The Theory of Distributions [1 ed.] 9781786309372
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Table of contents :
Cover
Title Page
Copyright Page
Contents
Preface
Introduction
Chapter 1. Topological Vector Spaces
1.1. Semi-norms
1.2. Topological vector space: definition and properties
1.2.1. Topology and semi-norms
1.2.2. Topological vector space
1.2.3. Locally convex topological vector space
1.2.4. Topological metrizable vector space
1.2.5. Convergence in a topological vector space
1.2.6. Applications and linear forms
1.3. Inductive limit topology
Chapter 2. Spaces of Test Functions
2.1. Multi-index notations
2.1.1. Leibniz formula
2.2. C∞ function with compact support
2.2.1. Support of a continuous function
2.2.2. Spaces of test functions
2.2.3. Convergence in D(Ω)
2.2.4. Convolution product
2.2.5. A few density results
2.3. Exercises with solutions
Chapter 3. Distributions on an Open Set of Rd
3.1. Definitions
3.1.1. Functional definition
3.1.2. Definition of order
3.1.3. Order of a distribution
3.2. Examples of distributions
3.2.1. Regular distributions
3.2.2. Non-regular distributions
3.2.3. Other examples
3.2.4. Radon measure
3.3. Convergence of sequences of distributions
3.3.1. Definition and examples
3.3.2. Other convergence results
3.4. Exercises with solutions
Chapter 4. Operations on Distributions
4.1. Multiplication by a C∞ function
4.1.1. Definition and some properties
4.1.2. Examples
4.1.3. Convergence properties
4.1.4. Solution of the equations xT = 0, xT = 1 and xT = S
4.2. Differentiation of a distribution
4.2.1. Definition and examples
4.2.2. Continuity of the differentiation operator
4.2.3. Solution of the equations T' = 0 and ∂xiT = 0
4.2.4. Jump formula in dimension 1
4.2.5. Differentiation/integration under the duality bracket
4.3. Transformations of distributions
4.3.1. Distribution translation
4.3.2. Distribution dilation
4.3.3. Distribution parity
4.3.4. Distribution homogeneity
4.4. Exercises with solutions
Chapter 5. Distribution Support
5.1. Distribution restriction and extension
5.1.1. Unit partitions
5.1.2. Distribution localization and recollement
5.2. Distribution support
5.2.1. Definition
5.2.2. Examples
5.2.3. Properties of the support
5.3. Compact support distributions
5.3.1. Definition and properties
5.3.2. Distributions with point support
5.4. Exercises with solutions
Chapter 6. Convolution of Distributions
6.1. Definition and examples
6.1.1. Convolution of two regular distributions
6.1.2. Convolution of a distribution and a function in D(Rd)
6.1.3. Density of D(Ω) in D'(Ω)
6.1.4. Convolution of two distributions
6.1.5. Some examples
6.2. Properties of convolution
6.2.1. Support of a convolution
6.2.2. Sequential continuity of the convolution product
6.2.3. Associativity and convolution
6.2.4. Differentiation and convolution
6.2.5. Translation and convolution
6.2.6. Algebraic study of D'+(R)
6.3. Exercises with solutions
Chapter 7. Schwartz Spaces and Tempered Distributions
7.1. S(Rd) Schwartz spaces
7.1.1. Definitions and examples
7.1.2. Topology and convergence in S(Rd)
7.1.3. First properties of S(Rd)
7.1.4. Operators in S(Rd)
7.2. Tempered distributions
7.2.1. Definition and examples
7.2.2. Convergence in S' (Rd)
7.2.3. First properties of S'(Rd)
7.2.4. Operators in S' (Rd)
7.3. Exercises with solutions
Chapter 8. Fourier Transform
8.1. Fourier transform in L1(Rd)
8.1.1. Definition and first properties
8.1.2. Fourier transform and operations
8.1.3. Fourier transform inversion
8.2. Fourier transform in S(Rd)
8.2.1. Definition and first properties
8.2.2. Fourier transform and operations
8.2.3. Fourier transform inversion
8.2.4. Fourier transform and convolution
8.3. Fourier transform in S'(Rd)
8.3.1. Definition and first properties
8.3.2. Fourier transform and differentiation
8.3.3. Fourier transform inversion
8.3.4. Fourier transform in E'(Rd)
8.3.5. Fourier transform and Poisson summation formula
8.3.6. Fourier transform and convolution
8.4. Exercises with solutions
Chapter 9. Applications to ODEs and PDEs
9.1. Partial Fourier transform
9.2. Tempered solutions of differential equations
9.3. Fundamental solutions of certain PDEs
9.3.1. Heat equation
9.3.2. Wave equation
Appendix
References
Index
EULA
Citation preview
The Theory of Distributions
Series Editor Nikolaos Limnios
The Theory of Distributions Introduction
El Mustapha Ait Ben Hassi
First published 2023 in Great Britain and the United States by ISTE Ltd and John Wiley & Sons, Inc.
Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Ltd 27-37 St George’s Road London SW19 4EU UK
John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA
www.iste.co.uk
www.wiley.com
© ISTE Ltd 2023 The rights of El Mustapha Ait Ben Hassi to be identified as the author of this work have been asserted by him in accordance with the Copyright, Designs and Patents Act 1988. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s), contributor(s) or editor(s) and do not necessarily reflect the views of ISTE Group. Library of Congress Control Number: 2023938463 British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library ISBN 978-1-78630-937-2
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Chapter 1. Topological Vector Spaces
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1.1. Semi-norms . . . . . . . . . . . . . . . . . . . . . . 1.2. Topological vector space: definition and properties 1.2.1. Topology and semi-norms . . . . . . . . . . . . 1.2.2. Topological vector space . . . . . . . . . . . . . 1.2.3. Locally convex topological vector space . . . . 1.2.4. Topological metrizable vector space . . . . . . . 1.2.5. Convergence in a topological vector space . . . 1.2.6. Applications and linear forms . . . . . . . . . . 1.3. Inductive limit topology . . . . . . . . . . . . . . . . Chapter 2. Spaces of Test Functions
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2.1. Multi-index notations . . . . . . . . . 2.1.1. Leibniz formula . . . . . . . . . . 2.2. C ∞ function with compact support . 2.2.1. Support of a continuous function 2.2.2. Spaces of test functions . . . . . . 2.2.3. Convergence in D(Ω) . . . . . . . 2.2.4. Convolution product . . . . . . . . 2.2.5. A few density results . . . . . . . 2.3. Exercises with solutions . . . . . . . .
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Chapter 3. Distributions on an Open Set of Rd . . . . . . . . . . . . . . 3.1. Definitions . . . . . . . . . . . . . . . . . . 3.1.1. Functional definition . . . . . . . . . . 3.1.2. Definition of order . . . . . . . . . . . . 3.1.3. Order of a distribution . . . . . . . . . . 3.2. Examples of distributions . . . . . . . . . . 3.2.1. Regular distributions . . . . . . . . . . 3.2.2. Non-regular distributions . . . . . . . . 3.2.3. Other examples . . . . . . . . . . . . . 3.2.4. Radon measure . . . . . . . . . . . . . . 3.3. Convergence of sequences of distributions 3.3.1. Definition and examples . . . . . . . . 3.3.2. Other convergence results . . . . . . . . 3.4. Exercises with solutions . . . . . . . . . . .
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37 37 38 39 39 39 41 46 46 48 48 52 55
Chapter 4. Operations on Distributions . . . . . . . . . . . . . . . . . . .
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75 76 77 77 78 81 82 87 87 91 98 100 100 102 102 103 103
Chapter 5. Distribution Support . . . . . . . . . . . . . . . . . . . . . . . . 123 5.1. Distribution restriction and extension . . . . 5.1.1. Unit partitions . . . . . . . . . . . . . . . 5.1.2. Distribution localization and recollement 5.2. Distribution support . . . . . . . . . . . . . . 5.2.1. Definition . . . . . . . . . . . . . . . . . 5.2.2. Examples . . . . . . . . . . . . . . . . . .
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123 123 125 126 126 127
Contents
5.2.3. Properties of the support . . . . 5.3. Compact support distributions . . . 5.3.1. Definition and properties . . . . 5.3.2. Distributions with point support 5.4. Exercises with solutions . . . . . . .
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Chapter 6. Convolution of Distributions
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6.1. Definition and examples . . . . . . . . . . . . . . . . . . . 6.1.1. Convolution of two regular distributions . . . . . . . . 6.1.2. Convolution of a distribution and a function in D(Rd ) 6.1.3. Density of D(Ω) in D (Ω) . . . . . . . . . . . . . . . . 6.1.4. Convolution of two distributions . . . . . . . . . . . . . 6.1.5. Some examples . . . . . . . . . . . . . . . . . . . . . . 6.2. Properties of convolution . . . . . . . . . . . . . . . . . . . 6.2.1. Support of a convolution . . . . . . . . . . . . . . . . . 6.2.2. Sequential continuity of the convolution product . . . . 6.2.3. Associativity and convolution . . . . . . . . . . . . . . 6.2.4. Differentiation and convolution . . . . . . . . . . . . . 6.2.5. Translation and convolution . . . . . . . . . . . . . . . 6.2.6. Algebraic study of D+ (R) . . . . . . . . . . . . . . . . 6.3. Exercises with solutions . . . . . . . . . . . . . . . . . . . .
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151 151 153 155 157 158 161 161 162 163 165 166 166 167
Chapter 7. Schwartz Spaces and Tempered Distributions . . . . . . . 179 7.1. S(Rd ) Schwartz spaces . . . . . . . . . . 7.1.1. Definitions and examples . . . . . . . 7.1.2. Topology and convergence in S(Rd ) 7.1.3. First properties of S(Rd ) . . . . . . . 7.1.4. Operators in S(Rd ) . . . . . . . . . . 7.2. Tempered distributions . . . . . . . . . . 7.2.1. Definition and examples . . . . . . . 7.2.2. Convergence in S (Rd ) . . . . . . . . 7.2.3. First properties of S (Rd ) . . . . . . . 7.2.4. Operators in S (Rd ) . . . . . . . . . . 7.3. Exercises with solutions . . . . . . . . . . Chapter 8. Fourier Transform 1
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8.2.1. Definition and first properties . . . . . . . . . . . . 8.2.2. Fourier transform and operations . . . . . . . . . . 8.2.3. Fourier transform inversion . . . . . . . . . . . . . . 8.2.4. Fourier transform and convolution . . . . . . . . . 8.3. Fourier transform in S (Rd ) . . . . . . . . . . . . . . . 8.3.1. Definition and first properties . . . . . . . . . . . . 8.3.2. Fourier transform and differentiation . . . . . . . . 8.3.3. Fourier transform inversion . . . . . . . . . . . . . . 8.3.4. Fourier transform in E (Rd ) . . . . . . . . . . . . . 8.3.5. Fourier transform and Poisson summation formula 8.3.6. Fourier transform and convolution . . . . . . . . . 8.4. Exercises with solutions . . . . . . . . . . . . . . . . . .
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Chapter 9. Applications to ODEs and PDEs . . . . . . . . . . . . . . . . 263 9.1. Partial Fourier transform . . . . . . . . . . . 9.2. Tempered solutions of differential equations 9.3. Fundamental solutions of certain PDEs . . . 9.3.1. Heat equation . . . . . . . . . . . . . . . 9.3.2. Wave equation . . . . . . . . . . . . . . .
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Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 References Index
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Preface
This book is a collection of course notes written in the spirit of presenting distributions in as clear a sense as possible, without overburdening them with even more interesting notions, and not to deviate from the main objective, which is to provide students with relatively simple course supporting material that contains the most essential and important results on distributions. The reader interested in more details on distributions will find a bibliographic list given at the end of the book. P.1. Summary This book is organized into nine chapters and an appendix, and almost all of them comprise a theoretical foundation in the form of definitions, propositions and theorems illustrated by examples, and a series of exercises with their answers. In Chapter 1, the essentials on topological vector spaces are presented, which will allow building topology on the space of test functions. Chapter 2 is dedicated to the presentation and study of Cc∞ : the space of test functions ends with some density results that will be useful later on. Chapter 3 introduces the notion of distributions through the presentation of equivalent definitions and various examples. Chapter 4 addresses some operations on distributions, namely multiplication by a C ∞ class function, differentiation of distributions, and classical transformations on distributions. Chapter 5 is dedicated to the notion of distribution support, its properties allowing a classification of distributions according to their support.
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In Chapter 6, the fundamental principle of convolution regularization is presented. More precisely, distribution convolution is defined and its properties are exposed. In order to define temperate distributions in Chapter 7, Schwartz spaces S(Rd ) are first defined and some properties are given that will be needed in the second part of this chapter, where the space of temperate distributions S (Rd ) is presented as well as its fundamental properties. Given the importance of the Fourier transform, Chapter 8 is dedicated to this notion in different functional spaces, namely the space L1 (Rd ), the Schwartz space S(Rd ) and finally the dual space S (Rd ) of temperate distributions. Chapter 9 presents an opportunity to apply all this theory exposed for the solution of some differential equations. An appendix concludes the book with a series of synthesis exercises that can be used for self-evaluation. P.2. Targeted audience This book is intended for master’s students in mathematics as well as for students of the physical sciences. It is also intended for students preparing for post-graduate degrees for the teaching of mathematics. It equips them with the most elementary definitions and notions that will help them familiarize with distributional calculus such as the notions of derivation, limit of a sequence or series of distributions as well as other operations on distributions. Through a range of well-chosen examples, they will be able to grasp the difference between what is classical for ordinary functions and what is related to distributions when considering what is usually referred to as “generalized functions”, which are themselves distributions. Finally, I would like to thank all the people who helped me, in one way or another, in the preparation of this book. I would especially like to thank my students and their contribution to preparing these notes. I will be grateful to those of my readers who will send me their comments about this first edition. April 2023
Introduction
The emergence of the theory of distributions can be traced back to 1945–1950 and is due to Laurent Schwartz. This theory provides a general setting with a pleasant formalism for studying functional spaces and partial differential equations. Since then, distributions are the natural framework that many analysts use basing themselves on Schwartz’s notations and ideas. A good familiarity with the language of distributions has become almost essential to any analyst. Schwartz himself was well aware that the main merit of his approach was not in the introduction of new tools, but in a clear and accessible synthesis of multiple recipes that had already been employed in various contexts. There are several reasons for introducing the concept of distribution. Some of them are purely physical (even experimental) reasons, while others are more mathematical ones; these latter, on the one hand, consist of giving meaning to some objects manipulated in physics, as for example the mysterious Dirac function introduced by Dirac (1929) which equals 0 anywhere, except at 0 where it equals +∞, and whose integral is equal to 1 (contrarily to all of the rules of Lebesgue’s theory of integration). Not only did Dirac use this function for formal computation purposes, but he would also derive it at will, simply observing that the successive derivatives were increasingly singular. On the other hand, they consist of exposing the operations, especially differentiation operations, carried out within the setting of partial differential equations. In order to motivate the generalization of the point aspect of functions and make it possible to move from the notion of function to the notion of functional, a classical physical example is given here related to the measurement of the temperature of a straight wire “at a given point”. Understandably and for obvious reasons, such a measurement is never perfectly feasible. Any thermometer, regardless of the physical principle used for the measurement, presents a spatial extension that cannot possibly
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be reduced to that of a point: what should be achieved to be able to measure the temperature at a point x0 . It can nonetheless be admitted that, in the case of a realistic temperature measurement, the thermometer takes into account all the temperatures in a “neighborhood”, of the point x0 according to a sensitivity function φ0 so that for a distribution function T (x) of the temperature along the bar (a function about which it is a priori not known what it is really equal to at a specific point of the bar), it can be said that the measured temperature T will be in fact T0 =
T (x)φ0 (x) dx. If the measurement would be taken at another point x1 , it would yield T1 = T (x)φ1 (x) dx, etc.
It can be seen that the measured temperature T , assuming that the functions T (x), φ0 , φ1 , etc., are sufficiently regular, appears as a linear expression in the sensitivity function φ. This expression can generously be expressed in the form: T, φ =
T (x)φ(x) dx
When the product x → T (x)φ(x) is Lebesgue integrable, this notation is perfectly justified. However, in many practical cases, the physical quantity being considered (which here would be T (x)) proves to be too singular for the written integral to have any meaning with a realistic choice of the function φ. The set of functions φ is called the set of test functions or trial functions. The measure of a physical quantity T is then represented by the “bracket”: T, φ independently or not of an integral form for this expression. The set of measurable quantities T by the test functions is then generically called a distribution. What should be the required minimum for the objects thus considered? 1) The set of test functions constitutes a vector space of functions. That is, any linear combination with complex coefficients of test functions is still a test function. 2) The distribution set is the set of continuous linear forms on the vector space of the test functions. 3) The result of the measure of T by φ is then the real or complex number T, φ. One of the other main interests of the theory of distributions is to allow the construction of a differential calculus that extends the ordinary differential calculus and for which any distribution is indefinitely differentiable. This theory has become an essential tool, especially in the study of partial differential equations. It has also allowed for a mathematical modernization for many physical phenomena.
1 Topological Vector Spaces
The purpose of this chapter is to present some of the tools needed to make sense of the notion of the distribution. In order to define a distribution as a continuous linear form on a space of test functions, it is necessary that a topology on this space be defined. To this end, a semi-norm is first defined, and then a topology on this space is deduced, namely the one constructed from a family of semi-norms. 1.1. Semi-norms D EFINITION 1.1.– Let E be a vector space over a field K = R, C. We call semi-norm on E any mapping P : E −→ R+ verifying: – ∀λ ∈ K, ∀x ∈ E, P (λx) = |λ|P (x); – ∀x, y ∈ E, P (x + y) ≤ P (x) + P (y). E XAMPLE 1.1.– Any norm on E is a semi-norm. E XAMPLE 1.2.– We consider KR the set of compact sets of R. Let K ∈ KR . – For f ∈ C(R, R), we consider PK (f ) = sup |f (x)| = max|f (x)|. The mapping x∈K
x∈K
PK defines a semi-norm on the space C(R, R).
– Let n ∈ N∗ . For f ∈ C n (R, R) , we consider Pn,K (f ) =
sup
|f (m) (x)| =
x∈K,m≤n
max |f (m) (x)|.
x∈K,m≤n
The mapping Pn,K defines a semi-norm on the space C n (R, R).
2
The Theory of Distributions
– The family (Pn,K )n∈N∗ is a family of semi-norms on the space C ∞ (R, R) . P ROPOSITION 1.1.– Let P be a semi-norm. We have ∀x, y ∈ E, |P (x) − P (y)| ≤ P (x − y). Proof. Let x, y ∈ E. From the triangle inequality, we have P (x) = P (x − y + y) ≤ P (x − y) + P (y). Therefore, P (x) − P (y) ≤ P (x − y). Similarly, we have P (y) = P (x − y + y) ≤ P (y − x) + P (x). Here, P (y) − P (x) ≤ P (y − x). Since P (y − x) = P (x − y), then P (y) − P (x) ≤ P (x − y). Consequently, |P (x) − P (y)| ≤ P (x − y). D EFINITION 1.2.– A family of semi-norms (Pα )α∈I is said to be separating if for all x ∈ E: x = 0 =⇒ ∃α ∈ I, Pα (x) = 0. E XAMPLE 1.3.– The families of semi-norms (PK )K∈KR , (Pn,K )K∈KR and (Pn,K )p∈N∗ ,K∈KR previously defined, respectively, on the spaces C (R, R) , C p (R, R) and C ∞ (R, R) are separating. D EFINITION 1.3.– A family of semi-norms (Pα )α∈I is said to be filtering if ∀α, β ∈ I, sup (Pα , Pβ ) ∈ Pγ , γ ∈ I . E XAMPLE 1.4.– The family (Pn,K )n∈N∗ ,K∈KR of semi-norms previously defined on the space C ∞ (R, R) is filtering. 1.2. Topological vector space: definition and properties In this section, a topology is defined from a family of semi-norms.
Topological Vector Spaces
3
1.2.1. Topology and semi-norms Let P := (Pα )α∈I be a family of semi-norms on E. D EFINITION 1.4.– We call a “P −ball” of center x0 and radius r > 0 the set Vn (x0 , r) := x ∈ E, Pi (x − x0 ) < r, for i ∈ Jn ⊆ I, Jn finite . We also write that Vn (x0 , r) = x ∈ E, maxPi (x − x0 ) < r, Jn ⊆ I, Jn finite . i∈Jn
D EFINITION 1.5.– We call “P −topology”, associated with the family of semi-norms (Pα )α∈I , the topology formed by any finite intersection or union of P −balls. This is the topology generated by the family (Pα )α∈I (i.e. the smallest one making the Pα continuous). D EFINITION 1.6.– A subset O of E is said to be open if it is empty or if, for each x0 ∈ O, there exists a finite non-empty subset Jn of I and a real r > 0 such that Vn (x0 , r) ⊂ O. Sets such as Vn (x0 , r) are open and their role is analogous to that of open balls in metric spaces. The subsets of E such as V n (x0 , r) := x ∈ E, maxPi (x − x0 ) ≤ r, Jn ⊆ I, Jn finite i∈Jn
are closed and their role is analogous to that of closed balls in metric spaces. R EMARK 1.1.– P −balls form a basis for the P −topology. T HEOREM 1.1.– If the family of semi-norms P = (Pα )α∈I is separating, then the associated P −topology is separated. In other words, ∀x, y ∈ E, if x = y, then there exists U an open set containing x and V an open set containing y such that U ∩ V = ∅. Proof. Let x, y ∈ E such that x = y. We have x = y, therefore x − y = 0. Then, ∃α ∈ I, r := Pα (x − y) > 0.
4
The Theory of Distributions
We consider U = z ∈ E, Pα (x − z) < 2r and V = z ∈ E, Pα (y − z) < 2r . U and V are two open sets of the P −topology. Let us assume that there exists z ∈ U ∩V. Then Pα (x−z)
0. For (x, y) ∈ E × E, we have Pi0 [(x + y) − (a + b)] ≤ Pi0 (x − a) + Pi0 (y − b). It follows that Pi0 [(x + y) − (a + b)] ≤ ε as soon as Pi0 (x − a) ≤ Pi0 (y − b) ≤ 2ε . Hence, the continuity of the addition at point (a, b).
ε 2
and
For the external multiplication (λ, x) −→ λx, let us establish continuity at a point (α, a) . We set an index i0 ∈ I and a real ε > 0. For (λ, x) ∈ K × E, we have Pi0 [(λx) − (αa)] ≤ |λ − α|Pi0 (x) + |α|Pi0 (x − a).
Topological Vector Spaces
5
We set a real r > 0 such that |α|r ≤ 2ε . It is observed that for x verifying Pi0 (x − a) ≤ r, we have Pi0 (x) ≤ Pi0 (a) + r and thus, |λ − α|Pi0 (x) ≤ |λ − α|(Pi0 (a) + r). Let us set a real η such that η Pi0 (a) + r ≤ 2ε . For (λ, x) ∈ K × E verifying |λ − α| < η and Pi0 (x − a) < r, we have Pi0 [(λx) − (αa)] ≤ ε, hence continuity of the multiplication at the point (α, a). It is deduced that – translations and homotheties, of ratios k = 0, are homeomorphisms of E; – the set V(a) of the neighborhoods of a point a ∈ E is the image by the translation τa of the set of neighborhoods V(0) of 0. The topology of a vector space is known as soon as the neighborhoods of 0 are known: V(a) = τa (V(0)). In other words, we have the following equivalence: V ∈ V(0) ⇐⇒ a + v ∈ V(a).
[1.1]
1.2.3. Locally convex topological vector space Reminder: let A be a subset of E. A is convex if, for every x, y ∈ A and for every λ ∈ [0, 1], we have λx + (1 − λ) y ∈ A. D EFINITION 1.8.– Let E be a topological vector space. E is locally convex if there exists a fundamental system (basis) of convex neighborhoods of the origin. T HEOREM 1.3.– Let E be a vector space. If the topology of E is defined by a family of semi-norms Pi i∈I , then E is a locally convex topological space.
6
The Theory of Distributions
Proof. If the topology of E is defined by a family of semi-norms Pi i∈I , then the sets Vn (ε) := x ∈ E, Pi (x) < ε, for i ∈ Jn ⊆ I, Jn finite , where 0 < ε < 1, form a fundamental system of convex neighborhoods of the origin. E XAMPLE 1.6.– Let Ω be an open set of Rd and 1 ≤ p < +∞. We denote by Lploc (Ω) the space of measurable functions p−locally integrable on Ω, that is, for any compact f (x)p < +∞. A semi-norm is defined by subset K of Ω, we have K
ï
f (x)p
PK (f ) =
ò p1 < +∞.
K
The family of semi-norms PK K∈KΩ determines a topology which makes Lploc (Ω) a locally convex vector space. 1.2.4. Topological metrizable vector space D EFINITION 1.9.– A metric d on a vector space E is said to be translation-invariant if, for each x, y, a ∈ E, we have d(x, y) = d x + a, y + a . Two translation-invariant metrics d and d on the same vector space E, which determine the same topology, are uniformly equivalent (i.e. the mappings IE : (E, d) −→ (E, d ) and IE : (E, d ) −→ (E, d) are uniformly continuous). When d and d are two translation-invariant and topologically equivalent metrics on E, then (E, d) is complete if and only if (E, d ) is complete. D EFINITION 1.10.– The topology T of a topological vector space E is said to be metrizable if there exists a translation-invariant metric on E which determines the topology T . T HEOREM 1.4.– The topology on a vector space determined by a separating sequence of semi-norms (i.e. I is at most countable) is metrizable. Proof. Let Pk k be a separating sequence of semi-norms on a vector space E.
Topological Vector Spaces +∞
7
Pk (x − y) . It is easy to verify that 1 + Pk (x − y) k=1 d is a translation-invariant distance on E and that this distance d determines the same topology as that defined by the sequence Pk k . For x, y ∈ E, we define d(x, y) =
2−k
E XAMPLE 1.7.– When Pk 1≤k≤m is a separating finite sequence of semi-norms on E, then max Pk and
1≤k≤m
m
r1 Pkr
, 1 ≤ r < +∞
k=1
are equivalent semi-norms that determine the same topology as the sequence Pk 1≤k≤m . D EFINITION 1.11.– We mention that E is a Frechet space if E is a locally convex topological vector space whose associated topology is metrizable and for the associated metric, E is complete. E XAMPLE 1.8.– The spaces C(R, R), C n (R, R) and C ∞ (R, R) are Frechet spaces. 1.2.5. Convergence in a topological vector space D EFINITION 1.12.– We state that a sequence xn n of E converges to an element x ∈ E if, for each neighborhood V of 0, there exists an integer m0 such that, for every integer m > m0 , we have xm − x ∈ V. Since the topology is separated, a convergent sequence possesses a single limit. In terms of semi-norms, we have the following equivalent definition. D EFINITION 1.13.– A sequence xn n of E converges to an element x ∈ E if, for each index i ∈ I and each ε > 0, there exists an integer m0 such that for each integer m > m0 , we have Pi (xm − x) ≤ ε. We can introduce the notion of Cauchy sequence in topological vector spaces. D EFINITION 1.14.– We say that a sequence xn n of E is a Cauchy sequence if, for each neighborhood V of 0, there exists an integer m0 such that, for each integer m, m > m0 , we have xm − xm ∈ V. This is expressed in terms of semi-norms as the following definition.
8
The Theory of Distributions
D EFINITION 1.15.– A sequence xn n of E is a Cauchy sequence if, for each index i ∈ I and each ε > 0, there exists an integer m0 such that for each integer m, m > m0 , we have Pi xm − xm ≤ ε. R EMARK 1.2.– Analogously, the topological notions met in undergraduate studies can be extended to the setting of vector spaces whose topology is determined by a separating family of semi-norms. 1.2.6. Applications and linear forms E and F denote two vector spaces respectively endowed with topologies determined by the families of semi-norms Pi i∈I and ql l∈L . D EFINITION 1.16.– A mapping f from E into F is continuous at point a ∈ E if, for each l ∈ L and each real ε > 0, there exists a finite non-empty subset Jn ⊂ I and a real r > 0 such that maxPi (x − a) < r =⇒ ql f (x) − f (a) < ε. i∈Jn
If f is a linear mapping, this definition can be reformulated as follows. T HEOREM 1.5.– Let f be a linear mapping from E into F . The following statements are equivalent: 1) f is continuous on E; 2) f is continuous at 0; 3) for all l ∈ L, there exists Jn ⊂ I, finite, and there exists c > 0, such that for each x ∈ E, we have ql (f (x)) ≤ c max Pi (x). i∈Jn
The characterization [1.1] of the neighborhood of a point a by the neighborhood of 0 makes it possible to easily see that f being continuous on E is equivalent to f being continuous at 0. When F = K and f is a linear form on E, we have the following result.
Topological Vector Spaces
9
T HEOREM 1.6.– Let f be a linear form on E. The following statements are equivalent: 1) f is continuous on E; 2) f is continuous at 0; 3) there exists finite Jn ⊂ I, and there exists c > 0, such that for each x ∈ E, we have |f (x)| ≤ c max Pi (x). i∈Jn
Moreover, if the family of semi-norms Pi i∈I is filtering, we have the following result. T HEOREM 1.7 (Fundamental theorem).– Let E be a locally convex topological vector space whose topology is defined by a separating and filtering family of semi-norms Pi i∈I . Let f be a linear form on E. The following statements are equivalent: 1) f is continuous on E; 2) f is continuous at 0; 3) there exists α ∈ I, ∃ c > 0, such that for each x ∈ E, we have |f (x)| ≤ cPα (x). E XERCISE 1.1.– Show the fundamental theorem 1.7. 1.3. Inductive limit topology This section addresses the notion of inductive limit topology from a point of view that will be used as another tool for defining the notion of distributions. Let Ei , Ti i∈N∗ be an increasing sequence of locally convex topological spaces such that the identity mapping id : Ei → Ei+1 is continuous for each i ∈ N∗ . We +∞ set E = Ei . Let us define on E the least fine-grained locally convex topology T i=1
making the identities id : Ei → Ei+1 continuous for i = 1, 2, · · · . It is called the inductive limit topology of E defined by the subspaces Ei . The space E endowed with this topology T is said to be inductive limit of spaces Ei i∈N∗ . For a convex set V to be a neighborhood of 0 for the inductive limit topology, it is necessary and sufficient that each intersection V ∩ Ei for all i = 1, 2, · · · is a
10
The Theory of Distributions
neighborhood of 0 in Ei . Therefore, a fundamental system of neighborhoods of the +∞ origin in E is obtained by taking all convex hulls of the form V = Conv Vi i=1
where each Vi belongs to the fundamental system of convex neighborhoods of each Ei , for all i ∈ N∗ . P ROPOSITION 1.2.– Let E be the inductive limit of Ei i∈N∗ and F a locally convex topological space. A linear map u : E −→ F is continuous if and only if the restriction ui = u|Ei of u is continuous from Ei to F for all i ∈ N∗ . Proof. If u is continuous, then every restriction ui is continuous since, by definition, the identity Ei → E is continuous. Conversely, we assume that each ui by: Ei −→ F is continuous. Let U be a convex neighborhood of 0 in F . If there exists a convex neighborhood of 0, denoted +∞ by Vi , in Ei such that ui (Vi ) ⊂ U, then, V = Conv( Vi ) is a neighborhood of 0 in i=1
E and we have u(V ) ⊂ U . Therefore, u is continuous from E into F . T HEOREM 1.8.– Let E, T be the union of an increasing sequence Ei , Ti i∈N∗ of locally convex topological spaces such that: – for all i, the identity Ei → Ei+1 is continuous; – the topology induced by Ei+1 on Ei coincides with the topology of Ei , for all i ∈ N∗ ; – Ei is a closed subspace of Ei+1 , for all i ∈ N∗ . Therefore, 1) the inductive limit topology of E induces on each Ei its original topology; 2) a subset A is bounded in the inductive topology of E if and only if there exists an index j such that A is contained and bounded in Ej . A completeness result for the inductive limit of complete spaces here follows. 1.3.– Let E, T be the inductive limit of an increasing sequence P ROPOSITION Ei , Ti i∈N∗ of locally convex separated topological vector spaces. If each Ei is complete, then E is also complete. Regarding the continuity of linear forms on an inductive limit space, we have the following result.
Topological Vector Spaces
11
1.4.– Let E, T be the inductive limit of an increasing sequence P ROPOSITION Ei , Ti i∈N∗ of locally convex topological vector spaces, and let T be a linear form on E. If each Ei is metrizable, then T is continuous if and only if it is sequentially continuous.
2 Spaces of Test Functions
Before discussing the notion of distributions, we give a few functional spaces on which are defined topological structures that will help making sense of the notions of limit and continuity within the context of distributions. In this chapter, Ω denotes an open set of Rd . 2.1. Multi-index notations A multi-index α is a d−uplet of natural numbers α = (α1 , α2 , · · · , αd ) ∈ Nd : αd 1 α2 ∀x ∈ Rd , xα = xα 1 x 2 · · · xd .
The length of a multi-index α is |α| = α1 + α2 + · · · + αd . Let α and β denote two multi-indexes. We have α ≤ β ⇐⇒ ∀i ∈ {1, · · · , d}, αi ≤ βi Å ã α! α = with α! = α1 !α2 ! · · · αd ! β β!(α − β)! ∂αϕ =
∂ α1 +α2 +···+αd ∂ α1 ∂ α2 ∂ αd ϕ. αd ϕ = α1 α2 α1 ◦ α2 ◦ · · · ◦ d ∂x1 ∂x2 · · · ∂xd ∂x1 ∂x2 ∂xα d
We also use the notation Dα ϕ which designates ∂ α ϕ, that is ∂ α ϕ = Dα ϕ.
14
The Theory of Distributions
2.1.1. Leibniz formula Binomial formula: (x + y)α =
Åαã xα−β y β β
0≤β≤α
Leibniz’s formula is an important one. Let k ≥ 1 and ϕ, ψ ∈ C k (Ω). Multinomial formula: for x ∈ Rd and k ∈ N, we have (x1 + x2 + · · · + xn )k =
k! xα . α!
|α|=k
For every multi-index α of length less than or equal to k, we have ∂ α (ϕψ) =
Åαã ∂ β ϕ∂ α−β ψ. β
0≤β≤α
. 2.2. C ∞ function with compact support This section begins by defining the support of a function. 2.2.1. Support of a continuous function D EFINITION 2.1.– Let f ∈ C 0 (Ω) be a continuous function on Ω. We call support of the function f the set supp (f ) := x ∈ Ω, f (x) = 0 . c Otherwise, supp (f ) is the largest open set where f is zero. R EMARK 2.1.– When working with Borel functions (not necessarily continuous), it is no longer reasonable to say that the support is the closure of the set of points where the function is non-zero. Actually, if we consider, for example, the function 1Q indicative of Q, it is zero almost everywhere, but as it has been defined before, its support will be R, whereas its support should reasonably be empty. This leads to redefining the support of a function, which is not necessarily continuous, as: the smallest closed set
Spaces of Test Functions
15
such that f = 0 a.e., on its complement. The support is then the complement of the largest cancellation open set almost everywhere. R EMARK 2.2.– If f and g coincide almost everywhere on Ω, then supp (f ) = supp(g). We can therefore refer to the support of an element of Lp (Ω) without having to specify which representative to choose in the equivalence class. E XAMPLE 2.1.– 1) supp 1[0,1] = supp 1[0,1[ = supp 1]0,1[ = [0, 1]; 2) supp 1Q∩[0,1] = ∅; 3) supp 1Q = ∅; 4) supp 1Qc = R; 5) supp sin = R; 6) supp (x −→ x) = R. P ROPOSITION 2.1.– Let f and g denote two continuous functions on Ω. We have – supp (f ) = ∅ ⇐⇒ f = 0 on Ω; – supp (f g) ⊂ supp (f ) ∩ supp (g); – if f ∈ C k (Ω), then for all α ∈ Nd , |α| ≤ k, supp (∂ α f ) ⊂ supp (f ). 2.2.2. Spaces of test functions We choose a space of test functions that possesses all the necessary physical and mathematical characteristics. D EFINITION 2.2.– We call space of test functions on Ω, denoted by D(Ω) or Cc∞ (Ω) the set of functions ϕ ∈ C ∞ (Ω) such that there exists a compact K ⊂ Ω verifying K = supp (ϕ): that is D(Ω) = {ϕ ∈ C ∞ , ∃K ⊂ Ω compact, K = supp (ϕ)} . ∞ (Ω) the set Let K ⊂ Ω be a fixed compact subset. We denote by DK (Ω) or CK
DK (Ω) = ϕ ∈ C ∞ (Ω), supp (ϕ) ⊂ K . Question: Is D(Ω) = ∅?
16
The Theory of Distributions
E XAMPLE 2.2.– We consider the function ρ defined by 1 ρ(x) = exp(− 1−x 2) ρ(x) = 0
if x < 1
if x ≥ 1.
· designates the Euclidean norm of Rd .
Figure 2.1. Function ρ, d = 1
Figure 2.2. Function ρ, d = 2
Spaces of Test Functions
17
We have ρ ∈ C ∞ (Rd ) and supp (ρ) = B(0, 1). Then ρ ∈ D(Rd ). Indeed, the last point is clear and the class of ρ is C ∞ for x = 1; since ρ depends on x only and that the class of x → x is C ∞ for x = 0, we merely have to prove that ρ is of class C ∞ in the case d = 1, and for x = 1 (by parity). 1 Pn (t) e t2 −1 for n ∈ N, |t| < 1, and 0 for |t| > 1, (t2 − 1)2n it can be seen by induction on n, that the class of ρ is C n and ρ(n) (1) = 0.
Since ρ(n) (t) has the form
R EMARK 2.3.– The function ρ is called the “Pic” function. nd ρ(x)dx. For n ∈ N∗ , we then set ρn (x) = ρ(nx), with I = I Rd P ROPOSITION 2.2.– The sequence of functions ρn n>0 verifies the following conditions: a) ρn ∈ D(Rd ) and ρn ≥ 0; b) supp (ρn ) ⊂ B 0, n1 ; c) ρn (x)dx = 1. Rd
D EFINITION 2.3.– Any sequence of functions satisfying (a), (b) and (c) of proposition 2.2 is called a regularizing sequence. E XAMPLE 2.3.– In this example, a method is given for constructing a C ∞ -class function with compact support. Let a, b, c and d be real numbers such that a < b < c < d. – We consider the function ψ defined by −1 ψ(x) = e t , ψ(x) = 0
t>0
otherwise.
It can be seen that ψ ∈ C ∞ (R). – We consider the function α : t −→ ψ(t)ψ(1 − t) > 0. t α(s)ds ≤ 1. – We consider the function β : t −→ 01 α(s)ds 0
18
The Theory of Distributions
ä Ä ä Ä d−x ∞ Then, the class of function ϕ : x −→ β x−a on R and equal to b−a β d−c is C 1 on [b, c] and with compact support contained in [a, d]. In addition, 0 ≤ ϕ ≤ 1. This function is called plateau function.
Figure 2.3. Plateau function. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
E XAMPLE 2.4.– If α is an indefinitely differentiable function on R and if ϕ is a function of D(R), then their product αϕ is a function of D(R). As a result, we have the following proposition. P ROPOSITION 2.3.– Let α be a function defined on R. We have the following equivalence: ∀ϕ ∈ D(R), αϕ ∈ D(R) ⇐⇒ α ∈ C ∞ (R). E XAMPLE 2.5.– Let ϕ be a function of D(R). Then, the translated function and the dilated function dλ ϕ are also in D(R) with τa ϕ defined by (τa ϕ) (x) = ϕ(x − a), where a is real and dλ ϕ is defined by (dλ ϕ) (x) = ϕ
x λ
,
where λ is a real number other than 0. E XAMPLE 2.6.– The test functions of D(R) and all their derivatives are integrable and bounded.
Spaces of Test Functions
R EMARK 2.4.–
1) We have D(Ω) =
19
DK (Ω).
K∈KΩ
2) The topology of DK (Ω) is defined by the family of semi-norms: Pn,K (ϕ) =
|∂ α ϕ(x)|,
max
x∈K,|α|≤n
|α| = α1 + α2 + · · · + αd .
L EMMA 2.1.– The space DK (Ω) is a Fréchet space. L EMMA 2.2 (Decomposition of Ω).– There exists a sequence Kn n≥1 of compacts of Ω such that: ◦
1) ∀n ≥ 1, Kn ⊂ K n+1 ; 2) Ω =
Kn =
n≥1
◦
K n;
n≥2
3) for any compact K of Ω, there exists n0 ≥ 1 such that K ⊂ Kn0 . Proof. It suffices to take 1 . Kn = x ∈ Rd , x ≤ n ∩ x ∈ Ω, d(x, Ωc ) ≥ n A family of semi-norms is then defined on D(Ω) as follows: Pn (ϕ) =
max
|∂ α ϕ(x)|.
x∈Kn ,|α|≤n
This sequence of semi-norms makes it possible to give D(Ω) a locally convex and metrizable topological vector space structure but which is not complete. 2
Counterexample: Let f be a function defined by f (x) = e−x . We consider the sequence of functions ψn n≥1 defined by
ψn (x) = 1, |x| ≤ n ψn (x) = 0 |x| > 2n.
The sequence of functions ψn f n≥1 is a Cauchy sequence in D(R); it simply converges to f but does not converge in D(R) because f does not belong to D(R) due to the fact that supp (f ) = R, which is not compact.
20
The Theory of Distributions
2.2.3. Convergence in D(Ω) D EFINITION 2.4.– It is said that a sequence ϕn n≥0 of elements of D(Ω) converges to ϕ ∈ D(Ω) if: i) there exists a compact subset K of Ω such that ∀n ∈ N, supp (ϕn ) ⊂ K; ii) ϕn and ∂ α ϕn uniformly converge on K, respectively, to ϕ and ∂ α ϕ for any multi-index α ∈ Nd . E XAMPLE 2.7.– Let ϕ ∈ D(R) be fixed. We consider the sequence ϕn n≥0 defined by ϕn (x) = ϕ x +
1 − ϕ(x), n+1
∀x ∈ R.
Let us show that ϕn −→ 0 in D(R). We have ϕ ∈ D(R), therefore there exists M > 0 such that supp (ϕ) [−M, M ] . Then ∀n ∈ N, supp (ϕn ) [−M − 1, M + 1] . Indeed, for all x ∈ R, if 1 x∈ / [−M − 1, M + 1], then ϕ(x) = 0 and ϕ(x + n+1 ) = 0. Consequently, we have ϕn (x) = 0. According to the finite increment theorem, we have (k) (x + ϕ(k) n (x) = ϕ
1 1 ) − ϕ(k) (x) ≤ supϕ(k+1) (x). n+1 n + 1 x∈R
Since supp ϕ(k) ⊂ supp ϕ, then, we have (k) x+ ϕ(k) n (x) = ϕ
1 − ϕ(k) (x) n+1
≤
1 supϕ(k+1) (x) n + 1 x∈R
≤
1 sup ϕ(k+1) (x). n + 1 x∈[−M,M ] (k)
Consequently, (ϕn ) uniformly converges to 0 on K = [−M − 1, M + 1]. Hence, ϕn −→ 0 in D(R).
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21
2.2.4. Convolution product 2.2.4.1. Definitions
Consider the measured space Rd , ML (Rd ), λd . We want to define the convolution product of two functions f and g by the formula ∀x ∈ R , f ∗ g(x) = d
Rd
f (x − t)g(t)dt.
In the case where the functions f and g are positive, their measurability is enough for this formula to make sense. Without assuming this positivity, the convolution product of f and g can still be defined provided that, in addition to their measurability, a regularity of order Lp is assumed. P ROPOSITION 2.4.– Let p ∈ [1, +∞]. Let f ∈ Lp (Rd ) and g ∈ L1 (Rd ). For almost any x ∈ Rd , the function y −→ f (x − y)g(y) is integrable. Then the convolution product f ∗ g is defined almost everywhere, moreover f ∗ g ∈ Lp (Rd ) and f ∗ gp ≤ f p g1 . In addition, we have f ∗ g = g ∗ f. Proof. The proof is given for the case where p = 1. We start it by applying the Fubini–Tonelli theorem and the change in variables theorem that the function (x, y) −→ f (x − y)g(y) is summable. In fact,
Rd
Rd
f (x − y)g(y)dxdy =
g(y) Rd
g(y)
= Rd
Rd
Rd
f (x − y)dx dy f (x)dx dy = f 1 g 1 . L L
The function (x, y) → f (x − y)g(y) is therefore summable on R2d , and Fubini’s theorem ensures that the function y −→ f (x − y)g(y) is summable for almost all x and that the function defined almost everywhere by f ∗ g(x) = is itself summable on Rd . Finally, we have f ∗ g 1 = L
Rd
f ∗ g (x)dx ≤
g(y) Rd
Rd
Rd
f (x − y)g(y)dy
f (x − y)dx dy = f 1 g 1 . L L
22
The Theory of Distributions
R EMARK 2.5.– 1) A sufficient condition for defining the convolution product f ∗ g is that the function y → f (x − y)g(y) be integrable on Rd . 2) To define the convolution product f ∗ g where f ∈ Lp (Rd ) and g ∈ Lq (Rd ), we simply have to assume that p1 + 1q > 1. A result generalizing proposition 2.4 is given by the following Hausdorf–Young inequality theorem. T HEOREM 2.1 (Hausdorf–Young inequality).– Let p, q, r ∈ [1, +∞] such that 1 1 1 1 + = 1 + , with the convention = 0, p q r ∞ and let f ∈ Lp (Rd ) and g ∈ Lq (Rd ). Then f and g are convolvable on Rd and f ∗ g ∈ Lr (Rd ). Moreover, f ∗ g verifies the inequality: f ∗ g r d ≤ f p d g q d . L (R ) L (R ) L (R ) R EMARK 2.6.– If f ∈ Lp (Rd ) and g ∈ Lq (Rd ) with p1 + 1q = 1, then f ∗g ∈ L∞ (Rd ). 2.2.4.2. Some properties of convolution The goal here is to give the main properties of convolution that are often used. 2.2.4.2.1. Translation of a convolution product Let a ∈ Rd and f a function. We denote by τa f the a−vector translation of f , that is, τa f (x) = f (x − a). P ROPOSITION 2.5.– Let f and g denote two functions such that (f ∗ g)(x) is defined for almost all x ∈ Rd and let a ∈ Rd . Therefore, τa (f ∗ g) = (τa f ) ∗ g = f ∗ (τa g). 2.2.4.2.2. Differentiation of a convolution product P ROPOSITION 2.6.– Let f ∈ L∞ (Rd ), g ∈ L1 (Rd ) and k ∈ N ∪ {∞}. It is assumed that f is of class C k and that its partial derivatives of any order are bounded. Then f ∗g is of class C k and for any α ∈ Nd such that |α| ≤ k, we have ∂ α (f ∗ g) = (∂ α f ) ∗ g.
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23
Proof. For almost every y ∈ Rd , the function x −→ f (x − y)g(y) is in C k (Rd ). Moreover, for all x ∈ Rd , α ∂ f (x − y)g(y) = ∂ α (f (x − y) g(y) ≤ ∂ α f ∞ |g(y)|. Since g ∈ L1 (Rd ), the theorem for the differentiation under the integral sign can be applied to obtain the result. C OROLLARY 2.1.– If f ∈ C p (Rd ) and g ∈ C q (Rd ), then f ∗ g ∈ C p+q (Rd ) and in addition: ∀α, β ∈ N∗ , |α| ≤ p, |β| ≤ q : ∂ α+β (f ∗ g) = (∂ α f ) ∗ (∂ β g). In particular, if f ∈ C ∞ (Rd ) and g ∈ C 0 (Rd ), then f ∗ g ∈ C ∞ (Rd ). It should be noted that the condition g ∈ L1loc (Rd ) is sufficient, as partial primitives are then continuous. 2.2.4.2.3. Convolution product support P ROPOSITION 2.7.– Let f and g be two functions such that (f ∗ g)(x) is defined for almost any x. Therefore, supp(f ∗ g) ⊂ supp(f ) + supp(g). Proof. Let x ∈ / supp(f ) + supp(g). Then x − supp(f ) ∩ supp(g) = ∅ and therefore, f ∗ g(x) =
Rd
f (x − y)g(y)dy
= (x−supp(f ))∩supp(g)
f (x − y)g(y)dy
=0 Therefore, f ∗ g(x) = 0 at any point not belonging to supp(f ) + supp(g) and therefore also at any point not belonging to its closure supp(f ) + supp(g), which gives the result. It should be recalled that, A + B = {x + y / x ∈ A, y ∈ B}, the sum of two closed sets is closed if one of them is compact and it is compact if both are compact.
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The Theory of Distributions
E XAMPLE 2.8.– Let us consider χ = 1[0,1] . We have supp (χ) = [0, 1] ,
supp (χ ∗ χ) = [0, 2] , supp (χ ∗ χ ∗ χ) = [0, 3]
supp χ ∗ . . . ∗ χ = [0, n] . ntimes
Figure 2.4. 1[0,1] → 1[0,1] ∗ 1[0,1] → 1[0,1] ∗ 1[0,1] ∗ 1[0,1]
2.2.4.2.4. Convolution and D(Rd ) P ROPOSITION 2.8.– Let f ∈ Lpc (Rd ) = {u ∈ Lp (Rd ), u = 0 outside a compact of Rd }. For all ϕ ∈ D(Rd ), we have ϕ ∗ f ∈ D(Rd ). Proof. Let a and b be two real numbers such that supp ϕ ⊂ B(0, a) and f (y) = 0 for |y| ≥ b. Let x ∈ Rd such that |x| > a + b. As in the integral which defines ϕ ∗ f , it can be assumed that x − y ∈ supp ϕ, from which |x − y| ≤ a. Then, |y| ≥ |x| − |x − y| > a + b − a = b. So, f (y) = 0. Therefore, (ϕ ∗ f )(x) = 0 for all x such that |x| > a + b. Consequently, supp (ϕ ∗ f ) ⊂ B (0, a + b) . The previous results will be used to show that it is possible to regularize a nonregular function by “convolving” it by a regular function. We start by considering a function Pic, ρ whose support is included in B(0, 1) and whose integral over Rd is 1. For ε > 0, we set ρε (x) = ε1d ρ xε . The sequence ρε is called an approximation of the identity. More precisely, the convolution operators by ρε converge to the identity operator. By convolving with the sequence (ρε ), the following approximation result is obtained:
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25
P ROPOSITION 2.9 (Bony 2001).–
1) If u ∈ Cck (Rd ) and k ∈ N, then for all α ∈ Nd such that |α| ≤ k, the sequence ∂ (ρε ∗ u) uniformly converges to ∂ α u on Rd when ε tends to 0. α
2) If u ∈ Lpc (Rd ), then the sequence ρε ∗ u converges to u in Lpc (Rd ) for all 1 ≤ p < +∞ when ε tends to 0. 2.2.5. A few density results Some density results now follow that will be useful later on. We endow Cc (Ω), the space of continuous functions with compact support in Ω, with the following topology: it is said that the sequence (un ) tends to u if for any ε > 0 there that for n ≥ n0 , we have supp(un ) ⊂ supp(u) + B(0, ε) exists n0 such and sup un (x) − u(x) < ε. x∈Ω
P ROPOSITION 2.10.– Let Ω be an open set of Rd . The space D(Ω) is dense in Cc (Ω). Proof. Let f ∈ Cc (Ω), by proposition 2.8, we indeed have ρε ∗ f ∈ D(Rd ). Moreover, since f ∈ Cc (Ω), then it is uniformly continuous and ρε ∗ f tends to f uniformly. In addition, by proposition 2.7, supp(ρε ∗ f ) ⊂ supp(f ) + B(0, ε). Therefore, ρε ∗ f indeed tends to f in Cc (Ω). The density of D(Ω) in Cc (Ω) is deduced thereof. T HEOREM 2.2.– Let Ω be an open set of Rd . The space D(Ω) is dense in Lp (Ω) for any 1 ≤ p < ∞. Proof. We consider Ωr = x, B(x, r) ⊂ Ω ∩ B(0, 1r ) the inner bounded neighborhood of Ω. Since Ω = Ωr , by Lebesgue’s theorem, it follows that: r>0
∀ε > 0, ∃r > 0,
Ω
f (x) − f (x)1Ωr p dx < εp .
Then we choose r < r such that f 1Ωr ∗ ρr − f 1Ωr Lp (Ω) < ε. This thus leads to f − f 1Ωr ∗ ρr Lp (Ω) < 2ε. Since r < r, propositions 2.7 and 2.8 imply that f 1Ωr ∗ ρr ∈ D(Ω). P ROPOSITION 2.11.– The space D(Ω) is dense in C k (Ω) for all k ∈ N.
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The Theory of Distributions
As an application of these two density results, lemma 2.3 is given, which is known as the Dubois–Reymond lemma, the proof of which is the subject of a corrected exercise. L EMMA 2.3 (Dubois-Reymond).– Let f ∈ L1loc (Ω). It is assumed that, for all ϕ ∈ D(Ω), f (x)ϕ(x)dx = 0. Then, f = 0 almost everywhere. Ω
Proof. See exercise 2.5. 2.3. Exercises with solutions E XERCISE 2.1.– Consider the sequence of functions fn (x) = x(1 − nx )n ln x χ[0,n] , for n ≥ 1 and the function f (x) = xe−x ln x. 1) Show that f is integrable on [0, +∞[. ∞ n x n 2) Show that lim x(1 − ) ln x dx = xe−x ln x dx. n →+∞ 0 n 0 S OLUTION 2.1.– 1) We have that f is continuous on ]0, 1] and extendable by continuity to the right at 0. Therefore,f is integrable on [0, 1]. On the other hand, f is continuous on [1, +∞[ x x and f (x) = O e− 2 . Since the function x → e− 2 is integrable on [1, +∞[, then f also is. Consequently, f is integrable on [0, +∞[. 2) Let n ∈ N∗ . For all x ∈ [0, n], we have (1 − nx )n ≤ e−x . Thereafter, ∀x ∈ [0, n], |fn (x)| ≤ |f (x)| . On the other hand, ∀x ∈ / [0, n], 0 = |fn (x)| ≤ |f (x)| . Hence: ∀x ∈ R, |fn (x)| ≤ f (x)χ[0,+∞[ . - The functions fn are integrable because they are upper bounded in absolute value by an integrable function. - We have ∀x ∈ R, |fn (x)| ≤ f (x)χ[0,+∞[ . - The function f χ[0,+∞[ is integrable on R because f is integrable on [0, +∞[. x - For all x ∈ R, we have lim (1 − )n = e−x and lim χ[0,n] = χ[0,+∞[ , n →+∞ n →+∞ n thus lim fn (x) = f (x)χ[0,+∞[ . n →+∞
Spaces of Test Functions
27
The assumptions of Lebesgue’s dominated convergence theorem are all verified. Consequently, it follows that
lim
n →+∞
R
fn (x)dx =
n
x(1 −
Hence, lim
n →+∞
0
R
f (x)χ[0,+∞[ dx.
x n ) ln xdx = n
∞
xe−x ln xdx.
0
E XERCISE 2.2.– Is there a function g integrable on R such that ∀n ∈ N, ∀x ∈ R, ne−n|x| ≤ g(x). S OLUTION 2.2.– We assume by contradiction that there exists a function g ∈ L1 (R) such that ∀n ∈ N, ∀x ∈ R, ne−n|x| ≤ g(x). Since the functions x → ne−n|x| are measurable and upper bounded by an integrable function, the dominated convergence theorem can be applied:
ne−n|x| dx =
lim
n→∞
R
Yet, ∀n ∈ N,
ne
−n|x|
lim ne−n|x| dx =
R n→∞
0 dx = 0. R
0
dx =
R
+∞
nx
ne dx + −∞
ne−nx dx = 1 + 1 = 2.
