Distributions, Partial Differential Equations, and Harmonic Analysis

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Universitext

Dorina Mitrea

Distributions, Partial Differential Equations, and Harmonic Analysis

Universitext

Universitext Series Editors: Sheldon Axler San Francisco State University Vincenzo Capasso Università degli Studi di Milano Carles Casacuberta Universitat de Barcelona Angus J. MacIntyre Queen Mary, University of London Kenneth Ribet University of California, Berkeley Claude Sabbah ´ CNRS, Ecole Polytechnique Endre Süli University of Oxford Wojbor A. Woyczynski Case Western Reserve University

Universitext is a series of textbooks that presents material from a wide variety of mathematical disciplines at master’s level and beyond. The books, often well class-tested by their author, may have an informal, personal even experimental approach to their subject matter. Some of the most successful and established books in the series have evolved through several editions, always following the evolution of teaching curricula, to very polished texts. Thus as research topics trickle down into graduate-level teaching, first textbooks written for new, cutting-edge courses may make their way into Universitext.

For further volumes: http://www.springer.com/series/223

Dorina Mitrea

Distributions, Partial Differential Equations, and Harmonic Analysis

123

Dorina Mitrea Department of Mathematics University of Missouri Columbia, MO, USA

ISSN 0172-5939 ISSN 2191-6675 (electronic) ISBN 978-1-4614-8207-9 ISBN 978-1-4614-8208-6 (eBook) DOI 10.1007/978-1-4614-8208-6 Springer New York Heidelberg Dordrecht London Library of Congress Control Number: 2013944653 Mathematics Subject Classification (2010): 35A08, 35A09, 35A20, 35B05, 35B53, 35B65, 35C15, 35D30, 35E05, 35G05, 35G35, 35H10, 35J05, 35J30, 35J45, 35J47, 35J48 © Springer Science+Business Media New York 2013 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)

Dedicat˘a cu drag lui Diana ¸si Adrian, Love, Mom

Preface

This book has been written from the personal perspective of a mathematician working at the interface between partial differential equations and harmonic analysis. Its aim is to offer, in a concise, rigorous, and largely self-contained form, a rapid introduction to the theory of distributions and its applications to partial differential equations and harmonic analysis. This is done in a format suitable for a graduate course spanning either a one-semester period, when the focus is primarily on the foundational aspects, or over a two-semester period, which allows for the proper amount of time to cover all intended applications as well. Throughout, a special effort has been made to develop the theory of distributions not as an abstract edifice but rather to give the reader a chance to see the rationale behind various seemingly technical definitions, as well as the opportunity to apply the newly developed tools (in the natural build-up of the theory) to concrete problems in partial differential equations and harmonic analysis, at the earliest opportunity. In addition to being suitable as a textbook for a graduate course, this book has been designed so that it may also be used for independent study since the presentation is reader-friendly, mostly self-sufficient (e.g., all auxiliary results originating outside the scope of this book have been carefully collected and presented in the appendix), and a large number of the suggested exercises have complete solutions. Columbia, MO, USA

Dorina Irena Rita Mitrea

vii

Contents Preface

vii

Introduction

xiii

Common Notational Conventions

xix

1 Weak Derivatives 1.1 The Cauchy Problem for a Vibrating 1.2 Weak Derivatives . . . . . . . . . . . 1.3 The Spaces E(Ω) and D(Ω) . . . . . 1.4 Additional Exercises for Chap. 1 . .

Infinite String . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . .

. . . .

Space D (Ω) of Distributions The Definition of Distributions . . . . . . . . . . . . . . The Topological Vector Space D (Ω) . . . . . . . . . . . Multiplication of a Distribution with a C ∞ Function . . Distributional Derivatives . . . . . . . . . . . . . . . . . The Support of a Distribution . . . . . . . . . . . . . . . Compactly Supported Distributions and the Space E (Ω) Tensor Product of Distributions . . . . . . . . . . . . . . The Convolution of Distributions in Rn . . . . . . . . . Distributions with Higher-Order Gradients Continuous or Bounded . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 Additional Exercises for Chap. 2 . . . . . . . . . . . . .

2 The 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9

. . . .

. . . .

. . . .

. . . .

. . . .

1 1 3 8 13

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

17 17 25 27 29 34 37 48 59

. . . . . . . . . .

72 80

3 The Schwartz Space and the Fourier Transform 89 3.1 The Schwartz Space of Rapidly Decreasing Functions . . . . . . . 89 3.2 The Action of the Fourier Transform on the Schwartz Class . . . . . . . . . . . . . . . . . . . . . . . . 99 3.3 Additional Exercises for Chap. 3 . . . . . . . . . . . . . . . . . . 106 4 The 4.1 4.2 4.3

Space of Tempered Distributions 109 Definition and Properties of Tempered Distributions . . . . . . . 109 The Fourier Transform Acting on Tempered Distributions . . . . 119 Homogeneous Distributions . . . . . . . . . . . . . . . . . . . . . 129 ix

CONTENTS

x 4.4 4.5 4.6 4.7

Principal Value Tempered Distributions . . . . . . . . . The Fourier Transform of Principal Value Distributions Tempered Distributions Associated with |x|−n . . . . . . A General Jump-Formula in the Class of Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . 4.8 The Harmonic Poisson Kernel . . . . . . . . . . . . . . . 4.9 Singular Integral Operators . . . . . . . . . . . . . . . . 4.10 Derivatives of Volume Potentials . . . . . . . . . . . . . 4.11 Additional Exercises for Chap. 4 . . . . . . . . . . . . .

. . . . . 135 . . . . . 141 . . . . . 146 . . . . .

. . . . .

. . . . .

. . . . .

. . . . .

151 162 167 175 184

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

189 189 191 195 200

6 Hypoelliptic Operators 6.1 Definition and Properties . . . . . . . . . . . . . . . . . 6.2 Hypoelliptic Operators with Constant Coefficients . . . 6.3 Integral Representation Formulas and Interior Estimates 6.4 Additional Exercises for Chap. 6 . . . . . . . . . . . . .

. . . .

. . . .

. . . .

. . . .

. . . .

201 201 203 210 216

5 The 5.1 5.2 5.3 5.4

Concept of a Fundamental Solution Constant Coefficient Linear Differential Operators A First Look at Fundamental Solutions . . . . . . The Malgrange–Ehrenpreis Theorem . . . . . . . . Additional Exercises for Chap. 5 . . . . . . . . . .

. . . .

. . . .

7 The Laplacian and Related Operators 7.1 Fundamental Solutions for the Laplace Operator . . . . . . 7.2 The Poisson Equation and Layer Potential Representation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Fundamental Solutions for the Bi-Laplacian . . . . . . . . . 7.4 The Poisson Equation for the Bi-Laplacian . . . . . . . . . 7.5 Fundamental Solutions for the Poly-Harmonic Operator . . 7.6 Fundamental Solutions for the Cauchy–Riemann Operator . 7.7 Fundamental Solutions for the Dirac Operator . . . . . . . 7.8 Fundamental Solutions for General Second-Order Operators 7.9 Layer Potential Representation Formulas Revisited . . . . . 7.10 Additional Exercises for Chap. 7 . . . . . . . . . . . . . . . 8 The 8.1 8.2 8.3 8.4 9 The 9.1 9.2 9.3

Heat Operator and Related Versions Fundamental Solutions for the Heat Operator . . . . . . . . The Generalized Cauchy Problem for the Heat Operator . . Fundamental Solutions for General Second Order Parabolic Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fundamental Solution for the Schrödinger Operator . . . .

217 . . . 217 . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

223 233 237 240 248 253 260 271 274

277 . . . 277 . . . 280 . . . 282 . . . 287

Wave Operator 289 Fundamental Solution for the Wave Operator . . . . . . . . . . . 289 The Generalized Cauchy Problem for the Wave Operator . . . . 307 Additional Exercises for Chap. 9 . . . . . . . . . . . . . . . . . . 308

CONTENTS

xi

10 The 10.1 10.2 10.3 10.4

Lam´ e and Stokes Operators General Remarks About Vector and Matrix Distributions . Fundamental Solutions and Regularity for General Systems Fundamental Solutions for the Lamé Operator . . . . . . . Mean Value Formulas and Interior Estimates for the Lamé Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 The Poisson Equation for the Lamé Operator . . . . . . . . 10.6 Fundamental Solutions for the Stokes Operator . . . . . . . 10.7 Additional Exercises for Chap. 10 . . . . . . . . . . . . . . .

11 More on Fundamental Solutions for Systems 11.1 Computing a Fundamental Solution for the Lamé Operator 11.2 Computing a Fundamental Solution for the Stokes Operator 11.3 Fundamental Solutions for Higher-Order Systems . . . . . . 11.4 Interior Estimates and Real-Analyticity for Null-Solutions of Systems . . . . . . . . . . . . . . . . . 11.5 Reverse Hölder Estimates for Null-Solutions of Systems . . 11.6 Layer Potentials and Jump Relations for Systems . . . . . . 11.7 Additional Exercises for Chap. 11 . . . . . . . . . . . . . . 12 Solutions to Selected Exercises 12.1 Solutions to Exercises from Sect. 1.4 . 12.2 Solutions to Exercises from Sect. 2.10 12.3 Solutions to Exercises from Sect. 3.3 . 12.4 Solutions to Exercises from Sect. 4.11 . 12.5 Solutions to Exercises from Sect. 5.4 . 12.6 Solutions to Exercises from Sect. 6.4 . 12.7 Solutions to Exercises from Sect. 7.10 . 12.8 Solutions to Exercises from Sect. 9.3 . 12.9 Solutions to Exercises from Sect. 10.7 . 12.10Solutions to Exercises from Sect. 11.7 .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

. . . . . . . . . .

13 Appendix 13.1 Summary of Topological and Functional Analytic . . . . . . 13.2 Summary of Basic Results from Calculus, Measure Theory, and Topology . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Custom-Designing Smooth Cut-Off Functions . . . . . . . . 13.4 Partition of Unity . . . . . . . . . . . . . . . . . . . . . . . 13.5 The Gamma and Beta Functions . . . . . . . . . . . . . . . 13.6 Surfaces in Rn and Surface Integrals . . . . . . . . . . . . . 13.7 Integration by Parts and Green’s Formula . . . . . . . . . . 13.8 Polar Coordinates and Integrals on Spheres . . . . . . . . . 13.9 Tables of Fourier Transforms . . . . . . . . . . . . . . . . .

309 . . . 309 . . . 314 . . . 319 . . . .

. . . .

. . . .

326 332 334 338

341 . . . 341 . . . 343 . . . 344 . . . .

. . . . . . . . . .

. . . .

. . . . . . . . . .

. . . .

356 361 365 373

. . . . . . . . . .

375 375 381 397 400 407 408 409 413 413 414

415 . . . 415 . . . . . . . .

. . . . . . . .

. . . . . . . .

424 428 429 434 435 437 437 445

xii

CONTENTS

Bibliography

449

Subject Index

455

Symbol Index

459

Introduction

It has long been recognized that there is a large overlap and intricate interplay among distribution theory (DT), partial differential equations (PDE), and harmonic analysis (HA). The purpose of this book is to guide a reader with a background in basic real analysis through the journey taking her/him to the stage when such connections become self-evident. Another goal of the present book is to convince the reader that traditional distinctions made among these branches of mathematics are largely artificial and are often simply a matter of choice in focus. Indeed, given the manner in which they complement, motivate, and draw inspiration from one another, it is not necessarily a stretch to attempt to pursue their development virtually simultaneously. Concerning the triumvirate DT, PDE, HA, while there exist a number of good reference texts available on the market, they are by and large conceived in such a way that they either emphasize more one of these topics, typically at the detriment of the others, or are simply not particularly well-suited for a nonspecialist. By way of contrast, not only is the present text written in a way that brings together and blends the aforementioned topics in a unified, coherent body of results, but the resulting exposition is also sufficiently detailed and reader-friendly so that it may be read independently, outside the formal classroom setting. Indeed, the book is essentially self-contained, presents a balanced treatment of the topics involved, and contains a large number of exercises (upwards of 200, more than half of which are accompanied by solutions), which have been carefully chosen to amplify the effect, and substantiate the power and scope, of the theory discussed here. While the topics treated are classical, the material is not entirely standard since a number of results are new even for a seasoned practitioner, and the overall architectural design of the monograph (including the way in which certain topics are covered) is original. Regarding its inception, the present monograph is an expanded version of the notes prepared for a course on distribution theory that I taught in the Spring of 2007 and the Spring of 2011 at the University of Missouri. My intention was to present the theory of distributions not as an abstract edifice but rather to give the student a chance to instantaneously see the justification and practical benefits of the multitude of seemingly technical definitions and results, as well xiii

INTRODUCTION

xiv

as give her/him the opportunity to immediately see how the newly introduced concepts (in the natural build-up of the theory) apply to concrete problems in PDE and harmonic analysis. Special care has been paid to the pedagogical aspect of the presentation of the material in the book. For example, a notable feature of the present monograph is the fact that fundamental solutions for some of the most basic differential operators in mathematical physics and engineering, including Laplace, heat, wave, poly-harmonic, Dirac, Lamé, Stokes, and Schrödinger, are systematically deduced starting from first principles. This stands in contrast with the more common practice in the literature in which one starts with a certain distribution (the origins of which are fairly obscure) and simply checks that the distribution in question is a fundamental solution for a given differential operator. Another feature is the emphasis placed on the interrelations between topics. For example, a clear picture is presented as to how DT vastly facilitates the computation of fundamental solutions, and the development of singular integral operators, tools which, in turn, are used to solve PDE as well as represent and estimate solutions of PDE. The presentation is also conceived in such a way as to avoid having to confront heavy-duty topology/functional analysis up front, in the main narrative. For example, the jargon associated with the multitude of topologies on various spaces of test functions and distributions is minimized by deferring to an appendix the technical details while retaining in the main body of the discussion only those consequences that are most directly relevant to the fluency of the exposition. While the core material I had in mind deals primarily with the theory of distributions, this book is ultimately devised in such a way as to make the present material a solid launching pad for a number of subsequent courses dealing with allied topics, including: • Harmonic analysis • Partial differential equations • Boundary integral methods • Sobolev spaces • Pseudodifferential operators For example, the theory of singular integral operators of convolution type in L2 (Rn ) is essentially developed here, in full detail, up to the point where more specialized tools from harmonic analysis (such as the Hardy–Littlewood maximal operator and the Calderón–Zygmund lemma) are typically involved in order to further extend this theory (via a weak-(1, 1) estimate, interpolation with L2 , and then duality) to Lp spaces with p ∈ (1, ∞), as in [6, 9, 16, 23, 47, 62–64], among others.

INTRODUCTION

xv

Regarding connections with PDE, the Poisson problem in the whole space, Lu = f

in

D (Rn ),

(0.0.1)

is systematically treated here for a variety of differential operators L, including the Laplacian, the bi-harmonic operator, and the Lamé system. While Sobolev spaces are not explicitly considered (the interested reader may consult in this regard [1, 2, 13, 40, 43, 44, 76], to cite just a small fraction of a large body of literature on this topic), they lurk just beneath the surface as their presence is implicit in estimates of the form

∂ α uLp (Rn ) ≤ C(L, p)f Lp(Rn ) ,

1 < p < ∞,

(0.0.2)

|α|=m

where m denotes the order of the elliptic differential operator L. Such estimates are deduced from an integral representation formula for u, involving a fundamental solution of L, and estimates for singular integral operators of convolution type in Rn . In particular, this justifies two features of the present monograph: (1) the emphasis placed on finding explicit formulas for fundamental solutions for a large number of operators of basic importance in mathematics, physics, and engineering; (2) the focus on the theory of singular integral operators of convolution type, developed along side the distribution theory. In addition, the analysis of the Cauchy problem formulated and studied for the heat and wave operators once more underscores the significance of the fundamental solutions for the named operators. As a whole, this material is designed to initiate the reader into the field of PDE. At the same time, it complements, and works well in tandem with, the treatment of this subject in [3, 12, 21, 30–32, 46, 68, 69, 75]. Whenever circumstances permit it, other types of problems are brought into play, such as the Dirichlet and Neumann problem in the upper-half space for the Laplacian, as well as more general second-order systems. In turn, the latter genre of boundary value problems motivates introducing and developing boundary integral methods, and serves as an opportunity to highlight the basic role that layer potential operators play in this context. References dealing with the latter topic include [34, 36, 46, 48, 74]. Analysis of the structure of the boundary layer potential operators naturally intervening in this context also points to the possibility of considering larger classes of operators where the latter may be composed, inverted, etc., in a stable fashion. This serves as an excellent motivation for the introduction of such algebras of operators as pseudodifferential and Fourier integral operators, a direction that the interested reader may then pursue in, for example, [24, 67, 69, 75], to name a few sources.

INTRODUCTION

xvi A brief description of this book’s contents is as follows.

Chapters 1–2 are devoted to the development of the most basic aspects of the theory of distributions. Starting from the discussion of the Cauchy problem for a vibrating infinite string as a motivational example, the notion of a weak derivative is introduced as a mean of extending the notion of solution to a more general setting, where the functions involved may lack standard pointwise differentiability properties. After touching upon classes of test functions, the space of distributions is then introduced and studied from the perspective of a topological vector space with various other additional features (such as the concept of support, and a partially defined convolution product). Chapter 3 contains material pertaining to the Schwartz space of functions rapidly decaying at infinity and the Fourier transform in such a setting. In Chap. 4 the action of the Fourier transform is further extended to the setting of tempered distributions, and several distinguished subclasses of tempered distributions are introduced and studied (including homogeneous and principal value distributions). The foundational material developed up to this point already has significant applications to harmonic analysis and PDE. For example, a general, higher-dimensional jump-formula is deduced in this chapter for a certain class of tempered distributions (that includes the classical harmonic Poisson kernel) which is latter used as the main tool in deriving information about the boundary behavior of layer potential operators associated with various partial differential operators and systems. Also, one witnesses here how singular integral operators of central importance to harmonic analysis (such as the Riesz transforms) naturally arise as an extension to L2 of the convolution product of tempered distributions of principal value type with Schwartz functions. The first explicit encounter with the notion of fundamental solution takes place in Chap. 5, where the classical Malgrange–Ehrenpreis theorem is presented. Subsequently, in Chap. 6, the concept of hypoelliptic operator is introduced and studied. In particular, here a classical result, due to L. Schwartz, is proved to the effect that a necessary and sufficient condition for a linear, constant coefficient differential operator to be hypoelliptic in the entire ambient space is that the named operator possesses a fundamental solution with singular support consisting of the origin alone. In Chap. 6 we also prove an integral representation formula and interior estimates for a subclass of hypoelliptic operators, which are subsequently used to show that null-solutions of these operators are real-analytic. One of the main goals in Chap. 7 is identifying (starting from first principles) all fundamental solutions that are tempered distributions for scalar elliptic operators. While the natural starting point is the Laplacian, this study encompasses a variety of related operators, such as the bi-Laplacian, the polyharmonic operator, the Cauchy–Riemann operator, the Dirac operator, as well as general second-order constant coefficient strongly elliptic operators. Having

INTRODUCTION

xvii

accomplished this task then makes it possible to prove the well-posedness of the Poisson problem (0.0.1) (equipped with a boundary condition at infinity), and derive qualitative/quantitative properties for the solution such as (0.0.2). Along the way, Cauchy-like integral operators are also introduced and their connections with Hardy spaces is brought to light in the setting of both complex and Clifford analysis. Chapter 8 has a twofold aim: determine all fundamental solutions that are tempered distributions for the heat operator and related versions (including the Schr¨ odinger operator), then use this as a toll in the solution of the generalized Cauchy problem for the heat operator. The same type of program is then carried out in Chap. 9, this time in connection with the wave operator. While the analysis up to this point has been largely confined to scalar operators, the final two chapters in the book are devoted to studying systems of differential operators. The material in Chap. 10 is centered around two such basic systems: the Lamé operator arising in the theory of elasticity, and the Stokes operator arising in hydrodynamics. Among other things, all their fundamental solutions that are tempered distributions are identified, and the well-posedness of the Poisson problem for the Lamé system is established. The former issue is then revisited in the first part of Chap. 11 from a different perspective, and subsequently generalized to the case of (homogeneous) constant coefficient systems of arbitrary order. In Chap. 11 we also show that integral representation formulas and interior estimates hold for null-solutions of homogeneous systems with non-vanishing full symbol. As a consequence, we prove that such nullsolutions are real-analytic and satisfy reverse Hölder estimates. The final topic addressed in Chap. 11 pertains to layer potentials associated with arbitrary constant coefficient second order systems in the upper-half space, and the relevance of these operators vis-a-vis to the solvability of boundary value problems for such systems in this setting. For completeness, a summary of topological and functional analysis results in reference to the description of the topology and equivalent characterizations of convergence in spaces of test functions and in spaces of distributions is included in the appendix (which also contains a variety of foundational results from calculus, measure theory, and special functions originating outside the scope of this book). One aspect worth noting in this regard is that the exposition in the main body of the book may be followed even without being fully familiar with all these details by alternatively taking, as the starting point, the characterization of convergence in the various topologies considered here (summarized in the main text under the heading Fact) as definitions. Such an approach makes the topics covered in the present monograph accessible to a larger audience while, at the same time, provides a full treatment of the topological and functional analysis background accompanying the theory of distributions for the reader interested in a more in-depth treatment.

xviii

INTRODUCTION

Finally, each book chapter ends with bibliographical references tailored to its respective contents under the heading Further Notes, as well as with a number of additional exercises, selectively solved in Chap. 12. Acknowledgments. The work on this project has been supported in part by the Simons Foundation grant # 200750, and by a University of Missouri Research Leave grant. The author wishes to express her gratitude to these institutions. In addition, the author thanks the referees for their useful comments which have lead to various improvements in the manuscript.

Common Notational Conventions

Throughout this book the set of natural numbers will be denoted by N, that is N := {1, 2, . . . }, while N0 := N ∪ {0}. For each k ∈ N set k! := 1 · 2 · · · (k − 1) · k, and make the convention that 0! := 1. The letter C will denote the set of complex numbers, and z denotes the complex conjugate of z ∈ C. Also the real and imaginary parts of a complex number z are denoted by Re z √ and Im z, respectively. The symbol i is reserved for the complex imaginary unit −1 ∈ C. The letter R will denote the set of real numbers and its n-fold Cartesian product of R with itself (where n ∈ N) is denoted by Rn . That is, (0.0.3) Rn := x = (x1 , . . . , xn ) : x1 , . . . , xn ∈ R considered with the usual vector space and inner product structure, that is, x + y := (x1 + y1 , . . . , xn + yn ),

cx := (cx1 , . . . , cxn ),

x · y :=

n

xj yj ,

j=1

∀ x = (x1 , . . . , xn ) ∈ Rn , ∀ y = (y1 , . . . , yn ) ∈ Rn , ∀ c ∈ R.

(0.0.4) The standard orthonormal basis of vectors in Rn is denoted by {ej }1≤j≤n , where we have set ej := (0, . . . , 0, 1, 0, . . . , 0) ∈ Rn with the only nonzero component on the j-th slot. We shall also consider the two canonical (open) half-spaces of Rn , denoted by Rn± := {x = (x1 , . . . , xn ) ∈ Rn : ± xn > 0}. Hence, Rn+ = Rn−1 × (0, ∞) and Rn− = Rn−1 × (−∞, 0). Given a multi-index α = (α1 , . . . , αn ) ∈ Nn0 , we set α! := α1 !α2 ! · · · αn ! and |α| :=

n

αj ,

(0.0.5)

j=1

∂ α := ∂1α1 · · · ∂nαn , xα :=

n

α

xj j

where

∂j :=

∂ for j = 1, . . . , n, ∂xj

for every x = (x1 , . . . , xn ) ∈ Cn .

(0.0.6)

(0.0.7)

j=1

Also if β = (β1 , . . . , βn ) ∈ Nn0 is another multi-index we shall write β ≤ α provided βj ≤ αj for each j ∈ {1, . . . , n}, in which case we set α − β := (α1 − β1 , . . . , αn − βn ). We shall also say that β < α if β ≤ α and β = α. Recall that the Kronecker symbol is defined by δjk := 1 if j = k and δjk := 0 if j = k. All functions in this monograph are assumed to be complex valued unless otherwise indicated. Derivatives of a function f defined on the real line are k going to be denoted using f , f , etc., or f (k) , or ddxfk . xix

xx

COMMON NOTATIONAL CONVENTIONS

Throughout the book, Ω denotes an arbitrary open subset of Rn . If A is an ˚ A, and ∂A denote its interior, its closure, and arbitrary subset of Rn , then A, its boundary, respectively. In addition, if B is another arbitrary subset of Rn , then their set theoretic difference is denoted by A \ B := {x ∈ A : x ∈ B}. In particular, the complement of A is Ac := Rn \ A. For any E ⊂ Rn we let χE stand for the characteristic function of the set E (i.e., χE (x) = 1 if x ∈ E and χE (x) = 0 if x ∈ Rn \ E). For k ∈ N0 ∪ {∞}, we will work with the following classes of functions that are vector spaces over C: C k (Ω) := ϕ : Ω → C : ∂ α ϕ continuous ∀ α ∈ Nn0 , |α| ≤ k , (0.0.8) (0.0.9) C k (Ω) := ϕΩ : ϕ ∈ C k (U ), U ⊆ Rn open set containing Ω , (0.0.10) C0k (Ω) := ϕ ∈ C k (Ω) : supp ϕ compact subset of Ω . As usual, for any Lebesgue-measurable (complex-valued) function f defined on a Lebesgue measurable set E ⊆ Rn and any p ∈ [1, ∞] we write

1/p p if 1 ≤ p < ∞, E |f | dx f Lp(E) := (0.0.11) ess-supE |f | if p = ∞, and denote by Lp (E) the Banach space of (equivalence classes of) Lebesguemeasurable functions f on E satisfying f Lp(E) < ∞. Also, we will work with locally integrable functions and with compactly supported integrable functions. For p ∈ [1, ∞] these are defined as Lploc (Ω) := f : Ω → C : f Lebesgue measurable such that f Lp(K) < ∞, ∀ K ⊂ Ω compact set , (0.0.12) and, respectively, as Lpcomp(Ω) := f ∈ Lp (Ω) : supp f compact subset of Ω ,

(0.0.13)

where supp f is defined in (2.5.10). Given a measure space (X, μ), a measurable set A ⊆ X with 0 < μ(A) < ∞, and a function f ∈ L1 (A, μ), we define the integral average of f over A by 1 − f dμ := f dμ. (0.0.14) μ(A) A A If E is a Lebesgue measurable subset of Rn , the Lebesgue measure of E is denoted by |E|. If x ∈ Rn and radius r > 0, set B(x, r) := {y ∈ Rn : |y −x| < r} for the ball of center x and radius R and its boundary is denoted by ∂B(0, r). The unit sphere in Rn centered at zero is S n−1 := {x ∈ Rn : |x| = 1} = ∂B(0, 1) and its surface measure is denoted by ωn−1 .

COMMON NOTATIONAL CONVENTIONS

xxi

For n, m ∈ N and R, an arbitrary commutative ring with multiplicative units, we denote by Mn×m (R) the collection of all n × m matrices with entries from R. If B ∈ Mn×m (R), then B denotes its transpose and if n = m, then det B denotes the determinant of the matrix B, while In×n denotes the identity matrix. Regarding semi-orthodox notational conventions, A := B stands for “A is defined as being B.” while A =: B stands for “B is defined as being A.” Also, the letter C when used as a multiplicative constant in various inequalities, is allowed to vary from line to line. Whenever necessary, its dependence on the other parameters a, b, . . . implicit in the estimate in question is stressed by writing C(a, b, . . .) or Ca,b,... in place of just C.

Chapter 1

Weak Derivatives 1.1

The Cauchy Problem for a Vibrating Infinite String

The partial differential equation ∂12 u − ∂22 u = 0

in R2

(1.1.1)

was derived by Jean d’Alembert in 1747 to describe the displacement u(x1 , x2 ) of a violin string as a function of time and distance along the string. Assuming that the string is infinite and that at time x2 = 0 the displacement is given by some function ϕ0 ∈ C 2 (R) leads to the following global Cauchy problem ⎧ u ∈ C 2 (R2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨∂12 u − ∂22 u = 0 ⎪ ⎪ u(·, 0) = ϕ0 ⎪ ⎪ ⎪ ⎪ ⎩ (∂2 u)(·, 0) = 0

in R2 , in R,

(1.1.2)

in R.

Thanks to the regularity assumption on ϕ0 , it may be checked without difficulty that the function u(x1 , x2 ) := 12 [ϕ0 (x1 + x2 ) + ϕ0 (x1 − x2 )],

for every (x1 , x2 ) ∈ R2 , (1.1.3)

is a solution of (1.1.2). This being said, the expression of u in (1.1.3) continues to be meaningful under much less restrictive assumptions on ϕ0 . For example, u is a well-defined continuous function in R2 whenever ϕ0 ∈ C 0 (R). While in this case expression (1.1.3) is no longer a classical solution of (1.1.2), it is natural to ask whether it is possible to identify a new (and possibly weaker) sense in which (1.1.3) would continue to satisfy ∂12 u − ∂22 u = 0. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6 1, © Springer Science+Business Media New York 2013

1

CHAPTER 1. WEAK DERIVATIVES

2

To answer this question, fix a function u ∈ C 2 (R2 ) satisfying ∂12 u − ∂22 u = 0 pointwise in R2 . If ϕ ∈ C0∞ (R2 ) is an arbitrary function and R ∈ (0, ∞) is a number such that supp ϕ ⊆ (−R, R) × (−R, R), then integration by parts gives 0= (∂12 u − ∂22 u)ϕ dx

R2

R

R

= −R

−

−R R

−R

= R2

∂12 u(x1 , x2 )ϕ(x1 , x2 ) dx1 dx2 R

−R

∂22 u(x1 , x2 )ϕ(x1 , x2 ) dx1 dx2

(∂12 ϕ − ∂22 ϕ)u dx.

(1.1.4)

Note that the condition R2 (∂12 ϕ − ∂22 ϕ)u dx = 0 for all ϕ ∈ C0∞ (R2 ) is meaningful even if u ∈ C 0 (R2 ), which suggests the following definition. Definition 1.1. A function u ∈ C 0 (R2 ) is called a weak (generalized) solution of the equation ∂12 u − ∂22 u = 0 in R2 if (∂12 ϕ − ∂22 ϕ)u dx = 0 for all ϕ ∈ C0∞ (R2 ). (1.1.5) R2

Returning to (1.1.3), let us now check that, under the assumption ϕ0 ∈ C 0 (R), the function u defined in (1.1.3) is a generalized solution of ∂12 u−∂22 u = 0 in R2 . Concretely, fix ϕ ∈ C0∞ (R2 ) and write (∂12 ϕ − ∂22 ϕ)u dx R2

=

1 2

R

R

∂12 ϕ(x1 , x2 ) − ∂22 ϕ(x1 , x2 ) ϕ0 (x1 + x2 ) + ϕ0 (x1 − x2 ) dx1 dx2

y1 + y2 y1 − y2 , ϕ0 (y1 ) + ϕ0 (y2 ) dy1 dy2 2 2 R R y1 + y2 y1 − y2 1 , − (∂22 ϕ) ϕ0 (y1 ) + ϕ0 (y2 ) dy1 dy2 , 4 R R 2 2

1 = 4

(∂12 ϕ)

(1.1.6)

where for the last equality in (1.1.6) we have made the change of variables 2 y1 −y2 for x1 + x2 = y1 , x1 − x2 = y2 . If we now let ψ(y1 , y2 ) := ϕ y1 +y 2 , 2 (y1 , y2 ) ∈ R2 , then y1 + y2 y1 − y2 y1 + y2 y1 − y2 1 1 ∂1 ψ(y1 , y2 ) = (∂1 ϕ) , , + (∂2 ϕ) , 2 2 2 2 2 2 (1.1.7)

1.2. WEAK DERIVATIVES and

3

y1 + y2 y1 − y2 y1 + y2 y1 − y2 1 , , − (∂2 ∂1 ϕ) 2 2 4 2 2 y1 + y2 y1 − y2 y1 + y2 y1 − y2 1 1 2 , , + (∂1 ∂2 ϕ) − (∂2 ϕ) 4 2 2 4 2 2 (1.1.8)

1 ∂2 ∂1 ψ(y1 , y2 ) = (∂12 ϕ) 4

which, when used in (1.1.6), give (∂12 ϕ − ∂22 ϕ)u dx = ∂1 ∂2 ψ(y1 , y2 )[ϕ0 (y1 ) + ϕ0 (y2 )] dy1 dy2 . (1.1.9) R2

R

R

Let R ∈ (0, ∞) be such that supp ϕ ⊂ (−R, R)×(−R, R). Then the support of ψ is contained in the set of points (y1 , y2 ) ∈ R2 satisfying −2R ≤ y1 +y2 ≤ 2R and −2R ≤ y1 −y2 ≤ 2R. Hence, if R > 2R we have supp ψ ⊂ (−R , R )×(−R , R ) and integration by parts yields (∂12 ϕ − ∂22 ϕ)u dx R2

R

= −R

ϕ0 (y1 )

−R R

= −R

−R

R

+

R

ϕ0 (y2 )

∂2 ∂1 ψ(y1 , y2 ) dy2 dy1

R

−R

∂1 ∂2 ψ(y1 , y2 ) dy1 dy2

ϕ0 (y1 ) ∂1 ψ(y1 , R ) − ∂1 ψ(y1 , −R ) dy1

R

+ −R

ϕ0 (y2 ) ∂2 ψ(R , y2 ) − ∂2 ψ(−R , y2 ) dy2 = 0.

(1.1.10)

In summary, this proves that for each ϕ0 ∈ C 0 (R) the function u defined as in (1.1.3) is a weak solution of the equation ∂12 u − ∂22 u = 0 in R2 .

(1.1.11)

We emphasize that in general, there is no reason to expect that u has any pointwise differentiability properties if ϕ0 is merely continuous.

1.2

Weak Derivatives

Convention. Unless otherwise specified, Ω denotes an arbitrary open subset of Rn . Before proceeding with the definition of weak (Sobolev) derivatives we discuss a phenomenon that serves as motivation for the definition. Let f ∈ C m (Ω),


4

where m ∈ N0 . Given an arbitrary ϕ ∈ C0∞ (Ω), consider a function F ∈ C0m (Rn ) that agrees with f in a neighborhood of supp ϕ. For example, we may take F = ψ f, where tilde denotes the extension by zero outside Ω and ψ ∈ C0∞ (Rn ) is identically one in a neighborhood of supp ϕ (see Proposition 13.26 for the construction of such a function). Also, pick R > 0 large enough so that supp ϕ ⊂ B(0, R). Then for each α ∈ Nn0 with |α| ≤ m, integration by parts (cf. Theorem 13.41 for a precise formulation) and support considerations yield α (∂ f )ϕ dx = (∂ α F )ϕ dx Ω

B(0,R) |α|

|α|

α

f ∂ α ϕ dx.

F ∂ ϕ dx = (−1)

= (−1)

B(0,R)

(1.2.1)

Ω

This computation suggests the following definition. Definition 1.2. If f ∈ L1loc (Ω) and α ∈ Nn0 , we say that ∂ α f belongs to L1loc (Ω) in a weak (Sobolev) sense provided there exists some g ∈ L1loc (Ω) with the property that gϕ dx = (−1)|α| f ∂ α ϕ dx for every ϕ ∈ C0∞ (Ω). (1.2.2) Ω

Ω

Whenever this happens, we shall write ∂ α f = g and call g the weak derivative of order α of f . The fact that the concept of weak derivative is unambiguously defined is then ensured by the next theorem.

Theorem 1.3. If g ∈ L1loc (Ω) and Ω gϕ dx = 0 for each ϕ ∈ C0∞ (Ω) then g = 0 almost everywhere on Ω. Proof. From the start, we note that the real and imaginary parts of g enjoy the same type of property as g itself (as seen by integrating g against real-valued test functions ϕ). Thus, without any loss of generality we may assume that g is real-valued.

We make the claim Ω gu dx = 0 for all u ∈ L∞ comp (Ω). To see that this is the case, pick an arbitrary u ∈ L∞ comp (Ω). By working with the extension of u by n zero outside Ω, we may assume that u ∈ L∞ comp (R ) and supp u ⊂ Ω, to begin with. To proceed, consider a function φ satisfying (see (13.3.3) for a concrete example) ∞ n φ(x)dx = 1, (1.2.3) φ ∈ C0 (R ), φ ≥ 0, supp φ ⊆ B(0, 1) and Rn

and for each ε > 0 define φε (x) :=

1 x , φ εn ε

for each x ∈ Rn .

(1.2.4)

1.2. WEAK DERIVATIVES

5

Then φε ∈ C0∞ (Rn ),

φε ≥ 0,

Let us now introduce

supp φε ⊆ B(0, ε),

and Rn

φε (x) dx = 1. (1.2.5)

uε (x) :=

Rn

u(y)φε (x − y) dy,

∀ x ∈ Rn ,

(1.2.6)

and set Kε := x ∈ Rn : dist (x, supp u) ≤ ε . Since u has compact support contained in Ω it follows that there exists ε0 > 0 such that (1.2.7) uε ∈ C ∞ (Rn ) and supp uε ⊆ Kε0 ⊂ Ω if 0 < ε ≤ ε0 . In particular, the restriction of uε to Ω satisfies uε Ω ∈ C0∞ (Ω) provided we select 0 < ε ≤ ε0 . Granted the current assumptions on g, it follows that guε dx = 0 whenever 0 < ε ≤ ε0 . (1.2.8) Ω

Next, using the properties of φε and the definition of uε , we may write |uε (x) − u(x)| = u(x − y)φε (y) dy − u(x) B(0,ε) = (1.2.9) [u(x − y) − u(x)]φε (y) dy B(0,ε) y 1 ≤ n dy |u(x − y) − u(x)|φ ε B(0,ε) ε c ≤ |u(x − y) − u(x)|dy −→ 0 for a.e. x ∈ Rn , |B(0, ε)| B(0,ε) ε→0+ ∞ n where c := ωn−1 n φL (R ) , and for the convergence in (1.2.9) we used Lebesgue’s differentiation theorem (cf. Theorem 13.11). Hence,

(1.2.10) uε → u as ε → 0+ pointwise for almost every x ∈ Rn .

In addition, since Rn φε (x−y) dy = 1 for every x ∈ Rn , it follows that |uε (x)| ≤ uL∞ (Rn ) for every x ∈ Rn . Together with (1.2.7), this implies |guε | = |uε ||g| ≤ uL∞ (Rn ) |g|χKε0 ∈ L1 (Ω).

(1.2.11)

Making use of (1.2.8) and Lebesgue’s dominated convergence theorem (cf. Theorem 13.12 that applies thanks to (1.2.10) and (1.2.11)), we therefore obtain 0 = lim+ guε dx = gu dx, (1.2.12) ε→0

Ω

finishing the proof of the claim made earlier.

Ω


6

To proceed, pick a sequence of compact sets {Kj }j∈N such that Ω =

Kj .

j∈N

For example, Kj := x ∈ Ω : dist(x, ∂Ω) ≥ 1j ∩ B(0, j),

j ∈ N,

(1.2.13)

will do. Also, for each j ∈ N, define uj := χ{x∈Kj : g(x)>0} − χ{x∈Kj : g(x) 0, x ≤ 0,

∀ x ∈ R.

(1.2.15)

Note that f is continuous on R but not differentiable at 0. Nonetheless, f has a weak derivative of order one that is equal to the Heaviside function 1, x > 0, H : R −→ R, H(x) := ∀ x ∈ R. (1.2.16) 0, x ≤ 0, Indeed, if ϕ ∈ C0∞ (R), then integration by parts yields ∞ ∞ ∞ f (x)ϕ (x) dx = − xϕ (x) dx = −xϕ(x) + − −∞

0

0

∞

ϕ(x) dx

0

∞

H(x)ϕ(x) dx,

=

(1.2.17)

−∞

which shows that

R

Hϕ dx = −

f ϕ dx,

∀ ϕ ∈ C0∞ (R).

R

(1.2.18)

Note that H ∈ L1loc (R) hence H is the weak (or Sobolev) derivative of order one of the function f in R. Example 1.5. Does there exist a function g ∈ L1loc (R) such that g is a weak derivative (of order one) of the Heaviside function? To answer this question, first observe that for each ϕ ∈ C0∞ (R) we have ∞ ∞ H(x)ϕ (x) dx = − ϕ (x) dx = ϕ(0). (1.2.19) − −∞

0

1.2. WEAK DERIVATIVES

7

1 Suppose that H has a weak derivative of order one,

and call this g ∈ Lloc (R). Then, by Definition 1.2 and (1.2.19), we have R g ϕ dx = ϕ(0) for all ϕ ∈

∞ C0∞ (R). This forces 0 gϕ dx = 0 for all ϕ ∈ C0∞ ((0, ∞)). In concert with Theorem 1.3, the latter yields g = 0 almost everywhere on (0, ∞). Similarly, we obtain that g = 0 almost everywhere on (−∞, 0), hence g = 0 almost everywhere on R. When combined with (1.2.19), this gives that 0 = R gϕ dx = ϕ(0) for all ϕ ∈ C0∞ (R), leading to a contradiction (as there are functions ϕ ∈ C0∞ (R) with ϕ(0) = 0). Thus, a weak (Sobolev) derivative of order one of H does not exist.

Having defined the notion of weak (Sobolev) derivatives for locally integrable functions, we return to the notion of weak solution considered in Definition 1.1 in a particular case, and extend this to more general partial differential equations. To set the stage, let P (x, ∂) be a linear partial differential operator of order m ∈ N of the form aα (x)∂ α , aα ∈ C |α| (Ω), α ∈ Nn0 , |α| ≤ m. (1.2.20) P (x, ∂) := |α|≤m

Also, suppose f ∈ C 0 (Ω) is a given function and that u is a classical solution of the partial differential equation P (x, ∂)u = f in Ω. That is, assume u ∈ C m (Ω) and the equation holds pointwise in Ω. Then, for each ϕ ∈ C0∞ (Ω), integration by parts gives f ϕ dx = ϕ aα (∂ α u) dx = (−1)|α| ∂ α (aα ϕ) u dx. (1.2.21) Ω

|α|≤m

Ω

Ω |α|≤m

Hence, if we define P (x, ∂)ϕ :=

(−1)|α| ∂ α (aα ϕ),

(1.2.22)

|α|≤m

and call it the transpose of the operator P (x, ∂), the resulting equation from (1.2.21) becomes f ϕ dx = [P (x, ∂)ϕ]u dx, ∀ ϕ ∈ C0∞ (Rn ). (1.2.23) Ω

Ω

Thus, any classical solution u of P (x, ∂)u = f in Ω satisfies (1.2.23). On the other hand, there might exist functions u ∈ L1loc (Ω) that satisfy (1.2.23) but are not classical solutions of the given equation. Such a scenario has been already encountered in (1.1.11) (cf. also the subsequent comment). This motivates the following general definition (compare with Definition 1.1 in the case when P (x, ∂) = ∂12 − ∂22 ). Definition 1.6. Let u, f ∈ L1loc (Ω) be given and assume that P (x, ∂) is as in (1.2.20). Then P (x, ∂)u = f is said to hold in the weak (or Sobolev) sense if f ϕ dx = [P (x, ∂)ϕ]u dx, ∀ ϕ ∈ C0∞ (Ω). (1.2.24) Ω

Ω


8

From the comments in the preamble to Definition 1.6 we know that if u is a classical solution of P (x, ∂)u = f in Ω for some f ∈ C 0 (Ω), then u is also a weak (Sobolev) solution of the same equation. Conversely, if u ∈ C m (Ω) is a weak solution of the partial differential equation P (x, ∂)u = f for some given function f ∈ C 0 (Ω), then by Definition 1.6 and integration by parts we obtain f ϕ dx = [P (x, ∂)ϕ]u dx = ϕ P (x, ∂)u dx. (1.2.25) Ω

Ω

Ω

C0∞ (Ω)

Since ϕ ∈ is arbitrary, Theorem 1.3 then forces f = P (x, ∂)u almost everywhere in Ω, hence ultimately everywhere in Ω, since the functions in question are continuous. In summary, the above discussion shows that the notion of weak solution of a partial differential equation is a natural, unambiguous, and genuine generalization of the concept of classical solution, in the following precise sense: • any classical solution is a weak solution, • any sufficiently regular weak solution is classical, • weak solutions may exist even in the absence of classical ones.

1.3

The Spaces E(Ω) and D(Ω)

A major drawback of Definition 1.2 is that while the right hand-side of (1.2.2) is always meaningful, it cannot always be written in the form given by the left-hand side of (1.2.2). In addition, it might be the case that some locally integrable function in Ω may admit weak (Sobolev) derivatives of a certain order and not of some intermediate lower order (see the example in Exercise 1.26). The remedy is to focus on the portion of (1.2.2) that always makes sense. Specifically, given f ∈ L1loc (Ω) and α ∈ Nn0 , define the mapping ∞ |α| f (∂ α ϕ) dx, ∀ ϕ ∈ C0∞ (Ω). (1.3.1) gα : C0 (Ω) → C, gα (ϕ) := (−1) Ω

The functional in (1.3.1) has the following properties. 1. gα is linear, i.e., gα (λ1 ϕ1 + λ2 ϕ2 ) = λ1 gα (ϕ1 ) + λ2 gα (ϕ2 ) for every λ1 , λ2 ∈ C, and every ϕ1 , ϕ2 ∈ C0∞ (Ω). 2. For each ϕ ∈ C0∞ (Ω) we may estimate α |gα (ϕ)| ≤ |f ||∂ ϕ| dx ≤ |f | dx sup |∂ α ϕ(x)|. Ω

supp ϕ

(1.3.2)

x∈suppϕ

The fact that the term supp ϕ |f |dx in (1.3.2) depends on ϕ is inconvenient if we want to considering the continuity of gα in some sense. Nonetheless, if a priori

1.3. THE SPACES E(Ω) AND D(Ω)

9

a compact set K ⊂ Ω is fixed and the requirement supp ϕ ⊆ K is imposed, then (1.3.2) becomes |f |dx sup |∂ α ϕ(x)| (1.3.3) |gα (ϕ)| ≤ x∈K

K

and, this time, K |f |dx is a constant independent of ϕ. This observation motivates considering an appropriate topology τ on C ∞ (Ω). For the exact definition of this topology see Sect. 13.1. We will not elaborate here more on this subject other than highlighting those key features of τ that are particularly important for our future investigations. To record the precise statements of these features, introduce (1.3.4) E(Ω) := C ∞ (Ω), τ , a notation that emphasizes that E(Ω) is the vector space C ∞ (Ω) equipped with the topology τ . We then have: Fact 1.7. A sequence {ϕj }j∈N ⊂ C ∞ (Ω) converges in E(Ω) to some ϕ ∈ C ∞ (Ω) as j → ∞ if and only if ∀ K ⊂ Ω compact, ∀ α ∈ Nn0 ,

we have

lim sup |∂ α (ϕj − ϕ)(x)| = 0,

j→∞ x∈K

(1.3.5) E(Ω)

in which case we use the notation ϕj −−−→ ϕ. j→∞

Fact 1.8. E(Ω) is a locally convex, metrizable, and complete topological vector space over C. It is easy to see that as a consequence of Fact 1.7 we have the following result. Remark 1.9. A sequence {ϕj }j∈N ⊂ C ∞ (Ω) converges in E(Ω) to ϕ ∈ C ∞ (Ω) as j → ∞ if and only if for any compact set K ⊂ Ω and any m ∈ N0 one has lim

sup

sup |∂ α (ϕj − ϕ)(x)| = 0.

j→∞ α∈Nn , |α|≤m x∈K 0

(1.3.6)

E(Ω)

Exercise 1.10. Prove that if ϕj −−−→ ϕ then the following also hold: j→∞

E(Ω)

(1) ∂ α ϕj −−−→ ∂ α ϕ for each α ∈ Nn0 ; j→∞

E(Ω)

(2) a ϕj −−−→ a ϕ for each a ∈ C0∞ (Ω). j→∞

A standard way of constructing a sequence of smooth functions in Rn that converges in E(Rn ) to a given f ∈ C ∞ (Rn ) is by taking the convolution of f with dilations of a function as in (1.2.3). This construction is discussed in detail next.


10

Example 1.11. Let f ∈ C ∞ (Rn ) be given. Then a sequence of functions from C ∞ (Rn ) that converges to f in E(Rn ) may be constructed as follows. Recall φ from (1.2.3) and define φj (x) := j n φ(jx)

for x ∈ Rn and each j ∈ N.

Clearly, for each j ∈ N we have φj ∈ C0∞ (Rn ),

supp φj ⊆ B(0, 1/j),

Now if we further set for each j ∈ N fj (x) := f (x − y)φj (y) dy = Rn

(1.3.7)

and Rn

φj dx = 1.

f (x − z/j)φ(z) dz,

(1.3.8)

for x ∈ Rn ,

B(0,1)

(1.3.9) then fj ∈ C ∞ (Rn ). Also, if K is an arbitrary compact set in Rn and α ∈ Nn0 , then |∂ α fj (x) − ∂ α f (x)| ≤ |∂ α f (x − z/j) − ∂ α f (x)| φ(z) dz B(0,1)

≤

1 max ∂ β f L∞ (K) j |β|=|α|+1

∀ x ∈ K,

(1.3.10)

n

E(R ) := {x ∈ Rn : dist (x, K) ≤ 1}. Hence fj − where K −−−→ f , as desired. j→∞

The previous approximation result may be further strengthened as indicated in the next two exercises. Exercise 1.12. Prove that C0∞ (Rn ) is sequentially dense in E(Rn ). That is, show that for each f ∈ C ∞ (Rn ) there exists a sequence {fj }j∈N ⊂ C0∞ (Rn ) with E(Rn )

the property that fj −−−−→ f . j→∞

Hint: Let ψ ∈ C0∞ (Rn ) be such that ψ(x) = 1 whenever |x| < 1. Then if f ∈ C ∞ (Rn ) define fj (x) := ψ(x/j)f (x), for each x ∈ Rn and each j ∈ N. Exercise 1.13. Prove that C0∞ (Ω) is sequentially dense in E(Ω). Kj = Ω and Hint: Recall the sequence of compacts from (1.2.13). Then j∈N

˚j+1 for each j ∈ N. For each j ∈ N, let ψj ∈ C0∞ (Ω) be such that Kj ⊂ K ψj = 1 on a neighborhood of Kj and supp ψj ⊆ Kj+1 (cf. Proposition 13.26). If f ∈ C ∞ (Ω), define fj := ψj f for every j ∈ N. Moving on, we focus on defining a topology on C0∞ (Ω) that suits the purposes we have in mind. Since C0∞ (Ω) ⊂ E(Ω), one option would be to consider the topology induced by this larger ambient on C0∞ (Ω). However, this topology has the distinct drawback of not preserving the property of being compactly supported under convergence. Here is an example to that effect.

1.3. THE SPACES E(Ω) AND D(Ω)

11

Example 1.14. Consider the function ⎧ 1 ⎪ ⎨ x− 1 2 − 1 2 4 e if 0 < x < 1, ϕ(x) := ⎪ ⎩ 0 if x ≤ 0 or x ≥ 1,

for each x ∈ R, (1.3.11)

that satisfies ϕ ∈ C ∞ (R), supp ϕ = [0, 1], and ϕ > 0 in (0, 1). For each j ∈ N define ϕj (x) := ϕ(x − 1) + 12 ϕ(x − 2) + · · · + 1j ϕ(x − j),

x ∈ R.

(1.3.12)

E(Rn )

Then ϕj ∈ C ∞ (R), supp ϕj = [1, j + 1], and ϕj −−−−→ ϕ where j→∞

ϕ(x) :=

∞

1 kϕ

x−j

for x ∈ R.

(1.3.13)

k=1

Clearly this limit function does not have compact support. The flaw just highlighted is remedied by introducing a different topology on C0∞ (Ω) that is finer than the one inherited from E(Ω). First, for each K ⊂ Ω compact, denote by DK (Ω) the vector space consisting of functions from C ∞ (Ω) supported in K endowed with the topology induced byE(Ω). Second, consider ∞ on C0 (Ω) the inductive limit topology of the spaces DK (Ω) K⊂Ω and decompact

note the resulting topological vector space by D(Ω). For precise definitions see Sect. 13.1. Two features that are going to be particularly important for our analysis are singled out below (see Sect. 13.1 in this regard). Fact 1.15. D(Ω) is a locally convex and complete topological vector space over C. Fact 1.16. A sequence {ϕj }j∈N ⊂ C0∞ (Ω) converges in D(Ω) to some function ϕ ∈ C0∞ (Ω) as j → ∞ if and only if the following two conditions are satisfied: (1) there exists a compact set K ⊂ Ω such that supp ϕj ⊆ K for all j ∈ N and supp ϕ ⊆ K; (2) for any α ∈ Nn0 we have lim supx∈K |∂ α (ϕj − ϕ)(x)| = 0. j→∞

D(Ω)

We abbreviate (1)–(2) by simply writing ϕj −−−→ ϕ. j→∞

In view of Fact 1.7 one obtains the following consequence of Fact 1.16. D(Ω)

Remark 1.17. ϕj −−−→ ϕ if and only if j→∞

(1) there exists a compact set K ⊂ Ω such that supp ϕj ⊆ K for all j ∈ N, and E(Ω)

(2) ϕj −−−→ ϕ. j→∞


12

If one now considers the identity map from D(Ω) into E(Ω), a combination of Remark 1.17, and Theorem 13.5 yields that this map is continuous. Hence, if we also take into account Exercise 1.12, it follows that D(Ω) is continuously and densely embedded into E(Ω).

(1.3.14)

Exercise 1.18. Suppose ω is an open subset of Ω and consider the map ϕ on ω, ∞ ∞ ι : C0 (ω) → C0 (Ω), ι(ϕ) := ∀ ϕ ∈ C0∞ (ω). (1.3.15) 0 on Ω \ ω, D(ω)

D(Ω)

j→∞

j→∞

Prove that if ϕj −−−→ ϕ then ι(ϕj ) −−−→ ι(ϕ). Use Theorem 13.5 to conclude that ι : D(ω) → D(Ω) is continuous.

Exercise 1.19. Let x0 ∈ Rn and consider the translation by x0 map defined as tx0 : D(Rn ) −→ D(Rn ) tx0 (ϕ) := ϕ(· − x0 ),

∀ ϕ ∈ C0∞ (Rn ).

(1.3.16)

Prove that tx0 is linear and continuous. Hint: Use Theorem 13.5. D(Ω)

Exercise 1.20. Prove that if ϕj −−−→ ϕ then the following also hold: j→∞

D(Ω)

(1) ∂ α ϕj −−−→ ∂ α ϕ for each α ∈ Nn0 ; j→∞

D(Ω)

(2) a ϕj −−−→ a ϕ for each a ∈ C0∞ (Ω). j→∞

E(Ω)

D(Ω)

j→∞

j→∞

Exercise 1.21. Prove that if ϕj −−−→ ϕ then a ϕj −−−→ a ϕ for each function a ∈ C0∞ (Ω). Also show that if ω is an open subset of Ω, a ∈ C0∞ (ω), and we E(Ω)

D(ω)

j→∞

j→∞

have ϕj −−−→ ϕ, then a ϕj −−−→ a ϕ. We have already noted that the topology D(Ω) is finer than the topology C0∞ (Ω) inherits from E(Ω), and an example of a sequence of smooth, compactly supported functions in Ω convergent in E(Ω) to a limit which does not belong to D(Ω) has been given in Example 1.14. The example below shows that even if the limit function is in D(Ω), one should still not expect that convergence in E(Ω) of a sequence of smooth, compactly supported functions in Ω implies convergence in D(Ω). Example 1.22. Let ϕ be as in (1.3.11) and for each j ∈ N set ϕj (x) := ϕ(x− j) for x ∈ R. Clearly, ϕj ∈ C ∞ (R) and supp ϕj = [j, j + 1] for all j ∈ N. If K is some compact subset of R, then there exists j0 ∈ N such that K ⊆ [−j0 , j0 ].

1.4. ADDITIONAL EXERCISES FOR CHAP. 1

13

(k) Consequently, supp ϕj ∩ K = ∅ for j ≥ j0 . Thus, trivially, sup ϕj (x) = 0 if x∈K

E(R)

j ≥ j0 which shows that ϕj −−−→ 0. Consider next the issue whether {ϕj }j∈N j→∞

converge in D(R). If this were to be the case, there would exist r ∈ (0, ∞) such ∞ that supp ϕj ⊆ [−r, r] for every j. However, supp ϕj = [1, ∞) which leads j=1

to a contradiction. Thus, {ϕj }j∈N does not converge in D(R). Convention. In what follows, we will often identify a function f ∈ C0∞ (Ω) with its extension by zero outside its support, which makes such an extension belong to C0∞ (Rn ). Further Notes for Chap. 1. The concept of weak derivative goes back to the pioneering work of the Soviet mathematician Sergei Lvovich Sobolev (1908– 1989). Although we shall later extend the scope of taking derivatives in a generalized sense to the larger class of distributions, a significant portion of partial differential equations may be developed solely based on the notion of a weak derivative. For example, this is the approach adopted in [12], where distributions are avoided all together. A good reference for the topological aspects that are most pertinent to the spaces of test functions considered here is [70], though there are many other monographs dealing with these issues. The interested reader may consult [10, 59, 71], and the references therein.

