A Course in Analysis: Vol. III: Measure and Integration Theory, Complex-Valued Functions of a Complex Variable [3, 1 ed.] 9813221593, 9789813221598

Table of contents :
Preface
Introduction
Contents
List of Symbols
Part 6: Measure and Integration Theory
1 A First Look at ˙-Fields and Measures
2 Extending Pre-Measures. Carath´eodory’s Theorem
3 The Lebesgue-Borel Measure and Hausdorff Measures
4 Measurable Mappings
5 Integration with Respect to a Measure — The Lebesgue Integral
6 The Radon-Nikodym Theorem and the Transformation Theorem
7 Almost Everywhere Statements, Convergence Theorems
8 Applications of the Convergence Theorems and More
9 Integration on Product Spaces and Applications
10 Convolutions of Functions and Measures
11 Differentiation Revisited
12 Selected Topics
Part 7: Complex-valued Functions of a Complex Variable
13 The Complex Numbers as a Complete Field
14 A Short Digression: Complex-valued Mappings
15 Complex Numbers and Geometry
16 Complex-Valued Functions of a Complex Variable
17 Complex Differentiation
18 Some Important Functions
19 Some More Topology
20 Line Integrals of Complex-valued Functions
21 The Cauchy Integral Theorem and Integral Formula
22 Power Series, Holomorphy and Differential Equations
23 Further Properties of Holomorphic Functions
24 Meromorphic Functions
25 The Residue Theorem
26 The ?-function, the -function and Dirichlet Series
27 Elliptic Integrals and Elliptic Functions
28 The Riemann Mapping Theorem
29 Power Series in Several Variables
Appendix I: More on Point Set Topology
Appendix II: Measure Theory, Topology and Set Theory
Appendix III: More on M¨obius Transformations
Appendix IV: Bernoulli Numbers
Solutions to Problems of Part 6
Solutions to Problems of Part 7
References
Mathematicians Contributing to Analysis (Continued)
Subject Index

Citation preview

A Course in Analysis (New) Vol. I Part 1 Part 2

Introductory Calculus Analysis of Functions of One Real Variable

Vol. II Part 3 Differentiation of Functions of Several Variables Part 4 Integration of Functions of Several Variables Part 5 Vector Calculus Vol. III Part 6 Measure and Integration Theory Part 7 Complex-valued Functions of a Complex Variable Vol. IV Part 8 Fourier Analysis Part 9 Ordinary Differential Equations Part 10 Introduction to the Calculus of Variations Vol. V Part 11 Functional Analysis Part 12 Operator Theory Part 13 Theory of Distributions Vol. VI Part 14 Partial Differential Equations: Classical Theory Part 15 Partial Differential Equations and Distributions Vol. VII Part 16 Differential Geometry of Curves and Surfaces Part 17 Differentiable Manifolds Part 18 Lie Groups

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Jacob, Niels. A course in analysis / by Niels Jacob (Swansea University, UK), Kristian P. Evans (Swansea University, UK). volumes cm Includes bibliographical references and index. Contents: volume 1. Introductory calculus, analysis of functions of one real variable ISBN 978-9814689083 (hardcover : alk. paper) -- ISBN 978-9814689090 (pbk : alk. paper) 1. Mathematical analysis--Textbooks. 2. Mathematics--Study and teaching (Higher) 3. Calculus--Textbooks. I. Evans, Kristian P. II. Title. QA300.J27 2015 515--dc23 2015029065

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2018 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher. ISBN 978-981-3221-59-8

Printed in Singapore

Preface A detailed description of the content of Volume III of our Course in Analysis will be provided in the introduction. Here we would like to take the opportunity to thank those who have supported us in writing this volume. We owe a debt of gratitude to James Harris who has typewritten the majority of the manuscript. Thanks for typewriting further parts are expressed to Saroj Limbu and James Morgan. Huw Fry, James Harris and Elian Rhind undertook a lot of proofreading for which we are grateful. We also want to thank the Department of Mathematics, the College of Science and Swansea University for providing us with funding for typewriting. It turned out that R. Schilling was working on the second edition of his book “Measure, Integration and Matingales” [75] while we were working on Part 6. This led to many interesting discussions between Dresden and Swansea from which we could benefit a lot and for which we are grateful. Finally we want to thank our publisher, in particular Tan Rok Ting and Ng Qi Wen, for a pleasant collaboration. Niels Jacob Kristian P. Evans Swansea, January 2017

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Introduction The third volume of our “Course in Analysis” covers two topics indispensable for every mathematician whether specializing in analysis or not. Part 6 discusses Lebesgue’s theory of integration and first consequences in the theory of a real-valued function of a real variable. Part 7 introduces the theory of complex-valued functions of one complex variable - traditionally just called the “the theory of functions”. The advanced theory of real-valued functions, Fourier analysis, functional analysis, the theory of dynamical systems, partial differential equations or the calculus of variations are all topics in analysis depending on Lebesgue’s integration theory. But in addition, probability theory, parts of geometry (fractal sets) and more applied subjects such as information theory or optimization needs a proper understanding of Lebesgue integration. Moreover, many mathematicians will agree that the structure of the real line is one of the most complicated structures, if not the most complicated structure of all, we have to deal with. Many problems leading to a better understanding of R and in fact leading to deep developments in the foundation of mathematics (mathematical logic and set theory) we encounter when relating the topological structure of R as induced by the Euclidean metric to the problem of defining and determining the size of subsets of R. Often the underlying model of set theory determines the relation between topology and measure theory. In our treatise we always assume ZF as underlying model of set theory but we do not spend much time on investigating the problems mentioned above, they are the topic of different and more advanced courses. We will describe the content of Part 6 in more detail below, however we would like to mention the influence of [11] in our presentation. The forerunner of H. Bauer’s monograph [11], i.e. [10], was the standard text book on measure theory in Germany for many decades. When [10] was split into two books the first named author was heavily involved in proof reading and discussing the material. The theory of functions is key to every other advanced theory in analysis and at the same time it is needed as a tool in many other applied mathematical disciplines such as mechanics or in fields such as electrical engineering or physics. But much more holds: holomorphic functions, i.e. complex differenvii

A COURSE IN ANALYSIS

tiable complex-valued functions of a complex variable enter into many fields in pure mathematics such as number theory, algebraic geometry, representation theory, differential geometry or combinatorics and many others. It is fair to say that without a proper knowledge of function theory no undergraduate education in pure or applied mathematics can be viewed as being complete. A more detailed discussion of Part 7 follows below. Before going into the details we have to add two remarks. Firstly, originally we also planned to include in our third volume Part 8: Fourier analysis. While writing Part 6 and 7 it emerged that a better strategy is to add more (advanced or specialized) material from the two theories covered in Part 6 and 7 already here and not in later parts where they will be needed, e.g. differentiability properties of real-valued functions, Sard’s theorem, dense subsets in Lp -spaces and the Friedrichs mollifier, or hypergeometric function, elliptic integrals and elliptic functions, just to mention a few. This will of course lead to some alteration of the previous plan of arranging the entire material. Secondly, the reader will notice a different mode of referring to the literature. We now sometimes deal with topics which admit quite different approaches and representations. Clearly we are influenced by authors who dealt with this material before and of course we want to and we have to give fair credit where appropriate. The fact that we have meanwhile reached more advanced material is also taken into account when in some but not many proofs we leave straightforward calculations to the reader, something we have strictly avoided in the first two volumes. Some of our problems are now more involved and some of the solutions are more brief, again a reflection of the fact that we now address more (mathematically) matured readers. In Chapter 1 we introduce σ-fields, their generators and measures, and Chapter 2 is devoted to the Carath´eodory extension theorem. A discussion of the Lebesgue-Borel measure and of the Hausdorff measure as well as the Hausdorff dimension follows in Chapter 3. In particular the Cantor set is treated in great detail. Measurable mappings are the topic of Chapter 4. The standard approach to define the Lebesgue integral with respect to a measure is developed in Chapter 5, and Chapter 6 starts to handle measures with densities. We prove the Radon-Nikodym theorem as we prove the transformation theorem for Lebesgue integrals. The role of sets of measure zero and almost everywhere statements are treated in Chapter 7 along with the main convergence results, especially the dominated convergence theorem. We also look viii

INTRODUCTION

at spaces of p-fold integrable functions. These results are then applied in Chapter 8 to prove typical theorems about the interchanging of limits such as the continuity or the differentiability of parameter dependent integrals. We prove Jensen’s inequality and discuss the relation between the Lebesgue integral and the Riemann integral. By this we fill in some of the gaps left by dealing with the Riemann integral in higher dimensions in Volume II. This discussion includes improper integrals in particular as well as the introduction of the Lp -spaces. Product integrations and most of all the theorems of Tonelli and Fubini are the main content of Chapter 9. As an important application we look at integration with respect to the distribution function and we give some examples of theoretical interest. We also provide a complete proof of Minkowski’s integral inequality. From our point of view the content of Chapters 1-9 forms the core of any module on measure and integration theory suitable for the purposes of analysis as well as probability theory. Chapters 10-12 deal with topics which are interesting by themselves but they are also major tools in areas such as Fourier analysis, functional analysis or (partial) differential equations. Chapter 10 is devoted to the convolution of functions and measures and we prove the density of continuous functions with compact support in the spaces Lp (Rn ), 1 ≤ p < ∞. Further we handle the Friedrichs mollifier which will turn out to become a first class tool in many later considerations. Finally, we have a first look at convolution operators. Lebesgue’s theory of differentiation is our topic in Chapter 11. After having proved the Vitali covering theorem we introduce absolutely continuous functions and functions of bounded variation and study their relations. The key results are a new version of the fundamental theorem of calculus and Lebesgue’s differentiation theorem, the proof of which is given by making use of the Hardy-Littlewood maximal function. Eventually in Chapter 12 we discuss three special results which will become important later: a version of Sard’s theorem on the measure of the critical points of a differentiable mapping, Lusin’s theorem which is followed by a first discussion of weak convergence and the Kolmogorov-Riesz theorem which characterises relative compact sets in Lp -spaces. Our treatment of complex-valued functions of a complex variable starts with a brief recollection of the complex numbers including convergence of sequences and series. In Chapter 14 we embark onto a small digression and summarize obvious properties of complex-valued functions defined on an arbitrary set and Chapter 15 is devoted to the geometry of the plane and complex ix

A COURSE IN ANALYSIS

numbers. We also include the Riemann sphere and the stereographic projection. A chapter on complex-valued functions of a complex variable follows this where we handle continuity, convergence and uniform convergence and we prove Abels’ theorem as well as the Cauchy-Hadamard theorem for power series. Our examples of power series include the binomial series and the Gaussian hypergeometric series. We close this chapter with a first look at M¨obius transformations. Complex differentiation is handled in Chapter 17 where all the standard results are shown including those related to the CauchyRiemann differential equations. The key notion is that of a holomorphic function and we also have a first encounter with biholomorphic functions as well as harmonic functions. Next is a discussion of some important holomorphic functions, e.g. the exponential function, trigonometrical functions and their inverses and hyperbolic functions and their inverses. Most important in Chapter 18 are the investigations into the logarithmic function. Topological notions such as connectivity, simple connectivity or homology play a crucial part in function theory and Chapter 19 treats some of the related questions before in Chapter 20 we introduce line integrals. For defining line integrals we can rely on our considerations in Chapter II.15 as we can when deriving their basic properties. Often we just need to change the notation, i.e. we use complex numbers. Once this is done the theory of complex-valued functions of a complex variable offers many new results when put in context of line integrals over (simply) closed curves. We introduce the notion of a primitive and relate properties of line integrals to the existence of primitives. Chapter 21 is a central chapter in the theory with a detailed discussion of Cauchy’s integral theorem, Cauchy’s integral formulae and many of their applications including standard estimates for derivatives of holomorphic functions and the relation to Taylor expansions. In Chapter 22 we continue our investigations by looking at power series, holomorphy and holomorphy domains, and applications to differential equations. An important result is Riemann’s theorem on removable singularities and the uniqueness theorem for holomorphic functions. We develop the theory further in Chapter 23 by discussing the boundary maximum principle, the Lemma of Schwarz, Liouville’s theorem and the approximation theorem of Runge. Meromorphic functions and the Laurent expansions are investigated in Chapter 24 including the classification of their singularities and a first version of the residue theorem as well as the argument principle and Rouch´e’s theorem. In Chapter 25 we give a general version of the residue theorem. This more theoretical consideration is followed by many very concrete applications, e.g. the evaluation of cerx

INTRODUCTION

tain (improper) integrals. We take up our discussion of the Γ-function from Volume I and now look at Γ(z) as a meromorphic function in Chapter 26. We also introduce Dirichlet series, in particular the Riemann ζ-function and prove some of their properties. This enables us to eventually formulate the Riemann hypothesis - we believe that the reader will understand that we do not provide its proof here. Mathematicians may encounter elliptic integrals only late in their education, physicists meet them quite early. Our discussion of elliptic integrals and elliptic functions in Chapter 27 starts with the integrals and then we move on to the Jacobi elliptic functions. We show that they are double periodic functions and we begin to investigate double periodic functions by proving the three Liouville theorems. Then we develop, up to a certain point, the theory of the Weierstrass ℘-function. This chapter serves as a first introduction but it covers more material (as the previous chapter) than is usually included in a first course on function theory. Chapter 28 provides a proof of the Riemann mapping theorem, a result which must be included in every course, however we do not discuss many applications to conformal mappings such as Schwarz-Christoffel mappings. The final chapter of Part 7 is devoted to power series in several complex variables. The aim is to point out that in several (real and complex) variables the domain of convergence of a power series is a much more delicate question as it is in one (real or complex) dimension. Two of the appendices handle some topological questions, partly in relation to measure theory. In Appendix III we discuss M¨obius transformations further and the geometry of the extended plane. The final appendix is devoted to Bernoulli numbers which we have encountered in several places before. As in the previous volumes we have provided solutions to all the ca. 275 problems. Moreover, in particular in Part 7, we needed quite a substantial number of figures (ca. 90). All these figures were done by the second named author using LaTeX. Finally a remark about referring to Volume I or II: Theorem II.4.20 means Theorem 4.20 in Volume II and (I.3.12) stands for formula (3.12) in Volume I. Problems marked with a ∗ are more challenging.

xi

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Contents Preface

v

Introduction

vii

List of Symbols

xvii

Part 6: Measure and Integration Theory

1

1 A First Look at σ-Fields and Measures

3

2 Extending Pre-Measures. Carath´ eodory’s Theorem

23

3 The Lebesgue-Borel Measure and Hausdorff Measures

45

4 Measurable Mappings

71

5 Integration with Respect to a Measure — The Lebesgue Integral

85

6 The Radon-Nikodym Theorem and the Transformation Theorem

109

7 Almost Everywhere Statements, Convergence Theorems

127

8 Applications of the Convergence Theorems and More

153

9 Integration on Product Spaces and Applications

173

10 Convolutions of Functions and Measures

195

11 Differentiation Revisited

215

12 Selected Topics

243 xiii

A COURSE IN ANALYSIS

Part 7: Complex-valued Functions of a Complex Variable

259

13 The Complex Numbers as a Complete Field

261

14 A Short Digression: Complex-valued Mappings

277

15 Complex Numbers and Geometry

283

16 Complex-Valued Functions of a Complex Variable

299

17 Complex Differentiation

319

18 Some Important Functions

335

19 Some More Topology

347

20 Line Integrals of Complex-valued Functions

365

21 The Cauchy Integral Theorem and Integral Formula

383

22 Power Series, Holomorphy and Differential Equations

407

23 Further Properties of Holomorphic Functions

425

24 Meromorphic Functions

445

25 The Residue Theorem

473

26 The Γ-function, the ζ-function and Dirichlet Series

505

27 Elliptic Integrals and Elliptic Functions

527

28 The Riemann Mapping Theorem

555

29 Power Series in Several Variables

569

Appendices

577

Appendix I: More on Point Set Topology

579

xiv

CONTENTS

Appendix II: Measure Theory, Topology and Set Theory

583

Appendix III: More on M¨ obius Transformations

589

Appendix IV: Bernoulli Numbers

593

Solutions to Problems of Part 6

599

Solutions to Problems of Part 7

653

References

741

Mathematicians Contributing to Analysis (Continued)

747

Subject Index

749

xv

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List of Symbols N N0 = N ∪ {0}, Nn0 Z Q R R+ = {x ∈ R | x ≥ 0}

R = R ∪ {−∞, ∞} D H Br (x) = {y ∈ R | kx − yk < r} Dρ (z0 ) = {z ∈ C | |z − z0 | < ρ} S n−1 = {x ∈ Rn | kxk = 1} Ar,R (z0 ) = Br (z0 )\Br (z0 ) A0,R = BR (z0 )\{z0 } a ∨ b = max(a, b) a ∧ b = min(a, b) ( 1, k = l δkl = 0, k 6= l

e = exp(1) i, i2 = −1 γ Bk

β2k = (−1)k−1 B2k n!   n k a k




=

a(a−1)(a−2)·...·(a−k+1) k!

natural numbers multi-indices (see more in Volume II) integers rational numbers real numbers

unit disc in C upper half plane in C

annulus in C punctured disc in C

Kronecker delta

Euler constant Bernoulli number n factorial binomial coefficient for n, k ∈ N general binomial coefficents xvii

A COURSE IN ANALYSIS

(a)n = a(a + 1)(a + 2) · . . . · (a + n − 1)

Pochhammer symbol

M(n; R) GL(n; R)

n × n-matrices with real elements general linear group over R in dimension n special linear group over R in dimension n orthogonal group in Rn special orthogonal group in Rn general linear group over C in dimension n special linear group over C in dimension n unitary group in Cn special unitary group in Cn automorphism group of a domain G⊂C

SL(n; R) O(n) SO(n) GL(n; C) SL(n; C) U(n) SU(n) Aut(G)

z = x + iy z = x − iy Re z = x Im z = y |z|

complex number conjugate complex number real part of z imaginary part of z absolute value or modulus of z

f, g, h

generic symbol for mappings or functions image of A under f pre-image of A0 underf positive part of f negative part of f

z α = z1α1 · . . . · znαn , z ∈ Cn , α ∈ Nn0

f (A) f −1 (A0 ) f+ = f ∨ 0 f − = (−f ) ∨ 0

xviii

LIST OF SYMBOLS

Γ(f ) supp f crit(f ) Λ(f ) sf Sing(f )

graph of f support of f critical points of f Lebesgue set of f singularity function of f singularities of a meromorphic function residue of f at z0 poles of f zeroes of f Jacobi matrix of f Jacobi determinant of f decomposition of f into real and imaginary parts convolution of f and g

res(f, z0 ) Pol(f ) N(f ) Jf det Jf f = u + iv f ∗g c T(a (z) = n)

∞ X k=0

ak (z − c)k

prj : A1 × · · · × AN → Aj

projection

∅ P(Ω)

empty set power set of Ω

A{

Ak ↑ A

complement of the set A [ Ak ⊂ Ak+1 and A = Ak

V/ ∼

equivalence classes in V with respect to the equivalence relation “∼”

(X, k · k) (X, d) (X, OX )

normed space metric space topological space

xix

A COURSE IN ANALYSIS

OX C(x) CX On Cn Kn Ir,n , Il,n

F (n)

B(n) , B(Rn ) (1)

topology, open sets in a topological space connectivity component of x in a topological space closed sets in a topological space open sets in R closed sets in Rn compact sets in Rn right open, left open intervals in Rn figures in Rn Borel σ-field in Rn

B , B(R) σ(E) δ(E)

Borel σ-field in R σ-field generated by E Dynkin system generated by E

˚ G

interior of a set G

G ∂G diam(A) = sup{d(x, y) | x, y ∈ A} dist(A, B) dist(A, x) = dist(A, {x}) dist(x, A) = dist({x}, A) voln (K) gT = det T ∗ T {f = g} = {x ∈ X | f (x) = g(x)} {f ≤ g} = {x ∈ X | f (x) ≤ g(x)} {f < g} = {x ∈ X | f (x) < g(x)} {f ≥ g} = {x ∈ X | f (x) ≥ g(x)} {f > g} = {x ∈ X | f (x) > g(x)}

closure of a set G boundary of a set G diameter of a set A distance of A and B

volume of a set K ⊂ Rn Gram determinant

xx

LIST OF SYMBOLS

A AΩ0 = Ω0 ∩ A (Ω, A) (Ω, A, µ) ˜ µ (Ω, A, ˜) Nµ

N O j=1

Aj = A1 ⊗ · · · ⊗ AN

A/A0 − measurable  f −1 (A0 ) = f −1 (A0 ) | A0 ∈ A0 Sx , Sy µ, ν w0  = 0 λ(n) (n)

generic σ-field trace σ-field measurable space measure space completion of (Ω, A, µ) with respect to µ null sets of (Ω, A, µ) product σ-field f : Ω → Ω0 measurable with respect to A, A0 for f : Ω → Ω0 x−, y−section of S ⊂ X × Y

generic measures unit measure, Dirac measure at w0 Dirac measure at 0 ∈ Rn

Lebesgue-Borel and Lebesgue measure in Rn

λG

restriction of the Lebesgue-Borel measure to G

βpN

binomial distribution

πα cα gσ,α µΩ0 (A) = µ (Ω0 ∩ A) T (µ) µ − a.e. ∼µ µ∗

Poisson distribution Cauchy distribution (density) Gauss distribution (density) trace of a measure image measure µ-almost everywhere µ-a.e. equivalence generic outer measure xxi

A COURSE IN ANALYSIS

µ⊗ν µ∗ν µν Hα dimH A

product measure convolution of measures absolute continuity of measures Hausdorff measure Hausdorff dimension of A

δ HA PX E(X)

outer Hausdorff measure distribution of a random variable expectation of a random measure

D+ , D − , D + , D − df (x) = f 0 (x) dx

Dini numbers or Dini derivatives derivative of a function of a real variable

k

d f = f (k) (x) dxk

higher order derivatives of a function of a real variable

df (z) = f 0 (z) dz

derivative of a function of a complex variable

k

d f (z) = f (k) (z) dz k ∂

α

f, ∂zα f

higher order derivative of a function of a complex variable partial derivates (α ∈ Nn0 )

∂u ∂x ∆ or ∆2 ∆n P (ζ, z)

Laplacian in R2 Laplacian in Rn Poisson kernel for the disc in R2 or C

γ γ −1 tr(γ)

generic curve (in C or R2 or Rn ) reversed curve trace of a curve

ux =

xxii

LIST OF SYMBOLS

γ1 ⊕ γ2 indγ int γ ext γ Z(t0 , . . . , tm )

sum of two curves index of a curve in C interior of a curve in C exterior of a curve in C partitions of [t0 , tm ] or [a, b], a = t0 and b = tm variation of a function or a curve with respect to the partition Z(t0 , . . . , tm ) (total) variation of a function or a curve length of a curve

VZ (g), VZ (γ) V (g), V (γ) lγ Z f (z) dz

line integral in C

γ

γ1 ∼ γ2 [γ] [γ1 ] ⊕ [γ2 ] := [γ1 ⊕ γ2 ]   [γ1 ]−1 := γ1−1 ˜ · · · +a ˜ M γM ˜ 2 γ2 + γ = a1 γ1 +a

equivalence of curves equivalence class of γ

Sx0 ,x1 , [x0 , x1 ], x0 x1 ∆(z1 , z2 , z3 )

line segment connecting x0 and x1 closed triangle with vertices z1 , z2 , z3 ∈ C boundary of ∆(z1 , z2 , z3 )

˜ · · · +(−a ˜ ˜ ' γ = (−a1 )γ1 +(−a M )γM 2 )γ2 +

∂∆(z1 , z2 , z3 ) k · k∞ kf k∞,G = sup |f (z)|

cycle

supremum norm

x∈G

kf k∞, tr(γ) = sup |f (z)| z∈ tr(γ)

k · k Lp = k · k p  p1 Z p |u(w)| µ(dw) Np (u) =

norm in Lp (Ω)

Ω

N∞ (u)

essential supremum of u xxiii

A COURSE IN ANALYSIS

Ar u(x) Mr (u)(x) M(f )(x) J (f )

average or mean value of u over Br (x) average operator Hardy-Littlewood maximal function Friedrichs mollifier applied to f

L = L(w1 , w2 )

periodicity lattice of a double periodic function fundamental parallelogram of a double periodic function order of an elliptic function polydisc in Cn distinguished boundary of a polydisc in Cn

P = P (w1 , w2 ) ord(f ) P (a, ~r) ∂d P (a, ~r)

S(Ω) H(G) AC([a, b]) BV ([a, b]) C(X), C(X; C) Cb (X), Cb (X; C) C0 (X), C0 (X; C) C∞ (Rn ) C k (X), C k (X; C) C ∞ (U) =

\

simple functions holomorphic functions absolutely continuous functions functions of bounded variation continuous (real-/complex-valued) functions on X bounded elements in C(X), C(X; C) elements with compact support in C(X), C(X; C) continuous functions on Rn vanishing at infinity k-times continuously differentiable elements in C(X), C(X; C)

C k (U)

k∈N

Cb∞ (U)

=

\

Cbk (U)

k∈N

xxiv

LIST OF SYMBOLS

C0∞ (U) =

\

C0k (U)

k∈N

Lp (Ω) L∞ (Ω) = Lp / ∼ µ, L1loc (Ω)

p-fold integrable functions 1≤p≤∞

locally integrable functions (equivalence classes)

χA exp sin cos tan cot arcsin arccos arctan arccot sinh cosh tanh coth arsinh arcosh artanh arcoth

characteristic function of the set A exponential function sine function cosine function tangent function cotangent function inverse sine function inverse cosine function inverse tangent function inverse cotangent function hyperbolic sine function hyperbolic cosine function hyperbolic tangent function hyperbolic cotangent function inverse hyperbolic sine function inverse hyperbolic cosine function inverse hyperbolic cotangent function inverse hyperbolic cotangent function

ln log z = ln |z| + iϕ

real logarithmic function ln : (0, ∞) → R principal branch of the complex logarithmic function

˜ + z) = ln(1

∞ X (−1)k−1 k=1

k

zk

xxv

A COURSE IN ANALYSIS

˜ − z) ˜ L(z) = ln(1   ˜ z La,b (z) = b + L a
1 y L(z) = ln x2 + y 2 + i arctan 2 z Ba (z) =

∞   X a k=0

B(z, w) Df (z) Eα Eαβ 2 F1 p Fq

k

zk

Gn = Gn (L) = Gn (w1 , w2 ) Jl am  sn   cn   dn

P(z) Γ(z) ζ(z) µ(n)

ψ(z) =

binomial series beta function Dirichlet series associated with f Mittag-Leffler function generalised Mittag-Leffler function Gauss hypergeometric series or functions generalised hypergeometric series or function Eisenstein series Bessel function of order l amplitude of an elliptic integral Jacobi elliptic functions Weierstrass ℘-function Γ-function ζ-function M¨obius function

Γ0 (z) Γ(z)

xxvi

Part 6: Measure and Integration Theory

1

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1

A First Look at σ-Fields and Measures

At this stage of our Course much has already been learnt about integration. In Part 1 we introduced the Riemann integral for real-valued functions of one real variable, in Part 2 we provided the underlying theory including all proofs. Then, in Part 4, we investigated integrals of real-valued functions of several real variables, followed by line and surface integrals of vector fields in Part 5. A motivation for introducing integrals was the problem to determine the area (volume) of a figure bounded by the graphs of certain functions. A further central topic was the fundamental theorem of calculus which links the derivative of a function to the integral. Powerful as these classical results are, eventually they lead to many open questions; here are some of these: • We can integrate certain discontinuous functions, but we cannot (yet) prove the fundamental theorem for them. It would be desirable to have necessary and sufficient conditions the fundamental theorem to hold. • We can define the length of a rectifiable curve and integrals over rectifiable curves, but we lack an analogous theory for surfaces or more generally hypersurfaces. • How do we define a volume integral for functions f : G → R, G ⊂ Rn , without depending on topological notions, i.e. ∂G? • Can we derive more satisfactory results for interchanging limits (continuity, differentiability, sequences) and integrals? As it turns out, within the Riemann (Darboux, Jordan) approach to define an integral we cannot resolve our problems in a convincing manner. Only after set theory was developed, and relying on the first ideas of E. Borel, the mathematician H. Lebesgue in [54] succeeded to construct a new theory of integration which allows us to resolve many of the problems we have with the Riemann integral. In addition this integral coincides with the Riemann integral for continuous functions defined on compact non-degenerate cells K ⊂ Rn , n ≥ 1, but it is defined on a much larger class of domains than the Riemann integral. The fundamental idea of Lebesgue was first to handle the problem of defining the volume (length, area) of a set. Starting with a nonempty set Ω he considered families A ⊂ P(Ω) of its subsets, P(Ω) denoting the power set of Ω, which satisfy certain “natural” conditions. Then he assigned a volume, or more generally a “mass” or a “measure” to each set A ∈ A. 3

A COURSE IN ANALYSIS

The elements of A are called measurable sets and the process of assigning a “measure to A” is formalized by introducing a mapping µ : A → [0, ∞] called the measure and µ(A) is then by definition the volume/mass/measure of A. Of course µ has to fulfil certain conditions. The first observation is the generality of this approach. In principle on one set Ω we can have quite a lot of different families of measurable sets, and in addition on one family A of measurable sets we can have different measures. It was ca. 25 years after Lebesgue had published his work that A. N. Kolmogorov [49] realised that Lebesgue’s thoery of measures and integrals provides the frame needed for probability theory. Thus the generality of the proposed approach was and is one of its advantages. At the same time it allows in the concrete setting of R or Rn to resolve most of the problems we encounter when using the Riemann integral. The definition of the Riemann integral (in Rn ) depends much on the topological structure of Rn induced by the Euclidean metric. In Chapter II.19 we have seen that Jordan measurable sets are characterised by properties of their boundary, and the boundary of a set is a topological notion. By separating the concept of measurable sets from topology, Lebesgue gained his advantage. However as a consequence we now have to investigate the relations of measurable sets belonging to A with the topological structure we have equipped Ω with. The surprise was that this is seriously non-trivial and deeply depends on the underlying model of set theory. This is a rather interesting topic but we cannot discuss it within our Course and refer to [9] or [46]. Before going into the details, a further general remark might be helpful. Due to the importance of measure theory in probability theory, nowadays many textbooks, e.g. [11], [15], [21], [75] or [85], on measure and integration concentrate on the general (abstract) theory and give some emphasis on those parts of relevance for probability theory. On the other hand, when studying real-valued functions other aspects of measure and integration theory are of more interest and there are also some good text books in this direction, e.g. [14], [47], [50], [59], [70], [71], [84], [87] or [94]. In our presentation of the topic, while still aiming to serve the needs of probabilists, we will emphasise the real-variable theory approach. In case of Rn we want to integrate functions, say continuous functions to n begin with, defined on a non-degenerate compact cell K := j=1 [aj , bj ], aj < bj . To achieve this within Lebesgue’s approach we need to define a

×

4

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

family of measurable sets which contains all non-degenerate compact cells K. Furthermore we want to have that the measure of aQcompact cell K = n [a , b ] is its natural “Euclidean” volume voln (K) = nj=1(bj −aj ). If for j=1 j j a moment Cn denotes all non-degenerate compact cells in Rn , we are searching for a family A of subsets of Rn containing all bounded Jordan measurable sets, hence in particular Cn , and for a Q mapping µ : A → [0, ∞] such that n µ|Cn = voln , i.e. µ(K) = voln (K) = nj=1 (bj − aj ) for K = j=1[aj , bj ]. From µ we would further expect

×

×

µ

µ(∅) = 0; µ(G) ≥ 0, G ⊂ Rn ; ! ∞ ∞ [ X Gj = µ(Gj ) if Gl ∩ Gk = ∅ for k 6= l;

j=1

j=1

µ(G + x) = µ(G), x ∈ R, G ⊂ Rn , (translation invariance); and µ(R(G)) = µ(G), G ⊂ Rn , R ∈ SO(n), (rotation invariance). In particular for G1 ∩ G2 = ∅ we expect µ(G1 ∪ G2 ) = µ(G1 ) + µ(G2 ) and this implies for G1 ⊂ G that µ(G\G1 ) = µ(G) − µ(G1 ) since G = (G\G1 ) ∪ G1 and (G\G1 ) ∩ G1 = ∅. For a single point c ∈ Rn we expect µ({c}) = 0 which implies, say for n = 1, that µ([a, b]) = µ([a, b)) = µ((a, b]) = µ((a, b)) = b − a. From this we deduce, since Q is countable, that µ(Q ∩ [a, b]) = 0 and µ([a, b]\Q) = µ([a, b]), i.e. we may take away from [a, b] an infinite, but countable set, and the “length” remains unchanged. It turns out that on P(Rn ) a mapping µ with 5

A COURSE IN ANALYSIS

all these properties does not exist. We refer to Appendix II where we will discuss some of the underlying problems. Let us collect some ideas for the conditions a family A of measurable subsets A ⊂ Ω, Ω 6= ∅ but otherwise arbitrary, should satisfy. It is natural to assume that Ω itself is a measurable set, i.e. Ω ∈ A. Moreover, if A1 , A2 ∈ A are two measurable sets we expect that A1 ∪ A2 is measurable too, which then clearly extends to finitely many sets, i.e. Aj ∈ A, j = 1, . . . , N, implies A1 ∪ . . . ∪ An ∈ A. Since Ω = A ∪ A{ , it is now natural to require also A{ ∈ A for A ∈ A. However we know that finite unions of cells or hyper-rectangles can never lead to balls, ellipsoids, etc, i.e. more general Jordan measurable sets. Lebesgue’s insight was S that denumerable unions will fit the bill, i.e. he added the condition that j∈N Aj ∈ A holds for (Aj )j∈N , Aj ∈ A (which we have already stated above). Before giving a formal definition, we recall that a set is called countable if it is the bijective image of a mapping with domain N, while a set is called denumerable if it is either finite or countable. By definition the empty set ∅ is finite. With these preparations we now start to develop Lebesgue’s theory. Definition 1.1. Let Ω 6= ∅ be a set. A family A of subsets of Ω is called a σ-field or σ-algebra in Ω if Ω ∈ A;

(1.1)

A ∈ A implies A{ ∈ A;

Aj ∈ A, j ∈ N, implies

[

j∈N

(1.2) Aj ∈ A.

(1.3)

The elements of A are called measurable sets and the pair (Ω, A) is called a measurable space. Remark 1.2. A. If we always choose Aj = AN in (1.3) for j ≥ N we find S { that A1 , . . . , AN ∈ A implies N j=1 Aj ∈ A. Since Ω = ∅ it follows also that ∅ ∈ A for every σ-field. B. In the case that we consider on Ω several σ-fields A1 , . . . , AM , we may, and may have to speak of Aj -measurable sets.

6

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

Example 1.3. A. The power set P(Ω) is obviously a σ-field as is the set A := {∅, Ω}. B. Let Ω 6= ∅ be any set and define n o A := A ∈ P(Ω) A or A{ is denumerable .

Then A is a σ-field in Ω, see Problem 1. C. Let (Ω, A) be a measurable space and Ω0 ⊂ Ω a non-empty subset. We define the trace or trace σ-field of A in Ω0 by  (1.4) AΩ0 := Ω0 ∩ A := Ω0 ∩ A A ∈ A .

Note that this construction is completely analogous to that of the trace or relative topology, see Proposition II.1.24, and the proof, which we leave as an exercise, goes analogously. The notation Ω0 ∩ A is quite common, hence we shall adopt this, but of course Ω0 ∩ A cannot be interpreted as an intersection of two sets. D. Let Ω and Ω0 be two non-empty sets and A0 a σ-field in Ω0 . Further let f : Ω → Ω0 be a mapping. For A0 ⊂ Ω0 we denote as usual by f −1 (A0 ) ⊂ Ω the pre-image of A0 under f . The family  f −1 (A0) := f −1 (A0 ) A0 ∈ A0 (1.5)

is a σ-field in Ω. Indeed, we have f −1 (Ω0 ) = Ω, thus Ω ∈ f −1 (A0 ) and { for A0 ∈ A0 it follows that (f −1 (A0 )) = Ω\f −1 (A0 ) = f −1 (Ω0 )\f −1 (A0 ) = −1 0 0 −1 0{ f (Ω \A ) = f (A ) and therefore A ∈ f −1 (A0 ) implies A{ ∈ f −1 (A0 ). Finally let (Aj )j∈N be a countable collection of elements of f −1 (A0 ). We can find A0j ∈ A0 such that Aj = f −1 (A0j ) and therefore we find [

j∈N

S

Aj =

[

f

−1

(A0j )

=f

−1

j∈N

[

j∈N

and since j∈N A0j ∈ A0 we conclude that f −1 (A0 ) is a σ-field.

S

j∈N

A0j

!

Aj ∈ f −1 (A0 ) implying that

Given a measurable space (Ω, A), since ∅ = Ω{ it follows that ∅∈A 7

(1.6)

A COURSE IN ANALYSIS

and for Aj ∈ A, j ∈ N, we find A{j ∈ A and therefore \

j∈N

Aj =

[

j∈N

A{j ∈ A

(1.7)

which also includes the statement that N \

j=1

Aj ∈ A for Aj ∈ A, j = 1, . . . , N.

(1.8)

Moreover, since A\B = A ∩ B { we have A\B ∈ A for A, B ∈ A.

(1.9)

Our first theorem gives a powerful tool to construct σ-fields. T Theorem 1.4. The intersection j∈I Aj of an arbitrary family (Aj )j∈I of σ-fields in Ω is a σ-field in Ω. Proof. First we note that \  Aj = A ∈ P(Ω) A ∈ Aj for all j ∈ I . j∈I

T Since Ω ∈ Aj for all j ∈ I it follows that Ω ∈ j∈I Aj . Furthermore, if T A ∈ j∈I Aj then A ∈ Aj for all j ∈ I, hence A{ ∈ Aj for all j ∈ I implying T Ak ∈ Aj for all j ∈ I and that A{ ∈ j∈I Aj . Eventually, suppose that T k ∈ N, which yields by (1.7) for all j ∈ I that k∈N Ak ∈ Aj and therefore T T j∈I Aj . k∈N Ak ∈

This theorem justifies

Definition 1.5. Let E ⊂ P(Ω), Ω 6= ∅. The σ-field \ σ(E) := A E ⊂ A and A is a σ-field in Ω

(1.10)

is called the σ-field generated by E and we call E a generator of σ(E). Note that since E ⊂ P(Ω) and P(Ω) is a σ-field we take an intersection over a non-empty system and therefore σ(E) is indeed a σ-field in Ω. 8

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

Example 1.6. A. If E is a σ-field in Ω then E = σ(E), in particular we have σ(σ(E)) = σ(E). B. For E = {A}, A ⊂ Ω, we find σ(E) = {∅, A, A{, Ω}. C. If E1 ⊂ E2 then σ(E1 ) ⊂ σ(E2 ). However it may happen that E1 ⊂ E2 , E1 6= E2 but σ(E1 ) = σ(E2 ) as we will see soon below in Corollary 1.8. Our next goal is to find a “good” σ-field on R which contains all intervals. An easy way to introduce such a σ-field is to take the σ-field generated by all intervals. However we want to give more emphasis on topological notions and start with Definition 1.7. The σ-field generated by all open subsets O1 of the real line is called the Borel σ-field in R and is denoted by B(1) or B(R), hence σ(O1 ) = B(1) .

(1.11)

The elements of B(1) are called the Borel sets of R. Corollary 1.8. The Borel σ-field B(1) is generated by all open intervals. Proof. From Example 1.6.C we deduce that the σ-field A generated by all open intervals is contained in B(1) . By Theorem I.19.27 we know that for every open set A ⊂ R there exists S a denumerable family Ij ⊂ R, j ∈ J ⊂ N, of open intervals such that A = j∈J Ij . This implies however for every open set A ⊂ R that A ∈ A which yields B(1) = σ(O1 ) ⊂ σ(A) = A ⊂ B(1) . The next step is to determine further generators of B(1) . For this let us collect some simple observations about generators of σ-fields, compare with Problem 2. Lemma 1.9. Let Ω 6= ∅ be a set and E, E1, E2 ⊂ P(Ω). A. If we denote by E { the system n o E { := E { E ∈ E

then we have

σ(E { ) = σ(E). 9

(1.12)

A COURSE IN ANALYSIS

n

B. For I := F =

S

o it follows that j∈J Ej Ej ∈ E, J ⊂ N σ(I) = σ(E).

(1.13)

σ(E1 ∪ E2 ) = σ(E1 ).

(1.14)

C. If σ(E1 ) = σ(E2 ) then

We use the following notations: On all open sets in Rn , Cn all closed sets in Rn , Kn all compact sets in Rn ,  Ir := [a, b) a < b, a, b ∈ R ,  Il := (a, b] a < b, a, b ∈ R , Ir,n := Irn , Il,n := Iln .

(1.15) (1.16) (1.17) (1.18) (1.19) (1.20)

Corollary 1.10. We have B(1) = σ(C1 ) = σ(K1 ).

(1.21)

Proof. Since C1 = O1{ the first equality follows from Lemma 1.9.A. Moreover,S since K1 ⊂ C1 we have σ(K1 ) ⊂ σ(C1 ). For C ∈ C1 we have C = k∈N (C ∩ [−k, k]), but C ∩ [−k, k] is compact implying by Lemma 1.9.B that σ(C1 ) ⊂ σ(K1 ). Corollary 1.11. The following holds B(1) = σ(Ir ) = σ(Il ).

(1.22)

Proof. For a < b we have [a, b) =

\

 1 a− ,b j

\



j∈N

and (a, b] =

j∈N

10

1 a, b + j

.

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

This implies σ(Ir ) ⊂ B(1) and σ(Il ) ⊂ B(1) .

Now for an open interval I = (a, b) we know that [ (a, b) = [c, d) [c, d) ⊂ (a, b) and c, d ∈ Q

and

(a, b) =

[ (c, d] (c, d] ⊂ (a, b) and c, d ∈ Q .

Since Q is countable, we deduce from Corollary 1.8

B(1) ⊂ σ(Ir ) and B(1) ⊂ σ(Il ). Remark 1.12. The proof of Corollary 1.11 yields even more, namely that B(1) is already generated by all half-open intervals (either from the right or from the left) with rational end points. For a, b ∈ Q, a < b, we have [a, b) = (−∞, b)\(−∞, a) which implies that

  B(1) = σ [a, b) ⊂ R a < b, a, b ∈ Q ⊂ σ (−∞, c) ⊂ R c ∈ Q . S However (−∞, c) = k∈N [−k, c) and therefore we deduce that −k 0 the measure µ in (1.34) is given as a finite sum N X µ= αkj ωkj . (1.37) j=0

An example of a corresponding probability measure is the Bernoulli or binomial distributions with parameter N and p ∈ [0, 1] defined by βpN note that

PN

k=0

N k




N   X N k := p (1 − p)N −k k , k k=0

(1.38)

pk (1 − p)N −k = (p + (1 − p))N = 1.

In the following let (Ω, A, µ) be an arbitrary measure space. For A, B ∈ A we always have A ∪ B = A ∪ (B\A) and A ∩ (B\A) = ∅ implying µ(A ∪ B) = µ(A) + µ(B\A) 18

(1.39)

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

and since B = (A ∩ B) ∪ (B\A) it follows that µ(B) = µ(A ∩ B) + µ(B\A).

(1.40)

Adding (1.39) and (1.40) we arrive at µ(A ∪ B) + µ(A ∩ B) + µ(B\A) = µ(A) + µ(B) + µ(B\A), and if µ(B\A) < ∞, which holds, for example, for every finite measure, we have µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B).

(1.41)

However, for µ(B\A) = ∞ both µ(A ∪ B) and µ(B) must be equal to +∞ and hence (1.41) holds in all cases. In particular we have for A ⊂ B µ(B) = µ(A ∩ B) + µ(B\A) = µ(A) + µ(B\A) implying the monotonicity of measures: A ⊂ B implies µ(A) ≤ µ(B).

(1.42)

Let us return to the problem of defining on Rn , better on B(n) , a measure λ(n) with the properties mentioned earlier on, in particular with the property n that for all compact cells K := j=1 [aj , bj ], aj < bj , we have

×

λ(n) (K) =

n Y (bj − aj ).

(1.43)

j=1

˚ but we can We can easily extend (1.43) to A ∈ Irn ∪ Iln or all open cells K, (n) not easily extend this definition to B . The general question we have to address is the following: given Ω 6= ∅ and a system G ⊂ P(Ω) of subsets. On G a mapping ν : G → [0, ∞] is defined. When can we extend ν to a measure on σ(G)? The answer is given by Carath´eodory’s extension theorem which we will discuss in the next chapter. 19

A COURSE IN ANALYSIS

Problems 1.

a) Prove that for a non-empty set Ω a σ-field is given by A := {A ∈ P(Ω)|A or A{ is denumerable}. b) Prove that the trace AΩ0 := Ω0 ∩ A as defined in (1.4) is a σ-field in Ω0 .

2. Prove Lemma 1.9. 3. Let (X, d) be a metric space and denote by B(X) = σ(X) the Borel σ-field in X, i.e. the σ-field generated by all open sets OX in X. Prove that B(X) is also generated by all closed sets in X. Let D = {Br (x) ⊂ X|r > 0 and x ∈ X} be the system of all open balls in X and suppose that X is separable, i.e. X has a countable dense subset. Does D generate B(X)? 4. Let (Ωj , Aj ), j = 1, 2, be two measurable spaces. Prove that in general A1 × A2 is not a σ-field in Ω1 × Ω2 . S 5. Let U ⊂ Rn be open. Prove that U = I⊂J,I⊂U I, where J is defined as in the proof of Corollary 1.21. 6. Solve Exercise 1.22. 7.

a) Let f : R → R be a bijective continuous mapping and choose on R the Borel σ-field B(1) . What can we say about σ(f )?

b) Consider the three measurable spaces (Ω1 , A1) := (R, P(R)), (Ω2 , A2 ) := (R, B(1) ) and (Ω3 , A3 ) := (R, {R, ∅, A, A{}) where A 6= ∅ is any set in R. Further, always consider on R the Euclidean topology. For 1 ≤ k, l ≤ 3 we can consider continuous mappings gkl : R → R. When is a generic map gkl Ak /Al -measurable. 8. Let (Ωj , Aj ), j = 1, 2, 3, be measurable spaces and f : Ω1 → Ω2 , g : Ω2 → Ω3 two measurable mappings. Prove that g ◦ f : Ω1 → Ω3 is measurable. Now prove that f : R → Rn is B(1) /B(n) -measurable if and only if fj : R → R is B(1) /B(1) -measurable, where f = (f1 , . . . , fn ). 9.

a) Let (Ω, A) be the measurable space from Example 1.3.B. Prove ( 0, A is denumerable is a probability that ν : A → [0, 1], ν(A) = 1, A{ is denumerable measure. 20

1

A FIRST LOOK AT σ-FIELDS AND MEASURES

P b) On (Z, P(Z)) consider µ : P(Z) → R, µ(A) = k∈Z k (A). Show that on P(Z) and µ(A + m) = µ(A) for all m ∈ Z and A ∈ P(Z).

10.

c) Let (Ω, P A, µ1j ), j ∈ N, be probability spaces. Prove that (Ω, A, µ) where µ = ∞ j=1 2j µj is a further probability space.

a) Let (Ω, A, µ) be a measure space and Ω0 ∈ A. Prove that µΩ0 defined by µΩ0 (A) := µ(Ω0 ∩ A) is a further measure on A. b) Let (Ω, A, µ) be a measure space and A˜ ⊂ A a σ-field. Show ˜ that µ| ˜ is a measure on A. A

21

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2

Extending Pre-Measures. Carath´ eodory’s Theorem

While in the last chapter we introduced the framework for eventually resolving our problem of defining a measure and integral for real-valued functions defined on suitable domains in Rn , we
still cannot apply this framework to this concrete situation, i.e. Rn , B(n) . The measure λ(n) we are seeking on B(n) should satisfy n Y λ(n) (K) = (bj − aj ) j=1

×

n [a , b ] j=1 j j

for every compact cell K = ⊂ Rn , but at the moment it is not defined on B(n) but only on (closed, open and half-open) hyper-rectangles. Thus we need to find a way to extend certain mappings defined on a subset of B(n) (or more generally of a σ-field A) to measures on B(n) (or the σ-field A). This problem will be solved by Carath´eodory’s theorem which we are going to prove in this chapter. First we want to explore further what we actually know about the set-function which we want to extend to the measure λ(n) on B(n) . In our discussion of the Riemann integral in Volume II we learnt that partitions are helpful and even important, however we cannot find partitions of non-degenerate compact cells into non-degenerate compact cells. This does not even work for compact intervals. Note that a partition of a set Ω consists of mutually disjoint subsets. For this reason we switch from non-degenerate compact cells to elements of Ir,n (we could also have chosen Il,n ). Recall that A ∈ Ir,n = (Ir )n if n A=

×[a , b ), j

j=1

j

aj < bj and aj , bj ∈ R, 1 ≤ j ≤ n.

First we note that for A ∈ Ir,n such that aj < bj for all 1 ≤ j ≤ n it follows that A is a non-degenerate compact cell and Q further that the “natural” volume for A and A is equal and given by nj=1 (bj − aj ), i.e. for λ(n) we will have λ(n) (A) = λ(n) (A). By switching from compact cells to elements of Ir,n we gain that we can find partitions of A into elements of Ir,n : each family of partitions of [aj , bj ) induces a partition of A into elements of Ir,n , compare with II.18. We now 23

A COURSE IN ANALYSIS

introduce the n-dimensional figures F (n) as finite unions of elements belonging to Ir,n : ( ) M [ (n) n F := F ∈ R F = Ak , Ak ∈ Ir,n . (2.1) k=1

Here are some examples of elements on F (2) :

A2

A1

F =A

F = A1 ∪ A2

Figure 2.1

Figure 2.2

A1 A3

A1

A2 A2

F = A1 ∪ A2 ∪ A3

F = A1 ∪ A2 Figure 2.4

Figure 2.3 24

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

Given F ∈ F (n) such that F = we can easily define

SM

Ak for mutually disjoint sets Ak ∈ Ir,n

k=1

λ(n) (F ) :=

M X

λ(n) (Ak ).

(2.2)

k=1

By standard arguments we often used in II.18 and which we can easily modify to this situation (see Problem 1) it is clear that the definition of λ(n) (F ) is independent of the choice of the partition (Ak )k=1,...,M . In order to study the structure of F (n) let us introduce some helpful notation. For a, b ∈ Rn we write a ≤ b if and only if aj ≤ bj for all 1 ≤ j ≤ n, and

n

[a, b) :=

×[a , b ). j

j

j=1

(0)

(1)

(kj )

Given partitions aj = aj < aj < . . . < aj [aj , bj ) =

kj −1 h

[

(l)

(l+1)

aj , aj

l=0

= bj of [aj , bj ), 

,

we obtain a partition of [a, b).  More precisely we find with α = (α1 , . . . , αn ), αj ∈ {0, . . . , kj − 1} and Q := α ∈ Nn0 αj ∈ {0, . . . , kj − 1} that [a, b) =

[

[a(α) , a(α+1) )

α∈Q

 h
 n (α ) (α +1) . By construction we have for α 6= where a(α) , a(α+1) := j=1 aj j , aj j 

0 0 α0 that a(α) , a(α+1) ∩ a(α ) , a(α +1) = ∅. We note further that [c, d) ⊂ [a, b) if and only if aj ≤ cj and dj ≤ bj for j = 1, . . . , n. Now let [c, d) ⊂ [a, b). Then a partition of [aj , bj ) is given by the points of {aj , cj , dj , bj } and therefore [
[a, b) = a(α) , a(α+1)

×

(0)

(1)

(2)

(3)

with aj = aj , aj = cj , aj = dj and aj = bj . Since [c, d) is one of the sets in this union it follows that [a, b)\[c, d) is a (disjoint) union of elements in Ir,n , hence [a, b)\[c, d) is a figure for [c, d) ⊂ [a, b). This is almost the proof of 25

A COURSE IN ANALYSIS

Lemma 2.1. For [a, b), [c, d) ∈ Ir,n it follows that [a, b) ∩ [c, d) ∈ F (n) and [a, b)\[c, d) ∈ F (n) . Proof. With a ∨ c := (max(a1 , c1 ), . . . , max(an , cn )) and b ∧ d := (min(b1 , d1 ), . . . , min(bn , dn )) we find that [a, b) ∩ [c, d) = [a ∨ c, b ∧ d) or otherwise [a, b) ∩ [c, d) = ∅, i.e. [a, b) ∩ [c, d) ∈ F (n) since ∅ = [a, a) ∈ F (n) . Since [a, b)\[c, d) = [a, b)\ ([a, b) ∩ [c, d)), and since we know by now that [a, b) ∩ [c, d) ∈ F (n) , in order to prove [a, b)\[c, d) ∈ F (n) we may assume that [c, d) ⊂ [a, b). But then we can apply our preceeding considerations. Lemma 2.2. Every figure F ∈ F (n) is a finite union of mutually disjoint elements belonging to Ir,n . S Proof. Let F = N k=1 Ak , Ak ∈ Ir,n . We choose A ∈ Ir,n and such that F ⊂ A. Now the intervals [aj,k , bj,k ), 1 ≤ j ≤ n and 1 ≤ k ≤ N, induce a partition n n of the intervals [aj , bj ) where Ak = j=1[aj,k , bj,k ) and A = j=1 [aj , bj ), see Figure 2.5.

×

×

b2

A

b2,2

A2 b2,1 b2,3

A1

a2,2

A3

a2,3 a2,1 a2 a1

a1,1

a1,2 b1,1

a1,3

b1,2

b1,3

b1

Figure 2.5 Hence we obtain a partition (Kα )α=1,...,M , Kα ∈ Ir,n of A. We observe that F =

N [

k=1

Ak =

M N [ [

k=1 α=1

26

(Kα ∩ Ak )

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

and Kα ∩ Ak ∈ Ir,n . Moreover, for α 6= α0 it follows that (Kα ∩ Ak ) ∩ (Kα0 ∩ Ak ) = ∅, and by the construction of the sets Kα , if (Kα ∩Ak )∩(Kα ∩Ak0 ) 6= ∅ then they must coincide which implies the lemma. Definition 2.3. Let Ω 6= ∅ be a set. We call R ⊂ P(Ω) a ring in Ω if ∅∈R A, B ∈ R implies A\B ∈ R; A, B ∈ R implies A ∪ B ∈ R.

(2.3) (2.4) (2.5)

If in addition Ω ∈ R then R is called an algebra in Ω. Remark 2.4. A. Since A ∩ B = A\(A\B) it follows for a ring R that A, B ∈ R implies A ∩ B ∈ R.

(2.6)

B. A family R ⊂ P(Ω) is an algebra if and only if Ω ∈ R, A ∈ R implies A{ ∈ R and A, B ∈ R implies A ∪ B ∈ R. Indeed, for an algebra these conditions are obvious since A{ = Ω\A. Conversely, since Ω{ = ∅ and A\B = A ∩ B { = (A{ ∪ B){ we deduce that these conditions imply (2.3) - (2.5). Now we claim Theorem 2.5. The n-dimensional figures F (n) form a ring. Proof. Since follows that ∅ ∈ F . Moreover, for F1 , F2 ∈ F (n) SM[a, a) = ∅ it S with F1 = k=1 Ak , F2 = N l=1 Bl , where Ak , Bl ∈ Ir,n , we obtain F1 ∪ F2 =

N M [ [

(Ak ∪ Bl ) ∈ F (n) .

k=1 l=1

In order to see that F1 \F2 ∈ F (n) for F1 , F2 ∈ F (n) we now assume that in the representations of F1 and F2 used above the families (Ak )1≤k≤M and (Bl )1≤l≤N consist of mutually disjoint sets Ak , Bl ∈ Ir,n , which in light of Lemma 2.2 can be done. This implies ! N M \ [ Ak \Bl F1 \F2 = k=1

TN

l=1

and it remains to prove that l=1 (Ak \Bl ) ∈ F (n) . We already know that Ak \Bl ∈ F (n) , hence we need to prove that the intersection of two, hence 27

A COURSE IN ANALYSIS

finitely many figures is a figure. But for two figures F1 and F2 with the given representation we have F1 ∩ F2 =

N M [ [

(Ak ∩ Bl )

k=1 l=1

and Ak ∩ Bl ∈ F (n) by Lemma 2.1, and the theorem follows. Since F (n) is not a σ-field, we cannot define a measure on F (n) . However it turns out that λ(n) is on F (n) a pre-measure in the sense of the following: Definition 2.6. Let R be a ring in Ω. A. We call µ : F (n) → [0, ∞] a content if µ(∅) = 0 and µ

N [

l=1

Al

!

=

N X

(2.7)

µ(Al )

for every finite family (Al )l=1,...,N of mutually disjoint sets Al ∈ R. B. If (2.7) holds for µ : F (n) → [0, ∞] and ! ∞ ∞ [ X Al = µ µ(Al ) l=1

(2.8)

l=1

(2.9)

l=1

for every countable family (Al )l∈N of mutually disjoint sets Al ∈ R such that S ∞ l=1 Al ∈ R then we call µ a pre-measure on R (or on (Ω, R)).

Exercise 2.7. If (µk )k∈N is a sequence of pre-measures on aP ring R and (ak )k∈N , ak ≥ 0, a sequence of non-negative numbers then µ := ∞ k=1 ak µk is a pre-measure on R too. Remark 2.8. A pre-measure on a σ-field is a measure, and if (Ω, A, µ) is a measure space and R ⊂ A a ring, then µ|R is a pre-measure. As in the case of a measure, compare with the end of Chapter 1, we can prove (2.10) - (2.12) in the following: 28

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

Lemma 2.9. Let R be a ring in Ω and µ : R → [0, ∞) a pre-measure. For A, B ∈ R we have µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B), A ⊂ B implies µ(A) ≤ µ(B), A ⊂ B and µ(A) < ∞ implies µ(B\A) = µ(B) − µ(A), and µ

∞ [

j=1

Aj

!

≤

∞ X

µ(Aj )

(2.10) (2.11) (2.12)

(2.13)

j=1

for any, not necessarily mutually disjoint, sequence (Aj )j∈N , Aj ∈ R such that S ∞ j=1 Aj ∈ R. Furthermore for A0 ∈ R and a sequence (Ak )k∈N , Ak ∈ R, we have ∞ [ X A0 ⊂ Ak implies µ(A0 ) ≤ µ(Ak ). (2.14) k∈N

k=1

S

Proof. It remains to prove (2.14). Since A0 = k∈N (AS0 ∩ Ak ) and since by (2.11) that A0 = k∈N Ak . With B1 := Sµ(A0 ∩ Ak ) ≤ µ(Ak ) we may assume S S k−1 Ak \ k∈N Bk , but Bk ∩ Bl = ∅ k∈N Ak = j=1 Aj , k > 1, we find A0 = for k 6= l, and therefore ! N N N ∞ [ X X X Bk = µ µ(Bk ) ≤ µ(Ak ) ≤ µ(Ak ). k=1

k=1

k=1

k=1

 S 
S N The sequence µ is increasing with upper bound µ k∈N Bk k=1 Bk k∈N
S = µ k∈N Ak implying now ! ∞ [ X Ak ≤ µ(A0 ) ≤ µ µ(Ak ). k∈N

k=1

Note that (2.14) yields in particular for all (Ak )k∈N , Ak ∈ R such that S k∈N Ak ∈ R, that ! ∞ X [ µ µ(Ak ). (2.15) Ak ≤ k=1

k∈N

29

A COURSE IN ANALYSIS

The next result deals with sequential continuity properties of a pre-measure and it will be needed to construct the Lebesgue pre-measure λ(n) on the ring F (n) . Theorem 2.10. Let µ be a pre-measure on the ring R. A. S For a sequence (Ak )k∈N , Ak ∈ R, such that Ak ↑ A, i.e. Ak ⊂ Ak+1 and k∈N Ak = A ∈ R, it follows lim µ(Ak ) = µ(A).

(2.16)

k→∞

B. Let T (Ak )k∈N , Ak ∈ R, be a sequence such that Ak ↓ A, i.e. Ak ⊃ Ak+1 and k∈N Ak = A. If µ(Ak ) < ∞ for all k ∈ N then lim µ(Ak ) = µ(A).

(2.17)

k→∞

Proof. A. With A0 := ∅ and Bk := Ak \Ak−1, k ∈ N, we now S have a sequence (Bk )k∈N of mutually disjoint sets Bk ∈ R such that A = k∈N Bk and Ak = Sk j=1 Bj . It follows that µ(A) =

∞ X

µ(Bk ) = lim

m→∞

k=1

m X

µ(Bk ) = lim µ(Am ).

k=1

m→∞

B. First we note that µ(A1 \Ak ) = µ(A1 ) − µ(Ak ) and Ak ↓ A yields (A1 ∩ Ak ) ↑ (A1 ∩ A) and all these sets belong to R. Now by part A we find µ(A1 \A) = lim (A1 \Ak ) = µ(A1 ) − lim µ(Ak ), k→∞

k→∞

and since A ⊂ Ak and µ(A) < ∞ we have µ(A1 \A) = µ(A1 ) − µ(A) implying the result. Corollary 2.11. Let µ be a pre-measure on the ring R. The statement of part B is equivalent to the following: if (Ak )k∈N is a sequence in R such that Ak ↓ ∅ and µ(Ak ) < ∞ for all k ∈ N then lim µ(Ak ) = 0

k→∞

holds. 30

(2.18)

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

Proof. Of course part B of Theorem 2.10 implies the statement of the corollary. Now let Ak ↓ A, µ(Ak ) < ∞. Then Ak \A ↓ ∅, and since Ak \A ⊂ Ak it follows that µ(Ak \A) < ∞. Hence, by (2.18) we have lim µ(Ak \A) = 0

k→∞

and from µ(Ak \A) = µ(Ak ) − µ(A) we now deduce that lim µ(Ak ) = µ(A).

k→∞

It is helpful to introduce some definitions: Definition 2.12. A. For a measure or a pre-measure we call the property ! ∞ ∞ [ X Ak = µ µ(Ak ) k=1

k=1

for every countable, mutually disjoint family (Ak )k∈N of sets for which µ is defined the σ-additivity of µ. B. We call a pre-measure µ on a ring R in Ω (a measure µ on a σ-field A in Ω) σ-finite if there exists S a sequence (Ak )k∈N , Ak ∈ R (Ak ∈ A), such that µ(Ak ) < ∞ and Ω = ∞ k=1 Ak = Ω. C. In the situation of Theorem 2.10.A we call µ continuous from below, and in the situation of Theorem 2.10.B we call µ continuous from above. If Corollary 2.11 holds we call µ continuous at ∅ or just ∅-continuous. Corollary 2.13. Let (Ω, A) be a measurable space and µ : A → [0, ∞] a content. In addition assume that for µ the following holds (continuity from below): for every sequence (Ak )k∈N , Ak ∈ A, such that Ak ↑ A it follows that limk→∞ µ(Ak ) = µ(A). Then µ is a measure. Proof. Clearly µ(∅) = 0. Now let (Bk )k∈N be a sequence S disjoint S of mutually sets. Define Al := B1 ∪· · ·∪Bl ∈ A, and note that Al ↑ k∈N Bk = j∈N Aj = A ∈ A. By our assumption and (2.8) we find µ(A) = lim µ(Al ) = lim l→∞

l→∞

l X k=1

31

µ(Bk ) =

∞ X k=1

µ(Bk ),

A COURSE IN ANALYSIS

i.e. µ

[

k∈N

Bk

!

=

∞ X

µ(Bk )

k=1

and we have proved that µ is σ-additive on A, hence a measure. In order to solve our major problem we are now going to prove two central results: on the ring F (n) we can define λ(n) as a pre-measure and every σfinite pre-measure µ ˜ on a ring R in Ω has a unique extension to a measure µ on σ(R). Both results require some lengthy proofs. Let F ∈ F (n) and let SM F = k=1 Ak be any representation of F by a finite family (Ak )k=1,...,M of mutually disjoint sets Ak ∈ Ir,n . We already know that λ(n) (F ) :=

M X

λ(n) (Ak )

k=1

is independent of the choice of the family (Ak )k=1,...,M , see the remark following (2.2) or Problem 1. Theorem 2.14. On the ring F (n) a pre-measure is given by λ(n) . Proof. (Following E. Behrends [13].) Clearly we have λ(n) (∅) = 0. We need (n) to show the σ-additivity (Fk )k∈N is a family of mutually disjoint S of λ : if(n) (n) figures Fk ∈ F with k∈N F ∈ F then ! ∞ [ X λ(n) Fk = λ(n) (Fk ). k∈N

k=1

S S With F = k∈N Fk and (Fk )k∈N as above we know that Gm :=TF \ m l=1 Fl ∈ F (n) . For the sequence (Gm )m∈N we have G1 ⊃ G2 ⊃ · · · and m∈N Gm = ∅, i.e. Gm ↓ ∅. Moreover we have λ(n) (Gm ) = λ(n) (F ) −

m X

λ(n) (Fl ).

l=1

Thus if we can prove limm→∞ λ(n) (Gm ) = 0 the σ-additivity of λ(n) on F (n) will follow. Note that we are aiming to prove that λ(n) is ∅-continuous on (n) (n) F (n) . From our construction it follows that
λ (Gk ) ≤ λ (Gl ) for l ≤ k and (n) since the decreasing sequence λ (Gk ) k∈N is bounded from below by 0 it 32

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

must have a limit δ ≥ 0, limk→∞ λ(n) (Gk ) = δ ≥ 0. The aim is to prove that δ = 0. We will prove that if δ > 0 then exists a sequence of non-empty compact sets (Km )m∈N suchTthat Km ⊃ Km+1 and Km ⊂ Gm . From Theorem T II.2.15 weTdeduce that m∈N Km 6= ∅, however by construction we have m∈N Km ⊂ m∈N Gm = ∅ which is a contradiction. (n) AssumeSnow that δ > 0. In this case SMleach Gm is a union of elements Fl ∈ F , ∞ Gm = l=m+1 Fl 6= ∅, and Fl = ν=1 Alν where we can assume that Alν is a non-degenerate compact cell since otherwise λ(n) (Alν ) = 0. By shrinking Alν to A0lν such that A0lν is still a non-degenerate compact cell we can shrink Gm to G0m such that G0m ⊂ Gm and λ(n) (G0m ) ≥ λ(n) (Gm ) −

δ . 2m

(2.19)

The set Km := G01 ∩ · · · ∩ G0m is compact, Km ⊂ Gm and K1 ⊃ K2 ⊃ · · · . We need to prove that Km 6= ∅. For this it is sufficient to prove G01 ∩· · ·∩G0m 6= ∅ or equivalently λ(n) (G01 ∩ · · · ∩ G0m ) > 0. We show now by induction that   1 (n) 0 0 (n) λ (G1 ∩ · · · ∩ Gm ) ≥ λ (Gm ) − δ 1 − m (2.20) 2 which implies λ(n) (G01 ∩ · · · ∩ G0m ) ≥

δ 2m

(2.21)

since λ(m) (Gm ) ≥ δ. For m = 1 estimate (2.20) is just (2.19). Suppose that (2.20) holds for m. Using (2.10) we obtain

λ(n) G01 ∩ · · · ∩ G0m+1 =λ(n) (G01 ∩ · · · ∩ G0m )+λ(n) (G0m+1 )−λ(n) (G01 ∩ · · · ∩ G0m ) ∪ G0m+1

and we observe λ

(n)

(G01

∩ ···∩

G0m )

≥λ

(n)



1 (Gm ) − δ 1 − m 2

by induction hypothesis, λ(n) (G0m+1 ) ≥ λ(n) (Gm+1 ) − 33

δ 2m+1



A COURSE IN ANALYSIS

by construction, and since (G01 ∩ · · · ∩ G0m ) ∪ G0m+1 ⊂ Gm we have

which implies (n)

λ

(G01

∩ ···∩


λ(n) (G01 ∩ · · · ∩ G0m ) ∪ G0m+1 ≤ λ(n) (Gm ),

G0m+1 )

  1 δ ≥ λ (Gm ) − δ 1 − m + λ(n) (Gm+1 ) − m+1 − λ(n) (Gm ) 2 2   1 = λ(n) (Gm+1 ) − δ 1 − m+1 , 2 (n)

and hence the theorem is proved. In order to prove our central extension and uniqueness theorem, Carath´eodory’s theorem, we need some further tools. At certain stages we need to decide of a given system of subsets of Ω extending a given ring R in Ω that it is already a σ-field, or even more whether it is the σ-field generated by R. As it turns out Dynkin systems are best suited for such a purpose. Definition 2.15. We call D ⊂ P(Ω), Ω 6= ∅, a Dynkin system in Ω if Ω∈D

(2.22)

A ∈ D implies A{ ∈ D,

(2.23)

and for every sequence (Ak )k∈N of mutually disjoint sets Ak ∈ D we have [

k∈N

Ak ∈ D.

(2.24)

Remark 2.16. A. Clearly ∅ = Ω{ ∈ D and finite unions of mutually disjoint sets belonging to D are elements of D. B. Every σ-field is a Dynkin system. C. For A, B ∈ D, A ⊂ B it follows that A ∩ B { = ∅ ∈ D and A ∪ B { ∈ D which yields  { B\A = B ∩ A{ = A ∪ B { ∈ D. (2.25) Theorem 2.17. A Dynkin system D in Ω is a σ-field in Ω if and only if A, B ∈ D implies that A ∩ B ∈ D. 34

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

Proof. (Compare with H. Bauer [11].) It remains to prove that if D is a Dynkin system and A, B ∈ D Simplies A ∩ B ∈ D, then D is a σ-field, i.e. Ω ∈ D, A{ ∈ D for A ∈ D and k∈N Ak ∈ D for a sequence (Ak )k∈N , Ak ∈ D. The first two assertions are trivial due to the definition of a Dynkin system. Now let (Ak )k∈N , Ak ∈ D. We define further A00 := ∅, A0k := A1 ∪ · · · ∪ Ak . For k 6= l this implies



A0k+1 \A0k ∩ A0l+1 \A0l = ∅.

Moreover, since for any two sets B1 , B2 ∈ D we have B1 \B2 ∈ D and B1 ∪ B2 = (B1 \B2 ) ∪ B2 and (B1 \B2 ) ∩ B2 = ∅ we deduce that a finite union of elements of D belongs to D, hence A0k+1 \A0k ∈ D which now yields [

Ak =

k∈N

[

k∈N


A0k+1 \A0k ∈ D

proving the theorem. The next construction is similar to that of generating a σ-field or a topology. Proposition 2.18. For every family E ⊂ P(Ω), Ω 6= ∅, the family δ(E) :=

\

D ⊂ P(Ω) E ⊂ D and D is a Dynkin system

(2.26)

is a Dynkin system called the Dynkin system generated by E. Proof. We refer to Problem 6. The usefulness of Dynkin systems is due to Theorem 2.19. Let E ⊂ P(Ω) be a family of subsets of Ω, Ω 6= ∅, such that A, B ∈ E implies A ∩ B ∈ E. Then we have δ(E) = σ(E),

(2.27)

i.e. the Dynkin system generated by E is already the σ-field generated by E. 35

A COURSE IN ANALYSIS

Proof. (Compare with H. Bauer [11].) It is trivial that δ(E) ⊂ σ(E). Thus we have only to prove the converse inclusion. For this we will employ Theorem 2.17 and show that D, E ∈ E implies that D ∩ E ∈ E. For D ∈ δ(E) we define  DD := Q ∈ P(Ω) Q ∩ D ∈ δ(E) (2.28) and claim that DD is a Dynkin system. Since Ω ∩ D = D it follows (2.22) and since D { ∈ δ(E) we find for Q ∈ DD    { Q{ ∩ D = Q{ ∪ D { ∩ D = (Q ∩ D){ ∩ D = (Q ∩ D) ∪ D { ∈ δ(E)

implying Q{ ∈ DD , i.e. (2.23). We now prove (2.24) for DD . For this let Ak ∈ DD , k ∈ N, such that Ak ∩Al = ∅ for k 6= l. It follows that Ak ∩D ∈ δ(E) and (Ak ∩ D) ∩ (Al ∩ D) = ∅ for k 6= l. Since δ(E) is a Dynkin system we find ! [ [ (Ak ∩ D) ∈ δ(E) Ak ∩ D = S

k∈N

k∈N

implying that k∈N Ak ∈ DD , i.e. DD is indeed a Dynkin system. For E ∈ E we know by our assumption that E ∩ E 0 ∈ E, for every E 0 ∈ E hence E ⊂ DE and therefore δ(E) ⊂ δ(DE ) = DE . Further, for D ∈ δ(E) and E ∈ E we have E ∩ D ∈ δ(E), i.e. E ⊂ DD and therefore δ(E) ⊂ DD for every D ∈ δ(E), which is just a reformulation of the statement we want to prove and the theorem is shown. Following R. Schilling [75], as a first application of Dynkin systems, we prove a uniqueness result for measures. Theorem 2.20. Let (Ω, σ(G)) be a measurable space where G ⊂ P(Ω), Ω 6= ∅. Assume that C1 , C2 ∈ G implies C1 ∩ C2 ∈ G and that there exists a sequence of sets (Bk )k∈N of sets Bk ∈ G such that Bk ↑ Ω. Suppose µ and ν are two measures on σ(G) such that µ(B) = ν(B) for all B ∈ G and that µ(Bk ) = ν(Bk ) < ∞ for all k. Then µ = ν, i.e. µ(A) = ν(A) for all A ∈ σ(G). Proof. For k ∈ N we claim that

 Dk := A ∈ σ(G) µ(Bk ∩ A) = ν(Bk ∩ A) 36

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

is a Dynkin system. Note that by assumption µ(Bk ∩A) < ∞. Since Ω∩Bk = Bk it is clear that Ω ∈ Dk . For A ∈ Dk we find   µ Bk ∩ A{ = µ(Bk \A) = µ(Bk ) − µ(Bk ∩ A) = ν(Bk ) − ν(Bk ∩ A) = ν(Bk \A) = ν(Bk ∩ A{ ),

i.e. A ∈ Dk implies A{ ∈ Dk . Now let (Aj )j∈N be a sequence of mutually disjoint sets Aj ∈ Dk . It follows that ! ! [ [ (Bk ∩ Aj ) µ Bk ∩ Aj = µ j∈N

j∈N

=

∞ X j=1

=ν

µ(Bk ∩ Aj ) = [

(Bk ∩ Aj )

j∈N

=ν

Bk ∩

S

[

j∈N

Aj

∞ X

!

!

j=1

ν(Bk ∩ Aj )

,

i.e. j∈N Aj ∈ Dk . By Theorem 2.17 we know that δ(G) = σ(G) which yields that G ⊂ Dk implies δ(G) = σ(G) ⊂ Dk for all k ∈ N. However Dk ⊂ σ(G) and we deduce for all k ∈ N that Dk = σ(G) which implies µ(Bk ∩ A) = ν(Bk ∩ A)

(2.29)

for all k ∈ N and all A ∈ σ(G). From Theorem 2.10.A we obtain now as k → ∞ that µ(A) = lim µ(Bk ∩ A) = lim ν(Bk ∩ A) = ν(A). k→∞

k→∞

Next we want to address the problem of how to extend a pre-measure µ defined on a ring R over Ω, Ω 6= ∅. As a first step we use µ to define a set function µ∗ on P(Ω) which we later will call the outer (Carath´eodory) 37

A COURSE IN ANALYSIS

measure induced by µ. For this let G ∈ P(Ω) be any subset of Ω. By U(G) we denote the family of all sequences (Ak )k∈N , Ak ∈ R, such that G is covered by (Ak )k∈N , i.e. [ G⊂ Ak . (2.30) k∈N

∗

Now we define µ : P(Ω) → [0, ∞] by ( P ∞ inf ∗ k=1 µ(Ak ) (Ak )k∈N ∈ U(G) , if U(G) 6= ∅ µ (G) := +∞, if U(G) = ∅,

(2.31)

note that µ∗ (G) ≥ 0 is trivial. Since the sequence (Ak )k∈N , Ak = ∅, belongs to U(∅) and µ(∅) = 0, we deduce µ∗ (∅) = 0.

(2.32)

Further, if G1 ⊂ G2 then U(G2 ) ⊂ U(G1 ), and therefore it follows that G1 ⊂ G2 implies µ∗ (G1 ) ≤ µ∗ (G2 ).

(2.33)

Finally we claim that for every sequence (Gl )l∈N , Gl ∈ P(Ω), we have ! ∞ X [ ∗ µ∗ (Gl ). (2.34) µ Gl ≤ l=1

l∈N

In order to prove (2.34) we may assume that µ∗ (Gl ) < ∞ for all l ∈ N, in particular we assume U(Gl ) 6= ∅ for all l ∈ N. For  > 0 and l ∈ N we find now a sequence (Alk )k∈N ∈ U(Gl ) such that ∞ X k=1

µ(Alk ) ≤ µ∗ (Ql ) +

The sequence (Alk )l,k∈N belongs to U ! ∞ [ X µ∗ Gl ≤ µ(Alk ) l∈N

S

l∈N

l,k=1

=

∞ ∞ X X l=1

µ(Alk )

k=1

38

!

 . 2l


Gl and therefore

≤

∞ X l=1

∞ X  µ (Gl ) + 2l l=1 ∗

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

or µ

∗

[

Gl

l∈N

!

≤

∞ X

µ∗ (Gl ) + 

l=1

for all  > 0 implying (2.34). Definition 2.21. A. Let Ω 6= ∅ and µ∗ : P(Ω) → [0, ∞] be a set function satisfying (2.32) - (2.34). Then µ∗ is called an outer measure on P(Ω). B. For an outer measure µ∗ on P(Ω) we call A ⊂ Ω a µ∗ -measurable set or outer measurable set with respect to µ∗ if µ∗ (G) = µ∗ (G ∩ A) + µ∗ (G ∩ A{ )

(2.35)

holds for all G ∈ P(Ω). Proposition 2.22. Let Ω 6= ∅ and R be a ring in Ω as well as µ a premeasure on R. Denote by µ∗ the outer measure associated with µ by (2.31). Then every A ∈ R is µ∗ -measurable and for A ∈ R we have µ∗ (A) = µ(A).

(2.36)

Proof. For A ∈ R we have to prove (2.35) and for this we may assume µ∗ (G) < ∞, and in particular U(G) 6= ∅. Since µ is finite additive we have for (Ak )k∈N ∈ U(G) µ(Ak ) = µ(Ak ∩ A) + µ(Ak \A) implying

∞ X

µ(Ak ) =

k=1

∞ X k=1

µ(Ak ∩ A) +

∞ X k=1

(Ak \A).

Furthermore (Ak ∩ A)k∈N ∈ U(G ∩ A) and (Ak \A)k∈N ∈ U(G\A) which yields ∞ X k=1

µ(Ak ) ≥ µ∗ (G ∩ A) + µ∗ (G\A)

for any such a sequence (Ak )k∈N and this implies µ∗ (G) ≥ µ∗ (G ∩ A) + µ∗ (G ∩ A{ ). 39

A COURSE IN ANALYSIS

Now we consider the sequence A1 = G ∩ A, A2 = G\A and Ak = ∅ for k > 2. Applying (2.34) to this sequence we find µ∗ (G) = µ∗ ((G ∩ A) ∪ (G\A)) ≤ µ∗ (G ∩ A) + µ∗ (G\A), and hence (2.35). In order to prove (2.36) for A ∈ R we note that the sequence (Ak )k∈N , A1 = A, Ak = ∅ for k > 1, is an element of U(A) and therefore we have µ(A) ≥ µ∗ (A). On the other hand, applying (2.14) to A and (Ak )k∈N we have µ(A) ≤ µ∗ (A), i.e. µ(A) = µ∗ (A), and the proposition is proven. So far we have seen that a pre-measure µ on a ring R in Ω 6= ∅ can be extended to an outer measure µ∗ on P(Ω) and all elements of R are µ∗ measurable. Our final step is to prove that the µ∗ -measurable sets form a σ-field in Ω. In this proof we will again make use of Dynkin systems. Theorem 2.23. Let µ∗ be an outer measure on P(Ω), Ω 6= ∅. The system A∗ of all µ∗ -measurable sets A ⊂ Ω is a σ-field and µ∗ |A∗ is a measure.

Proof. It is obvious that (2.35) holds for Ω since G ∩ Ω = G and G\Ω = ∅, and that A ∈ A∗ if and only if A{ ∈ A∗ . Next we prove that A∗ is an algebra, i.e. taking into account the previous observation, that A, B ∈ A∗ implies A ∪ B ∈ A∗ . We know that µ∗ (G) = µ∗ (G ∩ A) + µ∗ (G\A)

(2.37)

µ∗ (G) = µ∗ (G ∩ B) + µ∗ (G\A)

(2.38)

for G ∈ P(Ω) and A ∈ A∗ , thus we have also as well as

µ∗ (G ∩ A) = µ∗ ((G ∩ A) ∩ B) + µ∗ ((G ∩ A)\B) and µ∗ (G\A) = µ∗ ((G\A) ∩ B) + µ∗ ((G\A) ∩ B)

where we replaced in (2.38) G by G ∩ A and G\A, respectively. Combined with (2.37) this yields µ∗ (G) = µ∗ (G ∩ A) + µ∗ (G\A)

= µ∗ (G ∩ A ∩ B) + µ∗ (G ∩ A ∩ B { )

+ µ∗ (G ∩ A{ ∩ B) + µ∗ (G ∩ A{ ∩ B { ) 40

(2.39)

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

and replacing G by G ∩ (A ∪ B) we find µ∗ (G ∩ (A ∪ B)) = µ∗ (G ∩ A ∩ B) + µ∗(G ∩ A ∩ B {) + µ∗(G ∩ A{ ∩ B) (2.40) and with (2.39) µ∗ (G) = µ∗ (G ∩ (A ∪ B)) + µ∗ (G ∩ (A ∪ B){ ). Since G ∈ P(Ω) was arbitrary we deduce A ∪ B ∈ A∗ . Eventually we want to prove that A∗ is a Dynkin system stable under finite intersections which implies by Theorem 2.19 that A∗ is a σ-field. For S this let (Ak )k∈N be a sequence of mutually disjoint sets Ak ∈ A∗ and A := k∈N Ak . If we choose in (2.40) A = A1 and B = A2 we find using that A1 ∩ A2 = ∅ that µ∗ (G ∩ (A1 ∪ A2 )) = µ∗ (G ∩ A1 ) + µ∗ (G ∩ A2 ), and by induction we obtain µ

∗

G∩

N [

k=1

Ak

!

=

N X k=1

µ∗ (G ∩ Ak )

S ∗ for all N ∈ N and G ∈ P(Ω). We know already that BN := N k=1 Ak ∈ A ∗ ∗ and G\A ⊂ G\BN , i.e. µ (G\A) ≤ µ (G\BN ), therefore we get for all N ∈ N µ∗ (G) = µ∗ (G ∩ BN ) + µ∗ (G\BN ) ≥

N X k=1

µ∗ (G ∩ Ak ) + µ∗ (G\A)

and by (2.34) it follows that µ∗ (G) ≥

∞ X k=1

µ∗ (G ∩ Ak ) + µ∗ (G\A) ≥ µ∗ (G ∩ A) + µ∗ (G\A).

As in the proof of Proposition 2.22 we further conclude that µ∗ (G) =

∞ X k=1

µ∗ (G ∩ Ak ) + µ∗ (G\A) = µ∗ (G ∩ A) + µ∗ (G\A), 41

(2.41)

A COURSE IN ANALYSIS

S i.e. A = k∈N Ak ∈ A∗ . Thus we have proved that A∗ is a Dynkin system stable under finite intersections, i.e. it is a σ-field. If we choose in (2.41) now G = A then we find ! ∞ X [ ∗ ∗ µ∗ (Ak ), µ (A) = µ Ak = k=1

k∈N

i.e. µ∗ is on A∗ indeed a measure. Combining Theorem 2.23 with Theorem 2.20 and Proposition 2.22 we arrive at Carath´eodory’s theorem. Theorem 2.24. Every σ-finite pre-measure µ on a ring R in Ω, Ω 6= ∅, has a unique extension to a measure µ ˜ on σ(R). Proof. We know by Proposition 2.22 that R ⊂ A∗ , hence σ(R) ⊂ A∗ and therefore µ ˜ := µ∗ |σ(R) is a measure on σ(R). Since by assumption µ is σ-finite on R, R and µ have all properties required in Theorem 2.20. Finally we return to the pre-measure λ(n) on F (n) . We know from Theorem 2.14 that λ(n) is a σ-finite pre-measure on the ring F (n) . Hence according to Theorem 2.24 it has a unique extension as a measure on σ(F (n) ). For simplicity we will denote this extension again by λ(n) . Lemma 2.25. For n ∈ N we have σ(F (n) ) = B(n) . Proof. This result can be deduced from Exercise 1.22, but due to its importance we give the proof here (once again). Since F (n) ⊂ B(n) we have σ(F (n) ) ⊂ B(n) . In order to prove the converse inclusion we note first that every non-empty open set U ⊂ Rn is a countable ounion of elements belonn n ging to (a, b) ⊂ Rn (a, b) = j=1(aj , bj ), aj < bj . For example we may consider all sets (a, b) with aj , bj ∈ Q such that (a, b) ⊂ U. (This is purely a topological problem, see Problem 8, which we have encountered before several times). In addition we have  [ 1 (a, b) = a + h, b , h = (1, 1, . . . , 1) ∈ Rn , k

×

k∈N

which now implies On ⊂ σ(F (n) ), hence B(n) ⊂ σ(F (n) ) proving the lemma. 42

2

´ EXTENDING PRE-MEASURES. CARATHEODORY’S THEOREM

Theorem 2.26. On the Borel σ-field B(n) exist a unique measure λ(n) with the property that for all [a, b) ∈ Ir,n the following holds n

λ(n) ([a, b)) =

×(b − a ). j

j

j=1

Definition 2.27. The measure λ(n) is called the Lebesgue-Borel measure on B(n) or Rn . In the next chapter we also introduce a measure called Lebesgue measure. Later on we will (as do many authors) often speak about the Lebesgue measure when we actually mean the Lebesgue-Borel measure.

Problems (n) 1. Let F ∈ F (n) be an n-dimensional figure. Prove is indeSM that λ (F ) S pendent of the partition of F , i.e. if F = k=1 Ak and F = N l=1 Bl , PN (n) P (n) λ (B ). λ (A ) = Ak , Bl ∈ Ir,n , then M l k l=1 k=1

2. Let Ω be a non-denumerable set and define

R := {A ∈ P(Ω)|A or A{ is finite}. Prove that R is a ring in Ω and that µ : R → [0, ∞] defined on R by ( 0, A is finite µ(A) := 1, A{ is finite is a pre-measure on R. 3. Solve Exercise 2.7. 4. Let R be a ring in Ω and (µk )k∈N a sequence of pre-measures on R satisfying µk (A) ≤ µk+1 (A) for all A ∈ R. Prove that by µ(A) := supk∈N µk (A) a further pre-measure is defined on R. 5. Let R be a ring in Ω and µ a pre-measure on R. Further S let (Ak )k∈N , Ak ∈ R, be a sequence such that µ(Ak ) < ∞ and Ω = k∈N Ak , i.e. R ˜ ˜ is σ-finite with respect to µ. Construct S a sequence (Ak )k∈N , Ak ∈ R, such that A˜k ⊂ A˜k+1 , µ(A˜k ) < ∞ and k∈N A˜k = Ω. 43

A COURSE IN ANALYSIS

6. Prove Proposition 2.18. 7. Let Ω be a non-empty set and G ⊂ H ⊂ P(Ω). Show that δ(G) ⊂ δ(H) and δ(G) ⊂ σ(G). 8. Let U ⊂ Rn be open. Prove that U is the countable union of open rectangles with rational vertices contained in U. Hint: re-interpret Problem 5 of Chapter 1.

44

3

The Lebesgue-Borel Measure and Hausdorff Measures

The Lebesgue-Borel measure and its extension to the Lebesgue measure is a fundamental object of analysis and in this chapter we investigate more of its properties and extensions. Most of all we are interested in “geometric” properties, for example invariance properties. We need Lemma 3.1. Let B ∈ B(n) be a Borel set, x0 ∈ Rn and T ∈ O(n). Then x + B and T (B) are Borel sets too. Proof. From Corollary 1.18 we know that continuous mappings f : Rn → Rn are B(n) /B(n) -measurable. Since gx0 : Rn → Rn , g(y) = y − x0 is continuous and x0 + B = gx−1 (B), it follows that x0 + B ∈ B(n) . Further we know that 0 T ∈ O(n) is a homeomorphism implying that T −1 establishes a bijective correspondence of On with itself, i.e. T −1 U ∈ On for every U ∈ On , and every V ∈ On is of type V = T −1 U, U ∈ On . Since On generates B(n) we now deduce that B = T −1 (T (B)) is Borel-measurable for every B ∈ B(n) . Theorem 3.2. The Lebesgue-Borel measure λ(n) is translation invariant in the sense that λ(n) (x0 +B) = λ(n) (B) for all x0 ∈ Rn and B ∈ B(n) . Moreover, every translation invariant measure µ on B(n) for which µ ([0, 1)n ) is strictly positive and finite is a positive multiple of λ(n) , i.e. 0 < µ ([0, 1)n ) < ∞ implies for some constant cµ > 0 that µ(B) = cµ λ(n) (B) for all B ∈ B(n) .

×

×

n

n

Proof. For A = j=1 [aj , bj ) it follows that x0 + A = j=1[aj + x0j , bj + x0j ) Q Q and consequently λ(n) (x0 +A) = nj=1 (bj +x0j −(aj +x0j )) = nj=1 (bj −aj ) = λ(n) (A), which extends to all A ∈ F (n) . On the other hand µx0 : B(n) → [0, ∞] (n) (n) defined S by µx0 (B)S:= λ (x0 + B) is a measure(n)on B since x0 + ∅ = ∅ and the measures µx0 and λ(n) x0 + k∈N Bk = k∈N (x0 + Bk ). Since on F coincide it follows that µx0 = λ(n) on B(n) . In order to prove the second assertion we first note that it is sufficient to prove it for F (n) , hence for n Ir,n and thus for all A := j=1[aj , bj ), aj , bj ∈ Q. For each A we can find N, m1 , . .Q . , mn ∈ N such that Nmj = bj − aj and therefore we can cover A by m = nj=1 mj mutually disjoint half-open cubes with side length N1 . Each 

such a cube is a translation of the cube CN := 0, N1 × · · · × 0, N1 ⊂ Rn . Therefore we find by the translation invariance of µ and λ(n) that

×

µ(A) = m µ(CN ) and λ(n) (A) = m λ(n) (CN ) 45

A COURSE IN ANALYSIS

as well as µ(C1 ) = N n µ(CN ) and 1 = λ(n) (C1 ) = N n λ(n) (CN ). Thus

m Nn

= λ(n) (A) and this applies

m µ(C1 ) = µ(C1 )λ(n) (A) Nn proving the theorem with cµ = µ(C1 ).
Corollary 3.3. On Rn , B(n) the Lebesgue-Borel measure λ(n) is the unique translation invariant measure satisfying λ(n) (C1 ) = 1. µ(A) = m µ(CN ) =

Remark 3.4. We will use Theorem 3.2 in Appendix II to discuss the existence of non-Borel-measurable sets in Rn , i.e. of sets A ⊂ Rn , A ∈ / B(n) .

Invariance of measures is an important property. Before we prove that the Lebesgue-Borel measure is also invariant under the operation of O(n), we want to give an example of a “translation” invariant measure on Zn . Example 3.5. (Also see Problem 9.b) in Chapter 1) For a finite set A denote by #(A) the number of its elements. On Zn we consider P(Zn ) as σ-field and define the measure X µZn (A) := k (A). (3.1) k∈Zn

Clearly this is a measure with µZn (A) = #(A) for a finite set A ⊂ Zn and µZn (A) = +∞ otherwise. For l0 ∈ Zn fixed we can consider the translated set l0 + A for which we obtain X X X µZn (l0 + A) = k (l0 + A) = m (A), k−l0 (A) = k∈Zn

k−l0 ∈Zn

m∈Zn

i.e. µZn is on (Zn , P(Zn )) a measure invariant under the group translations of (Zn , +). Later we will discuss the existence of invariant measures on certain topological groups in more detail. Our intuition tells us that λ(n) is the “correct” object to measure the volume of (certain) subsets of Rn . Therefore for A ∈ B(n) the set T (A), T ∈ O(n), and the set A should have the same volume, i.e. we expect λ(n) (T (A)) = λ(n) (A)

(3.2)

for all A ∈ B(n) and all T ∈ O(n). We have already seen in Lemma 3.1 that T (A) ∈ B(n) for A ∈ B(n) and T ∈ O(n). 46

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Theorem 3.6. The Lebesgue-Borel measure λ(n) is invariant under the operation of O(n), i.e. (3.2) holds. Proof. Let T ∈ O(n) and A ∈ B(n) . Since T −1 ∈ O(n) it follows that T −1 (A) ∈ B(n) and we can define on B(n)
(n) (3.3) κT (A) := λ(n) T −1 (A) . (n)

Clearly, κT (∅) = 0 and for a sequence of mutually disjoint sets Ak ∈ B(n) , k ∈ N, we have ! !! [ [ (n) κT Ak = λ(n) T −1 Ak k∈N

k∈N

=λ

(n)

[

T

−1

(Ak )

k∈N

=

∞ X

!

=

∞ X

λ(n) T −1 (Ak )

k=1




(n)

κT (Ak ),

k=1

(n)

where we used T −1 (Ak )∩T −1 (Al ) = ∅ for Ak ∩Al = ∅. Hence κT is a measure (n) on B(n) . Next we prove that κT is translation invariant. For x0 ∈ Rn and A ∈ B we find (n)

κT (x0 + A) = λ(n) (T −1 (x0 + A))


= λ(n) (T −1 x0 ) + T −1 (A) = λ(n) (T −1 (A)) (n)

= κT (A),

where we used that T −1 is linear and λ(n) is translation invariant. For all R ∈ O(n) it follows R (Bρ (0)) = Bρ (0), Bρ (0) = {x ∈ Rn | kxk < ρ}, and with C1 = [0, 1) × · · · × [0, 1) ∈ Rn we have R(C1 ) ⊂ B√n (0). This implies
λ(n) (R(C1 )) ≤ λ(n) B√n (0) < ∞. Let y0 be the midpoint of C1 . Then B 1 (y0 ) ⊂ C1 and the translation invariance of λ(n) yields 4       0 < λ(n) B 1 (0) = λ(n) y0 + B 1 (0) = λ(n) B 1 (y0 ) ≤ λ(n) (C1 ). 4

4

4

(n)

Thus by Theorem 3.2 it follows that κT (A) = Cκ(n) λ(n) (A), but for A = T

(n)

B1 (0) we must have κT (B1 (0)) = λ(n) (B1 (0)) since B1 (0) is invariant under 47

A COURSE IN ANALYSIS (n)

T −1 ∈ O(n). It follows that Cκ(n) = 1, i.e. κT T replacing A by T (A) in (3.3) we obtain (3.2).

= λ(n) on B(n) and by

A second look at the proof of Theorem 3.6 reveals that (3.3) suggests a universal construction for a measure given a measure and a measurable mapping. Theorem 3.7. Let (Ω, A) and (Ω0 , A0) be two measurable spaces and f : Ω → Ω0 a measurable mapping. If µ is a measure on A then f (µ) : A0 → [0, ∞] defined by  (3.4) f (µ)(A0 ) := µ(f −1 (A0 )) = µ ω ∈ Ω f (ω) ∈ A0 , A0 ∈ A0, is a measure on A0 .

Proof. Clearly f (µ)(∅) = 0 and as in the proof of Theorem 3.6 we conclude for a sequence (A0k )k∈N of mutually disjoint sets A0k ∈ A0 , k ∈ N, that ! !! ∞ ∞ [ [ X X 0 −1 0 −1 0 f (µ) Ak = µ f Ak = µ(f (Ak )) = f (µ)(A0k ), k∈N

k∈N

k=1

k=1

proving the theorem. Definition 3.8. The measure f (µ) on A0 constructed in Theorem 3.7 by (3.4) is called the image of µ under f , or the image measure of µ under f . Instead of f (µ) we often will write µf . From Remark 1.16.C we deduce immediately Corollary 3.9. Let (Ωj , Aj ), j = 1, 2, 3, be three measurable spaces and f : Ω1 → Ω2 , g : Ω2 → Ω3 two measurable mappings. For a measure µ on A1 we have (g ◦ f )(µ) = g(f (µ))

where (g ◦ f )(µ) is a measure on A3 and f (µ) is a measure on A2 .

As usual we call the group generated by the group of translations on Rn and O(n) the group of motions on Rn . This is the group of all transformations on Rn leaving the Euclidean distance invariant. Combining Corollary 3.9 with Theorem 3.2 and Theorem 3.6 we have Corollary 3.10. The Lebesgue-Borel measure λ(n) is invariant under all motions on Rn . 48

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THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Next we calculate T (λ(n) ) for T ∈ GL(n; R). Theorem 3.11. For all A ∈ B(n) and T ∈ GL(n; R) the following holds λ(n) (T (A)) = |det T | λ(n) (A), i.e. T (λ(n) ) = Proof. First let A =

×

n [a , b ) j=1 j j



    Dk (r) =    

(3.5)

1 λ(n) . |det T |

(3.6)

and T = Dk (r), r > 0, 0 ≤ k ≤ n, where

1.



..



1 r 1.

0

..

0     

1

with r in position (k, k). It follows that k−1

Dk (r)(A) =

n

× j=1

[aj , bj ) × [rak , rbk ) ×

× [a , b ), j

j

j=k+1

and therefore λ(n) (Dk (r)(A)) = rλ(n) (A) = |det Dk (r)| λ(n) (A). For D(r1 , . . . , rn ) = D1 (r1 ) · · · Dn (rn ) = (rk δkl )k,l=1,...,n , rk > 0, we find λ(n) (D(r1 , . . . , rn )(A)) = r1 · . . . · rn λ(n) (A) = |det D(r1 , . . . , rn )| λ(n) (A). (3.7)

It follows that (3.7) holds for all A ∈ F (n) and hence for all A ∈ B(n) . By the polar decomposition theorem for T ∈ GL(n; R), Theorem II.A.I.28, we can find U ∈ O(n) and a symmetric positive definite matrix S such that T = US. Further, since S is symmetric, there exists V ∈ O(n) such that S = V −1 D(r)V with r1 , . . . , rn being the eigenvalues of S, rk > 0. Using the 49

A COURSE IN ANALYSIS

transitivity of the image measure, i.e. Corollary 3.9, and the invariance of λ(n) under the actions of O(n) we now get for all A ∈ B(n)
λ(n) (T (A)) = λ(n) UV −1 D(r)V (A) = λ(n) (D(r)V (A)) = |det D(r)| λ(n) (V (A))

= |det D(r)| λ(n) (A). Since

det T = det U · det V −1 det D(r) det V = det U det D(r)

and since |det U| = 1 we find |detT | = det D(r) = |det D(r)|, and (3.5) follows. Taking in (3.5) for A now T −1 (A) we arrive at (3.6). Remark 3.12. In Chapter II.16 when starting with our considerations on the Riemann integral in Rn we discussed parallelotops P (a1 , . . . , an ) defined for n linearly independent vectors aj ∈ Rn as ) ( n X n λj aj , λj ∈ [0, 1] , P (a1 , . . . , an ) := x ∈ R x = j=1

and we have seen that with T ∈ GL(n; R) defined by T ej = aj we always have P (a1 , . . . , an ) = T (P (e1 , . . . , en )),

but P (e1 , . . . , en ) = [0, 1] × · · · × [0, 1] ⊂ Rn . Thus we arrive again at (n) λ (P (a1 , . . . , an )) = |det T |λ(n) (P (e1 , . . . , en )) = |det T | = |det (a1 , . . . , an )| (3.8) where we used our old notation T = (a1 , . . . , an ). Clearly (3.8) is nothing but (II.16.8). Our notion of Jordan measurable sets which was key for developing the theory of Riemann integrals in Rn depends on the concept of Lebesgue null sets, see Definition II.19.6. Now we can put this concept in a more natural context. Definition 3.13. Let (Ω, A, µ) be a measure space. By Nµ we denote the sets of µ-measure zero, or just µ-zero sets, i.e.  Nµ := A ∈ A µ(A) = 0 . (3.9) A subset B of a set of µ-measure zero is called a µ-null set, compare with Definition I.32 and Definition II.19.6. 50

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Lemma 3.14. A. Clearly ∅ ∈ Nµ and if M ∈ A and M ⊂ N for some N ∈ Nµ then M ∈ Nµ . B. The union of a denumerable family S of µ-zero sets is a µ-zero set, i.e. for (Nk )k∈N , Nk ∈ Nµ , it follows that k∈N Nk ∈ Nµ . C. The union of a denumerable family of µ-null sets is a µ-null set. Proof. Part A is trivial, but note that the condition M ∈ A is essential, otherwise µ(M) may not be defined. Now, given a sequence (Nk )k∈N , Nk ∈ Nµ . For  > 0 we have µ(Nk ) < 2k and therefore ! ∞ ∞ [ X X 1 = Nk ≤ µ µ(Nk ) <  2k k∈N k=1 k=1 implying µ

S

k∈N


Nk = 0. Part C is obvious from Part B.

It is important to realise that two measures on one and the same σ-field A may have quite different sets of measure zero. Example 3.15. A. Let (Ω, A) be a measurable space and ω0 ∈ Ω. We consider the measure space (Ω, A, ω0 ). A set A ∈ A belongs to Nω0 if and only if ω0 ∈ / A, i.e. Nω0 = {A ∈ A | A ⊂ {ω0 }{ }. B. For every point x0 ∈ Rn we have λ(n) ({x0 }) = 0, i.e. {x0 } ∈ Nλ(n) . To see this, for  > 0 choose the cube C (x0 ) with centre x0 and side length . We have C (x0 ), {x0 } ∈ B(n) and λ(n) ({x0 }) ≤ λ(n) (C (x0 )) ≤ n implying {x0 } ∈ Nλ(n) . Combined with Lemma 3.14.B we now find that Nn , Zn , Qn ∈ Nλ(n) for n ∈ N. This is due to the fact that these are countable sets. C. When we now take in Part A the space (Rn , B(n) ) as the measurable space (Ω, A) and ω0 = x0 ∈ Rn we see immediately that Nx0 6= Nλ(n) since any non-degenerate compact cube K with x0 ∈ / K has positive Lebesgue measure, however x0 (K) = 0. Recall Definition II.19.6: A ⊂ Rn was said to be a Lebesgue null set if for everyP  > 0 exists a denumerable covering (Kj )j∈N of A by open cells such (n) (Kj ) < . In Problem 7 we will see that for sets A ∈ B(n) this that ∞ j=1 λ definition coincides with Definition 3.13. However in Definition II.19.6 the 51

A COURSE IN ANALYSIS

set A is not necessarily a Borel set. Thus we may have the situation that null sets in the sense of Definition II.19.6 are not elements in Nλ(n) since they do not belong to the Borel σ-field. As in general µ-null sets do not necessarily belong to A, in general they do not belong to Nµ . To cope with this problem we give Definition 3.16. A measure space (Ω, A, µ) is called complete if every subset of a set of measure zero belongs to A and hence is a set of measure zero. Theorem 3.17. Let (Ω, A, µ) be a measure space. Denote by Nµ all sets of measure zero and by N˜µ all subsets of Nµ , i.e. N ∈ N˜µ if there exists N0 ∈ Nµ such that N ⊂ N0 . The family o n A˜ := A ∪ N A ∈ A and N ∈ N˜µ

˜ Moreover µ is a σ-field in Ω and we have A ⊂ A. ˜ : A˜ → [0, ∞] defined by ˜ µ µ ˜(A ∪ N) := µ(A) is a measure defined on A˜ and (Ω, A, ˜) is a complete measure space with µ ˜|A = µ. ˜ µ Definition 3.18. The measure space (Ω, A, ˜) is called the completion of (Ω, A, µ).

Proof of Theorem 3.17. First we note that the representation A ∪ N for an element is not unique. For N0 ∈ Nµ we have (A\N0 ) ∪ (N ∪ N0 ) = A ∪ N, ˜ µ(N) ˜ = 0. In the following, when dealing with A\N0 ∈ A and N ∪ N0 ⊂ N, elements of A˜ we always start with a representation A ∪ N and we take care that results are independent of the special choice. Clearly A = A ∪ ∅ ∈ A˜ for all A ∈ A. Moreover for N ⊂ N0 ∈ Nµ we find with A ∈ A that   (A ∪ N){ = A{ ∩ N { = A{ ∩ N { ∩ N0{ ∪ N0     = A{ ∩ N { ∩ N0{ ∪ A{ ∩ N { ∩ N0     = A{ ∩ N0{ ∪ A{ ∩ N { ∩ N0 .

Since A{ ∩ N0{ ∈ A and A{ ∩ N { ∩ N0 ⊂ N0 ∈ Nµ it follows that (A ∪ N){ ∈ ˜ Now we consider a sequence (Ak ∪ Nk )k∈N with Ak ∪ Nk ∈ A˜ with A. corresponding sequence (N0,k )k∈N , Nk ⊂ N0,k ∈ Nµ . It follows that ! ! [ [ [ (Ak ∪ Nk ) = Nk Ak ∪ k∈N

k∈N

52

k∈N

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

S S Ak ∈ A as well as k∈N Nk ⊂ k∈N N0,k ∈ Nµ , hence we have ˜ ˜ k∈N (Ak ∪ Nk ) ∈ A and A is a σ-field containing A. Next we want to show that the definition of µ ˜ is independent of the representation. Suppose A ∪ N = B ∪ M with A, B ∈ A and N ⊂ N0 , M ⊂ M0 , N0 , M0 ∈ Nµ . By assumption we have and S

S

k∈N

A ⊂ A ∪ N = B ∪ M ⊂ B ∪ M0 and B ⊂ B ∪ M = A ∪ N ⊂ A ∪ N0 and the monotonicity of µ yields µ(A) ≤ µ(B ∪ M0 ) ≤ µ(B) + µ(M0 ) = µ(B) as well as µ(B) ≤ µ(A ∪ N0 ) ≤ µ(A) + µ(N0 ) = µ(A), i.e. µ(A) = µ(B) and hence µ ˜ is well defined, i.e. independent of the re˜ We claim that µ ˜ The fact that presentation of A˜ ∈ A. ˜ is a measure on A. µ ˜(∅) = 0 is trivial and for (Ak ∪ Nk )k∈N of mutually disjoint sets we find ! ! !! [ [ [ Ak ∪ (Ak ∪ Nk ) = µ ˜ µ ˜ Nk k∈N

k∈N

=µ

[

k∈N

Ak

!

k∈N

=

∞ X k=1

µ(Ak ) =

X k∈N

µ ˜(Ak ∪ Nk ),

S ˜ is a measure on A˜ extending where we used that k∈N Nk ⊂ N0 ∈ Nµ . Thus µ ˜ µ µ, i.e. µ ˜|A = µ. It remains to prove that (Ω, A, ˜) is complete. For this let ˜ ˜ ˜ ˜ ˜ We have to show that A = A ∪ N ∈ A such that µ ˜(A) = 0 and B ⊂ A. ˜ ∈ A. ˜ But µ ˜ = 0 implies µ(A) = 0 and we can write A˜ = ∅ ∪ (A ∪ N), B ˜(A) ˜ ⊂ A˜ = A ∪ N ⊂ N0 and B ˜ = ∅ ∪ B, ˜ ∅ ∈ A and A ∪ N ⊂ N0 ∈ Nµ . Thus B ˜ ⊂ N0 ∈ Nµ , i.e. B ˜ ∈ A. ˜ B Definition 3.19. The completion of B(n) is the σ-field of all Lebesgue sets and denoted by L(n) . The completion of λ(n) is called the Lebesgue mea˜ (n) . sure, however we will in general still write λ(n) instead of λ ˜ (n) is again translation invariant. Exercise 3.20. Show that λ 53

A COURSE IN ANALYSIS

We return to sets of measure zero, more precisely to sets of Lebesgue measure zero in (Rn , L(n) , λ(n) ). We can apply our results from Part 4 and find that in particular sets of Jordan content zero are Lebesgue sets of measure zero as are the graphs of continuous functions f : Rn−1 → R or certain boundaries of sets defined by equations, in particular this applies to hyperplanes in Rn as well as their subsets. (Compare with Remark II.19.7.C, Corollary II.19.13, Proposition II.19.19 and II.19.21, Lemma II.21.3, Proposition II.21.11 and Corollary II.21.12.) One of our first examples of a σ-field was the trace σ-field, Example 1.3.C. Let (Ω, A, µ) be a measure space and Ω0 ⊂ Ω a subset then the trace σ-field AΩ0 = Ω0 ∩ A was defined as {Ω0 ∩ A | A ∈ A}. We may ask whether we can construct a measure on Ω0 ∩ A starting with µ. In the case where Ω0 ∈ A, i.e. Ω0 is A-measurable, we can define for A ∈ Ω0 ∩ A µΩ0 (A) := µ(Ω0 ∩ A)

(3.10)

and it is trivial that µΩ0 is a measure on Ω0 ∩A. However for a non-measurable set Ω0 ⊂ Ω, i.e. for Ω0 ∈ / A, in general we can not define a corresponding measure on Ω0 ∩ A. (n)

Thus for every Borel set A ∈ B(n) the measure λA is well defined on A ∩ B(n) (n) as λA is well defined for every A ∈ L(n) . In particular, for a Lebesgue set N (n) (n) of measure zero (or a Borel set M of measure zero) the measure λN (and λM ) is defined, however by (3.10) it is identically zero. Thus this construction does not allow us to define a “nice” measure on a hyperplane or a C k -surface in Rn . For the graph of a measurable function we may try to work with the image measure. Let f : G → R, G ∈ B(n−1) , be a measurable function and consider (n−1) on G the measure λG . On Γ(f ) = {(x, f (x)) | x ∈ G} we may study (n−1) F (λG ) where F : G → Rn , F (x) = (x, f (x)). But it remains the general problem to find for Lebesgue measurable subsets of Rn a natural geometric measure. The Hausdorff measure we are going to study now will give such a measure. At the moment it mainly serves us as a further application of the Carath´eodory theorem, however we will make much use of it when returning to the question of how to define the area of a surface (or the area of a submanifold). Most of the results are best studied with context of metric spaces. We start with

54

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Definition 3.21. Let (X, d) be a metric space and µ∗ an outer measure on P(X). We call µ∗ a metric outer measure if µ∗ (A ∪ B) = µ∗ (A) + µ∗ (B)

(3.11)

holds for all A, B ∈ P(X) such that dist(A, B) > 0. Recall that in a metric space dist(A, B) is defined by  dist(A, B) = inf d(x, y) x ∈ A, y ∈ B , compare with (II.3.2).

The following theorem implies that a metric outer measure restricted to the Borel sets on X, i.e. to B(X) := σ(OX ) where OX denotes the metric topology in X, is a measure. Theorem 3.22. The Borel sets of a metric space are measurable with respect to any metric outer measure on X. Proof. We will prove that all closed sets are measurable with respect to µ∗ , i.e. that µ∗ (G) = µ∗ (G ∩ A) + µ∗ (G ∩ A{ )

holds for all closed sets A ⊂ X, and in fact we may assume that µ∗ (G) < ∞. Further it is clear that we only need to prove µ∗ (G) ≥ µ∗ (G ∩ A) + µ∗ (G ∩ A{ )

since the converse inequality follows from the sub-additivity of µ∗ . For k ∈  1 { N we define Gk := x ∈ A ∩ G | dist(x, A) > k ⊂ G. Clearly we have S Gk ⊂ Gk+1 and the closedness of A implies A{ ∩ G = k∈N Gk . Moreover dist(A ∩ G, Gk ) ≥ k1 and since µ∗ is an outer metric measure it follows that µ∗ (G) ≥ µ∗ ((A ∩ G) ∪ Gk ) = µ∗ (A ∩ G) + µ∗ (Gk ). If we can show that limk→∞ µ∗ (Gk ) = µ∗ (A{ ∩ G) then the result will follow. For this let Hk := Gk+1 ∩ G{k and consider   dist(Hk+1, Gk ) = dist Gk+1 ∩ G{k , Gk . 55

A COURSE IN ANALYSIS

For x ∈ Hk+1, in particular x ∈ G{k , and d(x, y) < d(y, A) ≤ d(x, y) + dist(x, A) < thus we have dist(Hk+1, Gk ) ≥

1 , k(k+1)

it follows that

1 1 1 + = , k(k + 1) k + 1 k 1 . k(k + 1)

(3.12)

Since H2l \G2l−1 ⊂ G2l+1 we arrive at µ∗ (G2l+1 ) ≥ µ∗ (H2l ∪ G2l−1 ) = µ∗ (H2l ) + µ∗ (G2l−1 ) where we used that µ∗ is a metric outer measure and (3.12). By induction we find now l X µ∗ (G2l+1 ) ≥ µ∗ (H2j ) j=1

and analogously

∗

µ (G2l ) ≥

l X

µ∗ (H2j−1).

j=1

∗ By µP (G) < ∞ implying the convergence of the series P∞ assumption ∞ ∗ ∗ µ (H ) and 2j j=1 µ (H2j−1 ), and hence j=1 ∗

∗

{

∗

µ (Gk ) ≤ µ (A ∩ G) ≤ µ (Gk ) +

∞ X

µ∗ (Hj )

j=k+1

which yields for k → ∞ that lim µ∗ (Gk ) = µ∗ (A{ ∩ G)

k→∞

and the theorem is proved. For a subset Y ⊂ X of a metric space (X, d) we define its diameter as  diam(Y ) := sup d(x, y) x, y ∈ Y ,

see Definition II.2.13. We note that if diam(Y ) ≤ ρ then diam(Y ) ≤ ρ, and further if Y ⊂ X then Y := {x ∈ X | d(x, Y ) < } is an open set and 56

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

diam Y ≤ diam Y + . Let A ⊂ X be any set and α > 0, δ > 0. We set (∞ ) X [ Hαδ (A) := inf (diam Yk )α A ⊂ Yk , diam Yk ≤ δ . k=1

(3.13)

k∈N

As δ decreases to 0 the function δ 7→ Hαδ (A) is increasing and therefore Hα∗ (A) := lim Hαδ (A)

(3.14)

δ→0

exists, maybe as improper limit +∞. In particular we have Hαδ (A) ≤ Hα∗ (A).

(3.15)

We claim Theorem 3.23. By Hα∗ a metric outer measure is defined on P(X). Proof. Clearly, Hαδ (∅) = 0, so Hα∗ (∅) = 0, and if A1 ⊂ A2 a covering of A2 is also a covering of A1 , so it follows that Hαδ (A1 ) ≤ Hαδ (A2 ) implying that Hα∗ (A1 ) ≤ Hα∗ (A2 ). Next we want to show the sub-additivity of Hα∗ . For this let (Ak )k∈N be a sequence in P(X) and (Yk,l )l∈N a covering of Ak by sets Yk,l such that diam(Yk,l ) < δ and ∞ X l=1

(diam(Yk,l ))α ≤ Hαδ (Ak ) =

 . 2k

S The family (Yk,l )k,l∈N is a covering of k∈N Ak by sets with diameter less than δ and therefore ! ∞ ∞ X X [  Hαδ Hαδ (Ak ) + Ak ≤ 2k k=1 k=1 k∈N ≤ which yields Hαδ

[

k∈N

Ak

∞ X k=1

!

≤

57

Hα∗ (Ak ) + , ∞ X k=1

Hα∗ (Ak ),

A COURSE IN ANALYSIS


P∞ S ∗ and for δ decreasing to 0 we obtain Hα∗ k=1 Hα (Ak ), i.e. the k∈N Ak ≤ ∗ ∗ subadditivity of Hα . This already implies that Hα is an outer measure. Finally we prove that Hα∗ is a metric outer measure. For this let A, B ⊂ X such that dist(A, B) > 0. We need to show that Hα∗ (A∪B) = Hα∗ (A)+Hα∗ (B), and since we already know the sub-additivity of Hα∗ it remains to verify Hα∗ (A ∪ B) ≥ Hα∗ (A) + Hα∗ (B).

(3.16)

Let 0 <  < dist(A, B) and let A ∪ B be covered by (Yk )k∈N with diam(Yk ) ≤ δ < . By A0k := A ∩ Yk and Bk0 := B ∩ Yk , k ∈ N, we now have disjoint covers of A and B, respectively. Therefore we find ∞ X

(diam A0k )α +

k=1

∞ X k=1

(diam Bk )α ≤

∞ X

(diam Yk )α .

k=1

We now take the infimum over all coverings and then we let δ go to zero to arrive at (3.16). Combining Theorem 2.23 with Theorem 3.22 and Theorem 3.23 we get Corollary 3.24. Restricted to B(n) the metric outer measure Hα∗ is a measure. Note that Hα∗ as a measure on B(n) allows a completion on L(n) , but at this stage we cannot decide whether for certain values of α the family of Hα∗ -measurable sets is larger than B(n) (or L(n) ). Definition 3.25. The measure Hα := Hα∗ B(n)

(3.17)

is called the α-dimensional Hausdorff measure on (X, d). Lemma 3.26. The α-dimensional Hausdorff measure on the n-dimensional Euclidean space Rn is translation invariant as well as invariant under the action of O(n), i.e. for x ∈ Rn and T ∈ O(n) Hα (x + A) = Hα (A) and Hα (T (A)) = Hα (A) holds for all A ∈ B(n) . 58

(3.18)

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Proof. This follows immediately since the metric induced by the Euclidean norm is invariant under translation and orthogonal transformations. Remark 3.27. Lemma 3.26 has a counterpart for a general metric space (X, d). The measure Hα is on B(n) (X) invariant under isometries of (X, d), i.e. if T : X → X satisfies d(T x, T y) = d(x, y), then for all A ∈ B(X) we have Hα (T (A)) = Hα (A). Corollary 3.28. There exists a constant γn > 0 such that Hn (A) = γn λ(n) (A)

(3.19)

holds for all A ∈ B(n) . Proof. The result will follow from Corollary 3.3 if we can prove for Cn = [0, 1] × · · · × [0, 1] ⊂PRn that 0 < Hn (Cn ) < ∞. Since for every covering n of Cn we have 1 ≤ ∞ k=1 (diam Yk ) the estimate 0 < Hn (Cn ) is√ trivial. If n we divide Cninto k cubes of side length k1 and choose δ > kn we find Hnδ (Cn ) ≤ k n

√

n k

n

n

= n 2 implying Hn (C) < ∞.

Remark 3.29. It can be shown with tools currently not at our disposal that n

1 π2 γn = n λ(n) (B1 (0)) = n n , 2 2 Γ( 2 + 1)

(3.20)

see [24]. The next result compares Hα∗ and Hβ∗ . Lemma 3.30. Let 0 < α < β < γ and assume that Hα∗ (A) < ∞, then Hβ∗ (A) = 0. If however Hγ∗ (A) > 0 then Hβ∗ (A) = +∞. Proof. Clearly, the second statement follows from the first. We note that if diam(Y ) < δ and β > α then (diam(Y ))β = (diam(Y ))β−α (diam(Y ))α ≤ δ β−α (diam(Y ))α which yields Hβδ (A) ≤ δ β−α Hαδ (A) ≤ δ β−α Hα∗ (A).

Now, if Hα∗ (A) > 0, since β − α > 0, for δ → 0 it follows that Hβ∗ (A) = 0. 59

A COURSE IN ANALYSIS

From this lemma we deduce immediately that in the case of Rn when dealing with Hα we can always assume that α ≤ n, otherwise Hα is identically zero. Example 3.31. Let U ⊂ Rn be an open set and α < n, then Hα (U) = ∞ since Hn (U) > 0. The latter statement follows since U must contain a non-degenerate cube. Now we can deduce that every Borel set of Rn with non-empty interior has for α < n an α-dimensional Hausdorff measure equal to zero. A further consequence of Lemma 3.30 is that for every Borel set A ∈ B(n) there exists a unique α ≥ 0 such that ( +∞, β < α, Hβ (A) = 0, α < β, i.e. we have   α = sup β > 0 Hβ (A) = ∞ = inf γ > 0 Hγ (A) = 0 .

(3.21)

Definition 3.32. For A ∈ B(n) the number α uniquely determined by (3.21) is called the Hausdorff dimension of A and we write α = dimH A. Remark 3.33. A. If A ∈ B(n) has Hausdorff dimension α, it is still possible that Hα (A) = 0 or Hα (A) = +∞. In the case where 0 < Hα (A) < ∞ some authors say that A has strict Hausdorff dimension α, see [84]. B. The Hausdorff dimension of a Borel set A ∈ B(n) can be a non-integer number. In such a case A is called a fractal. C. A priori there is no relation between the Hausdorff dimension of a Borel set and the dimensions of a subspace of Rn . However the situation will become more clear in the following examples. Before studying more examples we want to investigate the behaviour of the Hausdorff dimension under certain mappings. Lemma 3.34. For ρ > 0 define hρ : Rn → Rn by hρ (x) = ρx. If A ∈ B(n) and α > 0 then it follows that Hα (ρA) := Hα (hρ (A)) = ρα Hα (A). 60

(3.22)

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Proof. Let (Yk )k∈N , Yk ⊂ Rn , be a covering of A such that diam(Yk ) ≤ δ. Then (ρYk )k∈N is a covering of ρA and diam(ρYk ) ≤ ρδ which yields Hαρδ (ρA)

≤

∞ X

α

α

(diam(ρYk )) = ρ

k=1

∞ X k=1

(diam Yk )α ≤ ρα Hα∗ (A)

implying for Borel sets Hαρδ (ρA) ≤ ρα Hα (A) and now Hα (ρA) ≤ ρα Hα (A)follows for δ → 0. Replacing ρ by 1ρ and A by ρA we get Hα (A) = Hα ρ1 ρA ≤ 1 H (ρA) ρα α

or ρα Hα (A) ≤ Hα (ρA) and the result follows.

In Lemma 3.34 it was easy to see that hρ (A) is a Borel set since ρA = h−1 1 (A) ρ

and as a continuous mapping h 1 is measurable. For the following results we ρ need to know that the image of a Borel set under a continuous mapping is a Borel set, and this is a non-trivial result. A partial result is easy to get: since the image of a compact a continuous mapping is compact we
set under S S f = A can deduce using f k k∈N (Ak ) that the image of a countable k∈N union of compact sets under a continuous mapping is Borel measurable. For bijective continuous functions f : R → R we prove in Problem 11 that they map Borel sets into Borel sets. The proof of the general result needs knowledge about Suslin sets and goes far beyond what can be reasonably handled in a first course on measure theory. In fact there are many results in measure theory in topological spaces which depends on more advanced results of descriptive set theory. We refer the interested reader to H. Federer [25], K. Jacobs [42] and A. B. Kharazishvili [46], and T. Bartoszynski, H. Judah [9] where special attention is paid to the real line. For reference purpose we quote from [25] result 2.2.13 in the formulation suitable for our needs. Theorem 3.35. Denote on Rn the Euclidean metric by dE and suppose that (G, dE ), G ⊂ Rn , is a complete metric subspace of (Rn , dE ). Then for a continuous function f : G → Rm the image of a Borel set A ⊂ G is a Borel set in Rm . In particular every continuous mapping f : Rn → Rm maps Borel sets of Rn onto Borel sets in Rm . Of course we can take G just to be closed in Theorem 3.35, but we intended to be close to Federer’s formulation. Lemma 3.36. Let G ⊂ Rn be a closed set and f : G → Rm be a H¨ older continuous mapping with H¨ older exponent s ∈ (0, 1], i.e. kf (x) − f (y)k ≤ c kx − yks 61

(3.23)

A COURSE IN ANALYSIS

for all x, y ∈ G and some c > 0. For every Borel set A ∈ B(n) , A ⊂ G, we have α (3.24) H αs (f (A)) ≤ c s Hα (A). Proof. Let (Yk )k∈N , Yk ⊂ Rn , diam(Yk ) ≤ δ, be a covering of A. The H¨older condition (3.23) yields diam(f (A ∩ Yk )) ≤ c (diam(Yk ))s , i.e. (f (A ∩ Yk ))k∈N is a covering of f (A) by sets with diameter less or equal to cδ s . Since ∞ X k=1

we find

α s

(diam(f (A ∩ Yk ))) ≤ c

α s

∞ X

(diam(Yk ))α

k=1

α

s

δ α (f (A)) ≤ c s H (A) Hcδ s s

α

and in the limit δ → 0 we have H αs (f (A)) ≤ c s Hα (A).

Corollary 3.37. For a Lipschitz continuous mapping f : G → Rm and A ∈ B(n) , A ⊂ G, the following holds Hα (f (A)) ≤ cα Hα (A).

(3.25)

This estimate applies in particular if G is arcwise connected and f is a C 1 mapping with bounded derivative, since such a mapping is by the mean-value theorem Lipschitz continuous. Corollary 3.38. In the situation of Lemma 3.36 we have 1 dimH (f (A)) ≤ dimH A. s

(3.26)

α

Proof. By (3.34) we have H αs (f (A)) ≤ c s Hα (A) which implies for α > dimH A that H αs (f (A)) = 0, i.e. dimH f (A) ≤ αs for all α > dimH A, which in turn yields dimH f (A) ≤ 1s dimH A. In particular, for Lipschitz continuous mappings we have in the situation of Lemma 3.36 dimH f (A) ≤ dimH A. (3.27)

Corollary 3.39. Let G ⊂ Rn be closed and f : G → Rn , be a bi-Lipschitz mapping, i.e. for 0 < c1 ≤ c2 we have c1 kx − yk ≤ kf (x) − f (y)k ≤ c2 kx − yk

(3.28)

for all x, y ∈ G. If A ∈ B(n) , A ⊂ G, then we have the equality dimH f (A) = dimH A. 62

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Proof. First we note that f must be injective and so we can denote by f −1 the inverse of f : G → f (G). From (3.28) we deduce further that f −1 : f (G) → G is also Lipschitz continuous. Hence we have dimH f (A) ≤ dimH A as well as dimH A = dimH f −1 (f (A)) ≤ dimH f (A). Example 3.40. The embedding j : Rn → Rm , n < m, j(x1 , . . . , xn ) = (x1 , . . . , xn , 0, . . . , 0) is injective and its “inverse” in the sense of the proof of Corollary 3.39 is the projection π(n,m) : Rm → Rn , π(n,m) (y1 , . . . , ym ) = (y1 , . . . , yn ). Both mappings are trivially Lipschitz continuous and therefore the Hausdorff dimension of every Borel set A ∈ B(n) is equal to the Hausdorff ˚ ⊂ Rn dimension of j(A) ∈ B(m) . In particular if A has a non-empty interior A then dimH j(A) = n. This applies for example to non-degenerate cubes in Rn . The next proposition will eventually pave the way to define in Volume VI integrals over sub-manifolds. Analogous to Definition II.27.1.B we call for a linear mapping T : Rk → Rn gT := det(T ∗ T )

(3.29)

the Gram determinant of T . Since T ∗ T is a symmetric positive semi√ definite k × k-matrix gT is well defined. Proposition 3.41. Let k ≤ n and A ∈ B(k) . Further let T : Rk → Rn be a linear mapping. Then we have √ Hk (T (A)) = gT Hk (A). (3.30) Proof. (Following G. Folland [27]) In the case where k = n it follows that Hn = γn λ(n) , see Corollary 3.28, and gT = (det T )2 . Now (3.30) follows from Theorem 3.11. For k < n we can find a rotation R ∈ O(n) such that R(T (Rn )) ⊂ {(x1 , . . . , xn ) ∈ Rn | xk+1 = · · · = xn = 0} for which we write R(T (Rn )) ⊂ Rk × {0} ⊂ Rn . With S := RT we have S ∗ S = T ∗ R∗ RT = T ∗ T and the rotation invariance of the Hausdorff measure yields gS = gT as well as Hk (S(A)) = Hk (T (A)). Identifying Rk × {0} with Rk we observe that S maps Rk into itself and S ∗ S does not change under this identification. Hence we may apply the result already shown for the case k = n to obtain (3.30). We end this chapter by discussing a Borel set C ⊂ R which has fractional Hausdorff dimension. The set C is the Cantor set which we have already 63

A COURSE IN ANALYSIS

met in Chapter I.32.
We recall  its construction. Let C0 := [0, 1], define C1 := [0, 1]\ 31 , 23 = 30 , 31 ∪ 23 , 33 and now we continue in the following way: CN +1 is obtained by removing the 2N open middle intervals of length 3−N from the intervals forming CN . The Cantor set is then defined by C :=

∞ \

CN ,

(3.31)

N =0

which is by Theorem I.32.4 a compact non-denumerable null set with respect to λ(1) . The set CN is obtained by taking from [0, 1] away 2N − 1 open intervals which we denote by Jl , 1 ≤ l ≤ 2N − 1 and with Jl = (al , bl ) we assume that bl < al+1 . We set J 0 := {0} and J 2N := {1}. The set CN consists of 2N closed intervals Kk = [ck , dk ] and we order these intervals such that c1 = 0, ck = bk−1 and dk = ak , d2N = 1. We define now the function FN : [0, 1] → R in the following way  0,     l , N FN (x) = 2 gl (x),    1,

x=0 x ∈ J l , l = 1, . . . , 2N − 1 ˚l , l = 1, . . . , 2N x∈C x=0

1

(3.32)

1

F2

F1 3 4

1 2

1 2

1 4

1 3

2 3

1 9

1

2 9

1 3

2 3

7 9

8 9

1

Figure 3.2

Figure 3.1

64

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

where the graph Γ(gl ) is the line segment connecting (cl , FN (bl−1 )) with (dl , FN (al )). For N = 1 and N = 2 the functions F1 and F2 are drafted in Figure 3.1 and Figure 3.2 above. By construction FN : [0, 1] → [0, 1] is continuous and monotone increasing and further we have |FN +1 (x) − FN (x)| ≤

1 2N +1

.

(3.33)

The triangle inequality yields for M ∈ N that |FN +M (x) − FN (x)| ≤

M X j=1

|FN +j (x) − FN +j−1 (x)| ≤

M X j=1

1 2N +j

.

Taking the supremum over x ∈ [0, 1] we find kFN +M − FN k∞ ≤

1 , 2N

(3.34)

i.e. (FN )N ∈N is a Cauchy sequence in C([0, 1]), hence it has a limit F ∈ C([0, 1]), i.e. F (x) = lim FN (x) (uniform convergence). N →∞

(3.35)

Moreover F (0) = 0 and F (1) = 1, F is increasing since for y < x we have F (y) = limN →∞ FN (y) ≤ limN →∞ FN (x) = F (x), F is constant on each interval of the complement of C, and hence F maps C onto [0, 1]. Definition 3.42. The function F defined by (3.35) is called the CantorLebesgue function or by some authors the Lebesgue singular function. Lemma 3.43. The Cantor-Lebesgue function is H¨older continuous with 2 H¨older exponent s = ln . ln 3 Proof. The function F is the uniform limit of the functions FN which are piecewise linear and FN increases in an interval of length 31N by at most 21N .
N which yields Thus the slope of FN is bounded by 32  N 3 |FN (x) − FN (y)| ≤ |x − y|. 2 65

(3.36)

A COURSE IN ANALYSIS

Passing in (3.34) to the limit M → ∞ we find

1 . 2N Combining (3.36) and (3.37) gives for x, y ∈ [0, 1] |F (x) − FN (x)| ≤

(3.37)

|F (x) − F (y)| ≤ |FN (x) − FN (y)| + |F (x) − FN (x)| + |F (y) − FN (y)|  N
2 1 3 |x − y| + N = N 3N |x − y| + 2 . ≤ 2 2 2

For x, y ∈ [0, 1] fixed, x 6= y, we now choose N such that 1 ≤ 3N |x − y| ≤ 3 which is always possible since |x − y| ≤ 1. Hence we arrive at c |F (x) − F (y)| ≤ N 2 with some c ∈ [3, 5]. Since 3γ = 2 means eγ ln 3 = eln 2 , we find with s = that 3s = 2, and it follows that c |F (x) − F (y)| = sN 3

ln 2 ln 3

and taking into account that 3−N ≤ |x − y| we eventually get |F (x) − F (y)| ≤ c0 |x − y|s .

Now we can prove Theorem 3.44. The Hausdorff dimension of the Cantor set C is dimH (C) =

ln 2 . ln 3

(3.38)

T Proof. Recall that C = N ∈N CN and each CN is the union of 2N intervals of length 3−N . Given δ > 0 we choose M such that 3−M < δ and the intervals belonging to CM cover C. For α > 0 we have

For α =

ln 2 ln 3

Hαδ (C) ≤ 2M (3−M )α .

it follows that 2M (3−M )α = 1, i.e. H ln 2 (C) ≤ 1. ln 3

On the other hand, combining Lemma 3.36 with Lemma 3.43 and using that F (C) = [0, 1] we get H1 ([0, 1]) ≤ cH ln 2 (C), and both estimates imply ln 3 2 . dimH (C) = ln ln 3 66

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THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

Remark 3.45. A. Our arguments leading to the proof of (3.38) are much influenced by the presentation of E. M. Stein and R. Shakarchi [84]. B. We will meet the Cantor-Lebesgue function again in Chapter 11 when discussing differentiability and the fundamental theorem of calculus within Lebesgue’s theory of integration. C. Fractals are quite popular (and important) objects, partly due to the beautiful pictures that can be created using so called self-similar fractals. A good first reading is K. Falconer [23].

Problems 1.

a) On the set {1, . . . , n} we consider the power set as its σ-field P and we define the measure µ := nk=1 k . Prove that µ is invariant under the operation of the symmetric group Sn , i.e. µ(σ(A)) = µ(A) for every permutation σ : {1, . . . , n} → {1, . . . , n} and every subset A of {1, . . . , m}. b) Now consider on {1, . . . , n} the measure ν :=

X1 X 1 k + l 2 3 l≤n k≤n

k even

l odd

which has the power set as its σ-field. Prove that in general ν is not invariant under Sn . However, if σ ∈ Sn maps even numbers onto odd numbers, then σ leaves ν invariant. P 2. On (Z, P(Z)) consider the measure µ = k∈Z k . For T : Z → Z, k 7→ T (k) = k 2 find the image measure T (µ) on P(Z). 3. Consider the mapping T : R3 → R3 , T (x, y, z) = (ax, by, cz), where a > b > c > 0 are given and R3 is equipped with the Borel σ-field B(3) . Prove that T is measurable and find the image of B1 (0) ⊂ R3 under T as well as T (λ(3) ). Nowofind the volume of the ellipsoid E := n 2 2 2 (x, y, z) ∈ R3 x9 + y16 + z25 < 1 , i.e. find λ(n) (E).

4.

a) For the Bernoulli distribution βpN as well as for the Poisson distribution πα find all sets of measure zero. 67

A COURSE IN ANALYSIS

b) Give an example of a measurable mapping T : Rn → Rn and a Borel set A ∈ B(n) such that λ(n) (A) > 0 but λ(n) (T (A)) = 0. c) Let T : R → R be any mapping. Prove that λ(T (Q)) = 0.

5. Give an example of two measure spaces (Ω1 , A1 , µ1) and (Ω2 , A2 , µ2 ) and a measurable mapping h : Ω1 → Ω2 such that h maps a nontrivial set of µ1 -measure zero to a set of strictly positive µ2 -measure, i.e. A ∈ A1, A 6= ∅ and µ1 (A) = 0 but µ2 (h(A)) > 0. 6. The mapping ϕ : R → R2 , t 7→ ϕ(t) = (cos t, sin t) maps R onto the unit circle S 1 ⊂ R2 . Prove that ϕ(λ(1) )|S 1 is invariant under rotation. 7. Prove that a Borel set A ⊂ Rn which is a null set in the sense of Definition II.19.6 is a set of measure zero in the sense of Definition 3.13. (Note that the assumption that A is a Borel set, i.e. measurable, is crucial.) 8. Let (Ω, A, µ) be a complete measure space and Y1 , Y2 ∈ A such that Yj{ are sets of measure zero. Let f : Y1 → R be a measurable function such that for some g : Y2 → R we have f = g µ-a.e. Prove that g is measurable. ˜ (n) of the n9. Solve Exercise 3.20, i.e. prove that the completion λ dimensional Borel-Lebesgue measure λ(n) is translation invariant. 10. Let (Ω, A, µ) be a measure space and assume that S there exists a sequence of measurable sets (Bk )k∈N such that Ω = k∈N Bk and µ(Bk ) < ∞, i.e. the measure µ is σ-finite. A set N ∈ A has measure zero locally if µ(N ∩ C) = 0 for every C ∈ A, µ(C) < ∞. Prove that in the case of a σ-finite measure µ every set that has measure zero locally is a set of measure zero. 11. Prove that a bijective continuous function g : R → R maps Borel sets onto Borel sets. Hint: prove that {A ⊂ R|g(A) ∈ B(1) } is a σ-field. 12.

a) For an interval I ⊂ R with non-empty interior let γ : I → Rn be a C 1 -curve. Prove that for every interval I˜ ⊂ I with non-empty ˜ = 0 for α > 1. interior we have Hα (γ(I)) 68

3

THE LEBESGUE-BOREL MEASURE AND HAUSDORFF MEASURES

b) Let G ⊂ R2 be an open set and f : G → R3 a C 1 -mapping. Suppose that B1 (0) ⊂ G and consider the parametric surface f : B1 (0) → R3 . Prove that Hβ (f (B1 (0))) = 0 for β > 2.

c) The following theorem is essentially due to G. Peano and it had enormous influence in the development of many branches of mathematics: There exists a mapping P : [0, 1] → [0, 1] × [0, 1] which is continuous and surjective and satisfies the H¨older condition 1

|P (t) − P (s)| ≤ L|t − s| 2 . Such a curve is called a Peano curve or a space filling curve. We refer to [84] where a detailed proof and discussion is provided. Prove that it is not possible to have a continuous surjective mapping R : [0, 1] → [0, 1] × [0, 1] which is H¨older continuous for some exponent s > 12 . 13.

a) Let T : Rn → Rn be a symmetric positive definite linear mapping, where T is called positive definite if its matrix has this property. Prove that the Gram determinant of T is given by gT = det T . Given a parametric C 1 -surface f : G → R3 , x = (u, v) 7→  b)  f1 (x) f2 (x), where G ⊂ R2 is an open set. We denote by gf (x) the Gram f3 (x) determinant of dx f where dx f : R2 → R3 is the differential of f at x ∈ G. Find the expression of gf (x) in terms of E(x) := hfu (x), fu (x)i, F (x) := hfu (x), fv (x)i and G(x) := hfv (x), fv (x)i.

69

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4

Measurable Mappings

We have already introduced measurable mappings f : Ω → Ω0 , where (Ω, A) and (Ω0 , A0) are two measurable spaces. These are mappings having the property that the pre-image of A0 -measurable sets are A-measurable, i.e. f −1 (A0 ) ∈ A for A0 ∈ A0 , see Definition 1.15. In this chapter we want to investigate measurable mappings further, but first let us recollect what is already known to us: • the composition of measurable mappings is measurable (Remark 1.16.C); • f is measurable if and only if f −1 (E 0 ) ∈ A for a generator E 0 of A0 (Lemma 1.17); • for measurable spaces (Ωj , Aj ), j ∈ I, and mappings fj : Ω → Ωj exists a smallest σ-field in Ω such that all fj become measurable (Example 1.14 and Remark 1.16.A); • if f : Ω → Ω0 is measurable and µ is a measure on (Ω, A) then the image f (µ) is a well defined measure on (Ω0 , A0) (Theorem 3.7); • forming image measures is transitive (Corollary 3.9). We can derive more results if (Ω, A) and/or (Ω0 , A0 ) have additional structures respected by certain mappings. For example if (Ω, O) and (Ω0 , O0 ) are two topological spaces and A = B(Ω), A0 = B0 (Ω0 ) are the Borel σ-fields generated by the topologies, then every continuous mapping f : Ω → Ω0 is measurable, see Corollary 1.18 the proof of which extends without change to the general situation. In the following we want to discuss results when we have on Ω0 (and maybe on Ω) additional algebraic and topological structures. For example when Ω0 = Rn and A0 = B(n) then we can define f ± g and αf , and we may ask whether f ± g or αf are measurable if f and g are. The function f + g can be written as composition. We define add : Rn × Rn → Rn by add(x, y) = x + y, and x 7→ f (x) + g(x) is obtained as x 7→ (f (x), g(x)) 7→ add(f (x), g(x)) = f (x) + g(x). If we can prove that x 7→ (f (x), g(x)) and (x, y) 7→ add(x, y) are measurable, we can deduce that x 7→ f (x) + g(x) is measurable. Of general use is 71

A COURSE IN ANALYSIS

Lemma 4.1. Let (Ω, A) and (Ωj , Aj ), j = 1, 2, be measurable spaces. A mapping f : Ω → Ω1 × Ω2 is A/A1 ⊗ A2 -measurable if and only if the mappings prj ◦ f : Ω → Ωj are A/Aj -measurable, j = 1, 2. Proof. Since by the construction of A1 ⊗ A2 the projections are measurable, the measurability of f implies that of prj ◦ f . Conversely, since A1 × A2 is a generator of A1 ⊗ A2 and since for A1 × A2 ∈ A1 × A2 it follows that
−1 f −1 (A1 × A2 ) = f −1 pr−1 1 (A1 ) ∩ pr2 (A2 )

−1 = f −1 pr−1 pr−1 1 (A1 ) ∩ f 2 (A2 ) = (pr1 ◦ f )−1 (A1 ) ∩ (pr2 ◦ f )−1 (A2 ) ∈ A the measurability of prj ◦ f implies the measurability of f . In our motivating example it remains to study “add” but that the mapping (x, y) 7→ x + y is in Rn continuous, hence measurable, is well known to us. We want to handle this problem in a wider context for example we want to consider mappings with values in GL(n; R) considered as a group, or mappings defined on some measurable space mapping into the space of all continuous, real-valued functions defined on a compact space and this space we equip with the supremum norm. We need as preparation some additional background from topology. Since the examples most interesting to us are those where the topologies are induced by a metric, here we only discuss the case of metric spaces. We refer to Appendix I where we discuss the more general case. Definition 4.2. Let (X, dX ) and (Y, dY ) be two metric spaces and define for (x1 , y1), (x2 , y2) ∈ X × Y dX×Y ((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2).

(4.1)

Then (X × Y, dX×Y ) is called the product metric space of (X, dX ) and (Y, dY ) and dX×Y the product metric. Clearly, this definition extends to the case of N metric spaces (Xj , dXj ), j = 1, . . . , N, in the obvious way. Given two metrics d1 and d2 on a set X, we call d1 and d2 equivalent if for 0 < γ1 ≤ γ2 and all x, y ∈ X the following holds γ1 d1 (x, y) ≤ d2 (x, y) ≤ γ2 d1 (x, y). (4.2) 72

4

MEASURABLE MAPPINGS d

For the corresponding open balls Bρ j (x) := {y ∈ X | dj (x, y) < ρ} this implies B dρ1 (x) ⊂ Bρd2 (x) ⊂ B dρ1 (x) γ2

γ1

and hence equivalent metrics induce the same open sets in X, which yields that the corresponding Borel σ-fields are the same. Exercise 4.3. Prove that by dX×Y a metric on X × Y is given which is for 1 ≤ p < ∞ equivalent to each of the metrics dX×Y,p where 1

dX×Y,p ((x1 , y1 ), (x2 , y2 )) = (dpX (x1 , x2 ) + dpY (y1 , y2 )) p . Show further that the projections prX : X ×Y → X, prX (x, y) = x, and prY : X × Y → Y , prY (x, y) = y, are continuous. In addition give a formulation of these results in the case of N, N ∈ N, metric spaces (Xj , dXj ), 1 ≤ j ≤ N. In the following, when discussing metric spaces, we always consider them equipped with their metric topology, i.e. the topology induced by the metric, and the corresponding Borel σ-field, i.e. the σ-field generated by the open sets. On finite products of metric spaces we consider the topology induced by the product metric and the corresponding Borel σ-field which coincides with the product of the underlying σ-fields of each factor, see Problem 2. Definition 4.4. A. Let (G, ·) be a group and suppose that on G a metric dG is given. We call G (or more precisely (G, ◦, dG )) a metric group if the two mappings comp : G × G → G, comp(h1 , h2 ) = h1 ◦ h2 , and inv : G → G, inv(h) = h−1 , are continuous. B. Let (V, +, ·) be an Rn -vector space and let dV be a metric on V . We call V a metric vector space if + : V × V → V , (x, y) 7→ x + y, and · : R × V → V , (α, x) 7→ αx, are continuous, where on R we choose the topology induced by the absolute value. Remark 4.5. In Appendix I we discuss the product topology of two topological spaces. With this at hand we can immediately introduce the notion of a topological group and that of a topological vector space. If (G, ◦) is a group and on G a topology OG is given, then we call G (or (G, ◦, OG )) a topological group if the mappings comp and inv (as defined previously) are continuous. We call (V, +, ·) a topological vector space if on V a topology OV is given such that + : V × V → V and · : R × V → V are continuous where on R we choose as usual the Euclidean topology induced by the absolute value. 73

A COURSE IN ANALYSIS

Example 4.6. A. The vector space Rn equipped with the Euclidean metric is a metric vector space since we know that adding and multiplying with scalars is continuous. B. The general linear group GL(n; R) is a metric group. We can identify 2 GL(n; R) with an open subset of Rn , namely GL(n; R) = det−1 (R\{0}), and the group operations can be decomposed into a finite sequence of adding, subtracting, multiplying and dividing by a non-zero number. Since the 2 projections prj : Rn → R are continuous, all these operations are continuous too. C. The groups O(n) and SO(n) are metric groups. The continuity of the group operations is inherit from GL(n; R). D. Every normed vector space is a metric vector space. This example includes infinite dimensional vector spaces such as (Cb (K), k.k∞ ) where K is a metric space, or even a topological space. Corollary 4.7. A. Let (G, ◦) be a metric group. Then the mappings comp : G × G → G, comp(h1 , h2 ) = h1 ◦ h2 and inv : G → G, inv(h) = h−1 , are measurable. B. Let (V, +, ·) be a metric R-vector space. Then the vector space operations are measurable. Corollary 4.8. A. Let (Ω, A) be a measurable space and (G, ◦) a metric group. Further let u, v : Ω → G be two measurable mappings. Then the mappings u ◦ v : Ω → G, (u ◦ v)(ω) := u(ω) ◦ v(ω), and u−1 : Ω → G, u−1 (ω) := (u(ω))−1, are measurable. B. Let (Ω, A) be a measurable space and (V, +, ·) a metric vector space. Let f, h : Ω → V be two measurable mappings and α, β ∈ R. Then the mapping αf + βh : Ω → V , ω 7→ αf (ω) + βh(ω), is measurable too. Turning to real-valued functions f, h : Ω → R we can also consider the function f · h : Ω → R, ω 7→ f (ω)h(ω). Since multiplication in R is continuous we deduce that if f and h are measurable then f · h is measurable. (See also Problem 3.) The summary of all these considerations is simple: consider mappings from a measurable space (Ω, A) into a set H which carries an algebraic structure (group, vector space, etc) and a metric (or even a topology) such that the algebraic operations in H are continuous. Define the corresponding algebraic operations pointwisely for mappings f, g : Ω → H, (f ◦ g)(ω) = f (ω) ◦ g(ω), 74

4

MEASURABLE MAPPINGS

etc. Then these mappings are measurable when H is equipped with the Borel σ-field. Measures are certain mappings defined on a σ-field with values in [0, ∞]. When trying to define an integral with respect to a measure it makes sense to allow functions which may attain the values +∞ or −∞ too. The measurability of such functions will be investigated next. We have already introduced the arithmetic for R = [−∞, +∞], see (1.29) - (1.32). Now we want to find a σ-field on R extending B(1) . There are two equivalent routes. We can first consider R as a certain compactification of R and then we can work with the corresponding Borel σ-field. This will be discussed in Appendix I. Another (1) way is to define B directly by B

(1)

n o := A˜ = A ∪ S A ∈ B (1) and S ∈ {∅, {−∞}, {+∞}, {−∞, +∞}} . (4.3) (1)

(1)

It is easy to see that B is indeed a σ-field and that the trace σ-field BR = (1) R ∩ B is the Borel σ-field B(1) , see Problem 7. (1)

Lemma 4.9. Each of the following families is a generator of B :   [a, ∞] a ∈ R , [a, ∞] a ∈ Q ,   (b, ∞] b ∈ R , (b, ∞] b ∈ Q ,   [−∞, c) c ∈ R , [−∞, c) c ∈ Q ,   [−∞, d] d ∈ R , [−∞, d] d ∈ Q . (1)

(4.4) (4.5) (4.6) (4.7)

Proof. We prove that σ ({[a, ∞] | a ∈ R}) = B , the other cases can be proved in a similar way. In particular, switching from R to Q relies on arguments analogous to those employed in Chapter 1. Since [a, ∞] = [a, ∞)∪ (1) {+∞} and [a, ∞) ∈ B(1) it follows that [a, ∞] ∈ B , i.e. σ ({[a, ∞] | a ∈ R}) (1) ⊂ B . For b, c ∈ R we have [b, c) = [b, ∞]\[c, ∞] ∈ σ ({[a, ∞] | a ∈ R}) and since the right open intervals [b, c) generate B(1) we find in addition B(1) ⊂ T (1) σ ({[a, ∞] | a ∈ R}) ⊂ B . Finally we observe that {+∞} = k∈N [k, ∞] and T {−∞} = k∈N [−k, ∞]{ which yields {−∞}, {+∞} ∈ σ ({[a, ∞] | a ∈ R}) as well as
 A, A ∪ {+∞}, A ∪ {−∞}, A ∪ {−∞, ∞} ∈ σ [a, ∞] a ∈ R 
for all A ∈ B(1) , i.e. B(1) = σ [a, ∞] a ∈ R . 75

A COURSE IN ANALYSIS

Remark 4.10. Note that in Chapter 1 we have already seen that B(1) is generated by any of the systems {(c, ∞) | c ∈ K}, {[c, ∞) | c ∈ K} and {(−∞, c] | c ∈ K} where K ∈ {Q, R}. Definition 4.11. For any set G 6= ∅ we call f : G → R a numerical function. Thus saying that a certain numerical function is real-valued means to emphasise that it takes only elements of R as values, i.e. it does not attain the values −∞ or +∞. Combining Lemma 1.17 with Lemma 4.9 we find that for a measurable space (1) (Ω, A) the numerical function f : Ω → R is A\B -measurable if and only if one of the following equivalent conditions hold for all c ∈ K, K ∈ {Q, R},  f −1 ([c, ∞]) = ω  f −1 ((c, ∞]) = ω  f −1 ([−∞, c)) = ω  f −1 ([−∞, c]) = ω

∈ Ω f (ω) ≥ c ∈ A, ∈ Ω f (ω) > c ∈ A, ∈ Ω f (ω) < c ∈ A, ∈ Ω f (ω) ≤ c ∈ A.

(4.8) (4.9) (4.10) (4.11)

Note that the equivalence for (4.8) to (4.11) follows already from

and

   [ 1 , ω ∈ Ω f (ω) > c = ω ∈ Ω f (ω) ≥ c + k k∈N {   ω ∈ Ω f (ω) ≤ c = ω ∈ Ω f (ω) > c ,    [ 1 ω ∈ Ω f (ω) < c = ω ∈ Ω f (ω) ≤ c − k k∈N 

 { ω ∈ Ω f (ω) ≥ c = ω ∈ Ω f (ω) < c .

We make our life easier by introducing the notation

 {f ≤ g} := ω ∈ Ω f (ω) ≤ g(ω)

(4.12)

with {f = g}, {f 6= g}, {f < g}, {f ≥ g} and {f > g} being analogously 76

4

MEASURABLE MAPPINGS

defined and for example for {f ≥ h} ∩ {f ≤ g} we write {h ≤ f ≤ g}. Since {f < g} =

[

r∈Q

({f < r} ∩ {r < g}) ,

{f ≤ g} = {f > g}{ , {f = g} = {f ≤ g} ∩ {g ≤ f }, and

{f 6= g} = {f = g}{ ,

all these sets are measurable for measurable functions f, g : Ω → R. Constant functions are measurable and since for α, β ∈ R, β 6= 0, we have {α + βf ≥ c} = {f ≥ β1 (c − α)}, β > 0, and {α + βf ≥ c} = {f ≤ β1 (c − α)}, β < 0, we deduce that α + βf is measurable if f is, clearly the case β = 0 is trivial. Consequently for two numerical functions f and g such that f + g is defined we have {f + g ≥ c} = {f ≥ c − g} and hence f +√g is measurable. √ Moreover, whenever f 2 is defined we find {f 2 ≥ c} = {f ≥ c}∪{f ≤ − c} for c ≥ 0 and {f 2 ≥ c} = Ω for c < 0, hence f 2 is measurable if f is and from f · g = 41 (f + g)2 − 41 (f − g)2 we deduce that if f and g are measurable and f · g is defined, then f · g is measurable. Note that since we have on R no vector space or algebra structure we cannot use our previous arguments to deduce that f ± g, αf or f · g are measurable if f and g are. We know that the pointwise limit of a sequence of continuous functions need not be continuous. For a sequence of measurable functions the situation is different. Theorem 4.12. Let (Ω, A) be a measurable space and for k ∈ N let fk : Ω → R be a measurable numerical function. Suppose that for all ω ∈ Ω the limit f (ω) := limk→∞ fk (ω) exists (in R). Then the function f : Ω → R is (1) A/B -measurable. In particular if f is real-valued, i.e. f : Ω → R, then f is A/B(1) -measurable. Proof. The last remark is obvious. Since for ω ∈ Ω f (ω) = lim inf fk (ω) = lim sup fk (ω) k→∞

k→∞

77

A COURSE IN ANALYSIS

holds, it is sufficient to prove that lim inf fk or lim sup fk are measurable. Moreover, by definition, compare Definition I.19.19,   lim sup fk = inf sup fl k→∞

k∈N

and

l≥k





lim inf fk = sup inf fl . k→∞

k∈N

l≥k

Thus we need to prove that supk∈N fk and inf k∈N fk are measurable as functions from Ω → R. However for c ∈ R we have   \ sup fk ≤ c = {fk ≤ c} k∈N

k∈N

implying already the measurability of sup fk , but inf fk is given by inf k∈N fk = − supk∈N (−fk ) and the theorem follows. For reference purpose we note as corollary to the proof of Theorem 4.12. Corollary 4.13. For a sequence fk : Ω → R of measurable functions the functions supk∈N fk , inf k∈N fk , lim supk→∞ fk and lim inf k→∞ fk are measurable. We want to introduce a further helpful notation: For a, b ∈ R we write a ∧ b := min(a, b)

(4.13)

a ∨ b := max(a, b)

(4.14)

and with the extensions a1 ∧ · · · ∧ aN = min{a1 , . . . , aN } and a1 ∨ · · · ∨ aN = max{a1 , . . . , aN }. For numerical functions f, g : G → R we define further f ∧g and f ∨g pointwisely as (f ∧g)(ω) := f (ω)∧g(ω) and (f ∨g)(ω) := f (ω)∨g(ω). Corollary 4.14. For a finite number of measurable functions fk : Ω → R, 1 ≤ k ≤ N, the functions f1 ∧ · · · ∧ fN and f1 ∨ · · · ∨ fN are measurable. Proof. We apply Corollary 4.13 to the sequence (fk )k∈N , fl = fN for l ≥ N. 78

4

MEASURABLE MAPPINGS

Since f + = f ∨ 0, f − = −(f ∧ 0) and f = f + − f − as well as |f | = f + + f − it follows that together with f , f + , f − and |f | are also measurable. In fact, f is measurable if and only if f + and f − are measurable. Finally we want to investigate a special class of measurable functions, namely step functions, or as they are often called in the context of Lebesgue’s theory simple functions or elementary functions. Let (Ω, A) be a measurable space. We ( start with the observation that for the 1, ω ∈ A , of a set A ⊂ Ω we have characteristic function χA , χA (ω) = 0, ω ∈ /A   ∅, c ≥ 1 {χA > c} = A, 1 > c ≥ 0,   Ω, 0 > c

(4.15)

and therefore χA is measurable if and only if A ∈ A. Definition 4.15. Let (Ω, A) be a measurable space. We call a non-negative, real-valued measurable function u : Ω → [0, ∞) a simple function if it attains only finitely many values. The set of all simple functions on Ω we denote by S(Ω). Let u : Ω → R be a simple function and {γ1 , . . . , γN } ⊂ [0, ∞) its range. Then there exist measurable sets C1 , . . . , CN ∈ A such that u|Cj = γj and assuming γk 6= γl for k 6= l it follows that Ck ∩ Cl = ∅. Hence we can write u=

N X

γj χCj .

(4.16)

j=1

Indeed, for every partition of Ω into measurable sets B1 , . . . , BM ∈ A such that u|Bj is constant we can write u=

M X k=1

βk χBk , βk ∈ {γ1, . . . , γN }.

In this general case however we cannot assume that βk 6= βl for k 6= l. 79

(4.17)

A COURSE IN ANALYSIS

Remark 4.16. As mentioned before, simple functions are also called elementary functions. Note that step functions in the sense of Definition I.25.2 and Definition II.18.5 are simple functions if they attain only nonnegative values. But even on an interval a simple function is not necessarily a step function since the sets Cj in (4.16) (or Bk in (4.17)) are not necessarily intervals (or hyper-cubes in the case of Rn ). It turns out that it is not practical to always assume we have a representation of the form (4.16), i.e.γk 6= γl for k 6= l. For example if we want to have a representation of the sum of two simple functions as a simple function we need to use a representation as (4.17). Definition 4.17. Let u ∈ S(Ω). For any finite partition (Aj )j=1,...,N of Ω into measurable sets Aj ∈ A we call u=

N X

(4.18)

αj χAj

j=1

a normal representation of u. Let u, v ∈ S(Ω) and α ∈ [0, ∞). Then the following functions belong to S(Ω) too: u+ , u− , |u|, αu, u + v, u · v, u ∧ v, u ∨ v. (4.19)

Clearly each of these functions attains only finitely many values and they are non-negative. Moreover by our previous considerations these functions are all measurable. Let u, v ∈ S(Ω) and assume that they have the normal representation u=

N X

αj χAj and v =

j=1

M X

βk χBk .

(4.20)

k=1

Introducing Cjk := Aj ∩ Bk , j = 1, . . . , N and k = 1, . . . , M

(4.21)

we obtain a partition of Ω, i.e. Cjk ∩ Cnm = ∅ for (j, k) 6= (n, m) and Ω=

N [ M [

j=1 k=1

80

Cjk .

(4.22)

4

MEASURABLE MAPPINGS

In addition we have Aj =

M [

Cjk and Bk =

u=

M N X X

Cjk

(4.23)

j=1

k=1

as well as

N [

ξjk χCjk and v =

j=1 k=1

M N X X

ηjk χCjk

(4.24)

j=1 k=1

where ξjk = αj and ηjk = βk . The advantage of the representations (4.24) is that we can now easily reduce operations for u and v to operations on the numbers ξjk and ηjk . For example we find u+v =

N X M X

(ξjk + ηjk )χCjk ,

j=1 k=1

and since χCjk · χCnm = 0 for (j, k) 6= (n, m) we have (u · v) = =

N X M X

ξjk χCjk

j=1 k=1

N X M X

!

M N X X

n=1 m=1

ηnm χCnm

!

ξrs ηrs χCrs .

r=1 s=1

Finally we want to prove that measurable functions are pointwise limits of simple functions. Theorem 4.18. A non-negative numerical function f : Ω → [0, ∞] is measurable if and only if it is the increasing limit of simple functions uk ∈ S(Ω) such that uk ≤ uk+1 and limk→∞ uk (ω) = f (ω) for all ω ∈ Ω. Proof. Clearly, for every increasing sequence of simple functions the limit is non-negative and measurable. Conversely let f : Ω → [0, ∞] be a measurable function. For k ∈ N we define the sets  Ajk := j2−k ≤ f < (j + 1)2−k , j = 0, . . . , k2k − 1 and

Ak2k ,k = {k ≤ f } 81

A COURSE IN ANALYSIS

which are measurable and in addition for every k ∈ N they form a partition of Ω. Hence k2k X uk := j2−k χAjk (4.25) j=1

is a simple function in normal representation and by construction we have |f (ω) − uk (ω)| ≤ 2−k for ω ∈ {f < k}. Since uk+1|Ajk (ω) can attain only the values (2j)2−(k+1) and (2j + 1)2−(k+1) for j = 0, . . . , k2−k − 1, it follows that uk ≤ uk+1 . If f (ω) = +∞ we have uk (ω) = k for all k ∈ N, and for f (ω) < ∞ we find uk (ω) ≤ f (ω) ≤ uk (ω) + 2−k for all k > f (ω). Hence supk∈N uk (ω) = f (ω), ω ∈ Ω, and since the sequence is monotone increasing, i.e. uk (ω) ≤ uk+1 (ω) for all ω ∈ Ω and k ∈ N, the supremum is indeed a limit. If f : Ω → [−∞, ∞] is any numerical measurable function then we can decompose f according to f = f + − f − . Moreover for f + exists a sequence (uk )k∈N , uk ∈ S(Ω), converging pointwisely to f + , and for f − exists a sequence (vk )k∈N , vk ∈ S(Ω), converging pointwisely to f − . Consequently the sequence wk := uk − vk converges pointwisely to f and further we have |ωk | = uk + vk ≤ f + + f − = |f |. Thus we have proved Corollary 4.19. A numerical function f : Ω → R defined on a measurable space (Ω, A) is measurable if and only if there exists a sequence (wk )k∈N , wk ∈ S(Ω) − S(Ω), such that limk→∞ wk (ω) = f (ω) holds for all ω ∈ Ω. Recall that w ∈ S(Ω) − S(Ω) means that w = u − v, u, v ∈ S(Ω). There is a different way to phrase Corollary 4.19. We introduce span (S(Ω)), the span of S(Ω), and we denote by E(Ω) the set of all numerical measurable functions f : Ω → [−∞, ∞]. Then Corollary 4.19 states that span (S(Ω)) is sequentially dense in E(Ω) with respect to pointwise convergence. When we have to emphasise the σ-field A, we will write E(Ω, A) instead of E(Ω). Moreover, by E(Ω) (or E(Ω, A)) we denote all real-valued measurable functions f : Ω → R. 82

4

MEASURABLE MAPPINGS

Problems In the case of mappings between subsets on Rk and Rm we always consider the corresponding Borel σ-field except in some cases where we explicitly give the σ-field of interest. 1.

a) For arbitrary sets A, B, Ak ⊂ Ω, k ∈ N, prove the following: i) A ⊂ B implies χA ≤ χB ;

ii) χA{ = 1 − χA ;

iii) χSk∈N Ak = supk∈N χAk ; iv) χTk∈N Ak = inf k∈N χAk .

b) Let (Ω, A) be a measurable space.SSuppose that supk∈N χAk is measurable, Ak ⊂ Ω, k ∈ N. Deduce that k∈N Ak is measurable. Does this imply that all of the sets Ak are measurable, i.e. elements of A? 2.

a) Let (Xj , dj ), j = 1, . . . , N be metric spaces and (X, d) their product space. Denote by B(Xj ) and B(X) the corresponding Borel σ-fields and prove that ⊗N j=1 B(Xj ) = B(X). b) Solve Exercise 4.3.