0
It is a contradiction! E XERCISE 2.3.– Let f ∈ L1 (R). Consider the mapping: Tf : R −→ R 1 x → f (x − y) dy. 0
1) Show that if f is continuous and has compact support, Tf is continuous. 2) Let fn n∈N be a sequence of continuous functions with compact support that converges to f in L1 (R). Show that Tfn n∈N uniformly converges to Tf on R. Thereof, deduce that Tf is continuous on R. 3) Deduce that the convolution product on L1 (R) has no unit element.
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The Theory of Distributions
S OLUTION 2.3.– 1) Let f ∈ Cc (R) such that supp f ⊂ [a, b]. Then f ∈ L1 (R) and for all fixed x ∈ R, y → f (x − y) also is in L1 (R). So, Tf is well defined. Moreover, as f is continuous, so is x → f (x − y), we have ∀x ∈ R, ∀y ∈ [0, 1], | f (x − y) |≤ M 1[a−1,b+1] (y). Since y → M 1[a−1,b+1] (y) ∈ L1 (R) and this increase is uniform in x, then by the continuity theorem under the integral sign, Tf is continuous. 1 1 2) Since Cc (R) is dense in L (R), for any f ∈ L (R), there exists a sequence of functions fn n∈N of continuous functions with compact support that converges to f in L1 (R). Therefore,
∀x ∈ R, |Tfn (x) − Tf (x)| ≤
1 0
|fn (x − y) − f (x − y)| dy ≤ fn − f .
Moreover, sup |Tfn (x) − Tf (x)| ≤ fn − f −→ 0 if n tends to ∞. Therefore, x∈R Tfn converges uniformly to Tf on R. Since every Tfn is continuous, then Tf is continuous.
3) We have Tf = f ∗ 1[0,1] . We assume by contradiction that there exists u ∈ L1 (R) such that f ∗ u = f , then Tu = u ∗ 1[0,1] = 1[0,1] . However, according to Question 2, Tu is continuous on R, whereas 1[0,1] is not. It is a contradiction! E XERCISE 2.4.– Let h ∈ Rd . The translation operator is defined by h ∈ Rd , denoted by τh , acting on a function f : Rd → R by ∀x ∈ Rd , (τh f )(x) = f (x − h). The purpose of this exercise is to show the following result: if f ∈ Lp (Rd ) with 1 ≤ p < +∞, then lim τh f − f p = 0 (i.e. τh f tends to f in Lp (Rd ) when h tends h→0
to 0). Let 1 ≤ p < +∞. 1) Show that if f is continuous and compactly supported included in the ball B(0, M ) centered at 0 and of radius M and if | h |≤ 1, then ∀x ∈ Rn , | f (x − h) − f (x) |p ≤ 1B(0,M +1) (x)2p f p∞ . 2) Deduce that, for f lim τh f − f p = 0.
continuous with compact support,
we have
h→0
3) Prove the above stated result for any function in Lp (Rd ), 1 ≤ p < +∞. 4) What happens if p = ∞?
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29
S OLUTION 2.4.– 1) If f is a continuous function with compact support in the ball B(0, M ) and if | h |≤ 1, then ∀x ∈ Rd : | f (x − h) − f (x) |p ≤
| f (x − h) | + | f (x) |
p
p ≤ 2 f ∞ 1B(0,M +1)(x) ≤ 1B(0,M +1) (x)2p f p∞ .
2) Let f be a continuous function with compact support. Since f is continuous, then lim | f (x − h) − f (x) |= 0. h→0
Then, since x → 1B(0,M +1) (x)2p f p∞ is in L1 (Rd ) and dominates the functions fh : x →| f (x − h) − f (x) |p , Lebesgue’s dominated convergence theorem can be applied: lim τh f − f pp = lim
h→0
h→0
Rd
| f (x − h) − f (x) |p dx
lim | f (x − h) − f (x) |p dx
=
Rd h→0
=
0 dx Rd
= 0. 3) Let p ∈ [1, +∞[ and f ∈ Lp (Rd ). Since Cc (Rd ) is dense in Lp (Rd ) for the norm . p , for all ε > 0 there exists fε continuous with compact support such that f − fε p ≤ ε. Then, τh f − f p = τh (f − fε ) − (f − fε ) + τh fε − fε p ≤ τh (f − fε ) p + f − fε p + τh fε − fε p = 2 f − fε p + τh fε − fε p ≤ 2ε+ τh fε − fε p . Since fε ∈ Cc (Rd ), from Question 2, there exists δ > 0 such that | h |< δ =⇒ τh fε − fε p ≤ ε. Then | h |< δ ⇒ τh f − f p ≤ 3ε. Therefore, lim τh f − f p = 0. h→0
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The Theory of Distributions
4) The theorem does not hold for p = +∞. As a matter of fact, for f = 1B(0,1) , h = 0, we have τh f − f ∞ = 1. However, for h = 0, τh f − f ∞ = 0. As such, τh f − f ∞ does not tends to 0 when h tends to 0. E XERCISE 2.5 (Dubois-Reymond).– Let I be an open set of R and f ∈ L1loc (I). It is assumed that ∀ϕ ∈ D(I),
f (x)ϕ(x) dx = 0. Let [a, b] ⊂ I be an interval. I
1) Using a density argument, show that
0
∀g ∈ C (I), supp g ⊂ [a, b],
0
f (x)g(x) dx = 0 [a,b]
| f (x) − hε | dx ≤ ε. Show 2) Let ε > 0. Let hε ∈ C (]a, b[) be such that [a,b] that ∀g ∈ C 0 (]a, b[), hε (x)g(x) dx ≤ ε g ∞ . [a,b] 3) Properly chosing g ∈ C 0 (]a, b[), show that | hε (x) | dx ≤ ε. [a,b]
4) Derive therefrom that f = 0 almost everywhere on I. S OLUTION 2.5.– 1) The space D(I) = Cc∞ (I) is dense in Cc (I) for the infinite norm. More precisely, let g ∈ Cc (I) with supp g ⊂ [a, b] ⊂ I. Then, there exists a sequence ϕn ∈ D(I) with supp ϕn ⊂ [a , b ] ⊂ I; it can be specified that [a, b] ⊂ [a , b ], such that ϕn uniformly converges to g on [a, b]. Then, b f (x) g(x) − ϕn (x) dx ≤ g − ϕn |f (x)| dx is locally integrable on I, ∞ I a and finally f (x) g(x) − ϕn (x) dx ≤ M g − ϕn ∞ , which tends to 0 when n I f (x)ϕn (x)dx = 0, therefore tends to +∞. By hypothesis, for all n ∈ N, we have I f (x)g(x)dx = f (x)g(x)dx = 0. [a,b]
I 1
2) Since f ∈ L ([a, b]), for all ε > 0 there exists hε ∈ Cc (]a, b[) such that
b
|f (x) − hε (x)| dx ≤ ε. a
Spaces of Test Functions
31
Let g ∈ Cc (]a, b[). Then,
b
b
hε (x)g(x)dx = a
hε (x) − f (x) g(x)dx +
b
f (x)g(x)dx.
a
a
Therefore, b b hε (x)g(x)dx ≤ |hε (x) − f (x)| |g(x)| dx ≤ ε g ∞ . a a
3) We set gn (x) = gn ∞ ≤ 1.
hε |hε (x)| +
Let ψn (x) = hε (x)gn (x) =
1 n
. For all n ∈ N, we have gn ∈ Cc (]a, b[) and
h2ε . | hε (x) | + n1
We have lim ψn (x) =| hε (x) | and |ψn (x)| ≤ |hε (x)|. According to Lebesgue’s n→∞ ψn (x)dx = |hε (x)| dx. dominated convergence theorem, we have lim n→+∞
Hence the result.
[a,b]
[a,b]
4) From the definition and Question 3, it can be derived that
[a,b]
|f (x)| dx = ≤
[a,b]
[a,b]
|f (x) − hε (x) + hε (x)| dx |f (x) − hε (x)| dx +
[a,b]
|hε (x)| dx
≤ 2ε. Therefore, this integral is zero. This holds for all [a, b] ⊂ I, so f is zero almost everywhere on I. E XERCISE 2.6.– Let ϕ ∈ D(R) non-identically zero. For all n ∈ N∗ , and for all x ∈ R, we set ϕn (x) = n1 ϕ(nx). Study the convergence of the sequence (ϕn )n≥1 in D(R). S OLUTION 2.6.– It is easy to see that the sequence (ϕn )n≥1 simply converges to 0. If we prove that the sequence (ϕn )n≥1 converges in D(R), it can only be toward the zero function. It can also be observed that if ϕ has its support in [−M, M ], then any
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The Theory of Distributions
ϕn has its support in [−M/n, M/n] ⊂ [−M, M ]. Therefore, all the functions ϕn have their support contained in the same compact [−M, M ] . It will be interesting to (k) see whether for all k ∈ N, the sequence (ϕn )n≥1 converges uniformly to 0 on the compact [−M, M ] . By taking the first derivative, it can be seen that ϕn (x) = ϕ (nx), consequently ϕn ∞ = ϕ ∞ . It does not necessarily tend to 0 when n tends to +∞. E XERCISE 2.7 (Borel’s theorem).– Let (an )n∈N be a sequence of complex numbers. The goal of this exercise is to show that there exist functions f : R → C indefinitely differentiable and such that ∀n ∈ N, f (n) (0) = an . 1) Using a function φ ∈ D(R) equal to 1 in a neighborhood of 0, solve the problem an when the integer series xn has a non-zero convergence radius. n! n∈N 2) Let φ ∈ D(R) with support included in [−1, 1] and equal to 1 on − 12 , 12 . We define a sequence αn as αn = 1 if | an |≤ 1 and αn =| an | otherwise. We set, for each n ∈ N, fn (x) =
an n fn (x). x φ(αn x), f (x) = n! n∈N
a) Verify that the series defining f converges normally on R (bounds can be set for each nfn (x) by separating the cases | xαn |≥ 1 and | xαn |< 1). b) Show that f is C ∞ , and calculate f (k) (0). 3) Is it always possible to obtain a function verifying the same property if we require that f be integer? S OLUTION 2.7.– 1) Let R > 0 be the radius of convergence of the power series
an n∈N
n!
xn , and let
θ be a plateau functionelement of D(R), which is equal to 1 in the neighborhood of 0 and 0 outside − R2 , R2 . We set f (x) = θ(x)
an n∈N
n!
xn if | x |< R, f (x) = 0 otherwise.
The function f is clearly C ∞ in ]−R, R[, and is zero in R\ − R2 , R2 . It is thus of class C ∞ on R, and since it coincides with the power series in the neighborhood of 0, it correctly verifies that f (n) (0) = an .
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33
2) a) It should be noted that if |x|αn ≥ 1, then fn (x) = 0. On the other hand, if |x|αn ≤ 1, we have | fn (x) |≤
| an | φ ∞ , n! αnn
which is an inequality that is in fact verified for all x in R. Now |aαnn| ≤ 1 and thus n 1 | fn (x) |≤ n! , which is the general term of a convergent series. Consequently, the series fn (x) converges normally on R. n∈N
(k) b) It is simply necessary to show that for each k, the series fn converges normally on R. We set Mk = sup | φ(k) | (this exists because the function has x∈R
compact support). Using the Leibniz formula leads to fn(k) (x)
k Å ã xn−l k = a αk−l φ(k−l) (αn x). l (n − l)! n n l=0
Again, distinguishing between | x | αn ≤ 1 and | x | αn > 1, yields for n > k:
| fn(k) (x) |≤
Å ã k l . (n − 1)!
k Mk−l l=0
The left-hand side of this inequality is the general term of a convergent series. This proves that f is of class C inf ty , and that we have f (k) (x) =
fn(k) (x).
n≥0
f
(k)
In particular, since in the neighborhood of 0, fn (x) = (0) = ak , which correctly gives the desired function.
3) No, because if the function f is integer, the series all x.
an n n! x ,
we have
an xn must converge for
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The Theory of Distributions
E XERCISE 2.8.– 1) Let f ∈ D(Rn ), a, b ∈ Rn and m ≥ 0. Prove the Taylor formula with integral remainder (b − a)α ∂ α f (a)+ α! |α|≤m (b − a)α 1 (1 − t)m ∂ α f (a + t(b − a))dt. (m + 1) α! 0 f (b) =
|α|=m+1
2) Let f ∈ D(Rn ) which cancels out at the origin. Show that there exists functions g1 , · · · , gn ∈ D(Rn ) such that ∀x ∈ Rn , f (x) = x1 g1 (x) + · · · + xn gn (x). 3) Generalize to the case where f and all its (partial) derivatives up to order m − 1 cancel out at 0. S OLUTION 2.8.– 1) Let g be the function defined on R by g(t) = f (a + t(b − a)). The Taylor formula with integral remainder for g (in one variable) applied between 0 and 1 is written as g(1) =
m k g (0) k=0
k!
1
+ 0
(1 − t)m (m+1) (t)dt. g m!
We then show by induction on k that (b − a)α g k (t) = ∂ α f (a + t(b − a)). k! α! |α|=k
This is clear if k = 1. Let us assume that the formula is proved at rank k, and prove it at rank k + 1. We indeed have d α (bj − aj )∂ α+ej f (a + t(b − a)), ∂ f (a + t(b − a)) = dt j=1 n
where ej designates (0, · · · , 0, 1, 0, · · · , 0).
Spaces of Test Functions
35
We have g (k+1) (t) 1 d g (k) (t) = (k + 1)! k + 1 dt (k)! (b − a)α d = ∂ α f (a + t(b − a)) α! dt |α|=k
n (b − a)α (bj − aj )∂ α+ej f (a + t(b − a)) α! j=1
=
|α|=k
n (b − a)α
=
α!
|α|=k j=1
(bj − aj )∂ α+ej f (a + t(b − a)).
Let β such that |β| = k + 1. It should be noted that each term of the sum n (b − a)α |α|=k j=1
α!
(bj − aj )∂ α+ej f (a + t(b − a))
corresponds to expressing β in the form α + ek . By grouping the terms equal to (b − a)β ∂ β f (a + t(bã − a)) whose number is Å 1 1 1 the following result is obtained: + ··· + k + 1 (β1 − 1)! · · · βn ! β1 ! · · · (βn − 1)! (b − a)β g k+1 (t) = ∂ β f (a + t(b − 1)). (k + 1)! β! |β|=k+1
This completes the inductive reasoning. The requested formula results immediately therefrom. 2) Applying the previous formula at order 0! yields f (x) =
n i=1
xi
1 0
(1 − t)
Let us thus set hi (x) =
1 0
∂f (tx)dt. ∂xi
(1 − t)
∂f (tx)dt. ∂xi
Each function hi is clearly of class C ∞ . But caution should be taken, and there is no requirement that hi be compactly supported. Truncating with a plateau function n will help us achieve what we are looking for. Therefore, let [−A, A] containing
36
The Theory of Distributions n
the support of f , and let θ in D(Rn ) equal to 1 on [−A, A] . Finally, we set n gi = θhi ∈ D(Rn ). For x ∈ [−A, A] , the relation f (x) = x1 g1 (x) + · · · + xn gn (x) n is clearly satisfied. If x is not in [−A, A] , both members are equal to zero and the relation sought for is satisfied as well. 3) In this case, f is written as f (x) =
xα gα (x),
|α|=m
where the function gα is in D(Rn ).
3 Distributions on an Open Set of Rd
This chapter introduces the notion of distribution as a continuous linear form on D(Ω): the space of test functions is discussed in Chapter 2. After giving the order of a distribution, a series of examples illustrating this notion is presented. In this chapter, Ω denotes an open set of Rd . 3.1. Definitions Two equivalent definitions of the notion of distribution are given, one functional and theoretical in which continuity is sequentially expressed and the other effective in which continuity is directly expressed by estimates. 3.1.1. Functional definition D EFINITION 3.1.– We call distribution on the open set Ω, any linear form T : D(Ω) −→ K(= R or C) continuous in 0 (for any sequence ϕn n∈N of elements of D(Ω) that converges to 0, the sequence T, ϕn −→ 0 in K). We will denote by D (Ω) the set of distributions on Ω. The symbol T, ϕn refers here to a duality bracket, and it simply means the action of T on ϕn : T (ϕn ). The set D (Ω) is none other than the topological dual of the space D(Ω). Convergence in D(Ω) being a very restrictive condition, the continuity condition with respect to this topology is very strong and this will thus imply many properties for distributions. Another definition that can be deduced from this functional definition is as follows.
38
The Theory of Distributions
P ROPOSITION 3.1.– A linear form T on D(Ω) is a distribution on Ω if and only if for any compact subset K ∈ KΩ , the linear form T |K is continuous.
The proof is mainly based on the fact that D(Ω) =
DK (Ω).
K∈KΩ
These abstract definitions of distributions can be used for theoretical questions, but in order to show in practice that a linear form on D(Ω) is a distribution, the following characterization is preferable, which also gives an idea about the order of the distribution. 3.1.2. Definition of order P ROPOSITION 3.2.– A linear form T on D(Ω) is a distribution on Ω if and only if, for any compact subset K of Ω, there exists m ∈ N and CK > 0 such that, for any test function ϕ ∈ D(Ω) such that supp ϕ ⊂ K, | T, ϕ | ≤ CK
max
x∈K,|α|≤m
|∂ α ϕ(x)|.
N OTATION .– We keep denoting Pm (ϕ) =
max
x∈K,|α|≤m
|∂ α ϕ(x)|, which implicitly
depends on the compact subset K initially. Proof. We assume that T ∈ D (Ω) and will follow a reasoning by contradiction. For this, we assume that there exists a compact K ⊂ Ω on which ∀m ∈ N, ∀c > 0, ∃ ϕm ∈ D(Ω) such that supp (ϕ) ⊂ K and | T, ϕm | > cPm (ϕ). We consider that ϕm m ϕ˜m = T,ϕ . Since T is linear, then T, ϕ˜m = T,ϕ ˜m ⊂
T,ϕm = 1. Moreover, supp ϕ m K because supp ϕ˜m = supp ϕm . Consequently, ϕ˜m ∈ D(Ω). On the other hand, we have Pm (ϕ˜m ) =
Pm (ϕ) | T,ϕm |
Let k ∈ N be fixed, and let us show that Pk (ϕ˜m ) We have ∀m ≥ k, Pk (ϕ˜m ) ≤ Pm (ϕ˜m )
−→
0 such that, for any test function ϕ ∈ D(Ω), supp ϕ ⊂ K, we have |T, ϕ| ≤ CK
max
x∈K,|α|≤m
|∂ α ϕ(x)| .
The smallest possible integer m is called the order of the distribution T. 3.2. Examples of distributions In this section, some basic examples of distributions are presented. 3.2.1. Regular distributions One of the first things to verify is that the theory of distributions properly generalizes the theory of classical functions, typically integrable functions. We will therefore show how the functions L1loc (Ω) are injected in D (Ω). P ROPOSITION 3.3.– Let f ∈ L1loc (Ω). A distribution on D(Ω), denoted by Tf , can be associated with it, such that ∀ϕ ∈ D(Ω), Tf , ϕ = f (x)ϕ(x)dx. This distribution is of order 0.
Ω
40
The Theory of Distributions
Proof. 1) Linearity: The linearity of the mapping Tf : ϕ −→ trivial.
f (x)ϕ(x)dx is Ω
2) Continuity: Since f is in L1loc (Ω), its restriction to any compact set is integrable on this compact. Let K be a compact of subset Ω and let ϕ ∈ D(Ω), such that supp ϕ ⊂ K. The function f is integrable on K. Since f is bounded, because it is continuous on the compact K, it can be inferred that f (x)ϕ(x)dx ≤ max |ϕ(x)| |f (x)| dx. x∈K Ω
K
The linear form Tf : ϕ −→ of order 0.
f (x)ϕ(x)dx is thus a distribution, which is at most Ω
R EMARK 3.1.– The mapping f ∈ L1loc (Ω) −→ Tf ∈ D (Ω) is linear and injective. It can be written that L1loc (Ω) → D (Ω). E XAMPLE 3.1.– Let f and g be two locally integrable functions. If f and g are equal almost everywhere (i.e. f (x) = g(x) a.e.), then their associated distributions are equal (i.e. Tf = Tg ). Reciprocally, using the Dubois–Reymond lemma, we have the following result. T HEOREM 3.1.– Let f be a locally integrable function. We have Tf = 0 ⇐⇒ f = 0 a.e. D EFINITION 3.3.– A distribution T is regular if it is associated with a locally integrable function f . N OTATION .– When there is no ambiguity in the context, it is very often simpler to write Tf simply as f . On the other hand, the notation f (x), which is tolerated for a function f , is less so for the distribution Tf , because it gives the false impression that it is possible to know the value of a distribution at a precise point x. Some authors also write it as Tf = [f ] .
Distributions on an Open Set of Rd
41
E XAMPLE 3.2 (Constant distribution).– The constant (regular) distribution C is defined by ∀ϕ ∈ D(R),
C, ϕ =
Cϕ(x) dx = C R
ϕ(x) dx. R
In particular, C = 0 defines the (regular) null distribution. E XAMPLE 3.3 (Heaviside).– Similarly, the Heaviside function H(x) =
1,
if x ≥ 0
0,
otherwise
defines the Heaviside distribution H as
∀ϕ ∈ D(R),
H, ϕ =
H(x)ϕ(x) dx = R
ϕ(x) dx, R+
and the function 2H(x) − 1 = sign(x) defines a distribution and is denoted by signx . 3.2.2. Non-regular distributions D EFINITION 3.4.– Any distribution that is not regular is said to be singular or non-regular. 3.2.2.1. Dirac distribution E XAMPLE 3.4 (Dirac distribution).– Let a ∈ Ω. The linear form δa : D(Ω) −→ C defined by ∀ϕ ∈ D(Ω), δa , ϕ = ϕ(a) is a distribution on Ω of order 0. This distribution is called Dirac mass at point a. Proof. Let K be a compact of Ω. Let ϕ ∈ D(Ω), such that supp ϕ ⊂ K. Then, |δa , ϕ| = |ϕ(a)| ≤ max|ϕ(x)|. x∈K
The linear form δa is thus indeed a distribution, which is of order 0. The distribution δa is not associated with any function of L1loc (Ω).
42
The Theory of Distributions
As a matter of fact, let us assume that there exists f ∈ L1 (Ω) such that δa = Tf . f (x)ϕ(x)dx = ϕ(a), ∀ϕ ∈ D(Ω). That is, δa , ϕ = Ω
˜ = Ω − {a}. Then, δa , ϕ = We set Ω
Ω
˜ f (x)ϕ(x)dx = 0, ∀ϕ ∈ D(Ω).
Therefore, according to the Dubois–Reymond lemma, f = 0 almost everywhere in ˜ and thus almost everywhere in Ω. Therefore, Ω Ω
f (x)ϕ(x)dx = 0 for all ϕ ∈ D(Ω).
Consequently, ∀ϕ ∈ D(Ω), ϕ(a) = 0, which does not make sense. N OTE OF CAUTION .– Books about physics and electronics in their great majority employ the notation δ(x) for the distribution δ and δ(x − x0 ) for the distribution δx0 . This notation is a constant source of error in the calculations, since it implies that these distributions have a given value at a given point. To avoid errors, it is therefore strongly recommended that the computations are performed in the sense of the distributions, even if it requires presenting the final formulas as physicists and experts in electronics traditionally do. 3.2.2.2. The principal value of
1 x
1 1 E XAMPLE 3.5 (The principal value of ).– Consider the function x −→ from R in x x R. This function is not locally integrable on R; but it is on R∗ . We will see how it can be extended to R, the distribution that this function defines on R∗ . P ROPOSITION 3.4.– The linear form denoted Vp
1 defined by: x
≠ ∑ ϕ(x) ϕ(x) 1 Vp , ϕ = lim+ dx + dx x x x ε→0 xε ϕ(x) dx, ∀ϕ ∈ D(R), = lim+ x ε→0 |x|>ε is a first-order distribution. This distribution is called the principal value of
1 . x
Distributions on an Open Set of Rd
43
Proof. First method: Let ϕ ∈ D(R) be such that supp (ϕ) [−A, A]. There exists ψ ∈ C ∞ (R) such that ϕ(x) = ϕ(0) + xψ(x) with ψ(0) = ϕ (0). We have ≠
ñ − ô − A A Å ã ∑ 1 dx dx Vp ψ(x)dx + ψ(x)dx , ϕ = lim+ ϕ(0) + + x x →0 −A x −A A ψ(x)dx. = −A
Hence Å ã Vp 1 , ϕ ≤ 2A max |ψ| ≤ 2A max |ϕ |, x x∈[−A,A] x∈[−A,A] ϕ(x) − ϕ(0) because for x = 0, we have ψ(x) = = ϕ (θx), 0 < θ < 1 and x 1 ψ(0) = ϕ (0). Consequently, Vp is at most a first-order distribution on R. x Second method: It can be observed that by using a change in variable, we obtain ≠ Å ã ∑ ϕ(x) − ϕ(−x) 1 Vp , ϕ = lim+ dx. x x ε→0 x>ε We have |x|>ε
ϕ(x) dx = x =
A ε
ϕ(x) − ϕ(−x) dx x
ϕ(x) − ϕ(−x) ln x
A ε
A
−
ϕ (x) + ϕ (−x) ln xdx.
ε
But ϕ(x) = ϕ(0) + xψ(x), on the one hand, ϕ(ε) = ϕ(0) + εψ(ε) and ϕ(−ε) = ϕ(0) − εψ(−ε), on the other hand. Hence ϕ(ε) − ϕ(−ε) ln(ε) = ε ln(ε)ψ(ε) + ε ln(ε)ψ(−ε). Since ε ln(ε) −→ 0 and ψ(ε) −→ ψ(0), we have ε ln(ε)ψ(ε) −→ 0 and ε −→0
ε ln(ε)ψ(−ε) −→ 0. Therefore,
ε −→0
ε −→0
A A ≠ Å ã ∑ 1 Vp ϕ (x) ln |x|dx − ϕ (−x) ln |x|dx ,ϕ = − x 0 0 +A ϕ (x) ln |x|dx. =− −A
ε −→0
44
The Theory of Distributions
Let K be compact subset of R, and ϕ ∈ D(R) such that supp (ϕ) ⊂ K. We have ≠ Å ã ∑ 1 Vp ϕ (x) ln |x|dx. ,ϕ = − x K Hence, ≠ Å ã ∑ Vp 1 , ϕ ≤ max|ϕ (x)| ln |x|dx. x∈K x K
Then, Vp
1 is a first-order distribution, at most. x
Show that Vp
1 is not a zeroth-order distribution. x
R EMARK 3.2.– 1 is a first-order distribution, it can then never be regular. x 2) It is essential, in the definition, to specify that we integrate in the complementary of symmetrical neighborhoods of the origin. To prove this point, we simply consider 1 hε which takes 0 in the interval [−ε, ε2 ] and takes outside. It can be shown that the x sequence (hε ) does not have a D (R) limit. 1) Since Vp
1 3) In this example, the imparity of the function was used to define the x 1 distribution Vp as the limit, when ε −→ 0, of the distribution associated with the x 1 (x) . For a non-impair function, such a procedure does not work and function {x:|x|>ε} x this makes that a corrective (divergent) term has to be introduced to compensate for this defect by using the finite part method. E XAMPLE 3.6 (The finite part of Pf H(x) defined on D(R) by x
H(x) x ).–
Consider the linear form denoted by
¨ H(x) ∂ ∀ϕ ∈ D(R), Pf x , ϕ = lim ε−→0
Pf
H(x) x
+∞ ε
ϕ(x) dx + ϕ(0) log(ε) . x
is a first-order distribution.
1 1 E XAMPLE 3.7 (The finite part of 2 ).– The function x −→ 2 is not locally x x integrable on R, so a regular distribution cannot be associated therewith. To eliminate
Distributions on an Open Set of Rd
+∞
the divergent part of the integral 0
integral is defined by
ϕ(x) x2 dx
45
for ϕ ∈ D(R), the finite part of this
+∞ ≠ ∑
1 ϕ(x) ϕ(x) + ϕ(−x) − 2ϕ(0) ϕ(0) = Pf 2 , ϕ = lim dx − 2 dx. 2 ε−→0 x x ε x2 |x|≥ε 0 Pf
1 indeed defines a distribution. x2
E XAMPLE 3.8 (The finite part of xα ).– We can attempt to continue to integrate non-integrable functions, for example xα for −2 < α < −1 and x > 0. It is verified that +∞ +∞ +∞ xα ϕ(x)dx = xα ϕ(0)dx + xα+1 ϕ (0)dx + · · · . ∀ϕ ∈ D(R), ε
ε
ε
α+1
The first term is − εα+1 ϕ(0) and tends to +∞ when ε −→ 0. This is the infinite part of xα . More precisely, we have the following equality:
+∞
∀ϕ ∈ D(R), ε
εα+1 x ϕ(x)dx = − ϕ(ε) − α+1
+∞
α
ε
xα+1 ϕ (x)dx. α+1
xα+1 is integrable because α + 1 > −1; it α+1 therefore defines a distribution. The finite part thus becomes visible. On the other hand, the function
D EFINITION 3.5.– The finite part of xα , denoted by Pf(xα ), is the distribution defined by: Ç +∞ å εα+1 α α ∀ϕ ∈ D(R), Pf(x ), ϕ = lim x ϕ(x)dx + ϕ(ε) ε −→0 α+1 ε +∞ α+1 x =− ϕ (x)dx. α +1 0 Likewise, Pf(xα ) can be defined for α ∈] − n − 1, −n[, based on the following expression: ∀ϕ ∈ D(R),
Pf(x ), ϕ = (−1) α
+∞
n 0
xα+n ϕ(n) (x)dx. (α + 1)(α + 1) · · · (α + n)
In this case, Pf(xα ) defines a nth-order distribution.
46
The Theory of Distributions
3.2.3. Other examples This section starts by establishing the link between the theory of distributions and the theory of measures. Then another example of distributions is given, namely, the positive distribution. 3.2.4. Radon measure E XAMPLE 3.9 (Radon measure).– Let μ be a Radon measure on Ω. The linear form ϕ ∈ D(Ω) −→ ϕdμ is a distibution of order 0 on Ω. Ω
T HEOREM 3.2 (Admitted).– Let T ∈ D (Ω) be a distribution of order 0. Then, there exists a Radon measure μ on Ω such that ∀ϕ ∈ D(Ω),
T, ϕ =
ϕdμ. Ω
The proof of this theorem is based on the fact that, from the Riesz representation theorem, positive Radon measures on Ω identify with positive linear forms on Cc0 (Ω)
as ϕ : μ −→ ϕ(μ) with ϕ(μ) : f ∈ Cc0 (Ω) −→
f dμ. Ω
3.2.4.1. Positive distributions E XAMPLE 3.10 (Positive distribution).– A distribution T ∈ D (Ω) is said to be positive when: ∀ϕ ∈ D(Ω), ϕ ≥ 0 =⇒ T, ϕ ≥ 0. A positive distribution is a zeroth order distribution. Furthermore, let K a compact of Ω and let χ ∈ D(Ω), χ = 1 on K and 0 ≤ χ ≤ 1. If ϕ ∈ D(Ω), supp ϕ ⊂ K and ϕ is real, then the functions ψ± : x −→ χ(x)max|ϕ(x)| ± ϕ(x) x∈K
are in D(Ω) and are positive. Then, T, ψ± ≥ 0, from which |T, ϕ| ≤ |T, χ| max|ϕ(x)| = CK max|ϕ(x)|. x∈K
x∈K
This means that T is at most of zeroth order.
Distributions on an Open Set of Rd
47
3.2.4.2. Infinite order distribution E XAMPLE 3.11 (Infinite order distribution).– Let T be the linear form on D(R) +∞ ϕ(n) (n). T is a distribution. In fact, since a defined by ∀ϕ ∈ D(R), T, ϕ = n=0
compact of R contains only a finite number of integers n ∈ N, the above sum is a finite sum of non-zero terms. Let K be a compact of R and N = max(N ∩ K). Then, for any ϕ ∈ D(R) such that supp ϕ ⊂ K, we have N (n) ϕ (n) |T, ϕ| = n=0
≤
N n=0
max ϕ(n) (x) x∈K
≤ (N + 1)
max
x∈K,0≤n≤N
(n) ϕ (x)
≤ (N + 1) PN,K (ϕ). Hence T is a distribution. We can also show that T is a distribution of infinite order (see exercise 3.8). Let N ∈ N be fixed, and let K be a compact of R and ϕ ∈ D(R) such that N supp ϕ ⊂ K. We set TN , ϕ = ϕ(n) (n). Therefore, we can see that TN is linear. n=0
On the other hand, since supp ϕ(n) ⊂ supp ϕ, then |TN , ϕ| ≤
N n=0
max|ϕ(n) (x)| ≤ (N + 1) max |ϕ(n) (x)| = (N + 1)PN,K (ϕ). x∈K
x∈K,n≤N
Therefore, TN is at most a N th-order distribution. It can be observed that T, ϕ =
lim TN , ϕ .
N →+∞
R EMARK 3.3.– 1) Implicitly, we can see that D (Ω) is a K−vector space with K = R or C.
48
The Theory of Distributions
2) The above example leads to defining the notion of convergence in the space of distributions D (Ω). 3.3. Convergence of sequences of distributions As sequences of continuous linear mappings, distribution sequences behave in a very “simple” way. This is mainly due to the Banach–Steinhaus theorem, which is a result of the uniformization of bounds on families of continuous linear forms on a Banach space. We start by giving the definition of convergence in D (Ω). 3.3.1. Definition and examples In the space of distributions D (Ω), the weak topology is the topology generated by the family of semi-norms defined by Pϕ (T ) = |T, ϕ| , ϕ ∈ D(Ω), T ∈ D (Ω), making D (Ω) a locally convex space. The convergence for this topology is the simple convergence. More precisely, we have the following definition. D EFINITION 3.6.– We say that the sequence of distributions (Tn )n of D (Ω) converges to the distribution T if lim Tn , ϕ = T, ϕ for any test function ϕ in D(Ω). n→∞
R EMARK 3.4.– If this limiting distribution T exists, it is unique. E XAMPLE 3.12.– Let Tn n≥1 be a sequence of D (R) defined by: Tn = Teinx . We have Teinx −→ 0. D (R)
As a matter of fact, let ϕ ∈ D(R). There exists A > 0 such that supp ϕ [−A, A]. We have Tn , ϕ =
+∞
A
einx ϕ(x)dx
= ï =
einx ϕ(x)dx
−∞
−A
1 inx e ϕ(x) in
òA −A
−
1 in
A −A
einx ϕ (x)dx.
Distributions on an Open Set of Rd
1 Therefore, | Tn , ϕ | ≤ n ∀ϕ ∈ D(R), Tn , ϕ
−→
n−→+∞
A −A
ϕ (x) ≤
2 Aϕ ∞ n
−→
n−→+∞
49
0. Consequently,
D (R)
0. Then, Teinx −→ 0.
R EMARK 3.5.– D (R)
D (R)
1) As a result, we have Tcos nx −→ 0 and Tsin nx −→ 0. 2) We have just given an example of a sequence of functions (cos nx)n , which does not converge as a sequence of functions but which converges as a sequence of distributions. D (R)
E XAMPLE 3.13.– Let us show that δn −→ 0. Indeed, we have δn , ϕ = ϕ(n) and since ϕ has compact support, then lim
n−→+∞
δn , ϕ =
lim
n−→+∞
ϕ(n) = 0.
E XAMPLE 3.14 (Sequences of functions converging to δ).– Let fn be a sequence fn (x)dx = 1 and that fn (x) be zero of integrable positive functions such that R
outside an interval [−εn , +εn ] with lim εn = 0. Then, the sequence Tfn of regular distributions defined by fn tends to δ. Indeed, let ϕ ∈ D(R). We have
R
fn (x)ϕ(x)dx =
R
fn (x) (ϕ(x) − ϕ(0)) dx + ϕ(0)
R
fn (x)dx.
According to the finite increment theorem, since fn is zero outside an interval [−εn , +εn ], we have |ϕ(x) − ϕ(0)| ≤ |x| sup |ϕ (x)| ≤ εn sup |ϕ (x)| . |x|≤εn
|x|≤εn
We have lim εn = 0, thus there exists n0 ∈ N such that ∀n ≥ n0 , [−εn , +εn ] [−1, 1]. Hence, for all n ≥ n0 , we have sup |ϕ (x)| ≤ sup |ϕ (x)| . |x|≤εn
|x|≤1
Therefore, ∀n ≥ n0 , ∀x ∈ [−εn , + εn ], we have |ϕ(x) − ϕ(0)| ≤ εn C where
50
The Theory of Distributions
C = sup |ϕ (x)| is a constant independent of n. Since fn is a sequence of |x|≤1
positive functions, we then have fn (x) (ϕ(x) − ϕ(0)) dx ≤ R
[−εn ,+εn ]
fn (x) |ϕ(x) − ϕ(0)| dx
≤ Cεn
R
fn (x)dx = Cεn
−→
n−→+∞
0.
Similarly, by hypothesis on the sequence fn , we have ϕ(0) fn (x)dx = ϕ(0). R fn (x)ϕ(x)dx −→ ϕ(0) = δ, ϕ . Hence the desired result. It is deduced that R
n−→+∞
Using the same demonstration, the result of this example can be generalized as follows. P ROPOSITION 3.5.– If (fn ) is a sequence of functions such that i) fn (x) ≥ 0 for |x| ≤ C, C > 0 a fixed constant, ii) fn uniformly converges to 0 in any interval not containing the origin, a iii) for all a > 0 (fixed), lim fn (x)dx = 1, n−→+∞
−a
then the sequence (fn ) converges to δ the Dirac distribution. N OTE 3.1.– A sequence of regular distributions can converge to a regular distribution as it can converge to a singular distribution as can be seen in example 3.14. E XAMPLE 3.15 (Sequences of functions converging to δ).– Let f be an integrable f (x)dx = 1. Consider, for ε > 0, the regular distribution Tfε function verifying R
defined by the function fε = 1ε f ( xε ). Then Tfε tends to δ when ε −→ 0. E XAMPLE 3.16 (Dipole of moment μ = 1).– Let Tn n≥1 be a sequence of D (R) defined by: ∀n ∈ N∗ , Tn = n δ n1 − δ −1 . n
We have Tn −→ T with T defined by T, ϕ = 2ϕ (0). D (R)
Distributions on an Open Set of Rd
51
Indeed, let ϕ ∈ D(R). We have ¨ ∂ Tn , ϕ = n δ n1 − δ −1 , ϕ n
1 −1 = n ϕ( ) − ϕ( ) . n n
1
We recall that ϕ(x) = ϕ(0) + xψ(x) where ψ(x) =
ϕ (tx)dt. Therefore,
0
1 1 1 ϕ( ) = ϕ(0) + ψ( ) n n n
and
ϕ(
−1 1 −1 ) = ϕ(0) − ψ( ). n n n
Consequently, 1 −1 1 −1 n ϕ( ) − ϕ( ) = ψ( ) + ψ( ) −→ 2ψ(0) = 2ϕ (0). n n n n D (R)
Then, Tn , ϕ −→ T, ϕ . Hence, Tn −→ T. We will see in Chapter 4 that T is exactly T = −2δ0 . Analogously to the sequences of distributions, the convergence of a series of distributions can be defined. D EFINITION 3.7.– We consider the sequence of distributions Tn n≥1 . It is said that Tn defines a distribution T of D (Ω), if the sequence of the series of distributions n≥1
partial sums SN =
N
Tn converges in the sense of distributions to T .
n=1
In practice, we very often find series similar to
∞
Tn . In this case, the definition
n=−∞
of convergence is quite similar to the previous case: in fact, we merely have to consider N the convergence of the sequence of symmetric partial sums, SN = Tn . n=−N
E XAMPLE 3.17.– The sequence ΔN distribution denoted
nZ
=
N
δn converges in D (R) to a
n=−N
δn , which is called Cha and written III (a letter from the
Cyrillic alphabet). In physics, it is also called “Dirac comb”.
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The Theory of Distributions
3.3.2. Other convergence results When a sequence of distributions (Tfn ) is defined by locally integrable functions fn , we have to clearly distinguish the difference between convergence in the classical sense (simple convergence) and convergence in the sense of distributions. To prove this point, we only have to look at example 3.12. Theorem 3.3 characterizes the sequences of functions for which the simple limits coincide with the limits in D (Ω). T HEOREM 3.3.– Let fn be a sequence of locally integrable functions in Ω that converges to f almost everywhere in Ω, and let g ∈ L1loc (Ω) such that D (Ω)
∀n ∈ N, |fn | ≤ g, then fn converges to f in D (Ω), that is, Tfn −→ Tf . Proof. Let K ⊂ Ω be a compact and let ϕ ∈ D(Ω) be such that supp ϕ ⊂ K. We have Tfn , ϕ =
Ω
fn (x)ϕ(x)dx.
Since fn converges to f almost everywhere in Ω, fn ϕ therefore converges to f ϕ almost everywhere in Ω. Moreover, we have |fn ϕ| ≤ |gϕ|. Then, by application of Lebesgue’s dominated convergence theorem, we have lim
n−→+∞
Tfn , ϕ =
lim
n−→+∞
=
Ω
fn (x)ϕ(x)dx
lim (fn (x))ϕ(x)dx
Ω n−→+∞
=
f (x)ϕ(x)dx Ω
= T, ϕ . D (Ω)
Consequently, Tfn −→ Tf . N OTE 3.2.– The convergence almost everywhere on Ω does not imply the convergence in D (Ω). Counterexample: We consider the sequence fn n≥1 of L1loc (Ω) defined by fn : x −→
√
2
ne−nx .
a.e. We have for all x = 0 fn (x) −→ 0, that is fn −→ 0, but the sequence Tfn n≥1 √ converges in D (Ω) to πδ0 and not to the null distribution. Indeed, let ϕ ∈ D(Ω).
Distributions on an Open Set of Rd
53
We have Tfn , ϕ =
√
=
2
e−nx ϕ(x)dx
n R
2 y e−y ϕ( √ )dy. n R
Based on the dominated convergence theorem, we obtain
√ 2 y e−y ϕ( √ )dy −→ πϕ(0). n R
√ Hence, Tfn n≥1 converges in D (Ω) to πδ0 . Theorem 3.3 generalizes to the spaces Lploc (Ω), 1 ≤ p ≤ +∞ as follows. T HEOREM 3.4.– The convergence in Lploc (Ω), 1 ≤ p ≤ +∞ implies the convergence in D (Ω). In other words, Lp (Ω)
D (Ω)
−→ f ⇒ Tfn −→ Tf . fn loc Proof. Let fn be a sequence of functions in Lploc (Ω) that converges in Lploc (Ω) to a function f of Lploc (Ω). Let K ⊂ Ω denote a compact subset and ϕ ∈ D(Ω) such that supp ϕ ⊂ K. We have |Tfn , ϕ − Tf , ϕ| = |Tfn − Tf , ϕ| = fn − f (x)ϕ(x)dx Rn = fn − f (x)ϕ(x)dx K
ï
fn (x) − f (x)p dx
≤ K
where q is such that
1 p
+
1 q
ò p1
ϕ(x)Lq ,
= 1.
It should be noted that ϕ ∈ Lq (Ω) because it is in D(Ω).
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The Theory of Distributions
Lp (Ω)
ï
fn (x) − f (x)p dx
Since fn loc −→ f , then
ò p1
K
−→
n−→+∞
0.
D (Ω)
Moreover, we have ϕLq < ∞, then Tfn −→ Tf . d T HEOREM 3.5 (Uniform boundedness principle).– Let Ω an open subset of R and K ⊂ Ω a compact subset. Let Tn n≥0 a sequence distributions on Ω such that for all ϕ ∈ D(Ω) with support in K, the sequence (Tn , ϕ)n∈N admits a limit in K = R or C. Then ∃m ∈ N, ∃c > 0, such that
∀ϕ ∈ D(Ω) s.t. supp ϕ ⊂ K,
sup| Tn , ϕ | ≤ cPm,K (ϕ).
n∈N
R EMARK 3.6.– 1) The proof is based on the adaptation of the Banach–Steinhaus theorem. 2) The constant c and the integer m are independent of n. C OROLLARY 3.1.– Let Tn n≥0 be a sequence of distributions on Ω such that for all ϕ ∈ D(Ω) the sequence (Tn , ϕ)n∈N admits limits in K = R or C. Then the linear form T defined on D(Ω) by ∀ϕ ∈ D(Ω),
T, ϕ =
lim
n−→+∞
Tn , ϕ
is a distribution. Proof. Therefore, we simply have to verify that this linear form T is continuous on D(Ω). Let K ⊂ Ω be a compact subset. According to the principle of uniform boundedness, there exists an integer m ∈ N, and a constant cK > 0 such that ∀ϕ ∈ D(Ω)s.t. supp ϕ ⊂ K,
sup| Tn , ϕ | ≤ cK Pm,K (ϕ).
n∈N
By passing to the limit in the above inequality, it can be seen that | T, ϕ | ≤ cK Pm,K (ϕ) for any function ϕ ∈ D(Ω) s.t. supp ϕ ⊂ K, which ensures that T ∈ D (Ω).
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55
C OROLLARY 3.2 (Sequential continuity of the duality bracket).– Let (Tn )n≥0 be a sequence of distributions on Ω and (ϕn )n≥0 be a sequence of elements of D(Ω). If Tn
D (Ω)
−→
n−→+∞
T and ϕn
D(Ω)
−→
n−→+∞
ϕ, then Tn , ϕn
−→
n−→+∞
T, ϕ .
Proof. We have |Tn , ϕn − T, ϕ| = |Tn , ϕn − Tn , ϕ + Tn , ϕ − T, ϕ| = |Tn , ϕn − ϕ + Tn − T, ϕ| ≤ |Tn , ϕn − ϕ| + |Tn − T, ϕ| . We have |Tn , ϕn − ϕ| ≤ sup |Tn , ϕn − ϕ| n∈N
≤ cPm (ϕn − ϕ) On the other hand, |Tn − T, ϕ| Tn , ϕn
−→
n−→+∞
−→
n−→+∞
−→
n−→+∞
0.
0. Consequently,
T, ϕ .
3.4. Exercises with solutions E XERCISE 3.1.– Let ϕ ∈ D(R). Show that the following expressions define distributions whose order will be determined: 1) ϕ(x2 )dx;
R 2
ϕ (x)ex dx.
2) R
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The Theory of Distributions
S OLUTION 3.1.– 1) Let [a, b] ⊂ R. Let ϕ ∈ D(R) such that supp ϕ [a, b] . The mapping T : ϕ → ϕ(x2 )dx is a linear form on D(R); in addition, R
2 ϕ(x2 )dx = ϕ(x )dx R x∈R:x2 ∈[a, b] 2 ≤ ϕ(x ) dx x∈R:x2 ∈[a, b]
≤ ϕ∞
x∈R: x2 ∈[a, b]
dx
≤ Cϕ∞ . Therefore, T is at most a zeroth-order distribution. 2 2) The mapping T : ϕ → ϕ (x)ex dx is a linear form on D(R). R
Let ϕ ∈ D(R) and [a, b] ⊂ R such that supp ϕ [a, b] . Then, î ób 2 ϕ (x)ex dx = ϕ(x)ex2 − a R
b
≤
b a
ϕ(x)2xex dx 2
2 |ϕ(x)| 2xex dx
a
Ä 2 ä 2 ≤ ϕ∞ eb − ea = Cϕ∞ . So, T is at most a zeroth-order distribution. E XERCISE 3.2.– 1) Let ϕ ∈ D(R). Show that the expression first-order distribution T .
∞
ϕ (x) ln(x)dx defines at most a
0
2) Let ϕn denote a plateau function equal to 1 on 1 included in 2n ,2 .
1
n, 1
and whose support is
a) Apply a lower bound on T, ϕn . b) Deduce therefrom that T is exactly a first-order distribution.
Distributions on an Open Set of Rd
S OLUTION 3.2.–
1) The mapping T : ϕ →
∞ 0
57
ϕ (x) ln(x)dx is a linear form on D(R).
Let [a, b] ⊂ R and ϕ ∈ D(R) such that supp ϕ [a, b] . Then ϕ (x) ln(x)dx = ϕ (x) ln(x)dx + [a, b]∩R 0 | ln(x)|dx . ≤ ϕ ∞ [a, b]∩R+
|T, ϕ| =
∞
Therefore, T is at most a first-order distribution. 2)a) With ϕn as in the text, we have 2 2 ϕn (x) 2 ϕ (x) ln(x)dx = [ϕn (x) ln(x)] 1 − |T, ϕn | = dx 2n 1 n 1 x 2n 2n 1 1 ϕn (x) dx ≥ dx ≥ = ln(n). 1 1 dx x 2n n So, ∀n ≥ 1, | T, ϕn |≥ ln(n) −→ +∞. n→∞
b) We have ϕn ∞ = 1 for all n ≥ 1, thus, with [a, b] = [0, 2] , there exists [a, b] ⊂ R, such that for all c > 0 and all n ≥ 1, there exists ϕn ∈ D(R), supp ϕn ⊂ [0, 2] and |T, ϕn | ≥ cϕn ∞ = c (because ln(n) −→ +∞). n→∞
Therefore, T is not of zeroth order; it is then of first order. E XERCISE 3.3.– Let I ⊂ R be an open interval and let x0 ∈ I. Show that there is no function f ∈ L1loc (I) such that Tf = δx0 . S OLUTION 3.3.– Using a reasoning by contradiction, we assume that there exists f ∈ L1loc (I) such that δx0 = Tf . Then ∀ϕ ∈ D(I),
f (x)ϕ(x)dx = ϕ(x0 ). I
[3.1]
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The Theory of Distributions
In particular, for any ϕ ∈ D(I), if x0 ∈ supp ϕ, then
f (x)ϕ(x)dx = 0. I
Let J = I ∩ x ∈ R : x < x0 and let us assume that supp ϕ ⊂ J; then, f ϕ = 0. J
We have, for all ϕ ∈ D(J), f ϕ = 0. From exercise 2.5, we have f = 0 on J. J Similarly, f = 0 on J = I ∩ x ∈ R : x > x0 . Since I = J ∪ {x0 } ∪ J , then f = 0 almost everywhere I and ∀ϕ ∈ D(I),
f (x)ϕ(x)dx = 0.
[3.2]
I
However, if ϕ ∈ D(I) such that ϕ(x0 ) = 0, then equality [3.2] contradicts equality [3.1]. Therefore, a function f ∈ L1loc (I) such that δx0 = Tf does not exist. E XERCISE 3.4.– Let ϕ ∈ D(R). 1) Show that there exists ψ ∈ C ∞ (R) such that, ∀x ∈ R, ϕ(x) = ϕ(0) + xψ(x). ϕ(x) 2) Show that the limit lim dx exists. ε→0 |x|>ε x 3) Show that this expression defines at most a first-order distribution. This 1 1 distribution is called principal value of and is denoted by Vp( ). x x 1 4) Considering ϕn as in exercise 3.2, show that Vp( ) is exactly of first order. x S OLUTION 3.4.–
1) We have ∀x ∈ R, ϕ(x) = ϕ(0) + ∀x ∈ R, ϕ(x) = ϕ(0) + x
1
x
ϕ (t)dt. We set t = xu, then
0
ϕ (xu)du.
0
1 We set ψ(x) = 0 ϕ (xu)du, for all x ∈ R. Then, as ϕ ∈ D(R) and [0, 1] is compact, by the differentiation theorem under the integral sign we have ψ ∈ C ∞ (R) and ∀x ∈ R, ϕ(x) = ϕ(0) + xψ(x).
Distributions on an Open Set of Rd
59
2) Let > 0. Let [−a, a] ⊂ R and ϕ ∈ D(R) such that supp (ϕ) [−a, a]. We have ϕ(x) ϕ(x) ϕ(0) dx = dx = + ψ(x)dx x x |x|> a|x|> a|x|> x ϕ(0) ψ(x)dx) −→ ψ(x)dx = dx + →0 a|x|0 a|x|> x a|x|> =0 by imparity
(because ψ ∈ D(R)). ψ(x)dx.
Therefore, the limit exists and is a|x|0
3) Let ϕ ϕ → lim+ →0
∈
|x|>
D(R) such that supp(ϕ) [−a, a]. Then, the mapping ϕ(x) dx is a linear form on D(R) and x
a Å ã Vp 1 , ϕ = ψ(x)dx x −a Ç å 1 a ϕ (xu)du dx = −a 0 Ç å 1 a du dx ϕ ∞ −a 0 ≤ 2aϕ ∞ . Therefore, Vp( x1 ) is at most a first-order distribution. 4) Let n 1 be fixed and > 0 such that 2 ϕn (x) Vp( 1 ), ϕn dx x x 2 ϕn (x) dx ≥ 1 x 2n 1 ϕn (x) dx 1 x n 1 dx ≥ = ln(n). 1 x n
1 2n .
We have
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The Theory of Distributions
Therefore, ∀n 1, Vp( x1 ), ϕn ≥ ln(n). Then by making n tend to infinity, we obtain Vp( x1 ), ϕn −→ +∞; moreover, ϕn ∞ = 1, for all n 1. Therefore, n→∞
Vp( x1 ) is not of zeroth-order, it is thus a first-order distribution. E XERCISE 3.5.– Let T be the linear form on D(R) defined by T, ϕ = ϕ (0). 1) Show that T is a distribution of order less than or equal to 1. 2) We want to show that T is not a zeroth-order distribution. fn (0)
a) For any n ≥ 1, give an example of a function fn of class C ∞ such that = n and fn ∞ = 1.
b) Deduce, for all n ≥ 1, the existence of a function ϕn ∈ D(R), with support in [−1, 1] and such that ϕn (0) = n and ϕn ∞ ≤ 1. c) Conclude. S OLUTION 3.5.– 1) Let K be a compact of R and ϕ ∈ DK (R). We have |T, ϕ| = |ϕ (0)| ≤ max|ϕ (x)|, x∈K
which shows that T is indeed a distribution of order less than or equal to 1. 2) a) The function fn (x) = sin(nx) is appropriate. b) Let θ ∈ C ∞ (R) be a plateau function equal to 1 on [− 12 , 12 ] and supp(θ) [−1, 1]. Then, θ(0) = 1 and θ (0) = 0. Let ϕn = θfn . It is clearly visible that the support of ϕn is contained in that of θ and thus in [−1, 1] and that ϕn ∞ ≤ fn ∞ = 1. Moreover, ϕn (0) = θ (0)fn (0) + θ(0)fn (0) = n. c) We assume that T is of zeroth-order. Then, for the compact [−1, 1], there exists a constant C, such that any function ϕ ∈ D(R) with support in [−1, 1] verifies |T, ϕ| ≤ C max |ϕ(x)| . x∈[−1,1]
If this inequality is applied to the function ϕn defined just before, we will have, for all n ≥ 1, n = ϕn (0) ≤ C. This is a contradiction and the distribution is not of zeroth order.
Distributions on an Open Set of Rd
61
E XERCISE 3.6.– Let T be the linear form on D(R) defined by T, ϕ =
+∞
1 ϕ(n) ( ). n n=1
1) Show that the following formula S, ϕ = D (R∗ ).