1.4

Additional Exercises for Chap. 1

Exercise 1.23. Given ϕ0 ∈ C 2 (R), ϕ1 ∈ C 1 (R), and F ∈ C 1 (R2 ), show that the function u : R2 → R defined by x1 +x2 u(x1 , x2 ) := 12 (ϕ0 (x1 + x2 ) + ϕ0 (x1 − x2 ) + 12 ϕ1 (t) dt −

1 2

x2

x1 +(x2 −t)

F (ξ, t) dξ 0

x1 −x2

dt,

x1 −(x2 −t)

is a classical solution of the problem ⎧ u ∈ C 2 (R2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨∂12 u − ∂22 u = F in R2 , ⎪ ⎪ u(·, 0) = ϕ0 in R, ⎪ ⎪ ⎪ ⎪ ⎩ (∂2 u)(·, 0) = ϕ1 in R.

∀ (x1 , x2 ) ∈ R2 , (1.4.1)

(1.4.2)

Exercise 1.24. Determine the values of a ∈ R for which the function f : R → R x, if x ≥ a, defined by f (x) := for each x ∈ R, has a weak derivative. 0, if x < a,

14


Exercise 1.25. Does the function f : R → R defined by f (x) :=

1, if x ≥ 2, 0, if x < 2,

for each x ∈ R, have a weak derivative? Exercise 1.26. Let f : R2 → R2 be defined by f (x, y) := H(x) + H(y) for each (x, y) ∈ R2 , where H is the Heaviside function from (1.2.16). Prove that for α = (1, 1) and β = (1, 0) the weak derivatives ∂ α f and ∂ α+β f exist, while the weak derivative ∂ β f does not exist. Exercise 1.27. Compute the weak derivative of order one of f : (−1, 1) → R defined by f (x) := sgn(x) |x| for every x ∈ (−1, 1), where ⎧ ⎪ ⎨ 1 0 sgn(x) := ⎪ ⎩ −1

if x > 0, if x = 0,

(1.4.3)

if x < 0.

Exercise 1.28. Let f : R2 → R2 be the function defined by f (x, y) := x|y| for each (x, y) ∈ R2 . Prove that the weak derivative ∂12 ∂2 f exists, while the weak derivative ∂1 ∂22 f does not. Exercise 1.29. Let f : R2 → R2 be defined by f (x, y) := H(x) − sgn(y) for each (x, y) ∈ R2 , and let α = (α1 , α2 ) ∈ N20 . Prove that ∂ α f exists in the weak sense if and only if α1 ≥ 1 and α2 ≥ 1. Exercise 1.30. Let f : R → R be defined by f (x) := sin |x| for every x ∈ R. Does f exist in the weak sense? How about f ? Exercise 1.31. Let ω, Ω be open subsets of Rn with ω ⊆ Ω. Suppose f ∈ L1loc (Ω) is such that ∂ α f exists in the weak sense in Ω for some α ∈ Nn0 . Show that the weak derivative ∂ α (f |ω ) exists and equals (∂ α f )|ω almost everywhere in ω. Exercise 1.32. Let ε ∈ (0, 1) and consider the function f : Rn → R defined by |x|−ε if x ∈ Rn \ {0}, f (x) := ∀ x ∈ Rn . 1 if x = 0, Prove that ∂j f exists in the weak sense for each j ∈ {1, . . . , n} if and only if n ≥ 2. Also compute the weak derivatives ∂j f , j ∈ {1, . . . , n}, in the case when n ≥ 2. Exercise 1.33. Assume that a, b ∈ R are such that a < b. (a) Prove that if f ∈ L1loc (a, b) is such that the weak derivative f exists and is equal to zero almost everywhere on (a, b), then there exists some complex number c such that f = c almost everywhere on (a, b).

b Hint: Fix ϕ0 ∈ C0∞ (a, b) with a ϕ0 (t) dt = 1. Then each ϕ ∈

b C0∞ (a, b) is of the form ϕ = ϕ0 a ϕ(t) dt + ψ , for some ψ ∈ C0∞ (a, b) .


15

(b) Assume that g ∈ L1loc (a, b) and x0 ∈ (a, b). Prove that the function

x defined by f (x) := x0 g(t) dt, for x ∈ (a, b), belongs to L1loc ((a, b)) and has a weak derivative that is equal to g almost everywhere on (a, b). (c) Let f ∈ L1loc (a, b) be such that the weak derivative f (k) exists for some k ∈ N. Prove that all the weak derivatives f (j) exist for each j ∈ N with j < k.

x Hint: Prove that if g(x) := x0 h(t) dt where x0 ∈ (a, b) is a fixed point and h := f (k) , and if ϕ0 is as in the hint to (a), then f

(k−1)

=g−

b

b

(k−1)

k−1

g(t)ϕ0 (t) dt + (−1) a

f (t)ϕ0

(t)dt.

a

(d) Let f ∈ L1loc (a, b) such that f (k) = 0 for some k ∈ N. Prove that there k−1 exist a0 , a1 , . . . , ak−1 ∈ C such that f (x) = aj xj for almost every j=0

x ∈ (a, b). (e) If Ω is an open set in Rn for n ≥ 2 and f ∈ L1loc (Ω) is such that the weak derivative ∂ α f exists for some α ∈ Nn0 , does it follow that ∂ β f exists in a weak sense for all β ≤ α? Exercise 1.34. Let θ ∈ C0∞ (Rn ) and m ∈ N. Prove that the sequence ϕj (x) := e−j j m θ(jx),

∀ x ∈ Rn ,

j ∈ N,

converges in D(Rn ). Exercise 1.35. Let θ ∈ C0∞ (Rn ), h ∈ Rn \ {0}, and set ϕj (x) := θ(x + jh),

∀ x ∈ Rn ,

E(Rn )

D(Rn )

j→∞

j→∞

j ∈ N.

Prove that ϕj −−−−→ 0. Is it true that ϕj −−−−→ 0? Exercise 1.36. Let θ ∈ C0∞ (Rn ) be not identically zero, and for each j ∈ N define 1 ϕj (x) := θ(jx), ∀ x ∈ Rn . j Prove that the sequence {ϕj }j∈N does not converge in D(Rn ). Exercise 1.37. Let θ ∈ C0∞ (Rn ), h ∈ Rn \ {0}. Prove that the sequence ϕj (x) := e−j θ(j(x − jh)),

∀ x ∈ Rn ,

j ∈ N,

converges in D(Rn ) if and only if θ(x) = 0 for all x ∈ Rn .


16

Exercise 1.38. Consider θ ∈ C0∞ (Rn ) not identically zero, and for each j ∈ N define 1 x , ∀ x ∈ Rn . ϕj (x) := θ j j Does {ϕj }j∈N converge in D(Rn )? How about in E(Rn )? Exercise 1.39. Suppose that {ϕj }j∈N is a sequence in C0∞ (Ω) with the property that lim Rn f (x)ϕj (x) dx = 0 for every f ∈ L1loc (Ω). Is it true that j→∞

D(Rn )

ϕj −−−−→ 0? j→∞

Chapter 2

The Space D(Ω) of Distributions 2.1

The Definition of Distributions

Building on the idea emerging in (1.3.1), we now make the following definition that is central to all subsequent considerations. Definition 2.1. u : D(Ω) → C is called a distribution on Ω if u is linear and continuous. By design, distributions are simply elements of the dual space of the topological vector space D(Ω). Given a functional u : D(Ω) → C and a function ϕ ∈ C0∞ (Ω), we use the traditional notation u, ϕ in place of u(ϕ) (in particular, u, ϕ is a complex number). While any linear and continuous functional is sequentially continuous, the converse is not always true. Nonetheless, for linear functionals on D(Ω), continuity is equivalent with sequential continuity. This remarkable property, itself a consequence of Theorem 13.5, is formally recorded below. Fact 2.2. Let u : D(Ω) → C be a linear map. Then u is a distribution on Ω D(Ω)

if and only if for every sequence {ϕj }j∈N ⊂ C0∞ (Ω) with ϕj −−−→ ϕ for some j→∞

ϕ ∈ C0∞ (Ω), we have lim u, ϕj = u, ϕ (where the latter limit is considered in C).

j→∞

Remark 2.3. In general, if X, Y are topological vector spaces and Λ : X → Y is a linear map, then Λ is sequentially continuous on X if and only if Λ is sequentially continuous at the zero vector 0 ∈ X. When combined with Fact 2.2, D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6 2, © Springer Science+Business Media New York 2013

17

CHAPTER 2. THE SPACE D (Ω) OF DISTRIBUTIONS

18

this shows that a linear map u : D(Ω) → C is a distribution on Ω if and only if D(Ω)

lim u, ϕj = 0 for every sequence {ϕj }j∈N ⊂ C0∞ (Ω) with ϕj −−−→ 0.

j→∞

j→∞

Another important characterization of continuity of complex-valued linear functionals u defined on D(Ω) is given by the following proposition. Proposition 2.4. Let u : D(Ω) → C be a linear map. Then u is a distribution if and only if for each compact set K ⊂ Ω there exist k ∈ N0 and C ∈ (0, ∞) such that |u, ϕ| ≤ C sup |∂ α ϕ(x)| x∈K |α|≤k

for all ϕ ∈ C0∞ (Ω) with supp ϕ ⊆ K.

(2.1.1)

Proof. Fix u : D(Ω) → C linear and suppose that for each compact set K ⊂ Ω there exist k ∈ N0 and C ∈ (0, ∞) satisfying (2.1.1). To show that u is a D(Ω)

distribution, let ϕj −−−→ 0. Then, there exists a compact set K ⊆ Ω such that j→∞

supp ϕj ⊆ K for all j ∈ N, and ∂ α ϕj −−−→ 0 uniformly on K for any α ∈ Nn0 . j→∞

For this compact set K, by our hypotheses, there exist C > 0 and k ∈ N0 such that (2.1.1) holds, and hence, |u, ϕj | ≤ C sup |∂ α ϕj (x)| −−−→ 0,

(2.1.2)

j→∞

x∈K |α|≤k

which implies that u, ϕj −−−→ 0. From this and Remark 2.3 it follows that u j→∞

is a distribution in Ω. To prove the converse implication we reason by contradiction. Suppose that there exists a compact set K ⊆ Ω such that for every j ∈ N, there exists a function ϕj ∈ C0∞ (Ω) with supp ϕj ⊆ K and |u, ϕj | > j sup |∂ α ϕj (x)|.

(2.1.3)

x∈K |α|≤j 1 ϕj . Then for each j ∈ N we have ψj ∈ C ∞ (Ω), supp ψj ⊆ K, Define ψj := u,ϕ j and u, ψj = 1. On the other hand, from (2.1.3) we see that

sup |∂ α ψj (x)|
0 and ε > 0.

(2.1.9)


20

This is justified by starting with the elementary inequality ln t ≤ t for all t > 0, then replacing t by t±ε . In turn, for each R > 0 and ε ∈ (0, 1), estimate (2.1.9) gives ln |x| dx ≤ ε−1 (2.1.10) max |x|ε , |x|−ε dx < ∞. B(0,R)

B(0,R)

Let us now revisit the functional from (1.3.1). Example 2.10. For a given f ∈ L1loc (Ω) and multi-index α ∈ Nn0 , consider the functional gα : D(Ω) → C defined by f ∂ α ϕ dx for each ϕ ∈ C0∞ (Ω). (2.1.11) gα (ϕ) := (−1)|α| Ω D(Ω)

Clearly this is a linear mapping. Moreover, if ϕj −−−→ 0, then there exists K j→∞

compact subset of Ω such that supp ϕj ⊆ K for each j ∈ N, hence |gα (ϕj )| ≤ |f ||∂ α ϕj | dx K

≤ ∂ α ϕj L∞ (K) f L1 (K) → 0 as j → ∞.

(2.1.12)

By invoking Remark 2.3, it follows that gα is a distribution in Ω. Next we consider a set of examples of distributions that are not of function type. As a preamble, observe that f (x) := x1 for x = 0, is not locally integrable on R. Nonetheless, it is possible to associate to this function a certain distribution, not as in (2.1.6), but in the specific manner described below. Example 2.11. Consider the mapping P.V. x1 : D(R) → C defined by P.V.

1 (ϕ) := lim+ x ε→0

|x|≥ε

ϕ(x) dx, x

∀ ϕ ∈ C0∞ (R).

(2.1.13)

We claim that P.V. x1 is a distribution of order one in R. First, we prove that the mapping (2.1.13) is well-defined. Let ϕ ∈ C0∞ (R) and suppose R > 0 is such that supp ϕ ⊂ (−R, R). Fix ε ∈ (0, R) and observe that since x1 is odd on R \ {0}, we have |x|≥ε

ϕ(x) dx = x

ε≤|x|≤R

ϕ(x) dx = x

ε≤|x|≤R

ϕ(x) − ϕ(0) dx. x

(2.1.14)

≤ sup|y|≤R |ϕ (y)| for each x ∈ R\{0}. Thus, Lebesgue’s In addition, ϕ(x)−ϕ(0) x

dominated convergence theorem implies that lim+ |x|≥ε ϕ(x) x exists and is equal ε→0

to |x|≤R ϕ(x)−ϕ(0) dx. This shows that the mapping P.V. x1 is well-defined, and x

2.1. THE DEFINITION OF DISTRIBUTIONS

21

from definition it is clear that P.V. x1 is linear. Furthermore, it is implicit in the argument above that 1 (ϕ) ≤ 2R sup |ϕ (x)|, (2.1.15) ∀ ϕ ∈ C0∞ (−R, R) . P.V. x |x|≤R In concert with Proposition 2.4 this shows that P.V. x1 is a distribution in R of order at most one. We are left with showing that P.V. x1 does not have order 0. ∞ Consider the compact K = [0, 1] and for each j ∈ N let ϕ ∈ C j 0 (0, 1) be such 1 1 that 0 ≤ ϕj ≤ 1 and ϕj = 1 on j+2 , 1 − j+2 . Then from the very definition of P.V. x1 and the fact that ϕj vanishes near zero, 1 1− j+2 ! " 1 ϕ (x) 1 1 j dx ≥ dx = ln(j + 1), P.V. , ϕj = 1 x x x 0 j+2

∀ j ∈ N.

(2.1.16) Since supx∈K |ϕj (x)| ≤ 1 and lim ln(j+1) = ∞, the inequality in (2.1.16) shows j→∞

that there is no constant C ∈ (0, ∞) such that |P.V. x1 , ϕ| ≤ C supx∈K |ϕ(x)| for all ϕ ∈ C0∞ (R) with supp ϕ ⊆ K. This proves that the distribution P.V. x1 does not have order 0. Remark 2.12. An inspection of the proof in Example 2.11 shows that we have ! 1 " ϕ(x) − ϕ(0) ϕ(x) P.V. , ϕ = dx + dx, ∀ ϕ ∈ C0∞ (R). x x x |x|≤1 |x|>1 (2.1.17) Example 2.13. An important distribution is the Dirac distribution δ defined by δ(ϕ) := ϕ(0), ∀ ϕ ∈ C0∞ (Rn ). (2.1.18) It is not difficult to check that δ is a distribution in Rn of order 0. A natural question to ask is whether δ is a distribution of function type. To answer there exists f ∈ L1loc (Rn ) such that ϕ(0) = δ, ϕ =

this question, suppose ∞ f ϕ dx for all ϕ ∈ C0 (Rn ). The latter implies that Rn f ϕ dx = 0 for every Rn ϕ ∈ C0∞ (Rn ) satisfying supp ϕ ∩ {0} = ∅. Hence, by Theorem 1.3 we have f = 0 almost everywhere on Rn \{0}, thus f = 0 almost everywhere in Rn . Consequently, f ϕ dx = 0, ∀ ϕ ∈ C0∞ (Rn ), (2.1.19) ϕ(0) = δ, ϕ = Rn

which is false. This proves that the Dirac distribution is not of function type. Example 2.14. The Dirac distribution δ is sometimes referred to as having “mass at zero” since, for each x0 ∈ Rn , one may similarly define the map δx0 : C0∞ (Rn ) → C by setting δx0 (ϕ) := ϕ(x0 ). Then it is easy to see that δx0 is a distribution in Rn , called the Dirac distribution with mass at x0 , and the convention we make is to drop the subscript x0 if x0 = 0 ∈ Rn .

22


Example 2.15. Let μ be either a complex Borel measure on Ω, or a Borel positive measure on Ω that is locally finite (i.e., satisfies μ(K) < ∞ for every compact K ⊂ Ω). Consider ϕ dμ, ∀ ϕ ∈ C0∞ (Ω). (2.1.20) μ : D(Ω) → C, μ(ϕ) := Ω

The mapping in (2.1.20) is well-defined, linear, and if K is an arbitrary compact set in Ω, then ∀ ϕ ∈ C ∞ (Ω), supp ϕ ⊂ K.

|μ(ϕ)| ≤ |μ|(K) sup |ϕ(x)|,

(2.1.21)

x∈K

By Proposition 2.4 we have that μ induces a distribution in Ω. The estimate in (2.1.21) also shows that μ is a distribution of order zero. In the last part of this section we discuss the validity of a converse implication to the implication in Example 2.15. Proposition 2.16. Let u be a distribution in Ω of order zero. Then the distribution u extends uniquely to a linear map Λu : C00 (Ω) → C that is locally bounded, in the following sense: for each compact set K ⊂ Ω there exists CK ∈ (0, ∞) with the property that |Λu (ϕ)| ≤ CK sup |ϕ(x)|, x∈K

∀ ϕ ∈ C00 (Ω) with supp ϕ ⊆ K.

(2.1.22)

In addition, the functional Λu satisfies the following properties. ˚j+1 (i) Let {Kj }j∈N be a sequence of compact subsets of Ω satisfying Kj ⊂ K ∞ for j ∈ N and Ω = Kj . Then there exists a sequence of complex regular j=1

Borel measures μj on Kj , j ∈ N, with the following properties: ˚ , and each (a) μj (E) = μ (E) for each ∈ N, each Borel set E ⊂ K j ≥ ; (b) for each j ∈ N one has ϕ dμj , ∀ ϕ ∈ C00 (Ω) with supp ϕ ⊂ Kj . Λu (ϕ) =

(2.1.23)

Kj

(ii) Then there exist two Radon measures μ1 , μ2 , taking Borel sets from Ω into [0, ∞] (i.e., measures satisfying the regularity properties (ii)–(iv) in Theorem 13.17), such that Re Λu (ϕ) = ϕ dμ1 − ϕ dμ2 , ∀ ϕ ∈ C00 (Ω) real-valued. (2.1.24) Ω

Ω

Furthermore, a similar conclusion is valid for Im Λu .

2.1. THE DEFINITION OF DISTRIBUTIONS

23

Proof. Let u be a distribution in Ω of order zero and let K be a compact set contained in Ω. Fix ϕ ∈ C00 (Ω) such that supp ϕ ⊆ K. We claim that there exists a sequence of functions φj ∈ C0∞ (Ω), j ∈ N, supported in a fixed compact neighborhood K0 of K such that lim sup |φj (x) − ϕ(x)| = 0.

(2.1.25)

j→∞ x∈K0

To justify this claim consider φε as in (1.2.3)–(1.2.5) then define ϕε (x) := ϕ(y)φε (x − y) dy, ∀ x ∈ Rn ,

(2.1.26)

Rn

Then the same type of reasoning as in (1.2.9) (using that a continuous function on a compact set is uniformly continuous as a replacement for the use of Lebesgue’s differential theorem), then shows that φj := ϕ1/j do the job in (2.1.25). Because u is a distribution of order zero, there exists C = C(K0 ) ∈ (0, ∞) such that (2.1.1) holds with k = 0. The latter combined with (2.1.25) implies |u, φj − u, φk | ≤ C sup |φj (x) − φk (x)| −−−−−→ 0. j,k→∞

x∈K0

(2.1.27)

Hence, the sequence of complex numbers {u, φj }j∈N is Cauchy, thus convergent in C, which allows us to define Λu (ϕ) := lim u, φj .

(2.1.28)

j→∞

Proving that this definition is independent of the selection of {φj }j∈N is done by interlacing sequences. Specifically, if {φj }j∈N is another sequence of functions in C0∞ (Ω), supported in K0 and satisfying (2.1.25), then by considering the sequence {ψj }j≥2 defined by ψ2j+1 := φj and ψ2j := φj for every j ≥ 2, we obtain that {u, ψj }j≥2 is convergent, thus its subsequences {u, φj }j∈N and {u, φj }j∈N are also convergent and have the same limit. In turn, the independence of the definition of Λu (ϕ) on the approximating sequence {φj }j∈N readily implies that Λu : C00 (Ω) → C is a linear mapping. In addition, the fact that u is a distribution of order zero implies the existence of a finite constant C > 0 such that |u, φj | ≤ C sup |φj (x)|,

∀ j ∈ N.

(2.1.29)

x∈K0

Taking the limit as j → ∞ in (2.1.29) gives |Λu (ϕ)| ≤ C sup |ϕ(x)| on account x∈K

of (2.1.28). Let us now show that the linear extension Λu of u to C00 (Ω) is unique in the class of locally bounded mappings. By linearity, this comes down to proving that if Λ : C00 (Ω) → C is a linear locally bounded mapping that vanishes on C0∞ (Ω) then Λ is identically zero on C00 (Ω). To this end, pick an arbitrary ϕ ∈ C0∞ (Ω), set K := supp ϕ and, as before, let φj ∈ C0∞ (Ω), j ∈ N,


24

be a sequence of functions supported in a fixed compact neighborhood K0 of K such that (2.1.25) holds. Then |Λ(ϕ)| = |Λ(ϕ − φj )| ≤ C sup |ϕ(x) − φj (x)| −−−→ 0,

(2.1.30)

j→∞

x∈K0

proving that Λ(ϕ) = 0, as wanted. Moving on, fix a sequence of compacts as in the hypothesis of (i) (e.g., the sequence from (1.2.13) will do). Based on what we have proved so far and Riesz’s representation theorem for complex measures (see Theorem 13.18), it follows that there exists a sequence of complex regular Borel measures μj on Kj , j ∈ N, such that (2.1.23) holds. ˚ . If j ≥ , since the measures μj , Now fix ∈ N and a Borel set E ⊂ K μ are regular, in order to prove that μj (E) = μ (E) it suffices to show that μj (F ) = μ (F ) for every compact set F ⊆ E. Fix such a compact set and ˚ , with the property choose a sequence of compact sets {Fk }k∈N contained in K ∞ # ˚k for k ∈ N and that Fk+1 ⊂ F Fk = F . Applying Uryshon’s lemma (see k=1

Proposition 13.20) we obtain a sequence ϕk ∈ C00 (Ω), 0 ≤ ϕk ≤ 1, supp ϕk ⊆ Fk and ϕk = 1 on F , for each k ∈ N. Then lim ϕk (x) = χF (x) for x ∈ Ω and we k→∞

may write μj (F ) =

Kj

k→∞

Kj

=

ϕk dμj = lim u, ϕk = lim

χF dμj = lim

k→∞

χF dμ = μ (F ).

k→∞

ϕk dμ K

(2.1.31)

K

This completes the proof of (i). Finally, the claim in (ii) follows by invoking Riesz’s representation theorem for locally bounded functionals (cf. Theorem 13.19). Among other things, Proposition 2.16 is a useful ingredient in the following representation theorem for positive distributions. Theorem 2.17. Let u be a distribution in Ω such that u, ϕ ≥ 0 for every nonnegative function ϕ ∈ C0∞ (Ω). Then there exists a unique positive Borel regular measure μ on Ω such that u, ϕ = ϕ dμ, ∀ ϕ ∈ C0∞ (Ω). (2.1.32) Ω

Proof. First we prove that u has order zero. To do so, let K be a compact set in Ω and fix ψ ∈ C0∞ (Ω), ψ ≥ 0 and satisfying ψ ≡ 1 on K. Then, if ϕ ∈ C0∞ (Ω) has supp ϕ ⊆ K and is real valued, it follows that ( sup |ϕ(x)|)ψ ± ϕ ≥ 0, x∈K

(2.1.33)

2.2. THE TOPOLOGICAL VECTOR SPACE D (Ω)

25

hence by the positivity of u, we have ( sup |ϕ(x)|)u, ψ ± u, ϕ ≥ 0. x∈K

Thus, |u, ϕ| ≤ u, ψ supx∈K |ϕ(x)|, (2.1.34)

for all real valued ϕ ∈ C00 (Ω) with supp ϕ ⊆ K.

If now ϕ ∈ C0∞ (Ω) with supp ϕ ⊆ K is complex valued, say ϕ = ϕ1 + iϕ2 and ϕ1 , ϕ2 , are real valued, then by using (2.1.34) we may write |u, ϕ| = |u, ϕ1 |2 + |u, ϕ2 |2 ≤ u, ψ sup |ϕ1 (x)| + sup |ϕ2 (x)| x∈K

x∈K

≤ 2 u, ψ sup |ϕ(x)|.

(2.1.35)

x∈K

The fact that the distribution u has order zero now follows from (2.1.35). Having established this, we may apply Proposition 2.16 to conclude that u may be uniquely extended to a linear map Λu : C00 (Ω) → C that is locally bounded. We next propose to show that this linear map is positive. In this respect, we note that if ϕ ≥ 0 then the functions {φj }j∈N used in the proof of Proposition 2.16 (which are constricted by convolving ϕ with a nonnegative mollifier) may also be taken to be nonnegative. When combined with (2.1.28) and the fact that u is positive, this shows that the extension Λu of u to C00 (Ω) as defined in (2.1.28) is a positive functional. Consequently, Riesz’s representation theorem for positive functionals (see Theorem 13.17) may be may invoked to conclude that there exists a unique positive Borel regular measure μ on Ω such that Λu (ϕ) = ϕ dμ, ∀ ϕ ∈ C00 (Ω). (2.1.36) Ω

Now (2.1.32) follows by specializing (2.1.36) to ϕ ∈ C0∞ (Ω). This finishes the proof of the theorem.

2.2

The Topological Vector Space D (Ω)

The space of distributions in Ω endowed with the natural addition and scalar multiplication of linear mappings becomes a vector space over C. Indeed, if u1 , u2 , u are distributions in Ω and λ1 , λ2 , λ ∈ C, we define u1 + u2 : D(Ω) → C and λu : D(Ω) → C by setting ∀ ϕ ∈ C0∞ (Ω). (2.2.1) It is not difficult to check that u1 + u2 and λu are distributions in Ω. The topology we consider on the vector space of distributions in Ω is the weak∗-topology induced by D(Ω) (for details see Sect. 13.1), which makes it a (u1 + u2 )(ϕ) := u1 , ϕ + u2 , ϕ,

(λu)(ϕ) := λu, ϕ,


26

topological vector space over C and we denote this topological vector space by D (Ω). In particular, D (Ω) is a locally convex topological vector space over C. We record next an important equivalent characterization of convergence in D (Ω). Fact 2.18. A sequence {uj }j∈N in D (Ω) converges to some u ∈ D (Ω) as j → ∞ in D (Ω) if and only if uj , ϕ −−−→ u, ϕ for every ϕ ∈ C0∞ (Ω), in j→∞

D (Ω)

which case we use the notation uj −−−−→ u. j→∞

Moreover, the topological space D (Ω) is complete, in the following sense. If the sequence {uj }j∈N ⊂ D (Ω) is such that lim uj , ϕ exists (in C) for every j→∞

ϕ ∈ C0∞ (Ω) then the functional u : D(Ω) → C defined by u(ϕ) := lim uj , ϕ j→∞

for every ϕ ∈ C0∞ (Ω) is a distribution in Ω.

Note that from Fact 2.18 it is easy to see that if a sequence {uj }j∈N in D (Ω) is convergent, then its limit is unique. Indeed, if such a sequence would have two limits, say u, v ∈ D (Rn ), then it would follow that for each ϕ ∈ C0∞ (Rn ), the sequence of numbers {uj , ϕ}j∈N would converge to both u, ϕ and v, ϕ, thus u, ϕ = v, ϕ. Hence, u = v. Remark 2.19. Assume that we are given u ∈ D (Ω) and uε ∈ D (Ω) for each D (Ω)

ε ∈ (0, ∞). We make the convention that uε −−−−→ u is understood in the sense + ε→0

that for every sequence of positive numbers {εj }j∈N satisfying lim εj = 0 we j→∞

D (Ω)

have uεj −−−−→ u. j→∞

Example 2.20. Let φ be as in (1.2.3) and recall the sequence of functions {φj }j∈N from (1.3.7). Interpreting each φj ∈ L1loc (Rn ) as distribution in Rn , for each function ϕ ∈ C0∞ (Rn ) we have φj , ϕ = φj (x)ϕ(x) dx = φ(y)ϕ(y/j) dy, ∀ j ∈ N. (2.2.2) Rn

Rn

Thus, by Lebesgue dominated convergence theorem (cf. Theorem 13.12), φj , ϕ = φ(y)ϕ(y/j) dy −−−→ ϕ(0) φ(y)dy = ϕ(0) = δ, ϕ. (2.2.3) j→∞

Rn

Rn

D (Rn )

This proves that φj −−−−→ δ. j→∞

Exercise 2.21. Prove that if p ∈ [1, ∞] and a sequence of functions {fj }j∈N D (Ω)

from Lp (Ω) converges in Lp (Ω) to some f ∈ Lp (Ω) then fj −−−−→ f . j→∞

Hint: Use H¨ older’s inequality.

2.3. MULTIPLICATION OF A DISTRIBUTION WITH A C ∞ . . .

27

Exercise 2.22. (a) Assume that f ∈ L1 (Rn ) and for each ε > 0 define fε (x) := ε−n f (x/ε) for x ∈ Rn . Prove that for each g ∈ L∞ (Rn ) that is continuous at 0 ∈ Rn we have fε (x)g(x) dx −−−−→ f (x) dx g(0). (2.2.4) ε→0+

Rn

Rn

(b) Use part (a) to prove that if f ∈ L1 (Rn ) is given and for each j ∈ N we define fj (x) := j n f (jx) for each x ∈ Rn , then each fj belongs to L1 (Rn ) (hence fj ∈ D (Rn )) and D (Rn ) f (x) dx. (2.2.5) fj −−−−→ c δ where c := j→∞

Rn

Hint: To justify the claim in part (a), make a change of variables to write the integral Rn fε (x)g(x) dx as Rn f (y)g(εy) dy, then use Lebesgue’s dominated convergence theorem.

2.3

Multiplication of a Distribution with a C ∞ Function

The issue we discuss in this section is the definition of the multiplication of a distribution u ∈ D (Ω) with a smooth function a ∈ C ∞ (Ω). First we consider the case when u is of function type, i.e., u = uf for some f ∈ L1loc (Ω). In this particular case, af ∈ L1loc (Ω); thus, it defines a distribution uaf on Ω and uaf , ϕ = (af )ϕ dx = f (a ϕ) dx = f, a ϕ, ∀ ϕ ∈ C0∞ (Ω). (2.3.1) Ω

Ω

In the general case, this suggests defining au as in the following proposition. Proposition 2.23. Let u ∈ D (Ω) and a ∈ C ∞ (Ω). Then the mapping au : D(Ω) → C defined by

(au)(ϕ) := u, aϕ,

∀ ϕ ∈ C0∞ (Ω),

(2.3.2)

is linear and continuous, hence a distribution on Ω. Proof. The fact that au is linear is immediate from its definition. To show that au is also continuous we make use of Remark 2.3. To this end, consider a D(Ω)

D(Ω)

j→∞

j→∞

sequence ϕj −−−→ 0. By (2) in Exercise 1.20 we have a ϕj −−−→ 0. Since u is a distribution on Ω, the latter implies lim u, aϕj = 0. Moreover, from (2.3.2) j→∞

we have (au)(ϕj ) = u, aϕj for each j ∈ N. Hence, lim (au)(ϕj ) = 0 proving j→∞

that au is continuous.


28

Exercise 2.24. Let f ∈ L1loc (Ω) and a ∈ C ∞ (Ω). With the notation from (2.1.6), prove that auf = uaf in D (Ω). Remark 2.25. (1) When more information about u ∈ D (Ω) is available, (2.3.2) may continue to yield a distribution under weaker regularity demands on a than the current assumption that a ∈ C ∞ (Ω). In general, however, the condition a ∈ C ∞ (Ω) may not be weakened if (2.3.2) is to yield a distribution for arbitrary u ∈ D (Ω). (2) As observed later (see Remark 2.30), one may not define the product of two arbitrary distributions in a way that ensures associativity. (3) Based on (2.3.2) and (2.2.1), it follows that if u, u1 , u2 ∈ D (Ω), and if a, a1 , a2 ∈ C ∞ (Ω), then a(u1 + u2 ) = au1 + au2 , (a1 + a2 )u = a1 u + a2 u, and a1 (a2 u) = (a1 a2 )u, where the equalities are considered in D (Ω). Exercise 2.26. The following properties hold. E(Ω)

D (Ω)

j→∞

j→∞

1. If u ∈ D (Ω) and aj −−−→ a, then aj u −−−−→ au. D (Ω)

D (Ω)

j→∞

j→∞

2. If a ∈ C ∞ (Ω) and uj −−−−→ u, then auj −−−−→ au. Example 2.27. Recall the Dirac distribution defined in (2.1.18) and assume that some function a ∈ C ∞ (Ω) has been given. Then, for each ϕ ∈ C0∞ (Rn ) we may write $ % aδ, ϕ = δ, aϕ = (aϕ)(0) = a(0)ϕ(0) = a(0)δ, ϕ . This shows that aδ = a(0)δ

in D (Rn ) for each a ∈ C ∞ (Ω).

(2.3.3)

As a consequence, xm δ = 0

in D (R),

for every m ∈ N.

(2.3.4)

Example 2.28. The goal is to solve the equation xu = 1

in D (R).

(2.3.5)

Clearly, this equation does not have a solution u of function type. Recall the distribution defined in Example 2.11. Then for every ϕ ∈ C0∞ (R) we may write ! " 1 " ! 1 x ϕ(x) x P.V. , ϕ = P.V. , xϕ(x) = lim dx + x x x ε→0 |x|≥ε ϕ(x) dx = 1, ϕ. (2.3.6) = R

2.4. DISTRIBUTIONAL DERIVATIVES

29

Thus,

1 x P.V. = 1 in D (R). (2.3.7) x Given (2.3.4), it follows that u := P.V. x1 + c δ will also be a solution of (2.3.5) for any c ∈ C. We will see later (c.f. Remark 2.70) that in fact any solution of (2.3.5) is of the form P.V. x1 + c δ, where c ∈ C. Exercise 2.29. Let ψ ∈ C ∞ (R). Determine a solution u ∈ D (R) of the equation xu = ψ in D (R).

Remark 2.30. Suppose one could define the product of distributions as an associative operation, in a manner compatible with the multiplication by a smooth function. Considering then δ, x, and P.V. x1 ∈ D (R), one would then necessarily have 1 1 = δ x · P.V. = δ · 1 = δ in D (R), 0 = (δ · x) P.V. (2.3.8) x x thanks to (2.3.4) and (2.3.7), leading to the false conclusion that δ = 0 in D (R).

2.4

Distributional Derivatives

We are now ready to define derivatives of distributions. One of the most basic attributes of the class of distributions, compared with other classes of locally integrable functions, is that distributions may be differentiated unrestrictedly within this environment (with the resulting objects being still distributions), and that the operation of distributional differentiation retains some of the most basic properties as in the case of ordinary differentiable functions (such as, a suitable product formula, symmetry of mixed derivatives, etc.). In addition, the differentiation of distributions turns out to be compatible with the pointwise differentiation in the case when the distribution in question is of function type, given by a sufficiently regular function. To develop some sort of intuition, we shall start our investigation by looking first at a distribution of function type, and try to generalize the notion of weak (Sobolev) derivative from Definition 1.2. As noted earlier, if f ∈ L1loc (Ω), the mapping defined in (1.3.1) is a distribution on Ω. This suggests making the following definition. Definition 2.31. For each α ∈ Nn0 and u ∈ D (Ω), the distributional derivative (or the derivative in the sense of distributions) of order α of the distribution u is the mapping ∂ α u : D(Ω) → C defined by ∂ α u(ϕ) := (−1)|α| u, ∂ α ϕ,

∀ ϕ ∈ C0∞ (Ω).

(2.4.1)

Note that if u ∈ D (Ω) is of function type, say u = uf for some f ∈ L1loc (Ω), and if the weak derivative ∂ α f exists, that is, one can find g ∈ L1loc (Ω) such that (1.2.2) holds, then according to Definition 2.31 we have that the distributional derivative ∂ α u is equal to the distribution ug in D (Ω). In short,


30

∂ α uf = u∂ α f in this case. Thus, Definition 2.31 generalizes Definition 1.2. That the class of distributions in Ω is stable under taking distributional derivatives is proved in the next proposition. Proposition 2.32. For each α ∈ Nn0 and each u ∈ D (Ω) we have ∂ α u ∈ D (Ω). Proof. Fix α ∈ Nn0 and a distribution u ∈ D (Ω). The linearity of the mapping D(Ω)

∂ α u : D(Ω) → C is immediate. Regarding its continuity, let ϕj −−−→ 0. Then, j→∞

D(Ω)

since ∂ α ϕj −−−→ 0 (see (1) in Exercise 1.20) and u ∈ D (Ω), we may write j→∞

∂ α u(ϕj ) = (−1)|α| u, ∂ α ϕj −−−→ 0. j→∞

(2.4.2)

Remark 2.3 then shows that ∂ α u ∈ D (Ω). Exercise 2.33. Suppose that m ∈ N and f ∈ C m (Ω). Prove that for any α ∈ Nn0 satisfying |α| ≤ m, the distributional derivative of order α of uf is the distribution of function type given by the derivative, in the classical sense, of order α of f , that is, ∂ α (uf ) = u∂ α f in D (Ω). Proposition 2.34. Let m ∈ N0 and assume that P (x, ∂) := aα (x)∂ α , aα ∈ C ∞ (Ω), α ∈ Nn0 , |α| ≤ m.

(2.4.3)

|α|≤m

Also, suppose that u ∈ C m (Ω). Then P (x, ∂)u, computed in D (Ω), coincides as a distribution with the distribution induced by P (x, ∂)u, computed pointwise in Ω. Proof. This follows from (2.4.3), Exercises 2.33, and 2.24. Remark 2.35. Recall that a function f : Rn → C is called Lipschitz provided there exists M ∈ [0, ∞) such that |f (x) − f (y)| ≤ M |x − y|

∀ x, y ∈ Rn .

(2.4.4)

If f is Lipschitz, then the constant C := inf {M for which (2.4.4) holds} is called the Lipschitz constant of f . We will prove (see Theorem 2.104) that if f is Lipschitz then the distributional derivatives ∂k f , k = 1, . . . , n, belong to L∞ (Rn ). Consequently, f : Rn → R Lipschitz =⇒ ∂k (uf ) = u∂k f in D (Rn ) for k = 1, . . . , n. (2.4.5) Some basic properties of differentiation in the distributional sense are summarized below. Proposition 2.36. The following properties of distributional differentiation hold.


31

(1) Any distribution is infinitely differentiable (i.e. D (Ω) is stable under the action of ∂ α for any α ∈ N0 ). (2) If u ∈ D (Ω) and k, ∈ {1, . . . , n} then ∂k ∂ u = ∂ ∂k u in D (Ω). D (Ω)

D (Ω)

j→∞

j→∞

(3) If uj −−−−→ u and α ∈ Nn0 , then ∂ α uj −−−−→ ∂ α u. (4) For any u ∈ D (Ω) and any a ∈ C ∞ (Ω) we have ∂j (au) = (∂j a)u + a(∂j u) in D (Ω). Proof. The first property follows immediately from the definition of distributional derivatives. To prove the remaining properties, fix an arbitrary function ϕ ∈ C0∞ (Ω). Then, using (2.4.1) repeatedly and the symmetry of mixed partial derivatives for smooth functions (Schwarz’s theorem), we have ∂k ∂ u, ϕ = −∂ u, ∂k ϕ = u, ∂ ∂k ϕ = u, ∂k ∂ ϕ = −∂k u, ∂ ϕ = ∂ ∂k u, ϕ,

∀ k, ∈ {1, . . . , n},

which implies (2). Let now {uj }j∈N , u, and α satisfy the hypotheses in (3). Based on (2.4.1) and Fact 2.18 we may write ∂ α uj , ϕ = (−1)|α| uj , ∂ α ϕ −−−→ (−1)|α| u, ∂ α ϕ = ∂ α u, ϕ, j→∞

hence ∂ α uj converges to ∂ α u in D (Ω) as j → ∞ and the proof of (3) is complete. Finally, for u ∈ D (Ω) and a ∈ C ∞ (Ω), using (2.4.1) and Leibniz’s product formula for derivatives of smooth functions we can write ∂j (au), ϕ = −au, ∂j ϕ = −u, a(∂j ϕ) = −u, ∂j (aϕ) + u, (∂j a)ϕ $ % = ∂j u, aϕ + (∂j a)u, ϕ = a(∂j u) + (∂j a)u, ϕ , (2.4.6) from which (4) follows. Example 2.37. Recall the Heaviside function H from (1.2.16). This is a locally integrable function thus it defines a distribution on R that we denote also by H. Then, the computation in (1.2.19) implies H , ϕ = −H, ϕ = δ, ϕ, Hence,

H=δ

∀ ϕ ∈ C0∞ (R).

in D (R).

(2.4.7) (2.4.8)

Exercise 2.38. Prove that for every function a ∈ C ∞ (Ω) and every α ∈ Nn0 we have α! a(∂ α δ) = (−1)|β| (∂ β a)(0)∂ α−β δ in D (Ω). (2.4.9) β!(α − β)! β≤α

Hint: Use formula (13.2.4) when computing ∂ α (aϕ) for ϕ ∈ C0∞ (Ω).


32

Exercise 2.39. Prove that for every c ∈ R one has −c|x| e = −c e−cx H(x) + c ecx H(−x)

in

D (R).

(2.4.10)

Hint: Show e−c|x| = e−cx H(x) + ecx H(−x) in D (R) and then use part (4) in Proposition 2.36 and (2.4.8). Next, we look at the issue of existence of antiderivatives for distributions on open intervals. Proposition 2.40. Let I be an open interval in R and suppose u0 ∈ D (I). (1) The equation u = u0 in D (I) admits at least one solution. (2) If u1 , u2 ∈ D (I) are such that (u1 ) = u0 in D (I) and (u1 ) = u0 in D (I), then there exists c ∈ C such that u1 − u2 = c in D (I). Proof. Suppose I = (a, b), where a ∈ R ∪ {−∞} and b ∈ R ∪ {+∞}, and define the set A(I) := {ϕ : ϕ ∈ C0∞ (I)}. We claim that ϕ(x) dx = 0. (2.4.11) if ϕ ∈ C0∞ (I), then ϕ ∈ A(I) ⇐⇒ I

The left-to-right implication in (2.4.11) is immediate from the fundamental theorem of calculus. To prove the converse implication, suppose ϕ ∈ C0∞ (I) is

x such that I ϕ(x) dx = 0. Then the function ψ(x) := a ϕ(t) dt, x ∈ I, satisfies ψ ∈ C ∞ (I), supp ψ ⊆ supp ϕ, and ψ = ϕ on I, thus ϕ ∈ A(I). This finishes the justification of (2.4.11).

Next, fix ϕ0 ∈ C0∞ (I) with the property that I ϕ0 (x) dx = 1 and consider the map Θ : D(I) → D(I) defined for each ϕ ∈ C0∞ (I) by Θ(ϕ) := θϕ

where θϕ (x) := a

x

' & ϕ(x) dx ϕ0 (t) dt, ϕ(t) −

∀ x ∈ I.

I

(2.4.12) Since the integral of ϕ − I ϕ(x) dx ϕ0 over I is zero, our earlier discussion shows that Θ is well-defined. Since Θ is linear, Theorem 13.5 and Fact 1.15 imply that Θ is continuous if and only if it is sequentially continuous. The latter property may be verified from definitions. Finally, from (2.4.12) and the fundamental theorem of calculus we have

Θ(ϕ ) = ϕ for every ϕ ∈ C0∞ (I).

(2.4.13)

Suppose now that an arbitrary distribution u0 on I has been fixed. Define u := −u0 ◦ Θ, which is a distribution on I thanks to the properties of Θ. In concert with (2.4.13), this definition implies that u, ϕ = −u0 , ϕ for every function ϕ ∈ C0∞ (I), proving that u = u0 in D (I). This finishes the proof of the statement in (1).


33

Moving on, suppose u ∈ D (I) is such that u = 0 in D (I). Then if ϕ0 is as earlier in the proof, for any ϕ ∈ C0∞ (I) we may write " ! ϕ(x) dx ϕ0 + ϕ(x) dx u, ϕ0 u, ϕ = u, ϕ − I

I

$ % $ % = u, (θϕ ) + u, ϕ0 , ϕ = c, ϕ ,

(2.4.14)

where c := u, ϕ0 ∈ C. Hence, u = c in D(I). By linearity, this readily implies the statement in (2). The proof of the proposition is now complete. Proposition 2.41. Let I be an open interval in R, g ∈ C ∞ (I), and f ∈ C k (I) for some k ∈ N0 . If u ∈ D (I) satisfies u + gu = f in D (I), then u ∈ C k+1 (I). x

Proof. Fix a ∈ I and define F (x) := e a g0 (t) dt for x ∈ I. Then F ∈ C ∞ (I) and we may use (4) in Proposition 2.36 and the equation satisfied by u to write (keeping in mind that f is continuous) x F (t)f (t) dt = F u + F u − F f = 0 in D (I). (2.4.15) Fu − a

By Proposition 2.40, it follows that there exists some constant c ∈ C with the x x property that F u = a F (t)f (t) dt + c in D (I). Since a F (t)f (t) dt ∈ C k+1 (I)

x 1 1 c ∞ and F ∈ C (I), we conclude that u = F (x) a F (t)f (t) dt + F (x) ∈ C k+1 (I), as desired. We close this section by presenting a higher-degree version of the product formula for differentiation from part (4) in Proposition 2.36. Proposition 2.42 (Generalized Leibniz Formula). Suppose f ∈ C ∞ (Ω) and u ∈ D (Ω). Then for each α ∈ Nn0 one has ∂ α (f u) =

β≤α

α! (∂ β f )(∂ α−β u) β!(α − β)!

in

D (Ω).

(2.4.16)

Proof. The first step is to observe that for each j ∈ {1, . . . , n} and each k ∈ N0 we have ∂jk (f u) =

0≤≤k

k! (∂ f )(∂jk− u) in D (Ω), !(k − )! j

(2.4.17)

which is proved by induction on k making use of part (4) in Proposition 2.36. Hence, given any α = (α1 , α2 , . . . , αn ) ∈ Nn0 , via repeated applications of (2.4.17) we obtain ∂1α1 (f u) =

0≤β1 ≤α1

α1 ! (∂ β1 f )(∂1α1 −β1 u), β1 !(α1 − β1 )! 1

(2.4.18)


34 ∂1α1 ∂2α2 (f u) =

0≤β1 ≤α1 0≤β2 ≤α2

α2 ! α1 ! (∂ β1 ∂ β2 f )(∂1α1 −β1 ∂2α2 −β2 u), β1 !(α1 − β1 )! β2 !(α2 − β2 )! 1 2 (2.4.19)

and by induction ∂ α (f u) =

···

0≤β1 ≤α1

0≤βn ≤αn

αn ! α1 ! ··· × β1 !(α1 − β1 )! βn !(αn − βn )!

× (∂1β1 · · · ∂nβn f )(∂1α1 −β1 · · · ∂nαn −βn u) =

0≤β≤α

α! (∂ β f )(∂ α−β u), β!(α − β)!

(2.4.20)

as claimed.

2.5

The Support of a Distribution

In preparation to discussing the notion of support of a distribution, we first define the restriction of a distribution to an open subset of the Euclidean domain on which the distribution is considered. Necessarily, such a definition should generalize restrictions at the level of locally integrable functions. We start from the observation that if f ∈ L1loc (Ω) and ω is a non-empty set open subset of Ω, 1 then f ω ∈ Lloc (ω). Thus, f ω , ϕ = f ϕ dx = f ι(ϕ) dx, ∀ ϕ ∈ C0∞ (ω), (2.5.1) ω

Ω

where ι is the map from (1.3.15). Proposition 2.43. Let Ω be a nonempty open subset of Rn and suppose ω is a nonempty open subset of Ω. Also, recall the map ι from (1.3.15). Then for each u ∈ D (Ω), the mapping arising as the restriction of the distribution u to ω, i.e., uω : D(ω) → C defined by uω (ϕ) := u, ι(ϕ), ∀ ϕ ∈ C0∞ (ω), (2.5.2) is linear and continuous. Hence, uω ∈ D (ω). Proof. It is immediate that the map in (2.5.2) is well-defined and linear. To see that it is also continuous we use Proposition 2.4. Let K be a compact set contained in ω. Then K ⊂ Ω and, since u ∈ D (Ω), Proposition 2.4 applies and gives k ∈ N0 and C ∈ (0, ∞) such that (2.1.1) holds. In particular, for each ϕ ∈ C0∞ (ω) with supp ϕ ⊆ K, (2.5.3) u ω (ϕ) = u, ι(ϕ) ≤ C sup |∂ α ϕ(x)|. x∈K |α|≤k

2.5. THE SUPPORT OF A DISTRIBUTION

35

The conclusion that uω ∈ D (ω) now follows. For an alternative proof of the continuity of uω one may use Fact 2.2 and Exercise 1.18. Exercise 2.44. (1) Prove that the definition of the restriction of a distribution from (2.5.2) generalizes the usual restriction of functions. More specifically, using the notation introduced in(2.1.6), show that if ω is an open subset of Ω and f ∈ L1loc (Ω), then uf ω = uf |ω in D (ω). (2) Prove that the operation of differentiation of a distribution commutes with the operation of restriction of a distribution to open sets, that is, if ω is an open subset of Ω, then ∂ α uω = (∂ α u)ω , ∀ u ∈ D (Ω), ∀ α ∈ Nn0 . (2.5.4) The next proposition shows that a distribution is uniquely determined by its local behavior. Proposition 2.45. If u1 , u2 ∈ D (Ω) are such that for each x0 ∈ Ω there exists an open subset ω of Ω with x0 ∈ ω and satisfying u1 ω = u2 ω in D (ω), then u1 = u2 in D (Ω). Proof. Observe that this proposition may be viewed as a reconstruction problem; thus, it is meaningful to try to use a partition of unity. Let ϕ ∈ C0∞ (Ω) be arbitrary, fixed and set K := supp ϕ. The goal is to prove that u1 , ϕ = u2 , ϕ. From hypotheses it follows that for each x ∈ K there exists an open neighborhood ωx ⊂ Ω of x with the property that u1 ωx = u2 ωx . Based on the fact that K is compact, the cover {ωx}x∈K of K may be refined to a finite one, consisting of, say ω1 , . . . , ωN . These are open subsets of Ω and satisfy K⊂

N (

ωj

and u1 ωj = u2 ωj

for j = 1, . . . , N.

(2.5.5)

j=1

Consider a partition of unity {ψj : j = 1, . . . , N } subordinate to the cover ∞ {ωj }N j=1 of K, as given by Theorem 13.29. Hence, we have ψj ∈ C0 (Ω) with N ψj = 1 on K. Consequently, using supp ψj ⊂ ωj for each j = 1, . . . , N , and j=1

the linearity of distributions and (2.5.5), we obtain N N N ! " u1 , ϕ = u1 , ϕ ψj = u1 , ϕψj = u2 , ϕψj j=1

j=1

j=1

N " ! ϕψj = u2 , ϕ. = u2 , j=1

Thus, u1 , ϕ = u2 , ϕ and the proof of the proposition is complete.

(2.5.6)

36


Exercise 2.46. Let k ∈ N0 ∪ {∞} and suppose u ∈ D (Ω) is such that for each x ∈ Ω there exists a number rx > 0 and a function fx ∈ C k (B(x, rx )) such that B(x, rx ) ⊂ Ω and uB(x,rx ) = fx in D (B(x, rx )). Prove that u ∈ C k (Ω). Hint: Use Theorem 13.34 to obtain a partition of unity {ψj }j∈J subordinate to the cover {B(x, rx )}x∈Ω of Ω, then show that f := ψj fj is a function in j∈J

C k (Ω) satisfying u = f in D (Ω). Exercise 2.47. Let u, v ∈ D (Ω) be such that supp u ∩ supp v = ∅ and u + v = 0 in D (Ω). Prove that u = 0 and v = 0 in D (Ω). Hint: Note that v Ω\supp u = (u + v)Ω\supp u = 0 and v Ω\supp v = 0. Combine this with the fact that (Ω \ supp u) ∪ (Ω \ supp v) = Ω and Proposition 2.45 to deduce that v = 0. Now we are ready to define the notion of support of a distribution. Recall that if f ∈ C 0 (Ω) then its support is defined to be the closure relative to Ω of the set {x ∈ Ω : f (x) = 0}. However, the value of an arbitrary distribution at a point is not meaningful. The fact that Ω\supp f is the largest open set contained in Ω on which f = 0 suggests the introduction of the following definition. Definition 2.48. The support of a distribution u ∈ D (Ω) is defined as supp u := x ∈ Ω : there is no ω open such that x ∈ ω ⊆ Ω and uω = 0 . (2.5.7) Based on (2.5.7), it follows that Ω \ supp u = x ∈ Ω : ∃ ω open set such that x ∈ ω ⊆ Ω and uω = 0 , (2.5.8) which is an open set. Hence, supp u is relatively closed in Ω. Moreover, if we apply Proposition 2.45 to the distributions u and 0 ∈ D (Ω \ supp u) we obtain that ∀ u ∈ D (Ω). (2.5.9) uΩ\supp u = 0, In other words, Ω\supp u is the largest open subset of Ω on which the restriction of u is zero. Example 2.49. Recall the Dirac distribution from (2.1.18). We claim that supp δ = {0}. Indeed, if ϕ ∈ C0∞ (Rn \ {0}) it follows that δ, ϕ = ϕ(0) = 0. By Proposition 2.45, δ Rn \{0} = 0, thus supp δ ⊆ {0}. To prove the opposite inclusion, consider an arbitrary open subset ω of Ω such that 0 ∈ ω. Then there exists ϕ ∈ C0∞(ω) such that ϕ(0) = 1, and hence, δ, ϕ = 1 = 0, which in turn implies that δ ω = 0. Consequently, 0 ∈ supp δ as desired. Similarly, if x0 ∈ Rn , then supp δx0 = {x0 }, where δx0 is as in Example 2.14. Example 2.50. If f ∈ C 0 (Ω) then supp uf = supp f , where uf is the distribution from (2.1.6). Indeed, since f = 0 in Ω \ supp f , we have Ω f (x)ϕ(x) dx = 0 for every ϕ ∈ C0∞ (Ω \ supp f ), hence supp uf ⊆ supp f . Also, if x ∈ Ω \ supp uf then there exists an open neighborhood ω of x with ω ⊆ Ω and such that uf ω = 0. Thus, for every ϕ ∈ C0∞ (Ω) one has 0 = uf , ϕ = Ω f (x)ϕ(x) dx.

2.6. COMPACTLY SUPPORTED DISTRIBUTIONS ...

37

Invoking Theorem 1.3 we arrive at the conclusion that f = 0 almost everywhere in ω hence, ultimately, f = 0 in ω (since f is continuous in ω). Consequently, x ∈ supp f and this proves that supp f ⊆ supp uf . We propose to extend the scope of the discussion in Example 2.50 as to make it applicable to functions that are merely locally integrable (instead of continuous). This requires defining a suitable notion of support for functions that lack continuity, and we briefly address this issue first. Given an arbitrary set E ⊆ Rn and an arbitrary function f : E → C, we define the support of f as supp f := x ∈ E : there is no r > 0 so that f = 0 a.e. in B(x, r) ∩ E . (2.5.10) From this definition one may check without difficulty that ( E \ supp f = E ∩ B(x, rx )

(2.5.11)

x∈E\supp f

where for each x ∈ E \ supp f the number rx > 0 is such that f = 0 a.e. in B(x, rx ) ∩ E. Moreover, since Rn has the Lindelöf property, the above union can be refined to a countable one. Based on these observations, the following basic properties of the support may be deduced: supp f is a relatively closed subset of E,

(2.5.12)

f = 0 a.e. in E \ supp f,

(2.5.13)

supp f ⊆ F if F ⊆ E is relatively closed and f = 0 a.e. on E \ F ,

(2.5.14)

supp f = supp g if g : E → C is such that f = g a.e. on E.

(2.5.15)

In addition, if the set E ⊆ Rn is open and the function f : E → C is continuous, then supp f may be described as the closure in E of the set {x ∈ E : f (x) = 0}, which is precisely our earlier notion of support in this context. Exercise 2.51. If f ∈ L1loc (Ω), then supp uf = supp f , where uf is the distribution from (2.1.6). Hint: Use (2.5.10), (2.5.8), part (1) in Exercise 2.44, and the fact that the injection in (2.1.8) is one-to-one.