3. Prove that for M measurable functions gj : Rn → R the product g := QM n j=1 gj : R → R is measurable too. 4.

a) Let K ⊂ Rn be a convex set and h : K → R a concave function. Is h a B(n) /B(1) -measurable function?

b) Let g : (a, b) → R be differentiable. Prove that g 0 : (a, b) → R is measurable. (We do not assume that g ∈ C 1 ((a, b)), but only that g is differentiable.) 5.

a) For a measurable function f : [a, ∞) → R consider the function g : Rn → R, g(x) = f (||x||). Is f measurable?

b) Let f : Rn → R be a measurable function and σ ∈ Sn where Sn denotes the symmetric group. Is the function σ(f ) : Rn → R, σ(f )(x) = f (xσ(1) , . . . , xσ(n) ) measurable? 83

A COURSE IN ANALYSIS

6. Let f : Ω → R be a measurable function on a measurable space (Ω, A). Prove the existence of a sequence of functions ϕk ∈ S(Ω) − S(Ω) such that |ϕk (ω)| ≤ |ϕk+1(ω)| and limk→∞ ϕk (ω) = f (ω), ω ∈ Ω. (1)

7. Prove that BR = R ∩ B = B(1) . 8. We call u : R → R of Baire class zero if u is continuous. If u : R → R is the pointwise limit of functions of Baire class zero but not of Baire class zero itself we call u of Baire class 1. In general u : R → R is of Baire class k∈ N0 if u is the pointwise limit of functions of Baire class k − 1 but not of Baire class k − 1 itself. Prove that if u is of Baire class k it is measurable. For u defined by   1, |x| < 1, u(x) = 21 , |x| = 1,   0, |x| > 1, show that u is of Baire class 1. Hint: consider the sequence uk : R → R, uk (x) =

84

1 . 1+x2k

5

Integration with Respect to a Measure — The Lebesgue Integral

Let (Ω, A, µ) be a measure space and denote by S(Ω) the set of all simple functions u : Ω → R. Suppose that {γ1, . . . , γK } ⊂ [0, ∞) is the range of u and γj 6= γk for j 6= k. In this case the sets Cl := {u = γl }, l = 1, . . . , K, are mutually disjoint and we can define the (µ−)integral of u by Z K X u dµ := γl µ(Cl ) ∈ [0, ∞]. (5.1) l=1

Lemma 5.1. For u ∈ S(Ω) with normal representation u = have Z M X βk µ(Bk ) = u dµ. k=1

Proof. We show that for two normal representations u = P u= N j=1 αj χAj the following holds N X

αj µ(Aj ) =

j=1

M X

PM

k=1

βk χBk we (5.2)

PM

k=1 βk χBk

βk µ(Bk ).

k=1

With Cjk = Aj ∩ Bk , ξjk = αj and ηjk = βk we find as in (4.24) that N X M X

ξjk χCjk =

j=1 k=1

and therefore

M N X X

N X M X

ξjk µ(Cjk ) =

ξjk µ(Cjk ) =

j=1 k=1

and

N M X X k=1 j=1

ηjk µ(Cjk ) =

ηjk χCjk

k=1 j=1

N M X X

ηjk µ(Cjk ),

k=1 j=1

j=1 k=1

but

N M X X

N X

αj µ

M [

k=1

M X

N [

j=1

k=1

85

=

!

=

Cjk

j=1

βk µ

!

Cjk

N X

αj µ(Aj )

j=1

M X k=1

βk µ(Bk ).

and

(5.3)

A COURSE IN ANALYSIS

Before we investigate the µ-integral further, let us compare the situation with the Riemann integral where we started to define the integral on a hyperrectangle K ⊂ Rn for step functions f : K → R. In this case we first chose a partition of K into hyper-rectangles Kα and then we considered functions being constant on Kα , say f |Kα = cα . The volume or measure of Kα was taken as the classical volume as defined for hyper-rectangles in Rn . The last point was the serious and difficult restriction: only for hyper-rectangles we know the volume from classical geometry and consequently only for a partition of K into hyper-rectangles Kα we can form Z X f (x) dx := cα voln (Kα ). (5.4) K

α∈J

Now we start with a measurable function attaining (as a step function) only finitely many values, from this we obtain the partition of the domain into measurable sets and hence we can form (5.1). This allows us a priori to consider as the domain of the function any measurable set in a measurable space (Ω, A) and we are neither restricted to hyper-rectangles nor to Rn . Of course we have to pay a price: we have to introduce measure spaces and measurable mappings. In the case of the Riemann integral we just do this after having introduced an integral for functions we can approximate with step functions. It turns out that for n ≥ 2 this is a rather involved and less satisfactory approach, although it looks to be the more natural one. Proposition 5.2. Let (Ω, A, µ) be a measure space. For u, v ∈ S(Ω), and A ∈ A the following hold Z χA dµ = µ(A); Z Z (αu) dµ = α u dµ; Z Z Z (u + v) dµ = u dµ + v dµ; Z Z u ≤ v implies u dµ ≤ v dµ.

α≥0 (5.5) (5.6) (5.7) (5.8)

Proof. Of course, (5.5) and (5.6) are consequences of the definition. For (5.7) P α we start with normal representations of u and v, i.e. u = N j=1 j χAj and 86

5

INTEGRATION WITH RESPECT TO A MEASURE

P v= M k=1 βk χBk , and we switch to the common partition Cjk := Aj ∩ Bk ,1 ≤ j ≤ N, 1 ≤ k ≤ M, as in (4.24). Hence we have u=

N X M X

ξjk χCjk and v =

j=1 k=1

N M X X

ηjk χCjk

k=1 j=1

which yields u+v =

N X M X

(ξjk + ηjk )χCjk

j=1 k=1

and (5.7) follows from N X M X

(ξjk + ηjk )µ(Cjk ) =

j=1 k=1

N X M X

ξjk µ(Cjk ) +

j=1 k=1

=

N X

ηjk µ(Cjk )

k=1 j=1

αj µ(Aj ) +

j=1

N M X X

M X

βk µ(Bk ).

k=1

Moreover, since u ≤ v implies ξjk ≤ ηjk for 1 ≤ j ≤ N, 1 ≤ k ≤ M, we deduce (5.8) from N X j=1

αj µ(Aj ) =

N X M X j=1 k=1

ξjk µ(Cjk ) ≤

N M X X k=1 j=1

ηjk µ(Cjk ) =

M X

βk µ(Bk ).

k=1

P Exercise 5.3. Let u = N j=1 αj χAj for a partition of Ω into measurable sets R P u dµ but do not assume that αj 6= αk for j 6= k. Prove that N j=1 αj µ(Aj ) = still holds. Example 5.4. In Example I.25.12.A we have seen that the Dirichlet function χQ∩[0,1] : [0, 1] → R is not Riemann integrable. Since Q∩[0, 1] is a measurable set in the trace σ-field B(1) ([0, 1]) = [0, 1] ∩ B(1) , it follows that χQ∩[0,1] is an element of S([0, 1]), hence its λ(1) -integral is well defined. The set Q is R (1) countable and therefore λ(1) (Q) = 0 implying χQ∩[0,1] dλ|[0,1] = 0 where (1)

λ|[0,1] is the Lebesgue measure on R restricted to [0, 1]. The example already shows that not all λ(1) -integrable functions are Riemann integrable. 87

A COURSE IN ANALYSIS

Let u ∈ S(Ω) with ran(u) = {α1 , . . . , αN } ⊂ [0, ∞). The function ω 7→ up (ω) p is for every p > 0 well defined, has range {α1p , . . . , αN } and since {u ≥ α} = Ω 1 p for α ≤ 0 and {u ≥ α} = {u ≥ α p } for α > 0, it follows that up is measurable, hence up ∈ S(Ω). For p > 1 we define as usual the conjugate exponent q by 1p + 1q = 1. Consider u, v ∈ S(Ω) with the normal representation u=

N X

αj χAj and v =

j=1

N X

βj χAj .

j=1

Since αj , βj ≥ 0 and u · v, up , v q ∈ S(Ω) we deduce using H¨older’s inequality (for finite sums) Z

|u · v| dµ = =

N X j=1

N X j=1

≤ =

|αj · βj |µ(Aj ) 1

1

|αj |µ(Aj ) p |βj |µ(Aj ) q

N X j=1

Z

|αj |p µ(Aj ) p

|u| dµ

! p1

 p1 Z

N X j=1

q

|βj |q µ(Aj )

|v| dµ

! q1

 1q

and we have proved Lemma 5.5. For u, v ∈ S(Ω) and Z

|u · v| dµ ≤

Z

1 p

+ p

1 q

= 1, p > 1, H¨ older’s inequality

|u| dµ

 p1 Z

q

|v| dµ

 1q

(5.9)

holds. Corollary 5.6. For u, v ∈ S(Ω) and p ≥ 1 Minkowski’s inequality holds, i.e. Z  p1 Z  p1 Z  p1 p p p |u + v| dµ ≤ + (5.10) |u| dµ |v| dµ . 88

5

INTEGRATION WITH RESPECT TO A MEASURE

Proof. First we note that we do not need the absolute value in (5.10) since u, v ≥ 0. Further note that for p = 1 we even have by (5.7) equality in (5.10). For p > 1 it follows Z Z |u + v|p dµ = (u + v)(u + v)p−1 dµ Z Z p−1 = u(u + v) dµ + v(u + v)p−1 dµ, 1 p

and now H¨older’s inequality yields with Z

p

|u + v| dµ ≤

Z

p

u dµ +

 p1 Z

Z

+

1 q

=1

|u + v|

p

|v| dµ

q(p−1)

 p1 Z

However q(p − 1) = p and therefore we find Z or

p

Z

Z

p

|u + v| dµ ≤

p

u dµ

|u + v| dµ

 p1

Clearly, for u, v ∈ S(Ω) with Z

and for α < 0

R

 p1

≤

+

Z

v dµ

p

 p1

p

u dµ

u dµ < ∞ and

(u − v) dµ := Z

Z

Z

(αu) dµ := −

u dµ − Z

|u + v|

q(p−1)

 p1 ! Z

+

R

dµ

 1q

Z

p

dµ

 q1

p

.

|u + v| dµ

v dµ

 p1

 q1

.

v dµ < ∞ we can define

Z

v dµ

(−αu) dµ.

(5.11)

(5.12)

But we prefer to investigate this “linear” extension in a larger frame after having studied the behaviour of the integral on S(Ω) under monotone increasing limits. 89

A COURSE IN ANALYSIS

Theorem 5.7. Let (Ω, A, µ) be a measurable space and (ul )l∈N , ul ∈ S(Ω), a monotone increasing sequence, i.e. ul ≤ ul+1 . For u ∈ S(Ω) we have Z Z u ≤ sup ul implies u dµ ≤ sup ul dµ. (5.13) l∈N

l∈N

P Proof. Let u = N k=1 αk χAk be a normal representation of u and α ∈ (0, 1). It follows that Bl := {ul ≥ αu} ∈ A and ul ≥ αuχBl , implying for l ∈ N Z Z ul dµ ≥ α uχBl dµ. SinceS(ul )l∈N is monotone increasing and u ≤ supl∈N uS l we find Bl ⊂ Bl+1 and l∈N Bl = Ω as well as Ak ∩ Bl ⊂ Ak ∩ Bl+1 and l∈N (Ak ∩ Bl ) = Ak , k = 1, . . . , N. We know that µ is continuous from below, Theorem 2.10.A, and therefore we find Z Z sup ul dµ ≥ sup α uχBl dµ l∈N l∈N Z Z = α lim uχBl dµ = α u dµ. l→∞

Since α ∈ (0, 1) was arbitrary (5.13) follows. Corollary 5.8. For two monotone increasing sequences (un )n∈N and (vm )m∈N in S(Ω) it follows that Z Z sup un = sup vm implies sup un dµ = sup vm dµ. (5.14) n∈N

m∈N

n∈N

m∈N

Proof. For k, l ∈ N we have vl ≤ supn∈N un and uk ≤ supm∈N vm and Theorem 5.7 now implies Z Z Z Z vl dµ ≤ sup un dµ and uk dµ ≤ sup vm dµ. n∈N

m∈N

Now taking the supremum with respect to l and k, respectively, gives (5.14). With Theorem 4.18 in mind we can now give 90

5

INTEGRATION WITH RESPECT TO A MEASURE

Definition 5.9. Let (Ω, A, µ) be a measure space and f : Ω → [0, ∞] a nonnegative, numerical measurable function. The integral of f over Ω with respect to µ is defined by Z Z f dµ = sup uk dµ (5.15) k∈N

where (uk )k∈N is any monotone increasing sequence of functions in S(Ω) such that u = supk∈N uk = limk→∞ uk ≤ ∞. Remark 5.10. A. By Corollary 5.8 the definition is independent of the choice of the approximating sequence. R B. Sometimes we call f dµ the µ-integral if we want to emphasise the measure. From Proposition 5.2 we deduce (see Problem 5) Corollary 5.11. For f, g : Ω → [0, ∞] measurable and α ≥ 0, A ∈ A the following hold Z χA dµ = µ(A); (5.16) Z Z Z (f + g) dµ = f dµ + g dµ; (5.17) Z Z (αf ) dµ = α f dµ; (5.18) Z Z f ≤ g implies f dµ ≤ g dµ. (5.19) Let u, v : Ω → [0, ∞] be measurable functions such that u = supk∈N uk = limk→∞ uk and v = supl∈N vl = liml→∞ vl for monotone increasing sequences (uk )k∈N and (vl )l∈N of simple functions on Ω. It follows that u·v = limk→∞ uk · limk→∞ vk = limk→∞ (uk vk ) and for wk := uk vk we find wk = uk vk ≤ uk+1vk+1 = wk+1 , hence (uk · vk )k∈N is an increasing sequence of simple functions approximating u · v. Moreover we have up = |u|p = supk∈N |uk |p and v p = |v|p = supl∈N |vl |p . Consequently we can extend H¨ older’s inequality and Minkowski’s inequality to non-negative, numerical measurable functions. 91

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Corollary 5.12. For u, v : Ω → [0, ∞] measurable and p > 1, have  1q  p1 Z Z Z q p |v| dµ |u| dµ |uv| dµ ≤

1 p

+

1 q

= 1 we (5.20)

and for p ≥ 1 Z

p

|u + v| dµ

 p1

≤

Z

p

|u| dµ

 p1

+

Z

p

|v| dµ

 p1

.

(5.21)

The following result is due to B. Levi and gives an extremely powerful tool when studying the interchangeability of limits and integrals. It is either called Theorem of Beppo Levi or Monotone Convergence Theorem. Theorem 5.13. Let (Ω, A, µ) be a measure space and (fk )k∈N , fk : Ω → [0, ∞], an increasing sequence of measurable functions. Then supk∈N fk is a numerical, non-negative measurable function, hence µ-integrable, and we have Z Z sup fk dµ = sup fk dµ. (5.22) k∈N

k∈N

Proof. Set f := supk∈N fk . We prove the existence of a sequence of simple functions (vn )n∈N which is monotone increasing and satisfies supn∈N vn = f as well as vn ≤ fn . For such a sequence we obtain from (5.15) Z Z Z f dµ = sup vn dµ = sup vn dµ n∈N

and on the other hand we have Z

R

f dµ = sup n∈N

n∈N

vk dµ ≤

Z

R

fn dµ implying

vn dµ ≤ sup n∈N

Z

fn dµ,

R R R but f ≤ f further implies that f dµ ≤ f dµ, i.e. we get sup fn dµ ≤ n n n∈N R f dµ and (5.22) follows. Now we construct the sequence (vn )n∈N , vn ∈ S(Ω). For fk there exists a monotone increasing sequence (umk )m∈N of simple functions such that supm∈N umk = fk . Clearly we have vm := vm1 ∨ · · · ∨ umm ∈ S(Ω) and further vm ≤ vm+1 by construction. Since (fk )k∈N is increasing it follows that 92

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vm ≤ fm for all m ∈ N, implying supm∈N vm ≤ f . Moreover, for m ≥ k we find umk ≤ vm and therefore sup umk = fk ≤ sup vm ≤ f,

m∈N

m∈N

which yields supm∈N vm = supk∈N fk = f . Hence, (vm )m∈N has all the desired properties. Remark 5.14. The name “monotone convergence theorem” becomes more clear when we give it the following formulation: let (fk )k∈N be a monotone increasing sequence of non-negative, numerical functions on Ω converging pointwisely to f , i.e. limk→∞ fk = supk∈N fk = f . Then we have Z Z fk dµ = lim fk dµ. (5.23) lim k→∞

k→∞

Corollary 5.15. Let (fk )k∈N be any sequence of non-negative, numerical functions on the measure space (Ω, A, µ). Then the numerical function PN P∞ k=1 fk is measurable and k=1 fk = limN →∞ Z X ∞

fk dµ =

k=1

∞ Z X

fk dµ.

(5.24)

k=1

Proof. P We only have to apply Theorem 5.13 to the sequence of partial sums gN := N k=1 fk .

Remark 5.16. It is important to emphasise that (5.22), (5.23) or (5.24) are statements for non-negative numerical functions, hence the value +∞ is not excluded and convergence refers to convergence in [0, ∞]. Of different nature is the question whether we have convergence in R provided all functions fk are real-valued.

Example 5.17. Let (Ω, A) be any measurable space and for ω ∈ Ω let ω be the Dirac measure at ω, i.e. ω (A) = 1 for ω ∈ A and ω (A) = 0 for ω∈ / A. For any non-negative measurable numerical functions f : Ω → [0, ∞] we have Z f dω = f (ω).

93

(5.25)

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Indeed, for f ∈ S(Ω) with normal representation f = Z

f dω =

N X

PN

k=1

αk χAk it follows

αk ω (Ak ) = αk0 = f (ω)

k=1

for some αk0 ∈ {α1 , . . . , αN } and ω ∈ Ak0 . Now, if f : Ω → [0, ∞] is measurable and (fk )k∈N , fk ∈ S(Ω), is an increasing sequence converging to f we find Z Z Z fk dω = lim fk (ω) = f (ω). f dω = lim fk dω = lim k→∞

k→∞

k→∞

Example 5.18. Consider the measurable space (N, P(N)). Since a measure is σ-additive every measure on P(N) is determined by the sequence αk := µ({k}) ∈ [0, ∞], k ∈ N. A non-negative, numerical measurable function f : N → [0, ∞] can be described in the following way. Set fk := f (k)χ{k} . This is a measurable function and for f (k) < ∞ it is a P simple function, P∞ i.e. f = f (k) < ∞ implies fk ∈ S(Ω). Moreover f = limN →∞ N k k=1 fk . k=1 Therefore we find with Example 5.17 that Z

f dµ =

Z X ∞

fk dµ =

Z

f dµ =

k=1

i.e.

∞ X

f (k) αk =

k=1

∞ X

∞ X

f (k)µ({k}),

k=1

f (k)µ({k}).

(5.26)

k=1

(1) Example 5.19. A. For N ∈ N and p ∈ [0, 1], q := consider
k N −kon (R, B ) P1−p, N p N k . For the the Bernoulli distribution B(n, p) := βN := k=0 k p q non-negative measurable function f (x) = χ[0,∞) (x) · x we find

Z

f R

p dβN

Z N   X N k N −k = p q f dk k k=0 N   N   X X N k N −k N k N −k = p q f (k) = p q k k k k=0 k=1 = Np(p + q)N −1 = Np. 94

5

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B. For the Poisson distribution with parameter γ > 0, i.e. for the measure P −γ γ k  on (R, B(1) ), we find for f as in part A πγ := ∞ k=0 e k! k Z

f dπγ =

∞ X k=1

e−γ

γk k = γ. k!

(It would be a beneficial exercise to complete the remaining details in these two calculations.) Let [a, b], a < b be an interval and a = t0 < t1 < . . . < tN = b be a partition of [a, b]. Further let ϕ : [a, b] → R be a bounded, non-negative step function with values ϕ|(tk−1 ,tk ) = ck ≥ 0. This function is also a simple function when using on [a, b] ∈ [a, b] ∩ B(1) the partition into measurable sets (tk−1 , tk ), k = 1, . . . , N, and {tk }, k = 0, . . . , N. Since with respect to the measure (1) (1) λ[a,b] the sets {t} are negligible, i.e. λ[a,b] ({tk }) = 0, it follows that Z

(1)

ϕ dλ[a,b] = =

N X

k=1 N X

(1)

ck λ[a,b] ((tk−1 , tk )) +

N X k=0

(1) ck λ[a,b] ((tk−1 , tk ))

k=1

=

Z

(1)

ϕ(tk )λ[a,b] ({tk })

b

ϕ(x) dx

a

Rb (1) where λ[a,b] denotes the restriction of λ(1) to [a, b]∩B(1) and as usual a ϕ(x) dx denotes the Riemann integral of ψ. Thus we conclude that the Riemann (1) integral of a non-negative step function coincides with its λ[a,b] -integral (Lebesgue integral) of the corresponding simple function. But we already know that there are simple functions which are not step functions and which are not Riemann integrable. In the following we extend the integral to a class of real-valued functions which are not necessarily non-negative anymore. Before this it is maybe appropriate to acknowledge the influence of H. Bauer [11] on our presentation in parts of this in the following chapters. This is partly due to the fact that more than 25 years ago the first named author was much occupied with a critical proof -reading of a first draft of [11]. Definition 5.20. Let (Ω, A, µ) be a measure space and f : Ω → R be a numerical, measurable function. We call f (µ−)integrable if the integrals 95

A COURSE IN ANALYSIS

R

f + dµ and

R

f − dµ are real numbers, i.e. finite. The value Z Z Z + f dµ := f dµ − f − dµ

(5.27)

is called the (µ-)integral of f . Remark 5.21. A. For a measurable function f : Ω → [0, ∞] we have f + = f and f − = 0, hence (5.27) is consistent with our previous definitions of a (µ-)integral for non-negative numerical and measurable functions. B. The following notations are sometimes useful: Z Z Z f (ω) dµ(ω) := f (ω) µ(dω) := f dµ. (5.28) Our first result provides us with equivalent criteria for integrability. Theorem 5.22. Necessary and sufficient for the integrability of a measurable numerical function f : Ω → R is any of the following conditions: i) f + and f − have a finite integral; ii) f = u − v for two integrable functions u ≥ 0 and v ≥ 0 with finite integral; iii) there exists an integrable function g with finite integral such that |f | ≤ g; iv) |f | has a finite integral. Proof. Statement i) is just the definition of the integrability of f and with u = f + and v = f − it implies ii). Suppose now that ii) holds. Since u + v is integrable (with finite integral) and f = u − v ≤ u ≤ u + v and −f = v − u ≤ v ≤ u + v we deduce iii) from ii) with g := u + v. Further we note that (5.19) entails that iii) implies iv). Finally, if |f | has a finite integral, then f + ≤ |f | and f − ≤ |f | have finite integrals too and hence iv) yields i). Theorem 5.23. The set of all µ-integrable functions f : Ω → R form an R-vector space and the integral is a positivity preserving linear form on this 96

5

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vector space, i.e. for integrable functions f and g and for α, β ∈ R the function αf + βg is integrable too and the following hold: Z Z Z (αf + βg) dµ = α f dµ + β g dµ; (5.29) and f ≥ 0 and

Z

f dµ ≥ 0.

(5.30)

Proof. Since f = f + −f − and g = g + −g − we find f +g = (f + +g + )−(f − +g − ) implying the integrability of f + g. Further, for α ≥ 0 we have (αf )+ = αf + and (αf )− = αf − whereas for α ≤ 0 we have (αf )+ = |α|f − and (αf )− = |α|f + , hence αf = αf + − αf − , α ≥ 0, or αf = |α|f − − |α|f + for α ≤ 0. Thus αf is integrable and therefore we deduce that αf + βg is integrable. From f + g = (f + g)+ − (f + g)− = f + + g + − (f − + g − ) we deduce (f + g)+ + f − + g − = f + + g + + (f + g)− and using the additivity of the integral for non-negative functions we obtain Z Z Z Z Z Z + − + − + (f + g) dµ − (f + g) dµ = f dµ − f dµ + g dµ − g − dµ, i.e. the additivity of the integral. Further, for α ≥ 0 we find Z Z Z + αf dµ = (αf ) dµ − (αf )− dµ Z Z Z + − = α f dµ − α f dµ = α f dµ, and for α ≤ 0 we get Z Z Z + αf dµ = (αf ) dµ − (αf )− dµ Z  Z Z Z = |α| f − dµ − |α| f + dµ = −|α| f + dµ − f − dµ Z = α f dµ and (5.29) is proved. Finally, (5.30) follows from Remark 5.21.A and (5.19).

97

A COURSE IN ANALYSIS

Corollary 5.24. For two integrable functions f, g : Ω → R the functions f ∧ g and f ∨ g are integrable too. Furthermore f ≤ g implies Z Z f dµ ≤ g dµ (5.31) and for all f we have

Z Z f dµ ≤ |f | dµ.

(5.32)

In addition H¨ older’s inequality (5.20) and Minkowski’s inequality (5.21) hold for integrable functions f and g, i.e. for p > 1 and p1 + 1q = 1 we have Z  1p Z  q1 Z p q |f g| dµ ≤ |f | dµ |g| dµ (5.33) and for p ≥ 1  p1 Z Z  p1 Z  p1 p p p |f + g| dµ |f | dµ |g| dµ . ≤ +

(5.34)

Note that no assumption is made about the finiteness of the integrals in (5.33) and (5.34). Proof. Since f ∧ g = 12 (f + g − |f − g|) and f ∨ g = 21 (f + g + |f − g|) the integrability of f ∨ g and f ∧ g follows immediately. Moreover, (5.31) follows from (5.30) and (5.29) since 0 ≤ g − f . In order to see (5.32) we note that Z Z Z Z Z Z f dµ = f + dµ − f − dµ ≤ f + dµ + f − dµ = |f | dµ.

Since |f g| = |f ||g| H¨older’s inequality follows from (5.20). Noting that |f + g| ≤ |f | + |g| we derive from (5.21) Z  p1 Z  p1 p p |f + g| dµ ≤ (|f | + |g|) dµ

and now we may apply (5.21). Example 5.25. For (N, P(N), µ), µ(k) = αk ≥ 0, and a function f : N → R the integrability with respect to µ is equivalent to the condition ∞ X k=1

|f (k)|αk < ∞, 98

5

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and the µ-integral is given by Z

f dµ =

∞ X

f (k) αk .

k=1

Let (Ω, A, µ) be a measure space and A ∈ A. For a measurable function f : Ω → R we may consider the restriction f |A : A → R and we may ask whether fA is integrable. There are two possibilities to approach the question: we may consider f χA : Ω → R which coincides on A with f |A and is zero in the complement of A, or we may consider the measure space (A, A ∩ A, µ|A ) where µ|A denotes the restriction of µ to A ∩ A. Of course we expect that both approaches lead to the same result. We start with Definition 5.26. Let (Ω, A, µ) be a measure space and f : Ω → R be a measurable function which is either non-negative or integrable. For A ∈ A we define the (µ-)integral of f over A as Z Z f dµ := χA f dµ. (5.35) A

This definition is of course consistent with notation Z Z f dµ = f dµ.

(5.36)

Ω

The next lemma relates the two approaches described above. Lemma 5.27. For A ∈ A and every measurable function f : Ω → R which is either non-negative or integrable we have Z Z (5.37) f dµ = f |A dµ|A , A

where the integral on the right hand side is an integral in the measure space (A, A ∩ A, µ|A ). Proof. First consider the case f ≥ 0. Since A ∈ A it follows that fA : A → R is measurable (with A replaced by A∩A). For f we can find simple functions (uk )k∈N , uk ∈ S(Ω), such that uk ≤ uk+1 and limk→∞ uk = χA f , moreover uk |A ∈ S(A), uk |A ≤ uk+1|A and limk→∞ uk |A = f |A . Consequently we have Z Z Z Z f dµ = sup uk dµ and f |A dµ|A = sup uk |A dµ|A . (5.38) A

k∈N

k∈N

99

A COURSE IN ANALYSIS

Since 0 ≤ uk ≤ χA f we deduce that uk |A{ = 0, hence uk χA = uk |A and (5.38) implies the lemma for f ≥ 0. In the case where f : Ω → R and is integrable, we decompose f into positive and negative parts and apply the statement just proved for f ≥ 0. A trivial consequence is Corollary 5.28. The function f : A → R, A ∈ A, is µ-integrable if and only if fA defined by ( f (ω), ω ∈ A fA (ω) := 0, ω ∈ A{ is µ-integrable and in this case

R

A

f dµ =

R

fA dµ holds.

The following properties follow in a straightforward way from properties of the integral and those of characteristic functions. For a measure space (Ω, A, µ), sets A, B ∈ A and measurable functions f, g : Ω → R such that they are either non-negative or µ-integrable we have Z Z Z Z f dµ + f dµ = f dµ + f dµ, (5.39) A∩B

A∪B

A

B

in particular for A ∩ B = ∅ we have Z Z Z f dµ = f dµ + f dµ. A∪B

A

(5.40)

B

Moreover we have that f |A ≤ g|A implies

Z

A

f dµ ≤

Z

g dµ.

(5.41)

A

Lemma 5.29. Let (Ω, A, µ) be a measurable space and N ∈ A a set of measure zero, i.e. µ(N) = 0. Further let f : Ω R→ R be a measurable numerical function. Then f is integrable over N and N f dµ = 0. Moreover, if f is integrable then µ ({|f | = ∞}) = 0. R Proof. For k ∈ N the function uk := kχN is a simple function and uk dµ = kµ(N) = 0. Moreover the R sequence (uk )k∈N is increasing and limk→∞ uk = supk∈N uk implying that supk∈N uk dµ = 0. Since χN f ≤ supk∈N uk holds 100

5

INTEGRATION WITH RESPECT TO A MEASURE

for a Rnon-negative, measurable numerical function f : Ω → R, we conclude that N f dµ = 0. The general case now follows from the decomposition f = f + − f − and R the definition ofR the integral. Now suppose that f is integrable, implying f + dµ < ∞ and f − dµ < ∞. Hence we need to prove that for an integrable, non-negative numerical function f we have µ({f = ∞}) = 0. If µ({f = ∞}) > 0 then we have Z Z +∞ = +∞µ({f = ∞}) = (+∞)χ{f =∞} dµ ≤ f dµ < ∞ which is a contradiction. Example 5.30. A. For f ∈ Rn → R the support supp f is defined by supp f = {x ∈ Rn | f (x) 6= 0}, see (II.14.12). As a closed set supp f is Borel measurable and it follows that f is λ(n) -integrable if and only if f |supp f is λ(n) -integrable over supp f and the following holds Z Z f dµ. f dµ = supp f

B. Let f : Rn → R be a continuous function with compact support supp f . In this case |f | is bounded by M := maxx∈supp f |f (x)| and of course we have |f |supp f ≤ M. On the compact set supp f the constant function x 7→ M is a simple function, hence it is integrable and by Theorem 5.22.iii) we deduce that f |supp f is integrable and Z Z (n) f dλ = (χsupp f )f dλ(n) . (5.42) suppf

Thus all continuous functions f : Rn → R with compact support are λ(n) integrable as they are λ(n) -integrable over every measurable subset of their support. C. Let K ⊂ Rn be a compact set and h : K → R a bounded measurable function, (|h(x)| ≤ M for all x ∈ K. We extend h to H : Rn → R by h(x), x ∈ K (n) H(x) := . Since x 7→ M is λ|K -integrable it follows that h 0, x ∈ K{ (n)

is λ|K -integrable and Z Z Z (n) h dλ|K = H dλ(n) = HχK dλ(n) . 101

A COURSE IN ANALYSIS

In particular continuous functions defined on a compact set K ⊂ Rn are λ(n) -integrable over K and of its Borel subsets. Let us compare these results with the situation we have encountered in Chapter II.20 when introducing the Riemann integral for subsets of Rn not being a hyper-rectangle. The starting point was a bounded function f : G → R defined on a bounded Jordan measurable set G ⊂ Rn and with f we considered ( f (x), x ∈ G to Rn , see Definition II.20.1 its canonical extension f˜G (x) = 0, x∈ /G and note that H in Example 5.30.C is nothing but the canonical extension of h. We then defined the Riemann integral of f over G by Z Z f (x) dx := f˜G |K (x) dx (5.43) G

K

˚ see (II.20.2). The where K is any compact cell in Rn such that G ⊂ K, crucial point was the Jordan measurability of G which is essentially a property of ∂G, i.e. it depends on the topology. Now we can construct or define integrals over measurable subsets independent of any topological conditions. Of course, if the σ-field under consideration is related to a topology, e.g. if it is a Borel σ-field, then measurable sets will have in general relations to the elements of the topology, however usually in a less restricted sense as Jordan measurable sets do have. In Chapter 8 we will discuss cases where the Riemann approach to define volume integrals over a Jordan measurable subset of Rn coincides with the λ(n) -integral. In Definition 3.8 we have introduced the image measure: Let (Ω, A) and ˜ A) ˜ be two measurable spaces and µ a measure on (Ω, A). Further let (Ω, ˜ be a measurable mapping. Then the image measure T (µ) was T :Ω→Ω ˜ A) ˜ by defined as a measure on (Ω, ˜ = µ(T −1 (A)), ˜ ˜ T (µ)(A) A˜ ∈ A.

(5.44)

We now want to study integration with respect to an image measure. ˜ A) ˜ a measurable Theorem 5.31. Let (Ω, A, µ) be a measure space and (Ω, ˜ be a measurable mapping and f˜ : Ω ˜ → [0, ∞] a space. Further let T : Ω → Ω numerical, non-negative, measurable function. In this situation we have Z Z f˜ dT (µ) = (f˜ ◦ T ) dµ. (5.45) ˜ Ω

Ω

102

5

INTEGRATION WITH RESPECT TO A MEASURE

Proof. First we note that since f˜ ≥ 0 the integral on the left hand side is defined and the measurability of T implies that f˜ ◦ T is measurable and of course f˜ ◦ T ≥ 0, hence the integral on the right hand side is also defined. ˜ i.e. f˜ ∈ S(Ω), which yields that Now let f˜ be a simple function on Ω, f˜ =

N X

αk χA˜k

k=1

˜ k = 1, . . . , N. The sets for αk ≥ 0 and mutually disjoint sets A˜k ∈ A, −1 ˜ Ak := T (Ak ) belong to A and further we find f˜ ◦ T =

N X

αk χAk .

k=1

Taking into account that T (µ)(A˜k ) = µ(Ak ) we deduce N X

αk T (µ)(A˜k ) =

k=1

N X

αk µ(Ak ),

(5.46)

k=1

i.e. (5.45) holds for simple functions. The general result follows now by approximation. For f˜ ≥ 0 we can find a sequence of simple functions (˜ uk )k∈N , u˜k ≤ u˜k+1 and limk→∞ u˜k = f . The sequence (˜ uk ◦ T )k∈N is now a sequence in S(Ω), i.e. u˜k ◦ T ∈ S(Ω), u˜k ◦ T ≤ u˜k+1 ◦ T and limk→∞ u˜k ◦ T = f˜ ◦ T . Thus we may pass in (5.45) to the limit and (5.44) follows. ˜ → R and T as in Theorem 5.31 we have For f˜ : Ω (f˜ ◦ T )+ = f˜+ ◦ T and (f˜ ◦ T )− = f˜− ◦ T

(5.47)

and therefore we get immediately ˜ → R be a numeCorollary 5.32. In the situation of Theorem 5.31 let f˜ : Ω ˜ ˜ rical measurable function. If f is T (µ)-integrable then f ◦ T is µ-integrable and if f˜ ◦ T is µ-integrable then f˜ is T (µ)-integrable and in each case Z Z ˜ f˜ ◦ T dµ (5.48) f dT (µ) = ˜ Ω

Ω

holds. 103

A COURSE IN ANALYSIS

Corollary 5.33. If in the situation of Theorem 5.31 the mapping T : Ω → ˜ is bijective with a measurable inverse then for every numerical function Ω ˜ → R the T (µ)-integrability of f˜ is equivalent to the µ-integrability of f˜ : Ω ˜ f ◦ T. Proof. The key point is that if f˜ ◦ T is integrable then it is measurable and hence f˜ = f˜ ◦ T ◦ T −1 is measurable too.

The result of Theorem 5.31 and its corollaries is usually referred to as integration with respect to the image measure. Our first application of Theorem 5.31 relates to symmetry or invariance of a measure. Let (Ω, A, µ) be a measure space, T : Ω → Ω a measurable mapping. Suppose that T (µ) = µ and T (Ω) = Ω, and let f : Ω → R be µ-integrable (or f ≥ 0). From (5.48) we deduce that Z Z Z f dµ = f dT (µ) = f ◦ T dµ (5.49) or

Z

f (x) µ(dx) =

Ω

Z

f (T x)µ(dx).

(5.50)

Ω

In particular if µ is invariant with respect to a group G of measurable mappings T : Ω → Ω, i.e. T (µ) = µ for all T ∈ G, then (5.50) holds for T ∈ G and we say that the integral is invariant under G or the action of G. Example 5.34. By Theorem 3.2 the Lebesgue-Borel measure λ(n) is translation invariant, i.e. λ(n) (x0 +B) = λ(n) (B) for all B ∈ B(n) . With τx0 (x) = x0 + x we may rewrite this equality as τx0 (λ(n) )(B) = λ(n) (τx−1 (B)) = λ(n) (−x0 + 0 B) and therefore we find by Corollary 5.32 for an appropriate integrable function f : Rn → R (or f ≥ 0) that Z Z (n) (f ◦ τ∓x0 )(x)λ(n) (dx) f (x ∓ x0 )λ (dµ) = n R Rn Z Z (n) = f (x)τ∓x0 (λ )(dx) = f (x)λ(n) (dx). Rn

Rn

Example 5.35. According to Theorem 3.6 the Lebesgue-Borel measure λ(n) is invariant under the action of O(n), i.e. T (λ(n) ) = λ(n) for T ∈ O(n). Consequently we find for suitable functions f : Rn → R (or f : Rn → R) that Z Z (n) f (T x)λ (dx) = f (x)λ(n) (dx), T ∈ O(n). Rn

Rn

104

5

INTEGRATION WITH RESPECT TO A MEASURE

As a particular case we mention the reflection at the origin S : Rn → Rn , S(x) = −x, which yields for all suitable functions f Z

Z

f (−x)λ(n) (dx) =

f (x)λ(n) (dx).

Rn

Rn

Example 5.36. Consider the measure space (Zn , P(Zn ), µZn ), µZn (A) = P k∈Zn k (A), as introduced in Example 3.5. We have seen that µZn is invariant under the translation τl0 , τl0 : Zn → Zn , τl0 (k) = k + l0 , l0 ∈ Zn . For f : Zn → [0, ∞) we find Z X f dµZn = f (k), Zn

and Theorem 5.31 yields Z Z n f dµZ = Zn

or

X

f (k) =

k∈Zn

Zn

k∈Zn

f dτl0 (µ ) =

X

k∈Zn

Zn

Z

Zn

f ◦ τl0 dµZn ,

f (k + l0 ) for all l0 ∈ Zn ,

which is of course evident. Next we want to see how we can use Theorem 5.31 and its corollaries in pro˜ A) ˜ a measurable bability theory. Let (Ω, A, P ) be a probability space and (Ω, ˜ space. A measurable mapping X : Ω → Ω is called a random variable, thus we may speak of Rn -valued random variables, real-valued random variables, numerical random variables etc. The image measure of P under X is called the distribution of P under X and it is denoted by PX . Let X be a numerical random variable which we assume either to be non-negative or P -integrable. We call Z E(X) =

X dP

the expectation of X. Since P (Ω) = 1 we have 1 E(X) = P (X) 105

Z

X dP,

(5.51)

A COURSE IN ANALYSIS

i.e. E(X) has the interpretation of a mean with respect to P . Now let f : R → R be a measurable function integrable with respect to PX (or f ≥ 0). Theorem 5.31 or Corollary 5.32 yields Z E(f ◦ X) = f dPX . (5.52) In particular for the identity id(x) = x we find E(X) =

Z

xPX (dx).

(5.53)

Thus knowing the distribution PX of X allows us already to calculate its expectation and for this we need not to have any knowledge of Ω, A or even P. In light of these definitions and observations we can give Example 5.19 a probabilistic interpretation: P If a random variable X is Bernoulli-distributed according to βN , then its expectation is E(X) = Np,

if a random variable Y is Poisson-distributed with γ > 0, then its expectation is E(Y ) = γ. After we have introduced an integral with respect to a measure, we want to clarify the common fa¸con de parler. The integration theory discussed in this chapter is called Lebesgue-integration (theory) and hence the µ-integral for a measure µ is also called the integral with respect to µ in the sense of Lebesgue and often one speaks just about the Lebesgue integral with respect to the measure µ. At the same time it is common to understand as Lebesgue integral the integral constructed in this chapter for the Lebesgue or the Lebesgue-Borel measure λ(n) . In practice this will not cause any confusion. In the next chapter we will use the results on integration with respect to the image measure to derive a version of the change of variables formula or the transformation theorem we have discussed in Chapter II.21. 106

5

INTEGRATION WITH RESPECT TO A MEASURE

Problems 1. For Ω = {1, . . . , N} find R S(Ω) and for a measure µ on P(Ω) find for u ∈ S(Ω) the integral u dµ. (As usual, the σ-field in Ω is the power set P(Ω).) P 2. Let (Ω, A, µ) be a measure space and u = N j=1 αj χAj where (Aj )j=1,...,N is a partition of Ω into measurable sets, but it is not assumed that αj 6= αk for j 6= k. Prove that N X

αj µ(Aj ) =

j=1

Z

u dµ

still holds. 3. For 0 < r1 < r2 < · · · < rN define the measurable sets A1 := Br1 (0), Ak := Brk (0) \ Brk−1 (0), k = 2, . . . , N. Consider the function u = PN 1 1 k=1 γk χAk , γ1 := r1 , γk = rk −rk−1 , k = 2, . . . , N. Using the fact that R λ(2) (Br (0)) = πr 2 find udλ(2) . 4. On [0, 1) consider the sequence of elementary functions uk =

k X l−1 l=1

k

χ[ l−1 , l ) . k k

Prove that uk ∈ S([0, 1)), and find supk∈N (uk )k∈N increasing?

R

(1)

uk dλ[0,1) . Is the sequence

5. Give details of the proof of (5.17). R 6. Let ρ : R → R be a measurable function such that ρ > 0 and ρ dλ(1) < R ∞. Define on B(1) the mapping ν := ρλ(1) by ν(A) := A ρ(x)λ(1) (dx) and prove that ν is a measure on B(1) . Let g : R → R be a bounded measurable function. Prove that g is ν-integrable. 7. For g(x) = x2 χ[0,∞) (x) find the following integrals: R p p i) R g dβN , where βN is a Bernoulli distribution; R ii) R g dπγ , where πγ is a Poisson distribution. 107

A COURSE IN ANALYSIS

(

sin x , x

x 6= 0 is a continuous 1, x=0 function vanishing at infinity, i.e. lim|x|→∞ h(x) = 0. With the help of Theorem 5.22 prove that h is not integrable over R with respect to λ(1) . R ∞ sin x P∞ R (k+1)π sin x (1) λ (dx) Hint: use the decomposition 0 x dx = k=0 kπ x and now use integration with respect to the image measure to find Z (k+1)π Z π sin x (1) sin x (1) x λ (dx) = x + kπ λ (dx). kπ 0

8. Show that h : R → R defined by h(x) :=

(n)

9. Denote by W1 = [−1, 1]n ⊂ Rn the unit cube on Rn . Let T ∈ (n) GL(n; R) and g : W1 → R be a measurable function. Prove that Z Z 1 (n) (g ◦ T −1 )dλ(n) . g dλ = (n) (n) | det T | T (W1 ) W1   cos ϕ − sin ϕ . for ϕ = 0, π2 , π, 3π 10. Consider the mappings U(ϕ) := 2 sin ϕ cos ϕ (2) (2) Show that U(ϕ)W1 = W1 and deduce that Z Z (n) f (U(ϕ)x)λ (dx) = f (x)λ(n) (dx) (2)

(2)

W1

W1

(2)

for all continuous functions f : W1

108

→ R.

6

The Radon-Nikodym Theorem and the Transformation Theorem

In the last chapter we discussed integration with respect to an image measure, compare for example with the statement of Corollary 5.32, and we looked at measures being invariant under transformations and the behaviour of corresponding integrals. One purpose of this chapter is to study the behaviour of integrals over Borel sets G ⊂ Rn with respect to the measure λ(n) under suitable transformations. We start with Example 6.1. For T ∈ GL(n; R) we know from Theorem 3.11 that T (λ(n) ) =

1 λ(n) . | det T |

(6.1)

Now let f : Rn → R be an integrable function. According to Corollary 5.32 we have Z Z Z 1 dλ(n) = f dT (λ(n) ) = f ◦ T dλ(n) f | det T | or Z Z Z (n) (n) f dλ = (f ◦ T )| det T | dλ = f ◦ T | det JT | dλ(n) , (6.2) where JT denotes the Jacobi matrix (the differential) of T which is of course T itself. Our aim is to extend (6.2) to a diffeomorphism ϕ : G → H, G, H ⊂ Rn open, with T replaced by ϕ and | det JT | by | det Jϕ |. We start with the following observation: For a homeomorphism ϕ : G → H between two open sets G, H ⊂ Rn the mapping Φϕ : C(H) → C(G), Φϕ (h) = h ◦ ϕ, is an algebra homeomorphism, i.e. Φϕ is linear and in addition we have Φϕ (h1 h2 ) = Φϕ (h1 )Φϕ (h2 ). The compatibility with the algebraic structure is obvious, see also Problem 1. Moreover we have with Φϕ−1 : C(G) → C(H), Φϕ−1 (g) = g ◦ ϕ−1 , that Φϕ ◦ Φϕ−1 = idC(G) and Φϕ−1 ◦ Φϕ = idC(H) where as usual idA denotes the identity on A. Note that in general we have for open sets F, G, H ⊂ Rn and homeomorphisms ϕ : F → G and ψ : G → H the relation Φψ◦ϕ = Φϕ ◦ Φψ . 109

A COURSE IN ANALYSIS

Thus for integrable g ∈ C(Rn ), g = f ◦ T with f ∈ C(Rn ) integrable and T ∈ GL(n; Rn ), we have Z Z (n) f dλ = g| det T | dλ(n) (6.3) and we want to consider the right hand side of (6.3) as an integral of g with respect to the measure | det T |λ(n) . More generally for a measurable nonnegative function ρ : Rn → R we may investigate µ := ρλ(n) on Rn , or even better on B(n) . Indeed we have the following more general result: Proposition 6.2. Let (Ω, A) be a measurable space and ρ : Ω → [0, ∞] be a measurable function. For a measure µ we can define a new measure ν on A by Z Z ν(A) :=

ρ dµ =

χA ρ dµ.

(6.4)

A

Proof. Since χ∅ = 0 it follows that ν(∅) = 0 and since ρ ≥ 0 by assumption we find ν(A) ≥ 0 for all A ∈ A. Now letS(Ak )k∈N , Ak ∈ A, be P a sequence of mutually disjoint sets with union A = k∈N Ak . Since χA ρ = ∞ k=1 χAk ρ Corollary 5.15 yields ν(A) =

Z

χA ρ dµ =

Z X ∞

χAk ρ dµ =

k=1

∞ Z X k=1

χAk ρ dµ =

∞ X

ν(Ak ).

k=1

Definition 6.3. In the situation of Proposition 6.2 we call ρ the density of ν with respect to µ and we write ν = ρµ.

(6.5)

Example 6.4. By | det T |λ(n) , T ∈ GL(n; R), a measure is given on B(n) having density | det T | with respect to the Lebesgue-Borel measure λ(n) . We know that the Lebesgue-Borel measure λ(n) is determined on the set Ir,n = Irn , see Chapter 2, and hence on the n-dimensional figures F (n) . Moreover, for f : Rn → [0, ∞] measurable, a pre-measure is given on F (n) defined on Ir,n by Z ν(Q) :=

χQ f dλ(n) .

110

(6.6)

6

RADON-NIKODYM AND TRANSFORMATION THEOREMS

This pre-measure has a unique extension to B(n) implying that inequalities such as ν(Q) ≤ λ(n) (Q) or λ(n) (Q) ≤ ν(Q) for all Q ∈ Ir,n (hence all A ∈ F (n) ) extend to B(n) . Here are two simple consequences of Definition 6.3. Corollary 6.5. If ν = ρµ and κ = σν then κ = (ρσ)µ. Proposition 6.6. Let (Ω, A) be a measurable space and ρ : Ω → [0, ∞] be a measurable function. For a measure µ on A consider the measure ν having density ρ with respect to µ, i.e. ν = ρµ. For every measurable function f : Ω → [0, ∞] we have Z Z f dν =

f ρ dµ.

(6.7)

Moreover, for every measurable function f : Ω → R the ν-integrability is equivalent to the µ-integrability of f ρ and in the integrable case (6.7) holds. P Proof. For a simple function u = N k=1 αk χAk we find Z

u dν =

N X

αk ν(Ak ) =

k=1

N X k=1

αk

Z

χAk ρ dµ =

Z

uρ dµ,

(6.8)

and for an arbitrary non-negative measurable function f : Ω → [0, ∞] we can find an increasing sequence of simple functions (ul )l∈N converging to f . Passing for such a sequence in (6.8) to the limit l → ∞ gives the first statement. The second statement is derived from the first one by noting that Z Z Z ± ± f dν = f ρ dµ = (f ρ)± dµ. We return to our original problem and to (6.1). Taking advantage of the fact that T ∈ GL(n; R) has an image which is again measurable we deduce from T (λ(n) )(B) = λ(n) (T −1 (B)) =

1 λ(n) (B) | det T |

with A = T −1 (B) that λ(n) (A) =

1 λ(n) (T (A)) | det T | 111

A COURSE IN ANALYSIS

or λ(n) (T (A)) = | det T |λ(n) (A)

(6.9)

for all A ∈ B(n) . We want to find the analogue to (6.9) when T is replaced by a diffeomorphism. Let U, V ⊂ Rn be open sets and ϕ : U → V a C 1 -diffeomorphism. We denote, as usual, the differential of ϕ by dϕ which we represent at x0 ∈ U by its Jacobi matrix Jϕ (x0 ). We claim Theorem 6.7. Let ϕ : U → V be a C 1 -diffeomorphism between two open sets U, V ⊂ Rn and A ⊂ A ⊂ U a Borel set. Assume that Z (n) | det Jϕ (x)|λU (dx) < ∞ A

exists. Then ϕ(A) ∈ B(n) and the following holds Z (n) (n) | det Jϕ (x)|λU (dx). λV (ϕ(A)) =

(6.10)

A

Note that for ϕ = T ∈ GL(n; R) we recover (6.9) from (6.10) since in this case det Jϕ (x) = det T for all x ∈ Rn . Proof of Theorem 6.7. We will prove this theorem in several steps. First let ϕ : U → V be a Lipschitz continuous mapping, i.e. kϕ(x) − ϕ(y)k∞ ≤ Lkx − yk∞ ˚ 6= ∅, which can be for all x, y ∈ U. Further let Q ⊂ Q ⊂ U be a cube, Q open, half-open or closed. We denote the centre of Q by c = (c1 , . . . , cn ) and we assume that Q has side length 2d, d > 0. Thus we have n

Q=

×[c k=1

k

− d, ck + d].

Since Q = {y ∈ Rn | kc − yk∞ ≤ d} we find for the image of Q under ϕ = (ϕ1 , . . . , ϕn ) n

ϕ(Q) ⊂

×[ϕ (c) − Ld, ϕ (c) + Ld], k

k

k=1

112

6

RADON-NIKODYM AND TRANSFORMATION THEOREMS

˚ = λ(n) (Q) = λ(n) (Q) we obtain and since λ(n) (Q) λ(n) (ϕ(Q)) ≤ Ln λ(n) (Q). With the help of (6.11) we now prove for Q ∈ Ir,n , Q ⊂ U, Z (n) λ (ϕ(Q)) ≤ | det Jϕ (x)|λ(n) (dx).

(6.11)

(6.12)

Q

Let Q ∈ Ir,n , Q ⊂ U. On the compact set Q the mapping Jϕ is uniformly continuous and therefore for every  > 0 there exists δ > 0 such that  sup kJϕ (x) − Jϕ (x0 )k∞ < , (6.13) L x,x0 ∈Q kx−x0 k∞ ≤δ

where L := supx∈Q k(Jϕ (x))−1 k∞ . Here k · k∞ denotes the maximum norm in Rm , i.e. kzk∞ = max{|zl | | 1 ≤ l ≤ m}. Note that for a matrix A we 2 calculate kAk∞ by identifying A with an element in Rn . We also note that

sup (Jϕ (x))−1 ∞ ≤ sup k(Jϕ−1 )(y)k∞ . (6.14) x∈Q

y∈ϕ(Q)

SN

We now choose a partition of Q, Q = l=1 Ql into cubes Ql ∈ Ir,n such that the side length of each Ql is less that δ > 0, δ as in (6.13). The continuity of Jϕ implies that for every l = 1, . . . , N there exists xl ∈ Ql such that | det Jϕ (xl )| = inf | det Jϕ (x)|. x∈Ql

With Tl := Jϕ (xl ) ∈ GL(n; R) we now find using the chain rule that JT −1 ◦ϕ (x) = Tl−1 ◦ Jϕ (x) = idn + Tl−1 ◦ (Jϕ (x) − Jϕ (xl )) . l

From (6.13) and (6.14) we deduce for all l = 1, . . . , N



sup JT −1 ◦ϕ (x) ≤ 1 + L = 1 + , l ∞ L x∈Ql

(6.15)

where we used that Ql has side length less than δ. Estimate (6.15) implies the Lipschitz continuity of the mapping Tl−1 ◦ ϕ on the convex set Ql . Now it follows with (6.9)
λ(n) (ϕ(Ql )) = λ(n) (Tl ◦ Tl−1 ◦ ϕ)(Ql )
= | det Tl | λ(n) (Tl−1 ◦ ϕ)(Ql ) ≤ | det Tl | (1 + )n λ(n) (Ql ), 113

A COURSE IN ANALYSIS

where the last step follows from (6.11). Using the fact that the cubes Ql , l = 1, . . . , N, form a partition of Q and that by construction for l = 1, . . . , N we have | det Tl | ≤ | det Jϕ (x)| for all x ∈ Ql , we obtain λ

(n)

(ϕ(Q)) ≤

N X

λ(n) (ϕ(Ql ))

l=1

≤ (1 + )n ≤ (1 + )n = (1 + )n

N X

| det Tl | λ(n) (Ql )

l=1 N Z X

Zl=1 Q

Ql

| det Jϕ (x)| λ(n) (dx)

| det Jϕ (x)| λ(n) (dx).

If we now pass to the limit as  tends to 0 we arrive at (6.12). As mentioned before, we can extend (6.12) to all A ∈ B(n) , A ⊂ U, i.e. Z | det Jϕ (x)|λ(n) (dx) (6.16) λ(n) (ϕ(A)) ≤ A

(n)

holds for these Borel sets A ∈ B . Note that we allowed ourselves some (n) (n) imprecise notation, we should have always written λU = λ(n) |U and λV = λ(n) |V , i.e. (6.16) must be read as Z (n) (n) (6.17) | det Jϕ (x)|λU (dx) λV (ϕ(A)) ≤ A

and for the following it is helpful to be more strict with the notation. Since ϕ : U → V is a diffeomorphism we further have (n)

(n)

λV (ϕ(A)) = ϕ−1 (λV )(A).

For A ∈ U ∩ B(n) we have for some B ∈ V ∩ B(n) that A = ϕ−1 (B) and (6.17) yields Z (n) (n) χB (y)λV (dy) = λV (B) V Z (n) χϕ−1 (B) (x)| det Jϕ (x)|λU (dx) ≤ ZU (n) χB (ϕ(x))| det Jϕ (x)|λU (dx), = U

114

6

RADON-NIKODYM AND TRANSFORMATION THEOREMS

i.e. we have Z

V

(n) χB (y)λV (dy)

≤

Z

(n)

U

χB (ϕ(x))| det Jϕ (x)|λU (dx)

(6.18)

for all B ∈ V ∩ B(n) which extends to all non-negative, measurable functions u : V → [0, ∞]: Z Z (n) (n) u(y)λV (dy) ≤ (6.19) u(ϕ(x))| det Jϕ (x)|λU (dx). V

U

With (6.19) at our disposal, we can switch from ϕ : U → V to ϕ−1 : V → U and to the function v : U → R defined for some A = ϕ−1 (B) by v(x) := (χϕ(A) ◦ ϕ)(x)| det Jϕ (x)| replacing u. We obtain from (6.19) Z

U


(n) χϕ(A) ◦ ϕ (x)| det Jϕ (x)|λU (dx) Z
(n) ≤ (χϕ(A) ◦ ϕ)| det Jϕ | (ϕ−1 (y))| det Jϕ−1 (y)|λV (dy) ZV (n) = χϕ(A) (y)| det(Jϕ (ϕ−1 (y)) ◦ Jϕ−1 (y))|λV (dy) V Z (n) (n) = χϕ(A) (y)λV (dy) = λV (ϕ(A)). V

With (6.16) we now get (n) λV (ϕ(A))

=

Z

A

(n)

| det Jϕ (x)|λU (dx),

i.e. (6.10) and the theorem is proved. Since (6.10) is an equality of measures we can extend it (as we have done before with inequality (6.16)) to an equality for integrals. This is the famous and very important transformation theorem for integrals which was first noted by C. G. J. Jacobi. Theorem 6.8. Let U, V ⊂ Rn be open sets and ϕ : U → V be a C 1 diffeomorphism. Further let f : V → R be a numerical function. This (n) function is integrable with respect to λV if and only if the function (f ◦ ϕ) · 115

A COURSE IN ANALYSIS (n)

| det Jϕ | : U → R is integrable with respect to λU and in the case where these functions are integrable we find Z Z (n) (n) f (y)λV (dy) = (6.20) f (ϕ(x))| det Jϕ (x)|λU (dx), V

U

or compare with (II.21.5), we have Z Z (n) f (y)λ (dy) = (f ◦ ϕ)(x)| det Jϕ (x)|λ(n) (dx). ϕ(U )

(6.21)

U

Remark 6.9. A. All the proofs of Theorem 6.8 we know follow essentially the same idea, unfortunately some authors of textbooks leave serious gaps. In our presentation we have combined the discussions of R. Schilling [75] and [76]. B. The reader of Volume II might remember that for the Riemann integral (in Rn ) we only discussed the transformation theorem, Theorem II.21.8, but we postponed the proof to this part of our Course. At this stage we still need to identify in certain situations the Riemann integral with the Lebesgue integral in order to claim that Theorem 6.8 implies Theorem II.21.8 in these cases. Once this is done we can, in admissible situations, transfer the calculations made in the Riemannian context to Lebesgue integrals. C. We have already seen that sets of measure zero do not contribute to integrals. In the next chapter we will use this observation in a more precise sense to give an extension of Theorem 6.8. So far we have used the idea of a measure with density only in one case, namely to prove Theorem 6.8. It turns out that this notion has many more far reaching consequences and applications. Let (Ω, A, µ) be a measure space and ρR : Ω → [0, ∞] a measurable Rfunction with finite, non-zero integral, i.e. 0 < ρ dµ < ∞. With kρkL1 := ρ dµ it follows that 1 ρµ ν := kρkL1

is a probability measure and (Ω, A, ν) is a probability space. The most ˜ A, ˜ P) interesting case is when (Ω, A, µ) = (Rn , B(n) , λ(n) ). Suppose that (Ω, n ˜ is a probability space and X : Ω → R is a random variable with distribution PX which is by definition a probability measure on Rn (or better B(n) ). If PX 116

6

RADON-NIKODYM AND TRANSFORMATION THEOREMS

has a density p : Rn → [0, ∞] with respect to λ(n) , i.e. PX = pλ(n) , we call p the probability density or just the density of X with respect to λ(n) . When p is the probability density of X we find for the expectation of X Z Z XP (dω) = E(X) = xp(x)λ(n) (dx). (6.22) ˜ Ω

Rn

Hence the calculation of E(X) is reduced to the evaluation of an integral in Rn with respect to the Lebesgue-Borel measure (and we will see that often we can do this by using the corresponding Riemann integral). Interesting probability densities are for example for n = 1 1

gα,σ (x) = (2πσ 2 )− 2 e−

(x−α)2 2σ 2

(6.23)

which is the one-dimensional Gauss or normal distribution with mean α and standard deviation σ > 0, or the one-dimensional Cauchy distribution 1 α (6.24) π α2 + x2 and we will see in Problem 6 of Chapter 7 that a random variable which is Cauchy does not have a well defined expectation since the inteR distributed x (1) λ (dx) does not exist. gral R α2 +x 2 cα (x) =

However, recalling the Bernoulli or the Poisson distribution, compare with Example 1.25 or Example 5.19.A and B, we note that not all real-valued random variables admit a density with respect to λ(1) . The natural question which arises is when for two measures µ and ν on a measurable space (Ω, A) one measure has a density with respect to the other. A first observation yields the following necessary condition: Lemma 6.10. Suppose that ν = f µ on the measurable space (Ω, A) with a non-negative, numerical measurable density f : Ω → [0, ∞]. If A ∈ A is a set of measure zero with respect to µ, then A is also a set of measure zero with respect to ν, i.e. µ(A) = 0 implies ν(A) = 0. R Proof. Since ν(A) = A f dµ the result follows from Lemma 5.29.

Definition 6.11. We call a measure ν on the measurable space (Ω, A) absolutely continuous with respect to the measure µ on (Ω, A) if for every set A ∈ A with µ(A) = 0 it follows that ν(A) = 0. If ν is absolutely continuous with respect to µ we write ν  µ. (6.25) 117

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For a finite measure ν on (Ω, A) we can prove Theorem 6.12. A finite measure ν on (Ω, A) is absolutely continuous with respect to a measure µ on (Ω, A) if and only if for every  > 0 there exists δ > 0 such that for every A ∈ A with µ(A) ≤ δ it follows that ν(A) ≤ . Proof. If A ∈ A and µ(A) = 0 the assumptions made above imply that ν(A) ≤  for every  > 0, hence ν(A) = 0, and ν is absolutely continuous with respect to µ. Now assume that ν  µ. Suppose that there exists  > 0 such that for all δ > 0 there exists Aδ ∈ A such that µ(Aδ ) ≤ δ and ν(Aδ ) ≥ . We may choose for δ = 2−k−1 a set Ak ∈ A such that

For Bk :=

S∞

l=k

µ(Ak ) ≤ 2−k−1 and ν(Ak ) ≥ . Al ∈ A we deduce

µ(Bk ) ≤

∞ X l=k

2−l−1 = 2−k and ν(Bk ) ≥ ν(Ak ) ≥ .

Since the sequence (Bk )k∈N is decreasing, i.e. Bk+1 ⊂ Bk , and ν is a finite measure it follows by Theorem 2.10.B that ν

∞ \

k=1

Bk

!

= lim ν(Bk ) ≥ . k→∞

T∞ −k0 On the other hand we T∞know for every k0 that µ ( k=1 Bk ) ≤ µ(Bk0 ) ≤ 2 which yields that µ ( k=1 Bk ) = 0 and this is a contradiction. We will also need

Lemma 6.13. Let (Ω, A) be a measurable space and let µ and ν be two finite measures on A. For τ : A → R, τ := µ − ν, exists Ω0 ∈ A such that τ (Ω0 ) ≥ τ (Ω)

(6.26)

τ (A) ≥ 0 for all A ∈ Ω0 ∩ A.

(6.27)

and

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Proof. (See [11]) We start by proving that for every n ∈ N there exists Ωn ∈ A such that τ (Ωn ) ≥ τ (Ω) (6.28) and

1 for all A ∈ Ωn ∩ A. (6.29) n If τ (Ω) = 0 we choose Ωn = ∅, so we assume τ (Ω) > 0. If τ (Ω) > 0 and τ (A) > − n1 for all A ∈ A, we may choose Ωn := Ω. Thus we assume the existence of A1 ∈ A such that τ (A1 ) ≤ − n1 . It follows that τ (A) > −

τ (A{1 ) = τ (Ω) − τ (A1 ) ≥ τ (Ω) +

1 > τ (Ω) > 0. n

If τ (A) > − n1 for all A ∈ A{1 ∩ A we choose Ωn := A{1 . Otherwise we can find A2 ∈ A{1 ∩ A such that τ (A2 ) ≤ − n1 , and since A1 ∩ A2 = ∅ we have   2 τ (A1 ∪ A2 ){ ≥ τ (Ω) + > τ (Ω) > 0. n { S N We can iterate this process. If it stops after N steps we take ΩN := A j j=1 where the sets Aj are mutually disjoint. Suppose that this process does not stop. Then we can find a sequence (Ak )k∈N of mutually disjoint sets Ak ∈ A such that for all N ∈ N ! N [ 1 τ Ω\ Ak > τ (Ω) > 0 and τ (An ) ≤ − . n k=1 The measures µ and ν are finite additive and hence ! N N X [ N τ τ (Ak ) ≤ − , Ak = n k=1 k=1

P implying the divergence of ∞ k=1 τ (Ak ). However the σ-additivity of µ and ν yields ! ! ∞ ∞ ∞ [ X [ Ak − ν τ (Ak ) = µ Ak k=1

k=1

k=1

which is finite since µ and ν are finite measures. Hence, given n ∈ N there ex{ S N A ists mutually disjoint sets A1 , . . . , AN ∈ A such that with Ωn := k k=1 119

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condition (6.28) and (6.29) hold. We may apply S (6.28), (6.29) to Ωn , if necessary, we may assume Ωn+1 ⊂ Ωn . For Ω0 := n∈N Ωn we obtain from (6.28) now (6.26) and since in addition we have τ (A) ≥ − n1 for all A ∈ Ω0 ∩ A implying (6.27). With the help of Lemma 6.13 we can give a first answer to the question when ν has a density with respect to µ, the Radon-Nikodym theorem for finite measures. Theorem 6.14. For two finite measures µ and ν on the measurable space (Ω, A) the absolute continuity of ν with respect to µ is equivalent to the existence of a density f for ν with respect to µ, i.e. ν  µ if and only if ν = f µ. Proof. We already know that the existence of a density always, not only for finite measures, implies the absolute continuity. So we need to prove that the absolute continuity of ν implies the existence of a density. Consider the set   Z G := g : Ω → [0, ∞] g is measurable, g dµ ≤ ν(A) for all A ∈ A . A

This set is not empty since g(ω) = 0 for all ω ∈ Ω belongs to G. For g, h ∈ G we find further Z Z Z (g ∨ h) dµ = g dµ + h dµ A A∩{g≥h} A∩{g≥h}{   ≤ ν(A ∩ {g ≥ h}) + ν A ∩ {g ≥ h}{ = ν(A),

i.e. g, h ∈ G implies g ∨ h ∈ G. Since ν is a finite measure it follows that Z γ := sup g dµ ≤ ν(Ω) < ∞. g∈G

R Let (˜ gk )k∈N , gk ∈ G, be a sequence such that limk→∞ g˜k dµR = γ. The sequence gN := g˜1 ∨ · · · ∨ g˜N is increasing, g˜N ≤ gN and limk→∞ gk dµ = γ. The Rmonotone convergence theorem yields that f := supk∈N gk belongs to G and R f dµ = γ. Our claim is that ν = f µ. By construction we have f µ ≤ ν, i.e. A f dµ ≤ ν(A) for all A ∈ A, and therefore the measure τ := ν − f µ 120

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is absolutely continuous with respect to µ. If τ (A) = 0 for all A we are done. Suppose that τ (Ω) > 0 which also implies by absolute continuity that µ(Ω) > 0. For 1 τ (Ω) β := >0 2 µ(Ω) we have τ (Ω) = 2βµ(Ω) > βµ(Ω) and applying Lemma 6.13 to βµ instead of ν and τ as defined above we deduce the existence of Ω0 ∈ A such that τ (Ω0 ) − βµ(Ω0 ) ≥ τ (Ω) − βµ(Ω) > 0 and τ (A) ≥ βµ(A) for all A ∈ Ω0 ∩ A. The function f0 := f + βχΩ0 is measurable and non-negative implying that Z Z Z f0 dµ = f dµ + βµ(A ∩ Ω0 ) ≤ f dµ + τ (A ∩ Ω0 ) A A ZA ≤ f dµ + τ (A) = ν(A) A

for all A ∈ A. This estimate yields that f0 ∈ G and that Z Z f0 dµ = f dµ + βµ(Ω0 ) = γ + βµ(Ω0 ) > γ since τ (Ω0 ) > βµ(Ω0 ) and µ(Ω0 ) > 0 due to the absolute continuity of τ with respect to µ. Hence we have constructed a contradiction to the assumption τ (Ω) > 0 implying the theorem. The following example shows that we cannot expect the Radon-Nikodym theorem to hold for general, not necessarily finite measures.  Example 6.15. Let Ω be a non-denumerable set and A := A ∈ P (Ω) | A or A{ is denumerable the σ-field from Example 1.3.A. Further let ν be the measure on A given by ν(A) = 0 for A denumerable and ν(A) = +∞ for A non-denumerable. In addition we consider the measure µ on A defined for a finite set A by µ(A) = #(A), i.e. for a finite set µ gives the number of its elements, and µ(A) = +∞ for any infinite set. Since µ(A) = 0 if and only if A = ∅ it follows that ν is absolutely continuous with respect to µ. However, ν cannot have a density with respect to µ. Indeed, if ν = f µ with an A-measurable function f : Ω → [0, ∞], then for every ω ∈ Ω we find Z f dµ = f (ω)µ({ω}) = f (ω) 0 = ν({ω}) = {ω}

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implying f = 0, i.e. ν(A) = 0 for all A ∈ A which is a contradiction. As it turns out we can rescue the Radon-Nikodym theorem for σ-finite measures. Recall that by Definition 2.12.B a measure µ is σ-finite on (Ω, A) if there exists a sequence (Ak )k∈N S of measurable sets Ak ∈ A with finite measure µ(Ak ) < ∞ such that Ω = ∞ k=1 Ak . The following lemma links σ-finite measures to the existence of certain integrable functions. Lemma 6.16. Let (Ω, A, µ) be a measure space. The measure µ is σ-finite if there exists a µ-integrable function h : Ω → R such that 0 < h(ω) < ∞ for all ω ∈ Ω. Conversely, if on a measure space (Ω, A, µ) such a function h exists, then µ is σ-finite. Proof. Let (Ak )k∈NSbe a sequence of measurable sets Ak ∈ A such that Ak ⊂ Ak+1 , Ω = k∈N Ak and µ(Ak ) < ∞. Note that by Problem 5 to Chapter 2 such a sequence exists. We nowPchoose ck ∈ (0, 2−k ) such that ck µ(Ak ) < 2−k , k ∈ N. It follows that h := ∞ k=1 ck χAk is a strictly positive measurable function with h(ω) ≤ 1 for all ω ∈ Ω and Z

h dµ ≤

∞ X k=1

ck

Z

χAk dµ =

∞ X k=1

ck µ(Ak ) ≤ 1.

Conversely, suppose that on (Ω, A, µ) we can find R a measurable function h such that 0 < h(ω) < ∞ for all ω ∈ Ω and h dµ < ∞.  R With Ak := 1 h ≥ ∈ A we get χ ≤ kh and consequently µ(A ) ≤ k h dµ < ∞ and Ak k k S k∈N Ak = Ω. Theorem 6.17 (Radon-Nikodym). Let µ be a σ-finite measure on (Ω, A) and ν a measure. The existence of a density f for ν with respect to µ is equivalent to the absolute continuity of ν with respect to µ.

Proof. We first consider the case where µ is finite, i.e. µ(Ω) < ∞, but ν is not finite, i.e. ν(Ω) = +∞. We will prove that we can decompose Ω into a sequence (Ak )k∈N∪{∞} of measurable sets Ak ∈ A such that ν(Ak ) < ∞ for k ∈ N and if A ∈ A∞ ∩ A then ν(A) < ∞ implies ν(A) = 0. For this let   A(ν) := A ∈ A ν(A) < ∞ and a := sup µ(A) A ∈ A(ν) .

Since µ is finite a < ∞. We chooseSnow an increasing sequence A˜k ∈ A such that a = supk∈N µ(A˜k ). For A0 := k∈N A˜k it follows that µ(A0 ) = a and on 122

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A0 ∩ A the measure ν is σ-finite. Hence, S see Problem 5 of Chapter 2, we can represent A0 as disjoint union A0 = ∞ k=1 Ak , Ak ∈ A, and ν(Ak ) < ∞. We set A∞ := Ω\A0 and claim that for A ∈ A∞ ∩ A and ν(A) < ∞ it follows that µ(A) = 0. The following holds n o ˜ A˜ ∈ A(ν) a = sup µ(A˜k ) = sup µ(A) k∈N

≥ sup µ(A ∪ A˜k ) k∈N

since A ∪ A˜k ∈ A(ν). Thus, since A ∩ A˜k = ∅ we have a ≥ sup µ(A ∪ A˜k ) = sup(µ(A) + µ(A˜k )) k∈N

= µ(A) + a, and since 0 ≤ a < ∞ it follows µ(A) = 0. On Ak ∩ A the measure ν is finite too and according to Theorem 6.14 we can find a measurable function fk : Ak → [0, ∞] such that ν|Ak = fk µ|Ak . We define f : Ω → [0, ∞] by ( fk (ω), ω ∈ Ak , k ∈ N f (ω) := +∞, ω ∈ A∞ . Clearly, f is measurable and ν = f µ, since for B ∈ A we find the decompoS sition of B = k∈N (Ak ∩ B) ∪ (A∞ ∩ B) into mutually disjoint sets. Finally we drop the assumption that µ is finite but assume that µ is σ-finite. By Lemma 6.16 we can find a strictly positive function h : Ω → R with finite R µ-integral, i.e. 0 < h < ∞ and h dµ < ∞. Thus the measure hµ is on A a finite measure and hence ν admits a density f with respect to hµ, hence it follows that ν = f (hµ) = (f h)µ. The converse statement, i.e. that the existence of a density implies the absolute continuity has already been discussed in the proof of Theorem 6.14. Remark 6.18. Our proof is one of the standard proofs and we combined arguments from [11] and [13]. For a proof using martingale theory we refer to [75] or [56]. The latter proof is related to a “Hilbert space proof”, see [21], which is due to J. v. Neumann. 123

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A natural question is to which extent a density in ν = f µ is uniquely determined. From ν = f1 µ and ν = f2 µ we can deduce Z (f2 − f1 ) dµ = 0 for all A ∈ A, A

which however in general does not imply f1 R= f2 . If A is a set of measure zero with respect to µ, i.e. µ(A) = 0, then A g dµ = 0 for all measurable numerical functions g : Ω → [0, ∞]. How to handle properties on sets of measure zero will be studied in the next chapter.

Problems Although the transformation theorem is one of the central topics of this chapter we do not include problems related to the application of the transformation theorem. There are two reasons for this. Firstly, in Chapter II.21 (and further chapters in Volume II) we have already discussed many applications. Secondly, for interesting applications we need more tools such as Fubini’s theorem, Theorem 9.17, or we need criteria to identify the Lebesgue integral with the Riemann integral. Later on in this Course we will encounter again and again the transformation theorem and its applications. 1.

a) Let X and Y be two topological spaces and ϕ : X → Y a homeomorphism. Denote by C(X) the space of all continuous real-valued functions from X to R. Prove that C(X) and C(Y ) are isomorphic as algebras. b) Let G and H be open sets in Rn and ψ : G → H a diffeomorphism of class C k , k ∈ N0 . Consider the mapping Ψ : C k (H) → C k (G), Ψ(u) = u ◦ ψ, and prove that Ψ is a vector space isomorphism.

2. Prove that if g : Rn → Rn is Lipschitz continuous with respect to the norm || · ||p , 1 ≤ p ≤ ∞, i.e. ||g(x) − g(y)||p ≤ γp,p||x − y||p for all x, y ∈ Rn , then g is also Lipschitz continuous from (Rn , || · ||q ) to (Rn , || · ||r ), 1 ≤ q, r ≤ ∞, i.e. we have ||g(x) − g(y)||r ≤ γq,r ||x − y||q for all x, y ∈ Rn . 3.

a) Let G ⊂ Rn be a measurable set with λ(n) (G) > 0 and consider the measure ν := χG λ(n) . Find all Borel measurable functions f : Rn → R which are integrable with respect to ν. 124

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RADON-NIKODYM AND TRANSFORMATION THEOREMS s

b) For gs (x) := (1+|x|2)− 2 , s ∈ R, find a power growth condition for a continuous function u : R → R which implies that u is integrable with r respect to ν = gs λ(n) , i.e. find r ∈ R such that |u(x)| ≤ cu (1 + |x|2 ) 2 implies that u integrable with respect to ν. 4.

a) Let µj , j = 1, . . . , N be measures on the measurable space (Ω, A). Find a measure µ on (Ω, A) such that µj  µ for all j = 1, . . . , N.

b) Let µ, ν and π be measures on (Ω, A). Suppose that π  ν and ν  µ. Show that π is absolutely continuous with respect to µ. 5. Let g, h : Rn → [0, ∞) be continuous functions such that for all x ∈ Rn g(x) we have 0 < c0 ≤ h(x) ≤ c1 . Consider two measures ν1 := gλ(n) and ν2 := hλ(n) . Prove that ν2  ν1 and ν1  ν2 . 6. Let νk = gk µ, k ∈ N, and ν = gµ be absolutely continuous measures with respect to µ where µ is a measure on (Ω, A). Suppose that µ(A) < ∞ and limk→∞ ||gk − g||∞ = 0. Prove for every A ∈ A that limk→∞ νk (A) = ν(A). 7.

a) Is the Dirac measure x0 , x0 ∈ R, absolutely continuous with respect to the Borel-Lebesgue measure λ(1) ? Is λ(1) absolutely continuous with respect to x0 ? P P b) Let µ = k∈N bk k , ak , bk ≤ 0, be two k∈N ak k and ν = measures on B(1) . Give conditions on the sequences (ak )k∈N and (bk )k∈N which will imply that ν is absolutely continuous with respect to µ.

8. Let X be a real-valued random variable on the probability space (Ω, A, P ) which is Cauchy distributed. Prove that E(X) is not defined.

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Almost Everywhere Statements, Convergence Theorems

Let (Ω, A, µ) be a measure space and (uk )k∈N , uk : Ω → R, be a sequence of measurable functions. We want to discuss several different types of convergence of the sequence (uk )k∈N to a function u : Ω → R. As we will see later in this chapter we can often replace R by C, Rn or even a normed vector space (V, k.k), and sometimes we may replace R by R. Independent of A and µ we have of course the notion of pointwise convergence: a sequence of functions uk : Ω → R converges pointwisely to u : Ω → R if for every ω ∈ Ω and every  > 0 there exists N = N(, ω) ∈ N such that k ≥ N(, ω) implies |uk (ω) − u(ω)| < . In this definition we can obviously replace (R, |.|) by a normed R-vector space (V, k.k), i.e. uk : Ω → V converges pointwisely to u : Ω → V if for every ω ∈ Ω and every  > 0 there exists N = N(, ω) such that k ≥ N(, ω) implies kuk (ω) − u(ω)k < . We recall that Theorem 4.12 states that the pointwise limit of measurable functions uk : Ω → R is measurable. The following example shows that interchanging pointwise limits with integrals is in general a problem.   (1) Example 7.1. On (0, 1), B(1) ((0, 1)), λ|(0,1) we consider the sequence uk : (0, 1) → R, uk (x) := kχ(1− 1 ,1) (x). Each function uk is a simple function since k (1 − k1 , 1) and its complement in (0, 1) belongs to B(1) ((0, 1)) (= (0, 1) ∩ B(1) ). The integral of uk is   Z 1 1 (1) (1) uk (x) λ(0,1) (dx) = kλ 1 − ,1 =k· =1 k k R (1) for all k ∈ N and therefore limk→∞ uk dλ(0,1) = 1. We claim that on (0, 1) the sequence (uk )k∈N converges pointwisely to zero. Given x ∈ (0, 1) we can find N = N(x) ∈ N such that k ≥ N implies x < 1 − k1 and hence uk (x) = 0 for k ≥ N(x). Hence limk→∞ uk (x) = 0 for all x ∈ (0, 1). Thus we have constructed a sequence (uk )k∈N of measurable functions converging pointwisely to the measurable function u, u(x) = 0 for all x ∈ (0, 1), but while the sequence of corresponding integrals converges to 1 the limit function has integral 0. In the case of continuous functions and the Riemann integral we know that 127

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uniform convergence yields the desired result: Z Z   uk (x) dx. lim uk (x) dx = lim k→∞

k→∞

However for an arbitrary measure space (Ω, A, µ) the notion of continuity is not defined. Thus a natural question is to look for criteria as to when we can interchange limits of converging sequences of measurable functions with corresponding (converging) sequences of integrals. A key notion needed for this is that of almost everywhere convergence. Definition 7.2. Let (Ω, A, µ) be a measure space and (uk )k∈N , uk : Ω → R (or R) be a sequence of functions. We say that this sequence converges µ-almost everywhere (µ-a.e.) to the function u : Ω → R (or R) if there exists a set N ∈ A of measure zero, i.e. µ(N ) = 0, such that for every ω ∈ Ω\N the sequence (uk (ω))k∈N converges in R (or R) to u(ω). Remark 7.3. A. We can not change Definition 7.2 and require that the set of all ω ∈ Ω for which (uk (ω))k∈N does not converge to u(ω) is a set of µ-measure zero since this set is not necessarily measurable. However in the case where (Ω, A, µ) is a complete measure space, then this set will be a subset of a µ-null set and hence measurable. B. Clearly, if the sequence (uk )k∈N converges pointwisely to u, then u is measurable provided all uk are measurable and the convergence also holds almost everywhere. k Consider  the sequence uk : [0,1] → R, uk (t) = t , k ∈ N, on the measure (1) (1) space [0, 1], [0, 1] ∩ B(1) , λ[0,1] . This sequence converges λ[0,1] -a.e. to the function u : [0, 1] → R, u(t) = 0 for all t ∈ [0, 1]. Indeed, for t ∈ [0, 1) we (1) know that limk→∞ tk = 0 and {1} ⊂ [0, 1] has measure zero, i.e. λ[0,1] ({1}) = 0. We (note further that (uk )k∈N converges pointwisely to v : [0, 1] → R, 0, t ∈ [0, 1) (1) . Hence (uk )k∈N converges also λ[0,1] -a.e. to v and v 6= v(t) = 1, t = 1 u. Thus a µ-almost everywhere limit is in general not uniquely determined. For dealing with this and related non-uniqueness results we need

Definition 7.4. Let (Ω, A, µ) be a measure space and P (ω) a statement we can make for every ω ∈ Ω. We say that P (ω) holds µ-almost everywhere (µ-a.e.) if there exists a set N ∈ A, µ(N ) = 0, such that P (ω) holds for all ω ∈ Ω\N . 128

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Thus, if P (ω) is the statement: “(uk (ω))k∈N converges to u(ω)” then Definition 7.4 gives Definition 7.2. Statements such as f ≥ g µ-a.e. are now well defined. In particular we can consider µ-almost everywhere equality of functions u, v : Ω → Y where Y 6= ∅ is an arbitrary set. Definition 7.5. Let (Ω, A, µ) be a measure space and Y 6= ∅ be a set. We call u, v : Ω → Y µ-equivalent and write u ∼µ v, if there exists a set N ∈ A, µ(N ) = 0, such that u(ω) = v(ω) for all ω ∈ Ω\N . Corollary 7.6. The relation ∼µ is an equivalence relation on the set of all mappings u : Ω → Y . Proof. Clearly u ∼µ u, we just take N = ∅, and if u ∼µ v, i.e. u(ω) = v(ω) for all ω ∈ Ω\N , µ(N ) = 0, we have with the same set N that v ∼µ u. Finally, if u ∼µ v and v ∼µ z there exists sets N1 , N2 ∈ A, µ(N1) = µ(N2 ) = 0 such that u(ω) = v(ω) for ω ∈ Ω\N1 and v(ω) = z(ω) for ω ∈ Ω\N2 . Hence for ω ∈ Ω\(N1 ∪ N2 ) we have u(ω) = z(ω), and since µ(N1 ∪ N2 ) = 0 it follows that u ∼µ z. By [u]µ we denote the equivalence class of u : Ω → Y with respect to ∼µ . We can now reword the second statement of Lemma 5.29: every numerical integrable function is µ-a.e. finite. Proposition 7.7. Suppose that the sequence uk : Ω → R, k ∈ N, converges µ-almost everywhere to u : Ω → R. Then every further µ-a.e. limit v of (uk )k∈N belongs to [u]µ . Proof. Let N1 , N2 ∈ A, µ(N1 ) = µ(N2 ) = 0, and for ω ∈ Ω\N1 we have limk→∞ uk (ω) = u(ω) and for ω ∈ Ω\N2 we have limk→∞ uk (ω) = v(ω). This implies for ω ∈ Ω\(N1 ∪N2 ) that u(ω) = v(ω) holds and since µ(N1 ∪N2 ) = 0 we have u ∼µ v or v ∈ [u]µ . Theorem 7.8. If for a measurable function f : Ω → [0, ∞] on (Ω, A, µ) the integral is zero then f is µ-almost everywhere the zero function, i.e. Z f dµ = 0, f ≥ 0, implies f = 0 µ − a.e. (7.1) Conversely, f = 0 µ-a.e. yields

R

f dµ = 0. 129

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Proof. Since f is measurable it follows that {f > 0} ∈ A and (7.1) is proved 1 } and if we can show that µ({f > 0}) = 0. Since {f ≥ k1 } ⊂ {f ≥ k+1 S 1 1 k∈N {f ≥ k } = {f > 0} we need to prove that limk→∞ µ{f > k } = 0. Observing that f ≥ k1 χ{f ≥ 1 } we find k

0=

Z

f dµ ≥

Z

  1 1 1 ≥ 0, χ 1 dµ = µ f ≥ k {f ≥ k } k k

i.e. µ({f ≥ k1 }) = 0 implying f = 0 µ-a.e. The converse statement is just a consequence of Lemma 5.29. From Theorem 7.8 we can derive a few helpful consequences. Corollary 7.9. A. If two measurable functions f, g : Ω → [0, ∞] are µ -a.e. equal then they have the same integral. B. If two measurable functions f, g : Ω → R are µ-a.e. equal and f is µintegrable then g is µ-integrable and both integrals are equal. C. If f, g : Ω → R are measurable functions, |f | ≤ g µ-a.e. and g is µintegrable, then f is µ-integrable too. R R Proof. A. Since µ ({f 6= g}) = 0, by Lemma 5.29 {f 6=g} f dµ = {f 6=g} g dµ = 0 and we find Z Z Z Z f dµ f dµ = f dµ + f dµ = {f =g} {f =g} {f 6=g} Z Z Z Z = g dµ = g dµ + g dµ = g dµ. {f =g}

{f =g}

{f 6=g}

B. By assumption weRhave f + = gR+ µ-a.e. and f − = g − µ-a.e., hence by part R + R A we find f dµ = g + dµ and f − dµ = Rg − dµ andR the integrability of f implies now the integrability of g as well as f dµ = g dµ. C. The function g ∨ |f | is measurable and µ-a.e. we have g ∨ |f | = g. It follows that g ∨ |f | is integrable and from Theorem 5.22.iii) we deduce that f is integrable. Now we may return to the Radon-Nikodym theorem and discuss the uniqueness of densities. We do this in two steps combining the presentation in [13] and [11]. 130

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Theorem 7.10. Let (Ω, A) be a measurable space and µ a measure on A. A. If two non-negative measurable functions f, g : Ω → [0, ∞] are µ-a.e. equal, then the measures f µ and gµ are equal on A. B. If f or g is integrable and the measures f µ and gµ are equal on A then f = g µ-a.e. Proof. A. For A ∈ A it follows that f χA = gχA µ-a.e. and therefore Z Z g dµ = (gµ)(A). f dµ = (f µ)(A) = A

A

B. Suppose that f is integrable and f µ = gµ. We have to show that {f 6= g} is contained in a set of measure zero. By the second statement in Lemma 5.29 there exist sets Nf and Ng , µ(Nf ) = µ(Ng ) = 0 and f |Ω\Nf < ∞ as well as g|Ω\Ng < ∞. For n ∈ N consider the set   1 . Ωn := ω ∈ Ω\(Nf ∪ Ng ) |f (ω) − g(ω)| ≥ n

We claim that µ(Ωn ) = 0. Suppose µ(Ωn ) > 0. Then either Ωn,1 := {ω ∈ Ω\(Nf ∪ Ng ) | f (ω) − g(ω) ≥ n1 } or Ωn,2 := {ω ∈ Ω\(Nf ∪ Ng ) | f (ω) − g(ω) ≤ − n1 } must have positive measure. For Ωn,1 this would imply Z Z (f µ)(Ωn,1 ) = f dµ ≥ (g + χΩn,1 ) dµ Ωn,1

Ωn,1

= (gµ)(Ωn,1) + µ(Ωn,1 ) > (gµ)(Ωn,1),

and analogously for Ωn,2 we would obtain (f µ)(Ωn,2) < (gµ)(Ωn,2), both statements Scontradict f µ = gµ. Thus µ(Ωn ) = 0 and since {f 6= g} ⊂ Nf ∪ Ng ∪ ∞ n=1 Ωn it follows that f = g µ-a.e. Theorem 7.11. In the situation of Theorem 6.17 the density of f of ν with respect to µ is µ-a.e. unique.

Proof. Since µ is σ-finite there exists a strictly positive integrable function h on (Ω, A, µ). The measure hν = h(f µ) = f (hµ) has the density f with respect to the finite measure hµ, hence by Theorem 7.10 we have that f is a hµ-a.e. uniquely determined function. Since µ and hµ have the same sets of measure zero, recall that h > 0, it follows that f is µ-a.e. uniquely determined. We now return to discuss convergence results for which we give 131

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Definition 7.12. Let (Ω, A, µ) be a measurable space and (uk )k∈N , uk : Ω → R (or R) a sequence of measurable functions, and let u : Ω → R (or R) be a further measurable function. A. We say that (uk )k∈N converges in the pth mean, 1 ≤ p < ∞, to u if Z |uk − u|p dµ = 0. (7.2) lim k→∞

B. The sequence (uk )k∈N converges in measure to u if for every set A ∈ A with finite measure µ(A) < ∞ the following holds for every  > 0 that lim µ ({|uk − u| > } ∩ A) = 0.

k→∞

(7.3)

Remark 7.13. A. Convergence in the pth mean for p = 1 is called convergence in the mean and for p = 2 it is called convergence in the quadratic mean. B. In the case where (Ω, A, µ) is a finite measure space we can replace (7.3) by lim µ ({|uk − u| > }) = 0. (7.4) k→∞

C. If (Ω, A, P ) is a probability space convergence in measure is called stochastic convergence. Theorem 7.14. Let (Ω, A, µ) be a σ-finite measure space and suppose that the sequence (uk )k∈N of measurable mappings converges in measure to the measurable function u. Any further limit in measure v of (uk )k∈N belongs to [u]µ , i.e. u = v µ-a.e. Conversely, every measurable function z = u µ-a.e. is a further limit in measure of (uk )k∈N . Proof. The triangle inequality yields n o n o n o |u − v| ≥  ⊂ |uk − u| ≥ ∪ |uk − v| ≥ 2 2

implying for A ∈ A, µ(A) < ∞ n o  n  n  o o µ |u − v| ≥  ∩ A ≤ µ |uk − u| ≥ ∩ A +µ |uk − v| ≥ ∩A 2 2

and for k → ∞ we obtain µ

n o  |u − v| ≥  ∩ A = 0. 132

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S Since {u 6= v} ∩ A = n∈N {|u − v| ≥ n1 } ∩ A we deduce that u|A = v|A µ-a.e. Now we let A run through an S increasing sequence (Al )l∈N of measurable sets Al ⊂ Al+1 , µ(Al ) < ∞, and l∈N Al = Ω, to find that u = v µ-a.e. The converse statement is almost trivial since there exists a set Nz , µ(Nz ) = 0, such that on Nz{ we have u = z and therefore {|uk − u| ≥ } ∩ A and {|uk − z| ≥ } ∩ A differ only by a set of measure zero. In order to relate convergence in the pth mean to convergence in measure we prove the following, often very useful estimate: Lemma 7.15 (Chebyshev-Markov inequality). For every measurable function f : Ω → R defined on a measure space (Ω, A, µ) the following holds Z 1 µ ({|f | ≥ α}) ≤ p |f |p dµ (7.5) α for every α > 0 and p > 1. Proof. Since {|f | ≥ α} ∈ A we find Z Z p p α µ ({|f | ≥ α}) = α dµ ≤ {|f |≥α}

{|f |≥α}

p

|f | dµ ≤

Z

|f |p dµ.

Corollary 7.16. If the sequence (uk )k∈N of measurable functions converges in the pth mean to u then it converges in measure to u. Proof. For every A ∈ A we have µ ({|uk − u| ≥ } ∩ A) ≤ µ({|uk − u| ≥ }) and the Chebyshev-Markov inequality yields Z 1 µ ({|uk − u| ≥ } ∩ A) ≤ p |uk − u|p dµ α hence the result follows. We postpone the discussion of the converses of Theorem 7.14 and Corollary 7.16 as we do with the discussion of the relation of µ-a.e. convergence to convergence in measure or convergence in the pth means and we first study in more detail the latter notion.

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Let (uk )k∈N be a sequence of functions converging in the pth mean, p ≥ 1, to u and v. Using Minkowski’s inequality (5.34) we find Z

p

|u − v| dµ

 p1

≤

Z

p

|u − uk | dµ

 p1

+

Z

p

|uk − v| dµ

 p1

implying that |u − v| = 0 µ-a.e., hence u = v µ − a.e. and we have proved Corollary 7.17. Limits in the pth mean are µ-a.e. uniquely determined. Definition 7.18. Let (Ω, A, µ) be a measure space and p ≥ 1. By Lp (Ω) or Lp (Ω; µ) we denote the set of all measurable functions u : Ω → R for which Np (u) :=

Z

p

|u| dµ

 p1

(7.6)

is finite. Elements of Lp (Ω) are called p-fold integrable. Again we deduce from Lemma 5.29 that if for u : Ω → R and p ≥ 1 it follows that Np (u) < ∞ then u is finite µ-a.e. However, by assumption elements of Lp (Ω) are always real-valued, i.e. have on Ω finite values. Theorem 7.19. With the natural algebraic operations Lp (Ω) is an R-vector space. Furthermore, with u, v ∈ Lp (Ω) it follows that u ∨ v and u ∧ v belong to Lp (Ω) too. In addition u ∈ Lp (Ω) if and only if u+ , u− ∈ Lp (Ω). Proof. Note that elements in Lp (Ω) are real-valued and hence αu + βv is for all α, β ∈ R and u, v ∈ Lp (Ω) defined. Since for α ∈ R we have Np (αu) = |α|Np (u) and Minkowski’s inequality implies Np (u + v) ≤ Np (u) + Np (v) it is clear that Lp (Ω) is an R-vector space. Moreover, u ∈ Lp (Ω) implies |u| ∈ Lp (Ω) since Np (u) = Np (|u|) and from u ∨ v = 21 (u + v + |u − v|) and u ∧ v = 21 (u + v − |u − v|) it follows that u ∨ v, u ∧ v ∈ Lp (Ω) for u, v ∈ Lp (Ω). In particular we find u+ , u− ∈ Lp (Ω) for u ∈ Lp (Ω). Conversely, if u+ , u− ∈ Lp (Ω) then u = u+ − u− ∈ Lp (Ω). As a corollary to H¨older’s inequality (5.33) we obtain Corollary 7.20. Let p > 1 and p1 + 1q = 1. If u ∈ Lp (Ω) and v ∈ Lq (Ω) then u · v ∈ L1 (Ω). In particular if u, v ∈ L2 (Ω) then u · v ∈ L1 (Ω). 134

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Exercise 7.21. Prove that on L2 (Ω) by Z hu, vi := uv dµ

(7.7)

a symmetric, non-negative bilinear form is defined and hu, ui = 0 yields u = 0 µ-a.e. Corollary 7.22. Let (Ω, A, µ) be a finite measure space and 1 ≤ q ≤ p. If u ∈ Lp (Ω) then u ∈ Lq (Ω) and we have Nq (u) ≤ µ(Ω) Proof. We may take p q

1 r

=

applied to |u| and 1 = 1 Z

q

|u| dµ =

Z

p−q p

p p−q

q

to find

1 r

p−q pq

+

q p

Np (u).

(7.8)

= 1 and now H¨older’s inequality

gives

|u| · 1

p p−q

dµ ≤

Z

1 dµ

Z  p−q p

p

|u| dµ

 pq

which implies u ∈ Lq (Ω) and (7.8). In Problem 6 we will see that in Corollary 7.22 we cannot in general remove the condition, µ to be a finite measure. We want to add to the scale Lp (Ω), 1 ≤ p < ∞, a space which corresponds to bounded functions. Definition 7.23. Let (Ω, A, µ) be a measure space and f : Ω → R a measurable function. We call f essentially bounded if for some Mf ≥ 0 we have |f | ≤ Mf µ-a.e. and  N∞ (f ) := inf M ≥ 0 µ ({|f | > M}) = 0

is called the essential supremum of f . The set of all real-valued essentially bounded functions f : Ω → R is denoted by L∞ (Ω) or L∞ (Ω; µ). The triangle inequality implies that L∞ (Ω) is a vector space over R, and in Problem 8 we will see that N∞ (·) is a semi-norm. Moreover, for u ∈ Lp (Ω), 1 ≤ p ≤ ∞, and v ∈ L∞ (Ω) it follows that u · v ∈ Lp (Ω), in particular L∞ (Ω) is an algebra. Indeed we have Z Z Z p p |u · v|p dµ ≤ |u|p N∞ (v) dµ = N∞ (v) |u|p dµ, 135

A COURSE IN ANALYSIS

or Np (u · v) ≤ N∞ (v)Np (u).

(7.9)

On Lp (Ω), 1 ≤ p ≤ ∞, the mapping or functional Np : Lp (Ω) → R satisfies Np (u) ≥ 0, Np (αu) = |α|Np (u), Np (u + v) ≤ Np (u) + Np (v),

(7.10) (7.11) (7.12)

however Np (u) = 0 does not imply u = 0, but u = 0 µ-a.e. Hence Np is on Lp (Ω) a semi-norm in the sense of Definition II.20.11. Definition 7.24. Let V be an R-vector space and ρ : V → [0, ∞) be a seminorm on V . We call a sequence (xk )k∈N , xk ∈ V , convergent to x with respect to the semi-norm ρ if limk→∞ ρ(xk − x) = 0, i.e. for every  > 0 there exists K ∈ N such that k ≥ K implies ρ(xk − x) < . Proposition 7.25. If (xk )k∈N converges to x with respect to ρ then (xk )k∈N is bounded in the sense that supk∈N ρ(xk ) ≤ M < ∞. For two limits x and y of (xk )k∈N with respect to ρ it follows that ρ(x − y) = 0. Given two sequences (xk )k∈N and (yk )k∈N converging with respect to ρ to x and y respectively then the sequence (xk + yk )k∈N converges with respect to ρ to x + y and for α ∈ R the sequence (αxk )k∈N converges with respect to ρ to αx. Proof. We can essentially rely on the proofs for convergence with respect to a metric or a norm. Given  = 1 the convergence of (xk )k∈N to x implies for some K ∈ N that ρ(xk ) ≤ ρ(xk − x) + ρ(x) ≤ 1 + ρ(x) for k ≥ K and hence supk∈N ρ(xk ) ≤ max1≤k K, hence ρ(x − y) = 0. Again we employ the triangle inequality (7.12) for ρ to find ρ(xk + yk − (x + y)) ≤ ρ(xk − x) + ρ(yk − y) implying limk→∞ ρ(xk + yk − (x + y)) = 0. Finally we note that ρ(αxk − αx) = |α|ρ(xk − x) implying that limk→∞ ρ(αxk − αx) = 0. 136

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Corollary 7.26. If (uk )k∈N converges to u with respect to Np , 1 ≤ p ≤ ∞, then supk∈N Np (uk ) ≤ M < ∞ and two limits of (uk )k∈N with respect to Np are µ-a.e. equal, i.e. limk→∞(uk − u) = 0 and limk→∞ Np (uk − v) = 0 implies u = v µ-a.e. Further, for limk→∞ Np (uk − u) = 0 and limk→∞ Np (vk − v) = 0 it follows for all α, β ∈ R that limk→∞ Np ((αuk + βvk ) − (αu + βv)) = 0, i.e. if (uk )k∈N converges to u and (vk )k∈N converges to v with respect to Np then (αuk + βvk )k∈N converges to αu + βv with respect to Np . Furthermore we have Proposition 7.27. Let (uk )k∈N converge to u with respect to Np , 1 ≤ p ≤ ∞, and let (vk )k∈N converge to v in L∞ (Ω). Then (vk uk )k∈N converges in Lp (Ω) to vu. Moreover if 1 < p < ∞, p1 + 1q = 1, and (ωk )k∈N converges to ω in Lq (Ω) then (uk ωk )k∈N converges to uω in L1 (Ω). Proof. For the first statement let Mu be a bound for (Np (uk ))k∈N . It follows by (7.9) that Np (uk vk − uv) ≤ Np (uk (vk − v)) + Np ((uk − u)v) ≤ Np (uk )N∞ (vk − v) + Np (uk − u)N∞ (v) ≤ Mu N∞ (vk − v) + N∞ (v)Np (uk − u) implying that limk→∞ Np (uk vk − uv) = 0. By H¨older’s inequality we find uk ωk , uω ∈ L1 (Ω) and further since Nq (ωk ) ≤ Mω for all k ∈ N we get Z

N1 (uk ωk − uω) = |uk ωk − uω| dµ Z Z ≤ |uk ωk − uωk | dµ + |uωk − uω| dµ ≤

Z

p

|uk − u| dµ

 p1 Z

 p1 Z  1q Z  1q p q |u| dµ |ωk | dµ + |ωk − ω| dµ q

≤ Mω Np (uk − u) + Np (u)Nq (ωk − ω) implying limk→∞ N1 (uk ωk − uω) = 0. A further useful result is

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Lemma 7.28. Let (Ω, A, µ) be a measure space and (uk )k∈N a sequence of non-negative, numerical measurable functions. For 1 ≤ p < ∞ the following holds ! ∞ ∞ X X Np (uk ) (7.13) uk ≤ Np k=1

k=1

or

p ! p1  1p Z X ∞ Z ∞ X p |uk | dµ . ≤ uk dµ k=1

k=1

Proof. For the partial sum Np

(7.14)

N X

PN

k=1 uk

uk

k=1

!

P

≤

we find using the triangle inequality

N X k=1

Np (uk ) ≤





∞ X

Np (uk ).

(7.15)

k=1



PN N p are monotone and k=1 |uk | k=1 uk N ∈N P N ∈N P∞ PN N p = supN ∈N k=1 uk = k=1 |uk | k=1 uk and supN ∈N

Moreover the sequences

increasing with P∞ p |u | . The monotone convergence theorem, Theorem 5.13, now gives k k=1 ! ! N ∞ X X uk uk = sup Np Np k=1

N ∈N

k=1

and with (7.15) the lemma follows. Since the triangle inequality holds for a semi-norm ρ on a vector space V we can deduce with the standard arguments, |ρ(x) − ρ(y)| ≤ ρ(x − y), compare with Lemma II.1.4. For p = 1 this yields with A ∈ A that Z Z Z f dµ − g dµ ≤ |f − g| dµ ≤ N1 (f − g) A

A

(7.16)

(7.17)

A

for all f, g ∈ L1 (Ω), and for 1 < p < ∞ and f, g ∈ Lp (Ω), A ∈ A we find Z  p1  p1 Z p p − |g| dµ = |Np (χA f ) − Np (χA g)| |f | dµ A A ≤ Np (χA (f − g)) ≤ Np (f − g),

138

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ALMOST EVERYWHERE STATEMENTS, CONVERGENCE THEOREMS

i.e. |Np (χA f ) − Np (χA g)| ≤ Np (f − g).

(7.18)

From (7.17) and (7.18) we derive Proposition 7.29. If (fk )k∈N , fk ∈ L1 (Ω), converges in the mean to f ∈ L1 (Ω) then for all A ∈ A we have Z Z lim fk dµ = f dµ. (7.19) k→∞

A

A

For a sequence (gk )k∈N , gk ∈ Lp (Ω), converging in the pth mean to g ∈ Lp (Ω), 1 ≤ p < ∞, we have for all A ∈ A Z Z |g|p dµ. (7.20) |gk |p dµ = lim k→∞

A

A

To proceed further we need a result which is of great use in many situations: Lemma 7.30 (P. Fatou). For a sequence (fk )k∈N of non-negative measurable numerical functions fk : Ω → [0, ∞] on (Ω, A, µ) we always have Z Z lim inf fk dµ ≤ lim inf fk dµ. (7.21) k→∞

k→∞

Proof. Recall that lim inf k→∞ fk = supk∈N inf l≥k fl and that both inf l≥k fl ≥ 0 and f := lim inf k→∞ fk ≥ 0 are measurable. Moreover, the sequence (inf l≥k fl )k∈N is increasing and therefore by the monotone convergence theorem, Theorem 5.13, it follows   Z  Z Z  f dµ = sup inf fl dµ = lim inf fl dµ. l≥k

k∈N

k→∞

l≥k

Since inf l≥k fl ≤ fk we deduce  Z  Z inf fl dµ ≤ inf fl dµ l≥k

or

l≥k

Z Z   lim inf fk dµ ≤ lim inf fk dµ. k→∞

k→∞

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The following corollary to Fatou’s lemma is in particular in probability theory of great help. Corollary 7.31. For a sequence (Ak )k∈N , Ak ∈ A, the following holds ! ∞ ∞ \ [ (7.22) µ Al ≤ lim inf µ(Ak ) k→∞

k=1 l=k

and if µ is a finite measure we have ! ∞ ∞ [ \ µ Al ≥ lim sup µ(Ak ).

(7.23)

k→∞

k=1 l=k

Proof. For a proof of (7.22) we refer to Problem 9.c). We arrive at (7.23) when applying (7.22) to the sequence (A{k )k∈N :  ! !{  ∞ ∞ ∞ [ ∞ [ \ \ Al = µ  Al  µ(Ω) − µ k=1 l=k

k=1 l=k

=µ

∞ ∞ \ [

k=1 l=k

A{l

!

≤ lim inf µ(A{k ) k→∞

= µ(Ω) − lim sup µ(Ak ) k→∞

or µ

∞ ∞ [ \

k=1 l=k

Al

!

≥ lim sup µ(Ak ). k→∞

With the help of Fatou’s lemma we can prove the following convergence theorem due to F. Riesz: Theorem 7.32 (F. Riesz). Let (uk )k∈N , uk ∈ Lp (Ω), be a sequence converging µ-a.e. to u ∈ Lp (Ω), 1 ≤ p < ∞. If Z Z p lim |uk | dµ = |u|p dµ (7.24) k→∞

then (uk )k∈N converges to u in the pth mean, i.e. Z  p1 p = 0. lim |uk − u| dµ k→∞

140

(7.25)

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ALMOST EVERYWHERE STATEMENTS, CONVERGENCE THEOREMS

Proof. For a, b ≥ 0 and 1 ≤ p < ∞ we find (a + b)p ≤ (2(a ∨ b))p = 2p (ap ∨ bp ) ≤ 2p (ap + bp ) and since |a − b| ≤ a + b we have |a − b|p ≤ 2p (ap + bp ). Given (uk )k∈N as in the assumption, we deduce that vk := 2p (|uk |p + |u|p) − |uk − u|p ∈ L1 (Ω)

(7.26)

is a sequence of non-negative functions converging µ-a.e. to 2p+1 |u|p, hence lim inf k→∞ vk = 2p+1|u|p . Applying Fatou’s lemma to (vk )k∈N and taking into account (7.24) we get Z Z Z   p+1 p 2 |u| dµ = lim inf vk dµ ≤ lim inf vk dµ k→∞ k→∞ Z Z p+1 p =2 |u| dµ − lim sup |uk − u|p dµ, k→∞

or 0 ≤ lim sup k→∞

Z

|uk − u|p dµ ≤ 0.

By far the most important convergence result which in particular shows the advantage the Lebesgue integral has over the Riemann integral is the dominated convergence theorem due to H. Lebesgue. Theorem 7.33 (Dominated Convergence Theorem). Let (Ω, A, µ) be a measure space and (uk )k∈N , uk : Ω → R, be a sequence of integrable functions. Suppose that (uk )k∈N converges µ-a.e. to u : Ω → R and that for an integrable function g : Ω → [0, ∞] we have |uk | ≤ g µ-a.e. Then the function u is integrable, µ-a.e. bounded by g, and Z Z Z lim uk dµ = lim uk dµ = u dµ. (7.27) k→∞

k→∞

Proof. Since the µ-a.e. convergence of (uk )k∈N to u implies |u| ≤ g µ-a.e. and since g is integrable it follows from Theorem 5.22 that u is integrable. We first 141

A COURSE IN ANALYSIS

assume that uk , u and g are real-valued which implies that uk , u, g ∈ L1 (Ω). In this situation it is sufficient to prove that N1 (|uk − u|) converges to 0 in order to arrive at (7.27). For this set vk := |uk − u| and observe that 0 ≤ vk ≤ |uk | + |u| ≤ 2g,

hence vk is integrable. Applying the Lemma of Fatou to the sequence (|u| + g − vk )k∈N we get Z Z   lim inf (|u| + g − vk ) dµ ≤ lim inf (|u| + g − vk ) dµ k→∞ k→∞ Z Z = (|u| + g) dµ − lim sup vk dµ. k→∞

Since (|u|+g−vk )k∈N converges µ-a.e. to |u|+g it follows that lim inf k→∞ (|u|+ g − vk ) = |u| + g µ-a.e. which yields Z  Z  lim inf (|u| + g − vk ) dµ = (|u| + g) dµ k→∞

and therefore

Z

vk dµ ≤ 0. R But vk ≥ 0 µ-a.e. and we get limk→∞ |uk − u| dµ = 0 and hence (7.27). Now we turn to the general case. The integrability of uk , u and g implies the existence of a set N1 ∈ A of measure zero such that in N1{ we have lim sup k→∞

|uk (ω)| ≤ g(ω) < ∞ and |u(ω)| < ∞. Further, from the µ-a.e. convergence of (uk )k∈N to u follows the existence of N2 ∈ A, µ(N2 ) = 0, such that for ω ∈ N2{ we have limk→∞ uk (ω) = u(ω). Let N := N1 ∪ N2 . The first part of the proof yields Z Z (χN { uk ) dµ = (χN { u) dµ. lim k→∞

However N is a set of measure zero and therefore we have Z Z Z Z (χN { u) dµ = u dµ (χN { uk ) dµ = uk dµ and

and the theorem follows.

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Corollary 7.34. For 1 ≤ p < ∞ let (uk )k∈N be a sequence in Lp (Ω) converging µ-a.e. to u. Further assume that |uk | ≤ g holds for some g ∈ Lp (Ω). It follows that u ∈ Lp (Ω) and limk→∞ Np (uk − u) = 0. Proof. As in the proof of Theorem 7.33 we deduce uk , u, g ∈ Lp (Ω) and replacing vk by wk := |uk − u|p we have only to note that 0 ≤ wk ≤ (|uk | + |u|)p ≤ (|g| + |u|)p and with h := (|g| + |u|)p replacing |u| + g we can now follow the proof of Theorem 7.33 unchanged. For every semi-norm ρ on a vector space V the notion of a Cauchy sequence makes sense: (xk )k∈N , xk ∈ V , is called a Cauchy sequence with respect to ρ if for every  > 0 there exists N = N() ∈ N such that k, l ≥ N implies ρ(xk − xl ) < . From the triangle inequality which holds for ρ we deduce as usual that if (xk )k∈N converges with respect to ρ then it must be also a Cauchy sequence with respect to ρ. The converse is not necessarily true as we already know for norms (or metrics). The following result, often called the Fischer-Riesz theorem claims the convergence of Cauchy sequences in Lp (Ω), 1 ≤ p < ∞. Theorem 7.35. Every Cauchy sequence (uk )k∈N in Lp (Ω), 1 ≤ p < ∞, converges in the pth mean to some u ∈ Lp (Ω). Proof. The sequence (uk )k∈N admits a subsequence (ukl )l∈N such that for all l∈N Np (ukl+1 − ukl ) < 2−l holds. With vl := ukl+1 − ukl and v := we get by Lemma 7.28 Np (v) ≤

∞ X l=1

Np (vl ) ≤

∞ X l=1

∞ X

|vl | ≥ 0,

2−l = 1.

l=1

SincePv is measurable it is p-fold integrable, hence it is µ-a.e. real-valued, ∞ th partial sum of i.e. l=1 vl converges µ-a.e. absolutely. Since for the N 143

A COURSE IN ANALYSIS

P vl we find N l=1 vl = ukN+1 − uk1 it follows that (ukl )l∈N converges µ-a.e. Moreover we have

P∞

l=1

|ukl+1 | = |v1 + · · · + vl + uk1 | ≤ v + |uk1 | and v + |uk1 | ∈ Lp (Ω). Hence we may apply the dominated P∞ convergence theorem in form of Corollary 7.34 to (u ) with u = kl l∈N l=1 vl , i.e. u = P∞ p l=1 vl ∈ L (Ω) and lim Np (ukl − u) = 0. (7.28) l→∞

We claim that (7.28) also holds for the sequence (uk )k∈N which indeed follows from a standard argument. Since (uk )k∈N is a Cauchy sequence with respect to Np , given  > 0 there exists K1 () ∈ N such that k, l ≥ K1 () implies Np (uk − ul ) < 2 . Further, since (ukm )m∈N converges in the pth mean to u there exists m such that m ≥ K1 () implies Np (ukm − u) < 2 . Now the triangle inequality for Np yields for k ≥ K1 () Np (uk − u) ≤ Np (uk − ukm ) + Np (ukm − u) < .

For a sequence (uk )k∈N of measurable functions on a measure space (Ω, A, µ) we now have the following possible modes of convergence: • convergence in measure • µ-a.e. convergence • pointwise convergence • convergence in the pth mean • uniform convergence. So far we know the following relations uniform convergence ⇒ pointwise convergence ⇒ µ-a.e. convergence convergence in the pth mean ⇒ convergence in measure and none of these implications can be reversed. In the case where (Ω, A, µ) is a finite measure space, Corollary 7.22 implies 144

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that for 1 ≤ q < p convergence in the pth mean implies convergence in the q th mean. Moreover from Np (u) =

Z

p

|u| dµ

 p1

1

1

≤ µ(Ω) p sup |u(ω)| = µ(Ω) p kuk∞ ω∈Ω

which holds for a finite measure space (Ω, A, µ) and bounded measurable functions we deduce that for a finite measure space uniform convergence implies convergence in the pth mean for all p ≥ 1. Theorem 7.36. Let (uk )k∈N be a sequence of real-valued measurable functions on the same measure space (Ω, A, µ). A. If (uk )k∈N converges µ-a.e. to a real-valued measurable function u on Ω, then (uk )k∈N converges also in measure to u. B. If (uk )k∈N converges in measure to u then for every set A of finite measure the sequence (uk |A )k∈N has a subsequence converging µA -a.e. to u. Proof. A. Let A ∈ A, µ(A) < ∞,  > 0, and define the following sets:  B := ω ∈ A (uk (ω))k∈N does not converge to u(ω) ;  Bk () := ω ∈ A |uk (ω) − u(ω)| ≥  ; Rn () :=

∞ [

Bk ();

k=n

M() :=

\

Rn ().

n∈N

Since the functions uk and u are measurable all these sets belong to A. First we claim M ⊂ B. For ω0 ∈ / B it follows that limk→∞ uk (ω0 ) = u(ω0 ). Hence for every η > 0 there exists k0 ∈ N such that k ≥ k0 implies that |uk (ω0 ) − u(ω0 )| < η, i.e. ω0 ∈ / Bk for k ≥ k0 and therefore ω0 ∈ / M. Now µ|A is a finite measure and Rn () ⊃ Rn+1 () therefore by Theorem 2.10, we get µ|A (Rn ()) → µ|A (M), but Bn () ⊂ R n (),
which yields µ|A (Bn ()) → 0 implying limk→∞ µ ω ∈ Ω |uk − u| >  ∩ A = 0, i.e. uk → u in measure. B. Suppose that (uk )k∈N converges in measure to u and let A ∈ A, µ(A) < ∞. Switching to (A, A∩A, µ|A ) we may assume that (Ω, A, µ) is a finite measure space. For α > 0 and k, l ∈ N the triangle inequality yields n αo αo n ∪ |ul − u| ≥ . {|uk − ul | ≥ α} ⊂ |uk − u| ≥ 2 2 145

A COURSE IN ANALYSIS

SinceP (uk )k∈N converges in measure to u, for a sequence (ηj )j∈N , ηj > 0, such that ∞ j=1 ηj < ∞ we can find nj such that for k ≥ nj we have µ


 |uk − unj | ≥ ηj < ηj

and nj < nj+1 for all j ∈ N. The sets  Bj := ω ∈ Ω |unj+1 − unj | ≥ ηj are measurable and

∞ X j=1

µ(Bj ) ≤

∞ X j=1

ηj < ∞,

T∞ S∞ P∞ implying that limNT →∞ j=N Bj it follows N =1 j=N µ(Bj ) = 0. For B = P ∞ µ(B ) for all N, i.e. B and hence µ(B) ≤ further that B ⊂ ∞ j j j=N j=N { µ(B) = 0. For ω ∈ B the inequality |unj+1 (ω) − unj (ω)| ≥ ηj

P∞ can hold only for finitely many j. Further since j=1 ηj converges the seP∞ ries j=1(unj+1 (ω) − unj (ω)) converges µ-a.e. to a real-valued measurable function v : Ω → R. By part A v is also a limit in measure of (unj )j∈N and therefore we have u = v µ-a.e. Note we have proved that (uk )k∈N has a subsequence (unj )j∈N converging µ-a.e. on A ⊂ Ω, µ(A) < ∞, to u. Corollary 7.37. If (uk )k∈N , uk ∈ Lp (Ω), converges in the pth mean to u ∈ Lp (Ω), then for every measurable set A ∈ A, µ(A) < ∞, (uk )k∈N admits a subsequence converging on A µ-a.e. to u. Finally we want to discuss an interesting relation between µ-a.e. convergence and uniform convergence. Theorem 7.38 (D. F. Egorov). Let (Ω, A, µ) be a finite measure space and (uk )k∈N a sequence of real-valued measurable functions converging µ-a.e. to the real-valued (measurable) function u. For every δ > 0 there exists Aδ ∈ A such that µ(Aδ ) > µ(Ω) − δ, i.e. µ(A{δ ) < δ (7.29) and (uk |Aδ )k∈N converges uniformly to u|Aδ . 146

(7.30)

7

ALMOST EVERYWHERE STATEMENTS, CONVERGENCE THEOREMS

Proof. (Following [50]) We consider the measurable sets  \ 1 , Am,n := ω ∈ Ω |uk (ω) − u(ω)| < m k≥n

i.e. for m, n ∈ N fixed Am,n is the set of all ω ∈ Ω where |uk (ω) − u(ω)| < m1 for all k ≥ n. From the definition it follows that Am,n ⊂ Am,n+1 and we set [ Am := Am,n . n∈N

The continuity of µ, see Theorem 2.10, implies that for every m ∈ N and δ > 0 there exists n0 (m) ∈ N such that
δ µ Am \Am,n0 (m) < m . 2

We claim that

Aδ :=

\

Am,n0 (m)

m∈N

satisfies (7.29) and (7.30). First we note that for m ∈ N and ω ∈ Aδ given, we have for k > n0 (m) that |uk (ω) − u(ω)| < m1 which is however the uniform convergence of (uk |Aδ )k∈N to u|Aδ . It remains to prove (7.29). For ω0 ∈ Ω\Am we can find k sufficiently large such that |uk (ω0 ) − u(ω0 )| ≥ m1 , i.e. at ω0 (uk )k∈N does not converges to u, but since this sequence converges µ-a.e. to u we have µ(Ω\Am ) = 0 for every m ∈ N. This now implies

and further



δ µ Ω\Am,n0 (m) = µ Am \Am,n0 (m) < m 2

µ (Ω\Aδ ) = µ Ω\

\

Am,n0 (m)

m∈N

≤

∞ X

µ Ω\Am,n0 (m)

m=1

and the theorem is proved.




!

=µ

[

Ω\Am,n0 (m)

m∈N

∞ X δ =δ ≤ 2m m=1




!

We end this chapter by an obvious extension to the monotone convergence theorem, see [75]. 147

A COURSE IN ANALYSIS

Theorem 7.39 (General monotone convergence theorem). Let (Ω, A, µ) be a measure space and (uk )k∈N , uk ∈ L1 (Ω), an increasing sequence. If R supk∈N uk dµ is finite then u := supk∈N uk belongs to L1 (Ω) and the following holds Z Z sup uk dµ = sup uk dµ. (7.31) k∈N

k∈N

Proof. We may apply Theorem 5.13 to the sequence (uk − u1)k∈N which is non-negative, and we obtain Z Z 0 ≤ sup (uk − u1 ) dµ = sup(uk − u1) dµ. (7.32) k∈N

k∈N

R

Since supk∈N uk dµ is finite it follows that u = u1 + (u − u1 ) belongs to L1 (Ω) and now (7.32) implies (7.31). Remark 7.40. A. Note that u := supk∈N uk implies Z Z sup uk dµ = u dµ < ∞, k∈N

R and hence supk∈N uk dµ is finite if and only if u ∈ L1 (Ω). B. Changing the sign we can reformulate TheoremR 7.39 for decreasing sequences: (vk )k∈N , vk ∈ L1 (Ω), decreasing and inf k∈N vk dµ > −∞ implies Z Z inf vk dµ = inf vk dµ k∈N

and inf k∈N

R

k∈N

vk dµ > −∞ if and only if inf k∈N vk ∈ L1 (Ω).

Problems 1.

a) On the measurable PN space (Ω, P(Ω)), Ω = {1, . . . , N}, consider 1 the measure µ = N k=1 k . Let (uj )j∈N be a sequence of functions uj : Ω → R which converges µ-almost everywhere to u : Ω → R. Prove that the convergence of (uj )j∈N is in fact pointwise. P b) Now consider on (R, B(1) ) the measure ν = N1 N k=1 k and the sequence of functions (vj )j∈N , vj = χ[ 1 − 1 , 1 + 1 ] for j ≥ 5, and vj = j 4 j 2 j for j = 1, 2, 3, 4. Prove that (vj )j∈N converges pointwisely to χ[ 1 , 1 ] 4 2

148

7

ALMOST EVERYWHERE STATEMENTS, CONVERGENCE THEOREMS

and consequently ν-a.e. Prove further that (vj )j∈N converges also νalmost everywhere to the function v˜ = 0. Now consider the λ(1) -almost everywhere limit of (vj )j∈N . Is this λ(1) -almost everywhere limit equal to v˜? P 2. On the measure space (R, B(1) , ∞ k=1 ) consider Np (u), p ≥ 1, defined as in (7.6). Let ul , u : R → R, l ∈ N, be measurable functions such that liml→∞ Np (ul − u) = 0. Prove that this implies for all k ∈ N the convergence of (ul (k))l∈N to u(k). 3. Let f : R → R be a Lipschitz continuous function, i.e. |f (x) − f (y)| ≤ L|x − y| with L ≥ 0. Further let (uk )k∈N be a sequence of measurable functions uk : R → R converging λ(1) -almost everywhere to the measurable function u : R → R. Prove that (f (uk ))k∈N converges λ(1) -almost everywhere to f (u). 4.

a) Let g : Rn → R, g ≥ 0 and g ∈ Lq (Rn ). Consider on B(n) the measure ν = gλ(n) . For f ∈ Lp (Rn ), 1p + 1q = 1, prove ν({|f | ≥ α}) ≤

1 ||g||Lq ||f ||Lp . α

b) With g, q and (fk )k∈N , fk ∈ Lp (Rn ), be a sequence R p as above let(n) for which limk→∞ Rn |fk − h|g dλ = 0 holds for some h ∈ Lp (Rn ). Prove that (fk )k∈N converges in ν-measure to h. (k)

5. For k ∈ N we define on [0, 1) the functions fj , 1 ≤ j ≤ k, by ( 1, x ∈ [ j−1 , kj ) (k) k fj (x) := , kj ). 0, x ∈ [0, 1) \ [ j−1 k These functions form a sequence (gl )l∈N defined by (1)

(2)

(2)

(3)

(3)

g1 := f1 , g2 := f1 , g3 := f2 , g4 := f1 , g5 := f2 , . . . (1)

Prove that the sequence (gk )k∈N converges in λ|[0,1) -measure to zero but (1)

not λ|[0,1) -almost everywhere. (This is the standard counter example for a sequence converging in measure but not almost everywhere, see for example J.P.Natanson [59].) 149

A COURSE IN ANALYSIS 1 6. Prove that the function g : R → R, g(x) = 1+|x| , belongs to L2 (R) but not to L1 (R). Deduce that estimate (7.8) in Corollary 7.22 in general cannot hold for an unbounded set.

7.

a) Solve Exercise 7.21. P b) Prove that by p1 (u) := x∈[0,1] |u0(x)| a semi-norm is defined on Cb1 ([0, 1]) = {u ∈ Cb ([0, 1])|u0 ∈ Cb ([0, 1])} which is not a norm. For uk , u ∈ Cb1 ([0, 1]), k ∈ N, suppose that limk→∞ p1 (uk − u) = 0. Show that for every c ∈ R we also have limk→∞ p1 (uk − (u + c)) = 0. c) Prove that on Cb1 ([0, 1]) a further semi-norm is given by q1 (u) :=

Z

0

[0,1]

2 (1)

|u (x)| λ (dx)

 12

.

Characterise the functions u ∈ Cb1 ([0, 1]) for which q1 (u) = 0 holds. Suppose that (uk )k∈N , uk ∈ Cb1 ([0, 1]), and that for v ∈ L2 ([0, 1]) we have Z  21 0 2 (1) =0 lim |uk (x) − v(x)| λ (dx) k→∞

[0,1]

and assume that limk→∞ p1 (uk − u) = 0, u ∈ Cb1 ([0, 1]). What can we state about v? 8. Prove that by N∞ (·) a semi-norm is given on L∞ (Ω). 9.

a) Verify Fatou’s Lemma for the sequence uk : R → R, ( k 2 , 0 < x < k12 uk (x) = 0, otherwise.

b) Let (Ω, A, µ) be a measure space and vk , v : Ω → R, k ∈ N, are non-negative functions such that vk (x) ≤ v(x) and vkR → v µalmost everywhere. Use Fatou’s Lemma to prove that limk→∞ vk dµ = R v dµ. c) Prove (7.22).

150

7

ALMOST EVERYWHERE STATEMENTS, CONVERGENCE THEOREMS

10. Let (Ω, A, µ) be a finite measure space and u : Ω → R an essentially bounded, measurable function. Prove that lim

p→∞

Z

p

Ω

|u(ω)| µ(dω)

 p1

= ||u||∞.

11. Let (Ω, A, µ) be a finite measure space and let (fk )k∈N be a sequence of measurable functions converging on Ω in measure to the measurable function f . Show that a subsequence of (fk )k∈N converges µ-almost everywhere to f . 12. Use the result of Problem 10 to prove the following variant of the dominated convergence theorem: let (Ω, A, µ) be a finite measure space and (fk )k∈N be a sequence of measurable functions on Ω converging in measure to the measurable function f . If there exists an integrable function g such that |fk (ω)| ≤ g(ω) for all k ∈ N and ω ∈ Ω then Z Z fk dµ = f dµ. lim k→∞

Ω

13. Let (Ω, A, µ) be a finite measure space and (gk )k∈N , gk ∈ Lp (Ω), a sequence of functions such that limk→∞ Np (gk − g) = 0 for some g ∈ Lp (Ω). Further let (hk )k∈N be a sequence of measurable functions satisfying |hk (ω)| ≤ c < ∞ for all k ∈ N and ω ∈ Ω and which converges µ-almost everywhere to the measurable function h. Prove that lim Np (gk hk − gh) = 0.

k→∞

151

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8

Applications of the Convergence Theorems and More

In this chapter we collect various applications of the convergence results proved in the last chapter. Some are just useful tools, others are important statements in their own right. We will investigate the relations between the Lebesgue and the Riemann integral for domains in Rn . Finally we also discuss how we can pass from the spaces Lp (Ω) to the Banach spaces Lp (Ω) and therefore prepare some topics needed in functional analysis. The first topic we want to deal with is parameter dependent integrals. Already when discussing the Γ-function in I.28 we encountered integrals depending on a parameter, and in Part 4 we devoted in the context of the Riemann integral two chapters, II.17 and II.22 to these questions. We will now see that when using the Lebesgue integral certain results become clearer and more easy. We start with Theorem 8.1. Let (Ω, A, µ) be a measure space and (X, d) be a metric space. Further let u : X × Ω → R be a function satisfying i) for every x ∈ X the function u(x, ·) : Ω → R is µ-integrable; ii) for every ω ∈ Ω the function u(·, ω) : X → R is continuous at x0 ∈ X; iii) there exists a µ-integrable function h : Ω → R, h ≥ 0, such that |u(x, ω)| ≤ h(ω) for all (x, ω) ∈ X × Ω. Under these conditions the function g : X → R defined by Z g(x) := u(x, ω)µ(dω)

(8.1)

is continuous at x0 ∈ X. Proof. Let (xk )k∈N , xk ∈ X, be any sequence converging in (X, d) to x0 . Define the sequence uk : Ω → R by uk (ω) := u(xk , ω). By assumption it follows that |uk | ≤ h for all k ∈ N and since u is continuous at x0 the sequence 153

A COURSE IN ANALYSIS

(uk )k∈N converges pointwisely to u(x0 , ·). From the dominated convergence theorem, Theorem 7.33, we now deduce that Z lim g(xk ) = lim uk (ω)µ(dω) k→∞ k→∞ Z u(xk , ω)µ(dω) = lim k→∞ Z = lim u(xk , ω)µ(dω) k→∞ Z = u(x0 , ω) dµ = g(x0 ),

i.e. the continuity of g at x0 .

Corollary 8.2. If in the situation of Theorem 8.1 the function u(·, ω) : X → R is for every ω ∈ Ω continuous, then the function g defined by (8.1) is continuous on X. Note that we can apply Theorem 8.1 and its corollary in particular to the situation where X ⊂ Rm and d(x, y) = kx − yk, and (Ω, A, µ) = Y, Y ∩  (n) B(n) , λ|Y where Y ∈ B(n) . For example if X ⊂ Rm and Y ⊂ Rn are two open (or closed) sets and u : X × Y → R is continuous, thus x 7→ u(x, y) is continuous and y 7→ u(x, y) is continuous, hence measurable, then the (n) integrability of u(x, ·) with respect to λ|Y for all x ∈ X and the existence (n)

of an λ|Y -integrable function h : Y → R such that |u(x, y)| ≤ h(y) for all R (x, y) ∈ X × Y , entails the continuity of Y u(x, y)λ(n)(dy).

Example 8.3. Let u ∈ L1 (Rn ). The cosine-transform of u is defined for ξ ∈ Rn by Z −n coshx, ξiu(x) λ(n)(dx) (8.2) u˜(ξ) := (2π) 2

and u˜ : Rn → R is a continuous function. Indeed, we consider the function (x, ξ) 7→ coshx, ξiu(x) on Rn ×Rn . For every ξ ∈ Rn fixed this function is λ(n) integrable as product of a bounded measurable function with an integrable function, and for x ∈ Rn fixed this function is continuous since ξ 7→ coshx, ξi is continuous. Further we have uniformly in ξ the estimate | coshx, ξiu(x)| ≤ |u(x)|

and |u| ∈ L1 (Rn ). Hence by Theorem 8.1 the function u˜ is continuous on Rn . 154

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

We now turn to the differentiability of parameter dependent integrals. Once we understand the structure of the argument for functions u : I × Ω → R where I ⊂ R, ˚ I 6= ∅, is an interval, extending to functions v : X × Ω → R, X ⊂ Rm open, is almost obvious. Therefore we concentrate on the first mentioned case. Theorem 8.4. Let (Ω, A, µ) be a measure space and I ⊂ R, ˚ I 6= ∅, an interval. For the function u : I × Ω → R we assume i) for all x ∈ I the function u(x, ·) : Ω → R is µ-integrable; ii) for all ω ∈ Ω the function u(·, ω) : I → R is differentiable with deriva(·, ω); tive ∂u ∂x iii) for a µ-integrable function h : Ω → [0, ∞) we have on I × Ω ∂u (x, ω) ≤ h(ω). ∂x

Then the function

g(x) := is on I differentiable, ω 7→

Z

∂u (x, ω) ∂x

0

g (x) =

Z

u(x, ω)µ(dω)

(8.3)

(8.4)

is for all x ∈ I integrable and ∂u (x, ω)µ(dω) ∂x

(8.5)

holds. Proof. Let (xk )k∈N , xk ∈ I, xk 6= x0 , be an arbitrary sequence converging to x0 ∈ I and define on Ω the functions uk (ω) :=

u(xk , ω) − u(x0 , ω) . xk − x0

0 ,ω) These are µ-integrable functions and limk→∞ uk (ω) = ∂u(x for all ω ∈ Ω. ∂x By the mean value theorem we can find ξ = ξ(x, x0 , ω), x ∧ x0 ≤ ξ ≤ x ∨ x0 , such that uk (x, ω) − u(x0 , ω) ∂u = (ξ, ω). x − x0 ∂x

155

A COURSE IN ANALYSIS

For xk condition (8.3) implies with ξ = ξk as above ∂u |uk (ω)| = (ξk , ω) ≤ h(ω). ∂x

Again we can apply the dominated convergence theorem to derive that ∂u (x0 , ·) is integrable and that ∂x Z Z u(xk , ω) − u(x0 , ω) lim µ(dω) = lim uk (ω)µ(dω) k→∞ k→∞ xk − x0 Z Z ∂u = lim uk (ω)µ(dω) = (x0 , ω)µ(dω). k→∞ ∂x Remark 8.5. Of course, if we substitute in Theorem 8.4 the function u by ∂ m u(x,ω) and now assume the conditions i) − iii) modified accordingly, we can m ∂x obtain statements for higher order derivatives, see Problem 2. The following corollary is a version of Theorem 8.4 for functions u : U × Ω → R, U ⊂ Rm open, and again it extends to higher order partial derivatives.

Corollary 8.6. Let (Ω, A, µ) be a measure space and U ⊂ Rm an open set. Suppose that u : U × Ω → R satisfies i) for all x ∈ U the function u(x, ·) is µ-integrable; ii) for all ω ∈ Ω the partial derivative

∂ u(x, ω) ∂xj

exists, 1 ≤ j ≤ m;

iii) for a µ-integrable function h : Ω → [0, ∞) the following holds on U × Ω ∂u (8.6) ∂xj (x, ω) ≤ h(ω).

In this case the function v : U → R, Z v(x) := u(x, ω)µ(dω) has in U the partial derivative µ-integrable and we have

∂v , ∂xj

∂v (x) = ∂xj

Z

the function

∂u (x, ·) ∂xj

∂u (x, ω)µ(dω). ∂xj 156

is for all x ∈ U (8.7)

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

Proof. We can employ the proof of Theorem 8.4 when fixing all variables x1 , . . . , xj−1 , xj+1, . . . , xm . Of course, in Theorem 8.4 and its corollary we may take for (Ω, A, µ) the (n) measure space (V, V ∩ B(n) , λ|V ) for V ∈ B(n) and hence we can handle situations such as Z (n) α u(x, y)λ|V (dy). ∂x V

Example 8.7. Let u ∈ L1 (Rn ) be a function such that the function x 7→ |x|u(x) belongs to L1 (Rn ) too. In this case the cosine-transform u˜ has all first order partial derivatives and they are bounded. We consider the function defined on Rn × Rn by (x, ξ) 7→ coshx, ξiu(x). This function is with respect to x for all ξ integrable and since for 1 ≤ j ≤ n ∂ (coshxl , ξiu(x)) = −xj sinhx, ξiu(x) ∂ξj we have the estimate ∂ ∂ξj (coshx, ξiu(x)) ≤ |x| |u(x)|

and by assumption x 7→ |x||u(x)| belongs to L1 (Rn ), i.e. is integrable. It follows that Z ∂ u˜(ξ) = (−xj sinhx, ξiu(x))λ(n) (dx) ∂ξj and further

Z ∂ (n) ∂ξj u˜(ξ) ≤ |x||u(x)|λ (dx) = Mu < ∞.

Our next result is Jensen’s inequality which we have encountered already in several versions in Volume I and II. It is in particular of great use in probability theory and a more general variant than ours can be found in [75]. In Theorem I.23.5 we have proved that a convex function f : I → R, I ⊂ R an interval with end points a < b, is differentiable from the right and the proof yields f (x0 ) − f (x) ≤ f+0 (x0 ), (8.8) x0 − x 157

A COURSE IN ANALYSIS

where f+0 (x0 ) is the derivative from the right of f at x0 ∈ I. We can rewrite (8.8) as f (x) ≥ f+0 (x0 )(x − x0 ) + f (x0 ). (8.9) From (8.9) combined with Proposition I.23.10 we deduce, since x0 ∈ I, that f (x) = sup{f+0 (y)(x − y) + f (y)}.

(8.10)

y∈I

In particular, for u : Ω → R we get f (u(ω)) ≥ f+0 (y)(u(ω) − y) + f (y)

(8.11)

for every y ∈ I. With these preparations we can prove Theorem 8.8 (Jensen’s inequality). Let (Ω, A, µ) be a measure space and A ∈ A a set of finite measure µ(A) > 0. Let f : I → R be a convex function and u : A → R be a µ|A -integrable function such that u(A) ⊂ I. Then the following holds   Z Z 1 1 u dµ ≤ (f ◦ u) dµ. (8.12) f µ(A) A µ(A) A Proof. We take in (8.11) y = f



1 µ(A)

Z

A



u dµ +

1 µ(A)

f+0 (y)

R

A

u dµ ∈ I and we deduce

 u(ω) −

1 µ(A)

Z

A

u dµ



≤ f (u(ω)).

Integrating over A we find  Z  Z  Z Z 1 0 u dµ − u dµ ≤ (f ◦ u) dµ, u dµ + f+ (y) µ(A)f µ(A) A and dividing by µ(A) we obtain (8.12). Corollary 8.9. Let (Ω, A, P ) be a probability space and X : Ω → R a random variable with range contained in the interval I ⊂ R. For every convex function f : I → R we have f (E(X)) ≤ E(f ◦ X). 158

(8.13)

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

Example 8.10. For a finite R measure space (Ω, A, µ) we have for an integrable function u : Ω → R with u dµ = 1 the estimate Z 1 . (ln u) dµ ≤ µ(Ω) ln µ(Ω) Since (− ln)00 (x) = x12 the function − ln : (0, ∞) → R is convex and now (8.12) yields     Z Z 1 1 1 − ln = − ln u dµ ≤ (− ln u) dµ µ(Ω) µ(Ω) µ(Ω) or

Z

ln u dµ ≤ µ(Ω) ln

1 . µ(Ω)

We now want to compare the Riemann (Darboux-Jordan) intergration theory with the Lebesgue theory. We do this in three steps. First for functions defined on an interval I ⊂ R, then for functions defined on a non-degenerate cell or hyper-rectangle and then we turn to Jordan measurable sets. Our results will allow us to use in many cases the Riemann theory to evaluate integrals, while on the other hand they will provide us with a proof of the transformation theorem or the change of variable formula, Theorem II.21.8, for a large class of domains. When speaking in the following about Lebesgue integrable functions on subsets of Rn we always consider the Borel σ-field B(n) and the Lebesgue-Borel measure λ(n) restricted to this subset. Discussing integrability over a subset A ⊂ Rn we always assume A ∈ B(n) and for (n) simplicity we also write λ(n) for λ|A . We also adopt the notation of Volume I and II when dealing with the Riemann integral, see in particular Chapter I.25. Theorem 8.11. Let a < b and u : [a, b] → R be a Riemann integrable function. Then u is Lebesgue integrable over [a, b] and the Riemann and the Lebesgue integral of u coincide, i.e. Z

a

b

u(x) dx =

Z

u dλ(1) .

(8.14)

[a,b]

Proof. Let u : [a, b] → R be Riemann integrable, in particular u is bounded. By decomposing u = u+ − u− we may assume that u ≥ 0. Take a sequence 159

A COURSE IN ANALYSIS

(Zk )k∈N of partition of [a, b] with mesh(Zk ) tending to zero for k → ∞ and (k) (k) (k) (k) for an interval [xj , xj+1], xj , xj+1 ∈ Zk , let (k)

ξj

=

sup

  (k) (k) x∈ xj ,xj+1

u(x)

and

(k)

ηj

:=

 inf  u(x). (k) (k) x∈ xj ,xj+1

We assume further that Zk ⊂ Zk+1 , i.e. Zk+1 is a refinement of Zk . It follows that the Riemann sums S(u, Zk , ξ (k) ) and S(u, Zk , η (k) ) converge to the RieRb mann integral a u(x) dx. We can interpret S(u, Zk , ξ (k) ) and S(u, Zk , η (k) ) as Lebesgue integrals of the corresponding simple functions vk and ωk where (k) (k) vk |[x(k) ,x(k) ] = ξj and ωk |[x(k) ,x(k) ] = ηj . The sequence (vk )k∈N is decreasing j

j

j+1

j+1

whereas the sequence (ωk )k∈N is increasing and we have supk∈N ωk ≤ u ≤ inf k∈N vk . The general monotone convergence theorem, Theorem 7.39, now yields Z Z b (k) u(x) dx = lim S(u, Zk , η ) = lim ωk dλ(1) k→∞ k→∞ a Z = sup ωk dλ(1) , k∈N

as well as Z

a

b (k)

u(x) dx = lim S(u, Zk , ξ ) = lim k→∞ k→∞ Z = inf vk dλ(1) ,

Z

vk dλ(1)

k∈N

which implies supk∈N ωk , inf k∈N vk ∈ L1 (Ω) and  Z  0≥ inf vk − sup ωk dλ(1) = 0, k∈N

k∈N

i.e. ωk = inf k∈N vk = u λ(1) -a.e., i.e. u ∈ L1 (Ω) and R supk∈N u dλ(1) . [a,b]

Rb a

u(x) dx =

An analysis of the proof of Theorem 8.11 immediately gives

Corollary 8.12. Let K ⊂ Rn be a non-degenerate compact cell (a compact hyper-rectangle with voln (K) > 0) and u : K → R a measurable Riemann 160

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

integrable function. Then u is Lebesgue integrable and the Riemann and the Lebesgue integral of u over K coincide, i.e. Z Z u dλ(n) . (8.15) u(x) dx = K

K

Using the set-additivity S of both the Riemann and the Lebesgue integral we deduce that if G = N j=1 Kj is the union of finitely many compact nondegenerate cells Kj ⊂ Rn then a measurable Riemann integrable function u : G → R is also Lebesgue integrable and the two integrals of u over G coincide.

In Theorem II.19.25 we have seen that a bounded set G ⊂ Rn is Jordan measurable if ∂G has Lebesgue measure zero and we have mentioned that this condition indeed characterises Jordan measurable sets. Every compact set and every (bounded) open set in Rn is Borel measurable. Thus once we know that such a set is a Jordan measurable set, using the definition of the Riemann integral, i.e. Definition II.20.2, we can apply Corollary 8.12 to identify the corresponding Riemann and Lebesgue integral for a measurable and Riemann integrable function. A discussion when a given Borel measurable set in Rn is already Jordan measurable is provided in G. Folland [27]. The key problem is, when is it possible to “approximate” G ⊂ Rn by open cubes from the interior and by the compact cubes from the exterior (meaning the union of these cubes contains G) in such a way that the difference between the measures of the approximating sets tend to zero? See Figure 8.1 below.

Figure 8.1

161

A COURSE IN ANALYSIS

In Theorem II.20.4 we have proved that for a bounded function f : G → R on a bounded Jordan measurable set the Riemann integrability follows if the set of discontinuities has Jordan content zero. There are more general results, for example as stated in Theorem I.32.17 a bounded function u : [a, b] → R is Riemann integrable if and only if its set of discontinuity is a Lebesgue null set. We now want to provide a proof of Theorem I.32.17 which was not possible with the tools at our disposal in Volume I. Proof of Theorem I.32.17. As in the proof of Theorem 8.11 let (Zk )k∈N , Zk ⊂ Zk+1, be a sequence of partitions of [a, b] with mesh(Zk ) tending to 0 for (k) (k) k → ∞ and with ξj := supx∈[x(k) ,x(k) ) u(x), ηj := inf x∈[x(k) ,x(k) ) u(x) dej

j+1

j

j+1

noting the sequences (vk )k∈N and (wk )k∈N of simple functions corresponding S to S(u, Zk , ξ) and S(u, Zk , η), respectively. Note that k∈N Zk is a countable set, hence it has λ(1) -measure zero. The Riemann integrability of f implies that for  > 0 and x ∈ [a, b] there exists N = N(, x) ∈ N and tl−1 , tl ∈ ZN (,x) such that x ∈ (tl−1 , tl ) implies for k ≥ N(, x) that wk (x) − sup wk (x) + vk (x) − inf vk (x) ≤ . k∈N

k∈N

For x, y ∈ (tl−1 , tl ) we find

|u(x) − u(y)| ≤

sup u(t) −

t∈[tl−1 ,tl ]

inf

t∈[tl−1 ,tl ]

u(t)

= vN (,x) (x) − wN (,x) (x) ≤  + sup wk (x) − inf vk (x) . k∈N k∈N

We know from the proof of Theorem 8.11 that supk∈N wk = inf k∈N vk = u λ(1) a.e. The inequality just proven shows that {x ∈ [a, b] | u is discontinuous at S x} is contained in {x ∈ [a, b] | |supk∈N wk (x) − inf k∈N vk (x)| = 6 0} ∪ k∈N Zk , hence it is contained in a set of measure zero, i.e. a null set. On the other hand, we can deduce from the same inequality that   [  Zk . x ∈ [a, b] sup wk (x) = inf vk (x) ⊂ x ∈ [a, b] u is continuous at x ∪ k∈N

k∈N

k∈N

162

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APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

Thus, if the set of discontinuities of u is a null set, then the measurable set {x ∈ [a, b] | supk∈N wk (x) = inf k∈N vk (x)} is a set of measure zero and since Z Z Z
u = lim S u, Zk , η (k) = lim wk dλ(1) = sup wk dλ(1) k→∞

∗

and

Z

∗

k→∞

u = lim S u, Zk , ξ k→∞

this implies that

R

∗

u=

R∗

(k)




= lim

k→∞

k∈N

Z

vk dλ

(1)

=

Z

inf vk dλ(1)

k∈ N

u, i.e. the Riemann integrability of u.

Remark 8.13. In the proofs of Theorem 8.11 and Theorem I.32.17 we follow [76]. In Chapter I.28 and Chapter II.22 we discussed improper Riemann integrals which have a lot of important applications, we just want to mention the Γ-function or the Fourier transform which we will discuss in more detail in Part 8. Here we want to investigate the relations between improper Riemann integrals and the Lebesgue integral for n = 1, the higher dimensional case does not give a further insight. In Chapter I.28 we have discussed different types of improper Riemann integrals depending on whether we are working with bounded or unbounded domains and/or bounded or unbounded functions. For simplicity we have chosen here to treat in more detail the case of the half line [0, ∞) which covers for example (almost) the Γ-function. The corresponding results for the other cases we will state without proof. We also extend the definition of improper Riemann integrals slightly. While in Chapter I.28 and II.22 we worked with continuous functions only, we now assume that the functions under consideration are on every compact subinterval of their domain Riemann integrable. None of the results about the existence of improper Riemann integrals will change after this alteration. Theorem 8.14. Let f : [0, ∞) → R be a measurable function and suppose that f is improper Riemann integrable in the sense that f |[0,R] is for every R > 0 Riemann integrable and that Z R Z ∞ f (x) dx (8.16) f (x) dx := lim 0

R→∞

0

exists. The function f is Lebesgue integrable on [0, ∞) and the improper Riemann integral on [0, ∞) coincides with the Lebesgue integral of f over 163

A COURSE IN ANALYSIS

[0, ∞) if and only if |f | is improper Riemann integrable which means in particular that Z ∞ Z R |f (x)| dx (8.17) |f (x)| dx = lim R→∞

0

0

exists. In this case we also have Z Z ∞ |f (x)| dx = 0

[0,∞)

|f | dλ(1) .

Proof. If f is Riemann integrable over [0, R], R > 0, then f + and f − are also Riemann integrable over [0, R] and both functions are measurable by assumption. By Theorem 8.11 both, f + and f − , are Lebesgue integrable over [0, R], R > 0, and in each case the Lebesgue integral is equal to the Riemann integral, i.e. Z Z R + f (x) dx = χ[0,R] f + dλ(1) (8.18) 0

and

Z

R

−

f (x) dx = 0

Z

χ[0,R] f − dλ(1) .

(8.19)


For every sequence (RN )N ∈N , 0 < RN < RN +1 , the sequence χ[0,RN ] f + N ∈N
and χ[0,RN ] f − N ∈N are monotone increasing and if (8.16) and (8.17) hold the monotone R convergence theorem R applied to (8.18) and (8.19) yields the existence of χ[0,∞) f + dλ(1) and χ[0,∞) f − dλ(1) , hence f χ[0,∞) is Lebesgue integrable and we have Z Z Z ∞ (1) f (x) dx = χ[0,∞) f dλ = f dλ(1) . 0

[0,∞)

Conversely, if f ∈ L1 ([0, ∞)) then the functions f + , f − as well as gχ[0,R] , g ∈ {f, f + , f − }, are elements of L1 ([0, ∞)) and since f is by assumption improper Riemann integrable all these functions are Riemann integrals over [0, R], R > 0, and their Lebesgue and Riemann integrable over [0, R], R > 0, coincide. For every increasing sequence (Rj )j∈N , Rj > Rj−1 > 0, the monotone convergence theorem gives Z Z lim χ[0,Rj ] f + dλ(1) = χ[0,∞) f + dλ(1) j→∞

164

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

and lim

j→∞

Z

−

χ[0,Rj ] f dλ

(1)

=

Z

χ[0,∞) f − dλ(1) ,

which implies the existence of the limits (8.18) and (8.19), and hence (8.17). Example 8.15.R In Example I.28.19.B we have seen that the improper Rie∞ mann integral 0 sinx x dx exists. However the function g : [0, ∞) → R, g(x) = sinx x , x 6= 0, g(0) = 1, is not Lebesgue integrable over [0, ∞) as we have seen already in Problem 8 of Chapter 5. Remark 8.16. With the same type of arguments we can prove Theorem 8.14 for any type of improper Riemann integrals, compare with Definition I.28.1 and I.28.6. In the language of Definition I.28.14 we can state that an improper Riemann integral is a Lebesgue integral if it is absolutely convergent. In particular in the situation of Theorem I.28.17 an improper Riemann integral is a Lebesgue integral. Example 8.17. Since for x > 0 the function t 7→ tx−1 e−t is non-negative, Theorem 8.14 (or Remark 8.16) implies that the Γ-function Γ(x) R∞ = 0 tx−1 e−t dt can be looked at as a parameter dependent Lebesgue integral. Finally we want to return to the fact that for a measure space (Ω, A, µ) by
1 R Np (f ) = |f |p dµ p a semi-norm is given on Lp (Ω) which is in general not a norm. This fact has caused a number of problems in uniqueness statements, and moreover, since the theory of normed spaces, especially that of complete normed spaces, i.e. Banach spaces, gives much better tools to handle concrete problems such as solving equations, the question arises as to whether we can turn Lp (Ω) into a Banach space. The starting point is that for every semi-norm ρ on an R-vector space V an equivalence relation ∼ρ is given by x ∼p y if ρ(x − y) = 0. Indeed, since ρ(x−x) = ρ(0) = 0, ρ(x−y) = ρ(y−x) as well as ρ(x−z) ≤ ρ(x−y)+ρ(y−z), the fact that ∼ρ is an equivalence relation is trivial. As an equivalence relation ∼ρ induces on V a partition into equivalence classes which we denote by [x] = {y ∈ V | x ∼ρ y}. As usual we denote by V / ∼ρ the set of all equivalence classes induced by ∼ρ . With the operations [x] + [y] := [x + y], x, y ∈ V, 165

(8.20)

A COURSE IN ANALYSIS

and λ[x] := [λx], x ∈ V, λ ∈ R,

(8.21)

we can turn V / ∼ρ into a vector space. In particular the zero element 0 ∈ V / ∼ρ is [0V ] = {y ∈ V | 0V ∼ρ y} where for the moment 0V is the neutral element with respect to addition in V . Moreover we define on V / ∼ρ the mapping k.kρ : V / ∼ρ → R, k[x]kρ := ρ(x). (8.22) First we note that for x1 , x2 ∈ [x] it follows that ρ(x1 − x2 ) ≤ ρ(x1 − x) + ρ(x − x2 ) = 0, and therefore k[x1 ]kρ = ρ(x1 ) ≤ ρ(x1 − x2 ) + ρ(x2 ) = k[x2 ]kρ as well as k[x2 ]kρ = ρ(x2 ) ≤ ρ(x2 − x1 ) + ρ(x1 ) = k[x1 ]kρ i.e. k[x1 ]kρ = k[x2 ]kρ for all x1 , x2 ∈ [x] implying that k[x]kρ is independent of the representative and k.kρ is on V / ∼ρ well defined. Lemma 8.18. For every semi-norm ρ on the R-vector space V a norm is given on V / ∼ρ by k.kρ . Proof. Clearly, k[x]kρ ≥ 0 and k[x]kρ = 0 implies ρ(x) = 0, i.e. x ∈ [0V ] or [x] = 0 ∈ V / ∼ρ is the zero element. For λ ∈ R and [x] ∈ V / ∼ρ we have with x1 ∈ [x] kλ[x]kρ = k[λx]kρ = ρ(λx1 ) = |λ|ρ(x1 ) = |λ|k[x]kρ , and for [x], [y] ∈ V / ∼ρ with x1 ∈ [x] and y1 ∈ [y] it follows that k[x] + [y]kρ = k[x + y]kρ = ρ(x1 + y1 ) ≤ ρ(x1 ) + ρ(y1 ) = k[x]kρ + k[y]kp .
Thus V/∼ρ , k · kρ is a normed space. 166

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APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

Since for a measure space (Ω, A, µ) the term Np (u) = 1 ≤ p < ∞ a semi-norm on Lp (Ω) it follows that

R

Lp (Ω) := Lp (Ω)/ ∼Np

|u|p dµ


p1

is for (8.23)

is a normed space. We know that Np (u − v) = 0 if and only if u = v µ-a.e., 1 ≤ p < ∞, and hence we have Proposition 8.19. The space Lp (Ω) = Lp (Ω)/ ∼µ is a normed space with norm (8.24) k[u]kLp := k[u]kp := Np (u). We now want to lift the Fischer-Riesz theorem, Theorem 7.35, to Lp (Ω). Theorem 8.20. The normed space (Lp (Ω), k.kLp ), 1 ≤ p < ∞, is a Banach space where we have written kukLp for k[u]kLp . Proof. Let ([uk ])k∈N be a Cauchy sequence in Lp (Ω). For vk ∈ [uk ] it follows that (vk )k∈N is a Cauchy sequence in Lp (Ω) and this property is independent of the chosen representatives vk ∈ [uk ]. Hence by Theorem 7.35, (vk )k∈N determines µ-a.e. an element u ∈ Lp (Ω) satisfying Np (vk −u) → 0 as k → ∞. We claim that (8.25) lim k[uk ] − [u]kLp = 0. k→∞

We note that [uk ] − [u] = [uk − u] and consequently k[uk ] − [u]kLp = k[uk − u]kLp = Np (vk − u) and these equalities are independent from the representatives. Since limk→∞ Np (vk −u) = 0 it follows that (8.25) holds, i.e. in (Lp (Ω), k.kLp ) every Cauchy sequence has a limit and hence this is a Banach space. Theorem 8.21. Let (Ω, A, µ) be a measure space. The span of the set of all simple functions u ∈ S(Ω) with µ({x ∈ Ω | u(x) 6= 0}) < ∞ is dense in Lp (Ω), 1 ≤ p < ∞. Proof. Decompose f ∈ Lp (Ω) according to f = f + − f − . For f + we can find a sequence (uk )k∈N of simple functions such that 0 ≤ uk ≤ uk+1 ≤ f + and limk→∞ uk = f µ-a.e. This implies first of all that µ({x ∈ Ω | uk (x) 6= 0}) < ∞ and secondly uk ∈ Lp (Ω). Moreover we have |f + − uk |p ≤ |f |p and the dominated convergence theorem gives the result for f + which implies the general result. 167

A COURSE IN ANALYSIS

In Problem 11 we will discuss the space L∞ (Ω) := L∞ (Ω)/ ∼µ . The Banach spaces (Lp (Ω), k.kLp ) are from the structural point of view nice objects. However we must be very careful when working in Lp (Ω) since elements of Lp (Ω) are not any more functions but they are equivalence classes of functions. The common approach is to pretend that elements of Lp (Ω) are functions, i.e. to work with representatives, and to check that results are independent of the choice of the representatives. This needs some caution and experience. For example for u ∈ Lp (Ω) which is a pointwisely defined function u : Ω → R the statement u(x0 ) = max u(x) x∈Ω

makes sense. But for [u] ∈ Lp (Ω) such a statement does not make sense since [u](x0 ) is not defined independently of the representatives - it is not defined at all! Still we will later on follow the custom to write u ∈ Lp (Ω) when we should write [u] ∈ Lp (Ω). We leave the spaces Lp (Ω) here but return to them in Part 8 in this volume and in particular in Part 12 where we will discuss in more detail the properties of Lp (Ω) as Banach spaces.

Problems Where appropriate, results on the relationships between (improper) Riemann and Lebesgue integrals should be used to evaluate certain integrals below. 1.

a) For x ≥ 0 consider the integral Z −x(1+y2 ) e λ(1) (dy) ϕ(x) := 2 R 1+y and prove that ϕ : [0, ∞) → R is continuous with ϕ(0) = π. b) Show that the function k  Z sin x −ξx λ(1) (dx), k ≥ 2, e ψ(ξ) := x (0,∞)

is well defined and continuous on [0, ∞). 168

8

APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

2. Give precise conditions on u and µ which will imply for m ∈ N that Z ∂ m g(x) ∂u(x, ω) = µ(dω) m ∂x ∂xm R where g(x) = u(x, ω)µ(dω).

3. For the function in Problem 1 a) show

r π −x e , x > 0. ϕ (x) = − x 0

4.

x2

a) Prove that the function g : R → R, g(x) = e− 2 , solves the differential equation g 0(x) + xg(x) = 0.

(∗) b) Define g˜(ξ) = satisfies (∗).

√1 2π

R

R

x2

cos(ξx) e− 2 λ(1) (dx) and prove that g˜ also

c) Use the fact that two differentiable solutions h1 and h2 of (∗) differ only by a multiplicative constant, i.e. h1 = ch2 , to conclude that g˜ = g. R √ 2 Hint: use the fact that R e−x dx = π, see (I.28.29) or Theorem I.30.14. 5. Let µ be a bounded measure on B(1) . a) Show that

µ ˜(y) :=

Z

cos(yx) µ(dx)

R

is a continuous function µ ˜ : R → R. b) Find conditions on µ in terms of integrability properties of the functions x 7→ |x|k , k ∈ N, which will imply that µ ˜ is an N-times continuously differentiable function. P 6. Formulate Jensen’s inequality for the probability measure N1 N k=1 k on B(1) . 169

A COURSE IN ANALYSIS

7. For an integrable function u : [0, 1] → R such that |u(x)| ≤ 1 show that Z 8.

1 0

1 2

(1 − u2 (x)) dx ≤

1−

Z

1 0

2 ! 12 . u(x)dx

a) Let u, w : [a, b] → R be integrable functions and suppose that Rb w(x) ≥ 0 but a w(x)λ(1) (dx) > 0. Further assume that m ≤ u ≤ M. For a convex function ϕ : [m, M] → R prove ! Z 1 ϕ R u(x)w(x)λ(1) (dx) (1) (dx) w(x)λ [a,b] [a,b] Rb 1 (1) (dx). ≤ R w(x)λ (1) (dx) a ϕ(u(x))w(x)λ [a,b]

Hint: prove that µ(A) := [a, b] ∩ B(1) .

R

A

w(x)λ(1) (dx) is a bounded measure on

b) For a continuous function u : [0, 1] → R prove that Z

0

1

u(x) dx ≤ ln

Z

9. Prove that the Fresnel integrals Z ∞ cos(x2 )dx and 0

1 0

 eu(x) dx .

Z

∞

sin(x2 )dx

0

exist as improper Riemann integrals but not as Lebesgue integrals. Hint: use the transformation x2 = y andR investigate the transformed ∞ integrals. Now try to find the value of 0 sin(x2 )dx using methods from real analysis. We will also evaluate these two integrals later by using complex variable methods, see Chapter 21. R∞ α 10. Show that for every α > 0 the improper Riemann integral 0 e−x dx are Lebesgue integrals. 11. Discuss in detail the definition of L∞ (Ω) := L∞ (Ω)/ ∼µ as a Banach space with norm kuk∞ = esssup|u(x)|. 170

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APPLICATIONS OF THE CONVERGENCE THEOREMS AND MORE

12.

a) Let 1 < q and q ≤ r ≤ ∞. Prove that f ∈ Lq (Rn ) admits a decomposition f = g + h, g ∈ L1 (Rn ) and h ∈ Lr (Rn ).

b) For 1 < q < r ≤ ∞ show that L1 (Rn ) ∩ Lr (Rn ) ⊂ Lq (Rn ) and for u ∈ L1 (Rn ) ∩ Lr (Rn ) we have kukLq ≤ kukλL1 kuk1−λ Lr ,

λ=

r−q . q(r − 1)

13. This problem aims to remind us that we can also apply the convergence results in Lp spaces. The result itself is often used in the calculus of variations. Use Fatou’s Lemma to prove for a sequence (fk )k∈N , fk ∈ L1 (Rn ), g ≤ fRk for some g ∈ L1 (RnR), that fk → f λ(n) -almost everywhere implies f dλ(n) ≤ lim inf k→∞ fk dλ(n) .

171

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9

Integration on Product Spaces and Applications

Before introducing the Riemann integral for certain domains in Rn we studied in Chapter II.17 iterated integrals and then, in Chapter II.18 their relation to integrals over a hyper-rectangle in Rn , see Theorem II.18.19. We can interpret a hyper-rectangle as a product of intervals and an interesting question is whether we can find a natural measure on the product of two or finitely many measure spaces such that the integral with respect to this “product measure” will be for certain functions an iterated integral. In Chapter 1 we discussed product σ-fields and their generators and we will use the notations introduced there, see especially Definition 1.19 and Theorem 1.20. As a first step we will now construct a product measure. Let (Ωj , Aj , µj ), N j = 1, . . . , N, be measure spaces and let A := N j=1 Aj be the corresponding

×

N

product σ-field in Ω := j=1 Ωj . Further let Ej be a generator of Aj , j = 1, . . . , N. We are seeking for a measure µ on A such that for all E (j) ∈ Ej the following holds
µ E (1) × · · · × E (N ) = µ1 (E (1) ) · . . . · µN (E (N ) ). (9.1)

Having in mind the corresponding situation for a hyper-rectangle in Rn , n K := j=1 [aj , bj ], aj < bj , where we find

×

λ

(n)

(K) =

n Y j=1

(bj − aj ) =

n Y

λ(1) ([aj , bj ]),

(9.2)

j=1

it is natural to ask a condition such as (9.1) for a product measure. We want to establish the existence and uniqueness of a product measure µ satisfying (9.1) and it turns out that the uniqueness is quite easy to show. Theorem 9.1. For j = 1, . . . , N let (Ωj , Aj , µj ) be measure spaces and let Ej be a generator of Aj , i.e. σ(Ej ) = Aj . We assume that Ej is stable under (j) (j) (j) (j) finite intersections, i.e. E1 , E2 ∈ Ejimplies  that E1 ∩ E2 ∈ Ej , and we (j)

(j)

(j)

(j)

also require the existence of sequences Ek , Ek ∈ Ek , Ek ⊂ Ek+1 and k∈N S (j) (j) k∈N Ek = Ωj , such that µ(Ek ) < ∞, i.e. each measure space (Ωj , Aj , µj ) is σ-finite. Then A admits at most one measure µ such that (9.1) is satisfied. 173

A COURSE IN ANALYSIS

Proof. The σ-field A is generated by E := {E (1) × · · · × E (N ) | E (j) ∈ Ej , 1 ≤ j ≤ N}, and E is also stable under finite intersections. For the sequence S (N ) (1) (Ek )k∈N , Ek := Ek × · · · × Ek ∈ E, we find that Ek ⊂ Ek+1 and k∈N Ek = Ω1 × · · · × ΩN = Ω. Furthermore, we have     (N ) (1) < ∞. µ(Ek ) = µ1 Ek · . . . · µN Ek Now we apply Theorem 2.20 to obtain the uniqueness of a measure µ on A satisfying (9.1).