+∞
1 ϕ(n) ( ) defines an element of n n=1
2) We want to show that there is no distribution T ∈ D (R) such that, for any ϕ ∈ D(R∗ ), T, ϕ = S, ϕ. We reason by contradiction and assume that there is such a distribution T . a) Show that for any sequence ak k of complex numbers, there exists a 3 5 and such that f (j) (1) = aj for function f ∈ D(R) of support included in , 4 4 all j ∈ N∗ . b) We set fk (x) = f k 2 x − k + 1 . Show that the supports of the functions fk are pairwise disjoint. (n) 1 c) Compute fp n . m fp and ak = 1, show that T cannot be d) By taking K = 0, 54 , ϕm = p=1
continuous on D(R). S OLUTION 3.6.–
N
1 ϕ(n) ( ) is clearly a n n=1 distribution. For each ϕ ∈ D(R∗ ), the previous sequence converges to S, ϕ (the sum is in fact finite). According to the theorem on distribution sequences, S ∈ D (R∗ ). 1) The linear form on D(R) defined by SN , ϕ =
2) a) According to Borel’s theorem (see exercise 2.7), there exists a function g (n) of class C ∞ on R and such 3 that g (0) = an for each n. Let us take θ a plateau 5 function with support in 4 , 4 and equal to 1 in the vicinity of 1; the function f (x) = θ(x)g(x − 1) can be used. b) If x is in the support of fk , we have k2 x − k + 1 ∈
ò
ï ò ï −1 3 5 1 1 1 ⇔x∈ . + + , , 4 4 4k 2 k 4k 2 k
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The Theory of Distributions
If k ≥ 1, we have 1 1 2k 2 + 2k − 1 1 −1 = + − + > 0. 2 2 4k k 4(k + 1) k+1 4k 2 (k + 1)2 The supports are indeed disjoints. (n) 1 2n (n) (1) = n2n an . Since the supports c) If p = n, then we have fn n = n f (n) 1 are disjoint, then fp ( n ) = 0 if p = n. d) For K = [0, 54 ], there exists an integer j and a constant C such that, if ϕ has support in K, |T, ϕ| ≤ C max ϕ(α) (x) . α≤j,x∈K
Let us take the functions ϕm as shown. ϕm has a compact support contained in both R∗ and in K. We thus have T, ϕm =
+∞ m
m 1 p2p . fp(n) ( ) = n p=1 p=1 n=1
On the other hand, since the functions fp have disjoint supports, for α ≤ j, we have m m (α) max ϕ(α) (x) max (x) p2j , ≤ ≤ c f m p x∈R
p=1
x∈R
p=1
where c is a constant that depends only on f . Putting this together, we get m2m ≤
m
p2p ≤
p=1
m
p2j ≤ cm2j+1 .
p=1
This is impossible when m becomes large. E XERCISE 3.7.– Let ϕ ∈ D(R2 ). Show that the expression T, ϕ =
∞ 0
ã ã Å Å 1 ϕ 2 , sin t − ϕ(0, sin t) dt t
defines a distribution T of first order at most.
Distributions on an Open Set of Rd
63
S OLUTION 3.7.– T is a linear form on D(R2 ). 2
Let [−a, a] ⊂ R2 , a > 0, be a compact and ϕ ∈ D(R2 ) such that 2 supp ϕ [−a, a] . Therefore, ∞ Å Å ã ã 1 ϕ 2 , sin t − ϕ (0, sin t) dt |T, ϕ| = t 0 ∞ 1
s ds = ∂x ϕ 2 , sin t 2 dt 0 t t 0 ∞ 1
s dt ≤ ∂x ϕ 2 , sin t ds 2 1 t t √ 0 a ∞ dt ≤ ∂x ϕ∞ 1 t2 √ a √ √ ≤ a∂x ϕ∞ ≤ aP1 (ϕ). We deduce that T is at most a first-order distribution. E XERCISE 3.8.– Let ϕ ∈ D(R). Show that the expression T, ϕ =
ϕ(p) (p)
p=0
defines an infinite-order distribution on R. S OLUTION 3.8.– Let T be defined by: T, ϕ =
+∞
+∞
ϕ(p) (p).
p=0
It is easy to see that T is a linear form on D(R). Let us study the continuity of T . Let K ⊂ R be a compact. We have K ⊂ [−N, N ] where N ∈ N∗ . For any ϕ ∈ D(R) such that supp ϕ ⊂ K, we have |T, ϕ| ≤
N N (j) ≤ (j) max ϕ(j) (x) ≤ (N + 1)PN (ϕ). ϕ j=0
j=0
x∈K
Therefore, T is a distribution. Let us show that T is of infinite order. To this end, we follow on reasoning by contradiction.
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The Theory of Distributions
If T were of finite order, there would exist k ∈ N such that for all compact K of R we have |T, ϕ| ≤ C
k j=0
max ϕ(j) (x) .
[3.3]
x∈K
Let ψ0 ∈ D(R), such that ⎧ ⎪ ⎨ψ0 (x) = 1, if |x| ≤ ψ0 (x) = 0, if |x| ≥ ⎪ ⎩ 0 ≤ ψ0 (x) ≤ 1. Let us set ψ(x) =
1 4 1 2
xk+1 (k+1)! ψ0 (x).
The Leibniz formula shows that ψ (k+1) (0) = 1.
Let ϕ(x) = ψ (λ(x − (k + 1))) where λ is meant to tend to +∞. We have ™ ï ò ß 1 1 3 ⊂ k+ , k+ . supp ϕ ⊂ x ∈ R, |x − (k + 1)| ≤ 2λ 2 2 It can be deduced that ϕ(j) (j) = 0 if j
= k + 1 and k+1 (k+1) k+1 ϕ (k + 1) = λ ψ (0) = λ . On the other hand, for j ≤ k, (k+1)
(j) ϕ (x) ≤ λj max ψ (j) (y) ≤ λk max ψ (j) (x) . R
y∈R
Inequality [3.3] applied to ϕ would thus imply that λk+1 ≤ Cλk
k j=0
max ψ (j) (x) . R
This is absurd if λ is made to tend to +∞. E XERCISE 3.9.– Let ϕ ∈ D(R) and ε > 0. ∞ 1) For α ∈] − 2, −1[, show that xα ϕ(x)dx = Aϕ εα+1 + Rε , where Aϕ ∈ R 0
does not depend on ε and where Rε tends to a limit when ε tends to 0+ . 2) We set Pf(xα ), ϕ = lim+ Rε . Show that Pf(xα ) is at most a first-order distribution.
ε→0
Distributions on an Open Set of Rd
65
S OLUTION 3.9.– 1) Let [−a, a] ⊂ R, a > 0. Let ϕ ∈ D(R) with supp ϕ [−a, a]. For α ∈ ]−2, −1[ and > 0, we have
∞
a
α
x ϕ(x)dx =
x ϕ(x)dx =
1
with ψ(x) =
a
α
xα ϕ(0) + xα+1 ψ(x) dx,
ϕ (xu)du and thus ϕ(x) = ϕ(0) + xψ(x). Then,
0
+∞
a
α
x ϕ(x)dx =
a
α
xα+1 ψ(x)dx.
x ϕ(0)dx +
On the other hand, we have
ï
a
xα ϕ(0)dx =
xα+1 ϕ(0) α+1
ϕ(0) and R = We set Aϕ = − α+1
òa =
aα+1 α+1 ϕ(0) − ϕ(0). α+1 α+1
a
ϕ(0) α+1 . a α+1
xα+1 ψ(x)dx +
Since α ∈ ]−2, −1[ , x → x is integrable in the neighborhood of 0, thus as ψ belongs to D(R) and bounded, x → xα+1 ψ(x) is also integrable in the neighborhood of 0. α+1
Therefore, R has a limit when tends to 0+ given by
a
xα+1 ψ(x)dx + 0
∞
We have indeed
ϕ(0) α+1 . a α+1
xα ϕ(x)dx = Aϕ α+1 + R with R having a limit in 0+ .
ϕ(0) α+1 + a 2) We set Pf(x ), ϕ = α+1
a
α
x
α+1
0
1
ϕ (xu)du dx. Then,
0
|Pf(xα ), ϕ| ≤ C0 ϕ + C1 ϕ ≤ CP1 (ϕ), where C = 2 max(C0 , C1 ). Consequently, Pf(xα ) is a first-order distribution at most.
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The Theory of Distributions
E XERCISE 3.10.– For all n ∈ N, we denote by Tn the distribution associated with (nt) . Show that the sequence Tn n∈N the locally integrable function fn : t −→ sinπt converges in D (R) to the distribution δ0 .
∞
Hint: The identity 0
sin(t) π = can be used. t 2
S OLUTION 3.10.– Let ϕ ∈ D(R), such that supp(ϕ) [−A, A]. For all n ∈ N, we have Tn , ϕ =
A −A
sin(nt) dt. πt
ϕ can be decomposed into ϕ(t) = ϕ(0) + tψ(t) with ψ ∈ C ∞ (R). Then ∀n ∈ N, Tn , ϕ = ϕ(0)
A −A
sin(nt) 1 dt + πt π
A
ψ(t) sin(nt)dt. −A
According to the Riemann–Lebesgue lemma, the second integral tends to 0 when n tends to infinity. Then, using the change in variable u = nt in the first integral, we obtain
nA −nA
sin(u) 1 du −→ πu π
+∞ −∞
sin(u) du = 1. u
Therefore, lim Tn , ϕ = ϕ(0) = δ0 , ϕ , thus Tn converges to δ0 in D (R). n−→∞
E XERCISE 3.11.– Compute the limits, in D (R), of the following distribution sequences: 1) An = sin(nx); 2) Bn = n g(nx) where g ∈ L1 (R); n−1 1 δp ; n p=0 n Å ã 1 inx ; 4) En = e Vp x
3) Cn =
5) Dn (x) =
k=+n k=−n
eikx .
Distributions on an Open Set of Rd
67
S OLUTION 3.11.– 1) Let ϕ ∈ D(R) such that supp(ϕ) [−A, A]. Then for all n ∈ N, it follows that An , ϕ =
A
sin(nx)ϕ(x)dx −A
A
cos(nx) ϕ (x)dx. n
= −A
Therefore, 1 | An , ϕ |≤ n
A −A
| ϕ (x) | dx.
Then, An tends to 0 when n tends to infinity. R EMARK 3.7.– Using the same method, it can also be shown that cos(nx) and einx tend to 0 when n tends to infinity. 2) Let ϕ ∈ D(R) such that supp(ϕ) [−A, A]. Then for all n ∈ N, it follows that Bn , ϕ =
A
n.g(nx)ϕ(x)dx −A
nA
. u g(u)ϕ( )du n −nA u g(u)ϕ( )1[−nA,nA] du = n R
=
For all u ∈ R, we have u g(u)ϕ( )1[−nA,nA] −→ g(u)ϕ(0) n and u g(u)ϕ( )1[−nA,nA] ≤ |g(u)| ϕ ∞ . n So, according Å ã to the dominated convergence theorem, we obtain that Bn converges to g(u)du δ0 in D (R). R
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The Theory of Distributions
3) Let ϕ ∈ D(R). Then, Cn , ϕ =
1 n−1 1 p ϕ( ) −→ ϕ(x)dx = ϕ(x)1[0,1] dx. n p=0 n 0 R
Therefore, Cn converges to the distribution associated with 1[0,1] . 4) Let ϕ ∈ D(R) be such that supp(ϕ) [−A, A]. Then for all n ∈ N, we have ≠ ∑ 1 En , ϕ = einx Vp( ), ϕ x ∑ ≠ 1 inx = Vp( ), e ϕ(x) x einx ϕ(x) = lim dx −→0 |x|> x ñ ô ϕ(x) ϕ(x) cos(nx) sin(nx) dx + i dx . = lim −→0 x x A≥|x|> A≥|x|> For the real part: It can be written that ϕ(x) = ϕ(0) + xψ(x) with ψ ∈ C ∞ (R). We have cos(nx) A≥|x|>
ϕ(x) dx = ϕ(0) x
A≥|x|>
On the other hand, ϕ(0) A≥|x|>
cos(nx) dx+ x
cos(nx)ψ(x)dx. A≥|x|>
cos(nx) dx = 0 (by imparity). x
For the imaginary part: sin(nx) A≥|x|>
Therefore, lim
ϕ(x) dx = ϕ(0) x
A≥|x|>
sin(nx) dx+ x
sin(nx)ψ(x)dx. A≥|x|>
A A sin(nx) ϕ(x) sin(nx)ψ(x)dx. dx = ϕ(0) dx + −→0 A≥|x|> x x −A −A A A sin(nx) On the other hand, sin(nx)ψ(x)dx = π. dx + x −A −A A sin(nx)φ(x)dx = 0, then, En converges to iπδ0 in D (R). Since sin(nx)
−A
Distributions on an Open Set of Rd
69
5) A simple calculation makes it possible to see that ⎧ 1 ⎨ sin((n+ 2 )x) , x ∈ / 2πZ, x sin (2) Dn (x) = ⎩2n + 1, elsewhere. Let ϕ ∈ D(R) be a test function such that supp(ϕ) [a, b] ⊂ [−(2n + 1)π, (2n + 1)π] (for n large enough). We have n + 12 x ϕ(x)dx Dn , ϕ = sin x2 −∞ k=n (2k+1)π sin n + 1 x 2 = ϕ(x)dx sin x2 k=−n (2k−1)π
+∞
sin
n + 12 x ϕ(x + 2kπ)dx = sin x2 k=−n −π π sin n + 12 x x = ψ(x)dx, sin 2 −π k=n
k=n
with ψ(x) =
π
sin
ϕ(x + 2kπ). It can also be written that
k=−n
π n + 12 x sin n + 12 x x x ψ(0)dx + (ψ(x) − ψ(0)) dx sin 2 sin 2 −π −π π sin n + 12 x = ψ(0) dx sin x2 −π π ã ã ÅÅ x (ψ(x) − ψ(0)) 1 x + sin n+ dx 2 x sin x2 −π
π
Dn , ϕ =
= 2π
sin
k=n k=−n
ÅÅ
π
ϕ(2kπ) +
sin −π
ã ã x (ψ(x) − ψ(0)) 1 x n+ dx. 2 x sin x2
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The Theory of Distributions
This last integral tends to 0 by virtue of the Riemann–Lesbegue lemma. Then +∞ Dn , ϕ −→ 2π ϕ(2kπ). It should be recalled that the sum has a finite n−→+∞
k=−∞
support, since ϕ has a compact support. Conclusion: The sequence (Dn ) tends to the Dirac comb 2π
+∞ k=−∞
Figure 3.1. D4
δ2kπ .
Distributions on an Open Set of Rd
Figure 3.2. D40
E XERCISE 3.12.– Show that in the sense of the distributions, we have 1 ε 1) lim = δ; ε −→0+ π x2 + ε2 x2 1 2) lim + √ e− 4t = δ; ε −→0 2 πt 3)
lim
λ −→+∞
λe−λx H(x) = δ.
S OLUTION 3.12.– 1) Let ϕ ∈ D(R). We have ≠
∑ 1 ε 1 +∞ ε ,ϕ = ϕ(x)dx π x 2 + ε2 π −∞ x2 + ε2 1 +∞ 1 ϕ(εt)dt. = π −∞ t2 + 1
71
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The Theory of Distributions
Since
1 t2 +1 ϕ(εt)
1 t2 +1 ϕ(0),
−→
ε −→0+
then by Lebesgue’s dominated convergence
theorem (ϕ is bounded), we obtain 1 π
+∞ −∞
Therefore,
1 1 ϕ(εt)dt −→+ t2 + 1 π ε −→0
¨
ε 1 π x2 +ε2 , ϕ
∂
+∞ −∞
1 ϕ(0)dt = ϕ(0). t2 + 1
−→+ δ, ϕ. Consequently, lim +
ε −→0
ε −→0
1 ε = δ. 2 π x + ε2
2) Let ϕ ∈ D(R). We have ≠
+∞ ∑ x2 x2 1 1 √ e− 4t , ϕ = √ e− 4t ϕ(x)dx 2 πt 2 πt −∞ +∞ √ 2 1 e−u ϕ(2u t)du. =√ π −∞
√ 2 Since e−u ϕ(2u t)
−→
t −→0+
2
e−u ϕ(0), then by Lebesgue’s dominated
convergence theorem (ϕ is bounded), we obtain 1 √ π
+∞ −∞
√ 2 1 e−u ϕ(2u t)du −→+ √ π t −→0
+∞ −∞
Consequently, ≠
x2 1 √ e− 4t , ϕ 2 πt
∑ −→ δ, ϕ.
t −→0+
x2 1 Therefore, lim+ √ e− 4t = δ. t −→0 2 πt
3) Let ϕ ∈ D(R). We have −λx λe H(x), ϕ =
+∞
+∞
=
λe−λx H(x)ϕ(x)dx
−∞
λe−λx ϕ(x)dx
0 +∞
= 0
u e−u ϕ( )du. λ
2
e−u ϕ(0)du = ϕ(0).
Distributions on an Open Set of Rd
73
By Lebesgue’s dominated convergence theorem, we obtain
+∞
−u
e 0
u ϕ( )du −→ λ −→+∞ λ
+∞
e−u ϕ(0)du = ϕ(0).
0
Therefore, λe−λx H(x), ϕ −→+ δ, ϕ. Consequently, t −→0
lim
λ −→+∞
λe−λx H(x) = δ.
E XERCISE 3.13.– Show that in the sense of distributions, we have Å ã x 1 ; = Vp 1) lim + 2 2 x ε −→0 x + ε Å ã 1 1 − iπδ. 2) lim + = Vp x ε −→0 x + iε S OLUTION 3.13.– 1) Let ϕ ∈ D(R). We have ≠
∑ +∞ x x ,ϕ = ϕ(x)dx 2 2 2 2 x +ε −∞ x + ε +∞ x2 ϕ(x) − ϕ(−x) = dx. x2 + ε 2 x 0
2 ϕ(x)−ϕ(−x) ≤ However, x2x+ε2 ϕ(x)−ϕ(−x) and since the major function is x x integrable, then by Lebesgue’s dominated convergence theorem, we get 0
+∞
x2 ϕ(x) − ϕ(−x) dx −→+ 2 x + ε2 x ε −→0
Therefore,
¨
x x2 +ε2 , ϕ
∂
−→+ Vp
t −→0
Å ã x 1 = Vp lim . x ε −→0+ x2 + ε2
1 x
+∞ 0
≠ Å ã ∑ ϕ(x) − ϕ(−x) 1 ,ϕ . dx = Vp x x
, ϕ. Consequently,
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The Theory of Distributions
2) Let ϕ ∈ D(R). We have ≠
∑ +∞ x − iε 1 ϕ(x)dx ,ϕ = 2 2 x + iε −∞ x + ε +∞ +∞ x ε ϕ(x)dx − i ϕ(x)dx = 2 2 2 2 −∞ x + ε −∞ x + ε ∑ ≠ ∑ ≠ ε x ,ϕ − i ,ϕ . = x2 + ε2 x 2 + ε2
∂ ¨ x −→+ Vp x1 , ϕ . From exercise 3.13, it The first question gives x2 +ε 2,ϕ ε −→0 ∂ ¨ ε −→+ πδ, ϕ . follows that x2 +ε 2,ϕ Therefore,
¨
ε −→0
1 x+iε , ϕ
∂
−→+ Vp x1 − πδ, ϕ . Then,
ε −→0
Å ã 1 1 − iπδ. lim = Vp x ε −→0+ x + iε
4 Operations on Distributions
The multiplication of two distributions does not define an internal law in the set of distributions. However, if a is an indefinitely differentiable function and T an arbitrary distribution, the product aT of a by T can be defined as a distribution. Multiplication by an indefinitely differentiable function is then one of the important operations of distributions and will be the subject of section 4.1. Second, the notion of differentiation in the sense of distributions is presented and it is shown that the distributions are indefinitely differentiable and that their successive derivatives are also distributions. The chapter concludes with some properties concerning translation, scaling, parity and homogeneity of a distribution. Throughout this chapter, Ω denotes an open set of Rd . 4.1. Multiplication by a C ∞ function Multiplication or division of two distributions is something that is not always valid and can create problems, it is in general not defined; as can be seen in this example for f (x) = g(x) = √1x ∈ D (R) since they are locally integrable but their product f (x)g(x) = x1 ∈ / D (R). On the other hand, if f ∈ D (R) and ψ ∈ C ∞ (R), the product ψf can have a meaning and defines a distribution on D(R) as can be seen in section 4.1.1.
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The Theory of Distributions
4.1.1. Definition and some properties P ROPOSITION 4.1 (and definition).– Let T be a distribution of D (Ω) and a a C ∞ (Ω) function. The linear form S defined on D(Ω) by S, ϕ = T, aϕ , ∀ϕ ∈ D(Ω) is a distribution and is called the “product of a and T ”, and is denoted by S = aT . Proof. First of all, we actually have aϕ ∈ D(Ω). Therefore, the right-hand side member is well defined. Let K ⊂ Ω be a compact and let ϕ ∈ D(Ω) such that supp ϕ K. Since T ∈ D (Ω), there exists m ∈ N and C > 0 such that, |T, ϕ| ≤ CPm (ϕ). However, aϕ ∈ D(Ω) and supp (aϕ) ⊂ supp ϕ ⊂ K, then:Å|S, ϕ| = |T, aϕ| ≤ CPm (aϕ) αã By Leibniz’s formula, we have ∂ α (aϕ) = ∂ β a∂ α−β ϕ. Then, β 0≤β≤α
max|∂ α (aϕ)(x)| ≤ 2|α| x∈K
max
|∂ β1 a(x)|
x∈K,|β1 |≤|α|
max
x∈K,|β2 |≤|α|
|∂ β2 ϕ(x)|.
Hence, for any α such that |α| ≤ m, we have ˜ m (ϕ) max|∂ α (aϕ)(x)| ≤ CP x∈K
with C˜ = 2m
max
x∈K,|β1 |≤m
˜ m (ϕ). We finally have Then Pm (aϕ) ≤ CP ˜ m (ϕ) ≤ CS Pm (ϕ), |S, ϕ| = |T, aϕ| ≤ C CP which shows that S is a distribution on Ω. The following properties can be verified quite easily. P ROPOSITION 4.2.– For all and T, S ∈ D (Ω), we have i) (a + b)T = aT + bT ; ii) (ab)T = a(bT ); iii) a(T + S) = aT + aS.
|∂ β1 a(x)|.
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77
4.1.2. Examples E XAMPLE 4.1.– Let a ∈ C ∞ (Ω) and f ∈ L1loc (Ω). We have aTf = Taf . E XAMPLE 4.2.– Let α ∈ C ∞ (Ω) and x0 ∈ Ω. We have αδx0 = α(x0 )δx0 . In particular, in R we have αδ = α(0)δ and xδ0 = 0. This relation is often interpreted in physics by stating that δ is an eigen distribution of the multiplication operator by a function α, with the eigenvalue α(0). E XAMPLE 4.3.– We have xVp
1 x
= 1.
As it happens, if ϕ ∈ D(Ω), then ∑ ≠ Å ã ∑ ≠ Å ã 1 1 , ϕ = Vp , xϕ xVp x x xϕ(x) = lim+ dx x ε→0 |x|>ε ϕ(x) dx = lim+ ε→0
=
|x|>ε
ϕ(x) dx R
= 1, ϕ . Consequently, xVp
1 x
= 1.
R EMARK 4.1.– The relation xVp x1 = 1 is extremely interesting, because it says that the x in the sense of distributions is nothing else than the distribution inverse of 1 1 Vp x and not x , which obviously cannot define a regular distribution. 4.1.3. Convergence properties Another interesting result is that multiplication by a function C ∞ is a continuous operation. More precisely, we have the following theorem.
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The Theory of Distributions
T HEOREM 4.1.– Let T ∈ D (Ω) and a ∈ C ∞ (Ω). Let an n≥0 be a sequence which converges to a in C ∞ (Ω) and let Tn n≥0 be a sequence which converges to T in D (Ω). We then have i) an T ii) aTn
−→
aT ;
−→
aT ;
n−→+∞
n−→+∞
iii) an Tn
−→
n−→+∞
aT .
Convergence is in D (Ω). Proof. It is obvious that point (iii) has to be demonstrated.
ψn
Let ϕ ∈ D(Ω). We set for all n, ψn = an ϕ. By Leibniz’s formula, we have −→ aϕ. From corollary 3.2, we have n−→+∞
an Tn , ϕ = Tn , ψn
−→
n−→+∞
T, aϕ = aT, ϕ .
R EMARK 4.2.– Resulting from this theorem, the mapping (a, T ) ∈ C ∞ (Ω) × D (Ω) −→ aT ∈ D (Ω) is continuous. 4.1.4. Solution of the equations xT = 0, xT = 1 and xT = S These three equations will be studied in the one-dimensional case d = 1. A more general result will then be given in any dimension d. We start with a technical lemma that will be used later. L EMMA 4.1.– We have the following equivalence ψ = xθ, θ ∈ D(R) ⇐⇒ ψ ∈ D(R) and ψ(0) = 0. Proof. First of all, if ψ = xθ, with θ ∈ D(R), it is clear that ψ ∈ D(R) and that ψ(0) = 0.
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79
Conversely, let ψ ∈ D(R) such that ψ(0) = 0. Let us assume that supp ψ ] − A, A[. By the Taylor formula with integral remainder at first order, it can be written that
1
ψ(x) = ψ(0) + x
ψ (tx)dt = xθ(x),
0
1
where θ(x) = 0
ψ (tx)dt. Then, θ ∈ C ∞ (R), and if |x| > A, we have ψ(x) = 0
from which θ(x) = 0. Therefore, θ ∈ D(R). P ROPOSITION 4.3.– Let T ∈ D (R) denote a distribution. The following two assertions are equivalent: 1) xT = 0; 2) T = cδ0 where c is a constant. Proof. We have already seen that if T = cδ0 , then xT = cxδ0 = c0 = 0. Hence a first implication. For the other sense, suppose that xT = 0. For θ ∈ D(R), we have 0 = xT, θ = T, xθ . Therefore, T is null with all functions of the form xθ, θ ∈ D(R). According to lemma 4.1, the distribution T cancels out on all the functions ψ ∈ D(R) such that ψ(0) = 0. We set χ ∈ D(R) such that χ(x) = 1 for |x| ≤ 1. Let ϕ ∈ D(R). Let us set ψ = ϕ − ϕ(0)χ. Then ψ ∈ D(R) and ψ(0) = 0. Therefore, T, ψ = 0, or still: T, ϕ = ϕ(0) T, χ = c δ0 , ϕ with c = T, χ . Thereafter, T = cδ0 . Hence the other implication. A similar equation with a second member can then be studied. We start by looking at the equation xT = 1.
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The Theory of Distributions
P ROPOSITION 4.4.– Let T ∈ D (R) denote a distribution. The following two assertions are equivalent: 1) xT = 1; 2) T = cδ0 + Vp( x1 ) where c is a constant. Proof. We already saw, in example 4.3, that xVp( x1 ) = 1. So, if T is a solution of the equation xT = 1, we must have Å ã 1 x T − Vp( ) = 0. x By the previous proposition, 1 T − Vp( ) = cδ0 , x with c as a constant. A simple verification shows the opposite implication. Here, we come once more across the general principle of solving linear equations: the set of solutions is an affine space directed by the kernel of the linear application that defines the equation under consideration (i.e. the set of solutions of the associated homogeneous equation) and involving a particular solution of the equation. The equation xT = S can in fact be solved for any second member S ∈ D (R). P ROPOSITION 4.5.– Let S ∈ D (R) be a distribution. Then the equation xT = S admits a solution T ∈ D (R). Proof. Let ϕ ∈ D(R) and χ ∈ D(R) be such that χ(x) = 1 for |x| ≤ 1. Let us set ϕ˜ = ϕ − ϕ(0)χ. Then ϕ(0) ˜ = 0. There thus exists θ ∈ D(R) such that ϕ˜ = xθ. We define the distribution T by T, ϕ = S, θ . Then, xT, ϕ = T, xϕ = S, ϕ , from which xT = S. Actually, xϕ ˜ = xϕ − (xϕ)(0)χ = xϕ, and in this case θ = ϕ.
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81
All there is to do now is verify that the formula T, ϕ = S, θ defines correctly a distribution T ∈ D (R). First of all, if ϕ ∈ D(R), we have supp θ ⊂ supp ϕ ∪ supp χ, therefore θ ∈ C ∞ (R). Next, since S ∈ D (R), then |S, θ| ≤ CS Pm (θ).
1
However, θ = 0
(ϕ (tx) − ϕ(0)χ (tx)) dt, thus Pm (θ) ≤ CPm+1 (ϕ). As such,
|T, ϕ| ≤ CS CPm+1 (ϕ) = C Pm+1 (ϕ). Consequently, T ∈ D (R). E XAMPLE 4.4.– We have the following two identities: Å
1 xPf x2
ã
Å ã 1 = Vp x
and
Å
1 x Pf x2 2
ã = 1.
Indeed, let ϕ ∈ D(R). We have Ç å ≠ Å ã ∑ 1 ϕ(x) ϕ(0) Pf , ϕ = lim dx − 2 . 2 ε−→0 x2 ε |x|>ε x Then ∑ ≠ Å ã ∑ ≠ Å ã ∑ ≠ Å ã 1 ϕ(x) 1 1 , ϕ = Pf , xϕ = lim dx = Vp ,ϕ . xPf ε−→0 |x|>ε x2 x2 x x Consequently, Å xPf
1 x2
ã = Vp
Å ã 1 x
and
x2 Pf
Å
1 x2
ã = xVp
Å ã 1 = 1. x
4.2. Differentiation of a distribution In this section, we introduce the notion of derivation for distributions. A generalization of the notion of derivative of functions for distributions is given but keeping the same classical notion for ordinary functions. By a particular choice of a few examples, the difference between classical differentiation (i.e. differentiation of functions) and differentiation in the sense of distributions is discussed.
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The Theory of Distributions
4.2.1. Definition and examples 4.2.1.1. Definition Let f ∈ C 1 (R) and ϕ ∈ D(R). Then, an integration by parts gives
f , ϕ =
R
f (x)ϕ(x) dx = −
R
f (x)ϕ (x) dx = − f, ϕ ,
which motivates the following definition. D EFINITION 4.1.– We call derivative of a distribution T ∈ D (R) the distribution T ∈ D (R) defined by T , ϕ = − T, ϕ , ∀ϕ ∈ D(R). P ROPOSITION 4.6.– Any distribution T ∈ D (R) has derivatives of any order which are also distributions and we have, for all m ∈ N, ¨ ¨ ∂ ∂ T (m) , ϕ = (−1)m T, ϕ(m) ,
∀ϕ ∈ D(R).
Proof. Let m ∈ N be fixed and ϕ ∈ D(R). By reiterating the definition for successive derivatives of the distribution T , it can be seen that ¨ ¨ ∂ ∂ T (m) , ϕ = (−1)m T, ϕ(m) . Since ϕ ∈ D(R), then ∀j ∈ N, ϕ(j) ∈ D(R), which gives a meaning to the right-hand member of the above identity. We now show that, thus defined, the functional T (m) is indeed a distribution. The linearity is trivial, all that remains is to show continuity. Let K a compact of R and let ϕ ∈ D(R) be such that supp ϕ ⊂ K. Since T is a distribution, there exists CK > 0 and an integer n ∈ N such that |T, ϕ| ≤ CK Pn,K (ϕ).
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83
But ϕ(m) ∈ D(R), therefore ¨ ∂ T, ϕ(m) ≤ CK Pn,K (ϕ(m) ) ≤ CK Pn+m,K (ϕ). It can then be deduced that ¨ (m) ∂ , ϕ ≤ CK Pn,K (ϕ(m) ) ≤ CK Pn+m,K (ϕ). T Therefore, T (m) is indeed a distribution. R EMARK 4.3.– The proof given above leads to deducing that 1) if T is a distribution of order p, then T (m) is at most of order p + m; 2) any distribution is indefinitely differentiable, that is of C ∞ class. By reasoning about partial derivatives, the notion of differentiation of distributions can be generalized to any dimension and order. P ROPOSITION 4.7 (and definition).– Let T ∈ D (Ω) and α ∈ Nd be a multi-index. The αth-order derivative ∂ α T of T is the distribution defined by ∂ α T, ϕ = (−1)|α| T, ∂ α ϕ ,
∀ϕ ∈ D(Ω).
4.2.1.2. Examples We give a series of examples of differentiations in the sense of distributions. E XAMPLE 4.5.– The derivative of a distribution Tf with f ∈ C 1 (R) is the distribution Tf . This is the subject of the calculation carried out at the beginning of this section. E XAMPLE 4.6.– 1) Let H be the Heaviside function on R defined by
H(x) = 1, H(x) = 0,
x>0 x < 0.
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The Theory of Distributions
Then, H = δ. Indeed, ∀ϕ ∈ D(R), H , ϕ = − H, ϕ = − H(x)ϕ (x)dx =−
R
+∞
ϕ (x)dx
0
= ϕ(0) = δ, ϕ. 2) (sign(x)) = 2δ, where sign(x) = H(x) − H(−x). E XAMPLE 4.7.– The pth derivative of δ is the distribution δ (p) defined by ∂ ¨ δ (p) , ϕ = (−1)p ϕ(p) (0) for any test function ϕ. The distribution δ (p) is of order p. In fact, we simply have to take the sequence of functions ϕn (x) = xp ψ(nx) with ψ ∈ D(]−1, 1[) such that ψ = 1 close to 0. We have ¨ ∂ (p) δ , ϕn = p!, whereas for m < p, lim
sup (ϕn )
n−→+∞ x∈[−1,1]
(m)
(x) =
lim
sup (xp ψ(nx))
n−→+∞ x∈[−1,1]
(m)
= 0.
E XAMPLE 4.8.– The two distributions δ and δ are linearly independent, that is, if a and b are two arbitrary complex numbers, then we have aδ + bδ = 0 ⇐⇒ a = b = 0. The proof follows from the fact that the order of δ is 0, while that of δ is 1. E XAMPLE 4.9 (Differentiation of a product aT ).– If T is a distribution and a is a function of C ∞ class, then we have (aT ) = a T + aT .
Operations on Distributions
85
Indeed, for ϕ ∈ D(Ω), we have (aT ) , ϕ = − aT, ϕ = − T, aϕ = − T, (aϕ) − a ϕ = T , (aϕ) + T, a ϕ = aT , ϕ + a T, ϕ = aT + a T, ϕ . Generally speaking, Leibniz’s formula for the product of a distribution by a function of class C ∞ can be demonstrated and we have Åαã ∂ β a∂ α−β T ∂ (aT ) = β α
0≤β≤α
for T ∈ D (Ω) and a ∈ C ∞ (Ω). Application: For a ∈ C ∞ (Ω), we have: aδ = a(0)δ − a (0)δ. Special case: xδ = −δ
and ∀k > 1, xk δ = 0.
4.10.– Let f E XAMPLE f (x) = ln(|x|), x = 0, f (x) = 0,
be
the
function
defined
on
R
by
x = 0.
Since f ∈ L1loc (R), then it can be associated with a distribution Tf ∈ D(R). We then have (Tf ) = Vp( x1 ). To see this, the calculation made in example 3.5 should be revisited. x 1 u(t)dt. E XAMPLE 4.11.– Let u ∈ Lloc (R) and set, for x ∈ R, v(x) = 0
Then v is a continuous function on R and v = u in the sense of distributions. We begin by showing the continuity of v. Let b ∈ R and let (xn ) be a sequence that converges toward b. We have ∀n ≥ 0, v(xn ) =
xn
u(t)dt = 0
R
1(0,xn ) u(t)dt. By
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The Theory of Distributions
Lebesgue’s dominated convergence theorem, the sequence (v(xn ))n then converges 1(0,b) u(t)dt = v(b), hence the continuity of v in b, thus on R. to R
Let ϕ ∈ D(R) and assume that supp ϕ [−A, A]. We have v , ϕ = − v, ϕ = − v(x)ϕ (x)dx =− =− =−
R
A
v(x)ϕ (x)dx
−A
A −A A
u(t)dt ϕ (x)dx
0
0
x
x
0
u(t)ϕ (x)dtdx + −A
0
I1
0 x
u(t)ϕ (x)dtdx. I2
We use Fubini’s theorem to calculate the integral I1 . We have 0≤t≤x 0≤x≤A
⇐⇒
0≤t≤A t ≤ x ≤ A.
Then, I1 = −
A
u(t)
=−
0
A
ϕ (x)dx dt
t A
0
u(t)(ϕ(A) − ϕ(t))dt
A
u(t)ϕ(t)dt
= 0
(because ϕ(A) = 0).
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In the same way, we show that I2 =
v , ϕ =
A
0
u(t)ϕ(t)dt. Consequently, −A
0
u(t)ϕ(t)dt + 0
87
u(t)ϕ(t)dt −A
A
u(t)ϕ(t)dt
=
−A
=
u(t)ϕ(t)dt R
= u, ϕ . Indeed, we have Tv = Tu , otherwise v = u in D (R). 4.2.2. Continuity of the differentiation operator In this section, we give a simple but interesting result of the continuity of the derivative operator on the space of distributions. T HEOREM 4.2.– Let Tn be a sequence of distributions such that Tn −→ T in D (Ω). Then, for every fixed multi-index α ∈ Nd , we have ∂ α Tn −→ ∂ α T in D (Ω). Proof. Let ϕ ∈ D(R). We have ∂ α Tn , ϕ = (−1)|α| Tn , ∂ α ϕ. However, Tn −→ T , then Tn , ∂ α ϕ −→ T, ∂ α ϕ. Consequently, ∂ α Tn , ϕ = (−1)|α| Tn , ∂ α ϕ −→ (−1)|α| T, ∂ α ϕ = ∂ α T, ϕ . Hence, ∂ α Tn −→ ∂ α T in D (Ω). 4.2.3. Solution of the equations T = 0 and ∂xi T = 0 We start with a technical lemma that will be used to characterize a constant distribution. L EMMA 4.2.– We have the following equivalence:
ψ = ϕ , ϕ ∈ D(R) ⇐⇒ ψ ∈ D(R) and
ψ(x)dx = 0. R
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Proof. =⇒) : This direct direction is obvious since ϕ has a compact support. ⇐=) : We assume that ψ ∈ D(R) and ψ(x)dx = 0. R
x
We consider ϕ defined by ϕ(x) = −∞
ψ(t)dt and supp ψ [−M, M ] .
It is clear that ϕ ∈ C ∞ (R) and that ϕ (x) = ψ(x), ∀x ∈ R. Let us show that supp ϕ [−M, M ]. If x > M, then
x
ψ(t)dt
ϕ(x) =
−∞
M
ψ(t)dt +
= −∞
x
ψ(t)dt M =0
M
=
ψ(t)dt
−∞ M
ψ(t)dt
=
−M
=
ψ(t)dt R
= 0. Hence, x > M =⇒ ϕ(x) = 0. Similarly, we show that x < −M =⇒ ϕ(x) = 0. Then, supp ϕ [−M, M ]. Therefore, ϕ ∈ D(R).
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Consequently, there exists ϕ ∈ D(R) such that ψ = ϕ . It is concluded that
ψ = ϕ , ϕ ∈ D(R) ⇐⇒ ψ ∈ D(R) and
ψ(x)dx = 0. R
P ROPOSITION 4.8.– Let T ∈ D (R) denote a distribution on R. We have T = 0 ⇐⇒ T is constant. Proof. ⇐=) : It is assumed that T ≡ c. Let ϕ ∈ D(R) such that supp ϕ [−A, A]. We have T , ϕ = − T, ϕ = − cϕ (t)dt =−
R
A
= −c
cϕ (t)dt
−A A
ϕ (t)dt
−A
= −c ϕ(A) − ϕ(−A) = 0. Therefore, T , ϕ = 0, ∀ϕ ∈ D(R). Then T = 0. ⇒) : For the direct direction, we assume that T = 0. Then, ∀ϕ ∈ D(R), T , ϕ = − T, ϕ = 0. Therefore, T cancels out on all the functions ψ ∈ D(R) such as ψ = ϕ where ϕ ∈ D(R). According to lemma 4.2, we have
ψ = ϕ , ϕ ∈ D(R) ⇐⇒ ψ ∈ D(R) and
ψ(x)dx = 0. R
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From this equivalence, it is inferred that for all ψ ψ(x)dx = 0, we have T, ψ = 0.
∈
D(R) such that
R
Let χ ∈ D(R) be fixed such that
∀x ∈ R, ψ(x) = ϕ(x) −
R
χ(x)dx = 1. Let ϕ ∈ D(R). We set:
ϕ(x)dx χ(x). R
Then ψ ∈ D(R) and linearity of T , we obtain:
R
ψ(x)dx = 0. Consequently, T, ψ = 0. Then, by
T, ϕ = T, χ
R
ϕ(x)dx. = c 1, ϕ = c, ϕ with c = T, χ .
Therefore, T is constant and T = c. Conclusion: T = 0 ⇐⇒ T ∈ vect(1). The result holds in higher dimension, but its proof is more difficult. A more general result will be admitted from which the desired result will be immediately deduced. T HEOREM 4.3.– Let T ∈ D (Ω) be a distribution on Ω. It is assumed that there exists f1 , f2 , · · · , fd of the continuous functions on Ω such that ∀i ∈ {1, · · · , d}, ∂xi T = fi . Then there exists f ∈ C 1 (Ω), such that T = f. C OROLLARY 4.1.– Let T ∈ D (Ω) be a distribution on Ω a connected open set of Rd . It is assumed that ∀i ∈ {1, · · · , d}, ∂xi T = 0. Then, T is constant. Proof. We simply have to apply theorem 4.3 and then the classical result on functions of class C 1 with zero derivatives on a connected open (result based on finite increases).
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4.2.4. Jump formula in dimension 1 4.2.4.1. Piecewise C k -class function D EFINITION 4.2.– A function f : [a, b] −→ R is piecewise of class C k if there exists a subdivision a < a1 < a2 < . . . < an = b of [a, b] such that for all i = 0, · · · , n − 1, if we set gi = f |]ai ,ai+1 [ , then gi extends into a C k -class function, to the whole interval [ai , ai+1 ] . R EMARK 4.4.– If f is a piecewise C k -class function, then 1) the function f is of class C k on ]ai , ai+1 [ for all i = 0, · · · , n − 1; 2) for any integer j ≤ k and 1, ·· · ,n, the differentiated function at each ai ; i = (k) f (k) has a left limit f (k) a− and a right limit f a+ i i ; 3) the derivatives f (j) for j ≤ k exists in the ordinary sense everywhere except at points ai ; i = 0, · · · , n; 4) f is locally integrable on [a, b], so defines a distribution Tf ; 5) if now f is defined on an interval I of R which is not necessarily a segment anymore, then it is said that f is piecewise of class C k on I if it is piecewise of class Ck on any segment included in I. 4.2.4.2. Jump formula We will consider here the jump formula in dimension one only. To this end, we start by considering a piecewise C 1 -class function. P ROPOSITION 4.9 (Jump formula).– Let f be a piecewise C 1 -class function. The derivative of the distribution Tf determined by f is given by
(Tf ) = Tf +
n + f ai − f (a− i ) δ ai . i=1
This formula is called the jump formula.
[4.1]
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Proof. Let ϕ ∈ D(R), such that supp ϕ [−A, A], then for a discontinuity point a, we have (Tf ) , ϕ = − Tf , ϕ = − f (x)ϕ (x)dx =− =−
R
a −∞ a −A
f (x)ϕ (x)dx − f (x)ϕ (x)dx −
= −f (a− )ϕ(a) +
+∞ a A
a
f (x)ϕ(x)dx + f (a+ )ϕ(a) +
−A
(Tf ) , ϕ = f (a+ ) − f (a− ) ϕ(a) +
= f (a+ ) − f (a− ) ϕ(a) +
f (x)ϕ (x)dx
a
= f (a+ ) − f (a− ) ϕ(a) +
f (x)ϕ (x)dx
a
f (x)ϕ(x)dx +
−A A
A
f (x)ϕ(x)dx
a
A
f (x)ϕ(x)dx
a
f (x)ϕ(x)dx
−A
f (x)ϕ(x)dx R
f (a+ ) − f (a− ) δa , ϕ + Tf , ϕ = f (a+ ) − f (a− ) δa + Tf , ϕ .
=
Consequently, (Tf ) , ϕ = Tf + f (a+ ) − f (a− ) δa , ϕ , ∀ϕ ∈ D(R).
Conclusion: (Tf ) = Tf + (f (a+ ) − f (a− )) δa = Tf + ω(a)δa .
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Figure 4.1. Jump function. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
We now replace a by ai and carry out the sum i = 1, 2, . . . , n. If we assume that a1 < a2 < · · · < an , then for all ϕ ∈ D(R), we have
(Tf ) , ϕ = − Tf , ϕ a1 n−1 =− f (x)ϕ (x)dx − −∞
i=1
ai+1
f (x)ϕ (x)dx −
ai
+∞
f (x)ϕ (x)dx.
an
By integrations by parts, it can be easily seen that n + (Tf ) , ϕ = f a i − f a− ϕ(ai ) + i i=1
+∞
f (x)ϕ(x)dx,
−∞
∀ϕ ∈ D(R).
n n + − f ai −f ai δai = Tf + Consequently, (Tf ) = T + ω(ai )δai .
f
i=1
i=1
R EMARK 4.5.– The same thing can be obtained even with a countable number of +∞ discontinuity points (ai )i=1 :
that is, (Tf ) = Tf +
+∞ +∞ + + δ = T ω(ai )δai . f ai − f a− a f i i i=1
i=1
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The Theory of Distributions
E XAMPLE 4.12.– Using the jump formula, the following results can be found: 1) H = δ0
Figure 4.2. Heaviside function H. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
Figure 4.3. Derivative of H. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
Operations on Distributions
2) sign(x) = 2δ0
Figure 4.4. Function signx . For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
Figure 4.5. Derivative of signx . For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
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E XAMPLE 4.13.– Let E(x) be the integer part of the real x. The discontinuity points of this function are the points x = m with m ∈ Z. On each interval [m, m + 1[ , we have E(x) = m, thus E (x) = 0. The jump formula gives +∞
E =
δm .
m=−∞
The distribution
+∞
δm is denoted by 1 and called Dirac comb.
m=−∞
E XAMPLE 4.14.– Let f denote the 2π−periodic function defined by f (x) =
x 1 1− , 2 π
x ∈ [0, 2π].
The jump at discontinuity points x = ±2mπ, m ∈ N, is ω(x) = 1. Then by applying the jump formula, we obtain (Tf ) = −
+∞ 1 δ2mπ . + 2π m=−∞
E XAMPLE 4.15 (General case of a periodic function).– Let f denote the a−periodic function (a > 0) defined by f (x) =
x . a
Operations on Distributions
Figure 4.6. Function f (x) =
97
x a
Figure 4.7. The Dirac comb a
The jump at the discontinuity points xk = ±ka, k ∈ Z, is ω(xk ) = 1. Then, applying the jump formula, we get f =
+∞ 1 δka . − a
[4.2]
k=−∞
For a > 0, the distribution defined by a =
+∞
δma is called the Dirac comb.
m=−∞
The result from the jump formula can be extended to the successive derivatives by considering the jumps of f and those of its derivatives.
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The Theory of Distributions
P ROPOSITION 4.10 (Jump formula with C k ).– Let f denote a piecewise C k -class function. The successive derivatives of the distribution Tf determined by f are given by
(Tf )
(m)
= Tf (m) +
n
(m−1) − f (a+ i ) − f (ai ) δai
i=1
+
n
(m−2) − + ... f (a+ i ) − f (ai ) δai
n
(m−1) − (ai ) δai f (m−1) (a+ i )−f
i=1
+
i=1
for all m ∈ N∗ such that m ≤ k. 4.2.5. Differentiation/integration under the duality bracket The following two statements generalize, on the one hand, the differentiation theorem under the summation sign and, on the other hand, Fubini’s theorem. They will play an important part in differentiating the convolution product of two distributions. In the following two statements, the notation T, ϕ(., y) designates T, ϕy where ϕy ∈ D(Ω) is the function defined by ϕy (x) := ϕ(x, y). P ROPOSITION 4.11 (Differentiation under the bracket).– Let T ∈ D (Ω) be a distribution and ϕ a function of D(Ω × Rd ). Then the function F : y −→ T, ϕ(., y) is of class C ∞ on Rd and we have for all α ∈ Nd , ∂yα F (y) = T, ∂yα ϕ(., y) . Proof. We begin by demonstrating the result for |α| = 1. Let y0 ∈ Rd , and we will show that the differential of F at y0 is the mapping h −→ T, ∇y ϕ(., y0 ).h . Let (hn )n∈N a sequence of Rd converging to 0 and let us show that the sequence 1 (F (y0 + hn ) − F (y0 ) − T, ∇y ϕ(., y0 ).hn ) |hn |
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99
converges to 0 when n tends to infinity. Simply, we observe that the sequence of functions ϕn (x) :=
1 (ϕ(x, y0 + hn ) − ϕ(x, y0 ) − ∇y ϕ(., y0 ).hn ) |hn |
converges to the null function in D(Ω). So, we have T, ϕn
−→
n−→+∞
0,
so F is differentiable on Rd and its partial derivatives are ∂yj F (y) = T, ∂yj ϕ(., y) for all y ∈ Rd . The argument is iterated by induction on the differentiation order. R EMARK 4.6.– In the above text, the hypothesis ϕ ∈ D(Ω × Rd ) can be weakened by assuming only supp ϕ ⊂ K × Rd with K compact, that is, the support of ϕy is included in K for all y ∈ Rd . P ROPOSITION 4.12 (Integration under the bracket).– Let T ∈ D (Ω) denote a distribution and ϕ a function of D(Ω × Rd ). We have Rd
≠ T, ϕ(., y) dy = T,
∑ ϕ(., y)dy . Rd
Proof. We consider the case d = 1, the general case is then obtained by induction on d using Fubini’s theorem. Let R be a positive real such that ϕ have support in [−R, R]2 . Let ξ denote a C ∞ -class function on R of integral 1, supported in [−R, R]. Let ψ be the function defined by ψ(x, y) := ϕ(x, y) − ξ(y)
ϕ(x, u)du. R
∞ 2 yR] . Moreover, This function is of class C on Ω × R and has support in [−R, ψ(x, u)du is ψ(x, y)dy = 0, for all x ∈ Ω. Then the function Φ(x, y) := −∞
R
also of class C ∞ on Ω × R with support in [−R, R]2 . By proposition 4.12 on the differentiation under the bracket, we have d T, Φ(., y) = T, ∂y Φ(., y) = T, ψ(., y) . dy
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It is deduced that the mapping y −→ T, Φ(x, y) − constant. However, y −→ T, Φ(x, y) and y −→
y
−∞ y −∞
T, ψ(x, u) du is
T, ψ(x, u) du are
identically null for y < −R, therefore these two functions are equal. In particular, T, ψ(x, y) dy = 0. Then, R
0=
R
=
R
= R
T, ψ(x, y) dy ≠ T, ϕ(x, y) dy − T, ≠ T, ϕ(x, y) dy − T,
∑
y
ϕ(x, u)du −∞ y
∑ ϕ(x, u)du ,
ξ(y)dy R
−∞
hence the result. 4.3. Transformations of distributions This section presents a study on the effect of translation and dilation on distributions. 4.3.1. Distribution translation We first note that for a function f locally integrable in Rd and a ∈ Rd ; by change of variable, we have
Rd
f (x − a)ϕ(x)dx =
f (x)ϕ(x + a)dx. Rd
Denoting by τa g the translation by the vector a of the function g and which is defined by τa g(x) = g(x − a), ∀x ∈ Rd ,
we get Rd
τa f (x)ϕ(x)dx =
Rd
f (x)τ−a ϕ(x)dx. Otherwise,
τa f, ϕ = f, τ−a ϕ . The result can be generalized to any distribution.
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D EFINITION 4.3.– Let T ∈ D (Rd ) be a distribution and let a ∈ Rd . The translation of T by the vector a is the distribution denoted τa T and defined by τa T, ϕ = T, τ−a ϕ ,
∀ϕ ∈ D(R).
E XAMPLE 4.16.– We have τa δ = δa . As such, for any ϕ ∈ D(R), we have τa δ, ϕ = δ, τ−a ϕ = τ−a ϕ(0) = ϕ(a) = δa , ϕ . Then, τa δ = δa . D EFINITION 4.4.– Let T ∈ D (R) and a ∈ R∗+ . The distribution T is said to be periodic of period a, if τa T = T . E XAMPLE 4.17.– Let ϕ ∈ D(R). We have τ2π Tsin , ϕ = Tsin , τ−2π ϕ sin xϕ(x + 2π)dx =
R
=
R
=
sin(x − 2π)ϕ(x)dx sin(x)ϕ(x)dx
R
= Tsin , ϕ . Then, τ2π Tsin = Tsin . Consequently, Tsin is 2π−periodic. E XAMPLE 4.18.– The distribution T defined by T, ϕ = periodic of period a.
+∞
ϕ(na), a > 0 is
−∞
E XAMPLE 4.19.– Let α be an indefinitely differentiable function, and let T a distribution of D (R). For any real a, we have: τa (αT ) = (τa α)(τa T ).
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E XAMPLE 4.20.– For any real a, we have (x − a)T = 0 ⇐⇒ ∃k ∈ R, T = kδa .
4.3.2. Distribution dilation For λ ∈ R \ {0}, we recall that the dilation dλ g of a function g defined on Rd the function defined by x dλ g(x) = g( ), ∀x ∈ Rd . λ D EFINITION 4.5.– Let λ ∈ R \ {0}, and let T be a distribution of D (R). The dilation dλ T of the distribution T is the distribution defined by ¨ ∂ dλ T, ϕ = T, |λ|d λ1 ϕ , ∀ϕ ∈ D(R). E XAMPLE 4.21.– We have dλ δ = |λ|δ and dλ δx0 = |λ|δλx0 . 4.3.3. Distribution parity D EFINITION 4.6.– Let T denote a distribution of D (R). The distribution denoted by ˇ where ϕ(x) ˇ = ϕ(−x), is Tˇ := d−1 T and defined by ∀ϕ ∈ D(R), Tˇ, ϕ = T, ϕ
called the symmetric distribution of the distribution T . D EFINITION 4.7.– It is said that T is an even distribution if Tˇ = T and odd if Tˇ = −T . E XAMPLE 4.22.– The distribution of δ is even, as is δx0 + δ−x0 . E XAMPLE 4.23.– The distribution Vp
1 x
is odd, as is signx = 2H − 1.
E XAMPLE 4.24.– The distribution H has no well-defined parity, as is δx0 for x0 = 0. A PPLICATION EXERCISE .– Any distribution can be uniquely written as a sum of an even and an odd distribution: T =
T + Tˇ T − Tˇ + . 2 2
Apply this formula to the distributions H and δx0 .
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4.3.4. Distribution homogeneity The purpose of this section is to study the effect of a change of variable on a distribution. Consider the homothety hλ defined on Rd by: hλ (x) = λx, ∀x ∈ Rd . Let T denote a distribution of D (R). This distribution T is associated with a distribution 1 denoted by T ◦ hλ and defined by T ◦ hλ , ϕ = |λ| T, ϕ ◦ hλ , ∀ϕ ∈ D(R). R EMARK 4.7.– If T ◦ hλ = T , we say that the distribution T is invariant by hλ . D EFINITION 4.8.– We say that T is a homogeneous distribution of degree p ∈ N if T ◦ hλ = λp T,
∀λ > 0.