2.6

Compactly Supported Distributions and the Space E (Ω)

Next we discuss the issue of extending the action of a distribution u ∈ D (Ω) to a subclass of C ∞ (Ω) that is possibly larger than C0∞ (Ω). Observe that if f ∈ L1loc (Ω), the expression Ω f ϕ dx is meaningful for functions ϕ ∈ C ∞ (Ω)

38


such that supp f ∩ supp ϕ is a compact subset of

Ω. A particular case is when we have supp ϕ ∩ supp f = ∅ in which scenario Ω f ϕ dx = 0. This observation is the motivation behind the following theorem. Theorem 2.52. Let u ∈ D (Ω) and consider a relatively closed subset F of Ω satisfying supp u ⊆ F . Set MF := {ϕ ∈ C ∞ (Ω) : supp ϕ ∩ F is a compact set in Rn }.

(2.6.1)

Then there exists a unique linear map u : MF → C satisfying the following conditions: (i) u, ϕ = u, ϕ for every ϕ ∈ C0∞ (Ω), and (ii) u, ϕ = 0 for every ϕ ∈ C ∞ (Ω) with supp ϕ ∩ F = ∅, where, if ψ ∈ MF , then u, ψ denotes u (ψ). Moreover, extensions of u constructed with respect to different choices of F act in a compatible fashion. More precisely, if F1 , F2 ⊆ Ω are two relatively closed sets in Ω with the property that supp u ⊆ Fj , j = 1, 2, then u1 , ϕ = u2 , ϕ for every ϕ ∈ MF1 ∩ MF2 , where u 1 , u 2 , are the extensions of u constructed as above relative to the sets F1 and F2 , respectively. Before presenting the proof of this theorem, a few comments are in order. Remark 2.53. Retain the context of Theorem 2.52. (a) One has C0∞ (Ω) ⊆ MF and {ϕ ∈ C ∞ (Ω) : supp ϕ ∩ F = ∅} ⊆ MF . (b) MF is a vector subspace of C ∞ (Ω), albeit not a topological subspace of E(Ω). (c) If F = supp u we are in the setting discussed prior to the statement of Theorem 2.52. Also, if F1 ⊆ F2 then MF2 ⊆ MF1 . In particular, the largest MF corresponds to the case when F = supp u. (d) If supp u is compact and we take F = supp u then MF = C ∞ (Ω). In such a scenario, Theorem 2.52 gives an extension of u, originally defined as linear functional on C0∞ (Ω), to a linear functional defined on the larger space C ∞ (Ω). From a topological point of view, this extension turns out to be a continuous mapping of E(Ω) into C (as we will see later, in Theorem 2.59). Proof of Theorem 2.52. Fix a relatively closed subset F of Ω satisfying the condition supp u ⊆ F . First we prove the uniqueness statement in the first part of 2 : MF → C satisfy (i) and (ii). Fix ϕ ∈ MF and the theorem. Suppose u 1 , u consider ψ ∈ C0∞ (Ω) such that ψ ≡ 1 in an open neighborhood W of the set F ∩ supp ϕ. That such a function ψ exists is guaranteed by Proposition 13.26. Decompose ϕ = ϕ0 +ϕ1 where ϕ0 := ψϕ ∈ C0∞ (Ω) and ϕ1 := (1−ψ)ϕ ∈ C ∞ (Ω). In general, if A ⊆ Rn and f ∈ C 0 (Rn ), it may be readily verified that f = 0 on A if and only if supp f ⊆ (Ac ). Making use of this observation we obtain


39

c ˚ = W c . It follows that supp ϕ1 ⊆ W c ∩ supp ϕ, that supp (1 − ψ) ⊆ W c = W 1 and u 2 , we have hence supp ϕ1 ∩ F = ∅. Thus, by (i) and (ii) written for u u1 , ϕ = u1 , ϕ0 + u1 , ϕ1 = u, ϕ0 + 0 = u2 , ϕ0 = u2 , ϕ0 + u2 , ϕ1 = u2 , ϕ,

(2.6.2)

which implies that u 1 = u 2 . To prove the existence of an extension satisfying properties (i) and (ii), we make use of the decomposition of ϕ already employed in the proof of uniqueness. The apparent problem is that such a decomposition is not unique. Let ϕ ∈ MF and suppose that ϕ = ϕ0 +ϕ1 = ϕ0 +ϕ1 , with ϕ0 , ϕ0 ∈ C0∞ (Ω), ϕ1 , ϕ1 ∈ C ∞ (Ω) and supp ϕ1 ∩ F = ∅ = supp ϕ1 ∩ F. Thus, ϕ0 − ϕ0 = ϕ1 − ϕ1 , and since supp (ϕ1 − ϕ1 ) ∩ F = ∅, we also have supp (ϕ0 − ϕ0 ) ∩ F = ∅, which in turn implies supp (ϕ0 − ϕ0 ) ⊆ Ω \ supp u. The latter condition entails 0 = u, ϕ0 − ϕ0 = u, ϕ0 − u, ϕ0 . This suggests defining the extension u : MF −→ C,

u, ϕ := u, ψϕ for each ϕ ∈ MF and

each ψ ∈ C0∞ (Ω) with ψ ≡ 1 in a neighborhood of supp ϕ ∩ F .

(2.6.3)

Clearly u as in (2.6.3) is linear and, based on the previous reasoning, independent of the choice of ψ, thus well-defined. We claim that this extension also satisfies u, ϕ = (i) and (ii). Indeed, if ϕ ∈ C0∞ (Ω), we choose ψ ≡ 1 on supp ϕ. Then u, ϕ, so the extension in (2.6.3) satisfies (i). Also, if ϕ ∈ C ∞ (Ω) is such that supp ϕ ∩ F = ∅, we may choose ψ ∈ C0∞ (Ω) such that supp ψ ∩ F = ∅ which forces u, ϕ = u, ψϕ = 0, hence our extension satisfies (ii) as well. This proves the claim. We are left with proving the compatibility of extensions. Let F1 , F2 ⊆ Ω 2 be relatively closed sets in Ω each containing supp u. Denote by u 1 and u the linear extensions of u to MF1 and MF2 , respectively, constructed as above relative to the sets F1 and F2 . For ϕ ∈ MF1 ∩ MF2 let ψ ∈ C0∞ (Ω) be such that ψ = 1 on an open neighborhood of supp ϕ ∩ F1 and on an open neighborhood of supp ϕ ∩ F2 . Then by (2.6.3), u1 , ϕ = u, ψϕ = u2 , ϕ. The proof of the theorem is now complete. Remark 2.54. In the context of Theorem 2.52 consider u ∈ D (Ω), a ∈ C ∞ (Ω), and α ∈ Nn0 . Then the extension given in Theorem 2.52 satisfies the following properties:


40

(1) ) au, ϕ = u, aϕ for every ϕ ∈ MF ; α u, ϕ = (−1)|α| (2) ∂* u, ∂ α ϕ for every ϕ ∈ MF .

Indeed, since by Theorem 2.52 an extension with properties (i) and (ii) is unique, the statement in (1) above will follow if one proves that the actions of the linear functionals considered in the left- and right-hand sides of the equality in (1) coincide on C0∞ (Ω) and on C ∞ (Ω) functions with supports outside F , which are immediate from (2.6.3) and properties of distributions. A similar approach works for the proof of (2). We introduce the following notation Dc (Ω) := {u ∈ D (Ω) : supp u is a compact subset of Ω}.

(2.6.4)

By applying Theorem 2.52 to a distribution u ∈ Dc (Ω) and the set F := supp u, in which case we have MF = C ∞ (Ω), it follows that there exists a linear map u : C ∞ (Ω) → C satisfying (i) and (ii) in the statement of this theorem. In fact, this extension turns out to be continuous with respect to the topology E(Ω), an issue that we will address shortly. The dual of E(Ω) is the space {v : E(Ω) → C : v linear and continuous}.

(2.6.5)

Whenever v : E(Ω) → C is linear and continuous, and whenever ϕ ∈ C ∞ (Ω), we use the notation v, ϕ in place of v(ϕ). The following is an equivalent characterization of continuity for linear functionals on E(Ω) (see Sect. 13.1 for more details). Fact 2.55. A linear functional v : E(Ω) → C is continuous (for details see (13.1.22)) if and only if there exist a compact K ⊂ Ω, a number m ∈ N0 , and a constant C ∈ (0, ∞), such that |v(ϕ)| ≤ C

sup

sup |∂ α ϕ(x)| ,

x∈K α∈Nn 0 , |α|≤m

∀ ϕ ∈ C ∞ (Ω).

(2.6.6)

In the current setting, functionals on E(Ω) are continuous if and only if they are sequentially continuous. This can be seen by combining the general result presented in Theorem 13.14 with Fact 1.8. A direct proof, applicable to the specific case of linear functionals on E(Ω), is given in the next proposition. Proposition 2.56. Let v : E(Ω) → C be a linear map. Then v is continuous if and only if v is sequentially continuous. Proof. The general fact that any linear and continuous functional on topological vector spaces is sequentially continuous gives the left-to-right implication. To prove the converse implication, it suffices to check continuity at zero. This is done reasoning by contradiction. Assume that E(Ω)

v(ϕj ) −−−→ 0 whenever ϕj −−−→ 0, j→∞

j→∞

(2.6.7)


41

but that v is not continuous at 0 ∈ E(Ω). Then for each compact subset K of Ω and every j ∈ N, there exists ϕj ∈ E(Ω) such that |v(ϕj )| > j sup |∂ α ϕj (x)|.

(2.6.8)

x∈K |α|≤j

Consider now a nested sequence of compact sets {Kj }j∈N such that

∞

Kj = Ω.

j=1

For each j ∈ N, let ϕj be as given by (2.6.8) corresponding to K := Kj and ϕ define the function ψj := v(ϕjj ) which belongs to E(Ω). Then v(ψj ) = 1 and

sup x∈Kj , |α|≤j

|∂ α ψj (x)| ≤

1 for every j ∈ N. j

(2.6.9)

Thus, for each fixed α ∈ Nn0 and every compact subset K of Ω there exists some j0 ≥ |α| with the property that K ⊂ Kj0 and supx∈K |∂ α ψj (x)| < 1j for all E(Ω)

j ≥ j0 . The latter implies ψj −−−→ 0 which, in light of (2.6.7), further implies j→∞

v(ψj ) −−−→ 0. Since this contradicts the fact that v(ψj ) = 1 for every j ∈ N, j→∞

the proof is finished. The topology we consider on the dual of E(Ω) is the weak∗-topology, and we denote the resulting topological vector space by E (Ω) (see Sect. 13.1 for more details). A significant byproduct of this set up is singled out next. Fact 2.57. E (Ω) is a locally convex topological vector space over C, which is not metrizable, but is complete. In addition, we have the following important characterization of continuity in E (Ω). Fact 2.58. A sequence {uj }j∈N ⊂ E (Ω) converges to u ∈ E (Ω) as j → ∞ E (Ω)

in E (Ω), something we will indicate by writing uj −−−→ u, if and only if uj , ϕ −−−→ u, ϕ for every ϕ ∈ E(Ω).

j→∞

j→∞

We are now ready to state and prove a result that gives a complete characterization of the class of functionals that are extensions as in Theorem 2.52 of distributions u ∈ D (Ω) with compact support. Theorem 2.59. The spaces Dc (Ω) and E (Ω) are algebraically isomorphic. , where u is the Proof. Consider the mapping ι : Dc (Ω) → E (Ω), ι(u) := u extension of u given by Theorem 2.52 corresponding to F := supp u. Then Msupp u = C ∞ (Ω) and, to conclude that ι is well defined, there remains to show that the functional u is continuous on E(Ω). With this goal in mind, note that while in general the function ψ ∈ C0∞ (Ω) used in the construction of u as in (2.6.3) depends on ϕ, given that we are currently assuming that supp u


42

is compact, we may take ψ = 1 on a neighborhood of supp u (originally we only needed ψ = 1 on a neighborhood of supp u ∩ supp ϕ ⊆ supp u). Then u, ϕ = u, ψϕ for all ϕ ∈ C ∞ (Ω). Now let K0 := supp ψ. It follows that ϕψ ∈ C0∞ (Ω) and supp (ϕψ) ⊆ K0 , for all ϕ ∈ C ∞ (Ω). Fix ϕ ∈ C ∞ (Ω). Since u ∈ D (Ω), corresponding to the compact set K0 there exist k0 ∈ N0 and a finite constant C ≥ 0 such that | u, ϕ| = |u, ψϕ| ≤ C sup |∂ α (ψϕ)|.

(2.6.10)

x∈K0 |α|≤k0

Starting with Leibniz’s formula (13.2.4) applied to ψϕ, we estimate α! α α−β β ∂ |∂ (ψϕ)| = ψ∂ ϕ ≤ C sup |∂ β ϕ(x)|, x∈K0 β≤α β!(α − β)!

(2.6.11)

|β|≤k0

for some finite constant C = C (α, ψ) > 0. Combining (2.6.10) and (2.6.11), we obtain (2.6.12) | u, ϕ| ≤ C · C sup |(∂ β ϕ)(x)|, x∈K0 |β|≤k0

hence u ∈ E (Ω), proving that ι is well-defined. Moving on, it is clear that ι is linear, hence to conclude that it is injective, it suffices to show that if ι(u) = 0 for some u ∈ Dc (Ω), then u = 0. Consider u ∈ Dc (Ω) such that ι(u) = 0. Then by (i) in Theorem 2.52, u = ι(u)C ∞ (Ω) = 0, 0 as desired. Consider now the task of proving that ι is surjective. To get started, let v ∈ E (Ω) be arbitrary, and set u := v C ∞ (Ω) . Clearly u : D(Ω) → C is linear. 0 Since v ∈ E (Ω), Fact 2.55 ensures the existence of a compact set K ⊂ Ω, nonnegative integer k, and finite constant C > 0, such that |v, ϕ| ≤ C sup |∂ α ϕ(x)|,

∀ ϕ ∈ C ∞ (Ω).

(2.6.13)

x∈K |α|≤k

Then, for each compact subset A of Ω and ϕ ∈ C0∞ (Ω) with supp ϕ ⊆ A, by regarding ϕ as being in E(Ω) we may use (2.6.13) to write |u, ϕ| ≤ C sup |∂ α ϕ(x)| = C sup |∂ α ϕ(x)| ≤ C sup |∂ α ϕ(x)|. x∈K |α|≤k

x∈K∩A |α|≤k

(2.6.14)

x∈A |α|≤k

From (2.6.14) we may now conclude (invoking Proposition 2.4) that u ∈ D (Ω). Next, we claim that supp u ⊆ K. Indeed, if the function ϕ ∈ C0∞ (Ω) is such that supp ϕ ∩ K = ∅ then from (2.6.13) we obtain |u, ϕ| = 0, thus u = 0 on the set Ω \ K. Hence, the claim is proved which, in turn, shows that u ∈ Dc (Ω). To finish the proof of the surjectivity of ι, it suffices to show that ι(u)=v. Denote by u K the extension of u given by Theorem 2.52 with F := K. Then


43

reasoning as in the proof of the fact that ι is well-defined, we obtain that K C ∞ (Ω) = u. Also, if u K ∈ E (Ω). By part (i) in Theorem 2.52 it follows that u 0 ϕ ∈ C ∞ (Ω) satisfies the condition supp ϕ ∩ K = ∅, then by (ii) in Theorem 2.52 we have uK , ϕ = 0, while (2.6.13) implies v, ϕ = 0. Hence, the uniqueness result in Theorem 2.52 yields u K = v. On the other hand, since K and supp u are compact, we have MK = C ∞ (Ω) = Msupp u . Now the last conclusion in Theorem 2.52 gives u K = ι(u). Consequently, ι(u) = v, and the surjectivity of ι is proved. This finishes the proof of the theorem. In light of the significance of Dc (Ω), Theorem 2.59 provides a natural algebraic identification E (Ω) = u ∈ D (Ω) : supp u is a compact subset of Ω . (2.6.15) Remark 2.60. The spaces Dc (Ω) and E (Ω) are not topologically isomorphic since there exist sequences of distributions with compact support that converge in D (Ω) but not in E (Ω). For example, take the sequence {δj }j∈N ⊂ D (R) of Dirac distributions with mass at j ∈ N, that have been defined in Example 2.14. Then it is easy to check that the sequence {δj }j converges to 0 in D (R) but not in E (R). Theorem 2.59 nonetheless proves that the identity mapping is well-defined from E (Ω) into D (Ω). Keeping this in mind and relying on (1.3.14) and Proposition 13.3, we see that E (Ω) is continuously embedded into D (Ω).

(2.6.16)

This corresponds to the dual version of (1.3.14). In particular, the operation of restriction to an open subset ω of Ω is a well-defined linear mapping E (Ω) u → uω ∈ D (ω). (2.6.17) Moreover, uω ∈ E (ω) whenever the support of u ∈ E (Ω) is contained in ω. Remark 2.61. (1) In the sequel, we will often drop · from the notation of the extension (as defined in the proof of Theorem 2.59 or Proposition 2.63) of a compactly supported distribution. More precisely, if u ∈ Dc (Ω) we will simply use u for the extension of u to a functional in E (Ω), as well as for its extension to a functional in E (O), where O is an open subset of Rn containing Ω. (2) Whenever necessary, if u ∈ E (Ω), ϕ ∈ C0∞ (Ω), and ψ ∈ C ∞ (Ω), we will use the notation D u, ϕD for the action of u on ϕ as a functional in D (Ω), and the notation E u, ψE for the action of u on ψ as a functional in E (Ω). Proposition 2.62. Let u ∈ D (Ω) and ψ ∈ C0∞ (Ω). Then ψu ∈ E (Ω) and E ψu, ϕE

= D u, ψϕD ,

∀ ϕ ∈ E(Ω).

(2.6.18)


44

Proof. Since ψu ∈ D (Ω) and supp u ⊆ supp ψ, we have ψu ∈ Dc (Ω), thus ψu ∈ E (Ω) (i.e., ψu extends as an element in E (Ω)). Let φ ∈ C0∞ (Ω) be such that φ = 1 in a neighborhood of supp ψ. Then for every ϕ ∈ E(Ω), E ψu, ϕE

= E φψu, ϕE = E ψu, φϕE = D ψu, φϕD = D u, ψφϕD = D u, ψϕD

(2.6.19)

proving (2.6.18). Proposition 2.63. Let ω and Ω be open subsets of Rn such that ω ⊆ Ω. Then every u ∈ E (ω) extends to a functional u ∈ E (Ω) by setting u : E(Ω) → C,

u, ϕ := u, ψϕ,

∀ ϕ ∈ C ∞ (Ω),

(2.6.20)

where ψ ∈ C0∞ (ω) is such that ψ = 1 in a neighborhood of supp u. Proof. We first claim that the mapping in (2.6.20) is well-defined. Indeed, if ψj ∈ C0∞ (ω), ψj = 1 in a neighborhood of supp u, for j = 1, 2, then for each function ϕ ∈ C ∞ (Ω) we have (ψ1 − ψ2 )ϕ ∈ C0∞ (ω) and supp u ∩ supp [(ψ1 − ψ2 )ϕ] = ∅. Consequently, u, (ψ1 − ψ2 )ϕ = 0, so that the definition of u is independent of the choice of ψ with the given properties. The functional defined in (2.6.20) is also linear, its continuity is a consequence of Proposition 2.56 and Exercise 1.21. In addition, if ϕ ∈ C ∞ (ω) is given and if ψ ∈ C0∞ (ω) is such that ψ = 1 in a neighborhood of supp u, then supp [(1 − ψ)ϕ] ∩ supp u = ∅. Hence u, ϕ = u, ψϕ = u, (1 − ψ)ϕ + u, ϕ = u, ϕ,

(2.6.21)

proving that u is an extension of u. Exercise 2.64. In the context of Proposition 2.63 prove that αu ∂αu = ∂*

in D (Ω)

(2.6.22)

for every u ∈ E (ω) and every α ∈ Nn0 . D (Ω)

Exercise 2.65. Let uj −−−−→ u be such that there exists a compact K in Rn that j→∞

is contained in Ω and with the property that supp uj ⊆ K for every j ∈ N. Prove E (Rn )

E (Ω)

j→∞

j→∞

that supp u ⊆ K and uj −−−−→ u. Consequently, we also have uj −−−→ u. Exercise 2.66. Let k ∈ N0 and assume that cα ∈ C for α ∈ Nn0 with |α| ≤ k. Prove that cα ∂ α δ = 0 in D (Rn ) ⇐⇒ each cα = 0. (2.6.23) |α|≤k

Hint: Use (13.2.5).


45

Exercise 2.67. Prove that if u ∈ D (Rn ) and supp u ⊆ {a} for some a ∈ Rn , then u has a unique representation of the form cα ∂ α δa , (2.6.24) u= |α|≤k

for some k ∈ N0 and coefficients cα ∈ C. Sketch of proof: (I) Via a translation, reduce matters to the case a = 0. (II) Use Fact 2.55 to determine k ∈ N0 . (III) Fix ψ ∈ C0∞ (B(0, 1)) such that ψ = 1 on B(0, 12 ) and for ε > 0 define the function ψε (x) := ψ( xε ) for every x ∈ Rn . Prove that u = ψε u in D (Rn ). (IV) For ϕ ∈ C0∞ (Rn ) consider the kth order Taylor polynomial for ϕ at 0, i.e., 1 ϕk (x) := ∂ β ϕ(0) xβ , ∀ x ∈ Rn . (2.6.25) β! |β|≤k

Prove that for α ∈

Nn0

satisfying |α| ≤ k one has

∂ α (ϕ − ϕk ) = ∂ α ϕ − (∂ α ϕ)k−|α| . (V) Show that for each ϕ ∈ C0∞ (Rn ) there exists a constant c ∈ (0, ∞) such that |u, (ϕ − ϕk )ψε | ≤ c ε. (VI) Combine all the above to obtain that u, ϕ =

! (−1)|α| " u, xα ∂ α δ , ϕ α!

∀ ϕ ∈ C0∞ (Rn ).

(2.6.26)

|α|≤k

(VII) Prove that the representation in (VI) is unique. Example 2.68. Let m ∈ N. We are interested in solving the equation xm u = 0 in

D (R).

(2.6.27)

In this regard, assume that u ∈ D (R) solves (2.6.27) and note that if we have ϕ ∈ C0∞ (R \ {0}), then x1m ϕ ∈ C0∞ (R \ {0}). This observation permits us to write ! 1 " ∀ ϕ ∈ C0∞ (R \ {0}), (2.6.28) u, ϕ = xm u, m ϕ = 0, x which proves that supp u ⊆ {0}. In particular, u ∈ E (R). Applying ExerN cise 2.67 we conclude that there exists N ∈ N0 such that u = ck δ (k) in D (R), for some ck ∈ C, k = 0, 1, 2, . . . , N . We claim that c = 0

whenever m ≤ ≤ N.

k=0

(2.6.29)


46

To see why this is true, observe that since u ∈ E (R) it makes sense to apply u to any function in C ∞ (R). In particular, it is meaningful to apply u to any polynomial. Concerning (2.6.29), if N ≤ m − 1 there is nothing to prove, while in the case when N ≥ m for each ∈ {m, . . . , N } we may write 0 = xm u, x−m = u, x =

N

ck δ (k) , x

k=0

=

N

(−1)k ck

k=0

dk (x ) = (−1) ! c . dxk x=0

(2.6.30)

This proves (2.6.29) which, in turn, forces u to have the form u=

m−1

ck δ (k)

for some ck ∈ C, k = 0, 1, . . . , m − 1.

(2.6.31)

k=0

Conversely, one may readily verify that any distribution u as in (2.6.31) solves (2.6.27). In conclusion, any solution u of (2.6.27) is as in (2.6.31). Exercise 2.69. Let m ∈ N, a ∈ R, and u ∈ D (R). Prove u is a solution of the m−1 (k) equation (x − a)m u = 0 in D (R) if and only if u is of the form u = ck δa k=0

for some ck ∈ C, k = 0, 1, . . . , m − 1.

Remark 2.70. You have seen in Example 2.28 that P.V. x1 is a solution of the equation xu = 1 in D (R). Hence, if v ∈ D (R) is another solution of this equation, then the distribution v − P.V. x1 is a solution of the equation xu = 0 in D (R). By Example 2.68, it follows that v − P.V. x1 = c δ, where c ∈ C. Thus, the general solution of the equation xu = 1 in D (R) is u = P.V. x1 + c δ, for c ∈ C. Example 2.71. Let N ∈ N and aj ∈ Rn , j ∈ {1, . . . , N }, be distinct points. If u ∈ D (Rn ) is such that supp u ⊆ {a1 , a2 , . . . , aN }, then u has a unique representation of the form u=

N

cα,j ∂ α δaj ,

kj ∈ N0 ,

cα,j ∈ C.

(2.6.32)

j=1 |α|≤kj

To justify formula (2.6.32), fix a family of pairwise disjoint balls Bj := B(aj , rj ), j ∈ {1, . . . , N }, and for each j select a function ψj ∈ C0∞ (Bj ) satisfying ψj = 1 in a neighborhood of B(aj , rj /2). Then for each j ∈ {1, . . . , N } we have that ψj u ∈ E (Rn ) and supp (ψj u) ⊆ {aj }. Exercise 2.67 then gives that there exist cα,j ∂ α δaj in D (Rn ). In addition, kj ∈ N0 and cα,j ∈ C such that ψj u = since

N

|α|≤kj

ψj = 1 in a neighborhood of supp u, we have

j=1 N ! j=1

N N N " " ! uψj , ϕ = ψj u, ϕ = u, ψj ϕ = u, ϕ ψj = u, ϕ, (2.6.33) j=1

j=1

j=1

2.6. COMPACTLY SUPPORTED DISTRIBUTIONS ... for each ϕ ∈ C0∞ (Rn ). Hence, u =

N

47

ψj u in D (Rn ) which, given (2.6.33),

j=1

proves (2.6.32). Example 2.72. Let a, b ∈ R be such that a = b. We are interested in solving the equation (x − a)(x − b)u = 0 in D (R). (2.6.34) The first observation is that any solution u of this equation satisfies the condition supp u ⊆ {a, b}. Indeed, if we take an arbitrary ϕ ∈ C0∞ (R \ {a, b}), then the 1 ϕ belongs to the space C0∞ (R \ {a, b}) and function (x−a)(x−b) " ! 1 ϕ = 0. (2.6.35) u, ϕ = (x − a)(x − b)u , (x−a)(x−b) Hence, we may apply Example 2.71 to conclude that u=

N1

cj δa(j) +

j=0

N2

(j)

dj δb

in D (R),

(2.6.36)

j=0

where N1 , N2 ∈ N, {cj }0≤j≤N1 ⊂ C and {dj }0≤j≤N2 ⊂ C. Moreover, by dropping terms with zero coefficients, there is no loss of generality in assuming that cN1 = 0 and dN2 = 0.

(2.6.37)

In this scenario, we make the claim that N1 = N2 = 0. To prove this claim, suppose first that N1 ≥ 1. Then, using (2.6.36) and the hypotheses on u, we obtain % $ % $ 0 = (x − a)(x − b)u, (x − a)N1 −1 (x − b)N2 = u, (x − a)N1 (x − b)N2 +1 =

N1

N2 $ (j) % $ (j) % N1 N2 +1 + cj δa , (x − a) (x − b) dj δb , (x − a)N1 (x − b)N2 +1

j=0

=

N1

j=0

(−1)j cj

j=0

+

N2 j=0

dj N1 N2 +1 (x − a) (x − b) dxj x=a

(−1)j dj

dj N1 N2 +1 (x − b) (x − a) dxj x=b

= (−1)N1 cN1 N1 ! (a − b)N2 +1 .

(2.6.38)

Since by assumption a = b, from (2.6.38) we obtain cN1 = 0. This contradicts (2.6.37) and shows that necessarily N1 = 0. Similarly, we obtain that N2 = 0, hence any solution of (2.6.34) has the form u = c δa + d δb

in D (R),

c, d ∈ C.

(2.6.39)

Conversely, it is clear that any distribution as in (2.6.39) solves (2.6.34). To sum up, (2.6.39) describes all solutions of (2.6.34).


48

2.7

Tensor Product of Distributions

Let m, n ∈ N, U be an open subset of Rm , V be an open subset of Rn , and consider two complex-valued functions f ∈ L1loc (U ) and g ∈ L1loc (V ). Then the tensor product of the functions f and g is defined as f ⊗g : U ×V → C,

(f ⊗g)(x, y) := f (x)g(y)

for each (x, y) ∈ U ×V. (2.7.1)

In particular, it follows from (2.7.1) that f ⊗ g ∈ L1loc (U × V ). When f , g, and f ⊗ g are regarded as distributions, for each ϕ ∈ C0∞ (U × V ) we obtain that f ⊗ g, ϕ = f (x)g(y)ϕ(x, y) dx dy = f (x) g(y)ϕ(x, y) dy dx =

U×V

f (x)ϕ(x, y) dx dy

g(y) V

U

V

(2.7.2)

U

or, concisely, $ % $ % f ⊗ g, ϕ = f (x), g(y), ϕ(x, y) = g(y), f (x), ϕ(x, y) .

(2.7.3)

If, in addition, the test function ϕ has the form ϕ1 ⊗ ϕ2 , for some ϕ1 ∈ C0∞ (U ) and ϕ2 ∈ C0∞ (V ), then (2.7.3) becomes f ⊗ g, ϕ1 ⊗ ϕ2 = f, ϕ1 g, ϕ2 .

(2.7.4)

This suggests a natural way to define tensor products of general distributions granted the availability of the following density result for D(U × V ). Proposition 2.73. Let m, n ∈ N, U be an open subset of Rm , and V be an open subset of Rn . Then the set ⎧ ⎫ N ⎨ ⎬ ϕj ⊗ ψj : ϕj ∈ C0∞ (U ), ψj ∈ C0∞ (V ), N ∈ N C0∞ (U ) ⊗ C0∞ (V ) := ⎩ ⎭ j=1

(2.7.5) is sequentially dense in D(U × V ). Before proceeding with the proof of Proposition 2.73 we state and prove two lemmas. Lemma 2.74. Suppose that the sequence {fj }j∈N ⊂ E(Rn ) and f ∈ E(Rn ) are such that ∂ α fj − ∂ α f L∞ (B(0,j)) < E(Rn )

Then fj −−−−→ f . j→∞

1 j

for all α ∈ Nn0 satisfying |α| ≤ j.

(2.7.6)

2.7. TENSOR PRODUCT OF DISTRIBUTIONS

49

Proof. Suppose {fj }j∈N and f satisfy the current hypotheses, and let ε ∈ (0, ∞), α ∈ Nn0 , and a compact subset K of Rn be fixed. Then there exists j0 ∈ N such that K ⊂ B(0, j0 ). If we now fix some j ∗ ∈ N with the property that j ∗ > max{ 1ε , |α|, j0 }, it follows that for each j ≥ j ∗ we have |α| ≤ j ∗ ≤ j and ∂ α fj − ∂ α f L∞ (K) ≤ ∂ α fj − ∂ α f L∞ (B(0,j))
0 define the function |x−y|2 n ft (x) := (4πt)− 2 e− 4t f (y) dy, Rn

∀ x ∈ Rn .

(2.7.8)

The first goal is to prove that E(Rn )

ft −−−−→ f. t→0+

To this end, consider the function u defined by ft (x), x ∈ Rn , t > 0, ∀ (x, t) ∈ Rn × [0, ∞). u(x, t) := f (x), x ∈ Rn , t = 0,

(2.7.9)

(2.7.10)

From definition it is clear that u is continuous on Rn × (0, ∞). We claim that, n in fact, u is continuous √ on R × [0, ∞). Indeed, by making use of the change of variables x − y = 2 tz, we may write √ 2 −n 2 u(x, t) = π e−|z| f (x − 2 tz) dz, ∀ x ∈ Rn , ∀ t > 0. (2.7.11) Rn

Hence, for each x∗ ∈ Rn , Lebesgue dominated convergence theorem (cf. Theorem 13.12) gives 2 −n 2 lim u(x, t) = f (x )π e−|z| dz = f (x∗ ) = u(x∗ , 0), (2.7.12) ∗ x→x ∗

t→0+

Rn

proving that u is continuous at points of the form (x∗ , 0). Being continuous on Rn ×[0, ∞), u is uniformly continuous on every compact subset of Rn × [0, ∞), thus uniformly continuous on sets of the form K × [0, 1], where K ⊂ Rn is compact. Fix such a compact K and fix ε ∈ (0, 1) arbitrary. Then, there exists δ > 0 such that if (x1 , t1 ), (x2 , t2 ) ∈ K × [0, 1] satisfy


50

(x1 , t1 ) − (x2 , t2 ) ≤ δ then u(x1 , t1 ) − u(x2 , t2 ) ≤ ε. In particular, if x ∈ K and t ∈ (0, δ), then u(x, t) − u(x, 0) ≤ ε, that is ft − f L∞ (K) ≤ ε for all 0 < t < δ. This proves that lim+ ft (x) = f (x) uniformly on compact sets in Rn . t→0

The derivatives of ft enjoy the same type of properties as ft . More precisely, √ by a direct computation (involving also the change of variables x − y = 2 tz) we see that for each t > 0 and each α ∈ Nn0 we have |x−y|2 α −n 2 e− 4t (∂ α f )(y) dy, ∀ x ∈ Rn . (2.7.13) ∂x ft (x) = (4πt) Rn

In addition, as before, we obtain that lim (∂xα ft ) = ∂ α f uniformly on compact t→0+

sets in Rn . This completes the proof of (2.7.9). Next, recall that the Taylor expansion of the function ex , x ∈ R, about the ∞ j x origin is ex = j! with the series converging uniformly on compact subsets j=0

of R. In addition, for each N ∈ N, the remainder RN (x) := ex −

N j=0

xj j!

satisfies

N +1

C |RN (x)| ≤ eC (N +1)! whenever |x| ≤ C. Fix t > 0 and a compact subset K of 2

Rn . Then there exists C > 0 such that |x−y| ≤ C for every x ∈ K and every 4t y ∈ supp f , so |x − y|2 C N +1 lim RN − = 0. (2.7.14) ≤ lim eC N →∞ N →∞ 4t (N + 1)! Consequently, n

∂xα ft (x) = (4πt)− 2

j ∞ 1 |x − y|2 ∂ α f (y) dy, − j! 4t n R j=0

(2.7.15)

for each α ∈ Nn0 and each t > 0, and the series in (2.7.15) converges uniformly for x in a compact set in Rn . In addition, integrating by parts, we may write ∞ |x − y|2 j n (−1)|α| f (y) dy, (2.7.16) ∂xα ft (x) = (4πt)− 2 ∂yα − j 4t Rn j=0 where, for each t > 0 fixed, the series in (2.7.16) converges uniformly on compact sets in Rn . Hence, if for each t > 0 we define the sequence of polynomials j k 1 |x − y|2 −n 2 f (y) dy, ∀ x ∈ Rn , ∀ k ∈ N, Pt,k (x) := (4πt) − j! 4t n R j=0 E(Rn )

(2.7.17)

then the above proof implies that for each t > 0 we have Pt,k −−−−→ ft . k→∞

Next, we claim that there exists a sequence of positive numbers {tj }j∈N with the property that for each j ∈ N we have ∂ α ftj − ∂ α f L∞ (B(0,j))
0 and each j ∈ N there exists kt,j ∈ N such that ∂ α Pt,kt,j − ∂ α ft L∞ (B(0,j))
0 and j ∈ N. Since Pt,k −−−−→ ft and B(0, j) is a compact k→∞

subset of Rn , it follows that for each α ∈ Nn0 satisfying |α| ≤ j there exists kα∗ ∈ N such that 1 ∂ α Pt,k − ∂ α ft Lα (B(0,j)) < 2j , for k ≥ kα∗ . (2.7.22) If we now define kt,j := max kα∗ : α ∈ Nn0 , |α| ≤ j , then estimate (2.7.21) holds for this kt,j . This completes the proof the claim. Here is the end-game in the proof of the lemma. For each j ∈ N, let tj > 0 be as constructed above so that (2.7.18) holds, for this tj let ktj ,j be as defined above so that (2.7.21) holds, and set Pj := Ptj ,ktj ,j . Hence, for each j ∈ N and each α ∈ Nn0 satisfying |α| ≤ j we have

∂ α Pj −∂ α f L∞ (B(0,j)) = ∂ α Ptj ,ktj ,j −∂ α f L∞ (B(0,j)) ≤

1 1 2j + 2j

= 1j . (2.7.23)

E(Rn )

The fact that Pj −−−−→ f now follows from (2.7.23) by invoking Lemma 2.74. j→∞

Before turning to the proof of Proposition 2.73 we introduce some notation. For m, n ∈ N, if U is an open subset of Rm , V is an open subset of Rn , and A ⊆ U × V , the projections of A on U and V , respectively, are πU (A) := {x ∈ U : ∃ y ∈ V such that (x, y) ∈ A}, πV (A) := {y ∈ V : ∃ x ∈ U such that (x, y) ∈ A}.

(2.7.24)

We are ready to present the proof of the density result stated at the beginning of this section.


52

Proof of Proposition 2.73. Let ϕ ∈ C0∞ (U × V ). By Lemma 2.75, there exists a E(Rn+m )

sequence of polynomials {Pj }j∈N in Rn+m with the property that Pj −−−−−−→ ϕ. j→∞

Set K := supp ϕ, K1 := πU (K), and K2 := πV (K). Then K1 and K2 are compact sets in Rm and Rn , respectively. Fix a compact set L1 ⊂ U such that K1 ⊂ L˚1 and a compact set L2 ⊂ U such that K2 ⊂ L˚2 . Then there exists a function ϕ1 ∈ C0∞ (U ) with supp ϕ1 ⊆ L1 , ϕ1 = 1 in a neighborhood of K1 , and a function ϕ2 ∈ C0∞ (V ) satisfying supp ϕ2 ⊆ L2 and ϕ2 = 1 in a neighborhood of K2 . Consequently, ϕ1 ⊗ ϕ2 ∈ C0∞ (Rn+m ) and supp (ϕ1 ⊗ ϕ2 ) ⊆ L1 × L2 .

(2.7.25)

By (2) in Exercise 1.10 it follows that E(Rn+m )

(ϕ1 ⊗ ϕ2 )Pj −−−−−−→ (ϕ1 ⊗ ϕ2 )ϕ. j→∞

(2.7.26)

Hence, since supp [(ϕ1 ⊗ ϕ2 )Pj ] ⊆ L1 × L2 for every j ∈ N and since (ϕ1 ⊗ ϕ2 )ϕ = ϕ, we obtain D(U×V )

(ϕ1 ⊗ ϕ2 )Pj −−−−−− → ϕ. j→∞

(2.7.27)

Upon observing that (ϕ1 ⊗ϕ2 )Pj ∈ C0∞ (U )⊗C0∞ (V ) for every j ∈ N, the desired conclusion follows. The next proposition is another important ingredient used to define the tensor product of two distributions. Proposition 2.76. Let m, n ∈ N, U be an open subset of Rm , and V be an open subset of Rn . For each u ∈ D (U ) the following properties hold. (a) If for each ϕ ∈ C0∞ (U × V ) we define the mapping $ % ψ : V → C, ψ(y) := u(x), ϕ(x, y) ∀ y ∈ V,

(2.7.28)

then ψ ∈ C0∞ (V ). (b) The mapping D(U × V ) ϕ → ψ ∈ D(V ), with ψ as defined in (a), is linear and continuous. Remark 2.77. In the definition of ψ in part (a) of Proposition 2.76, the use of the notation u(x) does NOT mean that the distributions u%is evaluated at x since $ the latter is not meaningful. The notation u(x), ϕ(x, y) should be understood in the following sense: for each y ∈ V fixed, the distribution u acts on the function ϕ(·, y).


53

Proof of Proposition 2.76. Fix ϕ ∈ C0∞ (U × V ) and let K := supp ϕ that is a compact subset of U × V . Also, consider ψ as in (2.7.28) and recall the projections πU , πV from (2.7.24). Then clearly supp ψ ⊆ πU (K), thus ψ has compact support. Next we prove that ψ is continuous on V . Let {yj }j∈N be a sequence in V such that lim yj = y0 for some y0 ∈ V . Since u ∈ D (U ), based j→∞

on the definition of ψ and Fact 2.2, in order to conclude that lim ψ(yj ) = j→∞

D(U)

ψ(y0 ) it suffices to show that ϕ(·, yj ) −−−→ ϕ(·, y0 ). It is clear that for every j→∞

j ∈ N we have ϕ(·, yj ) ∈ C0∞ (U ) and supp ϕ(·, yj ) ⊆ πU (K). Moreover, since ϕ ∈ C ∞ (U × V ) it follows that ∂xα ϕ is continuous on K for every α ∈ Nm 0 , thus uniformly continuous on K. Consequently, (∂xα ϕ)(·, y j ) −−−→ (∂xα ϕ)(·, y 0 ) uniformly on πU (K). j→∞

This completes the proof of the fact that ψ is continuous on V . To continue, we claim that ψ is of class C 1 on V . Fix y ∈ V and some j ∈ {1, . . . , n}. Recall that ej is the unit vector in Rn with the jth component equal to 1, and let h ∈ R \ {0}. Since V is open, there exists ε0 > 0 such that if |h| < ε0 then y + hej ∈ V . Make the standing assumption that |h| < ε0 and set Rh (x, y) :=

∂ϕ ϕ(x, y + hej ) − ϕ(x, y) − (x, y), h ∂yj

∀ x ∈ U.

(2.7.29)

Then " ψ(y + hej ) − ψ(y) ! ∂ϕ − u(x), (x, y) = u(x), Rh (x, y), h ∂yj

∀ x ∈ U. (2.7.30)

Suppose lim Rh (·, y) = 0 in D(U ).

h→0

(2.7.31)

% $ Then lim u, Rh (·, y) = 0, which in view of (2.7.30) implies h→0

! ∂ϕ " ∂j ψ(y) = u, (·, y) . ∂yj

(2.7.32)

∂ϕ ∈ C0∞ (U × V ) by reasoning as in the proof of the continuity Moreover, since ∂y j of ψ on V , we also obtain that ∂j ψ is continuous on V . Hence, since j in {1, . . . , n} is arbitrary, to complete the proof of the claim, we are left with showing (2.7.31). Clearly supp [Rh (·, y)] ⊆ πU (K). Applying Taylor’s formula to ϕ in the variable y for each fixed x ∈ U we obtain

∂ϕ (x, y) + h2 ϕ(x, y + hej ) = ϕ(x, y) + h ∂yj

1

(1 − t) 0

∂ 2ϕ (x, y + thej ) dt. (2.7.33) ∂yj2


54

Hence, (2.7.29) and (2.7.33) imply 1 ∂2ϕ (1 − t) 2 (x, y + thej ) dt. Rh (x, y) = h ∂yj 0 Consequently, for every β ∈ Nm 0 , we have 1 ∂2ϕ ∂xβ Rh (x, y) = h (1 − t)∂xβ 2 (x, y + thej ) dt, ∂yj 0

∀ x ∈ U.

(2.7.34)

(2.7.35)

Since the integral in the right-hand side of (2.7.35) is bounded by a constant independent of h, x and y, it follows that lim ∂xβ Rh (·, y) = 0 uniformly on h→0

πU (K). Combined with the support information on Rh (·, y), this implies (2.7.31) and completes the proof of the claim that ψ ∈ C 1 (V ). By induction, we obtain ψ ∈ C ∞ (V ), completing the proof of the statement in part (a) of the proposition. The linearity of the mapping in part (b) is immediate since u is a linear mapping. To show that the mapping in (b) is also continuous, since D(V ) is locally convex, by Theorem 13.5 it suffices to prove that it is sequentially D(U×V )

continuous. To this end, let ϕj −−−−−−→ ϕ. In particular, there exists a compact j→∞

subset K of U × V such that supp ϕj ⊆ K for all j ∈ N and ∂ α ϕj −−−→ ∂ α ϕ j→∞

uniformly on K, for each α ∈ Nm+n . 0

(2.7.36)

$ % To proceed, for each y ∈ V set ψ(y) := u, ϕ(·, y) and ψj (y) := u, ϕj (·, y) for D(V )

each j ∈ N. The goal is to prove that ψj −−−→ ψ. Applying Proposition 2.4 to j→∞

the distribution u and compact πU (K) yields k ∈ N0 and C > 0 for which (2.1.1) holds with K replaced by πU (K). Then, for each β ∈ Nn0 , we have sup y∈πV (K)

β ∂ ψj (y) − ∂ β ψ(y)

=

sup y∈πV (K)

≤ =

u(x), ∂yβ ϕj (·, y) − ∂yβ ϕ(·, y)

sup

sup

y∈πV (K) x∈πU (K) |γ|≤k

C

γ β ∂x ∂y ϕj (x, y) − ∂xγ ∂yβ ϕ(x, y)

sup ∂xγ ∂yβ ϕj (x, y) − ∂xγ ∂yβ ϕ(x, y) −−−→ 0,

(x,y)∈K |γ|≤k

j→∞

(2.7.37)

where γ ∈ Nm 0 and the convergence to zero in (2.7.37) is due to (2.7.36) applied D(V )

for α := (γ, β). Thus, ψj −−−→ ψ and the proof of the statement in part (b) is j→∞

complete. We are now ready to define the tensor product of distributions. Theorem 2.78. Let m, n ∈ N, U be an open subset of Rm , and V be an open subset of Rn . Consider u ∈ D (U ) and v ∈ D (V ). Then the following statements are true.


55

(i) There exists a unique distribution u ⊗ v ∈ D (U × V ), called the tensor product of u and v, with the property that % $ u ⊗ v, ϕ1 ⊗ ϕ2 = u, ϕ1 v, ϕ2 , ∀ ϕ1 ∈ C0∞ (U ), ∀ ϕ2 ∈ C0∞ (V ). (2.7.38) (ii) The tensor product just defined satisfies u ⊗ v = v ⊗ u in D (U × V ). Proof. For each ϕ ∈ C0∞ (U × V ) consider the function ψ(y) := u(x), ϕ(x, y)

for y ∈ V.

By Proposition 2.76, we have ψ ∈ C0∞ (V ) and the mapping D(U × V ) ϕ → ψ ∈ D(V )

is linear and continuous.

(2.7.39)

Hence, v, ψ is meaningful and we may define u ⊗ v : D(U × V ) −→ C " u ⊗ v, ϕ := v(y), u(x), ϕ(x, y) for every ϕ ∈ C0∞ (U × V ). !

(2.7.40)

As defined, the mapping u ⊗ v is the composition of two linear and continuous mappings, linear and continuous. In addition, if ϕ1 ∈ C0∞ (U ) and ϕ2 ∈ C0∞ (V ), then ! " u ⊗ v, ϕ1 ⊗ ϕ2 = v(y) , u(x), ϕ1 (x)ϕ2 (y) " ! = v(y) , ϕ2 (y)u(x), ϕ1 (x) $ %$ % = v(y), ϕ2 (y) u(x), ϕ1 (x) , (2.7.41) thus the mapping u ⊗ v defined in (2.7.40) satisfies (2.7.38). To prove the uniqueness statement in part (i), suppose w1 , w2 ∈ D (U × V ) are such that wj , ϕ1 ⊗ ϕ2 = u, ϕ1 v, ϕ2 , j = 1, 2, (2.7.42) for every ϕ1 ∈ C0∞ (U ), ϕ2 ∈ C0∞ (V ). Then it follows that w1 , ϕ = w2 , ϕ for every ϕ ∈ C0∞ (U ) ⊗ C0∞ (V ), which in concert with Proposition 2.73 and the continuity of w1 and w2 implies w1 = w2 in D (U × V ). This completes the proof of the statement in (i). As for the statement in (ii), observe that based on (2.7.40) we have v ⊗ u : D(U × V ) → C and $ % v ⊗ u, ϕ = u(x), v(y), ϕ(x, y) for every ϕ ∈ C0∞ (U × V ).

(2.7.43)

In particular, for every ϕ1 ∈ C0∞ (U ) and ϕ2 ∈ C0∞ (U ) we have v ⊗ u, ϕ1 ⊗ ϕ2 = u, ϕ1 v, ϕ2 = u ⊗ v, ϕ1 ⊗ ϕ2 .

(2.7.44)

The uniqueness result from part (i) now implies u ⊗ v = v ⊗ u in D (U × V ).


56

Remark 2.79. If u ∈ D (U ) and v ∈ L1loc (V ), then the statement in part (ii) in Theorem 2.78 becomes " ! $ % v(y)ϕ(x, y) dy = v(y) u(x), ϕ(x, y) dy, ∀ ϕ ∈ C0∞ (U × V ). u(x), V

V

(2.7.45) The interpretation of (2.7.45) is that the distribution u commutes with the integral. We next establish a number of basic properties for the tensor products of distributions. Theorem 2.80. Let m, n ∈ N, U be an open subset of Rm , and V be an open subset of Rn . Assume that u ∈ D (U ) and v ∈ D (V ). Then the following properties hold. (a) supp u ⊗ v = supp u × supp v. n (b) ∂xα ∂yβ (u ⊗ v) = (∂xα u) ⊗ (∂yβ v) for every α ∈ Nm 0 and every β ∈ N0 .

(c) (f ⊗ g) · (u ⊗ v) = (f u) ⊗ (gv) for every f ∈ C ∞ (U ) and every g ∈ C ∞ (V ). (d) The mapping D (U ) × D (V ) (u, v) → u ⊗ v ∈ D (U × V ) is bilinear and separately continuous. (e) The tensor product of distributions is associative. Proof. We start by proving the set theoretic equality from (a). For the leftto-right inclusion, fix (x0 , y0 ) ∈ supp u × supp v. If C ⊆ U × V is an open neighborhood of (x0 , y0 ), then there exist an open set A ⊆ U containing x0 and an open set B ⊆ V containing y0 such that A × B ⊂ C. In particular, since x0 ∈ supp u and y0 ∈ supp v, there exist ϕ1 ∈ C0∞ (A) and ϕ2 ∈ C0∞ (B) with the property that u, ϕ1 = 0 and v, ϕ2 = 0. If we now set ϕ := ϕ1 ⊗ ϕ2 , then ϕ ∈ C0∞ (C) and u ⊗ v, ϕ = u, ϕ1 v, ϕ2 = 0. Hence (x0 , y0 ) ∈ supp (u ⊗ v), finishing the proof of the left-to-right inclusion in (a). To prove the opposite inclusion, observe that supp (u ⊗ v) ⊆ supp u × supp v is equivalent to (U × V ) \ (supp u × supp v) ⊆ (U × V ) \ supp (u ⊗ v).

(2.7.46)

Write the left hand-side of (2.7.46) as D1 ∪D2 , where D1 := (U \supp u)×V and D2 := U ×(V \supp v). Note that D1 and D2 are open sets in Rm ×Rn . Since the support of a distribution is the smallest relatively closed set outside of which the distribution vanishes, for (2.7.46) to hold it suffices to show that u ⊗ v, ϕ = 0 for every ϕ ∈ C0∞ (D1 ∪ D2 ). Fix such a function ϕ, set K := supp ϕ, and consider a partition of unity subordinate to the covering {D1 , D2 } of K, say ψj ∈ C0∞ (Dj ), j ∈ {1, 2},

ψ1 + ψ2 = 1 in a neighborhood of K.

(2.7.47)


57

Then ϕψ1 ∈ C0∞ (D1 ), ϕψ2 ∈ C0∞ (D2 ) (with the understanding that ψ1 and ψ2 have been extended by zero outside their supports), and ϕ = ϕψ1 + ϕψ2 on U × V . Since πU (D1 ) ∩ supp u = ∅ and πV (D2 ) ∩ supp v = ∅, we may write $ % u ⊗ v, ϕ = v(y), u(x), ϕ(x, y)ψ1 (x, y) % $ (2.7.48) + u(x), v(y), ϕ(x, y)ψ2 (x, y) = 0. This completes the proof of the equality of sets from part (a). n ∞ To prove the identity in (b), fix α ∈ Nm 0 , β ∈ N0 , and let ϕ1 ∈ C0 (U ), ∞ ϕ2 ∈ C0 (V ). Then starting with the definition of distributional derivatives and then using (2.7.38) we may write % % $ $ α β ∂x ∂y (u ⊗ v), ϕ1 ⊗ ϕ2 = (−1)|α|+|β| u ⊗ v, ∂xα ∂yβ (ϕ1 ⊗ ϕ2 ) = (−1)|α|+|β|u ⊗ v, ∂xα ϕ1 ⊗ ∂yβ ϕ2 = (−1)|α|+|β|u, ∂xα ϕ1 v, ∂yβ ϕ2 = ∂xα u, ϕ1 ∂yβ u, ϕ2 $ % = (∂xα u) ⊗ (ϕβy v), ϕ1 ⊗ ϕ2 .

(2.7.49)

By the uniqueness statement in part (i) of Theorem 2.78 we deduce from (2.7.49) that (∂xα u) ⊗ (ϕβy v) = ∂xα ∂yβ (u ⊗ v), completing the proof of the identity in (b). Moving on to the proof of the statement in (c), note that if f ∈ C ∞ (U ) and g ∈ C ∞ (V ), then f ⊗ g ∈ C ∞ (U × V ). The latter, combined with the definition of multiplication of a distribution with a smooth function and (2.7.38), permits us to write % $ (f ⊗ g) · (u ⊗ v), ϕ1 ⊗ ϕ2 = u ⊗ v, (f ⊗ g) · (ϕ1 ⊗ ϕ2 ) (2.7.50) $ % = u ⊗ v, (f ϕ1 ) ⊗ (gϕ2 ) = u, f ϕ1 v, gϕ2 % $ = f u, ϕ1 gv, ϕ2 = (f u) ⊗ (gv), ϕ1 ⊗ ϕ2 , for every ϕ1 ∈ C0∞ (U ), ϕ2 ∈ C0∞ (V ). The identity in (c) now follows from (2.7.50) by once again invoking the uniqueness result from part (i) in Theorem 2.78. The bilinearity of the mapping (u, v) → u ⊗ v is a consequence of the definition of u ⊗ v and (2.2.1). To prove that this mapping is also separately D (U)

continuous, let uj −−−−→ u and fix v ∈ D (V ). If ϕ ∈ C0∞ (U × V ), then j→∞

v(y), ϕ(x, y) ∈ C0∞ (U ) by Proposition 2.76, and we may use Fact 2.18 to write uj ⊗ v, ϕ = uj (x), v(y), ϕ(x, y) −−−→ u(x), v(y), ϕ(x, y) = u ⊗ v, ϕ. j→∞

(2.7.51)

D (v)

D (U×V )

j→∞

j→∞

Similarly, if u ∈ D (U ) is fixed and vj −−−→ v, then u ⊗ vj −−−−−−→ u ⊗ v. Upon recalling that the topological vector space D(U × V ) is locally convex, Theorem 13.5 may now be invoked to finish the proof of (d).


58

Finally, we are left with proving the associativity of the tensor product of distributions. To this end, let k ∈ N, W be an open subset of Rk , and w ∈ D (W ) be arbitrary. By Theorem 2.78, u ⊗ v ∈ D (U × V ), v ⊗ w ∈ D (V × W ) and, furthermore, (u ⊗ v) ⊗ w ∈ D (U × V × W ),

u ⊗ (v ⊗ w) ∈ D (U × V × W ). (2.7.52)

The goal is to prove that (u ⊗ v) ⊗ w = u ⊗ (v ⊗ w)

in

D (U × V × W ).

(2.7.53)

In this regard we first note that, for each ϕ ∈ C0∞ (U ), ψ ∈ C0∞ (V ), and η ∈ C0∞ (W ), we may write $

% (u ⊗ v) ⊗ w, (ϕ ⊗ ψ) ⊗ η = u, ϕv, ψw, η $ % = u ⊗ (v ⊗ w), ϕ ⊗ (ϕ ⊗ η) .

(2.7.54)

Define C0∞ (U ) ⊗ C0∞ (V ) ⊗ C0∞ (W ) as ⎧ ⎫ N ⎨ ⎬ ϕj ⊗ ψj ⊗ ηj : ϕj ∈ C0∞ (U ), ψj ∈ C0∞ (V ), ηj ∈ C0∞ (W ), N ∈ N , ⎩ ⎭ j=1

(2.7.55) and note that this set is sequentially dense in D(U × V × W ) (which can be proved by reasoning as in the proof of Proposition 2.73). Granted this, (2.7.53) is implied by (2.7.54), completing the proof of the theorem. Exercise 2.81. Let n, m ∈ N and x0 ∈ Rn , y0 ∈ Rm . Prove that δx0 ⊗ δy0 = δ(x0 ,y0 ) in D (Rn+m ). We close this section by revisiting the result proved in Proposition 2.76 and establishing a related version that is going to be useful later on. Proposition 2.82. Let m, n ∈ N, U be an open subset of Rm , and V be an open subset of Rn . Assume that u ∈ E (U ), ϕ ∈ C ∞ (U × V ), and define the function ψ : V → C,

ψ(y) := u(x), ϕ(x, y),

∀ y ∈ V.