Remark 9.2. From Theorem 9.1 we may already deduce that (9.2) uniquely determines λ(n) on B(n) given λ(1) on B(1) . Now we turn to the existence problem and we urge the reader to first recollect the definition of a ring and an algebra in a set Ω 6= ∅, see Definition 2.3. Let (Ω1 , A1, µ1 ) and (Ω2 , A2 , µ2 ) be two σ-finite measure spaces. We are searching for a measure τ on A := A1 ⊗ A2 with the property that for A × B ∈ A1 × A2 τ (A × B) = µ1 (A)µ2 (B)

(9.3)

holds. The system A1 × A2 ⊂ P(Ω1 × Ω2 ) is in some sense too small and it does not have the structure to apply general existence results such as Theorem 2.24 to τ . Therefore we introduce

A1,2 :=

 N [ 

j=1

  Aj × Bj Aj ∈ A1 , Bj ∈ A2 , Aj × Bj ∩ Al × Bl = ∅ for j 6= l, N ∈ N . 

(9.4)

Clearly ∅ and Ω1 × Ω2 belong to A1,2 . Furthermore we find for A1 × B1 , A2 × B2 ∈ A1 × A2 that (A1 × B1 ) ∩ (A2 × B2 ) = (A1 ∩ A2 ) × (B1 ∩ B2 ) and

(A1 × B1 ){ = (Ω1 × B1{ ) ∪ (A{1 × Ω2 )

implying that A1,2 is a ring, in fact an algebra and σ(A1,2 ) = A. For A×B ∈ A1 × A2 we set as expected τ (A × B) := µ1 (A)µ2 (B), 174

9

INTEGRATION ON PRODUCT SPACES AND APPLICATIONS

where we now use the convention 0 · ∞ = ∞ · 0 = 0.

S Consider A × B = k∈N Ak × Bk , Ak ∈ A1 , Bk ∈ A2 , where we suppose that the sets (Ak × Bk )k∈N are mutually disjoint. For the characteristic functions we find χA (ω1 )χB (ω2 ) = χA×B (ω1 , ω2 ) =

∞ X k=1

χAk ×Bk (ω1 , ω2 ) =

∞ X

χAk (ω1 )χBk (ω2 )

k=1

and the monotone convergence theorem in form of Corollary 5.15 yields Z χA (ω1 )χB (ω2 )µ1 (dω1 ) µ1 (A)χB (ω2 ) = Ω1

=

∞ Z X k=1

χAk (ω1 )χBk (ω2 )µ1 (dω1 ) =

Ω1

∞ X

µ1 (Ak )χBk (ω2 )

k=1

and further µ1 (A)µ2 (B) =

∞ X

µ1 (Ak )

Z

χBk (ω2 )µ2 (dω2 ) =

Ω2

k=1

∞ X

µ1 (Ak )µ2 (Bk ).

(9.5)

k=1

This allows us to define τ on A1,2 by τ (C) =

N X

µ1 (Aj )µ2 (Bj ),

j=1

C=

N [

j=1

Aj × Bj ,

(9.6)

where the sets (Aj × Bj )j=1,...,N are mutually disjoint. Indeed the above calculation shows that SM(9.5) is independent of the representation of C: if SN C = j=1 Aj × Bj = l=1 Dl × Fl then we can switch to the joint refinement ((Aj ∩ Dl ) × (Bj ∩ Fl ))j=1,...,N, l=1,...,M and find C=

N [ M [

((Aj ∩ Dl ) × (Bj ∩ Fl )).

(9.7)

j=1 l=1

The calculation leading to (9.5) also yields the σ-additivity of τ on A1,2: let S k Ck ∈ A1,2, Ck = N l=1 Ak,l × Bk,l , suchSthat Ck ∩ Cj = ∅ for k 6= j, and S consider C := k∈N Ck ∈ A, i.e. C = M i=1 Di × Fi with mutually disjoint 175

A COURSE IN ANALYSIS

sets Di × Fi , i ∈ M. Since C = conclude that τ (C) =

M X

S

k∈N

SNk

l=1

µ1 (Di )µ2 (Fi ) =

i=1

=

∞ X

Ak,l × Bk,l =

Nk ∞ X X

SM

i=1

Di × Fi we can

µ1 (Ak,l )µ2 (Bk,l )

k=1 l=1

τ (Ck ).

k=1

Finally we remark that ∅ = (Ω1 × Ω2 ){ ∈ A1,2 , i.e. τ (∅) = 0 and hence we have proved that τ is a pre-measure on the ring A1,2 . Using Theorem 2.24 we arrive finally at Theorem 9.3. Let (Ω1 , A1 , µ1 ) and (Ω2 , A2, µ2 ) be two σ-finite measure spaces. On the product (Ω1 × Ω2 , A1 ⊗ A2 ) there exists a unique σ-finite measure µ1 ⊗ µ2 such that for all sets A × B ∈ A1 × A2 we have µ(A × B) = µ1 (A)µ2 (B).

(9.8)

Remark 9.4. If in the situation of Theorem 9.3 both measure spaces are finite, then µ1 ⊗ µ2 is a finite measure on A1 ⊗ A2 . In addition for two probability spaces (Ω1 , A1 , P1 ) and (Ω2 , A2 , P2 ) the measure P1 ⊗ P2 is a probability measure too. The proof of Theorem 9.3 can be extended to the case of finitely many σ-finite measure spaces, i.e. we have Corollary 9.5. Let (Ωj , Aj , µj ), j = 1, . . . , N, be a finite family of σ-finite N measure spaces. On the product space (Ω, A), where Ω = j=1 Ωj and A = NN j=1 Aj , there exists a unique measure denoted by µ1 ⊗ · · · ⊗ µN such that for all Aj ∈ Aj the following holds

×

(µ1 ⊗ · · · ⊗ µN ) (A1 × · · · × AN ) = µ1 (A1 ) · . . . · µN (AN ).

(9.9)

Definition 9.6. The measure µ := µ1 ⊗ · · · ⊗ µN in Corollary 9.5 is called the product measure of the measures µj , j = 1, . . . , N, and (Ω, A, µ) is called the product (measure) space of the measure spaces (Ωj , Aj , µj ). 176

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The calculation leading to (9.5) allows us to immediately state Z Z Z f (ω1 )g(ω2)(µ1 ⊗ µ2 )(d(ω1 , ω2)) = f (ω1 )µ1 (dω1 ) g(ω2)µ2 (dω2 )  Z Z = f (ω1 )µ1 (dω1 ) g(ω2 )µ2 (dω2 )  Z Z = f (ω1 )g(ω2)µ1 (dω1 ) µ2 (dω2 )  Z Z = f (ω1 )g(ω2)µ2 (dω2 ) µ1 (dω1 ) for integrable functions f : Ω1 → R and g : Ω2 → R. Our next aim is to prove such a result for a function h : Ω1 × Ω2 → R which is integrable with respect to µ1 ⊗ µ2 , i.e. we want to find conditions when  Z Z Z h(ω1 , ω2)(µ1 ⊗ µ2 )(d(ω1 , ω2)) = h(ω1 , ω2)µ1 (dω1 ) µ2 (dω2 )  Z Z = h(ω1 , ω2)µ2 (dω2 ) µ1 (dω1 ) holds. In other words we want to prove that under certain conditions on h the integral with respect to a product measure is equal to an iterated integral and iterated integrals are independent of the order of integration. This needs some preparation. Definition 9.7. Let X and Y be two non-empty sets and S ⊂ X × Y . We define the x-section Sx ⊂ Y and the y-section S y ⊂ X of S by  Sx := y ∈ Y (x, y) ∈ S , x ∈ X (9.10) and

 S y := x ∈ X (x, y) ∈ S , y ∈ Y.

(9.11)

Lemma 9.8. Given two measurable spaces (Ω1 , A1) and (Ω2 , A2 ) with product σ-field A = A1 ⊗ A2 . For every S ∈ A1 ⊗ A2 and all ω1 ∈ Ω1 , ω2 ∈ Ω2 the ω1 -sections and the ω2 -sections are measurable sets, i.e. Sω 1 ∈ A 2

and 177

S ω2 ∈ A 1 .

(9.12)

A COURSE IN ANALYSIS

Proof. We prove that E := {S ∈ A1 ⊗ A2 | Sω1 ∈ A2 , ω1 ∈ Ω1 } is a σ-field and A ⊂ E. This of course implies the result for the ω1 -sections and the case of the ω2 -sections goes analogously. First we note that for S = A × B, A ∈ A1 , B ∈ A2 , it follows for ω1 ∈ Ω1 that Sω1 ∈ {∅, B} ⊂ A2 . Since σ(A1 × A2 ) = A1 ⊗ A2 it follows that A1 ⊗ A2 ⊂ E, and in particular we have Ω1 × Ω2 ∈ E. Moreover, if S ∈ E we find for all ω1 ∈ Ω1 that ((Ω1 × Ω2 )\S)ω1 = Ω2 \Sω1 ∈ A2 . Now let (Sk )k∈N be a sequence E. Since A2
S disjoint elements in S S of mutually and hence (S ) = S is a σ-field it follows that k ω k 1 k∈N Sk ∈ E k∈N k∈N ω1 implying that E is a σ-field and A1 ⊗ A2 ⊂ E, and hence the lemma is shown. Analogously to the ω1 - and ω2 -sections of sets we now define sections for mappings. Definition 9.9. Let f : X ×Y → Z be a mapping. The x-section fx : Y → Z and the y-section f y : X → Z of f are defined by fx (y) = f y (x) = f (x, y).

(9.13)

Example 9.10. For arbitrary sets X, Y 6= ∅ we find for the characteristic functions of S ⊂ X × Y , Sx ⊂ Y and S y ⊂ X (χS )x = χSx

and

(χS )y = χS y .

(9.14)

Corollary 9.11. Let (Ω1 , A1 ), (Ω2 , A2 ) and (Ω3 , A3 ) be three measurable spaces and let A = A1 ⊗ A2 . If f := Ω1 × Ω2 → Ω3 is A − A3 -measurable then for every ω1 ∈ Ω, the ω1 -section fω1 of f is A2 − A3 -measurable and for every ω2 ∈ Ω2 the ω2 -section f ω2 of f is A1 − A3 -measurable. Proof. First we note that for A ⊂ Ω3  (fω1 )−1 (A) = ω2 ∈ Ω2 (ω1 , ω2 ) ∈ f −1 (A) = (f −1 (A))ω1

and

(f ω2 )−1 (A) = (f −1 (A))ω2 ,

and now the measurability of f and Lemma 9.8 imply the result. 178

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Often we have used Dynkin systems to prove that a certain family of subsets of a given set form a σ-field. Sometimes a different way is more convenient, namely a monotone class argument. Definition 9.12. Let Ω 6= ∅ be a set and C ⊂ P(Ω). We call C a monotone class if it is closed under countable increasing unions and countable decreasing intersections, i.e. whenever for Ak , Bk ∈ C such that Ak ⊂ Ak+1 and Bk ⊃ Bk+1 it follows that [ \ Ak ∈ C and Bk ∈ C. k∈N

k∈N

Remark 9.13. Every σ-field is a monotone class as is the intersection of any family of monotone classes again a monotone class. Hence for E ⊂ P(Ω) we can look at all intersections of monotone classes containing E which gives the monotone class C(E) generated by E, see also Problem 5. Lemma 9.14 (Monotone class lemma). For every algebra A in Ω 6= ∅ we have σ(A) = C(A). (9.15) Proof. From Remark 9.13 it follows that C(A) ⊂ σ(A). We will prove that C(A) is a σ-field if A is an algebra which will imply σ(A) ⊂ C(A), i.e. (9.15). For A ∈ C(A) define  C(A) := B ∈ C(A) A\B, B\A, A ∩ B ∈ C(A) .

Since A is an algebra it follows that ∅ and A belong to C(A) and A ∈ C(B) if and only if B ∈ C(A). Moreover C(A) is itself a monotone class which is proved by inspection. For A ∈ A it follows that B ∈ C(A) for all B ∈ A since A is an algebra, implying that A ⊂ C(A), hence C(A) ⊂ C(A) for all A ∈ A. Further B ∈ C(A) yields B ∈ C(A) for all A ∈ A, i.e. A ∈ C(B) for all A ∈ A, which gives A ⊂ C(B) and therefore C(A) ⊂ C(B). Thus, A, B ∈ C(A) yields A\B ∈ C(A) and A ∩ B ∈ C(A), and since Ω ∈ A we have Ω ∈ C(A). Thus we have proved that C(A) is an algebra. For a sequence (Ak )k∈N of sets SNAk ∈ C(A) we can first use the fact that C(A) is an algebra to now, since C(A) is closed under countable deduce that k=1 Ak ∈ C(A) andS increasing unions we derive that k∈N Ak ∈ C(A), i.e. C(A) is a σ-field and the lemma is proved. 179

A COURSE IN ANALYSIS

The following result already links the product measure with integrals of sections. Theorem 9.15. Let (Ω1 , A1, µ1 ) and (Ω2 , A2 , µ2 ) be two σ-finite measure spaces and S ∈ A = A1 ⊗ A2 . For all ω1 ∈ Ω1 and all ω2 ∈ Ω2 the mappings ω1 7→ µ2 (Sω1 ) and ω2 7→ µ1 (S ω2 ) are measurable on (Ω1 , A1) and (Ω2 , A2 ), respectively, and the following holds Z Z (µ1 ⊗ µ2 )(S) = µ2 (Sω1 )µ1 (dω1 ) = (9.16) µ1 (S ω2 )µ2 (dω2 ). Ω1

Ω2

Proof. We prove the theorem first under the additional assumption that both measures are finite. We denote by C all sets in A = A1 ⊗ A2 for which the theorem holds. First we note that for S = A × B ∈ A1 × A2 it follows that µ2 (Sω1 ) = χA (ω1 )µ2 (B)

and

µ1 (S ω2 ) = µ1 (A)χB (ω2 )

implying that S ∈ C. Since µ1 and µ2 are additive, we deduce that finite unions of mutually disjoint “rectangles” S
k ∈ A1 × A2 also
k = {Ak × B { { belong to C as does (A × B) = Ω1 × B ∪ A × Ω2 for A × BS∈ A1 × A2 . Now let (Sk )k∈N be a sequence in C, Sk ⊂ Sk+1 and S = k∈N Sk , and consider the sequence uk : Ω2 → R defined by uk (ω2 ) = µ1 ((Sk )ω2 ). Each function uk is measurable since Sk ∈ C and further uk ≤ uk+1 with limk→∞ uk (ω2 ) = µ1 (S ω2 ). It follows that S ω2 is measurable and further, the monotone convergence theorem yields Z Z µ1 ((Sk )ω2 ) µ2 (dω) µ1 (S ω2 )µ2 (dω2 ) = lim Ω2

k→∞

Ω2

= lim (µ1 ⊗ µ2 )(Sk ) = (µ1 ⊗ µ2 )(S). k→∞

R Analogously we see that (µ1 ⊗ µ2 )(S) = Ω1 µ2 (Sω1 )µ1 (dω1 ) and hence S = S k∈N Sk ∈ C. Now let (Qk )k∈N , Qk ∈ C, be a decreasing sequence Qk ⊃ Qk+1 . Since µ1 and µ2 are finite measures, as assumed for the moment, we deduce that by vk : Ω2 → R, vk (ω2 ) = µ1 (Qk )ω2 , a decreasing sequence of functions vk ∈ L1 (Ω2 ) is defined, where the integrability follows from µ1 (Qk )ω2 ≤ µ1 (Q1 )ω1 ≤ µ1 (Ω1 ) < ∞. Applying the T dominated convergence theorem to this sequence we obtain that Q := k∈N Qk ∈ C. Thus C is a monotone class containing A1 × A2 , and by Lemma 9.14 it follows that C(A1 × A2 ) = A = A1 ⊗ A2 . 180

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Now we assume that both µ1 and µS Ak ∈ A1 , Bk ∈ A2 2 are σ-finite and letS be increasing sequences such that k∈N Ak = Ω1 and k∈N Bk = Ω2 and µ1 (Ak ) < ∞, µ2 (Bk ) < ∞. For S ⊂ A = A1 ⊗ A2 , we can apply the results of the theorem as well as the arguments of its proof to the sets S ∩ (Ak × Bk ) which yields (µ1 ⊗ µ2 )(S ∩ (Ak × Bk )) Z Z χAk (ω1 )µ2 (Sω1 ∩ Bk )µ(dω1 ) = = Ω1

Ω2

χBk (ω2 )µ1 (S ω2 ∩ Ak )µ2 (dω2 ),

and an application of the monotone convergence theorem finally induces the result. The following two theorems are cornerstones in the integration theory following Lebesgue’s approach and they have many important consequences. Theorem 9.16 (L. Tonelli). Let (Ω1 , A1 , µ1) and (Ω2 , A2 , µ2 ) be two σfinite measure spaces and f : Ω1 × Ω2 → [0, ∞] an A1 ⊗ A2 -measurable function. For every ω1 ∈ Ω1 and ω2 ∈ Ω2 the functions Z Z ω1 7→ fω1 dµ2 and ω2 7→ f ω2 dµ1 are A1 - and A2 -measurable, respectively. Moreover we have  Z Z Z ω2 f d(µ1 ⊗ µ2 ) = f dµ1 µ2 (dω2 )  Z Z = fω1 dµ2 µ1 (dω1 ).

(9.17)

Proof. For characteristic functions of sets S ∈ A1 ⊗ A2 this is just the statement of Theorem 9.15 and it can be extended immediately to simple functions by linearity (with non-negative scalars). Since f is measurable it is a pointwise limit of an increasing sequence (uk )k∈N of simple functions uk . From the monotone convergence theorem we deduce R first that (uk )ω1 increases to R ω1 7→ fω1 dµ2 and (uk )ω2 increases to ω2 7→ f ω2 dµ1 implying in particular that these functions are measurable. A further application of the monotone convergence theorem now yields   Z Z Z Z µ1 (dω1 ) uk dµ2 fω1 dµ2 µ1 (dω1 ) = lim k→∞ ω1 Z Z = lim uk d(µ1 ⊗ µ2 ) = f d(µ1 ⊗ µ2 ) k→∞

181

A COURSE IN ANALYSIS

and

Z Z

f

ω2



dµ1 µ2 (dω2 ) = lim

k→∞

= lim

k→∞

Z Z

ω2

µ2 (dω2 ) Z uk d(µ1 ⊗ µ2 ) = f d(µ1 ⊗ µ2 ).

Z

uk dµ1

Replacing in Tonelli’s theorem f by a function integrable with respect to µ1 ⊗ µ2 we obtain Theorem 9.17 (G. Fubini). Let (Ω1 , A1 , µ1 ) and (Ω2 , A2, µ2 ) be two σfinite measure spaces and f : Ω1 × Ω2 → R be an A-measurable function, A = A1 ⊗ A2 . Consider the following three integrals   Z Z Z Z Z |f | d(µ1 ⊗ µ2 ), |f | dµ1 dµ2 , |f | dµ2 dµ1 . (9.18) If one of these integrals is finite then all three are finite. In this case f ∈ L1 (Ω1 × Ω2 ) and further we have ω1 7→ f (ω1, ω2 ) is µ2 -a.e. ω2 7→ f (ω1, ω2 ) is µ1 -a.e. Z ω1 7→ f (ω1 , ω2 )µ2 (dω2 ) Z ω2 7→ f (ω1 , ω2 )µ1 (dω1 )

an element of L1 (Ω1 ) an element of L1 (Ω2 )

(9.19) (9.20)

is an element of L1 (Ω1 )

(9.21)

is an element of L1 (Ω2 ).

(9.22)

Moreover (9.17) holds. Proof. From Tonelli’s theorem we deduce that if one of the integrals in (9.18) is finite then they all are finite which in turn implies that f ∈ L1 (Ω1 × Ω2 ). The theorem of Tonelli also entails the measurability R of the functions ω1 7→ f + (ω1 R, ω2), ω1 7→ f − (ω1 , ω2 ) as well as of ω2 7→ f + (ω1 , ω2 )µ1 (dω1 ) and ω2 7→ f − (ω1 , ω2 )µ1 (dω1 ). Since f + ≤ |f | and f − ≤ |f | we deduce that Z Z + f (ω1 , ω2 )µ1 (dω1 ) ≤ |f (ω1 , ω2 )|µ1(dω1 ) (9.23) and

Z

f − (ω1 , ω2 )µ1 (dω1 ) ≤

Z

182

|f (ω1 , ω2 )|µ1(dω1 )

(9.24)

9

INTEGRATION ON PRODUCT SPACES AND APPLICATIONS

and the µ2 -integrability of the right hand side of (9.23) and (9.24) imply the µ2 -integrability of the left hand sides as well as the µ2 -a.e. integrability of ω1 7→ f + (ω1 , ω2 ) and ω1 7→ f − (ω1 , ω2 ) which yields (9.19) and (9.21). The statements (9.20) and (9.22) follow completely analogously. Finally, since (9.17) holds for f + and f − we conclude that it holds for f = f + − f − . Before we pay attention to some theoretical applications of Fubini’s theorem, some remarks are in order with respect to its role to evaluate integrals. So far we have no workable method to find the value of an integral, except in some very special cases. In one dimension when handling integrals with respect to the Lebesgue-Borel measure λ(1) our position is better: whenever we can identify the Lebesgue integral with the Riemann integral and the fundamental theorem is applicable to the corresponding Riemann integral, then we can take advantage of the theory developed in Volume I to evaluate integrals. Of course this applies to iterated integrals over hyper-rectangles in Rn . However, when combining the transformation theorem, Theorem 6.8, with Fubini’s theorem the situation improves dramatically: if the domain of integration is the (almost everywhere) image of a hyper-rectangle, then we can transform the integral to an integral over this hyper-rectangle (with respect to λ(n) ), which in turn by an application of Fubini’s theorem becomes an iterated integral, and now we may use the tools available for one-dimensional Riemann integrals. Our first application of Fubini’s theorem, or better of Tonelli’s theorem, is a reduction of an integral with respect to a measure µ to a one-dimension integral with respect to the Lebesgue measure λ(1) . We start with a detailed example. Consider the rectangle K := [a, b] × [c, d] ⊂ R2 , a < b, c < d. Its Lebesgue measure is of course λ(2) (K) = (b − a)(d − c). Since (2)

λ (K) =

Z

K

1 dλ

(2)

=

Z

R2

χK (z)λ(2) (dz), z = (x, y)

and since χK (z) = χK (x, y) = χ[a,b] (x)χ[c,d] (y) we find by Fubini’s or Tonelli’s theorem 183

A COURSE IN ANALYSIS

(2)

λ

(K) = =

Z Z R

Z

R

R

χK (x, y)λ

(1)

  Z Z (1) (1) (dy) λ (dx) = χK (x, y)λ (dx) λ(1) (dy)

χ[a,b] (x)λ(1) (dx)

Z

R

R

χ[c,d] (y)λ(1) (dy) =

Z

R b

a

1 λ(1) (dx)

Z

d

1 λ(2) (dy).

c

For a compact interval ∅ = 6 ˚ I ⊂ I = [a, b] we may now consider a nonnegative continuous function f : I → [0, ∞) and we interpret the integral Rb R f (x) dx = f dλ(1) as area of the set K ⊂ R2 bounded by I ×{0}∪Γ(f )∪ a I {a} × [0, f (a)] ∪ {b} × [0, f (b)], i.e.

  K = (x, y) ∈ R2 x ∈ [a, b], 0 ≤ y ≤ f (x) = (x, y) ∈ [a, b] × [0, ∞) y ≤ f (x) ,

see Figure 9.1.

b

Γ(f )

K b

x

a

b Figure 9.1

On the other hand we know that Z Z (2) (2) 1 dλ(2) , z = (x, y), χK (z)λ (dz) = λ (K) = R2

K

and we observe further that χK (x, y) = 1 if and only if x ∈ [a, b] and f (x) ≥ y 184

9

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as well as χ(0,f (x)] (y) = 1 if and only if x ∈ [a, b] and f (x) ≥ y. Therefore we find using the equivalence of the Lebesgue to the Riemann integral in our situation and Tonelli’s theorem ! Z f (x) Z Z Z b (1) 1 dy λ(1) (dx) f (x) dx = f dλ = a

Z

[a,b]

Z

[a,b]



1 λ (dy) λ(1) (dx)  Z Z (1) = χ(0,f (x)] (y)λ (dy) λ(1) (dx) [a,b] [0,∞) Z  Z (1) χK (x, y)λ (dx) λ(1) (dy) = [a,b] [0,∞) Z 
= λ(1) x ∈ [a, b] f (x) ≥ y λ(1) (dy) [0,∞) Z ∞ λ(1) ({f ≥ y}) λ(1) (dy), = =

[a,b]

(1)

0

(0,f (x)]

0

thus we have derived the formula Z Z b f (x) dx =

∞

0

a

λ(1) ({f ≥ y})λ(1) (dy).

(9.25)

Note that if f (x) = d for all x ∈ [a, b] then it follows that (  ∅, y>d x ∈ [a, b] f (x) = d ≥ y = [a, b], y ≤ d

and consequently we have

λ(1) ({f = d ≥ y}) = implying Z ∞ 0

(1)

(1)

λ ({f ≥ y})λ (dy) =

Z

(

0, y>d b − a, 0 < y ≤ d (1)

[0,d]

(b − a)λ (dy) +

Z

0λ(1) (dy)

(d,∞)

= (b − a)d = (b − a)(d − c), c = 0, 185

A COURSE IN ANALYSIS

i.e. we recover as expected λ(1) (K) for K = [a, b] × [0, d], and for general [a, b] × [c, d] we can use for example the invariance of the integral under translation. To illustrate the calculation we have a look at the set K from Figure 9.2, see below.

y5 A5

y4

b

Γ(f ) A4

A4

y3 b

A3

A3

K

y2 A2 y1 A1

x

a

b Figure 9.2

{f ≥ y3 }

{f {f {f {f {f

≥ y5 } = pr1 (∅) = ∅ ≥ y4 } = pr1 (A5 ) ≥ y3 } = pr1 (A4 ∪ A5 ) ≥ y2 } = pr1 (A3 ∪ A4 ∪ A5 ) ≥ y1 } = pr1 (A2 ∪ A3 ∪ A4 ∪ A5 ).

The important observation is now that the core result derived above holds for every σ-finite measure space and every non-negative measurable function. Definition 9.18. Let (Ω, A, µ) be a finite σ-finite measure space and let f : Ω → [0, ∞) a measurable function. We call the function defined on [0, ∞) by y 7→ µ({f ≥ y}) (9.26) 186

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the (measure theoretical) distribution function of f with respect to µ. Remark 9.19. A. The function y 7→ µ({f ≥ y}) is obviously a numeric function which is decreasing and the fact that µ is continuous implies the left-continuity of the distribution functions. B. In probability theory the function y 7→ µ({X ≤ y}) is of greater importance and it is called the distribution function of the random variable X. We will call it probabilistic distribution function to make a difference with the measure theoretical distribution function. Theorem 9.20. For a σ-finite measure space (Ω, A, µ) and a non-negative measurable function f : Ω → [0, ∞) we have Z Z f dµ = µ({f ≥ y})λ(1) (dy). (9.27) (0,∞)

Proof. We consider on Ω × [0, ∞) the measurable function F : Ω × [0, ∞) → R2 , (x, y) 7→ F (x, y) = (f (x), y). It follows that the set A := {(x, y) ∈ Ω × [0, ∞) | f (x) ≥ y} is measurable and χA (x, y) = χ(0,f (x)) (y), by the same argument used before. Now it follows again by using Tonelli’s theorem Z

f dµ =

Ω

Z

Ω

= =

Z

Z

f (x)

(1 dy) µ(dx) =

0

Ω×(0,∞) ∞

Z

0

!

Z

Ω

Z

(0,∞)

χA (x, y)(µ ⊗ λ(1) )(d(x, y)) =

(1)

χ(0,f (x)) (y)λ Z

(0,∞)

Z

Ω

!

(dy) µ(dx)

 χA (x, y)µ(dx) λ(1) (dy)

µ({f ≥ y})λ(1) (dy).

2

Example 9.21. The function g : Rn → R, g(x) = e−kxk , is Lebesgue integrable as well as improper Riemann integrable and by Example II.22.12 we know that Z n 2 e−kxk dx = π 2 . Rn

According to Theorem 9.20 we have Z Z ∞ −kxk2 λ(n) ({g ≥ y})λ(1) (dy). e dx = Rn

0

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A COURSE IN ANALYSIS

We first note that 
x ∈ Rn g(x) ≥ y n o 2 = λ(n) x ∈ Rn e−kxk ≥ y 
= λ(n) x ∈ Rn kxk2 ≤ − ln y ,

λ(n) ({g ≥ y}) = λ(n)

hence Z

−kxk2

e

dx =

Rn

Z

∞ 0

λ(n)


x ∈ Rn kxk2 ≤ − ln y λ(1) (dy)

and with − ln y = r, i.e. λ(1) (dy) = e−r λ(1) (dr) we find Z Z ∞ 
−kxk2 λ(n) x ∈ Rn kxk2 ≤ r e−r λ(1) (dr) e dx = Rn Z0 ∞
λ(n) B√r (0) e−r λ(1) (dr). = 0

We know by Proposition II.21.13, in particular (II.21.10), that

λ(n) B√r (0) = voln B√r (0) =

n

n

π2r2 . Γ( n2 + 1)

This implies now Z

n

−kxk2

e

Rn

π2 dx = n Γ( 2 + 1)

Z

∞

n

n

r 2 e−r dr = π 2

0

R∞ R∞ n since Γ(t) = 0 y t−1 e−y dy, i.e. Γ( n2 + 1) = 0 y 2 e−y dy. While this looks just like a further method to evaluate the Gauss integral given the volume of balls in Rn , the formula Z Z ∞
−kxk2 (9.28) e λ(n) B√r (0) e−r dr dx = Rn

0

is in itself of interest. It states that we can express the Gauss integral entirely with the help of the volumes of balls. We give two extensions of this observation. 188

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INTEGRATION ON PRODUCT SPACES AND APPLICATIONS

Example 9.22. A. Let g : [0, ∞) → [0, ∞) be aR bijective, continuous function with continuous derivative and suppose that Rn g(kxk) dx is a finite improper Riemann integral which then is equal to the Lebesgue integral of g. Then the following holds Z Z ∞ λ(n) ({g(kxk) ≥ y}) dy g(kxk) dx = Rn Z0 ∞   λ(n) Br{ (0) g 0(r) dr. = 0

B. Let f : [0, ∞) → [0, ∞) be a monotone increasing function and d(x, y) = 1 ψ 2 (x − y) a metric on Rn which generates on Rn the Euclidean topology. Here ψ is assumed to be a continuous function such that e−ψ is integrable. In this case we find Z Z Z 2 e−d (x,z) dx = e−ψ(x−z) dx = e−ψ(x) dx, Rn

Rn

Rn

and further Z

−ψ(x)

e

dx =

Rn

= = =

Z

∞

Z0 ∞ Z

0

∞

Z0 ∞ 0

λ(n)

 −ψ
e ≥ y dy

λ(n) ({ψ ≤ − ln y}) dy λ(n) ({ψ ≤ r}) e−r dr

  ψ λ(n) B√ (0) e−r dr, r

where BRψ (0) is the ball with centre 0 ∈ Rn and radius R > 0 in the metric 1 space (Rn , d) = (Rn , ψ 2 ). The formula Z Z ∞   ψ −ψ(x) λ(n) B√ e dx = (0) e−r dr (9.29) r Rn

0

is of some interest when discussing transition functions of certain L´evy processes, see [40]. The examples given above have in principle already provided the proof to 189

A COURSE IN ANALYSIS

Corollary 9.23. Let ϕ : [0, ∞) → [0, ∞) be a continuously differentiable 0 n function such that ϕ(0) = 0 and R ϕ (0) > 0. Further let f : R → [0, ∞) be a continuous function such that Rn (ϕ ◦ f )(x) dx exists. Then we have Z ∞ Z λ(n) ({f ≥ y})ϕ0(y) dy. (ϕ ◦ f )(x) dx = Rn

0

Proof. We apply Theorem 9.20 and use the substitution y = ϕ(s): Z ∞ Z λ(n) ({ϕ ◦ f ≥ y}) dy (ϕ ◦ f )(x) dx = 0 Rn Z ∞ λ(n) ({ϕ ◦ f ≥ ϕ(s)})ϕ0 (s) ds = 0 Z ∞ λ(n) ({f ≥ s})ϕ0 (s) ds. = 0

In Problem 10 we will give a further extension of Corollary 9.23. Example 9.24. Applying Corollary 9.23 to the function ϕ(y) = y p , p > 1, we find Z ∞ Z p p λ(n) ({|u| ≥ s})psp−1 ds |u(x)| dx = kukLp = Rn

0

and in Problem 10 we see that this formula holds for more general measure spaces (Ω, A, µ). Our final application in this chapter is known as Minkowski’s integral inequality. We follow closely the presentation in K. Kuttler [52] Theorem 9.25. Let (Ω1 , A1 , µ) and (Ω2 , A2 , ν) be two σ-finite measure spaces and f : Ω1 × Ω2 → R an A1 ⊗ A2 -measurable function. Further let p ≥ 1. Then Z

Ω2

Z

Ω1

p  p1 Z |f (ω1 , ω2 )|µ(dω1 ) ν(dω2 ) ≤

Ω1

Z

Ω2

 p1 µ(dω1 ). |f (ω1 , ω2 )|p ν(dω2 )

(9.30) Proof. Let (Ak )S k∈N be a sequence of sets Ak ∈ A1 such that µ(Ak ) < ∞, Ak ⊂ Ak+1 and k∈N Ak = Ω1 , and let (Bl )l∈N be a sequence of sets Bl ∈ A2 190

9

INTEGRATION ON PRODUCT SPACES AND APPLICATIONS

S such that ν(Bl ) < ∞, Bl ⊂ Bl+1 and l∈N Bl = Ω2 . Further, for m ∈ N we define ( f (x, y) if |f (x, y)| ≤ m fm (x, y) := 0 if |f (x, y)| > m. For k, l, m ∈ N it follows that Z Z Bl

|fm (ω1 , ω2 )|µ(dω1 )

Al

p

ν(dω2 )

 p1

0 the Marcinkiewicz integral is defined by MK,Q,ρ : R3 → R, Z (dist(K, y))ρ (3) λ (dy). MK,Q,ρ(x) := 3+ρ Q ||x − y|| Use Tonelli’s theorem to prove that MK,Q,ρ is λ(3) -almost everywhere finite in K and that x 7→ MK,Q,ρ(x) is integrable over K. Moreover prove the estimate Z 4π (3) MK,Q,ρ(x)λ(3) (dx) ≤ λ (Q \ K). ρ K Hint: use the fact that x 7→ dist(Q, x) is Lipschitz continuous and try R R λ(3) (dx) dr to prove K |x−y| 3+ρ ≤ 4π dist(Q,y) r 1+ρ .

8. Construct an example of a function f : R2 → R such that   Z Z Z Z (1) (1) (1) f (x, y)λ (dy) λ(1) (dx) f (x, y)λ (dx) λ (dy) = R

R

R

R

but f is not λ(2) -integrable. xy Hint: try f (x, y) = χ(−1,1)×(−1,1) (x, y) (x2 +y 2 )2 and use polar coordinates. 9.* In the situation of Example 9.22.B assume that for the measure of Bρψ (0) the following holds for c > 1 and ρ > 0 ψ λ(n) (Bcρ (0)) ≤ γ0 (c)λ(n) (Bρψ (0))

and γ0 (c) ≤ γ0 (1)cα for some α ≥ 0. Prove that there exists two constants κ0 > 0 and κ1 > 0 such that   Z   ψ ψ −tψ(x) (n) (n) (n) B √1 (0) . e λ (dx) ≤ κ1 λ B √1 (0) ≤ κ0 λ t

Rn

t

193

A COURSE IN ANALYSIS

Hint: first derive the equality Z Z −tψ(x) (n) e λ (dx) = t Rn

0

∞

ψ −tρ dρ. λ(n) (B√ ρ (0))e

10. Let (Ω, A, µ) be a σ-finite measure space and ϕ : [0, ∞) → [0, ∞) an increasing C 1 -function with ϕ(0) = 0. Prove that Z ∞ Z µ({u ≥ y})ϕ0(y)dy ϕ ◦ u dµ = Ω

0

holds for every measurable function u : Ω → [0, ∞). 11. Let k ∈ L2 ([a, b] × [a, b]) and u ∈ L2 ([a, b]), a < b. Use Minkowski’s integral inequality to show that Z

[a,b]

Z

[a,b]

! 21 2 k(x, y)u(y)λ(1)(dy) λ(1) (dx) ≤ kkkL2 kukL2 .

Now conclude that by Kop : L2 ([a, b]) → L2 ([a, b]) a linear mapping is defined which satisfies kKop ukL2 ≤ kkkL2 kukL2 . (Note that kukL2 is the L2 -norm in L2 ([a, b]), kkkL2 is the L2 -norm in L2 ([a, b] × [a, b]).)

194

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Convolutions of Functions and Measures

In Problem 10 of Chapter II.22 we briefly discussed the convolution for continuous functions with compact support, i.e. the function Z (u ∗ v)(x) := u(x − y)v(y) dy, (10.1) Rn

and in the lengthy Problem 11 of Chapter II.22 we investigated properties of the Friedrichs mollifier which we can interpret as a convolution operator. Convolutions of functions and measures are encountered in many situations, for example as solutions of partial differential equations with constant coefficients or as operators related to expectations of certain stochastic processes. In this chapter we will study convolutions rather systematically and by this provide tools that will be often employed in the following volumes. In order to handle integrals such as the one in (10.1) we need the additive group (Rn , +) and the invariance of the Lebesgue-Borel λ(n) under translations. In later volumes we will see that many results of this chapter extend to so called locally compact Abelian groups on which a unique (up to a normalization) translation invariant measure, the Haar measure exists. In this chapter we will only work in Rn and as long as we are dealing with the spaces Lp (Rn ) or Lp (Rn ) (or some of their subspaces) the underlying measure is the Lebesgue-Borel measure λ(n) . However, in the second part of this chapter we discuss the convolution µ ∗ ν of certain measures µ and ν on B(n) . Our first aim is to make sense of integrals such as the one in (10.1) for elements in Lp (Rn ) and even in Lp (Rn ). Recall that elements in Lp (Rn ) are not pointwisely defined functions but equivalence classes of elements from Lp (Rn ) under the equivalence relation “∼λ(n) ”, i.e. u ∼λ(n) v if u = v λ(n) -a.e. Thus for [u] ∈ L1 (Rn ) an expression such as [u](x−y) is undefined and [u(x−·)] needs an explanation. In order to simplify notation, in the following we write a.e. for λ(n) -a.e. and if no confusion can arise, we also write u for the equivalence class generated by u. Let u : Rn → R be a measurable function and let δ : Rn × Rn → Rn be defined by δ(x, y) := x − y. The mapping δ is continuous, hence measurable, and therefore the function ω := u ◦ δ : Rn × Rn → R, (u ◦ δ)(x, y) = u(δ(x, y)) = u(x − y) is measurable implying that for each x ∈ Rn fixed the function y 7→ u(x − y) is also measurable. Thus for two measurable functions 195

A COURSE IN ANALYSIS

u, v : Rn → R the function (x, y) 7→ u(x − y)v(y) is measurable as a function from Rn × Rn to R. The function on (x, y) 7→ |u(x − y)v(y)| is a non-negative measurable function defined on Rn × Rn and by Tonelli’s theorem we find Z

(n)

=

Rn

(n)

 (dy) λ(n) (dx)

|u(x − y)v(y)| λ ⊗ λ (dx, dy) = |u(x − y)| |v(y)| λ Rn Rn Z Z Z |v(y)| λ(n) (dy) |u(x − y)| λ(n) (dx) = |v(y)| λ(n) (dy) |u(x)| λ(n) (dx)

Rn ×Rn

Z

Z

Z

(n)

Rn

Rn

Rn

where in the last step we used that λ(n) is translation invariant. If we add as an assumption that both u and v are integrable, we then deduce by Fubini’s theorem that Z (10.2) |u(x − y)v(y)| λ(n) ⊗ λ(n) (dx, dy) = kvkL1 kukL1 < ∞, Rn ×Rn

and further that for almost all x ∈ Rn the integral Z |u(x − y)v(y)| λ(n)(dy)

(10.3)

Rn

is finite. Thus we can define for those x ∈ Rn the function Z (u ∗ v)(x) := u(x − y)v(y) λ(n)(dy)

(10.4)

Rn

which determines an equivalence class with respect to “∼λ(n) ”. This equivalence class we denote again by u ∗ v and from (10.2) we deduce u ∗ v ∈ L1 (Rn ) and (10.5) ku ∗ vkL1 ≤ kukL1 kvkL1 . Moreover, when replacing in (10.2) both u and v by functions u˜ and v˜ being a.e. equal to u and v, respectively, nothing will change in our arguments. In particular (10.4) will be a.e. equal to u˜ ∗ v˜. Thus we have proved Theorem 10.1. Let u, v ∈ L1 (Rn ). Then there exists a unique element u ∗ v ∈ L1 (Rn ) satisfying (10.5) and λ(n) -a.e. this element u ∗ v is equal to (10.4) where in (10.4) we may take any representative of u and v, respectively. Definition 10.2. The element u ∗ v from Theorem 10.1 is called the convolution of u and v. 196

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CONVOLUTIONS OF FUNCTIONS AND MEASURES

Remark 10.3. Once we have given a precise meaning of u ∗ v for u, v ∈ L1 (Rn ), we turn to the common fa¸con de parler and call u ∗ v the convolution of the functions u and v and we will also consider u ∗ v from time to time as a function. Corollary 10.4. For u, v, w ∈ L1 (Rn ) the following hold: u ∗ v = v ∗ u; (u ∗ v) ∗ w = u ∗ (v ∗ w).

(10.6) (10.7)

Proof. The first result follows from the change of variable z = x − y: Z Z (u ∗ v)(x) = u(x − y)v(y) λ(n)(dy) = u(z)v(x − z)λ(n) (dz) = (v ∗ u)(z). To see (10.7) we use (10.6) and Fubini’s theorem, and of course the fact that by Theorem 10.1 we have u ∗ v ∈ L1 (Rn ):   Z (n) u(y)v(· − y) λ (dy) ∗ w (x) ((u ∗ v) ∗ w) (x) =  Z Z (n) = u(y)v(x − z − y)w(z) λ (dy) λ(n) (dz)  Z Z (n) = u(y)v(x − y − z)w(z)λ (dz) λ(n) (dy)   Z (n) = u ∗ w(z)v(· − z) λ (dz) (x) = (u ∗ (v ∗ w)) (x).

Remark 10.5. From Theorem 10.1 and Corollary 10.4 we deduce that ∗ : L1 (Rn ) × L1 (Rn ) → L1 (Rn ) is an associative and commutative operation which we can interpret as multiplication, i.e. we may speak about the convolution product, which turns the vector space L1 (Rn ) into an algebra, also see Problem 2. Before we study the convolution for different Lp -spaces we want to change our point of view for understanding terms such as u(· − x) for x ∈ Rn and u ∈ L1 (Rn ). Recall that C0 (Rn ) denotes the vector space of all continuous 197

A COURSE IN ANALYSIS

functions u : Rn → R with compact support supp u, C∞ (Rn ) is the vector space of all continuous functions vanishing at infinity, and Cb (Rn ) is the space of all bounded continuous functions. We have the inclusions C0 (Rn ) ⊂ C∞ (Rn ) ⊂ Cb (Rn ) and kuk∞ := supx∈Rn |u(x)| is a norm on Cb (Rn ), hence it is also a norm on C0 (Rn ) and C∞ (Rn ). Let X ∈ {C0 (Rn ), C∞ (Rn ), Cb (Rn )}, x ∈ Rn , and A ∈ O(n). For every u ∈ X it follows that v defined by v(y) := (TA,x u)(y) = u(Ay + x) is again an element in X, see Problem 1. In particular u ∈ X implies u(x−·) ∈ X. Moreover the mapping TA,x : X → X, u 7→ TA,x u is linear and kTA,x uk∞ = kuk∞ . Now let 1 ≤ p < ∞ and u ∈ C0 (Rn ) ⊂ Lp (Rn ). It follows that Z  p1  p1 Z p p = kTA,x ukLp = |u(Ay + x)| dy |(TA,x u)(y)| dy Rn

Rn

and the substitution z := Ay + x yields Z  p1 Z p = |u(Ay + x)| dy Rn

p

Rn

|u(z)| dz

 p1

= kukLp ,

where we used that the absolute value of the Jacobi determinant of the underlying transformation is 1. Thus we have for all u ∈ C0 (Rn ) and all A ∈ O(n), x ∈ Rn , (10.8) kTA,x ukLp = kukLp , and (10.8) just reflects the invariance of the Lebesgue-Borel measure λ(n) under translations and O(n). We know by Theorem 8.20 that Lp (Rn ), 1 ≤ p < ∞, is a Banach space, hence continuity for a mapping T : Lp (Rn ) → Lp (Rn ) is well defined. In particular if with some constant cT the following holds for all u, v ∈ Lp (Rn ) kT u − T vkLp ≤ cT ku − vkLp

(10.9)

then T is even Lipschitz continuous. If T is linear then it follows that T u − T v = T (u − v) and the estimate kT wkLp ≤ cT kwkLp for all w ∈ Lp (Rn ) implies (10.9). Thus (10.8) means that TA,x |C0 (Rn ) is a continuous linear mapping from C0 (Rn ) to Lp (Rn ) when we choose on C0 (Rn ) and on Lp (Rn ) the norm k.kLp . Proposition 10.6. Let Z ⊂ Lp (Rn ) be a linear subspace which is dense in Lp (Rn ). If T : Z → Lp (Rn ) is a linear operator satisfying kT wkLp ≤ cT kwkLp 198

(10.10)

10

CONVOLUTIONS OF FUNCTIONS AND MEASURES

for all w ∈ Z, then T has a unique extension T˜ : Lp (Rn ) → Lp (Rn ) which is linear and satisfies (10.10) for all u ∈ Lp (Rn ). In particular T˜ is continuous. Remark 10.7. Recall that a set Z ⊂ X is dense in the metric space (X, d) if for every x ∈ X and every  > 0 there exists z ∈ Z such that d(z , x) < , or equivalently, for every x ∈ X there exists a sequence (zk )k∈N , zk ∈ Z, such that limk→∞ d(zk , x) = 0, see also Definition II.3.4. Proof of Proposition 10.6. Let u ∈ Lp (Rn ) and (uk )k∈N a sequence in Z converging in Lp (Rn ) to u. From (10.10) we deduce that kT uk − T ul kLp ≤ cT kuk − ul kLp , and therefore (T uk )k∈N is a Cauchy sequence in Lp (Rn ), hence it has a limit v := limk→∞ T uk . If (˜ uk )k∈N a further sequence in Z converging in Lp (Rn ) to u we find kT uk − T u˜k kLp = kT (uk − u˜k )kLp uk − ukLp , ≤ cT kuk − u˜k kLp ≤ cT kuk − ukLp + cT k˜ i.e. v := limk→∞ T uk = limk→∞ T u ˜k is independent of the approximating sequence. Therefore T˜u := lim T uk k→∞

(Lp -limit)

(10.11)

is well defined for all u ∈ Lp (Rn ) and for u ∈ Z we may choose as approximating sequence uk = u for all k ∈ N to find that T˜u = T u for u ∈ Z, thus T˜ |Z = T , i.e. T˜ is an extension of T . For u, v ∈ Lp (Rn ) and α, β ∈ R we choose sequences (uk )k∈N , (vk )k∈N , uk , vk ∈ Z, such that in Lp (Rn ) we have limk→∞ uk = u and limk→∞ vk = v. Thus limk→∞(αuk + βvk ) = αu + βv and we find T˜ (αu + βv) = lim T (αuk + βvk ) k→∞

= α lim T uk + β lim T vk = αT˜ u + β T˜ v, k→∞

k→∞

i.e. T˜ is linear on Lp (Rn ). Moreover, for (uk )k∈N as above we have kT˜ ukLp ≤ kT˜ (u − uk )kLp + kT uk kLp ≤ kT˜ (u − uk )kLp + cT kuk kLp . 199

A COURSE IN ANALYSIS

Since |kuk kLp − kukLp | ≤ kuk − ukLp and since limk→∞ kT˜ (u − uk )kLp = 0, we arrive at kT˜ ukLp ≤ cT kukLp . Finally suppose that T˜ and S˜ are two continuous and linear extensions of T . For u ∈ Lp (Rn ) with u = limk→∞ uk , uk ∈ Z, we find

˜ k − lim T˜ uk ˜ − T˜ ukLp = kSu

lim Su k→∞

k→∞

Lp

= lim kT uk − T uk kLp = 0, k→∞

i.e. S˜ = T˜. Remark 10.8. In the following, when no confusion may arise, we will denote the linear continuous extension of T again by T . Our next goal is to prove that C0 (Rn ) ⊂ Lp (Rn ) is dense. As a first preparation we need the following metric variant of the Lemma of Urysohn. Lemma 10.9. Let ∅ = 6 K ⊂ V ⊂ Rn where K is compact and V is open. For any open set U such that K ⊂ U ⊂ V and U is compact the function h : Rn → R defined by
dist x, U {
h(x) := (10.12) dist(x, K) + dist x, U {

is continuous with support supp h ⊂ U , h|K = 1, h|U { = 0 and 0 ≤ h(x) ≤ 1. Thus h ∈ C0 (Rn ) and χK ≤ h ≤ χU ≤ χV .
Proof. For x ∈ K it follows that dist x, U { > 0 and for x ∈ K { it follows that dist(x, K) > 0, i.e.
h is well defined on Rn . By Example II.3.28 the mappings x 7→ dist x, U { and x 7→ dist(x, K) are continuous, so h is condist(x,U { ) tinuous. Further, for x ∈ K we have h(x) = dist x,U { = 1 and for x ∈ U { ( )
we have dist x, U { = 0, i.e. h(x) = 0. Clearly 0 ≤ h(x) ≤ 1 and since U is compact, supp h ⊂ U is compact too. We will also need the next theorem, a proof of which we will discuss in Appendix II. Indeed it is worth discussing this result in a wider context as we will do in Part 12 when proving the Ri´esz representation theorem.

Theorem 10.10. Let A ∈ B(n) with λ(n) (A) < ∞. Then for every η > 0 we can find a compact set Kη and an open set Uη such that Kη ⊂ A ⊂ Uη , λ(n) (Uη ) < ∞ and λ(n) (Uη \Kη ) < η. 200

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With these preparations we can prove Theorem 10.11. For 1 ≤ p < ∞ the set C0 (Rn ) is a dense linear subspace of Lp (Rn ). Proof. We want to approximate elements in a linear space and we have achieved our aim if we can approximate elements whose linear combinations contain for every element in Lp (Rn ) a sequence converging in Lp (Rn ) to the given element. Since u = u+ − u− we first can reduce the problem to approximate non-negative elements in Lp (Rn ). Since these are measurable functions (or better, every representant is a λ(n) -a.e. non negative function) we can find an increasing sequence of simple functions (uk )k∈N such that 0 ≤ uk ≤ uk+1 ≤ u and limk→∞ uk = u λ(n) -a.e. The dominated convergence theorem applied to |uk −u|p now yields that limk→∞ kuk −ukLp = 0. Thus, if we can approximate in the norm k.kLp every characteristic function χA of a Borel measurable set A with λ(n) (A) < ∞ by continuous functions with compact support, the result will follow. Now we use Theorem 10.10: given A ∈ B(n) with λ(n) (A) < ∞ we can find for  > 0 a compact set K and an open set U such that λ(n) (U) < ∞ and Z   p , λ(n) (U\K) = χU \K dλ(n) < 2 or  kχU − χK kLp < . 2 n Further, by Lemma 10.9 we find h ∈ C0 (R ) such that χK ≤ h ≤ χU or 0 ≤ χU − h ≤ χU − χK which yields  kχU − hkLp < . 2

Noting that χK ≤ χA ≤ χU , we find also that kχU − χA kLp
0 and since x 7→ |v(x) − w(x)| is a continuous function there exists some ρ > 0 such that |v(x) − w(x)| > 0 for all x ∈ Bρ (x0 ). It follows that v 6= w on a set of positive measure, a contradiction to the fact that v = w a.e. which follows from our assumption. Hence u ∈ Lp (Rn ) can have at most one continuous representative. So far we have proved that u ∗ v is defined for u, v ∈ L1 (Rn ) and it is now our aim to study u ∗ v for u and v belonging to different spaces. The first result is the classical Young’s inequality Theorem 10.13. Let u ∈ L1 (Rn ) and v ∈ Lp (Rn ), 1 ≤ p ≤ ∞. Then (u ∗ v)(x) exists λ(n) -a.e., (u ∗ v) is an element of Lp (Rn ) and we have ku ∗ vkLp ≤ kukL1 kvkLp .

(10.13)

Proof. First we prove (10.13) for all v ∈ C0 (Rn ) ⊂ Lp (Rn ), 1 ≤ p < ∞. For this we note that (x, y) 7→ u(y)v(x − y) is λ(n) ⊗ λ(n) -measurable and furthermore, since u ∗ v = v ∗ u for u, v ∈ L1 (Rn ), see Corollary 10.4, we find using Minkowski’s integral inequaltiy, Theorem 9.25, p Z Z  1 Z Z 1 p p p ku ∗ vkLp = ≤ dx |u(y)v(x − y)| dy u(y)v(x − y) dy dx 1 Z 1 Z Z Z p p p p p ≤ |u(y)| |v(x − y)| dx dy = |u(y)| dy |v(x − y)| dx Z = |u(y)| kvkLp dy = kukL1 kvkLp ,

202

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where we used once more that the Lebesgue-Borel measure is translation invariant. If now v ∈ Lp (Rn ), we choose a sequence (vk )k∈N , vk ∈ C0 (Rn ), converging in Lp (Rn ) to v and (10.13) already proved for u ∈ L1 (Rn ) and vk ∈ C0 (Rn ) follows now by passing to the limit vk → v. Since we now know u ∗ v ∈ Lp (Rn ) it follows that u ∗ v must be λ(n) -a.e. finite and further by Fubini’s theorem we have λ(n) -a.e. that Z (u ∗ v)(x) = u(x − y)v(y) dy, recall that u(x − y) stands for (T−id,0 ◦ Tid,−x )u. The case p = ∞ is of course much simpler since in this case we have

Z Z



ku ∗ vk∞ = u(y)v(· − y) dy ≤ kvk∞ |u(y)| dy. ∞

Proposition 10.14. Let 1 ≤ p < ∞ and p1 + 1q = 1. If u ∈ Lp (Rn ) and v ∈ Lq (Rn ) then u ∗ v is defined and belongs to Cb (Rn ). Moreover we have ku ∗ vk∞ ≤ kukLp kvkLq .

(10.14)

Proof. Interpreting v(x − ·) ∈ Lq (Rn ) as mentioned above we find Z Z u(y)v(x − y) dy ≤ |u(y)| |v(x − y)| dy Rn

Rn

≤

Z

p

Rn

|u(y)| dy

= kukLp kvkLq

 p1 Z

Rn

q

|v(x − y)| dy

 q1

where we used once again the translation invariance of λ(n) . This estimate already implies that u∗v is almost everywhere defined and a bounded function satisfying (10.14). From the definition of u ∗ v it is obvious that (u, v) 7→ u ∗ v is a bilinear mapping, say from Lp (Rn ) × Lq (Rn ) to L∞ (Rn ). Further, for u, v ∈ C0 (Rn ) it follows that u ∗ v is continuous and bounded. Given u ∈ Lp (Rn ) and v ∈ Lq (Rn ) we can find a sequence (uk )k∈Rn , uk ∈ C0 (Rn ), 203

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converging in Lp (Rn ) to u, and we can find a sequence (vk )k∈N , vk ∈ C0 (Rn ), converging in Lq (Rn ) to v. It follows that k(u ∗ v) − (uk ∗ vk )k∞ ≤ ku ∗ (v − vk )k∞ + k(u − uk ) ∗ vk k∞ ≤ kukLp kv − vk kLq + ku − uk kLp kvk kLq . As a convergent sequence in Lq (Rn ) the sequence (vk )k∈N is in Lq (Rn ) bounded and we find k(u ∗ v) − (uk ∗ vk )k∞ ≤ M (kv − vk kLq + ku − uk kLp ) which means that u ∗ v is with respect to the sup-norm limit of a sequence of continuous functions, hence it is continuous. For a pointwisely defined function f : Rn → R (or f : G → R, G ⊂ Rn ) we can define the support f as  supp f := x ∈ Rn f (x) 6= 0 , (10.15)

compare with (II.14.12). This is a definition that we cannot transfer to the equivalence class [f ] of f with respect to “∼λ(n) ”. But there are other possibilities to characterise supp f , say for measurable functions which are integrable over every compact set K ⊂ Rn . Take x ∈ / supp f . Since {x} is compact and supp f is closed, f ) = δ > 0. Let ϕ ∈ C0 (Rn ), supp ϕ ⊂ Bδ (0). R dist(x, supp (n) It follows that Rn f ϕ dλ = 0. In fact for every ϕ ∈ C0 (Rn ) with supp ϕ ⊂ R (supp f ){ it follows that Rn f ϕ dλ(n) = 0. Given f we may consider the set Uf defined by Uf :=



x ∈ R for some ρ > 0 we have for all ϕ ∈ C0 (Bρ (x)) that n

Z

Rn

(n)

f ϕ dλ



=0 .

This is an open set since for x ∈ Uf and y ∈ B ρ2 (x) it follows for all ϕ ∈   R C0 B ρ2 (y) that ϕ ∈ C0 (Bρ (x)) and hence f ϕ dλ(n) = 0, i.e. B ρ2 (x) ⊂ Uf . Now for f ∈ Lp (Rn ) the set Uf is independent of the representative and we define supp f := Uf{ . For a continuous function f we recover the old definition. We already know for f continuous that (supp f ){ ⊂ Uf , i.e. Uf{ ⊂ supp f . Now let x ∈ supp f . For every  > 0 we can find y such that kx − yk <  and f (y) 6= 0. Suppose f (y) > 0, the case f (y) < 0 goes 204

10

CONVOLUTIONS OF FUNCTIONS AND MEASURES

analogously. Due to the continuity of f we can find η > 0R such that f (z) > 0 for all z ∈ Bη (y). However in this case we cannot have Rn f ϕ dλ(n) = 0 for all ϕ ∈ C0 (Rn ), supp ϕ ⊂ B η2 (y), hence x ∈ Uf{ , i.e. supp f ⊂ Uf{ . Proposition 10.15. Let u ∈ Lp (Rn ) and v ∈ Lq (Rn ) such that u ∗ v is defined and belong to some space Lr (Rn ). For the support of u ∗ v we have supp(u ∗ v) ⊂ supp u + supp v,

(10.16)

i.e. the support of u ∗ v is contained in the closure of supp u + supp v.

Proof. Recall that for A, B ⊂ Rn the set A + B is defined as A + B = {z ∈ Rn | z = x + y, x ∈ A, y ∈ B}. When both u and v are continuous we may argue pointwisely: if x ∈ / supp u + supp v then for any y ∈ supp u we find x−y ∈ / supp v and therefore u(y)v(x − y) = 0 implying that u ∗ v = 0, i.e. supp u ∗ v ⊂ supp u + supp v.R Now for the general case let ϕ ∈ C0 (Rn ). By our assumptions the integral ϕ(u ∗ v) dλ(n) exists and we have Z

Rn

ϕ(x)

Z

Rn

 Z u(y)v(x − y) dy dx =

supp v

Z

supp u

 ϕ(z + y)u(y)v(z) dy dz.

{

Thus, if supp ϕ ⊂ supp u + supp v then the integral vanishes and consequently we have supp(u ∗ v) ⊂ supp u + supp v. Using the Friedrichs mollifier which we have already introduced in Problem 11 of Chapter II.22 we will prove now that C0∞ (Rn ) is dense in Lp (Rn ), 1 ≤ p < ∞. With Z
−1 a := exp (kxk2 − 1)−1 dx Rn

we define j : R → R by n

( a exp ((kxk2 − 1)−1 ) , kxk < 1 (10.17) j(x) := 0, kxk ≥ 1. R It follows that j ∈ C0∞ (Rn ), j(x) dx = 1, j(x)
≥ 0, and supp j ⊂ B1 (0). For  > 0 we introduce further j (x) := −n j x and we know that j ∈ C0∞ (Rn ), R supp j ⊂ B (0), j (x) dx = 1 and j ≥ 0. For u ∈ Lp (Rn ) we can now define the convolution operator (the Friedrichs mollifier) Z J (u)(x) := (j ∗ u)(x) = j (x − y)u(y) dy. (10.18) Rn

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From Theorem 8.4 we deduce that J (u) ∈ C ∞ (Rn ), whereas Theorem 10.13 yields J (u) ∈ Lp (Rn ), i.e. J (u) ∈ C ∞ (Rn ) ∩ Lp (Rn ). Since C0∞ (Rn ) ⊂ T r n older’s inequality, p1 + 1q = 1, we find r≥1 L (R ), using H¨ p Z p u(x − y)j (y) dy |J (u)(x)| = n ZR  p  1 1 u(x − y)j (y) p j (y) q dy = Rn Z  pq Z p |u(x − y)| j (y) dy j (y) dy ≤ n Rn ZR = |u(x − y)|pj (y) dy. Rn

Integration with respect to x yields further  Z Z Z p p |u(x − y)| j (y) dy dx |J (u)(x)| dx ≤ n Rn Rn  Z ZR p |u(x − y)| dx dy = kukpLp , j (y) = Rn

Rn

thus we arrive at kJ (u)kLp ≤ kukLp .

(10.19)

Note that for p = 1 there is no need to use H¨older’s inequality to derive (10.19). Lemma 10.16. If u ∈ C0 (Rn ), then J (u) converges as  → 0 in Lp (Rn ) to u. Proof. For  > 0 and x ∈ Rn we find by the same arguments as above that p Z p j (y) (u(x − y) − u(x)) dy |J (u)(x) − u(x)| = n ZR ≤ j (y)|u(x − y) − u(x)|p dy. Rn

Since by assumption supp u is compact we can find R > 0 such that u|B{ (0) = R 0 and therefore, for kyk ≤ R it follows that Z Z p |u(x)|p dx = 0. |u(x − y)| dx ≤ kxk≥R

kxk≥2R

206

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In addition, since u is continuous, for η > 0 and  = (R, η) sufficiently small we have Z |u(x − y) − u(x)|p dx < η. sup kyk≤

kxk≤2R

For these  > 0 we get Z  Z Z p J (u)(x) − u(x) p dx ≤ j (y) |u(x − y) − u(x)| dx dy Rn Rn Rn Z |u(x − y) − u(x)|p dx ≤ sup n kyk≤ R Z  Z p p |u(x − y) − u(x)| dx + ≤ sup |u(x − y) − u(x)| dx kyk≤ kxk≥2R kxk≤2R Z = sup |u(x − y) − u(x)|p dx < η, kyk≤

kxk≤2R

which implies that lim>0 kJ (u) − ukLp = 0. Theorem 10.17. The space C0∞ (Rn ) of all arbitrarily often differentiable functions with compact support is dense in Lp (Rn ), 1 ≤ p < ∞. Proof. By Theorem 10.11 we know that C0 (Rn ) is dense in Lp (Rn ), 1 ≤ p < ∞. Hence, given η > 0 we can find for u ∈ Lp (Rn ) a function v ∈ C0 (Rn ) such that ku − vkLp < η2 . Further, by Lemma 10.16 we can find  > 0 such that kv −J (v)kLp < η2 and J (v) ∈ C ∞ (Rn ) with supp J (v) ⊂ supp v + B (0) by Proposition 10.15. Thus J (v) ∈ C0∞ (Rn ) and the triangle inequality gives ku − J (v)kLp ≤ ku − vkLp + kv − J (v)kLp < η. Let g : Rn → R be a non-negative integrable function. On B(n) we define the measure µ := gλ(n) and for u ∈ Lp (Rn ), 1 ≤ p < ∞, we may consider the convolution Z Z (n) u(x − y)µ(dy). (10.20) u(x − y)g(y)λ (dy) = Rn

Rn

Thus we can take (10.20) to extend the definition of the convolution of two functions to define the convolution of a measure with a suitable function. 207

A COURSE IN ANALYSIS

Suppose that u ≥ 0 is integrable and let B ∈ B(n) . It follows with µ = gλ(n) and ν = uλ(n) that Z

B

Z

Rn

(n)

u(x − y)g(y)λ

!

(dy) λ(n) (dx)

 χB (x)u(x − y)g(y)λ(n) (dy) λ(n) (dx) Rn Rn  Z Z (n) χB (x)u(x − y)λ (dx) g(y)λ(n) (dy) = Rn Rn  Z Z χB (y + z)u(z)λ(n) (dz) g(y)λ(n) (dy) = Rn Rn   Z Z Z Z χB (x + y)µ(dy) ν(dx). χB (x + y)ν(dx) µ(dy) = = =

Z

Z

Rn

Rn

The mapping B 7→

Rn

Z

Z

Rn (n)

is however a measure on B

(µ ∗ ν)(B) :=

Rn



χB (x + y)ν(dx) µ(dy)

, hence Z Z Rn

Rn

Rn



χB (x + y)ν(dx) µ(dy)

(10.21)

defines a measure on B(n) called the convolution of µ = gλ(n) and ν = uλ(n) . We have already seen that µ ∗ ν = ν ∗ µ holds.

Definition 10.18. Let µ and ν be two finite measures on B(n) . Their convolution µ ∗ ν is the measure defined on B(n) by  Z Z χB (x + y)ν(dx) µ(dy). (10.22) (µ ∗ ν)(B) = Rn

Rn

Lemma 10.19. For finite measures µ and ν on B(n) the following hold µ ∗ ν = ν ∗ µ;

(10.23)

(µ ∗ ν)(Rn ) = µ(Rn )ν(Rn ).

(10.24)

and In particular the convolution of two probability measures is a probability measure. Moreover, if µ = gλ(n) and ν = hλ(n) with non-negative integrable functions g and h, then µ ∗ ν has also a density with respect to λ(n) given by g ∗ h, i.e. µ ∗ ν = (g ∗ h)λ(n) . (10.25) 208

10

CONVOLUTIONS OF FUNCTIONS AND MEASURES

Proof. The commutativity relation (10.23) follows by applying Fubini’s theorem to the defining equation (10.22) and choosing B = Rn in (10.22) we find immediately  Z Z Z n (µ ∗ ν)(R ) = 1ν(dx) µ(dy) = ν(Rn )µ(dy) = ν(Rn )µ(Rn ). Rn

Rn

Rn

The final statement was already discussed:  Z Z χB (x + y)ν(dx) µ(dy) (µ ∗ ν)(B) = Rn Rn  Z Z (n) = h(x − y)g(y)λ (dy) λ(n) (dx) B Rn
= (g ∗ h)λ(n) (B). We want to change our point of view and give a different interpretation of µ P ∗ ν starting with (10.22). For a simple function f : Rn → [0, ∞), f= N k=1 αk χAk , we find  Z Z Z f d(µ ∗ ν) = f (x + y)µ(dy) ν(dx). (10.26) Rn

Rn

Further, if u is a (µ∗ν)-integrable non-negative function on Rn we may choose a sequence (fk )k∈N of non-negative measurable functions such that fk ≤ fk+1 and limk→∞ fk (x) = u(x) and it follows that (10.26) holds for u, i.e. for all non-negative (µ ∗ ν)-integrable functions. By Fubini’s theorem it now follows that f (x + ·) is µ- as well as ν-integrable. Next we apply our results on integration with respect to an image measure to the product measure µ ⊗ ν and the mapping add2 : Rn × Rn → Rn , add2 (x, y) = x + y, to find  Z Z Z (f ◦ add2 )(x, y)(µ ⊗ ν)(dy, dx) = f (x + y)µ(dy) ν(dx) Rn ×Rn Rn Rn Z f (z)(µ ∗ ν)(dz), = Rn

i.e. we have proved Lemma 10.20. The convolution µ ∗ ν is the image of the measure µ ⊗ ν under the mapping add2 . 209

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With Lemma 10.20 in mind we may introduce for finite measures µ1 , . . . , µN on B(n) the measure µ1 ∗ · · · ∗ µN by µ1 ∗ · · · ∗ µN := addN (µ1 ⊗ · · · ⊗ µN )

(10.27)

where addN : Rn × · · · × Rn → Rn , (x1 , . . . , xN ) 7→ x1 + · · · + xN . Let BN +1 : Rn(N +1) → R2n , BN +1 (x1 , . . . , xN +1 ) = (x1 + · · · + xN , xN +1 ) and DN +1 : Rn(N +1) → R2n , DN +1 (x1 , . . . , xN +1 ) = (x1 , x2 + · · · + xN +1 ). Both mappings are continuous, hence measurable, and we have BN +1 (µ1 ⊗ · · · ⊗ µN +1 ) = addN (µ1 ⊗ · · · ⊗ µN ) ⊗ µN +1 as well as DN +1 (µ1 ⊗ · · · ⊗ µN +1 ) = µ1 ⊗ addN (µ2 ⊗ · · · ⊗ µN +1 ). Moreover we have addN +1 = add2 ◦ BN +1 = add2 ◦ DN +1 , implying µ1 ∗ · · · ∗ µN +1 = addN +1 (µ1 ⊗ · · · ⊗ µN +1 ) = add2 ((µ1 ∗ · · · ∗ µN ) ⊗ µN +1 ) = (µ1 ∗ · · · ∗ µN ) ∗ µN +1 as well as µ1 ∗ · · · ∗ µN +1 = addN +1 (µ1 ⊗ · · · ⊗ µN +1 ) = add2 (µ1 ⊗ (µ2 ∗ · · · ∗ µN +1 )) = µ1 ∗ (µ2 ∗ · · · ∗ µN +1 ). In particular we have that iterating our initial definition is consistent with (10.27) and convolution of finite measures is associative, i.e. (µ1 ∗ µ2 ) ∗ µ3 = µ1 ∗ (µ2 ∗ µ3 ).

(10.28)

Example 10.21. A. A trivial consequence of the properties of integrals is that for finite measures µ1 , µ2 and µ3 on B(n) and for α ≥ 0 we have µ1 ∗ (µ2 + µ3 ) = µ1 ∗ µ2 + µ1 ∗ µ3

(10.29)

(αµ1) ∗ µ2 = µ1 ∗ (αµ2 ) = α(µ1 ∗ µ2 ).

(10.30)

and

210

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CONVOLUTIONS OF FUNCTIONS AND MEASURES

B. For a finite measure µ on B(n) and the translation Tx0 (y) = x0 + y we always find that 
µ(B − x0 ) = µ(T−x0 (B)) = µ y ∈ Rn y ∈ −x0 + B 
= µ y ∈ Rn y + x0 ∈ B = Tx0 (µ)(B)

and therefore we find for the Dirac measure at x0 Z (µ ∗ x0 )(B) = (x0 ∗ µ)(B) = µ(B − x)x0 (dx) = µ(B − x0 ) = Tx0 (µ)(B),

i.e. we have µ ∗ x0 = x0 ∗ µ = Tx0 (µ).

(10.31)

x1 ∗ x2 = x1 +x2

(10.32)

µ ∗ 0 = µ = 0 ∗ µ,

(10.33)

In particular we find and i.e. in the set of all finite measures on B(n) with convolution as operation 0 behaves as a unit element. C. Consider for t, s > 0 the Poisson distributions πt =

∞ X

k −t t

e

k!

k=0

k

and

πs =

∞ X l=0

sl e−s l . l!

These are probability measures on B(1) and therefore their convolution πt ∗ πs is well defined and we find using Part A and B and the binomial theorem πt ∗ πs =

∞ ∞ X X

e−(t+s)

k=0 l=0 −(t+s)

=e

∞ X

m=0 ∞ X

tk sl k ∗ l k! l!

m X tm k=1

sm−k k! (m − k)!

!

(t + s)m m = πt+s . = e−(t+s) m! m=0 211

m

A COURSE IN ANALYSIS

Once we have mastered the basic properties of the Fourier transform in Part 8, we will understand this example in a much broader context. We will encounter convolutions of functions or of measures, and later on of distributions, in many places in the rest of these treatises. Often they appear as “linear operators” in our considerations. Example 10.22. With j ,  > 0, and j as in (10.17) we can define a linear mapping or operator J : Lp (Rn ) → Lp (Rn ), 1 ≤ p < ∞ satisfying the estimate (10.19), hence J is continuous with respect to the norm k.kLp , 1 ≤ p < ∞.

More generally let k : Rn → R be an element of L1 (Rn ) and define on Lp (Rn ), 1 ≤ p < ∞, the term Kop u := k ∗ u, (10.34) or

(Kop u)(x) =

Z

k(x − y)u(y) dy.

(10.35)

Young’s inequality, i.e. Theorem 10.13, yields that kKop ukLp ≤ kkkL1 kukLp ,

which implies that Kop : Lp (Rn ) → Lp (Rn ) is a linear and (Lipschitz) continuous operator. Operators of this type are called convolution operators ˜ y) := k(x − y) are called the kernel of the operator Kop , soand k or k(x, metimes k is called the convolution kernel whereas k˜ is called the kernel. We claim that every convolution operator Kop is translation invariant, i.e. for x0 ∈ Rn we have (10.36) Tx0 (Kop u) = Kop (Tx0 u) or Tx0 ◦ Kop = Kop ◦ Tx0 ,

(10.37)

i.e. for the commutator [Kop , Tx0 ] = Kop ◦ Tx0 − Tx0 ◦ Kop it follows that [Kop , Tx0 ] = 0. Indeed we have for u ∈ Lp (Rn ) Z Tx0 (Kop u)(x) = (Kop u)(x0 + x) = k(x + x0 − y)u(y) dy n R Z Z k(x − z)(Tx0 u)(z) dz k(x − z)u(x0 + z) dz = = Rn

Rn

= Kop (Tx0 u)(x).

212

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We will see much later that all translation invariant operators (satisfying some type of continuity condition) are of convolution type, but Lp (Rn ) must be replaced by certain spaces of generalised functions.

Problems  1. Let X ∈ C0 (Rn ), C∞ (Rn ), Cb(Rn ), C k (Rn ) and define for u ∈ X the function v(y) := (TA,x u)(y) = u(Ay + x), where A ∈ O(n) and x ∈ Rn . Prove that v ∈ X. 2. Prove that L1 (Rn , +, ∗) is a commutative algebra over R. 3.

a) Let a ∈ Cb (Rn ) and define on C0 (Rn ) the linear operator A : C0 (Rn ) → Lp (Rn ), 1 ≤ p < ∞, by Au := au. Prove that A has ˜ Lp ≤ a continuous extension A˜ : Lp (Rn ) → Lp (Rn ) satisfying kAuk p kak∞ kukL . Hint: Theorem 10.17 may be used. d : C 1 ([0, 1]) → L2 ([0, 1]), u 7→ du , does b) Prove that the operator dx dx

2 ≤ ckukL2 . not have a linear continuous extension satisfying du dx L Hint: investigate the functions uk (x) = sin 2πkx.

4. Recall the definition of the Schwartz space S(Rn ), see Problem 3 of Chapter II.20: S(Rn ) = {u ∈ C ∞ (Rn )|pα,β (u) < ∞ for all α, β ∈ Nn0 } where pα,β (u) = supx∈Rn |xβ ∂ α u(x)|. Prove that for u, v ∈ S(Rn ) it follows that u ∗ v ∈ S(Rn ). Hint: first prove that for u ∈ S(Rn ) and all m1 , m2 ∈ N0 we have m2 P α supx∈Rn (1 + kxk2 ) 2 |α|≤m1 |∂ u(x)| < ∞. Further recall Peetre’s inequality (I.23.14). 5. Prove the following C ∞ -version of Urysohn’s lemma: let K ⊂ G ⊂ Rn where K is compact and G is an open bounded set. Then there exists a function ϕ ∈ C0∞ (Rn ) such that supp ϕ ⊂ G, 0 ≤ ϕ ≤ 1 and ϕ|K = 1. 6.* (C∞ -partition of unity) Let K ⊂ Rn be a compact set and Gj ⊂ Rn , S j = 1, . . . , N, be bounded open sets such that K ⊂ N j=1 Gj . Then there P n ∞ exist functions ϕj ∈ C0 (R ), supp ϕj ⊂ Gj , such that N j=1 ϕj (x) = 1 for all x ∈ K. 213

A COURSE IN ANALYSIS

7. For t > 0 and a ≥ 0 define on (R, B(1) ) the measures µt := e−at 0 and prove that µt ∗ µs = µt+s for all t, s > 0. 8. Let (µt )t>0 be a family of probability measures on Rn having the property that µt ∗ µs = µt+s for all t, s > 0. See Example 10.21.C or Problem 7 for examples. For u ∈ C∞ (Rn ) and t > 0 define Z Tt u(x) := u(x − y)µt(dy), and take for granted that Tt u ∈ C∞ (Rn ) for all t > 0, a proof of this will be provided in Part 8. Thus Tt : C∞ (Rn ) → C∞ (Rn ) by assumption. Prove that Tt is a linear operator satisfying kTt uk∞ ≤ kuk∞ for all t > 0 and u ∈ C∞ (Rn ). Furthermore, prove for t, s > 0 that Ts+t = Ts ◦ Tt , i.e. Ts+t u = Ts (Tt u). 9. For 0 < α < 1 the function kα (x) = χ[−1,1] |x|−α , x 6= 0, kα (0) = 0, belongs to L1 (R). (Why is this the case?) Define on C0 (R) the integral operator Kop by Kop u := kα ∗ u (such an operator is called a convolution operator). Prove that for 1 ≤ p < ∞ the operator Kop has a continuous extension from Lp (R) to Lp (R). Now switch to Rn and give conditions on γ > 0 such that with kγ,n (kxk) := χB1 (0) (x)kxk−γ , γ > 0, kγ,n (0) = 0, the operator (n) Kop u(x) :=

Z

kγ,n (kx − yk)u(y)λ(n)(dy)

originally defined on C0 (Rn ) extends to a continuous linear operator from Lp (Rn ) to Lp (Rn ).

214

11

Differentiation Revisited

Let f : [a, b] → R be a Borel measurable function. We want to investigate questions such as: for which points x ∈ [a, b] is f differentiable, or if f is R integrable, when is x 7→ [a,x] f (t)λ(1) (dt) differentiable, and does the fundamental theorem hold? Before we start with our studies we want to provide a result seemingly unrelated to our problems, but as it will turn out, it is a key technical ingredient for deriving central results. Definition 11.1. Let A ⊂ R be a Borel measurable set. A family V := (Ij )j∈J , J 6= ∅, of closed intervals Ij ⊂ R, ˚ Ij 6= ∅, is called a Vitali covering of A if for each x ∈ A and  > 0 there exists an interval I = Ix, ∈ V such that x ∈ I and λ(1) (I) < . Theorem 11.2 (Vitali’s covering theorem). Let A ⊂ R be a Borel measurable set and V a Vitali covering of A. Then we can find a denumerable family (Ijk )k∈N of mutually disjoint intervals Ijk ∈ V such that 

λ(1) A ∩

[

Ij k

k∈N

!{ 

 = 0.

(11.1)

Moreover, if λ(1) (A) < ∞ then for every  > 0 we can find a finite family of mutually disjoint intervals Ij1 , . . . , IjN ∈ V such that 

λ(1) A ∩

N [

k=1

Ij k

!{ 

 < .

(11.2)

Proof. (Following [35]) Suppose first that λ(1) (A) < ∞ and take U ⊂ R open such that A ⊂ U and λ(1) (U) < ∞, recall Theorem 10.10. Now consider V0 := {I ⊂ V | I ⊂ U} which is a further Vitali covering of A. Pick any I1 ∈ V0 . In the case that A ⊂ I1 it follows that A∩I1{ = ∅, i.e. λ(1) (A∩I1{ ) = 0 and the assertion is proved. Assume that A ∩ I1{ 6= ∅, i.e. A is not contained in I1 . We now set A1 := I1 , U1 := U ∩ A{1 215

A COURSE IN ANALYSIS

and we note that A1 is closed, U1 is open and U1 ∩ A 6= ∅. We choose now I2 ∈ V0 such that I2 ⊂ U1 , in particular I1 ∩ I2 = ∅, and such that λ(1) (I2 ) > 12 δ1 where  δ1 = sup λ(1) (I) I ∈ V0 , I ⊂ U1 ≤ λ(1) (U) < ∞. If A ⊂ I1 ∪ I2 we are done. Otherwise we continue this construction of intervals Ij . Thus suppose that we have already selected mutually disjoint { S N intervals I1 , . . . , IN , Ij ∈ V0 and A ∩ 6= ∅. We set k=1 Ik AN :=

N [

UN := U ∩ A{N

Ik ,

k=1

and we note again that AN is closed and UN is open as well as UN ∩ A 6= ∅. Let  δN := sup λ(1) (I) I ∈ V0 , I ⊂ UN (11.3)

and choose IN +1 ∈ V0 such that IN +1 ⊂ UN , λ(1) (IN +1 ) > 21 δN . If this process S 0 stops for some N0 , i.e. A ⊂ N k=1 Ik then the theorem is already proved. Otherwise we obtain an infinite sequence (Ik )k∈N of mutually disjoint closed S intervals Ik ∈ V0 . We need to prove in this case with D := k∈N Ik that
λ(1) A ∩ D { = 0. For k ∈ N we denote by Jk the closed interval with the same midpoint as Ik and satisfying λ(1) (Jk ) = 5λ(1) (Ik ). By our assumptions we have ! ∞ ∞ [ X λ(1) Jk ≤ 5 λ(1) (Ik ) k=1

k=1 (1)

= 5λ (D) ≤ 5λ(1) (U) < ∞.

From Corollary 2.13 we deduce that lim λ

M →∞

(1)

∞ [

k=M

Jk

!

= 0.

(11.4)

S We claim that A ∩ D { ⊂
∞ k=M Jk holds for every M ∈ N which implies by (11.3) that λ(1) A ∩ D { = 0, i.e. (11.2). We fix M ∈ N and pick 216

11

DIFFERENTIATION REVISITED

x ∈ A ∩ D { . It follows that x ∈ A ∩ A{M ⊂ UM . Hence we can find I ∈ V0 such that x ∈ I ⊂ UM . Clearly δM < 2λ(1) (IM +1 ) and by (11.4) we have limk→∞ λ(1) (Ik ) = 0, i.e. for some k0 ∈ N we have δk0 < λ(1) (I). On the other hand, by (11.4) there exists l0 ∈ N such that I is not a subset of Ul0 , and therefore there is a smallest M0 ∈ N with this property. Therefore we must have M < M0 , and consequently we find I ∩ AM0 6= ∅

and

I ∩ AM0 −1 = ∅,

hence I ∩ IM0 6= ∅.

(11.5)

λ(1) (I) ≤ δM0 −1 < 2λ(1) (IM0 ).

(11.6)

Since I ⊂ UM0 −1 we get

Using that λ(1) (IM0 ) = 5λ(1) (IM0 ), (11.5) and (11.6) we deduce that I ⊂ JM 0 ⊂

∞ [

Jk ,

k=M


S (1) A ∩ D { = 0. Given  > 0 we i.e. x ∈ ∞ k=M Jk , which eventually yields λ can find M ∈ N such that ∞ X

λ(1) (Ik ) < 

k=M +1

which implies by A ∩ A{M that λ

(1)

  ⊂ A ∩ D{ ∪

  { A ∩ AM ≤ λ(1) (1)

∞ [

k=M +1

Ik

!

∞ [

Ik

k=M +1

=

∞ X

!

λ(1) (Ik ) < ,

k=M +1

i.e. the theorem is proved for λ (A) < ∞. Now let λ(1) (A) = ∞. For m ∈ Z we define A(m) := A ∩ (m, m + 1) and Vm := {I ∈ V | I ⊂ (m, m + 1)} which is a Vitali covering of A(m) and λ(1) (A(m) ) ≤ 1. Thus we can apply the result of the first part to find a 217

A COURSE IN ANALYSIS

denumerable family Im ⊂ Vm of mutually disjoint
intervals such that with S (1) (m) { Dm = {K |K ∈ I } we have λ A ∩ D m m = 0, m ∈ Z. The family S I = k∈Z Ik is denumerable and S consists of mutually disjoint elements of V. Moreover we have with D∞ = m∈Z Dm that  [ { { A ∩ D∞ ⊂Z∪ A(m) ∩ Dm m∈Z

implying     X { { = 0. ≤ λ(1) (Z) + λ(1) A(m) ∩ Dm λ(1) A ∩ D∞ m∈Z

Before we now turn to differentiability questions we recollect some results from Chapter I.32. Let a < b and f : [a, b] → R be a bounded increasing function. We know that f can have at most countably many jumps on [a, b]. By [f ](x0 ) := f (x0 +) − f (x0 −) (11.7) we denote the jump of f at x0 where f (x0 +) = x→x lim f (x) and f (x0 −) = x→x lim f (x). 0

(11.8)

0

x>x0

x 0 such that for all m ∈ N and any choice ofP mutually disjoint open intervals Pm (aj , bj ) ⊂ (b − a ) < δ implies [a, b], j = 1, . . . , m, the estimate m j j=1 |h(bj ) − j=1 j h(aj )| < . Any Lipschitz continuous function is absolutely continuous and every absolutely continuous function is of bounded variation. Definition 11.3. By AC([a, b]) we denote the set of all absolutely continuous functions on [a, b]. From Problem 6 of Chapter I.32 we know that AC([a, b]) is an algebra. We have already seen relation between BV - and AC-functions and integrals: Z

•

f (t) dt ∈ BV ([a, b]) and V (f ) = a Z • f (t) dt ∈ AC([a, b]). f ∈ C([a, b]) ∪ BV ([a, b]) implies F (·) = f ∈ C([a, b]) implies F (·) =

Z

b

f (t) dt, (11.13)

a

(11.14)

a

Thus there are indicators that BV ([a, b]) and AC([a, b]) are related to differentiability properties and the fundamental theorem.

219

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Let f : [a, b] → R be a function. The Dini derivatives or Dini numbers of f at x ∈ [a, b] are defined as f (x + h) − f (x) , h f (x + h) − f (x) , D− f (x) = lim inf h→0− h f (x + h) − f (x) , D + f (x) = lim sup h h→0+ f (x + h) − f (x) D − f (x) = lim sup . h h→0− D+ f (x) = lim inf h→0+

(11.15) (11.16) (11.17) (11.18)

Clearly, if f is differentiable at x then all four Dini derivatives are equal to f 0 (x). For x = a we cannot define D− f (a) and D − f (a) while for x = b we cannot define D+ f (b) and D + f (b). We call D+ f (x) the lower right derivative of f at x, D + f (x) the upper right derivative of f at x, and further D− f (x) is the lower left derivative of f at x whereas D − f (x) is the upper left derivative of f at x. Note that the values +∞ and −∞ are not excluded. Definition 11.4. We say that f : [a, b] → R has right derivative at x ∈ [a, b] or is differentiable from the right at x if D + f (x) = D+ f (x), and in this case we write f+0 (x) := D + f (x) = D+ f (x).

(11.19)

If D − f (x) = D− f (x) we call f differentiable from the left or we say that f has a left derivative at x, and we write f−0 (x) := D − f (x) = D− f (x).

(11.20)

We also note that in (11.15) - (11.18) we can switch to sequences, e.g. D+ f (x) = lim inf n→∞

f (x + hn ) − f (x) hn

where hn > 0 and limn→∞ hn = 0. This implies immediately that for f measurable the functions D+ f, D + f, D− f and D − f are measurable. Thus for 0 < p < q, p, q ∈ Q the sets  Ap,q (f ) := x ∈ [a, b] D+ f (x) < p < q < D + f (x) (11.21) 220

11

DIFFERENTIATION REVISITED

and [  A(f ) := x ∈ [a, b] D+ f (x) < D + f (x) = Ap,q (f )

(11.22)

p,q∈Q 0 0 such that 0 0 such that [x, x + δ] ⊂ U ∩ [a, b] and f (x + δ) − f (x) < pδ.

(11.23)

If we let x vary over all points in Ap,q (f ) and allow all δ > 0 with the above property we obtain a Vitali covering V of Ap,q (f ) consisting of closed intervals. By Theorem 11.2 there exists a finite subfamily of mutually disjoint intervals [x1 , x1 + δ1 ], . . . , [xN , xN + δN ] belonging to V, such that  !{  N [ [xk , xk + δk ]  < . λ(1) Ap,q (f ) ∩ k=1

For the open set V :=

SN

k=1 (xk , xk

+ δk ) we also find   λ(1) Ap,q (f ) ∩ V { < 

and since V ⊂ U it follows that N X k=1

δk = λ(1) (V ) ≤ λ(1) (U) < α + . 221

(11.24)

A COURSE IN ANALYSIS

Now (11.23) yields N X k=1

(f (xk + δk ) − f (xk )) ≤ p

N X

δk < p(α + ).

(11.25)

k=1

Moreover, for y ∈ Ap,q (f ) ∩ V there exists η > 0 such that [y, y + η] ⊂ V and f (y + η) − f (y) > qη,

(11.26)

and we get a Vitali covering of Ap,q (f ) ∩ V by closed intervals as collection of all these closed intervals for y ∈ Ap,q (f ) ∩ V and η > 0 such that (11.26) holds. A further application of Vitali’s covering theorem allows us to pick a finite family [y1 , y1 + η1 ], . . . , [yM , yM + ηM ] of mutually disjoint intervals of this covering such that  !{  M [ λ(1) Ap,q (f ) ∩ V ∩ [yj , yj + ηj ]  < . (11.27) j=1

Combining (11.24) with (11.27) yields   α = λ(1) (Ap,q (f )) ≤ λ(1) Ap,q (f ) ∩ V { + λ(1) (Ap,q (f ) ∩ V ) ! M X < + + ηj ,

(11.28)

j=1

which yields with (11.26) q(α − 2) < q By construction we have increasing we deduce M X j=1

M X

ηj
f−0 (x) .

For x ∈ A1 we choose r(x) ∈ Q such that f+0 (x) < r(x) < f−0 (x). Further we pick p(x), q(x) ∈ Q such that a < p(x) < q(x) < b and f (y) − f (x) > r(x) for p(x) < y < x y−x

(11.31)

and

f (y) − f (x) < r(x) for x < y < q(x) (11.32) y−x hold. These two inequalities imply for y 6= x and p(x) < y < q(x) that f (y) − f (x) < r(x)(y − x).

(11.33)

We define g : A1 → Q3 , g(x) = (r(x), p(x), q(x)), and we are going to show that g is an injective mapping, hence A1 must be a denumerable set. Let x, y ∈ A1 , x 6= y, and suppose that g(x) = g(y). It follows that (p(y), q(y)) = (p(x), q(x)) (as open intervals) and x, y ∈ (p(y), q(y)) = (p(x), q(x)). From (11.33) we deduce now that f (y) − f (x) < r(x)(y − x) and f (x) − f (y) < r(y)(x − y). 223

(11.34)

A COURSE IN ANALYSIS

By assumption we have r(x) = r(y), and hence by adding the two inequalities from (11.34) we obtain 0 < 0, i.e. a contradiction. Thus we have to conclude that g is injective and therefore f+0 (x) = f−0 (x) λ(1) -a.e. The case of points belonging to A2 goes analogously. Finally we prove that the set B = {x ∈ (a, b) | f 0 (x) = ∞} has measure zero, i.e. λ(1) (B) = 0. We follow the ideas of the proof of Lemma 11.5. Let R > 0. For x ∈ B there exists δ > 0 such that [x, x + δ] ⊂ (a, b) and f (x + δ) − f (x) > Rδ.

(11.35)

The family of all these closed intervals is a Vitali covering of B. By Theorem 11.2 there exists a countable subfamily of mutually disjoint intervals ([xk , xk + δk ])k∈N of this covering such that  !{  [ λ(1) B ∩ [xk , xk + δ]  = 0, k∈N

which together with (11.35) implies (1)

Rλ (B) ≤ R

∞ X

δk
0, i.e. λ(1) (B) = 0, and the theorem is proved. Remark 11.7. A. For proving Lemma 11.5 and Theorem 11.6 we have adopted the discussion in E. Hewitt and K. Stromberg [35], originally Theorem 11.6 was proved by H. Lebesgue. B. Note that the proof of Theorem 11.6 yields the following interesting result: let f : (a, b) → R be any function, then the set of all points x ∈ (a, b) for which f+0 (x) and f−0 (x) exist but are not equal is at most countable. Corollary 11.8. Every function f : [a, b] → R of bounded variation, and hence every absolutely continuous function f : [a, b] → R is λ(1) -a.e. differentiable with finite derivative. Moreover the derivative f 0 of f ∈ BV ([a, b]) belongs to L1 ([a, b]) and if f is increasing the following holds Z b f 0 (x) dx ≤ f (b) − f (a). (11.36) a

224

11

DIFFERENTIATION REVISITED

In addition, if f is increasing then f 0 ≥ 0 a.e. and if f is decreasing then f 0 ≤ 0. Proof. The first part of the corollary is trivial in light of Theorem 11.6 since f = f1 −f2 and f1 , f2 are increasing functions. In order to prove the remaining part it is sufficient that for f : [a, b] → R increasing it follows f 0 ∈ L1 ([a, b]) and f 0 ≥ 0 λ(1) -a.e. Consider the sequence (fk )k∈N defined by
    f x + k1 − f (x) 1 =k f x+ 0 ≤ fk (x) := − f (x) . 1 k k By Theorem 11.6 we know that λ(1) -a.e. we have limk→∞ fk (x) = f 0 (x), hence f 0 ≥ 0 λ(1) -a.e. Using Fatou’s lemma we find further 0≤

Z

a

b

0

f (x) dx ≤ lim inf k→∞

≤ lim inf k→∞

Z

k

fk (x) dx

Z

b

b+ k1

f (x) dx − k

Z

a

a+ k1

!

f (x) dx

≤ f (b) − f (a).

This proves that f 0 ∈ L1 ([a, b]) for f increasing as well as (11.36). Since a general function belonging to BV ([a, b]) is the difference between two increasing functions the corollary follows. Consider now f ∈ L1 ([a, b]) and note that f |[a,x] ∈ L1 ([a, x]) for every x ∈ [a, b]. Therefore we can define a function F : [a, b] → R by Z x f (t) dt. (11.37) F (x) := a

Note Rthat the valueRF (x) is pointwisely defined, i.e. if f1 = f2 λ(1) -a.e. we x x have a f1 (t) dt = a f2 (t) dt. Of course we would like to obtain a result such as F 0 (x) = f (x) λ(1) -a.e. In preparation for a proof of such a statement we show the following proposition which we can interpret as the absolute continuity of the integral. Proposition 11.9. Let (Ω, A, µ) be a measure space and f ∈ L1 (Ω). Then Rfor every  > 0 there exists δ > 0 such that A ∈ A and µ(A) < δ implies |f | dµ < . A 225

A COURSE IN ANALYSIS

Proof. If kf k∞ < ∞, i.e. f is λ(1) -a.e. bounded then we have for δ := (the case kf k∞ = 0 is trivial) the estimate Z |f | dµ ≤ kf k∞ µ(A) < ,

 kf k∞

A

i.e. the assertion holds. Now let f ∈ L1 (Ω) and define gk (ω) := |f (ω)| ∧ k ≤ |f (ω)|. The sequence (gk )k∈N consists of measurable functions and kgk k∞ ≤ k. Moreover gk → |f | holds pointwisely. By the dominated convergence theorem we conclude that Z Z |f | dµ = lim |gk | dµ, k→∞

or for every  > 0 exists N ∈ N such that k ≥ N implies that Z  (|f (ω)| − gk (ω)|) µ(dω) < . 2

For 0 < δ
0 and any P choice of mutually disjoint intervals (a S1M, b1 ), . . . , (aM , bM ) ⊂ M [a, b] such that j=1 (bj − aj ) < δ we have with A = j=1 (aj , bj ) Z Z M Z X (1) f dλ(1) ≤ |f | dλ = A

A

j=1

bj

aj

|f (x)|λ(1) (dx) < .

Corollary 11.10. For f ∈ L1 ([a, b]) the function F defined by (11.37) is absolutely continuous and hence λ(1) -a.e. differentiable. 226

11

DIFFERENTIATION REVISITED

Proof. Since for a < x < y < Z by we have Z f (t) dt − F (y) − F (x) =

Z

x

f (t) dt =

f (t) dt

x

a

a

y

and we find for a finite number of mutually disjoint intervals (x1 , y1 ), . . . , (xM , yM ) ⊂ [a, b] that M M Z yj X X (F (yj ) − F (xj )) = f (t) dt. j=1

xj

j=1

Hence PM by Proposition 11.9, for  > 0 there exists δ > 0 such that whenever j=1 (yj − xj ) < δ it follows that M M Z yj X X |f (t)| dt < , |F (yj ) − F (xj )| ≤ j=1

j=1

xj

i.e. F is absolutely continuous, hence λ(1) -a.e. differentiable. Rx Proposition 11.11. If f ∈ L1 ([a, b]) and F (x) := a f (t) dt = 0 for all x ∈ [a, b] then f = 0 λ(1) -a.e.

Proof. Suppose that for a set A ∈ B([a, b]), λ(1) (A) > 0, we have f |A > 0 (the case f |A < 0 goes analogously). By Theorem 10.10 we can find a compact set K ⊂ A such that λ(1) (K) >S0. The set U := (a, b)\K is open, hence by Theorem I.19.27 we have U = M k=1 (ak , bk ) where the open intervals (ak , bk ) are mutually disjoint and M ∈ N ∪ {∞}. We handle the case M = ∞, the other cases are then trivial. Since Z Z b Z f (t) dt = 0 = F (b) = f (t) dt + f (t) dt a

K

Z

we find

U

f (t) dt < 0,

U

hence 0 6=

Z

U

f (t) dt =

a

bk

f (t) dt

k=1 ak Rb have akk0 f (t) dt 0

and therefore for some k0 we must Z Z bk Z bk 0 0 f (t) dt − f (t) dt = ak0

∞ Z X

ak0

a

which is a contradiction. 227

6= 0. However we have

f (t) dt = F (bk0 ) − F (ak0 ) = 0,

A COURSE IN ANALYSIS

Proposition 11.12. For a bounded measurable function f : [a, b] → R the derivative of F as defined by (11.37) is f , i.e. F 0 = f , λ(1) -a.e. Proof. Since F is absolutely continuous it is λ(1) -a.e. differentiable, see Corollary 11.10. Suppose that |f (x)| ≤ M and consider the sequence gk of functions defined by
    F x + k1 − F (x) 1 gk (x) = − F (x) =k F x+ 1 k k where we agree to set F (y) = F (b) for y ≥ b. It follows that gk (x) = k

Z

x+ k1

f (t) dt

x

implying that |gk (x)| ≤ M. Furthermore (gk (x))k∈N converges λ(1) -a.e. to F 0 (x) and the dominated convergence theorem yields that for any c ∈ [a, b] we have Z c Z c 0 gk (x) dx F (x) dx = lim k→∞ a a   Z c  1 F x+ − F (x) dx = lim k k→∞ k a ! Z a+ 1 Z c+ 1 k k F (x) dx − F (x) dx = lim k k→∞

c

a

= F (c) − F (a) where for the last step we use the fact that F is absolutely continuous, hence continuous and the mean-value theorem for Riemann integrals applies. Thus we arrive at Z c Z c 0 f (x) dx F (x) dx = F (c) − F (a) = a

a

for all c ∈ [a, b] and by Proposition 11.11 it follows that F 0 = f λ(1) -a.e. With the help of Proposition 11.12 we eventually can show

Theorem 11.13. For f ∈ L1 ([a, b]) and F defined as in (11.37) we have F 0 = f λ(1) -a.e. 228

(11.38)

11

DIFFERENTIATION REVISITED

Proof. The basic idea is to approximate f by bounded functions. Since we can decompose f into positive and negative parts, we may assume that f ≥ 0. We define the approximating sequence of functions hk ∈ L1 ([a, b]) by hk (x) := f (x) ∧ k.

(11.39)

Clearly f − hk ≥ 0 and |hk (x)| ≤ k. The function Hk : [a, b] → R Z x (f (t) − hk (t)) dt Hk (x) := a

is increasing, λ(1) -a.e. differentiable and Hk0 (x) ≥ 0 λ(1) -a.e. By Proposition 11.12 we have Z x d hk (t) dt = hk (x) λ(1) -a.e. dx a and therefore λ(1) -a.e. 0

F (x) =

Hk0 (x)

d + dx

Z

a

x

hk (x) dx ≥ hk (x).

Since k was arbitrary we find that F 0 (x) ≥ f (x) λ(1) -a.e. It follows that Z b Z b f (x) dx = F (b) − F (a), F 0 (x) dx ≥ a

a

and (11.36) in Corollary 11.8 yields now Z b Z b 0 f (x) dx F (x) dx = F (b) − F (a) = a

a

as well as

Z

a

b

(F 0 (x) − f (x)) dx = 0.

Since F 0 (x) ≥ f (x) λ(1) -a.e. it follows now that F 0 (x) = f (x) λ(1) -a.e. and the theorem is proved. Corollary 11.14. For f ∈ L1 ([a, b]) and Fa defined by Z x f (t) dt Fa (x) = F (a) + a

it follows that F 0 = f λ(1) -a.e. 229

A COURSE IN ANALYSIS

Recall that according to Definition I.12.6 a differentiable function F : [a, b] → R is a primitive of f : [a, b] → R if F 0 = f . The fundamental theorem of calculus, Theorem I.12.7, states that if f is continuous then it admits a primitive F and Z b

a

f (t) dt = F (b) − F (a).

We will now show that a function F is a primitive if and only if it is absolutely continuous. For this we need

Lemma 11.15. An absolutely continuous function f : [a, b] → R with derivative f 0 = 0 λ(1) -a.e. is constant. Proof. We want to prove that f (a) = f (c) for every c ∈ [a, b]. By assumption we can find a measurable set A ⊂ (a, c) such that f 0 (x) = 0 for all x ∈ A. Let  > 0. For every x ∈ A there exists y such that [x, y] ⊂ [a, c] and  (y − x) since f 0 (x) = 0 for x ∈ A. We choose now δ > 0 |f (y) − f (x)| < 2(c−a) such that δ is determined by the absolute continuity of f with  replaced by  . This δ we now use in Vitali’s covering theorem, Theorem 11.2 as , to 2 get that we can find mutually disjoint intervals [x1 , y1 ], . . . , [xM , yM ] ⊂ [a, c], xk < xk+1 , k = 1, . . . , M − 1, such that  |f (yk ) − f (xk )| < (yk − xk ) (11.40) 2(c − a) and



λ(1) A\

M [

[xk , yk ]

k=1

!{ 

 < δ.

(11.41)

With y0 = a and xM +1 = c it follows from (11.41) that M X k=0

|xk+1 − yk | < δ

and further using again the absolute continuity of f and (11.40) M M X X |f (c) − f (a)| = (f (xk+1 ) − f (yk )) + (f (yk ) − f (xk )) k=0


0 was arbitrary it follows that f (c) = f (a). 230

11

DIFFERENTIATION REVISITED

Now we can prove the extension of the fundamental theorem of calculus: Rx Theorem 11.16. Let f ∈ L1 ([a, b]) then F (x) := a f (t) dt is absolutely continuous on [a, b] and F 0 = f holds in L1 ([a, b]) as well as Z x f (t) dt. (11.42) F (x) − F (a) = a

Conversely, if F : [a, b] → R is an absolutely continuous function then there Rx 1 exists f ∈ L ([a, b]) such that F (x) = a f (t) dt for all x ∈ [a, b].

Proof. We know already from Corollary 11.10 and Theorem 11.13 that for f ∈ L1 ([a, b]) the function F is absolutely continuous and that F 0 = f holds. Since F (a) = 0 we also have (11.42). Now let F be absolutely continuous, hence F ∈ BV ([a, b]) and with two increasing functions F1 and F2 we have F (x) = F1 (x) − F2 (x). It follows that F 0 exists λ(1) -a.e. and |F 0(x)| ≤ F10 (x) + F20 (x), and therefore Z

b a

|F 0 (x)| dx ≤ F1 (b) + F2 (b) − F1 (a) − F2 (a)

by (11.36) in Corollary 11.8. This shows already that F 0 ∈ L1 ([a, b]). We consider now the two absolutely continuous functions Z x F 0 (t) dt and g := F − H. H(x) := a

By Theorem 11.13 we have g 0 = F 0 − H 0 = 0 λ(1) -a.e., hence by Lemma 11.15 the function g is constant implying that Z x F 0 (t) dt + F (a) F (x) = a

and the second part of the theorem is proved with f := F 0 . The fundamental theorem allows us to generalise the integration by parts formula: 231

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Theorem 11.17. Let f, g ∈ L1 ([a, b]) and denote by F and G the absolutely continuous functions Z x Z x g(t) dt. f (t) dt and G(x) = F (x) = a

a

The following formula Z b Z b F (t)g(t) dt f (t)G(t) dt = F · G|ba −

(11.43)

a

a

holds. Proof. We also know that F · G is absolutely continuous, hence λ(1) -a.e. differentiable and Z b (F · G)0 (t) dt = F (b)G(b) − F (a)G(a) = F · G|ba , a

while on the other hand by Leibniz’s rule Z b Z b Z b (F g)(t) dt, (f G)(t) dt + (F · G)0 (t) dt = a

a

a

implying (11.43). Corollary 11.18. For two absolutely continuous functions f, g : [a, b] → R we have Z b Z b 0 b f 0 (t)g(t) dt. (11.44) f (t)g (t) dt = f g|a − a

a

Remark 11.19. Our presentation of this part of “differentiation theory” is much influenced by [35], [70], also see [14]. In Chapter 6 we introduced measures ν having a density with respect to given measure µ and where led to the notion of absolutely continuous measures, see Definition 6.11: ν is absolutely continuous with respect to µ if µ(A) = 0 implies that ν(A) = 0. If ν = f λ(1) , f ∈ RL1 ([a, b]), i.e. R · ν is absolutely · continuous with respect to λ(1) , then F (·) = a f (t)dt = a 1 dν is absolutely continuous and hence both notions are related.

The property that f : A → R, A ∈ B(1) , maps sets of measure zero onto sets of measure zero is often called the Lusin property. The relation of the 232

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Lusin property to other properties of f such as being of bounded variation, Lipschitz continuous etc. is discussed in detail in [6]. Without proof we state the Banach-Zaretzky theorem, see [14]: Theorem 11.20. A function F : [a, b] → R of bounded variation is absolutely continuous if and only if for every A ∈ B(1) ([a, b]) it follows that λ(1) (A) = 0 implies λ(1) (F (A)) = 0. We now want to study two examples. Example 11.21. In Example I.32.13 we have seen that the function f : [0, 1] → R defined by f (x) =

(

x sin x1 , x ∈ (0, 1] 0, x=0

is not of bounded variation, hence not absolutely continuous. We extend this example in the following direction: For α, β > 0 define ( xα sin x1β , x ∈ (0, 1] gα,β (x) := 0, x = 0. We claim that if 0 < β < α then gα,β is absolutely continuous while for 0 < α ≤ β it is not. For 0 < β < α we find for x > 0 that 0 gα,β (x) = αxα−1 sin

1 1 − βxα−β−1 cos β β x x

and this function belongs to L1 ([0, 1]), thus we have gα,β (x) =

Z

0

x

  1 1 α−β−1 α−1 cos β dy, αy sin β − βy y y

i.e. gα,β is for 0 < β < α absolutely continuous. Of course, for 0 < α ≤ β the 0 0 expression of gα,β does not change, but now gα,β does not belong any more 1 to L ([0, 1]) and hence by Corollary 11.8 the function gα,β is not of bounded variation. 233

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Example 11.22. Let F : [0, 1] → R be the Cantor-Lebesgue function as introduced in Definition 3.42. By Lemma 3.43 we know that F is H¨older 2 continuous with exponent α = ln , furthermore F 0 = 0 λ(1) -a.e. by its ln 3 construction, recall that F is constant on the intervals forming C { and λ(1) (C) = 0. Thus FR cannot be absolutely continuous, since otherwise we x would have F (x) = 0 F 0 (t) dt = 0 for all x ∈ [0, 1]. This example shows that neither continuous monotone functions nor H¨older continuous functions need to be absolutely continuous. Let f : [a, b] → R be a continuous function and x0 ∈ (a, b). It follows for h sufficiently small that Z x0 +h Z x0 +h 1 1 (f (t) − f (x0 ))dt = f (t) dt − f (x0 ) 2h x0 −h 2h x0 −h 1 (F (x0 + h) − F (x0 − h)) − f (x0 ) = 2h Rx where F (x) = a f (t) dt. Thus we find 1 h→0 2h lim

or

Z

x0 +h

x0 −h

1 lim h→0 2h

Z

(f (t) − f (x0 )) dt = 0

(11.45)

x0 +h

f (t) dt = f (x0 ).

(11.46)

x0 −h

This result has a far reaching extension to Rn known as Lebesgue’s differentiation theorem which we are going to discuss next. In doing so we will also introduce a new, rather powerful tool, namely the Hardy-Littlewood maximal function. As preparation we need a further covering result, sometimes called Wiener’s covering lemma. Lemma 11.23. For every collection B0 = {B1 , . . . , BN } of open balls Bj ⊂ Rn we can find a subcollection Bj1 , · · · , BjM of B0 such that ! N M [ X (n) λ Bk ≤ 3n λ(n) (Bjl ). (11.47) k=1

l=1

Proof. Let Bj1 = Br1 (x1 ) ∈ B0 be a ball with largest radius and consider now B1 := {Bk ∈ B0 | Bk ∩ Bj1 = ∅}. Since Bj1 has the largest radius, by the 234

11

DIFFERENTIATION REVISITED

triangle inequality it follows that [

B∈B0 \B1

B ⊂ B3r1 (x1 ).

We now apply the same procedure to B1 and thus obtain an open ball Bj2 = Br2 (x2 ) ∈ B1 and we introduce B2 := {Bk ∈ B1 | Bk ∩Br2 = ∅}. After M ≤ N steps we have a subcollection of mutually disjoint balls Bj1 , . . . , BjM ∈ B0 S which covers N k=1 Bk and therefore we have λ

(n)

N [

k=1

Bk

!

≤λ ≤

M [

(n)

M X

B3rl (xl )

l=1

λ

(n)

!

(B3rl (xl )) = 3

n

M X

λ(n) (Bjl ).

l=1

l=1

Definition 11.24. A. We call a Rmeasurable function u : Rn → R locally integrable with respect to λ(n) if K |u(x)|λ(n) (dx) < ∞ for every compact set K ⊂ Rn . The space of all locally integrable functions u we denote by L1loc (Rn ). (As usual L1loc (Rn ) consists of equivalence classes induced by the equivalence relation “∼λ(n) ”). B. For u ∈ L1loc (Rn ) we define its average over Br (x) ⊂ Rn by Z 1 u(y) dy. (11.48) (Ar u) (x) := (n) λ (Br (x)) Br (x) Lemma 11.25. For u ∈ L1loc (Rn ) the mapping (r, x) 7→ (Ar u)(x) is continuous from (0, ∞) × Rn to R. Proof. Recall that λ(n) (Br (x)) = ωn r n , ωn =

n

π2 , +1) Γ( n 2

see (II.21.11), and that

∂Br (x) is a set of measure zero for λ(n) . In addition we have on Rn \∂Br0 (x0 ) pointwisely lim χBr (x) (y) = χBr0 (x0 ) (y), (r,x)→(r0 ,x0 )

for r < r0 + 21 and i.e. lim(r,x)→(r0 ,x0) χBr (x) = χBr0 (x0 ) λ(n) -a.e. Furthermore, 1 kx0 − xk < 2 the triangle inequality yields χBr (x) (y) ≤ χBr0 +1 (x0 ) (y). Now 235

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the dominated convergence theorem implies Z Z χBr (x) (y)u(y) dy u(y) dy = lim lim (r,x)→(r0 ,x0 ) Rn (r,x)→(r0 ,x0 ) Br (x) Z Z = χBr0 (x0 ) (y)u(y) dy = u(y) dy, Rn

Br0 (x0 )

and consequently lim

(r,x)→(r0 ,x0 )

(Ar u)(x) = (Ar0 u)(x0 ),

proving the lemma. Definition 11.26. For f ∈ L1loc (Rn ) the Hardy-Littlewood maximal function, in short the maximal function, of f is defined by Z 1 M(f )(x) := sup(Ar |f |)(x) = sup (n) |f (y)| dy. (11.49) (Br (x)) Br (x) r>0 r>0 λ Lemma 11.27. The Hardy-Littlewood maximal function is measurable on Rn . Proof. From Lemma 11.25 we know that for a > 0 the pre-image of (a, ∞) under Ar |f | : Rn → R, i.e. (Ar |f |)−1(a, ∞), is open. Since M(f )−1(a, ∞) = S −1 r>0 (Ar |f |) (a, ∞) is open it follows that M(f ) is measurable. The following theorem, often called maximal or Hardy-Littlewood maximal theorem, is of great importance and has many applications in real analysis. Theorem 11.28. The estimate λ

(n)


3n x ∈ Rn M(f )(x) > α ≤ α

Z

Rn

|f (t)| dt =

3n kf kL1 α

(11.50)

holds for all α > 0 and f ∈ L1 (Rn ).

Proof. Let Aα := {x ∈ Rn | M(f )(x) > α}. From the definition of M(f ) for x ∈ Aα we can find a ball Bx := Br(x) (x) such that Z 1 |f (y)| dy > α. λ(n) (Bx ) Bx 236

11

DIFFERENTIATION REVISITED

For a compact set K ⊂ Aα the family {Bx | x ∈ K} is an open covering of S B K, hence we can cover K by finitely many of such balls, i.e. K ⊂ M k=1 k , Bk ∈ {Bx | x ∈ K}. Applying Lemma 11.23 we can find a further subcovering of mutually disjoint balls Bk1 , . . . , BkN ∈ {B1 , . . . , BM } such that M [

λ(n)

Bk

k=1

!

≤ 3n

N X

λ(n) (Bkj ).

(11.51)

j=1

For each Bkj we have λ

(n)

1 (Bkj ) < α

Z

Bkj

|f (y)| dy.

(11.52)

Now it follows that λ

(n)

(K) ≤ λ

(n)

M [

k=1

Bk

!

≤ 3n

N X

λ(n) (Bkj )

j=1

Z N Z 3n 3 X |f (y)| dy = |f (y)| dy ≤ α j=1 Bkj α SNj=1 Bkj Z 3n |f (y)| dy, ≤ α Rn n

i.e. for every compact set K ⊂ Aα λ(n) (K) ≤

3n kf kL1 α

holds, implying (11.50). Eventually we can show Lebesgue’s differentiation theorem: Theorem 11.29. For every f ∈ L1 (Rn ) the limit Z 1 lim f (y) dy = f (x) (n) (B) λ(n) (B)→0 λ B

(11.53)

x∈B

exists λ(n) -a.e. where B ⊂ Rn denotes an open ball B ⊂ Rn containing x. 237

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Proof. Suppose that we can prove for every α > 0 that the set     Z   1 n f (y) dy − f (x) > 2α lim sup (n) Aα := x ∈ R   (B) B   λ(n) (B)→0 λ x∈B

has measure zero, i.e. λ(n) (Aα ) = 0. It follows that the set A = {

(n)

S

n∈N

A1

n

also has measure zero and (11.53) holds for x ∈ A , i.e. λ -a.e. Let α > 0 be fixed and let  > 0. By Theorem 10.11 we can find g ∈ C0 (Rn ) such that kf − gkL1 < . For g we have Z 1 g(y) dy = g(x). lim (n) (B) λ(n) (B)→0 λ B x∈B

Indeed, this we can see as follows: first we note that Z Z 1 1 g(y) dy − g(x) = (n) (g(y) − g(x)) dy. λ(n) (B) B λ (B) B

Since g is continuous, given η > 0 there exists δ > 0 such that kx − yk < δ implies |g(x) − g(y)| < η. For every ball B ⊂ Rn with x ∈ B and diameter δ we find now Z Z 1 1 g(y) dy − g(x) ≤ (n) |g(y) − g(x)| dy < η. λ(n) (B) λ (B) B B We return to f and write 1 (n) λ (B)

Z

Z 1 f (y) dy − f (x) = (n) (f (y) − f (x)) dy λ (B) B B Z Z 1 1 (f (y) − g(y)) dy + (n) (g(y) − g(x)) dy + g(x) − f (x). = (n) λ (B) B λ (B) B

Since for λ(n) (B) → 0, x ∈ B, the second term tends to 0 we get Z 1 lim sup (n) f (y) dy − f (x) (B) B λ(n) (B)→0 λ x∈B Z 1 |f (y) − g(y)| dy + |g(x) − f (x)| ≤ lim sup (n) (B) B λ(n) (B)→0 λ x∈B

≤ M(f − g)(x) + |g(x) − f (x)|. 238

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We observe that   Aα ⊂ x ∈ Rn M(f − g)(x) > α ∪ x ∈ Rn |f (x) − g(x)| > α . By the Chebyshev-Markov inequality, Lemma 7.15, we find λ(n)


1 x ∈ Rn |f (x) − g(x)| > α ≤ kf − gkL1 , α

and the maximal theorem yields λ(n) or


3n x ∈ Rn M(f − g)(x) > α ≤ kf − gkL1 , α  n  3 3n + 1 1 (n) λ (Aα ) ≤ kf − gkL1 ≤ +  α α α

by our choice of g. Since  > 0 was arbitrary the theorem follows. Remark 11.30. A. Theorem 11.29 extends to f ∈ L1loc , see for example [27] or the remark [84]. Our proof is a combination of the arguments in [27], [84] and [14]. B. In [27] it is also discussed that we can replace the balls in Theorem 11.29 by more general families of sets shrinking to x. C. For f ∈ L1loc (Rn ) the Lebesgue set of f , Λ(f ), is defined by   Z 1 |f (y) − f (x)| dy = 0 , (11.54) Λ(f ) := x ∈ Rn lim (n) r→0 λ (Br (x)) Br (x)
and in [27] it is proved that λ(n) Λ(f ){ = 0 for f ∈ L1loc (Rn ), i.e. we can strengthen Theorem 11.29 to Z 1 lim |f (y) − f (x)| dy = 0, λ(n) -a.e. r→0 λ(n) (Br (0)) B (x) r for f ∈ L1loc (Rn ).

Problems 1. Let f, g : [a, b] → R be two absolutely continuous functions and g(x) 6= 0 for all x ∈ [a, b]. Prove that fg is also an absolutely continuous function. 239

A COURSE IN ANALYSIS

2.

a) Let f : [a, b] → R be an absolutely continuous function with f ([a, b]) = [c, d]. Further let ϕ : [c, d] → R be a Lipschitz continuous function. Prove that ϕ ◦ f : [a, b] → R is an absolutely continuous function. b) Let f, g : [a, b] → R be absolutely continuous and g in addition a monotone function. Show that f ◦ g is also an absolutely continuous function.

3. Let f : [a, b] → R be an absolutely continuous function and N ⊂ [a, b] be a set of measure zero. Show that f (N) is of measure zero too. 4. Let g ∈ Lp ([a, b]), 1 ≤ p < ∞, and extend g to be 0 in [a, b]{ . For h > 0 define the function Z x+h 1 g(t)dt. gh (x) := 2h x−h Prove that gh is a continuous function and kgh kLp ≤ kgkLp . 5. We call a non-constant continuous function s : [a, b] → R which is of bounded variation a singular function if s0 = 0 λ(1) |[a,b] -almost everywhere. a) Prove that if f : [a, b] → R is a continuous function with bounded variation then f admits a unique decomposition f = ϕ + s where ϕ is absolutely continuous with ϕ(a) = f (a), and a function s which is either identically zero or a singular function. b) For a monotone function f with decomposition f = ϕ + s as in part a) show that ϕ and s are monotone too. ( x2 cos xπ2 , x ∈ (0, 1] 6. Consider u : [0, 1] → R, u(x) = , and prove that 0, x=0 u has a (finite) derivative u0 (x) everywhere, but u0 ∈ / L1 ([0, 1]). 7.* Let g : (a, b) → R be a convex function. Prove that then g 0 (x) exists for all x ∈ (a, b) \ A, where A ⊂ (a, b) is at most a countable set. Furthermore g 0 is an increasing function. Hint: recall Theorem I.23.5. 240

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8.* For f ∈ L1loc (R) denote by M(f ) its Hardy-Littlewood maximal function. For 1 < p < ∞ prove that f ∈ Lp (R) implies M(f ) ∈ Lp (R) and that kM(f )kLp ≤ ckf kLp . Hint: Example 9.24 may be used to state for M(f ) that Z ∞ p λ(1) ({M(f ) > s})sp−1ds. kM(f )kLp = p 0

9.* In this problem we construct a continuous function which is at no point differentiable. The first example of such a function is due to Weierstrass, his example was modified by many authors. We have chosen the function discussed in [45]. Let g : R → R be the periodic extension from x 7→ |x| defined on [−1, 1], i.e. g(x) = g(x + 2) and g|[−1,1] = |x|.
P 3 k g(4k x) Prove that the Weierstrass function x 7→ f (x) := ∞ k=0 4 is continuous on R and at no point x ∈ R differentiable.

241

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12

Selected Topics

In this chapter we want to discuss three topics with some of their applications: the theorem of Sard; the theorem of Lusin; and Kolmogorov’s version of the Arzela-Ascoli theorem for Lp -spaces. In some discussions we will need auxilliary results (mainly from point set topology) which we will quote but not prove. The main idea of this chapter is to present some useful results and to indicate that the theory developed so far has far reaching consequences in quite different, and maybe unexpected fields of mathematics. We start with a version of Sard’s theorem. In Chapter II.9 we introduced the notion of a critical point of a differentiable function f : G → R, G ⊂ Rn , as a point x0 ∈ G for which grad f (x0 ) = 0. For differentiable mappings f : G → Rn , G ⊂ Rn , we can extend this definition to points x0 ∈ G where det Jf (x0 ) = 0, i.e. x0 ∈ G is called a critical point if the Jacobi determinant of f at x0 vanishes. From here it is an easy step to Definition 12.1. Let G ⊂ Rn be an open set and f : G → Rm a differentiable mapping. We call x0 ∈ G a critical point of f if dx0 f = Jf (x0 ) has no maximal rank. The set of all critical points of f we denote by crit(f ). The set f (crit(f )) is the set of critical values and its complement in f (G) is the set of all regular values. The version of Sard’s theorem we are going to prove is Theorem 12.2. For a C 1 -mapping f : G → Rn , G ⊂ Rn open, the set f (crit(f )) has Lebesgue measure zero, i.e. f (crit(f )) is a Borel set and λ(n) (f (crit(f ))) = 0. Remark 12.3. In Volume VI we will extend the definition of critical points even to differentiable mappings between differentiable manifolds, and if these manifolds have a countable base, Sard’s theorem still holds, compare with [58]. For C ∞ -mappings f : G → Rm , G ⊂ Rn open, a complete proof of Sard’s theorem is given in [57] stating that the measure of the set f ({x ∈ G | rank of Jf (x) < m}) is zero. A different extension of Sard’s theorem is given in [78] and it reads as follows: for G ⊂ Rn and f : G → Rn continuously differentiable the set f (A) is measurable if A ⊂ G is measurable and we have Z (n) |det Jf (x)| dx. (12.1) λ (f (A)) ≤ A

Clearly (12.1) implies Theorem 12.2.

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For the proof of Theorem 12.2 we need an easy extension of Theorem II.9.1: For f : G → Rn , G ⊂ Rn convex and f continuously differentiable f (x) − f (y) − (df (y))(x − y) =

Z

1

(df (y + t(x − y)) − df (y)) (x − y) dt

0

(12.2)

holds for all x, y ∈ G. In addition the proof requires Lemma 12.4. Let G ⊂ Rn be an open set. Then G is the union of countably many compact cubes. We do not prove this result which can be derived from general results for separable, locally compact and σ-compact spaces, compare with [65], otherwise we would have to derive a more lengthy concrete proof for the special case of G ⊂ Rn . Proof of Theorem 12.2. (Compare with [19] or [74]). As stated in Lemma 12.4, G is a countable union of compact cubes Qj , j ∈ N. Once we can prove the result for a compact cube Q ⊂ G, the result will follow since [ f (crit(f )) = f (crit(f |Qj )). j∈N

Let Q ⊂ G a compact cube with side length r > 0. Since by assumption df (x) is continuous on G it is bounded and uniformly continuous on Q. Thus √ kdf (x)k ≤ M and for every  > 0 there exists l ∈ N such that with δ := lnr the following holds kx − yk < δ implies kdf (x) − df (y)k < . Note that for x and f (x) the norm kxk and kf (x)k is the norm in Rn whereas 2 for df (x) the norm kdf (x)k is the norm (matrix norm) in Rn . For x, y ∈ Q such that kx − yk < δ we now find kf (x) − f (y) − (df (y))(x − y)k ≤

Z

1 0

kdf (y + t(x − y)) − df (y)k kx − yk dy

≤ kx − yk.

244

12

SELECTED TOPICS

Let N = ln and divide Q into N cubes Qj , j = 1, . . . , N, with diameter δ, i.e. side length √δn = rl . For x, y ∈ Qj we have f (x) = f (y) + (df (y))(x − y) + R(x, y),

kR(x, y)k ≤ δ.

Now let xj ∈ Qj ∩ crit(f ) and set D := (df )(xj ) and g(y) := f (xj + y) − f (xj ), y ∈ Qj − xj , which gives ˜ ˜ g(y) = Dy + R(y), kR(y)k = kR(xj + y, xj )k ≤ δ. We note that det(Jf (xj )) = det(df (xj )) = 0 implies that D(Qj − xj ) is contained in an (n − 1)-dimensional subspace of Rn , say the subspace characterised by hz, b1 i = 0 for all z ∈ D(Qj − xj ) with some b1 ∈ Rn , kb1 k = 1. Let {b1 , . . . , bn } be an orthogonal basis of Rn obtained by extending {b1 }. It follows that n X g(y) = hg(y), bk ibk k=1

and

|hg(y), b1i| ≤ |hDy, b1i| + |hR(xj + y, xj ), b1 i| ≤ kR(xj + y, xj )k ≤ δ,

(12.3)

whereas for 2 ≤ k ≤ n we find |hg(y), bk i| ≤ kDk kyk + kR(xj + y, xj )k ≤ kDkδ + δ.

(12.4)

Using the definition of g we deduce that f (Qj ) is contained in some cube Wj with centre f (xj ) such that λ(n) (Wj ) = (2(kDkδ + δ))n−1 2δ = 2n (kDk + )n−1 δ n . 245

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Consequently we have f (critf |Q ) ⊂ and N X j=1

λ

(n)

(Wj ) ≤

N X

N [

Wj

j=1

(2(kdf (xj )k + ))n−1 2δ n 

j=1

√ ≤ N2n (M + )n−1 ( nr)n ,

S implying that N j=1 Wj is a set of measure zero, and hence f (critf |Q ) is a null set, hence a Lebesgue set of measure zero. Since f (crit(f )) is the union of countably many compact set it is also a Borel set. Corollary 12.5. Let G ⊂ Rn be an open set and f : G → Rm a continuously differentiable mapping. If n < m then f (G) is a Borel set and λ(m) (f (G)) = 0. Proof. Let pr : Rm → Rn be the canonical projection and consider the mapping f ◦ pr : pr−1 (G) → Rm . Since d(f ◦ pr) = df ◦ d pr the Jacobi determinant of f ◦ pr is identically zero, however (f ◦ pr)(pr−1 (G)) = f (G). Sard’s theorem has many applications in particular when discussing the critical or regular values of mappings between differentiable manifolds, but it is also possible to give a proof of the Brouwer fixed point theorem based on Sard’s theorem. Of particular importance are applications in the degree theory of mappings, compare for example with [78] or [19]. The next theorem we want to discuss is Lusin’s theorem. Although it holds for a much more general situation, see Appendix II, we state and prove it here for the case Rn and the Lebesgue measure. For the proof we need once again a result relating measurable sets to closed sets, namely Lemma 12.6. Let A ∈ B(n) . For every  > 0 we can find a closed set C ⊂ A such that λ(n) (A\C) ≤ . Of course, this lemma is closely related to Theorem 10.10 and we will discuss it in Appendix II. 246

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Theorem 12.7 (Lusin). For A ∈ B(n) , λ(n) (A) < ∞, let f : A → R be a measurable function. Given  > 0 we can find a closed set C ⊂ A such that λ(n) (A\C ) <  and f |C is continuous. Proof. (Adopted from [84]). Since f is measurable it is the λ(n) -a.e. limit of simple functions fk : A → R. For each fk we can find a measurable set Ak ⊂ A such that λ(n) (Ak ) < 21k and f |A{ is continuous, see Problem 5. k Further by Egorov’s theorem, Theorem 7.38, there exists a measurable set
B 3 such that B 3 ⊂ A, λ(n) A\B 3 < 3 and on B 3 the sequence (fk )k∈N converges uniformly to f . For  > 0 we can always find N ∈ N such that P∞ −k ≤ 3 and with this N we define k=N 2 [ Ak . C˜ := B 3 \ k≥N


Since fk |C˜ is continuous and fk |C˜ k∈N converges on C˜ uniformly to f , it follows that f |C˜ is continuous. Clearly C˜ is measurable and therefore we  can  (n) C˜ \C <  find by Lemma 12.6 a closed set C such that C ⊂ C˜ and λ 3

implying that λ

(n)

(A\C ) <  and the theorem is proved.

Remark 12.8. A. As is emphasised in [84], and rarely in other books, the statement of Lusin’s theorem is that f |C is continuous, but we cannot expect that f : A → R (as a function defined on A) is continuous at all points C ⊂ A. We also want to note that in general we have no possibility to identify C . B. In [94] the equivalence of measurability and the property stated in Lusin’s theorem is discussed. C. For justifiable reasons some authors call Lusin’s theorem the Vitali-Lusin theorem. We want to use Lusin’s theorem to characterise a new, rather important type of convergence in Lp (G), G ∈ B(n) , λ(n) (G) < ∞. First we give Definition 12.9. Let G ⊂ B(n) be any measurable set and 1 < p < ∞. We call a sequence (gk )k∈N , gk ∈ Lp (G), weakly convergent to g ∈ Lp (G) if for all ϕ ∈ Lq (G), p1 + 1q = 1, we have lim

k→∞

Z

gk ϕ dλ

(n)

=

G

Z

G

247

gϕ dλ(n) .

(12.5)

A COURSE IN ANALYSIS

Corollary 12.10. Every sequence (gk )k∈N converging in Lp (G) to g, i.e. limk→∞ kgk − gkLp = 0, converges weakly to g. Moreover the weak limit of a sequence is uniquely determined. Proof. From H¨older’s inequality we deduce Z Z (n) gk ϕ dλ(n) − gϕ dλ ≤ kgk − gkLp kϕkLq G

G

implying that limk→∞ kgk − gkLp entails weak convergence. Furthermore for two weak limits g and h of (gk )k∈N we find Z Z Z (g − h)ϕ dλ(n) ≤ lim (g − gk )ϕ dλ(n) + lim (gk − h)ϕ dλ(n) = 0, k→∞

G

k→∞

G

G

i.e. for all ϕ ∈ Lq (G) the following holds Z (g − h)ϕ dλ(n) = 0. G

˜ of h. Since (p − 1)q = p the function Let g˜ be any representant of g and h ( ˜ g˜−h ˜ p−1, g˜ = ˜ g − h| 6 h ˜ |˜ ϕ˜g˜,h˜ (x) = |˜g−h| ˜ 0, g˜ = h is measurable and |ϕ˜g˜,h˜ |q is integrable. It follows that Z Z (n) ˜ ˜ p dλ(n) 0 = (˜ g − h)ϕ˜g˜,h˜ dλ = |˜ g − h| G

G

˜ λ(n) -a.e., hence g = h in Lp (G). implying that g˜ = h Example 12.11. In L2 ([0, 2π]) consider the function gk (x) = sin kx. Since for k 6= l Z 2π Z 2π Z 2π 2 (1) 2 (1) 2 gk gl dλ(1) gl dλ − 2 gk dλ + kgk − gl kL2 = 0 0 0 Z 2π Z 2π Z 2π 2 2 sin kx sin lx dx (sin lx) dx − 2 (sin kx) dx + = 0

0

= 2π,

248

0

12

SELECTED TOPICS

the sequence (gk )k∈N is not a Cauchy sequence in L2 ([0, 2π]), hence it is not convergent. On the other hand, for ϕ = χ[a,b] , [a, b] ⊂ [0, 2π], we have as k→∞ Z 2π 1 (sin kx)ϕ(x) dx = (cos ak − cos bk) → 0 k 0 P and this extends to all linear combinations N j=1 γj χ[aj ,bj ] , [aj , bj ] ⊂ [0, 2π], γj ∈ R, and hence by Theorem 8.21 to all ϕ ∈ L2 ([0, 2π]) implying that the sequence (gk )k∈N converges weakly in L2 ([0, 2π]) to 0, but it does not converge in L2 ([0, 2π]). Remark 12.12. A. In order to have a simpler fa¸con de parler, a sequence (gk )k∈N converging in Lp (G) to g is also called strongly convergent or norm convergent. B. In light of Corollary 12.10 we know that if (gk )k∈N , gk ∈ Lp (G), has a weak limit g ∈ Lp (G), then the only candidate for a strong limit of (gk )k∈N is g. With the help of Egorov’s and Lusin’s theorem we can prove Proposition 12.13. Let G ∈ B(n) , λ(n) (G) < ∞, and let (gk )k∈N , gk ∈ Lp (G), 1 < p < ∞, be a sequence of functions converging weakly in Lp (G) to g. In addition assume that (gk )k∈N converges also λ(n) -a.e. to h. It follows that g = h λ(n) -a.e. Proof. Combining the statements of Egorov’s and Lusin’s theorem, for  > 0 we can find a closed set G ⊂ G such that λ(n) (G\G ) <  and (gk )k∈N converges on G uniformly to h and g|G is continuous and bounded. This implies also that (h − g)χG ∈ L∞ (G), i.e. (h − g)χG ∈ Lq (G) for all q ≥ 1. The uniform convergence of (gk )k∈N on G to h yields Z Z (n) lim gk (h − g) dλ = h(h − g) dλ(n) , k→∞

G

G

while the weak convergence of (gk )k∈N to g implies Z Z (n) lim gk (h − g) dλ = lim gk (h − g)χG dλ(n) k→∞ G k→∞ G  Z = g(h − g) dλ(n) . G

249

A COURSE IN ANALYSIS

Subtracting these lines gives for every  > 0 Z (h − g)2 dλ(n) 0= G

which in turn implies h = g λ(n) -a.e. Combining Proposition 12.13 with Theorem 7.35 we get Corollary 12.14. Let G ∈ B(n) , λ(n) (G) < ∞. If a sequence (gk )k∈N , gk ∈ Lp (G), 1 < p < ∞, converges weakly to g ∈ Lp (G) and in measure to h then g = h λ(n) -a.e. Results such as in Proposition 12.13 or its corollary are important tools in the analysis of (non-linear) partial differential equations when combined with certain compactness results. Our next aim is to characterise pre-compact sets in the spaces Lp (G), G ∈ B(n) . The space Lp (Rn ), 1 ≤ p < ∞, is a Banach space, hence a complete metric space and compact subsets of Lp (Rn ) are defined. We are longing to find a characterisation of all relatively compact sets K ⊂ Lp (Rn ), i.e. sets whose closures are compact, compare with Definition II.3.15.A. An important tool to check relative compactness is the notion of totally bounded sets (see Definition II.3.15.B): a subset Y of a metric space (X, d) is called totally bounded if for every  > 0 there exists a finite covering (Uj )j=1,...,M , M ∈ N, of Y with open sets Uj ⊂ X all having diameter diam Uj < . We can reformulate this definition in the following way: for S every  > 0 exist finitely many points x1 , . . . , xM such that Y ⊂ M j=1 B (xj ). The set {x1 , . . . , xM } is often called an -net for Y . For every -net for Y we have for all x ∈ Y  d(x, {x1 , . . . , xM }) = inf d(x, xj ) j = 1, . . . , M < . (12.6) Now we can state Theorem II.3.18 in the following way

Theorem 12.15. For a subset Y of a metric space (X, d) the following are equivalent i) Y is compact; ii) every infinite sequence (xk )k∈N , xk ∈ Y , has at least one accumulation point x ∈ Y ; 250

12

SELECTED TOPICS

iii) Y is complete (as metric subspace) and totally bounded; iv) Y is complete and for every  > 0 there exists an -net for Y . The metric space we are interested in is Lp (Rn ), 1 ≤ p < ∞, which is a Banach space, hence complete, but it is an infinite dimensional space. We have already seen a compactness result for an infinte dimensional Banach space, namely the Arzela-Ascoli theorem, Theorem II.14.25. We will use the Arzela-Ascoli in our proof for a compactness result for Lp (Rn ). Recall that equi-continuity, Definition II.14.24, was one condition a relatively compact subset of C(Y ), Y a compact metric space, must satisfy. For Lp -spaces we need a similar condition and we start by investigating the continuity of the translation operator Th in Lp (Rn ) as well as the averaging operator Z 1 (Th u)(x) dh, (12.7) (MR u)(x) := (n) λ (BR (0)) BR (0) R > 0 and u ∈ Lp (Rn ). In Chapter 10 we have seen that for u ∈ Lp (Rn ) the translation operator (Th u)(x) = u(x + h), h ∈ Rn , is well defined and hence it makes sense to consider the linear mapping Th : Lp (Rn ) → Lp (Rn ). The translation invariance of λ(n) yields of course Z Z p p (n) |u(x)|p λ(n) (dx) = kukpLp . (12.8) |u(x + h)| λ (dx) = kTh ukLp = Rn

Rn

Theorem 12.16. Let 1 ≤ p < ∞ and Th as well as MR be defined as above. A. The operator Th maps to Lp (Rn ) boundedly into itself and is continuous with respect to h as h → 0 in the sense that lim kTh u − ukLp = 0.

(12.9)

h→0

B. For R > 0 fixed the operator MR satisfies for all u ∈ Lp (Rn ), 1 ≤ p < ∞, the estimate (12.10) kMR u − ukLp ≤ sup kTh u − ukLp khk≤R

as well as kMR uk∞ ≤



1 kukpLp (n) λ (BR (0)) 251

 p1

.

(12.11)

A COURSE IN ANALYSIS

Proof. A. For  > 0 we can find ϕ ∈ C0 (Rn ) such that ku − ϕkLp < 3 , recall C0 (Rn ) is dense in Lp (Rn ). Suppose that supp ϕ ⊂ BR (0). The function ϕ is uniformly continuous and therefore we can find δ, 0 < δ < 1, such that khk < δ implies  |ϕ(x + h) − ϕ(x)| < 3



1 (n) λ (BR (0))

 p1

.

For h ∈ Rn such that khk < δ we find Z

Rn

p

|ϕ(x + h) − ϕ(x)| dx =

Z

BR (0)

|ϕ(x + h) − ϕ(x)|p dx
0 fixed we have with V (R) := λ(n) (BR (0)) and 1 + 1p = 1 that q

kMR u − ukLp = =

≤ ≤

p ! p1 1 Z ((Th u)(x) − u(x)) dh dx V (R) n R BR (0) ! p1 Z Z p 1 (u(x + h) − u(x)) dh dx Rn V (R) BR (0) p ! p1 Z Z 1 1 · (u(x + h) − u(x)) dh dx V (R) Rn BR (0)  p ! !  p1 Z Z Z q 1  q p 1 dh |u(x + h) − u(x)| dh dx , V (R) Rn BR (0) BR (0) Z

where in the last step we used H¨older’s inequality. Thus we arrive by using 252

12

SELECTED TOPICS

Fubini’s theorem at kMR u − ukLp ≤ V (R) = V (R)

1 −1 q

− p1

1

=

Z

Z

Rn

Z

BR (0)

BR (0) 1 p

  p1 |u(x + h) − u(x)| dh dx p

Z

1 p |u(x + h) − u(x)| dx dh n p

R

1

λ(n) (BR (0)) p sup kTh u − ukLp

λ(n) (BR (0)) = sup kTh u − ukLp ,

khk≤R

khk≤R

which proves (12.10). In order to derive (12.11) we can use parts of the calculations made above to find Z 1 |u(x + h)| dh |(MR u)(x)| ≤ V (R) BR (0) Z  p1 1 1 p ≤ |u(x + h)| dh (V (R)) q V (R) BR (0)  p1 Z 1 −1 p |u(x + h)| dh ≤ V (R) q =



Rn

1 kukpLp V (R)

 p1

.

Note that the proof ofR (12.10) and (12.11) for p = 1 follows when we agree to consider in this case |uv| dλ(n) ≤ kuk∞ kvkL1 , u ∈ L∞ (Rn ) and v ∈ L1 (Rn ) also as H¨older’s inequality with p = 1 and q = ∞. We can now prove the Kolmogorov-Riesz theorem which characterises relatively compact sets in Lp (Rn ), 1 ≤ p < ∞. Theorem 12.17. A subset K ⊂ Lp (Rn ), 1 ≤ p < ∞, is relatively compact in Lp (Rn ) if and only if the following three conditions hold: i) K is bounded in Lp (Rn ), i.e. sup kukLp ≤ M < ∞;

u∈K

253

(12.12)

A COURSE IN ANALYSIS

ii) K is equi-continuous in Lp (Rn ), i.e. lim sup kTh u − ukLp = 0;

(12.13)

h→0 u∈K

iii) lim sup

R→∞ u∈K

Z

{ (0) BR

|u(x)|p λ(n) (dx)

! p1

= 0.

(12.14)

Proof. Assume first that K is relatively compact. Since K is compact in Lp (Rn ) it must be bounded, hence K ⊂ K must be bounded. Furthermore K is totally bounded, i.e. for  > 0 we can find an 2 -net {u1 , . . . , uN } for K such that for every u ∈ K there exists j ∈ {1, . . . , N} such that ku − uj kLp < 2 . Now let ϕ1 , . . . , ϕN ∈ C0 (Rn ) be such that kuj − ϕj kLp < 2 , since C0 (Rn ) is dense in Lp (Rn ), 1 ≤ p < ∞, such functions exists. We may also assume that supp ϕj ⊂ BR (0) for some R > 0 and all j = 1, . . . , N. For u ∈ K it follows that Z

!1

p

p

{ (0) BR

|u(x)| dx

≤ =

Z

{ (0) BR

Z

{ (0) BR

!1

p

p

|u(x) − ϕj (x)| dx

+

!1

Z

{ (0) BR

!1

p

p

|ϕj (x)| dx

p

|u(x) − ϕj (x)|p dx

≤ ku − ϕj kLp ≤ ku − uj kLp + kuj − ϕj kLp < ,

implying that lim sup

R→∞ u∈K

Z

{ (0) BR

! p1

|u(x)|p dx

= 0,

i.e. iii) holds. We now prove ii). For this we will use Theorem 12.16.A. Note that we cannot just apply this theorem to every single function u ∈ K since we long for uniformity with respect to u ∈ K in (12.13). Given  > 0 let {u1 , . . . , uN } be now an 3 -net of K. It follows that kTh u − ukLp ≤ kTh u − Th uj kLp + kTh uj − uj kLp + kuj − ukLp ≤ 2ku − uj kLp + kTh uj − uj kLp , 254

12

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where we used that kTh vkLp = kvkLp for every v ∈ Lp (Rn ) and Th u − Th uj = Th (u − uj ). In light of Theorem 12.16.A we may choose δ > 0 such that khk < δ implies for all j = 1, . . . , N that kTh uj − uj kLp < 3 and we arrive at sup kTh u − ukLp < 

u∈K

for all khk < δ, i.e. ii) is shown. We are now going to prove the more important converse statement. Condition ii) requires limh→0 kTh u − ukLp = 0 uniformly for u ∈ K, hence by (12.10) we have limh→0 kMR u − ukLp = 0 uniformly for u ∈ K. Fix R > 0 and pick x1 , x2 ∈ Rn . It follows with V (R) := λ(n) (Br (0)) that Z 1 |MR u(x1 ) − MR u(x2 )| ≤ |u(x1 + h) − u(x2 + h)| dh V (R) BR (0)  p1 Z 1 p |u(x1 + h) − u(x2 + h)| dh ≤ 1 BR (0) V (R) p Z  p1 1 p ≤ |u(x1 + h) − u(x2 + h)| dh 1 Rn V (R) p 1 p = 1 kTx1 u − Tx2 ukL V (R) p 1 p p ≤ 1 (kTx1 u − ukL + kTx2 u − ukL ) , λ(n) (BR (0)) p and now (12.13) implies the equi-continuity of the function MR u for R > 0 fixed. In particular MR is continuous. Moreover (12.11) yields that the family {MR u | u ∈ K}, R > 0 fixed, is bounded with respect to the sup-norm k.k∞ . We consider the set o n DR := v = χBR (0) · MR u u ∈ K .

  This set can be considered as a subset of C BR (0) and in the Banach     space C BR (0) , k.k∞ this set DR is bounded and equi-continuous. Since

BR (0) is compact, by Arzela-Ascoli theorem, Theorem II.14.25, the set DR is 255

A COURSE IN ANALYSIS

relatively compact in C(BR (0)). Now we prove that K is relatively compact in Lp (Rn ), 1 ≤ p < ∞. Given  > 0. By iii) we can find R0 > 0 such that ! p1 Z  (12.15) |u(x)|p dx < , { (0) 3 BR 0

and further we can find R1 > 0 such that for all u ∈ K  kMR1 u − ukLp < (12.16) 3   holds. Moreover, in C BR0 (0) the set DR0 is relative compact. Hence there exist functions u1 , . . . , uM ∈ K such that for every u ∈ K there exists j ∈ {1, . . . , M} such that for all x ∈ BR0 (0) we have |MR1 u(x) − MR1 uj (x)| ≤

1  . 3 V (R0 ) p1

For j = 1, . . . , M we define gj := χBR (0) MR1 uj and we claim that {g1 , . . . , 0 gM } is an -net for K. Let u ∈ K and pick uj , 1 ≤ j ≤ M, as above. Then it follows Z

ku − gj kLp = ≤ 

+



≤ ≤

Z

{ (0) BR

{ (0) BR 0

Z

3

0

|u(x)| dx +

Z

p

BR0 (0)

! p1

|u(x) − MR1 uj (x)| dx

|u(x)|p dx

BR0 (0)

  p

p

! p1

|u(x) − MR1 u(x)|p dx 

+ ku − MR1 ukLp +

+

  p 3

Z

BR0 (0)

1 V (R0 )

 p  p1   1 2 (1 + 2p ) p  p  < , + = 3 3 3

Z

! p1 p ! p1

|MR1 u(x) − MR1 uj (x)|p dx

BR0 (0)

! p1 p  p1 1 dx  



i.e. the set {g1 , . . . , gM } is indeed an -net for K and the theorem is proved. Remark 12.18. Our proof of Theorem 12.17 is a modification and an adaption of the proof given in [90] for the case p = 2. 256

12

SELECTED TOPICS

Example 12.19. Consider on L2 (R) the sequence of functions uk (x) := χ[0,2π] (x) sin kx. From Example 12.11 we deduce that kuk − ul k = 2π, hence no subsequence of (uk )k∈N will ever be a Cauchy sequence, and hence (uk )k∈N has no accumulation point, i.e. {uk | k ∈ N} is not relatively compact in L2 (R). With the arguments of Example 12.11 it still follows that (uk )k∈N 2 converges weakly in L√ (R) to 0. The calculations of Example 12.11 also yields that kuk kL2 = 2π, i.e. the set {uk | k ∈ N} is uniformly bounded in L2 (R) and since uk |[0,2π]{ = 0, condition (12.14) in Theorem 12.17 holds. However the equi-continuity of {uk | k ∈ N} fails to hold since Z

2

R

|uk (x + h) − uk (x)| dx =

Z

2π

0

| sin k(x + h) − sin kx|2 dx

= 2π − 2π cos hk

√ and therefore lim suph→0 (supk∈N kTh uk − uk kL2 ) = 2 π.

Problems 1. Let G ⊂ Rn C 1 -mapping. critical point have at most

be a bounded open set, y0 ∈ Rn , and f : G → Rn a Consider the equation f (x) = y0 and suppose that no of f solves this equation. Prove that then f (x) = y0 can finitely many solutions in G.

2. Let h := [0, 1] → R be a continuously differentiable function and denote by Ex(h) the set of all local extreme values of h, i.e. Ex(h) = {h(x)|x ∈ [0, 1], h has a local extreme value at x}. Clearly we have Ex(h) ⊂ h(crit(h|(0,1) ))∪{h(0), h(1)}. Prove the existence of a function h ∈ C 1 ([0, 1]) such that Ex(h) is finite but h(crit(h|(0,1) )) is nondenumerable. Hint: consider for the Cantor set C the R xfunction g : R → R, g(x) := dist(x, C) and then investigate h(x) := 0 g(t)dt, x ∈ [0, 1]. 3. Why is the trace of every parametrized C 1 -surface f : G → R3 , G ⊂ R2 open, a set of λ(3) -measure zero?

4. Let A ∈ B(n) be a bounded set and s : A → R a simple function. Prove that for every  > 0 there exists a closed set C ⊂ A such that λ(n) (A \ C ) ≤  and s|C is continuous. 257

A COURSE IN ANALYSIS

5. Let f : A → R, A ∈ B(n) . Suppose that f has the following property (compare with Lusin’s theorem, Theorem 12.7): for every  > 0 there exists a closed set C ⊂ A such that λ(n) (A \ C ) <  and f |C is continuous. Prove that f is a measurable function. 6. Prove that if a subset Y of a metric space (X, d) has the property that for every SN δ > 0 there exist closed balls Bδ (x1 ), . . . , Bδ (xN ) such that Y ⊂ j=1 Bδ (xj ) then Y admits for every  > 0 a finite -net. 7.

a) Prove that (a, b) ⊂ R has for every  > 0 a finite -net.

b) Consider the Banach space (C([0, 1]), k · k∞) and its subset Y := {u ∈ C([0, 1])| 0 = u(0) ≤ u(t) ≤ u(1) = 1}. Prove that Y ⊂ C([0, 1]) is bounded and that for  = 21 the set Y admits a finite 12 -net. Hint: use Problem 6 and consider the function u0 ∈ C([0, 1]), u0 (t) = 21 for all t ∈ [0, 1]. Further show that for 0 <  < 21 the set Y does not have an -net. Hint: suppose that for 0 <  < 21 the set {u1 , . . . , um} is a finite -net for Y . Investigate the function u : [0, 1] → R defined by u(0) = 0, u(1) = 1 (
1   j 1, uj m+1 ≤ j
12 , := u j m+1 0, uj m+1 > 2
j j+1 for j = 1, . . . , m, and for t ∈ m+1 , j = 0, . . . , m, we define u as , m+1



j j+1 the linear function connecting j, u m+1 and j + 1, u m+1 . This problem is taken from [7].

258

Part 7: Complex-valued Functions of a Complex Variable

259

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13

The Complex Numbers as a Complete Field

Although we assume that the reader is familiar with the complex numbers C from studies about algebra or geometry, we want to discuss in some detail the algebraic properties of C as well as the fact that C equipped with the modulus or absolute value is metric in a complete metric space. In fact C is also a complete algebraic field in the sense that every polynomial factorises into linear factors, but for this we will give a proof much later in Chapter 23 and our proof will be analytic not algebraic. This chapter also serves to fix notations and to make a start with getting used to the traditional fa¸con de parler in the theory of complex functions of a complex variable. A complex number z ∈ C is written as z = x + iy, x, y ∈ R

(13.1)

x = Re z and y = Im z

(13.2)

where

are the real and the imaginary part of z, respectively. Thus we can identify the complex numbers with pairs (x, y) of real numbers x, y, hence we can identify C with R2 using the bijective mapping j : C 7→ R2 , z = x + iy 7→ j(z) = (x, y).

(13.3)

This identification will soon become very useful when discussing the topology of C. However to handle the algebraic structure of C another identification is more useful, namely  n o x −y h : C → H := x, y ∈ R ⊂ M(2, R) (13.4) y x   x −y . z = x + iy → 7 h(z) = y x

Let us study the algebraic structure of H with respect to the standard matrix 261

A COURSE IN ANALYSIS

operations. It is an easy exercise to see  ad the standard  matrix  that H with x −y −x y and is inverse to dition forms an Abelian group where −y −x y x  0 0 . Moreover, the neutral element with respect to addition is of course 0 0   o n 0 0 is also an Abelian group with respect to matrix multiplicaH\ 0 0       1 0 x −y x y 1 is the neutral and is inverse to tion where x2 +y2 0 1 y x −y x element with respect to multiplication. Since matrix multiplication is in general not commutative ofH that   it  is worth noting for two elements   x −y u −v xu − yv −xv − yu u −v x −y , = = y x v u xv + yu xu − yv v u y x i.e. restricted to H the multiplication of matrices is commutative. We also have the law of distributivity, i.e. !      c −d a −b x −y + d c b a y x =

!   a −b x −y + b a y x

!   c −d x −y , d c y x

which is easily verified by inspection. Thus H with these operations is a commutativefield.  ! x −y = x + iy = z, we now find with zk = xk + iyk = Using h−1 y x  ! xk −yk −1 that h yk xk z1 + z2 = (x1 + x2 ) + i(y1 + y2 )

(13.5)

or Re(z1 + z2 ) = Re z1 + Re z2 and Im(z1 + z2 ) = Im z1 + Im z2 ,

(13.6)

as well as z1 + z2 = z2 + z1 , z1 · z2 = z2 · z1 ,

(z1 + z2 ) + z3 = z1 + (z2 + z3 ), (z1 · z2 )z3 = z1 · (z2 · z3 ), 262

(13.7) (13.8)

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

and z1 (z2 + z3 ) = z1 z2 + z1 z3 . The neutral element with respect to addition is given by !  0 0 −1 0 := 0 + i 0 = h 0 0 and the neutral element with respect to multiplication is given by !  1 0 . 1 := 1 + i 0 = h−1 0 1

(13.9)

(13.10)

(13.11)

We will not introduce different symbols for the zero element (neutral element with respect to addition) in R, C or H and usually we also take 0 as the symbol for the zero element in R2 . Similarly 1 denotes the neutral element with respect to multiplication in R and C, however the corresponding element in H (or M2 (R)) we often denote id or idH . For z1 · z2 we find i.e.

z1 · z2 = x1 x2 − y1 y2 + i(x2 y1 + x1 y2 ),

(13.12)

Re z1 · z2 = x1 x2 − y1 y2 , Im z1 · z2 = x2 y1 + x1 y2 .

(13.13)

For z1 6= 0 it follows that z1−1

x1 −y1 1 = 2 +i 2 = 2 z1 x1 + y1 x1 + y12

!

=

x21

x1 y1 −i 2 2 + y1 x1 + y12

(13.14)

and we have z2 = z2 · z1−1 = z1−1 · z2 . z1

(13.15)

The most interesting relations are of course −1

i = 0 + i.1 = h

263

!  0 −1 1 0

(13.16)

A COURSE IN ANALYSIS

implying i2 = −1

(13.17)

where we adopted the notation z1k = z1 · ......· zk , for k ∈ Z, with the convention z10 = 1.    x 0 By Re H := x ∈ R a subfield of H is given and we can identify 0 x Re H via h−1 with R, hence we can identify R with a subfield of C and we simply consider R as a subset, i.e. R ⊂ C, with the interpretation that  z ∈R x 0 . means z = x + i 0 or z is identified with (x, 0), or as we started with 0 x When comparing or identifying C with R2 , i.e. using the mapping j, we can use the additive group operation on C and the multiplication with scalars from R ⊂ C to consider C as a two dimensional R-vector space, but we do not have a multiplication on R2 . From now on, as in the beginning, we write z = x + iy for a complex number, and for z1 · z2 we write simply z1 z2 . We will rarely make explicit use of the identification of C by H, and the identification of C by R2 will be used and is in fact of importance when dealing with the geometry and the topology of C, but again, in the algebraic context it is rarely used. On C we have a further algebraic operation, we may consider the conjugate complex number z¯ := x + iy = x − iy,

(13.18)

Re z¯ = Re z, Im z¯ = − Im z.

(13.19)

(¯ z ) = z,

(13.20)

i.e. Obviously we have i.e. the mapping z 7→ z¯ is an involution and we can look at C as an algebra with involution. The following rules follow from the definition and are easily verified by inspection z1 + z2 = z¯1 + z¯2 ,

(13.21)

z1 · z2 = z¯1 · z¯2 ,

(13.22)

264

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

z1 z¯1 ) = , z2 6= 0, z2 z¯2

(13.23)

z· z¯ = (Re z)2 + (Im z)2 = x2 + y 2 ,

(13.24)

(

Re z =

z + z¯ z − z¯ , Im z = , 2 2i

(13.25)

see also Problem 3. Note that for z 6= 0 we have z¯ z· z¯ = z· = 1, z· z¯ z· z¯ i.e. z −1 =

y x z¯ −i 2 , = 2 2 z· z¯ x + y x + y2

(13.26)

where the last equality is of course (13.14). It is important to become familiar in manipulating complex numbers. In particular, as we will suggest in some problems, i.e. Problems 1 and 2, to get experience with handling fractions of complex numbers in an efficient way. Definition 13.1. The modulus or absolute value |z| of z = x + iy ∈ C is defined by |z| :=

√

z¯ z=

Note that (13.27) implies

p

(Re z)2 + (Im z)2 =

p x2 + y 2 .

(13.27)

|Re z| ≤ |z| and |Im z| ≤ |z|

(13.28)

|z|2 = z¯ z ≥ 0.

(13.29)

|z1 · z2 | = |z1 ||z2 |

(13.30)

and

Furthermore we have which follows from (z1 z2 )(z1 z2 ) = (z1 z¯1 )(z2 z¯2 ), and for z2 6= 0 we find z1 |z1 | = (13.31) z2 |z2 | . 265

A COURSE IN ANALYSIS

For z ∈ R, i.e. Im z = 0, we have of course |z| =

√

x2 = |x|

which justifies the name absolute value. Identifying C with R2 , i.e. z = x + iyp with (x, y), we note that |z| is nothing but the Euclidean norm ||(x, y)||2 = (x2 + y 2). Consequently we have |z| ≥ 0 and |z| = 0

if and only if z = 0

(13.32)

and the triangle inequality

|z1 + z2 | ≤ |z1 | + |z2 |

(13.33)

holds. Moreover, the R-positive homogeneity of the Euclidean norm extends to multiplication with complex numbers, i.e. the equality ||λ(x, y)||2 = |λ|k(x, y)k2 or |λz| = |λ||z|

for λ ∈ R and (x, y) ∈ R2 , z = x + iy, extends to

|z1 · z2 | = |z1 ||z2 |, z1 , z2 ∈ C.

(13.34)

Definition 13.2. Let V be a vector space over the field C. A norm on V is a mapping ||.|| : V → R satisfying i) ||w|| ≥ 0 and ||w|| = 0 if and only if w = 0, w ∈ V , ii) ||zw|| = |z|||w||, z ∈ C, w ∈ V , iii) ||w1 + w2 || ≤ ||w1|| + ||w2||. Whenever ||.|| is a norm on V the converse triangle inequality holds |||w1|| − ||w2 ||| ≤ ||w1 − w2 ||.

(13.35)

d(w1 , w2) := ||w1 − w2 ||

(13.36)

Moreover by a metric d is given on V , see Problem 6. Thus the modulus |.| is a norm on C (which is a of course a C-vector space of complex dimension 1) and hence it induces a metric d on C. For z1 , z2 ∈ C we find p d(z1 , z2 ) = |z1 − z2 | = Re(z1 − z2 )2 + Im(z1 − z2 )2 266

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

or with z1 = (x1 , y1 ) and z2 = (x2 , y2 ) we have p d(z1 , z2 ) = (x1 − x2 )2 + (y1 − y2 )2 ,

(13.37)

i.e. d is just the Euclidean metric of R2 . We can now consider convergence in the metric space (C, d) which must correspond to convergence in the Euclidean space R2 . Since we fix on C the metric d, we will write just C instead of (C, d) when meaning the metric space (C, d). In order to get a better understanding of how to handle complex numbers, the following considerations are much more detailed than needed given our identification of the metric space C with the metric space R2 . We start with Definition 13.3. Let (zn )n∈N , zn ∈ C, be a sequence of complex numbers and z ∈ C. We call z the limit of (zn )n∈N and we write lim zn = z

n→∞

(13.38)

if for every  > 0 there exists N = N() such that n ≥ N() implies |zn − z| = d(zn , z) < . Since in metric space every limit is unique, z is uniquely determined. Moreover, we call a sequence (zn )n∈N bounded if |zn | ≤ M for all n ∈ N and some M ≥ 0, and it follows that every convergent sequence is bounded. Lemma 13.4. Let (zn )n∈N , zn = xn + iyn ∈ C, and z = x + iy be the limit of (zn )n∈N . Then we have lim xn = x and lim yn = y.

n→∞

n→∞

(13.39)

Proof. First we recall (13.28), i.e. for every complex number w ∈ C the following holds p |Re w| ≤ (Re w)2 + (Im w)2 = |w|

and

|Im w| ≤ Now we find

p

(Re w)2 + (Im w)2 = |w|.

|xn − x| ≤ |zn − z| 267

(13.40)

A COURSE IN ANALYSIS

as well as |yn − y| ≤ |zn − z|.

(13.41)

lim xn = x and lim yn = y.

(13.42)

Hence, given  > 0 and chosen N() such that n ≥ N() implies |zn − z| <  it follows also for n ≥ N() that |xn − x| <  and |yn − y| < , i.e. n→∞

n→∞

Suppose now that two sequences (xn )n∈N , (yn )n∈N of real numbers xn , yn ∈ R are given and that with x, y ∈ R we have in R that limn→∞ xn = x and limn→∞ yn = y. For the sequence (zn )n∈N of complex numbers zn := xn + iyn we find |zn − z|2 = |xn − x|2 + |yn − y|2 or

|zn − z| ≤ (|xn − x| + |yn − y|).

Thus, given  > 0 we can find N = N() such that n ≥ N() implies |xn − x| < 2 as well as |yn − y| < 2 and consequently |zn − z| < , i.e. limn→∞ zn = z. Hence, we have proved. Proposition 13.5. A sequence of complex numbers (zn )n∈N converges to a complex number z if and only if the sequence (Re zn )n∈N converges to Re z and the sequence (Im zn )n∈N to Im z. From here it is easy to derive the basic statements for the algebra of limits in C. Proposition 13.6. Let (zn )n∈N and (wn )n∈N be two sequences of complex numbers zn = xn + iyn ∈ C and wn = un + ivn ∈ C. Suppose that (zn )n∈N converges to z and that (wn )n∈N converges to w. Then we have lim (zn ± wn ) = z ± w,

(13.43)

lim zn · wn = z· w,

(13.44)

n→∞

n→∞

and if w 6= 0 it follows that

zn z lim = . n→∞ wn w 268

(13.45)

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

Proof. We will reduce each statement to a statement involving only the sequences of the real and imaginary parts and then we may apply our results to sequences of real numbers. Of course we have zn ± wn = (xn ± un ) + i(yn ± vn )

(13.46)

and since (xn ± un )n∈N converges to x ± u, and (yn ± vn )n∈N converges to y ± v, the first statement follows. Now zn · wn = xn · un − yn vn + i(un yn + xn vn ) and since

lim (xn un − yn vn ) = xu − yv = Re(z· w)

n→∞

and lim (yn un + xn vn ) = yu + xv = Im(z· w)

n→∞

the second statement follows. Finally we observe for wn 6= 0

zn w¯n zn = wn |wn |2

xn un + yn vn + i(yn un − xn vn ) u2n + vn2 xn un + yn vn yn un − xn vn = + i( ). u2n + vn2 u2n + vn2 =

Since |w|2 = u2 + v 2 6= 0 and limn→∞ (u2n + vn2 ) = u2 + v 2 we deduce first that wn 6= 0 for n sufficiently large, hence u2n + vn2 6= 0 for n sufficiently large, and now it follows that z zn = . lim n→∞ wn w Corollary 13.7. Let (zn )n∈N , zn = xn +iyn , be a sequence of complex numbers converging to z = x + iy ∈ C. Then we have lim z¯n = z¯

(13.47)

lim |zn | = |z|.