E XAMPLE 4.25.– 1) The distribution |x| is homogeneous of degree 1. 2) The distribution signx is homogeneous of degree 0. 3) The distribution Vp( x1 ) is homogeneous of degree −1. 4.4. Exercises with solutions E XERCISE 4.1.– Let T ∈ D (R). Show that sin x T = 0 if and only if there exists a sequence cn n∈Z of complex numbers such that T = cn δnπ . n∈Z
S OLUTION 4.1.– ï ò 3π 3π i) Let On = − + nπ, + nπ . Then On n∈Z is a locally finite covering 4 4 of R. Let (χn )n be a unit partition associated with On n∈N . Let ϕ ∈ Cc∞ (R). We decompose ϕ into: ϕ = ϕχn := ϕn . According to the Taylor formula with n∈Z
n∈Z
integral remainder, each ϕn is written as ∀x ∈ R, ϕn = ϕ(nπ) + (x − nπ)ψn (x)
with
ψn ∈ C ∞ (R).
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x − nπ is of class C ∞ and has compact sin(x) support in On . Finally, if νn is a plateau function equal to 1 on On and with support in ]−π + nπ, π + nπ[, it can then be written that, for all x ∈ R, Moreover, it can be seen that x −→
ϕn (x) = ϕ(nπ)νn (x) + sin(x)
x − nπ ψn (x)νn (x) sin(x)
= ϕ(nπ)νn (x) + sin(x)θn (x). So, T, ϕ =
[T, νn ϕ(nπ) + (sin(x)T ), θn ]
n∈Z
=
T, νn ϕ(nπ).
n∈Z
If we set cn = T, νn , we correctly have: T = ii) Conversely, let us assume that T =
cn δnπ .
n∈Z
cn δnπ . Since sin(nπ) = 0, we have
n∈Z
indeed sin(x)T = 0. sin x T = 0 if and only if there exists a sequence cn n∈Z of complex numbers such that T = cn δnπ . Conclusion:
n∈Z
E XERCISE 4.2.– For a = b, solve, in D (R), the following equations: 1) (x − a)(x − b)T = 0; 2) (x − a)(x − b)T = 1. S OLUTION 4.2.– 1) We recall that f δa = f (a)δa for any function f indefinitely differentiable. Therefore, (x − a)δa = 0. Let ϕ ∈ D(R). According to the Taylor formula with integral remainder, there exists a function ψ indefinitely differentiable on R such that ∀x ∈ R, ϕ(x) = ϕ(a) + (x − a)ψ(x). We then have T, ϕ = ϕ(a) T, 1 + T, (x − a)ψ = ϕ(a) T, 1 + (x − a)T, ψ = T, 1 δa , ϕ + (x − a)T, ψ .
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Therefore, (x − a)T = 0 ⇐⇒ T ∈ Rδa . Returning to our equation, we get (x − a)(x − b)T = 0 ⇐⇒ (x − b)T ∈ Rδa . We then solve the equation (x − b)T = δa .
Ä δa = δa , thus (x−b) T − It is known that (x−b) a−b where c is a real constant. It is deduced that
δa a−b
ä
= 0. Then T =
δa a−b +cδb
(x − a)(x − b)T = 0 =⇒ T ∈ Vect(δa , δb ). Since the reciprocal is trivial, we obtain the following equivalence: (x − a)(x − b)T = 0 ⇐⇒ T ∈ Vect(δa , δb ), for a = b. We have (x − a)2 T = 0 =⇒ (x − a)T ∈ Rδa and we are looking for a particular solution of (x − a)T = δa . To this end, we should observe that (x − a)δa = δa . Therefore, T − δa ∈ Rδa . As such, (x − a)2 T = 0 =⇒ T ∈ Vect(δa , δa ). Since the reciprocal is trivial, the following is deduced: (x − a)2 T = 0 ⇐⇒ T ∈ Vect(δa , δa ). 2) If we recall Question 1, solving the equation (x − a)(x − b)T = 1 is equivalent to Ä finding ä a particular solution to this equation. We then simply observe that 1 (x − a)Vp x−a = 1. A small decomposition into simple elements gives 1 1 = (x − a)(x − b) (b − a)
Å
ã 1 1 − . x−b x−a
Let us set T0 =
Å Å ã Å ãã 1 1 1 Vp − Vp . (b − a) x−b x−a
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We have Å Å ã 1 1 (x − a)(x − b)Vp (b − a) x−b ãã Å 1 − (x − b)(x − a)Vp x−a 1 = (x − a) − (x − b) (b − a)
(x − a)(x − b)T0 =
= 1. Consequently, (x − a)(x − b)T = 1 ⇐⇒ T ∈ T0 + Vect(δa , δb ). Å ã 1 . E XERCISE 4.3.– Solve in D (R) the equation: xT = Vp x S OLUTION 4.3.– The general solution of xT = Å 0ãis T = cδ, where c is a constant. 1 simply has to be obtained. From A particular solution of the equation xT = Vp Å ã Å ã x 1 1 example 4.4, we have xPf . Therefore, the general solution to the = Vp 2 x Å ã x 1 proposed equation is T = cδ + Pf , where c is a constant. x2 E XERCISE 4.4.– Let H be a Heaviside function (i.e. indicative of R+ ). 1)a) Show that H = δ0 in D (R). b) Show that (ln(| x |)) = Vp x1 in D (R). c) Show that in D (R), Ç å Å ã Å ã 1 ϕ(x) 2ϕ(0) 1 Vp . = lim − = Pf 2 2 −→0 x x x |x|> 2) Let m ∈ N. Calculate the successive derivatives in D (R) of the function x −→
xm H(x). m!
Operations on Distributions
S OLUTION 4.4.– 1)a) Let ϕ ∈ D(R). Then, H , ϕ = − H, ϕ = ϕ(0) = δ0 , ϕ hence the result is sought for. b) Let ϕ ∈ D(R). Then, (ln | x |) , ϕ = − ln | x |, ϕ = − lim −→0
ln | x | ϕ (x)dx
A≥|x|≥
ϕ(x) x |x|≥ ≠ ∑ 1 = Vp( ), ϕ . x
= lim
−→0
The result is then proven. c) Let ϕ ∈ D(R), supp(ϕ) [−A, A], A ≥ 0. Then, Æ
Å ã ∏ ∑ ≠ Å ã 1 1 , ϕ Vp , ϕ = − Vp x x ϕ (x) = − lim −→0 A≥|x|≥ x Ç å ϕ(x) ϕ(−) ϕ() = − lim − . − + 2 −→0 A≥|x|≥ x
Now, ϕ() = ϕ(0) + ψ1 () with ψ1 () −→ 0. −→0
And ϕ(−) = ϕ(0) − ψ2 () with ψ2 () −→ 0. −→0
Here, ϕ() + ϕ(−) = 2ϕ(0) + (ψ1 () − ψ2 ()). Thereafter, −
ϕ(−) ϕ() 2ϕ(0) − + = (ψ1 () − ψ2 ()) −→ 0. −→0
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The Theory of Distributions
Consequently, Ç å Å ã Å ã 1 ϕ(x) 2ϕ(0) 1 . Vp = lim − = Pf 2 −→0 x x2 |x|> x 2) By integration by successive parts, for ϕ ∈ D(R) and for all k ≤ m, we have ≠ m ∑ ∑ ≠ m (k) x x , ϕ = (−1)k H(x) H(x), ϕ(k) m! m! +∞ m x = (−1)k ϕ(k) (x)dx m! 0 +∞ xm−k k k = (−1) .(−1) ϕ(x)dx. (m − k)! 0 Hence, – for k ≤ m, we have (k) xm xm−k = H(x) H(x); m! (m − k)! – for k = m + 1, we have xm (m+1) = δ0 ; H(x) m! – for k ≥ m + 2, we have xm (k) = δ0k−m−1 . H(x) m! E XERCISE 4.5.– Consider the function f defined by
f (x) = cos x, x > 0, f (x) = 0, x ≤ 0.
Determine the derivative in the sense of the distributions of the function f and express the result using the Heaviside function H(x). S OLUTION 4.5.– Since f is piecewise of C 1 class, the jump formula is applicable. We then obtain f = δ0 − sin xH(x).
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109
E XERCISE 4.6.– Show that the sequence of distributions Tn n∈N∗ defined by ∀n ∈ N∗ , Tn = n δ1/n − δ−1/n , converges in D (R). Is the order of the limit of a distribution of order m always m? S OLUTION 4.6.– Let ϕ 1 ψ(x) = ϕ (xu)du.
∈
D(R). We have ϕ(t)
=
ϕ(0) + tψ(t) with
0
Then for all n ∈ N, we have Å ã 1 −1 Tn , ϕ = n ϕ( ) − ϕ( ) n n ã Å 1 −1 1 1 ) = n ϕ(0) + ψ( ) − ϕ(0) − ψ( n n n n 1 ã Å −1 1 −→ 2ψ(0) = 2 ϕ (0)dt = 2ϕ (0) = −2δ0 , ϕ . = ψ( ) + ψ n n 0 Therefore, Tn converges to −2δ0 in D (R). Now for all n ∈ N, Tn is of zeroth order and δ0 is of first order; then in general, there is no connection between the order of the elements of a sequence of distributions and the order of the limit distribution. E XERCISE 4.7.– Determine the limit, when α −→ 0, of the distribution defined by Ta =
Å ã 1 1 1 Vp . − Vp 2α x−α x+α
S OLUTION 4.7.– Let ϕ ∈ D(R). We have Å ã ∑ 1 1 1 Vp ,ϕ Ta , ϕ = − Vp 2α x−α x+α ≠ ≠ ∑ ∑ 1 1 1 1 Vp Vp = ,ϕ − ,ϕ . 2α x−α 2α x+α ≠
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1 Now, the distribution Vp x±α is the translate of Vp therefore,
1 x
by the translation: τ±α ,
≠ Å ã ∑ ≠ Å ã ∑ 1 1 1 1 Vp , ϕ(x + α) − Vp , ϕ(x − α) 2α x 2α x ∑ ≠ Å ã ϕ(x + α) − ϕ(x − α) 1 , = Vp x 2α ≠ Å ã ∑ 1 = Vp , Φα . x
Ta , ϕ =
where Φα =
ϕ(x + α) − ϕ(x − α) ∈ D(R). We have 2α 1 α−→0 2
Å
lim Φα = lim
α−→0
ϕ(x + α) − ϕ(x) ϕ(x − α) − ϕ(x) + α −α
ã
= ϕ (x).
By reiterating the same procedure for the derivatives of Φα , we show that Φα tends toward ϕ in the sense of the convergence in D(R). Since Vp x1 is continuous on D(R), then Å ã ∑ 1 , Φα α−→0 x ∑ ≠ Å ã 1 , ϕ = Vp x ÆÅ Å ãã ∏ 1 =− Vp ,ϕ x ≠ ∑ 1 = Pf 2 , ϕ . x ≠
lim Ta , ϕ = lim
α−→0
Vp
Consequently, lim Ta = Pf( α−→0
1 ). x2
E XERCISE 4.8.– Solve in D (R), the differential equation: xT + T = 0. S OLUTION 4.8.– We have (xT ) = xT + T . Therefore, xT + T = 0 ⇔ (xT ) = 0. 1 Then, there exists λ ∈ R such that xT = λ. However, λVp( ) is a solution of the x latter equation, so Å Å ãã 1 = 0. xT = λ ⇐⇒ x T − λVp x
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111
Å ã 1 Therefore, there exists λ, α ∈ R, such that: T = λVp + αδ0 . x E XERCISE 4.9.– Solve in the space of distributions the following differential equations: 1) u + xu = δ; 2) u + u = H; 3) u + e−x u = δ. S OLUTION 4.9.– 1) We have x2
x2
u + xu = δ ⇐⇒ (u + xu) e 2 = e 2 δ x2 d x2 e2 u =e2 δ ⇐⇒ dx d x2 e2 u =δ ⇐⇒ dx ⇐⇒ e
x2 2
u(x) = H(x) + C
⇐⇒ u(x) = e
−x2 2
H(x) + Ce
−x2 2
with C a real constant. 2) We have u + u = H ⇐⇒ (u + u) ex = ex H(x), ∀x ∈ R d x (e u) = ex H(x), ∀x ∈ R dx x ⇐⇒ ex u = et H(t)dt + C, ∀x ∈ R ⇐⇒
−∞
Å ⇐⇒ u(x) = with C a real constant.
x
ã
e H(t)dt + C e−x , ∀x ∈ R t
−∞
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The Theory of Distributions
⎧ x ⎪ ⎪ et H(t)dt = 0, ⎪ ⎪ ⎨ −∞ However,
⎪ ⎪ ⎪ ⎪ ⎩
x < 0,
x
−∞
et H(t)dt = ex − 1,
x ≥ 0,
then u + u = δ ⇐⇒ u(x) = e−x (H(x) (ex − 1) + C) , ∀x ∈ R. 3) We have −x
−x
u + e−x u = δ ⇐⇒ (u + u) e−e = e−e δ −x d Ä −e−x ä ⇐⇒ u = e−e δ . e dx Since a(x)δ = −a (0)δ + a(0)δ , then ee
−x
δ = −e−1 δ + e−1 δ . Consequently,
d Ä −e−x ä u = −e−1 δ + e−1 δ e dx −x −e−1 H(x) + e−1 δ + C , ⇐⇒ u(x) = ee
u + e−x u = δ ⇐⇒
with C a real constant. E XERCISE 4.10.– In the R2 plane, we call Heaviside function to the function H H(x, y) = 1, x > 0 and y > 0, defined by: H(x, y) = 0, elsewhere. In the sense of distributions, we have
∂2H = δ(0,0) . ∂x∂y
.
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113
S OLUTION 4.10.– Let ϕ ∈ D(R2 ). We have ≠
∑ ∑ ≠ ∂2H ∂2ϕ , ϕ = H, ∂x∂y ∂x∂y ∂2ϕ = H(x, y) (x, y)dxdy ∂x∂y R2 +∞ +∞ 2 ∂ ϕ (x, y)dxdy = ∂x∂y 0 0 å +∞ Ç +∞ 2 ∂ ϕ = (x, y)dx dy ∂x∂y 0 0 +∞ ∂ϕ =− (0, y)dy ∂y 0 = ϕ(0, 0) = δ(0,0) , ϕ .
Consequently,
∂2H = δ(0,0) . ∂x∂y
E XERCISE 4.11.– 1) Let p and q represent two natural numbers. Compute the distribution: T = xp δ (q) where δ (i) is the ith derivative of the Dirac measure on R. 2) Solve the equation xU = 0 in D (R). 3) Solve the equation xm U = 1 in D (R). S OLUTION 4.11.– 1) Since xp ∈ C ∞ (R), then xp δ (q) makes sense. Let ϕ ∈ D(R): ¨ ∂ ∂ ¨ (q) (q) xp δ (q) , ϕ = (−1)q δ, (xp ϕ) = (−1)q (xp ϕ) (0). According to the Leibniz rule, (xp ϕ)
(q)
=
q Å ã q i=0
i
(xp )(i) ϕ(q−i) =
q i=0
Fi,q (x).
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The Theory of Distributions
Two cases can be distinguished: – If P > q, then (xp ϕ)
(q)
(0) = 0. Actually,
(xp )(i) = p(p − 1) . . . (p − i + 1)xp−i and p − i is strictly positive. – If P ≤ q, then q
Fi,q (x) =
i=0
p−1 i=0
Fi,q (x) +
q
Fi,q (x).
i=p
The first of the two sums cancels out at the origin for the same reason as above and the second is written as q i=p
Å ã q Fi,q (x) = (xp )(p) ϕ(q−p) p
because (xp )(i) (0) = 0 for i > p + 1. In this case, Å ã ¨ ∂ (−1)p q! ¨ (q−p) ∂ p (q) q q x δ , ϕ = (−1) p!ϕ(q−p) (0) = ,ϕ . δ p (q − p)! Conclusion:
xp δ (q)
⎧ ⎨0, p > q, = (−1)p q! (q−p) ⎩ ,p ≤ q δ (q − p)!
.
2) We must have u = cδ with c a constant. Let H denote the Heaviside function, then H = δ, that is to say, H is a particular solution to the equation u = δ. The 2 general solution of xu = 0 is cH + dδ, (c, d) ∈ R . 3) It is already known that the solutions of the equation xm u = 0 are given by
m−1 k=0
ck δ (k) . We should simply find a particular solution to the equation: xm u = 1. If
m = 1, we know that vp( x1 ) is appropriate. In general, we can construct a solution
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115
of xm+1 u = 1 from a solution um to xm u = 1. We derive the equation xm um = 1, which gives mxm−1 um + xm um = 0, then multiply by x, which yields 0 = mxm um + xm+1 um = m + xm+1 um . 1 Then um+1 = − m um is a particular solution to the equation xm+1 u = 1.
In summary, we find as a particular solution 1 u1 = Vp( ) x Å ã 1 u2 = − Vp( ) x
um
(−1)m−1 = (m − 1)!
Å ã 1 (m−1) . Vp( ) x
Knowing that (ln |x|) = Vp( x1 ), we can also write um =
(−1)m−1 (m) . (ln |x|) (m − 1)!
E XERCISE 4.12.– Compute, in the sense of distributions, the following derivatives: ã Å d − λ H(x).eλx ; 1) dx Ç 2 å d H(x) sin(ωx) 2 ; 2) +ω 2 dx ω 3)
dm H(x)xm−1 , for m ∈ N∗ . dxm (m − 1)!
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The Theory of Distributions
S OLUTION 4.12.– 1) Let ϕ ∈ D(R). We have ≠
∑ d H(x)eλx , ϕ = − H(x)eλx , ϕ dx +∞ =− H(x)eλx ϕ (x)dx =−
−∞
+∞
eλx ϕ (x)dx
0
+∞
= ϕ(0) + λ
eλx ϕ(x)dx.
0
Consequently: Å
ã d − λ H(x).eλx = δ. dx
Otherwise, by simply applying the differentiation formula we obtain d d H(x).eλx = (H(x)) .eλx + λH(x)eλx = δ.eλx + λH(x)eλx . dx dx 2) Let ϕ ∈ D(R). We have Æ
d2 H(x) sin(ωx) ,ϕ dx2 ω
∏
∑ H(x) sin(ωx) = ,ϕ ω 1 +∞ = sin(ωx)ϕ (x)dx ω 0 +∞ sin(ωx)ϕ(x)dx = ϕ(0) − ω ≠
0
= δ, ϕ − ω Consequently, Ç
d2 + ω2 dx2
å
H(x) sin(ωx) = δ. ω
2
≠
∑ H(x) sin(ωx) ,ϕ . ω
Operations on Distributions
3) For m = 1, we have
117
d H(x) = δ. dx
For m ≥ 2, we have xm−1 xm−2 xm−2 d H(x)xm−1 = δ+ H(x) = H(x). dx (m − 1)! (m − 1)! (m − 2)! (m − 2)! Set by set, we get xm−3 dm−2 H(x)xm−1 d2 H(x)xm−1 = H(x), . . . , = xH(x). dx2 (m − 1)! (m − 3)! dxm−2 (m − 1)! Then, dm−1 H(x)xm−1 = xδ + H(x). dxm−1 (m − 1)! Consequently, dm H(x)xm−1 = δ. dxm (m − 1)! It should be noted that a more general result is obtained in exercise 4.4 using the definition. E XERCISE 4.13.– Show that the function f defined by f (x) = 0, x < 0, f (x) = √1x , x > 0 is a distribution, then calculate its derivative. S OLUTION 4.13.– The function f is locally integrable, therefore it defines a 1 3 distribution. Its quasi-derivative equals 0 on ]−∞, 0[ , and − x− 2 on ]0, +∞[ , 2 which is not locally integrable. We are thus going to regularize it. We have f , ϕ = − f, ϕ +∞ ϕ (x) √ dx =− x 0 +∞ ϕ (x) √ dx. = − lim ε−→0 ε x
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The Theory of Distributions
If we integrate by parts,
+∞
− ε
ϕ(ε) ϕ (x) √ dx = √ + x ε
+∞
−ϕ(x) 3
2x 2
ε
dx.
Since ϕ(ε) = ϕ(0) + O(ε) when ε −→ 0, we finally have å Ç +∞ −ϕ(x) ϕ(0) √ + f , ϕ = lim dx . 3 ε−→0 ε 2x 2 ε E XERCISE 4.14.– For ϕ ∈ D(R2 ), we consider two linear forms on D(R2 ) defined +∞ ϕ(x, x)dx and T + , ϕ = ϕ(x, x)dx. by: T, ϕ = R
0
2
1) a) Show that T ∈ D (R ) and that T
+
∈ D (R2 ).
b) Calculate ∂x T + ∂t T and ∂x T + + ∂t T + . 2) Let D = (x, t) ∈ R2 / t ≥| x | . We denote by E the regular distribution associated with 1D indicative of D. Calculate ∂t E − ∂x E; then, deduce therefrom ∂t2 E − ∂x2 E. S OLUTION 4.14.– 1) Let A > 0. 2
a) For any ϕ ∈ D(R2 ) such that supp ϕ [−A, A] , we have |T, ϕ| ≤ 2Aϕ∞
and
+ T , ϕ ≤ 2Aϕ∞
Consequently, T ∈ D (R2 ) and T + ∈ D (R2 ); moreover, they are of zeroth order. 2
b) Let ϕ ∈ D(R2 ) such that supp ϕ [−A, A] , we have ∂x T, ϕ = − T, ∂x ϕ A =− ∂x ϕ(x, x)dx =−
−A
A −A
∂x (ϕ(x, x)) dx +
= 0 + T, ∂t ϕ = − ∂t T, ϕ .
A −A
∂t ϕ(x, x)dx
Operations on Distributions
Then, ∂x T + ∂t T = 0. In the same way, we show that ∂x T + + ∂t T + = δ(0,0) . In reality, ∂x T + , ϕ = − T + , ∂ x ϕ A =− ∂x ϕ(x, x)dx =−
0
A
∂x (ϕ(x, x)) dx +
0
= ϕ(0, 0) + T + , ∂t ϕ = ϕ(0, 0) − ∂t T + , ϕ = δ(0,0) − ∂t T + , ϕ .
A 0
∂t ϕ(x, x)dx
The result sought for can be deduced therefrom. 2
2) Let A > 0. For any ϕ ∈ D(R2 ) such that supp ϕ [−A, A] , we have ∂t E − ∂x E, ϕ = − E, ∂t ϕ − ∂x ϕ ∂t ϕ(t, x) − ∂x ϕ(t, x)dtdx =−
t≥|x|
=−
x∈R
+∞ |x|
∂t ϕ(t, x)dt dx −
ϕ |x|, x dx −
=−
x∈R
= 2 T +, ϕ .
t≥0
+∞ 0
t −t
∂x ϕ(t, x)dx dt
(ϕ(t, t) − ϕ(t, −t)) dt
Therefore, ∂t E − ∂x E = 2T + . On the other hand, we have 2 ∂t E − ∂x2 E, ϕ = ∂t E − ∂x E, (∂t + ∂x ) ϕ = −2 T + , (∂t + ∂x ) ϕ = 2 (∂t + ∂x ) T + , ϕ = 2δ(0,0) , ϕ . Consequently, ∂t2 E − ∂x2 E = 2δ(0,0) .
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E XERCISE 4.15.– 1 is an element of D (R2 ). x + iy Å ã 1 ∂ ∂ ˜ . Show that 2) The Cauchy–Riemann operator is defined by ∂ = +i 2 ∂x ∂y ˜ = πδ. ∂f 1) Show that the function f defined by f (x, y) =
S OLUTION 4.15.– 1) First of all, |f (x, y)| =
1 (x2
+
1 y2 ) 2
=
1 |X|
X = (x, y).
where
1 is integrable in the neighborhood of the origin in |X|α d R if α < d. Here, α = 1 and d = 2 therefore f is L1loc (R2 ), which implies that f defines an element of D (R2 ). We know that the function
2) Let ϕ ∈ D(R2 ). We have 1 ∂f, ϕ = − f, ∂ϕ = − 2
R2
1 x + iy
Å
∂f ∂f +i ∂x ∂y
ã dxdy.
If we shift to polar coordinates: x = r cos θ and y = r sin θ. We have dxdy = rdrdθ ∂ ∂ sin θ ∂ = cos θ − ∂x ∂r r ∂θ ∂ ∂ cos θ ∂ = sin θ + . ∂y ∂r r ∂θ Then 1 ∂f, ϕ = − 2
2π 0
+∞ 0
e−iθ r
where ϕ(r, ˜ θ) = ϕ(r cos θ, r sin θ).
Ç
e−iθ ∂ ϕ˜ e +i ∂r r ∂θ ˜ iθ ∂ ϕ
å rdrdθ,
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121
According to Fubini’s theorem,
1 ∂f, ϕ = − 2
2π
Ç
0
+∞ 0
∂ ϕ˜ dr ∂r
å dθ −
i 2
+∞ 0
1 r
2π 0
Å
ã ∂ ϕ˜ dθ dr. ∂θ
Since ϕ(0, ˜ 0) = ϕ(0, 0) and θ −→ ϕ(r, ˜ θ) is 2π−periodic, we get 1 ∂f, ϕ = − × 2π × (−ϕ(0, 0)) = πϕ(0, 0) = πδ, ϕ . 2 Consequently, ∂f = πδ. E XERCISE 4.16.– In the sense of distributions, solve the following equations: 1) T − λT = δ; 2) T − 3T + 2T = δ; 3) T − w2 T = δ and T + w2 T = δ. S OLUTION 4.16.– 1) We get T − λT = δ ⇐⇒ e−λx (T − λT ) = e−λx δ, ∀x ∈ R ⇐⇒
d −λx T = δ, ∀x ∈ R e dx
⇐⇒ e−λx T = H + c, ∀x ∈ R ⇐⇒ T = eλx H + ceλx , ∀x ∈ R, where c is a constant. 2) The operator D2 − 3D + 2I = (D − 2I) ◦ (D − I) prompts us to introduce S = T − T . Then S − 2S = δ, therefore S(x) = e2x (H(x) + c). We then obtain T − T = e2x (H(x) + c) ⇐⇒ e−x (T − T ) = ex (H(x) + c) ⇐⇒
d −x (e T ) = ex (H(x) + c) . dx
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The Theory of Distributions
Let us look for a primitive of ex H(x) = H(x) + (ex − 1)H(x). A primitive of H(x) is xH(x). A primitive of the continuous function (ex − 1)H(x) is x (et − 1)H(t)dt = 0 if x < 0 and ex − x − 1 if x ≥ 0 −∞
In summary, a primitive of ex H(x) is the function f defined by
f (x) = 0,
x < 0,
f (x) = e − 1, x
x ≥ 0.
Otherwise, f = (ex − 1) H(x). Therefore, e−x T = (ex − 1)H(x) + cex + d. Consequently, T = e2x − ex H(x) + ce2x + dex , ∀x ∈ R, where c and d are constants. 3) These are second-order equations with constant coefficients. Let us look for a particular solution to the equation T − w2 T = δ in the form T (x) = H(x)f (x) where f is a C 2 -class function to be determined. We have T − w2 T = δ ⇐⇒ H f − w2 f + f (0)δ + f (0)δ = δ ⇐⇒ f − w2 f = 0, f (0) = 1, f (0) = 0 sh(ωx) Then, f (x) = . The thing that remains is to add the general solution of the ω associated homogeneous equation: T (x) = A ch(ωx) + B sh(ωx) + H(x)
sh(ωx) , ∀x ∈ R. ω
The same procedure is followed for the other equation.
5 Distribution Support
We do not know how to make sense of the convolution product for any pair of distributions unless at least we have a compact support, hence the interest in defining the support of a distribution and giving its fundamental properties. In this chapter, Ω denotes an open set of Rd . 5.1. Distribution restriction and extension 5.1.1. Unit partitions We start by recalling a lemma ensuring the existence of a plateau-type function. L EMMA 5.1.– Let a ∈ Rd and r > 0. There exits a function ϕ ∈ D(Ω) such that i) supp ϕ ⊂ B(a, 2r); ii) 0 ≤ ϕ ≤ 1; iii) ϕ|B(a,r) = 1. D EFINITION 5.1.– Let K be a compact such that K ⊂ Ω and K ⊂
p
Ωj with
j=1
(Ωj )1≤j≤p a finite family of open sets included in Ω. A partition of the unit p Ωj of K is a sequence of functions (χj )1≤j≤p in subordinated to the open cover D(Ω) such that
j=1
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The Theory of Distributions
i) 0 ≤ χj ≤ 1,
∀j ∈ {1, · · · , p};
ii) supp (χj ) ⊂ Ωj ,
∀j ∈ {1, · · · , p};
iii) there exists a neighborhood V of K such that ∀x ∈ V,
p
χj (x) = 1.
j=1
We give a technical lemma that will be very useful later on; it is the lemma of unit partitions. This lemma is a tool that allows us shifting from local to global. L EMMA 5.2.– For all K a compact of Ω and for any open cover exists a partition of the unit subordinate to the cover
p
p
Ωj of K, there
j=1
Ωj .
j=1
Proof. First, if S is a compact included in an open set U of Rd , then there exists an open subset V ⊂ V that S ⊂ Vand V is compact. We actually simply have ⊂ U such to construct V = x ∈ U d(x, S) < δ/2 where δ = min d(x, U c ). x∈S
Let S1 = K\(Ω2 ∪ · · · ∪ Ωp ). Then, S1 is closed, bounded and therefore compact. Moreover, S1 ⊂ Ω1 . Therefore, there exists an open subset V1 of compact adherence such that S1 ⊂ V1 and V1 ⊂ Ω. Consequently, K ⊂ S1 ∪ Ω2 ∪ · · · ∪ Ωp , from which K ⊂ V1 ∪ Ω2 ∪ · · · ∪ Ωp . We set S2 = K\(Ω1 ∪ Ω3 ∪ · · · Ωp ) ⊂ Ω2 and in the same way we built V2 an open neighborhood of compact adhesion of S2 such that V2 ⊂ Ω2 . Then, K ⊂ Ω1 ∪ V2 ∪ Ω3 ∪ · · · ∪ Ωp , from where by intersection, K ⊂ V1 ∪ V2 ∪ Ω3 ∪ · · · ∪ Ωp . Therefore, we construct by recurrence a sequence of open subsets V1 , · · · Vp with compact adherence such that for all j, K ⊂ V1 ∪ · · · ∪ Vj ∪ Ωj+1 ∪ · · · ∪ Ωp . For any j ∈ {1, ..., p}, let χ˜j ∈ D(Ω) such that supp (χ˜j ) ⊂ Ωj , χ˜j =1 on Vj and 0 ≤ χ˜j ≤ 1. Finally, we set χ1 = χ˜1 , χ2 = χ˜1 (1−χ˜2 ) and χj = 1−χ˜2 · · · 1−χ˜j . The functions χj prove useful because we have, in addition, 1−
p j=1
χj =
p ! j=1
(1 − χ˜j ) = 0 on K.
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125
5.1.2. Distribution localization and recollement D EFINITION 5.2.– Let w ⊂ Ω be an open subset and let T ∈ D (Ω). For ϕ ∈ D(w), we define ϕ˜ ∈ D(Ω) which is equal to ϕ on w and at 0 on Ω \ w. Then, the linear form T |w defined on D(w) by ˜ ∀ϕ ∈ D(w), T |w , ϕ = T, ϕ is a distribution on w called the restriction of T to w.
Figure 5.1. Extension of the function ϕ ∈ D(w). For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
It is clear by coupling, since supp ϕ ⊂ w ⊂ Ω is compact, that ϕ˜ ∈ D(Ω). Therefore, the definition is well established. D EFINITION 5.3.– Let T ∈ D (Ω), and let w ⊆ Ω be an open subset. It is said that T is null in w if T |w = 0. R EMARK 5.1.– We can also write T |w = 0 ⇐⇒ T, ϕ = 0, ∀ϕ ∈ D(Ω) such that supp ϕ ⊂ w.
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E XAMPLE 5.1.– 1) Let T = δ ∈ D (R) and w =]1, 2[⊂ R. Let ϕ ∈ D(R) such that supp ϕ ⊂ w. We have δ, ϕ = ϕ(0) = 0. Therefore, T |w = 0. 2) Let T = H ∈ D (R) is the Heaviside distribution. We have T |R∗− = 0. 3) Let f ∈ D(R) such that supp f ⊂ [a, b]. We have Tf |]−∞,a[ = Tf |]b,+∞[ = 0. We then have a result from shifting from local to global for this notion of local nullity for a distribution. L EMMA 5.3.– Let (wi )i∈I be a family of open sets of Ω and w := Let T ∈ D (Ω) such that for all i ∈ I, T |wi = 0. Then T |w = 0.
wi their union.
i∈I
Proof. We must show that for any ϕ ∈ D(w), T, ϕ = 0. Let thus ϕ ∈ D(w) and K = supp ϕ. Since K ⊂ wi , we can extract from i∈I
this open cover of K a finite sub-cover indexed by finite J ⊂ I (Borel–Lebesgue of the unit relative to the property). Let then (χi )i∈J be a partition cover (wi )i∈J of χi (x) = 1 for x ∈ wi . Since ϕ has K. Then for all i ∈ J, χi ∈ D(wi ) and support in K ⊂
wi , we have ϕ =
i∈J
T, ϕ =
i∈J χj ϕ and thus,
i∈J
j∈J
T, χi ϕ .
j∈J
Since, for all i ∈ J, χi ϕ ∈ D(wi ) and T |wi = 0, then T, χi ϕ = 0 and therefore T, ϕ = 0. A consequence of this lemma is the following proposition. P ROPOSITION 5.1.– For any distribution T∈ D (Ω), there exists a larger open subset U ⊂ Ω such that T |U = 0. 5.2. Distribution support 5.2.1. Definition By proposition 5.1, for any distribution T∈ D (Ω), there exists a larger subset where T is null. This result allows us to define the support of a distribution as follows.
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D EFINITION 5.4.– Let T ∈ D (Ω) be a distribution. We call support of T the complementary of the largest open set where T is null. It is denoted by supp T . R EMARK 5.2.– As the complementary of an open set, supp T is always a closed set. By translating the definition, the following properties that characterize the support of a distribution can be obtained. P ROPOSITION 5.2.– Let T ∈ D (Ω). We have 1) x0 ∈ / supp T ⇐⇒ ∃ Vx0 , an open neighborhood of x0 such that ∀ϕ ∈ D Vx0 , T, ϕ = 0; c 2) supp T = x ∈ Ω/ T null near x ;
3) x0 ∈ supp T ⇐⇒ ∀ Vx0 open neighborhood of x0 , ∃ϕ ∈ D Vx0 , T, ϕ = 0;
4) supp T ⊂ F ⇐⇒ T = 0 in F c . 5.2.2. Examples In this section, some examples of distribution supports are given. Others will be presented in more detail in exercises with solutions. E XAMPLE 5.2.– Let f be a continuous function on Ω. Then, supp Tf = supp f where supp f is the support in the classical sense of the continuous function f . Actually, if x0 ∈ / supp f , there exists Vx0 an open neighborhood of x0 such that, for all x ∈ Vx0 , f (x) = 0. So, for all ϕ ∈ D Vx0 , we have (ϕf )(x) = 0 in Ω. It follows that Tf , ϕ =
Ω
f (x)ϕ(x)dx = 0. Therefore, x0 ∈ / supp Tf . Hence the first
inclusion: supp Tf ⊂ supp f . Conversely, if x0 ∈ / supp Tf , there exists Vx0 an open neighborhood of x0 such that ∀ϕ ∈ D Vx0 ,
f (x)ϕ(x)dx = 0. Ω
By the Dubois–Reymond lemma, f is zero almost everywhere (so everywhere / supp f , hence the second inclusion. since f is continuous) in Vx0 . Therefore, x0 ∈ Consequently, supp Tf = supp f .
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E XAMPLE 5.3.– For any a ∈ Ω, we have supp δa = {a}. Indeed, let ϕ ∈ D(Ω{a}), we have δa , ϕ = ϕ(a) = 0, thus supp δa ⊂ {a}. Conversely, let Va be an open neighborhood of a, and let ρ be a peak function in the neighborhood of a, such that ρ(a) = 1. Then, δa , ρ = ρ(a) = 0. Therefore, a ∈ supp δa , hence the converse inclusion. Consequently, supp δa = {a}. (n)
E XAMPLE 5.4.– For any a ∈ Ω and for any n ∈ N, we have supp δa = {a}. ¨ (n) ∂ Actually, let ϕ ∈ D(Ω/{a}), we have δa , ϕ = (−1)n ϕ(n) (a) = 0, thus (n)
supp δa
⊂ {a}.
Conversely, let Va be an open neighborhood of a, and let χ ∈ D(Va ) a plateau function in the neighborhood of a, such that χ(a) = 1 in the neighborhood a. We n consider the function ϕ defined by ϕ(x) = (x−a) χ(x). By application of Leibniz’s n! ¨ (n) ∂ (n) rule, it is clear that δa , ϕ = 1 = 0. Therefore, a ∈ supp δa , hence the converse (n)
inclusion. Consequently, supp δa
= {a}.
E XAMPLE 5.5.– We have supp Vp( x1 ) = R. Actually, let x0 = 0. Let us assume that x0 > 0, and the demonstrationïis the same ò x0 3x0 . for x0 < 0. Let ρx0 be a peak function centered in x0 and supported on , 2 2 x0 For any ε > 0 such that ε < , we have 2 |x|≥ε
ρx0 (x) dx = x
x≥ε
ρx0 (x) dx = x
3x0 2 x0 2
ρx0 (x) dx = C > 0. x
Hence, Vp( x1 ), ρx0
= 0 and x0 ∈ supp Vp( x1 ). Therefore, R∗ ⊂ 1 supp Vp( x ) ⊂ R. Since the support of a distribution is closed, we necessarily have supp Vp( x1 ) = R. 5.2.3. Properties of the support A useful practical result is given in the following manner. P ROPOSITION 5.3.– For any distribution T ∈ D (Ω), we have supp T = ∅ ⇐⇒ T = 0.
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129
Proof. If T = 0, then clearly, by definition supp T = ∅. Conversely, if supp T = ∅, then for all x ∈ Ω, there exists an open set wx ∈ Ω containing x such that T |wx = 0. By lemma 5.3, T = 0 because wx = Ω. x∈Ω
The corollary of this proposition is the following localization principle. C OROLLARY 5.1.– Let T ∈ D (Ω). It is assumed that T is locally a C k -class function for 0 ≤ k ≤ ∞ ( i.e. ∀x ∈ Ω, ∃wx open neighborhood of x, ∃fx ∈ C k (wx ), T |wx = Tfx ). Then there exists f ∈ C k (Ω) such that T = Tf . Proof. Since Ω =
wx , we can choose for all x ∈ Ω, fx ∈ C k (wx ) such that
x∈Ω
T |wx = Tfx . On wx ∩ wy , we have fx = fy because Tfx |wx ∩wy = T |wx ∩wy = Tfy |wx ∩wy , then we use the continuity of fx and fy to deduce that fx = fy everywhere on wx ∩ wy (and not only almost everywhere ). Then, we can legitimately set f : Ω → C defined by f (z) = fx (z) if z ∈ wx . The function f is of class C k on Ω because it is of class C k in the neighborhood of any x ∈ Ω and we have ∀x ∈ Ω, (T − Tf )|wx = 0. By definition of the support, supp (T − Tf ) = ∅, therefore by proposition 5.3, T = Tf . In the following, some results are given that reflect the result of the operations on the supports of the distributions. We start with the effect of multiplication. P ROPOSITION 5.4.– Let T ∈ D (Ω) and a ∈ C ∞ (Ω). We have supp (aT ) ⊂ supp a ∩ supp T. c
c
Proof. Let x0 ∈ (supp a) ∪ (supp T ) . c
If x0 ∈ (supp a) , there exists Vx0 a neighborhood of x0 such that a(x) = 0 for all x ∈ Vx0 . Then if ϕ ∈ D(Vx0 ), we have for all x ∈ Ω, a(x)ϕ(x) = 0. / supp (aT ). Hence aT, ϕ = T, aϕ = 0 and therefore, x0 ∈ If x0 ∈ (supp T )c , there exists Vx0 such that T, ψ = 0 for all ψ ∈ D(Vx0 ). Let ϕ ∈ D(Vx0 ). We have aϕ ∈ D(Vx0 ), thus aT, ϕ = T, aϕ = 0. / supp (aT ). This completes the proof. Then x0 ∈
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For differentiation, the following result is obvious. P ROPOSITION 5.5.– Let T ∈ D (Ω). For all α ∈ Nd , supp ∂ α T ⊂ supp T. Proof. Let O the largest open set where T = 0 and W the largest open set where ∂ α T = 0. By definition, we have supp T = Oc and supp ∂ α T = W c . Since T = 0 on O, then ∂ α T = 0 on O. As such, W ⊇ O, consequently, W ⊆ Oc . Hence, the desired result. c
R EMARK 5.3.– – The inclusion in proposition 5.5 can be strict. As a matter of fact, let us consider the non-null constant distribution T = c. We have supp T = R, while supp T (n) = ∅, for all n ∈ N∗ . (n)
– By applying this property, it can be found that supp δa
= {a}, for all n ∈ N∗ .
P ROPOSITION 5.6.– Let T ∈ D (Ω) and let ϕ ∈ D(Ω). We have supp T ∩ supp ϕ = ∅ =⇒ T, ϕ = 0. Proof. The proof is mainly based on the Borel–Lebesgue property applied to an open cover of the compact supp ϕ. N OTE 5.1.– Let T ∈ D (Ω) and ϕ ∈ D(Ω). It is possible that ϕ = 0 on supp T but T, ϕ =
0. Counter example: We consider T = δ . Let ϕ ∈ D(R). We have supp T = {0}. On the other hand, T, ϕ = δ , ϕ = − δ, ϕ = −ϕ (0). Is there ϕ ∈ D(R) such that ϕ (0) = 0 and ϕ(0) = 0? Let θ ∈ D(R) be a plateau function that takes the value 1 on [−1, 1] and null outside of [−2, 2]. We now consider ϕ = xθ. We have ϕ ∈ D(R), ϕ (0) = 1 = 0 and ϕ(0) = 0. The function ϕ thus constructed answers the question. A result giving sufficient conditions for T, ϕ = 0 is the following.
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131
P ROPOSITION 5.7.– Let T ∈ D (Ω) be a mth-order distribution and ϕ ∈ D(Ω) null on the support of T as well as its partial derivatives of order less than or equal to m. Then T, ϕ = 0. Proof. Let K be a compact neighborhood of the support of ϕ and K := K ∩ supp T . Let (ψn )n∈N a sequence of functions of D (Ω) with values in [0, 1], null if d(x, K) > 3 n such that – ψn (x) = 1 if d(x, K) ≤ – ∀α ∈ Nd ,
3 n;
|∂ α ψn (x)| ≤ c n|α| .
Such a sequence does exist; in practice, we simply have to choose ρ ∈ D(Rd ) positive, with support in B(0, 1) and of integral 1 and we set ψn := 1Kn ∗ ρn where ρn (x) = nd ρ(nx) and Kn := x ∈ Ω, d(x, K) ≤ n2 . The function ϕ (1 − ψn ) has its support in K and is disjoint from K, thus disjoint from supp T . By definition of the support of T , we thus have T, ϕ (1 − ψn ). It follows that T, ϕ = T, ϕψn . Now, T is of order m, so there exists M > 0 such that |T, ϕψn | ≤ M sup sup |∂ α (ϕψn )(x)| . |α|≤m x∈Ω
Let us show that for |α| ≤ m, we have sup |∂ α (ϕψn )(x)| ≤ Cn|α|−m−1
[5.1]
x∈Ω
for n large enough and with C a constant independent of n, which will show the desired result. It should be noted that if x ∈ supp (ϕψn ), then x ∈ K in which case ∂ β (ϕ)(x) = 0 for all |β| ≤ m, or x ∈ supp (ϕψn )\K and then the Taylor formula (applied at a point a ∈ K such that d(x, K) = |x − a|) implies that for all |β| ≤ m, we have β m+1−|β| ≤ Cn|β|−m−1 . ∂ ϕ(x) ≤ C |x − a|
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The Theory of Distributions
Then, by the Leibniz rule, we have ∂ α (ϕψn )(x) =
Åαã β≤α
β
∂ α−β ψn (x)∂ β ϕ(x).
Therefore, if |α| ≤ m, we have |∂ α (ϕψn )(x)| ≤ c
n|α|−|β|
β≤α
sup
3 d(x,K)≤ n
β ∂ ϕ(x) .
So for n large enough, we have |∂ α (ϕψn )(x)| ≤ c ε
n|α|−|β| n|α|−m−1 ≤ Cn|α|−m−1 .
β≤α
We have thus demonstrated inequality [5.1], which ends the demonstration. P ROPOSITION 5.8.– Let T ∈ D (Ω) and ϕ ∈ D(Ω) be null on the support of T as well as its partial derivatives of any order. Then T, ϕ = 0. Proof. Let (Kn )n∈N be an exhaustive sequence of compacts Ω and for all n ∈ N, let ψn ∈ D(Ω) is equal to 1 in the neighborhood of Kn . Then ψn T is a distribution with compact support. However, then ψn T is of finite order mn . Moreover, the support of ψn T is included in the support of T . Hence, it can be deduced that ϕ and all its partial derivatives up to order mn are null on the support of ψn T and thus by proposition 5.8, ∀n ∈ N, ψn T, ϕ = 0. But there exists n0 such that the support of ϕ is included in Kn0 , thus ψn0 ϕ = ϕ and then T, ϕ = T, ψn0 ϕ = ψn0 T, ϕ = 0. Hence, the result is looked for. P ROPOSITION 5.9.– Let T, S ∈ D (Ω) denote two distributions. We have supp (T + S) ⊂ supp T ∪ supp S. 5.3. Compact support distributions Distributions with compact support have an important place in distribution theory. A distribution with compact support is actually a finite sum of (multiple) derivatives in the sense of distributions of continuous functions. This somehow confirms that the notion of distribution is the minimal extension of the notion of continuous function, which can be differentiable as many times as desired.
Distribution Support
133
5.3.1. Definition and properties D EFINITION 5.5.– Let T ∈ D (Ω) be a distribution. It is said that T is compactly supported when supp T is compact. The set of distributions with compact support is denoted by E (Ω).
Obviously, E (Ω) is a vector subspace of D (Ω) on R (or C). A priori, a distribution with compact support is a continuous linear form on the space of functions of class C ∞ with compact supports. However, since the support of the distribution is itself compact, the test functions outside the support of the distribution can be ignored. This makes it possible to extend the duality and to consider distributions with compact support as continuous linear forms on the space of functions C ∞ (Ω) endowed with the topology of uniform convergence on all compact derivatives of all orders; this is exactly the subject of section 5.3.1.1. 5.3.1.1. Topology of C ∞ (Ω) Let ϕ ∈ C ∞ (Ω). Based on the decomposition lemma, lemma 2.2, a family of semi-norms (Pn )n can be defined on C ∞ (Ω) by Pn (ϕ) =
max
|α|≤n,x∈Kn
|∂ α ϕ(x)|.
T HEOREM 5.1.– The space C ∞ (Ω) endowed with the P −topology associated with the family of semi-norms (Pn )n is a Fréchet space. D EFINITION 5.6.– Let T : C ∞ (Ω) −→ K = (R or C) a linear form. T is continuous ⇐⇒ ∃c > 0, ∃n0 ∈ N, ∀ϕ ∈ C ∞ (Ω), |T, ϕ| ≤ cPn0 (ϕ). An equivalent sequential definition is given by definition 5.7. D EFINITION 5.7.– It is said that a sequence (ϕn ) of C ∞ (Ω) converges to ϕ in C ∞ (Ω) if for any multi-index α ∈ Nd , ∂ α ϕn converges to ∂ α ϕ uniformly in any compact of Ω. R EMARK 5.4.– 1) Unlike convergence in D(Ω), no assumption is done here on the supports. 2) Continuity and linearity for a functional on C ∞ (Ω) has been defined in the same way as in D(Ω); the dual of C ∞ (Ω) is denoted C ∞ (Ω) , which is the space of continuous linear functionals defined on C ∞ (Ω).
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5.3.1.2. Extension of a distribution with compact support The following density result is recalled. P ROPOSITION 5.10.– The space D(Ω) is dense in C ∞ (Ω). The following result is intended to extend a compactly supported distribution. P ROPOSITION 5.11.– A distribution E (Ω) can be extended to C ∞ (Ω). Proof. Let K be the compact support of T ∈ E (Ω). Let K be compact containing K and K a compact containing K such that there exists χ a plateau function, which is equal to 1 on K and that is supported in K . We verify that, for ϕ ∈ D(Ω), we have T, ϕ = T, χϕ + T, (1 − χ)ϕ , since (1 − χ)ϕ is supported in a compact contained in the complementary of the support of T , T, (1 − χ)ϕ = 0. Hence, ∀ϕ ∈ D(Ω), T, ϕ = T, χϕ .
Figure 5.2. χ function. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
[5.2]
Distribution Support
135
Given the density of D(Ω) in C ∞ (Ω), we can define on C ∞ (Ω) a functional T˜ extending T by ¨ ∂ ∀ϕ ∈ C ∞ (Ω), T˜, ϕ = T, χϕ .
[5.3]
Since χϕ ∈ D(Ω), T, χϕ is well defined, in addition, ¨ ∂ ∃c1 > 0, ∃n ∈ N, T˜, ϕ = |T, χϕ| ≤ c1 max
max
|α|≤n x∈supp (χϕ)
|∂ α (χϕ)(x)| .
On the other hand, we have |∂ α (χϕ)| ≤ c2 max max ∂ β ϕ(x) . Consequently, |β|≤|α| x∈K
¨ ∂ ∃c > 0, ∃n ∈ N, T˜, ϕ ≤ c max max ∂ β ϕ(x) . |β|≤n x∈K
Finally, ¨ ∂ ∃c > 0, ∃n ∈ N, ∀ϕ ∈ C ∞ (Ω), T˜, ϕ ≤ cPn (ϕ). It is deduced that T˜ ∈ C ∞ (Ω) . Consequence: We have just shown that for T ∈ E (Ω) there exists a linear mapping T˜ from C ∞ (Ω) to C verifying: ¨ ∂ i) T˜, ϕ = T, ϕ , ∀ϕ ∈ D(Ω); ¨ ∂ ii) T˜, ϕ = 0, ∀ϕ ∈ C ∞ (Ω), supp ϕ ∩ supp T = ∅; iii) there exists n ∈ N, a compact K (neighborhood of K0 = supp T ) and c > 0 such that ¨ ∂ ∀ϕ ∈ C ∞ (Ω), T˜, ϕ ≤ c max max ∂ β ϕ(x) . |β|≤n x∈K
R EMARK 5.5.– 1) The expression [5.3] defining T˜ is independent of the function χ. Actually, let χ1 and χ2 ∈ D(Ω) such that χ1 |V1 = 1 and χ2 |V2 = 1 where V1 and V2 are two open neighborhoods of K in Ω. Then χ1 − χ2 |V1 ∩V2 = 0.
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Since V1 ∩ V2 is an open of supp (T ) in Ω, for any function neighborhood ϕ ∈ D(Ω), the function χ1 − χ2 ϕ, which is of class C ∞ on Ω, verifies supp (χ1 − χ2 ) ϕ ∩ supp T = ∅. It follows that T, (χ1 − χ2 ) ϕ = 0. Consequently, T, χ1 ϕ = T, χ2 ϕ. ∞ 2) From to thisextension, E (Ω) can be identified with (C (Ω)) just as we already ∞ saw for D (Ω) = Cc (Ω) . In fact, we have the following result.
P ROPOSITION 5.12.– The mapping Φ from E (Ω) to C ∞ (Ω) , which associates with T ∈ E (Ω) the continuous linear form T˜ ∈ C ∞ (Ω) , is a bijective linear mapping. It makes it possible to identify E (Ω) with C ∞ (Ω) . An important property of compact support distributions is the following. P ROPOSITION 5.13.– Any distribution with compact support is of finite order. Proof. Let K be the compact support of T ∈ E (Ω). Consider K a compact contained K and K a compact containing K , so that there exists χ a plateau function, which is equal to 1 on K and which is supported in K . We recall that ∀ϕ ∈ D(Ω), T, ϕ = T, χϕ + T, (1 − χ) ϕ = T, χϕ . Considering the compact K and since T ∈ D (Ω), then there exists an integer m0 and a constant C0 , such that ∀φ ∈ DK (Ω), |T, φ| ≤ C0 max max |∂ α φ(x)| . |α|≤m0 x∈K
On the other hand, since ϕχ ∈ DK (Ω), (i.e. supported in K ), then |T, ϕ| ≤ C0 max max |∂ α (χϕ) (x)| . |α|≤m0 x∈K
Applying Leibniz’s rule yields ∂ α (χϕ) =
α β
∂ γ χ∂ β ϕ and the uniform
γ+β=α
upper bounds, for |γ| ≤ m0 of |∂ γ χ| by χm0 = max max |∂ γ χ(x)| and of |γ|≤m0 x∈K α |α| results in obtaining a constant C such that = 2 β β+γ=α
C0 max max |∂ α (χϕ)| ≤ C max |α|≤m0 x∈K
max |∂ α ϕ(x)|,
|α|≤m0 x∈supp ϕ
Distribution Support
137
and |T, ϕ| ≤ C max
max |∂ α ϕ(x)|.
|α|≤m0 x∈supp ϕ
This proves that T is of order m0 at most, therefore it is of finite order. N OTE 5.2.– Theconverse of this property is not always true. Simply consider the 1 distribution Vp . This distribution is a distribution of finite order, but x 1 supp Vp x = R is not compact. 5.3.2. Distributions with point support We have already seen in the examples above that the support of the Dirac (n) derivatives is supp δa = {a}. Conversely, distributions whose support is a singleton admit a very simple characterization. T HEOREM 5.2.– Let T ∈ D (Ω) and let x0 ∈ Ω. Let us assume that supp T = {x0 }. Then there exists an integer m and some complex number (aα )|α|≤m such that ∀ϕ ∈ D(Ω),
T, ϕ =
aα ∂ α ϕ(x0 ),
|α|≤m
which can also be written as T =
a ˜α ∂ α δx0 where a ˜α = (−1)|α| aα .
|α|≤m
For the proof of this theorem, see exercise 5.12. 5.4. Exercises with solutions E XERCISE 5.1.– Consider the linear form T defined on T defined on D(R) by T : ϕ −→
+∞ 1 1 1 √ ϕ( ) − ϕ(− ) . k k k k=1
1) Show that T is a distribution of order 1 at most. 2) Show that T is of order 1. 3) Show that the support of T is S =
1 /k∈Z ∩ 0 . k
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The Theory of Distributions
S OLUTION 5.1.– 1) Let A > 0 and ϕ ∈ D(R) such that supp ϕ [−A, A]. By Taylor’s formula, there exists ψ ∈ C ∞ (R) such that ϕ(x) = ϕ(0) + xψ(x). Then, for all N ∈ N∗ , we have N N 1 1 1 1 √ ϕ( ) − ϕ(− ) ≤ 2 ϕ ∞ 3 . k k k k2 k=1
k=1
1 The series 3 is convergent as a Riemann series. Evaluating the limit when k2 N −→ +∞, we obtain |T, ϕ| ≤
2
+∞ 1 3
k2
k=1
ϕ ∞ ≤
2
+∞ 1 3
k=1
k2
P1 (ϕ).