(2.7.56)

Then ψ ∈ C ∞ (V ) and for every α ∈ Nn0 we have $ % ∂ α ψ(y) = u(x) , ∂yα ϕ(x, y) ,

∀ y ∈ V.

(2.7.57)

Proof. Fix some η ∈ C0∞ (U ) that satisfies η = 1 in a neighborhood of supp u. Then for each θ ∈ C0∞ (V ) we may write

2.8. THE CONVOLUTION OF DISTRIBUTIONS IN RN

59

$ % $ % (θψ)(y) = (ηu)(x), θ(y)ϕ(x, y) = u(x), (η ⊗ θ)(x, y)ϕ(x, y) $ % = u(x), (η ⊗ θ)ϕ (x, y) , ∀ y ∈ V. (2.7.58) Given that (η ⊗ θ)ϕ ∈ C0∞ (U × V ), Proposition 2.76 applies and gives that the right-most side of (2.7.58) depends in a C ∞ manner on the variable y ∈ V . Hence, θψ ∈ C ∞ (V ) and since θ ∈ C0∞ (V ) has been arbitrarily chosen we deduce that ψ ∈ C ∞ (V ). This takes care of the first claim in the statement of the proposition. As regards (2.7.57), observe that it suffices to prove this formula in the case |α| = 1, since the general case then follows by iteration. With this in mind, fix j ∈ {1, . . . , n} and pick an arbitrary point y ∗ ∈ V . Also, select some function θ ∈ C0∞ (V ) such that θ = 1 near y ∗ . These properties of θ permit us to compute ∂yj (η ⊗ θ)ϕ (x, y ∗ ) = η(x) (∂j θ)(y ∗ )ϕ(x, y ∗ ) + θ(y ∗ )(∂yj ϕ)(x, y ∗ ) = η(x)(∂yj ϕ)(x, y ∗ ),

∀ x ∈ U.

Making use of (2.7.32), (2.7.58), and (2.7.59), we may then write $ % ∂j ψ(y ∗ ) = ∂j (θψ)(y ∗ ) = u(x), ∂yj (η ⊗ θ)ϕ (x, y ∗ ) $ % = u(x), η(x)(∂yj ϕ)(x, y ∗ ) $ % = u(x), (∂yj ϕ)(x, y ∗ ) .

(2.7.59)

(2.7.60)

This corresponds precisely to formula (2.7.57) written at the point y = y ∗ and for the multi-index α = (0, . . . , 0, 1, 0, . . . , 0) ∈ Nn0 with the nonzero component on the jth slot. As remarked earlier, this suffices to finish the proof.

2.8

The Convolution of Distributions in Rn

Recall that if f, g ∈ L1 (Rn ) then, as a consequence of Fubini’s theorem, the function h : Rn × Rn → C, defined by h(x, y) := f (x − y)g(y) for every (x, y) ∈ Rn × Rn , is absolutely integrable on Rn × Rn and |h(x, y)| dx dy = |f (x − y)g(y)| ddx dy Rn ×Rn

Rn

=

Rn

Rn

|g(y)|

Rn

|f (x − y)| dx dy

= f L1 (Rn ) gL1(Rn ) . Hence, the convolution of f and g defined as n f ∗ g : R → C, (f ∗ g)(x) := f (x − y)g(y) dy Rn

(2.8.1)

for each x ∈ Rn , (2.8.2)

60


satisfies f ∗ g ∈ L1 (Rn ) (and a natural estimate). We would like to extend this definition to functions that are not necessarily in L1 (Rn ). Specifically, assume that f, g ∈ L1loc (Rn ) have the property that MB(0,r) := (x, y) ∈ supp f × supp g : x + y ∈ B(0, r) (2.8.3) is a compact set in Rn × Rn for every r ∈ (0, ∞). In this scenario, consider the function G : Rn → [0, ∞] defined by |f (x − y)| |g(y)| dy for each x ∈ Rn . G(x) :=

(2.8.4)

Rn

Note that for every r ∈ (0, ∞), by making a natural change of variables and using the fact that f ⊗ g ∈ L1loc (Rn × Rn ), we obtain G(x) dx = |f (z)||g(y)| dy dz < ∞. (2.8.5) |x|≤r

MB(0,r)

Thus, the function G is locally integrable, hence finite almost everywhere in Rn . In conclusion, whenever f, g ∈ L1loc (Rn ) satisfy (2.8.3), for almost every x ∈ Rn the integral (f ∗ g)(x) := Rn f (x − y)g(y) dy is absolutely convergent and f ∗ g ∈ L1loc (Rn ). Furthermore, having fixed an arbitrary ϕ ∈ C0∞ (Rn ) we may write f (x − y)g(y)ϕ(x) dy dx f ∗ g, ϕ = Rn

Rn

f (z)g(y)ϕ(z + y) dy dz.

= Rn

(2.8.6)

Rn

To proceed, observe that the function ϕΔ defined by ϕΔ : Rn × Rn → C,

ϕΔ (x, y) := ϕ(x + y) for every x, y ∈ Rn , (2.8.7)

satisfies ϕΔ ∈ C ∞ (Rn × Rn ) though, in general, the support of ϕΔ is not compact. Formally, the last double integral in (2.8.6) has the same expression as f ⊗ g, ϕΔ . However, under the current assumptions on ϕ, f , and g, it is not clear that this pairing may be interpreted in the standard distributional sense. Indeed, even though f ⊗ g is a well-defined distribution in D (Rn × Rn ) (cf. Theorem 2.78), the function ϕΔ does not belong to C0∞ (Rn × Rn ), as it lacks the compact support property. Nonetheless, (2.8.3) implies supp ϕΔ ∩ supp (f ⊗ g) is a compact set in Rn × Rn .

(2.8.8)

Theorem 2.52 applies with F := supp (f ⊗ g) and allows us to uniquely extend the action of the distribution f ⊗ g to the set of functions ψ ∈ C ∞ (Rn × Rn ) satisfying the property that supp ψ ∩ supp (f ⊗ g) is a compact set in Rn × Rn . Denote this unique extension by f ⊗ g. Then f ⊗ g, ϕΔ is well-defined, and it Δ is meaningful to set f ∗ g, ϕ := f ⊗ g, ϕ . This discussion justifies making the following definition.


61

Definition 2.83. Suppose u, v ∈ D (Rn ) are such that for each compact subset K of Rn the set MK := {(x, y) ∈ supp u × supp v : x + y ∈ K} is compact in Rn × Rn . (2.8.9) Granted this, define the convolution of the distributions u and v as the functional u ∗ v : D(Rn ) → C whose action on each ϕ ∈ C0∞ (Rn ) is given by u ∗ v, ϕ := u ⊗ v, ϕΔ

(2.8.10)

where ϕΔ (x, y) := ϕ(x+y) for every x, y ∈ Rn , and u ⊗ v is the unique extension of u ⊗ v obtained by applying Theorem 2.52 with F := supp (u ⊗ v). Remark 2.84. Retain the context of Definition 2.83. (1) If ϕ ∈ C0∞ (Rn ) and ψ ∈ C0∞ (Rn × Rn ) is such that ψ = 1 on a neighborhood of Msupp ϕ then % $ (2.8.11) u ∗ v, ϕ = u ⊗ v, ψϕΔ . (2) If (2.8.9) holds for the compacts Kj = B(0, j), j ∈ N, then (2.8.9) holds for arbitrary compact sets K ⊂ Rn . (3) Condition (2.8.9) is always satisfied if either u or v is compactly supported. The issue of continuity of the convolution map introduced in Definition 2.83 is discussed next. Theorem 2.85. If u, v ∈ D (Rn ) are such that (2.8.9) holds, then one has u ∗ v ∈ D (Rn ). In particular, the convolution between two distributions in Rn , one of which is compactly supported, is always well-defined and is a distribution in Rn . Proof. Let u, v ∈ D (Rn ) be such that (2.8.9) is satisfied. From Theorem 2.52, D(Rn )

we have that u ⊗ v is linear, hence u ∗ v is linear as well. Let ϕj −−−−→ 0. j→∞

Then there exists a compact subset K of Rn such that supp ϕj ⊆ K for each j ∈ N, and lim ∂ α ϕj = 0 uniformly on K, for every α ∈ Nn0 . In particular, j→∞

Msupp ϕj ⊆ MK for every j ∈ N. Hence, if we fix ψ ∈ C0∞ (Rn × Rn ) such that ψ = 1 in a neighborhood of MK , then part (1) in Remark 2.84 gives % $ ∀ j ∈ N. (2.8.12) u ∗ v, ϕj = u ⊗ v, ψϕΔ j , Moreover, we claim that

D(Rn ×Rn )

ψ ϕΔ −−−−−−→ 0. j − j→∞

(2.8.13)

To prove this claim, note that supp (ψ ϕΔ j ) ⊆ supp ψ for each j ∈ N and for each α1 , α2 , β1 , β2 ∈ Nn0 we have α1 β1 sup (∂x ∂y ψ)(x, y)(∂xα2 ∂yβ2 ϕΔ j )(x, y) (x,y)∈supp ψ


62 ≤

α β ∂x 1 ∂y 1 ψ(x, y)∂ α2 +β2 ϕj L∞ (Rn )

sup (x,y)∈supp ψ

≤ ∂xα1 ∂yβ1 ψL∞ (supp ψ) ∂ α2 +β2 ϕj L∞ (K) −−−→ 0. j→∞

(2.8.14)

Hence, (2.8.13) follows by combining (2.8.14) with Leibniz’s formula (13.2.4). Since u ⊗ v ∈ D (Rn × Rn ), from (2.8.12) and (2.8.13) we deduce % $ −−−→ 0. (2.8.15) u ∗ v, ϕj = u ⊗ v, ψϕΔ j j→∞

On account of Remark 2.3, this proves that u ∗ v ∈ D (Rn ). Finally, the last statement of the theorem is a consequence of what we proved so far and part (3) in Remark 2.84. Remark 2.86. A combination of Theorem 2.85 and the discussion regarding (2.8.6) yields the following result. If f, g ∈ L1loc (Rn ) are such that for each compact K ⊂ Rn the set (x, y) : x ∈ supp f, y ∈ supp g, x + y ∈ K is compact, then uf ∗ ug = uf ∗g in D (Rn ). That is, f ∗ g ∈ L1loc (Rn ), the convolution between the distributions uf and ug [recall (2.1.6)] is well-defined, and uf ∗ ug is a distribution of function type that is equal to the distribution uf ∗g . In particular, ⎫ f ∈ L1loc (Rn ), g ∈ L1comp(Rn ) ⎪ ⎬ or =⇒ uf ∗ ug = uf ∗g . (2.8.16) ⎪ 1 n 1 n ⎭ f ∈ Lcomp(R ), g ∈ Lloc (R ) The main properties of the convolution of distributions, whenever meaningfully defined, are stated and proved in the next theorem. Recall that for any A, B ⊆ Rn the set A ± B is defined as {x ± y : x ∈ A, y ∈ B}. Theorem 2.87. The following statements are true. (a) If u, v ∈ D (Rn ) are such that (2.8.9) is satisfied, then supp (u ∗ v) ⊆ supp u + supp v.

(2.8.17)

(b) If u, v ∈ D (Rn ) are such that (2.8.9) is satisfied, then u ∗ v = v ∗ u. (c) If u, v, w ∈ D (Rn ) are such that ⎧ n ⎪ ⎨ for each compact subset K of R the set := {(x, y, z) ∈ supp u × supp v × supp w : x + y + z ∈ K} (2.8.18) MK ⎪ ⎩ is compact in Rn × Rn × Rn , then (u ∗ v) ∗ w and u ∗ (v ∗ w) are well-defined, belong to D (Rn ), and are equal.


63

(d) Let u ∈ D (Rn ), α ∈ Nn0 . Then (∂ α u) ∗ δ = ∂ α u = u ∗ ∂ α δ. In particular, u ∗ δ = u. (e) If the distributions u, v ∈ D (Rn ) are such that (2.8.9) is satisfied and α ∈ Nn0 , then ∂ α (u ∗ v) = (∂ α u) ∗ v = u ∗ (∂ α v). Proof. Let u, v ∈ D (Rn ) be such that (2.8.9) holds. Since supp u + supp v is closed, the inclusion in (a) will follow as soon as we show that u ∗ v Rn \(supp u+supp v) = 0. Pick some arbitrary function ϕ ∈ D(Rn \ (supp u + supp v)). Then necessarily supp ϕΔ ∩ supp (u ⊗ v) = ∅, hence u ⊗ v, ϕΔ = 0 which, in turn, implies u ∗ v, ϕ = 0, as wanted. The statement in (b) is an immediate consequence of Definition 2.83 and the fact that the tensor product is commutative (see (ii) in Theorem 2.78). To prove the statement in (c), suppose u, v, w ∈ D (Rn ) satisfy (2.8.18). Define the functional u ∗ v ∗ w : D(Rn ) → C by setting $ % v ⊗ w, ϕΔ , u ∗ v ∗ w, ϕ := u ⊗ (2.8.19) for each ϕ ∈ C0∞ (Rn ), v⊗w where ϕΔ (x, y, z) := ϕ(x + y + z) for each x, y, x ∈ Rn , and where u ⊗ is the unique extension of u ⊗ v ⊗ w obtained by applying Theorem 2.52 for the set F := supp (u ⊗ v ⊗ w). The mapping in (2.8.19) is well-defined since if ϕ ∈ C0∞ (Rn ) then ϕΔ ∈ C ∞ (Rn × Rn × Rn ) and, based on (2.8.18), the set

Δ Msupp ϕ = supp (u ⊗ v ⊗ w) ∩ supp ϕ

is compact in Rn × Rn × Rn . (2.8.20)

Reasoning as in the proof of Theorem 2.85, it follows that u ∗ v ∗ w ∈ D (Rn ) and $ % for ϕ ∈ C0∞ (Rn ) and for each u ∗ v ∗ w, ϕ = u ⊗ v ⊗ w, ψϕΔ (2.8.21) ψ ∈ C0∞ (Rn × Rn × Rn ) with ψ = 1 in a neighborhood of Msupp ϕ. Given the freedom in selecting ψ as in (2.8.21), we choose to take ψ as foln lows. Let πj : Rn × Rn × Rn → R , j = 1, 2, 3, be the projections defined by π1 (x, y, z) := x, π2 (x, y, z) := y, π3 (x, y, z) := z, for all x, y, z ∈ Rn . Given ϕ ∈ C0∞ (Rn ), fix ψj ∈ C0∞ (Rn ) with ψj = 1 near πj (Msupp ϕ ), j = 1, 2, 3,

then choose

(2.8.22)

(2.8.23) ψ := ψ1 ⊗ ψ2 ⊗ ψ3 ∈ C0∞ (Rn × Rn × Rn ). The next two claims are designed to proving that u ∗ (v ∗ w) exists.

Claim 1. For every compact K in Rn the set NK := {(y, z) ∈ supp v × supp w : y + z ∈ K} is compact in Rn × Rn .


64

To see why this is true, start by observing that, for every x0 ∈ supp u, the set K + x0 is compact in Rn and B := {(x, y, z) ∈ {x0 } × supp v × supp w : x + y + z ∈ K + x0 }

(2.8.24)

. Since (2.8.18) ensures that MK+x is compact is closed and contained in MK+x 0 0 in Rn × Rn × Rn , it follows that B is compact in Rn × Rn × Rn . In addition, the mapping θ : Rn × Rn × Rn → Rn × Rn , defined by θ(x, y, z) := (y, z) for x, y, z ∈ Rn , is continuous and θ(B) = NK . Therefore, NK must be compact, and Claim 1 is proved. The latter ensures that v ∗ w exists.

Claim 2. For every K ⊂ Rn compact, the set PK := {(x, z) ∈ supp u × supp (v ∗ w) : x + z ∈ K} is compact in Rn × Rn . By part (a) in the theorem, we have supp (v ∗ w) ⊆ supp v + sup w. Thus PK ⊆ {(x, z) ∈ supp u × (supp v + supp w) : x + z ∈ K} = (x, y + t) : x ∈ supp u, y ∈ supp v, t ∈ supp w, x + y + t ∈ K (2.8.25) and the last set in (2.8.25) is closed in Rn × Rn . If we now set σ : Rn × Rn × Rn → Rn × Rn , σ(x, y, t) := (x, y + t) for every x, y, t ∈ Rn ,

(2.8.26)

then σ is continuous and PK ⊆ σ(MK ). By (2.8.18), the set MK is compact in n n n n n R × R × R , hence σ(MK ) is compact in R × R . The set PK being closed in Rn × Rn , we may conclude that PK is compact in Rn × Rn . This proves Claim 2 and, as a consequence, the fact that u ∗ (v ∗ w) exists.

With an eye toward proving u ∗ (v ∗ w) = u ∗ v ∗ w, we dispense with two more claims. Claim 3. For each K ⊂ Rn compact, A := (supp v + supp w) ∩ (K − supp u) is a compact set in Rn . Rewrite A as A = {t ∈ supp v + supp w : t = ω − x, for some ω ∈ K, x ∈ supp u}. Then if t ∈ A, it follows that there exist y ∈ supp v, z ∈ supp w, ω ∈ K and x ∈ supp u such that t = y+z = ω−x. Hence, (x, y, z) ∈ supp u×supp v×supp w and A ⊆ ν(MK ), where and x + y + z = ω, which implies that (x, y, z) ∈ MK ν : Rn × Rn × Rn → Rn , ν(x1 , x2 , x3 ) := x2 + x3

for every x1 , x2 , x3 ∈ Rn .

(2.8.27)


65

We may now conclude that A is compact since the map ν is continuous, MK is compact and A is closed. This proves Claim 3.

Claim 4. Fix ϕ ∈ C0∞ (Rn ) and set K := supp ϕ. Also, let A be as in Claim 3 corresponding to this K, and suppose η ∈ C0∞ (Rn ) is such that η = 1 in a neighborhood of A. Then, with ψ1 as in (2.8.22), we have ψ1 ⊗ η = 1

on

(supp u × supp (v ∗ w)) ∩ supp ϕΔ

(2.8.28)

where, as before, ϕΔ (x, y) = ϕ(x + y) for x, y ∈ Rn . ) and η = 1 on A, it suffices to show To prove this claim, since ψ1 = 1 on π1 (MK that supp u × supp (v ∗ w) ∩ supp ϕΔ ⊆ π1 (MK ) × A. (2.8.29)

To justify the latter inclusion, let x ∈ supp u, y ∈ supp v, and z ∈ supp w, be such that (x, y + z) ∈ supp ϕΔ . Then x + y + z ∈ K which forces x ∈ π1 (MK ) as well as y + z ∈ K − supp u ⊆ A. Thus, (x, y + z) ∈ π1 (MK ) × A and this completes the proof of Claim 4. Consider now an arbitrary function ϕ ∈ C0∞ (Rn ) and set K := supp ϕ. Also, assume that ψ1 , ψ2 , ψ3 are as in (2.8.22), and let η be as in Claim 4. Making use of the definition of the convolution and tensor products, and keeping in mind (2.8.28), we may write " $ % ! (2.8.30) u ∗ (v ∗ w), ϕ = u(x) ⊗ (v ∗ w)(t), ψ1 (x)η(t)ϕ(x + t) " ! $ % = u(x), (v ∗ w)(t), ψ1 (x)η(t)ϕ(x + t) ! $ %" = u(x), v(y) ⊗ w(z), ψ1 (x)η(y + z)ϕ(x + y + z)ψ2 (y)ψ3 (z) . A few words explaining the origin of the last equality are in order. According to the definition of the convolution, passing from v ∗ w to v ⊗ w requires that we consider the set C := (supp v × supp w) ∩ supp η Δ ∩ supp[ϕ(x + ·)]Δ .

(2.8.31)

) × π3 (MK ), it follows that C is Since C is closed and satisfies C ⊂ π2 (MK compact. Now, the fact that ψ2 ⊗ ψ3 = 1 in a neighborhood of C justifies the presence of ψ2 ⊗ ψ3 in the last term in (2.8.30). Going further, since ψ1 (x)ψ2 (y)ψ3 (z) = ψ1 (x)ψ2 (y)ψ3 (z)η(y + z) for (x, y, z) near MK , (2.8.32)

referring to (2.8.21) and (2.8.23) allows us to rewrite (2.8.30) in the form ! %" $ u ∗ (v ∗ w), ϕ = u(x), v(y) ⊗ w(z), ψ1 (x)ψ2 (y)ψ3 (z)ϕ(x + y + z) " ! = u ⊗ v ⊗ w, (ψ1 ⊗ ψ2 ⊗ ψ3 )ϕΔ = u ∗ v ∗ w, ϕ. (2.8.33)

66


Since ϕ ∈ C0∞ (Rn ) was arbitrary, it follows that u ∗ (v ∗ w) = u ∗ v ∗ w. Similarly, it can be seen that (u ∗ v) ∗ w = u ∗ v ∗ w and this completes the proof of the statement in (c). Moving on to (d), fix u ∈ D (Rn ) and α ∈ Nn0 . Since the Dirac distribution δ has compact support, by Theorem 2.85 it follows that both δ ∗ u and (∂ α δ) ∗ u are well-defined and belong to D (Rn ). Let ϕ ∈ C0∞ (Rn ) be arbitrary and consider ψ ∈ C0∞ (Rn × Rn ) such that ψ = 1 on a neighborhood of the set ({0} × supp u) ∩ supp ϕΔ . Starting with Definition 2.83, then using (2.4.1) combined with Proposition 2.32, then Leibniz’s formula (13.2.4), then (2.1.18), the support condition for ψ and then (2.4.1) again, we may write $ α % $ % ∂ δ ∗ u, ϕ = (∂xα δ) ⊗ u(y), ψ(x, y)ϕ(x + y) ! %" $ = u(y), ∂ α δ(x), ψ(x, y)ϕ(x + y) ! $ %" = (−1)|α| u(y), δ(x), ∂xα (ψ(x, y)ϕ(x + y)) ! ! = (−1)|α| u(y), δ(x),

β α−β α! ϕ)(x β!(α−β)! ∂x ψ(x, y)(∂x

"" + y)

β≤α

! = (−1)|α| u(y) ,

"

β α−β α! ϕ)(y) β!(α−β)! (∂x ψ)(0, y)(∂x

β≤α

% $ = (−1)|α| u(y), ψ(0, y)∂ α ϕ(y) $ % $ % = (−1)|α| u, ∂ α ϕ = ∂ α u, ϕ .

(2.8.34)

In particular, if |α| = 0, the above implies δ ∗ u = u. When combined with (b), this finishes the proof of the statement in (d). Finally, by making use of the results from (d) and (c) we have ∂ α (u ∗ v) = ∂ α (δ ∗ (u ∗ v)) = ∂ α δ ∗ (u ∗ v) = (∂ α δ ∗ u) ∗ v = (∂ α u) ∗ v. (2.8.35) A similar argument also shows that ∂ α (u ∗ v) = u ∗ (∂ α v). The proof of the theorem is now complete. Exercise 2.88. Prove that for a distribution u ∈ D (Rn ), u = δ ⇐⇒ u ∗ f = f

for each

f ∈ C0∞ (Rn ).

(2.8.36)

Hint: For the right-to-left implication use f = φj , where φj is as in Example 2.20, and let j → ∞. Next, we extend the translation map (1.3.16) to distributions. Proposition 2.89. For each x0 ∈ Rn and each u ∈ D (Rn ) fixed, the translation mapping D(Rn ) ϕ → u, t−x0 (ϕ) ∈ C is linear and continuous. Denoting this map by tx0 u thus yields a distribution in Rn that satisfies tx0 u, ϕ = u, t−x0 (ϕ),

∀ ϕ ∈ C0∞ (Rn ).

(2.8.37)


67

Proof. This follows by observing that the mapping in question is the composition u ◦ t−x0 where the latter translation operator is consider in the sense of Exercise 1.19. Exercise 2.90. Fix x0 ∈ Rn and recall the distribution δx0 from Example 2.14. Prove that δx0 = tx0 δ in D (Rn ). Also show that δx0 ∗ u = tx0 u for every u ∈ D (Rn ). In particular, if x1 ∈ Rn is arbitrary, then δx0 ∗ δx1 = δx0 +x1 in D (Rn ). Remark 2.91. (1) If u, v ∈ E (Rn ) then u ∗ v ∈ E (Rn ). This is an immediate consequence of (2.8.17). (2) There exists a sequence {uj }j∈N of compactly supported distributions in R that converges to some u ∈ D (R) and such that uj ∗ v does not necessarily converge to u ∗ v in D (R) for each v ∈ D (R). To see an example in this regard, consider the sequence of compactly D (R)

supported distributions {δj }j∈N that satisfies δj −−−− → 0. Then if 1 denotes j→∞

the distribution on R given by the constant function 1, Exercise 2.90 gives δj ∗ 1 = 1 for each j, and the constant sequence {1}j∈N ⊂ D (R) does not converge in D (R) to 0 ∗ 1 = 0. This shows that sequential continuity for convolution of distributions cannot be expected in general. (3) Condition (2.8.18) is necessary for the operation of convolution of distributions to be associative. To see this, consider the distributions 1, δ , and H on R. Then we have supp δ = {0}, supp 1 = R, supp H = [0, ∞). If K is a compact set in Rn , the set = (x, 0, z) : x ∈ R, z ∈ [0, ∞), x + z ∈ K (2.8.38) MK is not compact in R × R × R, thus (2.8.18) does not hold. Furthermore, 1 ∗ δ = 1 ∗ δ = 0 ∗ δ = 0 so (1 ∗ δ ) ∗ H = 0, while 1 ∗ (δ ∗ H) = 1 ∗ (δ ∗ H ) = 1 ∗ (δ ∗ δ) = 1 ∗ δ = 1 and clearly (1 ∗ δ ) ∗ H = 1 ∗ (δ ∗ H) in D (R). Proposition 2.92. The following statements are true. D (Rn )

D (Rn )

j→∞

j→∞

(1) If u ∈ E (Rn ) and vj −−−−→ v, then u ∗ vj −−−−→ u ∗ v. D (Rn )

(2) If uj −−−−→ u and there exists K ⊂ Rn compact with supp uj ⊆ K for j→∞

D (Rn )

every j ∈ N, then uj ∗ v −−−−→ u ∗ v for every v ∈ D (Rn ). j→∞

Proof. To see why (1) is true, fix ϕ ∈ C0∞ (Rn ). Then by definition, for each j ∈ N we have u ∗ vj , ϕ = u ⊗ vj , ψj ϕΔ for any smooth compactly supported


68

function ψj with ψj = 1 on a neighborhood of (supp u × supp vj ) ∩ supp ϕΔ . Note that (supp u × supp vj ) ∩ supp ϕΔ ⊆ supp u × (supp ϕ − supp u)

(2.8.39)

and supp u × (supp ϕ − supp u) is compact since both supp u and supp ϕ are compact. Hence, if we fix ψ ∈ C0∞ (Rn × Rn ) such that ψ = 1 in a neighborhood of the set supp u × (supp ϕ − supp u), then u ∗ vj , ϕ = u ⊗ vj , ψϕΔ for every j ∈ N. Based on (d) in Theorem 2.80, we may write u ∗ vj , ϕ = u ⊗ vj , ψϕΔ −−−→ u ⊗ v, ψϕΔ = u ∗ v, ϕ, j→∞

(2.8.40)

and the desired conclusion follows. Assume next the hypotheses in part (2) of the proposition. In particular, these entail supp u ⊆ K. Let ϕ ∈ C0∞ (Rn ) and note that (supp uj × supp v) ∩ supp ϕΔ ⊆ K × (supp ϕ − K),

∀ j ∈ N.

(2.8.41)

Then if ψ ∈ C0∞ (Rn × Rn ) is a function with the property that ψ = 1 in a neighborhood of K × (supp ϕ − K), we have uj ∗ v, ϕ = uj ⊗ v, ψϕΔ for every j ∈ N. Hence, uj ∗ v, ϕ = uj ⊗ v, ψϕΔ −−−→ u ⊗ v, ψϕΔ = u ∗ v, ϕ, j→∞

(2.8.42)

where for the convergence in (2.8.42) we used part (d) in Theorem 2.80. When convolving an arbitrary distribution with a distribution of function type given by a compactly supported smooth function, the resulting distribution is of function type. This fact is particularly useful in applications and we prove it next. Proposition 2.93. If u ∈ D (Rn ) and g ∈ C0∞ (Rn ), then the distribution u ∗ g is of function type given by the function f : Rn → C,

f (x) := u(y), g(x − y),

∞

∀ x ∈ Rn ,

(2.8.43)

that satisfies f ∈ C (R ). Moreover, if u is compactly supported then so is f . In short, n

D (Rn ) ∗ C0∞ (Rn ) ⊆ C ∞ (Rn )

and

E (Rn ) ∗ C0∞ (Rn ) ⊆ C0∞ (Rn ). (2.8.44)

Proof. Let φ : Rn × Rn → C be defined by φ(x, y) := g(x − y) for each (x, y) ∈ Rn × Rn . Then φ ∈ C ∞ (Rn × Rn ) and φ(x, ·) ∈ C0∞ (Rn ) for each x ∈ Rn . This shows that the function f in (2.8.43) is well-defined. To prove that f is of class C ∞ in Rn , pick an arbitrary point x∗ ∈ Rn and pick a function ψ ∈ C0∞ (Rn ) with the property that ψ = 1 in a neighborhood of B(x∗ , 1). In addition, select η ∈ C0∞ (Rn ) such that η = 1 near B(x∗ , 1)−supp g. Then $ % $ % f ∗ (x) = u(y), g(x − y) = u(y), ψ(x)g(x − y) B(x ,1)

$ % = (ηu)(y), ψ(x)g(x − y) ,

∀ x ∈ B(x∗ , 1).

(2.8.45)


69

Since ηu ∈ E (Rn ) and the function Rn × Rn (x, y) → ψ(x)g(x − y)∈ C is of class C ∞ , we may further invoke Proposition 2.82 to conclude that f B(x∗ ,1) ∈ C ∞ B(x∗ , 1) . Given that x∗ ∈ Rn has been arbitrarily chosen, it follows that f ∈ C ∞ (Rn ). We now turn to the task of showing that the distribution u ∗ g is of function type and is given by f . To this end, fix ϕ ∈ C0∞ (Rn ) and consider a function ψ ∈ C0∞ (Rn × Rn ) such that ψ = 1 in a neighborhood of the set (x, y) : x ∈ supp u, y ∈ supp g, x + y ∈ supp ϕ .

(2.8.46)

Then, starting with Definition 2.83 we write $ % u ∗ g, ϕ = u ⊗ g, ψϕΔ ! $ %" = u(x), g(y), ψ(x, y)ϕ(x + y)

!

" g(y)ψ(x, y)ϕ(x + y) dy

= u(x) , Rn

! = u(x) ,

" g(y)ϕ(x + y) dy

Rn

! = u(x) ,

Rn

= Rn

g(z − x)ϕ(z) dz

"

u(x), g(z − x)ϕ(z) dz

= f, ϕ.

(2.8.47)

For the third equality in (2.8.47) we have used the fact that g is a distribution of function type, for the forth equality we used condition (2.8.46), the fifth equality is based on a change of variables, the sixth equality follows from (2.7.45), while the last equality is a consequence of the definition of f . To complete the proof of the proposition there remains to notice that, by part (2.8.17), we have supp f ⊆ supp u + supp g. In particular, if u is compactly supported, then so is f . Exercise 2.94. Prove that if u ∈ E (Rn ) and g ∈ C ∞ (Rn ) then the distribution u ∗ g is of function type given by the function f : Rn → C,

f (x) := u(y), g(x − y),

∀ x ∈ Rn ,

(2.8.48)

that satisfies f ∈ C ∞ (Rn ). In short, E (Rn ) ∗ C ∞ (Rn ) ⊆ C ∞ (Rn ). Hint: Use Proposition 2.82 to show that f ∈ C ∞ (Rn ), then reason as in Proposition 2.93 to take care of the remaining claims.


70

Exercise 2.95. Let Ω be a bounded open set in Rn . Suppose u ∈ L1 (Ω) and define u to be the extension by zero outside Ω of u. Also assume that a function ∈ D (Rn ), it satisfies supp u ⊆ Ω, hence g ∈ L1loc (Rn ) is given. Prove that u n u ∗ g is well-defined in D (R ), and that the distribution u ∗ g is of function type given by the function ( u ∗ g)(x) = g(x − y)u(y) dy, ∀ x ∈ Rn . (2.8.49) Ω

Hint: Apply (2.8.16) with f = u . D (Rn )

Exercise 2.96. Prove that if u ∈ D (Rn ) and ϕj −−−−→ ϕ, then one has j→∞

E(Rn )

u ∗ ϕj −−−−→ u ∗ ϕ. j→∞

Next, we prove that distributions of function type given by smooth, compactly supported functions are dense in the class of distributions. First we treat the simpler case when the distributions are considered in Rn . The case when the distributions are considered on arbitrary open subsets of Rn needs to be handled with a bit more care. Theorem 2.97. The set C0∞ (Rn ) is sequentially dense in D (Rn ). Proof. First we will show that C ∞ (Rn ) is sequentially dense in D (Rn ). Let φ be as in (1.2.3) and recall the sequence of functions {φj }j∈N from (1.3.7). In particular, we have that supp φj ⊆ B(0, 1) for every j ∈ N. Recall from D (Rn )

Example 2.20 that φj −−−−→ δ. j→∞

Let u ∈ D (Rn ) be arbitrary and define uj := u ∗ φj

in D (Rn ),

∀ j ∈ N.

(2.8.50)

By Proposition 2.93 we have uj ∈ C ∞ (Rn ) for all j ∈ N. Also, by part (2) in Proposition 2.92 and part (d) in Theorem 2.87 we obtain D (Rn )

uj = u ∗ φj −−−−→ u ∗ δ = u. j→∞

(2.8.51)

This completes the proof of the fact that C ∞ (Rn ) is sequentially dense in D (Rn ). Moving on, and keeping the notation introduced so far, fix ψ ∈ C0∞ (Rn ) satisfying ψ(x) = 1 if |x| < 1 and set wj (x) := ψ(x/j)(u ∗ φj )(x)

∀ x ∈ Rn ,

∀ j ∈ N.

(2.8.52)

It is immediate that wj ∈ C0∞ (Rn ). Moreover, if ϕ ∈ C0∞ (Rn ) is given, then there exists j0 ∈ N with the property that supp ϕ ⊂ B(0, j0 ). Therefore, for all j ≥ j0 we may write

2.8. THE CONVOLUTION OF DISTRIBUTIONS IN RN wj , ϕ =

ψ

x

Rn

j

71

(u ∗ φj )(x)ϕ(x) dx =

Rn

(u ∗ φj )(x)ϕ(x) dx

= u ∗ φj , ϕ −−−→ u, ϕ.

(2.8.53)

j→∞

D (Rn )

This shows that wj −−−−→ u. Hence, ultimately, C ∞ (Rn ) is sequentially dense j→∞

in D (Rn ), finishing the proof of the theorem. The same type of result as in Theorem 2.97 actually holds in arbitrary open subsets of the Euclidean ambient. Theorem 2.98. The set C0∞ (Ω) is sequentially dense in D (Ω). Proof. Fix u ∈ D (Ω) and recall the sequence of compact sets introduced in ˚j+1 for every j ∈ N. For each j ≥ 2 (1.2.13). Then Kj = Ω and Kj ⊂ K j∈N

consider a function ψj ∈ C0∞ (Ω), ψj = 1 on a neighborhood of Kj−1 , supp ψj ⊆ Kj ,

(2.8.54)

and define uj := ψj u ∈ D (Ω). Since supp uj ⊆ Kj , Proposition 2.63 states that each uj may be extended to a distribution in E (Rn ), which we continue to denote by uj . If we now set εj :=

1 dist (Kj , ∂Kj+1 ) > 0, 4

∀ j ∈ N,

(2.8.55)

then the definition of the compacts in (1.2.13) forces εj 0 as j → ∞. Having

fixed some φ ∈ C0∞ (Rn ) satisfying supp φ ⊆ B(0, 1) and Rn φ(x) dx = 1, define φj (x) := ε−n j φ(x/εj ),

∀ x ∈ Rn , ∀ j ∈ N.

(2.8.56)

Thus, φj ∈ C0∞ (Rn ),

supp φj ⊆ B(0, εj ) ⊆ B(0, 1),

∀ j ∈ N,

(2.8.57)

and reasoning as in Example 2.20 we see that D (Rn )

φj −−−−→ δ. j→∞

(2.8.58)

Finally, for each j ≥ 2 introduce wj := uj ∗ φj . By combining part (1) in Remark 2.91 with Proposition 2.93 we obtain that wj ∈ C0∞ (Rn ) for every j ≥ 2. In addition, (2.8.17) and (2.8.55) imply that supp wj ⊆ Kj + B(0, εj ) ⊂ Kj+1 , so in fact we have wj ∈ C0∞ (Ω). D (Ω)

To complete the proof of the theorem it suffices to show that wj −−−−→ u. j→∞

To this end, fix ϕ ∈ C0∞ (Ω) and observe that based on (1.2.13) there exists j0 ∈ N such that supp ϕ ⊆ Kj0 . Then for j > j0 + 1 we may write wj , ϕ = uj ∗ φj , ϕ = (uj ∗ φj )(x)ϕ(x) dx Ω


72

uj (y), φj (x − y)ϕ(x) dx =

= Ω

u(y), ψj (y)φj (x − y)ϕ(x) dx Ω

u(y), ψj0 +2 (y)φj (x − y)ϕ(x) dx = uj0 +2 ∗ φj , ϕ.

=

(2.8.59)

Ω

For the second equality in (2.8.59) we used the fact that uj ∗ φj ∈ C0∞ (Rn ) for j ≥ 2, for the third equality we used Proposition 2.93, while for the forth the fact that uj = ψj u for every j ∈ N. These observations also give the last equality in (2.8.59). As for the penultimate equality in (2.8.59), observe that if j > j0 + 1, x ∈ supp ϕ ⊆ Kj0 and x − y ∈ supp φj ⊆ B(0, εj ), then y ∈ Kj0 − B(0, εj ) ⊆ Kj0 − B(0, εj0 ) ⊂ Kj0 +1 , thus ψj (y) = 1 = ψj0 +2 (y). If we now combine (2.8.58) with (2.8.57) and part (2) in Proposition 2.92, it D (Rn )

follows that uj0 +2 ∗ φj −−−−→ uj0 +2 ∗ δ = uj0 +2 . Hence, if we return with this j→∞

in (2.8.59), we may write lim wj , ϕ = lim uj0 +2 ∗φj , ϕ = uj0 +2 , ϕ = ψj0 +2 u, ϕ = u, ϕ, (2.8.60)

j→∞

j→∞

since ψj0 +2 = 1 in a neighborhood of the support of ϕ. This finishes the proof of the theorem.

2.9

Distributions with Higher-Order Gradients Continuous or Bounded

We have seen that if u ∈ C m (Rn ) for some m ∈ N, then its distributional derivatives up to order m are distributions of function type, each given by the corresponding pointwise derivative of u. A more subtle question pertains to the possibility of deducing regularity results for distributions with distributional derivatives of a certain order that are of function type, and these functions exhibit a certain amount of smoothness. In this section we prove two main results in this regard. In the first result (see Theorem 2.102), we show that if a distribution u ∈ D (Ω) has all distributional derivatives of order m ∈ N continuous, then in fact u ∈ C m (Ω). In the second main result (see Theorem 2.104) we prove that a distribution in D (Rn ) is of function type given by a Lipschitz function if and only if all its first-order distributional derivatives are bounded functions in Rn . We start by proving a weaker version of Theorem 2.102. Proposition 2.99. If u ∈ D (Rn ) and there exists m ∈ N0 such that for each multi-index α ∈ Nn0 satisfying |α| ≤ m the distributional derivative ∂ α u belongs to C 0 (Rn ), then u ∈ C m (Rn ). Proof. Recall the sequence of distributions {φj }j∈N from Example 2.20 and for u satisfying the hypothesis of the proposition let {uj }j∈N be as in (2.8.50). In particular, (2.8.51) holds, thus the distributional and classical derivatives of each uj coincide.

2.9. DISTRIBUTIONS WITH HIGHER-ORDER GRADIENTS...

73

Next we make the following claim: lim ∂ α uj = ∂ α u uniformly on compact sets in Rn

j→∞

for every multi-index α ∈ Nn0 satisfying |α| ≤ m.

(2.9.1)

To prove this claim, observe that by part (e) in Theorem 2.87, for each j ∈ N we have ∂ α uj = (∂ α u) ∗ φj for every α ∈ Nn0 . Fix α ∈ Nn0 satisfying |α| ≤ m. Since by the current hypotheses the distributional derivative ∂ α u is continuous, by invoking Proposition 2.93 we may write ∂ α uj (x) = ((∂ α u) ∗ φj )(x) = ∂ α u(y), φj (x − y) α (∂ u)(y)φj (x − y) dy = (∂ α u)(x − z)φj (z) dz = Rn

=

Rn

Rn

(∂ α u) x − y/j φ(y) dy,

∀ x ∈ Rn .

(2.9.2)

Fix a compact set K in Rn . Making use of (2.9.2) and the properties of φ (recall (1.2.3)) we estimate α α α α sup |∂ uj (x) − ∂ u(x)| = sup [(∂ u) x − y/j − (∂ u)(x)]φ(y) dy x∈K

x∈K

≤

Rn

sup (∂ α u) x − y/j − (∂ α u)(x).

(2.9.3)

x∈K y∈B(0,1)

Since ∂ α u is continuous in Rn it follows that ∂ α u is uniformly continuous on compact subsets of Rn , thus lim sup |∂ α uj (x) − ∂ α u(x)| ≤ lim sup (∂ α u) x − z/j − (∂ α u)(x) = 0, j→∞ x∈K

j→∞

x∈K y∈B(0,1)

(2.9.4) completing the proof of the claim. With (2.9.1) in hand, we may invoke Lemma 2.100 below and proceed by induction on |α| to conclude that, as desired, u ∈ C m (Rn ). Lemma 2.100. Suppose the functions {uj }j∈N and u are such that: (i) uj ∈ C 1 (Ω) for every j ∈ N, (ii) lim uj = u uniformly on compact subsets of Rn contained in Ω, and j→∞

(iii) for each k ∈ {1, . . . , n} there exists a function vk ∈ C 0 (Ω) with the property that lim ∂k uj = vk uniformly on compact subsets of Rn contained in Ω. j→∞

Then u ∈ C 1 (Ω) and ∂k u = vk for each k ∈ {1, . . . , n}. Proof. From the start, since uniform convergence on compact sets preserves continuity, we have that u ∈ C 0 (Ω). Fix x ∈ Ω and k ∈ {1, 2, . . . , n} and let t0 > 0 be such that x + tek ∈ Ω whenever t ∈ [−t0 , t0 ] where, as before, ek is


74

the unit vector in Rn where the kth component is equal to 1. Then, for each t ∈ [−t0 , t0 ], we may write t d uj (x + s ek ) ds u(x + tek ) − u(x) = lim [uj (x + tek ) − uj (x)] = lim j→∞ j→∞ 0 ds t t = lim (∂k uj )(x + s ek ) ds = vk (x + s ek ) ds. (2.9.5) j→∞

0

0

The first equality in (2.9.5) is based on (ii) while the last one is a consequence of (iii). Next, since vk is continuous on Ω, by the mean value theorem for integrals it follows that there

t exists some st , belonging to the interval with endpoints 0 and t, such that 0 vk (x + s ek ) ds = tvk (x + st ek ). Hence, u(x + tek ) − u(x) = lim vk (x + st ek ) = vk (x), t→0 t→0 t lim

(2.9.6)

which proves that (∂k u)(x) exists and is equal to vk (x). Thus, ∂k u = vk ∈ C 0 (Ω) for every k, which shows that u ∈ C 1 (Ω). Lemma 2.101. Let u ∈ D (Ω) be such that for each j ∈ {1, . . . , n}, the distributional derivatives ∂j u are of function type and belong to C 0 (Ω). Then u ∈ C 0 (Ω).

1 Proof. Since ∇u ∈ [C 0 (Ω)]n the function v(x) := 0 (∇u)(tx) · x dt for x ∈ Ω (where “·” denotes the dot product of vectors) is well-defined and belongs to C 0 (Ω). Given the current goals, by Exercise 2.46 it suffices to show that for each x0 ∈ Ω there exists an open set ω ⊂ Ω such that x0 ∈ ω and uω = v ω in D (ω). To this end, fix x0 ∈ Ω and let r ∈ (0, dist(x0 , ∂Ω)). Then B(x0 , r) ⊂ Ω. Without loss of generality in what follows we may assume that x0 = 0 (since translations interact favorably, in a reversible manner, with both hypotheses and conclusion). Let ϕ ∈ C0∞ (Ω) be such that supp ϕ ⊂ B(0, r) and fix some j ∈ {1, . . . , n}. Then we have 1 ∂j v, ϕ = −v, ∂j ϕ = − (∇u)(tx) · x ∂j ϕ(x) dt dx =−

1

Ω

0

(∇u)(tx) · x ∂j ϕ(x) dx dt

Ω 1

1

= − lim

ε→0+

ε

= − lim

ε→0+

= − lim

ε→0+

0

ε

n Ω k=1

(∇u)(tx) · x ∂j ϕ(x) dx dt Ω

y

∇u(y) · Ω

tn+1

∂k u(y) ε

1

(∂j ϕ)(y/t) dy dt

yk tn+1

(∂j ϕ)(y/t) dt dy

2.9. DISTRIBUTIONS WITH HIGHER-ORDER GRADIENTS... n ! = lim+ u(y), ε→0

k=1

1

ε

∂yk

y " k (∂j ϕ)(y/t) dt . n+1 t

75

(2.9.7)

For the fourth equality in (2.9.7) we have used Lebesgue’s dominated convergence theorem (cf. Theorem 13.12) and for the fifth one the change of variables tx = y (note that since supp ϕ ∈ B(0, r) and t ∈ (0, 1], we have supp ϕ(·/t) ⊂ B(0, r) ⊂ Ω). Furthermore, for each ε > 0, differentiating with respect to y and then integrating by parts in t, gives n 1 y k ∂yk n+1 (∂j ϕ)(y/t) dt t ε k=1

n 1

=

ε k=1

1 tn+1

(∂j ϕ)(y/t) +

yk tn+2

(∂k ∂j ϕ)(y/t) dt

n 1 d (∂ (∂ ϕ)(y/t) − ϕ)(y/t) dt j j tn+1 tn dt ε t=1 1 1 = − n (∂j ϕ)(y/t) = −∂j ϕ(y) + n (∂j ϕ)(y/ε). t ε t=ε 1

=

By combining (2.9.7) and (2.9.8) we obtain ! ∂j v, ϕ = −u, ∂j ϕ + lim+ u(y), ε→0

= ∂j u, ϕ + lim

ε→0+

! ∂j u(y),

1

∂y εn−1 j

" ϕ(y/ε)

" ϕ(y/ε) . n−1 1

ε

(2.9.8)

(2.9.9)

Since ∂j u ∈ C 0 (Ω), the pairing under the limit in the right-most term in (2.9.9) is given by an integral in which we make the change of variables x = y/ε to further compute " ! 1 1 (∂j u)(y)ϕ(y/ε) dy lim ∂j u(y), n−1 ϕ(y/ε) = lim+ n−1 ε ε→0+ ε→0 ε Ω (2.9.10) = lim+ ε (∂j u)(εx)ϕ(x) dx = 0 ε→0

Ω

given that, by the continuity of ∂j u at 0 ∈ Ω, lim (∂j u)(εx)ϕ(x) dx = ∂j u(0) ϕ(x) dx. ε→0+

Ω

(2.9.11)

Ω

In summary, from (2.9.9)–(2.9.11), the fact that ϕ is an arbitrary element in D(B(0, r)), and that j is arbitrary in {1, . . . , n}, we conclude that necessarily ∇v B(0,r) = ∇uB(0,r) in D (B(0, r)). By Exercise 2.140 there exists c ∈ C such that uB(0,r) − v B(0,r) = c in D (B(0, r)). Since v ∈ C 0 (Ω), the latter implies that uB(0,r) belongs to C 0 (B(0, r)) as desired. This completes the proof of the lemma.

76


After these preparations, we are ready to state and prove our first main result. Theorem 2.102. Let u ∈ D (Ω) and suppose that there exists m ∈ N0 such that for each α ∈ Nn0 satisfying |α| = m the distributional derivative ∂ α u is continuous on Ω. Then u ∈ C m (Ω). Proof. We prove the theorem by induction on m. For m = 0 there is nothing to prove. Suppose m = 1. Applying Lemma 2.101, we obtain that u ∈ C 0 (Ω). To prove that u ∈ C 1 (Ω), it suffices to show that u is of class C 1 in a neighborhood of any point x0 ∈ Ω. Fix x0 ∈ Ω, take r > 0 such that B(x0 , r) ⊂ Ω and a function ψ ∈ C0∞ (Rn ) such that supp ψ ⊂ B(x0 , r) and ψ ≡ 1 on B(x0 , r/2). Then ) ∈ E (Rn ). Also, by Proposition 2.63 the distribution ψu extends to v := ψu v ∈ C 0 (Rn ) since u ∈ C 0 (Ω) and ψ is compactly supported in Ω. Using (2.5.4) and part (4) in Proposition 2.36 we obtain 0 n * ∂j v = ∂ j (ψu) = (∂j ψ)u + ψ ∂j u ∈ C (R ).

(2.9.12)

1 n Applying Proposition 2.99 since with m = 1 then gives v ∈ C (R ). In addition, by design v B(x ,r/2) = u B(x ,r/2) , we conclude that u B(x ,r/2) ∈ C 1 (B(x0 , r/2)), 0 0 0 as wanted. Assume now that the theorem is true for all nonnegative integers up to, and including, some m ∈ N. Take u ∈ D (Ω) with the property that ∂ α u ∈ C 0 (Ω) for all α ∈ Nn0 satisfying |α| = m + 1 and fix j ∈ {1, . . . , n}. Then ∂j u ∈ D (Ω) satisfies ∂ α (∂j u) ∈ C 0 (Ω) for all α ∈ Nn0 with |α| = m. By the induction hypothesis, it follows that ∂j u ∈ C m (Ω). In particular, ∂j u ∈ C 0 (Ω). This being true for all j ∈ {1, . . . , n}, what we already proved for m = 1 implies u ∈ C 1 (Ω). Thus, u is a C 1 function in Ω with first-order partial derivatives that are of class C m in Ω. Then necessarily u ∈ C m+1 (Ω) as wanted. The proof of the theorem is finished.

Exercise 2.103. Let u ∈ D (Rn ) be such that for some N ∈ N0 we have ∂ α u = 0 in D (Rn ) for each α ∈ Nn0 with |α| > N . Prove that u is a polynomial of degree less than or equal to N . Hint: Use Theorem 2.102 to conclude that u ∈ C ∞ (Rn ) then invoke Taylor’s formula (13.2.6). Moving on to the second issue discussed at the beginning of this section we recall (cf. Remark 2.35) that a function f : Ω → C is called Lipschitz in Ω provided there exists some constant M ∈ [0, ∞) such that |f (x) − f (y)| ≤ M |x − y|,

∀ x, y ∈ Ω,

(2.9.13)

and that the Lipschitz constant of f is the smallest M for which (2.9.13) holds. Our next task is to prove the following theorem. Theorem 2.104. For f ∈ D (Rn ) and a number M ∈ [0, ∞) the following two statements are equivalent:


77

(i) f is given by a Lipschitz function in Rn with Lipschitz constant less than or equal to M ; (ii) for each k ∈ {1, . . . , n}, the distributional derivative ∂k f belongs to L∞ (Rn ) and ∂k f L∞ (Rn ) ≤ M . Proof. Fix a distribution f ∈ D (Rn ) and let φ be as in (1.2.3). Consider the sequence of functions {φj }j∈N from (1.3.7) that satisfies the properties listed in (1.3.8). Furthermore, set fj := f ∗ φj

in D (Rn ),

∀ j ∈ N.

(2.9.14)

By Proposition 2.93 we have fj ∈ C ∞ (Rn ) and fj (x) = f, φj (x − ·),

∀ j ∈ N.

(2.9.15)

D (Rn )

Also, since (as proved in Example 2.20) one has φj −−−−→ δ, by part (2) in j→∞

Proposition 2.92 and part (d) in Theorem 2.87 one obtains D (Rn )

fj = f ∗ φj −−−−→ f ∗ δ = f. j→∞

(2.9.16)

Next we proceed with the proof of (i) =⇒ (ii). Suppose f is Lipschitz with Lipschitz constant ≤ M . In particular, the formula in (2.9.15) becomes fj (x) = f (y)φj (x − y) dy = f (x − y)φj (y) dy, ∀ x ∈ Rn , ∀ j ∈ N. Rn

Rn

(2.9.17)

We claim that for each j ∈ N one has |fj (x) − fj (y)| ≤ M |x − y|,

∀ x, y ∈ Rn

and ∇fj L∞ (Rn ) ≤ M. (2.9.18)

Indeed, if j ∈ N is fixed, then from (2.9.17) we obtain |fj (x) − fj (y)| ≤ |f (x − z) − f (y − z)|φj (z) dz ≤ M |x − y|,

(2.9.19)

Rn

for every x, y ∈ Rn . In turn, since fj is smooth, we have fj (x + hek ) − fj (x) , h→0 h

(∂k fj )(x) = lim

∀ x ∈ Rn , ∀ k ∈ {1, . . . , n}. (2.9.20)

In combination with (2.9.19) this implies ∇fj L∞ (Rn ) ≤ M , completing the proof of the claims made in (2.9.18). Next, fix k ∈ {1, . . . , n} and ϕ ∈ C0∞ (Rn ). Then based on (2.9.16) we may write ∂k f, ϕ = −f, ∂k ϕ = − lim fj , ∂k ϕ = lim ∂k fj , ϕ j→∞

j→∞


78

= lim

j→∞

Rn

(∂k fj )(x)ϕ(x) dx.

Using the second estimate in (2.9.18) we obtain n (∂k fj )(x)ϕ(x) dx ≤ ∇fj L∞ (Rn ) ϕL1 (Rn ) ≤ M ϕL1 (Rn ) .

(2.9.21)

(2.9.22)

R

Hence, from (2.9.22) and (2.9.21) it follows that (∂k fj )(x)ϕ(x) dx ≤ M ϕL1(Rn ) . ∂k f, ϕ ≤ lim sup

(2.9.23)

Rn

j→∞

Consequently, the linear assignment C0∞ (Rn ) ϕ → ∂k f, ϕ ∈ R

(2.9.24)

is continuous in the L1 -norm and has norm less than or equal to M . Since C0∞ (Rn ) is dense in L1 (Rn ), the linear functional in (2.9.24) extends to a linear, bounded Λk : L1 (Rn ) → C with norm less or equal to M . Thus, 1functional n ∗ Λk ∈ L (R ) = L∞ (Rn ) has norm less than or equal to M , which implies that there exists a unique gk ∈ L∞ (Rn )

with gk L∞ (Rn ) ≤ M

(2.9.25)

and such that Λk (h) =

Rn

gk (x)h(x) dx,

∀ h ∈ L1 (Rn ).

(2.9.26)

Granted (2.9.26) and keeping in mind that Λk is an extension of the linear assignment in (2.9.24), we arrive at the conclusion that ∂k f, ϕ = gk (x)ϕ(x) dx, ∀ ϕ ∈ C0∞ (Rn ). (2.9.27) Rn

The identity in (2.9.27) yields ∂k f = gk in D (Rn ), which proves (ii) in view of (2.9.25). Conversely, suppose (ii) is true. Fix j ∈ N and note that from (2.9.14) and part (e) in Theorem 2.87, for every k ∈ {1, . . . , n} one has ∂k fj = (∂k f ) ∗ φj in D (Rn ). Thus, using Proposition 2.93, the current assumptions on f , and the properties of φj , we have |(∂k fj )(x)| = |∂k f, φj (x − ·)| = (∂k f )(y)φj (x − y) dy Rn φj (x − y) dy ≤ M, ∀ x ∈ Rn , (2.9.28) ≤ ∂k f L∞ (Rn ) Rn


79

for each k ∈ {1, . . . , n}. Now fix x0 ∈ Rn and consider the sequence of functions {gj }j∈N given by gj (x) := fj (x) − fj (x0 ),

for x ∈ Rn .

(2.9.29)

Then gj ∈ C ∞ (Rn ) and based on the mean value theorem, (2.9.29), and (2.9.28) we also obtain |gj (x) − gj (y)| ≤ M |x − y|, |gj (x)| ≤ M |x − x0 |,

∀ x, y ∈ Rn ,

and

∀ x ∈ Rn .