(13.48)

n→∞

and n→∞

269

A COURSE IN ANALYSIS

Proof. We need only to note that z¯n = xn − iyn and that |zn | = and that the square root is on R+ a continuous function.

p

x2n + yn2

As in R or in any metric space we can introduce the notion of a Cauchy sequence of complex numbers. Definition 13.8. Let (zn )n∈N be a sequence of complex numbers zn ∈ C. We call (zn )n∈N a Cauchy sequence if for every  > 0 there exists N = N() ∈ N such that n, m ≥ N() implies |zn − zm | < . Corollary 13.9. Let (zn )n∈N be a sequence of complex numbers zn = xn +iyn . The sequence (zn )n∈N is a Cauchy sequence if and only if the sequences (xn )n∈N and (yn )n∈N are Cauchy sequences. Proof. We just have to note the inequalities |Rez| ≤ |z|, |Imz| ≤ |z|, |z| ≤ (|Rez| + |Imz|) to deduce that |zn − zm | <  implies |xn − xm | <  and |yn − ym | <  as well as |xn − xm |
N implies | Proof. Let Sp :=

Pp

k=1

n X

k=m

zk | < .

zk be the pth partial sum. It follows that Sn − Sm−1 = 271

n X

k=m

zk ,

A COURSE IN ANALYSIS

and the criterion is nothing but the Cauchy criterion for the sequence of partial sums. As in the case of series of real numbers, see Theorem I.18.2, we have P Theorem 13.12. If the series ∞ k=1 zk converges, then limn→∞ zn = 0. P∞ Proof. P If k=1 zk converges then by Theorem 13.11, for any  > 0 it follows that | nk=m zk | <  for n ≥ m ≥ N. Taking n = m we find |zn | <  for n ≥ N, i.e. limn→∞ zn = 0. Of great importance is that we can extend the result about the limit of the geometric series from R to C: P k Theorem 13.13. For z ∈ C, |z| < 1, the geometric series ∞ k=0 z converges 1 to 1−z , i.e. ∞ X 1 zk = , |z| < 1. (13.50) 1−z k=0 Moreover, the geometric series converges absolutely.

Proof. As in the real case we find compare with Theorem I. 16.4 n X

zk =

k=0

1 − z n+1 . 1−z

(13.51)

Now we note that lim |z n+1 | = |z| lim |z|n = 0

n→∞

n→∞

since |z| < 1. This implies however that limn→∞ z n+1 = 0 and passing in (13.51) to the limit gives the convergence. Further we note that ∞ X k=0

|z k | =

∞ X k=0

|z|k =

1 , |z| < 1, 1 − |z|

which implies the absolute convergence. More generally, as in the real case, see Theorem I.18.11, we have Theorem 13.14. Any absolutely convergent series of complex numbers is convergent. 272

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

P converges absolutely. According to the Cauchy Proof. Suppose that ∞ k=1 zk P criterion applied to the series ∞ k=1 |zk |, for  > 0 there exists N() ∈ N such that n ≥ m ≥ N() implies n X |zk | < . k=m

Now the triangle inequality yields for n ≥ m ≥ N() |

n X

k=m

zk | ≤

n X

k=m

|zk | < ,

P∞

zk implying convergence. P Theorem 13.15 (Comparison test). Let ∞ k=1 ck be a convergent series of non-negative real numbers ck ≥ 0. Further let (zk )k∈N , zk ∈ C, be a series of complex P numbers such that |zk | ≤ ck for k ∈ N(or k ≥ N0 ). Then the series ∞ k=1 zk converges absolutely.

i.e. the Cauchy criterion holds also for

k=1

Proof. Given  > 0 there exists N() ∈ N such that |

n X

ck | =

n X

|zk | ≤

k=m

n X

ck <  f or n ≥ m ≥ N().

n X

ck <  f or n ≥ m ≥ N()

k=m

Therefore we find

k=m

k=m

which proves the result. Finally we prove the ratio test for the complex series, see Theorem I.18.14 for the real case. P∞ Theorem 13.16. Let k=0 zk be a series of complex numbers such that zn 6= 0 for all n ≥ N0 . Suppose that there exists ν, 0 < ν < 1, such that | Then the series

P∞

k=0

zk+1 | ≤ ν f or all k ≥ N0 . zk

zk converges absolutely. 273

A COURSE IN ANALYSIS

P Proof. The convergence of the series ∞ k=0 zk does not depend on the first N0 terms. Now zk+1 | ≤ ν f or k ≥ N0 | zk P n implies |zN0 +n | ≤ |zN0 |ν n . Since 0 < ν < 1 the series ∞ n=N0 ν converges, and by Theorem 13.15 the result follows.

Problems 1. Simplify the following expressions: 4i ; 3+i

a)

3+5i 2i−7

b)

(2i)8 −128 ; (1+2i)(1−2i)

c)

1−i 2+3i 2+4i 6−2i

d) 2. Verify: a)



+

;

− 12

+

√ 2 3 i . 2

(a+ib)2 −2iab i(a−b)

= −i(a + b), a, b ∈ R, a 6= b;

q 2 2 b) For p, q ∈ R, p4 < q, the numbers z1 = − p2 + i q − p4 and q 2 z2 = − p2 − i q − p4 are solutions to the equation z 2 + pz + q = 0.

3. Verify the following for complex numbers z1 , z2 and z = x + iy: a) z1 · z2 = z1 · z2 ;

b) z · z¯ = x2 + y 2 ; c) Re z =

z+¯ z 2

and Im z =

z−¯ z . 2i

4. Formulate and indicate the proof of the binomial theorem for two complex numbers. 5. Prove the following inequalities: a) ||z| − |w|| ≤ |z − w| ∧ |z + w|, recall that a ∧ b = min{a, b}; b)

|x|+|y| √ 2

≤ |z| ≤ |x| + |y|, for z = x + iy. 274

13

THE COMPLEX NUMBERS AS A COMPLETE FIELD

6.

a) Prove that if V is a vector space over C and k · k is a norm on V then d(f, g) := kf − gk, f, g ∈ V , is a metric on V . 1 P n 2 2 b) For z = (z1 , . . . , zn ) ∈ Cn define kzk := |z | and prove j j=1 n that this gives a norm on C .

7. Find the following limits: 3n2 − 5in ; n→∞ n2 i (k − ik)2 (2k + 5i) b) lim ; k→∞ k 2 + (3 + ik)3

a) lim

c) lim

N →∞

N X k=4

z k , |z| < 1, N ≥ 4.

8. Prove the convergence of the following power series for the range of z ∈ C as indicated. You may use any criterion for convergence we have discussed. ∞ X zk a) , z ∈ C; k! k=0 b)

∞ X k=0

c)

∞ X

(−1)k

z 2k , z ∈ C; (2k)!

(−1)k−1

k=1

d)

∞ X

(−1)k−1

zk , |z| < 1; k

(−1)k−1

z 2k−1 , |z| < 1. 2k − 1

k=1

e)

∞ X k=1

z 2k−1 , z ∈ C; (2k − 1)!

P 9. Let k=0 ak and ∞ k=0 bk be two absolutely convergent P∞ series of complex numbers. Define their Cauchy product as n=0 cn where cn := Pn k=0 an−k bk . Prove that the Cauchy product converges absolutely and find its value. Hint: compare with Theorem I.29.21. P∞

275

A COURSE IN ANALYSIS

10. Define for z ∈ C the number e := z

∞ X zk

which is justified by Problem k! 8 a). For z1 , z2 ∈ C prove with the help of Problem 4 and Problem 9 the functional equation of the exponential function: ez1 +z2 = ez1 ez2 . k=0

11. Let QN (ak )k∈N be a sequence of complex numbers ak 6= 0 and set PN := ak . In the case where P := limN →∞ PN exists, we write P = Qk=1 ∞ k=1 ak and call P the infinite product of the sequence (ak )k∈N . Prove: Q a) If ∞ k=1 ak converges then limk→∞ ak = 1; Q b) If the product ∞ k=1 ak converges then for every  > 0 there exists N = N() ∈ N such that n > m > N() implies n Y ak − 1 < . k=m+1

(Compare with Proposition I.30.7.)

12. Let (ak )k∈N , ak 6= −1,Qbe a sequence of complex numbers. We say that the infinite product ∞ k=1 (1 + ak ) converges absolutely if the product Q ∞ (1 + |a |) converges. k k=1 Q a) Prove that if ∞ k=1 (1+ak ) converges absolutely, then it converges. Q∞ if b) Prove P∞that k=1 (1 + ak ) converges absolutely if andPonly ∞ a |a | converges, i.e. if and only if the series the series k=1 k k=1 k converges absolutely. Hint: compare with Proposition I.30.9 and Proposition I.30.10.

276

14

A Short Digression: Complex-valued Mappings

Part 7 is essentially devoted to functions defined on some subset D ⊂ C with values in C. The fact that C is a field allows us to consider difference (z0 ) and hence complex differentiation which is significantly quotients f (z)−f z−z0 different from differentiating functions g : G → R2 , G ⊂ R2 . However, before turning to this theory, in this chapter we want to discuss mappings f : X → C where X might be just a non-empty set or a subset of Rn , n ≥ 1, or a metric space, etc. We want to see how the additional algebraic structure of C gives new results or allows to reformulate results for mappings f : X → R2 . Clearly, the set of all mappings f : X → C, X 6= ∅, equipped with the natural operations (f ± g)(x) = f (x) + g(x) (λf )(x) = λf (x) (f · g)(x) = f (x)g(x)

(addition of complex numbers) (λ ∈ C) (multiplication of complex numbers),

is an algebra over C which allows the involution f 7→ f . Since for x ∈ X a complex number is given by f (x) we can introduce its real and imaginary part u(x) and v(x), respectively, and we write f (x) = u(x) + i v(x)

(14.1)

where u, v : X → R are real-valued mappings. For two mappings fk : X → C, fk = uk + i vk , we find Re (f1 + f2 ) = u1 + u2 and Im (f1 + f2 ) = v1 + v2 .

(14.2)

Moreover, the modulus of f = u + iv is given at x ∈ X by |f (x)| = (u2 (x) + v 2 (x))1/2 .

(14.3)

For a sequence (fk )k∈N , fk : X → C, we can define pointwise convergence to a function f : X → C by lim fk (x) = f (x) for all x ∈ X,

k→∞

(14.4)

and uniform convergence on X to f by lim (sup |fk (x) − f (x)|) = lim ||fk − f ||∞ = 0,

k→∞ x∈X

k→∞

277

(14.5)

A COURSE IN ANALYSIS

where we have introduced the sup-norm of f : X → C as ||f ||∞ := sup |f (x)|.

(14.6)

x∈X

In Problem 1 we will discuss norms and scalar products on vector spaces over C and see that ||f ||∞ is indeed a norm on the vector space of all mappings f : X → C satisfying that ||f ||∞ < ∞. Depending on additional structures on X we may carry over further properties and ideas from the real case to the complex case. The first case we want to discuss is the case where on X a σ - field and a measure is given. Instead of writing X we now deal with a measure space (Ω, A , µ) and look at complexvalued mappings f : Ω → C. If not stated otherwise we consider on C the Borel σ - field B(2) induced when identifying C with the Euclidean space R2 . It follows immediately that f : Ω → C, f = u + iv, is A/B(2) measurable if and only if u and v are A/B(1) measurable. For this we only need to note that the projections pr1 : C → R, z 7→ Re z, pr2 : C → R, z → Im z, are measurable since after identification of C with R2 they coincide with the projections prk : R2 → R, k = 1, 2. In the case where f , hence u and v, are measurable it is natural to extend integrability, more precisely µ - integrability, of f by Z Z Z f (w) µ(dw) := u(w) µ(dw) + i v(w) µ(dw), (14.7)

R which we define for µ-integrable function u and v. Note that f dµ is now a complex number. All µ-integrable complex-valued functions f : Ω → C form a vector space and f is µ-integrable if and only if |f | is µ-integrable. We can further define the (semi-)norms ||f ||Lp :=

note that ||f ||Lp =

Z

Z

p

|f (w)| µ(dw)

2

1/p

2 p/2

(|u(w)| + |v(w)| )

,

µ(dw)

(14.8) 1/p

where f = u + iv. This leads to the complex vector spaces Lp (Ω ; C) and Lp (Ω ; C), and if no confusion may arise we often will write just as before Lp (Ω) and Lp (Ω), respectively. As we will see in Problem 2, typical inequalities such as H¨older’s inequality or the Minkowski’s inequality carry over to 278

14

A SHORT DIGRESSION: COMPLEX-VALUED MAPPINGS

the case of complex-valued mappings. Moreover, on L2 (Ω ; C) a scalar product is given by < f, g >=

Z

f (w) g(w) µ(dw).

(14.9)

Indeed, we can also prove that for 1 ≤ p < ∞ the spaces Lp (Ω ; C) are Banach spaces, i.e. complete with respect to the norm ||.||Lp . Next we assume that X carries some topology OX . In this case continuity of f : X → C is well defined when on C we have as usual chosen the 2 dimensional Euclidean topology induced by the identification of C with R2 . As expected we find that f = u + i v is continuous if and only if u and v are continuous. In the following assume that (X, OX ) is a Hausdorff space so that compactness is well defined. If K ⊂ X is compact and f : X → C is continuous then f |K is bounded, i.e. |f (x)| ≤ M for all x ∈ K and some M < ∞. Further, if X is even a metric space and K ⊂ X compact, then f |K is uniformly continuous. If X is a Hausdorff space and fk : X → C is a sequence of functions we call (fk )k∈N locally uniform or uniformly on compact sets convergent to f : X → C if for every compact set K ⊂ X the following holds lim sup |fk (x) − f (x)| = 0. (14.10) k→∞ X∈K

Locally uniform convergence will become a very important notion later on when studying complex-differentiable mappings f : D → C, D ⊂ C. In general we denote by C(X; C) the set of all continuous functions f : X → C which is of course a vector space over C, and as before, if no confusion may arise we just write C(X). The space Cb (X; C) is the subspace of all bounded continuous function on X and equipped with the sup-norm ||.||∞ it is again a Banach space. For a Hausdorff space (X, OX ) we denote by C0 (X; C) the space of all continuous functions f : X → C with compact support, i.e. supp f := {x ∈ X|f (x) 6= 0} is compact for f ∈ C0 (X; C). In the topological space (X, OX ) we can always consider the Borel σ - field B(X) := σ(OX ). Now the properties of OX will determine the relations of the spaces Lp (X; C), C(X; C), Cb(X; C) or C0 (X; C) and we do not want to discuss these details now, some will be discussed in Part 12, i.e. Volume IV. In the special case of Rn equipped with the Euclidean topology we can 279

A COURSE IN ANALYSIS

also consider the space C∞ (Rn ; C) of all continuous complex-valued funtions f : Rn → C vanishing at infinity, i.e. functions for which we have lim |f (x)| = 0.

||x||→∞

˚ 6= ∅, we can consider functions f : G → C with real Finally, if G ⊂ Rn , G partial derivatives, i.e. for α ∈ Nn0 we may look at D α f (x) = D α u(x) + i D α v(x).

(14.11)

Thus spaces such as C m (G; C) or Cbm (G; C) as well as C0m (G; C), C ∞ (G; C) or C0∞ (G; C) are all well defined. But even in the case n = 2 we make and have to make a strict distinction between functions having some real partial derivatives at a point (or in G) and functions admitting a complex derivative as we will introduce it soon. Once more, if H(X; C) is a certain space of functions f : X → C and no confusion may arise, we will write simply H(X).

Problems 1. P a) For z, w ∈ Cn define the unitary scalar product by hz, wi := n k=1 zk wk and prove the Cauchy-Schwarz inequality |hz, wi| ≤ kzkkwk

where kzk = (

Pn

k=1

1

|zk |2 ) 2 .

b) Show that for a compact set K ⊂ Rn norms are given on C(K; C) R
1 by kuk∞ := supx∈K |u(x)| and kukL2 := K |u(x)|2 dx 2 .

2. Let f, g ∈ C(K; C) where K ⊂ Rn is a compact set. Sketch a proof of the H¨older and the Minkowski inequalities, i.e. Z 1 1 |f (x)g(x)|dx ≤ kf kLp kgkLq , 1 < p < ∞ and + = 1, p q K and kf + gkLp ≤ kf kLp + kgkLp , 1 ≤ p < ∞. 280

14

A SHORT DIGRESSION: COMPLEX-VALUED MAPPINGS

3. Let (Ω, A) be a measurable space and f : Ω → C a function. Prove that f is measurable if and only if Re f and Im f are measurable. 4. Let ∅ = 6 G ⊂ Rn be an open set and f : G → C, f = u + iv, a function. Define differentiability of f at x0 as follows: there exists a ∈ Cn and ϕx0 (x) = 0 and ϕx0 defined in a neighbourhood of x0 such that lim x→x0 kx − x0 k f (x) = f (x0 ) = ha, (x − x0 )i + ϕx0 (x) in the domain of ϕx0 . a) Prove that ak :=

∂f (x0 ), 1 ∂xk

≤ k ≤ n.

b) Prove that f :G →C is differentiable at x0 ∈ G if and only if u(x) is differentiable at x0 . F : G → R2 , F (x) = x(x)

281

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15

Complex Numbers and Geometry

The preparations of Chapter 13 allow us to start immediately with the investigation of complex-valued functions defined on a subset G ⊂ C and nowadays many textbooks on this topic follow the same approach. However complex analysis is also a rather geometric theory and we prefer to collect early in a more or less coherent manner interpretations and results linking geometry and algebra in C. This is a more classical approach and we refer to the classical text of C. Carath´eodory [16] or the book of E. Peschl [64]. Complex numbers can be identified as points in the plane, see Figure 15.1, and by its very definition, the addition of two complex numbers z1 and z2 is vector addition in R2 , see Figure 15.2. Moreover, mapping z onto z is just a reflection in the real axis, see Figure 15.3.

y = Im z (imaginary axis)

Im z

z1 + z2

z = x + iy b

z2

y

z1

x

x = Re z (real axis)

Figure 15.1

283

Re z

Figure 15.2

A COURSE IN ANALYSIS

Im z

z = x + iy b

Re z

b

z¯ = x − iy

Figure 15.3

The multiplication of two complex numbers z1 , z2 ∈ C, zj 6= 0, can   be interx1 −y1 , preted in the following way: we write z1 as a matrix, i.e. h(z1 ) = y1 x1   x −y2 and z2 as a matrix h(z2 ) = 2 , and then z1 · z2 corresponds to y2 x2    x1 −y1 x2 −y2 h(z1 · z2 ) = y1 x1 y2 x2 =

(x21

The matrix U =

+

y12 )1/2

x1 (x21 +y12 )1/2 y1 (x21 +y12 )1/2

x1 (x21 +y12 )1/2 y1 (x21 +y12 )1/2 −y1 (x21 +y12 )1/2 x1 (x21 +y12 )1/2

−y1 (x21 +y12 )1/2 x1 (x21 +y12 )1/2

!

!

x2 −y2 y2 x2



has determinant 1 and therefore

U ∈ SO(2), i.e. it corresponds to a rotation. Thus it looks as if complex multiplication is related to a rotation followed by a dilation. In order to come to a firm statement we introduce polar coordinates. We have already encountered polar coordinates (r, ϕ) in the plane R2 , see Chapter II.12, in particular Example II.12.3. We want to treat polar coordinates in C and we will make use of Euler’s relation eiϕ = cos ϕ + i sin ϕ, ϕ ∈ R, which we 284

15

COMPLEX NUMBERS AND GEOMETRY

already needed in Chapter II.17, also see Problem 6 of Chapter II.17. Using the ratio test we have seen in Problem 8 of Chapter 13 that for every z ∈ C the series ∞ X zk , (15.1) exp z := k! k=0 cos z :=

∞ X

(−1)l

l=0

and sin z :=

∞ X

(−1)m−1

m=1

z 2l (2l)!

z 2m−1 (2m − 1)!

(15.2)

(15.3)

converge for all z ∈ C and for z = i ϕ, ϕ ∈ R, we find using i2 = −1 that iϕ

e

= =

∞ X (iϕ)k k=0 ∞ X l=0

= =

k!

∞

(iϕ)2l X (iϕ)2m−1 + (2l)! (2m − 1)! m=1

∞ 2l X i ϕ2l l=0 ∞ X

(2l)!

(−1)l

l=0

+

∞ 2m−1 2m−1 X i ϕ l=0

(2m − 1)! ∞

X ϕ2m−1 ϕ2l +i (−1)m−1 (2l)! (2m − 1)! l=0

= cos ϕ + i sin ϕ. With polar coordinates

y r = |z| = (x2 + y 2 )1/2 and ϕ = arctan , x

(15.4)

where some care is needed in defining ϕ, compare with Chapter II.12, and see also Figure 15.4 below, we get for z 6= 0 z = r (cos ϕ + i sin ϕ) = reiϕ , where ϕ is only determined modulo 2π, see Figure 15.4. 285

(15.5)

A COURSE IN ANALYSIS

Im z

y1 = Im z1 b

r1

ϕ2 z2

r2

y2 = Im z2

x2 = Re z2

ϕ3

b

z1

ϕ1 Re z x3 = Re z3

x1 = Re z1

r3

y3 = Im z3

b

z3

Figure 15.4 The function tan is π -periodic, not 2π - periodic and it has the natural real domain ( −π , π ). In order to use arg z = arctan xy for all z = x + iy we use 2 2 the following (justifiable) convention  y  x > 0 and y ≥ 0 arctan x ,    π  , x = 0 and y > 0  2   y π  + arctan |x| , x < 0 and y > 0 y arctan := 2 |y|  x π + arctan |x| , x < 0 and y ≤ 0     3π  , x = 0 and y < 0  2    3π + arctan |y| , x > 0 and y < 0. 2

x

We have already mentioned that |z| is called the modulus of z and arg z := ϕ is called the argument of z. In order to have arg z uniquely defined we reduce the range of ϕ to a half-open interval of length 2π. Traditionally in complex analysis the choice is −π < ϕ ≤ π, the modern choice which we will 286

15

COMPLEX NUMBERS AND GEOMETRY

follow is 0 ≤ ϕ < 2π. So far we may summarize our consideration in Corollary 15.1. Every complex number z ∈ C\{0} has a unique representation z = reiϕ (15.6) with r = |z| > 0 and ϕ = arg z ∈ [0, 2π). This is called the polar representation of z. Now we find for complex multiplication of z1 , z2 ∈ C\{0} z1 · z2 = r1 eiϕ1 r2 eiϕ2 = r1 r2 ei(ϕ1 +ϕ2 )

= |z1 ||z2 |ei arg(z1 z2 ) = |z1 ||z2 |(cos(arg (z1 · z2 )) + i sin(arg (z1 · z2 )),

where arg (z1 · z2 ) ≡ (ϕ1 + ϕ2 ) mod 2π, arg (z1 · z2 ) ∈ [0, 2π). This also leads to Corollary 15.2. The inverse of z ∈ C\{0} with respect to multiplication is given by 1 1 z −1 = = e −iϕ . z r The reader may recall from algebra or number theory that for integers k, l, m ∈ Z the meaning of k ≡ l mod m (read k is congruent to l modulus m) is that k − l is divisible by m. This notation is extended to real numbers in the sense that a ≡ b mod c means that a−b = k for some k ∈ Z. Thus c arg (z1 · z2 ) ≡ (ϕ1 + ϕ2 )

mod 2π

(15.7)

means that arg (z1 · z2 ) = ϕ1 + ϕ2 + 2πk, k ∈ Z, 0 ≤ arg (z1 · z2 ) < 2π.

(15.8)

Applying now the polar coordinate representation to the matrix representation we find with zj = rj eiϕj that   cos ϕ1 − sin ϕ1 iϕ2 iϕ1 h(z2 ), h(z1 · z2 ) = h (r1 e r2 e ) = r1 sin ϕ1 cos ϕ1 287

A COURSE IN ANALYSIS

and our geometric interpretation becomes more clear: 2 by z1  multiplying z cos ϕ1 − sin ϕ1 corresponds to a rotation by U(arg z1 ) = U(ϕ1 )= follosin ϕ1 cos ϕ2 wed by a dilation of size r1 . This observation leads to some interesting geometric consequences. Consider a subset A ⊂ C. For w ∈ C, |w| = 1, we can study the new set wA := {ζ = wz|z ∈ A}. This set is obtained from A by a rotation around the origin with an angle arg w. Hence invariance of A under a certain rotation can be expresses by the equality A = wA. We will see further cases where geometry in the plane is best described algebraically with the help of complex numbers. Using the law of the exponential function, compare with Problem 10 of Chapter 13, we now find the formula of de Moivre Lemma 15.3. For z = reiϕ the following holds zn = r n ei n ϕ = r n (cos nϕ + i sin nϕ).

(15.9)

This lemma allows us to determine the nth roots of unity, n ∈ N, i.e. all complex numbers z with z n = 1 or z n − 1 = 0. We take for granted the funN P damental theorem of algebra stating that a polynomial p(z) = ak z k , ak ∈ k=0

C, z ∈ C, of degree N, i.e. aN 6= 0, has N complex roots (taking multiplicity into account). An analytic proof of the fundamental theorem of algebra will also be given later in Chapter 23, Theorem 23.15. Now, if z = r eiϕ satisfies z n = 1 it follows that r n einϕ = 1 = 1ei 0 . Thus we must have r = 1 and ϕ = ϕk = 2π k, k = 0, ......, n − 1. Indeed for r = 1 and ϕ = 2π k, k = 0, ......, n − 1, we n n have 2π z n = (1 e n ϕk i)n = e2π k i = 1. In Figure 15.5 and Figure 15.6 we have shown the positions on the unit circle S 1 of the 5th and the 12th roots of unity, respectively. These figures also demonstrate nicely that the associated polygons are invariant under certain rotations which can be expressed by showing that multiplying by the corresponding root of unity does not change the set. 288

15

COMPLEX NUMBERS AND GEOMETRY

e e

2π 5 i

4π 5 i

2π 5 b

e

0

e

2π 5 0i

=1

6π 5 i

e

8π 5 i

Figure 15.5

π

e2i e

e

2π 3 i

5π 6 i

π

e6i

2π 12

eπi = −1

e

π

e3i

2π

e 12 0i = 1

b

0

7π 6 i

e

e

e

4π 3 i

e

3π 2 i

11π 6 i

5π 3 i

Figure 15.6

Thus the number of the nth roots of unity is exactly n and they have the repre2π sentation zk = e n k i , k = 0, ....., n−1. Of course for other values k ∈ Z we still have zkn = 1, but zk can not anymore be considered as given in polar coordina2π tes with argument in [0, 2π). Indeed if zm = e n m i , m ∈ Z\{0, 1, ......, n − 1}, n we have zm = 1 and we can find a unique km ∈ {0, 1, ......, n − 1} and lm ∈ Z 289

A COURSE IN ANALYSIS

such that

Consider

m − km = n or m = km mod n. lm 2π

E(n) := {e n k i |k = 0, 1, ...., n − 1} and

2π

C(n) := {e n k i |k ∈ Z}.

Both sets are subsets of S 1 , and S 1 with complex multiplication is a subgroup of C\{0}. It follows that both E(n) and C(n) are subgroups of S 1 which as point sets are equal. However we can identify C(n) with the infinite group n Z = {n k|k ∈ Z} with the group operation being addition in Z, whereas E(n) we can identify with the finite group, the group Z/n formed by the system of representation {0, 1, ....., n−1} and the corresponding group operation and E(n) is considered as the group of nth unit roots. We urge the student to get used to the idea that with groups we can associate geometric objects which are invariant under the action or operation of the group under consideration. First let us give Definition 15.4. Let X 6= ∅ be a set and G a group. A. We say that G operates or acts on X if for every g ∈ G there exists a bijective mapping Tg : X → X. We call the mapping g 7→ Tg from G into M(X) a representation of G if Tg− 1 = Tg−1 and Tg1 g2 = Tg1 · Tg2 . As usual M(X) denotes the set of all mappings of X into itself. B. A set Y ⊂ X is called invariant under Tg if Tg Y = Y . C. If R : G → M(X) is a representation of G and Y ⊂ X is invariant under all Tg := R(g) then we call Y invariant under the representation R of G and if no confusion may arise we say that Y is invariant under G. Example 15.5. Let X = C and G = C\{0} the multiplicative group of C. For every w ∈ G we define Tw : C → C by Tw z = w · z. Note that the inverse of Tw is given by Tw−1 = Tw− 1 and that Tw2 · Tw1 as well as T1 = idC . We may consider S 1 as a subgroup of G and it follows for w ∈ S 1 and z ∈ S 1 that we have Tw z = w z = S 1 , i.e. S 1 as a subgroup of C\{0} leaves S 1 as subset of C invariant. 290

15

COMPLEX NUMBERS AND GEOMETRY

Example 15.6. Consider the regular n-gon Pn ⊂ C with vertices zk = 2π e n k i , k = 0, ......., n − 1, as subset of C. Thus the vertices of Pn are just the elements of E(n), for n = 5 and n = 12 compare with Figure 15.5 and Figure 15.6, respectively. Since multiplication with an nth root of unity corresponds to a rotation around the origin by an angle ϕk ∈ {0, 2nπ , ......., 2nπ (n − 1)}, it follows that Pn ⊂ C is invariant under the action of the representation of E(n) as multiplication operators. We now want to describe some simple geometric objects (subsets) in the plane with the help of complex numbers. Let us start with the following observation. We can identify C with R2 , i.e. z = x + iy is identified with (x, y). If we can characterise a geometric object in the plane R2 by a relation h(x, y) = 0 or an inequality g(x, y) ≥ 0 with some functions h, g : R2 → R, then we may switch to complex numbers using the relations x = Re z =

z−z z+z and y = Im z = . 2 2i

, z−z ) = 0 and similarly g(x, y) ≥ 0 becomes Thus h(x, y) = 0 becomes h ( z+z 2 2i z+z z−z g( 2 , 2i ) ≥ 0. If we now can find H and G such that   z+z z−z H(z, z) = h , 2 2i

and G(z, z) = g



 z+z z−z . , 2 2i

We can give a description of the corresponding geometric object with the help of complex numbers. However, in concrete situations direct (geometric) considerations are often easier. A straight line in C we can give in its parametric form by z = z1 + t (z2 − z1 ), t ∈ R,

(15.10)

      x1 x x2 − x1 = +t y1 y y2 − y1

(15.11)

which corresponds to

291

A COURSE IN ANALYSIS

    x1 x2 and which passes through z1 , i.e. , and z2 , i.e. . y1 y2 1 ∈ S 1 ⊂ C we find With c := |zz22 −z −z1 | z − z1 = c t, |z2 − z1 |

(15.12)

where c, interpreted as vector in R2 , gives the direction of the line. If we rewrite (15.12) as z − z1 c = t, (15.13) |z2 − z1 |

note c c = 1, we observe that the left hand side in (15.13) must be a real number for all z on the straight line given by (15.10). Thus for points on this line Im (c (z − z1 )) = 0 (15.14) must hold and it is easy to derive from (15.14) that {z ∈ C | Im (c (z − z1 )) = 0}, |c| = 1, must be a straight line passing through z1 and z2 = c + z1 . In Problem 8 we will also find the line orthogonal to the line (15.10). The normal form of (15.10) is given by d z + d z − 2 p = 0, d = i c, p = Re (dz1 ).

(15.15)

Next consider for a, b ∈ R and c ∈ C the equation azz + cz + cz + b = 0

(15.16)

D := |c|2 − ab > 0.

(15.17)

under the constraint For a = 0 it follows from (15.15) that (15.16) gives a straight line. In the case where a 6= 0 we can rewrite (15.16) as    c |c|2 − ab D c z+ = = 2, z+ 2 a a a a or with R2 =

D a2

it follows from (15.16) that c |z + | = R a 292

(15.18)

15

COMPLEX NUMBERS AND GEOMETRY

must hold, whereas for D < 0 and a 6= 0 no z ∈ C can solve (15.16). For % > 0 the set {z ∈ C||z − z0 | = %} is however a circle with centre z0 and radius %: with z = x + i y and z0 = x0 + i y0 we have indeed |z − z0 |2 = (x − x0 )2 + (y − y0 )2 = %2 .

(15.19)

Thus (15.18) gives a circle with centre z0 = − ac and radius R. In polar coordinates we can rewrite the equation |z − z0 | = % as z = z0 + % eiϕ , ϕ ∈ [0, 2π). Note that when we choose in (15.18) the value c = 0 we have a circle with the origin 0 as centre. For every w ∈ C, |w| = 1, we have of course |wz| = |z| = R if z lies on the circle |z| = R. This nicely demonstrates our geometric interpretation of the multiplication with a complex number of modulus 1: a circle with centre 0 is of course invariant under all rotations around the origin. We can identify the open disc with centre z0 and radius % > 0 with D% (z0 ) = {z ∈ C||z − z0 | < %}

(15.20)

for which we more often write as before B% (z0 ) = B% ((x, y)), z = x + iy, see Figure 15.7. The closed disc with centre z0 and radius % is of course D% (z0 ) = {z ∈ C||z − z0 | ≤ %} = B% (z0 ). However we also use D for B1 (0). We also note that the inequality Im z > 0

(15.21)

characterises the open upper half plane, see Figure 15.8 H := {z ∈ C| Imz > 0} = {(x, y) ∈ R2 |y > 0},

(15.22)

and the closed upper half plane H as well as the open and closed lower half planes H− and H− are defined analogously. 293

A COURSE IN ANALYSIS

Im z

Re z

ρ b

z0 Dρ (z0 ) = {z ∈ C||z − z0 | < %}

Figure 15.7

Im z

H = {z ∈ C|Im z > 0}

Re z

Figure 15.8

294

15

COMPLEX NUMBERS AND GEOMETRY

On R we have a natural order structure and limit points +∞ and −∞ were introduced. In the plane this needs a different approach: there are too many directions leading to “infinity”. We overcome this problem by mapping S 2 \{N}, where S 2 ⊂ R3 is the unit sphere and N = (0, 0, 1) is the “north pole”, bijectively onto the complex plane C. Then we extend this mapping to S 2 and consider the image of N as “point at infinity” for C. We consider the sphere S 2 ⊂ R3 with centre (0, 0, 0) and radius 1 in the ξ, η, ζ coordinate system, S 2 = {(ξ, η, ζ) ∈ R3 | ξ 2 + η 2 + ζ 2 = 1}, and we consider the ξ − η - plane also as z = x + iy plane on which we want to project S 2 , see Figure 15.9.

ξ

N = (0, 0, 1)

ζ

ρ

P (ξ, η, ζ)

η ξ

η, y r z = x + iy

ξ, x

2

2

r = (x + y )

1 2

Figure 15.9

The point z ∈ C, z = x + iy, is connected with the north pole of the sphere by a straight line. This line hits the sphere in the point P = (ξ, η, ζ). Now we want to determine z = x + iy as function of ξ, η, ζ and then we want to determine the inverse mapping. Using Figure 15.9 we get the following set of equations ξ 2 + η2 + ζ 2 = 1 295

(15.23)

A COURSE IN ANALYSIS

and

y r 1 x = = = , ξ η ρ 1−ζ

implying

ξ η , y= , 1−ζ 1−ζ

(15.25)

ξ + iy ξ − iη and z = . 1−ζ 1−ζ

(15.26)

x= and therefore z= Furthermore we find

(15.24)

1 − ζ2 1+ζ ξ 2 + η2 = = , 2 2 (1 − ζ) (1 − ζ) 1−ζ 1−ζ 2ζ 1+ζ − = , zz − 1 = 1−ζ 1−ζ 1−ζ 2 , zz + 1 = 1−ζ zz =

and therefore (15.25) gives

ξ = 2x

2x z+z 1−ζ = = 2 zz + 1 zz + 1

2y (z − z) =i . zz + 1 zz + 1 Thus we have the mapping and its inverse: η=

z= and

ξ + iη 1−ζ

(15.27)

z+z i(z − z) zz − 1 , η= , ζ= . (15.28) zz + 1 zz + 1 zz + 1 The mapping S : S 2 \{N} → C, (ξ, η, ζ) 7→ S(ξ, η, ζ) = z = ξ+iη , is called 1−ζ the stereographic projection. Our problem is to extend S to S 2 , so the question is to which point shall we map N? We introduce the “ideal” point ∞, ∞ ∈ / C, hence we extend C to ˆ := C ∪ {∞}, and now we extend S to a mapping S˜ : S 2 → C ˆ = C ∪ {∞} C where ( S(ξ, η, ζ) , (ξ, η, ζ) 6= N = (0, 0, 1) S˜ (ξ, η, ζ) = ∞ , (ξ, η, ζ) = N = (0, 0, 1). ξ=

296

15

COMPLEX NUMBERS AND GEOMETRY

ˆ the closed or extended complex plane or the Riemann sphere. We call C ˆ First we extend arithmetic operations to C z + ∞ = ∞ for z ∈ C,

(15.29)

z = 0 for z ∈ C, ∞

(15.30)

z · ∞ = ∞ for z ∈ C\{0}, z = ∞ for z ∈ C, 0

(15.31) (15.32)

∞ however we do not define ∞ + ∞, 0 · ∞, ∞ or 00 . Let (zn )n∈N be a sequence of complex numbers. If for every R > 0 there exists N(R) ∈ N such that n ≥ N(R) implies |zn | ≥ R we say that (zn )n≥N converges to ∞ and we write

lim zn = ∞.

n→zn

(15.33)

ˆ In Appendix I we will give a further description of the topology on C.

Problems 1. Find the polar representations of: √ a) z = 4 + 4 3i; b) z = −1 + i; √ √ c) z = − 24 − 8i. 2.

a) Express cos(5ϕ) as a P polynomial in cos ϕ, i.e. find N ∈ N and k ak ∈ C such that cos(5ϕ) = N k=0 ak cos ϕ. b) Use Euler’s formula to verify

8 cos4 (2ϕ) − 3 = cos(8ϕ) + 4 cos(4ϕ). c) Find the value of



√ 12 1+ 31 3i √ . 1 1− 3 3i

297

A COURSE IN ANALYSIS

3.

P a) Let zP of the polynomial p(z) = nk=0 ak z k . 1 , . . . , zn be the roots Q and nk=1 zk = (−1)n aan0 . (For n = 2 these Prove that nk=1 zk = − an−1 an formulae are known as the formulae of Vi´ eta.) b) Show that the sum of all nth roots of unity is equal to 0 provided n ≥ 2.

4.

a) Sketch the positions of all 8th roots of unity. b) Prove that if n divides m then E(n) is a subgroup of E(m).

5. Show that the annulus Ar,R = {z ∈ C|0 < r < |z| < R} is invariant under the action of S 1 . (Compare with Example 15.5.) 6. Find the subset E ⊂ C determined by the equation |z −3|+|z +3| = 10. 7. Find the non-parametric equation of the line passing through z1 = −2 − i and z2 = 3 + 5i. 8. Let z = z1 + t(z2 − z1 ), t ∈ R, be a straight line L in C given in parametric form. Find a non-parametric equation for the line passing through z1 and orthogonal to L. 9. Find the image of a line L ⊂ C passing through z0 = 0 on the Riemann sphere. 10. Let n ∈ N and zn := z0n , |z0 | > 1. Prove that limn→∞ zn = ∞. Compare this result with the analogous result in the real case.

298

16

Complex-Valued Functions of a Complex Variable

In Chapter 14 we discussed complex-valued functions defined on an arbitrary non-empty set. When specifying this set to a subset D ⊂ C we can of course apply the earlier results. We denote points in D by z = x + iy and it follows that every f : D → C admits a decomposition f (z) = Re f (z) + i Im f (z).

(16.1)

It is very helpful to consider Re f (z) and Im f (z) also as real-valued functions of the two variables (x, y) and we write f (z) = u (x, y) + i v (x, y).

(16.2)

Since D equipped with the metric d(z1 − z2 ) = |z1 − z2 | is a metric space as is C, continuity of f : D → C is well defined. Using the results of Chapter II. 2 we know: A function f : D → C, D ⊂ C, is continuous at z0 ∈ D if for every  > 0 there exists δ > 0 such that for all z ∈ D with 0 < |z − z0 | < δ it follows that |f (z) − f (z0 )| < . If D contains a sequence (zn )n∈N , zn 6= z0 , then this is equivalent to the fact that for every sequence (zn )n∈N , zn ∈ D, converging to z0 it follows that (f (zn ))n∈N converges to f (z0 ), i.e. lim zn = z0 implies lim f (zn ) = f (z0 )(= f ( lim zn )). n→∞

n→∞

n→∞

If f : D → C is continuous for all z ∈ D we call f continuous in D. The continuous functions in D form a vector space, in fact an algebra, over C which we denote by C(D, C) or if no confusion may arise simply by C(D). We have already noted in Chapter 14 that f is continuous (at z0 or in D) if and only if u and v are continuous (at (x0 , y0) or in D considered as subset of R2 ). If D ⊂ C is compact then a continuous function f : D → C is uniformly continuous. Since on C we do not have a natural order structure the notion of a monotone function does not make sense and we call f : D → C bounded if |f (z)| ≤ M for all z ∈ D. Continuous functions on compact sets are bounded and for some z1 , z2 ∈ D the following hold: |f (z1 )| = sup |f (z)| and |f (z2 )| = inf |f (z)|. z∈D

z∈D

299

(16.3)

A COURSE IN ANALYSIS

This follows from the fact that real-valued functions which are continuous on a compact set are bounded and attain their supremum and infimum and the observation that |f (z)| = (u2 (x, y) + v 2 (x, y))1/2 . If f : D → C is continuous and g : f (D) → C is continuous too, then g ◦ f : D → C is continuous. Using explicitly the fact that C is a field for two functions f, g : D → C we can define on {z ∈ D | g(z) 6= 0} the quotient f : D\{z ∈ C | g(z) = 0} → C and if both, f and g are continuous, then fg is g continuous too. So far we can deduce that every polynomial p(z) :=

N X k=0

ak z k , N ∈ N ∪ {0}, ak ∈ C,

(16.4)

is continuous on C and hence on every subset D ⊂ C. As usual we call the largest k0 ∈ {0, 1, ......., N} with ak0 6= 0 the degree of p and usually, if we N P write p(z) = ak z k we assume that p is of degree N. k=0

Moreover we know that rational functions are continuous on their natural domain, i.e. if with two polynomials p and q we have r(z) :=

p(z) q(z)

(16.5)

then r : C\{z ∈ C | q(z) = 0} → C is continuous. Note that r might have a continuous extension to a larger set: 2 0) is defined and continuous on C\{z0 } but The rational function r(z) = (z−z z−z0 it admits a continuous extension (or continuation) to C by h(z) = z − z0 . We will discuss such problems for so called meromorphic function later on in Chapters 22 and 24. We next want to discuss whether it is possible to extend elementary transcendental functions defined on subsets of R, for example exp, cos, arctan, etc., to subsets of C. Having in mind that we have derived power series expansions for these functions, namely their Taylor series, we may try to extend the domain of convergence of such a series (which is an interval in R) to some complex domain. In fact we have already seen that for example the expo∞ k P z nential series converges for all z ∈ C and uniformly on compact sets, k! k=0

300

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

hence we can extend exp : R → R to exp : C → C using the power series. Thus we were led to investigate power series in C. We start with recollecting some results on uniform convergence. Definition 16.1. A sequence fk : D → C, D ⊂ C, of complex-valued functions converges uniformly to f : D → C if for every  > 0 there exists N = N() ∈ N such that k ≥ N() implies sup | fk (z) − f (z)| < .

(16.6)

z∈D

Remark 16.2. A. We know that uniform convergence is convergence with respect to the supremum norm ||f ||∞ := supz∈D |f (z)|. B. By Theorem II.2.32 the uniform limit of a sequence of continuous functions is continuous. ∞ P C. We call a series fk of functions fk : D → C absolutely convergent

if the series

∞ P

k=0

k=0

|fk (z)| converges for all z ∈ D.

Next we prove a complex version of the Weierstrass M-test, compare with Theorem I.29.1. Theorem 16.3. Let (fk )k∈N0 , fk : D → C, D ⊂ C, be a sequence of functions ∞ ∞ P P and suppose that ||fk ||∞ < ∞. Then the series fk converges absolutely k=0

k=0

and uniformly on D to a function f : D → C. If all functions fk are continuous then f is continuous too.

Proof. First we show that

∞ P

k=0

fk (z), z ∈ D, converges pointwisely to some

function f : D → C. Since |fk (z)| ≤ ||fk ||∞ the series absolutely by the comparison test, Theorem 13.15. Thus for z ∈ D we can define f (z) :=

∞ X

∞ P

fk (z) converges

k=0

fk (z)

(16.7)

k=0

which gives a function f : D → C. Now we prove that this convergence is N P uniformly, i.e. the sequence of partial sums (S N )N ∈N0 , S N (z) := fk (z), k=0

301

A COURSE IN ANALYSIS ∞ P

converges uniformly to f . Given  > 0 the convergence of the existence of N0 () ∈ N such that N ≥ N0 () implies Now it follows for N ≥ N0 () that

||fk ||∞ implies

k=0 ∞ P

k=N +1

||fk ||∞ < .

||S N − f ||∞ = sup |S N (z) − f (z)| z∈D

= sup | z∈D

=

∞ X

∞ X

k=N +1

k=N +1

fk (z)| ≤

∞ X

k=N +1

sup |fk (z)| z∈D

||fk ||∞ < ,

i.e. the uniform convergence of (S N )N ∈N0 to f . It follows further from Remark 16.2.A that the continuity of all fk , k ∈ N0 , implies the continuity of f. Using the arithmetic in C we find immediately by first looking at partial sums Corollary 16.4. Let (fk )k∈N0 and (gk )k∈N0 be two sequences of functions ∞ ∞ P P fk , gk : D → C, D ⊂ C. If the series fk and gk converge pointwise k=0

k=0

(absolutely, uniformly) to f and g respectively, then for a, b ∈ C we have ∞ X

(a fk + b gk ) = af + b g

(16.8)

k=0

and

∞ X

fk = f

(16.9)

k=0

where the convergence is pointwise (absolutely, uniformly). We now turn to power series and the reader may compare our discussion with Chapter I.29. 302

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Definition 16.5. Let (ak )k∈N0 be a sequence of complex numbers and let c ∈ C. We call ∞ X c T(a (z) := ak (z − c)k , z ∈ C, (16.10) k) k=0

the (formal) power series associated with the sequence (ak )k∈N0 and centre c. The numbers ak are called the coefficients of the power series.

c Of course, the most urgent questions is: for which z ∈ C does T(a (z) conk) verge?

Theorem 16.6 (N.H.Abel). Let (ak )k∈N0 be a sequence of complex numbers c and c ∈ C. If T(a (z) converges for some z1 6= c, then it converges for all k) c z ∈ C such that |z − c| ≤ % < |z1 − c|, i.e. T(a (z) converges in B% (z). k) In addition, on B% (z) the convergence is uniform and the same statement ∞ P holds for T˜ c (z) := k ak (z − c)k . In particular on B% (z) the functions (ak )

z 7→

c T(a (z) k)

k=1

c and z 7→ T˜(a (z) are continuous. k)

c Proof. We set fk (z) := ak (z − c)k , and formally we find with f (z) := T(a (z) k) ∞ ∞ P P fk (z1 ) converges by assumption there exists M ≥ 0 fk . Since that f = k=0

k=0

such that |fk (zn )| ≤ M for all k ∈ N0 . For z ∈ [c − %, c + %] it follows with 0 < % < |z1 − c| that k k k z − c |fk (z)| = |ak (z − c) | = |ak (z1 − c) ≤ M ϑk , z1 − c where ϑ :=

% z1 −c

< 1. Thus we have

||fk ||∞, B% (c) = sup |fk (z)| ≤ M ϑk z∈B% (c)

implying that

∞ X k=0

||fk ||∞, Bϑ (c) ≤ M

Now Theorem 16.3 implies that the series

∞ P

k=0

and uniformly. 303

1 . 1−ϑ

fk converges on B% (c) absolutely

A COURSE IN ANALYSIS

In order to prove the second statement we define gk (z) := k ak (z − c)k−1 and ∞ P g := gk . As before we can derive k=1

||gk ||∞, B% (c) ≤ k M ϑk−1

and the ratio test, Theorem 13.16, implies the convergence of Note that

(k+1)M ϑk kM ϑk−1

∞ P

k M ϑk−1 .

k=1

=

k+1 k

ϑ < 1 for k large since ϑ < 1 and lim

k→∞

k+1 = 1. k

Theorem 16.3 gives the convergence result for the series

∞ P

k ak (z−c)k−1 .

k=1

Remark 16.7. Since Theorem 16.6 gives uniform and absolute convergence for all B% (c), % < |z1 − c|, we obtain pointwise absolute convergence in the open disc B%0 (c), %0 = |z1 − c|, i.e. B%0 (c) =

[

B% (c).

% 0 such that T(ak ) (z) converges for all z ∈ B%0 (c) the radius of c convergence of T(a (z). k)

Of interest is Corollary 16.9. If the radius of convergence %0 > 0 of a power series ∞ P c ak (z − c)k is finite, then for z ∈ C such that |z − c| > %0 , T(a (z) = k) k=0 {

i.e. z ∈ B% (c) , we have

sup |ak ||z − c|k = ∞.

k∈N0

304

(16.11)

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Proof. Suppose that z1 ∈ B%0 (c){ and | ak (z1 − c)k | ≤ M < ∞ for all k ∈ N0 . c The proof of Theorem 16.6 yields that the power series T(a (z) must converge k) for all z ∈ C such that |z − c| < |z1 − c|. However for z ∈ C in the annulus %0 < |z − c| < |z1 − c| we get a contradiction to the maximality of %0 . The next result tells us how we can (in principle) determine %0 . Theorem 16.10 (Cauchy and Hadamard). The radius convergence %0 of ∞ P the power series ak (z − c)k is given by k=0

1 1 p %0 = lim inf p = . k k→∞ |ak | lim supk→∞ k |ak |

Proof. We know that

∞ P

k=0 k

(16.12)

ak (z − c)k converges for |z − c| = r < %0 which

implies that lim |ak | r = 0. Thus there exists N0 ∈ N such that |ak | r k < 1 k→∞

1 for k ≥ N0 or r < √ k

|ak |

for k ≥ N0 . This implies of course that r ≤

1 lim inf √ , but r < %0 was arbitrary, hence k

k→∞

|ak |

1 . %0 ≤ lim inf p k k→∞ |ak |

1 When %0 = ∞ it follows, of course, that lim inf k→∞ √ k

|ak |

= ∞ = %0 . If

%0 < ∞ and |z − c| = r > %0 then the sequence (|ak | r k )k∈N0 is by Corollary 1 , for an infinite 16.9 unbounded. Therefore we have |ak | r k > 1, i.e. r > √ k |ak |

1 number of values of k ∈ N0 . This implies r ≥ lim inf k→∞ √ k

since r was arbitrary but larger than %0 we arrive at

and the result follows.

|ak |

and again

1 %0 ≥ lim inf p , k k→∞ |ak |

For power series, using the Cauchy-Hadamard result, we can obtain an improved ratio test, see R.Remmert [67]. 305

A COURSE IN ANALYSIS

Proposition 16.11. Let (ak )k∈N0 , ak ∈ C and ak 6= 0 for all k ∈ N0 , and c ∈ C. Suppose that %0 > 0 is the radius of convergence of the power series ∞ P ak (z − c)k . Then

k=0

lim inf k→∞

|ak | |ak | ≤ % ≤ lim sup |ak+1 | k→∞ |ak+1 | |ak | k→∞ |ak+1 |

holds and in particular in the case that lim

(16.13)

= A exists we have %0 = A.

k| Proof. Let 0 < σ < lim inf k→∞ |a|ak+1 . It follows the existence of l ∈ N | ak such that | ak+1 | > σ, i.e. |ak+1 | σ < |ak |, for all k ≥ l. With α := |al | σ l it follows that |al+m | σ l+m ≤ α for all m ≥ 0. Hence (|ak | σ k )k∈N is a bounded sequence and Corollary 16.9 implies that σ < %0 . Now assume that 0 < k| lim supk→∞ |a|ak+1 < τ < ∞. | a n In this case we can find m ∈ N such that an+1 > τ or |an+1 | τ > |an | for n ≥ m. This yields for 0 < β := |am | τ m that |an+m | τ n+m ≥ β for all n ≥ 0 and therefore the sequence (|an | τ n )n∈N cannot converge to 0, implying that ∞ P |ak | τ k cannot converge, hence τ ≥ %0 and the proposition is proved.

k=0

Finally we can link the convergence of power series with real coefficients and a real variable to that of the power series with the same coefficients but a complex variable.

Lemma 16.12. Let (ak )k∈N0 , ak ∈ R, and c ∈ R. Suppose that the power ∞ P series ak (x − c)k converges for all x ∈ (c − %0 , c − %0 ), %0 > 0. Then the k=0

power series

∞ P

k=0

ak (z − c)k converges for all z ∈ C such that |z − c| < %0 , i.e.

z ∈ D%0 (c) = B%0 (c) ⊂ C.

Proof. Let 0 < r < %0 . We know that

∞ P

k=0

|ak | r k converges, hence for all

z ∈ C, |z − c| ≤ r it follows the convergence of convergence of

∞ P

∞ P

k=0

|ak | |z − c|k implying the

ak (z − c)k for |z − c| ≤ r. Since r < %0 was arbitrary we

k=0 ∞ P

have proved that

k=0

ak (z − c)k converges for all z ∈ C, |z − c| < %0 . 306

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Corollary 16.13. Let I ⊂ R be an interval and f : I → R a function which is arbitrary often differentiable in ˚ I. Suppose that for some c ∈ ˚ I the function f has the convergent Taylor expansion f (x) =

∞ X f (k) (c)

k!

k=0

(x − c)k

in the interval (c − %0 , c + %0 ) ⊂ ˚ I, %0 > 0. Then the power series ∞ X f (k) (c) k=0

k!

(z − c)k

converges for all z ∈ C, |z − c| < %0 , to a continuous function extending f to the open disc with centre c and radius %0 . Thus we obtain the following continuous complex-valued functions by extending the corresponding Taylor series of the known real-valued function: ∞ X zk

exp(z) =

k=0

k!

,

∞ X (−1)k

cos (z) =

k=0

(2k)!

z 2k ,

∞ X (−1)k+1 2k−1 z , sin (z) = (2k − 1)! k=1

∞ X (−1)k 2k+1 arctan(z) = z , 2k + 1 k=0

˜ + z) = ln(1

∞ X (−1)k+1 k=1

1 = 1−z Jl =

∞ X k=0

∞ X

k

zk ,

zk ,

k=0

(−1)k z 2n+l , 22k+l k!(k + l)! 307

|z| < ∞;

(16.14)

|z| < ∞;

(16.15)

|z| < ∞;

(16.16)

|z| < 1;

(16.17)

|z| < 1;

(16.18)

|z| < 1;

(16.19)

|z| < ∞,

(16.20)

A COURSE IN ANALYSIS

where Jl is the Bessel function of order l, see Problem 7 to Chapter I.29. We will discuss some more examples below and in the problems. Some of these and other complex-valued functions we want to study now in more detail. The following three examples demonstrate a very important point: extending functions from subsets of R to subsets of C may dramatically change their properties. Example 16.14. The function g2N +1 : R → R, g(x) = x2N +1 is for all N ∈ N −1 a strictly monotone increasing bijective function with inverse g2N +1 (y) = 1 2N+1 . The complex extension h2N +1 of g2N +1 defined by h2N +1 : sgn (y) |y| C → C, h2N +1 (z) = z 2N +1 is for no N ∈ N injective. In fact for every real number R > 0 we find 2N + 1 distinct pre-images of R namely 1

2π i

zk = R 2N+1 e 2N+1 k ,

k = 0, ......, 2N,

or interpreting h−1 2N +1 (R) as the pre-image of {R} ⊂ C 1

2N+1 E(2N + 1) h−1 2N +1 (R) = R

where E(2N + 1) is again the group of (2N + 1)th roots of unity. (Note that for G ⊂ C and a ∈ C we write G = {z = ag | g ∈ G}). Indeed, the equation h2N +1 (z) = z 2N +1 = R is equivalent to |z|2N +1 e(2N +1)i ϕ = Rei 0 1

which yields |z| = R 2N+1 and ϕ ∈ E(2N + 1). Example 16.15. The complex exponential function exp (z) =

∞ P

k=0

zk k!

is a

periodic function with period τ = 2πi. To see this we first note that exp (z1 + z2 ) = exp (z1 ) exp (z2 ) also holds for all z1 , z2 ∈ C, see Problem 10 of Chapter 13. Now it follows for z = x + iy that exp (z + 2πi) = exp (x + i(y + 2π)) = ex (cos (y + 2π) + i sin (y + 2π)) = ex (cos y + i sin y) = exp (x + iy) = exp (z). Thus the strictly monotone, hence injective function exp : R → R is extended to a periodic function exp : C → C. 308

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Example 16.16. Using the power series for exp, cos and sin we find cos z =

exp (iz) + exp (−iz) 2

exp (iz) − exp (−iz) . 2i For z = iy, i.e. a purely imaginary number, we find sin z =

cos (iy) =

(16.21) (16.22)

e−y − ey e−y + ey and sin(iy) = , 2 2i

implying that | cos (iy)| and | sin (iy)| are unbounded. While cos and sin as functions from R to R are bounded functions, their complex extensions to C are not. We want to extend the binomial formula n   X n k n (1 + z) = z (16.23) k k=0

to non-integer values. For this we define for a ∈ C and n ∈ N the binomial coefficients     n a(a − 1)· .....· (a − n + 1) Y a − k + 1 a a := 1. (16.24) = , := 0 n n! k k=1 An easy calculation shows that     a−n a a = , n+1 n+1 n

a ∈ C and n ∈ N0 .

(16.25)

Definition 16.17. For a ∈ C we define the binomial series by Ba (z) :=

∞   X a k=0

k

zk .

(16.26)

Clearly, for a ∈ N formula (16.26) reduces to (16.23). For a ∈ C\N we have Proposition 16.18. For all a ∈ C\N the binomial series has radius of convergence 1. 309

A COURSE IN ANALYSIS

  a 6 0 and with (16.25) we find = Proof. For a ∈ C\N we always have k    1 + k1 k+1 a a =− , k ∈ N, = k+1 k a−k 1 − ka    a a = 1 and therefore by Theorem 16.11 it implying lim k+1 k k→∞ follows that the radius of convergence %0 is equal to 1. In Problem 8 we will use the formula       a a−1 a−1 , k ≥ 1, = + k k−1 k

(16.27)

to derive (1 + z) Ba−1 (z) = Ba (z).

(16.28)

For the next example we need some notational preparation. For a real number α ∈ R the Pochhammer symbols are defined by (α)0 := 1, (α)n = α (α + 1)· .....· (α + n − 1), n ∈ N.

(16.29)

It follows that (1)n = n! , and further (−k)n = 0 for k ∈ N and k ≤ n − 1.

Theorem 16.19. Let α, β ∈ R and γ ∈ R\(−N0 ). The series 2 F1 (α, β; γ; z) :=

∞ X (α)k (β)k k=0

(γ)k k!

zk

converges for |z| < 1. Proof. With ak :=

(α)k (β)k (γ)k k!

we find

ak+1 z k+1 |(α)k+1 (β)k+1 (γ)k k!| ak z k = |(α)k (β)k (γ)k+1 | (k + 1)! |z| |(α + k)(β + k)| |z|. = |(γ + k)(k + 1)| Since lim

k→∞

the result follows.

|(α + k)(β + k)| =1 |(γ + k)(k + 1)| 310

(16.30)

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Definition 16.20. The series 2 F1 (α, β; γ; z) is called the hypergeometric series associated with α, β, γ. Example 16.21. The following hold: 2 F1 (1, 1; 1; z) =

∞ X

zk ;

(16.31)

k=0

(α, 1; 1; z) = (1 − z)−α ; α 2 F1 (−α, 1; 1; −z) = (1 + z) ; z 2 F1 (1, 1; 2; −z) = ln (1 + z);   3 2 1 1+z 1 ; z 2 F1 ( , 1; ; z ) = ln 2 2 2 1−z 1 3 z 2 F1 ( , 1; ; −z 2 ) = arc tan z. 2 2 2 F1

(16.32) (16.33) (16.34) (16.35) (16.36)

Thus we can identify many well-known functions as (modified) hypergeometric series. There are many questions to answer on the behaviour of the complex extension of functions such as exp, cos, sin, ln etc. In particular, since exp is periodic on C we can not expect ln to be the inverse, thus the relation ln ◦ exp and exp ◦ ln are in C at first glance meaningless. In order to investigate these functions it is helpful to introduce and discuss complex derivatives of complex-valued functions of a complex variable. This we will do in the next chapter. In the remaining part of this chapter we want to return to more geometric considerations by discussing the mapping properties of so called linear transformations or M¨obius transformation. Definition 16.22. Let a, b, c, d ∈ C such that   a b = ad − bc 6= 0. det c d

(16.37)

We call the expression w = w(z) :=

az + b , cz + d

z ∈ C,

(16.38)

the M¨ obius transformation or linear transformation associated with a, b, c, d. 311

A COURSE IN ANALYSIS

If c = 0 then it follows from (16.37) that a 6= 0 and d 6= 0. In particular, in the case where w is defined for all z ∈ C. If c 6= 0 then w is defined for all z ∈ C \ {− dc }. Now, if c = 0, then (16.38) yields w=

b d b a z + and z = w − , d d a d

(16.39)

i.e. by (16.38) a bijective mapping from C into C is defined. The classical and still used fa¸con de parler is that for c = 0 a one-to-one mapping of the z-plane onto the w-plane is defined by (16.38). In the case that c 6= 0 and z 6= − dc we find by a simple calculation w=

az + b −dw + b and z = , cz + d cw − a

(16.40)

and w attains all values of the complex plane except w0 = ac . For z0 ∈ C we call C\{z0 } the complex plane punctured at z0 . Our considerations so far yield az+b maps the puncProposition 16.23. The M¨obius transformation w = cz+d tured plane C\{− dc } bijectively and continuously with a continuous inverse onto the punctured plane C\{ ac }.

Next we show Lemma 16.24. The composition of two M¨obius transformations w = is again a M¨ obius transformation given by and ω = αz+β γz+δ W (z) = ω(w(z)) =

Az + B Cz + D

az+b cz+d

(16.41)

where A = α a + β c, B = α b + β d, C = γ a + δ c, D = γ b + δ d,

(16.42)

    a b α β A B . Moreover the M¨obius transforma= i.e. we have c d γ δ C D tion W maps the punctured plane C\{− D } onto the punctured place C\{ CA }. C 

312

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

Proof. First we note that   A B = AD − BC = (a d − b c)(α δ − β γ) 6= 0, det C D and since by assumption a d − b c 6= 0 and α δ − β γ 6= 0 it follows that by (16.41) a M¨obius transformation with the claimed mapping properties is given. Moreover we have W (z) = ω(w(z)) = = =

α w (z) + β γ w (z) + δ z+d z+b ) + β ( cc z+d ) α ( ac z+d z+b ) + δ ( cc z+d ) γ ( ca z+d z+d

(α a + β c)z + α b + β d (γ a + δ c)z + γ b + δ d

implying (16.41) and (16.42). Corollary 16.25. For the two M¨obius transformations w = dz+b − cz−a we find W (z) = ω (w (z)) = z.

az+b cz+d

and ω =

Proof. This follows immediately from (16.42). Remark 16.26. It seems that the combined content of Proposition 16.23 and Corollary 16.25 is that the M¨obius transformation form a group. However there is a problem with the domain of definition and we will resolve this shortly. Remark 16.27. The reader might have noticed a change in some formulations and notations, for example we write w = az+b and z = −dw+b , i.e. we use cz+b cw−a w as a symbol for a mapping, the image of one point under a mapping and as independent variable of a new mapping, as we do with z. This is due to the fact that many traditional notations in the theory of complex variables are still used and no adaption to more modern notation rarely took place. In order to develop the students ability to read the standard literature we shall stick to this common notation. Every M¨obius transformation associated with a, b, c, d ∈ C can be written as a composition of simple transformations. For c 6= 0 we have −c (cz + d) 1 a , w2 = , w = w1 + , w1 = c w2 ad − bc 313

(16.43)

A COURSE IN ANALYSIS

and for c = 0 we find

b a w1 , w1 = z + . (16.44) d a The transformations w = α z, α 6= 0, and w = z + β map straight lines onto straight lines, which is trivial, and they map circles onto circles. Indeed, if |z − z0 | = %0 , then we find w=

1 | w − z0 | = % α

or

|w − α z0 | = |α|%0,

and |w1 − β − z0 | = |w1 − (z0 + β)| = %0 .

However the transformation w = z1 has the same property. Consider equation (15.16) which gives circles and straight lines, i.e. looking at A z z + Cz + Cz + B = 0, A, B ∈ R, C ∈ C, we find with w =

(16.45)

1 z

A

1 1 1 1 + C + C + B = 0, ww w w

or A + C w + C w + B w w = 0,

(16.46)

i.e. the image of the set of all z satisfying (16.45) is a set of points in the w-plane satisfying a similar equation. Now, if A = 0, then (16.46) yields that the straight line defined by (16.45) is mapped onto a circle containing the point w = 0, and if B = 0, then a circle in the z-plane passing through z = 0 is mapped onto a straight line in the w-plane. In all other cases we find that circles are mapped onto circles. Combining these results we obtain az+b Proposition 16.28. Every M¨obius transformation w = cz+d maps a straight line or a circle in the z-plane onto a straight line or a circle in the w-plane. In particular, if a circle in the z-plane is given by (16.45) with A 6= 0 and B 6= 0, then w maps this circle onto a circle in the w-plane.

Thus we may look at complex-valued mappings of a complex variable also from the following point of view: they map certain nice domains onto nice domains and by this certain properties of geometric objects are left invariant, i.e. unchanged. Eventually we will see in Chapter 28 that a corner stone of 314

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

classical complex analysis, the Riemann mapping theorem, is exactly concerned with such a statement. In Appendix III we will discuss the M¨obius transformation further, in particular how we can extend them to the Riemann sphere and how to interpret this extension geometrically.

Problems 1.

P 1 a) Prove that the series ∞ n=1 n4 +z 4 converges uniformly for 1 < |z| < 2. P sin nz for |z| < 1? Comb) Does the series ∞ n=1 n3 converge P∞ sinuniformly nx pare with the result for the series n=1 n3 , x ∈ R.

2. Find the radius of convergence of the following power series: P (z−i)n a) ∞ n=0 (n+2)(n+3) ; P (−1)n−1 (n+1)(z+i)n ; b) ∞ 1 n=1 n 2 c)

P∞

3 (n +1) 2

(n!)2 n n=0 (2n)! z .

Hint to part c): rewrite the Sterling formula as n! = nn e−n rn with 1

limn→∞ rnn = 1. P zk 3. Prove that the Mittag-Leffler function Eα (z) := ∞ k=0 Γ(αk+1) , α ∈ N converges in C, but for α = 0 it converges in |z| < 1. Verify that 1 , 1−z E(z 2 ) = cosh x, E(−z 2 ) = cos z. E0 (z) =

4. Prove that the generalised Mittag-Leffler function Eα,β (z) := P∞ zk k=0 Γ(αk+β) , α, β ∈ N, converges in C and verify that: E1,1 (z) = ez , ez − 1 , E1,2 (z) = z sinh z E2,2 (z 2 ) = . z 315

A COURSE IN ANALYSIS

Remark: once we have extended the Γ-function as a meromorphic function to C we may increase the range of the parameters to Re α > 0, β ∈ C, compare with [31]. 5. Assume that for |z| < 2π the equality (∗)

∞

X Bk z = zk ez − 1 k=0 k!

holds where the Bernoulli numbers are defined by this equality. (This can be thought of as expanding x 7→ exx−1 into a Taylor series about 0 for x ∈ (−2π, 2π), then the corresponding Taylor coefficients are Bk!k .) a) For N ∈ N, N > 1, prove B0 B1 BN −1 + +···+ =0 0!N! 1!(N − 1)! (N − 1)!1! as well as       N N N B0 + B1 + · · · + BN −1 = 0. 0 1 N −1 Further find B1 , B2 , B3 and B4 . b) Prove that BN = 0 if N > 1 isPan odd number. B2k 2k Hint: show that g(x) = exx−1 + x2 = ∞ k=0 (2k)! x .

6. Use that

z ez −1

+

z 2

=

P∞

B2k 2k k=0 (2k)! z

πz cos t(πz) := πz

to show that ∞

(2π)2k 2k cos(πz) X = z . (−1)k sin(πz) (2k)! k=0


7. Prove that for a ∈ N it follows that ak = 0 for k > a. Moreover show that the binomial series extends the binomial theorem to (1+x)a , a ∈ R, x ∈ R, |x| < 1, by expanding x 7→ (1 + x)a into a Taylor series about 0. From the latter result derive the formula Ba (x)Bb (x) = Ba+b (x) for |x| < 1, a, b ∈ R. 316

16

COMPLEX-VALUED FUNCTIONS OF A COMPLEX VARIABLE

8. For a ∈ C and k ≥ 1 prove the identity       a−1 a−1 a + = k k−1 k P∞ a
k and show that with Ba (z) := k=0 k z it follows that (1+z)Ba−1 (z) = Ba (z). 9.

a) Denote by B(α, β) the beta-function and prove that for α, β > 0 we have: Γ(α + n) B(α + n, β − α) (α)n (α)n = = and . Γ(α) (β)n B(α, β − α) b) Verify: 2 F1 (1, 1; 1; z)

=

∞ X

zk ,

k=0

and

˜ + z). z 2 F1 (1, 1; 2; −z) = ln(1

10. Using the series representation of 2 F1 prove that for |z| < 1 we have a) 2 F1 (α, β; γ; z) = 2 F1 (β, α; γ; z);

b) (γ − α − β)2 F1 (α, β; γ; z) + α(1 − z)2 F1 (α + 1, β; γ; z) − (γ − β)2 F1 (α, β − 1; γ; z) = 0. 11.

a) Let z1 , z2 , z3 ∈ C be three distinct points and w1 , w2 , w2 ∈ C be three further distinct points. Find the M¨obius transformation which maps the point zk onto wk , 1 ≤ k ≤ 3. Thus we have to find azk +b , k = 1, 2, 3. a, b, c, d, ac − bd 6= 0 such that wk = cz k +d

b) Find the M¨obius transformation mapping 0 onto i, −i onto 1 and 1 onto −i.

12. Find the image of R ⊂ C under the following two M¨obius transformations: a) w(z) = b) w(z) =

z−i ; z+i z−i ,z z+i

6= −1.

13. Let w(z) = az+b be a M¨obius transformation with c 6= 0. Find conditicz+d ons for a, b, c, d such that w admits exactly two fixed points ζ1 , ζ2 ∈ C, i.e. w(ζk ) = ζk for k = 1, 2. 317

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17

Complex Differentiation

Let f : D → C be a function defined on a non-empty open subset of C. For z, z0 ∈ D, z 6= z0 , the difference quotient f (z) − f (z0 ) z − z0

(17.1)

is well defined. This observation makes the main difference between complex analysis, i.e. the analysis of a complex-valued function depending on a complex variable and real analysis of mappings from an open set in R2 to R2 . For the latter mapping the difference quotient cannot be defined. We can divide by a non-zero element in the field C, but we cannot divide by a non-zero element in the vector space R2 . Once (17.1) makes sense we may try to pass in (17.1) to the limit z → z0 as we do this in the case of a real-valued function defined on an open set in R. Before starting to investigate the limit of (17.1) as z tends to z0 let us agree to the following conventions commonly used in complex analysis: A non-empty open set D ⊂ C is called a domain in C; A connected domain is called a region in C. Note that C always carries the topology generated by the metric induced by the modulus |.|, hence connectivity is well-defined, compare with Definition II.3.30. If not otherwise stated we assume functions to be defined on a domain. Definition 17.1. Let f : D → C be defined on the domain D ⊂ C. A. We call f (complex) differentiable of z0 ∈ D, or we say that f has the (complex) derivative f 0 (z0 ) at z0 if df f (z) − f (z0 ) (z0 ) := f 0 (z0 ) := lim z→z0 dz z − z0

(17.2)

exists. B. We call f holomorphic in the domain D if f is for every z ∈ D differentiable. If f is holomorphic on C we call f an entire function. 319

A COURSE IN ANALYSIS

Obviously, if f (z) = c ∈ C is a constant function then f 0 (z) = 0 for all z ∈ C. The following results are proved by using results on limits in metric spaces, more precisely in C, and do not differ in their structure from those given for the case of real-valued functions of a real variable. Corollary 17.2. If f : D → C is differentiable of z0 ∈ C then f is continuous at z0 . In particular, holomorphic functions are continuous. Proposition 17.3. For functions f, g : D → C defined on a domain D ⊂ C and differentiable at z0 ∈ D the functions f ± g, af (a ∈ C), and f · g are differentiable at z0 and the following hold: (f ± g)0 (z0 ) = f 0 (z0 ) ± g 0 (z0 );

(17.3)

(af )0 (z0 ) = af 0 (z0 );

(17.4)

(f · g)0(z0 ) = f 0 (z0 ) g(z0) + f (z0 ) g 0(z0 ) (Leibniz rule).

(17.5)

In addition, if g(z0 ) 6= 0 then fg is defined in an open neighbourhood of z0 and differentiable at z0 with derivative  0 f 0 (z0 ) g(z0) − f (z0 ) g 0(z0 ) f (z0 ) = g g(z0 )2

(quotient rule).

(17.6)

We deduce from Proposition 17.3 that the set of all holomorphic functions on a domain is with the natural operations an algebra over C which we denote by H(D). A further consequence of Proposition 17.3 is that all polynomials N P p : C → C, p(z) = ak z k , are (complex) differentiable. For this we need k=0

only to note that for f (z) = z we have lim

z→z0

z − z0 f (z) − f (z0 ) = lim = 1. z→z0 z − z0 z − z0

Further, rational functions, i.e. quotients h = fg of polynomials f and g are differentiable on the set {z ∈ C | g(z) 6= 0}. Since g is continuous it follows that {z ∈ C | g(z) 6= 0} = g −1 (C\{0}) is open, i.e. a domain. The proof of the chain rule for real-valued functions of a real variable also extends to the complex case: 320

17

COMPLEX DIFFERENTIATION

Proposition 17.4. Let D1 , D2 ⊂ C be two domains and f : D1 → C and g : D2 → C be two functions. Suppose that f (D1 ) ⊂ D2 and f is differentiable at z0 ∈ D1 as well as g is differentiable at ω0 := f (z0 ) ∈ D2 . Then g ◦ f : D1 → C is well defined, differentiable at z0 and we have (g ◦ f )0 (z0 ) = g 0(f (z0 )) f 0 (z0 ).

(17.7)

We now want to discuss the difference between complex differentiability and differentiability of mappings h : U → R2 , U ⊂ R2 open. For this we write z = x + i y and identify sometimes C with R2 . For a function f : D → C, D ⊂ C, we obtain two new functions u, v : D → R by decomposing f into real and imaginary part. f (z) = u(z) + i v(z),

u = Ref, v = Imf.

We also may write f (x, y) = u (x, y) + i v(x, y) or even f (z) = u (x, y) + i v(x, y). for z = x + i y ∈ D. This allows us for example to take partial derivatives of u and v with respect to x and y. Now let f : D → C be complex differentiable at z ∈ D. Since D is open we can find a disc Dr ( z) ⊂ D for r > 0 sufficiently small. Let h ∈ R such that z + h, z + i h ∈ D. Thus we can consider the two difference quotients f (z + h) − f (z) f (x + h, y) − f (x, y) = h h

(17.8)

and

f (x, y + h) − f (x, y) f (z + i h) − f (z) = . ih ih Passing in (17.8) to the limit as h tends to 0 we obtain f (z + h) − f (z) f (x + h, y) − f (x, y) = lim h→0 h→0 h h u (x + h, y) − u (x, y) + i (v(x + h, y) − v (x, y)) = lim h→0 h = ux (x, y) + i vx (x, y),

f 0 (z) = lim

321

(17.9)

A COURSE IN ANALYSIS

while passing in (17.9) to the limit as h tends to 0 we find f (z + i h) − f (z) f (x, y + h) − f (x, y) = lim h→0 ih ih (u (x, y + h) − u (x, y)) + i (v (x, y + h) − v (x, y)) = lim h→0 ih 1 = (uy (x, y) + i vy (x, y)) = vy (x, y) − i uy (x, y). i

f 0 (z) = lim

h→0

By calculating these limits we have used the fact that a limit in C exists if and only if the corresponding limits for the real and imaginary part exists. Further we used our standard notation of partial derivatives, i.e. ux = ∂u ∂x etc. From the considerations made above we get Theorem 17.5. If f : D → C, f = u + i v, is differentiable at z = x + i y then the Cauchy-Riemann differential equations ux (x, y) = vy (x, y) and vx (x, y) = −uy (x, y)

(17.10)

must hold at z = x + iy. Proof. We know that ux (x, y) + i vx (x, y) = f 0 (z) = vy (x, y) − i uy (x, y), but the real and imaginary part of the complex number f 0 (z) is uniquely determined. Since f : D → C can be viewed as a function defined on a subset D ⊂ R2 with values in R2 and its differential at (x, y) ∈ D would be the Jacobi matrix   ux (x, y) uy (x, y) . vx (x, y) vy (x, y) In order that this matrix represents a complex number we need to have (as discussed in Chapter 13) ux (x, y) = vy (x, y) and vx (x, y) = −uy (x, y), i.e. the Cauchy-Riemann differential equations must hold. Thus complex differentiability is a much more restrictive notion compared with differentiability for functions f : D → R2 , D ⊂ R2 . Indeed we have 322

17

COMPLEX DIFFERENTIATION

Theorem 17.6. Let f : D → C be a mapping. Then f is complex differen  u : D → R2 is real differentiable tiable at z = x + iy ∈ D if and only if v at (x, y) ∈ D and the Cauchy-Riemann differential equations hold. Proof. We have seen that complex differentiability implies   differentiability in u at (x, y) represents R2 . We have also seen that the Jacobi matrix of v a complex number if and only if the Cauchy-Riemann differential equations hold. We know in the real-valued case that the existence of partial derivatives does not imply the differentiability of function f : D → R2 , D ⊂ R2 . However if all first order partial derivatives are continuous, then the function is differentiable. This yields Theorem 17.7. Let u, v : D → R be functions having continuous first order partial derivatives and suppose that u, v satisfy the Cauchy-Riemann differential equations. Then the function f : D → C, D ⊂ C, f (z) = u(z) + iv(z) is complex differentiable. We want to study some examples for the Cauchy-Riemann differential equation. Example 17.8. Consider the function f : C → C, f (z) = z. Thus with u(x, y) = x and v(x, y) = y we have f (z) = u(z) + iv(z), or f (x, y) = u(x, y) + iv(x, y), and moreover ux (x, y) = 1, vx (x, y) = 0, uy (x, y) = 0, vy (x, y) = 1 and we find as expected vx = 0 = −uy .

ux = 1 = vy ,

Now let us consider f (z) = z = x − iy. The real part is again u(x, y) = x while the imaginary part is v˜ (x, y) = −y. We also find ux (x, y) = 1, v˜x (x, y) = 0, uy (x, y) = 0, v˜y (x, y) = −1 and consequently ux (x, y) 6= v˜y (x, y). Therefore the function   z 7→ z is u : D → R2 not complex differentiable. Nonetheless the real function v˜ 323

A COURSE IN ANALYSIS

  1 0 . One consequence of this obis differentiable with differential 0 −1 servation is that while C and the set C(D) of all continuous functions f : D → C, D ⊂ C, are algebras with involutions, namely z 7→ z and f 7→ f , respectively, for the algebra of holomorphic functions H (D) the mapping f 7→ f is not an involution since in general f ∈ / H (D) for f ∈ H (D). Example 17.9. For the complex exponential function we know the equality, z = x + iy, exp (z) = ex (cos y + i sin y) = ex cos y + i ex sin y. With u (x, y) = ex cos y and v (x, y) = ex sin y we find ux (x, y) = ex cos y, uy (x, y) = −ex sin y, vx (x, y) = ex sin y, vy (x, y) = ex cos y, and it follows that ux (x, y) = ex cos y = vy (x, y) vx (x, y) = ex sin y = −uy (x, y). Thus exp : C → C is complex differentiable. Example 17.10. Since the line {z = x + iy ∈ C | Rez = x = 0} is closed the set C\{z ∈ C | Re z = 0} is a domain on which we can define the function L(z) :=

y 1 ln (x2 + y 2 ) + i arc tan = u(x, y) + i v(x, y), 2 x

(17.11)

where we still use ln and arc tan as defined in Volume I as inverse functions to the real-valued exponential function ξ 7→ eξ , ξ ∈ R, and inverse to the tangent function restricted to (− π2 , π2 ). Taking partial derivatives of u and v we obtain x y ux (x, y) = 2 , uy (x, y) = 2 2 x +y x + y2 as well as vx (x, y) = −

y , x2 + y 2

vy =

x . x2 + y 2

Hence we find ux = vy and vx = −vy on C\{z ∈ C | Re z = 0} implying that L is on this domain complex differentiable. 324

17

COMPLEX DIFFERENTIATION

Example 17.11. Let G ⊂ C be a region and f : G → C, f = u + i v, a function. Assume that Ref = u = c ∈ R in G, i.e. u is a constant function. If f is not constant then f is not complex differentiable. Indeed, if f was complex differentiable then the Cauchy-Riemann differential equations imply ux = vy and uy = −vx , but since u = c it follows that ux = uy = vx = vy = 0. Since G is connected this implies that v must be constant too, see Problem 14 of Chapter II.6. Note that if G is not a region but a domain, then f must be constant on every connectivity component of G but on each component the constant may have a different value. Such a function is locally constant in the sense that every point of its domain has an open neighbourhood on which it is constant. Exercise 17.12. Let f : G → C be a function which on the region G ⊂ C satisfies either Imf = c ∈ R or |f | = c ∈ R. Prove that if f is complex differentiable then f must be constant. Example 17.13. Let f : D → C, D ⊂ C,  be  a function with f = u + iv. u from D ⊂ R2 to R2 . The We can also interpret f as a mapping g = v Jacobi determinant of g has some geometric meaning, for example when we think of the transformation theorem for integrals. Assuming that f is complex differentiable, hence u and v satisfy the Cauchy-Riemann differential equations we find   ux uy det Jg (x) = det = ux vy − uy vx . vx vy = u2x + u2y = vx2 + vy2 .

Using further that u2x = vy2 and u2y = vx2 and our calculation leading to the Cauchy-Riemann differential equations we find |f 0(z)|2 = f 0 (z) f 0 (z) = (ux + i vx )(ux − i vx ) = u2x + vx2 = u2x + u2y . Thus we arrive at the remarkable fact that   ux (x, y) uy (x, y) 0 2 ≥ 0. |f (z)| = det vx (x, y) vy (x, y) 325

(17.12)

A COURSE IN ANALYSIS

Polar coordinates are quite helpful in many situations and if f : D → C, D = B% (0) = {z ∈ C | |z| < %} = {(x, y) ∈ R2 | x2 + y 2 < %2 }, is complex differentiable with real part u and imaginary part v we may consider u and v as functions of the polar coordinates (r, ϕ). The Cauchy-Riemann differential equations in polar coordinates read as ∂u 1 ∂v ∂v ∂u = and =− , ∂r r ∂ϕ ∂r ∂ϕ

(17.13)

see Problem 6. We want to enlarge the class of examples of complex differentiable functions by looking at power series with postive radius of convergence. Recall that by ∞ P ak (z − c)k in B% (c) enAbel’s theorem, Theorem 16.6 the convergence of k=0

∞ P

tails the convergence of the series

k=1

ak k(z −c)

k−1

in B% (c) and in both cases

the convergence is uniform if % < %0 where %0 is the radius of convergence ∞ P ak (z − c)k . By the Cauchy-Hadamard theorem, Theorem 16.10, we of k=0 √ 1 . It is worth noting that lim k k = 1 further know that %0 = lim inf √ k |ak |

k→∞

implies that

k→∞

1 = lim inf p , k k→∞ |ak |k k→∞ |ak | ∞ P i.e. %0 is also the radius of convergence of the series ak k(z − c)k−1. Since lim inf p k

d dz

1

k=1

(ak (z − c)k ) = ak k (z − c)k−1 we are led to conjecture that

∞ ∞ ∞ X X d X d k k (ak (z − c) ) = ak (z − c) = ak k (z − c)k−1 , dz k=0 dz k=1 k=0

which we are now going to prove. For every complex differentiable function   d df (z − c) = f 0 (z − c) f (z − c) = dz dz holds and therefore we may assume without loss of generality that c = 0. Theorem 17.14. Let T(ak ) (z) =

∞ P

ak z k have radius of convergence %0 > 0.

k=0

Then the function z 7→ T(ak ) (z) is in B%0 (0) ⊂ C complex differentiable and 326

17

COMPLEX DIFFERENTIATION

we have 0 T(a k)

(z) =

∞ X

ak k z k−1 .

(17.14)

k=1

0 Moreover, the series T(a (z) has again %0 as radius of convergence. k)

Remark 17.15. We already know this result for power series with real coefficients of a real variable, compare with Corollary I.29.7. However the proof of this corollary indirectly depends on the fundamental theorem of calculus which is now not at our disposal. Hence we need a different proof which we have adapted from [67]. Proof. We := T(ak ) (z) and P set f (z) k−1 k a z . We have to prove that f 0 (z) = g(z) for z ∈ B%0 (0). g(z) := ∞ k k=1 Fix z0 ∈ B%0 (0) and define for k ∈ N gk (z) := z k−1 + z k−2 z0 + .... + z k−j z0j−1 + ... + z0k−1 ,

z ∈ C.

Note that gk (z0 ) = k z0k−1 . It follows that f (z) − f (z0 ) =

∞ X k−1

ak (z k − z0k ) ∞ X

= (z − z0 )

ak gk (z),

k=1

With f1 (z) :=

∞ X

z ∈ B%0 (0).

ak gk (z)

k=1

we find and

f (z) = f (z0 ) + (z − z0 ) f1 (z), z ∈ B%0 (0), f1 (z0 ) =

∞ X

k ak z0k−1 = g(z0 ).

k=1

Suppose for a moment that f1 is continuous at z0 . Then, since f (z) − f (z0 ) = f1 (z) z − z0 327

A COURSE IN ANALYSIS

we find that f 0 (z0 ) = lim

z→z0

f (z) − f (z0 ) = f1 (z0 ) = g(z0 ) z − z0

and the result was proved. The continuity of f1 is shown as follows. For B% (0), |z0 | < % < %0 it follows that ||ak gk (.)||∞,B% (0) ≤ |ak |k%k−1

and therefore

∞ X k=1

||ak gk (.)||∞,B% (0) ≤

∞ X k=1

k|ak |%k−1 < ∞

P∞ by our previous considerations. Hence the series k=1 ak gk (z) converges uniformly in B% (0) to a continuous function. Since z0 ∈ B% (0) the result follows. Suppose that f : D → C is holomorphic in D. Then f 0 : D → C defines a new complex-valued function in D which might be complex differentiable. Thus as in the case of real-valued functions of a real variable we can define for complex-valued functions of a complex variable higher order derivatives k which we denote with ddz fk (z) or f (k) (z) or for k small by f 00 (z), f 000 (z), etc. As we can derive Corollary I.29.8 from Corollary I.29.7, we can now derive from Theorem 17.14. ∞ P Corollary 17.16. Let f (z) = ak (z − c)k be a power series with radius of k=0

convergence %0 > 0. The function f is arbitrarily complex differentiable in B% (c) and for the coefficients ak ∈ C we find ak =

1 (k) f (c). k!

(17.15)

Proof. We may iterate the arguments of the proof of Theorem 17.14 to find for z ∈ B%0 (0) and m ∈ N that f (m) (z) =

∞ X dm k(k − 1)· .....· (k − m + 1) ak (z − c)k−m (17.16) f (z) = dz m k=m

which implies already that f is arbitrarily often differentiable and setting in (17.16) z = c formula (17.15) follows. 328

17

COMPLEX DIFFERENTIATION

Using the definition of the binomial coefficients we can rewrite (17.16) as   ∞ X k m f (z) = a (z − c)k−m . (17.17) m! m k k=m

Remark 17.17. In Chapter 22 we will prove that if f : D → C is holomorphic then for every z0 ∈ D there exists a radius ρ > 0 such that in Bρ (z0 ) the function f has a representation as a convergent power series. This implies that any holomorphic function is arbitrary often differentiable, i.e. one time differentiable implies arbitrarily often differentiable. Such a result is clearly false in the case of real-valued functions.

We now can deduce that all functions defined in (16.14)-(16.20) and (16.30) are in B%0 (0), %0 being the radius of convergence, complex differentiable, infact they are arbitrarily often complex differentiable in B%0 (0). For exp, cos, sin and the Bessel functions Jl , l ∈ N0 , this implies that they are entire functions. At the same time the question arises whether a given power series c f (z) = T(a (z) with radius of convergence %0 > 0 has a holomorphic extenk) ∞ P sions to a larger domain. For example the power series z k has radius of convergence 1 and

∞ P

k=0

zk =

k=0

1 1−z

holds for |z| < 1. The function z 7→

1 1−z

is a

rational function being defined and complex differentiable for all z ∈ C\{1}. 1 Thus we may consider z 7→ 1−z as a holomorphic extension or continua∞ P k tion of the power series z . We will return to the problem of the existence k=0

of a holomorphic extension of a given holomorphic function at several occasions. Once we know that a certain function f : D → C, D ⊂ C, is complex differentiable we can apply our rules from Proposition 17.3 and Proposition 17.4 to calculate the complex derivatives. In particular we obtain for f extending ∞ P ak (x − c)k , ak , c ∈ R, converging in the disc B% (c) ⊂ C a power series k=0

0

that f (z) is the complex extension of

immediately that

d ( dx

∞ P

(x − c)k ). Therefore we find

k=0

d exp(z) = exp(z), dz d cos(z) = − sin(z), dz 329

(17.18) (17.19)

A COURSE IN ANALYSIS

d sin(z) = cos(z) dz

(17.20)

hold for all z ∈ C. Definition 17.18. Let G1 and G2 be two regions and f : G1 → G2 a holomorphic mapping. We call f biholomorphic if f is bijective with holomorphic inverse f −1 : G2 → G1 . Two regions G˜1 and G˜2 are called holomorphically equivalent if there exists a biholomorphic mapping f˜ : G˜1 → G˜2 . The Riemann mapping theorem, see Chapter 28, will classify all regions holomorphically equivalent to the unit disc B1 (0). This theorem is one of the major results of complex analysis. For the moment we note a further result which is essentially obtained by carrying over the proof of the corresponding result for a real-valued function of a real variable. Proposition 17.19. A mapping f : G1 → G2 between two regions is biholomorphic if and only if f is holomorphic, bijective, f −1 is continuous and f 0 (z) 6= 0 for all z ∈ G1 . In this case we have (f −1 )0 (w) =

1 , f 0 (z)

i.e. (f −1 )0 (w) =

w = f (z),

1 . f 0 (f −1 (w))

(17.21)

(17.22)

For a proof we refer to Problem 9. We want to return to the Cauchy-Riemann differential equations. Let f : D → C, f = u + iv, be a complex differentiable, i.e. a holomorphic function. If f has higher order complex derivatives, for example in the case where f is given by a convergent power series, then u and v must also have higher order partial derivatives. In light of Remark 17.17, eventually the assumption that u and v have continuous second order partial derivatives uxx , uxy = uyx , uyy and vxx , vx,y = vy,x , vyy will be no restriction. Recall that w ∈ C 2 (G), G ⊂ 2 2 R2 , is called harmonic in G if ∆2 w = ddxw2 + ddyw2 = wxx + wyy = 0 in G, compare with Chapter II.9. Since in this part of the Course we deal essentially only with functions of one complex variable or two real variables we will write here ∆ for ∆2 . 330

17

COMPLEX DIFFERENTIATION

Theorem 17.20. We assume that the real part and the imaginary part of the holomorphic function f : D → C, f = u + iv, have all second order partial derivatives in D and that they are continuous. Both, u and v are then harmonic functions in D, i.e. in D we have ∆u = uxx + uyy = 0 and ∆v = vxx + vyy = 0.

(17.23)

Proof. Since f is holomorphic in D the Cauchy-Riemann differential equations hold in D, i.e. ux = vy and uy = −vx . Further, by assumption all second order partial derivatives of u and v exist and are continuous and therefore it follows that uxx = vyx , uxy = vyy and uyy = −vxy , uyx = −vxx and in addition uxy = uyx and vxy = vyx . Thus we find uxx + uyy = vyx − vxy = 0 and vxx + vyy = −uyx + uxy = 0. Example 17.21. A. Since z 7→ exp (z) is an entire function Re exp and Im exp are harmonic functions in R2 . Hence Re exp (x, y) = ex cos y and Im exp (x, y) = ex sin y are harmonic on R2 . B. From Example 17.10 we deduce that (x, y) 7→

1 y log (x2 + y 2) and (x, y) 7→ arctan , x 6= 0, 2 x

are harmonic functions. N P C. If p (z) = ak z k , ak ∈ C, is a polynomial then Re p(x, y) and Im p (x, y) k=0

are polynomials in two real variables which are on R2 harmonic functions. For p (z) = z k , k = 1, 2, 3, 4 we find the following harmonic polynomials: P1r (x, y) = x, P1i (x, y) = y,

(17.24)

P2r (x, y) = x2 − y 2, P2i (x, y) = 2xy,

(17.25)

P3r (x, y) = x3 − 3xy 2, P3i (x, y) = −y 3 + 2x2 y, 4

4

2 2

3

3

P4r (x, y) = x + y − 6x y , P4i (x, y) = 4x y − 4xy . 331

(17.26) (17.27)

A COURSE IN ANALYSIS

As it turns out there is a close relation between harmonic functions of two real variables and holomorphic functions. We will discuss this relation in much more detail once we can resolve the following central question: given a harmonic function u on the open set D ⊂ R2 . Can we find a second harmonic function v in D such that f (z) = u(z) + i v(z) = u(x, y) + i v(x, y) is holomorphic in D? When this is true we call v a conjugate harmonic function to u in D. Before we continue to develop further the general theory of holomorphic functions, in the next chapter we will discuss in more detail some concrete functions f : D → C. This is followed by deepening our understanding of some topological concepts in Chapter 19.

Problems In the following G ⊂ C is always a region. Where appropriate we consider G as a subset of R2 , z = x + iy ∈ G, (x, y) ∈ G. 1. Let f, g : G → C be m-times complex differentiable functions. Prove that f · g : G → C is also m-times complex differentiable and we have m   X dm m (m−l) (f · g)(z) = f (z)g (l) (z). m dz l l=0

2. Solve Exercise 17.12. 3. Let f : [0, ∞) → R be a C ∞ -function and consider the function g : C → C, g(z) := f (|z|2 ). Is g complex differentiable? 4. Let f : G → C be a complex differentiable function. Denote by G∗ the set G∗ := {x ∈ C|z = w, ¯ w ∈ G}. Sketch the set G∗ and prove that ∗ z ) is complex differentiable. H : G → C, h(z) := f (¯ 2

5. Decompose f : C → C, f (z) = ez + z 3 , into its real and imaginary parts and verify the Cauchy-Riemann differential equations for f . 6. Find the Cauchy-Riemann differential equations in polar coordinates, i.e. verify (17.13). 332

17

COMPLEX DIFFERENTIATION

7. Prove that the following functions are complex differentiable: P einz a) ∞ n=1 n4 , Im z > 0; P 2n b) ∞ n=1 z n , |z| > 2. 8. Prove that w : C \ {−1} → C \ {1}, w(z) = mapping.

z−1 , z+1

is a biholomorphic

9. Provide a proof for Proposition 17.19. 10. For z = x + iy ∈ B1 (0) we define with ζ ∈ ∂B1 (0) fixed the Poisson kernel P (ζ, z) by 1 1 − |z|2 . P (ζ, z) := 2π |ζ − z|2 With z = reiϕ , ζ = eiϑ , prove that P (ζ, z) =

1 1 − r2 . 2π 1 + r 2 − 2r cos(ϑ − ϕ)

Moreover, show that for ζ ∈ ∂B1 (0) fixed P (ζ, z) = P (ζ, (x, y)) is a harmonic function in B1 (0) ⊂ R2 . ζ+z . Hint: consider the real part of z 7→ ζ−z 11. Let f : G1 → G2 be a holomorphic function and consider G1 and G2 also as subsets of R2 . Let h : G2 → R be a harmonic function. Is the function h ◦ f : G1 → R a harmonic function where with f = u + iv and z = x + iy the meaning of h ◦ f is (h ◦ f )(z) = (h ◦ f )(x + iy) = h(u(x, y), v(x, y))?

333

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18

Some Important Functions

In this chapter we want to discuss some of the so called elementary transcendental functions defined on complex domains. We start with the exponential function exp defined by the power series z

exp(z) := e :=

∞ X zk k=0

k!

(18.1)

which converges in C and hence exp is an entire function. By differentiating in (18.1) term by term we find that exp satisfies the differential equation d z e = ez dz

(18.2)

and further we have exp (0) = 1. Since d z −z (e e ) = ez e−z − ez e−z = 0 dz and since C is connected it follows that ez e−z = 1 or (ez )−1 = e−z ,

(18.3)

which implies that e 6= 0 for all z ∈ C. Moreover, with the same arguments as given in the proof of Lemma I.9.7 we obtain the well-known functional equation for the exponential function, but now for complex values of the argument: ez+w = ez ew . (18.4) z

We have already seen that on C the exponential function is periodic with purely imaginary period 2πi: ez+2πi = ez for all z ∈ C.

(18.5)

Thus, while on R the function x 7→ exp (x) is injective with range (0, ∞), on C the function z 7→ exp (z) is not injective and hence we cannot expect to extend ln : (0, ∞) → R to become the inverse function of the complex exponential function. Indeed the logarithmic function is a much more difficult object on C which we start to investigate now. 335

A COURSE IN ANALYSIS

Definition 18.1. Let G ⊂ C\{0} be a region. A continuous function f : G → C is called a branch of the logarithmic function if for all z ∈ G we have ef (z) = z. (18.6) Remark 18.2. If f is a branch of the logarithmic function then exp |f (G) is the inverse to f , in particular f is injective. Furthermore, the functions fk : G → C, fk (z) = f (z) + 2kπi, k ∈ Z, are branches of the logarithmic function on G. Corollary 18.3. If f : G → C is a branch of the logarithmic function then all other branches of the logarithmic function on G are of the type fk (z) = f (z) + 2kπi, k ∈ Z.

Proof. Let g : G → C be a further branch of the logarithmic function on G. For all z ∈ G it follows that ef (z) = z = eg(z) , i.e. ef (z)−g(z) = 1 implying that f (z) − g(z) = 2πi h(z) with h(z) ∈ Z. But h must be a continuous function on the region G, hence it must be constant. Theorem 18.4. Every branch f : G → C, G ⊂ C\{0} a region, is on its domain a holomorphic function and we have 1 f 0 (z) = , z

z ∈ G.

(18.7)

Conversely, if f : G → C is holomorphic and satisfies (18.7) and if ef (z0 ) = z0 holds for some z0 ∈ G, then f is a branch of the logarithmic function in G. Proof. The mapping f : G → f (G) is bijective with inverse exp : f (G) → G. For z, z0 ∈ G, z 6= z0 , we set w = f (z) and w0 = f (z0 ) and we observe that w − w0 f (z) − f (z0 ) = w = z − z0 e − ew0

1 ew −ew0 w−w0

.

(18.8)

Since f is continuous, z → z0 implies w → w0 and therefore the limit z → z0 in (18.8) exists and it follows that lim

z→z0

f (z) − f (z0 ) 1 1 = w0 = . z − z0 e z0

We now prove the converse statement. For this let g(z) := z e−f (z) which is in G holomorphic and satisfies g 0(z) = e−f (z) − z f 0 (z) e−f (z) = 0 since z f 0 (z) = 1. Since G is a region it follows that g = c ∈ C\{0}, i.e. z = c ef (z) in G. By assumption we have z0 = ef (z0 ) implying that c = 1. 336

18

SOME IMPORTANT FUNCTIONS

Note that so far we do not know whether for some region G ⊂ C\{0} a branch of the logarithmic function exists. This problem is solved by the following existence result: Theorem 18.5. On B1 (1) = {z ∈ C | |z −1| < 1} a branch of the logarithmic ∞ P (−1)k−1 ˜ : B1 (1) → C, L(z) ˜ function is given by L = (z − 1)k . k=1

∞ ˜ + z) = P Proof. We know that ln(1

k=1

(−1)k−1 k

k

z k converges in B1 (0), hence it

is holomorphic in B1 (0), and differentiating term by term yields ∞

∞

X X 1 d ˜ ln (1 + z) = . (−1)k−1 z k−1 = (−z)k = dz 1 + z k=1 k=0 ˜ (z −1), z ∈ B1 (1), which is holomorphic ˜ we find L(z) ˜ For the function L = ln 1 0 ˜ ˜ (z) = . In addition we have L(1) = 0 or eL(1) = 1. in B1 (1) and satisfies L z Now the second part of Theorem 18.4 yields the result. Exercise 18.6. For a ∈ C\{0} define La,b : B|a| (a) → C by La,b (z) := ˜ z ) where b is a logarithm of a, in particular eb = a. Prove that La,b is b + L( a a branch of the logarithmic function. Our aim is to now introduce the principal branch of the logarithmic function as the mapping log : C\(−∞, 0] → C. For this we represent z ∈ C\(−∞, 0] as z = |z| ei ϕ , −π < ϕ < π. Note that the domain of ϕ is now different to the case of polar coordinates and it is chosen such that the representation of z is unique for every z ∈ C\(−∞, 0]. Moreover the mapping z 7→ ln |z| is continuous from C\(−∞, 0] to R as the mapping z 7→ i ϕ is continuous from C\(−∞, 0] to {i ϕ | ϕ ∈ (−π, π)}. The set C\(−∞, 0] is often called the cut plane. Theorem 18.7. The function log : C\(−∞, 0] → C |z| ei ϕ , −π < ϕ < π, by log z := ln |z| + i ϕ

defined for z = (18.9)

where ln : (0, ∞) → R is the natural logarithm as defined in Chapter I.9 is a ˜ : B1 (1) → C branch of the logarithmic function which extends the function L from Theorem 18.5. 337

A COURSE IN ANALYSIS

Proof. The continuity of z 7→ ln |z| and z 7→ i ϕ, z = |z| ei ϕ with −π < ϕ < π, implies that log z is continuous from C\(−∞, 0] to C. Furthermore elog z = e(ln |z|+iϕ) = eln |z| eiϕ = |z|eiϕ = z holds, hence log is in the region C\(−∞, 0] a branch of the logarithmic ∞ P (−1)k−1 ˜ function. Since B1 (1) ⊂ C\(−∞, 0] and L(z) = (z − 1)k is a k k=1

˜ branch of the logarithmic function we know that log z = L(z) + c, c ∈ C, ˜ for all z ∈ B1 (1). However for z = 1 we find log 1 = L(1) = 0 implying that c = 0. Definition 18.8. The function log : C\(−∞, 0] → C defined by (18.9) is called the principal branch of the logarithmic function. Remark 18.9. A. We will use the following conventions: ln : (0, ∞) → R denotes the natural logarithm; log : C\(−∞, 0] → C denotes the principal branch of the logarithmic function; ˜ : B1 (0) → C ln

˜ + z) = P∞ (−1)k−1 z k . Further we use is defined as ln(1 k=1 k ˜ − z). ˜ : B1 (1) → C, L(z) ˜ L = ln(1

B. A consequence of (18.9) is the equality log i =

1 π i, 2

(18.10) π

since |i| = 1, i.e. ln |i| = 0, and i = cos π2 + i sin π2 = e 2 i . C. Further branches of the logarithmic function on C\(−∞, 0] are given by z 7→ log z + 2π i k, k ∈ Z. 338

18

SOME IMPORTANT FUNCTIONS

Next we want to study the functional equation for log. Let z1 , z2 ∈ C\(−∞, 0] such that z1 · z2 ∈ C\(−∞, 0] with representation z1 = |z1 |ei ϕ1 , z2 = |z2 |ei ϕ2 and z1 · z2 = |z1 · z2 |ei ϕ3 , −π < ϕj < π, j = 1, 2, 3. For ϕ3 we find ϕ3 = ϕ1 + ϕ2 + α where α ∈ {−2π, 0, 2π} and consequently we have log (z1 · z2 ) = ln |z1 ||z2 | + i ϕ3 = (ln |z1 | + i ϕ1 ) + (ln |z2 | + i ϕ2 ) + i α, i.e. log (z1 · z2 ) = log z1 + log z2 + i α.

(18.11)

, π ), hence ϕ1 + ϕ2 ∈ (−π, π) and For Re zj > 0, j = 1, 2, we have ϕj ∈ ( −π 2 2 α = 0 implying log (z1 · z2 ) = log z1 + log z2 ,

Re zj > 0.

(18.12)

Thus the functional equation for ln is in general to be replaced by (18.11) and for Re z1 , Re z2 > 0 it reduces to the one for ln. Next we want to study the relation log (exp (z)). First of all the definition of log requires that exp z 6= (−∞, 0] which means z = x + iy with x ≤ 0 and y = 0 must be excluded. Since ez = ex cos y + i ex sin y these conditions imply that y = (2k + 1)π, k ∈ Z, must be excluded. Otherwise, in G := C\{z ∈ C | Im z = (2k + 1)π, k ∈ Z} the term log (exp (z)) is defined, but note that G is not a region since it is not connected. Indeed we have Gk ∩ Gl = ∅ for k 6= l and [ G= Gk , Gk := {z ∈ C | (2k − 1)π < Im z < (2k + 1)π}. k∈Z

However each Gk , k ∈ Z, is a region, namely a strip in the complex plane parallel to the real axis and with width 2π. For z ∈ Gk we have ez = ex eiy = ex ei(y−2k π) with y − 2 k π ∈ (−π, π). This implies for z ∈ Gk that log(ez ) = ln ex + i (y − 2k π) = x + iy − 2k π i = z − 2π k i. In particular, for k = 0 we have log (ez ) = z, Hence we have proved the result. 339

z ∈ G0 .

A COURSE IN ANALYSIS

Lemma 18.10. The exponential functions maps the region G0 = {z ∈ C | − π < Im z < π} biholomorphically onto the cut plane C\(−∞, 0] with the principal branch of the logarithmic function being the inverse. Once the logarithmic function is at our disposal we can turn to power functions, i.e. we want to define z 7→ z α , α ∈ C, for z belonging to a region of C. Of course for n ∈ Z we have no problem to define z n . But already for a > 0 and b ∈ R we need for the definition of ab the function ln, recall ab := eb ln a .

(18.13)

Since exp is an entire function, whenever log z is defined, for β ∈ C we can use (18.13) as guide and define z β := eβ log z ,

(18.14)

but we know that some care is needed to deal with the logarithmic function and to have the relation elog z = z. The general and proper definition of a power function is therefore Definition 18.11. Let G ⊂ C\(−∞, 0] be a region and L : G → C be a branch of the logarithmic function. Further let β ∈ C. We define a branch of the power function z 7→ z β as pβ : G → C, pβ (z) := z β := eβ L(z) . The functional equation of the exponential function yields z α+β = e(α+β) L(z) = eα L(z) eβ L(z) = z α · z β ,

(18.15)

and further we have d βL(z) β d pβ (z) = e = β L0 (z) eβ L(z) = z β , dz dz z i.e. we have d pβ (z) = β z β−1 , z ∈ G, L : G → C, G ⊂ C\(−∞, 0]. dz

(18.16)

For β = n ∈ N we observe that pn has of course the holomorphic continuation to C given by z 7→ z n , and for k ∈ Z, k < 0, we can extend pk to C\{0} 1 , −k ∈ N, holomorphically. The key point to note is that by z 7→ z k = z −k the general power function is not defined on all of C, but as the logarithmic 340

18

SOME IMPORTANT FUNCTIONS

function it has branches, in fact for every β every branch of the logarithmic function can be used to define a branch of z 7→ z β . For certain values of β ∈ C the function pβ has extensions onto larger sets than the region on which its defining branch of the logarithmic function is originally given, recall the cases β ∈ N or β ∈ Z, β < 0. An interesting special result, an observation made already by L. Euler, is the following: Since i = 0 + i, i.e. Re i = 0 and Im i = 1 ∈ (−π, π), it follows that i ∈ G0 . Hence i is in the domain of the principal branch of the logarithmic function and we have ii = ei log i . Using (18.10) we find further 1

ii = ei log i = ei ( 2 πi) = e

−π 2

,

(18.17)

and the surprising fact is that ii is a real number. Of special interest are roots or root functions, i.e. power functions of type p 1 , n ∈ N. We choose a fixed branch L : G → C of the logarithmic function n

2πi

1

and observe that with ζn := e n further branches of p 1 (z) = e n L(z) are given n by p 1 ,k (z) = ζnk p 1 (z), k = 1, ....., n−1. Thus given L we obtain in the region n n where L is defined at least n branches of p 1 . n

Now we can better understand our definition of the square root function on (0, ∞). We√have picked the principal branch of the logarithmic function to define x 7→ x on (0, ∞) and by this we made the square root of a positive 2π i real number uniquely defined. For n = 2 we have e 2 = eπ i = −1 =√ζ21 and p 1 ,1 (z) = −p 1 (z). Thus for x ∈ (0, ∞) we obtain now p 1 ,1 (x) = − x. In 2 n 2 Problem 3 we will discuss the formula ∞   X β k β (1 + z) = z , z ∈ B1 (0), β ∈ C. (18.18) k k=0

The functions sin and cos can be easily extended to C by their power series representations and the relation ei z = cos z + i sin z

(18.19)

follows for all z ∈ C. Moreover the two symmetries cos z = cos (−z) and sin z = − sin (−z) continue to hold, hence we have e−i z = cos z − i sin z 341

(18.20)

A COURSE IN ANALYSIS

which with (18.19) yields cos z =

ei z − e−i z ei z + ei z and sin z = . 2 2i

(18.21)

From (18.21) it is easy to extend typical addition theorem for trigonometrical functions defined on R to C. Furthermore we find with z = x + iy and taking into account the power series representations of sinh and cosh that Re (cos z) = cos x cosh y,

Im (cos z) = − sin x sinh y

(18.22)

Im(sin z) = cos x sinh y.

(18.23)

and Re (sin z) = sin x cosh y, For z ∈ C\{(k +

1 )π | k 2

∈ Z} the extension of tan is given by tan z =

sin z 1 e2iz − 1 = cos z i e2iz + 1

(18.24)

and for z ∈ C\{k π | k ∈ Z} we get cot z =

e2i z + 1 cos z = i 2i z . sin z e −1

(18.25)

Again, making use of the exponential function and the power series representations of sin, cos, sinh and cosh we find sinh z =

ez − e−z 1 = sin(i z) 2 i

(18.26)

ez + e−z = cos (i z) 2

(18.27)

and cosh z = as well as

1 sinh(i z) and cos z = cosh(i z). (18.28) i More complicated is the discussion of the inverse functions for trigonometrical and hyperbolic functions, therefore we only want to discuss arctan in more detail. One message we get from our considerations made above on concrete (holomorphic) functions such as M¨obius transformations, the exponential function, the logarithmic function, etc., is that much care is needed to study their mapping properties. For example the discussion of the principal branch sin z =

342

18

SOME IMPORTANT FUNCTIONS

of the logarithmic function requires some knowledge of the image of G0 := {z ∈ C | − π < Im z < π} under the exponential mapping. In case of w = tan z, using (18.24), it is formally easy to find z as a function of w. We start with 1 e2iz − 1 w = tan z = i e2iz + 1 2i z and with v := e we find 1 v−1 w= i v+1 or 1 + iw v= 1 − iw and therefore 1 + iw 1 log . z= 2i 1 − iw Thus the formula for arctan suggested by this calculation is arctan w =

1 1 + iw log , 2i 1 − iw

1+iw . Note 1−iw 

that w = 1i v−1 v+1  1 −1 with deteris a M¨obius transformation with corresponding matrix i i   i 1 1+iw . minant 2 i 6= 0 and inverse v = 1−iw with corresponding matrix −i 1 From Problem 8 of Chapter I.29 we know that for |x| < 1 but we know that there are problems to define log

arctan x =

∞ X

(−1)k

k=0

x2k+1 2k + 1

and from Problem 6 of Chapter I.29 we know that for |x| < 1 ∞

1 1 + x X x2k+1 ln = . 2 1 − x k=0 2k + 1 Both power series converge absolutely and uniformly for |z| < 1 and with z = ix, |x| < 1, we find ∞ X (ix)2k+1 k=0

2k + 1

=i

∞ X k=0

343

(−1)k

x2k+1 , 2k + 1

A COURSE IN ANALYSIS

or arctan(ix) =

1 1 + ix ln , 2 1 − ix

x ∈ (−1, 1).

We want to extend the equality to z ∈ B1 (0) ⊂ C. Consider the M¨obius . In Problem 5 we will show that w(B1 (0)) = transformation w(z) = 1+z 1−z {ζ ∈ C|Re ζ > 0}. This implies that for the principal part of the logarithmic function log we can define W (z) := log w(z),

z ∈ B1 (0),

and W is in B1 (0) a holomorphic function with W 0 (z) =

z w 0 (z) = . w(z) 1 − z2

With arctan z defined for z ∈ B1 (0) by the convergent power series we consider F (z) := W (iz) − 2i arctan z which is holomorphic in B1 (0) and has the derivative F 0 (z) = iW 0 (iz) + 2i(arctan z)0 =

2i 2i − = 0. 2 1+z 1 + z2

Hence F must be constant on B1 (0) and since F (0) = 0 it follows that arctan z =

1 1 + iz log , 2i 1 − iz

z ∈ B1 (0).

(18.29)

Note that (18.29) is, at the moment, an equality for power series converging in B1 (0), but in Chapter 22 we will identify the power series on the left hand side with a unique holomorphic extension of arctan as defined for a real-valued argument. With similar arguments one can first obtain “algebraic” formulae for arcsin, arccos, arccot as well as arcsinh, arccosh and arctanh and then one can also 344

18

SOME IMPORTANT FUNCTIONS

determine the domain of the corresponding principal branch: √ 1 arcsin w = log (iw + 1 − w 2 ) on C\{w ∈ C | |Re w| ≥ 1}, i √ 1 arccos w = log (iw + w 2 − 1) on C\{w ∈ C | |Re w| ≤ 1}, i   w+i 1 on C\{w ∈ C | |Im w| ≤ 1}, log arccot w = 2i w−i √ arcsinh w = log (w + w 2 + 1) on C\{w ∈ C | |Im w| ≥ 1}, √ arccosh w = log (w + w 2 − 1) on C\{w ∈ C | Re w ≤ 1},   1+w 1 on C\{w ∈ C | |Re w| ≥ 1}. arctanh w = log 2 1−w

(18.30) (18.31) (18.32) (18.33) (18.34) (18.35)

At this point it is worth reminding the reader that there are many tools available to find results such as the ones listed. Of course there are fast developing (and changing) internet resources. More classical resources are books, for example the one edited by M.Abramowitz and F.A.Stegun [1]. In order to learn new mathematics proofs, and typical examples must be worked through. Working mathematicians cannot do every calculation by themselves, they need to use tools from sources they can rely on - of course they must quote the resources they have used.

Problems √ 1. Use the principal branch of the logarithmic function to find ( 21 2e2π + √ 1 2e2π i)i . 2 2. Prove that in general for the principal branch of the logarithmic function we have log(z1 · z2 ) 6= logz1 + logz2 . 3. For x ∈ R, |x| < 1, and β ∈ C, β = s + it, show that ∞   X β k x = (1 + x)s (cos(t ln(1 + x)) + i sin(t ln(1 + x))). k k=0 Hint: use Problem 7 of Chapter 16.

4. Solve Exercise 18.6. 5. Prove that w(z) :=

1+z 1−z

maps B1 (0) onto the set {ζ ∈ C|Re ζ > 0}. 345

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Some More Topology

Line integrals of holomorphic and related functions turn out to be very helpful tools in investigating these functions. In fact they lead to completely new insights, including, surprisingly, about the relation between analysis and topology. For understanding these relations we first need to study in more detail the notion of a connected set in the plane. However, since many ideas and results hold for more general topological spaces and since we want to make use of these results when we study differential geometry or differentiable manifolds later on, we discuss them here either in the context of topological spaces or at least in the context of metric spaces. But there will also be a few results especially related to the plane. Note that from the topological or metrical point of view C and R2 are identical, but sometimes we may prefer to use complex numbers or complex-valued functions for the formulation of topological statements in the plane. In the first part of this chapter we mainly recollect definitions and results from Volume I and especially from Volume II. We start with Definition 19.1. Let (X, O) be a topological space. A. A pair (O1 , O2 ) of non-empty open subsets of X is called a splitting of X if O1 ∪ O2 = X and O1 ∩ O2 = ∅. B. A topological space is called connected if it does not admit a splitting. C. A subset of a topological space is called connected if it is connected as a topological space equipped with the relative topology induced by X. D. A region G ⊂ X is a non-empty open subset which is connected. We have seen these definitions before in I.19.24 for the real line and in II.3.30 for general metric spaces. Since only topological notions are involved in these definitions they extend easily to topological spaces. In addition many of the results proved in Chapter II.3 do hold with the same proof in topological spaces. Here is a collection of such results. Theorem 19.2. Let (X, O) be a topological space. A. The space is connected if and only if X and ∅ are the only sets which are both, open and closed. (See Remark II.3.3.1.) B. The image of a connected space under a continuous mapping is connected, or in other words, the image of a connected subset of a topological space under a continuous mapping is connected. (See Theorem II.3.32.)T C. Let (Ak )k∈I be a family of connected subsets of X. If k∈I Ak 6= ∅ then 347

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S

Ak is connected too. (See Theorem II.3.34.)

k∈I

D. If Y ⊂ X is connected and Y ⊂ Z ⊂ Y then Z is connected. In particular the closure Y of a connected set Y is connected. (See Proposition II.3.37.) We have already seen examples of connected sets: Example 19.3. A. The connected sets in R are the intervals (see Theorem I.19.25) and a set in R is open if and only if it is the denumerable union of mutually disjoint open intervals. B. Every star-shaped set in Rn is connected. (See Problem 12.b in Chapter II.3.) SRecall that S ⊂ Rn is called star-shaped if for some x0 ∈ Rn we have S= Sx,x0 where Sx,x0 denotes the closed line segment connecting x0 with

x.

x∈S

Using Theorem 19.2.C we can introduce the notion of the (connectivity) component of a point. Definition 19.4. In a topological space (X, O) the union C(x) of all connected sets containing the point x ∈ X is called the connectivity component of x. (Compare with Corollary II.3.36.) Clearly C(x) is the largest connected subset of X which contains x. If y ∈ C(x) then C(x) = C(y), and if y ∈ / C(x) then C(x) ∩ C(y) = ∅. Thus the components (C(x))x∈X form a partition of X. A further notion we have already introduced before is that of a pathwise or arcwise connected set. Since later on we will work in C and since it is customary to use in the plane the name arcwise connected set we will do the same. We need some preparations, again we can easily extend some definitions from metric spaces (or even Rn ) to general topological spaces. Definition 19.5. Let (X, O) be a topological space. A. For a closed interval I ⊂ R we call a continuous mapping γ : I → X an arc or a path or a parametric curve in X. If I = [a, b], a < b, then γ(a) is called the initial point of γ and γ(b) is called the terminal point of γ. The image of I under γ is called the trace of γ and sometimes denoted by tr(γ). The interval I is called the parameter interval of γ and t ∈ I is the arc or curve parameter. B. We say that γ : [a, b] → X connects the points x, y ∈ X if x, y ∈ 348

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{γ(a), γ(b)}, i.e. x is either initial or terminal point of γ and then y is terminal or initial point of γ, respectively. C. We call γ : [a, b] → X closed if γ(a) = γ(b). If γ is closed and in addition γ|[a,b) is injective we call γ a simply closed curve in X. These definitions are completely analogous to those given in Chapter II.4 (Definition II.4.11, Definition II.4.12) and we refer also to examples given there. Further, following Definition II.4.16 we give Definition 19.6. Let (x, O) be a topological space and G ⊂ X. We call G pathwise or arcwise connected if every pair of points x, y ∈ G can be connected by an arc, i.e. there exists a continuous mapping γ : [a, b] → X such that γ(a) = x and γ(b) = y. The proofs given in Chapter II.4 for the case G ⊂ Rn carry over to the general situation: Proposition 19.7. A. Every arcwise connected subset G ⊂ X is connected. (See Proposition II.4.18.) B. Let G ⊂ X and x0 ∈ G. The set [ Gx0 := {tr (γ) | γ : [a, b] → G is an arc with γ(a) = x0 } is arcwise connected. (See Lemma II.4.19.)

Note that Proposition 19.7.B implies that a star-shaped set is arcwise connected. However, Theorem II.4.20 does not hold for arbitrary topological spaces, therefore we state it here once more as in Chapter II.4: Theorem 19.8. If G ⊂ Rn is a non-empty open set which is connected then G is arcwise connected, i.e. regions in Rn are arcwise connected. We state explicitly Corollary 19.9. A region G ⊂ C is arcwise connected. Since intervals are connected sets in R the trace tr (γ) of an arc (a path, a parametric set) is by Proposition 19.7.A connected. In Problem 5 we prove Proposition 19.10. A continuous function between two topological spaces maps arcwise connected sets onto arcwise connected sets. 349

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We invite the reader to reconsider Example II.4.21 - II.4.23 to get a better understanding of all these notions and results. The careful reader will have noticed that the proofs of some of the results stated above will need the notion of a sum of two arcs (paths or curves) which we will discuss now in more detail. We start with a modification of Definition II.4.17. Definition 19.11. Given a topological space (X, O) and two paths γj : [aj , bj ] → X such that γ1 (b1 ) = γ2 (a2 ), i.e. the terminal point of γ1 is the initial point of γ2 . We define the sum γ1 ⊕ γ2 of γ1 and γ2 as the mapping γ1 ⊕ γ2 : [a1 , b1 + b2 − a2 ] → X, ( γ1 (t) for t ∈ [a1 , b1 ] (γ1 ⊕ γ2 )(t) = γ2 (t + a2 − b1 ) for t ∈ [b1 , b1 + b2 − a2 ].

(19.1)

Obviously we have that tr (γ1 ⊕ γ2 ) = tr (γ1 ) ∪ tr(γ2 ), however it is not obvious that γ1 ⊕ γ2 is a path, i.e. continuous. If X were a metric space this statement is easy to prove by using the characterisation of continuity with the help of limits of sequences. For the general situation we need Proposition 19.12. Let (X, OX ) be a topological space and A1 , ....., AN ⊂ X N S closed sets such that X = Ak . A mapping f : X → Y into a topological k=1

space (Y, OY ) is continuous if and only if the mappings f |Ak : Ak → Y, k = 1, ...., N, are continuous.

Proof. Of course, if f : X → Y is continuous, each of the mappings f |Ak is continuous. Now let C ⊂ Y be any closed set. It follows that (f |Ak )−1 (C) = Ak ∩ f (−1) (C) is closed in Ak equipped with the relative topology induced by X. Since Ak is closed it follows that Ak ∩ f (−1) (C) is closed in X. Using that N N S S (f |Ak )−1 (C) and this is a Ak we deduce further that f −1 (C) = X = k=1

k=1

closed set in X. Thus we have proved that the pre-image of every closed set in Y is under f closed in X, i.e. f is continuous.

Corollary 19.13. The sum of two paths γj : [aj , bj ] → X is a path, i.e. a continuous mapping γ1 ⊕ γ2 : [a1 , b1 + b2 − a2 ] → X. Clearly we can extend Definition 19.11 to a finite number of paths γ1 , ......., γN whenever the terminal point of γj is the initial point of γj+1, j = 1, ....., N −1, and in Problem 6 we will see that the operation ⊕ is associative. We also need a generalisation of Definition II.4.15. 350

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Definition 19.14. Let γ : [γ, β] → X be a path in the topological space (X, θ). A. We call a strictly increasing continuous function ϕ : [a, b] → [α, β] with ϕ(a) = α and ϕ(b) = β an orientation preserving parameter transformation for γ. B. A strictly decreasing continuous function ψ : [a, b] → [α, β] with ψ(a) = β and ψ(b) = α is called an orientation reversing parameter transformation. C. The paths γ ◦ ϕ and γ ◦ ψ are said to be obtained from γ by a change or transformation of the parameter. If ϕ is an orientation preserving parameter transformation for γ then γ ◦ ϕ : [a, b] → X is a path with tr (γ ◦ ϕ) = tr (γ) and (γ ◦ ϕ)(a) = γ(α) as well as (γ ◦ ϕ)(b) = γ(β), i.e. the initial and the terminal points of γ and γ ◦ ϕ coincide. If however ψ is an orientation reversing parameter transformation then we still have tr (γ ◦ ψ) = tr (γ) but (γ ◦ ψ)(a) = γ(β) and (γ ◦ ψ)(b) = γ(α), i.e. the initial point of γ ◦ ψ is the terminal point of γ and the terminal point of γ ◦ ψ is the initial point of γ. In other words on γ ◦ ψ we are running through tr (γ) in the reversed direction namely from γ(β) to γ(α). Definition 19.15. We call two paths γj : [aj , bj ] → X, j = 1, 2, in a topological space (X, O) equivalent paths if there exists an orientation preserving parameter transformations ϕ : [a2 , b2 ] → [a1 , b1 ] such that γ2 = γ1 ◦ ϕ. Proposition 19.16. The equivalence of paths is an equivalence relation which we denote by “∼γ 00 or just by “ ∼ ”. Proof. Since the identity id : [a, b] → [a, b] satisfies all conditions of an orientation preserving parameter transformation it follows always γ ∼ γ, and since the inverse of an orientation preserving parameter transformation is such a mapping as well as the composition of two orientation preserving parameter transformations is of this type, it follows γ1 ∼ γ2 implies γ2 ∼ γ1 and γ1 ∼ γ2 together with γ2 ∼ γ3 yields γ1 ∼ γ3 . We can now lift the definition of the sum γ1 ⊕ γ2 of two paths γ1 and γ2 to the set of equivalence classes by setting [γ1 ] ⊕ [γ2 ] := [γ1 ⊕ γ2 ].

(19.2)

Clearly, we have to check that this definition is independent of the parameterization which we will do in Problem 7. Of course we can extend (19.2) 351

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to finite number of paths γ1 , ....., γN with the terminal point of γj being the initial point of γj+1 for j = 1, ....., N − 1. For the extension (19.2) of (19.1) we can again prove the associative law, see Problem 7, i.e. we have ([γ1 ] ⊕ [γ2 ]) ⊕ [γ3 ] = [γ1 ] ⊕ ([γ2 ] ⊕ [γ3 ]).

(19.3)

For x0 ∈ X we denote by γx0 a path γx0 : [a, b] → X, γx0 (t) = x0 for all t ∈ [a, b]. Moreover, for γ : [a, b] → X we can define γ −1 : [a, b] → X, γ(t) = γ(a+ b−t), which starts at γ(b) and terminates at γ(a). Note that ψ : [a, b] → [a, b], ψ(t) = a + b − t is a strictly decreasing continuous function with ψ(a) = b and ψ(b) = a. Hence γ −1 is obtained from γ by applying an orientation reversing parameter transformation. Here is an illustration of these operations

γ1

b

γ2

x0

X

Figure 19.1

[γx0 ] b

[γ1 ]

x0

Figure 19.2

b

x0

Figure 19.3

352

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[γ2 ] [γ1−1 ] b

b

x0

x0

Figure 19.4 Figure 19.5

[γ1 ] ⊕ [γ2 ] b

x0

Figure 19.6

We have seen that it is possible to define some algebraic operations for (equivalence classes of) paths. This idea will now become more important when discussing homotopies and fundamental groups, concepts needed to study connected sets. The basic idea is to study the continuous deformation of continuous mappings. For example we may ask whether we can shrink a simply closed curve to a point. In C we may consider the circle h : [0, 2π] → C, h(t) = eit , and the point 0 ∈ C. The family H : [0, 2π] × [0, 1] → C, H(t, r) = r eit is continuous as function on [0, 2π] × [0, 1] and H(t, 1) = h(t), H(t, 0) = h0 : [0, 2π] → C, h0 (t) = 0 for all t ∈ [0, 2π]. Thus by H(· , r), r ∈ [0, 1], a family of curves is given shrinking continuously to the point, i.e. the constant curve 0, see Figure 19.7. 353

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C

H(t, r) = reit , 0 < r < 1

H(t, 0) = 0

b

0

−1

1

h(t) = eit = H(t, 1)

Figure 19.7

Now we consider the set C\B 1 (0) and start again with h : [0, 2π] → C, h(t) = 2 eit . In this case it seems that we cannot anymore shrink h continuously to a point, the set B 1 (0) is in our way, see Figure 19.8. 2

C \ B 1 (0) 2

H(t, r) = reit , 0 < r < 1

H(t, 0) = 0

b

−1

− 21

0

1 2

1

h(t) = eit = H(t, 1)

Figure 19.8

354

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In the following we use much from the discussion in H.Schubert [77]. Definition 19.17. Let (X, OX ) and (Y, OY ) be two topological spaces and f, g : Y → X two continuous mappings. We call f homotopic to g if there exists a continuous mapping H : Y × [0, 1] → X such that H(· , 0) = f and H(· , 1) = g. The mapping H is called a homotopy between f and g. Remark 19.18. A. It is important to note that H is supposed to be continuous on Y × [0, 1], i.e. with respect to the product topology. It is not sufficient to assume that H (· , t) and H (y, · ) are separately continuous. For an example of a separately continuous function which is not continuous we refer to Example II.4.1. B. If H : Y × [0, 1] → X is a homotopy then for every fixed y ∈ Y the mapping H (y, · ) : [0, 1] → X is a path in X. C. We may in particular consider the case where Y = [a, b]. In this case f and g are two paths in X and a homotopy between f and g deforms continuously the path f = H (· , 0) through a family of paths H (· , t), 0 < t < 1, into the path g = H (· , 1), see Figure 19.9. Of special interest is the case where all paths H (· , t), t ∈ [0, 1], have the same initial point x0 = H (a, t) for t ∈ [0, 1] and the same terminal point x1 = H (b, t) for all t ∈ [0, 1], see Figure 19.10. The latter case includes also closed paths with H (a, t) = H (b, t) for all t ∈ [0, 1], see Figure 19.11. f (b) b

f

f (a)

b b

b

H(·, t) : [a, b] → X b

b b b b b

g(b) b

b b

g(a)

b

Figure 19.9 f b

f (a) = g(a)

f (b) = g(b)

b

H(·, t) : [a, b] → X

Figure 19.10 355

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H(·, t) : [a, b] → X

b

f (a) = f (b) = g(a) = g(b)

g

Figure 19.11 Theorem 19.19. On the set of all continuous mappings f : Y → X homotopy is an equivalence relation. Proof. Denote by πY the projection πY : Y × [0, 1] → Y, πY (y, t) = y. Since πY is continuous it follows that H : Y × [0, 1] → X, H = f ◦ πY is continuous and H (y, t) = f (y) for all t ∈ [0, 1], i.e. f is homotopic to itself. Let f, g : Y → X be two continuous mappings which are homotopic and H : Y × [0, 1] → X be a homotopy between f and g. The mapping τ : Y × [0, 1] → Y × [0, 1], τ (y, t) = (y, 1 − t) is continuous implying that ˜ : Y × [0, 1] → X, H ˜ := H ◦ τ, (y, t) 7→ H(y, ˜ t) = H(y, 1 − t), is continuous H ˜ 0) = g holds as well as H(y, ˜ 1) = f . Thus if f is homotopic and further H(y, to g then g is homotopic to g. Finally suppose that f is homotopic to g and that g is homotopic to h with homotopies H1 and H2 respectively. We define H3 : Y × [0, 1] → X by ( H1 (y, 2t), 0 ≤ t ≤ 21 H3 (y, t) := 1 ≤ t ≤ 1. H2 (y, 2t − t), 2 First we note that H3 (y, 0) = f (y) and H3 (y, 1) = h(y) and in addition H3 (y, 12 ) = g(y). Further, since H3 is continuous on Y ×[0, 12 ] and on Y ×[ 12 , 1], and both sets are closed in Y ×[0, 1], Proposition 19.12 implies the continuity of H3 on Y × [0, 1]. We denote the equivalence relation “f and g are homotopic” by f ∼H g and we call the equivalence class of f the homotopy class of f . 356

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Example 19.20. Let (X, O) be a topological space and Y = {y0} be the topological space having just one point y0 . Every (continuous) mapping f : Y → X can be identified with the point xf := f (y0). We note further that {y0 } × [0, 1] and [0, 1] are homomorphic topological spaces and it follows that f ∼H g if and only if we can join xf with xg by a path γ : [0, 1] → X. Indeed using a homotopy H between f and g we may choose γ(t) = H (y0 , t). Conversely, given γ we consider H (y0 , t) := γ(t) as homotopy. Definition 19.21. A continuous mapping f : Y → X is called nullhomotopic if it is homotopic to a constant mapping g : Y → X, g(y) = x0 for all y ∈ Y . Definition 19.22. Two topological spaces X1 and X2 are called homotopically equivalent or of the same homotopy type if and only if there exist continuous mappings f : X1 → X2 and g : X2 → X1 such that g ◦ f is homotopic to the identity on X1 and f ◦ g is homotopic to the identity on X2 . Remark 19.23. In the case that X1 and X2 are of the same homotopy type it does not follow that f and g are uniquely determined. Moreover, if f1 , f2 : X1 → X2 both induce a homotopy equivalence of X1 with X2 it need not follow that they are homotopic. In Problem 8 we will see that homotopy equivalence for topological spaces is an equivalence relation. S Example 19.24. Let S = Sx,x0 ⊂ Rn be a star-shaped set as in Example x∈S

19.3.B. We claim that S with the relative topology from Rn is homotopically equivalent to any topological space Z = {z}. Points in S can be described by affine coordinates, or using the standard basis {e1 , ....., en } every x ∈ S is n P given by x0 + ξj ej where ξ = (ξ1 , ...., ξn ) are the coordinates of x − x0 . j=1

We consider the two continuous mappings f : S → Z, f (x) = z for all x ∈ S, and g : Z → S, g(z) = x0 . It follows that f ◦ g : Z → Z, (f ◦ g)(z) = z, is the identity on Z, and for g ◦f : S → S we have (g ◦f )(x) = x0 for all x ∈ S. We define the mapping H : S ×[0, 1] → S by H(x, t) = (t ξ1 +x0,1 , ...., t ξn +x0,n ). This mapping is clearly continuous, H(x, 0) = x0 , i.e. H(· , 0) = g ◦ f , and H(x, 1) = x, i.e. H(· , 1) = idS . Since f ◦ g : Z → Z is the identity on Z it follows that S is indeed homotopically equivalent to Z = {z}. 357

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The mappings we are interested in are paths which we can always assume to be parametrized by the interval [0, 1]. So let X be a topological space and γ : [0, 1] → X be a path. The mapping H : [0, 1] × [0, 1] → X defined by H(s, t) = γ (s − st) is continuous with H(s, 0) = γ(s) and H(s, 1) = γ(0). Thus every path is null-homotopic in the sense of Definition 19.21 and therefore homotopy for paths is not a far reaching notion. Matters change if we assume that the initial and the terminal points must be fixed under the action of a homotopy. Definition 19.25. A. Let F ⊂ Y be a subset of the topological space Y and let f, g : Y → X be two continuous mappings from Y into the topological space X. We call f and g homotopic relative to F if there exists a homotopy H : Y × [0, 1] → X between f and g such that H(y, t) = H(y, 0) for all y ∈ F and t ∈ [0, 1], i.e. H(y, t) = f (y) for y ∈ F and t ∈ [0, 1]. B. We call two paths γ1 , γ2 : [0, 1] → X homotopic relative to their intial and terminal points if there exists a homotopy H : [0, 1] × [0, 1] → X between γ1 and γ2 such that H(0, t) = γ1 (0) and H(1, t) = γ1 (1) for all t ∈ [0, 1]. In particular we must have γ1 (0) = γ2 (0) and γ1 (1) = γ2 (1), see Figure 19.10. C. We call two closed paths γ1 ,γ2 : [0, 1] → X homotopic relative to x0 if γ1 (0) = γ1 (1) = γ2 (0) = γ2 (1) = x0 and if γ1 and γ2 are homotopic with respect to their initial and terminal points, see Figure 19.11. D. If a closed path γ : [0, 1] → X is homotopic to the constant path γx0 , x0 = γ(0), we call γ null-homotopic. Remark 19.26. In order to have a shorter fa¸con de parler we call two paths γ1 and γ2 homotopic paths if they are homotopic relative to their initial and terminal points. It is easy to check that homotopy relative to F is again an equivalence relation, as is homotopy of paths or homotopy of closed paths relative to a point x0 . Furthermore we have Proposition 19.27. Let ϕ : [0, 1] → [0, 1] be an orientation preserving parameter transformation and γ : [0, 1] → X a path. The two paths γ and γ ◦ ϕ are homotopic. Proof. The mapping τ : [0, 1] × [0, 1] → [0, 1], τ [s, t] = s (1 − t) + t ϕ(s), is continuous from [0, 1] × [0, 1] to [0, 1]. We define H : [0, 1] × [0, 1] → X by H (s, t) = (γ ◦ τ ) (s, t) = γ (s (1 − t) + t ϕ (s)). This is a continuous mapping 358

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from [0, 1] × [0, 1] to X. For all t ∈ [0, 1] we have H (0, t) = γ (t ϕ (0)) = γ (0) as well as H (1, t) = γ (1 − t + t ϕ (1)) = γ (1). Moreover we have H (s, 0) = γ (s) and H (s, 1) = γ (ϕ (s)) proving the proposition. For a path γ : [0, 1] → X we consider as before the path γ −1 → X, γ −1 (s) = γ (1 − s). Lemma 19.28. If γ1 and γ2 are two homotopic paths then γ1−1 and γ2−1 are homotopic too. Proof. Let H : [0, 1] × [0, 1] → X be a homotopy between γ1 and γ2 . Then ˜ : [0, 1] × [0, 1] → X, H(s, ˜ t) = H (1 − s, t) is a homotopy the mapping H −1 −1 between γ1 and γ2 . We want to study the sum of paths in light of their behaviour under homotopies. Since we now let all paths be defined on [0, 1] (which after a parameter transformation is always possible) we have to adapt Definition 19.11 accordingly. If γ1 , γ2 : [0, 1] → X are two paths with γ1 (1) = γ2 (0) we define γ1 ⊕ γ2 : [0, 1] → X by ( γ1 (s), 0 ≤ s ≤ 21 (19.4) (γ1 ⊕ γ2 )(s) = 1 ≤ s ≤ 1, γ2 (2s − 1), 2 and as above we set γ1−1 : [0, 1] → X, γ1−1 (s) = γ1 (1 − s). It follows that (γ1 ⊕ γ2 )−1 = γ2−1 ⊕ γ1−1 . Now, if γ3 : [0, 1] → X is a third path with γ2 (1) = γ3 (0) we find   0 ≤ s ≤ 41 γ1 (4 s), 1 ((γ1 ⊕ γ2 ) ⊕ γ3 ) (s) = γ2 (4 s − 1), ≤ s ≤ 12 4   1 ≤s≤1 γ3 (2 s − 1), 2   γ1 (2 s), (γ1 ⊕ (γ2 ⊕ γ3 )) (s) = γ2 (4 s − 1),   γ2 (4 s − 3),

0 ≤ s ≤ 21 1 ≤ s ≤ 34 2 3 ≤ s ≤ 1, 4

(19.5)

(19.6)

(19.7)

and these are different paths having however the same trace. This implies that ⊕ as introduced by (19.4) is not giving rise to an associative operation. 359

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We may resolve this problem by looking at equivalence classes with respect to orientation preserving parameter transformations, however we prefer to look at equivalence classes with respect to homotopy of paths. Since (19.6) and (19.7) are obtained from each other by an orientation preserving parameter transformation they are homotopic if we impose the correct conditions on their initial and terminal points, see Proposition 19.26. For a path γ : [0, 1] → X with x0 = γ(0) and x1 = γ(1) we can always form γx0 ⊕ γ and γ ⊕ γx1 and from Proposition 19.26 we deduce that γx0 ⊕ γ, γ and γ ⊕ γx1 are homotopic paths. Furthermore we can always form γ ⊕ γ −1 according to ( γ (2 s), 0 ≤ s ≤ 12 −1 (19.8) (γ ⊕ γ ) (s) = 1 ≤ s ≤ 1. γ −1 (2 s − 1) = γ (2 − 2s), 2 Clearly, γ ⊕ γ −1 is a closed path. Consider the continuous mapping H : [0, 1] × [0, 1] → X defined by ( γ (2 s (1 − t)), 0 ≤ s ≤ 21 H(s, t) := 1 γ (2 (1 − s) (1 − t)), ≤ s ≤ 1. 2 For t = 0 we find H(· , 0) = γ ◦ γ −1 and for t = 1 we have H(· , 1) = γ(0). Thus γ ◦ γ −1 is null-homotopic to γx0 . Next let γ1 , γ2 : [0, 1] → X be two paths such that γ1 (1) = γ2 (0) and let γ˜j be homotopic to γj , j = 1, 2. Denote by Hj a homotopy between γj and γ˜j and define ( H1 (2 s, t), 0 ≤ s ≤ 21 (19.9) H (s, t) := 1 ≤s≤1 H2 (2 s − 1, t), 2 which turns out to be a homotopy between γ1 ⊕ γ2 and γ˜1 ⊕ γ˜2 . With this preparation we can lift the operation ⊕ to equivalence classes of homotopic paths. The definition is of course the obvious one and we do not change the notation: [γ1 ] ⊕ [γ2 ] := [γ1 ⊕ γ2 ] (19.10) and

[γ]−1 := [γ −1 ].

(19.11)

From our previous investigation we conclude: [γ1 ] ⊕ ([γ2 ] ⊕ [γ3 ]) = ([γ1 ] ⊕ [γ2 ]) ⊕ [γ3 ], 360

(19.12)

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[γ] + [γx0 ] = [γ] = [γ] ⊕ [γx1 ], γ(0) = x0 and γ(1) = x1 ,

(19.13)

and [γ] ⊕ [γ −1 ] = [γx0 ], [γ −1 ] ⊕ [γ] = [γx1 ].

(19.14)

We denote by Γx0 ,x1 the set of all homotopy classes of paths η : [0, 1] → X with η (0) = x0 and η (1) = x1 , and by Γx0 we denote the set Γx0 ,x0 , i.e. all homotopy classes of closed path η : [0, 1] → X with η (0) = x0 . Theorem 19.29. Let γ : [0, 1] → X be a fixed path with initial point γ(0) = x0 and terminal point γ(1) = x1 . Then the mapping σ : Γx0 ,x → Γx0 , σ([η]) := [η]⊕[γ −1 ] is a bijection with inverse τ : Γx0 : Γx0 x , τ (w) := [w]⊕[γ]. Proof. For [η] ∈ Γx0 ,x1 it follows that (τ ◦ σ)([η]) = τ (σ [η]) = τ ([η] ⊕ [γ −1 ]) = ([η] ⊕ [γ −1 ]) ⊕ [γ] = [η], and with ω ∈ Γx0 we have (σ ◦ τ )([ω]) = σ (τ ([ω])) = σ ([ω] ⊕ [γ]) = ([ω] ⊕ [γ]) ⊕ [γ −1 ] = [ω], i.e. σ and τ are inverse to each other which implies the theorem. Corollary 19.30. Let γ1 and γ2 be two paths with γ1 (0) = γ2 (0) = x0 and γ1 (1) = γ2 (1) = x1 . These two paths are homotopic if and only if γ1 ⊕ γ2−1 is null-homotopic. In other words, the investigations on homotopy of paths can be reduced to the study of the homotopy of closed paths. From our preparations we now deduce Theorem 19.31. The set Γx0 with the operation ⊕ forms a group with neutral element [γx0 ] and inverse to [γ] given by [γ]−1 = [γ −1 ]. Definition 19.32. The group (Γx0 , ⊕) is called the first fundamental group of X with respect to x0 and is denoted by π1 (X, x0 ). Theorem 19.33. For an arcwise connected topological space X the groups π1 (X, x0 ) and π1 (X, x1 ), x0 , x1 ∈ X, are isomorphic. 361

A COURSE IN ANALYSIS

Proof. Let γ : [0, 1] → X be a path connecting x0 with x1 , i.e. γ(0) = x0 and γ(1) = x1 . We define the mapping hγ : π1 (X, x0 ) → π1 (X, x1 ), hγ ([η]) := [γ] ⊕ [η] ⊕ [γ]−1 , and claim that hγ is an isomorphism. For [η1 ], [η2 ] ∈ π1 (X, x0 ) we find hγ ([η1 ] ⊕ [η2 ]) = [γ] ⊕ [η1 ] ⊕ [η2 ] ⊕ [γ]−1 = [γ] ⊕ [η1 ] ⊕ [γ]−1 ⊕ [γ] ⊕ [η2 ] ⊕ [γ]−1 = hγ ([η1 ]) ⊕ hγ ([η2 ]), i.e. hγ is a homomorphism. Now we can use [γ]−1 to define hγ −1 : π1 (X, x1 ) → π1 (X, x0 ) by hγ −1 ([η]) = [γ]−1 ⊕ [η] ⊕ [γ] and it follows that hγ −1 ◦ hγ = idπ1 (x,x0 ) and hγ ◦ hγ −1 = idπ1 (x,x1 ) , i.e. hγ (and hence hγ −1 ) is an isomorphism. Of central importance is Definition 19.34. A pathwise connected topological space X is said to be simply connected if for one point x0 ∈ X the first fundamental group π1 (X, x0 ) is trivial, i.e. consist only of the neutral element. Remark 19.35. From Theorem 19.32 we deduce that all groups π1 (X, x), x ∈ X, are trivial if π1 (X, x0 ) is trivial. Hence the notion of simple connectivity of a space does not dependent on the special point x0 in Definition 19.33. Example 19.36. Every convex set K ⊂ Rn is simply connected. For γ1 , γ2 : [0, 1] → K we define H (s, t) = (1 − t) γ1 (s) + t γ2 (s) which is continuous and due to the convexity of K well defined. Moreover we have H(· , 0) = γ1 and H(· , 1) = γ2 . Thus if we choose x0 ∈ K and γ to be any closed path in K with initial point x0 then we find that γ is null-homotopic to γx0 , i.e. π1 (K, x0 ) is trivial. A classical result related to (simply) connected sets in the plane is the Jordan curve theorem. Although we will not make use of it, we want to state it in the refined version due to Schoenflies and for a proof we refer to [8], [33], [69] or [83]. Theorem 19.37. Let γ : [0, 1] → C be a simply closed arc. The trace tr (γ) divides C into three mutually disjoint sets Γ+ , Γ− and tr (γ), C = Γ+ ∪ Γ− ∪ tr (γ), where Γ+ is bounded, Γ− is unbounded and ∂Γ+ = ∂Γ− = tr (γ). The sets Γ+ and Γ− are connected and Γ+ is simply connected. 362

19

SOME MORE TOPOLOGY

For our final result in this chapter we need Definition 19.38. An arc γ : [a, b] → C is called a polygonal arc (or curve or path) if its trace consists of a finite number of line segments. If all these line segments are parallel either to the real axis or to the imaginary axis we call γ an axis parallel polygonal arc (or curve or path). By definition for a polygonal arc γ : [a, b] → C there exists a partition of [a, b], a = t0 < t1 < ...... < tn = b such that the mappings γ |[tj−1 , tj ] , j = 1, ....., n, have as traces line segments. For example we may look at t 7→ t−t tj −t γ (tj−1) + tj −tj−1 γ (tj ), t ∈ [tj−1 , tj ], as a further parametrization of tj −tj−1 j−1 γ |[tj−1 ,tj ] . Proposition 19.39. In a region G ⊂ C we can connect two points z1 , z2 ∈ G always by a polygonal arc. Proof. Let γ : [a, b] → G be a curve connecting z1 and z2 . Since G is open and tr (γ) ⊂ G is compact we can cover tr (γ) by a finite number of balls Bj (wj ), j = 1, ...., N, wj ∈ tr (γ), w1 = z1 and wN = z2 . Each pair of points in Bj (wj ) can be joined by a line segment since these are convex sets. Further we have Bj (wj ) ∩ Bj+1 (wj+1 ) 6= ∅ when we choose the enumeration of the balls appropriately. Denote by wk, k+1 a point in Bk (wk ) ∩ Bk+1 (wk+1). The line segments joining z1 = w1 with w1,2 , w1,2 with w2,3 , w2,3 with w3,4 , ...., wN −1,N with wN = z2 form a polygonal arc in G connecting z1 and z2 .

Problems 1.

a) Let (X, O) be a discrete topological space, i.e. O = P(X). When is (X, O) connected?

b) Let (X, OX ) be a connected topological space and (Y, OY ) a topological space with at least two connectivity components. Does there exist a surjective mapping f : X → Y which is continuous? 2. Let A ⊂ Rn and B ⊂ Rm be two connected non-empty open sets. Is the set A × B ⊂ Rn × Rm connected? 3. Consider the simply closed curve γ in C the trace tr(γ) of which is given by the figure below: 363

A COURSE IN ANALYSIS

1 + 2i b

2 + 2i b

2+i b

b

b

b

0

1

2

b

3+i

Find a piecewise continuous differentiable, anticlockwise parametrization of γ, γ : [0, 1] → C, such that γ(0) = 2 = γ(1). 4. Find a bijective continuous mapping f : R2 \ {0} → R2 \ {0} with continuous inverse which maps ∂B2 (0) onto ∂B1 (0) = S 1 . 5. Show that a continuous mapping between two topological spaces maps arcwise connected sets onto arcwise connected sets. 6. Prove that for three paths γj : [aj , bj ] → X, j = 1, 2, 3, such that γ1 (b1 ) = γ2 (a2 ) and γ2 (b2 ) = γ3 (a3 ) we have (γ1 ⊕ γ2 ) ⊕ γ3 = γ1 ⊕ (γ2 ⊕ γ3 ). 7. Let γ1 and γ2 be two paths in the topological space X such that γ1 ⊕ γ2 is defined. Prove that the definition [γ1 ]⊕[γ2 ] := [γ1 ⊕γ2 ] is independent of the paratmetrization. Now use Problem 6 to deduce that ([γ1 ] ⊕ [γ2 ]) ⊕ [γ3 ] = [γ1 ] ⊕ ([γ2 ] ⊕ [γ3 ]) holds. 8. Prove that homotopy equivalence for topological spaces is an equivalence relation. 9. Is π1 (A 1 ,2 (0), 1) where A 1 ,2 (0) = {z ∈ C| 21 < |z| < 2}, trivial? 2

2

364

20

Line Integrals of Complex-valued Functions

For a parametrized curve γ : [a, b] → C with tr γ ⊂ D, D ⊂ C a domain, we want to define for complex-valued functions f : D → C the line inteR gral f (z) dz. While in the case of holomorphic functions the properties of γ R f (z) dz will turn out to be quite different to those of general line integrals γ R (in Rn ), the basic definition of the line integral f (z) dz is similar, in fact γ

it is an adaptation of the theory developed in Chapter II.15 where we now identify R2 with C. Therefore we start by recollecting definitions and results from Chapter II.15 but now we will use complex notation. From this chapter on we will also use the standard notation from complex variable theory and we will often write γ for the curve (as mapping) as well as for its trace.

Definition 20.1. (II.15.1) Let γ : [a, b] → C, a < b, be a continuous parametrized curve and Z = Z (t0 , ..., tm ), t0 = a, tm = b, be a partition of [a, b]. A. We call m−1 X VZ (γ) := | γ (tk+1 ) − γ (tk )| (20.1) k=0

the Z-variation of γ and

V (γ) := sup VZ (γ)

(20.2)

Z

the total variation of γ where the supremum is taken over all partitions Z of [a, b]. B. If V (γ) < ∞ we call γ rectifiable and define its length as lγ := V (γ).

(20.3)

Proposition 20.2. (II.15.2) A continuous parametric curve γ : [a, b] → C, γ = Re γ + i Im γ, is rectifiable if and only if the functions Re γ and Im γ are of bounded variations. Lemma 20.3. (II.15.5) If γ1 and γ2 are rectifiable and γ1 ⊕ γ2 is defined then γ1 ⊕ γ2 is rectifiable too. 365

A COURSE IN ANALYSIS

Definition 20.4. (II.15.7) A continuous curve γ : [a, b] → C is piecewise continuously differentiable if there exists a partition Z = Z (t0 , ....., tm ) of [a, b] such that γ | [tj , tj+1 ] , j = 0, ....., m − 1, is continuously differentiable. Corollary 20.5. (II.15.9) A piecewise continuously differentiable curve γ : [a, b] → C is rectifiable. We know from Example II.15.8 that every polygon is a piecewise continuously differentiable curve and if γ1 and γ2 are two piecewise continuously differentiable curves for which γ1 ⊕ γ2 is defined, then γ1 ⊕ γ2 is piecewise continuously differentiable too. Of a more technical nature is Theorem 20.6. (II.15.11) Let γ : [a, b] → C be a rectifiable curve. For every  > 0 there exists δ > 0 such that for every partition Z of [a, b] with mesh size less than δ it follows that V (γ) − VZ (γ) < . If (Z (ν) )ν∈N is a sequence of partition of [a, b] with mesh size η ν and lim η (ν) = 0 then lim VZ (ν) (γ) = ν→∞

ν→∞

V (γ) = lγ . Theorem 20.7. (II.15.12) For a C 1 curve γ : [a, b] → C we have lγ = V (γ) =

Z

a

b

|γ˙ (t)| dt =

Z

b

1

((Re γ˙ (t))2 + (Im γ˙ (t))2 ) 2 dt.

(20.4)

a

Corollary 20.8. (II.15.13) Let γ : [a, b] → C be a piecewise continuously differentiable curve, i.e. with Z = Z (t0 , ...., tm ) we have that γ | [tj , tj+1 ] , j = 0, ....., m − 1, is a C 1 curve. Then γ is a rectifiable curve and lγ =

m−1 X j=0

lγ | [tj , tj−1 ] =

m−1 X Z tj+1 j=0

tj

| γ˙ (t)| dt.

(20.5)

Now we can develop the theory of line integrals for complex-valued functions. In the remaining part of this chapter we deal with the more elementary theory using the books [26] and [67] a bit as a guide. In analogy to Definition II.15.15 we give Definition 20.9. Let D ⊂ C be a domian and f : D → C be a continuous complex-valued function. Further let γ : [a, b] → D be a piecewise continuously differentiable curve, γ = γ1 ⊕ ....... ⊕ γN , where γj : [tj−1 , tj ] → D, a = 366

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

t0 < ..... < tN = b, is a C 1 - curve. The line integral of f along γ is defined by Z

f (z) dz :=

γ

N Z X

f (z) dz

(20.6)

j=1 γ

j

:=

Ztj N X

f (γ (t)) γ˙ (t) dt =

j=1 t j−1

Ztj N X

f (γj (t)) γ˙ (t) dt.

j=1 t j−1

Remark 20.10. Sometimes it has an advantage to write z(t) for the curve and then to look at Ztj f (z(t)) z 0 (t) dt. tj−1

With this notation we can rewrite (20.4) as lγ =

Zb a

|z 0 (t)| dt.

(20.7)

Furthermore we note that for γ : [a, b] → C, γ(t) = t, we find Z

f (z)dz =

γ

Zb

f (γ(t))γ(t)dt ˙ =

a

Zb

f (t)dt,

a

i.e. our notations are consistent. R Quite a few properties of f (z)dz are straightforward to prove: γ

If f : [a, b] → C, [a, b] ⊂ R is continuous then Re

Zb a

f (t)dt =

Zb

Ref (t)dt and Im

a

Zb

f (t)dt =

a

Zb

Imf (t)dt,

(20.8)

a

as well as Zb a

f (t)dt =

Zb a

367

f (t)dt.

(20.9)

A COURSE IN ANALYSIS

Next let f, g : D → C be continuous functions and γ, γ1 and γ2 be curves satisfying each of the conditions of Definition 20.9 and assume that γ1 ⊕ γ2 is defined. Further let c ∈ C. The following hold: Z Z Z (f ± g)(z)dz = f (z)dz + g(z)dz; (20.10) γ

γ

Z

γ

cf (z)dz = c

γ

and

Z

Z

f (z)dz;

(20.11)

Z

(20.12)

γ

f (z)dz =

γ1 ⊕γ2

Z

f (z)dz +

γ1

If tr γ = {z0 }, i.e. γ(t) = z0 for all t, then

R

f (z)dz.

γ2

f (z) dz = 0. If γ −1 is the inverse

γ

to γ, i.e. γ −1 : [a, b] → C, γ −1 (t) = γ (a + b − t), then γ˙ −1 (t) = −γ(t) ˙ and it follows that Z

f (z) dz =

γ −1

Zb

f (γ −1 (t)) (γ˙ −1 ) (t) dt

a

=−

Zb

f (γ (a + b − t)) γ˙ (a + b − t) dt

a

=

Zb a

and we have proved

Z

γ −1

f (s) γ(s) ˙ ds = −

f (z) dz = −

Z

Zb

f (s) γ(s) ˙ ds,

a

f (z) dz.

(20.13)

γ

Example 20.11. Let γ : [0, 2π] → C, γ(t) = z0 + r eit , be the circle with 1 is defined centre z0 and radius r > 0. Since z0 ∈ / tr γ the function z 7→ z−z 0 and continuous in a neighbourhood of γ and we find Z

|z−z0 |=r

1 dz = z − z0

Z2π 0

368

1 ir eit dt = 2πi, r eit

(20.14)

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

or

1 2πi

Z

1 dz = 1. z − z0

|z−z0 |=r

(20.15)

Note that this integral is independent of z0 and r. In (20.14) and (20.15) we have used the rather common notation Z Z f (z) dz := f (z) dz for γ(t) = z0 + r eit . (20.16) γ

|z−z0 |=r

Example 20.12. The upper half circle with radius 1 can be parametrized by γ : [0, π] → C, γ(t) = ei (π−t) . The initial point of γ is γ(0) = ei π = −1 and the terminal point of γ is γ(π) = ei 0 = 1, see Figure 20.1. For f (z) = |z| we find Zπ Z Z ˙ dt = γ(π) − γ(0) = 2. f (z) dz = |z| dz = 1 γ(t) γ

0

γ

We may also integrate f (z) = |z| along the interval [−1, 1], i.e. the line segment σ connecting (−1, 0) and (1, 0), σ : [−1, 1] → C, σ(t) = t, see Figure 20.2. For this curve we get Z

f (z) dz =

σ

Thus in general

R

Z

|z| dz =

σ

Z1

−1

|t| dt = −

Z0

−1

t dt +

Z1

t dt = 1.

0

f (z) dz will depend on γ and not only on the initial and

γ

terminal points of γ.

γ

σ −1

0

1

−1

Figure 20.1

369

0

1

Figure 20.2

A COURSE IN ANALYSIS

The following proposition gives basic estimates for line integrals. Proposition 20.13. A. For a continuous function f : [a, b] → C the following holds b Zb Z f (t) dt ≤ |f (t)| dt. (20.17) a

a

B. If for a domain D the curve γ : [a, b] → D, D ⊂ C, is piecewise continuously differentiable and f : D → C is continuous on tr γ then Z f (z) dz ≤ ||f ||∞,tr γ lγ (20.18) γ

where ||f ||∞,tr γ = sup |f (z)|. z∈tr γ

Proof. A. If f is real-valued, (20.17) is just a consequence of the triangle Rb inequality and is well known to us. The case where f (t) dt = 0 is trivial. a

For

Rb a

Rb Rb f (t) dt 6= 0 we can find s ∈ R such that eis f (t) dt = Re (eis f (t)) dt a

as well as

a

b b Z Zb Z is is f (t) dt = e f (t) dt = Re (e f (t)) dt a

a

≤

Zb a

1

is

|Re (e f (t))| dt ≤

Zb a

|f (t)| dt.

B. Since γ is continuous and [a, b] is compact, tr γ is compact too and f |tr γ is bounded, i.e. ||f ||∞,tr γ is finite. Moreover, using part A we have |

Z γ

f (z) dz| = |

Zb a

f (γ(t)) γ(t) ˙ dt| ≤

≤ ||f ||∞,tr γ

Zb

Zb a

|f (γ(t))|| γ(t)| ˙ dt

|γ(t)| ˙ dt = ||f ||∞,tr γ lγ .

a

370

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

R

Next we want to prove that

f (z) dz is independent of the parametrization,

γ

i.e. it is unchanged under an orientation preserving parameter transformation or change of parameter. Since we are working with piecewise continuously differentiable curves we can reformulate Definition 19.14.A as Definition 20.14. Let I := [a, b] and J := [c, d] be non-degenerate compact intervals and ϕ : [c, d] → [a, b] a strictly increasing piecewise continuously differentiable function, i.e. ϕ0 > 0 wherever it is defined, with ϕ(c) = a and ϕ(d) = b. Then we call ϕ a piecewise continuously differentiable orientation preserving parameter transformation or just a change of parameter. It is easy to see that if ϕ and ψ are changes of parameter then the same holds for ϕ ◦ ψ and ϕ−1 . Let γ : I → C be a piecewise continuously differentiable curve and ϕ : J → I a change of parameter. Then γ ◦ ϕ : J → C is a piecewise continuously differentiable curve such that tr γ = tr γ ◦ ϕ and the initial and terminal points of both curves are the same. We say that γ ◦ ϕ is obtained from γ by a change of parameter. Theorem 20.15. Let D ⊂ C be a domain and γ1 : [a, b] → D be a piecewise continuously differentiable curve and f : D → C a function such that f |tr γ1 is continuous. If γ2 : [c, d] → C is obtained from γ1 by a change of parameter ϕ : [c, d] → [a, b] then we have Z Z f (z) dz = f (z) dz. (20.19) γ1

γ2

Proof. By a straightforward calculation we get Z

f (z) dz =

Zb

f (γ1 (t)) γ˙1 (t) dt

=

Zd

f ((γ1 ◦ ϕ) (s)) γ˙1 (ϕ (s)) ϕ0(s) ds

Zd

f (γ2 (s)) γ˙2 (s) ds =

γ1

a

c

=

c

Z

γ2

371

f (z) dz.

A COURSE IN ANALYSIS

The central result of calculus of real-valued function of a real variable is the fundamental theorem and for this the notion of a primitive is essential. Definition 20.16. A holomorphic function F : D → C defined as a domain D ⊂ C is called a primitive of the continuous function f : D → C if F 0 = f . In Chapter 18 we have seen that holomorphic functions might have more complicated domains and therefore it is worth giving Definition 20.17. A continuous function f : D → C has a local primitive if for every z0 ∈ D there exists an open neighbourhood U(z0 ) ⊂ D such that f |U (z0 ) has a primitive. One formulation of the fundamental theorem is, see Theorem I.26.4, that for f : [a, b] → R continuous with primitive F we have Zb

f (x) dx = F (b) − F (a).