Therefore, T is at most a first-order distribution. 2) Let (ϕn )n≥1 be a sequence of positive functions of D(R) such that ó ó 1 ∪ [2, +∞[ ϕn (x) = 1 if x ∈ n1 , 1 and ϕn (x) = 0 if x ∈ −∞, n+1 For any n ≥ 1, supp ϕn ⊂ K := [0, 2] and ϕn ∞ = 1. If we assume that T is of order 0, then there will exist a constant C > 0 such that ∀n ∈ N∗ ,
|T, ϕn | ≤ C.
However, T, ϕn = because
n 1 √ k k=1
−→
n 1 √ , thus ∀n ∈ N∗ , k k=1
n−→+∞
n 1 √ ≤ C. This is impossible k k=1
+∞. Therefore, T is a first-order distribution.
1 3) Let k ∈ N∗ . For any open set V containing the compact k , there exists î ó 1 1 a “peak” function, ψk such that supp ψk ⊂ k+1 , k−1 ∩ V, ψk k1 = 1 and 0 ≤ ψk ≤ 1. For any k ∈ N∗ , we have T, ψk = √1k = 0. Then Similarly, we also show that ∀k ∈ Z∗− , k1 ∈ supp T .
1 k
∈ supp T, ∀k ∈ N∗ .
Distribution Support
It follows that
1
k,k
ß S = {0} ∪
139
∈ Z∗ ⊂ supp T . Since supp T is a closed set, then
1 , k ∈ Z∗ k
™ ⊂ supp T.
/ S then, there exists an open neighborhood V of x0 such that Conversely, if x0 ∈ V ∩ S = ∅ and such that for any ϕ ∈ D(V ), we have T, ϕ = 0. Therefore, x0 ∈ / supp T . It is deduced that supp T ⊂ S. Consequently, supp T = k1 , k ∈ Z∗ ∪ {0}. E XERCISE 5.2.– For all n ≥ 0, we set Tn =
n
δk , where δk is the Dirac distribution
k=0
at point x = k.
1) Show that Tn is a distribution with compact support. 2) What is its support? S OLUTION 5.2.– 1) The distribution Tn =
n
δk is a sum of a finite number of distributions with
k=0
compact support. Since the space E (R) of compactly supported distributions is a vector space, then Tn is itself with compact support. 2) Let ϕ ∈ D(R) with support in the complementary of {0, 1, .......n}. We have Tn , ϕ =
n k=0
δk , ϕ =
n
ϕ(k) = 0.
k=0
Therefore, the support of Tn is contained in {0, 1, . . . , n}. It should be noted that proposition 5.9 allows us to easily obtain the inclusion of the support of Tn in {0, 1, . . . , n}. Let k ∈ {0, 1, . . . , n}. So, Vk be an open neighborhood of k. It is considered that ϕ ∈ D(R) the “peak” function with support in ]k − 12 , k + 12 [∩Vk and which equals 1 in k.
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The Theory of Distributions
We have Tn , ϕ =
n
δj , ϕ
j=0
= δk , ϕ. = ϕ(k) = 1 = 0. Then k ∈ supp Tn . As a result, we get the other inclusion. Finally, supp Tn = {0, 1, . . . , n}. E XERCISE 5.3.– Consider the linear form T defined on D(R) by T : ϕ −→
lim
n−→+∞
nϕ(0) −
n k=1
1 ϕ( 2 ) . k
1) Show that T is a distribution. 2) Determine the support of T . S OLUTION 5.3.– 1) For all n ∈ N∗, we set Tn , ϕ = nϕ(0) −
n k=1
1 1 ) = uk , where uk = ϕ(0) − ϕ( 2 ). k2 k n
ϕ(
k=1
ï 1 The increment theorem implies that there exists ck ∈ 0, 2 such that k uk = −ϕ (ck ) k12 . In addition, |ϕ (ck )| ≤ sup |ϕ (x)|. The series with general ò
0≤x≤1
term uk is thus upper bounded by a convergent Riemann sum (since of index 2). It is therefore convergent with limit T, ϕ. Given that each Tn is linear in ϕ, the same +∞ uk , which is thus a linear form on can be said of the limit T defined by T, ϕ = k=1
Distribution Support
141
D(R). Moreover, for any compact K ⊂ R and any test function ϕ ∈ D(R) whose support is contained in K, we have |T, ϕ| ≤ C sup |ϕ (x)| , where C := x∈K
+∞ 1 . k2
k=1
It is then deduced that T is a distribution of order at most equal to 1. 2) Let ϕk ∈ D(R) verifying (for k ≥ 2): ϕk (
ï ò 1 1 1 . ) = 1 and supp ϕ ⊂ , k k2 (k + 1)2 (k − 1)2
1 ) = 1. Therefore, k12 ∈ supp T . Since the support of k2 ß ™ 1 ∗ ∪ {0} ⊂ supp T . , k ∈ N T is closed, this implies that H := k2 We obtain T, ϕk = ϕk (
Then let ϕ ∈ D(R) verifying supp ϕ ∩ H = ∅. Since Tn , ϕ = 0, we recover at the limit T, ϕ = 0, which yields supp T = H. E XERCISE 5.4.– Let f be a function indefinitely differentiable on R and T a distribution and show that {x ∈ R : f (x) = 0} ∩ supp T ⊂ supp (f T ) ⊂ supp f ∩ supp T . S OLUTION 5.4.– We set: A = {x ∈ R : f (x) = 0}, supp f = A =: B, supp T = F and supp (f T ) = E. We know that, for any ϕ ∈ D(R), if supp ϕ ⊂ F c , then T, ϕ = 0, and if supp ϕ ⊂ E c , then f T, ϕ = 0. – Let us show that A ∩ F ⊂ E. To this end, we have to show that if u ∈ E c and u ∈ A, then u ∈ F c . Since E c is an open set and u ∈ E c , there exists a neighborhood of u, denoted V (u), such that V (u) ⊂ E c , f (u) = 0, f T, ϕ = 0, for any function ϕ ∈ D(R), having its support in V (u). We know that f ϕ ∈ D(R) and henceforth, 0 = f T, ϕ = T, f ϕ .
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The Theory of Distributions
Consequently, V (u) ⊂ F c and thus u ∈ F c . – Let us show that E ⊂ B ∩ F . To this end, we merely have to show that if u ∈ B c or u ∈ F c , then u ∈ E c . Let u ∈ B c , then f (u) = 0 and f (u)ϕ(u) = 0, for all ϕ ∈ D(R). Henceforth, 0 = T, f ϕ = f T, ϕ, and thus u ∈ / E. Recall that u ∈ F if and only if T is not zero on any neighborhood V (u) of u. Therefore, if u ∈ F c , then T, ϕ = 0, for any function ϕ ∈ D(R), having its support in V (u). Hence,T, f ϕ = f T, ϕ = 0, and thus u ∈ E c . Conclusion: We have just shown that A ∩ F ⊂ E and that E ⊂ B ∩ F . Consequently, A ∩ F ⊂ E ⊂ B ∩ F , that is, {x ∈ R : f (x) = 0} ∩ supp T ⊂ supp (f T ) ⊂ supp f ∩ supp T. E XERCISE 5.5.– For a ∈ R∗ , we consider Ta = cos(ax)Vp( x1 ),
∀x ∈ R∗ .
1) Show that Ta defines a distribution on R. 2) Determine supp Ta . S OLUTION 5.5.– 1) It is known that Vp
1 x
is a distribution on R.
Since cos(ax) ∈ C ∞ (R), then Ta = cos(ax)Vp x1 is indeed a distribution on R. ß ™ π kπ 2) Let A = {x ∈ R : cos(ax) = 0} = x ∈ R : x = + , k ∈ Z . From 2a a exercise 5.4, we have A ∩ supp Vp
1 x
⊂ supp Ta ⊂ supp (cos(ax)) ∩ supp Vp
1 x
.
Now supp Vp x1 = R and supp (cos(ax)) = A = R, thus A ⊂ supp Ta ⊂ R. Therefore, A ⊂ supp Ta ⊂ R, then supp Ta = R. Consequently, supp Ta = R.
Distribution Support
E XERCISE 5.6.– For all ϕ ∈ D(R), we set T, ϕ =
e−x
2
−y 2
143
ϕ(sin(xy))dxdy.
R2
1) Show that T is a distribution. 2) What is the support of T ? S OLUTION 5.6.–
e−x
1) The integral =
2
−y 2
R2
ϕ(sin(xy))dxdy exists for all ϕ ∈ D(R) since ϕ is
bounded. Therefore, T is indeed a functional, and its linearity is obvious. On the other hand, for ϕ ∈ D(R) such that suppϕ [−a, a], we have |T, ϕ| =
2
−y 2
R2
≤ ϕ∞
e−x e−x
2
−y 2
|ϕ(sin(xy))| dxdy
dxdy
R2
≤ πϕ∞ . Then T is a zeroth-order distribution. 2) Consider the open set U =] − ∞, −1[∪]1, +∞[. If a function ϕ has a support included in U , then we have ϕ(t) = 0 for all t ∈ [−1, 1]. For such a function ϕ, we have T, ϕ = 0. The distribution T is null in U . The support of T is therefore included in [−1, 1]. Let us show that supp T = [−1, 1]. We denote by Ω the complementary of the support of T . We assume that there exists u ∈ Ω such that −1 < u < 1. Let us choose ρ > 0 such that ]u − ρ, u + ρ[ ⊂ Ω ∩ ]−1, 1[ . We consider a"positive function ϕ ∈ D(R), with support included in ]u − ρ, u + ρ[ ρ ρ# and equal to 1 in u − , u + . The subset W of R2 defined by 2 2 ρ W = (x, y) : | sin(xy) − u| < 2 is open (because the function which associates with (x,y) the real sin(xy) is a continuous function on R2 ). If (x, y) ∈ W , we have ϕ(sin(xy)) = 1. Therefore, T, ϕ ≥ W
e−x
2
−y 2
dxdy > 0.
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The Theory of Distributions
However, the support of ϕ is included in Ω and T, ϕ = 0. This leads to a contradiction because such a real u does not exist. Since the support of T is closed, we correctly have supp T = [−1, 1]. 2 E XERCISE 5.7.– Let T be the mapping from D(R ) into R defined by
T (φ) = R
φ(x, −x)dx.
1) Show that T is a zeroth-order distribution on R2 . 2) Determine the support of T . S OLUTION 5.7.– 2 1) Let support in the compact K ⊂ ϕ ∈ D(R ) with K1 = x ∈ R : (x, −x) ∈ K . Then K1 is compact and we have
T, ϕ = ϕ(x, −x)dx ≤ R
sup ϕ(x, y)
(x,y)∈K
R2 . We set
1dx ≤ C.ϕ∞ . K1
Therefore, T is a distribution and, in addition, of zeroth order. 2) Let us show that the support of T is the line D := (x, −x), x ∈ R . By definition, it is clear that supp T ⊂ D (because T ≡ 0 on R2 \ D). Moreover, for all P0 ∈ D and > 0, there exists ϕ ∈ D(R2 ) such that supp(ϕ) ⊂ B(P0 , ) and T, ϕ = 0 (choose ϕ positive, for example). The statement is deduced thereof. E XERCISE 5.8.– We are given the function g ∈ L1 (R) and u ∈ C 1 (R). We consider g(x)ϕ(x, u(x))dx, ∀ϕ ∈ D(R2 ). the mapping T defined on D(R2 ) by T, ϕ = R
1) Show that T is a distribution. 2) Show that the support of T is contained in the support of a R2 curve to be determined. S OLUTION 5.8.– 1) Clearly, the mapping T is a linear form. For any compact K ⊂ R2 and for all ϕ ∈ D(R2 ) with compact support contained in K, we have T, ϕ ≤ gL1 (R) sup |ϕ(x, y)|. (x,y)∈K
Therefore, T is a distribution of order 0.
Distribution Support
145
2) Let Γ denote the R2 curve of class C 1 determined by Γ := {(x, u(x)), x ∈ R}. The set Γ is a closed set of R2 . Its complementary Ω := Γc is an open set R2 . Let ϕ ∈ D(R2 ) with support in Ω. Then, ∀x ∈ R, ϕ(x, u(x)) = 0 such that T, ϕ = 0. This proves that the support of T is contained in Γ. E XERCISE 5.9.– Let T be the distribution defined by T, ϕ =
∞ 0
Å Å ã ã 1 ϕ 2 , sin t − ϕ (0, sin t) dt, ∀ ϕ ∈ D(R2 ). t
Determine the support of T . S OLUTION 5.9.– Let us show that if S = (x, y) ∈ R2 /y = sin √1x , x > 0 , then supp T = S = S ∩ {0} × [−1, 1] . c
Let x0 ∈ / S. Then, there exists > 0 such that B(x0 , ) ⊂ S . Let ϕ ∈ D(R2 ) be such that supp ϕ ⊂ B(x0 , ). / B(x0 , ) and of T, we have T, ϕ = 0 because ∀t > 0, t12 , sin t ∈ By definition / supp T and as such, we obtain supp T ⊂ S. 0, sin t ∈ / B(x0 , ). Therefore, x0 ∈ Conversely, let x0 ∈ S and let > 0 such that B(x0 ,) ∩ ({0} × [−1, 1]) = ∅. Let ϕ ∈ D(R2 ) such that suppT ⊂ B(x0 , ), ϕ = 1 on B x0 , 2 and ϕ ≥ 0. Then, T, ϕ =
∞ 0
1 , sin t dt ≥ (t2 − t1 ) > 0, 2 t
1 t1 verifies the equality x0 = , sin t1 and 2 t1 ß ™ 1 / B(x0 , 2 ) . Then, T is non-null in the neighborhood t2 = inf t > t1 , 2 , sin t ∈ t of x0 and S ⊂ supp T . where
Since supp T is closed, S ⊂ supp T and finally, supp T = S. E XERCISE 5.10.– Let T ∈ D (R) be a distribution such that supp T ⊂ [a, +∞[, and let ϕ ∈ C ∞ (R) such that supp ϕ ⊂] − ∞, b]. 1) We consider the following expression: T, ϕ = T, θϕ,
[5.4]
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The Theory of Distributions
where θ ∈ D(R) is a function equal to 1 on [−M, M ] the interval that contains a and b. Show that this expression indeed defines a distribution. 2) Show that the function ψ(t) = Tx , ϕ(x + t) is defined for all t ∈ R and that supp ψ ⊂] − ∞, b − a]. S OLUTION 5.10.– 1) Since ϕ ∈ C ∞ (R) and θ ∈ D(R), then θϕ ∈ D(R). Hence, the expression T, θϕ where θ ∈ D(R) indeed makes sense. Let us show that it is independent of θ. We assume that there exists θ1 ∈ D(R), which equals 1 on [−M, M ] and that verifies expression [5.4]. We have that (θ − θ1 )ϕ cancels out on ] − M, +∞[, thus supp (θ − θ1 )ϕ ⊂] − ∞, −M ]. On the other hand, supp T ⊂ [a, +∞[, then supp (θ − θ1 )ϕ ∩ supp T = ∅. Consequently, T, (θ − θ1 )ϕ = 0. 2) For any t ∈ R, the function τ−t ϕ belongs to C ∞ (R). In addition, we have supp τ−t ϕ ⊂] − ∞, b − t]. According to the first question, the bracket Tx , ϕ(x + t) thus makes sense. The function ψ cancels out if supp τ−t ϕ ∩ supp T = ∅ that is, if b − t < a. Consequently, supp ψ ⊂] − ∞, b − a]. E XERCISE 5.11.– Let T be a distribution on Rd , and let f be a function of class C ∞ on Rd , with values in R. 1) Show that, if f T = 0, then the support of T is included in the set Z(f ) = x ∈ Rd , f (x) = 0 . 2) In addition, it is assumed that T is of order 0. Show that theconverse is true, if the support of T is included in the set Z(f ) = x ∈ Rd , f (x) = 0 then f T = 0.
3) Taking T = δ , show that the converse does not hold in general if T is not of zeroth order.
4) Characterize the functions f of class C ∞ on R such that f δ = 0. S OLUTION 5.11.– c
d 1) Let us assume that f T = 0. Let¨ϕ ∈ D(R ∂ ) such that supp ϕ ⊂ (Z(f )) . Then,c ϕ ϕ d ∈ D(R ) and we have T, ϕ = f T, f = 0. Then, T cancels out on (Z(f )) f and therefore supp T ⊂ Z(f ).
2) Let T ∈ D (Rd ) such that T is of zeroth order and supp T ⊂ Z(f ). Let K be a compact of Rd . Since T is of zeroth order, then there exists a positive constant c such that for any ϕ ∈ D(Rd ) such that supp ϕ ⊂ K, we have |T, ϕ| ≤ c ϕ∞ . Since for
Distribution Support
147
any ϕ ∈ D(Rd ) such that supp ϕ ⊂ K, then we have f ϕ ∈ D(Rd ) and supp f ϕ ⊂ K, thus |f T, ϕ| = |T, f ϕ| ≤ c f ϕ∞ . Since supp T ⊂ Z(f ), then f ϕ cancels out on supp T and thus f T, ϕ = 0. Consequently, f T = 0.
3) Let us assume that T = δ which is not of zeroth order. We take f such that f (x) = x. Then supp T ⊂ Z(f ), because supp T = {0}. However, if we take ϕ ∈ D(Rd ) such that ϕ(0) = 0, then xδ , ϕ = − δ, (xϕ) = −(xϕ) (0) = −ϕ(0) = 0. Therefore, xδ is not a null distribution. 4) Let ϕ ∈ D(Rd ). We have f δ , ϕ = −f (0)ϕ(0) − f (0)ϕ (0). Then f δ , ϕ is null for any ϕ if f (0) = f (0) = 0. Therefore, f = x2 g with g ∈ D(R). E XERCISE 5.12.– Let T be a distribution on R such that supp(T ) = {0}. 1) Justify that T is of finite order. We will denote by m its order further in the text. 2) Let ϕ ∈ D(R) such that ϕ(x) = o(xm ) in the neighborhood of 0. Let ρ be a plateau function equal to 1 in the neighborhood of 0 and 0 outside ] − 1, 1[. We denote, for r > 0 and x ∈ R, ρr (x) = ρ(x/r). a) Show that, if l ≤ m, then lim sup |(ρr ϕ)(l) (x)| = 0. r→0 |x|≤r
b) Therefrom, deduce that T, ϕ = 0. 3) Let ϕ ∈ D(R). Show that ϕ is written in the form ∀x ∈ R, ϕ(x) =
m ϕ(k) (0) k=0
k!
xk + ψ(x),
where ψ is a function of class C ∞ verifying ψ(x) = o(xm ) in the neighborhood of 0. 4) Deduce the existence of complex numbers α0 , ..., αm such that, for all m m (k) (−1)k αk δ0 . αk ϕ(k) (0), or still, T = ϕ ∈ D(R), T, ϕ = k=0
k=0
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The Theory of Distributions
S OLUTION 5.12.– 1) Since T has compact support, it is of finite order. 2) a) The Leibniz rule gives
∀ ≤ m,
(ρr ϕ)
( )
=
C k (ρr )
( −k)
k
(ϕ) .
k=0
ϕ
(k)
Let r > 0. Since ϕ(x) = o(xm ) in the neighborhood of 0, then (x) = o(xm−k ), by uniqueness of the limited expansion. Furthermore, ∀x ∈ R,
(ρr )
( −k)
(x) =
1 ( −k) x ρ ( ). r −k r
For r small enough and for |x| < r, we have rm−k ( −k) k (x) (ϕ) (x) ≤ −k ρ( −k) (ρr ) ∞ r ≤ rm− ρ( −k) . ∞
( ) By combination, we obtain lim sup (ρr ϕ) = 0. r−→0 |x|≤r
b) Since T is a distribution having {0} for support and that ρr = 1 in the neighborhood of 0, then ∀r > 0, ∀ϕ ∈ D(R), T, ϕ = T, ρr ϕ. Since T is of mth order and ρr ϕ is a function with support in [−r, r] with [−r, r] ⊂ [−1, 1] which is a fixed compact. From question (a), we have lim |T, ρr ϕ| = 0. Consequently, r−→0
T, ϕ = lim |T, ρr ϕ| = 0. r−→0
3) By the Taylor rule with integral remainder, we have ∀x ∈ R,
ϕ(x) =
m ϕ(k) (0) k=0
k!
xk +
xm+1 m!
1 0
(1 − u) ϕ(m+1) (xu)du. m
xm+1 1 m (1 − u) ϕ(m+1) (xu)du and using the derivation m! 0 under the integral sign, we have that ψ is of class C ∞ and ψ(x) = o(xm ) in the neighborhood of 0. Setting ψ(x) =
Distribution Support
149
4) Let ϕ ∈ D(R). From the previous question, there exists ψ of class C ∞ such that ψ(x) = o(xm ) in the neighborhood of 0 and ∀x ∈ R,
ϕ(x) =
m ϕ(k) (0)
k!
k=0
xk + ψ(x).
We choose a function ρ as in the second question, then ρϕ has a compact support; in addition, ϕ and ρϕ are equal in the neighborhood of 0. Since supp T = {0}, then $ T, ϕ = T, ρϕ =
T,
m ϕ(k) (0)
k!
k=0
% ρx
k
+ T, ρψ(x) .
However, ρψ is of class C ∞ , has compact support and ρψ(x) = o(xm ) in the neighborhood of 0, then, by Question 2, T, ρψ = 0. Finally, T, ϕ =
m ϕ(k) (0) k=0
k!
T, ρx
k
m m T, ρxk (k) = αk ϕ(k) (0), ϕ (0) = k! k=0
T, ρxk with αk = . Finally, k! T =
m k=0
(−1)k αk δ (k) =
m k=0
ak δ (k) where ak = (−1)k αk .
k=0
6 Convolution of Distributions
The importance of the convolution product lies essentially in its regularization role. In this chapter, we introduce the convolution product in the space of distributions. To this end, we consider the convolution product of two integrable functions and then generalize to any two distributions by means of regular distributions. 6.1. Definition and examples In this section, we define the convolution product of two distributions. To illustrate this definition, various examples are given. 6.1.1. Convolution of two regular distributions Recall that according to proposition 2.4, the convolution product of two integrable functions on Rd can be defined almost everywhere. Actually, for f, g ∈ L1 (Rd ), f ∗ g is defined almost everywhere; in addition, f ∗ g ∈ L1 (Rd ) and in particular we obtain f ∗ g ∈ D (Rd ). In order to introduce the convolution product of two distributions, it is most natural to define it for regular distributions, namely locally integrable functions. For this purpose, we add an additional condition, which is sufficient to give a meaning to the convolution product within this framework. More precisely, we consider f and g two locally integrable functions on Rd such that at least one of the functions has a compact support. The fact that one of the functions has a compact support K that makes it possible to restrict the computation of the integral to the compact K and
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The Theory of Distributions
thus to address it as for integrable functions. The convolution product of f and g is well defined and given by (f ∗ g)(x) =
Rd
f (x − y)g(y)dy =
f (x − y)g(y)dy. K
In addition, we have
(f ∗ g)(x) =
Rd
f (x − y)g(y)dy =
Rd
f (y)g(x − y)dy = (g ∗ f )(x).
Hence, f ∗ g = g ∗ f . Let ϕ ∈ D(Rd ). We have Tf ∗g , ϕ =
Rd
(f ∗ g)(x)ϕ(x)dx Å
ã
=
Rd
Rd
=
Rd
Rd
=
f (x − z)g(z)dz ϕ(x)dx
f (x − z)g(z)ϕ(x)dzdx f (y)g(z)ϕ(y + z)dzdy
Rd
Rd
=
f (x)g(y)ϕ(x + y)dxdy
Rd
Rd
Å
ã g(y)ϕ(x + y)dy dx
f (x)
= Rd
Rd
= Tf , Tg , ϕ(x + y) . Then ∀ϕ ∈ D(Rd ), Tf ∗g , ϕ = Tf , Tg , ϕ(x + y) .
[6.1]
Note that it can also be written that Tf ∗g , ϕ := f ⊗ g, ϕ(x + y) where f ⊗ g is the tensor product of f and g and verifies
f ⊗ g, ψ =
f (x)g(y)ψ(x, y)dxdy, Rd
Rd
∀ψ ∈ D(R2 × R2 ).
Convolution of Distributions
153
Consequently, ∀ϕ ∈ D(Rd ), Tf ∗g , ϕ = Tf , Tg , ϕ(x + y) = f ⊗ g, ϕ(x + y) .
[6.2]
D EFINITION 6.1.– Let f and g be two locally integrable functions on Rd such that at least one of the functions has compact support. The convolution product of Tf and Tg is the distribution denoted by Tf ∗ Tg and defined by ∀ϕ ∈ D(Rd ), Tf ∗ Tg , ϕ = Tf , Tg , ϕ(x + y) . Otherwise, ∀ϕ ∈ D(Rd ), Tf ∗ Tg , ϕ = f ⊗ g, ϕ(x + y). 6.1.2. Convolution of a distribution and a function in D(Rd ) The definition of the convolution product can be extended with no difficulty to any problem to the convolution product of a distribution by a function of class C ∞ with compact support. D EFINITION 6.2.– Let T ∈ D (Rd ) be a distribution and let ϕ ∈ D(Rd ). The convolution product of T by ϕ is the function T ∗ ϕ defined by T ∗ ϕ(x) = T, ϕ(x − ·) , ∀x ∈ Rd . P ROPOSITION 6.1 (Support of T ∗ ϕ).– Let T ∈ D (Rd ) be a distribution and let ϕ ∈ D(Rd ). We have supp (T ∗ ϕ) ⊂ supp T + supp ϕ. & Proof. Let x ∈ Rd (supp T + supp ϕ). We have supp ϕ(x − .) = {x} − supp ϕ. Thus, supp ϕ(x − .) ∩ supp T = ∅. This results & in T ∗ ϕ(x) = T, ϕ(x − ·) = 0. Consequently, T ∗ ϕ = 0 on the open set Rd (supp T + supp ϕ), which shows that this open set is included in the complementary support of T ∗ ϕ. It is deduced that supp (T ∗ ϕ) ⊂ supp T + supp ϕ. As for the convolution of a locally integrable function by a test function, the product of convolution of a distribution by a test function can be derived by obtaining the derivative of any of the two factors. P ROPOSITION 6.2 (Regularity of T ∗ ϕ).– Let T ∈ D (Rd ) be a distribution and let ϕ ∈ D(Rd ). The convolution product T ∗ ϕ of T by ϕ is a function of class C ∞ on Rd ; in addition, we have ∀α ∈ N d , ∂ α (T ∗ ϕ) = (∂ α T ) ∗ ϕ = T ∗ ∂ α (ϕ).
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The Theory of Distributions
Proof. Let α ∈ Nd . We have (∂ α T ) ∗ ϕ(x) = ∂ α T, ϕ(x − ·) = (−1)|α| T, ∂yα (ϕ(x − ·)) . Since (−1)|α| ∂yα (ϕ(x − ·)) = (∂ α ϕ)(x − ·), then (∂ α T ) ∗ ϕ(x) = T ∗ ∂ α ϕ(x). On the other hand, let x0 ∈ Rd and let χ ∈ D(Rd ) be the plateau function such that χ = 1 on B(x0 , 1). The function Φ defined by Φ : (x, y) −→ χ(x)ϕ(x − y) is a function of class C ∞ on Rd × Rd and supp Φ ⊂ supp (χ) × (supp (χ) + supp (ϕ)). According to proposition 4.11, addressing differentiation under the bracket, the function x −→ T, χ(x)ϕ(x − ·) = T, Φ(x, y) is of class C ∞ on Rd and we have ∂ α T, χ(x)ϕ(x − ·) = T, ∂xα (χ(x)ϕ(x − ·)) . The function x −→ T, Φ(x, y) coincide with T ∗ ϕ on B(x0 , 1), then T ∗ ϕ is of class C ∞ on B(x0 , 1) and we have T ∗ ∂ α ϕ(x) = (∂ α T ) ∗ ϕ(x) = ∂ α (T ∗ ϕ) (x) for all x ∈ B(x0 , 1). We conclude by observing that the result found is valid for all x0 ∈ Rd . R EMARK 6.1.– The convolution product of a distribution with compact support by a function of class C ∞ is defined by the same formula as that in definition 6.2. P ROPOSITION 6.3 (A different notation).– Let T ∈ D (Rd ) and ϕ ∈ D(Rd ). For all ψ ∈ D(Rd ), we have T ∗ ϕ, ψ = T, ϕˇ ∗ ψ = T, ϕ, ψ(x + y). Proof. Let ψ ∈ D(Rd ). Using the integration property, under the bracket, of proposition 4.12, we have T ∗ ϕ, ψ =
Rd
= Rd
T ∗ ϕ(x)ψ(x)dx Ty , ϕ(x − y) ψ(x)dx
≠ = Ty ,
∑ Rd
ϕ(x − y)ψ(x)dx
= Ty , ϕˇ ∗ ψ .
Convolution of Distributions
155
Then, T ∗ ϕ, ψ = T, ϕˇ ∗ ψ. By establishing that z = x − y, we get
Rd
ϕ(x − y)ψ(x)dx =
Rd
ϕ(z)ψ(y + z)dz = ϕ, ψ(y + ·)x .
It follows that T ∗ ϕ, ψ = Ty , ϕ, ψ(y + ·)x y = T, ϕ, ψ(y + x) . 6.1.3. Density of D(Ω) in D (Ω) Using proposition 6.2, a distribution can be approximated by a sequence of functions of class C ∞ . This is the goal of the following proposition. P ROPOSITION 6.4 (Density of C ∞ (Rd ) in D (Rd )).– Let T ∈ D(Rd ) be a distribution and ρε be a regularizing sequence (i.e. ρε (x) = ε−d ρ xε where
ρ ∈ C ∞ (Rd ) with support in B(0, 1), ρ ≥ 0 and
ρ(x)dx = 1). We set Rd d
Tε = T ∗ ρε . Then Tε ∈ C ∞ (Rd ) and Tε −→ T in D (R ). ε−→0
Proof. The fact that Tε ∈ C ∞ (Rd ) directly follows from proposition 6.2. In addition to proposition 6.2, the rest of the proof is based on the property of integration under the duality bracket seen in proposition 4.12. R EMARK 6.2.– Obviously, the above regularization property is easily localized in an open set of Rd . By combining the truncation principle and the regularization principle, a distribution can be approximated by a sequence of functions of class C ∞ with compact support. This is the concern of the following proposition. P ROPOSITION 6.5 (Density of D(Ω) in D (Ω)).– Let T ∈ D (Ω) be a distribution. Then there exists a sequence (Tn )n≥1 of functions of class C ∞ with compact supports such as Tn −→ T in D (Ω). n−→+∞
In proposition 6.6, we show the effect of convergence on a sequence of convoluted distributions. P ROPOSITION 6.6 (Convolution and convergence).– Let ϕ ∈ D(Rd ) denote a test function and (Tn )n≥1 a sequence of distributions on Rd converging to T in the sense of distributions. Then, for any multi-index α ∈ Nd , ∂ α (Tn ∗ ϕ) −→ ∂ α (T ∗ ϕ) n−→+∞
uniformly on any compact Rd .
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The Theory of Distributions
Proof. Without restricting generality, it can be assumed that T = 0. On the other hand, we simply have to show that (Tn ∗ ϕ) −→ 0 and to apply this statement by n−→+∞
recurrence by changing Tn into ∂ α Tn since, we have ∂ α (Tn ∗ ϕ) = ∂ α (Tn ) ∗ ϕ. By hypothesis, we have Tn , ϕ(x − ·)
−→
n−→+∞
∀x ∈ Rd .
0,
Let us set K = B(0, 1) + supp (ϕ) q where ϕ(x) q = ϕ(−x), which is compact in R . According to the principle of uniform boundedness (proposition 3.5), there exists C > 0 and p ∈ N such that d
sup |Tn ∗ ϕ(x)| = sup |Tn , ϕ(x − ·)| ≤ C sup sup |∂ α ϕ(x)| .
|x|≤R
|x|≤R
|α|≤p x∈K
Let us show that sup |Tn ∗ ϕ(x)| |x|≤R
−→
n−→+∞
0. Otherwise, there would exist η > 0
and a sequence xn ∈ B(0, 1) such that |Tn ∗ ϕ(xn )| > η, ∀n ≥ 1. Since B(0, 1) is compact, there exists nk ≥ 1 such that the sub-sequence xnk −→ x∗ ∈ B(0, 1),when k −→ +∞. Then |Tnk ∗ ϕ(x∗ )| ≥ |Tnk ∗ ϕ(xnk )| − |Tnk ∗ ϕ(xnk ) − Tnk ∗ ϕ(x∗ )| ≥ η − N |x∗ − xnk | max sup |Tnk ∗ ∂j ϕ(y)| 1≤j≤N |y|≤R
≥ η − N |x∗ − xnk | max (C max sup |∂ α ∂j ϕ(z)|) 1≤j≤N
≥ η − C |x∗ − xnk | ≥
|α|≤p z∈K
η . 2
Å ã α with C = N max C max sup |∂ ∂j ϕ(z)| and where the second inequality
1≤j≤N
|α|≤p z∈K
above follows from the increment theorem, while the third is a consequence of the uniform estimation on Tn ∗ ϕ obtained above. η > 0, but this contradicts the fact that 2 Tn ∗ ϕ(x) −→ 0 for all x ∈ Rd when n −→ +∞. It is deduced that the above assumption concerning the existence of the real η is false, that is, Therefore,
lim
k−→+∞
|Tnk ∗ ϕ(x∗ )| ≥
sup |Tn ∗ ϕ(x)|
|x|≤R
−→
n−→+∞
0.
Convolution of Distributions
Now, this means that Tn ∗ ϕ
−→
n−→+∞
157
0 uniformly B(0, R) for all R > 0.
6.1.4. Convolution of two distributions In a similar way, the notion of the tensor product and the convolution product can be generalized to distributions that are not necessarily regular. The derivation property under the bracket already seen in proposition 4.11 supports and gives a meaning to the notation defining this tensor product. T HEOREM 6.1.– Let T ∈ D (Ω) be a distribution on Ω and S ∈ D (Ω ) a distribution on Ω . Then, there exists a unique distribution on Ω × Ω denoted T ⊗ S, called tensor product such that ∀ϕ ∈ D(Ω), ∀ψ ∈ D(Ω ), T ⊗ S, ϕ ⊗ ψ = T, ϕ S, ψ . In addition, for any ϕ in D(Ω × Ω ) we have ¨ ∂ T ⊗ S, ϕ = Tx , Sy , ϕy = Sy , Tx , ϕx y . x
Based on proposition 6.3, the convolution product can be generalized to two distributions, of which at least one has compact support. D EFINITION 6.3.– Let T and S be two distributions in D (Rd ); it is assumed that at least one has compact support. The convolution T ∗ S is a distribution on Rd defined by ∀ϕ ∈ D(Rd ), T ∗ S, ϕ = Tx ⊗ Sy , ϕ(x + y) = Tx , Sy , ϕ(x + y) . We simply write ∀ϕ ∈ D(Rd ), T ∗ S, ϕ = T, S, ϕ(x + y) .
[6.3]
R EMARK 6.3.– 1) According to lemma 4.2, the function Φ : x −→ S, ϕ(x + y) is of class C ∞ on Rd , in addition if S ∈ D (Rd ) has support compact, then Φ ∈ D(Rd ), which indeed gives a meaning to the convolution product defined by formula [6.3]. 2) By generalizing the convolution product given in proposition 6.3 to the case of two distributions of which at least one is compactly supported, it can be written that ∀ϕ ∈ D(Rd ), T ∗ S, ϕ = T, Sˇ ∗ ϕ .
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The Theory of Distributions
R EMARK 6.4.– Why assume that at least one of the distributions has compact support? In order to simplify our development, only the case in which d = 1 is considered here. 1) We defined a test function on R2 by ψ : (x, y) −→ ψ(x, y) = ϕ(x + y) where ϕ ∈ D(R). Let (a, b) ∈ R2 such that supp ϕ ⊂ [a, b], we have ψ(x, y) = ϕ(x + y), is constant for x + y = c, with arbitrary c ∈ [a, b]. Therefore, supp ψ contains all lines of slope −1 that intersect the y−axis (as well as the x−axis) at a point c ∈ supp ϕ, as represented here below. As such, ψ does not have compact support, so it is not a test function. However, the action of extended distributions to test functions ϕ ∈ / D(R) can be used.
Figure 6.1. supp ψ is unbounded. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
Let us assume that the distribution S has a compact support. Then let a and b be such that supp S ⊂ [a, b]; in Figure 6.2, it can be seen that the test function ψ has compact support. In this figure, supp ϕ ⊂ [α, β]. 2) The definition of the convolution product T ∗ S of two distributions remains valid for two distributions T and S, “causal”, having two left bound supports: that is, supp T ⊂ [a, +∞[ and supp S ⊂ [b, +∞[. 3) The result remains true if the two distributions have their supports limited to the right: supp T ⊂] − ∞, a] and supp S ⊂] − ∞, b]. 6.1.5. Some examples E XAMPLE 6.1.– Let T be a distribution in D (Rd ).
Convolution of Distributions
Figure 6.2. First example in which supp ψ is bounded. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
Figure 6.3. Second example in which supp ψ is bounded. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
1) We have T ∗ δ0 = T . Indeed, for any ϕ in D(Rd ), we have ∂ ¨ T ∗ δ, ϕ = Tx , δ y , ϕ(x + y)y
x
= Tx , ϕ(x + 0) = Tx , ϕ(x) = T, ϕ .
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The Theory of Distributions
Therefore, T ∗ δ = T . Similarly, we have δ ∗ T = T . 2) We have δa ∗ T = τa T . For any ϕ in D(Rd ), we actually have δa ∗ T, ϕ = δa ⊗ T, ϕ(x + y) ¨ ∂ = δax , Ty , ϕ(x + y)y
x
= Ty , ϕ(a + y) = T, τ−a ϕ = τa T, ϕ . 3) We have δa ∗ δb = δa+b . Furthermore, for any ϕ in D(Rd ), we have δa ∗ δb , ϕ = δa ⊗ δb , ϕ(x + y) ¨ ∂ = δax , δby , ϕ(x + y)y
x
=
δby , ϕ(a
+ y)
= ϕ(a + b) = δa+b , ϕ . R EMARK 6.5.– 1) The convolution product «∗ » is an internal composition law on E (Rd ) of neutral element δ = δ0 . 2) The convolution product «∗ » is not an internal law on D (Rd ). E XAMPLE 6.2.– Let T be a distribution in D (R). We have δ ∗ T = T . Indeed, for any ϕ in D(R), we have δ ∗ T, ϕ = δ ⊗ T, ϕ(x + y) ¨ ∂ = δx , Ty , ϕ(x + y)y
x
=−
d (Ty , ϕ(x + y)) |x=0 dx
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161
∑ ≠ d = − Ty , (ϕ(x + y)|x=0 (we use differentiation under the bracket) dx = − Ty , ϕ (0 + y) = − T, ϕ = T , ϕ . D+ (R) = T ∈ D (R), supp T ⊂ [a, +∞[ ,a ∈ R E XAMPLE 6.3.– We consider (respectively, D− (R) = T ∈ D (R), supp T ⊂ ]−∞, b] , b ∈ R ) the set of distributions with left-hand limited supports “also called causal” (respectively, right-hand limited). The set D+ (R) (respectively, D− (R)) is a real vector space. In addition, the convolution product “∗” is well defined on D+ (R) (respectively, on D− (R)).
6.2. Properties of convolution In this section, we give some properties of the convolution of two distributions. 6.2.1. Support of a convolution P ROPOSITION 6.7.– Let S and T be two distributions on Rd of which at least one has compact support. We have supp (T ∗ S) ⊂ supp T + supp S. Proof. We introduce the mapping s : supp T × supp S −→ Rd (x, y) −→ x + y. Let ϕ denote the mapping ϕ : (x, y) −→ ϕ(x + y). Let x ∈ / supp T + supp S. Since supp T + supp S is a closed set, there exists δ > 0 such that B(x, δ) ∩ (supp T + supp S) = ∅. Then, for ϕ ∈ D B(x, δ) , we have supp ϕ ∩ (supp T × supp S) = s−1 (supp ϕ) ⊂ s−1 B(x, δ) = ∅. / supp (T ∗ S). Therefore, T ∗ S, ϕ = T ⊗ S, ϕ = 0 and x ∈ Consequently, supp (T ∗ S) ⊂ supp T + supp S.
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The Theory of Distributions
R EMARK 6.6.– The reciprocal inclusion does not hold in general. Two locally integrable functions u and v should be taken with u = 1 and v(x)dx = 0. We Rd
then have u ∗ v = 0, hence supp (Tu ∗ Tv ) = ∅ while supp Tu + supp Tv = Rd since supp Tu = Rd . However, we can have a form of reciprocal inclusion by considering distributions with compact support only and convex envelopes. If A ⊂ Rd , we denote Conv(A) its convex envelope, that is, the smallest convex Rd set that contains A. We then have the following result. T HEOREM 6.2.– Let S,T ∈ E (Rd) be two distributions on Rd with compact supports. We have Conv supp (T ∗ S) = Conv supp T + Conv supp S . 6.2.2. Sequential continuity of the convolution product T HEOREM 6.3.– Let Tn n≥1 be a sequence of D (Rd ) and Sn n≥1 a sequence of E (Rd ) such that Tn −→ T and Sn −→ S in D (Rd ). Furthermore, let us n→+∞
n→+∞
assume that there exists a compact K ⊂ Rd such that supp(Sn ) ⊂ K, for all n ≥ 1. Then, Tn ∗ Sn −→ T ∗ S in D Rd . n→+∞
Proof. Let ϕ ∈ D(Rd ). We have Tn ∗ Sn , ϕ = Tn , Sˇn ∗ ϕ . By hypothesis for all n ≥ 1, we have supp Sˇn ∗ ϕ ⊂ supp Sˇn + supp ϕ ⊂ supp ϕ − K. Therefore, supp Sˇn ∗ ϕ is compact as a closed set included in the sum of two compacts. By proposition 6.6, for any multi-index α ∈ Nd , we have ∂ α Sˇn ∗ ϕ = ∂ α Sˇn ∗ ϕ
−→
n−→+∞
∂ α Sˇ ∗ ϕ = ∂ α Sˇ ∗ ϕ
uniformly on any compact of Rd . Otherwise, Sˇn ∗ ϕ
−→
n−→+∞
Sˇ ∗ ϕ,
in D(Rd ).
Corollary 3.2 gives Tn ∗ Sn , ϕ = Tn , Sˇn ∗ ϕ
−→
n−→+∞
T, Sˇ ∗ ϕ = T ∗ S, ϕ .
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163
The test function ϕ ∈ D(Rd ) being arbitrary, it is deduced that Tn ∗ S n
−→
n−→+∞
T ∗ S in D (Rd ).
6.2.3. Associativity and convolution
The convolution product “∗” is not associative in D (Rn ) (see the counterexample below). However, we have the following result. P ROPOSITION 6.8.– Let R, S and T ∈ D (Rd ), verifying one of the following two conditions, namely, i) at least two of them have compact support; ii) or they are all causal. Then we have R ∗ S ∗ T = R ∗ S ∗ T . Proof. By virtue of commutativity, the problem can always be reduced to the case where S and T have compact supports. If we take r > 0 and large enough, we will assume that supp S, supp T ⊂ B(0, r). We consider the regularizing sequence ρn defined by ρn (x) = nd ρ(nx) where ρ(x)dx = 1. ρ ∈ D(Rd ) is a function such that supp ρ ⊂ B(0, 1), ρ ≥ 0, Rd
We set Tn = T ∗ ρn and Sn = S ∗ ρn . Let us verify that R ∗ Sn ∗ Tn = R ∗ Sn ∗ Tn . We have R ∗ Sn ∗ Tn (x) = R, (Sn ∗ Tn )(x − ·) ≠ ∑ = R, Sn ((x − ·) − z) Tn (z)dz . Rd
The proposition of integration under the duality bracket makes it possible to have R ∗ Sn ∗ Tn (x) =
Rd
= Rd
R, Sn ((x − ·) − z) Tn (z) dz R ∗ Sn (x − z) Tn (z)dz
= (R ∗ Sn ) ∗ Tn (x).
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The Theory of Distributions
Furthermore, R ∗ Sn ∗ Tn = R ∗ Sn ∗ Tn . By construction, for all n ≥ 1, we have supp Sn ⊂ B(0, r + 1)
and
supp Tn ⊂ B(0, r + 1)
and Sn
−→
n−→+∞
S,
Tn
−→
n−→+∞
T
in D (Rd ).
Theorem 6.3 of sequential continuity of the convolution product ensures, on the one hand, that R∗Sn −→ R∗S, then R∗Sn ∗Tn −→ R∗S ∗T in D (Rd ). n−→+∞
n−→+∞
On the other hand, it is similarly shown that Sn ∗Tn
−→
n−→+∞
S ∗T
and supp Sn ∗Tn ⊂ supp Sn + supp Tn ⊂ B(0, 2r +2).
By sequential continuity of the convolution product, it is further deduced that R ∗ Sn ∗ T n
−→
n−→+∞
R ∗ S ∗ T in D (Rd ).
Consequently, R ∗ S ∗ T = R ∗ S ∗ T . R EMARK 6.7.– The convolution product of n distributions of which (n − 1) at least have compact support is associative and commutative. We should bear in mind that this result may not hold if only one of the three distributions has compact supports, although the right- and left-hand members of the previous equality make sense. Here follows an example of this phenomenon. E XAMPLE 6.4 (Counterexample of non-associativity).– We consider the distributions H, δ and 1. The products H ∗ δ ∗ 1 and H ∗ δ ∗ 1 do exist but are different. As a matter of fact, H ∗ δ ∗ 1 = H ∗ δ ∗ 1 = δ ∗ 1 = 1 H ∗ δ ∗ 1 = H ∗ δ ∗ 1 = H ∗ 1 = H ∗ 0 = 0. It should be observed that H, δ and 1 do not satisfy the conditions ensuring associativity.
Convolution of Distributions
165
6.2.4. Differentiation and convolution In this section, we present the action of differentiation on the convolution of two distributions. P ROPOSITION 6.9.– Let S and T in D (R). If S ∗ T exists, then we have S ∗ T = S ∗ T = S ∗ T . In general, we have (p+q) ∀(p, q) ∈ N2 , S (p) ∗ T (q) = S (p+q) ∗ T = S ∗ T (p+q) = S ∗ T . Proof. Let S and T in D (R) such that S ∗ T exists, that is, either they are both causal, or at least one of them has compact support, then δ , S and T satisfy the previous conditions (actually, in the first case, they are all causal, including δ , and in two have indeed compact support, in particular δ ). We thus have the second, δ ∗ S ∗ T = S ∗ δ ∗ T = δ ∗ S ∗ T . From example 6.2, we obtain S ∗ T = S ∗ T = S ∗ T . Reasoning by recurrence leads to easily generalizing the formula to any order, and we can write that (p+q) . ∀(p, q) ∈ N2 , S (p) ∗ T (q) = S (p+q) ∗ T = S ∗ T (p+q) = S ∗ T R EMARK 6.8.– To derive a convolution product, one of the factors has simply to be derived. The result is given for distributions on R, but it remains valid for distributions defined on Rd . A generalization in the space is given by the following proposition. P ROPOSITION 6.10.– Let S and T in D (Rd ) and α ∈ Nd . If S ∗ T exists, we then have 1) ∂ α T = (∂ α δ) ∗ T ; 2) ∂ α (S ∗ T ) = ∂ α S ∗ T = S ∗ ∂ α T . (n)
A PPLICATION EXERCISE.– Compute xm δ0
(p)
∗ x p δ0 .
E XAMPLE 6.5.– Let T ∈ D (R) be a causal distribution, then a primitive P of T is simply written as P = H ∗ T . In fact, P = H ∗ T = δ ∗ T = T .
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The Theory of Distributions
6.2.5. Translation and convolution P ROPOSITION 6.11.– Let S, T ∈ D (Rd ) denote two distributions such that S ∗ T exists. For all a ∈ Rd , we have τa S ∗ T = S ∗ τa T = τa (S ∗ T ). for applying associativity. Proof. The δa ,S and T verify the conditions distributions We have δa ∗ S ∗ T = S ∗ δa ∗ T = δa ∗ S ∗ T . Using the fact that δa ∗ T = τa T already seen in example 6.1, we get τa S ∗ T = S ∗ τa T = τa S ∗ T . 6.2.6. Algebraic study of D+ (R) (R). P ROPOSITION 6.12.– The convolution produit “∗” is an internal law in D+ (R), such that supp T ⊂ [a, +∞[ and supp S ⊂ [b, +∞[, Proof. For any T, S ∈ D+ the convolution product T ∗ S is well defined. By proposition 6.7 on the support of the convoluted of two distributions, we have supp (T ∗ S) ⊂ supp T + supp S. Therefore, supp (T ∗ S) ⊂ [a + b, +∞[ . Consequently, T ∗ S ∈ D+ (R). P ROPOSITION 6.13.– D+ (R) is a commutative unitary algebra for the convolution product.
Proof. The mapping (T, S) −→ T ∗ S is bilinear. The commutativity follows from the fact that ϕ(x+y) = ϕ(y +x). The neutral element is δ. Associativity is the subject of proposition 6.8. R EMARK 6.9.– 1) The proposition holds D+ (R) ∩ D− (R) = E (R).
for
D− (R)
and
consequently
for
2) An interesting special case: the spaces D0,+ (R) = T ∈ D (R), supp T ⊂ [0, +∞[ , D0,− (R) = T ∈ D (R), supp T ⊂ ]−∞, 0] and D0 (R) = D0,+ (R) ∩ D0,− (R) are commutative unitary algebras for the convolution product. Any element T of D0 (R) has point support {0}. By theorem 5.2, the distribution T is uniquely written as a differential polynomial in δ. Then, the following result is deduced.
Convolution of Distributions
167
P ROPOSITION 6.14.– D0 (R) endowed with the law “∗” is isomorphic to the algebra of polynomials. Proof. We consider the mapping Θ :
ak δ (k) −→
0≤k≤m
ak X k .
0≤k≤m
R EMARK 6.10.– Since the algebra of polynomials is integral, D0 (R) is also integral and admits a fraction field. The latter is identified with a subalgebra of D0,+ (R). (R) generated by D0 (R) and distributions T HEOREM 6.4.– The subalgebra of D0,+ λx similar to H(x)e with λ ∈ C is a commutative field for the law “∗”. It is the fraction field of D0 (R). Moreover, it is the direct sum of D0 (R) and the space of distributions N similar to H(x) Pj eλj x (the λj ∈ C and the Pj are polynomials). j=0
This field is isomorphic to the field C(X) of rational fractions. 6.3. Exercises with solutions E XERCISE 6.1.– Let a, b ∈ Rd . Compute the convolution product in each of the following cases: 1) δa ∗ δb ; 2) δa ∗ δb ; 3) δa ∗ u where u ∈ D (Rd ). S OLUTION 6.1.–
1) Let ϕ ∈ D Rd . By definition, we have δa ∗ δb , ϕ = δa , δb , ϕ(x + y) = δa , ϕ(x + b) = ϕ(a + b)
which shows that δa ∗ δb = δa+b . 2) Let us recall that ∂ α (u ∗ v) = (∂ α u) ∗ v = u ∗ (∂ α v). Thus, δa ∗ δb = (δa ∗ δb ) = δa+b .
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The Theory of Distributions
3) We have δa ∗ u = δa ∗ u , and a calculation similar to that of the first question shows that δa ∗ u = τa (u ). E XERCISE 6.2.– Calculate the convolution product (H(x) cos x) ∗ (δ + H(x)) , where H(x) is the Heaviside function. S OLUTION 6.2.– We have (H(x) cos x) ∗ (δ + H(x)) = (H(x) cos x) ∗ δ + (H(x) cos x) ∗ H(x)
= (H(x) cos x) ∗ δ + H(x) ∗ (H(x) cos x)
= (H(x) cos x) + H(x) ∗ (H(x) cos x) = H (x) cos x − H(x) sin x + H(x) ∗ (H(x) cos x) = δ cos x − H(x) sin x + H(x) ∗ (H(x) cos x) = δ − H(x) sin x + H(x) ∗ (H(x) cos x). We compute H(x) ∗ (H(x) cos x). For ϕ ∈ D(R), we have H(x) ∗ (H(x) cos x), ϕ = H(x), (H(y) cos y), ϕ(x + y) ∏ Æ +∞ =
cos yϕ(x + y)dy
H(x),
+∞
0
+∞
=
cos yϕ(x + y)dydx
0
0
+∞
Ç
=
ϕ(x + y)dx cos ydy. 0
å
+∞
0
+∞
We set ψ(y) =
ϕ(x + y)dx. Then 0
H(x) ∗ (H(x) cos x), ϕ =
+∞
ψ(y) cos ydy 0 +∞
= [ψ(y) cos y]0 =−
+∞ 0
−
+∞
0
ψ (y) sin ydy.
ψ (y) sin ydy
Convolution of Distributions
+∞
By setting u = x + y, we have ψ(y) =
169
ϕ(u)du. Then, ψ (y) = −ϕ(y). It
y
follows that H(x) ∗ (H(x) cos x), ϕ =
+∞
+∞
ϕ(y) sin ydy = 0
H(y) sin yϕ(y)dy. −∞
Consequently, H(x) ∗ (H(x) cos x), ϕ = H(x) sin x, ϕ. Therefore, H(x) ∗ (H(x) cos y) = H(x) sin x. Finally, (H(x) cos x) ∗ (δ + H(x)) = δ. E XERCISE 6.3.– Let S, T ∈ D+ (R) from the space add two distributions and a ∈ C be a complex number. Show that eax (S ∗ T ) = eax S ∗ eax T .
S OLUTION 6.3.– By definition, for any test function ϕ ∈ D(R), we have eax S ∗ eax T, ϕ = eax Sx , eay Ty , ϕ(x + y) = Sx , eax eay Ty , ϕ(x + y) = Sx , Ty , ea(x+y) ϕ(x + y) = S ∗ T, eax ϕ = eax (S ∗ T ), ϕ. Consequently, eax (S ∗ T ) = eax S ∗ eax T . E XERCISE 6.4.– Let H be a Heaviside distribution, and let H ∗n denote the convolution product of H by itself n−times: H ∗n = H ∗ H ∗ · · · ∗ H. 1) Show that for any natural number n ≥ 1, we have H ∗n =
xn−1 H. (n − 1)!
2) Therefrom, deduce the derivative of order n of H ∗n , for all n ≥ 1. S OLUTION 6.4.– 1) The relation is obvious for n = 1. By induction, assume that H ∗(n−1) =
xn−2 H. (n − 2)!
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The Theory of Distributions
For any test function ϕ ∈ D(R), we have H ∗n , ϕ = Hx , Hy∗(n−1) , ϕ(x + y). By induction hypothesis, we have
Hy∗(n−1) , ϕ(x + y)
+∞
= 0
y n−2 ϕ(x + y)dy = (n − 2)!
+∞ x
(s − x)n−2 ϕ(s)ds. (n − 2)!
As a result, it follows that H ∗n , ϕ =
+∞
0
+∞
x
(s − x)n−2 ϕ(s)dsdx. (n − 2)!
By inverting the order of the integrals, we obtain H ∗n , ϕ =
+∞
Å
0
0 +∞
= 0
=
s
ã (s − x)n−2 dx ϕ(s)ds (n − 2)!
sn−1 ϕ(s)ds (n − 1)!
xn−1 H, ϕ. (n − 1)!
The proof is thus complete. 2) A simple calculation gives (H ∗n )
(n)
= δ, for all n ≥ 1.
