Complex Analysis: A Functional Analytic Approach 9783110417241, 9783110417234

Table of contents :
Preface
Contents
1. Complex numbers and functions
2. Cauchy’s Theorem and Cauchy’s formula
3. Analytic continuation
4. Construction and approximation of holomorphic functions
5. Harmonic functions
6. Several complex variables
7. Bergman spaces
8. The canonical solution operator to ∂̄
9. Nuclear Fréchet spaces of holomorphic functions
10. The ∂̄-complex
11. The twisted ∂̄-complex and Schrödinger operators
Bibliography
Index

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Friedrich Haslinger Complex Analysis De Gruyter Graduate

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Friedrich Haslinger

Complex Analysis |

A Functional Analytic Approach

Mathematics Subject Classification 2010 30-01, 30H20, 32-01, 32A36, 32W05, 35J10 Author Prof. Dr Friedrich Haslinger Fakultät für Mathematik Universität Wien Oskar-Morgenstern-Platz 1 A-1090 Wien Austria [email protected]; http://www.mat.univie.ac.at/˜has/

ISBN 978-3-11-041723-4 e-ISBN (PDF) 978-3-11-041724-1 e-ISBN (EPUB) 978-3-11-042615-1 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2018 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck Cover image: Friedrich Haslinger ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface This book provides a thorough introduction to complex analysis in one variable and features special topics of several variables connected with the Cauchy–Riemann differential equation from a functional analysis point of view. Chapters 1 and 2 are basic and cover the classical material of the complex plane and of holomorphic functions, their power series and the Cauchy integral formula with the consequences for the behavior of zeroes and the maximum principle. The homology and the homotopy version of Cauchy’s theorem are related to important topological concepts and will be crucial for the characterization of simply connected domains in Chapter 4. The calculus of residues of meromorphic functions is used to evaluate real integrals and Fourier transforms without computing antiderivatives. There are only 2 sections which do not belong to a standard treatment: holomorphic parameter integrals, where the knowledge of basic properties of the Lebesgue integral is assumed, and the inhomogeneous Cauchy formula, where Stokes’ integral theorem from real analysis is used. This is the first place where we refer to real methods in complex analysis. Chapter 3 is devoted to analytic continuation and the monodromy theorem. In Chapter 4 we continue with real methods such as the 𝒞∞ -partition of unity and the inhomogeneous Cauchy–Riemann equations. This is also true for the general treatment of Runge’s approximation theorem using the Hahn–Banach theorem. Now we are able to prove Mittag-Leffler’s theorem on the existence of global meromorphic functions with prescribed singularities and, by solutions of the inhomogeneous Cauchy–Riemann equations, the cohomology version of Mittag-Leffler’s theorem. Infinite products are introduced in order to obtain Weierstraß’ factorization theorem on the existence of holomorphic functions with prescribed zeroes. This gives the tools to show that each domain in the complex plane is a domain of holomorphy, i.e. has a holomorphic function which cannot be analytically extended beyond any point of the boundary of the domain. In addition, normal families of holomorphic functions are studied, a concept which refers to compactness in spaces of holomorphic functions and provides the main idea in the proof of the Riemann mapping theorem, which states that each simply connected domain not equal to ℂ is biholomorphic equivalent to the unit disc. Chapter 4 ends with a characterization of simply connected domains by means of properties of holomorphic functions. In Chapter 5 we study harmonic functions – a concept of real analysis, solve the Dirichlet problem constructing a continuous function on the closure of a domain which is harmonic in the interior and coincides with a given continuous function on the boundary of the domain. Furthermore, we collect the basics of subharmonic functions. We would like to emphasize that this presentation of the classical theory for one complex variable was inspired by the first 30 pages of Lars Hörmander’s book “An Introduction to Complex Analysis in Several Variables” [41], which, after more than 50 years, is still singular in its elegance and importance. https://doi.org/10.1515/9783110417241-201

VI | Preface Chapter 6 describes the main differences between the univariate and multivariate theories with emphasis on the inhomogeneous Cauchy–Riemann differential equations. We explain the concept of pseudoconvexity without giving a full proof of the characterization of domains of holomorphy by pseudoconvexity. In the following chapters we treat basic functional analytic results on Hilbert spaces and spectral theory of operators on Hilbert spaces for bounded and unbounded operators in order to provide the tools for Bergman spaces of holomorphic functions (Hilbert spaces with the reproducing property) and for a thorough discussion of the canonical solution operator to 𝜕. In Chapter 8 the solution operator to 𝜕 restricted to holomorphic L2 -functions in one complex variable is investigated, pointing out that the Bergman kernel of the associated Hilbert space of holomorphic functions plays an important role. We investigate operator properties like compactness and Schatten-class membership (nuclear and Hilbert–Schmidt operators). At this place we have the prerequisites to study the space of all holomorphic functions on a domain endowed with the topology of uniform convergence on all compact subsets of the domain. Chapter 9 is devoted to the study of these spaces, which turn out to be complete metric spaces with the Montel property (nuclear Fréchet spaces). In addition, we describe the dual spaces of these Fréchet spaces again as certain spaces of holomorphic functions and represent them as sequence spaces (Köthe sequence spaces). In Chapter 10 we consider the general 𝜕-complex and derive properties of the complex Laplacian on L2 -spaces of bounded pseudoconvex domains. For this purpose we first concentrate on basic results about distributions, Sobolev spaces, and unbounded operators on Hilbert spaces. The key result is the Kohn–Morrey formula, which is presented in different versions. Using this formula the basic properties of the 𝜕-Neumann operator – the bounded inverse of the complex Laplacian – are proved. It turns out to be useful to investigate an even more general situation, namely the twisted 𝜕-complex, where 𝜕 is composed with a positive twist factor. In this way one obtains a rather general basic estimate, from which one gets Hörmander’s L2 -estimates for the solution of the Cauchy–Riemann equation together with results on related weighted spaces of entire functions. The last chapter contains an account of the application of the 𝜕-methods to Schrödinger operators, Pauli and Dirac operators and to Witten–Laplacians. We use the 𝜕-methods and some spectral theory to settle the question whether certain Schrödinger operators with magnetic field on ℝ2 have compact resolvent. Most of the material in this book is self-contained with the exception of some parts in Chapters 6 and 11. Each single chapter contains exercises enhancing and reinforcing the material discussed in the text. Notes at the end of each chapter refer to the literature and more advanced results. The prerequisites for reading this book are: real analysis, basic measure theory and point set topology. I was partially supported by the FWF-grant P 28154 of the Austrian Science Fund. Vienna February 2017

Friedrich Haslinger

Contents Preface | V 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12

Complex numbers and functions | 1 Complex numbers | 1 Some topological concepts in ℝn | 3 Holomorphic functions | 6 The Cauchy–Riemann equations | 7 A geometric interpretation of the complex derivative | 13 Uniform convergence | 15 Power series | 18 Line integrals | 21 Primitive functions | 24 Elementary functions | 28 Exercises | 33 Notes | 38

2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14

Cauchy’s Theorem and Cauchy’s formula | 39 Winding numbers | 39 The theorem of Cauchy–Goursat and Cauchy’s formula | 43 Important consequences of Cauchy’s Theorem | 50 Isolated singularities | 53 The maximum principle and Cauchy’s estimates | 56 Open mappings | 61 Holomorphic parameter integrals | 64 Complex differential forms | 66 The inhomogeneous Cauchy formula | 68 General versions of Cauchy’s Theorem and Cauchy’s formula | 70 Laurent series and meromorphic functions | 80 The residue theorem | 84 Exercises | 95 Notes | 102

3 3.1 3.2 3.3 3.4 3.5

Analytic continuation | 103 Regular and singular points | 103 Analytic continuation along a curve | 105 The Monodromy Theorem | 107 Exercises | 110 Notes | 110

VIII | Contents 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12

Construction and approximation of holomorphic functions | 111 A partition of unity | 111 The inhomogeneous Cauchy–Riemann differential equations | 113 The Hahn–Banach Theorem | 117 Runge’s approximation theorems | 120 Mittag-Leffler’s Theorem | 130 The Weierstraß Factorization Theorem | 136 Some applications of the Mittag-Leffler and Weierstraß Theorems | 142 Normal families | 144 The Riemann Mapping Theorem | 146 Characterization of simply connected domains | 151 Exercises | 152 Notes | 154

5 5.1 5.2 5.3 5.4 5.5 5.6

Harmonic functions | 155 Definition and important properties | 155 The Dirichlet problem | 157 Jensen’s formula | 161 Subharmonic functions | 165 Exercises | 167 Notes | 168

6 6.1 6.2 6.3 6.4 6.5

Several complex variables | 169 Complex differential forms and holomorphic functions | 169 The inhomogeneous CR equations | 174 Domains of holomorphy | 180 Exercises | 187 Notes | 188

7 7.1 7.2 7.3 7.4 7.5

Bergman spaces | 191 Elementary properties | 191 Examples | 199 Biholomorphic mappings | 203 Exercises | 206 Notes | 207

8 8.1 8.2 8.3 8.4

The canonical solution operator to 𝜕 | 209 Compact operators on Hilbert spaces | 209 The canonical solution operator to 𝜕 restricted to A2 (𝔻) | 222 (0, 1)-forms with holomorphic coefficients | 226 Exercises | 228

Contents | IX

8.5

Notes | 229

9 9.1 9.2 9.3 9.4

Nuclear Fréchet spaces of holomorphic functions | 231 General properties of Fréchet spaces | 231 The space ℋ(DR (0)) and its dual space | 232 Exercises | 237 Notes | 238

10 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9

The 𝜕-complex | 239 Unbounded operators on Hilbert spaces | 239 Distributions and Sobolev spaces | 255 Friedrichs’ lemma | 261 A finite dimensional analog | 266 The 𝜕-Neumann operator | 267 Density in the graph norm | 275 Properties of the 𝜕-Neumann operator | 280 Exercises | 287 Notes | 288

11 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11

The twisted 𝜕-complex and Schrödinger operators | 291 An exact sequence of unbounded operators | 291 The twisted basic estimates | 292 Hörmander’s L2 -estimates | 295 The 𝜕-Neumann operator on weighted (0, q)-forms | 297 Weighted spaces of entire functions | 305 Spectral analysis of self-adjoint operators | 310 Real differential operators | 315 Dirac and Pauli operators | 322 Compact resolvents | 323 Exercises | 327 Notes | 329

Bibliography | 331 Index | 335

1 Complex numbers and functions This chapter is devoted to the complex plane, its field structure and its topological properties. The differential calculus is extended to the complex plane, leading to the concept of a holomorphic function. Power series are introduced in order to obtain further examples of holomorphic functions. After explaining the basic facts about line integrals and primitive functions, we describe the most common elementary complex functions, such as exponential, logarithmic and trigonometric functions.

1.1 Complex numbers The quadratic equation x2 + 1 = 0 has the two formal solutions x1,2 = ±√−1, and Euler1 writes in 1777: “…formulam √−1 littera i in posterum designabo.” Definition 1.1. We consider ℂ ∶= ℝ2 as a vector space over ℝ, with multiplication in ℂ: (a, b) ∈ ℂ, a, b ∈ ℝ, (c, d) ∈ ℂ, c, d ∈ ℝ, (a, b)(c, d) = (ac − bd, bc + ad). In this way ℂ becomes a commutative field with zero element (0, 0) and unit element (1, 0); for (a, b) ≠ (0, 0), we have (a, b)−1 = (

a −b , ). a2 + b2 a2 + b2

For (a, b) ∈ ℂ, we also write a+ib, this means that (a, 0) corresponds to the real number a and (0, 1) to the imaginary unit i. The multiplication law from above stems from the formal multiplication (a + ib)(c + id) = ac − bd + i(bc + ad). For z = (x, y), we will write z = (x, y) = (x, 0) + (0, 1)(y, 0) = x + iy, where x = ℜz is called the real part of z and y = ℑz the imaginary part of z. For z = (x, y) = x + iy ∈ ℂ, the complex number z = (x, −y) = x − iy

1 Euler, Leonhard (1707–1783). https://doi.org/10.1515/9783110417241-001

2 | 1 Complex numbers and functions is called the complex conjugate of z, and we have the following rules for z, w ∈ ℂ: (z + w)− = z + w,

zw = z w,

z = z.

We define |z|2 = zz; |z| is the absolute value of z. We have |z| = √x 2 + y2 , for z = x + iy, and |z| = |z|. In addition, 1 1 ℜz = (z + z), ℑz = (z − z), 2 2i |zw| = |z||w|, |z + w| ≤ |z| + |w|, ||z| − |w|| ≤ |z − w|. Polar representation Let z = x + iy ≠ 0. We set x = r cos θ and y = r sin θ, where r = |z| is the absolute value of z and θ = arg z the argument of z. We have z = r(cos θ + i sin θ) = reiθ , which will become clear after introducing the complex exponential function, see Section 1.10. Since cos θ = cos(θ + 2kπ) and sin θ = sin(θ + 2kπ), for k ∈ ℤ, there are infinitely many values of θ corresponding to a single z. The principal argument Arg z of z is to be taken −π < Arg z ≤ π, in this way the polar representation of z becomes uniquely determined. Examples. (a) z = 2 + 2i, r = |z| = 2√2, Arg z = π4 ; (b) z = 2 − 2i, r = |z| = 2√2, Arg z = − π4 . Product of complex numbers Let z = r(cos θ + i sin θ) and w = s(cos ϕ + i sin ϕ). Then zw = rs[(cos θ cos ϕ − sin θ sin ϕ) + i(cos θ sin ϕ + sin θ cos ϕ)] = rs(cos(θ + ϕ) + i sin(θ + ϕ)), where we used the addition rules for the cosine and sine function. Hence one has to multiply the absolute values and to add the angles. Concerning the principal argument, one has to be careful in this connection as the following example shows: let z = −1 and w = i. Then Arg(−1) = π and Arg(i) = π/2, but Arg((−1)i) = Arg(−i) = −π/2 ≠ Arg(−1) + Arg(i). de Moivre’s formula Let n ∈ ℕ. Then z n = r n (cos nθ + i sin nθ) = r n einθ , which follows by induction. Roots of complex numbers Let n ∈ ℕ, z ≠ 0, z = r(cos θ + i sin θ). Using de Moivre’s formula, we get an nth root of z by

1.2 Some topological concepts in ℝn

| 3

z 1/n = r 1/n (cos θ/n + i sin θ/n). As z = r(cos(θ + 2kπ) + i sin(θ + 2kπ)), for k ∈ ℤ, the expressions r 1/n (cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)) are also nth roots of z. There are n different nth roots of z, with arguments θ θ + 2π θ + 2(n − 1)π , ,…, . n n n Riemann sphere and stereographic projection Let S2 = {(x1 , x2 , x3 ) ∈ ℝ3 ∶ x12 + x22 + x32 = 1} be the unit sphere in ℝ3 with N = (0, 0, 1) as the north pole on S2 . In this context, S2 is also called Riemann2 sphere. We consider ℂ as the equator plane of S2 . A point z ∈ ℂ is associated with the intersection point P of S2 with the ray joining the north pole with z ∈ ℂ. In this way we obtain a homeomorphism ϕ between S2 ⧵ {N} and ℂ, given by ϕ(x1 , x2 , x3 ) =

1 (x + ix2 ), 1 − x3 1

with the inverse ϕ−1 (x1 + ix2 ) =

1 (2x , 2x , x 2 + x22 − 1). x12 + x22 + 1 1 2 1

One-point compactification By continuous continuation of ϕ to the whole of S2 , we obtain a topological homeomorphism between the compact space S2 and the so-called extended plane (one-point compactification of ℂ) ℂ: ϕ(N) ∶= ∞,

ϕ−1 (∞) ∶= N,

ϕ(S2 ) = ℂ.

1.2 Some topological concepts in ℝn We explain some topological concepts and basic results, which will be important later on. For this purpose we consider the complex vector space ℂn of all n-tuples of complex numbers with the standard inner product n

⟨z, w⟩ = ∑ zj wj , j=1

z = (z1 , … , zn ), w = (w1 , … , wn )

2 Riemann, Georg Friedrich Bernhard (1826–1866).

4 | 1 Complex numbers and functions and the Euclidean norm n

|z|2 ∶= ∑ |zj |2 . j=1

If we forget about the “complex structure”, we identify ℂn ≅ ℝ2n . We will use the notations DR (a) ∶= {z ∈ ℂ ∶ |z − a| < R}, n

BR (a) ∶= {z ∈ ℂ ∶ |z − a| < R},

a ∈ ℂ, R > 0, n = 1, a ∈ ℂn , R > 0, n > 1.

We assume some knowledge about basic topological concepts in ℂn , such as that each Cauchy3 sequence in ℂn is convergent (ℂn is complete). The following topological concepts are formulated for sets in ℝn , they have analogous definitions in ℂ ≅ ℝ2 and ℂn ≅ ℝ2n . Definition 1.2. Let G ⊆ ℝn . G is open, if for each x ∈ G there exists ϵ > 0 such that Bϵ (x) ⊆ G. A set U is a neighborhood of the set M, if there exists an open set V such that M ⊂ V ⊂ U. A ⊆ ℂ is called closed, if ℂ ⧵ A is open. The union of arbitrarily many open sets is open; the intersection of finitely many open sets is open; the union of finitely many closed sets is closed; the intersection of arbitrarily many closed sets is closed. Let M ⊆ ℝn be an arbitrary set in ℝn . The set M ∘ ∶= ⋃{U ∶ U ⊆ M, U open} is called the interior of M; M ∘ is an open set, it is the largest open set which is contained in M. The set M ∶= ⋂{A ∶ A ⊇ M, A closed} is called the closure of M; M is a closed set, it is the smallest closed set which contains M. The set 𝜕M ∶= M ⧵ M ∘ is called the boundary of M. Let M ⊆ ℝn be an arbitrary set in ℝn . A subset U ⊆ M is called relatively open in M, if there is an open set O in ℝn such that U = O ∩ M. A set X ⊆ ℝn is called (pathwise) connected, if any two points can be joined by a continuous curve in G. If X ⊆ ℝn and x ∈ X, we denote by Ex the largest connected set in X containing the point x. The set Ex is called the connected component of x. 3 Cauchy, Augustin Louis (1789–1857).

1.2 Some topological concepts in ℝn

| 5

A domain G in the complex plane is called simply connected, if its complement with respect to the extended plane, ℂ ⧵ G, is connected. Let N ⊆ M ⊆ ℝn . We say that N is dense in M, if N ⊇ M. Let N ⊆ M ⊆ ℝn . We say that N is discrete in M, if for each z ∈ M there exists a neighborhood U (an open ball with center z) such that U ∩ N contains at most finitely many elements of N. A subset K ⊆ ℝn is called compact, if each open cover of K has a finite subcover. K is compact in ℝn , if and only if K is closed and bounded in ℝn (i.e. there exists C > 0 such that |x| ≤ C ∀x ∈ K). Let M ⊆ ℝn be a subset of ℝn . U is relatively compact in M (we write U ⊂⊂ M), if U ⊆ M and U is compact. Definition 1.3. Let G ⊆ ℂ and f ∶ G ⟶ ℂ a function. We separate f (z) into its real and imaginary part: f (z) = u(z) + iv(z) = ℜf (z) + iℑf (z), where u, v ∶ ℂ ⟶ ℝ are realvalued functions. The set {(z, w) ∶ f (z) = w, z ∈ G} ⊆ ℂ2 is called the graph of f . The sets {z ∶ ℜf (z) = const.}, {z ∶ ℑf (z) = const.}, {z ∶ |f (z)| = const.} are the level lines of f . Examples. (a) f (z) = z 2 , ℜf (z) = x2 − y2 , ℑf (z) = 2xy. The level lines are circles and hyperbolas. (b) f (z) = az, a ∈ ℂ, a ≠ 0. This is a rotation–dilation. We write a = α + iβ, then we have az = αx − βy + i(βx + αy). If one considers f as a mapping from ℝ2 to ℝ2 , one obtains x αx − βy α ( )↦( )=( y βx + αy β

−β x cos γ ) ( ) = (α2 + β2 )1/2 ( α y sin γ

− sin γ x )( ), cos γ y

where cos γ =

(α2

α , + β2 )1/2

sin γ =

(α2

β . + β2 )1/2

Definition 1.4. Let O ⊆ ℂ be an open subset of ℂ, and let f ∶ O ⟶ ℂ be a function. f is called continuous at z0 ∈ O, if for each ϵ > 0 there exists δ > 0 such that |f (z) − f (z0 )| < ϵ

for |z − z0 | < δ.

f is called continuous on a set M if f is continuous at each point of M. f is continuous at z0 , if and only if for each sequence (zn )∞ n=1 with limn→∞ zn = z0 we have limn→∞ f (zn ) = f (limn→∞ zn ) = f (z0 ). If f and g are continuous, then f + g, f ⋅ g, gf (g ≠ 0) are continuous. f is continuous, if and only if ℜf and ℑf are continuous.

6 | 1 Complex numbers and functions If f is continuous, then |f | is continuous. If f is continuous on a compact set K, then sup|f (z)| = max|f (z)| = |f (z1 )| for some z1 ∈ K, z∈K

z∈K

inf |f (z)| = min|f (z)| = |f (z2 )| for some z2 ∈ K.

z∈K

z∈K

If, in addition, f ≠ 0 on K, then there exists δ > 0 such that |f (z)| ≥ δ

∀z ∈ K.

1.3 Holomorphic functions Definition 1.5. Let U ⊆ ℂ be an open set and f ∶ U ⟶ ℂ a function. f is called complex differentiable at z0 ∈ U, if there exists a function Δ ∶ U ⟶ ℂ, which is continuous at z0 , such that f (z) = f (z0 ) + (z − z0 )Δ(z),

z ∈ U.

f is called holomorphic on U, if f is complex differentiable at each point of U; we write f ∈ ℋ(U). f is holomorphic at z0 ∈ U, if there exists an open neighborhood U0 of z0 such that f is holomorphic on U0 . Remark. If f is complex differentiable at z0 , we have f (z) − f (z0 ) = Δ(z); z − z0

lim

z→z0

f (z) − f (z0 ) = Δ(z0 ) = f ′ (z0 ). z − z0

The following theorem has the same proof as in the real case. Theorem 1.6. Let f and g be complex differentiable at z0 . Then f + g and f ⋅ g are complex differentiable at z0 . In addition, λf is complex differentiable at z0 , where λ ∈ ℂ, and the following rules are valid: (f + g)′ = f ′ + g ′ ,

(λf )′ = λf ′ ,

(f ⋅ g)′ = f ′ ⋅ g + f ⋅ g ′ .

If g(z0 ) ≠ 0, then f ′ f ′ ⋅ g − f ⋅ g′ ( ) = . g g2 Let w0 = f (z0 ) and let h be complex differentiable at w0 . Then (h ∘ f )′ (z0 ) = h′ (f (z0 ))f ′ (z0 )

(chain rule).

1.4 The Cauchy–Riemann equations | 7

We will need some results about holomorphic functions derived from the Cauchy theorem to show an inverse function theorem for holomorphic functions. Theorem 1.7. Suppose f is holomorphic on a domain G, z0 ∈ G, and f ′ (z0 ) ≠ 0. Then there exists an open neighborhood U of z0 with U ⊂ G such that f is injective on U, the image V = f (U) of U is open, and the inverse function f −1 ∶ V ⟶ U is holomorphic on V and satisfies (f −1 )′ (f (z)) = 1/f ′ (z),

z ∈ U.

For the proof, see Theorem 2.40. Examples. (a) f (z) = z n , n ∈ ℕ. f ′ (z0 ) = lim

z→z0

z n − z0n (z − z0 )(z n−1 + z n−2 z0 + ⋯ + zz0n−2 + z0n−1 ) = lim z − z0 z→z0 z − z0

= nz0n−1 . (b) f (z) = z.

First we take the limit z → z0 parallel to the real axis: z − z0 = h ∈ ℝ, h → 0, z − z0 h = = 1; z − z0 h and now parallel to the imaginary axis: z − z0 = ih, h ∈ ℝ, h → 0, z − z 0 −ih = = −1. z − z0 ih Hence f (z) = z is nowhere complex differentiable.

1.4 The Cauchy–Riemann equations Here we explain the relationship between real and complex differentiable functions, which is expressed by the Cauchy–Riemann equations. Definition 1.8. Let U ⊆ ℂ be open and g ∶ U ⟶ ℝ. g is real differentiable at z0 ∈ U, if there exist functions Δ1 , Δ2 ∶ U ⟶ ℝ, continuous at z0 , such that g(z) = g(z0 ) + (x − x0 )Δ1 (z) + (y − y0 )Δ2 (z), where z = x + iy and z0 = x0 + iy0 .

(1.1)

8 | 1 Complex numbers and functions (z ) = gx (z0 ), which is the partial derivative with respect We have that Δ1 (z0 ) = 𝜕g 𝜕x 0 to x, and Δ2 (z0 ) = 𝜕g (z ) = g (z y 0 ) is the partial derivative with respect to y. In order to 𝜕y 0 see this, we first choose z = x + iy0 , then (1.1) implies Δ1 (z) =

g(z) − g(z0 ) , x − x0

and putting x → x0 , we obtain the assertion about the partial derivative with respect to x. Choosing z = x0 + iy, we get the assertion about the partial derivative with respect to y. In addition, we have that each real differentiable function at z0 is also continuous at z0 . Example 1.9. Let z = x + iy and { xy2 , if z ≠ 0, u(z) = { |z| 0, if z = 0. { Then u fails to be continuous at z = 0 since 1/n2 1 = ≠ u(0, 0) = 0. n→∞ 2/n2 2

lim u(1/n, 1/n) = lim

n→∞

But the partial derivatives exist and ux (0, 0) = uy (0, 0) = 0, since ux (0, 0) = lim

h→0

u(h, 0) − u(0, 0) = 0. h

Using the mean value theorem from real analysis, one can show the following result: if ux and uy are continuous at z0 , then u is real differentiable at z0 . Now we take a function f which is complex differentiable at z0 . For h ∈ ℝ both of the limits lim

h→0

f (z0 + h) − f (z0 ) h

and

lim

h→0

f (z0 + ih) − f (z0 ) ih

exist and are equal. Splitting f into real and imaginary parts, f (z) = u(x, y) + iv(x, y), we obtain u(x0 + h, y0 ) − u(x0 , y0 ) v(x0 + h, y0 ) − v(x0 , y0 ) + i lim h→0 h h u(x0 , y0 + h) − u(x0 , y0 ) v(x0 , y0 + h) − v(x0 , y0 ) = lim + i lim . h→0 h→0 ih ih

f ′ (z0 ) = lim

h→0

This implies that ux (z0 ) + ivx (z0 ) = 1/i(uy (z0 ) + ivy (z0 )) and, again after taking real and imaginary parts, ux (z0 ) = vy (z0 ),

vx (z0 ) = −uy (z0 ).

1.4 The Cauchy–Riemann equations | 9

This system of two partial differential equations for the functions u and v is called the Cauchy–Riemann differential equations. Hence we have shown the following Theorem 1.10. Let f be complex differentiable at z0 and split the function into real and imaginary parts, f = u + iv. Then ux (z0 ) = vy (z0 ),

vx (z0 ) = −uy (z0 ).

In addition, f ′ (z0 ) = ux (z0 ) + ivx (z0 ) = vy (z0 ) − iuy (z0 ). Now we consider complex-valued functions. Definition 1.11. Now let f ∶ U ⟶ ℂ be a complex-valued function. We say that f is real differentiable at z0 , if there are functions Δ1 , Δ2 ∶ U ⟶ ℂ, both continuous at z0 , such that f (z) = f (z0 ) + (x − x0 )Δ1 (z) + (y − y0 )Δ2 (z),

z ∈ U.

We have again Δ1 (z0 ) =

𝜕f (z ) = fx (z0 ) 𝜕x 0

and Δ2 (z0 ) =

𝜕f (z ) = fy (z0 ). 𝜕y 0

Lemma 1.12. The following assertions are equivalent: (1) f ∶ U ⟶ ℂ is real differentiable at z0 . (2) Real and imaginary part of f = u + iv are real differentiable at z0 . (3) There exist functions A1 , A2 ∶ U ⟶ ℂ, both continuous at z0 , such that f (z) = f (z0 ) + (z − z0 )A1 (z) + (z − z0 )− A2 (z), where

1 A1 (z0 ) = (fx (z0 ) − ify (z0 )) and 2

1 A2 (z0 ) = (fx (z0 ) + ify (z0 )). 2

Proof. For f = u + iv we have f (z) = f (z0 ) + (x − x0 )Δ1 (z) + (y − y0 )Δ2 (z)

= u(z0 ) + (x − x0 )ℜΔ1 (z) + (y − y0 )ℜΔ2 (z)

+ i[v(z0 ) + (x − x0 )ℑΔ1 (z) + (y − y0 )ℑΔ2 (z)].

This shows that (1) and (2) are equivalent and that fx (z0 ) = ux (z0 ) + ivx (z0 )

and fy (z0 ) = uy (z0 ) + ivy (z0 ).

10 | 1 Complex numbers and functions An easy computation shows that f (z) = f (z0 ) + (x − x0 )Δ1 (z) + (y − y0 )Δ2 (z) 1 = f (z0 ) + [(x − x0 ) + i(y − y0 )] (Δ1 (z) − iΔ2 (z)) 2 1 + [(x − x0 ) − i(y − y0 )] (Δ1 (z) + iΔ2 (z)) 2 = f (z0 ) + (z − z0 )A1 (z) + (z − z0 )− A2 (z), where

1 A1 (z) = (Δ1 (z) − iΔ2 (z)) and 2

1 A2 (z) = (Δ1 (z) + iΔ2 (z)). 2

The expressions for the functions A1 and A2 lead to the following Definition 1.13. The Wirtinger-derivatives4 are defined by 𝜕f 1 𝜕f 𝜕f = ( − i ), 𝜕z 2 𝜕x 𝜕y

𝜕f 1 𝜕f 𝜕f = ( +i ) 𝜕z 2 𝜕x 𝜕y

Theorem 1.14. The function f ∶ U ⟶ ℂ satisfies the Cauchy–Riemann differential 𝜕f equations at z0 ∈ U if and only if 𝜕z (z0 ) = 0. Proof. Splitting f into real and imaginary part, one obtains

Hence

𝜕f 1 𝜕f 𝜕f 1 = ( + i ) = [ux + ivx + i(uy + ivy )] 𝜕z 2 𝜕x 𝜕y 2 1 = [ux − vy + i(vx + uy )]. 2 𝜕f =0 𝜕z

⟺

ux = vy ,

uy = −vx .

Remark. The advantage of this concept is that the system of partial differential equations ux (z0 ) = vy (z0 ),

vx (z0 ) = −uy (z0 )

can be written as one equation, namely 𝜕f (z ) = 0. 𝜕z 0 Theorem 1.15. Let f ∶ U ⟶ ℂ and z0 ∈ U. Then f is complex differentiable at z0 , if and 𝜕f 𝜕f only if f is real differentiable at z0 and 𝜕z (z0 ) = 0. In this case we have f ′ (z0 ) = 𝜕z (z0 ). 4 Wirtinger, Wilhelm (1865–1945).

1.4 The Cauchy–Riemann equations | 11

Proof. First suppose that f (z) = f (z0 ) + (z − z0 )Δ(z), where Δ is continuous at z0 . We set A1 = Δ, A2 = 0. Then f (z) = f (z0 ) + (z − z0 )A1 (z) and, by Lemma 1.12, we have that f is real differentiable. Since A2 = 0, the Cauchy–Riemann differential equations are satisfied. For the other direction, let f be real differentiable at z0 . Then, by Lemma 1.12, there exist functions A1 and A2 , both continuous at z0 , such that f (z) = f (z0 ) + (z − z0 )A1 (z) + (z − z0 )− A2 (z), and, by assumption, we have A2 (z0 ) =

𝜕f (z ) = 0. 𝜕z 0

0) { A2 (z)(z−z z−z0 ̃ Δ(z) = { 0 {

−

Let

for z ≠ z0 ,

for z = z0 . −

(z−z ) ̃ = 0. Since A2 is continuous at z0 , A2 (z0 ) = 0 and | z−z0 | = 1, we have limz→z0 Δ(z) 0 Hence Δ̃ is continuous at z0 . Now let Δ = A1 + Δ.̃ Then Δ is continuous at z0 and

f (z) = f (z0 ) + (z − z0 )A1 (z) + (z − z0 )− A2 (z)

A2 (z)(z − z0 )− ] z − z0 ̃ = f (z0 ) + (z − z0 )(A1 (z) + Δ(z)) = f (z0 ) + (z − z0 )[A1 (z) +

= f (z0 ) + (z − z0 )Δ(z).

Hence f is complex differentiable at z0 . The assumptions about differentiability can considerably be weakened: Theorem 1.16 (Looman–Menchoff). Let f = u + iv ∶ U ⟶ ℂ. Suppose that all partial derivatives ux , uy , vx , vy exist and satisfy the Cauchy–Riemann differential equations on U. Then f is complex differentiable on U. For a proof see, for instance, [57]. Examples. (1) f = u+iv, u(x, y) = x2 −y2 , v(x, y) = 2xy. f is real differentiable on ℂ and since ux = 2x, uy = −2y, vx = 2y, vy = 2x, the Cauchy–Riemann differential equations are satisfied. Hence f is complex differentiable on ℂ and holomorphic on ℂ. We have f (z) = z 2 = x2 − y2 + 2ixy. (2) f = u + iv, u(x, y) = x3 − 3xy2 , v(x, y) = 3x2 y − y3 . Also in this case, the Cauchy– Riemann differential equations are satisfied and f is holomorphic on ℂ; we have f (z) = z 3 .

12 | 1 Complex numbers and functions (3) f = u + iv, u(x, y) = ex cos y, v(x, y) = ex sin y. The Cauchy–Riemann differential equations are satisfied and f is holomorphic on ℂ. Later on (in Section 1.8) we will see that f is the complex exponential function. We have f (z) = ez = ex+iy = ex (cos y + i sin y). (4) f = u + iv, u(x, y) = x3 y2 , v(x, y) = x2 y3 . f is real differentiable on ℂ and ux = 3x 2 y2 , vy = 3x2 y2 , uy = 2x3 y, vx = 2xy3 . It follows that the Cauchy–Riemann differential equations are satisfied if and only if (x 2 + y2 )xy = 0. These are exactly the points on the coordinate axes. Hence f is complex differentiable there but not holomorphic. In the following we explain some formulas of the so-called Wirtinger calculus. Remark. (1) We consider complex-valued functions which can be expressed by the complex conjugate variables z and z. When computing the Wirtinger derivatives 𝜕 𝜕 and 𝜕z of such functions, one can take z and z as independent variables. 𝜕z 𝜕f 𝜕f Example. f (z) = |z|2 = zz, 𝜕z = z, 𝜕z = z. f is complex differentiable at z = 0, but fails to be holomorphic at z = 0. 𝜕 𝜕 (2) 𝜕z , 𝜕z are ℂ-linear operators, i.e. for real differentiable functions f , g ∶ U ⟶ ℂ and for c, d ∈ ℂ we have 𝜕f 𝜕g 𝜕 (cf + dg) = c + d , 𝜕z 𝜕z 𝜕z

𝜕 𝜕f 𝜕g (cf + dg) = c + d . 𝜕z 𝜕z 𝜕z

(3) 𝜕 𝜕f 𝜕 1 𝜕f 𝜕f 1 𝜕 𝜕 𝜕f 𝜕f ( ) = ( ( − i )) = ( + i )( − i ) 𝜕z 𝜕z 𝜕z 2 𝜕x 𝜕y 4 𝜕x 𝜕y 𝜕x 𝜕y 1 𝜕2 f 𝜕2 f 𝜕2 f 𝜕2 f = ( 2 −i +i + 2) 4 𝜕x 𝜕x𝜕y 𝜕y𝜕x 𝜕y

1 𝜕2 f 𝜕2 f = ( 2 + 2) 4 𝜕x 𝜕y 1 = Δf . 4

The differential operator Δ is called Laplace operator.5 (4) Let f , φ ∶ U ⟶ ℂ real differentiable functions. The equation 𝜕f =φ 𝜕z is called the inhomogeneous Cauchy–Riemann differential equation. It corresponds to the system of the two partial differential equations 1/2(ux − vy ) = ℜφ, 5 Laplace, Pierre Simon (1749–1827).

1/2(uy + vx ) = ℑφ.

1.5 A geometric interpretation of the complex derivative

| 13

1.5 A geometric interpretation of the complex derivative Definition 1.17. Let U ⊆ ℂ and let f ∶ U ⟶ ℂ be real differentiable on U. We can consider f as a mapping from a subset of ℝ2 to ℝ2 , if we separate f into real and imaginary parts, f (z) = u(x, y) + iv(x, y). Let Jf = |

ux vx

uy | = ux vy − uy vx . vy

Then Jf is called the Jacobi6 determinant of f . Remark. If f is complex differentiable, then the Cauchy–Riemann differential equations are valid: ux = vy , uy = −vx . Hence Jf = u2x + vx2 . By Theorem 1.10, we have f ′ = ux + ivx , which implies that |f ′ |2 = u2x + vx2 and Jf = |f ′ |2 . Definition 1.18. Let f ∶ U ⟶ ℂ be complex differentiable at z0 ∈ U. Then f (z0 + h) − f (z0 ) − hf ′ (z0 ) = 0. h→0 h lim

Let (Tf (z0 ))(h) ∶= f ′ (z0 )h,

h ∈ ℂ.

(Tf (z0 )) ∶ ℂ ⟶ ℂ is called the tangential map of f at the point z0 . It is a ℂ-linear map. Remark. If f is only real differentiable at z0 , then by Lemma 1.12 we have f (z) = f (z0 ) + (z − z0 )A1 (z) + (z − z0 )− A2 (z). Take for h = z − z0 and set (Tf (z0 ))(h) =

𝜕f 𝜕f (z0 )h + (z0 )h. 𝜕z 𝜕z

This mapping (Tf (z0 )) ∶ ℂ ⟶ ℂ is in general only ℝ-linear. It is easy to show that f is complex differentiable at z0 , if and only if (Tf (z0 )) ∶ ℂ ⟶ ℂ is ℂ-linear. 6 Jacobi, Carl Gustav (1804–1851).

14 | 1 Complex numbers and functions Definition 1.19. Let 𝒯 ∶ ℂ ⟶ ℂ be a bijective ℝ-linear map. For z = x + iy and w = u + iv let ⟨z, w⟩ = ℜ(zw) = xu + yv be the Euclidean scalar product of the vector space ℂ = ℝ2 over ℝ. 𝒯 is an angle preserving map, if |z||w|⟨𝒯z, 𝒯w⟩ = |𝒯z||𝒯w|⟨z, w⟩

∀z, w ∈ ℂ.

Remark. If φ is the angle between z and w, then cos φ =

⟨z, w⟩ , |z||w|

hence the upper assumption means that the angle between 𝒯z and 𝒯w coincides with the original angle. Definition 1.20. Let f ∶ U ⟶ ℂ be real differentiable on U. f is angle preserving at z0 ∈ U, if the tangential map (Tf (z0 )) ∶ ℂ ⟶ ℂ is angle preserving. f is called angle preserving on U, if f is angle preserving at each point of U. Theorem 1.21. Let f ∶ U ⟶ ℂ be a holomorphic function on U and suppose that f ′ (z) ≠ 0 ∀z ∈ U. Then f is angle preserving on U. Proof. By Definition 1.18, we have (Tf (z))(h) = f ′ (z)h. We have to show that |h||k|⟨(Tf (z))(h), (Tf (z))(k)⟩ = |(Tf (z))(h)||(Tf (z))(k)|⟨h, k⟩

∀h, k ∈ ℂ.

The left-hand side is equal to |h||k|⟨f ′ (z)h, f ′ (z)k⟩ = |h||k|ℜ(f ′ (z)hf ′ (z)k) = |h||k||f ′ (z)|2 ℜ(hk), and the right-hand side is |f ′ (z)h||f ′ (z)k|ℜ(hk) = |f ′ (z)|2 |h||k|ℜ(hk). Let γ ∶ [a, b] ⟶ U ⊆ ℂ be a curve in ℂ (see Chapter 2, Section 2.1). We split into real and imaginary parts, t ↦ γ(t) = x(t) + iy(t),

t ∈ [a, b].

γ is differentiable at s ∈ (a, b) if the derivatives x ′ (s) and y′ (s) exist. We set γ ′ (s) = x′ (s) + iy′ (s). Let z = γ(s) and suppose that γ ′ (s) ≠ 0. The map t ↦ z + γ ′ (s) t, is the tangent to the curve γ at z = γ(s).

t ∈ ℝ,

1.6 Uniform convergence

| 15

If f ∶ U ⟶ ℂ is a holomorphic function, we can consider the image curve: f ∘ γ ∶ [a, b] ⟶ ℂ. We have t ↦ f (γ(t)) = u(x(t), y(t)) + iv(x(t), y(t)),

γ(t) = z = x(t) + iy(t).

Using the chain rule, we obtain (f ∘ γ)′ (s) = ux (z)x′ (s) + uy (z)y′ (s) + i[vx (z)x′ (s) + vy (z)y′ (s)]

= 1/2[ux (z) + ivx (z) − i(uy (z) + ivy (z))](x′ (s) + iy′ (s))

+ 1/2[ux (z) + ivx (z) + i(uy (z) + ivy (z))](x′ (s) − iy′ (s))

𝜕f 𝜕f (z)(x ′ (s) + iy′ (s)) + (z)(x′ (s) − iy′ (s)) 𝜕z 𝜕z = (Tf (z))(γ ′ (s)) (by Definition 1.18). =

(by Definition 1.13)

If (f ∘ γ)′ (s) ≠ 0, then the image curve has the tangent t ↦ f (z) + (f ∘ γ)′ (s) t = f (z) + (Tf (z))(γ ′ (s)) t,

t∈ℝ

at the point f (z). Now let γ1 and γ2 be two curves through the point z. The angle between the two curves at the point z is given by the angle between the corresponding tangents at this point. The direction vectors of the tangents are γ1′ (s) and γ2′ (s). Hence the angle between the two curves at the point z is given by ∢(γ1′ (s), γ2′ (s)). The angle between the image curves at the point f (z) is ∢((Tf (z))(γ1′ (s)), (Tf (z))(γ2′ (s))). We suppose that f is holomorphic at z and f ′ (z) ≠ 0. Then, by Theorem 1.21, we get ∢(γ1′ (s), γ2′ (s)) = ∢((Tf (z))(γ1′ (s)), (Tf (z))(γ2′ (s))). Example. f (z) = z 2 , f ′ (z) = 2z, f ′ (z) ≠ 0, ∀z ∈ ℂ∗ = ℂ ⧵ {0}. Then we have f = u + iv with u(x, y) = x2 − y2 and v(x, y) = 2xy. The parallels to the y-axis with equation x = a are mapped by f to the parabolas v2 = 4a2 (a2 − u) with the common focus (0, 0). The parallels to the x-axis with equation y = b are mapped by f to the parabolas v2 = 4b2 (b2 + u) with common focus (0, 0). This follows by using the above formulas for u and v, plugging in the equations for the parallels and finally eliminating x and y. The angles between the curves (90∘ ) are preserved by the map f .

1.6 Uniform convergence The concept of uniform convergence will play an important role in the investigation of power series and limits of sequences of holomorphic functions.

16 | 1 Complex numbers and functions Definition 1.22. Let fn ∶ U ⟶ ℂ be a sequence of functions and A ⊆ U. (fn )∞ n=1 converges uniformly on A to a function f , if for ϵ > 0 there exists nϵ ∈ ℕ such that |fn (z) − f (z)| < ϵ

∀n > nϵ and ∀z ∈ A.

N A series ∑∞ n=1 fn converges uniformly on A, if the sequence of its partial sums (∑n=1 fn )N converges uniformly on A. We write |f |A ∶= supz∈A |f (z)|.

Theorem 1.23. Let fn ∶ U ⟶ ℂ be a sequence of functions. The following assertions are equivalent: (1) (fn )n converges uniformly on A ⊆ U. (2) (fn )n is a Cauchy sequence on A, i.e. ∀ϵ > 0 ∃nϵ ∈ ℕ such that |fn − fm |A < ϵ

∀m, n > nϵ .

Proof. (1) ⇒ (2). If (fn ) → f converges uniformly on A, then |fn − fm |A ≤ |fn − f |A + |f − fm |A
nϵ .

For a fixed z ∈ A, the sequence (fn (z))n is a Cauchy sequence in ℂ. Since ℂ is complete, the limit of this sequence exists: lim fn (z) = f (z)

n→∞

(pointwise convergence). For an arbitrary z ∈ A, we choose m0 = m(z) such that |fm (z) − f (z)| < ϵ ∀m > m0 . Then |fn (z) − f (z)| ≤ |fn (z) − fm (z)| + |fm (z) − f (z)| < ϵ + ϵ ∀n > nϵ , ∀m > m0 and for arbitrary z ∈ A. Theorem 1.24. ∑∞ n=1 fn converges uniformly on A, if and only if for each ϵ > 0 there exists nϵ ∈ ℕ such that |fm+1 (z) + ⋯ + fn (z)| < ϵ for all n > m ≥ nϵ and for all z ∈ A.

1.6 Uniform convergence

| 17

Proof. Since n

m

k=1

k=1

fm+1 (z) + ⋯ + fn (z) = ∑ fk (z) − ∑ fk (z), everything follows from Theorem 1.23. Theorem 1.25 (Weierstraß7 majorant criterion). Let fn ∶ U ⟶ ℂ be a sequence of functions with sup|fn (z)| = |fn |A ≤ Mn ,

Mn ≥ 0.

z∈A

∞ Suppose that ∑∞ n=1 Mn < ∞. Then the series ∑n=1 fn converges uniformly on A.

Proof. Let ϵ > 0. There exists nϵ with n

n

n

k=m+1

k=m+1

k=m+1

| ∑ fk (z)| ≤ ∑ |fk (z)| ≤ ∑ Mk < ϵ, ∀n > m ≥ nϵ and ∀z ∈ A; now we can apply Theorem 1.24. Examples. (1) fn (z) = z n , n ∈ ℕ. fn → 0 converges uniformly on each compact subset of 𝔻 = {z ∶ |z| < 1}. If K ⊂ 𝔻 is a compact subset of 𝔻, then there exists 0 < r < 1 with K ⊂ Dr (0) = {z ∶ |z| < r}, and we have |fn (z)| = |z n | ≤ r n → 0 ∀z ∈ K. 1 n (2) ∑∞ n=0 z = 1−z with uniform convergence on all compact subsets of 𝔻. Let K be as above in example (1), then ∞

∞

∞

n=0

n=0

n=0

| ∑ z n | ≤ ∑ |z n | ≤ ∑ r n < ∞, ∀z ∈ K, and we can apply Theorem 1.25. zn (3) ∑∞ n=0 n! converges uniformly on all compact subsets of ℂ. If K is a compact subset in ℂ, then there exits an N ∈ ℕ with K ⊂ DN (0). We assume to know the Taylor series expansion of the real exponential function and obtain zn zn Nn | ≤ ∑| | ≤ ∑ = eN , n! n! n! n=0 n=0 n=0 ∞

∞

|∑

∀z ∈ K. Now apply again Theorem 1.25. 7 Weierstraß, Karl Theodor Wilhelm (1815–1897).

∞

18 | 1 Complex numbers and functions

1.7 Power series Definition 1.26. Let z0 ∈ ℂ be a fixed point and an ∈ ℂ for n ∈ ℕ0 = ℕ ∪ {0}. The expression ∞

∑ an (z − z0 )n

n=0

is called the formal power series at z0 with coefficients an . For two formal power series ∞

P = ∑ an (z − z0 )n n=0

and

∞

Q = ∑ bn (z − z0 )n , n=0

we define the sum ∞

P + Q = ∑ (an + bn )(z − z0 )n , n=0

and the product ∞

P ⋅ Q = ∑ cn (z − z0 )n , n=0

where cn = a0 bn + a1 bn−1 + ⋯ + an−1 b1 + an b0 ;

the last expression is also called the Cauchy product. Theorem 1.27. If s, M > 0 are two constants with the property that |an |sn ≤ M

∀n ∈ ℕ0 ,

n then the power series ∑∞ n=0 an (z − z0 ) converges uniformly and absolutely on each compact subset of Ds (z0 ) = {z ∶ |z − z0 | < s}.

Proof. If K is a compact subset of Ds (z0 ) then there exists a positive r < s with K ⊂ Dr (z0 ). Set q = r/s < 1. Then we have sup|an (z − z0 )n | ≤ sup |an (z − z0 )n | ≤ |an |r n = |an |sn z∈K

z∈Dr (z0 )

rn ≤ Mqn , sn

n since ∑∞ n=0 Mq < ∞, the assertion follows from Theorem 1.25.

Remark. If |an |sn ≤ M ∀n ∈ ℕ0 , then the sequence (|an |r n )n converges to 0 for each positive r < s. n Definition 1.28. Let ∑∞ n=0 an (z − z0 ) be a power series and

R ∶= sup{t ≥ 0 ∶ (|an |t n )n is a bounded sequence}. n R is called radius of convergence of the power series ∑∞ n=0 an (z − z0 ) .

1.7 Power series | 19

n Theorem 1.29. Let ∑∞ n=0 an (z − z0 ) be a power series with radius of convergence R. Then: n (1) ∑∞ n=0 an (z − z0 ) is uniformly convergent on each compact subset of DR (z0 ). ∞ (2) ∑n=0 an (z − z0 )n fails to be convergent in ℂ ⧵ DR (z0 ).

Proof. (1) If R = 0, then the assertion is trivial. If R > 0, then we have for an arbitrary positive s < R that the sequence (|an |sn )n is bounded. By Theorem 1.27, the power series n ∑∞ n=0 an (z − z0 ) converges uniformly on each compact subset of Ds (z0 ), and as s was an arbitrary positive number strictly smaller than R, the first assertion of the theorem follows. (2) If |z − z0 | > R, then the sequence (|an | |z − z0 |n )n is unbounded and the power n series ∑∞ n=0 an (z − z0 ) fails to be convergent. In the following theorem we show that the radius of convergence can be computed in terms of the coefficients of the power series. n Theorem 1.30 (Cauchy–Hadamard8 ). Let ∑∞ n=0 an (z − z0 ) be a power series with radius of convergence R. Then −1

R = (lim sup |an |1/n ) n→∞

(1/0 = ∞; 1/∞ = 0).

Proof. Let L = (lim supn→∞ |an |1/n )−1 . We show that L = R. Let ϵ > 0 be an arbitrary positive number. For almost all n ∈ ℕ we have |an |1/n ≤ 1/(L − ϵ). Hence |an |(L − ϵ)n ≤ 1, which implies that the sequence (|an |(L − ϵ)n )n is bounded. So we have L − ϵ ≤ R and L ≤ R, as ϵ was arbitrary. Now suppose that L < ∞. In order to show that R ≤ L, it suffices to prove that R ≤ s ∀s > L. Let L < s < ∞. Then s−1 < L−1 = lim supn→∞ |an |1/n , hence there exists an infinite subset M ⊆ ℕ such that s−1 < |am |1/m

∀m ∈ M.

This implies |am |sm > 1 ∀m ∈ M and (|an |sn )n fails to be a null-sequence. Therefore we must have s ≥ R, because s < R would imply that (|an |sn )n is a null-sequence (see the remark from above). If L = ∞, then we get L = R from the first step of the proof. n n 1/n Examples. (a) ∑∞ = n and R = 0. n=0 n z , |an |

8 Hadamard, Jacques Solomon (1865–1963).

20 | 1 Complex numbers and functions n 1/n (b) ∑∞ = 1 and R = 1. n=0 z n , |an | ∞ z (c) ∑n=0 nn , |an |1/n = 1/n and R = ∞. zn (d) ∑∞ n=0 n! , for this example we use the following n Theorem 1.31. Let ∑∞ n=0 an (z − z0 ) be a power series with radius of convergence R, and suppose that an ≠ 0 for almost all n. Then

lim inf| n→∞

an a | ≤ R ≤ lim sup| n |. an+1 n→∞ an+1

a

If the limit limn→∞ | a n | exists, then n+1

R = lim | n→∞

Example. ∑∞ n=0

a zn , n n! an+1

=

(n+1)! n!

an |. an+1

= n + 1 and R = ∞.

a

Proof. Let S = lim infn→∞ | a n |. If S = 0, we have S ≤ R. If S > 0, it suffices to show that n+1 s ≤ R for each 0 < s < S. Since S is a lim inf, there exists an l ∈ ℕ such that |

an | > s ∀n ≥ l. an+1

Hence |an |s−1 > |an+1 | ∀n ≥ l. Let A = |al |sl . Then the last inequality implies that |al+1 |sl+1 < |al |s−1 sl+1 = |al |sl = A, iterating this argument one obtains |al+m |sl+m ≤ A ∀m ∈ ℕ. Hence the sequence (|an |sn )n is bounded. Now we get from the definition of the radius of convergence that s ≤ R. a Now let T = lim supn→∞ | a n |. If T = ∞, we have R ≤ T. If T < ∞, it remains to n+1 show that t ≥ R for each t > T. We get again an l ∈ ℕ with |

an | |an |t −1 ∀n ≥ l. One can choose l such that B = |al |t l > 0. Then |al+1 |t l+1 > |al |t −1 t l+1 = |al |t l = B, and by iteration |am+l |t l+m > B ∀m ∈ ℕ. Hence the sequence (|an |t n )n fails to be a nullsequence. Therefore we have t ≥ R.

1.8 Line integrals | 21

1.8 Line integrals Here we take up the complex integral calculus. We describe the basic properties of line integrals and its relationship to complex primitives for holomorphic functions. Definition 1.32. Let γ ∶ [a, b] ⟶ U ⊆ ℂ be a continuous map. We call γ a curve in ℂ. We use γ to denote the map and γ ∗ for the set γ ∗ = {γ(t) ∶ t ∈ [a, b]}. γ(a) is called initial point and γ(b) end point of the curve. [a, b] is the parameter interval. If γ(a) = γ(b), then γ is called a closed curve. Let a = s0 < s1 < s2 < ⋯ < sn = b and γ|[sj−1 ,sj ] for j = 1, … , n continuously differen-

tiable (we will also say a 𝒞1 -function). This is a piecewise continuously differentiable curve, which will be called a path. Let f ∶ U ⟶ ℂ be a continuous function and γ ∶ [a, b] ⟶ U a path in U. Then b

n−1

a

j=0 sj

∫ f (z) dz = ∫ f (γ(t))γ ′ (t) dt ∶= ∑ ∫ γ

b

sj+1

f (γ(t))γ ′ (t) dt

b

= ∫ ℜ(f (γ(t))γ ′ (t)) dt + i ∫ ℑ(f (γ(t))γ ′ (t)) dt a

a

is the line integral of f along γ. Remark 1.33. (a) Let φ ∶ [a1 , b1 ] ⟶ [a, b] be a bijective 𝒞1 function such that φ′ > 0 everywhere on [a1 , b1 ], let γ ∶ [a, b] ⟶ U and γ1 ∶= γ ∘ φ ∶ [a1 , b1 ] ⟶ U be paths in U. Since we supposed that φ′ > 0, we have φ(a1 ) = a and φ(b1 ) = b, which means that the orientation in γ and γ1 coincides. For each continuous function f ∶ U ⟶ ℂ we have b1

b1

a1

a1

∫ f (z) dz = ∫ f (γ1 (t))γ1′ (t) dt = ∫ f (γ(φ(t)))γ ′ (φ(t))φ′ (t) dt γ1

b

= ∫ f (γ(s))γ ′ (s) ds = ∫ f (z) dz, a

γ

where we substituted s = φ(t), ds = φ′ (t) dt. We say that the curves γ1 and γ are equivalent. The line integrals are independent of the parametrization. (b) Let γ1 ∶ [a, b] ⟶ U and γ2 ∶ [b, c] ⟶ U be paths in U with γ1 (b) = γ2 (b). We define a new path γ ∶ [a, c] ⟶ U by γ|[a,b] = γ1 and γ|[b,c] = γ2 (composition of paths). Then we have ∫ f (z) dz = ∫ f (z) dz + ∫ f (z) dz. γ

γ1

γ2

(c) Let γ ∶ [0, 1] ⟶ U be a path. We define the inverse path γ1 to γ by γ1 (t) ∶= γ(1−t), t ∈ [0, 1]. Then γ ∗ = γ1∗ and ∫ f (z) dz = − ∫ f (z) dz. γ1

γ

22 | 1 Complex numbers and functions This follows by the substitution 1 − t = s, dt = −ds in the integral 1

0

∫ f (z) dz = ∫ f (γ(1 − t))(−γ ′ (1 − t)) dt = ∫ f (γ(s))(−γ ′ (s))(−1) ds γ1

0

1

1

= − ∫ f (γ(s))γ ′ (s) ds = − ∫ f (z) dz. 0

γ

We write γ1 = γ −1 . (d) Let γ ∶ [a, b] ⟶ U be a path and f ∶ U ⟶ ℂ a continuous function. The length L(γ) of the path γ is given by b

L(γ) = ∫ |γ ′ (t)| dt. a

We have |∫ f (z) dz| ≤ L(γ) max |f (z)|, ∗ z∈γ

γ

which is shown by the following estimate: b

b

a

a

|f (z)|. |∫ f (z) dz| = |∫ f (γ(t))γ ′ (t) dt| ≤ ∫ |f (γ(t))| |γ ′ (t)| dt ≤ L(γ) max ∗ γ

z∈γ

Example 1.34. (1) Let a ∈ ℂ, r > 0 and γ(t) = a + r(cos t + i sin t), t ∈ [0, 2π] the positively oriented circle with center a and radius r (once passed through). Let f ∶ U ⟶ ℂ be a continuous function and suppose that Dr (a) ⊂ U. Then 2π

∫ f (z) dz = r ∫ f (a + r(cos t + i sin t))(− sin t + i cos t) dt. 0

γ

Special cases are as follows: – a = 0, r = 1, f (z) = z. Here 2π

∫ z dz = i ∫ (cos 2t + i sin 2t) dt = 0; γ

–

0

f (z) = z. Here 2π

2π

0

0

∫ z dz = i ∫ (cos2 t + sin2 t) dt = i ∫ γ

–

dt = 2πi;

f (z) = 1/z. Here ∫

γ

2π − sin t + i cos t 1 dz = ∫ dt = 2πi. z cos t + i sin t 0

1.8 Line integrals | 23

–

For arbitrary a ∈ ℂ and r > 0, one has ∫

γ

2π 2π 1 dz − sin t + i cos t =r∫ dt = ir ∫ dt = 2πi. z−a 0 a + r(cos t + i sin t) − a 0 r

(2) Let a, b ∈ ℂ, a ≠ b. The path γ(t) = a + (b − a)t, t ∈ [0, 1] describes the straight line segment [a, b] joining the points a and b, f ∶ U ⟶ ℂ, where γ ∗ ⊂ U. Then 1

∫ f (z) dz = (b − a) ∫ f (a + (b − a)t) dt. 0

γ

If a = −1, b = 1 and f (z) = z, then 1

∫ z dz = 2 ∫ (−1 + 2t) dt = 2(−t + t 2 )|10 = 0. 0

γ

If γ1 is the positively oriented semicircle between −1 and 1, then 2π

∫ z dz = i ∫ (cos 2t + i sin 2t) dt = 0. π

γ1

Taking the function f (z) = z, we see that the line integral does not depend on the path between −1 and 1. But for f (z) = z we have 1

∫ z dz = 2 ∫ (−1 + 2t) dt = 0, 0

γ

and 2π

∫ z dz = i ∫ (cos2 t + sin2 t) dt = iπ. γ1

π

(3) Let a, b, c ∈ ℂ and Δ = Δ(a, b, c) be the triangle with vertices a, b, c, with the orientation a → b → c → a. Then ∫ f (z) dz = ∫ 𝜕Δ

[a,b]

f (z) dz + ∫

[b,c]

f (z) dz + ∫

[c,a]

f (z) dz.

If Δ′ = Δ(a, c, b) is the triangle with the orientation a → c → b → a, then ∫

𝜕Δ′

f (z) dz = ∫

[a,c]

f (z) dz + ∫

= −∫ [c,a]

[c,b]

f (z) dz − ∫

= − ∫ f (z) dz. 𝜕Δ

f (z) dz + ∫

[b,c]

[b,a]

f (z) dz − ∫

f (z) dz

[a,b]

f (z) dz

24 | 1 Complex numbers and functions

1.9 Primitive functions In the following we state an analogue to the fundamental theorem of calculus and show that the second statement of the fundamental theorem holds for holomorphic functions on convex domains. Definition 1.35. Let U ⊆ ℂ be an open set and f ∶ U ⟶ ℂ a continuous function. f has a primitive function on U, if there exists a holomorphic function F on U such that F ′ = f on U. F is called a primitive function for f . Remark. If U ⊆ ℂ is a connected set and f ∶ U ⟶ ℂ has two primitive functions F1 and F2 on U, then we have F1 = F2 + C, where C is a constant. This follows from F1′ − F2′ = f − f = 0. Theorem 1.36. Suppose that the continuous function f ∶ U ⟶ ℂ has a primitive function F on U. Let z0 , z1 ∈ U and γ an arbitrary path in U from z0 to z1 . Then ∫ f (z) dz = F(z1 ) − F(z0 ). γ

Proof. Let γ ∶ [a, b] ⟶ U, a = t0 < t1 < ⋯ < tn = b and γ a 𝒞1 map on [tk−1 , tk ] for k = 1, … , n. Then n

∫ f (z) dz = ∑ ∫ γ

tk

k=1 tk−1 n tk

n

f (γ(t))γ ′ (t) dt = ∑ ∫ n

tk

k=1 tk−1

F ′ (γ(t))γ ′ (t) dt

= ∑ ∫ (F ∘ γ)′ (t) dt = ∑ [F(γ(tk )) − F(γ(tk−1 ))] k=1 tk−1

= F(z1 ) − F(z0 ).

k=1

Corollary 1.37. Suppose that the continuous function f ∶ U ⟶ ℂ has a primitive on U, then ∫ f (z) dz = 0, γ

for each closed path γ in U. Remark. If F ∈ ℋ(U) and F ′ is continuous (later we will see that we do not need the last assumption), then ∫ F ′ (z) dz = 0, γ

for each closed path γ in U. Example 1.38. (a) The function f (z) = z n , n ∈ ℤ, n ≠ −1 has F(z) = on ℂ for n ≥ 0 and on ℂ∗ for n < −1. We have

z n+1 n+1

as a primitive

1.9 Primitive functions |

∫ [z0 ,z1 ]

z n dz =

1 (z n+1 − z0n+1 ), n+1 1

25

∫ z n dz = 0, γ

for n ≠ −1 and for each closed path γ with 0 ∉ γ ∗ . ak k+1 The complex polynomial p(z) = ∑nk=0 ak z k has ∑nk=0 k+1 z as a primitive. (b) For the function f (z) = z, the line integral ∫γ f (z) dz depends not only on the initial

and endpoint of the path (see Example 1.34 (2)), hence, by Theorem 1.36, f has no primitive on any open subset of ℂ. (c) If γ(t) = cos t + i sin t, t ∈ [0, 2π], then ∫

γ

dz = 2πi, z

so, by Corollary 1.37, the function f (z) =

1 z

has no primitive on ℂ∗ .

Now we will prove the converse of Corollary 1.37. Theorem 1.39. Let G ⊆ ℂ be a domain (open and connected) and f ∶ G ⟶ ℂ a continuous function. Suppose that for each closed path γ in G we have ∫ f (z) dz = 0. γ

Then f has a primitive on G. Proof. Fix a ∈ G. For an arbitrary z ∈ G, we choose a path γz in G from a to z and we set F(z) = ∫ f (ζ ) dζ . γz

We will show that F (z0 ) = f (z0 ) for an arbitrary z0 ∈ G. If z is sufficiently close to z0 , then [z0 , z] ⊂ G and the path γ, which is composed by γz0 , [z0 , z] and γz−1 , is a closed path in G. Hence, by assumption, we have ′

0 = ∫ f (ζ ) dζ = ∫ f (ζ ) dζ + ∫ γ

γz0

[z0 ,z]

f (ζ ) dζ − ∫ f (ζ ) dζ . γz

This implies F(z) − F(z0 ) = ∫ f (ζ ) dζ − ∫ f (ζ ) dζ = ∫ γz

γz0

1

[z0 ,z]

f (ζ ) dζ

= ∫ f (z0 + t(z − z0 ))(z − z0 ) dt = (z − z0 )A(z), 0

1

where A(z) = ∫0 f (z0 + t(z − z0 )) dt, A(z0 ) = f (z0 ). Hence A(z) =

F(z) − F(z0 ) . z − z0

26 | 1 Complex numbers and functions In order to show F ′ (z0 ) = f (z0 ), it suffices to prove that A is continuous at z0 . For this we estimate |A(z) − A(z0 )| ≤ max |f (z0 + t(z − z0 )) − f (z0 )|. t∈[0,1]

Since f is continuous at z0 , we get now the same for A. Hence F is a primitive of f on G. We remark that the definition of the function F does not depend on the choice of the path γz . If γ̃ z is a different path from a to z, then γz γ̃ −1 z is a closed path in G and hence ∫ f (ζ ) dζ − ∫ f (ζ ) dζ = ∫ γz

γ̃ z

γz γ̃ −1 z

f (ζ ) dζ = 0,

which implies that ∫ f (ζ ) dζ = ∫ f (ζ ) dζ . γz

γ̃ z

If we assume something more about the domain G, we can considerably weaken the assumptions in Theorem 1.39. Definition 1.40. The domain G ⊆ ℂ is convex, if whenever two points z0 , z1 belong to G, then the straight line segment [z0 , z1 ] joining the two points is contained in G. Theorem 1.41. Let G be a convex domain in ℂ and f ∶ G ⟶ ℂ a continuous function. Suppose that for each triangle Δ ⊆ G ∫ f (z) dz = 0. 𝜕Δ

Then f has a primitive on G. Proof. Fix a ∈ G and let F(z) = ∫

[a,z]

f (ζ ) dζ ,

for z ∈ G. By assumption, the straight line segment [a, z] ⊂ G. If z0 ∈ G, then the triangle Δ with vertices a, z, z0 is contained in G and hence 0 = ∫ f (ζ ) dζ = ∫ 𝜕Δ

[a,z0 ]

f (ζ ) dζ + ∫

[z0 ,z]

f (ζ ) dζ − ∫

[a,z]

f (ζ ) dζ .

Now we can continue as in the proof of Theorem 1.39. Remark. If G is not convex, the assertions of Theorem 1.41 are true at least in each convex neighborhood Uz ⊆ G of an arbitrary point z ∈ G.

1.9 Primitive functions |

27

In the following we study the interchange of limit processes where uniform limits of sequences of functions and line integrals are involved. Theorem 1.42. Let γ be a path in ℂ and let (fn )n be a sequence of continuous functions on γ ∗ . Suppose that the sequence (fn )n converges uniformly on γ ∗ to a function f . Then lim ∫ fn (z) dz = ∫ lim fn (z) dz = ∫ f (z) dz.

n→∞ γ

γ n→∞

γ

Proof. By Remark 1.33 (d), we have |∫ fn (z) dz − ∫ f (z) dz| = |∫ (fn (z) − f (z)) dz| ≤ L(γ) max |fn (z) − f (z)|, ∗ γ

γ

z∈γ

γ

which implies the assertion. ∗ Remark. If the series ∑∞ n=0 fn of continuous functions fn converges uniformly on γ , then ∞

∞

∫ ( ∑ fn (z)) dz = ∑ ∫ fn (z) dz. γ

n=0 γ

n=0

n Theorem 1.43. Let P(z) = ∑∞ n=0 an (z − z0 ) be a power series with radius of convergence R > 0. Then P is holomorphic on DR (z0 ) and ∞

P ′ (z) = ∑ nan (z − z0 )n−1 n=1

(we can interchange summation and differentiation). Proof. First we show the following assertion. Let R′ be the radius of convergence of n−1 the power series Q(z) = ∑∞ . Then R′ ≥ R. n=1 nan (z − z0 ) n Without loss of generality, we can assume that z0 = 0. The power series ∑∞ n=0 nz n has radius of convergence 1, i.e. ∀ρ ∈ [0, 1) the sequence (nρ )n is bounded. Now we have |an z1n | ≤ M, ∀n ∈ ℕ (M > 0), for an arbitrary z1 ∈ DR (0), in addition we have |nan z2n−1 | =

z n M n z2 n | | |an z1n | ≤ n| 2 | , |z2 | z1 z1 |z2 |

z

for 0 < |z2 | < |z1 |. Since | z2 | < 1, the sequence (|nan z2n−1 |)n is bounded. As |z2 | < |z1 | < R 1 were chosen arbitrarily, we get from the definition of the radius of convergence (see Definition 1.28) that R′ ≥ R. Now let γ be an arbitrary closed path in DR′ (0). We will show that Q has P as a primitive on DR′ (0): ∞

∞

n=1

n=1

∫ Q(z) dz = ∫ ( ∑ nan z n−1 ) dz = ∑ nan ∫ z n−1 dz = 0, γ

γ

γ

28 | 1 Complex numbers and functions here we interchanged integration and summation (see Theorem 1.42) and used Example 1.38 (a). By Theorem 1.39, this implies that there exists a primitive of Q on DR′ (0). A primitive of Q is ∫ [0,z]

Q(ζ ) dζ = ∫

[0,z] ∞

∞

∞

n=1 n

n=1

( ∑ nan ζ n−1 ) dζ = ∑ nan ∫

= ∑ nan n=1

z = ∑ a zn , n n=1 n ∞

[0,z]

ζ n−1 dζ

where we used again Theorem 1.42 and Example 1.38 (a). Hence also ∞

P(z) = a0 + ∑ an z n n=1

is a primitive of Q on DR′ (0). We have also seen that P converges on DR′ (0), hence R′ = R. Remark. We can apply Theorem 1.43 to P ′ to see that P ′ is holomorphic on DR (z0 ). Iterating this argument, we obtain the existence of derivatives of P of arbitrary order and the formula ∞

P (k) (z) = ∑ n(n − 1) ⋯ (n − k + 1)an (z − z0 )n−k , n=k

an =

P (n) (z0 ) . n!

1.10 Elementary functions Here we define the complex elementary functions by its power series and derive the most important properties including a definition of π and a proof of the Eulerian identity e2πi = 1. Definition 1.44. We define the complex exponential function by the power series zn exp z = ez = ∑∞ n=0 n! . The series converges uniformly on all compact subsets of ℂ. By Theorem 1.43, we have nz n−1 z n−1 zn =∑ =∑ = ez , n=1 n! n=1 (n − 1)! n=0 n! ∞

(ez )′ = ∑

∞

∞

i.e. (ez )′ = ez . Theorem 1.45. For z, w ∈ ℂ we have ez+w = ez ew . Proof. Let f (z) = e−z ez+w . Then f ′ (z) = −e−z ez+w + e−z ez+w = 0

∀z ∈ ℂ.

1.10 Elementary functions | 29

f is holomorphic in ℂ, and since f ′ = ux + ivx = vy − iuy = 0, we obtain ux = uy = vx = vy = 0. Hence f = const., and e−z ez+w = f (0) = ew , setting w = 0, we get e−z ez = e0 = 1. This implies (ez )−1 = e−z

and ez+w = ez ew .

Remark. We know from the last proof that e−z ez = e0 = 1 for all z ∈ ℂ. Hence the exponential function has no zeroes. Next we define the complex sine and cosine function again by power series. Definition 1.46. (−1)n 2n+1 z , n=0 (2n + 1)! ∞

sin z = ∑

(−1)n 2n z , n=0 (2n)! ∞

cos z = ∑

the series converge uniformly on all compact subsets of ℂ. Using the power series, it is easy to show the following formulas: cos z + i sin z = eiz , cos(−z) = cos z, sin(−z) = − sin z, 1 1 cos z = (eiz + e−iz ), sin z = (eiz − e−iz ), 2 2i (sin z)′ = cos z, (cos z)′ = − sin z. For z = x + iy, we have by Theorem 1.45 ez = ex+iy = ex eiy = ex (cos y + i sin y), since |eiy |2 = eiy eiy = (cos y + i sin y)(cos y − i sin y) = eiy e−iy = 1, we get |eiy | = 1 and cos2 y + sin2 y = 1 for all y ∈ ℝ. In addition ℜez = ex cos y,

ℑez = ex sin y,

|ez | = |ex (cos y + i sin y)| = ex = eℜz ,

since | cos y + i sin y| = 1. The expression ez = ex (cos y + i sin y)

30 | 1 Complex numbers and functions can be interpreted as a polar representation of the exponential function, and we have arg ez = ℑz. Next we will determine the zeroes of the sine and cosine functions. First we claim that there are positive real numbers x such that cos x = 0. Suppose that this is not true. Then, since cos 0 = 1, we have cos x > 0 for each x > 0. Hence we have for the derivative sin′ x = cos x > 0 and the function sin would be strictly increasing. As sin 0 = 0, we would have sin x > 0 for x > 0. This would imply that for 0 < x1 < x2 , x2

(x2 − x1 ) sin x1 < ∫ sin t dt = cos x1 − cos x2 ≤ 2. x1

Since we have sin x > 0, the last inequality gives a contradiction if x2 is sufficiently large. The zero set of the continuous function cos is closed and cos 0 ≠ 0, hence there exists a smallest positive number x0 with cos x0 = 0. We define the number π by π ∶= 2x0 . Then cos(π/2) = 0 and since cos2 x + sin2 x = 1 this implies sin(π/2) = ±1. As cos x > 0 in (0, π/2), the function sin is increasing in (0, π/2). Hence sin(π/2) = 1. This implies exp(

π i) = i. 2

By Theorem 1.45, we get exp(πi) = −1,

exp(2πi) = 1.

In this way we get that all real zeroes of the function cos are of the form {(2k + 1)π/2 ∶ k ∈ ℤ}, and all real zeroes of the function sin are of the form {kπ ∶ k ∈ ℤ}. Suppose that cos z = 0. Set z = x + iy and recall that 2 cos z = eiz + e−iz . It follows that e2iz + 1 = 0, which implies e−2y cos 2x = −1, e−2y sin 2x = 0. Hence cos z = 0 implies that z must be real and cos has no other zeroes. Also sin has no other zeroes. Remark. (a) The mapping φ ∶ t ↦ eit is a homomorphism between the additive group (ℝ, +) and the multiplicative group S1 = {z ∈ ℂ ∶ |z| = 1}, since we have φ(t1 + t2 ) = φ(t1 )φ(t2 ),

t1 , t2 ∈ ℝ.

(b) The mapping ψ ∶ z ↦ ez is a homomorphism between the additive group (ℂ, +) and the multiplicative group (ℂ∗ = ℂ ⧵ {0}, ⋅): ψ(z1 + z2 ) = ψ(z1 )ψ(z2 )

∀z1 , z2 ∈ ℂ.

1.10 Elementary functions | 31

ψ is surjective. To see this, let w ∈ ℂ∗ , then each z = x + iy with x = log |w| and y = arg w is the pre-image of w, since we have ez = ex eiy = elog |w| ei arg w = |w|ei arg w , where the last expression corresponds to the polar representation of w. ψ fails to be injective. Ker ψ = {z ∶ ψ(z) = 1} = {2πik ∶ k ∈ ℤ} = 2πiℤ, since 1 = ez = ex (cos y + i sin y) implies x = 0 and y = 2πk, k ∈ ℤ. The exponential function is periodic with period 2πi, i.e. exp(z + 2πi) = exp(z). (c) Each strip {z ∈ ℂ ∶ a ≤ ℑz < a + 2π}, a ∈ ℝ, is mapped onto ℂ∗ by the exponential function. This follows from (b). In the following we introduce some other elementary functions related to the exponential function. Definition 1.47.

sin z , cos z cos z cot z = , sin z

tan z =

We have

z ≠ (k + 1/2)π, z ≠ kπ,

k∈ℤ

k ∈ ℤ.

1 e2iz − 1 e2iz + 1 , cot z = i , i e2iz + 1 e2iz − 1 1 1 (tan z)′ = , (cot z)′ = − 2 . cos2 z sin z tan z =

Definition 1.48.

We have

1 cosh z = (ez + e−z ), 2

cosh z = cos iz,

sinh z =

1 sin iz, i

1 sinh z = (ez − e−z ). 2 cos z = cosh iz,

sin z =

1 sinh iz. i

Now we investigate the inverse functions of the elementary functions introduced above. Definition 1.49. For each z ∈ ℂ∗ there are infinitely many w ∈ ℂ with ew = z. Each of these values w is called a logarithm of z. Each logarithm has the form w = log |z| + i arg z. Two logarithms of z differ by an entire multiple of 2πi. Let ℂ− = ℂ ⧵ {z ∈ ℂ ∶ ℑz = 0, ℜz ≤ 0} be the slit plane. For each z ∈ ℂ− we have a unique representation of z = |z|eiφ with −π < φ < π. We define Log z ∶= log |z| + iφ, as the principal branch of the logarithm.

z ∈ ℂ−

32 | 1 Complex numbers and functions Example 1.50. Log i = iπ/2. Theorem 1.51. Let G0 = {z ∈ ℂ ∶ −π < ℑz < π}. Then the exponential function exp ∶ G0 ⟶ ℂ− is holomorphic and bijective, the inverse function is the principal branch of the logarithm, it is a map from ℂ− onto G0 , which is also holomorphic and bijective. Proof. For z = x + iy ∈ G0 , we observe that w = ez = ex eiℑz belongs to the domain of the principal branch of the logarithm, since |w| = ex > 0 and arg w = ℑz. Then we have Log(exp z) = x + iℑz = x + iy = z. If w ∈ ℂ− , then exp(Log w) = exp(log |w| + i arg w) = |w|ei arg w = w. Hence exp and Log are inverses to each other, and so both of them are bijective. Since (exp z)′ = exp z ≠ 0 ∀z ∈ ℂ, it follows from the differentiation rule for inverse functions (see Theorem 1.7) that the principal branch of the logarithm Log is holomorphic on ℂ− , and we have (Log w)′ =

1 , w

w ∈ ℂ− .

Remark. One has to be careful when using the functional equation which is valid for the real logarithm. The following lines show that one must stay in the domain of the branch of the logarithm when using the functional equation: log(−1) = log(i ⋅ i) = log i + log i = iπ/2 + iπ/2 = iπ, but 0 = log 1 = log((−1)(−1)) = log(−1) + log(−1) = 2 log(−1) ⇒ log(−1) = 0. Definition 1.52. Let z ∈ ℂ− and a ∈ ℂ. We define z a ∶= exp(a Log z). The chain rule (see Theorem 1.6) shows that (z a )′ = az a−1 . Example 1.53. (i) 1a = exp(a Log 1) = e0 = 1, ∀a ∈ ℂ. (ii) The function z ↦ z 1/2 = exp(1/2 Log z) for z ∈ ℂ− is the principal branch of the complex root function.

1.11 Exercises | 33

(iii) In addition, we have ii = exp(i Log i) = exp(i2 π/2) = exp(−π/2) = 0.208 … and √i = i1/2 = exp(1/2 Log i) = exp(iπ/4) =

√2 (1 + i). 2

Also here one has to be careful when using arithmetic rules like (z α )β = z αβ , which is to be understood as an equality of sets. Otherwise it can lead to absurd conclusions. Indeed, let z ≠ 0 and set z = eα for a certain α ∈ ℂ; let β = α/2πi. Then z = eα = e2πiβ = (e2πi )β = 1β = 1.

1.11 Exercises 1.

Determine the real and imaginary part and the absolute value of 2 , 1 − 3i

(1 + i√3)6 ,

(

1+i 5 ) , 1−i

(

1 + i √3 4 ) . 1−i

2. Determine all complex numbers z, for which z = z 2 . 3. Show that for |z| = r > 0: 1 r2 ℜz = (z + ), 2 z

ℑz =

1 r2 (z − ). 2i z

4. Prove the identity |z1 + z2 |2 + |z1 − z2 |2 = 2(|z1 |2 + |z2 |2 ), 5.

and explain its geometric meaning. Determine the absolute value and the principal argument of 2.718 − 3.010i,

3 + 2i , 5i − 4

3(cos

4π 4π + i sin ). 3 3

3+i 6. Let z0 = −1−i . Compute z0123 . 7. Compute all eighth roots of 256(cos 80∘ + i sin 80∘ ). 8. Determine all real numbers x and y, for which the following equations hold: (i) 2(x + iy) = (x + iy)2 , (ii) |2 − (x − iy)| = x + iy, x−iy (iii) x+iy = i. √

34 | 1 Complex numbers and functions 9.

Give a geometric description of the sets of all points determined by the following relations: (i) |z − 2 + 3i| < 5, (ii) ℑz ≥ ℜz, (iii) |z − i| + |z − 1| = 2, (iv) ℜz = |z − 2|. 10. Let |w| < 1. Show that |

z−w | < 1, wz − 1

for |z| < 1,

|

z−w | = 1, wz − 1

for |z| = 1.

and

11. Let f (z) = U(r, θ) + iV(r, θ), where z = r(cos θ + i sin θ). Let f be complex differentiable at z0 ≠ 0. Show that at z0 we have 𝜕U 1 𝜕V = 𝜕r r 𝜕θ

𝜕V 1 𝜕U =− . 𝜕r r 𝜕θ

and

These are the Cauchy–Riemann differential equations in polar coordinates. 12. Let f (z) = u(x, y) + iv(x, y) and suppose that f satisfies the Cauchy–Riemann differential equations in an open set G ⊆ ℂ. Let G the set in ℂ obtained by the reflection of G with respect to the real axis, i.e. (x, y) ∈ G, if (x, −y) ∈ G. Define a function g on G by g(z) = f (z),

z ∈ G.

Show that g satisfies the Cauchy–Riemann differential equations in G. 13. Suppose that a function f satisfies the Cauchy–Riemann differential equations for |z| < R. Define g by g(z) = f (R2 /z),

14.

15. 16. 17.

|z| > R.

Show that g satisfies the Cauchy–Riemann differential equations for |z| > R. (Use polar coordinates!) Let a, b, c ∈ ℝ. For x, y ∈ ℝ and z = x + iy set P(z) = ax2 + 2bxy + cy2 . Find a necessary and sufficient condition for the existence of a holomorphic function f ∈ ℋ(ℂ) such that ℜ(f ) = P. Suppose that f is holomorphic in ℂ and real-valued. Show that f is constant. Let f (z) = |z|4 + (ℑz)2 . Compute fz and fz . Let f be a real differentiable function. Prove that 𝜕f 𝜕f = , 𝜕z 𝜕z

𝜕f 𝜕f = . 𝜕z 𝜕z

1.11 Exercises | 35

18. Let f be a real-valued, real-differentiable function. Show 𝜕f 𝜕f = . 𝜕z 𝜕z 19. Prove the following chain rules: 𝜕(g ∘ f ) 𝜕g 𝜕f 𝜕g 𝜕f = + , 𝜕z 𝜕w 𝜕z 𝜕w 𝜕z 𝜕(g ∘ f ) 𝜕g 𝜕f 𝜕g 𝜕f = + . 𝜕z 𝜕w 𝜕z 𝜕w 𝜕z 20. Let φ be a differentiable function of the real variable t. Show that d(f ∘ φ) 𝜕f dφ 𝜕f dφ = + . dt 𝜕z dt 𝜕z dt 21. Let 1 1 f (z) = (z + ), 2 z

z ∈ ℂ ⧵ {0}.

Show that f is angle preserving on ℂ ⧵ {−1, 0, 1}. Determine the images under f of the circles |z| = r < 1 and of the rays z = ct, 0 < t < 1, |c| = 1 and use these image curves in order to check the angle preservation of f . 22. Let fn (z) =

1 , 1 + az n

a ≠ 0.

Show that the sequence (fn ) converges to 1 uniformly on each compact subset of the open unit disc D1 (0) and that (fn ) converges to 0 uniformly on each compact subset of ℂ ⧵ Dr (0), for r > 1. 23. Show that the series zn n n+1 ) n=1 (1 − z )(1 − z ∞

∑

converges to z(1 − z)−2 uniformly on each compact subset of D1 (0) and that it converges to (1 − z)−2 uniformly on each compact subset of ℂ ⧵ D1 (0). 24. Where is the series zn n n=1 1 − z ∞

∑

uniformly convergent?

36 | 1 Complex numbers and functions 25. Let R1 and R2 be the radii of convergence of the power series ∞

∑ an z n

∞

∑ bn z n .

and

n=0

n=0

Show that the radius of convergence R of ∞

∑ an bn z n

n=0

satisfies R ≥ R1 R2 ; the radius of convergence R′ of an n z , b n=0 n ∞

∑

bn ≠ 0,

n ∈ ℕ0

satisfies R′ ≤ R1 /R2 ; and the radius of convergence R0 of ∞

∑ (an b0 + an−1 b1 + ⋯ + a0 bn )z n

n=0

satisfies R0 ≥ min(R1 , R2 ). 26. Determine the radius of convergence of the following power series ∞

∑ (−1)n (z + i)n ,

n=0 ∞ n n

2 z , ∑ (2n)! n=1 ∞

n

∑ 2−n z 2 ,

n=0 ∞

∞

∞

∑ (−1)n 2n z 2n+2 ,

n=0

n! ∑ n zn , n n=1 ∞

(2n)! n z , ∑ 2 n=0 (n!) ∞

n=2

∞

n=1

∞

∑ 2log n z n ,

√(2n)! 2n z , ∑ n! n=0

∞

∑ n−1/2 z n ,

∑ (n + an )z n ,

n=0 n

∑ n−1 z 3 ,

∞

∑[

n=1

n=1

n n2 ] zn . (n + 1)(n + 2)

27. Suppose that ∞

f (z) = ∑ an z n n=0

has radius of convergence R > 0. Prove that 1 2π ∫ |f (reiθ )|2 dθ = ∑ |an |2 r 2n , 2π 0 n=0 ∞

Suppose that f is also bounded on DR (0), i.e. |f (z)| ≤ M,

z ∈ DR (0),

0 < r < R.