(20.20)

a

Interpreting [a, b] as trace of a curve with initial point a and terminal point b we may seek to generalise (20.20) to Z

f (z) dz = F (z1 ) − F (z0 )

(20.21)

γ

where now f : D → C, γ : [a, b] → C with γ(a) = z0 and γ(b) = z1 , and F is a primitive of f . Theorem 20.18. Let f : D → C be a continuous function which has a primitive on the domain D. Further let γ : [a, b] → D be a piecewise continuously differentiable curve such that γ(a) = z0 and γ(b) = z1 then we indeed have (20.21), i.e. Z f (z) dz = F (z1 ) − F (z0 ). γ

In particular the integral depends only on the initial and terminal point of γ. 372

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

Proof. Let γ : [a, b] → D and a = t0 < t1 < ...... < tN = b be a partition of [a, b] such that γj = γ|[tj−1 ,tj ] , j = 1, ...., N, is continuously differentiable. It follows that Z

f (z) dz =

γ

Zb

f (γ(t)) γ˙ (t) dt =

t N Zj X

0

F (γ(t)) γ˙ (t) dt =

j=1 t j−1

=

f (γ(t)) γ˙ (t) dt

j=1 t j−1

a

=

t N Zj X

N X j=1

t N Zj X j=1 t j−1

(F ◦ γ)0 (t) dt

((F ◦ γ)(tj ) − (F ◦ γ)(tj−1)) = F (z1 ) − F (z0 ).

Corollary 20.19. Suppose that f : D → C is continuous and has a primitive. If γ : [a, b] → D is a closed curve satisfying the assumption of Theorem 20.18 we have Z f (z) dz = 0. (20.22) γ

Example 20.20. A. For n ∈ N the function f : C → C, f (z) = z n , has the 1 z n+1 and for every piecewise continuously differentiable primitive F (z) = n+1 curve γ with initial point z0 and terminal point z1 we have Z 1 z n dz = (z n+1 − z0n+1 ). n+1 1 γ

This implies for every polynomial p(z) =

N P

j=0

Z γ

aj z j , aj ∈ C, that

N N X X aj j+1 aj j+1 z1 − z . p(z) dz = j+1 j+1 0 j=0 j=0

B. For the entire functions exp, cos and sin we have exp0 (z) = exp(z), cos0 (z) = − sin z and sin0 (z) = cos(z) 373

A COURSE IN ANALYSIS

implying for any piecewise continuously differentiable curve γ : [a, b] → C with γ(a) = z0 and γ(b) = z1 Z ez dz = ez1 − ez0 , Z

γ

cos(z) dz = sin z1 − sin z0 ,

γ

Z

sin(z) dz = − cos z1 + cos z0 .

γ

C. Let F (z) =

∞ P

k=0

ak (z − c)k be a power series converging in Br (c), r > 0.

We know by Theorem 17.14 that F 0 (z) =

∞ P

k=1

ak k (z − c)k converges in Br (c)

too, hence it converges uniformly on every compact subset. Thus F is in ∞ P ak k (z − c)k−1 . From Theorem 18.5 and its proof Br (c) a primitive of k=1

we now find in B1 (1) that for a branch of the logarithmic function given by ˜ (z − 1), L ˜ : B1 (1) → C, L(z) ˜ ˜ 0 (z) = 1 , and therefore we have for a L = ln z piecewise continuously differentiable curve γ with tr γ ⊂ B1 (1) and initial point z0 as well as terminal point z1 the formula Z 1 ˜ (z1 − 1) − ln ˜ (z0 − 1), dz = ln z γ

∞ ˜ (1 + z) = P where as before, see (16.18), ln

k=1

(−1)k k

z k , |z| < 1.

Example 20.21. A. In Example 20.12 we have seen that for z 7→ |z| the R line integral |z| dz depends on γ and not only on the initial and terminal γ

point of γ. Consequently z 7→ |z| has no primitive. Note that this differs completely from the situation of real-valued functions. If g : [a, b] → R is Rt continuous then G(t) := g(s) ds is a primitive of g. a R 1 B. We know by Example 20.11 that dz = 2πi and therefore by Corolz

lary 20.12 the function z 7→

1 z

|z|=1

defined for all C\{0} can not have a primitive 374

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

in C\{0}. However we have seen in Example 20.20.C that it has a primitive (for example) in B1 (1). We want to find conditions which imply the converse to Corollary 20.19. The main constraint is of topological nature: D must be a region, i.e. connected. Theorem 20.22. Let G ⊂ C be a region and f : G → C a continuous R function. If f (z) dz = 0 for every piecewise continuously differentiable and γ

closed curve γ with tr γ ⊂ D, then f has a primitive in G.

Proof. We fix a point a ∈ G and for z ∈ G let γaz be a piecewise continuously differentiable curve connecting a and z. Note that by Proposition 19.38 such a curve exists. We define on G the function F by Z

z 7→ F (z) :=

f (w) dw

(20.23)

γaz

and note first of all that F is independent of the choice of γaz . Indeed if γ˜az is a further piecewise continuously differentiable curve connecting a and z −1 −1 with corresponding reversed curve γ˜az we find that γaz ⊕ γ˜az is a piecewise continuously differentiable closed curve, hence Z

f (z) dz = 0,

−1 γaz ⊕˜ γaz

or 0=

Z

−1 γaz ⊕˜ γaz

i.e.

f (z) dz =

Z

f (z) dz +

Z

f (z) dz =

−1 γ ˜az

γaz

Z

γaz

f (z) dz =

Z

γaz

Z

f (z) dz −

Z

f (z) dz,

γ ˜az

f (z) dz.

γ ˜az

We claim that F is a primitive of f . Let z0 ∈ G such that for some r > 0 we have z0 ∈ Br (z) ⊂ G. We denote that line segment, connecting z0 and z by 375

A COURSE IN ANALYSIS

[z0 , z] and we consider F (z) − F (z0 ) =

Z

f (w) dw −

γaz

=

Z

f (w) dw

γaz0

Z

f (w) dw

[z0 ,z]

=

Z1

f (z0 + t(z − z0 )) (z − z0 ) dt,

0

or F (z) − F (z0 ) − f (z0 ) = z − z0

Z1

f (z0 + t(z − z0 )) dt − f (z0 ).

0

The function f is continuous. Thus, given  > 0 there exists δ > 0 such that |z − z0 | < δ implies |f (z) − f (z0 )| < . For 0 ≤ t ≤ 1 we have |z0 +t(z−z0 )−z0 | = t|z−z0 | ≤ |z−z0 | < δ implying |f (z0 +t(z−z0 ))−f (z0 )| <  and further sup |f (z0 + t(z − z0 )) − f (z0 )| < . 0≤t≤1

Thus we have proved that F (z) − F (z0 ) = f (z0 ), z→z0 z − z0

F 0 (z0 ) = lim i.e. F is a primitive of f .

In order to apply Theorem 20.22 we need to verify

R

f (z) dz = 0 for all pie-

γ

cewise continuously differentiable curves to decide whether f has a primitive. A natural R question is whether we can obtain the same conclusion if the condition f (z) dz = 0 holds for a smaller set of curves. It turns out that in the γ

case where R the region G has additional geometric properties it is sufficient to verify f (z) dz = 0 for boundaries of triangles only. Since triangles and γ

their boundaries will enter more and more into our considerations we first give some preparations. For z1 , z2 ∈ C we denote by [z1 , z2 ] the line segment connecting z1 and z2 376

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

and also the curve t 7→ z1 + t(z2 − z1 ), t ∈ [0, 1]. A triangle ∆(z1 , z2 , z3 ) with vertices z1 , z2 , z3 ∈ C is the convex hull of these three points in the plane, compare with Theorem II.13.24, i.e. ∆(z1 , z2 , z3 ) = {z ∈ C | z = t1 z1 + t2 z2 + t3 z3 , t1 , t2 , t3 ≥ 0, t1 + t2 + t3 = 1} (20.24) = {z ∈ C | z = z1 + s(z2 −z1 )+ t(z3 −z1 ), s ≥ 0, t ≥ 0, s+t ≤ 1} The boundary ∂∆(z1 , z2 , z3 ) is first of all the topological boundary of ∆(z1 , z2 , z3 ) ⊂ C and we can write it as ∂∆(z1 , z2 , z3 ) = [z1 , z2 ] ∪ [z2 , z3 ] ∪ [z3 , z1 ].

(20.25)

Interpreting however [zj , zk ] as a parametrized curve as mentioned above we can and will also interpret ∂∆(z1 , z2 , z3 ) as a closed, piecewise continuously differentiable curve being the sum of the three curves [z1 , z2 ], [z2 , z3 ] and [z3 , z1 ], i.e. as a curve we consider ∂∆(z1 , z2 , z3 ) as ∂∆(z1 , z2 , z3 ) = [z1 , z2 ] ⊕ [z2 , z3 ] ⊕ [z3 , z1 ],

(20.26)

compare with Figure 20.3: z3 [z3 , z1 ]

z1

[z2 , z3 ]

z2

[z1 , z2 ] ∂∆(z1 , z2 , z3 ) = [z1 , z2 ] ⊕ [z2 , z3 ] ⊕ [z3 , z1 ]

Figure 20.3 R The starting point to reduce the set of curves γ for which f (z) d(z) = 0 γ

must hold should be the definition of the primitive F , i.e. Z F (z) = f (w) dw, γaz

377

A COURSE IN ANALYSIS

see (20.23). If every z ∈ G can be connected by a straight line [z0 , z] to a fixed point z0 ∈ G and then define Z F (z) = f (w) dw (20.27) [z0 ,z]

(ζ) we are a step closer to our goal. Next we note that in order to form F (z)−F z−ζ and pass to the limit z → ζ we only need to consider values of z with |z − ζ| being small. So if z0 is fixed as above, ζ ∈ G and r > 0 such that Br (ζ) ⊂ G, for all z ∈ Br (ζ) the curve [z0 , ζ] ⊕ [ζ, z] ⊕ [z, z0 ] is a piecewise continuously differentiable curve which is closed and its trace is entirely contained in G. However [z0 , ζ] ⊕ [ζ, z] ⊕ [z, z0R] = ∂∆(z0 , ζ, z). Thus if for every triangle f (z) dz = 0 then the proof of Theorem ∆(z0 , z1 , z2 ) ⊂ G we have ∂∆(z0 , z1 , z2 )

20.22 works and establishes that F defined by (20.27) is a primitive of f . It remains to recall the definition of a star-shaped domain to conclude

Theorem 20.23. Let G ⊂ C be a star-shaped domain with respect to z0 ∈ G. Further let f : G → C be a continuous function such that for every triangle R ∆(z0 , z1 , z2 ) ⊂ G we have f (z) dz = 0. Then by ∂∆(z0 , z1 , z2 )

F (z) :=

Z

f (w) dw

(20.28)

[z0 , z]

a primitive of f is given and

R

f (z) dz = 0 for all closed, piecewise continu-

γ

ously differentiable curves in G. Corollary R 20.24. If G ⊂ C is a convex set, f : G → C a continuous function f (z) dz = 0 holds for every triangle ∆(z1 , z2 , z3 ) ⊂ G then f and ∂∆(z1 , z2 , z3 )

has a primitive F in G.

Proof. Every convex set is star-shaped with respect to each of its points. Remark 20.25. Note that a star-shaped set need not be convex, see Figure 20.4 or Figure 20.5. 378

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

z0 b

b

z0

Star-shaped but non-convex domain

Star-shaped but non-convex domain Figure 20.5

Figure 20.4

Corollary 20.26. Let D ⊂ C be a domain and suppose that f : D → C is a continuous function R such that for every piecewise continuously differentiable curve γ we have f (z) dz = 0. Then f has a local primitive. γ

Proof. Every point z ∈ D has a convex open neighbourhood entirely contained in D, for example B% (z), % > 0 sufficiently small.

Remark 20.27. Since the relies on Theorem 20.23, R R proof of Corollary 20.26 f (z) dz for all triangles we can therefore replace f (z) dz = 0 by γ

∂∆(z1 , z2 , z3 )

∆(z1 , z2 , z3 ) ⊂ D.

Problems it 1. Let γ : [0, 2πk] → C, k ∈ N, be the curve R n γ(t) = e . When is γ a simply closed curve? Find the integral γ z dz for n ∈ Z.

2. With γ and σ as in Example 20.12 find the integrals: Z Z Z Z Re zdz; Re zdz; Im zdz; Im zdz. γ

σ

γ

σ

3. Consider the triangle ABC with A = −2, B = 2, C
= 2i. Parametrize
∂ABC by γ : [0, 1] → C such that γ(0) = A, γ 13 = B, γ 32 = C, γ(1) = A and γ is piecewise linear, see the figure below: 379

A COURSE IN ANALYSIS Im z

b

C = 2i ∂ABC = tr(γ)

b

b

A = −2

B=2

Now find by a direct calculation that 4.

R

γ

Re z

|z|dz = 0.

Rb a) In R we have for every interval [a, b], a < b, the equality a 1dx = b − a, i.e. the integral of the constant function x 7→ 1 over [a, b], gives R the interval length. Can we expect that γ 1dz = lγ holds?

b) Let R ⊂ C be a region. For k ∈ N let γk : [0, 1] → R be a piecewise continuously differentiable curve. We suppose that limk→∞ lγk = 0. Prove that for every continuous function f : R → C we have Z 1 lim sup f (z)dz < ∞ lγk γk k→∞

provided tr γk ⊂ K for all k ∈ N where K ⊂ D is a compact set.

5. Consider the curve γ R: [0, 2π] → C, γ(t) = 4eit + e−it . Find the trace of γ and the integral γ z 3 dz.

6. Let f : BR (z0 ) → C be given by a convergent power series. Show that for every 0 < r < R the function F |Br (z0 ) admits a primitive.

7. Find the following integrals: R a) [−1−2πi,1+2πi] e2z dz, where [−1 − 2πi, 1 + 2πi] is the line segment connecting −1 − 2πi and 1 + 2πi; 380

20

LINE INTEGRALS OF COMPLEX-VALUED FUNCTIONS

R b) γ sinh(z)dz, where the trace of γ is the arc of the circle with centre 0 and radius 2 connecting the points 2 and −2i; R c) γ p(z)ez dz where p is a polynomial and γ is the curve from Problem 5. Hint: prove that z 7→ p(z)ez has a power series representation in C. 8.

a) Prove that the functions z 7→ Re z, z 7→ Im z and z 7→ |z| do not have a primitive. R R b) With γ and σ as in Example 20.12 find γ f (|z|)dz and σ f (|z|)dz where f : [0, ∞) → R is a continuous function. Does z 7→ f (|z|) have a primitive when f (r) = r k , k ∈ N.

381

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21

The Cauchy Integral Theorem and Integral Formula

The results in this Chapter can be viewed as the starting point as well as the essence of Cauchy’s approach to complex variable theory. We start with a more or less heuristic (but still correct) consideration. Let G ⊂ C ∼ = R2 be a domain for which Green’s theorem in the form of (II.26.13) holds. In particular ∂ G is the trace of a parametric curve and G can be considered as a normal domain with respect to both coordinate axes. Since we identify R2 with C the canonical basis in R2 can be identified with {1, i}, a vector z has components z = x + iy and f : G → C is interpreted as vector field f (z) = u (x, y) + i v (x, y). Suppose that f is in a neighbourhood of G holomorphic, i.e. f 0 (z) exists. We assume in addition that f 0 is continuous implying that ux , uy , vx and vy are continuous too. Now Green’s theorem, i.e. (II.26.13) yields Z Z f (z) dz = (u + iv)(dx + i dy) ∂G

∂G

= =

Z

∂G Z 

u dx −

Z

v dy + i

∂G

∂v ∂u − − ∂x ∂y

G





Z

v dx +

∂G

dx dy + i

Z  G

Z

u dy



∂G  ∂u ∂v dx dy. − ∂x ∂y

Since f is holomorphic the Cauchy-Riemann differential equations hold, i.e. ux = vy and vx = −uy , implying that Z f (z) dz = 0. (21.1) ∂G

Thus for a function f holomorphic in a neighbourhood of G and continuous complex derivative f 0 we expect that (21.1) holds for at least all domains for which Green’s theorem holds. We want to prove (21.1) without the assumption that f 0 is continuous and we start with Goursat’s Theorem which some authors refer to as Goursat’s lemma. Theorem 21.1 (E. Goursat). Let ∆ ⊂ C be a closed triangle and U ⊂ C an open neighbourhood of ∆. For a holomorphic function f : U → C it 383

A COURSE IN ANALYSIS

follows that

Z

f (z) dz = 0,

(21.2)

∂∆

where ∂ ∆ is interpreted as trace of a piecewise continuously differentiable parametric curve. Proof. The following proof is standard and essentially identical in every book about complex analysis. Let ∆ = ∆(a, b, c) and U be as in the assumptions, see Figure 21.1. c U ∆ = ∆(a, b, c) b a

Figure 21.1 (1)

(4)

We now divide ∆ into four triangles ∆1 , ......, ∆1 by connecting the midpoints of the sides of ∆, see Figure 21.2. c

b

z3

z2

b

b b

a

z1 Figure 21.2

If the midpoints are z1 , z2 , z3 we can consider ∆(z1 , z2 , z3 ) and assume that ∂∆(z1 , z2 , z3 ) interpreted as parametric curve has anticlockwise orientation. 384

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

We also assume that the boundaries of the other three triangles interpreted as parametric curves have anticlockwise orientation. If the line segment [w1 , w2 ] (again interpreted as oriented curve) connects two midpoints (k) w1 , w2 ∈ {z1 , z2 , z3 }, then [w1 , w2 ] is a side in a triangle ∆1 and (the trace (l) of) the reversed curve [w2 , w1 ] is a side of triangle ∆1 for suitable k, l ∈ {1, 2, 3, 4}. Since Z Z f (z) dz = −

[w1 ,w2 ]

we find

Z

[w2 ,w1 ]

f (z) dz =

Z 4 X k=1

∂∆

and it follows that

f (z) dz

f (z) dz,

(21.3)

(k)

∂∆1

Z Z f (z) dz ≤ 4 max 1≤k≤4 ∂∆

(k) ∂∆1

(1)

Denote by ∆max a triangle in the set {∆1 1 Z f (z) dz = max

f (z) dz .

(2)

(3)

(21.4)

(4)

, ∆1 , ∆1 , ∆1 } such that Z (21.5) f (z) dz . 1≤k≤4

∂∆max 1

(k)

∂∆1

we repeat the procedure, see Figure 21.3. For ∆max 1 c

b

z3 b

z2 b

b

b b

b

a

z1

Figure 21.3

Thus we get a triangle ∆max for which we find 2 Z Z ≤ 4 f (z) dz f (z)dz ∂∆max 1

∂∆max 2

385

(21.6)

A COURSE IN ANALYSIS

and therefore

Z Z f (z) dz ≤ 42

∂∆max 2

∂∆

f (z) dz .

(21.7)

Now we iterate this process to obtain triangles

∆ = ∆max ⊃ ∆max ⊃ ∆max ⊃ ∆max ⊃ .............. 0 1 2 3 such that

Z Z f (z) dz ≤ 4n

f (z) dz

∂∆

∂∆max n

l∂∆max = n

1 = .......... = 2−n l∂∆ , l∂∆max n−1 2

holds as well as

(21.8)

(21.9)

max where l∂∆max denotes the length of the curve ∂∆max is compact k . Since ∆n k max ) → 0 as n → ∞ there exists a unique point z and diam (∆ 0 such that n T max ∆k . As f is complex differentiable at z0 there exists a continuous {z0 } = k∈N

function h such that lim h(z) = 0 and z→z0

f (z) = f (z0 ) + (z − z0 ) f 0(z0 ) + (z − z0 ) h(z).

(21.10)

Furthermore we know that z 7→ f (z0 ) + (z − z0 ) f 0(z0 ) has a primitive which yields for all n ∈ N Z (f (z0 ) + (z − z0 ) f 0 (z0 )) dz = 0. (21.11) ∂∆max n

Now we find

Z

∂∆max n

f (z) dz =

Z

∂∆max n

(z − z0 ) h(z) dz

max (|z − z0 ||h(z)|) ≤ l∂ ∆max n max z∈∂ ∆n

)2 max |h(z)|. ≤ (l∂∆max n max z∈∆n

386

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

By (21.8) and (21.9) it follows that Z f (z) dz ≤ l2 max |h(z)|. ∂∆ max

(21.12)

z∈∆n

∂∆

From lim h(z) = 0 we deduce that for  > 0 we can find N ∈ N such that z→z0

n ≥ N implies max |h(z)| < max z∈∆n

 2 l∂∆



Z

which yields that for every  > 0

∂∆

holds, i.e. we have

Z

f (z) dz <  f (z) dz = 0,

(21.13)

∂∆

and the theorem is proved. It turns out that a small generalisation of Theorem 21.1 will have enormous implications. Theorem 21.2. Let ∆ ⊂ C be a closed triangle and z0 ∈ ∆. If U ⊂ C is an open neighbourhood of ∆ and f : U → C a continuous function which is holomorphic in U\{z0 } then we have Z f (z) dz = 0. (21.14) ∂∆

Proof. We will show the result in three steps. First suppose that z0 is a vertex of ∆, see Figure 21.4 c

ζ2 b

∆1 a

∆2

∆3 b

b

ζ1 Figure 21.4 387

A COURSE IN ANALYSIS

We pick a point ζ1 on a side of ∆(a, b, c) with vertex z0 and determine ζ2 on the other side of ∆(a, b, c) with vertex z0 by drawing the parallel to the side of ∆(a, b, c) opposite to z0 . Using Theorem 21.1 and its proof we get with ∆ = ∆(a, b, c) Z Z Z Z Z f (z) dz = f (z) dz + f (z) dz + f (z) dz = f (z) dz (21.15) ∂∆

∂∆1

∂∆2

∂∆3

∂∆1

where ∆1 has the vertices z0 , ζ1 and ζ2 , ∆2 has the vertices ζ1 , ζ2 and one vertex z1 6= z0 of ∆(a, b, c), and ∆3 has the vertices z1 , z2 of ∆(a, b, c) not equal to z0 and one of the points ζ1 or ζ2 . This gives a decomposition of ∆(a, b, c) into triangles ∆1 , ∆2 , ∆3 and to ∆2 , ∆3 we may apply Theorem 21.1, i.e. we get (21.15). Furthermore we have Z f (z) dz ≤ l∂∆1 max |f (z)| ≤ l∂∆1 ||f ||∞,∆ , z∈∆1

∂∆1

where ||f ||∞,∆ = maxz∈∆ |f (z)|. The compactness of ∆ together with the continuity of f implies that ||f ||∞,∆ < ∞. Since ζ1 was arbitrary but [ζ1 , ζ2 ] is parallel to the side of ∆(a, b, c) opposite to z0 it follows that ζ1 → z0 implies l∂∆1 → 0 and therefore Z lim f (z) dz = 0 ζ1 →z0 ∂∆1

which yields the result for the first case. In the second case we assume that z0 ∈ ∂∆(a, b, c) but z0 ∈ / {a, b, c}, i.e. z0 is not a vertex of ∆(a, b, c). We now decompose ∆(a, b, c) into two triangles ∆1 and ∆2 , where in each triangle z0 is a vertex, see Figure 21.5. c

∆2

∆1

b

b

z0 a

Figure 21.5 388

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

Since for ∆1 and ∆2 we can apply the first case it follows that Z Z Z f (z) dz = f (z) dz + f (z) dz = 0. ∂∆

∂∆1

∂∆2

˚ i.e. z0 is point in the interior of ∆. By Finally we assume that z0 ∈ ∆, drawing a line passing through z0 and one vertex of ∆(a, b, c) we decompose ˜ 1 and ∆ ˜ 2 , see Figure 21.6. ∆(a, b, c) into two triangles ∆ c

b

˜1 ∆

˜2 z0 ∆ b Figure 21.6

a

˜ 1 and ∆ ˜ 2 the second case applies and we deduce For the two triangles ∆ Z Z Z f (z) dz = f (z) dz + f (z) dz = 0 ∂∆

˜1 ∂∆

˜2 ∂∆

and the theorem is proved. Combining Theorem 21.2 with Theorem 20.23 we obtain Theorem 21.3. Let G ⊂ C be a star-shaped domain with respect to some point ζ ∈ G. Further let f : G → C be a continuous function which is holomorphic in G or in G\{z0 }, z0 ∈ G. Then f admits a primitive in G. As a final consequence of these considerations we derive Theorem 21.4 (Cauchy’s Integral Theorem). Let G ⊂ C be a starshaped domain and f : G → C a continuous function which is with the possible exception of one point z0 ∈ G holomorphic in G. For every closed piecewise continuously differentiable curve γ : [a, b] → G the following holds: Z f (z) dz = 0. (21.16) γ

389

A COURSE IN ANALYSIS

Proof. Since f admits a primitive in G and γ is closed the theorem follows. Example 21.5. Let z = r eiϕ ∈ C\(−∞, 0]. We can reach z starting from 1 by following the line segment [1, z], but also by following γ = γ1 ⊕ γ2 where γ1 is the line segment from 1 to r = |z| and γ2 is the arc of the circle with centre 0 and radius r connecting r and z = r eiϕ , see Figure 21.7: γ2 z = reiϕ b

[1, z]

b

b

0

1

b

γ1

r

Figure 21.7 The set C\(−∞, 0] is star-shaped with respect to 1 and therefore z 7→

R

[1,z]

1 ζ

dζ

is a primitive to 1z in C\(−∞, 0]. Since by Cauchy’s integral theorem applied to ζ1 and the curve γ1 ⊕ γ2 ⊕ [z, 1] it follows that Z Z Z 1 1 1 dζ = dζ + dζ ζ ζ ζ [1,z]

=

γ1

γ2

Zr

Zϕ

1 dt + t

1

i r eit dt r eit

0

= ln r + i ϕ = log z, i.e. the function z 7→

R

[1,z]

1 ζ

dζ gives in C\(−∞, 0] the principal branch of the

logarithmic function, compare with Definition 18.8. Example 21.6. In Problem 9 of Chapter 8 we calculated the Fresnel integrals using real variable methods, in particular a type of mollifying technique. We will now evaluate these integrals with the help of Cauchy’s integral 390

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

theorem. This does not give only a new method for finding these two particular integrals, but will lead eventually to a complete new tool for calculating integrals, which we will discuss in great detail in Chapter 25 once we have proved the residue theorem. 2

We start the observation that the function z 7→ eiz is an entire function, R with 2 hence eiz dz = 0 for every piecewise continuously differentiable, simply γ

closed curve γ : [a, b] → C. For r > 0 we consider the curve γ = γ1 ⊕ γ2 ⊕ γ3 , see Figure 21.8, where γ1 is the line segment connecting 0 with r, γ2 is the π arc of the circle with centre 0 and radius r connecting r and rei 4 , whereas π γ3 is the line segment connecting rei 4 and 0. y = Im z

π

re 4 i b

π 4 b

x = Re z

r

0

Figure 21.8 It follows that

0=

Z γ

iz 2

e

dz =

Z

γ1

iz 2

e

dz +

Z

iz 2

e

γ2

dz +

Z

2

eiz dz, z = x + iy.

γ3

On γ1 we have z = x, on γ2 we have z = reiϕ , 0 ≤ ϕ ≤ π4 , and on γ3 we have π z = % e 4 i where % moves from r to 0. For the corresponding integrals this 391

A COURSE IN ANALYSIS

yields Z

iz 2

e

dz =

γ1

Zr

e

dx =

0

Zr

2

cos x dx + i

0

γ2

Z

Z0

iz 2

e

dz =

e

dz =

γ3

sin x2 dx,

0

ir 2 e2iϕ

e

Z

iϕ

ire dϕ =

0

iz 2

Zr

π 4

π 4

Z

Z

ix2

eir

2

cos 2ϕ−r 2 sin 2ϕ

ireiϕ dϕ,

0

i%2

e e

r

π i 2

e

π i 4

π i 4

d% = − e

Zr

2

e−% d%,

0

or Zr

2

cos x dx + i

0

Zr

1√ sin x dx = 2 2 2

0

Zr

−%2

e

i√ 2 d% + 2

0

Zr

2

e−% d%

(21.17)

0

π 4

−

Z

eir

2

cos 2ϕ−r 2 sin 2ϕ

i reiϕ dϕ.

0

Next we observe π

π

Z4 Z4 2 cos 2ϕ−r 2 sin 2ϕ 2 ir iϕ e ir e dϕ ≤ e−r sin 2ϕ r dϕ 0

0

π

r = 2

Z2 0

−r 2

e

r sin ϕ dϕ ≤ 2

π

Z2 0

e−r

2ϕ π

dϕ

π 2 = (1 − e−r ), 4r where we used first the change of variable 2ϕ → ϕ and then the estimate 2ϕ ≤ sin ϕ for 0 ≤ ϕ ≤ π2 . The latter estimate we see as follows: on [0, π2 ) we π have 1 d 1≤ = tan y 2 cos y dy or Zϕ Zϕ d ϕ = 1dy ≤ tan y dy = tan ϕ dy 0

0

392

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA


ϕ d sin ϕ implying ϕ cos ϕ ≤ sin ϕ on [0, π2 ). Since dϕ = ϕ cos ϕ−sin , we derive ϕ ϕ2 sin ϕ 2ϕ sin ϕ π 2 that ϕ is on [0, 2 ] a decreasing function with ϕ ≥ π , i.e. π ≤ sin ϕ for ϕ ∈ [0, π2 ]. Now we pass in (21.17) to the limit as r tends to ∞ to obtain Z∞

2

cos x dx + i

0

and since

1√ 2 sin x dx = 2 2

R∞

2

e−% d% =

√

Z∞

−%2

e

i√ 2 d% + 2

0

0

0

as well as

Z∞ π 2

we eventually have

Z∞ 0

r √ π π 1√ 2 = cos x dx = 2 2 8

Z∞

r √ π π 1√ sin x dx = = . 2 2 2 8

0

Z∞

2

e−% d%,

0

2

2

Looking more closely at our calculation we see the power of complex integration combined with the Cauchy theorem: in order to evaluate certain line integrals (including integrals of functions defined on a segment of the real axis) we may replace the original integral by an integral more convenient to handle. In Example 20.11 a simple calculation has shown that Z 1 1 dz = 1. 2πi z − z0 |z−z0 |=r

This result has the following extension Lemma 21.7. For z ∈ Br (z0 ) we have Z 1 1 dζ = 1, 2πi ζ −z

(21.18)

|ζ−z0 | = r

where as usual the circle |ζ − z0 | = r is parametrized by t 7→ z0 + reit , t ∈ [0, 2π]. 393

A COURSE IN ANALYSIS

Proof. For z ∈ Br (z0 ) we find q := |z−z0 | r −1 ∈ [0, 1) implying for z ∈ Br (z0 ) and ζ ∈ ∂Br (z0 ) that   z − z0 n = q n , n ∈ N. max ζ∈∂ Br (z0 ) ζ − z0 z−z0 , ζ−z0

With w :=

z ∈ Br (z0 ) and ζ ∈ ∂ Br (z0 ) this implies

k ∞ ∞  1 1 1 1 X k 1 X z − z0 = = w = ζ −z ζ − z0 1 − w ζ − z0 k=0 ζ − z0 k=0 ζ − z0 and this geometric series converges absolutely and uniformly in ζ. Now it follows for z ∈ Br (z0 ) k Z Z ∞  1 1 1 X z − z0 1 dζ = dζ 2πi ζ −z 2πi ζ − z0 k=0 ζ − z0 |ζ−z0 |=r

∞ X

|ζ−z0 |=r

1 = (z − z0 ) 2πi k=0

Z

k

|ζ−z0 |=r

1 dζ. (ζ − z0 )k+1

For k + 1 = 1, i.e. k = 0, we know that the value of the integral above is 1, while for k + 1 > 1 it follows that Z

|ζ−z0 |=r

1 dζ = (ζ − z0 )k+1

Z2π

1 r k+1 e(k+1) it

ireit dt

0

1 = k r

Z2π

e−ikt dt = 0,

0

which yields 1 2πi

Z

|ζ−z0 |=r

Z ∞ X 1 1 1 dζ = dζ (z − z0 )k k+1 ζ −z 2πi (ζ − z 0) k=0 |ζ−z0 |=r Z 1 1 = dζ = 1. 2πi ζ − z0 |ζ−z0 |=r

394

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THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

With the help of Lemma 21.7 and the Cauchy integral theorem we can obtain a formula expressing a holomorphic function in the interior of a disc by its values on the boundary of the disc. Theorem 21.8 (Cauchy’s Integral Formula). Let G ⊂ C be a region and f : G → C a holomorphic function. For every z ∈ Br (z0 ) ⊂ G, r > 0, z0 ∈ G, we have Z 1 f (ζ) f (z) = dζ (21.19) 2πi ζ −z |ζ−z0 |=r

where the circle |ζ − z0 | = r is parametrized by γ(t) = z0 + reit , t ∈ [0, 2π]. Proof. We choose  > 0 such that Br+ (z0 ) ⊂ G. Since Br+ (z0 ) is an open, convex, hence star-shaped neighbourhood of Br (z0 ) we can apply Cauchy’s integral theorem to the function g : Br+ (z0 ) → C defined by g(ζ) :=

(

f (ζ)−f (z) , ζ−z 0

ζ 6= z ζ =z

f (z),

(21.20)

since g is continuous in Br+ (z0 ) and at least holomorphic in Br+ (z0 )\{z}. It follows that Z Z f (ζ) − f (z) 0= g(ζ) dζ = dζ ζ −z |ζ−z0 |=r

=

Z

|ζ−z0 |=r

|ζ−z0 |=r

f (ζ) dζ − f (z) ζ −z

Z

|ζ−z0 |=r

1 dz, ζ −z

and by Lemma 21.7 we find 1 f (z) = 2πi

Z

|ζ−z0 |=r

f (ζ) dζ. ζ −z

Using the parametrization γ(t) = z0 + reit formula (21.19) yields

395

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Corollary 21.9 (Mean-value Theorem). Under the assumptions of Theorem 21.8 the following hold 1 f (z0 ) = 2π

Z2π

f (z0 + reit ) dt;

(21.21)

0

and |f (z0 )| ≤ ||f |∂Br (z0 ) ||∞ = ||f ||∞,∂Br (z0 ) .

(21.22)

Example 21.10. A. We want to find the values of the two integrals Z Z 1 1 dz and dz, 2 2 z +1 z +1 |=1 |z+ 3i 2

|z− 2i |=1

where in each case the circle is parametrized as in Theorem 21.8. First we 1 1 · z−i and this function is holomorphic in C\{i, −i}. note that z 21+1 = z+i 1 The function z+i is holomorphic in C\{−i} and therefore we can apply the 1 in the disc |z − 2i | = 1 to find Cauchy integral formula to g(z) = z+i   Z Z 1 1 g(z) = π. dz = dz = 2πi g(i) = 2πi 2 z +1 z−i 2i |z− 2i |=1

|z− 2i |=1

1 Analogously we find for h(z) = z−i which is holomorphic in C\{i} that   Z Z 1 1 h(z) = −π. dz = dz = 2πi h(−i) = 2πi z2 + 1 z+i −2i |z+ 3i |=1 2

|z+ 3i |=1 2

B. Since the function z 7→ cos z is an entire function we find for the circle |z| = 3, given with the usual parametrization that Z cos z dz = 2πi cos i = π i(ei − e−i ). z−i |z|=3

For developing the theory further we need to reformulate our results on the continuity and differentiability of parameter dependent integrals. The proof of the following results are the same as in the cases discussed in Chapter II.17 or in Chapter 8 of this volume. 396

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THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

Theorem 21.11. Let γ : [a, b] → C be a piecewise continuously differentiable curve, G ⊂ Rn and f : tr (γ) × G → C a continuous function. A. The function Z F : G → C, F (x) := f (z, x) dz (21.23) γ

is on G continuous. ∂f exists and is B. If G is open and if on tr (γ) × G the partial derivative ∂x j continuous, then F has a continuous partial derivative with respect to xj and Z ∂F ∂f (x) = (z, x) dz. (21.24) ∂xj ∂xj γ

C. Now let G ⊂ C be open and assume that for every z ∈ tr (γ) the function ζ 7→ f (z, ζ) is complex differentiable with continuous complex (partial) derivative fζ (z, ζ). Then F is holomorphic on G and we have Z 0 F (ζ) = fζ (z, ζ) dz. (21.25) γ

We will now apply Theorem 21.11 to (21.19). Taking in (21.19) the derivative under the integral sign which is by Theorem 21.11.C justified we find Z f (ζ) 1 0 dζ. (21.26) f (z) = 2πi (ζ − z)2 |ζ−z0 |=r

f (ζ) We observe that z 7→ (ζ−z) 2 again satisfies all assumptions of Theorem 21.11.C, so we can iterate the process. In fact we obtain the generalised Cauchy integral formula:

Theorem 21.12. Let G ⊂ C be a region and f : G → C a holomorphic function. Then f has all higher order complex derivates and for n ∈ N, Br (z0 ) ⊂ G, z ∈ Br (z0 ) it follows Z f (ζ) n! (n) dζ, (21.27) f (z) = 2πi (ζ − z)n+1 |ζ−z0 |=r

where we use the parametrization t 7→ z0 + reit , t ∈ [0, 2π] for the circle |ζ − z0 | = r. 397

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We leave the formal, short proof by induction of Theorem 21.12 to the reader, see Problem 8. Remark 21.13. Theorem 21.12 is quite a remarkable result as it states that if a function f : G → C is one time complex differentiable, then all complex derivatives exist in G and hence they are holomorphic too. As a consequence we deduce that u = Ref and v = Imf are in G ⊂ C ∼ = R2 arbitrary often differentiable. This raises the following two questions: 1. Suppose u, v : G → R are one time partially differentiable functions in a region G ⊂ R2 and satisfy the Cauchy-Riemann differential equations. Does this imply that u and v are arbitrarily differentiable in G? 2. Suppose that 2 2 u : G → R have the second order partial derivative ∂∂xu2 and ∂∂yu2 and is 2

2

harmonic in the region G ⊂ R2 , i.e. ∂dxu2 + ∂∂yu2 = 0. Does this imply that u is arbitrarily often differentiable in G? We will later see that both questions have an affirmative answer which is well understood within the theory of linear elliptic partial differential equations. As in the case of Theorem 21.8 we can use the generalised Cauchy integral formulae to evaluate certain line integrals. Example 21.14. A. Consider the integral Z

2

eζ dζ (ζ + 1)4

(21.28)

|ζ|=4 2

with the parametrization t 7→ 4 eit , t ∈ [0, 2π]. Since ζ 7→ eζ is an entire 2 function, when taking in (21.27) as f the function f (ζ) = eζ , n = 3 and z = −1 we get Z 2 eζ 3! (3) f (−1) = dζ. 2πi (ζ + 1)4 |ζ|=4

2

Since f (3) (ζ) = (12 ζ + 8 ζ 3) eζ we find f (3) (−1) = −20 e and therefore Z

2

eζ 20 eπi dζ = − . 4 (ζ + 1) 3

|ζ|=4

398

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

B. For k ≥ 1 and r ≥ 8 the following holds with the parametrization of |ζ| = r given as ζ = reiϕ , ϕ ∈ [0, 2π], Z 1 eζ 1 = dζ. (21.29) Γ(k + 1) 2πi (ζ − 2πi)k+1 |ζ|=r

Since r ≥ 8 it follows that 2πi ∈ {z ∈ C | |z| < r}, further we have and with ζ = 2πi we find e2πi = 1. Thus by (21.27) we have Z eζ k! dζ 1= 2πi (ζ − 2πi)k+1

dk d ζk

eζ = eζ

|ζ|=r

and since k! = Γ(k + 1) formula (21.29) follows. Let G ⊂ C be a region and let f : G → C be a continuous function. Let γj : [aj , bj ] → G, j = 1, 2, be two simply closed piecewise continuously differentiable curves with disjoint traces tr (γ1 ) ∩ tr (γ2 ) = ∅. Since tr (γ1 ) and tr (γ2 ) are compact, dist (tr (γ1 ), tr (γ2 )) > 0. Pick z1 ∈ tr (γ1 ) and z2 ∈ tr (γ2 ) and assume that for the line segment [z1 , z2 ] connecting z1 and z2 we have [z1 , z2 ] ∩ tr (γ1 ) = {z1 } and [z1 , z2 ] ∩ tr (γ2) = {z2 }, see Figure 21.9.

γ1

G

z1

z2

γ2−1

γ = γ1 ⊕ [z1 , z2 ] ⊕ γ2−1 ⊕ [z2 , z1 ]

Figure 21.9 If necessary by changing the parametrization we may assume that z1 = γ1 (a1 ) and z2 = γ2 (a2 ). Denote by γ2−1 the curve inverse to γ2 and consider the 399

A COURSE IN ANALYSIS

simply closed, piecewise continuously differentiable curve γ := γ1 ⊕ [z1 , z2 ] ⊕ γ2−1 ⊕ [z2 , z1 ], see again Figure 21.9. It follows that Z Z Z Z Z f (z) dz + f (z) dz = f (z) dz + f (z) dz + γ

γ1

=

Z

γ1

since

R

[z1 ,z2 ]

cular for

R

γ2−1

[z1 ,z2 ]

f (z) dz −

f (z) dz = −

R

Z

f (z) dz

[z2 ,z1 ]

f (z) dz,

γ2

R

f (z) dz and

γ2−1

[z2 ,z1 ]

f (z) dz = 0 we conclude

f (z) dz = −

R

f (z) dz. In parti-

γ2

γ

Z

f (z) dz =

γ1

Z

f (z) dz,

(21.30)

γ2

i.e. we can replace γ1 by γ2 to evaluate

R

f (z) dz.

γ1

Let us have again a look at

Z

1 dz z

(21.31)

γ

where tr (γ) = S 1 . For example we may take γ = γk : [0, 2πk] → C, γk (t) = eit , k ∈ N. From the very definition of the line integral we find that Z

1 dz = z

Z2πk

1 it ie dt = 2πk i. eit

0

γk

Thus this integral depends onR γ, i.e. the parametrization, and not only on 1 1 dz counts how often γk is winding around tr(γ). We also note that 2πi z γk

z0 = 0 ∈ C. This and related observation will soon become of more importance. From Theorem 21.12 we have learnt that if f : G → C is holomorphic, i.e. has a complex derivative f 0 , then it has complex derivation f (k) of all order. 400

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

We even have a formula for the derivative f (k) (z) provided z ∈ Br (z0 ) ⊂ G, namely (21.27). We want to derive some useful estimates, called standard estimates or Cauchy’s inequalities for f (k) (z). Theorem 21.15. Let G ⊂ C be a region and f : G → C a holomorphic function. Let z0 ∈ G and Br (z0 ) ⊂ G. For every d, 0 < d ≤ r, and all z ∈ Br−d (z0 ), we find for k ∈ N0 |f (k) (z)| ≤

r k! d dk

max |f (ζ)|.

|ζ−z0 |=r

(21.32)

Proof. Using Cauchy’s integral formula (21.27) we find for |z − z0 | < r that Z f (ζ) k! (k) dζ. f (z) = 2πi (ζ − z)k+1 |ζ−z0 |=r

For |z − z0 | ≤ r − d it follows that d = r + d − r ≤ |ζ − z0 | − |z − z0 | ≤ |ζ − z|, 1 |ζ−z|k+1

i.e. |ζ − z| ≥ d or |f

(k)

≤

1 , dk+1

k! (z)| ≤ 2π

which implies Z

|ζ−z0 |=r

f (ζ) dζ (ζ − z)k+1

k! 1 ≤ 2πr k+1 max |f (ζ)| |ζ−z0 |=r 2π d r k! max |f (ζ)|. = d dk |ζ−z0 |=r

Corollary 21.16. Under the assumptions of Theorem 21.15 the following holds k! |f (k) (z0 )| ≤ k max |f (ζ)| (21.33) r |ζ−z0|=r and for z ∈ B 2r (z0 ) we find a constant C independent of f such that |f (k) (z)| ≤ C

k! max |f (ζ)|. r k |ζ−z0 |=r 401

(21.34)

A COURSE IN ANALYSIS

Proof. In the first case we take d = r, while the second estimate follows for d = 2r . From (21.33) we deduce for every z ∈ C such that |z − z0 | = % < r that (k)  k f (z0 ) % k k! (z − z0 ) ≤ ||f ||∞, ∂Br (z0 ) r

and since 0
0 such that in every disc Br (z0 ) ⊂ BR (z0 ) ⊂ G the function f has a Taylor expansion f (z) =

∞ X k=0

ak (z − z0 )k

(21.39)

representing f and for the Taylor coefficients ak we find for 0 < % < R Z f (k) (z0 ) 1 f (ζ) ak = dζ. (21.40) = k! 2πi (ζ − z0 )k+1 |ζ−z0 |=%

The coefficients ak , k ∈ N0 , are uniquely determined and satisfy the estimate |ak | ≤

1 rk

max |f (ζ)|.

|ζ−z0 |=r

(21.41)

Proof. It remains to prove the uniqueness of the coefficients and estimate (21.41). The uniqueness problem is solved by differentiating and evaluating f (k) at z0 , whereas the estimate is a consequent of Corollary 21.16. As first consequence of Theorem 21.17 is Corollary 21.18. Let z0 T(a (z) := k)

∞ X k=0

ak (z − z0 )k , z0 ∈ C, ak ∈ C, 403

(21.42)

A COURSE IN ANALYSIS

be a convergent power series with radius of convergence R > 0. For every z0 z1 ∈ BR (z0 ) we can expand the holomorphic function z 7→ T(a (z) into a k) convergent power series T(bz1k ) (z)

=

∞ X k=0

bk (z − z1 )k

(21.43)

where the radius of convergence of T(bz1k ) (z) is at least R − |z1 − z0 |. z0 (· ) is a holomorphic function we just need to apply Theorem Proof. Since T(a k) 21.17.

From Corollary 21.18 we deduce further Corollary 21.19. If a holomorphic function is defined as a convergent power z0 z0 series, i.e. f (z) = T(a (z) with radius of convergence R > 0, then T(a (z) k) k) is also the Taylor series of f in B% (z0 ), % < R. In Chapter 16 and 17 we have already encountered several well-known holomorphic functions defined by power series. The next chapter will give a further insight to holomorphic functions and power series.

Problems 1. Suppose that f, g : G → C are complex differentiable functions defined on a region G ⊂ C and let γ : [a, b] → R be a piecewise continuously differentiable curve. Prove that Z Z 0 f (z)g (z)dz = f (γ(b))g(γ(b)) − f (γ(a))g(γ(a)) − f 0 (z)g(z)dz γ

γ

and deduce for a simply closed curve that Z Z 0 f (z)g (z)dz = − f 0 (z)g(z)dz. γ

γ

2.

a) For the parametric curve γ(t) aeit + be−it , t ∈ [0, R 2π] and 0 < P= M b < a and every polynomial p(z) = k=1 ak z k prove that γ p(z)dz = 0.

b) Sketch the curve γ : [0, 2π] → C definedPby γ|[0,π] (t) = t+i sin t and γ[π,2π] (t) = 2π −t. Let Q(ζ, η) = |α|≤k aα ζ α1 η α2 , aα ∈ C, 404

21

THE CAUCHY INTEGRAL THEOREM AND INTEGRAL FORMULA

(α1 , α2 ) ∈ N20 and ζ, η ∈ C, be a polynomial in two complex variables. Find Z Q(cos z, sin z)dz. γ

c) Let g : B1 (0) → C be a continuous function and h : B1 (0) → C a holomorphic function. For 0 < ρ < 1 prove that Z (g(ζ) + h(ζ))dζ ≤ 2π||g||∞,∂Bρ(0) . |ζ|=ρ

3.

a) Let G ⊂ C be a region and γ : [a, b] → G be a curve such that with a < c < b the curves γ1 := γ|[a,c] and γ2 := γ|[c,b] are piecewise continuously differentiable curves which are closed and further assume that Rtrγ1 ∩ trγ2 = {γ(c)}. For a holomorphic function h : G → C show that γ h(z)dz = 0. b) Consider the set Γ := ∂B1 (0)∪∂B1 (2). Find a curve η : [0, 4π] → C which has trace Γ and with c = 2π satisfies all requirements made for γ in part a). Find the integral Z 1 1 dz. 2πi η z − 23

4. Find the integrals 1 2πi and

Z

|ζ−4|=1

ζ−

π 4




sin ζ + (ζ − π) cos ζ

ζ2 −

5π ζ 4

−

π2 4

dζ


Z ζ − π4 sin ζ + (ζ − π) cos ζ 1 dζ π2 2πi |ζ+ 21 |= 52 ζ − ζ 2 − 5π 4 4

where in both cases we use the standard parametrization of the corresponding circles. 5. Prove that

Z ∞ X 1 e−ζ dζ = 1. 2 2πi 2 ζ − ln(k + k) |ζ|=k k=1

Hint: use Example I.16.3.

405

A COURSE IN ANALYSIS

6.* Prove that for f as in Corollary 21.9 the estimate |f (z)| ≤ kf k∞,∂Br (z0 ) holds for all z ∈ Br (z0 ).

  k1 1 Hint: apply (21.19) to f k and use limk→∞ rk ζ−z k∞,∂Br (z0 ) = 1.

7. Let G ⊂ C be a region.

a) Prove that for a holomorphic function f = u + iv in G the real part and the imaginary part satisfy the following mean-value equation: Z 2π 1 w(z0 + reit )dt w(z0 ) = 2π 0

for every Br (z0 ) ⊂ G.

b) Let fk : G → C be a sequence of holomorphic functions converging on every compact subset K ⊂ G uniformly to a function f : G → C. Prove that f must satisfy the mean-value property, i.e. Z 2π 1 f (z0 ) = f (z0 + reit )dt. 2π 0 8. Prove the generalised Cauchy integral formula, i.e. Theorem 21.12. 9. Use the generalised Cauchy integral formula to evaluate the following integrals: R cos√ζ 2 dζ; a) |ζ−z|=1 (ζ− π)2 b) For the hypergeometric function 2 F1 find Z 1 2 F1 (z) dz, n ∈ N. 2πi |z|= 21 z n+1

10. Let f be a holomorphic function in a neighbourhood of B1 (z0 ). Suppose in addition that for all z ∈ B1 (z0 ) we have Z 1 + ζ − z + (ζ − z)2 1 f (ζ)dζ = 0. 2πi |ζ−z0|=1 (ζ − z)4 Prove that in B1 (z0 ) the function f satisfies the differential equation 000

00

f (z) + 3f (z) + 6f 0(z) = 0. 406

22

Power Series, Holomorphy and Differential Equations

Let us summarize what we know about holomorphic functions so far. Theorem 22.1. Let G ⊂ C be a region. The following statements are equivalent: i) f : G → C is holomorphic; ii) f has locally a primitive; iii) for every z0 ∈ G the function f admits an expansion into a convergent power series with strictly positive radius of convergence which represents f , this power series is the Taylor series of f about z0 ; iv) for every closed triangle 4 ⊂ G the following holds Z f (z) dz = 0.

(22.1)

∂4

Proof. It remains only to prove that iv) implies i) which however follows from Remark 20.27. Remark 22.2. A. The result that statement iv) implies holomorphy is often called the Theorem of Morera. B. We have seen in the beginning of Chapter 21 that if two functions u, v : G → R, G ⊂ C ∼ = R2 , satisfy the Cauchy-Riemann differential equations and their partial derivatives are continuous then Cauchy’s theorem holds for f = u + iv, and hence f is holomorphic. It takes some more effort to prove that the Cauchy-Riemann differential equations without the continuity condition on the partial derivatives already imply the holomorphy of f . Furthermore, given a harmonic function u˜ in G, i.e. ∆˜ u = 0, then one can find a conjugate harmonic function v˜ in G, i.e. ∆˜ v = 0, such that f˜ = u˜ + i˜ v is holomorphic. We will deal with these results later, in particular when studying partial differential equations. Let us consider the following 407

A COURSE IN ANALYSIS

Example 22.3. Given the two power series ∞ X 1 k z T1 (z) = k+1 2 k=0

and T2 (z) =

∞ X k=0

(22.2)

1 (z − i)k . (2 − i)k+1

(22.3)

For T1 we find in the disc |z| < 2 that

1 X  z k 1 1 T1 (z) = = 2 k=0 2 21− ∞

and for T2 it follows in the disc |z − i| < ∞

1 X T2 (z) = 2 − i k=0 =

√



z 2

=

1 2−z

z−i < 1, that 5, i.e. 2−i

z−i 2−i

k

1 1 1 . z−i = 2 − i 1 − 2−i 2−z

1 Both, T1 and T2 , are Taylor series and they represent the function z 7→ 2−z in different domains, but these domains overlap. We can put this differently: Suppose we start with the holomorphic function f : B2 (0) → C, f (z) := T1 (z). Then the function g : B2 (0) ∪ B√5 (i) → C defined by ( T1 (z), z ∈ B2 (0) g(z) := (22.4) T2 (z), z ∈ B√5 (i)

is a holomorphic extension or continuation of f to B2 (0) ∪ B√5 (i). The above example shows that a holomorphic function f : G → C might have a holomorphic continuation to a larger domain and we are of course interested to know whether we can determine the maximal holomorphic continuation of f . The latter problem is much more far reaching than it initially looks, eventually one is led to Riemannian surfaces. A further problem which arises is the question of the uniqueness of a holomorphic continuation. As the following example shows we cannot expect uniqueness of C ∞ -extensions of C ∞ -functions defined on open sets of the real line. 408

22

POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

Example 22.4. The two functions f1 , f2 : R → R, ( 1 ( 1 e− x2 , x 6= 0 e− x2 , x > 0 , f2 (x) := f1 (x) := 0, x≤0 0, x=0 (k)

(k)

are C ∞ -functions, f1 |[0,∞) = f2 |[0,∞) , f1 (0) = f2 (0) = 0 for all k ∈ N0 . Hence f1 is an extension of f2 |[0,∞) as f2 is an extension of f1 |[0,∞). Both functions have even the same Taylor series about 0 which however does not represent any of them. Thus C ∞ -extensions of C ∞ -functions need not be unique at all. The absolute value |.| : R → R is a continuous function and its restriction to R\{0} is even a C ∞ -function. For a continuous function f : G → C defined on a region G ⊂ C it is not possible for f |G\{z0 } , z0 ∈ G, to be holomorphic when f is not, a result we prove next. We need Definition 22.5. Let (X, d) be a metric space and M ⊂ X, M 6= ∅. We call M a discrete subset of X if every point x ∈ M has a neighbourhood U(x) such that U(x) ∩ (M\{x}) = ∅. Points in a discrete set M are in some sense isolated points. If y ∈ M { then for every x ∈ M we have d(x, y) > 0 and hence y has a neighbourhood completely contained in M { , i.e. M { is open. Therefore a discrete set is closed. Moreover, every discrete subset M of a compact metric space is a finite set.

Theorem 22.6. Let G ⊂ C be a region and M ⊂ G a discrete set. Suppose that f : G → C is continuous and f |G\M is holomorphic. Then f is holomorphic in G. Proof. Let z0 ∈ M and choose B (z0 ) ⊂ G such that B (z0 ) ∩ (M\{z0 }) = ∅. Theorem 21.3 implies that f has a primitive in B (z0 ) implying that f is holomorphic in z0 . The next result allows us to remove certain singularities for holomorphic functions in the sense that originally f is defined and holomorphic on some set D\{z0 } but admits a holomorphic continuation to z0 . Theorem 22.7 (Riemann’s theorem on removable singularities). Let G ⊂ C be a region and z0 ∈ G. Further let f : G\{z0 } → C be a holomorphic function which is bounded on Br (z0 )\{z0 } ⊂ G. Then there exists a holomorphic function h : G → C such that h|G\{z0 } = f . 409

A COURSE IN ANALYSIS

Proof. We define on G the function ( (z − z0 )f (z), z ∈ G\{z0 } g(z) := 0, z = z0

(22.5)

which is holomorphic in G\{z0 } and due to the boundedness of f in Br (z0 )\{z0 } it is continuous at z0 , hence continuous on G. Now we can apply Theorem 22.6 to deduce that g is holomorphic in G, i.e. complex differentiable. It follows the existence of a function h : G → C such that g(z) = g(z0 ) + h(z)(z − z0 ), where h is continuous at z0 . On G\{z0 } we have h = f , in particular h is holomorphic on G\{z0 }. A further application of Theorem 22.6 yields that h : G → C is holomorphic and we know already that h|G\{z0 } = f . Theorem 22.6 and in particular Theorem 22.7 tell us that at certain (isolated) points we can extend a holomorphic function and the extension is holomorphic too. Our next goal is to understand how many values we need to know from f and its derivatives to determine f . As Example 22.4 shows, for C ∞ -functions this question does not make much sense. For holomorphic functions the situation is quite different. We start with Definition 22.8. A. Let D ⊂ C be a domain and f : D → C in a neighbourhood of z0 ∈ D a complex differentiable function. We call z0 a zero of order n of f if f (k) (z0 ) = 0

for k = 0, 1, . . . , n − 1

(22.6)

and f (n) (z0 ) 6= 0.

(22.7)

B. A complex number w ∈ C is a value of order n for f at z0 ∈ D if f − w has at z0 a zero of order n. C. The function f has a zero of order ∞ at z0 if f (k) (z0 ) = 0 for all k ∈ N0 . We can transfer Definition 22.8 easily and in the obvious way to function f : I → R where I ⊂ R is an open interval. The functions f1 and f2 in 410

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POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

Example 22.4 are now examples of C ∞ -functions having at 0 a zero of order ∞. On the other hand, if the Taylor series T z0 (z) =

∞ X f (k) (z0 ) k=0

k!

(z − z0 )k

has a positive radius of convergence and represents a function not identically 0, then z0 cannot be a zero of order ∞. This observation prepares our main result. It turns out that if a holomorphic function f has a zero at z0 then its Taylor expansion about z0 allows us to determine the order of the zero at z0 . More precisely we have Lemma 22.9. A holomorphic function f : G → C has a zero of order n at P f (k) (z0 ) z0 ∈ G if and only if its Taylor series about z0 is given by ∞ (z−z0 )k k=n k! where f (n) (z0 ) 6= 0. This holds if and only if we can find a neighbourhood of z0 and a holomorphic function g in this neighbourhood such that g(z0 ) 6= 0 and f (z) = (z − z0 )n g(z). Proof. If f has a zero of order n at z0 then f (k) (z0 ) = 0 for k = 0, . . . , n − 1, and hence the Taylor series of f about z0 will start with the nth term since f (n) (z0 ) 6= 0. Conversely, if the Taylor series of f about z0 starts with the nth term (assumed to be non-zero) then by differentiating the series we obtain f (k) (z0 ) = 0 for k = 0, 1, . . . , n − 1. Further, if the Taylor series of f about P f (k) (z0 ) z0 starts with the nth term, the function g(z) := ∞ (z − z0 )k−n is k=n k! holomorphic in a neighbourhood of z0 and f (z) = (z −z0 )n g(z). On the other hand, if f (z) = (z − z0 )n g(z) then Leibniz’s rule yields that f (k) (z0 ) = 0 for k = 0, 1, . . . , n − 1 implying that the Taylor series of f starts with the nth term. Let us introduce a rather convenient notation. If ϕ : X → Y is any mapping and y ∈ Y we write ϕ(x) ≡ y (22.8)

if ϕ(x) = y for all x ∈ X, i.e. ϕ is the constant mapping x 7→ y. Now we prove

Theorem 22.10. Let G ⊂ C be a region and f : G → C a holomorphic function. The function f is identically zero, i.e. f (z) ≡ 0 if and only if f has in G a zero of order ∞. Equivalent to this condition is the existence of a set A ⊂ G which is not discrete such that f (z) = 0 for all z ∈ A. 411

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Proof. Clearly, if f (z) ≡ 0 then there exists a non-discrete set A ⊂ G such that f |A = 0. Just take A = G. Now let A ⊂ G be a non-discrete set such that f |A = 0. In this case there exists z0 ∈ G and a sequence (zl )l∈N , zl ∈ A, such that zl 6= z0 and liml→∞ zl = z0 . We now prove that z0 must be a zero of order ∞ of f . Consider the Taylor series of f about z0 , P f (k) (z0 ) (z − z0 )k . By continuity of f we find that f (z0 ) = i.e. f (z) = ∞ k=0 k! liml→∞ f (zl ) = 0, i.e. f (0) (z0 ) = 0. Now assume that we have already proved that f (0) (z0 ) = · · · = f (n−1) (z0 ) = 0. We want to show that then f (n) (z0 ) = 0 too, and by induction this would imply that z0 is a zero of order ∞. In a neighbourhood of z0 we have ∞ X f (n) (z0 ) f (k) (z0 ) n f (z) = (z − z0 ) + (z − z0 )k , n! k! k=n+1

and for z = zl (if necessary we have to assume that l is sufficiently large) it follows that ∞ X f (k) (z0 ) f (n) (z0 ) n (zl − z0 ) + (zl − z0 )k , 0 = f (zl ) = n! k! k=n+1

or after dividing by (zl − z0 )n 6= 0 ∞ X f (n) (z0 ) f (k) (z0 ) 0= + (zl − z0 )k−n n! k! k=n+1

=

∞ X f (n) (z0 ) f (k) (z0 ) + (zl − z0 ) (zl − z0 )k−n−1 . n! k! k=n+1

P f (k) (z0 ) (zl − z0 )k−n−1 is continuous in a neighbourhood of Since z 7→ ∞ k=n+1 k! z0 , in fact it is holomorphic, it follows that f

(n)

∞ X f (k) (z0 ) (zl − z0 )k−n−1 = 0, (z0 ) = −(n!) lim (zl − z0 ) l→∞ k! k=n+1

proving that z0 is a zero of order ∞ of f . Next suppose that z0 ∈ G is a given zero of order ∞. We want to prove that f (z) ≡ 0. For this we consider the set  M := z ∈ G f (k) (z) = 0 for all k ∈ N0 . 412

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POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

We will show that M is both closed and open. Since z0 ∈ M it is not an empty set and therefore the connectivity of G implies M = G, hence f (z) ≡ 0. First we prove that M is closed. Let (zl )l∈N , zl ∈ M, be a sequence in M converging to z 0 ∈ G. Since f is holomorphic it follows for all k ∈ N0 that 0 = lim f (k) (zl ) = f (k) (z 0 ), l→∞

i.e. zl ∈ M, implying that M is closed. Now we show that M is open. Let z1 ∈ M. In this case it follows that f (z) =

∞ X f (k) (z0 ) k=0

k!

(z − z1 )k = 0

for all z in a neighbourhood U of z1 . Thus all derivatives of f vanish in U, i.e. U ⊂ M, and therefore M is open which proves the theorem. Corollary 22.11. The number 0 does not belong to the range of the exponential function, i.e. ez 6= 0 for all z ∈ C. Proof. Since (ez )0 = ez , if ez0 = 0 then z0 would be a zero of order ∞. An application of Theorem 22.10 to f − g gives the following uniqueness theorem for holomorphic functions. Theorem 22.12. Let f, g : G → C be holomorphic functions on the region G ⊂ C. The following statements are equivalent: i) f (z) = g(z) for all z ∈ G, i.e. f = g; ii) there exists z0 ∈ G such that f (k) (z0 ) = g (k) (z0 ) for all k ∈ N0 ; iii) there exists a non-discrete set A ⊂ G such that f (z) = g(z) for all z ∈ A. P∞ P z2 z1 k (z) = ∞ Corollary 22.13. Let T(a k=0 bk (z− k=0 ak (z−z1 ) and T(bk ) (z) = k) k z2 ) be two power series converging in Br1 (z1 ) and Br2 (z2 ), respectively, r1 , r2 > 0. Suppose that Br1 (z1 ) ∩ Br2 (z2 ) 6= ∅. Then there exists a holomorz1 phic function f : G → C, Br1 (z1 ) ∪ Br2 (z2 ) ⊂ G, such that f |Br1 (z1 ) = T(a k) z2 and f |Br2 (z2 ) = T(bk ) . 413

A COURSE IN ANALYSIS

Of course it is possible to extend the procedure suggested in Corollary 22.13 to several power series. We may choose a sequence (T zν )ν∈A , A ∈ {Nn | n ∈ N} ∪ {N}, Nn = {1, . . . , n}, of power series with positive radii rν > 0 of convergence such that Brν (zν ) ∩ Brν+1 (zν+1 ) 6= ∅, Brν (zν ) ∩ Brµ (zµ ) = ∅ for |ν − µ| ≥ 2, and T zν |Brν (zν )∩Brν+1 (zν+1 ) = T zν+1 |Brν (zν )∩Brν+1 (zν+1 ) , see Figure 22.1 and Figure 22.2.

Allowed configuration

Figure 22.1

Forbidden configuration 414

Figure 22.2

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POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

S In this case there exists a holomorphic function g : ν∈A Brν (zν ) → C such that g|Brν (zν ) = T zν . Note S that the condition Brν (zν ) ∩ Brµ (zµ ) = ∅ for |ν − µ| ≥ 2 assures that ν∈A Brν (zν ) is simply connected. However some care is still needed when trying to identify g with a known function, a problem we in fact encounter already with a single power series: the power series T (z) := log(−1 + i) +

∞ X (−1)k−1

k

k=1

√

1 (z + 1 − i)k (−1 + i)k

converges for |z + 1 − i| < 2 but (C\{Re z ≤ 0}) ∩ B√2 (−1 + i) is not a connected set, see Figure 22.3. Im z

G1

b

i

−1 + i −1

Re z

0

G2 Figure 22.3   For G1 := z ∈ B√2 (−1 + i)|Imz > 0 and G2 = z ∈ B√2 (−1 + i)|Imz < 0 the functions T |G1 and T |G2 correspond to different branches of log. In Chapter 17 we have seen that if a C ∞ -function g : (−r + a, a + r) → R, r > 0, has a representation as convergent power series about a, i.e. g(x) =

∞ X f (k) (a) k=0

k!

(x − a)k , 415

|x − a| < r,

A COURSE IN ANALYSIS

then we can extend g to a holomorphic function f defined by f (z) :=

∞ X f (k) (a) k=0

k!

(z − a)k ,

|z − a| < r.

Since (−r + a, a + r) ∩ Br (a) ⊂ C is a non-discrete set in Br (a), we conclude that f is uniquely determined by g. This observation allows us to transfer certain results holding for g to f . For example the exponential function on R satisfies exp(s + t) = exp(s) exp(t), s, t ∈ R. (22.9)

For t ∈ R fixed we consider s 7→ exp(s + t) and s 7→ exp(s) exp(t), both defined on R. From (22.9) and the uniqueness theorem we deduce that exp(z + t) = exp(z) exp(t)

(22.10)

for all z ∈ C. Now we consider z ∈ C to be fixed and conclude by the same argument that exp(z + w) = exp(z) exp(w) for all z, w ∈ C. In a similar way we can prove that cos or sin are 2π-periodic. But note that the holomorphic extension f of g : (−r + a, a + r) → R might have additional properties. For example we know that the complex exponential function is also periodic with period 2πi. It is obvious that a Taylor series expansion of a holomorphic function f : G → C about a point z0 ∈ G of its domain is a powerful tool to study its local behaviour. However it is in general not easy (k) to determine all coefficients f k!(z0 ) of the expansion. There are some tools at our disposal which might be helpful in concrete situations. We first give Theorem 22.14. Let G ⊂ C be a region and γ : [a, b] → G be a piecewise continuously differentiable curve. Further let h : tr(γ) × G → C be a continuous function such that for every t ∈ [a, b] the function z 7→ h(γ(t), z) is holomorphic in G. Then the function f (z) :=

Z

h(ζ, z) dξ =

Z

a

γ

is holomorphic in G. 416

b

h(γ(t), z)γ(t) ˙ dt

(22.11)

22

POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

Proof. Let 4 ⊂ G be a closed triangle and note that  Z Z Z h(ζ, z) dζ dz f (z) dz = γ ∂4 ∂4  Z Z = h(ζ, z) dz dζ = 0, γ

∂4

where we have used our standard results for interchanging the order of iterated integrals of continuous functions over the product of compact R sets, and the fact that z 7→ g(ζ, z) is for ζ ∈ tr(γ) fixed holomorphic, i.e. ∂∆ g(ζ, z) dz = 0 for every closed triangle 4 ⊂ G. From Theorem 22.1 we deduce now that f is holomorphic in G. The reader should note the difference between this and Theorem 21.11.C where we have assumed that fζ (ζ, z) is continuous which we do not need anymore. Returning to our problem, we now find that if Z f (z) = h(ζ, z) dζ

(22.12)

γ

or

dk g(z) , (22.13) dz k and we know the Taylor expansion about z0 ∈ G of either z 7→ h(ζ, z) for every ζ ∈ tr(γ) or the one of z 7→ g(z), then we may use (22.12) or (22.13) to find the Taylor expansion of f about z0 . f (z) =

Example 22.15. For example we know that ∞   X a z k , |z| < |a|. = a a−z k=0
d a a Since dz = − (a−z) 2 we find a−z   ∞ X 1 d  z k 1 d a = − = − (a − z)2 a dz a − z dz a k=0

implying

∞ X k k−1 1 = − z . 2 k (a − z) a k=1

417

A COURSE IN ANALYSIS

We will discuss similar problems in Problem 8. We want to study how we can use the results of this and previous chapters to find solutions of certain differential equations. More precisely, for some linear differential equations we try to obtain a solution as a convergent power series. In Problem 7 to Chapter I.29 we encountered for l ∈ N0 the Bessel function of order l ∞ ∞ X X (−1)k  x 2k+1 (−1)k l Jl (x) = x2k (22.14) =x 2k+l k!(k + 1)! k!(k + 1)! 2 2 k=0 k=0 as power series converging for all x ∈ R and solving the Bessel differential equation x2 u00 (x) + xu0 (x) + (x2 − l2 )u(x) = 0, x ∈ R. (22.15) From the results developed in this chapter we deduce that Jl has a holomorphic continuation to an entire function Jl with now complex power series representation Jl (z) = z l

∞ X k=0

(−1)k z 2k , z ∈ C, 22k+l k!(k + 1)!

(22.16)

and this entire function solves in C the equation z 2 u00 (z) + zu0 (z) + (z 2 − l)u(z) = 0.

(22.17)

In the above mentioned problem of Chapter I.29 the series was just given and the task was to check the convergence as well as the fact that Jl solves (22.15). By using the hypergeometric differential equation z(1 − z)u00 (x) + (γ − (α + β + 1)z)u0 (z) − αβu(z) = 0,

(22.18)

α, β, γ ∈ R, as an example we want to study now a method, namely a (power) series Ansatz, which allows us to find solutions for differential equations such as (22.17) or (22.18). We try to obtain a solution to (22.18) as a series, more precisely we start with ∞ X u(z) = z ρ ck z k (22.19) k=0

418

22

POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

where ρ and the coefficients ck , k ∈ N0 , are to be determined. Since (22.18) is a linear homogeneous equation, if u solves this equation, every scalar multiple of u will do the same. We use this freedom to normalise in (22.19) the coefficient c0 by setting it equal to 1, i.e. c0 := 1. Now we substitute the series (22.19) into the equation (22.18) and try to determine ρ and ck , k ∈ N. Our approach is to use first formal calculations and once ρ and ck , k ∈ N, are determined we prove that the series converges and verify that it is indeed a solution of (22.18). Differentiating in (22.19) yields 0

u (z) =

∞ X

ck (k + ρ)z k+ρ−1 ,

k=0

u00 (z) =

∞ X k=0

leading to

ck (k + ρ)(k + ρ − 1)z k+ρ−2 ,

0 = z(1 − z)u00 (z) + (γ − (α + β + 1)z)u0 (z) − αβu(z)

= zu00 (z) + γu0 (z) − z 2 u00 (z) − (α + β + 1)zu0 (x) − αβu(z) ∞ ∞ X X k+ρ−1 ck (k + ρ)(k + ρ − 1)z + ck γ(k + ρ)z k+ρ−1 = k=0 ∞ X

−

k=0

k=0

ck (k + ρ)(k + ρ − 1)z k+ρ −

∞ X k=0

ck (α + β + 1)(k + ρ)z k+ρ −

∞ X

ck αβz k+ρ .

k=0

We are longing for a solution which has no singularity at z = 0 and therefore we require ρ(γ + ρ − 1) = 0 (22.20) and

(k + ρ)(ρ + k + γ − 1)ck = (ρ + k + α − 1)(ρ + k + β − 1), k ∈ N. (22.21) For γ ∈ / −N0 we choose ρ = 0 which gives ck =

(k + α − 1)(k + β − 1) ck−1 k(k + γ − 1)

and with c0 = 1 we arrive at ck =

α(α + 1) · . . . · (α + k − 1)β(β + 1) · . . . · (β + k − 1) , k!γ(γ + 1) · . . . · (γ + k − 1) 419

(22.22)

A COURSE IN ANALYSIS

or with the Pochhammer symbol notation, see (16.29), we get ck =

(α)k (β)k , k ∈ N0 , k!(γ)k

which yields u(z) =

∞ X (α)k (β)k k=0

k!(γ)k

(22.23)

zk .

(22.24)

However, this series we know already as the hypergeometric series associated with α, β, γ, compare with Definition 16.20. Theorem 22.16. The hypergeometric series 2 F1 (α, β; γ; z)

:=

∞ X (α)k (β)k k=0

k!(γ)k

zk

(22.25)

has radius of convergence 1 and satisfies in B1 (0) the differential equation (22.18). Proof. The convergence of 2 F1 (α, β; γ; z) was already investigated in Theorem 16.19. We will verify that 2 F1 (α, β; γ; z) indeed solves (22.18) in Part 9. In the case where γ ∈ / N\{1} we can also choose ρ = 1 − γ to find c˜k =

(α − γ + 1)k (β − γ + 1)k k!(2 − γ)k

(22.26)

and it follows that u˜(z) = z 1−γ 2 F1 (α − γ + 1, β − γ + 1; 2 − γ; z)

(22.27)

is a further solution to (22.18) if γ ∈ / N\{1}, again see Part 9. This method to solve the hypergeometric differential equation works for a much larger class of linear differential equations, for example for equations with polynomials as coefficients and we will return to these questions in Volume IV. Here it is important to learn that solutions of certain ordinary differential equations in the plane are holomorphic functions which we can 420

22

POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

find by a series Ansatz, u(z) = z ρ

P∞

k=0 ck z

k

, c0 = 1.

There is an obvious way to generalise the hypergeometric series, and more importantly, this generalisation is quite useful. For real numbers αj , 1 ≤ j ≤ p, and γl , 1 ≤ l ≤ q, we define p Fq (α1 , . . . , αp ; γ1 , . . . , γq ; z) :=

∞ X (α1 )k · . . . · (αp )k k z k!(γ1 )k · . . . · (γq )k

(22.28)

k=0

and call p Fq for (p, q) 6= (2, 1) the generalised hypergeometric series whereas 2 F1 is often called the Gauss hypergeometric series and sometimes just denoted by F . Theorem 22.17. Let p, q ∈ N0 , α1 , . . . , αp ∈ R and γ1 , . . . , γq ∈ R\{−n | n ∈ N0 }. i) For p > q + 1 the series p Fq converges only at z = 0. ii) For p = q + 1 the series p Fq converges in B1 (0), i.e. 1 is its radius of convergence. iii) For p ≤ q the radius of convergence of p Fq is ∞. Proof. With ak :=

(α1 )k ·...·(αp )k k!(γ1 )k ·...·(γq )k

we find

ak+1 z k+1 (α1 + k) · . . . · (αp + k) ak z k = (γ1 + k) · . . . · (γq + k)(k + 1) |z| k p |(1 + σ(k))| |z|, = q+1 k |(1 + τ (k))|

(22.29)

where σ(x) is a polynomial of degree at most p − 1 and τ (x) is a polynomial of degree at most q. Hence, if p > q + 1, i.e. p − 1 > q, then the quotient (22.29) becomes unbounded, hence p Fq (α1 , . . . , αp ; γ1, . . . , γq ; z) diverges for ak+1 z k+1 p > q + 1 and z 6= 0. If p = q + 1 it follows that limk→∞ ak z k = |z| and therefore for |z| < 1 the series p Fp+1 (α1 , . . . , αp , γ1 , . . . , γp+1; z) converges in ak+1 z k+1 B1 (0). Finally, if p ≤ q then for every z ∈ C we have limk→∞ ak zk = 0 implying that p Fq (z) represents for p ≤ q an entire function. 421

A COURSE IN ANALYSIS

Here are some examples 0 F0 (z) =

∞ X zk k=0

1 F0 (α, z)

=

k!

= ez ;

∞ X (α)k

k!

k=0

zk =

(22.30) 1 . (1 − z)α

(22.31)

Using the solution to Problem 11, i.e. the identities 1 1 1 1 = k 1
= k 3
and (2k)! (2k + 1)! 4 2 k k! 4 2 k k!

we find further



 1 1 2 ;− z ; 2 4   1 1 2 cosh z = 0 F1 ; z ; 2 4   3 1 2 ;− z ; sin z = z 0 F1 2 4   3 1 2 sinh z = z 0 F1 ; z . 2 4 cos z = 0 F1

Moreover, for l ∈ N0 we obtain Jl (z) =


z l 2 l!

0 F1



 1 2 l + 1; − z . 4

(22.32)

(22.33) (22.34) (22.35) (22.36)

(22.37)

As 2 F1 (z) satisfies a differential equation, namely the hypergeometric differential equation (22.18), it can be shown that p Fq is also a solution of a certain differential equation. However this will go far beyond an introduction to complex variables. For a good theoretical treatment we refer to [12], however it is worthwhile to become familiar with highly valuable sources such as [1] or [55].

Problems 1. Let G ⊂ C be a bounded domain and M ⊂ G a discrete set. Prove that M is a finite set. 422

22

POWER SERIES, HOLOMORPHY AND DIFFERENTIAL EQUATIONS

2. Let G ⊂ C be a bounded region and M ⊂ G a discrete set. Let f : G \ M → C be a holomorphic function which is bounded on G \ M. Prove that f has a holomorphic extension to G. Hint: use Theorem 22.7 and the uniqueness result for holomorphic functions. P∞ 2k+1 be a power series converging in Br (0), 3. Let f (z) = k=0 a2k+1 z r > 0. Consider on Br (0) \ {0} the function g(z) := f (z) . Prove that g z has a removable singularity at z = 0. Deduce that z 7→ sinz z is an entire function if appropriately extended at z0 = 0. 4. Let h : G → C be a holomorphic function which has at z0 ∈ G a zero of order k ≥ 1. Obviously it follows that h2 (z) = (h(z))2 has a zero at z0 too. Find the order of the zero of h2 . 5. Let p(z) be a polynomial of order k. Prove that the entire function z 7→ p(z)eaz , a ∈ C, cannot have a zero of order k + 1. 6. Use the fact that for all x, y ∈ R the equality cos(x − y) = cos x cos y + sin x sin y holds to derive that this equality also holds for all z, w ∈ C. 7. Let G ⊂ C be a region and γ : [0, 1] → G a piecewise continuously differentiable curve. Further let f : G → C be a continuous function. Suppose that for every z0 ∈ tr(γ) there exists rz0 > 0 such that in Brz0 (z0 ) ⊂ G the function f has a representation as a convergent Taylor P k series, i.e. f (z) = ∞ k=0 ak (z0 )(z − z0 ) , z ∈ Brz0 (z0 ). Prove that there exists a domain D ⊂ G such that f |D is holomorphic. Hint: note that tr(γ) is compact and use the uniqueness result for holomorphic functions. P∞ 2k 8. a) Use the series k=0 z , |z| < 1, to find the Taylor series of z+6z 2 h(z) = (1−z 2 )3 .
b) Use the equality x2 F1 21 , 1; 23 , −x2 = arctan x, |x| < 1, to find 1 a power series expansion of x 7→ 1+x 2. 9. Let u : (a, b) → R, a < b, be a twice continuously differentiable functions satisfying the differential equation u00 (x) + x2 u0 (x) + x4 u(x) = 0 in (a, b). Suppose that for some x0 ∈ (a, b) the Pfunction u hask the convergent Taylor series representation u(x) = ∞ k=0 ak (x − x0 ) , ak ∈ R, 423

A COURSE IN ANALYSIS

x ∈ (−η + x0 , x0 + η) ⊂ (a, b), η > 0. Show that there exists a holomorphic function h : Bη (x0 ) → C such that h satisfies h00 (z) + z 2 h0 (z) + z 4 h(z) = 0 in Bη (x0 ). 10.* For l ∈ N prove the integral representation of the Bessel function Jl (z) :  z l Z 1 1 1

Jl (z) = (1 − t2 )l− 2 cos zt dt. 1 1 2 Γ l+ 2 Γ 2 −1

Hint: use the integral representation of the beta-function and the power series expansion of Jl , as well as the doubling formula for the Γ-function, i.e. (I.31.35).

11. Prove the identities (22.32) and then derive (22.33) and (22.34).

424

23

Further Properties of Holomorphic Functions

In this chapter we want to discuss several properties of holomorphic functions, including some of their global mapping properties, as well as properties of their range, the question of how we can approximate holomorphic functions, and infinite products of holomorphic functions. We start with some mapping properties and we need the result below as preparation. Lemma 23.1. Let D ⊂ C be a domain, z0 ∈ D and Br (z0 ) ⊂ D, r > 0. Further let f : D → C be a holomorphic function. If |f (z0 )| < min |f (z)| |z−z0 |=r

(23.1)

then f has a zero in Br (z0 ). Proof. Suppose that f has no zero in Br (z0 ). Then the function z 7→ g(z) := 1 is holomorphic in some neighbourhood of Br (z0 ) since (23.1) implies f (z) f (z) 6= 0 for |z − z0 | = r. The Cauchy inequalities, Theorem 21.15, for k = 0 now imply |g(z0 )| = max |g(z)|, or

which yields

|z−z0 |=r

1 1 1 ≤ max = |f (z0 )| |z−z0|=r |f (z)| min|z−z0|=r |f (z)| |f (z0 )| ≥ min |f (z)| > 0, |z−z0 |=r

hence a contradiction. Our first main result is surprising when compared with the situation of realvalued C ∞ -functions. We have proved, for example in Problem 5 of Chapter 10, the existence of a C ∞ -function ϕ : R → R such that 0 ≤ ϕ(t) ≤ 1, ϕ|(−∞,0] = 0 and ϕ|[1,∞) = 1. In particular ϕ(R) = [0, 1] is a closed set. Non-constant holomorphic functions however will always map regions onto regions: Theorem 23.2. The image of a region G ⊂ C under a non-constant holomorphic function f : G → C is a region, i.e. f (G) is open and connected. 425

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Proof. Since f is continuous it follows that f (G) is connected. The strong statement of the theorem is that f (G) is open which we are going to prove now. Let w0 ∈ f (G), i.e. w0 = f (z0 ) for some z0 ∈ G. Since f is not constant we can find some r > 0 such that Br (z0 ) ⊂ G and f (z) 6= w0 for z ∈ Br (z0 )\{z0 }. Otherwise the uniqueness theorem, Theorem 22.12, would imply that f (z) = w0 for all z ∈ G. For z ∈ ∂Br (z0 ) we can now deduce that |f (z) − w0 | ≥ 3 > 0.

(23.2)

Our aim is to prove that B (w0 ) ⊂ f (G), i.e. for w˜ ∈ B (w0 ) there exists z˜ ∈ G such that f (˜ z ) = w, ˜ from which the openness of f (G) will follow. Now, for |w − w0 | <  and |z − z0 | = r, r as above, it follows that |f (z) − w| ≥ |f (z) − w0 | − |w − w0 | ≥ 3 −  = 2, i.e. min |f (z) − w0 | ≥ 2.

|z−z0 |=r

However for z = z0 we find |f (z0 ) − w| = |w0 − w| < , i.e. |f (z0 ) − w| = |w0 − w| < , hence |f (z0 ) − w| < min |f (z) − w|, |z−z0 |=r

implying that z 7→ f (z) − w has at least one zero in Br (z0 ), i.e. for every w, |w − w0 | < , there exists z ∈ Br (z0 ) such that w = f (z) which means that B (w0 ) ⊂ f (G) implying that f (G) is open. Corollary 23.3. Let f : G → C be a holomorphic function defined on a region G ⊂ C. If |f |, Re f or Im f is constant then f must be constant too. Proof. In each case f (G) cannot be open. In Chapter 17 we introduced biholomorphic functions and in Proposition 17.19 we characterised biholomorphic function f : G1 → G2 between two regions. Now we want to prove a holomorphic version of the inverse function theorem, Theorem II.10.12. 426

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Definition 23.4. Let D ⊂ C be a domain and f : D → C a holomorphic function. We call f locally biholomorphic at z0 ∈ D if there exists an open neighbourhood U = U(z0 ) ⊂ D such that f |U : U → f (U) is biholomorphic. We aim to prove that f 0 (z0 ) 6= 0 implies f to be locally biholomorphic at z0 . For this we need some auxillary results. P k Lemma 23.5. Let g(z)P = ∞ k=0 ck (z − z0 ) be a power series converging in ∞ k−1 Br (z0 ), r > 0. If |c1 | > k=2 k|ck |r , then g : Br (z0 ) → C is injective.

Proof. Let w1 , w2 ∈ Br (z0 ) such that g(w1) = g(w2 ), and define z1 := w1 −z0 , z2 := w2 − z0 . It follows that ∞ X k=1


ck z1k − z2k = 0.


Since z1k − z2k = (z1 − z2 ) z1k−1 + z1k−2 z2 + · · · + z1 z2k−2 + z2k−1 we find 0=

∞ X k=1

ck z1k − z2k

= (z1 − z2 )c1 +




∞ X k=2


ck (z1 − z2 ) z1k−1 + z1k−2 z2 + · · · + z1 z2k−2 + z2k−1 ,

or for z1 6= z2 −c1 =

∞ X k=2


ck z1k−1 + z1k−2 z2 + · · · + z1 z2k−2 + z2k−1 .

Recall that w1 , w2 ∈ Br (z0 ), i.e. |z1 |, |z2 | < r, and we conclude now |c1 | ≤

∞ X k=2

|ck |kr k−1,

which is a contradiction. Corollary 23.6. Let D ⊂ C be a domain and f : D → C a holomorphic function. If f 0 (z0 ) 6= 0 for some z0 ∈ D then there exists a neighbourhood U = U(z0 ) ⊂ D such that f |U is injective. 427

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P f (k) (z0 ) (z − z0 )k of f Proof. We consider the Taylor expansion f (z) = ∞ k=0 k! about z0 which represents f in some open ball Br (z0 ), r > 0. Since f 0 (z0 ) 6= 0 P |f (k) (z0 )| k−1 and since the function ρ 7→ ∞ ρ is continuous at zero with k=2 k k! value 0 for ρ = 0, it follows the existence of r1 < r such that |f 0 (z0 )| >

∞ X |f (k) (z0 )| k−1 k r , k! k=2

and Lemma 23.5 implies that f |Br (z0 ) is injective. Now we can prove Theorem 23.7. Let D ⊂ C be a domain and f : D → C a holomorphic function with f 0 (z0 ) 6= 0 for some z0 ∈ D. Then there exists an open neighbourhood U = U(z0 ) ⊂ D of z0 such that f |U : U → f (U) is biholomorphic, i.e. f is locally biholomorphic at z0 . Proof. We know that we can find an open neighbourhood U = U(z0 ) ⊂ D of z0 ∈ D which we can choose to be a region such that f |U is injective, and we know further that f (U) is a region too. An injective mapping f |U : U → f (U) is bijective. Shrinking U if necessary we can use Theorem II.3.38 to deduce that f −1 is continuous. Now Proposition 17.19 implies that f |U is biholomorphic from U to f (U). A further consequence of Theorem 23.2 is the following maximum principle. Theorem 23.8 (Boundary Maximum Principle). Let f : G → C be a holomorphic function defined on a region G ⊂ C. If |f | has a local maximum at z0 ∈ G then f is constant on G. If G is bounded and f is defined and continuous on G then |f | attains its maximum on ∂G, i.e. for all z ∈ G we have |f (z)| ≤ max |f (ζ)|. (23.3) ζ∈∂G

Proof. Let U ⊂ G be a neighbourhood of z0 such that |f (z0 )| ≥ |f (z)| for z ∈ U. The image of U under f is a subset of {w | |w| ≤ |f (z0 )|}, i.e. f (U) ⊂ {w | |w| ≤ f (z0 )}, but {w | |w| ≤ f (z0 )} is not a neighbourhood of f (z0 ). Thus by Theorem 23.2 f must be constant equal to f (z0 ) in a neighbourhood of z0 , and hence by the uniqueness theorem for holomorphic functions f must be constant in G. 428

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FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

The second part is seen as follows. The continuous function |f | : G → R attains its maximum on G since G is now compact. The first part implies that for a non-constant function this maximum cannot be attained in G, so it must be attained on ∂G. Corollary 23.9. Let G ⊂ C be a bounded region and f, g : G → C two continuous functions such that f, g : G → C are holomorphic and f |∂G = g|∂G . Then f = g in G. Proof. The function f − g : G → C is continuous and holomorphic on G. By the boundary maximum principle it follows that |f (z) − g(z)| ≤ max |f (ζ) − g(ζ)| = 0

(23.4)

ζ∈∂G

for all z ∈ G. We can also derive from Theorem 23.8 a minimum principle. Theorem 23.10 (Boundary Minimum Principle). Let G ⊂ C be a region and f : G → C be a holomorphic function. A. If |f | has a local minimum at z0 ∈ G then either f (z0 ) = 0 or f is constant. B. Now suppose that G is compact and f has a continuous extension to G, i.e. f : G → C is continuous and f : G → C is holomorphic. Then f has either a zero in G or |f | attains its minimum on ∂G, i.e. |f (z)| ≥ min |f (ζ)|

(23.5)

ζ∈∂G

for all z ∈ G. Proof. If f (z) 6= 0 in G we can consider the function g(z) = satisfies all assumptions of Theorem 23.8.

1 f (z)

which

As an application of the maximum principle we prove the Lemma of Schwarz which we will later on, see Chapter 28, find to be very useful. Theorem 23.11. For a holomorphic function f : B1 (0) → B1 (0) with f (0) = 0 the estimates |f (z)| ≤ |z| for all z ∈ B1 (0) (23.6)

and

hold.

|f 0 (0)| ≤ 1 429

(23.7)

A COURSE IN ANALYSIS

Proof. f (0) = 0 we have for f the series representation f (z) = P∞powerk−1 P∞ Since k c z is a further holomorphic c z and g defined by g(z) = k=1 k k=1 k function defined on B1 (0). Clearly f (z) = zg(z) for all z ∈ B1 (0) as well as g(0) = c1 = f 0 (0). Since f (B1 (0)) ⊂ B1 (0) it follows that |f (z)| < 1 for all z ∈ B1 (0), hence r max|z|=r |g(z)| ≤ 1 for every r ∈ (0, 1). Now the maximum principle gives |g(z)| ≤ 1r for all z ∈ Br (0), 0 < r < 1. For r → 1 this implies |g(z)| ≤ 1 for all z ∈ Br (0) and therefore we find |f (z)| ≤ |z| for all z ∈ B1 (0) as well as |f 0(0)| = |g(0)| ≤ 1. For later purposes we add Corollary 23.12. In the situation of Theorem 23.11 assume the existence of a point a ∈ B1 (0)\{0} such that |f (a)| = |a| or that |f 0(0)| = 1. Then we can find γ ∈ S 1 = ∂B1 (0) such that f (z) = γz for all z ∈ B1 (0), i.e. f is a rotation of B1 (0), in fact of C. Proof. With the notations of the proof of Theorem 23.11 we find for a ∈ B1 (0)\{0} with |f (a)| = |a| or in the case that |f 0 (0)| = 1 that either |g(a)| = 1 or |g(0)| = 1. This however means that f attains its supremum in B1 (0) and hence it must be a constant. Next we want to use our knowledge of the growth of |f | for a holomorphic function f to study, in some cases, the range of f . Our first goal is to prove the fundamental theorem P of algebra which states in its original version that every polynomial p(z) = nk=0 ak z k of order n, i.e. an 6= 0, with complex coefficients ak ∈ C has n zeroes z1 , . . . , zn counted according to their multiplicity. In other words, given w ∈ C we can always solve the equation p(z) = w and the set of solutions has at most n distinct elements. Another way of wording the result is that the range of a polynomial function is C, and with arguments employed later on one easily derives the fundamental theorem of algebra in its classical form. We start by estimating the growth of a polynomial. P k Lemma 23.13. Let p(z) = ∞ k=0 ak z , ak ∈ C, an 6= 0, be a polynomial of degree n. For |z| ≥ 1 the estimate ! n X |p(z)| ≤ |ak | |z|n (23.8) k=0

430

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FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

holds. Furthermore, for every  ∈ (0, 1) there exists ρ() such that |z| ≥ ρ() implies (1 − )|an ||z|n ≤ |p(z)| ≤ (1 + )|an ||z|n . (23.9) Proof. For |z| ≥ 1 we find immediately that |p(z)| ≤ i.e. (23.8). With q(z) = |z| ≥ 1 we get

n X k=0

k

|ak ||z| ≤

Pn−1 k=0

|q(z)| ≤ For  ∈ (0, 1) we choose

n X k=0

!

|ak | |z|n ,

ak z k we have p(z) = an z n + q(z) and for n−1 X k=0

!

|ak | |z|n−1 .

) n−1 1 X |ak | . ρ() := max 1, |an | k=0 (

Now |z| ≥ ρ() implies that n−1

|q(z)| ≤

1 X |ak ||z|n |z| k=0

!

(23.10)

≤ |an ||z|n ,

which yields (1 − )|an ||z|n ≤ |an ||z|n − |q(z)| ≤ |p(z)| ≤ |an ||z|n + |q(z)| ≤ (1 + )|an ||z|n proving (23.9). P Corollary 23.14. nThe zeroes of the polynomial p(z) =oo nk=0 ak z k , an 6= 0, n Pn−1 must lie in the set z ∈ C | |z| ≤ max 1, |a1n | k=0 |ak | . Proof. From (23.9) we deduce that the zeroes of p(z) must belong for every  ∈ (0, 1) to Bρ() (0), ρ() as in (23.10), and passing to the limit as  tends to 1 yields the result.

Theorem 23.15 (Fundamental Theorem of Algebra). Every nonconstant polynomial with complex coefficients has a zero in C. 431

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P 6
0, of degree n ≥ 1 we Proof. For the polynomial p(z) = nk=0 ak zk , an = choose  = 12 in Lemma 23.13 and R ≥ max 1, ρ 12 such that |p(0)| ≤

|an | n R . 2

Now (23.9) implies that |p(0)| ≤ min|z|=R |p(z)| and the boundary minimum principle, Theorem 23.10, yields that p must have a zero in BR (0). P Corollary 23.16. A polynomial p(z) = nk=0 ak z k , an 6= 0, factorises into linear factors, i.e. there exists complex numbers z1 , . . . , zn ∈ C (not necessarily all different) such that with some c ∈ C p(z) = c

n Y (z − zj )

(23.11)

j=1

holds. Proof. According to the fundamental theorem there exists z1 ∈ C such that p(z1 ) = 0, hence p(z) = (z − z1 )q(z) where q(z) is a polynomial of degree n − 1. Now we can iterate the argument. The next result is a further one which relates growth properties of an entire function to its structure. Theorem 23.17. Let f : C → C be an entire function and suppose that for some n ∈ N there exists R > 0 and M > 0 such that |f (z)| ≤ M|z|n ,

|z| ≥ R.

Then f is a polynomial of degree less than or equal to n. Proof. We consider the Taylor expansion of f about z0 = 0, f (z) =

∞ X f (k) (0) k=0

k!

zk .

Using the Cauchy inequalities from Theorem 21.15 we find for r ≥ R that |f (k) (0)| 1 ≤ k max |f (z)| ≤ r −k Mr n = Mr n−k . k! r |z|=r For r → ∞ it follows that if k > n then f (k) (0) = 0, i.e. f (z) = 432

Pn

k=0

f (k) (0) k z . k!

23

FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

Corollary 23.18 (Theorem of Liouville). A bounded entire function is constant. Proof. Apply Theorem 23.17 for n = 0. We now turn to convergence and approximation properties of holomorphic functions. For continuous functions we know that uniform convergence is needed to assure that in general the limit function of a sequence of continuous functions is continuous too, however see also the remark following Theorem 23.20. For real-valued differentiable functions we need the pointwise convergence of the sequence and the uniform convergence of the sequence of derivatives, compare with Theorem I.24.6 and Theorem I.26.19. In order to assure that limit functions also have a higher order degree of differentiability appropriate conditions of the approximating sequence and the mode of convergence of the functions and their derivatives are needed. While this suggests strong conditions to assure that the limit function of a sequence of holomorphic functions is holomorphic too, we need in fact a less strong mode of convergence. Definition 23.19. Let D ⊂ C be a domain and fk : D → C, k ∈ N, a sequence of functions. We say that (fk )k∈N converges locally uniform to f : D
→ C if every point z0 ∈ D has a neighbourhood U(z0 ) such that fk |U (z0 ) k∈N converges uniformly to f |U (z0 ) .

Theorem 23.20. Let G ⊂ C be a region and fk : G → C, k ∈ N, be a sequence of holomorphic functions converging locally uniform to a function f : G → C. Then f is holomorphic and for every n ∈ N the sequence (n) fk converges locally uniformly to f (n) . k∈N

Proof. Clearly f is continuous since for every point z0 ∈ G there exists a ball B (z0 ) ⊂ G,  > 0, such that (fk )k∈N converges on B (z0 ) uniformly to f . Let 4 ⊂ G be a triangle and ∂4 anti-clockwise oriented. The uniform convergence of (fk |∂4 )k∈N to f |∂4 follows from the compactness of ∂4 ⊂ G and it implies Z Z fk (z) dz = 0. f (z) dz = lim ∂4

k→∞

∂4

Now the theorem of Morera, compare with Theorem 22.1, yields the holomorphy of f . In order to prove the second statement it is sufficient to show that (fk0 )k∈N converges locally uniform to f 0 , the general case follows then by 433

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iterating the argument. Let r > 0 and Br (z0 ) ⊂ G. Applying the Cauchy inequalities to the holomorphic function f − fk we find for all z ∈ B 2r (z0 ) that 1 max |f (ζ) − fk (ζ)| r |ζ−z0|=r c ≤ kf − fk k∞,Br (z0 ) , r

|f 0(z) − fk0 (z)| ≤ c

implying the uniform convergence of (fk0 )k∈N on B 2r (z0 ) to f 0 and the result follows. Remark 23.21. Note that the proof of Theorem 23.20 yields that already the locally uniform limit of continuous functions is continuous, and this holds at least for every metric space once we modify Definition 23.19 in the obvious way. We can even apply such a kind of argument to derivatives, we still need convergence of the derivatives which we do not need in the case of Theorem 23.20, i.e. holomorphic functions. The Weierstrass approximation theorem, Theorem II.14.3, assures that the restriction of polynomials on the real line are dense in (C([0, 1]; R), k.k∞ ). Extensions such as Stone’s approximation theorem, Theorem II.14.9, give further results on the approximation of continuous functions defined on a compact metric space. We are longing for some approximation results for holomorphic functions on compact sets. For this we first need Definition 23.22. Let K ⊂ C be a compact set. We call f : K → C holomorphic on K if there exists an open neighbourhood U of K, K ⊂ U, and a holomorphic function g : U → C such that f = g|K . In the following, when dealing with holomorphic functions f on K, if no confusion may arise, we denote the function g in Definition 23.22 with f too, i.e. we assume that f is already given as holomorphic function f : U → C, K ⊂ U. The first idea to approximate a holomorphic function on K, K compact, by polynomials turns out to be not always successful. Consider the function h, h(z) = 1z , defined and holomorphic on C\{0}. In particular h is holomorphic on the compact set S 1 . For theR unit circle S 1 we have when using 1 1 dz = 1. Now suppose that our standard parametrization that 2πi |z|=1 z 434

23

FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

there exists a sequence of polynomials converging locally uniformly in a neighbourhood of S 1 to h. It follows that (pk )k∈N converges uniformly on S 1 to R 1 h, but 2πi p (z) dz = 0. Since by uniform convergence of (pk )k∈N on S 1 |z|=1 k to h we get lim

k→∞

Z

pk (z) dz =

Z

h(z) dz

|z|=1

|z|=1

we have a contradiction. Thus, in general, to try to approximate holomorphic functions on a compact set by polynomials seems not to be a valid option. Instead we try to use rational functions and we will prove Runge’s approximation theorem. As preparation we prove

Theorem 23.23. Let K ⊂ C be a compact set and U ⊂ C an open neighbourhood of K, K ⊂ U. Then there exists finitely many axis-parallel line segments γ1 , . . . , γN such that γ := γ1 ⊕ · · · ⊕ γN is a simply closed, positive oriented, piecewise continuously differentiable curve with tr(γ) ⊂ U\K and 1 f (z) = 2πi

Z

γ

f (ζ) dζ ζ −z

(23.12)

holds for every z ∈ K and every holomorphic function f : U → C. Remark 23.24. It is important to observe that γ, i.e. γj , j = 1, . . . , N, depends on K and U but it is independent of f .

Proof of Theorem 23.23. Since K is compact and U is open it follows that δ := dist(K, ∂U) > 0. We cover √ U with axis-parallel closed square of side length h where we assume that 2h < δ, see Figure 23.1. 435

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U

K

Qk

γ

Figure 23.1 Since K is compact we can find a finite number Q1 , . . . , QM of these squares with the property that M [ K⊂ Qk ⊂ U. (23.13) k=1

The choice of h in relation to δ assures that for z0 ∈ K ∩ Qk we have √ Bδ (z0 ) ⊂ U and since diam Qk = 2h < δ it follows that Qk ⊂ U, hence (23.13) holds. We consider further ∂Qk as positive oriented simply closed parametric curves and we denote by γ1 , . . . , γN those segments which are part of the boundary of some Qk but never of two adjacent squares Qk and Ql . The enumeration of γ1 , . . . , γN is chosen such that γ := γ1 ⊕ · · · ⊕ γN is a simply closed positive oriented paramteric curve. By construction we also find that tr(γ) ⊂ K { ∩ U. Indeed, if γl ∩ K 6= ∅ then γl would be part of the boundary of two adjacent squares, see again Figure 23.1. Now let f : U → C be a holomorphic function and z ∈ K, hence z ∈ Qk0 for some k0 . Suppose ˚k0 . In this case we have that z ∈ Q Z 1 f (ζ) f (z) = dζ. 2πi ∂Qk0 ζ − z 436

23

FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

Since for k 6= k0 we have now

1 2πi

R

f (ζ) ∂Qk ζ−z

dζ = 0 it follows that

Z Z M X 1 f (ζ) f (ζ) 1 dζ = dz, f (z) = 2πi ∂Qk0 ζ − z 2πi γ ζ − z k=1 where for the last equality we used that if the trace of a line segment γl belongs to the boundary of two adjacent squares we integrate the same function over γl and γl−1 and hence the sum of these two integrals vanishes. This proves (23.12) for z ∈ Qk0 . If z ∈ ∂Qk0 it is in the intersection of two squares Qk0 / tr(γ). Hence by approximating z by a sequence (zj )j∈N , and Ql0 , hence z ∈ ˚ zj ∈ Qk0 ∩ K, the result will follow. With the help of Theorem 23.23 we can now prove the approximation theorem of Runge. Theorem 23.25 (Runge). Let K ⊂ C be a compact set and f : K → C a holomorphic function. For every  > 0 we can find a rational function (z) , q (w) 6= 0 for w ∈ K, such that |f (z) − r (z)| <  for all z ∈ K, r (z) = pq(z) i.e. kf − r k∞,K < . Proof. Let f : K → C be holomorphic. Choose U ⊂ C open such that K ⊂ U and f = g|K for a holomorphic function g : U → C, which we now denote by f too. Now we determine the curve γ according to Theorem 23.23 (ζ) and we note that on the compact set tr(γ) × K the function (ζ, z) 7→ fζ−z is continuous. As a continuous function on a compact set this function is uniformly continuous. Hence for  > 0 there exists δ > 0 such that for z ∈ K and z1 , z2 ∈ tr(γ) with |z1 − z2 | < δ it follows that f (z1 ) 2π f (z ) 2 z1 − z − z2 − z < lγ

where as usual lγ denotes the length of γ. By construction γ is rectifiable, in fact it is an axis-parallel polygonal curve. Therefore we can divide γ into segments κl , lκl < δ, l = 1, . . . , M, and then choose points ζl ∈ tr(κl ) such that for all z ∈ K Z Z 1  f (ζ) f (ζl ) 1 dζ − dζ < lκl . 2πi 2πi κl ζl − z lγ κl ζ − z 437

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The function M

1 X f (ζl ) r (z) := 2πi l=1 ζl − z

Z

M

1 X f (ζl ) lκ 1 dζ = 2πi l=1 ζl − z l κl

is a rational function which is not holomorphic at the points ζl ∈ κl ⊂ K { , l = 1, . . . , M, only. Note that of course the points ζl depend indirectly on . Now we observe that M Z M X 1 Z f (ζ) X 1 f (ζ ) l |f (z) − r (z)| = dζ − dζ 2πi ζ − z 2πi ζ − z κl κl l l=1 l=1 M X  lκ = , < lγ l l=1

and the theorem is proven. Remark 23.26. In proving Runge’s theorem we found the presentations in [30] and [83] quite helpful. We now turn to infinite products of holomorphic functions and infinite product of holomorphic functions. For a polynomial p(z) = Pn representations k a z , a = 6 0, we have a finite product representation given by (23.11). k n k=0 In Chapter I.30 and I.31 we discussed infinite products of real numbers as well as of real-valued functions. For example in Theorem I.31.14 we have proved for x ∈ R the following representation of the sine function  ∞  Y x2 sin πx = πx 1 − 2 , x ∈ R. (23.14) k k=1 A natural question is whether (23.14) extends to z ∈ C. We have also obtained a product representation of the Γ-function, namely the Weierstrass representation, Theorem I.31.2, x ∞ e−γx Y e k Γ(x) = , x > 0, x k=1 1 + xk

(23.15)

 P N 1 − ln N is the Euler constant, also see Theorem where γ = limN →∞ k=1 k I.18.24. Now we may ask whether we can extend the product on the right 438

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FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

hand side to some domain in C which could allow us to extend the Γ-function to subsets of C larger than R+ \{0}. We start by looking at the definition of infinite products of sequences of complex numbers. However it is more convenient for our purposes to start with (ak )k∈N , ak ∈ C, and look at ∞ Y

(1 + ak ).

(23.16)

k=1

In Problem 11 we will handle of Chapter I.30.

Q∞

k=1 ck ,

ck ∈ C, in light of some of the results

Definition 23.27. For a sequence (ak )k∈N of complex numbers ak ∈ C the infinite product (23.16) converges if lim

N →∞

N Y

(1 + ak )

(23.17)

k=1

exists. In this case the limit is denoted by (23.16). Analogously to Proposition I.30.10 we have Proposition 23.28. Let (ak )k∈N be a sequence of complex Q numbers for which P∞ ∞ k=1 (1 + ak ) conk=1 ak converges absolutely. Then the infinite product verges. It diverges to 0 if and only if one of its factors is 0. Q Remark 23.29. Definition I.30.2 it was defined that if limN →∞ ∞ k=0 ck = QIn c divergent to 0. In order to be consistent we use in 0 then we call ∞ k=0 k Proposition 23.28 now the wording “diverges to 0” instead of “converges to 0” as for example in [83]. P∞ Proof of Proposition 23.28. Since k=1 |ak | converges there exists M ∈ N such that |ak | < 21 for k ≥ M. Since finitely many terms do not affect the convergence we assume |ak | < 21 for all k ∈ N. Now we can use the branch of ˜ + z) = P∞ (−1)k−1 , the logarithmic function given as in Chapter 18 by ln(1 k=1 k |z| < 1, and we obtain in particular ˜

1 + z = eln(1+z) , 439

|z| < 1.

A COURSE IN ANALYSIS

This implies further that N Y

(1 + ak ) =

k=1

Since |z|
0 we note the estimate kz 2 ≤ Rk2 , z ∈ BR (0), and since the series  Q  P∞ R2 N z2 converges locally k=1 k 2 converges, we conclude that k=1 1 − k 2 N ∈N  Q∞  2 uniform on C. This implies that πz k=1 1 − kz 2 is an entire function which coincides by Theorem I.31.14 on R with sin x. Hence by the uniqueness theorem, Theorem 22.12, we deduce (23.18).

Theorem 23.34. Let D ⊂ C be a domain and the sequence fk : D → C, k ∈ N, of holomorphic functions is assumed to satisfy the assumptions of Theorem 23.31. Suppose further that for all k ∈ N the function fk has no zero, i.e. fk (z) 6= 0 for all z ∈ D. If we denote by f (z) the infinite product of hk (z) = fk (z) − 1, then we have ∞

∞

f 0 (z) X fk0 (z) X h0k (z) = = . f (z) fk (z) 1 + hk (z) k=1 k=1

(23.19)

Proof. Once we can assure the uniform convergence of the products and series entering in (23.19), the formula itself follows as inthe proof of Theorem QN I.30.19. Now, from Theorem 23.31 we deduce that conk=1 fk (z) N ∈N

verges locally uniformly to f (z) 23.20 it follows also the and by Theorem 0  QN locally uniform convergence of to f 0 (z). Let K ⊂ D be k=1 fk (z) N ∈N  Q 0  QN N converge a compact set, hence and k=1 fk (z) k=1 fk (z) N ∈N

441

N ∈N

A COURSE IN ANALYSIS

Q N



is uniformly bounk=1 fk (z) N ∈N Q N ded away from 0, i.e. for z ∈ K we have k=1 fk (z) ≥ η(K) > 0 for all N ∈ N. From this we can deduce that Q 0 N f (z) k k=1 f 0 (z) → uniformly on K, QN f (z) k=1 fk (z) uniformly on K and on K the sequence

hence for all z ∈ D. Now it remains to argue as in Theorem I.30.19 that Q N

k=1

QN

fk (z)

0

k=1 fk (z)

and the theorem follows.

N X f 0 (z) = fk (z) k=1

We will return to product representatives, especially to the Weierstrass representation of the Γ-function, once we have a better understanding of so called meromorphic functions.

Problems 1.

a) Give a counter example to the statement: if f 0 (z) 6= 0 for all z ∈ C, then the holomorphic function f maps C biholomorphically onto its range. b) Let g : BR (0) → C be a holomorphic function such that g(0) 6= 0. Prove that the function f : BR (0) → C, f (z) = g(z) sin z, is locally biholomorphic at 0.

2. Revisit the proof of Theorem 23.2 and prove the following refined statement: let BR (z0 ) ⊂ G and δ := 12 minz∈∂BR (z0 ) |f (z) − f (z0 )| > 0. Then it follows that BR (f (z)) ⊂ f (BR (z0 )). 3. Give a counter example to the boundary maximum principle in the case where G is unbounded. Hint: you may consider on G = {z ∈ C|Re z > 0, Im z > 0} the 4 function z 7→ f (z) := eiz . 442

23

FURTHER PROPERTIES OF HOLOMORPHIC FUNCTIONS

4. Let (fk )k∈N , fk : G → C, be a sequence of holomorphic functions on the bounded region G with continuous extension to G. Suppose that there exists a continuous function f : G → C such that fk (z) → f (z) for all z ∈ G as k tends to infinity and limk→∞ ||fk − f ||∞,∂G = 0. Prove that limk→∞ ||fk − f ||∞,G = 0 Is f |G a holomorphic function? 5. Let D ⊂ C be a domain and BR (0) ⊂ D. On the set F := {f |BR (0) |f : D → C is holomorphic} define kf kF := maxz∈∂BR (0) |f (z)|. Prove that (F , k · kF ) is a Banach space. 6. Let f : BR (0) → C be a holomorphic function and M : [0, R) → R be defined by M(r) := kf k∞,∂Br (0) . Prove that M is monotone increasing. 7. (This problem is taken from [67]). Let G ⊂ C be a relatively compact region and fk : G → C, k ∈ N, be a sequence of holomorphic functions converging locally uniform to f : G → C. Then for every z0 ∈ G we can find N(z0 ) ∈ N and a sequence (zn )n ≥ N(z0 ) such that limn→∞ zn = z0 and fn (zn ) = f (z0 ) for all n ≥ N(z0 ). Hint: we may assume that f (z0 ) = 0. Now deduce that f has no further zero in a neighbourhood of z0 = 0 and use the boundary minimum principle. 8. Prove that the range of a non-constant polynomial is closed and now use Theorem 23.2 to deduce the fundamental theorem of algebra. 9. Does there exist a biholomorphic function f : C → B1 (0)? 10. Let f : C → C be a continuous function and a, b ∈ C be linearly independent over R, i.e. a and b span a parallelogram in the plane. Suppose that for all z ∈ C we have f (a+z) = f (z) and f (b+z) = f (z). Prove that if f is holomorphic then f is constant. 11.

a) Prove that if

Q∞

k=1 ck

= c 6= 0 then limk→∞ ck = 1, ck ∈ C.

Q∞b) Formulate and prove the Cauchy criterion for infinte products k=1 ck , ck ∈ C. Hint: follow the ideas of Proposition I.30.7. 443

A COURSE IN ANALYSIS πz 12. For π cot πz := π cos defined on {z ∈ C| sin πz 6= 0} prove the product sin πz representation ∞ 1 X 2z . π cot πz = + z z2 − k2 k=1

444

24

Meromorphic Functions

P Rational functions R = Q , P and Q are polynomials, are defined on C\{z ∈ C | Q(z) = 0} and in this set they are holomorphic. Since Q has only a finite number of zeroes, in the case where Q is of degree m the set {z ∈ C | Q(z) = 0} has by the fundamental theorem of algebra at most m elements, the function R is holomorphic in C except at a finite number of points. For reasons which will become clear later we write

Sing(R) := {z ∈ C | Q(z) = 0} .

(24.1)

The fact that R is holomorphic in C\ Sing(R) has immediate consequences for the Taylor expansion of R. Let ζ0 ∈ C\ Sing(R), hence δ := dist(ζ0 , Sing(R)) > 0. In every disc Br (ζ0 ), r < δ, the function R is holomorphic and admits a Taylor expansion about every point of Br (ζ0 ). However we cannot expect that a Taylor expansion of R about a point ζ ∈ Br (ζ0 ) has radius of convergence greater than η := dist(ζ, Sing(R)) otherwise R would be holomorphic in a neighbourhood of a point of Sing(R). The question arises whether we can obtain a local and convergent series expansion of R in some set A ⊂ C\{z0 } where R|A is holomorphic and z0 ∈ Sing(R). Of course, this expansion in general cannot be a Taylor or power series. We may try to obtain a series expansion which also contains terms of negative powers in (z − z0 ), and certainly we shall expect such a series not to converge at z0 nor to represent R at z0 since R is not defined at z0 . Definition 24.1. For 0 ≤ r < R ≤ ∞ the open annulus with centre z0 ∈ C is defined by  (24.2) Ar,R (z0 ) := z ∈ C r < |z − z0 | < R , see Figure 24.1. For r = 0 we obtain the punctured disc A0,R (z0 ) = BR (z0 )\{z0 }, see Figure 24.2 445

(24.3)

A COURSE IN ANALYSIS

R

r b

z0

Ar,R (z0 )

Figure 24.1

R z0

A0,R (z0 ) = BR (z0 ) \ {z0 }

Figure 24.2 Obviously we have Ar,R (z0 ) = BR (z0 )\Br (z0 )

(24.4)

Ar,∞ (z0 ) = Br (z0 ){ .

(24.5)

and

446

24

MEROMORPHIC FUNCTIONS

There are many holomorphic functions on Ar,R (z0 ) which are not holomorphic on Br (z0 ). Just look at z 7→ (z−z1 0 )k , k ∈ N, or even at z 7→ (z−z1 1 )k , z1 ∈ Br (z0 ), and now take linear combinations of such functions, e.g. z 7→ P N ck k=1 (z−zk )k , zk ∈ Br (z0 ). A further example, not being a rational function is z 7→ sin1 z , z0 = 0, r = 12 and R = π. In general we shall not expect that a function holomorphic in Ar,R (z0 ) has a holomorphic extension to BR (z0 ). However in the case A0,R (z0 ) we can rely on Theorem 22.7 and we know that if h : A0,R (z0 ) → C is a holomorphic function bounded on some set Bρ (z0 )\{z0 }, 0 < ρ < R, then h admits a holomorphic extension to BR (z0 ). The next Theorem 24.2 shows that every holomorphic function f : Ar,R (z0 ) → C has a decomposition f = f1 + f2 where f1 is holomorphic in Ar,∞ (z0 ) and f2 is holomorphic in BR (z0 ), see Figure 24.3.

Ar,∞ (z0 )

r

R

b

BR (z0 )

z0

Ar,R (z0 )

Figure 24.3 447

A COURSE IN ANALYSIS

For its proof we need the following considerations: let G ⊂ C be a region and h : G → C a holomorphic function. For z0 ∈ C, note that z0 need not belong to G, and 0 < r2 < r1 assume that Ar2 ,r1 (z0 ) ⊂ G, see Figure 24.4.

Ar2 ,r1 (z0 )

b

b

z0

G

(Re z0 + r1 ) + iIm z0 b

(Re z0 + r2 ) + iIm z0

Figure 24.4

Let γ1 (ϕ) = z0 + r1 eiϕ and γ2 (ϕ) = z0 + r2 eiϕ , ϕ ∈ [0, 2π], the standard
parametrization of ∂Br1 (z0 ) and ∂Br2 (z0 ), respectively. For  ∈ − π8 , π8 let z1,± = r1 e±i and z2,± = r2 e±i . By γj, , j = 1, 2, we denote the curves γj, : [, 2π−] → C, γj, (ϕ) = γj (ϕ), ϕ ∈ [, 2π−]. Further we denote by σ± the line segments connecting z1, with z2, and z1,− with z2,− , respectively, see Figure 24.5. 448

24

MEROMORPHIC FUNCTIONS

γ1, Ar2 ,r1 (z0 )

z

b

z1,

σ−1 z2, b

2 b

z0

b

b

Ir2 ,r1 b

(Re z0 + r2 ) + iIm z0

z2,−

(Re z0 + r1 ) + iIm z0

b

σ− b

z1,−

−1 γ2,

Figure 24.5

Now we consider the curve −1 γ := γ1, ⊕ σ− ⊕ γ2, ⊕ σ−1 ,

(24.6)

compare again with Figure 24.5. The curve γ is a simply closed, piecewise continuously differentiable curve the trace of which belongs entirely to a domain on which h is holomorphic and it is the boundary of a set Ar2 ,r1 , (z0 ) ⊂ Ar2 ,r1 (z0 ). Therefore the Cauchy integral formula implies for z ∈ Ar2 ,r1 , (z0 ) 1 h(z) = 2πi 1 = 2πi

Z

γ

Z

h(ζ) dζ ζ −z

γ1 ,

(24.7) !

Z Z Z h(ζ) h(ζ) h(ζ) h(ζ) dζ + dζ + dζ + dζ −1 ζ − z ζ −z σ− ζ − z γ2, σ−1 ζ − z

We introduce, see Figure 24.5,  Ir1 ,r2 := w ∈ C Rez0 + r2 ≤ Re w ≤ Re z0 + r1 , Im w = Im z0 449

.

A COURSE IN ANALYSIS

and note that for every z ∈ Ar2 ,r1 (z0 )\Ir1 ,r2 there exists 0 = 0 (z) > 0 such that z ∈ Ar2 ,r1 , (z0 ) for all  < 0 . Therefore it follows for all z ∈ Ar2 ,r1 , (z0 )\Ir1 ,r2 ,  < 0 , that Z Z h(ζ) h(ζ) dζ = dζ (24.8) lim →0 γ ζ −z γ1 ζ − z 1, and lim

→0

Z

−1 γ2,

h(ζ) dζ = ζ −z

Z

h(ζ) dζ = − ζ −z

γ2−1

Z

γ2

h(ζ) dζ. ζ −z

We also claim for these values of z that Z  Z h(ζ) h(ζ) lim dζ + dζ = 0. →0 σ− ζ − z σ−1 ζ − z

(24.9)

(24.10)

In order to prove (24.10) we introduce the curves ηj, : [−, ] → C, ηj, (ϕ) = rj eiϕ , ϕ ∈ [−, ], see Figure 24.6, and consider the simply closed, piecewise continuously differentiable curve −1 η := σ−1 ⊕ η1, ⊕ σ− ⊕ η2, .

b

z

z1,

σ−1 z2,

b

−1 η1,

b

η2, b

z0

b

Ir2 ,r1 b

(Re z0 + r2 ) + iIm z0

z2,−

(Re z0 + r1 ) + iIm z0

b

σ− b

z1,−

Figure 24.6

450

24

MEROMORPHIC FUNCTIONS

For z ∈ Ar2 ,r1 (z0 )\Ir1 ,r2 and  < 0 (z) we now deduce Z h(ζ) dζ 0= ζ −z η and consequently Z Z Z Z h(ζ) h(ζ) h(ζ) h(ζ) dζ + dζ = dζ − dζ, σ− ζ − z σ−1 ζ − z η1, ζ − z η2, ζ − z or

Z

σ−

h(ζ) dζ + ζ −z

Z

σ−1


h(ζ) dζ ≤ M(0 ) lη1, + lη2 , , ζ −z

h(ζ) where we used that for z ∈ Ar2 ,r1 ,0 (z0 ) and ζ ∈ tr(η ) the term ζ−z is bounded and the bound is independent of , but may depend on 0 . Since for  → 0 both, lη1 , and lη2 , tend to zero, we arrive at Z Z 1 1 h(ζ) h(ζ) h(z) = dζ − dζ (24.11) 2πi γ1 ζ − z 2πi γ2 ζ − z for all z ∈ Ar2 ,r1 (z0 )\Ir1 ,r2 . By an analogous argument we obtain (24.11) for all z ∈ Ar2 ,r1 (z0 )\I˜r1 ,r2 where  I˜r1 ,r2 := w ∈ C − Re z0 − r1 ≤ Re w ≤ −Re z0 − r2 , Im w = Imz0 , and hence we have proved (24.11) to hold for all z ∈ Ar2 ,r1 (z0 ).

In Chapter 25 we will have a closer look at this type of result and argument after we have developed the concept of null-homologous cycles and related integrals. Now we prove Theorem 24.2. For a holomorphic function f : Ar,R (z0 ) → C there exists a holomorphic function f1 : Ar,∞ (z0 ) → C and a holomorphic function f2 : BR (z0 ) → C such that on Ar,R (z0 ) we have f = f1 + f2 .

(24.12)

If in addition we require lim|z|→∞ |f1 (z)| = 0 then the decomposition (24.12) is unique. 451

A COURSE IN ANALYSIS

Proof. We start by defining f2 . For z ∈ Bρ (z0 ), r < ρ < R, we set Z f (ζ) 1 dζ f2,ρ (z) := 2πi |ζ−z0|=ρ ζ − z

(24.13)

which is a holomorphic function on Bρ (z0 ). The Cauchy integral yields that for r < ρ1 < ρ2 < R we have on Bρ1 (z0 ) the equality f2,ρ1 = f2,ρ2 . It follows that a holomorphic function f2 is defined on BR (z0 ) by Z f (ζ) 1 dζ (24.14) f2 (z) := 2πi |ζ−z0 |=ρ ζ − z provided max{r, |z−z0 |} < ρ < R. Furthermore, for r < σ < min{R, |z−z0 |} a holomorphic function f1 is defined on Ar,∞ (z0 ) by Z 1 f (ζ) f1 (z) := − dζ. (24.15) 2πi |ζ−z0 |=σ ζ − z From the Cauchy inequalities, Theorem 21.15, we derive that lim |f1 (z)| = 0,

|z|→∞

(24.16)

note that for |z| → ∞ we can choose a sequence (σl )l∈N with liml→∞ σl = σ. We are now in a situation where we can apply (24.11) to f , γ1 is the circle |z − z0 | = ρ and γ2 is the circle |z − z0 | = σ, r < σ < |z − z0 | < ρ < R. Thus we obtain Z Z 1 f (ζ) f (ζ) 1 dζ − dζ = f2 (z) + f1 (z), f (z) = 2πi |ζ−z0|=ρ ζ − z 2πi |ζ−z0|=σ ζ − z i.e. the decomposition of f is established. It remains to prove the uniqueness result. Suppose that f = g1 + g2 is a second decomposition of f with lim|z|→∞ |g1 (z)| = 0. Then we find that f1 − g1 = g2 − f2 on Ar,R (z0 ). Thus ( f1 − g1 on Ar,∞ (z0 ), h(z) := f2 − g2 on BR (z0 ) is an entire function vanishing at infinity, i.e. lim|z|→∞ |h(z)| = 0. Hence h is bounded and by Louiville’s theorem h must be constant and therefore identically zero. 452

24

MEROMORPHIC FUNCTIONS

Let f : Ar,R (z0 ) → C be as in Theorem 24.2 admitting the decomposition f = f1 + f2 . The function f2 has in BR (z0 ) an expansion into a Taylor series representing f2 , i.e. ∞ X f2 (z) = ck (z − z0 )k (24.17) k=0

where

1 f (k) (z0 ) = ck = k! 2πi

Z

f (ζ) dζ, r < ρ < R, k ∈ N0 . (24.18) k+1 |ζ−z0 |=ρ (ζ − z0 )

In order to handle f1 we consider the biholomorphic function h : B 1 (0)\{0} → r

C\Br (z0 ), h(w) = z0 + w1 . We leave it as an exercise to prove that h is indeed a biholomorphic function, see Problem 1. Note that B 1 (0)\{0} = A0, 1 (0) and r

r

C\Br (z0 ) = Ar,∞ (z0 ), so h maps biholomorphically A0, 1 (0) onto Ar,∞ (z0 ). r Since f1 has the property that lim|z|→∞ f1 (z) = 0 we conclude that for f1 ◦ h it follows that limw→0(f1 ◦ h)(w) = 0. Therefore, by Theorem 22.7 we may extend f1 ◦ h to a holomorphic function on B 1 (0) which admits in B 1 (0) a r r Taylor representation ∞ X (f1 ◦ h)(w) = γl w l , (24.19) l=1

where we used that limw→0 (f1 ◦ h)(w) = 0 implies γ0 = 0. The convergence of the series in (24.19) is uniform and absolute on every disc B 1 (0), ρ > r. ρ

1 Now we recall that for z = h(w) it follows that w = z−z and we can rewrite 0 (24.19) as ∞ ∞ X X 1 , (24.20) f1 (z) = γl (z − z0 )−l = γl l (z − z 0) l=1 l=1

and due to the mapping properties of h we know that this series converges in C\Bρ (z0 ) uniformly for every ρ > r. We define k∈N

(24.21)

ck (z − z0 )k .

(24.22)

c−k := γk , and we arrive at f1 (z) =

−∞ X

k=−1

453

A COURSE IN ANALYSIS

Combining (24.17) with (24.21) we find for z ∈ Ar,R (z0 ) f (z) = f1 (z) + f2 (z) =

−∞ X

k=−1

k

ck (z − z0 ) +

∞ X k=0

ck (z − z0 )k ,

(24.23)

where the first series converges locally uniformly in Ar,∞ (z0 ) and the second in BR (z0 ), i.e. both converge locally uniformly in Ar,R (z0 ) and the sum of both series in Ar,R (z0 ) is f . We claim further that (24.18) also holds for k ∈ −N. Indeed, for r < ρ < R and k ∈ Z we find (z − z0 )

−k−1

f (z) =

∞ X

l=−1

l

cl+k+1 (z − z0 ) +

∞ X l=0

cl+k+1 (z − z0 )l

and these series converge uniformly on |z − z0 | = ρ. Thus we may integrate term by term which yields Z Z dz f (ζ) dζ = ck = 2πick , (24.24) k+1 |z−z0 |=ρ z − z0 |z−z0 |=ρ (z − z0 ) as usual in (24.24) we used the standard parametrization of the circle |z − z0 | = ρ. Thus we have proved Theorem 24.3. A holomorphic function f : Ar,R (z0 ) → C, 0 ≤ r < R ≤ ∞, admits the representation (24.23) with coefficients given by Z 1 f (ζ) ck = dζ, r < ρ < R. (24.25) 2πi |ζ−z0 |=ρ (ζ − z0 )k+1 Furthermore we can differentiate the series in (24.23) term by term. Definition 24.4. A. A series ∞ X

k=−∞

k

ck (z − z0 ) =

−∞ X

k=−1

k

ck (z − z0 ) +

∞ X k=0

ck (z − z0 )k ,

(24.26)

is called a Laurent series about z0 ∈ C. It converges at z ∈ C if both series in (24.26) converge for z. B. The first series in (24.26) is called the principal part of the series (24.26). 454

24

MEROMORPHIC FUNCTIONS

Remark 24.5. A. Now we can read Theorem 24.3 as follows: a holomorphic function f : Ar,R (z0 ) → C admits a locally uniform convergent Laurent series expansion in Ar,R (z0 ) and the principal part of this expansion is given by the expansion of f1 according to (24.22). Hence we can also call f1 the principal part of f . Consider the series (24.26) and assume that it converges locally uniformly in some annulus Ar,R (z0 ). For every k ∈ Z\{−1} the function z 7→ (z − z0 )k has a primitive in Ar,R (z0 ), a fact which we indirectly used to derive (24.25). This gives the coefficient c−1 a special role. Let γ : [a, b] → Ar,R (z0 ) be a simply closed, piecewise continuously differentiable curve which is anticlockwise oriented. We assume in addition that the set bounded by tr(γ) ⊂ Ar,R (z0 ) contains Br (z0 ), see Figure 24.7. Note that due to the Jordan curve theorem, Theorem 19.37, such an assumption is justified.

R

γ

> >

>

r b

z0

Ar,R (z0 )

Br (z0 )

>

Figure 24.7 Since tr(γ) is compact the locally uniform convergence of (24.25) in Ar,R (z0 ) allows us to interchange integration and summation and we find for such a curve that Z Z X ∞ k (24.27) ck (z − z0 ) dz = c−1 (z − z0 )−1 dz = 2πic−1 . γ k=−∞

γ

455

A COURSE IN ANALYSIS

In particular if f : Ar,R (z0 ) → C is holomorphic with Laurent series (24.23) then we have for every curve as described above Z Z 1 1 f (z) dz = f1 (z) dz. (24.28) c−1 = 2πi γ 2πi γ Of course, for (24.27) or (24.28) it is important that z0 lies in the set bounded by γ. Definition 24.6. Given the punctured disc A0,R (z0 ) ⊂ C and a holomorphic function f : A0,R (z0 ) → C. The number Z 1 res(f, z0 ) := f (ζ) dζ, 0 < ρ < R, (24.29) 2πi |ζ−z0|=ρ is called the residue of f at z0 . Hence res(f, z0 ) is the first coefficient of the principal part of the Laurent expansion of f in A0,R (z0 ). Remark 24.7. From the considerations preceding Definition 24.6 it is clear that res(f, z0 ) is independent of the choice of ρ. In fact we may choose instead |ζ −z0 | = ρ a simply closed piecewise continuously differentiable curve γ with tr(γ) ⊂ A0,R (z0 ) and z0 in the interior of γ. Before we continue with our theoretical considerations we want to look at some examples. Example 24.8. A. The function z 7→ z1 is holomorphic in C\{0} and its Laurent series about z0 = 0 is just the function itself, the same applies to z 7→ z1l , l ∈ N. B. Let h : C → C be an entire function with Taylor expansion h(z) =

∞ X k=0

ck (z − z0 )k

  1 about z0 ∈ C. The function f (z) := h z−z is holomorphic in A0,∞ (z0 ) and 0 its Laurent expansion about z0 is given by f (z) =

∞ X k=1

ck (z − z0 )−k + c0 . 456

24

MEROMORPHIC FUNCTIONS 1

In particular we find for the Laurent expansion of e z , z 6= 0, 1

ez =

∞ X 1 −k z + 1. k! k=1

C. Let
us consider once more the function z 7→ 1z but this
time in the annulus 1 1 A1,2 2 . Clearly the function is holomorphic in A1,2 2 and we want to find its Laurent series about 12 . For this we write 1 1 = z z − 21 +

1 2

2
1 + 2 z − 21  k ∞ X 1 k =2 (−2) z − 2 k=0 =

which converges for z − 21 < 21 . Hence the Laurent expansion of z1 about 12 is a Taylor expansion and the principal part vanishes. However the residue
res 1z (0) = 1 and not equal to 0. For determining the residue of a function f at z0 we must take its Laurent expansion about z0 , not about a different point. D. In this example, essentially taken from [26] we want to calculate the 1 Laurent expansion for h : C\{0, i} → C, h(z) = z(z−i) 2 , for three different values of z0 and corresponding annuli: A0,1 (0), A1,∞ (1) and A0,1 (i). In each case a reduction to a geometric or related series is essential, and in fact this is a major tool to find Taylor as well as Laurent expansions. i) In A0,1 (0) we find ∞  z k 1 1 1 1X (k + 1) = − = −
z(z − i)2 z 1− z 2 z i k=0 i ∞ ∞  z k X 1 1 Xk+2 k =− +i z . (k + 2) =− + k−1 z i z i k=0 k=0

ii) In A1,∞ (1) we have −∞ −∞ X X 1 1 1 k+2 k −k−1 k = z . = −i (k + 2)z =
2 2 3 i z(z − i) z 1− ik+2 k=−3 k=−3 z

457

A COURSE IN ANALYSIS

iii) In A0,1 (i) we obtain −i 1 1 1 −i i 1 = + + − = + 2 2 2 z(z − i) (z − i) z−i z (z − i) z − i 1 − i(z − 1) ∞ X 1 −i i(i(z − i))k + + = 2 (z − i) z−i k=0

=

−i 1 + + 2 (z − i) z−i

∞ X k=0

ik+1 (z − i)k .

P∞ P k k Proposition 24.9. A. Let ∞ k=−∞ ck (z −z0 ) be two k=−∞ ak (z −z0 ) and Laurent series representing on some annulus Ar,R (z0 ) a holomorphic function h, then ak = ck for all k P ∈ Z. k B. Suppose that f (z) = ∞ k=−∞ ck (z − z0 ) , z ∈ Ar,R (z0 ) and let r < ρ < R. It follows for all k ∈ Z that |ck | ≤ ρ−k sup |f (z)|.

(24.30)

|z−z0 |=ρ

Proof. A. Fix k0 ∈ Z and note that (z − z0 )−k0 −1

∞ X

k=−∞

ak (z − z0 )k = (z − z0 )−k0 −1

∞ X

k=−∞

ck (z − z0 )k .

Integrating this equality over the circle |z − z0 | = ρ, r < ρ < R, yields ak0 = ck0 . B. The coefficients ck are given by (24.25) and it follows Z Z 1 |f (z)| 1 1 dζ, dζ ≤ sup |f (z)| |ck | ≤ 2π |z−z0|=ρ ρk+1 2πρk+1 |z−z0|=ρ |z−z0 |=ρ i.e. |ck | ≤ ρ−k sup|z−z0|=ρ |f (z)|. Let f : A0,R (z0 ) → C be a holomorphic function with a non-trivial principal part. In this case f cannot be bounded in a neighbourhood of z0 and z0 cannot be a removable singularity of f , compare with Theorem 22.7. We now want to study certain singularities of holomorphic functions. Definition 24.10. Let z0 ∈ C and U ⊂ C a neighbourhood of z0 . Further let f : U\{z0 } → C be a holomorphic function. We call z0 an isolated singularity of f . 458

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Note that Definition 24.10 also covers the case where f is bounded in a neighbourhood of z0 and hence admits a holomorphic extension f˜ to U by Theorem 22.7. Clearly f˜ will be holomorphic in U, i.e. will have no isolated singularity. In such a case having an isolated singularity is mainly a problem of the domain and not a property of the function. We can extend Definition 24.10 in the following sense: if M ⊂ D is a discrete set in a domain D ⊂ C and f : D\M → C is a holomorphic function, then we call the points of M the isolated singularities of f . Here are some examples of holomorphic functions with isolated singularities. Example 24.11. A. For z0 ∈ C and k ∈ N the function f (z) = (z−z1 0 )k has an isolated singularity at z0 and is holomorphic in C\{z0 }. P has isolated singularities for all points B. The rational function R = Q belonging to Sing(R) as defined by (24.1). a C. The function z 7→ e zk , a 6= 0, k ∈ N, has an isolated singularity at z0 = 0, otherwise it is holomorphic on C\{0}. D. The function z 7→ sinz z has an isolated singularity at z0 = 0 as does the z function 1−cos . z2 E. The set Z is the set of isolated singularities of the function cot : C\Z → C, πz . cot πz = cos sin πz The natural programme to follow is to classify isolated singularities and to study the behaviour of a holomorphic function in a neighbourhood of an isolated singularity. Of course, Theorem 22.7 on removable singularities is such a result. In our discussion of singularities we follow partly [26]. Definition 24.12. Let z0 be an isolated singularity of the holomorphic function f : U\{z0 } → C where U ⊂ C is a neighbourhood of z0 . A. If there exists a neighbourhood V ⊂ U of z0 such that f |V \{z0 } is bounded then we call z0 a removable singularity of f . B. Suppose that limz→z0 |f (z)| = ∞, then we call z0 a pole of f . C. In the case that z0 ∈ C is an isolated singularity of the holomorphic function f : U\{z0 } → C which is neither removable nor a pole, then we call z0 an essential singularity of f . We do already understand removable singularities. Now we turn to poles. Let f : U\{z0 } → C be a holomorphic function where U is a neighbourhood of z0 and assume that z0 is a pole of f . Since in this case limz→z0 |f (z)| = ∞, we can find a neighbourhood V of z0 such that |f (z)| ≥ 1 for z ∈ 459

A COURSE IN ANALYSIS 1 V \{z0 }. In particular f |V \{z0 } does not vanish implying that z 7→ f (z) is a 1 1 holomorphic function on V \{z0 } and limz→z0 f (z) = 0. Thus f is bounded in a neighbourhood of z0 and hence z0 is a removable singularity of f1 : V \{z0 } →   C and we can extend f1 to V by defining f1 (z0 ) = 0. By Theorem 22.7 this extension is holomorphic. Thus there exists a holomorphic function g : V → C and n ∈ N such that

1 = (z − z0 )n g(z) f (z)

(24.31)

holds and g does not vanish on V . Therefore we can find a neighbourhood 1 W ⊂ U of z0 and a holomorphic function h(z) = g(z) not vanishing at z0 such that in W \{z0 } we have f (z) = (z − z0 )−n h(z).

(24.32)

The number n ∈ N in (24.32) is called the order or the multiplicity of the pole z0 : since |f (z)| = |z − z0 |−n |h(z)| and h does not vanish in W \{z0 } it follows that limz→z0 |f (z)| = ∞, i.e. if (24.32) holds, then f has a pole (of order n) at z0 . Corollary 24.13. A. Let g : U → C be a holomorphic function and suppose that its set of zeroes N(f ) := {z ∈ U | g(z) = 0} is a discrete set. In this case the function g1 : U\N → C is holomorphic and has poles at N. The order of a pole z0 ∈ N of g equals the order of the zero z0 of f . P B. Let R = Q be a rational function and Sing(R) defined as before, i.e. Sing(R) = {w ∈ C | Q(w) = 0}. Then Sing(R) is a subset of the poles of R. Part B of Corollary 24.13 stimulates a closer study of the isolated singularities P and P has order n, Q has order m. of rational functions. Suppose that R = Q According to the fundamental theorem of algebra we have the factorizations p(z) = c

N Y j=1

(z − zj )αj and Q(z) = d

M Y (z − wl )βl l=1

where zj , 1 ≤ j ≤ N, are the zeroes of P with multiplicity αj , and wl , P 1 ≤ l ≤ M, are the zeroes of Q with multiplicity βl . Hence N j=1 αj = n and PM l=1 βl = m. Thus we can write R on C\{w1 , . . . , wn } as R(z) = cd−1 (z − z1 )α1 · . . . · (z − zN )αN (z − w1 )−β1 · . . . · (z − wM )−βM . 460

24

MEROMORPHIC FUNCTIONS

In the case where {z1 , . . . , zN } ∩ {w1 , . . . , wM } = ∅, the function R has poles of order βl at wl . However, whenever zj0 = wl0 for some j0 and l0 we have to have a closer look at (z − zj0 )αj0 (z − wl0 )−βl0 = (z − wl0 )αj0 −βl0 . If αj0 − βl0 ≥ 0, then R has a removable singularity at wl0 , whereas if αj0 − βl0 < 0 then R has a pole of order βl0 − αj0 at wl0 . Hence for a rational function we know that all singularities are isolated singularities or removable singularities. They are located at the zeroes of Q(z) and their type is determined by the relations of the zeroes of Q(z) to the zeroes of P (z): If Q(z) and P (z) have no common zeroes, then R has at every zero of Q(z) a pole the order of which is the multiplicity of the corresponding zero of Q(z). For a common zero z0 of P (z) and Q(z) the function R has a removable singularity of the order of z0 as zero of P (z) is larger or equal to the order of z0 as zero of Q(z). In the other case R has a pole of order equal to the absolute value of the difference of the multiplicities. We can characterise a pole in terms of estimates. Theorem 24.14. Let f : U\{z0 } → C be a holomorphic function, where U is a neighbourhood of z0 . The isolated singularity z0 of f is a pole of order n if and only if in some neighbourhood V ⊂ U of z0 the estimates κ0 |z − z0 |−n ≤ |f (z)| ≤ κ1 |z − z0 |−n

(24.33)

hold for constants κ0 , κ1 > 0. Proof. Suppose that z0 is a pole of order n. By (24.32) we can find κ0 , κ1 > 0 such that 0 < κ0 ≤ |h(z)| ≤ κ1 which immediately implies (24.33). Now suppose that (24.33) holds. In this case h(z) := (z − z0 )n f (z) is bounded in a neighbourhood of z0 , hence z0 is for h a removable singularity. Since by (24.33) the function h is bounded from below by κ0 it follows that h(z0 ) 6= 0 and we obtain for f the representation (24.32) which yields that f has a pole of order n at z0 . Definition 24.15. A complex-valued function f : D → C defined on a domain D ⊂ C is called meromorphic in D if there exists a discrete subset Pol(f ) ⊂ D such that f |D\ Pol(f ) is a holomorphic function and every z ∈ Pol(f ) is a pole of f . 461

A COURSE IN ANALYSIS

Example 24.16. A. If Pol(f ) = ∅, which is not excluded by our definition, then f is holomorphic in D, i.e. holomorphic functions are meromorphic. B. The rational functions are meromorphic on C but in general they are not holomorphic. C. We claim that the function z 7→ cot πz is meromorphic on C. Since πz cot πz = cos the integers Z forms the set of isolated singularities of cot πz. sin πz For z0 = 0 we find πz cos πz π(z − z0 ) cot πz = πz cot πz = sin πz and for z → z0 , i.e. z → 0, it follows

 πz  · cos πz z→0 sin πz limz→0 cos πz = = 1, limz→0 sinπzπz

lim πz cot πz = lim

z→0

thus cot πz has a pole of order 1 at z0 = 0. Using either the periodicity of cot πz or the Taylor expansion of sin πz about kπ, k ∈ Z, we deduce that every k ∈ Z is a pole of order 1 of cot πz. Example 24.16.C is of particular interest since it shows firstly that there exists meromorphic functions which are not rational functions. Secondly it suggests that meromorphic functions are quotients of holomorphic functions. At least locally we can already prove Theorem 24.17. Let f : D → C, D ⊂ C a domain, be a meromorphic function. For every z0 ∈ D we can find a neighbourhood U ⊂ D of z0 and g(z) . two holomorphic functions g, h : U → C such that on V we have f (z) = h(z) Proof. Let z0 ∈ D. If z0 is not a singularity of f we choose U = D, g = f and h = 1. In the case where z0 is an isolated singularity it must be a pole, say of order n. Now we may use (24.32), i.e. the representation of f in some neighbourhood U of z0 as f (z) = (z − z0 )−n g(z), i.e. we take as h(z) the function z 7→ (z − z0 )n . Remark 24.18. The general global results, i.e. the statement that every meromorphic function f : G → C defined on a region G ⊂ C admits a representation on G as quotient of two holomorphic functions g, h : G → C, g(z) i.e. f (z) = h(z) for all z ∈ D we do not prove in our treatise and we refer to [26] or [67]. 462

24

MEROMORPHIC FUNCTIONS

We turn now our attention to essential singularities. Theorem 24.19 (F. Casorati and K. Weierstrass). Let z0 ∈ D and D ⊂ C be a domain. Further let f : D\{z0 } → C be a holomorphic function with isolated singularity at z0 . This isolated singularity is an essential singularity if and only if for every w0 ∈ C there exists a sequence (zk )k∈N , zk ∈ D\{z0 }, converging to z0 and limk→∞ f (zk ) = w0 . In other words, the image f (U\{z0 }) is for every neighbourhood U ⊂ D of z0 dense in C, i.e. f (U\{z0 }) = C. Proof. The “only if” part immediately implies that z0 can neither be a removable singularity nor a pole, hence it must be an essential singularity. Now let z0 be an essential singularity of f . Suppose that there exists a neighbourhood U ⊂ D of z0 such that f (U\{z0 }) is not dense in C. In this case we can find some w0 ∈ C and  > 0 such that B (w0 ) ∩ f (U\{z0 }) = ∅, i.e. 1 |f (z) − w0 | >  for all z ∈ U\{z0 }. It follows that z 7→ g(z) := f (z)−w is on 0 1 U\{z0 } holomorphic and bounded by  , hence g has a removable singularity 1 has a removable at z0 . If now limz→z0 g(z) 6= 0, then z 7→ f (z) = w0 + g(z) 1 singularity at z0 , while if limz→z0 g(z) = 0 then z 7→ f (z) = w0 + g(z) has a pole at z0 . Both cases contradict our assumption that z0 is an essential singularity of f . Now we combine the classifications of isolated singularities with the Laurent series expansion of a holomorphic function. Theorem 24.20. Let f : D\{z0 } → C be a holomorphic function which has an isolated singularity at z0 ∈ D, D ⊂ C is a domain. Let  > 0 such that the punctured P disc A0, (z0 ) iska subset of D\{z0 }, i.e. A0, (z0 ) ⊂ D\{z0 } and let f (z) = ∞ k=−∞ ck (z − z0 ) be the Laurent expansion of f about z0 converging in A0, (z0 ). i) The singularity z0 is removable if and only if ck = 0 for k < 0. ii) The singularity z0 is a pole of order n if and only if c−n 6= 0 and ck = 0 for k < n. iii) The singularity z0 is an essential singularity if ck 6= 0 for infinitely many k ∈ Z, k < 0. 463

A COURSE IN ANALYSIS

Proof. If z0 is a removable singularity the Taylor series of its holomorphic extension to B (z0 ) must coincide in A0, (z0 ) with the Laurent expansion of f , hence ck = 0 for k < 0. Conversely, if all ck , k < 0, are zero, then the principal part of the Laurent expansion of f about z0 , vanishes identically, implying that f is bounded in a neighbourhood of z0 , hence z0 is removable. Suppose next that z0 is a pole of order n. In this case we can write f (z) = (z−z0 )−n h(z) where h is a holomorphic function in B (z0 ) not vanishing at z0 , i.e. h(z0 ) 6= 0. For the Laurent expansion of f about z0 thisP implies ck = 0 for k k < −n and c−n 6= 0. The converse is trivial, i.e. if f (z) = ∞ k=−n ck (z −z0 ) , c−n 6= 0, then f has a pole of order n at z0 . Finally, if neither i) or ii) hold f must have an essential singularity at z0 and at the same time infinitely many coefficients ck , k < 0, cannot vanish. Example 24.21. A. Consider the function f (z) = Using the Taylor expansion of cos we find ! ∞ 1 X (−1)k 2k cos z − 1 = 2 z −1 z2 z (2k)! k=0

cos z−1 z2

defined on C\{0}.

∞ ∞ 1 X (−1)k 2k X (−1)k 2(k−1) = 2 z = z z (2k)! (2k)! k=1

k=1

∞ X (−1)k+1 2k z , = 2(k + 1)! k=0

and it follows that f has a removable singularity at z0 = 0. B. Using the results from Example 24.8.D we find that z 7→ pole of order 2 at i. C. Since for z 6= 0 and n ∈ N 1

e zn =

1 z(z−i)2

has a

∞ X 1 1 l! z nl l=0

1

it follows that z 7→ e zn has an essential singularity at z0 = 0. We want to explore the structure of the set of all meromorphic functions f on a region (or a domain with finitely many connectivity components) G ⊂ C. For f there exists a discrete set Pol(f ) ⊂ G and w0 ∈ Pol(f ) is a pole of f . In G\ Pol(f ) the function f is holomorphic and since G\ Pol(f ) is a region, 464

24

MEROMORPHIC FUNCTIONS

if f is not identically equal to 0, then by Theorem 22.10 f has no zero of order ∞ and the set N(f ) of all zeroes is discrete. It follows that Pol(f ) ∪ N(f ) is a discrete set in G and the functions f1 : G\ (Pol(f ) ∪ N(f )) → C is holomorphic. In a neighbourhood of a pole w0 ∈ Pol(f ) we have the representation f (z) = (z − w0 )−n g(z) where n ∈ N is the order of the pole w0 and g is holomorphic with g(w0) 6= 0. In a neighbourhood of a zero z0 ∈ N(f ) we have f (z) = (z − z0 )m h(z) where m ∈ N is the multiplicity of order of the zero z0 and h is a holomorphic function with h(z0 ) 6= 0. It follows that f1 is meromorphic in G and     Pol f1 = N(f ). Moreover we have that N f1 = Pol(f ), i.e. f1 has a holomorphic extension to G\N(f ). Now we may look at algebraic operations for meromorphic functions defined on a region G ⊂ C. It is important to recall that a meromorphic function on G is a holomorphic function defined on G\X where X ⊂ G is a discrete set. Let fj , j = 1, 2, be meromorphic functions with sets of poles Pol(fj ). On G\(Pol(f1 ) ∪ Pol(f2 )) we can define f1 + f2 , f1 − f2 and f1 · f2 and these are holomorphic functions on the given domain, hence meromorphic functions on G since Pol(f1 )∪ Pol(f2 ) is a discrete set in G. In addition we can define for a meromorphic function f not identically zero in G with set of poles Pol(f ) and set of zeroes N(f ) the holomorphic function f1 : G\(Pol(f )∪N(f )) → C, then 1 is a meromorphic function on G. Note that Pol(f1 ±f2 ) and Pol(f1 ·f2 ) are in f general subsets of Pol(f1 ) ∪ Pol(f2 ) since singularities may cancel or become  1 removable. We already know that Pol f = N(f ). We agree that when operating with meromorphic functions we always extend these functions into their removable singularities as holomorphic functions. The following result is now easy to check although it takes some time and we leave this to the reader. Theorem 24.22. The meromorphic functions on G ⊂ C, where G is a region or a domain with finitely many connectivity components, form with the operations introduced above a field containing the holomorphic functions. 465

A COURSE IN ANALYSIS

The neutral element with respect to addition is the function 0, i.e. z 7→ 0 for all z ∈ G, and the neutral element with respect to multiplication is the function 1, i.e. z 7→ 1 for all z ∈ G. This theorem determines the algebraic structure of the meromorphic functions on G. There are other operations we may apply to obtain a new meromorphic function from a given one. Differentiation is such an operation. Lemma 24.23. For a meromorphic function f on a domain D ⊂ C the derivative f 0 is a meromorphic function on G too and Pol(f 0 ) = Pol(f ). Proof. On D\ Pol(f ) the function f , hence f 0 , is holomorphic and it remains to prove that z0 ∈ Pol(f ) is also a pole of f 0 . For some  > 0 we have in A0, (z0 ) the Laurent series representation f (z) =

∞ X

k=−n

ck (z − z0 )k

(24.34)

where n ∈ N is the order of the pole z0 , c−n 6= 0. Since we can differentiate this series term by term we find in A0, (z0 ) that 0

f (z) =

∞ X

k=−n

kck (z − z0 )

k−1

=

∞ X

(k + 1)ck+1(z − z0 )k

(24.35)

k=−n−1

and it follows that f 0 has a pole of order −n − 1 at z0 . Thus we conclude that Pol(f 0 ) = Pol(f ) and the lemma is proved. Since with f also f 0 is meromorphic, under the assumption that f is not 0 identically zero we may consider on G the term ff . Suppose that f has a zero z0 ∈ G of order n, i.e. with some holomorphic function g, g(z0 ) 6= 0, we have in a neighbourhood of z0 the equality f (z) = (z − z0 )n g(z), hence f 0 (z) = n(z − z0 )n−1 g(z) + (z − z0 )n g 0 (z), implying f 0 (z) n(z − z0 )n−1 g(z) + (z − z0 )n g 0(z) = f (z) (z − z0 )n g(z) n g 0 (z) = + . (24.36) z − z0 g(z)  0  0 Thus ff has a pole of order 1 at z0 and res ff , z0 = n. Now, if w0 is a pole of order m of f then we find in a neighbourhood of w0 the equality 466

24

MEROMORPHIC FUNCTIONS

f (z) = (z−w0 )−m h(z) where h is a holomorphic function such that h(w0 ) 6= 0. Now we find h0 (z) −m f 0 (z) + = , f (z) z − w0 h(z)

(24.37)

  0 0 i.e. ff has a pole of order 1 at w0 and res ff , w0 = −m. We have proved that if f is meromorphic in G ⊂ C with set of poles Pol(f ) and set of zeroes 0 N(f ), then ff is meromorphic in G having poles of order 1 for all points in Pol(f ) ∪ N(f ). In order to exploit this observation further we first need to extend (24.29) which leads to a first version of the residue theorem.

Theorem 24.24. Let D ⊂ C be a domain and Br (a) ⊂ D. Assume that f is meromorphic in D and Pol(f ) = {w1 , . . . , wn } ⊂ Br (a). Then we have 1 2πi

Z

f (z) dz = ∂Br (a)

N X

res(f, wk ),

(24.38)

k=1

where we used for the circle ∂Br (a) our standard parametrization.

Proof. Consider Figure 24.8. With an argument similar to that leading to (24.11) we obtain that

1 2πi

Z

Z N X 1 f (z) dz = f (z) dz, 2πi ∂Bk (wk ) ∂Br (a) k=1

(24.39)

where the k ’s are chosen such that Bk (wk ) does not contain any further pole of f , and all circles are parametrized in the standard way. Now we can R apply (24.29) to each integral ∂B (wk ) f (z) dz and (24.38) follows. k

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1 w1

D r

2 a

b

w2

w3

3

Figure 24.8

In the next chapter we will provide many applications of Theorem 24.24 and its generalisation. We will also see that ∂Br (a) in (24.38) can be replaced by a more general curve γ. This applies also for the following result, the argument principle: Theorem 24.25. For a domain D ⊂ C and Br (a) ⊂ D let f be a meromorphic function in D. We denote the zeroes of f in Br (a) by z1 , . . . , zM and zk has multiplicity αk . The poles of f in Br (a) are w1 , . . . , wN and wl has order βl . If f has no zero and no pole on ∂Br (a) then we have 1 2πi

Z

M

∂Br (a)

N

X X f 0 (z) dz = αk − βl . f (z) k=1 l=1

(24.40)

Proof. We need only combine (24.36) and (24.37) with Theorem 24.24. We end this chapter with an interesting application of the argument principle. 468

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Theorem 24.26 (Theorem of E. Rouch´ e). Let D ⊂ C be a domain and Br (a) ⊂ D. Suppose that f, g : D → C are holomorphic functions and that |f (z)| > |g(z)| on ∂Br (a). In this situation the functions f and f ± g have the same number of zeroes in Br (a). Proof. For s ∈ [0, 1] we consider the family fs := f + sg of holomorphic functions in D, f0 = f and f1 = f + g. Denote by n(s) the number of zeroes of fs in Br (a) counted according to their multiplicity. On ∂Br (a) we have |fs (z)| = |f (z) + sg(z)| ≥ |f (z)| − s|g(z)| > 0 since |f (z)| > |g(z)| on ∂Br (a). By the argument principle we obtain Z 1 f 0 (z) n(s) = dz 2πi ∂Br (a) f (z) where we use the standard parametrization for ∂Br (a). Since n(s) ∈ Z for every s ∈ [0, 1], once we have proved that n(·) is continuous it follows that n(·) is constant, i.e. n(s) = n(0) for all s implying that n(1) = n(0), 0 i.e. the theorem will follow. The function (z, s) 7→ ffss (z) is continuous on (z) ∂Br (a) × [0, 1] since fs (z) 6= 0 for all z ∈ ∂Br (a). Now the compactness of ∂Br (a) implies, see Theorem 21.11, the continuity of n(·). Remark 24.27. In Problem 11 we will see how we can use the Theorem of Rouch´e to localise zeroes of holomorphic functions. We need a better “calculus for parametric curves” and this will be developed in the following chapter.

Problems 1. Prove that h : B 1 (0) \ {0} → C \ Br (z0 ), h(w) = z0 + w1 , is a biholor morphic function. 2. Find Laurent expansions of the following functions and identify the type of singularity: 3

a)

z− z6 −sin z z5

at z0 = 0;

1 b) (z − 4) sin z+3 at z0 = −3;

c)

cos 2z (z−1)3

at z0 = 1.

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A COURSE IN ANALYSIS

3. Expand f (z) =

1 (z+2)(z+4)

as a Laurent series for:

a) z ∈ A2,4 (0);

b) z ∈ A4,∞ (0);

c) z ∈ A1,2 (−2); d) z ∈ B2 (0).

4. Determine the singularities of each of the following functions: a) f (z) = b) h(z) =

1 ; (4 sin z−2)2 z

1

e 2z −1

.

5. Give an example of a meromorphic function which is not a rational function and has a pole of order 2 at z0 = 1. 6. Let f : U \ {z0 } → C be a holomorphic function where U is a neighbourhood of z0 . Suppose that in some neighbourhood V ⊂ U of z0 we have the estimate 1

κ0 e |z−z0 | ≤ |f (z)|,

κ0 > 0.

What type of singularity does f have at z0 ?
7. Consider on B 1 (0) \ {0} the holomorphic function h(z) = exp z1 . Let 2 w0 ∈ C \ {0}. Prove that there exists a countable set {zk | k ∈ N} ⊂ B 1 (0) \ {0} such that h(zk ) = w0 . 2

8. Let f, g : U \ {z0 } → C be holomorphic functions and assume that f as well as g has a pole of order N at z0 . Prove that for some neighbourhood V ⊂ U of z0 the function fg is defined in V \ {z0 } and has a removable singularity at z0 . Further prove that ( f (z), z ∈ V \ {z0 } h(z) := ag−N , z = z0 b−N is a holomorphic extension of fg : V \ {z0 } → C, where a−N and b−N are the coefficients of the leading terms in the Laurent expansion of f and g about z0 . 470

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MEROMORPHIC FUNCTIONS

9. Give a further proof of the fundamental theorem of algebra by using the argument principle. Hint: for p(z) = z n + a1 z n−1 + · · · + an , n ≥ 1, choose R > 0 such that |p(z)| ≥ 1 for z ∈ BR{ (0) and consider Z 1 p0 (z) dz. 2πi ∂BR (0) p(z) 10. Use the Theorem of Rouch´e to prove that all zeroes of the polynomial z 5 − 2z 3 + 10 belong to A1,2 (0). Hint: consider f (z) = 10 and g(z) = z 5 − 2z 3 to prove that there are no zeroes in B1 (0). Then consider f (z) = z 5 and g(z) = −2z 3 + 10 to prove that all zeroes are in B2 (0).

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25

The Residue Theorem

So far when working with closed curves we emphasised that they were simply closed, in particular for the circle |z0 − z| = r we used the standard (anticlockwise) parametrization t 7→ z(t) = z0 + reit , t ∈ [0, 2π]. A very useful observation was that with this parametrization we have Z 1 1 dζ = 1, z ∈ Br (z0 ), (25.1) 2πi |ζ0 −z0 |=r ζ − z compare with Lemma 21.7. In preparing the proof of the Runge theorem, Theorem 23.25, we gave already some extension R dζof (25.1), see Theorem 23.23. We now want to investigate the integral γ ζ−z for general (i.e. piecewise continuously differentiable) closed curves γ : [a, b] → C. Proposition 25.1. For a piecewise continuously differentiable closed curve γ : [a, b] → C and z ∈ / tr(γ) the integral Z 1 1 dξ (25.2) indγ (z) := 2πi γ ζ − z has the following properties: i) indγ (z) ∈ Z; ii) indγ (z1 ) = indγ (z2 ) for z1 , z2 belonging to the same open connectivity component of tr(γ){ ; iii) for z in the unbounded component of tr(γ){ it follows that indγ (z) = 0. Proof. For t ∈ [a, b] and z ∈ / tr(γ) fixed consider the integral hz (t) :=

Z

t

a

γ(t) ˙ dt. γ(t) − z

Since γ is piecewise continuously differentiable we find with the possible exception of finitely many values of t ∈ [a, b] that h˙ z (t) =

γ(t) ˙ . γ(t) − z

473

A COURSE IN ANALYSIS

With the
possible exception of finitely many values the function t 7→ (γ(t) − z)e−hz (t) is continuously differentiable and we find  
d −hz (t) −hz (t) −hz (t) ˙ = γ(t)e ˙ + (γ(t) − z) −hz (t)e (γ(t) − z)e dt   γ(t) ˙ = γ(t) ˙ − (γ(t) − z) e−hz (t) = 0. γ(t) − z It then follows that (γ(·) − z)e−hz (·) must be equal to a constant A 6= 0, note that γ(t) 6= z for all t ∈ [a, b] by our assumptions. Using the fact that γ(a) = γ(b) we get 1 1 (γ(a) − z) = (γ(b) − z)ehz (b) , A A or hz (b) = 2πik, k ∈ Z. Thus we arrive at Z 1 1 indγ (z) = dζ 2πi γ ζ − z Z b γ(t) ˙ 1 1 dt = hz (b) ∈ Z, = 2πi a γ(t) − z 2πi 1 = ehz (a) =

and i) is proved. The function z 7→ indγ (z) is continuous in C\tr(γ) and an integer-valued continuous function on an (open) connected set must be constant which implies ii). Finally we note that Z 1 1 lim indγ (z) = lim dζ = 0, z→∞ z→∞ 2πi γ ζ − z which now yields iii). Definition 25.2. The integer indγ (z) is called the winding number or the index of γ with respect to z ∈ C\tr(γ). Example 25.3. A. Let γ be the circle |ζ − z0 | = r with our standard parametrization. Then we have indγ (z) = 1 for z ∈ Br (z0 ) and indγ (z) = 0 {

for z ∈ Br (z0 ) . B. Let γk : [0, 2πk] → C, γk (t) = z0 + reit , k ∈ N. The trace of γk is again the circle with centre z0 and radius r, i.e. ∂Br (z0 ), and for z ∈ Br (z0 ) we find Z 2πk Z 2πk ireit 1 1 i dt = k. dt = indγk (z) = indγk (z0 ) = 2πi 0 z0 + reit − z0 2πi 0 474

25

THE RESIDUE THEOREM

In the case where we use the parametrization ηk (t) = z0 + re−it , t ∈ [0, 2πk], i.e. ηk = γk−1 we find indηk (z) = indηk (z0 ) =

1 2πi

Z

2πk

(−i) dt = −k, z ∈ Br (z0 ).

0

Note that we have used in both calculations the fact that indγ (·) is constant on Br (z0 ). Further note that indγ −1 (z) = −indγk (z), for z ∈ C\∂Br (z0 ), and this is k indeed a general result. Lemma 25.4. Let γj : [aj , bj ] → C be two closed piecewise continuously differentiable curves such that γ1 (b1 ) = γ2 (a2 ). For z ∈ / tr(γ1 ) ∪ tr(γ2 ) we have (25.3) indγ1 ⊕γ2 (z) = indγ1 (z) + indγ2 (z), and indγ1 (z) = −indγ1−1 (z).