E XERCISE 6.5.– Let P ∈ C[X] be a polynomial of degree d and let u ∈ E (R). Show that u ∗ P is also a polynomial of degree less than or equal to d. S OLUTION 6.5.– Since v −→ u ∗ v is a linear mapping, it can be assumed that: P = X k , with 0 ≤ k ≤ d. We have (u ∗ X k )(x) = u, (x − y)k k Å ã k xi (−y)k−i = u, i i=0
=
k Å ã k i=0
i
xi u, (−y)k−i .
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Since the u, (−y)k−i are constants in C, it is deduced that u ∗ P is a polynomial of degree ≤ d. E XERCISE 6.6.– 1) Let T ∈ D (R). Show that there exists a distribution E ∈ D (R), with compact support, such that E ∗ T = T (k) . 2) Let T and S in D (R), andS is assumed to have compact support. For n ∈ N, we denote by X n the function, from R in R, x −→ xn . Demonstrate the following n Å ã n n (X k T ) ∗ X n−k S . formula: X (T ∗ S) = k k=0
S OLUTION 6.6.– (k)
1) The distribution E = δ0 is suitable. It happens that if we derive k-times the (k) identity δ0 ∗ T = T , we will obtain δ0 ∗ T = T (k) . 2) Let ϕ ∈ D(R). If we set ϕ (x, y) = ϕ(x + y), we have X n (T ∗ S), ϕ = T ∗ S, X n ϕ = T ⊗ S, (x + y)n ϕ (x, y) n Å ã n = T ⊗ S, xk y n−k ϕ (x, y) k k=0
=
n Å ã n
k=0
=
k
(xk T ) ⊗ (y n−k S), ϕ (x, y)
n Å ã n (X k T ) ∗ (X n−k S, ϕ . k k=0
Consequently, X (T ∗ S) = n
n Å ã n k=0
k
(X k T ) ∗ X n−k S .
(R) = T ∈ D (R), supp(T ) ⊂ R+ . E XERCISE 6.7.– We write D0,+ Let χ ∈ C ∞ (R) such that, χ = 1 on − 12 , +∞ and χ = 0 on − ∞, −1 . 1) a) Let ϕ ∈ D(R). Show that the mapping ϕ : (x, y) → χ(x)χ(y)ϕ(x + y) is in D(R2 ).
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b) Let T, S ∈ D0,+ (R). We define: T ∗ S, ϕ = Tx ⊗ Sy , ϕ . Show that T ∗ S is independent of the choice of χ. Then deduce that T ∗ S ∈ D0,+ (R). (R) is invertible, if there exists S ∈ D0,+ (R) such that 2) We say that T ∈ D0,+ T ∗ S = δ0 . We write S = T −1 . a) Show that δ0 is invertible and compute its inverse.
b) Show that if T ∈ D(]0, +∞[), then T is not invertible. S OLUTION 6.7.– 1) a) Since ϕ has compact support, there exists a > 0 such that supp(ϕ) ⊂ [−a, a]. It can be observed that if χ(x)χ(y)ϕ(x + y) = 0, then we have x, y ≥ −1 and x + y ≤ a. Hence, x, y ∈ [−1, a + 1]. Therefore, we obtain supp(ϕ ) ⊂ [−1, a + 1]2 , and it follows that ϕ ∈ D(R2 ). b) According to the previous question, the distribution T ∗ S is well defined. Let then χ1 denote another function verifying the same assumptions as χ. We have Tx ⊗ Sy , ϕ (x, y) − Tx ⊗ Sy , χ1 (x)χ1 (y)ϕ(x + y) = Tx ⊗ Sy , (χ(x) − χ1 (x)) χ(y)ϕ(x + y) + Tx ⊗ Sy , χ1 (x) (χ(y) − χ1 (y)) ϕ(x + y) . Since supp (χ − χ1 ) ⊂ −∞, − 21 , then supp (T ) ∩ supp (χ − χ1 ) = ∅, which implies that the first bracket of the second member of the equality is zero. In addition, since supp(S) ∩ supp (χ − χ1 )∂ = ∅, the second bracket is also zero. ¨ Therefore, Tx ⊗ Sy , ϕ = Tx ⊗ Sy , ϕ (corresponding to χ1 ) and T ∗ S does 1 not depend on the choice of χ. 2) Let γ ∈ 0, 12 and let χγ ≡ 1 on ] − γ, +∞[ and χγ ≡ 0 on ] − ∞, −2γ[. From Question (a, b), we have ∀ϕ ∈ Cc∞ (R), T ∗ S, ϕ = T ⊗, ϕ(x + y)χγ (x)χγ (y) . We choose ϕ with support in ] − ∞, 0[, then there exists a > 0 such that: supp(ϕ) ⊂] − ∞, −a[. Therefore, (x, y) −→ ϕ(x + y)χγ (x)χγ (y) is non-zero only if x + y ≤ −a and x, y ≥ −2γ, or still x, y ≤ −a + 2γ.
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However, for γ small enough, we can have −a + 2γ < 0. Since supp (T ) ⊂ [0, +∞[ and supp(S) ⊂ [0, +∞[, we then obtain: T ∗ S, ϕ = 0. As such, supp (T ∗ S) ⊂ [0, +∞[. 3) a) Let H denote a Heaviside distribution, that is the indicator function of ]0, +∞[. Then, H ∈ D0,+ (R) and we have H ∗ δ0 = H ∗ δ0 = δ0 ∗ δ0 = δ0 . Therefore, H is the inverse of δ0 in D0,+ (R). b) We assume by contradiction that there exists S ∈ D0,+ (R) such that T ∗ S = δ0 .
Since T ∈ D (]0, +∞[), we should have that T ∗ S is the distribution associated with a function of D (]0, +∞[). However, δ0 is not. It is a contradiction! We have the following result: if T ∈ D (]0, +∞[), then T is not invertible in D0,+ (R). (R), the following equation: E XERCISE 6.8.– We consider, in D+
(E) :
(δ + δ) ∗ X = δ.
We set X = H(t)Z(t). 1) Verify that in order to solve (E), we simply have to solve the differential equation (E ) Z + Z = 0 with Z(0) = 0 and Z (0) = 1. 2) Solve (E ), then deduce X. S OLUTION 6.8.– 1) We have (δ + δ) ∗ X = δ ⇐⇒ δ ∗ X + δ ∗ X = δ ⇐⇒ δ ∗ X (t) + δ ∗ X(t) = δ ⇐⇒ X (t) + X(t) = δ ⇐⇒ H (t)Z(t) + 2H (t)Z (t) + Z (t)H(t) + H(t)Z(t) = δ ⇐⇒ δ Z(t) + 2δZ (t) + Z (t)H(t) + H(t)Z(t) = δ ⇐⇒ (δZ(t)) + δZ (t) + H(t)Z (t) + H(t)Z(t) = δ ⇐⇒ Z(0)δ + Z (0)δ + (Z (t) + Z(t))H(t) = δ ⇐⇒ Z (t) + Z(t) = 0, Z(0) = 0, Z (0) = 1.
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We point out here that shifting from the penultimate equivalence to the last one is due to the fact that the three distributions H, δ and δ form a free family. Then, to solve (E) the differential equation (E ): Z + Z = 0 with Z(0) = 0 and Z (0) = 1 should be solved.
2) The solution of the equation (E ) is Z(t) = αe−it + βeit . Since Z(0) = 0 and 1 1 Z (0) = 1, then α + β = 0 and (−α + β)i = 1. Hence, α = − 2i and β = − 2i . It follows that Z(t) = sin t.
We have Z(t) = sin t, thus X(t) = H(t) sin t is a solution to the equation (E). Conclusion: From equation (E), it is deduced that (δ + δ) is invertible in D+ (R) −1 and (δ + δ) = H(t) sin t. (R), of the following distributions: E XERCISE 6.9.– Compute the inverses, in D+
1) δ ; 2) δ + H. S OLUTION 6.9.– 1) By definition, the inverse of δ in D+ (R) is obtained by solving the equation δ ∗ X = δ.
We have (δ ∗ X) = δ, d’où X = δ and consequently, X = H. Finally, in D+ , we have (δ )−1 = H. 2) As previously, we look for X in D+ (R) such that (δ + H) ∗ X = δ.
We have ((δ + H) ∗ X) = δ , hence (δ + H) ∗ X = δ . Therefore, (δ + δ) ∗ X = δ , thus X = (δ + δ)−1 ∗ δ . From exercise 6.8, we have (δ + δ)
−1
= H(t) sin t, thus
X = H sin t ∗ δ = (H sin t ∗ δ) = (H sin t) = H sin t + H cos t = δ sin t + H cos t. Consequently, X = H(t) cos t. Finally, in D+ (R), we have
(δ + H)
−1
= H(t) cos t.
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E XERCISE 6.10.– In the following, we study the solutions to the equation T − T − 2T = F. I) The homogeneous equation F = 0. 1) Give the set of solutions to the homogeneous equation y − y − 2y = 0.
[6.4]
2) Give the unique solution to the homogeneous equation [6.4] such that y(0) = 0, y (0) = 1. This solution will be denoted by y0 . 3) Specify the set of distributions solutions to the equation T − T − 2T = 0. II) Search for a particular solution. 1) Let us assume that F is a function-type distribution, for example F = [x → e3x ]. Find, by application of the variation of the constant, a particular solution to the equation: y − y − 2y = e3x . Deduce a distribution which is solution of: T − T − 2T = [x → e3x ]. 2) Let us examine the more general case of a distribution F with compact support. a) Recall the definition of the support of a distribution. b) Give the support of each of the following distributions: δ1 and [1[−1,1] ]. c) Consider E = [y0 H]. After calculating E and E , verify that E is a solution to: T − T − 2T = δ0 . d) What is F ∗ δ0 equal to? e) It should be noted that since F has compact support, E∗F is well defined. Verify that E ∗ F is a particular solution to the equation T − T − 2T = F. f) Give a particular solution for F = [1[0,1] ]. III) The conclusion. Let S be a solution distribution of T − T − 2T = F . Show that S − F ∗ E is a solution to the homogeneous equation: T − T − 2T = 0. Then conclude the solution set.
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IV) Example. Solve: T − 2T = δ0 + 1[0,1] . S OLUTION 6.10.– I) The homogeneous equation F = 0. 1) The characteristic equation is: r2 − r − 2 = 0. It has two possible roots r1 = −1, r2 = 2. The real solutions of the homogeneous equation are given by SH = t −→ c1 e−t + c2 e2t |c1 , c2 ∈ R . 2) Let us look for the solution y verifying the conditions y(0)=0 (∗) : y (0) = 1. C1 + C2 = 0 We have (∗) ⇐⇒ −C1 + 2C2 = 1
1 and therefore C1 = −C2 = − . 3
Hence, y0 : t −→
1 2t e − e−t . 3
3) The induced distributions [y] with y ∈ SH simply have to be considered. II) Search for a particular solution. 1) Let us look directly for a solution in the form yp : t −→ ce3t (since 3 is not root). Thus: yp (t) − yp (t) − 2yp (t) = (9c − 3c − 2c)e3t = e3t =⇒ c =
1 . 4
Finally, [yp ] is a solution distribution. 2) a) The support of a distribution is the smallest closed set F such that for any ϕ with support in R \ F , we have T, ϕ = 0. b) Here we have supp(δ1 ) = {1} and supp(1[−1,1] ) = [−1, 1]. c) Using the fact that for a function such that f (0) = 0, we have f δ0 = f (0)δ0 = 0, the derivatives are E = [y0 H] = y0 [H] + y0 [H] = y0 [H] + y0 (0)δ0 = y0 [H]
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E = [y0 H] = [y0 [H]] = y0 [H] + y0 (0)δ0 = y0 [H] + δ0 . By summation and using the fact that y0 − y0 − 2y0 = 0, we have E − E + 2E = (y0 − y0 − 2y0 )[H] + δ0 = δ0 . E is therefore indeed a solution. d) Consider two distributions T and S, then (T ∗ S) = T ∗ S = T ∗ S . By applying this property of the convolution operator, E − E + 2E = δ0 =⇒ F ∗ (E − E + 2E) = F ∗ δ0 . We have F ∗δ0 = F and F ∗(E −E +2E) = (F ∗E) −(F ∗E) +2(F ∗E). Since F ∗ E = E ∗ F , it is deduced that this distribution, E ∗ F , is indeed a solution to our problem. e) A particular solution is given by the distribution: [y0 H] ∗ [1[0,1] ] = [y0 H1[0,1] ], that is, the distribution induced by the function 1 x −→ 3
min(x+1,0) min(x−1,0)
e2t − e−t dt.
f) We simply have to achieve the difference of the two following equations: S − S + 2S = F, (F ∗ E) − (F ∗ E) + 2(F ∗ E) = F. Then (S − F ∗ E) − (S − F ∗ E) + 2(S − F ∗ E) = F − F = 0 III) Conclusion. From the above, if S is a solution distribution, then S − F ∗ E is a solution to the homogeneous equation. Now, we have already described the space of solutions of the homogeneous equation. We know that S − F ∗ E is of [yH ] type with yH ∈ SH . The set of solutions is therefore S = F ∗ E + [t → c1 e−t + c2 e2t ]|(c1 , c2 ) ∈ R2 . The space is said to possess an affine space structure.
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IV) Example. The characteristic equation is r2 − 2 = 0.√The set of solutions of the homogeneous √ equation is S = F ∗ E + [t → c1 e 2t + c2 e− 2t ]|(c1 , c2 ) ∈ R2 . The √ 1 √ fundamental solution is y0 (t) = √ e 2t − e− 2t . A particular solution is given 2 2 by [y0 H] ∗ δ1 + [1[0,1] ] . Now, for any function f ∈ L1loc , [f ] ∗ δ1 = [t → f (t − 1)] ⇒ [δ0 H] ∗ δ1 = [t → (t − 1)e
[δ0 H] ∗ [1[−1,1] ] = [δ0 H] ∗ 1[−1,1]
1 = x→ √ 2 2
√ 2(t−1)
1[1,+∞[ (t)]
min(x+1,0) √ 2t
e
min(x,0)
− e−
√ 2t
dt .
Finally, the solutions to the equation are given by the particular solution (found with the fundamental solution) and the solutions to the homogeneous equation √ √ S = [y0 H] ∗ δ1 + [1[0,1] ] + t → c1 e 2t + c2 e− 2t | (c1 , c2 ) ∈ R2 .
7 Schwartz Spaces and Tempered Distributions
So far, we have defined the distributions on the space of test functions D(Rd ). There are, however, particular distributions that are defined as linear continuous functionals on spaces larger than D(Rd ). Among these spaces, the Schwartz spaces are introduced here, which will be denoted by S(Rd ). These spaces will play an important role in defining a restricted class of distributions called tempered distributions, which are very useful for solving physical problems. 7.1. S(Rd ) Schwartz spaces Several definitions can be found in the literature on Schwartz spaces and they are all equivalent. This section starts by presenting some of these definitions. 7.1.1. Definitions and examples D EFINITION 7.1.– Let ϕ be a Rd definite function with complex values. The function ϕ is said to be rapidly decreasing if for any α ∈ Nd , there exists a constant cα > 0 such that: ∀ x ∈ Rd , |xα ϕ(t)| ≤ cα . R EMARK 7.1.– It can be seen that a function is rapidly decreasing if and only if its product by any polynomial tends to 0 at infinity. E XAMPLE 7.1.– 1) The function ϕ1 : x −→ exp(−x2 ) is a rapidly decreasing function.
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2) The function ϕ2 : x −→ exp(−|x|) is a rapidly decreasing function but not indefinitely differentiable. 3) A non-zero polynomial does not define a rapidly decreasing function. 4) The function
1 1+x2
is of class C ∞ but not rapidly decreasing.
R EMARK 7.2.– Contrary to what its name might suggest, definition 7.1 implies no monotonicity for f , even in a neighborhood of infinity as inf (x) = e−|x| sin x.
Figure 7.1. Function f (x) = e−|x| sin x
P ROPOSITION 7.1.– Let ϕ be a Rd definite function with complex values. The function ϕ is rapidly decreasing if only if, for all m ∈ N, there exists a constant 1 d cm > 0 such that: |ϕ(x)| ≤ cm (1+|x|) m,∀x ∈ R . | · | denotes the Euclidean norm of Rd .
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181
d Proof. Let us assume √ that ϕ αis rapidly decreasing. Let m ∈ N be fixed. For all x ∈ R , we have |x| ≤ d max |x |. On the other hand, by a convexity argument of the |α|=1
function x → xm , there exists am > 0 such that ∀t ≥ 0, (1 + t)m ≤ am (1 + tm ). m m Therefore, (1 + |x|) ≤ am (1 + d 2 max |xα |). Hence, |α|=m
m
m
(1 + |x|) |ϕ(x)| ≤ am (1 + d 2 max |xα |) |ϕ(x)| . |α|=m
Given that ϕ is rapidly decreasing, then there exists a constant bm > 0 such that m
∀x ∈ Rd , (1 + d 2 max |xα |) |ϕ(x)| ≤ bm . |α|=m
m
Consequently, ∀x ∈ Rd , (1 + |x|) |ϕ(x)| ≤ cm = am bm . For the reciprocal, for all α ∈ Nd , we have ∀x ∈ Rd , |xα | ≤ |x||α| ≤ |α| |α| (1 + |x|) . Furthermore, ∀x ∈ Rd , |xα | |ϕ(x)| ≤ |x||α| ≤ (1 + |x|) |ϕ(x)| < ∞. This completes the proof. R EMARK 7.3.– It has just been shown that m
∀α ∈ Nd , sup |xα | |ϕ(x)| < ∞ ⇐⇒ ∀m ∈ N, sup (1 + |x|) |ϕ(x)| < ∞ x∈Rd
[7.1]
x∈Rd
D EFINITION 7.2.– Let ϕ be a Rd definite function with complex values. The function ϕ is said to be tempered or at most polynomially growing, if there exists N ∈ N, and N a constant cN > 0 such that |ϕ(x)| ≤ cN (1 + |x|) , ∀ x ∈ Rd . E XAMPLE 7.2.– 1) Any polynomial function is at most a polynomially growing function. 2) The two functions f1 : x −→ 1 + t2 cos(t) and f2 : x −→ sin(exp(t)) are tempered. D EFINITION 7.3.– – A complex-valued function ϕ defined on Rd is said to be a Schwartz function if it is indefinitely differentiable and if it is, as well as all its derivatives rapidly decreasing. – The Schwartz space S(Rd ) is the set of functions of C ∞ (Rd ) which are as well as all their derivatives rapidly decreasing. We write ® S(Rd ) =
´ ϕ ∈ C ∞ (Rd ) ∀α, β ∈ Nd , Nα,β (ϕ) = sup xα ∂ β ϕ(x) < ∞ . x∈Rd
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An equivalent definition of the Schwartz space is given in the following manner. P ROPOSITION 7.2 (Equivalent definitions).– We have S(Rd ) =
⎧ ⎨ ⎩
ϕ ∈ C ∞ (Rd ) ∀p ∈ N, Np (ϕ) =
|α|,|β|≤p
⎫ ⎬ sup xβ ∂ α ϕ(x) < ∞ ⎭ x∈Rd
´ p β ϕ ∈ C (R ) ∀p ∈ N, Qp (ϕ) = sup sup (1 + |x|) ∂ ϕ(x) < ∞ .
® =
∞
d
|β|≤p x∈Rd
Proof. – For the first equality, it should be observed that Np (ϕ) is a finite sum of Nα,β (ϕ). Nα,β (ϕ). More precisely, we have Np (ϕ) = |α|,|β|≤p
– For the second equality, we use the equivalence [7.1] for the functions ∂ β ϕ.
R EMARK 7.4.– – The two expressions of Np and Qp , respectively, define two equivalent seminorms. – In the expression of Np , can be replaced by sup . |α|,|β|≤p
|α|,|β|≤p
Conversely, in the expression of Qp , sup can be replaced by |β|≤p
.
|β|≤p
– Let ϕ ∈ C ∞ (R). We have ϕ ∈ S(R) ⇐⇒ ∀n, m ∈ N, xn ϕ(m) (x) −→ 0 . |x|−→+∞
– Any function ϕ ∈ S(Rd ) is obviously bounded and summable. E XAMPLE 7.3.– 1) D(Rd ) = Cc∞ (Rd ) ⊂ S(Rd ). 2
2) ϕ(x) = e−|x| ∈ S(Rd ). 2
3) ϕ(x) = P (x)e−α|x| ∈ S(Rd ) with P a polynomial and α > 0. 4) No rational function (other than the null function) belongs to S(Rd ).
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7.1.2. Topology and convergence in S(Rd ) 7.1.2.1. S(Rd ) topology On the space S(Rd ), several families of semi-norms have been defined; the three types that are often used are given. First type Nα,β (ϕ) = sup |xβ ∂ α ϕ(x)|,
See (Zuily 1988) ∀α, β ∈ Nd .
x∈Rd
Nn,p (ϕ) =
Nα,β (ϕ),
∀n, p ∈ N.
Nα,β (ϕ),
∀n, p ∈ N.
|α|≤n,|β|≤p
Nn,p (ϕ) =
sup
|α|≤n,|β|≤p
sup |xβ ∂ α ϕ(x)|,
Np (ϕ) =
|α|,|β|≤p
∀p ∈ N.
x∈Rd
Second type See (Khoan 1972) n sup 1 + |x|2 ∂ β ϕ(x)|, ∀n, p ∈ N.
mn,p (ϕ) =
x∈Rd ,|β|≤p
Mn,p (ϕ) =
|β|≤p
Mp (ϕ) =
n
p
sup 1 + |x|2 x∈Rd
1 + |x|2
sup
|∂ β ϕ(x)|,
∀n, p ∈ N.
|∂ α ϕ(x)|,
∀p ∈ N.
x∈Rd ,|α|n
Third type sup (1 + |x|)n ∂ β ϕ(x)|,
qn,p (ϕ) =
See (Guillopé 2008) ∀n, p ∈ N.
x∈Rd ,|β|≤p
Qn,p (ϕ) =
|β|≤p
Qp (ϕ) =
sup (1 + |x|)n |∂ β ϕ(x)|,
∀n, p ∈ N.
x∈Rd
sup
(1 + |x|)p |∂ α ϕ(x)|, ,
∀p ∈ N.
x∈Rd ,|α|p
Table 7.1. Several types of semi-norms
The topology of the S(Rd ) space is defined by the countable family of norms, defined by Np (ϕ) = sup xβ ∂ α ϕ(x). |α|,|β|≤p x∈R
d
R EMARK 7.5.– 1) The topology of the S(Rd ) space can also be defined by the countable family of norms Qp (ϕ) =
sup x∈Rd ,|α|p
p
(1 + |x|) |∂ α ϕ(x)|
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or Mp (ϕ) =
sup x∈Rd ,|α|p
1 + |x|2
p
|∂ α ϕ(x)|.
2) For any p ∈ N, we have Qp (ϕ) ≤ Np (ϕ) and Mp (ϕ) ≤ N2p (ϕ). 3) Endowed with the family of semi-norms (Np )p∈N (or (Mp )p∈N , (Qp )p∈N ), the Schwartz space is a Fréchet space. 4) For the introduction of Sobolev spaces, some authors use the family (Mp )p∈N , as basis family (see (Bony 2001), for instance). With the family of norms (Np ), a distance can be defined on S(Rd ) by d(f, g) =
+∞ p=0
2−p
Np (f − g) . 1 + Np (f − g)
(S(Rd ), d) is a complete metric space whose distance d is translation invariant. It is said that S(Rd ) is a metrizable space. 7.1.2.2. Convergence in S(Rd ) With the family of norms (Np ), the convergence of a sequence in S(Rd ) can be defined as follows. D EFINITION Let ϕn be a sequence of elements of S(Rd ). It is said that the 7.4.– sequence ϕn converges to ϕ ∈ S(Rd ) in the space S(Rd ) if Np ϕn − ϕ −→ 0, ∀p ∈ N. N OTE 7.1.– There is no norm N on S(Rd ) which is adequate to express this convergence itself, that is, such that ϕn −→ ϕ in S(Rd ) if and only if, N (ϕn − ϕ) −→ 0. This shows why the topology of S(Rd ) is defined by the family of norms and not by a single norm. 7.1.3. First properties of S(Rd ) P ROPOSITION 7.3.– For all q ∈ [1, +∞], we have: S(Rd ) ⊂ Lq (Rd ). More precisely, for any ϕ ∈ S(Rd ), we have α β x ∂ ϕ
1
Lq
1
≤ Cq (Np (ϕ))1− q (Np+d+1 (ϕ)) q , ∀ |α|, |β| ≤ p.
[7.2]
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185
Proof. It should be noted that for α = β = 0, the inequality [7.2] makes it possible to deduce that S(Rd ) ⊂ Lq (Rd ). Let α, β ∈ Nd such that |α|, |β| ≤ p. We set g(x) = xα ∂ β ϕ(x). Then, α β q x ∂ ϕ q ≤ L
Rd
|g(x)|q dx
≤ sup |g(x)|q−1 x∈Rd
Rd
|g(x)|dx
≤ sup |g(x)|q−1 sup (1 + |x|)d+1 |g(x)| x∈Rd
x∈Rd
Rd
1 dx (1 + |x|)d+1
≤ Cd N0 (g)q−1 Nd+1 (g) ≤ Cd (Np (ϕ))q−1 Np+d+1 (ϕ). 1 1 Consequently, xα ∂ β ϕLq ≤ Cq (Np (ϕ))1− q (Np+d+1 (ϕ)) q , ∀ |α|, |β| ≤ p. We have Np (ϕ) ≤ Np+d+1 (ϕ), then α β x ∂ ϕ
Lq
≤ Cq Np+d+1 (ϕ), ∀ |α|, |β| ≤ p.
Special case: If q = ∞, then xα ∂ β ϕL∞ ≤ CNp (ϕ). Using the density results given in section 2.2.5 and the fact that D(Rd ) ⊂ S(Rd ) ⊂ Lp (Rd ), the following density results can be shown. P ROPOSITION 7.4.– We have the following properties: 1) D(Rd ) = S(Rd ); 2) S(Rd ) = Lp (Rd ) for all p ∈ [1, +∞[; 3) S(Rd ) = C 0 (Rd ). Proof. 1) Let χ ∈ D(Rd ), with support in B(0, 2), with values in [0, 1], and equal to 1 on B(0, 1). For all n ≥ 1, we set χn = χ nx . Let ϕ ∈ S(Rd ) and we set ϕn = χn ϕ.
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Then ϕn ∈ D(Rd ) has support in B(0, 2n) and coincides with ϕ sur B(0, 1). Let p denote a fixed integer and α, β ∈ Nd such that |α| ≤ p and |β| ≤ p. Based on the Leibniz rule, it can be easily shown that
∂ (ϕ − ϕn ) (x) = (1 − χn (x)) ∂ ϕ(x)+ β
α
γ≤β,|γ|>1
Ç å β 1 γ x β−γ ∂ ∂ χ ϕ(x). γ n|γ| n
Then, α β x ∂ (ϕ − ϕn ) (x) ≤ xα (1 − χn (x)) ∂ β ϕ(x) Ç å 1 β + sup |∂ γ χ (z)| sup xα ∂ β−γ ϕ(x) γ z∈Rd n d x∈R γ≤β,|γ|≥1
C ≤ sup xα ∂ β ϕ(x) + sup xα ∂ β−γ ϕ(x) n d |x|>n x∈R γ≤β
C ≤ sup xα ∂ β ϕ(x) + Np (ϕ). n |x|>n with C = 2p
sup ∂ β χ(x).
d |γ|≤p R
On the other hand, 1 1 sup xα ∂ β ϕ(x) ≤ 2 sup |x|2 xα ∂ β ϕ(x) ≤ 2 Np+2 (ϕ). n |x|>n n |x|>n It can be deduced that 1 C sup xα ∂ β (ϕ − ϕn ) (x) ≤ 2 Np+2 (ϕ) + Np (ϕ). n n |α|, |β|≤p x∈Rd max
Then, Np (ϕ − ϕn ) ≤
1 n2 Np+2 (ϕ)
p ∈ N; it is deduced that ϕn
−→
n−→+∞
+
C n
|Np (ϕ)|
−→
n−→+∞
0. This result holds for
ϕ in S(Rd ), which expresses the result sought
for, namely D(Rd ) = S(Rd ). 2) This result is derived from the inclusion D(Rd ) ⊂ S(Rd ) ⊂ Lp (Rd ) and from theorem 2.2. 3) We apply proposition 2.11.
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7.1.4. Operators in S(Rd ) 7.1.4.1. Differentiation and multiplication operators P ROPOSITION 7.5 (Derivative and multiplication operators).– Let ϕ ∈ S(Rd ), and we have: 1) for any β ∈ Nd , ∂ β ϕ ∈ S(Rd ); 2) for any function ρ ∈ C ∞ (Rd ) whose derivatives are all polynomially increasing (tempered), ρϕ ∈ S(Rd ). Proof. 1) For any p, q ∈ N, and for any α, β ∈ Nd such that |α|, |β| ≤ q, we have Np (xα ∂ β ϕ) =
sup xλ ∂ μ xα ∂ β ϕ(x) d
|λ|,|μ|≤p x∈R
≤ Np+q (ϕ). For all β ∈ Nd , there exists q ∈ N such that |β| ≤ q. By applying the above estimate for α = 0, we obtain ∀p ∈ N, Np (∂ β ϕ) ≤ Np+q (ϕ) < ∞. because ϕ ∈ S(Rd ). Hence, ∂ β ϕ ∈ S(Rd ). 2) This point follows from the Leibniz rule and the fact that ρ ∈ C ∞ (Rd ) is a function whose derivatives are all polynomially increasing. The Leibniz rule is used to obtain the stability of the S(Rd ) space by multiplication. P ROPOSITION 7.6.– If f, g ∈ S(Rd ), then f g ∈ S(Rd ). N OTE 7.2.– ρ ∈ C ∞ (Rd ) and f ∈ S(Rd ) does not always imply that ρf ∈ S(Rd ). 2 In reality, we simply have to take f (x) = e−x ∈ S(R) and 2 ρ(x) = e2x ∈ C ∞ (R). 7.1.4.2. Convolution in S(Rd ) D EFINITION 7.5.– Let ϕ1 , ϕ2 ∈ S(Rd ). Their convolution product is the function d ϕ1 (x − t)ϕ2 (t)dt. ϕ1 ∗ ϕ2 defined by ∀x ∈ R , ϕ1 ∗ ϕ2 (x) = Rd
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The Theory of Distributions 2
E XAMPLE 7.4.– For the Gaussian γ1 such that γ1 (t) = e−t , ∀t ∈ R, we have γ1 ∗ γ1 (t) =
»
2
π − t2 2e
.
P ROPOSITION 7.7.– Let ϕ1 , ϕ2 , ϕ3 ∈ S(Rd ). We have i) ϕ1 ∗ ϕ2 is a Schwartz function; that is ϕ1 ∗ ϕ2 ∈ S(Rd ); ii) ϕ1 ∗ ϕ2 = ϕ2 ∗ ϕ1 ; that is the law “∗” is commutative in S(Rd ); iii) (ϕ1 ∗ ϕ2 ) ∗ ϕ3 = ϕ1 ∗ (ϕ2 ∗ ϕ3 ); that is the law “∗” is associative in S(Rd ). Proof. 1) The convolution product ϕ1 ∗ ϕ2 is well defined because ϕ1 , ϕ2 ∈ L1 (Rd ). Let p ∈ N and α, β ∈ Nd with |α|, |β| ≤ p. For all x ∈ Rd , we have α β α β ϕ1 (x − y)ϕ2 (y)dy. x ∂ (ϕ1 ∗ ϕ2 ) (x) = x ∂x Rd
Since the function y −→ ∂xβ ϕ1 (x−y)ϕ2 (y) is integrable for all x and is dominated by sup |∂xβ ϕ1 ||ϕ2 | ∈ L1 (Rd ), we can derive under the integral sign to obtain xα ∂ β (ϕ1 ∗ ϕ2 ) (x) = xα ∂xβ ϕ1 (x − y)ϕ2 (y)dy Rd
xα ∂xβ ϕ1 (x − y)ϕ2 (y)dy
=
Rd
= Rd
(x − y + y)α ∂xβ ϕ1 (x − y)ϕ2 (y)dy
Ç å α = (x − y)γ (y)α−γ ∂xβ ϕ1 (x − y)ϕ2 (y)dy γ d R γ≤α Ç å α (x − y)γ ∂xβ ϕ1 (x − y)y α−γ ϕ2 (y)dy = γ d R γ≤α Ç å α = y γ ∂ β ϕ1 ∗ y α−γ ϕ2 (x). γ
γ≤α
According to Young’s inequality, we have α β x ∂ (ϕ1 ∗ ϕ2 )
L∞
≤
γ≤α
Ç å α γ β y ∂ ϕ1 1 y α−γ ϕ2 L∞ . L γ
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189
It follows that Np (ϕ1 ∗ ϕ2 ) ≤ CNp+d+1 (ϕ1 )Np (ϕ2 ). Consequently, ϕ1 ∗ ϕ2 ∈ S(Rd ). P ROPOSITION 7.8.– Let T ∈ E (Rd ). If ϕ ∈ S(Rd ), then ϕ ∗ T ∈ S(Rd ). Moreover, there exists m ∈ N such that ∀p ∈ N, ∃ Cp > 0, Np (ϕ ∗ T ) ≤ Cp Np+m (ϕ). Proof. Since ϕ ∈ C ∞ (Rd ), then ϕ ∗ T is an indefinitely differentiable function and we have xα ∂ β (ϕ ∗ T )(x) = Ty , xα ∂ β ϕ(x − y). On the other hand, since T is compactly supported, then T is of finite order m, and there exists C > 0 such that, if supp T ⊂ B(0, R), we have α β sup ∂yμ xα ∂xβ ϕ(x − y) x ∂ (ϕ ∗ T )(x) ≤ C |μ|≤m
≤C
|μ|≤m
≤C
|μ|≤m
|y|≤R
sup xα ∂ β+μ ϕ(x − y)
|y|≤R
α sup (x − y + y) ∂ β+μ ϕ(x − y)
|y|≤R
≤ C Np+m (ϕ) (binomial formula). Hence, for all α, β ∈ Nd such that |α|, |β| ≤ p, we have α β x ∂ (ϕ ∗ T )(x) ≤ C Np+m (ϕ). Therefore, there exists m ∈ N such that ∀p ∈ N, ∃ Cp > 0, Np (ϕ ∗ T ) ≤ Cp Np+m (ϕ).
7.2. Tempered distributions This section begins with the definition of a tempered distribution followed by examples illustrating this definition.
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7.2.1. Definition and examples D EFINITION 7.6.– A tempered distribution is a definite linear and continuous form on S(Rd ), that is, a linear form T on S(Rd ) such that there exists p ≥ 0 and there exists c > 0 such that ∀ϕ ∈ S(Rd ),
| T, ϕ | ≤ c Np (ϕ).
[7.3]
The set of tempered distributions is a vector space that we denote by S (Rd ). R EMARK 7.6.– 1) Given the density of D(Rd ) in S(Rd ), it suffices that inequality [7.3] is verified for ϕ ∈ D(Rd ). 2) In the definition, the family of norms (Np ) can be replaced by the previously defined family (Qp ). 3) A linear form T on S(Rd ) is a tempered distribution if and only if, for any sequence (ϕn ) of elements of S(Rd ), S(Rd )
ϕn −→ 0 leads to T, ϕn −→ 0 in C. P ROPOSITION 7.9.– For a distribution T to belong to S (Rd ), it is necessary and sufficient that T be a continuous linear form on D(Rd ) for the convergence induced by that on S(Rd ). Proof. The necessary condition is immediate. The converse is true because since T is continuous linear on D(Rd ) which is dense in S(Rd ), it uniquely extends into a continuous linear form on S(Rd ). It should be noted that here we are talking about a density extension of a continuous linear form on D(Rd ) that is dense in S(Rd ), which is not the case for Banach spaces. For more details, see Bony (2001).
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191
Figure 7.2. Spaces and their duals. For a color version of this figure, see www.iste.co.uk/aitbenhassi/distribution.zip
E XAMPLE 7.5.– We have δa ∈ S (Rd ). Actually, ∀ϕ ∈ S(Rd ), |δa , ϕ| = |ϕ(a)| sup |ϕ(x)| ≤ N0 (ϕ). x∈Rd
E XAMPLE 7.6.– Any polynomial function defines a tempered distribution (that is to say R[X] ⊂ S (Rd ). We simply have to show the result for a function such as f (x) = xα where α ∈ Nd . A linear combination will make it possible to conclude. Let α ∈ Nd such that |α| ≤ N and let ϕ ∈ S(Rd ). We have |f, ϕ| ≤ ≤ ≤
Rd
Rd
|f (x)ϕ(x)| dx |xα ϕ(x)| dx 1
Rd
(1 + |x|)
d+1
≤ sup (1 + |x|) Ç
Rd
d+1
|xα ϕ(x)| dx
d+1
x∈Rd
≤
(1 + |x|)
1 (1 + |x|)
|xα ϕ(x)| å
1 Rd
(1 + |x|)
dx NN +d+1 (ϕ) d+1
≤ C NN +d+1 (ϕ). Hence, f defines a tempered distribution.
d+1
dx
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The Theory of Distributions
E XAMPLE 7.7.– 1) If f is a measurable function and if there exists a polynomial function P such that |f (x)| |P (x)|, then f ∈ S (Rd ). 2) The principal value Vp
1 is a tempered distribution (see exercise 7.2). x
3) We have E (Rd ) ⊂ S (Rd ). 4) The distribution defined by T
=
an δna , a
>
0 is a tempered
n∈Z
distribution if the sequence (an )n∈Z is at most polynomially growing. (i.e. ∃ p > 0, ak ≤ (1 + |k|)p ), ∀k ∈ Z) (see exercise 7.5). P ROPOSITION 7.10.– Any function in Lp (Rd ) defines a tempered distribution (i.e. Lp (Rd ) ⊂ S (Rd ), ∀p ∈ [1, ∞]). Proof. Let f ∈ Lp (Rd ). For any ϕ ∈ S(Rd ), we have
f, ϕ =
f (x)ϕ(x)dx = Rd
Rd
f (x) (1 + |x|)m ϕ(x)dx. (1 + |x|)m
– For 1 < p < +∞: let m be large enough so that ä Ä q such that p1 + 1q = 1 . We have
Rd
dx < +∞ with (1 + |x|)qm
|f (x)| dx sup (1 + |x|)m |ϕ(x)| m x∈Rd Rd (1 + |x|) 1 q sup (1 + |x|)m |ϕ(x)| f Lp m
|f, ϕ|
(1+|x|)
L
x∈Rd
1 q Qm (ϕ) ≤ f Lp (1+|x|) m L ≤ CNm (ϕ). – For p = 1: (the result is obvious). – For p = ∞: the result is derived from example 7.6. P ROPOSITION 7.11.– Any continuous function f at most polynomially growing defines a tempered distribution on Rd .
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193
Proof. Since f is continuous and at most polynomially growing, then there exists N N ∈ N and it exists C > 0 such that |f (x)| ≤ C (1 + |x|) , ∀ x ∈ Rd . Let ϕ ∈ S(Rd ). We have |f, ϕ| ≤
Rd
|f (x)ϕ(x)| dx N
≤
Rd
C (1 + |x|) |ϕ(x)| dx
≤ C sup (1 + |x|)
N +d+1
x∈Rd
Ç ≤C
å
1 (1 + |x|)
Rd
|ϕ(x)|
d+1
1 Rd
(1 + |x|)
(d+1)
dx
dx NN +d+1 (ϕ)
≤ CN NN +d+1 (ϕ). Consequently, f defines a tempered distribution. E XAMPLE 7.8 (Counterexamples).– 2
1) The function f (x) = ex defines a distribution but it is not a tempered distribution. As a matter of fact, for ϕ(x) = e−
x2 2
∈ S(R), we have
f, ϕ =
f (x)ϕ(x)dx
R 2
ex e−
=
x2 2
dx
R
=
e
x2 2
dx
R
= +∞. 2) The distributions Tex , Tshx and Tchx are not tempered distributions (see exercise 7.3).
7.2.2. Convergence in S (Rd )
The set of tempered distributions S (Rd ) is naturally endowed with the weak dual topology, defined by the weak convergence ∗ of sequences.
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The Theory of Distributions
D EFINITION 7.7.– We say that a sequence (Tn )n≥1 of tempered distributions converges to T ∈ S (Rd ) if, Tn , ϕ −→ T, ϕ , ∀ϕ ∈ S (Rd ). n→+∞
P ROPOSITION 7.12.– Let (Tn )n≥1 be a sequence of tempered distributions. If for any ϕ ∈ S(Rd ), the sequence (Tn , ϕ)n≥1 converges, then there exists T ∈ S (Rd ) such that (Tn )n≥1 converges to T in S (Rd ). R EMARK 7.7.–
1) If fn −→ f in Lp (Rd ), then fn −→ f in S (Rd ).
2) If Tn −→ T in S (Rd ), then Tn −→ T in D (Rd ). The converse does not hold. Furthermore, for any real sequence (an ), the sequence (an δn ) −→ 0 in D (R), whereas it converges in S (R), only if there exists c > 0 and p ∈ N such that |an | ≤ c(1 + n)p . 7.2.3. First properties of S (Rd ) A first property of tempered distributions is the following.
P ROPOSITION 7.13.– The space S (Rd ) injects into D (Rd ).
Proof. Let T ∈ S (Rd ) be a tempered distribution. By definition, there exists m ∈ N and a constant C > 0 such that: |T, ϕ| ≤ CNm (ϕ), ∀ϕ ∈ S(Rd ). In particular, for any compact K ⊂ Rd and for all ϕ ∈ D(Rd ) with support in K, we have sup |xα ∂ β ϕ| ≤ CK,α sup |∂ β ϕ|. Therefore, x∈K
x∈Rd
|T, ϕ| ≤ C(
CK,α )
|α|≤m
|β|≤m
sup |∂ β ϕ|
x∈K
≤ C Pm (ϕ),
which shows that T|D(Rd ) ∈ D (Rd ). Therefore, the mapping Ψ : T −→ T|D(Rd ) is well defined. Moreover, by density of D(Rd ) in S(Rd ), if T|D(Rd ) = 0, then T = 0 on S(Rd ). Hence the injection of the mapping Ψ. C OROLLARY 7.1.– Any tempered distribution is of finite order.
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195
Proof. Let T ∈ S (Rd ) be a tempered distribution. In the proof of proposition 7.13, it has been shown that there exists m ∈ N such that for any compact K ⊂ Rd , there exists a constant C > 0 such that: |T, ϕ| ≤ C Pm (ϕ), ∀ϕ ∈ D(Rd ), supp ϕ ⊂ K. Thus, T is at most an mth-order distribution.
Application: In exercise 3.8, it has been shown that the distribution
+∞
δn(n) is of
n=0
infinite order; it is then deduced that this distribution can never be tempered.
7.2.4. Operators in S (Rd ) All the operations already seen in the space S(Rd ) are transmitted by duality to the space S (Rd ). 7.2.4.1. Differentiation and multiplication operators The properties of the derivation and multiplication operators in S(Rd ) are transposed to its dual space S (Rd ).
P ROPOSITION 7.14.– Let T ∈ S (Rd ), and we have
1) for all α ∈ Nd , ∂ α T ∈ S (Rd ); 2) for any function ρ at most polynomially growing and all its derivatives, we have ρT ∈ S (Rd ).
C OROLLARY 7.2.– Let T ∈ S (Rd ). For all α, β ∈ Nd , we have xα ∂ β T ∈ S (Rd ). E XAMPLE 7.9 (Infinite growth and tempered character).– To see if a distribution defined by a function, for example, is tempered or not, studying the infinity growth of the modulus of this function does not suffice. Consider, for example, the function x x x −→ iex eie . We have |iex eie | = ex , which has the same infinite growth as the above counterexamples. However, x
x
iex eie = ddx (eie ), x
and since the function x −→ eie is a function of class C 1 bounded on R, it defines a tempered distribution on R. On the other hand, its derivative in the sense of distributions coincides with the distribution defined by its derivative function in the usual sense, so that, according to x proposition 7.14 this, derivative function x −→ iex eie properly defines an element of S (Rd ).
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The Theory of Distributions x
Intuitively, the fast oscillations of the function x −→ eie in the neighborhood of x +∞ are the underlying cause that annihilate the growth of x −→ eie for x in the neighborhood of +∞.
R EMARK 7.8.– In practice, to decide if a distribution T ∈ D (Rd ) is tempered, it will be obviously necessary to find out if it belongs to the classes of examples above, namely distributions with compact support, functions of Lp , functions with polynomial growth, etc. If this is not the case, we will then have to find out whether the distribution under consideration is a derivative (of any order) of a distribution, which is already known to be tempered (this is obviously the case in the previous example). If none of these approaches allows us to conclude, we must then return to the definition of temperate distributions, and verify the continuity property. In fact, we have the following characterization of temperate distributions.
T HEOREM 7.1 (Characterization of S (Rd ) distributions).– Any distribution T ∈ S (Rd ) is of the form T = ∂xα (1 + |x|2 )n f in the sense of distributions where d α ∈ N is a natural number, n is a natural number, and f is a continuous function on Rd .
7.2.4.2. Convolution in S (Rd )
P ROPOSITION 7.15 (Convolution).– Let T ∈ S (Rd ) be a tempered distribution and S ∈ E (Rd ) a distribution with compact support. We have T ∗ S ∈ S (Rd ). Proof. Let ϕ ∈ D(Rd ). There exists a sequence (ψn ) ∈ D(Rd ) such that ψn −→ S in D (Rd ). Therefore, T ∗ ψn , ϕ = T, ψˇn ∗ ϕ. Since T ∗ ψn −→ T ∗ S in D (Rd ) and ψˇn ∗ ϕ −→ Sˇ ∗ ϕ in D(Rd ), then T ∗ S, ϕ = lim T, ψˇn ∗ ϕ = lim T ∗ ψn , ϕ = T, Sˇ ∗ ϕ. n →+∞
n →+∞
Since T ∈ S (Rd ), according to proposition 7.8, there exists C, C > 0 and p, m ∈ N such that T, Sˇ ∗ ϕ ≤ CNp (Sˇ ∗ ϕ) ≤ C Np+m (ϕ).
Consequently, T ∗ S ∈ S (Rd ). 7.3. Exercises with solutions E XERCISE 7.1.– Let f be the real function defined by 2
2
f (x) = e−x cos(e2x ), ∀x ∈ R.
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197
1) Show that f is a rapidly decreasing function. 2) Does f belong to the Schwartz space S(R)? S OLUTION 7.1.– 1) From the definition of a rapidly decreasing function, we have to show that for any index α ∈ N, we have lim |xα f (x)| = 0. Since, on the one hand, |x| →+∞ 2 2x2 | cos(e )| ≤ 1 and, on the other hand, lim xα e−x = 0, then |x| →+∞
lim
|x| →+∞
Therefore,
2 α −x2 cos(e2x ) = 0. x e lim
|x| →+∞
|xα f (x)| = 0. Consequently, f is a rapidly decreasing
function. 2) Through a direct and easy calculation, we obtain Ä ä 2 2 2 2 f (x) = −2x e−x cos(e2x ) + 2ex sin(e2x ) . Considering the sequence (xk ) defined by xk = lim |f (xk )| = +∞ and
k →+∞
Then, it is impossible to have
»
1 2
ln 2kπ + π2 , we have
lim |xk | = +∞.
k →+∞
lim
|x| →+∞
|xα f (x)| = 0. Consequently, f is not a
rapidly decreasing function, so f does not belong to the Schwartz space S(R). E XERCISE 7.2.– 1) Show that a Dirac mass is a tempered distribution. 2) Show that a polynomial function is a tempered distribution. 3) Let f be a measurable function verifying x → (1 + |x|2 )−m f (x) ∈ Lp (Rd ) for a specific integer m ∈ N and a specific p ∈ [1, +∞]. Show that f is a temperate distribution. 4) Show that the function x −→ ln |x| ∈ Lploc (R) is a tempered distribution and compute its derivative. What can be deduced about the principal value of x1 ? Indication: For the differentiation, ϕ ∈ D(R) will be considered and we will integrate the function x −→ ln |x|ϕ (x) by parts on a domain such as {x ∈ R : ε ≤ |x| ≤ R} before making R tend to +∞ then ε to 0.
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The Theory of Distributions
S OLUTION 7.2.– 1) For any rapidly decreasing function ϕ ∈ S(Rd ) and any a ∈ Rd , we have |δa , ϕ| = |ϕ(a)| ϕ∞ . Consequently, δa is therefore indeed a tempered distribution. It can also be observed that δa is a compact support distribution, so it is tempered. 2) If P is a polynomial function, then there exists m ∈ N such that m |P (x)| 1 d P (x) = O 1 + |x|2 . Let ϕ ∈ S(Rd ). Since x −→ (1+|x| 2 )m+d+1 ∈ L (R ), ∞ then ´ ® m+d+1 |P (x)| dx sup 1 + |x|2 |ϕ(x)| d P (x)ϕ(x)dx m+d+1 2 x∈Rd R Rd (1 + |x| ) ≤ CMm+d+1 (ϕ) ≤ CN2(m+d+1) (ϕ). Consequently, P is thus indeed a tempered distribution. m 3) We denote Qm (x) := 1 + ||x||2 . We have in this case, for any rapidly decreasing function ϕ ∈ S(Rd ), and using the Hölder inequality if p ∈]1, ∞[, and simply the inequality of the mean if p ∈ {1, ∞},
f (x) |ϕ(x)Qm (x)| dx f (x)ϕ(x)dx Rd Rd Qm (x) f 1 1 Q ϕQm p ( p + p = 1). m p
−d−1 actually belongs to all spaces Lp (Rd ), we finally Since qd :−→ 1 + ||x||2 have f qd sup 1 + |x|2 m 1 + |x|2 d+1 |ϕ(x)| f (x)ϕ(x)dx p Qm p x∈Rd Rd d+m+3 C sup 1 + |x|2 |ϕ(x)| x∈Rd
≤ CMm+d+3 (ϕ) ≤ CN2(m+d+3) (ϕ). Hence, f is thus a tempered distribution.
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199
4) First, x −→ ln |x| is indeed a tempered distribution, since for instance −2 x −→ ln |x| 1 + x2 is integrable, so that the criterion of the previous question applies. Then, if ϕ ∈ D(R), we follow the indication given in the text and it follows that IR := ln |x|ϕ (x)dx |x|R
=
− [ln(−x)ϕ(x)]−R
+
[ln(x)ϕ(x)]R
−
− −R
ϕ(x) dx − x
R
ϕ(x) dx, x
where we recall that the function x −→ ln |x| is differentiable on R∗ , and where its 1 derivative is x −→ . We thus have x IR
:= ln(R) [ϕ(R) − ϕ(−R)] + ln() [ϕ(−) − ϕ()] − |x|R
ϕ(x) dx. x
We first evaluate the limit for R −→ +∞. The factor term of ln(R) is zero for R diam{supp(ϕ)} (support which is compact). Since ϕ ∈ D(R), then, by dominated convergence, it is obtained that |x|
ln |x|ϕ (x)dx = ln() [ϕ(−) − ϕ()] −
|x|
ϕ(x) dx. x
[7.4]
We now evaluate the limit −→ 0. By the finite increment theorem, we have |ϕ(−) − ϕ()| |ϕ(−) − ϕ(0)| + |ϕ(0) − ϕ()| 2||ϕ ||∞ , so that the term in factor of ln() tends to 0 (by domination). Finally, by dominated convergence, the left member of the equality [7.4], ln |x|ϕ (x)dx tends to |x|
ln |x|, ϕ 1, and by definition of the principal value, the right-hand side member tends to − Vp( x ), ϕ . By definition of the derivative in the sense of distributions, we finally have that the derivative of x −→ ln |x| is precisely the principal value. It is known that S (R) is stable by differentiation, so Vp( x1 ) ∈ S (R).
E XERCISE 7.3.– 1) Show that the locally integrable function x → ex is not a tempered distribution. Hint: Set a non-identically zero positive test function θ with support in ] − 1, 1[ and study the family of square brackets Texp , τa θ for a ∈ R.
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The Theory of Distributions
2) Let f (x) := ex cos(ex ). a) Show that f is not bounded by any polynomial. b) Show, however, that f is a tempered distribution. Hint: Give a primitive of f . S OLUTION 7.3.– 1) We follow the hint given above. If Texp were tempered, then there would exist a constant C > 0 and integersm, p ∈ N such that for any test function ϕ ∈ D(R), we m have |Texp , ϕ| Csup sup 1 + x2 |ϕ(k) (x)|. Let θ ∈ D (] − 1, 1[) be a positive kp x∈R
function and a ∈ R. Since τa θ has support in Ia =]a − 1, a + 1[, we have m |Texp , τa θ| Csup sup 1 + x2 |(τa θ)(k) (x)| kp x∈Ia
m sup sup 1 + x2 |τa (θ)(k) (x)| kp x∈Ia
m ≤ sup sup 1 + x2 |θ(k) (x − a)|. kp x∈Ia
m m is bounded by 2 + (1 + a)2 on Ia , so that However, x −→ 1 + x2 m finally it would follow that |Texp , τa θ| C 2 + (1 + a)2 sup sup |θk |, but kp x∈R
on the other hand, using a change of variable u = x − a, it can be noted that ex θ(x − a)dx=ea Texp , θ such that the previous inequality is Texp , τa θ = R m written as ea |Texp , θ| C 2 + (1 + a)2 sup sup |θ(k) |, and since θ = 0, we kp x∈R
have Texp , θ = 0, and by dividing it follows that m ea Cθ 2 + (1 + a)2 sup sup |θ(k) | kp x∈R
that is, a polynomial bound (at a) for the exponential function, which is obviously contradictory: exp ∈ / S (R). We can do better without following the hint! Simply taking the function ϕ(x) =
+∞ −∞
ex dx = +∞, cosh(x)
/ S(R). which shows that ex ∈
1 ∈ S(R), we then obtain cosh(x)
Schwartz Spaces and Tempered Distributions
201
2)a) If f were bounded by a polynomial, by domination, we would have cos (ex )
−→
x−→+∞
0,
which is not true, as can be seen by considering the sequence (xk ) defined by xk = ln(2πk), which tends to +∞ when k tends to +∞, whereas cos (exk ) = 1. b) A primitive of f is the function F : x −→ sin(ex ) ∈ L∞ (R). Since L (R) ⊂ S (R) and that the latter space is stable by differentiation, we thus indeed have f ∈ S (R). ∞
E XERCISE 7.4.– Show that if f is locally summable and if it is square-summable, then it determines a tempered distribution. S OLUTION 7.4.– Let ϕ ∈ S(R). From Schwarzt’s inequality, we have Ç
å2
+∞
f (x)ϕ(x)dx −∞
≤
−∞ +∞
2
+∞
ϕ2 (x)dx
−∞ +∞
f (x)dx −∞
≤
2
f (x)dx
≤
+∞
−∞
+∞
−∞
≤π
ϕ2 (x) 1 + |x|2 dx 1 + |x|2
f 2 (x)dx π sup |ϕ(x)| x∈R
+∞ −∞
2
1 + |x|2
2 f 2 (x)dx sup |ϕ(x)| 1 + |x|2 . x∈R
Then: |f, ϕ| ≤
√
πf 2 sup ϕ(x) 1 + |x|2 . x∈R
Otherwise, we have |f, ϕ| ≤ tempered distribution.