1.11 Exercises | 37

for some constant M > 0. Show that in this case ∞

∑ |an |2 R2n ≤ M 2 .

n=0

Let r < R and M(r) = sup0≤θ≤2π |f (reiθ )|. Prove that |an | ≤ r −n M(r),

n ∈ ℕ0 .

28. Compute the following line integrals: ∫ z dz, ∫ z dz, γ1

dz , ∫ γ4 z

γ2

∫ z dz, γ3

∫ z dz, γ5

where γ1 is the polygon from (1, 0) to (0, 0) and then to (0, 1), and γ2 is the polygon from (0, 0) to (1, 1). γ3 is the square with vertices (0, 0), (1, 0), (1, 1), (0, 1) passed through once in positive direction, γ4 is the unit circle |z| = 1 passed through once in positive direction, and γ5 is a quarter of a circle with vertices (0, 0), (2, 0), (0, 2) passed through once in positive direction. 29. Let Log denote the principal branch of the complex logarithm. Determine for which z, w ∈ ℂ one has Log(zw) = Log z + Log w. 30. Let z ≠ 0. Determine α, β ∈ ℂ such that (z α )β = z αβ . 31. Show that the complex sine function sin maps the strip {z = x + iy ∶ −π/2 < x < π/2, y ∈ ℝ} bijectively onto the region G obtained from the plane by deleting the two intervals (−∞, −1] and [1, ∞). Determine the inverse function of sin on G in terms of the principal branch of the complex logarithm. 32. Determine the real and imaginary part of tan z and cot z. 33. Show that for z = x + iy one has | tan z|2 =

sin2 x + sinh2 y cos2 x + sinh2 y

and | tanh z|2 =

sinh2 x + sin2 y . sinh2 x + cos2 y

34. Compute the sum of the geometric series n

∑ ekiθ

k=0

and conclude from the result the sums of n

∑ sin kθ

k=1

and

n

1 + ∑ cos kθ. 2 k=1

38 | 1 Complex numbers and functions

1.12 Notes We assume basic knowledge of real analysis and topology. The interested reader is referred to other introductory textbooks for more details: [1, 4, 12, 11, 15, 21, 23, 26, 53, 63]. For interesting historical remarks, see [60].

2 Cauchy’s Theorem and Cauchy’s formula The first section is devoted to winding numbers of closed paths. This is an essentially topological concept but has important implications for the behavior of holomorphic functions. Section 2.1 also contains a holomorphic parameter integral which will be frequently used in the sequel. The theorem of Cauchy–Goursat1 is presented with the classical proof of A. Pringsheim.2 One of the most important consequences is Cauchy’s integral formula, which is a first example of a reproducing formula. This gives the tools required in Sections 2.3, 2.4, 2.5 and 2.6 to prove the most important local properties of holomorphic functions, like the local Taylor series expansion, behavior of zeroes of holomorphic functions and the maximum principle. A more general holomorphic parameter integral is discussed in Section 2.7. The rest of the chapter is devoted to generalizations of Cauchy’s integral formula – the inhomogeneous Cauchy formula, the homology and the homotopy version of it – and to meromorphic functions, their residues and applications of the residue theorem to compute real integrals and Fourier3 transforms.

2.1 Winding numbers The boundary of a domain in ℂ may be unusually complicated. A natural assumption is that the boundary consists of one or several closed curves. For this purpose the Jordan Curve Theorem will be relevant. We recall some topological concepts. Remark. Let U ⊆ ℂ be an open set, p, q ∈ U. We say p is equivalent to q (p ∼ q), if there exists a curve in U joining p with q. It is easily seen that ∼ defines an equivalence relation, the associated equivalence classes are the connected components of U. For p ∈ U let Up denote the equivalence class containing p. Up is the largest connected subset of U containing p. Theorem 2.1 (Jordan4 Curve Theorem). Let γ ∶ [a, b] ⟶ ℂ be a closed Jordan curve in ℂ, i.e. γ(a) = γ(b), but γ(s) ≠ γ(t) for any s, t ∈ (a, b), s ≠ t. Then the open set ℂ ⧵ γ ∗ has two connected components, a bounded one and an unbounded one. The bounded component is called the interior of γ, the unbounded one is called the exterior of γ. The bounded one is simply connected and γ ∗ is the boundary of each of the components. 1 2 3 4

Goursat, Édouard Jean Baptiste (1858–1936). Pringsheim, Alfred (1850–1941). Fourier, Joseph (1768–1830). Jordan, Camille (1838–1921).

https://doi.org/10.1515/9783110417241-002

40 | 2 Cauchy’s Theorem and Cauchy’s formula The proof is lengthy and difficult, see [19]. The next result leads to the concept of the winding number (or index). Theorem 2.2. Let γ be a closed path and let Ω = ℂ ⧵ γ ∗ . Let Indγ (z) ∶=

1 dζ , ∫ 2πi γ ζ − z

z ∈ Ω.

Then Indγ is an integer-valued function on Ω, this means Indγ (Ω) ⊆ ℤ, and Indγ is constant on each connected component of Ω, in addition, Indγ (z) = 0 for each z belonging to the unbounded connected component of Ω. Indγ (z) is called the winding number of γ with respect to z. Remark. The set γ ∗ is compact, hence there exists a disc D such that γ ∗ ⊂ D. The set ℂ ⧵ D is connected and is therefore contained in a connected component of Ω. Hence Ω has exactly one unbounded connected component. In order to prove the theorem above, we need the following Lemma 2.3. Let ϕ, ψ ∶ [a, b] ⟶ ℂ be two curves in ℂ. Let Ω ⊂ ℂ be an open subset of ℂ such that Ω ∩ ϕ∗ = ∅. Let b

f (z) = ∫

a

ψ(t) dt, ϕ(t) − z

z ∈ Ω.

Then f is a holomorphic function on Ω. Remark. Later on we will prove another general result on a holomorphic parameter integral, see Theorem 2.43. Proof. We consider an arbitrary w ∈ Ω. There exists r > 0 such that Dr (w) ⊆ Ω. We will show that ∞

f (z) = ∑ cn (z − w)n , n=0

where the sum is uniformly convergent on each compact subset of Dr (w). Then, by Theorem 1.43, the function f is holomorphic on Dr (w) and as w ∈ Ω was an arbitrary point, we conclude that f is holomorphic on Ω. For this aim we take z ∈ Dr/2 (w) and observe that the denominator in the integral can be written as 1 1 ϕ(t) − w 1 1 = = z−w ϕ(t) − z ϕ(t) − w ϕ(t) − w − (z − w) ϕ(t) − w 1 − ϕ(t)−w 1 z−w n (z − w)n ) =∑ , ∑( n+1 ϕ(t) − w n=0 ϕ(t) − w n=0 (ϕ(t) − w) ∞

=

∞

2.1 Winding numbers | 41

since |ϕ(t) − w| > r, ∀t ∈ [a, b] and Ω ∩ ϕ∗ = ∅, the last sum converges uniformly for all t ∈ [a, b]. Hence b (z − w)n (z − w)n ]ψ(t) dt = ψ(t) dt ∑ ∫ n+1 n+1 n=0 (ϕ(t) − w) n=0 a (ϕ(t) − w)

b

∞

∞

f (z) = ∫ [ ∑ a ∞

= ∑ [∫ n=0

b

a

ψ(t) dt](z − w)n = ∑ cn (z − w)n , (ϕ(t) − w)n+1 n=0 ∞

where b

|cn | = |∫

a

ψ(t) b−a dt| ≤ n+1 max |ψ(t)|. t∈[a,b] (ϕ(t) − w)n+1 r

Now the series ∞

f (z) = ∑ cn (z − w)n n=0

converges uniformly for z in an arbitrary compact subset of Dr (w), because we can estimate ∞

∞

n=0

n=0

∑ |cn | |(z − w)n | ≤ C ∑ ρn ,

for some ρ < 1. Proof of Theorem 2.2. Since Indγ (z) =

dζ 1 b γ ′ (s) 1 = ds, ∫ ∫ 2πi γ ζ − z 2πi a γ(s) − z

z ∈ Ω,

we have Indγ (z) ∈ ℤ ⟺ Φ(b) = 1, where t

Φ(t) = exp[∫

a

γ ′ (s) ds], γ(s) − z

and we used the fact that w ∈ℤ 2πi

⟺

ew = 1,

see Definition 1.46. Now we compute the logarithmic derivative of Φ(t): Φ′ (t) γ ′ (t) = , Φ(t) γ(t) − z excluding the finitely many points a = s0 < s1 < ⋯ < sn = b, where γ possibly fails to be differentiable. In addition, we have Φ′ (t)(γ(t) − z) − Φ(t)γ ′ (t) = 0,

42 | 2 Cauchy’s Theorem and Cauchy’s formula hence (

Φ(t) ′ Φ′ (t)(γ(t) − z) − Φ(t)γ ′ (t) ) = = 0, γ(t) − z (γ(t) − z)2

except for finitely many points. On the other hand, the function Φ(t) γ(t) − z

t↦

is continuous on [a, b], which follows from the fundamental theorem of calculus, hence it must be constant on [a, b]. As Φ(a) = e0 = 1, it follows that Φ(a) Φ(t) 1 = = , γ(a) − z γ(a) − z γ(t) − z

t ∈ [a, b]

and Φ(t) =

γ(t) − z , γ(a) − z

t ∈ [a, b].

Since γ is a closed path, we have γ(a) = γ(b), and hence Φ(b) = 1, and Indγ (z) ∈ ℤ. By Lemma 2.3, the function z ↦ Indγ (z) =

1 b γ ′ (s) ds ∫ 2πi a γ(s) − z

is holomorphic on Ω. Since the image of connected sets under continuous maps is again connected, we obtain that the function Indγ is constant on the connected components of Ω. From Indγ (z) =

1 b γ ′ (s) ds ∫ 2πi a γ(s) − z

we conclude that | Indγ (z)| < 1, if |z| is sufficiently large. Hence Indγ (z) = 0 on the unbounded connected component of Ω. In the following we explain why Indγ (z) is called a winding number. For this purpose let λ(t) = ∫

t

a

γ ′ (s) ds. γ(s) − z

Then λ(b) = 2πi Indγ (z) and hence ℑλ(b) = 2π Indγ (z). Using the same notation as in the proof from above, we get exp(λ(t)) = Φ(t) =

γ(t) − z γ(a) − z

2.2 The theorem of Cauchy–Goursat and Cauchy’s formula | 43

and arg Φ(b) = ℑλ(b) = 2π Indγ (z). Now we set z = 0. We have arg Φ(a) = 0 and arg Φ(t) = arg γ(t) − arg γ(a). This means, if t runs through the interval [a, b], the expression Indγ (0) counts how many times γ turns around z = 0. For a circle, we get Theorem 2.4. Let γ(t) = re2πit + a, t ∈ [0, 1], a ∈ ℂ. Then {1 Indγ (z) = { 0 {

for |z − a| < r, for |z − a| > r.

Proof. From (1) in Example 1.34 we know that 1 dζ = 1. ∫ 2πi γ ζ − a Now we can apply Theorem 2.2 to finish the proof.

2.2 The theorem of Cauchy–Goursat and Cauchy’s formula The following result was first formulated by C. Goursat in full generality with a small gap in its proof (see [27]), which was closed by A. Pringsheim one year later [58]. Pringsheim’s proof, which uses the method of shrinking triangles, became the classical one, still usual nowadays. Later on, we will prove a more general result, which is derived from Stokes’ Theorem from real analysis (see Section 2.9). The theorem of Cauchy– Goursat and Cauchy’s Formula yield powerful methods for the study of holomorphic functions, having no similar counterparts in real analysis. Theorem 2.5. Let Ω ⊆ ℂ be an open subset and fix a point p ∈ Ω. Let f ∶ Ω ⟶ ℂ be a continuous function and suppose that f ∈ ℋ(Ω ⧵ {p}). Let Δ be an open triangle with Δ ⊂ Ω. Then ∫ f (z) dz = 0. 𝜕Δ

Proof. First we suppose that p ∉ Δ. We divide Δ into 4 equal area triangles Δ1 , Δ2 , Δ3 , Δ4 as in Figure 2.1. Then 4

|∫ f (z) dz| = |∑ ∫ 𝜕Δ

j=1 𝜕Δj

f (z) dz| ≤ 4 max|∫ 1≤j≤4

𝜕Δj

f (z) dz|.

The paths in the interior of the large triangle cancel each other out.

44 | 2 Cauchy’s Theorem and Cauchy’s formula

Figure 2.1

We denote the triangle where the maximum is attained by Δ(1) and repeat the procedure from above for the triangle Δ(1) instead of Δ obtaining a triangle Δ(2) , and so on. In this way we get a sequence of triangles Δ ⊃ Δ(1) ⊃ Δ(2) ⊃ ⋯ and after n steps the inequality |∫ f (z) dz| ≤ 4n |∫ 𝜕Δ

𝜕Δ(n)

f (z) dz|.

Let L(𝜕Δ(n) ) be the length of the circumference of 𝜕Δ(n) . Then we have L(𝜕Δ(n) ) = 2−1 L(𝜕Δ(n−1) ) = ⋯ = 2−n L(𝜕Δ). (n) Next we show that there exists a uniquely determined point z0 ∈ ⋂∞ n=1 Δ . (n) For this aim we choose a sequence (zn )n with zn ∈ Δ , n ∈ ℕ. Since Δ is compact, there exists a limit point z0 ∈ Δ of the sequence (zn )n , in addition, z0 also belongs to (m) Δ(m) , m ∈ ℕ, which follows from the fact that the sequence (zn )∞ . Hence we n=m+1 ⊂ Δ ∞ (n) (n) have z0 ∈ ⋂n=1 Δ . z0 is uniquely determined, since L(𝜕Δ ) → 0 as n → ∞. The function f is complex differentiable in z0 . Hence, by Definition 1.5, we can write

f (z) = f (z0 ) + (z − z0 )(f ′ (z0 ) + B(z)), where B is continuous at z0 and B(z0 ) = 0. The function z ↦ f (z0 ) + (z − z0 )f ′ (z0 ) has a primitive (by formal integration), hence, by Corollary 1.37, ∫

𝜕Δ(m)

(f (z0 ) + (z − z0 )f ′ (z0 )) dz = 0 ∀m ∈ ℕ.

2.2 The theorem of Cauchy–Goursat and Cauchy’s formula

| 45

Now we have |∫

𝜕Δ(m)

f (z) dz| = |∫

(z − z0 )B(z) dz| ≤ L(𝜕Δ(m) ) max [|z − z0 | |B(z)|]

𝜕Δ(m) (m)

≤ [L(𝜕Δ

z∈𝜕Δ(m)

2

)] max |B(z)|, z∈𝜕Δ(m)

where we estimated |z − z0 |, z ∈ 𝜕Δ by the length of the circumference L(𝜕Δ(m) ). From this we get for the original integral (m)

|∫ f (z) dz| ≤ 4m (2−m )2 [L(𝜕Δ)]2 max |B(z)| = [L(𝜕Δ)]2 max |B(z)|. z∈𝜕Δ(m)

𝜕Δ

z∈𝜕Δ(m)

As B is continuous at z0 and B(z0 ) = 0, the expression maxz∈𝜕Δ(m) |B(z)| tends to 0 as m → ∞, and hence ∫ f (z) dz = 0. 𝜕Δ

If p is a corner point of Δ, for instance, p = a, then we divide Δ as in Figure 2.2.

Figure 2.2

For the integral we obtain ∫ =∫ 𝜕Δ

+∫+⋯,

𝜕{a,x,y}

where we only have to handle the first summand, because the other integrals do not contain the point p and are therefore 0 by the first step of the proof. For the first summand we can approach x and y arbitrarily close to a, by which ∫ 𝜕{a,x,y}

f (z) dz → 0

as f is continuous. If p lies in the interior of Δ, we can use the decomposition of the triangle in Figure 2.3 to reduce everything to the second step of the proof.

46 | 2 Cauchy’s Theorem and Cauchy’s formula

Figure 2.3

Corollary 2.6. Let Ω be a convex open subset of ℂ, p ∈ Ω and let f ∶ Ω ⟶ ℂ be a continuous function with f ∈ ℋ(Ω ⧵ {p}). Then f has a primitive on Ω and ∫ f (z) dz = 0, γ

for each closed path γ in Ω. Proof. By Theorem 2.5, we have ∫ f (z) dz = 0 𝜕Δ

for each triangle Δ in Ω. Now, by Theorem 1.41, there exists a primitive of f on Ω and finally, by Corollary 1.37. ∫ f (z) dz = 0 γ

for each closed path in Ω. Definition 2.7. An open subset M ⊆ ℂ is called a star-shaped domain, if there exists a point z1 ∈ M (called a center), such that for each z ∈ M the straight line segment [z1 , z] joining the points z1 and z is contained in M. Example 2.8. (a) Figure 2.4 shows a star-shaped domain with center z1 . (b) Consider ℂ− . Here the center is to be chosen on the positive semi-axis. Theorem 2.9. Let G be a star-shaped domain and let f ∈ ℋ(G). Then f has a primitive on G and ∫ f (z) dz = 0, γ

for each closed path γ in G.

2.2 The theorem of Cauchy–Goursat and Cauchy’s formula | 47

Figure 2.4

Proof. Theorem 1.41 is also valid for star-shaped domains as is easily seen. Hence we can use the same proof as in Corollary 2.6. Example 2.10. We consider ℂ− with 1 as center, we write z = reiϕ ∈ ℂ− , and take the closed path γ, as shown in Figure 2.5.

Figure 2.5

By Theorem 2.9, we have ∫

γ

In addition, we can define z ↦ ∫[1,z] ∫ [1,z]

dζ ζ

dζ = 0. ζ

as a primitive for z ↦ 1/z on ℂ− . Then

r dt ϕ ireit dζ =∫ +∫ dt = log r + iϕ. it ζ 1 t 0 re

In this way we have found the principal branch of the logarithm. In the following we study integral representations of holomorphic functions. Cauchy’s formula is the prototype of an integral representation. It will allow us to estimate the size of the holomorphic function involved, to show that all derivatives of holomorphic functions are again holomorphic and to obtain power series expansions of holomorphic functions.

48 | 2 Cauchy’s Theorem and Cauchy’s formula Theorem 2.11 (Cauchy’s formula). Let Ω be a convex domain in ℂ, let γ be a closed path in Ω, and let z ∈ Ω, z ∉ γ ∗ . Suppose that f ∈ ℋ(Ω). Then f (z) Indγ (z) =

1 f (ζ ) dζ . ∫ 2πi γ ζ − z

Proof. Let { f (ζ )−f (z) , g(ζ ) = { ζ −z f ′ (z), {

if ζ ≠ z, if ζ = z.

Then g is continuous on Ω and g ∈ ℋ(Ω ⧵ {z}). Hence g satisfies the assumptions of Corollary 2.6. So we have 1 ∫ g(ζ ) dζ = 0. 2πi γ Finally, we get 1 1 1 f (ζ ) − f (z) f (ζ ) dζ dζ = dζ − f (z) = 0. ∫ ∫ ∫ 2πi γ ζ − z 2πi γ ζ − z 2πi γ ζ − z We mention an important special case. Let γ(t) = z + reit , t ∈ [0, 2π], f ∈ ℋ(DR (z)), R > r. Then Indγ (z) = 1 and f (z) =

1 f (ζ ) dζ . ∫ 2πi γ ζ − z

In addition, we obtain an important estimate |f (z)| =

1 f (ζ ) 1 f (ζ ) |∫ dζ | ≤ 2πr max | | = max |f (ζ )|. ∗ ζ ∈γ ζ − z ζ ∈γ ∗ 2π γ ζ − z 2π

Plugging in the definition of the line integral, we obtain f (z) =

1 1 2π f (z + reit ) it 1 2π f (ζ ) dζ = ire dt = ∫ ∫ ∫ f (z + reit ) dt, it 2πi γ ζ − z 2πi 0 re 2π 0

which is the mean value property of holomorphic functions. Example 2.12. Using Cauchy’s formula, we compute the following line integral: I =∫

γ

(a) If G = D3/4 (0), f (z) = lary 2.6.

ez , (z−1)(z−2)

ez dz. (z − 1)(z − 2)

f ∈ ℋ(G), γ(t) =

eit , 2

t ∈ [0, 2π], then I = 0, by Corol-

2.2 The theorem of Cauchy–Goursat and Cauchy’s formula | 49

(b) If G = D11/6 (0), f (z) = we have

ez , z−2

f (1) =

f ∈ ℋ(G), γ(t) =

3eit , 2

t ∈ [0, 2π], then by Theorem 2.11

1 f (ζ ) 1 eζ dζ = dζ . ∫ ∫ 2πi γ ζ − 1 2πi γ (ζ − 2)(ζ − 1)

Hence I = −2πie. (c) If G = D4 (0), f (z) = ez , f ∈ ℋ(G), γ(t) = 3eit , t ∈ [0, 2π], then by Theorem 2.11 we have I =∫

γ

ez ez f (z) f (z) dz − ∫ dz = ∫ dz − ∫ dz = 2πi(e2 − e). z−2 γ z−1 γ z−2 γ z−1

Theorem 2.13 (Taylor5 series expansion). Let Ω ⊆ ℂ be an open set and let a ∈ Ω, R > 0 be such that DR (a) ⊆ Ω. Suppose that f ∈ ℋ(Ω). Then ∞

f (z) = ∑ an (z − a)n , n=0

where the sum converges uniformly on all compact subsets of DR (a) and an =

f (n) (a) , n!

n = 0, 1, 2, … ,

an are the Taylor coefficients of f in the expansion around the point a. Proof. Let γ(t) = a + reit , r < R, t ∈ [0, 2π]. Then Indγ (z) = 1, ∀z ∈ Dr (a) (see Theorem 2.4). By Theorem 2.11, we have f (z) =

1 f (ζ ) 1 2π f (γ(t))γ ′ (t) dζ = dt ∫ ∫ 2πi γ ζ − z 2πi 0 γ(t) − z

∀z ∈ Dr (a).

As in the proof of Lemma 2.3 we expand the last integral into a power series ∞

f (z) = ∑ an (z − a)n , n=0

where an =

1 f (ζ ) 1 2π f (γ(t))γ ′ (t) dt = dζ ∫ ∫ n+1 2πi 0 (γ(t) − a) 2πi γ (ζ − a)n+1

and the series converges uniformly on all compact subsets of Dr (a). Since r was an arbitrary number such that r < R, we obtain the same assertion for DR (a). By Theorem 1.43, we obtain the formula for the Taylor coefficients. 5 Taylor, Brook (1685–1731).

50 | 2 Cauchy’s Theorem and Cauchy’s formula The proof of the last theorem contains the following important results: Theorem 2.14. If f ∈ ℋ(Ω), then f (n) ∈ ℋ(Ω) ∀n ∈ ℕ, f has complex derivatives of arbitrary order. Theorem 2.15. Under the same assumptions as in Theorem 2.13 and for γ(t) = a + reit , r < R, t ∈ [0, 2π], we have f (n) (a) =

f (ζ ) n! dζ , ∫ 2πi γ (ζ − a)n+1

n ∈ ℕ.

We are now able to prove the converse of Cauchy’s Theorem. Theorem 2.16 (Morera’s6 Theorem). Let G be an open set in ℂ and let f ∶ G ⟶ ℂ be continuous on G. Suppose that ∫ f (z) dz = 0 𝜕Δ

for all solid triangles Δ in G. Then f ∈ ℋ(G). Proof. Let V ⊆ G be an arbitrary convex subset of G. By Theorem 1.41, there exists a primitive F of f on V , we have that F ∈ ℋ(V) and, by Theorem 2.14, also that F ′ = f ∈ ℋ(V). Since V was an arbitrary convex subset of G, we conclude that f ∈ ℋ(G).

2.3 Important consequences of Cauchy’s Theorem In the following we study the zero sets of holomorphic functions. Also in this context there is no analogue in the theory of real differentiable functions. Theorem 2.17. Let Ω ⊆ ℂ be a domain and f ∈ ℋ(Ω). We define the zero set of f by Z(f ) = {a ∈ Ω ∶ f (a) = 0}. Then Z(f ) = Ω or Z(f ) is a discrete subset of Ω (i.e. Z(f ) has no limit point in Ω). In the second case, for each a ∈ Z(f ) there exists a uniquely determined integer ma ∈ ℕ such that f (z) = (z − a)ma g(z), where g ∈ ℋ(Ω) and g(a) ≠ 0. We say that a is a zero of order ma of f . In addition, the zero set Z(f ) is at most countably infinite. 6 Morera, Giacinto (1856–1909).

2.3 Important consequences of Cauchy’s Theorem

| 51

Proof. Let A be the set of all limit points of Z(f ) in Ω. Since f is continuous, we have A ⊆ Z(f ). Fix a ∈ Z(f ). There exists r > 0, such that Dr (a) ⊆ Ω. By Theorem 2.13, we can expand f into a Taylor series ∞

f (z) = ∑ an (z − a)n , n=0

z ∈ Dr (a).

Now we can distinguish between two cases: Case 1. All Taylor coefficients an = 0, which implies that f ≡ 0 in Dr (a). Hence Dr (a) ⊆ A, i.e. a is an interior point of A. Case 2. There exists a minimal m ∈ ℕ with am ≠ 0. Now we define a function g by {(z − a)−m f (z) for z ∈ Ω ⧵ {a}, g(z) = { a for z = a. { m Then we have f (z) = (z − a)m g(z), for all z ∈ Ω ⧵ {a} and g ∈ ℋ(Ω ⧵ {a}). But as f (z) = n ∑∞ n=0 an (z − a) in Dr (a), we get ∞

g(z) = ∑ an+m (z − a)n , n=0

z ∈ Dr (a).

Hence g ∈ ℋ(Dr (a)) and, by the definition of g, we obtain g ∈ ℋ(Ω). Since g(a) = am ≠ 0 and since g is continuous, there exists an open neighborhood U of a such that g(z) ≠ 0 for all z ∈ U, which implies that f (z) ≠ 0, for all z ∈ U ⧵ {a}. Now we have shown that a is an isolated point in Z(f ), i.e. there exists a neighborhood U of a, containing no other point of Z(f ). In summary, we can state the following: if a ∈ A, then all an = 0, because otherwise a would be an isolated point (by Case 2); so, by Case 1, we have Dr (a) ⊆ A and A is an open set. But A is the set of all limit points of Z(f ) and is therefore also closed in Ω, and the set B = Ω ⧵ A is open. We have Ω = A ∪ B, which is a union of two disjoint open sets. We assumed that Ω is connected, hence Ω = A or A = ∅. If A = Ω we have f ≡ 0 on Ω; if A = ∅ we get that Z(f ) is discrete in Ω. In this case, there are at most finitely many points of Z(f ) in each compact subset of Ω, otherwise a limit point of Z(f ) would belong to this compact subset and A ≠ ∅. We can write Ω as a countable union of compact subsets of Ω: ∞

Ω = ⋃ (Dn (0) ∩ Ω1/n )− , n=1

where Ω1/n = {z ∈ Ω ∶ dist(z, 𝜕Ω) > 1/n}, hence Z(f ) is at most countably infinite.

52 | 2 Cauchy’s Theorem and Cauchy’s formula The following results are easy but important consequences, which are also referred to as the identity principles. Theorem 2.18. Let Ω be a domain in ℂ, let f , g ∈ ℋ(Ω) and suppose that f = g on a subset M of Ω, which has a limit point in M. Then f = g on the whole of Ω. Proof. Let ϕ = f − g. Then ϕ(z) = 0 for all z ∈ M and Theorem 2.17 implies that ϕ(z) = 0 on the whole of Ω. Remark. If f = g on a non-empty open subset of Ω, then f = g on the whole of Ω. If Ω is not connected, we cannot draw this conclusion: let Ω = D1 (0) ∪ D1 (3) and {1, f (z) = { 0, {

z ∈ D1 (0), z ∈ D1 (3).

Then f ∈ ℋ(Ω), f = 0 on the open subset D1 (3) of Ω, but f is not identically zero on Ω. Corollary 2.19. Let G be a domain in ℂ, let f ∈ ℋ(G) and suppose that f (n) (a) = 0, n = 0, 1, 2, …, for some a ∈ G. Then f ≡ 0 on G. Proof. Let r > 0 be such that Dr (a) ⊆ G. Then, by Theorem 2.13, f can be expanded into a Taylor series f (n) (a) (z − a)n n! n=0 ∞

f (z) = ∑

for z ∈ Dr (a). Since f (n) (a) = 0, n = 0, 1, 2, …, we have f ≡ 0 on Dr (a). Now Theorem 2.18 implies that f ≡ 0 on G. Remark. The above result is false for 𝒞∞ -functions on ℝ. Let 2

{e−1/x , f (x) = { 0, {

x ∈ ℝ, x ≠ 0, x = 0.

It follows that f ∈ 𝒞∞ (ℝ) and f (n) (0) = 0, n = 0, 1, 2, …, but f is not identically zero on ℝ. Definition 2.20. Let U ⊆ ℂ be open and f ∈ ℋ(U). We say that f has a holomorphic logarithm on U, if there exists a function g ∈ ℋ(U) such that f = exp(g) on U. We say that f has a holomorphic mth root on U, if there exists a function q ∈ ℋ(U) such that qm = f on U.

2.4 Isolated singularities | 53

Remark. If exp(g) = f , then f ≠ 0 on U. In addition, we have f ′ = g ′ exp(g) = g ′ f . Hence g ′ = f ′ /f . This expression is called the logarithmic derivative of f . Theorem 2.21. Let G be a domain in ℂ, and f ∈ ℋ(G) with f ≠ 0 on G. Then the following assertions are equivalent: (1) f has a holomorphic logarithm on G. (2) f ′ /f has a primitive on G. Proof. Suppose (1) is true. Then we have g ′ = f ′ /f . Hence f ′ /f has a primitive on G. Now suppose that (2) holds. Let F ∈ ℋ(G) be a primitive of f ′ /f on G. Set h = f exp(−F). Then h′ = f ′ exp(−F) − fF ′ exp(−F) = f ′ exp(−F) −

ff ′ exp(−F) = 0. f

Since h ∈ ℋ(G) and h′ = 0 on G, it follows that h is constant on G, i.e. f = a exp(F) for some a ∈ ℂ, a ≠ 0. Now there exists b ∈ ℂ with eb = a. We define ϕ = F + b, and observe that ϕ ∈ ℋ(G) and exp(ϕ) = exp(F + b) = a exp(F) = f . Theorem 2.22. Let G be a star-shaped domain in ℂ, let f ∈ ℋ(G), f ≠ 0 on G. Then f has a holomorphic logarithm and a holomorphic mth root on G. Proof. By Theorem 2.21, it suffices to show that f ′ /f has a primitive on G. Now we use Theorem 2.9: if c is a center for G, then g(z) = ∫

[c,z]

f ′ (ζ ) dζ + b, f (ζ )

f (c) = eb

is a primitive of f ′ /f on G and we have exp(g) = f . Then q = exp(g/m) is the desired holomorphic mth root.

2.4 Isolated singularities Definition 2.23. Let Ω ⊆ ℂ be an open set and a ∈ Ω. We say that a function f ∈ ℋ(Ω ⧵ {a}) has an isolated singularity at a. If one can define f at a such that f ∈ ℋ(Ω), we say that a is a removable singularity of f . Theorem 2.24. Let f ∈ ℋ(Ω ⧵ {a}) and denote by D′r (a) = {z ∶ 0 < |z − a| < r} the punctured disc. Suppose that there exists a constant M > 0 such that |f (z)| ≤ M for all z ∈ D′r (a). Then f has a removable singularity at a.

54 | 2 Cauchy’s Theorem and Cauchy’s formula Proof. Let {(z − a)2 f (z), h(z) = { 0, {

z ∈ Ω ⧵ {a}, z = a.

We claim that h ∈ ℋ(Ω). For this aim we compute (z − a)2 f (z) h(z) − h(a) = lim = lim (z − a)f (z) = 0, z→a z→a z→a z−a z−a

h′ (a) = lim

where we used that f is bounded on D′r (a). By Theorem 2.13, we can expand h into a Taylor series in Dr (a), ∞

h(z) = ∑ an (z − a)n , n=0

z ∈ Dr (a).

Now we define f (a) = a2 . Since h(z) = (z − a)2 f (z) in Dr (a), we get ∞

f (z) = ∑ an+2 (z − a)n , n=0

where the series converges in Dr (a). Hence f ∈ ℋ(Dr (a)), and also that f ∈ ℋ(Ω). We can now characterize all possible isolated singularities of holomorphic functions. Theorem 2.25. Let Ω ⊆ ℂ be an open set, let a ∈ Ω, and f ∈ ℋ(Ω ⧵ {a}). We distinguish between three possible cases: (a) f has a removable singularity at a. (b) There exist c1 , … , cm ∈ ℂ, m ∈ ℕ, cm ≠ 0, such that m

ck (z − a)k k=1

f (z) − ∑

has a removable singularity at a; in this case a is called a pole of order m of f and limz→a |f (z)| = +∞: m

ck k k=1 (z − a) ∑

is called the principal part of f . (c) For all r > 0 with Dr (a) ⊆ Ω, the image f (D′r (a)) is dense in ℂ, i.e. for all w ∈ ℂ and for all ϵ > 0 there exists z ∈ D′r (a) such that |f (z)−w| < ϵ; equivalently, for all w ∈ ℂ there exists a sequence (zn )n in Ω ⧵ {a} with limn→∞ zn = a, such that limn→∞ f (zn ) = w; in this case, a is called an essential singularity of f .

2.4 Isolated singularities | 55

Remark. The assertion from (c) is known as the Casorati7 –Weierstraß Theorem. Proof. We will show that case (a) or (b) is valid, if (c) fails. Suppose that (c) fails, then there exist r > 0 and δ > 0 and there exists some w ∈ ℂ such that |f (z) − w| > δ, for all z ∈ D′r (a). Set g(z) =

1 , f (z) − w

z ∈ D′r (a).

Then g ∈ ℋ(D′r (a)) and |g(z)|
0 with |g(z)| ≥ ρ, for all z ∈ Ds (a). Hence |f (z)| = |

1 1 + w| ≤ + |w|, g(z) ρ

for all z ∈ D′s (a), which means that f is bounded on D′s (a). By Theorem 2.24, f has a removable singularity at a. Case 2. g has a zero of order m ∈ ℕ at a (see Theorem 2.17). Therefore, g(z) = (z − a)m g1 (z)

∀z ∈ Dr (a),

where g1 ∈ ℋ(Dr (a)) and g1 (a) ≠ 0. Since g(z) =

1 = (z − a)m g1 (z) f (z) − w

on D′r (a), we even have that g1 (z) ≠ 0 ∀z ∈ Dr (a). Now we set h = 1/g1 . Then h ∈ ℋ(Dr (a)) and f (z) − w = (z − a)−m h(z), for all z ∈ D′r (a). 7 Casorati, Felice (1835–1890).

56 | 2 Cauchy’s Theorem and Cauchy’s formula We expand h into a Taylor series on Dr (a) ∞

h(z) = ∑ bn (z − a)n , n=0

where b0 = h(a) ≠ 0. Setting ck = bm−k , k = 1, … , m, we have cm = b0 ≠ 0 and f (z) − w = (z − a)−m (cm + cm−1 (z − a) + ⋯ + c1 (z − a)m−1 + bm (z − a)m + ⋯) cm cm−1 c = + + ⋯ + 1 + bm + bm+1 (z − a) + ⋯ . m m−1 (z − a) (z − a) z−a Hence m

ck = w + bm + bm+1 (z − a) + ⋯ , k k=1 (z − a)

f (z) − ∑

where the right-hand side is a holomorphic function for z ∈ Dr (a). Example 2.26. (a) Let f (z) = sin z/z, z ∈ ℂ ⧵ {0}. Then f ∈ ℋ(ℂ ⧵ {0}). Since sin 0 = 0 and cos 0 = (sin z)′ |z=0 , we get from Theorem 2.17 that there exists g ∈ ℋ(ℂ) such that sin z = zg(z). Hence f has a removable singularity at z = 0 and f ∈ ℋ(ℂ). Since 3 5 2 4 sin z = z − z3! + z5! − ⋯, we have f (z) = 1 − z3! + z5! − ⋯. 1 (b) Let f (z) = sin z . Then f ∈ ℋ(D′1 (0)). Since sin z = zg(z), g ∈ ℋ(ℂ), g(0) ≠ 0, there exists r > 0 such that g(z) ≠ 0 for all z ∈ Dr (0). Setting h(z) = 1/g(z), we get h ∈ ℋ(Dr (0)). Hence f (z) =

h(z) 1 1 = = ( ∑ bn z n ), sin z z z n=0 ∞

where we used the Taylor series expansion of h on Dr (0) and that h(0) = b0 ≠ 0. Therefore, f (z) −

b0 = b1 + b2 z + ⋯ , z

where the right-hand side is a holomorphic function on Dr (0). This means that f (z) = sin1 z has a pole of order 1 at z = 0. (c) Let f (z) = exp(1/z). Then f ∈ ℋ(ℂ∗ ). For n ∈ ℕ we get that f (1/n) → ∞ as n → ∞. But |f (i/n)| = 1, for all n ∈ ℕ. Hence the assertions (a) and (b) from Theorem 2.25 are not valid. Hence f has an essential singularity at z = 0. (d) In a similar way one shows that f (z) = exp(−1/z 2 ) has an essential singularity at z = 0. But f |ℝ ∈ 𝒞∞ (ℝ).

2.5 The maximum principle and Cauchy’s estimates In this section we prove some useful inequalities related to the absolute value of a holomorphic function and its derivatives.

2.5 The maximum principle and Cauchy’s estimates | 57

n Theorem 2.27. Let f (z) = ∑∞ n=0 cn (z − a) be a holomorphic function on DR (a). Let 0 < r < R. Then ∞

∑ |cn |2 r 2n =

n=0

1 2π ∫ |f (a + reiθ )|2 dθ. 2π 0

Proof. We write z − a = reiθ . Then 1 2π 1 2π ∫ |f (a + reiθ )|2 dθ = ∫ ( ∑ c r n einθ )( ∑ cn r n e−inθ ) dθ 2π 0 2π 0 n=0 n n=0 ∞

∞

∞

= ∑ |cn |2 r 2n . n=0

Definition 2.28. A function f ∈ ℋ(ℂ) is called an entire function. Theorem 2.29 (Liouville’s8 Theorem). Each bounded entire function is constant. n Proof. Let f (z) = ∑∞ n=0 cn z be a bounded entire function, i.e. there exists a constant M > 0 such that |f (z)| ≤ M, for all z ∈ ℂ. Then, by Theorem 2.27, we have ∞

∑ |cn |2 r 2n ≤ M 2 ,

n=0

for all r > 0, hence cn = 0 for all n ∈ ℕ. Theorem 2.30 (Maximum Principle). Let Ω be a domain in ℂ and f ∈ ℋ(Ω). Let a ∈ Ω be an arbitrary point in Ω and r > 0 such that Dr (a) ⊂⊂ Ω, i.e. Dr (a) ⊂ Ω. Then |f (a)| ≤ max{|f (a + reiθ )| ∶ 0 ≤ θ ≤ 2π}, with equality if and only if f is constant on Ω. Remark. The assertion from above implies that the function z ↦ |f (z)| has no local maximum in Ω. Proof. If max{|f (a + reiθ )| ∶ 0 ≤ θ ≤ 2π} ≤ |f (a)|, n then Theorem 2.27 implies that the function f (z) = ∑∞ n=0 cn (z − a) satisfies ∞

∑ |cn |2 r 2n =

n=0

1 2π ∫ |f (a + reiθ )|2 dθ ≤ |f (a)|2 = |c0 |2 . 2π 0

Hence cn = 0 for all n ∈ ℕ, and f ≡ f (a) on Dr (a). Since Ω is connected, we get from Theorem 2.18 that f ≡ f (a) on the whole of Ω. 8 Liouville, Joseph (1809–1882).

58 | 2 Cauchy’s Theorem and Cauchy’s formula Corollary 2.31. Let Ω be a bounded domain in ℂ, and let f ∈ ℋ(Ω) be a non-constant function, which is continuous on Ω. Then the function |f | attains its maximum on 𝜕Ω. Proof. Suppose that there exists z0 ∈ Ω with maxz∈Ω |f (z)| = |f (z0 )|. Then there exists r > 0 such that Dr (z0 ) ⊂⊂ Ω. By Theorem 2.30, we conclude that |f (z0 )|< max |f (z0 + reiθ )|, 0≤θ≤2π

where we used that f is non-constant. So we arrive at a contradiction. At this point we can give a proof of the fundamental theorem of algebra. Theorem 2.32 (Fundamental Theorem of Algebra). Let n ∈ ℕ and let p(z) = z n + an−1 z n−1 + ⋯ + a1 z + a0 be a polynomial of degree n with complex coefficients a0 , a1 , … , an−1 ∈ ℂ. Then p has exactly n zeros (some of which may be counted according to their multiplicities). Proof. First we prove the following assertion: |p(0)| = |a0 | < |p(reiθ )| for each θ ∈ [0, 2π] and each r > 1 + 2|a0 | + |a1 | + ⋯ + |an−1 |. This assertion implies, that all zeros of p are contained in the disk DR (0) where R = 1 + 2|a0 | + |a1 | + ⋯ + |an−1 |. In order to prove this assertion, we take r > R. Then r > 1, and we get 2|a0 | + |a1 |r + ⋯ + |an−1 |r n−1 ≤ r n−1 (2|a0 | + |a1 | + ⋯ + |an−1 |) < r n−1 r = r n .

(2.1)

In addition, we have r n − |p(reiθ )| ≤ | |r n einθ | − |p(reiθ )| | ≤ |r n einθ − p(reiθ )| = |r n einθ − r n einθ − an−1 r n−1 ei(n−1)θ − ⋯ − a0 | ≤ |an−1 |r n−1 + ⋯ + |a1 |r + |a0 |,

hence r n − (|an−1 |r n−1 + ⋯ + |a0 |) ≤ |p(reiθ )|. Now we use (2.1) and (2.2) to obtain |a0 | = 2|a0 | + |a1 |r + ⋯ + |an−1 |r n−1 − (|a0 | + |a1 |r + ⋯ + |an−1 |r n−1 ) < r n − (|a0 | + |a1 |r + ⋯ + |an−1 |r n−1 ) ≤ |p(reiθ )|.

(2.2)

2.5 The maximum principle and Cauchy’s estimates | 59

To prove the fundamental theorem of algebra, we suppose that p has no zero. Then f = 1/p is an entire non-constant function and the assertion we just proved implies that |f (reiθ )| < |f (0)| for all sufficiently large r, which contradicts Theorem 2.30. Hence there exists z1 ∈ ℂ with p(z1 ) = 0, and by Theorem 2.17 we have p(z) = (z − z1 )m q(z) for some m ∈ ℕ and a polynomial q with grad(q) < grad(p). (Expand p into a Taylor series around the point z1 .) Induction on the degree of the polynomial concludes the proof. The following estimate is related to the absolute value of a holomorphic function and its derivatives. Theorem 2.33 (Cauchy’s estimates). Let Ω ⊆ ℂ be an open subset, a ∈ Ω, r > 0, Dr (a) ⊂⊂ Ω, f ∈ ℋ(Ω). Then for n ∈ ℕ we have |f (n) (a)| ≤

n! M (f ), r n r,a

where Mr,a (f ) = maxz∈𝜕Dr (a) |f (z)|. Proof. By Theorem 2.15, we have f (n) (a) =

f (ζ ) n! dζ ∫ 2πi 𝜕Dr (a) (ζ − a)n+1

and hence |f (n) (a)| ≤

M (f ) n! 2πr r,a . 2π r n+1

In the following we address the problem under which conditions the limit of a sequence of holomorphic functions is again holomorphic. We point out that we use the whole theory we have developed up to now. This result will also be fundamental for the study of topological vector spaces of holomorphic functions and gives a first occasion to use the terminology of functional analysis. Theorem 2.34 (Weierstraß’ Theorem). Let Ω be an open subset of ℂ, fn ∈ ℋ(Ω), n ∈ ℕ. Suppose that the sequence (fn )n converges uniformly on all compact subsets of Ω to a function f . Then f ∈ ℋ(Ω) and, for each k ∈ ℕ, the sequence fn(k) converges to f (k) uniformly on all compact subsets of Ω.

60 | 2 Cauchy’s Theorem and Cauchy’s formula If fn ∈ ℋ(Ω) is a Cauchy sequence in the sense of uniform convergence on all compact subsets of Ω, i.e. for each compact subset K ⊂ Ω and for each ϵ > 0 there exists Nϵ,K > 0 such that sup|fn (z) − fm (z)| < ϵ z∈K

for all n, m > Nϵ,K . Then there exists a holomorphic function f ∈ ℋ(Ω) with limn→∞ fn = f uniformly on all compact subsets of Ω. We say that the space ℋ(Ω) is complete in the sense of uniform convergence on all compact subsets of Ω. Proof. f is the uniform limit of continuous functions and therefore again continuous. Let Δ be a solid triangle in Ω. We get from Theorem 2.5 that ∫ fn (z) dz = 0, 𝜕Δ

and, since 𝜕Δ is compact, ∫ f (z) dz = 0. 𝜕Δ

Apply Theorem 2.16, to get that f ∈ ℋ(Ω). If K is a compact subset of Ω, then s = dist(K, Ωc ) > 0 and, for r = s/3, we have K ⊂ ⋃ Dr (z) = U. z∈K

The closure U is a compact subset of Ω and we can apply Theorem 2.33 to obtain |fn′ (z) − f ′ (z)| ≤

1 max|f (w) − f (w)| r w∈U n

∀z ∈ K,

hence fn′ converges to f ′ uniformly on K. If fn ∈ ℋ(Ω) is a Cauchy sequence, there exists a limit function f , which is at least continuous on Ω (see real analysis). The first part of the theorem implies that f ∈ ℋ(Ω). Remark. There are sequences of 𝒞∞ -functions on ℝ, which converge uniformly to nowhere differentiable functions (see [43]). Remark. We already mentioned that the space ℋ(Ω), endowed with the topology of uniform convergence on compact subsets of Ω, is a complete topological vector space (Theorem 2.34). The topology on ℋ(Ω) can be described by a metric. Indeed, consider a so-called compact exhaustion of Ω, which is a sequence (Kj )j of compact subsets of Ω such that Kj ⊂ K̊ j+1 , j ∈ ℕ and ⋃∞ j=1 Kj = Ω.

2.6 Open mappings | 61

For instance, one can take Kj = {z ∶ dist(z, Ωc ) ≥ 1/j} ∩ Dj (0),

j ∈ ℕ.

Now one defines a sequence of seminorms ‖f ‖j = sup|f (z)|, z∈Kj

f ∈ ℋ(Ω),

j ∈ ℕ,

and we have ‖f ‖j ≤ ‖f ‖j+1 . The metric on ℋ(Ω) is now defined by ∞

d(f , g) = ∑ 2−j j=1

‖f − g‖j

1 + ‖f − g‖j

,

f , g ∈ ℋ(Ω).

The metric d generates the original topology of uniform convergence on all compact subsets of Ω on ℋ(Ω), see Exercises. ℋ(Ω) is a Fréchet space, i.e. a complete, metrizable topological vector space.

2.6 Open mappings Definition 2.35. Let U ⊆ ℂ be an open set and ϕ ∶ U ⟶ ℂ a function. ϕ is called open, if ϕ(V) is open, for each open subset V ⊆ U. Remark. Recall that ϕ ∶ U ⟶ ℂ is continuous, if ϕ−1 (O) is open, for each open subset O ⊆ ℂ. If ϕ is open and invertible, then ϕ−1 is continuous. Example 2.36. Let ϕ ∶ ℝ ⟶ ℝ, ϕ(x) = x2 . We have ϕ((−1, 1)) = [0, 1). Hence ϕ is not open. Theorem 2.37 (Minimum Principle). Let U ⊆ ℂ be open and f ∈ ℋ(U), let c ∈ U and let V a disc with center c and V ⊂ U. Suppose that min|f (z)| > |f (c)|. z∈𝜕V

Then f has a zero in V . Proof. Suppose that f has no zero in V . Our assumption implies that there exists an open neighborhood V1 of V such that V1 ⊆ U, and that f has no zero on V1 . Set g(z) = 1/f (z), z ∈ V1 . Then g ∈ ℋ(V1 ), and by Theorem 2.30 we have −1

|f (c)|−1 = |g(c)| ≤ max|g(z)| = [min|f (z)|] , z∈𝜕V

and we arrive at a contradiction.

z∈𝜕V

62 | 2 Cauchy’s Theorem and Cauchy’s formula Theorem 2.38 (Open Mapping Theorem). Let U ⊆ ℂ be an open set and f ∈ ℋ(U). Suppose that there is no open subset of U where f is constant. Then f is an open mapping. Proof. Let O ⊆ U be open and c ∈ O. We have to show that f (O) contains an open disc with center f (c). Without loss of generality, we can suppose that f (c) = 0, otherwise we could consider the function z ↦ f (z) − f (c) instead of f . We supposed that f is not constant in any neighborhood of c. We claim that there exists a disc V with center c such that V ⊂ O and 0 ∉ f (𝜕V). (If for each disk V with center c and V ⊂ O there exists z0 ∈ 𝜕V with f (z0 ) = 0, then, by Theorem 2.18, f ≡ 0 in some neighborhood of c, which contradicts our assumption on f .) Now we set 2δ = minz∈𝜕V |f (z)| > 0 and D = Dδ (0). We will show that D ⊆ f (O). For this aim let b ∈ D be an arbitrary point. As |b| < δ we have |f (z) − b| ≥ |f (z)| − |b| > δ, for all z ∈ 𝜕V , so we get min|f (z) − b| ≥ δ > |b| = |f (c) − b|. z∈𝜕V

Now we can apply the minimum principle Theorem 2.37 to the function z ↦ f (z) − b, and get a point z ′ ∈ V with f (z ′ ) − b = 0. Hence f (z ′ ) = b, and finally b ∈ f (O). Remark. Let πm (z) = z m , m ∈ ℕ. Then πm is an open mapping. Each w ≠ 0 is the image under πm of exactly m different points zk , k = 1, … , m, i.e. πm (zk ) = w, k = 1, … , m. For w = reiθ we have zk = r 1/m ei(θ+2kπ)/m , k = 1, … , m. The point w = 0 has only z = 0 as its preimage, it is a so-called branching point. We will show that each non-constant holomorphic function is locally of the form πm ∘ ϕ + c, where ϕ is an invertible holomorphic function and c is a constant. Lemma 2.39. Let Ω ⊆ ℂ be open and f ∈ ℋ(Ω). Define { f (z)−f (w) , z ≠ w, z, w ∈ Ω, g(z, w) = { z−w f ′ (z), z = w ∈ Ω. { Then g ∶ Ω × Ω ⟶ ℂ is continuous on Ω × Ω. Proof. It suffices to show continuity on the diagonal {(a, a) ∶ a ∈ Ω}. For this aim, fix a ∈ Ω and let r > 0 be such that Dr (a) ⊆ Ω and |f ′ (ζ ) − f ′ (a)| < ϵ

∀ζ ∈ Dr (a).

For z, w ∈ Dr (a) we set ζ (t) = (1 − t)z + tw, t ∈ [0, 1], which describes the straight line from z to w. It is clear that ζ (t) ∈ Dr (a) ∀t ∈ [0, 1]. Now we compute the integral

2.6 Open mappings | 63 1

1 1 df dζ 1 1 dt ∫ f ′ (ζ (t))(−z + w) dt = ∫ −z + w 0 −z + w 0 dζ dt 1 f (z) − f (w) = f (ζ (t))|10 = . −z + w z−w

∫ f ′ (ζ (t)) dt = 0

Hence we get

|g(z, w) − g(a, a)| = |

1 f (z) − f (w) − f ′ (a)| = |∫ [f ′ (ζ (t)) − f ′ (a)] dt| z−w 0

≤ sup |f ′ (ζ (t)) − f ′ (a)| ≤ ϵ. t∈[0,1]

First we prove a result about invertible holomorphic functions. Theorem 2.40. Let Ω ⊆ ℂ be open, ϕ ∈ ℋ(Ω), z0 ∈ Ω and ϕ′ (z0 ) ≠ 0. Then there exists an open neighborhood V of z0 , V ⊆ Ω such that (1) ϕ|V is injective, (2) the function ψ ∶ ϕ(V) ⟶ V defined by ψ(ϕ(z)) = z, z ∈ V , is holomorphic on W = ϕ(V), ϕ has a holomorphic inverse on V . Proof. By Lemma 2.39, there exists an open neighborhood V ⊆ Ω of z0 such that 1 |ϕ(z1 ) − ϕ(z2 )| ≥ |ϕ′ (z0 )||z1 − z2 | 2

∀z1 , z2 ∈ V,

for this aim one has to choose V in such a way that |

ϕ(z1 ) − ϕ(z2 ) |ϕ′ (z0 )| |≥ . z1 − z2 2

If z1 , z2 ∈ V, z1 ≠ z2 , then ϕ(z1 ) ≠ ϕ(z2 ) and ϕ is injective on V . Since ϕ′ (z0 ) ≠ 0, we can choose V also such that ϕ′ (z) ≠ 0, ∀z ∈ V . By assertion (1), each w ∈ W = ϕ(V) has a uniquely determined z ∈ V with ϕ(z) = w. Now let z, z1 ∈ V and w, w1 ∈ W be chosen such that ϕ(z) = w, ϕ(z1 ) = w1 and ψ(w) = z, ψ(w1 ) = z1 . Then we have z − z1 ψ(w) − ψ(w1 ) = ; w − w1 ϕ(z) − ϕ(z1 ) if w → w1 , then z → z1 and so the left-hand side converges to ψ′ (w1 ) as w → w1 and the right-hand side converges at the same time to 1/ϕ′ (z1 ). Hence we have ψ′ (w1 ) =

1 ϕ′ (z1 )

and since ϕ′ ≠ 0 on V , we obtain ψ ∈ ℋ(W). Now we are able to show the local form of a holomorphic function as indicated above.

64 | 2 Cauchy’s Theorem and Cauchy’s formula Theorem 2.41. Let Ω be a domain in ℂ, f ∈ ℋ(Ω) non-constant, z0 ∈ Ω and w0 = f (z0 ). Let m be the order of the zero z0 of the function z ↦ f (z) − w0 . Then there exists an open neighborhood V ⊆ Ω of z0 and a function ϕ ∈ ℋ(V) such that (1) f (z) = w0 + [ϕ(z)]m ∀z ∈ V ; (2) ϕ is invertible on V . Remark. On V , we have f − w0 = πm ∘ ϕ, where πm (z) = z m . Hence f is an m–to–1 mapping on V ⧵ {z0 }. Proof. We can take a convex open neighborhood V of z0 such that f (z) ≠ w0 , ∀z ∈ V ⧵ {z0 }, otherwise we could find a sequence (zn )n in V such that limn→∞ zn = z0 and f (zn ) = w0 ∀n ∈ ℕ, then, by Theorem 2.18, we would get that f ≡ w0 on V , which is excluded by our assumption on f . Now we can apply Theorem 2.17 to obtain that f (z) − w0 = (z − z0 )m g(z) ∀z ∈ V , where g ∈ ℋ(V) and g ≠ 0 on V . By Theorem 2.22, g has a holomorphic logarithm on V , i.e. there exists h ∈ ℋ(V) such that exp(h) = g on V . Now we set ϕ(z) = (z − z0 ) exp(h(z)/m). Then [ϕ(z)]m = (z − z0 )m exp(h(z)) = (z − z0 )m g(z) = f (z) − w0 . In addition, we have that ϕ′ (z) = exp(h(z)/m) + (z − z0 ) exp(h(z)/m)h′ (z)/m, and since exp(h(z0 )/m) ≠ 0, we get ϕ′ (z0 ) ≠ 0. By taking a possibly smaller V we can also get that ϕ′ ≠ 0 on V . The rest of the proof now follows from Theorem 2.40. Theorem 2.42. Let Ω be a domain in ℂ, f ∈ ℋ(Ω). Suppose that f is injective on Ω. Then f ′ ≠ 0 on Ω and f has a holomorphic inverse. Proof. If f ′ (z0 ) = 0 for some z0 ∈ Ω, then f is an m–to–1 mapping in a punctured neighborhood of z0 , see Theorem 2.41, where m > 1. As f ′ (z0 ) = 0, we arrive at a contradiction. Remark. The converse of the last theorem is false. For example, f (z) = ez , f ′ (z) = ez ≠ 0 on ℂ. But the exponential function is not injective on ℂ.

2.7 Holomorphic parameter integrals In this section we prove a result about holomorphic parameter dependence of integrals, which will be very useful for many applications later on. A first result in this

2.7 Holomorphic parameter integrals | 65

direction was already used for the Taylor series expansion of holomorphic functions, see Lemma 2.3. Now we use general properties of L1 -functions; see, for instance, [20]. Theorem 2.43. Let Ω ⊂ ℂ be open and let (X, μ) be a measure space with a positive measure μ. Let L1 (μ) = {g ∶ X ⟶ ℂ measurable ∶ ∫ |g| dμ < ∞}. X

Suppose that the function f ∶ Ω × X ⟶ ℂ has the following properties: (i) f (z, ⋅) ∈ L1 (μ) for all z ∈ Ω; (ii) for all x ∈ X, the function f (⋅, x) ∶ Ω ⟶ ℂ is holomorphic; (iii) for each disc K ⊂ Ω there exists an integrable non-negative function gK on X, such that for all z ∈ K we have |f (z, ⋅)| ≤ gK μ-almost everywhere. Then the function F ∶ Ω ⟶ ℂ, defined by F(z) = ∫ f (z, x) dμ(x), X

z ∈ Ω,

is holomorphic on Ω. For all integers n ≥ 0, the function 𝜕n f (z, ⋅) 𝜕z n is integrable on X, and for z ∈ Ω one has F (n) (z) = ∫

X

𝜕n f (z, x) dμ(x). 𝜕z n

Proof. Let a ∈ Ω and choose r > 0 such that K ∶= D2r (a) ⊂ Ω. Then, by Cauchy’s formula (see Theorem 2.11), we have for all z ∈ D2r (a) f (z, x) =

1 f (ζ , x) dζ . ∫ 2πi 𝜕D2r (a) ζ − z

Hence, for all z, w ∈ Dr (a), z ≠ w, we obtain F(z) − F(w) 1 f (ζ , x) =∫ dζ dμ(x). ∫ z−w 2πi (ζ − z)(ζ − w) X 𝜕D2r (a) Now let (wk )k be a sequence in Dr (a) with limk→∞ wk = z, where wk ≠ z for all k; define φk (z, x) ∶=

1 f (ζ , x) dζ . ∫ 2πi 𝜕D2r (a) (ζ − z)(ζ − wk )

Then |φk (z, ⋅)| ≤

4πrgK (⋅) 2 = gK (⋅) 2πr 2 r

since |ζ − a| = 2r and wk , z ∈ Dr (a).

μ-almost everywhere,

66 | 2 Cauchy’s Theorem and Cauchy’s formula In addition, we have φk (z, ⋅) =

f (z, ⋅) − f (wk , ⋅) , z − wk

since f (ζ , x) f (ζ , x) (z − wk )f (ζ , x) − = , ζ −z ζ − wk (ζ − z)(ζ − wk ) and we can apply Cauchy’s formula to φk . Hence φk (z, ⋅) is a measurable function. Considering the limit wk → z, we observe that f (ζ , x)/[(ζ −z)(ζ −wk )] tends to f (ζ , x)/(ζ −z)2 uniformly for ζ ∈ 𝜕D2r (a). Hence we can interchange limit and integration and obtain lim φk (z, x) =

k→∞

𝜕f 1 f (ζ , x) dζ = (z, x), ∫ 2 2πi 𝜕D2r (a) (ζ − z) 𝜕z

where we used Cauchy’s formula for the first derivative of f with respect to z. Now we apply the dominated convergence theorem (see [20]) and get the desired assertion for n = 1. Using Cauchy’s formula for the higher derivatives, we obtain the general result.

2.8 Complex differential forms Here we recall Stokes’ Theorem for 1-forms on ℝ2 ; see, for instance, [59]. As ℝ2 is identified with ℂ, we use a complex notion of differential forms which is compatible with the Wirtinger derivatives. Definition 2.44. We define the differential dx as a linear mapping from ℝ2 ≅ ℂ to ℝ, dx ∶ z ↦ (dx)z = x,

z = x + iy,

analogously (dy)z = y. If f ∶ M ⊆ ℂ ⟶ ℂ is a real differentiable function, we define (df )z0 ∶=

𝜕f 𝜕f (z ) dx + (z0 ) dy 𝜕x 0 𝜕y

as the complete differential of f at the point z0 . More general, we say that α = f dx + g dy is a 1-form, where f , g are functions. If h is another function, we define hα ∶= hf dx + hg dy. In particular, we have for f (z) = z and f (z) = z: dz = dx + i dy

and dz = dx − i dy.

2.8 Complex differential forms | 67

Recall the Wirtinger derivatives with respect to z and z: 𝜕 1 𝜕 𝜕 = ( −i ) 𝜕z 2 𝜕x 𝜕y

and

𝜕 1 𝜕 𝜕 = ( + i ), 𝜕z 2 𝜕x 𝜕y

and simple computation shows that df =

𝜕f 𝜕f 𝜕f 𝜕f dx + dy = dz + dz. 𝜕x 𝜕y 𝜕z 𝜕z

𝜕f 𝜕f If f is a holomorphic function, we have df = 𝜕z dz, since 𝜕z = 0. 2 2 The 2-form dx ∧ dy ∶ ℝ × ℝ → ℝ is the alternating 2-linear form

ξ η ξ dx ∧ dy (( 1 ) , ( 1 )) = | 1 ξ2 η2 ξ2

η1 | = ξ1 η2 − ξ2 η1 . η2

If f is a function, ω = f (dx ∧ dy) is a general 2-form. The following rules are valid: (f dx + g dy) ∧ (f1 dx + g1 dy) = (fg1 − gf1 ) dx ∧ dy,

dx ∧ dy = −dy ∧ dx,

dz ∧ dz = −2i dx ∧ dy = −2i dλ(z),

where dλ is the Lebesgue measure on ℂ ≅ ℝ2 , dx ∧ dx = dy ∧ dy = dz ∧ dz = dz ∧ dz = 0. The differential of a 1-form α = f dx + g dy is defined by dα ∶= df ∧ dx + dg ∧ dy = (

𝜕g 𝜕f − ) dx ∧ dy. 𝜕x 𝜕y

If α = F dz + G dz, we have dα = (

𝜕G 𝜕F − ) dz ∧ dz. 𝜕z 𝜕z

In addition, d(df ) = 0 and if f is real differentiable, then d(f dz) = −

𝜕f dz ∧ dz. 𝜕z

Example 2.45. Let z be a fixed point and let f be a real differentiable function, let ω be the following 1-form ω(ζ ) =

1 f (ζ ) dζ 2πi ζ − z

for ζ ≠ z.

68 | 2 Cauchy’s Theorem and Cauchy’s formula Then dω(ζ ) = − =− since ζ ↦

1 ζ −z

1 1 𝜕f /𝜕ζ 𝜕 [ + f (ζ ) (ζ ↦ )] dζ ∧ dζ 2πi ζ − z ζ −z 𝜕ζ 1 𝜕f /𝜕ζ dζ ∧ dζ , 2πi ζ − z

is a holomorphic function.

Let G be open in ℂ and k ∈ ℕ ∪ {∞}. Then 𝒞k (G) denotes the space of k times continuously differentiable (in the real sense) functions on G. We also write 𝒞(G) instead of 𝒞0 (G). The space 𝒞k (G) is the space of all functions f in 𝒞k (G) such that all derivatives of f up to order k extend continuously to G. Theorem 2.46 (Stokes’ Theorem). Let G be a domain in ℂ. Let 𝜕G consist of a positively oriented path and let ω ∈ 𝒞1 (G) be a continuously differentiable 1-form on G. Then ∫ ω = ∫ dω. 𝜕G

G

For a proof, see [59].

2.9 The inhomogeneous Cauchy formula Now we apply Stokes’ Theorem to prove a more general version of Cauchy’s formula, which will be a useful tool for the study of the inhomogeneous Cauchy–Riemann equations. Theorem 2.47. Let G be a bounded domain in ℂ with piecewise smooth positively oriented boundary 𝜕G. Let f ∈ 𝒞1 (G). Then for z ∈ G we have f (z) =

1 f (ζ ) 1 (𝜕f /𝜕ζ )(ζ ) dζ + dζ ∧ dζ . ∫ ∫ 2πi 𝜕G ζ − z 2πi G ζ − z

Remark. If f ∈ ℋ(G), we have (𝜕f /𝜕ζ )(ζ ) = 0 ∀ζ ∈ G and hence f (z) =

1 f (ζ ) dζ . ∫ 2πi 𝜕G ζ − z

In this sense, Theorem 2.47 yields a generalization of Cauchy’s formula. Proof. Fix z ∈ G and choose r > 0 such that Dr (z) ⊆ G. We remove the disc Dr (z) from G and define Gr = G ⧵ Dr (z), the boundary of Gr consists of the positively oriented boundary of G and of the negatively oriented circle κr = 𝜕Dr (z). Walking on 𝜕Gr , the domain Gr lies always on the left-hand side.

2.9 The inhomogeneous Cauchy formula

| 69

For ζ ∈ Gr we define the 1-form ω(ζ ) =

1 f (ζ ) dζ , 2πi ζ − z

we can apply Stokes’ Theorem (see Theorem 2.46) and obtain from Theorem 2.45 1 f (ζ ) 1 (𝜕f /𝜕ζ )(ζ ) dζ = − dζ ∧ dζ . ∫ ∫ 2πi 𝜕Gr ζ − z 2πi Gr ζ −z Now we take the limit r → 0. First we show that the integral ∫

Ds (0)

1 dz ∧ dz |z|

exists. For this purpose we use polar coordinates and get ∫

Ds (0)

s 2π 1 1 1 dz ∧ dz = −2i ∫ dx ∧ dy = −2i ∫ ∫ r dr dϕ, |z| 0 0 r Ds (0) |z|

where the last integral exists. This implies that the function ζ ↦ the integral ∫

Gr

(𝜕f /𝜕ζ )(ζ ) dζ ∧ dζ ζ −z

(𝜕f /𝜕ζ )(ζ ) ζ −z

tends to

is absolutely integrable on G. Hence

∫

G

(𝜕f /𝜕ζ )(ζ ) dζ ∧ dζ , ζ −z

as r → 0. Here we used again the dominated convergence theorem. Now we consider the line integral of Stokes’ Theorem and get 1 f (ζ ) 1 f (ζ ) 1 f (ζ ) dζ = dζ + dζ . ∫ ∫ ∫ 2πi 𝜕Gr ζ − z 2πi 𝜕G ζ − z 2πi κr ζ − z We have

1 1 f (ζ ) f (z) f (ζ ) − f (z) 1 dζ = dζ + dζ ∫ ∫ ∫ 2πi κr ζ − z 2πi κr ζ − z 2πi κr ζ −z 1 1 1 f (ζ ) − f (z) = f (z) ⋅ dζ + dζ ∫ ∫ 2πi κr ζ − z 2πi κr ζ −z 1 f (ζ ) − f (z) = −f (z) + dζ , ∫ 2πi κr ζ −z

and hence |∫

κr

f (ζ ) − f (z) f (ζ ) − f (z) dζ | ≤ 2πr max | | = 2π max |f (ζ ) − f (z)| → 0, ∗ ζ ∈κr ζ ∈κr∗ ζ −z ζ −z

as r → 0, since f is continuous. So the line integral 1 f (ζ ) dζ ∫ 2πi 𝜕Gr ζ − z

tends to

as r → 0, and we arrive at the desired result.

1 f (ζ ) dζ − f (z), ∫ 2πi 𝜕G ζ − z

70 | 2 Cauchy’s Theorem and Cauchy’s formula

2.10 General versions of Cauchy’s Theorem and Cauchy’s formula It will be convenient to consider integrals over sums of paths. This leads to the concepts of chains and cycles. Definition 2.48. Let Ω ⊆ ℂ be open and let γ1 , … , γn be paths in Ω. Let Γ ∗ = ⋃nj=1 γj∗ and define γ̃ j (f ) ∶= ∫ f (z) dz, γj

for f ∈ 𝒞(Γ ∗ ). γ̃ j ∶ 𝒞(Γ ∗ ) ⟶ ℂ can be seen as a linear functional on 𝒞(Γ ∗ ). We set Γ̃ = γ̃ 1 + ⋯ + γ̃ n , and denote by Γ = γ1 + ⋯ + γn the formal sum of the paths γ1 , … , γn . We define n

̃ ) = ∑ ∫ f (z) dz, ∫ f (z) dz = Γ(f j=1 γj

Γ

for f ∈ 𝒞(Γ ∗ ). Γ is called a chain in Ω. If all paths γ1 , … , γn are closed, we call Γ a cycle in Ω. Remark. (a) Chains and cycles can be represented as sums of paths in many ways. (b) By −Γ we denote the cycle, where each path γj , j = 1, … , n is replaced by its opposite path, for f ∈ 𝒞(Γ ∗ ) we have ∫ f (z) dz = − ∫ f (z) dz. −Γ

Γ

(c) If Γ1 and Γ2 are chains or cycles, we can form the sum Γ = Γ1 + Γ2 and have ∫ f (z) dz = ∫ f (z) dz + ∫ f (z) dz, Γ

Γ1

Γ2

f ∈ 𝒞(Γ1∗ ∪ Γ2∗ ).

Definition 2.49. Let Γ = γ1 + ⋯ + γn be a cycle in Ω and α ∉ Γ ∗ . We define the index of α with respect to Γ by IndΓ (α) ∶=

dz 1 . ∫ 2πi Γ z − α

Obviously, we have n

IndΓ (α) = ∑ Indγj (α). j=1

Theorem 2.50 (Homology-version of Cauchy’s Theorem). Let Ω ⊆ ℂ be an arbitrary open set, f ∈ ℋ(Ω), let Γ be a cycle in Ω such that IndΓ (α) = 0 for every α ∉ Ω. Then f (z) IndΓ (z) =

1 f (w) dw ∫ 2πi Γ w − z

∀z ∈ Ω ⧵ Γ ∗ ;

2.10 General versions of Cauchy’s Theorem and Cauchy’s formula

| 71

in addition, ∫ f (w) dw = 0. Γ

If Γ0 and Γ1 are cycles in Ω such that IndΓ0 (α) = IndΓ1 (α)

∀α ∉ Ω,

then ∫ f (z) dz = ∫ f (z) dz. Γ0

Γ1

Proof. Let { f (z)−f (w) , z ≠ w, z, w ∈ Ω, g(z, w) = { z−w f ′ (z), z = w ∈ Ω. { By Lemma 2.39, g ∶ Ω × Ω ⟶ ℂ is continuous. The first assertion of the theorem is equivalent to the statement h(z) = 0 ∀z ∈ Ω ⧵ Γ ∗ , where h(z) =

1 ∫ g(z, w) dw, 2πi Γ

because 1 1 dw 1 f (z) − f (w) f (w) dw = f (z) ∫ − dw ∫ ∫ 2πi Γ z − w 2πi 2πi Γ z − w Γ z−w f (w) 1 dw. = −f (z) IndΓ (z) + ∫ 2πi Γ w − z First we show that h ∈ ℋ(Ω). g is uniformly continuous on every compact subset of Ω × Ω. Hence, if z ∈ Ω and zn → z in Ω, then g(zn , w) → g(z, w) uniformly for w ∈ Γ ∗ , which is a compact subset. So we get 1 lim ∫ g(zn , w) dw 2πi n→∞ Γ 1 1 = ∫ lim g(z , w) dw = ∫ g(z, w) dw = h(z), 2πi Γ n→∞ n 2πi Γ

lim h(zn ) =

n→∞

and h is continuous on Ω; the limit and integral can be interchanged because of uniform convergence. Now let Δ be an arbitrary closed triangle in Ω. Then Fubini’s theorem implies ∫ h(z) dz = 𝜕Δ

1 1 ∫ (∫ g(z, w) dw) dz = ∫ (∫ g(z, w) dz) dw. 2πi 𝜕Δ Γ 2πi Γ 𝜕Δ

72 | 2 Cauchy’s Theorem and Cauchy’s formula For w fixed, the function z ↦ g(z, w) has a removable singularity at z = w (see Theorem 2.24), so it is holomorphic on Ω and, by Theorem 2.5, we have ∫ g(z, w) dz = 0

∀w ∈ Ω,

𝜕Δ

hence ∫ h(z) dz = 0, 𝜕Δ

and, by Theorem 2.16, h ∈ ℋ(Ω). Next we prove that h(z) = 0 ∀z ∈ Ω ⧵ Γ ∗ . For this aim let Ω1 = {z ∈ ℂ ∶ IndΓ (z) = 0}. If we define h1 (z) =

1 f (w) dw, ∫ 2πi Γ w − z

then h1 ∈ ℋ(Ω1 ), and for z ∈ Ω ∩ Ω1 h(z) =

dw 1 f (w) 1 f (w) 1 f (z) ∫ − dw = 0 + dw = h1 (z). ∫ ∫ 2πi 2πi Γ z − w 2πi Γ w − z Γ z−w

Hence the function {h(z), ϕ(z) = { h (z), { 1

z ∈ Ω, z ∈ Ω1

is holomorphic on Ω ∪ Ω1 . By assumption, we have IndΓ (α) = 0 ∀α ∉ Ω, hence Ωc ⊆ Ω1 and Ω ∪ Ω1 = ℂ. So ϕ ∈ ℋ(ℂ) is an entire function. The set Ω1 contains the unbounded connected component of ℂ ⧵ Γ ∗ , since the index is always zero there (see Theorem 2.2). Therefore lim ϕ(z) = lim h1 (z) =

|z|→∞

|z|→∞

1 f (w) lim ∫ dw = 0, 2πi |z|→∞ Γ w − z

and Liouville’s theorem (see Theorem 2.29) implies ϕ ≡ 0, in particular h(z) = 0 ∀z ∈ Ω ⧵ Γ∗ . Finally, we have to show that ∫Γ f (z) dz = 0. Let a ∈ Ω ⧵ Γ ∗ and F(z) ∶= (z − a)f (z). From the first assertion of the theorem, we get for F and z = a that 0 = F(a) IndΓ (a) =

1 F(w) 1 dw = ∫ ∫ f (w) dw. 2πi Γ w − a 2πi Γ

For the last assertion of the theorem, we pick two cycles Γ0 and Γ1 with IndΓ0 (α) = IndΓ1 (α)

∀α ∉ Ω.

Define Γ = Γ0 − Γ1 , then IndΓ (α) = 0 ∀α ∉ Ω and, by the first part of the theorem, 0 = ∫ f (z) dz = ∫ f (z) dz − ∫ f (z) dz. Γ

Γ0

Γ1

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Remark. (a) If IndΓ (α) = 0 ∀α ∉ Ω, the cycle Γ is called null-homologous in Ω. (b) Theorem 2.50 is a generalization of Corollary 2.11: if Ω is convex γ a closed path in Ω, then w ↦ 1/(w − α) is holomorphic on Ω for every α ∉ Ω and, by Corollary 2.6, we obtain Indγ (α) =

1 dw = 0. ∫ 2πi γ w − α

Therefore, the assumptions of Theorem 2.50 are satisfied, hence Theorem 2.50 implies Corollary 2.11. Example 2.51. Let Ω = ℂ ⧵ (D1/2 (−2) ∪ D1/2 (0) ∪ D1/2 (2)), and let γ1 (t) = −2 + 43 eit , γ2 (t) = 3 it e , γ3 (t) = 2 + 43 eit , Γ(t) = 6eit , t ∈ [0, 2π]. Define γ = γ1 + γ2 + γ3 . Then 4 Indγ (α) = IndΓ (α)

∀α ∉ Ω,

hence, by Theorem 2.50, we have ∫ f (z) dz = ∫ f (z) dz, γ

Γ

for each f ∈ ℋ(Ω). In the following we discuss another topological concept which is important for Cauchy’s Theorem. Definition 2.52. Let Ω ⊆ ℂ and γ0 , γ1 ∶ [0, 1] ⟶ Ω be closed curves. γ0 and γ1 are Ω-homotopic, if there is a continuous mapping H ∶ [0, 1] × [0, 1] ⟶ Ω such that H(s, 0) = γ0 (s)

∀s ∈ [0, 1];

H(s, 1) = γ1 (s)

H(0, t) = H(1, t)

∀t ∈ [0, 1].

∀s ∈ [0, 1]

and

Set γt (s) = H(s, t) for a fixed t ∈ [0, 1] and s ∈ [0, 1]. Since H(0, t) = H(1, t), the curves γt are also closed. We get a one-parameter family γt , t ∈ [0, 1] of closed curves connecting γ0 and γ1 . If a closed curve γ0 is Ω-homotopic to a constant curve (consisting of just one point), we say that γ0 null-homotopic in Ω. A domain Ω is said to be simply connected, if every closed curve in Ω is nullhomotopic in Ω.

74 | 2 Cauchy’s Theorem and Cauchy’s formula Example 2.53. Let Ω be a convex domain and let γ be a closed curve in Ω. Fix z0 ∈ Ω. Define H(s, t) = (1 − t)γ(s) + tz0 ,

s, t ∈ [0, 1].

Then H(s, 0) = γ(s) and H(s, 1) = z0 ∀s ∈ [0, 1]. Since γ(0) = γ(1), we have H(0, t) = (1 − t)γ(0) + tz0 = (1 − t)γ(1) + tz0 = H(1, t)

∀t ∈ [0, 1].

For a fixed s, the expression H(s, t) = (1 − t)γ(s) + tz0 , t ∈ [0, 1] describes the straight line from γ(s) to z0 , which is contained in Ω, as Ω is convex. Hence H(s, t) ∈ Ω ∀s, t ∈ [0, 1] and γ is null-homotopic in Ω. Therefore, Ω is simply connected. Lemma 2.54. Let γ0 and γ1 be closed paths in ℂ. Let α ∈ ℂ be a complex number such that |γ0 (s) − γ1 (s)| < |α − γ0 (s)|

∀s ∈ [0, 1].

Then Indγ0 (α) = Indγ1 (α). Proof. The assumption implies that α ∉ γ0∗ and α ∉ γ1∗ . We set γ(s) =

γ1 (s) − α , γ0 (s) − α

s ∈ [0, 1],

then we get for the derivatives γ ′ (s) =

(γ0 (s) − α)γ1′ (s) − (γ1 (s) − α)γ0′ (s) (γ0 (s) − α)2

and γ ′ (s) (γ0 (s) − α)γ1′ (s) − (γ1 (s) − α)γ0′ (s) γ0 (s) − α = γ(s) (γ0 (s) − α)2 γ1 (s) − α ′ γ1 (s) γ0′ (s) = − . γ1 (s) − α γ0 (s) − α By assumption we have |1 − γ(s)| = |

γ0 (s) − γ1 (s) | < 1, γ0 (s) − α

hence γ ∗ ⊂ D1 (1) and, by Theorem 2.2, Indγ (0) = 0. This implies 0 = Indγ (0) = =

1 dz 1 1 γ ′ (s) = ds ∫ ∫ 2πi γ z 2πi 0 γ(s)

1 1 γ1′ (s) 1 1 γ0′ (s) ds − ds = Indγ1 (α) − Indγ0 (α). ∫ ∫ 2πi 0 γ1 (s) − α 2πi 0 γ0 (s) − α

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| 75

Theorem 2.55 (Homotopy version of Cauchy’s Theorem). Let Ω be a domain in ℂ, let Γ0 and Γ1 closed paths in Ω, which are Ω-homotopic. Let α ∉ Ω. Then IndΓ0 (α) = IndΓ1 (α) and, by Theorem 2.50, ∫ f (z) dz = ∫ f (z) dz Γ0

∀f ∈ ℋ(Ω).

Γ1

Proof. Let H be a homotopy function between Γ0 and Γ1 . The difficulty of the proof relies on the fact that the one-parameter family Γt (s) = H(s, t) does not consist of necessarily piecewise differentiable curves. We will approximate the curves Γt by suitable paths, in our case by polygonal closed paths. Fix α ∉ Ω. Since H is uniformly continuous on the compact set [0, 1] × [0, 1], there exists ϵ > 0 such that |α − H(s, t)| > 2ϵ

∀s, t ∈ [0, 1];

(2.3)

and there exists n ∈ ℕ such that |H(s, t) − H(s′ , t ′ )| < ϵ/2,

(2.4)

if |s − s′ | + |t − t ′ | < 1/n. Now we define the approximating polygonal paths: for k = 0, 1, … , n and s ∈ [0, 1] let j−1 k j k , )(j − ns), γk (s) = H( , )(ns + 1 − j) + H( n n n n for j − 1 ≤ ns ≤ j, j = 1, … , n. It is easily seen that the curves γ0 , γ1 , … , γn are closed. They are also piecewise differentiable, since the variable s only appears as a linear term, and not as an argument of the function H. By (2.4) and from the definition of γk , we get k |γk (s) − H(s, )| < ϵ, n for s ∈ [0, 1] and k = 0, 1, … , n. If j − 1 ≤ ns ≤ j, we have j − ns ≤ 1, hence k j k j−1 k k |γk (s) − H(s, )| = |H( , )(ns + 1 − j) + H( , )(j − ns) − H(s, )| n n n n n n j k j−1 k j k k ≤ (j − ns)|H( , ) − H( , )| + |H( , ) − H(s, )| n n n n n n n < ϵ/2 + ϵ/2 = ϵ.

(2.5)

76 | 2 Cauchy’s Theorem and Cauchy’s formula In particular, for k = 0 |γ0 (s) − H(s, 0)| = |γ0 (s) − Γ0 (s)| < ϵ and for k = n |γn (s) − H(s, 1)| = |γn (s) − Γ1 (s)| < ϵ. Now we get from (2.3) and (2.5) |α − γk (s)| = |α − H(s, k/n) − (γk (s) − H(s, k/n))| ≥ |α − H(s, k/n)| − |γk (s) − H(s, k/n)| > 2ϵ − ϵ = ϵ,

for s ∈ [0, 1] and k = 0, 1, … , n. Using (2.4) and the definition of γk , we get |γk−1 (s) − γk (s)| < ϵ, for s ∈ [0, 1] and k = 1, … , n. If j − 1 ≤ ns ≤ j, we have ns + 1 − j ≤ 1, therefore j k j k−1 ) − H( , )| |γk−1 (s) − γk (s)| ≤ (ns + 1 − j)|H( , n n n n j−1 k−1 j−1 k + (j − ns)|H( , ) − H( , )| n n n n < ϵ/2 + ϵ/2 = ϵ. Since |α − γk (s)| > ϵ and |γk−1 (s) − γk (s)| < ϵ, we obtain |γk−1 (s) − γk (s)| < |α − γk (s)|, ∀s ∈ [0, 1] and k = 1, … , n. Similarly, |γ0 (s) − Γ0 (s)| < |α − Γ0 (s)| and |γn (s) − Γ1 (s)| < |α − Γ1 (s)|, ∀s ∈ [0, 1]. Now we can apply Lemma 2.54 to the pairs (Γ0 , γ0 ), (γ0 , γ1 ), … , (γn−1 , γn ), (γn , Γ1 ) and get IndΓ0 (α) = Indγ0 (α) = Indγ1 (α) = ⋯ = Indγn−1 (α) = Indγn (α) = IndΓ1 (α).

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| 77

Corollary 2.56. (a) Let Ω be a simply connected domain in ℂ and let γ be a closed path in Ω. Then ∫ f (z) dz = 0, γ

for each f ∈ ℋ(Ω). (b) Let Ω ⊆ ℂ be an open set and let a ∈ Ω. Let γ be a closed path in Ω ⧵ {a}, which is null-homotopic in Ω. Then f (n) (a) Indγ (a) =

f (z) n! dz, ∫ 2πi γ (z − a)n+1

n ∈ ℕ,

for each f ∈ ℋ(Ω). Proof. (b) For n = 1, the assertion follows from Theorem 2.50. Finally, differentiation inside the integral with respect to the variable a gives the desired result. Remark. If Ω is a domain and γ a closed null-homotopic path in Ω, then ∫

γ

dz =0 z−α

∀α ∉ Ω,

this follows from Theorem 2.55, since the function z ↦ 1/(z − α) is holomorphic and γ is null-homotopic in Ω. This implies that γ is also null-homologous. Every nullhomotopic path is null-homologous. The converse is false (see Exercises). But if every closed path in Ω is null-homologous, then every closed path is also null-homotopic and Ω is simply connected (see Chapter 4). We conclude this chapter with two useful results about the index of a path. Definition 2.57. Let G be a domain in ℂ and γ ∶ [0, 1] ⟶ ℂ a path. We say that γ runs in G from boundary to boundary, if (1) there exists t1 , t2 ∈ [0, 1], t1 < t2 , γ(t1 ), γ(t2 ) ∈ 𝜕G, γ(t1 ) ≠ γ(t2 ); (2) γ(t) ∈ G, for t1 < t < t2 ; (3) γ(t) ∉ G, for t ∈ [0, 1] but t ∉ [t1 , t2 ]; (4) G ⧵ γ ∗ has exactly two connected components and γ ∗ ∩ G belongs to the boundary of both of these components. Remark. If γ is injective and smooth and z0 ∈ γ ∗ is an arbitrary point on γ, there exists a neighborhood U of z0 such that γ runs in U from boundary to boundary. Theorem 2.58. Let γ be a closed path in ℂ and let D be a disc. Suppose that γ runs in D from boundary to boundary. Let t1 , t2 ∈ [0, 1] with t1 < t2 and a = γ(t1 ), b = γ(t2 ), a, b ∈ 𝜕D,

78 | 2 Cauchy’s Theorem and Cauchy’s formula and γ|[t1 ,t2 ] = γ0 . Let D1 , D2 denote the two connected components of D ⧵ γ ∗ . Suppose that D1 lies to the left of γ. Then Indγ (z1 ) = Indγ (z2 ) + 1, for z1 ∈ D1 and z2 ∈ D2 . Remark. Let γ in ℂ be an arbitrary path. The index Indγ (z) = 0 for z in the unbounded component of ℂ ⧵ γ ∗ . Now one can use the theorem above, in order to compute the indices of γ one after the other. Proof. We separate γ = γ0 + γ1 , and 𝜕D into κ1 and κ2 (positively oriented), κ1∗ ⊂ 𝜕D1 , κ2∗ ⊂ 𝜕D2 , see Figure 2.6.