(2.9.30) (2.9.31)

By a corollary of the classical Arzel´ a–Ascoli theorem (see Theorem 13.16), there exists a subsequence {gj }∈N that converges uniformly on any compact subset of Rn to some function g ∈ C 0 (Rn ). As such, for each ϕ ∈ C0∞ (Rn ) we may write gj (x)ϕ(x) dx = g(x)ϕ(x) dx = g, ϕ lim gj , ϕ = lim →∞

→∞

supp ϕ

supp ϕ

(2.9.32) D (Rn )

which goes to show that gj −−−−→ g. The latter, (2.9.16), and (2.9.29), state →∞

that whenever ψ ∈ C0∞ (Rn ) is such that Rn ψ(x) dx = 0 we have g, ψ = lim gj , ψ = lim gj (x)ψ(x) dx →∞

= lim

→∞

Rn

→∞

Rn

fj (x)ψ(x) dx = lim fj , ψ = f, ψ. →∞

(2.9.33)

Thus, we may apply Exercise 2.139 to conclude that there exists some constant c ∈ C with the property that f = g + c in D (Rn ). This proves that the distribution f is of function type and is given by the function g + c. Moreover, writing the estimate in (2.9.30) with j replaced by j and then taking the limit as → ∞ (recall that {gj } converges pointwise to g) leads to |g(x) − g(y)| ≤ M |x − y|,

∀ x, y ∈ Rn ,

(2.9.34)

which in concert with the fact that f (x) − f (y) = g(x) − g(y) proves that f is a Lipschitz function with Lipschitz constant ≤ M . Now the proof of (ii) =⇒ (i) is finished.

Further Notes for Chap. 2. The material in this chapter is at the very core of the theory of distributions since it provides a versatile calculus for distributions that naturally extends the scope of the standard calculus for ordinary functions. The definition of distributions used here is essentially that of the French mathematician Laurent Schwartz (1915–2002), cf. [60], though nowadays there are many books dealing


80

at length with the classical topics discussed here. These include [8, 15, 17–20, 24, 25, 30, 31, 59, 65, 68, 70–72], and the reader is referred to these sources for other angles of exposition. In particular, in [24, 31, 68], distributions are defined on smooth manifolds, while in [53] the notion of distributions is adapted to rough settings.

2.10


Exercise 2.105. Consider the mapping u : D(R) → C by setting u(ϕ) :=

∞ 1 ϕ 2 − ϕ(0) , j j=1

∀ ϕ ∈ C0∞ (R).

(2.10.1)

Prove that u is well-defined. Is u a distribution? If yes, what is the order of u? Exercise 2.106. Prove that there exists u ∈ D (Ω) for which the following statement is false: For each compact K contained in Ω there exist C > 0 and k ∈ N0 such that |u, ϕ| ≤ C

sup x∈K, |α|≤k

|∂ α ϕ(x)|,

∀ ϕ ∈ C0∞ (Ω). (2.10.2)

Exercise 2.107. Prove that |x|N ln |x| ∈ L1loc (Rn ) whenever N is a real number satisfying N > −n. Thus, in particular, |x|N ln |x| ∈ D (Rn ) when N > −n. Exercise 2.108. Suppose n ≥ 2 and given ξ ∈ S n−1 define f (x) := ln |x · ξ| for each x ∈ Rn \ {0}. Prove that f ∈ L1loc (Rn ). In particular, ln |x · ξ| ∈ D (Rn ). Exercise 2.109. Prove that (ln |x|) = P.V. x1 in D (R), where P.V. x1 is the distribution defined in (2.1.13). Exercise 2.110. Let f : R → R be defined by f (x) = x ln |x| − x if x ∈ R \ {0} and f (0) = 0. Prove that f ∈ C 0 (R) and the distributional derivative of f in D (R) equals ln |x|. Exercise 2.111. Suppose n ≥ 2. Prove that ∂j (ln |x|) = every j ∈ {1, . . . , n}.

xj |x|2

in D (Rn ) for

Exercise 2.112. Let θ ∈ C0∞ (R) be supported in the interval (−1, 1) and such

that R θ(t) dt = 1. For each j ∈ N, define

j

j θ(jx − jt) dt,

ψj (x) :=

∀ x ∈ R.

1/j D (R)

Prove that ψj −−−−→ H, where H is the Heaviside function. j→∞


81 ∞

Exercise 2.113. Let u : D(R) → C defined by u(ϕ) :=

ϕ(j) (j) for each

j=1

ϕ ∈ D(R). Prove that u ∈ D (R) and that this distribution does not have finite order. Exercise 2.114. For each j ∈ N define fj (x) :=

1

for x ∈ Rn \ {0}.

1

j|x|n− j

D (Rn )

Prove that fj −−−−→ ωn−1 δ, where ωn−1 is the area of the unit sphere in Rn . j→∞

Exercise 2.115. For each ε > 0 define fε (x) :=

ε 1 π x2 + ε2

for x ∈ R.

D (R)

Prove that fε −−−−→ δ. + ε→0

Exercise 2.116. For each ε > 0 define n

fε (x) := (4πε)− 2 e−

|x|2 4ε

for x ∈ Rn .

D (Rn )

Prove that fε −−−−→ δ. ε→0+

Exercise 2.117. Recall that i = fε± (x) :=

√

−1 ∈ C and, for each ε ∈ (0, ∞), define

1 x±iε

for x ∈ R.

Also, recall the distribution from (2.1.13). Prove that 1 1 −→ ∓iπδ + P.V. x±iε x

as

This is the so-called Sokhotsky’s formula. Exercise 2.118. Prove that the sequence as j → ∞.

ε → 0+ in D (R).

sin(jx) πx

j∈N

(2.10.3)

converges to δ in D (R)

Exercise 2.119. In each case determine if the given sequence of distributions in D (R) indexed over j ∈ N converges√and determine its limit whenever convergent. Below m ∈ N is fixed and i = −1. √ 2 j (a) fj (x) := √ e−jx for x ∈ R; π (b) fj (x) ::= j m cos(jx)

for x ∈ R;


82 (c) fj (x) =

2j 3 x2 π(1 + j 2 x2 )2

for x ∈ R;

(d) uj := (−1)j δ 1j ; (e) uj :=

j δ 1j − δ− 1j ; 2

(f ) fj (x) :=

1 1 χ x |x|≥ j

(g) fj (x) :=

1 sin2 (jx) jπ x2

for x ∈ R \ {0}; for x ∈ R \ {0};

(h) fj (x) := j m eijx for x ∈ R; jeijx if x > 0, (j) fj (x) := 0 if x ≤ 0,

for x ∈ R.

Exercise 2.120. Let a ∈ R. Compute (H(· − a)) in D (R). Exercise 2.121. Consider the function x f : R −→ R defined by f (x) := 0

if x > a, if x ≤ a,

∀ x ∈ R,

(2.10.4)

where a ∈ R is fixed. Compute (uf ) in D (R), where uf is defined as in (2.1.6). Exercise 2.122. Let f : R → R be defined by f (x) := sin |x| for every x ∈ R. Compute (uf ) and (uf ) in D (R). Exercise 2.123. Let I ⊆ R be an open interval, x0 ∈ I, and f ∈ C 1 (I \ {x0 }) be such that f ∈ L1loc (I) (here f is the pointwise derivative of f in I \ {x0 }). Prove that the one-sided limits lim f (x), lim f (x) exist and are finite, that f∈

L1loc (I),

x→x+ 0

x→x− 0

and that

(uf ) = uf +

lim f (x) − lim f (x) δx0

x→x+ 0

x→x− 0

in

D (I).

Remark 2.124. Prove that there exist pointwise differentiable functions on R for which the distributional derivative does not coincide with the classical derivative. You may consider the function f defined by f (x) := x2 cos x12 for x = 0 and f (0) := 0, and show that f ∈ C 1 (R \ {0}) and also f is differentiable at the origin, while f ∈ L1loc (R). Exercise 2.125. Let I ⊆ R be an open interval, x0 ∈ I, and let m ∈ N. Suppose that f ∈ C ∞ (I \ {x0 }) is such that the pointwise derivatives f , f , . . . , f (m) , belong to L1loc (I). Prove that for each k ∈ {0, , 1, . . . , m−1} the limits lim+ f (k) (x) x→x0


83

and lim− f (k) (x) exist, are finite, and that x→x0

(m)

uf

lim f (x) − lim f (x) δx(m−1) 0

=uf (m) +

+

x→x− 0

lim+ f (x) − lim− f (x) δx(m−2) + ··· 0

+

x→x+ 0

x→x0

x→x0

lim f (m−1) (x) − lim f (m−1) (x) δx0

x→x+ 0

in D (I).

x→x− 0

Exercise 2.126. Let I ⊆ R be an open interval and consider a sequence {xk }k∈N ⊂ I with no accumulation point in I. Suppose f ∈ C 1 (I \ {xk : k ∈ N}) is such that the pointwise derivative f belongs to L1loc (I). Prove that for each k ∈ N the limits lim± f (x) exist and are finite, that f ∈ L1loc (I), and that x→xk

(uf ) = uf +

∞ k=1

lim+ f (x) − lim− f (x) δxk

x→xk

in

x→xk

D (I).

Exercise 2.127. Let f : R → R be the function defined by f (x) := x for each x ∈ R, where x is the integer part of x. Determine (uf ) . Exercise 2.128. Let Σ ⊂ Rn be a surface of class C 1 as in Definition 13.36, and denote by σ the the mapping δΣ : C0∞ (Rn ) → C

surface measure on Σ. Define ∞ n by δΣ (ϕ) := Σ ϕ(x) dσ(x) for each ϕ ∈ C0 (R ). Prove that δΣ ∈ D (Rn ), it has order zero, and supp δΣ = Σ. Also show that if g ∈ L∞ (K ∩ Σ) for each compact set K in Rn , and if we define gδΣ (ϕ) := g(x)ϕ(x) dσ(x), ∀ ϕ ∈ C0∞ (Rn ), (2.10.5) Σ

then gδΣ ∈ D (R ). n

Exercise 2.129. Let Ω ⊂ Rn be a domain of class C 1 (recall Definition 13.40) and denote by ν = (ν1 , . . . , νn ) its outward unit normal. Denote by δ∂Ω the distribution defined as in Exercise 2.128 corresponding to Σ := ∂Ω. Set Ω+ := Ω and Ω− := Rn \ Ω. Suppose that f ∈ L1loc (Rn ) has the property that the distributional derivative ∂k f belongs to L1loc (Rn ) for each number k ∈ {1, 2 . . . , n}. In addition, assume that the restrictions f± := f |Ω± satisfy f± ∈ C 1 (Ω± ) and that they may be extended so that f± ∈ C 0 (Ω± ). Prove that for each k ∈ {1, 2 . . . , n} the following equality holds: ∂k uf = u∂k f + s∂Ω (f )νk δ∂Ω

in D (Rn ),

where s∂Ω (f ) : ∂Ω → C is defined by s∂Ω (f )(x) := f− (x) − f+ (x) =

lim

Rn \Ω y→x

f (y) − lim f (y) Ω y→x

for every x ∈ ∂Ω.

(2.10.6)


84

Exercise 2.130. Let Ω ⊂ Rn be a bounded domain of class C 1 with outward unit normal ν = (ν1 , . . . , νn ). Prove that for each k ∈ {1, 2 . . . , n} one has ∂k χΩ = −νk δ∂Ω in D (Rn ). Exercise 2.131. Suppose R ∈ (0, ∞) and let u ∈ D (Rn ) be such that (|x|2 − R2 )u = 0

in D (Rn ).

(2.10.7)

Prove that u has compact support. Give an example of a distribution u satisfying condition (2.10.7). Exercise 2.132. Let f ∈ C 0 (Ω) be such that uf ∈ E (Ω). Prove that f has compact support and supp uf = supp f . Exercise 2.133. Compute the derivatives of order m ∈ N of each distribution on R given below. (a) |x|

(b) sgn x

(c) cos x H

(d) sin x H

(e) x2 χ[−1,1]

Exercise 2.134. Consider the set A := {(x, y) ∈ R2 : |x−2|+|y−1| < 1} ⊂ R2 . Compute (∂12 − ∂22 )χA in D (R2 ). Exercise 2.135. Let f : R2 → R be defined by f (x, y) := χ[0,1] (x − y) for x, y ∈ R. Compute ∂1 (uf ), ∂2 (uf ) in D (R2 ). Prove that ∂12 (uf ) − ∂22 (uf ) = 0 in D (R2 ). Exercise 2.136. Let ψ ∈ C ∞ (Ω) be such that ψ(x) = 0 for every x ∈ Ω. Prove that for each v ∈ D (Ω) there exists a unique solution u ∈ D (Ω) of the equation ψu = v in D (Ω). Exercise 2.137. Let ψ ∈ C ∞ (Ω) and suppose u1 , u2 ∈ D (Ω) are such that u1 = u2 and ψu1 = ψu2 in D (Ω). Prove that the set {x ∈ Ω : ψ(x) = 0} is not empty. Exercise 2.138. Suppose {Ωj }j∈I is an open cover of the open set Ω ⊆ Rn and there exists a family of distributions {uj }j∈I such that uj ∈ D (Ωj ) for each j ∈ I and uj Ωj ∩Ω = uk Ωj ∩Ω in D (Ωj ∩ Ωk ) for every j, k ∈ I such that k k Ωj ∩ Ωk = ∅. Prove that there exists a unique distribution u ∈ D (Ω) with the property that uΩj = uj in D (Ωj ) for every j ∈ I. Exercise 2.139. Let u ∈ D (Rn ) be such that u, ϕ = 0 for every ϕ ∈ C0∞ (Rn )

satisfying Rn ϕ(x) dx = 0. Prove that there exists c ∈ C such that u = c in D (Rn ). Exercise 2.140. Let Ω ⊆ Rn be open and connected and let u ∈ D (Ω) be such that ∂1 u = ∂2 u = · · · = ∂n u = 0 in D (Ω). Prove that there exists c ∈ C such that u = c in D (Ω). Exercise 2.141. Let u ∈ D (Rn ) be such that xn u = 0 in D (Rn ). Prove that there exists v ∈ D (Rn−1 ) such that u(x , xn ) = v(x ) ⊗ δ(xn ) in D (Rn ).


85

Exercise 2.142. Let u ∈ D (Rn ) be such that x1 u = · · · = xn u = 0 in D (Rn ). Determine the expression for u. Exercise 2.143. Let u ∈ D (Rn ) be such that ∂n u = 0 in D (Rn ). Prove that there exists v ∈ D (Rn−1 ) such that u(x , xn ) = v(x ) ⊗ 1 in D (Rn ), where 1 denotes the constant function (equal to 1) in R. Exercise 2.144. Let v, w ∈ D (R) and define the distribution u(x1 , x2 ) := 1 ⊗ v(x2 ) + w(x1 ) ⊗ 1

in

D (R2 ),

where 1 denotes the constant function (equal to 1) in R. Prove that ∂1 ∂2 u = 0 in D (R2 ). Exercise 2.145. Let u(x1 , x2 , x3 ) := H(x1 ) ⊗ δ(x2 ) ⊗ δ(x3 ) in D (R3 ), where H is the Heaviside function on the real line. Compute ∂1 u, ∂2 u, ∂3 u in D (R3 ). Exercise 2.146. Consider the sequence eix·ξ dξ, fj (x) := (2π)−n

∀ x ∈ Rn , ∀ j ∈ N.

(2.10.8)

[−j,j]×···×[−j,j] D (Rn )

Prove that fj −−−−→ δ. j→∞

Exercise 2.147. Solve each equation in D (R) for u. (1) (x − 1)u = δ; (2) xu = a, where a ∈ C ∞ (R); (3) xu = v, where v ∈ D (R). Exercise 2.148. Prove that the given convolutions exist and then compute them. (a) H ∗ H (b) H(−x) ∗ H(−x) (c) x2 H ∗ (sin x H) (d) χ[0,1] ∗ (xH) (e) |x|2 ∗ δ∂B(0,r) where r > 0 and δ∂B(0,r) is as defined in Exercise 2.128 corresponding to the surface Σ := ∂B(0, r). Exercise 2.149. Let a ∈ Rn \ {0}, uj := δja , vj := δ−ja , for each j ∈ N. Determine lim uj , lim vj , lim (uj ∗ vj ), in D (Rn ). j→∞

j→∞

j→∞


86

Exercise 2.150. For each j ∈ N, consider the functions defined by j j fj (x) : = (−1) 2 χ 1 1 (x) and gj (x) := (−1) , for every x ∈ R. Determine if −j,j

the given limits exist in D (R). (a) lim fj j→∞

(b) lim gj j→∞

(c) lim (fj ∗ gj ) j→∞

Exercise 2.151. Let u ∈ D (Rn ) and consider the map Λ : D(Rn ) → E(Rn ) given by Λ(ϕ) := u ∗ ϕ, for every ϕ ∈ C0∞ (Rn ). Prove that Λ is a well-defined, linear and continuous map. Also prove that Λ commutes with translations, that is, if x0 ∈ Rn and ϕ ∈ C0∞ (Rn ), then tx0 Λ(ϕ) = Λ tx0 ϕ , where tx0 is the map from (1.3.16). Exercise 2.152. Suppose Λ : D(Rn ) → E(Rn ) is a linear, continuous map that commutes with translations (in the sense explained in Exercise 2.151). Prove that there exists a unique u ∈ D (Rn ) such that Λ(ϕ) = u ∗ ϕ for every function ϕ ∈ C0∞ (Rn ). Exercise 2.153. Let u ∈ E (Rn ) be such that u, xα = 0 for every α ∈ Nn . Prove that u = 0 in E (Rn ). Exercise 2.154. Let u : E(R) → C be the functional defined by ⎡ ⎤ k 1 u(ϕ) := lim ⎣ ϕ ∀ ϕ ∈ C ∞ (R). − kϕ(0) − ϕ (0) ln k ⎦ , j

k→∞

j=1

Prove that u ∈ E (R) and determine supp u. Exercise 2.155. For each j ∈ N consider the function fj : R → R defined by fj (x) :=

j 2

if |x| ≤

1 j

E (R)

and fj (x) := 0 if |x| > 1j . Prove that fj −−−→ δ. j→∞

Exercise 2.156. For each j ∈ N consider the function fj : R → R defined by fj (x) := 1j if |x| ≤ j and fj (x) := 0 if |x| > j. Prove that the sequence {fj }j∈N converges in D (R) but not in E (R).

Exercise 2.157. Let ψ ∈ C0∞ (Rn ) be such that Rn ψ(x) dx = 1 and for each j ∈ N define fj : Rn → C by fj (x) := j n ψ(jx) for each x ∈ Rn . Prove that E (Rn )

fj −−−−→ δ. j→∞

Exercise 2.158. Let {xj }j∈N be a sequence of points in Rn . Prove that {xj }j∈N is convergent in Rn if and only if {δxj }j∈N is convergent in E (Rn ).


87

Exercise 2.159. Let a ∈ R and k ∈ N0 be given. Prove that (x + a)δa(k) = 2a δa(k) + kδa(k−1)

in

D (R),

(x2 − a2 )δa(k) = −2k a δa(k−1) + k(k − 1)δa(k−2) (−m)

with the convention that δa

(2.10.9) in

D (R),

:= 0 ∈ D (R) for each m ∈ N.

(2.10.10)

Chapter 3

The Schwartz Space and the Fourier Transform 3.1

The Schwartz Space of Rapidly Decreasing Functions

Recall that if f ∈ L1 (Rn ) then the Fourier transform of f is the mapping f3 : Rn → C defined by e−ix·ξ f (x) dx for each ξ ∈ Rn , (3.1.1) f3(ξ) := Rn

√ where i := −1 ∈ C. Note that under the current assumptions the integral in (3.1.1) is absolutely convergent (which means that f3 is well-defined pointwise in Rn ) and one has sup |f3(ξ)| ≤ f L1 (Rn )

ξ∈Rn

and f3 ∈ C 0 (Rn ).

(3.1.2)

where the second condition is seen by applying Lebesgue’s dominated convergence theorem. Hence, the mapping F : L1 (Rn ) → {g ∈ C 0 (Rn ) : g is bounded},

F f := f3,

∀ f ∈ L1 (Rn ), (3.1.3)

called the Fourier transform, is well-defined. Besides being continuous, functions belonging to the image of F also vanish at infinity. This property is proved next. Proposition 3.1. If f ∈ L1 (Rn ), then lim f3(ξ) = 0. |ξ|→∞

D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6 3, © Springer Science+Business Media New York 2013

89

CHAPTER 3. THE SCHWARTZ SPACE AND THE FOURIER...

90

Proof. First, consider the case when f ∈ C0∞ (Rn ). In this scenario, integrating n 4 (ξ) for each ξ ∈ Rn \ {0}, where Δf := ∂ 2 f . by parts gives |ξ|2 f3(ξ) = −Δf j=1

j

Hence, |f3(ξ)| ≤

4 (ξ)| Δf L1 (Rn ) |Δf ≤ , 2 |ξ| |ξ|2

∀ ξ ∈ Rn \ {0},

(3.1.4)

from which it is clear that lim f3(ξ) = 0 in this case. |ξ|→∞

Consider now the case when f is an arbitrary function in L1 (Rn ). Since C0∞ (Rn ) is dense in the latter space, for each ε > 0 fixed there exists g ∈ C0∞ (Rn ) such that f − gL1 (Rn ) ≤ 2ε . Also, from what we proved so far, there exists R > 0 with the property that |3 g (ξ)| ≤ ε2 whenever |ξ| > R. Keeping this in mind and using the linearity of the Fourier transform as well as the estimate in (3.1.2) we may write |f3(ξ)| ≤ |(f − g)(ξ)| + |3 g (ξ)| ≤ f − gL1 (Rn ) + From this, the desired conclusion follows. therefore complete.

ε ≤ ε, 2

if |ξ| > R.

(3.1.5)

The proof of the proposition is

We are very much interested in the possibility of extending the action of the Fourier transform from functions to distributions, though this is going to be accomplished later. For now, we note the following consequence of Fubini’s theorem: f (x)3 g (x) dx, ∀ f, g ∈ L1 (Rn ). (3.1.6) f3(ξ)g(ξ) dξ = Rn

Rn

Identity (3.1.6) might suggest defining the Fourier transform of a distribution based on duality. However, there is a serious impediment in doing so. Specif3 ∈ C ∞ (Rn ) (as may be seen ically, while for every ϕ ∈ C0∞ (Rn ) we have ϕ directly from (3.1.1)) and ϕ 3 decays at infinity (as proved in Proposition 3.1), we nonetheless have F (C0∞ (Rn )) ⊂ C0∞ (Rn ).

(3.1.7)

ϕ ∈ C0∞ (Rn ) and ϕ 3 ∈ C0∞ (Rn ) =⇒ ϕ = 0.

(3.1.8)

In fact, we claim that

3 has compact To see that this is the case, suppose ϕ ∈ C0∞ (Rn ) is such that ϕ support in Rn , and pick an arbitrary point x∗ = (x∗1 , . . . , x∗n ) ∈ Rn . Define the function Φ : C → C by setting n ∗ Φ(z) := e−izx1 + j=2 xj xj ϕ(x1 , . . . , xn ) dx1 · · · dxn , for z ∈ C. (3.1.9) Rn

Then Φ is analytic in C and Φ(t) = ϕ(t, 3 x∗2 , . . . , x∗n ) for each t ∈ R. Given that ϕ 3 has compact support, this implies that Φ vanishes on R \ [−R, R] if

3.1. THE SCHWARTZ SPACE OF RAPIDLY DECREASING...

91

R > 0 is suitably large. The identity theorem for ordinary analytic functions of one complex variable then forces Φ = 0 everywhere in C. In particular, ϕ(x 3 ∗ ) = Φ(x∗1 ) = 0. Since x∗ ∈ Rn has been chosen arbitrarily, we conclude that ϕ 3 = 0 in Rn . However, as we will see in the sequel, the Fourier transform is injective on C0∞ (Rn ), so (3.1.8) follows. To overcome the difficulty highlighted in (3.1.7), we introduce a new (topological vector) space of functions, that contains C0∞ (Rn ), is invariant under F , and that has a dual that is a subspace of D (Rn ). This is the space of Schwartz functions, named after the French mathematician Laurent–Mo¨ıse Schwartz (1915–2002) who pioneered the theory of distributions and first considered this space in connection with the Fourier transform. Before presenting the definition of Schwartz functions, we introduce some notation, motivated by the observation that each time a partial derivative of ϕ 3 is taken, the exponential introduces i as a multiplicative factor. To adjust for this factor, it is therefore natural to re-normalize the ordinary partial differentiation operators as follows: Dj := 1i ∂j ,

j = 1, . . . , n,

Dα := D1α1 · · · Dnαn ,

D := (D1 , . . . , Dn ),

∀ α = (α1 , . . . , αn ) ∈ Nn0 .

(3.1.10)

At times, we will also use subscripts to specify the variable with respect to which the differentiation is taken. For example, Dxα stands for Dα with the additional specification that the differentiation is taken with respect to the variable x ∈ Rn . Fix now α, β ∈ Nn0 and observe that for each ϕ ∈ C0∞ (Rn ) integration by parts implies (−Dx )β (e−ix·ξ ) (−x)α ϕ(x) dx 3 = ξ β e−ix·ξ (−x)α ϕ(x) dx = ξ β Dξα ϕ(ξ) Rn

=

Rn

Hence,

e−ix·ξ Dxβ (−x)α ϕ(x) dx.

3 ≤ sup ξ β Dξα ϕ(ξ)

ξ∈Rn

Rn

(3.1.11)

Rn

|Dxβ (xα ϕ(x))| dx < ∞.

(3.1.12)

The conclusion from (3.1.12) is that derivatives of any order of ϕ 3 decrease at ∞ faster than any polynomial. This suggests making the following definition. Definition 3.2. The Schwartz class of rapidly decreasing functions is defined as (3.1.13) S(Rn ) := ϕ ∈ C ∞ (Rn ) : sup |xβ ∂ α ϕ(x)| < ∞, ∀ α, β ∈ Nn0 . x∈Rn

We shall simply say that ϕ is a Schwartz function if ϕ ∈ S(Rn ). Obviously, C0∞ (Rn ) ⊂ S(Rn ) ⊂ C ∞ (Rn )

(3.1.14)

though both inclusions are strict. An example of a Schwartz function that is not compactly supported is provided below.


92

Exercise 3.3. Prove that for each fixed number a ∈ (0, ∞), the function f , 2 defined by f (x) := e−a|x| for each x ∈ Rn , belongs to S(Rn ) and has the property that supp f = Rn . Other elementary observations pertaining to the Schwartz class from Definition 3.2 are recorded below. Remark 3.4. One has S(Rn ) = ϕ ∈ C ∞ (Rn ) : sup |xβ Dα ϕ(x)| < ∞, ∀ α, β ∈ Nn0 ,

(3.1.15)

and if ϕ ∈ C ∞ (Rn ), then ϕ ∈ S(Rn ) if and only if ∀ m ∈ N0 , ∀ α ∈ Nn0 , |α| ≤ m. sup (1 + |x|)m |∂ α ϕ(x)| < ∞,

(3.1.16)

x∈Rn

x∈Rn

Indeed, (3.1.15) is immediate from Definition 3.2. Also, the second claim in the remark readily follows from the observation that for each m ∈ N0 there exists C ∈ [1, ∞) with the property that C −1 |x|m ≤ |xγ | ≤ C|x|m , ∀ x ∈ Rn . (3.1.17) |γ|=m

In turn, the second inequality in (3.1.17) is seen by noting that for each α ∈ Nn0 and x = (x1 , . . . , xn ) ∈ Rn , we have |xα | = |x1 |α1 · · · |xn |αn ≤ |x|α1 · · · |x|αn = |x||α| .

(3.1.18) |xγ | To justify the first inequality in (3.1.17), consider the function g(x) := |γ|=m

for x ∈ Rn . Then its restriction to S n−1 attains a nonzero minimum, and the desired inequality follows by rescaling. Exercise 3.5. Prove that if f ∈ S(Rn ) then for each α, β ∈ Nn0 and N ∈ N there exists C = Cf,N,α,β ∈ (0, ∞) such that α β x ∂ f (x) ≤

C (1 + |x|)N

for each x ∈ Rn .

(3.1.19)

Use this to deduce that S(Rn ) ⊂ Lp (Rn ) for every p ∈ [1, ∞]. allows us to In particular, S(Rn ) ⊂ L1 (Rn ) which, in concert with (3.1.3), consider the Fourier transform on S(Rn ). Also, F C0∞ (Rn ) ⊆ S(Rn ) as seen from the computation in (3.1.12). Clearly, S(Rn ) is a vector space when endowed with the canonical operations of addition of functions and multiplication by complex numbers. For a detailed discussion regarding the topology we consider on S(Rn ) see Sect. 13.1. We continue to denote by S(Rn ) the respective topological vector space and we single out here a few important facts that are useful for our analysis.


93

Fact 3.6. S(Rn ) is a Frechét space, that is, S(Rn ) is a locally convex, metrizable, complete, topological vector space over C. Fact 3.7. A sequence {ϕj }j∈N ⊂ S(Rn ) converges in S(Rn ) to some ϕ ∈ S(Rn ) provided ∀ α, β ∈ Nn0 , (3.1.20) sup xβ ∂ α ϕj (x) − ϕ(x) −−−→ 0, j→∞

x∈Rn

S(Rn )

in which case we use the notation ϕj −−−−→ ϕ. j→∞

S(Rn )

Exercise 3.8. Use (3.1.17) to prove that ϕj −−−−→ ϕ if and only if j→∞

sup

x∈Rn α∈Nn 0 , |α|≤m

(1 + |x|)m ∂ α [ϕj (x) − ϕ(x)] −−−→ 0,

∀ m ∈ N0 .

j→∞

(3.1.21)

It is useful to observe that the Schwartz class embeds continuously into Lebesgue spaces. Exercise 3.9. Prove that if p ∈ [1, ∞] and a sequence of functions {ϕj }j∈N in S(Rn ) converges in S(Rn ) to some ϕ ∈ S(Rn ) then {ϕj }j∈N also converges in Lp (Rn ) to ϕ. Hint: Use Exercise 3.8. Definition 3.10. The space of slowly increasing functions in Rn is defined as L(Rn ) := a ∈ C ∞ (Rn ) : ∀ α ∈ Nn0 ∃ k ∈ N0 such that sup (1 + |x|)−k |∂ α a(x)| < ∞ . (3.1.22) x∈Rn

Note that an immediate consequence of Definition 3.10 is that L(Rn ) is stable under differentiation (i.e., if a ∈ L(Rn ) then ∂ α a ∈ L(Rn ) for every α ∈ Nn0 ). Also, (3.1.23) S(Rn ) ⊂ L(Rn ), though L(Rn ) contains many additional functions that lack decay as, for example, the class of polynomials (other examples are contained in the two exercises below). 2

Exercise 3.11. Prove that the function f (x) := ei|x| , x ∈ Rn , belongs to L(Rn ). Exercise 3.12. Prove that for each k ∈ R the function f (x) := (1 + |x|2 )k , x ∈ Rn , belongs to L(Rn ). Some other basic properties of the Schwartz class are collected in the next theorem.


94

Theorem 3.13. The following statements are true. (a) For each a ∈ L(Rn ), the mapping S(Rn ) ϕ → aϕ ∈ S(Rn ) is welldefined, linear, and continuous. (b) For each α ∈ Nn0 , the mapping S(Rn ) ϕ → ∂ α ϕ ∈ S(Rn ) is well-defined, linear, and continuous. (c) D(Rn ) → S(Rn ) → E(Rn ) and the embeddings are continuous. (d) C0∞ (Rn ) is sequentially dense in S(Rn ). Also, the Schwartz class S(Rn ) is sequentially dense in E(Rn ). (e) If m, n ∈ N and f ∈ S(Rm ), g ∈ S(Rn ), then f ⊗ g ∈ S(Rm × Rn ) and the mapping S(Rm ) × S(Rn ) (f, g) → f ⊗ g ∈ S(Rm × Rn )

(3.1.24)

is bilinear and continuous. (f ) If f, g ∈ S(Rn ) then f ∗ g ∈ S(Rn ) and the mapping S(Rn ) × S(Rn ) (f, g) → f ∗ g ∈ S(Rn )

(3.1.25)

is bilinear and continuous. Proof. Clearly, the mappings in (a) and (b) are linear. By Fact 3.6 and Theorem 13.14, their continuity is equivalent with sequential continuity at 0, something that can be easily checked using Fact 3.7. Moving on to the statement in (c), we first prove that D(Rn ) embeds continuously into S(Rn ). Let ι : D(Rn ) → S(Rn ) be defined by ι(ϕ) := ϕ for each ϕ ∈ C0∞ (Rn ). From (3.1.14) this is a well-defined and linear mapping. To see that ι is sequentially continuous D(Rn )

at 0 ∈ D(Rn ), consider ϕj −−−−→ 0. Then there exists a compact set K ⊂ Rn j→∞

with the property that supp ϕj ⊆ K for every j ∈ N, and lim sup |∂ α ϕj | = 0 j→∞ x∈K

for every α ∈ Nn0 . Hence, for any α, β ∈ Nn0 , sup xβ ∂ α ϕj (x) = sup xβ ∂ α ϕj (x) ≤ C sup ∂ α ϕj (x) −−−→ 0, x∈Rn

x∈K

x∈K

j→∞

(3.1.26)

proving that ι is sequentially continuous at the origin. Recalling now Fact 3.6 and Theorem 13.5, we conclude that ι is continuous. Our next goal is to show that S(Rn ) embeds continuously in E(Rn ). From (3.1.14) we have that the identity ι : S(Rn ) → E(Rn ) given by ι(f ) := f , for each f ∈ S(Rn ), is a well-defined linear map. By Fact 3.6, Fact 1.8, and Theorem 13.14, ι is continuous if and only if it is sequentially continuous at S(Rn )

zero. However, if fj −−−−→ 0, then for any compact set K ⊂ Rn and any α ∈ Nn0 ,

j→∞

lim sup |∂ α fj (x)| ≤ lim sup |∂ α fj (x)| = 0.

j→∞ x∈K

j→∞ x∈Rn

(3.1.27)


95

This shows that ι is sequentially continuous at zero, finishing the proof (c). Next, we prove the statement in (d). Let f ∈ S(Rn ) be arbitrary and, for some fixed ψ ∈ C0∞ (Rn ) satisfying ψ = 1 in a neighborhoodofB(0, 1), define the sequence of functions fj : Rn → C by setting fj (x) := ψ xj f (x) for every x ∈ Rn and every j ∈ N. Then fj ∈ C0∞ (Rn ) and fj = f on B(0, j) for each j ∈ N. We claim that S(Rn )

fj −−−−→ f. j→∞

(3.1.28)

n To see that this is the case, if α, β ∈ N x0 are arbitrary, by making use of Leibniz’s formula (13.2.4) and the fact that ψ j = 1 for each x ∈ B(0, j), we may write x β β α α! sup x ∂ fj (x) − f (x) = sup x ∂ γ f (x)∂ α−γ ψ − 1 γ!(α − γ)! j x∈Rn x∈Rn γ≤α x α! ≤ sup xβ ∂ γ f (x)∂ α−γ ψ γ!(α − γ)! j |x|≥j γ 0, depending only on ψ and α, such that α−γ x C ≤ , sup ∂ ψ ∀ γ ∈ Nn0 , γ < α, ∀ j ∈ N. (3.1.30) j j |x|≥j

Also, since f ∈ S(Rn ), we may invoke (3.1.19) to conclude that there exists some C = Cf,α,β ∈ (0, ∞) such that C x − 1 ≤ 1 + ψL∞ (Rn ) . sup xβ ∂ α f (x) ψ (3.1.31) j j |x|≥j Combining (3.1.29)–(3.1.31), and keeping in mind that f ∈ S(Rn ), we obtain sup |xβ ∂ α (fj (x) − f (x))|

x∈Rn

(3.1.32)

β C C α! γ sup x ∂ f (x) + −−−→ 0. ≤ j x∈Rn γ!(α − γ)! j j→∞ γ≤α S(Rn )

This shows that fj −−−−→ f and completes the proof of the fact that C0∞ (Rn ) j→∞

is sequentially dense in S(Rn ). The sequential continuity of S(Rn ) in E(Rn ) is a consequence of Exercise 1.13 and (3.1.14). This completes the proof of (d). The claims in part (e) follow using the observation that (x, y)(α,β) ∂xγ ∂yμ (f ⊗ g)(x, y) = xα ∂ γ f (x)y β ∂ μ g(y), (3.1.33)


96

for every (x, y) ∈ Rm × Rn , for every f ∈ S(Rm ), g ∈ S(Rn ), and every n α, γ ∈ Nm 0 , β, μ ∈ N0 . Consider now the statement in (g). From Exercise 3.5 we know that S(Rn ) ⊂ 1 L (Rn ), thus S(Rn ) ∗ S(Rn ) is meaningful. To see that S(Rn ) ∗ S(Rn ) ⊂ S(Rn ), fix some arbitrary f, g ∈ S(Rn ) and α, β ∈ Nn0 . Then, making use of the binomial theorem (cf. Theorem 13.7) as well as Exercise 3.5, we may estimate β α β α sup x ∂x (f ∗ g)(x) = sup ((x − y) + y) ∂x f (x − y)g(y)dy x∈Rn

x∈Rn

≤ sup

x∈Rn

≤ Cα,β ≤ Cα,β

γ≤β

β! γ!(β − γ)!

Rn

Rn

|(x − y)β−γ (∂ α f )(x − y)||y γ g(y)| dy (3.1.34)

sup (1 + |z|)|β| ∂ α f (z)

z∈Rn

Rn

(1 + |y|)|β| |g(y)| dy

sup (1 + |y|)|β|+n+1 g(y) < ∞. sup (1 + |z|)|β| ∂ α f (z)

z∈Rn

y∈Rn

This implies f ∗ g ∈ S(Rn ). The fact that the mapping in (e) is bilinear is immediate from definitions. As regards its continuity, we may invoke again Theorem 13.14 and Fact 3.6 to reduce matters to proving sequential continuity instead. However, the latter is apparent from the estimate in (3.1.34). This finishes the proof of the theorem. n Exercise 3.14. Assume ψ ∈ S(R ) isn given and, for each j ∈ N, define xthat the function ψj (x) := ψ j for every x ∈ R . Prove that S(Rn )

ψj f −−−−→ ψ(0)f j→∞

for every f ∈ S(Rn ).

(3.1.35)

Hint: Adapt estimates (3.1.29) and (3.1.30) to the current setting and, in place of (3.1.31), this time use the mean value theorem for the term ψ(x/j) − ψ(0) to get a decay factor of the order 1/j. Proposition 3.15. Let m, n ∈ N. Then C0∞ (Rm ) ⊗ C0∞ (Rn ) is sequentially dense in S(Rm × Rn ). Proof. Since the topology on S(Rm × Rn ) is metrizable (recall Fact 3.6), there exists a distance function d : S(Rm × Rn ) × S(Rm × Rn ) → [0, ∞) that induces its topology. Hence, S(Rm ×Rn )

fj −−−−−−−→ f j→∞

if and only if

lim d(fj , f ) = 0.

j→∞

(3.1.36)

Now let f ∈ S(Rm × Rn ). Then by part (d) in Theorem 3.13, there exists a sequence {fj }j∈N ⊂ C0∞ (Rm × Rn ) with the property that d(fj , f ) < 1j for every j ∈ N. Furthermore, by Proposition 2.73, for each fixed number j ∈ N, there


97

D(Rm ×Rn )

exists a sequence {gjk }k∈N ⊂ C0∞ (Rm ) ⊗ C0∞ (Rn ) such that gjk −−−−−−−→ fj . k→∞

In particular, by (c) in Theorem 3.13, S(Rm ×Rn )

gjk −−−−−−−→ fj k→∞

for each j ∈ N,

(3.1.37)

thus lim d(gjk , fj ) = 0

k→∞

for each j ∈ N.

(3.1.38)

Condition (3.1.38) implies that for each j ∈ N there exists kj ∈ N with the property that d(gjkj , fj ) < 1j . Now the sequence {gjkj }j∈N ⊂ C0∞ (Rm ) ⊗ C0∞ (Rn ) satisfies d(gjkj , f ) ≤ d(gjkj , fj ) + d(fj , f )
M + n and note that this ensures Rn (1 + |x − y|)M (1 + |y|)−N dy < ∞, proving the first claim in the statement. The fact that f ∗ g ∈ C ∞ (Rn ) is now seen in a similar fashion given that ∂ α f continues to be in L(Rn ) for every α ∈ Nn0 . n

Exercise 3.18. Prove that Lp (Rn ) ∗ S(Rn ) ⊆ C ∞ (Rn ) for every p ∈ [1, ∞]. Hint: Use the blueprint as in the proof of Proposition 3.17, using Hölder’s inequality in place of estimates for slowly growing functions, and arrange matters so that all derivatives fall on the Schwartz function. We conclude this section with an integration by parts formula that will be useful shortly. Lemma 3.19. If f ∈ L(Rn ) and g ∈ S(Rn ), then for every α ∈ Nn0 the following integration by parts formula holds: (∂ α g)(x)f (x) dx = (−1)|α| g(x)(∂ α f )(x) dx. (3.1.43) Rn

Rn

Proof. Fix f ∈ L(Rn ) and g ∈ S(Rn ). Since the classes L(Rn ) and S(Rn ) are stable under differentiation, it suffices to show that for each j ∈ {1, . . . , n} we have (∂j g)(x)f (x) dx = − g(x)(∂j f )(x) dx, (3.1.44) Rn

Rn

since (3.1.43) then follows by iterating (3.1.44). To this end, fix some integer j ∈ {1, . . . , n} along with some arbitrary R ∈ (0, ∞). The classical integration by parts formula in the bounded, smooth, domain B(0, R) ⊂ Rn then reads (cf. (13.7.4)) (∂j g)(x)f (x) dx = − g(x)(∂j f )(x) dx B(0,R)

B(0,R)

+

g(x)f (x)(xj /R) dσ(x).

(3.1.45)

∂B(0,R)

From part (a) in Theorem 3.13 we know that f g ∈ S(Rn ). Based on this and Exercise 3.5, it follows that g(x)f (x)(xj /R) dσ(x) ≤ ωn−1 Rn−1 sup |(f g)(x)| ∂B(0,R)

|x|=R

≤ CR−1 −−−−→ 0. R→∞

(3.1.46)

3.2. THE ACTION OF THE FOURIER TRANSFORM ...

99

On the other hand, (∂j g)f, g∂j f ∈ S(Rn ) ⊂ L1 (Rn ). As such, passing to limit R → ∞ in (3.1.45) yields (3.1.44) on account of Lebesgue’s dominated convergence theorem and (3.1.46).

3.2

The Action of the Fourier Transform on the Schwartz Class

Originally, we defined the Fourier transform in (3.1.3) as a mapping acting on functions from L1 (Rn ). Since S(Rn ) is contained in L1 (Rn ), it makes sense to consider the Fourier transform acting on the Schwartz class. In this section we study the main properties of the Fourier transform in this setting. The reader is advised that we use the symbols F and 3· interchangeably to denote this Fourier transform. To state our first major result pertaining to the Fourier transform in this setting, recall the notation introduced in (3.1.10). Theorem 3.20. The following statements are true. α f (ξ) = ξ α f3(ξ) for every 5 (a) If f ∈ S(Rn ) and α ∈ Nn0 are arbitrary, then D n ξ∈R . α f (ξ) = (−D)α f3(ξ) for (b) If f ∈ S(Rn ) and α ∈ Nn0 are arbitrary, then x5 n every ξ ∈ R .

(c) The Fourier transform, originally introduced in the context of (3.1.3), induces a mapping F : S(Rn ) → S(Rn ) that is linear and continuous. (d) If m, n ∈ N, f ∈ S(Rm ) and g ∈ S(Rn ), then f ⊗ g = f3 ⊗ 3 g. Proof. Fix f ∈ S(Rn ) and α ∈ Nn0 . Then the decay of f (cf. (3.1.19)) ensures that we may differentiate under the integral sign in (3.1.1) and obtain e−ix·ξ (−x)α f (x) dx Dα f3(ξ) = Rn

α f (ξ), = (−x)

∀ ξ ∈ Rn .

(3.2.1)

From this, the statement in (b) readily follows. Also, if β ∈ Nn0 is arbitrary, then using the first identity in (3.2.1), the fact that ξ β e−ix·ξ = (−Dx )β (e−ix·ξ ), and the integration by parts formula from Lemma 3.19, we obtain (−Dx )β (e−ix·ξ )(−x)α f (x) dx ξ β Dα f3(ξ) = Rn

=

Rn

e−ix·ξ Dxβ (−x)α f (x) dx,

∀ ξ ∈ Rn .

(3.2.2)


100

The formula in (a) follows by specializing (3.2.2) to the case α = (0, . . . , 0) ∈ Nn0 . In addition, starting with (3.2.2) we may estimate sup |ξ D f3(ξ)| ≤ β

2 −n

α

ξ∈Rn

Rn

(1 + |x| )

× sup

x∈Rn

dx ×

(1 + |x|2 )n Dxβ xα f (x) < ∞,

where the finiteness condition is a S(Rn ). Clearly, (3.2.1) also implies with (3.2.3), shows that f3 ∈ S(Rn ). The fact that this mapping is linear S(Rn )

(3.2.3)

consequence of the membership of f to that f3 ∈ C ∞ (Rn ) which, in combination Hence, the mapping in (c) is well-defined. is immediate from definition. In addition,

if fj −−−−→ 0, then based on the first inequality in (3.2.3) we have that, for each j→∞

m, k ∈ N0 , sup

ξ∈Rn |α|≤m, |β|≤k

|ξ β ∂ α f3j (ξ)|

≤C

sup

n

(1 + |x|2 )n ∂ β xα fj (x) → 0 as j → ∞.

(3.2.4)

x∈R |α|≤m, |β|≤k S(Rn )

In view of Exercise 3.8, this proves F fj −−−−→ 0. The latter combined with j→∞

Fact 3.6 and Theorem 13.14 then implies that F is continuous from S(Rn ) into S(Rn ). At this stage we are left with proving the statement in (d). To this end, fix f ∈ S(Rm ) and g ∈ S(Rn ). Then by part (e) in Theorem 3.13, we obtain that f ⊗ g ∈ S(Rm × Rn ), so F (f ⊗ g) is well-defined. Furthermore, by applying Fubini’s theorem, we may write e−ix·ξ−iy·η (f ⊗ g)(x, y) dy dx f ⊗ g(ξ, η) = Rm

Rn

e

=

−ix·ξ

e−iy·η g(y) dy

f (x) dx

Rm

Rn

= f3(ξ)3 g (η) = (f3 ⊗ g3)(ξ, η),

∀ ξ ∈ Rm , ∀ η ∈ Rn .

(3.2.5)

This finishes the proof of the theorem. Example 3.21. Suppose λ ∈ C satisfies Re(λ) > 0 and if λ = reiθ for r > 0 √ 2 1 and −π/2 < θ < π/2, set λ 2 := reiθ/2 . Consider the function f (x) := e−λ|x| for x ∈ Rn . Then f ∈ S(Rn ) and f3(ξ) =

π n2 λ

e−

|ξ|2 4λ

for each ξ ∈ Rn .

(3.2.6)


101

Proof. Fix λ ∈ C satisfying the given hypotheses. Then Exercise 3.3 ensures 2 2 that f is a Schwartz function. Also, f (x) = e−λx1 ⊗ · · · ⊗ e−λxn for each x = (x1 , . . . , xn ) ∈ Rn . As such, part (d) in Theorem 3.20 shows that it suffices 2 to prove (3.2.6) when n = 1, in which case f (x) = e−λx for every x ∈ R. Suppose that this is the case and observe that f satisfies f + 2λxf = 0 in R. By taking the Fourier transform of both sides of this differential equation, and using (a) and (b) in Theorem 3.20, we arrive at ξ f3+ 2λ f3 = 0. The solution to this ξ2 latter ordinary differential equation is f3(ξ) = f3(0)e− 4λ for ξ ∈ R. There remains

1 2 to show that f3(0) = πλ 2 . Since by definition, f3(0) = R f (x) dx = R e−λx dx, we are left with showing that π 12 2 e−λx dx = whenever λ ∈ C has Re(λ) > 0. (3.2.7) λ R 1

2 Corresponding to the case when λ ∈ R+ , the identity R e−λx dx = πλ 2 is a standard exercise in basic calculus. To extend this to complex λ’s, observe that the function π 12 2 e−zx dx − for z ∈ C with Re(z) > 0, h(z) := z R is analytic and equal to zero for every z ∈ R+ . This forces h(z) = 0 for all z ∈ C 1 with Re(z) > 0. Thus, f3(0) = π 2 , as desired. λ

Exercise 3.22. Let a ∈ (0, ∞) and b ∈ R be fixed. Show that if x ∈ R then F (e−ax

2

+ibx

)(ξ) =

π 12 a

Hint: First prove that F (e−ax then use Example 3.21.

2

e−

+ibx

(ξ−b)2 4a

for every ξ ∈ R.

(3.2.8)

2

)(ξ) = (F (e−ax ))(ξ − b) for every ξ ∈ R,

Exercise 3.23. Prove that if A ∈ Mn×n (R) is such that det A = 0, then ϕ ◦ A−1 = | det A| (ϕ 3 ◦ A ),

∀ ϕ ∈ S(Rn ),

(3.2.9)

where A−1 and A denote, respectively, the inverse and the transpose of the matrix A. Next we note a consequence 3.20 of basic importance. As moti of Theorem vation, suppose P (D) = aα Dα is a differential operator of order m ∈ N, |α|≤m

with constant coefficients aα ∈ C, for every α ∈ Nn0 with |α| ≤ m. Furthermore, assume that f ∈ S(Rn ) has been given. Then any solution u ∈ S(Rn ) of the differential equation P (D)u = f in Rnalso satisfies P (ξ)3 u(ξ) = f3(ξ) for each n α aα ξ . In particular, if P (ξ) has no ξ ∈ R , where we have set P (ξ) := zeros, then u 3(ξ) =

f(ξ) P (ξ) ,

|α|≤m

for every ξ ∈ Rn . This gives a formula for the Fourier


102

transform of u. In order to find a formula for u itself, the natural question that arises is whether we can reconstruct u from u 3. The next theorem provides a positive answer to this question in the class of Schwartz functions. Theorem 3.24. The mapping F : S(Rn ) → S(Rn ) is an algebraic and topologic isomorphism, that is, it is bijective, continuous, and its inverse is also continuous. In addition, its inverse is the operator F −1 : S(Rn ) → S(Rn ) given by the formula −1 −n (F g)(x) = (2π) eix·ξ g(ξ) dξ, ∀ x ∈ Rn , ∀ g ∈ S(Rn ). (3.2.10) Rn

Proof. The proof of the fact that the mapping F −1 : S(Rn ) → S(Rn ) defined as in (3.2.10) is well-defined, linear, and continuous is similar to the proof of part (c) in Theorem 3.20. There remains to show that F −1 ◦ F = I = F ◦ F −1 on S(Rn ), where I is the identity operator on S(Rn ). To proceed, observe that the identity F −1 ◦ F = I is equivalent to −n eix·ξ f3(ξ) dξ = f (x), (3.2.11) ∀ x ∈ Rn , ∀ f ∈ S(Rn ). (2π) Rn

As regards (3.2.11), fix some function f ∈ S(Rn ) along with some point x ∈ Rn . Recall (cf. (3.1.1)) that e−iy·ξ f (y) dy, ∀ ξ ∈ Rn . (3.2.12) f3(ξ) = Rn

As such, one is tempted to directly replace f3(ξ) in (3.2.11) by the right-hand side of (3.2.12) and then use Fubini’s theorem to reverse the order of integration in the variables ξ and y. There is, however, a problem in carrying out this approach, since the function ei(x−y)·ξ f (y), considered jointly in the variable (ξ, y) ∈ Rn × Rn , does not belong to L1 (Rn × Rn ), hence Fubini’s theorem is not necessarily applicable. To remedy this problem, we introduce a “convergence factor” in the form of a suitable family of Schwartz functions ψ ε , indexed by ε > 0 (to be specified shortly), designed to provide control in the variable ξ, thus ensuring the applicability of Fubini’s theorem.

Hence, in place of Rn eix·ξ f3(ξ) dξ the idea is to consider Rn eix·ξ ψ ε (ξ)f3(ξ) dξ, for which we may write (granted that ψ ε ∈ S(Rn )) eix·ξ ψ ε (ξ)f3(ξ) dξ = eix·ξ ψ ε (ξ) e−iy·ξ f (y) dy dξ Rn

Rn

Rn

e−i(y−x)·ξ ψ ε (ξ)f (y) dy dξ

= Rn ×Rn

= Rn

=

Rn

e−i(y−x)·ξ ψ ε (ξ) dξ f (y) dy

Rn

4ε (y − x)f (y) dy = ψ

4ε (z)f (x + z) dz. ψ

Rn

(3.2.13)


103

Given the goal we have in mind (cf. (3.2.11)), as well as the format of the current identity, we find it convenient to define ψ ε by setting ψ ε (ξ) := ϕ(ε ξ) for each ξ ∈ Rn and ε > 0,

(3.2.14)

where ϕ ∈ S(Rn ) is to be specified momentarily. The rationale behind this 4ε (x) are reasonably easy to compute. choice is that both lim ψ ε (ξ) and lim ψ ε→0+

ε→0+

This will eventually allow us to deduce (3.2.11) from (3.2.13) by letting ε → 0+ . Concretely, from (3.2.14) we have lim+ ψ ε (ξ) = ϕ(0), while from the definition ε→0

of the Fourier transform it is immediate that z 4ε (z) = ε−n ϕ ψ 3 ε (z) for every z ∈ Rn . = ϕ 3 ε

(3.2.15)

Keeping this in mind and employing part (a) in Exercise 2.22 we obtain 4ε (z)f (x + z) dz = lim ϕ 3 ε (z)f (x + z) dz ψ lim ε→0+

ε→0+

Rn

=

Rn

Rn

ϕ(z) 3 dz f (x).

(3.2.16)

Also, on account of (3.2.14) and the fact that f3 ∈ S(Rn ) ⊂ L1 (Rn ), Lebesgue’s dominated convergence theorem gives eix·ξ ψ ε (ξ)f3(ξ) dξ = lim+ eix·ξ ϕ(ε ξ)f3(ξ) dξ lim+ ε→0

ε→0

Rn

Rn

= ϕ(0)

eix·ξ f3(ξ) dξ.

(3.2.17)

Rn n In summary,

(3.2.13), (3.2.16), and (3.2.17), show that whenever ϕ ∈ S(R ) is such that Rn ϕ(z) 3 dz = 0 we have eix·ξ f3(ξ) dξ = f (x), (3.2.18) ∀ x ∈ Rn , ∀ f ∈ S(Rn ), C Rn

where the normalization constant C is given by ϕ(0) . ϕ(z) 3 dz Rn

C :=

(3.2.19)

As such, (3.2.11) will follow as soon as we prove that C = (2π)−n . For this task, we have the freedom of choosing the function ϕ ∈ S(Rn ) and a candidate that springs to mind is the Schwartz function from Example 3.21 (say, in the partic2 ular case when λ = 1). Hence, if ϕ(x) := e−|x| for each x ∈ Rn , formula (3.2.6) gives n

ϕ(ξ) 3 = π 2 e−

|ξ|2 4

for each ξ ∈ Rn .

(3.2.20)


104

Consequently, n ϕ(ξ) 3 dξ = π 2 Rn

e−

|ξ|2 4

n

dξ = π 2

Rn

e−

|t|2 4

n dt

n

n

= π 2 (4π) 2 = (2π)n .

R

(3.2.21)

where the second equality is simply Fubini’s theorem, while the third equality is provided by (3.2.7) with λ := 1/4. Since in this case we also have ϕ(0) = 1, it follows that C = (2π)−n , as wanted. This finishes the justification of the identity F −1 ◦ F = I on S(Rn ). The same approach also works for the proof of F ◦ F −1 = I, completing the proof of the theorem. In what follows, for an arbitrary function f : Rn → C we define f ∨ (x) := f (−x),

∀ x ∈ Rn .

(3.2.22)

Exercise 3.25. Prove that the mapping S(Rn ) f → f ∨ ∈ S(Rn )

(3.2.23)

is well-defined, linear, and continuous. Hint: Use Fact 3.6 and Theorem 13.14. Recall that z denotes the complex conjugate of z ∈ C. Exercise 3.26. Let f ∈ S(Rn ). Then the following formulas hold: (1) f4∨ = (f3 )∨ . 4∨ (2) f3 = f . 3 (3) f3 = (2π)n f ∨ .