(25.4)

Proof. From our assumptions we deduce that γ1 ⊕ γ2 is a well defined closed piecewise continuously differentiable curve. Since 1 2πi

Z

γ1 ⊕γ2

1 1 dζ = ζ −z 2πi

Z

γ1

1 1 dζ + ζ −z 2πi

Z

γ2

1 dζ, ζ −z

we have proved (25.3) whereas (25.4) follows from the fact that 1 2πi

Z

γ1

1 1 dζ = − ζ −z 2πi

Z

γ1−1

1 dζ. ζ −z

For a simply closed curve we can rely on the Jordan curve theorem, Theorem 19.37, to define the interior and the exterior of its trace, where the interior is bounded and simply connected. Such a result cannot be expected for an arbitrary closed curve, see Figure 25.1 below, 475

A COURSE IN ANALYSIS

Figure 25.1 Therefore we give Definition 25.5. Let γ : I → C be a closed, piecewise continuously differentiable curve. The interior of γ is defined as  int γ := z ∈ C\tr(γ) indγ (z) 6= 0 , (25.5)

and the exterior of γ is defined as  ext γ := z ∈ C\tr(γ) indγ (z) = 0 .

(25.6)

With these definitions we find the decomposition

C = int γ ∪ tr(γ) ∪ ext γ,

(25.7)

and we have Lemma 25.6. For a closed, piecewise continuously differentiable curve γ : I → C the sets int γ and ext γ are open and ∂(int γ) ⊂ tr(γ) as well as ∂(ext γ) ⊂ tr(γ). Moreover, int γ ⊂ BR (z0 ) and C\BR (z0 ) ⊂ ext γ for some open disc such that tr(γ) ⊂ BR (z0 ). Proof. The first part is trivial since indγ is locally constant and tr(γ) is compact. Furthermore, since ∅ = 6 BR (z0 ){ is a connected set the function indγ must be constant on BR (z0 ){ , but for z → ∞ we have indγ (z) → 0, so BR (z0 ){ ⊂ ext γ which implies by (25.7) that int γ ⊂ BR (z0 ). 476

25

THE RESIDUE THEOREM

Remark 25.7. If tr(γ) is also the trace of a simply connected curve, then we have of course the Jordan curve theorem at our disposal. The following figure shows that we cannot expect ∂(ext γ) = tr(γ).

Figure 25.2 In previous chapters we often have replaced a given path of integration by another one without changing the value of the integral, see for example (24.11) when preparing the proof of Theorem 24.2. Of course we can interpret the Cauchy integral theorem, Theorem 21.4, in this sense: we replace ∂4 by γ. We want to study integrals over finite systems of curves γk , k = 1, . . . , M, in such a way that the “arithmetic” for integrals is reflected in operations in systems of curves. We start with Definition 25.8. Let G ⊂ C be a domain and let γk : Ik → G, k = 1, . . . , M, be a finite family of piecewise continuously differentiable curves. The formal sum ˜ · · · +a ˜ M γM , ak ∈ Z, γ := a1 γ1 + (25.8)

is called a chain of curves in G with integer coefficients.

Remark 25.9. A. As long as no integrals are involved, we can allow in all considerations below to work just with continuous curves. B. It is important ˜ in (25.8) has no meaning, we just use it to define to note that so far “+” a new class of objects. In particular one shall not mix up (25.8) with the operation η1 ⊕η2 for two curves with the terminal point of η1 being the initial point of η2 . We now want to introduce an operation on the set of all chains in G. Let ˜ · · · +a ˜ M γM . We agree that γ = a1 γ1 + ˜ · · · +a ˜ σ(M ) γσ(M ) ˜ · · · +a ˜ M γM = aσ(1) γσ(1) + a1 γ1 + 477

A COURSE IN ANALYSIS

˜ · · · +a ˜ M γM and η = b1 γ1 + ˜ ··· for any permutation σ ∈ SM . For γ = a1 γ1 + ˜ M γM we define +b ˜ := (a1 + b1 )γ1 + ˜ · · · +(a ˜ M + bM )γM , γ +η ˜ = η +γ, ˜ ˜ ··· which immediately yields γ +η and with −γ := −a1 γ1 + ˜ ˜ ˜ ˜ +(−a )γ we have γ +(−γ) = 0γ + · · · +0γ , and furthermore we have M M 1 M ˜ ˜ ˜ ˜ ˜ ˜ (γ +η)+τ = γ +(η +τ ) where τ = c1 γ1 + · · · +cM γM . We will write 0 := ˜ · · · +0γ ˜ M and γ −η ˜ for γ +(−η), ˜ ˜ = 0−γ. ˜ 0γ1 + in particular we write −γ ˜ · · · +a ˜ M γM and η = b1 η1 + ˜ · · · +b ˜ N ηN we define Next, for γ = a1 γ1 + ˜ = a1 γ1 + ˜ · · · +a ˜ M γM +b ˜ 1 η1 + ˜ · · · +b ˜ N ηN , γ +η

(25.9)

and we note that if γj = ηl for some j and l, then we can rewrite (25.9) as ˜ = a1 γ1 + ˜ · · · +a ˜ j−1 γj−1 +(a ˜ j − bl )γj +a ˜ j+1 γj+1+ ˜ · · · +γ ˜ M γ +η ˜ 1 η1 + ˜ · · · +b ˜ l−1 ηl−1 +b ˜ l+1 ηl+1 + ˜ · · · +b ˜ N ηN . +b With these definitions we have introduced on the set of all chains in G the structure of an Abelian group. ˜ · · · +a ˜ M γM we define its trace as union of the traces For a chain γ = a1 γ1 + of the curves γk , 1 ≤ k ≤ M, i.e. [  tr(γ) = tr(γk ), Z = l ∈ {1, . . . , M} al 6= 0 . k∈Z

Now let f : tr(γ) → C be a continuous function. We set Z

f (z) dz =

γ

M X

ak

Z

f (z) dz,

(25.10)

γk

k=1

˜ · · · +a ˜ M γM . where γ = a1 γ1 + If for example all curves γk are simply closed and G ⊂ C starshaped as well as f holomorphic on G then we have by the Cauchy integral theorem Z

γ

f (z) dz =

M X

ak

k=1

478

Z

γk

f (z) dz = 0.

25

THE RESIDUE THEOREM

˜ · · · +γ ˜ M In the case where we can form γ1 ⊕· · ·⊕γM then it follows for γ = γ1 + that Z Z f (z) dz =

f (z) dz.

γ1 ⊕···⊕γM

γ

(0)

For a curve η1 we denote its initial point by η1 and its terminal point by P (1) ˜ · · · +a ˜ M γM we can form η1 . For a chain γ = a1 γ1 + (0) ak which counts k,z=γk how often z, taking multiplicity into P account, is an initial point of some curve γk of the chain γ. Analogously k,z=γ (1) ak counts, again multiplicity taken k into account, how often z is a terminal point of γ. For z ∈ C not an initial point (terminal point) P of any of the  curves γk of the chain we have of course P (0) ak = 0 (1) ak = 0 . k,z=γ k,z=γ k

k

˜ · · · +a ˜ M γM in G closed or a Definition 25.10. We call a chain γ = a1 γ1 + cycle in G if for all z ∈ C we have X X ak = ak , (25.11) (0)

k,z=γk

(1)

k,z=γk

i.e. every z ∈ C is as often initial as terminal point (multiplicity taken into account) of some γk . Example 25.11. In the following, by curve we always mean a piecewise continuously differentiable curve. A. Every curve is of course a chain and every simply closed curve is a cycle. Moreover, if γ is a curve for which every point is an m-fold point in the sense of Definition II.7.12.B, then γ is a cycle. The last example has some further interesting features. Consider the simply closed curve γ : [0, 1] → G ⊂ C, and now extend γ periodically to γ (m) : [0, m] → G, γ (m) (t + 1) = γ(t), t ∈ [0, m − 1]. The curve γ (m) is a cycle in G. The chain mγ we may write ˜ · · · +γ ˜ and we may interpret this as running m-times through as mγ = γ + γ, which is the same as running once through γ (m) or through γ ⊕ · · · ⊕ γ (m terms). ˜ −1 is a cycle. B. If γ is a curve, hence a chain, then γ +γ (1) (0) C. Let γk , 1 ≤ k ≤ M, be curves such that γk+1 = γk , k = 1, . . . , M − 1, (0) (1) ˜ · · · +γ ˜ M is a cycle. and γM = γ1 . Then γ1 + As a first application we can now extend Theorem 20.22. 479

A COURSE IN ANALYSIS

Theorem 25.12. A continuous function f : G → C defined on the region G ⊂ C admits a primitive in G if and only if for all cycles γ in G we have Z f (z) dz = 0. (25.12) γ

˜ · · · +a ˜ M γM a cycle. It follows Proof. Let F be a primitive of f and γ = a1 γ1 + that Z Z M M      X X (0) (1) f (z) dz = ak f (z) dz = ak F γk − F γk γ

k=1

=

γk



k=1



X X X  ak  F (z), ak −  z

(1)

k,z=γk

(0)

k,z=γk

where the summation is over all z ∈ C which are initial or terminal points of some γk . Since γ is a cycle the inner sum is however equal to zero. To see the converse direction, we just need to note that simply closed piecewise continuously differentiable curves are cycles and therefore Theorem 20.22 yields the existence of a primitive. To proceed further we need ˜ · · · +a ˜ M γM be a cycle. For z ∈ Definition 25.13. A. Let γ = a1 γ1 + / tr(γ) we define its index as Z dζ 1 . (25.13) indγ (z) := 2πi γ ζ − z B. We call a cycle γ in a domain G ⊂ C null-homologous if for every z ∈ G{ the index of γ at z is zero, i.e. indγ (z) = 0 for z ∈ G{ .

(25.14)

C. If the difference of two cycles γ and η in G is null-homologous we call γ and η homologous. ˜ · · · +a ˜ M γM we have Since for every cycle γ = a1 γ1 + 1 2πi

Z

M

γ

X dζ = ak ζ −z k=1 480

Z

γk

dζ ζ −z

25

THE RESIDUE THEOREM

we deduce that ind−γ ˜ (z) = −indγ (z), and indγ (z) =

M X

ak indγk (z),

k=1

z∈ / tr(γ) z∈ / tr(γ).

(25.15)

(25.16)

˜ that From (25.16) we deduce further for two cycles γ and η and z ∈ / tr(γ +η) indγ +η ˜ (z) = indγ (z) + indη (z).

(25.17)

The formulae (25.15) - (25.17) allow us to immediately transfer results holding for the index of a curve to the index of a cycle. Proposition 25.14. The index of a cycle is an integer-valued function and on every connectivity component C\ tr(γ) it is constant. Moreover, if R > 0 is such that tr(γ) ⊂ BR (0), then indγ | { is equal to the zero function, i.e. BR (0)

{

indγ (z) = 0 for z ∈ BR (0) . ˜ · · · +a ˜ M γM , and consider the cycle −γ ˜ = Let γ be a cycle, γ = a1 γ1 + ˜ ˜ (−a1 )γ1 + S · · · +(−aM )γM . Both cycles have the same trace namely tr(γ) = ˜ = M tr(−γ) k=1 tr(γ), and for z ∈ C\ tr(γ) we find ind−γ ˜ (z) = =

M X

(−ak )indγk (z) =

k=1

M X

M X k=1

ak ind−γ ˜ k (z)

ak indγ1−1 (z) = indγ −1 (z),

k=1

−1 ˜ · · · +a ˜ M γM with γ −1 = a1 γ1−1 + . In this sense we can identify for calculating ˜ with γ −1 . indices of cycles −γ

Example 25.15. A. For γ given as γ(t) = z0 + reikt , t ∈ [0, 2π], we find for / tr(γ) = ∂Br (z0 ), that indγ (z) = k since indγ (z0 ) = k as is z ∈ Br (z0 ), z ∈ shown by a simple calculation. B. Let γj , j = 1, 2, be the standard parametrization of the circle |ζ − zj | = rj ˜ and suppose Br1 (z1 ) ⊂ Br2 (z2 ). We consider the cycle γ := γ2 +(−γ 1 ) and −1 . Therefore we find for z ∈ tr(γ) we know that indγ = indγ2 +γ ˜ 1 i) indγ (z) = 0 for z ∈ Br{2 (z2 ); 481

A COURSE IN ANALYSIS

ii) indγ (z) = indγ2 (z) + indγ1−1 (z) = 1 for z ∈ Br2 (z2 )\Br1 (z1 ), since {

in this case z ∈ Br1 (z1 ) and hence indγ1−1 (z) = −indγ1 (z) = 0, but indγ2 (z) = 1. iii) indγ (z) = indγ2 (z) − indγ1 (z) = 0 for z ∈ Br (z1 ), since now indγ2 (z) = indγ1 (z) = 1. C. From part B we deduce that in C\{z0 } the cycles γ1 , γ1 (t) = z0 + r1 eit , t ∈ [0, 2π], and γ2 , γ2 (t) = z0 + r2 eit , t ∈ [0, 2π], r1 > r2 > 0, are homologous ˜ 2 = γ1 +γ ˜ 2−1 is null-homologous in C\{z0 }. since γ1 −γ Lemma 25.16. On the set of all cycles in G homology is an equivalence relation. Proof. Clearly γ is homologous to itself since indγ −γ ˜ (z) = indγ +γ ˜ −1 (z) = 0, z ∈ G{ . Further, since indγ −η ˜ (z) = indγ +η ˜ −1 (z) = indγ (z) − indη (z) = − (indη (z) − indγ (z)) = −indη−γ ˜ (z), we deduce indγ −η homology ˜ (z) = 0 if and only if indη−γ ˜ (z) = 0, i.e. { ˜ ˜ be is a symmetric relation. Now let z ∈ G and γ −η as well as η −τ null-homologous cycles, i.e. indγ +η ˜ −1 (z) = 0 and indη+τ ˜ −1 (z) = 0. For indγ −τ (z) = ind (z) we find −1 −1 ˜ ˜ γ +τ indγ +τ ˜ −1 = indγ (z) − indτ (z) = indγ (z) − indη (z) + indη (z) − indτ (z) = indγ +η ˜ −1 (z) = 0, ˜ −1 + indη+τ i.e. homology is transitive and the lemma is proved. We are now in a position to prove one of the main results in complex analysis, the general Cauchy integral formula for null-homologous cycles. Theorem 25.17. Let γ be a null-homologous cycle in the domain D ⊂ C and f : D → C a holomorphic function. For all k ∈ N0 and all z ∈ D\ tr(γ) we have Z f (ζ) k! (k) dζ. (25.18) indγ (z)f (z) = 2πi γ (ζ − z)k+1 482

25

THE RESIDUE THEOREM

First we want to draw some consequences of Theorem 25.17 and then we will provide the proof. Theorem 25.18. Under the assumptions of Theorem 25.17 we find Z f (ζ) dζ = 0. (25.19) γ

Proof. Let z1 ∈ D\ tr(γ) and consider h(z) = f (z)(z − z1 ). This function is holomorphic on D and h(z1 ) = 0. Now, using (25.18) we get Z Z 1 h(ζ) 1 dζ = f (ζ) dζ. 0 = indγ (z1 )h(z1 ) = 2πi γ ζ − z1 2πi γ Theorem 25.19. For two homologous cycles γ and η in D and every holomorphic function f : D → C we have Z Z f (ζ) dζ = f (ζ) dζ. (25.20) γ

η

˜ is null-homologous and therefore we have by TheProof. By assumption γ −η orem 25.17 Z Z Z 0= f (z) dz = f (z) dz − f (z) dz. ˜ γ −η

γ

η

Note that (25.20) is exactly the type of statement we are searching for: we replace the closed curve (the cycle) and the integral remains unchanged for all holomorphic functions. Proof of Theorem 25.17. (Following [26]). First we note that if (25.18) holds for k = 0 it holds for all k ∈ N0 . Indeed we can differentiate in Z 1 f (ζ) indγ (z)f (z) = dζ, (25.21) 2πi γ (ζ − z) on the right hand side k-times to obtain the right hand side in (25.18) and on the left hand side we only need to note that indγ (z) is constant on connectivity components. Thus we aim to prove (25.21). Since Z 1 f (z) indγ (z)f (z) = dζ, 2πi γ ζ − z 483

A COURSE IN ANALYSIS

we can rewrite (25.21) as Z f (ζ) − f (z) dζ = 0, ζ −z γ

z ∈ D\ tr(γ).

We first study the function g(ζ, z) :=

(

f (ζ)−f (z) , ζ−z 0

f (z),

ζ 6= z, ζ=z

as function defined on D ×D. For (ζ0 , z0 ) ∈ D ×D and ζ0 6= z0 the function g is continuous as function of two variables at (ζ0 , z0 ). Now let ζ0 = z0 and δ > 0 to the determined later. We study the function (ζ, z) 7→ g(ζ, z) − g(ζ0, z0 ) on Bδ (z0 ) × Bδ (z0 ) where we assume that Bδ (z0 ) ⊂ D. In the case that ζ = z (∈ Bδ (z0 )) we note that g(ζ, z) − g(ζ0, z0 ) = g(z, z) − g(z0 , z0 ) = f 0 (z) − f 0 (z0 ), whereas for ζ 6= z we have f (ζ) − f (z) − f 0 (z0 ) g(ζ, z) − g(ζ0, z0 ) = g(ζ, z) − g(ζ0, z) = ζ −z Z 1 = (f 0 (w) − f 0 (z0 )) dw. ζ − z [z,ζ] Since f is holomorphic, f 0 is continuous, in particular at z0 . Given  > 0 we can now choose δ > 0 in such a way that for w ∈ Bδ (z0 ) it follows that |f 0(w) − f 0 (z0 )| < . This implies in the case where ζ = z that |g(z, z) − g(z0 , z0 )| = |f 0(z) − f 0 (z0 )| < , and for ζ 6= z we find |g(ζ, z) − g(ζ0 , z0 )| ≤

1 |ζ − z| sup |f 0 (w) − f 0 (z0 )| < . |ζ − z| w∈[z,ζ]

Thus we have proved that g is on D × D continuous. Since tr(γ) is compact we immediately find that the function Z h0 (z) := g(ζ, z) dζ γ

484

25

THE RESIDUE THEOREM

is on D continuous. For a triangle 4 ⊂ D with anticlockwise orientated boundary ∂4 ⊂ D we find   Z Z Z Z Z h0 (z) dz = g(ζ, z) dζ dz = g(ζ, z) dz dζ = 0, ∂4

∂4

γ

γ

∂4

since for all ζ fixed z → 7 g(ζ, z) is holomorphic in D\{ζ}, but since g is continuous on D, this is a removable singularity, hence we have indeed by Goursat’s theorem Z h0 (z) dz = 0

∂4

˜ := for every such a triangle implying that h0 is holomorphic in D. Let D { ˜ {z ∈ C | indγ (z) = 0}. Since γ is null-homologous it follows that D ⊂ D ˜ ˜ and therefore we have D ∪ D = C. On D ∩ D we have Z 0 f (ζ) dζ h0 (z) = γ ζ −z ˜ { we conclude that and since tr(γ) ⊂ D Z f (ζ) dζ h1 (z) := ζ −z γ ˜ holomorphic. Thus by is on D ( h0 (z), z ∈ D, h(z) := ˜ h1 (z), z ∈ D an entire function is given and if tr(γ) ⊂ BR (0) ⊂ BR (0) ⊂ D, then we have {

for z ∈ BR (0) the representation

h(z) =

Z

γ

f (ζ) dζ. ζ −z

(25.22)

We can estimate h(z) in (25.22) according to |h(z)| ≤

1 lγ kf k∞, tr(γ) dist(z, γ) 485

(25.23)

A COURSE IN ANALYSIS

P ˜ · · · +a ˜ M γM we have lγ = M where for γ = a1 γ1 + k=1 |ak |lγk . On the compact set BR (0) the holomorphic function h is bounded and by (25.23) it is {

also bounded on BR (0) . Hence h is a bounded entire function and by Liouville’s theorem h is a constant. Our aim is to proveR that this constant is zero, which will imply h0 is identically zero, hence γ g(ζ, z) dζ = 0 for z∈ / tr(γ) and therefore (25.18) will follow. Again the compactness of tr(γ) {

helps. For a sequence (zk )k∈N of complex numbers such that zk ∈ BR (0) and limk→∞ |zk | = ∞ the estimate (25.23) implies limk→∞ |h(zk )| = 0. We can prove the residue theorem for null-homologous cycles.

Theorem 25.20. Let G ⊂ C be a region and f a function which is on G holomorphic with the exception of isolated singularities. For every nullhomologous cycle γ in G on the trace of which are no singularities of f we have Z X 1 f (ζ) dζ = indγ (z) res(f, z). (25.24) 2πi γ z∈G Remark 25.21. Since tr(γ) is compact, hence tr(γ) ⊂ BR (0) for some R > 0, outside of the compact set BR (0) we have indγ (z) = 0 which implies that in the sum of the right hand side of (25.24) only finitely many terms do not vanish, i.e. the sum is a finite one. Proof of Theorem 25.20. Let γ be a null-homologous cycle in G. We decompose the set sing(f ) of singularities of f according to   sing(f ) = sing(f, γ) ∪ sing f, γ { where and

 sing(f, γ) := z ∈ sing(f ) indγ (z) 6= 0

 sing(f, γ { ) := z ∈ sing(f ) indγ (z) = 0 .

By our assumptions it follows that sing(f, γ) is a finite set, say sing(f, γ) = {z1 , . . . , zM }. For zk ∈ sing(f, γ) we denote the principal part of its Laurent expansion about zk by fzk . On C\{zk } the function fzk is holomorphic and 486

25

THE RESIDUE THEOREM


P { therefore f − M k=1 fzk is holomorphic in G\ sing f, γ
. Since γ is nullhomologous in G it is also null-homologous in G\ f, γ { and by Theorem 25.17 (or Theorem 25.18) we find ! Z M X fzk (ζ) dζ f− 0= γ

or

Z

k=1

M Z X

f (ζ) dζ = γ

k=1

fzk (ζ) dζ. γ

By compactness of tr(γ) the Laurent series fzk (z) = verges uniformly on tr(γ) and consequently 1 2πi

Z

fzk (ζ) dζ =

γ

∞ (k) Z X a−l l=1

(k)

= a−1

2πi

1 2πi

Z

γ

γ

P∞

l=1

(k)

a−l (z − zl )−l con-

(ζ − zk )−l dζ 1 dζ = indγ (zk ) res(f, zk ) ζ − zk

implying the theorem. With Theorem 25.17 and Theorem 25.20 we have proved some of the main results of classical complex analysis in their most general form. Of course there is a need to explore the notion of homology, in particular of nullhomology in more detail. In particular the relation of simply connected sets, i.e. sets with a trivial first fundamental group, and null-homology is of interest. This leads to more topological discussions and we will postpone this to our final volume. We will now apply the residue theorem to evaluate certain integrals. For this we need first of all an easier way to find residues. By its very definition the residue of f at z0 is the coefficient c−1 of the principal part of the Laurent expansion of f about z0 , provided z0 is an isolated singularity of f and we have Z 1 res(f, z0 ) = c−1 = f (ζ) dζ, (25.25) 2πi |ζ−z0|=ρ for every circle |ζ − z| = ρ, such that z0 is the only singularity of f in Bρ (z0 ). Since we want to use residues to calculate integrals we are looking for a different way to find res(f, z0 ). Note that finding the Laurent expansion 487

A COURSE IN ANALYSIS

might also be troublesome. Here comes an easier way for poles. First suppose that the pole is of first order. Then it follows that in some annulus A0,r (z0 ) we have ∞ X f (z) = ck (z − z0 )k (25.26) k=−1

and (z − z0 )f (z) =

∞ X

k=−1

ck (z − z0 )k+1

implying lim (z − z0 )f (z) =

z→z0

∞ X

k=−1

ck lim (z − z0 )k+1 = c−1 . z→z0

(25.27)

In the case that z0 is a pole of order n we find f (z) =

∞ X

k=−n

ck (z − z0 )k

(25.28)

and therefore (z − z0 )n f (z) =

∞ X

k=−n

ck (z − z0 )k+n =

∞ X k=0

ak−n (z − z0 )k .

This is a holomorphic function in Br (z0 ) and therefore we get ∞ X dn−1 n ((z − z0 ) f (z)) = ak−n (n − 1)!(z − z0 )k−n+1 , n−1 dz k=n−1

implying Proposition 25.22. Let the holomorphic function f have a pole of order n at z0 . For the residue res(f, z0 ) we find dn−1 res(f, z0 ) = lim n−1 z→z0 dz



 1 n (z − z0 ) f (z) . (n − 1)! 488

(25.29)

25

THE RESIDUE THEOREM

z Example 25.23. A. The rational function R(z) = (z−1)(z+1) 2 has the pole z0,1 = 1 of order 1 and the pole z0,2 = −1 of order 2. The corresponding residues are   1 z = res(R, 1) = lim (z − 1) 2 z→1 (z − 1)(z + 1) 4

and



  1 d 1 z 2 res(R, −1) = lim =− . (z + 1) 2 z→−1 1! dz (z − 1)(z + 1) 4
1 B. The functions z 7→ exp − z has an essential singularity at z0 = 0, so we cannot apply Proposition 25.22. However we know its series about PLaurent P ∞ k 1 −k k 1 −k (−1) z with principal part z . This (−1) z0 = 0 which is ∞ k=1 k=0 k! k! implies that     1 , 0 = −1. res exp − z A further useful rule with trivial proof is Proposition 25.24. Let the holomorphic f have a pole of first order at z0 and let g be holomorphic in a neighbourhood of z0 with g(z0 ) 6= 0. In this case we have res((f · g), z0 ) = g(z0 ) res(f, z0 ). (25.30)

Proof. First we note that in this case f · g has at z0 a pole of first order and it follows that lim (z − z0 )(f (z)g(z)) = lim g(z) lim (z − z0 )f (z) = g(z0 ) res(f, z0 ).

z→z0

z→z0

z→z0

We now will evaluate quite a few integrals of (mainly) real-valued functions defined on an interval using the residue theorem. We do not long for a systematical treatment which can be found in [26] or [67], to mention just some sources, but we want to provide some of the basic ideas. Convention: in all examples, if not otherwise stated, circles or parts of a circle are parametrized in the standard way, i.e. |z − z0 | = ρ is parametrized by γ(t) = z0 + ρeit , t ∈ [0, 2π] (or a subinterval). Furthermore, whenever we consider a simply closed curve its parametrization is anti-clockwise oriented. Moreover, as before [z1 , z2 ] stands for the line segment connecting z1 with z2 usually parametrized as z1 + t(z2 − z1 ), t ∈ [0, 1]. 489

A COURSE IN ANALYSIS

Example 25.25. A. We want to show that Z

2π 0

! √ 1 + (2 − 3)4 √ √ −8 . 2 3(2 − 3)2

cos 2ϑ dϑ = π 2 − cos ϑ z+z −1 2

2

(25.31)

−2

and cos 2ϑ = z −z and it follows that 2  −1 Z 2π Z cos 2ϑ 1 z + z −1 z 2 + z −2 2− dϑ = dz 2 − cos ϑ 2 2 iz 0 |z|=1 Z Z i(z 4 + 1) i(z 4 + 1) √ √ dz. = dz = 2 2 2 3)(z − 2 + 3) |z|=1 z (z − 4z + 1) |z|=1 z (z − 2 −

With z = eiϑ we find cos ϑ =

On the circle |z| = 1 the integral has no singularity and in the interior of |z| = 1, i.e. in√B1 (0), we have a pole √ of order 2 at z1 =√0 and a pole of order 1 at z2 = 2 − 3. Note that 2 + 3 > 1, so z3 = 2 + 3 is a pole of order 1 outside of B1 (0). The residues can be determined by using Proposition 25.22 4 +1) and we find with f (z) = z 2 (zi(z2 −4z+1)     d 1 2 d i(z 4 + 1) res(f, 0) = lim = 4i z f (z) = lim z→0 dz z→0 dz 1! z 2 − 4z + 1 and res(f, 2 −

√

3) = =

which yields Z

2π 0

lim√ (z z→2− 3

−2+

√

3)f (z) = lim√ z→2− 3 !

√ 1 + (2 − 3)4 √ − √ 2 3(2 − 3)2



i(z 4 + 1) √ z 2 (z − 2 − 3)

i

 √  cos 2ϑ dϑ = 2πi res(f, 0) + res(f, 2 − 3) 2 − cos ϑ √ 1 + (2 − 3)4 √ − 8π. =π √ 2 3(2 − 3)2

B. The following holds for a > |b| Z 2π dϕ 2π . =√ a + b sin ϕ a2 − b2 0 490



25

THE RESIDUE THEOREM −1

Again we substitute z = eiϕ and now with sin ϕ = z−z2i we arrive at Z Z Z 2π 1 2 dϕ

= dz. dz = 2 z 2 + 2ai − b z−z −1 a + b sin ϕ b |z|=1 iz a + b |z|=1 0 2i

The integrand has poles of first order at

a i√ 2 z1,2 = − i ± a − b2 b b and |z1 | < 1 whereas |z2 | > 1 which follows from |b| < a. With h(z) = 2 we find b2 z 2 +2ai−b Z 2π Z dϕ 2 = dz 2 2 a + b sin ϕ 0 |z|=1 b z + 2ai − b = 2πi res(h, z1 )

= 2πi lim ((z − z1 )h(z)) = 2πi z→z1

2π . =√ a2 − b2



1 √ i a2 − b2



The integrals in Example 25.25 are of the type Z 2π R(cos ϕ, sin ϕ) dϕ 0

where R(x, y) is a rational function in two real variables. With the substitution z = eiϕ , ϕ ∈ [0, 2π], they transform to  Z 2π 
1 iϕ
1 iϕ −iϕ −iϕ R dϕ e +e e −e , 2 2i 0      Z Z 1 dz 1 1 1 1 = z+ , z− = R h(z) dz. i |z|=1 2 z 2i z z |z|=1 Applying the residue theorem we find Z 2π X R(cos ϕ, sin ϕ) dϕ = 2π res(h, z) 0

(25.32)

|z| 1

Z

[−R,R]⊕κ+ R

  1 1 dz = 2πi res ,i 1 + z2 1 + z2   1 = 2πi lim (z − i) = π. z→i (z − i)(z + i)

On the other hand we find Z

[−R,R]⊕κ+ R

Z

Z 1 1 dz + dz 2 2 + [−R,R] 1 + z κR 1 + z Z R Z π 1 Rieit = dx + dt. 2 it 2 −R 1 + x 0 1 + (Re )

1 dz = 1 + z2

For R → ∞ the first integral tends to integral we use the estimate Rieit 1 + (Reit )2 =

R∞

1 −∞ 1+x2

R (1 +

R4

+

1

2R2 cos 2t) 2

≤

dx, while for the second

R (1 +

R4

−

which holds for R > 1. This estimate implies however

lim

R→∞

Z

0

R

Rieit dt = 0. 1 + (Reit )2 492

1

2R2 ) 2

=

R −1

R2

25

THE RESIDUE THEOREM Im z κ+ R

0

−R

Re z

R

Figure 25.3 Example 25.27. We claim Z ∞ cos x π −α e , α > 0. (25.33) dx = 2 2 α +x 2α 0 Due to the symmetry of the integrand we need to prove that Z Z ∞ Z ∞ eix 1 ∞ cos x 1 π −α cos x Re e . dx = dx = dx = 2 2 2 2 2 2 α +x 2 −∞ α + x 2 2α −∞ α + x 0

For R > α we consider the simply closed, piecewise continuously differentiable curve γR := [−R, R] ⊕ [R, R + iR] ⊕ [R + iR, −R + iR] ⊕ [−R + iR, −R] see Figure 25.4 Im z

−R + iR

−R

b

iR

R + iR

0

R

Re z

Figure 25.4 iz

e With h(z) = (z−iα)(z+iα) it follows that Z Z π eiz eiz dz = dz = 2πi res(h, iα) = e−α , 2 2 α γR (z − iα)(z + iα) γR α + z

493

A COURSE IN ANALYSIS

where we used that iα is the only pole (and of first order) of h in the interior of γR and therefore eiz e−α = . z→iα z + iα 2iα

res(h, iα) = lim ((z − iα)h(z)) = lim z→iα

Now we note that Z

R

−R

eiz dz = α2 + z 2

or lim Re

R→∞

Z

R

−R

Z

R −R

cos x + i sin x dx, α2 + x2

eiz dz = α2 + z 2

Z

∞

−∞

cos x dx. α2 + x2

We to show that for R → ∞ the remaining three integrals, i.e. R want eiz dz, j = 2, 3, 4, vanish. For γ2 , γ2 (t) = R + itR, t ∈ [0, 1], we γj α2 +z 2 have Z 1 Z ei(R+itR) eiz dz = (iR) dt. 2 2 2 2 0 α + (R + itR) γ2 α + z Since for R > α it follows that |α2 + (R(1 + it)2 )| ≥ |R(1 + it)|2 − α2 = R2 (1 + t2 ) − α2 ≥ R2 − α2 , we find Z

Z 1 Re−tR eit ≤ dz dt 2 2 2 2 2 γ2 α + z 0 |α + R (1 + it) | Z 1 1 − e−R Re−tR dt = , ≤ 2 2 R2 − α 2 0 R −α R eiz which tends to 0 for R → ∞. Analogously we find that γ4 α2 +z 2 dz tends to 0 as R → ∞. For the final integral we have Z

γ3

eiz dz = α2 + z 2

Z

1

ei(R+iR−2tR) (−2R) dt 2 2 0 α + (R + iR − 2tR) Z 1 ei(R−2tR) e−R R = −2 dt. 2 2 2 0 α + R (1 − 2t + i) 494

25

THE RESIDUE THEOREM

Now we use that |R2 (1 − 2t + i)2 + α2 | ≥ R2 |1 − 2t + i|2 − α2 ≥ R2 − α2 to obtain Z Z 1 Re−R Re−R eiz ≤ 2 dz dt = 2 → 0 as R → ∞. 2 2 2 2 R2 − α 2 0 R −α γ3 α + z

Hence we have proved lim Re

R→∞

Z

γR

eiz dz = α2 + z 2

Z

∞

−∞

π cos x dx = e−α 2 2 α +x α

implying (25.33). The insight we get from Example 25.26 and 25.27 is the following: let g : I → R be a function defined on an unbounded interval which is improperly Riemann integrable. Suppose that there exists a region G ⊂ C such that with the exception of finitely many isolated singularities z1 , . . . , zN the function g has a holomorphic extension to G. For simplicity we denote this extension again by g. Suppose further that for every R ≥ R0 we can find a simply closed piecewise continuously differentiable curve γR such that the interior of γR contains all the singularities of g and tr(γR ) ⊂ G. Moreover assume that Z Z lim g(z) dz = g(x) dx. R→∞

I

γR

In this case we have Z

I

g(x) dx = 2πi

∞ X

res(g, zj ).

(25.34)

k=1

Of course, the problem is to find a good family of curves γR , but classical books in complex analysis, see [36] or [93] just to mention two of them, provide a lot of results in this direction. It is also possible to use the residue theorem in some cases for integrals having a singularity on the interval or the endpoints of the interval of integration. 495

A COURSE IN ANALYSIS

Example 25.28. Using the residue theorem we will prove Z ∞ π sin x dx = . x 2 0

(25.35)

The integral suggests to look at an integral over some curve for the function iz z 7→ ez . While limx→0 sinx x = 1 assures that x 7→ sinx x is at x = 0 a continuous z function, for z 7→ ez the value z0 = 0 is a pole and it is located at the boundary of [0, ∞) (or (0, ∞)) and hence it causes at first glance a problem. + −1 + For  > 0 and R > 0 we consider the curve γ,R = κ+ R ⊕ [−R, −] ⊕ (κ ) + [, R], see Figure 25.5, where as before κR is the upper half-circle with centre iϕ 0 and radius R parametrized as κ+ R (ϕ) = Re , ϕ ∈ [0, π]. Im z κ+ R

−1 (κ+  )

−

−R



Re z

R

Figure 25.5 iz

Since z → 7 ez is in C\{0} holomorphic and 0 is not a point on γ,R we find R iz e dz = 0, or γ,R z Z



R

eix dx + x

leading to

Z

R



i.e.

Z



∞

Z

−

−R

Z Z eix eiz eiz dx = − dz − dz −1 x z κ+ (κ+ ) z R Z

eix − e−ix dx = x

1 sin x dx = x 2i

κ+ 

Z

κ+ 

eiz dz − z

496

Z

eiz dz − z

κ+ R

Z

κ+ R

eiz dz z

eiz dz z

!

.

25

THE RESIDUE THEOREM

Now we observe that 1 2i

Z

κ+ 

Z π ieiϕ e 1 eiz ieiϕ dϕ dz = z 2πi 0 eiϕ Z 1 π ieiϕ = e dϕ 2 0

and in the limit  → 0 we get 1 lim →0 2π

Z

z dz = π. z

1 lim R→∞ 2i

Z

eiz dz = 0. z

κ+ 

It remains to prove that

κ+ R

For this we note Z Z iϕ 1 eiz 1 π eiRe iϕ iRe dϕ dz = iϕ + 2i 0 Re 2i κR z Z π iR cos ϕ−R sin ϕ e 1 iϕ ≤ Re dϕ iϕ 2 0 Re Z Z π 1 −R sin ϕ = e dϕ = e−R sin ϑ dϑ, π 2 0 2 where we used the symmetry of the sine function with respect to the axis passing through π2 and being orthogonal to the ϑ-axis. Using again the for 0 ≤ ϑ ≤ π2 , see Example 21.6, we find estimate sin ϑ ≥ 2ϑ π Z

0

π 2

−R sin ϑ

e

dϑ ≤

Z

π 2

ϑ

e−2R π dϑ =

0

and we have indeed proved that 1 lim R→∞ 2i

Z

κ+ R

ez dz = 0 z

implying eventually (25.35). 497

π (1 − e−R ) 2R

A COURSE IN ANALYSIS

Finally we want to discuss an integral of the type Z 1 √ e−ixξ g(x) dx. 2π R

(25.36)

We will learn in Part 8 that (25.36) is the Fourier transform of g. In Example II.22.19 we have already seen that Z Z 2 ξ2 x2 1 1 −ixξ − x2 √ e e cos xξ e− 2 dx = e− 2 . dx = √ 2π R 2π R Now we want to find the Fourier transform of f (x) =

1 . x2 +a2

1 Example 25.29. The Fourier transform of f (x) = x2 +a 2 , a > 0, is the function r −|ξ|a Z 1 πe 1 −ixξ √ . (25.37) dx = e 2 2 x +a 2 a 2π R −ixξ

−izξ

We can extend fξ (x) := xe2 +a2 , x ∈ R, to the function fξ (z) = ze2 +a2 = e−izξ which is holomorphic in C\{−ia, ia} and at z1 = ia, z2 = −ia (z−ia)(z+ia) we have poles of first order. However we have a problem with the growth of z 7→ e−izξ depending on ξ. With z = x + iy we find e−izξ = eyξ e−ixξ . In order to proceed further we must note that eyξ need not be bounded, but it will be so long as yξ < 0. Therefore we discuss two cases, first let ξ < 0 and y = Im z > 0. In the upper half plane Im z > 0 we have the pole ia and we choose the curve γR := [−R, R] ⊕ κ+ R , R > ia. It follows that Z Z √ e−izξ e−izξ 1 1 √ √ 2πi res(fξ , ia) dz = dz = 2π γR z 2 + a2 2π γR (z − ia)(z + ia) and

 res(fξ , ia) = lim (z − ia) z→ia

e−izξ (z − ia)(z + ia)

or, for ξ < a



=

eξa e−|ξ|a = 2ia 2ia

√ Z e−izξ 1 π −|ξ|a √ √ e . dz = 2 + a2 z 2π γR 2a On the other hand we find Z Z R −ixξ Z 1 1 1 e e−izξ e−izξ √ √ √ dz = dx + dz 2π γR z 2 + a2 2π −R x2 + a2 2π κ+R z 2 + a2 498

25

THE RESIDUE THEOREM

and further with arguments as in Example 25.28 Z Z π e−iRξ cos ϕ+ξR sin ϕ e−izξ iϕ = dz iRe dϕ κ+R z 2 + a2 (Reiϕ )2 + a2 0 Z π Z π −|ξ|R sin ϕ 2 ϑ R e R dϕ ≤ 2 e−2|ξ|R π dϑ ≤ 2 2 2 R −a R −a 0 0
1 R −|ξ|R · , 1 − e = 2 R − a2 2|ξ|R i.e. lim

Z

Z

e−ixξ dx = x2 + a2

R→∞

κ+ R

e−izξ dz = 0, z 2 + a2

implying for ξ < 0 that 1 √ 2π

R

r

π e−|ξ|a . 2 a

For ξ > 0 we choose the pole −ia and a similar calculation now yields 1 √ 2π

Z

R

e−ixξ dx = x2 + a2

r

π e−ξa = 2 a

r

π e−|ξ|a 2 a

and (25.37) is proved. This results should be compared with Example 25.27. In Part 8 we will find further Fourier transforms by applying the residue theorem.

Problems 1. Let K ⊂ C be a convex set such that ∂K is a polygon with vertices A1 , . . . , AN . Prove that we can interpret ∂K as a cycle. 2. In the following all line segments are parametrized with the help of the unit interval [0, 1]. Consider the following figure 499

A COURSE IN ANALYSIS

C F

E

D

B

A Why can we interpret this figure as the trace of a chain? Is this chain a cycle?

3. Consider the following figure in G = B2 (z0 ):

G = B2 (z0 )

C

B b

√ 2 2

z0

D

κ√2 (z0 )

A

where ABCD = tr(η), η = [A, B]⊕[B, C]⊕[C, D]⊕[D, A]. Are κ√2 (z0 ) and η cycles? Are they homologous? (For all curves appearing in the figure we use our standard parametrizations.)

4. Let ∆ = ∆(A, B, C) ⊂ C be the open triangle with boundary ABC = tr(γ) = tr([A, B] ⊕ [B, C] ⊕ [C, A]), see the figure below. 500

25

THE RESIDUE THEOREM

C

B A For z ∈ C \ ∂∆ find indγ (z). 5. Denote by τa : C → C, τa (z) = z + a, a ∈ C, the translation by a. Let g : U \ {z0 } → C be a holomorphic function. Define the function h := g ◦ τa and prove that res(g, z0 ) = res(g ◦ τa , z0 − a). 6. Find the residue of a) f (z) =

z 3 −z 2 +2z (z+1)2 (z 2 +4)

b) g(z) =

e2z sin2 z

at z1 = −1 and z2,3 = ±2i;

at z = kπi, k ∈ Z.

7. Prove the following: PM PN l k Proposition. Let p(z) = l=0 bl z be two k=0 ak z and q(z) = polynomials with M ≥ N + 2. Suppose that q has no real zeroes and c1 , . . . , ck are the poles of f := pq in the upper plane Im z > 0. Then we have Z ∞ K X f (x)dx = 2πi Res(f, cj ). −∞

8. Find the integral

j=1

Z

0

∞

x6

1 dx. +1

9. Show that for k ∈ N we have Z ∞ π (2k − 2)! 1 ds = 2k−2 . 2 k 2 ((k − 1)!)2 −∞ (s + 1) 10. For µ, ν ∈ Z, 0 ≤ µ < ν, verify Z ∞ t2µ π dt = π . 2ν ) ν sin((2µ + 1) 2ν −∞ 1 + t 501

A COURSE IN ANALYSIS

11. Use formula (25.33) to find for all r ∈ R \ {1, −1} the integral Z 2π dϕ . 1 − 2r cos ϕ + r2 0 12. For 0 < α < 1 prove that Z

∞

π xα dx = . 1+x sin απ

0

Hint: consider the following curve γ and use the fact that for x ∈ [a, b] ⊂ R,  > 0 and β ∈ R we have lim(x + i)β = xβ and lim(x − i)β = xβ e2πiβ . →0

→0

y

C R

D



H −1

r

0

A

B

G

F

x

E

13. Use the residue theorem to prove for η ∈ R that Z ∞ 1 e−iyη 1 p π
dx = pπ
. √ cosh y η 2π −∞ cosh 2 2 Hint: first transform the identity to Z ∞ −2πixξ e 1 dx = . cosh πξ −∞ cosh πx 502

25

THE RESIDUE THEOREM

Now integrate along the following curve: Im z

2i b

3i 2 i 2

−R

0

503

γR R

Re z

26

The Γ-function, the ζ-function and Dirichlet Series

This chapter is devoted to some special functions which are of central importance in many branches of mathematics. In Volume I we encountered the Γ-function and related functions such as the beta-function, but on that occasion we considered these functions as real-valued depending on real variables. Now we want to study them as complex-valued functions of a complex variable. We start with defining for a > 0 the function az by z 7→ az := ez ln a .

(26.1)

The usual rules for dealing with exponentials still apply and further we have d z a = ln aez ln a = (ln a)az . dz

(26.2)

We also note that for z = x + iy az = e(x+iy) ln a = ex ln a eiy ln a = ax aiy . For x > 0 the Γ-function was defined as the improper integral Z ∞ tx−1 e−t dt, Γ(x) =

(26.3)

(26.4)

0

and in Lemma I.28.21 the convergence of the integral in (26.4) was proved. We extend this result first in Lemma 26.1. For all z ∈ C, Re z > 0, the integral Z ∞ tz−1 e−t dt Γ(z) = 0

converges and we have Γ(¯ z ) = Γ(z). Proof. With z = x + iy, x > 0, we find for t > 0 tz−1 e−t = tx−1 e−t e−iy ln t 505

(26.5)

A COURSE IN ANALYSIS

and therefore

z−1 −t t e = tx−1 e−t .

This implies already the convergence of Γ(z) in the half plane Re z > 0, and using the proof of Lemma I.28.21 we can also deduce that in every plane Re z ≥ x0 > 0 the convergence is uniform, indeed this follows from z−1 −t t e ≤ tx0 −1 e−t for t < 1, whereas for t > 1 the exponential term takes care for the convergence.

Theorem 26.2. The Γ-function is on Re z > 0 a holomorphic function. Proof. Since

d z−1 −t
= (ln t)tz−1 e−t , t e dz for  > 0 and R >  we find Z R Z R d z−1 −t (ln t)tz−1 e−t dt t e dt = dz  

and the integral on the right hand side converges uniformly in z for Re z ≥ RR z−1 −t x0 > 0 and by Theorem 22.14 the function z 7→  (ln t)t e dt is holomorphic in Re z ≥ x0 > 0 for every x0 > 0, hence Rin Re z > 0. Now R uniform convergence in Re z ≥ x0 > 0 of  (ln t)tz−1 e−t dt to Rwe∞ use the (ln t)tz−1 e−t dt which follows from the fact that at 0 the term tz−1 con0 trols ln t as does e−t at ∞. This however implies that Γ(z) is differentiable for Re z > 0 and Z ∞ (ln t)tz−1 e−t dt. (26.6) Γ0 (z) = 0

Using the integration by parts argument as in the proof of Theorem I.28.23 we arrive at Lemma 26.3. For Re z > 0 Γ(z + 1) = zΓ(z)

(26.7)

holds and for n ∈ N, Re z > 0 we have Γ(z + n) = z(z + 1) · . . . · (z + n − 1)Γ(z). 506

(26.8)

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

From Lemma 26.3 we deduce as in Chapter I.31. Theorem 26.4. The gamma function is a meromorphic function on C\(−N0 ) with simple poles at −n, n ∈ N0 .

Proof. For −n, n ∈ N0 , we define for Re z > −n but z 6= 0, −1, . . . , −n, using (26.8) Γ(z + n) . (26.9) Γ(z) = z(z + 1) · . . . · (z + n − 1) We note that for z = −k, k ∈ N0 , 0 ≤ k ≤ n, we have Γ(z + n) = Γ(n − k) = (n − k − 1)! 6= 0, hence Γ(z) has poles of first order at −k, k ∈ N0 . Corollary 26.5. The residue of Γ(z) at −k is

(−1)k , k ∈ N0 . k! Proof. Since −k is a pole of order 1 we have res(Γ, −k) =

(26.10)

res(Γ, −k) = lim (z + k)Γ(z) z→−k

Γ(z + k + 1) (−k)(−k + 1) · . . . · (−1) (−1)k Γ(1) = , = (−k)(−k + 1) · . . . · (−1) k!

= lim

z→−k

where we used (26.8). Using our uniqueness result for meromorphic functions, Theorem I.31.13 implies immediately Theorem 26.6. For all z ∈ C\Z we have

π . (26.11) sin πz Formula (26.11) has some interesting consequences. We can write for z ∈ C\Z 1 sin πz = Γ(1 − z) . (26.12) Γ(z) π The function z 7→ Γ(1 − z) has poles of first order for z ∈ N0 and is holomorphic on C\N0 . However z 7→ sinππz has simple zeroes, i.e. zeroes of multiplicity 1, at z ∈ N0 . Hence the singularities of z 7→ Γ(1 − z) sinππz are all removable and this entails Γ(z)Γ(1 − z) =

507

A COURSE IN ANALYSIS 1 Theorem 26.7. The function z 7→ Γ(z) is an entire function and the Γfunction has no zero, i.e. Γ(z) 6= 0 for all z ∈ C\(−N0 ).

Next we turn to the product representation (I.31.12). For all x > 0 we have ∞  Y x −x 1 1+ = xeγx e k Γ(x) k k=1

where γ is the Euler constant γ = limN →∞ Lemma 26.8. For every n ∈ N we have

P N

1 k=1 k

(26.13)

 − ln N . First we prove

∞  X · −·

< ∞. e k

1 − 1 + k ∞,Bn (0) k=1

Proof. For a ∈ C we have

(26.14)

       1 1 1 1 1 1− + a + ··· + al−1 + · · · − − 2! 2! 3! l! (l + 1)!  ∞  X 1 1 al−1 . − = a2 l! (l + 1)!

1 − (1 − a)ea = a2

l=1

Since

1 l!

−

1 (l+1)!

> 0 it follows for |a| ≤ 1 that a

|1 − (1 − a)e | ≤ |a|

2

∞  X 1 l=1

1 − l! (l + 1)!



= |a|2 .

For a = − kz and |z| ≤ k we find   z  − z |z|2 e k ≤ 2 1− 1+ k k

implying now

 X 1 z  − z sup 1 − 1 + e k ≤ n2 0 the uniqueness result for holomorphic functions yields Theorem 26.9. For

1 , Γ(z)

z ∈ C, the following product representation

∞  Y 1 z  −z γz 1+ = ze e k Γ(z) k

(26.15)

k=1

holds uniformly on compact subsets on C. Finally we deduce from (26.15) the product representations for the Γ function. Theorem 26.10. For z ∈ C\(−N0 ) the Γ-function has the presentation z ∞ e−γz Y e k , Γ(z) = z 1 + kz

(26.16)

k=1

where the convergence is uniform on compact subsets of C\(−N0 ). Proof. With   1 1 1 z  −z z  −z := zez (1+ 2 +···+ k −ln k) 1 + e 1 ·...· 1 + e k Γk (z) 1 k we have as k → ∞ formula (26.15). Since z

z

1 1 1 e1 Γk (z) = e−z (1+ 2 +···+ k −ln k) z 1+

z 1

ek · ...· 1 + kz

we find that as k → ∞ formula (26.16) follows. In Definition I.31.9 we introduce the beta-function for x, y > 0 as Z 1 tx−1 (1 − t)y−1 dt, B(x, y) =

(26.17)

0

and we proved in Theorem I.31.11 that B(x, y) =

Γ(x)Γ(y) . Γ(x + y)

509

(26.18)

A COURSE IN ANALYSIS

Since this is a function of two real variables we can only try to find an extension of B(x, y) as a separately holomorphic or meromorphic function B(z, w), i.e. as a function which is holomorphic in z (in w) for w(z) being fixed. Since Γ(z) is holomorphic in the half plane Re z > 0, formula (26.18) allows us to identify (z, w) 7→ B(z, w) :=

Γ(z)Γ(w) Γ(z + w)

(26.19)

as a separately holomorphic function on {Re z > 0} × {Re w > 0}. In Problem 3 we will see that in {Re z > 0} × {Re w > 0} the integral Z 1 tz−1 (1 − t)w−1 dt (26.20) 0

converges on compact sets uniformly and absolutely to a separate holomorphic function representing B(z, w). Now we turn to the Riemann ζ-function and in close relation with it to Dirichlet series. The importance of these functions lies in their relation to number theory. In particular the Riemann ζ-function has deep connections with prime numbers. Once we have studied the ζ-function we can also formulate the best known among all challenging open problems in Mathematics, the Riemann conjecture which links the (non-trivial) zeroes of the ζ-function with the distribution of prime numbers. While the results we are going to discuss are interesting in their own right, for us they serve mainly to see complex analysis in action. But we need some basic facts from elementary number theory to proceed and we collect them first. A good reference is for example [5] in which the proofs are also given. Number theory is concerned with the structure of N and we begin with the following two classical results: Theorem 26.11. A. (Euclid) There exists infinitely many prime numbers. B. (Fundamental Theorem of Arithmetic) Every n ∈ N, n > 1, admits a unique representation as a product of prime numbers when the order of the factors is not taken into account. Note that our formulation of the fundamental theorem of arithmetic allows and must allow that some prime numbers occur as a factor more than once 510

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

in the product representations of n, i.e. we have 1 r · . . . · pa(r) n = pa(1) =

r Y

a

j p(j) ,

j=1

with aj ∈ N and the numbers p(1) , . . . , p(r) are the distinct prime factors. As number theory is concerned with the set N, we expect functions f : N → C will play a role in number theory and they are called arithmetical functions. But the reader should note that this is just a new name for a sequence (f (n))n∈N of complex numbers. Definition 26.12. Let f be an arithmetical function not identically equal to zero. A. We call f multiplicative if f (mn) = f (m)f (n)

for m and n relative prime.

(26.21)

B. If f (mn) = f (m)f (n)

for all m, n ∈ N

(26.22)

we call f completely multiplicative. Recall that m, n ∈ N are called relatively prime if their greatest common divisor is equal to 1. Example 26.13. For n ∈ N and s = σ + it, σ, t ∈ R, we have ns = nσ ei ln t and for the arithmetical function fs (n) = ns we find fs (mn) = (mn)s = (mn)σ eit ln(mn) = mσ nσ eit(ln m+ln n)

= mσ eit ln m nσ eit ln n = ms ns = fs (m)fs (n),

i.e. ns , hence n−s , is completely multiplicative.

In Example 26.13 we have already picked up the standard notation from analytical number theory and in particular the theory of Dirichlet series: complex numbers are in some typical situations, for example in terms as n−s , denoted by s = σ + it, s, t ∈ R. 511

A COURSE IN ANALYSIS

Let f be an arithmetical function and s = σ + it ∈ C. If Re s = σ ≥ σ0 we find (n) f |f (n)| ns ≤ nσ0 P P∞ f (n) |f (n)| implying that if ∞ n=1 nσ0 converges then n=1 ns converges absolutely in the half plane Re s ≥ σ0 and in fact this convergence is also uniform. Proposition 26.14. Let f be an arithmetical function such that the seP∞ f (n) ries n=1 ns neither converges for all s ∈ C nor diverges for all s ∈ C. P f (n) Then we can find σa ∈ R called the abscissa of convergence of ∞ n=1 ns P∞ f (n) such that for Re s > σa the series n=1 ns converges absolutely whereas for Re s < σa the series does not converge absolutely. n o P f (n) Proof. Consider the set Af := σ ∈ R ∞ diverges, s = σ + it . s n=1 n Since by assumption our series does not converge absolutely for all s ∈ C the set Af is not empty. On the other hand, since the series converges absolutely for at least one σ0 , by our preceding consideration, Af is bounded from above. For σa := sup Af the result follows: if σ < σa then σ ∈ Af , hence the series does not converge absolutely, if σ > σa , then σ ∈ Af and the absolute convergence follows. Definition 26.15. A. Let f : N → C be an arithmetical function. We call s 7→

∞ X f (n) n=1

ns

the Dirichlet series associated with or generated by f . B. The Dirichlet series associated with the constant function f (n) = 1 is by definition the Riemann ζ-function, i.e. ∞ X 1 . ζ(s) = ns n=1

(26.23)

Corollary 26.16. A. The abscissa of convergence of the Riemann ζ-function is 1, i.e. for σ > 1, s = σ + it, the series (26.23) converges absolutely while it does not converge absolutely for σ ≤ 1. B. An arbitrary Dirichlet series converges absolutely in the half plane Re s = σ > σa with the understanding that σa = −∞ if the series converges for all σ ∈ R absolutely and σa = +∞ if the series converges nowhere absolutely. 512

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

Proof.PPart B follows from Proposition 26.14. In order to see Part A we note 1 that ∞ n=1 nσ converges for σ > 1 and diverges for σ ≤ 1. Let us agree to write

Df (s) :=

∞ X f (n) n=1

ns

s = σ + it ∈ C,

,

(26.24)

with the understanding that Df (s) denotes the formal series as well as the corresponding function in the half plane Re s > σa , σa < ∞. Corollary 26.17. If the abscissa of absolute convergence σa of the Dirichlet series Df (s) satisfies σa < ∞ then Df (s) is in the half-plane Re s > σa a holomorphic function and for its derivative we find Df0 (s) = −

∞ X f (n) ln n n=1

ns

.

(26.25)

f (n) Proof. Let σ0 > σa . From ns ≤ |fn(n)| we deduce that Df (s) converges σ0 in each half-plane Re s ≥ σ0 > σa uniformly, hence locally uniformly and from Theorem 23.20 we deduce that Df (s) is in the open half-plane Re s > σa holomorphic. The formula (26.25) follows now since we are allowed to differentiate the series term by term. Corollary 26.18. The Riemann ζ-function is in the half-plane Re s > 1 holomorphic with derivative 0

ζ (s) = −

∞ X ln n n=2

ns

.

(26.26)

Proposition 26.19. For Df (s) as in (26.24) assume σa < ∞. Then we have uniformly for t ∈ R lim Df (σ + it) = f (1).

σ→∞

(26.27)

Proof. We have to prove that lim

σ→∞

∞ X f (n) n=2

ns

513

= 0.

(26.28)

A COURSE IN ANALYSIS

For σ ≥ σ0 > σa we find with s = σ + it ∞ ∞ X f (n) X |f (n)| ≤ ns nσ n=2

n=2

≤2

and since

P∞

|f (n)| n=2 nσ0

−(σ−σ0 )

∞ X |f (n)| n=2

nσ0

,

< ∞ we arrive at (26.28).

As in the case of power series the “coefficients” f (n) determine a Dirichlet series. Theorem 26.20. Let Df1 (s) and Df2 (s) be two Dirichlet series both converging absolutely for Re s = σ > σa . If Df1 (s) = Df2 (s) in the half-plane Re s > σa then f1 (n) = f2 (n) for all n ∈ N. Proof. By linearity we need to prove that if the Dirichlet series Df (s) converges absolutely for Re s = σ > σa to the zero function, then f (n) = 0 for all n ∈ N. Suppose not all values f (n), n ∈ N, are zero. Let n0 be the smallest number such that f (n0 ) 6= 0. It follows that  −s  −s ∞ ∞ X X n n s = f (n0 ) . f (n) 0 = Df (s)n0 = f (n) n0 n0 n=n +1 n=n 0

0

n n0

> 1 for n ≥ n0 + 1 The convergence of the series and the fact that  −σ P∞ implies that limσ→∞ n=n0 +1 f (n) nn0 = 0, i.e. f (n0 ) = 0 which is a contradiction. Exercise 26.21. Prove the following refinement of the uniqueness result, i.e. Theorem 26.20. Two Dirichlet series Df1 (s) and Df2 (s) converging absolutely in the half-plane Re σ > σa are equal if and only if there exists a sequence (sk )k∈N , Re sk = σk > σa and limk→∞ σk = ∞, such that Df1 (sk ) = Df2 (sk ). Hint: Prove that these conditions imply f1 (n) = f2 (n) for all n ∈ N. For this the following result may be used: for k ≥ 1 and σ ≥ c ≥ σa the following holds: ∞ ∞ X X −(σ−c) −s |g(n)|n−c g(n)n ≤ k n=k n=k P∞ g(n) where the Dirichlet series k=1 ns has the abscissa of convergence σa . 514

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

Using the result of Exercise 26.21 we find Corollary 26.22. Let the Dirichlet series Df (s) converge absolutely in Re s = σ > σa and assume that for some s0 , Re s0 > σa , we have Df (s0 ) 6= 0. Then there exists σ0 ≥ σa such that Df (s) 6= 0 for all s such that Re s > σ0 . Proof. Assume that no such a half-plane exists. In this case, for k ∈ N we can pick sk ∈ C, Re sk = σk > k such that Df (sk ) = 0. Since limk→∞ σk = ∞ the uniqueness result in form of Exercise 26.21 yields for f (n) = 0 for all n ∈ N, i.e. Df (s) = 0 for all s, which of course contradicts the assumption. Our next goal is to prove a product representation of certain Dirichlet series, in particular we want to show that ζ(s) =

Y p

1 , 1 − p−s

Re s = σ > 1,

where the product is taken over all prime numbers. We need some preparations. Our presentation of the following theorem and proof follows closely [5]. Theorem P∞ 26.23. Let f be a multiplicative arithmetical function and assume that n=1 |f (n)| converges. It follows that ! ∞ ∞ X Y X
k (26.29) f (n) = f p n=1

p

k=0

where the product is taken over all prime numbers p. If in addition f is completely multiplicative then we have ∞ X

f (n) =

n=1

Y p

1 . 1 − f (p)

Proof. For x ∈ R we define P (x) :=

∞ Y X

p≤x

k=0

515

f p


k

!

(26.30)

.

(26.31)

A COURSE IN ANALYSIS


P f pk converge absolutely This is a finite product and all the series ∞ k=0P which follows from thePabsolute convergence of ∞ k=1 f (n). Thus the finite
∞ k can be taken in any order and we may product of the series k=0 f p arrange the sum as we need. A typical term in P (x) can be written as f (pa11 ) f (pa22 ) · . . . · f (par r ) = f (pa11 · . . . · par r ) ,

where we used the fact that f is multiplicative and we assume p1 < p2 < · · · < pr . Denote by A(x) the set of all n ∈ N with all prime factors being less or equal to x and let B(x) be the set of all n ∈ N with at least one prime factor strictly larger than x. By the fundamental theorem of arithmetic it follows that X f (n) P (x) = n∈A(x)

and therefore

∞ X X X |f (n)| ≤ |f (n)|. f (n) − P (x) ≤ n=1 n>x n∈B(x) P P Since ∞ n>x |f (n)| = 0 implying n=1 |f (n)| converges we find that limx→∞ that ! ! ∞ ∞ ∞ Y X X Y X

k k = lim f (n). = f p f p p

k=0

x→∞

p≤x

n=1

k=0

In order to see theQabsolute convergence of the product in (26.29) P we note ∞ that the product n=1 (1 + cn ) converges absolutely if and only if ∞ n=1 cn converges absolutely. Since ∞ ∞ ∞ X X
X X
X k k ≤ f p |f (n)| f p ≤ p≤x k=1

p≤x k=1

n=2

the absolute convergence of the product follows. Finally we
if
P note that k = f p fPis completely multiplicative we have f pk = f (p)k and ∞ k=0 ∞ 1 k k=1 f (p) . Since this geometric series converges its limit is 1−f (p) and (26.30) is proved. P f (n) Corollary 26.24. Suppose that Df (s) = ∞ n=1 ns converges absolutely for Re s = σ > σa . For f being multiplicative we have
! ∞ ∞ X f (n) Y X f pk , Re s > σa , (26.32) = ns pks n=1 p k=0 516

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

and if f is completely multiplicative ∞ X f (n) n=1

ns

=

Y p

1 1 − f (p)p−s

(26.33)

holds. Corollary 26.25. For the Riemann ζ-function we have the Euler product representation ∞ X Y 1 1 = , ns 1 − ps n=1 p

ζ(s) =

Re s = σ > 1,

(26.34)

which implies in particular ζ(s) 6= 0 for Re s > 0. A remark to further examples seems to be appropriate. Of course we can start to “play” with arithmetic functions f , i.e. complex sequences (f (n))n∈N to construct Dirichlet series. However it is much more important to note that there are classical arithmetic functions such that ϕ(n) µ(n) λ(n) χ(n)

the the the the .. .

Euler ϕ-function M¨obius function Liouville function Dirichlet character

and each give rise to a Dirichlet series which in turn is an important tool in number theory. For example we have ∞

or

X µ(n) Y
1 −s = = , 1 − p ζ(s) n=1 ns p L(s, χ) :=

∞ X χ(n) n=1

ns

=

Y p

1 , 1 − χ(p)p−s

Re s = σ > 1,

Re s = σ > 1.

The latter function is the Dirichlet L-function which is of great importance when studying for example arithmetic progression. For the definitions and the properties of these classical arithmetical functions as well as for their 517

A COURSE IN ANALYSIS

applications to number theory we refer to standard textbooks in number theory, [5] or the classics by G. Hardy and E. Wright [32]. In the following we only deal further with the Riemann ζ-function. Our first result is an integral representation for the ζ-function which gives also a first link to the Γ-function. We remind the reader on Lemma 26.1 stating that Z ∞ y s−1e−y dy (26.35) Γ(s) = 0

holds for all s ∈ C with Re s > 0. Theorem 26.26. For s ∈ C, Re s > 1, we have Z ∞ s−1 Z ∞ s−1 −x x x e 1 1 dx. dx = ζ(s) = Γ(s) 0 ex − 1 Γ(s) 0 1 − e−x

(26.36)

Proof. First we show that for σ = Re s > 1 the integral exists as an improper, absolutely convergent integral. For 0 < η ≤ 1 < R we have for s = σ + it Z R s−1 Z R σ−1 it ln x x e x = dx x x e −1 η η e −1 Z R σ−1 Z 1 σ−1 Z R σ−1 x x x ≤ dx = dx + dx. x−1 x−1 x−1 e e e η η 1 Since σ − 2 > −1 and limx→0 exx−1 = 1 we find that Z

0

Furthermore, from Z

1

R

1

xσ−1 dx = lim η→0 ex − 1

1 1−e−x

≤

xσ−1 dx = lim R→∞ ex − 1

1 1−e−1

Z

R

1

Z

1 η

xσ−1 dx < ∞. ex − 1

for x ≥ 1 we deduce

xσ−1 dx = lim R→∞ ex − 1

R∞

xs−1

Z

∞ 1

e−x xσ−1 dx < ∞. 1 − e−x

Hence for σ > 1 the integral 0 ex −1 dx exists both as an improper complexvalued Riemann integral and as a Lebesgue integral. Now we use in (26.35) the substitution y = nx to find Z ∞ s xs−1 e−nx dx Γ(s) = n 0

518

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

or −s

n

1 = Γ(s)

which yields N X k=1

−s

n

Z

∞

xs−1 e−nx dx,

0

N Z 1 X ∞ s−1 −nx = x e dx Γ(s) k=1 0
Z ∞ −x 1 − e−N x 1 s−1 e x dx = Γ(s) 0 1 − e−x Z ∞ 1 1 − e−N x = xs−1 x dx, Γ(s) 0 e −1

where we have used that Γ(s) has no zeroes and N X n=1

e−nx =

N X

e−x

n=1


n

=

N X

e−x

n=0


n

−1


e−x 1 − e−N x 1 − e−(N +1)x −1 = . = 1 − e−x 1 − e−x

We can split the integral into two absolutely convergent integrals to obtain N X n=1

−s

n

1 = Γ(s)

Z

0

∞

xs−1 dx − ex − 1

Z

0

∞

 e−N x xs−1 dx . ex − 1

For the second integral we find for c > 0 that Z ∞ −N x s−1 Z c −N x s−1 Z ∞ −N x s−1 e x e x e x dx = dx + dx. x x x e −1 e −1 e −1 0 0 c We note that for Re s = σ > 1 we have Z c −N x s−1 Z c −N σ−1 Z c σ−1 e x e x x dx ≤ dx < dx x x x e −1 e −1 0 0 0 e −1

and therefore, given  > 0 we can find c0 ≤ 1 such that c ≤ c0 implies Z c −N x s−1 Z c σ−1 e x x dx < dx < . x x e −1 0 0 e −1 519

A COURSE IN ANALYSIS

For the second integral we observe that for Re s = σ > 1 Z ∞ −N x s−1 Z ∞ −N x σ−1 e x e x dx ≤ dx x x e −1 e −1 c c Z ∞ σ−1 x −N x dx ≤e ex − 1 c

R ∞ σ−1 and since c exx −1 dx is finite for σ > 1 and c ≥ 0 it follows as N → ∞ that for s = σ + it, Re s = σ > 1, we have Z ∞ s−1 N X x 1 1 = dx. ζ(s) = lim s N →∞ n Γ(s) 0 ex − 1 n=1

Remark 26.27. In [61] the proof of (26.36) first given for s ∈ R, s > 1, R ∞is s−1 and then it is shown that the integral s 7→ 0 exx −1 dx admits a holomorphic extension to the half-plane Re s > 1. Now the uniqueness result for holomorphic functions entails that if (26.36) holds for s ∈ R, s > 1, then it holds also for σ ∈ C, Re σ > 1. 1 By Theorem 26.7 we know that s 7→ Γ(s) is an entire function. Hence if we R ∞ xs−1 R ∞ xs−1 e−x can extend s 7→ 0 ex −1 dx = 0 1−e−x dx to a larger subset of C than the half-plane Re s = σ > 1 as a holomorphic function we can extend the ζ-function. We start with formula (26.36)

Γ(s)ζ(s) =

Z

∞ 0

xs−1 dx = ex − 1

Z

1 0

xs−1 dx + ex − 1

Z

∞

1

xs−1 dx ex − 1

which holds for Re s = σ > 1. For R > 1 we find that Z R s−1 Z R (s−1) ln x x e s 7→ dx = dx x ex − 1 1 e −1 1 is a holomorphic function and for s in a compact set we have uniformly lim

R→∞

Z

R 1

xs−1 dx = ex − 1 520

Z

∞ 1

xs−1 dx ex − 1

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

R ∞ s−1 implying that s 7→ 1 exx −1 dx is an entire function. Now we turn to the first integral in the decomposition and observe that x 7→ exx−1 admits a Taylor expansion about x = 0 given by ∞ X x (−1)k−1 βk 2k 1 = 1 − x + x ex − 1 2 (2k)! k=1

(26.37)

where the βk ’s are related to the Bernoulli numbers by B2k = (−1)k−1 βk , also see Problem 5 in Chapter 16. The radius of convergence of this series is 2π. Indeed, for z 7→ ezz−1 we have poles at 2πim, m ∈ Z\{0}, so 2π must be a bound for the radius of convergence. For the following the actual values of the Bernoulli numbers are not important. However, since there is some nice part of mathematics to learn we discuss the expansion (26.37) and the Bernoulli numbers in some detail in Appendix IV. We multiply (26.37) with xs−2 , Re s > 1, to get ∞

xs−1 1 s−1 X (−1)k−1βk 2k+s−1 s−2 = x − x + x ex − 1 2 (2k)! k=1

and integration over [0, 1] yields Z 1 s−1 ∞ X x (−1)k−1 βk 1 1 1 dx = − + . x s − 1 2s k=1 (2k)! 2k + s 0 e −1

(26.38)

For s0 6= 1, 0, −1, −3, −5, . . . we can always find a ball Br (s0 ), r > 0, such that the right hand side in (26.38) is in Br (s0 ) a holomorphic function in s, hence the right hand side is a meromorphic function in C with isolated singularities at s0 = 1, 0, −1, −3, . . ., which are all poles of first order. 1 is an entire function with simple zeroes at 0, −1, −2, −3 . . .. We know that Γ(s) This implies now

Theorem 26.28. The ζ-function originally defined for Re s > 0 can be extended to a meromorphic function on C by ζ(s) =

1 Γ(s)

Z

0

1

xs−1 1 dx + ex − 1 Γ(s)

Z

∞ 1

xs−1 1 dx = ex − 1 Γ(s)

Z

0

∞

xs−1 dx ex − 1

(26.39)

which has only one singularity at s0 = 1. This is a pole of first order with residue res(ζ, 1) = 1. 521

A COURSE IN ANALYSIS

In order to study ζ(s) in more detail we need the following functional equation for the ζ-function, ζ(s) = 2(2π)s−1 sin

πs Γ(1 − s)ζ(1 − s) 2

or with s replaced by 1 − s ζ(1 − s) = 21−s π −s cos

πs Γ(s)ζ(s). 2

(26.40)

(26.41)

There are different ways to derive this functional equation, one uses Fourier analysis and we will provide this proof in Part 8 of our Course. Other proofs are given in [20], [33], [61] or [73]. Once (26.40) or (26.41) are established we can draw further consequences about the Riemann ζ-function. So far we know that the ζ-function is a meromorphic function with a pole of first order at 1 with residue 1. Further we know that for Re s > 1 the ζ-function has no zeroes, a fact which follows from the product representation given in Corollary 26.25. The functional equations are identities for meromorphic functions hence they must hold everywhere (except at the pole of ζ at 1). For Re s = σ < 0 the function s 7→ Γ(1 − s) is holomorphic as is the function s 7→ ζ(1 − s) and both functions are non-zero in the half-plane Re s < 0. The same holds for s 7→ (2π)−s which now implies that in the half-plane Re s < 0 the ζ-function has the same zeroes as s 7→ sin πs , i.e. we 2 have Corollary 26.29. In the half-plane Re s < 0 the Riemann ζ-function has simple zeroes for s = −2, −4, . . ., i.e. s ∈ −2N. These zeroes of the ζfunctions are called the trivial zeroes of ζ. Thus we know that outside the strip 0 ≤ σ ≤ 1, Re s = σ, the ζ-function has only the trivial zeroes. The strip 0 ≤ σ ≤ 1 is called the critical strip of the Riemann ζ-function. It needs further efforts to prove that on Re s = 1 and Re s = 0 the ζ-function has no zeroes, however in the critical strip there are infinitely many zeroes of the ζ-function. We can now formulate the Riemann Conjecture or as it should be called historically correct the Riemann Hypothesis: All non-trivial zeroes of ζ(s) lie on the critical line Re s = 21 . 522

26

THE Γ-FUNCTION, THE ζ-FUNCTION AND DIRICHLET SERIES

We refer to [20], [39], [61], [63], [73], [82] or [86] where partly details of the above mentioned results and partly the relation of the non-trivial zeroes of the Riemann ζ-function to the distribution of prime numbers are discussed. A fascinating topic which however goes beyond our Course.

Problems 1.

a) Find the value Γ −k + b) Prove |Γ(iy)|2 =

π y sinh πy




for k ∈ N.
2 and Γ 12 + iy =

1 2

π . cosh πy

Hint: use (26.11), Γ(¯ z ) = Γ(z), −i sin iy = sinh y and cos iy = cosh y. 2. Prove the Legendre duplication formula (I.31.35) for z ∈ C, Re z > 0:   1 22z−1 Γ(2z) = √ Γ(z)Γ z + . 2 π

3. For Re z > 0 and Re w > 0 justify B(z, w) =

Z

1

tz−1 (1 − t)w−1 dt.

0

4. For Re z > 0 prove Γ(z) = 2

Z

∞

2

t2z−1 e−t dt.

0 0

(z) of the Γ-function. 5. Consider the logarithmic derivative ψ(z) := ΓΓ(z) Find its singularities and in some natural domain prove the functional equation 1 ψ(z + 1) = + ψ(z) z and ψ(1 − z) − ψ(z) = π cot πz.

Hint: use the functional equation Γ(z + 1) = zΓ(z) and Γ(z)Γ(1 − z) = sinππz , respectively. 523

A COURSE IN ANALYSIS

6. With ψ as in Problem 5 prove in a natural domain  ∞  1 X 1 1 ψ(z) = −γ − − . − z k=1 z + k k Hint: use the product representation of the Γ-function. 7. The M¨ obius function µ : N → R is defined as follows. For n = 1 we set µ(1) = 1. Now let n ∈ N, n ≥ 2, have the prime number decomposition u = pν11 · . . . · pνkk . We define µ(n) = (−1)k if ν1 = · · · = νk = 1, µ(n) = 0 otherwise. a) Prove that µ(n) = 0 if and only if n has a square factor larger than 1. b) Prove that µ is a multiplicative arithmetic function which is not completely multiplicative. 8. Solve Exercise 26.21. 9.* Let (ak )k∈N be a bounded sequence of complex numbers. Prove for σ > 1 that 2 Z R X ∞ ∞ X |an |2 a 1 n dt = . lim R→∞ 2R −R nσ+it n2σ n=1 n=1 (This problem is taken from [83].)

10.* Finding special values of ζ(z) is non-trivial, we will discuss some ideas in Appendix IV and in Part 8. The strategy as indicated in the following hint to prove π2 ζ(2) = 6 is taken from [43]. Hint. Prove: (i) For 0 < x
1). Lemma 27.8. For sn, cn and dn we have   ˜ = 1, sn K + iK k   ˜ ˜ = −i k , cn K + iK k and

  ˜ = 0. dn K + iK

(27.31) (27.32) (27.33)

Proof. As already mentioned (27.31) follows from (27.30) when noting Z w dt p sn(z) = w if z = . (1 − t2 )(1 − k 2 t2 ) 0

For (27.32) we note that r   p ˜ ˜ = 1 − sn2 (K + iK) = 1 − 1 = −i k , cn K + iK k2 k and (27.33) follows from (27.31) and the definition of dn: q ˜ = 1 − k 2 sn2 (K + iK) ˜ = 0. dn (K + iK) Combining the addition formulae with Lemma 27.8 we find

      ˜ cn K + iK ˜ dn K + iK ˜   2 sn K + iK ˜ =   = 0, sn 2K + 2iK ˜ 1 − k 2 sn4 K + iK

(27.34)

      ˜ − sn2 K + iK ˜ dn2 K + iK ˜   cn2 K + iK ˜ =   = 1, cn 2K + 2iK (27.35) ˜ 1 − k 2 sn4 K + iK

and

      ˜ − k 2 sn2 K + iK ˜ cn2 K + iK ˜   dn2 K + iK ˜ =   = −1. dn 2K + 2iK (27.36) ˜ 1 − k 2 sn4 K + iK

535

A COURSE IN ANALYSIS

We may iterate the arguments which led to (27.34) - (27.36) and then we get   ˜ = 0, sn 4K + 4iK (27.37)   ˜ = 1, cn 4K + 4iK (27.38) and

  ˜ = 1. dn 4K + 4iK

Now we can prove

(27.39)

Theorem 27.9. The functions sn, cn and dn are double periodic functions and we have ˜ are the periods of sn 4K and 2iK (27.40) ˜ are the periods of cn 4K and 2K + 2iK (27.41) ˜ are the periods of dn. 2K and 4iK (27.42) ˜ are given by (27.22) and (27.23) respectively. Here K and K Proof. We have already seen that 4K is a period of sn and cn as well as 2K is a period of dn. Using the addition formulae and our previous results on special values of sn, cn and dn we find for sn ˜ = sn(z − 2K + 2K + 2iK) ˜ sn(z + 2iK) ˜ dn(2K + 2iK) ˜ + sn(2K + 2iK) ˜ cn(z − 2K) dn(z − 2K) sn(z − 2K) cn(2K + 2iK) = ˜ 1 − k 2 sn2 (z − 2K) sn2 (2K + 2iK) = −sn(z − 2K) = −sn(z + 2K) = sn(z).

For cn it now follows ˜ = cn(z + 2K + 2iK)

˜ − sn(z) sn(2K + 2iK) ˜ dn(z) dn(2K + 2iK) ˜ cn(z) cn(2K + 2iK) ˜ 1 − k 2 sn2 (z) sn(2K + 2iK)

= cn(z),

and the calculations for dn yields ˜ = dn(z − 4K + 4K + 4iK) ˜ dn(z + 4iK) ˜ cn(z − 4K) cn(4K + 4iK) ˜ ˜ − k 2 sn(4K + 4iK) dn(z − 4K) dn (4K + 4iK) = 2 2 2 ˜ 1 − k sn (z − 4K) sn (4K + 4iK) = dn(z − 4K) = dn(z − 2K) = dn(z).

536

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

Remark 27.10. We appreciate very much the useful and very explicit calculations in [80] which we employed to derive Theorem 27.9, and which are much easier to follow than those in [41]. The theory of elliptic integrals and elliptic functions had changed completely when first F. G. M. Eisenstein realised that starting with double periodic meromorphic functions is an alternative approach. Without noting or at least without mentioning Eisenstein’s work Weierstrass developed the whole theory largely depending on his ℘-function. In the following we give some introduction to this theory. Before we turn to double periodic functions we want to make a few remarks about simply periodic functions. Let f be a meromorphic function on C satisfying for some w 6= 0 f (z + w) = f (z) for all z ∈ C.

(27.43)

In this case we call w a period of f . Clearly with w every number kw, k ∈ Z\{0}, is a further period of f . In [61] it is proved that the moduli of all periods have an infimum bounded away from zero and that periods have no accumulation point in C. Hence for a periodic meromorphic function there exists a period w0 of minimal modulus, |w0 | = 6 0. Of course, with w0 the number −w0 is a further period of minimal modulus, otherwise w0 is uniquely determined. Clearly, all periods of the type kw0 , k ∈ Z\{0}, are on the straight line g satisfying z ∈ g if and only if arg z = arg w0 . On g we do not have any further period w1 of f . Suppose w1 ∈ g is such a period lying between kw0 and (k + 1)w0 , k ∈ Z. It follows that w1 − kw0 is a further period of f on the line g with |w1 − kw0 | < |w0| which is a contradiction. We call w0 a primitive period of f . Definition 27.11. A meromorphic function f on C which admits only one primitive period is called simply periodic. We refer again to [61] where the following theorem is proved: Theorem 27.12. Suppose that f is a simply periodic function with primitive period w0 = 2πi. If f satisfies the growth condition |f (z)| ≤ Meκ|x| , |x| ≥ x0 , z = x + iy, then there exists a rational function R such that f (z) = R(ez ). 537

(27.44)

A COURSE IN ANALYSIS

We now turn to meromorphic functions on C with exactly two periods w1 6= 0 and w2 6= 0, i.e. for all z ∈ C we have f (z + w1 ) = f (z) and f (z + w2 ) = f (z).

(27.45)

The case of interest is where w1 and w2 are linearly independent over R. If w1 = λw2 , λ ∈ R, then both w1 and w2 lie on the same line and depending on whether λ is rational or irrational the function f is either simply periodic or constant. (An exercise with hints leading to this result can be found in [83]). In the following we assume that w1 and w2 are linearly independent over R, hence they span a parallelogram P = P (w1, w2 ) in C, which we call the fundamental parallelogram, i.e.  (27.46) P = P (w1, w2 ) := z ∈ C z = λ1 w1 + λ2 w2 0 ≤ λ1 , λ2 < 1 . In order to establish some type of uniqueness for w1 and w2 we assume that they have minimal modulus among all periods lying on the line they generate. Further we assume that w1 belongs to the first quadrant, i.e. Re w1 ≥ 0 and Im w1 ≥ 0, and that {w1 , w2 } is positive (anticlockwise) orientated and that w2 belongs to the upper half plane, see Figure 27.1 below. Im z

ω1 + ω2 ω2

P (ω1 , ω2 ) ω1 0

Re z

Figure 27.1

538

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

With w1 , w2 ∈ C as above we also introduce the (periodicity) lattice L = L(w1 , w2) generated by w1 and w2 as  L = L(w1 , w2 ) = mw1 + nw2 m, n ∈ Z ,

(27.47)

see Figure 27.1. Some authors define the fundamental parallelogram to be P (w1, w2 ) which has sometimes an advantage, however our definition has the advantage that we obtain a cover of C by mutually disjoint sets with the help of the family (Pm,n (w1 , w2 ))m,n∈Z , where  Pm,n (w1 , w2 ) = z = λ1 w1 + λ2 w2 m ≤ λ1 < m + 1, n ≤ λ2 < n + 1 .

It is clear that for w ∈ L and z in the domain of f we have f (z + w) = f (z).

(27.48)

Since f is meromorphic it might have poles on a discrete set Pol(f ). For w ∈ L and a ∈ Pol(f ) we find that a + w ∈ Pol(f ), thus (27.48) holds for all z ∈ C and w ∈ L. We can now give the definition of an elliptic function as it is used nowadays. Definition 27.13. An elliptic function associated with a periodicity lattice L = L(w1 , w2 ) is a meromorphic function on C with the property (27.48), i.e. for all w ∈ L and all z ∈ C we have f (z + w) = f (z). Remark 27.14. A. Instead of f (z + w) = f (z) for all w ∈ L and z ∈ C it is sufficient to require f (z + w1 ) = f (z) and f (z + w2 ) = f (z) for all z ∈ C and the generators w1 and w2 of L. B. Elliptic functions are sometimes called double periodic functions. C. If f is double periodic then f 0 is double periodic too. The following three results are often referred to as the three Liouville theorems. Theorem 27.15 (1st Liouville theorem). An elliptic function f without pole is constant. 539

A COURSE IN ANALYSIS

Proof. Suppose that f has no pole. Then f is continuous and on the compact set P (w1, w2 ) the continuous function |f | is bounded. If z ∈ C then we can find integers m, n ∈ Z such that z − mw1 − nw2 ∈ P (w1 , w2) implying |f (z)| = |f (z − mw1 − nw2 )| ≤

max w∈P (w1 ,w2 )

|f (w)| < ∞.

This yields however that supz∈C |f (z)| ≤ maxw∈P (w1,w2 ) |f (w)|, i.e. f is a bounded entire function and by Liouville’s theorem, Corollary 23.18, f must be a constant. Theorem 27.16 (2nd Liouville theorem). Let w1 and w2 generate the lattice L and let f be a non-constant elliptic function with periodicity lattice L. Taking multiplicity into account, the total numbers of poles f has in P (w1, w2 ) is strictly larger than 1 and their residues add up to zero. Proof. First we note that in the compact set P (w1 , w2 ) the meromorphic function f can have only a finite number of poles, hence there are only finitely many poles in P (w1, w2 ). We want to integrate f over ∂P (w1 , w2 ) and to apply the residue theorem. However some of the poles of f may lie on ∂P (w1 , w2 ). In this case, since we are dealing only with finitely many poles we can translate P (w1, w2 ) by some z0 ∈ C such that on ∂(z0 + P (w1 , w2 )) there are no poles of f . Hence we may assume that on ∂P (w1 , w2) are no poles and now by the residue theorem we find Z

f (z) dz = 2πi

∂P (w1 ,w2 )

X

res(f, zν ),

ν

where the sum goes over all residues of f in P (w1, w2 ). We claim that the integral on the left hand side is zero. Taking the periodicity of f into account this is almost trivial. We may write ∂P (w1 , w2 ) = [0, w1 ] ⊕ [w1 , w1 + w2 ] ⊕ [w1 + w2 , w2 ] ⊕ [w2 , 0] and we note that [0, w1 ] and [w1 + w2 , w2 ] as well as [w1 , w1 + w2 ] and [w2 , 0] have opposite orientation whereas f |[0,w1 ] = f |[w1+w2 ,w2] and f |[w1,w1 +w2 ] = f |[w2,0] , see Figure 27.2 below. 540

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

Im z [ω1 + ω2 , ω2 ]

ω1 + ω2

ω2

[ω1 , ω1 + ω2 ] [ω2 , 0] [0, ω1 ]

ω1 Re z

0

Figure 27.2 Thus we have proved that a single pole of first order.

P

ν

res(f, zν ) = 0 which excludes that f has only

For the third Liouville theorem we need Definition 27.17. The order ord(f ) of an elliptic function f with periodicity lattice L = L(w1 , w2 ) is the number of its poles in P (w1, w2 ) counted according to their multiplicity. Theorem 27.18 (3rd Liouville theorem). Suppose that the elliptic function f has order ord(f ) = M. When we take multiplicity into account then f also has M zeroes in P (w1, w2 ), i.e. the number of poles and the number of zeroes (each counted with its multiplicity) of an elliptic function in its fundamental parallelogram is the same. Proof. First we note that the argument principle, Theorem 24.25, can be proved for boundaries of parallelograms. Secondly we can argue as in the proof of Theorem 27.16 and assume that on ∂P (w1 , w2 ) there are no poles and no zeroes of f . Denoting again by N(f ) the number of zeroes of f in P (w1 , w2 ), of course counted according to their multiplicity, we find by the argument principle that Z

∂P (w1 ,w2 )

f 0 (z) dz = 2πi(N(f ) − ord(f )). f (z) 541

(27.49)

A COURSE IN ANALYSIS 0

Since ff has also L(w1 , w2 ) as periodicity lattice, with the same argument as in the proof of Theorem 27.16 we find that the integral in (27.49) must be zero and the theorem follows. For a given periodicity lattice L = L(w1 , w2 ) we want to determine, if possible, all associated elliptic functions. In Problem 9b P of Chapter 1 and in Example 5.36 we discussed the counting measure µZ = k∈Z k and we have seen its invariance under the action Z. This observation is helpful to construct periodic functions. If the series X G(z) := g(z + kw) (27.50) k∈Z

converges on C and hence represents a function then we find X X G(z + w) = g(z + w + kw) = g(z + (k + 1)w) k∈Z

=

X

k∈Z

g(z + lw) = G(z),

l∈Z

i.e. G is periodic with period w. This construction works of course for two periods w1 and w2 : the function X G(z) = g(z + k1 w1 + k2 w2 ) (27.51) (k1 ,k2 )∈Z2

has two periods w1 and w2 . The problem is of course the convergence of a series such as (27.51), in particular if we need to construct a function with poles. Consider the series

X

(k1 ,k2 )∈Z2 (k1 ,k2 )6=(0,0)

α ∈ R. We define the partial sums Sα,N :=

(k12

X

(k1 ,k2 )∈KN

1 , + k22 )α

(k12

(27.52)

1 + k22 )α

 where KN := (k1 , k2 ) ∈ Z2 \{(0, 0)} max {|k1 |, |k2|} < N . The limit of (Sα,N )N ∈N , if it exists, is also denoted by (27.52). Since all terms in (27.52) 542

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

are non-negative, convergence of (27.52) will be absolute convergence and hence any other way of forming partial sums will converge to the same limit. Note that this is essentially depending on a “one-dimensional” proof working with different enumerations of Z2 , see also Theorem I.18.27. The following lemma, extending the one-dimensional result, i.e. Theorem I.18.21, gives a helpful criterion to decide on the convergence of (27.52). Lemma 27.19. The sequence (Sα,N )N ∈N converges if and only if the inte R dx dy 2 gral K (x2 +y2 )α , K = (x, y) ∈ R k(x, y)k∞ ≥ 1 , converges as improper Riemann integral. This proof works as that of Theorem I.18.21 by approximating the integral with special R Riemann sums and is left to the reader. Of course the convergence of K (x2dx+ydy2 )α is equivalent to the convergence of the integral R dx dy and now with the help of polar coordinates we get B { (0) (x2 +y 2 )α 1

Z

B1{ (0)

dx dy = 2π (x2 + y 2 )α

Z

∞

1

1 r dr = r 2α

Z

∞

r 1−2α dr

1

which is finite if and only if 2α > 2, i.e. α > 1. The function z 7→

X

1 (z − w)2 w∈L

is a good candidate (see below) for an elliptic function with a pole of order 2 at z = 0 and periodicity lattice L = L(w1 , w2 ). The summation is to be understood as X 1 . (27.53) (z − mw1 − nw2 )2 m,n∈Z

However, for w1 = 1 and w2 = i we have with z = 0 that 1 1 (z − mw1 − nw2 )2 = m2 + n2

and from Lemma 27.19 we deduce that (27.53) will not converge absolutely on C. In general we have

Lemma 27.20. Let L = L(w1 , w2 ) be a periodicity lattice and α > 2. Then P the series w∈L\{0} |w|−α converges where 0 stands for the lattice point (0, 0). 543

A COURSE IN ANALYSIS

Proof. The function h(ξ, η) = h(tξ, tη) =

|ξw1 +ηw2 |2 ξ 2 +η2

is homogeneous of degree 0 since

|ξw1 + ηw2|2 |tξw1 + tηw2 |2 = = h(ξ, η). t2 ξ 2 + t2 η 2 ξ 2 + η2

As a continuous function it obtains its minimum κ on S 1 . Since w1 and w2 are linearly independent over R, for (ξ, η) ∈ S 1 we have |ξw1 + ηw2 |2 > 0, implying that h(ξ, η) ≥ κ > 0. The homogeneity of h now implies |mw1 + nw2 |2 ≥ κ(m2 + n2 )

1 1 1 for all m, n ∈ Z2 which yields |mw1 +nw 2 ≤ κ m2 +n2 and from Lemma 27.19 2| we deduce for α > 2 that X X 1 1 |w|−α ≤ α < ∞. 2 κ (m + n2 ) 2 w∈L\{0} m,n∈Z2 \{0}

Of greatest interest is now Lemma 27.21. Let L = L(w1 , w2 ) be a periodicity lattice. The series  X  1 1 (27.54) − (z − w)2 w 2 w∈L\{0}

converges absolutely and locally uniformly in every subset of C\(L\{0}). Proof. Since

we have

1 1 w 2 − (z − w)2 −z(z − 2w) − = = 2 2 2 2 (z − w) w (z − w) w (z − w)2w 2 |z| |z − 2w| 1 1 (z − w)2 − w 2 = |z − w|2|w|2 .

Now let z ∈ Br (0) and |w| ≥ 2r. It follows that |z| ≤ |z| |z − 2w| ≤ 3r|w| and |z − w| ≥ |w| − |z| ≥ |w| , i.e. 2

|w| 2

and therefore

|z| |z − 2w| ≤ 12r|w|−3, |z − w|2|w|2 which implies the locally uniform and absolute convergence of (27.54) for all z ∈ C\(L\{0}). 544

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

Definition 27.22. The Weierstrass ℘-function associated with the periodicity lattice L = L(w1 , w2 ) is the function defined by  X  1 1 1 , z∈ / L. (27.55) − ℘(z, L) = ℘(z) = 2 + z (z − w)2 w 2 w∈L\{0}

For z ∈ L we may define ℘(z) = ∞ when we understand by ∞ the point at infinity obtained by the one point compactification of C. Our considerations so far are summarised in Theorem 27.23. Given a periodicity lattice L = L(w1 , w2 ). The associated Weierstrass ℘-function is a function meromorphic on C with poles of second order at the lattice points. In the complement of L it is a holomorphic function which is even, i.e. ℘(z) = ℘(−z) and its Laurent expansion about 0 is of the type ∞ X 1 a2k z 2k . (27.56) ℘(z) = 2 + z k=1 Proof. We only need a remark as to why ℘ is an even function, which however follows when we replace z by −z in (27.55) and note that − (L\{0}) = L\{0}. Since we can differentiate (27.55) in C\ (L\{0}) term by term we obtain Corollary 27.24. The derivative of ℘ as in Theorem 27.23 is given by ℘0 (z) = −2

X

1 . (z − w)3 w∈L

(27.57)

This is meromorphic function on C with poles of order 3 at the lattice points and holomorphic in C\L. In addition ℘0 is an odd function, i.e. ℘0 (−z) = −℘0 (z). Our goal is to prove that ℘ and ℘0 are elliptic functions and every elliptic function f with periodicity lattice L = L(w1 , w2) is of the type f = R1 (℘) + ℘0 R2 (℘)

(27.58)

where R1 and R2 are rational functions. Hence the whole theory of elliptic functions is encoded in the functions ℘. It is of some interest to note that 545

A COURSE IN ANALYSIS

Eisenstein was the first who noted that such an approach might work, his starting point was the series (27.53). He was well aware of the convergence problems and could manage related problems as A. Weil [91] has pointed out. Nonetheless, his work was not picked up and it is one of the merits of Weierstrass to have developed the theory of elliptic functions as we handle them today. Theorem 27.25. The function ℘ is an elliptic function of order 2 and ℘0 is an elliptic function of order 3. Proof. It remains to prove the periodicity of ℘ and ℘0 and this is easier for ℘0 . Indeed, for w0 ∈ L we have X 1 ℘0 (z + w0 ) = −2 = ℘0 (z) 3 (z + w − w) 0 w∈L since −w 7→ w0 − w is a bijection of L onto itself. Now we prove that for every w0 ∈ L the function z 7→ ℘(z + w0 ) − ℘(z) is constant and equal to zero. Since ℘0 (z + w0 ) = ℘0 (z) this function must be constant, note that C\L is arcwise connected. Take w0 = w1 , recall L = L(w1 , w2 ), i.e. w1 is one of / L and the generators of the periodicity lattice. It follows that − w21 ∈  w   w  w   w  1 1 1 1 ℘ − =℘ −℘ − =0 + w1 − ℘ − 2 2 2 2 where we used that ℘ is an even function. Hence ℘(z + w0 ) = ℘(z).

Let w ∈ {w1 , w2 }. It follows that  w w  w w = ℘0 = −℘0 − w = ℘0 − ℘0 2 2 2 2

0 w2 0 w1 implying that ℘ 2 = ℘ 2 = 0. Moreover, for w = w1 + w2 we have       w1 + w2 w1 + w2 w1 + w2 ℘0 = ℘0 , − (w1 + w2 ) = −℘0 2 2 2
2 = 0 and we have found in P (w1 , w2) three mutually distinct i.e. ℘ w1 +w 2 2 zeroes of ℘0 : w21 , w22 and w1 +w . Since by the 3rd Liouville theorem, Theorem 2 0 27.18, ℘ has exactly three zeroes in P (w1, w2 ) we know all zeroes of ℘0 . The values   w  w  w1 + w2 1 2 e1 := ℘ (27.59) , e2 := ℘ , e3 = ℘ 2 2 2 546

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

are playing some important role in the theory of elliptic functions. We want to determine the coefficients of the Laurent expansion of ℘ about z0 = 0. For this we note first that by Lemma 27.20 for n ≥ 3 the series X Gn = Gn (L) = Gn (w1 , w2) := w −n (27.60) w∈L\{0}

converges absolutely. Moreover for n odd it follows that X X X Gn = w −n = (−w)n = − w −n w∈L\{0}

w∈L\{0}

w∈L\{0}

implying G2n+1 = 0

for all n ∈ N.

(27.61)

Theorem 27.26. The Laurent expansion of the Weierstrass ℘-function associated with the periodicity lattice L = L(w1 , w2 ) is given by ∞

X 1 (2k + 1)G2(k+1) z 2k . ℘(z) = 2 + z k=1

(27.62)

Proof. We know already from Theorem 27.23 that ∞

X 1 ℘(z) = 2 + a2k z 2k . z k=1 Consider the function

1 z2 which must have Taylor coefficients a0 = 0 and g(z) = ℘(z) −

a2k =

g (2k) (0) . (2k)!

For l > 1 we have g (l) (z) = (−1)l (l + 1)!

X

w∈L\{0}

1 , (z − w)l+2

(27.63)

i.e. g (l) (z) = (−1)l (l + 1)! Gl+2. For l odd we know by (27.61) that Gl+2 = 0, i.e. g (l) (0) = 0. For l = 2k we find g (2k) (0) = (2k + 1)! G2(k+1) 547

A COURSE IN ANALYSIS

implying that a2k =

(2k + 1)! G2(k+1) g (2k) (0) = = (2k + 1)G2(k+1) . (2k)! (2k)!

Remark 27.27. The series Gn are called Eisenstein series and they are of great importance in several branches of Mathematics. Recall that Gn depends on L, i.e. Gn = Gn (L). We can now prove the first main result in Weierstrass’ theory of elliptic functions. Theorem 27.28. Let L = L(w1 , w2 ) be a periodicity lattice and f an elliptic function with periodicity lattice L. Then there exists two rational functions R1 and R2 such that with ℘(z) = ℘(z, L) we have f (z) = R1 (℘(z)) + ℘0 (z)R2 (℘(z)).

(27.64)

Proof. We prove this theorem in several steps similar to the structure of the proof given in [28]. 1. Assume that f is not constant, even and that all poles of f are contained in L. This implies that f must have a pole at 0 and the Laurent expansion of f about 0 is of the form (n)

(n)

f (z) = a−2n z −2n + a2(n−1) z −2(n−1) + · · · . The Laurent expansion of ℘(z)n about 0 is of the form ℘(z)n = z −2n + · · · . (n)

This implies that the function f − a−2n ℘n is an even elliptic function with poles only on L which have order strictly less than ord(f ). Hence for some m < n we have that (n) (m) f − a−2n ℘n − a−2m ℘m (m)

is an elliptic function with poles only on L where a−2m is determined from (n)

(m)

(m)

f − a−2n ℘n = a−2m z −2m + a−2(m−1) z −2(m−1) + · · · . 548

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

Thus, after at most n steps we find that f−

n X k=1

(k)

a−2k ℘k

is an even elliptic function with no pole, i.e. it must be constant. Hence we have proved: If f is an even elliptic function with poles only on L then f is a polynomial ) in ℘(·, L) of order ord(f . 2 2. Suppose that f is still even and not constant but f may have poles not only on L. Let a ∈ / L be such a pole of f and consider h(z) = (℘(z) − ℘(a))N f (z),

where N is determined such that the singularity of h at a is removable. Since f can have only a finite number of poles in P (w1, w2 )\L, say a1 , . . . , aM , we deduce that with suitable numbers Nj ∈ N g(z) := f (z)

M Y j=1

(℘(z) − ℘(aj ))Nj

has no pole outside of L. This implies that g is a polynomial in ℘ and f is a rational function of ℘, i.e. f (z) = R(℘(z)) for some rational function R. 3. Now let f be an arbitrary non-constant elliptic function with periodicity lattice L and ℘ = ℘(·, L) the Weierstrass ℘-function associated with L. We decompose f = feven + fodd into its even and odd part where feven (z) = 1 (f (z) + f (−z)). Both are elliptic functions with periodicity lattice L. For 2 f even we have already a representation feven (z) = R1 (℘(z)) with a rational function R1 . Now, ℘0 is an odd function and therefore fodd ℘0 is an even function which is an elliptic function with periodicity lattice L. = R2 (℘) which Hence we can find a rational function R2 such that fodd ℘0 eventually yields f (z) = feven (z) + fodd (z) = R1 (℘(z)) + ℘0 (z)R2 (℘(z)).

549

A COURSE IN ANALYSIS

As the Jacobi elliptic functions sn, cn and dn are double periodic functions ˜ Lcn = Lcn (4K, 2K + 2iK) ˜ with periodicity lattice Lsn = Lsn (4K, K + 2iK), ˜ and Ldn = Ldn (2K, 4iK), respectively, to each of these functions corresponds a Weierstrass ℘-function with the respective periodicity lattice. This allows us to represent the Jacobi elliptic function with the help of the ℘-function, its derivative and two rational functions in ℘. We do not want to enter into a more detailed discussion and refer for example to [88]. Finally we want to derive a differential equation for ℘. For this consider the function H(z) = (℘(z) − e1 ) (℘(z) − e2 ) (℘(z) − e3 ) (27.65)  2 and where ej is as in (27.59). Since ℘0 (w) ˜ = 0 for w˜ = w21 , w22 , w1 +w 2 w1 +w2 w1 w2 0 these are simple roots of ℘ , it follows that 2 , 2 and 2 are zeroes of multiplicity two of H. Moreover (℘0 )2 has roots of order two at w21 , w22 and w1 +w2 . In addition H has poles of order 6 at w1 , w2 and w1 +w2 as does (℘0 )2 . 2 0 2 This implies that the elliptic function (℘H) which has periodicity lattice L is in fact holomorphic, hence constant. By using the Laurent expansion about zero, i.e. 1 2 ℘(z) = 2 + · · · , ℘0 (z) = − 3 + · · · , z z we deduce that

(℘0 (z))2 H(z)

= 4, and we have proved

Theorem 27.29. The Weierstrass ℘-function satisfies the differential equation (℘0 )2 = 4 (℘ − e1 ) (℘ − e2 ) (℘ − e3 ) . (27.66) Note that solving the differential equation (g 0 )2 = 4(g − e1 )(g − e2 )(g − e3 ) for a function g = g(t), t ∈ R, leads by the method of separating variables, formally to Z dg p = t − t0 2 (g − e1 )(g − e2 )(g − e3 )

and we encounter an elliptic integral!

550

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

There is another way to express (27.66). For this we use a very formal but justifiable way of calculation. Given the Laurent expansion about 0 of two meromorphic functions f (z) = a−k z −k + a−(k−1) z −(k−1) + · · · and g(z) = b−m z −m + b−(m−1) z −(m−1) + · · · . We obtain the Laurent expansion of the product by multiplying these two series and then arranging the product series according to increasing powers: (f · g)(z) = a−k b−m z −k−m + (a−k b−(m−1) + a−(k−1) b−m )z −k−m−1 + · · · . Thus we find for ℘ and ℘0 ℘(z) = z −2 + 3G4 z 2 + 5G6 z 4 + · · · ℘0 (z) = −2z −3 + 6G4 z + 20G6 z 3 + · · · ℘(z)2 = z −4 + 6G4 + 10G6 z 2 + · · · ℘(z)3 = −z −6 + 9G4 z −2 + 15G6 + · · · and ℘0 (z)2 = 4z −6 − 24G4 z −2 − 80G6 + · · · which implies that ℘0 (z) − 4℘(z)3 + 60G4 ℘(z) + 140G6 is an elliptic function with periodicity lattice L which is holomorphic, hence it must be constant. Since for z = 0 this holomorphic function is 0, we arrive at Theorem 27.30. The Weierstrass ℘-function associated with the periodicity lattice L = L(w1 , w2 ) satisfies the differential equation (℘0 )2 = 4℘3 − 60G4 ℘ − 140G6 .

(27.67)

Comparing (27.66) with (27.67) allows us to establish relations between e1 , e2 and e3 and the Eisenstein series G4 and G6 . Finally we state a fundamental result linking certain elliptic integrals with elliptic functions. 551

A COURSE IN ANALYSIS

Theorem 27.31. Let p(t) be a polynomial of degree 3 or 4 without multiple zeroes. Then there exists a non-constant elliptic function f and an open set D ⊂ C such that f |D admits an inverse g : f (D) → C and (with an appropriate choice of the branch of the square root we have g 0(z) = R z function) 1 1 1 √ , i.e. g is a primitive of √ , or g(z) = a √ dt. p(z)

p(·)

p(t)

Here we end our discussion of elliptic integrals and elliptic functions, but we emphasise once again that the theory and its application just starts with these results. In addition to the sources already mentioned, e.g. [12], [28], [83] and [88] we recommend [53] as a further reading.

Problems 1. For 0 < b < a let γ : [0, 2π] → R2 , γ(t) = (a sin t, b sin t), be a parame2 2 trization of the ellipse xa2 + yb2 = 1. Find the length lγ of the ellipse in terms of an elliptic integral. 2. The lemniscate is the algebraic curve in the plane determined by the equation (x2 + y 2 )2 = x2 − y 2, see the corresponding figure in the solution. In polar coordinates the lemniscate is given by the equation r 2 = 2 cos 2ϕ. Find the length of an arc of the lemniscate starting at 0 and terminating at a point z with arg(z) < π4 . Use symmetry and a consideration for the limit ϕ → π4 to find the length l of the lemniscate. 3. Prove formula (27.6). 4. With K(k) as in (27.22) prove Z

π 2

0

√ dx √ = 2 sin x

Z

0

2π

q

dt 1 − 21 sin2 t

=

√

  1 . 2K 2

5. Let a, b ∈ R and a+b 6= 0 and K(k) as in (27.22). Using the substitution ϑ = 2ϕ prove J :=

Z

0

π

dϑ 2 √ = 2 2 a+b a + 2ab cos ϑ + b 552

Z

π 2

0

q

dϕ 1−

4ab (a+b)2

sin2 ϕ

.

27

ELLIPTIC INTEGRALS AND ELLIPTIC FUNCTIONS

Now apply the substitution a sin(ϑ − ψ) = b sin ψ to derive Z π Z π dϑ 2 2 1 √ q dψ. = 2 2 2 a 0 a + 2ab cos ϑ + b 0 1 − b sin2 ψ a2

Deduce

or with k =

√ 2 ab a+b

√ !   b 2 2 ab K , = K a a+b a+b √ and k 0 = 1 − k 2   1 − k0 2 . K K(k) = 1 + k0 1+k

6. Prove the equation and

0

dn2 z − k 2 cos2 z = k 2 sn2 z =

1 − cos 2z . 1 + dn2z

7. Verify (27.16) - (27.18). Hint: work through the solution we have provided as a proof of (27.16). 8. For the Jacobi elliptic functions sn, cn and dn derive the following differential equations:  2 dsnz = (1 − sn2 z)(1 − k 2 sn2 z); dz  ccnz 2 0 = (1 − cn2 z)(k 2 + k 2 cn2 z); dz  2 ddnz 0 = −(1 − dn2 z)(k 2 − dn2 z). dz 9. Give a more detailed proof of Lemma 27.19. Formulate an extension of this lemma to Rn and indicate the proof. 10. Consider the Weierstrass ℘-function ℘(z, L) as a function of z and w1 , w2 , L = L(w1 , w2 ), i.e. write ℘(z, L) = ℘(z; w1 , w2 ). Prove that (z, w1 , w2) 7→ ℘(z; w1 , w2 ) is homogeneous of degree −2. 553

A COURSE IN ANALYSIS

11. Let L = L(1, τ ), Im τ > 0. For the corresponding Eisenstein series Gn , n ≥ 3, prove with Gn (τ ) := Gn (1; τ ) that   1 −n . Gn (τ + 1) = Gn (τ ) and Gn (τ ) = τ Gn − τ

554

28

The Riemann Mapping Theorem

Our approach to complex-valued functions of a complex variable is in some sense an analytic approach putting properties of holomorphic functions on the centre stage. However on several occasions we encountered the need to think more (complex-) geometrical. In fact, once basic notions and tools have been introduced, we can look more at topological and most of all geometric or metric mapping properties of these functions. Here the aspect of mapping sets onto sets becomes more prominent. One such result is Theorem 23.2 stating that holomorphic functions are open mappings, i.e. they map open sets onto open sets, provided they are not constant. In this chapter we want to look at simply connected open sets in the plane and study their biholomorphic images. Geometry has entered somewhat hidden into our considerations in some other context: it is a problem to find “good” domains of definition for so called multi-valued functions such as z 7→ z α , α ∈ / Z, or z → log z, etc. Our way of dealing with these functions was to shrink their domains in such a way that they become well defined holomorphic functions admitting a primitive in this domain. This led to the introduction of branches of functions, for example in the case of the logarithmic function. Of similar nature was our problem when we dealt with elliptic integrals and Jacobi elliptic functions and tried to determine their periods. B. Riemann suggested (in our modern interpretation) that instead of shrinking the domains of a multi-valued function to enlarge the domains and go beyond subsets of the complex plane (or its one-point-compactification, the Riemann sphere). This resulted in the theory of Riemannian surfaces, thanks to H. Weyl [92] in the theory of differentiable manifolds, but also in the theory of uniformisation of (algebraic) functions. Eventually it turns out that both problems are closely connected and Koebe’s uniformisation theorem, according to L. Ahlfors [4] “perhaps the single most important theorem in the whole theory of analytic [=holomorphic] functions of one variable”, does for Riemann surfaces what the Riemann mapping theorem does for regions in C. The Koebe theorem and the theory of uniformisation will not be discussed here, see for example [60], and Riemann surfaces we will briefly handle in our final volume when dealing with differentiable manifolds. However the Riemann mapping theorem and some of its consequences we can treat and 555

A COURSE IN ANALYSIS

this is the aim of this Chapter. By Definition 17.18 two regions G1 and G2 are holomorphically equivalent if there exists a biholomorphic function f : G1 → G2 . By Proposition 17.19 a mapping f : G1 → G2 is biholomorphic if and only if f is holomorphic, bijective, f −1 is continuous and f 0 (z) 6= 0 for all z ∈ G1 . Since the identity is biholomorphic and since with f : G1 → G2 the function f −1 : G2 → G1 is biholomorphic it follows that biholomorphy is an equivalence relation in the set of all regions. From Liouville’s theorem we deduce that there cannot exist a biholomorphic mapping from C onto the unit disc D := B1 (0). Indeed, if f : C → D were biholomorphic it would be a bounded entire function, hence constant. Thus the problem to classify all biholomorphically equivalent regions is a non-trivial one. Since biholomorphic mappings are also homeomorphisms we restrict ourselves to the topologoical simplest situation namely to simply connected regions. Before we turn to the Riemann mapping theorem and its proof we first want to look at some concrete examples and then we want to look at biholomorphic mappings from a simply connected region into itself and discuss the underlying group structure. It is helpful to use in this chapter as notation for special sets  D := B1 (0) and H := z ∈ C Im z > 0

(28.1)

which are both simply connected regions. We next use our results on M¨obius transformations from Chapter 16. In Proposition 16.23 we have proved the following:

 az+b maps the punctured plane C\ − dc biThe M¨obius transform w(z) = cz+d jectively  and continuously with continuous inverse onto the punctured plane C\ ac .

For z 6= − dc the function w is holomorphic. Further we find   a b det c d 0 6= 0 (28.2) w (z) = (cz + d)2   a b az+b since for a M¨obius transform cz+d by assumption det 6= 0. Thus, by c d Proposition 17.19 the M¨obius transform w is a biholomorphic mapping bet556

28

THE RIEMANN MAPPING THEOREM

  ween the two simply connected regions C\ − dc and C\ ac . As a corollary we obtain Corollary 28.1. Two punctured planes are biholomorphic equivalent.  We can exploit Corollary 28.1 further. Let G1 ⊂ C\ − dc be a simply conaz+b nected region and w(z) = cz+d a M¨obius transformation. Then w(G1 ) and G1 are biholomorphic equivalent provided ac ∈ / w(G1). i−z corresponding to the maConsider the M¨obius transformation W1 (z) = i+z   −1 i . with determinant −2i and inverse W1−1 (w) = V1 (w) = i−iw trix 1+w 1 i   a b then Recall that if a M¨obius transformation is associated with c d   −d b . We know already that W1 its inverse is associated with c −a maps C\{−i} biholomorphically onto C\{−1}. For z ∈ H it follows that |i − z| < |z + i|, hence W1 (H) ⊂ D. On the other hand for w = u + iv ∈ D it follows that     1 − u − iv i − i(u + iv) = Re Im V1 (w) = Im i + u + iv 1 + u + iv   (1 − u − iv)(1 + u − iv) = Re (1 + u)2 + v 2 1 − u2 − v 2 = > 0, (1 + u2 ) + v 2

thus V1 (D) ⊂ H and we have proved

maps the upper half-plane H biLemma 28.2. The mapping W1 (z) = i−z i+z holomorphically onto the unit disc D with inverse mapping V1 (w) = i−iw . i+w In Problem 1 we will see that W1 maps ∂H ∼ = R continuously onto ∂D \{−1}, 1 ∼ recall that ∂D = S . Exercise 28.3. Prove that the M¨obius transformation w(z) =  D ∩ H onto the first quadrant u + iv ∈ C u > 0, v > 0 .

For a ∈ D we define the M¨obius transform z−a Wa (z) := . 1 − az 557

1+z 1−z

maps

(28.3)

A COURSE IN ANALYSIS

The matrix associated with Wa is for a ∈ D.



1 −a −a 1



with determinant 1−|a|2 > 0

Lemma 28.4. For a ∈ D the M¨obius transformation Wa is a bijective mapping from D onto itself as well as from S 1 onto itself. The inverse transformation to Wa is W−a and we have Wa0 (0) = 1 − |a|2 ,

Wa0 (a) =

1 . 1 − |a|2

(28.4)

Proof. The to Wa is the M¨obius transformation corresponding to the  inverse  −1 −a matrix and hence is given by −a −1 z 7→ Since

−z − a z+a = = W−a (z). −az − 1 1 + az 

1 −a det −a 1 Wa0 (z) = (1 − az)2



=

1 − |a|2 , (1 − az)2

1 we find for z = 0 that Wa0 (0) = 1 − |a|2 and Wa0 (a) = 1−|a| 2. it Now let z = e , t ∈ [0, 2π]. It follows that it e − a eit − a = |Wa (z)| = 1 − ae−t e−it − a |eit − a| = 1, = eit − a

i.e. Wa (S 1 ) ⊂ S 1 . The same argument applies to W−a implying that Wa (S 1 ) = S 1 . Moreover, the boundary maximum principle, Theorem 23.8, yields now Wa (D) ⊂ D as well as W−a (D) ⊂ D which again gives that Wa (D) = D. Lemma 28.4 provides us with an example of a biholomorphic mapping from D onto itself D. We want to have a closer look at such mappings. First we note a simple general result. 558

28

THE RIEMANN MAPPING THEOREM

Lemma 28.5. Let G ⊂ C be a simply connected region. The set of all biholomorphic mappings from G onto itself forms a group under the usual compositions of mappings. Proof. Since with f and g also f ◦ g is a biholomorphic mapping from G onto G, and since f −1 is biholomorphic if f is, we only need add the remark that idG is a biholomorphic mapping. Definition 28.6. The automorphism group of a simply connected region G ⊂ C is the group of all biholomorphic mappings from G onto itself and is denoted by Aut(G). We want to determine Aut(D). Clearly, if ϕ ∈ R then z 7→ eiϕ z is an automorphism of D. Combined with Lemma 28.4 we find that for every ϕ ∈ R and a ∈ D an automorphism of D is given by z 7→ eiϕ Wa (z).

(28.5)

For the following it is more convenient  to consider Va (z) := −Wa (z). The −1 a matrix corresponding to Va is and now the matrix corresponding −a 1   −1 a to Va−1 becomes , i.e. Va = Va−1 . In light of (28.5) changing the −a 1 sign of Wa (z) can be compensated substituting ϕ by ϕ + π. The surprising result is that all automorphisms of D are of type (28.5) or equivalently of type ei(ϕ+π) Va (·). Theorem 28.7. For every automorphism W of D there exists ϑ ∈ R and a ∈ D such that a−z W (z) = eiϑ (28.6) 1 − az holds. Proof. The proof will make much use of the Lemma of Schwarz, i.e. Theorem 23.11. Let W be an automorphism of D. Denote by a ∈ D the unique point with W (a) = 0 and consider the automorphism U := W ◦ Va , a−z . We find that Va (0) = a, hence U(0) = 0. According to Va (z) = 1−az Theorem 23.11 it follows that |U(z)| ≤ |z| 559

for z ∈ D

(28.7)

A COURSE IN ANALYSIS

and |U 0 (0)| ≤ 1.

Since U −1 is also an automorphism of D and U −1 (0) = 0, we find by a further application of Schwarz’ lemma that |U −1 (w)| ≤ |w|

for all w ∈ D,

and with w = U(z) we arrive at |z| ≤ |U(z)|

for all z ∈ D,

|U(z)| = |z|

for all z ∈ D.

(28.8)

i.e. we have Now we use Corollary 23.12, the refinement to the Lemma of Schwarz, to conclude that for some ϑ ∈ R (or equivalently some γ ∈ S 1 ) we must have U(z) = eiϑz (or U(z) = γz). With z replaced by Va (z) and using the fact that Va ◦ Va = id, we eventually have eiϑ Va (z) = U(Va (z)) = W (Va ◦ Va (z)) = W (z).

Corollary 28.8. If an element of Aut(D) has the origin as a fixed point it is a rotation. Lemma 28.9. The groups Aut(D) and Aut(H) are isomorphic. Proof. We define the mapping Φ : Aut(D) → Aut(H) by Φ(W ) := W1−1 ◦ W ◦ W1 where W1 (z) = i−z is the mapping from Lemma 28.2, i.e. W1 maps i+z H biholomorphically onto D. This implies that Φ(W ) is holomorphic and bijective with inverse Φ(V )−1 = W1 ◦ V ◦ W1−1 which is again a holomorphic mapping, hence Φ(W ) is a biholomorphic mapping. Moreover we have Φ(W ◦ V ) = W1−1 ◦ (W ◦ V ) ◦ W1 = W1−1 ◦ W ◦ W1 ◦ W1−1 ◦ V ◦ W1 = Φ(W ) ◦ Φ(V ) and Φ(idD ) = idH . Thus we have proved that Φ is an isomorphism. 560

28

THE RIEMANN MAPPING THEOREM

Remark 28.10. Note that the proof of Lemma 28.9 yields that whenever G1 and G2 are biholomorphic equivalent then their automorphism groups Aut(G1 ) and Aut(G2 ) are isomorphic. It is now possible to determine the structure of elements of Aut(H).   a b ∈ SL(2; R) = Theorem 28.11. For V ∈ Aut(H) there exists M = c d {A ∈ GL(2; R) | det A = 1} such that V (z) =

az + b . cz + d

(28.9)

Conversely, every mapping V as in (28.9) belongs to Aut(H). We refer to [83] for a detailed proof, also compare with Problem 4. The next preparation for the proof of the Riemann mapping theorem is a type of compactness result for subsets of holomorphic functions. The reader should compare this result, which is sometimes called Montel’s theorem, with the theorem of Arzela-Ascoli, Theorem II.14.25, which we will use in its proof. Definition 28.12. Let G ⊂ C be a region and H = H(G) be a family of holomorphic functions f : G → C. We call H a normal family or a normal set if every sequence in H contains a subsequence which converges uniformly on compact sets of G. The limit of this subsequence is of course holomorphic but does not necessarily belong to H. Theorem 28.13. Let H be a family of holomorphic functions f : G → C, where G ⊂ C is a region. If H is uniformly bounded on every compact subset K of G, i.e. sup kf k∞,K ≤ MK < ∞ (28.10) f ∈H

where MK is independent of f ∈ H, then H is a normal family. Proof. The first observation is that if (Kn )n∈NS, Kn ⊂ G compact, is a com˚n+1 and pact exhaustion of G, i.e. Kn ⊂ K n∈G Kn = G, and if we can prove that every sequence in H has a subsequence converging uniformly on each set Kn , then the result follows by a standard diagonal argument. Next we note that the Arzela-Ascoli theorem, Theorem II.14.25 also holds 561

A COURSE IN ANALYSIS

for complex-valued functions where in the proof we only need to replace the absolute value in R by the modulus in C, see also [38] which we used in our proof of Theorem II.14.25. Moreover, condition (28.10) implies that H(z) := {f (z) | f ∈ H} is relative compact in C as long as z belongs to a compact set K or Kn , respectively. Thus, if we can prove that H|Kn = {f |Kn | f ∈ H} is equi-continuous, then by the Arzela-Ascoli theorem every sequence in H|Kn would have a subsequence converging uniformly on Kn implying the theorem. Now we prove the equi-continuity condition. For Kn there exists δn > 0 such that for z ∈ Kn , we have B2δn (z) ⊂ Kn+1 . Let z1 , z2 ∈ Kn such that |z1 − zn | < δn . Let γ(t) = z1 + 2δn eit , t ∈ [0, 2π]. Since 1 z1 − z2 1 − = ζ − z1 ζ − z2 (ζ − z1 )(ζ − z2 ) the Cauchy integral formula yields for f ∈ H that z1 − z2 f (z1 ) − f (z2 ) = 2πi

Z

γ

f (ζ) dζ. (ζ − z1 )(ζ − z2 )

(28.11)

Since |ζ − z1 | = 2δn and |ζ − z2 | > δn for ζ ∈ tr(γ) by our assumptions, we deduce from (28.11) |f (z1 ) − f (z2 )| ≤

MKn+1 |z1 − z2 | δn

for all f ∈ H, i.e. H|Kn is equi-continuous and H is a normal family. Remark 28.14. Our proof of the equi-continuity of H|Kn follows [71] where a proof independent of the Arzela-Ascoli theorem is given. More precisely, once the equi-continuity of H|Kn is proved, a proof of the Aszela-Ascoli theorem is in this special case is provided in [71]. Our final preparations for the proof of the Riemann mapping theorem states the existence of the (complex) square root of certain holomorphic functions. For this we need the existence of a branch of the logarithmic functions first. Theorem 28.15. Let G ⊂ C be a simply connected region and f : G → C be a holomorphic function not vanishing in G. Then there exists a branch g of the logarithmic function in G such that f (z) = eg(z) . 562

28

THE RIEMANN MAPPING THEOREM

Proof. Fix z0 ∈ G and choose c0 ∈ C such that ec0 = f (z0 ). For z ∈ G let γ be a continuously differentiable path in G connecting z0 with z. Consider the integral Z 0 f (w) dw + c0 . g(z) = γ f (w) 0

Since ff is holomorphic in G, g(z) is independent of the choice of γ, recall that G is simply connected. We can now argue similarly as in the proofs leading to Theorem 20.23 and Theorem 21.4 to deduce that g is a holomorphic function. Using
d f (z)e−g(z) = (f 0 (z) − f (z)g 0 (z)) e−g(z) = 0 dz we derive that z 7→ f (z)e−g(z) is constant in the connected set G. Since f (z0 )e−g(z0 ) = 1 it follows that f (z) = eg(z) . Corollary 28.16. Let G ⊂ C be a simply connected region and f p :G→C be a holomorphic function not vanishing in G. Then by h(z) = f (z) = 1 1 (f (z)) 2 = e 2 g(z) a holomorphic branch of the square root of f in G is defined, i.e. h2 (z) = f (z) in G. Now we can prove a version of the Riemann mapping theorem. Theorem 28.17. A simply connected region G ⊂ C, G 6= C, is biholomorphically equivalent to D. Proof. First we prove the existence of an injective holomorphic mapping W : G → D. For this let w0 ∈ G{ . By Corollary 28.16 there exists a holomorphic function V : G → C such that V 2 (z) = z − w0 . This mapping is injective since V (z1 ) = V (z2 ) implies that z1 − w0 = z2 − w0 , i.e. z1 = z2 . Furthermore suppose that for z1 , z2 ∈ G we have V (z1 ) = −V (z2 ), i.e. V 2 (z1 ) = V 2 (z2 ). As before we deduce now that z1 = z2 . As a holomorphic mapping V is open, compare with Theorem 23.2, i.e. V (G) is a region. By our previous considerations we can find a ∈ V (G) and r, 0 < r < |a|, such that Br (a) ⊂ V (G) and Br (−a) ∩ V (G) = ∅. We define now the mapping r W : G → C by W (z) = V (z)+a . Since −a ∈ / V (G) it follows that W is a holomorphic mapping and since the distance of V (z) to −a is by construction larger than r we have r < 1. |W (z)| = V (z) + a 563

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The injectivity of V implies that W : G → D is an injective holomorphic mapping. Next let W : G → D be an injective holomorphic mapping such that W (G) 6= D, i.e. there exists some w ∈ D not belonging to W (G). We claim that in this case for z0 ∈ G there exists W1 : G → D holomorphic and injective such that |W 0 (z0 )| < |W10 (z0 )|. (28.12) z−a Let Wa (z) = 1−az be the M¨obius transformation (28.3) considered in Lemma 28.4. We know that Wa is a biholomorphic mapping from D into itself with inverse W−a . If W : G → D is an injective holomorphic mapping and a ∈ D ∩ W (D){ then Wa ◦ W : G → D is again an injective mapping and in addition (W0 ◦ W )(z) 6= 0 for all z ∈ G. Again by Corollary 28.16 we can find a holomorphic mapping H : G → C such that for z ∈ G we have H 2 (z) = (Wa ◦ W )(z). With arguments along the lines used above we deduce that H is also injective and maps G into D. We set b := H(z0 ) and define W1 := Wb ◦ H : G → D which is an injective holomorphic mapping. With S : C → C, S(w) := w 2 , it follows now that

W = W−a ◦ S ◦ H = W−a ◦ S ◦ W−b ◦ W1 . With F := W−a ◦ S ◦ W−b the chain rule yields W 0 = F 0 (W1 )W10 and since W1 (z0 ) = 0 we find W 0 (z0 ) = F 0 (0)W1 (z0 ). The mapping F maps D into itself but since S is not injective F is not injective either. We can apply the lemma of Schwarz in form of Corollary 23.12 to F and we deduce that |F 0(0)| < 1, or |W 0 (z0 )| < |W10 (z0 )|. Note that since W is injective we have W 0 (z0 ) 6= 0. For z0 ∈ G fixed we set  m := sup |W 0(z0 )| W : G → D is injective and holomorphic . (28.13)

In view of (28.12) if we can find an injective holomorphic mapping W0 : G → D for which the supremum in (28.13) is attained then W0 must be biholomorphic mapping. The family of all injective mappings W : G → D is by the Montel theorem, Theorem 28.13, a normal family. Note that sup |W (z)| ≤ 1 for all W : G → D and z ∈ G. By the definition of the supremum there 564

28

THE RIEMANN MAPPING THEOREM

must exist a sequence Wk : G → D of injective holomorphic mappings with limk→∞ Wk0 (z0 ) = m, and this sequence must have a subsequence, again denoted by (Wk )k∈N , which converges uniformly on compact subsets of G to some mapping J, hence |J 0 (z0 )| = m. We know that the set of all injective holomorphic mappings W : G → D is not empty and that m > 0. Thus J is not a constant mapping and since Wk (G) ⊂ D it follows that J(G) ⊂ D where we used that by Theorem 23.2 the holomorphic mapping J is open. Our aim is now to prove that J is injective. Let z1 , z2 ∈ G, z1 6= z2 , and put a := J(z1 ), ak := Wk (z1 ), and choose r > 0 such that z1 ∈ / Br (z2 ) and such that J − a has no zeroes on ∂Br (z2 ). Note that in Br (z2 ) the function J − a can have only finitely many zeroes. Then if we reduce r a little if necessary we can always achieve that J − a has no zeros on ∂Br (z2 ). The sequence (Wk − a)k∈N converges uniformly on Br (z2 ) to J −a, and the functions Wk −a do not have a zero in Br (z2 ) since they are injective and Wk (z1 ) − a = 0. Now, as seen in Problem 6, we can apply the theorem of Rouch´e, Theorem 24.26, to conclude that J − a has no zero in Br (z2 ), which yields in particular that J(z1 ) 6= J(z2 ). Remark 28.18. Our proof of Theorem 28.17 is the standard proof used nowadays and which goes back to C. Carath´eodory and P. Koebe. We followed essentially the presentation in [71], but we have to make some adaptations to our different approach to some auxillary results. Suppose that G 6= C is a simply connected region and let W : G → D be biholomorphic mapping. This mapping induces a bijective mapping −1 : H(G) → H(D), HW : H(D) → H(G) by f 7→ f ◦ W with inverse HW −1 HW = HW −1 . Thus, once we understand H(D) we understand up to biholomorphic mappings all spaces H(G), G ⊂ C, G 6= C and a simply connected region. However there might be severe problems at the boundaries. So far we have not even established that we can map G onto D with the help of a continuous extension of a biholomorphic mapping W : G → D. In fact this is not necessarily true. The discussion of the boundary regularity of biholomorphic mappings W : G1 → G2 is a very interesting, challenging and important subject, but it goes beyond the scope of our Course which gives a first encounter with complex variable theory. Finally we want to give an application of biholomorphic mappings to the theory of harmonic functions.

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Let f : G2 → G1 be a holomorphic function f = u+iv and with z = x+iy we write f (z) = u(x, y) + iv(x, y). For u and v the Cauchy-Riemann differential equations hold, i.e. ux = vy and uy = −vx . This implies in particular that |f 0(z)|2 = |ux + ivx |2 = u2x + u2y = vx2 + vy2 , compare with Example 17.13. Moreover we know that u and v must be harmonic, i.e. ∆u = ∆v = 0, see Theorem 17.20 and add that we know meanwhile that u and v must be C ∞ -functions. Now let h : G1 → R be a harmonic function and with w = f (z) = u + iv consider the function g : G2 → R, g = h ◦ f . A straightforward calculation yields for g(x, y) = h(u(x, y, v(x, y))) that ∂2h 2 ∂2h ∂ 2 h 2 ∂h ∂h ∂2g = u + 2 v + v u + uxx + vxx x x x ∂x2 ∂u2 ∂u∂v ∂v 2 x ∂u ∂v and

∂2g ∂ 2 h 2 ∂h ∂h ∂2h 2 ∂2h v u + uyy + vyy = u vy + + 2 y y y 2 2 2 ∂y ∂u ∂u∂v ∂v ∂u ∂v which yields
∂2h 2
∂2h 2 ∂2g ∂2g 2 2 + = u + u + v + v x y x y ∂x2 ∂y 2 ∂u2 ∂v 2 ∂h ∂h ∂2h + ∆u + ∆v + 2 (vx ux + vy uy ) ∂u  ∂v ∂u∂v  ∂2h ∂2h 2 = |f 0 (z)| . + ∂u2 ∂v 2 Since h is by assumption harmonic we have proved Proposition 28.19. If f : G2 → G1 is holomorphic and h : G1 → R is harmonic, then g = h ◦ f : G2 → R is harmonic too. Many two-dimensional problems in physics or mechanics, but also differential geometry lead to the task to find a harmonic function with special properties, for example prescribed behaviour at the boundary. Harmonic functions on D or H are easy to construct as we will see in our Course when dealing with partial differential equations. Thus if we can map a complicated (simply connected) region onto a simpler one, say D or H, by a biholomorphic mapping which also behaves “nice” at the boundary we can reduce the original problem from physics or mechanics or geometry to a much simpler one. For such a strategy the Riemann mapping theorem is a good starting point.

566

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In this chapter we dealt with biholomorphic mappings f : G1 → G2 and by their very definition they are bijective, hence injective. Moreover their derivative f 0 never vanishes. A more classical way and also still very important approach is to start with conformal mappings. These are by definition holomorphic functions h : G1 → G2 with h0 (z) 6= 0 for all z ∈ G1 . This condition yields only local injectivity. The importance of conformal mappings lies in the fact that in the small they preserve angles. Conformal mappings have a large domain of real-world applications as they have many interesting mathematical properties. Again, we will not handle the theory of conformal mappings in our Course and have to refer to the literature, e.g. [3], [4], [17], [20], [37], [61], [68], [80] or [83].

Problems 1. Consider W1 : H → D, W1 (z) = i−z , as in Lemma 28.2 and prove that i+z W1 maps ∂H continuously onto a subset of ∂D. 2. Solve Exercise 28.3. 3. Prove that if G1 and G2 are two biholomorphically equivalent regions then Aut(G1 ) is isomorphic to Aut(G2 ).   a b ∈ SL(2; R) induces by 4. Prove that A = c d fA (z) :=

az + b cz + d

an automorphism of H. 5. Let G ⊂ C be a region and H a normal family of holomorphic functions f : G → C. Prove that for every k ∈ N the family H := {f (k) |f ∈ H} is normal too. Hint: use the standard estimates for derivatives. 6. Fill the gaps at the end of the proof of the Riemann mapping theorem, i.e. use the theorem of Rouch´e to prove that J − a has no zero in D. 567

A COURSE IN ANALYSIS

7. Verify that h(x, y) = x2 −y 2 is in R2 ∼ = C harmonic. Find the harmonic function g : {(x, y) ∈ R2 |y > 0} → R where g = h ◦ W1 and W1 is from Problem 1.

568

29

Power Series in Several Variables

In this chapter we discuss some properties of power series of several variables, complex or real. We do not intend to give an introduction to the theory of complex-valued functions of several variables, this is beyond our Course. However already when looking at real-valued functions of several real variables, we stopped short of treating Taylor series which should be of course power series. We have seen the very close relationship between power series of one real variable and of one complex variable. Now we will see some similarity in several variables, but convergence domains (and results) will be quite different. This is best understood in several complex variables although in [51] a readable treatment of the real case is available, and it also indicates why the theory of differentiable functions from some set G ⊂ Cn to C should be expected to have quite different features than the one-dimensional theory. In our presentation we make use of [29] and [66]. We will use multi-index notation as we are used to from Volume II, in particular for z = (z1 , . . . , zn ) ∈ Cn and α ∈ Nn0 we will write z α := z α1 · . . . · znαn .

(29.1)

While dimension there is a natural way partial sums of a series PNto form P∞ in one k k c z , N ∈ N0 , this is not the c z , c ∈ C, i.e. we consider S := k N k=0 k k=0 k n case in several variables. Every enumeration of N0 will give rise to a sequence of partial sums and there is no reason to consider one as superior to the other. Since we know that for absolutely convergent series every rearrangement will lead to the same limit, see Theorem I.18.27 for the real case which easily extends to the complex case, we discuss only absolutely convergence series in case of several variables. In order to proceed we need some definitions. Definition 29.1. A. Any bijective mapping ϕ : N0 → Nn0 we call an enumeration of Nn0 . P B. For a sequence (cα )α∈Nn0 of complex numbers we call the series α∈Nn cα 0 P∞ cϕ(k) = convergent if for every enumeration ϕ of Nn0 the series k=0 P limN →∞ N k=0 cϕ(k) converges. P Remark 29.2. Convergence of α∈Nn cα is always absolute convergence, i.e. 0 P∞ P∞ k=0 cϕ(k) k=0 cϕ(k) converges. Since by Theorem I.18.27 the limit of is of the order of summation it follows that the convergence of Pindependent ∞ c k=0 ϕ(k) for one enumeration will already imply the convergence for all 569

A COURSE IN ANALYSIS n other enumerations of NP 0 . A different question is whether for all enumerations the limit limN →∞ N k=1 cϕ(k) is the same. The result holds and can be proved along the lines of the proof of Theorem I.18.27, see Problem 1.

Definition 29.3. A. The open ball in Cn with centre a ∈ Cn and ran dius r > 0 is denoted by Br (a) := z ∈ C kz − ak < r , where kzk = 1

(|z1 |2 + . . . + |zn |2 ) 2 . B. The polydisc P (a, ~r) ⊂ Cn with centre a ∈ Cn and multi-radius ~r = (r1 , . . . , rn ), rj ≥ 0, is the set  (29.2) P (a, ~r) = z ∈ Cn |zj − aj | < rj , j = 1, . . . , n .

C. The distinguished boundary ∂d P (a, ~r) is the set  ∂d P (a, ~r) = z ∈ Cn |zj − aj | = rj , j = 1, . . . , n .

(29.3)

Remark 29.4. The reader can easily check that

P (a, ~r) = Br1 (a1 ) × · · · × Brn (an )

(29.4)

∂d P (a, ~r) = ∂Br1 (a1 ) × · · · × ∂Brn (an ).

(29.5)

and In particular for a = 0 ∈ Cn and ~1 = (1, . . . , 1) we can identify ∂d P (0, ~1) with the n-dimensional torus Tn = S 1 × · · · × S 1 ⊂ Cn . The topology on Cn induced by kzk is of course the Euclidean topology on R2n and the family of all open balls Br (a) as well as the family of all polydiscs P (a, ~r) form a base of this topology.  Definition 29.5. The set Rn+ = x ∈ Rn xj ≥ 0 is called the absolute space (of Cn ) and τ : Cn → Rn+ , τ (z) = τ (z1 , . . . , zn ) := (|z1 |, . . . , |zn |) is called the natural projections of Cn onto Rn+ . Note that for ~r ∈ Rn+ the pre-image under τ is the set ∂d P (0, ~r). Definition 29.6. An open connected set R ⊂ Cn is called a Reinhardt domain if z ∈ R implies that ∂d P (0, τ (z)) ⊂ G. Note that ∂d P (0, τ (z)) = τ −1 (τ (z)). If G is Reinhardt domain then G = τ −1 (τ (G)). 570

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POWER SERIES IN SEVERAL VARIABLES

Definition 29.7. A. A Reinhardt domain G ⊂ Cn is called proper if 0 ∈ G. B. A complete Reinhardt domain G ⊂ Cn is a Reinhardt domain with the property that for all z ∈ G ∩ (C\{0})n it follows that P (0, τ (z)) ⊂ G. Note that G ⊂ Cn is a complete Reinhardt domain if and only if G=

[

P (0, τ (z)).

z∈G

We want to discuss some examples. Of course even for n = 2 there is no chance to sketch an open set G ⊂ C2 . However, for n = 2 the absolute space is a subset of R2 and we can at least draft τ (G) for n = 2. For Br (0) ⊂ C2 we find that τ (Br (0)) is the quarter disc in R2 , see Figure 29.1, and for P (0, (r1, r2 )) ⊂ C2 the natural projection τ (P (0, (r1 , r2 ))) is a square, see Figure 29.2. |z2 |

|z2 |

r r2 τ (Br (0)) τ (P (0, (r1 , r2 ))) 0

r

|z1 |

0

Figure 29.1

r1

|z1 |

Figure 29.2

Note that in both cases the dotted lines do not belong to the set. It is not hard to see that polydiscs and open balls with centre 0 ∈ Cn are complete and proper Reinhardt domains. For a complete Reinhardt domain G ⊂ C2 the natural projection τ (G) looks as in Figure 29.3 while Figure 29.4 features the natural projection of a possible Reinhardt domain which however is not complete. 571

A COURSE IN ANALYSIS |z2 |

0

|z2 |

|z1 |

|z1 |

0

Figure 29.3

Figure 29.4

Example 29.8. Let a = (a1 , a2 ) ∈ C2 , |aj | > 1. For every t ∈ R it follows that τ (eit a) = (|eit a1 | , |eit a2 |) = (|a1 |, |a2|) = τ (a). We claim that we can find some  > 0 such that P (a, (, )) is not a Reinhardt domain. For this we note that |eit a − a| = |eit − 1| |a| > |eit − 1| since |aj | > 1 which implies for t suitably chosen that |eit a − a| ∈ / P (a, (, )). We now turn to power series about the origin which is sufficient for our purposes: X Tcα (z) = cα z α . (29.6) α∈Nn 0

Definition 29.9. The domain of convergence of Tcα is the interior of the set of all z ∈ Cn for which (29.6) converges (in the sense of Definition 29.1). P k The key result for power series ∞ k=0 ak (z − c) in one dimension was Abel’s theorem, Theorem 16.6, with Corollary 16.9. If the series converges for some z0 6= c then it converges in the open set Bρ (c), 0 < ρ < |z0 |. Combined with the Hadamard-Cauchy theorem, Theorem 16.10, we know that indeed there exists a maximal R ≥ 0 called the radius of convergence such that locally on {

compact sets of Br (c) the series converges uniformly and in BR (c) the series diverges.

P Consider the series α∈Nn α1α1 z1α1 z2α2 . It is trivial that this series converges 0 on the complex line z2 = 0. However it diverges for every z1 6= 0 if z2 6= 0 572

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POWER SERIES IN SEVERAL VARIABLES

implying that its domain of convergence is empty. Thus convergence of a series of the type Tcα (z) for some point z0 ∈ C n will in general not help us to find a subset of its domain of convergence. The analogous result to Abel’s theorem in several variables, often called Abel’s lemma, is Theorem 29.10. Let P (0, r~0) ⊂ Cn be a polydisc with r~0 = (r0,1 , . . . , r0,n ), r0,j > 0, and let P (0, ~r) a further polydisc with ~r = (r1 , . . . , rn ), 0 < rj < r0,j for j = 1, . . . , n. If the power series X Tcα (z) = cα z α (29.7) α∈Nn 0

converges for one point z0 in the distinguished boundary of P (0, r~0), i.e. z0 ∈ ∂d P (0, r~0) then Tcα (z) converges for all z ∈ P (0, ~r). P Proof. Let z0 ∈ ∂d P (0, r~0) and suppose that α∈Nn cα z0α converges. It follows 0 that the sequence (|cα z0α |)α∈Nn must be bounded, i.e. |cα z0α | < K. Since 0 0 < rj < r0,j for 1 ≤ j ≤ n, we can find ηj ∈ (0, 1) such that for all z ∈ P (0, ~r) we have |zj | < ηj |z0,j |, which implies that |cα z α | ≤ η α , η = (η1 , . . . , ηn ), or supz∈P (0,~r) |cα z α | ≤ η α . Now we observe for η that

and we find that

P

X

|α|≤N

α∈Nn 0

α

η ≤

n X N Y

α ηj j

j=1 αj =1

≤

n Y

1 1 − ηj j=1

cα z α converges for z ∈ P (0, ~r).

The proof of Theorem 29.10 shows more, namely that the convergence in P (0, ~r) is uniform. A preferred notion of convergence in this context is that P f of normal convergence: a series ofP functions ∞ k=0 k , fk : G → C, is said kf k to be normally convergent in G if ∞ k ∞,G converges. Furthermore, k=0 since the open polydiscs form a base of the topology in Cn , we have Corollary 29.11. Let P (0, r~0) ⊂ Cn and suppose that the power series (29.7) converges for some point z0 ∈ ∂d P (0, r~0). Then Tcα (z) converges uniformly on every compact subset of P (0, r~0). The importance of Reinhardt domains is seen from the following Theorem 29.12. The domain D of convergence of a power series of type (29.7) is a complete Reinhardt domain. 573

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Proof. Let z0 ∈ D. Then there exists a polydisc P (z0 , ~) ⊂ D, ~ = (, . . . , ), i.e. P (0, ~) = B (z0,1 ) × · · · × B (z0,n ). We pick z1 = (z1,1 , . . . , z1,n ), z1,j ∈ B (z0,j ), such that |z1,j | > |z0,j |. It follows that z1 ∈ D and z0 ∈ P (0, τ (z1 )). For every z0 we fix such a point z1 . We know that Tcα (z1 ) converges, but z1 ∈ ∂d P (0, τ (z1 )), i.e. P (0, τ (z1 )) ⊂ D. Since P (0, τ (z0 )) ⊂ P (0, τ (z1 )) and ∂d P (0, τ (z0 )) ⊂ P (0, τ (z1 )) we deduce that D is a complete Reinhardt domain. Note that this result alreadyP indicates implications for the domain of convergence of a real power series α∈Nn aα xα , aα ∈ R, x ∈ Rn . When passing the 0 complexification xj 7→ zj = xj + iyj , we must end up in Cn with a Reinhardt domain. As it turns out, being a Reinhardt domain is not sufficient to be a domain of convergence. We need Definition 29.13. A Reinhardt domain G is called logarithmically convex if ln (τ (G ∩ (C\{0})n )) is a convex set in Rn . Here ln ~r = (ln r1 , . . . , ln rn ) and ln A stands for the image of the set A under ln. Theorem 29.14. The domain D of convergence of a power series of the type (29.7) is logarithmically convex. Proof. Let z1 , z2 ∈ D ∩ (C\{0})n and set x1 := ln(τ (z1 )) as well as x2 = ln(τ (z2 )). Since D is open, for η > 0 sufficiently small ηzj ∈ D ∩ (C\{0})n , hence Tcα (ηz1 ) and Tcα (ηz2 ) converge implying for some M that for all α ∈ Nn0 we have |cα |η |α| |z1α | ≤ M and |cα |η α |z2α | ≤ M. For 0 ≤ λ ≤ 1 this implies for every α ∈ Nn0 that |cα |η |α| |z1α |λ |z2α |1−λ ≤ M. From Theorem 29.10 we deduce that Tcα (·)
converges in a neighbourhood of zλ := |z1,1 |λ |z2,1 |1−λ , . . . , |z1,n |λ |z2,n |1−λ , i.e. zλ ∈ D and λx1 + (1 − λ)x2 = ln(τ (zλ )), or ln τ (zλ ) ∈ ln (τ (G ∩ (C\{0})n )) proving the theorem. Thus only logarithmically convex Reinhardt domains are convergence domains of power series of type (29.7). It turns out that the analogous result holds for real power series, details are given in [51]. One way of defining holomorphy in Cn , n ≥ 1, is Definition 29.15. Let D ⊂ Cn be an open set. We call f : D → C holomorphic if for every point z0 ∈ D there P exists an open neighbourhood U(z0 ) ⊂ D and a power series Tczα0 (z) = α∈N0 cα (z − z0 )α converging in U(z0 ) to f (z). 574

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In some sense Reinhardt domains then turn out to be maximal holomorphy domains. But here we have to stop.

Problems n 1. Let ϕ : N0 → Nn0 be an enumeration  of Nn0 and let  (cα )α∈N0 be a P converges. sequence of complex numbers for which |α|≤N |cα | N ∈N0 P P∞ Prove that k=0 cϕ(k) = α∈Nn cα . 0 P 2. a) Prove that the power series α∈N2 z α converges in B1 (0)×B1 (0). 0 P z1 and b) For z1 ∈ C and z2 ∈ B1 (0) prove that (1,α2 )∈N2 z α = 1−z 2 0 deduce that the power series converges in C × B1 (0). P c) Show that for |z1 z2 | < 1 the power series ν∈N0 z1ν z2ν converges.

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Appendices

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Appendix I: More on Point Set Topology In this appendix we want to collect some further results from topology, in particular the construction of a product topology as well as the construction of compactifications. Let (X, OX ) be a topological space. A family (Uj )j∈I of non-empty open sets is called a base of the topology (or the open sets) of X if every non-empty open set U ⊂ X is the union of a sub-family of (Uj )j∈I , see Definition II.3.7. Now let (Xk , OXk ), k = 1, . . . , N, be a finite family of topological spaces Q and X := N k=1 Xk the Cartesian product. We want to construct on X a topology such that all projections prk : X → Xk are continuous and which is minimal. It turns out, see [21], or [65] that the family ) ( N \ U ∈ X U = pr−1 (Oj ), Oj ∈ Oj j

j=1

is a base of this (uniquely determined) topology on X and this topology is called the product topology on X. In particular if (Xj , dj ), j = 1, . . . , N, are metric spaces then the product Q topology on X = N j=1 Xj induced by the metric topologies Odj (on Xj ) and P the topology on X induced by the metric d = N j=1 dj are equal. Within this frame where we can now define a topological group (G, OG , ◦) as a topological space (G, OG ) which carries a group operation comp : G×G → G, comp(g1 , g2 ) = g1 ◦ g2 , which together with the inverse is continuous, i.e. comp : G × G → G, comp(g1 , g2 ) = g1 ◦ g2 is continuous from (G × G, OG×G ) to (G, OG ), as the inverse, inv : G → G, g 7→ g −1, is continuous from (G, OG ) into itself. Examples of topological groups are GL(n; R), GL(n; C) as well as the subgroups SL(n; R), SL(n; C), O(n; R), U(n, C), SO(n; R) or SU(n; C). Similarly we can introduce topological vector spaces over R or C. The additional requirement to the vector space axioms is that vector addition and its inverse as well as scalar multiplication are continuous. Next we turn to the problem of finding a compactification of a given (noncompact) toplogical space. We first discuss the one-pont compactification or Alexandrov-compactification of a locally compact space (X, OX ). Recall that (X, OX ) is a locally compact space if it is a Hausdorff space and if every point x ∈ X has a compact neighbourhood. Here we call (X, OX ) 579

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a Hausdorff space if for x 6= y, x, y ∈ X, there exist (open) neighbourhoods U(x) and U(y) of x and y, respectively, such that U(x) ∩ U(y) = ∅. Every metric space (X, d) is a Hausdorff space since Br (x) ∩ Br (y) = ∅ for r = 21 d(x, y) > 0. The aim is to find for (X, OX ) a compact space (Y, OY ) such that Y contains a topological subspace Y1 , which is homeomorphic to X. If (X, OX ) is compact itself we can of course choose (Y, OY ) = (X, OX ). Theorem A.I.1 (Alexandrov compactification). Let (X, OX ) be a locally compact, non-compact Hausdorff space and let ∞ be a point not belonging to X, i.e. ∞ ∈ / X. Then the set Y := X ∪ {∞} admits a topology OY such that (Y, OY ) is a compact space. Moreover there exists a topological subspace Y1 ⊂ Y such that X is homeomorphic to Y1 , Y \Y1 = {∞}, and Y1 is dense in Y . The point ∞ in Theorem A.I.1 is called the point at infinity. A detailed proof of Theorem A.I.1 is given in [47] or [65]. We only want to indicate the construction of OY : U ∈ OY if either U ∈ OX or U = Y \K where K is a compact set in X. ˆ n of Rn can It is not difficult to see that the one-point compactification R n−1 be identified with the sphere S . In particular for R we obtain the circle S 1 ⊂ R2 , and for R2 we obtain the sphere S 2 ⊂ R3 . When identifying topologically C with R2 we now understand that topologically the Riemann sphere is the one-point compactification of C and that the north pole (0, 0, 1) ∈ S 2 ⊂ R3 can be looked at as the point at infinity. A further remark is helpful. If u ∈ C∞ (Rn ), i.e. u : Rn → R is a continuous function which vanishes at infinity, then u has a continuous extension to ˆ n ) when R ˆ n is considered as one-point compactification of Rn , we just C(R define u(∞) := 0 where ∞ denotes the point at infinity. It follows that ˆ n ). C∞ (Rn ) can be considered as a subspace of C(R The one-point compactification of R is not helpful when we want to maintain the natural order structure on R, i.e. when dealing with [−∞, ∞] = R. The solution is to pass to a two-point compactification by adding to R the two points −∞, ∞ ∈ / R, −∞ = 6 ∞. A natural way to define on R the topology OR is the following: choose as a base for OR all open sets belonging to R (or just all open intervals) and in addition all sets of the type {x ∈ R | x > a}, 580

APPENDIX I: MORE ON POINT SET TOPOLOGY

a ≥ 1, and {x ∈ R | x < b}, b ≤ −1. This compactification of R is the one (1) we need to handle B as Borel σ-field in Chapter 4.

We close this appendix by extending our discussion of the Lemma of Urysohn, see [47]. A topological space (X, OX ) is called normal topological space if for any pair of disjoint closed subsets C1 , C2 ⊂ X there exist disjoint open sets U1 , U2 ⊂ X such that C1 ⊂ U1 and C2 ⊂ U2 . The space Rn with the Euclidean topology is a normal space as is every metric space. Theorem A.I.2. For two disjoint closed sets C1 , C2 in a normal space (X, OX ) we can find a continuous function f : X → [0, 1] such that f |C1 = 0 and f |C2 = 1. Corollary A.I.3. Theorem A.I.2 holds for compact Hausdorff spaces. Corollary A.I.4. If (X, OX ) is a locally compact Hausdorff space and C1 , C2 ⊂ X are two disjoint closed sets then there exists a continuous function f : X → [0, 1] with compact support such that f |C1 = 0 and f |C2 = 1. Recall that the support of a continuous function is defined as  supp f = x ∈ X f (x) 6= 0 .

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Appendix II: Measure Theory, Topology and Set Theory When starting to study analysis in most investigations we depend on the notation of set theory but we really do not use much set theory. At the beginning we do not need to understand its foundations, its axioms, more precisely the system of axioms of a model of set theory. The standard one would be ZF , i.e. the Zermelo-Fraenkel system of axioms. Nonetheless when coming to more subtle topics we need more set theory and usually a second course on measure theory, functional analysis or point set topology on a graduate level starts with some deeper discussion of set theory. There are good reasons for this, and we shall not forget that set theory emerged from the efforts to understand the convergence of trigonometrical series, i.e. problems from analysis were the starting point of Cantor’s investigations leading to set theory. Measure theory starts by introducing σ-fields, a proper theory of continuous functions starts by introducing topologies. Both are systems of subsets of a given set closed under certain operations. In the case of σ-fields only certain countable operations are permitted, for topologies we allow arbitrary unions of open sets. We can use topologies to generate σ-fields (Borel σ-fields) and now it becomes apparent that a more detailed study of Borel sets (for example) will depend on properties of sets obtained from such operations, i.e. unions or intersections of certain families of sets, hence on the underlying model of set theory. A highly non-trivial ingredient from set theory needed to understand the relation of measure theory and topology is the axiom of choice. Let (Yj )j∈I , I 6= ∅, be an arbitrary family of sets and denote by j∈I Yj the set of all mapping f with domain I such that f (j) ∈ Yj . We call f a choice function for the family (Yj )j∈I and f (j) the j th coordinate of f .

×

Axiom of Choice (1st version) If I 6= ∅ and Yj 6= ∅ for all j ∈ I then there exists at least one choice function for the family (Yj )j∈I . We can rephrase the axiom of choice as follows Axiom of Choice (2nd version) Let I = 6 ∅ and Yj 6= ∅ for all j ∈ I. Suppose S that the sets Yj are mutually disjoint. Then there exists a set M ⊂ j∈I Yj 583

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which contains exactly one element from each set Yj , j ∈ I. It is remarkable that this axiom is independent of the other axioms of set theory, a result due to P. Cohen. The axiom of choice is equivalent to other surprising statements such as the, • Hausdorff maximality principle: every non-empty partially ordered set contains a maximal chain, • Lemma of Zorn: every partially ordered set in which each chain has an upper bound has a maximal element, • Well-ordering theorem of Zermelo: every set can be well-ordered. We do not need these results or notions here and therefore we refer to [35] or the nice general introduction to set theory [79]. The reader might be interested to note that the axiom of choice is in its equivalent form of Zorn’s lemma is also key in linear algebra: it is needed to prove that every vector space admits an algebraic basis, i.e. a maximal linearly independent set. Following closely R. Schilling [75] we prove Theorem A.II.1. Under the assumption that the axiom of choice holds there exists a non-Lebesgue measurable subset of [0, 1). Recall that the Lebesgue σ-field is the completion of the Borel σ-field with respect to the Lebesgue-Borel measure. Proof of Theorem A.II.1. We call x, y ∈ [0, 1) equivalent and write x ∼ y if x − y ∈ Q. This yields for the equivalence class [x] that [x] = (x + Q) ∩ [0, 1). These equivalence classes form a partition (Mj )j∈J = ([xj ])j∈J of [0, 1), i.e. S [0, 1) = j∈J Mj and Mj ∩ Ml = ∅ for j 6= l. By the axiom of choice (in the second version) there exists a set M ⊂ [0, 1) containing exactly one element mj ∈ Mj , j ∈ J. We claim that M is not Lebesgue measurable. Suppose that M is Lebesgue measurable. Then we can find for x ∈ [0, 1) some j0 ∈ J such that [x] ∩ M = [mj0 ], hence the existence of q ∈ Q follows such that x = mj0 + q, −1 < q < 1. We deduce that [0, 1) ∩ M + (Q ∩ (−1, 1)) ⊂ [0, 1) + (−1, 1) = [−1, 2), or [0, 1) ⊂

[

(q + M) ⊂ [−1, 2).

q∈Q∩(−1,1)

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By the definition of the equivalence relation “∼” we conclude further that (q + M) ∩ (r + M) = ∅ for q 6= r. Since λ(1) is σ-additive we now find λ(1) ([0, 1)) ≤

X

q∈Q∩(−1,1)

λ(1) (q + M) ≤ λ(1) ([−1, 2)) = 3.

The translation invariance of the Lebesgue measure yields 1≤

X

q∈Q∩(−1,1)

λ(1) (M) ≤ 3

which cannot hold and therefore we have a contradiction. Hence M is not Lebesgue measurable. With further efforts, compare with [75], Appendix D, we can now prove • the existence of Lebesgue measurable sets which are not Borel sets; • there exists a subset of R (Rn ) which is not a Borel set. The proofs of these statements use comparisons of the cardinality of sets such as R, B(1) and the Lebesgue sets of R. Although the following discussion extends to more general topological spaces we restrict ours to Rn or even to R. Let O(n) denote the Euclidean topology, C (n) the closed sets and B(n) the Borel σ-field in Rn . All elements of O(n) and C (n) belong to B(n) , the converse is of course not true. Moreover, for a T (n) U sequence (Uk )k∈N , Uk ∈ O , we can in general not expect that k∈N k is T Analogously, for a an element of O(n) , but k∈N Uk is always a Borel set. S (n) sequence (Ck )k∈N , Ck ∈ C , we cannot expect that k∈N Ck is a closed set but it is always a Borel set. This leads to Definition A.II.2. A set E ⊂ Rn is called a Gδ -set if it is a countable intersection of open sets. We call F ⊂ Rn an Fσ -set if it is a countable union of closed set. As already mentioned, Gδ - and Fσ -sets are Borel sets. In the study of realvalued functions on Rn , especially on R, Fσ - and Gδ -sets play an important role. Here are without proof some results highlighting the importance of Fσ and Gδ -sets and indirectly the interplay of measurability and topology. 585

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Theorem A.II.3. A. Let f : R → R be a function. Its set of discontinuities {x ∈ R | f is discontinuous at x} is an Fσ -set. Conversely, to every Fσ -set A ⊂ R there exists a function fA such that the set of discontinuities of fA is A. B. Let f : R → [0, 1] be a continuous function with compact support. For every r > 0 the set {f ≥ r} is a compact Gδ -set. If r ≥ 0 then the set {f > r} is an open Fσ -set. C. Let F ⊂ Rn be an Fσ -set and g : F → Rm be an H¨older continuous mapping in the sense that kg(x) − g(y)k∞ ≤ ckx − ykα∞ holds for all x, y ∈ F and some 0 < α ≤ 1. Then for every Fσ -set F˜ ⊂ Rn the set g(F ∩ F˜ ) is an Fσ -set in Rm . A proof of part A can be found in [6] or [14], part B is proved in [47], and a proof of part C is given in [75]. We now turn to Theorem 10.10. A full proof we will provide in Part 12. Any proof of this theorem is either lengthy or needs more preparation of a different type, i.e. the construction of the Lebesgue measure with the help of the Riesz representation theorem. The statement of Theorem 10.10 is best understood in the context of regular measures on locally compact metrizable Hausdorff spaces, however we stay in the context of Rn . We call a Borel measure µ on B(n) a regular measure if it is finite on the compact sets and if for every Borel set A ∈ B(n) we have  (A.II.1) µ(A) = sup µ(K) K ⊂ A, K compact  = inf µ(U) A ⊂ U, U open .

Note that (A.II.1) implies for every A ∈ B(n) such that µ(A) < ∞ that for  > 0 there exists a compact set K and an open set U such that K ⊂ A ⊂ U and µ (U \K ) < . However this statement is needed to prove the regularity of µ. One way to prove (A.II.1) is to prove that the sets of all A satisfying (A.II.1) form a σ-field containing the compact set. This can be done by using Dynkin systems and explicit proofs are given in [11] or [47]. In [75] a proof is given by using Carath´eodory construction of the Lebesgue measure, in [27] or [71] a proof is given using the construction of Borel measures with the help of the Riesz representation theorem. Finally we mention a result a proof of which can be found in [75]. 586

APPENDIX II: MEASURE THEORY, TOPOLOGY AND SET THEORY

Lemma A.II.4. Let A ∈ B(n) be a Borel set in Rn . Then we can find an Fσ set F and a Gδ -set G such that F ⊂ A ⊂ G and λ(n) (F ) = λ(n) (A) = λ(n) (G). Of course, this lemma implies Theorem 10.10, but again Theorem 10.10 is needed to prove this lemma.

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Appendix III: More on M¨ obius Transformations In Chapter 16 we introduced M¨obius transformations az + b w(z) := (A.III.1) cz + d   a b 6= 0. They have served us first of all as exwith a, b, c, d ∈ C, det c d ample of complex-valued functions of a complex variable, however we could use them later to construct certain biholomorphic mappings, for example from D to H. Furthermore we have seen in Proposition 16.28 that M¨obius transformations map straight lines or circles onto straight lines or circles. In addition we could determine the inverse of a M¨obius transformation and prove that the composition of two M¨obius transformations is again a M¨obius transformation. In this appendix we want to briefly collect some further geometry related properties of M¨obius transformations. For this topic classical texts such as [16] or [61] are still some of the best sources. A further good reference is [2]. First ˆ we extend every M¨obius transformation to the one-point compactification C of C by ( az+b , z 6= − dc . (A.III.2) w(z) := cz+d ∞, z = − dc ˆ and we obtain We can now invert w on C ( w −1(ζ) =

−dζ+b , cζ−a

∞,

ζ 6= ζ=

a c a c

,

(A.III.3)

and it follows that the family of all extended M¨obius transformations on Cˆ (∼ = S 2 ) from a non-abelian group. A first result is ˆ be given triples of disTheorem A.III.1. Let (z1 , z2 , z3 ), (w1 , w2 , w3 ) ∈ C ˆ →C ˆ such tinct points. Then exists a unique M¨obius transformation w : C that w(zj ) = wj , j = 1, 2, 3. It is convenient to call the set of all circles and straight lines on C the set of all generalised circles. Let ∂Br (z0 ), z0 ∈ C, be a circle in C. The mapping 589

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Rz0 ,r : C\{z0 } → C\{z0 } defined by Rz0 ,r (z) = z0 +

r2 z − z0

(A.III.4)

is called the reflection at the circle ∂Br (z0 ) and we have Theorem A.III.2. Every reflection at a circle maps a generalised circle into a generalised circle. With the help of Theorem A.III.2 one can deduce from Theorem A.III.1. Corollary A.III.3. Three distinct points z1 , z2 , z3 ∈ C always lie on a unique generalised circle. Proof. (Sketch) Transform the points {z1 , z2 , z3 } by a M¨obius transformation onto the points {0, ∞, 1} which must lie by Theorem A.III.2 on a generalised circle. Corollary A.III.4. Two generalised circles can be transformed onto each other by a M¨obius transformation. To proceed further we need ˆ Their cross ratio is defined by Definition A.III.5. Let z1 , z2 , z3 , z4 ∈ C. (z1 : z2 ; z3 : z4 ) :=

z1 −z3 z2 −z3 z1 −z4 z2 −z4

.