Figure 2.6

Then Ind−κ1 +γ1 (z1 ) = Ind−κ1 +γ1 (z2 ) and hence Indκ1 (z1 ) − Indκ1 (z2 ) = Indγ1 (z1 ) − Indγ1 (z2 ), where we used the notation Indκj (zj ) = also for paths which are not closed.

1 dζ , ∫ 2πi κj ζ − zj

2.10 General versions of Cauchy’s Theorem and Cauchy’s formula

| 79

z2 belongs to the unbounded component of ℂ ⧵ (κ1 + γ0 )∗ , so we have Indκ1 +γ0 (z2 ) = 0. Similarly, Indκ2 −γ0 (z1 ) = 0. This gives Indγ (z1 ) − Indγ (z2 ) = Indγ0 (z1 ) − Indγ0 (z2 ) + Indγ1 (z1 ) − Indγ1 (z2 )

= Indγ0 (z1 ) − Indγ0 (z2 ) + Indκ1 (z1 ) − Indκ1 (z2 )

= Indγ0 (z1 ) + Indκ1 (z1 )

= Indγ0 (z1 ) + Indκ1 (z1 ) + Indκ2 −γ0 (z1 )

= Indκ1 +κ2 (z1 ) = 1.

Lemma 2.59. Let A ⊂ ℂ be a compact subset and U ⊃ A an open set. Then there exists a cycle Γ in U ⧵ A such that IndΓ (a) = 1

∀a ∈ A

and

IndΓ (z) = 0 ∀z ∉ U.

Proof. Case 1. First we suppose that A is connected. Let δ > 0 be such that 0 < 2δ < dist(A, 𝜕U). We use a lattice parallel to the axes with mesh width δ and positively oriented lattice squares. Since A is compact, there exist finitely many lattice squares Q1 , … , Qn with Qj ∩ A ≠ ∅, j = 1, … , n. Let Γj be the boundary cycle of Qj and define Γ = Γ 1 + ⋯ + Γn . Let a ∈ A be an arbitrary point in the compact set A. Without loss of generality, we can suppose that a ∈ Q∘1 . Then n

IndΓ (a) = ∑ IndΓj (a) = IndΓ1 (a) = 1. j=1

If a belongs to 𝜕Qj for some j, then a belongs to the interior of four adjacent squares of the lattice, and we get the same result IndΓ (a) = 1. Now we modify Γ in the following way: we take only line segments [p, q], which are line segments of exactly one square of our collection, these are line segments [p, q] with [p, q] ∩ A = ∅; line segments with non-empty intersection with A are passed through in both directions and drop out. The modified boundary cycle is again denoted by Γ. Now we have Γ ∗ ∩ A = ∅ and, by the choice of the mesh width, that Γ ∗ ⊂ U. Hence ∗ Γ ⊂ U ⧵ A. As A is connected, we get from Theorem 2.2 that IndΓ (a) = 1 ∀a ∈ A.

80 | 2 Cauchy’s Theorem and Cauchy’s formula If z ∉ U, it follows that IndΓj (z) = 0, j = 1, … , n and IndΓ (z) = 0. Case 2. A has finitely many connected components A1 , … , AN . Now we choose points aj ∈ Aj , j = 1, … , N and a lattice parallel to the axes with mesh width δ > 0, where 2δ < min{dist(A, 𝜕U), |aj − ak | j ≠ k}, and we also choose the lattice such that different aj belong to the interior of different squares of the lattice. Then IndΓ (aj ) = 1 and hence IndΓ (a) = 1 for every each a ∈ Aj , j = 1, … , N. Everything else can now be reduced to the first case. Case 3. In the general case, A could have infinitely many connected components. For every z ∈ A there exists an open square Q(z) with line segments parallel to the axes such that z ∈ Q(z) ⊂⊂ U; since A is compact, finitely many of these squares cover A. We denote them by Q1 , … , Qm . Now let m

A0 = ⋃ Qj ⊂ U. j=1

A0 is compact and has only finitely many connected components. Since A ⊂ A0 ⊂ U, it suffices to prove the assertion for A0 , which follows from Case 2.

2.11 Laurent series and meromorphic functions We now study holomorphic functions on annuli and obtain their canonical representations as Laurent series. Theorem 2.60. Let Dr,R (a) = {z ∈ ℂ ∶ r < |z − a| < R} be an annulus, we define D0,R (a) = DR (a) ⧵ {a} and Dr,∞ (a) = {z ∶ |z − a| > r}. Let f ∈ ℋ(Dr,R (a)), and define U1 = Dr,∞ (a), U2 = DR (a). Then there exist functions f1 ∈ ℋ(U1 ) and f2 ∈ ℋ(U2 ) such that f = f1 + f2 ,

on U1 ∩ U2 = Dr,R (a).

The function f1 can be chosen with the property lim|z|→∞ f1 (z) = 0. In this way f1 and f2 are uniquely determined. Proof. Let r < ρ < R and define f2,ρ (z) =

1 f (ζ ) dζ , ∫ 2πi γρ ζ − z

where γρ (t) = a + ρe2πit , t ∈ [0, 1]. The function in the integral is holomorphic for z in Dρ (a), hence f2,ρ ∈ ℋ(Dρ (a)). By Theorem 2.55, we have f2,ρ (z) = f2,ρ′ (z) on Dρ (a) for r < ρ < ρ′ < R.

2.11 Laurent series and meromorphic functions |

81

For z ∈ U2 = DR (a) and max{r, |z − a|} < ρ < R, we define f2 (z) =

1 f (ζ ) dζ , ∫ 2πi γρ ζ − z

where the integral is independent of ρ as long as r < ρ < R. Hence f2 ∈ ℋ(U2 ). For z ∈ U1 = {z ∶ |z − a| > r} and r < σ < min{R, |z − a|}, we define f1 (z) = −

1 f (ζ ) dζ . ∫ 2πi γσ ζ − z

Similarly, we get f1 ∈ ℋ(U1 ), and from the definition of f1 we derive immediately that lim|z|→∞ |f1 (z)| = 0. For z ∈ Dr,R (a), we choose ρ and σ such that r < σ < |z − a| < ρ < R and define the cycle Γ = γρ − γσ . It follows that IndΓ (α) = 0 ∀α ∉ Dr,R (a) and IndΓ (z) = 1. Now apply Theorem 2.50 and obtain f (z) =

1 f (ζ ) 1 f (ζ ) 1 f (ζ ) dζ = dζ − dζ = f1 (z) + f2 (z). ∫ ∫ ∫ 2πi Γ ζ − z 2πi γρ ζ − z 2πi γσ ζ − z

It remains to show uniqueness of the representation f = f1 + f2 . For this let f = g1 + g2 be another representation with g1 ∈ ℋ(U1 ) and g2 ∈ ℋ(U2 ), as well as lim|z|→∞ |g1 (z)| = 0. Then we have f1 − g1 = g2 − f2 on U1 ∩ U2 . We define {f − g1 h={ 1 g −f { 2 2

on U1 ,

on U2 ,

and get a holomorphic function on U1 ∪ U2 = ℂ. Hence h ∈ ℋ(ℂ) and lim|z|→∞ |h(z)| = 0. So h is a bounded entire function. By Liouville’s Theorem (see Theorem 2.29), it follows that h ≡ 0, and f1 = g1 as well as f2 = g2 . Remark. f1 is called the principal part of f . Since f2 is holomorphic on DR (a), it can be expanded as a Taylor series ∞

f2 (z) = ∑ an (z − a)n , n=0

z ∈ DR (a).

Let F(w) = a + 1/w. Then F is a biholomorphic mapping (i.e. holomorphic in both directions) from D′1/r (0) = {w ∶ 0 < |w| < 1/r} to U1 = {w ∶ |w − a| > r}. Hence, f1 ∘ F ∈ ℋ(D′1/r (0)).

82 | 2 Cauchy’s Theorem and Cauchy’s formula Since lim|z|→∞ |f1 (z)| = 0, we have limw→0 (f1 ∘ F)(w) = 0. Hence f1 ∘ F has a removable singularity at w = 0 (see Theorem 2.24), in addition, f1 ∘ F ∈ ℋ(D1/r (0)) and we can expand it as a Taylor series around w = 0 to get ∞

(f1 ∘ F)(w) = ∑ bn wn , n=1

where the series converges uniformly on D1/ρ (0), ρ > r. For w = 1/(z − a), we have F(w) = z and −∞

f1 (z) = ∑ an (z − a)n , n=−1

where a−n = bn and the series converges uniformly on ℂ ⧵ Dρ (a), ρ > r. Theorem 2.61. Let f ∈ ℋ(Dr,R (a)). Then f can be represented in the form −∞

∞

n=−1

n=0

f (z) = ∑ an (z − a)n + ∑ an (z − a)n , which is the Laurent9 series of f in Dr,R (a), and the series converges uniformly on all compact subsets of Dr,R (a). The Laurent coefficients an are given by an =

1 f (z) dz, ∫ 2πi γρ (z − a)n+1

n∈ℤ

where γρ (t) = a + ρe2πit , t ∈ [0, 1], r < ρ < R. Proof. It remains to prove the formula for the Laurent coefficients an . We have −∞

∞

k=−1

k=0

(z − a)−n−1 f (z) = ∑ ak+n+1 (z − a)k + ∑ ak+n+1 (z − a)k with uniform convergence on ∫

γρ

γρ∗ .

Hence integration term by term yields

f (z) dz dz = an ∫ = 2πian , (z − a)n+1 γρ z − a

because all summands are zero, except for k = −1. Example 2.62. (1) Let f (z) = 9 Laurent, Pierre Alphonse (1813–1854).

1 . z(z − i)2

2.11 Laurent series and meromorphic functions |

83

(a) Its Laurent series expansion in D′1 (0) = {z ∶ 0 < |z| < 1} is 1 z n n+2 1 1 1 1 =− = − ∑ (n + 1)( ) = − + i ∑ n z n . 2 2 z(z − i) z (1 − z/i) z n=0 i z n=0 i ∞

∞

(b) Its Laurent series expansion in D1,∞ (0) = {z ∶ |z| > 1} is 1 1 1 = 3 = ∑ i−n−1 (n + 2)z n . 2 z(z − i) z (1 − i/z)2 n=−3 −∞

(c) Its Laurent series expansion in D0,1 (i) = {z ∶ 0 < |z − i| < 1} is 1 −i 1 1 −i 1 i = + − = + − = ⋯, 2 2 2 z(z − i) (z − i) z − i z (z − i) z − i 1 − i(z − i) where the last term can be written as the sum of a geometric series. (2) Consider the Laurent series ∞

zn n+1 n=0 2 ∞

∑ z −n + ∑

n=1

1 with infinitely many negative powers of z. The first summand converges to z−1 for 1 |z| > 1, and the second summand converges to 2−z for |z| < 2. So the whole series 1 1 converges to the function f (z) = z−1 + 2−z on the annulus D1,2 (0). The function f is holomorphic at z = 0, although the Laurent series in the annulus D1,2 (0) has infinitely many negative powers of z.

Remark. We return to the characterization of isolated singularities in Theorem 2.25, and can easily show that a function f has a pole of order k in a point a if and only if the Laurent expansion of f in the punctured disc D′r (a) has the form f (z) = a−k (z − a)−k + ⋯ + a−1 (z − a)−1 + a0 + a1 (z − a) + ⋯ , where a−k ≠ 0. In addition, it follows that f has an essential singularity in a if and only if the Laurent expansion of f in the punctured disc D′r (a) has infinitely many terms of the form a−k (z − a)−k , where k > 0 and a−k ≠ 0. Definition 2.63. Let U ⊆ ℂ be an open set. A function f is called meromorphic on U, if there exists a discrete subset P of U such that f ∈ ℋ(U ⧵ P) and f has poles in P. ℳ(U) denotes the set of all meromorphic functions on U. Example 2.64. (1) Let U = D1 (0) and f (z) = 1/z. Then f ∈ ℳ(U). Every rational function p/q, where p and q are polynomials, belongs to ℳ(ℂ). sin z (2) tan z = cos belongs to ℳ(ℂ). It is easily seen that tan z has infinitely many poles. z (3) Let f , g ∈ ℋ(U) and suppose that g ≢ 0 on every connected component of U. Then f /g ∈ ℳ(U) (see Theorem 2.17).

84 | 2 Cauchy’s Theorem and Cauchy’s formula Theorem 2.65. Let f ∈ ℳ(U). Then, for each a ∈ U, there exists an open neighborhood V of a, and g, h ∈ ℋ(V) such that f = g/h on V . Proof. If a is not a pole of f , we put g = f and h ≡ 1 and take V = U ⧵ Pf , where Pf denotes the set of all poles of f . Then f = g/h on V and g, h ∈ ℋ(V). If a is a pole of order m > 0 of f , then, by Theorem 2.25, there exist complex numbers c1 , … , cm (cm ≠ 0) such that the function m

ck = ϕ(z) (z − a)k k=1

f (z) − ∑

has a removable singularity in a. Hence m

m

1 ck = [(z − a)m ϕ(z) + ∑ ck (z − a)m−k ], m k (z − a) (z − a) k=1 k=1

f (z) = ϕ(z) + ∑

where the expression in brackets is holomorphic in a neighborhood of a. Denote this expression by g, then we have g(a) = cm ≠ 0 and f (z) = g(z)/(z −a)m in a suitable neighborhood of a. Later we will be able to show that f = g/h globally on U. Theorem 2.66. Let f ∈ ℳ(U) and let a be a pole of f . Then lim |f (z)| = ∞,

z→a

i.e. for each compact subset K ⊂ ℂ there exists δ > 0 with f (D′δ (a)) ⊆ ℂ ⧵ K. Proof. From the proof of the last theorem we get f (z) =

m

1 [(z − a)m ϕ(z) + ∑ ck (z − a)m−k ], (z − a)m k=1

where cm ≠ 0. Taking the limit z → a yields the desired result. Theorem 2.67. Let Ω be a domain in ℂ. Then ℳ(Ω) is a field with respect to pointwise addition and multiplication of functions.

2.12 The residue theorem The residue theorem is not only a generalization of the homology version of Cauchy’s theorem when the function f has singularities, it enables us also to evaluate definite real integrals which are certainly not solvable by methods of real analysis and to count the number of zeroes and poles of meromorphic functions.

2.12 The residue theorem

| 85

Definition 2.68. Let U ⊆ ℂ be open and f a holomorphic function on U except for isolated singularities. Let a be an isolated singularity of f . Then there exists r > 0, such that f can be expanded in D′r (a) as a Laurent series −∞

∞

n=−1

n=0

f (z) = ∑ cn (z − a)n + ∑ cn (z − a)n , the coefficient c−1 = Res(f ; a) is called the residue of f at a. Remark. If one computes the line integral ∫ f (z) dz, γs

where γs (t) = a + seit , t ∈ [0, 2π] and 0 < s < r, one observes that the integral can be computed term by term in the Laurent series expansion and that only one term is left, namely ∫

γs

c−1 dz = 2πic−1 , z−a

which motivates the notion of a residue. Theorem 2.69 (Residue Theorem). Let U ⊆ ℂ be open and f a holomorphic function on U except for isolated singularities. Denote by Sf the set of all singularities of f in U. Let Γ be a cycle in U ⧵ Sf such that IndΓ (α) = 0 ∀α ∉ U. Then 1 ∫ f (z) dz = ∑ Res(f ; a) IndΓ (a). 2πi Γ a∈S f

Proof. First we show that the set B = {a ∈ Sf ∶ IndΓ (a) ≠ 0} is finite, which implies that the sum in the theorem is a finite sum. For this let W = ℂ ⧵ Γ ∗ . The index IndΓ is constant on every connected component V of W . If V is unbounded or if V ∩ (ℂ ⧵ U) ≠ ∅, then IndΓ (α) = 0 ∀α ∈ V , by our assumption that IndΓ (α) = 0 ∀α ∉ U. Sf has no limit point in U, the limit points of Sf can only be on the boundary of U, therefore limit points can belong to the unbounded component of W or to a component V with V ∩ (ℂ ⧵ U) ≠ ∅. We have dist(Γ ∗ , 𝜕U) > 0, hence B must be finite. Let B = {a1 , a2 , … , an }, and let Qj be the principal parts of f in the Laurent expansion around aj , j = 1, … , n. Define g = f − (Q1 + Q2 + ⋯ + Qn ), (if B = ∅, set g = f ) then g has removable singularities at the points of B, and g ∈ ℋ(U0 ), where U0 = U ⧵ (Sf ⧵ B). Now IndΓ (α) = 0 ∀α ∉ U0 , and, by Theorem 2.50, we have ∫ g(z) dz = 0 Γ

86 | 2 Cauchy’s Theorem and Cauchy’s formula and, by the definition of g, we get n

n

1 1 ∑ ∫ Q (z) dz = ∑ Res(Qk ; ak ) IndΓ (ak ) ∫ f (z) dz = 2πi Γ 2πi k=1 Γ k k=1 n

= ∑ Res(f ; ak ) IndΓ (ak ). k=1

Remark. If U is open and convex and Γ is a closed, positively oriented path without double points in U and f is holomorphic in U except for isolated singularities, then 1 ∫ f (z) dz = ∑ Res(f ; ak ), 2πi Γ where the sum is taken over all singularities of f in the interior of Γ. For the applications it will be convenient to know about some simple rules how to compute residues. Theorem 2.70. Let U ⊆ ℂ be open and let f and g be holomorphic on U except for isolated singularities. Then (a) Res(f + g; a) = Res(f ; a) + Res(g; a) and for α1 , α2 ∈ ℂ Res(α1 f + α2 g; a) = α1 Res(f ; a) + α2 Res(g; a). (b) If z0 is a pole of first order of f , we have Res(f ; z0 ) = lim [(z − z0 )f (z)]. z→z0

(c) If g is holomorphic at z0 and f has a pole of first order at z0 , we have Res(fg; z0 ) = g(z0 ) Res(f ; z0 ). (d) If h is holomorphic at z0 and z0 is a simple zero of h, we have Res(1/h; z0 ) = 1/h′ (z0 ). (e) If z0 is a pole of order n of f , we have Res(f ; z0 ) =

1 dn−1 lim { n−1 [(z − z0 )n f (z)]}. (n − 1)! z→z0 dz

Proof. (a) Follows from the Laurent expansion of f and g around a.

2.12 The residue theorem

| 87

(b) The Laurent expansion of f around z0 has the form c−1 + ∑ c (z − z0 )n , z − z0 n=0 n ∞

f (z) = hence

∞

(z − z0 )f (z) = c−1 + ∑ cn (z − z0 )n+1 , n=0

taking the limit z → z0 , the infinite series disappears. (c) We have ∞

g(z) = g(z0 ) + ∑ bn (z − z0 )n , n=1

f (z) =

c−1 + ∑ c (z − z0 )n , z − z0 n=0 n ∞

this implies f (z)g(z) =

g(z0 )c−1 + ∑ dn (z − z0 )n . z − z0 n=0 ∞

(d) 1/h has a pole of first order at z0 . Hence, by (b), Res(1/h; z0 ) = lim

z→z0

z − z0 z − z0 = lim = 1/h′ (z0 ). h(z) z→z0 h(z) − h(z0 )

(e) We have f (z) =

c−n c + ⋯ + −1 + ∑ ck (z − z0 )k (z − z0 )n z − z0 k=0 ∞

which implies that ∞

(z − z0 )n f (z) = c−n + c−n+1 (z − z0 ) + ⋯ + c−1 (z − z0 )n−1 + ∑ ck (z − z0 )k+n . k=0

Differentiating (n − 1) times we get the desired result. Theorem 2.71 (Rouché’s10 Theorem). Let Ω be a domain in ℂ and f ∈ ℳ(Ω) a meromorphic function. Let γ be a closed, null-homologous path in Ω. Suppose that f has no zeros and no poles on γ ∗ and that Indγ (α) = 1 or = 0 ∀α ∈ ℂ ⧵ γ ∗ . Let Ω1 = {z ∈ Ω ∶ Indγ (z) = 1} and let Nf the number of zeros of f in Ω1 and Pf the number of poles of f in Ω1 . Then Nf − Pf = where Γ = f ∘ γ. 10 Rouché, Eugéne (1832–1910).

1 f ′ (z) dz = IndΓ (0), ∫ 2πi γ f (z)

88 | 2 Cauchy’s Theorem and Cauchy’s formula In addition, let g1 , g2 ∈ ℋ(Ω) be holomorphic functions on Ω such that |g1 (z) − g2 (z)| < |g1 (z)|

∀z ∈ γ ∗ .

Then Ng2 = Ng1 . Proof. Let ϕ = f ′ /f . Then ϕ ∈ ℳ(Ω). Now let a be a zero of f of order m(a). Then, by Theorem 2.17, f (z) = (z − a)m(a) h(z), where h is holomorphic in a neighborhood of a and h ≠ 0 there. We obtain ϕ(z) =

m(a)(z − a)m(a)−1 h(z) + (z − a)m(a) h′ (z) m(a) h′ (z) = + , (z − a)m(a) h(z) z − a h(z)

where the second summand is holomorphic in a neighborhood of a. Hence Res(ϕ; a) = m(a). If b is a pole of f of order p(b), we obtain from the Laurent expansion of f around b that f (z) = (z − b)−p(b) k(z), where k is holomorphic in a neighborhood of b and k ≠ 0 there. An analogous computation as above shows that Res(ϕ; b) = −p(b). Let A = {a ∈ Ω1 ∶ f (a) = 0} and B the set of all poles of f in Ω1 . Then, by Theorem 2.69, 1 f ′ (z) dz = ∑ Res(ϕ; a) + ∑ Res(ϕ; b) = ∑ m(a) − ∑ p(b) = Nf − Pf . ∫ 2πi γ f (z) a∈A a∈A b∈B b∈B The chain rule implies that IndΓ (0) =

1 1 1 f ′ (γ(s)) ′ 1 dz f ′ (z) = γ (s) ds = dz = Nf − Pf . ∫ ∫ ∫ 2πi Γ z 2πi 0 f (γ(s)) 2πi γ f (z)

Our assumption |g1 (z) − g2 (z)| < |g1 (z)| ∀z ∈ γ ∗ implies that g2 has no zeros on γ ∗ . Let Γ1 = g1 ∘ γ and let Γ2 = g2 ∘ γ. Then |Γ1 (s) − Γ2 (s)| < |Γ1 (s)|

∀s ∈ [0, 1].

Now we apply Lemma 2.54 and the first part of the proof to obtain Ng1 = IndΓ1 (0) = IndΓ2 (0) = Ng2 .

2.12 The residue theorem

| 89

Example 2.72. Let g(z) = z 4 − 4z + 2. How many zeros does g have in D1 (0)? On |z| = 1 we have |z|4 = 1 < 2 ≤ |−4z + 2|. Set f (z) = −4z + 2, then on |z| = 1 we have |f (z) − g(z)| = |−4z + 2 − z 4 + 4z − 2| = |z 4 | < |−4z + 2| = |f (z)|. f has exactly one zero in D1 (0), namely z0 = 1/2, hence, by Theorem 2.71, we get that g also has exactly one zero in D1 (0). Example 2.73 (Applications of the Residue Theorem). First we compute the real definite integrals ∞

∫ −∞

cos x dx x2 − 2x + 2

and

∞

∫ −∞

sin x dx. x 2 − 2x + 2

For this purpose we consider the line integral ∫

γR

z2

R eix eiz eiz dz = ∫ dx + ∫ 2 dz 2 − 2z + 2 −R x − 2x + 2 ΓR z − 2z + 2

(2.6)

where γR is the path consisting of the line segment on the real axis from −R to R and the semicircle ΓR from R to −R in the upper half-plane, we take R > 3. The function eiz /(z 2 − 2z + 2) has poles of the first order at the points 1 + i and 1 − i. The point 1 + i lies in the interior of γR , and we use Theorem 2.70 (b) to compute the residue of this function at 1 + i: eiz (z − 1 − i) eiz ie−1+i = lim = − . z→1+i z 2 − 2z + 2 z→1+i z − 1 + i 2 lim

By the Residue Theorem (see Theorem 2.69), we have ∫

γR

z2

eiz dz = 2πi(−ie−1+i )/2 = πe−1+i . − 2z + 2

(2.7)

If z = x + iy lies on the semicircle ΓR from R to −R in the upper half-plane, we have y ≥ 0, hence |eiz | = e−y ≤ 1, and we can estimate |

eiz 1 |≤ 2 , z 2 − 2z + 2 R − 2R − 2

for z ∈ ΓR∗ . This implies |∫ ΓR

z2

eiz πR dz| ≤ 2 , − 2z + 2 R − 2R − 2

and we obtain ∫ ΓR

eiz dz → 0 z 2 − 2z + 2

as R → ∞.

90 | 2 Cauchy’s Theorem and Cauchy’s formula By (2.6) and (2.7), we get lim ∫

R

R→∞ −R

x2

π(cos 1 + i sin 1) eix dx = πe−1+i = . − 2x + 2 e

Taking real and imaginary parts on each side and using the dominated convergence theorem, we finally obtain ∞

∫ −∞

x2

cos x π cos 1 dx = − 2x + 2 e

and

∞

∫ −∞

sin x π sin 1 dx = . x2 − 2x + 2 e

In the next example we consider two polynomials P and Q with grad Q ≥ grad P +2. Suppose that Q(x) ≠ 0 ∀x > 0 and that Q(0) = 0 is a simple zero. Let 0 < α < 1 and R = P/Q. Using the Residue Theorem we will compute the real definite integral ∞

∫ xα R(x) dx, 0

where xα = exp(α log x). For this purpose we choose the star-shaped domain Ω = ℂ ⧵ {x ∈ ℝ ∶ x ≥ 0}. By Theorem 2.22, there exists a branch g of the logarithm on Ω (it is not the principal branch) such that for fixed x > 0, lim g(x + iy) = log x,

y→0+

lim g(x − iy) = log x + 2πi.

y→0+

For δ, ϵ > 0 small and ρ > 0 large, choose the following closed path γ in Ω from Figure 2.7.

Figure 2.7

2.12 The residue theorem

| 91

By the Residue Theorem (see Theorem 2.69), we have 1 ∫ eαg(z) R(z) dz = ∑ Res(f ; a), 2πi γ a

(2.8)

where f (z) = eαg(z) R(z) and the sum is taken over all poles of f in the interior of γ. Observe that |eαg(z) | = eαℜg(z) = eα log |z| = |z|α and |R(z)| ≤

M , |z|

M>0

in a neighborhood of zero, since we supposed that Q has a zero of order ≤ 1 at zero, and choose δ > 0 small enough that the last estimate holds on C2∗ . Now we have |∫ f (z) dz| ≤ 2πδ max (|z|α |R(z)|) ≤ 2πδδα M/δ = 2πMδα . ∗ z∈C2

C2

Hence lim |∫ f (z) dz| = 0.

δ→0

(2.9)

C2

On the other hand, |R(z)| ≤

M′ , |z|2

M′ > 0

for |z| large enough, since grad Q ≥ grad P + 2. Hence we get for ρ > 0 large enough that (|z|α |R(z)|) ≤ 2πρα+1 M ′ /ρ2 = 2πM ′ ρα−1 , |∫ f (z) dz| ≤ 2πρ max ∗ z∈C1

C1

and since 0 < α < 1, we obtain lim |∫ f (z) dz| = 0.

ρ→∞

(2.10)

C1

Now fix δ and ρ, and take the limit ϵ → 0 to get ρ

ρ

∫ f (z) dz + ∫ f (z) dz → ∫ eα log x R(x) dx − ∫ eα(log x+2πi) R(x) dx L1

L2

δ

ρ

δ

= (1 − e2πiα ) ∫ eα log x R(x) dx. δ

92 | 2 Cauchy’s Theorem and Cauchy’s formula Hence we obtain from (2.9) and (2.10), using the dominated convergence theorem, ∞

∫ f (z) dz = ∫ + ∫ + ∫ + ∫ → (1 − e2πiα ) ∫ eα log x R(x) dx, γ

C1

L1

L2

C2

0

where we first took the limit as ϵ → 0, then as δ → 0, and finally as ρ → ∞. Taking these limits, the sum on the right-hand side of (2.8) must now we taken over all poles of f in Ω. This implies ∞

∫ xα R(x) dx = 0

2πi ∑ Res(f ; a). 1 − e2πiα a∈Ω

(2.11)

Now let R(x) = 1/(1 + x2 ). By (2.11) we obtain ∞

∫

0

xα 2πi dx = [Res(z α /(1 + z 2 ); i) + Res(z α /(1 + z 2 ); −i)]. 1 + x2 1 − e2πiα

Using Theorem 2.70, we can compute the residues iα 1 zα ] = = ei(α−1)π/2 , z→i (z + i)(z − i) 2i 2 α (−i)α 1 i(α−1)3π/2 z ]= = e Res(z α /(1 + z 2 ); −i) = lim [(z + i) , z→−i (z + i)(z − i) −2i 2 Res(z α /(1 + z 2 ); i) = lim[(z − i)

and get finally ∞

∫

0

2πi 1 i(α−1)π/2 1 π xα dx = [e + ei(α−1)3π/2 ] = . 1 + x2 1 − e2πiα 2 2 cos(απ/2)

Example 2.74. We compute the inverse Fourier transform of x ↦ compute A

lim ∫

A→∞ −A

The function z ↦

sin z z

sin x itx e dx. x

has a removable singularity at 0, therefore ψ(z) =

sin z izt 1 eiz(1+t) − eiz(−1+t) e = z 2i z

is an entire function. Hence, by Corollary 2.6, we obtain A

∫

−A

sin x itx e dx = ∫ ψ(z) dz, x ΓA

where the path ΓA is shown in Figure 2.8. Now set 1 1 eisz ϕA (s) = dz. ∫ π 2πi ΓA z

sin x . x

For t ∈ ℝ we

2.12 The residue theorem

| 93

Figure 2.8

Then A

∫

−A

sin x itx e dx = ϕA (t + 1) − ϕA (t − 1). x

The function z ↦ eisz /z has a pole of the first order at 0 with residue 1. Hence, by the Residue Theorem (see Theorem 2.69), we have 1 eisz dz = 0, ∫ 2πi γ z where the path γ is shown in Figure 2.9.

Figure 2.9

In addition, 1 eisz dz = 1, ∫ 2πi δ z where δ is the path in Figure 2.10. Hence we get 1 eisz 1 1 −π iAeiθ dz = ϕA (s) + ∫ ∫ exp(isAeiθ ) iθ dθ = 0, 2πi γ z π 2πi 0 Ae which implies 1 1 0 ϕA (s) = ∫ exp(isAeiθ ) dθ, π 2π −π

(2.12)

94 | 2 Cauchy’s Theorem and Cauchy’s formula

Figure 2.10

and eisz 1 1 π 1 dz = ϕA (s) + ∫ ∫ exp(isAeiθ ) dθ = 1, 2πi δ z π 2π 0 and finally, 1 1 π ϕA (s) = 1 − ∫ exp(isAeiθ ) dθ. π 2π 0

(2.13)

If s and sin θ have the same signature, | exp(isAeiθ )| = exp(−sA sin θ) → 0, as A → ∞. By the dominated convergence theorem, we obtain from (2.12) and (2.13) {π, lim ϕA (s) = { A→∞ 0, {

s > 0, s < 0.

Again from (2.12) or (2.13) we get ϕA (0) = π/2. Hence A

lim ∫

A→∞ −A

sin x itx e dx = lim [ϕA (t + 1) − ϕA (t − 1)], A→∞ x

where the last expression is the function π, { { { χ(t) = {π/2, { { {0,

−1 < t < 1, t = ±1, |t| > 1.

The Fourier transform of χ is 1 ∞ sin x . ∫ χ(t)e−itx dt = 2π −∞ x

2.13 Exercises | 95

2.13 Exercises 35. Compute the following line integrals: ∫ (z − a)k dz, γ1

where a ∈ ℂ, k ∈ ℤ and γ1 is the unit circle |z| = 1 passed through once in the positive direction. 36. Let γ(t) = eit , 0 ≤ t ≤ 2π. Compute the following line integrals and compare the results with the assertions of Cauchy’s Theorem: ∫ (z)2 dz, γ

∫ z −2 dz, γ

where the annulus A = {z ∶ 21 < |z| < 2} is the corresponding domain. 37. Let γ(t) = reit , r > 0, 0 ≤ t ≤ 2π, and let a, b ∈ ℂ such that |a| < r < |b|. Show that 2πi dz = . (z − a)(z − b) a − b

∫

γ

38. Let γ, a, b be as in Exercise 37, m, n ∈ ℕ. Compute ∫

γ

dz

(z

− a)m (z

− b)n

.

39. Let γ(t) = eit , 0 ≤ t ≤ 2π. Compute ∫

γ

eiz dz, z2

sin z dz. z3

∫

γ

40. Let γ(t) = eit , 0 ≤ t ≤ 2π and n ∈ ℕ. Compute ∫

γ

ez − e−z dz, zn

∫

γ

dz . (z − 1/2)n

41. Let γ(t) = 1 + 21 eit , 0 ≤ t ≤ 2π, n ∈ ℕ0 . Compute Log(z) dz. zn

∫

γ

42. Let γ(t) = −1 + eit , 0 ≤ t ≤ 2π. Compute ∫

γ

dz . (z + 1)(z − 1)3

43. Let γ(t) = 2eit , 0 ≤ t ≤ 2π. Compute ∫

γ

sin z dz. z+i

96 | 2 Cauchy’s Theorem and Cauchy’s formula 44. Let γ(t) = 21 eit , 0 ≤ t ≤ 2π. Compute ∫

γ

e1−z

z 3 (1 − z)

dz.

45. Compute ∫

γ

z2 + 1 dz, z(z 2 + 4)

where γ(t) = reit , 0 ≤ t ≤ 2π, first for 0 < r < 2 and then for 2 < r < ∞. 46. Prove that 2π

∫ cos(cos θ) cosh(sin θ) dθ = 2π. 0

Hint: use the mean value property for the function f (z) = cos z. 47. (a) Let U ⊆ ℂ be an open and L a straight line, suppose that f ∶ U ⟶ ℂ is continuous and holomorphic on U ⧵ L. Show that f is holomorphic on the whole of U (Use Morera’s Theorem!). (b) Let G be a domain in ℂ, which is symmetric with respect to the real axis, i.e. if z ∈ G then z ∈ G. Let f ∶ {z ∈ G ∶ ℑz ≥ 0} ⟶ ℂ be continuous, suppose that f is holomorphic on {z ∈ G ∶ ℑz > 0} and has real values on {z ∈ G ∶ ℑz = 0}. Show that {f (z), for ℑz ≥ 0, f ̃(z) = { f (z), for ℑz < 0 { is a holomorphic function on G (Schwarz’s reflection principle). 48. Examine which functions can be holomorphically extended into the point 0: z cot z,

z , ez − 1

1 z 2 sin . z

49. If f is holomorphic in {z ∶ |z| > R} and f ( z1 ) has an isolated singularity at 0, it is said that f has an isolated singularity at ∞. Determine the type of isolated singularities (possibly also at ∞) of the following functions:

exp(

z ), 1−z

z5 ez 1 − ez , , , (1 − z)2 1 + z2 1 + ez 1 1 1 −1 (ez − 1)−1 exp( ), exp(tan ), sin(cos ) . 1−z z z 1 , z − z3

2.13 Exercises | 97

50. Let a ∈ ℂ, R > 0 and f ∈ ℋ(D′R (a)) and suppose that a is an essential singularity of f . Let g be a non-constant entire function. (i) Show that the closure of g(ℂ) equals to ℂ. (ii) Prove that a is an essential singularity of g ∘ f . 51. Let a ∈ ℂ, R > 0 and f ∈ ℋ(D′R (a)) such that ℜf (z) ≥ 0 for each z ∈ D′R (a). (i) Show that a is not an essential singularity of f . (ii) Prove that f can in fact be extended to a holomorphic function on DR (a). 52. Expand the following functions as power series around z0 : ez ,

z0 = πi;

(z 2

2z + 1 , + 1)(z + 1)2

z0 = 0;

1 , (z − i)3

z0 = −i

and determine the radius of convergence. 53. Do the same as in Exercise 52 for: (cosh z)2 , ∫ [0,z]

2

eζ dζ ,

1 , b ≠ 0, z0 = 0; az + b sin ζ z0 = 0; ∫ dζ , z0 = 0. [0,z] ζ

z0 = 0;

z

54. Let f (z) = z/(e − 1). Expand f as a power series around z0 = 0, and set ak k z . k=0 k! ∞

f (z) = ∑

Determine the radius of convergence and show that 0 = a0 + (

n+1 n+1 )an . )a1 + ⋯ + ( n 1

Use the fact that f (z) + 21 z is an even function in order to show that ak = 0 for k odd and k > 1. 55. Let an , n ∈ ℕ0 be as in Exercise 54. The numbers B2n = (−1)n−1 a2n , n ≥ 1, are called the Bernoulli numbers. Compute B2 , B4 , … , B10 . n 56. Let f (z) = ∑∞ n=0 an z be a function in ℋ(D1 (0)) such that |f (z)|(1 − |z|) ≤ 1,

z ∈ D1 (0).

Prove that for n ∈ ℕ, 1 n |an | ≤ (1 + ) (n + 1) < e(n + 1). n 57. Let f be an entire function such that |f (z)| ≤ A + B|z|k , for z ∈ ℂ, where A, B, k are positive constants. Show that f is a polynomial.

98 | 2 Cauchy’s Theorem and Cauchy’s formula 58. An entire function is called transcendental, if it has an essential singularity at ∞. Let f be a transcendental entire function and let M(r) = max{|f (z)| ∶ |z| = r}. Show that lim

r→∞

log M(r) = ∞. log r

59. Let a ∈ ℂ, R > 0 and f ∈ ℋ(D′R (a)). Suppose that a is a pole of f . Let g be a transcendental entire function. Show that a is an essential singularity of g ∘ f . 60. Let f ∈ ℋ(DR (0)) be non-constant. Show that the function r ↦ M(r) = sup|z|=r |f (z)| is strictly increasing for r ∈ (0, R). 61. Let f be an entire function, α a zero of f and z ∈ ℂ. Show that |f (z)| ≤ 2|z − α| sup{|f (w)| ∶ |z − w| = 1}, for all z ∈ ℂ. 62. Let f be a holomorphic on Dr1 ,r2 (0) = {z ∶ r1 < |z| < r2 }, suppose that f is continuous on Dr1 ,r2 (0), let Mk = sup|z|=rk |f (z)|, k = 1, 2, and M(r) = sup|z|=r |f (z)|. Show that log M(r) ≤

log r − log r1 log r2 − log r log M1 + log M2 . log r2 − log r1 log r2 − log r1

(Hadamard’s Three Circle Theorem). We say that log M(r) is a convex function of log r. Hint: consider the function [f (z)]p z −q , where p, q are integers, and use the maximum principle. 63. Which of the following domains are simply connected: ℂ ⧵ {0};

ℂ ⧵ [0, 1];

ℂ ⧵ {x ∶ x ≤ 0} ?

64. Which of the following domains are simply connected: (1) ℂ ⧵ {(x, y) ∶ x = 0, |y| ≤ 1} ⧵ {(x, y) ∶ x > 0, y = sin x1 }; (2) the complement of an Archimedean spiral around 0, ℂ∗ ⧵ {z ∶ z = et(1+i) , t ∈ ℝ}; 1 1 (3) G = {(x, y) ∶ 0 < x < 1, 0 < y < 1} ⧵ ⋃∞ n=1 {(x, y) ∶ x = n , 0 < y ≤ 2 }; 1 (4) {z ∶ |z| < 1} ⧵ { n ∶ n = 2, 3, …} ⧵ {0}? 65. Find a suitable domain G and a path in G, which is null-homologous but not nullhomotopic in G. 66. Determine the Laurent expansion of the function

f (z) = in

(z 2 − 1) (z + 2)(z + 3)

2.13 Exercises | 99

(1) {z ∶ 2 < |z| < 3}; (2) {z ∶ |z| > 3}. 67. Determine the Laurent expansion of the function 1 , (z − a)(z − b)

f (z) =

0 < |a| < |b|,

in (1) {z ∶ |a| < |z| < |b|}; (2) {z ∶ |z| > b}. 68. Determine the Laurent expansion of the function f (z) = [

1/2 z ] , (z − 1)(z − 2)

ℑf (3/2) > 0

in the annulus {z ∶ 1 < |z| < 2}. 69. Compute the residues of the following functions at the given points: z (2 − 3z)(4z + 3)

sin z 1 − 2 cos z

at

π ; 3

2 3 at , − ; 3 4

cos2 z (2π − z)3

ez−1 ez − 1

at 2π;

at 0;

z tan z

at

eiπz 16 − z 4

π ; 2

at 2;

z+1 (z 2 + 4)2

at 2i.

70. Let G be a domain in ℂ, which is symmetric with respect to the real axis. Let f ∈ ℳ(G) and suppose that f is real-valued on the real axis. Show that Res(f ; z) = Res(f ; z),

z ∈ G.

71. Compare the residue of the function f at a simple pole z = a ≠ 0 with the residue of the function zf (z 2 ) at the point z = a1/2 . 72. Suppose that the function f has an isolated residue at ∞. The residue of f at ∞ is defined by 2πi Res(f ; ∞) = ∫ f (z) dz, γ

where γ is a negatively oriented circle containing all other singularities f , this means that ∞ lies to the left of γ. Prove that if f is holomorphic on ℂ except for isolated singularities, then the sum of all residues of f is zero. 73. Suppose that the function f has an isolated singularity at ∞. Let g(z) = −z −2 f (1/z). Show that Res(f ; ∞) = Res(g; 0). 74. Compute the residues at ∞ of the following functions: f (z) = z n ,

n ∈ ℤ;

g(z) =

z2 + 3 ; 5z 4 − 7z 2 + 6z

h(z) =

2z − 3 . z2

100 | 2 Cauchy’s Theorem and Cauchy’s formula 75. How many zeros does the function f (z) = z 8 − 4z 5 + z 2 − 1 have in D = {z ∶ |z| < 1}? 76. How many zeros does the function g(z) = 2iz 2 + sin z have in the rectangle R = {(x, y) ∶ |x| ≤ π/2, |y| ≤ 1}? 77. Prove the following theorem (Hurwitz’11 Theorem): Let G be a domain in ℂ and (fn )n a sequence of holomorphic functions on G without zeros in G, which converges uniformly on all compact subsets of G to f ∈ ℋ(G). Then f ≡ 0 or f has no zeros in G. Hint: use Rouché’s Theorem. What are the properties of the sequence fn (z) = ez /n with respect to this theorem? 78. Let G be a domain in ℂ and let f ∈ ℋ(G) be the limit of a sequence fn ∈ ℋ(G), uniformly on all compact subsets of G. Show that the zeros of f are limits of sequences of zeros of the functions fn . Find an example of a limit point of zeros of the functions fn at the boundary of G which is not necessarily a zero of f . 79. Let R(x, y) be a rational function of two variables such that R(cos t, sin t) is defined for all t ∈ ℝ. Show that 2π 1 1 1 1 1 ∫ R(cos t, sin t) dt = 2π ∑ Res( R( (ζ + ), (ζ − )); z). ζ 2 ζ 2i ζ 0 |z| 1;

a ∈ ℝ;

2π

∫

0

π/2

dt ; 1 + sin2 t cos2 2t dt, 1 − 2a cos t + a2 ∫

0

−1 < a < 1.

80. Let R(z) be a rational function having no poles on ℝ and suppose that the degree of the denominator is larger than the degree of the numerator. Then +∞

∫ −∞

R(x)eix dx = 2πi ∑ Res(R(ζ )eiζ ; z). ℑz>0

Hint: choose positive r1 , r2 , s so large that all poles of R in the upper half-space lie in the rectangle [r2 , r2 + is, −r1 + is, −r1 , r2 ] and use the Residue Theorem for integration along the boundary of this rectangle. Finally, take the limits as r1 , r2 → ∞. Compute the following definite integrals using the formula from above: ∞

∫

0

cos x dx, a2 + x 2

11 Hurwitz, Adolf (1859–1919).

a > 0;

∞

∫

0

cos ax dx, (x 2 + b2 )2

a, b > 0.

2.13 Exercises | 101

81. Compute ∞

∫ −∞

eitx dx, 1 + x2

t ∈ ℝ.

82. Let α be a complex number, |α| ≠ 1. Compute 2π

∫

0

dθ , 1 − 2α cos θ + α2

by integration of (z − α)−1 (z − 1/α)−1 along the unit circle. 83. Let G be a domain in ℂ and f ∈ ℋ(G), let z1 , z2 , … ∈ G, set ω0 (z) ≡ 1 and k

ωk (z) = ∏(z − zj ). j=1

Let γ be a closed path in G without double points and such that the points z1 , … , zn belong to the interior of γ. Prove that ℒn−1 (z) =

1 f (ζ ) ωn (ζ ) − ωn (z) dζ , ∫ 2πi γ ωn (ζ ) ζ −z

z ∉ γ⋆

is a polynomial of degree n − 1 with the property ℒn−1 (zj ) = f (zj ),

j = 1, … , n.

Use the Residue Theorem to show that n

f (zj ) ωn (z) ′ ω j=1 n (zj ) z − zj

ℒn−1 (z) = ∑

(Lagrange12 interpolation). Put Rn (z) = f (z) − ℒn−1 (z) and show that Rn (z) =

1 f (ζ ) ωn (z) dζ . ∫ 2πi γ ζ − z ωn (ζ )

Finally, prove that n−1

ℒn−1 (z) = ∑ [ j=0

(Newton13 interpolation). 12 Lagrange, Joseph-Louis (1736–1813). 13 Newton, Isaac (1643–1727).

1 f (ζ ) dζ ]ωj (z), ∫ 2πi γ ωj+1 (ζ )

z∈G

102 | 2 Cauchy’s Theorem and Cauchy’s formula

2.14 Notes Further details on the topics of this chapter can be found in [26, 11, 60, 61]. A thorough treatment of the applications of the Residue Theorem is given in [56]. For more details on Lagrange and Newton interpolation methods the reader should consult [53].

3 Analytic continuation We called a function f holomorphic at a point z0 if it is complex differentiable in an open neighborhood of z0 . We can choose an open disc DR (z0 ) for this neighborhood and can ask whether for another point z1 ∈ DR (z0 ) we obtain a disc DR1 (z1 ) where f is also complex differentiable. We will see that the disc DR1 (z1 ) may protrude beyond the original disc DR (z0 ). This process yields an extension of the domain of holomorphy of f . In this chapter we shall describe it in more detail.

3.1 Regular and singular points Definition 3.1. Let f ∈ ℋ(DR (a)), a ∈ ℂ, R > 0. A point z0 ∈ 𝜕DR (a) is called a regular boundary point for f , if there exists r > 0 and g ∈ ℋ(Dr (z0 )) such that f (z) = g(z) ∀z ∈ DR (a) ∩ Dr (z0 ). Points in 𝜕DR (a), which are not regular are called singular boundary points for f . Remark. If z0 is a regular boundary point for f , then {g(z), z ∈ Dr (z0 ), h(z) = { f (z), z ∈ DR (a) { is holomorphic on DR (a) ∪ Dr (z0 ). Example 3.2. Let f (z) = 1/(1 − z). Then f ∈ ℋ(D1 (0)). We consider the boundary point −1 and expand f around the point −1: 1 1 1 1 = = = ∑ 2−n−1 (z + 1)n . 1 − z 2 − (z + 1) 2 1 − (z + 1)/2 n=0 ∞

The radius of convergence of this power series is 2. Hence f ∈ ℋ(D2 (−1)) and the point −1 is regular for f . The point 1 is obviously singular for f . Remark. It is easily seen that the set of regular boundary points for f is an open subset of the boundary which can be empty. The set of singular boundary points for f is closed. Example 3.3. We construct a function f ∈ ℋ(D1 (0)) having no regular boundary points. Let ∞

n

f (z) = ∑ z 2 . n=0

https://doi.org/10.1515/9783110417241-003

104 | 3 Analytic continuation The radius of convergence of this power series is 1, hence f ∈ ℋ(D1 (0)). Now we consider the 2m th roots of unity {exp(2πik/2m ) ∶ k, m ∈ ℕ}. They form a dense set in 𝕋 = 𝜕D1 (0) since for arbitrary α ∈ [0, 1] and given ϵ > 0 there exist k, m ∈ ℕ such that |α − k/2m | < ϵ. In addition, we have for 0 < r < 1, m

m−1

m

n

f (re2πik/2 ) = ∑ r 2 e(2πik/2

)2n

n=0

n

+ ∑ r2 , n≥m

hence m

lim f (re2πik/2 ) r→1

m

is unbounded and e2πik/2 is a singular point for f . Since the set of all 2m th roots of unity is dense in 𝕋 and the set of all singular points is closed, all points of 𝕋 must be singular for f . n Theorem 3.4. Let f (z) = ∑∞ n=0 an z be a power series with radius of convergence 1. Then f has at least one singular point on 𝕋.

Proof. Suppose that all points in 𝕋 are regular for f . Since 𝕋 is compact, there exist finitely many disks D1 , … , DN with centers on 𝕋 such that (1) 𝕋 ⊆ ⋃Nk=1 Dk , (2) ∃ gk ∈ ℋ(Dk ) with gk = f on D1 (0) ∩ Dk , k = 1, … , N. If Dk ∩ Dj ≠ ∅, set Vk,j = Dk ∩ Dj ∩ D1 (0). Then Vk,j ≠ ∅, since the line connecting the centers of Dj and Dk passes through Dk ∩ Dj and belongs to D1 (0). On Vk,j we have gk = f = gj . Now define N

Ω = D1 (0) ∪ ⋃ Dk k=1

and {f (z), h(z) = { g (z), { k

z ∈ D1 (0),

z ∈ Dk .

Then h is holomorphic on Ω and h = f on D1 (0). Ω is open and contains D1 (0). Hence there exists ϵ > 0 such that D1+ϵ ⊆ Ω. By Theorem 2.13, the radius of convergence of f must be at least 1 + ϵ, and we arrive at a contradiction. n Corollary 3.5. Let f (z) = ∑∞ n=0 an (z − z0 ) be a power series with radius of convergence R > 0. Let ζ ∈ 𝜕DR (z0 ) and let z1 be a point on the line connecting z0 and ζ , suppose that z1 ≠ z0 , z1 ≠ ζ . Let Δ = R − |z1 − z0 |.

3.2 Analytic continuation along a curve

| 105

(a) If Δ = [lim sup | n→∞

1/n −1

f (n) (z1 ) | n!

] ,

then ζ is a singular point for f . (b) If 1/n −1

f (n) (z1 ) Δ < [lim sup | | n! n→∞

] ,

then ζ is a regular point for f . Proof. To show (a), let ρ = [lim sup | n→∞

1/n −1

f (n) (z1 ) | n!

]

be the radius of convergence of f , if f is expanded around z1 . By Theorem 3.4, there exists at least one singular point for f on 𝜕Dρ (z1 ). If Δ = ρ, this point must be ζ . (b) follows immediately.

3.2 Analytic continuation along a curve Example 3.6. Let f1 be the principle branch of the logarithm on G1 , f1 (z) = log |z| + i arg ϕ,

z = |z|eiϕ , −π < ϕ < π.

On G2 there exists a branch f2 of the logarithm with f1 = f2 on G1 ∩ G2 – take, for instance, f2 (z) = log |z| + i arg ϕ,

z = |z|eiϕ , 0 < ϕ < 2π.

And there exists a branch f3 of the logarithm with f1 = f3 on G1 ∩ G3 , choose f3 = f1 , see Figure 3.1. But on G3 ∩ G2 we have f2 = f3 + 2πi. This means that analytic continuation along a full circle around 0 can lead to a different result. A similar behavior shows up for z 1/k = exp(1/k log z). Definition 3.7. Let γ ∶ [0, 1] ⟶ ℂ be a curve connecting z0 and z1 , i.e. γ(0) = z0 and γ(1) = z1 , let ϵ, η > 0 and f ∈ ℋ(Dϵ (z0 )) as well as g ∈ ℋ(Dη (z1 )).

106 | 3 Analytic continuation

Figure 3.1

We say that g arises from f by analytic continuation along γ, if (1) there exists a subdivision 0 = t0 < t1 < ⋯ < tn = 1 and there exist open neighborhoods Uν of γ([tν−1 , tν ]) and functions fν ∈ ℋ(Uν ), ν = 1, … , n such that fν = fν+1 on the connected component of γ(tν ) in Uν ∩ Uν+1 for ν = 1, … , n − 1; (2) f = f1 in a neighborhood of z0 ; (3) fn = g in a neighborhood of z1 . In this case we say that f has an analytic continuation along γ. Remark. (a) The function f (z) = 1/(1 − z) has no analytic continuation along a curve which passes through the point 1. (b) If f has an analytic continuation along γ, the result g is uniquely determined by f and γ, it is independent of the subdivision and of the functions fν and the neighborhoods Uν . This follows from the identity principle of holomorphic functions, Theorem 2.18. (c) But the result g depends in general on the route of γ and not only on the endpoint of γ, see Example 3.6. (d) In stead of the neighborhoods Uν one can also choose discs Kν such that the center of Kν+1 lies in Kν , for ν = 1, … , n − 1. The functions fν can be taken as power series around γ(tν ). Example 3.8. (a) We are looking for a branch of the logarithm which has the value log 2 + (2k + 1)πi at the point z1 = −2, where k ∈ ℕ. For this purpose we start with the principle branch of the logarithm in a suitable neighborhood of z0 = 1. We choose a path starting at z0 which circles around the origin k times in the positive direction and ends at the point z1 = −2 coming from

3.3 The Monodromy Theorem | 107

the upper half-plane (compare with Example 3.6). In this way we get the desired branch of the logarithm. (b) Let G ⊆ ℂ be a domain, f ∈ ℋ(G). If V ⊆ G is a convex subset, then f has a primitive F ∈ ℋ(V) on V , i.e. F ′ = f on V . Let γ be an arbitrary curve in G from z0 to z1 . We choose a suitable subdivision of the parameter interval and discs Dν as associated neighborhoods. Since all Dν are convex, there exist primitive functions Fν of f on Dν (see Corollary 2.6). By the identity principle (see Theorem 2.18), we have Fν = Fν+1 on Dν ∩ Dν+1 for ν = 1, … , n − 1. Hence, local primitive functions have an analytic continuation along each curve in G.

3.3 The Monodromy Theorem As already mentioned before, different paths between two fixed points may give different results for the analytic continuation along them. We will now consider sufficient conditions for equality of the results. For this purpose we need the concept of homotopy. Definition 3.9. Let U ⊆ ℂ be an open set and let γ0 , γ1 denote two curves in U from z0 ∈ U to z1 ∈ U. γ0 is homotopic to γ1 in U, if there exists a continuous function H ∶ [0, 1] × [0, 1] ⟶ U with the following properties: (1) H(t, 0) = γ0 (t), H(t, 1) = γ1 (t) ∀t ∈ [0, 1]; (2) H(0, s) = z0 , H(1, s) = z1 ∀s ∈ [0, 1]. H is called a homotopy from γ0 to γ1 . For fixed s ∈ [0, 1], we define γs (t) = H(t, s), t ∈ [0, 1]. Then γs is also a curve in U from z0 to z1 . If γ0 is homotopic to γ1 in U, we also write γ0 ∼U γ1 . Remark. ∼U is an equivalence relation: reflexivity is clear; if H0 is a homotopy from γ0 to γ1 , then H(t, s) = H0 (t, 1 − s) is a homotopy from γ1 to γ0 ; hence ∼U is symmetric. If γ0 ∼U γ1 and γ1 ∼U γ2 with corresponding homotopies H0 and H1 , then {H (t, 2s), H(t, s) = { 0 H (t, 2s − 1), { 1

0 ≤ s ≤ 1/2, 1/2 ≤ s ≤ 1

defines a homotopy from γ0 to γ2 , and we have H(t, 0) = H0 (t, 0) = γ0 (t) and H(t, 1) = H1 (t, 1) = γ2 (t), in addition H is continuous and its image belongs to U. Hence we have γ0 ∼U γ2 , and ∼U is transitive.

108 | 3 Analytic continuation

Figure 3.2

We denote the equivalence class by [γ]U = {δ curve in U ∶ δ ∼U γ}. These classes are invariant under parameter transformations. Indeed, if γ is a curve in U and ϕ ∶ [0, 1] ⟶ [0, 1] is bijective, continuous and monotone increasing, we get for δ = γ ∘ ϕ that δ ∈ [γ]U . By H(t, s) = γ((1 − s)t + sϕ(t)) we have a homotopy from γ to δ since H(t, 0) = γ(t) and H(t, 1) = γ(ϕ(t)) = δ(t), as well as H(0, s) = γ(0) and H(1, s) = γ(1) ∀s ∈ [0, 1]. Theorem 3.10 (Monodromy Theorem). Let U ⊆ ℂ, and let H ∶ [0, 1] × [0, 1] ⟶ U be a homotopy of curves in U with common starting point z0 and common endpoint z1 . Let f ∈ ℋ(Dϵ (z0 )), ϵ > 0, and suppose that f has an analytic continuation along all the curves γs = H(⋅, s), s ∈ [0, 1]. Then the analytic continuation along any curve γs leads to the same function g (the analytic continuations gs along γs all coincide in a neighborhood of z1 ). Example 3.11. Let U = ℂ. Consider the curves γ0 and γ1 from z0 to z1 in Figure 3.2. Each homotopy from γ0 to γ1 passes through the origin. The logarithm has no analytic continuation to a neighborhood of the origin. The assumptions of the monodromy theorem are not satisfied for any branch of the logarithm. Analytic continuation along γ0 and along γ1 yields a different result for each branch of the logarithm (see Example 3.6). Proof. (a) Suppose that the continuation of f along γs0 yields gs0 . First we show the following: if s is sufficiently close to s0 , the continuation along γs also yields gs0 . For the continuation along γs0 we choose the subdivision points tν , corresponding neighborhoods Uν and holomorphic functions fν ∈ ℋ(Uν ), ν = 1, … , n. Let 𝒰 = ⋃nν=1 Uν . Then 𝒰 is open. Since H is continuous, H −1 (𝒰) is open in [0, 1] × [0, 1]. 𝒰 contains γs∗0 . Hence [0, 1] × {s0 } ⊂ H −1 (𝒰). Since H −1 (𝒰) is open, there exists

3.3 The Monodromy Theorem | 109

an open neighborhood W of s0 in [0, 1] such that [0, 1] × W ⊂ H −1 (𝒰). Let Vν be the connected component of γs0 (tν ) in Uν ∩ Uν+1 . Taking a possibly smaller W we can accomplish that γs (tν ) ∈ Vν ∀s ∈ W , ν = 1, … , n − 1. Hence all tν , Uν and fν yield the same result gs0 taking analytic continuation of f along γs , for each s ∈ W . (b) Let g be the result of the analytic continuation of f along γ0 , s = 0. Let J = {s ∈ [0, 1] ∶ continuation of f along γσ yields g for 0 ≤ σ ≤ s}. By (a), the set J is open [0, 1] and 0 ∈ J. Suppose that J ≠ [0, 1]. Then sup J = s0 ∉ J, otherwise we would get a contradiction to the openness of J. Apply part (a) for this s0 . Now we get a contradiction to the assertion that s0 is the supremum of J. Hence we obtain J = [0, 1]. Theorem 3.12. Let G be a simply connected domain in ℂ. Let z0 ∈ G, f ∈ ℋ(Dϵ (z0 )). Suppose that f has an analytic continuation along each curve in G with starting point z0 . Then there exists F ∈ ℋ(G) with f = F on Dϵ (z0 ). Proof. In a simply connected domain, each two curves γ0 and γ1 with the same start and end points are homotopic (see Definition 2.52). γ = γ0 − γ1 is a closed curve, which is homotopic to a constant curve. Let H be the corresponding homotopy. Using H one can easily construct a homotopy H0 from γ0 to γ1 (see Exercises). Now let z ∈ G and let γz be a curve in G from z0 to z. Let gz denote the result of the analytic continuation of f along γz . The function gz is holomorphic in a neighborhood U(z) ⊂ G of z. By Theorem 3.10, gz is independent of γz . Let F(w) = gz (w) for w ∈ U(z). In order to show that F ∈ ℋ(G), it suffices to show the following assertion: if z1 ≠ z2 and U(z1 ) ∩ U(z2 ) ≠ ∅, then gz1 = gz2 on U(z1 ) ∩ U(z2 ). For this purpose let z∗ ∈ U(z1 ) ∩ U(z2 ). Then we obtain gz∗ by analytic continuation along γz1 + [z1 , z∗ ], and also by analytic continuation of gz1 along the straight line [z1 , z∗ ] which lies in U(z1 ). By the identity principle, we obtain that gz1 = gz∗ on U(z1 ) ∩ U(z∗ ). But we also get that gz∗ = gz2 on U(z∗ ) ∩ U(z2 ). Hence gz1 = gz2 on U(z1 ) ∩ U(z2 ). Remark. By the last theorem one can construct a global holomorphic function starting with a locally defined holomorphic function (a germ of a holomorphic function) and using analytic continuation. If G is a simply connected domain which does not contain the origin, one can extend the logarithm to a globally defined holomorphic function F ∈ ℋ(G). If G is an annulus around the origin, there exists no global holomorphic function as an extension of the logarithm.

110 | 3 Analytic continuation

3.4 Exercises 84. Let ∞

f (z) = ∑ ak z k k=0

be a power series with radius of convergence 1, and where all coefficients ak are non-negative real numbers. Prove that z = 1 is a singular point for f . Hint: use Corollary 3.5. 85. Let f be the restriction of the principle branch of the function √z to the disc D1 (1). Determine the analytic continuations of f along the curves γ(t) = exp(2πit) and σ(t) = exp(4πit) for 0 ≤ t ≤ 1. 86. Let G be a simply connected domain in ℂ. Show that each two curves with common start and end points are homotopic in G. 87. Let Ω be a domain in ℂ and choose a fixed w ∉ Ω. Let D be a disc in Ω. Show that there exists a function f ∈ ℋ(D) such that exp[f (z)] = z − w and f ′ (z) = (z − w)−1 in D. Prove that f ′ has an analytic continuation along each curve in Ω, which starts at the center of D, and show that it is necessary to assume simple connectivity in Theorem 3.12.

3.5 Notes The fact that analytic continuation along a closed path can lead to a different function than the one started with, leads to the concept of a Riemann surface, see [15]. The more algebraic aspects of the topic are treated in [46].

4 Construction and approximation of holomorphic functions In this chapter we discuss the question in which domains G ⊆ ℂ holomorphic functions can be approximated by polynomials or by functions which are holomorphic in a larger domain than G (Runge’s Theorem, Section 4.4). We use the Hahn–Banach Theorem (Section 4.3) to prove a preferably general assertion which will be helpful later on. In addition, we construct holomorphic functions with a prescribed zero set and given multiplicities (Weierstraß’ Factorization Theorem, Section 4.6) and we prove the existence of a meromorphic function with prescribed principle parts (Mittag-Leffler’s Theorem, Section 4.5). For all of these problems the solution of the inhomogeneous Cauchy–Riemann equation plays an important role (Section 4.2). First 𝒞∞ -functions with the desired properties are constructed, which are finally corrected by suitable solutions of the inhomogeneous Cauchy–Riemann equation to become holomorphic functions. This method has its origin in the theory of holomorphic functions of several variables, which will be explained in more detail in the second part of this book. This chapter ends with the Riemann Mapping Theorem showing the existence of biholomorphic mappings between simply connected domains G ≠ ℂ and the unit disc D1 (0) (Section 4.9) and with a characterization of simply connected domains in terms of properties of holomorphic function on these domains (Section 4.10).

4.1 A partition of unity We start with an important method in real analysis, which enables us to glue together locally defined functions with special properties. Definition 4.1. Let Ω ⊆ ℝn be open and let 𝒰 = {Ui ∶ i ∈ I} be an open cover of Ω, i.e. Ω ⊆ ⋃i∈I Ui , where the sets Ui are open. A 𝒞∞ -partition of unity with respect to the cover 𝒰 is a family {Φi ∶ i ∈ I} of 𝒞∞ -functions Φi ∶ Ω ⟶ ℝ with the following properties: (1) Φi ≥ 0 on Ω and the support of Φi supp(Φi ) = {x ∈ ℝn ∶ Φi (x) ≠ 0} ⊆ Ui

∀i ∈ I;

(2) {supp(Φi ) ∶ i ∈ I} is locally finite, i.e. the set {i ∈ I ∶ supp(Φi ) ∩ K ≠ ∅} is finite for each compact subset K ⊂ Ω. (3) ∑i∈I Φi ≡ 1 on Ω (by (2) this sum is finite ∀x ∈ Ω). Lemma 4.2. Let Ω ⊆ ℝn be open and let K ⊂ Ω be a compact subset. Then there exists a function Φ ∈ 𝒞∞ 0 (Ω) (Φ has compact support) with Φ(x) > 0 ∀x ∈ K. https://doi.org/10.1515/9783110417241-004

112 | 4 Construction and approximation of holomorphic functions Proof. Let 1 {exp(− 1−t ), ψ(t) = { 0, {

t < 1, t ≥ 1.

Then ψ ∈ 𝒞∞ (ℝ). Now we set χ(x1 , … , xn ) = ψ((x12 + ⋯ + xn2 )/2). n n 2 2 Then χ ∈ 𝒞∞ 0 (ℝ ) with supp χ ⊆ {x ∈ ℝ ∶ x1 + ⋯ + xn ≤ 2}, and χ(0) > 0. c Choose δ > 0 such that δ < dist(K, Ω ). Let a ∈ K be an arbitrary point and define ). Then we have Φa ∈ 𝒞∞ Φa (x) = χ( x−a 0 and Φa (a) = χ(0) > 0 as well as supp Φa ⊆ Ω. δ We define Va = {x ∈ Ω ∶ Φa (x) > 0}. The sets Va are open and, since Φa (a) > 0 ∀a ∈ K, we get

K ⊂ ⋃ Va . a∈K

Since K is compact, there exist finitely many points a1 , … , ap ∈ K such that p

K ⊂ ⋃ Vaj j=1

and we set Φ = Φa1 + ⋯ + Φap , then we finally obtain Φ(x) > 0 ∀x ∈ K. Lemma 4.3. Let Ω ⊆ ℝn be open and let 𝒰 = {Ui ∶ i ∈ I} be an open cover of Ω. Then there exists a 𝒞∞ -partition of unity with respect to 𝒰. Proof. Ω ⊆ ℝn is paracompact (see [19]), i.e. 𝒰 has a locally finite refinement, this is a locally finite open cover {Vj ∶ j ∈ J} of Ω with the property that ∀j ∈ J ∃i ∈ I with Vj ⊆ Ui . Hence there exists a mapping τ ∶ J ⟶ I such that Vj ⊆ Uτ(j) . In addition, there exist compact subsets Kj ⊂ Vj such that ⋃j∈J Kj = Ω (see [19]). By Lemma 4.2, there exist ψj ∈ 𝒞∞ 0 (Vj ) such that ψj (x) > 0 ∀x ∈ Kj . Now we set ψ = ∑ ψj . j∈J

Since the cover {Vj ∶ j ∈ J} is locally finite, the sum from above is always a finite sum and ψ ∈ 𝒞∞ . We also have ψ > 0 on Ω, since ψj > 0 on Kj and ⋃j∈J Kj = Ω. Let χj = ψj /ψ. The family {χj ∶ j ∈ J} is a 𝒞∞ -partition of unity with respect to {Vj ∶ j ∈ J} because we have ∑ χj = ∑ j∈J

j∈J

ψj ψ

=

1 ∑ ψ = 1. ψ j∈J j

Set Ji = τ−1 ({i}) for a i ∈ I and define Φi = ∑j∈J χj . i Then {Φi ∶ i ∈ I} is the desired 𝒞∞ -partition of unity with respect to the original cover 𝒰 = {Ui ∶ i ∈ I}.

4.2 The inhomogeneous Cauchy–Riemann differential equations | 113

Theorem 4.4. Let Ω ⊆ ℝn be an open set, let U be an open subset of Ω and X a closed subset of U, i.e. X ⊂ U ⊂ Ω. Then there exists Φ ∈ 𝒞∞ (Ω) with 0 ≤ Φ ≤ 1 such that Φ|X = 1 and Φ|Ω⧵U = 0. Proof. Let V = Ω ⧵ X. Then V is open and {U, V} is an open cover of Ω. By Lemma 4.3, there exists a corresponding 𝒞∞ -partition of unity, denoted by {ΦU , ΦV }. Then we have ΦU + ΦV = 1 on Ω and supp ΦU ⊆ U, supp ΦV ⊆ V . Now we set Φ = ΦU . Then Φ = 0 outside of U, and since ΦV = 0 outside of V , we must have ΦU = 1 on X. Theorem 4.5. Let Ω ⊆ ℝn be open and let X1 and X2 be two disjoint closed subsets of Ω. Let Φ1 , Φ2 ∈ 𝒞∞ (Ω). Then there exists Φ ∈ 𝒞∞ (Ω) such that Φ|X1 = Φ1 |X1

and Φ|X2 = Φ2 |X2 .

Proof. By Theorem 4.4, there exists α ∈ 𝒞∞ (Ω) with 0 ≤ α ≤ 1 such that α|X1 = 1 and α|X2 = 0. Now set Φ = αΦ1 + (1 − α)Φ2 . Then Φ has the desired properties.

4.2 The inhomogeneous Cauchy–Riemann differential equations We apply methods from complex analysis to solve the inhomogeneous Cauchy– Riemann differential equation 𝜕u = f. 𝜕z In the second part of this book we will reconsider this problem using functional analysis tools such as unbounded operators on Hilbert spaces. Theorem 4.6. Let G be a bounded domain in ℂ with a piecewise smooth boundary and let f be a function in a neighborhood U of G, which has continuous first real derivatives. Then u(z) =

f (ζ ) 1 dζ ∧ dζ , ∫ 2πi G ζ − z

z∈G

is a continuously differentiable function on G such that 𝜕u = f, 𝜕z on G. The function u is a solution of the inhomogeneous Cauchy–Riemann differential equation with right-hand side f .

114 | 4 Construction and approximation of holomorphic functions Remark. For the proof we will use the inhomogeneous Cauchy formula 2.47. Suppose = f , then we get from 2.47 that u is continuously differentiable on G and a solution of 𝜕u 𝜕z u(z) =

1 u(ζ ) 1 f (ζ ) dζ + dζ ∧ dζ , ∫ ∫ 2πi 𝜕G ζ − z 2πi G ζ − z

where the boundary of G has positive orientation. The first term is a holomorphic function at z ∈ G. Differentiation with respect to z gives 1 𝜕 u(ζ ) 1 𝜕 f (ζ ) 𝜕u = ( ) dζ + [∫ dζ ∧ dζ ] ∫ 𝜕z 2πi 𝜕G 𝜕z ζ − z 2πi 𝜕z G ζ − z 1 𝜕 f (ζ ) = [∫ dζ ∧ dζ ]. 2πi 𝜕z G ζ − z By this computation we get a hint for the solution, namely u(z) =

1 f (ζ ) dζ ∧ dζ . ∫ 2πi G ζ − z

Proof. First we show that u is continuously differentiable on G. Let z0 ∈ G be an arbitrary point and let r > 0 be such that D2r (z0 ) ⊂⊂ G. By Theorem 4.4, there exists ϕ ∈ 𝒞∞ (ℂ) with ϕ(z) = 1 for z ∈ Dr (z0 ) and ϕ(z) = 0 for z ∈ ℂ ⧵ D2r (z0 ). Now we separate u into two summands u(z) =

1 ϕ(ζ )f (ζ ) 1 (1 − ϕ(ζ ))f (ζ ) dζ ∧ dζ + dζ ∧ dζ = u1 (z) + u2 (z). ∫ ∫ 2πi G ζ − z 2πi G ζ −z

Let z ∈ Dr/2 (z0 ). Since 1 − ϕ(ζ ) = 0 for ζ ∈ Dr (z0 ), we have u2 (z) =

1 (1 − ϕ(ζ ))f (ζ ) dζ ∧ dζ , ∫ 2πi G⧵Dr (z0 ) ζ −z

the term in the integral is continuously differentiable on G ⧵ Dr (z0 ) as a function of the variable z. Hence we can interchange integration and differentiation with respect to z and obtain 𝜕u2 1 𝜕 (1 − ϕ(ζ ))f (ζ ) = ( ) dζ ∧ dζ = 0, ∫ 𝜕z 2πi G⧵Dr (z0 ) 𝜕z ζ −z therefore u2 is even holomorphic on Dr/2 (z0 ). The term u1 is handled in a different way: we have ϕf ≡ 0 on G ⧵ D2r (z0 ), if we extend f on the whole of ℂ in an arbitrary way, the last assertion remains true and we can write u1 as u1 (z) =

1 ϕ(ζ )f (ζ ) dζ ∧ dζ . ∫ 2πi ℂ ζ − z

4.2 The inhomogeneous Cauchy–Riemann differential equations | 115

We substitute ζ = z + w and obtain u1 (z) =

ϕ(z + w)f (z + w) 1 dw ∧ dw. ∫ 2πi ℂ w

Now the integrand is a continuously differentiable function of the variables z and z; all derivatives with respect to z and z are uniformly bounded by integrable functions of the variable w (see Theorem 2.43 and the proof of Theorem 2.47). Hence one can interchange integration and differentiation with respect to z and z. Therefore u1 is continuously differentiable, as well as u = u1 + u2 . It remains to show that u is a solution of the inhomogeneous Cauchy–Riemann differential equation. Let z ∈ Dr/2 (z0 ). By the first part of the proof we have 𝜕u 1 𝜕 ϕ(ζ )f (ζ ) 𝜕u (z) = 1 (z) = [∫ dζ ∧ dζ ] 𝜕z 𝜕z 2πi 𝜕z G ζ − z ϕ(z + w)f (z + w) 1 𝜕 [∫ dw ∧ dw] = 2πi 𝜕z ℂ w 1 𝜕 ϕ(z + w)f (z + w) = [ ] dw ∧ dw. ∫ 2πi ℂ 𝜕z w For a differentiable function h, we can write 𝜕h 𝜕h (z + w) = (ζ ), 𝜕z 𝜕ζ where ζ = z + w. Now set z = x + iy and ζ = ξ + iη, then 𝜕h h(z + w + t) − h(z + w) (z + w) = lim t∈ℝ,t→0 𝜕x t and so 𝜕h h(ζ + t) − h(ζ ) (ζ ) = lim ; t∈ℝ,t→0 𝜕ξ t analogous expressions are obtained for the derivative with respect to η. Now we can continue the computation of 𝜕u (z) and get 𝜕z 𝜕u 1 𝜕 ϕ(z + w)f (z + w) (z) = [ ] dw ∧ dw ∫ 𝜕z 2πi ℂ 𝜕z w =

1 𝜕/𝜕ζ [ϕ(ζ )f (ζ )] dζ ∧ dζ . ∫ 2πi ℂ ζ −z

(4.1)

We use Theorem 2.47 for ϕf and for a disc D with D2r (z0 ) ⊂⊂ D. Since ϕf = 0 on 𝜕D, we obtain for z ∈ Dr/2 (z0 ) f (z) = ϕ(z)f (z) = =

1 ϕ(ζ )f (ζ ) 1 𝜕/𝜕ζ [ϕ(ζ )f (ζ )] dζ + dζ ∧ dζ ∫ ∫ 2πi 𝜕D ζ − z 2πi D ζ −z

1 𝜕/𝜕ζ [ϕ(ζ )f (ζ )] 𝜕u dζ ∧ dζ = (z) ∫ 2πi ℂ ζ −z 𝜕z

using (4.1).

116 | 4 Construction and approximation of holomorphic functions Corollary 4.7. (a) If f ∈ 𝒞∞ 0 (ℂ) and u(z) =

1 f (ζ ) dζ ∧ dζ , ∫ 2πi ℂ ζ − z

then u ∈ 𝒞∞ (ℂ) and 𝜕u = f on ℂ. 𝜕z ∞ (b) If f ∈ 𝒞0 (ℂ), then f (z) =

1 𝜕f /𝜕ζ (ζ ) dζ ∧ dζ on ℂ. ∫ 2πi ℂ ζ − z

Both assertions follow from the proof of Theorem 4.6. Theorem 4.8. Let G ⊆ ℂ be open and f ∈ ℋ(G), let Kbe a compact subset of G. Let α ∈ 𝒞∞ 0 (G) be such that α = 1 on K (see Theorem 4.4). Then f (z) =

1 𝜕α f (ζ ) dζ ∧ dζ ∫ 2πi G 𝜕ζ ζ − z

∀z ∈ K.

Proof. Let {f (z)α(z), z ∈ G, Φ(z) = { 0, z ∉ G. { Then Φ ∈ 𝒞∞ 0 (ℂ). By Lemma 2.59, there exists an open subset G1 ⊂ G such that supp(α) ⊂ G1 and 𝜕G1 is piecewise smooth. We apply Theorem 2.47 and get Φ(z) =

1 Φ(ζ ) 1 𝜕Φ/𝜕ζ dζ + dζ ∧ dζ . ∫ ∫ 2πi 𝜕G1 ζ − z 2πi G1 ζ − z

Now 𝜕Φ 𝜕f 𝜕α 𝜕α = α + f = f, 𝜕ζ 𝜕ζ 𝜕ζ 𝜕ζ since f is holomorphic. Hence we get for z ∈ K, 1 Φ(ζ ) 1 𝜕α f (ζ ) dζ + dζ ∧ dζ ∫ ∫ 2πi 𝜕G1 ζ − z 2πi G1 𝜕ζ ζ − z 1 𝜕α f (ζ ) = dζ ∧ dζ , ∫ 2πi G 𝜕ζ ζ − z

f (z) = Φ(z) =

where the first summand vanishes since 𝜕G1 ⊂ (supp(α))c .

4.3 The Hahn–Banach Theorem

| 117

4.3 The Hahn–Banach Theorem For the prove of our general approximation result we will need functional analytic methods, in particular, we will use a characterization of dense subspaces of a normed space in terms of properties of continuous linear functionals – an important version of the Hahn–Banach Theorem. We consider a vector space X over ℝ or over ℂ, together with a mapping ‖ ⋅ ‖ ∶ X ⟶ ℝ + with the properties ‖cx‖ = |c|‖x‖ for c ∈ ℝ (or ℂ) and ‖x + y‖ ≤ ‖x‖ + ‖y‖ for x, y ∈ X. We call ‖ ⋅ ‖ a seminorm. If, in addition, ‖x‖ = 0 implies x = 0, the mapping ‖ ⋅ ‖ is a norm on X. Theorem 4.9 (Hahn1 –Banach2 Theorem). Let X be a vector space over ℝ with a seminorm ‖ ⋅ ‖. Let Y be a subspace of X. Suppose that l ∶ Y ⟶ ℝ is a linear functional on Y such that l(x) ≤ C‖x‖

∀x ∈ Y,

where C > 0 is a constant. Then there exists a linear extension L of l on the whole of X, i.e. L|Y = l, such that L(x) ≤ C‖x‖

∀x ∈ X.

Proof. Let x0 ∉ Y and Z = ⟨Y, x0 ⟩ be the linear span of Y and x0 . If x ∈ Z, then x = y +cx0 , y ∈ Y , c ∈ ℝ, and the representation of x is uniquely determined. We set ϕ(x) = l(y) + ca0 for a0 ∈ ℝ. Then ϕ is an extension of l on Z. Now we show that a0 can be chosen such that ϕ(x) ≤ C‖x‖ ∀x ∈ Z, i.e. ca0 ≤ C‖y + cx0 ‖ − l(y).

(4.2)

For c = 0 we always have (4.2). If c > 0, then (4.2) is equivalent to y y a0 ≤ C‖ + x0 ‖ − l( ); c c if c < 0, then (4.2) is equivalent to y y a0 ≥ −C‖− − x0 ‖ − l( ). c c Therefore we have to show that there exists a0 such that − C‖−y − x0 ‖ − l(y) ≤ a0 ≤ C‖y + x0 ‖ − l(y) 1 Hahn, Hans (1879–1934). 2 Banach, Stefan (1892–1945).

∀y ∈ Y.

(4.3)

118 | 4 Construction and approximation of holomorphic functions For this purpose take y1 , y2 ∈ Y . Then l(y2 ) − l(y1 ) = l(y2 − y1 ) ≤ C‖(y2 + x0 ) + (−y1 − x0 )‖ ≤ C(‖y2 + x0 ‖ + ‖−y1 − x0 ‖), which implies −C‖−y1 − x0 ‖ − l(y1 ) ≤ C‖y2 + x0 ‖ − l(y2 )

∀y1 , y2 ∈ Y,

hence we can find a0 ∈ ℝ with (4.3). Therefore we have found a linear extension ϕ of l to Z such that ϕ(x) ≤ C‖x‖

∀x ∈ Z.

Now let ℰ be the set of all linear extensions λ of l, which are dominated by C‖ ⋅ ‖ on its domain of definition dom(λ). We define a partial order on ℰ: for λ1 , λ2 ∈ ℰ let λ1 ≤ λ2

⟺

λ2 is an extension of λ1 , dom(λ1 ) ⊆ dom(λ2 ).

Then we have transitivity: if λ1 ≤ λ2 and λ2 ≤ λ3 , then λ1 ≤ λ3 ; and we have reflexivity: for each λ ∈ ℰ we have λ ≤ λ. If λ1 ≤ λ2 and λ2 ≤ λ1 , then λ1 = λ2 . Now let ℱ be a totally ordered subset of ℰ, i.e. for each pair λ1 , λ2 ∈ ℱ we have λ1 ≤ λ2 or λ2 ≤ λ1 . We define a linear functional U ∶ ⋃ dom(λ) ⟶ ℝ λ∈ℱ

by U(x) = λ(x), if x ∈ dom(λ). U is well-defined since ℱ is totally ordered. And U is also en extension of l with U(x) ≤ C‖x‖

∀x ∈ dom(U).

Hence U ∈ ℰ and U is an upper bound of ℱ. Zorn’s3 Lemma. Let ℰ be a non-empty partially ordered set, in which each totally ordered subset has an upper bound. Then ℰ contains at least one maximal element. We indicate that Zorn’s Lemma is equivalent to the Axiom of Choice (see [63]). The assumptions of Zorn’s Lemma are satisfied in our case. Hence there exists a maximal element L in ℰ. If dom(L) = M ≠ X, then there exists x0 ∈ X ⧵M and an extension of L to ⟨M, x0 ⟩ with the required properties and we arrive at a contradiction to the maximality of L. 3 Zorn, Max August (1906–1993).

4.3 The Hahn–Banach Theorem

| 119

Theorem 4.10. Let X be a vector space over ℂ with a seminorm ‖⋅‖. Let Y be a subspace of X. Let l ∶ Y ⟶ ℂ be a linear functional on Y with |l(x)| ≤ C‖x‖

∀x ∈ Y,

where C > 0 is a constant. Then there exists a linear extension L of l to the whole X, i.e. L|Y = l, with |L(x)| ≤ C‖x‖

∀x ∈ X.

Proof. Let l1 (x) = ℜl(x). l is now complex linear. We have il(x) = l(ix) = l1 (ix) + iℑl(ix) and, on the other hand, il(x) = i(l1 (x) + iℑl(x)) = −ℑl(x) + il1 (x), hence ℑl(x) = −l1 (ix). l1 is a real linear functional on Y and the assumption implies |l1 (y)| ≤ C‖y‖

∀y ∈ Y.

By Theorem 4.9, there exists a real linear extension L1 of l1 to the whole of X with |L1 (x)| ≤ C‖x‖ ∀x ∈ X. Inspired by the computation at the beginning of the proof, we set now L(x) = L1 (x) − iL1 (ix),

x ∈ X.