(4) Rn f (x) dx = f3(0) and Rn f3(ξ) dξ = (2π)n f (0). Proposition 3.27. Let f, g ∈ S(Rn ). Then the following identities hold:

g (x) dx = Rn f3(ξ)g(ξ) dξ. (a) Rn f (x)3 (b)

f (x)g(x) dx = (2π)−n Rn f3(ξ) g3(ξ) dξ an identity referred to in the literature as Parseval’s identity. Rn

(c) f ∗ g = f3 · 3 g. (d) f5 · g = (2π)−n f3 ∗ g3. Proof. The identity in (a) follows via a direct computation using Fubini’s the4 ∨ g3 = g4 = (2π)n g which, when orem. Also, based on Exercise 3.26, we have 3


105

combined with (a) gives (b). The identity in (c) follows using Fubini’s theorem. Specifically, for each ξ ∈ Rn we may write e−ix·ξ (f ∗ g)(x) dx = e−ix·ξ f (x − y)g(y) dy dx f ∗ g(ξ) =

Rn

g(y)

=

Ru

Rn

Rn

e−ix·ξ f (x − y) dx dy

e−iy·ξ g(y)

=

Rn

e−iz·ξ f (z) dz dy = f3(ξ)3 g (ξ),

(3.2.24)

Rn

R

as wanted. Next, the identity from (c) combines with Exercise 3.26 to yield 3 f3 ∗ g3 = f3 · 3 g = (2π)2n f ∨ · g ∨ = (2π)2n (f · g)∨ . 3

(3.2.25)

Applying now the Fourier transform to the most extreme sides of (3.2.25) and once again invoking Exercise 3.26, we obtain ∨

−n

(2π)

∨ · g)∨ = f5 · g. f3 ∗ g3 = (2π)−2n f3 ∗ 3 g = (f

(3.2.26)

This completes the proof of the proposition. Remark 3.28. (1) It is not difficult to see via a direct computation that we also have f3(ξ)g(ξ) dξ, ∀ f ∈ L1 (Rn ), ∀ g ∈ S(Rn ). f (x)3 g (x) dx = Rn

Rn

(3.2.27)

(2) Parseval’s identity written for f = g ∈ S(Rn ) becomes f3L2 (Rn ) = (2π) 2 f L2(Rn ) . n

(3.2.28)

As a consequence, since C0∞ (Rn ) is dense in L2 (Rn ), the Fourier transform F may be extended to a linear operator from L2 (Rn ) into itself, and the latter identity continues to hold for every f ∈ L2 (Rn ). In summary, this extension of F , originally considered as in part (c) of Theorem 3.20, satisfies F : L2 (Rn ) → L2 (Rn ) is linear and continuous and n

F f L2 (Rn ) = (2π) 2 f L2 (Rn ) ,

∀ f ∈ L2 (Rn ).

(3.2.29)

Based on this, part (3) in Exercise 3.26, the continuity of the linear mapping L2 (Rn ) f → f ∨ ∈ L2 (Rn ), and the density of Schwartz functions in L2 (Rn ), we further deduce that (3.2.30) F F f = (2π)n f ∨ , ∀ f ∈ L2 (Rn ).

106

CHAPTER 3. THE SCHWARTZ SPACE AND THE FOURIER... Combined with (3.2.29), this proves that F : L2 (Rn ) → L2 (Rn ) is a linear, continuous, isomorphism, ∨ and F −1 f = (2π)−n F f ∨ = (2π)−n F f , ∀ f ∈ L2 (Rn ). (3.2.31) We will continue to use the notation f3 for F f whenever f ∈ L2 (Rn ). The identity in (3.2.29) is called Planch´ erel’s identity. The same type of density argument shows that formula from part (b) of Proposition 3.27 extends to −n f (x)g(x) dx = (2π) ∀ f, g ∈ L2 (Rn ), f3(ξ) g3(ξ) dξ, Rn

Rn

(3.2.32)

to which we continue to refer as Parseval’s identity. (3) An inspection of the computation in (3.2.24) reveals that the identity f ∗ g = f3 · 3 g remains valid if f, g ∈ L1 (Rn ).

Exercise 3.29. Prove that Rn f (x)3 g (x) dx = Rn f3(ξ)g(ξ) dξ for every f, g ∈ L2 (Rn ). Hint: Use part (a) in Proposition 3.27, (3.2.29), and the fact that C0∞ (Rn ) is sequentially dense in L2 (Rn ) to first prove the identity for f ∈ L2 (Rn ) and g ∈ S(Rn ). Further Notes for Chap. 3. The basic formalism associated with the Fourier transform goes back to the French mathematician and physicist Joseph Fourier (1768–1830) in a more or less precise form. A distinguished attribute of this tool, of fundamental importance in the context of partial differential equations, is the ability to render the action of a constant coefficient differential operator simply as ordinary multiplication by its symbol on the Fourier transform side. As the name suggest, the Schwartz space of rapidly decreasing functions has been formally introduced by Laurent Schwartz who was the first to recognize its significance in the context of the Fourier transform. Much of the elegant theory presented here is due to him.

3.3


Exercise 3.30. Prove that if f ∈ L1comp(Rn ) then f3 ∈ C ∞ (Rn ). Exercise 3.31. Prove that if f ∈ L1 (Rn ) is real-valued and odd, then so is f3. Exercise 3.32. Let ϕ ∈ C0∞ (Rn ) be such that ϕ = 0 and for each j ∈ N set ϕj (x) := e−j ϕ(x/j), S(Rn )

∀ x ∈ Rn .

Prove that ϕj −−−−→ 0 but the sequence {ϕj }j∈N does not converge in D(Rn ). j→∞


107

Exercise 3.33. Let ϕ ∈ C0∞ (Rn ) be such that ϕ = 0 and for each j ∈ N set ϕj (x) :=

1 ϕ(x/j), j

∀ x ∈ Rn .

E(Rn )

Prove that ϕj −−−−→ 0 but the sequence {ϕj }j∈N does not converge in S(Rn ). j→∞

Exercise 3.34. Let θ ∈ C0∞ (R) be such that θ(x) = 1 for |x| ≤ 1, and let ψ ∈ C ∞ (R) be such that ψ(x) = 0 for x ≤ −1 and ψ(x) = e−x for x ≥ 0. For each j ∈ N then set ϕj (x) :=

1 ψ(x)θ(x/j), j

∀ x ∈ R.

Prove that the sequence {ϕj }j∈N converges in S(R). Exercise 3.35. Determine which of the following functions belongs to S(Rn ). 2

2

(a) e−(x1 +x2 +···+xn ) (b) (x21 + x22 + · · · + x2n )n! e−|x|

2

n

(c) (1 + |x|2 )−2 2

(d)

sin(e−|x| ) 1+|x|2

(e)

cos(e−|x| ) (1+|x|2 )n

2

2

2

(f ) e−|x| sin(ex1 ) (g) e−(Ax)·x , where A ∈ Mn×n (R) is symmetric and satisfies (Ax) · x > 0 for all x ∈ Rn \ {0} (as before, “·” denotes the dot product of vectors in Rn ). Exercise 3.36. Let A ∈ Mn×n (R) be symmetric and such that (Ax) · x > 0 for every x ∈ Rn \ {0}. Prove that if f (x) := e−(Ax)·x for x ∈ Rn , then n (A−1 ξ)·ξ for every ξ ∈ Rn . f3(ξ) = √π 2 e− 4 det A

2

2

Exercise 3.37. Prove that f : R2 → R defined by f (x1 , x2 ) := e−(x1 +x1 x2 +x2 ) for (x1 , x2 ) ∈ R2 belongs to S(R2 ), then compute its Fourier transform. Exercise 3.38. If P (x) is a polynomial in Rn , compute the Fourier transform 2 of the function defined by f (x) := P (x)e−|x| for each x ∈ Rn . Exercise 3.39. If a ∈ (0, ∞) and x0 ∈ Rn are fixed, compute the Fourier 2 transform of the function defined by f (x) := e−a|x| sin(x · x0 ) for each x ∈ Rn . Exercise 3.40. Let ϕ ∈ S(R). Prove that the equation ψ = ϕ has a solution ψ ∈ S(R) if and only if R ϕ(x) dx = 0.

108

CHAPTER 3. THE SCHWARTZ SPACE AND THE FOURIER... 2

Exercise 3.41. Does the equation ψ = e−x have a solution in S(R)? Exercise 3.42. Fix x0 ∈ Rn . Prove that the translation map tx0 from (1.3.16) extends linearly and continuously as a map from S(Rn ) into itself. More precisely, show that the translation map tx0 : S(Rn ) → S(Rn ) defined by tx0 (ϕ) := ϕ(· − x0 ) for every ϕ ∈ S(Rn ), is linear and continuous. Also, prove that for every ϕ ∈ S(Rn ) the following identities hold in S(Rn ) F tx0 (ϕ) (ξ) = e−ix0 ·ξ ϕ(ξ) 3 = F eix0 ·x ϕ . 3 and tx0 ϕ (3.3.1)

Chapter 4

The Space of Tempered Distributions 4.1

Definition and Properties of Tempered Distributions

The algebraic dual of S(Rn ) is the vector space u : S(Rn ) → C : u is linear and continuous .

(4.1.1)

Functionals u belonging to this space are called tempered distributions (a piece of terminology justified a little later). An important equivalent condition for a linear functional on S(Rn ) to be a tempered distribution is stated next (see (13.1.32)). Fact 4.1. A linear functional u : S(Rn ) → C is continuous if and only if there exist m, k ∈ N0 , and a finite constant C > 0, such that ∀ ϕ ∈ S(Rn ). |u(ϕ)| ≤ C sup sup xβ ∂ α ϕ(x) , (4.1.2) x∈Rn α,β∈Nn 0 , |α|≤m, |β|≤k

From Fact 3.6 and Theorem 13.14 we also know that any f : S(Rn ) → C (linear or not) is continuous if and only if it is sequentially continuous. As a consequence, we have the following characterization of tempered distributions. Proposition 4.2. Let u : S(Rn ) → C be linear. Then u is a tempered distribuS(Rn )

tion if and only u(ϕj ) −−−→ 0 whenever ϕj −−−−→ 0. j→∞

j→∞

As discussed in Example 2.7, to any locally integrable function f in Rn one can associate an “ordinary” distribution uf ∈ D (Rn ). This being said, it is not always the case that uf is actually a tempered distribution (for more on this, see Remark 4.15). This is, however, true if the locally integrable function f becomes integrable at infinity after being tempered by a polynomial. We elaborate on this issue in the next example. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6 4, © Springer Science+Business Media New York 2013

109

110

CHAPTER 4. THE SPACE OF TEMPERED DISTRIBUTIONS

Example 4.3. Let f ∈ L1loc (Rn ) be such that there exist m ∈ [0, ∞) and R ∈ (0, ∞) with the property that |x|≥R

|x|−m |f (x)| dx < ∞.

(4.1.3)

We claim that the distribution of function type defined by f is a tempered distribution, that is, the mapping uf : S(Rn ) → C, uf (ϕ) := f ϕ dx, ∀ ϕ ∈ S(Rn ), (4.1.4) Rn

is a well-defined tempered distribution. To see that this is the case, pick N ∈ N0 such that N ≥ m and, for an arbitrary ϕ ∈ S(Rn ), estimate Rn

|f ϕ| dx ≤

|x| 0) gives lim Γ2

f3(ξ) =

R→∞

π −a|ξ| ae

for ξ ∈ (0, ∞).

Case 3: Assume ξ = 0. It is immediate from (4.2.9) that 1 π f3(0) = dx = . 2 2 a R x +a In summary, we have proved 1 π F 2 (ξ) = e−a|ξ| for every ξ ∈ R, as well as in S (R). 2 x +a a

(4.2.20)

In the next example we compute the Fourier transform of a tempered distribution induced by a slowly growing function that is not absolutely integrable in Rn . 2

Example 4.24. Let a ∈ (0, ∞) and consider the function f (x) := e−ia|x| , x ∈ Rn . It is not difficult to check that f ∈ L(Rn ); thus, by Exercise 4.6 the function f may be regarded as a tempered distribution (identifying it with uf ∈ S (Rn ) associated with f as in (4.1.4)). We claim that −ia|x|2 = e

π n2 ia

ei

|ξ|2 4a

in

S (Rn ).

(4.2.21)

To prove (4.2.21), fix ϕ ∈ S(Rn ) and starting with the definition of the Fourier transform on S (Rn ) write ! " $ % 2 2 −ia|x|2 −ia|x| e ,ϕ = e ,ϕ 3 = e−ia|ξ| ϕ(ξ) 3 dξ = lim+ ε→0

= lim

ε→0+

Rn

Rn

$ 2 2 % e−(ia+ε)|ξ| ϕ(ξ) 3 dξ = lim+ ϕ 3 , e−(ia+ε)|ξ| ε→0

! $ n2 − |ξ|2 " 2 % π ϕ , F e−(ia+ε)|x| = lim ϕ , ia+ε e 4(ia+ε) ε→0+

n2 |ξ|2 π e− 4(ia+ε) ϕ(ξ) dξ ia + ε ε→0+ Rn π n2 |ξ|2 = e− 4ia ϕ(ξ) dξ ia Rn " ! π n2 |ξ|2 e i 4a , ϕ . = ia = lim

(4.2.22)

Above, the first equality is based on (4.2.2), while the second equality is a con2 sequence of the way in which the slowly growing function e−ia|x| is regarded as a tempered distribution. The third and second-to-last equalities are based on Lebesgue’s dominated convergence theorem. In the fourth equality we interpret the Schwartz function ϕ 3 as being in S (Rn ) and rely on the fact that

4.2. THE FOURIER TRANSFORM ACTING ON TEMPERED...

123

2

e−(ia+ε)|x| ∈ S(Rn ) for each ε > 0 (as noted in Example 3.21). The fifth equality once uses Remark 4.22 and once again (4.2.2). The sixth equality is a consequence of formula (3.2.6), while the rest is routine. Theorem 4.25. The following statements are true. (a) The mapping F : S (Rn ) → S (Rn ) defined by F (u) := u 3, for every u ∈ S (Rn ), is bijective and continuous, and its inverse is also continuous. αu = ξα u 5 3 for all α ∈ Nn0 and all u ∈ S (Rn ). (b) D α u = (−D)α u (c) x5 3 for all α ∈ Nn0 and all u ∈ S (Rn ).

(d) If u ∈ S (Rm ) and v ∈ S (Rn ), then u ⊗v =u 3 ⊗ v3. Proof. Recall from Theorem 3.24 that the map F : S(Rn ) → S(Rn ) is linear, continuous, and bijective, and its inverse is also continuous. Since the transpose of this map [in the sense of (13.1.10)] is precisely the Fourier transform in the context of part (a) of the current statement, from Propositions 13.1 and 13.2 it follows that F : S (Rn ) → S (Rn ) is also well-defined, linear, continuous, and bijective, and its inverse is also continuous. Consider next u ∈ S (Rn ) and α ∈ Nn0 . Then for every ϕ ∈ S(Rn ), using part (c) in Theorem 4.13, part (a) in Theorem 3.20, (4.2.2), and part (b) in Theorem 4.13, we may write α u, ϕ = D α u, ϕ αϕ 5 D 3 = (−1)|α| u, Dα ϕ 3 = u, ξ5

= 3 u, ξ α ϕ = ξ α u 3, ϕ,

(4.2.23)

and α u, ϕ = xα u, ϕ αϕ x5 3 = u, xα ϕ 3 = u, D

= 3 u, Dα ϕ = (−D)α u 3, ϕ.

(4.2.24)

This proves the claims in (b) and (c). We are left with the proof of the statement in (d). Fix u ∈ S (Rm ) and v ∈ S (Rn ). Then based on (d) in Theorem 4.13 we have u ⊗ v ∈ S (Rm × Rn ). Hence, starting with (4.2.2), then using (d) in Theorem 3.20 and (4.1.13), we may write u ⊗ v, ϕ ⊗ ψ = u ⊗ v, ϕ ⊗ ψ = u ⊗ v, ϕ 3 ⊗ ψ3

(4.2.25)

= u, ϕ 3 v, ψ3 = 3 u, ϕ3 v , ψ = 3 u ⊗ v3, ϕ ⊗ ψ,

∀ ϕ ∈ S(Rm ), ∀ ψ ∈ S(Rn ).

3 ⊗ v3C ∞ (Rm )⊗C ∞ (Rn ) , which in combiConsequently, u ⊗ v C ∞ (Rm )⊗C ∞ (Rn ) = u 0 0 0 0 nation with Proposition 3.15 proves the statement in (d).


124

Exercise 4.26. Recall (4.1.4) and (4.1.8). Prove that u 4f = uf in S (Rn ) for 2 n 1 n each f ∈ L (R ) and each f ∈ L (R ). Hint: Use (3.2.27) and Exercise 3.29. Lemma 4.27 (Riemann–Lebesgue Lemma). If f ∈ L1 (Rn ), then the tempered distribution uf satisfies u 4f ∈ C 0 (Rn ). Proof. This is a consequence of Exercise 4.26 and (3.1.3). Exercise 4.28. Prove that for every f, g ∈ L2 (Rn ) one has f ∗ g = f3 · g3 in n S (R ). Hint: Use a density argument based on the formula from part (c) of Proposition 3.27, Exercise 4.11, Young’s inequality (cf. Theorem 13.13) in the particular case when p = q = 2, and the fact that the Fourier transform is continuous both on L2 (Rn ) and on S (Rn ). Example 4.29. Let a ∈ (0, ∞). We are interested in computing the Fourier x transform of the bounded function x2 +a 2 , viewed as a distribution in S (R). 1 In this vein, consider the auxiliary function f (x) := x2 +a2 for x ∈ R, and recall from (4.2.20) that f3(ξ) = πa e−a|ξ| in S (R). This formula, part (c) in Theorem 4.25, and (2.4.10), allows us to write x d 3 d π −a|ξ| (ξ) = F xf (ξ) = i e f (ξ) = i F 2 2 x +a dξ dξ a πi = − a e−aξ H(ξ) + a eaξ H(−ξ) a = −πi(sgn ξ)e−a|ξ|

in

S (R).

(4.2.26)

Proposition 4.30. For each u ∈ S (Rn ) define the mapping u∨ : S(Rn ) → C, ∨

Then u ∈ S (R ) and

u∨ (ϕ) := u, ϕ∨ ,

∀ ϕ ∈ S(Rn ).

(4.2.27)

n

3 u 3 = (2π)n u∨ .

(4.2.28)

Proof. Fix u ∈ S (R ). Then (4.2.27) is simply the composition of u with the mapping from Exercise 3.25. Since both are linear and continuous, it follows that u∨ ∈ S (Rn ). Formula (4.2.28) then follows by combining (4.2.2) with the identity in part (3) of Exercise 3.26. n

Exercise 4.31. Show that 3 1 = (2π)n δ

in

S (Rn ).

(4.2.29)

Exercise 4.32. Prove that for any tempered distribution u the following equivalence holds: u∨ = −u

in S (Rn ) ⇐⇒ (3 u )∨ = −3 u in S (Rn ).

(4.2.30)


125

One suggestive way of summarizing (4.2.30) is to say that a tempered distribution is odd if and only if its Fourier transform is odd. Theorem 4.33. The following statements are true. ∗a = 3 au 3 in S (Rn ), where 3 a is (a) If a ∈ S(Rn ) and u ∈ S (Rn ), then u n viewed as an element from S(R ). (b) If u ∈ E (Rn ) then the 3 is of function type given by $ tempered %distribution u the formula u 3(ξ) = u(x), e−ix·ξ for every ξ ∈ Rn , and u 3 ∈ L(Rn ). (c) If u ∈ E (Rn ), v ∈ S (Rn ) then u5 ∗v = u 33 v , where u 3 is viewed as an element in L(Rn ). Proof. By (d) in Theorem 4.18 we have u ∗ a ∈ S (Rn ), hence u ∗ a exists and a ∈ S(Rn ), by (3.1.23) and (b) in Theorem 4.13 belongs to S (Rn ). Also, since 3 it follows that 3 au 3 ∈ S (Rn ). Then, we may write $

% $ % $ % $ % u ∗ a, ϕ = u ∗ a, ϕ 3 = u, a∨ ∗ ϕ 3 = (2π)−n u, 3 3 a∗ϕ 3 $ % $ % $ % = u, 3 a4ϕ = u 3, 3 aϕ = 3 au 3, ϕ , ∀ ϕ ∈ S(Rn ).

(4.2.31)

For the second equality in (4.2.31) we used (4.1.34), for the third we used part (3) in Exercise 3.26, while for the forth we used (d) in Proposition 3.27. This proves the statement in (a). n Moving $ on to the %proof of (b), fix some u ∈ E (R ) and introduce the function f (ξ) := u(x), e−ix·ξ for every ξ ∈ Rn . From Proposition 2.82 it follows that f ∈ C ∞ (Rn ) and ! " ∂ α f (ξ) = u(x) , ∂ξα [e−ix·ξ ] , ∀ ξ ∈ Rn , for every α ∈ Nn0 . (4.2.32) In addition, since u ∈ E (Rn ), there exist a compact subset K of Rn , along with a constant C ∈ (0, ∞), and a number k ∈ N0 , such that u satisfies (2.6.6). Combining all these facts, for each α ∈ Nn0 , we may estimate α ∂ f (ξ) = u(x), ∂ξα e−ix·ξ ≤ C sup ∂xβ ∂ξα e−ix·ξ x∈K, |β|≤k

≤C

sup x∈K, |β|≤k

α β x ξ ≤ C(1 + |ξ|)k .

(4.2.33)

From (4.2.33) and the fact that f is smooth we deduce that f ∈ L(Rn ). Hence, if we now recall Exercise 4.6, it follows that f ∈ S (Rn ). We are left with proving that u 3 = f as tempered distributions. To this end, fix θ ∈ C0∞ (Rn ) such that θ = 1 in a neighborhood of supp u. Then, for every ϕ ∈ C0∞ (Rn ) one has " $ % ! 3 u, ϕ = u, ϕ 3 = u(ξ) , θ(ξ)ϕ(ξ) 3 = u(ξ), e−ix·ξ θ(ξ)ϕ(x) dx . (4.2.34) Rn

126


At this point recall Remark 2.79 (note that the function e−ix·ξ θ(ξ)ϕ(x) belongs to C0∞ (Rn ×Rn ) and one may take v = 1 in (2.7.45)) which allows one to rewrite the last term in (4.2.34) and conclude that $ % 3 u, ϕ = u(ξ), e−ix·ξ ϕ(x) dx = f, ϕ, ∀ ϕ ∈ C0∞ (Rn ). (4.2.35) Rn

Hence the tempered distributions u 3 and f coincide on C0∞ (Rn ). By (4.1.25), we n therefore have u 3 = f in S (R ), completing the proof of the statement in (b). Regarding the formula in part (c), while informally this is similar to the formula proved in part (a), the computation in (4.2.31) through which the latter has been deduced no longer works in the current case, as various ingredients used to justify it break down (given that u 3 is now only known to belong to L(Rn ) and n not necessarily to S(R )). This being said, we may employ what has been established in part (a) together with a limiting argument to get the desired result. Specifically, assume that u ∈ E (Rn ) and v ∈ S (Rn ). Then u ∗ v ∈ S (Rn ) (recall statement (a) in Theorem 4.18) hence, u ∗ v is well-defined in S (Rn ). n 3 v3 is meaningful and u 3 v3 ∈ S (Rn ). To Also, from (b) we have u 3 ∈ L(R ), thus u proceed, recall the sequence {φj }j∈N from Example 1.11. In particular, (1.3.8) D (Rn )

holds and φj −−−−→ δ (see Example 2.20). Thus, the statement from part (b) j→∞

S (Rn )

in Theorem 4.18 applies and gives that u ∗ φj −−−−→ u ∗ δ = u. Moreover, j→∞

since u ∈ E (Rn ), by (2.8.17) it follows that u ∗ φj ∈ E (Rn ) and supp (u ∗ φj ) ⊆ supp u + B(0, 1) for every j ∈ N (by part (1) in Remark 2.91). Hence, one may S (Rn )

apply (b) in Theorem 4.18 to further conclude that (u ∗ φj ) ∗ v −−−−→ u ∗ v. j→∞

Given (a) in Theorem 4.25, the latter implies (recall Fact 4.10) $ % lim F (u ∗ φj ) ∗ v , ϕ = u ∗ v, ϕ, ∀ ϕ ∈ S(Rn ). j→∞

(4.2.36)

Note that (2.8.44) gives u ∗ φj ∈ C0∞ (Rn ) for every j ∈ N. Hence, based on what we have proved already in part (a), we obtain (keeping in mind that u 3 ∈ L(Rn )) 4j u 3 ∗ φj v3 = φ 3 v3 = φ(·/j) u 3 v3 in S (Rn ), ∀j ∈ N. F (u ∗ φj ) ∗ v = u (4.2.37) In concert, (4.2.36) and (4.2.37) yield that for each ϕ ∈ S(Rn ) $ % $ % $ % $ % 3 3 ϕ = u 33 v , φ(·/j) ϕ = u 3 v3, φ(0) 3 v3, ϕ (4.2.38) u ∗ v, ϕ = lim u j→∞

S(Rn )

3 ϕ by Exercise 3.14, and φ(0) 3 3 = 1. This proves the since φ(·/j) ϕ −−−−→ φ(0) j→∞

statement in (c) and finishes the proof of the theorem. Example 4.34. If a ∈ (0, ∞) then χ[−a,a] , the characteristic function of the interval [−a, a], belongs to E (R) and by statement (b) in Theorem 4.33 we have ⎧ a ⎨ 2 sin(aξ) for ξ ∈ R \ {0}, ξ −ixξ (ξ) = e dx = χ (4.2.39) [−a,a] ⎩ −a 2a for ξ = 0.


127

Exercise 4.35. Use Exercise 2.65 and statement (b) in Theorem 4.33 to prove 3 ⊆ {a} for some a ∈ Rn , then there exist that if u ∈ S (Rn ) is such that supp u k ∈ N0 and constants cα ∈ C, for α ∈ Nn0 with |α| ≤ k, such that u= cα xα eix·a in S (Rn ). (4.2.40) |α|≤k

In particular, if a = 0 then u is a polynomial in Rn . Example 4.36. Let a ∈ R and consider the function f (x) := sin(ax) for x ∈ R. Then f ∈ C ∞ (R) ∩ L∞ (R), hence f ∈ S (R). Suppose a = 0. We shall compute the Fourier transform of f in S (R) by making use of a technique relying on the ordinary differential equation that f satisfies. More precisely, since f +a2 f = 0 in R, the same equation holds in S (R), thus by (a) and (b) in Theorem 4.25 we have (ξ 2 − a2 )f3 = 0 in S (R). By Example 2.72 this implies f3 = C1 δa + C2 δ−a in S (R) for some C1 , C2 ∈ C. Applying again the Fourier transform to the last equation, using (4.2.28) and part (b) from Theorem 4.33, we have −ixξ = C1 δ3a + C2 δ5 + C2 δ−a (ξ), e−ixξ 2π sin(−ax) = sin(ax) −a = C1 δa (ξ), e = C1 e−iax + C2 eiax = (C1 + C2 ) cos(ax) + i(C1 − C2 ) sin(−ax).

(4.2.41)

The resulting identity in (4.2.41) forces C1 = −iπ and C2 = iπ. Plugging these constants back in the expression for f3 yields = −iπδa + iπδ−a sin(ax)

in S (R).

(4.2.42)

It is immediate that (4.2.42) also holds if a = 0. Example 4.37. Let a, b ∈ R. Then the function g(x) := sin(ax) sin(b x) for x ∈ R satisfies g ∈ L∞ (R), thus g ∈ S (R) (cf. (4.1.8)). Applying the Fourier transform to the identity in (4.2.42) and using (4.2.28) we obtain sin(ax) =

i3 i δa − δ5 −a 2 2

in S (R).

(4.2.43)

Also, making use of (4.2.43), part (c) in Theorem 4.33, and then Exercise 2.90, we may write i i i i δ3a − δ5 δ3b − δ5 sin(ax) sin(bx) = −a −b 2 2 2 2 1 = − F δa+b − δa−b − δb−a + δ−a−b in S (R). (4.2.44) 4 Hence, another application of the Fourier transform gives (relying on (4.2.28)) π F sin(ax) sin(bx) = − δa+b − δa−b − δb−a + δ−a−b in S (R). (4.2.45) 2


128

2

Example 4.38. Let a ∈ (0, ∞) and consider the function f (x) := e ia|x| for x ∈ Rn . Then f ∈ L∞ (Rn ); thus, f ∈ S (Rn ) by (4.1.8). The goal is to compute the Fourier transform of f in S (Rn ). The starting point is the observation that 2

2

2

f (x) = e iax1 ⊗ e iax2 ⊗ · · · ⊗ e iaxn

∀ x = (x1 , . . . , xn ) ∈ R.

(4.2.46)

Invoking part (d) in Theorem 4.25 then reduces matters to computing the Fourier transform of f in the case n = 1. 2 Assume that n = 1, in which case f (x) = e iax for x ∈ R. Then f satisfies the differential equation: f −2ia xf = 0 in S (R). Taking the Fourier transform in S (R) and using the formulas from (b)–(c) in Theorem 4.25 we obtain (f3) +

i 3 ξ f = 0 in S (R). 2a

(4.2.47)

The format of (4.2.47) suggests that we consider the ordinary differential equai ξy = 0 in R, in the unknown y = y(ξ). One particular solution of tion y + 2a ξ2

this o.d.e. is y(ξ) = e−i 4a . Note that both y and 1/y belong to L(Rn ). In particular, it makes sense to consider the tempered distribution u := (1/y)f3, with the derivative u = (1/y) f3 + (1/y)(f3) = −(y /y 2 )f3 + (1/y)(f3) i i ξ /y f3 − ξ(1/y)f3 = 0 in S (R). = 2a 2a

(4.2.48)

From this and part (2) in Proposition 2.40 we then deduce that there exists c ∈ C such that u = c in S (R) which, given the significance of u and y, forces ξ2 f3 = c e−i 4a in S (R) for some c ∈ C. We are therefore left with determining the constant c. This may be done by choosing a suitable Schwartz function and ξ2 then computing the action of f3 on it. Take ϕ(ξ) := e− 4a for ξ ∈ R. Since √ 2 ϕ(x) 3 = 4aπe−ax for x ∈ R (recall Example 3.21), we may write 2 2 √ 2 2 −i ξ4a − ξ4a 3 c e dξ = f , ϕ = f, ϕ 3 = 4aπ e iax −ax dx. (4.2.49) R

R

The two integrals in (4.2.49) may be computed using formula (3.2.7) with λ := 1+i 4a and λ := a(1 − i), respectively. After some routine algebra (i.e., computing these integrals, replacing their values in (4.2.49), then solving for c), we find π c = πa e i 4 . In summary, this analysis proves that |ξ| nπ ia|x|2 (ξ) = π 2 e i 4 e−i 4a e a n

2

in S (Rn ).

(4.2.50)

Partial Fourier Transforms In the last part of this section we define partial Fourier transforms. To set the stage, fix m, n ∈ N. We shall denote by x, ξ generic variables in Rn , and by

4.3. HOMOGENEOUS DISTRIBUTIONS

129

y, η generic variables in Rm . The partial Fourier transform with respect to the 3x or Fx ϕ, is defined by variable x of a function ϕ ∈ S(Rn+m ), denoted by ϕ e−ix·ξ ϕ(x, y) dx, ∀ ξ ∈ Rn , ∀ y ∈ Rm . (4.2.51) ϕ 3x (ξ, y) := Rn

Reasoning in a similar manner as in the proof of Theorem 3.24, it follows that Fx : S(Rn+m ) → S(Rn+m ) is bijective,

(4.2.52)

continuous, with continuous inverse and its inverse is given by (Fx−1 ψ)(x, η)

−n

eix·ξ ψ(ξ, η) dξ,

:= (2π)

Rn

for all (x, η) ∈ R

(4.2.53)

and all ψ ∈ S(R

n+m

n+m

).

Furthermore, analogously to Proposition 4.20, the partial Fourier transform Fx extends to S (Rn+m ) as a continuous map by setting ∀ u ∈ S (Rn+m ),

Fx u, ϕ := u, Fx ϕ,

∀ ϕ ∈ S(Rn+m ),

(4.2.54)

and this extension is an isomorphism from S (Rn+m ) into itself, with continuous inverse denoted by Fx−1 . Moreover, the action of Fx enjoys properties analogous to those established for the “full” Fourier transform in Theorem 3.20, Exercise 3.26, Theorem 4.25, and Proposition 4.30. Exercise 4.39. Let 3· denote the full Fourier transform in Rn+m . Prove that for each function ϕ ∈ S(Rn × Rm ) we have Fx Fy ϕ(x, y) (ξ, η) = Fy Fx ϕ(x, y) (ξ, η) = ϕ(ξ, 3 η) ∀ (ξ, η) ∈ Rn × Rm . (4.2.55) Also, show that Fx Fy u = Fy Fx u = u 3 in S (Rn+m ),

∀ u ∈ S (Rn × Rm ).

Exercise 4.40. Prove that Fx δ(x) ⊗ δ(y) = 1(ξ) ⊗ δ(y)

4.3

in S (Rn+m ).

(4.2.56)

(4.2.57)

Homogeneous Distributions

Let A ∈ Mn×n (R) be such that det A = 0. Then for every f ∈ L1 (Rn ) one has f ◦ A ∈ L1 (Rn ); thus, f, f ◦ A ∈ S (Rn ) by (4.1.8). Moreover, f (Ax)ϕ(x) dx = | det A|−1 f (y)ϕ(A−1 y) dy f ◦ A, ϕ = Rn

Rn

−1

= | det A|

−1

f, ϕ ◦ A

,

∀ ϕ ∈ S(Rn ).

(4.3.1)

This and Exercise 3.16 justifies extending the operator of composition with linear maps to S (Rn ) as follows.

130


Proposition 4.41. Let A ∈ Mn×n (R) be such that det A = 0. For each u ∈ S (Rn ), define the mapping u ◦ A : S(Rn ) → C by setting $ % (4.3.2) ∀ ϕ ∈ S(Rn ). u ◦ A (ϕ) := | det A|−1 u, ϕ ◦ A−1 , Then u ◦ A ∈ S (Rn ). Proof. This is an immediate consequence of (3.1.40). Exercise 4.42. Let A, B ∈ Mn×n (R) be such that det A = 0 and det B = 0. Then the following identities hold in S (Rn ): (1) (u ◦ A) ◦ B = u ◦ (AB) for every u ∈ S (Rn ). (2) u ◦ (λA) = λu ◦ A for every u ∈ S (Rn ) and every λ ∈ R. (3) (u + v) ◦ A = u ◦ A + v ◦ A for every u, v ∈ S (Rn ). In the next proposition we study how the Fourier transform interacts with the operator of composition by an invertible matrix. Recall that the transpose of a matrix A is denoted by A . Proposition 4.43. Assume that A ∈ Mn×n (R) is such that det A = 0. Then for each u ∈ S (Rn ), −1 u ◦ A = | det A|−1 u 3 ◦ A . (4.3.3) Proof. For each ϕ ∈ S(Rn ), based on (4.2.2), (4.3.2), and (3.2.9), we may write $ % 3 ◦ A−1 u ◦ A, ϕ = u ◦ A, ϕ 3 = | det A|−1 u, ϕ $ % $ % = u, ϕ ◦ A = u 3, ϕ ◦ A $ −1 % 3 ◦ A = | det A|−1 u ,ϕ . (4.3.4) This proves (4.3.3). The mappings in (3.1.40) and (4.3.2) corresponding to A := tIn×n , for some number t ∈ (0, ∞), are called dilations and will be denoted by τt . More precisely, for each t ∈ (0, ∞) we have τt : S(Rn ) → S(Rn ), and

(τt ϕ)(x) := ϕ(tx),

∀ ϕ ∈ S(Rn ), ∀ x ∈ Rn ,

(4.3.5)

τt : S (Rn ) → S (Rn ), τt u, ϕ := t−n u, τ 1t ϕ,

∀ u ∈ S (Rn ), ∀ ϕ ∈ S(Rn ).

(4.3.6)

Exercise 4.44. Prove that for each t ∈ (0, ∞) the following are true: F (τt ϕ) = t−n τ 1t F (ϕ)

in

S(Rn ),

∀ ϕ ∈ S(Rn ),

(4.3.7)

F (τt u) = t−n τ 1t F (u)

in

S (Rn ),

∀ u ∈ S (Rn ).

(4.3.8)


131

Hint: Use (3.2.9) with A = 1t In×n to prove (4.3.7), then use (4.3.7) and (4.3.6) to prove (4.3.8). To proceed, we make a couple of definitions. Definition 4.45. A linear transformation A ∈ Mn×n (R) is called orthogonal provided A is invertible and A−1 = A . Some of the most basic attributes of an orthogonal matrix A are (A )−1 = A,

|det A| = 1,

|Ax| = |x| for every x ∈ Rn .

(4.3.9)

Definition 4.46. A distribution u ∈ S (Rn ) is called invariant under orthogonal transformations provided u = u ◦ A in S (Rn ) for every orthogonal matrix A ∈ Mn×n (R). Proposition 4.47. Let u ∈ S (Rn ). Then u is invariant under orthogonal transformations if and only if u 3 is invariant under orthogonal transformations. Proof. This is a direct consequence of (4.3.3) and the fact that any orthogonal matrix A satisfies (4.3.9). Next we take a look at homogeneous functions to gain some insight into how this notion may be defined in the setting of distributions. Definition 4.48. (1) A nonempty open set O in Rn is called a cone-like region if tx ∈ O whenever x ∈ O and t ∈ (0, ∞). (2) Given a cone-like region O ⊆ Rn , call a function f : O → C positive homogeneous of degree k ∈ R if f (tx) = tk f (x) for every t > 0 and every x ∈ O. Exercise 4.49. Prove that if O ⊆ Rn is a cone-like region, N ∈ N, and f ∈ C N (O) is positive homogeneous of degree k ∈ R on O, then ∂ α f is positive homogeneous of degree k − N on O for every α ∈ Nn0 with |α| ≤ N . Exercise 4.50. Prove that if f ∈ C 0 (Rn \{0}) is positive homogeneous of degree 1 − n on Rn \ {0}, and g ∈ C 0 (S n−1 ), then g(x/R)f (x) dσ(x) = g(x)f (x) dσ(x), ∀ R ∈ (0, ∞). ∂B(0,R)

S n−1

(4.3.10) Exercise 4.51. Prove that if f ∈ C 0 (Rn \{0}) is positive homogeneous of degree k ∈ R, then |f (x)| ≤ f L∞ (S n−1 ) |x|k for every x ∈ Rn \ {0}. x k Hint: Write f (x) = f |x| |x| for every x ∈ Rn \ {0}.

132


Exercise 4.52. Show that if f ∈ C 0 (Rn \{0}) is positive homogeneous of degree k ∈ R with k > −n, then f ∈ S (Rn ). Hint: Make use of Exercise 4.51 and the result discussed in Example 4.3. After this preamble, we are ready to extend the notion of positive homogeneity to tempered distributions. Definition 4.53. A distribution u ∈ S (Rn ) is called positive homogeneous of degree k ∈ R provided τt u = tk u in S (Rn ) for every t > 0. Exercise 4.54. Prove that δ ∈ S (Rn ) is positive homogeneous of degree −n. Exercise 4.55. Let f ∈ L1loc (Rn ) be such that (4.1.3) is satisfied for some m ∈ N and some R ∈ (0, ∞) and let k ∈ R. Show that the tempered distribution uf is positive homogeneous of degree k, if and only if f is positive homogeneous of degree k. Exercise 4.56. Prove that if u ∈ S (Rn ) is positive homogeneous of degree k ∈ R then for every α ∈ Nn0 the tempered distribution ∂ α u is positive homogeneous of degree k − |α|. Deduce from this that for every α ∈ Nn0 the tempered distribution ∂ α δ is positive homogeneous of degree −n − |α|. Proposition 4.57. Let k ∈ R. If u ∈ S (Rn ) is positive homogeneous of degree k, then u 3 is positive homogeneous of degree −n − k. Proof. Let u ∈ S (Rn ) be positive homogeneous of degree k, and fix t > 0. Then (4.3.8) and the assumption on u give 3 = t−n F τ 1t u = t−n F t−k u = t−n−k u 3 in S (Rn ), (4.3.11) τt u hence u 3 is positive homogeneous of degree −n − k. Proposition 4.58. If u ∈ S (Rn ), uRn \{0} ∈ C ∞ (Rn \ {0}), and uRn \{0} is positive homogeneous of degree k, for some k ∈ R, then u 3 n ∈ C ∞ (Rn \{0}). R \{0}

Proof. Fix u satisfying the hypotheses of the proposition. By (c) in Proposi α u in S (Rn ). Also, it is not tion 4.25, for each α ∈ Nn0 one has Dξα u 3 = (−x) α n difficult to check that (−x) u ∈ S (R ) continues to satisfy the hypotheses of the proposition with k replaced by k + |α|. Hence, the desired conclusion follows once we prove that u 3Rn \{0} is continuous on Rn \ {0}. To this end, assume first that k < −n and fix ψ ∈ C0∞ (Rn ) such that ψ = 1 on B(0, 1). Use this to decompose u = ψu + (1 − ψ)u. Since ψu ∈ E (Rn ) part 4 ∈ C ∞ (Rn ). Furthermore, (1 − ψ)u vanishes near (b) in Theorem 4.33 gives ψu x x = |x|k u |x| . Given the origin while outside supp ψ becomes u(x) = u |x| |x| the current assumption on k, this behavior implies (1 − ψ)u ∈ L1 (Rn ), hence (1 − ψ)u ∈ C 0 (Rn ) by Lemma 4.27. To summarize, this analysis shows that u 3 ∈ C 0 (Rn ) whenever k + n < 0.

(4.3.12)


133

To treat the case k+n ≥ 0, let α ∈ Nn0 be arbitrary and set vα := Dα u ∈ S (Rn ). Since uRn \{0} ∈ C ∞ (Rn \ {0}) differentiating u(tx) = tk u(x) yields t|α| (Dα u)(tx) = tk (Dα u)(x) for x ∈ Rn \ {0} and t > 0. Given that vα Rn \{0} ∈ C ∞ (R \ {0}), the latter translates into

(4.3.13)

vα (tx) = tk−|α| vα (x) for x ∈ Rn \ {0} and t > 0. (4.3.14) Hence, vα Rn \{0} is homogeneous of degree k − |α|. Based on what we proved 0 n earlier (cf. (4.3.12)), it follows that v4 α ∈ C (R ) whenever k − |α| < −n. In terms of the original distribution u this amounts to saying that ξα u 3 ∈ C 0 (Rn ),

∀ α ∈ Nn0

with

|α| > k + n.

(4.3.15)

The end-game in the proof is then as follows. Given an arbitrary k ∈ R, pick a natural number N with the property that 2N > k + n. Writing (cf. (13.2.1)) for each ξ ∈ Rn N! ξ 2α , |ξ|2N = (4.3.16) α! |α|=N

we obtain |ξ|2N u 3=

N! ξ 2α u 3 in S (Rn ). α!

(4.3.17)

|α|=N

Collectively, (4.3.17), (4.3.15), and the assumption on N imply |ξ|2N u 3 ∈ C 0 (Rn ).

(4.3.18)

1 2N

belongs to C ∞ (Rn \ {0}), condition (4.3.18) further implies |ξ| Rn \{0}0 n that u 3Rn \{0} ∈ C (R \ {0}). This completes the proof of the proposition. Since

An inspection of the proof of Proposition 4.58 shows that several other useful versions could be derived, two of which are recorded below. Exercise 4.59. If u ∈ S (Rn ), uRn \{0} ∈ C ∞ (Rn \ {0}) and uRn \{0} is positive

homogeneous of degree k, for some k ∈ R satisfying k < −n, then u 3 ∈ C k0 (Rn ), where k0 := max{j ∈ N0 , j + k < −n}. Exercise 4.60. Assume that u ∈ S (Rn ), uRn \{0} ∈ C N (Rn \{0}) where N ∈ N is even, and u n is positive homogeneous of degree k, for some k ∈ R R \{0}

satisfying k < N − n. Then u 3 ∈ C m (Rn \ {0}) for every m ∈ N0 satisfying m < N − n − k. Next we take on the task of computing the Fourier transform of certain homogeneous tempered distributions that will be particularly important later in applications. Recall the gamma function Γ from (13.5.1).

134


−λ n Proposition 4.61. Let λ ∈ (0, n)∞andnset fλ (x) := |x| , for each x ∈ R \{0}. n 3 Then fλ ∈ S (R ), fλ Rn \{0} ∈ C (R \ {0}), and n−λ n Γ 2 |ξ|λ−n for every ξ ∈ Rn \ {0}. f3λ (ξ) = 2n−λ π 2 (4.3.19) Γ λ2

Proof. Fix λ ∈ (0, n). Exercise 4.4 then shows that fλ ∈ S (Rn ). Clearly, |x|−λ is invariant under orthogonal transformations and is positive homogeneous of degree −λ. Hence, by Proposition 4.47 and Proposition 4.57, it follows that and positive homogeneous of f4 λ is invariant under orthogonal transformations ∞ n degree −n + λ. In addition, f4 λ Rn \{0} ∈ C (R \ {0}) by Proposition 4.58. Fix ξ ∈ Rn \ {0} and choose an orthogonal matrix A ∈ Mn×n (R) with the property that Aξ = (0, 0, . . . , 0, |ξ|) (such a matrix may be obtained by ξ completing the vector vn := |ξ| to an orthonormal basis {v1 , . . . , vn } in Rn and then taking A to be the matrix mapping each vj into ej for j = 1, . . . , n). Then λ−n 4 4 , f4 λ (ξ) = fλ (Aξ) = fλ (0, . . . , 0, |ξ| ) = cλ |ξ|

(4.3.20)

where cλ := f4 λ (0, . . . , 0, 1) ∈ C. As such, we are left with determining the value |x|2 of cλ . We do so by apply f3λ to the particular Schwartz function ϕ(x) := e− 2 n

|ξ|2

for x ∈ Rn . From Example 3.21 we know that ϕ(ξ) 3 = (2π) 2 e− 2 for every n 3 may ξ ∈ R . Based on this formula and (4.3.20), the identity f4 λ , ϕ = fλ , ϕ be rewritten as |ξ|2 |x|2 n λ−n − 2 2 cλ |ξ| e dξ = (2π) |x|−λ e− 2 dx. (4.3.21) Rn

Rn

The two integrals in (4.3.21) may be computed simultaneously, by adopting a slightly more general point of view, as follows. For any p > −n use polar coordinates (cf. (13.8.9)) and a natural change of variables to write (for the definition and properties of the gamma function Γ see (13.5.1) and the subsequent comments) ∞ 2 ρ2 p − |ξ| 2 |ξ| e dξ = ωn−1 ρp+n−1 e− 2 dρ Rn

0

= ωn−1 2

p+n−2 2

∞

t

p+n−2 2

e−t dt

0

=2

p+n−2 2

ωn−1 Γ

p + n 2

.

(4.3.22)

When used with p := λ − n and p := −λ, formula (4.3.22) allows us to rewrite (4.3.21) as λ −λ + n λ−n+n−2 −λ+n−2 n 2 = (2π) 2 2 2 ωn−1 Γ . (4.3.23) ωn−1 Γ cλ 2 2 2 n

This gives cλ = 2n−λ π 2

Γ( n−λ 2 ) Γ( λ 2)

, finishing the proof of (4.3.19).

4.4. PRINCIPAL VALUE TEMPERED DISTRIBUTIONS

135

A remarkable consequence of Proposition 4.61 is singled out below. Corollary 4.62. Assume that n ∈ N, n ≥ 2, and fix λ ∈ [0, n − 1). Then for each j ∈ {1, . . . , n}, we have Γ n−λ ξj xj n−λ−1 n 2 n−λ π 2 λ in S (Rn ). (4.3.24) = −i 2 F λ+2 |x| Γ 2 + 1 |ξ| In particular, corresponding to the case when λ = n−2, formula (4.3.24) becomes xj ξj F (4.3.25) = −i ωn−1 2 in S (Rn ). |x|n |ξ| Proof. Fix an integer n ≥ 2, and suppose first that λ ∈ (0, n−1). In this regime, both (4.3.19) and (4.1.32) hold. In concert with part (b) in Theorem 4.25, these give xj −λ F = F (∂j fλ ) = iξj f3λ (ξ) |x|λ+2 Γ n−λ n−λ n 2 2 ξj |ξ|λ−n in S (Rn ). π (4.3.26) = i2 Γ λ2 Hence, whenever λ ∈ (0, n − 1), xj F = Cλ ξj |ξ|λ−n |x|λ+2

in S (Rn ),

(4.3.27)

where we have set Cλ := −i 2

n−λ−1

π

n 2

Γ

n−λ

2λ λ Γ 2 2

= −i 2

n−λ−1

Γ n−λ 2 π Γ λ2 + 1 n 2

(4.3.28)

and the last equality follows from (13.5.2). This proves formula (4.3.24) in the case when λ ∈ (0, n − 1). The case when λ = 0 then follows from what we have just proved, by passing to limit λ → 0+ in (4.3.24) and observing that all quantities involved depend continuously on λ, in an appropriate sense. Finally, (4.3.25) is a direct consequence of (4.3.24) and (13.5.6).

4.4

Principal Value Tempered Distributions

Recall the distribution P.V. x1 ∈ D (R) from Example 2.11. As seen from Exercise 4.103, we have P.V. x1 ∈ S (R). The issue we address in this section is the generalization of this distribution to higher dimensions. The key features of the function Θ(x) := x1 , x ∈ R \ {0}, that allowed us to define P.V. x1 as a tempered distribution on the real line are as follows: first, Θ ∈ C 0 (R \ {0}), second, Θ is positive homogeneous of degree −1, and third, Θ(1) + Θ(−1) = 0.

136


Moving from one dimension to Rn , this suggests considering the class of functions satisfying Θ ∈ C 0 (Rn \ {0}), positive homogeneous of degree −n,

Θ dσ = 0. S n−1

(4.4.1) It is worth noting that a function Θ as above typically fails to be in L1loc (Rn ) [though obviously Θ ∈ L1loc (Rn \ {0})]. As such, associating a distribution with Θ necessarily has to be a more elaborate process than the one identifying functions from L1loc (Rn ) with distributions in Rn . This has already been the case when defining P.V. x1 and here we model the same type of definition in higher dimensions. Specifically, given Θ as in (4.4.1) consider the linear mapping P.V. Θ : S(Rn ) → C, P.V. Θ (ϕ) := lim Θ(x)ϕ(x) dx, ε→0+

|x|≥ε

∀ ϕ ∈ S(Rn ).

(4.4.2)

That this definition does the job is proved next. Proposition 4.63. Let Θ be a function satisfying (4.4.1). Then the map P.V. Θ considered in (4.4.2) and is a tempered distribution in Rn . In is well-defined addition, P.V. Θ Rn \{0} = Θ Rn \{0} in D (Rn \ {0}). Before proceeding with the proof of Proposition 4.63 we recall a definition and introduce a class of functions that will be used in the proof. Definition 4.64. A function ψ : Rn → C is called radial if ψ(x) depends only on |x| for every x ∈ Rn , that is, ψ(x) = f (|x|) for all x ∈ Rn , where f : R → C. Next, consider the class of functions ⎧ ψ : Rn → C, ψ ∈ C 1 (Rn ), ⎪ ⎪ ⎨ ψ is radial and ψ(0) = 1, ⎪ ⎪ ⎩ ∃ εo ∈ (0, ∞) such that ψ decays like |x|−εo at ∞, (for example ψ(x) = e−

|x|2 2

(4.4.3)

, x ∈ Rn , satisfies (4.4.3)) and set

Q := ψ : ψ satisfies (4.4.3) . Now we are ready to return to Proposition 4.63.

(4.4.4)

4.4. PRINCIPAL VALUE TEMPERED DISTRIBUTIONS

137

Proof of Proposition 4.63. Fix an arbitrary ψ satisfying (4.4.3). Then, making use of formula (13.8.9) and the properties of Θ and ψ, for each ε ∈ (0, ∞) we have x Θ( |x| ) Θ(x)ψ(x) dx = ψ(x) dx n |x|≥ε |x|≥ε |x| ∞ ψ(ρ) Θ(ω) dσ(ω) dρ = 0. (4.4.5) = ρ ε S n−1 Hence, Θ[ϕ − ϕ(0)ψ] ∈ L1 (Rn ) for every ϕ ∈ S(Rn ), which when combined with (4.4.5) and Lebesgue dominated convergence theorem 13.12 yields P.V. Θ (ϕ) = Θ(x)[ϕ(x) − ϕ(0)ψ(x)] dx, ∀ ϕ ∈ S(Rn ). (4.4.6) Rn

Note that because of (4.4.5), the right-hand side in (4.4.6) is independent of the choice of ψ. Estimating the right hand-side of (4.4.6) (using Exercise 4.51, the decay at infinity of functions from Q and the Schwartz class, and the mean value theorem near the origin) shows that there exists a constant C ∈ (0, ∞) independent of ϕ with the property that α β P.V. Θ (ϕ) ≤ C x ∂ ϕ(x), ∀ ϕ ∈ S(Rn ). (4.4.7) sup x∈Rn , |α|≤1, |β|≤1

Since from (4.4.6) we see that P.V. Θ is linear, in light of Fact 4.1 estimate (4.4.7) implies P.V. Θ ∈ S (Rn ) as wanted. The fact that the restriction in the distributional sense of P.V. Θ to Rn \ {0} is equal to the restriction of the function Θ to Rn \ {0} is immediate from definitions. Remark 4.65. (1) As already alluded to, if n = 1 and Θ(x) := x1 , x ∈ R \ {0}, then we have P.V.Θ = P.V. x1 . (2) Suppose Θ is as in (4.4.1). Since identity (4.4.6) holds for any ψ ∈ Q, we may select ψ ∈ Q that also satisfies ψ = 1 on S n−1 and observe that for this choice of ψ we have Θ(x)[ϕ(x) − ϕ(0)ψ(x)] dx (4.4.8) Rn

=

Θ(x) ϕ(x)−ϕ(0) dx+

|x|≤1

Hence,

|x|>1

P.V. Θ, ϕ =

|x|≤1

Θ(x)ϕ(x) dx, ∀ ϕ ∈ S(Rn ).

Θ(x) ϕ(x) − ϕ(0) dx

Θ(x)ϕ(x) dx,

+ |x|>1

∀ ϕ ∈ S(Rn ).

(4.4.9)

138

CHAPTER 4. THE SPACE OF TEMPERED DISTRIBUTIONS x

j Example 4.66. If j ∈ {1, . . . , n}, the function Θ defined by Θ(x) := |x|n+1 for xj n each x ∈ R \ {0} satisfies (4.4.1). By Proposition 4.63 we have P.V. |x|n+1 ∈ n n S (R ) and part (2) in Remark 4.65 gives that for every ϕ ∈ S(R )

! P.V.

" xj xj ϕ(x) , ϕ = lim dx n+1 n+1 + |x| ε→0 |x|≥ε |x| = |x|≤1

xj (ϕ(x) − ϕ(0)) dx + |x|n+1

(4.4.10) |x|>1

xj ϕ(x) dx. |x|n+1

The next proposition elaborates on the manner in which principal value tempered distributions convolve with Schwartz functions. Proposition 4.67. Let Θ be a function the conditions in (4.4.1). satisfying Then for each ϕ ∈ S(Rn ) one has that P.V. Θ ∗ ϕ ∈ S (Rn ) ∩ C ∞ (Rn ) and P.V. Θ ∗ ϕ (x) = lim+ ε→0

|x−y|≥ε

Θ(x − y)ϕ(y) dy,

∀ x ∈ Rn .

(4.4.11)

n Proof. Fix an arbitrary ϕ ∈ S(Rn ) and note that since P.V. Θ ∈ S (Rn ) (c.f. Proposition 4.63), part (e) in Theorem 4.18 gives P.V. Θ ∗ ϕ ∈ S (R ). Let ψ ∈ C0∞ (Rn ) be such that ψ = Then 1 − ψ ∈ L(Rn ) and it 1 near the origin. n makes sense to consider (1−ψ) P.V. Θ in S (R ) (cf. part (b) in Theorem 4.13). Hence, we may decompose P.V. Θ = u + v where

u := ψ P.V. Θ ∈ E (Rn ) and v := (1 − ψ) P.V. Θ ∈ S (Rn ).

(4.4.12)

The last part in Proposition 4.63 also permits us to identify v = (1 − ψ)Θ in L1loc (Rn ). By Exercise 4.51 we have v ∈ Lp (Rn ) for every p ∈ (1, ∞) which, in combination with Exercise 3.18, allows us to conclude that v ∗ ϕ ∈ C ∞ (Rn ) and (v ∗ ϕ)(x) =

Rn

(1 − ψ(y))Θ(y)ϕ(x − y) dy,

∀ x ∈ Rn .

(4.4.13)

Since the above integral is absolutely convergent, by Lebesgue’s dominated convergence theorem we further express this as (v ∗ ϕ)(x) = lim (1 − ψ(y))Θ(y)ϕ(x − y) dy, ∀ x ∈ Rn . (4.4.14) ε→0+

|y|≥ε

Thanks to Exercise 2.94 we also have u ∗ ϕ ∈ C ∞ (Rn ) and (u ∗ ϕ)(x) = ψ P.V. Θ, ϕ(x − ·)

for each x ∈ Rn .