(A.III.5)

Here we use the conventions (∞ : z2 ; z3 : z4 ) := or

z2 − z4 z2 − z3

z4 − z2 , z4 − z1 etc, i.e. we pass in (A.III.5) to the corresponding limit as zj → ∞. Clearly, the cross ratio is translation invariant on C. Furthermore we have (z1 : z2 ; ∞ : z4 ) :=

ˆ are distinct points and w : C ˆ →C ˆ is Theorem A.III.6. If z1 , z2 , z3 , z4 ∈ C a M¨obius transformation then we have (w(z1 ) : w(z2 ); w(z3) : w(z4 )) = (z1 : z2 ; z3 : z4 ). 590

(A.III.6)

¨ APPENDIX III: MORE ON MOBIUS TRANSFORMATIONS

As a corollary we obtain ˆ lie on a generalised circle Corollary A.III.7. Four points z1 , z2 , z3 , z4 ∈ C if and only if their cross ratio is a real number. Proof. (Sketch) We map z1 , z2 , z3 by a M¨obius transformation w onto 0, ∞, 1 and note that now (z1 : z2 ; z3 : z4 ) = (0 : ∞; 1; w(z4)) =

1 . w(z4 )

It is now possible to use M¨obius transformations more extensively as a tool ˆ (∼ in studying geometric objects in the plane or in C = S 2 ), a beautiful and classical topic in geometry but we refer to the sources mentioned above and end our considerations here.

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Appendix IV: Bernoulli Numbers In Chapter 26 we introduced the Bernoulli numbers B2k = (−1)k−1 β2k , k ≥ 1, with the help of the Taylor expansion of x 7→ exx−1 , i.e. ∞

X (−1)k−1 β2k x 1 = 1 − x + x2k . ex − 1 2 (2k)! k=1

(A.IV.1)

Starting with the Ansatz ∞

X Bk x = xk x e − 1 k=0 k!

(A.IV.2)

and using the Taylor expansion of the exponential function   x x2 x2 x3 x + +··· = x 1+ + +··· e −1 =x+ 2! 3! 2! 3! we find

x = ex − 1 1+

For x ∈ R such that

P∞

|x|l l=1 (l+1)!

x 1  P = ex − 1 1− − ∞ l=1

∞ X

x 2!

1 . 3 + x3! + · · ·

< 1 we deduce further

xl (l+1)!

=

∞ X

(−1)

k

k=0

∞ X l=1

xl (l + 1)!

!k

Bn n 1 x , =1− x+ 2 n! n=2

where the last equality now defines the Bernoulli numbers Bn , n ≥ 2, and further we set B0 := 1 and B1 := − 21 . Since   x x 1 1 e +1 + x= x x ex − 1 2 2 e −1

and

ex + 1 1 + ex e−x + 1 = − = e−x − 1 1 − ex ex − 1 593

A COURSE IN ANALYSIS

it follows that

∞

X Bn 1 x + x = xn ex − 1 2 n! n=2

is an even function and therefore Bn = 0 for n ≥ 2 and odd. Moreover, from (A.IV.2) we deduce by using the Cauchy product of series ! ∞ ! ∞ ∞ k X Bn X X x B n xn = xn x = (ex − 1) n! k! n! n=0 n=0 k=1 =

∞ X

cm xm

m=1

with cm =

m−1 X l=0

Bl l!(m − l)!

implying by comparing coefficients that m−1 X l=0

 m Bl = 0, l

(A.IV.3)

thus we have a recursion formula for the Bernoulli numbers Bk . Here is a list of the non-trivial Bernoulli numbers up to k = 12. Note that by (A.IV.3) all Bernoulli numbers are rational numbers: 1 1 1 B0 = 1, B1 = − , B2 = , B4 = − , 2 6 30 1 5 691 1 . B6 = , B8 = − , B10 = , B12 = − 42 30 66 2730 P Bn n Since the series ∞ n=2 n! x has positive radius of convergence, namely 2π, it follows that there exists infinitely many non-zero Bernoulli numbers. We can also prove that the series (B2k )k∈N is alternating, i.e. sgn B2k = −sgn B2(k+1) . Note that since Bn = 0 for n ≥ 2 and odd some authors call the numbers βk := (−1)k−1 B2k , k ∈ N, together with β0 = B0 = 1 and β1 = 21 the Bernoulli numbers. Bernoulli were first introduced by Jacob I Bernoulli when sumP numbers n k . Since then they have plenty appearances in Mathematics, a ming N k=0 594

APPENDIX IV: BERNOULLI NUMBERS

classical book dealing entirely with Bernoulli numbers in N. Nielsen’s “Trait´e ´el´ementaire des nombres de Bernoulli”, [62], but still new results on Bernoulli numbers are being discovered. We want to prove next the formula N X

n

n! Bn−j+1 (N + 1)j (n − j + 1)!j! j=1  n+1  1 X n+1 = Bn−j+1(N + 1)j . n + 1 j=1 j

k =

k=1

n+1 X

(A.IV.4)

For N = 1, 2, 3 we find N X

k=

k=1 N X

k2 =

k=1

N(N + 1) , 2 N(N + 1)(2N + 1) , 6

and N X

k3 =

k=1

(N(N + 1))2 . 4

In order to derive (A.IV.4) we first note that N X k=0

ekx

∞ N X N ∞ X X X 1 n = kn (kx) = n! n=0 k=0 n=0 k=0

!

xn , n!

(A.IV.5)

but in addition we have for x 6= 0 N X

kx

e

=

k=0

N X

(ex )k =

k=0

e(N +1)x − 1 . ex − 1

Combining (A.IV.5) and (A.IV.6) we find N ∞ X X n=0

k=0

kn

!


xn+1 x = x e(N +1)x − 1 n! e −1 595

(A.IV.6)

A COURSE IN ANALYSIS

and with (A.IV.2) we arrive at ! N ∞ X X xn+1 kn = n! n=0 k=0 =

∞ X Bl

l=0 ∞ X

l!

xl

!

∞ X (N + j)j j x j! j=1

!

cn xn+1

n=0

with cn =

n+1 X j=1

1 Bn−j+1 (N + 1)j . (n − j + 1)!

Finally, by comparing the coefficients in ! ∞ ∞ N X X xn+1 X n = cn xn+1 k n! n=0 n=0 k=0

we obtain (A.IV.4). In our argument we followed [22] closely. It was Euler who could prove that ζ(2k) =

∞ 2k X 1 k+1 (2π) = (−1) B2k 2k n 2(2k)! n=1

(A.IV.7)

holds for k ∈ N. Hence the Bernoulli numbers are also linked to the Riemann ζ-function which in Euler’s times was not yet studied as a function in the complex plane. Euler had also obtained Taylor expansions for trigonometrical series with the help of the Bernoulli numbers, see [67], cot z = tan z =

∞

1 X 4j + B2j z 2j−1 , (−1)j 2 j=1 (2j)!

∞ X

(−1)j−1

j=1

or

4j (4j − 1) B2j z 2j (2j)!

∞

X 22j − 2 z = B2j z 2j . (−1)j−1 sin z (2j)! j=0 596

(A.IV.8) (A.IV.9)

(A.IV.10)

APPENDIX IV: BERNOULLI NUMBERS

Finally we want to mention that there exists integral representations of the Bernoulli numbers such as Z ∞ 4n t −1 n+1 dt, (A.IV.11) B2n = 8n(−1) 2πt − 1 e 0 which is taken from [93]. More formulae of this type as well as a table of Bernoulli numbers can be found in [1].

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Solutions to Problems of Part 6 Chapter 1 1.

a) By definition ∅ is denumerable, hence Ω ∈ A, and if A ∈ A then A or
{ A is denumerable, hence A{ belongs to A since A{ = A. Finally we note S for a sequence (Aj )j∈N , Aj ∈ A, that if all sets Aj are denumerable, then Aj is {

j∈N

denumerable. However if one of the sets, say Aj0 . is not denumerable, then we find [

Aj

j∈N

and A{j0 is denumerable, hence

S

{

=

Aj

{

j∈N

\

j∈N

A{j ⊂ A{j0

is denumerable implying

S

j∈N

Aj ∈ A.

b) Since Ω0 = Ω0 ∩Ω and Ω ∈ A it follows that Ω0 ∈ AΩ0 . Moreover, for A0 ∈ AΩ0 we can find A ∈ A such that A0 = Ω0 ∩ A. Now we have to find the complement of A0 in Ω0 which is of course Ω0 \ A0 = Ω0 ∩(Ω0 ∩A){ = Ω0 ∩(Ω0{ ∪A{ ) = Ω0 ∩A{ ∈ AΩ0 since A{ ∈ A. For a sequence (A0j )j∈N , A0j ∈ A0Ω , it holds that A0j = Ω0 ∩ Aj for some Aj ∈ A . S  S 0 S 0 S Therefore Aj = (Ω ∩ Aj ) = Ω0 ∩ Aj . Since Aj ∈ A it follows j∈N

j∈N

j∈N

j∈N

eventually that AΩ0 is a σ-field.

2. Proof of Lemma 1.9.A: We note that if A is a σ-field in Ω such that E ⊂ A then E { ∈ A and hence ∼ ∼ E { ⊂ A. Conversely, if A is a σ-field containing E { then A also contains E. Thus ∼

∼

∼

{A | E ⊂ A and A is a σ-field in Ω} = {A | E { ⊂ A and A is a σ field in Ω} implying σ(E) = σ(E { ). Proof of Lemma 1.9.B: Since S E ⊂ I it follows that σ(E) ⊂ σ(I). On the other side, if Ej ∈ E, j ∈ J ⊂ N, then Ej ∈ σ(E), i.e. I ⊂ σ(E) implying σ(I) ⊂ σ(E). j∈J

Proof of Lemma 1.9.C: Obviously we have σ(E1 ) ⊂ σ(E1 ∪E2 ). Since E1 ∪E2 ⊂ σ(E1 ) it also follows that σ(E1 ∪ E2 ) ⊂ σ(E1 ).

{ 3. Since OX = CX where CX denotes the closed sets in X the first part follows from Lemma 1.9.A. Now let (X, d) be a separable metric space. Denote by xk , k ∈ N, the points of a dense subset of X, i.e. X = {xj | j ∈ N}. As countable union of countable sets the set Dd := {B n1 (xj ) | n ∈ N, xj ∈ X} ∪ {Bn (xj ) | n ∈ N, xj ∈ X}

is countable and every open set in X is a union of sets from Dd . This implies that σ(Dd ) = σ(D) = σ(OX ) = B(X).

4. Consider the Cartesian product B 1 × B 1 = {A = B1 × B2 | Bj ∈ B (1) }. For [0, 1] × [0, 1] ∈ B (1) × B (1) it follows that

599

A COURSE IN ANALYSIS ([0, 1] × [0, 1]){ = (−∞, 0) × R ∪ (1, ∞) × R ∪ (0, 1) × (−∞, 0) ∪ (0, 1) × (1, ∞) which is a set not belonging to B 1 × B 1 , so B 1 × B 1 can not be a σ-field. √ 5. The key observation is that the estimates kxk∞ ≤ kxk2 ≤ n kxk∞ imply for all r > 0 and z ∈ Rn (∞) B r (z) ⊂ Br(2) (z) ⊂ Br(∞) (z) n

(2) Bρ (y)

(∞)

where = {x ∈ R | kx − yk2 < ρ} and Bρ (y) = {x ∈ Rn | kx − yk∞ < ρ}. S (2) S (∞) Now, if U ∈ On then we have U = Bρ(x) (x) and therefore B ρ(x) (x), however √ n x∈U x∈U S (∞) S (∞) (x) ⊂ U. Thus we have U = (x) and B ρ(x) it is trivial that U ⊂ B ρ(x) √ √ n n x∈U x∈U S (∞) B ρ(x) (x) ∈ J which now gives U = I. √ n

I∈J,I⊂U

n

6. From Problem 3 it follows that σ(Cn ) = B (n) . Further we know that σ(KSn ) ⊂ σ(Cn ) since Kn ⊂ Cn . For C ∈ Cn the set Cj := C ∩Bj (0) is compact and C = Cj , hence j∈N

Cn ⊂ σ(Kn ), i.e. σ(Cn ) = σ(Kn ). Thus we have already proved B (n) = σ(On ) = σ(Cn ) = σ(Kn ). In order to prove σ(Ir,n ) = σ(Il,n ) = B (n) we can use the considerations of Corollary 1.11. It is helpful to introduce first some notation. For a = (a1 , ..., an ) and b = (b1 , ..., bn ) we set n

(a, b) :=

n

×

(ak , bk ),

[a, b] :=

k=1

as well as

k

k

k=1

n

[a, b) :=

×[a , b ], n

×

[ak , bk ),

k=1

(a, b] :=

×(a , b ], k

k

k=1

and we call these sets the n-dimensional open, closed and half-open intervals, respectively. With c = (1, ..., 1) ∈ Rn it follows that [a, b) =

\

j∈N

\ 1 1 (a, b + c), (a − c, b) and (a, b] = j j j∈N

hence σ(Ir,n ) ⊂ B (n) and σ(Il,n ) ⊂ B (n) . On the other hand we have (a, b) = and (a, b) =

[ {[c, d] | [c, d) ⊂ (a, b), c, d ∈ Qn },

[

{(c, d] | (c, d] ⊂ (a, b), c, d ∈ Qn }.

The countability of Q now yields that

B n ⊂ σ(Ir,n ) and B n ⊂ σ(Il,n ).

600

SOLUTIONS TO PROBLEMS OF PART 6

The remaining part consists of two problems: firstly to pass from half-open ndimensional intervals to open or closed n-dimensional intervals, and secondly to pass from arbitrary real endpoints, to rational endpoints. Since we can approximate real numbers by rational numbers the step from n-dimensional intervals with real endpoints to those with rational endpoints is clear: replace the real endpoints a, b ∈ Rn by sequences of approximating rational endpoints say (ak )k∈N , (bk )k∈N (and ∼

∼

( a k )k∈N , ( b k )k∈N ) and now consider the intersection and union respectively, i.e. (a, b) =

[

(ak , bk ) and [a, b] =

k∈N

\

∼

∼

[ a k , b k ].

k∈N

The proof of B (n) = σ((Ir,n ) implies already that we can use the n-dimensional  S ∼ ∼ { open intervals to generate B (n) . Finally we remark that [a, b] = [ a k , b k ]{ to k∈N

conclude that B (n) is also generated by the n-dimensional closed intervals.

7.

a) If we know that f maps open intervals onto open intervals the results is proved: in this case we have {(a, b) | a < b} ⊂ σ(f ) = {f −1 (V ) | V ∈ B (1) } but we have B (1) = σ({(a, b) | a < b}) ⊂ σ(B) ⊂ B (1) . Since f is bijective f and f −1 are strictly monotone (either increasing or decreasing) and since f is continuous f ((a, b)) is an interval with midpoints A and B. By monotonicity and continuity we must have A, B ∈ {f (a), f (b)}. Suppose A ∈ J. Then f −1 (A) must belong to (a, b), i.e. a or b belongs to (a, b) which is a contradiction. The same argument applies to B. Hence J = (A ∨ B, A ∧ B) is open and the result follows.

b) The mappings g11 , g12 , g13 are A1 /A3 -measurable since for every A ⊂ R it −1 (A) ∈ P(R) = A1 . follows that g1j The mapping g21 is in general not B 1 /P(R)-measurable, just take the identity and consider the pre-image of a non-Borel measurable set. Since continuous mappings are Borel-measurable g22 is measurable. If A ∈ B (1) then g23 is B (1) /A3 -measurable, otherwise in general it is not. If B ∈ / A3 then id : R → R is not A3 /P(R)measurable, hence in general g31 is not A3 /P(R)-measurable and with the same type of argument we see that g32 is in general not A3 /B (1) -measurable. However, since for a constant function g : R → R, g(x) = c for all x, we have for every set B ⊂ R that g −1 (B) ∈ {∅, R}, these functions are both A3 /P(R)-measurable and A3 /B (1)-measurable. Finally, g33 is in general not A3 /A3 -measurable since −1 (A) = A 6= a. g33 := id + a, a ∈ R, yields for any set g33 8. Since (g ◦ f )−1 (A) = f −1 (g −1 (A)) it follows that if A ∈ A3 then g −1 (A) ∈ A2 and consequently f −1 (g −1 (A)) ∈ A1 , i.e. for every A ∈ A3 the set (g ◦ f )−1 (A) belongs to A1 , in other words g ◦ f is A1 /A3 . Now since prj : Rn → R, 1 ≤ j ≤ n, is B (1) /B (1)-measurable it follows for a B (1) /B (n)-measurable mapping f : R → Rn that fi = prj ◦ f is B (1) /B (1)measurable. On the other hand if all mappings fj , 1 ≤ j ≤ n, are B (1) /B (1)Tn (−1) measurable and A ∈ B (n) then f (−1) (A) = j=1 fj (A) belongs to B ( 1) implying that f is B (1) − B (n) -measurable.

601

A COURSE IN ANALYSIS

9.

a) Since ∅ is by definition finite, hence denumerable, it follows that ν(O) = 0. Let (Aj )j∈N be a sequence of mutually disjoint sets belonging to it. If all sets Aj S are denumerable we find that Aj is denumerable and therefore j∈N

0=ν

[

j∈N

∞  X Aj = ν(Aj ) = 0. j=1

Now suppose that some Aj0 of the family (Aj )j∈N is not denumerable, i.e. A{j0 is denumerable. We claim that no other set Aj , j 6= j0 , can be denumerable. Indeed if Ajn , jn 6= j0 was denumerable then A{j0 and A{jn were denumerable and hence A{j0 ∪ A{jn . Consequently (A{j0 ∪ A{jn ){ was non-denumerable, but (A{j0 ∪ A{jn ){ = Aj0 ∩ Ajn = ∅ by assumption. Thus at most one set of (Aj )j∈NS, Aj ∩ Al = ∅, can be non(Aj ) is non-denumerable denumerable. If Aj0 is such a set we find now since j∈N

ν

[

j∈N

 X X ν(Aj ) = (Aj ) = 1 = ν(Aj0 ) = ν(Aj0 ) + ν(Aj ), j6=j0

j∈N

implying that ν is a measure. Finally we observe that ν(Ω) = 1 since Ω is denumerable, so ν is a probability measure. b) Clearly X k (∅) = 0 k∈Z

and for Aj ⊂ Z, j ∈ N, with Aj ∩ Al = ∅ if j 6= l it follows that [  X [  XX   X µ (Aj ) = k (Aj ) = k Aj ) = µ(Aj ). j∈N

k∈Z

j∈N

j∈N k∈Z

j∈N

This proves already that µ is a measure on (Z, P(Z)). If A ⊂ Z is finite and has N elements and if m ∈ Z then A + m has also N elements and therefore X X k (A) = N = k (A + m). k∈Z

k∈Z

If A is not finite then A + m is not finite either and X X k (A) = ∞ = k (A + m). k∈Z

k∈Z

Thus for all A ∈ P(Z) and all m ∈ Z we have µ(A) = µ(A + m). c) That µ is a measure follows from µ(∅) =

∞ X 1 µ (∅) = 0 j j 2 j=1

602

SOLUTIONS TO PROBLEMS OF PART 6

and µ

[

k∈N

∞ ∞ ∞ ∞ X ∞ ∞  X X X 1 1 [  X 1 X µj µj (Ak )= µj (Ak )= Ak = µ(Ak ) Ak = j j j 2 2 2 j=1 j=1 j=1 k=1

k∈N

k=1

k=1

for every sequence (Ak )k ∈ N, Ak ∈ A, of mutually disjoint sets. Since ∞ X 1 =1 j 2 j=1 it follows further that µ(Ω) = 1, i.e. µ is a probability measure. 10.

a) Obviously we have µΩ0 (∅) = µ(Ω0 ∩ ∅) = 0. If (Ak )k∈N , Ak ∈ A, is S µ(∅) = S a sequence of mutually disjoint sets then Ω0 ∩ Ak = (Ω0 ∩ Ak ) and the sets k∈N

(Ω0 ∩ Ak ), k ∈ N, are mutually disjoint too. Hence we have µΩ0

[

k∈N

k∈N

∞ ∞ ∞    X [ [  X X Ak =µ Ω0 ∩ (Ω0 ∩ Ak ) = Ak = µ(Ω0 ∩ Ak )= µΩ0 (Ak ), µ k∈N

k∈N

k=1

k=1

k=1

i.e. µ0 is a measure. ∼

∼

b) Since A is a σ-field, ∅ ∈ A and therefore we find µ| ∼ (∅) = µ(∅) = 0. Now

∼

∼

∼

∼

∼

A

let Ak ∈ A, k ∈ N, such that Ak ∩ Al = ∅ for k 6= l. Since A ⊂ A it follows that µ∼

A

[

k∈N

∞ ∞  [ ∼  X X ∼ ∼ Ak = µ Ak = µ(Ak ) = µ ∼ (A), k∈N

k=1

k=1

Ak

∼

hence µ ∼ is a measure on (Ω, A). A

Chapter 2 1. We must first recollect the way we worked with the partitions in Volume II Chapter 18. If Q ⊂ Rn is any non-degenerate hyper-rectangle, we learnt that partitions of the generating intervals led to a partition of Q. If Aα , α ∈ I, denotes the hyperrectangle obtained from such a partition, then we have X λ(1) (Q) = λ(1) (Aα ). α∈I

We also note that for this equality it does not matter whether Aα is open, closed ◦

or half-open, since λ(n) (Aα ) = λ(n) (Aα ). Next we observe that F =

M [ N [

(Ak ∩ Bl ) =

k=1 l=1

603

M [

k=1

Ak =

N [

l=1

Bl

A COURSE IN ANALYSIS

and Ak ∩ Bl is an element in Ir,n or Ak ∩ Bl is a set contained in a hyperplane of Rn and then λ(1) (Ak ∩ Bl ) = 0. Thus we have λ(n) (F ) =

M X N X k=1 l=1

λ(n) (Ak ∩ Bl ).

On the one hand we have N M X X

k=1 l=1

λ(n) (Ak ∩ Bl ) =

=

M X

λ(n)

k=1

M X

k=1

N [

N M  X  [
 Bl (Ak ∩ Bl ) = λ(n) Ak ∪

l=1

k=1

M X

λ(n) (Ak ∩ F ) =

k=1 l=1

λ(n) (Ak ∩ Bl ) =

=

N X l=1

N X

λ(n)

and hence we have shown that M X

λ(n) (Ak ) =

k=1

M  [

k=1

l=1

λ(n) (F ∩ Bl ) =

λ(n) (Ak )

k=1

while on the other hand we have M X N X

l=1

N X

N X


Ak ∩ Bl

λ(n) (Bl ),

l=1

λ(n) (Bl ).

l=1

2. We start by proving that R is a ring. Since ∅ is by definition finite ∅ ∈ R. Now let A, B ∈ R. If A is finite then A\ B is finite. In general we have A\ B = A∩B { . Thus if A is infinite and B is infinite then B { is finite and A∩B { too, hence A\B is finite. Finally, if A is infinite and B is finite, then A{ ∪ B is finite, but A{ ∪ B = (A \ B){ , thus (A \ B){ is finite and hence A \ B ∈ R in each case. Moreover, for A and B finite it follows that A ∪ B is finite and if A is finite and B is infinite then A ∪ B must be infinite but (A ∪ B){ = A{ ∩ B { is finite since B { is finite. The same argument holds for A infinite and B finite. If however both, A and B are infinite then (A ∪ B){ = A{ ∩ B { is finite. Thus R is indeed a ring. Clearly we have µ(∅) = 0 since ∅ is finite. Now letS(Ak )k∈N , Ak ∈ R, be a sequence of mutually disjoint elements of R such that k∈N Ak ∈ R. Suppose that T S { all Ak are finite and assume S that k∈N Ak is not S finite. Then k∈N Ak must be finite  k∈N Ak ∈ R. But k∈N Ak is denumerable, hence  S since by  assumption T { must be denumerable which is a contradiction to A A Ω= ∪ k∈N k k∈N k S our assumption. Thus if all Ak are finite then k∈N Ak must be finite too. { T S On the other hand, if Ak0 is infinite then A{k0 is finite and = k∈N A{k ⊂ k∈N Ak

604

SOLUTIONS TO PROBLEMS OF PART 6 A{k0 is finte. Suppose Ak1 , k1 6= k0 , is infinite too. Then A{k0 and A{k1 are finite and  {  { so is A{k0 ∪ A{kn . Consequently A{k0 ∪ A{k1 is not finite, but A{k0 ∪ A{k1 = Ak0 ∩ Ak1 = ∅. Thus only one of the sets Ak , k ∈ N, can be not finite. For µ this means that: µ(∅) = 0 since ∅ is finite. If all Ak , k ∈ N, are finite then 0=µ

∞  X Ak = µ(Ak ) = 0

[

k∈N

k=1

and if Ak0 is infinite then 1=µ

[

k∈N

∞  X X Ak = µ(Ak0 ) = µ(Ak0 ) + µ(Ak ) = µ(Ak ). k=1

k6=k0

3. Obviously we find for ∅ that µ(∅) =

∞ X

ak µk (∅) = 0.

k=1

Now let (Al )l∈N be a sequence of mutually disjoint sets Al ∈ R such that R. Now we find ∞ [  X [  Al = Al µ a k µk l∈N

=

∞ X

ak

∞ X ∞ X

µk (Al ) =

l=1

k=1

l∈N

Al ∈

l∈N

k=1

∞ X

S

ak µk (Al )

l=1 k=1

=

∞ X

µk (Al ).

l=1

4. For ∅ we have obviously µ(∅) = supk∈N µk (∅) S = 0. Now let (Al )l∈N , Al ∈ R, be a sequence of mutually disjoint sets such that l∈N Al ∈ R. Then we have [  [  Al = sup µk Al µ k∈N

l∈N

= sup

∞ X

µk (Al ) = sup sup

N X

N ∈N k→∞ l=1

= sup N ∈N

N X l=1

N X

k∈N N ∈N l=1

k∈N l=1

= sup lim

l∈N

µk (Al ) = sup

N X

µk (Al )

lim µk (Al )

N ∈N l=1 k→∞

sup µk (Al ) = sup

N ∈N

k∈N

605

N X l=1

µ(Al ) =

∞ X l=1

µ(Al ).

A COURSE IN ANALYSIS

5. We start with B1 := A1 and by induction we define Bk := Ak \ (A1 ∪ ... ∪ Ak−1 ). First we note that all sets Bk belong to R. Further we have by construction that ∼ ∼ ∼ SN Bk ∩ Bl = ∅ for k 6= l. Now we set AN := j=n Bj . Clearly we have AN ⊂ AN +1 ∼ S S S and n∈N AN = j∈N Bj = k Ak = Ω. Finally, since Bk ∩ Bl = ∅ for k 6= l we find ∼

µ(AN ) = µ

N [

j=n

N  X Bj = µ(Bj ) < ∞ j=1

since µ(Bj ) ≤ µ(Aj ).

6. Since Ω ∈ D for all Dynkin systems D it follows that Ω ∈ δ(E). Suppose that A ∈ δ(E). Then A belongs to all Dynkin system D containing E, hence A{ belongs to all Dynkin systems containing E, i.e. A{ ∈ δ(E). Finally let (Ak )k∈N , Ak ∈ δ(E), be a sequence of mutually disjointed sets. S These sets also belong Dynkin systems D such that E ⊂ D and henceStheir union k∈N Ak is an element of every Dynkin systems S D with E ⊂ D. Thus k∈N Ak is in the intersection of all Dynkin systems, i.e. k∈N Ak ∈ δ(E), proving that δ(E)is a Dynkin system.

7. Of course we have g ⊂ H ⊂ δ(H). In addition δ(g) is the minimal Dynkin system containing g and δ(H) is a Dynkin system containing g, thus δ(g) ⊂ δ(H). We know that every σ-field is a Dynkin system and since δ(g) is the smallest Dynkin system containing g we must have δ(g) ⊂ σ(g).

8. We can use essentially the proof of Problem 5 in Chapter 1. An open hyper-rectangle n (ai , bj ) and (a, b) has rational vertices if ai , bj ∈ Q for all is of the type (a, b) = Xj=1 j = 1, ..., n. If U ⊂ Rn is open then for every x ∈ U exists an open hyper-rectangle (2) (2) R(x) ⊂ U since there exists an open ball Bρ (x) ⊂ U and Bρ (x) contains such an open hyper-rectangle. The density of Q in R allows us to now replace R(x) by an open hyper-rectangle Q(x) ⊂ U with rational vertices. S S Clearly we have x∈U Q(x) ⊂ U and U ⊂ x∈U Q(x) is trivial.

Chapter 3 1.

Pn a) The measure µ = k=1 k counts the elements of a subset A ⊂ {1, ..., n}, i.e. µ(A) = #(A). Since for all σ ∈ Sn the sets A and σ(A) have the same number of elements we have for all σ ∈ Sn and all A ⊂ {1, ..., n} that µ(σ(A)) = #(σ(A)) = #(A) = µ(A). b) Let A ⊂ {1, ..., n} and A1 ⊂ A be the subset of A which contains all even numbers belonging to A and A2 = A{1 is the set containing all odd numbers belonging to A. We set #(A1 ) = n1 and #(A2 ) = n2 and we have 0 ≤ nj ≤ n, n1 + n2 = n. We now find n2 n1 + . ν(A) = 2 3 For n = 7 and σ ∈ S7 with σ(1) = 1, σ(2) = 3, σ(3) = 5, σ(4) = 7, σ(5) = 2, σ(6) = 4, σ(7) = 6, and the set A = {1, 2, 3, 4} it follows that σ(A) = {1, 3, 5, 7} and further ν(A) =

606

5 2 2 + = 2 3 3

SOLUTIONS TO PROBLEMS OF PART 6

but

4 0 4 + = , 2 3 3 i.e. µ(A) 6= ν(σ(A)) which means that in general ν is not invariant under Sn . However if σ maps even numbers onto even numbers and odd numbers onto odd numbers, then for A = A1 ∪A2 and σ(A) = σ(A)1 ∪σ(A)2 , where σ(A)1 contains the even numbers of σ(A) whereas σ(A)2 contains the odd numbers of σ(A). Moreover we have #(A1 ) = #(σ(A)1 ) and #(A2 ) = #(σ(A)2 ) implying that ν(A) = ν(σ(A)). ν(σ(A)) =

2. By definition we have T (µ)(A) = µ{T −1 (A)} = µ{k ∈ Z | T (k) ∈ A}. Let A ⊂ Z, A = A1 ∪ A2 where A1 = {l ∈ A | l ≥ 0} and A2 = {l ∈ A | l < 0}. −1 First we note that for l < 0 it follows = ∅ since T (k) = k 2 ≥ 0. Thus  Sthat T ({l}) S −1 −1 −1 −1 {l}. we have T (A) = T (A1 ) = T l∈A1 T l∈A1 {l} =

Moreover, for l1 , l2 ∈ A1 , l1 6= l2 it follows that T {l1} ∩ T {l2 } = ∅ and therefore X µ(T −1 ({l})). T (µ)(A) = µ(T −1 (A)) = µ(T −1 (A1 )) = l∈A1

Furthermore we have T

−1

(l) =



∅, if l 6= k 2 , k ∈ Z {−k, k}, if l = k 2 , k ∈ Z

This now implies that T (µ) =

X

l ∈ An l = k2 ∈ Z

(E−√l + E√l ).

Thus T (µ)(A) is twice the number of squares of integers belonging to A. 3. Since T is continuous it is Borel measurable. For (x, y, z) ∈ B1 (0) let (ξ, η, ζ) =  2 T (x, y, z) = (ax, by, cz), i.e. ξ = ax, η = by, ζ = cz. Now it follows that aξ +  2  2 η + ζc = x2 + y 2 + z 2 < 1, and therefore b o n  ξ 2  η 2  ζ 2 + + b > c. Since T is a linear transformation with determinant det T = abc > 0,by Theorem 3.11 we have according to (3.6) T (λ(3) ) =

607

1 (3) λ . abc

A COURSE IN ANALYSIS 2

2

2

For E = {(x, y, z) ∈ R3 | x9 + y16 + z25 < 1} we find further with T(3,4,5) (x, y, z) = (3x, 4y, 5z) that E = T(3,4,5) (B1 (0)) and (3.5) in Theorem 3.11 now yields 4 λ(3) (E) = k det T(3,4,5) kλ(3) (B1 (0)) = 60λ(3) (B1 (0) = 60. π = 80π. 3 4.

a) The Bernoulli distribution βpN is a measure on B (1) , so we are seeking all Borel sets A ⊂ R with βpN (A) = 0 where βPN (A)

N   X N κ = p (1 − p)N −k k (A). k k=0

βPN (A)

It follows that = 0 for all Borel sets A not containing a subset of {0, 1, ..., N }. Whereas if A contains a subset of {0, 1, ..., N } then βPN (A) 6= 0. P∞ k Similarly, we find for the Poisson distribution πα = k=0 e−α αk! k that πα (A) = 0 for all Borel sets A ⊂ (N0 ){ where N0 = N ∪ {0}, but for a Borel set A ⊂ R such that A ∩ N0 6= 0 it follows πα (A) 6= 0. b) For example every constant mapping will fit the bill. Take Tc : Rn → Rn , Tc x = c ∈ Rn . Then for every non-empty set A ⊂ Rn we have Tc (A) = {c} and λ(1) ({c}) = 0. Thus if we choose A ∈ B (n) such that λ(n) (A) > 0, every open set A ⊂ Rn will have this property, then we have constructed a measurable mapping Tc and a measurable set A ⊂ Rn such that λ(n) (Tc (A)) = 0 but λ(n) (A) > 0. c) For every mapping T : R → R the set T (Q) is denumerable, hence measurable and λ(1) (T (Q)) = 0. 5. Consider (Ω1 , A1 , µ1 ) as any measure space admitting a non-empty set of measure zero. Now take (Ω2 , A2 , µ2 ) as the space with Ω2 = {a1 , a2 }, a1 = 6 a2 , A2 = {∅, {a1}, {a2 }.{a1 , a2 }} = P({a1 , a2 }) and define µ2 on A2 by µ2 (∅) = 0, µ2 (a1 ) = 0, µ(a2 ) = 1, µ2 (a1 , a2 ) = 1. Clearly, µ2 is a measure on (Ω2 , A2 ). Now we have as a mapping h the constant mapping h : Ω1 → Ω2 , h(w) = a2 for all ω ∈ Ωn . It follows for every A ⊂ Ω1 , A 6= ∅, that h(A) = {a2 } and µ2 (h(A)) = 1. This holds in particular for every non-empty set of measure zero in A1 .

6. We note that ϕ(R) = S 1 and we recall that

ϕ(λ(1) )(A) = λ(1) {t ∈ R | ϕ(t) ∈ A} and therefore we may assume A!⊂ S 1 , A measurable. cos r − sin r Let U = U (r) = ∈ SO(2). It follows that sin r cos r ϕ(λ(1) )|S 1 (U (r)A) = λ(1) {t ∈ R | ϕ(t) ∈ U (r)A}

608

SOLUTIONS TO PROBLEMS OF PART 6

= λ(1) {t ∈ R |

cos t sin t

∈ U (r)A}

cos t = λ(1) {t ∈ R | U −1 (r) ∈ A} sin t !   cos r − sin r cos t (1) ∈ A} = λ {t ∈ R | sin r cos r sin t  cos(r + t) ∈ A} sin(r + t)   cos s ∈ A} = λ(1) {s − r ∈ R | sin s   cos s ∈ A} = λ(1) {s ∈ r + R | sin s   cos s ∈ A} = λ(1) {s ∈ R | sin s

= λ(1) {t ∈ R |



= ϕ(λ(1) )|S 1 (A),

hence ϕ(λ(1) )|S 1 (U (r)A) = ϕ(λ(1) )|S 1 (A), i.e. we have proved the invariance of ϕ(λ(1) )|S 1 under rotations. 7. First note that a Lebesgue null set in the sense of Definition II.19.6 need not belong to B (n) , i.e. need not be on Borel measurable set. However, if A ⊂ Rn is a Borel set and a Lebesgue null set in the sense of Definition II.19.6, 0 we can find open cells Kj ⊂ Rn , j ∈ N, such that S then for every P∞  > (n) A ⊂ j∈N Kj and j=1 λ (Kj ) < . Since now λ(n) (A) is defined it follows that λ(n) (A) ≤

∞ X

λ(n) (Kj ) < .

j=1

for every  > 0, implying that λ(n) (A) = 0, i.e. A is indeed a set of measure zero. 8. Consider the set A = {x ∈ Y1 ∩ Y2 | f (x) = g(x)}. It follows from our assumptions that µ(A{ ) = 0 and all subsets of A{ belong to A. Furthermore we find for any a ∈ R that g −1 ((a, ∞)) = (g −1 ((a, ∞)) ∩ A) ∪ (g −1 ((a, ∞)) ∩ A{ ) = (f −1 ((a, ∞)) ∩ A) ∪ (g −1 ((a, ∞)) ∩ A{ ).

Since f is measurable, f −1 ((a, ∞)) ∩ A ∈ A, and since g −1 ((a, ∞)) ∩ A{ ⊂ A{ it follows that g −1 ((a, ∞)) ∩ A{ ⊂ A too. Hence g −1 ((a, ∞)) ⊂ A for all a ∈ R and Remark 1.12 yields the result.

609

A COURSE IN ANALYSIS

9. Let A ∈ l(n) , i.e. A = B ∪ N where B ∈ B (n) and N ⊂ K ∈ B (n) with λ(n) (K) = 0. For x ∈ Rn it follows that x+A = (x+B)∪(x+N ), x+B ∈ B (n) and x+N ⊂ x+K, but λ(n) (x + N ) = λ(n) (K) = 0. Thus we find λ(n) (x + A) = λ(n) ((x + B) ∪ (x + N )) = λ(n) (x + B = λ(n) (B) = λ(n) (A).
S 10. Let zero. Since N = N ∩ Ω = N ∩ k∈N Bk = P∞ S N ∈ A be locally of measure k∈N (N ∩ Bk ) we have µ(A) ≤ k=1 µ(N ∩ Bk ). By assumption µ(Bk ) < ∞ and hence µ(N ∩ Bk ) = 0 for all k ∈ N implying that µ(A) = 0. Note that in general a set which is locally of measure zero need not be a set of measure zero. 11. Since g is bijective it is strictly monotone and for two sets A, B ⊂ R we have g(A ∩ B) = g(A) ∩ g(B) as well as g(A \ B) = g(A) \ g(B). Now consider A := {A ⊂ R | g(A) ∈ B (1) } and we claim that it is a σ-field. For A ∈ A it follows that g(A{ ) = g(R \ A) = R \ g(A), hence A{ ∈ A, S and R
∈ ASis trivial since g(R) = R. If (A ) is a sequence in A then g k k∈R k∈N g(Ak ) and k∈N Ak = S therefore k∈N Ak ∈ A. The continuity of g implies that compact sets belong to A which in turn yields that B (1) ⊂ A, i.e. the image of every Borel set in R under g is a Borel set. 12.

a) We can apply Corollary 3.37 to obtain first ·

·

·

Hα (γ(I)) ≤ (sup ||γ||)α Hα (I), ·

t∈I

·

·

where we used that γ is bounded on I and that ||γ(t) − γ(s)|| = ||

Z

s

t

·

·

γ(v)dv|| ≤ sup ||γ(v)||(t − s) v∈[s,t]

·

·

holds for all t, s ∈ I, s < t. Now limH (I) = 1 implying by Lemma 3.30 that ·

Hα (γ(I)) = 0 for α > 1.

b) Our argument goes along the same lines as in part a): f |B1 (0) is Lipschitz continuous and we have the estimate  β Hβ (f (B1 (0)) ≤ sup || grad f (x)|| Hβ (B1¯(0)), x∈B1¯(0)

but H2 (B1 (0)) = 2, hence Hβ (f (B1¯(0))) = 0 for β > 2. By Lemma 3.36 the following must hold

α

H αs ([0, 1] × [0, 1]) ≤ L s Hα ([0, 1]). For α = 1 we obtain H αs ([0, 1] × [0, 1]) < ∞ implying that

610

α s

=

1 s

≥ 2, i.e. s ≤ 1s .

SOLUTIONS TO PROBLEMS OF PART 6

13.

a) Since T is symmetric and positive definite we find that det(T ∗ T ) = (det T )2 = Πnj=1 κ2j where κj , 1 ≤ j ≤ n, denotes the j th eigenvalue of T (multiplication taken into account). By assumption we have κj > 0 and det T = Πnj=1 κj . This now implies gT =

n Y  21 p κ2j det(T ∗ T ) = = Πnj=1 κj = det T. j=1

b) First note that dx f is given by dx f =

where f (x) = f (u, v) =

! f1,v (x) f2,v (x) , f3,v (x)

f1,u (x) f2,u (x) f3,u (x)

! f1 (u, v) f2 (u, v) = f3 (u, v)

f1 (x) f2 (x) f3 (x)

!

and fj,u =

∂fj ∂u

as well as fj,v =

∂fj ∂v .

This yields for gf2 (x): gf2 (x) = gdx f = (det(dx f )(dx f )) f (x) = det 1,u f1,v (x)

f2,u (x) f2,v (x)

hfu , fu i(x) = det hfv , fu i(x) i.e. we have gf (x) = Chapter 4

f3,u (x) f3,v (x)

hfu , fv i(x) hfv , fv i(x)

!

! f (x) 1,u f2,u (x) f3,u (x)

f1,v (x)! f2,v (x) f3,v (x)

= E(x)G(x) − F 2 (x),

p E(x)G(x) − F 2 (x).

1. Recall χA (ω) =



1, ω ∈ A 0, ω ∈ /A

i) Since A ⊂ B means ω ∈ A implies ω ∈ B it follows for ω ∈ A that χA (ω) = χB (ω) = 1. For ω ∈ B \ A we have χA (ω) = 0 but χB (ω) = 1, and for ω ∈ B { it follows that χA (ω) = χB (ω) = 0. Hence we have indeed χA ≤ χB . ii) Note that

χA{ (ω) =



611

1, ω ∈ A{ 0, ω ∈ /A

A COURSE IN ANALYSIS

=1−



0, ω ∈ A{ 1, ω ∈ /A

1 − χA (ω).

S

iii) For ω ∈ k∈N Ak there exists k0 ∈ N such that ω ∈ Ak0 and therefore T S supk∈N χAk (x) = 1. If ω ∈ / k∈N Ak then ω ∈ k∈N A{k and χAk (ω) = 0 for all k ∈ N, i.e. supk∈N χAk (ω) = 0, and we have proved that χSk∈N Ak = sup χAk . k∈N

Ak then ω ∈ Ak for all k ∈ N and χAk (ω) = 1 for all k ∈ N, T { hence inf k∈N χAk (ω) = 1. In the case that ω ∈ we have ω ∈ A{k1 for k∈N Ak some k1 ∈ N and χAk1 = 0, hence inf k∈N χAk (ω) = 0. b) From (4.15) it follows that χA is measurable if and only S if A is measurable. Thus if sup χAk is measurable, by iii) in part a) it follows that k∈N Ak is measurable. Of course this does not imply that all sets Ak are measurable. Just take a decomposition of Ω = B1 ∪B2 with B1 not measurable, hence B2 is not measurable. Define S A1 := B1 and Ak := B2 for k > 2. None of the sets Ak , k ∈ N, is measurable but k∈N Ak = Ω is. iv) If ω ∈

2.

T

k∈N

a) The product space (X,d) is given by X=

M Y

Xj and dX =

j=1

M X

dj ,

j=1

and we know that dX is equivalent to any of the metrics dp :=

M X

dpXj

j=1

 p1

, 1 ≤ p ≤ ∞,

which follows from the inequalities cp,q ||x||q ≤ ||x||p ≤ Cp,q ||x||q , where ||x||p = 1 P M p p . In addition we know that the projection prj : X → Xj is contij=1 |xj | nuous, hence measurable. Now Theorem 1.20 yields the solution when noting that the open sets generate the σ-fields. b) The key observation is the validity of the estimate   p1 N N N X X X |xj | (∗) cp |xj |p  ≤ Cp |xj | ≤  j=1

j=1

j=1

which holds for all x = (x1 , . . . , xN ) ∈ RN , compare with Problem 6 of Chapter II.1. If we substitute in (∗) xj by dXj (ξj , ηj ), ξj , ηj ∈ Xj , the equivalence of the metrics  p1 P PN QN N p d d1 := j=1 dXj and dp := on X = j=1 Xj follows immediately. j=1 Xj The fact that dp is a metric on X can be seen as follows:

612

SOLUTIONS TO PROBLEMS OF PART 6

(i) dp (ξ, η) = 0 is equivalent to dXj (ξj , ηj ) = 0 for all j = 1, . . . , N ; (ii) dp (ξ, η) = dp (η, ξ) since dXj (ξj , ηj ) = dXj (ηj , ξj ) for all j = 1, . . . , N ; (iii) The triangle inequality for the Euclidean metric yields  p1    p1 N N X X
p  dpXj (ξj , ηj ) ≤  dXj (ξj , ζj ) + dXj (ζj , ηj )  j=1

j=1



≤

N X j=1

 p1



dpXj (ξj , ζj ) + 

N X j=1

 p1

dpXj (ζj , ηj )p  .

The continuity of the projection prj = prXj : X → Xj , prj (ξ) = ξj follows  
(k) (k) from the fact that for every sequence ξ (k) k∈N = ξ1 , . . . , ξN the (k)

convergence limk→∞ ξ (k) = ξ implies limk→∞ ξj

k∈N

= ξj .

3. The mapping multM : RM → R defined for x = (x1 , ..., xM ) ∈ RM by multM (x) = x1 · ... · xM = pr1 (x) · ... · prM (x)

4.

is continuous, hence measurable. Further,
with gj , 1 ≤ j ≤ M, also the mapping G : Rn → Rn , G(x) = g1 (x), ..., gm (x) is measurable. Finally we note g(x) = multM G(x).

a) If h is concave then −h is convex and by Theorem II.13.27 it follows that −h, hence h, is continuous. This implies in turn the measurability of h. b) By definition we have for every x ∈ R the existence of the limit g 0 (x) = lim

h→0

For h =

1 n

g(x + h) − g(x) . h

we find

  1 g 0 (x) = lim n g(x + ) − g(x) . h→0 n We define gn : R → R by gn (x) = h(g(x + n1 ) − g(x)) which is continuous, hence measurable. Now we see that g 0 is the pointwise limit of the measurable functions gn , thus g 0 is measurable. 5.

n a) This result is an easy consequence of the fact that || · || : R → R is continuous, note ||x|| − ||y|| ≤ ||x − y||, and that the composition of measurable functions is measurable. Note that if we replace (Rn , || · ||) by any normed space (X, || · ||) and take on X the Borel σ-field generated by the norm topology, then the result still holds. b) Our problem is solved if we can prove that σ(idn ) : Rn → Rn , (x1 , ..., xn ) 7→ (xσ(1) , ..., xσ(n) ) is measurable. Let A = A1 × ... × An ∈ B (n) , Aj ∈ B (1) . It follows that n \ −1 σ(idn )−1 (A) = prσ(j) (Aj ) j=1

613

A COURSE IN ANALYSIS −1 (Aj ) is in B (n) the result follows from Theorem 1.20. and since each set prσ(j)

6. This result is essentially a refinement of Corollary 4.19 by looking closer to its proof. With f = f + − f − and (uk )k∈N , (vk )k∈N ∈ S(Ω) converging to f + and f − , respectively we define ϕk := uk − vk . By Theorem 4.18 we may assume that uk ≤ uk+1 and vk ≤ vk+1 . It follows that limk→∞ ϕk (w) = f (w) and further, using the fact that both sequence (uk )k∈N and (vk )k∈N are increasing we find with − ϕ+ k = uk and ϕk = vk that |ϕk (w)| = uk (w) + vk (w) ≤ uk+1 (w) + vk+1 (w) = |ϕk (w)|. Note further that |ϕk (w)| ≤ |f (w)| holds for w ∈ Ω and k ∈ N since uk ≤ f + and vk ≤ f − . (n)

7. The first equality is just notational, B R and R ∩ B σ-field B

(1)

(n)

both denote the trace of the

in R. By definition of the trace σ-field we have R∩B

(1)

= {R ∩ A | A ∈ B

(1)

(1)

}.

(1)

But for A ∈ B it follows that R ∩ A ∈ B , i.e. R ∩ B (1) (1) (1) inclusion B ⊂ R ∩ B is trivial since R ∈ B .

(1)

⊂B

(1)

, whereas the

8. The functions uk are continuous, hence measurable and of Baire class zero. The function u is obviously not continuous, hence it is not of Baire class zero. However we have limk→∞ uk (x) = u(x) for all x ∈ R. Indeed, if |x| < 1 then 1 2k limk→∞ x2k = 0 and consequently limk→∞ 1+x =1 2k = 1. For |x| = 1 we have x 1 1 2k and limk→∞ 1+1 = 2 , and finally, for |x| > 0 it follows that limk→∞ x = ∞ implying that 1 limk→∞ 1+x 2k = 0. Chapter 5 1. In Ω = {1, ..., N } we choose of course the power set P(Ω) and σ-field and hence every subset of Ω is measurable. We can represent A ⊂ Ω as A = {a1 } ∪ ... ∪ {ak }, aj ∈ {1, ..., N }, 1 ≤ j ≤ k ≤ N, aj 6= al for j 6= l. It follows that for every A we Pk have χA = j=1 χ{aj } and therefore every simple function function u ∈ S(Ω) PN is of type u = k=1 γk χ{h} , γk ≥ 0. Consequently we have a bijective mapping PN k : (R+ ∪{0})N → S(Ω) defined by γ1 , ..., γN ) 7→ k=1 γk χ{k} . Every measure µ on R PN P(Ω) is determined by its values µ({k}) and therefore we find udµ = k=1 µ({k}). 2. Let {αj1 , ...αjM } be the maximal subset of {α1 , ..., αN } consisting of elements which are mutually distinct, i.e. αjl 6= αjm for l 6= m. The value αjk is attained by u on the (k) (k) (k) set Bk = A1 ∪ ... ∪ AL(jk ) for certain sets Al , 1 ≤ l ≤ L(jk ), and by assumption (k)

Al

(k)

∩ A1 Am = ∅ for l 6= m. Thus it follows that u=

N X

αj χAj =

j=1

M X

k=1

614

αjk χBk

SOLUTIONS TO PROBLEMS OF PART 6

and therefore N X

αj µ(Aj ) =

j=1

M X

αjk µ(Bk ) =

j=1

Z

udµ.

3. The sets Ak , k = 1, ..., N, are Borel sets and we have λ(2) (Br1 (0)) = πr12 , λ(2) (Ak ) = 2 2 πrk2 − πrk−1 = π(rk2 − rk−1 ), k = 2, ..., N. It follows that Z

udλ(2) =

N X

N

γk λ(2) (Ak ) =

k=2

k=1

= πr1 +

1 1 2 X πr + λ(2) (Ak ) r1 1 rk − rk−1

X

N

π

X 1 2 (rk2 − rk−1 ) = 2π rk − rN . rk − rk−1 k=1

4. The following is an illustration of the situation: h  h  h m l−1 l is Borel measurable, l−1 ∩ m−1 = ∅ for l 6= m and Since k , k k k , k Sk h l−1 l  = [0, 1), the function uk belongs to S([0, 1)) and is given in norl=1 k , k R mal representation. The integral udλ(1) is given by Z

uk dλ(1) =

k X l−1 l=1

k

λ(1)

k h l − 1 l  X l−1 , = k k k2 l=1

k k+1 1 k−1 1 X (l − 1) = − = . k2 2k k 2k l=1

k ≥ k−1 Since k ≥ k − 1 = (k + 1)(k − 1) we have k+1 k and therefore we find that   (k+1)−1 k−1 k−1 is a monotone increasing sequence and we find 2(k+1) ≥ 2k , i.e. 2k 2

2

k∈N

sup k∈N

Z

uk dλ(1) = sup k∈N

1 k−1 k−1 = lim = . k→∞ 2k 2k 2

 h 1 the functions uk The sequence (uk )k∈N is not monotone increasing. On 0, k+1  h 1 1 and uk+1 are equal (with value 0), on k+1 , k the function uk still has the value h  i h 1 1 1 0, but uk+1 equals on k+1 , k1 to k+1 , so uk ≤ uk+1 on k+1 , k1 . However on i 2 1 [ k1 , k+1 we find uk = k1 and uk+1 = k+1 , i.e. uk+1 ≤ uk . Thus the standard R1 approximation of the integral 0 xdx by the Riemann sums of the function uk will not lead to a monotone approximation needed for our new theory. Of course the functions vk := u1 ∨ ... ∨ of S(Ω), the sequence (vk )k∈N is monotone R uk are elements R increasing and supk∈N vk dλ(1) = xλ(1) (dx).

615

A COURSE IN ANALYSIS

5. The key observation is that by Proposition 5.2 the statement (5.17) holds for simple functions. Now let (uk )k∈N and (vk )k∈N be sequences of simple functions increasing to f and g, respectively, i.e. uk ≤ uk+1 and f = supk∈N uk as well as vk ≤ vk+1 and g = sup vk . The sequence (uk + vk )k∈N is increasing to f + g and we have supk∈N (uk + vk ) = supk∈N uk + supk∈N vk = f + g. It follows that Z Z Z Z Z  sup (uk + vk )dµ = sup uk dµ + vk dµ = sup uk dµ + sup vk dµ k∈N

k∈N

= as well as sup k∈N

Z

Z

k∈N

sup uk dµ +

Z

sup vk dµ =

(uk + vk )dµ =

Z

sup(uk + vk )dµ =

k∈N

k∈N

Z

k∈N

f dµ + Z

k∈N

Z

gdµ

(f + g)dµ.

Thus f + g is µ-integrable and Z Z Z (f + g)dµ = f dµ + gdµ.

R 6. Clearly we have ν(A) = A ρ(x)λ(1) (dx) ≥ 0 for all A ∈ B (1) . Furthermore we have R ν(∅) = 0 since ∅ ρ(x)λ(1) (dx) = 0, and for a mutually disjoint collection of Borel sets (Ak )k∈N we find ν

[

k∈N



Ak =

Z

(1)

S

ρ(x)λ

k∈N

(dx) =

Ak

∞ Z X k=1

ρ(x)λ(1) (dx) = Ak

∞ X

ν(Ak ).

k=1

Thus ν is a measure on B (1) . It is sufficient to prove that if g ≥ 0 is measurable and bounded then g is ν-integrable. Let uk be a simple function, then uk ρ ≥ 0 is B (1) -integrable. Moreover, if (uk )k∈N is a sequence of simple function increasing to g, then (uk )k∈N is a sequence of non-negative measurable function increasing to gρ. By the monotone convergence theorem, Theorems 5.13, we find that Z Z Z   sup uk ρdλ(1) = sup uk ρ dλ(1) = gρdλ(1) , k∈N

k∈N

which we can re-interpret as Z Z Z sup uk dν = sup uk dν = gdν, k∈N

k∈N

i.e. g is ν-integrable. 7. We proceed as in Example 5.19. i) Z Z N   X N k N −k p g g dβNP = g dk k R k=0

616

SOLUTIONS TO PROBLEMS OF PART 6

=

N   N   X X N k N −k 2 N k N −k p q k p q g(k) = k k k=0

k=0

= Np

N X

k

k=1

ii)

Z =

∞ X

k=0

e−γ

N − 1 pk−1 q N −k = N p(N p + q). k−1

g dπγ =

∞ X

e−γ

k=0

γt k!

Z

g dk

∞

γ t 2 X −γ γ k = (k − 1 + 1) e k! (k − 1)! k=1

= γ 2 + γ.

p Note that if we consider βN and πγ as distributions of random variables X and Y respectively, then the variance of X is given by

V (X) := E((X − E(X))2 ) = E(X 2 ) − E(X)2 = Npq and the variance of Y is V (Y ) = E(Y 2 ) − E(Y )2 = γ.

1 8. Since limx→0 sinx x = 1 the function h is continuous on R and from sinx x ≤ |x| we deduce that h vanishes at infinity. The function h is even, hence it is integrable over R if and only if it is integrable over [0, ∞). Now we can apply some ideas of the solution to Problem 8 of Chapter 1.28. The function h is integrable over [0, ∞) if and only if |h| is integrable over [0, ∞), see Theorem 5.22. Next we note that Z ∞ ∞ Z (k+1)π X sin x (1) sin x (1) λ (dx) = λ (dx). x x kπ 0 k=0

With Tk : [0, π] → [kπ, (k + 1)π], Tk x = x + kπ, we find using Theorem 5.31 and the translation invariance of λ(1) that Z (k+1)π sin x Tk (λ(1) )(dx) x kπ Z π Z π sin(x + kπ) (1) sin x (1) = λ (dx) = λ (dx). x + kπ x + kπ 0 0

Since

Z

0

and

Z

0

π

π

Z π sin x 1 sin x dx dx ≥ x + kπ (k + 1)π 0

sin x dx ≥

Z

5π 6

π 6

617

sin x dx ≥

π 1 4π · = 2 6 3

A COURSE IN ANALYSIS

we arrive at ∞ Z X

k=0

(k+1)π

kπ

∞ ∞ sin x πX 1X1 1 (1) ≥ = ∞, λ (dx) ≥ x 3 (k + 1)π 3 k k=0

k=1

i.e. the function h is not integrable over R.

9. We use Corollary 5.32 and we note that λ(n) in the formula we want to prove stands for the restriction of the Lebesgue-Borel measure λ(n) onto the appropriate sets, i.e. (n) (n) (n) W1 and T (W1 ), respectively. With T ∈ GL(n, R) we consider T −1 : T (W1 ) → (n) (n) (n) W1 where we also consider on T (W1 ) the measure T (λ ). Applying (5.47) to (n) g : W1 → R we get Z Z Z (g ◦ T (−1) )d(T (λ(n) )). g d(T −1 ◦ T (λ(n) ) = g dλ(n) = (n)

(n)

(n)

T (W1

W1

W1

By Theorem 3.11 we have that T (λ(n) ) = Z

(n)

gdλ(n) =

W1

1 | det T |

1 (n) , | det T | λ

Z

(n)

T (W1

)

and hence we arrive at

(g ◦ T −1 )dλ(n) .

10. The mapping U (ϕ) is a rotation around the origin by the angle ϕ. For ϕ ∈ 2 {0, π2 , π, 3π 2 } such a rotation leaves every square with centre 0 ∈ R and sides (2) (2) parallel to the coordinate axes invariant, i.e. U (ϕ)W1 = W1 . Moreover, since U (ϕ) ∈ SO(2) we have U (ϕ)(λ(2) = λ(2) . (2)

With (5.49) we find now for a continuous, hence integrable function f : W1 that Z Z Z (f ◦ U (ϕ))dλ(2) . f dU (ϕ)(λ(2) ) = f dλ(2) = (2)

W1

→R

(2)

(2)

W1

W1

Chapter 6 1.

a) We define Φ : C(Y ) → C(X) by Φ(u) = u ◦ ϕ. For u ∈ C(Y ) it follows that Φ(u) ∈ C(X). Since ϕ is a homomorphism it is injective implying that u 6= v yields u ◦ ϕ 6= v ◦ ϕ, and for g ∈ C(X) the function g ◦ ϕ−1 belongs to C(Y ) and it follows that Φ(g ◦ ϕ−1 ) = g ◦ ϕ ◦ ϕ−1 = g, i.e. Φ is also surjective, hence Φ is bijective. For u, v ∈ C(Y ) and α, β ∈ R we find (?) Φ(αu + βv) = (αu + βv) ◦ ϕ = (αu) ◦ ϕ + (βv) ◦ ϕ = αu ◦ ϕ + βv ◦ ϕ = αΦ(u) + βΦ(v), i.e. Φ is linear. Furthermore ϕ(u.v) = (u.v) ◦ ϕ = (u ◦ ϕ) · (v ◦ ϕ) = Φ(u) · Φ(v),

618

SOLUTIONS TO PROBLEMS OF PART 6

The inverse to Φ is of course Φ−1 : C(X) → C(Y ), Φ−1 (g) = g ◦ ϕ−1 , which has the same properties as Φ. b) First we note that with u ∈ C k (H) the function Ψ(u) = u ◦ ψ is an element in C k (G) and with the same arguments as in part a) we see that Ψ−1 : C k (G) → C k (H), Ψ−1 (g) = g ◦ ψ −1 is inverse to Ψ. The calculation (?) imply now that Ψ is indeed a vector space isomorphism. 2. We have seen already at several occasions that there exists constants ca,b > 0 and Ca,b such that for all x ∈ Rn we have ca,b ||x||b ≤ ||x||a ≤ Ca,b ||x||b , 1 ≤ a, b < ∞, for example compare with Problem 6 of Chapter 1 in Volume II. Thus given p, q, r we find constants Cp,r and Cp,q such that ||g(x) − g(y)||r ≤ Cp,r ||g(x) − g(y)||p ≤ Cp,r γp,p ||x − y||p ≤ Cp,r γp,p Cp,q ||x − y||q . 3.

a) The function f : Rn → R is integrable with respect to ν if i.e. is finite. Since ν = χG λ(n) it follows that Z Z Z |f |dν = |f |χG dλ(n) = |f |dλ(n) . Rn

R

R

Rn

|f |dν exists,

G

Thus f is integrable over Rn with respect to ν = χG λ(n) if and only if f |G is (n) integrable over G with respect to λ|G . R b) We need to assure that R |u(x)|gs (x)λ(1) (dx) is finite. From our assumption we deduce −s r |u(x)|gs (x) ≤ cu (1 + |x|2 ) 2 (1 + |x|2 ) 2 . Thus we need that

Z

R

(1 + |x|2 )

r−s 2

λ(1) (dx) < ∞.

We look at the decomposition. Z Z Z r−s r−s (1 + |x|2 ) 2 λ(1) (dx) + (1 + |x|2 ) 2 λ(1) (dx) = R

|x|≤1

|x|>1

(1 + |x|2 )

r−s 2

λ(1) (dx).

The first integrand is always finite since the integral is a bounded continuous function on a set of finite Borel-Lebesgue measure. Now we handle the second r−s integral. We note that x 7→ (1 + |x|2 ) 2 is an even function, hence we only need R∞ R∞ r−s r−s to investigate 1 (1 + |x|2 ) 2 λ(1) (dx). If r − s ≥ 0 then 1 (1 + |x|2 ) 2 λ(1) (dx) R r−s ∞ is not finite since in this case 1 ≤ (1 + |x|2 ) 2 and the integral 1 1λ(1) (dx) is not finite. So let r − s < 0. For |x| ≥ 1 we now have 1 1 1 ≤ s−r ≤ 2s−r |x|s−r |x|s−r (1 + |x|2 ) 2

619

A COURSE IN ANALYSIS

and therefore Z ∞ Z ∞ Z ∞ 1 (1) 1 1 (1) 2 r−s 2 λ(1) (dx) ≤ λ (dx) ≤ λ (dx). (1 + x ) s−r 2s−r 1 xs−r x 1 1 Further we have Z

∞

1

∞ Z X

1

λ(1) (dx) = xs−r

∞ X

X 1 ≤ s−r (k + 1)

k+1

1 (1) λ (dx) xs−r

k

k=1

and this yields ∞ X

k=2

1 k s−r

=

k=1

∞

k=1

Z

k+1

k

1

λ(1) (dx) ≤ xs−r

∞ X

k=1

1 k s−r

,

R ∞ 1 (1) λ (dx) diverges whereas for which implies that for s − r < 1 the integral 1 xs−r s − r > 1 this integral converges. Thus we conclude that u is ν-integrable over R if r < s − 1. P 4. a) With µj , j = 1, ..., N, a further measure on (Ω, A) is given by µ := N j=1 µj and for every A ∈ A we have µj (A) ≤ µ(A), i.e. µ(A) = 0 implies µj (A) = 0 for j = 1, ..., N, and hence µj continuous with respect to µ. b) Let A ∈ A and µ(A) = 0. It follows that ν(A) = 0 since ν continuous with respect to µ. But now the absolute continuity of π to ν implies that π(A) = 0, i.e. µ(A) = 0 implies π(A) = 0, i.e. π continuous with respect to µ. g(x) h(x)

5. From ν1 = gλ(n) and 0 < c0 ≤ ν1 (A) =

Z

(n)

g(x)λ

A

is absolutely is absolutely with respect is absolutely

≤ c1 we conclude for A ∈ B (n) that Z

(dx) ≤ c1

h(x)λ(n) (dx) = c1 ν2 (A)

A

and if ν2 (A) = 0 then ν1 (A) = 0, i.e. ν2 is absolutely continuous with respect to ν1 . Furthermore we have Z Z Z 1 1 1 (n) (n) g(x)λ (dx) = g(x)λ(n) (dx) = γ1 (A). ν2 (A) = h(x)λ (dx) ≤ c c c 0 A 0 A A 0 i.e. ν2 is absolutely continuous with respect to ν1 . 6. For A ∈ A the following holds |νk (A) − ν(A)| = | =|

Z

A

Z

A

(gk − g)dµ| ≤

620

gk dµ − Z

A

Z

gdµ|

A

|gk − g|dµ

SOLUTIONS TO PROBLEMS OF PART 6

≤

Z

Ω

||gk − g||∞ dµ ≤ ||gk − g||∞

Z

1dµ Ω

= µ(Ω)||gk − g||∞ . Thus limk→∞ ||gk −g||∞ = 0 implies limk→∞ |νk (A)−ν(A)| = 0 or limk→∞ νk (A) = ν(A) for all A ∈ A. 7.

a) The set {x0 } has Borel-Lebesgue measure zero, i.e. λ(1) ({x0 }) = 0 but x0 ({x0 }) = 1, so x0 is not absolutely continuous with respect to λ(1) . Conversely, every open interval (a, b) ⊂ R has Borel-Lebesgue measure λ((a, b)) = b − a 6= 0, but for x0 ∈ / (a, b) it follows that x0 ((a, b)) = 0, and therefore λ(1) is not absolutely continuous with respect to 0 . b) In order that ν is absolutely continuous with respect to µ it must follow for A ∈ B (1) and µ(A) = 0 that ν(A) = 0. Let A ∈ B (1) . Only points in A ∩ N will contribute to ν(A) and µ(A), i.e. X X µ(A) = ak µ(A) = ak k∈N

and µ(A) =

X

k∈A∩N

bk ν(A) =

k∈N

X

bk .

k∈A∩N

If µ(A) = 0 then ak = 0 for all k ∈ A ∩ N and in order that ν(A) = 0 holds we must have bk = 0 for all k ∈ A ∩ N. This implies that ν is absolutely continuous with respect to µ if and only if bk = 0 for all ak = 0. 8. This problem is essentially solved with the calculation made with the solution to Problem 3.b). The expectation of X is given by Z x α λ(1) (dx), α > 0. E(X) = π R α2 + x2

R (1) (dx) exists, i.e. is finite. This Now, this integral exists if and only if R α2|x| +x2 λ Rα x R ∞ x (1) (1) requires 0 α2 +x2 λ (dx) and α α2 +x2 λ (dx) to be finite. However we have 1 2

Z

∞

α

1 (1) λ (dx) ≤ x

Z

∞

α

x λ(1) (dx) α2 + x2

and we know that the integral on the left hand side does not exist (as a finite integral), compare the solution to Problem 3.b). Chapter 7 1.

a) We determine the sets of µ-measure zero. Let A ∈ P(Ω), i.e. A ⊂ Ω, and PN suppose that µ(A) = 0. This means that N1 k=1 k (A) = 0, or for all k ∈ {1, ..., N } it must hold k ∈ / A, i.e. A = ∅. Thus only the empty set has µ-measure zero implying that every µ-a.e. statement is in fact an everywhere statement.

621

A COURSE IN ANALYSIS

b) First we note that for every consideration of convergence the first finitely many terms T∞of a sequence do not count. Thus we only have to look at (vj )j≥5 . Since j=5 [ 41 − 1j + 21 ] = [ 41 , 12 ] it follows that for all x ∈ R we have limj→∞ vj (x) = χ[ 41 , 12 ] (x), i.e the pointwise convergence of (vj )j∈N to χ[ 41 , 12 ] . From Remark 7.3.B we further deduce that (vj )j∈N converges ν-almost everywhere to X[ 14 , 12 ] (x). For χ ∈ [ 14 , 21 ]{ = {X < 14 } ∪ {X > 12 } the function χ[ 41 , 12 ] (x) is identiPN cally zero. However with respect ν = N1 k=1 k the set [ 41 , 21 ] is a set of ν-measure ∼ zero since [ 14 , 12 ] ∩ N = ∅. Thus the functions χ[ 14 , 12 ] and v = 0 only differ on a set of ν-measure zero and therefore (vj )j∈N converges also ν-almost everywhere to ∼ v = 0. Again, from the pointwise convergence of (vj )j∈N to χ[ 14 , 12 ] it follows that (vj )j∈N converges λ(1) -almost everywhere to χ[ 41 , 12 ] . However the set [ 41 , 21 ] has measure λ(1) ([ 41 , 12 ]) = 41 > 0. ∼ Therefore with respect to λ(1) the two functions χ[ 14 , 12 ] and v are not almost everyw∼

here equal and convergence v cannot be an λ(1) -almost everywhere limit of (vj )j∈N . 2. From the assumption lim Np (ul − u) = lim

l→∞

l→∞

Z

R

|ul (x) − u(x)|p

∞ X

k=1

 p1 =0 k (dx)

we deduce Z

R

|ul (x)−u(x)|p

∞ X

k=1

k (dx) =

∞ Z X

k=1

R

Since |ul (x) − u(x)| ≤

|ul (x)−u(x)|p k (dx) =

∞ X k=1

|ul (k) − u(k)|p

 p1

∞ X

k=1

|ul (k)−u(k)|p → 0.

→0

the result follows. 3. Since (uj )j∈N converges λ(1) -almost everywhere to u there exists a set N ∈ B (1) of λ(1) -measure zero, i.e. λ(1) (N ) = 0, such that for x ∈ N { we have limk→∞ uk (x) = u(x). For x ∈ N { it follows |f (uk (x)) − f (u(x))| ≤ L|uk (x) − u(x)| → 0,


i.e. f (uk ) k∈N converges in the complement of a λ(1) -measure set of zero to f (u), i.e. it converges λ(1) -almost everywhere to f (u). 4.

a) We first use the Chebychev-Markov inequaltiy for f and the measure ν : Z Z 1 1 (?) ν({|f | ≥ α}) ≤ |f |dν = |f |g dλ(1) . α α

622

SOLUTIONS TO PROBLEMS OF PART 6

Since δ ∈ Lp (Rn ) and g ∈ Lq (Rn ), 1p + inequality to obtain

1 q

ν({|f | ≥ α}) ≤

= 1, we can apply the Cauchy-Schwarz 1 ||g||Lq ||f ||Lq . α

b) Replacing in (?) f by fk − h we find for every α > 0 Z 1 ν({|fk − h| ≥ α}) ≤ |fk − h|gdλ(n) α implying lim ν({|fk − h| ≥ α}) = 0

k→∞

for every α > 0, i.e. (fk )k∈N converges in ν-measure to h. (k)

5. Let gn = fj

be given and α > 0. It follows that {x ∈ [0, 1) | |gn | ≥ α} =

hj − 1 j  , k k

and therefore we have λ(1) ({x ∈ [0, 1) | |gn | ≥ α}) = k1 which implies λ(1) ({x ∈ (1) [0, 1) | |gn | ≥ α}) → 0 as n → ∞, i.e. (gn )n∈N converges in λ[0,1) -measure to zero. However for any x ∈ [0, 1) we have limhk→∞ gk(x) = 0. Given x0 ∈ [0, 1) we can (k) j find for every k some j such that x0 ∈ j−1 k , k , i.e. gn (x0 ) = fj (x0 ) = 1. Thus

for every x0 ∈ [0, 1) the sequence (gn (x0 ))n∈N contains a subsequence (gnl (x0 ))l∈N with gkl (x0 ) = 1.  2 1 1 1 6. Since 1+|x| = 1+2|x|+|x| 2 ≤ 1+|x|2 we find N2



Z Z  Z 1 (1) 1 1 2 (1) 1  (1) λ (dx) + λ (dx) = λ (dx) ≤ 2 2 1+|·| 1 + |x| 1 + |x| |x| |x|≥1 R |x|≤1 =2

Z

0

1

∞

X 1 λ(1) (dx) + 2 2 1 + |x|

k=1

Z

k+1

k

1 (1) λ (dx) |x|2

∞ X 1 M }) = 0} = |α|N∞ (f ). The triangle inequality follows for f, g ∈ L∞ (Ω) as follows: N∞ (f ) < ∞ means that |f (ω)| ≤ Cf < ∞ for all ω ∈ Ω \ Nf , µ(Nf ) = 0. We call Cf an essential bound of f . Clearly we have N∞ (f ) = inf{Cf |Cf is an essential bound of f }. Thus |f (ω)| ≤ Cf µ-a.e. and |g(ω)| ≤ Cg µ-a.e. yields |f (ω) + g(ω)| ≤ Cf + Cg ≤ N∞ (f ) + N∞ (g)µ − a.e. which gives N∞ (f + g) ≤ N∞ (f ) + N∞ (g). R R 9. a) For k ∈ N we have uk λ(1) (dx) = 1, hence limk→∞ uk λ(1) (dx) = 1. On the other hand we have for all x ∈ R that limk→∞ uk (x) = 0. This is trivial for x ∈ (0, 1){ . For x ∈ (0, 1) we can find N such that x > N12 implying uk (x) = 0 for k > N. Thus Z Z Z 0= lim uk (x)λ(1) (dx) = lim inf uk (x)λ(1) (dx) ≤ lim inf uk (x)λ(1) (dx) = 1. R k→∞

R k→∞

k→∞

R

R

R

R vdµ hence limk→∞ vk dµ   and R vk dµ exists as limit of the monotone and bounded sequence . The Lemma b) From vk (x) ≤ v(x) we deduce that

vk dµ ≤

k∈N

of Fato yields further Z Z Z Z vdµ = lim vk dµ ≤ lim inf vk dµ ≤ vdµ, k→∞

hence we have limk→∞

R

vk dµ =

R

k→∞

vdµ.

625

A COURSE IN ANALYSIS

c) This follows when taking in the lemma of Fatou the sequence fk = χAk and when recalling the definition of lim inf which yields ∞ χ∪ ∞ = lim inf χAk k=1 ∩l=k Al

k→∞

and now we just have to use the fact that µ(A) = R  p1 1 p ≤ ||u||∞ µ(Ω) p it follows that 10. Since |u| dµ Ω lim sup p→∞

Z

Ω

|u|p dµ

 p1

R

χA dµ.

1

≤ ||u||∞ lim µ(Ω) p = ||u||∞ , p→∞

1

recall that for a > 0 we have limp→∞ a p = 1. Suppose that ||u||∞ > 0, otherwise the statement is trivial. For 0 <  ≤ ||f ||∞ we find δ := µ({ω ∈ Ω | |f (ω)| > ||f ||∞ − }) > 0 which implies that Z  p |f |p dµ ≥ ||f ||∞ −  δ Ω

and therefore

lim inf p→∞

Z

Ω

|f |p dµ

 p1

1

≥ (||f |||∞ − ) lim inf δ p = ||f ||Ω − . p→∞

For  → 0 if follows that Z  p1 Z  p1 ≥ ||f ||∞ ≥ lim sup |f |p dµ , |f |p dµ lim inf p→∞

p→∞

Ω

i.e. lim

p→∞

Z

Ω

|f |p dµ

 p1

Ω

= ||f ||∞ .

11. Since (fk )k∈N converges on Ω in measure to f, for m ∈ N there exists km such that µ({ω ∈ Ω | |fkm (ω) − f (ω)| >

1 1 } ≤ m. m 2

Without loss of generality we assume that the sequence (km )m∈N is strictly increasing. Define ∞ [ 1 Ωl := Ω \ {ω ∈ Ω | |fkm (ω) − f (ω)| > }. m m=l

It follows that

µ(Ω \ Ωl ) ≤ S

∞ X 1 = 21−l 2m

m=l

1 . and therefore µ(Ω \ Ωl ) = 0. For m ≥ l and ω ∈ Ωl we haveS|fkm (ω) − f (ω)| < m Ω , but as already (ω)) to f (ω) for ω ∈ This implies the convergence of (f l m∈N k m l∈N S stated, µ(Ω \ l∈N Ωl ) = 0.

626

SOLUTIONS TO PROBLEMS OF PART 6

12. The estimate |fk (ω)| ≤ g(ω) implies that all functions are fk are integrable. From Problem 10 we deduce that a subsequence (fkm )k∈N converges µ-almost everywhere to f. To this subsequence we may apply the dominated convergence theorem to find first that f is integrable and secondly that Z Z f dµ = lim fkm dµ. m→∞

R Now we consider the sequence (γk )k∈N , γk := fk dµ. Since Z Z Z |γk | = | fu dµ| ≤ |fk |dµ ≤ gdµ

the R sequence (γk )k∈N is bounded. Suppose that (γk )k∈N does not converge to γ := f dµ. Then it must have a subsequence (γkl )l∈N which does not converge to γ, but by the Bolzano-Weierstrass theorem a subsequence of this subsequence must have ∼ some limit γ 6= γ. We denote this subsequence again with (γkl )l∈N . However the corresponding sequence (fkl )l∈N has a subsequence converging µ-almost everywhere to f and the dominated convergence theorem applied to this subsequence of (fkl )l∈N yields the convergence of the corresponding integrals to γ which is a contradiction. 13. Since |hk | ≤ c it follows that |hk g|p ≤ cp |g|p and the dominated convergence theorem implies that Z Z |hk g|p dµ = |hg|p dµ. lim k→∞

Ω

Ω

From Theorem 7.31 we deduce that for the sequence (hk g)k∈N that lim Np (hk g − hg) = lim

k→∞

k→∞

Z

  p1 = 0. |hk g − hg|p dµ

In addition we find Z Z Z  p p p p |hk gk − hg| dµ ≤ 2 |hk | |gk − g| dµ + |hk g − hg|p dµ   ≤ 2p cp Npp (gk − g) + Npp (g(hk − h) ,

since Np (gk − g) tends for k to infinity to 0 by our assumption and since we have proved before that Np (hk g − hg) tends to 0 too, the result is proved. Chapter 8 1.

−x(1+y2 )

a) We use Theorem 8.1. The function g(x, y) = e 1+y2 defined on [0, ∞) × R is continuous, in particular for y ∈ R fixed the function x 7→ g(x, y) is continuous on [0, ∞). Furthermore, for x ∈ [0, 1] and y ∈ R we have 2

e−x(1+y 1 + y2

)

≤

e−x|y| 1 ≤ 1 + y2 1 + y2

627

A COURSE IN ANALYSIS

and for x ≥ 1 and y ∈ R we have 2

e−x(1+y 1 + y2

)

≤

e−x|y| ≤ e−|y| . 1 + y2

Thus for all (x, y) ∈ [0, ∞] × R it follows that |g(x, y)| = g(x, y) ≤

1 + e−|y| 1 + y2

1 −|y| and the function g 7→ 1+y is integrable over R which we can deduce for 2 + e example from Theorem 8.14. Thus by Theorem 8.1 we find that ϕ is a continuous function. In addition it follows that Z R Z ∞ R 1 1 dy = lim dy = lim arctan y = π. ϕ(0) = 2 R→∞ −R 1 + y 2 R→∞ −R −∞ 1 + y

b) Consider the function hk : [0, ∞] × [0, ∞) → R defined for k ≥ 2 by hk (ξ, x) :=

(

 k e−ξx sinx x , ξ ∈ [0, ∞) × (0, ∞) 1, ξ ∈ [0, ∞) × {0}

Since limx→0 sinx x = 1 it follows that hk is on [0, ∞) × [0, ∞) continuous and Z Z  sin x k  sin x k λ(1) (dx) = λ(1) (dx). eξx e−ξx x x [0,∞) (0,∞)

 k For x ≥ 1 we find sinx x ≤ x1k ≤ 1 and further we know that the continuous k  is bounded on [0, 1]. Thus it follows that function x 7→ sinx x   M −ξx sin x k , ≤ e x 1 + xk

and Theorem 8.1 implies for k ≥ 2 the continuity of ψ since k ≥ 2.

R∞ 0

1 dx 1+xk

< ∞ for

2. The proof of the extension of Theorem 8.4 follows by induction. So we fix m ∈ N and we assume that ∂ k u(x ,·) i) for all x ∈ I the functions ∂xkj : Ω → R, 0 ≤ k ≤ m − 1, are µ-integrable; m−1

u(·,ω) ii) for all ω ∈ Ω the function ∂ ∂xm−n : I → R is partial differentiable with respect to x; iii) for µ-integrable functions hk : Ω → [0, ∞) the following holds on I × Ω

∂k k u(x, ω) ≤ hk (ω), 1 ≤ k ≤ m. ∂x

628

SOLUTIONS TO PROBLEMS OF PART 6 R Then g(x) := u(x, ω)µ(dω) is on I m-times differentiable, ω 7→ k ≤ m, is for all x ∈ I integrable and Z k ∂k ∂ u g(x) = (x, ω)µ(dω), k ≤ m. ∂xk ∂xk

∂ku (x, ω), ∂xk

1≤

These conditions are such that we can take Theorem 8.4 with m = 1 as starting ∂k point of the induction and then we replace in the induction steps u by ∂x k u(x, ω). 3. Note that we claim the differential equation holds in the open interval (0, ∞). For proving that ϕ is in (0, ∞), differentiable we fix x0 > 0 and note that 2 2 ∂  e−x(1+y )  x |x0 = e−x0 (1+y ) ≤ e−x0 |y| ∂x 1 + y2

which implies by Theorem 8.4 for x > x0 that ϕ0 (x)=

d dx

Z

R

Z Z 2 2 2 d  e−x(1+y )  (1) e−x(1+y ) (1) e−x(1+y ) λ(1) (dy). λ (dy)= λ (dy)= 2 1 + y2 dx 1 + y R R

Since x0 > 0 was arbitrary we obtain Z 2 e−x(1+y ) λ(1) (dy) for all x > 0, ϕ0 (x) = R

and we can identify this integral as the improper Riemann integral √ Using the substitution ξ := y x we find Z

∞

2

e−x(1+y ) dy = lim

R→∞

−∞

= lim

R→∞

4.

Z

√

R x √

e

−x −ξ 2

e

−R x

Z

1 1 √ dξ = √ e−x x x

R

−∞

2

e−x(1+y ) dy.

2

e−x(1+y ) dy

−R

Z

R∞

∞

e

−ξ 2

dξ =

−∞

r

π −x e , x > 0. x

a) The following holds x2 d − x2 d g(x) = e 2 = −xe− 2 = −xg(x) dx dx

or g 0 (x) + xg(x) = 0. x2

x2

∼

b) Since |(cos ξx)e− 2 | ≤ e− 2 we can differentiate g under the integral and we can identify Lebesgue integrals with improper Riemann integrals. Now we find Z x2 1 d∼ (cos ξx)e− 2 λ(1) (dx) g(ξ) = √ dξ 2π R Z Z ∞ 1 1 ∂ ∂ x2 x2 = √ ( cos ξx)e− 2 λ(1) (dx) ( cos ξx)e− 2 λ(1) (dx) = √ 2π R ∂ξ 2π −∞ ∂ξ

629

A COURSE IN ANALYSIS Z R   d 1 x2 dx = √ e− 2 dx (sin ξx) lim dx 2π R→∞ −R −R Z R    R 2 2 1 d x x = √ − sin ξx e− 2 dx lim (sin ξx)e− 2 −R 2π R→∞ −R dx Z ∞ x2 1 ∼ ξ(cos ξx)e− 2 dx = −ξ g(ξ) = −√ 2π −∞

1 = √ lim 2π R→∞

Z

∼

R

(−x sin ξx)e−

x2 2

∼

∼

implying that g(ξ) + ξ g (ξ) = 0, i.e. (?) holds for g. ∼

∼

c) We can now conclude that g(x) = c g(x) which yields that g(0) = c g(0). R R 2 ∼ x However g(0) = 1 and g(0) = √12π R e− 2 λ(1) (dx) = 1 which follows from R e−x λ(1) √ (dx) = π and the substitution y = √x2 . Thus c = 1 and eventually we have proved ∼

g = g. 5.

a) Since | cos yx| ≤ 1 and x 7→ 1 is integrable against the bounded measure µ ∼ it follows that µ is continuous. b) We want to use Theorem 8.4. A formal application yields that Z dl ∼ dl µ(y) = (cos yx)µ(dx). l dy l R dy l d l Since dy l cos yx| ≤ |x| we can prove the following result after introducing a further R common definition. For a measure ν the number Ml := |x|l ν(dx) is called the lth absolute moment of ν. If a bounded measure µ on B (1) has all absolute moments up to order N then the ∼ function µ is N -times continuously differentiable and for 1 ≤ k ≤ N we have Z  k  d dk ∼ µ(y) = cos yx µ(dy). k k dy R dy

6. For every probability measure P , every integrable random variable X and convex functions f Jensen’s inequality states, see (8.13), (E(X)) ≤ E(f ◦ X) or f For (R, B (1) , N1

and

PN

k=1 k

Z



R

Z

Z

XdP



≤

Z

f ◦ XdP.

and X = g : R → R measurable we find

XdP =

(f ◦ g)

Z

R

g

N N 1 X 1 X dk = g(k) N N k=1

k=1

N N 1 X 1 X dk = f (g(k)) N N k=1

630

k=1

SOLUTIONS TO PROBLEMS OF PART 6

which eventually yields f

N N 1 X  1 X g(k) ≤ f (g(k)). N N k=1

k=1

7. The statement can be read as − 1−

Z

1

0

2 u(x)dx

! 12

≤

Z

0

1

1

(−(1 − u2 ) 2 )dx

1

and will follow once ϕ(t) = −(1 − t2 ) 2 is verified to be a convex function. For t ∈ (0, 1) we have 1 (−ϕ0 )(t) = t(1 − t2 )− 2 and

1

3

1

(−ϕ00 )(t) = (1 − t2 )− 2 + t2 (1 − t2 )− 2 = implying the result.

3

(1 − t2 )− 2

>0

a) On [a, b] equipped with the σ-field [a, b] ∩ B(1) a measure is given by µ(A) := R w(x)λ(1) (dx). Indeed, for A = ∅ it follows that µ(∅) = ∅ w(x)λ(1) (dx) = 0, and A for a collection (Ak )k∈N of mutually disjoint sets Ak ∈ [a, b] ∩ B(1) we have ∞ Z ∞ [  Z X X w(x)λ(1) (dx) = w(x)λ(1) (dx) = µ Ak = S µ(Ak ).

8. R

k∈N

k∈N

Ak

k=1

Ak

k=1

R

By assumption we have µ[a, b] = w(x)λ(1) (dx) < ∞, i.e. µ is a finite measure. Now Theorem 8.8 implies Z Z   1 1 u(x)µ(dx) ≤ (ϕ ◦ u)(x)µ(dx) ϕ µ([a, b]) [a,b] µ([a, b])

or

Z  1 (1) u(x)w(x)λ (dx) (1) (dx) [a,b] ([a,b]] w(x)λ Z 1 ≤R (ϕ ◦ u)(x)w(x)λ(1) (dx). (1) (dx) w(x)λ [a,b] [a,b]  ϕ R

b) The function t 7→ et is convex on [0, 1] and with w(x) = 1 we find for the (1) measure µ in part a) that µ = λ|[0,1] . Further, in our situation the Riemann and the Lebesgue integral coincide and it follows that Z 1 R1 u(x)dx 0 eu(x) dx ≤ e 0

or

Z

0

1

u(x)dx ≤ ln

631

Z

0

1

 eu(x) dx .

A COURSE IN ANALYSIS

9. The suggested substitution leads for the improper Riemann integrals to Z

1

cos(x2 )dx =

0

and

Z

∞

sin(x2 )dx =

0

1 2

Z

1 2

Z

∞

cos y √ dy y

∞

sin y √ dy. y

0

0

For the improper Riemann integral to be a Lebesgue integral we need the integray| y| | sin y| √ √ √ bility of | cos and | sin y y over [0, ∞). Suppose that y → y was integrable over [0, ∞). It follows that Z

Z

∞

X | sin y| √ dy = y

∞

0

kπ

k=0

≥

∞ X

k=0

P∞

∞

X | sin y| √ dy = y

(k+1)π

1 p (k + 1)π

Z

k=0

Z

π

0

| sin(y + kπ| √ dy y + kπ

∞ 2 X 1 √ , | sin(y + kπ)|dy = √ π k k=1

π 0

√1 diverges. A similar argument holds for the second integral. kR ∞ find 0 sin(x2 )dx (as improper integral) and we will make use

but the series

k=1

Now we want to of the calculations in [44]. Note that the following calculation is in the frame of Riemann’s theory, not of Lebesgue’s theory. This refers in particular for applications √ R∞ 2 of the change of variable and the interchanging of integrals. From 0 e−x dx = 2π R∞ 2 we deduce that √1u = √2π 0 e−uy dy. Now consider for  > 0 the integral. Z

∞

0

sin u −u 2 √ e du = √ u π 2 = √ π 2 =√ π

Z

Z

0

0

Z

∞

(sin u)e−u

0

∞Z

∞

Z

∞

Z

0

(sin u)e−(+y

2

)u

0 ∞

(sin u)e−(+y

2

0

)u

∞

 2 e−uy dy du

 dy du

 du dy.

The inner integral we can calculate explicitly using integration by parts twice: Z ∞ Z ∞ 2 2 (sin u)e−(+y )u du = 1 − ( + y 2 )2 (sin u)e−(+y )u du 0

0

or

Z

∞

(sin u)e−(+y

2

)u

du =

0

1 . 1 + ( + y 2 )2

Therefore we find Z

0

∞

(sin u) −u 2 √ e du = √ u π

632

Z

0

∞

1 dy. 1 + ( + y 2 )2

SOLUTIONS TO PROBLEMS OF PART 6

We are allowed to pass to the limit as  tends to 0 to get Z ∞ Z ∞ (sin u) 2 1 √ du = √ dy. u π 0 1 + y4 0 R∞ R∞ u) √ du = 0 sin(x2 )dx exists At this stage we already know that the integral 0 (sin u 1 as an improper Riemann integral. Moreover, a primitive of g(y) = 1+y 4 is the function √  √ √ 1 y 2 + 2y + 1 1  √ G(y) = √ ln + √ arctan( 2y + 1) + arctan( 2y − 1) . 2 4 2 y − 2y + 1 2 2 This yields

π lim G(y) − lim G(y) = √ y→0 2 2

y→∞

and eventually we have

Z

∞

sin(x2 )dx =

0

r

π . 8

α

α

10. For all x ≥ 0 and α > 0 we have e−x ≥ 0 thus x 7→ e−x is improper Riemann α integrable if and only if it is Lebesgue integrable. Since limx→∞ (x2 e−x ) = 0 we α can find R ≥ 1 such that x ≥ R implies e−x ≤ x12 and therefore Z

∞ 0

e

−xα

dx ≤

Z

0

R

e

−xα

dx +

Z

∞

R

e

−xα

dx ≤

Z

R

e

−xα

dx +

0

Z

∞

R

1 dx. x2

Both integrals on the right hand side are finite, the first as it of a R ∞is an integral 1 . continuous function over a compact set, and the second since R x−2 dx = R

11. We define L∞ (Ω) to consist of all real-valued functions f : Ω → R which are essentially bounded, i.e. |f | ≤ Mf < ∞ µ-almost everywhere and further we define N∞ (f ) := inf{M ≥ 0 | µ({|f |} > µ}) = 0}. Clearly N∞ (f ) ≥ 0 and if λ ∈ R and f ∈ L∞ (Ω) then |λf | = |λ||f | ≤ |λ|Mf µ-almost everywhere, i.e. λf is also essentially bounded and N∞ (λf ) ≤ |λ|N∞ (f ). If f, g ∈ L∞ (Ω) and |f | ≤ Mf , |g| ≤ Mg µ-almost everywhere then we have |f + g| ≤ |f | + |g| ≤ Mf + Mg µ-almost everywhere and it follows that N∞ (f + g) ≤ N∞ (f ) + N∞ (g). Thus N∞ is a semi-noun on the vector space L∞ . Now let (fj )j∈N be a Cauchy sequence with respect to N∞ , i.e. for every  > 0 there exists N () ∈ N such that k, l ≥ N () implies N∞ (fk − fl ) < . Consider the sets S

Ak,l = {|fk | ≥ N∞ (fk )} ∪ {|fk − fl | > N∞ (f− fl )}

and A = k,l∈N Ak,l . From the definition it follows that µ(Ak,l ) = 0 and µ(A) = 0 implying that N∞ (χA fj ) = 0 for all j ∈ N. On A{ the sequence (fj )j∈N converges

633

A COURSE IN ANALYSIS with respect to the supremums norm to a bounded function: for ω ∈ A{ we have uniformly |fk (ω) − fl (ω)| <  , k, l ≥ N (), hence for every ω ∈ A{ the limit limk→∞ fk (ω) =: f (ω) exists and the convergence is again uniform with respect to ω ∈ A{ . In particular f is measurable and bounded on A{ . Hence f extends to a µ-almost everywhere bounded and measurable function on Ω and limj→∞ N∞ (fj − f χA{ ) = 0. Thus L∞ (Ω) is complete in the same that every Cauchy sequence with respect to N∞ has a limit. When now passing to the quotient space L∞ (Ω) = L∞ (Ω)\∼µ we obtain a Banach space, the space L∞ (Ω). In L∞ (Ω) we often write ||u||∞ for the norm instead of N∞ (u) or ess supω∈Ω |u(ω|. 12.

a) For f ∈ Lq (Rn ) we consider the set {|f | > 1} and the decomposition of f according to f = g + h = f χ{|f |>1} + f χ{|f |>1}{ . Since for 1 < q we have |g| ≤ |f |χ{|f |>1} it follows that g ∈ L1 (Rn ) and for q ≤ r we have |h|r = |f |r χ{|f |≥1} ≤ |f |q χ{|f |>1}{ hence h ∈ Lr (RN ). In the case of r = ∞ we find ||h||∞ ≤ 1.

(1−λ)q r 1 and (1−λ)q have the property that λq = b) For r < ∞ the numbers λq 1 + r 1, i.e. they are conjugate indices in H¨ older’s inequality and therefore it follows that Z (1−λ)q r 1 |||u| ||u||qLq = |u||λq ||u||(1−λ)q dλ(n) ≤ |||u|λq || λq || (1−λ) q L

R

= which yields

Z

(n)

|u|dλ

λq  Z

|u|r dλ(n)

 (1−λ)q r

L

(1−λ)q

≤ ||u||λq L1 ||u||Lr

(1−λ)

||u||Lq ≤ ||u||λL1 ||u||Lr

.

13. The sequence (fk − g)k≥0 is a sequence of non-negative integrable functions converging λ(1) -almost everywhere to f − g. By Fato’s lemma we get Z Z lim inf (fk − g)dλ(n) ≤ lim inf (fk − g)dλ(n) k→∞

k→∞

but lim inf (fk − g) = f − g k→∞

and it follows that Z

(n)

f dλ

−

Z

gdλ

≤ lim inf k→∞

or

Z

(n)

Z

≤ lim inf k→∞

fk dλ(n) −

f dλ(n) ≤ lim inf k→∞

634

Z

Z

Z

(fk − g)dλ(n) gdλ(n)

fk dλ(n) .

SOLUTIONS TO PROBLEMS OF PART 6

Chapter 9 1. The system E := {C × D | C ∈ A1 ∩ A1 and D ∈ A2 ∩ A2 } = {C × D |C ∈ A1 , C ⊂ A1 and D ∈ A2 , D ⊂ A2 } is the natural choice for a generator of (A1 × A2 ) ∩ (A1 ⊗ A2 ) = (A1 ∩ A2 ) ⊗ A2 ∩ A2 . Note that we have (A1 × A2 ) ∩ (A1 ⊗ A2 ) = (A1 ∩ A2 ) ⊗ (A2 ∩ A2 ). 2. We know that B(n) = ⊗nj=1 B (1) (n copies of B (1) ) and for x ∈ Rn , A ∈ B (n) we have x (A) =



1, x ∈ A 0, x ∈ /A

Furthermore we know that B (1) × ... × B (1) (n copies of B (1) ) is a generator of B (n) which determines x . Let A = A1 × ... × Ak , Aj ∈ B (1) , and x = (x1 , ..., xn ) ∈ Rn . Since x ∈ A if and only if xj ∈ Aj it follows that x (A) = x1 (A1 ) · ... · xn (An ), in other words x = x1 ⊗ ... ⊗ xn . 3.

a) Again we note that A1 × ... × AN = {A1 × ... × AN | Aj ∈ Aj } is a generator of A1 ⊗ ... ⊗ AN and ν1 ⊗ ... ⊗ νN as well as µ1 ⊗ ... ⊗ µN are determined on A1 ⊗ ... ⊗ AN . It follows with ω = (ω1 , ..., ωN ), ωj ∈ R, that (ν1 ⊗ ... ⊗ νn )(A1 ⊗ ... ⊗ AN ) = ν1 (A1 ) · ... · νN (AN ) =

Z

A1

=

Z

g1 (ω1 )µ1 (dω1 ) · ... ·

A1 ⊗...⊗AN

Z

gN (ωN )µN (dωN ) AN

g1 (ω1 ) · ... · gN (ωN )(µ1 ⊗ ... ⊗ µN )(dω),

and therefore ν1 ⊗ ... ⊗ νN = g1 · ... · gN µ1 ⊗ ... ⊗ µN . Since by assumption all functions gj are integrable we have (ν1 ⊗ ... ⊗ νN )(Ω1 ⊗ ... ⊗ ΩN ) =

N Z Y

j=1

Ωj

gj (ωj )µj (dωj ) < ∞,

i.e. ν1 ⊗ ... ⊗ νN is finite measure. b) Since A1 ⊗ A2 is generated by A1 × A2 it follows for A ∈ A1 and N ∈ N2 that A × N ∈ A1 × A2 . Moreover we have (µ1 ⊗ µ2 )(A × N ) = µ1 (A)µ2 (N ) = 0, or A × N ∈ N. 4.

a) Consider the following figure

635

A COURSE IN ANALYSIS y

6 2 n E = (x, y) ∈ R2 x4 +

Sx

Sy

≤1

o

x

x 2

−2

y2 36

y

−6

Since Ex := {y ∈ R | (x, y) ∈ E} for x ∈ (−∞, −2)∪(2, ∞) we have obviously Ex = ∅. √ √ 2 2 If x ∈ [−2, 2] then (x, y) ∈ E means x4 + y36 ≤ 1 or y ∈ [−3 4 − x2 , 3 4 − x2 ], i.e. Ex =

√ √ [−3 4 − x2 , 3 4 − x2 ], x ∈ [−2, 2] ∅, otherwise.



Analogously we find Ey =



[− 31 ∅,

p p 36 − y 2 , 31 36 − y 2 ], y ∈ [−6, 6] otherwise

b) We use the figure below y C = (1, 1) b

x Sy

y Sx b

A = (0, 0)

y

x

636

b

B = (1, 0)

x

SOLUTIONS TO PROBLEMS OF PART 6 By definition we have Sx = {y ∈ R | (x, y) ∈ S = ABC} which yields for x ∈ [0, 1]{ that Sx = ∅. For x ∈ [0, 1] we find that Sx = {y ∈ R | 0 ≤ y ≤ 1 − x}, hence  [0, 1 − x], x ∈ [0, 1] Sx = ∅, otherwise, and analogously Sy =



[y, 1], y ∈ [0, 1] ∅, otherwise.

It follows that λ(1) (Sx ) = 1 − x and λ(1) (Sy ) = 1 − y implying Z Z 1 (1 − x)dx = λ(1) (Sx )λ(1) (dx) = 2 [0,1] R and

Z

R

λ(1) (Sy )λ(1) (dy) =

Z

[0,1]

(1 − y)dy =

Since λ(2) (S) = (λ(1) ⊗ λ(1) )(S) = area(ABC) = (λ(1) ⊗ λ(1) )(S) =

1 = 2

Z

R

1 2

1 . 2

we have indeed verified (g.16)

λ(1) (Sx )λ(1) (dx) =

Z

R

λ(1) (Sy )λ(1) (dy).

T 5. Let Cj ⊂ P(Ω), j ∈ I, be a collection of monotone classes in Ω and C := j∈I Cj . Further let (Ak )k∈N and (Bk )k∈N be sequences of S elements Ak , Bk ∈TC such that Ak ⊂ Ak+1 and T Bk ⊃ Bk+1 . We have to show that k∈N Ak ∈ C and k∈N S Bk ∈ C. Since A ∈ C it follows that A ∈ C for all j ∈ I and therefore k j k j j∈N k∈N Ak ∈ T S T C we deduce that A ∈ C. Analogously, since B ∈ C = C , i.e. j k k j j∈I k∈N j∈I T B ∈ C for all j ∈ I which yields B ∈ C for all j ∈ I and therefore k j k j k∈N T k∈N Bk ∈ C and we have proved that C is a monotone class.

6. Suppose that f is measurable and consider the function F : Rn+1 → R defined by F (x, y) = y − f (x). It follows that {F (x, y) ≤ 0} is measurable as is {y ≥ 0} hence Γs (f ) = {F (x, y) ≤ 0} ∩ {y ≥ 0} is measurable. Now suppose that Γs (f ) is measurable. This implies that Γs (f )x = {y ∈ R | (x, y) ∈ Γs (f )} = [0, f (x)] is measurable and x 7→ λ(1) ([0, f (x)]) = f (x) is measurable too by Theorem 9.15. Finally, by (9.16) we have Z λ(n+1) (Γs (f )) = χΓs (f ) (x, y)λ(n+1) (dxdy) =

Z

Rn

λ(1) (Γs (f )x )λ1 (dx) =

Z

Rn

f (x)λ(n) (dx).

The interpretation of this result is that the volume of the body bounded from below Rn × {0} and from above by the graph of f, i.e. Γ(f ), is given by Rby the plane (n) f (x)λ (dx) as expected. Rn

637

A COURSE IN ANALYSIS

7. First we note that Z Z Z  (dist(K, y))ρ (3) MK,Q,ρ (x)λ(3) (dx) = λ (dy) λ(3) (dx) 3+ρ Q ||x − y|| K K Z Z  (dist(K, y))ρ (3) λ (dx) λ(3) (dy) = 3+ρ K ||x − y|| Q Z Z  1 = (dist(K, y))ρ λ(3) (dx) λ(3) (dy) 3+ρ Q K ||x − y|| Z Z  1 (3) = (dist(K, y))ρ λ (dx) λ(3) (dy) 3+ρ Q\K K ||x − y||

Since dist(K, y) = 0 for y ∈ K, for y ∈ Q \ K and x ∈ K we have ||x − y|| ≥ dist(K, y) > 0 since K is closed, see also Example II. 3.28. This implies now for y ∈ Q \ K fixed and x ∈ K Z Z 1 1 (3) λ (dy) ≤ λ(3) (dx) 3+ρ 3+ρ ||x−y||≥dist(K,y) ||x − y|| K ||x − y|| Z 1 λ(3) (dz) = 3+ρ ||z|| ||z||≥dist(K,y) Z ∞ 1 2 ≤ 4π r dr 3+ρ r dist(K,y) Z ∞ 1 4π = 4π dr = . 1+ρ ρ(dist(K, y))ρ dist(K,y) r Now we find Z Z (3) MK,Q,ρ (x)λ (dx) ≤ K

=

(dist(K, y))ρ Q\K

4π λ(3) (dy) ρ(dist(K, y))ρ

4π (3) λ (Q \ K). ρ

In our calculation we needed that x 7→ dist(K, x) is measurable which follows from its continuity, compare with Example II.3.28. However it is helpful to know that dist(K, ·) is Lipschitz continuous and this we see as follows: for x, y ∈ Rn fixed assume that dist(K, x) ≥ dist(K, y). Hence we can find for  > 0 some z1 ∈ K such that dist(K, y) ≥ ||y − z1 || − , and therefore   0 ≤ dist(K, x) − dist(K, y) ≤ inf ||x − z|| − ||y − z1 || +  z∈K

≤ ||x − z1 || − ||y − z1 || +  ≤ ||x − y|| + ,

and since  > 0 was arbitrary we arrive at

| dist(K, x) − dist(K, y)| ≤ ||x − y||, i.e. x 7→ dist(K, x) is Lipschitz continuous with constant 1.

638

SOLUTIONS TO PROBLEMS OF PART 6 xy xy 8. The functions x 7→ (x2 +y 2 )2 and y 7→ (x2 +y 2 )2 are on (−1, 1) odd functions or identically zero and therefore we have Z 1 Z 1 xy xy (1) λ (dx) = λ(1) (dy) = 0. 2 2 2 2 2 2 −1 (x + y ) −1 (x + y )

Suppose that f is Lebesgue integrable. Then |f | is Lebesgue integrable too as it was Riemann integrable and using polar coordinates we find Z Z 1 Z 2π (r| cos ϕ|r| sin ϕ| 2) |f (x, y)|λ (dxdy) ≥ rdϕdr r4 0 0 (−1,1)×(−1,1) Z 1 Z 2π 1 | cos ϕ|| sin ϕ|dϕ = dr. 0 r 0 R1 1 However the integral 0 r dr is not finite, hence |f |, i.e. f, cannot be Lebesgue integrable.

9. With the arguments given in Example 9.22.B we get Z ∞ Z n o λ(n) e−tψ ≥ y dy e−tψ(x) λ(n) (dx) = =

Z

Rn

∞

0

0

Z ∞ 1 λ(n) ({ψ ≤ − ln y})dy = t λ(n) ({ψ ≤ ρ})e−tρ ds t 0 Z ∞ ψ −tρ λ(n) (B√ =t dρ, ρ (0))e 0

which yields further Z

Rn

e−tψ(x) λ(n) (dx) =

Z

∞

Now we observe that Z Z ∞ ψ −r (0))e dr ≥ λ(n) (B√ r t

0

ψ (0)) ≥ λ(n) (B√ 1 t

Z

ψ λ(n) (B√ (0))e−r dr. r t

0

∞

∞

ψ λ(n) (B√ (0))e−r dr r t

1

e−r dr =

1

1 (n) ψ λ (B√ 1 (0)), e t

which gives the lower bound with κ0 = 1e . For the upper bound we note that Z 1 Z ∞ Z ∞ ψ ψ ψ (n) −r −r √ λ (B λ(n) (B√ (0))e dr = (0))e dr + (0))e−r dr λ(n) (B√ r r r t

0



≤ 1−

t

0

1 e

λ(n) (B√ 1 (0)) + λ(n) (B√ 1 (0)) t

t

= κ1 λ(n) (B√ 1 (0)),

Z

t

1

∞

α

γ0 (1)r 2 e−r dr

1

t

which is the upper bound. Here we that for α ≥ 0 the integral  have used R∞ α is finite and we set κ1 := 1 − 1e + γ0 (1) 1 r 2 e−r dr.

639

R∞ 1

α

r 2 e−r dr

A COURSE IN ANALYSIS

10. The proof is quite similar to that of Corollary 9.23. We apply again Theorem 9.20 and use the substitution y = ϕ(s) to find Z ∞ Z µ({ϕ ◦ u ≥ y})dy (ϕ ◦ u) dµ = 0

Ω

=

Z

∞ 0

µ({ϕ ◦ u ≥ ϕ(s)})ϕ0 (s)ds =

Z

∞

0

µ({u ≥ y})ϕ0 (y)dy.

11. The following holds Z

||Kop u||L2 = ≤

[a,b]

≤

(?) = (??) ≤

[a,b]

Z

Z

Z

Z

[a,b]

[a,b]

[a,b]

Z

[a,b]

Z

Z

[a,b]

[a,b]

 21 k(x, y)u(y)dy|2 dx


2  12 |k(x, y)u(y)|dy dx

[a,b]

Z

Z

|

 21 |k(x, y)|2 |u(y)|2 dx dy

 12 |k(x, y)|2 dx |u(y)|dy


 12  |k(x, y)| dy 2

[a,b]

Z

[a,b]

|u(y)|2 dy

= ||k||L2 ||u||L2 ,

 21

where we used in (?) Minkowski’s integral inequality and in (??) the CauchySchwarz inequality. Chapter 10 1. Since for A ∈ O(n) and x ∈ Rn the mapping y 7→ T y := Ay + x is arbitrary often differentiable it follows that u ◦ T is continuous for u continuous and u ◦ T ∈ C k (Rn ) for u ∈ C k (Rn . Furthermore, if u ∈ Cb (Rn , i.e. ||u||∞ = M < ∞ then ||u ◦ T ||∞ ≤ M, i.e. u ◦ T ∈ Cb (RN ). We note that T is bijective with inverse T −1 z = A−1 (z − x) = A−1 z − A−1 x and that T (Br (0)) = A(Br (0)) + x = Br (x) since A ∈ O(n). This implies that if |u(y)| <  in the complement of some compact set K, then |(u ◦ T )(y)| <  in the complement of some compact set KT , i.e. u ∈ C∞ (Rn ) yields u ◦ T ∈ C∞ (R4 ). Finally, if supp u ⊂ BR (0) then we have supp(u ◦ T ) ⊂ BR (x), i.e. u ∈ C0 (Rn ) implies u ◦ T ∈ C0 (Rn ). 2. We know that (L1 (R), +) is a vector space of R and by Young’s inequality we have u ∗ v ∈ L1 (Rn ) for u, v ∈ L1 (Rn ). Moreover, by (10.6) and (10.7) it follows that convolution is an associative and commutative operation on L1 (Rn ). Furthermore, from the definition of convolution we get for α, β ∈ R and u, v, w ∈ L! (Rn ) Z Z (αu ∗ βv)(x) = αu(x − y)βv(y)λ(n) (dy) = αβ u(x − y)v(y)λ(n) (dy) Rn

Rn

640

SOLUTIONS TO PROBLEMS OF PART 6

= αβ(u ∗ v)(x), and

=

Z

((u + v) ∗ w)(x) =

Rn

u(x − y)w(y)λ(n) (dy) +

Z

Rn

Z

Rn

(u + v)(x − y)w(y)λ(n) (dy)

v(x − y)w(y)λ(n) (dy) = (u ∗ w)(x) + (v ∗ w)(x).

Hence (L1 (Rn ), +, ∗) is a commutative R-algebra. 3.

a) For a ∈ Cb (Rn ) and u ∈ C0 (Rn ) it follows that au ∈ C0 (Rn ) ⊂ Lp (Rn ). Moreover we have  Z   p1 Z  p1 = sup |a(x)| |u(x)|p λ(n) (dx) ||au||Lp = |a(x)u(x)|p λ(n) (dx) Rn

Rn

x∈Rn

= ||a||∞ ||u||Lp .

Now Proposition 10.6 combined with the density of C0 (Rn ) in Lp (Rn ) yields the result. Note that C0∞ (Rn ) ⊂ C0 (Rn ) ⊂ Lp (Rn ) and since by Theorem 10.17 C0∞ (Rn ) is dense in Lp (Rn ) it follows that C0 (Rn ) is dense in Lp (Rn ) too. b) A function u ∈ C0∞ ((0, 1)) we can extend by zero outside of (0, 1) and we obtain a C ∞ -function on R with support in [0, 1]. Therefore we have C0∞ ((0, 1)) ⊂ C 1 ([0, 1]) ⊂ L2 ([0, 1]) and C 1 ([0, 1]) is dense in L2 ([0, 1]). Consider now the sequence d (uk )k∈N , uk (x) = sin(2πkx), of functions in L2 ([0, 1]). It follows that dx uk (x) = duk 2 (2πk) cos(2πkx), hence dx ∈ L ([0, 1]). We find further

whereas

Z 1 du 2 k |2πk cos(2πkx)|2 dx = 2π 2 k 2 , 2 = dx L 0 ||uk ||2L2 =

If

d dx

Z

1

0

| sin(2πkx)|2 dx =

1 . 2

was continuous in L ([0, 1]) satisfying with some c > 0 the estimate du dx 2

L2

c||u||L2 , then we must have

du √ 1√ k 2 = c||uk ||L2 , = 2πk ≤ c dx L2 2

≤

d has no continuous extension from L2 (Rn ) which is of course not possible. Thus dx 2 n to L (R ). Remark. Note that we used for linear operators defined in a Banach space the equivalence of continuity and boundedness in the sense of ||T u||X ≤ c||u||X . That a linear continuous operator is bounded can be seen as follows: for  = 1 there exists δ > 0 such that ||u||X < δ implies ||T u||X = ||T u − T 0||X ≤ 1. If u 6= 0 it follows 2 δ δu δu 2||u|| ∈ Bδ (u) and therefore T 2||u||X X = 2||u||X ||T u||X ≤ 1, or ||T u||X ≤ δ ||u||X which is trivial for u = 0.

641

A COURSE IN ANALYSIS

4. For m1 , m2 ∈ N0 we find by rather rough estimates X X m2 |∂ α u(x)| ≤ (1 + ||x||2 ) 2 (1 + ||x||2 )m2 |∂ α u(x)| |α|≤m1

≤

|α|≤m1

m2  m2    X X X X m2 m2 (x21 + · · · + x2n )m2 −1 |∂ α u(x)|, ||x||m2 −l |∂ α u(x)| = l l |α|≤m1 l=0

|α|≤m1 l=0

and it follows (1 + ||x||2 )

X

m2 2

|α|≤m1

|∂ α u(x)| ≤ C

X

X

|x2β ∂ α u(x)|

|α|≤m1 |β|≤2m2

which is finite since pαβ (u) is for all α, β ∈ Nn0 finite. Thus, for u ∈ J (Rn ) we have for every α ∈ Nn0 and every m ∈ N0 the estimate m

|∂ α u(x)| ≤ Cα,m (u)(1 + ||x||2 )− 2 .

(?)

For u, v ∈ J (Rn ) we find now using our results on differentiating parameter dependent integrals, note that (?) implies ∂ α u ∈ L1 (Rn ) ∩ L∞ (Rn ) for all α ∈ Nn0 and u ∈ J (Rn ), the estimate Z u(x − y)v(y)dy| xβ ∂xα (u ? v)(x)| = |xβ ∂xα

≤

Z

≤

Rn

≤C

Z

Rn

Rn

|xβ ∂xα u(x − y)v(y)|dy |β| 2

(1 + ||x||2 )

Z

Rn α

|(∂ α u)(x − y)||v(y)|dy

(1 + ||x||2 )

(1 + ||x −

|β| 2

y||2 )

k

|β| 2

(1 + ||y||2 )− 2 dy,

where we used (?) for (∂ u) and v and k will determined later. From Peetre’s inequality we deduce (1 + ||x||2 )

(1 + ||x −

|β| 2

y||2 )

|β| 2

≤2

which yields |xβ ∂xα (u ∗ v)(x)| ≤ C 0

Z

|β| 2

(1 + ||y||2 )

(1 + ||y||2 )

|β| 2

|β|−k 2

dy

and for k > n + |β| we find Pαβ (u ∗ v) < ∞, hence u ∗ v ∈ J (Rn ). 5. Since K ∩ ∂G = ∅ it follows that dist(K, ∂G) > 0. We choose  > 0 such that  < 21 dist(K, ∂G). For K := {x ∈ Rn | there exists y ∈ K such that ||x − y|| < } = K + B (0)

642

SOLUTIONS TO PROBLEMS OF PART 6

we consider J (χK ), i.e. Z Z (n) j (x − y)χK (x − y)λ (dy) = J (χK )(x) = Rn

B (0)

χK (x − y)j (y)λ(n) (dy),

which is nothing but the Friedrichs Mollifier applied to the characteristic functions   ∞ 4 of K . Thus J (χK ) ∈ C (R ), 0 ≤ J χK ≤ 1 and since  < 21 dist(K, ∂G) we have   supp J χK ⊂ B (0) + K ⊂ B2 (0) + K ⊂ G.

Finally,  for x ∈ K and y ∈ B (0) we find that χK (x − y) = 1 implying that J χK (x) = 1.

6. Consider the sets Gj, := {x ∈ Gj | inf y∈∂Gj ||x − y|| > }, see the Figure below.

Gj

Gj,

 There exists 0 > 0 such that for 0 <  < 0 , the sets Gj, are non-empty and open. Furthermore Gj, is compact and Gj, ⊂ Gj . We claim now the existence of 1 > 0 SN such that 0 <  < 1 ≤ 0 implies K ⊂ j=1 Gj, . If this was not the case then SN there would be a sequence (xl )l∈N , xl ∈ K, such that xl ∈ j=1 Gj, 1l . Since K is compact (xl )l∈N has a convergent subsequence which we denote again by (xl )l∈N S and which has a limit x ∈ K. Since K ⊂ N j=1 Gj there exists j0 such that x ∈ Gj0 and hence x ∈ Gj0 , k1 for k ≥ k0 . This however implies for l ≥ k0 that xl ∈ Gj0 , k1 SN and hence xl ∈ j=1 Gj, k1 which is a contradiction. Hence for 0 <  < 1 we have SN K ⊂ j=1 Gj, . We now fix  > 0 with these properties. Since K is compact and SN j=1 Gj, is open, there exists η > 0 such that {x ∈ Rn | there exists y ∈ K such that ||x − y|| < η} := Kη ⊂

N [

Gj, .

j=1

Let ψj ∈ C0∞ (Gj ) ⊂ C0∞ (Rn ) such that supp ψj ⊂ Gj , 0 ≤ ψj ≤ 1 and ψ|Gj, = 1. P  N By Problem 5 such a function ψj does exist. It follows that j=1 ψj |K ≥ 1.

643

A COURSE IN ANALYSIS

n For ψ ∈ C0∞ (Kη ) ⊂ L∞ 0 (R ) such that ψ|K = 1 we define the functions ϕj (x) := ψ (x) j ∞ ∞ n P∞ ψl (x) ψ(x), which belong to C0 (Gj ) ⊂ C0 (R ) and for x ∈ K it follows that Pl=1 N j=1 ψj (x) = ψ(x) = 1. (Our solution follows closely [89].)

7. For A ∈ B (n) we have

(µt ∗ µs )(A) = Z Z

=

R

R

=

Z Z R

 χA (x + y)µs (dx) µt (dy)

R

 χA (x + y)e−as 0 (dx) e−at 0 (dy) Z

R

e−as χA (y)e−at 0 (dy)

= e−a(s+t) χA (0) =

Z

R

χA (z)e−a(s+t) 0 (dz)

= µt+s (A). 8. Under our assumptions the following calculation is justified Z Z   u(x − z − y)µt (dy) µs (dz) (Ts ◦ Ts u (x) = =

Rn

Rn

Z

Rn

u(x − z)(µt ∗ µs )(dz) =

Z

Rn

u(x − z)µt+s (dz)

= Ts+t u(x).

9. Since kα is even and non-negative we find Z Z ∞ kα (x)dx = 2 χ[0,1] (x)|x|−α dx R

=2

Z

0

1

0

|x|−α dx =

1 2 2 |x|1−α = < ∞, 1−α 1−α 0

hence kα ∈ L1 (R) and by Young’s inequality we find for u ∈ C0 (R) ⊂ Lp (R) ||Kop u||Lp ≤

2 ||u||Lp 1−α

which allows us to obtain a continuous extension of Kop to Lp (R) which satisfies the same estimate. Now we turn to kγ,n (||x||) = χB1 (0) (x)||x||−γ in Rn . It follows with the help of spherical coordinates that Z Z χB1 (0)\{0} (x)||x||−γ dx kγ,n (||x||)dx = Rn

Rn

= Cn

Z

0

1

r−γ rn−1 dr = Cn

Z

0

644

1

rn−1−γ dr = Cn

1 rn−γ |10 n−γ

SOLUTIONS TO PROBLEMS OF PART 6

=

Cn , n−γ

provided n − γ > 1 i.e. γ < n. Note that by (II.12.26) we have Z π Z π  sinn−2 ϑ1 · ... · sin ϑn−2 dϑn−2 ...dϑ1 . ... Cn = 2π 0

0

(n)

Now we may argue as before to find that Kop has a continuous extension to Lp (Rn ) satisfying Cn (n) ||Kop u||Lp ≤ ||u||Lp . n−γ Chapter 11 1. Since g is continuous and g(x) 6= 0 on [a, b] it follows for some c0 that |g(x)| ≥ c0 > 0 for all x ∈ [a, b] implying that 1 1 |g(x) − g(y)| − . ≤ g(x) g(y) c20

This estimate however implies the absolute continuity of that 2.

f g

1 g

and therefore it follows

is absolutely continuous too.

a) For ϕ there exists a constant L > 0 such that |ϕ(x) − ϕ(y)| ≤ L|x − y| holds for all x, y ∈ [a, b]. This implies |(ϕ ◦ f )(x) − (ϕ ◦ f )(y)| ≤ L|f (x) − f (y)| for all x, y ∈ [a, b] and again the absolute continuity of f implies that of ϕ ◦ f. b) Let  > 0 and δ > 0 such that for every finite number of intervals (xk , yk ) ⊂ PN PN [a, b], k = 1, ..., N, with k=1 (yk −xk ) < δ it follows that k=1 |f (xk )−f (yk )| < . Suppose that g is an increasing function. Since it is by assumption absolutely continuous we can find ν < 0 such that for every finite collection of intervals (sj , tj ) ⊂ PM P [a, b], j = 1, ..., M, with N j=1 (g(tj ) − g(sj )) < δ. This j=1 (tj − sj ) < ν it follows implies of course M X |f (g(tj )) − f (g(sj ))| < . j=1

The case of a decreasing function g goes analogously. 3. Let N ⊂ (a, b) be a set of measure zero and  > 0. We can find δ > 0 such that for every pairwise disjoint, finite family of intervals (xk , yk ) ⊂ [a, b], 1 ≤ k ≤ N, with PN PN k=1 (yk −xk ) < δ it follows that k=1 |f (xk )−f (yk )| < . For the set N ⊂ (a, b) of measure zero we can find an open set U ⊂ (a, b) with N ⊂ U and λ(1) (U ) < δ. Since every open set in R is the the denumerable union of open intervals, compare with Theorem S I.19.27, there exist P∞pairwise disjoint open intervals (sj , tj ), j ∈ N, such and U ⊂ j∈N (sj , tj ) and j∈N (tj − sj ) < δ. On [sj , tj ] the continuous function f

645

A COURSE IN ANALYSIS

attains its minimum, say at αj , and its maximum, say at βj , with aj := αj ∧ βj and bj = αj ∨ βj it follows thatf ([sj , tj ]) = [f (aj ), f (bj )]. For every N ∈ N we have PN PN and consequently j=1 |f (aj ) − f (bj )| < , which yields in the j=1 (bj − aj ) < δ P limit N → ∞ that ∞ j=1 |f (aj ) − f (bj )| < . However [ [ f (N ) ⊂ f (U ) = f ((sj , tj )) ⊂ f ([sj , tj ]), j∈N

j∈N

implying

λ(1) (f (N )) ≤

∞ X j=1

λ(1) (f ([sj , tj ])) ≤

∞ X j=1

|f (aj ) − f (bj )| ≤ ,

i.e. λ(1) (f (N )) = 0. Note that since λ(1) ({a}) = λ(1) ({b}) = 0 the assumption N ⊂ (a, b) is no restriction.

4. Since [a, Rb] is of finite measure it follows that Lp ([a, b]) is a subset of L1 ([a, b]). With x G(x) := a−h g(t)dt we find that gh (x) :=

 1  G(x + h) − G(x − h) 2h

and by Corollary 11.10 the function G, hence gk , is a continuous function. For p = 1 we have Z x+h 1 |g(t)|dt, |gh (x)| ≤ 2h x−h and for 1 ≤ p ≤ ∞,

1 p

+

1 q

= 1, H¨ older’s inequality gives

|gh (x)|p ≤

1  (2h)p =

Z

1 2h

x+h

1dt x−h

Z

x+h

x−h

 pq  Z

x+h x−h

|g(t)|p dt



|g( t)|p dt.

Now we find for 1 ≤ p < ∞ that Z b Z b  Z x+h  1 p |gh (x)| dx ≤ 1|g(t)|p dt dx 2h a a x−h Z b Z h Z h Z b   1 1 |g(t + x)|p dt dx = |g(t + x)|p dx dt = 2h a 2h −h −h a Z h  Z b+t  1 = |g(s)|p ds dt. 2h −h a+t

Since g|[a,b]{ = 0 we get further 1 2h

Z

h

−h

Z

b+t

a+t

Z h Z b   1 |g(s)|p ds dt ≤ |g(s)|p ds dt, 2h −h a

646

SOLUTIONS TO PROBLEMS OF PART 6

or

Z

a

b

|gk (x)|p dx ≤

Z

b

a

|g(s)|p ds,

implying that ||gh ||Lp ≤ ||g||Lp . Rx 5. If we set ϕ(x) = f (a) + a f 0 (t)dt and s(x) = f (x) − ϕ(x), it follows that ϕ is absolutely continuity with ϕ(a) = f (a), and for s it follows that s is either identically zero or s is a singular function. For two decomposition of f, ∼

∼

f (x) = ϕ(x) + s(x) = ϕ(x) + s(x) ∼

∼

we find ϕ(x) − ϕ(x) = s(x) − s(x), which implies by the absolute continuously ∼ ∼ ∼ of ϕ and ϕ that (ϕ − ϕ)0 = 0 almost everywhere, hence (ϕ − ϕ) = constant, and ∼ ∼ ∼ from ϕ(a) = ϕ(a) − f (a) we deduce ϕ = ϕ which in turn implies s = s, i.e. the decomposition of f is unique. 6. Before turning to the calculation let us understand the meaning of this result: there are functions having a derivative everywhere, but still the fundamental theorem in Rx the form of u(x) − u(y) = y u0 (t)dt does not necessarily hold. The function u is bounded on [0, 1] and we have |u(x)| ≤ x2 , u(0) = 0. We extend u as an even function to [−1, 1] or directly modify the solution to Problem 7 of Chapter I.21 to see that u is differentiable at x = 0. The differentiability at all other points follows from the chain rule. For 0 < a < b < b ≤ 1 we can apply the fundamental theorem to find Z b π π u0 (t)dt = b2 cos 2 − a2 cos 2 . b a a q q 2 1 and bn := 2n it follows that Taking an := 4n+1 Z

bn

u0 (t)dt =

an

1 . 2u

Now, [an , bn ] ∩ [al , bl ) = ∅ for k 6= l and therefore we find with A := [0, 1] that Z ∞ X 1 = +∞, |u0 (t)|dt ≥ 2n A n=1

S

n∈N [an , bn ]

⊂

i.e. u0 ∈ / L1 ([0, 1]).

7. As a convex function g iscontinuous.  By the proof of Theorem I.23.4 we know that h 7→ g(x+h)−g(x) is decreasing in h. We also know that the = − g(x)−g(x+h) h h 0 0 derivatives g+ (x) from the right and g− (x) from the left exist and 0 g+ (x) =

lim

h→0,h>0

647

g(x + h) − g(x) h

A COURSE IN ANALYSIS

as well as 0 g− (x) =

lim

h→0,h>0

g(x) − g(x − h) h

hold. Moreover, we have g(x + h) − g(x) g(x) − g(x − h) ≤ , h > 0, h h 0 0 (x) and both are finite numbers. Now we claim that (x) ≤ g+ implying that g− 0 0 (x). This follows from (I.23.4) which yields (y) ≤ g− a < y < x < b implies that g+ 0 g+ (y) ≤

Thus we have (?)

g(x) − g(y) 0 ≤ g− (x). x−y

0 0 0 g+ (y) ≤ g− (x) ≤ g+ (x).

0 0 implying that g+ is monotone increasing. The proof that g− is monotone increasing 0 0 have at and g− is similar. By Problem 6 to Chapter I.20 we know that both g+ 0 the most countable many points of discontinuity. For a point of continuity x of g+ 0 0 estimates (?) implies g+ (x) = g− (x). Thus g 0 exists at every point of continuity of 0 , in particular λ(1) -almost everywhere, and is monotone increasing. g+

8. For simplicity we set µ(s) := λ(1) ({M(f ) > s}) and  f (x), |f (x)| ≥ 2s g(x) := 0, otherwise. It follows that |f (x)| ≤ |g(x)| +

s 2

1 2h h>0

M(f )(x) ≤ sup

and consequently

Z

x+h x−h

|g(y)|dy +

s s = M(g)(x) + . 2 2

Moreover, since {x ∈ R | M(f )(x) > s} ⊂ {x ∈ R | M(g) > 2s } we find by Theorem 11.28 that s 6 µ(s) ≤ λ(1) ({M(g) > }) ≤ ||g||L1 2 s Z 6 |f (y)|dy. = s |f |≥ 2s Now we find

≤p

Z

0

∞

Z

s

p−1

6 Z s

R

|M(f )|p λ(1) = p

|f |≥ 6s

Z

∞

µ(s)sp−1 ds

0

Z  Z |f (y)|dy ds = 6p |f (y)| R

2|f (y)|

0

where in the last step we used Tonelli’s theorem. Since Z 2|f (y)| (2|f (y)|)p−1 , p > 1, sp−2 ds = (?) p−1 0

648

 sp−2 ds dy,

SOLUTIONS TO PROBLEMS OF PART 6

we arrive at

Z

R

|M(f )|p dλ(1) ≤ 2p

i.e. ≤ ||M||Lp ≤ 2

3p p−1

Z

R

|f |p dλ(1) ,

 3p  p1 ||f ||Lp . p−1

A remark is needed to (?). Since f ∈ Lp (R), f is λ(1) -almost everywhere finite, hence (?) holds almost everywhere only, which however is sufficient for our purpose. It is now easy to extend the result to Rn , a detailed proof which we need for our case is given in [90].
k P∞ 9. The function g is bounded, i.e. |g(x)| ≤ 1 for all x ∈ R, and since k=0 43 = 4, the Weierstrass test implies the continuity of f. Now let x ∈ R and m ∈ N. The intervals (4m x, 4m x + 21 ) and (4m x − 21 , 4m x) cannot both contain an integer. Thus with νm := ± 12 4−m chosen approximately, in the interval with endpoints 4m x and 4m (x + νm ) there is no integer. Note that for m ∈ N fixed the term g(4m (x + νm )) − g(4m x)) , n ∈ N0 , νm has always the same sign and moreover we have g(4m (x + ν )) − g(4m x))  0, m = 4n , νm

n>m 0 ≤ n ≤ m.

Here we need that for n > m we have 4n νm = ± 21 4n−m = ±2l and g has period 2. It follows that for f that ∞ f (x + ν ) − f (x) X 3
k g(4k (x + νm )) − g(4k x)) m = νm 4 νm k=0

m X 3
k g(4k (x + νm )) − g(4k x)) = 4 νm k=0

=

m X 3m+1 − 1 3
k k ·4 = . 4 2 k=0

This implies however that for νm → 0, i.e. m → ∞, the limit limm→∞ cannot exist, i.e. f 0 (x) does not exist.

f (x+νm )−f (x) νm

Chapter 12 1. Let x1 , x2 ∈ G be two solutions of f (x) = y0 . It follows that 0 = f (x2 ) − f (x1 ) = 2 −x1 )|| Jf (x1 )(x2 − x1 ) + ϕ(x2 − x1 ) and limx2 →1 ||ϕ(x ||x2 −x1 || = 0. Since x1 is not a critical

649

A COURSE IN ANALYSIS

point, the symmetric matrix Jf (x1 ) has full rank and we can find some ν > 0 such that ||Jf (x1 )z|| ≥ ν||z|| for all z ∈ Rn . This implies ν||x2 − x1 || ≤ ||Jf (x1 )(x2 − x1 )|| + ||ϕ(x2 − x1 )|| or 0 0 for x ∈ R \ C. This implies that h : [0, 1] → R x defined by h(x) := 0 g(t)dt is a strictly monotone increasing function which is continuously differentiable and h0 (x) = 0 for x ∈ C. Since f is injective the set h(crit(h|(0,1) )) ∪ {h(0), h(1)} = h(C) is non-denumerable, whereas h has only the R1 (local) extreme values h(0) = 0 and h(1) = 0 g(t)dt. Thus Ex(h) has two elements but h(crit(h|(0,1) )) is non-denumerable. 3. We only need to apply Corollary 12.5. 4. Let {α1 , ..., αN } ⊂ R be the range of s and set Aj = {x ∈ A|s(x) = αj }. Clearly, Aj is measurable. By Lemma 12.6, given  > 0 we can find closed sets Cj, ⊂ A such SN that λ(n) (Aj \ Cj, ) ≤ N , j = 1, ..., N. with C := j=1 Cj, we find λ(n) (A\ C ) ≤ . Since dist(Cj, , Cl, ) > 0 and s|Cj, = αj it follows that s|C is continuous. 5. For k ∈ N we can find closed sets Ck ⊂ A such that λ(n) (A\C) < k1 and fk := f |Ck is S continuous. The set C := k∈N Ck belongs to A and λ(n) (A \ C) = 0. Furthermore, for a ∈ R we have {x ∈ A | f (x) > a} = {x ∈ C | f (x) > a} ∪ {x ∈ A \ C | f (x) > a} [ = {x ∈ Ck | fk (x) > a} ∪ {x ∈ A \ C | f (x) > a}. k∈N

On Ck the functions fk is continuous and therefore measurable, which yields that {x ∈ Ck | fk (x) > a} is measurable and since {x ∈ A \ C | f (x) > a} is a set of measure zero it follows that {x ∈ A | f (x) > a} is measurable, hence f is measurable. 6. For  > 0 choose ν > 0S such that δ :=  − ν > 0. Then we can find points N x1 , ..., xN such that Y ⊂ j=1 Bδ (xj ), but Bδ (xj ) ⊂ B (xj ) and therefore we have SN Y ⊂ j=1 B (xj ), i.e. Y admits a finite -net. Clearly, if Y admits for every S SM  > 0 a finite -net, i.e. Y ⊂ M j=1 B (yj ), then we find Y ⊂ j=1 B (yj ), i.e. the assumptions of Problem 6 are fulfilled.

650

SOLUTIONS TO PROBLEMS OF PART 6

7.

a) Given  > 0, choose N > b−a  . Then {a + , a + 2, ..., a + N } is a finite -net for (a, b). b) Since v ∈ Y implies 0 ≤ v(t) ≤ 1 for all t ∈ [0, 1] it follows that ||v||∞ ≤ 1 for all v ∈ Y, i.e. Y is a bounded set in C([0, 1]), in fact it is a subset of the unit ball B1 (0), where 0 denotes the zero function on [0, 1]. Since for all v ∈ Y we have supt∈[0,1] |v(t) − 21 | ≤ 21 it follows from the solution of Problem 6 that Y admits a finite 12 -net. The following figure indicates how u may look like:

1

1 By construction we have u ∈ Y and ||u − uj ||∞ ≥ implying that {u1 , ...um } cannot be an -net for Y.

651

1 2

>  for all j ∈ {1, ..., m}

This page intentionally left blank

Solutions to Problems of Part 7 Chapter 13 1.

a) 4i (3 + 5i)(−7 − 2i) 4i(3 − i) 3 + 5i + = + 2i − 7 3 + i (−7 + 2i)(−7 − 2i) (3 + i)(3 − i) −11 − 41i 4 + 12i = + 53 10 51 + 113i ; = 265 b) 28 i8 − 128 256 − 128 (2i)8 − 128 = = = 16 (2 + 2i)(2 − 2i) 4+4 8

note i8 = i2 · i2 · i2 · i2 = (−1)(−1)(−1)(−1) = 1; c) 1−i 2+3i 2+4i 6−2i

= =

d)

2.

1 − i 6 − 2i 4 − 8i 2 − 4i · = = 2 + 3i 2 + 4i −8 + 10i −4 + 5i (2 − 4i)(−4 − 5i) −28 + 6i = ; 16 + 25 41

√ ! √ ! √ √ 1 1 3 3 3 3 3 1 − + i i = − i− i− − + 2 2 2 2 4 4 4 4 √ 1 3 =− − i. 2 2

a) a2 + 2iab − b2 − 2iab (a + ib)2 − 2iab = i(a − b) i(a − b) a2 − b 2 = = −i(a + b); i(a − b) b) With z+ = z1 , z− = z2 we find

!2 r p p2 = − ±i q− 2 2 r p2 p2 − q ∓ ip q − = 2 2 r p2 p2 pz± = − ± ip q − 2 4 2 z±

653

A COURSE IN ANALYSIS

and therefore 2 z± + pz± + q =

r r p2 p2 p2 p2 − q ∓ ip q − − ± ip q − + q = 0. 2 4 2 4

a) With zj = xj + iyj we have z j = xj − iyj and it follows

3.

z1 · z2 = (x1 + iy1 )(x2 + iy2 )

= (x1 x2 − y1 y2 ) + i(x1 y2 + x2 y1 )

= x1 x2 − y1 y2 − i(x1 y2 + x2 y1 ) and

z 1 · z 2 = (x1 − iy1 )(x2 − iy2 ) = x1 x2 − y1 y2 + i(−x1 y2 − x2 y1 ) = x1 x2 − y1 y2 − i(x1 y2 + x2 y1 ),

implying the assertion. b) z · z = (x + iy)(x − iy) = x2 − ixy + ixy + y 2 = x2 + y 2 . c) x + iy + x − iy 2x z+z = = = Re z 2 2 2 and z−z x + iy − x + iy 2iy = = = Im z. 2i 2i 2i 4. The binomial theorem, see Theorem I.3.9, has for C the same form as for R: for z, w ∈ C and n ∈ N we have n   X n n−k k (z + w)n = z w . k k=0

There is also no need to modify the proof: the manipulations of terms involving z and w use only the axioms of a field and the manipulations of the binomial coefficients are unchanged. 5.

a) The triangle inequality yields |z| ≤ |z − w| + |w|

or

|z| − |w| ≤ |z − w|

as well as |z| ≤ |z + w| + | − w|

or

|z| − |w| ≤ |z + w|,

and therefore |z| − |w| ≤ |z − w| ∧ |z + w|. Changing the role of z and w we find − (|z| − |w|) = |w| − |z| ≤ |z − w| ∧ |z + w|,

654

SOLUTIONS TO PROBLEMS OF PART 7

which gives eventually ||z| − |w|| ≤ |z − w| ∧ |z + w|. b) For z = x + iy and |z| = x2 + y 2 Since


12

we derive immediately |z| ≤ |x| + |y|.

|x|2 + 2|x||y| + |y|2 |x|2 + |y|2 |x|2 + |y|2 |x|2 + |y|2 = + |x||y| ≤ + 2 2 2 2 or

2

(|x| + |y|) ≤ |x|2 + |y|2 2 implying |x| + |y| √ ≤ |z|. 2 6.

a) For f, g ∈ V we have of course d(f, g) = kf − gk ≥ 0 and d(f, g) = 0 implies kf − gk = 0, i.e. f = g. Moreover it follows that d(f, g) = kf − gk = k − 1(g − f )k = | − 1|kg − f k = kg − f k = d(f, g). Eventually we observe that d(f, g) = kf − gk = kf − h + h − gk ≤ kf − hk + kh − gk = d(f, h) + d(h, g) and we have proved that d is a metric. b) Clearly kzk ≥ 0 and kzk = 0 implies |zj | = 0 for j = 1, . . . , n, i.e. z = 0. For λ ∈ C we find that  21   21  21   n n n X X X |λzj | =  kλzk +  |λ|2 |zj |2  = |λ|  |zj |2  = |λ|kzk. j=1

j=1

j=1

In order to prove the triangle inequality we first rewrite kzk. Let zj = xj + ixn+j , 1 ≤ j ≤ n, xl ∈ R, 1 ≤ l ≤ 2n. It follows that |zj |2 = x2j + x2n+j and  

n X j=1

|zj |

 21

2

=

2n X l=1

x2l

! 21

,

in other words kzk is the Euclidean norm of the vector (x1 , . . . , x2n ). With wj = yj + iyn+j , 1 ≤ j ≤ n, yl ∈ R, 1 ≤ l ≤ 2n, we find now kz + wk = k(x1 + y1 , . . . , x2n + y2n )k where the norm on the right hand side is the Euclidean norm in R2n and therefore we get kz + wk ≤ k(x1 , . . . , x2n )k + k(y1 , . . . , y2n )k ≤ kzk + kwk.

655

A COURSE IN ANALYSIS

7.

a) Note that (3n2 − 5in)(−i) −3n2 i − 5n 5 3n2 − 5in = = = − − 3i n2 i n2 n2 n implying

3n2 − 5in = −3i. n→∞ n2 i lim

b) We have −4ik 3 + 10k 2 (k − ik)2 (2k + 5i) = 2 3 k + (3 + ik) 27 − 8k 2 + i(27k − k 3 ) −4i + 10 k = 27 8 27i k3 − k + k − i and it follows that

(k − ik)2 (2k + 5i) = 4. k→∞ k 2 + (3 + ik)3 lim

c) First we note that N X

zk =

k=4

N X

k=0

=

z k − (1 + z + z 2 + z 3 )

1 − zN − (1 + z + z 2 + z 3 ). 1−z

Since |z| < 1 it follows that limN →∞ z N = 0 and therefore lim

N →∞

8.

a) Since

N X

k=4

k−1 z k! 1 (k + 1)! z k = |z| k + 1 → 0

by the ratio test the series b) Since

it follows that

z4 1 − (1 + z + z 2 + z 3 ) = . 1−z 1−z

zk =

P∞

zk k=0 k!

as k → ∞,

converges for all z ∈ C.

2k 2k (−1)k z ≤ |z| (2k)! (2k)!

∞ ∞ 2k X X |z|l (−1)k z ≤ = e|z| (2k)! l! l=0

k=0

and by the comparison test the convergence of

656

P∞

k z 2k k=0 (−1) (2k)!

for all z ∈ C follows.

SOLUTIONS TO PROBLEMS OF PART 7

c) We argue as in part b) and observe first 2k−1 2k−1 (−1)k−1 z ≤ |z| (2k − 1)! (2k − 1)!

to conclude that

∞ 2k−1 X ≤ e|z| (−1)k z (2k − 1)! k=1

P∞

2k−1

z and the convergence of k=1 (−1)k (2k−1)! for all z ∈ C follows again by the comparison test. d) For |z| < 1 we find k+1 z k k k + 1 · z = k + 1 |z| ≤ |z| < 1 P∞ k and the ratio test yields the convergence of k=1 zk for |z| < 1. P∞ |z|k e) From part d) we know that for |z| < 1 the series P k=1 k converges (which ∞ follows also from the convergence of the geometric series k=0 |z|k ). Since X ∞ ∞ 2k−1 X |z|l ≤ (−1)k z 0 we can find N = N () such that k ≥ N () implies |Bk − B| <  and for n > N () it follows that N n n X X X |an−k ||Bk − B| + |an−k ||Bk − B| an−k (Bk − B) ≤ k=0

k=0

k=N +1

≤ max |Bk − B| k∈N

≤ max |Bk − B| k∈N

657

N X

k=0 N X

k=0

|an−k | + 

|an−k | + 

n X

k=N +1 ∞ X

k=0

|an−k |

|ak |.

A COURSE IN ANALYSIS P∞ For n → ∞ it follows that |an | → 0 since k=0 |ak | converges. Therefore we find for every fixed N N X lim |an−k | = 0 n→∞

implying that

k=0

∞ n X X |ak |, 0 ≤ lim sup an−k (Bk − B) ≤  n→∞ k=0

k=0

P∞ i.e. we find k=0 ck = A · B. P∞ Now we show the absolute convergence of k=0 ck : n M M X M X n X X X |ck | = an−k bk ≤ |an−k ||bk | n=0 n=0 k=0 n=0 k=0 ! ! ∞ ! ! M ∞ M X X X X |bk | . |an | |bk | ≤ |an | = n=0

10. We form the Cauchy product of ez1 ez2 = =

∞ X

n=0 ∞ X

n=0

k=0

P∞

z1k

P∞

k=0

z2k

and k=0 k! to find ! ! n   ∞ n X X 1 X n n−k k z1n−k z2k z z2 = k 1 (n − k)! k! n! n=0 k=0 k!

k=0

k=0

1 (z1 + z2 )n = ez1 +z2 , n! n=0

where we have used the binomial theorem, see Problem 4. 11.

a) Since ak =

Pk Pk−1

Pk Pk −1 = 1. But this is limk→∞ Pk k = lim limk→∞ PPk−1 k→∞ Pk−1

we need to prove that limk→∞

trivial

= 1. Q∞ b) We sketch once more the proof of Proposition I.30.7. Suppose that k=1 Q  ak N a converges to a 6= 0. By the Cauchy criterion applied to the sequence k=1 k

since limk→∞ Pk = limk→∞ Pk−1 implying that

N ∈N

we can find for  > 0 and η > 0 a number N = N (, η) ∈ N such that n > m > N (, η) implies n m Y Y ak − ak < η k=1

or

Qm

k=1

n Y η ak − 1 < Q m . | k=1 ak | k=m+1

Since limm→∞ k=1 ak = a 6= 0 it follows for m > N0 that | 2 therefore, if n > m > max(N0 , N (, η)), with η = |a| we have n Y ak − 1 < . k=m+1

658

Qm

k=1

ak | ≥

|a| 2

and

SOLUTIONS TO PROBLEMS OF PART 7

By inspecting the proof of Proposition I.30.7 that, as in the real case, the converse of the statement also holds, i.e. if the conclusion of the problem holds then the product converges. 12. Again we can argue as in the real case. a) We note the estimate |(1 + a1 ) · . . . · (1 + an ) − 1| ≤ (1 + |a1 |)(1 + |a2 |) · . . . · (1 + |an |) − 1, which follows as in the proof of Proposition I.30.9: For n = 1 we have |(1+a1 )−1| = |a1 | = (1 + |a1 |) − 1, and now we find |(1 + a1 )(1 + a2 ) · . . . · (1 + an )(1 + an+1 ) − 1| = |(1 + a1 )(1 + a2 ) · . . . · (1 + an + an+1 + an an+1 ) − 1|

≤ |(1 + |a1 |)(1 + |a2 |) · . . . · (1 + |an + an+1 + an an+1 |) − 1|,

but 1 + |an + an+1 + an an+1 | ≤ (1 + |an |)(1 + |an+1 |). Thus we have n n Y Y (1 + ak ) − 1 ≤ (1 + |ak |) − 1 k=m+1

k=m+1

Q∞ and by the Cauchy Q∞ criterion the convergence of k=1 (1 + ak ), follows from the convergence of k=1 (1 + |ak |). b) We argue as in the proof of Proposition I.30.10. Since |a1 | + · · · + |an | ≤ (1 + |a1 |)(1 + |a2 |) · . . . · (1 + |an |) Q∞ the absolute of the P∞ convergence of k=1 (1 + ak ) implies the absolute convergence series k=1 ak . On the other hand, for x > 0 we know that 1+x ≤ ex and therefore

(1 + |a1 |)(1 + |a2 |) · . . . · (1 + |an |) ≤ e|a1 |+···+|an | . P Q∞ If the series ∞ k=1 |ak |converges then it follows that k=1 (1 + |ak |) converges too QN since (1 + |a |) is an increasing sequence bounded from above. k k=1 N ∈N

Chapter 14 1.

Pn Pn 2 a) Since hz, zi = j=1 |zj | it follows that hz, zi ≥ 0 and if j=1 zj z j = hz, zi = 0 we must have zj = 0 for all j = 1, . . . , n, i.e. z = 0 ∈ Cn . Moreover we find hz, wi = =

n X

zj w j =

j=1

z j wj

j=1

j=1

n X

n X

wj z j = hw, zi.

659

A COURSE IN ANALYSIS

Finally we note for ζ1 , ζ2 ∈ C that hζ1 z + ζ2 w, vi = = ζ1

n X

(ζ1 zj + ζ2 wj )v j

j=1 n X

zj v j + ζ2

n X

wj v j

j=1

j=1

= ζ1 hz, vi + ζ2 hw, vi,

and we have proved that hz, wi is an unitary scalar product. Now we prove the Cauchy-Schwarz inequality. For hz, wi = 0 nothing remains to be proved. From the definition of hz, wi we conclude for ζ ∈ C that 0 ≤ hz − ζw, z − ζwi

= hz, zi − ζhw, zi − ζhz, wi + ζζhw, wi = kzk2 − 2Re (ζhw, zi) + |ζ|2 kwk2 .

With ζ =

kzk2 hw,zi ,

hw, zi 6= 0, it follows that 0 ≤ kzk2 − 2kzk2 +

kzk4 kwk2 |hw, zi|2

or |hz, wi|2 kzk2 ≤ kzk4kwk2 , i.e. |hz, wi| ≤ kzkkwk.

b) First let us make explicit the meaning of kuk∞ . For u : K → C we have the decomposition u = a + ib where a, b : K → R are continuous functions. Therefore
1 |u(x)| = a2 (x) + b2 (x) 2 and
1 kuk∞ = sup a2 (x) + b2 (x) 2 . x∈K

Clearly kuk∞ ≥ 0 and kuk∞ = 0 implies a(x) = 0 for all x ∈ K and b(x) = 0 for all x ∈ K, thus u = 0. For ζ ∈ C we find first |ζu(x)| = |ζ||u(x)| and therefore kζuk∞ = sup |ζu(x)| = |ζ| sup |u(x)| = |ζ|kuk∞ . x∈K

x∈K

Finally we derive the triangle inequality for u, v ∈ C(K; C): ku + vk∞ = sup |u(x) + v(x)| ≤ sup |u(x)| + sup |v(x)| = kuk∞ + kvk∞ , x∈K

x∈K

x∈K

where we used that the triangle inequality holds for the modulus of complex numbers.

660

SOLUTIONS TO PROBLEMS OF PART 7

Next we turn to kukL2 for u ∈ C(K; C). Note that for u = a + ib, a, b ∈ C(K; R) we have  21 Z
. |a(x)|2 + |b(x)|2 λ(n) (dx) kukL2 = K

Again it is easy to seeR that kukL2 ≥ 0 and R kζukL2 = |ζ|kukL2 . Suppose kukL2 = 0. Then both integrals K a2 (x) dx and K b2 (x) dx must vanish. Now a and b are continuous functions, then a2 ≥ 0 and b2 ≥ 0 are continuous functions and for a R continuous function f : K → R, f ≥ 0, we know that K f (x)λ(n) (dx) = 0 implies that f (x) = 0 for all x ∈ K, see Proposition I.20.10. The triangle inequality can either be derived as in the next problem or it can be reduced to the real case: for u, v ∈ C(K, C) we find Z

K

Z  12  21 2 ≤ ||u| + |v|| dx |u + v|2 dx K

and using the Cauchy-Schwarz inequality for real-valued functions we get Z Z Z |v|(|u| + |v|) dx |u|(|u| + |v|) dx + (|u| + |v|)2 dx = K

≤

Z

K

K

K

 12 Z  12 Z  21  12 Z + |v|2 dx (|u| + |v|)2 dx (|u| + |v|)2 dx |u|2 dx K

K

K

implying Z

K

 12 Z  12 Z  21 (|u| + |v|)2 dx ≤ + , |u|2 dx |v|2 dx K

K

which yields ku + vkL2 ≤ kukL2 + kvkL2 . 2. It is possible to derive the H¨ older inequality along the lines we proved in the previous problem the Cauchy-Schwarz inequality as we can reduce the new case to the case of real-valued functions which was treated in Chapter 5. However, since this chapter shall help to become some routine with complex-valued functions, we sketch a more direct proof here. We prove first H¨ older’s inequality Cn , then we indicate the approximation process how to come from sums to integrals, and finally we use the standard argument to deduce from H¨ older’s inequality Minkowski’s inequality. Recall that by Lemma I.23.11 for p, q ∈ (1, ∞), p1 + q1 = 1, and A, B ≥ 0 we have 1

1

Ap B q ≤

A p

+

B q.

Now assume for z, w ∈ CN that at least one zj0 and one wk0 , 1

1

1 ≤ j0 , k0 ≤ N is not zero. For 1 ≤ j ≤ N we replace in A p B q ≤ PN

(

|zj |

k=1

1

|zk |p ) p

P N

and B by

k=1

PN

PN

(

|wj |

k=1

|zj ||wj |  p1 P N p j=1

|zk |

k=1

1

|wk |q ) q

|wk |q

A p

+

B q

now A by

to find

 q1

PN PN p q 1 j=1 |wj | 1 j=1 |zj | + PN =1 ≤ PN p k=1 |zk |p q k=1 |wk |q

661

A COURSE IN ANALYSIS

implying for z, w ∈ CN H¨ older’s inequality

  q1  p1  N N N X X X |wj |q  |zj |p   zj wk ≤  j=1 j=1 j=1

as well as

  q1  p1  X N N X X N q p |wj |  . zj wj ≤  |zj |   j=1 j=1 j=1

Minkowski’s inequality in CN follows now with the standard argument using H¨ older’s inequality N X j=1

|zj + wj |p ≤

N X j=1



≤

|zj + wj |p−1 |zj | +

N X j=1



+

PN

|zj + wj |

N X j=1

N X j=1

|zj + wj |p−1 |wj |

 q1 

q(p−1) 



N X j=1

 q1 

|zj + wj |q(p−1)  

|zj |

N X j=1

 p1

p

 p1

|wj |p  .

|zj + wj |p 6= 0 we now get Minkowski’s inequality on CN by dividing 1 P PN N q(p−1) q through . The case where j=1 |zj + wj |p = 0 is trivial. j=1 |zj + wj | Now we turn to the corresponding inequalities for integrals. For complex-valued step functions we get immediately N Z X f (x)g(x) dx = (αj βj )λ(n) (Kj ) K j=1    N N   q1   p1 X X |βj | λ(n) (Kj )  |αj | λ(n) (Kj )   ≤

For

j=1

j=1

j=1

 p1   q1 N N X X |αj |p λ(n) (Kj )  ≤ |βj |q λ(n) (Kj ) 

=

j=1

Z

K

j=1

 q1  p1 Z q p |g(x)| dx |f (x)| dx K

P where we assume that f = j=1 αj χKj and g = N j=1 βj χKj are already given with respect to the joint partition (Kj )j=1,...,N of K. The rest of the proof is PN

662

SOLUTIONS TO PROBLEMS OF PART 7

now done by a standard approximation argument. In the case where K is Jordan measurable, i.e. λ(n) (∂K) = 0, we can approximate the integrals by Riemann sums. In the other case, i.e. the Borel set K is not Jordan measurable, we have to work with decompositions of f and g into real and imaginary parts and then decompose these into positive and negative parts. For step functions Minkowski’s inequality follows now from Minkowski’s inequality in CN and then again in the general case by approximation. 3. On C we take of course the Borel σ-field and since C ∼ = R2 topologically the σfield is B (2) . Thus  when we are concerned with measurability we can switch from  u . Now the statement is that f is measurable if and only if f = u + iv to f = v pr1 (f ) = u and pr2 (f ) = v are measurable which is of course the case as proved in Lemma 4.1. 4.

a) If we choose x = x0 + hek we find f (x0 + hek ) − f (x0 ) = ha, hek i + ϕx0 (x0 + hek ) or ϕx (x0 + hek ) f (x0 + hek ) − f (x0 ) = ak + 0 h h ∂f (x0 ) = ak where we use that the limit on the and therefore as h → 0 we get ∂x k ∂f ∂u ∂v right hand side exists by assumption. Note that ∂x (x0 ) = ∂x (x0 ) + i ∂x (x0 ). k k k

A =   F is differentiable at x0 . Then there exists a matrix  b) Suppose that ψx0,1 a11 (x0 ) . . . an1 (x0 ) 2 defined and a function ψx0 : U (x0 ) → R , ψx0 = ψx0,2 a21 (x0 ) . . . a2n (x0 ) kψ

(x)k

x0 = 0 and F (x) − F (x0 ) = in a neighbourhood of x0 such that limx→x0 kx−x 0k A(x − x0 ) + ψx0 (x). With ak := a1k (x0 ) + ia2k (x0 ) as well as a = (a1 , . . . , an ) and ϕx0 (x) := ψx0,1 (x) + iψx0,2 (x) we can rewrite F (x) − F (x0 ) = A(x − x0 ) +

ϕ

(x)

x0 ψx0 (x) as f (x) − f (x0 ) = ha, (x − x0 )i + ϕx0 (x) and limx→x0 kx−x = 0, i.e. 0k the differentiability of F implies the differentiability of f . But now the converse is obvious: if f is differentiable define a1k := Re ak , a2k := Im ak as well as ψx0,1 := Re ϕx0 and ψx0,2 := Im ϕx0 .

Chapter 15 1.

1

a) We have r = |z| = (16 + 16 · 3) 2 = √ π arctan 3 = π3 which yields z = 8ei 3 . 1

b) It follows that r = |z| = (1 + 1) 2 = √ 3π z = 2ei 4 .

√ √ 64 = 8 and ϕ = arctan 4 4 3 =

√ 2 and ϕ =

π 2

arctan 1 =

3π 4 ,

i.e.

√ √ 1 c) √ Now we find r = |z| = (24 + 8) 2 = 32 = 4 2 and further ϕ = π + √ 8 i 7π 6 . = π + arctan √13 = 7π arctan √24 6 which yields z = 4 2e

663

A COURSE IN ANALYSIS

a) Our starting point is the equality eiϕ

2.


5

= e5ϕi or

cos 5ϕ + i sin 5ϕ = (cos ϕ + i sin ϕ)5

(∗)

and by the binomial theorem we find     5 5 (cos ϕ + i sin ϕ)5 = cos5 ϕ (i sin ϕ)0 + cos4 ϕ (i sin ϕ)1 0 1       5 5 5 3 2 2 3 + cos ϕ (i sin ϕ) + cos ϕ (i sin ϕ) + cos ϕ (i sin ϕ)4 2 3 4   5 + (cos ϕ)0 (i sin ϕ)5 5 = cos5 ϕ − 10 cos3 ϕ sin2 ϕ + 10 cos ϕ sin4 ϕ

+ i(5 cos4 ϕ sin ϕ − 10 cos2 ϕ sin3 ϕ + sin5 ϕ).

With the help of (∗) we now find by comparing real and imaginary parts cos5 ϕ = cos5 ϕ − 10 cos3 ϕ sin2 ϕ + 10 cos ϕ sin4 ϕ

= cos5 ϕ − 10 cos3 ϕ(1 − cos2 ϕ) + 10 cos ϕ(1 − cos2 ϕ)2

= 16 cos5 ϕ − 20 cos3 ϕ + 5 cos ϕ. 1 8

b) We note that 8 cos4 (2ϕ)−3 = cos(8ϕ)+4 cos(4ϕ) is equivalent to cos4 (2ϕ) = cos(8ϕ) + 21 cos(2ϕ) + 83 . Now we make use of 

4 e2ϕi + e−2ϕi cos (2ϕ) = 2
1 8ϕi e + 4e4ϕi + 6 + 4e−4ϕi + e−8ϕi = 16    1 e8ϕi + e−8ϕi 1 e4ϕi + e−4ϕi 3 = + + 8 2 2 2 8 1 3 1 = cos 8ϕ + cos 4ϕ + . 8 2 8 √ √ π 4π c) Since 1 + 13 3i = √23 ei 6 and 1 − 31 3i = √23 ei 3 we find 4

1+ 1− 3.

√ !12

1 3 3i √ 1 3 3i

=

π √2 ei 6 3 4π √2 ei 3 3

!12

a) Let z1 , . . . , zn be the roots of p(z) =

 7π 12 = e−14πi = 1. = e−i 6 Pn

k=0

ak z k . It follows that

an (z − z1 )(z − z2 ) · . . . · (z − zn ) = 0 or


an z n − (z1 + · · · + zn )z n−1 + · · · + (−1)n z1 · · · zn = 0

664

SOLUTIONS TO PROBLEMS OF PART 7 Pn

implying that −an (z1 + · · · + zn ) = an−1 , i.e. Qn an (z1 · · · zn ) = a0 or k=1 zk = (−1)n aan0 .

k=1 zk

n = − an−1 an , and (−1)

b) The nth roots of unity are the roots of z n − 1. In this polynomial the coefficient an−1 is equal to 0, recall that we assume n ≥ 2. Hence by part a) the sum of all nth roots of unity for n ≥ 2 is equal to 0. 4.

a) The locations of the 8th roots of unity are shown in the figure below: Im z

z2 = b

z3 =

z4 = π

3π 4

b

z1 =

b

b

z5 =

π 2

b

b

0 5π 4

π 4

z0 = 1

b

b

b

z6 =

z7 =

Re z

7π 4

3π 2

2πk

b) The elements of E(m) are the numbers e m i , k = 0, . . . , m−1. Suppose that 2πk 2πk n < m divides m, i.e. m = pn. We write e m i = e np i , k = 0, . . . , pn − 1. Consider the quotient kp for k = 0, . . . , pn−1. For the values k = 0, p, 2p, . . . , p(n−1) < pn−1 it follows that 2πk 2πl e np i = e n i , l = 0, . . . , n − 1, 2πl

and the numbers e n i , l = 0, . . . , n − 1 are exactly the elements of E(n), hence E(n) ⊂ E(m). Since for every group E(j) the multiplication is multiplication in C it follows that E(n) is a subgroup of E(m). 5. Let z ∈ Ar,R and w ∈ S 1 . With Tw as in Example 15.5 we find Tw z = wz and |Tw z| = |z| implying that if |z| = r0 , r < r0 < R, i.e. z ∈ Ar,R , then |Tw z| = r0 (= |z|) and therefore Tw z ∈ Ar,R , or Tw (Ar,R ) ⊂ Ar,R . Since we can find for every ζ ∈ Ar,R a number z ∈ Ar,R such that Tw z = ζ, namely z = w−1 ζ, it even follows that Tw (Ar,R ) = Ar,R . 6. The subset E consists of all points in the plane with the property that their distance to the point (3, 0) and their distance to the point (−3, 0) always add up to a constant, namely 10. From this we deduce that E must be an ellipse with foci (3, 0) and (−3, 0). Replacing z by −z and z by z we find further that this ellipse must be symmetric in the real axis as well as in the imaginary axis, hence its midpoint must be (0, 0). For z ∈ R, z = x, we now find x = 5 or x = −5 depending whether

665

A COURSE IN ANALYSIS

we are on the right of (3, 0) or on the left of (−3, 0). For z = iy, y ∈ R we find 1 |iy − 3| + |iy + 3| = 2(y 2 + y) 2 = 10 or y = ±4. Thus we can also give the equation 2 2 of the ellipse with respect to the real coordinates x and y as x25 + y16 = 1. −z1 7. With c := |zz22 −z the equation according to (15.14) is given by Im (c(z − z1 )) = 0. 1| For z1 = −2 − i and z2 = 3 + 5i this yields z2 − z1 = 3 + 5i + 2 + i = 5 + 6i and √ . Therefore we find c = 5+6i 61

0 = Im



 1 5 − 6i √ (z + 2 + i) = √ Im ((5 − 6i)z + 16 − 7i) , 61 61

i.e. the equation for the line can be written as Im ((15 − 6i)z + 16 − 7i) = 0. If we write further z = x + iy we find 0 = Im ((15 − 6i)(x + iy) + 16 − 7i) = 15y − 6x − 7 and we can give the equation of the line in Cartesian coordinates as y = 25 x +

7 15 .

2 8. Orthogonality in the plane refers of course to orthogonality of two vectors inR . x 2 If z = x + iy and w = u + iv represent two vectors in R , z corresponds to y   u and w corresponds to , then orthogonality means xu + yv = 0, or Re z Re w + v   x2 − x1 , zj = xj + iyj , is of course Im z Im w = 0. The direction orthogonal to y2 − y1   −(y2 − y1 ) (or any non-zero real multiple of it). This vector corresponds to x2 − x1 i(z2 − z1 ), indeed

i(z2 − z1 ) = i((x2 − x1 ) + i(y2 − y1 )) = i(x2 − x1 ) − (y2 − y1 )

  −(y2 − y1 ) . Thus a complex parametric form so i(z2 − z1 ) is associated with x2 − x1 of the line orthogonal to z = z1 + t(z2 − z1 ) and passing through z1 is w = z1 + t(i(z2 − z1 )). This is however not the form (15.10) but may be rewritten as w = z1 + t(z1 + iz2 − iz1 − z1 ). If we now replace c by ic in (15.14)
we find a further 1 form of the equation for the orthogonal line as Im ic(z − z1 ) = 0, c = |zz22 −z −z1 | .