L is real linear, and we have L(ix) = L1 (ix) − iL1 (−x) = i[L1 (x) − iL1 (ix)] = iL(x), hence L is also complex linear. For y ∈ Y , we get L(y) = L1 (y) − iL1 (iy) = l1 (y) − il1 (iy) = l(y), therefore L is an extension of l. Since we have L(x) = |L(x)|eiθ for some θ ∈ ℝ, we obtain finally |L(x)| = e−iθ L(x) = L(e−iθ x) = L1 (e−iθ x) ≤ C‖e−iθ x‖ = C‖x‖. Let (X, ‖ ⋅ ‖) be a normed vector space. A linear functional L ∶ X ⟶ ℂ is continuous if there exists a constant C > 0 such that |L(x)| ≤ C‖x‖ for all x ∈ X. Theorem 4.11. Let (X, ‖ ⋅ ‖) be a normed vector space over ℂ, x0 ∈ X and Y a subspace of X. Then x0 belongs to Y , if and only if there is no continuous linear functional L on X such that L(x) = 0 ∀x ∈ Y but L(x0 ) ≠ 0. Proof. If x0 ∈ Y and L is a continuous linear functional on X with L(x) = 0 ∀x ∈ Y , then, by the continuity of L, we have also that L(x0 ) = 0.

120 | 4 Construction and approximation of holomorphic functions If x0 ∉ Y , then there exists δ > 0 such that ‖x − x0 ‖ > δ ∀x ∈ Y . Let Z = ⟨Y, x0 ⟩ and define l(x + cx0 ) = c for x ∈ Y and c ∈ ℂ. Then we have δ|c| ≤ |c|

‖x0 + c−1 x‖ = ‖cx0 + x‖,

hence |l(x + cx0 )| = |c| ≤ δ−1 ‖x + cx0 ‖, i.e. |l(z)| ≤ δ−1 ‖z‖ ∀z ∈ Z. By definition we have l(x) = 0 ∀x ∈ Y and l(x0 ) = 1. Finally, by Theorem 4.10, there exists an extension L of l to the whole of X such that |L(x)| ≤ δ−1 ‖x‖

∀x ∈ X,

this means that L is also continuous. Corollary 4.12. Let (X, ‖ ⋅ ‖) be a normed vector space over ℂ and Y a subspace of X. The subspace Y is dense in X, if and only if each continuous linear functional on X, which vanishes on Y , also vanishes on the whole of X.

4.4 Runge’s approximation theorems If f ∈ ℋ(DR (z0 )), then f can be approximated by polynomials uniformly on all compact subsets of DR (z0 ) (Taylor series expansion). In this section we will discuss approximation of holomorphic functions on general domains by polynomials or by rational functions. For this purpose we will need the Hahn–Banach theorem in a suitable form. As a warm-up we first point out that the Cauchy integral formula suggests how to approximate holomorphic functions by rational functions. Theorem 4.13. Let γ be a path in ℂ and f be the holomorphic function on ℂ ⧵ γ ∗ defined by the Cauchy integral f (z) =

1 g(ζ ) dζ , ∫ 2πi γ ζ − z

z ∉ γ∗ ,

where g is a continuous function on γ ∗ . Let K be a compact subset of ℂ such that K ∩ γ ∗ = ∅ and let ϵ > 0 be arbitrary. Then there is a rational function R with simple poles in γ ∗ such that sup|f (z) − R(z)| = |f − R|K < ϵ. z∈K

4.4 Runge’s approximation theorems | 121

Proof. Let γ ∶ [a, b] ⟶ ℂ. The function t ↦ g(γ(t))/(γ(t) − z) is uniformly continuous. Hence there exists a partition a = t0 < t1 < ⋯ < tn = b such that |

2πϵ g(γ(t)) g(γ(tj )) − |< , γ(t) − z γ(tj ) − z M(b − a)

tj ≤ t ≤ tj+1 ∀z ∈ K, j = 0, 1, … n − 1,

where M = supt∈[a,b] |γ ′ (t)|. Now we define the rational function R(z) =

n−1 g(γ(tj )) 1 (γ(tj+1 ) − γ(tj )). ∑ 2πi j=0 γ(tj ) − z

Then we obtain the desired approximation |

n−1 t j+1 1 g(γ(t)) g(γ(tj )) ′ g(ζ ) 1 dζ − R(z)| = | − )γ (t) dt| ∑∫ ( ∫ 2πi γ ζ − z 2πi j=0 tj γ(t) − z γ(tj ) − z

≤

1 2πϵ M (b − a) = ϵ. 2π M(b − a)

In the following we get approximation by rational functions using the homologyversion of Cauchy’s Theorem (see Theorem 2.50). Theorem 4.14. Let Ω ⊆ ℂ be open and K ⊂ Ω a compact subset. Then each function f ∈ ℋ(Ω) can be uniformly approximated on K by rational functions that have their poles at points of Ω ⧵ K. Proof. By Lemma 2.59, there exists a cycle Γ in Ω such that IndΓ (z) = 1, for each z ∈ K and IndΓ (z) = 0 for all z ∉ Ω. Using this cycle and the homology-version of Cauchy’s Theorem (see Theorem 2.50), we have f (z) =

1 f (ζ ) dζ , ∫ 2πi Γ ζ − z

for all z ∈ K. Let ϵ > 0. From Theorem 4.13, we get a rational function R with poles in Γ ∗ such that |f − R|K < ϵ. Now we prove a more general approximation result which is remarkable in particular for the fact that we prove equivalence of a purely topological assertion to properties of holomorphic functions. For a similar result using a duality principle see Theorem 9.3. Theorem 4.15 (Runge’s4 Theorem). Let Ω ⊆ ℂ be open and K ⊂ Ω a compact subset. Then the following assertions are equivalent: 4 Runge, Carl David Tolmé (1856–1927).

122 | 4 Construction and approximation of holomorphic functions (a) Each function f , being holomorphic on neighborhood of K, can be uniformly approximated on K by functions in ℋ(Ω), i.e. ∀ϵ > 0 ∃g ∈ ℋ(Ω) such that |f − g|K < ϵ; (b) The open set Ω ⧵ K has no connected component, which is relatively compact in Ω, i.e. there is no connected component O of Ω ⧵ K, which is bounded and O ⊂ Ω; (c) ∀z ∈ Ω ⧵ K ∃f ∈ ℋ(Ω) with sup|f (w)| = |f |K < |f (z)|, w∈K

(4.4)

a function f satisfying the above condition is a called a peaking function. Corollary 4.16. Let K ⊂ ℂ be a compact set. Then the following assertions are equivalent: (a) Each function being holomorphic in a neighborhood of K can be uniformly approximated on K by polynomials; (b) ℂ ⧵ K is connected. This follows immediately from Theorem 4.15, taking Ω = ℂ. Remark. More generally, if ℂ ⧵ K is connected, each function in A(K) can be approximated by polynomials uniformly on K, where ̊ A(K) = {f ∈ 𝒞(K) ∶ f holomorphic on K}. (Mergelyan Theorem, see [63]). Proof of Theorem 4.15. (c) ⟹ (b). Suppose that (b) is not valid. Then there exists a connected component O of Ω ⧵ K such that O is compact and O ⊂ Ω. We claim that 𝜕O ⊂ K. In order to show this, suppose that there exists a point a ∈ 𝜕O with a ∉ K. Then, since O ⊂ Ω, we have a ∉ 𝜕Ω. Hence a belongs to the open set Ω ⧵ K, and there exists r > 0 such that Dr (a) ⊂ Ω ⧵ K. Since a ∈ O, we get Dr (a) ∩ O ≠ ∅ and the set Dr (a) ∪ O ⊂ Ω ⧵ K is connected, and therefore a larger connected subset of Ω ⧵ K than O. We arrive at a contradiction, hence 𝜕O ⊂ K. Now we use the maximum principle (see Corollary 2.31). Let f ∈ ℋ(Ω). Then sup|f (z)| = sup |f (z)| ≤ sup|f (z)|, z∈O

z∈𝜕O

z∈K

which contradicts (c). (a) ⟹ (b). Suppose that (b) is not valid. Then there exists a relatively compact connected component O of Ω ⧵ K and, by the first step of the proof, we have 𝜕O ⊂ K. By the maximum principle, we get sup|f (z)| ≤ sup|f (z)| z∈O

z∈K

∀f ∈ ℋ(Ω).

(4.5)

4.4 Runge’s approximation theorems | 123

Let ζ ∈ O and f (z) = 1/(z − ζ ). Then f is holomorphic in a neighborhood of K. By assertion (a), there exists a sequence of functions fn ∈ ℋ(Ω) such that |fn − f |K → 0 as n → ∞. Therefore we have sup|fn (z)(z − ζ ) − 1| → 0, z∈K

as n → ∞.

(4.6)

On the other hand, |fn − fm |K < ϵ

∀n, m > Nϵ ,

and by (4.5) sup|fn (z) − fm (z)| < ϵ z∈O

∀n, m > Nϵ .

By Theorem 2.34, there exists F ∈ ℋ(O), being continuous on O, such that fn → F uniformly on O. Using (4.6), we obtain F(z)(z − ζ ) = 1

∀z ∈ O,

choosing z = ζ , we get 0 = 1. Contradiction! (b) ⟹ (a). Let K ⊂ ℂ be a compact subset and let 𝒞(K) be the space of all continuous functions ψ ∶ K ⟶ ℂ, endowed with the norm ‖ψ‖ = supz∈K |ψ(z)|. 𝒞(K) is a Banach space. Let L ∈ 𝒞(K)′ be a continuous linear functional on 𝒞(K) with L(f ) = 0 for each f ∈ ℋ(Ω). We will show this implies that L(g) = 0, for each function g being holomorphic in a neighborhood of K. By the Hahn–Banach Theorem (see Theorem 4.12), this assertion is equivalent to (a). So let L ∈ 𝒞(K)′ and L(f ) = 0 for each f ∈ ℋ(Ω). Let ζ ∈ ℂ ⧵ K. Define fζ (z) = 1/(z − ζ ). Then fζ ∈ 𝒞(K), f is even holomorphic in a neighborhood of K. We define ϕ(ζ ) = L(fζ ). We claim that ϕ ≡ 0 on ℂ ⧵ K. First we show ϕ ∈ ℋ(ℂ ⧵ K). For this we consider ϕ(ζ1 ) − ϕ(ζ2 ) 1 1 = [L(fζ1 ) − L(fζ2 )] = L(f − f ). ζ1 − ζ2 ζ1 − ζ2 ζ1 − ζ2 ζ1 ζ2 We get fζ1 (z) − fζ2 (z) =

1 1 ζ1 − ζ2 − = . z − ζ1 z − ζ2 (z − ζ1 )(z − ζ2 )

For Fζ1 ,ζ2 (z) = 1/[(z − ζ1 )(z − ζ2 )], we obtain from (4.7) that ϕ(ζ1 ) − ϕ(ζ2 ) 1 = L((ζ1 − ζ2 )Fζ1 ,ζ2 ) = L(Fζ1 ,ζ2 ), ζ1 − ζ2 ζ1 − ζ2

(4.7)

124 | 4 Construction and approximation of holomorphic functions where we used that L is linear. Now we let ζ1 → ζ2 and get that Fζ1 ,ζ2 → Fζ2 ,ζ2 uniformly on K, and we have Fζ2 ,ζ2 (z) = 1/(z − ζ2 )2 . Since L is continuous, we obtain lim

ζ1 →ζ2

ϕ(ζ1 ) − ϕ(ζ2 ) = L(Fζ2 ,ζ2 ), ζ1 − ζ2

hence ϕ ∈ ℋ(ℂ ⧵ K). In a similar way one can show that the higher derivatives of ϕ can be expressed as 1 ). (z − ζ )k+1

ϕ(k) (ζ ) = k!L(z ↦

If ζ ∈ ℂ ⧵ Ω, then the function z ↦ (z − ζ )−(n+1) is holomorphic on Ω; hence, by assumption, L(z ↦ (z − ζ )−(n+1) ) = 0, i.e. ϕ(k) (ζ ) = 0

∀k ∈ ℕ,

and ϕ = 0 on each connected component of ℂ ⧵ K, which has non-empty intersection with ℂ ⧵ Ω. In addition we have L(z ↦ z n ) = 0 ∀n ∈ ℕ0 . If |ζ | > supz∈K |z|, then the series 1 = ∑ a zn z − ζ n=0 n ∞

converges uniformly on K and we have ∞

∞

n=0

n=0

ϕ(ζ ) = L(z ↦ 1/(z − ζ )) = L(z ↦ ∑ an z n ) = ∑ an L(z ↦ z n ) = 0, hence ϕ = 0 on the unbounded connected component of ℂ ⧵ K. Now we know that ϕ = 0 on the unbounded connected component of ℂ ⧵ K and on each connected component of ℂ ⧵ K, which has non-empty intersection with ℂ ⧵ Ω. Let U be a bounded connected component of ℂ ⧵ K with U ∩ (ℂ ⧵ Ω) = ∅. Then U ⊆ Ω and U ⊆ Ω ⧵ K. We know from the proof of (c) ⟹ (b), applied for Ω = ℂ, that 𝜕U ⊆ K. Hence U ⊆ Ω, contradicting (b). Therefore there is no connected component like U. This means that ϕ = 0 on the whole of ℂ ⧵ K. Now let g be holomorphic in a neighborhood ω ⊃ K. We have to show that L(g) = 0. We choose a compact set K1 such that ω ⊃ K1 ⊃ K. By Theorem 4.8, there exists ψ ∈ 𝒞∞ 0 (ω) with ψ = 1 on K1 and 1 𝜕ψ g(ζ ) (ζ ) dζ ∧ dζ ∀z ∈ K, ∫ 2πi ω 𝜕ζ ζ −z 1 𝜕ψ g(ζ ) = (ζ ) dζ ∧ dζ , ∫ 2πi ω⧵K1 𝜕ζ ζ −z

g(z) =

4.4 Runge’s approximation theorems | 125

since on K1 we have

𝜕ψ 𝜕ζ

= 0. No we apply the functional L to this formula. The integral ∫

g(ζ ) 𝜕ψ (ζ ) dζ ∧ dζ ζ −z 𝜕ζ

lim ∑

𝜕ψ g(ζ ) (ζν ) ν m(Rν ), ζ 𝜕ζ ν −z

ω⧵K1

can be written as a limit ν

where Rν are rectangles covering ω ⧵ K1 , the points ζν ∈ Rν and where m(Rν ) is the measure of Rν . Taking the limit means that the diameters of Rν tend to zero. Furthermore, the limit occurs uniformly for all z ∈ K. Since L is continuous, we can interchange the limit with the application of L and obtain 𝜕ψ g(ζ ) (ζ ) dζ ∧ dζ ) ζ −z ω⧵K1 𝜕ζ 𝜕ψ 1 (ζ )g(ζ )L(z ↦ ) dζ ∧ dζ = 0, ζ − z 𝜕ζ

L(g) = L(z ↦ ∫ =∫

ω⧵K1

since ϕ(ζ ) = L(fζ ) = 0 ∀ζ ∈ ℂ ⧵ K. We have shown that (c) ⟹ (b) ⟺ (a). It remains to show that (b) ⟹ (c). For this purpose let z ∈ Ω ⧵ K. We have to prove that there exists f ∈ ℋ(Ω) such that |f |K < |f (z)|. We choose a disc D around z such that D ⊂ Ω ⧵ K. The connected components of Ω ⧵ (K ∪ D) coincide with the connected components of Ω ⧵ K except for those components where D was removed. As we assume (b), there is no connected component in Ω ⧵ K being relatively compact in Ω; in other words, all connected components of Ω ⧵ K come arbitrarily close to the boundary 𝜕Ω. This property is also true for the connected components of Ω ⧵ (K ∪ D). Hence, if we take the compact set K ∪ D in stead of K, property (b) is also satisfied. Now we use the conclusion (b) ⟹ (a) for the compact set K ∪ D ⊂ Ω. We have K ∩ D = ∅, since D ⊂ Ω ⧵ K. Now let g = 0 in a neighborhood of K and g = 1 in a neighborhood of D, where the neighborhoods are chosen with empty intersection. Then g is holomorphic in a neighborhood of K ∪ D. By (a), there exists f ∈ ℋ(Ω) such that |f | = |f − g| < 1/2 on K and |f − 1| = |f − g| < 1/2 on D. If z ∈ D, then we have |f (z)| ≥ 1 − |f (z) − 1| > 1 − 1/2 = 1/2 > |f |K . Definition 4.17. Let Ω be open in ℂ and K ⊂ Ω a compact subset. The set K̂ Ω ∶= {z ∈ Ω ∶ |f (z)| ≤ |f |K , ∀f ∈ ℋ(Ω)} is called the holomorphically convex hull of K with respect to Ω.

126 | 4 Construction and approximation of holomorphic functions Remark 4.18. (a) Taking the holomorphically convex hull of a compact set K, the holes of K are plugged in a certain sense, which corresponds to the maximum principle. (b) We have dist(K, Ωc ) = dist(K̂ Ω , Ωc ). Since K ⊆ K̂ Ω , we get dist(K, Ωc ) ≥ dist(K̂ Ω , Ωc ). Considering the function f (z) = 1/(z − ζ ), ζ ∈ Ωc , we get for z ∈ K̂ Ω that −1 1 1 ≤ sup| | = ( inf |w − ζ |) , w∈K |z − ζ | w∈K w − ζ

since f ∈ ℋ(Ω). Hence infw∈K |w − ζ | ≤ |z − ζ |. Therefore we obtain dist(K, Ωc ) ≤ dist(K̂ Ω , Ωc ). (c) A subset M ⊆ ℝ2 is called convex, if for any two points in M the whole straight line segment between the two points also lies in M. An equivalent formulation is the following: for any point (x0 , y0 ) ∈ ℝ2 ⧵ M there exists a straight line g = {(x, y) ∈ ℝ2 ∶ l((x, y)) = ax + by + c = 0} such that (x0 , y0 ) ∈ g and M ⊂ {(x, y) ∈ ℝ2 ∶ l((x, y)) < 0}, which is also called the separation property. The convex hull K̃ of K is the smallest convex set, which contains K. One can show that K̃ = {z ∈ ℂ ∶ |eαz | ≤ sup|eαw | ∀α ∈ ℂ}. w∈K

Since the functions z ↦ e

αz

are holomorphic on Ω, it follows that K̂ Ω ⊆ K,̃

hence the notation “holomorphically convex.” ̂ = K̂ Ω . (d) From that definition we get immediately (K̂ Ω ) Ω ̂ (e) In addition, we have that K Ω is bounded: since |w| ≤ C ∀w ∈ K, C > 0, we get for the holomorphic function f (z) = z on Ω that |z| = |f (z)| ≤ |f |K ≤ C and hence that |z| ≤ C ∀z ∈ K̂ Ω . Furthermore K̂ Ω is closed in Ω: if z ∈ Ω ⧵ K̂ Ω , then there exists f ∈ ℋ(Ω) such that |f (z)| > |f |K , since |f | is continuous, the same inequality holds on a whole neighborhood U(z) ⊂ Ω, which is therefore contained in Ω ⧵ K̂ Ω . Hence Ω ⧵ K̂ Ω is open. Next we claim that K̂ Ω ⊂ Ω: if there is a point z0 ∈ K̂ Ω ⧵ Ω, then z0 belongs to 𝜕K̂ Ω ∩ 𝜕Ω. But then the function f (z) = 1/(z − z0 ) is holomorphic in Ω. If (zn )n is a sequence in K̂ Ω with limn→∞ zn = z0 , then, by Definition 4.17, |f (zn )| ≤ |f |K ∀n ∈ ℕ, which yields a contradiction to the fact that the sequence (f (zn ))n is unbounded.

4.4 Runge’s approximation theorems | 127

Finally, we get K̂ Ω = K̂ Ω . If (zn )n is a sequence in K̂ Ω such that limn→∞ zn = z0 , then the assertion from above yields z0 ∈ Ω. Hence we have |f (z0 )| = limn→∞ |f (zn )| ≤ |f |K ∀f ∈ ℋ(Ω), which implies that z0 ∈ K̂ Ω . The holomorphically convex hull K̂ Ω is itself a compact subset of Ω and always satisfies condition (c) of Theorem 4.15: ∀z ∈ Ω ⧵ K̂ Ω ∃f ∈ ℋ(Ω) with |f (z)| > |f |K . Hence, functions being holomorphic in a neighborhood of K̂ Ω can be approximated by functions in ℋ(Ω) uniformly on K. (f) There exists a compact exhaustion (Kj )j of Ω (see Section 2.5), which has the prop̂ = Kj j ∈ ℕ: we choose an arbitrary exhaustion (Kj∗ )j of Ω and set K1 = erty (Kj ) Ω ̂ . Now let K1 , … , Kn−1 be already constructed. There exists ̂ , then K1 = (K1 ) Ω (K1∗ ) Ω ∗ ∗ ̂ . Then Kn is a compact subset of λ(n) ∈ ℕ with Kn−1 ⊂ (Kλ(n) )∘ . We set Kn = (Kλ(n) )Ω ∗ ∘ ̂ and we have Kn ⊂ (Kλ(n+1) )∘ ⊂ Kn+1 Ω such that Kn = (Kn ) Ω . Finally, ∞

∞

∞

n=1

n=1

n=1

∗ ∗ ̂ ⊇ ⋃ Kλ(n) ⋃ Kn = ⋃ (Kλ(n) )Ω = Ω.

(g) Given M, ϵ > 0 and p ∈ Ω ⧵ K̂ Ω , there is f ∈ ℋ(Ω) with |f |K < ϵ and |f (p)| > M. To prove this, observe that p ∈ Ω ⧵ K̂ Ω implies that there is h ∈ ℋ(Ω) with |h|K < |h(p)|. After multiplying with a suitable constant, one may assume that |h|K < 1 < |h(p)|, now take f = hN with N > 0 sufficiently large. Theorem 4.19. Let Ω ⊆ ℂ be open and let K ⊂ Ω be a compact subset. Then K̂ Ω is the union of K with all connected components of Ω ⧵ K, which are relatively compact in Ω. Proof. Let O be a connected component of Ω ⧵ K, which is relatively compact in Ω. By the first step of the proof of Theorem 4.15, we get 𝜕O ⊆ K and, by the maximum principle, sup|f (z)| ≤ sup|f (z)| z∈O

z∈K

∀f ∈ ℋ(Ω).

Hence O ⊆ K̂ Ω = {z ∈ Ω ∶ |f (z)| ≤ |f |K ∀f ∈ ℋ(Ω)}. Let K1 denote the union of K with all connected components of Ω ⧵ K being relatively compact in Ω. We have already shown that K1 ⊆ K̂ Ω . Furthermore, Ω ⧵ K1 is open, since this set is the union of open connected components of Ω ⧵ K. So K1 ⊂ Ω is compact and satisfies condition (b) of Theorem 4.15. Hence ̂ . Since K ⊆ K1 , K1 also satisfies condition (c) in Theorem 4.15, and we get K1 = (K1 ) Ω we finally get ̂ = K1 . K̂ Ω ⊆ (K1 ) Ω Remark. Conditions (a), (b) and (c) of Runge’s approximation Theorem 4.15 are satisfied for K ⊂ Ω if and only if K = K̂ Ω . In this case K is called holomorphically convex.

128 | 4 Construction and approximation of holomorphic functions Now we consider two open sets Ω1 and Ω2 in ℂ such that Ω1 ⊆ Ω2 . We have ℋ(Ω1 ) ⊇ ℋ(Ω2 ) and we can ask the question under which conditions each function f ∈ ℋ(Ω1 ) can be approximated by functions in ℋ(Ω2 ) uniformly on all compact subsets of Ω1 . In other words, when is ℋ(Ω2 ) dense in ℋ(Ω1 )? Theorem 4.20. Let Ω1 and Ω2 be two open subsets in ℂ with Ω1 ⊆ Ω2 . Then the following conditions are equivalent: (a) Each f ∈ ℋ(Ω1 ) can be approximated by functions in ℋ(Ω2 ) uniformly on all compact subsets of Ω1 , i.e. ℋ(Ω2 ) is dense in ℋ(Ω1 ). (b) If Ω2 ⧵ Ω1 = L ∪ F, where F ⊂ Ω2 is closed, L is compact and L ∩ F = ∅, then L = ∅. (c1 ) ∀K ⊂ Ω1 compact, one has K̂ Ω2 = K̂ Ω1 . (c2 ) ∀K ⊂ Ω1 compact, one has K̂ Ω2 ∩ Ω1 = K̂ Ω1 . (c3 ) ∀K ⊂ Ω1 compact, the set K̂ Ω2 ∩ Ω1 is compact. Proof. (a) ⟹ (c2 ). Let K ⊂ Ω1 . By definition we have K̂ Ω2 ∶= {z ∈ Ω2 ∶ |f (z)| ≤ |f |K ∀f ∈ ℋ(Ω2 )} and K̂ Ω1 ∶= {z ∈ Ω1 ∶ |f (z)| ≤ |f |K ∀f ∈ ℋ(Ω1 )}. First we claim that K̂ Ω1 ⊃ K̂ Ω2 ∩ Ω1 . For z ∈ K̂ Ω2 ∩ Ω1 , ϵ > 0 arbitrary and f ∈ ℋ(Ω1 ), we use (a) to find g ∈ ℋ(Ω2 ) such that |f − g|K∪{z} < ϵ. Then we have |f (z)| ≤ |g(z)| + |f (z) − g(z)| ≤ |g(z)| + ϵ ≤ |g|K + ϵ ≤ |f − g|K + |f |K + ϵ ≤ |f |K + 2ϵ, which implies K̂ Ω1 ⊃ K̂ Ω2 ∩ Ω1 . Since ℋ(Ω1 ) ⊇ ℋ(Ω2 ), we have K̂ Ω1 ⊆ K̂ Ω2 ∩ Ω1 , and we get (c2 ). (c2 ) ⟹ (c3 ). This follows from Remark 4.18(e). (c3 ) ⟹ (a). Let K ′ = K̂ Ω2 ∩ Ω1 and K ″ = K̂ Ω2 ∩ Ωc1 . Then K ′ ∪ K ″ = K̂ Ω2 and K ′ ∩ ″ K = ∅. We have K ′ ⊂ Ω1 and K ′ ∪ K ″ ⊂ Ω2 . By (c3 ), the set K ′ is compact and, by Remark 4.18 (e), also K ″ is compact. Let h ∈ ℋ(Ω1 ) and define {h in a neighborhood of K ′ , g={ 1 in a neighborhood of K ″ , {

4.4 Runge’s approximation theorems | 129

where we chose disjoint neighborhoods. Then g is holomorphic in a neighborhood of the compact set K ′ ∪ K ″ = K̂ Ω2 . Using Theorem 4.15 (actually, condition (b) of Theorem 4.15 is satisfied), we can approximate g uniformly on K ′ ∪ K ′ by functions in ℋ(Ω2 ). In particular, h can be approximated uniformly on K by functions in ℋ(Ω2 ), and we have shown (a). If we choose h = 0, then {0 in a neighborhood of K ′ , g={ 1 in a neighborhood of K ″ , { and, again by Theorem 4.15, there exists an approximating function ϕ ∈ ℋ(Ω2 ) such that |ϕ(z)| > |ϕ(ζ )|

∀z ∈ K ″ ∀ζ ∈ K ′ .

But now K ⊂ K ′ and K ′ ∪ K ″ = K̂ Ω2 = {z ∈ Ω2 ∶ |ϕ(z)| ≤ |ϕ|K ∀ϕ ∈ ℋ(Ω2 )}, hence K ″ = ∅ and we get K̂ Ω2 = K̂ Ω2 ∩ Ω1 . We know already that condition (c3 ) implies condition (c2 ), hence we obtain K̂ Ω2 = K̂ Ω2 ∩ Ω1 = K̂ Ω1 . Therefore we have proved that (c3 ) ⟹ (c1 ). (c1 ) ⟹ (c2 ). By assumption we have K̂ Ω1 = K̂ Ω2 . Since always K̂ Ω1 ⊆ K̂ Ω2 ∩ Ω1 ⊆ K̂ Ω2 , we get immediately K̂ Ω1 = K̂ Ω2 ∩ Ω1 . So far we have shown the equivalences of (a), (c1 ), (c2 ) and (c3 ). Now we still prove that (b) ⟺ (c1 ), then we are done. (c1 ) ⟹ (b). For a compact set K ⊂ Ω1 , we have by assumption that K̂ Ω1 = K̂ Ω2 . Let Ω2 ⧵ Ω1 = L ∪ F, where L is compact, F is closed in Ω2 and L ∩ F = ∅; we have to show that L = ∅. For this let ω ⊃ L be an open set, which is relatively compact in Ω2 and for which ω ∩ F = ∅. We have 𝜕ω ⊂ Ω2 and 𝜕ω ∩ (Ω2 ⧵ Ω1 ) = 𝜕ω ∩ (L ∪ F) = (𝜕ω ∩ L) ∪ (𝜕ω ∩ F) = ∅. Hence 𝜕ω ⊂ Ω1 . Furthermore, the maximum principle implies that ̂ = {z ∈ Ω2 ∶ |f (z)| ≤ |f |𝜕ω = |f |ω ∀f ∈ ℋ(Ω2 )}, (𝜕ω) Ω 2 ̂ and since L ⊂ ω, we can use condition (c1 ) for the compact set therefore ω ⊆ (𝜕ω) Ω 2 𝜕ω to show that ̂ = (𝜕ω) Ω ̂ ⊆ Ω1 . L ⊂ (𝜕ω) Ω 2 1 Hence L ⊂ Ω1 ; but since Ω2 ⧵ Ω1 = L ∪ F, we get L = ∅.

130 | 4 Construction and approximation of holomorphic functions (b) ⟹ (c1 ). Let K ⊂ Ω1 be compact. One always has K̂ Ω1 ⊆ K̂ Ω2 . Hence we still have to show that K̂ Ω1 ⊇ K̂ Ω2 . For this we use Theorem 4.19. Let O be a connected component of Ω2 ⧵ K, which is relatively compact in Ω2 . We will show that O ⊆ Ω1 . Then we are done by Theorem 4.19. From the first part of the proof of Theorem 4.15, we know that 𝜕O ⊆ K ⊂ Ω1 . Now let L = O ∩ (Ω2 ⧵ Ω1 ). Then L is compact, and we have L = (O ∪ 𝜕O) ∩ (Ω2 ⧵ Ω1 ) = (O ∩ Ω2 ⧵ Ω1 ) ∪ (𝜕O ∩ Ω2 ⧵ Ω1 ) = O ∩ Ω2 ⧵ Ω1 , since 𝜕O ⊆ Ω1 . Let F = Oc ∩ (Ω2 ⧵ Ω1 ). Then F is closed in Ω2 and L ∪ F = Ω2 ⧵ Ω1 . Here we assume condition (b) and get L = ∅. Hence O ⊆ Ω1 . Corollary 4.21. Let Ω ⊆ ℂ be an open set. Then the polynomials are dense in ℋ(Ω) (i.e. each f ∈ ℋ(Ω) can be approximated by polynomials uniformly on all compact subsets of Ω), if and only if ℂ ⧵ Ω is connected. Proof. Let ℂ ⧵ Ω be connected. Taylor series expansion implies that the polynomials are dense in ℋ(ℂ). Set Ω2 = ℂ, Ω1 = Ω and use Theorem 4.20: we write Ω2 ⧵ Ω1 = ℂ ⧵ Ω = L ∪ F, where L is compact, F is closed in ℂ and L ∩ F = ∅. We will show that L = ∅. For this aim let F̃ = F ∪ {∞}; then we have F̃ ∩ L = ∅ and ℂ ⧵ Ω = F̃ ∪ L, where F̃ ≠ ∅. Since both sets F̃ and L are closed, we must have L = ∅, otherwise we would get a contradiction to the assumption that ℂ ⧵ Ω is connected. Now we apply Theorem 4.20 and get that ℋ(ℂ) is dense in ℋ(Ω). To prove the other direction, suppose that ℂ ⧵ Ω is not connected. It suffices to show that condition (b) from Theorem 4.20 is not satisfied. Now we can write ℂ ⧵ Ω as the union L ∪ F,̃ where L is compact in ℂ, the set and F̃ is closed in ℂ and ∞ ∈ F,̃ L ≠ ∅ as well as L ∩ F̃ = ∅. Let F = F̃ ⧵ {∞}. Then F is closed in ℂ and ℂ ⧵ Ω = L ∪ F, where L ∩ F = ∅. Therefore condition (b) from Theorem 4.20 is not satisfied. Remark. Compare the assertions of Corollary 4.16 and Corollary 4.21! Example 4.22. Let Ω = {z ∈ ℂ ∶ −1 < ℑz < 1}. Then ℂ ⧵ Ω is not connected, whereas ℂ ⧵ Ω is connected.

4.5 Mittag-Leffler’s Theorem Let Ω ⊆ ℂ be an open subset and let (zj )j be a discrete sequence of points in Ω being different from each other. Furthermore, let mj

fj (z) = ∑

Ajk

k k=1 (z − zj )

be principle parts in zj .

4.5 Mittag-Leffler’s Theorem

| 131

We want to construct a meromorphic function f ∈ ℳ(Ω) which has the given principle parts in zj . Taking the sum of all principle parts ∑∞ j=1 fj raises the question of convergence of this sum. We will construct terms which leave the given principle parts unchanged, but ensure convergence. For this purpose we will use Runge’s approximation theorem. Theorem 4.23 (Mittag-Leffler’s5 Theorem). Let Ω ⊆ ℂ be an open subset and let (zj )j be a discrete sequence of points in Ω being different from each other. Let fj be functions meromorphic in a neighborhood of zj , j ∈ ℕ. Then there exists a meromorphic function f ∈ ℳ(Ω) such that f − fj is holomorphic in a neighborhood of zj , j ∈ ℕ, i.e. f − fj has a removable singularity at zj . We call f a solution of the Mittag-Leffler problem. Proof. By Theorem 2.65, we can assume that mj

fj (z) = ∑

Ajk

k=1

(z − zj )k

j ∈ ℕ.

,

We will construct functions uj ∈ ℋ(Ω) such that the series ∞

f (z) = ∑(fj (z) − uj (z)) j=1

converges. Since uj ∈ ℋ(Ω), the principle parts of f will then exactly be the given fj . ̂ = Kj , For this purpose we choose a compact exhaustion (Kj )j of Ω such that (Kj ) Ω j ∈ ℕ (see Remark 4.18), furthermore we can assume that zk ∉ Kj ∀k ≥ j. The functions fj ̂ = Kj . Hence, are holomorphic in a neighborhood of Kj , since zj ∉ Kj , and we have (Kj ) Ω by Theorem 4.15, there exist uj ∈ ℋ(Ω) with |fj (z) − uj (z)| < 2−j

∀z ∈ Kj .

Then the series ∞

∑(fj (z) − uj (z))

j=k

converges uniformly on Kk and defines a holomorphic function in the interior of Kk , since we have the estimate ∞

∞

j=k

j=k

∑|fj (z) − uj (z)| < ∑ 2−j < ∞

and all fj are holomorphic in the interior of Kk ∀j ≥ k. 5 Mittag-Leffler, Magnus Gösta (1846–1927).

∀z ∈ Kk

132 | 4 Construction and approximation of holomorphic functions If K ⊂ Ω is an arbitrary compact subset, there exists k ∈ ℕ with K ⊆ Kk such that k−1

∞

j=1

j=k

f (z) = ∑ (fj (z) − uj (z)) + ∑(fj (z) − uj (z)) where the second series converges uniformly on K. Therefore, f has the desired properties. Example 4.24. We are looking for a meromorphic function on ℂ having the principle parts (z − k)−2 for k ∈ ℤ. Let f (z) = ∑k∈ℤ (z − k)−2 . If |z| ≤ R and |k| > 2R, then |z − k| ≥ |k| − |z| > |k| − R > |k| − |k|/2 = |k|/2. Hence |z − k|−2 ≤ 4k −2 , and the series defining f converges uniformly on compact subsets of ℂ which do not contain the poles k ∈ ℤ. The functions f (z) and π 2 / sin2 (πz) have the same poles and principle parts. Hence f (z) − π 2 / sin2 (πz) must be an entire function. It is easily seen that f (z + 1) = f (z). The same is true for π 2 / sin2 (πz), so that f (z) − π 2 / sin2 (πz) is periodic with period 1. It suffices to consider the periodic strip {z = x + iy ∶ 0 ≤ x ≤ 1, y ∈ ℝ}. In this strip we have |

1 1 1 |= 2 ≤ , 2 2 (z − k) y + (k − x) (k − 1)2

for |k| ≥ 2 and |y| ≥ 1. The series for f (z) converges uniformly and f (z) is uniformly bounded for |y| ≥ 1 and 0 ≤ x ≤ 1. Each summand of f (z) tends to 0 as |y| → ∞. Now the uniform convergence implies that f (x + iy) → 0 as 0 ≤ x ≤ 1 and |y| → ∞. As we have | sin z|2 = | sin x|2 + | sinh y|2 , we see that the function π 2 / sin2 (πz) is bounded for |y| ≥ 1 and tends to 0 as |y| → ∞. Hence the entire function f (z)−π 2 / sin2 (πz) is bounded for 0 ≤ x ≤ 1 and |y| ≥ 1 and also bounded on the entire vertical strip 0 ≤ x ≤ 1. By periodicity, this function is bounded on ℂ. By Liouville’s Theorem (see Theorem 2.29) it is constant and this constant is zero, since both f (z) and π 2 / sin2 (πz) tend to 0 as |y| → ∞. Therefore we have π2 1 =∑ , sin2 (πz) k∈ℤ (z − k)2

(4.8)

and the series converges uniformly on compact subsets of ℂ which do not contain the poles k ∈ ℤ. The series is called the partial fractions decomposition of π 2 / sin2 (πz). Example 4.25. Let Ω = ℂ, and let the points a0 = 0, a1 , a2 , … ∈ ℂ be different from each other and suppose that limn→∞ |an | = ∞. Furthermore, let (cn )∞ n=0 be arbitrary sequence in ℂ.

4.5 Mittag-Leffler’s Theorem

| 133

We are looking for a meromorphic function f ∈ ℳ(ℂ) having simple poles at the points an with residues cn , i.e. a meromorphic function with principle parts cn /(z − an ) at the points an for n = 0, 1, … . Let ϵj > 0 such that ∑∞ j=1 ϵj < ∞. The series 1 1 z k =− ∑( ) z − an an k=0 an ∞

converges uniformly on all compact subsets of {z ∶ |z| < |an |}. Now choose ln ∈ ℕ such that l

|

1 1 n z k ϵ + ∑( ) | < n , z − an an k=0 an |cn |

for |z| ≤ 2n and n = 0, 1, … . Define |a |

f (z) =

l

n c0 1 zk + ∑ cn ( + ∑ k+1 ). z n=1 z − an k=0 an

∞

Since the expression in the brackets has absolute value ≤ ϵn /|cn | for |z| ≤ 2n , we have f ∈ ℳ(ℂ). Now we choose ak = k, k ∈ ℤ and ck = 1, k ∈ ℤ. In this case it suffices to take lk = 0 because |a |

|

1 z 2R 1 + |=| |≤ 2, z−k k k(z − k) k

if |z| ≤ R and |k| > 2R. We have then |z − k| ≥ |k| − |z| > |k| − R > |k| − |k|/2 = |k|/2 and the estimate | ∑ ( |k|>2R

1 2R 1 + )| ≤ ∑ 2 < ∞, z−k k |k|>2R k

for |z| ≤ R. Hence the function f (z) =

1 1 1 + ∑ ( + ) z k∈ℤ⧵{0} z − k k

(4.9)

is meromorphic on ℂ with simple poles at ak = k ∈ ℤ and residues ck = 1, k ∈ ℤ. Using Example 4.24, we can identify the function f (z) as π cot(πz): for this aim we integrate the partial fractions decomposition (4.8) term by term, this is justified on account of the uniform convergence on compact subsets of ℂ which do not contain the poles. We have ∫ [0,z]

dζ 1 1 + ), = −( 2 (ζ − k) z−k k

134 | 4 Construction and approximation of holomorphic functions for k ≠ 0, and (

∫ [0,z]

π2 1 1 − ) dζ = −π cot(πz) + . z sin2 (πζ ) ζ 2

Finally, we get from (4.8) 1 1 1 + ∑ ( + ). z k∈ℤ⧵{0} z − k k

π cot(πz) =

(4.10)

In the following we will prove a generalization of Mittag-Leffler’s Theorem which is of importance for the theory of functions of several complex variables. For this purpose we use solutions of the inhomogeneous Cauchy–Riemann differential equations, also in a more general version than known so far. Theorem 4.26. Let Ω ⊂ ℂ be open and f ∈ 𝒞∞ (Ω). Then there exists u ∈ 𝒞∞ (Ω) such that 𝜕u = f (compare with Corollary 4.7). 𝜕z ̂ = Kj . Choose ψj ∈ 𝒞∞ Proof. Let (Kj )j be a compact exhaustion of Ω with (Kj ) Ω 0 (Ω) such that ψj = 1 in a neighborhood of Kj , j ∈ ℕ, define ϕ1 = ψ1 , and ϕj = ψj − ψj−1 , ∀ j > 1. Then ϕj = 0 in a neighborhood of Kj−1 ∀j > 1. By construction we have ∞

∑ ϕj = 1 j=1

on Ω.

∞ Observe that fϕj ∈ 𝒞∞ 0 (Ω) and define fϕj = 0 on ℂ ⧵ Ω. Then fϕj ∈ 𝒞0 (ℂ), j ∈ ℕ ∞ and, by Corollary 4.7, there exist smooth functions uj ∈ 𝒞 (ℂ) such that

𝜕uj 𝜕z

= fϕj ,

j ∈ ℕ.

Since fϕj = 0 in a neighborhood of Kj−1 , j > 1, the functions uj are holomorphic in a neighborhood of Kj−1 . By Theorem 4.15, there exist holomorphic functions vj ∈ ℋ(Ω) such that |uj − vj | < 2−j on Kj−1 , j > 1. (For v1 we take an arbitrary holomorphic function on Ω.) Now we set ∞

u = ∑(uj − vj ) j=1

and observe that for each l > 1 the series ∞

∑ (uj − vj )

j=l+1

4.5 Mittag-Leffler’s Theorem

| 135

consists of terms which are holomorphic in a neighborhood of Kl . Furthermore, the series converges uniformly on Kl . Therefore it represents a holomorphic function in the interior of Kl . Now we get that the series ∞

u = ∑(uj − vj ) j=1

converges uniformly on all compact subsets of Ω, which is also true for all derivatives of u. Hence u ∈ 𝒞∞ (Ω) and we can interchange differentiation and summation to obtain ∞ ∞ ∞ 𝜕uj 𝜕vj 𝜕u = ∑( − ) = ∑ fϕj = f ∑ ϕj = f . 𝜕z j=1 𝜕z 𝜕z j=1 j=1

Theorem 4.27. Let Ω = ⋃∞ j=1 Ωj be an open cover of Ω. Let fj ∈ ℳ(Ωj ), j ∈ ℕ, be meromorphic functions such that fj − fk ∈ ℋ(Ωj ∩ Ωk ) for each pair (j, k) where Ωj ∩ Ωk ≠ ∅. Then there exists a meromorphic function f ∈ ℳ(Ω) such that f − fj ∈ ℋ(Ωj ) ∀j ∈ ℕ. Remark. Theorem 4.27 implies Theorem 4.23: if (zj )j is a discrete sequence in Ω and if the functions fj are meromorphic in a neighborhood of zj , then we choose Ωj to be neighborhoods of zj such that zk ∉ Ωj ∀k ≠ j. We choose Ω0 to cover the rest of Ω. In addition, we can assume that fj ∈ ℋ(Ωj ⧵ {zj }). Then fk − fj ∈ ℋ(Ωj ∩ Ωk ). Hence, by Theorem 4.27, there exists a meromorphic function f ∈ ℳ(Ω) such that f − fj ∈ ℋ(Ωj ), which is the assertion of Theorem 4.23. Theorem 4.28. Let Ω = ⋃∞ j=1 Ωj be an open cover of Ω. Let gjk ∈ ℋ(Ωj ∩ Ωk ) such that gjk = −gkj and gjk + gkl + glj = 0 for all triples (j, k, l), where Ωj ∩ Ωk ∩ Ωl ≠ ∅. (We call (gjk ) a cocycle.) Then there exist holomorphic functions gj ∈ ℋ(Ωj ) such that gjk = gk − gj for all pairs (j, k) where Ωj ∩ Ωk ≠ ∅. (We say that the first cohomology group disappears.) Remark. Theorem 4.28 implies Theorem 4.27: if fj ∈ ℳ(Ωj ), we put gjk = fk − fj ∈ ℋ(Ωj ∩ Ωk ). Then (gjk ) satisfies the cocycle conditions gkj = fj − fk = −(fk − fj ) = −gjk and gjk + gkl + glj = fk − fj + fl − fk + fj − fl = 0. Hence, by Theorem 4.28, there exist gj ∈ ℋ(Ωj ) such that gjk = gk − gj on Ωk ∩ Ωj , hence fk − fj = gk − gj and fk − gk = fj − gj on Ωk ∩ Ωj , where fk − gk ∈ ℳ(Ωk ) and fj − gj ∈ ℳ(Ωj ). Now we set f = fj − gj on Ωj and we have a meromorphic function on the whole of Ω satisfying f − fj = −gj ∈ ℋ(Ωj ). This is the assertion of Theorem 4.27. Proof of Theorem 4.28. Let (ϕj )j be a 𝒞∞ -partition of unity with respect to the cover ∞ (Ωj )j of Ω; one has ϕj ∈ 𝒞∞ 0 (Ωj ) and ∑j=1 ϕj = 1 on Ω. For k ∈ ℕ fixed, we set hk = ∑ ϕj gjk , j

136 | 4 Construction and approximation of holomorphic functions where the sum is taken over all j for which Ωk ∩ Ωj ≠ ∅. Then the cocycle conditions on Ωk ∩ Ωl imply hk − hl = ∑ ϕj (gjk − gjl ) = ∑ ϕj (gjk + glj ) = ∑ ϕj glk = glk ∑ ϕj = glk , j

j

j

j

by construction of the 𝒞∞ -partition of unity (see Lemma 4.3). The functions hk satisfy glk = hk − hl , but they are still not holomorphic. Using the solutions of the inhomogeneous Cauchy–Riemann differential equations, we construct a term u such that gk = hk + u is holomorphic on Ωk . Since glk ∈ ℋ(Ωl ∩ Ωk ), we have 𝜕hk 𝜕hl = 𝜕z 𝜕z

on Ωl ∩ Ωk .

Setting ψ ∶= 𝜕zk on Ωk , k ∈ ℕ, we obtain a function in 𝒞∞ (Ω). By Theorem 4.26, there exists u ∈ 𝒞∞ (Ω) such that 𝜕u = −ψ. 𝜕z Now let gk = hk + u. Then gk − gl = hk + u − hl − u = glk and 𝜕h

𝜕gk 𝜕hk 𝜕u 𝜕hk = + = − ψ = ψ − ψ = 0, 𝜕z 𝜕z 𝜕z 𝜕z hence gk ∈ ℋ(Ωk ).

4.6 The Weierstraß Factorization Theorem In order to construct holomorphic functions with prescribed zeroes, we define infinite products first of complex numbers and then of holomorphic functions. The definition of convergent infinite products is suggested by the definition of convergent series. Definition 4.29. (1) Let an ∈ ℂ ⧵ {0}, n ∈ ℕ. The product ∏∞ n=1 an is convergent, if N limN→∞ ∏n=1 an exists and is not zero. (2) Let (an )n be an arbitrary sequence in ℂ. The product ∏∞ n=1 an is convergent, if almost all an ≠ 0 and if the product ∞

∏ an

n=1,an ≠0

exists in the sense of (1). If one an = 0, we set ∏∞ n=1 an = 0. ∞ Remark. If ∏∞ n=1 an converges, then limn→∞ an = 1; if ∏n=1 (1 + un ) converges, then limn→∞ un = 0.

Lemma 4.30. Let un ≠ −1 ∀n ∈ ℕ. If one can choose a value of log(1 + un ) for each n ∈ ℕ ∞ such that ∑∞ n=1 log(1 + un ) is convergent, then the product ∏n=1 (1 + un ) is convergent.

4.6 The Weierstraß Factorization Theorem

| 137

Proof. We use the functional equation and the continuity of the exponential function. Since ∑∞ n=1 log(1 + un ) is convergent, we have N

N

N

lim ∏(1 + un ) = lim ∏ exp log(1 + un ) = lim exp( ∑ log(1 + un ))

N→∞

n=1

N→∞

N→∞

n=1 ∞

n=1

= exp( ∑ log(1 + un )). n=1

Lemma 4.31. Let Log be the principle branch of the logarithm. Then ∑∞ n=1 Log(1 + un ) ∞ converges absolutely if and only if ∑n=1 un converges absolutely. Proof. We claim that for |u| ≤ 1/2 one has |u| 3|u| ≤ |Log(1 + u)| ≤ , 2 2 from this one gets the assertion of the lemma. For |u| < 1 one has Log(1 + u) = u − u2 /2 + u3 /3 − ⋯, hence |1 −

Log(1 + u) | = |1 − 1 + u/2 − u2 /3 + ⋯| u |u|

≤

|u| (1 + |u| + |u|2 + ⋯) = 2 2 1 − |u|

since |u| < 1.

If |u| ≤ 1/2, then |1 −

1 Log(1 + u) |≤ . u 2

Remark. Let fn ∈ ℋ(U), n ∈ ℕ. The product ∏∞ n=1 (1 + fn ) converges absolutely and uniformly on each compact subset K of U, if for each compact subset K of U there exists n0 ∈ ℕ such that ∑∞ n=n0 Log(1 + fn ) converges absolutely and uniformly on K. The limit function of the product is again holomorphic on U, see Lemma 4.30 and Lemma 4.31. Lemma 4.32. Let U ⊆ ℂ be open and suppose that the points a, b belong to one and the same connected component of ℂ ⧵ U. Then there exists f ∈ ℋ(U) such that exp(f (z)) =

z−a , z−b

z ∈ U,

i.e. there exists a holomorphic logarithm on z ↦ (z − a)/(z − b) on U. Proof. Let Gk , k ∈ ℕ, denote the connected components of U. The set U is the union of the domains Gk . If a, b belong to one and the same connected component of ℂ ⧵ U, then a, b belong to one and the same connected component of ℂ ⧵ Gk ∀k. We consider the function

138 | 4 Construction and approximation of holomorphic functions z−a on Gk . In order to show that g has a holomorphic logarithm on Gk , it suffices g(z) = z−b to show that g ′ /g has a primitive on Gk , see Theorem 2.21. By Theorem 1.39, it suffices to show that

∫

γ

g ′ (z) dz = 0, g(z)

for each closed path γ in Gk . We have 1 z−a 1 1 g ′ (z) z − b = ( − − )= 2 g(z) z − a z − b (z − b) z−a z−b hence ∫

γ

g ′ (z) dz dz dz = ∫ −∫ = 2πi(Indγ (a) − Indγ (b)). g(z) z − a z γ γ −b

a, b belong to one and the same connected component of ℂ ⧵ Gk and therefore also to one and the same connected component of ℂ ⧵ γ ∗ . By Theorem 2.2, the index is locally constant, hence Indγ (a) − Indγ (b) = 0. This gives a function fk ∈ ℋ(Gk ) such that exp(fk (z)) =

z−a , z−b

z ∈ Gk .

Now define f ∈ ℋ(U) by f (z) = fk (z), z ∈ Gk , k ∈ ℕ. Now we are ready to construct holomorphic functions with prescribed zeroes of prescribed order. Theorem 4.33 (Weierstraß Factorization Theorem). Let Ω ⊆ ℂ be open and let (zj )j be a discrete sequence of points in Ω, being different from each other. Furthermore, let nj ∈ ℤ, j ∈ ℕ. Then there exists a meromorphic function f on Ω, which is holomorphic and ≠ 0 except at the points zj , and such that for all j ∈ ℕ the function f (z)(z − zj )−nj is holomorphic in a neighborhood of zj and does not vanish there. This means f has a zero of order nj at the points zj , when nj > 0, and f has a pole of order |nj | at the points zj , when nj < 0. Remark. Before we start with the proof, we recall some properties of holomorphically convex sets which will be important for the construction of the desired meromorphic function. Let K ⊂ Ω be a compact set such that K̂ Ω = K. Then we have the following assertions: (1) ∀z ∈ 𝜕Ω there exists a neighborhood U(z) of z and a connected component U of Ω ⧵ K such that U ∩ U(z) ≠ ∅.

4.6 The Weierstraß Factorization Theorem | 139

(2) For each connected component U of Ω⧵K there exists z0 ∈ 𝜕Ω such that U ∩U(z0 ) ≠ ∅, for each neighborhood U(z0 ) of z0 . (Assume this is not the case, then U must be relatively compact in Ω.) Proof of Theorem 4.33. Consider the product ∏(1 − j

z nj ) . zj

We will construct holomorphic factors which do not vanish on Ω and generate convergence of the infinite product. ̂ = Kj , j ∈ ℕ. Let (Kj )j be a compact exhaustion of Ω such that (Kj ) Ω We will choose successively rational functions fj having the prescribed zeros and poles in Kj and we will choose functions gj ∈ ℋ(Ω) such that |

fj+1 fj

exp(gj ) − 1| < ϵj

(4.11)

on Kj , where ϵj > 0, j ∈ ℕ and ∑∞ j=1 ϵj < ∞. Suppose we have already found f1 , … , fj and g1 , … , gj−1 with the desired properties. Let h(z) = ∏(z − zν )mν , ν

for finitely many zν ∈ {zj ∶ j ∈ ℕ} and mν ∈ {nj ∶ j ∈ ℕ}, be a rational function with the prescribed zeros and poles in Kj+1 (we have only finitely many points zν ∈ Kj+1 ). Then we can write h(z) = c ∏ (z − wν )mν fj (z) ν∈M

(4.12)

where c ∈ ℂ, c ≠ 0 and M ⊂ ℕ is a finite set, wν ∈ Kjc and wν ∈ {zj ∶ j ∈ ℕ}, for ν ∈ M. Since no connected component of Ω ⧵ Kj is relatively compact in Ω, it follows from c Remark 4.6 that for each ν ∈ M one can find a point wν′ ∈ Kj+1 which belongs to the same connected component of Ω ⧵ Kj as wν . Recall that each connected component of Ω ⧵ Kj has points close to 𝜕Ω. Let wν be a point in a connected component of Ω ⧵ Kj . This component comes close to 𝜕Ω, but in this vicinity there is also a connected component of Ω ⧵ Kj+1 , and we choose wν′ in this connected component of Ω ⧵ Kj+1 . Now we set fj+1 (z) = h(z) ∏ (z − wν′ )−mν . ν∈M

Then fj+1 is a rational function and has the prescribed zeros and poles in Kj+1 . By (4.12), we get fj+1 (z) fj (z)

=c

h(z) z − wν mν ) . ∏ (z − wν′ )−mν = c ∏ ( ′ fj (z) ν∈M ν∈M z − wν

140 | 4 Construction and approximation of holomorphic functions The last function is holomorphic in a neighborhood of Kj . The points wν , wν′ belong to one and the same connected component of Ω ⧵ Kj , hence, by Lemma 4.32, there exists a holomorphic logarithm of c ∏( ν∈M

z − wν mν ) z − wν′

in a neighborhood of Kj . We set exp(ϕν (z)) = (

z − wν mν ) , z − wν′

ν∈M

and write ϕν (z) = mν log(ν)

z − wν . z − wν′

Now we set log(fj+1 (z)/fj (z)) ∶= log c + ∑ mν log(ν) ν∈M

z − wν , z − wν′

and observe that this function is holomorphic in a neighborhood of Kj . Hence, by Theorem 4.15, there exists gj ∈ ℋ(Ω) such that |log(fj+1 /fj ) + gj | < log(1 + ϵj ) on Kj . From this inequality we obtain (4.11). Now we set N

∞

j=1

j=1

f = lim fN+1 ∏ exp(gj ) = f1 ∏[(fj+1 /fj ) exp(gj )]. N→∞

By Lemma 4.31, the product ∏∞ j=l [(fj+1 /fj ) exp(gj )] converges uniformly on Kl to a function being holomorphic in the interior of Kl which does not vanish there. The factor l−1

f1 ∏[(fj+1 /fj ) exp(gj )] j=1

has the prescribed zeros and poles in Kl . Remark 4.34 (Classical version of the Weierstraß Factorization Theorem). Let Ω = ℂ and 0 = |a0 | < |a1 | < ⋯, limj→∞ |aj | = ∞, nj ∈ ℕ. We want to construct an entire function u with prescribed zeros in aj of order nj , j ∈ ℕ.

4.6 The Weierstraß Factorization Theorem

| 141

For this purpose we first solve the Mittag-Leffler problem for the principle parts nj /(z − aj ) and obtain a meromorphic function h ∈ ℳ(ℂ) with poles in aj and the corresponding principle parts nj /(z − aj ), j ∈ ℕ. By Example 4.25, the function h can be written as ∞

h(z) = ∑ nj [ j=1

k

1 1 j z k + ∑ ( ) ] = ∑ hj (z), z − aj aj k=0 aj j=1 ∞

where the indices kj are chosen in such a way that the series converges uniformly on compact subsets which contain no poles. Now let k

n

j j z z k+1 1 ) exp( ∑ ( ) )] . aj k=0 k + 1 aj

uj (z) = [(1 −

A simple computation shows that u′j /uj = hj . We set ∞

u(z) = z n0 ∏ uj (z). j=1

Let R > 0 and choose j0 so large that |aj | > R for j ≥ j0 . Then uj has no zeros in DR (0) for j ≥ j0 . Let z ∈ DR (0) and set vj (z) = ∫

u′j (ζ )

[0,z]

uj (ζ )

dζ = ∫

[0,z]

hj (ζ ) dζ .

Then exp(vj ) = uj (see Theorem 2.22) and ∑ vj = ∑ log uj .

j≥j0

j≥j0

By construction, the series ∑j≥j hj converges uniformly on DR (0) and the same is true 0 for ∑j≥j vj and also for ∑j≥j log uj . 0 0 By Lemma 4.30, we get uniform convergence of ∏j≥j uj on DR (0). 0 Hence ∞

u(z) = z n0 ∏ uj (z) j=1

is the desired entire function. Example 4.35. We are looking for an entire function, which has simple zeroes at the integer points. From (4.9) we get that z

z ∏ (1 − ) ez/k k k∈ℤ⧵{0}

142 | 4 Construction and approximation of holomorphic functions is convergent and has the desired properties. As the function sin(πz) has the same zeros, there exists an entire function g such that z sin(πz) = z ∏ (1 − ) ez/k eg(z) . k k∈ℤ⧵{0} Taking the logarithmic derivative and using the partial fraction decomposition of π cot(πz) (see (4.10)), it follows that g ′ (z) = 0 and so that g is constant. Since lim

z→0

sin(πz) = π, z

we have eg(z) = π and we obtain z sin(πz) = πz ∏ (1 − ) ez/k , k k∈ℤ⧵{0} which can also be written in the form ∞

sin πz = πz ∏(1 − j=1

z2 ). j2

Similarly, one can show that cos πz = ∏(1 − j∈ℤ

z z/aj )e , aj

where aj = j + 1/2, j ∈ ℤ.

4.7 Some applications of the Mittag-Leffler and Weierstraß Theorems First we solve an interpolation problem for holomorphic functions taking preassigned values in a discrete sequence of interpolation points. Theorem 4.36 (Interpolation). Let Ω ⊆ ℂ be open and let (aj )j be a discrete sequence of points in Ω being different from each other and let (bj )j be an arbitrary sequence of complex numbers. Then there exists a function f ∈ ℋ(Ω) such that f (aj ) = bj , j ∈ ℕ. Proof. By Theorem 4.33, there exists a function g ∈ ℋ(Ω) such that g(aj ) = 0, j ∈ ℕ, where all zeros have order 1, i.e. g ′ (aj ) ≠ 0. Now we solve the Mittag-Leffler problem for the principle parts bj 1 , g ′ (aj ) z − aj

j ∈ ℕ.

4.7 Some applications of the Mittag-Leffler and Weierstraß Theorems | 143

For the solution h to this problem, we have h ∈ ℋ(Ω ⧵ {aj ∶ j ∈ ℕ}). Now we set f = gh, then f ∈ ℋ(Ω ⧵ {aj ∶ j ∈ ℕ}) and in the points aj we have f (aj ) = lim g(z)h(z) = lim [ z→aj

z→aj

g(z) − g(aj ) z − aj

(z − aj )h(z)] = g ′ (aj )

bj ′ g (a

j)

= bj ,

hence f has a removable singularity in aj , j ∈ ℕ. Next we show that a meromorphic function can globally be written as the quotient of two holomorphic functions. Theorem 4.37. Let Ω ⊆ ℂ be open and h ∈ ℳ(Ω) a meromorphic function. Then there exist f , g ∈ ℋ(Ω) such that h = f /g on Ω (compare with Theorem 2.65). Proof. Let (aj )j be the poles of h with the orders nj . By Theorem 4.33, there exists a function g ∈ ℋ(Ω) with zeroes of order nj in the points aj . Now define f ∶= hg. Then f ∈ ℋ(Ω) and h = f /g. The next result is specific for the complex analysis of one variable, it has no analogue in several variables. It states that each domain has holomorphic functions which cannot be extended holomorphically to a larger domain. Theorem 4.38. Let G ⊆ ℂ be a domain.Then there exists a holomorphic function f ∈ ℋ(G) such that each boundary point of G is singular for f , i.e. f cannot be extended along any curve from the interior of G to the boundary of G. Proof. First we construct a sequence (aj )j which has limit points close to each boundary point of 𝜕G and then we will find a function f ∈ ℋ(G) with f ≢ 0 and f (aj ) = 0, j ∈ ℕ. Finally, we will show that this function has the desired properties. For this aim we construct a sequence of open discs (Dn )n having the following properties: Dn ⊂ G; (Dn )n is locally finite (i.e. each compact subset of G has non-empty intersection with only finitely many Dn ); ⋃∞ n=1 Dn = G; the radii rn of Dn tend to 0 as n → ∞. Let (Kp )p be a compact exhaustion of G. We choose the discs Dn in the following way: K1 ⊂ D1 ∪ ⋯ ∪ Dn1 and Kp+1 ⧵ K̊ p ⊂ Dnp +1 ∪ ⋯ ∪ Dnp+1 ; furthermore, Dn ∩ Kp = ∅ for n ≥ 1 + np+1 . Let the radius of Dn be < 1/p for np < n ≤ np+1 , p ∈ ℕ. Existence of such discs follows from the fact that the sets Kp+1 ⧵ K̊ p are compact. Choose points aj ∈ Dj such that aj ≠ ak for j ≠ k. The sequence (aj )j has no limit point in G since (Dn )n is locally finite. By Theorem 4.33, there exists a function f ∈ ℋ(G) such that f (aj ) = 0, j ∈ ℕ, and f ≠ 0 elsewhere. We claim that this function f has the desired properties.

144 | 4 Construction and approximation of holomorphic functions Suppose there exists a point a ∈ 𝜕G and a path γ ∶ [0, 1] ⟶ G with γ([0, 1)) ⊂ G and γ(1) = a, such that there exists F ∈ ℋ(Dρ (a)), ρ > 0, which arose from analytic continuation of f along γ (see Definition 3.7). Then we can find ϵ > 0 such that γ(t) ∈ Dρ (a) for 1 − ϵ ≤ t ≤ 1. Let U be the connected component of G ∩ Dρ (a), which contains the set {γ(t) ∶ 1 − ϵ ≤ t < 1}. We have ′ F|U = f |U . Denote D′ = Dρ/2 (a). By construction we have D′ ∩ U ⊆ ⋃∞ n=1 Dn , but D ∩ U ⊄ N ⋃n=1 Dn ∀N ∈ ℕ. Hence there exists a sequence (nk )k of natural numbers such that Dnk ∩ (D′ ∩ U) ≠ ∅, we can even assume that the radii of the discs Dnk are all < ρ/8, for this purpose we possibly have to take a further subsequence. Then Dnk ⊆ Dρ (a) ∩ G ∀k and Dnk ∩ U ≠ ∅. Since U and Dnk are connected, we must have Dnk ⊆ U ∀k. Since F|U = f |U , we also have F(ank ) = 0 ∀k. Finally, we get ank ∈ Dnk ⊆ D3ρ/4 (a) ⊂⊂ Dρ (a). Hence F has infinitely many zeroes in a relatively compact subset of Dρ (a), there exists a limit point of the zeroes of F in Dρ (a) and, by Theorem 2.18, F ≡ 0 on Dρ (a). This implies f ≡ 0 on U and, again by Theorem 2.18, f ≡ 0 on G, which yields a contradiction to the construction of f . Remark. For holomorphic functions on ℂ2 we have a completely different result. Let ‖z‖ = (|z1 |2 + |z2 |2 )1/2 and G = {z = (z1 , z2 ) ∶ 1/2 < ‖z‖ < 1}. Then G is a domain in ℂ2 with the following properties: for each f ∈ ℋ(G) there exists F ∈ ℋ(U), U = {z = (z1 , z2 ) ∶ ‖z‖ < 1}, with F|G = f |G , see Example 6.12. Domains in ℂn for which there exist holomorphic functions with the properties in Theorem 4.38 are called domains of holomorphy. Theorem 4.38 says that each domain in ℂ is a domain of holomorphy (see [59] or [51]).

4.8 Normal families The concept of a normal family arises from the aim to find an appropriate description of compact subsets in the space ℋ(Ω) endowed with the topology of uniform convergence on compact subsets of Ω. Montel’s Theorem, the main result of this section, corresponds to the Heine–Borel Theorem which characterizes compact subsets of ℝn . Montel’s Theorem will be crucial for the proof of the Riemann Mapping Theorem in the next section. Definition 4.39. Let Ω ⊆ ℂ be a domain and ℱ a family of functions in ℋ(Ω). The family ℱ is called normal in Ω, if each sequence of elements of ℱ has a subsequence, which is uniformly convergent on the compact subsets of Ω.

4.8 Normal families |

145

A family ℱ is a bounded family on Ω, if for each compact subset K ⊂ Ω there exists a constant M(K) > 0 such that sup|f (z)| ≤ M(K) z∈K

∀f ∈ ℱ.

Remark. Normality corresponds to the topological concept of sequential compactness. Boundedness for a family in ℋ(Ω) is the appropriate concept of a bounded subset in metrizable topological vector spaces; see Chapter 9. Theorem 4.40 (Montel’s6 Theorem). Let Ω ⊆ ℂ be a domain. Then each bounded family is normal. Proof. Let (Kn )n be a compact exhaustion of Ω. We can find δn > 0 such that D2δn (z) ⊆ Kn+1 ∀z ∈ Kn , n ∈ ℕ. Let z ′ , z ″ ∈ Kn be such that |z ′ − z ″ | < δn ; furthermore, let γ(t) = z ′ + 2δn e2πit , t ∈ [0, 1]. For arbitrary f ∈ ℱ, we have f (z ′ ) − f (z ″ ) = =

f (ζ ) 1 f (ζ ) 1 dζ − dζ ∫ ∫ ′ 2πi γ ζ − z 2πi γ ζ − z ″ 1 (z ′ − z ″ )f (ζ ) dζ . ∫ 2πi γ (ζ − z ′ )(ζ − z ″ )

(4.13)

If ζ ∈ γ ∗ , then |ζ − z ′ | = 2δn and |ζ − z ″ | > δn . By assumption we have sup |f (z)| ≤ M(Kn+1 )

z∈Kn+1

∀f ∈ ℱ,

and we get from (4.13) |f (z ′ ) − f (z ″ )| ≤

1 1 1 M(Kn+1 ) ′ 4πδn |z ′ − z ″ |M(Kn+1 ) = |z − z ″ |, 2π 2δn δn δn

(4.14)

this is true for z ′ , z ″ ∈ Kn with |z ′ − z ″ | < δn and for each f ∈ ℱ. We have shown that ℱ is equicontinuous on Kn , i.e. ∀ϵ > 0 ∃δ > 0 with |f (z ′ ) − f (z ″ )| < ϵ ∀f ∈ ℱ, if z ′ , z ″ ∈ Kn and |z ′ − z ″ | < δ. For a given ϵ > 0, we take δ=

ϵδn , M(Kn+1 )

then (4.14) implies |f (z ′ ) − f (z ″ )|
0. Choose δ > 0 like in the first part of the proof. Since Kn is compact, we can find points z1 , … , zp ∈ Kn ∩ E such that Kn ⊂ ⋃pj=1 Dδ (zj ). Furthermore, there exists N > 0 such that |fr,r (zj ) − fs,s (zj )| < ϵ,

j = 1, … , p; r, s > N,

which follows from the construction of the diagonal sequence. For each z ∈ Kn , there exists a point zj ∈ E (1 ≤ j ≤ p) such that |z − zj | < δ, and we obtain |fr,r (z) − fs,s (z)| ≤ |fr,r (z) − fr,r (zj )| + |fr,r (zj ) − fs,s (zj )| + |fs,s (zj ) − fs,s (z)| < 3ϵ, ∀z ∈ Kn and r, s > N, where we used the equicontinuity in the first and third term. Hence (fm,m )m is a uniform Cauchy sequence on Kn , which converges uniformly on Kn by Theorem 2.34.

4.9 The Riemann Mapping Theorem We are now ready to prove that there exists a biholomorphic mapping ϕ ∶ Ω ⟶ D1 (0), where Ω ⊊ ℂ is a simply connected domain. To find conditions implying uniqueness of this mapping, we need the following lemma. Lemma 4.41 (Schwarz7 Lemma). Let f ∈ ℋ(D1 (0)) and supz∈D1 (0) |f (z)| ≤ 1, as well as f (0) = 0. Then we have: (a) |f (z)| ≤ |z| ∀z ∈ D1 (0); (b) |f ′ (0)| ≤ 1. If we have equality in (a) for some z ∈ D1 (0) ⧵ {0}, or if we have equality in (b), then f (z) = λz for some λ ∈ ℂ with |λ| = 1. 7 Schwarz, Hermann Amandus (1843–1921).

4.9 The Riemann Mapping Theorem

| 147

Proof. Define g(z) = f (z)/z, z ∈ D1 (0) ⧵ {0}. Since f (0) = 0, the function g has a removable singularity in 0 (see Theorem 2.24), and f ′ (0) = lim

z→0

f (z) − f (0) f (z) = lim = g(0). z→0 z z

For each ϵ > 0 and z ∈ D1−ϵ (0), we have |g(z)| ≤ max |g(z)| = max | z∈D1−ϵ (0)

z∈𝜕D1−ϵ (0)

f (z) 1 |≤ , z 1−ϵ

where we have used the maximum principle. Since ϵ > 0 was arbitrary, we get |g(z)| ≤ 1 ∀z ∈ D1 (0), and |f (z)| ≤ |z| ∀z ∈ D1 (0). Furthermore, we have |f ′ (0)| = |g(0)| ≤ 1. If there exists z0 ∈ D1 (0) ⧵ {0} such that |f (z0 )| = |z0 |, then |g(z0 )| = 1 and, again by the maximum principle, we get g(z) = λ, where λ is a constant with |λ| = 1. If |f ′ (0)| = 1, then |g(0)| = 1, and the maximum principle yields the same result as before. Before we prove the Riemann Mapping Theorem, we investigate holomorphic automorphisms of the unit disc. Theorem 4.42. Let α ∈ D1 (0) and ϕα (z) =

z−α , 1 − αz

z ∈ D1 (0).

Then ϕα maps 𝕋 bijectively onto 𝕋. Furthermore, ϕα (D1 (0)) = D1 (0), ϕα is injective on D1 (0) and ϕα (α) = 0. The inverse of ϕα is ϕ−α . And we have ϕ′α (0) = 1 − |α|2 as well as ϕ′α (α) = 1/(1 − |α|2 ). Remark. Each holomorphic automorphism of the unit disc has the form λϕα , where λ ∈ ℂ is a constant with |λ| = 1 and α ∈ D1 (0). This follows from the Schwarz lemma (see also Exercises). Proof of Theorem 4.42. ϕα is holomorphic except for a pole at the point 1/α ∉ D1 (0). One has ϕ−α (ϕα (z)) =

z−α 1−αz

+α

z−α 1 + α 1−αz

=

z − α + (1 − αz)α z − |α|2 z = = z, 1 − αz + αz − αα 1 − |α|2

hence ϕα is injective on D1 (0) and has the inverse ϕ−α . Let z0 ∈ 𝕋, z0 = eit , t ∈ ℝ. Then |ϕα (eit )| = |

eit − α eit − α | = | | = 1, 1 − eit α e−it − α

hence ϕα (𝕋) ⊆ 𝕋 and ϕ−α (𝕋) ⊆ 𝕋, so we have ϕα (𝕋) = 𝕋. Now it follows from the maximum principle that ϕα (D1 (0)) ⊆ D1 (0) as well as ϕ−α (D1 (0)) ⊆ D1 (0) and therefore ϕα (D1 (0)) = D1 (0). The assertions about the derivative of ϕα are easy to check.

148 | 4 Construction and approximation of holomorphic functions Theorem 4.43 (Riemann Mapping Theorem). Let Ω ⫋ ℂ be a simply connected domain. Then there exists a biholomorphic mapping h ∶ Ω ⟶ D1 (0), i.e. h ∈ ℋ(Ω) and h−1 ∈ ℋ(D1 (0)). Proof. Since Ω ≠ ℂ, there exists w0 ∈ ℂ, w0 ∉ Ω. Let Σ = {ψ ∈ ℋ(Ω) ∶ ψ ∶ Ω ⟶ D1 (0) injective}. We will show that there exists a ψ ∈ Σ, which is also surjective. (a) Suppose Σ ≠ ∅. Let f (z) = z − w0 . Then f (z) ≠ 0 ∀z ∈ Ω and the function z ↦ f ′ (z)/f (z) = 1/(z − w0 ) is holomorphic on Ω. Since Ω is simply connected, we can use Corollary 2.56 to get ∫

γ

f ′ (z) dz = 0 f (z)

for each closed path γ in Ω. By Theorem 1.39, the function f ′ /f has a primitive on Ω and by Theorem 2.21, the function f has a holomorphic square root ϕ on Ω, i.e. ϕ2 (z) = z − w0 ∀z ∈ Ω. We claim that ϕ is injective. Indeed, if ϕ(z1 ) = ϕ(z2 ), then ϕ2 (z1 ) = ϕ2 (z2 ), hence z1 − w0 = z2 − w0 , and z1 = z2 . ϕ is non-constant. Hence, by Theorem 2.38, ϕ is an open mapping. Furthermore, we have ϕ ≠ 0 on Ω. Let z0 ∈ Ω and ϕ(z0 ) = a ≠ 0. There exists r > 0 with 0 < r < |a| such that Dr (a) ⊆ ϕ(Ω). Now we define ψ(z) = r/(ϕ(z) + a), z ∈ Ω. If ϕ(z) = −a for some z ∈ Ω, then ϕ2 (z) = 2 a = ϕ2 (z0 ), and we have z = z0 , which is a contradiction. So ψ ∈ ℋ(Ω) and ψ is injective, since ϕ is. Next we claim that Dr (−a) ∩ ϕ(Ω) = ∅. Suppose ∃ζ ∈ ϕ(Ω) such that |ζ + a| < r. Then there exists z1 ∈ Ω with ϕ(z1 ) = ζ and |ϕ(z1 ) + a| < r, hence | − ϕ(z1 ) − a| < r and so −ϕ(z1 ) ∈ Dr (a) ⊆ ϕ(Ω). But there are no two points z1 , z2 ∈ Ω such that ϕ(z1 ) = −ϕ(z2 ). Contradiction! So we have |ϕ(z) + a| > r ∀z ∈ Ω and |ψ(z)| =

r r < =1 |ϕ(z) + a| r

∀z ∈ Ω,

this means that ψ ∶ Ω ⟶ D1 (0) and ψ ∈ Σ. We remark that for ψ ∈ Σ with ψ(ζ0 ) = α, the composition ϕα ∘ ψ ∈ Σ (see Theorem 4.42) and one has ϕα (ψ(ζ0 )) = ϕα (α) = 0. Hence Σ contains functions having a zero at a given point ζ0 ∈ Ω. (b) Fix z0 ∈ Ω and define Σ0 = {ψ ∈ ℋ(Ω) ∶ ψ(z0 ) = 0, ψ ∶ Ω ⟶ D1 (0) injective}.

4.9 The Riemann Mapping Theorem

| 149

Let ψ ∈ Σ0 and suppose that ψ(Ω) ≠ D1 (0). We claim that there exists a ψ1 ∈ Σ0 such that |ψ′1 (z0 )| > |ψ′ (z0 )|. For this aim let ψ ∈ Σ0 and α ∈ D1 (0) such that α ∉ ψ(Ω). Take the function ϕα as in Theorem 4.42. Then we have ϕα ∘ ψ ∈ Σ and since ϕα vanishes only at z = α, the function ϕα ∘ ψ has no zero in Ω. As in (a) one shows that there exists a g ∈ ℋ(Ω) such that g 2 = ϕα ∘ ψ. The function g is again injective so that g ∈ Σ. Let β = g(z0 ) and ψ1 = ϕβ ∘ g. Then we have ψ1 ∈ Σ0 , since ψ1 (z0 ) = ϕβ (g(z0 )) = ϕβ (β) = 0. Consider the square function s(w) = w2 . Now we have ψ = ϕ−α ∘ s ∘ g = ϕ−α ∘ s ∘ ϕ−β ∘ ψ1 = F ∘ ψ1 , where we took F = ϕ−α ∘ s ∘ ϕ−β . By the chain rule, we have ψ′ (z0 ) = F ′ (ψ1 (z0 ))ψ′1 (z0 ) = F ′ (0)ψ′1 (z0 ). Furthermore, we have F(D1 (0)) ⊆ D1 (0) and F(0) = F(ψ1 (z0 )) = ψ(z0 ) = 0; since the function s is involved we know that F fails to be injective. We can apply the Schwarz Lemma 4.41 to get that |F ′ (0)| < 1. This implies |ψ′ (z0 )| < |ψ′1 (z0 )| and we have proved our claim. We remark that ψ′ (z0 ) ≠ 0, since ψ is injective (see Theorem 2.42). (c) Fix z0 ∈ Ω and let η = sup{|ψ′ (z0 )| ∶ ψ ∈ Σ0 }. We claim that there exists an h ∈ Σ0 such that η = |h′ (z0 )|. By (b), such a function h must be surjective. For this claim we will use Montel’s Theorem (see Theorem 4.40). The family Σ0 is bounded, since |ψ(z)| < 1 ∀z ∈ Ω ∀ψ ∈ Σ0 . Hence, by Definition 4.39, Σ0 is normal. From the definition of η, we get a sequence (ψn )n in Σ0 such that lim |ψ′n (z0 )| = η.

n→∞

Hence there exists a subsequence, again denoted by (ψn )n which converges uniformly on all compact subsets of Ω to a function h ∈ ℋ(Ω) and |h′ (z0 )| = η. We have η > 0 hence h is non-constant. So h is an open mapping, in particular, h(Ω) is open. Since ψn (Ω) ⊆ D1 (0) ∀n ∈ ℕ, we have h(Ω) ⊆ D1 (0) and even h(Ω) ⊆ D1 (0), as h(Ω) is open. It remains to show that h is injective. Let z1 , z2 ∈ Ω, z1 ≠ z2 . Let α = h(z1 ) and αn = ψn (z1 ). By Theorem 2.18, there exists a disc D ⊂ Ω with center z2 , such that z1 ∉ D and such that h − α has no zero on the boundary 𝜕D. Since ψn − αn converges uniformly on D to h − α, we have |ψn − αn − (h − α)| < |h − α| on 𝜕D, if n is large enough. Since all functions ψn are injective, the function ψn − αn has no zeroes in D. By Theorem 2.71, we know that the number of zeroes of ψn − αn in D coincides with the number of zeroes of h − α in D. Hence h − α has no zeroes in D and h(z1 ) ≠ h(z2 ).

150 | 4 Construction and approximation of holomorphic functions Corollary 4.44. Let Ω ⫋ ℂ be a simply connected domain, and let z0 ∈ Ω. Then there exists a biholomorphic mapping g ∶ Ω ⟶ D1 (0) such that g(z0 ) = 0 and such that g ′ (z0 ) is real and g ′ (z0 ) > 0. The mapping g is uniquely determined by the above properties. Proof. The existence of g follows immediately from the proof of Theorem 4.43, by multiplication with a suitable λ ∈ ℂ, |λ| = 1 one can attain that g ′ (z0 ) > 0. To show uniqueness, we apply the Schwarz Lemma (see Lemma 4.41): let h be another function with the above properties. Define ϕ = h ∘ g −1 ∶ D1 (0) ⟶ D1 (0). Then ϕ(0) = h(g −1 (0)) = h(z0 ) = 0. By Lemma 4.41, we get |ϕ(z)| ≤ |z|

∀z ∈ D1 (0).

Let z = g(w). Then |h(g −1 (g(w)))| ≤ |g(w)| and |h(w)| ≤ |g(w)|. Interchanging the roles of g and h, one gets |g(w)| ≤ |h(w)|. Hence |h(w)| = |g(w)| ∀w ∈ D1 (0), therefore |ϕ(z)| = |z| on D1 (0) and, again by Lemma 4.41, ϕ(z) = λz for some λ ∈ ℂ with |λ| = 1. Furthermore, we have ϕ′ (0) = h′ (g −1 (0))(g −1 )′ (0) = h′ (z0 )

1 > 0. g ′ (z0 )

Since ϕ′ (z) = λ, we finally have λ = 1, and ϕ(z) = z on D1 (0). Hence h = g. Remark. A holomorphic function f ∶ D1 (0) ⟶ ℂ is called univalent, if f is injective, f ′ (0) = 1 and f (0) = 0. The Taylor series expansion of a univalent function f has the form f (z) = z + a2 z 2 + a3 z 3 + ⋯ . Bieberbach’s conjecture (1914) was that |an | ≤ n ∀n ∈ ℕ. Bieberbach8 himself showed that |a2 | ≤ 2. K. Löwner proved |a3 | ≤ 3 and |a4 | ≤ 4. Bieberbach’s conjecture was proved by Louis De Branges in 1984; for a proof, see [40]. Example 4.45. Let H = {z ∶ ℑz > 0} the upper half-space. The functions h(z) = eit

z−i , z+i

t∈ℝ

are biholomorphic mappings from the upper half-space onto D1 (0) such that h(i) = 0. 8 Bieberbach, Ludwig Moses (1886–1982).

4.10 Characterization of simply connected domains | 151

The proof of the Riemann mapping theorem is not constructive, which means that it is impossible to find nice formulas in most cases. There are powerful approximation methods which have interesting applications, for instance, in fluid mechanics; see [40]. Later we will discuss another approach using Hilbert space methods and the Bergman kernel.

4.10 Characterization of simply connected domains Using the whole theory developed so far, we can now characterize simply connected domains – a purely topological concept – by properties of holomorphic functions. Theorem 4.46. Let Ω ⊆ ℂ be a domain. The following assertions are equivalent: (a) Ω is homeomorphic to D1 (0), i.e. there exists a bijective mapping ϕ ∶ Ω ⟶ D1 (0) which is continuous in both directions; (b) Ω is simply connected; (c) Indγ (α) = 0 for each closed path γ in Ω and for each α ∈ ℂ ⧵ Ω; (d) ℂ ⧵ Ω is connected; (e) Each f ∈ ℋ(Ω) can be approximated by polynomials uniformly on all compact subsets, i.e. for each compact subset K ⊂ Ω and for each ϵ > 0 there exists a polynomial P such that |f − P|K < ϵ; (f) ∀f ∈ ℋ(Ω) and for each closed path γ in Ω one has ∫ f (z) dz = 0; γ

(g) ∀f ∈ ℋ(Ω) ∃F ∈ ℋ(Ω) such that F ′ = f , i.e. f has a primitive on Ω; (h) ∀f ∈ ℋ(Ω) such that f (z) ≠ 0 ∀z ∈ Ω there exists g ∈ ℋ(Ω) such that eg = f , i.e. f has a holomorphic logarithm on Ω; (i) ∀f ∈ ℋ(Ω) such that f (z) ≠ 0∀z ∈ Ω there exists h ∈ ℋ(Ω) such that h2 = f , i.e. f has a holomorphic root on Ω. Proof. (a) ⟹ (b). Let ϕ ∶ Ω ⟶ D1 (0) be a homeomorphism and let γ a closed curve in Ω. Define H(s, t) = ϕ−1 (tϕ(γ(s))),

s, t ∈ [0, 1].

Then H ∶ [0, 1] × [0, 1] ⟶ Ω is continuous and one has that H(s, 0) = ϕ−1 (0), s ∈ [0, 1] is constant and H(s, 1) = γ(s), s ∈ [0, 1]; furthermore, H(0, t) = H(1, t), t ∈ [0, 1]. Hence γ is null-homotopic and Ω is simply connected. (b) ⟹ (c). Theorem 2.55. (c) ⟹ (d). Suppose that ℂ ⧵ Ω is not connected. Then there are two non-empty closed, disjoint subsets H, K of ℂ ⧵ Ω such that H ∪ K = ℂ ⧵ Ω. If ∞ ∈ H, then K is

152 | 4 Construction and approximation of holomorphic functions compact in ℂ. Let W = ℂ ⧵ H. Then W = Ω ∪ K. By Lemma 2.59, there exists a cycle Γ in W ⧵ K = Ω such that IndΓ (α) = 1 ∀α ∈ K, which yields a contradiction to (c). (d) ⟹ (e). Corollary 4.21. (e) ⟹ (f). Let f ∈ ℋ(Ω), γ be a closed path in Ω, and let (pn )n be a sequence of polynomials, which converges to f uniformly on γ ∗ . Polynomials always have primitives. Hence ∫ pn (z) dz = 0 γ

∀n ∈ ℕ,

which implies ∫ f (z) dz = 0. γ

(f) ⟹ (g). Theorem 1.39. (g) ⟹ (h). If f ∈ ℋ(Ω) and f (z) ≠ 0 on Ω, then f ′ /f ∈ ℋ(Ω) and f ′ /f has a primitive on Ω. Now use Theorem 2.21. (h) ⟹ (i). Trivial. (i) ⟹ (a). If Ω = ℂ, then ϕ(z) =

z , 1 + |z|

z∈ℂ

gives the desired homeomorphism from ℂ to D1 (0). Finally, let Ω ≠ ℂ. In the proof of Theorem 4.43, we only use assertion (i) to get the biholomorphic mapping ϕ ∶ Ω ⟶ D1 (0).