(4.4.15)

On the other hand, the definition of the principal value gives that for each x ∈ Rn

4.4. PRINCIPAL VALUE TEMPERED DISTRIBUTIONS $ % $ % ψ P.V. Θ, ϕ(x − ·) = P.V. Θ, ψ(·)ϕ(x − ·) = lim Θ(y)ψ(y)ϕ(x − y) dy. ε→0+

139

(4.4.16)

|y|≥ε

Collectively, these arguments show that, for each x ∈ Rn , P.V. Θ ∗ ϕ (x) = (u ∗ ϕ)(x) + (v ∗ ϕ)(x) = lim+ Θ(y)ψ(y)ϕ(x − y) dy ε→0

|y|≥ε

+ lim

ε→0+

|y|≥ε

(1 − ψ(y))Θ(y)ϕ(x − y) dy

= lim

ε→0+

|x−y|≥ε

Θ(x − y)ϕ(y) dy,

(4.4.17)

proving (4.4.11). The next example discusses a basic class of principal value tempered distributions arising naturally in applications. Example 4.68. Let Φ ∈ C 1 (Rn \ {0}) be positive homogeneous of degree 1 − n. Then for each j ∈ {1, . . . , n} it follows that ∂j Φ satisfies the conditions in (4.4.1). Consequently, P.V.(∂j Φ) is a well-defined tempered distribution. To see why this is true fix j ∈ {1, . . . , n} and note that ∂j Φ ∈ C 0 (Rn \ {0}) and ∂j Φ is positive homogeneous of degree −n (cf. Exercise 4.49). Moreover, using Exercise 4.50, then integrating by parts based on (13.7.4), and then using (13.8.5), we obtain xj 0= Φ(x) dσ(x) − Φ(x)xj dσ(x) = ∂j Φ(x) dx 2 |x|=2 |x|=1 1 0

(itself a consequence of the fact that Θ has mean value zero over S n−1 ). At this stage, we wish to invoke Lebesgue’s dominated convergence theorem in order to absorb the limit inside the integral. To see that this theorem is applicable in the current context, we first note that for each ξ ∈ Rn \ {0} and each ω ∈ S n−1 such that ξ · ω = 0, formulas (4.11.5) and (4.11.6) give R −i(ω·ξ)ρ) dρ e − cos ρ lim ρ ε→0+ ε R→∞

6

= lim

ε→0+

R→∞

ε

R

7 R cos (ω · ξ)ρ − cos ρ sin (ω · ξ)ρ dρ − i dr ρ ρ ε

= − ln |(ω · ξ)| − i

π sgn (ω · ξ) = − log i(ω · ξ) . 2

(4.5.17)


144

This takes care of the pointwise convergence aspect of Lebesgue’s theorem. To verify the uniform domination aspect, based on (4.11.7) and (4.11.8) we first estimate

Rn

|ϕ(ξ)|

S n−1

|Θ(ω)| sup

≤

Rn

|ϕ(ξ)|

01 |x| |x|≤1 (4.6.21) is a tempered distribution, F wχB(0,1) ∈ S (Rn ) is of function type, and there exists a constant c ∈ C such that (4.6.22) F wχB(0,1) (ξ) = −ωn−1 ln |ξ| + c, ∀ ξ ∈ Rn \ {0}.

150


In the last part of this section we compute the constant c from (4.6.22) in the case n = 1. Before doing so we recall that Euler’s constant γ is defined by ⎤ ⎡ k 1 − ln k ⎦ . γ := lim ⎣ (4.6.23) k→∞ j j=1 The number γ plays an important role in analysis, as it appears prominently in a number of basic formulas. Two such identities that are relevant for us here are as follows: √ ∞ ∞ π −x −x2 (γ + 2 ln 2). (4.6.24) e ln x dx = −γ, e ln x dx = − 4 0 0 Example 4.75. We claim that F wχ(−1,1) (ξ) = −2 ln |ξ| − 2γ

in S (R),

(4.6.25)

where wχ(−1,1) is defined in (4.6.21). To see why this is true, note that (4.6.22) written for n = 1 (in which case ωn−1 = 2) becomes F wχ(−1,1) (x) = −2 ln |x|+ c in S (R). Consequently, to obtain (4.6.25) it remains to prove that c = −2γ when n = 1. The idea for determining the value of c is to apply F wχ(−1,1) to the 2

Schwartz function e−x . First, from (4.6.22) we have " ! 2 2 F wχ(−1,1) , e−x = −2 ln |x| + c, e−x .

(4.6.26)

Second, we compute the term in the left hand-side of (4.6.26). Specifically, based on the definition of the Fourier transform of a tempered distribution, (3.2.6), and (4.6.21) we may write " $ ! % $ √ 2 2 % 2 F wχ(−1,1) , e−x = wχ(−1,1) , F (e−x ) = wχ(−1,1) , πe−x /4 =

√ π

2

|x|>1

√ =2 π

∞ 1/2

√ e−x /4 dx + π |x| 2

√ e−x dx + 2 π x

e−x

1/2

0

/4

−1

|x|

|x|≤1

2

dx

2

e−x − 1 dx =: I + II. x (4.6.27)

Going further, integrating by parts then changing variables gives ∞ √ √ 2 I = 2 πe−1/4 ln 2 + 4 π (ln x)e−x x dx √ √ = 2 πe−1/4 ln 2 + π

1/2 ∞

√ (ln x)e 1/ 2

−x

dx.

(4.6.28)

4.7. A GENERAL JUMP-FORMULA IN THE CLASS . . .

151

Regarding II, for each ε > 0 an integration by parts and then a change of variables yield

1/2

ε

2

2 e−x − 1 dx = −(e−1/4 − 1) ln 2 − (e−ε − 1) ln ε + 2 x

2

= −(e−1/4 − 1) ln 2 − (e−ε

1 − 1) ln ε + 2

1/2

2

(ln x)e−x x dx

ε

√ 1/ 2

√

(ln x)e−x dx.

ε

(4.6.29) Observe that since 2

lim+ (e−ε − 1) ln ε =

ε→0

ln ln y − ln(y − 1) 1 lim = 0, 2 y→∞ y

(4.6.30)

by taking the limit as ε → 0+ in (4.6.29) we may conclude that √ √ II = −2 π(e−1/4 − 1) ln 2 + π

√ 1/ 2

(ln x)e−x dx.

(4.6.31)

0

Hence, from (4.6.28), (4.6.31), and the first identity in (4.6.24), it follows that ∞ √ √ √ √ I + II = 2 π ln 2 + π (ln x)e−x dx = 2 π ln 2 − πγ. (4.6.32) 0

This takes care of the term in the left-hand side of (4.6.25). To compute the term in the right-hand side of (4.6.25), write 2 −x2 −x2 −2 ln |ξ| + c, e = −2 (ln |x|)e dx + c e−x dx R

= −4

∞

R

√ 2 (ln x)e−x dx + c π

0

√ √ = π(γ + 2 ln 2) + c π.

(4.6.33)

For the last equality in (4.6.33) we used the second identity in (4.6.24). A combination of (4.6.26), (4.6.27), (4.6.32), and (4.6.33), then implies c = −2γ, as desired.

4.7

A General Jump-Formula in the Class of Tempered Distributions

The aim in this section is to prove a very useful formula expressing the limits of certain sequences of tempered distributions {Φε }ε=0 as the index parameter ε ∈ R \ {0} approaches the origin from either side. The trademark feature (which also justifies the name) of this formula is the presence of a jump-term of


152

the form ±Cδ (where the sign is correlated to sgn ε) in addition to a suitable principal value tempered distribution. A conceptually simple example of this phenomenon has been presented in Exercise 2.117, which our theorem contains as a simple special case (see Remark 4.79). With the notational convention that for points x ∈ Rn we write x = (x , t), where x = (x1 , . . . , xn−1 ) ∈ Rn−1 and t ∈ R, our main result in this regard reads as follows. Theorem 4.76. If Φ ∈ C 4 (Rn \ {0}) is odd and positive homogeneous of degree 1 − n, then i 3 lim Φ(x , ε) = ± Φ(0 , 1) δ(x ) + P.V. Φ(x , 0) 2

ε→0±

in S (Rn−1 ).

(4.7.1)

A few comments before presenting the proof of this result are in order. First, above we have employed the earlier convention of writing u(x ) for a distribution u in Rn−1 simply to stress that the test functions to which u is applied are considered in the variable x ∈ Rn−1 . Second, for each fixed ε ∈ R \ {0}, applying Exercise 4.51 yields −(n−1)/2 |Φ(x , ε)| ≤ ΦL∞ (S n−1 ) |x |2 + ε2

for each x ∈ Rn−1 .

(4.7.2)

Having observed this, the discussion in Example 4.3 then shows Φ(·, ε) ∈ S (Rn−1 ). Third, it is worth recalling an earlier convention to the effect that given a family of tempered distributions uε , indexed by ε ∈ I, I = (a, b) ⊆ R open interval, we say that uε → u ∈ D (Rn ) in S (Rn ) as I ε → a+ provided uεj → u in S (Rn ) for every sequence {εj }j∈N ⊆ I such that εj → a+ as j → ∞. In particular, this interpretation is in effect for (4.7.1). Fourth, from Exercise 4.52 we know that Φ ∈ S (Rn ). In addition, Exer3 n 3 , 1) cise 4.60 applied to Φ shows that Φ ∈ C 1 (Rn \{0}). In particular, Φ(0 R \{0} is meaningfully defined. Fifth, Φ(·, 0) (viewed as a function in Rn−1 \ {0 }) is continuous and posn−1 \ {0 } and, being odd, satisfies

itive homogeneous of degree −(n − 1) in R S n−2 Φ(·, 0) dσ = 0. As such, the conditions in (4.4.1) are satisfied (with n − 1 in place of n), hence Proposition 4.63 ensures that P.V. Φ(·, 0) is a well-defined tempered distribution in Rn−1 . Proof of Theorem 4.76. Assume Φ ∈ C 4 (Rn \ {0}) is odd and positive homogeneous of degree 1 − n, and let ϕ ∈ S(Rn−1 ). Then, for any fixed ε > 0, write Φ(x , t)ϕ(x ) dx = lim+ Φ(x , t)ϕ(x ) dx lim+ t→0

Rn−1

t→0

|x |>ε

+ lim+ t→0

|x |0 |x−y|≥ε Θ(x − y)f (y) dy , for x ∈ Rn ). Having dealt with (4.9.45) in the case p = 2, its proof for p ∈ (1, ∞) proceeds as follows. In a first step, a suitable version of (4.9.45) is established for p = 1 which, when combined with the case p = 2 already treated yields (via a technique called interpolation) (4.9.45) for p ∈ (1, 2). In a second step, one uses duality to handle the case p ∈ (2, ∞). There are two important factors at play p ∈ (2, ∞), here: that the dual of Lp (Rn ) with p ∈ (1, 2) is Lp (Rn ) with p = p−1 and the formula for the adjoint of TΘ we deduced in (4.9.42). In particular, Theorem 4.97 gives that, given p ∈ (1, ∞), the jth Riesz transform Rj , j ∈ {1, . . . , n}, originally considered as in (4.9.9) extends by density so that Rj : Lp (Rn ) −→ Lp (Rn )

is linear and bounded,

and for each f ∈ Lp (Rn ) we have xj − yj f (y) dy (Rj f )(x) = lim+ |x − y|n+1 ε→0 |x−y|≥ε

for a.e. x ∈ Rn .

(4.9.46)

(4.9.47)

Of course, the same type of result is true for the Hilbert transform on the real line.

4.10. DERIVATIVES OF VOLUME POTENTIALS

4.10

175

Derivatives of Volume Potentials

One basic integral operator in analysis is the so-called Newtonian potential, given by 1 −1 f (y) dy, x ∈ Rn , (4.10.1) (NΩ f )(x) := (n − 2)ωn−1 Ω |x − y|n−2 where Ω is an open set in Rn , n ≥ 3, and f ∈ L∞ (Ω) with bounded support. In this regard, an important issue is that of computing ∂j ∂k NΩ f in the sense of distributions in Rn , where j, k ∈ {1, . . . , n}. First, we will show (cf. Theorem 4.101) that NΩ f belongs to C 1 (Rn ) and we have xk − yk 1 f (y) dy, x ∈ Rn . (4.10.2) ∂k (NΩ f )(x) = ωn−1 Ω |x − y|n

zk 1 n Hence, ∂k (NΩ f )(x) = Ω Φ(x − y)f (y) dy where Φ(z) := ωn−1 |z|n for z ∈ R \ {0}. Note that for x fixed, Φ(x − ·) is locally integrable, though this is no longer the case for (∂j Φ)(x − ·). This makes the job of computing ∂j ∂k (NΩ f ) considerably more subtle. There is a good reason for this since, as it turns out, the latter distributional derivative involves (as we shall see momentarily) SIO. We note that the function Φ considered above is C ∞ and positive homogeneous of degree 1 − n in Rn \ {0}. These are going to be key features in our subsequent analysis. Our most general results encompassing the discussion about the Newtonian potential introduced above are contained in Theorems 4.100 and 4.101. We begin by first considering the case when Ω = Rn and f is a Schwartz function. Proposition 4.98. Let Φ ∈ C 1 (Rn \ {0}) be a function

that is positive homogeneous of degree 1 − n and let f ∈ S(Rn ). Then Rn Φ(x − y)f (y) dy is differentiable in Rn and for each j ∈ {1, . . . , n} we have Φ(x − y)f (y) dy = Φ(ω)ωj dσ(ω) f (x) (4.10.3) ∂xj Rn

S n−1

+ lim+ ε→0

|y−x|≥ε

(∂j Φ)(x − y)f (y) dy

∀ x ∈ Rn .

Proof. The key ingredient in the proof of (4.10.3) is formula (4.4.19). To see how (4.4.19) applies, first note that by using Exercise 4.51 we have for each R > 0, Φ(x − y)f (y) dy Rn

≤ ΦL∞ (S n−1 )

Rn

|f (y)| dy |x − y|n−1


176

≤ ΦL∞ (S n−1 ) f L∞ (Rn )

|y−x|≤R

dy |x − y|n−1

+ ΦL∞ (S n−1 ) (1 + |y|) f L∞ (Rn ) 2

thus

Φ(x − y)f (y) dy is well-defined. Second, since

Rn

Rn

Φ(x − y)f (y) dy =

we have that ∂xj

|y−x|>R

dy < ∞, (1 + |y|2 )|x − y|n−1 (4.10.4)

Rn

Rn

Rn

Φ(y)f (x − y) dy

∀ x ∈ Rn ,

(4.10.5)

Φ(x − y)f (y) dy is differentiable and

Φ(x − y)f (y) dy =

Rn

Φ(y)(∂j f )(x − y) dy

∀ x ∈ Rn .

(4.10.6)

Third, for each x ∈ Rn , with tx as in (2.8.37) (and recalling (3.2.22)), we may write Φ(x − y)ϕ(y) dy = tx (Φ∨ ), ϕ, ∀ ϕ ∈ S(Rn ). (4.10.7) Rn

Now fix j ∈ {1, . . . , n}. Then for x ∈ Rn we have Φ(x − y)f (y) dy ∂xj Rn

= tx (Φ∨ ), ∂j f = Φ∨ , t−x (∂j f ) = Φ∨ , ∂j [t−x f ] = −∂j [Φ∨ ], t−x f =−

S n−1

! " ωj Φ∨ (ω) dσ(ω) δ, t−x f − P.V. ∂j (Φ∨ ) , t−x f

= f (x) S n−1

= f (x) S n−1

ωj Φ(ω) dσ(ω) + lim+

(∂j Φ)(−y)f (x + y) dy

ωj Φ(ω) dσ(ω) + lim

(∂j Φ)(x − y)f (y) dy.

ε→0

ε→0+

|y|≥ε

|y|≥ε

(4.10.8) The first equality in (4.10.8) uses (4.10.6) and (4.10.7), the fifth uses (4.4.19), the sixth the fact that ∂j (Φ∨ ) = −(∂j Φ)∨ , and the last one a suitable change of variables. This completes the proof of the corollary. Next, we present a version of Proposition 4.98 when the function f is lacking any type of differentiability properties. Here we make use of the basic Calder´ on– Zygmund result recorded in Theorem 4.97.


177

Theorem 4.99. Let Φ ∈ C 1 (Rn \{0}) be a function that is positive homogeneous of degree 1 − n and let f ∈ Lp (Rn ) for some p ∈ (1, n). Then Φ ∗ f ∈ L1loc (Rn ) and for each j ∈ {1, . . . , n} we have T∂j Φ f ∈ Lp (Rn ) and Φ(ω)ωj dσ(ω) f + T∂j Φ f in D (Rn ), (4.10.9) ∂j (Φ ∗ f ) = S n−1

where T∂j Φ is the operator from (4.9.45) with ψ replaced by ∂j Φ. Proof. Fix p ∈ (1, n), f ∈ Lp (Rn ), and R ∈ (0, ∞). Then we write |f (y)| 1 dy dx = |f (y)| dx dy n−1 |x − y| |x − y|n−1 n B(0,R) R |y|≤2R B(0,R)

+ |y|>2R

|f (y)| B(0,R)

1 dx dy= : I+II. |x − y|n−1 (4.10.10)

If y ∈ B(0, 2R) and x ∈ B(0, R), then |x − y| ≤ 3R, thus 1 |f (y)| dx dy I≤ n−1 |y|≤2R |x−y|≤3R |x − y| ≤

1 dz |B(0, 2R)|1− p f Lp(B(0,2R) < ∞,

1

|z|≤3R

|z|n−1

(4.10.11)

where for the second inequality in (4.10.11) Hölder’s inequality has been used. Also, if y ∈ Rn \ B(0, 2R) and x ∈ B(0, R), then |y| ≤ |y − x| + |x| ≤ |y − x| + R ≤ |y − x| +

|y| , 2

which implies |y − x| ≥ |y|/2. Using this, II is estimated as 1 II ≤ 2n−1 |f (y)| dx dy n−1 |y| |y|>2R |x|≤R ≤2

n−1

|B(0, R)|

|y|>2R

|f (y)| dy |y|n−1

≤ 2n−1 |B(0, R)|f Lp(Rn )

dy |y|>2R

|y|

(n−1)p

1/p

< ∞,

(4.10.12)

p where p := p−1 is the H¨ older conjugate exponent for p. In the third inequality in (4.10.12) the assumption p ∈ (1, n) has been used to ensure that (n−1)p > n. A combination of (4.10.10), (4.10.11), and (4.10.12) yields the following conclusion: for every R ∈ (0, ∞) there exists some finite positive constant C, depending on R, n, and p, such that |f (y)| dy dx ≤ Cf Lp(Rn ) , ∀ f ∈ Lp (Rn ). (4.10.13) |x − y|n−1 n B(0,R) R


178

In turn, (4.10.13) entails several useful conclusions. Recall that the assumptions on Φ imply |Φ(x)| ≤ ΦL∞ (S n−1 ) |x|1−n for each x ∈ Rn \{0} (cf. Exercise 4.51). The first conclusion is that for each f ∈ Lp (Rn ) and each R ∈ (0, ∞), Rn

|Φ(x − y)||f (y)| dy < ∞ for a.e.

x ∈ B(0, R).

(4.10.14)

This shows that (Φ ∗ f )(x) is well-defined for a.e. x ∈ Rn . Second, from what we have just proved and (4.10.13) we may conclude that Φ ∗ f ∈ L1loc (Rn ). Next, recall that C0∞ (Rn ) is dense in Lp (Rn ) for each p ∈ (1, ∞). As such there exists a sequence {fk }k∈N of function in C0∞ (Rn ) with the property that lim fk = f in Lp (Rn ). Based on (4.10.13) we may conclude that the sequence k→∞

{Φ ∗ fk }k∈N converges in Lp (Rn ) to Φ ∗ f . Therefore, by Exercise 2.21, D (Rn )

Φ ∗ fk −−−−→ Φ ∗ f.

(4.10.15)

k→∞

Moving on, fix j ∈ {1, . . . , n} and note that, by Exercise 4.49, ∂j Φ is positive homogeneous of degree −n. The function Φ ∗ fk belongs to C ∞ (Rn ) and Proposition 4.98 applies and implies that Φ(ω)ωj dσ(ω) fk + T∂j Φ fk , ∀ k ∈ N, (4.10.16) ∂j Φ ∗ fk = S n−1

pointwise in Rn . In particular, from (4.10.16) we infer that T∂j Φ fk ∈ C ∞ (Rn ) for all k ∈ N, and that the equality in (4.10.16) also holds in D (Rn ). Since Lp (Rn )

→ T∂j Φ f , based on Exercise 2.21 it follows Theorem 4.97 gives that T∂j Φ fk −−−−− k→∞

that D (Rn )

T∂j Φ fk −−−−→ T∂j Φ f k→∞

and

D (Rn )

fk −−−−→ f.

(4.10.17)

k→∞

D (Rn )

Since an immediate consequence of (4.10.15) is that ∂j (Φ∗ fk ) −−−−→ ∂j (Φ∗ f ), k→∞

combining this, (4.10.17), and the fact that (4.10.16) holds in D (Rn ), we obtain (4.10.9). We are prepared to state and prove our most general results regarding distributional derivatives of volume potentials. In particular, the next two theorems contain solutions to the questions formulated at the beginning of this section. Theorem 4.100. Let Φ ∈ C 1 (Rn \ {0}) be a function that is positive homogeneous of degree 1 − n. Consider a measurable set Ω ⊆ Rn and assume that f ∈ Lp (Ω) for some p ∈ (1, n). Then Ω Φ(x−y)f (y) dy is absolutely convergent for a.e. x ∈ Rn and belongs to L1loc (Rn ) as a function of the variable x. Moreover, if f denotes the extension of f by zero to Rn , then for each j ∈ {1, . . . , n} we have


Φ(x − y)f (y) dy =

∂xj

S n−1

Ω

179

Φ(ω)ωj dσ(ω) f(x)

+ lim+ ε→0

(∂j Φ)(x − y)f (y) dy (4.10.18) y∈Ω\B(x,ε)

where the derivative in the left-hand side is taken in D (Rn ) and the equality is also understood in D (Rn ). In particular, formula (4.10.18) holds for any function f belonging to some Lq (Ω), with q ∈ (1, ∞), that vanishes outside of a measurable subset of Ω of finite measure. Proof. The main claim in the statement follows by applying Theorem 4.99 to the function f in Ω, f := (4.10.19) 0 in Rn \ Ω, upon observing that f ∈ Lp (Rn ). Moreover, if f is as in the last claim in the statement then H¨ older’s inequality may be invoked to show that f belongs to Lp (Ω) for some p ∈ (1, n). Recall that if a ∈ R, the integer part of a is denoted by a and is by definition the largest integer that is less than or equal to a. To state the next theorem we introduce a if a ∈ Z, a := (4.10.20) a − 1 if a ∈ Z. Hence, a is the largest integer strictly less than a (thus, in particular, a < a for every a ∈ R). Theorem 4.101. Let Φ ∈ C ∞ (Rn \ {0}) be a function that is positive homogeneous of degree m ∈ R where m > −n. Define the generalized volume n potential associated with Φ by setting for each f ∈ L∞ comp (R ) (ΠΦ f )(x) :=

Rn

Φ(x − y)f (y) dy,

∀ x ∈ Rn .

(4.10.21)

n m+n Then for each f ∈ L∞ (Rn ) and comp (R ) one has ΠΦ f ∈ C

∀ α ∈ Nn0 with |α| ≤ m + n. (4.10.22) (Rn ), the Moreover, if α ∈ Nn0 is such that |α| = m + n, then for each f ∈ L∞ comp distributional derivative ∂ α ΠΦ f is of function type and satisfies ∂ α (ΠΦ f ) = Π∂ α Φ f

pointwise in Rn ,


180

∂ ΠΦ f (x) =

α

S n−1

(∂ β Φ)(ω)ωj dσ(ω) f (x)

+ lim

ε→0+

Rn \B(x,ε)

(∂ α Φ)(x − y) f (y) dy

in D (Rn ), (4.10.23)

for any β ∈ Nn0 and j ∈ {1, . . . , n} such that β + ej = α. n n Proof. Fix f ∈ L∞ comp (R ) and let K := supp f which is a compact set in R . By Exercises 4.49 and 4.51,

we have that ∂ α Φ ∈ C ∞ (Rn \ {0}),

for every α ∈ Nn0

∂ α Φ is positive homogeneous of degree m − |α|, and α ∂ Φ(x − y) ≤ ∂ α ΦL∞ (S n−1 ) |x − y|m−|α| ∀ x, y ∈ Rn , x = y.

(4.10.24)

Fix now α ∈ Nn0 such that |α| ≤ m + n. Then m − |α| ≥ m + n > −n, hence (4.10.24) further yields α |(∂ Φ)(x − y)f (y)| dy ≤ Cf L∞ (Rn ) |x − y|m−|α| dy < ∞. (4.10.25) Rn

K

This proves that Π∂ α Φ f is well-defined. Next, we focus on proving Π∂ α Φ f ∈ C 0 (Rn ).

(4.10.26)

To see this, fix x0 ∈ Rn and pick an arbitrary sequence {xk }k∈N of points in Rn satisfying lim xk = x0 . Consider the following functions defined a.e. in Rn : k→∞

vk := (∂ α Φ)(xk − ·)f,

∀ k ∈ N,

and v := (∂ α Φ)(x0 − ·)f.

(4.10.27)

To conclude that Π∂ α Φ f is continuous at x0 , it suffices to show that lim vk (y) dy = v(y) dy. (4.10.28) k→∞

K

K

The strategy for proving (4.10.28) is to apply Vitali’s theorem (cf. Theorem 13.21) with X := K and μ being the restriction to K of the Lebesgue measure in Rn . Since K is compact we have μ(X) < ∞ and from (4.10.25) we know that vk ∈ L1 (X, μ) for all k ∈ N. Clearly, |v(x)| < ∞ for μ-a.e. x ∈ K. Also, lim vk (y) = v(y) for μ-almost every y ∈ K. Hence, in order to obtain k→∞

(4.10.28), the only hypothesis left to verify in Vitali’s theorem is that the sequence {vk }k∈N is uniformly integrable in (X, μ). With this goal in mind, let ε > 0 be fixed and consider a μ-measurable set A ⊂ K such that μ(A) is sufficiently small, to a degree to be specified later. Then for every k ∈ N, based on (4.10.24), we have


181

α vk (x) dμ(x) ≤ ∂ ΦL∞ (S n−1 ) f L∞ (Rn ) |xk − y|m−|α| dy A

A

|x|m−|α| dx,

=C

(4.10.29)

A−xk

where C := ∂ α ΦL∞ (S n−1 ) f L∞(Rn ) . Note that μ(A − xk ) = μ(A) for k ∈ N. Also, since the sequence {xk }k∈N and the set K are bounded, ∃ R ∈ (0, ∞) such that A − xk ⊂ K − xk ⊂ B(0, R) ∀ k ∈ N.

(4.10.30)

Given that m − |α| > −n, we have |x|m−|α| ∈ L1 (B(0, R)) and we may invoke Proposition 13.22 to conclude that there exists δ > 0 such that for every Lebesgue measurable set E ⊂ B(0, R) with Lebesgue measure less than δ we

have E |x|m−|α| dx < ε/C. At this point return with the latter estimate in (4.10.29) to conclude that (4.10.31) if μ(A) < δ then vk (x) dμ(x) < ε ∀ k ∈ N. A

This proves that the sequence {vk }k∈N is uniformly integrable in (X, μ) and finishes the proof of the fact that Π∂ α Φ f is continuous at x0 . Since x0 was arbitrary in Rn the membership in (4.10.26) follows. Our next goal is to show (4.10.22). First note that based on what we proved so far we have ΠΦ f, Π∂ α Φ f ∈ L1loc (Rn ), thus they define distributions in Rn . We claim that the distribution Π∂ α Φ f is equal to the distributional derivative ∂ α [ΠΦ f ]. To see this, fix ϕ ∈ C0∞ (Rn ) and using the definition of distributional derivatives and the definition of ΠΦ f write $ $ α % % ∂ ΠΦ f, ϕ = (−1)|α| ΠΦ f, ∂ α ϕ Φ(x − y)f (y) dy ∂ α ϕ(x) dx = (−1)|α| Rn

|α|

= (−1)

Rn

f (y)

Rn

Rn

Φ(x − y)∂ α ϕ(x) dx dy.

(4.10.32)

Based on (4.10.24), the assumptions on ϕ, and the fact that m − |α| > −n, we may use Lebesgue’s dominated convergence theorem and formula (13.7.4) repeatedly to further write

Rn

Φ(x − y)∂ α ϕ(x) dx = lim

ε→0+

Rn \B(y,ε)

= lim

ε→0+

+

(−1)|α|

cjβγ

α=β+γ+ej

Φ(x − y)∂ α ϕ(x) dx

Rn \B(y,ε)

(∂ α Φ)(x − y)ϕ(x) dx

⎫ ⎬ x − y j j γ ∂ ϕ(x) dσ(x) , (∂ β Φ)(x − y) ⎭ ε ∂B(y,ε) (4.10.33)


182

where cjβγ are suitable constants independent of ε. Note that, for each β, γ, and j such that α = β + γ + ej , in light of (4.10.24) we have xj − yj γ β ∂ ϕ(x) dσ(x) (∂ Φ)(x − y) ∂B(y,ε) ε ≤ |(∂ β Φ)(x − y)| |∂ γ ϕ(x)| dσ(x) ∂B(y,ε)

≤ ∂ ϕL∞ (Rn ) ∂ ΦL∞ (S n−1 ) γ

|x − y|m−|β| dσ(x)

β

∂B(y,ε)

= Cεm−|β|+n−1 −−−−→ 0, ε→0+

(4.10.34)

since m−|β|+n−1 ≥ m+n−|α| > 0. Returning with (4.10.34) to (4.10.33), and applying one more time Lebesgue’s dominated convergence theorem, it follows that Φ(x − y)∂ α ϕ(x) dx = (−1)|α| (∂ α Φ)(x − y)ϕ(x) dx. (4.10.35) Rn

Rn

This, when used in (4.10.32) further yields $ α % ∂ ΠΦ f, ϕ = f (y) (∂ α Φ)(x − y)ϕ(x) dx dy Rn

Rn

= Rn

Rn

(∂ α Φ)(x − y)f (y) dy ϕ(x) dx

% $ = Π∂ α Φ f, ϕ .

(4.10.36)

Since ϕ ∈ C0∞ (Rn ) is arbitrary, from (4.10.36) we conclude ∂ α ΠΦ f = Π∂ α Φ f

in

D (Rn ),

∀ α ∈ Nn0 with |α| ≤ m + n.

(4.10.37)

Upon observing that (4.10.26) and (4.10.37) hold for every α ∈ Nn0 with the property that |α| = m + n, we may invoke Theorem 2.102 to infer that ΠΦ f ∈ C m+n (Rn ) and that (4.10.22) is valid. We are left with proving the very last statement in the theorem. To this end, suppose there exists α ∈ Nn0 satisfying |α| = m + n. In particular, we have m ∈ Z and |α| ≥ 1. Hence, there exists j ∈ {1, . . . , n} with the property that αj ≥ 1. Set β := (α1 , . . . , αj−1 , αj − 1, αj+1 , . . . , αn ), so that α = β + ej

and |β| = m + n − 1 = m + n.

(4.10.38)

Based on what we proved earlier, we have ∂ β [ΠΦ f ] = Π∂ β Φ f pointwise in Rn . Also, ∂ β Φ is of class C ∞ and positive homogeneous of degree 1 − n in Rn \ {0}. Thus, we may apply Theorem 4.100 with Φ replaced by ∂ β Φ and Ω replaced by K and obtain


183

∂ α ΠΦ f (x) = ∂j ∂ β [ΠΦ f ] (x) = ∂j Π∂ β Φ f (x) (∂ β Φ)(x − y)f (y) dy = ∂x j

K

= S n−1

+ lim+ ε→0

(∂ β Φ)(ω)ωj dσ(ω) f (x) Rn \B(x,ε)

(∂ α Φ)(x − y) f (y) dy,

(4.10.39)

where the derivative in the left hand-side of (4.10.39) is taken in D (Rn ) and the equality is understood in D (Rn ). This proves (4.10.23) and finishes the proof of the theorem. Exercise 4.102. In the context of Theorem 4.101 prove directly, without relying n m+n on Theorem 2.102, that whenever f ∈ L∞ (Rn ). comp (R ) one has ΠΦ f ∈ C Hint: Show that for each x ∈ Rn and each j ∈ {1, . . . , n} fixed, one has (4.10.40) lim (ΠΦ f )(x + hej ) − (ΠΦ f )(x) = (Π∂j Φ f )(x) h→0

The proof of (4.10.40) may be done by using Vitali’s theorem (cf. Theorem 13.21) in a manner analogous to the proof of (4.10.26). Once (4.10.40) is established, iterate to allow higher-order partial derivatives. Further Notes for Chap. 4. The significance of the class of tempered distributions stems from the fact that this class is stable under the action of the Fourier transform. The topics discussed in Sects. 4.1–4.6 are classical and a variety of expositions is present in the literature, though they differ in terms of length and depth, and the current presentation is no exception. For example, while the convolution product of distributions is often confined to the case in which one of the distributions in question is compactly supported, that is, E (Rn ) ∗ D (Rn ), in Theorem 4.18 we have seen that S(Rn ) ∗ S (Rn ) continues to be meaningfully defined in S (Rn ). For us, extending the action of the convolution product in this manner is motivated by the discussion in Sect. 4.9, indicating how SIO may be interpreted as multipliers. The main result in Sect. 4.7, Theorem 4.76, appears to be new at least in the formulation and the degree of generality in which it has been presented. This result may be regarded as a far-reaching generalization of Sokhotsky’s formula (2.10.3); cf. Remark 4.79 for details. Theorem 4.76 has a number of remarkable consequences, and we use it to offer a new perspective on the treatment of the classical harmonic Poisson kernel in Sect. 4.8. Later on, in Sect. 11.6, Theorem 4.76 resurfaces as the key ingredient in the study of boundary behavior of layer potential operators in the upper-half space. The treatment of the SIO from in Sect. 4.9 highlights the interplay between distribution theory, harmonic analysis, and partial differential equations. Specifically, first the action of a SIO of the form TΘ on a Schwartz function ϕ is interpreted as

P.V. Θ ∗ϕ (which, as pointed out earlier, is a well-defined object in S(Rn )∗S (Rn ) ⊂ S (Rn )). Second, the Fourier analysis of tempered distributions is invoked to conclude

184


that T F P.V. Θ which shifts the focus of understanding the properties of Θϕ = ϕ

the SIO TΘ to clarifying the nature of the tempered distribution mΘ := F P.V. Θ . Third, it turns out that the distribution mΘ is of function type and, in fact, a suitable pointwise formula may be deduced for it. In particular, it is apparent from this formula that mΘ ∈ L∞ (Rn ). Having established this, in a fourth step we may return to the original focus of our investigation, namely the SIO TΘ , and use the fact that along with Planchérel’s formula and the boundedness of mΘ to evenT Θ ϕ = mΘ ϕ, tually conclude (via a density argument) that TΘ extends to a linear and bounded operator on L2 (Rn ). In turn, SIO naturally intervene in the derivatives of volume potentials discussed in Sect. 4.10, and the boundedness result just derived ultimately becomes the key tool in obtaining estimates for the solution of the Poisson problem, treated later. The discussion in the previous paragraph also sheds light on some of the main aims of the theory of SIO, as a subbranch of harmonic analysis. For example, one would like to extend the action of SIO to Lp (Rn ) for any p ∈ (1, ∞), rather than just L2 (Rn ). Also, it is desirable to consider SIO which are not necessarily of convolution type, and/or in a setting in which Rn is replaced by a more general type of ambient (e.g., some type of surface in Rn+1 ). The execution of this program, essentially originating in the work of A. P. Calder´ on, A. Zygmund, and S. G. Mikhlin in the 1950s, stretches all the way to the present day. The reader is referred to the exposition in the monographs [6, 7, 16, 23, 47, 62, 63], where references to specific research articles may be found. Here, we only wish to mention that, in turn, such progress in harmonic analysis has led to significant advances in those areas of partial differential equations where SIO play an important role. In this regard, the reader is referred to the discussion in [28, 36, 47, 50].

4.11


Exercise 4.103. Prove that P.V. x1 ∈ S (R). Exercise 4.104. Prove that |x|N ln |x| ∈ S (Rn ) if N is a real number satisfying N > −n. Exercise 4.105. Prove that (ln |x|) = P.V. x1 in S (R). Exercise 4.106. Let a ∈ (0, ∞) and recall the Heaviside function H from (1.2.16). Prove that the function eax H(x), x ∈ R, does not belong to S (R). Do any of the functions e−ax H(−x), eax H(−x), or e−ax H(x), defined for x ∈ R, belong to S (R)? Exercise 4.107. In each case below determine if the given sequence of tempered distributions indexed over j ∈ N converges in S (R). Whenever convergent, determine its limit. (a) fj (x) =

x x2 +j −2 ,

(b) fj (x) =

1 j

·

x∈R

1 x2 +j −2 ,

x∈R

4.11. ADDITIONAL EXERCISES FOR CHAP. 4 (c) fj (x) =

1 sin jx π x ,

185

x ∈ R \ {0}

2

(d) fj (x) = ej δj Exercise 4.108. For each j ∈ N let fj (x) := j n θ(jx), x ∈ Rn , where θ ∈ S(Rn ) is fixed and satisfies Rn θ(x) dx = 1. Does the sequence {fj }j∈N converge in S (Rn )? If yes, determine its limit. 2

Exercise 4.109. For each j ∈ N consider the function fj (x) := ex χ[−j,j] (x) defined for x ∈ R. Prove that the sequence {fj }j∈N converges in D (R) but not in S (R). Exercise 4.110. For each j ∈ N let fj (x) := χ[j−1,j] (x) for x ∈ R. Prove that the sequence {fj }j∈N does not converge in Lp (R) for any p ≥ 1 but it converges to zero in S (R). Exercise 4.111. For each j ∈ N consider the tempered distribution uj ∈ S (R) D (R)

defined by uj := χ[−j,j] ex sin(ex ). Prove that uj −−−−→ ex sin(ex ). Does the j→∞

sequence {uj }j∈N converge in S (R)? If yes, what is the limit? Exercise 4.112. Let m ∈ N. Prove that any solution u of the equation xm u = 0 in D (R) satisfies u ∈ S (R). Use the Fourier transform to show that the general m−1 solution to this equation is u = ck δ (k) in S (R), where ck ∈ C, for every k = 0, 1, . . . , m − 1.

k=0

Exercise 4.113. Prove that for any f ∈ S (R) there exists u ∈ S (R) such that xu = f in S (R). 2

Exercise 4.114. Does the equation e−|x| u = 1 in S (Rn ) have a solution? Exercise 4.115. Prove that the Heaviside function H belongs to S (R) and compute its Fourier transform in S (R). Exercise 4.116. Compute the Fourier transform of P.V. x1 in S (R). Exercise 4.117. Compute the Fourier transform in S (R) of each of the following tempered distribution (all are considered in S (R) and recall the definition of the sgn function from (1.4.3)): (a) sgn x. (b) |x|k for k ∈ N. (c)

sin(ax) x

for a ∈ R \ {0}.


186 (d)

sin(ax) sin(bx) x

for a, b ∈ R \ {0}.

(e) sin(x2 ). (f ) ln |x|. Exercise 4.118. Let n = 3 and R ∈ (0, ∞). Compute the Fourier transform of the tempered distribution δ∂B(0,R) in S (R3 ), where δ∂B(0,R) the distribution defined as in Exercise 2.128 corresponding to Σ := ∂B(0, R). Exercise 4.119. Let k ∈ N0 and suppose f ∈ L1 (Rn ) has the property that xα f ∈ L1 (Rn ) for every α ∈ Nn0 satisfying |α| ≤ k. Prove that f3, the Fourier transform of f in S (Rn ) satisfies f3 ∈ C k (Rn ). Exercise 4.120. Show that χ[−1,1] ∈ S (R) and compute χ [−1,1] in S (R). Do 1 we have χ [−1,1] ∈ L (R)?

Exercise 4.121. Fix x0 ∈ Rn and set tx0 : S (Rn ) → S (Rn ) tx0 u, ϕ) := u, t−x0 (ϕ),

∀ ϕ ∈ S(Rn ),

(4.11.1)

where t−x0 (ϕ) is understood as in Exercise 3.42. Prove that the map in (4.11.1) is well-defined, linear, and continuous, and is an extension of the map from Exercise 3.42 in the sense that, if f ∈ S(Rn ), then tx0 uf = utx0 (f ) in S (Rn ). Also show that F tx0 u = e−ix0 ·(·) u 3 in S (Rn ) for every u ∈ S (Rn ). (4.11.2) Exercise 4.122. Let a ∈ R and consider the function g(x) := cos(ax) for every x ∈ R. Prove that g ∈ S (R) and compute g3 in S (R). Exercise 4.123. Let m ∈ N and suppose P is a polynomial in Rn of degree 2m no real roots. Prove that P1 ∈ S (Rn ), and that if 2m > n then 1that has 2m−n−1 F P ∈C (Rn ). Exercise 4.124. Let P be a polynomial in Rn and suppose k ∈ N. Prove that if P is homogeneous of degree −k then P ≡ 0. Exercise 4.125. Let ζ, η ∈ Rn , n ≥ 2, be two unit vectors such that ζ · η = 1, and consider the linear mapping R : Rn → Rn define by Rξ := ξ −

ξ · (η + ζ) ξ · [(1 + 2η · ζ)ζ − η] ζ+ η, 1+η·ζ 1+η·ζ

∀ ξ ∈ Rn .

(4.11.3)

Show that this is an orthogonal transformation in Rn that satisfies Rζ = η

and

R η = ζ.

(4.11.4)


187

Exercise 4.126. Prove that for every c ∈ R \ {0} the following formulas hold

R

lim

ε→0+

R→∞

ε

lim

ε→0+

R→∞

ε

R

cos(cρ) − cos ρ dρ = − ln |c|, ρ

(4.11.5)

π sin(cρ) dρ = sgn c, ρ 2

(4.11.6)

R cos(cρ) − cos ρ sup dρ ≤ 2 ln |c|, ρ 0ε

Proof. Apply Corollary 4.78 to the function Φ : R2 \ {(0, 0)} → C given by −1 Φ(x, y) := 2πi(x+iy) for all (x, y) ∈ R2 \ {(0, 0)}. Note that Φ is C ∞ , odd, and homogeneous of degree −1 and, under the canonical identification R2 ≡ C, takes −1 3 for z ∈ C \ {0}. Proposition 7.33 then gives that Φ(ξ) = 1ξ the form Φ(z) = 2πiz 3 1) = −i. Having established for all ξ ∈ C \ {0} which, in particular, yields Φ(0, this, (7.6.32) follows directly from (4.7.45). Remark 7.38. Upon recalling formula (4.9.30) for the Hilbert transform H on the real line, we may recast the version of Plemelj jump-formula (7.6.32) corresponding to considering the Cauchy operator in the upper half-plane in the form ver i 1 Hϕ in R, ∀ ϕ ∈ S(R), (7.6.33) Cϕ 2 = ϕ + 2 2π ∂R+

7.7. FUNDAMENTAL SOLUTIONS FOR THE DIRAC...

253

where the “vertical limit” of C ϕ to the boundary of the upper half-plane is understood as in (4.8.20). In turn, formula (7.6.33) suggests the consideration of the operator (with I denoting the identity) P :=

i 1 I+ H. 2 2π

(7.6.34)

From Corollary 4.95 it follows that P is a well-defined, linear and bounded operator on L2 (R). Using the fact that H 2 = −π 2 I and H ∗ = −H on L2 (R) (again, see Corollary 4.95), we may then compute 1

i 2 1 i 2 i H2 H = I+ H+ 2 2π 4 2π 2π i 1 1 i 1 H+ I= I+ H = I+ 4 2π 4 2 2π

P2=

I+

= P,

(7.6.35)

and P∗ =

i ∗ i 1 1 I− H = I+ H = P. 2 2π 2 2π

(7.6.36)

Any linear and bounded operator on L2 (R) satisfying these two properties (i.e., P 2 = P and P ∗ = P ) is called a projection. Then one may readily verify that i I − P = 12 I − 2π H is also a projection and if we introduce (what are commonly referred to as Hardy spaces on the real line) H±2 (R) :=

1 2

I±

i H f : f ∈ L2 (R) , 2π

(7.6.37)

then any complex-valued function f ∈ L2 (R) may be uniquely decomposed as f = f+ + f− with f± ∈ H±2 (R) and, moreover, any two functions f± ∈ H±2 (R) are orthogonal, in the sense that f+ (x)f− (x) dx = 0. (7.6.38) R

7.7

Fundamental Solutions for the Dirac Operator

In a nutshell, Dirac operators are first order differential operators factoring the Laplacian. When n = 1, Δ = d2 /dx2 , hence if we set D := i(d/dx), then D2 = −Δ. We seek a higher-dimensional generalization of the latter factorization formula. The natural context in which such a generalization may be carried out is the Clifford algebra setting.

254

CHAPTER 7. THE LAPLACIAN AND RELATED OPERATORS

The Clifford algebra with n generators Cn is the associative algebra with unit Cn , !, +, 1 freely generated over R by the family {ej }1≤j≤n , of the standard orthonormal base in Rn , now called imaginary units, subject to the following axioms: ej ! ek + ek ! ej = −2δjk , ∀ j, k ∈ {1, . . . , n}. (7.7.1) Hence, ej ! ek = −ek ! ej if 1 ≤ j = k ≤ n, and ej ! ej = −1

for j ∈ {1, . . . , n}.

(7.7.2)

The first condition above indicates that Cn is noncommutative if n > 1, while the second condition justifies calling {ej }1≤j≤n imaginary units. Elements in the Clifford algebra Cn can be uniquely written in the form a=

n

aI e I

(7.7.3)

l=0 |I|=l

with aI ∈ C, where eI stands for the product ei1 ! ei2 ! · · · ! eil whenever I = (i1 , i2 , . . . , il ) with 1 ≤ i1 < i2 < · · · < il ≤ n,e∅ := 1 ∈ R (which plays the role of the multiplicative unit in Cn ) and indicates that the |I|=l

sum is performed over strictly increasingly ordered indexes I with l components (selected from the set {1, . . . , n}). In the writing (7.7.3) we shall refer to the numbers aI ∈ C as the scalar components of a. Exercise 7.39. Given any a, b ∈ Cn with scalar components aI , bI ∈ C define (a, b) :=

n

aI b I ,

a :=

n

l=0 |I|=l

aI e I ,

(7.7.4)

l=0 |I|=l

and abbreviate Ma b := a ! b. Prove that Ma b = Ma b

and

(Ma b , c) = (b , Ma c)

for every

a, b, c ∈ Cn .

(7.7.5)

Clifford algebra-valued functions defined in an open set Ω ⊆ Rn may be defined naturally. Specifically, any function f : Ω → Cn is an object of the form f=

n

f I eI ,

(7.7.6)

l=0 |I|=l

where each fI is a real-valued function defined in Ω. Given any k ∈ N0 ∪{∞}, we shall denote by C k (Ω, Cn ) the collection of all Clifford algebra-valued functions f whose scalar components fI are of class C k in Ω. In a similar manner we may define C0∞ (Ω, Cn ), L1loc (Ω, Cn ), Lp (Ω, Cn ), etc. In fact, we may also consider Clifford algebra-valued distributions in an open set Ω ⊆ Rn . Specifically, write u ∈ D (Ω, Cn ) provided u=

n l=0 |I|=l

u I eI

with uI ∈ D (Ω) for each I.

(7.7.7)


255

Much of the theory originally developed for scalar-valued distributions naturally extends to this setting. For example, if u ∈ D (Ω, Cn ) is as in (7.7.7), we may define n α ∂ α u I eI , ∀ α ∈ Nn0 , (7.7.8) ∂ u := l=0 |I|=l

and f u :=

n n

f J u I eJ ! eI ,

(7.7.9)

=0 |J|= k=0 |I|=k

for any f =

n k=0 |J|=k

fJ eJ ∈ C ∞ (Ω, Cn ).

We are ready to introduce the Dirac operator D associated with Cn . Specifically, given a Clifford algebra-valued distribution u ∈ D (Ω, Cn ) as in (7.7.7), we define Du ∈ D (Ω, Cn ) by setting Du :=

n n

∂j uI ej ! eI .

(7.7.10)

j=1 l=0 |I|=l

In other words, D :=

n

Mej ∂j ,

(7.7.11)

j=1

where Mej denotes the operator of Clifford algebra multiplication by ej from the left. Proposition 7.40. Let Ω ⊆ Rn be an open set. Then the Dirac operator D satisfies D2 = −Δ in D (Ω, Cn ),

(7.7.12)

where Δ is the Laplacian. Proof. Pick an arbitrary u ∈ D (Ω, Cn ), say u =

n

l=0 |I|=l

uI eI with uI ∈ D (Ω)

for each I. Then using (7.7.10) twice yields D2 u = D(Du) =

n n

∂k ∂j uI ek ! ej ! eI .

(7.7.13)

j,k=1 l=0 |I|=l

Observe that, on the one hand,

n

1≤j=k≤n l=0 |I|=l

∂k ∂j uI ek ! ej ! eI = 0,

(7.7.14)

256


since for each I, we have ∂k ∂j uI = ∂j ∂k uI and ek ! ej ! eI = −ej ! ek ! eI whenever 1 ≤ j = k by the first formula in (7.7.2). On the other hand, corresponding to the case when j = k n n

∂k2 uI ek ! ek ! eI = −

k=1 l=0 |I|=l

n n

∂k2 uI eI = −Δu,

(7.7.15)

k=1 l=0 |I|=l

since ek ! ek = −1 by the second formula in (7.7.2). Exercise 7.41. Consider the embedding R → Cn , n

R x = (xj )1≤j≤n ≡ n

n

xj ej ∈ Cn ,

(7.7.16)

j=1

which identifies vectors from Rn with elements in the Clifford algebra Cn . With this identification in mind, show that x ! x = −|x|2

for any x ∈ Rn ,

x ! y + y ! x = −2x · y

for any x, y ∈ Rn .

(7.7.17) (7.7.18)

For example, in light of the embedding described in (7.7.16) we may regard the assignment Rn \ {0} x → |x|xn as a Clifford algebra-valued function. Proposition 7.40 combined with Remark 5.6 yields the following result. Theorem 7.42. The Clifford algebra-valued function E(x) := −

x 1 ∈ L1loc (Rn , Cn ) ∩ S (Rn , Cn ) ωn−1 |x|n

(7.7.19)

is a fundamental solution for the Dirac operator D in Rn , i.e., DE = δ

in S (Rn , Cn ).

(7.7.20)

Moreover, any u ∈ S (Rn , Cn ) satisfying Du = δ in S (Rn , Cn ) is of the form E + P where E as in (7.7.19) and P is a Clifford algebra-valued function whose components are polynomials and such that DP = 0 in Rn . Proof. Let EΔ be the fundamental solution for Δ as described in (7.1.12) for n ≥ 2. Then − (DEΔ )(x) = −

x 1 , ωn−1 |x|n

x ∈ Rn \ {0},

(7.7.21)

becomes, thanks to (7.7.12) and Remark 5.6, a fundamental solution for the Dirac operator D in Rn . To justify the last claim in the statement of the theorem, let F ∈ S (Rn , Cn ) be an arbitrary fundamental solution for the Dirac operator D in Rn . Then the tempered distribution P := F − E satisfies DP = 0 in S (Rn , Cn ) and, on the


257

Fourier transform side, we have ξ ! P3 = 0 in S (Rn , Cn ). Multiplying (in the Clifford algebra sense) this equality with the Clifford algebra-valued function with polynomial growth ξ then yields −|ξ|2 P3 = 0 in S (Rn , Cn ) (cf. (7.7.17)). The latter implies supp P3 ⊆ {0}, hence the components of P are polynomials in Rn (cf. Exercise 4.35). There are other versions of the Dirac operator D from (7.7.11) that are ∂ more in line with the classical Cauchy–Riemann operator ∂z := 12 (∂1 + i∂2 ). n+1 n+1 set x = (x0 , x1 , x2 , . . . , xn ) ∈ R and consider Specifically, in R ±

D := ∂0 ± D = ∂0 ±

n

Mej ∂j ,

(7.7.22)

j=1

Note that in the case when n = 1 the Dirac operator D− corresponds to a ∂ . A reasoning similar to constant multiple of the Cauchy–Riemann operator ∂z that used above for D also yields fundamental solutions for D± . We leave this as an exercise for the interested reader. Exercise 7.43. Let Ω ⊆ Rn+1 be an open set and let x = (x0 , x1 , x2 , . . . , xn ) ∈ Rn+1 . Then the Dirac operators D± satisfy D+ D− = D− D+ = Δn+1 in D (Ω, Cn ),

(7.7.23)

where Δn+1 is the Laplacian in Rn+1 . Moreover, use a reasoning similar to the one used in the proof of Theorem 7.42 to show that the functions

1 E (x) : = ωn +

1 E (x) : = ωn −

x0 +

n

xj ej

j=1 |x|n+1

x0 −

n

∈ L1loc (Rn+1 , Cn ) → D (Rn+1 , Cn )

(7.7.24)

∈ L1loc (Rn+1 , Cn ) → D (Rn+1 , Cn )

(7.7.25)

xj ej

j=1 |x|n+1

are fundamental solution for the Dirac operators D− and D+ , respectively, in Rn+1 . We next introduce the Cauchy-Clifford operator and discuss its jump formulas in the upper and lower half-spaces. Theorem 7.44. For each Cn -valued function ϕ ∈ S(Rn−1 ) define the Cauchy– Clifford operator n−1 1 j=1 (xj − yj )ej + xn en ! en ! ϕ(y ) dy , (7.7.26) (C ϕ)(x) := − x − (y , 0)n ωn−1 Rn−1 for each x = (x1 , . . . , xn ) ∈ Rn with xn = 0. Then for each x ∈ Rn−1 one has


258

lim (C ϕ)(x , xn )

xn →0±

1 1 = ± ϕ(x ) + lim+ 2 ε→0 ωn−1

y ∈Rn−1

n−1

j=1 (yj |x −

− xj )ej y |n

! en ! ϕ(y ) dy .

|x −y |>ε

(7.7.27) Proof. Consider the Clifford algebra-value function Φ : Rn \ {0} → Cn given by n j=1 xj ej Φ(x) := − ! en for each x = (x1 , . . . , xn ) ∈ Rn \ {0}. (7.7.28) ωn−1 |x|n Then Φ is C ∞ , odd, and positive homogeneous of degree 1 − n in Rn \ {0}. As such, Corollary 4.78 may be applied (to each component of Φ). In this regard note from Corollary 4.62 that ⎡ ⎤ ⎤ ⎡ n n xj 1 ⎣ ξj 3 Φ(ξ) =− F e ⎦ ! en (7.7.29) (ξ) ej ⎦ ! en = i ⎣ 2 j ωn−1 j=1 |x|n |ξ| j=1 in S (Rn ). In particular, since en ! en = −1, we obtain 3 , 1) = i en ! en = −i. Φ(0

(7.7.30)

Given that, as seen from (7.7.26) and (7.7.28), we have (C ϕ)(x) = Φ(x − y , xn ) ! ϕ(y ) dy ,

(7.7.31)

Rn−1

for each x = (x1 , . . . , xn ) ∈ Rn with xn = 0, the jump formulas for the Cauchy– Clifford operator in (7.7.27) follow from (4.7.45) and (7.7.30). In parallel with the discussion in Remark 7.38, in the higher-dimensional setting we have the following connection between the Cauchy–Clifford operator and Riesz transforms. Remark 7.45. Let Rj , j ∈ {1, . . . , n − 1}, be the Riesz transforms in Rn−1 (i.e., singular integral operators defined as in (4.9.10) with n − 1 in place of n). Also, recall the definition of the “vertical limit” of a function defined in Rn+ to the boundary of the upper half-space from (4.8.20). Then we may express the version of the jump-formula (7.7.27) corresponding to considering the Cauchy– Clifford operator in the upper half-space as n−1 ver 1 1 ej ! en ! (Rj ϕ) Cϕ n = ϕ − 2 ωn−1 j=1 ∂R+

in

Rn−1

(7.7.32)

for each Cn -valued function ϕ ∈ S(Rn−1 ), where the Riesz transforms act on ϕ componentwise.


259

The format of the jump formula displayed in (7.7.32) suggests considering the operator acting on Cn -valued functions according to P :=

n−1 1 1 ej ! en ! Rj I− 2 ωn−1 j=1

1 1 I+ en ! ej ! Rj , 2 ωn−1 j=1 n−1

=

(7.7.33)

where I stands for the identity operator. In the second line of (7.7.33) the change in sign is due to the formula (cf. (7.7.2)) ej ! en = −en ! ej

for each j ∈ {1, . . . , n − 1}.

(7.7.34)

Theorem 4.93 then gives that P is a well-defined, linear, and bounded operator on L2 (Rn−1 , Cn ). Moreover, since each Rj has a real-valued kernel, its action commutes with multiplication by elements from Cn (i.e., for each a ∈ Cn we have Rj Ma = Ma Rj , in the notation from Exercise 7.39). Keeping this in mind and relying on (7.7.34) and the fact that e2n = −1 (cf. (7.7.2)), we may then write P2 =

1 2

I+

1 ωn−1

en !

n−1

ej ! Rj

2

j=1

1 2 2 1 1 I+ en ! ej ! Rj + ej ! Rj . 4 ωn−1 ωn−1 j=1 j=1 n−1

=

n−1

(7.7.35)

Furthermore, using the fact that, as proved in Theorem 4.93, the Riesz transn−1 2 2 forms commute with one another and satisfy Rj = − ωn−1 I on L2 (Rn−1 ), 2 j=1

we may expand n−1

ej ! Rj

2 =

j=1

n−1

ej ! ek ! Rj Rk

j,k=1

=

n−1

e2j Rj2 +

j=1

=−

n−1 j=1

ej ! ek ! Rj Rk

1≤j=k≤n−1

Rj2 =

ωn−1 2 2

I,

(7.7.36)

where the source of the cancellation taking place in the third equality above is the observation that ej ! ek ! Rj Rk = −ek ! ej ! Rk Rj whenever we have that 1 ≤ j = k ≤ n − 1. Combining (7.7.35)–(7.7.36) and recalling (7.7.33) then yields (7.7.37) P 2 = P on L2 (Rn−1 , Cn ).