9. The line L we consider in parametric form as z(t) = z0 t, |z0 | = 1. In general for a point z ∈ C we find the position on the Riemann sphere by ξ=

z+z , 2z + 1

η=

i(z − z) , zz + 1

666

ζ=

zz − 1 zz + 1

SOLUTIONS TO PROBLEMS OF PART 7

which gives for z = z(t) = z0 t (z0 + z 0 )t 2 Re z0 t = 2 , t2 + 1 t +1 2 Im z0 t i(z 0 − z0 )t = 2 , η= t2 + 1 t +1 2 2 z0 z 0 t − 1 t −1 ζ= = 2 . t2 + 1 t +1 ξ=

Thus on S 2 ⊂ R3 we have with T denoting the inverse of S˜   2 Re z0 t 1  2 Im z0 t . T (z(t)) = 1 + t2 t2 − 1   − Im z0 Since with ~n :=  Re z0  we have 0 +   * −Im z0 2 Re z0 t 2 Im z0 t ,  Re z0  = 0 0 t2 − 1

it a plane orthogonal to ~n. Furthermore we have (z(0)) = T  follows  that T (L)lies in  0 Re z0 0  0 , T (z(1)) = Im z0  and limt→∞ T (z(t)) = limt→−∞ T (z(t)) = 0. Iden1 0 −1 tifying T (z(1)) with z0 we have determined  T (L) is  and it followsthat   theplane 0 Re z0 the circle on S 2 passing through z0 ∼ = Im z0  the north pole 0 and the 1 0   0 south pole  0 . −1 h i 10. Given R > 1 and choose N = N (R) = lnln|zR0 | + 1. For n > N we find ln R

|z0 |n = en ln |z0 | ≥ e ln |z0 |

ln |z0 |

=R

implying that limn→∞ zn = limn→∞ z0n = ∞. Now in the case of real numbers the situation is quite different. For x0 > 1 we again find convergence for (xn0 )n∈N , namely to +∞, i.e. limn→∞ xn0 = +∞. However for x0 < −1 we have limn→∞ xn0 = limn→∞ (−1)n |x0 |n does not exist. This looks like a contradiction. In both cases “∞” denotes something different. In the complex case “∞” is the point at infinity of the one-point compactification, see

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A COURSE IN ANALYSIS

Appendix I. However in the case of the real line we have with “+∞” and “−∞” due to the order structure two points at infinity and the extended real-line R is not a one-point compactification of R or a subset of the one-point compactification of C. Chapter 16 1.

a) Since |n4 + z 4| ≥ n4 − |z|4 ≥ n4 − 14 n4 = 34 n4 holds for n ≥ 4 and 1 ≤ |z| ≤ 2 we deduce that 4 1 ≤ 4 |n4 + z 4 | 3n

for n ≥ 4 and 1 ≤ |z| ≤ 2

and the Weierstrass M-test implies that the series |z| ≤ 2 uniformly and absolutely. b) With z = x + iy we find

P∞

1 n=1 n4 +z 4

converges in 1 ≤

 sin nz 1  i(x+iy)n e − e−i(x+iy)n = 3 3 n 2in eny e−ixn e−ny eixn − = 3 2in 2in3 implying

sin nz en|y| e−n|y| en|y| 1 n3 ≥ 2n3 − 2n3 ≥ 2n3 − 2n3 ,

P∞ eny but n=1 2n3 diverges for |y| 6= 0, note for this that for a > 0 we have that ena limn→∞ 2n 3 = ∞. However, for x ∈ R we find sin nx 1 n3 ≤ n3 P∞ and the Weierstrass test implies the uniform and absolute convergence of n=1 sinn3nx on R. Note that sin is bounded on R but not on C.  p 2. a) We set an := n (n + 2)(n + 3) − 1 > 0 and find n   X n k (n + 2)(n + 3) = (1 + an ) = a k n k=0   n 3 ≥1+ a 3 n n

which gives

3(n2 + 5n + 5) n2 + 5n + 5
, = n (n − 2)(n − 1)n 3 p i.e. limn→∞ an = 0 implying limn→∞ n (n + 2)(n + 3) = 1 and by the HadamardCauchy theorem it follows that the radius is 1. a3n ≤

668

SOLUTIONS TO PROBLEMS OF PART 7

b) We first estimate (−1)n−1 (n + 1) n+1 1 3n (n2 + 1) 2 ≤ 3n (n2 + 1) 12

and now study the series

P∞

k=0

s n

n+1

1

3n (n2 +1) 2

z n by looking at

1 1 n+1 1 = n 2 3 (n + 1) 2 3

As in part a) we see that limn→∞

lim

n→∞

q

n+1

n

s n

1

(n2 +1) 2

s

n+1

n

1

(n2 + 1) 2

.

= 1 which first yields

1 1 n+1 = 3n (n2 + 1) 12 3

and then by the Hadamard-Cauchy theorem that the radius of convergence is 3. 1

c) Following the hint we write n! = nn e−n rn with limn→∞ rn2 = 1. This implies 

Since limn→∞ rn2 is 4.


n1

(n!)2 (2n)!

 n1



 n1 n2n e−2n rn2 (2n)2n e−2n r2n   n1 1  rn2 1 rn2 n = = . 22n r2n 4 r2n =

1

= limn→∞ (r2n ) n = 1 we find that the radius of convergence 1

Now let us prove n! = nn e−n rn , limn→∞ rnn = 1, using Stirling’s formula, Theorem n I.31.7. First we note that by Problem 10 of Chapter I.15 we have limn→∞ (2π) 2 = 1 √ n and by Example I.15.12 we know that limn→∞ n = 1. Now using (I.39.29) we find ! n1 √ √ 1 (n!) n 2π n(n!) √ √ 1 = 2π n(nn e−n ) (nn e−n ) n which yields that limn→∞

(n!) 1

(nn e−n ) n

= 1, but

n! = nn e−n rn , 3. Since

rn =

n! . nn e−n

z k+1 Γ(αk + 1) = |z| (αk)! = |z| , Γ(αk + 2) zk (αk + 1)! αk + 1

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A COURSE IN ANALYSIS

the ratio test implies the uniform and absolute convergence in C for every α ∈ N while for α = 0 we have uniform and absolute convergence for |z| < 1. For α = 0 we find Γ(αk + 1) = Γ(1) = 1, i.e. E0 (z) =

∞ X

zk =

k=0

1 , 1−z

while for α = 2 we have Γ(2k + 1) = (2k)! and therefore E2 (w) =

∞ X wk . (2k)! k=0

With w = z 2 we obtain

∞
X z 2k E2 z 2 = = cosh z (2k)! k=0

and


k ∞ ∞ X
X −z 2 z 2k E2 −z 2 = = = cos z. (−1)k (2k)! (2k)! k=0

k=0

Also see the remark in the solution to Problem 4.

4. Again we use the ratio test and find first Γ(αk + β) 1 (αk + β − 1)! z k+1 ≤ |z| = |z| Γ(αk + α + β) k z (αk + α + β − 1)! αk + β

since α, β ∈ N. Now it follows the convergence of Eαβ (z) for all z ∈ C. If we choose α = β = 1 we have E1,1 (z) =

∞ X

k=0

∞

X zk zk = = ez . Γ(k + 1) k! k=0

For α = 1 and β = 2 we have for z 6= 0 E1,2 (z) =

∞ X

k=0 ∞ X

1 = z

k=0 z

=

∞

X zk zk = Γ(k + 2) (k + 1)! k=0

1 z k+1 = (k + 1)! z

e −1 . z

z

Since limz→0 e z−1 = 1 we have E1,2 (z) = it follows that for z 6= 0 ∞
X E2,2 z 2 = k=0

ez −1 z

∞ X zk k=0

k!

−1

!

for all z ∈ C. Finally for α = β = 2

∞ 1 X z 2k+1 z 2k = Γ(2k + 2) z (2k + 1)! k=0

1 = sinh(z), z

670

SOLUTIONS TO PROBLEMS OF PART 7

which extends continuously to C. Remark. Denoting natural numbers by α or β is uncommon in our treatise. In fact the definitions of Eα (z) and Eα,β (z) extends to a much larger range of α and β. The functions Eα (z) can be defined for all α > 0 and they are convergent in the plane C as are the functions Eα,β (z) for Re α > 0 and β ∈ C, however many books also prefer to restrict Eα,β (z) to the range α > 0 and β > 0. In order to prove the convergence we need the asymptotic expansion z b−a


1 Γ(z + a) (a − b)(a + b − 1) 1 a − b ∼1+ + 3(a + b − 1)2 − a + b − 1 2 + · · · , Γ(z + b) 2z 12 z 2

see [1], (6, 1.47) on page 257. Note that Mittag-Leffler functions play a crucial role in fractional calculus, see [31]. P zk 5. a) Since ez = ∞ k=0 k! we find z z = P∞ ez − 1 k=1

which implies

1=

∞ X

k=0

zk k!

zk (k + 1)!

= P∞

1

zk k=0 (k+1)!

!

∞ X Bl l=0

l!

z

l

!

,

.

Taking the Cauchy product of the two power series we get ! ∞ n X X Bk z n, 1= (n − k + 1)!k! n=0 k=0

which yields

N X

k=0

Bk = k!(N − k + 1)!

(

1, N = 0 , 0, N > 0

i.e. for N > 0, summing up to N − 1, we get 0=

N −1 X k=0

B0 B1 BN −1 Bk = + + ···+ k!(N − k)! 0!N ! 1!(N − 1)! (N − 1)!1!

and multiplying by N ! we get 0=

N −1 X k=0

    N −1   X N! N N N Bk = Bk = B0 + · · · + BN −1 . k k!(N − k)! 0 N −1 k=0

For B0 we find of course the value B0 = 1 and now it follows B1 B0 + =0 0!2! 1!1! B1 B2 B0 + + =0 0!3! 1!2! 2!1! B0 B1 B2 B3 + + + =0 0!4! 1!3! 2!2! 3!1!

or or or

671

B0 1 B1 = − =− ; 2 2  1 1 1 B2 = 2 − + = ; 6 4 6   1 1 1 B3 = 6 − + − = 0; 24 12 24

A COURSE IN ANALYSIS

and B0 B1 B2 B3 B4 + + + + =0 0!5! 1!4! 2!3! 3!2! 4!1!

  1 1 1 1 B4 = 24 − + − =− . 120 48 72 30

or

b) Since (∗) holds for all |z| < 2π it holds for all x ∈ R, |x| < 2π. We consider the odd part of the function g(x) = exx−1 + x2 , i.e.   x x −x x + − − ex − 1 2 ex − 1 2   −x −1 + e + ex − 1 =x+x = 0. 1 − ex + e−x − 1

2(g(x) − g(−x)) =

Thus g is an even function implying that for |x| < 2π ∞ X Bk k=0

k!

xk +

∞

x x X Bk k x =1− + x + 2 2 k! 2 k=2

∞ X Bk k =1+ x k! k=2

must be an even function which yields Bk = 0 for k = 2l + 1, l ∈ N.

6. In the solution to Problem 5 we have seen that for |x| < 2π, x ∈ R, we have ∞

ex

x X B2k 2k x + = x . −1 2 (2k)! k=0

From Theorem 22.12 we can conclude that this equality must hold also for all z ∈ C, |z| < 2π. Thus we will justify the use of this identity for |z| < 2π when proving Theorem 22.12. We look now at the function f (z) = and we find f (z) =

z 2



z +1 ez − 1

z z + ez − 1 2



With z = 2πiξ it follows that

z

=

z

z ez + 1 z e 2 + e− 2 = z . z z 2e −1 2 e 2 − e− 2

z cos z2 f (πξ) = f (2πiz) = − i = πξ cot(πξ) 2 sin z2 and replacing again ξ by z we arrive at (πz) cot(πz) =

∞ ∞ X X (2π)2k z 2k B2k (2πiz)k = . (−1)k (2k)! (2k)! k=0

k=0

672

SOLUTIONS TO PROBLEMS OF PART 7
7. The first part is easy. By the definition of ak , see (16.24) we have   a a(a − 1) · . . . · (a − k + 1) = k k! and in the case where a ∈ N0 it follows that the numerator becomes zero for k > α. The second part is the claim that for |x| < 1, x ∈ R, and a ∈ R the Taylor expansion ∞   X a n a (1 + x) = x n n=0 holds. If a ∈ N, this is just the binomial theorem a   X a n (1 = x)a = x n n=0 in light of the first conclusion, i.e. the fact that the k th derivative of x 7→ (1 + x)a , a ∈ R, we find




a n

= 0 for n > a if a ∈ N0 . For

f (k) (x) = a(a − 1) · . . . · (a − k + 1)(1 + x)a−k   a = k! (1 + x)a−k . k The Taylor coefficients are given by is indeed given by

f (n) (0) n!

(1 + x)a =

implying that the formal Taylor series

∞   X a

n=0

n

xn .

We determine the domain of convergence by applying the ratio test and find a

a n+1 a − n n+1
x n+1
= |x| = |x| qn := a n a n + 1 n x n

implying that lim qn = |x| < 1 for |x| < 1 which yields the convergence for |x| < 1. Now the equality Ba+b (x) = (1 + x)a+b = (1 + x)a (1 + x)b = Ba (x)Bb (x) which holds for |x| < 1 implies the final
Pclaim. ∞ Note that the absolute convergence of n=0 na xn , |x| < 1, x ∈ R, also implies the absolute convergence of the series for |z| < 1, z ∈ C. We will see later in Theorem 22.12, that the identity Ba+b (x) = Ba (x)Bb (x) for all x ∈ (−1, 1) will extend to the identity Ba+b (z) = Ba (z)Bb (z) for all z ∈ C, |z| < 1.

a 8. Since ak = a−(k−1) k−1 we find k       a−k a−1 a−1 (a − 1) − (k − 1) a − 1 = = k k k−1 k−1 k

673

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and therefore         a−1 a−1 a−1 a−k a−1 + + = k k−1 k−1 k k−1      a−k a−1 a a−1 = +1 = k k k−1 k−1 a(a − 1) · . . . · (a − 1 − (k − 1) + 1) = k(k − 1)!   a(a − 1) · . . . · ((a − k) + 1) a = = . k! k Furthermore, for |z| < 1 we have

 ∞  X a−1 k (1 + z)Ba−1 (z) = (1 + z) z k k=0     ∞  X a−1 k a − 1 k+1 = z + z k k k=0    ∞ ∞ X a−1 X a − 1 = zk + zk k k−1 k=0 k=1      ∞  a−1 0 X a−1 a−1 = z + + zk 0 k k−1 k=1   ∞   ∞   a−1 0 X a k X a k = z + z = z = Ba (z), 0 k k k=1

9.

k=0

where we used the result of the previous consideration and the fact that
a 0 = 1. a) By definition we have

(α)n = α(α + 1) · . . . · (α + n − 1) and the functional equation of the Γ-function yields Γ(α + n) = (α + n − 1)Γ(α + n − 1)

= (α + n − 1)(α + n − 2)Γ(α + n − 2) .. . = (α + n − 1)(α + n − 2) · . . . · (α + 1)αΓ(α),

i.e.

Γ(α + n) = (α)n . Γ(α)

674




a−1 0

=

SOLUTIONS TO PROBLEMS OF PART 7

This leads now to (α)n Γ(α + n) Γ(β) = (β)n Γ(α) Γ(β + n) Γ(β) Γ(α + n)Γ(β − α) · = Γ(n + β) Γ(α)Γ(β − α) B(α + n, β − α) . = B(α, β − α) b) Since (1)k = k! we have 2 F1 (1, 1; 1; z) =

=

∞ X (1)k (1)k

k=0 ∞ X

k=0

(1)k k!

zk ∞

(1)k k X k z = z . k! k=0

Moreover, since (2)k = 2 · 3 · . . . · (2 + k − 1) = (k + 1)! it follows that z 2 F1 (1, 1; 2; −z) = =

∞ X

k=0 ∞ X

(−1)k

(1)k (1)k k+1 z (2)k k!

(−1)k

X z k+1 k! (−1)k z k+1 = (k + 1)! k+1 k=0

k=0

=

∞ X

k=1

10.

∞

zk ˜ + z). (−1)k+1 = ln(1 k

a) Recalling the definition of 2 F1 (α, β; γ; z) we find 2 F1 (α, β; γ; z)

= =

∞ X (α)k (β)k

k=0 ∞ X k=0

(γ)k k!

zk

(β)k (αk ) k z = 2 F1 (β, α; γ; z). (γ)k k!

b) We have (γ − α − β) 2 F1 (α, β; γ, z) + α(1 − z)2 F1 (α + 1, β; γ; z) − (γ − β)2 F1 (α, β − 1; γ; z) =

∞ X

k=1

=

∞ X

(α)k (β)k (α + 1)k (β)k (α)k (β − 1)k +α − (γ − β) (γ)k k! (γ)k ! (γ)k k! ! (α + 1)k−1 (β)k−1 −α zk (γ)k−1 (k − 1)! (γ − α − β)

Ak z k ,

k=1

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A COURSE IN ANALYSIS

where for the coefficients Ak we find Ak =

11.

(α)k (β)k  (γ − α − β)(β + k − 1) + (α + k)(β + k − 1) − (γ − β)(β − 1) (γ)k k!  − (γ + k − 1)k = 0.

a) For k = 1, 2, 3 we have az + b azk + b − cz + d czk + d (ad − bc)(z − zk ) = (cz + d)(czk + d)

w − wk =

and we find therefore (ad − bc)(z − z1 ) (cz + d)(cz1 + d) (ad − bc)(z − z3 ) w − w3 = (cz + d)(cz3 + d) (ad − bc)(z2 − z1 ) w2 − w1 = (cz2 + d)(cz1 + d) (ad − bc)(z2 − z3 ) . w2 − w3 = (cz2 + d)(cz3 + d) w − w1 =

Since we must assume that ad − bc 6= 0 we get further (w − w1 )(w2 − w3 ) (z − z1 )(z2 − z3 ) = (w − w3 )(w2 − w1 ) (z − z3 )(z2 − z1 ) which yields w=

az + b cz + d

where a = (w2 − w3 )w1 (z2 − z1 ) + (w1 − w2 )(z3 − z2 ) b = (w2 − w3 )w1 (z2 − z1 ) + (w2 − w1 )z1 (z2 − z3 ) c = (w2 − w3 ) − (z2 − z3 ) d = (w3 − w2 )z3 (z2 − z1 ) + z1 (z2 − z3 ).

b) Using part a) we find w(z) =

i(1 + i)z + 1 + i (2 − 2i)z + i(1 + i)

676

SOLUTIONS TO PROBLEMS OF PART 7

a) With z = t ∈ R we find

12.

w(t) =

t2 − 1 2t t−i = 2 − i t+i t + 1 t2 + 1

and 

2 t2 − 1 4t2 + 2 2 t +1 (t + 1)2 4 2 t − 2t + 1 + 4t2 t4 + 2t2 + 1 = = (t2 + 1)2 (t2 + 1)2 = 1.

|w(t)|2 =

2

2 2 Thus w(R) ⊂ S 1 . Clearly, 1 ∈ / w(R) since otherwise 1 = tt2 −1 +1 or t + 1 = t − 1.
π π 1 Suppose that w ∈ S \{1}. We can write w as w = − cos
2ϕ + i sin 2ϕ, ϕ ∈ − 2 , 2 , and expressing cos 2ϕ and sin 2ϕ by tan ϕ, ϕ ∈ − π2 , π2 , we find

tan2 ϕ − 1 2 tan ϕ − i. 1 + tan2 ϕ 1 + tan2 ϕ
Thus the parametrization t = tan ϕ, ϕ ∈ − π2 , π2 , allows us to find for every 2 −1 1 w ∈ S 1 \{1} a t = tan ϕ ∈ R such that w = tt2 +1 − t22t +1 i, i.e. w(R) = S \{1}. w=

b) For z = x ∈ R we have w(x) = x−1 x+1 ∈ R\{1}. Indeed if for some x0 we x0 −1 have x0 +1 = 1, then this implies x0 − 1 = x0 + 1. Let y ∈ R\{1} and consider the   y+1 y+1 equation x−1 = y i.e. x+1 = y with solution x = 1−y , y 6= 1. It follows that w 1−y

w(R\{−1}) = R\{1}.

13. We have to solve the equation

aζ + b =ζ cζ + d

under the conditions ac − bd 6= 0 and c 6= 0. This leads to the quadratic equation cζ 2 + dζ − aζ − b = 0 or ζ2 +

b d−a ζ− =0 c c

which has two distinct solutions for (d − a)2 + 4bc 6= 0. These solutions are given by s r 2 d−a d−a d−a b (d − a)2 + 4bc ζ1,2 = ± ± . + = 2c 2c c 2c 4c2

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A COURSE IN ANALYSIS

Chapter 17 1. We can use the solution to Problem 2 of Chapter I.21. We use mathematical induction and the product rule, i.e. Leibniz’s rule. For m = 1 it follows that d (f · g)(z) = f 0 (z)g(z) + f (z)g 0 (z) dz     1 (1) 1 (0) = f (z)g (0) (z) + f (z)g (1) (z). 0 1 Now suppose that for some k ≥ 1

k   X dk k (k−l) (f · g)(z) = f (z)g (l) (z) dz k l l=0

holds and consider ! k   X k (k−l) (l) f (z)g (z) l l=0 k    X k d  (k−l) f (z)g (l) (z) = l dz l=0  k  X k  = f (k+1−l) (z)g (l) (z) + f (k−l) (z)g (l+1) (z) l l=0 k   X k (k+1−l) (k+1) =f (z)g(z) + f (z)g (l) (z) l l=1  k  X k + f (k−(l−1)) (z)g (l) (z) + f (z)g (k+1) (z) l−1 l=1 k+1 X k + 1 = f (k+1−l) (z)g (l) (z), l

dk+1 d (f · g)(z) = k+1 d dz

l=0

where we used the calculation of the solution to Problem 2 of Chapter I.21. 2. Suppose first that for f = u + iv we have Im f = v = c. The Cauchy-Riemann differential equations ux = vy and uy = −vx imply ux = 0 and uy = 0. Since G is connected it follows that u must be constant on G, hence f = u + iv must be constant on G. 1 Now assume that |f | = (u2 + v 2 ) 2 = c. If c = 0 it follows that u = v = 0. If c 6= 0 ux + vx 3

(u2 + v 2 ) 2

=0

and

uy + vy 3

(u2 + v 2 ) 2

= 0.

This implies that 0 = u2x + 2ux vx + vx2

678

(∗)

SOLUTIONS TO PROBLEMS OF PART 7

and 0 = u2y + 2uy vy + vy2 .

(∗∗)

Using the Cauchy-Riemann differential equations we find 2ux vx = −2uxuy and 2uy vy = 2ux uy . Thus adding (∗) and (∗∗) yields u2x + vx2 + u2y + vy2 = 0 which implies ux = uy = vx = vy = 0 and the result follows as before since G is connected. 3. Since g(z) = u(z) + iv(z) = f (|z|2 ) we find v(z) = 0 for all z ∈ C and by Problem 2 it follows that g is complex differentiable if and only if g, hence f , is constant. 4. Since passing from w to w is a reflection in the x-axis (or the real axis), G∗ is obtained by reflecting G in the x-axis, see the figure below

Im z

G w b

Re z

b

z=w ¯ G∗

G∗ = {z ∈ C|z = w, ¯ w ∈ G} Now we look at h(z) := f (z), z ∈ G∗ . For z0 ∈ G∗ and z ∈ G∗ there exist w0 , w ∈ G

679

A COURSE IN ANALYSIS

such that w = z and w0 = z 0 and we find f (z) − f (z 0 ) h(z) − h(z0 ) = z − z0 z − z0

f (w) − f (w0 ) w − w0 f (w) − f (w0 ) = w − w0   f (w) − f (w0 ) . = w − w0 =

Using the continuity of z 7→ z it follows for z → z0 that, h0 (z0 ) = h0 (w 0 ) = f 0 (w0 ),

in particular h is differentiable. 5. For z = x + iy we have z 2 = x2 − y 2 + 2ixy and z 3 = x3 − 3xy 2 + i(−y 3 + 3x2 y) which gives 2

ez + z 3 = ex

2

2

−y 2 i2xy

+ x3 − 3xy 2 + i(−y 3 + 3x2 y)

e

2

2

2

= ex −y cos 2xy + iex −y sin 2xy + x3 − 3xy 2 + i(−y 3 + 3x2 y)  2 2   2 2  = ex −y cos 2xy + x3 − 3xy 2 + i ex −y sin 2xy − y 3 + 3x2 y

=: u(x, y) + iv(x, y). Now we find ux (x, y) = 2xex

2

−y 2

uy (x, y) = −2yex vx (x, y) = 2xex

2

vy (x, y) = −2ye

2

−y 2

−y 2

cos 2xy − 2yex

2

x −y

−y 2

cos 2xy − 2xex

sin 2xy + 2yex 2

2

2

sin 2xy + 2xe

2

sin 2xy + 6x2 − 3y 2 ;

−y 2

−y

2

sin 2xy − 6xy;

cos 2xy + 6xy;

x2 −y 2

cos 2xy − 3y 2 + 6x2

implying ux = vy and uy = −vx .

6. We want to show for polar coordinates (r, ϕ) that 1 ∂v ∂u = ∂r r ∂ϕ

and

∂v ∂u =− ∂r ∂ϕ 1

where f (z) = u(z) + iv(z), z = x + iy. With x = r cos ϕ, y = r sin ϕ, r = (x2 + y 2 ) 2 and ϕ = arctan xy we find first ∂r x ∂r y = = cos ϕ, = = sin ϕ ∂x r ∂y r −y ∂ϕ x sin ϕ cos ϕ ∂ϕ = 2 , = 2 =− = ∂x x + y2 r ∂y x + y2 r

680

SOLUTIONS TO PROBLEMS OF PART 7

and now ∂u(r cos ϕ, r sin ϕ) ∂u = = ∂x ∂x ∂u sin ϕ ∂u = cos ϕ − ; ∂r r ∂ϕ ∂u(r cos ϕ, r sin ϕ) ∂u = = ∂y ∂y ∂u cos ϕ ∂u + ; = sin ϕ ∂r r ∂ϕ ∂v(r cos ϕ, r sin ϕ) ∂v = = ∂x ∂x ∂v sin ϕ ∂v = cos ϕ − ; ∂r r ∂ϕ ∂v(r cos ϕ, r sin ϕ) ∂v = = ∂y ∂y cos ϕ ∂v ∂v + . = sin ϕ ∂r r ∂ϕ

∂u ∂r ∂u ∂ϕ + ∂r ∂x ∂ϕ ∂x

∂u ∂r ∂u ∂ϕ + ∂r ∂y ∂ϕ ∂y

∂v ∂r ∂u ∂ϕ + ∂r ∂x ∂ϕ ∂x

∂v ∂r ∂v ∂ϕ + ∂r ∂y ∂ϕ ∂y

Using the Cauchy-Riemann differential equations we deduce that     1 ∂u ∂v ∂u 1 ∂v − + cos ϕ − sin ϕ = 0 ∂r r ∂ϕ ∂r r ∂ϕ and



∂u 1 ∂v − ∂r r ∂ϕ



sin ϕ +



∂v 1 ∂u + ∂r r ∂ϕ



cos ϕ = 0.

(∗)

(∗∗)

If we multiply (∗) with cos ϕ and (∗∗) with sin ϕ and then add the resulting equalities we find   ∂u 1 ∂v ∂u 1 ∂r 0= − − . (cos2 ϕ + sin2 ϕ) = ∂r r ∂ϕ ∂r r ∂ϕ

Multiplying (∗) with − sin ϕ and (∗∗) with cos ϕ and adding these new equalities we get  
∂v 1 ∂u ∂v 1 ∂u + + , sin2 ϕ + cos2 ϕ = 0= ∂r r ∂ϕ ∂r r ∂ϕ and the result follows.

7.

a) For y = Im z > 0 we have the estimate inz inx −ny e e e ≤ 1 n4 = n4 n4 P∞ inz implying that the series n=1 en4 converges absolutely and uniformly, hence pointwise, for Im z > 0. Consider now the series ∞ ∞ ∞ X X X in inz einx e−ny d einz = e = i . dz n4 n4 n3 n=1 n=1 n=1

681

A COURSE IN ANALYSIS

For Im z > 0 we have the estimate inx −ny e e ≤ 1 n3 n3

implying that the differentiated series converges absolutely and uniformly for Im z > 0. Hence we can justify the interchanging of summation and differentiation and it P einz follows that for Im z > 0 the function z 7→ ∞ n=1 n4 is holomorphic. P∞  2 n 2 b) For |z| > 2, i.e. |z| < 1 we have the convergent geometric series n=1 |z| and therefore we find ∞ X 1 2 2n =1− 2 = 2−z n |z| 1 − z n=1 which is holomorphic for z 6= 2, in particular for all z with |z| > 2.

z−1 8. We know that for z 6= −1 the rational function z 7→ w(z) = z+1 is holomorphic. z−1 −1 We want to determine w for which we solve the equation ζ = z+1 , which leads ζ+1 provided ζ 6= 1. Thus the image of C\{−1} under w is C\{1} and to z = ζ−1 ζ+1 is holomorphic and inverse to w. Thus on C\{1} the function w−1 (ζ) = ζ−1 w : C\{−1} → C\{1} is a biholomorphic mapping.

9. If f is biholomorphic then f is holomorphic, bijective and since f −1 is also holomorphic it is continuous. Furthermore we have z = f −1 (f (z)),

z ∈ G1

which implies by the chain rule that
0 1 = f −1 (f (z)) · f 0 (z).

Hence f 0 (z) 6= 0 for z ∈ G1 and with w = f (z) we have
0 f −1 (w) =

1 . f 0 (z)

Now let f be holomorphic, bijective and assume that f −1 is continuous as well as that f 0 (z) 6= 0 for all z ∈ G1 . We want to show that g := f −1 : G2 → G1 is holomorphic. With w := f (z), z ∈ G1 , and w0 = f (z0 ), z0 ∈ G1 , we find g(w) − g(w0 ) = w − w0

1 w−w0 g(w)−g(w0 )

=

1 f (z)−f (z0 ) z−z0

and since f 0 is continuous and f 0 (z0 ) 6= 0 we find a neighbourhood of z0 where f (z)−f (z0 ) 6= 0. Thus we may pass to the limit as z tends to z0 and we find z−z0 g 0 (w0 ) =
0 in particular f −1 (w0 ) exists.

682

1 , f 0 (z0 )

SOLUTIONS TO PROBLEMS OF PART 7

10. Clearly we have |z| = r2 and



|ζ − z|2 = (ζ − z)(ζ − z) = e−iϑ − re−iϕ eiϑ − reiϕ   = 1 + r2 − r e−i(ϑ−ϕ) − e−(ϕ−ϑ) = 1 + r2 − 2r cos(ϑ − ϕ)

implying that P (ζ, z) = We know that the function z 7→ its real part is harmonic. Since

1 1 − r2 . 2 2π 1 + r − 2r cos(ϑ − ϕ)

ζ+z ζ−z ,

ζ ∈ ∂B1 (0), z ∈ B1 (0), is holomorphic, hence

ζ +z (ζ + z)(ζ − z) ζζ + zζ − ζz − zz = = 2 ζ −z |ζ − z| |ζ − z|2 2 1−r 2r sin(ϑ − ϕ) = +i , 1 − r2 − 2r cos(ϑ − ϕ) 1 − r2 − 2r cos(ϑ − ϕ) where we used that zζ − ζz = (2r sin(ϑ − ϕ))i, it follows that P (ζ, (x, y)) as a function of (x, y) is a harmonic function. 11. We have to find

It follows that



∂2 ∂2 + ∂x2 ∂y 2



h(u(x, y), v(x, y)).

∂h ∂h ∂u ∂h ∂v = + , ∂x ∂u ∂x ∂v ∂x 2 2 ∂ 2 h ∂v ∂u ∂h ∂ 2 u ∂ h ∂u ∂u ∂ h + + = ∂x2 ∂u2 ∂x ∂x ∂v∂u ∂x ∂x ∂u ∂x2 2 2 ∂ h ∂v ∂u ∂h ∂ 2 v ∂ h ∂v ∂v + + + 2 ∂v ∂x ∂x ∂v∂u ∂x ∂x ∂v ∂x2 = huu ux ux + 2huv ux vx + huv vx vx + hu uxx + hv vxx , and analogously ∂h ∂u ∂h ∂v ∂h = + , ∂y ∂u ∂y ∂v ∂y ∂2h = huu uy uy + 2huv uy vy + hvv vy vy + hu uyy + hv vyy , ∂y 2 which yields ∂2h ∂2h + 2 = huu (u2x + u2y ) + 2huv (vx ux + vy uy ) + hvv (vx2 + vy2 ) ∂x2 ∂y + hu (uxx + uyy ) + hv (vxx + vyy ).

683

A COURSE IN ANALYSIS

Since f is holomorphic, u and v are harmonic, i.e. uxx + uyy = 0 and vxx + vyy = 0, and we find ∂2h ∂2h + 2 = huu (u2x + u2y ) + 2huv (vx ux + vy uy ) + hvv (vx2 + vy2 ). ∂x2 ∂y The Cauchy-Riemann differential equations give ux = vy and uy = −vx and therefore we have vx ux + vy uy = −ux uy + ux uy = 0, and it follows that ∂2h ∂2h + 2 = huu (u2x + u2y ) + hvv (vx2 + vy2 ). ∂x2 ∂y However, again by the Cauchy-Riemann differential equations, we note that u2x + u2y = vx2 + vy2 , i.e. 

∂2 ∂2 + ∂x2 ∂y 2



h (u(xy), v(x, y)) = (huu + hvv ) u2x + u2y

and since by assumption h is harmonic we deduce that   2 ∂2 ∂ + 2 h (u(x, y), v(x, y)) = 0. ∂x2 ∂y




This result is interesting: we can obtain harmonic functions in G1 once we know harmonic functions in G2 and a holomorphic mapping from G1 onto G2 . In the case where f is biholomorphic, hence u2x +u2y 6= 0, we obtain a one-to-one correspondence of harmonic functions on G1 and harmonic functions on G2 . Chapter 18 1. For z = reiϕ , −π < ϕ < π, we have z i = ei log z = ei(ln r+iϕ) = e−ϕ (cos ln r + i sin ln r). √ √ π Since 21 2e2π + 21 2e2π i = e2π ei 4 it follows with r = e2π and ϕ =  2. With z1 = e

2πi 3

1 √ 2π 1 √ 2π 2e + 2e i 2 2 and z2 = e

3πi 4

i

π

π

= e− 4 (cos 2π + i sin 2π) = e− 4 .

we find

log z1 =

2πi 3

π 4

and

leading to log z1 + log z2 =

684

log z2 = 17πi . 12

3πi 4

that

SOLUTIONS TO PROBLEMS OF PART 7

On the other hand we have z1 · z2 = e

2πi 3

e

3πi 4

= e−

and log(z1 · z2 ) = −

7πi 12

7πi . 12

Note however that 17πi = log z1 + log z2 . 12

log(z1 · z2 ) + 2πi =

3. From Problem 7 in Chapter 16 we know that for x ∈ (−1, 1) ∞   X β k x = (1 + x)β k k=0

holds and since for x ∈ (−1, 1) the number 1 + x belongs to C\(−∞, 0] we find ∞   X β k x = (1 + x)β = eβ log(1+x) = es log(1+x) eit log(1+x) k k=0

= (1 + x)s (cos(t ln(1 + x)) + i sin(t ln(1 + x))) .
P β k |z| < 1 and we know By Proposition 16.18 the series ∞ k=0 k z converges for all
P∞ from Problem 7 of Chapter 16 that for x ∈ (−1, 1) we have k=0 βk xk = (1 + x)β . The uniqueness theorem for holomorphic functions which we will prove in Chapter 22, Theorem 22.16, now implies ∞   X β k z = (1 + z)β for z ∈ B1 (0) and β ∈ C. k k=0



k k−1 z ˜ a = b+ P∞ (−1) converges in B|a| (a) 4. We know that La,b (z) := b+ L k=1 z k a −1 and is continuous, in fact holomorphic. We want to use Theorem 18.4 in order to show that La,b is a branch of the logarithmic function. For this we note that ! ∞ k  X d ˜ z (−1)k−1  z d 0 −1 −1 ln = b+ La,b (z) = dz k a dz a k=1

1 1 1 = · z = . a a z

In addition it follows that eLa,b (a) = eb = a since by assumption b is a logarithm of a and La,b (a) = b +

∞ X

(−1)k

k=1

685

a

a

−1

k

= b.

A COURSE IN ANALYSIS

5. The inverse of w(z) =

1+z 1−z

is ζ(w) =

w−1 w+1 .

For |z| < 1 we find with z = x + iy

1+z (1 + z)(1 − z) 1 − (x2 + y 2 ) 2y = = +i 1−z (1 − x)2 + y 2 (1 − x2 ) + y 2 (1 − x)2 + y 2   implying Re 1+z 1−z > 0 for |z| < 1. On the other hand for w = u+iv the condition Re w = u > 0 implies |w−1| < |w+1| w−1 or w+1 < 1. Thus we have proved that the bijective mapping w : C\{1} → C\{−1} maps B1 (0) into {ζ ∈ C | Re ζ > 0} and its inverse maps {ζ ∈ C | Re ζ > 0} into B1 (0), i.e. w(B1 (0)) = {ζ ∈ C | Re ζ > 0}. Chapter 19 1.

a) Since in a discrete topological space every set is open and closed, a discrete topological space is connected if and only if its underlying set X has one element. b) Such a mapping cannot exist. A continuous mapping must map the connected space X onto a connected subset of Y . Since by assumption f is surjective, i.e. f (X) = Y , and Y is not connected such f cannot be continuous.

2. Take (a1 , b1 ), (a2 , b2 ) ∈ A × B. The set {a1 } × B is homeomorphic to B and the set A × {b2 } is homeomorphic to A, hence A × {b2 } and {a1 } × B are connected in Rn+m . Furthermore we have {a1 } × B ∩ A × {b2 } = {a1 , b2 }, implying that {a1 } × B ∪ A × {b2 } is connected in Rn+m and therefore (a1 , b1 ) and (a2 , b2 ) belong to the same connectivity component of A × B. However (a1 , b1 ) and (a2 , b2 ) are arbitrary points, hence A × B is connected. 3. We decompose tr(γ) into four curves γ1 , γ2 , γ3 , γ4 such that   • γ1 : 0, 41 → C, tr(γ
1 ) is the half circle with centre 2 + i and radius 1 and γ1 (0) = 2 and γ1 14 = 2 + 2i;

  • γ2 :
14 , 12 → C, tr(γ2 ) is
the line segment connecting 2 + 2i and 1 + 2i, i.e. γ2 41 = 2 + 2i and γ2 12 = 1 + 2i;

  • γ3 :
12 , 34 → C, tr(γ3 )
is the line segment connecting 1 + 2i and 0, i.e. γ3 21 = 1 + 2i and γ3 34 = 0;   • γ4 : 34 , 1 → C, tr(γ4 ) is the line segment connecting 0 with 2, i.e. γ4 and γ4 (1) = 2.

686

3 4




=0

SOLUTIONS TO PROBLEMS OF PART 7

The corresponding curves are γ1 (t) = (2 + i) − ie4πit , γ2 (t) = −4t + 3 + 2i, γ3 (t) = (−4 − 8i)t + 3 + 6i, γ4 (t) = 8t − 6i,

  1 ; t ∈ 0, 4   1 1 t∈ , ; 4 2   1 3 t∈ , ; 2 4   3 ,1 . t∈ 4

4. We identify R2 with C and we introduce polar coordinates (r, ϕ). Every z ∈ C\{0} ∼ = R2 \{0} corresponds to z = reiϕ , r > 0, ϕ ∈ [0, 2π). For ρ > 0 the mapping Rρ : C\{0} → C\{0}, z 7→ Rρ (z) = ρreiϕ , z = reiϕ , is bijective, continuous with continuous inverse R ρ1 . Finally, f = R 12 maps ∂B2 (0) onto ∂B1 (0) since

for z = 2eiϕ we have R 21 2eiϕ = eiϕ and conversely 2eiϕ = R2 eiϕ .

5. Let C ⊂ X be an arcwise connected set and f : X → Y a continuous mapping. For y1 , y2 ∈ f (C) we choose x1 , x2 ∈ C such that f (x1 ) = y1 and f (x2 ) = y2 . Now let γ : [0, 1] → C be a continuous curve connecting x1 and x2 , i.e. γ(0) = x1 and γ1 = x2 . We consider the arc f ◦ γ : [0, 1] → f (C) which is continuous and f ◦ γ(0) = f (x1 ) = y1 as well as (f ◦ γ)(1) = f (x2 ) = y2 . Hence f (C) is arcwise connected. 6. Let γj : [aj , bj ] → X, j = 1, 2, 3, with γ(b1 ) = γ2 (a2 ) and γ2 (b2 ) = γ3 (a3 ). It follows that γ1 ⊕ γ2 : [a1 , b1 + b2 − a2 ] → X, (γ1 ⊕ γ2 )(t) =

(

γ1 (t), γ2 (t + a2 − b1 ),

t ∈ [a1 , b1 ] t ∈ [b1 , b1 + b2 − a2 ]

and (γ1 ⊕ γ2 ) ⊕ γ3 : [a1 , b1 + b2 − b3 − a2 − a3 ] → X where (

(γ1 ⊕ γ2 )(t), t ∈ [a1 , b1 + b2 − a2 ] γ3 (t + a3 + a2 − b1 − b2 ), t ∈ [b1 + b2 − a2 , b1 + b2 + b3 − a2 − a3 ]   t ∈ [a1 , b1 ] γ1 (t), = γ2 (t + a2 − b1 ), t ∈ [b1 , b1 + b2 − a2 ]   γ3 (t + a2 + a3 − b1 − b2 ), t ∈ [b1 + b2 − a2 , b1 + b2 + b3 − a2 − a3 ].

((γ1 ⊕ γ2 ) ⊕ γ3 ) =

On the other hand we have γ2 ⊕ γ3 : [a2 , b2 + b3 − a3 ] → X with (γ2 ⊕ γ3 )(t) =

(

γ2 (t), γ3 (t + a3 − b2 ),

687

t ∈ [a2 , b2 ] t ∈ [b2 , b2 + b3 − a3 ]

A COURSE IN ANALYSIS

and γ1 ⊕ (γ2 ⊕ γ3 ) : [a1 , b1 + b2 − a2 − a3 ] → X where ( γ1 (t), t ∈ [a1 , b1 ] (γ1 ⊕ (γ2 ⊕ γ3 ))(t) = (γ2 ⊕ γ3 )(t + a2 − b1 ), t ∈ [b1 , b1 + b2 + b3 − a2 − a3 ]   t ∈ [a1 , b1 ] γ1 (t), = γ2 (t + a2 − b1 ), t ∈ [b1 , b1 + b2 − a2 ]   γ3 (t + a2 + a3 − b1 − b2 ), t ∈ [b1 + b2 − a2 , b1 + b2 + b3 − a3 ],

i.e. we indeed have (γ1 ⊕ γ2 ) ⊕ γ3 = γ1 ⊕ (γ2 ⊕ γ3 ).

7. Let γ1 ∼ γ˜1 and γ2 ∼ γ˜2 , we have to prove that γ1 ⊕ γ2 ∼ γ˜1 ⊕ γ˜2 . With ϕ1 : [˜ a1 , ˜b1 ] → [a1 , b1 ] and ϕ2 : [˜ a2 , ˜b2 ] → [a2 , b2 ] we have γ˜1 = γ1 ◦ ϕ1 and γ˜2 = γ2 ◦ ϕ2 . Furthermore it follows that γ1 ⊕ γ2 : [a1 , b1 + b2 − a2 ] → X where ( γ1 (t), t ∈ [a1 , b1 ] (γ1 ⊕ γ2 )(t) = γ2 (t + a2 − b1 ), t ∈ [b1 , b1 + b2 − a2 ], and γ˜1 ⊕ γ˜2 : [˜ a1 , ˜b1 + ˜b2 − a ˜2 ] → X where ( γ˜1 (t), (˜ γ1 ⊕ γ˜2 )(t) = γ˜2 (t + a ˜2 − ˜b1 ),

t ∈ [˜ a1 , ˜b1 ] ˜ t ∈ [b1 , ˜b1 + ˜b2 − a ˜2 ].

˜2 ] → [a1 , b1 + b2 − a2 ] defined by With ϕ : [˜ a1 , ˜b1 + ˜b2 − a ( ϕ˜1 (˜ r ), r˜ ∈ [˜ a1 , ˜b1 ] ϕ(˜ r) = ˜ ˜ ϕ(˜ ˜ r+a ˜2 − b1 ), r˜ ∈ [b1 , ˜b1 + ˜b2 − a ˜2 ] it follows that γ^ ˜1 ⊕ γ˜2 , and therefore 1 ⊕ γ2 = (γ1 ⊕ γ2 ) ◦ ϕ = (γ1 ◦ ϕ) ⊕ (γ2 ◦ ϕ) = γ the definition of [γ1 ] ⊕ [γ2 ] is independent of the parametrization. With the help of Problem 6 we now find ([γ1 ] + [γ2 ]) ⊕ [γ3 ] = [γ1 ⊕ γ2 ] ⊕ [γ3 ] = [γ1 ⊕ γ2 ⊕ γ3 ]

= [γ1 ] ⊕ [γ2 ⊕ γ3 ] = [γ1 ] ⊕ ([γ2 ] ⊕ [γ3 ]) .

8. With f = g = idX it follows that every topological space X is homotopically equivalent to itself. If X and Y are two topological spaces and X is homotopically equivalent to Y then there exists mappings f : X → Y and g : Y → X such that g◦f is homotopic to the identity idX on X and f ◦ g is homotopic to the identity idY on Y . This implies that Y is homotopically equivalent to X just by switching the roles of X and Y as well as f and g. Now let X be homotopically equivalent to Y and Y homotopically equivalent to Z. Thus there exist continuous mappings fX : X → Y and gY : Y → X such that gY ◦ fX is homotopic to idX and fX ◦ gY is homotopic to idY . Further there exists continuous mappings fY : Y → Z and gZ : Z → Y such that gZ ◦ fY is homotopic to idY and fY ◦ gZ is homotopic to idZ . We consider the mappings fY ◦ fX : X → Z and gY ◦ gZ : Z → X which are continuous and from Theorem 19.19 we deduce that (gY ◦ gZ ) ◦ (fY ◦ fX ) = gY ◦ (gZ ◦ fY ) ◦ fX is homotopic to idZ . Hence X is homotopically equivalent to Z.

688

SOLUTIONS TO PROBLEMS OF PART 7

9. Let x0 ∈ A 21 ,2 (0), |x0 | = 1. The path γ : [0, 1] → A 12 ,2 (0) γ(t) = e2πit , has its trace in A 12 ,2 (0), 1 ∈ tr(γ), but γ is not homotopic to the constant path γ1 : [0, 1] → A 21 ,2 (0), γ1 (t) = 1 for all t ∈ [0, 1]. Chapter 20 1. For n 6= −1 we find Z

z n dz =

Z

2πk

eint ieit dt = i

=

2πk

ei(n+1)t dt

0

0

γ

Z

 1  2πk(n+1) e − 1 = 0. n+1

However for n = −1 we have Z Z z −1 dz = γ

2πk

0

1 it ie dt = i eit

Z

2πk

1 dt = 2πki.

0

2. Recall that γ : [0, π] → C, γ(t) = ei(π−1) , and σ : [−1, 1] → C, σ(t) = t. Note that γ(0) = σ(−1) = −1 and γ(π) = γ(1) = 1. Further we have γ(t) ˙ = −iei(π−t) and it it σ(t) ˙ = 1. On γ we find Re z = Re e = cos t and Im z = Im e = sin t whereas on σ we have Re z = t and Im z = 0. This yields Z Z π   (cos t) −iei(π−t) dt Re z dz = 0

γ

= −i Z =i

π =i 2 and

Z

Z π cos t sin(π − t) dt cos t cos(π − t) dt + 0 0 Z π π cos t sin t dt cos2 t dt −

Z

0

π

0

Re z dz =

σ

Moreover we have Z

γ

Im z dz =

Z

π

0

= −i =− and

Z

1

t dt = −1

t2 1 = 0. 2 −1

  sin t −iei(π−t) dt

Z

π 2

Z

0

π

sin t cos(π − t) dt −

Im z dz =

σ

Z

σ

689

0 dz = 0.

Z

0

π

sin2 t dt

A COURSE IN ANALYSIS

3. The parametrization is given by γ : [0, 1] → C with     t ∈ 0, 31 12t − 2, 1 2 γ(t) = −(6 − 6i)t + 4 − 2i, t ∈ 3 , 3     −(6 + 6i)t + 4 + 6i, t ∈ 23 , 1 with corresponding derivative

    t ∈ 0, 13 12,   γ(t) ˙ = −(6 − 6i), t ∈ 13 , 32     −(6 + 6i), t ∈ 23 , 1 . Now we find with γ1 = γ 1 , γ2 = γ 1 2 and γ3 = γ 2 that [3,3] [ 3 ,1] [0, 3 ] Z Z Z Z z dz z dz + z dz + z dz = γ

γ3

γ2

γ1

and further Z Z

z dz =

Z

1 3

0

γ1

(12t − 2) · 12 dt

13 t2 = 144 − 24t = 0 2 0 Z 32 (−(6 − 6i)t + (4 − 2i)) (−(6 − 6i)) dt z dz = 1 3

γ2

= (−6 + 6i)2 = −4

32 t2 + (4 − 2i)(−6 + 6i)t 1 2 3

and Z

z dz =

γ3

Z

1 2 3

((−6 − 6i)t + 4 + 6i) (−6 − 6i) dt

= (−6 − 6i)2 = 4. Thus we find indeed Z Z z dz = γ

4.

z dz + γ1

Z

1 t2 − (4 + 6i)(6 + 6i)t 2 2 3

z dz +

γ2

Z

γ3

z dz = 0 − 4 + 4 = 0.

a) Since for every simply closed, piecewise continuously differentiable curve γ R we have γ 1 dz = 0 such a result is not possible.

690

SOLUTIONS TO PROBLEMS OF PART 7

b) From estimate (20.18) we deduce that Z 1 lγ

γk

k

f (z) dz ≤ kf k∞,tr γk ≤ kf k∞,K .

  R Hence the sequence lγ1 γk f (z) dz is bounded and therefore its lim sup must k k∈N be finite.

  4 cos t + 5. With z = x + iy we find when identifying C with R that γ(t) = 4 sin t     cos t 5 cos t = and for every t ∈ [0, 2π] we find − sin t 3 sin t 2

 x 2 5

+

 y 2 3

= cos2 t + sin2 t = 1.

Hence tr(γ) is the ellipse with centre 0 and principal axes being the real- and imaginary axes, respectively, passing through (5, 0), (−5, 0), (0, 3) and (0, −3). Now we find with γ(t) ˙ = 4ieit − ie−it that Z

z 3 dz = γ

=

Z

Z

2π

4eit + e−it

0 2π 0

= 0,


3


4ieit − ie−it dt


256ie4it + 128ie2it − 8ie2it − ie−4it dt

where we used that for n ∈ N we have

R 2π 0

enit dt = 0.

P k 6. Let f (z) = ∞ of f in BR (z0 ) k=0 ak (z − z0 ) be the power series representation P∞ and let 0 < r < R. Since for r < R it follows that k=0 |ak |rk converges, we P∞ |ak | k P∞ |ak | k+1 deduce that k=0 k+1 r and hence k=0 k+1 r converges. This implies that P ak k+1 converges for z ∈ Br (z0 ), 0 < r < R, the power series F (z) := ∞ k=0 k+1 (z − z0 ) absolute and uniformly and therefore we find that d F (z) = dz 0

∞ X ak (z − z0 )k+1 k+1

k=0

!

=

∞ X

k=0

ak (z − z0 )k = f (z),

i.e. F is primitive of f in every disc Br (z0 ), 0 < r < R. 7.

a) We use the standard parametrization γ : [0, 1] → C, γ(t) = (2+4πi)t−1−2πi

691

A COURSE IN ANALYSIS

with γ(t) ˙ = 2 + 4πi. Now we find Z

e2z dz =

[−1−2πi,1+2πi]

Z

1

e2(2+4πi)t−2−4πi (2 + 4πi) dt

0

Z 1 = (2 + 4πi)e−2 e(4+8πi)t dt 0   2 + 4πi −2 (4+8πi)t 1 e e = 4 + 8πi 0  
1 1 1 4+8πi 2 = 2 e −1 = e − 2 . 2e 2 e

  → C, γ(t) = 2eit that γ(t) ˙ = b) It follows with the parametrization γ : 0, 3π 2 it 2ie and Z Z 3π 2 sinh |z| dz = sinh 2eit 2ieit dt γ

0

Z

3π 2

eit dt = 2i sinh 2 0  3π  = 2 sinh 2 e 2 i − 1 = 2 sinh 2(−i − 1)

= −(2 + 2i) sinh 2. c) Once we know that z 7→ p(z)ez admits aR primitive in a neighbourhood of the closure of the interior of γ, it follows that γ p(z)ez dz = 0. We prove that z → p(z)ez has a convergent power series representation in C which in turn implies PN that the primitive exists, see Problem 6. Now let p(z) = k=0 ak z k and consider the Taylor expansion of the expontential function with centre 0, i.e. ez =

∞ X zl l=0

We find that p(z)ez =

N X

ak

k=0

l!

.

∞ X z k+l l=0

l!

P∞ and the result follows if we can prove that for every k ∈ N0 the series l=0 converges in C. Since m−1 ∞ ∞ z X X (m − k)! |z| zm z k+l = and · = , l! (m − k)! (m + 1 − k)! |z m | m+1−k l=0

z k+l l!

m=k

we deduce from the ratio test that the series

692

P∞

l=0

z k+l l!

converges for all z ∈ C.

SOLUTIONS TO PROBLEMS OF PART 7

8.

a) From Example 20.21 we know that z 7→ |z| has no primitive and Problem 2 yields that neither z 7→ Re z nor z 7→ Im z has a primitive. b) With γ : [0, π] → C, γ(t) = ei(π−t) we find Z π Z f (|γ(t)|)γ(t) ˙ dt f (|z|) dz = 0 γ Z π γ(t) ˙ dt = f (1) 0

= f (1) (γ(π) − γ(0)) = 2f (1).

On the other hand, with σ : [−1, 1] → C, σ(t) = t, we have Z Z 0 Z Z 1 f (−t) dt + f (|t|) dt = f (|z|) dz = −1

−1

σ

1

f (t) dt.

0

Now for f (r) = rk the first calculation yields Z |z|k dz = 2. γ

For k odd we find f (−t) = −f (t) and Z Z Z 0 f (t) dt + f (|z|) dz = − −1

σ

whereas for k even it follows Z Z f (|z|) dz = 2 In all cases we find has no primitive.

γ

f (t) dt = 2

f (|z|) dz 6=

R

σ

f (t) dt = 0,

0

Z

0

0

σ

R

1

1

1

tk dt =

2 . k

f (|z|) dz and consequently z 7→ f (|z|) = |z|k

Chapter 21 d 1. Since f · g is a primitive of dz (f · g) we find Z d (f (z)g(z)) dz = f (γ(b))g(γ(b)) − f (γ(a))g(γ(a)) dz γ

and on the other hand we have Z Z Z d f (z)g 0 (z) dz + f 0 (z)g(z) dz (f (z)g(z)) dz = γ dz γ γ implying the result. Since for a simply closed curve we have in addition γ(b) = γ(a) the second statement follows too. We have seen, compare with Theorem 21.12,Rthat f g 0 and f 0 g are complex differenR 0 tiable too, hence we always have γ f g dz = γ gf 0 dz = 0 and the second statement does not contain much useful new information.

693

A COURSE IN ANALYSIS

2.

a) The curve γ is a simply closed differentiable curve. In fact it is the ellipse with centre 0 ∈ C, principal axes in the coordinate axes, i.e. the real and the imaginary axes passing through the points ±(a + b) and ±(a − b)i, as we see from the representation γ(t) = (a + b) cos t + (a − b)i sin t = x + iy and y2 x2 + = 1. (a + b)2 (a − b)2

R Since every polynomial is an entire function it follows that γ p(z) dz = 0. b) For t ∈ [0, π] the trace of γ is just the graph of sin [0,2π] while for t ∈ [π, 2π] we deal with the line segment connecting π with 0 considered as points on the real axis. Im z

γ(t) = t + i sin t

1

b

π

0

Re z

Since γ is a simply closed piecewise continuously differentiable curve we find, again by using the fact that polynomials as well as sin and cos are entire functions, that Z Q(cos z, sin z) dz = 0. γ

This implies in particular Z

2π

π

Z 2π Q(cos t, − sin t) dt Q(cos(2π − t), sin(2π − t)) dt = π Z π Q(cos(t + i sin t), sin(t + i sin t))(1 + i cos t) dt. =− 0

R R c) Since |ζ|=ρ (g(ζ) + h(ζ)) dζ = |ζ|=ρ g(ζ) dζ due to the holomorphy of h the estimate follows immediately from Proposition 20.13.B. 3.

a) Here is a sketch of the geometric situation

694

SOLUTIONS TO PROBLEMS OF PART 7

γ(a) = γ(c) = γ(b)

γ|[a,c]

a

γ|[c,b]

c

b

From our assumptions it follows that Z Z Z h(z) dz + h(z) dt =

h(z) dz.

γ2

γ1

γ

But each integral on the right hand side must be zero since it is an integral of a holomorphic function over a simply closed piecewise continuously differentiable curve. b) First we look at Γ:

b

−1

η : [0, 4π] → C,

b

0

1

η1 (t) := η [0,2π] (t) = eit

3

2

and

η2 (t) := η [2π,4π] (t) = 2 − eit .

It follows that η(0) = 1 = η(2π) = η(4π) and further we find that z 7→ holomorphic in B 54 (0). Thus it follows that 1 2πi

Z

η

1 z−

3 2

Z 1 1 2πi η1 z − Z 1 1 = 2πi η2 z −

dz =

695

3 2 3 2

dz + dz.

1 2πi

Z

η2

1 z−

3 2

dz

1 z− 23

is

A COURSE IN ANALYSIS

We cannot apply directly Theorem 21.8 since η2 is not given with the parametrization required. For this we have to change the orientation of the parametrization of η2 and hence we arrive at Z 1 1 dz = −1. 2πi η z − 32 4. We note that ζ2 − which yields ζ−

π 4

 π2 π 5π ζ− = ζ− (ζ − π) 4 4 4




sin ζ + (ζ − π) cos ζ sin ζ cos ζ = + π ζ − π ζ − π4 ζ 2 − 5π − 4 4


/ B1 (4) as well as π ∈ / B 52 − 21 but and next we note that π ∈ B1 (4), π4 ∈
B 52 − 21 . By the Cauchy integral formula we find therefore 1 2πi

Z

|z−4|=1

ζ−

π 4




sin ζ + (ζ − π) cos ζ

ζ2

−

5π 4 ζ

− 

π2 4

π 4

∈

dζ

 Z 1 sin ζ cos ζ dζ = + 2πi |z−4|=1 ζ − π ζ − π4 Z 1 sin ζ = dζ = sin 0 = 0, 2πi |z−4|=1 ζ − π

cos ζ where we have used that in B 54 (4) the function ζ 7→ ζ− π is holomorphic and for the 4 remaining integral we applied the Cauchy integral formula. By a similar argument we find
Z Z ζ − π4 sin ζ + (ζ − π) cos ζ 1 cos ζ 1 dζ = π dζ π2 5 ζ − 1 2πi |z+ 21 |= 25 2πi ζ 2 − 5π ζ − = z+ | 2| 2 4 4 4 π 1√ = cos = 2. 4 2

5. Note that for all k ∈ N we have 0 < ln(k 2 + k) < k 2 . Since z 7→ e−z is an entire function we find Z 2 1 1 e−ζ 1 dζ = e− ln(k +k) = ln(k2 +k) = . 2πi |ζ|=k2 ζ − ln(k 2 + k) k(k + 1) e Using the result of Example I.16.3 we now derive Z ∞ ∞ X X e−ζ 1 1 dζ = = 1. 2πi |ζ|=k2 ζ − ln(k 2 + k) k(k + 1) k=1

k=1

696

SOLUTIONS TO PROBLEMS OF PART 7

6. Since f , in fact f k , satisfies the assumption of Theorem 21.8 we have for k ∈ N and z ∈ Br (z0 ) Z f k (ζ) 1 dζ f k (z) = 2πi |ζ−z0 |=r ζ − z implying by (20.18) k

k

|f (z)| = |f (z)| ≤ ≤ or |f (z)| ≤



1

r · − z

∞,∂Br (z0 )



1

r · − z

∞,∂Br (z0 )



1

r

· − z

∞,∂Br (z0 )

! k1

! !

k

f ∞,∂Br (z0 ) kf k∞,∂Br (z0 )

kf k∞,∂Br (z0 ) .

The hint now implies the estimate |f (z)| ≤ kf k∞,∂Br (z0 ) for all z ∈ Br (z0 ). This nice argument is due to E. Landau, we refer to [67]. 7.

a) Since we have for f the mean-value equality f (z0 ) =

Z

1 2π

2π

f (z0 + reit ) dt

0

we may pass to the real and the imaginary part. Note that u = Re f and v = Im f are harmonic functions and we will see in Part 10 that harmonic functions are indeed characterised by the mean-value property. b) For every k ∈ N and every ball Br (z0 ) ⊂ Br (z0 ) ⊂ G we have fk (z0 ) =

1 2π

Z

2π

0


fk z0 + reit dt.

(∗)

Since we have uniform convergence to f on all compact subsets of G we may pass in (∗) to the limit and obtain f (z0 ) =

1 2π

Z

8. Given (21.19), i.e. f (z) =

1 2πi

2π

0

Z


f z0 + reit dt.

|ζ−z0 |=r

f (ζ) dζ, ζ −z

we have a starting point for our induction. Now suppose that for all z ∈ Br (z0 ) Z f (ζ) n! dζ. f (n) (z) = 2πi |ζ−z0 |=r (ζ − z)n+1

697

A COURSE IN ANALYSIS

We are allowed to differentiate under the integral sign and therefore we find Z f (ζ) n! d dζ 2πi dz |ζ−z0 |=r (ζ − z)n+1 Z f (ζ) d n! dζ = 2πi |ζ−z0 |=r dz (ζ − z)n+1 Z n! n+1 = f (ζ) dζ 2πi |ζ−z0 |=r (ζ − z)n+1 Z (n + 1)! f (ζ) = dζ. n+1 2πi |ζ−z0 |=r (ζ − z)

f (n+1) (z) =

a) With f (z) = cos z 2 the Cauchy integral formula for n = 2 reads

9.

f (2) (z) =

2! 2πi

Z

|ζ−2|=1

1 f (ζ) √ 2 dζ = πi (ζ − π)

Z

|ζ−2|=1

√ √ where we used that 1 < π < 2, i.e. π ∈ B1 (2). Since − 4ζ 2 cos ζ 2 it follows that Z

|ζ−2|=1

cos ζ 2 √ dζ, (ζ − π)2 d2 dz 2


cos ζ 2 = −2 sin ζ 2

cos ζ 2 √ dζ = πi(4π 2 ) = 4π 3 i. (ζ − π)2

b) Recall that for α, β ∈ R, γ ∈ R\ (−N0 ) 2 F1 (z)

=

∞ X (α)k (β)k k=0

(γ)k k!

zk

converges for |z| < 1. Consequently we have ∞ X dn (α)n (β)n F (z) = k(k − 1) · · · (k − n + 1)z k−n 2 1 n dz (γ)n k! k=n

and (α)n (β)n dn . 2 F1 (0) = dz n (γ)n Now we find 1 2πi

Z

|z|= 21

2 F1 (z) z n+1

dz =

(α)n (β)n 1 dn 2 F1 (0) = n! dz n (γ)n n!

which is of course the nth Taylor coefficient as expected.

698

SOLUTIONS TO PROBLEMS OF PART 7

10. The generalised Cauchy integral formula gives Z 1 1 f (ζ) dζ f 0 (z) = 2πi |ζ−z0 |=1 (ζ − z)2 Z 2 1 f 00 (z) = f (ζ) dζ 2πi |ζ−z0 |=1 (ζ − z)3 Z 6 1 f 000 (z) = f (ζ) dζ 2πi |ζ−z0 |=1 (ζ − z)4 which yields 00



 1 1 1 f (ζ) dz + + (ζ − z)2 (ζ − z)3 (ζ − z)4 |ζ−z0 |=1   Z 6 1 + ζ − z + (ζ − z)2 = f (ζ) dζ = 0. 2πi |ζ−z0 |=1 (ζ − z)4

6 f (z) + 3f (z) + 6f (z) = 2πi 000

Z

Chapter 22 1. We have seen that a discrete set M is closed, hence a bounded discrete set is compact. Further, for every x ∈ M we can find rx > 0 such that Br (x)∩(M \{x}) = ∅. Thus each x ∈ M is exactly in one of these balls. They form an open covering of M and the compactness of M implies that finitely many of these balls will cover M , hence M itself must be finite. 2. We know from Problem 1 that M is a finite set, say M = {z1 , . . . , zn } ⊂ G. We choose Br1 (z1 ) ⊂ G such that Br1 (z1 ) ∩ (M \{z1}) = ∅. By Theorem 22.7 we can extend f as a holomorphic function f1 to Br1 (z1 ). The uniqueness theorem, Theorem 22.12, implies that f1 Br (z1 )\{z1 } = f and hence we obtain a holomorphic 1 function f˜1 : G\{z2 , . . . , zN } → C with the property that f˜1 G = f and f˜1 is bounded in a neighbourhood of each point zk , 2 ≤ k ≤ N . This process can be iterated. If f˜k : G\{zk+1 , . . . , zN } → C, k ≤ N , is already constructed as a holomorphic function such that f˜k G = f and f˜k is bounded in a neighbourhood of each zl , k + 1 ≤ l ≤ N , then we choose rk+1 > 0 such that Brk+1 (zk+1 ) ⊂ G and Brk+1 (zk+1 )∩{zk+2 , . . . , zn } = ∅. We can extend f˜k to a holomorphic function fk+1 defined Brk+1 (zk+1 ) and then we apply once more the uniqueness theorem to find that this gives rise to a holomorphic function f˜k+1 : G\{zk+2 , . . . , zN } extending to f . Eventually the result follows. P∞ 2k+1 3. Since converges uniformly and absolutely in Br (0) it follows that k=0 a2k+1 z P∞ 2k a (2k + 1)z converges uniformly and absolutely in PBr (0), i.e. 2k+1 k=0 P∞ ∞ 2k converges uniformly in Br (0) implying that k=0 |a2k+1 | k=0 |a2k+1 |(2k + 1)|z| P ∞ |z|2k also converges uniformly in Br (0), hence k=0 a2k+1 z 2k converges uniformly and absolutely in Br (0). Now, on Br (0)\{0} we have g(z) =

∞

f (z) X = a2k+1 z 2k z k=0

699

A COURSE IN ANALYSIS

which is bounded in a neighbourhood of z = 0. Therefore g has at z = 0 a removable singularity and can be extended to Br (0) by g(0) = a1 . P∞ z 2k+1 for all z ∈ C we first conclude that z 7→ sinz z must Since sin z = k=0 (−1)k (2k+1)! have an entire extension and at z = 0 this extension has the value 1. 4. We know that h(z0 ) = 0 as well as h(l) (z0 ) = 0 for l = 1, . . . , k − 1, but h(k) (z0 ) 6= 0. We claim that h2 has at z0 a zero of order 2k. The Leibniz rule gives
dl dl h2 (z) = l (h(z)h(z)) l dz dz l   X l (l−j) = h (z)h(j) (z). j j=0 If l < 2k−1 at least one term in the product h(l−j) (z0 )h(j) (z0 ) must vanish, however, (2k−j) (j) (k) (k) if l = 2k all products term  l h  (z0 )h (z0 ) vanish exceptthe  h (z0 )h (z0 ). 2k d d 2 2 Therefore we have dz (z0 ) = 0 for l ≤ 2k − 1, but dz (z0 ) 6= 0 implying lh 2k h that h2 has at z0 a zero of order 2k.

5. We know that z 7→ eaz , a ∈ C, has no zeroes. This implies that p(z)eaz = 0 if and only if p(z) = 0 and as a polynomial of order k we know that p(z) can have zeroes only of order less than or equal to k. Indeed, if p(z) = ak z k + q(z), ak 6= 0, and Pk−1 dk q(z) = l=0 al z l , we find dz k p(z) = k!ak 6= 0.

6. We proceed as in our considerations leading to ez+w = ez ew . The functions z 7→ cos z and z 7→ sin z are entire functions. For y ∈ R fixed we consider the two functions x 7→ cos(x − y) and x 7→ cos z cos y + sin x sin y. Both functions have a holomorphic extension to C and by the uniqueness theorem we must have for all z ∈ C that cos(z − y) = cos z cos y + sin z sin y. Now we fix z ∈ C and consider the functions y 7→ cos(z −y) and y 7→ cos z cos y +sin z sin y. Again, both functions have a holomorphic extension to C and we find once more by the uniqueness theorem cos(z − w) = cos z cos w + sin z sin w for all z, w ∈ C. 7. Since tr(γ) is compact we can cover tr(γ) with finitely many balls Brz0j (z0j ), j = 1, . . . , M , with the property that f Br (z0j ) has a convergent power series z0j SM represenation. Hence the function f has for every point w ∈ j=1 Brz0j (z0j ) a P∞ convergent power series representation k=0 bk (z − w)k with some positive radius SM of convergence. Therefore f is in j=0 Brz0j (z0j ) a holomorphic function. Note that since f is a priori given in G we cannot run into problems as we could in the consideration following Corollary 22.13. 8.

a) The series is a geometric series, namely ∞ X

k=0

z 2k =

∞ X

k=0

700

z2


k

=

1 . 1 − z2

SOLUTIONS TO PROBLEMS OF PART 7

We note further that

and

d dz



1 1 − z2



=

2z (1 − z 2 )2

d2 dz 2



1 1 − z2



=

2 + 6z 2 . (1 − z 2 )3

It follows that

∞ 2 + 6z 2 d2 X 2k z = (1 − z 2 )3 dz 2 k=0

∞ d X 2kz 2k−1 = dz k=0

=

∞ X

(2k)(2k − 1)z 2k−2

k=0

=

∞ X

(2k + 2)(2k + 1)z 2k .

k=1

b) First note that (1)k = k! and further   2k − 1 1 3 5 1 = · · ····· 2 k 2 2 2 2 and

  3 2k + 1 3 5 7 . = · · · ···· 2 k 2 2 2 2

Therefore we find x 2 F1



3 1 , 1; ; −x2 2 2



=x

∞ X

k=0

=

∞ X

k=0

Since




1 2 k

k!


k (1)k
−x2

3 2 k

1 (−1)k x2k+1 = arctan x. 2k + 1

d 1 arctan x = dx 1 + x2

it follows that ∞

∞

k=0

k=0

1 d X (−1)k 2k+1 X k x = = (−1) x2k , 2 1+x dx 2k + 1 a result which we of course expect when looking at series.

701

1 1+x2

as limit of a geometric

A COURSE IN ANALYSIS P∞ 9. Since u(x) = k=0 ak (x − x0 )k converges uniformly P∞ in (−η + x0 , x0 + η), we can extend u to a holomorphic function h(z) := k=0 ak (z − x0 )k with a uniformly and absolutely convergent power series in Bη (x0 ) ⊂ C. We claim that h satisfies h00 (z)+ z 2 h0 (z)+ z 4 h(z) = 0 in Bη (x0 ). The function z 7→ w(z) := h00 (z)+ z 2 h0 (z)+ z 4 h(z) is in Bη (x0 ) holomorphic since h is. On the open interval (−η + x0 , x0 + η) this function coincides with x 7→ u00 (x) + x2 u0 (x) + x4 u(x), but u00 (x) + x2 u0 (x) + x4 u(x) = 0 for all x ∈ (a, b), in particular for all x ∈ (−η + x0 , x0 + η). Now the uniqueness theorem for holomorphic functions implies w(z) = 0 for all z ∈ Bη (x0 ), i.e. h00 (z) + z 2 h0 (z) + z 4 h(z) = 0 for all z ∈ Bη (x0 ).

10. We start to prove the identity for x ∈ R. Recall the definition of the beta-function, (I.31.31), Z 1 ta−1 (1 − t)b−1 dt, a > 0, b > 0, B(a, b) = 0

and the relation to the Γ-function, Theorem I.31.11, Γ(a)Γ(b) . Γ(a + b)

B(a, b) = It follows that



Z 1 Γ k + 12 Γ l + 21 1 1 sk− 2 (1 − s)l− 2 ds = Γ(k + l + 1) 0 Z 1 1 t2k−1 (1 − t2 )l− 2 t dt =2 0

=2 =

Z

Z

1

0 1

−1

1

t2k (1 − t2 )l− 2 dt 1

t2k (1 − t2 )l− 2 dt.

Thus we have 1 1

= 1 Γ(k + l + 1) Γ k + 2 Γ l + 21

For the Bessel function Jl , l ∈ N, this yields

Z

1

−1

1

t2k (1 − t2 )l− 2 dt.


l+2k (−1)k x2 Jl (x) = Γ(k + 1)Γ(k + l + 1) k=0
Z 1 ∞ X (−1)k x l+2k 1 1 2

= t2k (1 − t2 )l− 2 dt Γ(k + 1) Γ k + 21 Γ l + 21 −1 k=0
Z 1 ∞ x l X (xt)2k (−1)k 2 l− 12 2

dt. (1 − t ) = 1 2k 2 Γ l + 2 −1 Γ(k + 1)Γ k + 12 ∞ X

k=0

702

SOLUTIONS TO PROBLEMS OF PART 7

Now we use Theorem I.31.12, i.e. the doubling formula for the Γ-function       1 1 1 =Γ Γ(2k + 1) = Γ (2k)! 22k Γ(k + 1)Γ k + 2 2 2 and we find
x l 2

Jl (x) =


1

Z

1

1

(1 − t2 )l− 2

∞ X (−1)k (xt)2k
dt 2k 2 Γ(k + 1)Γ k + 12

Γ 1 + 2 −1 k=0
Z 1 ∞ x l X (−1)k (xt)2k 2 l− 12 2

= dt (1 − t ) 1 Γ l + 2 −1 Γ 21 Γ(2k + 1) k=0
Z 1 ∞ x l X (−1)k 1 2
(xt)2k dt, = (1 − t2 )l− 2 1 (2k)! Γ l + 2 −1 k=0

or for all x ∈ R

Jl (x) =

Γ


1

2


x l 2

Γ l+


1

2

Z

1

−1

1

(1 − t2 )l− 2 cos(xt) dt.

(∗)

We know R 1 that Jl is 1an entire function. The complex differentiability of the function z 7→ −1 (1 − t2 )l− 2 cos(zt) dt for z ∈ BR (0), R > 0, follows by using standard arguments for differentiating parameter depending integrals. Hence (∗) holds for all z ∈ BR (0), R > 0, therefore (∗) holds in C. Note that the result also holds for R 1 l = 0, but 1for this case we must be a bit more careful when arguing that z 7→ −1 (1 − t2 )− 2 cos(zt) dt is complex differentiable due to the (weak) singularity of the integral at t = ±1. 1 (2k)!

11. We prove 1

4· 21 ·1

=

1 2.

=

1 4k ( 12 ) k!

by induction. For k = 1 we have

k

=

1 2!

!

=

1 41 ( 21 ) 1!

Now if the statement holds for k we find

1 1 = = k 2(k + 1)! (2k)!(2k + 1)(2k + 2) 4

4k




1 2 k

1 ! = k+1 k!(2k + 1)(2k + 2) 4

1 1 ! = (2k + 1)(2k + 2) 4(k + 1)


where we used that 12 k+1 = 21 k 21 + k . Since 2




1 2 k

and we long to prove

or

1 2




1 k!(2k + 1)(2k + 2)

1

1 2 k+1

1 2

+k

(2k + 1)(2k + 2) = 4k + 6k + 2 = 4(k + 1)

703

(k + 1)!




 1 +k , 2

1

=

A COURSE IN ANALYSIS

the equality follows. Analogously the second identity follows: For k = 1 we have 1 ! 1 1 1
= . = = 6 3! 6 4 23 1

Furthermore we find assuming the statement for k that 1 1 1 = = (2(k + 1) + 1)! (2k + 3)! (2k + 1)!(2k + 2)(2k + 3) 1 = k 3
4 2 k k!(2k + 2)(2k + 3) 1 ! = k+1 3
, 4 2 k+1 (k + 1)!

which reduces to the claim

1 1 ! = (2k + 2)(2k + 3) 4(k + 1)

3 2

+k




which follows from (2k + 2)(2k + 3) = 4k 2 + 10k + 6 = 4(k + 1) identities we find now  X k   ∞ 1 1 2 1 1 2
; − z z = − F 0 1 2 4 4 k! 12 k

3 2


+ k . With these

k=0

=

∞ X

k=0

(−1)k 2k
z 1

4k k!

2 k

∞ X (−1)k 2k = z = cos z, (2k)! k=0

and analogously 0 F1



1 1 2 ; z 2 4



=

∞ X z 2k = cosh z. (2k)!

k=0

Chapter 23 1.

a) The exponential function z 7→ ez is an entire function without any zeroes and (ez )0 = ez . However it is a periodic function with periodic 2πi and therefore it is not globally bijective. Indeed all our problems of defining and understanding the logarithmic function are due to the non-injectivity of z 7→ ez . b) Since f 0 (z) = g 0 (z) sin z + g(z) cos z we have f 0 (0) = g(0). If g(0) 6= 0 then by Theorem 23.7 there exists an open neighbourhood U of 0 such that f : U → f (U ) is biholomorphic, i.e. f is locally biholomorphic at 0.

704

SOLUTIONS TO PROBLEMS OF PART 7

2. For w ∈ Bδ (f (z0 )), i.e. |w − f (z0 )| < δ, it follows for all z ∈ ∂BR (z0 ) that |f (z) − w| ≥ |f (z) − f (z0 )| − |w − f (z0 )| > δ and therefore min

z∈∂BR (z0 )

|f (z) − w| > |f (z0 ) − w|.

By applying Lemma 23.1 to z 7→ f (z) − w, we conclude that there exists some z1 ∈ BR (z0 ) such that f (z1 ) = w, i.e. w ∈ f (BR (z0 )). 4

3. We consider on G = {z ∈ C | Re z > 0, Im z > 0} the function f (z) = e−iz . Since ∂G = ([0, ∞) + 0i) ∪ (0 + i[0, ∞)) i.e. ∂G is the union of the closed positive real and positive imaginary axis, we find for f on ∂G that z = x + iy : z ∈ [0, ∞) + i0 4 implies z 4 = x4 and f (z) = eix , hence |f (z)| = 1. Then for z ∈ 0 + i[0, ∞) we have 4 z 4 = (iy)4 = y 4 and it follows that f (z) = eiy and we find again |f (z)|  = 1. Thus √  √ √
5 − i 86 supz∈∂G |f (z)| = 1. However for z = r 12 2 + i 21 3 we have z 4 = r4 − 16 √

6r4

√ 4 6r

5r4

and therefore f (z) = e 8 e− 16 i implying that |f (z)| = e 8 and for r tending to +∞ it yields that |f (z)| is unbounded, i.e. the estimate |f (z)| ≤ supz∈∂G |f (z)| = 1 cannot hold. This example shows that the boundedness of G is essential for (23.3) to hold.
4. Since limk→∞ kfk −f k∞,∂G = 0 it follows that the sequence fk ∂G k∈N is a Cauchy
sequence and the boundary maximum principle implies that fk G k∈N is a Cauchy sequence too, i.e. liml,k→∞ kfk − fl k∞,G = 0. Since (fk (z))k∈N converges for all z ∈ G to f (z) we find that (fk )k∈N converges uniformly on G to f , i.e. limk→∞ kfk − f k∞,G = 0, also compare with Theorem II.14.2. As a uniform limit of holomorphic functions fk the function f is on G holomorphic too.

5. Since g, h ∈ F implies for λ, µ ∈ C that λg + µh ∈ F , i.e. F is vector space over C. Clearly kf kF ≥ 0 and if kf kF = 0 then by the boundary maximum principle we have |f (z)| ≤ maxz∈∂BR (0) |f (z)| = kf kF = 0 for all z ∈ BR (0) which yields f ≡ 0. For λ ∈ C we find further that kλf kF =

max

z∈∂BR (0)

|λf (z)| = |λ|

max

z∈∂BR (0)

|f (z)| = |λ|kf kF

and the triangle inequality follows from kf + gkF =

z∈∂BR (0)

max

|f (z) + g(z)|

≤

z∈∂BR (0)

max

(|f (z) + |g(z)|)

≤ kf kF + kgkF . Thus (F , k.kF ) is a normed space. Now let (fk )k∈N be a sequence in F which is a Cauchy sequence  with  respect to k.kF . By the boundary maximum principle it follows that fk BR (0) is a Cauchy sequence with respect to the norm k.k∞,BR (0) .

705

A COURSE IN ANALYSIS   From Theorem II.14.2 we deduce that (fk )k∈N has a limit in C BR (0) and now Problem 4 implies that f must be holomorphic in BR (0). Hence (F , k.kF ) is complete. 6. Suppose that for 0 < r2 < r1 < R we have M (r1 ) < M (r2 ) i.e. maxz∈∂Br1 (0) |f (z)| < maxz∈Br2 (0) |f (z)|. By the maximum boundary principle this implies max |f (z)| ≤

z∈Br1 (0)

max

z∈∂Br1 (0)

|f (z)|
0 there exists n N = N () ∈ N such that n > m > N () implies k=m+1 ck − 1 < . Q∞ Proof. Suppose first that k=1 ck = c 6= 0. The Cauchy criterion applied to   QN states: for every  > 0 and η > 0 there exists N = N (, η) ∈ N k=1 ck N ∈N

such that n > m > N (η, ) implies n m Y Y ck − ck < η, k=1

or

|c| 2

k=m+1

Qm = c 6= 0 it follows that for m ≥ N0 we have  | k=1 ck | ≥ 2 2 implies 6 0, hence we have for η = |c| = that n > m > max N0 , N , |c| n 2 Y |c|  ≤ . ck − 1 < Qm | k=1 ck |

Since limm→∞

Qm

k=1

n Y η . ck − 1 < Qm | k=1 ck |

k=1 ck

k=m+1

707

A COURSE IN ANALYSIS

Now we prove the converse. For  = 12 there exists N1 ∈ N such that n > m > N1 implies n Y 1 ck − 1 < 2 k=m

which yields

n 1 Y 3 < ck < . 2 2 k=m

In particular cl 6= 0 for all l > N1 . For N > N1 fixed and 0 <  < assumption N () > N such that n > m > N () implies Y Qn n k=N ck = Qm − 1 c − 1 k k=N ck

1 2

there exists by

k=m

= |cm+1 · cm+2 · . . . · cn − 1|
r, and w =   h B r1 (0)\{0} .

we find w ∈ B r1 (0)\{0} and z = z0 + w1 , i.e. C\Br (z0 ) ⊂

a) We have

z−

z3 6

− sin z

z5

= =

z− 5 − z5!

z3 6

 − z−

+

z7 7!

− z5

z9 9!

z3 3!

+

z5 5!

z5 ± ···

−

z7 7!

+ ···



1 z2 z4 + − ± ··· 5! 7! 9! ∞ X z 2k = (−1)k+1 (2k + 5)!

=−

k=0

3 z− z6

−sin z

and therefore z 7→ has a removable singularity at z0 = 0. z5 b) With z + 3 = w, i.e. z = w − 3, it follows that

1 1 = (w − z) sin z+3 w   1 1 1 = (w − z) − + ± · · · w 3!w3 5!w5 1 7 1 7 7 + + − ± ··· =1− − w 3!w2 3!w3 5!w4 5!w5 1 7 7 1 7 − =1− + + − ± ··· 2 3 4 z + 3 3!(z + 3) 3!(z + 3) 5!(z + 3) 5!(z + 3)5

(z− 4) sin

1 has an essential singularity at z0 = −3. and therefore z 7→ (z − 4) sin z+3 π c) With z − 4 = w, i.e. z = w + π4 , we find
cos π2 + 2w cos 2z − sin 2w = =
3 π 3 w w3 z− 4

and further

  (2w)5 (2w)7 sin 2w 8 (2w)3 − sin 2w + − ± · · · = −8 = − 2w − w3 (2w)3 (2w3 ) 3! 5! 7! 2 4 8 8(2w) 8(2w) 8 =− + ± ··· + − (2w)2 3! 5! 7! ∞ X 8 (2w)2k =− + 8 (−1)k 2 (2w) (2k + 3)! k=0

709

A COURSE IN ANALYSIS

or
2k ∞ X 2z − π2 cos 2z 8 k
=− (−1)
2 + 8 (2k + 3)! z − π4 2z − π2 k=0
2k ∞ 2k X z − π4 2 k2 + 8 (−1) =−
2 (2k + 3)! z−π k=0

4

implying that z 7→

cos 2z

(z− π4 )3

has a pole of order 2 at z0 =

3. First we note that we can decompose f (z) =

1 (z+2)(z+4)

π 4.

by using partial fractions:

1 1 1 = − . (z + 2)(z + 4) 2(z + 2) 2(z + 4) a) For A2,4 (0) = B4 (0)\B2 (0) we find: since |z| > 2 ∞ 1 X 1 1 2k
= = (−1)k k 2 2(z + 2) 2z z 2z 1 + z k=0

and since |z| < 4

∞

1X 1 zk 1
= (−1)k k = z 2(z + 4) 8 4 8 1+ 4 k=0

it follows for the Laurent series of f (z) = f (z) = =

∞ X

k=0 ∞ X

(−1)k

1 (z+2)(z+4)

in A2,4 (0) that

∞

zk 1X 2k−1 (−1)k k − k+1 z 8 4 k=0

(−1)k

k=0

k−1

2 + z k+1

∞ X

(−1)k+1

k=0

zk . 2 · 4k+1

b) For z ∈ A4,∞ (0) we can use a partial result from the previous part to find f (z) =

∞

X 1 1 1 2k−1 − = . (−1)k k+1 − 2(z + 2) 2(z + 4) z 2(z + 4) k=0

But for |z| > 4 we have ∞ 4k 1 1 X 1
= (−1)k k = 4 2(z + 4) 2z z 2z 1 + z k=0

=

∞ X

(−1)k

k=0

710

2 · 4k−1 , z k+1

SOLUTIONS TO PROBLEMS OF PART 7

and therefore f (z) = =

∞ X (−1)k 2k−1

k=0 ∞ X

z k+1

(−1)k

k=0

−

∞ X (−1)k 2 · 4k−1 k=0
k

z k+1

2k−1 1 − 2 z k+1

for z ∈ A4,∞ (0). c) Note that z ∈ A1,2 (−2) means 1 < |z + 2| < 2 and with z + 2 = w we find 1 1 1
= = (z + 2)(z + 4) w(w + 2) 2w 1 + w2 ∞ ∞ X 1 X wk 1 (z + 2)k−1 = (−1)k k = + (−1)k 2w 2 2(z + 2) 2k+1 k=0

k=1

∞

X (z + 2)k 1 + (−1)k+1 k+2 . = 2(z + 2) 2 k=0

1 1 d) Again using the decomposition f (z) = 2(z+2) − 2(z+4) we get for z ∈ B2 (0), i.e. |z| < 2, that ∞ 1 1 1X
(−1)k z k , = = 2(z + 2) 4 4 1 + z2 k=0

and for the second term we may use the result of part a), i.e. ∞

1 zk 1X (−1)k k , = 2(z + 4) 8 4 k=0

which yields in B2 (0) ∞ ∞ 1X zk 1X (−1)k z k − (−1)k k 4 8 4 k=0 k=0   ∞ 1 1X zk (−1)k 1 − = 4 2 · 4k k=0   ∞ X 2 · 4k − 1 k zk, = (−1) 2 · 4k+1

f (z) =

k=0

hence in B2 (0) we obtain the Taylor series. 4.

a) We have to look at the zeroes of z 7→ 4 sin z − 2, i.e. we have to solve the equation sin z = 21 . For z real we obtain the points π4 + 2kπ and 5π 4 + 2kπ, k ∈ Z, and since sin0 = cos these are simple zeroes. Consequently f has at the points π4 + 2kπ and 5π 4 + 2kπ, k ∈ Z, a pole of order 2. Using the representation

711

A COURSE IN ANALYSIS
1 eiz − e−iz we first deduce that sin z = 12 cannot have a purely imaginary sin z = 2i solution iy, and in the general case, i.e. z = x + iy, for sin z = 12 we must have sin x = e−y1+ey > 1. b) We first look at the equation ew − 1, w ∈ C. With w = u + iv it follows that eu eiv = 1 must hold, or eu = e−iv . Since e−iv = 1 we deduce that eu = 1 or u = 0. Thus ew = 1, w = u+iv implies w = iv, and we have to solve cos v+i sin v = 1 which 1 1 1 implies v = 2kπ. With w = 2z we now deduce zk = 4πik = − 4πk i, k ∈ Z\{0}. So for these values of zk the function 1z has a pole of first order. Whereas for e 2z −1

z = 0 we obtain that Laurent expansion z 1

e 2z − 1

=

∞ X 1 1 −k+1 z k! 2k k=1

implying that the function has an essential singularity at z0 = 0. sin

π

z

π 2 5. Consider the function g(z) = ((cos π(z−1))−1) 2 . For z0 = 1 we have sin 2 z0 = 1 and cos(π(z0 − 1)) − 1 = 0. Moreover z0 = 1 is a simple zero of cos π(z − 1) − 1, hence it is a zero of order two for (cos π(z − 1) − 1)2 implying that g has a pole of order two at z0 = 1. 1

6. If z → z0 then |z − z0 | → 0 and therefore e |z−z0 | → ∞. This implies that f is not bounded in a neighbourhood of z0 , hence it cannot have a removable singularity. If f had a pole at z0 we must have for some N ∈ N with κ0 > 0 that 1

κ0 e |z−z0 | ≤ |f (z)| ≤ κ1

1 |z − z0 |N

holds in a neighbourhood of z0 . This however yields that 1

|z − z0 |N e |z−z0 | ≤

κ1 , κ0

1

1

i.e. lim supr→0 rN e r < ∞, but we know that limr→0 rN e r = ∞. Hence f must have an essential singularity at z0 . 7. We consider the equation exp z1 = w0 , w0 6= 0, for 0 < |z| < 12 . With ζ = z1 this is equivalent to exp ζ = w0 , w0 6= 0, 2 < |ζ| < ∞. With ζ = u + iv and w0 = reiϕ it follows now that eu eiv = reiϕ . With u = ln r we arrive at ei(v−ϕ) = 1, i.e. cos(v − ϕ) + i sin(v − ϕ) = 1 or v − ϕ = 2kπ, k ∈ Z. Hence we have in 0 < |z| < 21 1 1 , k ∈ Z, of e z = w0 . the countable many zeroes zk = ζ1k = ln r+i(ϕ+2πk) 8. If f and g have a pole of order N at z0 we can find an open neighbourhood of z0 such that with a−N 6= 0 and b−N 6= 0 we have f (z) = a−N (z − z0 )−N +

N −1 X l=1

712

a−l (z − z0 )−l +

∞ X

k=0

ak (z − z0 )k

SOLUTIONS TO PROBLEMS OF PART 7

and

N −1 X

g(z) = b−N (z − z0 )−N +

l=1

or −N

f (z) = a−N (z − z0 )

1+

N −1 X l=1

b−l (z − z0 )−l +

∞ X

k=0

bk (z − z0 )k

∞

X ak a−l (z − z0 )N −l + (z − z0 )k+N a−N a−N k=0

!

and −N

g(z) = b−N (z − z0 )

1+

∞

N −1 X

X bk b−l (z − z0 )N −l + (z − z0 )k+N b−N b−N k=0

l=1

which yields for z 6= z0 that a−N f (z) = g(z) b−N =

1+ 1+

P∞

ak k=−N +1 a−N (z P∞ bk k=−N +1 b−N (z

a−N F (z) . · b−N G(z)

− z0 )k+N

− z0 )k+N

!

,

!

For z → z0 it follows that limz→z0 F (z) = 1 and limz→z0 G(z) = 1 which first of all (z) implies that fg(z) is bounded in a neighbourhood of z0 , hence the singularity at z0 is removable, and further we find lim

z→z0

f (z) a−N = . g(z) b−N

9. First we choose R > 0 such that |p(z)| ≥ 1 for |z| ≥ R. Now we note that nz n−1 + a1 (n − 1)z n−1 + · · · + an−1 p0 (z) = p(z) z n + a1 z n−1 + · · · + a0 n = + h(z), z ν where h(z) is a sum of terms of the type A z ν , ν ≥ 2, as we can deduce by applying polynomial division. Integrating over ∂BR (0) given in standard parametrization yields, recall that for ν ≥ 2 we have Z 1 dz = 0, ν z ∂BR (0)

that

1 2πi

Z

p0 (z0 ) 1 dz = p(z) 2πi

Z

n dz = n. z P { Since p(z) has no pole in BR (0) the argument principle implies M k=1 αk = n where α1 , . . . , αM are the multiplicities of the zeroes of p. ∂BR (0)

713

∂BR (0)

A COURSE IN ANALYSIS

10. First we decompose h(z) = z 5 − 2z 3 + 10 according to h(z) = f (z) + g(z) where f (z) = 10 and g(z) = z 5 − 2z 3 . We note that |g(z)| = z 5 − 2z 3 ≤ |z|5 + 2|z|3 ≤ 3 < 10

for z ∈ B1 (0). This implies by Rouch´e’s theorem that f (z) + g(z) = z 5 − 2z 2 + 10 and f (z) = 10 have the same number of zeroes in B1 (0), but f (z) has no zero. Now we decompose h(z) = f (z) + g(z) with f (z) = z 5 and g(z) = 10 − 2z 3 . On ∂B2 (0) we find |g(z)| = |10 − 2z 3 | ≤ 10 + 2|z|3 ≤ 26 < 25 = 32.

Thus h(z) = z 5 −2z 3 +10 has the same numbers of zeroes inside B2 (0) as f (z) = z 5 , i.e. all zeroes of h(z) are inside B2 (0) implying now that all zeroes of h(z) = z 5 − 2z 3 + 10 belong to A1,2 (0). Chapter 25 1. We can consider ∂K as the trace of the N line segments [Aj , Aj+1 ], j = 1, . . . , N , AN +1 = A1 , i.e. we have ∂K = tr ([A1 , A2 ] ⊕ [A2 , A3 ] ⊕ · · · ⊕ [AN −1 , AN ] ⊕ [AN , A1 ]) , which implies that ∂K is a cycle provided that the line segments are parametrized in the standard way. 2. Using standard parametrization we introduce the two curves γ : [A, B] ⊕ [B, C] ⊕ [C, A] and η := [D, E] ⊕ [E, F ] ⊕ [F, D]. These are two simply closed curves which ˜ we can add to a chain γ +η. Since tr(γ) ∩ tr(η) = ∅ and both curves are simply ˜ is indeed a cycle. closed γ +η 3. We denote the simply closed curve with trace κ√2 (z0 ) by γ1 and that with trace being the rectangle ABCD by γ2 . We already know that both curves are cycles. Moreover, both are null-homologous in G = B2 (z0 ) and since {

indγ1 −γ2 (z) = indγ1 (z) − indγ2 (z)

we have for z ∈ G that indγ1 −γ2 (z) = 0 implying that γ1 and γ2 are homologous. {

1 7 4. For z ∈ ∆ we already know that indγ (z) = 0 since in this case ζ → z−ζ is holomorphic in a neighbourhood of ∆. Next let z ∈ ∆ and consider the figure below C

C0 b

A0

z

B0

A B

714

SOLUTIONS TO PROBLEMS OF PART 7

The Cauchy integral theorem yields Z Z 1 1 dζ = dz = 1. ζ − z ζ − z ∂Br (z) γ 5. Let z0 be a pole of order k. Then we have by the translation invariance of the differentiation
dk−1 1 (z − z0 )k g(z) z→z0 (k − 1)! dz k−1
1 dk−1 = lim (z + a − z0 )k g(z + a) z→z0 (k − 1)! dz k−1 = res(g ◦ τa , z0 − a).

res (g, z0 ) = lim

6.

a) We have a simple pole at z = 2i, z = −2i and a pole of order two at z = −1. z 3 −z 2 +2z Therefore we find with f (z) = (z+1) 2 (z 2 +4)  z 3 − z 2 + 2z z→2i (z + 1)2 (z − 2i)(z + 2i) 3 2 z − z + 2z 1−i = lim = , z→2i (z + 1)2 (z + 2i) −4 − 3i   z 3 − z 2 + 2z res(f, −2i) = lim (z + 2i) z→−2i (z + 1)2 (z − 2i)(z + 2i) 3 2 z − z + 2z 1+i = lim = z→−2i (z + 1)2 (z − 2i) −4 + 3i   1 d z 3 − z 2 + 2z res(f, −1) = lim (z + 1)2 z→−1 1! dz (z + 1)2 (z 2 + 4)  3  d z − z 2 + 2z = lim z→−1 dz z2 + 4 2 (3z − 2z + 2)(z 2 + 4) − (z 3 − z 2 + 2z)(2z) = lim z→−1 (z 2 + 4)2 34 = . 25 res(f, 2i) = lim

  (z − 2i)

b) At z = kπ, k ∈ Z, we have poles of order two and with g(z) =

e2z sin2 z

we have

  2z 1 d 2 e (z − kπ) res(g, kπ) = lim k→kπ 1! dz sin 2z   2 (z − kπ) sin z + 2(z − kπ) sin z − 2(z − kπ)2 cos z = lim 2e2z z→kπ sin3 z

715

A COURSE IN ANALYSIS

and the substitution w = z − kπ yields   2 w sin w + 2w sin w − 2w2 cos w res(g, kπ) = 2 lim e2w+2kπ w→0 sin3 w  2   w sin w + 2w sin w − 2w2 cos w w3 = 2e2kπ lim w→0 w3 sin3 w w2 sin w + 2w sin w − 2w2 cos w = 2e2kπ lim w→0 w3 where we used that limw→0 we find

sin w w

= 1. Applying the rules of l’Hospital three times

(2w2 + 2w + 2) sin w + (w2 − 2w) cos w w2 sin w + 2w sin w − 2w2 cos w = lim 3 w→0 w→0 w 3w2 2 2 (−w + 6w + 2) sin w + (2w + 4w) cos w = lim w→0 6w (−2w2 + 2w + 6) sin w + (−w2 + 10w + 6) cos w = lim w→0 6 = 1, lim

hence we have res(g, kπ) = 2e2kπ . 7. Consider the path γR = κ+ r (0) ⊕ [−R, R] as below Im z iR

γR = κ+ R + [−R, R]

0

−R

R

Re z

where R > 0 and all poles c1 , . . . , cK of f = pq lie in the interior of γR . By the residue theorem we have Z Z K X p(z) dz = 2πi cj . f (z) dz = γR q(z) γR j=1 We split the integral according to Z Z f (z) dz = γR

κ+ R (0)

716

f (z) dz +

Z

R

−R

f (z) dz.

SOLUTIONS TO PROBLEMS OF PART 7

For the first integral on the right hand side we find Z π Z
f Reiϕ · iReiϕ dϕ f (z) dz = κ+ R (0)

0

and since M ≥ N + 2 we can find a constant d > 0 such that
f Reiϕ iReiϕ ≤ d R

implying

Z dπ . f (z) dz ≤ κ+ R R (0)

Now we find 2πi

K X j=1

cj = lim

R→∞

= lim

R→∞

Z

and since lim

R→∞

Z

κ+ R (0)

κ+ R (0)

Z

f (z) dz +

f (z) dz +

κ+ R (0)

Z

Z

R

f (z) dz

−R

∞

!

f (z) dz

−∞

f (z) dz = 0

the result follows. 8. The integral is of the type discussed in Problem 7. The poles of z 7→ z61+1 are the zeroes of z 6 + 1 = 0 which yields that the poles in the upper half plane Im z > 0 πi πi 5πi are at the points e 6 , e 2 and e 6 , and these are simple poles. Now let z0 be any zero of z 6 + 1. It follows that z − z0 1 = P5 5−k z k z6 + 1 0 k=0 z

and therefore we find for the residues of z 7→ res



1 , z0 z6 + 1



= lim

z→z0

 (z − z0 )

1 z6 + 1

This now implies Z

0

∞

1 z 6 +1



by Proposition 25.22

= lim P5 z→z0

1

5−k z k 0 k=0 z

=

1 . 6z05

  −5πi Z 25πi 1 1 ∞ 1 1 e 6 − 5πi 2 + e− 6 dx = dx = · 2πi + e x6 + 1 2 −∞ x6 + 1 2 6 π = . 3

717

A COURSE IN ANALYSIS

1 9. The meromorphic function f (z) = (z2 +1) k has in the upper half-plane Im z > 0 the pole z0 = i only and this pole is of order k. Therefore we have

 1 1 (z − i)k 2 (k − 1)! (z + 1)k   k−1 d (z − i)k 1 = lim k−1 z→i dz (k − 1)! (z − i)k (z + i)k   k−1 1 1 d = lim k−1 z→i dz (k − 1)! (z + i)k −i (2k − 2)! , = 2k−1 2 ((k − 1)!)2 dk−1 z→i dz k−1

res (f, i) = lim



which implies by the residue theorem (in form of Problem 7) Z

∞

−∞

(2k − 2)! π 1 ds = 2k−2 . (s2 + 1)k 2 ((k − 1)!)2

We note the first four integrals explicitly: Z

∞

1 ds = π, 2 −∞ s + 1 Z ∞ 3π 1 , ds = 2 + 1)3 (s 8 −∞

Z

∞

1 π ds = , 2 + 1) 2 1 5π . ds = (s2 + 1)4 16

(s2

−∞ Z ∞

−∞

10. Again we can apply Problem 7. The polynomial q(z) = 1 + z 2ν has in Im z > 0 the (2l+1)πi simple zeroes zl = e 2ν , l = 0, . . . , ν − 1. Using Proposition 25.22, also compare with Problem 8, we find

res



z 2µ ; zl 1 + z 2ν



  z 2µ = lim (z − zl ) z→zl 1 + z 2ν 1 1 2µ−2ν+1 z = − zl2µ+1 = 2ν l 2ν

718

SOLUTIONS TO PROBLEMS OF PART 7 where in the last step we used that zl−2ν = −1. Thus we find Z

∞

−∞

ν−1

2πi X 2µ+1 t2µ zl dt = − 2ν 1+t 2ν l=0 ν−1 X

=−

iπ ν

=−

ν−1 iπ (2ν1 )πi X  πi(2µ+1) l e ν e 2ν ν

=−

iπ (2µ+1)πi e 2ν ν

πi

e 2ν (2l+1)(2µ+1)

l=0

l=0 ν−1 X

e

πi(2µ+1) ν

l=0

l

iπ (2µ+1)πi 1 − e(2µ+1)πi = − e 2ν (2µ+1)πi ν 1−e ν i π iπ    = . =− (2µ+1)π ν sin (2µ+1)π ν sin 2ν

R 2π

ν

dϕ 1−2r cos ϕ+r 2

is of the type (25.32) and with R(cos ϕ, sin ϕ) = 1 we find h(z) = (z−r)(1−rz) which has in the unit circle one simple 1 pole, either r if |r| < 1 or r if |r| > 1. Using Proposition 25.22 to calculate the corresponding residue we arrive at ( Z 2π 2π |r| < 1 dϕ 2, = 1−r 2π 2 1 − 2r cos ϕ + r , |r| > 1. 2 0 r −1

11. The integral 1 1−2r cos ϕ+r 2

0

12. We consider the integral

R

z α−1 γ 1+z

dz along the curve given in the hint where 0 <  < 1

and 0 < r < 1 < R. In the interior of γ the function g(z) = at z0 = −1 and the residue at z0 is   z α−1 = e(α−1)πi . lim (z + 1) z→−1 1+z This implies

Z

γ

BF

[A,B]

has a simple pole

z α−1 dz = 2πie(α−1)πi . 1+z

We now split the integral and we obtain Z Z Z Z g(z) dz + _ g(z) dz + g(z) dz = γ

z α−1 1+z

[F,G]

g(z) dz +

Z

_

g(z) dz,

GA

where as usual [A, B] and [F, G] are the line segments from A to B and from F _

to G, respectively, and BF denotes the arc of the circle with centre 0 and radius

719

A COURSE IN ANALYSIS _

R that goes through B, C, D, E and F , whereas GA is the arc of the circle with centre 0 and radius r that goes through G, H and A. First we note that with B = Reiϕ Z Z 2π−ϕ α−1 iϕ(α−1) R e iReiϕ = dϕ g(z) dz _ iϕ 1 + Re BF ϕ Z Z 2π−ϕ 2π Rα Rα dϕ ≤ dϕ = iϕ |1 + Re | R−1 0 ϕ implying that

Z

lim lim

→0 R→0

_

g(z) dz = lim lim

R→∞ →0

BF

Z

_

g(z) dz = 0.

BF

Furthermore we have A = reiψ that Z ψ Z α−1 iψ(α−1) iϕ r e ire dϕ _ g(z) dz = iϕ 1 + re GA 2π−ψ Z 2π−ψ α Z 2π α r r ≤ dϕ ≤ dϕ 1 − r 1 −r ψ 0 which yields lim lim

→0 r→0

Z

_

g(z) dz = lim lim

r→0 →0

GA

Now we use the second part of the hint, i.e. lim (x + i)β = xβ

and

→0

Z

_

g(z) dz = 0.

GA

lim (x − i)β = xβ e2πiβ ,

→0

x > 0, which simply follows from z α = eα log z and lim→0 log(x + i) = ln x and lim→0 log(x − i) = ln x + 2πi. For x ∈ [r, R] we find that Z R α−1 Z x z α−1 dz = dx, lim →0 [A,B] 1 + z r 1+x and lim

→0

or lim lim lim

R→∞ r→0 →0

Z

Z

[A,B]

[F,G]

z α−1 dz = −e2πiα 1+z

z α−1 dz + 1+z

Z

[F,G]

Z

R r

z α−1 dz 1+z

!

xα−1 dx, 1+x

= 1−e

2πiα

Finally we arrive at −2πieαπi = e(α−1)πi = or

Z

0

∞

Z

γ


z α−1 dz = 1 − e2πiα 1+z

Z

0

∞

∞
xα−1 dx. 1+x 0

xα−1 dx 1+x

2πieαπi 2πi π xα−1 dx = − . = − −απi = 1+x 1 − e2πiα e − eαπi sin απ

720

SOLUTIONS TO PROBLEMS OF PART 7

13. It turns out that some calculations become easier when first noting that Z

∞

−∞

e−2πixξ 1 dx = cosh πx cosh πξ

√ √ will imply our result. This follows from the substitutions 2πx = y and 2πξ = η. −2πizξ In the remaining part we argue along the lines of [83]. We integrate f (z) = ecosh πz over the curve indicated in the limit. The denominator of this function vanishes πz −πz = 0, i.e. eπz = −e−πz or e2πz = −1 implying that in the if cosh πz = e +e 2 i 3i interior of γR we have two  i simple poles 2 and 2 . We now use Proposition 25.22 to 3i find the residue at z0 ∈ 2 , 2 : 2(z − z0 ) eπz + e−πz 1 = 2e−2πizξ eπz e2πz −e−2πz0 ,

(z − z0 )f (z) = e−2πizξ

z−z0

where we have used that eπi = e2π( 2 ) = e−2π( 2 ) = 1. We note further that i

1

lim

=

2πz −e2πz0 z→z0 e z−z0

3i

1 d 2πz dz e

Therefore it follows that lim (z − z0 )f (z) = lim

z→z0

z→z0

=

2e




(z0 )

−2πizξ πz

e

1 −2πiz0 ξ −πz0 , e e π

1 . 2πe2πz0

=




!

1 e2πz −e−2πz0 z−z0

which yields 

i res f, 2 So far we have proved Z

e



i = − eπξ , π

−2πizξ

γR

Now we split the integral Z

γR

f (z) dz =

Z

R+2i

R

R

γR



3i res f, 2



=

i 3πξ e . π

  i πξ i 3πξ 1 = 2πi − e + e cosh πz π π
πξ 3πξ =2 e −e .

f (z) dz in the following way:

f (z) dz +

Z

−R+2i R+2i

721

f (z) dz +

Z

−R

−R+2i

f (z) dz +

Z

R

−R

f (z) dz.

A COURSE IN ANALYSIS

We observe that πz e + e−πz 1 πz ≥ |e | − e−πz | cosh πz| = 2 2

and for z = R + iy, 0 ≤ y ≤ 2 we get

| cosh πz| ≥ whereas


1 πR e − e−πR 2

−2πizξ −2πiRξ 2πyξ e = e ≤ e4πξ e

implying

−2πi(R+iy)ξ e = 0, lim R→∞ cosh π(R + iy)ξ

i.e.

lim

R→∞

Z

for 0 ≤ y ≤ 2,

R+2i

f (z) dz = 0.

R

Replacing z = R + iy by z = −R + iy, 0 ≤ y ≤ 2, yields similar estimates and consequently Z −R f (z) dz = 0. lim R→∞

−R+2i

−2πi(x+2i)

On the other hand we find e = e−2πixξ e4πξ and cosh(πx) = cosh(π(x+2i)) since cosh(z + 2πi) = cosh z. It follows that Z

R+2i

e−2πizξ

−R+2i

1 = cosh πz

Z

=e

R+2i

e−2πi(x+2i)ξ

−R+2i Z R 4πξ

e−2πixξ

−R

or

Z

−R+2i

e−2πizξ

R+2i

This implies Z −R+2i e−2πizξ R+2i

1 dz = −e4πξ cosh πz

1 dz + cosh πz

Z

R

e−2πixξ

−R

Z

R −R

1 dx cosh π(x + 2i)

1 dx, cosh πx e−2πixξ

1 dx. cosh πx


1 dx = 1 − e4πξ cosh πx

Z

R

f (z) dz

−R

and for R → ∞ we finally arrive at Z Z

∞ −2πixξ 1 1 e−2πizξ 2 eπξ − e3πξ = lim dz = 1 − e4πξ dx e R→∞ γ cosh πz cosh πx −∞ R or

Z

∞ −∞

e−2πxξ


2 eπξ − e3πξ 1 dx = . cosh πx 1 − e4πξ

722

SOLUTIONS TO PROBLEMS OF PART 7

Since 1 − e4πξ = −e2πξ eπξ − e−πξ and eπξ − e3πξ = −e2πξ eπξ − e−πξ we eventually obtain Z ∞ 1 2 1 e−2πixξ = πξ . = cosh πx e − e−πξ cosh πz −∞ Chapter 26 a) From (26.11) we deduce for k ∈ N that   π 1 π 1

. = (−1)k = Γ −k + 2 Γ k + 21 sin π(k + 1) Γ k + 21

1.

Furthermore we know that Γ

1 2




=

√ π and a short induction yields

    1 1 · 3 · 5 · . . . · (2k − 1) 1 Γ = . Γ k+ 2 2k 2 Indeed, for k = 0 the result is known, and assuming the statement for k we find          1 1 1 1 Γ k+1+ =Γ k+ +1 = k+ Γ k+ 2 2 2 2     1 1 1 · 2 · 3 · . . . · (2k − 1)(2k + 1) 2k + 1 Γ k+ Γ = . = k+1 2 2 2 2 Note that

(2k)! 1 · 3 · 5 · . . . · (2k − 1) = k 2k 4 k!

and therefore we find   1 (−1)k (2k)! √ Γ −k + = π. 2 4k k! b) From (26.11) we deduce π = Γ(z)Γ(1 − z) = −zΓ(z)Γ(−z) sin πz or Γ(z)Γ(−z) = −

π . z sin πz

Since by Lemma 26.1 we have Γ(z) = Γ(z) we find for z = iy, y ∈ R, that Γ(iy)Γ(−iy) = Γ(iy)Γ(iy) = |Γ(iy)| implying that 2

|Γ(iy)| = −

π π = iy sin π(iy) y sinh y

723

2

(∗)

A COURSE IN ANALYSIS

where we used that −i sin(iy) = sinh y. Analogously we now find when taking z +

1 2

instead of z in (∗) that

    π 1 1 π
−z = Γ Γ z+ 1 = cos πz 2 2 sin π z + 2

and with z = iy, y ∈ R, it follows that Γ



     2 1 1 1 π π + iy Γ − iy = Γ + iy = = , 2 2 2 cos πiy cosh πy

since cos iw = cosh w, w ∈ R.

2. We know from (I.31.35) that this identity holds for all x > 0 and that Γ is in Re z > 0 a holomorphic function. The uniqueness theorem for holomorphic functions implies now that the identity must hold for all z ∈ C, Re z > 0. 3. By Theorem I.31.11 we have the equality B(x, y) =

Z

0

1

tx−1 (1 − t)y−1 dt =

Γ(x)Γ(y) Γ(x + y)

Γ(x)Γ(y) Γ(x+y)

for all x, y > 0. We can extend y 7→

for fixed x > 0 to a holomorphic R1 on Re w > 0 as we can extend 0 tx−1 (1 − t)y−1 dt for x > 0 function w 7→ R1 fixed to a holomorphic function w 7→ 0 tx−1 (1 − t)w−1 dt on Re w > 0. By the uniqueness theorem for holomorphic functions it follows that for all x > 0 and w ∈ C, Re w > 0, we have Γ(x)Γ(w) Γ(x+w)

Z

0

1

tx−1 (1 − y)w−1 dt =

Γ(x)Γ(w) . Γ(x + w)

(∗)

Now we fix w, Re w > 0, and extend both sides in (∗) to holomorphic functions in R1 Re z > 0, i.e. z 7→ 0 tz−1 (1 − t)w−1 dt and z 7→ Γ(z)Γ(w) Γ(z+w) , and again the uniqueness result for holomorphic function yields that (∗) must hold for Re z > 0 and Re w > 0. 4. Once we have proved Γ(x) = 2

Z

∞

2

t2x−1 e−t dt

0

for all x > 0 we can argue as in Problem 3 to extend this equality to Re z > 0. The substitution s = t2 gives Z ∞ Z ∞
x−1 −t2 Γ(x) = sx−1 e−s ds = e 2t dt t2 0 0 Z ∞ 2 t2x−1 e−t dt. =2 0

724

SOLUTIONS TO PROBLEMS OF PART 7

5. We know that z 7→

1 Γ(z)

is an entire function and Γ is a meromorphic function in 0

(z) is a meromorphic C\ (−N0 ) with simple poles at −n, n ∈ N0 . Thus ψ(z) := ΓΓ(z) function with poles at −k, k ∈ N0 . Since about −k we have the Laurent expansion (compare with (26.10))

Γ(z) =

(−1)k 1 + T (z + b) k! z + k

where T is the regular part, we find Γ0 (z) =

(−1)k+1 1 + T 0 (z + k), k! (z + k)2

0

(z) has a simple pole at −k. Using the functional equation and it follows that z 7→ ΓΓ(z) of the Γ-function for x > 0 we find

d d ln Γ(x + 1) = ln (xΓ(x)) dx dx or d d 1 ln Γ(x + 1) = (xΓ(x)) dx xΓ(x) dx 1 1 Γ0 (x) 1 = (Γ(x) + xΓ0 (x)) = + = + ψ(x). xΓ(x) x Γ(x) x

ψ(x + 1) =

This identity extends to the half plane Re z > 0. Similarly we have x ∈ (0, 1) d  π  d ln (Γ(x)Γ(1 − x)) = ln dx dx sin πx or 1 d 1 d π (Γ(x)Γ(1 − x)) = π . Γ(x)Γ(1 − x) dx sin πx dx sin x For the left hand side we find

d Γ0 (x)Γ(1 − x) − Γ(x)Γ0 (1 − x) 1 (Γ(x)Γ(1 − x)) = Γ(x)Γ(1 − x) dx Γ(x)Γ(1 − x) Γ0 (x) Γ0 (1 − x) = − = ψ(x) − ψ(1 − x), Γ(x) Γ(1 − x) and the right hand side gives 1 π sin πx

π d = π cot πx. dx sin πx

Thus we have for x ∈ (0, 1) ψ(x) − ψ(1 − x) = π cot πx which by the uniqueness theorem extends to the holomorphy domain of both sides.

725

A COURSE IN ANALYSIS

6. We use the product representation of Γ(z), i.e. (26.16), and for x > 0 we obtain x ∞ e−γx Y e k x 1 + xk

d d ψ(x) = ln Γ(x) = ln d dx

k=1

!

.

We note that ln

x N eγx Y e k x 1 + kx

k=1

!

= −γx − ln x +

N X x

k=1

k

−

N X

k=1

 x ln 1 + k

and differentiation yields x N e−γx Y e k x 1 + xk

d ln dx

k=1

!

N N 1 X1 X 1 + − x k x+k k=1 k=1   N 1 X 1 1 = −γ − − − . x x+k k

= −γ −

k=1

PN Since k=1 we arrive at



1 x+k

−

1 k



=−

PN

x k=1 k2 +kx

we have convergence for x > 0 and hence ∞

1 X ψ(x) = −γ − − x k=1



1 1 − x+k k



which again extends to the half plane Re z > 0. 7.

a) This is trivial. From the definition we deduce that if some νj > 1 then α ν νj = 2 + αj , αj ≥ 0, and pj j = p2j pj j .

b) Let m and n be relative prime. In the case where m or n has a prime square factor then mn also has a prime square factor. Hence µ(m · n) = 0 as well as µ(m)µ(n) = 0. If both m and n have no square factor, then we can write m = p1 ·. . .·pr and n = q1 ·. . .·qs with distinct primes p1 , . . . , pr , q1 , . . . , qs . Consequently we have µ(m) = (−1)r , µ(n) = (−1)s and since m · n = p1 · · · pr q1 · · · qs we find µ(mn) = (−1)r+s = µ(m)µ(n), i.e. µ is multiplicative. The example µ(2)µ(2) = 1 but µ(4) = 0 shows that µ is not completely multiplicative.

8. (Following [5]) With f = f1 · f2 we have to prove that f (n) = 0 for all n ∈ N. Let n0 be the smallest natural number with f (n) 6= 0. It follows that Df (s) =

∞ ∞ X X f (n0 ) f (n) f (n) = + s s n n ns 0 n=n n=n +1 0

0

or f (n0 ) = ns0 Df (s) − ns0

726

∞ X

n=n0

f (n) . ns +1

SOLUTIONS TO PROBLEMS OF PART 7

For s = sk we have Df (sk ) = 0, hence f (n0 ) = −ns0k

∞ X

n=n0

f (n) . nsk +1

Using the estimate from the hint we find for k such that σk > c > σa |f (n0 )| ≤ nσ0 k (n0 + 1)−σk −c ≤



n0 n0 + 1

σk

∞ X

n=n0 +1

|f (n)|n−c

κ,

 σk 0 where κ is independent of k. Since limk→∞ n0n+1 = 0 we arrive at f (n0 ) = 0 which is a contradiction, i.e. f (n) = 0 for all n ∈ N. Finally we prove the estimate given in the hint. We have ∞ ∞ X g (n) X |g(n)| −(σ−c) n ≤ s n nc n=k

n=k

≤ k −(σ−c)

∞ X |g(n)| . nc

n=k

9. Note that ns = nσ nit = eσ ln n eit ln n for s = σ + it. We start with 2 Z R X Z R X N X N N an ak an dt dt = s n ns k s −R −R n=1 k=1

n=1

=

Z

R

N X N X

ak ak e−σ ln n e−σ ln k eit ln n e−it ln k dt

−R n=1 k=1

N X N X

=

an ak e−σ ln n e−σ ln k

Z

R

eit ln n e−t ln k dt.

−R

n=1 k=1

For n = k it follows that an ak e−σ ln n e−σ ln k

Z

R

−R

eit ln n e−i ln k dt = 2T |an |2 e−2σ ln =

However for n 6= k we find Z R eit(ln n−ln k) dt = −R

n n
1 eiR ln k − e−iR ln k i ln nk 2 sin R ln nk . = ln nk

727

2R|an |2 . n2σ

A COURSE IN ANALYSIS

Thus we find 2 N N X 1 X 2R|an |2 an dt = s 2R n=1 n2σ −R n=1 n

Z

1 2R

R

+

N 1 X an ak 2 sin R ln nk 2R nσ k σ ln nk n6=k




N

N X

|an |2 X an ak sin R ln nk
+ = n2σ nσ k σ R ln nk n=1 n6=k





.

Since σ > 1 and |an | ≤ M we obtain in the limit N → ∞ and R → ∞ that 1 lim R→∞ 2R

∞ ∞ X a 2 X |an |2 n dt = . σ+it n2σ −R n=1 n n=1

Z

R

10. For 0 < x < π2 we have sin x < x < tan x or cot x < x < sin1 x , which gives kπ cot2 x < x12 < sin12 x = 1 + cot2 x. For x = 2m+1 , k = 1, . . . , m, it follows that m X

2

cot

k=1



kπ 2m + 1

We claim that

For y ∈ 0,

π 2






  m m X (2m + 1)2 X 1 kπ 2

n 2.

10. Using (27.55) we find for λ > 0 that X

1

2 + (λz) w∈L\{0}  X 1 1 = 2 2+ λ z

℘ (λz; λw1 , λw2 ) =

w∈L\{0}

=

1 (λz − λw)

2

−

2

(λw)    1  1 − 2 (z − w)2 w

1 ℘(z; w1 , w2 ) = λ−2 ℘(z; w1 , w2 ). λ2

735

1

!

A COURSE IN ANALYSIS

11. Note that in our situation with w1 = 1 and w2 = τ we have X 1 Gn (τ ) = Gn (1, τ ) = n. (k1 + k2 τ )

(∗)

(k1 ,k2 )∈Z\{0}

Since k1 + k2 (τ + 1) = k1 + k2 + k2 τ a rearrangement of the summation in (∗) with k1 + k2 replacing k1 we see that Gn (τ ) has period 1, i.e. Gn (τ + 1) = Gn (τ ). In addition we have   n −1 n k1 + k2 = τ −1 (k1 τ − k2 ) τ and replacing in (∗) the summation over (k1
, k2 ) by the summation over (−k2 , k1 ) we obtain the relation Gn (τ ) = τ −n Gn − τ1 .

Chapter 28

1. The mapping W1 is continuous on C\{−i}, in particular it is continuous when restricted to ∂H and ∂H can be identified by R, recall H = {z ∈ C | Im z > 0}. For |i−x| = 1 implying that z = x ∈ R we have |x − i| = |x + i| which implies |W1 (x)| = |i+w| W1 (∂H) ⊂ ∂D. Note that with some extra effort, see [83], it is possible to show that W1 maps ∂H bijectively onto ∂D\{−1}. 2. We decompose w(z) = we have

1+z 1−z

into its real and imaginary parts, i.e. with z = x + iy

(1 + z)(1 − z) |1 − z|2
1 − x2 + y 2 2y = + i. (1 − x)2 + y 2 (1 − x)2 + y 2  For z ∈ D∩H = x + iy ∈ C | x2 + y 2 < 1, y > 0 we find Re w(z) = w(z) = w(x, y) =

and Im w(z) =

2y (1−x)2 +y 2

1−(x2 +y 2 ) (1−x)2 +y 2

>0

> 0, i.e. w(D ∩ H) ⊂ K := {x + iy ∈ C | x > 0, y > 0}.

The inverse M¨ obius transformation of w : C\{1} → C\{−1} is v(ζ) = ζ−1 ζ+1 and since for ζ = ξ + iη, ξ > 0 and η > 0, we have |ζ + 1| > |ζ − 1|, see the figure below, we conclude that |v(ζ)| < 1, i.e. v(K) ⊂ D.

K |ζ + 1|

b

ζ = ξ + iη

|ζ − 1|

−1

1

736

SOLUTIONS TO PROBLEMS OF PART 7

For v(ζ) we have the decomposition v(ξ, η) = v(ζ) =

ξ 2 + η2 − 1 2η (ζ − 1)(ζ + 1) = + i 2 2 2 |ζ + 1| (ξ + 1) + η (ξ + 1)2 + η 2

implying that Im v(ζ) > 0 for (ξ, η) ∈ K. Hence we have proved that v(K) ⊂ D ∩ H and together with the first part we conclude that w(D ∩ H) = K. It remains to remark that w : C\{1} → C\{−1} and v : C\{−1} → C\{1} are holomorphic functions. 3. We can adopt the arguments from the proof of Lemma 28.9. Let g : G1 → G2 be a biholomorphic mapping with biholomorphic inverse g −1 : G2 → G1 and define for W ∈ Aut(G1 ) Φ : Aut(G1 ) → Aut(G2 )

Φ(W ) := g ◦ W ◦ g −1 .

We claim that Φ is a group isomorphism. First we note for W, V ∈ Aut(G1 ) that Φ (W ◦ V ) = g ◦ (W ◦ V ) ◦ g −1

= g ◦ W ◦ g −1 ◦ g ◦ V ◦ g −1 = Φ(W ) ◦ Φ(V ),

and further Φ (idG1 ) = g ◦ idG1 ◦ g −1 = g ◦ g −1 = idG2 . Next we note that Φ(W )−1 = g ◦ W ◦ g −1


−1

= g −1 ◦ W −1 ◦ g

and both Φ(W ) as well as Φ(W )−1 are biholomorphic mappings. Hence Φ is a group isomorphism with inverse Φ−1 : Aut(G2 ) → Aut(G1 ) given by Φ−1 (V ) = g −1 ◦V ◦g. 4. We decompose fA (z) into its real and imaginary part and find fA (z) =

ac|z|2 + (bc + ad) Re z + bd (ad − bc) Im z +i |cz + d|2 |cz + d|2

and since ad − bc = 1 we get Im fA (z) > 0 for z ∈ H, i.e. fA (H) ⊂ H. Since this holds for all A ∈ SL(2; R) it follows already that fA (H) = H. Next we note that by Lemma 16.24 we have for A, A0 ∈ SL(2; R) the relations fA ◦ fA0 = fA◦A0 , and fA−1 = (fA )−1 . Hence fA is for every A ∈ SL(2; R) an automorphism of H. 5. According to Theorem 28.13 it is sufficient to show for every compact set K ⊂ G that



≤ MK,k < ∞. (∗) sup f (b) f ∈H

∞,K

737

A COURSE IN ANALYSIS

Of course (∗) will follows if we can prove the estimate for all closed balls B2r (ζ) ⊂ G. Since H is normal we can find MB2r (ζ) such that sup kf k∞,B2r (ζ) ≤ MB2r (ζ) .

f ∈H

The standard estimates, i.e. Theorem 21.15, imply

(k)

f

∞,Br (ζ)

for all f ∈ H, hence



sup f (k)

≤

∞,Br (ξ)

f ∈H

2MB2r (ζ) k! rk

≤

2MB2r (ζ) k! rk

and Theorem 28.13 implies the normality of Hk .

6. Since for all k ∈ N the functions fk := Wk − a are injective and Wk (z1 ) = a, we can find η > 0 such that with z2 as in the proof of Theorem 28.17 it follows that inf

inf

k∈N z∈∂Br (z2 )

|fk (z)| ≥ η > 0.

Since gk := (J − a) − (Wk − a) converges uniformly on Br (z2 ) to zero we can find k0 ∈ N such that |gk0 (z)| < |fk0 (z)|

for all z ∈ ∂Br (z2 ).

This implies by the theorem of Rouch´e, Theorem 24.26, that fk0 = Wk0 − a and fk0 + gk0 = J − a have in Br (z2 ) the same number of zeroes, i.e. J − a has no zero in Br (z2 ). 7. With W1 (z) = W1 (x, y) = u(x, y) + iv(x, y) we find g(x, y) = h(u(x, y), v(x, y)) = u2 (x, y) − v 2 (x, y). Since W1 (z) =

i−z i+z

we find further

W1 (x, y) = u(x, y) + iv(x, y) =

2x 1 − (x2 + y 2 ) +i 2 2 2 x + (y + 1) x + (y + 1)2

implying that g(x, y) =


!2
2 1 − x2 − y 2 x2 + y 2 − 6x2 − 2y 2 + 1 4x2 = − x2 + (y + 1)2 (x2 + (y + 1)2 )2 (x2 + (y + 1)2 )2

 is harmonic in (x, y) ∈ R2 | y > 0 .

738

SOLUTIONS TO PROBLEMS OF PART 7

Chapter 29

P

 P |c | implies the convergence of |α|≤N cα = C. α |α|≤N N ∈N PM We have to prove that cϕ(k) = C. For  > 0 we know the existence PlimM→∞ k=0 of N0 ∈ N such that |α|≥N0 |cα | < 2 which yields

1. The convergence of

X X X  C − c c = |c|α| | < . α α ≤ 2 |α|≤N0 −1 |α|≥N0 |α|≥N0

Now we take N such that {α | |α| ≤ N0 − 1} ⊂ {ϕ(β) | |β| ≤ N }. For m ≥ N we find m m X X X X cϕ(k) − C ≤ cϕ(k) − cα + cα − C |α|≤N0 −1 k=0 k=0 |α|≤N0 −1 X X ≤ |cα | + cα − C |α|≤N0 −1 |α|≥N0   + = 2 2 and the result is proved.

P P P P α2 α1 and 2. a) We have α∈N2 z α = α1 ,α2 ∈N0 z1α1 z2α2 = α2 ∈N0 z2 α1 ∈N0 z1 0 P P α1 α2 1 1 for |z1 | < 1, |z2 | < 1 we find α1 ∈N0 z1 = 1−z1 as well as α2 ∈N0 z2 = 1−z 2 implying that X 1 zα = (1 − z )(1 − z2 ) 1 2 α∈N0

for (z1 , z2 ) ∈ B1 (0) × B1 (0). P P P b) We note that (1,α2 )∈N2 z α = α2 ∈N0 z1 z2α2 = z1 α2 ∈N0 z2α2 and therefore 0 we obtain for (z1 , z2 ) ∈ C × B1 (0) that X

zα =

(1,α)∈N20

z1 . 1 − z2

P P ν c) Since ν∈N0 z1ν z2ν = ν∈N0 (z1 z2 ) we deduce that for |z1 z2 | < 1 the series converges and has limit 1−z11 z2 , i.e. X

z1ν z2ν =

ν∈N0

739

1 . 1 − z1 z2

This page intentionally left blank

References [1] Abramowitz, M., Stegun, J. A., (eds.), Handbook of Mathematical Functions. 7th printing. Dover Publications, New York 1970. [2] Agricola, J., Friedrich, T., Elementary Geometry. American Mathematical Society, Providence R.I. 2008. [3] Ahlfors, L.V., Complex Analysis. 2nd ed. McGraw-Hill Book Company, New York 1966. [4] Ahlfors, L., Conformal Invariants. Topics in Geometric Function Theory. McGrawHill Book Company, New York 1973. [5] Apostol, T.M., Introduction to Analytic Number Theory. Springer Verlag, Berlin 1976. [6] Appell, J., Analysis in Beispielen und Gegenbeispielen. Springer Verlag, Berlin · Heidelberg 2009. [7] Appell, J., V¨ ath, M., Elemente der Funktionalanalysis. Vieweg Verlag, Wiesbaden 2005. [8] Armstrong, M.,A., Basic Topology. McGraw-Hill Book Company (U.K.), London 1979. [9] Bartoszynski, T., Judah, H., Set Theory. On the Structure of the Real Line. A. K. Peters, Wellesly MA 1995. [10] Bauer, H., Wahrscheinlichkeitstheorie und Grundz¨ uge der Maßtheorie. 2. Aufl. Walter de Gruyter Verlag, Berlin 1974. English: Probability Theory and Elements of Measure Theory. 2nd ed. Academic Press, London · New York, 1981. [11] Bauer, H., Maß-und Integrationstheorie. Walter de Gruyter Verlag, Berlin 1990. English: Measure and Integration Theory. Walter de Gruyter Verlag, Berlin 2001. [12] Beals, R., and Wong, R., Special Functions. A Graduate Text. Cambridge University Press, 2010. [13] Behrends, E., Maß-und Integrationstheorie. Springer Verlag, Berlin 1987. [14] Benedetto, J.J., Czaja, W., Integration and Modern Analysis. Birkh¨ auser Verlag, Boston · Basel · Berlin 2009.

[15] Billingsley, P., Probability and Measure. 3rd ed. John Wiley & Sons, New York 1995. [16] Carath´eodory, C., Funktionentheorie. 1. Band. 2. Aufl. Birkh¨ auser Verlag, Basel · Stuttgart 1960. English: Theory of Functions of a Complex Variable Volume 1. 2nd ed. Chelsea Publishing Company, New York 1964.

[17] Carath´eodory, C., Funktionentheorie. 2. Band. 2. Aufl. Birkh¨ auser Verlag, Basel · Stuttgart 1961. English: Theory of Functions of a Complex Variable. Volume 2. 2nd ed. Chelsea Publishing Company, New York 1960.

741

A COURSE IN ANALYSIS

[18] Cerny, J., Foundations of Analysis in the Complex Domain. Academia, Praha 1992. [19] Deimling, K., Nichtlineare Gleichungen und Abbildungsgrade. Springer Verlag, Berlin 1974. [20] Dinghas, A., Vorlesungen u ¨ber Funktionentheorie. Springer Verlag, Berlin 1961. [21] Dudley, R.M., Real Analysis and Probability. Wadsworth & Brooks, Pacific Grove CA 1989. [22] Duren, P., Invitation to Classical Analysis. American Mathematical Society, Providence R.I. 2012. [23] Falconer, K.J., Fraktale Geometrie. Spektrum Akademischer Verlag, Heidelberg 1993. English: Fractal Geometry. Mathematical Foundations and Applications. John Wiley & Sons, New York 1990. [24] Falconer, K.J., The Geometry of Fractal Sets. Cambridge University Press, paperback reprint, Cambridge 1995. [25] Federer. H., Geometric Measure Theory. Reprint: Classics in Mathematics. SpringerVerlag, Berlin 1996. [26] Fischer, W., Lieb, I., Funktionentheorie. 5. Aufl. Vieweg Verlag, Braunschweig · Wiesbaden 1988. [27] Folland, G.B., Real Analysis. Modern Techniques and Their Applications. 2nd ed. John Wiley & Sons, New York 1999. [28] Freitag, E., Busam, R., Funktionentheorie 1. 4th ed. Springer Verlag, Berlin 2006. [29] Fritsche, K., Grauert, H., From Holomorphic Functions to Complex Manifolds. Springer Verlag, Berlin 2002. [30] Gaier, D., Vorlesungen u ¨ber Approximation im Komplexen. Birkh¨ auser Verlag, Basel · Boston · Stuttgart 1980. [31] Gorenflo, R., Kilbas, A.A., Mainardi, F., Rogosini, S.V., Mittag-Leffler Functions, Related Topics and Applications. Springer Verlag, Berlin 2014. [32] Hardy, G.H., Wright, E.M., An Introduction to the Theory of Numbers. 5th ed. Clarendon Press, Oxford 1979, several corrected reprints. [33] Heins, M., Selected Topics in the Classical Theory of Functions of a Complex Variable. Holt, Rinehart and Winston, New York 1962. [34] Heins, M., Complex Function Theory. Academic Press, New York 1968. [35] Hewitt, E., Stromberg, K., Real and Abstract Analysis. Springer Verlag, 1978. [36] Hille, E., Analytic Function Theory. Volume I. Ginn and Company, Boston MA 1959. [37] Hille, E., Analytic Function Thoery. Volume II. Ginn and Company, Boston MA 1962.

742

REFERENCES

[38] Hirzebruch, F., Scharlau, W., Einf¨ uhrung in die Funktionalanalysis. B. I.Wissenschaftsverlag, Mannheim · Wien · Z¨ urich 1971. [39] Ingham, A.E., The Distribution of Prime Numbers. Reprinted in “Cambridge Mathematical Library”, Cambridge University Press, Cambridge 1995. [40] Jacob, N., Knopova, V., Landwehr, S., Schilling, R.L., A Geometric Interpretation of the Transition Density of a Symmetric L´evy Process. Sci. China Math. 55(2012), 1099-1126. [41] Jacobi, C.G.J., Fundamenta nova theoria functionum elliptarium. Gebr¨ uder Borntr¨ ager, K¨ onigsberg 1829. [42] Jacobs, K., Measure and Integral. Academic Press, New York 1978. [43] Kaczor, W.J., and Nowak, M.T., Problems in Mathematical Analysis I. American Mathematical Society, Providence R.I., 2000. [44] Kaczor, W.J., and Nowak, M.T., Problems in Mathematical Analysis II. American Mathematical Society, Providence R.I., 2001. [45] Kaczor, W.J., and Nowak, M.T., Problems in Mathematical Analysis III. American Mathematical Society, Providence RI 2003. [46] Kharazishvili, A.B., Strange Functions in Real Analysis. 2nd ed. Chapman & Hall, Boca Raton FL 2006. [47] Knapp, A. W., Basic Real Analysis. Birkh¨ auser Verlag, Boston · Basel · Berlin 2005. [48] Koecher, M., Krieg, A., Elliptische Funktionen und Modulformen. Springer Verlag, Berlin 1998. [49] Kolmogoroff, A.N., Grundbegriffe der Wahrscheinlichkeitsrechnung. Springer Verlag, Berlin 1933. [50] Kolmogorov, A.N., Fomin, S.V., Reelle Funktionen und Funktionalanalysis. VEB Deutscher Verlag der Wissenschaften, Berlin 1975. [51] Krantz, S.G., Parks, H.R., A Primer of Real Analytic Functions. Birkh¨ auser Verlag, Basel · Boston · Berlin 1992. [52] Kuttler, K.L., Modern Analysis. CRC Press, Boca Raton FL 1998. [53] Lang, S., Elliptic Functions. 2nd ed. Springer Verlag, Berlin 1987. [54] Lebesgue, H., Int´egrale, Longueur, Aire. These Paris 1902. Also: Ann. Mat. Pura Appl. 7(1902), 231-359. [55] Magnus, W., Oberhettinger, F., Soni, R.P., Formulas and Theorems for Special Functions of Mathematical Physics. 3rd ed. Springer Verlag, Berlin 1966. [56] Malliavin, P., Integration and Probability. Springer Verlag, Berlin 1995. [57] Milnor, J.W., Topology from the Differential Viewpoint. The University Press of Virginia, Charlottesville, 5th printing, 1978.

743

A COURSE IN ANALYSIS

[58] Narasimhan R., Analysis on Real and Complex Manifolds. North-Holland Publishing Company, Amsterdam, 3rd printing, 1985. [59] Natanson, J.P., Theorie der Funktionen einer reelen Ver¨ anderlichen. 4. Aufl. Verlag Harri Deutsch, Z¨ urich · Frankfurt · Thun 1975. [60] Nevanlinna, R., Uniformisierung. Springer Verlag, Berlin 1953. [61] Nevanlinna, R., Paatero, V., Einf¨ uhrung in die Funktionentheorie. Birkh¨ auser Verlag, Basel · Stuttgart 1965. English: Introduction to Complex Analysis. Addison-Wesley Publishing Company, Reading MA 1969. [62] Nielsen, N., Trait´e ´el´ementaire des nombres de Bernoulli. Gauthier-Villars, Paris 1923. [63] Patterson, S.J., An Introduction to the Theory of the Riemann Zeta-Function. Cambridge University Press, Paperback Edition, Cambridge 1995. [64] Peschl, E., Funktionentheorie. Band 1. B.I.-Verlag, Mannheim · Wien · Z¨ urich 1967. [65] Querenburg, Boto von, Mengentheoretische Topologie. Springer Verlag, Berlin 1973. [66] Range, R.M., Holomorphic Functions and Integral Representations in Several Complex Variables. Springer Verlag, Berlin 1986. [67] Remmert, R., Funktionentheorie 1. 3. Aufl. Springer Verlag, Berlin 1992. English: Theory of Complex Functions. Springer Verlag, Berlin 1991. [68] Remmert, R., Funktionentheorie 2. 2. Aufl. Springer Verlag, Berlin 1995. English: Classical Topics in Complex Function Theory. Springer Verlag, Berlin 1998. [69] Rinow, W., Topologie. VEB Deutscher Verlag der Wissenschaften, Berlin 1975. [70] Royden, H.L., Real Analysis. Macmillan Company, New York 1963. [71] Rudin, W., Real and Complex Analysis. 2nd ed. McGraw-Hill Book Company, New York 1974. [72] R¨ uhs, F., Funktionentheorie. 2 Aufl. VEB Deutscher Verlag der Wissenchaften, Berlin 1971. [73] Saks, S., Zygmund, A., Analytic Functions. Nakladem Polskiego Towarzystwa Matematycznego, Warszawa 1952. [74] Schechter, M., An Introduction to Nonlinear Analysis. Cambridge University Press, Cambridge 2004. [75] Schilling, R.L., Measures, Integrals and Martingales. Cambridge University Press, Cambridge 2005. (2nd ed. to be published in 2017.) [76] Schilling, R.L., Mass und Integral. Walter de Gruyter Verlag, Berlin 2015. [77] Schubert, H., Topologie. 4. Aufl. Teubner Verlag, Stuttgart 1976.

744

REFERENCES

[78] Schwartz, J., Nonlinear Functionalanalysis. Gordon and Breach Science Publishers, New York 1969. [79] Shen, A., Vereshchagin, N.K., Basic Set Theory. American Mathematical Society, Providence R.I. 2002. [80] Spiegel, M.R., Theory and Problems of Complex Variables. (SI(metric) edition.) McGraw-Hill Book Company, New York 1974. [81] Srivastara, S.M., A Course on Borel Sets. Springer Verlag, Berlin 1998. [82] Stein, E.M., Shakarchi, R., Princeton Lectures in Analysis I. Fourier Analysis: An Introduction. Princeton University Press, Princeton NJ 2003. [83] Stein, E.M., Shakarchi, R., Princeton Lectures in Analysis II. Complex Analysis. Princeton University Press, Princeton NJ 2003. [84] Stein, E.M., Shakarchi, R., Princeton Lectures in Analysis III. Real Analysis. Princeton University Press, Princeton NJ 2005. [85] Stroock, D.W., A Concise Introduction to the Theory of Integration. 3rd ed. Birkh¨ auser Verlag, Boston · Basel · Berlin 1999. [86] Titchmarsh, E.C., The Theory of the Riemann Zeta-Function. Oxford at the Clarendon Press, Oxford 1951. [87] Torchinsky, A., Real Variables. Perseus Books, Reading MA 1995. [88] Tricomi, F., Krafft, M., Elliptische Funktionen. Akademische Verlagsgesellschaft, Geest & Partig, Leipzig 1948. [89] Triebel, H., H¨ ohere Analysis. VEB Deutscher Verlag der Wissenschaften, Berlin 1972. English: Higher Analysis. J. A. Barth Verlag, Leipzig 1992. [90] Voigt, A., Wloka, J., Hilbertr¨ aume und elliptische Differentialoperatoren. B.-I Verlag, Mannheim · Wien · Z¨ urich 1975. [91] Weil, A., Elliptic Functions According to Eisenstein and Kronecker. Springer Verlag, Berlin 1976. (Reprinted as Springer Classics in Mathematics 1999.) [92] Weyl, H., Die Idee der Riemannschen Fl¨ ache. Teubner Verlag 1913, 2nd corrected ed. 1923. [93] Whittaker, E.T., Watson, G.N., A Course of Modern Analysis. 4th ed., reprinted in “Cambridge Mathematical Library”, Cambridge University Press, Cambridge 1996. [94] Wheeden, R.L., Zygmund, A., Measure and Integral. An Introduction to Real Analysis. Marcel Dekker Inc., New York · Basel 1977.

745

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Mathematicians Contributing to Analysis (Continued) Ahlfors, Lars Valerian (1907-1996). Alexandrov, Pawel Sergejevich (1896-1982). Bernoulli, Johann I (1667-1748). Carath´ eodory, Constantin (1873-1950). Casorati, Felice (1835-1890). Chebyshev, Pafnuti Lwowich (1821-1894). Dynkin, Eugene Borisovich (1924-2014). Egorov, Dmitri Fedorovich (1869-1931). Eisenstein, Ferdinand Gotthold Max (1823-1852). Fatou, Pierre Joseph Louis (1878-1829). Federer, Herbert (1920-2010). Fischer, Ernst (1875-1954). Fresnel, Augustin Jean (1788-1827). Friedrichs, Kurt-Otto (1901-1982). Goursat, Edouard Jean-Baptiste (1858-1936). Grauert, Hans (1930-2011). Hardy, Godefrey Harold (1877-1947). Kolmogorov, Andrej Nikolajevich (1903-1987). Laurent, Pierre Alphonse (1813-1854). Levi, Beppo (1875-1961). Liouville, Joseph (1809-1882). Littlewood, John Edensor (1885-1977). Lusin, Nikolaj Nikolajevich (1883-1950).

747

A COURSE IN ANALYSIS

M¨ obius, August Ferdinand (1790-1868). Marcinkiewicz, Jozef (1910-1940). Markov, Andrej Andrejevich (1856-1922). Mittag-Leffler, Magnus G¨ osta (1846-1927). Montel, Paul Antoine Aristide (1876-1975). Morera, Giacinto (1856-1909). Nevanlinna, Rolf Hermann (1895-1980). Nielsen, Niels (1865-1931). Nikodym, Otton Martin (1887-1974). Pochhammer, Leo August (1841-1920). Radon, Johann Karl August (1887-1956). Reinhardt, Karl August (1895-1941). Riesz, Frigyes (1880-1956). Rouch´ e, Eugene (1832-1910). Runge, Carl David Tolm´e (1856-1927). Sard, Arthur (1909-1980). Tonelli, Leonida (1885-1946). Tricomi, Francesco Giacomo Filippo (1897-1978). Urysohn, Pawel Samuilovich (1898-1924). Vi` ete (Vieta), Fran¸cois (1540-1603). Vitali, Giuseppe (1875-1932). Weil, Andr´e (1906-1998). Weyl, Hermann Klaus Hugo (1885-1955). Wiener Norbert (1894-1964). Zygmund Antoni (1900-1992).

748

Subject Index averaging operator, 251 axioms of choice, 583 axis parallel polygonal arc, 363

α-dimensional Hausdorff measure, 58 ∅-continuous, 31 -net, 250 µ-almost everywhere convergence, 128 µ-equivalent, 129 µ-integrable, 95 µ-integral, 85, 91, 96, 99 µ-null set, 50 σ-additivity, 31 σ-algebra, 6 σ-field, 6 Borel, 9, 12 generator, 8 product, 14 σ-finite, 31 x-section, 177

Baire class, 84 Banach-Zaretzky theorem, 233 base of a topology, 579 Bernoulli distribution, 18, 94 Bernoulli numbers, 316, 593 Bessel differential equation, 418 Bessel function, 418 bi-Lipschitz mapping, 62 biholomorphic mapping, 330 binomial coefficients, 309 binomial distribution, 18 binomial series, 309 Borel σ-field, 9, 12 Borel sets, 9, 12 boundary maximum principle, 428 boundary minimum prnciple, 429 bounded, 267, 299 bounded variation, 219 branch of the logarithmic function, 336 branch of the power function, 340

Abel’s lemma, 573 Abel’s theorem, 303 abscissa of convergence, 512 absolute continuity of the integral, 225 absolute convergence, 271, 301, 440 absolute moment, 630 absolute space, 570 absolute value, 265 absolutely continuous, 219 absolutely continuous measure, 117 addition formula for elliptic functions, 531 Alexandrov-compactification, 579 algebra, 27 amplitude, 531 annulus, 445 arc, 348 axis parallel polygonal, 363 arc parameter, 348 arcwise connected set, 348 arcwise connected topological space, 349 argument, 286 argument principle, 468 arithmetical functions, 511 automorphism group, 559 average, 235

Cantor set, 63 Cantor-Lebesgue function, 65 Casorati and Weierstrass’ theorem, 463 Cauchy and Hadamard’s theorem, 305 Cauchy criterion, 271 Cauchy distribution, 117 Cauchy integral formula (generalised), 397 Cauchy integral formula for null homologous cycles, 482 Cauchy product, 275 Cauchy sequence, 270 Cauchy’s inequalities, 401 Cauchy’s integral theorem, 389 Cauchy-Riemann equations, 322 chain closed, 479 cycle, 479

749

A COURSE IN ANALYSIS

chain of curves, 477 chain rule, 320 change of parameter, 371 Chebyshev-Markov inequality, 133 choice function, 583 closed chain, 479 coefficients, 303 comparison test, 273 complementary modulus, 531 complete elliptic integral of the first kind, 532 complete elliptic integral of the second kind, 532 complete measure space, 52 complete Reinhardt domain, 571 completely multiplicative function, 511 completion, 52 complex derivative, 319 Cauchy-Riemann equations, 322 chain rule, 320 Leibniz rule, 320 quotient rule, 320 complex differentiable, 319 complex number absolute value, 265 argument, 286 conjugate, 264 imaginary part, 261 modulus, 265, 286 polar representation, 287 real part, 261 conformal mapping, 567 conjugate, 264 conjugate harmonic function, 332 connected topological space, 347 connectivity component, 348 content, 28 continuous, 31 continuous from above, 31 continuous from below, 31 continuous function, 299 convergence µ-almost everywhere, 128 absolute, 301, 440 local uniform, 433

norm, 249 normal, 573 pointwise, 127 stochastic, 132 strong, 249 uniform, 301, 440 weak, 247 with respect to a semi-norm, 136 convergence in measure, 132 convergence in the pth mean, 132 convergence in the mean, 132 convergence in the quadratic mean, 132 converse triangle inequality, 266 convolution, 196, 208 convolution kernel, 212 convolution operator, 212, 214 cosine-transform, 154 cosines amplitudines, 531 countable set, 6 critical point, 243 critical strip of the Riemann ζ-function, 522 critical values, 243 cross ratio, 590 curve arc, 348 arc parameter, 348 chain, 477 exterior of, 476 index, 474, 480 initial point, 348 interior of, 476 parameter interval, 348 parametric, 348 piecewise continuously differentiable, 366 rectifiable, 365 simply closed, 349 sum, 350 terminal point, 348 total variation, 365 trace, 348 winding number, 474 Z-variation, 365 cut plane, 337

750

SUBJECT INDEX

cycle, 479 null-homologous, 480

Eisenstein series, 548 elementary function, 80 elementary transcendental functions, 335 elliptic function, 539 elliptic function in the sense of Jacobi, 529 elliptic integral, 528 entire function, 319 enumeration of Nn0 , 569 equi-continuous, 254 equivalent metrics, 72 equivalent paths, 351 essential bound, 625 essential singularity, 459 essential supremum, 135 essentially bounded function, 135 Euclid’s theorem, 510 extended complex plane, 297 exterior of a curve, 476

de Moivre’s formula, 288 degree, 300 density, 110, 117 denumerable set, 6 derivative complex, 319 Dini, 220 left, 220 lower left, 220 lower right, 220 right, 220 upper left, 220 upper right, 220 difference quotient, 319 differentiable, 319 differentiable from the left, 220 differentiable from the right, 220 differential equation Bessel, 418 hypergeometric, 418 Dini derivatives, 220 Dini numbers, 220 Dirac measure, 17 Dirichlet series, 512 discrete subset, 409 distinguished boundary, 570 distribution, 105 Bernoulli, 18, 94 binomial, 18 Cauchy, 117 Gauss, 117 normal, 117 Poisson, 18, 95 distribution function, 187 domain, 319 complete Reinhardt, 571 of convergence, 572 Reinhardt, 570 domain of convergence, 572 dominated convergence theorem, 141 double periodic function, 531, 539 Dynkin system, 34

finite measure, 16 finite measure space, 16 first fundamental group, 361 Fischer-Riesz theorem, 143 fractal, 60 Fresnel integrals, 390 Friedrichs mollifier, 205 Fubini’s theorem, 182 function µ-almost everywhere convergence, 128 µ-equivalent, 129 absolutely continuous, 219 arithmetical, 511 Bessel, 418 bounded, 299 bounded variation, 219 branch of the logarithmic function, 336 branch of the power function, 340 Cantor-Lebesgue, 65 choice, 583 completely multiplicative, 511 complex differentiable, 319 conjugate harmonic, 332

751

A COURSE IN ANALYSIS

continuous, 299 convergence in measure, 132 convergence in the pth mean, 132 convergence in the mean, 132 convergence in the quadratic mean, 132 convolution, 196 critical point, 243 density, 110 distribution, 187 double periodic, 531, 539 elementary, 80 elementary transcendental, 335 elliptic, 539 elliptic in the sense of Jacobi, 529 entire, 319 essential singularity, 459 essential supremum, 135 essentially bounded, 135 Hardy-Littlewood maximal, 236 harmonic, 330 holomorphic, 319 holomorphic on a compact set, 434 isolated singularity, 458 jump, 218 Lebesgue singular, 65 line integral, 367 local primitive, 372 locally biholomorphic, 427 locally constant, 325 locally integrable, 235 M¨ obius, 524 meromorphic, 461 Mittag-Leffler, 315 multiplicative, 511 numerical, 76 orientation preserving parameter transformation, 351 orientation reversing parameter transformation, 351 pointwise convergence, 127 pole, 459 primitive, 372 primitive period, 537 principal part of the logarithmic

function, 338 probabilistic distribution, 187 rational, 300 removable singularity, 459 residue, 456 Riemann ζ, 512 root, 341 simple, 79 simply periodic, 537 singular, 240 support, 581 variation, 219 Weierstrass, 241 Weierstrass ℘-function, 545 fundamental parallelogram, 538 fundamental theorem of algebra, 431 fundamental theorem of arithmetic, 510 fundamental theorem of calculus, 231 Gauss distribution, 117 Gauss hypergeometric series, 421 generalised circles, 589 generalised hypergeometric series, 421 generator of a σ-field, 8 Goursat’s theorem, 383 Gram determinant, 63 group automorphism, 559 first fundamental, 361 metric, 73 topological, 73 H¨ older continuous mapping, 61 H¨ older exponent, 61 H¨ older’s inequality, 88, 91, 98 Hardy-Littlewood maximal function, 236 Hardy-Littlewood maximal theorem, 236 harmonic, 330 Hausdorff dimension, 60 Hausdorff maximality principle, 584 Hausdorff space, 580 holomorphic, 319, 574 holomorphic extension, 329 holomorphic function on a compact set, 434

752

SUBJECT INDEX

holomorphically equivalent, 330, 556 homotopic mapping, 355 homotopic path, 358 homotopically equivalent topological spaces, 357 homotopy, 355 relative, 358 homotopy class, 356 homotopy type, 357 hypergeometric differential equation, 418 hypergeometric series, 311, 420

kernel, 212 Kolmogorov-Riesz theorem, 253 lattice, 539 Laurent series, 454 Lebesgue integral, 106 Lebesgue integration, 106 Lebesgue measure, 53 Lebesgue set, 53, 239 Lebesgue singular function, 65 Lebesgue’s differentiation theorem, 237 Lebesgue-Borel measure, 43 left derivative, 220 Leibniz rule, 320 lemma Abel’s, 573 Schwarz’, 429 Urysohn’s, 200 Zorn’s, 584 lemniscate, 552 limit, 267 linear transformation, 311 Liouville’s theorem, 433 Liouville’s theorem (1st ), 539 Liouville’s theorem (2nd ), 540 Liouville’s theorem (3rd ), 541 local primitive, 372 local uniform convergence, 279, 433 locally biholomorphic function, 427 locally constant function, 325 locally integrable, 235 logarithmically convex, 574 lower left derivative, 220 lower right derivative, 220 Lusin property, 232 Lusin’s theorem, 247

image measure, 48, 102 imaginary part, 261 index of a curve, 474, 480 inequality Chebyshev-Markov, 133 converse triangle, 266 H¨ older’s, 88, 91, 98 Jensen’s, 158 Minkowski’s, 88, 91, 98 Minkowski’s integral, 190 Young’s, 202 infinite product, 276 initial point, 348 integrable locally, 235 p-fold, 134 integral, 91, 96, 99 absolute continuity, 225 complementary modulus, 531 elliptic, 528 Fresnel, 390 Lebesgue, 106 modulus, 531 transformation theorem, 115 integration by parts, 231 interior of a curve, 476 invariant, 290 involution, 264 isolated singularity, 458

M¨ obius function, 524 M¨ obius transformation, 311 mapping arc, 348 bi-Lipschitz, 62 biholomorphic, 330 conformal, 567 H¨ older continuous, 61 homotopic, 355

Jensen’s inequality, 158 Jordan curve theorem, 362 jump function, 218

753

A COURSE IN ANALYSIS

involution, 264 measurable, 12 measure, 16 null-homotopic, 357 path, 348 representation, 290 Marcinkiewicz integral, 193 mean-value theorem, 395 measurable mapping, 12 measurable sets, 4, 6 measurable space, 6 measure, 4, 16 α-dimensional Hausdorff, 58 absolutely continuous, 117 continuous, 31 continuous from above, 31 continuous from below, 31 convolution, 208 Dirac, 17 finite, 16 image, 48, 102 integral, 85, 91, 96 Lebesgue, 53 Lebesgue-Borel, 43 measure zero locally, 68 metric outer, 55 monotonicity, 19 outer, 39 probability, 16 product, 176 regular, 586 measure space, 16 complete, 52 completion, 52 finite, 16 meromorphic function, 461 metric equivalent, 72 product, 72 metric group, 73 metric outer measure, 55 metric space product, 72 metric vector space, 73 Minkowski’s inequality, 88, 91, 98

Minkowski’s integral inequality, 190 Mittag-Leffler function, 315 modulus, 265, 286, 531 monotone class, 179 monotone class lemma, 179 monotone convergence theorem, 92, 148 monotonicity of measures, 19 Montel’s theorem, 561 Morera’s theorem, 407 motions, 48 multiplicative function, 511 multiplicity of a pole, 460 natural projection, 570 norm, 266 norm convergent, 249 normal convergence, 573 normal distribution, 117 normal family, 561 normal representation, 80 normal set, 561 normal topological space, 581 null-homologous cycle, 480 null-homotopic mapping, 357 null-homotopic path, 358 numerical function, 76 one-point compactification, 579 open ball, 570 open disc, 293 operator convolution, 212, 214 kernel, 212 order of a pole, 460 outer measurable set, 39 outer measure, 39 p-fold integrable, 134 parameter interval, 348 parametric curve, 348 partial sum, 271 partition of unity, 213 path, 348 equivalent, 351 homotopic, 358 null-homotopic, 358

754

SUBJECT INDEX

pathwise connected set, 348 pathwise connected topological space, 349 Peano curve, 69 piecewise continuously differentiable curve, 366 Pochhammer symbols, 310 point at infinity, 580 pointwise convergence, 127, 277 Poisson distribution, 18, 95 Poisson kernel, 333 polar coordinates, 285 polar representation, 287 pole, 459 multiplicity, 460 order, 460 polydisc, 570 polygonal arc, 363 polynomial, 300 degree, 300 power series, 303 coefficients, 303 continuation, 329 holomorphic extension, 329 radius of convergence, 304 pre-measure, 28 primitive, 372 primitive period of a function, 537 principal part, 454 principal part of the logarithmic function, 338 probabilistic distribution function, 187 probability density, 117 probability measure, 16 probability space, 16 product σ-field, 14 product (measure) space, 176 product measure, 176 product metric, 72 product metric space, 72 product topology, 579 projection natural, 570 punctured disc, 445

radius of convergence, 304 Radon-Nikodym’s theorem, 120, 122 random variable, 105 ratio test, 273 rational function, 300 real part, 261 rectifiable curve, 365 reflection at the circle, 590 region, 319, 347 regular measure, 586 regular values, 243 Reinhardt domain, 570 relative homotopy, 358 relatively compact sets, 250 relatively prime numbers, 511 removable singularity, 459 representation, 290 residue of a function, 456 residue theorem, 467 residue theorem for null-homologous cycles, 486 Riemann ζ-function, 512 Riemann conjecture, 522 Riemann hypothesis, 522 Riemann mapping theorem, 563 Riemann sphere, 297 Riemann’s theorem on removable singularities, 409 right derivative, 220 ring, 27 root function, 341 Rouch´e’s theorem, 469 Runge’s approximation theorem, 437 Sard’s theorem, 243 Schwarz’ lemma, 429 sequence bounded, 267 Cauchy, 270 convergence with respect to a semi-norm, 136 infinite product, 276 limit, 267 local uniform, 279 series absolute convergence, 271

quotient rule, 320

755

A COURSE IN ANALYSIS

binomial, 309 Cauchy criterion, 271 Cauchy product, 275 comparison test, 273 Dirichlet, 512 Eisenstein, 548 Gauss hypergeometric, 421 generalised hypergeometric, 421 hypergeometric, 311, 420 Laurent, 454 partial sum, 271 principal part, 454 ratio test, 273

splitting, 347 star-shaped set, 348 stereographic projection, 296 stochastic convergence, 132 strongly convergent, 249 sum of curves, 350 support, 101 support of a function, 581 Taylor expansion, 403 terminal point, 348 theorem 1st Liouville theorem, 539 2nd Liouville theorem, 540 3rd Liouville theorem, 541 Abel’s, 303 Banach-Zaretzky’s, 233 Beppo Levi’s, 92 Casorati and Weierstrass’, 463 Cauchy and Hadamard’s, 305 Cauchy’s integral, 389 dominated convergence, 141 Euclid’s, 510 Fischer-Riesz’, 143 Fubini’s, 182 fundamental theorem of algebra, 431 fundamental theorem of arithmetic, 510 Goursat’s, 383 Hardy-Littlewood’s maximal, 236 Jordan curve, 362 Kolmogorov-Riesz’, 253 Lebesgue’s differentiation, 237 Liouville’s, 433 Lusin’s, 247 mean-value, 395 monotone convergence, 92, 148 Montel’s, 561 Morera’s, 407 Radon-Nikodym’s, 120, 122 residue, 467 residue theorem for null-homologous cycles, 486 Riemann mapping, 563

set µ-measure zero, 50 µ-null, 50 arcwise connected, 348 Borel, 9, 12 Cantor, 63 countable, 6 critical values, 243 denumerable, 6 discrete, 409 equi-continuous, 254 invariant, 290 Lebesgue, 53, 239 measurable, 4, 6 normal, 561 outer measurable, 39 region, 347 regular values, 243 relatively compact, 250 splitting, 347 star-shaped, 348 totally bounded, 250 sets of µ-measure zero, 50 simple function, 79 simply closed curve, 349 simply periodic function, 537 sines amplitudines, 531 singular function, 240 space Hausdorff, 580 normal topological, 581 space filling curve, 69

756

SUBJECT INDEX

Riemann’s theorem on removable singularities, 409 Rouch´e’s, 469 Runge’s approximation theorem, 437 Sard’s, 243 Tonelli’s, 181 uniqueness theorem for holomorphic functions, 413 Vitali’s covering, 215 Zermelo’s well-ordering theorem, 584 topological group, 73 topological space arcwise connected, 349 connected, 347 homotopically equivalent, 357 homotopy type, 357 pathwise connected, 349 topological vector space, 73 topology base, 579 product, 579 total variation, 365 totally bounded, 250 trace, 7, 348 transformation theorem for integrals, 115 trivial zero of the ζ-function, 522

uniform convergence, 277, 301, 440 uniqueness theorem for holomorphic functions, 413 unit mass, 17 upper half plane, 293 upper left derivative, 220 upper right derivative, 220 value of order n, 410 variation of a function, 219 vector space metric, 73 topological, 73 Vi´eta’s formula, 298 Vitali covering, 215 Vitali’s covering theorem, 215 weakly convergent, 247 Weierstrass ℘-function, 545 Weierstrass function, 241 well-ordering theorem of Zermelo, 584 Wiener’s covering lemma, 234 winding number, 474 Young’s inequality, 202 Z-variation, 365 zero of order ∞, 410 zero of order n, 410

757