4.11 Exercises 88. Let Ω = {z ∶ |z| < 1 and |2z − 1| > 1}, and let f ∈ ℋ(Ω). (1) Is there a sequence of polynomials Pn such that Pn → f uniformly on all compact subsets of Ω? (2) Is there such a sequence converging to f uniformly on Ω? (3) Is the answer for (2) different, if it is assumed that f is holomorphic in an open set containing the closure of Ω? 89. Show that there exists a sequence (pn )n of polynomials such that {1, lim pn (z) = { 0, {

n→∞

z ∈ ℝ, z ∉ ℝ.

Hint: Use Corollary 4.16 for the compact sets Kn = {z = x + iy ∶ |z| ≤ n, y = 0 or 1/n ≤ |y| ≤ n}

4.11 Exercises | 153

and consider the characteristic functions 1Un of the open sets Un = {z = x + iy ∶ |x| < n + 1, |y| < 1/3n}, which are holomorphic in a neighborhood of Kn . 90. Prove that a subset M ⊆ ℝ2 is convex if and only if for each point (x0 , y0 ) ∈ ℝ2 ⧵ M there exists a straight line g = {(x, y) ∈ ℝ2 ∶ l((x, y)) = ax + by + c = 0} such that (x0 , y0 ) ∈ g and M ⊂ {(x, y) ∈ ℝ2 ∶ l((x, y)) < 0} (separation property). 91. The convex hull K̃ of K ⊂ ℝ2 is the smallest convex set, which contains K. Show that K̃ = {z ∈ ℂ ∶ |eαz | ≤ sup|eαw | ∀α ∈ ℂ}. w∈K

92. Let U = {z ∶ |z| < 1}, V = {z ∶ 0 < |z| < 1}, K = {z ∶ |z| ≤ 1/2} and L = {z ∶ |z| = 1/2}. Determine K̂ U , K̂ ℂ , L̂ U , L̂ V and L̂ ℂ . Discuss the consequences for Runge’s Theorem! 93. Prove that a domain G in ℂ is simply connected if and only if for each compact subset K of G one has K̂ G = K̂ ℂ . 94. Let Ω be a domain in ℂ and let (ak )k∈ℕ be a discrete sequence in Ω. For each k ∈ ℕ let Rk (z) be the principle part of a function being meromorphic in a neighborhood of ak ; furthermore, let Dk be a sequence of disjoint discs with center ak and let ϕk ∈ 𝒞∞ 0 (Dk ). Define ∞

u = ∑ ϕ k Rk . k=1

Use the solution of a suitable inhomogeneous Cauchy–Riemann equation to find a correction term for u such that the result gives a solution to the corresponding Mittag-Leffler problem, i.e. a function being holomorphic on Ω ⧵ {ak ∶ k ∈ ℕ} and having Rk as principle part at ak , k ∈ ℕ. 95. Where are the following infinite products absolutely convergent: ∞

∞

∏(1 − z n );

∏ cos z/n;

n=1

n=1

∞

∏(1 − n=1

1 ); nz

sin z/n ; n=1 z/n ∞

∏

∞

n

∏(1 + z 2 )? n=1

96. Use Example 4.24 to show that 1 π2 = . 2 6 k=1 k ∞

∑

97.

Construct a meromorphic function with poles at the points zk = ik 2 , k ∈ ℕ and corresponding principle parts k , z − ik 2

k ∈ ℕ.

154 | 4 Construction and approximation of holomorphic functions 98. Find the partial fraction decompositions of the functions f (z) =

1 , sin z

g(z) =

1 , cos z

h(z) =

1 , 1 − ez

i.e. represent the functions from above as the sum of their principle parts, where one has to take care for possible convergence generating summands in the sense of the Mittag-Leffler Theorem. 99. Show that each holomorphic automorphism ϕ of the unit disc D1 (0) has the form ϕ(z) = λ

z−α , 1 − αz

where α ∈ D1 (0) and λ ∈ ℂ is a constant with |λ| = 1. Hint: use the Schwarz Lemma 4.41. 100. Let f ∈ ℋ(D1 (0)) and suppose that |f (z)| < 1 for |z| < 1. Show that |f ′ (z)| ≤

1 − |f (z)|2 , 1 − |z|2

for |z| < 1

(Pick’s Lemma). Hint: fix z0 ∈ D1 (0) and set w0 = f (z0 ). Use Lemma 4.41 for the function h ∘ f ∘ g, where g(z) =

z + z0 1 + z0 z

and

h(w) =

w − w0 . 1 − w0 w

4.12 Notes The proof of the second version of Runge’s Theorem (Theorem 4.20) and the proof of Weierstraß’ Factorization Theorem (Theorem 4.33) is based on Hörmander’s presentation [41]. We also refer to Narasimhan’s monograph [57], where the so-called real method in complex analysis to use the inhomogeneous Cauchy–Riemann equation is explained for functions of one complex variable. This powerful method has its origin in the theory of several variables and will be discussed in detail in the last chapters of the book in combination with L2 -methods. The generalization of Mittag-Leffler’s Theorem (Theorem 4.28) is inspired by the solution of the Cousin I problem for meromorphic functions of several variables, where singularities are never isolated; see, for instance, [59] or [51].

5 Harmonic functions Harmonic functions in ℝ2 appear as real and imaginary parts of holomorphic functions. Their properties are crucial for the theory of partial differential equations. Harmonic functions and the Laplace operator describe many important phenomena in mathematical physics, such as fluid mechanics, thermodynamics, and electricity. The solution of the Dirichlet problem in Section 5.2 is a striking example how methods of complex analysis can be applied for a problem in real partial differential equations. Therefore the relationship between harmonic and holomorphic functions is of great interest. Jensen’s formula (Theorem 5.16) stands as an example for the other direction, where properties of harmonic functions are used to explain the zero distribution of holomorphic functions. Finally, we discuss the basic properties of subharmonic functions.

5.1 Definition and important properties Definition 5.1. Let G ⊆ ℝ2 be a domain and let u ∶ G ⟶ ℝ be a function being twice continuously differentiable in the real sense. u is called harmonic on G, if Δu = 0 on G 𝜕2 𝜕2 𝜕2 where Δ = 𝜕x 2 + 𝜕y 2 or Δ = 4 𝜕z𝜕z . Example 5.2. (a) Linear polynomials in the two real variables x and y are always harmonic. f (x, y) = x2 fails to be harmonic. But u(x, y) = x2 − y2 and v(x, y) = 2xy are harmonic. If p(z) = z 2 , z = x + iy, then ℜp(x, y) = x2 − y2 and ℑp(x, y) = 2xy. (b) If f ∈ ℋ(G) and f = u + iv, then the Cauchy–Riemann differential equations are valid: ux = vy , uy = −vx . Differentiation with respect to x and y yields uxx = vyx , uyy = −vxy , hence uxx + uyy = 0. Similarly, we have vxx + vyy = 0. Hence u and v are harmonic on G. x (c) Let G = ℝ2 ⧵ {(0, 0)} and let h(x, y) = 1/2 log(x2 + y2 ). We have hx = x2 +y 2 and hxx =

y2 − x2 , (x2 + y2 )2

hyy =

x2 − y2 . (x 2 + y2 )2

Hence Δh = 0 on G. Definition 5.3. Let G ⊆ ℝ2 be a domain and let u ∶ G ⟶ ℝ be harmonic on G. A harmonic function v ∶ G ⟶ ℝ is a harmonic conjugate to u, if f = u + iv ∈ ℋ(G) (where G is taken to be a domain in ℂ). Remark. If f ∈ ℋ(G), then ℜf and ℑf are conjugate harmonic functions. Theorem 5.4. Let G ⊆ ℝ2 be a domain. G is simply connected if and only if each harmonic function u on G has a harmonic conjugate function v on G. https://doi.org/10.1515/9783110417241-005

156 | 5 Harmonic functions Proof. (⟹) First suppose that G is ℝ2 or D1 (0). If u ∶ G ⟶ ℝ is harmonic, we define y

x

0

0

v(x, y) = ∫ ux (x, t) dt − ∫ uy (s, 0) ds, then vy (x, y) = ux (x, y) and y

y

0

0

vx (x, y) = ∫ uxx (x, t) dt − uy (x, 0) = − ∫ uyy (x, t) dt − uy (x, 0) = −uy (x, y) + uy (x, 0) − uy (x, 0) = −uy (x, y). The Cauchy–Riemann differential equations are valid, hence f = u + iv ∈ ℋ(G). If G ≠ ℂ is simply connected, then, by Theorem 4.43, there exists a biholomorphic mapping h ∶ G ⟶ D1 (0). Let u1 = u ∘ h−1 ∶ D1 (0) ⟶ ℝ. Then u1 is harmonic, as is easily seen. By the first part of the proof, there exists a harmonic function v1 ∶ D1 (0) ⟶ ℝ, which is harmonic conjugate to u1 . Hence f1 = u1 + iv1 ∈ ℋ(D1 (0)). Let f = f1 ∘ h ∶ G ⟶ ℂ. Then f ∈ ℋ(G) and ℜf = u. Set v = ℑf , then v is harmonic conjugate to u on G. (⟸) Let f ∈ ℋ(G) and f (z) ≠ 0 ∀z ∈ G. We claim that f has a holomorphic logarithm on G. Then we get from Theorem 4.46 that G is simply connected. Let f = u + iv and define U(x, y) = 1/2 log(u2 (x, y) + v2 (x, y)) = log|f (z)|. Then U ∶ G ⟶ ℝ and, by Example 5.2 (c), U is harmonic on G. By assumption there exists a harmonic function V ∶ G ⟶ ℝ, which is harmonic conjugate to U. Hence g = U + iV ∈ ℋ(G). Let h = exp(g). Then |

f (z) |f (z)| |f (z)| f (z) |= = =| |=1 h(z) exp(ℜg(z)) exp(log |f (z)|) f (z)

∀z ∈ G.

In addition f /h ∈ ℋ(G). Since |f /h| = 1 on G, the function f /h cannot be open and, by Theorem 2.38, f = ch, where c ∈ ℂ is a constant. Hence f has a holomorphic logarithm on G. Theorem 5.5 (Mean value property). Let G ⊆ ℝ2 be open and let h ∶ G ⟶ ℝ be a harmonic function. Let a ∈ G and R > 0 such that DR (a) ⊂ G. Then h(a) =

1 2π ∫ h(a + Reit ) dt. 2π 0

Proof. We choose R′ > R with DR′ (a) ⊂ G. By Theorem 5.4, there exists a harmonic function g on DR′ (a), which is harmonic conjugate to h. Hence the function f = h + ig ∈ ℋ(DR′ (a)).

5.2 The Dirichlet problem

| 157

Now let γR (t) = a + Reit , t ∈ [0, 2π]. Then, by Cauchy’s integral formula, f (a) =

1 f (ζ ) 1 2π dζ = ∫ ∫ f (a + Reit ) dt. 2πi γR ζ − a 2π 0

Taking real and imaginary parts, we get the desired result. Remark. Later we will see that a certain form of the converse is also true. Example 5.6. We consider the idealized flow of a fluid with velocity vector v(z) = v1 (x, y) + iv2 (x, y), z = (x, y) ∈ G ⊆ ℝ2 , where G is simply connected. We suppose that the flow in G is sourceless and irrotational, i.e. 𝜕v1 𝜕v2 + = 0, 𝜕x 𝜕y 𝜕v 𝜕v rot v = 2 − 1 = 0. 𝜕x 𝜕y

div v =

(5.1) (5.2)

From (5.2) we get that there exists Φ ∶ G ⟶ ℂ (velocity potential) such that grad Φ = v, i.e. 𝜕Φ = v1 , 𝜕x

𝜕Φ = v2 . 𝜕y

By (5.1), we have ΔΦ = 0, hence Φ is harmonic. Conversely, each solution of ΔΦ = 0 can be interpreted as the velocity potential of a sourceless and irrotational flow. By Theorem 5.4, there exists a Φ, which is harmonic conjugate to Ψ on G. We call F = Φ + iΨ ∈ ℋ(G) the complex velocity potential. Using Definition 1.13 and the Cauchy–Riemann differential equations, we obtain F ′ (z) =

𝜕Φ 𝜕Ψ +i = v1 − iv2 , 𝜕x 𝜕x

hence v = F ′ .

5.2 The Dirichlet problem Let g ∶ 𝕋 ⟶ ℝ be a continuous function. We want to find a continuous function u on D1 (0), which is harmonic in the interior of D1 (0) and which coincides with the given function g on 𝕋. We say that u is a solution of the Dirichlet1 problem with boundary values g. 1 Dirichlet, Gustav Peter Lejeune (1805–1859).

158 | 5 Harmonic functions Definition 5.7. Let a ∈ ℂ, R > 0, z ∈ DR (a). Then Pa,R (z, t) = ℜ(

Reit + (z − a) ), Reit − (z − a)

t∈ℝ

is called the Poisson2 kernel of DR (a). Let Φ ∶ 𝜕DR (a) ⟶ ℝ be a continuous function. The expression Pa,R (Φ)(z) =

1 2π ∫ P (z, t)Φ(a + Reit ) dt, 2π 0 a,R

z ∈ DR (a)

is called the Poisson integral of Φ. Remark. Let a = 0 and R = 1. We set P(z) = P0,1 (z, 0) = ℜ(

1+z ), 1−z

|z| < 1.

For z = reiθ , we have Pr (θ) = P(reiθ ) = ℜ( =

1 + r cos θ + ir sin θ 1 − r 2 + 2ri sin θ ) = ℜ( ) 1 − r cos θ − ir sin θ 1 − 2r cos θ + r 2

1 − r2 . 1 − 2r cos θ + r 2

For |z| < 1, we have 1+z = (1 + z) ∑ z n = 1 + 2 ∑ z n , 1−z n=0 n=1 ∞

∞

hence Pr (θ) = ℜ(

1+z ) = 1 + 2 ∑ r n cos nθ = ∑ r |n| einθ . 1−z n=1 n∈ℤ ∞

For a z ∈ Dr (a) such that z = a + reiθ , r < R, we get Pa,R (z, t) = P(r/R ei(θ−t) ). Lemma 5.8. The Poisson kernel Pa,R (z, t) is harmonic as a function of z ∈ DR (a), where t is fixed. Furthermore, we have Pa,R (z, t) > 0 ∀z ∈ DR (a), ∀t ∈ ℝ. For each r with 0 ≤ r < R, one has 1 2π ∫ P (a + reiθ , t) dθ = 1. 2π 0 a,R 2 Poisson, Siméon Denis (1781–1840).

5.2 The Dirichlet problem

| 159

Proof. Pa,R (z, t) is the real part of the holomorphic function z↦

Reit + (z − a) Reit − (z − a)

for z ∈ DR (a) and hence harmonic there. Set ρ = r/R, then Pa,R (z, t) = P(ρei(θ−t) ) = Pρ (θ − t) =

1 − ρ2 . 1 − 2ρ cos(θ − t) + ρ2

Since 1 − 2ρ cos(θ − t) + ρ2 ≥ 1 − 2ρ + ρ2 = (1 − ρ)2 > 0, we have Pa,R (z, t) > 0 ∀z ∈ DR (a), ∀t ∈ ℝ. In addition, we have 2π 1 1 2π 1 2π = 1. ∑ ρ|n| e−int ∫ einθ dθ = ∫ Pρ (θ − t) dθ = 2π 0 2π n∈ℤ 2π 0

Lemma 5.9. Let Pr (θ) =

1−r 2 1−2r cos θ+r 2

and 0 < δ < π/2. If δ ≤ θ ≤ 2π − δ, then

0 < Pr (θ) ≤

1 − r2 1 − cos2 δ

∀r, 0 ≤ r < 1.

In particular, Pr (θ) tends to 0 as r → 1, even uniformly in θ for δ ≤ θ ≤ 2π − δ. Proof. If π/2 ≤ θ ≤ 3π/2, then cos θ ≤ 0 and hence 1 − 2r cos θ + r 2 ≥ 1, so we have Pr (θ) ≤ 1 − r 2 ≤

1 − r2 . 1 − cos2 δ

If δ ≤ θ ≤ π/2, we have 0 ≤ cos θ ≤ cos δ and 1 − 2r cos θ + r 2 ≥ 1 − 2r cos δ + r 2 = (1 − cos2 δ) + (r − cos δ)2 ≥ 1 − cos2 δ. Therefore, Pr (θ) ≤

1 − r2 . 1 − cos2 δ

If 3π/2 ≤ θ ≤ 2π − δ, we have 0 ≤ cos θ ≤ cos(2π − δ) = cos δ and hence again Pr (θ) ≤

1 − r2 . 1 − cos2 δ

Theorem 5.10 (Solution of the Dirichlet problem). Let Φ ∶ 𝕋 ⟶ ℝ be a continuous function. Let z ∈ D1 (0), z = reiθ , u(z) =

1 2π ∫ P (θ − t)Φ(eit ) dt, 2π 0 r

and let u(eit ) = Φ(eit ) on 𝕋. Then u ∶ D1 (0) ⟶ ℝ is a solution of the Dirichlet problem with boundary values Φ.

160 | 5 Harmonic functions Proof. First we show that u is harmonic on D1 (0). We can write u in the following form: u(z) =

2π eit + z 1 2π eit + z 1 )Φ(eit ) dt = ℜ(∫ Φ(eit ) dt). ∫ ℜ( it it 2π 0 e −z 2π 0 e −z

The last integral is a holomorphic parameter integral with respect to z ∈ D1 (0). Hence u is the real part of a holomorphic function and therefore harmonic on D1 (0). By Lemma 5.8, we have 1 2π 1 2π ∫ Pr (θ − t)Φ(eit ) dt − ∫ P (θ − t)Φ(eit0 ) dt 2π 0 2π 0 r 1 2π = ∫ P (θ − t)(Φ(eit ) − Φ(eit0 )) dt. 2π 0 r

u(z) − Φ(eit0 ) =

(5.3)

Now we consider the limit as z → eit0 . Φ is uniformly continuous on 𝕋, hence, for each ϵ > 0 there exists δ > 0 such that |Φ(eit ) − Φ(eit0 )| < ϵ, if |eit − eit0 | < δ, where δ is independent of t and t0 . Let z = reiθ . If |eit − eit0 | ≥ δ, then |ei(t−θ) − 1| ≥ δ/2, if θ is sufficiently close to t0 , because we have ei(t−θ) − 1 = e−iθ (eit − eiθ ), therefore |ei(t−θ) − 1| = |eit − eiθ |. Now we decompose (5.3) into two summands: 1 2π ∫ P (θ − t)(Φ(eit ) − Φ(eit0 )) dt 2π 0 r 1 = (∫ P (θ − t)(Φ(eit ) − Φ(eit0 )) dt 2π |eit −eit0 |≥δ r

u(z) − Φ(eit0 ) =

+∫

|eit −eit0 | 0, which depends only on δ (we know that |ei(t−θ) − 1| ≥ δ/2, if θ is sufficiently close to t0 ). Then

5.3 Jensen’s formula

|I| ≤

| 161

1 C(δ)(1 − r 2 ) ∫ |Φ(eit ) − Φ(eit0 )| dt 2π |eit −eit0 |≥δ 2π

≤ C ′ (δ)(1 − r 2 ) ∫ |Φ(eit ) − Φ(eit0 )| dt 0

≤ C ′ (δ)(1 − r 2 )M, where the constant M > 0 depends only on Φ. Taking the limit as z = reiθ → eit0 , we observe that r → 1, hence I tends to 0. So we have lim u(z) = Φ(eit0 ),

z→eit0

and u is continuous on D1 (0). Corollary 5.11. If Φ ∶ 𝜕DR (a) ⟶ ℝ is continuous, then u(z) = Pa,R (Φ)(z), z ∈ DR (a), and u(z) = Φ(z), z ∈ 𝜕DR (a), is a solution of the Dirichlet problem on DR (a) with boundary values Φ. Remark 5.12. If G is a simply connected domain, G ≠ ℂ, whose boundary is a piecewise smooth curve without double points, then the biholomorphic mapping ψ ∶ G ⟶ D1 (0) has a continuous extension to the boundary (again denoted by ψ) ψ ∶ G ⟶ D1 (0) such that ψ(𝜕G) = 𝕋 (see [22]). If g ∶ 𝜕G ⟶ ℝ is a continuous function, then g ∗ = g ∘ ψ−1 ∶ 𝕋 ⟶ ℝ is continuous on 𝕋. For g ∗ as boundary value, we can solve the Dirichlet problem (Theorem 5.10): u∗ (w) =

1 2π ∫ P (w, t)g ∗ (eit ) dt, 2π 0 0,1

w ∈ D1 (0),

then u(z) = u∗ (ψ(z)), z ∈ G, is harmonic on G, and for z ∈ 𝜕G we have u(z) = u∗ (ψ(z)) = g ∗ (ψ(z)) = g(ψ−1 (ψ(z))) = g(z), hence u is a solution of the Dirichlet problem on G with boundary values g. Let G be the upper half-space H = {ζ = σ + iτ ∶ τ > 0} and let w ∶ ℝ ⟶ ℝ be a bounded continuous function. We want to find a continuous function W ∶ H ⟶ ℝ, which is harmonic on H and coincides with w on 𝜕H = ℝ. The Riemann mapping func−i , ζ ∈ H. Using the procedure from above, tion ψ ∶ H ⟶ D1 (0) has the form ψ(ζ ) = ζζ +i one finally gets W(ζ ) =

1 ∞ τ dt, ∫ w(t) π −∞ (t − σ)2 + τ2

ζ = σ + iτ.

5.3 Jensen’s formula Jensen’s formula describes the relationship between the growth behavior and the distribution of zeros of a holomorphic function. It is a consequence of the mean value

162 | 5 Harmonic functions property of harmonic functions and has important applications in various fields of analysis. We start with some further properties of harmonic functions. Definition 5.13. Let u ∶ Ω ⟶ ℝ be a continuous functions on the open set Ω ⊆ ℂ. We say that u has the mean value property, if for each z ∈ Ω there exists a sequence (rn )n of positive real numbers with limn→∞ rn = 0 such that u(z) =

1 2π ∫ u(z + rn eit ) dt 2π 0

∀n ∈ ℕ.

Remark. By Theorem 5.5, we know that harmonic functions have the mean value property. In the following we will prove the converse of this assertion. Theorem 5.14. Let u ∶ Ω ⟶ ℝ be a continuous function with the mean value property. Then u is harmonic on Ω. Proof. Let a ∈ Ω and R > 0 with DR (a) ⊂ Ω. By Corollary 5.11, the Poisson integral h(z) =

Reit + (z − a) 1 2π )u(a + Reit ) dt ∫ ℜ( it 2π 0 Re − (z − a)

defines a function which is continuous on DR (a), harmonic on DR (a) and which coincides with u on 𝜕DR (a). Now let v = u − h and m = sup{v(z) ∶ z ∈ DR (a)}. Suppose that m > 0, then the set E = {z ∈ DR (a) ∶ v(z) = m} is a compact subset of DR (a) since v = 0 on 𝜕DR (a). Hence there exists a z0 ∈ E with |z0 −a| ≥ |z −a| ∀z ∈ E. If r > 0 is sufficiently small, at least half of the disk Dr (z0 ) does not belong to E. The function v has the mean value property. But the mean value of v over 𝜕Dr (z0 ) is smaller than v(z0 ) = m, and we have a contradiction. Hence m = 0 and v ≤ 0 on DR (a). The same reasoning for −v yields v ≥ 0. Hence u = h on DR (a). Therefore u is harmonic on Ω. Lemma 5.15. 1 2π ∫ log|1 − eiθ | dθ = 0 2π 0 Proof. Let Ω = {z ∶ ℜz < 1}. Then 1 − z ≠ 0 ∀z ∈ Ω. Since Ω is simply connected, we get from Theorem 4.46 that there exists a holomorphic function h ∈ ℋ(Ω) such that exp h(z) = 1 − z ∀z ∈ Ω. Setting h(0) = 0, the function h becomes uniquely determined. We have 1 − z = eℜh(z) (cos(ℑh(z)) + i sin(ℑh(z))), and since ℜ(1 − z) > 0 ∀z ∈ Ω, we obtain ℜh(z) = log |1 − z| and

|ℑh(z)| < π/2.

5.3 Jensen’s formula

| 163

For δ > 0 small, let Γ be the path defined by Γ(t) = eit , δ ≤ t ≤ 2π − δ and let γ the circular line from eiδ to e−iδ in Ω, see Figure 5.1.

Figure 5.1

Cauchy’s formula implies that 0 = h(0) =

1 h(z) h(z) (∫ dz − ∫ dz). 2πi Γ z γ z

Hence 1 2π−δ 1 h(z) 1 h(z) log |1 − eiθ | dθ = ℜ( dz) = ℜ( dz). ∫ ∫ ∫ 2π δ 2πi Γ z 2πi γ z The length of γ is smaller than πδ, and we can estimate as follows: |ℜ(

h(z) 1 h(z) 1 h(z) 1 dz)| ≤ | dz| ≤ πδ sup| | ∫ ∫ 2πi γ z 2πi γ z 2π z∈γ∗ z

≤ Cδ[(log|1 − 1 − δeit |)2 + (π/2)2 ]1/2 ≤ C ′ (δ| log δ| + δ),

the last expression tends to 0 as δ → 0. So we get the desired result.

164 | 5 Harmonic functions Theorem 5.16 (Jensen’s3 formula). Let f ∈ ℋ(DR (0)), f (0) ≠ 0, 0 < r < R, and let α1 , … , αN be the zeros of f in Dr (0), repeated according to multiplicities. Then N

r 1 2π = exp{ ∫ log|f (reiθ )| dθ}. 2π 0 n=1 |αn |

|f (0)| ∏

Remark. If f (0) = 0 of order k, then consider the function z ↦ f (z)/z k instead of f . Proof. Arrange the zeros αj such that α1 , … , αm ∈ Dr (0) and |αm+1 | = ⋯ = |αN | = r. Now define m

N

r 2 − αn z αn , ∏ r(α − z) α n n=1 n=m+1 n − z

g(z) = f (z) ∏

then g ∈ ℋ(D), where D = Dr+ϵ (0) for some ϵ > 0. We can choose ϵ > 0 such that g has no zero in D. We know from Section 5.1 that log |g| is harmonic on D, and Theorem 5.5 implies log|g(0)| =

1 2π ∫ log|g(reiθ )| dθ. 2π 0

(5.4)

In addition, we have m

r . |α n=1 n |

|g(0)| = |f (0)| ∏

(5.5)

A simple computation shows that for 1 ≤ n ≤ m and |z| = r one has |

r 2 − αn z | = 1. r(αn − z)

Now let αn = reiθn for m < n ≤ N. From the definition of g we have N

log|g(reiθ )| = log|f (reiθ )| − ∑ log|1 − ei(θ−θn ) |. n=m+1

Integrating (5.6) with respect to θ from 0 to 2π and using Lemma 5.15, we obtain 2π

2π

0

0

∫ log|g(reiθ )| dθ = ∫ log|f (reiθ )| dθ. Inserting this into (5.4) and using (5.5), we obtain Jensen’s formula. 3 Jensen, Johan Ludwig William Valdemar (1859–1925).

(5.6)

5.4 Subharmonic functions | 165

Corollary 5.17. Let f ∈ ℋ(DR (0)) and 0 < ρ < R, suppose that f (0) = 0 of order λ (where λ = 0 means that f (0) ≠ 0). Let n(ρ) be the number of zeros of f in Dρ (0), where multiplicities are included and let M(ρ) = max |f (ρeiθ )| = sup |f (z)|. 0≤θ≤2π

z∈Dρ (0)

Then M(ρ) ρn(ρ) ≤ . |α1 | ⋯ |αn(ρ) | |f (λ) (0)/λ!|ρλ Proof. From Theorem 5.16 for f (z)/z λ instead of f , we obtain n(ρ)

log |f (λ) (0)/λ!| + log(ρλ ∏ k=1

1 2π ρ )= ∫ log|f (ρeiθ )| dθ ≤ log M(ρ). |αk | 2π 0

Remark. The last result describes an important relationship between the distribution of the zeroes and the growth behavior of a holomorphic function, which is of special interest for entire functions, see [11].

5.4 Subharmonic functions We discuss elementary properties of subharmonic functions in the complex plane. The generalization to several complex variables, so-called plurisubharmonic functions, plays a fundamental role in many areas of complex analysis. Definition 5.18. Let U be an open set in ℂ. A real-valued function v, continuous on U is said to be subharmonic on U if it satisfies v(a) ≤

1 2π ∫ v(a + Reit ) dt, 2π 0

(5.7)

for each a ∈ U and R > 0 such that DR (a) ⊂ U. It is easy to see that |h| is subharmonic, if h is harmonic. Hence if f is holomorphic, then |f | is subharmonic. Furthermore, if v1 and v2 are subharmonic functions on U, then v1 + v2 is subharmonic, as well as cv1 , if c > 0, and max{v1 , v2 }. Subharmonic, and therefore also harmonic, functions fulfill the maximum principle: Theorem 5.19 (Maximum principle for subharmonic functions). Let v be a subharmonic function on a domain G and suppose that v(z) ≤ C for all z ∈ G, where C is a constant. If there exists a point z0 ∈ G such that v(z0 ) = C, then v(z) = C for all z ∈ G.

166 | 5 Harmonic functions Proof. Let DR (z0 ) ⊂ G. We consider (5.7) taking a radius r with 0 ≤ r ≤ R. Then we multiply both sides by r and integrate with respect to r from 0 to R. So we get R

∫ rv(z0 ) dr ≤ 0

1 R 2π ∫ ∫ v(z0 + reit )r dr dt. 2π 0 0

On the left-hand side we have R2 v(z0 )/2, whereas on the right-hand side we have polar coordinates of the Lebesgue measure dλ in ℝ2 . Hence v(z0 ) ≤

1 v(z) dλ(z). ∫ πR2 DR (z0 )

(5.8)

This implies 1 (v(z) − v(z0 )) dλ(z) ≥ 0, ∫ πR2 DR (z0 ) the integrand satisfies v(z) − v(z0 ) ≤ 0, as v(z0 ) = C, hence the integrand must be identically zero on DR (z0 ). When we say that a function satisfies the maximum principle, we refer to the property in the last theorem, that is, we suppose that it does not assume a maximum value in G unless it is constant. Theorem 5.20. Let G be a domain and v ∶ G ⟶ ℝ a continuous function. Then v is subharmonic on G if and only if for every domain G1 ⊂ G and every harmonic function u1 on G1 , the function v − u1 satisfies the maximum principle on G1 . Proof. If v is subharmonic and G1 and u1 are as in the statement, then v − u1 is subharmonic and satisfies the maximum principle by Theorem 5.19. For the converse, let DR (a) ⊂ G. By Corollary 5.11, there exists a continuous function u ∶ DR (a) ⟶ ℝ which is harmonic in DR (a) and u(z) = v(z) for z ∈ 𝜕DR (a). By hypothesis, v − u satisfies the maximum principle on DR (a). As v − u = 0 on 𝜕DR (a), we obtain v ≤ u on DR (a). Hence v(a) ≤ u(a) =

1 2π 1 2π ∫ u(a + Reit ) dt = ∫ v(a + Reit ) dt, 2π 0 2π 0

and v is subharmonic on G. Immediately we get the following corollary which explains the name subharmonic. Corollary 5.21. Let G be a domain and v ∶ G ⟶ ℝ a continuous function. Then v is subharmonic on G if and only if for every bounded domain G1 such that G1 ⊂ G and for every continuous function u1 ∶ G1 ⟶ ℝ which is harmonic in G1 and satisfies v ≤ u1 on 𝜕G1 , one has v ≤ u1 on G1 .

5.5 Exercises | 167

Remark 5.22. For many important applications it is necessary to consider a more general concept. A function v ∶ G ⟶ ℝ ∪ {−∞} is called subharmonic if v is upper semicontinuous, i.e. {z ∈ G ∶ v(z) < c} is open for every c ∈ ℝ, and if for every compact set K ⊂ G and every function h ∈ 𝒞(K) which is harmonic on the interior of K and satisfies v ≤ h on 𝜕K it follows that v ≤ h on K. For this more general concept one also has the maximum principle and submean value property. In addition, if {vα ∶ α ∈ A} is a family of subharmonic functions on G such that v = supα vα is finite and upper semicontinuous, then v is subharmonic; if {vj ∶ j ∈ ℕ} is a decreasing sequence of subharmonic functions on G, then v = limj→∞ vj is subharmonic. Important examples of subharmonic functions are |f |c for c > 0 and log |f |, where f ∈ ℋ(G) is a holomorphic function on G. If one supposes that v ∈ 𝒞2 (G), then v is subharmonic on G if and only if Δv ≥ 0 on G. For further details, see [59].

5.5 Exercises 101. Let u be a harmonic function on a domain G ⊆ ℝ2 ≅ ℂ. Show that ux and uy are harmonic on G and that f = ux − iuy is holomorphic on G. 102. Let u and v be real-valued harmonic functions on a domain G ⊆ ℝ2 . Under what conditions is uv harmonic on G? Show that u2 is harmonic on G if and only if u is constant. 103. Let u be harmonic on a domain G and suppose DR (a) ⊂⊂ G. Show that u(a) =

1 u(x, y) dλ(x, y). ∫ πR2 DR (a)

104. Let u ∶ G ⟶ ℝ be a function with continuous partial derivatives up to order 2. Define U(r, θ) ∶= u(r cos θ, r sin θ). Suppose that 0 ∉ G. Show that u is harmonic in G if and only if r

𝜕2 U 𝜕 𝜕U (r ) + 2 = 0. 𝜕r 𝜕r 𝜕θ

105. Suppose that 0 ∉ G and that u depends only on |z|, i.e. u(z) = φ(|z|). Show that u is harmonic on G if and only if u(z) = a log |z| + b for some constants a, b ∈ ℝ. 106. Let u ∶ DR (a) ⟶ ℝ be a continuous function, harmonic in DR (a). Suppose that u ≥ 0 on DR (a). Show that for 0 ≤ r < R and all θ ∈ [0, 2π] we have R−r R+r u(a) ≤ u(a + reiθ ) ≤ u(a) R+r R−r (Harnack’s inequality). Hint: use the integral representation in the proof of Theorem 5.14.

168 | 5 Harmonic functions 108. Let f be an entire function, let M(ρ) and n(ρ) be as in Corollary 5.17. Suppose that f (0) = 1 and show that n(ρ) log 2 ≤ log M(2ρ). 109. Let f ∈ ℋ(G). Show that |f |c for c > 0 and log |f | are subharmonic on G. 110. Suppose that v ∈ 𝒞2 (G). Show that v is subharmonic on G, if and only if Δv ≥ 0 on G.

5.6 Notes A thorough treatment of the relationship between the growth behavior and the distribution of zeroes of entire functions can be found in several textbooks, see [53, 11, 15]. Subharmonic functions and a general treatment of the Dirichlet problem are basic tools in potential theory, see [11, 15, 26].

6 Several complex variables To think that the analysis of several complex variables is more or less the one variable theory with some more indices turns out to be incorrect. Completely new phenomena appear which will be exploited in the following. Many differences between the univariate and multivariate theories originate from the Cauchy–Riemann differential equations which constitute an overdetermined system of partial differential equations for several complex variables. We start with the basic definitions and complex differential forms. Section 6.1 also presents the main differences between univariate and multivariate analysis, such as the Identity Theorem and Hartogs phenomenon. Section 6.2 provides another important example for this difference, namely in the analysis of the inhomogeneous Cauchy–Riemann differential equations. In addition, the concept of the tangential Cauchy–Riemann equation is introduced. This gives the tools required for the famous Lewy example of a partial differential operator without a solution. In Section 6.3 we discuss pseudoconvex domains and plurisubharmonic functions and explain the concept of a domain of holomorphy.

6.1 Complex differential forms and holomorphic functions Let Ω ⊆ ℂn be an open subset and let f ∶ Ω ⟶ ℂ be a 𝒞1 -function. We write zj = xj + iyj and consider for P ∈ Ω the differential n

dfP = ∑( j=1

𝜕f 𝜕f (P)dxj + (P)dyj ). 𝜕xj 𝜕yj

We use the complex differentials dzj = dxj + idyj ,

dz j = dxj − idyj

and the derivatives 𝜕 1 𝜕 𝜕 = ( − i ), 𝜕zj 2 𝜕xj 𝜕yj

𝜕 1 𝜕 𝜕 = ( +i ) 𝜕z j 2 𝜕xj 𝜕yj

to rewrite the differential dfP in the form n

dfP = ∑( j=1

𝜕f 𝜕f (P)dzj + (P)dz j ) = 𝜕fP + 𝜕fP . 𝜕zj 𝜕z j

A general differential form is given by ω=

∑

′

|J|=p,|K|=q https://doi.org/10.1515/9783110417241-006

aJ,K dzJ ∧ dz K ,

170 | 6 Several complex variables where ∑

′ |J|=p,|K|=q

denotes the sum taken only over all increasing multiindices J =

(j1 , … , jp ), K = (k1 , … , kq ) and

dzJ = dzj1 ∧ ⋯ ∧ dzjp ,

dz K = dz k1 ∧ ⋯ ∧ dz kq .

We call ω a (p, q)-form and write ω ∈ 𝒞k(p,q) (Ω), if ω is a (p, q)-form with coefficients belonging to 𝒞k (Ω). The derivative dω of ω is defined by dω =

′

∑

|J|=p,|K|=q

daJ,K ∧ dzJ ∧ dz K =

′

∑

|J|=p,|K|=q

(𝜕aJ,K + 𝜕aJ,K ) ∧ dzJ ∧ dz K ,

and we set 𝜕ω =

′

∑

|J|=p,|K|=q

𝜕aJ,K ∧ dzJ ∧ dz K

and 𝜕ω =

∑

′

|J|=p,|K|=q

𝜕aJ,K ∧ dzJ ∧ dz K .

We have d = 𝜕 + 𝜕 and, since d2 = 0, it follows that 0 = (𝜕 + 𝜕) ∘ (𝜕 + 𝜕)ω = (𝜕 ∘ 𝜕)ω + (𝜕 ∘ 𝜕 + 𝜕 ∘ 𝜕)ω + (𝜕 ∘ 𝜕)ω, 2

which implies 𝜕2 = 0, 𝜕 = 0 and 𝜕 ∘ 𝜕 + 𝜕 ∘ 𝜕 = 0, by comparing the types of the differential forms involved. Before we proceed, we mention important domains in ℂn and some basic facts about them. Definition 6.1. A polydisc with center a = (a1 , … , an ) ∈ ℂn and multiradius r = (r1 , … , rn ), rj > 0 is the set P(a, r) = {z ∈ ℂn ∶ |zj − aj | < rj , 1 ≤ j ≤ n}. A ball with center a = (a1 , … , an ) ∈ ℂn and radius r > 0 is defined by n

B(a, r) = {z ∈ ℂn ∶ ∑ |zj − aj |2 < r 2 }. j=1

We write 𝔹 for the unit ball B(0, 1). The Siegel1 upper half-space 𝕌 in ℂn , n ≥ 2, is defined by n−1

𝕌 = {z ∈ ℂn ∶ ℑzn > ∑ |zj |2 }. j=1

In the sequel we will use the symbol bΩ for the boundary of a domain Ω in ℂn . The symbol 𝜕 is now reserved for differential forms. 1 Siegel, Carl Ludwig (1896–1981).

6.1 Complex differential forms and holomorphic functions | 171

Definition 6.2. A domain Ω ⊂ ℝn , n ≥ 2, is said to have 𝒞k (1 ≤ k ≤ ∞) boundary at the boundary point p if there exists a real-valued function ρ defined in some open neighborhood U of p such that ρ ∈ 𝒞k (U) and U ∩ Ω = {x ∈ U ∶ ρ(x) < 0}, bΩ ∩ U = {x ∈ U ∶ ρ(x) = 0}, and dρ(x) ≠ 0 on bΩ ∩ U. The function ρ is called a 𝒞k local defining function for Ω near p. If U is an open neighborhood of Ω, then ρ is called a global defining function for Ω, or simply a defining function for Ω. In the following we consider the relationship between two defining functions. Lemma 6.3. let ρ1 and ρ2 be two local defining functions of Ω of class 𝒞k in a neighborhood U of p ∈ bΩ. Then there exists a positive 𝒞k−1 function h on U such that ρ1 = hρ2 on U and dρ1 (x) = h(x)dρ2 (x) for x ∈ U ∩ bΩ. Proof. Since dρ2 ≠ 0 on the boundary near p, we may assume that p = 0, xn = ρ2 (x) and U ∩ bΩ = {x ∈ U ∶ xn = 0}, after a 𝒞k change of coordinates. Let x ′ = (x1 , … , xn−1 ). Then we have ρ1 (x′ , 0) = 0 and by the fundamental theorem of calculus 1

ρ1 (x ′ , xn ) = ρ1 (x′ , xn ) − ρ1 (x′ , 0) = xn ∫

0

𝜕ρ1 ′ (x , txn ) dt. 𝜕xn

Hence ρ1 = hρ2 for some 𝒞k−1 function on U. If k − 1 ≥ 1, we get dρ1 (x) = h(x)dρ2 (x) for x ∈ U ∩ bΩ, as ρ2 (x) = 0 for x ∈ U ∩ bΩ. If k = 1, we get the same conclusion from the fact that for a function f differentiable at 0 ∈ ℝn such that f (0) = 0 and for a function h continuous at 0, one has that f ⋅ h is differentiable at 0 and d(hf )0 = h(0)df0 . Finally, as dρ1 (x) ≠ 0 and dρ2 (x) ≠ 0 for x ∈ U ∩ bΩ, we get h(x) ≠ 0 for x ∈ U ∩ bΩ. In addition, since h > 0 on U ⧵ bΩ, and h is continuous, we obtain h > 0 on U. Definition 6.4. Let Ω ⊆ ℂn be open. A function f ∶ Ω ⟶ ℂ is called holomorphic on Ω if f ∈ 𝒞1 (Ω) and f satisfies the system of partial differential equations 𝜕f (z) = 0 𝜕z j

for 1 ≤ j ≤ n and z ∈ Ω,

(6.1)

equivalently, if f satisfies 𝜕f = 0. We remark that there is no biholomorphic mapping between a polydisc and a ball in ℂn , n ≥ 2, see [59]. The Siegel upper half-space 𝕌 is biholomorphic to the unit ball 𝔹, by the so-called Cayley transform, so 𝕌 is an unbounded realization of a bounded symmetric domain. The boundary ℍ of 𝕌 carries the structure of the Heisenberg group, see Exercises for more details. Next we establish a Cauchy integral formula for holomorphic functions on polydiscs.

172 | 6 Several complex variables Theorem 6.5. Let P = P(a, r) be a polydisc in ℂn . Suppose that f ∈ 𝒞1 (P) and that f is holomorphic on P, i.e. for each z ∈ P and 1 ≤ j ≤ n, the function ζ ↦ f (z1 , … , zj−1 , ζ , zj+1 , … , zn ) is holomorphic on {ζ ∈ ℂ ∶ |ζ − aj | < rj }. Then f (z) =

1 f (ζ ) dζ1 ⋯ dζn , ∫ ⋯∫ n (2πi) γ1 γn (ζ1 − z1 ) ⋯ (ζn − zn )

(6.2)

for z ∈ P, where γj (t) = aj + rj eit , for t ∈ [0, 2π] and j = 1, … , n. Proof. We use induction over n. For n = 1, one has the classical Cauchy Formula, see Theorem 2.47. Suppose that the theorem has been proven for n − 1 variables. For z ∈ P fixed, we apply the inductive hypothesis with respect to (z2 , … , zn ) and obtain f (z1 , z2 , … , zn ) =

1 f (z1 , ζ2 , … , ζn ) dζ2 ⋯ dζn . ∫ ⋯∫ (2πi)n−1 γ2 (ζ γn 2 − z2 ) ⋯ (ζn − zn )

For ζ2 , … , ζn fixed, we get from the 1-dimensional case f (z1 , ζ2 , … , ζn ) =

1 f (ζ1 , … , ζn ) dζ1 ∫ 2πi γ1 ζ1 − z1

which can be substituted to the formula above to obtain (6.2). Like in the case n = 1, we get also here that multivariate holomorphic functions are 𝒞∞ functions, and all complex derivatives of holomorphic functions are again holomorphic, by differentiating under the integral sign in (6.2). In addition, we get the Cauchy estimates: for f ∈ ℋ(P(a, r)) and α = (α1 , … , αn ) ∈ α α n ℕ0 ∶ let |α| = α1 + ⋯ + αn and α! = α1 ! ⋯ αn !; furthermore, set r α = r1 1 ⋯ rn n , then |Dα f (a)| = |

𝜕|α| f α! α (a)| ≤ α sup{|f (z)| ∶ z ∈ P(a, r)}. r ⋯ 𝜕zn n

α 𝜕z1 1

(6.3)

Next we show that every holomorphic function can be represented locally by a convergent power series: Theorem 6.6. Let f ∈ ℋ(P(a, r)). Then the Taylor series of f at a converges to f uniformly on all compact subsets of P(a, r), that is, f (z) = ∑

α∈ℕn0

for z ∈ P(a, r).

Dα f (a) (z − a)α , α!

(6.4)

6.1 Complex differential forms and holomorphic functions | 173

Proof. Use the same method as in the proof of Theorem 2.13 for each of iterated integrals in (6.2). From this we get the following: let Ω ⊆ ℂn be a domain and f ∈ ℋ(Ω), suppose that there is a ∈ Ω such that Dα (a) = 0 for all α ∈ ℕn0 , then f (z) = 0 for z ∈ Ω. In particular, if there is a nonempty open set U ⊂ Ω such that f (z) = 0 for z ∈ U, then f ≡ 0 on Ω (Identity Theorem). But Theorem 2.18 is not valid for n > 1 ∶ let f (z1 , z2 ) = z1 . Then this function is zero on {(0, z2 ) ∶ z2 ∈ ℂ}, but f is not identically zero. The following result is also an easy consequence of the corresponding one variable result. Theorem 6.7. Let Ω be a domain in ℂn and suppose that f ∈ ℋ(Ω) is not constant. Then f is an open mapping. Proof. We refer to Theorem 2.38. It is enough to show that for any ball B(a, r) = {z ∈ ℂn ∶ ∑nj=1 |zj − aj |2 < r 2 } the image f (B(a, r)) is a neighborhood of f (a). The restriction of f to B(a, r) is not constant, otherwise f would have to be constant on Ω. Choose p ∈ B(a, r) with f (p) ≠ f (a) and define g(ζ ) = f (a + ζp) for ζ ∈ D1 (0). Then g is nonconstant and holomorphic on D1 (0). By Theorem 2.38, g(D1 (0)) contains a neighborhood of g(0). As g(0) = f (a) and g(D1 (0)) ⊂ f (B(a, r)), the image f (B(a, r)) is a neighborhood of f (a). The maximum principle follows from this result as for n = 1: if Ω ⊆ ℂn is a domain and f ∈ ℋ(Ω) such that |f | has a local maximum at a point a ∈ Ω, then f is constant on Ω; if Ω is a bounded domain in ℂn and f ∈ ℋ(Ω) ∩ 𝒞(Ω), then |f (z)| ≤ |f |bΩ for all z ∈ Ω. We remark that Weierstraß’ Theorem (see Theorem 2.34) and Montel’s Theorem (see Theorem 4.40) also hold for holomorphic functions of several variables with analogous proofs. A striking difference between univariate and multivariate analysis appears in the next result, which gives a domain in ℂn , n > 1, with the property that each holomorphic function can be analytically extended to a larger domain; compare with Theorem 4.38. Theorem 6.8 (Hartogs2 ). Let n ≥ 2 and suppose that 0 < rj < 1 for j = 1, … , n. Then every function f holomorphic on the domain H(r) = {z ∈ ℂn ∶ |zj | < 1 for j < n, rn < |zn | < 1} ∪ {z ∈ ℂn ∶ |zj | < rj for j < n, |zn | < 1},

see Figure 6.1, has a unique holomorphic extension f ̃ to the polydisc P(0, 1). 2 Hartogs, Friedrich Moritz (1874–1943).

174 | 6 Several complex variables

Figure 6.1: The Hartogs domain H(r) in absolute space.

Proof. The extension is unique because of the Identity theorem. Fix δ with rn < δ < 1. Then we define f ̃(z ′ , zn ) =

1 f (z ′ , ζ ) dζ , ∫ 2πi γδ ζ − zn

(6.5)

where z ′ = (z1 , … , zn−1 ) and γδ (t) = δeit for t ∈ [0, 2π]. In this way we defined a function holomorphic on the polydisc P(0, (1′ , δ)), where (1′ , δ) = (1, … , 1, δ). For z ′ ∈ P(0, r ′ ) the function f (z ′ , ⋅) is holomorphic on |zn | < 1, hence (6.5) implies that f ̃(z ′ , zn ) = f (z ′ , zn ) for (z ′ , zn ) ∈ P(0, (r ′ , δ)). The Identity Theorem implies f ̃ = f on H(r) ∩ P(0, (1′ , δ)), so f ̃ is the desired extension of f to the polydisc P(0, 1). The reason for this phenomenon can be better understood by studying the inhomogeneous Cauchy–Riemann differential equations in several complex variables (CR equations).

6.2 The inhomogeneous CR equations Let Ω ⊆ ℂn be a domain and let

n

g = ∑ gj dz j j=1

1

be a (0, 1)-form with coefficients gj ∈ 𝒞 (Ω), for j = 1, … , n. We want to find a function f ∈ 𝒞1 (Ω) such that 𝜕f = g,

(6.6)

6.2 The inhomogeneous CR equations | 175

in other words, 𝜕f = gj , 𝜕z j

j = 1, … , n.

(6.7)

f is called a solution to the inhomogeneous CR equation 𝜕f = g. 2 Since 𝜕 = 0, a necessary condition for solvability of (6.6) is that the right-hand side g satisfies 𝜕g = 0. So, the (0, 2)-form 𝜕g satisfies n

n

𝜕g = ∑ ∑ k=1 j=1

𝜕gj 𝜕z k

dz k ∧ dz j = 0,

which means that 𝜕gj

=

𝜕z k

𝜕gk , 𝜕z j

j, k = 1, … , n.

Theorem 6.9. Let n ≥ 2 and let g = ∑nj=1 gj dz j be a (0, 1)-form with coefficients gj ∈

𝒞k0 (ℂn ), j = 1, … , n, where 1 ≤ k ≤ ∞ and suppose that 𝜕g = 0. Then there exists f ∈ 𝒞k0 (ℂn ) such that 𝜕f = g. We shall see that this result enables us to explain the Hartogs phenomenon in a rather general setting. For n = 1 the above theorem is false: Suppose that ∫ℂ g(ζ ) dλ(ζ ) ≠ 0 and that there is a compactly supported solution f 𝜕f of the equation 𝜕z = g. Then there exists R > 0 such that f (ζ ) = 0 for |ζ | ≥ R. Applying Stokes’ Theorem (see Example 2.45), we obtain for γ(t) = Reit , t ∈ [0, 2π],

0 = ∫ f (ζ ) dζ γ

=∫

DR (0)

= 2i ∫ ≠ 0,

𝜕f dζ ∧ dζ 𝜕ζ

DR (0)

g(ζ ) dλ(ζ )

whenever DR (0) contains the support of g. This is a contradiction. Proof of Theorem 6.9. Define f on ℂn by f (z1 , … , zn ) = By Corollary 4.7 (a), f ∈ 𝒞k (ℂn ) and 𝜕gk . Since g1 𝜕z 1

1 g (ζ , z2 , … , zn ) dζ ∧ dζ . ∫ 1 2πi ℂ ζ − z1 𝜕f 𝜕z 1

= g1 . Now let k > 1. By hypothesis we have

(6.8) 𝜕g1 𝜕z k

=

has compact support, we can interchange differentiation and integration

176 | 6 Several complex variables when we take the derivative of (6.8) with respect to z k and get 𝜕f 1 = ∫ 𝜕z k 2πi ℂ 1 = ∫ 2πi ℂ

𝜕g1 (ζ , z2 , … , zn ) 𝜕z k ζ 𝜕gk (ζ , z2 , … , zn ) ζ 𝜕ζ

= gk (z1 , … , zn ),

1 dζ ∧ dζ − z1 1 dζ ∧ dζ − z1

where we used Corollary 4.7 (b) for the last equality. Hence 𝜕f = g. We still have to show that f is with compact support. Choose R > 1 such that gk (z) = 0 for ∑nj=1 |zj |2 ≥ R, k = 1, … , n. Then f is holomorphic on the domain Ω = {z ∈ ℂn ∶ ∑nj=1 |zj |2 > R}. Since n ≥ 2, we can fix z2 such that |z2 | > R. Then g1 (ζ , z2 , … , zn ) = 0 for all ζ ∈ ℂ and all z3 , … , zn ∈ ℂ. From the definition of f , it follows that f (z) = 0, if |z2 | > R. The set {z ∈ Ω ∶ |z2 | > R} is a nonempty open subset of the domain Ω. Since f ∈ ℋ(Ω), the Identity Theorem yields that f ≡ 0 on Ω. Therefore f has compact support. Now we are able to describe the Hartogs phenomenon in a more general way. Theorem 6.10. Let Ω be a bounded open set in ℂn such that ℂn ⧵ Ω is connected. Suppose that n > 1. Let U be an open neighborhood of the boundary bΩ = Ω ⧵ Ω. Then there exists an open set V with bΩ ⊂ V ⊂ U having the following property: if f ∈ ℋ(U), then there exists F ∈ ℋ(Ω) such that for the restriction to V ∩ Ω one has f |V∩Ω = F|V∩Ω . Proof. Let W be an open neighborhood of bΩ such that W ⊂⊂ U. Choose α ∈ 𝒞∞ 0 (U) such that α = 1 on W . For f ∈ ℋ(U) we define {αf g={ 0 {

on U ∩ Ω, on Ω ⧵ U.

Since α = 0 in a neighborhood of bU, we have g ∈ 𝒞∞ (Ω). Next, define on Ω,

{ 𝜕g ϕk = { 𝜕z k 0 {

on ℂn ⧵ Ω.

𝜕g 𝜕f Since 𝜕z = 𝜕z on W ∩ Ω, we have ϕk ∈ 𝒞∞ (ℂn ). Furthermore ϕk = 0 on (ℂn ⧵ Ω) ∪ W , k k n which implies that supp(ϕk ) ⊂ Ω and ϕk ∈ 𝒞∞ 0 (ℂ ). n Next we claim that the (0, 1)-form ϕ = ∑j=1 ϕj dz j satisfies 𝜕ϕ = 0. We have to show that

𝜕ϕj 𝜕z k

=

𝜕ϕk , 𝜕z j

6.2 The inhomogeneous CR equations | 177

for all j, k = 1, … , n. Both ϕj and ϕk are zero on (ℂn ⧵ Ω) ∪ W , so on Ω we have 𝜕ϕj 𝜕z k

=

𝜕2 g 𝜕ϕk = . 𝜕z j 𝜕z k 𝜕z j

n By Theorem 6.9, there exists u ∈ 𝒞∞ 0 (ℂ ) such that 𝜕u = ϕ. Now we set F = g − u. We have

𝜕F 𝜕g = − ϕk = 0 𝜕z k 𝜕z k on Ω, for k = 1, … , n. Hence F ∈ ℋ(Ω). Let Ω0 be the connected component of (ℂn ⧵ Ω) ∪ W containing ℂn ⧵ Ω. Define V ∶= Ω0 ∩ U. We claim that f |V∩Ω = F|V∩Ω . Since V ∩ Ω ⊆ W ∩ Ω and α = 1 on W so that 𝜕f g = f on V ∩ Ω, it suffices to show that u|Ω0 = 0. Since ϕk = 0 on ℂn ⧵ Ω and ϕk = 𝜕z =0 k

𝜕u on W , we have 𝜕z = 0 on Ω0 . Hence u ∈ ℋ(Ω0 ). And since supp(u) is compact and k Ω is bounded, ℂn ⧵ Ω must intersect ℂn ⧵ supp(u). In particular, the open set Ω1 = Ω0 ∩ (ℂn ⧵ supp(u)) ≠ ∅, and u|Ω1 = 0. Since u ∈ ℋ(Ω0 ), the Identity Theorem implies that u = 0 on Ω0 .

Corollary 6.11. Let Ω be a bounded open set in ℂn such that ℂn ⧵ Ω is connected. Suppose that n > 1. Let U be an open neighborhood of the boundary bΩ = Ω ⧵ Ω. Furthermore, suppose that U ∩ Ω is connected. If f ∈ ℋ(U), then there exists G ∈ ℋ(Ω ∪ U) such that G|U = f . Proof. If F is as in Theorem 6.10, and Ω ∩ U is connected, the Identity Theorem implies that F|Ω∩U = f |Ω∩U , and we may define G by G|Ω = F and G|U = f . Example 6.12. Let |z|2 ∶= |z1 |2 + ⋯ + |zn |2 for z ∈ ℂn . Let Ω = {z ∈ ℂn ∶ |z| < 1} and let U = {z ∈ ℂn ∶ 1/2 < |z| < 3/2}. Then each f ∈ ℋ(U) has a unique holomorphic extension to Ω ∪ U = {z ∈ ℂn ∶ |z| < 3/2}, see Figure 6.2 in absolute space. It is even possible to extend certain functions on the boundary of a domain to holomorphic functions in the interior of the domain. Theorem 6.13. Let Ω be a bounded open set in ℂn , n > 1. Suppose that ℂn ⧵ Ω is connected and bΩ ∈ 𝒞4 , i.e. there exists a real-valued defining function ρ ∈ 𝒞4 (ℂn ) such that ρ vanishes precisely on bΩ and dρ ≠ 0 on bΩ. If u ∈ 𝒞4 (Ω) and 𝜕u ∧ 𝜕ρ = 0 on bΩ, one can then find a function U ∈ 𝒞1 (Ω) such that U ∈ ℋ(Ω) and U = u on bΩ. Remark 6.14. The condition 𝜕u ∧ 𝜕ρ = 0 on bΩ, can also be stated as n

∑ tj j=1

𝜕u =0 𝜕z j

on bΩ,

178 | 6 Several complex variables

Figure 6.2 𝜕ρ for all (t1 , … , tn ) ∈ ℂn with ∑nj=1 tj 𝜕z = 0 on bΩ. We say that u satisfies the tangential j Cauchy–Riemann equations and that u is CR-function. Using Lemma 6.3, one easily sees that this definition does not depend on the choice of the defining function ρ (see Exercises).

Proof of Theorem 6.13. First we construct U0 ∈ 𝒞2 (Ω) such that U0 = u on bΩ and 𝜕U0 = ρ2 v where v is a (0, 1)-form with 𝒞1 coefficients on bΩ. First we claim that 𝜕u = h0 𝜕ρ + ρh1 , where h0 ∈ 𝒞3 (Ω) and h1 ∈ 𝒞2(0,1) (Ω). For this aim we consider the coefficients of the (0, 1)-form 𝜕u. Using the assumption that 𝜕u ∧ 𝜕ρ = 0 on bΩ, we see that there exists h0 ∈ 𝒞3 (Ω) such that 𝜕u 𝜕ρ − h0 =0 𝜕z j 𝜕z j

on bΩ, j = 1, … , n.

From the proof of Lemma 6.3, we get that there exist h1,j ∈ 𝒞2 (Ω), j = 1, … , n such that 𝜕ρ 𝜕u − h0 = ρh1,j , 𝜕z j 𝜕z j

j = 1, … , n.

Now define the (0, 1)-form h1 = ∑nj=1 h1,j dz j . Then 𝜕u = h0 𝜕ρ + ρh1 . Next we get 𝜕(u − 2

h0 ρ) = ρ(h1 − 𝜕h0 ) = ρh2 , where h2 ∈ 𝒞2(0,1) (Ω). Since 0 = 𝜕 (u − h0 ρ) = 𝜕(ρh2 ) = 𝜕ρ ∧ h2 + ρ𝜕h2 , we have 𝜕ρ ∧ h2 = 0 on bΩ. As in the first part of the proof, we can again write h2 = h3 𝜕ρ + ρh4 , where h3 ∈ 𝒞2 (Ω) and h4 ∈ 𝒞1(0,1) (Ω). Now set U0 = u − h0 ρ − h3 ρ2 /2. An easy computation shows that 𝜕U0 = ρ2 (h4 − 𝜕h3 /2),

6.2 The inhomogeneous CR equations | 179

which completes the construction of U0 . Next we define the (0, 1)-form {𝜕U f ={ 0 0 {

on Ω,

on ℂn ⧵ Ω.

Since f = ρ2 v on bΩ we have f ∈ 𝒞1(0,1) (ℂn ) and f has compact support. By Theorem 6.9 we can find a function V ∈ 𝒞10 (ℂn ) with compact support such that 𝜕V = f . The definition of f implies that V is holomorphic in the connected set ℂn ⧵ Ω and, as V has compact support, that V = 0 on ℂn ⧵ Ω. The function U = U0 − V is therefore equal to U0 = u on bΩ, and 𝜕U = 𝜕U0 − 𝜕V = f − f = 0 in Ω. The tangential Cauchy–Riemann equations for the Siegel upper half-space 𝕌 are of special interest. Let n = 2. The function ρ(z1 , z2 ) = − 2i1 (z2 − z 2 ) + z1 z 1 is a defining function for b𝕌. The boundary can be identified with ℍ2 = ℂ × ℝ via the mapping π ∶ (z1 , t + i|z1 |2 ) ↦ (z1 , t), where z2 = t + is. We call ℍ2 the Heisenberg group, see Exercises. If 𝜕u ∧ 𝜕ρ = 0 on b𝕌, we have for a function u ∈ 𝒞1 (𝕌) 1 𝜕u 𝜕u − z1 =0 2i 𝜕z 1 𝜕z 2 on b𝕌. This means we have to consider the differential operator L=

𝜕 𝜕 𝜕 + i − 2i(x + iy) 𝜕x 𝜕y 𝜕t

on ℍ2 . This operator has a special property giving a partial differential operator without solution. Theorem 6.15 (H. Lewy3 ). Let f be a continuous real-valued function depending only on t. If there is a 𝒞1 -function u on (x, y, t) ∈ ℍ2 satisfying Lu = f in some neighborhood of the origin, then f is analytic at t = 0, i.e. can be expanded into a convergent Taylor series in a neighborhood of t = 0. So if one takes a continuous function f which is not analytic at 0, the partial differential equation Lu = f has no solution. Proof. Suppose Lu = f in the set where x2 + y2 < R2 and |t| < R, R > 0. Let γ(θ) = reiθ , θ ∈ [0, 2π], 0 < r < R. Consider the line integral 2π

V(r, t) = ∫ u(x, y, t) dz = ir ∫ u(r cos θ, r sin θ, t)eiθ dθ. γ

3 Lewy, Hans (1904–1988).

0

180 | 6 Several complex variables By Example 2.45 and Stokes’ Theorem (see Theorem 2.46), V(r, t) = i ∫

(

Dr (0) r 2π

𝜕u 𝜕u + i )(x, y, t) dλ(z) 𝜕x 𝜕y

= i∫ ∫ ( 0

0

𝜕u 𝜕u + i )(σ cos θ, σ sin θ, t)σ dσdθ, 𝜕x 𝜕y

where we used polar coordinates dλ(z) = σ dσdθ. Hence 2π 𝜕u 𝜕u 𝜕V = i ∫ ( + i )(r cos θ, r sin θ, t)r dθ 𝜕r 𝜕x 𝜕y 0 𝜕u 𝜕u dz = ∫ ( + i )(x, y, t)r . 𝜕x 𝜕y z γ

Now set s = r 2 and use Lu = f to get 𝜕V 1 𝜕V 𝜕u 𝜕u dz = = ∫ ( + i )(x, y, t) 𝜕s 2r 𝜕r 𝜕y 2z γ 𝜕x 𝜕u dz = i∫ (x, y, t) dz + ∫ f (t) 2z γ 𝜕t γ 𝜕V + iπf (t). =i 𝜕t τ

Now we set F(t) = ∫0 f (τ) dτ, and U(t, s) = V(t, s) + πF(t), Then 𝜕U 𝜕U +i = 0, 𝜕t 𝜕s which is the Cauchy–Riemann equation. Hence U is a holomorphic function of w = t + is for 0 < s < R2 , and |t| < R, in addition, U is continuous up to the line s = 0, and V = 0 when s = 0, therefore U(t, 0) = πF(t) is real-valued. We can apply the Schwarz’ reflection principle (see Exercise 47 (b)): the definition U(t, −s) = U(t, s) gives a holomorphic continuation of U to a full neighborhood of the origin. In particular, U(t, 0) = πF(t) is analytic in t, hence so is f = F ′ .

6.3 Domains of holomorphy In this section we describe domains for which the Hartogs extension phenomenon does not occur; these are the so-called domains of holomorphy. First we study holomorphically convex domains, a concept which was of importance for the Runge type theorems and which serves as an interesting concept where the difference between one and several complex variables becomes apparent. It turns out that another generalization of convexity, so-called pseudoconvexity, is the appropriate geometric concept to characterize domains of holomorphy in ℂn , n ≥ 2. It is beyond the level of this book to give all the details in this context, and we refer to textbooks on several complex variables for a thorough treatment; see [41, 59, 51].

6.3 Domains of holomorphy | 181

Definition 6.16. Let Ω be a domain in ℂn . A holomorphic function f on Ω is completely singular at p ∈ bΩ if for every connected neighborhood U of p there is no h ∈ ℋ(U) which agrees with f on some connected component of U ∩ Ω. Ω is called a weak domain of holomorphy if for every p ∈ bΩ there is f ∈ ℋ(Ω) which is completely singular at p, and Ω is called a domain of holomorphy if there exists f ∈ ℋ(Ω) which is completely singular at every boundary point p ∈ bΩ. We already know that every domain in ℂ is a domain of holomorphy, see Theorem 4.38. For n ≥ 2 we already know examples of domains which fail to be domains of holomorphy. The concept of weak domain of holomorphy is convenient at the introductory level; it is, in fact, equivalent to the concept of domain of holomorphy, but this result is not elementary. Lemma 6.17. Every convex domain Ω in ℂn is a weak domain of holomorphy. Proof. Let p ∈ bΩ. The convexity implies that we can find an ℝ-linear function l ∶ ℂn ⟶ ℝ such that the hyperplane {z ∈ ℂn ∶ l(z) = l(p)} separates Ω and p, i.e. we may assume that l(z) < l(p) for z ∈ Ω. We can write n

n

j=1

j=1

l(z) = ∑ αj zj + ∑ βj z j , where αj , βj ∈ ℂ. Since l is real-valued, we have βj = αj , for j = 1, … , n. Set h(z) = 2 ∑nj=1 αj zj . Then h is complex-linear and l(z) = ℜh(z). Now the function fp (z) ∶=

1 h(z) − h(p)

is holomorphic on Ω and completely singular at p. In the following we consider the concept of holomorphically convex domains in ℂn in order to get further examples of domains of holomorphy. This concept was already introduced in Chapter 4 for a general treatment of the Runge approximation theorem in one complex variable. Definition 6.18. A domain Ω in ℂn is called holomorphically convex, if K̂ Ω is relatively compact in Ω for every compact set K ⊂ Ω, where K̂ Ω = {z ∈ Ω ∶ |f (z)| ≤ |f |K for all f ∈ ℋ(Ω)}. We call K holomorphically convex (ℋ(Ω)-convex), if K = K̂ Ω . Remark. A domain in ℂ is always holomorphically convex (see Remark 4.18 (e)). The situation is different in higher dimensions. Let Ω = {z ∈ ℂn ∶ 1/2 < |z| < 2} and K = {z ∈ ℂn ∶ |z| = 1}. Then K̂ Ω = K, if n = 1, but if n > 1, Corollary 6.11 implies that every f ∈

182 | 6 Several complex variables ℋ(Ω) extends to a holomorphic function f ̃ on B(0, 2). It follows from the maximum principle applied to f ̃ that for 1/2 < |z| ≤ 1, one has |f (z)| = |f ̃(z)| ≤ |f ̃|K = |f |K , hence {z ∈ Ω ∶ |z| ≤ 1} ⊂ K̂ Ω , and K̂ Ω is not relatively compact in Ω. Lemma 6.19. Let Ω be a holomorphically convex domain in ℂn . Then there is a compact exhaustion (Kj )j of Ω by holomorphically convex sets Kj . Proof. Since Ω is holomorphically convex, one can use Remark 4.18 (f). This can be used to construct unbounded holomorphic functions. Lemma 6.20. Let (Kj )j be a compact exhaustion of Ω by holomorphically convex sets Kj . Suppose that pj ∈ Kj+1 ⧵ Kj for j = 1, 2, … . Then there exists f ∈ ℋ(Ω) such that limj→∞ |f (pj )| = ∞. Proof. The desired function f is constructed as the limit of a series ∑m fm , where fm ∈ ℋ(Ω) is chosen such that |fm |Km < 2−m ,

m = 1, 2, … ,

(6.9)

and j−1

|fj (pj )| > j + 1 + ∑ |fm (pj )|, m=1

j = 2, 3, … .

(6.10)

We construct the sequence (fm )m inductively: set f1 = 0, and if k ≥ 2, suppose that f1 , … , fk−1 have already been found such that (6.9) and (6.10) hold. By Remark 4.18 (g), ̂ , there exists fk ∈ ℋ(Ω) with |fk |K < 2−k and such that (6.10) holds. since pk ∉ (Kk ) Ω k Now (6.9) implies that f = ∑∞ j=1 fj converges uniformly on all compact subsets of Ω. Hence f ∈ ℋ(Ω). Furthermore (6.10) implies |f (pj )| ≥ |fj (pj )| − ∑ |fm (pj )| > j + 1 − ∑ |fm (pj )|, m≠j

m>j

j ≥ 2.

Then (6.9) implies that ∑m>j |fm (pj )| < ∑m>j 2−m ≤ 1, and hence that |f (pj )| > j. It is now easy to show that a domain Ω is holomorphically convex, if and only if for every sequence (pj )j in Ω without limit point in Ω there is f ∈ ℋ(Ω) with supj |f (pj )| = ∞. In addition, one can now use Lemma 6.17 to show that every convex domain in ℂn is holomorphically convex (see Exercises).

6.3 Domains of holomorphy | 183

Now we introduce a class of domains which generalize the polydiscs. Definition 6.21. An open set Ω ⊂⊂ ℂn is called an analytic polyhedron if there are a neighborhood U of Ω and finitely many functions f1 , … , fk ∈ ℋ(U) such that Ω = {z ∈ U ∶ |f1 (z)| < 1, … , |fk (z)| < 1}. Example 6.22. Let 0 < q < 1. Then Ω = {(z1 , z2 ) ∈ ℂ2 ∶ |z1 | < 1, |z2 | < 1, |z1 z2 | < q} is an analytic polyhedron which is not a convex domain, see Figure 6.3.

Figure 6.3

Theorem 6.23. Every analytic polyhedron is holomorphically convex. Proof. Let Ω be like in Definition 6.21. If K ⊂ Ω is compact, then rj ∶= |fj |K < 1, for j = 1, … , k. It follows that K̂ Ω ⊂ {z ∈ U ∶ |f1 (z)| ≤ r1 , … , |fk (z)| ≤ rk }, and the set on the right-hand side is relatively compact in Ω. It is relatively easy to show that holomorphically convex domains are domains of holomorphy. For this we need some preparations which are similar to the proof of Theorem 4.38.

184 | 6 Several complex variables Lemma 6.24. Let Ω be a domain in ℂn . Let U be a connected neighborhood of p ∈ bΩ and let Ω1 ⊂ U ∩ Ω be a nonempty connected component of U ∩ Ω. Then bΩ1 ∩ (U ∩ bΩ) ≠ ∅. Proof. Since Ω1 is a component of the open set U ∩ Ω, it follows that Ω1 is open in ℂn and closed in U ∩ Ω. Since U is connected and Ω1 ≠ U, one has that Ω1 cannot be closed in U. Hence there exists q ∈ (bΩ1 ∩ U) ⧵ Ω1 . Since Ω1 ⊂ Ω and Ω1 is closed in U ∩ Ω, we have q ∈ bΩ, and so q ∈ bΩ1 ∩ (U ∩ bΩ). Lemma 6.25. Let (Km )m be a compact exhaustion of the domain Ω in ℂn . Then there are a subsequence (mj ) of ℕ and a sequence (pj )j of points in Ω such that (a) pj ∈ Kmj+1 ⧵ Kmj , for j = 1, 2, … , and (b) for every p ∈ bΩ and every connected neighborhood U of p, each component Ω1 of U ∩ Ω contains infinitely many points from (pj )j . Proof. Let (ak )k be an enumeration of the points of Ω with rational coordinates. Let rk = dist(ak , bΩ). Then the balls Bk = B(ak , rk ) are contained in Ω. Let (Qj )j be a sequence of such balls Bk which contains each Bk infinitely many times; for example, the sequence B1 , B1 , B2 , B1 , B2 , B3 , B1 , … . Now take Km1 = K1 and use induction: assume that l > 1 and p1 , … , pl−1 and Km1 , … , Kml have been chosen so that (a) holds for j = 1, … , l − 1. Since Ql is not contained in any compact subset of Ω, we may choose pl ∈ Ql ⧵ Kml and ml+1 such that pl ∈ Kml+1 . Then (a) holds for all j = 1, 2, … . We claim that the points (pj )j satisfy (b): given Ω1 as in (b), there is a point q ∈ bΩ1 ∩ (U ∩ bΩ), see Lemma 6.24. Hence there is aν ∈ Ω1 with rational coordinates sufficiently close to q so that Bν ⊂ Ω1 . Since Bν occurs infinitely many times in the sequence (Qj )j , and pj ∈ Qj for j = 1, 2, … , the ball Bν contains infinitely many points of the sequence (pj )j , and we are done. Theorem 6.26. Every holomorphically convex domain Ω in ℂn is a domain of holomorphy. Proof. We can choose a compact exhaustion (Kj )j of Ω with by holomorphically convex sets Kj . We apply Lemma 6.20 to the sequences (pj )j and (Kmj )j given by Lemma 6.25 to get f ∈ ℋ(Ω) with limj→∞ |f (pj )| = ∞. We claim that f is completely singular at every point p ∈ bΩ. If Ω1 is a component of U ∩ Ω, where U is a connected neighborhood of p, suppose there exists h ∈ ℋ(U) with f |Ω1 = h|Ω1 . Now we replace U by U ′ ⊂⊂ U and we replace Ω1 by a component Ω′1 of U ′ ∩ Ω which meets Ω1 , then we may assume that |h|Ω′1 ≤ |h|U ′ < ∞. Hence f would have to be bounded on Ω′1 , and this contradicts Lemma 6.25 (b) and limj→∞ |f (pj )| = ∞. Using Example 6.22, we have an example of a domain of holomorphy which is not convex. We now introduce the suitable generalization of convexity to characterize domains of holomorphy.

6.3 Domains of holomorphy | 185

Definition 6.27. A 𝒞2 real valued function φ on Ω is plurisubharmonic, if n

𝜕2 φ (z)tj t k ≥ 0, j,k=1 𝜕zj 𝜕z k

i𝜕𝜕φ(t, t)(p) ∶= ∑ for all t = (t1 , … , tn ) ∈ ℂn and all z ∈ Ω. φ is strictly plurisubharmonic, if n

𝜕2 φ (z)tj t k > 0, j,k=1 𝜕zj 𝜕z k

i𝜕𝜕φ(t, t)(p) ∶= ∑ for all t ∈ ℂn , t ≠ 0.

Remark 6.28. (a) A 𝒞2 real-valued function φ on Ω is plurisubharmonic, if and only if for every a ∈ Ω and w ∈ ℂn the function u ↦ φ(a + uw) is subharmonic on {u ∈ ℂ ∶ a + uw ∈ Ω}, see Exercises. For technical reasons it is convenient to include upper semicontinuous functions and to admit the value −∞ in the definition of plurisubharmonic functions, where one has to take the general definition of subharmonicity (see Remark 5.22). (b) Suppose r ∈ 𝒞2 (U) is a defining function for a domain Ω ⊂ ℂn , where U is a neighborhood of a point p ∈ bΩ. One can write the Taylor expansion of r at p in complex form: r(p + t) = r(p) + 2ℜ(𝜕rp (t) + Qp (r; t)) + i𝜕𝜕r(t, t)(p) + o(|t|2 ),

(6.11)

where t = (t1 , … , tn ) ∈ ℂn , n

𝜕r (p)tj , 𝜕z j j=1

𝜕rp (t) = ∑ Qp (r; t) =

n

1 𝜕2 r (p)tj tk . ∑ 2 j,k=1 𝜕zj 𝜕zk

(6.12) (6.13)

Definition 6.29. A bounded domain Ω in ℂn is called strictly pseudoconvex if there are a neighborhood U of bΩ and a strictly plurisubharmonic function r ∈ 𝒞2 (U) such that Ω ∩ U = {z ∈ U ∶ r(z) < 0}. The simplest example of a strictly pseudoconvex domain is a ball B(p, R), the function r(z) = |z − p|2 − R2 is strictly plurisubharmonic, and B(p, R) = {z ∈ ℂn ∶ r(z) < 0}. In the following we shall show that a strictly pseudoconvex domain is (at least) locally a domain of holomorphy. Lemma 6.30. Let U be open in ℂn and suppose r ∈ 𝒞2 (U) is strictly plurisubharmonic on U. If W ⊂⊂ U, then there are positive constants c > 0 and ϵ > 0 such that the function F (r) (ζ , z) defined on U × ℂn by

186 | 6 Several complex variables n

n

𝜕r 𝜕2 r 1 (ζ )(ζj − zj ) − ∑ (ζ )(ζj − zj )(ζk − zk ) 2 j,k=1 𝜕ζj 𝜕ζk j=1 𝜕ζj

F (r) (ζ , z) = ∑

(6.14)

satisfies the estimate 2ℜF (r) (ζ , z) ≥ r(ζ ) − r(z) + c|z − ζ |2

(6.15)

for ζ ∈ W and |z − ζ | < ϵ. Proof. From (6.11) with p = ζ ∈ U and t = z − ζ , we obtain the Taylor expansion of r(z) at ζ : r(z) = r(ζ ) − 2ℜF (r) (ζ , z) + i𝜕𝜕r(z − ζ , z − ζ )(ζ ) + o(|z − ζ |2 ).

(6.16)

Since r is strictly plurisubharmonic, we have i𝜕𝜕r(t, t)(p) ≥ κ|t|2 where κ = min{i𝜕𝜕r(t, t)(p) ∶ |t| = 1} is positive. So, if 0 < c < κ, the continuity of the second derivatives of r implies i𝜕𝜕r(t, t)(z) ≥ c|t|2 , for t ∈ ℂn and all z in some neighborhood of p. As W ⊂ U is compact there is c > 0 such that i𝜕𝜕r(z − ζ , z − ζ )(ζ ) ≥ 2c|z − ζ |2 for ζ ∈ W and z ∈ ℂn . Now we use Taylor’s theorem and the uniform continuity on W of the derivatives of r up to order 2 to show that there exists ϵ > 0 such that the error term o(|z − ζ |2 ) in (6.16) can be estimated in the form o(|z − ζ |2 ) ≤ c|z − ζ |2 uniformly for ζ ∈ W and if |z − ζ | < ϵ. The desired estimates (6.15) now follow from (6.16). Theorem 6.31. Let Ω be a strictly pseudoconvex domain. Then every point p ∈ bΩ has a neighborhood V such that V ∩ Ω is a (weak) domain of holomorphy. Proof. Let r ∈ 𝒞2 (U) be strictly plurisubharmonic in a neighborhood U of bΩ so that Ω∩U = {z ∈ U ∶ r(z) < 0}. Choose c, ϵ as in Lemma 6.30 such that (6.15) holds for ζ ∈ bΩ. For ζ ∈ bΩ we have r(ζ ) = 0, and (6.15) implies that ℜF (r) (ζ , z) > 0 for z ∈ Ω with |z − ζ | < ϵ (choose ϵ so small that B(ζ , ϵ) ⊂ U for ζ ∈ bΩ). If p ∈ bΩ is fixed, set V = B(p, ϵ/2). We claim that V ∩ Ω is a weak domain of holomorphy: for ζ ∈ V ∩ bΩ, the function fζ (z) ∶=

1 F (r) (ζ , z)

is holomorphic on V ∩ Ω and completely singular at ζ ; for any of the remaining boundary points ζ ∈ bV ∩ Ω of V ∩ Ω the convexity of V implies that there is g ∈ ℋ(V) which is completely singular at ζ , see Lemma 6.17. Remark 6.32. We mention different types of pseudoconvexity: Let Ω be a bounded domain in ℂn with n ≥ 2, and let r be a 𝒞2 defining function for Ω. Ω is called Levi pseudoconvex at p ∈ bΩ, if the Levi form n

𝜕2 r (p)tj t k ≥ 0 j,k=1 𝜕zj 𝜕z k

i𝜕𝜕r(t, t)(p) ∶= ∑

6.4 Exercises | 187

for all n

t ∈ Tp1,0 (bΩ) = {t = (t1 , … , tn ) ∈ ℂn ∶ ∑ tj (𝜕r/𝜕zj )(p) = 0}, j=1

where Tp1,0 (bΩ) is the space of type (1, 0) vector fields which are tangent to the boundary at the point p. The domain Ω is said to be strictly Levi pseudoconvex at p, if the Levi form is strictly positive for all such t ≠ 0. Ω is called a Levi pseudoconvex domain if Ω is Levi pseudoconvex at every boundary point of Ω. A bounded domain Ω in ℂn is pseudoconvex if Ω has a 𝒞2 strictly plurisubharmonic exhaustion function φ ∶ Ω ⟶ ℝ, i.e. the sets {z ∈ Ω ∶ φ(z) < c} are relatively compact in Ω, for every c ∈ ℝ. (Here there is no assumption on the boundary of Ω.) It turns out that for bounded domains with 𝒞2 boundary the concepts of (strictly) Levi pseudoconvex and (strictly) pseudoconvex domains coincide. Furthermore, the following assertion holds: Let Ω be a domain in ℂn . The following are equivalent: (1) Ω is pseudoconvex. ∞ (2) The equation 𝜕u = f always has a solution u ∈ 𝒞∞ (p,q) (Ω) for any form f ∈ 𝒞(p,q+1) (Ω) with 𝜕f = 0, q = 0, 1, … , n − 1. (3) Ω is a domain of holomorphy. The proof is beyond the scope of this book. The most difficult part is the solution of the Levi problem, to prove that a pseudoconvex domain is a domain of holomorphy, see [41, 59, 51].

6.4 Exercises 111. Show that the Cayley transform Φ(z1 , … , zn ) = (w1 , … , wn ), where wj = zj /(1 + zn ) for 1 ≤ j ≤ n − 1 and wn = i(1 − zn )/(1 + zn ) is a biholomorphic map from 𝔹 ⟶ 𝕌. 112. Let n > 1. Show that the boundary b𝕌 = {(z ′ , t + i|z ′ |2 ) ∶ z ′ ∈ ℂn−1 , t ∈ ℝ} of the Siegel upper half-space can be identified with ℂn−1 × ℝ. Show that the multiplication (z ′ , t) ⋅ (ζ ′ , τ) = (z ′ + ζ ′ , t + τ + 2ℑ⟨z ′ , ζ ′ ⟩) turns b𝕌 into a group which is non-Abelian. This group is called the Heisenberg4 group. 4 Heisenberg, Werner (1901–1976).

188 | 6 Several complex variables 113. Let f be holomorphic in a neighborhood of the closed polydisc P(0, r) ⊂ ℂn , where n > 1, with the possible exception of the origin (0, … , 0) ∈ ℂn . Suppose that not all zj , where 1 ≤ j ≤ n − 1, are zero. Prove that f (z1 , … , zn ) =

1 f (z1 , … , zn−1 , ζn ) dζn , ∫ 2πi |ζn |=rn ζn − zn

and show that the integral on the right-hand side depends holomorphically on z1 , … , zn for all z = (z1 , … , zn ) ∈ P(0, r). Therefore holomorphic functions of several variables do not have isolated singularities. 114. Let Ω be a bounded domain in ℂn , n > 1, with a defining function ρ ∈ 𝒞2 which vanishes precisely on bΩ and dρ ≠ 0 on bΩ. Show that the condition 𝜕u ∧ 𝜕ρ = 0 on bΩ can also be stated as n

∑ tj j=1

𝜕u =0 𝜕z j

on bΩ,

𝜕ρ for all (t1 , … , tn ) ∈ ℂn with ∑nj=1 tj 𝜕z = 0 on bΩ. We say that u is a CR-function. j Show that this definition does not depend on the choice of the defining function ρ. 115. Use Lemma 6.17 and Lemma 6.20 to show that every convex domain in ℂn is holomorphically convex. 116. Let 0 < q < 1. Define

Ω = {(z1 , z2 ) ∈ ℂ2 ∶ |z1 | < 1, |z2 | < 1, |z1 z2 | < q}. Show that Ω is a domain of holomorphy, which is not convex. 117. Let Ω be a domain in ℂn . Show that a 𝒞2 real-valued function φ on Ω is plurisubharmonic, if and only if for every a ∈ Ω and w ∈ ℂn the function u ↦ φ(a + uw) is subharmonic on {u ∈ ℂ ∶ a + uw ∈ Ω}. 118. Let Ω be a domain in ℂn and let f ∈ ℋ(Ω). Show that |f |α , α > 0, and log |f | are plurisubharmonic on Ω. 119. Let Ω ⊆ ℂn and G ⊆ ℂm be domains and let F ∶ G ⟶ Ω be a holomorphic map. Suppose that u ∈ 𝒞2 (Ω) is plurisubharmonic on Ω. Show that u ∘ F is plurisubharmonic on G.