√
πf 2 N2 (ϕ). Consequently, f defines a
E XERCISE 7.5.– Let (an )n∈Z be at most a polynomially growing sequence (i.e. an δna , ∃ p > 0, an ≤ (1 + |n|)p , ∀n ∈ Z). We consider T defined by T = n∈Z
a > 0. Show that T is a tempered distribution.
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S OLUTION 7.5.– Let M > 0 and ϕ ∈ D(R) such that supp ϕ [−M, M ]. Let N ∈ N such that N ≥ M . We have |T, ϕ| ≤
N
|ak ϕ(ka)|
−N
≤ ϕ∞
N
|ak |
−N
≤ C ϕ∞ . Therefore, T is indeed a zeroth-order distribution. Let ϕ ∈ S(R). We have |ak δka , ϕ| = |ak ϕ(ka)| ≤ C(1 + |k|)p |ϕ(ka)| ≤ C(1 + |k|)p |ϕ(ka)| ≤
|ϕ(ka)| C(1 + |k|)p (1 + |ka|)p+2 . (1 + |ka|)p (1 + |ka|)2
Since ϕ ∈ S(R), then (1 + |ka|)p+2 |ϕ(ka)| ≤ sup(1 + |x|)p+2 |ϕ(x)| ≤ NP +2 (ϕ). xR
In addition, we have
(1 + |x|)p C(1 + |k|)p ≤ C sup = C1 . Then p p (1 + |ka|) xR (1 + |xa|)
|ak δka , ϕ(k)| ≤ C1
NP +2 (ϕ) . (1 + |ka|)2
Consequently, |T, ϕ| ≤ C2 NP +2 (ϕ) where C2 = C1 conclude that T ∈ S (R).
n∈Z
1 . We then (1 + |ka|)2
Consequence: In the special case when ak = 1, we find that the Dirac comb a = δna is a tempered distribution. n∈Z
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203
E XERCISE 7.6.– For all nN∗ , consider the function fn defined by 1 − cos(2nπx) . πx
∀x ∈ R∗ , fn (x) =
1) Show that for all n ∈ N∗ , fn defines a tempered distribution that will be denoted by fn . 2) Show that the limit of the sequence of tempered distributions (fn ) is converging in S (R) to the distribution π1 Vp( x1 ). S OLUTION 7.6.– 1) For any n ≥ 1, the function fn is continuous in R∗ as the quotient of two continuous functions in R∗ , since the function of the denominator does not cancel out in R∗ . In the neighborhood of 0, by a limited expansion of the function cos, we have 1 − cos(2nπx) =
(2nπx)2 (2nπx)4 − + o(x4 ). 2 4!
We deduce that lim fn (x) = lim
x −→0
x −→0
1 − cos(2nπx) = 0. πx
The function fn is thus extendable by 0 in the neighborhood of 0, and thus defines a continuous function on R. The function fn is bounded on [−1, 1] as a continuous function on a compact. On the other hand, for all x ∈ R such that |x| > 1, we have |fn (x)| ≤ π2 . fn is a function of L1loc (R) and at most polynomially growing, so fn ∈ S (R). 2) Let n ∈ N∗ and ϕ ∈ S(R). We start by verifying that ≠ ∑ ϕ(x) − ϕ(0) ϕ(x) 1 Vp( ), ϕ = dx − dx. x x x [−1,1] |x|≥1 Then, we show that ≠ fn , ϕ −
1 1 Vp( ), ϕ π x
∑
[−1,1]
cos(2nπx)(ϕ(x) − ϕ(0)) dx πx
|x|≥1
cos(2nπx)ϕ(x) dx. πx
= −
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By using the Taylor formula at the first order at the point 0 applied to the function ϕ at x, we have ϕ(x) − ϕ(0) = xψ(x) with ψ ∈ C ∞ . Therefore, ∑ cos(2nπx)ϕ(x) 1 1 1 cos(2nπx)ψ(x)dx− fn , ϕ− Vp( ), ϕ = dx. π x π [−1,1] πx |x|≥1 ≠
Showing that both integrals tend to 0 when n tends to +∞, we get ≠ lim
n−→+∞
fn , ϕ =
∑ 1 1 Vp( ), ϕ , for all ϕ ∈ S(R). π x
This allows us to conclude.
8 Fourier Transform
The Fourier transform is a particularly important tool in mathematics, either from the fundamental point of view, or from the point of view of its applications (in signal theory for example, or in everything related to wave phenomena). There are therefore many varied motivations underlying its study. Because of the inversion theorem, the Fourier transform will make it possible to write an integrable function and more generally a tempered distribution as a superposition of complex exponential functions. This is a very powerful tool for studying convolution equations, its resolution being reduced to division problems. As for the most important applications of the Fourier transform, the reader will find them in the study of partial differential equations, which is the subject of Chapter 9 of this book. It is important to be able to use this transformation under the most general conditions possible. We define it within the too restricted framework of integrable functions L1 (Rd ). To extend it to a larger space, it is restricted to the S(Rd ) space, which is a much smaller space than L1 (Rd ). By duality, this transformation can be finally extended to S (Rd ), which presents good invariance properties for the Fourier transform. Unfortunately, it is not possible to define the Fourier transform of any distribution. 8.1. Fourier transform in L1 (Rd ) We begin by recalling the definition and some properties of the Fourier transform in the L1 (Rd ) space.
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8.1.1. Definition and first properties D EFINITION 8.1.– Let f ∈ L1 (Rd ), its Fourier transform is the function denoted by e−i2πxy f (y)dy, with xy = x, y, f* (or Ff ) and is defined by ∀x ∈ Rd , f*(x) = which denotes the scalar product in Rd .
Rd
R EMARK 8.1.– 1) We should point out that some authors adopt other definitions and the reader should bear them in mind. For example, we cite the following formulas: f*(x) =
e
−iπxy
f (y)dy
or
Rd
1 f*(x) = √ 2π
e−iπxy f (y)dy. Rd
2) Note that the Fourier transform does not always exist. For example, the function f (x) = x does not have a Fourier transform because ye−i2πxy dy does not exist for any value of x. the integral R
3) In order to avoid overburdening the book with further details, the Fourier transform of f is indifferently expressed by f* or by Ff independently of the space on which it is defined. For more details concerning these two notations, the interested reader can refer to Dalmasso and Witomski (1996). E XAMPLE 8.1.– The rectangular function denoted by Π is defined by Π(t) = 1 Π(t) = 0
if t ∈ − 12 , 12 if t ∈ / − 12 , 12 .
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207
Figure 8.1. Rectangular function Π
Given that Π is even, then if s = 0, we have F(Π)(s) = 2
1 2
ï Π(t) cos (2πst) dt = 2
0
sin 2πst 2πs
ò 12 = 0
sin πs . πs
If s = 0, then F(Π)(0) = 1. The function F(Π) is extendable by continuity at 0. Therefore, the Fourier transform of the rectangular function, Π, is the function defined on R by ∀s ∈ R, F(Π)(s) =
sin πs . πs
This function is called cardinal sine function and is denoted by sinc. Its graphical representation can be found in Figure 8.2.
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The Theory of Distributions
Figure 8.2. Sinc function: F (Π)
R EMARK 8.2.– The rectangular function Π is an example of an integrable function whose Fourier transform is not integrable! In other words, the Fourier transform does not operate in L1 (Rd ). E XAMPLE 8.2 (Poisson density type function).– Consider the function f defined by f (y) = e−a|y| where a > 0. We have F(f )(s) =
2a . a2 + 4πs2
In reality, the function f is even, so F(f )(s) = 2 double integration by part gives F(f )(s) =
2a . a2 + 4πs2
+∞ 0
e−ay cos(2πsy)dy. A
Fourier Transform
Figure 8.3. The function e−|x|
Figure 8.4. The function F (e−|x| )
The first natural property of the Fourier transform is linearity.
209
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The Theory of Distributions
P ROPOSITION 8.1 (Linearity).– Let f, g ∈ L1 (Rd ) and α, β ∈ C. Then, F (αf + βg) = αF (f ) + βF (g) . Proof. The proof is derived from the linearity of the integral. Among the properties of the Fourier transform, we give a result often used, which is directly derived from the Riemann–Lebesgue lemma. P ROPRIETY 8.1.– Let f ∈ L1 (Rd ), then f*, the Fourier transform of f , is a continuous function, and in addition lim f*(x) = 0. x→+∞
Proof. f* is well defined, since |f*(x)| f 1 , then f* makes sense. Continuity: Let x0 ∈ Rd be fixed and let (sn )n≥1 be a sequence converging to d * x0 ∈ R . We have f (sn ) = e−i2πsn y f (y)dy. Since sn → x0 , then Rd
e−i2πsn y f (y) → e−i2πx0 y f (y) a.e. y ∈ Rd . Moreover, ∀y ∈ Rd , e−i2πsn y f (y) |f (y)|. Now f ∈ L1 (Rd ), according to Lebesgue’s dominated convergence theorem, we have f*(sn ) −→ f*(x0 ). Hence, f* is continuous at x0 . Consequently, f* is continuous n→+∞
on Rd . To show that f*(x) tends to 0 when x → +∞, only the scalar case is considered. Furthermore, the space of piecewise continuous functions with bounded support on R or that of step functions are dense in L1 (R). Any function f ∈ L1 (R) can be approximated in the sense of the norm of L1 (R) by a sequence of step functions with bounded support. Therefore, a stepped function ϕ can be found; this means that there exist bounded intervals [ak−1 , ak ] and values c1 , ..., cn such that ϕ=
n k=1
ck 1[ak−1 ,ak ] ,
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211
where 1[ak−1 ,ak ] designates the characteristic function of [ak−1 , ak ]. As such, ϕ is written as a finite linear combination of interval characteristic functions and its Fourier transform ϕ * can be computed by linearity. More precisely, we have ϕ(x) * =
=
n +∞ −∞ k=1
n
=
k=1
ak
ck
e−i2πxy dy
ak−1
k=1 n
ck 1[ak−1 ,ak ] (y)e−i2πxy dy
ick −i2πxak e − e−i2πxak−1 . 2πx
Then 1 |ck |. π|x| n
|ϕ(x)| * ≤
k=1
Let ε > 0. For |x| → +∞, we have |ϕ(x)| * ≤
ε . Consequently, 2
* * * + |ϕ(x)| * ≤ ε. f (x) ≤ f (x) − ϕ(x) Then
lim
x→+∞
f*(x) = 0. This completes the proof.
R EMARK 8.3.– 1) To show that
lim
x→+∞
f*(x) = 0, we could directly use the Riemann–Lebesgue
theorem. The proof given here implicitly reuses the proof of the Riemann–Lebesgue theorem. 2) Although the Fourier transform of a function f ∈ L1 (Rd ) has meaning and is even continuous, in general it is not in L1 (Rd ). 3) Application of the Fourier transform F : L1 (Rd ), · 1 −→ C0 (Rd ), · ∞ is a linear and continuous mapping.
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It should be noted that C0 (Rd ) designates the set of continuous functions on Rd which tends to 0 when x→ +∞. We have C0 (Rd ) = {f ∈ C(Rd ),
f*(x) = 0}.
lim
x→+∞
8.1.2. Fourier transform and operations We now present some highly relevant properties of the Fourier transform. 8.1.2.1. Fourier transform and differentiation We start with the scalar case d = 1. 8.1.2.1.1. Scalar case We give a lemma which will be useful in the proof of the following theorem. L EMMA 8.1.– Let f ∈ L1 (R). If f ∈ L1 (R), then lim f (x) = 0. x→±∞
Proof. For all x ∈ R, we have f (x) = f (0) + Therefore, lim f (x) = f (0) + x→+∞
Let us assume that =
|L| , we get 2
+∞
x
f (s)ds.
0
f (s)ds. Hence lim f (x) does exist. x→+∞
0
lim f (x) = 0 and we set L =
x→+∞
∃A > 0, ∀x ∈ R, x > A =⇒ |f (x)|>
lim f (x). If we take
x→+∞
|L| , 2
which is non-sense because f ∈ L1 (R). Then, lim f (x) = 0. x→+∞
Similarly, we also show that lim f (x) = 0. x→−∞
Now we will study the Fourier transform of a derivative.
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213
T HEOREM 8.1.– Let f ∈ L1 (R). We assume that f is differentiable and that f ∈ L1 (R). Then F(f )(x) = (2iπ)xFf (x) = (2iπ)xf*(x). If, in addition, f admits derivatives up to order n that are in L1 (R), then: F(f (n) )(x) = (2iπ)n xn Ff (x). Proof. We have F(f )(x) =
e−2iπxy f (y)dy R
= f (x)e−2iπxy
+∞
+ (2iπ)x −∞
e−2iπxy f (y)dy. R
By hypothesis, f ∈ L1 (R) and f ∈ L1 (R). According to lemma 8.1, we have lim f (x) = 0. It follows that lim f (x)e−2iπxy = 0. Then,
x→±∞
x→±∞
F(f )(x) = (2iπ)x
e−2iπxy f (y)dy = (2iπ)xF(f )(x). R
Since f admits derivatives up to order n then, by repeating the above process, we obtain F(f (n) )(x) = (2iπ)n xn Ff (x). The following result shows that the more differentiable the function f is, with derivatives in L1 (R), the more rapidly its Fourier transform f* decreases at infinity. C OROLLARY 8.1.– Let f ∈ L1 (R). If f admits derivatives up to order n which are in C L1 (R), then there exists C > 0 such that ∀x ∈ R∗ , f*(x) ≤ |x| n. Proof. From the formula F(f (n) (x)) = (2πx)n f*(x), it is immediately deduced that ∀x ∈ R , f*(x) ≤ ∗
1 with C = (2π)n
+∞ −∞
1 (2π|x|)n
+∞ −∞
C (n) , f (y) dy ≤ |x|n
(n) f (y) dy. This completes the proof.
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The Theory of Distributions
Observation: The corollary shows that the more differentiable f is and the more integrable its derivatives are, the more rapidly f* decreases at infinity. The following result is a result that gives conditions which ensure the regularity of the Fourier transform of a function if this latter multiplied with a polynomial is integrable. P ROPOSITION 8.2.– Let f ∈ L1 (R). If yf (y) ∈ L1 (R), then f* is differentiable and we have Ä ä f* (x) = F ((−2iπx)f ) . Moreover, if y n f (y) ∈ L1 (R), then f* is n times differentiable and we have Ä ä(n) = F ((−2iπ x)n f ) . f* Proof. We have (−2iπy)f (y)e−2iπxy = 2π |yf (y)|. Since, by hypothesis, yf (y) ∈ L1 (R), then according to the differentiation theorem under the integral sign, we have Ä ä * f (x) = (−2iπ) e−2iπxy yf (y)dy. R
Consequently, Ä ä f* (x) = F ((−2iπx)f ) . Since y n f (y) ∈ L1 (R), then repeating the above process gives the formula Ä ä(n) = F ((−2iπ x)n f ) . f* C OROLLARY 8.2.– Let f ∈ L1 (R). If y n f (y) ∈ L1 (R), then Ä ä (n) n * ∀x ∈ R, f (x) ≤ (2π) |y n f (y)| dy. R
Observation: The corollary reflects the fact that the faster f decreases at infinity, the more differentiable f* is and its derivatives are bounded.
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215
8.1.2.1.2. Vector case The properties given in the two previous propositions generalize to the multidimensional case (d ≥ 1). T HEOREM 8.2 (Fourier transform and differentiation).– Let f ∈ L1 (Rd ) and let k ≥ 1 be an integer. i) If ∂ α f exists for |α| k and ∂ α f ∈ L1 (Rd ), then we have F(∂ α f )(x) = (2iπ)|α| xα Ff (x). ii) If for any α : |α| k, xα f ∈ L1 (Rd ), then Ff is of class C k and we have α
∂ α Ff = F ((−2iπ)x) f ) . 8.1.2.2. Fourier transform and translation P ROPOSITION 8.3 (Fourier transform and translation).– Let f ∈ L1 (Rd ) and a ∈ Rd . We have i) the Fourier transform of the function τa (f ) = f ◦ τ−a : x → f (x − a) is F(τa (f ))(ζ) = e−2iπζa F(f )(ζ), ∀ζ ∈ Rd . ii) We also have τa (F(f )(ζ)) = F(f )(ζ − a) = F(e2iaπx f )(ζ), ∀ζ ∈ Rd . Proof. Let ζ ∈ Rd . We have F (τa (f )) (ζ) =
Rd
e−2iπxζ f (x − a)dx e−2iπ(y+a)ζ f (y)dy
= Rd
= e−2iπaζ
e−2iπyζ f (y)dy Rd
= e−2iπaζ F(f )(ζ). Hence, ∀ζ ∈ Rd , F(τa (f ))(ζ) = e−2iπζa F(f )(ζ). Similarly, we show property (ii).
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The Theory of Distributions
8.1.2.3. Fourier transform and dilation P ROPOSITION 8.4.– Let f ∈ L1 (Rd ) and λ ∈ R∗ . We have i) if g(t) = f ( λt ), then Fg(x) = |λ|d Ff (λx); ˇ ˇ ii) f“= f“ with fˇ(x) = f (−x). Proof. i) We have g*(x) = =
e−i2πxy g(y)dy Rd
y e−i2πxy f ( )dy. λ Rd
Using the change in variable y = λu, we obtain dy = |λ|d du. As a result, g*(x) =
Rd
|λ|d e−2iπλxu f (u)du
= |λ|d f*(λx). Conclusion: F(f ( λt ))(x) = |λ|d Ff (λx). ii) We have ˇ f“(x) = f*(−x) e2iπxy f (y)dy =
Rd
e−2iπxy f (−y)dy
= Rd
ˇ(x). = f“ We can also apply the first point of the proposition by taking λ = −1. Observation: For a small λ, f ( λt ) is very concentrated while its Fourier transform is very spread out.
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217
C OROLLARY 8.3.– Let f ∈ L1 (Rd ). We have – if f is even, then f* is also even; – if f is odd, then f* is also odd. Proof. ˇ ˇ = f“ – If f is even, then fˇ = f . As a result, f* = f“ . Consequently, f*is even. – The proof is similar to that of the first point. 8.1.2.4. Exchange formula P ROPOSITION 8.5 (Transfer formula).– Let f, g ∈ L1 (Rd ). We have
f*(x)g(x)dx.
f (x)* g (x)dx = Rd
Rd
Proof. The proof is mainly based on the Fubini theorem. Furthermore, according to the Fubini–Tonelli theorem, we have
Rd
f (x)g(y)e−2iπxy dxdy = Rd
Rd
|f (x)| dx
Rd
|g(y)| dy = f 1 g1 < ∞.
Fubini’s theorem then gives
Å
f (x)* g (x)dx = Rd
ã g(y)e−2iπxy dy dx
f (x)
Rd
Å
Rd
g(y)
=
Rd
f (x)e
−2iπxy
ã dx dy
Rd
f*(y)g(y)dy.
= Rd
Hence the exchange formula, also called the duality formula. One of the applications of the exchange (or transfer) formula is the theorem on the injection of the Fourier transform on L1 (Rd ). T HEOREM 8.3.– (El Amrani 2008) The Fourier transform F : L1 (Rd ) −→ C0 (Rd ) is an injective application.
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The Theory of Distributions
8.1.2.5. Fourier transform and convolution P ROPOSITION 8.6 (Fourier transform and convolution).– Let f, g ∈ L1 (Rd ). We have f ∗ g ∈ L1 (Rd ) and ∀x ∈ Rd , F (f ∗ g) (x) = F (f ) (x)F (g) (x). In other words, the Fourier transform of a convolution product is equal to the ordinary product of the Fourier transforms of each factor. Proof. In order to provide a simple proof, the proposition is presented for the scalar case (Fubini will allow generalizing the proof to the vector case). +∞ We have ∀x ∈ R, f ∗ g(x) = −∞ f (t)g(x − t)dt. Hence, for all x ∈ R, we have F (f ∗ g) (x) =
+∞ −∞ +∞
f ∗ g(y)e−2iπxy dy
+∞
=
−∞ +∞
−∞
+∞
= −∞
−∞
f (t)g(y − t)e−2iπxy dtdy f (t)e−2iπxt g(y − t)e−2iπx(y−t) dtdy.
We set u = t and v = y − t. From the change in variables theorem in multiple integrals, we have F (f ∗ g) (x) = Ç with J = det
∂t ∂u ∂y ∂u
∂t ∂v ∂y ∂v
+∞
−∞
å
+∞
−∞
f (t)e−2iπxt g(v)e−2iπxv |J|dtdv,
1 0 = 1. = 1 1
Then ∀x ∈ R, F (f ∗ g) (x) =
+∞
f (t)e
−2iπxt
+∞
dt
−∞
Consequently, ∀x ∈ R, F (f ∗ g) (x) = F (f ) (x)F (g) (x).
g(v)e−2iπxv dv.
−∞
The operations on the Fourier transform are summarized in Table 8.1. Function Fourier transform
f (x) F (f )(ζ) (2iπ)
∂ α f (x) |α|
(f ∗ g)(x)
f (x + a)
xα F (f )(ζ) F (f )(ζ)F (g)(ζ) ei2πa·ζ F (f )(ζ)
Table 8.1. Operations and Fourier transform
f (ax) 1 |a|d
ζ F (f )( a )
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219
8.1.3. Fourier transform inversion Although the Fourier transform is not bijective onto L1 (Rd ) (it is only injective), when it exists its inverse is defined by the following. D EFINITION 8.2.– Let f ∈ L1 (Rd ). The conjugate Fourier transform (or inverse) of the function f is the function denoted by Ff and it is defined by ∀x ∈ Rd , Ff (x) =
e2iπxy f (y)dy. Rd
T HEOREM 8.4.– Let f ∈ L1 (Rd ). If Ff ∈ L1 (Rd ) and f is a continuous function, then F (Ff ) = f.
[8.1]
Proof. To simplify, the theorem is shown for d = 1. Let x ∈ R. We have F (F(f )) (x) = R
e2iπy(x−t) e−ε|y| f (t)dtdy
= lim
ε−→0
R
= lim
ε−→0
Now, limε−→0
R
e2iπxy e−ε|y| f*(y)dy
= lim
ε−→0
e2iπxy f*(y)dy
R
e−2iπy(t−x) e−ε|y| dy.
f (t)dt R
R
F(e−ε|y| )(s)
=
2ε , ε2 + 4π 2 s2
thus
F (F(f )) (x)
=
2ε f (t) 2 dt. By change in variable t = x + εu, we obtain ε + 4π 2 (t − x)2 R 1 ε−→0 π
F (F(f )) (x) = lim
f (x + εu) R
1 du. 1 + u2
We have that f is continuous, so lim f (x + εu) = f (x). By Lebesgue’s ε−→0 dominated convergence theorem, we obtain F (F(f )) (x) = f (x)
1 π
R
1 1 +∞ du = f (x) [arctan(u)]−∞ = f (x). 2 1+u π
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Consequently, F (Ff ) = f . R EMARK 8.4.– 1) The identity [8.1] shows that the Fourier transform can be inverted: there exists therefore an inverse transform which is F, which could also be written as F −1 . 2) If f ∈ L1 (R) is a function that is not continuous in t, then F (F(f )) (t) =
f (t+ ) + f (t− ) . 2
3) These results are particularly significant when the operator F is used for the solution of partial differential equations. 8.2. Fourier transform in S(Rd ) As pointed out in theorem 8.3, the Fourier transform is injective on L1 (Rd ) but is not surjective. To overcome this shortcoming, the starting space is reduced to the subspace S(Rd ) ⊂ L1 (Rd ) while hoping to ensure the bijection of the Fourier transform on this new space S(Rd ) and preserving all the properties established in the L1 (Rd ) space. 8.2.1. Definition and first properties We start with a lemma which will be useful to establish a stability result of the Schwartz space with respect to the Fourier transform. L EMMA 8.2.– Let (fn ) be a sequence of elements of S(Rd ). If fn −→ 0 in S(Rd ), then fn −→ 0 in L1 (Rd ). n→+∞
n→+∞
Proof. Let (fn ) be a sequence of elements of S(Rd ) such that fn −→ 0 in S(Rd ). n→+∞
We have: fn −→ 0 in S(Rd ), then ∀p ∈ N, Qp (fn ) −→ 0. n→+∞
n→+∞
Fourier Transform
Let p ∈ N be such that Rd
|fn (x)|dx = ≤
Rd
221
dx < +∞. We have (1 + |x|)p
Rd
|fn (x)| ((1 + |x|)p )dx (1 + |x|)p
Rd
dx × sup |(1 + |x|)p fn (x)| (1 + |x|)p x∈Rd
Rd
dx × Qp (fn ) (1 + |x|)p
Rd
dx Np (fn ) −→ 0. n→+∞ (1 + |x|)p
Then, fn −→ 0 in L1 (Rd ). n→+∞
We have S(Rd ) ⊂ Lp (Rd ), ∀p ∈ [1, +∞], in particular, S(Rd ) ⊂ L1 (Rd ). Then, for all f ∈ S(Rd ), f* exists. T HEOREM 8.5.– The Fourier transform F is continuous on S(Rd ) and moreover, F(S(Rd )) ⊂ S(Rd ). Proof. Let f ∈ S(Rd ). Ff ∈ S(Rd ): We have f*(x) =
e−2iπxy f (y)dy. According to the derivation Rd
property of the Fourier transform of a function, we have ∀β ∈ Nd , ∂ β f*(x) = (−2iπ)|β| F(xβ f ). On the other hand, we know that if f ∈ S(Rd ) and P ∈ R[X], then P f ∈ S(Rd ). Hence, ∀β ∈ Nd , ∂ β f* exists. As a result, f* ∈ C ∞ (Rd ). Let us show that sup xα ∂ β f*(x) < +∞. x∈Rd
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The Theory of Distributions
We have xα ∂ β f*(x) = = xα (−2iπ)|β| F(xβ f) and F(∂ α (xβ f )) |α| α β α β* (2iπ) x (F(x f )). Hence, x ∂ f = c.F ∂ α (xβ f ) where c is a constant. We have f ∈ S(Rd ) =⇒ xβ f ∈ S(Rd ) =⇒ ∂ α (xβ f ) ∈ S(Rd ) ⊂ L1 (Rd ) =⇒ xα ∂ β f* is bounded =⇒ sup |xα ∂ β f*(x)|< +∞. x∈Rd
Hence f* ∈ S(Rd ). Consequently, ∀f ∈ S(Rd ), F(f ) ∈ S(Rd ). Conclusion: F is a linear mapping that sends S(Rd ) in S(Rd ). Continuity: Let (fn ) be a sequence of elements of S(Rd ) such that fn −→ 0. Let us show that: ∀α ∈ N , ∀β ∈ N d
d
F(fn )
, xα ∂ β f+ n
=
f+ n
n→+∞
−→
n→+∞ d
0 in S(R ), d
that is,
→ 0 uniformly on R .
(−2iπ)|β| F(∂ α (xβ fn )). We have fn −→ 0 in n→+∞ (−2iπ)|α| 0 in S(Rd ). As such, ∂ α (xβ fn ) −→ 0 in S(Rd ).
We recall that xα ∂ β f+ n = S(Rd ). Then, xβ fn
−→
n→+∞ α
According to lemma 8.2, ∂ (xβ fn ) −→ 0 in L1 (Rd ). Since
n→+∞
n→+∞
sup |F(∂ α (xβ fn ))(y)| ∂ α (xβ fn )L1 (Rd −→ 0, n→+∞
y∈Rd
d then sup |F(∂ α (xβ fn ))(y)| −→ 0. As a result, xα ∂ β f+ n −→ 0 uniformly on R . y∈Rd
n→+∞
n→+∞
Conclusion: F is a continuous linear mapping from S(Rd ) to S(Rd ). 8.2.2. Fourier transform and operations Here are some elementary but very important properties of the Fourier transform on S(Rd ).
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223
8.2.2.1. Fourier transform and differentiation P ROPOSITION 8.7 (Fourier transform and differentiation).– Let ϕ be a function of S(Rd ). We have the following: i) the function F(ϕ) is of class C 1 (Rd ), and for all j = 1, ..., d, we have ∂ξj F(ϕ)(ξ) = −2iπF(xj ϕ)(ξ), ∀ξ ∈ Rd ; ii) for j = 1, ..., d, we have F(∂xj ϕ)(ξ) = 2iπξj F(ϕ)(ξ), ∀ξ ∈ Rd . Proof. According to theorem 8.5, the conditions of theorem 8.2 are verified by the function ϕ ∈ S(Rd ). The function F(ϕ) is even of class C ∞ and the differentiation formulas follow therefrom. Application: Consider the “Gaussian kernel” function f , defined by 2 f (x) = e−πx . We know that f ∈ S(Rd ), and we show that f* = f . +∞ 2 * We set F (x) := f (x) = e−πy e−2iπxy dy and it can be noted that −∞ +∞ −πy 2 e dy = 1 (a classical integral that can be found through a little F (0) = −∞
calculation in polar coordinates). A legitimate differentiation with respect to x of the integral in the expression of F (x) gives
+∞
F (x) = −∞
+∞
=i
2
(−2iπy)e−πy e−2iπxy dy =if (y)
f (y)e−2iπxy dy.
−∞
A new reading of property (ii) in proposition 8.7 allows us to write F (x) = i(2iπx)f*(x) = −2πxF (x). This ordinary differential equation is easily integrated and we obtain 2 2 F (x) = ce−πx . Yet, F (0) = 1, thus c = 1. As a result, F (x) = e−πx = f (x). Consequently, f* = f .
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8.2.2.2. Fourier transform and translation P ROPOSITION 8.8 (Fourier transform and translation).– Let ϕ be a function of S(Rd ). We have i) for all a ∈ Rd , the function τa (ϕ) = ϕ ◦ τ−a : x → ϕ(x − a) has the Fourier transform F(τa (ϕ))(ξ) = e−2iπξa F(ϕ)(ξ); ii) for all a ∈ Rd and for all ξ ∈ Rd , we have F(e2iaπx ϕ)(ξ) = τa (F(ϕ)(ξ)) = F(ϕ)(ξ − a). 8.2.2.3. Fourier transform and dilation P ROPOSITION 8.9.– Let f ∈ S(Rd ) and λ ∈ R∗ . We have i) F(f ( λt ))(ζ) = |λ|d Ff (λζ); ˇ ˇ ii) f“= f“ with fˇ = f (−x). R EMARK 8.5.– The Fourier transform F substitutes differentiation and multiplication with x, as we have just seen. Consequently, F exchanges regularity and decreasing at infinity, that is, the more differentiable a function is (with integrable derivatives on Rd ), the more rapidly its Fourier transform decreases at infinity. This fact makes it possible to understand the interest of the Schwartz class with respect to the Fourier transform: since any function of S(Rd ) is of class C ∞ with derivatives which are all integrable, it can be deduced that its Fourier transform is rapidly decreasing. And since any function of S(Rd ) is rapidly decreasing, it is deduced that its Fourier transform is of class C ∞ . Therefore, it can be seen that the Schwartz class is invariant by Fourier transform, which is one of the advantages of this space. 8.2.3. Fourier transform inversion This section shows that the Fourier transform is invertible from S(Rd ) to S(Rd ). T HEOREM 8.6 (Fourier inversion).– Let f ∈ S(Rd ). We have f (x) = Rd
e2iπxy f*(y)dy, ∀x ∈ Rd ,
that is, F(F(f )) = f .
[8.2]
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225
t 2 Proof. Let λ > 0. We set ϕ(x) = e−|x| and fλ (t) = f ( ). λ According to the exchange formula, we have Rd
f*λ (x)ϕ(x)dx =
Rd
fλ (x)ϕ(x)dx * =
Rd
λd f (u)ϕ(λu)du * (u =
x ). λ
Since f*λ (x) = λd f*(λx), then
Rd
λd f (u)ϕ(λu)dx * =
Rd
f (u)+ ϕλ (u)du =
Rd
f (x)+ ϕλ (x)dx.
On the other hand, we have
Rd
fλ (x)+ ϕλ (x)dx =
Rd
f*(x)ϕλ (x)dx =
x f*(x)ϕ( )dx. λ Rd
We then obtain
x f*(x)ϕ( )dx = λ d R
Rd
fλ (x)ϕ(x)dx. *
When λ → +∞, Lebesgue’s dominated convergence theorem implies that
f*(x)dx = f (0)
Rd
Rd
Then, Rd
get
e−π
2
ϕ(x)dx * =
x2
R
R
(π) 2 e−π d
2
(x21 +...+x2d )
Rd
e−y
dx =
ϕ(x)dx. *
2
dx1 dx2 ...dxd . Setting y = πx, we
dy 1 =√ . π π 1
f*(x)dx. Let d π2 Rd ϕ ∈ S(Rd ) and we set f = τ−x ϕ ∈ S(Rd ). We have f (z) = ϕ(z + x) and f (0) = ϕ(x). Therefore, d
Hence,
Rd
ϕ(x)dx * = (π) 2 .
= 1. As a result, f (0) =
f (0) = Rd
F(τ−x ϕ(y))dy = ϕ(x).
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Consequently, ϕ(x) = Rd
demonstration.
e2iπxy ϕ(y)dy, * ∀ϕ ∈ S(Rd ). This completes the
Physical +∞ Interpretation: Physically, in one dimension, we often write [8.2] as f (t) = −∞ e2iπνt f*(ν)dν. t denotes time and ν designates frequency. The term e2iπνt is called the harmonic component and f*(ν)dν is called the weighting factor. These harmonic components e2iπνt are trigonometric functions of frequency ν and are the complex equivalent of sine waves. This is the reason why it is sometimes said that any function f (t) (admitting a Fourier transform) can be written as a continuous sum of sinusoids of frequencies ν ranging from −∞ to +∞. The term f*(ν) represents the weight of the harmonic component of frequency ν in the expansion of f . R EMARK 8.6.– We have F ◦ F = IS(Rd ) . T HEOREM 8.7.– F is a bijection from S(Rd ) onto S(Rd ): (thatis ∀g ∈ S(Rd ), ∃!f ∈ S(Rd ) : Ff = g). “ Proof. Let f ∈ S(Rd ). Let us first show that f“ = fˇ (we should recall that fˇ(x) = f (−x)). “ “ e−i2πxy f*(y)dy. According to the inverse Fourier formula, We have f (x) = Rd “ e−i2πxy f*(y)dy. Since f (−x) = fˇ(x), thus f“(x) = fˇ(x). It we have f (−x) = Rd
“ “ “ “ “ “ ˇ = f . Consequently, follows that f“ = fˇ. We have f“ = fˇ, thus f“ = f“ F ◦ F ◦ F ◦ F = IdS(Rd ) . It can be deduced that F is bijective and F −1 = F 3 . Conclusion: Since F is also continuous from S(Rd ) to S(Rd ), then F is an isomorphism from S(Rd ) to S(Rd ). We complete this series of properties of the Fourier transform on the Schwartz class by the Plancherel formula, which can be seen as a trivial consequence of the inversion formula.
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227
T HEOREM 8.8 (Plancherel).– The Fourier transform F is extended by continuity into an isometry from L2 (Rd ) to L2 (Rd ). That is to say, there exists an extension that is denoted again by F : L2 (Rd ) → L2 (Rd ) verifying, in addition, the identity f, gL2 = Ff, FgL2 , ∀f, g ∈ L2 (Rd ), and in particular, f L2 = Ff L2 , ∀f ∈ L2 (Rd ). Proof. Let f, g ∈ S(R ). We have f, g = d
f (x)g(x)dx. Now Rd
g ∈ S(Rd ) =⇒ g(x) =
Rd
e2iπxy g*(y)dy =
Rd
e−2iπxy g*(y)dy,
thus
f, g =
f (x) Rd
Rd
e−2iπxy g*(y)dydx.
According to Fubini’s theorem, we have f, g =
Å Rd
=
Rd
g*(y)
ã e−2iπxy f (x)dx dy Rd
g*(y)f*(y)dy g (y)dy f*(y)*
= Rd
= f*, g*. Conclusion: ∀f, g ∈ S(Rd ), f, g = Ff, FgL2 ∀f ∈ S(Rd ), f L2 = Ff L2 . Since S(Rd )|L2 = L2 (Rd ), then F is extended by continuity on L2 (Rd ) and again this extension is denoted by F. Moreover, we have ∀f, g ∈ L2 (Rd ), f, gL2 = Ff, FgL2 , f L2 = Ff L2 .
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R EMARK 8.7.– The extension F verifying f L2 = Ff L2 is and isomorphism from L2 (Rd ) to L2 (Rd ) and F 4 = IdL2 (Rd ) . 8.2.4. Fourier transform and convolution Along with the Fourier transform and in connection with it, the convolution product is an essential tool for the study of physical phenomena. Many systems actually operates as filters, in the sense that an input signal is modified after traveling through the system. Under very general assumptions of linearity and time invariance, the output signal is expressed as a convolution product between the input signal and a function that depends on the system under study. The Fourier transform exchanges convolution product and usual multiplication. T HEOREM 8.9.– Let ϕ, ψ ∈ S(Rd ). We have F(ϕ ∗ ψ) = Fϕ · Fψ
and
F(ϕψ) = Fϕ ∗ Fψ
Proof. The use of Fubini’s theorem can be easily justified, so we have 1) For all x ∈ Rd , we have F(ϕ ∗ ψ)(x) =
Rd
e−2iπxy (ϕ ∗ ψ)(y)dy
=
Rd
Rd
e
=
e−2iπxy ϕ(s)ψ(y − s)dsdy
−2iπxs
ï ϕ(s)
Rd
e
−2iπx(y−s)
Rd
* e−2iπxs ϕ(s)ψ(x)ds
= Rd
* = ψ(x)
e−2iπxs ϕ(s)ds Rd
* = (ϕ *ψ)(x) = (Fϕ · Fψ)(x). Conclusion: F(ϕ ∗ ψ) = Fϕ · F ψ.
ò ψ(y − s)dy ds
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229
2) For all x ∈ Rd , we have F(ϕψ)(x) =
e−2iπxy (ϕψ)(y)dy = Rd
e−2iπxy ϕ(y)ψ(y)dy. Rd
* we have ϕ(y) = F −1 (Φ)(y) = Let Φ = ϕ * and Ψ = ψ, Therefore, F(ϕψ)(x) =
e
−2iπxy
Rd
=
Φ(s)
Rd
e2iπsy Φ(s)ds. Rd
e2iπsy Φ(s)ds ψ(y)dy Rd
Rd
=
e−2iπ(x−s)y ψ(y)dy ds Rd
Φ(s)Ψ(x − s)ds
= Φ ∗ Ψ(x) = Fϕ ∗ Fψ(x). Conclusion: F(ϕψ) = Fϕ ∗ Fψ. P ROPOSITION 8.10.– Let T ∈ E (Rd ). For any ϕ ∈ S(Rd ), we have F(ϕ ∗ T ) = F(ϕ)F(T ). Proof. Let ϕ ∈ S(Rd ) and ζ ∈ Rd . Using integration under the bracket, we get F(ϕ ∗ T )(ζ) =
Rd
e−2iπxζ Ty , ϕ(x − y)dx
= Ty , = Ty ,
= Ty , e
Rd
e−2iπxζ ϕ(x − y)dx e−2iπ(y+z)ζ ϕ(z)dx
Rd −2iπyζ
F(ϕ)(ζ)
= F(T )(ζ)F(ϕ)(ζ). Consequently, F(ϕ ∗ T ) = F(ϕ)F(T ).
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The Theory of Distributions
At this point, a justification should be given about how we move from the first line to the second one in the above calculation because the function (x, y) → ϕ(x−y) does not necessarily have a compact support if we take ϕ ∈ S(Rd ). To this end, consider (ϕn ) a sequence of D(Rd ) which converges to ϕ in S(Rd ). According to the previous calculation, we have F(ϕn ∗ T ) = F(ϕn )F(T ). According to the first point of the proposition, ϕn ∗ T −→ ϕ ∗ T in S(Rd ). Since the Fourier transform is continuous S(Rd ), then F(ϕn ∗ T ) −→ F (ϕ ∗ T ) in S(Rd ). Finally, we have F(ϕn )(ζ)F(T )(ζ) −→ F(ϕn )(ζ)F(T )(ζ) in S(Rd ). It completes the demonstration. 8.3. Fourier transform in S (Rd ) In section 8.2, the Fourier transform has been defined on the Schwartz class S(Rd ). Unfortunately, this cannot be considered as sufficient, because there are far too few functions in S(Rd ). We will now extend its definition to the space S (Rd ) of tempered distributions by applying the same duality strategy. 8.3.1. Definition and first properties To make sense of the Fourier transform of a tempered distribution, we start with the following result. P ROPOSITION 8.11.– Let T ∈ S (Rd ). The relation FT, ϕ = T, ϕ, * ∀ϕ ∈ S(Rd ) defines a tempered distribution. The relation FT, ϕ = T, ϕ, * ∀ϕ ∈ S(Rd ) also defines a tempered distribution. Proof. Let ϕn −→ 0 in S(Rd ). Let us show that FT, ϕn −→ 0. n→+∞
Since ϕn
n→+∞
−→ 0 in S(Rd ), the continuity of the Fourier transform on S(Rd )
n→+∞
leads to ϕ +n −→ 0 in S(Rd ). As a result, T, ϕ +n −→ 0. On the other hand, we n→+∞
n→+∞
have T, ϕ +n = FT, ϕn . Hence, FT, ϕn −→ 0. Consequently, FT ∈ S (Rd ). n→+∞
Similarly, we show that FT ∈ S (Rd ). D EFINITION 8.3.– Let T ∈ S (Rd ). Then the Fourier transform FT and the inverse Fourier transform are defined by * ∀ϕ ∈ S(Rd ). FT, ϕ = T, ϕ * and FT, ϕ = T, ϕ,
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231
E XAMPLE 8.3.– 1) Let f ∈ L1 (R). We have Tf ∈ S (R), donc F(Tf ) ∈ S (R). Moreover, we have F(Tf ) = Tf*. As a matter of fact, Let ϕ ∈ S(R), we have F(Tf ), ϕ = Tf , F(ϕ) = f (x)F(ϕ)(x)dx
R
Å f (x)
=
e
R
−2iπxt
ã ϕ(t)dt dx.
R
The function (x, t) → f (x)ϕ(t) is integrable on R2 . Then, according to Fubini’s theorem, we have F(Tf ), ϕ =
Å
ã e−2iπxt f (x)dx dt
ϕ(t)
R
R
ϕ(t)f*(t)dt
= R
= Tf*, ϕ . +f = T . Hence, T f* 2) We have δ0 ∈ S (R) because δ0 ∈ E (R). In addition, δ“0 = 1. In reality, for any ϕ ∈ S(R), we have F(δ), ϕ = δ, F(ϕ) = F(ϕ)(0) = ϕ(t)dt R
= 1, ϕ = T1 , ϕ. Finally, δ“0 = 1. 3) We have δ“a (x) = e−2iπxa .
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The Theory of Distributions
Furthermore, for all ϕ ∈ S(R), it follows that Fδa , ϕ = δa , Fϕ = F(ϕ)(a) = e−2iπta ϕ(t)dt R
= e−2iπta , ϕ = Te−2iπta , ϕ. Hence, δ“a = e−2iπ.a : t → e−2iπta . R EMARK 8.8.– By duality, the translation property and the dilation property of F on S(R) remain valid on S (R). 8.3.2. Fourier transform and differentiation Given its importance, we reconsider the property of the differentiation of the Fourier transform of a tempered distribution. T HEOREM 8.10.– Let T ∈ S (Rd ) and α ∈ Nd . Then, we have i) ∂ α (FT ) = F[(−2iπx)α T ] = (−2iπ)|α| F[xα T ]; ii) F(∂ α T ) = (2iπx)α FT . Proof. i) Let ϕ ∈ S(Rd ). We have ∂ α (FT ), ϕ = (−1)|α| FT, ∂ α ϕ = (−1)|α| T, F(∂ α ϕ). Now, F(∂ α ϕ) = (2iπ)|α| xα ϕ, * then ∂ α (FT ), ϕ = (−1)|α| T, (2iπx)α ϕ * = F((−2iπx)α T ), ϕ. Hence, the formula ∂ α (FT ) = F[(−2iπx)α T ].
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233
ii) Let ϕ ∈ S(Rd ). We have F(∂ α T ), ϕ = ∂ α T, ϕ * = (−1)|α| T, ∂ α (Fϕ) = (−1)|α| T, (−2iπ)|α| F(xα ϕ) = (2iπ)|α| T, F(xα ϕ) = F((2iπ)|α| T ), xα ϕ = xα F((2iπ)|α| T ), ϕ = (2iπx)α T*, ϕ. Consequently, F(∂ α T ) = (2iπx)α FT . R EMARK 8.9.– As on the Schwartz class, the Fourier transform F on the space S (R) of the tempered distributions substitute differentiation and multiplication with x up to a factor ±2π i. All the interest of the Fourier transform for tempered distributions lies in this simple fact: on the one hand, the framework of the distributions makes it possible to differentiate as many times as necessary; on the other hand, after Fourier transform, any partial derivative of any order corresponds to the multiplication by a monomial. 8.3.3. Fourier transform inversion T HEOREM 8.11.– F is a continuous bijection of S (Rd ) to S (Rd ) and we have: F −1 = F 3 = F. Proof. – Bijection: We have FFT, ϕ = T, FF ϕ = T, ϕ. Thus, FFT = T and FFT = T. – Continuity: Consider a sequence (Tj ), Tj ∈ S (Rd ) such that Tj S (Rd ). We have S (Rd )
Tj −→ 0 ⇐⇒ ∀ϕ ∈ S(Rd ), Tj , ϕ −→ 0. j→+∞
j→+∞
−→ 0 in
j→+∞
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The Theory of Distributions
Thus, +j , ϕ = Tj , ϕ T * −→ 0 because ϕ * ∈ S(Rd ). j→+∞
d
(R ) +j S− → 0. Consequently, F is continuous (even bicontinuous), that Therefore, T j→+∞
is, F −1 = F is continuous. E XAMPLE 8.4.– The constant function 1 belongs neither to L1 (R) nor to L2 (R) and its Fourier transform in the sense of functions does not exist. Its transform in the sense of distributions is given by: F1 = δ, which can be easily obtained from the fact that F(δ) = 1 and with Fourier inverse. Observation: Plancherel’s theorem can be obtained by at least three methods: – the first method utilizes the Fourier transform on L2 (Rd ), built by density of L (Rd ) ∩ L1 (Rd ) in the Hilbert space L2 (Rd ), and the Plancherel formula which can be demonstrated for step functions with compact support; 2
– the second method is the one that appears in theorem 8.8 and which consists of extending the identity provided by the Fourier transform defined on S(Rd ) to the Hilbert space L2 (Rd ); – the third method consists of showing that L2 (Rd ) is a subspace of S (Rd ) that is stable under the action of the Fourier transform F. T HEOREM 8.12 (Plancherel).– The Fourier transform F defined on S (Rd ) induces a C−vector space isomorphism from L2 (Rd ) in itself, verifying in addition the identity f, gL2 = Ff, FgL2 , ∀f, g ∈ L2 (Rd ), and in particular, f L2 = Ff L2 , ∀f ∈ L2 (Rd ). 8.3.4. Fourier transform in E (Rd ) T HEOREM 8.13.– For any distribution T ∈ E (Rd ), the tempered distribution FT is the function defined by Rd ξ → T, e−ξ , with e−ξ : x → e−2iπxξ . This function g : ξ → F (T )(ξ) is of class C ∞ (Rd ) and polynomially increasing as well as all its derivatives.
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235
We also write F(T )(ξ) = g(ξ) = Tx , e−2iπxξ . Proof. Since the distribution T has compact support, then the expression Tx , e−2iπxξ exists and defines an indefinitely differentiable function that extends into a holomorphic function on the complex plane. We have F(T ), ϕ = T, Fϕ Æ ∏ +∞ −2iπxy = Tx , e ϕ(y)dy −∞
= Tx , ϕ(y), e−2iπxy = Tx ϕ(y), e−2iπxy = ϕ(y), Tx , e−2iπxy +∞ = ϕ(y) Tx , e−2iπxy dy −∞
= g, ϕ . This completes the demonstration. E XAMPLE 8.5.– We find that the Fourier transform of the Dirac distribution δ is δ* = 1. In fact, the support of δ being {0}, so δ ∈ E (Rd ) and we have immediately F(δ)(ζ) = g(ζ) = δ, e−2iπxζ = 1. E XAMPLE 8.6.– The Fourier transform of the nth derivative of the Dirac distribution δ is Ä ä n F δ (n) = (2iπx) . R EMARK 8.10.– We recall that, as mentioned above, the transform F exchanges regularity and decreasing at infinity. A function of L1 (Rd ) being a less irregular object than the Dirac mass, it is not surprising that its Fourier transform tends to 0 at infinity, whereas Fδ0 = 1 does not tend to 0 at infinity. On the other hand, the Dirac mass tends to 0 at infinity as fast as possible, since its support (reduced to {0}) is as small as possible: it is therefore not surprising that the Fourier transform of δ0 is a constant function, constants being the most regular functions possible. In reality, the Fourier transform of a distribution with compact support is much better than of class C ∞ : since it is an analytic function. More precisely, if T ∈ E (Rd ), then FT is analytic.
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T HEOREM 8.14 (Paley-Wiener).– (Dalmasso and Witomski 1996) Let T ∈ E (R) and M > 0 be such that supp T [−M, M ]. The function g = FT ∈ C ∞ (R) is extended to a function g˜ holomorphic on C. In addition, ∃m ∈ N, ∃C > 0, such that for all z ∈ C, we have m |˜ g (z)| ≤ C 1 + |z|2 2 e2πM |mz| . Proof. We know that g(ξ) = F(T )(ξ) is a function of class C ∞ defined by g(ξ) = Tx , e−2iπxξ . For all z ∈ C, we set g˜(z) = Tx , e−2iπxz . A direct calculation shows that g˜ is holomorphic on C and the inequality on the modulus of g˜ follows from the continuity of T . R EMARK 8.11.– A generalization of theorem 8.14 to vectors is given by the Paley– Wiener–Schwartz theorem, which consists of extending the Fourier transform of any distribution T ∈ E (Rd ) into an analytic function on C d . A very important consequence of theorem 8.14 is the following proposition. P ROPOSITION 8.12.– The Fourier transform of a distribution T with compact support (T = 0) cannot have compact support. Proof. Let T ∈ E (R) and let g = FT . The function g is analytic on R as a restriction of a holomorphic function on C. If g has compact support, then it will cancel out on a non-empty open set of R. Since it is analytic, it will be identically zero. This contradicts the fact that T = 0. 8.3.5. Fourier transform and Poisson summation formula We give a remarkable identity that is often used in Fourier analysis. T HEOREM 8.15 (Poisson summation formula).– The distribution T = to S (R) and its Fourier transform is F
δk =
k∈Z
Proof. Consider the distribution (Dirac comb), T =
δk belongs
k∈Z
δk in S (R).
k∈Z
δk .
k∈Z
The distribution T is tempered because the ϕ −→ δk , ϕ = ϕ(k) is linear and continuous from S(R) to R. k∈Z
k∈Z
mapping
Fourier Transform
237
A more detailed proof is exposed in exercise 7.5. The Fourier transform F is a continuous operation on S (R), therefore F
δk (ξ) = F (δk ) (ξ) = e−i2πkξ =: R(ξ).
k∈Z
k∈Z
k∈Z
This distribution is tempered and periodic of period 1. Let us determine R(ξ). We have ei2πξ R(ξ) = ei2πξ e−i2πkξ = e−i2πξ(k−1) = e−i2πkξ = R(ξ). k∈Z
k∈Z
k∈Z
Hence, g(ξ)R(ξ) = 0 avec g(ξ) = ei2πξ − 1. The function g is indefinitely differentiable and in addition g(ξ) = 0 if and only if, ξ ∈ Z with g (ξ) = 0. ak δk (ξ). The coefficients ak are all equal to a same constant Therefrom, R(ξ) = k∈Z
c. The fact that the distribution T is 1−periodic means that τ1 T = T where the distribution τ1 T is the translate of T by the translation 1. Therefore, R(ξ) = cδk (ξ). We take for ϕ the “rectangular” function defined by k∈Z ϕ(x) = 1, |x| < 12 , We have ϕ(x) = 0, |x| ≥ 12 . δk (ξ)), ϕ(ξ) R(ξ), ϕ(ξ) = c k∈Z
=c
δk (ξ)), ϕ(ξ)
k∈Z
= cϕ(0) = c. Now R(ξ), ϕ(ξ) =
=
1 2
− 12
k∈Z
k∈Z
ei2πkξ dξ
1 2
− 12
ei2πkξ dξ = 1,
because
1 2
− 12
e
i2πkξ
dξ =
1, k = 0 0, k = 0.
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The Theory of Distributions
Then, c = 1. Therefore, R(ξ) =
δk (ξ). Consequently,
k∈Z
F
δk = δk in S (R).
k∈Z
k∈Z
R EMARK 8.12.– For all a ∈ R∗+ , we have F
δak = δak .
k∈Z
k∈Z
P ROPOSITION 8.13 (Poisson summation formula).– If ϕ ∈ S(R), then
ϕ(k) =
k∈Z
ϕ(k). *
[8.3]
k∈Z
Proof. We first observe that these series converge absolutely. As such, we set c1 = sup x2 |ϕ(x)| , c2 = sup x2 |ϕ (x)| , c3 = sup ξ 2 |ϕ(ξ)| * and C = max(c1 , c2 , c3 ). x∈R
x∈R
ξ∈R
Now, ϕ ∈ S(R), thus ϕ * ∈ S(R) and we have C < +∞. As a result, |ϕ(k)| ≤
C k2
and
|ϕ(k)| * ≤
C . k2
C Since 2 is the general term of an absolutely convergent series, the convergence k of series in identity [8.3] derives thereof. According to theorem 8.15, we have F
δk = δk in S (R).
k∈Z
k∈Z
δk , ϕ = δk , ϕ . Consequently, Then, for any ϕ ∈ S(R), we have F ∀ϕ ∈ S(R),
k∈Z
ϕ(k) =
k∈Z
k∈Z
ϕ(k). *
k∈Z
R EMARK 8.13.– For all a ∈ R∗+ , and for any function ϕ ∈ S(R), the Poisson 1 k ϕ( ) = ϕ(ka). * summation formula is written as a a k∈Z
k∈Z
Fourier Transform
239
8.3.6. Fourier transform and convolution The transformation of the convolution product into the classical product, and vice versa, by the Fourier transform on the convolution products of the elements of S(Rd ) is also valid on the convolution products of distributions, assuming that they are well defined. This relation, reinforced by the Fourier inversion formula in S (Rd ), F(S ∗ T ) = FS · FT, can be used to provide the definition of the convolution product of certain tempered distributions whose product of Fourier transforms can be defined as a distribution in S (Rd ). P ROPOSITION 8.14.– Let S and T denote two distributions with compact supports. Then, the support of the convolution product S ∗ T is compact and we have F(S ∗ T ) = (FS).(FT ). Proof. It should be recalled that the convolution product of two distributions S and T exists if one of the supports (supp S or supp T ) is compact. In this case, supp (S ∗ T ) is also compact because it is included in the adhesion of the sum of the support of S and the support of T . By theorem 8.13, F(S ∗ T ) is a function defined by F(S ∗ T )(ξ) = S ∗ T, e−2iπxξ ∂ ¨ = Sx ⊗ Ty , e−2iπξ(x+y) = Sx , e−2iπxξ Ty , e−2iπyξ = (FS)(ξ).(FT )(ξ). The demonstration is complete. R EMARK 8.14.– The above result is valid with more general assumptions. On the other hand, the product (FS).(FT ) cannot be defined to most cases. If S is a distribution with compact support and T any tempered distribution, then the product (FS).(FT ) must be interpreted as the product of a distribution by a function of class C ∞ . It should also be noted that if S and T are tempered distributions and have lower bounded supports, then the product S ∗ T is well defined but the product (FS).(FT ) is generally not defined. To better understand this, the case of the Heaviside distribution should be considered.