6.5 Notes In Section 6.2 we have followed the expositions of L. Hörmander [41] and R.M. Range [59]. For a thorough treatment of Lewy’s Theorem (see Theorem 6.15) including interesting consequences for Hardy spaces, the reader should consult E. Stein [67]. Pseudoconvexity is also crucial for the inhomogeneous Cauchy–Riemann equations, as well as plurisubharmonic functions, see Chapters 10 and 11. The Levi problem to construct

6.5 Notes | 189

a holomorphic function on a pseudoconvex domain which is completely singular at the boundary is solved by means of integral representations in [59]. Another proof uses the powerful method of global solutions and estimates for the inhomogeneous Cauchy–Riemann equations [41], this method will be discussed and exploited in more details in the following chapters.

7 Bergman spaces In the following we study topological vector spaces of holomorphic functions to demonstrate the interplay between functional analysis and complex analysis. The first example for this will be a Hilbert space of holomorphic functions which has a striking property – the so-called reproducing property. The corresponding integral kernel – the Bergman kernel – is computed for the ball and polydisc, as well as for the Fock space of entire functions (Section 7.2). In addition we prove the important transformation formula for the Bergman kernel and discuss its relationship to the Riemann Mapping Theorem.

7.1 Elementary properties Definition 7.1. Let Ω ⊆ ℂn be a domain. The Bergman1 space A2 (Ω) is defined by A2 (Ω) = {f ∶ Ω ⟶ ℂ holomorphic ∶ ‖f ‖2 = ∫ |f (z)|2 dλ(z) < ∞}, Ω

where λ is the Lebesgue measure of ℂn . The inner product is given by (f , g) = ∫ f (z)g(z) dλ(z), Ω

for f , g ∈ A2 (Ω). For sake of simplicity, we first restrict to domains Ω ⊆ ℂ. We consider special continuous linear functionals on A2 (Ω), the point evaluations. Let f ∈ A2 (Ω) and fix z ∈ Ω. By Cauchy’s integral theorem, we have f (z) =

1 f (ζ ) dζ , ∫ 2πi γs ζ − z

where γs (t) = z + seit , t ∈ [0, 2π], 0 < s ≤ r and Dr (z) = {w ∶ |w − z| < r} ⊂ Ω. Using polar coordinates and integrating the above equality with respect to s between 0 and r, we get f (z) =

1 Bergman, Stefan (1895–1977). https://doi.org/10.1515/9783110417241-007

1 f (w) dλ(w). ∫ πr 2 Dr (z)

(7.1)

192 | 7 Bergman spaces Then, by Cauchy–Schwarz inequality, |f (z)| ≤ ≤ ≤ ≤

1 1 ⋅ |f (w)| dλ(w) ∫ πr 2 Dr (z)

1/2 1/2 1 (∫ 12 dλ(w)) (∫ |f (w)|2 dλ(w)) 2 πr Dr (z) Dr (z)

1

π 1/2 r 1

π 1/2 r

(∫ |f (w)|2 dλ(w))

1/2

Ω

‖f ‖.

If K is a compact subset of Ω, there is an r(K) > 0 such that for any z ∈ K we have Dr(K) (z) ⊂ Ω and then get sup|f (z)| ≤ z∈K

1 ‖f ‖. π 1/2 r(K)

n

If K ⊂ Ω ⊂ ℂ , we can find a polycylinder P(z, r(K)) = {w ∈ ℂn ∶ |wj − zj | < r(K), j = 1, … , n} such that for any z ∈ K we have P(z, r(K)) ⊂ Ω. Hence by iterating the above Cauchy integrals, we get Theorem 7.2. Let K ⊂ Ω be a compact set. Then there exists a constant C(K), only depending on K, such that sup|f (z)| ≤ C(K)‖f ‖, z∈K

(7.2)

for any f ∈ A2 (Ω). Theorem 7.3. A2 (Ω) is a Hilbert space. Proof. If (fk )k is a Cauchy sequence in A2 (Ω), by (7.2), it is also a Cauchy sequence with respect to uniform convergence on compact subsets of Ω. Hence the sequence (fk )k has a holomorphic limit f with respect to uniform convergence on compact subsets of Ω. On the other hand, the original L2 -Cauchy sequence has a subsequence, which converges pointwise almost everywhere to the L2 -limit of the original L2 -Cauchy sequence (see, for instance, [63]), and so the L2 -limit coincides with the holomorphic function f . Therefore A2 (Ω) is a closed subspace of L2 (Ω) and itself a Hilbert space. In the sequel we present basic facts about Hilbert2 spaces and their consequences for the Bergman spaces. 2 Hilbert, David (1862–1943).

7.1 Elementary properties | 193

Theorem 7.4. Let E be a non-empty, convex, closed subset of the Hilbert space H, i.e. for x, y ∈ E one has tx + (1 − t)y ∈ E, for each t ∈ [0, 1]. Then E contains a uniquely determined element of minimal norm. Proof. The parallelogram rule says that ‖x + y‖2 + ‖x − y‖2 = 2‖x‖2 + 2‖y‖2 ,

x, y ∈ H.

Let δ = inf{‖x‖ ∶ x ∈ E}. For x, y ∈ E we have 21 (x + y) ∈ E, hence 1/4‖x − y‖2 = 1/2‖x‖2 + 1/2‖y‖2 − ‖1/2(x + y)‖2 , implies that ‖x − y‖2 ≤ 2‖x‖2 + 2‖y‖2 − 4δ2 . So, if ‖x‖ = ‖y‖ = δ, then x = y (uniqueness). By the definition of δ, there exists a sequence (yk )k in E such that ‖yk ‖ → δ if k → ∞. The estimate ‖yk − ym ‖2 ≤ 2‖yk ‖2 + 2‖ym ‖2 − 4δ2 implies that (yk )k is a Cauchy sequence in H. Since H is complete there exists x0 ∈ H with ‖yk − x0 ‖ → 0 and, as E is closed, we have x0 ∈ E; the mapping x ↦ ‖x‖ is continuous and therefore ‖x0 ‖ = limk→∞ ‖yk ‖ = δ. Theorem 7.5. Let M be a closed subspace of the Hilbert space H. Then there exist uniquely determined mappings P ∶ H ⟶ M,

Q ∶ H ⟶ M⟂

such that (1) x = Px + Qx ∀x ∈ H (2) for x ∈ M we have Px = x, hence P 2 = P and Qx = 0; for x ∈ M ⟂ we have Px = 0, Qx = x, and Q2 = Q. (3) The distance of x ∈ H to M is given by inf{‖x − y‖ ∶ y ∈ M} = ‖x − Px‖. (4) For each x ∈ H, we have ‖x‖2 = ‖Px‖2 + ‖Qx‖2 . (5) P and Q are continuous, linear, self-adjoint operators. P and Q are the orthogonal projections of H onto M and M ⟂ .

194 | 7 Bergman spaces Proof. For each x ∈ H, the set x + M = {x + y ∶ y ∈ M} is convex. Hence, by Proposition 7.4, there exists a uniquely determined element of minimal norm in x + M, which is denoted by Qx. We set Px = x − Qx and see that Px ∈ M, since Qx ∈ x + M. Now we claim that Qx ∈ M ⟂ . We have to show that (Qx, y) = 0 ∀y ∈ M: we can suppose that ‖y‖ = 1, then we have (Qx, Qx) = ‖Qx‖2 ≤ ‖Qx − αy‖2 = (Qx − αy, Qx − αy)

∀α ∈ ℂ,

by the minimality of Qx. Therefore we get 0 ≤ −α(y, Qx) − α(Qx, y) + |α|2 , setting α = (Qx, y), we obtain 0 ≤ −|(Qx, y)|2 and (Qx, y) = 0; hence Q ∶ H ⟶ M ⟂ . If x = x0 +x1 with x0 ∈ M and x1 ∈ M ⟂ , then x0 −Px = Qx −x1 , and since M ∩M ⟂ = {0} we obtain x0 = Px and x1 = Qx, therefore P and Q are uniquely determined. In a similar way, we get that P(αx + βy) − αPx − βPy = αQx + βQy − Q(αx + βy). The left-hand side belongs to M, the right-hand side belongs to M ⟂ , hence both sides are 0, which proves that P and Q are linear. Property (3) follows by the definition of Q, whereas property (4) by the fact that (Px, Qx) = 0

∀x ∈ H.

In addition, we have ‖Q(x − y)‖ = inf{‖x − y + m‖ ∶ m ∈ M} ≤ ‖x − y‖, hence Q and P = I − Q are continuous. For x, y ∈ H, we have (Px, y) = (Px, Py + Qy) = (Px, Py)

and (x, Py) = (Px + Qx, Py) = (Px, Py)

hence (Px, y) = (x, Py), and so P is self-adjoint. Corollary 7.6. If M ≠ H is a closed, proper subspace of the Hilbert space H, then there exists an element y ≠ 0 with y ⟂ M. Proof. Let x ∈ H such that x ∉ M. Set y = Qx. Then x ≠ Px implies y ≠ 0. The next result is the Riesz3 representation theorem. 3 Riesz, Frigyes (1880–1956).

7.1 Elementary properties | 195

Theorem 7.7. Let L be a continuous linear functional on the Hilbert space H. Then there exists a uniquely determined element y ∈ H such that Lx = (x, y) ∀x ∈ H. Proof. If L(x) = 0 ∀x ∈ H, then we set y = 0. Otherwise we define M = {x ∶ Lx = 0}. Then, by the continuity of L, the subspace M of H is closed. By Corollary 7.6, we have M ⟂ ≠ 0. Lz ⟂ Let z ∈ M ⟂ with z ≠ 0. Then Lz ≠ 0. Now set y = αz, where α = ‖z‖ and 2 . Then y ∈ M Ly = L(αz) =

Lz |Lz|2 Lz = = (y, y) = |α|2 (z, z). ‖z‖2 ‖z‖2

For x ∈ H, we define x′ = x −

Lx y (y, y)

and x″ =

Lx y. (y, y)

Then we obtain Lx′ = 0 and x′ ∈ M, hence (x′ , y) = 0 and (x, y) = (x″ , y) = (

Lx y, y) = Lx. (y, y)

If (x, y) = (x, y′ ) ∀x ∈ H, then we get (x, y − y′ ) = 0 ∀x ∈ H, in particular (y − y′ , y − y′ ) = 0. Therefore y = y′ , which shows that y is uniquely determined. Corollary 7.8. Let H be a Hilbert space and L ∈ H ′ a continuous linear functional. Then the dual norm ‖L‖ = sup{|Lx| ∶ ‖x‖ ≤ 1} can be expressed in the form ‖L‖ = sup{|(x, y)| ∶ ‖x‖ ≤ 1} = ‖y‖, where y ∈ H corresponds to L. For fixed z ∈ Ω, (7.2) also implies that the point evaluation f ↦ f (z) is a continuous linear functional on A2 (Ω), hence, by the Riesz representation theorem (see Theorem 7.7), there is a uniquely determined function kz ∈ A2 (Ω) such that f (z) = (f , kz ) = ∫ f (w)kz (w) dλ(w). Ω

(7.3)

We set K(z, w) = kz (w). Then w ↦ K(z, w) = kz (w) is an element of A2 (Ω), hence the function w ↦ K(z, w) is anti-holomorphic on Ω and we have f (z) = ∫ K(z, w)f (w) dλ(w), Ω

f ∈ A2 (Ω).

196 | 7 Bergman spaces The function of two complex variables (z, w) ↦ K(z, w) is called Bergman kernel of Ω and the above identity represents the reproducing property of the Bergman kernel. Now we use the reproducing property for the holomorphic function z ↦ ku (z), where u ∈ Ω is fixed: ku (z) = ∫ K(z, w)ku (w) dλ(w) Ω

= ∫ kz (w)K(u, w) dλ(w) Ω

−

= (∫ K(u, w)kz (w) dλ(w)) Ω

= kz (u), hence we have ku (z) = kz (u), or K(z, u) = K(u, z). It follows that the Bergman kernel is holomorphic in the first variable and antiholomorphic in the second variable. Theorem 7.9. The Bergman kernel is uniquely determined by the properties that it is an element of A2 (Ω) in z and that it is conjugate symmetric and reproduces A2 (Ω). Proof. To see this, let K ′ (z, w) be another kernel with these properties. Then we have K(z, w) = ∫ K ′ (z, u)K(u, w) dλ(u) Ω

= (∫ K(w, u)K ′ (u, z) dλ(u))

−

Ω

= K ′ (w, z) = K ′ (z, w). Now let ϕ ∈ L2 (Ω). Since A2 (Ω) is a closed subspace of L2 (Ω) there exists a uniquely determined orthogonal projection P ∶ L2 (Ω) ⟶ A2 (Ω), see Theorem 7.5. For the function Pϕ ∈ A2 (Ω), we use the reproducing property and obtain Pϕ(z) = ∫ K(z, w)Pϕ(w) dλ(w) = (Pϕ, kz ) = (ϕ, Pkz ) = (ϕ, kz ); Ω

(7.4)

where we still have used that P is a self-adjoint operator and that Pkz = kz . Hence Pϕ(z) = ∫ K(z, w)ϕ(w) dλ(w). Ω

(7.5)

P is called the Bergman projection. In order to compute the Bergman kernel for some special cases, we need the concept of a complete orthonormal basis.

7.1 Elementary properties | 197

A subset {uα ∶ α ∈ A} of a Hilbert space is called orthonormal, if (uα , uβ ) = δαβ for each α, β ∈ A. If (xk )k is a linearly independent sequence in H, there is a standard procedure, called the Gram–Schmidt process, for converting (xk )k into an orthonormal sequence (uk )k such that the linear span of (uk )Nk=1 equals the linear span of (xk )Nk=1 for all N ∈ ℕ. We start with u1 = x1 /‖x1 ‖. Having defined u1 , … , uN−1 , we set N−1

vN = xN − ∑ (xN , uj )uj . j=1

The element vN is nonzero because xN is not in the linear span of x1 , … , xN−1 and hence of u1 , … , uN−1 . So we can set uN = vN /‖vN ‖. It is now clear that (uk )Nk=1 has the desired properties. Next we prove Bessel’s4 inequality. Theorem 7.10. If {uα ∶ α ∈ A} is an orthonormal set in the Hilbert space H, then for any u∈H ∑ |(u, uα )|2 ≤ ‖u‖2 .

(7.6)

α∈A

In particular, the set {α ∶ (u, uα ) ≠ 0} is countable. Proof. It suffices to show that ∑α∈F |(u, uα )|2 ≤ ‖u‖2 , for any finite set F ⊂ A. We use the property ‖uα + uβ ‖2 = ‖uα ‖2 + ‖uβ ‖2 , for α, β ∈ A, α ≠ β. 2

0 ≤ ‖u − ∑ (u, uα )uα ‖ α∈F

2

2

= ‖u‖ − 2ℜ(u, ∑ (u, uα )uα ) + ‖ ∑ (u, uα )uα ‖ α∈F

α∈F

= ‖u‖2 − 2 ∑ |(u, uα )|2 + ∑ |(u, uα )|2 α∈F

= ‖u‖2 − ∑ |(u, uα )|2 .

α∈F

α∈F

Theorem 7.11. If {uα ∶ α ∈ A} is an orthonormal set in the Hilbert space H, then the following conditions are equivalent: (1) (Completeness) If (u, uα ) = 0 for all α ∈ A, then u = 0. (2) (Parseval’s5 equation) ‖u‖2 = ∑α∈A |(u, uα )|2 for all u ∈ H. (3) u = ∑α∈A (u, uα )uα for each u ∈ H, where the sum has only countably many nonzero terms and converges in norm to u no matter how these terms are ordered. 4 Bessel, Friedrich Wilhelm (1784–1846). 5 Parseval, Marc-Antoine (1784–1846).

198 | 7 Bergman spaces Proof. (Claim (1) implies Claim (3)) If u ∈ H, let α1 , α2 , … be any enumeration of those 2 α’s for which (u, uα ) ≠ 0. By Bessel’s inequality, the series ∑∞ j=1 |(u, uαj )| converges, so M

2

M

‖ ∑ (u, uα )uα ‖ = ∑ |(u, uα )|2 → 0 j=m

as m, M → ∞.

j=m

∞ Hence the series ∑∞ j=1 (u, uαj )uαj converges in H. If v = u − ∑j=1 (u, uαj )uαj , then (v, uα ) = 0 for all α ∈ A, so by Claim (1), v = 0. (Claim (3) implies Claim (2)) As in the proof of Bessel’s inequality, we have 2

m

m

j=1

j=1

‖u‖2 − ∑|(u, uαj )|2 = ‖u − ∑(u, uαj )uαj ‖ → 0

as m → ∞.

Finally, that Claim (2) implies Claim (1) is obvious. An orthonormal set having the properties of Theorem 7.11 is called an orthonormal basis of H. Remark. An application of Zorn’s Lemma shows that the collection of orthonormal sets in a Hilbert space, ordered by inclusion, has a maximal element. Maximality is equivalent to Claim (1) of Theorem 7.11, hence for each orthonormal set there exists an orthonormal basis, which contains the given orthonormal set. If H is a separable Hilbert space, it has a countable orthonormal basis. Theorem 7.12. Let K ⊂ Ω be a compact subset and {ϕj } be an orthonormal basis of A2 (Ω). Then the series ∞

∑ ϕj (z)ϕj (w) j=1

sums uniformly on K × K to the Bergman kernel K(z, w). Proof. For the proof of this statement, we use the duality for the sequence space l2 to get ∞

sup(∑|ϕj (z)|2 ) z∈K

j=1

1/2

∞

∞

j=1

j=1

= sup{|∑ aj ϕj (z)| ∶ ∑ |aj |2 = 1, z ∈ K} = sup{|f (z)| ∶ ‖f ‖ = 1, z ∈ K} ≤ C(K),

(7.7)

where we have used (7.2) in the last inequality. Now Cauchy–Schwarz inequality gives ∞

∞

2

∑|ϕj (z)ϕj (w)| ≤ (∑|ϕj (z)| ) j=1

j=1

1/2

∞

2

1/2

(∑|ϕj (w)| ) j=1

7.2 Examples | 199

with uniform convergence at z, w ∈ K. In addition it follows that (ϕj (z))j ∈ l2 and the function ∞

w ↦ ∑ ϕj (z)ϕj (w) j=1

belongs to A2 (Ω). Let the sum of the series be denoted by K ′ (z, w). Notice that K ′ (z, w) is conjugate symmetric and that for f ∈ A2 (Ω) we get ∞

∫ K ′ (z, w)f (w) dλ(w) = ∑ ∫ f (w)ϕj (w) dλ(w) ϕj (z) = f (z) j=1 Ω

Ω

with convergence in the Hilbert space A2 (Ω). But (7.7) implies uniform convergence on compact subsets of Ω, hence f (z) = ∫ K ′ (z, w)f (w) dλ(w), Ω

for all f ∈ A2 (Ω), so K ′ (z, w) is a reproducing kernel. By the uniqueness of the Bergman kernel, we obtain K ′ (z, w) = K(z, w). We notice that (7.7) implies K(z, z) = sup{|f (z)|2 ∶ f ∈ A2 (Ω), ‖f ‖ = 1}.

(7.8)

7.2 Examples (a) The functions ϕn (z) = √ n+1 z n , n = 0, 1, 2, … constitute an orthonormal basis in π A2 (𝔻), 𝔻 = {z ∈ ℂ ∶ |z| < 1}. This follows from 2π

1

0

0

∫ z n z m dλ(z) = ∫ ∫ r n einθ r m e−imθ r dr dθ = 𝔻

2π δ . n + m + 2 n,m

n For each f ∈ A2 (𝔻) with Taylor series expansion f (z) = ∑∞ n=0 an z , we get 1

2π

0

0

(f , z n ) = ∫ f (z)z n dλ(z) = ∫ ∫ f (reiθ )r n e−inθ r dr dθ 𝔻 1

2π

0

0

=∫ ∫

1 f (reiθ ) a iθ 2n+1 re dθr dr = 2πa r 2n+1 dr = π n , ∫ n n +1 r n+1 ei(n+1)θ 0

where we used the fact that an =

1 f (z) ∫ n+1 dz, 2πi γr z

200 | 7 Bergman spaces for γr (θ) = reiθ . Hence, by the uniqueness of the Taylor series expansion, we obtain that (f , ϕn ) = 0, for each n = 0, 1, 2, … implies f ≡ 0. This means that (ϕn )∞ n=0 constitutes an orthonormal basis for A2 (𝔻) and we get ∞

‖f ‖2 = ∑ |(f , ϕn )|2 , n=0

which is equivalent to |an |2 , n=0 n + 1

∞

∞

f (z) = ∑ an z n .

‖f ‖2 = π ∑

n=0

Hence each f ∈ A2 (𝔻) can be written in the form f = ∑∞ n=0 cn ϕn , where the sum converges in A2 (𝔻), but also uniformly on compact subsets of 𝔻. For the coefficients cn we have cn = (f , ϕn ). Now we compute an explicit formula for the Bergman kernel K(z, w) of 𝔻. The function z ↦ K(z, w), with w ∈ 𝔻 fixed, belongs to A2 (𝔻). Hence we get from the above formula that ∞

K(z, w) = ∑ cn ϕn (z), n=0

where cn = (K(⋅, w), ϕn ), in other words, cn = (ϕn , K(⋅, w)) = ∫ ϕn (z)K(w, z) dλ(z) = ϕn (w), 𝔻

by the reproducing property of the Bergman kernel. This implies that the Bergman kernel is of the form ∞

K(z, w) = ∑ ϕn (z)ϕn (w), n=0

(7.9)

where the sum converges uniformly in z on all compact subsets of 𝔻. (This is true for any complete orthonormal system, as is shown above.) A simple computation now gives ∞

K(z, w) = ∑ ϕn (z)ϕn (w) = n=0

1 1 1 . ∑ (n + 1)(zw)n = π n=0 π (1 − zw)2 ∞

Hence for each f ∈ A2 (𝔻) we have f (z) =

1 1 f (w) dλ(w). ∫ π 𝔻 (1 − zw)2

(7.10)

7.2 Examples | 201

(b) Theorem 7.13. Let Ωj ⊂ ℂnj , j = 1, 2 be two bounded domains with Bergman kernels KΩ1 and KΩ2 . Then the Bergman kernel KΩ of the product domain Ω = Ω1 × Ω2 is given by KΩ ((z1 , z2 ), (w1 , w2 )) = KΩ1 (z1 , w1 )KΩ2 (z2 , w2 )

(7.11)

for (z1 , z2 ), (w1 , w2 ) ∈ Ω1 × Ω2 . Proof. In order to show this, let F denote the function on the right-hand side of (7.11). It is clear that (z1 , z2 ) ↦ F((z1 , z2 ), (w1 , w2 )) belongs to A2 (Ω) for each fixed (w1 , w2 ) ∈ Ω and that F is anti-holomorphic in the second variable. The reproducing property f (z1 , z2 ) = ∫

Ω1 ×Ω2

F((z1 , z2 ), (w1 , w2 ))f (w1 , w2 ) dλ(w1 , w2 )

is a consequence of Fubini’s theorem and the corresponding reproducing properties of KΩ1 and KΩ2 . Hence, by the uniqueness property of the Bergman kernel, Theorem 7.9, we obtain F = KΩ . From this we get that the Bergman kernel of the polycylinder 𝔻n is given by K𝔻n (z, w) =

n

1 1 . ∏ π n j=1 (1 − zj wj )2

(7.12)

(c) For the computation of the Bergman kernel K𝔹n of the unit ball in ℂn , we use the Beta and Gamma functions 1

∫ xk (1 − x)m dx = B(k + 1, m + 1) = 0

Γ(k + 1)Γ(m + 1) , Γ(k + m + 2)

where k, m ∈ ℕ and that for 0 ≤ a < 1,

∫

√1−a2

0

x2k+1 (1 −

1 x2 m+1 1 2 k+1 ) dx = (1 − a ) yk (1 − y)m+1 dy ∫ 1 − a2 2 0 1 = (1 − a2 )k+1 B(k + 1, m + 2) 2 1 Γ(k + 1)Γ(m + 2) . = (1 − a2 )k+1 2 Γ(k + m + 3) α

α

Now we can normalize the orthogonal basis {z α = z1 1 ⋯ zn n } in A2 (𝔹n ) to obtain ‖z α ‖2 = ∫ |z1 |2α1 ⋯ |zn |2αn dλ(z) 𝔹n

=

π ∫ |z |2α1 ⋯ |zn−1 |2αn−1 (1 − |z1 |2 − ⋯ − |zn−1 |2 )αn +1 dλ αn + 1 𝔹n−1 1

202 | 7 Bergman spaces

=

π ∫ |z |2α1 ⋯ |zn−2 |2αn−2 (1 − |z1 |2 − ⋯ − |zn−2 |2 )αn +1 αn + 1 𝔹n−1 1

αn +1 |zn−1 |2 ) dλ 1 − |z1 |2 − ⋯ − |zn−2 |2 π πΓ(αn−1 + 1)Γ(αn + 2) = αn + 1 Γ(αn + αn−1 + 3)

× |zn−1 |2αn−1 (1 −

×∫

𝔹n−2

|z1 |2α1 ⋯ |zn−2 |2αn−2 (1 − |z1 |2 − ⋯ − |zn−2 |2 )αn +αn−1 +2 dλ

π πΓ(αn−1 + 1)Γ(αn + 2) πΓ(α1 + 1)Γ(αn + ⋯ + α2 + n) ⋯ αn + 1 Γ(αn + αn−1 + 3) Γ(αn + ⋯ + α1 + n + 1) π n α1 ! ⋯ αn ! = . (αn + ⋯ + α1 + n)! =

Hence the Bergman kernel of the unit ball is given by K𝔹n (z, w) = ∑ α

=

(αn + ⋯ + α1 + n)! α α z w π n α1 ! ⋯ αn !

1 (α + ⋯ + α1 + n)! α α z w ∑ ∑ n π n k=0 |α|=k α1 ! ⋯ αn ! ∞

1 ∑ (k + n)(k + n − 1) ⋯ (k + 1)(z1 w1 + ⋯ + zn wn )k π n k=0 1 n! . = n π (1 − (z1 w1 + ⋯ + zn wn ))n+1 ∞

=

2

(d) In the sequel we will also consider the Fock space A2 (ℂn , e−|z| ) consisting of all entire functions f such that ∫ |f (z)|2 e−|z| dλ(z) < ∞. 2

ℂn

It is clear, that the Fock space is a Hilbert space with the inner product 2

(f , g) = ∫ f (z)g(z)e−|z| dλ(z). ℂn

2

Similar as in beginning of this chapter, setting n = 1, we obtain for f ∈ A2 (ℂ, e−|z| ) that |f (z)| ≤ ≤

2 2 1 e|w| /2 |f (w)|e−|w| /2 dλ(w) ∫ πr 2 Dr (z)

1/2 1/2 1 2 −|w|2 |w|2 (∫ e dλ(w)) (∫ |f (w)| e dλ(w)) πr 2 Dr (z) Dr (z)

≤ C(∫ |f (w)|2 e−|w| dλ(w)) 2

ℂ

≤ C‖f ‖,

1/2

7.3 Biholomorphic mappings | 203

where C is a constant only depending on z. This implies that the Fock space 2 A2 (ℂn , e−|z| ) has the reproducing property. The monomials {z α } constitute an orthogonal basis and the norms of the monomials are 2

2

‖z α ‖2 = ∫ |z1 |2α1 e−|z1 | dλ(z1 ) ⋯ ∫ |zn |2αn e−|zn | dλ(zn ) ℂ

ℂ

n

∞

= (2π) ∫ r 0

2α1 +1 −r 2

e

∞

2

dr ⋯ ∫ r 2αn +1 e−r dr 0

= π n α1 ! ⋯ αn !. 2

Hence the Bergman kernel of A2 (ℂn , e−|z| ) is of the form K(z, w) = ∑ α

1 z α wα 1 z α wα = = exp(z1 w1 + ⋯ + zn wn ). ∑ ∑ ‖z α ‖2 π n k=0 |α|=k α1 ! ⋯ αn ! π n ∞

(7.13)

7.3 Biholomorphic mappings Here we describe the behavior of the Bergman kernel under biholomorphic mappings. Theorem 7.14. Let F ∶ Ω1 ⟶ Ω2 be a biholomorphic map between bounded domains in ℂn . Let f1 , … , fn be the components of F and F ′ (z) = ( Then

𝜕fj (z) n ) . 𝜕zk j,k=1

KΩ1 (z, w) = det F ′ (z)KΩ2 (F(z), F(w))det F ′ (w),

(7.14)

for all z, w ∈ Ω1 . Proof. Notice that, by the Cauchy–Riemann equations for fj = uj + ivj , j = 1, … , n, the determinant of the Jacobian of the mapping F equals to | det F ′ (z)|2 . The substitution formula for integrals implies that for g ∈ L2 (Ω2 ) we have ∫ |g(ζ )|2 dλ(ζ ) = ∫ |g(F(z))|2 |det F ′ (z)|2 dλ(z). Ω2

Ω1

(7.15)

Hence the map TF ∶ g ↦ (g ∘ F) det F ′ establishes an isometric isomorphism from L2 (Ω2 ) to L2 (Ω1 ), with inverse map TF −1 , which restricts to an isomorphism between A2 (Ω1 ) and A2 (Ω2 ). Now let f ∈ A2 (Ω1 ) and apply the reproducing property of KΩ2 to the function TF −1 f = (f ∘ F −1 ) det(F −1 )′ . Setting F(z) = u, we get ∫ KΩ2 (u, v)TF −1 f (v) dλ(v) = TF −1 f (u) = f (z)(det F ′ (z))−1 . Ω2

(7.16)

Since TF is an isometry, ∫ TF −1 f (v)[KΩ2 (v, u)]− dλ(v) = ∫ f (w)[TF KΩ2 (⋅, u)(w)]− dλ(w). Ω2

Ω1

(7.17)

204 | 7 Bergman spaces From (7.16) and (7.17), we obtain f (z) = ∫ det F ′ (z)KΩ2 (F(z), F(w))det F ′ (w)f (w) dλ(w), Ω1

which means that the right-hand side of (7.14) has the required reproducing property, belongs to A2 (Ω1 ) in the variable z and is anti-holomorphic in the variable w, and hence must agree with KΩ1 (z, w). Finally, we derive a useful formula for the corresponding orthogonal projections Pj ∶ L2 (Ωj ) ⟶ A2 (Ωj ),

j = 1, 2.

Theorem 7.15. For all g ∈ L2 (Ω2 ) one has P1 (det F ′ g ∘ F) = det F ′ (P2 (g) ∘ F).

(7.18)

Proof. The left-hand side of (7.18) can be written in the form P1 (TF (g)), hence, by (7.5), we obtain P1 (TF (g))(z) = ∫ KΩ1 (z, w)TF (g)(w) dλ(w), Ω1

z ∈ Ω1 .

Now (7.14), together with (7.17), implies that KΩ1 (w, z) = [TF (KΩ2 (⋅, F(z)))(w)]det F ′ (z), so, since TF is an isometric isomorphism, we get P1 (TF (g))(z) = det F ′ (z) ∫ TF (g)(w)[TF (KΩ2 (⋅, F(z)))(w)]− dλ(w) Ω1

= det F ′ (z) ∫ g(v)[KΩ2 (v, F(z))]− dλ(v) Ω2

= det F (z)(P2 (g))(F(z)), ′

which proves (7.18). If n = 1 and Ω ⊊ ℂ is a simply connected domain, there is an interesting connection between the Bergman kernel KΩ of Ω and the Riemann mapping F ∶ Ω ⟶ 𝔻, see Corollary 4.44. Theorem 7.16. Let Ω ⊊ ℂ be a simply connected domain, and let KΩ be the Bergman kernel of Ω. Let F ∶ Ω ⟶ 𝔻 be the Riemann mapping with the uniqueness properties F(a) = 0, F ′ (a) > 0 for some a ∈ Ω. Then F ′ (z) = √

π K (z, a), KΩ (a, a) Ω

z ∈ Ω.

(7.19)

7.3 Biholomorphic mappings | 205

Proof. By (7.15), the transformation TF establishes an isometry between L2 (Ω) and L2 (𝔻) which restricts to be an isometry between A2 (Ω) and A2 (𝔻). Therefore we have (TF u, TF u)Ω = (u, u)𝔻 ,

u ∈ L2 (𝔻),

where (⋅, ⋅)Ω denotes the inner product in L2 (Ω) and (⋅, ⋅)𝔻 denotes the inner product of L2 (𝔻). Similarly, for v ∈ L2 (Ω) and G = F −1 we have (TG v, TG v)𝔻 = (v, v)Ω . The polarization identity 1 i (u1 , u2 ) = (‖u1 + u2 ‖2 − ‖u1 − u2 ‖2 ) − (‖u1 + iu2 ‖2 − ‖u1 − iu2 ‖2 ) 4 4 in an inner product space over ℂ yields (TF u1 , TF u2 )Ω = (u1 , u2 )𝔻 ,

u1 , u2 ∈ L2 (𝔻),

(7.20)

(TG v1 , TG v2 )𝔻 = (v1 , v2 )Ω ,

v1 , v2 ∈ L2 (Ω).

(7.21)

and

Since TF TG is the identity operator, we obtain from (7.20) and (7.21) (TF u, v)Ω = (TF u, TF (TG v))Ω = (u, TG v)𝔻 .

(7.22)

Now let h ∈ A2 (𝔻) and observe that, by (7.1) (h, 1)𝔻 = πh(0). By (7.22) we get for f ∈ A2 (Ω) (f , F ′ )Ω = (G′ (f ∘ G), 1)𝔻 = πG′ (0)f (G(0)) =

π f (a). F ′ (a)

It follows that the function k(z, a) = F π(a) F ′ (z) has the reproducing property and belongs to A2 (Ω), so, by Proposition 7.9, k(z, a) = KΩ (z, a) and we get ′

F ′ (z) =

π K (z, a). F ′ (a) Ω

Setting z = a, we obtain F ′ (a)2 = πKΩ (a, a), which proves (7.19).

206 | 7 Bergman spaces

7.4 Exercises 120. Let H be a vector space over ℂ with inner product (⋅, ⋅). Show that 4(x, y) = ‖x + y‖2 − ‖x − y‖2 + i‖x + iy‖2 − i‖x − iy‖2 , for all x, y ∈ H. 121. Let H be a Hilbert space, x0 ∈ H and let M be a closed subspace of H. Show that min{‖x − x0 ‖ ∶ x ∈ M} = max{|(x0 , y)| ∶ y ∈ M ⟂ , ‖y‖ = 1}. 122. Let f ∈ A2 (𝔻). Determine a polynomial p of degree ≤ n such that ‖f − p‖ is minimal. 123. Let H be a Hilbert space. Show that x1 , x2 , … , xn ∈ H are linearly independent, if and only if the Gram determinant G(x1 , x2 , … , xn ) = det((xj , xk ))nj,k=1 ≠ 0. 124. Let H be a Hilbert space and let y, x1 , x2 , … , xn ∈ H be linearly independent. Let M be the linear span of x1 , x2 , … , xn . Show that (dist(y, M))2 =

G(y, x1 , x2 , … , xn ) . G(x1 , x2 , … , xn )

125. For n ∈ ℕ0 let Hn (x) = (−1)n exp(x2 ) exp(n) (−x2 ) (Hermite polynomials). Determine H0 , H1 , H2 and H3 . Show that the Hermite6 functions φn (x) = exp(−x2 /2)Hn (x) constitute an orthogonal system in L2 (ℝ). 2 ″ ″ Hint: show first that φ″ n (x) = (x − 2n − 1)φn (x) and φn φm − φn φm = 2(m − n)φn φm . 126. Compute ‖φn ‖2 for n ∈ ℕ0 , where ‖ ⋅ ‖2 is the norm in L2 (ℝ). Hint: show first that Hn′ = 2nHn−1 for n ∈ ℕ, then use partial integration. 127. Show that the normalized Hermite functions constitute a complete orthonormal system in L2 (ℝ). Hint: let f ∈ L2 (ℝ) with (f , φn ) = 0 for n ∈ ℕ0 ; define F(z) =

1 ∫ exp(−izx − x2 /2)f (x) dx, √2π ℝ

z ∈ ℂ,

and show that F is an entire function. Compute F (n) (0) for n ∈ ℕ0 and use the uniqueness of the Fourier transform. 6 Hermite, Charles (1822–1901).

7.5 Notes | 207

128. Let z ∈ 𝔻. Show that 1 1 1 dλ(w) = . ∫ 4 π 𝔻 |1 − zw| (1 − |z|2 )2 Hint: consider the function f (w) = 1/(1 − wz)2 , and use the reproducing property of the Bergman kernel. 129. Let Ω = {(z1 , z2 ) ∶ |z1 | < |z2 | < 1} ⊂ ℂ2 be the Hartogs triangle. Show that every holomorphic function on a neighborhood of Ω continues to the bidisc (see Theorem 6.10) and find an orthonormal basis for A2 (Ω). Hint: use Laurent series expansion for the variable z2 .

7.5 Notes The basics on Hilbert spaces can be found, for instance, in [70]. There are numerous good texts on different aspects of Bergman spaces and their reproducing kernels, the reader may consult the books by S.-C. Chen and M.-C. Shaw [13], St. Krantz [51], and M. Jarnicki and P. Pflug [44]. The transformation formula (see Theorem 7.14) is of great importance in the study of the boundary behavior of biholomorphic mappings on strictly pseudoconvex domains (Fefferman’s Theorem, see [59]). The theorem about the Riemann mapping and the Bergman kernel is contained in St. Bell’s book [7]. Further results regarding the Bergman kernel and the solution to 𝜕 will be discussed in the next chapters.

8 The canonical solution operator to 𝜕 In this chapter we will use properties of the Bergman kernel to solve the inhomo𝜕 𝜕 𝜕 geneous Cauchy–Riemann equation 𝜕u = g, where 𝜕z = 21 ( 𝜕x + i 𝜕y ), z = x + iy and 𝜕z

g ∈ A2 (𝔻). It turns out that the Bergman kernel plays an important role for this problem, especially to find the so-called canonical solution in L2 (𝔻) which is perpendicular to the corresponding space of holomorphic functions. For this purpose we develop the spectral theory of compact operators on Hilbert spaces together with basics on the ideals of Schatten class operators (Section 8.1). In Section 8.2 it is shown that the canonical solution operator to 𝜕 is a Hilbert–Schmidt operator on A2 (𝔻), whereas on A2 (𝔹), for the ball 𝔹 ⊂ ℂn , n ≥ 2, the corresponding operator is compact but fails to be Hilbert–Schmidt (Section 8.3 together with the exercises in Section 8.4).

8.1 Compact operators on Hilbert spaces Let H1 and H2 be separable Hilbert spaces and A ∶ H1 ⟶ H2 a bounded linear operator. The operator A is compact, if the image A(B1 ) of the unit ball B1 in H1 is a relatively compact subset of H2 , since H2 is complete this is equivalent to the concept of a totally bounded set, i.e. for each ϵ > 0 there exists a finite number of elements v1 , … , vm ∈ H2 such that m

A(B1 ) ⊂ ⋃ B(vj , ϵ), j=1

where B(vj , ϵ) = {v ∈ H2 ∶ ‖v − vj ‖ < ϵ}. Another equivalent definition of compactness is the following: for each bounded sequence (uk )k in H1 the image sequence (A(uk ))k has a convergent subsequence in H2 . Let ℒ(H1 , H2 ) denote the space of all bounded linear operators from H1 to H2 endowed with the topology generated by the operator norm ‖A‖ = sup{‖Au‖ ∶ ‖u‖ ≤ 1}. In this way ℒ(H1 , H2 ) becomes a Banach space. Let 𝒦(H1 , H2 ) denote the subspace of all compact operators from H1 to H2 . The following characterization of compactness is useful for the special operators in the text, see, for instance, [17]: Theorem 8.1. Let H1 and H2 be Hilbert spaces, and assume that S ∶ H1 → H2 is a bounded linear operator. The following three statements are equivalent: – S is compact. https://doi.org/10.1515/9783110417241-008

210 | 8 The canonical solution operator to 𝜕 –

For every ϵ > 0 there are a C = Cϵ > 0 and a compact operator T = Tϵ ∶ H1 → H2 such that ‖Sv‖ ≤ C‖Tv‖ + ϵ‖v‖.

–

(8.1)

For every ϵ > 0 there are a C = Cϵ > 0 and a compact operator T = Tϵ ∶ H1 → H2 such that ‖Sv‖2 ≤ C‖Tv‖2 + ϵ‖v‖2 .

(8.2)

Proof. First we show that (8.1) and (8.2) are equivalent. Suppose that (8.2) holds. Write (8.2) with ϵ and C replaced by their squares to obtain ‖Sv‖2 ≤ C 2 ‖Tv‖2 + ϵ2 ‖v‖2 ≤ (C‖Tv‖ + ϵ‖v‖)2 , which implies (8.1). Now suppose that (8.1) holds. Choose η with ϵ = 2η2 and apply (8.1) with ϵ replaced by η to get ‖Sv‖2 ≤ C 2 ‖Tv‖2 + 2ηC‖v‖‖Tv‖ + η2 ‖v‖2 . It is easily seen (small constant – large constant trick) that there is C ′ > 0 such that 2ηC‖v‖‖Tv‖ ≤ η2 ‖v‖2 + C ′ ‖Tv‖2 , hence ‖Sv‖2 ≤ (C 2 + C ′ )‖Tv‖2 + 2η2 ‖v‖2 = C ″ ‖Tv‖2 + ϵ‖v‖2 . To prove the theorem it therefore suffices to show that (8.1) is equivalent to compactness. When S is known to be compact, we choose T = S and C = 1, and (8.1) holds for every positive ϵ. For the converse let (vn )n be a bounded sequence in H1 . We want to extract a Cauchy subsequence from (Svn )n . From (8.1) we have ‖Svn − Svm ‖ ≤ C‖Tvn − Tvm ‖ + ϵ‖vn − vm ‖.

(8.3)

Given a positive integer N, we may choose ϵ sufficiently small in (8.3) so that the second term on the right-hand side is at most 1/(2N). The first term can be made smaller than 1/(2N) by extracting a subsequence of (vn )n (still labeled the same) for which (Tvn )n converges, and then choosing n and m large enough. Let (vn(0) )n denote the original bounded sequence. The above argument shows that, for each positive integer N, there is a sequence (vn(N) )n with the property that

8.1 Compact operators on Hilbert spaces | 211

(vn(N) )n is a subsequence of (vn(N−1) )n , and for any pair v and w in (vn(N) )n we have ‖Sv − Sw‖ ≤ 1/N. Let (wk )k be the diagonal sequence defined by wk = vk(k) . Then (wk )k is a subsequence of (vn(0) )n and the image sequence under S of (wk )k is a Cauchy sequence. Since H2 is complete, the image sequence converges and S is compact. Theorem 8.2. 𝒦(H1 , H2 ) is a closed subspace of ℒ(H1 , H2 ) endowed with the operator norm. Proof. Let A ∈ ℒ(H1 , H2 ). Suppose, for each ϵ > 0, there is a compact operator Aϵ such that ‖A − Aϵ ‖ ≤ ϵ. Then for each u ∈ H1 we have ‖Au − Aϵ u‖ ≤ ϵ‖u‖. Now we get ‖Au‖ = ‖Au − Aϵ u + Aϵ u‖

≤ ‖Au − Aϵ u‖ + ‖Aϵ u‖

≤ ϵ‖u‖ + ‖Aϵ u‖. Theorem 8.1 implies that A is compact.

Theorem 8.3. Suppose that A ∈ 𝒦(H1 , H2 ), S ∈ ℒ(H1 , H1 ) and that T ∈ ℒ(H2 , H2 ) is a bounded operator on H2 . Then both AS and TA are compact. Proof. If (uk )k is a bounded sequence in H1 , then (S(uk ))k is also bounded, because S is a bounded operator. A is compact, so (A(S(uk )))k has a convergent subsequence. Thus AS is compact. To show that TA is compact we use Theorem 8.1: ‖TAu‖ ≤ ‖T‖ ‖Au‖ ≤ ‖T‖(ϵ‖u‖ + C‖Au‖) ≤ ϵ‖T‖ ‖u‖ + C‖T‖ ‖Au‖. Corollary 8.4. Let H be a Hilbert space. 𝒦(H, H) forms a two-sided, closed ideal in ℒ(H, H). Theorem 8.5. Let A ∶ H1 ⟶ H2 be a bounded linear operator. The following properties are equivalent: (i) A is compact. (ii) The adjoint operator A∗ ∶ H2 ⟶ H1 is compact. (iii) A∗ A ∶ H1 ⟶ H1 is compact. Proof. Suppose that A is compact, then, by Theorem 8.3, AA∗ is also compact. Given ϵ > 0, it follows that there is a constant C such that

212 | 8 The canonical solution operator to 𝜕 ‖A∗ u‖2 = (A∗ u, A∗ u) = (u, AA∗ u) ≤ ‖u‖ ‖AA∗ u‖ ≤ ϵ‖u‖2 + C‖AA∗ u‖2 . Therefore, by Theorem 8.1, A∗ is compact. Since A∗∗ = A, (i) and (ii) are equivalent. (i) implies (iii) (Theorem 8.3), so it leaves us to show that (iii) implies (i): let A∗ A be compact. Given ϵ > 0 there is again a constant C such that ‖Au‖2 = (Au, Au) = (u, A∗ Au) ≤ ϵ‖u‖2 + C‖A∗ Au‖2 , so Theorem 8.1 implies that A is compact. Theorem 8.6. A bounded operator A ∶ H ⟶ H is compact, if and only if there exists a sequence (Ak )k of linear operators with finite-dimensional range such that ‖A − Ak ‖ → 0 as k → ∞. Proof. An operator with finite-dimensional range is compact, because any bounded sequence in a finite-dimensional Hilbert space has a convergent subsequence. The limit of compact operators in the operator norm is again compact, so we proved one direction. The converse will follow from the next two results. The following theorem is the spectral theorem for compact, self-adjoint operators: Theorem 8.7. Let A ∶ H ⟶ H be a compact, self-adjoint operator on a separable Hilbert space H. Then there exists a real zero-sequence (μn )n and an orthonormal system (en )n in H such that for x ∈ H ∞

Ax = ∑ μn (x, en )en , n=0

where the sum converges in the operator norm, i.e. N

sup ‖Ax − ∑ μn (x, en )en ‖ → 0,

‖x‖≤1

n=0

as N → ∞. Proof. First we collect some elementary properties of self-adjoint operators: Claim (a). Let A ∈ ℒ(H, H) = ℒ(H) and suppose that (Ax, x) = 0 for all x ∈ H. Then A = 0. We have 0 = (A(x + y), x + y) − (A(x − y), x − y) = 2[(Ax, y) + (Ay, x)],

8.1 Compact operators on Hilbert spaces | 213

hence (Ax, y) + (Ay, x) = 0, now replace x by ix, then i(Ax, y) − i(Ay, x) = 0, therefore (Ax, y) = 0 for all x, y ∈ H, which implies A = 0. Claim (b). A = A∗ , if and only if (Ax, x) ∈ ℝ for all x ∈ H. A = A∗ implies (Ax, x)− = (x, Ax) = (Ax, x) ∈ ℝ. If (Ax, x) = (Ax, x)− for all x ∈ H, then (Ax, x) = (x, A∗ x)− = (A∗ x, x), and so ((A − A∗ )x, x) = 0 for all x ∈ H. By Claim (a) we have A = A∗ . Now we show that for a self-adjoint operator A ∈ ℒ(H) the norm of A is given by ‖A‖ = sup |(Ax, x)|. ‖x‖=1

(8.4)

For this let NA = sup‖x‖=1 |(Ax, x)|. Then NA ≤ ‖A‖. We also have 2[(Ax, y) + (Ay, x)] = (A(x + y), x + y) − (A(x − y), x − y), as A∗ = A, and by the parallelogram rule, |4ℜ(Ax, y)| ≤ NA (‖x + y‖2 + ‖x − y‖2 ) = 2NA (‖x‖2 + ‖y‖2 ). There is a θ ∈ ℝ such that e−iθ (Ax, y) = |(Ax, y)|. Now replace y by e−iθ y, then 2|(Ax, y)| ≤ NA (‖x‖2 + ‖y‖2 ). For a t > 0 we replace x by tx and y by y/t. Then one obtains 2|(Ax, y)| ≤ NA (t 2 ‖x‖2 +

1 ‖y‖2 ). t2

We consider the right-hand side of this inequality as a function of t and get after differentiation with respect to t 2t‖x‖2 −

2 ‖y‖2 , t3

so the right-hand side of the inequality will be minimal, if t 2 = ‖y‖/‖x‖. Hence we obtain 2|(Ax, y)| ≤ 2NA ‖x‖ ‖y‖, and ‖A‖ ≤ NA , which proves (8.4). Next we show that, if A ≠ 0, there exists an eigenvector x0 ∈ H of A such that ‖x0 ‖ = 1 and |(Ax0 , x0 )| = ‖A‖ = sup |(Ax, x)|, ‖x‖=1

(8.5)

and we show that the corresponding eigenvalue λ0 is real and satisfies |λ0 | = ‖A‖. By (8.4) we have ‖A‖ = sup‖x‖=1 |(Ax, x)|. Hence there is a sequence (xn )n in H with ‖xn ‖ = 1 and limn→∞ |(Axn , xn )| = ‖A‖. The inner products (Axn , xn ) are real, so there

214 | 8 The canonical solution operator to 𝜕 exists a subsequence, which is again denoted by (xn )n , such that limn→∞ (Axn , xn ) = λ0 . Then we have λ0 = ‖A‖ or λ0 = −‖A‖. Now we get 0 ≤ ‖Axn − λ0 xn ‖2 = ‖Axn ‖2 − 2λ0 (Axn , xn ) + λ02 ‖xn ‖2 ≤ ‖A‖2 − 2λ0 (Axn , xn ) + ‖A‖2 ,

and letting n → ∞ lim ‖Axn − λ0 xn ‖ = 0.

n→∞

Since A is compact and (xn )n is a bounded sequence, there exists a subsequence (xnk )k such that limk→∞ Axnk = x. Hence limk→∞ λ0 xnk = x, and as λ0 ≠ 0, we also have limk→∞ xnk = x0 with λ0 x0 = x. Since A is continuous, this implies lim Axnk = Ax0 ,

k→∞

and hence x = λ0 x0 = Ax0 . So x0 is an eigenvector of A with eigenvalue λ0 . This proves (8.5). Now, each eigenvalue of A is real: Ax = λx and x ≠ 0 imply λ(x, x) = (Ax, x) = (x, Ax) = (x, λx) = λ(x, x). Therefore λ ∈ ℝ. Let Hλ = {x ∈ H ∶ Ax = λx} be the eigenspace of the eigenvalue λ. Since A is continuous, Hλ is closed. There exists an orthonormal basis in Hλ consisting of eigenvectors of A. If λ1 ≠ λ2 are eigenvalues, then Hλ1 ⊥ Hλ2 , because for x ∈ Hλ1 and y ∈ Hλ2 we have λ1 (x, y) = (Ax, y) = (x, Ay) = (x, λ2 y) = λ2 (x, y), and hence (x, y) = 0. We claim that the eigenspace Hλ of an eigenvalue λ ≠ 0 is always finite-dimensional: if Hλ is of infinite dimension, we would have an infinite orthonormal system {xα }α∈A with Axα = λxα and since the restriction of A to Hλ is also compact, we could find a convergent subsequence of (Axα )α , ending up in a contradiction to ‖λxα1 − λxα2 ‖2 = 2|λ|2 for α1 ≠ α2 . Now we have two possible cases: (i) there are finitely many eigenvalues ≠ 0; (ii) there are infinitely many eigenvalues ≠ 0. (i) In this case we have H = Hλ0 ⊕ ⋯ ⊕ Hλk−1 ⊕ M, where M = (Hλ0 ⊕ ⋯ ⊕ Hλk−1 )⟂ . We claim that A(M) ⊆ M. Indeed, for y ∈ M we have (Ay, x) = (y, Ax) = 0, for all x ∈ Hλ0 ⊕ ⋯ ⊕ Hλk−1 , because Ax ∈ Hλ0 ⊕ ⋯ ⊕ Hλk−1 , hence Ay ∈ M. The restriction AM of A to M is again compact and self-adjoint: for x, y ∈ M we have (AM x, y) = (Ax, y) = (x, Ay) = (x, AM y), because A(M) ⊆ M. We claim that AM = 0. Suppose AM ≠ 0, then, by (8.4), there is a λ ≠ 0 with AM x = λx for some x ≠ 0, hence we would get x ∈ Hλ , which is a contradiction.

8.1 Compact operators on Hilbert spaces | 215

Let Pj ∶ H ⟶ Hλj be the orthogonal projection. Then one obtains x = P0 x + P1 x + ⋯ + Pk−1 x + y, where Ay = 0, therefore Ax = λ0 P0 x + λ1 P1 x + ⋯ + λk−1 Pk−1 x. (ii) For each ϵ > 0, the set {λi ∶ λi eigenvalue of A with |λi | ≥ ϵ} is a finite set, otherwise we would get an infinite orthonormal system {xi } with Axi = λi xi and the sequence (Axi )i , would have to contain a convergent subsequence. But ‖Axk − Axl ‖2 = ‖λk xk − λl xl ‖2 = |λk |2 + |λl |2 ≥ 2ϵ2 , which is a contradiction. Let λ0 , … , λk−1 be the eigenvalues with |λi | ≥ ϵ and let M = (Hλ0 ⊕ ⋯ ⊕ Hλk−1 )⟂ . The restriction AM of A to M is self-adjoint and compact, hence, by (8.4), there is an eigenvector x with AM x = λx and |λ| = ‖AM ‖, and hence Ax = λx. Therefore x ∈ (Hλi )⟂ and |λ| = ‖AM ‖ < ϵ. So we have Ax = Ax1 + AM x2 ,

x1 ∈ M ⟂ , x2 ∈ M.

Since Ax1 = λ0 P0 x + ⋯ + λk−1 Pk−1 x we obtain ‖AM x2 ‖ = ‖Ax − λ0 P0 x − ⋯ − λk−1 Pk−1 x‖ < ϵ‖x‖. Letting ϵ → 0 we get ∞

Ax = ∑ λj Pj x. j=0

Now let {ψ1 , … , ψk } be an orthonormal basis of Hλi . Then k

APi (x) = λi Pi (x) = λi ∑(x, ψj )ψj . j=1

The sequence of the eigenvalues can be ordered in the following way |λ0 | ≥ |λ1 | ≥ ⋯ , because for each ϵ > 0 there are only finitely many λi with |λi | ≥ ϵ. Finally, the orthogonality of the eigenvectors xk and Bessel’s inequality (7.6) give n

2

∞

2

‖Ax − ∑ λk (x, xk )xk ‖ = ∑ |λk (x, xk )|2 ≤ (‖x‖ sup |λk |) , k=1

k=n+1

k>n

and since (λk )k≥1 is a zero-sequence, the above series converges to A in the operator norm.

216 | 8 The canonical solution operator to 𝜕 Now we drop the assumption of self-adjointness and obtain Theorem 8.8. Let A ∶ H1 ⟶ H2 be a compact operator. There exists a decreasing zerosequence (sn )n in ℝ+ and orthonormal systems (en )n≥0 in H1 and (fn )n≥0 in H2 such that ∞

Ax = ∑ sn (x, en )fn n=0

∀x ∈ H1 ,

where the sum converges again in the operator norm. Proof. In order to show this, one applies the spectral theorem for the self-adjoint, compact operator A∗ A ∶ H1 ⟶ H1 and gets ∞

A∗ Ax = ∑ s2n (x, en )en ,

(8.6)

n=0

where s2n are the eigenvalues of A∗ A. If sn > 0, we set fn = s−1 n Aen and get (fn , fm ) =

1 s2 1 (Aen , Aem ) = (A∗ Aen , em ) = n (en , em ) = δn,m . sn sm sn sm sn sm

For y ∈ H1 with y ⟂ en for each n ∈ ℕ0 , we have by (8.6) that ‖Ay‖2 = (Ay, Ay) = (A∗ Ay, y) = 0. Hence we have ∞

∞

Ax = A(x − ∑ (x, en )en ) + A( ∑ (x, en )en ) ∞

n=0

∞

n=0

= ∑ (x, en )Aen = ∑ sn (x, en )fn . n=0

n=0

Similarly as in the last theorem, we get n

2

∞

2

‖Ax − ∑ sk (x, ek )fk ‖ = ∑ |sk (x, ek )|2 ≤ (‖x‖ sup |sk |) , k=1

k=n+1

k>n

which implies that the series converges in the operator norm. Remark 8.9. From the last statement we get the missing direction in the proof of Theorem 8.6. The numbers sn are called the s-numbers of A. They are uniquely determined by the operator A, since they are the square roots of the eigenvalues of A∗ A. Let 0 < p < ∞. The operator A belongs to the Schatten1 -class Sp , if its sequence (sn )n of s-numbers belongs to lp . The elements of the Schatten class S1 are called nu1 Schatten, Robert (1911–1977).

8.1 Compact operators on Hilbert spaces | 217

clear operators, the elements of the Schatten class S2 are called Hilbert–Schmidt2 operators. Theorem 8.10. Let A ∶ H1 ⟶ H2 be a bounded linear operator between Hilbert spaces. The following conditions are equivalent: (i) There is an orthonormal basis (ei )i∈I of H1 such that ∑i∈I ‖Aei ‖2 < ∞. (ii) For each orthonormal basis (fj )j∈J of H1 one has ∑j∈J ‖Afj ‖2 < ∞. (iii) A is a Hilbert–Schmidt operator. Proof. (i) implies (ii). Let (ei )i∈I be as in (i) and (gl )l∈L an orthonormal basis of H2 . Then, by Parseval’s equality (see Theorem 7.11) ∑‖A∗ gl ‖2 = ∑ ∑|(A∗ gl , ei )|2 = ∑ ∑|(gl , Aei )|2 = ∑ ‖Aei ‖2 < ∞. l∈L i∈I

l∈L

i∈I l∈L

i∈I

If (fj )j∈J is an arbitrary orthonormal basis of H1 , we get again ∑ ‖Afj ‖2 = ∑‖A∗ gl ‖2 = ∑ ‖Aei ‖2 < ∞. j∈J

i∈I

l∈L

(ii) implies (iii). Let (ei )i∈I be an orthonormal basis of H1 and M a finite subset of I. We set PM x = ∑i∈M (x, ei )ei . Then ‖(A − APM )x‖ = ‖A(I − PM )x‖ = ‖ ∑ (x, ei )Aei ‖ 1/2

≤ ( ∑ ‖Aei ‖2 ) i∈I⧵M

i∈I⧵M

1/2

( ∑ |(x, ei )|2 ) i∈I⧵M

1/2

≤ ( ∑ ‖Aei ‖2 ) i∈I⧵M

‖x‖,

where we used Bessel’s inequality (7.6). By assumption, ∑i∈I ‖Aei ‖2 < ∞, hence we can approximate A in the operator norm by operators with finite range. Therefore A is compact and, by Theorem 8.8, can be written as ∞

Ax = ∑ sn (x, xn )fn . n=0

By Remark 7.1, there exists an orthonormal basis (ξj )j∈J , which contains the orthonormal system (xn )n≥0 . Then we have ∞

2

∞

∑ ‖Aξj ‖2 = ∑‖ ∑ sn (ξj , xn )fn ‖ = ∑ s2n < ∞, j∈J

j∈J n=0

n=0

so A is a Hilbert–Schmidt operator. (iii) implies (i). If A is a Hilbert–Schmidt operator, A can be written as above and we obtain (i). 2 Schmidt, Erhard (1876–1959).

218 | 8 The canonical solution operator to 𝜕 Corollary 8.11. Let A ∶ H1 ⟶ H2 be a Hilbert–Schmidt operator. Then for each orthonormal basis (ei )i∈I of H1 we have ∞

ν2 (A)2 ∶= ∑ sn (A)2 = ∑ ‖Aei ‖2 ≥ ‖A‖2 . n=0

i∈I

In particular, ν2 is a norm in the space S2 (H1 , H2 ) of all Hilbert–Schmidt operators between H1 and H2 . Proof. We have 1/2

‖Ax‖ = ‖∑(x, ei )Aei ‖ ≤ (∑ ‖Aei ‖2 ) i∈I

i∈I

‖x‖,

hence ν2 (A) ≥ ‖A‖. It is easily seen that A ↦ (∑i∈I ‖Aei ‖2 )1/2 is a norm. If H1 and H2 are separable Hilbert spaces, A ∈ ℒ(H1 , H2 ), and (en )n≥1 an orthonormal basis of H1 and (fn )n≥1 an orthonormal basis of H2 , then for each x ∈ H1 we have ∞

Ax = ∑ (x, en )Aen n=1

and hence for all j ∈ ℕ ∞

(Ax, fj ) = ∑ (x, en )(Aen , fj ). n=1

The coefficients of Ax with respect to (fj )j≥1 can be computed from the coefficients of x with respect to (en )n≥1 by means of the infinite matrix aj,n = (Aen , fj ),

j, n ∈ ℕ.

Parseval’s equation implies ∞

∞

∞

j,n=1

j,n=1

n=1

∑ |aj,n |2 = ∑ |(Aen , fj )|2 = ∑ ‖Aen ‖2 ,

and we obtain Corollary 8.12. A ∈ ℒ(H1 , H2 ) is a Hilbert–Schmidt operator, if and only if the matrix (aj,n )j,n≥1 of A with respect to arbitrary orthonormal bases satisfies ∞

∑ |aj,n |2 < ∞.

j,n=1

In this case ∞

ν2 (A)2 = ∑ |aj,n |2 . j,n=1

8.1 Compact operators on Hilbert spaces | 219

Corollary 8.13. A linear operator A ∶ l2 ⟶ l2 is a Hilbert–Schmidt operator, if and only 2 2 if there exists a matrix (aj,n )j,n≥1 with ∑∞ j,n=1 |aj,n | < ∞ such that for all x = (xn )n≥1 ∈ l one has ∞

Ax = ( ∑ aj,n xn ) n=1

.

j≥1

Proof. If A has the given form, one can apply Cauchy–Schwarz inequality to get ∞

‖Ax‖22 ≤ ( ∑ |aj,n |2 )‖x‖22 , j,n=1

hence everything follows from Corollary 8.12. Theorem 8.14. Let S ⊆ ℝn and T ⊆ ℝm be open sets and A ∶ L2 (T) ⟶ L2 (S) a bounded linear operator. A is a Hilbert–Schmidt operator, if and only if there exists K ∈ L2 (S × T) such that Af (s) = ∫ K(s, t)f (t) dt, T

f ∈ L2 (T).

In this case, one has ν2 (A) = (∫

S×T

1/2

|K(s, t)|2 ds dt)

.

Proof. Let (gk )k≥1 and (fj )j≥1 be orthonormal bases of L2 (T) and L2 (S), respectively. Then (hj,k )j,k≥1 , defined by hj,k (s, t) ∶= fj (s)gk (t),

(s, t) ∈ S × T,

is an orthonormal basis of L2 (S × T). This can be shown as follows: if F ∈ L2 (S × T), then t ↦ F(s, t)fj (s) is for almost every s ∈ S in L2 (T). If F ⟂ hj,k for all j, k ∈ ℕ, it follows that the functions t ↦ F(s, t)fj (s) vanish in L2 (T) for all j ∈ ℕ almost everywhere with respect to s. This implies F = 0 and (hj,k )j,k≥1 is an orthonormal basis. If A is a Hilbert–Schmidt operator and (aj,k )j,k≥1 is its matrix with respect to (gk )k≥1 and (fj )j≥1 , then, by Corollary 8.12, K(s, t) ∶= ∑ aj,k hj,k (s, t) = ∑ aj,k fj (s)gk (t) j,k≥1

j,k≥1

belongs to L2 (S × T) and ‖K‖22 = ∑j,k≥1 |aj,k |2 = ν2 (A)2 . For f ∈ L2 (T) we have Af (s) = ∑ aj,k (f , gk )fj (s) = ∫ ∑ aj,k fj (s)gk (t)f (t) dt = ∫ K(s, t)f (t) dt. j,k≥1

T j,k≥1

T

220 | 8 The canonical solution operator to 𝜕 If A is given by the kernel function K ∈ L2 (S × T), then A is continuous because ‖Af ‖2 ≤ (∫

S×T

|K(s, t)|2 ds dt) ∫ |f (t)|2 dt. T

For the matrix (aj,k )j,k≥1 of A, we get aj,k = (Agk , fj ) = ∫ (∫ K(s, t)gk (t) dt)fj (s) ds = (K, hj,k ), S

T

for all j, k ∈ ℕ. Bessel’s inequality and Corollary 8.12 imply that A is a Hilbert–Schmidt operator. Lemma 8.15. Let H be a separable Hilbert space and let A ∶ H ⟶ H be a positive, compact operator (i.e. (Ax, x) ≥ 0 ∀x ∈ H) with spectral decomposition ∞

Ax = ∑ λn (x, xn )xn , n=0

where λ0 ≥ λ1 ≥ ⋯. Then (i) λ0 = max x∈H

(ii)

(Ax, x) , (x, x)

where the maximum is attained by an eigenvector with eigenvalue λ0 . λj = min max ⟂ L∈Nj x∈L

(Ax, x) , (x, x)

j ≥ 1,

where Nj is the set of all j-dimensional subspaces of H. The minimum is attained by the subspace L = Lj = ⟨x0 , … , xj−1 ⟩, i.e. λj = max ⟂ x∈Lj

(Ax, x) . (x, x)

Proof. (i) follows from (8.4). For j ≥ 1 we have λj = (Axj , xj )/(xj , xj ). The assertion follows, if we can show that for j

each j-dimensional subspace L there exists z0 ⟂ L with z0 ≠ 0 and z0 = ∑k=0 (z0 , xk )xk , because then we have j

(Az0 , z0 ) ∑k=0 λk |(z0 , xk )|2 = ≥ λj , j (z0 , z0 ) ∑k=0 |(z0 , xk )|2 as λj ≤ λi for 0 ≤ i ≤ j and we get max ⟂ x∈L

(Ax, x) ≥ λj . (x, x)

The existence of z0 follows from the fact that for a basis {yk ∶ k = 0, … , j − 1} of L the system of equations

8.1 Compact operators on Hilbert spaces | 221 j

∑ ai (xi , yk ) = 0,

i=0

k = 0, … , j − 1,

j

has a non-trivial solution. Set z0 = ∑i=0 ai xi , then z0 ⟂ L. Theorem 8.16. Let A ∶ H1 ⟶ H2 be a compact operator. The s-numbers s0 (A) ≥ s1 (A) ≥ ⋯ have the following properties: (a) s0 (A) = ‖A‖; (b) If A∗ = A, then sk (A) = |λk |; (c) sk (λA) = |λ|sk (A); (d) sk (A∗ ) = sk (A); (e) sk (BA) ≤ ‖B‖sk (A) for B ∈ ℒ(H1 ); (f) sk (AB) ≤ ‖B‖sk (A) for B ∈ ℒ(H2 ); (g) sj (A) = minK∈𝒦j ‖A − K‖, where 𝒦j is the set of operators with j-dimensional image. Proof. Properties (a), (b) and (c) follow from (8.4) and Theorem 8.8. ∞ ∗ (d) If Ax = ∑∞ k=0 sk (A)(x, ek )fk , then A x = ∑k=0 sk (A)(x, fk )ek . The canonical representation of a compact operator in Theorem 8.8 is uniquely determined: if Ax = ∑∞ j=0 tj (x, χj )ωj is another representation with t0 ≥ t1 ≥ ⋯ and orthonormal systems (χj )j and (ωj )j , then we have ∞

∞

∞

i=0

j=0

i=0

A∗ Ax = ∑ ti ( ∑ tj (x, χj )ωj , ωi )χi = ∑ ti2 (x, χi )χi , and ti2 are the eigenvalues of A∗ A. (e) Let A1 , A2 ∶ H ⟶ H be compact operators such that A1 ≥ 0 and A2 − A1 ≥ 0. Then 0 ≤ A1 ≤ A2 , and for each j-dimensional subspace L of H we have max ⟂ x∈L

(A1 x, x) (A2 x, x) ≤ max x∈L⟂ (x, x) (x, x)

and, by Lemma 8.15 (ii), λj (A1 ) = min max ⟂ L∈Nj x∈L

(A2 x, x) (A1 x, x) ≤ min max = λj (A2 ), L∈Nj x∈L⟂ (x, x) (x, x)

for j = 0, 1, 2, …. It follows that (A∗ B∗ BAx, x) = ‖BAx‖2 ≤ ‖B‖2 ‖Ax‖2 = (‖B‖2 A∗ Ax, x), which implies 0 ≤ A∗ B∗ BA ≤ ‖B‖2 A∗ A, hence λj (A∗ B∗ BA) ≤ λj (‖B‖2 A∗ A) = ‖B‖2 λj (A∗ A), and sj (BA) ≤ ‖B‖sj (A) for j ≥ 0. (f) By (d) and (e), we have sj (AB) = sj ((AB)∗ ) = sj (B∗ A∗ ) ≤ ‖B∗ ‖sj (A∗ ) = ‖B‖sj (A).

222 | 8 The canonical solution operator to 𝜕 (g) By Lemma 8.15, we have s2j (A) = λj (A∗ A) = min max ⟂ L∈Nj x∈L

(A∗ Ax, x) , (x, x)

therefore sj (A) = min max ⟂ L∈Nj x∈L

‖Ax‖ . ‖x‖

If K ∈ 𝒦j , then K ∗ ∈ 𝒦j and we have (im K ∗ )⟂ = ker K, since (K ∗ x, y) = 0 ∀x

⟺

(x, Ky) = 0 ∀x

⟺

Ky = 0.

Hence sj (A) ≤ max

x∈ker K

‖Ax‖ . ‖x‖

For x ∈ ker K we have Ax = (A − K)x and ‖Ax‖ ≤ ‖A − K‖ ‖x‖, hence sj (A) ≤ ‖A − K‖. For j−1

Kx = Kj x = ∑ sk (A)(x, fk )ek k=0

we have, on the other hand, (A − Kj )x = ∑ sk (A)(x, fk )ek . k≥j

This implies ‖A − Kj ‖ = supk≥j sk (A) = sj (A), hence sj (A) = min ‖A − K‖. K∈𝒦j

8.2 The canonical solution operator to 𝜕 restricted to A2 (𝔻) We return to the inhomogeneous Cauchy–Riemann equation on the disc 𝔻 and use the notations of Chapter 7. Let S(g)(z) = ∫ K(z, w)g(w)(z − w)− dλ(w). 𝔻

Using the Bergman projection P ∶ L2 (𝔻) ⟶ A2 (𝔻)

(8.7)

8.2 The canonical solution operator to 𝜕 restricted to A2 (𝔻)

| 223

and (7.5), we get ̃ S(g)(z) = zg(z) − P(g)(z), ̃ where g(w) = wg(w). We claim that S(g) is a solution of the inhomogeneous Cauchy– Riemann equation: 𝜕 𝜕z 𝜕g 𝜕P(g)̃ S(g)(z) = g(z) + z + = g(z), 𝜕z 𝜕z 𝜕z 𝜕z because g and P(g)̃ are holomorphic functions, therefore 𝜕S(g) = g. In addition, we have S(g) ⟂ A2 (𝔻), because for arbitrary f ∈ A2 (𝔻) we get ̃ f ) = (g,̃ f ) − (P(g), ̃ f ) = (g,̃ f ) − (g,̃ Pf ) = (g,̃ f ) − (g,̃ f ) = 0. (Sg, f ) = (g̃ − P(g), The operator S ∶ A2 (𝔻) ⟶ L2 (𝔻) is called the canonical solution operator to 𝜕. Now we want to show that S is a compact operator. For this purpose we consider the adjoint operator S∗ and prove that S∗ S is compact, which implies that S is compact (Theorem 8.5). For g ∈ A2 (𝔻) and f ∈ L2 (𝔻), we have (Sg, f ) = ∫ (∫ K(z, w)g(w)(z − w)− dλ(w))f (z) dλ(z) 𝔻

𝔻

−

= ∫ (∫ K(w, z)(z − w)f (z) dλ(z)) g(w) dλ(w) = (g, S∗ f ), 𝔻

𝔻

hence S∗ (f )(w) = ∫ K(w, z)(z − w)f (z) dλ(z). 𝔻

(8.8)

Now set cn2 = ∫ |z|2n dλ(z) = 𝔻

π , n+1

n

and ϕn (z) = z /cn , n ∈ ℕ0 , then the Bergman kernel K(z, w) can be expressed in the form z k wk . 2 k=0 ck ∞

K(z, w) = ∑ Next we compute

z k wk wn z k−1 wk wn c z n−1 P(ϕ̃ n )(z) = ∫ ∑ 2 w dλ(w) = ∑ 2 ∫ dλ(w) = n 2 , cn cn−1 𝔻 k=0 ck k=1 ck−1 𝔻 cn ∞

∞

hence we have S(ϕn )(z) = zϕn (z) −

cn z n−1 , 2 cn−1

n ∈ ℕ.

224 | 8 The canonical solution operator to 𝜕 Now we apply S∗ and get zz n cn z n−1 wk z k (z − w)( − 2 ) dλ(z). 2 cn cn−1 𝔻 k=0 ck ∞

S∗ S(ϕn )(w) = ∫ ∑

The last integral is computed in two steps: first, multiplication by z gives wk z k zz n+1 cn z n ( − 2 ) dλ(z) 2 cn cn−1 𝔻 k=0 ck ∞

∫ ∑

z n+1 wk z k+1 c wk z k dλ(z) − 2n ∫ z n ∑ 2 dλ(z) ∑ 2 cn k=0 ck cn−1 𝔻 k=0 ck ∞

=∫ 𝔻

∞

wn wn = 3 ∫ |z|2n+2 dλ(z) − 2 ∫ |z|2n dλ(z) cn 𝔻 cn−1 cn 𝔻 =(

2 c cn+1 − 2n )wn . cn3 cn−1

Next, multiplication by w yields wk z k zz n cn z n−1 − 2 ) dλ(z) ( 2 cn cn−1 𝔻 k=0 ck ∞

w∫ ∑

=w∫

𝔻

= w( = 0.

wk z k+1 cn z n−1 wk z k zn dλ(z) − w dλ(z) ∑ ∑ ∫ 2 2 cn k=0 ck2 𝔻 cn−1 k=0 ck ∞

∞

cn wn−1 cn wn−1 − 2 ) 2 cn−1 cn−1

Then it follows that S∗ S(ϕn )(w) = (

2 cn+1 cn2 − )ϕn (w), 2 cn2 cn−1

n = 1, 2, … ,

and for n = 0 an analogous computation shows S∗ S(ϕ0 )(w) =

c12 ϕ (w). c02 0

Finally, we get Theorem 8.17. Let S ∶ A2 (𝔻) ⟶ L2 (𝔻) be the canonical solution operator for 𝜕 and (ϕk )k the normalized monomials. Then S∗ Sϕ = for each ϕ ∈ A2 (𝔻).

2 cn+1 cn2 c12 (ϕ, ϕ )ϕ + ( − )(ϕ, ϕn )ϕn ∑ 0 0 2 2 c02 cn−1 n=1 cn ∞

(8.9)

8.2 The canonical solution operator to 𝜕 restricted to A2 (𝔻)

| 225

Since 2 c2 1 cn+1 − 2n = →0 2 cn cn−1 (n + 2)(n + 1)

as n → ∞,

it follows that S∗ S is compact, and so is S. c2 We have also shown that the s-numbers of S are ( cn+1 2 − n

∞

∑(

n=1

cn2 1/2 2 ) cn−1

and since

2 cn2 cn+1 − ) 0 ∶ x ∈ tU} is a continuous seminorm on X; we call ‖ ⋅ ‖U the Minkowski functional of U. One can explain the topology of a locally convex vector space X in a different way: a family 𝒰 of 0-neighborhoods is a fundamental system of 0-neighborhoods, if for https://doi.org/10.1515/9783110417241-009

232 | 9 Nuclear Fréchet spaces of holomorphic functions each 0-neighborhood U there exist V ∈ 𝒰 and ϵ > 0 such that ϵV ⊂ U. A family (pα )α∈A of seminorms is called a fundamental system of seminorms, if the sets Uα = {x ∈ X ∶ pα (x) < 1} constitute a fundamental system of 0-neighborhoods of X. We will write (X, (pα )α∈A ) to refer to that. Let X and Y be locally convex vector spaces with fundamental systems (pα )α∈A and (qβ )β∈B of seminorms. A linear mapping T ∶ X ⟶ Y is continuous, if and only if for each β ∈ B there exist α ∈ A and a constant C > 0 such that qβ (Tx) ≤ Cpα (x), for all x ∈ X. A linear functional x′ on X is continuous, if and only if there exist α ∈ A and a constant C > 0 such that |x′ (x)| ≤ Cpα (x) for all x ∈ X. We indicate that the consequences of the Hahn–Banach Theorem (Theorem 4.11 and Corollary 4.12) are also true for locally convex vector spaces, one has to replace the norm in the proof of Theorem 4.11 by one of the seminorms defining the topology of a locally convex vector space; a subspace Y of a locally convex vector space X is dense in X, if and only if each continuous linear functional on X, which vanishes on Y , also vanishes on the whole of X. The appropriate concept of a bounded subset in X reads as follows: a subset B of a locally convex vector space is said to be bounded if to every 0-neighborhood U in X corresponds a number s > 0 such that B ⊂ tU for every t > s. It is easily seen that B is bounded, if and only if supx∈B pα (x) < ∞ for all α ∈ A, where (pα )α∈A is a fundamental system of seminorms for the topology of X; compare with Definition 4.39. Now let X ′ be the space of all continuous linear functionals on a locally convex vector space (X, (pα )α∈A ). We endow the dual space X ′ with the topology of uniform convergence on all bounded subsets of X; which can be expressed in the following way: (X ′ , (pB )B∈ℬ ), where pB (x′ ) = supx∈B |x′ (x)| and ℬ denotes the family of all bounded subsets of X. It is called the strong topology on X ′ .

9.2 The space 𝓗(DR (0)) and its dual space Our main example is the space ℋ(Ω) of all holomorphic functions on a domain Ω ⊆ ℂn endowed with the topology of uniform convergence on all compact subsets of Ω. Let (Km )m∈ℕ be a compact exhaustion of Ω. The topology of ℋ(Ω) can be described by the increasing system of norms |f |m ∶= supz∈Km |f (z)|, for f ∈ ℋ(Ω). The system of norms (| ⋅ |m )m∈ℕ is a fundamental system of (semi)norms. Let f , g ∈ ℋ(Ω) and define 1 |f − g|m . m m=1 2 1 + |f − g|m ∞

d(f , g) = ∑

(9.1)

It is easily seen that d(⋅, ⋅) is a metric which generates the original topology of uniform convergence on all compact subsets of Ω.

9.2 The space ℋ(DR (0)) and its dual space | 233

By Weierstraß’ Theorem (see Theorem 2.34), ℋ(Ω) is a complete metric vector space. These spaces are called Fréchet spaces. Montel’s Theorem (see Theorem 4.40) indicates that all closed bounded subsets of ℋ(Ω) are compact subsets of ℋ(Ω). The topology of ℋ(Ω) does not stem from a single norm, but from a countable system of norms, see Exercises, Section 9.3. In sake of simplicity, we describe the following properties of ℋ(Ω) for 1-dimensional discs DR (0), most of the results can be generalized to arbitrary domains in ℂn using standard functional analysis methods. Take an increasing sequence rm ↗ R and define |f |m ∶= sup|z|≤rm |f (z)| for f ∈ ℋ(DR (0)). Using (7.2) we find that for each m ∈ ℕ there exist an ℓ ∈ ℕ and a constant C, depending only on m and ℓ, such that |f |m ≤ C(∫

1/2

Drℓ (0)

|f (z)|2 dλ(z))

,

(9.2)

≤ C ′ |f |m

(9.3)

for each f ∈ ℋ(DR (0)); the inequality ‖f ‖m ∶= (∫

Drm (0)

1/2

|f (z)|2 dλ(z))

is clear. Hence, using the Hilbert norms ‖ ⋅ ‖m , the space (ℋ(DR (0)), (‖ ⋅ ‖m )m∈ℕ ) carries the original topology of uniform convergence on all compact subsets of DR (0). Now we consider the Bergman spaces A2 (Drm (0)) endowed with the norm ‖ ⋅ ‖m , see Section 7.1. If rm < rℓ < R, we have the inclusions ℋ(DR (0)) ⊂ A2 (Drℓ (0)) ⊂ A2 (Drm (0)); and we can show that the natural embedding ιℓ,m ∶ A2 (Drℓ (0)) ↪ A2 (Drm (0)) is a Hilbert–Schmidt operator. Fix m ∈ ℕ and set ϕℓn (z) ∶= √

n + 1 zn , π rℓn+1

2 for n = 0, 1, 2, … . Then (ϕℓn )∞ n=0 constitutes an orthonormal basis in A (Drℓ (0)), see Section 7.2. By Theorem 8.10, we have to show that ∞

∑ ‖ιℓ,m (ϕℓn )‖2m < ∞.

n=0

An easy computation shows that ‖ιℓ,m (ϕℓn )‖2m = (rm /rℓ )2n+2 , and as rℓ > rm we get (9.4).

(9.4)

234 | 9 Nuclear Fréchet spaces of holomorphic functions We just showed that for each m ∈ ℕ there exists an ℓ ∈ ℕ such that the natural embedding ιℓ,m ∶ A2 (Drℓ (0)) ↪ A2 (Drm (0)) is a Hilbert–Schmidt operator; we say that ℋ(DR (0)) is a nuclear Fréchet space. Using the Taylor series expansion and its uniqueness property, it is shown that the spaces ℋ(DR (0)) are topologically isomorphic to certain sequence spaces (Köthe1 sequence spaces): Theorem 9.1. Let rm ↗ R be an increasing sequence of positive numbers. Define ∞

∞ n ΛR = {(ξn )∞ n=0 ∶ pm ((ξn )n=0 ) ∶= ∑ |ξn |rm < ∞ ∀m ∈ ℕ}. n=0

Then the spaces (ℋ(DR (0)), (| ⋅ |m )m∈ℕ ) and (ΛR , (pm )m∈ℕ ) are topologically isomorphic, where the isomorphism T ∶ ΛR ⟶ ℋ(DR (0)) is given by ∞

n T((ξn )∞ n=0 )(z) = ∑ ξn z , n=0

z ∈ DR (0),

and T −1 (f ) = (

f (n) (0) ∞ ) , n! n=0

f ∈ ℋ(DR (0)).

Proof. For (ξn )∞ n=0 ∈ ΛR , we have ∞

∞

n n ∞ |T((ξn )∞ n=0 )|m = sup | ∑ ξn z | ≤ ∑ |ξn |rm = pm ((ξn )n=0 ). |z|≤rm n=0

n=0

On the other side, we get from Cauchy’s estimates (see Theorem 2.33) that |

|f | f (n) (0) | ≤ nℓ , n! rℓ

hence, if rℓ > rm , we get ∞

pm (T −1 (f )) = ∑ | n=0 ∞

f (n) (0) n |rm n!

|f |ℓ n r n m n=0 rℓ

≤∑

n rm )|f |ℓ . n n=0 rℓ ∞

= (∑

1 Köthe, Gottfried (1905–1989).

9.2 The space ℋ(DR (0)) and its dual space | 235

In a similar way, we can describe the dual space of ℋ(DR (0)). Recall that L is a continuous linear functional on (ℋ(DR (0)), (| ⋅ |m )m∈ℕ ), if and only if there exist an m ∈ ℕ and a constant C > 0 such that |L(f )| ≤ C|f |m for each f ∈ ℋ(DR (0)). Theorem 9.2. Let rm ↗ R be an increasing sequence of positive numbers. Define ∞ Λ′R = {(ηn )∞ n=0 ∶ ∃m ∈ ℕ with qm ((ηn )n=0 ) ∶= sup n

|ηn | < ∞}. n rm

Then the dual space ℋ′ (DR (0)) is isomorphic to the sequence space (Λ′R , (qm )m∈ℕ ). Proof. We indicate that the seminorms (qm )m∈ℕ are decreasing in m, and that (Λ′R , (qm )m∈ℕ ) is not a metric space, but a dual metric space. Let L ∈ ℋ′ (DR (0)). Then there exist an m ∈ ℕ and a constant C > 0 such that |L(f )| ≤ C|f |m for each f ∈ ℋ(DR (0)), in particular applying L to the monomials z ↦ z n we obtain a sequence ηn ∶= L(z ↦ z n ) such that n |ηn | ≤ Crm ,

n = 0, 1, 2, … .

′ This implies that (ηn )∞ n=0 ∈ ΛR . ′ ∞ If, for the other direction, (ηn )∞ n=0 ∈ ΛR is given with qm ((ηn )n=0 ) = supn we define a linear functional on ℋ(DR (0)) by ∞

L(f ) = ∑ ηn n=0

f for an arbitrary function f (z) = ∑∞ n=0 mates (see Theorem 2.33), we obtain

(0) n z n!

(n)

∞

|L(f )| ≤ ∑ |ηn | n=0

|ηn | n rm

< ∞,

f (n) (0) , n! in ℋ(DR (0)). Again, from Cauchy’s esti-

|f |m+1 n rm+1

n rm |ηn | n n r r n=0 m m+1 ∞

= |f |m+1 ∑

∞

≤ qm ((ηn )∞ n=0 )|f |m+1 ∑

n rm

n n=0 rm+1

.

Hence L ∈ ℋ′ (DR (0)). ′ Furthermore, we associate to each sequence (ηn )∞ n=0 ∈ ΛR a function F which is holomorphic in a neighborhood of ∞, i.e. in a set {w ∈ ℂ ∶ |w| > ℓ}, and with the property F(∞) = limz→0 F(1/z) = 0. This is done in the following way: suppose that |η | supn rnn < ∞, then ℓ ∶= lim supn→∞ |ηn |1/n ≤ rm < R. Hence the function m

ηn , n+1 n=0 w ∞

F(w) = ∑

236 | 9 Nuclear Fréchet spaces of holomorphic functions is holomorphic in {w ∶ |w| > ℓ} and satisfies F(∞) = 0. We know from the last proof that the expression ∞

L(f ) = ∑ ηn n=0

f (n) (0) , n!

f ∈ ℋ(DR (0))

represents an arbitrary continuous linear functional on ℋ(DR (0)). Let ℓ < ρ < R and γρ (t) = ρeit , t ∈ [0, 2π]. Then 1 ηn 1 f (w) dw ∫ F(w)f (w) dw = ∫ ∑ n+1 2πi γρ 2πi γρ n=0 w ∞

= ∑ ηn

f (w) 1 dw ∫ 2πi γρ wn+1

= ∑ ηn

f (n) (0) n!

∞

n=0 ∞ n=0

= L(f ). Given L ∈ ℋ′ (DR (0)), we obtain the corresponding holomorphic function F, representing L as before, by Cauchy’s integral formula L(w ↦

1 1 F(w) )= dw = F(z), ∫ z−w 2πi γρ z − w

(9.5)

1 F (k) (z) , )= k+1 k! (z − w)

(9.6)

and L(w ↦

where |z| > ρ. Let H0 (R) be the space of all functions holomorphic in an open neighborhood of {z ∈ ℂ ∶ |z| ≥ R} such that F(∞) = 0. We have just shown that the dual space ℋ′ (DR (0)) can be identified with H0 (R), a space of holomorphic functions in a neighborhood of the complement of DR (0), which is known as the Köthe duality. This will now be used, together with the Hahn–Banach Theorem, to give a simple proof of a Runge type approximation theorem; compare with the proof of Theorem 4.15 where a similar method was used. For this purpose we have to explain the concept of a subset W ⊂ ℂ with multiplicity m ∶ W ⟶ {1, 2, 3, … } ∪ {∞}. By a limit point of (W, m) we mean an ordinary limit point of W or a point w ∈ W with m(w) = ∞. Given a set (W, m) with multiplicity, let ℛ(W) denote the following collection of functions: if w ∈ W , w ≠ ∞ and m(w) < ∞, then z ↦ 1/(z − w) belongs to ℛ(W); if w ≠ ∞ and m(w) = ∞, then the functions z ↦ 1/(z − w)k , for k = 1, 2, … belong to ℛ(W); if ∞ ∈ W and m(∞) = ∞, then the functions z ↦ z k , for k = 0, 1, 2, … belong to ℛ(W).

9.3 Exercises | 237

Theorem 9.3. If W ⊂ ℂ ⧵ DR (0) is a set with multiplicity which has a limit point in ℂ ⧵ DR (0), then the linear span of ℛ(W) is dense in ℋ(DR (0)). Proof. To show that the linear span of ℛ(W) is dense in ℋ(DR (0)), we take a continuous linear functional L ∈ ℋ′ (DR (0)) which vanishes on the linear span of ℛ(W). Using Corollary 4.12, we will be done if we can show that L vanishes on ℋ(DR (0)). The assumptions on ℛ(W) imply that the holomorphic function F corresponding to L by (9.5) vanishes on a set with a limit point or, using (9.6), has the property that F (k) (ζ ) = 0, for k = 0, 1, 2, … and some ζ ∈ ℂ ⧵ DR (0). In both cases, the Identity Theorems (see Theorem 2.18 and Corollary 2.19) imply that F ≡ 0, and hence L = 0 on ℋ(DR (0)).

9.3 Exercises 140. Let (Km )m∈ℕ be a compact exhaustion of the domain Ω ⊆ ℂn and let |f |m ∶= supz∈Km |f (z)|, for f ∈ ℋ(Ω). Show that 1 |f − g|m m 1 + |f − g| 2 m m=1 ∞

d(f , g) = ∑

defines a translation invariant metric on ℋ(Ω). 141. Show that the metric d(⋅, ⋅) generates the original topology of uniform convergence on all compact subsets of Ω. 142. Let X be a locally convex vector space and let U be an absolutely convex 0-neighborhood in X. Show that the Minkowski functional ‖x‖U = inf{t > 0 ∶ x ∈ tU} is a continuous seminorm on X. 143. Let X be a locally convex vector space. A collection Λ of neighborhoods of a point x ∈ X is called a local base at x, if every neighborhood of x contains a member of Λ. A set B ⊂ X is called balanced if cB ⊂ B for every c ∈ ℂ with |c| ≤ 1. Show that X has a local base consisting of balanced convex sets. 144. Let X be a locally convex vector space and let U be a 0-neighborhood in X. Let (rj )j a strictly increasing sequence of positive numbers with rj → ∞ as j → ∞. Show that ∞

X = ⋃ rj U. j=1

145. Let X be a locally convex vector space. Show that every compact subset K ⊂ X is bounded. Hint: Choose a 0-neighborhood U and a balanced 0-neighborhood W such that W ⊂ U and K ⊂ ⋃∞ n=1 nW . 146. Let (sj )j be a strictly decreasing sequence of positive numbers such that limj→∞ sj = 0 and V be a bounded subset of the locally convex vector space X. Show that the collection {sj V ∶ j ∈ ℕ} is a local base for X.

238 | 9 Nuclear Fréchet spaces of holomorphic functions 147. Show that each finite dimensional subspace of a locally convex subspace is closed. 148. X is locally compact if 0 has a neighborhood whose closure is compact. Show that each locally compact locally convex vector space has finite dimension. Hint: take a 0-neighborhood V whose closure is compact, since V is also bounded, the sets 2−n V , n ∈ ℕ form a local base for X. The compactness of V shows that there exist x1 , … , xm ∈ X such that 1 1 V ⊂ (x1 + V) ∪ ⋯ ∪ (xm + V). 2 2 Let Y be the vector space spanned by x1 , … , xm . Show that Y = X. 149. Let Ω ⊆ ℂn be a domain. Show that ℋ(Ω) endowed with the topology of uniform convergence on all compact subsets of Ω is not normable, i.e. has no bounded 0-neighborhood. Hint: if U is a bounded 0-neighborhood, Montel’s Theorem (see Theorem 4.40) implies that ℋ(Ω) is locally compact, now use Exercise 148. 150. Show that the system of seminorms pr (f ) ∶= sup |ak |r k , 0≤k 0 such that the ball B(4r, y0 ) = {y ∈ Y ∶ ‖y − y0 ‖ < 4r} is contained in T(B1 ). Pick y1 = Tx1 ∈ T(B1 ) such that ‖y1 − y0 ‖ < 2r. Then B(2r, y1 ) ⊂ T(B1 ), so if ‖y‖ < 2r, we have y = Tx1 + (y − y1 ) ∈ T(x1 + B1 ) ⊂ T(B2 ). Dividing both sides by 2, we obtain that there exists r > 0 such that if ‖y‖ < r then y ∈ T(B1 ). If we could replace T(B1 ) by T(B1 ), the proof would be complete. Dilation invariance implies that if ‖y‖ < r2−n then y ∈ T(B2−n ). Suppose that ‖y‖ < r/2; we can find x1 ∈ B1/2 such that ‖y − Tx1 ‖ < r/4, and proceeding inductively, we can find xn ∈ B2−n such that n

‖y − ∑ Txj ‖ < r2−n−1 . j=1

(10.1)

∞ −n We have ∑∞ = 1 and, since X is complete, the absolutely convergent n=1 ‖xn ‖ < ∑n=1 2 ∞ series ∑n=1 xn is also convergent, say to x = ∑∞ n=1 xn ∈ X, and, by (10.1), we have y = Tx. This means that T(B1 ) contains all y with ‖y‖ < r/2.

Corollary 10.12. Let T ∶ X ⟶ Y be a bijective bounded linear operator between Banach spaces. Then T −1 is also bounded. Proof. By Theorem 10.11, T is open, hence T −1 is bounded. Now it is easy to prove the closed graph theorem. Theorem 10.13. Let X and Y be Banach spaces. Let T ∶ X ⟶ Y be a closed operator with dom(T) = X. Then T is bounded. Proof. Let π1 and π2 be the projections of 𝒢(T) onto X and Y . They are both continuous. X × Y is also complete, 𝒢(T) is a closed subspace of X × Y and therefore also complete. π1 is a bijection from 𝒢(T) to X. By Corollary 10.12, π1−1 is bounded. But T = π2 ∘ π1−1 , therefore T is bounded. Before we proceed in the theory for unbounded operators on Hilbert spaces, we still prove the uniform boundedness principle again as an application of the Baire category theorem.

244 | 10 The 𝜕-complex Theorem 10.14. Suppose X and Y are normed spaces and 𝒜 is a subset of ℒ(X, Y). (i) If supT∈𝒜 ‖Tx‖ < ∞ for all x in a nonmeager subset of X, then sup ‖T‖ < ∞.

T∈𝒜

(ii) If X is a Banach space and supT∈𝒜 ‖Tx‖ < ∞ for all x ∈ X, then sup ‖T‖ < ∞.

T∈𝒜

Proof. Let En = {x ∈ X ∶ sup ‖Tx‖ ≤ n} = ⋂ {x ∈ X ∶ ‖Tx‖ ≤ n}. T∈𝒜

T∈𝒜

Then the sets En are closed, so under the hypothesis of (i), some En must contain a nontrivial closed ball B(r, x0 ). Then E2n ⊃ B(r, 0), for if ‖x‖ ≤ r then x − x0 ∈ En and ‖Tx‖ ≤ ‖T(x − x0 )‖ + ‖Tx0 ‖ ≤ 2n. In other words, ‖Tx‖ ≤ 2n whenever T ∈ 𝒜 and ‖x‖ ≤ r, so sup ‖T‖ ≤ 2n/r.

T∈𝒜

(ii) follows from (i) by Theorem 10.10. The following characterization of compactness uses the uniform boundedness principle and the concept of weak convergence. Definition 10.15. A sequence (xk )k in a Hilbert space H is a weak null-sequence, if (xk , x) → 0 for each x ∈ H. A sequence (xk )k converges weakly to x0 , if (xk − x0 )k is a weak null-sequence. Remark 10.16. A weakly convergent sequence (xk )k in a Hilbert space is always bounded: we have sup|(xk − x0 , x)| < ∞, k

for all x ∈ H, then, by Theorem 10.14, sup ‖xk − x0 ‖ = sup sup |(xk − x0 , x)| < ∞ k

k

‖x‖≤1

and therefore ‖xk ‖ ≤ ‖xk − x0 ‖ + ‖x0 ‖ < ∞, for all k ∈ ℕ. In the same way we can show that each weak Cauchy sequence is bounded. If A ∈ ℒ(H1 , H2 ) and (xk )k is a weakly convergent sequence in H1 , then (Axk )k converges weakly in H2 , which follows from (Axk − Ax0 , y)2 = (xk − x0 , A∗ y)1 , where y ∈ H2 .

10.1 Unbounded operators on Hilbert spaces | 245

Theorem 10.17. Let A ∈ ℒ(H1 , H2 ) be a bounded operator between Hilbert spaces. A is compact, if and only if (Axk )k converges to 0 in H2 for each weak null-sequence (xk )k in H1 . Proof. Let A be compact and (xk )k be a weak null-sequence in H1 . Then (Axk )k is a weak null-sequence in H2 . Suppose that (‖Axk ‖)k does not converge to 0. Then there exists ϵ > 0 and a subsequence (yk )k of (xk )k such that ‖Ayk ‖ ≥ ϵ for each k ∈ ℕ. By Remark 10.16, the sequence (‖yk ‖)k is bounded. As A is compact, there exists a subsequence (zk )k of (yk )k such that (Azk )k is convergent in H2 . Since (Azk )k converges weakly to 0, we have ‖Azk ‖ = sup |(Azk , y)| → 0, ‖y‖≤1

contradicting ‖Ayk ‖ ≥ ϵ. For the opposite direction we need a number of prerequisites: first we claim the following assertion: if (xk )k is a bounded sequence in a Hilbert space H and if ((xk , y))k is a Cauchy sequence in ℂ for each y in a dense subset M of H, then (xk )k is a weak Cauchy sequence in H. To show this, let x ∈ H be an arbitrary element and choose y ∈ M such that ‖x − y‖ < ϵ. Then |(xk − xm , x)| ≤ |(xk , x − y)| + |(xk − xm , y)| + |(xm , y − x)|. Since (xk )k is a bounded sequence, we get that ((xk , x))k is a Cauchy sequence in ℂ for each x ∈ H. Next we claim that each weak Cauchy sequence in H is also weakly convergent: a weak Cauchy sequence is bounded, therefore we have |(xk , x)| ≤ C‖x‖, for some constant C > 0 and for each k ∈ ℕ. Hence F(x) ∶= lim (x, xk ) k→∞

defines a continuous linear functional, for which there exists x0 ∈ H such that (x, x0 ) = lim (x, xk ), k→∞

which means that (xk )k converges weakly to x0 . Next we show that each bounded sequence (xk )k contains a weakly convergent subsequence: the sequence ((xk , x))k is bounded in ℂ, hence we can find subsequences (xk(m) )k of (xk(m−1) )k for m ∈ ℕ with (xk )k = (xk(0) )k such that ((xk(m) , xm ))k converges. The diagonal sequence (xk(k) )k has the property that the sequence ((xk(k) , xm ))k is convergent for each m ∈ ℕ. This implies that ((xk(k) , x))k converges for each x in the linear span L of the elements {xℓ ∶ ℓ ∈ ℕ}. Since the sequence (xk(k) )k is bounded, we have from above that ((xk(k) , x))k is convergent for each x ∈ L. It is clear that ((xk(k) , x)) = 0 for each x ∈ L⊥ , and therefore we get that (xk(k) )k is a weak Cauchy sequence, which converges weakly, by our assumptions.

246 | 10 The 𝜕-complex Now we can finish the prove of the opposite direction: let (xk )k be a bounded sequence in H1 . We know that it contains a weakly convergent subsequence xkℓ → x0 , hence the sequence (xkℓ − x0 )ℓ is a weak null sequence. By assumption we obtain that Axkℓ → Ax0 in H2 . So A is a compact operator. Now we continue to derive further important properties of unbounded operators on Hilbert spaces which will later be used for the 𝜕-operator. Lemma 10.18. Let T ∶ H1 ⟶ H2 be a bounded linear operator. T(H1 ) is closed, if and only if T|(ker T)⟂ is bounded from below, i.e. ‖Tf ‖ ≥ C‖f ‖

∀f ∈ (ker T)⟂ .

Proof. If T(H1 ) is closed, then the mapping T ∶ (ker T)⟂ ⟶ T(H1 ) is bijective and continuous and, by the open-mapping theorem (see Theorem 10.11), also open. This implies the desired inequality. To prove the other direction, let (fn )n be a sequence in H1 with Tfn → y in H2 . We have to show that there exists h ∈ H1 with Th = y. Decompose fn = gn + hn , where gn ∈ ker T and hn ∈ (ker T)⟂ . By assumption we have ‖hn − hm ‖ ≤ C‖Thn − Thm ‖ = C‖Tfn − Tfm ‖ < ϵ, for all sufficiently large n and m. Hence (hn )n is a Cauchy sequence. Let h = limn→∞ hn . Then we have ‖Tfn − Th‖ = ‖Thn − Th‖ ≤ ‖T‖ ‖hn − h‖, and therefore y = lim Tfn = Th. n→∞

Lemma 10.19. Let T be as before. T(H1 ) is closed, if and only if T ∗ (H2 ) is closed. Proof. Since T ∗∗ = T, it suffices to show one direction. We will show that the closedness of T(H1 ) implies, that (ker T)⟂ = im T ∗ ; since (ker T)⟂ is closed, we will be done. Let x ∈ im T ∗ . Then there exists y ∈ H2 with x = T ∗ y. Now we get for x ′ ∈ ker T that (x, x′ ) = (T ∗ y, x′ ) = (y, Tx′ ) = 0, hence im T ∗ ⊆ (ker T)⟂ . For x′ ∈ (ker T)⟂ we define a linear functional λ(Tx) = (x, x ′ )

10.1 Unbounded operators on Hilbert spaces | 247

on the closed subspace T(H1 ) of H2 . We remark that λ is well-defined, since Tx = T x̃ implies that x − x̃ ∈ ker T, hence (x − x,̃ x′ ) = 0 and (x, x ′ ) = (x,̃ x ′ ). The operator T ∶ H1 ⟶ T(H1 ) is continuous and surjective. Since T(H1 ) is closed, the open-mapping theorem implies ‖v‖ ≤ C‖Tv‖ for all v ∈ (ker T)⟂ where C > 0 is a constant. Set y = Tx and write x = u + v, where u ∈ ker T and v ∈ (ker T)⟂ . Then we obtain |λ(y)| = |(x, x′ )| = |(v, x ′ )| ≤ ‖v‖‖x′ ‖ ≤ C‖Tv‖‖x′ ‖ = C‖Tx‖‖x′ ‖ = C‖y‖‖x′ ‖. Hence λ is continuous on im T. By Theorem 7.7, there exists a uniquely determined element z ∈ im T with λ(y) = (y, z)2 = (x, x′ )1 . This implies (y, z)2 = (Tx, z)2 = (x, T ∗ z)1 = (x, x′ )1 , for all x ∈ H1 , and hence x ′ = T ∗ z ∈ im T ∗ . Lemma 10.20. Let T ∶ H1 ⟶ H2 be a densely defined closed operator. im T is closed in H2 , if and only if T|dom(T)∩(ker T)⟂ is bounded from below, i.e. ‖Tf ‖ ≥ C‖f ‖

∀f ∈ dom(T) ∩ (ker T)⟂ .

̃ , Tf )) = Tf and get a bounded Proof. On the graph 𝒢(T) we define the operator T((f linear operator T̃ ∶ 𝒢(T) ⟶ H2 , since ̃ , Tf ))‖ = ‖Tf ‖ ≤ (‖f ‖2 + ‖Tf ‖2 )1/2 = ‖(f , Tf )‖; ‖T((f and im T̃ = im T. By Lemma 10.18, im T is closed, if and only if T|̃ (ker T)̃ ⟂ is bounded from below. We have ker T̃ = ker T ⊕ {0}, and it remains to show that T|̃ (ker T)̃ ⟂ is bounded from below, if and only if T|dom(T)∩(ker T)⟂ is bounded from below. But this follows from ̃ , Tf ))‖ = ‖Tf ‖ ≥ C(‖f ‖2 + ‖Tf ‖2 )1/2 , ‖T((f and hence, for 0 < C < 1, ‖Tf ‖2 ≥

C2 ‖f ‖2 . 1 − C2

248 | 10 The 𝜕-complex Lemma 10.21. Let P, Q ∶ H ⟶ H be orthogonal projections on the Hilbert space H. Then the following assertions are equivalent: (i) im(PQ) is closed; (ii) im(QP) is closed; (iii) im(I − P)(I − Q) is closed; (iv) P(H) + (I − Q)(H) is closed. Proof. (i) and (ii) are equivalent by Lemma 10.19 and the fact that QP = Q∗ P ∗ = (PQ)∗ . Suppose (ii) holds and let (fn )n and (gn )n be sequences in H with Pfn + (I − Q)gn → h. Then Q(Pfn + (I − Q)gn ) = QPfn → Qh. By assumption, im(QP) is closed, hence there exists f ∈ H with QPf = Qh; it follows that Qh = Pf − (I − Q)(Pf ) and h = Qh + (I − Q)h = Pf − (I − Q)(Pf ) + (I − Q)h = Pf + (I − Q)(h − Pf ) ∈ P(H) + (I − Q)(H), which yields (iv). If (iv) holds and (fn )n is a sequence in H with QPfn → h, we get QPfn = Pfn − (I − Q)Pfn ∈ P(H) + (I − Q)(H). Hence there exist f , g ∈ H with h = Pf + (I − Q)g; and it follows that Qh = Q( lim QPfn ) = lim Q2 Pfn = h, n→∞

n→∞

and h = Pf + (I − Q)g = Qh = QPf ∈ im(QP), therefore (ii) holds. Finally, replace P by I − P and Q by I − Q. Then, using the assertions proved so far, we obtain the equivalence im(I − P)(I − Q) closed

⟺

(I − P)(H) + Q(H) closed,

which proves the remaining assertions. At this point, we are able to prove Lemma 10.19 for densely defined closed operators. Theorem 10.22. Let T ∶ H1 ⟶ H2 be a densely defined closed operator. im T is closed, if and only if im T ∗ is closed.

10.1 Unbounded operators on Hilbert spaces | 249

Proof. Let P ∶ H1 × H2 ⟶ 𝒢(T) be the orthogonal projection of H1 × H2 on the closed subspace 𝒢(T) of H1 × H2 , and let Q ∶ H1 × H2 ⟶ {0} × H2 be the canonical orthogonal projection. Then im T ≅ im QP and, since I − Q ∶ H1 × H2 ⟶ H1 × {0} and I − P ∶ H1 × H2 ⟶ 𝒢(T)⟂ = V(𝒢(T ∗ )) ≅ 𝒢(T ∗ ), we obtain the desired result from Lemma 10.21. Theorem 10.23. Let T ∶ H1 ⟶ H2 be a densely defined closed operator and G a closed subspace of H2 with G ⊇ im T. Suppose that T ∗ |dom(T ∗ )∩G is bounded from below, i.e. ‖f ‖ ≤ C‖T ∗ f ‖ for all f ∈ dom(T ∗ ) ∩ G, where C > 0 is a constant. Then G = im T. Proof. We have ker T ∗ = (im T)⟂ . Since im T ⊆ G, it follows that ker T ∗ ⊇ G⟂ . If G⟂ is a proper subspace of ker T ∗ , then G ∩ ker T ∗ ≠ {0}, which is a contradiction to the assumption that T ∗ |dom(T ∗ )∩G is bounded from below. Hence ker T ∗ = G⟂ and G = G⟂⟂ = (ker T ∗ )⟂ = im T ⟂⟂ = (im T). In addition, we have T ∗ |dom(T ∗ )∩G = T ∗ |dom(T ∗ )∩(ker T ∗ )⟂ and, by Lemma 10.20, we obtain that im T ∗ is closed. By Theorem 10.22, im T is also closed and we get that G = im T. Remark 10.24. The last theorem also holds in the other direction: if T ∶ H1 ⟶ H2 is a densely defined closed operator and G is a closed subspace of H2 with G = im T, then T ∗ |dom(T ∗ )∩G is bounded from below. Since in this case G = im T, we have that im T is closed and hence, by Lemma 10.22, im T ∗ is also closed. Therefore, Lemma 10.20 and the fact that G = (ker T ∗ )⟂ implies that T ∗ |dom(T ∗ )∩G is bounded from below. Theorem 10.25. Let T ∶ H1 ⟶ H2 be a densely defined closed operator and let G be a closed subspace of H2 with G ⊇ im T. Suppose that T ∗ |dom(T ∗ )∩G is bounded from below. Then for each v ∈ H1 with v ⟂ ker T there exists f ∈ dom(T ∗ ) ∩ G with T ∗ f = v and ‖f ‖ ≤ C‖v‖. Proof. We have ker T = (im T ∗ )⟂ , hence v ∈ (ker T)⟂ = im T ∗ . In addition, G⟂ ⊆ (im T)⟂ = ker T ∗ , and therefore im T ∗ |dom(T ∗ )∩G = im T ∗ , which means that im T ∗ is closed and that for v ∈ (ker T)⟂ = im T ∗ there exists an f ∈ dom(T ∗ ) ∩ G with T ∗ f = v. The desired norm-inequality follows from the assumption that T ∗ |dom(T ∗ )∩G is bounded from below.

250 | 10 The 𝜕-complex In the following we introduce the fundamental concept of an unbounded selfadjoint operator, which will be crucial for both spectral theory and its applications to complex analysis. Definition 10.26. Let T ∶ dom(T) ⟶ H be a densely defined linear operator. T is symmetric, if (Tx, y) = (x, Ty) for all x, y ∈ dom(T). We say that T is self-adjoint, if T is symmetric and dom(T) = dom(T ∗ ). This is equivalent to requiring that T = T ∗ and implies that T is closed. We say that T is essentially self-adjoint, if it is symmetric and its closure T is self-adjoint. Remark 10.27. (a) If T is a symmetric operator, there are two natural closed extensions. We have dom(T) ⊆ dom(T ∗ ) and T ∗ = T on dom(T). Since T ∗ is closed (Lemma 10.5), T ∗ is a closed extension of T; it is the maximal closed extension. T is also closable, by Lemma 10.2, therefore T exists; it is the minimal closed extension. (b) If T is essentially self-adjoint, then its self-adjoint extension is unique. To prove this, let S be a self-adjoint extension of T. Then S is closed and, being an extension of T, it is also an extension of its smallest extension T. Hence T ⊂ S = S∗ ⊂ (T)∗ = T and S = T. Lemma 10.28. Let T be a densely defined, symmetric operator. (i) If dom(T) = H, then T is self-adjoint and T is bounded. (ii) If T is self-adjoint and injective, then im(T) is dense in H, and T −1 is self-adjoint. (iii) If im(T) is dense in H, then T is injective. (iv) If im(T) = H, then T is self-adjoint, and T −1 is bounded. Proof. (i) By assumption dom(T) ⊆ dom(T ∗ ). If dom(T) = H, it follows that T is selfadjoint, therefore also closed (Lemma 10.5) and continuous by the closed graph theorem. (ii) Suppose y⊥ im(T). Then x ↦ (Tx, y) = 0 is continuous on dom(T), hence y ∈ dom(T ∗ ) = dom(T), and (x, Ty) = (Tx, y) = 0 for all x ∈ dom(T). Thus Ty = 0 and, since T is assumed to be injective, it follows that y = 0. This proves that im(T) is dense in H. T −1 is therefore densely defined, with dom(T −1 ) = im(T), and (T −1 )∗ exists. Now let U ∶ H × H ⟶ H × H be defined by U((x, y)) = (−y, x). It easily follows that U 2 = −I and U 2 (M) = M for any subspace M of H × H, and we get 𝒢(T −1 ) = U(𝒢(−T)) and U(𝒢(T −1 )) = 𝒢(−T). Being self-adjoint, T is closed; hence −T is closed and T −1 is closed. By Lemma 10.6 applied to T −1 and to −T, we get the orthogonal decompositions H × H = U(𝒢(T −1 )) ⊕ 𝒢((T −1 )∗ )

10.1 Unbounded operators on Hilbert spaces | 251

and H × H = U(𝒢(−T)) ⊕ 𝒢(−T) = 𝒢(T −1 ) ⊕ U(𝒢(T −1 )). Consequently, 𝒢((T −1 )∗ ) = [U(𝒢(T −1 ))]⊥ = 𝒢(T −1 ), which shows that (T −1 )∗ = T −1 . (iii) Suppose Tx = 0. Then (x, Ty) = (Tx, y) = 0 for each y ∈ dom(T). Thus x⊥ im(T), and therefore x = 0. (iv) Since im(T) = H, (iii) implies that T is injective, dom(T −1 ) = H. If x, y ∈ H, then x = Tz and y = Tw, for some z ∈ dom(T) and w ∈ dom(T), so that (T −1 x, y) = (z, Tw) = (Tz, w) = (x, T −1 y). Hence T −1 is symmetric. (i) implies that T −1 is self-adjoint (and bounded), and now it follows from (ii) that T = (T −1 )−1 is also self-adjoint. Lemma 10.29. Let T be a densely defined closed operator, dom(T) ⊆ H1 and T ∶ dom(T) ⟶ H2 . Then B = (I + T ∗ T)−1 and C = T(I + T ∗ T)−1 are everywhere defined and bounded, ‖B‖ ≤ 1, ‖C‖ ≤ 1; in addition, B is self-adjoint and positive. Proof. Let h ∈ H1 be an arbitrary element and consider (h, 0) ∈ H1 × H2 . From the proof of Lemma 10.7, we get H1 × H2 = 𝒢(T) ⊕ V −1 (𝒢(T ∗ )),

(10.2)

which implies that (h, 0) can be written in a unique way as (h, 0) = (f , Tf ) + (−T ∗ (−g), −g), for f ∈ dom(T) and g ∈ dom(T ∗ ), which gives h = f + T ∗ g and 0 = Tf − g. We set Bh ∶= f and Ch ∶= g. In this way we get two linear operators B and C defined everywhere on H1 . The two equations from above can now be written as I = B + T ∗ C,

0 = TB − C,

which gives C = TB

and I = B + T ∗ TB = (I + T ∗ T)B.

The decomposition in (10.2) is orthogonal, therefore we obtain ‖h‖2 = ‖(h, 0)‖2 = ‖(f , Tf )‖2 + ‖(T ∗ g, −g)‖2 = ‖f ‖2 + ‖Tf ‖2 + ‖T ∗ g‖2 + ‖g‖2 ,

(10.3)

252 | 10 The 𝜕-complex and hence ‖Bh‖2 + ‖Ch‖2 = ‖f ‖2 + ‖g‖2 ≤ ‖h‖2 , which implies ‖B‖ ≤ 1 and ‖C‖ ≤ 1. For each u ∈ dom(T ∗ T) we get ((I + T ∗ T)u, u) = (u, u) + (Tu, Tu) ≥ (u, u) hence, if (I + T ∗ T)u = 0 we get u = 0. Therefore (I + T ∗ T)−1 exists, and (10.3) implies that (I + T ∗ T)−1 is defined everywhere and B = (I + T ∗ T)−1 . Finally, let u, v ∈ H1 . Then (Bu, v) = (Bu, (I + T ∗ T)Bv) = (Bu, Bv) + (Bu, T ∗ TBv) = (Bu, Bv) + (T ∗ TBu, Bv) = ((I + T ∗ T)Bu, Bv) = (u, Bv) and (Bu, u) = (Bu, (I + T ∗ T)Bu) = (Bu, Bu) + (TBu, TBu) ≥ 0, which proves the lemma. At this point we can describe the concept of the core of an operator, which will be very useful later for spectral analysis. Definition 10.30. Let T be a closable operator with domain dom(T). A subspace D ⊂ dom(T) is called a core of the operator T, if the closure of the restriction T|D is an extension of T. Remark 10.31. If T is a closed operator, then T|D = T. Lemma 10.32. Let T be a densely defined closed operator, dom(T) ⊆ H1 and T ∶ dom(T) ⟶ H2 . Then dom(T ∗ T) is a core of the operator T. Proof. We have to show that 𝒢(T) = 𝒢(T|dom(T ∗ T) ). For this purpose we consider elements (x, Tx) in the graph of T. We suppose that (x, Tx) ⟂ (y, Ty) for each y ∈ dom(T ∗ T). Then (x, (I + T ∗ T)y) = (x, y) + (Tx, Ty) = ((x, Tx), (y, Ty)) = 0, and, as im(I + T ∗ T) = H1 (Lemma 10.29), we conclude that x = 0, which means that 𝒢(T|dom(T ∗ T) ) is dense in 𝒢(T). Finally, we describe a general method to construct self-adjoint operators associated with Hermitian sesquilinear forms. This leads to a self-adjoint extension of an unbounded operator, which is known as the Friedrichs1 extension. 1 Friedrichs, Kurt Otto (1901–1982).

10.1 Unbounded operators on Hilbert spaces | 253

Definition 10.33. Let (𝒱, ‖ ⋅ ‖𝒱 ) and (H, ‖ ⋅ ‖H ) be Hilbert spaces such that 𝒱 ⊂ H,

(10.4)

and suppose that there exists a constant C > 0 such that for all u ∈ 𝒱 we have ‖u‖H ≤ C‖u‖𝒱 .

(10.5)

We also assume that 𝒱 is dense in H. In this situation the space H can be imbedded into the dual space 𝒱′ : for h ∈ H the mapping L(u) = (u, h)H ,

u∈𝒱

is continuous on 𝒱, this follows from (10.5): |L(u)| ≤ ‖u‖H ‖h‖H ≤ C‖h‖H ‖u‖𝒱 . Hence there exists a uniquely determined vh ∈ 𝒱′ such that vh (u) = (u, h)H ,

u ∈ 𝒱,

and the mapping h ↦ vh is injective, as 𝒱 is dense in H. Definition 10.34. A form a ∶ 𝒱 × 𝒱 ⟶ ℂ is sesquilinear, if it is linear in the first component and anti-linear in the second component. The form a is continuous, if there exists a constant C > 0 such that |a(u, v)| ≤ C‖u‖𝒱 ‖v‖𝒱

(10.6)

for all u, v ∈ 𝒱, and it is Hermitian, if a(u, v) = a(v, u) for all u, v ∈ 𝒱. The form a is called 𝒱-elliptic, if there exists a constant α > 0 such that |a(u, u)| ≥ α‖u‖2𝒱

(10.7)

for all u ∈ 𝒱. Theorem 10.35. Let a be a continuous, 𝒱-elliptic form on 𝒱 × 𝒱. Using (10.6) and the Riesz representation theorem (Theorem 7.7), we can define a linear operator A∶𝒱⟶𝒱 such that a(u, v) = (Au, v)𝒱 . This operator A is a topological isomorphism from 𝒱 onto 𝒱.

(10.8)

254 | 10 The 𝜕-complex Proof. First we show that A is injective: (10.8) and (10.7) imply that for u ∈ 𝒱 we have ‖Au‖𝒱 ‖u‖𝒱 ≥ |(Au, u)𝒱 | ≥ α‖u‖2𝒱 , hence ‖Au‖𝒱 ≥ α‖u‖𝒱 ,

(10.9)

which implies that A is injective. Now we claim that A(𝒱) is dense in 𝒱. Let u ∈ 𝒱 be such that (Av, u)𝒱 = 0 for each v ∈ 𝒱. Taking v = u, we get a(u, u) = 0 and, by (10.7), u = 0, which proves the claim. Next we observe that (10.8) implies a(u, Au) = ‖Au‖2𝒱 , therefore, using (10.6), we obtain ‖A(u)‖𝒱 ≤ C‖u‖𝒱 , hence A ∈ ℒ(𝒱). If (vn )n is a Cauchy sequence in A(𝒱) and Aun = vn , we derive from (10.9) that (un )n is also a Cauchy sequence. Let u = limn→∞ un . We know already that A is continuous, therefore limn→∞ Aun = Au, which shows that limn→∞ vn = v = Au and A(𝒱) is closed. As we have already shown that A(𝒱) is dense in 𝒱, we conclude that A is surjective. Finally, (10.9) yields that A−1 is continuous. Theorem 10.36. Let a be a Hermitian, continuous, 𝒱-elliptic form on 𝒱 × 𝒱 and suppose that (10.4) and (10.5) hold. Let dom(S) be the set of all u ∈ 𝒱 such that the mapping v ↦ a(u, v) is continuous on 𝒱 for the topology induced by H. For each u ∈ dom(S) there exists a uniquely determined element Su ∈ H such that a(u, v) = (Su, v)H

(10.10)

for each v ∈ 𝒱 (by the Riesz representation theorem (Theorem 7.7)). Then S ∶ dom(S) ⟶ H is a bijective densely defined self-adjoint operator and S−1 ∈ ℒ(H). Moreover, dom(S) is also dense in 𝒱. Proof. First we show that S is injective. For each u ∈ dom(S), we get from (10.7) and (10.5) that α‖u‖2H ≤ Cα‖u‖2𝒱 ≤ C|a(u, u)|

= C|(Su, u)H | ≤ C‖Su‖H ‖u‖H ,

which implies that α‖u‖H ≤ C‖Su‖H ,

(10.11)

for all u ∈ dom(S), therefore S is injective. Now let h ∈ H and consider the mapping v ↦ (h, v)H for v ∈ 𝒱. Then, by (10.5), we obtain |(h, v)H | ≤ ‖h‖H ‖v‖H ≤ C‖h‖H ‖v‖𝒱 ,

10.2 Distributions and Sobolev spaces | 255

which implies that there exists a uniquely determined w ∈ 𝒱 such that (h, v)H = (w, v)𝒱 for all v ∈ 𝒱. Now we apply Theorem 10.35 and get from (10.8) that a(u, v) = (w, v)𝒱 , where u = A−1 w. Since a(u, v) = (h, v)H for each v ∈ 𝒱, we conclude that u ∈ dom(S) and that Su = h, which shows that S is surjective. Suppose that (u, h)H = 0 for each u ∈ dom(S). As S is surjective, there is v ∈ dom(S) such that Sv = h and we get that (u, Sv)H = 0 for each u ∈ dom(S). Using the 𝒱-ellipticity (10.7), we get for u = v that 0 = (Sv, v)H = a(v, v) ≥ α‖v‖2𝒱 , which implies that v = 0, and consequently h = 0. Therefore we have shown that dom(S) is dense in H. As a(u, v) is Hermitian, we get for u, v ∈ dom(S) that (Su, v)H = a(u, v) = a(v, u) = (Sv, u)H = (u, Sv)H . Hence S is symmetric and dom(S) ⊂ dom(S∗ ). Let v ∈ dom(S∗ ). Since S is surjective, there exists v0 ∈ dom(S) such that Sv0 = S∗ v. This implies (Su, v0 )H = (u, Sv0 )H = (u, S∗ v)H = (Su, v)H , for all u ∈ dom(S). Using again the surjectivity of S, we derive that v = v0 ∈ dom(S). This implies that dom(S) = dom(S∗ ) and that S is self-adjoint. By Lemma 10.28 (iv), S−1 ∈ ℒ(H). Finally, we show that dom(S) is dense in 𝒱. Let h ∈ 𝒱 be such that (u, h)𝒱 = 0, for all u ∈ dom(S). By Theorem 10.35, there exists f ∈ 𝒱 such that Af = h. Then 0 = (u, h)𝒱 = (u, Af )𝒱 = (Af , u)𝒱 = a(f , u) = a(u, f ) = (Su, f )H . S is surjective, therefore we obtain f = 0 and h = Af = 0.

10.2 Distributions and Sobolev spaces In our context the unbounded operator T will mainly be the 𝜕-operator. In order to achieve an appropriate closed extension, we will have to consider the derivatives 𝜕z𝜕 k in the sense of distributions. Therefore we will now briefly summarize elementary definitions and results of distribution theory. ∞ Definition 10.37. Let Ω ⊆ ℝn an open subset and 𝒟(Ω) = 𝒞∞ 0 (Ω) the space of 𝒞 -functions with compact support (test functions). A sequence (ϕj )j tends to 0 in 𝒟(Ω), if there exists a compact set K ⊂ Ω such that supp ϕj ⊂ K for every j and

256 | 10 The 𝜕-complex 𝜕|α| ϕj

α 𝜕x1 1

α

⋯ 𝜕xn n

→0

uniformly on K for each α = (α1 , … , αn ). A distribution is a linear functional u on 𝒟(Ω) such that for every compact subset K ⊂ Ω there exist a k ∈ ℕ0 = ℕ ∪ {0} and a constant C > 0 with |u(ϕ)| ≤ C ∑ sup| |α|≤k x∈K

𝜕|α| ϕ(x) α |, α 𝜕x1 1 ⋯ 𝜕xn n

for each ϕ ∈ 𝒟(Ω) with support in K. We denote the space of distributions on Ω by 𝒟′ (Ω). It is easily seen that u ∈ 𝒟′ (Ω), if and only if u(ϕj ) → 0 for every sequence (ϕj )j in 𝒟(Ω) converging to 0 in 𝒟(Ω). Example 10.38. (1) Let f ∈ L1loc (Ω), where L1loc (Ω) = {f ∶ Ω ⟶ ℂ measurable ∶ f |K ∈ L1 (K)

∀K ⊂ Ω, K compact}.

The mapping Tf (ϕ) = ∫Ω f (x)ϕ(x) dλ(x), ϕ ∈ 𝒟(Ω), is a distribution. (2) Let a ∈ Ω and δa (ϕ) ∶= ϕ(a), which is the point evaluation in a. The distribution δa is called Dirac delta distribution. In the sequel, certain operations for ordinary functions, such as multiplication of functions and differentiation, are generalized to distributions. Definition 10.39. Let f ∈ 𝒞∞ (Ω) and u ∈ 𝒟′ (Ω). The multiplication of u with f is defined by (fu)(ϕ) ∶= u(fϕ) for ϕ ∈ 𝒟(Ω). Notice that fϕ ∈ 𝒟(Ω). For u ∈ 𝒟′ (ℝn ) and f ∈ 𝒟(ℝn ) the convolution of u and f is defined by (u ∗ f )(x) ∶= u(y ↦ f (x − y)). If u = Tg for some locally integrable function g, it is the usual convolution of functions (Tg ∗ f )(x) = ∫ g(y)f (x − y) dλ(y) = (g ∗ f )(x). Ω

Let Dk =

𝜕 𝜕xk

and Dα =

α 𝜕x1 1

𝜕|α| α , ⋯ 𝜕xn n

where α = (α1 , … , αn ) is a multiindex. The partial derivative of a distribution u ∈ 𝒟′ (Ω) is defined by (Dk u)(ϕ) ∶= −u(Dk ϕ),

ϕ ∈ 𝒟(Ω);

higher order mixed derivatives are defined as (Dα u)(ϕ) ∶= (−1)|α| u(Dα ϕ),

ϕ ∈ 𝒟(Ω).

10.2 Distributions and Sobolev spaces | 257

This definition stems from integrating by parts: ∫ (Dk f )ϕ dλ = − ∫ f (Dk ϕ) dλ, Ω

Ω

where f ∈ 𝒞1 (Ω) and ϕ ∈ 𝒟(Ω). Definition 10.40. Let Ω1 ⊆ Ω be an open subset and u ∈ 𝒟′ (Ω). The restriction of u to Ω1 is the distribution u|Ω1 defined by u|Ω1 (ϕ) = u(ϕ),

ϕ ∈ 𝒟(Ω1 ) ⊆ 𝒟(Ω).

The support supp(u) of u ∈ 𝒟′ (Ω) consists of x ∈ Ω such that there is no open neighborhood U of x with u|U = 0. We denote the space of distributions with compact support by ℰ′ (Ω). A fundamental solution of a differential operator P(D) is a distribution E ∈ ℰ′ (ℝn ) with P(D)E = δ0 . In what follows we will find a fundamental solution for 𝜕 and we will investigate regularity properties of the 𝜕-operator. For this purpose we recall Stokes’ Theorem in the following way: let Ω ⊂ ℂ be a bounded domain with piecewise 𝒞1 -boundary γ (positively oriented), let f ∈ 𝒞1 (Ω) and ζ ∈ Ω; for 0 < ϵ < dist(ζ , Ωc ), let Ωϵ ∶= {z ∈ Ω ∶ |z − ζ | > ϵ}, where the boundary of Ωϵ consists of γ and the negatively oriented circle (z) ρϵ = {z ∶ |z − ζ | = ϵ}. Consider the differential form ω = fz−ζ dz, then by Stokes’ Theorem ∫ dω = ∫ ω + ∫ ω = ∫ γ

Ωϵ

ρϵ

γ

2π f (z) dz − ∫ if (ζ + ϵeiθ ) dθ. z−ζ 0

(10.12)

We have dω =

𝜕 f (z) 𝜕f 1 ( ) dz ∧ dz = − dz ∧ dz, 𝜕z z − ζ 𝜕z z − ζ

2π

and by continuity ∫0 if (ζ + ϵeiθ ) dθ → 2πif (ζ ) as ϵ → 0. Hence we obtain f (ζ ) =

1 1 f (z) 𝜕f 1 dz + dz ∧ dz; ∫ ∫ 2πi γ z − ζ 2πi Ω 𝜕z z − ζ

(10.13)

see Theorem 2.47. In particular, if ϕ ∈ 𝒟(Ω), it follows that ϕ(ζ ) = −

1 𝜕ϕ 1 dλ(z), ∫ π Ω 𝜕z z − ζ

where we used the fact that dz ∧ dz = (−2i) dx ∧ dy = (−2i) dλ(z).

(10.14)

258 | 10 The 𝜕-complex This can be interpreted in the sense of distributions as δζ (ϕ) = ϕ(ζ ) = −

1 𝜕ϕ 1 𝜕 1 dλ(z) = ∫ (z ↦ )ϕ(z) dλ(z), ∫ π Ω 𝜕z z − ζ π(z − ζ ) Ω 𝜕z

which means that z ↦

1 πz

(10.15)

is a fundamental solution to 𝜕.

Theorem 10.41. Let Ω ⊆ ℂ be a domain and u ∈ 𝒟′ (Ω) a distribution satisfying 𝜕u = 0. Then there exists a holomorphic function f on Ω such that u = Tf . Proof. Let E be the fundamental solution to 𝜕 given in (10.15). We have 𝜕E = δ0 . Now let Ω′ ⊂⊂ Ω be a relatively compact subset and choose g ∈ 𝒟(Ω) such that g = 1 in some neighborhood of Ω′ . Let S ∶= 𝜕(gu). Then S ∈ ℰ′ (Ω) with supp(S) ⊂ ℂ ⧵ Ω′ , hence gu = δ0 ∗ (gu) = (𝜕E) ∗ (gu) = E ∗ 𝜕(gu) = E ∗ S.

(10.16)

Now pick z0 ∈ Ω′ and ϵ > 0 such that Vϵ = {z ∶ dist(z, ℂ ⧵ Ω′ ) > ϵ} is a neighborhood of z0 . Let ψϵ ∈ 𝒟(ℂ) such that ψϵ (z) = 1 for |z| ≤ ϵ/2 and ϕϵ (z) = 0 for |z| ≥ ϵ. We can write (10.16) in the form gu = (ψϵ E) ∗ S + ((1 − ψϵ )E) ∗ S.

(10.17)

Since supp((ψϵ E) ∗ S) ⊆ supp(ψϵ E) + supp(S) ⊆ {z ∶ dist(z, supp(S)) ≤ ϵ}, we have that (ψϵ E) ∗ S has to vanish on Vϵ . This implies u|Vϵ = (gu)|Vϵ = (((1 − ψϵ )E) ∗ S)|V . ϵ

(10.18)

The function (1 − ψϵ )E is in 𝒞∞ (ℂ) and supp(S) is compact, therefore ((1 − ψϵ )E) ∗ S is smooth. Hence, using (10.18), we obtain that u is smooth in a neighborhood of z0 , and, z0 being arbitrary, u is smooth in Ω′ and therefore also in Ω. This implies that u is holomorphic, since it also solves the Cauchy–Riemann equation. For an appropriate description of the appearing phenomena, we will use further Hilbert spaces of differentiable functions – the Sobolev spaces. Definition 10.42. If Ω is a bounded open set in ℝn , and k is a nonnegative integer we define the Sobolev2 space W k (Ω) = {f ∈ L2 (Ω) ∶ 𝜕α f ∈ L2 (Ω), |α| ≤ k}, where the derivatives are taken in the sense of distributions and endow the space with the norm 1/2

‖f ‖k,Ω = [ ∑ ∫ |𝜕α f |2 dλ] |α|≤k Ω

2 Sobolev, Sergej Lvovich (1908–1989).

,

(10.19)

10.2 Distributions and Sobolev spaces | 259

where α = (α1 , … , αn ) is a multiindex, |α| = ∑nj=1 αj and 𝜕α f =

𝜕|α| f α . ⋯ 𝜕xn n

α 𝜕x1 1

k ∞ W0k (Ω) denotes the completion of 𝒞∞ 0 (Ω) under W (Ω)-norm. Since 𝒞0 (Ω) is dense in L2 (Ω), it follows that W00 (Ω) = W 0 (Ω) = L2 (Ω). Using the Fourier transform, it is also possible to introduce Sobolev spaces of non-integer exponent; see [2, 25].

In general, a function can belong to a Sobolev space, and yet be discontinuous and unbounded. Example 10.43. Take Ω = 𝔹 the open unit ball in ℝn , and u(x) = |x|−α ,

x ∈ 𝔹, x ≠ 0.

We claim that u ∈ W 1 (𝔹) if and only if α < n−2 . 2 First note that u is smooth away from 0, and that uxj (x) =

−αxj

,

x ≠ 0.

|α| , |x|α+1

x ≠ 0.

|x|α+2

Hence |∇u(x)| =

Now, recall the Gauß–Green Theorem: for a smoothly bounded ω ⊆ ℝn , we have ∫ ∇ ⋅ F(x) dλ(x) = ∫ (F(x), ν(x)) dσ(x), ω

bω

where ν(x) = ∇r(x) is the normal to bω at x, and F is a 𝒞1 vector field on ω, and n

∇ ⋅ F(x) = ∑ j=1

𝜕Fj 𝜕xj

;

see [24]. Let ϕ ∈ 𝒞∞ 0 (𝔹) and let 𝔹ϵ be the open ball around 0 with radius ϵ > 0. Take ω = 𝔹 ⧵ 𝔹ϵ and F(x) = (0, … , 0, uϕ, 0, … , 0), where uϕ appears at the jth component. Then ∫ 𝔹⧵𝔹ϵ

u(x)ϕxj (x) dλ(x) = − ∫

𝔹⧵𝔹ϵ

uxj (x)ϕ(x) dλ(x) + ∫

b𝔹ϵ

u(x)ϕ(x)νj (x) dσ(x),

260 | 10 The 𝜕-complex where ν(x) = (ν1 (x), … , νn (x)) denotes the inward pointing normal on b𝔹ϵ . If α < n − 1, then |∇u(x)| ∈ L1 (𝔹), and we obtain |∫

b𝔹ϵ

u(x)ϕ(x)νj (x) dσ(x)| ≤ ‖ϕ‖∞ ∫

b𝔹ϵ

ϵ−α dσ(x)

≤ Cϵn−1−α → 0,

as ϵ → 0. Thus ∫ u(x)ϕxj (x) dλ(x) = − ∫ uxj (x)ϕ(x) dλ(x) 𝔹

for all

ϕ ∈ 𝒞∞ 0 (𝔹).

𝔹

As |∇u(x)| =

α ∈ L2 (𝔹), |x|α+1

if and only if 2(α + 1) < n, we get that u ∈ W 1 (𝔹), if and only if α
0, choose a continuous function g with compact support such that ‖f − g‖2 < ϵ/3. Then also ‖fx − gx ‖2 < ϵ/3, so ‖fx − f ‖2 ≤ ‖fx − gx ‖2 + ‖gx − g‖2 + ‖g − f ‖2 < ‖gx − g‖2 + 2ϵ/3. For |x| sufficiently small, ‖gx − g‖2 < ϵ/3, hence ‖fx − f ‖2 < ϵ. n Let χ ∈ 𝒞∞ 0 (ℝ ) be a function with support in the unit ball such that χ ≥ 0 and

∫ χ(x) dλ(x) = 1. ℝn

10.3 Friedrichs’ lemma

| 263

We define χϵ (x) = ϵ−n χ(x/ϵ) for ϵ > 0. Let f ∈ L2 (ℝn ) and define for x ∈ ℝn fϵ (x) = (f ∗ χϵ )(x) = ∫ f (x′ )χϵ (x − x ′ ) dλ(x′ ) ℝn

= ∫ f (x − x′ )χϵ (x′ ) dλ(x′ ) ℝn

= ∫ f (x − ϵx ′ )χ(x′ ) dλ(x′ ). ℝn

In the first integral we can differentiate under the integral sign to show that fϵ ∈ 𝒞∞ (ℝn ). The family of functions (χϵ )ϵ is called an approximation to the identity. Lemma 10.48. ‖fϵ − f ‖2 → 0 as ϵ → 0. Proof. fϵ (x) − f (x) = ∫ [f (x − ϵx′ ) − f (x)]χ(x ′ ) dλ(x′ ). ℝn

We use Minkowski’s inequality (10.20) to get ‖fϵ − f ‖2 ≤ ∫ ‖f−ϵx′ − f ‖2 |χ(x′ )| dλ(x′ ). ℝn

But ‖f−ϵx′ − f ‖2 is bounded by 2‖f ‖2 and tends to 0 as ϵ → 0 by Lemma 10.48. Now set Fϵ (x′ ) = ‖f−ϵx′ − f ‖2 χ(x′ ). Then Fϵ (x′ ) → 0 as ϵ → 0 and |Fϵ (x′ )| ≤ 2‖f ‖2 χ(x ′ ), and we can apply the dominated convergence theorem to get the desired result. n If u ∈ 𝒞∞ 0 (ℝ ), we have

Dj (u ∗ χϵ ) = (Dj u) ∗ χϵ , where Dj = 𝜕/𝜕xj . This also true, if u ∈ L2 (ℝn ) and Dj u is defined in the sense of distributions. We will show even more using these methods for approximating a function in a Sobolev space by smooth functions. Let Ω ⊆ ℝn be an open subset and let Ωϵ = {x ∈ Ω ∶ dist(x, bΩ) > ϵ}. Lemma 10.49. Let u ∈ W k (Ω) and set uϵ = u ∗ χϵ in Ωϵ . Then (i) uϵ ∈ 𝒞∞ (Ωϵ ), for each ϵ > 0, (ii) Dα uϵ = Dα u ∗ χϵ in Ωϵ , for |α| ≤ k.

264 | 10 The 𝜕-complex Proof. (i) has already been shown. (ii) means that the ordinary αth partial derivative of the smooth functions uϵ is the ϵ-mollification of the αth weak partial derivative of u. To see this, we take x ∈ Ωϵ and compute Dα uϵ (x) = Dα ∫ u(y)χϵ (x − y) dλ(y) Ω

= ∫ Dαx χϵ (x − y)u(y) dλ(y) Ω

= (−1)|α| ∫ Dαy χϵ (x − y)u(y) dλ(y). Ω

For a fixed x ∈ Ωϵ the function ϕ(y) ∶= χϵ (x − y) belongs to 𝒞∞ (Ω). The definition of the αth weak partial derivative implies ∫ Dαy χϵ (x − y)u(y) dλ(y) = (−1)|α| ∫ χϵ (x − y)Dα u(y) dλ(y). Ω

Ω

Thus Dα uϵ (x) = (−1)|α|+|α| ∫ χϵ (x − y)Dα u(y) dλ(y) α

Ω

= (D u ∗ χϵ )(x), which proves (ii). We are now ready to prove Lemma 10.50 (Friedrichs’ Lemma). If v ∈ L2 (ℝn ) is a function with compact support and a is a 𝒞1 -function in a neighborhood of the support of v, it follows that ‖aDj (v ∗ χϵ ) − (aDj v) ∗ χϵ ‖2 → 0

as ϵ → 0,

where Dj = 𝜕/𝜕xj and aDj v is defined in the sense of distributions. n Proof. If v ∈ 𝒞∞ 0 (ℝ ), we have

Dj (v ∗ χϵ ) = (Dj v) ∗ χϵ → Dj v,

(aDj v) ∗ χϵ → aDj v,

with uniform convergence. We claim that ‖aDj (v ∗ χϵ ) − (aDj v) ∗ χϵ ‖2 ≤ C‖v‖2 ,

(10.22)

where v ∈ L2 (ℝn ) and C is some positive constant independent of ϵ and v. Since n 2 n 𝒞∞ 0 (ℝ ) is dense in L (ℝ ), the lemma will follow like in the proof of Lemma 10.48 from (10.22) and the dominated convergence theorem.

10.3 Friedrichs’ lemma

| 265

To show (10.22), we may assume that a ∈ 𝒞10 (ℝn ), since v has compact support. We n have for v ∈ 𝒞∞ 0 (ℝ ), a(x)Dj (v ∗ χϵ )(x) − ((aDj v) ∗ χϵ )(x) = a(x)Dj ∫ v(x − y)χϵ (y) dλ(y) − ∫ a(x − y) = ∫(a(x) − a(x − y))

𝜕v (x − y)χϵ (y) dλ(y) 𝜕xj

= − ∫(a(x) − a(x − y))

𝜕v (x − y)χϵ (y) dλ(y) 𝜕yj

= ∫(a(x) − a(x − y))v(x − y) − ∫(

𝜕v (x − y)χϵ (y) dλ(y) 𝜕xj

𝜕 χ (y) dλ(y) 𝜕yj ϵ

𝜕 a(x − y))v(x − y)χϵ (y) dλ(y). 𝜕yj

Let M be the Lipschitz constant for a such that |a(x) − a(x − y)| ≤ M|y|, for all x, y ∈ ℝn . Then |a(x)Dj (v ∗ χϵ )(x) − ((aDj v) ∗ χϵ )(x)| ≤ M ∫|v(x − y)|(χϵ (y) + |y

𝜕 χ (y)|) dλ(y). 𝜕yj ϵ

By Minkowski’s inequality (10.20), we obtain ‖aDj (v ∗ χϵ ) − (aDj v) ∗ χϵ ‖2 ≤ M‖v‖2 ∫(χϵ (y) + |y

𝜕 χ (y)|) dy 𝜕yj ϵ

= M(1 + mj )‖v‖2 , where mj = ∫|y

𝜕 𝜕 χϵ (y)| dy = ∫|y χ(y)| dλ(y). 𝜕yj 𝜕yj

n ∞ n 2 n This shows (10.22) when v ∈ 𝒞∞ 0 (ℝ ). Since 𝒞0 (ℝ ) is dense in L (ℝ ), we have proved (10.22) and the lemma.

Lemma 10.51. Let n

L = ∑ aj Dj + a0 j=1

be a first order differential operator with variable coefficients where aj ∈ 𝒞1 (ℝn ) and a0 ∈ 𝒞(ℝn ). If v ∈ L2 (ℝn ) is a function with compact support and Lv = f ∈ L2 (ℝn ) where n Lv is defined in the distribution sense, the convolution vϵ = v ∗ χϵ is in 𝒞∞ 0 (ℝ ) and 2 n vϵ → v, Lvϵ → f in L (ℝ ) as ϵ → 0.

266 | 10 The 𝜕-complex Proof. Since a0 v ∈ L2 (ℝn ), we have lim a0 (v ∗ χϵ ) = lim (a0 v ∗ χϵ ) = a0 v

ϵ→0

ϵ→0

in L2 (ℝn ). Using Friedrichs’ Lemma (see Lemma 10.50), we have Lvϵ − Lv ∗ χϵ = Lvϵ − f ∗ χϵ → 0 in L2 (ℝn ) as ϵ → 0. The lemma follows easily since f ∗ χϵ → f in L2 (ℝn ).

10.4 A finite dimensional analog We demonstrate the method for the 𝜕-Neumann problem first in its finite dimensional analog: let E, F, G denote finite dimensional vector spaces over ℂ with inner product. We consider an exact sequence of linear maps S

T

E ⟶ F ⟶ G, which means that im S = ker T, hence TS = 0. Given f ∈ im S = ker T, we want to solve Su = f with u ⟂ ker S, then u will be called the canonical solution. For this purpose we investigate S

T

⟵ ∗

⟵ ∗

E⟶F⟶G S

T

and observe that ker T = (im T ∗ )⟂ and ker T ∗ = (im T)⟂ . We claim that the operator SS∗ + T ∗ T ∶ F ⟶ F is bijective. Let (SS∗ + T ∗ T)g = 0, then SS∗ g = −T ∗ Tg, which implies SS∗ g ∈ im T ∗ ∩ im S = im T ∗ ∩ ker T = im T ∗ ∩ (im T ∗ )⟂ = {0}, hence SS∗ g = T ∗ Tg = 0, but this gives S∗ g ∈ ker S ∩ im S∗ = ker S ∩ (ker S)⟂ = {0}, and g ∈ ker S∗ = (im S)⟂ ; from T ∗ Tg = 0 we get Tg ∈ ker T ∗ ∩ im T = (im T)⟂ ∩ im T = {0} and g ∈ ker T = im S, therefore we obtain g ∈ im S ∩ (im S)⟂ = {0}. So SS∗ + T ∗ T is injective and as F is finite dimensional SS∗ + T ∗ T is bijective. Let N = (SS∗ + T ∗ T)−1 . We claim that u = S∗ Nf is the canonical solution to Su = f . So we have to show that SS∗ Nf = f and S∗ Nf ⟂ ker S. The latter easily follows from the fact that S∗ Nf ∈ im S∗ = (ker S)⟂ .

10.5 The 𝜕-Neumann operator

| 267

We have f = SS∗ Nf + T ∗ TNf , therefore the assumption Tf = 0 implies 0 = Tf = TSS∗ Nf + TT ∗ TNf = TT ∗ TNf , since TS = 0. From here we obtain 0 = (TT ∗ TNf , TNf ) = (T ∗ TNf , T ∗ TNf ) and T ∗ TNf = 0, hence SS∗ Nf = f and we are done. In the following we will use this method for the 𝜕-operator.

10.5 The 𝜕-Neumann operator Let Ω = {z ∈ ℂn ∶ r(z) < 0}, where r is a real valued 𝒞1 -function with ∇z r ∶= (

𝜕r 𝜕r ,…, )≠0 𝜕z1 𝜕zn

on bΩ = {z ∶ r(z) = 0}. Then r is called a defining function for Ω. Without loss of generality, we can suppose that |∇z r| = |∇r| = 1 on bΩ. For u, v ∈ 𝒞∞ (Ω) and (u, v) = ∫ u(z)v(z) dλ(z) Ω

we have (uxk , v) = −(u, vxk ) + ∫ u(z)v(z)rxk (z) dσ(z), bΩ

where dσ is the surface measure on bΩ. This follows from the Gauß–Green Theorem: for ω ⊆ ℝn we have ∫ ∇ ⋅ F(x) dλ(x) = ∫ (F(x), ν(x)) dσ(x), ω

bω

where ν(x) = ∇r(x) is the normal to bω at x, and F is a 𝒞1 vector field on ω, and n

∇ ⋅ F(x) = ∑ j=1

𝜕Fj 𝜕xj

.

For k = 1 and F = (uv, 0, … , 0), one gets (ux1 , v) = −(u, vx1 ) + ∫ u(z)v(z)rx1 (z) dσ(z), bΩ

similarly, one obtains (

𝜕u 𝜕v 𝜕r , v) = −(u, ) + ∫ u(z)v(z) (z) dσ(z). 𝜕zk 𝜕z k 𝜕zk bΩ

(10.23)

268 | 10 The 𝜕-complex Definition 10.52. Let n

L2(0,1) (Ω) ∶= {u = ∑ uj dz j ∶ uj ∈ L2 (Ω), j = 1, … , n} j=1

be the space of (0, 1)-forms with coefficients in L2 (Ω). For u, v ∈ L2(0,1) (Ω), we define the inner product by n

(u, v) = ∑(uj , vj ). j=1

In this way L2(0,1) (Ω) becomes a Hilbert space. (0, 1)-forms with compactly supported 𝒞∞ coefficients are dense in L2(0,1) (Ω). Definition 10.53. Let f ∈ 𝒞∞ 0 (Ω) and set n

𝜕f dz j , 𝜕z j j=1

𝜕f ∶= ∑ then

2 𝜕 ∶ 𝒞∞ 0 (Ω) ⟶ L(0,1) (Ω).

𝜕 is a densely defined unbounded operator on L2 (Ω). It does not have a closed graph. Definition 10.54. The domain dom(𝜕) of 𝜕 consists of all functions f ∈ L2 (Ω) such that 𝜕f , in the sense of distributions, belongs to L2(0,1) (Ω), i.e. 𝜕f = g = ∑nj=1 gj dz j , and for each ϕ ∈ 𝒞∞ 0 (Ω) we have ∫ f( Ω

𝜕ϕ − ) dλ = − ∫ gj ϕ dλ, 𝜕zj Ω

j = 1, … , n.

(10.24)

2 It is clear that 𝒞∞ 0 (Ω) ⊆ dom(𝜕), hence dom(𝜕) is dense in L (Ω). Since differentiation is a continuous operation in distribution theory we have

Lemma 10.55. 𝜕 ∶ dom ⟶ L2(0,1) (Ω) has closed graph and ker 𝜕 is a closed subspace of L2 (Ω). Proof. We use the arguments of the proof of Theorem 10.45: let (fk )k be a sequence in dom(𝜕) such that fk → f in L2 (Ω) and 𝜕fk → g in L2(0,1) (Ω). We have to show that 𝜕f = g. From the proof of Theorem 10.45 we know that 𝜕fk → 𝜕f as distributions. As 𝜕fk → g in L2(0,1) (Ω), it follows that f ∈ dom(𝜕) and 𝜕f = g. Now we can apply Lemma 10.9 and get that ker 𝜕 is a closed subspace of L2 (Ω). ker 𝜕 coincides with the Bergman space A2 (Ω) of all holomorphic functions on Ω 𝜕f belonging to L2 (Ω). This is due to the fact that 𝜕z = 0 in the sense of distributions k

10.5 The 𝜕-Neumann operator

| 269

implies that f is already a holomorphic function (regularity of the Cauchy–Riemann operator, see Theorem 10.41). More generally, for q ≥ 1, 𝜕 ∶ L2(0,q) (Ω) ⟶ L2(0,q+1) (Ω) with domain as before is again a densely defined, closed operator (see below). In this case ker 𝜕 is a closed subspace of L2(0,q) (Ω), which does not mean that all coefficients are holomorphic functions. The (0, 1)-form f (z1 , z2 ) = z 2 dz 1 + z 1 dz 2 satisfies 𝜕f = 0, but has non-holomorphic coefficients. Theorem 10.56. Let Ω be a smoothly bounded domain in ℂn , with defining function r such that |∇r(z)| = 1 on bΩ. We set 𝒞∞ (Ω) for the restriction of 𝒞∞ (ℂn ) to Ω and 𝒟0,1 = ∗ n ∞ 𝒞∞ (0,1) (Ω) ∩ dom(𝜕 ). A (0, 1)-form u = ∑j=1 uj dz j with coefficients in 𝒞 (Ω) belongs to

𝒟0,1 , if and only if ∑nj=1

𝜕r u 𝜕zj j

= 0 on bΩ.

Proof. For ψ ∈ 𝒞∞ (Ω) ⊂ dom(𝜕) we have n

∑(− j=1

𝜕uj 𝜕zj

n

, ψ) = ∑(uj , j=1

n

𝜕ψ 𝜕r ) − ∑ ∫ uj ψ dσ 𝜕z j 𝜕z j j=1 bΩ n

= (u, 𝜕ψ) − ∑ ∫ uj ψ j=1 bΩ

𝜕r dσ, 𝜕zj

if ψ has in addition compact support in Ω, we have (𝜕 u, ψ) = (u, 𝜕ψ). ∗

Since the compactly supported smooth function are dense in L2 (Ω), we must have n

∑ ∫ uj ψ j=1 bΩ

𝜕r dσ = 0, 𝜕zj

for any ψ ∈ 𝒞∞ (Ω). This implies n

∑ uj

𝜕r =0 𝜕zj

𝜕

𝜕

j=1

on bΩ. Now we consider the 𝜕-complex 𝜕

𝜕

L2 (Ω) ⟶ L2(0,1) (Ω) ⟶ ⋯ ⟶ L2(0,n) (Ω) ⟶ 0,

(10.25)

where L2(0,q) (Ω) denotes the space of (0, q)-forms on Ω with coefficients in L2 (Ω). The 𝜕-operator on (0, q)-forms is given by n

𝜕( ∑ aJ dz J ) = ∑ ∑ ′

J

j=1 J

′

𝜕aJ dz ∧ dz J , 𝜕z j j

(10.26)

where ∑′ means that the sum is only taken over strictly increasing multiindices J = (j1 , … , jq ).

270 | 10 The 𝜕-complex The derivatives are taken in the sense of distributions, and the domain of 𝜕 consists of those (0, q)-forms for which the right-hand side belongs to L2(0,q+1) (Ω). So 𝜕 is a densely defined closed operator, and therefore has an adjoint operator from L2(0,q+1) (Ω) ∗ into L2(0,q) (Ω) denoted by 𝜕 . We consider the 𝜕-complex 𝜕

𝜕

⟵ ∗

⟵ ∗

L2(0,q−1) (Ω) ⟶ L2(0,q) (Ω) ⟶ L2(0,q+1) (Ω), 𝜕

(10.27)

𝜕

for 1 ≤ q ≤ n − 1. ∗ We remark that a (0, q + 1)-form u = ∑′J uJ dz J belongs to 𝒞∞ (0,q+1) (Ω) ∩ dom(𝜕 ), if and only if n

∑ ukK

k=1

𝜕r =0 𝜕zk

(10.28)

on bΩ for all K with |K| = q. To see this, let α ∈ 𝒞∞ (0,q) (Ω) n

(u, 𝜕α) = ( ∑ uJ dz J , ∑ ∑ ′

′

j=1|K|=q

|J|=q+1 n ′

𝜕αK dz ∧ dz K ) 𝜕z j j

𝜕α = ∑ ∑ ∫ ujK K dλ 𝜕zj j=1|K|=q Ω n

′

= −∑ ∑ ∫

j=1|K|=q Ω n ′

= ( ∑ (− ∑ |K|=q

𝜕ujK 𝜕zj 𝜕ujK 𝜕zj

j=1

n

αK dλ + ∑ ∑ ∫ ujK αK ′

j=1|K|=q bΩ

𝜕r dσ 𝜕zj

n

) dz K , ∑ αK dz K ) + ∑ ∫ (∑ ujK ′

′

|K|=q bΩ

|K|=q

j=1

𝜕r )α dσ 𝜕zj K

= (ϑu, α) − ∫ ⟨θ(ϑ, dr)u, α⟩ dσ, bΩ

where n

ϑu = ∑ (− ∑ ′

|K|=q

j=1

𝜕ujK 𝜕zj

) dz K

(10.29)

and n

∑ ∫ (∑ ujK ′

|K|=q bΩ

j=1

𝜕r )α dσ = − ∫ ⟨θ(ϑ, 𝜕r)u, α⟩ dσ, 𝜕zj K bΩ

(10.30)

hence we have (ϑu, α) = (u, 𝜕α) + ∫ ⟨θ(ϑ, 𝜕r)u, α⟩ dσ, bΩ

(10.31)

where θ(ϑ, dr)u denotes the symbol of ϑ in the 𝜕r direction. Note that for u ∈ dom(𝜕 ) ∗ we have 𝜕 u = ϑu. ∗

10.5 The 𝜕-Neumann operator | 271

∞ Similarly, we have for u ∈ 𝒞∞ (0,q) (Ω) and α ∈ 𝒞(0,q+1) (Ω),

(𝜕u, α) = (u, ϑα) + ∫ ⟨𝜕r ∧ u, α⟩ dσ,

(10.32)

bΩ

where n

∫ ⟨𝜕r ∧ u, α⟩ dσ = ∑ ∑ ∫ uK ′

bΩ

|K|=q k=1 bΩ

𝜕r α dσ. 𝜕z k kK

(10.33)

Theorem 10.57. The complex Laplacian □ = 𝜕 𝜕 + 𝜕 𝜕, defined on the domain ∗ ∗ ∗ dom(□) = {u ∈ L2(0,q) (Ω) ∶ u ∈ dom(𝜕) ∩ dom(𝜕 ), 𝜕u ∈ dom(𝜕 ), 𝜕 u ∈ dom(𝜕)} acts as an unbounded, densely defined, closed and self-adjoint operator on L2(0,q) (Ω), for 1 ≤ q ≤ n, which means that □ = □∗ and dom(□) = dom(□∗ ). ∗

∗

Proof. dom(□) contains all smooth forms with compact support, hence □ is densely ∗ defined. Showing that □ is closed depends on the fact that both 𝜕 and 𝜕 are closed: note that (□u, u) = (𝜕 𝜕 u + 𝜕 𝜕u, u) = ‖𝜕u‖2 + ‖𝜕 u‖2 , ∗

∗

(10.34)

∗

for u ∈ dom(□). We have to prove that for every sequence uk ∈ dom(□) such that uk → u in L2(0,q) (Ω) and □uk converges, we have u ∈ dom(□) and □uk → □u. It follows from (10.34) that (□(uk − uℓ ), uk − uℓ ) = ‖𝜕(uk − uℓ )‖2 + ‖𝜕 (uk − uℓ )‖2 , ∗

which implies that 𝜕uk converges in L2(0,q+1) (Ω) and that 𝜕 uk converges in L2(0,q−1) (Ω). ∗ ∗ Since 𝜕 and 𝜕 are closed operators, we get u ∈ dom(𝜕) ∩ dom(𝜕 ) and 𝜕uk → 𝜕u in ∗ ∗ 2 2 L(0,q+1) (Ω) and 𝜕 uk → 𝜕 u in L(0,q−1) (Ω). ∗ ∗ ∗ ∗ To show that 𝜕u ∈ dom(𝜕 ) and 𝜕 u ∈ dom(𝜕), we first notice that 𝜕 𝜕 uk and 𝜕 𝜕uk are orthogonal, which follows from ∗

2 ∗

(𝜕 𝜕 uk , 𝜕 𝜕uk ) = (𝜕 𝜕 uk , 𝜕uk ) = 0. ∗

∗

Therefore the convergence of □uk = 𝜕 𝜕 uk + 𝜕 𝜕uk implies that both 𝜕 𝜕 uk and 𝜕 𝜕uk ∗ ∗ converge. Now use again that 𝜕 and 𝜕 are closed operators to obtain that 𝜕 𝜕 uk → ∗ ∗ ∗ 𝜕 𝜕 u and 𝜕 𝜕uk → 𝜕 𝜕u. This implies that □uk → □u. Hence □ is closed. In order to show that □ is self-adjoint we use Lemma 10.29. Define ∗

∗

∗

∗

R = 𝜕𝜕 + 𝜕 𝜕 + I ∗

∗

on dom(□). By Lemma 10.29, both (I + 𝜕 𝜕 )−1 and (I + 𝜕 𝜕)−1 are bounded, self-adjoint operators. Consider ∗

∗

L = (I + 𝜕 𝜕 )−1 + (I + 𝜕 𝜕)−1 − I. ∗

∗

272 | 10 The 𝜕-complex Then L is bounded and self-adjoint. We claim that L = R−1 . Since (I + 𝜕 𝜕 )−1 − I = (I − (I + 𝜕 𝜕 ))(I + 𝜕 𝜕 )−1 = −𝜕 𝜕 (I + 𝜕 𝜕 )−1 , ∗

∗

∗

∗

∗

we have that the range of (I + 𝜕 𝜕 )−1 is contained in dom(𝜕 𝜕 ). Similarly, we have that ∗ ∗ the range of (I + 𝜕 𝜕)−1 is contained in dom(𝜕 𝜕) and get ∗

∗

L = (I + 𝜕 𝜕)−1 − 𝜕 𝜕 (I + 𝜕 𝜕 )−1 . ∗

∗

∗

2

Since 𝜕 = 0, we have that the range of L is contained in dom(𝜕 𝜕) and ∗

∗

∗

∗

𝜕 𝜕L = 𝜕 𝜕(I + 𝜕 𝜕)−1 . Similarly, we have that the range of L is contained in dom(𝜕 𝜕 ) and ∗

𝜕 𝜕 L = 𝜕 𝜕 (I + 𝜕 𝜕 )−1 . ∗

∗

∗

This implies that the range of L is contained in dom(□). In addition, we have RL = 𝜕 𝜕 (I + 𝜕 𝜕 )−1 + 𝜕 𝜕(I + 𝜕 𝜕)−1 + L = I. ∗

∗

∗

∗

If Ru = 0, we get □u = −u and 0 ≤ (□u, u) = −(u, u), which implies that u = 0. Hence R is injective, and we have that L = R−1 . By Lemma 10.29, we know that L is self-adjoint. Apply Lemma 10.28 to get that R is self-adjoint. Therefore □ = R − I is self-adjoint. The two boundary conditions u ∈ dom(𝜕 ) and 𝜕u ∈ dom(𝜕 ) which appear in the definition of dom(□) are called the 𝜕-Neumann boundary conditions. They amount to a Dirichlet boundary condition on the normal component of u and to the normal component of 𝜕u, respectively, see [68] for more details. ∗

∗

Example 10.58. Let Ω be a smoothly bounded domain in ℂn with 0 ∈ bΩ. Assume that for some neighborhood U of 0 Ω ∩ U = {z ∈ ℂn ∶ ℑzn = yn < 0} ∩ U. Let u = ∑nj=1 uj dz j ∈ 𝒞2(0,1) (Ω) and suppose that the support of u lies in U ∩ Ω. Then u ∈ dom(□), if and only if un = 0 𝜕uj 𝜕z n

=0

on bΩ ∩ U,

(10.35)

on bΩ ∩ U, j = 1, … , n − 1.

(10.36)

Equation (10.35) follows from (10.28), which means that u ∈ dom(𝜕 ), and 𝜕u ∈ ∗ dom(𝜕 ) is equivalent to ∗

𝜕uj 𝜕z n again by (10.28). Since

𝜕un 𝜕z j

−

𝜕un =0 𝜕z j

on bΩ ∩ U, j = 1, … , n − 1,

= 0 on bΩ ∩ U, j = 1, … , n − 1, we get (10.36).

10.5 The 𝜕-Neumann operator |

273

It is the second boundary condition which makes the system non-coercive, see [49]. We continue investigating the boundary conditions: Theorem 10.59. Let Ω be a smoothly bounded domain in ℂn , with defining function r such that |∇r(z)| = 1 on bΩ. Then, if u ∈ 𝒟0,1 , we have n

‖𝜕u‖2 + ‖𝜕 u‖2 = ∑ ‖ ∗

j,k=1

𝜕uj 𝜕z k

2

n

𝜕2 r uj uk dσ. bΩ j,k=1 𝜕zj 𝜕z k

‖ +∫

∑

(10.37)

Proof. For u ∈ 𝒟0,1 , we have 𝜕u = ∑( j 0. ∗ First we will show that (10.48) implies that 𝜕 and 𝜕 have closed range. ∗

(10.48)

10.7 Properties of the 𝜕-Neumann operator |

281

Theorem 10.66. Let Ω ⊂ ℂn be a smoothly bounded pseudoconvex domain. Then 𝜕 and ∗ 𝜕 have closed range. Proof. We notice that ker 𝜕 = (im 𝜕 )⟂ , which implies that ∗

(ker 𝜕)⟂ = im 𝜕 ⊆ ker 𝜕 . ∗

∗

If u ∈ ker 𝜕 ∩ ker 𝜕 , we have by (10.48) that u = 0. Hence ∗

(ker 𝜕)⟂ = ker 𝜕 . ∗

(10.49)

If u ∈ dom(𝜕) ∩ (ker 𝜕)⟂ , then u ∈ ker 𝜕 , and (10.48) implies ∗

1 ‖u‖ ≤ ‖𝜕u‖. c Now we can apply Lemma 10.20 to conclude that im 𝜕 is closed. Finally, Theorem 10.22 ∗ gives that im 𝜕 is also closed. The next result describes the implication of the basic estimates (10.48) for the □-operator. Theorem 10.67. Let Ω ⊂ ℂn be a smoothly bounded pseudoconvex domain. Then □ ∶ dom(□) ⟶ L2(0,q) (Ω) is bijective and has a bounded inverse N ∶ L2(0,q) (Ω) ⟶ dom(□). N is called 𝜕-Neumann4 operator. In addition, 1 ‖Nu‖ ≤ ‖u‖. c

(10.50)

Proof. Since (□u, u) = ‖𝜕u‖2 + ‖𝜕 u‖2 , it follows that for a convergent sequence (□un )n we get ∗

‖□un − □um ‖ ‖un − um ‖ ≥ (□(un − um ), un − um ) ≥ c‖un − um ‖2 , which implies that (un )n is convergent and, since □ is a closed operator, we obtain ∗ that □ has closed range. If □u = 0, we get 𝜕u = 0 and 𝜕 u = 0 and by (10.48) that u = 0, hence □ is injective. By Lemma 10.28 (ii), the range of □ is dense, therefore □ is surjective. We showed that □ ∶ dom(□) ⟶ L2(0,q) (Ω) 4 Neumann, Carl (1832–1925).

282 | 10 The 𝜕-complex is bijective and therefore, by Lemma 10.28 (iv), has a bounded inverse N ∶ L2(0,q) (Ω) ⟶ dom(□). For u ∈ L2(0,q) (Ω), we use (10.48) for Nu to obtain c‖Nu‖2 ≤ ‖𝜕Nu‖2 + ‖𝜕 Nu‖2 ∗

= (𝜕 𝜕Nu, Nu) + (𝜕𝜕 Nu, Nu) ∗

∗

= (u, Nu) ≤ ‖u‖ ‖Nu‖, which implies (10.50). For u ∈ L2(0,q) (Ω) and v ∈ dom(𝜕) ∩ dom(𝜕 ), we get ∗

(u, v) = (□Nu, v) = ((𝜕 𝜕 + 𝜕 𝜕)Nu, v) = (𝜕 Nu, 𝜕 v) + (𝜕Nu, 𝜕v). ∗

∗

∗

∗

(10.51)

Now we discuss a different approach to the 𝜕-Neumann operator, which is related to the quadratic form Q(u, v) = (𝜕u, 𝜕v) + (𝜕 u, 𝜕 v). ∗

∗

For this purpose we consider the embedding j ∶ dom(𝜕) ∩ dom(𝜕 ) ⟶ L2(0,q) (Ω), ∗

where dom(𝜕) ∩ dom(𝜕 ) is endowed with the graph-norm ∗

u ↦ (‖𝜕u‖2 + ‖𝜕 u‖2 )1/2 . ∗

The graph-norm stems from the inner product Q(u, v) = (u, v)Q = (□u, v) = (𝜕u, 𝜕v) + (𝜕 u, 𝜕 v). ∗

∗

The basic estimates (10.48) imply that j is a bounded operator with operator norm ‖j‖ ≤

1 . √c

By (10.48) it follows in addition that dom(𝜕) ∩ dom(𝜕 ) endowed with the graph-norm ∗ u ↦ (‖𝜕u‖2 + ‖𝜕 u‖2 )1/2 is a Hilbert space. Since (u, v) = (u, jv), we have that (u, v) = (j∗ u, v)Q . Equation (10.51) suggests that as ∗ an operator to dom(𝜕) ∩ dom(𝜕 ),N coincides with j∗ and as an operator to L2(0,q) (Ω), N should be equal to j ∘ j∗ . For this purpose set ∗

Ñ = j ∘ j∗ ,

(10.52)

10.7 Properties of the 𝜕-Neumann operator |

283

∗ and note that Ñ = (j ∘ j∗ )∗ = j ∘ j∗ = Ñ , i.e. Ñ is self-adjoint (of course, also bounded). We claim that the range of Ñ is contained in dom(□). To show this we use an approach ̃ ∈ dom(□∗ ) due to F. Berger (see [8]): since □ is self-adjoint, it suffices to show that Nu ̃ is bounded for all u ∈ L2(0,q) (Ω), which means to show that the functional v ↦ (□v, Nu) on dom(□): ∗ ∗ ̃ ̃ ̃ + (𝜕∗ v, 𝜕∗ Nu)| ̃ |(□v, Nu)| = |((𝜕 𝜕 + 𝜕 𝜕)v, Nu)| = |(𝜕v, 𝜕Nu)

= |Q(v, j∗ u)| = |(jv, u)| = |(v, u)| ≤ ‖v‖ ‖u‖. For v ∈ dom(𝜕) ∩ dom(𝜕 ) we have ∗

̃ v) = (Nu, ̃ v)Q = (j∗ u, v) = (u, jv) = (u, v), (□Nu, Q ̃ = u. In a similar way we obtain, for u ∈ dom(□), hence □Nu ̃ ̃ = (u, Nv) ̃ Q = (u, j∗ v) = (ju, v) = (u, v), (N□u, v) = (□u, Nv) Q ̃ which implies that N□u = u. Altogether we obtain that N = Ñ . Remark 10.68. Similarly one can use Theorem 10.36 in order to prove the existence and the main properties of the 𝜕-Neumann operator. For this purpose let H = L2(0,q) (Ω) ∗ ∗ and 𝒱 = dom(𝜕) ∩ dom(𝜕 ), endowed with the graph-norm u ↦ (‖𝜕u‖2 + ‖𝜕 u‖2 )1/2 . The corresponding sesquilinear form a ∶ 𝒱 × 𝒱 ⟶ ℂ is given by a(u, v) = Q(u, v) = (𝜕u, 𝜕v) + (𝜕 u, 𝜕 v), ∗

∗

for u, v ∈ dom(𝜕) ∩ dom(𝜕 ). The basic estimates (10.48) show that 𝒱 is continuously embedded into H, which gives (10.5), and, by definition, the sesquilinear form a(u, v) is 𝒱-elliptic, see (10.7). The domain of the operator S in Theorem 10.36 coincides with the domain of the □-operator because the mapping v ↦ a(u, v) is continuous in the L2 -norm exactly, if we can write ∗

|a(u, v)| = |(□u, v)| ≤ ‖□u‖ ‖v‖, see Exercise 164. We have S = □ and, by Theorem 10.36, we get immediately that □ ∶ dom(□) ⟶ L2(0,q) (Ω) is bijective and self-adjoint and has a bounded inverse. Theorem 10.69. The operators 𝜕N ∶ L2(0,q) (Ω) ⟶ L2(0,q+1) (Ω) and 𝜕 N ∶ L2(0,q) (Ω) ⟶ L2(0,q−1) (Ω) ∗

are both bounded.