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The Theory of Distributions
P ROPOSITION 8.15.– Let S and T denote two distributions with compact supports. Then we have: F(ST ) = (FS) ∗ (FT ). Proof. By setting FS = U and FT = V , we know from the Fourier reciprocity formulas that FU = F FS = S, FV = F FT = T, and we will have F(ST ) = FS ∗ FT. We conclude this section with Table 8.2, which gives the Fourier transform of some tempered distributions. T T (k) (−2iπx)k T τa T e2iπxa T δa e2iπxa (k) δa xk signx Vp( x1 )
F
−→ F(T ) −→ (2iπζ)k F(T ) −→ (F(T ))(k) −→ e−2iπζa F(T ) −→ τa F(T ) −→ e−2iπζa −→ δa −→ (2iπζ)k −→ (−2iπx)k δak 1 1 −→ iπ Vp( ζ ) −→ −iπsign
Table 8.2. The Fourier transform of a tempered distribution
8.4. Exercises with solutions E XERCISE 8.1.– Build a function f ∈ S(R) non-identically zero such that, for all k ∈ N,
xk f (x)dx = 0. R
S OLUTION 8.1.– For all k ∈ N, we have F xk f (0) =
xk f (x)dx = 0. R
dk dk Now (−2iπ)k F xk f (0) = k F(f )(0), thus ∀k ∈ N, k F(f )(0) = 0. The dζ dζ condition imposed on f is expressed by the fact that all derivatives of F(f ) cancel out at 0.
Fourier Transform
241
We now consider the plateau function χ with support in ]−2, 2[ and which takes 1 on [−1, 1]. Let g ∈ S(R). We take h : ζ → h(ζ) = (1 − χ(ζ)) g(ζ). Since g ∈ S(R), then h ∈ S(R). Moreover, all derivatives of h are zero at 0. The function f = F −1 (h) answers the question. E XERCISE 8.2.– Solve in D (R) the differential equation u + u = δ0 . What are the tempered solutions? S OLUTION 8.2.– We start by determining the classical solutions of the associated homogeneous equation u + u = 0. We find ∀x ∈ R, u(x) = ce−x , c ∈ R. Then, we look for a distribution S, which is a particular solution to u + u = δ0 as S = ex T , where T is the solution of u + u = δ0 . We then have S = (ex T ) = ex (T + T ) = δ0 ex = δ0 e0 = δ0 . Therefore, S = H + c, c ∈ R (H is the Heaviside distribution). Hence, T = He−x + ce−x in D(R). However, the distribution associated with x −→ e−x is not in S (R) (see the behavior at −∞), then the only solution of u + u = δ0 which is in S (R) is the distribution T = He−x . E XERCISE 8.3.– Compute the Fourier transform of the tempered distributions on R associated with the following functions: 1) eiax (a ∈ R); 2) cos(x); 3) x sin(x); 4) e−a|x| (a > 0); 5) |x|e−a|x| (a > 0). S OLUTION 8.3.– 1) Let a ∈ R. We have F(δa ) = e−i2πaζ . Then, by Fourier inversion, FF(δa ) = F(e−i2πaζ ) with FF = σ où σ(f )ζ = f (−ζ). Hence, δa = σF(e−i2πaζ ). Therefore, F(ei2πax ) = δa . By change in variable, we obtain a . F(eiax ) = δ 2π Special case: F(δ) = 1 and F(1) = δ. 2) We have ∀x ∈ R, cos x =
eix + e−ix . 2
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The Theory of Distributions
Hence, by linearity of F and from Question 1, we have F(cos x) =
1 1 1 + δ −1 . F(eix ) + F(e−ix ) = δ 2π 2π 2 2
It should noted that we also have F(sin x) =
1 1 1 − δ −1 . F(eix ) − F(e−ix ) = δ 2π 2π 2i 2i
3) We have F(x sin x) =
i 1 i 1 ∂ζ F(sin x) = ∂ζ δ 1 − δ −1 = δ 1 + δ −1 . 2π 2π 2π 2π 2i 2π 4π 2π
4) Since x −→ e−a|x| ∈ S, for a > 0, we then have F e−a|x| (ζ) =
+∞
−∞ 0
−∞
0
1 = e(a−i2πζ)x a − i2πζ = =
+∞
eax−i2πxζ dx +
= ï
e−a|x| e−i2πxζ dx
ò0 −∞
e−ax−i2πxζ dx
ï + −
1 e−(a+i2πζ)x a − +2πζ
1 1 + a − i2πζ a + i2πζ a2
2a 2 . + 2πζ
Observation: This issue was addressed in example 8.2. 5) Similarly, we have F |x|e−a|x| (ζ) =
+∞
−∞ 0
= −∞
ï = −x
|x|e−a|x| e−xi2πζ dx
−xeax−i2πxζ dx +
1 e(a−i2πζ)x a − iζ
ò0
+∞ 0
xe−ax−i2πxζ dx
0
+ −∞
−∞
e(a−i2πζ)x dx a − i2πζ
ò+∞ 0
Fourier Transform
243
ò+∞ +∞ −(a+i2πζ)x e −x −(a+i2πζ)x + + e dx a + i2πζ a + i2πζ 0 0 ñ ô0 ô∞ ñ e−(a+i2πζ)x e(a−i2πζ)x +0+ =0+ (a − i2πζ)2 −∞ (a + i2πζ)2 0 ï
=
1
1
2 a + i2πζ 2 2 a + i2πζ + a − i2πζ = 2 2 a2 + 2πζ 2 2 a2 − 2πζ = 2 2 . a2 + 2πζ a − i2πζ
2 +
−a|x|
Therefore: F(|x|e
2 2 a2 − 2πζ )= 2 2 . a2 + 2πζ
E XERCISE 8.4.– Compute the Fourier transform of the tempered distributions on R associated with the following functions: sin(x) ; x 2) sin(|x|). 1)
S OLUTION 8.4.– 1) Direct method: If u = F i 2π u
= F(sin x).
sin x sin x , since sin x = × x, then we have x x
sin x Hence, u = F − i2πx and according to Question 2 of exercise 8.3, we x obtain 1 − δ −1 . u = −π δ 2π
[8.4]
2π
So u |]−∞, −1 [ = 0 ⇒ u|]−∞, −1 [ = a 2π
u |] −1 , 2π
1 2π [
2π
= 0 ⇒ u|] −1 , 2π
1 2π [
=b
1 1 and u |] 2π ,+∞[ = 0 ⇒ u|] 2π ,+∞[ = c.
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The Theory of Distributions
⎧ −1 ⎪ ⎨a if x < 2π , If we set u ˜ = b if −1 2π < x < ⎪ ⎩ 1 c if x > 2π ,
1 2π ,
then by the jump formula, we have
1 , u ˜ = (b − a)δ −1 + (c − b)δ 2π 2π
and we want a, b, c such that b − a = π and c − b = −π; this leads us into taking c = a and b = π + a. Then u ˜ = a + π 1] −1 , 2π
1 2π [
is a solution to equation [8.4].
Hence, the general solution to equation [8.4] is u=u ˜ + constant = A + π1] −1 , −1 [ with A ∈ R. 2π
Yet,
2π
sin x ∈ L2 (R), then by Plancherel’s theorem, F( sinx x ) ∈ L2 (R). x
So, since the constant A is not in L2 (R), unless it is zero, we must have A = 0. Finally, F( sinx x ) = π 1] −1 , −1 [ . 2π
2π
Inverse Fourier method: We have Ä F 1] −1 , 2π
ä 1 2π [
(ζ) =
1 2π −1 2π
e−i2πxζ dx =
2 sin(ζ) . 2πζ
ä Ä Then, FF 1] −1 , 1 [ (x) = π1 F( sinx x ) = 1] −1 , 1 [ (x). 2π 2π 2π 2π sin x Hence, F x = π 1] −1 , 1 [ . Using the same calculation, we get the following 2π 2π result: Å F
sin(2πRx) x
ã = π 1]−R,R[ .
2) Since sin 0 = 0, we have sin |x| = sin x 1x>0 − sin x 1x0 = lim sin x e−εx 1x>0 in S (R) ε−→0
(∗).
Fourier Transform
245
Moreover, ∀ε > 0, we have
F sin x e
−εx
1x>0 (ζ) =
+∞
0
1 = 2i
sin xe−εx−i2πxζ dx
+∞
e
x(i−ε−2πiζ)
0
1 dx − 2i
+∞
e−x(i+ε+i2πζ) dx
0
ï ò 1 1 1 − 2i i − ε − 2πiζ −(i + ε + i2πζ) ï ò −1 1 1 = . − 2 2πζ − 1 − iε 2πζ + 1 − iε =
A classical result on Vp( x1 ), in D (R), allows us to deduce that 1 1 −→ Vp( ) + iπδa , y + a − iε y+a when ε tends to 0+ , ∀a ∈ R. Therefore, ï Å ã Å ãò 1 1 iπ −1 F sin x e−εx 1x>0 −→ Vp − Vp + δ1 −δ−1 , 2 2πζ − 1 2πζ + 1 2 when ε tends to 0+ . Similarly, F (sin x e
εx
ï Å ã Å ãò 1 1 iπ 1 Vp − Vp + 1x0 ) − F(sin x eεx 1x 0) the Gaussian of a π + variable, we have fα = α f π2 . α
On Rd , we have
π d2 ä Ä 2 2 π2 F e−αx (ζ) = e− α ζ . α E XERCISE 8.7.– Consider the two functions Π and defined by Π(x) =
1 if |x| ≤ 0 if |x| >
1 2 1 2
and (x) =
1 − |x| if |x| ≤ 1 0 if |x| > 1.
1) Compute the Fourier transform of the function Π. 2) Compute the Fourier transform of the function . 3) Compute and give (x) as a function of Π(x). 4) Using Question 3, find the result of Question 2. 5) Determine Π ∗ Π, and then deduce the result of Question 2. S OLUTION 8.7.– 1) In example 8.1, it has been shown that ∀ζ ∈ R, F(Π)(ζ) =
sin(πζ) . πζ
2) For any ζ ∈ R, we have F()(ζ) =
+∞ −∞
(x)e−2iπxζ dx
+∞
=2 0
1
=2 0
(x) cos(2iπxζ)dx
(1 − x) cos(2iπxζ)dx.
By integration by parts, it is obtained that ∀ζ ∈ R, F()(ζ) =
sin2 (πζ) 1 − cos(2πζ) = . π2 ζ 2 π2 ζ 2
Fourier Transform
249
3) It is easily shown that ⎧ ⎪ ⎨−1 if 0 < x ≤ 1 (x) = 1 if − 1 ≤ x < 0 ⎪ ⎩ 0 if |x| > 1. Otherwise, we have (x) = Π(x + 12 ) − Π(x − 12 ). 4) We have (x) = Π(x + 12 ) − Π(x − 12 ). Therefore, 1 1 F( )(ζ) = F(Π(x + ))(ζ) − F(Π(x − ))(ζ) 2 2 = eπiζ F(Π(x))(ζ) − e−πiζ F(Π(x))(ζ) = 2i sin(πζ)F(Π(x))(ζ) = 2i
sin2 (πζ) . πζ
On the other hand, from the Fourier transform property and the derivation, it is known that F( )(ζ) = 2iπζF((x))(ζ). Consequently, F()(ζ) =
sin2 (πζ) . π2 ζ 2
5) For any ζ ∈ R, we have Π ∗ Π(x) = = =
+∞ −∞ 1 2
− 12
Π(t)Π(x − t)dt
Π(x − t)dt
x− 12 x+ 12
Π(s)ds
⎧ ⎪ ⎨1 − x if 0 ≤ x ≤ 1 = 1 + x if − 1 ≤ x ≤ 0 ⎪ ⎩ 0 if |x| > 1 = (x).
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The Theory of Distributions
We know that F(Π ∗ Π) = F(Π)F(Π), thus F()(ζ) =
sin2 (πζ) . π2 ζ 2
E XERCISE 8.8.– 1) Determine the Fourier transform of the function f defined by f (x) = 2) Deduce the Fourier transform of the function g(x) =
1 . 1 + x2
x . (1 + x2 )2
S OLUTION 8.8.– 1) Using the residual method, we obtain F(f )(ζ) =
R
e−2iπxζ dx = πe−2π|ζ| . 1 + x2
2) We have Å F
x (1 + x2 )2
ã
Ç
Å ã å 1 (ζ) = F − (ζ) 2 1 + x2 ÇÅ ã å 1 1 (ζ) =− F 2 1 + x2 ã Å 1 (ζ) = −iπζ F 1 + x2 ã Å 1 (ζ) = −iπζF 1 + x2 = −π 2 iζe−2π|ζ| .
Å Then, F
x (1 + x2 )2
ã
(ζ) = −π 2 iζe−2π|ζ| .
E XERCISE 8.9.– Let k > 0 and u ∈ S (R) be such that Show that
d4 u + ku ∈ L2 (R). dx4
dj u ∈ L2 (R) for 0 ≤ j ≤ 4. dxj
d4 u + ku ∈ L2 (R), we have S OLUTION 8.9.– By the Plancherel theorem, since dx4 (i2πζ)4 + k u *(ζ) ∈ L2 (R). Therefore, u * is measurable. Since k > 0, then we have 4 4 i2πζ + k = k + 2πζ > 0. We set ξ = 2πζ.
Fourier Transform
251
– If |ξ| 1, then ∀j ∈ {0, ..., 4}, |ξ|j 1 C1 k if C1 k1 . Therefore, ∃C1 > 0, |ξ|j C1 k C1 (k + ξ 4 ). – If |ξ 1, then for C2 1, |ξ|j C2 |ξ|4 = C2 ξ 4 . Hence, ∃C2 1, |ξ|j C2 (ξ 4 + k). – If C = max(C1 , C2 ), we have just shown that in all cases, ∃C > 0, ∀ξ ∈ R, |ξ|j C(k + ξ 4 ), thus |ξ|j |* u| C(k + ξ 4 )|* u|, that is, |2πζ|j |* u| k + (2πζ)4 u *, for all j ∈ {0, ..., 4}. * ∈ L2 (R) Placing an upper bound, since k + (2πζ)4 u * ∈ L2 (R), then (2πζ)j u for all j ∈ {0, ..., 4}. By Plancherel,
dj u ∈ L2 (R), ∀j ∈ {0, ..., 4}. dxj
E XERCISE 8.10.– Let P (ξ) be a polynomial in Rd not identically zero. Let P (D) designate the differential operator with constant coefficients associated with P (ξ). Show that if u ∈ E (Rd ) is such that P (D)u = 0, then u = 0. * is a S OLUTION 8.10.– First, we should observe that since u has compact support, u function C ∞ on Rn . Let us apply F to the equation P (∂)u = 0, we obtain P (2πiζ)* u(ζ) = 0, ∀ζ ∈ Rn . *(ζ) = 0. We set Z = {ζ ∈ Rn , P (2πiζ) = 0}. Then ∀ζ ∈ Rn \Z, u However, since P is a polynomial, Z is a closed set with empty interior because P is non-identically null. Therefore, Rn \Z is dense in Rn . Since u * is continuous and null on the dense set Rn \Z, u * is null everywhere. By inverse Fourier transform, we obtain that u = 0. E XERCISE 8.11.– Let P (D) =
aα ∂ α be a differential operator on Rd with
|α|≤m
constant coefficients such that: Σ :=
ξ ∈ Rd ; P (ξ) =
|α|≤m
aα ξ α = 0
Show that the kernel of P (D) in S (Rd ) consists of polynomials.
= {0}.
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The Theory of Distributions
S OLUTION 8.11.– Let u ∈ S (Rn ), u ∈ ker P (D). Then P (D)u = 0. Since u ∈ S (Rn ) and does not necessarily have a compact support, the argument of exercise 8.10 should be adapted. Since u ∈ S (Rn ) and P (D)u = 0, by Fourier transform, P (2πiζ)* u = 0 in S (Rn ). Therefore, u * is null in Rn \Σ.
Since Σ = {0}, we deduce that supp u * ⊂ Σ = {0}. Hence u * = By inverse Fourier transform, u =
(α)
aα δ0 .
|α|m α
aα (2πix) . Consequently, u is indeed a
|α|m
polynomial. E XERCISE 8.12 (Dirac comb Fourier transform).– We consider the Dirac comb +∞ a = δka , where a > 0. k=−∞
1) Show that +∞
a =
δka =
k=−∞
+∞ 1 2iπk x a. e a
[8.5]
k=−∞
2) Compute the Fourier transform of a . 3) Deduce that F(a ) = a1 a1 . S OLUTION 8.12.– 1) Considering the periodic function f of period a > 0 defined on [0, a[ by f (x) = xa , we recall that identity [4.2] gives f (x) =
+∞ 1 δka . − a k=−∞
A Fourier series expansion of the function f gives f (x) =
i 1 2iπk x 1 a. + e 2 2π k k=0
[8.6]
Fourier Transform
253
Obviously, the convergence of this series is in L2 (0, a), but it can be shown that it also takes place in D (R). By continuity of the derivation operator, we can differentiate term by term: f (x) = −
1 2iπk x a. e a
[8.7]
k=0
The two formulas [8.6]–[8.7] ensure that a =
+∞
δka =
k=−∞
+∞ 1 2iπk x a. e a
[8.8]
k=−∞
2) From exercise 7.5, a is a tempered distribution. The continuity of the Fourier +∞ transform on the space S (R) leads to F(a ) = F(δka ). We know that k=−∞
+∞
∀ζ ∈ R, F(δka )(ζ) = e2iπkaζ , then F(a )(ζ) =
e2iπkaζ .
k=−∞
3) According to identity [8.8], we have
+∞
e2iπkaζ =
k=−∞
F(a ) =
1 1 . Consequently, a a
1 1 . a a
Special case: If a = 1, again we find F
+∞
+∞ δka = δka .
k=−∞
k=−∞
E XERCISE 8.13.– In the following, we will denote by δathe Dirac distribution at point a. We define by induction the sequence of distributions Tk k>0 as follows: T1 =
1 (δ1 + δ−1 ) and ∀k 2, Tk = Tk−1 ∗ T1 . 2
1) Express Tk as a finite linear combination of distributions with point support. 2) Compute the Fourier transform T*k of the distribution with compact support Tk . ξ 3) For k 1, we set fk (ξ) = T*k √ . Show that fk ∈ S (R) and the sequence k fk k1 converges in S (R) to a distribution that we will determine.
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The Theory of Distributions
4) Let gk denote the distribution of which fk is the Fourier transform. Show that the sequence gk k1 converges in S (R) to a distribution that we will determine. 2
Hint: The fact that for a > 0, F(e−ax ) =
√ π 2 x2 √π e− a a
can be used.
S OLUTION 8.13.– 1) We first point out that since supp(T ∗ S) ⊂ supp(T ) + supp(S), given that T1 has compact support, we show by induction that all Tk also have compact support. Recall also that for a and b in R, for all ϕ ∈ Cc∞ (R), we have δa ∗ δb , ϕ = δa , δb , ϕΔ = δa , ϕ(x + b) = ϕ(a + b) = δa+b , ϕ . Therefore, δa ∗ δb = δa+b . We can then calculate Tk . Let k 2: Tk = T1 ∗ ..... ∗ T1 = T1∗k =
1 2k
δ1 + δ−1
∗k
.
Now, the law ∗ is commutative and associative, so we can apply Newton’s binomial formula to obtain k Å ã 1 k ∗j ∗(k−j) Tk = k δ ∗ δ−1 2 j=0 j 1
=
k Å ã 1 k δj ∗ δj−k 2k j=0 j
=
k Å ã 1 k δ2j−k . 2k j=0 j
Therefore, ∀k 2, Tk =
k Å ã 1 k δ2j−k . 2k j=0 j
2) Since Tk = T1∗k , we have T*k (ζ) = (T*1 (ζ))k . Now, T*1 (ζ) = 12 (δ*1 + δ*−1 ) = 12 (e−i2πζ + ei2πζ ) = cos(2πζ), thus k T*k (ζ) = cos(2πζ) and T*k is indeed of class C ∞ (which we already knew because Tk ∈ E (R)).
Fourier Transform
255
k 3) According to Question 2, we have ∀k 1, ∀ζ ∈ R, fk (ζ) = cos √ζk . Then, fk ∈ L∞ (R) ⊂ S (R), thus fk ∈ S (R). To show the convergence of the sequence (fk ) in S (R), we set ϕ ∈ S(R) and look for the possible limit of the sequence fk , ϕ. Let ϕ ∈ S(R), we have ∀k 1, fk , ϕ = fk (ζ)ϕ(ζ)dζ. R
ä Ä π2 and cos 2π √ζk 0. We set ζ ∈ R. For k 16|ζ|2 , we have ä Ä 2ä Ä 2 On the other hand, we have cos 2π √ζk = 1 − 2π 2 ζk + o ζk and we have |ζ| 2π √ k
ãã Å Å ζ ln fk (ζ) = k ln cos 2π √ k Å ã 2π 2 ζ 2 ζ2 = k ln 1 − + o( ) k k = −2π 2 ζ 2 + o(ζ 2 ). Now, ∀ζ ∈ R, |fk (ζ)ϕ(ζ)| |ϕ(ζ)| ∈ L1 (R), so by the dominated convergence theorem, we get lim
k−→+∞
fk , ϕ =
lim
k−→+∞
R
=
lim
R k−→+∞
e−2π
=
2 2
ζ
fk (ζ)ϕ(ζ)dζ fk (ζ)ϕ(ζ)dζ ϕ(ζ)dζ
R
¨ ∂ 2 2 = e−2π ζ , ϕ . 2 2 Hence, fk k1 converges in S (R) to e−2π ζ . 4) Let gk ∈ S (R) such that fk = F(gk ). Then F(fk ) = σ(gk ) and gk = (F ◦ σ)(fk ), ∀k 1.
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By continuity of F in S (R), from the convergence in S (R) of it is deduced (fk )k1 , the convergence of gk k1 in S (R). Moreover, gk k1 converges in S (R) to (F ◦ σ)(e−2π
2 2
ζ
∀x ∈ R, (F ◦ σ)(e
), and we have
−2π 2 ζ 2
Therefore, gk : x −→
… )(x) = 2
x √1 e− 2 2π
1 − x2 e 2. 2π
in S (R).
E XERCISE 8.14.– For ψ differentiable on R, we write Dx ψ =
1 dψ . i dx
2
For ϕ ∈ S(R) and s ∈ R, we set e−isΔ ϕ = F −1 (eisξ Fϕ). 1) Demonstrate that for any ψ ∈ S(R) and all t ∈ R, e−itΔ eitΔ ψ = ψ. 2) Let ψ ∈ S(R). Show that for all t ∈ R, we have eitΔ xe−itΔ ψ = 4π 2 xψ + 2itDx ψ. 3) Thereof deduce that if P (x) = we have ∀t ∈ R, eitΔ P (x)e−itΔ ψ =
m k=0 m
ak xk with ak ∈ C, then for all ψ ∈ S(R), ak (4π 2 x + 2itDx )k ψ.
k=0
S OLUTION 8.14.– 1) Let t ∈ R and ϕ ∈ S(R). We have 2 F e−itΔ eitΔ ψ = FF −1 eitζ F eitΔ ψ 2 2 = eitζ F F −1 e−itζ F(ψ) 2
2
= eitζ e−itζ F(ψ) = F(ψ). Then, by the inverse Fourier formula, we obtain: e−itΔ eitΔ ψ = ψ. 2) By definition of eisΔ , for all ψ ∈ S(R) and any real t we have 2 F(eitΔ xe−itΔ ψ) = FF −1 eitζ F(xe−itΔ ψ) 2
= e−itζ F(xe−itΔ ψ).
Fourier Transform
257
Then F xe−itΔ ψ = F(x) ∗ F(e−itΔ ψ) 2 = i2πDζ δ0 ∗ F F −1 eitζ Fψ 2
= i2πδ0 ∗ eitζ Fψ(ζ). Hence 2 2 * F eitΔ xe−itΔ ψ = i2πe−itζ δ0 ∗ eitζ ψ(ζ) 2 2 * = i2πe−itζ δ0 ∗ eitζ ψ(ζ) 2 2 * = i2πe−itζ eitζ ψ(ζ) 2 2 * + eitζ 2 d ψ(ζ) * = i2πe−itζ 2itζeitζ ψ(ζ) dζ * + d ψ(ζ) * = i2π 2itζ ψ(ζ) . dζ Now, F(Dx ψ) = −i2πζF(ψ) and F(xψ) =
i 2π Dζ F(ψ),
thus
F eitΔ xe−itΔ ψ = 2itF(Dx ψ) + 4π 2 F(xψ) = F 2itDx ψ + 4π 2 xψ (from the linearity of F). Hence, by inverse Fourier transform, we obtain ∀ψ ∈ S(R), ∀t ∈ R, eitΔ xe−itΔ ψ = 2itDx ψ + 4π 2 xψ. 3) By linearity, it suffices to show that for all k 0, we have k ∀t ∈ R, ∀ψ ∈ S(R), eitΔ xk e−itΔ ψ = 2itDx + 4π 2 x ψ.
[8.9]
For k = 0, this is Question 1. For k = 1, this is Question 2. We can proceed by induction on k, the initialization at k = 0 or k = 1 being done. We assume that the result holds at rank k 1.
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The Theory of Distributions
Then, by Question 1, as eitΔ e−itΔ = Id, for all t ∈ R and for all ψ ∈ S(R), we have eitΔ xk+1 e−itΔ ψ = eitΔ xe−itΔ eitΔ xk e−itΔ ψ = eitΔ x e−itΔ (2itDx + 4π 2 x)k ψ (by assumption of induction) k = 2itDx + 4π 2 x x − 2tDx ψ (by Question 2) k+1 = 2itDx + 4π 2 x ψ. Therefore, identity [8.9] holds at rank k + 1. By induction, we obtain k ∀k 0, ∀t ∈ R, ∀ψ ∈ S(R), eitΔ xk e−itΔ ψ = 2itDx + 4π 2 x ψ. Then, if P (x) = have
m
ak xk , by linearity, for all ψ ∈ S(R) and for all t ∈ R, we
k=0
eitΔ P (x)e−itΔ ψ =
m
ak eitΔ xk e−itΔ ψ
k=0
=
m
k ak 2itDx + 4π 2 x ψ
k=0
= P 2itDx + 4π 2 x ψ. E XERCISE 8.15.– For ϕ ∈ D(Rd ) and t > 0, we denote by ϕt the function defined by ϕt (x) = ϕ( xt ). Let T ∈ S (Rd ). It is assumed that T* ∈ L∞ . Using the fact that T = F −1 FT , show that, ∀t > 0, | T, ϕt | ≤ T*L∞ ϕ * L1 . S OLUTION 8.15.– Let ψ ∈ D(Rd ) and t > 0. For T ∈ S (Rd ) such that T* ∈ L∞ (Rd ), we have | T, ϕt | = F −1 FT, ϕt = |FT, (F ◦ σ)ϕt | T*(ζ)(F ◦ σ)ϕt )(ζ)dζ = Rd
T* L∞ (Rd ) ×
Rd
|((F ◦ σ)ϕt )(ζ)| dζ.
Fourier Transform
259
Now (F ◦ σ)ϕt (ζ) =
x ϕ( )e2πixζ dx = t Rd
Hence |((F ◦ σ)ϕt )(ζ)| dζ = Rd
Rd
Rd
ϕ(y)e2πityζ td dy = td ϕ(−tζ). *
d * t ϕ(−tζ) dζ =
Rd
|ϕ(−ζ)| * dζ = ϕ * L1 (Rd ) .
* L1 (Rd ) . Consequently, |T, ϕt | T* L∞ (Rd ) ϕ
E XERCISE 8.16 (Diagonal distribution).– Consider the distribution T of D (R2 ) defined by T : D(R2 ) −→ R ϕ(t, t)dt. ϕ −→ R
1) Show that T is tempered. 2) Show that ∂x1 T + ∂x2 T = 0. 3) We wish to calculate the Fourier transform of T . 2 2 a) We set, for ϕ ∈ S(R ), I (ϕ) := e−t ϕ(t, t)dt. R
¨ ∂ + Show that T*, ϕ = lim I (ϕ). →0
b) Applying the Fubini theorem √ and an appropriate change in variable, deduce ¨ ∂ 2 1 ε ϕ(x, that T*, ϕ = lim √ z − x)e−z dxdz and conclude. →0 π π R2 2
Hint:,By denoting fα : x → e−αx (α > 0) the Gaussian of a variable, we π + have fα = α f π2 . α
S OLUTION 8.16.– 1) It should be noted that if ϕ ∈ D(R2 ), T, ϕ is well defined, since the integrand is regular and has compact support in t. Then, we have for example, Å ã 1 ϕ(t, t)dt ≤ dt sup 1 + t2 |ϕ(t, t)| 2 1 + t t∈R R R Å ã 1 ≤ dt sup 1 + x21 + x22 |ϕ(x1 , x2 )| . 2 (x1 ,x2 )∈R2 R 1+t
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The Theory of Distributions
So T is indeed a tempered distribution. 2) For ϕ ∈ D(R2 ), we set ψ(t) = ϕ(t, t). We have ψ (t) = ∂x1 ϕ(t, t) + ∂x2 ϕ(t, t). Since ψ has compact support, we thus have
ψ (t)dt
0=
R
= R
∂x1 ϕ(t, t)dt +
R
∂x2 ϕ(t, t)dt = T, ∂x1 ϕ + T, ∂x2 ϕ .
= − ∂x1 T + ∂x2 T, ϕ Finally, ∂x1 T + ∂x2 T = 0. 3) a) Let ϕ ∈ S(R2 ). We have t −→ ϕ(t, t) ∈ L1 (R), because −1 ∈ L1 (R). Therefore, the expression of T defined on the |ϕ(t, t)| ≤ C 1 + t2 test functions extends naturally (and in a way that is unique) to rapidly decreasing 2 2 functions. Then, since e−εt ϕ(t, t) ≤ |ϕ(t, t)|, and e−εt converges simply to 1 on R, the expected convergence results from Lebesgue’s theorem of dominated convergence (since the Fourier transform sends S(R2 ) onto itself). 2
b) Using the notation of the statement fα : x −→ e−αx , we denote Lε (ϕ) := Iε (ϕ) * =
ï R
fε (t)
ϕ(x, y)e
−i2πt(x+y)
ò dx dy dt.
R2
The function (x, y, t) −→ fε (t) |ϕ(x, y)| ∈ L1 (R3 ); by application of Fubini’s theorem, we get Lε (ϕ) =
ï ϕ(x, y)
R2
= R2
R
ò fε (t)e−i2πt(x+y) dt dx dy
ϕ(x, y)f“ε (x + y)dx dy.
and according to the hint in the text, we have f“ε = … Lε (ϕ) =
π ε
ϕ(x, y)e− R2
π 2 (x+y)2 ε
dx dy.
,π
ε f πε
2
, then
Fourier Transform
However, for a fixed x, by change in variable z = …
π ε
−
ϕ(x, y)e
π 2 (x+y)2 ε
R
…
π(x+y) √ , ε
261
we have
√ Å √ ã ε ε −z 2 dy = ϕ x, z−x e dz π π R Å √ ã 2 ε 1 ϕ x, z − x e−z dz. =√ π π R π ε
Finally, 1 Lε (ϕ) = √ π
Å √ ã 2 ε ϕ x, z − x e−z dz dx. π R2
¨ ∂ Moreover, we have seen in the previous question that T*, ϕ = lim Lε (ϕ). ε−→0 ä Ä √ Since, on the other hand, ϕ x, πε z − x −→ ϕ(x, −x) and, on the other hand, ε−→0
Å √ ã ϕ x, ε z − x e−z2 ≤ C 1 e−z2 ∈ L1 (R2 ). π 1 + x2 then using the fact that f“1 (0) = theorem yields
√
πfπ (0) =
√
π, Lebesgue’s dominated convergence
Lε (ϕ) −→
ε−→0
R
ϕ(x, −x)dx.
Consequently, T* is the tempered distribution with support on the second bisector defined as follows: T* : S(R2 ) −→ R ϕ(x, −x)dx. ϕ −→ R
“ “ = T , since T is an even In particular, the inversion formula is verified: T distribution. E XERCISE 8.17 (Convolution equation).– 1) Let f, g ∈ S(Rd ) denote two functions such that f ∗ g = 0. Can we say that f = 0 or g = 0?
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The Theory of Distributions
2) Same question if f ∈ D(R) and g ∈ D(R)? S OLUTION 8.17.– 1) By applying the Fourier transform to the equality f ∗ g = 0, this equality becomes f*g* = 0; this thus just means that the supports of f* and g* are disjoint, and it d cannot be deduced that f = 0 or g = 0: the ring S(R ), +, ∗ is not integral. If we are looking for an explicit counterexample, we simply have to take ϕ ∈ D(Rd ) nonnull with support in B(0, 1) and ψ ∈ D(Rd ) non-null with support in {x : |x| > 1}, for instance, and to consider f := F −1 (ϕ) and g := F −1 (ψ), we have in particular f, g ∈ S(Rd ) (since this space is stable by inverse Fourier transform and contains D(Rd ). We then have f‘ ∗ g = f*g* = ϕψ, therefore f ∗ g = 0 by injectivity of the Fourier transform, without having f = 0 or g = 0 (since their Fourier transforms are non-zero). 2) In this case, we know that f* and g* both admit two extensions holomorphic to C, and their zeros are therefore isolated (unless they are zero). If these two functions are non-zero, this implies in particular that the set of their zeros is countable, which contradicts the fact that the product of these two functions is identically null on R.
9 Applications to ODEs and PDEs
In a number of situations, and especially for the study of evolutionary partial differential equations, it proves to be helpful to perform a partial Fourier transform, which do not involve all the variables, but only some of them. This process will be briefly described, using the framework of evolutionary problems, which is its main application. ODEs are Ordinary Differential Equations. PDEs are Partial Differential Equations. 9.1. Partial Fourier transform D EFINITION 9.1 (Partial Fourier transform on S(Rt × Rdx )).– Let ϕ : (t, x) −→ ϕ(t, x) ∈ S(Rt × Rdx ). The partial Fourier transform at x of the e−i2πξx ϕ(t, x)dx. function ϕ is defined by the formula Fx ϕ(t, ξ) = Rd
This partial Fourier transform is endowed of the same continuity and invertibility properties as the usual Fourier transform. Consider now the case of distributions. D EFINITION 9.2 (Partial Fourier transform on S (Rt × Rdx )).– Let T ∈ S (Rt × Rdx ) be a distribution. The partial Fourier transform at x of the distribution T is defined by the formula Fx T, ϕ = T, Fx ϕ , ∀ϕ ∈ S(Rt × Rdx ). The task of verifying that the right-hand side of the above equality defines a tempered distribution on Rt × Rdx will be left to the reader.
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The Theory of Distributions
9.2. Tempered solutions of differential equations Let ω > 0 and let g be a distribution with compact support. The following differential equation, where we are looking for y ∈ S (R): y − ω 2 y = g
[9.1]
can be written in terms of convolution as follows: (δ − ω 2 δ) ∗ y = g.
[9.2]
To solve it, it is therefore sufficient to have a distribution E such that E − ω 2 E = δ.
[9.3]
As a matter of fact, by convolving the two members of [9.2] by E on the left, and assuming that associativity is permissible, we obtain E ∗g = E ∗ (δ −ω 2 δ)∗y = E ∗(δ −ω 2 δ) ∗y = E −ω 2 E ∗y = δ ∗y = y. We verify that yg = E ∗ g is indeed a solution of equation [9.1]. D EFINITION 9.3.– The distribution E ∈ S (R) as a solution of equation [9.3] is called elementary solution of the differential equation y − ω 2 y = g. The equation E − ω 2 E = δ of unknown E ∈ S (R) is equivalent to “ − ω2 E “ = 1. −4π 2 x2 E “ = − ω 2 + 4π 2 x2 −1 . Therefore, E −ω|t|
As a result, by inverse transform, we have E(t) = − e 2ω , and it is the unique solution in S (R) of the equation E − ω 2 E = δ. The solution of equation [9.1] is thus, if g is a function, yg (t) = −
R
e−ω|s−t| g(s)ds. 2ω
Applications to ODEs and PDEs
265
9.3. Fundamental solutions of certain PDEs aα (x)∂ α be a differential operator on Rd , with coefficient Let P (x, ∂) = |α|≤m
aα ∈ C ∞ (Rd ). If the coefficients aα are constants with respect to x, it is said that the aα ∂ α . A differential operator has constant coefficients and we denote P (∂) = |α|≤m
distribution E ∈ D (Ω) is said to be an elementary solution of P (∂) if it verifies P (∂)E = δ, in D (Ω). It should be noted that an elementary solution, when it exists, is not unique. To understand it better, we merely have to add a solution of the homogeneous equation P (∂)T = 0, in D (Ω). The distribution E+T is still an elementary solution. It is necessary that conditions be imposed to characterize one of the solutions and to prove the uniqueness of the solution. The existence of the solution is ensured by the following famous result. T HEOREM 9.1 (Malgrange-Ehrenpreis).– Any differential operator with constant coefficients P (∂) on Rd admits an elementary solution E in D (Rd ). The notion of elementary solution is employed to find the solutions to some partial differential equations in the following way. T HEOREM 9.2.– Let P (∂) be a differential operator with constant coefficients on Rd , and let E ∈ D (Rd ) be an elementary solution to P (∂). Then, for any f ∈ E (Rd ), the equation P (∂)u = f has at least one solution u ∈ D (Rd ) of the form u = E ∗ f . Proof. Since f has compact support, the expression for u makes sense. We then have P (∂)u = P (∂) E ∗ f = P (∂)E ∗ f = δ ∗ f = f.
9.3.1. Heat equation The heat equation is the partial differential equation ∂u ∂2u , x ∈ R, t ∈ R+ . = ∂t ∂x2
266
The Theory of Distributions
The associated Cauchy problem consists in finding the solution u(x, t) to this equation which verifies the initial condition u(x, 0) = f (x). The solution, in the sense of distributions, is u ∈ Dx,t (R2 ). We set
D=
∂ ∂2 . − ∂t ∂x2
Let us verify that the regular distribution associated with locally integrable the 2 function E : R2 −→ R defined by E(x, t) = 2√1πt exp − x4t is an elementary solution. We must then verify that DE = δ(0,0) . Actually, ≠Å
∂ ∂2 − ∂t ∂x2
ã
∑ ≠ ∑ ∑ ≠ ∂ϕ ∂2ϕ E, ϕ = − E, − E, 2 ∂t ∂x +∞ Å Å 2ã ã ∂ϕ x 1 √ exp − = − lim dt dx ε −→0 −∞ 4t ∂t t>ε 2 πt +∞ Å Å 2ã 2 ã ∂ ϕ x 1 √ exp − dt dx − lim ε −→0 −∞ 4t ∂x2 t>ε 2 πt = ϕ(0, 0) = δ(0,0) , ϕ .
In the general case, let x = (x1 , · · · , xd ) ∈ Rd and t ∈ R be the time variable. The ∂ Laplace operator is D = − Δ. It is also called the heat operator. The distribution ∂t E(x, t) =
|x|2 1 d √ H(t) exp − 4t 2 πt
is a fundamental solution to D. 9.3.2. Wave equation The wave equation is the partial differential equation ∂2u ∂2u = , x ∈ R, t ∈ R+ . ∂t2 ∂x2 The associated Cauchy problem consists of finding the solution u(x, t) to this equation that verifies the initial conditions: u(x, 0) = f0 (x),
∂u (x, 0) = f1 (x). ∂t
Applications to ODEs and PDEs
267
where f0 ∈ C 2 (R) and f1 ∈ C 1 (R). The elementary solution to the Cauchy problem is the regular distribution gt associated with the locally integrable function gt (x) =
1 H(x + t) − H(x − t) , 2
∂gt and we have lim gt = 0 and lim = δ. The general solution to the wave equation t −→0 t −→0 ∂t is written as ut = f 0 ∗
∂gt + f1 ∗ gt . ∂t
In fact, the solution to the wave equation is analogous to u(x, t) = f (x + t) + g(x − t), where f and g are continuous functions. Furthermore, for any ϕ ∈ D(R), ≠
∑ ∂2u ∂2 f (x + t) + g(x − t) ϕ(x)dx ,ϕ = 2 ∂t2 ∂t R = f (u)ϕ”(u − t)du + g(u)ϕ”(u + t)du
R
=
R
f (x + t)ϕ”(x)dx + R
≠ ∑ ∂2ϕ = u, 2 ∂x ≠ 2 ∑ ∂ u = , ϕ . ∂x2
R
g(x − t)ϕ”(x)dx
Appendix Evaluations
This appendix provides exercises for student self-assessment. E XERCISE A.1.– 1) Let fn n≥1 be the sequence of functions from R into R defined by ∀n ≥ 0, ∀x ∈ R, fn (x) = n cos(nx). a) Justify that for all n ≥ 0, fn ∈ L1loc (R). We denote by Tfn the distribution associated with fn . b) Show that Tfn n≥1 converges in D (R) and determine its limit. 2) Consider the linear form T defined on D(R) by T : ϕ −→
Å ãã +∞ Å Å ã 1 1 ϕ −ϕ − . k k+1
k=1
a) Show that T is a distribution. b) Determine the order of T . 3) Let f be the function defined on R by: f (x) = x sin(x). a) Justify that the distribution associated with f is tempered. b) Determine the Fourier transform in S (R) of the distribution associated with the function f . 4) a) Solve in D (R) the differential equation T + T = 0.
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The Theory of Distributions
b) Solve in D (R) the differential equation T + T = δ0 . c) Deduce the set of tempered solutions of the differential equation T +T = δ0 . E XERCISE A.2.– For all n ≥ 1, we consider un the function defined on R by ∀x ∈ R, un (x) =
x2 if x ≤ n n2 if x > n.
1) Show that un defines a distribution Tun . Let Tn = Tun be the distribution associated with un .
(3)
2) Calculate Tn , Tn and Tn . (3)
3) What is the order of Tn ? 4) Determine
lim
n−→+∞
Tn in D (R).
E XERCISE A.3.– 1) Let ϕ ∈ D(R). a) Show that there exists a function ψ ∈ C ∞ (R) such that ∀x ∈ R, ϕ(x) = ϕ(0) + xϕ (0) + x2 ψ(x). b) Therefrom, deduce the existence of the limit lim +
ε−→0
ϕ(x) |x|>ε
x2
−2
ϕ(0) dx. ε
We denote this limit as ≠ Å ã ∑
ϕ(x) 1 ϕ(0) Pf dx. , ϕ := lim − 2 x2 x2 ε ε−→0+ |x|>ε 2) Justify that the linear form on D(R), Pf x12 : ϕ −→ Pf x12 , ϕ is a distribution of order 2 at most. 1 3) Let Vp x1 be the principal value distribution of defined by x ≠ Å ã ∑ 1 ϕ(x) ∀ϕ ∈ D(R), Vp , ϕ := lim + dx. x x ε−→0 |x|>ε
Appendix
a) Show that in the sense of the distributions, xVp
271
Å ã 1 = 1. x
Å ã 1 ∈ S (R). b) Justify briefly that Vp x c) Show that Vp x1 is an odd distribution. d) Deduce that the Fourier transform F Vp x1 is an odd distribution. = −2iπδ0 . e) Show that F Vp x1 1 f) Then deduce that F Vp x = −2iπH + C, where H is the distribution associated with the Heaviside function and C is a constant (which by abuse of notation is confused with the distribution associated with it). 4) Show that, in D (R), Vp x1 = Pf x12 . 5) Can we say that Pf x12 ∈ S (R)? 6) What can it be said about the parity of Pf x12 ? E XERCISE A.4.– 1) Let f : R −→ R be the function defined by ∀x ∈ R, f (x) =
−2 if x ≤ 0 2 if x > 0.
a) Justify that the function f defines a distribution on R that will be denoted by Tf . b) Compute the derivative of Tf in D (R). 2) Let fn n≥1 be the sequence of functions from R into R defined by ∀n ≥ 0, ∀x ∈ R, fn (x) = cos2 (nx). a) Justify that, for all n ≥ 0, fn ∈ L1loc (R). We denote by Tfn the distribution associated with fn . b) Show that Tfn n≥1 converges in D (R) and determine its limit. 3) Consider the linear form T defined on D(R) by Å Å ã ã +∞ 1 1 T : ϕ −→ ϕ − ϕ(0) . k k k=1
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The Theory of Distributions
a) Show that T is a distribution of order 1 at most. b) Show that T is of order 1. 4) Show that a Dirac mass δa is a tempered distribution. 5) a) Show that the function x −→ ln |x| defines a tempered distribution and compute its derivative. b) Is Vp x1 (the principal value of x1 ) a tempered distribution? Justify. 6) a) Let f be the function defined on R by f (x) = cos(2x). Justify that the distribution associated with f is tempered. b) Determine the Fourier transform in S (R) of the distribution associated with the function f . E XERCISE A.5.– For all n ≥ 0, we set Tn =
n
δk , where δk is the Dirac distribution
k=0
at the point x = k.
1) a) Show that Tn is a distribution with compact support in R. b) What is its support? c) Explain why Tn is tempered. +∞ δk is a tempered distribution. 2)a) Verify that T = k=0
b) Show that the sequence of distributions Tn n converges in S (R) to T . 3) Show that the sequence of convoluted Tn ∗ Tn n converges in S (R) to T ∗ T. (Hint: We can use the properties of the Fourier transform and its inverse and obtain the result without calculating Tn ∗ Tn ). 4) We assume that T ∗ T =
+∞
(k + 1)δk .
k=0
a) Compute F(Tn )(λ) the Fourier transform of Tn in λ ∈ R. b) Compute F(T ∗ T )(λ), λ ∈ R. c) Deduce the multiplication of series formula +∞ k=0
e
−2iπkλ
+∞ k=0
e
−2iπkλ
=
+∞
(k + 1)e−2iπkλ .
k=0
Appendix
273
E XERCISE A.6.– 1) Let ψ ∈ C ∞ (R) and let Y = 1R+ be the indicator of R+ . We set E = TY ψ . a) Verify that E ∈ D (R). b) Show that for all p ≥ 1, (p−1)
E (p) = TY ψ(p) + ψ (p−1) (0)δ0 + ψ (p−2) (0)δ0 + ... + ψ(0)δ0 2) Let ψ be a solution to
k
.
aj ψ (j) (x) = 0 for x ∈ R, where aj ∈ R and ak = 0,
j=0
satisfying ψ(0) = ψ (0) = ... = ψ k−2 (0) = 0 a) Give the expression of
k
and
ψ k−1 (0) =
1 . ak
aj E (j) where E = TY ψ .
j=0
b) Compute ψ and then deduce a solution in D (R) of the equation E + E + E = δ0 .
References
Bony, J-M. (2001). Cours d’analyse. Théorie des distributions et analyse de Fourier. Ellipses, Paris. Boumaza, H. (2018). Théorie des distributions. Duplicate copy, Université Sorbonne Paris Nord. Dalmasso, R. and Witomski, P. (1996). Analyse de Fourier et applications. Masson, Paris. El Amrani, M. (2008). Analyse de Fourier dans les espaces fonctionnels. Ellipses, Paris. Gasquet, C. and Witomski, P. (2001). Analyse de Fourier et applications. Dunod, Paris. Guillopé, L. (2008). Analyse fonctionnelle. Duplicate copy, École polytechnique de l’Université de Nantes. Hirsch, F. and Lacombe, G. (1999). Éléments d’analyse fonctionnelle. Dunod, Paris. Khoan, V. (1972). Distributions analyse de Fourier opérateurs aux derivées partielles. Durand, Paris. Lesfari, A. (2012). Distributions, analyse de Fourier et transformation de Laplace. Ellipses, Paris. Pathak, R.S. (2001). A Course in Distribution Theory and Applications. Alpha Science, London. Schwartz, L. (1966). Théorie des distributions. Hermann, Paris. Schwartz, L. (1983). Méthodes mathématiques pour les sciences physiques. Hermann, Paris.
276
The Theory of Distributions
Strichartz, R. (2003). A Guide to Distribution Theory and Fourier Transforms. World Scientific Publishing Co. Inc., Singapore. Zuily, C. (1988). Éléments de distributions et d’équations aux dérivées partielles. Hermann, Paris. Zuily, C. (2002). Problems in Distributions and Partial Differential Equations. Dunod, Paris.
Index
A, B algebra, 166, 167 associativity, 164, 166, 264 Banach, 48, 54, 190 basis, 3, 5, 184 bilinear, 166 Borel, 14, 32, 61, 126, 130
C closed, 3, 10, 15, 23, 124, 127, 128, 139, 141, 144, 145, 161, 162, 176, 251 commutativity, 163, 166 compacts, 19, 39, 132, 162 complete, 6, 7, 10, 19, 170, 184, 226, 239 connected, 90 continuous, 3–5, 8–15, 23, 25–29, 37, 38, 40, 48, 54, 61, 77, 78, 85, 90, 110, 121, 127, 132, 133, 136, 143, 179, 190, 192, 193, 196, 203, 210–212, 219–222, 226, 230, 233, 234, 236, 237, 251, 267 converges uniformly, 32 convex envelope, 162
convolution, 7, 8, 10, 21, 22, 24, 27, 98, 123, 151–154, 157, 158, 160–169, 177, 187, 188, 205, 218, 228, 239, 264
D density, 5, 9, 25, 26, 30, 134, 135, 185, 190, 194, 208, 234 derivation, 10, 81, 148, 157, 195, 221, 249, 253 differentiable, 12, 18, 32, 75, 83, 99, 101, 104, 132, 141, 180, 181, 189, 199, 213, 214, 224, 235, 237, 256 Dirac, 11, 41, 50, 51, 70, 96, 97, 113, 137, 139, 197, 202, 235, 236, 252, 253, 272 comb, 51, 70, 96, 97, 202, 236, 252 direct sum, 167 distribution, 6, 7, 9–13, 18, 37–48, 50–52, 54, 56–66, 68, 75–77, 79–85, 87, 89–91, 93–103, 109, 113, 117, 118, 123, 125–128, 130–134, 136–148, 153–155, 157–160, 165, 166, 169, 171–173, 175–177, 189–203, 205, 230, 232, 234–237, 239–241, 245, 246, 253, 254, 259–261, 263–267, 269–272
278
The Theory of Distributions
duality, 6, 37, 55, 98, 133, 155, 163, 195, 205, 217, 230, 232
F formula, 5, 6, 8, 14, 21, 33–35, 61, 64, 76, 78, 79, 81, 85, 91, 94, 96–98, 102–104, 108, 116, 131, 138, 154, 157, 165, 171, 189, 204, 213, 214, 217, 225, 226, 232, 234, 236, 238, 239, 244, 254, 256, 261, 263, 272 Fourier transform, 7, 8, 10, 205–224, 226–228, 230, 232–237, 239–241, 243, 245–252, 254, 257, 259, 260, 262, 263, 269, 271, 272 function, 5–7, 9, 11, 12, 14, 16–19, 21–24, 26–29, 31–36, 38–42, 44, 45, 48, 50, 53, 54, 57, 58, 60, 61, 66, 69, 73, 75–77, 83–85, 91, 93, 94, 96, 98–102, 104, 106, 108, 112, 114, 117, 120–123, 125, 127–132, 134–136, 138, 139, 141–143, 146–149, 153–155, 157, 158, 163, 168–173, 175–182, 187–189, 191–193, 195–201, 203–211, 213–215, 219–221, 223, 224, 226, 228, 230, 231, 234–241, 246–248, 250–252, 258, 260, 263, 264, 266, 267, 269–272 functional, 10, 11, 13, 37, 82, 133, 135, 143 fundamental system, 5, 6, 10
H, I Hausdorf–Young, 22 Heaviside, 41, 83, 94, 106, 108, 112, 114, 126, 168, 169, 173, 239, 241, 245, 246, 271 homogeneous, 80, 103, 122, 175–178, 241, 265
identity, 9, 10, 24, 66, 82, 171, 220, 227, 234, 236, 238, 252, 253, 258 inductive limit, 9–11 isomorphic, 167
L, M law, 75, 160, 166, 167, 188, 254 Lebesgue, 11, 12, 25, 27, 29, 31, 52, 66, 72, 73, 85, 126, 130, 210, 211, 219, 225, 260, 261 Leibniz, 5, 14, 33, 64, 76, 78, 85, 113, 128, 132, 136, 148, 186, 187 linear application, 80 form, 5, 8–13, 37–42, 44, 46–48, 54, 56, 57, 59–61, 63, 76, 118, 125, 133, 136, 137, 140, 144, 190, 269–271 locally convex, 5–7, 9–11, 19, 48 metrizable, 5–7, 11, 19, 184 multi-indexes, 13
N, O neighborhoods, 5, 6, 10, 44, 135 neutral, 160, 166 norm, 13, 16, 29, 30, 180, 184, 210 open, 3, 4, 6, 13–15, 25, 30, 37, 54, 57, 75, 90, 123–130, 135, 136, 139, 141, 143, 145, 153, 155, 236 order, 6, 10, 11, 13, 21, 22, 34, 35, 37–47, 55–60, 62–65, 79, 82–84, 99, 109, 118, 122, 131, 132, 136–138, 141, 143, 144, 146–148, 151, 158, 165, 169, 170, 173, 189, 194–196, 202, 204, 206, 213, 218, 233, 269, 270, 272
P, R partial derivatives, 22, 83, 99, 131, 132 periodic, 96, 101, 121, 237, 252
Index
principal value, 42, 58, 192, 197, 199, 272 rapidly decreasing, 179–181, 197, 198, 224, 260 regular, 6, 7, 12, 24, 40, 41, 44, 49, 50, 77, 118, 151, 157, 235, 259, 266, 267 regularizing, 17, 163
S semi-norm, 2–9, 19, 48, 133, 183, 184 separating, 2–4, 6–9, 32 sequence, 6–11, 17, 19, 20, 24–26, 28, 30–32, 37–39, 44, 48–55, 61, 66, 70, 78, 84, 85, 87, 98, 99, 103, 104, 109, 123, 124, 131–133, 138, 155, 156, 162, 163, 184, 190, 192, 194, 196, 201, 203, 210, 220, 222, 230, 233, 253–255, 269, 271, 272 series, 9, 10, 32, 33, 37, 51, 83, 138, 140, 226, 238, 252, 253, 272 singular, 11, 12, 41, 50 summable, 21, 182, 201, 246
279
support, 5–7, 9, 14, 15, 17, 18, 23–25, 27–29, 31–33, 36, 49, 54, 60–62, 69, 88, 99, 104, 123, 126–129, 131–134, 136, 137, 139–146, 148, 149, 151, 153–155, 157, 158, 161–166, 171, 172, 175, 176, 185, 186, 194, 196, 198–200, 210, 230, 234–236, 239, 241, 251–254, 259–262, 264, 265, 272
T, U topology, 2–10, 19, 25, 37, 48, 133, 183, 184, 193 translated, 18 translation-invariant, 6, 7 uniformly continuous, 6, 25 unit, 27, 103, 123, 124, 126
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