284 | 10 The 𝜕-complex Proof. From the above considerations on N, we get ‖𝜕Nu‖2 + ‖𝜕 Nu‖2 = (j∗ u, j∗ u)Q ≤ ‖j∗ ‖2 ‖u‖2 , ∗

for u ∈ L2(0,q) (Ω), which implies the result. Theorem 10.70. Let Nq denote the 𝜕-Neumann operator on L2(0,q) (Ω). Then Nq+1 𝜕 = 𝜕Nq ,

(10.53)

Nq−1 𝜕 = 𝜕 Nq ,

(10.54)

on dom(𝜕)(𝜕) and ∗

∗

on dom(𝜕 ). ∗ In addition, we have that 𝜕 Nq is zero on (ker 𝜕)⟂ . ∗

Proof. For u ∈ dom(𝜕) we have 𝜕u = 𝜕𝜕 𝜕Nq u and ∗

Nq+1 𝜕u = Nq+1 𝜕 𝜕 𝜕Nq u = Nq+1 (𝜕 𝜕 + 𝜕 𝜕)𝜕Nq u = 𝜕Nq u, ∗

∗

∗

which proves (10.53). In a similar way we get (10.54). Now let k ∈ (ker 𝜕)⟂ and u ∈ dom(𝜕), then (𝜕 Nq k, u) = (Nq k, 𝜕u) = (k, Nq 𝜕u) = (k, 𝜕Nq−1 u) = 0, ∗

since 𝜕Nq−1 u ∈ ker(𝜕), which gives 𝜕 Nq k = 0. ∗

Since we already know that both operators 𝜕Nq and 𝜕 Nq are bounded, we can ∗ extend both operators Nq+1 𝜕 and Nq−1 𝜕 to bounded operators on L2(0,q) (Ω). ∗

Theorem 10.71. Let α ∈ L2(0,q) (Ω), with 𝜕α = 0. Then u0 = 𝜕 Nq α is the canonical solution of 𝜕u = α, which means that 𝜕u0 = α, u0 ⊥ ker 𝜕, and ∗

‖𝜕 Nq α‖ ≤ c−1/2 ‖α‖.

(10.55)

∗

Proof. For α ∈ L2(0,q) (Ω) with 𝜕α = 0, we get α = 𝜕 𝜕 Nq α + 𝜕 𝜕Nq α. ∗

(10.56)

∗

If we apply 𝜕 to the last equality, we obtain 0 = 𝜕α = 𝜕𝜕 𝜕Nq α, ∗

and since 𝜕Nq α ∈ dom(𝜕 ), we have ∗

0 = (𝜕 𝜕 𝜕Nq α, 𝜕Nq α) = (𝜕 𝜕Nq α, 𝜕 𝜕Nq α) = ‖𝜕 𝜕Nq α‖2 . ∗

∗

∗

∗

(10.57)

| 285

10.7 Properties of the 𝜕-Neumann operator

Finally, we set u0 = 𝜕 Nq α and derive from (10.56) and (10.57) that for 𝜕α = 0, ∗

α = 𝜕u0 , and we see that u0 ⊥ ker 𝜕, since for h ∈ ker 𝜕 we get (u0 , h) = (𝜕 Nq α, h) = (Nq α, 𝜕h) = 0. ∗

It follows that ‖𝜕 Nq α‖2 = (𝜕 𝜕 Nq α, Nq α) ∗

∗

= (𝜕 𝜕 Nq α, Nq α) + (𝜕 𝜕Nq α, Nq α) ∗

∗

= (α, Nq α) ≤ ‖α‖ ‖Nq α‖,

and using (10.50) we obtain ‖𝜕 Nq α‖ ≤ c−1/2 ‖α‖. ∗

We showed that the canonical solution operator Sq for 𝜕 coincides with 𝜕 Nq as operator on ∗

L2(0,q) (Ω) ∩ ker 𝜕 and is a bounded operator. The 𝜕-Neumann operator N can be expressed in terms of the canonical solution operators: Theorem 10.72. ∗ Nq = Sq∗ Sq + Sq+1 Sq+1 .

(10.58)

Proof. We use (10.53) and (10.54) to show that 𝜕 Nq = 𝜕 Nq∗ = (Nq 𝜕)∗ ∗

∗

and (𝜕 Nq )∗ = Nq 𝜕, ∗

and 𝜕Nq = 𝜕 Nq∗ = (Nq 𝜕 )∗ = (𝜕 Nq+1 )∗ ∗∗

∗

∗

and

𝜕 Nq+1 = (𝜕Nq )∗ = Nq 𝜕 , ∗

∗

where we applied Lemma 10.3. Hence it follows that for u ∈ L2(0,q) (Ω) we have Nq u = Nq (𝜕 𝜕 + 𝜕 𝜕)Nq u ∗

∗

= (Nq 𝜕)(𝜕 Nq )u + (Nq 𝜕 )(𝜕Nq )u ∗

∗

= (𝜕 Nq )∗ (𝜕 Nq )u + (𝜕 Nq+1 )(𝜕 Nq+1 )∗ u ∗

∗

∗ = Sq∗ Sq u + Sq+1 Sq+1 u.

∗

∗

286 | 10 The 𝜕-complex Theorem 10.73. Let Pq ∶ L2(0,q) (Ω) ⟶ ker 𝜕 denote the orthogonal projection, which is the Bergman projection for q = 0. Then Pq = I − 𝜕 Nq+1 𝜕,

(10.59)

∗

on dom(𝜕). Proof. First we show that the range of the right-hand side of (10.59), which we denote by P,̃ coincides with ker 𝜕. Indeed, for u ∈ dom(𝜕), we have 𝜕u − 𝜕 𝜕 Nq+1 𝜕u = 𝜕u − □Nq+1 𝜕u + 𝜕 𝜕Nq+1 𝜕u = 𝜕u − 𝜕u = 0, ∗

∗

where we used (10.53) to show that 𝜕Nq+1 𝜕u = Nq+2 𝜕 𝜕u = 0, and since u − 𝜕 Nn+1 𝜕u = u ∗

for u ∈ ker 𝜕, we have shown the first claim. Now we obtain ∗ ∗ ∗∗ ∗ P̃ = (I − 𝜕 Nq+1 𝜕)∗ = I − 𝜕 Nq+1 𝜕 = P,̃

and 2 ̃ − 𝜕∗ Nq+1 𝜕Pu ̃ P̃ u = Pu

̃ − 𝜕 Nq+1 𝜕u + 𝜕 Nq+1 𝜕 𝜕 Nq+1 𝜕u = Pu ∗

∗

∗

̃ − 𝜕∗ Nq+1 𝜕u + 𝜕∗ Nq+1 (□ − 𝜕∗ 𝜕)Nq+1 𝜕u = Pu ̃ = Pu. This means that P̃ coincides with Pq on dom(𝜕). Finally, we remark that P̃ can be extended to a unique bounded operator on which coincides with Pq . Indeed, for u ∈ dom(𝜕), we have by (10.53) that ∗ ∗ ∗ = 𝜕 𝜕Nq u and u = □Nq u = 𝜕 𝜕 Nq u + 𝜕 𝜕Nq u is an orthogonal decomposition, which follows from

L2(0,q) (Ω), ∗ 𝜕 Nq+1 𝜕u

(𝜕 𝜕 Nq u, 𝜕 𝜕Nq u) = (𝜕 𝜕 𝜕 Nq u, 𝜕Nq u) = 0. ∗

∗

∗

Hence ‖𝜕 Nq+1 𝜕u‖ = ‖𝜕 𝜕Nq u‖ ≤ ‖u‖, ∗

∗

u ∈ dom(𝜕),

which proves the claim since dom(𝜕) is dense in L2(0,q) (Ω).

10.8 Exercises | 287

10.8 Exercises 152. Let H be a Hilbert space. Show that for each ϵ > 0 there exists a constant Cϵ > 0 such that |(x, y)| ≤ ϵ‖x‖2 + Cϵ ‖y‖2

∀x, y ∈ H.

153. Show that a densely defined operator T ∶ dom(T) ⟶ H2 between Hilbert spaces is closable, if and only if the following holds: if (fn )n is a sequence in dom(T) such that limn→∞ fn = 0, and the sequence (Tfn )n in H2 is convergent, then we have limn→∞ Tfn = 0. 154. Let T be a densely defined operator. For f , g ∈ dom(T), define a scalar product by (f , g)T ∶= (f , g)1 + (Tf , Tg)2 . Show that T is closed if and only if (dom(T), (⋅, ⋅)T ) is a Hilbert space. 155. Let X and Y be Fréchet spaces and T ∶ X ⟶ Y a continuous linear operator which is surjective. Show that T is open. Hint: use Theorem 10.10 and proceed similarly as in Theorem 10.11. 156. Let H be a Hilbert space and X be a Banach space. Show that an operator A ∈ ℒ(H, X) is compact, if and only if for each orthonormal sequence (en )n in H one has that limn→∞ Aen = 0. 157. Let A ∈ ℒ(H1 , H2 ) be a bounded operator between Hilbert spaces. Show that A is compact, if and only if for arbitrary orthonormal sequences (en )n in H1 and (fn )n in H2 one has that limn→∞ (Aen , fn ) = 0. 158. Let A ∈ ℒ(H) be a compact operator on the Hilbert space H. Show that I − A has closed range. Hint: use Theorem 10.11 for the operator I − A. 159. Show that every symmetric operator T on the Hilbert space H is closable and T is also symmetric. 160. Let Ω ⊂ ℂn be a bounded domain, n ≥ 2. Show that the kernel of 𝜕 in L2 (Ω) coincides with the Bergman space A2 (Ω). Show that the kernel of 𝜕 in L2(0,q) (Ω) for 1 ≤ q ≤ n − 1 is larger than the space of (0, q)-forms with coefficients in A2 (Ω). 161. Let Ω ⊂ ℂn be a bounded pseudoconvex domain. Consider the operator T ∶ L2(0,1) (Ω) ⟶ L2 (Ω) defined by Tf (z) = ∫ K(z, w)⟨f (w), z − w⟩ dλ(w), Ω

where f = ∑nk=1 fk dz k ∈ L2(0,1) (Ω) and n

⟨f (w), z − w⟩ = ∑ fk (w)(z k − wk ). k=1

288 | 10 The 𝜕-complex Show that T can be written as a sum of commutators n

Tf = ∑ [M k , P]fk , k=1

where M k v(z) = z k v(z), v ∈ L2 (Ω), k = 1, … , n, and P is the Bergman projection. 162. Let T ∶ L2(0,1) (Ω) ⟶ L2 (Ω) be as in 161. Let 𝒫 ∶ L2(0,1) (Ω) ⟶ A2(0,1) (Ω) be the orthogonal projection on the space of (0, 1)-forms with holomorphic coefficients belonging to A2 (Ω). Show that Tf = T𝒫f ,

f ∈ L2(0,1) (Ω).

Conclude from this property that the canonical solution operator S from (8.10) is compact as an operator from A2(0,1) (Ω) to L2 (Ω), if and only if the operator T is compact as an operator from L2(0,1) (Ω) to L2 (Ω). ∗ 163. Show that 𝜕 N restricted to A2(0,1) (Ω) is compact, if and only if the commutators [P, Mk ] ∶ L2 (Ω) ⟶ L2 (Ω) are compact, for k = 1, … , n, where Mk v(z) = zk v(z), v ∈ L2 (Ω), k = 1, … , n, and P is the Bergman projection. Hint: use Exercise 162 and the fact that [M k , P]∗ = [P, Mk ]. 164. Prove that dom(S) in Remark 10.68 coincides with dom(□) and that S = □. Hint: in order to show that dom(S) ⊆ dom(□), take f ∈ dom(S) and decompose ∗ ∗ u ∈ dom(𝜕)̄ in the form u = u1 + u2 , where u1 ∈ ker 𝜕̄ ∩ dom(𝜕)̄ and u2 ∈ (ker 𝜕̄ )⊥ , ∗ ∗ ̄ 𝜕f̄ ) and show that 𝜕f̄ ∈ dom(𝜕̄ ). Similarly, consider (𝜕̄ w, 𝜕̄ ∗ f ), now consider (𝜕u, where w = w1 + w2 ∈ ker 𝜕̄ ∩ dom(𝜕̄ ) ⊕ (ker 𝜕)̄ ⊥ , ∗

̄ and show that 𝜕̄ f ∈ dom(𝜕̄ ) = dom(𝜕). ∗

∗∗

10.9 Notes The basics on functional analysis, such as the open-mapping, the closed-graph theorem and the uniform boundedness principle were taken from [24]. A thorough treatment of unbounded operators on Hilbert spaces can be found in [70]. The construction of self-adjoint operators by means of sesquilinear forms acting on different Hilbert spaces is mainly based on the assumption of the 𝒱-ellipticity, sometimes also known as the Lax–Milgram situation, see [10, 34]. Basic facts about pseudoconvex domains are given, for instance, in [41, 51, 44]. We point out that Theorem 10.67, and the following results on the 𝜕-Neumann operator, also hold for general bounded pseudoconvex domains without regularity conditions for the boundary of the domain, see [13, 68]. Theorems 10.71 and 10.73 are due to J.J. Kohn [47, 48]. The approach to the 𝜕-Neumann operator N using the quadratic form Q can be found in [68]. Later on, the formula N = j ∘ j∗ will be very important to handle the problem of compactness

10.9 Notes | 289

of N. This approach is also inspired by the general study of quadratic forms associated with non-negative self-adjoint operators and their quadratic roots, see [18]. We will explain some of these ideas, which are related to the spectral theorem in more details in Chapter 11. It turns out that the question of compactness of N is crucial for regularity of N and the Sobolev theory for N. For a thorough treatment of compactness of N, the reader should consult [68, 30].

11 The twisted 𝜕-complex and Schrödinger operators We will consider the twisted 𝜕-complex T

S

L2 (Ω) ⟶ L2(0,1) (Ω) ⟶ L2(0,2) (Ω)

(11.1)

for operators T = 𝜕 ∘ √τ and S = √τ ∘ 𝜕, where τ ∈ 𝒞2 (Ω) and τ > 0 on Ω (Section 11.2). It turns out that this approach leads to generalized basic estimates which are seminal for the applications. In Section 11.3 we derive the basic estimates (10.48) from the theorem on the twisted 𝜕-complex and the density result (Theorem 10.61). Next we investigate the basic properties of the 𝜕-operator on weighted L2 -spaces on the whole of ℂn (Section 11.4) and apply the results to construct entire functions with a given growth behavior at infinity (Section 11.5). It will turn out that the behavior of the 𝜕-operator essentially depends on the properties of the Levi-matrix of the weight function. For n = 1, there is an interesting connection between 𝜕 and the theory of Schrödinger operators with magnetic fields, for n ≥ 2, the corresponding concept is the Witten Laplacian (Section 11.7). The corresponding density result from the last chapter is much easier to handle in the weighted case, therefore also the basic estimate and existence of the 𝜕-Neumann operator. Nevertheless, we need some facts from spectral analysis of unbounded self-adjoint operators, in particular consequences of the spectral theorem, and Persson’s Lemma about the bottom of the essential spectrum, where all the details and proofs are beyond the scope of this book. We collect all necessary information for this in Section 11.6.

11.1 An exact sequence of unbounded operators First we prove a general result about operators like T and S from (11.1). Theorem 11.1. Let H1 , H2 , H3 be Hilbert spaces and T ∶ H1 ⟶ H2 and S ∶ H2 ⟶ H3 densely defined closed linear operators such that S(T(f )) = 0, for each f ∈ dom(T), and let P ∶ H2 ⟶ H2 be a positive invertible operator such that ‖Pu‖22 ≤ ‖T ∗ u‖21 + ‖Su‖23 ,

(11.2)

for all u ∈ dom(S) ∩ dom(T ∗ ), where dom(T ∗ ) = {u ∈ H2 ∶ |(u, Tf )2 | ≤ C‖f ‖1 for all f ∈ dom(T)}. Suppose (11.2) holds and let α ∈ H2 be such that Sα = 0. Then there exists σ ∈ H1 such that (i) T(σ) = α and (ii) ‖σ‖21 ≤ ‖P −1 α‖22 . Proof. Since P is positive, it follows that P = P ∗ . Now let α ∈ H2 be such that Sα = 0. We consider the linear functional T ∗ u ↦ (u, α)2 for u ∈ dom(T ∗ ). We show that this linear https://doi.org/10.1515/9783110417241-011

292 | 11 The twisted 𝜕-complex and Schrödinger operators functional is well-defined and continuous. If u ∈ ker S, then |(u, α)2 | = |(Pu, P −1 α)2 | ≤ ‖Pu‖2 ‖P −1 α‖2 ≤ (‖T ∗ u‖21 + ‖Su‖23 )1/2 ‖P −1 α‖2 = ‖T ∗ u‖1 ‖P −1 α‖2 ,

if u⊥2 ker S, then (u, α)2 = 0. In addition, we have that T ∗ w = 0 for all w⊥2 ker S, this follows from the assumption that Tf ∈ ker S, so 0 = (w, Tf )2 ≤ C‖f ‖1 , which means that w ∈ dom(T ∗ ) and T ∗ w = 0, since (T ∗ w, f )1 = (w, Tf )2 = 0 for all f ∈ dom(T). If T ∗ u = 0, it follows from the above estimate that (u, α)2 = 0, which implies that the linear functional T ∗ u ↦ (u, α)2 for u ∈ dom(T ∗ ) is well-defined and continuous. We apply the Hahn–Banach Theorem, where we keep the constant for the estimate of the functional and the Riesz representation theorem to get σ ∈ H1 such that (T ∗ u, σ)1 = (u, α)2 , which implies that σ ∈ dom(T ∗∗ ). By Lemma 10.7, we have T = T ∗∗ , so we get (u, Tσ)2 = (u, α)2 , for all u ∈ dom(T ∗ ). Since dom(T ∗ ) is dense in H2 , we obtain Tσ = α and, again by the above estimate ‖σ‖1 ≤ ‖P −1 α‖2 .

11.2 The twisted basic estimates Let Ω be a smoothly bounded pseudoconvex domain in ℂn , with defining function r such that |∇r(z)| = 1 on bΩ. Let τ ∈ 𝒞2 (Ω) and τ > 0 on Ω. For f ∈ 𝒞∞ (Ω), we define n

𝜕 (√τf ) dz k , 𝜕z k k=1

Tf = (𝜕 ∘ √τ)f = ∑

(11.3)

and for u = ∑nj=1 uj dz j with coefficients uj in 𝒞∞ (Ω), we will write u ∈ Λ0,1 (Ω), we define Su = ∑ √τ( j 0 on Ω and let u ∈ 𝒟0,1 . Then ‖√τ + A 𝜕φ u‖2φ + ‖√τ 𝜕u‖2φ ≥ ‖√τu‖2φ,z + ∫ Θ(u, u)e−φ dλ + ∫ τi𝜕𝜕r(u, u)e−φ dσ, ∗

Ω

bΩ

where Θ(u, u) = τi𝜕𝜕φ(u, u) − i𝜕𝜕τ(u, u) −

|⟨𝜕τ, u⟩|2 . A

Proof. Like in the case without twists, we get ‖√τ 𝜕u‖2φ = ∑‖√τ( j 0 are constants. Hence we get that Nφ is a continuous linear operator from L2(0,1) (ℂn , φ) into itself (see also [41] or [13]). The rest is clear from the remarks made for the 𝜕-Neumann operator without weights in Section 10.7. Remark 11.12. In this case one can also show that Nφ = jφ ∘ jφ∗ , where jφ ∶ dom(𝜕) ∩ dom(𝜕φ ) ⟶ L2(0,1) (ℂn , e−φ ) ∗

is the embedding and dom(𝜕) ∩ dom(𝜕φ ) is endowed with the graph norm ∗

u ↦ (‖𝜕u‖2φ + ‖𝜕φ u‖2φ )1/2 . ∗

Theorem 11.13. Let 1 ≤ q ≤ n and suppose that the sum sq of any q (equivalently, the smallest q) eigenvalues of Mφ satisfies (11.21). Then there exists a uniquely determined bounded linear operator Nφ,q ∶ L2(0,q) (ℂn , e−φ ) ⟶ L2(0,q) (ℂn , e−φ ), such that □φ ∘ Nφ,q u = u, for any u ∈ L2(0,q) (ℂn , e−φ ).

304 | 11 The twisted 𝜕-complex and Schrödinger operators Proof. Let μφ,1 ≤ μφ,2 ≤ ⋯ ≤ μφ,n denote the eigenvalues of Mφ and suppose that Mφ is diagonalized. Then, in a suitable basis, n

n

𝜕2 φ ′ ujK ukK = ∑ ∑ μφ,j |ujK |2 𝜕z 𝜕z j k |K|=q−1 j,k=1 |K|=q−1 j=1 ∑

′

∑

=

∑

(μφ,j1 + ⋯ + μφ,jq )|uJ |2

′

J=(j1 ,…,jq )

≥ sq |u|2 . The last equality results as follows: for J = (j1 , … , jq ) fixed, |uJ |2 occurs precisely q times in the second sum, once as |uj1 K1 |2 , once as |uj2 K2 |2 , etc. At each occurrence, it is multiplied by μφ,jℓ . For the rest of the proof, we use (11.20) and proceed as in the proof of Theorem 11.11. ̄ Remark 11.14. For the 𝜕-Neumann operator Nφ,q on (0, q)-forms, one obtains in a sim∗ ilar way as in Theorem 11.9 that Nφ,q = jφ,q ∘ jφ,q , where jφ,q ∶ dom(𝜕) ∩ dom(𝜕φ ) ⟶ L2(0,q) (ℂn , e−φ ), ∗

and dom(𝜕) ∩ dom(𝜕φ ) is endowed with the graph norm (‖𝜕u‖2φ + ‖𝜕φ u‖2φ )1/2 . ∗ It also follows in this case that both 𝜕 and 𝜕φ have closed range, see Theorem 10.66. ∗

∗

In Theorem 11.38, we will see that we can replace (11.21) by the weaker condition lim inf sq (z) > 0.

(11.23)

|z|→∞

For this purpose we will have to use a result about the bottom of the essential spectrum of an elliptic operator (Persson’s Lemma), together with the Witten Laplacian, which is a unitarily equivalent form of □φ ; for n = 1, it is a Schrödinger2 operator with magnetic field. Theorem 11.15. Let Ω be a smoothly bounded pseudoconvex domain in ℂn and let φ ∶ Ω ⟶ ℝ be a real valued function in 𝒞2 (Ω) such that n

n

𝜕2 φ(z) wj w k , j,k=1 𝜕zj 𝜕z k

c(z) ∑ |wj |2 ≤ ∑ j=1

z ∈ Ω, w ∈ ℂn ,

where c is a positive continuous function on Ω. If g ∈ L2(0,1) (Ω, e−φ ) and 𝜕g = 0, it follows that one can find f ∈ L2 (Ω, e−φ ) with 𝜕f = g and ∫ |f (z)|2 e−φ(z) dλ(z) ≤ 2 ∫ |g(z)|2 Ω

provided that the right-hand side is finite. 2 Schrödinger, Erwin (1887–1961).

Ω

e−φ(z) dλ(z), c(z)

(11.24)

11.5 Weighted spaces of entire functions | 305

Proof. We use Theorem 11.2 with τ = A = 1 and get n

𝜕2 φ ∗ uj uk e−φ dλ ≤ ‖𝜕u‖2φ + 2‖𝜕φ u‖2φ , Ω j,k=1 𝜕zj 𝜕z k

∫ ∑

for u ∈ 𝒟0,1 . Let P ∶ L2(0,1) (Ω, e−φ ) ⟶ L2(0,1) (Ω, e−φ ) be the multiplication operator by the function √c. Then it follows from the assumption that ‖Pu‖2φ ≤ ‖𝜕u‖2φ + 2‖𝜕φ u‖2φ . ∗

By Theorem 11.1, we get a function f ∈ L2 (Ω, e−φ ) with 𝜕f = g and ‖f ‖φ ≤ √2‖P −1 g‖φ . For a positive ψ = ∑nj,k=1 ψj,k dzj ∧ dz k ∈ Λ1,1 (Ω) and α = ∑nj=1 αj dz j ∈ Λ0,1 (Ω) we set n

|α|2ψ ∶= ∑ ψj,k αj αk , j,k=1

(11.25)

where ψj,k = (ψℓ,m )−1 j,k . Then we get a more general result Theorem 11.16. Let Ω be a smoothly bounded pseudoconvex domain in ℂn and let φ ∶ Ω ⟶ ℝ a strictly plurisubharmonic function belonging to 𝒞2 (Ω). If α ∈ L2(0,1) (Ω, e−φ ) satisfies 𝜕α = 0, then one can find u ∈ L2 (Ω, e−φ ) such that 𝜕u = α and ∫ |u(z)|2 e−φ(z) dλ(z) ≤ 2 ∫ |α(z)|2i𝜕𝜕φ e−φ(z) dλ(z), Ω

Ω

(11.26)

provided that the right-hand side is finite.

11.5 Weighted spaces of entire functions In this part Hörmander’s L2 -estimates are used to construct nontrivial entire functions with given growth behavior at infinity. Theorem 11.17. Let Ω be a smoothly bounded pseudoconvex domain in ℂn and let φ ∶ Ω ⟶ ℝ be a plurisubharmonic function in 𝒞2 (Ω). For every g ∈ L2(0,1) (Ω, e−φ ) with 𝜕g = 0 there is a solution u ∈ L2 (Ω, loc) of the equation 𝜕u = g such that ∫ |u(z)|2 e−φ(z) (1 + |z|2 )−2 dλ(z) ≤ ∫ |g(z)|2 e−φ(z) dλ(z). Ω

Ω

(11.27)

Proof. We apply Theorem 11.15 with φ replaced by φ + 2 log(1 + |z|2 ) and use that n

∑ wj wk

j,k=1

𝜕2 log(1 + |z|2 ) = (1 + |z|2 )−2 (|w|2 (1 + |z|2 ) − |(w, z)|2 ) 𝜕zj 𝜕z k ≥ (1 + |z|2 )−2 |w|2 ,

so we can take c(z) = 2(1 + |z|2 )−2 to obtain the desired result.

306 | 11 The twisted 𝜕-complex and Schrödinger operators Theorem 11.18. Let Ω ⊆ ℂn be a smoothly bounded pseudoconvex domain in ℂn and let φ ∶ Ω ⟶ ℝ be a plurisubharmonic function in 𝒞2 (Ω). If z0 ∈ Ω, one can find a holomorphic function u in Ω such that u(z0 ) = 1 and ∫ |u(z)|2 e−φ(z) (1 + |z|2 )−3n dλ(z) < ∞. Ω

(11.28)

Proof. We may assume that z0 = 0. Choose a polydisc D = {z ∶ |zj | < r, j = 1, … , n} ⊂ Ω. The function e−φ is integrable there. Define Ωk ∶= {z ∈ Ω ∶ |zj | < r for j > k}, for k = 0, 1, … , n. We shall prove inductively that for every k there is a holomorphic function uk in Ωk with uk (z0 ) = 1 and ∫ |uk (z)|2 e−φ(z) (1 + |z|2 )−3k dλ(z) < ∞. Ωk

When k = 0, we can take u0 (z) ≡ 1, and un will have the desired properties. Assume that 0 < k ≤ n and that uk−1 has already been constructed. Choose ψ ∈ 𝒞∞ (ℂ) so that ψ(zk ) = 0 when |zk | > r/2 and ψ(zk ) = 1 when |zk | < r/3, and set 0 uk (z) ∶= ψ(zk )uk−1 (z) − zk v(z), notice that ψ(zk )uk−1 (z) = 0 in Ωk ⧵ Ωk−1 . To make uk holomorphic we must choose v as a solution of the equation 𝜕v = zk−1 uk−1 𝜕ψ = f . By the inductive hypothesis, we have ∫ |f (z)|2 e−φ(z) (1 + |z|2 )−3(k−1) dλ(z) < ∞. Ωk

Hence it follows from Theorem 11.17 that v can be found such that ∫ |v(z)|2 e−φ(z) (1 + |z|2 )1−3k dλ(z) < ∞. Ωk

Together with the inductive hypothesis on uk−1 this implies that ∫ |uk (z)|2 e−φ(z) (1 + |z|2 )−3k dλ(z) < ∞. Ωk

Since 𝜕v = 0 in a neighborhood of 0, v is a 𝒞∞ -function there, and we have uk (0) = uk−1 (0) = 1, so uk has the required properties. Lemma 11.19. Let ζ ∈ ℂn and K > 0 and define g(z) = log(1 + K|z − ζ |2 ). Then for each w ∈ ℂn we have

11.5 Weighted spaces of entire functions | 307

K K |w|2 ≤ i𝜕𝜕g(w, w)(z) ≤ |w|2 , 2 2 (1 + K|z − ζ | ) 1 + K|z − ζ |2

(11.29)

where we use notation (11.5). Proof. An easy computation shows K 2 (z j − ζ j )(zk − ζk ) Kδjk 𝜕2 g (z) = − + 2 2 𝜕zj 𝜕z k (1 + K|z − ζ | ) 1 + K|z − ζ |2 =

K [(1 + K|z − ζ |2 )δjk − K(z j − ζ j )(zk − ζk )]. (1 + K|z − ζ |2 )2

This implies i𝜕𝜕g(w, w)(z) =

K [(1 + K|z − ζ |2 )|w|2 − K|(w, z − ζ )|2 ], (1 + K|z − ζ |2 )2

and hence K [(1 + K|z − ζ |2 )|w|2 − K|w|2 |z − ζ |2 ] (1 + K|z − ζ |2 )2 K ≤ i𝜕𝜕g(w, w)(z) ≤ |w|2 . 1 + K|z − ζ |2 We are now able to show that weighted spaces of entire functions are of infinite dimension if the weight function has an appropriate behavior at infinity. Theorem 11.20. Let W ∶ ℂn ⟶ ℝ be a 𝒞∞ function and let μ(z) denote the smallest eigenvalue of the Levi matrix i𝜕𝜕W(z) = (

𝜕2 W(z) n ) . 𝜕zj 𝜕zk j,k=1

Suppose that lim |z|2 μ(z) = ∞.

|z|→∞

Then the Hilbert space A2 (ℂn , e−W ) of all entire functions f such that ∫ |f (z)|2 exp(−W(z)) dλ(z) < ∞, ℂn

is of infinite dimension. Proof. Assumption (11.30) implies that there exists a constant K > 0 such that i𝜕𝜕W(w, w)(z) ≥ −K|w|2 , for all z, w ∈ ℂn , and that i𝜕𝜕W(z) is strictly positive for large |z|.

(11.30)

308 | 11 The twisted 𝜕-complex and Schrödinger operators From Lemma 11.19, we have i𝜕𝜕g(w, w)(z) ≥

4K |w|2 , (1 + 4K|z − ζ |2 )2

where g(z) = log(1 + 4K|z − ζ |2 ). Hence, for |z − ζ | ≤ 1/√4K, we have i𝜕𝜕g(w, w)(z) ≥ K|w|2 . Since i𝜕𝜕W(w, w)(z) is negative at most on a compact set in ℂn , there exist finitely many points ζ1 , … , ζM ∈ ℂn such that this compact set is covered by the balls {z ∶ |z − ζl | < 1/√4K}. Hence M

̃ φ(z) ∶= W(z) + ∑ gl (z) l=1

is strictly plurisubharmonic, where gl (z) = log(1 + 4K|z − ζl |2 ), l = 1, … , M. ̃ be the least eigenvalue of i𝜕𝜕φ.̃ Then, by assumption (11.30), we have Let μ(z) ̃ = ∞. lim |z|2 μ(z)

|z|→∞

For each N ∈ ℕ there exists R > 0 such that ̃ ≥ μ(z)

N +M+1 , |z|2

for |z| > R.

̃ ∶ |z| ≤ R}. Then μ̃ 0 > 0. Set Let μ̃ 0 ∶= inf{μ(z) κ=

μ̃ 0 2(N + M)

and M

φ(z) ∶= W(z) + ∑ gl (z) − (N + M) log(1 + κ|z|2 ). l=1

It follows that e−φ is locally integrable. Next we claim that φ is strictly plurisubharmonic. Notice that ̃ − i𝜕𝜕φ(w, w)(z) ≥ |w|2 (μ(z)

(N + M)κ ). 1 + κ|z|2

For |z| ≤ R, we have ̃ − μ(z)

(N + M)μ̃ 0 μ̃ 0 (N + M)κ ≥ μ̃ 0 − (N + M)κ = μ̃ 0 − = > 0, 2 1 + κ|z| 2(N + M) 2

11.5 Weighted spaces of entire functions | 309

and for |z| > R, ̃ − μ(z)

(N + M)κ N + M + 1 N + M 1 ≥ − = 2, 1 + κ|z|2 |z|2 |z|2 |z|

which implies that φ is strictly plurisubharmonic. Therefore we can apply Theorem 11.18 and get an entire function f with f (0) = 1 and ∫ |f (z)|2 (1 + |z|2 )−3n e−φ(z) dλ(z) < ∞. ℂn

Now we set Ñ = N − 3n and get ∫ |f (z)|2 (1 + |z|2 )N e−W(z) dλ(z) ̃

ℂn

2 ∏M ̃ l=1 (1 + 4K|z − ζl | ) |f (z)|2 (1 + |z|2 )N e−φ(z) dλ(z) 2 N+M (1 + κ|z| )

=∫

ℂn

≤ sup { z∈ℂn

2 (1 + |z|2 )N ∏M l=1 (1 + 4K|z − ζl | ) } (1 + κ|z|2 )N+M

× ∫ |f (z)|2 (1 + |z|2 )−3n e−φ(z) dλ(z) ℂn

< ∞. Hence fp ∈ A2 (ℂn , e−W ) for any polynomial p of degree < Ñ , and since N = Ñ + 3n was arbitrary, we are done. The following example in ℂ2 shows that Theorem 11.20 is not sharp. Let φ(z, w) = |z|2 |w|2 + |w|4 . In this case we have that A2 (ℂ2 , e−φ ) contains all the functions fk (z, w) = wk for k ∈ ℕ, since ∞

∞

0

0

2 2

4

∞

∞

0

0

2 2

4

∫ ∫ r22k e−(r1 r2 +r2 ) r1 r2 dr1 dr2 = ∫ (∫ r1 r22 e−r1 r2 dr1 )r22k−1 e−r2 dr2 4 4 1 1 ∞ = ∫ ( ∫ e−s ds)r22k−1 e−r2 dr2 = ∫ r22k−1 e−r2 dr2 < ∞. 2 2 0 0 0

∞

∞

The Levi matrix of φ has the form i𝜕𝜕φ = (

|w|2 wz

zw ) |z|2 + 4|w|2

hence φ is plurisubharmonic and the smallest eigenvalue has the form 1 μφ (z, w) = (5|w|2 + |z|2 − (9|w|4 + 10|z|2 |w|2 + |z|4 )1/2 ) 2 16|w|4 = , 2 2 2(5|w| + |z| + (9|w|4 + 10|z|2 |w|2 + |z|4 )1/2 )

310 | 11 The twisted 𝜕-complex and Schrödinger operators hence lim |z|2 μφ (z, 0) = 0,

|z|→∞

which implies that condition (11.30) of Theorem 11.20 is not satisfied.

11.6 Spectral analysis of self-adjoint operators Definition 11.21. The spectrum σ(T) of an operator T ∈ ℒ(H) is the set of all λ ∈ ℂ, such that λI − T has no inverse in ℒ(H). The complement ρ(T) = ℂ ⧵ σ(T) is called the resolvent set. If λ ∈ ρ(T) the operator (λI − T)−1 ∈ ℒ(H) is called the resolvent of T at λ and is denoted by RT (λ). We have an operator-valued function RT ∶ ρ(T) ⟶ ℒ(H). An operator T ∈ ℒ(H) is called normal if TT ∗ = T ∗ T. Theorem 11.22. Let T ∈ ℒ(H). Then the spectrum σ(T) of T is a compact subset of ℂ and |λ| ≤ ‖T‖, for every λ ∈ σ(T). Proof. First we show that ρ(T) is open. Let λ ∈ ρ(T). Then α = ‖RT (λ)‖−1 > 0. Let μ ∈ ℂ with |μ| < α. We will show that (λ + μ)I − T has a bounded inverse. Then we proved that ρ(T) is open. We have (λ + μ)I − T = λI − T + μI = (λI − T)[I + μ(λI − T)−1 ] = (λI − T)(I + μRT (λ)). Formally, ∞

(I + μRT (λ))−1 = I + ∑ (−1)k (μRT (λ))k , k=1

but as |μ| < α, we have ∞

∞

∞

k=1

k=1

‖ ∑ (−1)k (μRT (λ))k ‖ ≤ ∑ |μ|k ‖RT (λ)‖k = ∑ (|μ|/α)k < ∞, k=1

k k ∑∞ k=1 (−1) (μRT (λ))

therefore the partial sums of Since ℒ(H) is complete, we obtain that ∞

form a Cauchy sequence in ℒ(H).

∑ (−1)k (μRT (λ))k ∈ ℒ(H)

k=1

and ρ(T) is open.

11.6 Spectral analysis of self-adjoint operators | 311

If η ∈ ℂ with |η| > ‖T‖, then I − T/η has a bounded inverse, since ∞

(I − T/η)−1 = I + ∑ (T/η)k . k=1

This implies that ηI − T has a bounded inverse. Hence, if λ ∈ σ(T), then |λ| ≤ ‖T‖, and σ(T) is a bounded set. The resolvent has the following properties: Lemma 11.23. If λ, μ ∈ ρ(T), then RT (λ) − RT (μ) = (μ − λ)RT (λ)RT (μ).

(11.31)

If λ ∈ ρ(T) and |λ − μ| < ‖RT (λ)‖−1 , then ∞

RT (μ) = ∑ (λ − μ)k [RT (λ)]k+1 , k=0

(11.32)

therefore one says that RT is a holomorphic operator-valued function. Proof. Equation (11.31) follows from RT (λ) = RT (λ)(μI − T)RT (μ) = RT (λ)[(λI − T) + (μ − λ)I]RT (μ) = RT (μ) + (μ − λ)RT (λ)RT (μ).

Equation (11.32) follows immediately from the proof of Theorem 11.22. Theorem 11.24. Let T be a bounded self-adjoint operator. Then σ(T) is real and is contained in [m, M], where m = inf

u≠0

(Tu, u) (u, u)

and

M = sup u≠0

(Tu, u) . (u, u)

Moreover, m and M belong to the spectrum of T. Proof. If ℑλ ≠ 0, then |ℑλ| ‖u‖2 = |ℑ((T − λI)u, u)| ≤ ‖(T − λI)u‖ ‖u‖, for all u ∈ H, where we used the fact that (Tu, u) = (u, Tu) is real. This implies that T − λI is injective and has closed range. As (im(T − λI))⟂ = ker(T − λI) and this kernel reduces to {0}, we obtain that T − λI is bijective.

(11.33)

312 | 11 The twisted 𝜕-complex and Schrödinger operators This follows also from Theorem 10.35, once one has observed that |((T − λI)u, u)| ≥ |ℑλ| ‖u‖2 .

(11.34)

Hence, the spectrum is real. If λ is real and λ > M, we can apply Theorem 10.35 for the sesquilinear form (u, v) ↦ λ(u, v) − (Tu, v), to see that λ ∉ σ(T). To show that M ∈ σ(T), we apply Cauchy–Schwarz inequality to the scalar product (u, v) ↦ M(u, v) − (Tu, v). We get |(Mu − Tu, v)| ≤ (Mu − Tu, u)1/2 (Mv − Tv, v)1/2 . In particular, ‖Mu − Tu‖ ≤ ‖MI − T‖1/2 (Mu − Tu, u)1/2 .

(11.35)

Now let (un )n be a sequence in H such that ‖un ‖ = 1 for each n ∈ ℕ and (Tun , un ) → M as n → ∞. By (11.35), lim (MI − T)un = 0,

n→∞

and this implies M ∈ σ(T), otherwise un = (MI − T)−1 (MI − T)un would tend to 0 in contradiction to ‖un ‖ = 1. Corollary 11.25. Let T ∈ ℒ(H) be a self-adjoint operator such that σ(T) = {0}. Then T = 0. Proof. By Theorem 11.24, we have m = M = 0, hence (Tu, u) = 0, for each u ∈ H. As (Tu, v) can be written as a linear combination of terms of the type (Tw, w), we obtain T = 0. More general, we introduce the following concepts: Definition 11.26. The resolvent set of a linear operator T ∶ dom(T) ⟶ H is the set of all λ ∈ ℂ such that λI − T is an injective mapping of dom(T) onto H whose inverse belongs to ℒ(H). The spectrum σ(T) of T is the complement of the resolvent set of T.

11.6 Spectral analysis of self-adjoint operators | 313

First we collect some facts about the spectrum of an unbounded operator. Lemma 11.27. If the spectrum σ(T) of an operator T does not coincide with the whole complex plane ℂ, then T must be a closed operator. The spectrum of a linear operator is always closed. Moreover, if ζ ∉ σ(T) and c ∶= ‖RT (ζ )‖ = ‖(ζI − T)−1 ‖, then the spectrum σ(T) does not intersect the ball {w ∈ ℂ ∶ |ζ − w| < c−1 }. The resolvent operator RT is a holomorphic operator-valued function. Proof. For ζ ∉ σ(T) let S = (ζI − T)−1 which is a bounded operator. Let xn ∈ dom(T) with limn→∞ xn = x and limn→∞ Txn = y and set un = (ζI − T)xn . Then lim un = lim (ζxn − Txn ) = ζx − y,

n→∞

n→∞

therefore S(ζx − y) = lim Sun = lim xn = x. n→∞

n→∞

This implies x ∈ dom(T) and (ζI − T)x = ζx − y, or Tx = y. Hence T is closed. The remainder of the proof is similar to the case when T is bounded, see Lemma 11.23. Theorem 11.28. The spectrum σ(T) of any self-adjoint operator T is non-empty. Proof. Suppose T has empty spectrum. Then T −1 is a bounded self-adjoint operator. We claim that σ(T −1 ) = {0}. For λ ≠ 0, we can write the inverse of T −1 − λI in the form (T −1 − λI)−1 = λ−1 T(λ−1 I − T)−1 = −λ−1 I + λ−2 (λ−1 I − T)−1 , which is a bounded operator. Now Corollary 11.25 gives that T −1 = 0, which contradicts that T ∘ T −1 = I. In the following we describe how to generalize the spectral decomposition (Theorem 8.7) of a compact self-adjoint operator. Definition 11.29. Let Ω be a subset of ℂ and M be a σ-algebra in Ω and let H be a Hilbert space. A resolution of the identity is a mapping E ∶ M ⟶ ℒ(H) of M to the algebra ℒ(H) of bounded linear operators on H with the following properties: (a) E(∅) = 0 (zero-operator), E(Ω) = I (identity on H). (b) For each ω ∈ M the image E(ω) is an orthogonal projection on H. (c) E(ω′ ∩ ω″ ) = E(ω′ )E(ω″ ).

314 | 11 The twisted 𝜕-complex and Schrödinger operators (d) If ω′ ∩ ω″ = ∅, then E(ω′ ∪ ω″ ) = E(ω′ ) + E(ω″ ). (e) For every x ∈ H and y ∈ H, the set function Ex,y defined by Ex,y (ω) = (E(ω)x, y) is a complex measure on M. The main result in spectral analysis of unbounded self-adjoint operators is the spectral theorem: Theorem 11.30. Let T be an unbounded self-adjoint operator (T = T ∗ and dom(T) = dom(T ∗ )). Then there exists a uniquely determined resolution of the identity E on the Borel subsets of ℝ such that (Tx, y) = ∫

∞

−∞

t dEx,y (t),

x ∈ dom(T), y ∈ H.

(11.36)

Moreover, E is concentrated on the spectrum σ(T) ⊂ ℝ of T in the sense that E(σ(T)) = I. Equation (11.36) can be viewed as a generalization of the spectral decomposition (Theorem 8.6) of a compact, self-adjoint operator. As an important consequence, we mention that a positive self-adjoint operator has a uniquely determined square root which is again a positive self-adjoint operator. Theorem 11.31. Let T be a self-adjoint operator on H. (a) (Tx, x) ≥ 0 for every x ∈ dom(T) (briefly T ≥ 0), if and only if σ(T) ⊂ [0, ∞). (b) If T ≥ 0, there exists a unique self-adjoint operator S ≥ 0 such that dom(S) ⊇ dom(T) and S2 = T on dom(T). Furthermore, x ∈ dom(T), if and only if x ∈ dom(S) and also Sx ∈ dom(S). In addition, dom(T) is a core of S. (c) If T is bounded, positive and self-adjoint, then the square root S has the same properties as T. We will also need a more detailed study of the spectrum of a self-adjoint operator. Definition 11.32. Let T be a self-adjoint operator on H. The discrete spectrum σd (T) of T is the set of all eigenvalues λ of finite multiplicity which are isolated in the sense that the intervals (λ − ϵ, λ) and (λ, λ + ϵ) are disjoint from the spectrum for some ϵ > 0. The non-discrete part of the spectrum of T is called the essential spectrum of T, and is denoted by σe (T). Theorem 11.33. Let T be an unbounded self-adjoint operator on H which is nonnegative in the sense that σ(T) ⊆ [0, ∞). Then the following conditions are equivalent: (i) The resolvent operator (I + T)−1 is compact. (ii) σe (T) = ∅.

11.7 Real differential operators | 315

(iii) There exists a complete orthonormal system of eigenvectors {xn }n of T with corresponding eigenvalues μn ≥ 0 which converge to +∞ as n → ∞. Theorem 11.34. Let T be a positive, self-adjoint operator, which is also injective. Suppose that the bottom of the essential spectrum satisfies inf σe (T) > 0. Then T has a bounded inverse, i.e. T −1 ∈ ℒ(H). This is also a consequence of the spectral theorem; see [70, Theorem 8.17].

11.7 Real differential operators We consider differential operators H(A, V) of the form H(A, V) = −ΔA + V,

(11.37)

where V ∶ ℝn ⟶ ℝ is the electric potential and n

Aj ∶ ℝn ⟶ ℝ,

A = ∑ Aj dxj , j=1

j = 1, … , n

is a 1-form, and n

ΔA = ∑( j=1

2 𝜕 − iAj ) . 𝜕xj

The 2-form B = dA = ∑( j 0 is a positive constant. Let dom(H(A, V)) = 𝒞∞ 0 (ℝ ). Then H(A, V) is a sym2 n metric, semibounded operator on L (ℝ ). n Proof. For u ∈ 𝒞∞ 0 (ℝ ), we have

(H(A, V)u, u) = ∫ (−ΔA u + Vu)u dλ ℝn

n

= ∫ ∑ |Xj u|2 dλ + ∫ V|u|2 dλ ℝn j=1

ℝn

≥ −C‖u‖2 . Using Theorem 10.36, we obtain Theorem 11.36. Let H(A, V) be as in Theorem 11.35. Then H(A, V) admits a self-adjoint extension. Proof. Define a(u, v) ∶= (H(A, V)u, v) + (C + 1)(u, v) n and define 𝒱 to be the completion of 𝒞∞ 0 (ℝ ) with respect to the inner product a(u, v). Then one can apply Theorem 10.36 to get the desired result.

Remark 11.37. The operator H(A, V) of Theorem 11.36 is essentially self-adjoint (see n [34]) on 𝒞∞ 0 (ℝ ). This implies that its self-adjoint extension is uniquely determined. In our first application of methods from complex analysis we will concentrate on Schrödinger operators with magnetic field on ℝ2 of the form −ΔA + B, where the 2-form dA is given by dA = B dx ∧ dy. So, in this case, the Schrödinger operator −ΔA + B = H(A, B) is completely determined by the vector field A.

11.7 Real differential operators | 317

Let φ be a subharmonic 𝒞2 -function. We consider the inhomogeneous 𝜕-equation 𝜕u = f for f ∈ L2 (ℂ, e−φ ). The canonical solution operator to 𝜕 gives a solution with minimal L2 (ℂ, e−φ )-norm. We substitute v = ue−φ/2 and g = fe−φ/2 and the equation becomes Dv = g, where D = e−φ/2

𝜕 φ/2 e . 𝜕z

(11.40)

u is the minimal solution to the 𝜕-equation in L2 (ℂ, e−φ ), if and only if v is the solution to Dv = g which is minimal in L2 (ℂ). ∗ 𝜕 −φ/2 The formal adjoint of D is D = −eφ/2 𝜕z e . We define dom(D) = {f ∈ L2 (ℂ) ∶ Df ∈ L2 (ℂ)} and likewise for D . Then D and D are closed unbounded linear operators from L2 (ℂ) to itself. Further we define ∗

∗

dom(D D ) = {u ∈ dom(D ) ∶ D u ∈ dom(D)} ∗

∗

∗

and we define D D as D ∘ D on this domain. Any function of the form eφ/2 g, with ∗ ∗ g ∈ 𝒞20 (ℂ) belongs to dom(D D ) and hence dom(D D ) is dense in L2 (ℂ). Since ∗

∗

D=

𝜕 1 𝜕φ + 𝜕z 2 𝜕z

and

D =− ∗

𝜕 1 𝜕φ + , 𝜕z 2 𝜕z

we see that 𝒮 = DD = − ∗

𝜕2 1 𝜕φ 𝜕 1 𝜕φ 𝜕 1 𝜕φ 2 1 𝜕2 φ − + + | | + . 𝜕z𝜕z 2 𝜕z 𝜕z 2 𝜕z 𝜕z 4 𝜕z 2 𝜕z𝜕z

(11.41)

It follows that 4𝒮 = −(

𝜕2 𝜕2 i 𝜕φ 𝜕 i 𝜕φ 𝜕 1 𝜕φ 2 𝜕φ 2 1 + ) + − + (( ) + ( ) ) + Δφ, 𝜕x2 𝜕y2 2 𝜕x 𝜕y 2 𝜕y 𝜕x 4 𝜕x 𝜕y 2

which implies that 𝒮 is a Schrödinger operator with magnetic field, namely 1 𝒮 = (−ΔA + B), 4

(11.42)

where the 1-form A = A1 dx + A2 dy is related to the weight φ by A1 = −𝜕y φ/2, ΔA = (

A2 = 𝜕x φ/2, 2

(11.43) 2

𝜕 𝜕 − iA1 ) + ( − iA2 ) , 𝜕x 𝜕y

(11.44)

318 | 11 The twisted 𝜕-complex and Schrödinger operators and the magnetic field Bdx ∧ dy satisfies 1 B(x, y) = Δφ(x, y). 2 Both D D and D D are non-negative, self-adjoint operators, see Lemma 10.28 and Lemma 10.29. By Theorem 11.36 and Remark 11.37, we now know that the operator 𝒮 with its domain described above is the uniquely determined self-adjoint extension of the Schrödinger operator with magnetic field 41 (−ΔA + B). ∗ Since 4D D = −ΔA + 21 Δφ, it follows that ((−ΔA + 21 Δφ)f , f ) ≥ 0, for f ∈ 𝒞20 (ℂ). Sim∗ ilarly one shows that 4D D = −ΔA − 21 Δφ, and this implies ∗

∗

1 −2ΔA ≥ −ΔA + Δφ ≥ −ΔA . 2 It follows that D D = e−φ/2 𝜕φ 𝜕eφ/2

(11.45)

D D = e−φ/2 𝜕 𝜕φ eφ/2 ,

(11.46)

∗

∗

and ∗

∗

𝜕 where 𝜕φ = − 𝜕z + ∗

𝜕φ . 𝜕z

For n = 1, we have ∗

□φ = 𝜕 𝜕φ , which means that D D = e−φ/2 □φ eφ/2 . ∗

For several complex variables, the situation is more complicated. Let φ ∶ ℂn ⟶ ℝ be a 𝒞2 -weight function. We consider the 𝜕-complex 𝜕

𝜕

L2 (ℂn , e−φ ) ⟶ L2(0,1) (ℂn , e−φ ) ⟶ L2(0,2) (ℂn , e−φ ). For v ∈ L2 (ℂn ), let n

D1 v = ∑ ( k=1

𝜕v 1 𝜕φ + v) dz k 𝜕z k 2 𝜕z k

and for g = ∑nj=1 gj dz j ∈ L2(0,1) (ℂn ), let n

D1 g = ∑( ∗

j=1

𝜕gj 1 𝜕φ gj − ), 2 𝜕zj 𝜕zj

where the derivatives are taken in the sense of distributions.

(11.47)

11.7 Real differential operators | 319

It is easy to see that 𝜕u = f for u ∈ L2 (ℂn , e−φ ) and f ∈ L2(0,1) (ℂn , e−φ ), if and only if D1 v = g, where v = ue−φ/2 and g = fe−φ/2 . It is also clear that the necessary condition 𝜕f = 0 for solvability holds, if and only if D2 g = 0 holds. Here n

D2 g = ∑ ( j,k=1

𝜕gj 𝜕z k

+

1 𝜕φ g ) dz k ∧ dz j 2 𝜕z k j

and n

D2 h = ∑ ( ∗

j,k=1

𝜕hkj 1 𝜕φ hkj − ) dz j 2 𝜕zk 𝜕zk

for a suitable (0, 2)-form h = ∑′|J|=2 hJ dz J . We consider the corresponding D-complex: D1

D2

⟵ ∗

⟵ ∗

L2 (ℂn ) ⟶ L2(0,1) (ℂn ) ⟶ L2(0,2) (ℂn ). D1

D2

The so-called Witten Laplacians (see [33]) Δ(0,0) and Δ(0,1) are defined by φ φ Δ(0,0) = D1 D1 , φ ∗

Δ(0,1) = D1 D1 + D2 D2 . φ ∗

∗

(11.48)

A computation shows that n

D1 D1 v = ∑( ∗

j=1

1 𝜕φ 𝜕v 1 𝜕φ 𝜕φ 1 𝜕φ 𝜕v 1 𝜕2 φ 𝜕2 v + v− − v− ) 2 𝜕zj 𝜕z j 4 𝜕zj 𝜕z j 2 𝜕z j 𝜕zj 2 𝜕zj 𝜕z j 𝜕zj 𝜕z j

and n

n

(D1 D1 + D2 D2 )g = ∑ [∑( ∗

∗

k=1 j=1

−

1 𝜕φ 𝜕gk 1 𝜕φ 𝜕φ + g 2 𝜕zj 𝜕z j 4 𝜕zj 𝜕z j k

1 𝜕φ 𝜕gk 1 𝜕2 φ 𝜕2 gk 𝜕2 φ − gk − + g )] dz k . 2 𝜕z j 𝜕zj 2 𝜕zj 𝜕z j 𝜕zj 𝜕z j 𝜕zj 𝜕z k j

(11.49)

𝜕φ 𝜕φ More generally, we set Zk = 𝜕z𝜕 + 21 𝜕z , Zk∗ = − 𝜕z𝜕 + 21 𝜕z and consider (0, q)-forms k k k k h = ∑′|J|=q hJ dz J , where ∑′ means that we sum up only over increasing multiindices J = (j1 , … , jq ) and where dz J = dz j1 ∧ ⋯ ∧ dz jq . We define n

Dq+1 h = ∑ ∑ Zk (hJ ) dz k ∧ dz J ′

k=1|J|=q

and n

Dq h = ∑ ∑ Zk∗ (hJ ) dz k ⌋ dz J , ∗

′

k=1|J|=q

320 | 11 The twisted 𝜕-complex and Schrödinger operators where dz k ⌋ dz J denotes the contraction, or interior multiplication by dz k , i.e. we have ⟨α, dz k ⌋ dz J ⟩ = ⟨dz k ∧ α, dz J ⟩ for each (0, q − 1)-form α. The complex Witten Laplacian on (0, q)-forms is then given by Δ(0,q) = Dq Dq + Dq+1 Dq+1 , φ ∗

(11.50)

∗

for q = 1, … , n − 1. The general D-complex has the form Dq

Dq+1

⟵ ∗

⟵ ∗

L2(0,q−1) (ℂn ) ⟶ L2(0,q) (ℂn ) ⟶ L2(0,q+1) (ℂn ). Dq

Dq+1

It follows that Dq+1 Δ(0,q) = Δ(0,q+1) Dq+1 φ φ

and

Dq+1 Δ(0,q+1) = Δ(0,q) φ φ Dq+1 . ∗

∗

We remark that n

n

Dq h = ∑ ∑ Zk∗ (hJ ) dz k ⌋ dz J = ∑ ∑ Zk∗ (hkK ) dz K . ∗

′

′

k=1|J|=q

|K|=q−1 k=1

In particular, we get for a function v ∈ dom(Δ(0,0) φ ) that n

∗ Δ(0,0) φ v = D1 D1 v = ∑ Zj Zj (v), ∗

j=1

and for a (0, 1)-form g = ∑nℓ=1 gℓ dz ℓ ∈ dom(Δ(0,1) φ ) we obtain Δ(0,1) φ g = (D1 D1 + D2 D2 )g ∗

∗

n

= ∑ {Zj (Zk∗ (gℓ )) dz j ∧ (dz k ⌋ dz ℓ ) + Zk∗ (Zj (gℓ )) dz k ⌋ (dz j ∧ dz ℓ )} j,k,ℓ=1 n

= ∑ {Zk∗ (Zj (gℓ ))(dz j ∧ (dz k ⌋ dz ℓ ) + dz k ⌋ (dz j ∧ dz ℓ )) j,k,ℓ=1

+ [Zj , Zk∗ ](gℓ ) dz j ∧ (dz k ⌋ dz ℓ )} n

n

𝜕2 φ gℓ δkℓ dz j j,k,ℓ=1 𝜕z j 𝜕zk

= ∑ Zj∗ Zj (gℓ ) dz ℓ + ∑ j,ℓ=1

= (Δ(0,0) ⊗ I)g + Mφ g, φ where we used that for (0, 1)-forms α, a and b we have α ⌋ (a ∧ b) = (α ⌋ a) ∧ b − a ∧ (α ⌋ b),

11.7 Real differential operators | 321

which implies dz j ∧ (dz k ⌋ dz ℓ ) + dz k ⌋ (dz j ∧ dz ℓ )

= dz j ∧ (dz k ⌋ dz ℓ ) + (dz k ⌋ dz j ) ∧ dz ℓ − dz j ∧ (dz k ⌋ dz ℓ ) = (dz k ⌋ dz j ) ∧ dz ℓ = δkℓ dz ℓ ,

and where we set n

n

𝜕2 φ gk ) dz j k=1 𝜕zk 𝜕z j

Mφ g = ∑( ∑ j=1

and n

(Δ(0,0) ⊗ I)g = ∑ Δ(0,0) φ φ gk dz k . k=1

We will now explain how to prove the following theorem, for all details see [31, 30]. Theorem 11.38. Let φ ∶ ℂn ⟶ ℝ be a 𝒞2 -plurisubharmonic function and suppose that the least eigenvalue μφ of the Levi matrix Mφ of φ satisfies lim inf μφ (z) > 0. |z|→∞

(11.51)

Then the operator Δ(0,1) has a bounded inverse on L2(0,1) (ℂn ) φ −1 −φ/2 Nφ eφ/2 , (D1 D1 + D2 D2 )−1 = (Δ(0,1) φ ) =e ∗

∗

(11.52)

where Nφ = □−1 φ . For n = 1, condition (11.51) reads as lim inf Δφ(z) > 0. |z|→∞

(11.53)

Proof. For g ∈ dom(Δ(0,1) φ ), we have (0,0) Δ(0,1) ⊗ I)g + Mφ g, φ g = (Δφ

(11.54)

and, as Δ(0,0) is a positive operator, obtain φ (Δ(0,1) φ g, g) ≥ (Mφ g, g),

(11.55)

which implies that n

2 (Δ(0,1) φ g, g) ≥ ∫ ∑ μφ (z)|gℓ (z)| dλ(z). ℂn ℓ=1

(11.56)

322 | 11 The twisted 𝜕-complex and Schrödinger operators Recall that μφ ≥ 0. If Δ(0,1) φ g = 0, (11.56) together with the assumption (11.51) implies that μφ is nonzero at ∞, hence g = 0 on a non-empty open set. By the uniqueness result for elliptic differential operators (see [5]), it follows that g = 0 everywhere. So Δ(0,1) is injective. φ Now we apply Persson’s Lemma (see [3]) to Δ(0,1) and get φ (0,1) ∞ n inf σe (Δ(0,1) φ ) = sup inf{(Δφ g, g) ∶ ‖g‖ = 1, g ∈ 𝒞0 (ℂ ⧵ K)}, K

(11.57)

where the supremum is taken over all compact subsets K ⊂ ℂn . Using (11.56) together with the assumption (11.51), we now conclude that the bottom of the essential spectrum of Δ(0,1) is strictly positive and Theorem 11.34 implies that Δ(0,1) has a bounded φ φ inverse. Remark 11.39. Under the same assumptions as before, we have that also Nφ exists and is bounded, Nφ is self-adjoint and positive. So we get from Theorem 11.31 that Nφ has a bounded square root Nφ1/2 . Let u ∈ dom(□φ ). Then there exists a w ∈ L2(0,1) (ℂn , e−φ ) such that Nφ w = u. Define v ∶= Nφ1/2 w. It follows that Nφ1/2 v = u, and we have ‖u‖2φ = ‖Nφ1/2 v‖2φ ≤ C‖v‖2φ = C(Nφ1/2 w, Nφ1/2 w)φ = C(w, Nφ w)φ = C(□φ u, u)φ

= C(‖𝜕u‖2φ + ‖𝜕φ u‖2φ ), ∗

which is the basic estimate (11.22). In addition, we get from Theorem 11.31 that □φ has a self-adjoint square root 1/2 ̄ ̄∗ □φ with dom(□1/2 φ ) = dom(𝜕) ∩ dom(𝜕φ ) and that we can write for the corresponding Dirichlet form 1/2 1/2 ̄ ̄∗ ̄∗ Qφ (f , g) = (𝜕f̄ , 𝜕g) φ + (𝜕φ f , 𝜕φ g)φ = (□φ f , □φ g)φ ;

Theorem 11.31 implies that dom(□φ ) is a core of □1/2 φ . If we suppose that lim|z|→∞ μφ (z) = +∞, then Nφ is a compact operator (see [31, 30]). For the complex 1-dimensional case, see Section 11.9.

11.8 Dirac and Pauli operators There is an interesting connection to Dirac3 and Pauli4 operators: recall (11.44) and define the Dirac operator 𝒟 by 𝒟 = (−i

𝜕 𝜕 − A1 )σ1 + (−i − A2 )σ2 = 𝒜1 σ1 + 𝒜2 σ2 , 𝜕x 𝜕y

3 Dirac, Paul (1902–1984). 4 Pauli, Wolfgang (1900–1958).

(11.58)

11.9 Compact resolvents | 323

where σ1 = (

0 1 ), 1 0

σ2 = (

0 i

−i ). 0

Hence we can write 𝒟=(

0 𝒜1 + i𝒜2

𝒜1 − i𝒜2 ). 0

We remark that i(𝒜2 𝒜1 − 𝒜1 𝒜2 ) = B and hence it turns out that the square of 𝒟 is diagonal with the Pauli operators P± on the diagonal: 𝒟2 = ( =(

𝒜21 − i(𝒜2 𝒜1 − 𝒜1 𝒜2 ) + 𝒜22 0 P− 0

0 ) 𝒜21 + i(𝒜2 𝒜1 − 𝒜1 𝒜2 ) + 𝒜22

0 ), P+

where P± = (−i

2 2 𝜕 𝜕 − A1 ) + (−i − A2 ) ± B = −ΔA ± B. 𝜕x 𝜕y

By Lemma 10.28 and Lemma 10.29, the Pauli operators P± are non-negative and selfadjoint. It follows that 4𝒮 = P+ is the Schrödinger operator with magnetic field and that 4Δ(0,0) = P− . φ

(11.59)

In addition, we obtain that 𝒟2 is self-adjoint and likewise 𝒟 by Theorem 11.31.

11.9 Compact resolvents In the final section we demonstrate additional examples for the interplay between real and complex analysis, mainly to discuss compactness for the 𝜕-Neumann operator. It turns out that compactness of the 𝜕-Neumann operator is related to the question of whether corresponding Bergman spaces of entire functions are of infinite dimension. First we give a sufficient condition for compactness of the 𝜕-Neumann operator Nφ on L2 (ℂ, e−φ ). For this purpose we use a criterion for magnetic Schrödinger operators with compact resolvent. Then we describe a characterization of compactness of the 𝜕-Neumann operator Nφ on L2 (ℂ, e−φ ) as it is done in [31] using various methods from real analysis.

324 | 11 The twisted 𝜕-complex and Schrödinger operators Theorem 11.40. Let H(A, V) = −ΔA + V be a magnetic Schrödinger operator, where A = ∑nj=1 Aj dxj , with Aj ∈ L2loc (ℝn ), j = 1, … , n and V ∈ L2loc (ℝn ) and V(x) ≥ 0, for all x ∈ ℝn . Then H(A, V) has compact resolvent (i.e. the resolvent (H(A, V) − iI)−1 is a compact operator on L2 (ℝn )), if and only if there exists a real-valued continuous function ψ on ℝn such that lim|x|→∞ ψ(x) = ∞ and (H(A, V)u, u) ≥ ∫ ψ(x)|u(x)|2 dλ(x) ℝn

(11.60)

n for all u ∈ 𝒞∞ 0 (ℝ ).

For the proof, see [42]. Theorem 11.41. Let φ be a subharmonic 𝒞2 -function such that Δφ(z) → +∞ as |z| → ∞. Then the 𝜕-Neumann operator Nφ is compact on L2 (ℂ, e−φ ). Proof. Suppose that Δφ(z) → ∞ as |z| → ∞. From (11.47) we know that □φ = ∗ ∗ eφ/2 D D e−φ/2 and that D D = − 41 ΔA + 81 Δφ. We also proved that −ΔA ≥ 21 Δφ, which implies that − 41 ΔA + 81 Δφ ≥ 41 Δφ and hence for f ∈ 𝒞∞ 0 (ℂ) we obtain (□φ f , f )φ = (eφ/2 D D e−φ/2 f , f )φ ∗

= (e−φ/2 D D e

∗ −φ/2

= (D D e

∗ −φ/2

f,e

(11.61)

f,f)

(11.62)

f ),

(11.63)

−φ/2

and setting g = e−φ/2 f we get 1 1 ∗ (□φ f , f )φ = (D D g, g) ≥ (Δφ g, g) = (Δφ f , f )φ . 4 4

Then we can apply Theorem 11.40 to see that Nφ is compact. In our case the resol∗ ∗ vent (D D )−1 exists and is compact, since (D D − iI)−1 is compact and we can use (11.31). For the characterization of compactness in the complex one-dimensional case, we need the following concepts: The reverse Hölder5 class B2 (ℝ2 ) consists of L2 positive and almost nonzero everywhere functions V for which there exists a constant C > 0 such that 1

2 1 1 ( ∫ V 2 dλ) ≤ C( ∫ V dλ) |Q| Q |Q| Q

for any ball Q in ℝ2 . Note that any positive (nonzero) polynomial is in B2 (ℝ2 ). 5 Hölder, Otto (1859–1937).

(11.64)

11.9 Compact resolvents | 325

Theorem 11.42. Let φ be a subharmonic 𝒞2 -function on ℝ2 such that Δφ ∈ B2 (ℝ2 ).

(11.65)

Then the 𝜕-Neumann operator Nφ is compact on L2 (ℂ, e−φ ), if and only if lim ∫

|z|→∞ D (z) 1

Δφ(w) dλ(w) = +∞,

(11.66)

where D1 (z) = {w ∈ ℂ ∶ |w − z| < 1}. That (11.66) is necessary for compactness follows from a result of Iwatsuka. The sufficiency of (11.66) is derived form the diamagnetic inequality (see [30]) and a special form of the Fefferman–Phong inequality (see [31]). Example 11.43. Let z = x + iy. Define 1 1 1 φ(x, y) = (x 2 + y2 ) + (x4 y2 + x2 y4 ) − (x6 + y6 ). 4 24 360 Then Δφ(x, y) = 1 + x2 y2 and φ is subharmonic. It is clear that lim|z|→∞ Δφ(z) does not exist. But condition (11.66) is satisfied (see Exercise 171). We return to the Dirac and Pauli operators related with the weight function φ: 𝒟 = (−i where A1 = − 21 𝜕φ , A2 = 𝜕y

1 𝜕φ 2 𝜕x

𝜕 𝜕 − A1 )σ1 + (−i − A2 )σ2 , 𝜕x 𝜕y

and σ1 = (

0 1 ), 1 0

σ2 = (

0 i

−i ). 0

The square of 𝒟 is diagonal with the Pauli operators P± on the diagonal: 𝒟2 = (

P− 0

0 ), P+

where P± = (−i

2 2 𝜕 𝜕 − A1 ) + (−i − A2 ) ± B = −ΔA ± B, 𝜕x 𝜕y

and B = 21 Δφ. Definition 11.44. A non-negative Borel measure μ on ℂ is doubling, if there exists a constant C > 0 such that for any z ∈ ℂ and any r > 0 μ(D(z, r)) ≤ Cμ(D(z, r/2)).

(11.67)

326 | 11 The twisted 𝜕-complex and Schrödinger operators It can be shown that μ(D(z, 2r)) ≥ (1 + C −3 )μ(D(z, r)),

(11.68)

for each z ∈ ℂ and for each r > 0; in particular μ(ℂ) = ∞, unless μ(ℂ) = 0 (see [67]). For example, if p(z, z) is a polynomial on ℂ of degree d, then dμ(z) = |p(z, z)|a dλ(z),

a>−

1 d

is a doubling measure on ℂ, see [67]. Theorem 11.45. Let φ ∶ ℂ ⟶ ℝ + be a subharmonic 𝒞2 -function. Suppose that dμ = Δφ dλ is a non-trivial doubling measure. Then the weighted space of entire functions A2 (ℂ, e−φ ) = {f entire ∶ ‖f ‖2φ = ∫ |f |2 e−φ dλ < ∞} ℂ

is of infinite dimension. Proof. We know that μ(ℂ) = ∫ Δφ dλ = ∞. ℂ

(11.69)

Let Kφ (z, w) denote the Bergman kernel of A2 (ℂ, e−φ ). So, if f ∈ A2 (ℂ, e−φ ), then f (z) = ∫ Kφ (z, w)f (w)e−φ(w) dλ(w). ℂ

(11.70)

We claim that ∞

∫ Kφ (z, z)e−φ(z) dλ(z) = ∑ ‖fk ‖2φ , k=1

ℂ

(11.71)

for any complete orthonormal basis (fk )k of A2 (ℂ, e−φ ), finite or infinite. The partial sums N

FN (z) = ∑ |fk (z)|2 k=1

form a pointwise non-decreasing sequence: FN (z) ≤ FN+1 (z), for each z ∈ ℂ and for each N ∈ ℕ. In addition, we have ∞

lim FN (z) = ∑ fk (z)fk (z) = Kφ (z, z),

N→∞

k=1

11.10 Exercises | 327

see (7.9). The monotone convergence theorem implies that ∞

∞

k=1

ℂ k=1

∑ ‖fk (z)‖2φ = ∫ ∑ |fk (z)|2 e−φ(z) dλ(z) = ∫ Kφ (z, z)e−φ(z) dλ(z). ℂ

(11.72)

Now we use the following pointwise inequality for the Bergman kernel which is due to Marzo and Ortega-Cerda [54]: there exists a constant C > 0 such that C −1 ρ(z)−2 ≤ Kφ (z, z)e−φ(z) ≤ Cρ(z)−2 ,

(11.73)

for each z ∈ ℂ, where μ(Dρ(z) (z)) = 1 (see [54]), to conclude that A2 (ℂ, e−φ ) is of infinite dimension. Theorem 11.46. Suppose that φ is a subharmonic function such that dμ = Δφ dλ is a non-trivial doubling measure. Then the corresponding Dirac operator 𝒟 has noncompact resolvent. Proof. Since (I + 𝒟2 ) = (iI + 𝒟)(−iI + 𝒟), 𝒟2 has compact resolvent, if and only if 𝒟 has compact resolvent. Suppose that 𝒟 has compact resolvent. Since 𝒟2 = (

P− 0

0 ), P+

this would imply that both P± have compact resolvent. We know from (11.45) that P− = 4D D = 4e−φ/2 𝜕φ 𝜕eφ/2 ∗

∗

and that P− is a non-negative self-adjoint operator. It follows from Theorem 11.45 that the space of entire functions A2 (ℂ, e−φ ) is of infinite dimension. This means that 0 belongs to the essential spectrum of P− . Hence, by Theorem 11.33, P− fails to have compact resolvent, and we arrive at a contradiction.

11.10 Exercises 165. Let H be a Hilbert space and A ∈ ℒ(H) a compact self-adjoint operator. Suppose that μ ∉ σ(A) ∪ {0}. Take an arbitrary y ∈ H and show that the equation (A − μI)x = y has a uniquely determined solution x ∈ H.

328 | 11 The twisted 𝜕-complex and Schrödinger operators 166. Let A, A′ ∈ 𝒞2 (ℝn , ℝn ) be such that dA = dA′ . Suppose that V ∈ L2loc (ℝn ) and V ≥ 0. Show that H(A′ , V) = eig H(A, V)e−ig , where g ∈ 𝒞1 (ℝn ) and that σ(H(A, V)) = σ(H(A′ , V)). This is called the gauge invariance of the spectrum of H(A, V). Hint: use the Poincaré lemma to show that A − A′ = dg. 167. Consider the weighted space L2 (ℂ, e−φ ), where φ(z) = |z|2 . Show that □φ u = 𝜕 𝜕φ u = − ∗

𝜕2 u 𝜕u +z + u, 𝜕z𝜕z 𝜕z

for u ∈ 𝒞∞ 0 (ℂ). 2 168. Show that the Bergman space A2 (ℂ, e−|z| ) is a subspace of the eigenspace to the eigenvalue 1 of the operator □φ , where φ(z) = |z|2 , and that 1 ∈ σe (□φ ). 169. Consider the weighted space L2 (ℂ, e−φ ), where φ(z) = |z|α , α > 0 and set β = α/2. Let uk (z) = z k , k = 0, 1, …. Show that ‖uk ‖2φ =

π k 1 Γ( + ) β β β

and k ∗ ‖𝜕φ uk ‖2φ = πβΓ( + 1). β 170. Let φ(z1 , z2 ) = φ1 (z1 ) + φ2 (z2 ). Consider the space L2 (ℂ2 , e−φ ). Show that □φ,1 u = (−

𝜕2 u1 𝜕2 u1 𝜕φ 𝜕u 𝜕φ 𝜕u 𝜕2 φ1 − + 1 1+ 2 1+ u ) dz 1 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z1 𝜕z 1 1

+ (−

𝜕2 u2 𝜕2 u2 𝜕φ 𝜕u 𝜕φ 𝜕u 𝜕2 φ2 − + 1 2+ 2 2+ u ) dz 2 , 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z2 𝜕z 2 2

2 for u = u1 dz 1 + u2 dz 2 and u1 , u2 ∈ 𝒞∞ 0 (ℂ ), and that □φ,2 V equals to

(−

𝜕2 v 𝜕2 v 𝜕φ 𝜕v 𝜕φ2 𝜕v 𝜕2 φ1 𝜕2 φ2 − + 1 + + v+ v) dz 1 ∧ dz 2 , 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2 𝜕z1 𝜕z 1 𝜕z2 𝜕z 2

2 where V = v dz 1 ∧ dz 2 and v ∈ 𝒞∞ 0 (ℂ ). 171. Let φ(z1 , z2 ) = |z1 |2 + |z2 |2 and consider the corresponding 𝜕-Neumann operator

Nφ,1 ∶ L2(0,1) (ℂ2 , e−φ ) ⟶ L2(0,1) (ℂ2 , e−φ ). Show that Nφ,1 fails to be compact.

Hint: let uk (z1 , z2 ) = ψk (z1 ) dz 2 , where ψk (z1 ) = □φ,1 uk = uk .

z1k √πk! ,

for k ∈ ℕ, and show that

11.11 Notes | 329

172. Show that the weight function 1 1 1 φ(x, y) = (x 2 + y2 ) + (x4 y2 + x 2 y4 ) − (x6 + y6 ) 4 24 360 satisfies the condition lim ∫

|z|→∞ D (z) 1

Δφ(w) dλ(w) = +∞.

11.11 Notes Theorem 11.20 is due to I. Shigekawa [66]. A. Iwatsuka [42] and J. Avron, I. Herbst and B. Simon [6] were the first, who obtained results on the problem of compact resolvent of Schrödinger operators with magnetic field. Basic facts about Schrödinger operators with magnetic fields can be found in [16, 34, 35, 37]. The relationship to the 𝜕-equation appears in [9, 28, 31]. M. Christ [14] initiated the study of weighted spaces of entire functions, where the weight φ has the property that dμ = Δφ dλ is a non-trivial doubling measure. Witten Laplacians, as they appear in the context of semiclassical analysis, are studied in [33, 36, 45]. The Witten Laplacians were introduced by E. Witten on a compact manifold in connection with a Morse function, which is a function with nondegenerate critical points. The main idea of E. Witten was that the dimension of the kernels of these Laplacians, which are related by the Hodge theory to the Betti numbers, can be estimated from above by the dimension of the eigenspace associated with the small eigenvalues of these operators (see [72, 73, 33, 38, 39, 45]). A thorough treatment of Dirac and Pauli operators can be found in [16, 37, 69, 62].

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Index A2 (Ω) 191 Tp1,0 (bΩ) 187 Δ(0,0) 319 φ Δ(0,q) 320 φ 𝜕-Neumann operator 281, 298, 303 𝒦(H1 , H2 ) 209 ℒ(H1 , H2 ) 209 𝒱-elliptic form 253 𝜕-complex 269, 270, 297, 318 □ 271, 281, 282 □φ 297 ℂ 3 ℋ(U) 6 absolute value 2 absolutely convex 231 adjoint operator 223, 239, 297 analytic continuation 106, 108 analytic polyhedron 183 angle preserving 14 approximation to the identity 263 argument 2 Arzela–Ascoli Theorem 146 automorphism 147 Baire category theorem 242 balanced set 237 basic estimate 281 Bergman kernel 196, 200, 226 Bergman projection 196 Bergman space 191 Bernoulli numbers 97 Bessel’s inequality 197, 215 Bieberbach’s conjecture 150 biholomorphic mapping 81, 203 boundary 4 bounded family 145 bounded subset 232 branching point 62 canonical solution operator 223, 226, 284, 317 Casorati–Weierstraß Theorem 55 Cauchy product 18 Cauchy sequence 4 Cauchy’s estimates 59 Cauchy’s formula 48 Cauchy–Goursat Theorem 43

Cauchy–Riemann differential equation – inhomogeneous 12 Cauchy–Riemann equations 7 Cayley transform 187 chain 70 closable operator 239 closed graph 268 closed graph theorem 242, 243 closed operator 239, 270, 313 closed range 281 closed set 4 closure 4 closure of an operator 239 cocycle 135 cohomology-group 135 compact exhaustion 60 compact operator 209, 212, 223 compact resolvent 323, 327 compact set 5 complete differential 66 complex conjugate 2 complex differentiable function 6 complex Laplacian 271, 297 complex measure 314 complex velocity potential 157 conjugate harmonic function 155 connected component 4 connected set 4 continuous form 253 continuous function 5 continuous linear functional 119 convergent product 136 convex domain 26 convex set 126 core of the operator 252, 322 CR-function 178 curve 14 cycle 70 De Branges, L. 150 de Moivre’s formula 2 defining function 171 dense set 5 derivative of a distribution 256 diamagnetic inequality 325 differential 66

336 | Index

Dirac Delta distribution 256 Dirac operator 322, 325, 327 Dirichlet form 298 Dirichlet problem 157, 159 discrete set 5 discrete spectrum 314 distribution 255, 256 domain of holomorphy 144, 181 doubling measure 325 dual metric space 235 dual norm 195 electric potential 315 elliptic partial differential operator 315 entire function 57 equivalence relation 39 essential singularity 54 essential spectrum 314, 327 essentially self-adjoint operator 250 exact sequence 266 exhaustion function 187 exponential function 28 extended plane 3 Fefferman–Phong inequality 325 Fock space 202 formal power series 18 Fourier transform 39, 92 Fréchet space 61, 233 Friedrichs extension 252 Friedrichs’ Lemma 264 fundamental solution 257 fundamental system 232 Fundamental Theorem of Algebra 58 gauge invariance 328 Gauß–Green Theorem 259, 267 germ of a holomorphic function 109 Gram determinant 206 Gram–Schmidt process 197 graph of a function 5 Hadamard’s Three Circle Theorem 98 Hahn–Banach Theorem 117 harmonic function 155 Harnack’s inequality 167 Hartogs domain 173, 174 Hartogs triangle 207 Heisenberg group 171, 187

Hermite functions 206 Hermitian form 253 Hilbert–Schmidt operator 217, 233 holomorphic function 6, 171 holomorphic inverse 63 holomorphic logarithm 52 holomorphically convex domain 181 holomorphically convex hull 125, 127 homotopic curves 73 homotopy 107 Hörmander, L. 296 Hurwitz’ Theorem 100 identity principle 52 Identity Theorem 173 imaginary part 1 index of a cycle 70 inhomogeneous Cauchy formula 68 interior 4 interior multiplication 320 interpolation 142 irrotational flow 157 isolated singularity 53 Jacobi determinant 13 Jensen’s formula 164 Jordan Curve Theorem 39 Kohn, J.J. 288 Köthe duality 236 Köthe sequence space 234 Lagrange interpolation 101 Laplace operator 12 Laurent coefficients 82 Laurent series 82 level lines 5 Levi pseudoconvex domain 186 line integral 21 Liouville’s Theorem 57 local base 237 locally compact set 238 locally convex vector space 231 locally finite cover 111 logarithm 31 Löwner, K. 150 magnetic field 315, 317 majorant 17

Index | 337

maximum principle 57, 173 meager set 242 mean value property 48, 156, 162 Mergelyan’s Theorem 122 meromorphic function 83 Minimum Principle 61 Minkowski functional 231 Minkowski’s inequality 263 Mittag-Leffler’s Theorem 131 Monodromy Theorem 108 Montel’s Theorem 145 Morera’s Theorem 50 neighborhood 4 Newton interpolation 101 non-coercive system 273 norm 117 normal family 144 normal operator 310 nowhere dense set 242 nuclear Fréchet space 234 nuclear operator 217 null-homologous cycle 73 null-homotopic curve 73 one-parameter family 73 one-point compactification 3 open mapping 61 open set 4 open-mapping theorem 242, 246 operator norm 209 order of a zero 50 orthogonal projection 193, 313 orthonormal basis 197, 198 paracompact set 112 parallelogram rule 193, 213 parameter integral 64 Parseval’s equation 197 partial fractions decomposition 132 partition of unity 111 path 21 – inverse 21 pathwise connected set 4 Pauli operator 322, 325 peaking function 122 Persson’s Lemma 304, 322 Pick’s Lemma 154

plurisubharmonic function 185, 297, 302, 305, 306, 308, 309, 321 pointwise convergence 16 Poisson integral 158 Poisson kernel 158 polar representation 2 polarization identity 205 pole of order m 54 polydisc 170 power series 18 primitive function 24 principal argument 2 principal branch 31 principal part 81 product of complex numbers 1 pseudoconvex domain 274, 275, 280, 288, 292, 296, 304–306 punctured disc 53 quadratic form 282 radius of convergence 18 real differentiable function 7 real part 1 reflection principle 96 regular boundary point 103 relatively compact set 5 relatively open set 4 removable singularity 53 reproducing property 196 residue 85 residue theorem 85 resolution of the identity 313 resolvent of an operator 310 resolvent set 310, 312 restriction of a distribution 257 reverse Hölder class 324 Riemann mapping 204 Riemann Mapping Theorem 148 Riemann sphere 3 Riesz representation theorem 194 root function 32 rotation–dilation 5 Rouché’s Theorem 87 Runge’s Theorem 121 s-number 216 Schatten-class 216 Schrödinger operator 291, 317 Schrödinger operator with magnetic field 315

338 | Index

Schwarz Lemma 146 self-adjoint operator 196, 212, 239, 250, 252, 271, 313, 314, 327 seminorm 117 sesquilinear form 253 set of the second category 242 Siegel upper half-space 170 simply connected domain 5, 73, 151 singular boundary point 103 slit plane 31 Sobolev space 258 sourceless flow 157 spectral theorem 314 spectrum 310, 312, 313 starshaped domain 46 stereographic projection 3 Stokes’ Theorem 68 strictly Levi pseudoconvex domain 187 strictly plurisubharmonic function 185 strictly pseudoconvex domain 185 strong topology 232 subharmonic function 165, 317 support 111 support of a distribution 257 symmetric operator 250 tangential Cauchy–Riemann equations 178 tangential map 13

Taylor coefficient 49 Taylor series expansion 49 test function 255 topological vector space 231 totally bounded set 209 transcendental entire function 98 twist factor 292 twisted 𝜕-complex 291 uniform boundedness principle 243 uniform convergence 16 univalent function 150 upper halfspace 150 velocity potential 157 weak null-sequence 244 Weierstraß Factorization Theorem 138 Weierstraß’ Theorem 59 winding number 40 Wirtinger calculus 12 Wirtinger-derivatives 10 Witten, E. 329 Witten Laplacian 291, 319 zero set 50 Zorn’s Lemma 118, 198