260

Note that this is in agreement with the result obtained in (7.6.35) in the case of the two-dimensional setting. Let us also consider the higher-dimensional analogue of (7.6.36). In this regard, we first observe that based on Exercise 7.39 for any f, g ∈ L2 (Rn−1 , Cn ) we may write Rn−1

n−1 Mej (Rj f )(x ) , g(x ) dx Men j=1

n−1

=

Rn−1 j=1

=−

(Rj f )(x ) , Mej Men g(x ) dx

n−1

Rn−1 j=1

= Rn−1

f (x ) , Mej Men (Rj g)(x ) dx

n−1 Mej (Rj g)(x ) dx . f (x ) , Men

(7.7.38)

j=1

From this and (7.7.33) it follows that for every f, g ∈ L2 (Rn−1 , Cn ) we have (7.7.39) (P f )(x ), g(x ) dx = f (x ), (P g)(x ) dx , Rn−1

Rn−1

a condition that we shall interpret simply as P∗ = P

on L2 (Rn−1 , Cn ).

(7.7.40)

In summary, the above analysis shows that the operator P defined as in (7.7.33) is a projection on L2 (Rn−1 , Cn ). Starting from this result, a corresponding higher-dimensional Hardy space theory may be developed in the Clifford-algebra setting as well.

7.8

Fundamental Solutions for General Second-Order Operators

Consider a constant, complex coefficient, homogeneous, second-order differential n ajk ∂j ∂k in Rn . In the first stage, our goal is to find necessary operator L = j,k=1

and sufficient conditions, that can be expressed without reference to the theory of distributions, guaranteeing that a function E ∈ L1loc (Rn ) is a fundamental solution for L. Several necessary conditions readily present themselves. First, it is clear that LE is the zero distribution in Rn \ {0}. In addition, if L is elliptic then this necessarily implies that E ∈ C ∞ (Rn \ {0}). In the absence of ellipticity as a hypothesis for L, we may wish to assume that E is reasonably regular, say E ∈ C 2 (Rn \ {0}). Second, if the fundamental solution E is a priori

7.8. FUNDAMENTAL SOLUTIONS FOR GENERAL...

261

3 known to be a tempered distribution, then −L(ξ)E(ξ) = 1 in S (Rn ). If L is elliptic and n ≥ 3, this forces E = E0 + P where E0 := −F −1 L(ξ)−1 is homogeneous of degree 2 − n and P is a polynomial that is annihilated by L. Hence, working with E0 in place of E, there is no loss of generality in assuming that E is homogeneous of degree 2 − n. The case n = 2 may also be included in this discussion by demanding that ∇E is homogeneous of degree 1 − n. In summary, it is reasonable to restrict our search for a fundamental solution for L in the class of functions satisfying E ∈ C 2 (Rn \ {0}) ∩ L1loc (Rn ) with the property that ∇E is positive homogeneous of degree 1 − n in Rn \ {0}. However, these conditions do not rule out such trivial candidates as the zero distribution. In Theorem 7.46 we identify the key nondegeneracy property (7.8.2), guaranteeing that E is in fact a fundamental solution. Theorem 7.46 is then later used to find an explicit formula for such a fundamental solution, under a strong ellipticity assumption on L (cf. Theorem 7.54). Theorem 7.46. Assume that n ≥ 2 and consider L=

n

ajk ∂j ∂k ,

ajk ∈ C.

(7.8.1)

j,k=1

Then for a function E ∈ C 2 (Rn \ {0}) ∩ L1loc (Rn ) with the property that ∇E is positive homogeneous of degree 1 − n in Rn \ {0}, the following statements are equivalent: (1) When viewed in L1loc (Rn ), the function E is a fundamental solution for L in Rn ; (2) One has LE = 0 pointwise in Rn \ {0} and n j,k=1

S n−1

ajk ωj ∂k E(ω) dσ(ω) = 1.

(7.8.2)

Remark 7.47.

(i) In the partial differential equation parlance, the integrand n in (7.8.2), that is, the expression ajk ωj ∂k E(ω), is referred to as the j,k=1

conormal derivative of E on S n−1 . (ii) One remarkable aspect of Theorem 7.46 is that the description of a fundamental solution from part (2) is purely in terms of ordinary calculus (i.e., without any reference to the theory of distributions). Proof of Theorem 7.46. Let E ∈ C 2 (Rn \ {0})∩L1loc(Rn ) with the property that ∇E is positive homogeneous of degree 1 − n in Rn \ {0}. Exercise 4.51 then implies that E ∈ L1loc (Rn ). Fix an arbitrary f ∈ C0∞ (Rn ). Then making use of (4.4.19) and Proposition 4.67 (applied to each ∂k E) we obtain

262


(LE) ∗ f =

n

ajk ∂j (∂k E) ∗ f

j,k=1 n

ajk

S n−1

j,k=1

n

=

S n−1 j,k=1

n ∂k E(ω)ωj dσ(ω) (δ ∗ f ) + ajk P.V.(∂j ∂k E) ∗ f j,k=1

ajk ∂k E(ω)ωj dσ(ω) f

+ lim+ ε→0

n

|y−· |>ε j,k=1

ajk (∂j ∂k E)(· − y)f (y) dy

(7.8.3)

in D (Rn ),

where we have also used the fact that δ ∗ f = f . Having proved this, we now turn in earnest to the proof of the equivalence in the statement of the theorem. First, assume that E is solution for L in Rn . Then LE= δ in a fundamental n D (R ) implies that L E Rn \{0} = 0 in D (Rn ). Since by assumption E Rn \{0} belongs to C 2 (Rn \ {0}), we arrive at the conclusion that LE = 0 pointwise in Rn \ {0}. Explicitly, n

∀ x ∈ Rn \ {0}.

ajk ∂j ∂k E(x) = 0,

(7.8.4)

j,k=1

This proves the first claim in (2). Next, for each function f ∈ C0∞ (Rn ) we may write f = δ ∗ f = (LE) ∗ f in D (Rn ) which, in light of (7.8.3) and (7.8.4), forces f=

n

S n−1 j,k=1

ajk ∂k E(ω)ωj dσ(ω) f.

(7.8.5)

Since f ∈ C0∞ (Rn ) was arbitrary, (7.8.2) follows. This finishes the proof of (1 ) ⇒ (2 ). Conversely, suppose that E ∈ C 2 (Rn \ {0}) ∩ L1loc (Rn ) with the property that ∇E is positive homogeneous of degree 1 − n in Rn \ {0}, such that LE = 0 pointwise in Rn \ {0}, and (7.8.2) holds. Then for each f ∈ C0∞ (Rn ) formula (7.8.3) simply reduces to (LE) ∗ f = f . Now Exercise 2.88 may be invoked to conclude that LE = δ in D (Rn ), as wanted. Next, we turn to the task of finding all fundamental solutions that are tempered distributions for general homogeneous, second-order, constant coefficient operators that are strongly elliptic. We begin by defining this stronger (than originally introduced in Definition 6.13) notion of ellipticity. Let A = ajk 1≤j,k≤n ∈ Mn×n (C) and associated to such a matrix A, consider the operator LA := LA (∂) :=

n j,k=1

ajk ∂j ∂k .

(7.8.6)


263

This is a homogeneous, second-order, constant coefficient operator for which the ellipticity condition reads n

LA (ξ) :=

ajk ξj ξk = 0,

∀ ξ = (ξ1 , . . . , ξn ) ∈ Rn \ {0}.

(7.8.7)

j,k=1

As a trivial consequence of the Malgrange–Ehrenpreis theorem (cf. Theorem 5.9), any elliptic operator has a fundamental solution. The goal is to obtain explicit formulas for such fundamental solutions for a subclass of homogeneous, elliptic, second-order, constant coefficient operators satisfying a stronger condition than (7.8.7). Definition 7.48. Call an operator LA as in (7.8.6) strongly elliptic, if there exists a constant C ∈ (0, ∞) such that Re

n

ajk ξj ξk ≥ C|ξ|2 ,

∀ ξ = (ξ1 , . . . , ξn ) ∈ Rn .

(7.8.8)

j,k=1

By extension, call a matrix A = ajk 1≤j,k≤n ∈ Mn×n (C) strongly elliptic provided there exists some C ∈ (0, ∞) with the property that (7.8.8) holds. Remark 7.49. (1) It is obvious that any operator LA as in (7.8.6) that is strongly elliptic is elliptic. (2) Up to changing L to −L, any elliptic, homogeneous, second-order, constant coefficient differential operator L with real coefficients is strongly elliptic. To see why this is the case let A ∈ Mn×n (R) and suppose that LA is ellipn ajk ξj ξk tic. Consider the function f : S n−1 → R defined by f (ξ) := j,k=1

for ξ ∈ S n−1 . Then f is continuous and since LA is elliptic the number 0 is not in the image of f . The unit sphere S n−1 being compact and connected, it is mapped by f in a compact, connected, subset of R, not containing 0. This forces the image of f to be a compact interval that does not contain 0. Hence, there exists c ∈ (0, ∞) with the property that either f (ξ) ≥ c for every ξ ∈ S n−1 or −f (ξ) ≥ c for every ξ ∈ S n−1 . This implies that either f (ξ/|ξ|) ≥ c for every ξ ∈ Rn \ {0} or −f (ξ/|ξ|) ≥ c for every ξ ∈ Rn \ {0}, or equivalently, that either LA or −LA is strongly elliptic. (3) Consider the operator L = ∂12 + i∂22 in R2 . Then L is a homogeneous, second-order, constant coefficient differential operator with L(ξ) = ξ12 +iξ22 , ξ = (ξ1 , ξ2 ) ∈ R2 . Clearly L(ξ) = 0 if ξ = 0, so L is elliptic. However, L is not strongly elliptic since Re [L(ξ)] = ξ12 , which cannot be bounded from below by a constant multiple of |ξ|2 since the latter blows up if |ξ2 | → ∞.


264

Fix A = ajk 1≤j,k≤n ∈ Mn×n (C) and consider the operator LA as in (7.8.6). Due to the symmetry of mixed partial derivatives in the sense of distributions, it is immediate that LA = LAsym ,

where Asym :=

A + A . 2

(7.8.9)

As such, any fundamental solution for LAsym is also a fundamental solution for LA . Also, since (Asym ξ) · ξ = (Aξ) · ξ for each ξ ∈ Rn , we have that LA

is strongly elliptic if and only if

LAsym

is strongly elliptic.

(7.8.10)

Consequently, when computing the fundamental solution for LA we may assume without loss of generality that A is symmetric, that is, A = A . For further reference we summarize a few basic properties of symmetric matrices (throughout, the symbol dot denotes the real inner product of vectors with complex components). A ∈ Mn×n (R), A = A =⇒ (Aζ) · ζ ∈ R,

∀ ζ ∈ Cn ,

(7.8.11)

A ∈ Mn×n (C), A = A =⇒ Re A = (Re A) and Im A = (Im A) , (7.8.12) A ∈ Mn×n (C), A = A =⇒ Re (Aζ) · ζ = (Re A)ζ · ζ, ∀ ζ ∈ Cn . (7.8.13) It is easy to see that (7.8.11)–(7.8.12) hold, while (7.8.13) follows from (7.8.11)– (7.8.12) after writing A = Re A + i Im A. Also, recall that a matrix A ∈ Mn×n (C) is said to be positive definite provided (Aζ) · ζ is real and strictly positive for each ζ ∈ Cn \ {0}.

(7.8.14) It is easy to see that any positive definite matrix A ∈ Mn×n (C) satisfies A = A, and there exists c ∈ (0, ∞) such that (Aζ) · ζ ≥ c|ζ|2 ,

∀ ζ ∈ Cn .

(7.8.15)

Remark 7.50. Fix a matrix A ∈ Mn×n (C) that is symmetric and satisfies (7.8.8). Then, for each ζ ∈ Cn we have (with C ∈ (0, ∞) as in (7.8.8)) Re [(Aζ) · ζ ] = Re A(Re ζ + i Im ζ) · (Re ζ − i Im ζ) = Re (ARe ζ) · Re ζ + (A Im ζ) · Im ζ ≥ C|Re ζ|2 + C|Im ζ|2 = C|ζ|2 .

(7.8.16)

The second equality in (7.8.16) uses the fact that A is symmetric, while (7.8.8) is used for the inequality in (7.8.16). Thus, combining (7.8.16) with the Cauchy– Schwarz inequality, we obtain |Aζ| ≥ C|ζ|

for every

ζ ∈ Cn ,

(7.8.17)


265

proving that the linear map A : Cn → Cn is injective, thus invertible. In particular, detA = 0. Also, thanks to (7.8.13) and (7.8.16) we have that Re A is a positive definite matrix. More precisely, with C as in (7.8.8), we have ∀ ζ ∈ Cn . (7.8.18) (Re A)ζ · ζ ≥ C|ζ|2 From Remark 7.50 and definitions we see that, given any A ∈ Mn×n (C), the following implications hold: Re A positive definite =⇒ A strongly elliptic, A strongly elliptic ⇐⇒ Asym strongly elliptic, A strongly elliptic and symmetric =⇒ Re A positive definite.

(7.8.19) (7.8.20) (7.8.21)

Remark 7.51. Assume that A ∈ Mn×n (C) is symmetric and satisfies (7.8.8). From Remark 7.50 it follows that A is invertible. Moreover, if we define A := sup |Aζ| : ζ ∈ Cn , |ζ| = 1 , (7.8.22) then (7.8.17) ensures that A > 0. We claim that Re (A−1 ζ) · ζ ≥

C |ζ|2 A2

∀ ζ ∈ Cn ,

(7.8.23)

where C is as in (7.8.8). To justify this, first note that |ζ|2 = |AA−1 ζ|2 ≤ A2 |A−1 ζ|2 for every ζ ∈ Cn . In turn, this and (7.8.16) permit us to estimate Re [(A−1 ζ) · ζ ] =Re [ (A−1 ζ) · ζ] = Re (A−1 ζ) · (AA−1 ζ) ≥ C|A−1 ζ|2 ≥

C |ζ|2 A2

∀ ζ ∈ Cn .

(7.8.24)

This proves (7.8.23). In particular, (7.8.23) yields A−1 |ξ|2 ≥ |(A−1 ξ) · ξ| ≥

C |ξ|2 A2

∀ ξ ∈ Rn .

Remark 7.52. Consider the set M := A ∈ Mn×n (C) : A = A , Re A is positive definite .

(7.8.25)

(7.8.26)

Since the n × n symmetric matrices A = (ajk )1≤j,k≤n with complex entries are uniquely determined by the elements ajk with 1 ≤ j ≤ k ≤ n, we may naturally identify M with an open convex subset of Cn(n+1)/2 . Throughout, this identification is implicitly assumed. Moreover, every A ∈ M satisfies det A = 0, since if Aζ = 0 for some ζ ∈ Cn , then (7.8.13) and the fact that Re A is positive definite force ζ = 0. The fact that M is convex, implies that there is a unique 1 1 analytic branch of the mapping M A → (detA) 2 ∈ C such that (detA) 2 > 0 1 when A is real. Thus (detA) 2 is unambiguously defined for A ∈ M.


266

To proceed with the discussion regarding determining a fundamental solution for a strongly elliptic operator LA we first analyze the case when A has real entries. The case when A is real, symmetric and satisfies (7.8.8). Since the matrix A is real, symmetric and positive definite, A is diagonalizable with A = U −1 DU for some orthogonal matrix U ∈ Mn×n (R) and some diagonal n × n matrix D where the entries on the main diagonal are strictly positive real 1 numbers. Hence, D 2 is meaningfully defined as the n × n diagonal matrix with the entries on the diagonal being equal to the square roots of the entries on the 1 1 main diagonal in D. In addition, A 2 √ := U −1 D 2 U is well-defined, symmetric, 1 1 1 invertible, A 2 A 2 = A and det(A 2 ) = detA. Next, fix a function u ∈ L1loc (Rn ) such that

∂k u ∈ L1loc (Rn )

for each k ∈ {1, . . . , n}, (7.8.27)

where the derivatives are taken in the sense of distributions. Also, consider ϕ ∈ C0∞ (Rn ). Then, using the fact that A is symmetric, we obtain LA u, ϕ =

n

n $ $ % % ajk ∂j ∂k u, ϕ = − ajk ∂k u, ∂j ϕ

j,k=1

=−

Rn

j,k=1

∇u(x) · A∇ϕ(x) dx.

(7.8.28) 1

In the integral in (7.8.28) we make the change of variables x = A 2 y. Since for every invertible matrix B ∈ Mn×n (R) and any function f the chain rule gives (∇f )(By) = (B )−1 ∇[f (By)] for each y ∈ Rn

(7.8.29)

with the property that f is differentiable at By, we obtain 1 (∇u)(x) · A(∇ϕ)(x) dx = |detA 2 | Rn

=

√ detA

Rn

Rn

1 1 (∇u)(A 2 y) · A(∇ϕ)(A 2 y) dy

1 ∇ u(A 2 y)

1 1 1 · (A 2 )−1 A(A 2 )−1 ∇ ϕ(A 2 y) dy √ = detA ∇v(y) · ∇ψ(y) dy Rn √ = − detA v(y)Δψ(y) dy,

(7.8.30)

Rn

where we have set 1

v := u ◦ A 2

1

and ψ := ϕ ◦ A 2 ∈ C0∞ (Rn ),

(7.8.31)


267

and for the last equality in (7.8.30) we integrated by parts one more time. Hence, (7.8.28) and (7.8.30) imply √ LA u, ϕ = detA v(y)Δψ(y) dy. (7.8.32) Rn

In particular, if we now choose 1 1 u(y) := √ EΔ ◦ A− 2 (y) for y ∈ Rn \ {0}, detA

(7.8.33)

where EΔ is the fundamental solution for the Laplacian from (7.1.12), then u satisfies the conditions listed in (7.8.27), and the function v from (7.8.31) becomes 1 EΔ (y) for y ∈ Rn \ {0}. (7.8.34) v(y) = √ detA 1

Since A 2 (0) = 0, we may write 1 ϕ(0) = ϕ ◦ A 2 (0) = ψ(0) = δ, ψ = ΔEΔ , ψ = EΔ , Δψ √ √ 1 √ EΔ (y)Δψ(y) dy = detA = detA v(y)Δψ(y) dy detA Rn Rn % 1 1 $ LA EΔ ◦ A− 2 , ϕ , (7.8.35) = LA u, ϕ = √ detA where for the second to the last equality in (7.8.35) we used (7.8.32). Since (7.8.35) holds for every ϕ ∈ C0∞ (Rn ), we may conclude that √

1 1 EΔ ◦ A− 2 detA

is a fundamental solution for LA in Rn .

(7.8.36) 1

Denoting by EA and keeping in mind that |A− 2 x|2 = − 1 this− fundamental −1 solution 1 A 2 x · A 2 x = A x · x for every x ∈ Rn , we obtain that the function defined by ⎧ 1 −1 ⎪ √ · if n ≥ 3, ⎪ n−2 ⎪ −1 ⎨ (n − 2)ωn−1 detA [(A x) · x] 2 (7.8.37) EA (x) := ⎪ ⎪ ⎪ ⎩ √1 ln (A−1 x) · x if n = 2, 4π detA for every x ∈ Rn \ {0} is a fundamental solution for LA in the current case. Moreover, as is apparent from (7.8.37), (7.8.25), and Exercise 4.5, the function EA is locally integrable and a tempered distribution in Rn . In preparation for dealing with the case of matrices with complex entries, we state and prove the following useful complex analysis result.

268


Lemma 7.53. Let N ∈ N and assume O is an open and convex subset of CN with the property that O ∩RN = ∅ (where RN is canonically embedded into CN ). Also, suppose f, g : O → C are two functions that are separately holomorphic (i.e., in each scalar complex component in CN ) such that (7.8.38) f O∩RN = g O∩RN . Then f = g in O. Proof. Fix an arbitrary point (x1 , . . . , xN ) ∈ O ∩ RN and consider O1 := z1 ∈ C : (z1 , x2 , . . . , xN ) ∈ O .

(7.8.39)

Then O1 is an open convex subset of C, which contains x1 , hence O1 ∩ R = ∅. Define the functions f1 , g1 : O1 → C by f1 (z) := f (z, x2 , . . . , xN ) and g1 (z) := g(z, x2 , . . . , xN ),

∀ z ∈ O1 . (7.8.40)

Then f1 and g1 are holomorphic functions in O1 that coincide on O1 ∩ R. Since the latter contains an accumulation point in the convex (hence connected) set O1 , it follows that f1 = g1 on O1 by the coincidence theorem for holomorphic functions of one complex variable. Since (x1 , . . . , xN ) ∈ O ∩ RN was arbitrary, we may conclude that (7.8.41) f O∩(C×RN −1) = g O∩(C×RN −1 ) . Next, fix (z1 , x2 , . . . , xN ) ∈ O ∩ (C × RN −1 ) and define O2 := z2 ∈ C : (z1 , z2 , x3 , . . . , xN ) ∈ O .

(7.8.42)

Once again, O2 is an open convex subset of C, which contains x2 , hence O2 ∩R = ∅. If we now define the functions f2 , g2 : O2 → C by ∀ z ∈ O2 , (7.8.43) it follows that f2 , g2 are holomorphic in O2 , which, by (7.8.41), coincide on O2 ∩ R. Given that the latter set contains an accumulation point in the convex set O1 , we deduce that f2 = g2 on O2 by once again invoking the coincidence theorem for holomorphic functions of one complex variable. Upon recalling that (z1 , x2 , . . . , xN ) ∈ O ∩ (C × RN −1 ) was arbitrary, we conclude that (7.8.44) f O∩(C2 ×RN −2 ) = g O∩(C2 ×RN −2 ) . f2 (z) := f (z1 , z, x3 , . . . , xN ) and g2 (z) := g(z1 , z, x3 , . . . , xN ),

Continuing this process inductively, we arrive at the conclusion that f = g in O. After this preamble, we are ready to consider the general case. The case when A has complex entries, is symmetric and satisfies (7.8.8). As observed in Remark 7.50, under the current assumptions, A continues to be


269

invertible. Also, (7.8.25) holds. In addition, under the current assumptions we 1 have that A ∈ M and (detA) 2 is unambiguously defined (see in Remark 7.52). These comments show that the function EA from (7.8.37) continues to be well-defined under the current assumption on A if ln is replaced by the principal branch of the complex log (defined for points z ∈ C\(−∞, 0] so that z a = ea log z for each a ∈ R). In addition, EA continues to belong to L1loc (Rn ) and we have EA ∈ C ∞ (Rn \ {0}). Furthermore, from (7.8.25) and Exercise 4.5 it follows that the function EA continues to be a tempered distribution in Rn . The goal is to prove that this expression is a fundamental solution for LA in the current case. First, observe that since A−1 is symmetric, for each j, k ∈ {1, . . . , n} we have ∂k A−1 x · x = 2(A−1 x)k and ∂j A−1 x k = (A−1 )kj . (7.8.45) Hence, for every x ∈ Rn \ {0}, differentiating pointwise we obtain n 2−n −n LA A−1 x · x 2 = ajk ∂j (2 − n) A−1 x · x 2 (A−1 x)k j,k=1

= −n(2 − n)

n

−n−2 ajk A−1 x · x 2 (A−1 x)j (A−1 x)k

j,k=1

+ (2 − n)

n

−n ajk A−1 x · x 2 (A−1 )kj = 0. (7.8.46)

j,k=1

Similarly, LA log (A−1 x) · x = 0 for x ∈ R2 \ {0}. Thus, we may conclude that ∀ x ∈ Rn \ {0}.

LA EA (x) = 0

(7.8.47)

Second, by making use of (7.8.45) and the expression for EA we obtain ∇EA (x) =

A−1 x 1 √ · n ωn−1 detA [(A−1 x) · x] 2

∀ x ∈ Rn \ {0}

(7.8.48)

which, in particular, shows that ∇EA is positive homogeneous of degree 1 − n in Rn \ {0}. Furthermore, for each x ∈ S n−1 we have n n

ajk xj ∂k EA (x) =

j,k=1

= =

1 √

ωn−1 detA

ωn−1

1 √ 1 √

·

ajk xj (A−1 x)k

j,k=1 n

[(A−1 x) · x] 2

(A x) · (A−1 x) n detA [(A−1 x) · x] 2

ωn−1 detA

· ·

1 [(A−1 x)

where the last equality uses the fact that |x| = 1.

n

· x] 2

,

(7.8.49)

270


Invoking Theorem 7.46 we conclude that EA is a fundamental solution for LA in Rn if and only if √ dσ(x) ωn−1 detA = . (7.8.50) −1 x) · x] n 2 n−1 [(A S The fact that we already know that EA is a fundamental solution for LA in Rn in the case when A ∈ Mn×n (R) satisfies A = A and condition (7.8.8), implies that formula (7.8.50) holds for this class of matrices. We make the claim that in fact (7.8.50) actually holds for the larger class of matrices A ∈ Mn×n (C) satisfying A = A and condition (7.8.8). To see why this is true, recall the open subset M of Cn(n+1)/2 (7.8.26) and consider from the functions f, g : M → C defined for every A = ajk 1≤j,k≤n ∈ M by √ (7.8.51) f (ajk )1≤j≤k≤n := ωn−1 detA, dσ(x) (7.8.52) g (ajk )1≤j≤k≤n := n . −1 x) · x] 2 S n−1 [(A Then f and g are analytic (as functions of several complex variables) on M, which is an open convex set in Cn(n+1)/2 . If A ∈ M has real entries, then A satisfies (7.8.8), and (7.8.50) holds for such A, hence f = g on M ∩ Mn×n (R). Invoking Lemma 7.53 we may therefore conclude that f = g on M. Thus, (7.8.50) holds for every A ∈ Mn×n (C) satisfying A = A and condition (7.8.8). Finally, we note that thanks to Proposition 5.7 and the current strong ellipticity assumption, any other fundamental solution of LA belonging to S (Rn ) differs from EA by a polynomial that LA annihilates. In summary, the above analysis proves the following result. Theorem 7.54. Suppose A = ajk 1≤j,k≤n ∈ Mn×n (C) and consider the operator LA associated to A as in (7.8.6). If LA is strongly elliptic, then the function defined by ⎧ 1 1 ⎪ ⎪ − · if n ≥ 3, n−2 ⎪ ⎪ −1 (n − 2)ω detA ⎪ x) · x] 2 n−1 sym [((Asym ) ⎨ EA (x) := ⎪ ⎪ ⎪ 1 ⎪ ⎪ log ((Asym )−1 x) · x if n = 2, ⎩ 4π detAsym (7.8.53) for x ∈ Rn \{0} belongs to L1loc (Rn )∩S (Rn )∩C ∞ (Rn \{0}) and is a fundamental solution for LA in Rn . Above, Asym := 12 (A + A ), detAsym is defined as in Remark 7.52, and log denotes the principal branch of the complex logarithm (defined for complex numbers z ∈ C \ (−∞, 0] so that z a = ea log z for each a ∈ R). Moreover, (7.8.54) u ∈ S (Rn ) : LA u = δ in S (Rn ) = EA + P : P polynomial such that LA P = 0 in Rn .

7.9. LAYER POTENTIAL REPRESENTATION FORMULAS...

271

We conclude this section with a couple of related exercises about fundamental solutions for second-order, constant coefficient, differential operators. n Exercise 7.55. Let n ≥ 2 and consider a differential operator L = ajk ∂j ∂k j,k=1

with complex coefficients. Assume that E ∈ C 2 (Rn \{0})∩L1loc(Rn ) is a function with the property that ∇E is positive homogeneous of degree 1 − n in Rn \ {0}. In addition, suppose that n ajk ωj ∂k E(ω) dσ(ω) = 0. (7.8.55) λ := j,k=1

S n−1

Prove that λ−1 E is a fundamental solution for L in Rn . Use this result to find the proper normalization for the standard fundamental solution for the Laplacian in Rn , starting with E(x) := |x|2−n when n ≥ 3, and with E(x) := ln |x| for n = 2. Exercise 7.56. Let n ≥ 2 and suppose that E ∈ C 2 (Rn \ {0}) ∩ L1loc (Rn ) has the property that ∇E is odd and positive homogeneous of degree 1 − n in Rn \ {0}. In addition, assume that the function E is a fundamental solution for n ajk ∂j ∂k in Rn . the complex constant coefficient differential operator L = j,k=1

Prove that for every ξ ∈ S n−1 one has n 1 ajk ωj ∂k E(ω) dσ(ω) = . 2 ω∈S n−1

(7.8.56)

j,k=1

ω·ξ>0

7.9

Layer Potential Representation Formulas Revisited

The goal here is to derive a layer potential representation formula generalizing the identity from Proposition 7.17 for the Laplacian. We begin by describing the setting in which we intend to work. Given a matrix A = ajk 1≤j,k≤n ∈ Mn×n (C), we associate the homogeneous second-order differential operator LA =

n

ajk ∂j ∂k

in Rn ,

(7.9.1)

j,k=1

and for every unit vector ν = (ν1 , . . . , νn ) and any complex-valued function u of class C 1 , define the conormal derivative of u associated with the matrix A (along ν) as n νj ajk ∂k u. (7.9.2) ∂νA u := j,k=1

272


Theorem 7.57. Suppose n ≥ 2 and let Ω ⊂ Rn be a bounded domain of class ν and surface measure σ. In addition, assume C 1 , with outward unit normal that the matrix A = ajk 1≤j,k≤n ∈ Mn×n (C) is such that the operator LA associated with A as in (7.9.1) is strongly elliptic, and recall the fundamental solution EA for LA defined in (7.8.53). Then for every complex-valued function u ∈ C 2 (Ω ) one has (LA u)(y)EA (x − y) dy − EA (x − y)(∂νA u)(y) dσ(y) Ω

−

∂Ω

u(y)(∂νA E)(x − y) dσ(y) u(x), x ∈ Ω, = 0, x ∈ Rn \ Ω, ∂Ω

(7.9.3)

where is the conormal derivative associated with the matrix A (along ν). Proof. When x ∈ Rn \ Ω, it clear from (7.8.53) that EA (x − ·) ∈ C ∞ (Ω ). Also, since EA is a fundamental solution for LA in Rn we have that LA [EA (x − ·)] = 0 in Rn \ {x}, hence (LA EA )(x − ·) = 0 in Ω. Based on these and repeated use of (13.7.4) we may then write Ω

(LA u)(y)EA (x−y) dy =

n Ω

j,k=1

=

n Ω

j,k=1

ajk (∂k u)(y)(∂j EA )(x − y) dy

n

+

ajk (∂j ∂k u)(y)EA (x − y) dy

j,k=1

= Ω

∂Ω

ajk νj (y)(∂k u)(y)EA (x−y) dσ(y)

u(y)(LA EA )(x−y) dy+

+ ∂Ω

∂Ω

u(y)(∂νA EA )(x − y) dσ(y)

EA (x−y)(∂νA u)(y) dσ(y).

(7.9.4)

Upon recalling that (LA EA )(x − ·) = 0 in Ω, the last solid integral drops out and the resulting identity is in agreement with (7.9.3). Consider now the case when x ∈ Ω. Since Ω is open, there exists r > 0 such that B(x, r) ⊆ Ω. For each ε ∈ (0, r) define Ωε := Ω \ B(x, ε) which is a bounded domain of class C 1 . Since EA (x − ·) ∈ C ∞ (Ωε ) and (LA EA )(x − ·) = 0 in Ωε , the same type of reasoning as above gives (keeping in mind that ∂Ωε = ∂Ω ∪ ∂B(x, ε)) Ωε

(LA u)(y)EA (x−y) dy =

∂Ω

u(y)(∂νA EA )(x−y) dσ(y)

−

∂B(x,ε)

u(y)(∂νA EA )(x − y) dσ(y)

7.9. LAYER POTENTIAL REPRESENTATION FORMULAS...

+ ∂Ω

EA (x−y)(∂νA u)(y) dσ(y)

−

273

∂B(x,ε)

EA (x−y)(∂νA u)(y) dσ(y) =: I+II+III+IV. (7.9.5)

As seen from (7.8.53), we have |IV | ≤ CA ∇uL∞ (Ω) ε max 1, ln |ε| , from which we deduce that lim+ IV = 0. Next, split II = II + u(x)II where ε→0

II := −

[u(y) − u(x)](∂νA EA )(x − y) dσ(y),

(7.9.6)

∂B(x,ε)

and observe that II := −

(∂νA EA )(x − y) dσ(y) ∂B(x,ε)

n

=−

∂B(x,ε) j,k=1

=

n

S n−1 j,k=1

= S n−1

n

ajk

yk − xk (∂j EA )(x − y) dσ(y) ε

ajk ωk (∂j EA )(ω) dσ(ω)

(A )jk ωj (∂k EA )(ω) dσ(ω) = 1.

(7.9.7)

j,k=1

Above, the first equality defines II , the second equality uses the definition of the conormal derivative and the outward unit normal to the ball, the third equality is based on the change of variables ω = (x − y)/ε and the fact that ∇EA is positive homogeneous of degree 1 − n. Finally, in the fourth equality we have interchanged j and k in the summation and used the identities EA = EA , LA = LA , while the last equality is due to (7.8.2). In addition, |II | ≤ CA ∇uL∞ (Ω) ε, hence lim II = 0, and ε→0+

(LA u)(y)EA (x − y) dy =

lim

ε→0+

Ωε

(LA u)(y)EA (x − y) dy Ω

by Lebesgue’s dominated convergence theorem (recall that EA (x − ·) ∈ L1 (Ω) since Ω is bounded). Collectively, the results deduced in the above analysis yield (7.9.3) in the case when x ∈ Ω, finishing the proof of the theorem.

Further Notes for Chap. 7. As evidenced by the treatment of the Poisson problem for the Laplacian and bi-Laplacian (from Sects. 7.2 and 7.4, respectively), fundamental solutions play a key role both for establishing integral representation formulas and for deriving estimates for the solution. This type of application to partial differential

274


equations amply substantiate the utility of the tools from distribution theory and harmonic analysis derived in Sect. 4.10 (dealing with derivatives of volume potentials) and Sect. 4.9 (dealing with singular integral operators). The aforementioned Poisson problems serve as a prototype for other types of boundary value problems formulated for other differential operators and with the entire space Rn replaced by an open set Ω ⊂ Rn . In the latter scenario one specifies boundary conditions on ∂Ω in place of ∞ as in the case of Rn (note that ∞ plays the role of the topological boundary of Rn regarded as an open subset of its compactification Rn ∪ {∞}). In Sect. 7.7 the Dirac operator has been considered in the natural setting of Clifford algebras. For more information pertaining to this topic, the interested reader is referred to the monographs [4, 22, 49]. The last two references also contain a discussion of Hardy spaces in the context of Clifford algebras (a topic touched upon in Sect. 7.7). A classical reference to Hardy spaces in the ordinary context of C (which appeared at the end of Sect. 7.6) is the book [27].

7.10


Exercise 7.58. Prove that there exists a unique E ∈ S (Rn ) with the property that ΔE − E = δ in S (Rn ). Exercise 7.59. Does there exist E ∈ L1 (Rn ) such that ΔE = δ in D (Rn )? Exercise 7.60. Prove that for every u ∈ D (Ω) we have div(∇u) = Δu

D (Ω).

in

(7.10.1)

Exercise 7.61. Prove that if f ∈ C ∞ (Ω) and u ∈ D (Ω), then Δ(f u) = (Δf )u + 2(∇f ) · (∇u) + f Δu

in

D (Ω).

(7.10.2)

Exercise 7.62. Suppose n ≥ 2 and denote by EΔ the fundamental solution for the Laplacian operator Δ given in (7.1.12). Without making use of Corollary 4.62, prove that for each j ∈ {1, . . . , n} one has F (∂j EΔ ) = −i

ξj |ξ|2

In turn, use (7.10.3) to show that xj ξj F = −iωn−1 2 |x|n |ξ| and F −1

ξj |ξ|2

=

S (Rn ).

in

in

i xj · ωn−1 |x|n

in

S (Rn ),

S (Rn ).

(7.10.3)

(7.10.4)

(7.10.5)

Exercise 7.63. Suppose n ≥ 3 and denote by EΔ2 the fundamental solution for the bi-Laplacian operator Δ2 given in (7.3.8). Prove that for each j, k ∈ {1, . . . , n} one has F (∂j ∂k EΔ2 ) = −

ξj ξk |ξ|4

in

S (Rn ).

(7.10.6)


275

Consequently, δjk 1 xj xk ξj ξk 1 −1 · − · F = |ξ|4 2(n − 2)ωn−1 |x|n−2 2ωn−1 |x|n

and F

xj xk |x|n

= ωn−1

δjk ξj ξk − 2ωn−1 4 |ξ|2 |ξ|

in

in

S (Rn ).

S (Rn ), (7.10.7)

(7.10.8)

Exercise 7.64. Let P (D) be a nonzero linear constant coefficient operator of order m ∈ N0 . Prove that P (D) is elliptic if and only if there exist two constants C, R ∈ (0, ∞) such that |P (ξ)| ≥ C|ξ|m for every ξ ∈ Rn \ B(0, R). Exercise 7.65. Reprove Theorem 7.54 without making any appeal to Theorem 7.46.

Chapter 8

The Heat Operator and Related Versions Throughout this chapter we use the notation (x, t) := (x1 , . . . , xn , t) ∈ Rn+1 . n ∂x2j and will be The heat operator1 is then defined as L := ∂t − Δx = ∂t − j=1

at the center of our investigations in this chapter. The focus is on determining all fundamental solutions for the heat operator that are tempered distributions. As an application of the latter, we also discuss the generalized Cauchy problem for the heat operator. In addition, at the end of this section we compute a fundamental solution for the Schr¨ odinger operator 1i ∂t − Δx .

8.1

Fundamental Solutions for the Heat Operator

The starting point in determining all fundamental solutions for the heat operator L that are tempered distributions is Theorem 5.13, which guarantees that such fundamental solutions do exist. As we have seen in the case of the Laplace operator, the Fourier transform is an important tool in determining explicit expressions for fundamental solutions that are tempered distributions. We will continue to make use of this tool in the case of the heat operator with the adjustment that, this time, we work with the partial Fourier transform Fx discussed at the end of Sect. 4.2. Let E ∈ S (Rn+1 ) be a fundamental solution for L. Thus, in view of Exercise 2.81 and (4.1.25), we have ∂t E − Δx E = δ(x) ⊗ δ(t)

in S (Rn+1 ).

(8.1.1)

1 First considered in 1809 for n = 1 by Laplace (cf. [39]) and then for higher dimensions by Poisson (cf. [56] for n = 2.)


277

278

CHAPTER 8. THE HEAT OPERATOR AND RELATED VERSIONS

3x , and using Exercise 4.40, it Applying Fx to (8.1.1), denoting Fx (E) by E follows that 3x + |ξ|2 E 3x = 1(ξ) ⊗ δ(t) in ∂t E

S (Rn+1 ).

(8.1.2)

In particular, for each fixed ξ ∈ Rn , we have

3x = δ(t) ∂t + |ξ|2 E

in D (R).

(8.1.3)

3x (ξ, t) := H(t)e−|ξ|2 t is a solution of Applying Example 5.12, we obtain that E (8.1.3), where as before, H denotes the Heaviside function from (1.2.16). Note 3x (ξ, t) ∈ S (Rn+1 ). Also, if ϕ ∈ C ∞ (Rn+1 ) then, integration by that we have E 0 parts yields 2

2

x +|ξ|2 E x , ϕ = −H(t)e−|ξ| t , ∂t ϕ(ξ, t) +H(t)e−|ξ| t , |ξ|2 ϕ(ξ, t) ∂t E ∞ ∞ 2 2 =− e−|ξ| t ∂t ϕ(ξ, t) dt dξ+ |ξ|2 e−|ξ| t ϕ(ξ, t) dξ dt

Rn

= Rn

0

0

Rn

ϕ(ξ, 0) dξ = 1(ξ) ⊗ δ(t), ϕ ,

(8.1.4)

3x + |ξ|2 E 3x = 1(ξ) ⊗ δ(t) in D (Rn+1 ). Invoking (4.1.25), 3x verifies ∂t E hence E 3x (ξ, t) = H(t)e−|ξ|2 t verifies (8.1.2). We remark here that it follows that E 2 while the distribution −H(−t)e−|ξ| t satisfies (8.1.3), based on our earlier discussion pertaining to the nature of the function in (4.1.27), it does not belong to S (Rn+1 ), thus we cannot apply Fx−1 to it. Starting from the identity Fx (Fx E(x, t)) = (2π)n E(−x, t) (which is easy to check), we may write 3x (ξ))(x) = (2π)−n H(t)Fξ (e−t|ξ| )(x) E(−x, t) = (2π)−n Fξ (E π n2 |x|2 |x|2 n = (2π)−n H(t) e− 4t = H(t)(4πt)− 2 e− 4t in S (Rn+1 ), t (8.1.5) 2

where for the third equality in (8.1.5) we used Remark 4.22 and (3.2.6). Hence, we may conclude that the tempered distribution from (8.1.5) is a fundamental solution for the heat operator. n Note that, with the notation from (3.1.10), we have ∂t −Δ = iDn+1 + Dj2 , j=1

hence the heat operator satisfies the hypothesis of Proposition 5.7. Consequently, if u ∈ S (Rn+1 ) is an arbitrary fundamental solution for the heat operator in Rn+1 , then u − E = P (x, t) in S (Rn+1 ), for some polynomial P (x, t) in Rn+1 satisfying (∂t − Δx )P (x, t) = 0 in Rn+1 . As a final remark, we claim that E as in (8.1.5) satisfies E ∈ L1loc (Rn+1 ). Indeed, if K is a compact subset of Rn × [−R, R] for some R ∈ (0, ∞), then

8.1. FUNDAMENTAL SOLUTIONS FOR THE HEAT OPERATOR

0≤

R

E(x, t) dx dt ≤

Rn

0

K

n

(4πt)− 2 e−

n

= π− 2

R

dx dt

2

Rn

0

|x|2 4t

279

e−|y| dy dt = R < ∞.

(8.1.6)

In summary, we proved the following theorem. Theorem 8.1. The function defined as n

E(x, t) := H(t)(4πt)− 2 e−

|x|2 4t

∀ (x, t) ∈ Rn+1 ,

,

(8.1.7)

belongs to S (Rn+1 )∩L1loc (Rn+1 )∩C ∞ (Rn+1 \{0}) and is a fundamental solution for the heat operator ∂t − Δx in Rn+1 . Moreover, u ∈ S (Rn+1 ) : (∂t − Δx )u = δ(x) ⊗ δ(t) in S (Rn+1 ) (8.1.8) = E + P : P polynomial in Rn+1 satisfying (∂t − Δx )P (x, t) = 0 . Corollary 8.2. The heat operator ∂t − Δx is hypoelliptic in Rn+1 . Proof. This is a consequence of Theorem 6.8 and Theorem 8.1. Exercise 8.3. Let E be the function defined in (8.1.8) and set F (x, t) := −E(x, −t), for (x, t) ∈ Rn+1 . Prove that F is a fundamental solution for the operator ∂t + Δ in Rn+1 . n

|x|2

Remark 8.4. Given the expression E(x, t) = H(t)(4πt)− 2 e− 4t defined for each (x, t) ∈ Rn+1 , then one may check via a direct computation that this is a fundamental solution for the heat operator L = ∂t − Δx . First, the computation in (8.1.6) gives that E ∈ L1loc (Rn+1 ), which in turn implies E ∈ D (Rn+1 ). Second, if ϕ ∈ C0∞ (Rn+1 ) is arbitrary, then using integration by parts we may write LE, ϕ = −E, ∂t ϕ + Δx ϕ ' & ∞ E(x, t)(∂t ϕ(x, t) + Δx ϕ(x, t)) dx dt = − lim+ ε→0

ε

&

= lim+ ε→0

= lim+ ε→0

= lim+ ε→0

Rn

∞

Rn

LE(x, t)ϕ(x, t) dx dt ε

n

(4πε)− 2 e−

|x|2 4ε

'

E(x, ε)ϕ(x, ε) dx + Rn

ϕ(x, ε) dx

Rn

√ 2 n π − 2 e−|y| ϕ(2 εy, ε) dy

Rn

= ϕ(0) = δ, ϕ.

(8.1.9)

For the fourth equality in (8.1.9) we have used the fact that LE = 0 pointwise in Rn × (0, ∞), for the fifth a suitable change of variables, while for the sixth

280


equality we applied Lebesgue’s dominated convergence theorem. This proves that LE = δ in D (Rn+1 ), thus E is a fundamental solution for L. In closing we record a Liouville-type theorem for the operator ∂t − Δ, which is a particular case of Theorem 5.3. Theorem 8.5 (Liouville’s theorem for the heat operator). Any bounded function in Rn+1 that is a null solution of the heat operator is constant.

8.2

The Generalized Cauchy Problem for the Heat Operator

Let F ∈ C 0 Rn × [0, ∞) , f ∈ C 0 (Rn ) and suppose u is a solution of the Cauchy problem for the heat operator: ⎧ u ∈ C 0 Rn × [0, ∞) , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨∂t u, ∂ 2 u ∈ C 0 Rn × (0, ∞) , j = 1, . . . , n, xj (8.2.1) ⎪ ⎪ ∂t u − Δx u = F in Rn × (0, ∞), ⎪ ⎪ ⎪ ⎩ u(·, 0) = f in Rn . Denote by u and F the extensions by zero of u and F to the entire space Rn+1 . Then, if ϕ ∈ C0∞ (Rn+1 ), integrating by parts and using Lebesgue’s dominated convergence theorem we obtain ∞ $ % $ % (∂t − Δx ) u, ϕ = − u , ∂t u + Δx ϕ = − lim u(∂t ϕ + Δx ϕ) dx dt & = lim+ ε→0

∞

ε

=

ε→0+

∞ Rn

Rn

ε

(∂t u − Δx u)ϕ dx dt +

Rn

' u(x, ε)ϕ(x, ε) dx

Rn

F (x, t)ϕ(x, t) dx dt + 0

f (x)ϕ(x, 0) dx

% $ % $ = F, ϕ + f (x) ⊗ δ(t), ϕ(x, t) .

Rn

(8.2.2)

This proves that (∂t − Δx ) u = F + f (x) ⊗ δ(t) in D (Rn+1 ) and suggests the definition made below. As a preamble, we introduce the notation Rn+1 := (x, t) ∈ Rn+1 : t ≥ 0 (8.2.3) + and the space D+ (Rn+1 ) := {u ∈ D (Rn+1 ) : supp u ⊆ Rn+1 }. + D+ (Rn+1 )

(8.2.4)

Definition 8.6. Given F0 ∈ and f ∈ D (R ), a distribution u ∈ (Rn+1 ) is called a solution of the generalized Cauchy problem for the heat D+ operator for the data F0 and f , if u verifies (∂t − Δx )u = F0 + f (x) ⊗ δ(t)

n

in D (Rn+1 ).

(8.2.5)

8.2. THE GENERALIZED CAUCHY PROBLEM FOR THE HEAT . . . 281 The issue of solvability of Eq. (8.2.5) fits into the framework presented in Remark 5.5. More precisely, let E be the fundamental solution for the heat (Rn+1 ) and f ∈ D (Rn ) are such operator as given in (8.1.7). Then, if F0 ∈ D+ that u := E ∗ [F0 + f ⊗ δ] exists in D (Rn+1 )

(8.2.6)

the distribution u satisfies (8.2.5). For this u to be a solution of the generalized Cauchy problem for the heat operator, we would also need supp u ⊆ Rn+1 + . Since it is not difficult to check that supp E ⊆ Rn+1 and supp (f ⊗ δ) ⊆ Rn+1 + + , and since by assumption supp F0 ⊆ Rn+1 + , it follows that whenever condition (8.2.6) is verified, the distribution u defined in (8.2.6) also satisfies supp u ⊆ Rn+1 (by + (2.8.17)). While the convolution of two arbitrary distributions in D+ (Rn ) does not always exist (you may want to check that an exception is the case n = 1), under the additional assumptions f ∈ E (Rn ) and F0 ∈ E (Rn+1 ) condition (8.2.6) is verified. The above discussion is the reason why we analyze in detail the following setting. Retain the notation introduced at the beginning of the section, and assume that F ∈ C0∞ (Rn+1 ) and f ∈ C0∞ (Rn ). (8.2.7) Then u := E ∗ F + E ∗ f (x) ⊗ δ(t)

exists, belongs to D+ (Rn+1 ),

(8.2.8)

and is a solution of (8.2.5). We proceed by rewriting the expression for u in a more explicit form. First, by Proposition 2.93 and the fact that E ∈ L1loc (Rn+1 ), we have E ∗ F ∈ C ∞ (Rn+1 ) and $ % $ % (E ∗ F )(x, t) = E(y, τ ), F (x − y, t − τ ) = E(x − y, t − τ ), F (y, τ ) ∞ − n (x−y)2 = H(t − τ ) 4π(t − τ ) 2 e− 4(t−τ ) F (y, τ ) dy dτ 0

t

Rn

= 0

Rn

E(x − y, t − τ )F (y, τ ) dy dτ.

(8.2.9)

To compute E ∗ (f ⊗ δ), fix an arbitrary compact set K ⊂ Rn+1 , consider a function ϕ ∈ C0∞ (Rn+1 ) such that supp ϕ ⊆ K, and pick some ψ ∈ C0∞ (R2n+2 ) with the property that ψ = 1 in a neighborhood of the set (x, t, y, 0) ∈ Rn × R × Rn × R : y ∈ supp f and (x + y, t) ∈ K . (8.2.10) Relying on the definition of convolution we have $ % ! $ %" E ∗ (f ⊗ δ), ϕ = E(x, t), f (y) ⊗ δ(τ ), ψ(x, t, y, τ )ϕ(x + y, t + τ ) 9 8 f (y)ϕ(x + y, t)dy = E(x, t), Rn

282


=

E(x, t)f (y)ϕ(x + y, t) dx dy dt R

Rn

R

Rn

Rn

= Rn

E(z − y, t)f (y)ϕ(z, t) dz dy dt.

Hence, E ∗ f (x) ⊗ δ(t) is given by the function E(x − y, t)f (y) dy ∈ C, Rn × R (x, t) →

(8.2.11)

(8.2.12)

Rn

whose restriction to Rn × (0, ∞) is of class C ∞ . In summary, this analysis proves the following result.

Proposition 8.7. Let f ∈ C0∞ (Rn ) and assume that F ∈ C 0 Rn × [0, ∞) is such that its extension F by zero to Rn+1 satisfies F ∈ C0∞ (Rn+1 ). Also let E be the fundamental solution for the heat operator ∂t − Δx as given in (8.1.7). Then the generalized Cauchy problem for the heat operator for the data F and f has a solution u ∈ D+ (Rn+1 ) that is of function type, whose restriction n ∞ to R × (0, ∞) is of class C , and has the expression t E(x − y, t)f (y) dy + E(x − y, t − τ )F(y, τ ) dy dτ (8.2.13) u(x, t) = Rn

0

Rn

for every x ∈ R and t ∈ (0, ∞). n

Note that the integrals in (8.2.13) are meaningfully defined under weaker assumptions on F and f . In fact, starting with u as in (8.2.13) one may prove that this is a solution to a version of (8.2.1) (corresponding to finite time, i.e., t ∈ (0, T ), for some T > 0) under suitable yet less stringent conditions on F and f .

8.3

Fundamental Solutions for General Second Order Parabolic Operators

Let A = ajk 1≤j,k≤n ∈ Mn×n (C) and associated to such a matrix A, consider the parabolic operator LA := LA (∂) := ∂t −

n

ajk ∂j ∂k .

(8.3.1)

j,k=1

The goal is to obtain explicit formulas for all tempered distributions that are fundamental solutions for LA under the additional assumption that there exists a constant C ∈ (0, ∞) such that the matrix A satisfies the strict positiveness condition n Re ajk ξj ξk ≥ C|ξ|2 , ∀ ξ = (ξ1 , . . . , ξn ) ∈ Rn . (8.3.2) j,k=1

8.3. FUNDAMENTAL SOLUTIONS FOR GENERAL . . .

283

The approach is an adaptation to the parabolic setting of the ideas used in Sect. 7.8 for the derivation of (7.8.53). In a first stage, we note that, via the same reasoning as in Sect. 7.8, when computing the fundamental solution for LA we may assume without loss of generality that A is symmetric, that is, A = A . Also, we treat first the case when A has real entries. The case when A is real, symmetric,, and satisfies (8.3.2). 1 Since A is real, symmetric and positive definite, √ A 2 is well-defined, real, sym1 1 1 metric, invertible, A 2 A 2 = A, and det(A 2 ) = detA. Throughout this discussion, we agree to denote by ∇x,t the global gradient in Rn+1 , and reserve ∇ for the gradient in the variable x only. As in the past, Δ refers to the Laplacian in the variable x. Fix such that ∂k u ∈ L1loc (Rn+1 ) for each k ∈ {1, . . . , n},

u ∈ L1loc (Rn+1 )

(8.3.3) Also, fix some ϕ ∈ C0∞ (Rn+1 ). Then, using the fact that A is symmetric we obtain n

LA u, ϕ = ∂t u, ϕ −

n $ $ % % ajk ∂j ∂k u, ϕ = −u, ∂t ϕ + ajk ∂k u, ∂j ϕ

j,k=1

j,k=1

=−

Rn+1

u(x, t)∂t ϕ(x, t) dx dt +

=−

R

Rn

n j,k=1

Rn+1

∂j u(x, t)ajk ∂k ϕ(x, t) dx dt

u(x, t)∂t ϕ(x, t) − ∇u(x) · A∇ϕ(x, t) dx dt.

(8.3.4)

In the last integral over Rn+1 in (8.3.4) we make the change of variables 1 x = A 2 y. Since for every invertible matrix B ∈ Mn×n (R) and any function f the chain rule gives (∇f )(By) = (B )−1 ∇[f (By)]

(8.3.5)

for each y ∈ Rn such that f is differentiable at By, we obtain (the analogue of (7.8.30)) u(x, t)∂t ϕ(x, t) − ∇u(x, t) · A∇ϕ(x, t) dx dt R

Rn

=

√ detA R

=

√ detA

Rn

R

Rn

v(y, t)∂t ψ(y, t) − ∇v(y, t) · ∇ψ(y, t) dy dt v(y, t) ∂t ψ(y, t) + Δψ(y, t) dy dt,

(8.3.6)

where we have set 1

v(x, t) := u(A 2 x, t)

1

and ψ(x, t) := ϕ(A 2 x, t)

for (x, t) ∈ Rn+1 . (8.3.7)

284


Note that ψ ∈ C0∞ (Rn+1 ). Hence, (8.3.4) and (8.3.6) imply √ LA u, ϕ = − detA v(y, t)[∂t ψ(y, t) + Δψ(y, t)] dy dt.

(8.3.8)

Rn+1

Choose u to be the function u(y, t) := √

1 detA

1

E∂t −Δ (A− 2 y, t) for

(y, t) ∈ Rn+1 ,

(8.3.9)

where E∂t −Δ is the fundamental solution for the heat operator from (8.1.7). In particular, u satisfies (8.3.3). For this choice of u, the function v from (8.3.7) becomes 1 v(y, t) = √ E∂t −Δ (y, t) for (y, t) ∈ Rn+1 . detA

(8.3.10)

1

Since A 2 (0) = 0, we may write $ % 1 ϕ(0) = ϕ(A 2 0, 0) = ψ(0) = δ(x) ⊗ δ(t), ψ $ % $ % = (∂t − Δ)E∂t −Δ , ψ = E∂t −Δ , −∂t ψ − Δψ √ 1 √ = − detA E∂t −Δ (y, t) ∂t ψ(y, t) + Δψ(y, t) dy dt detA Rn+1 √ v(y, t) ∂t ψ(y, t) + Δψ(y, t) dy dt = − detA Rn+1

= LA u, ϕ = √

% $ 1 1 LA E∂t −Δ (A− 2 x, t) , ϕ(x, t) , detA

(8.3.11)

where the penultimate equality in (8.3.11) uses (8.3.8). From (8.3.11) we may conclude that the function 1 1 E∂t −Δ (A− 2 x, t) EA (x, t) := √ detA

for (x, t) ∈ Rn+1 ,

(8.3.12)

n is a fundamental solution Rn+1 . Keeping − 1for LA in −1 in mind that for each x ∈ R 1 − 12 2 − we have |A x| = A 2 x · A 2 x = A x · x, the function EA may be rewritten as (A−1 x)·x n 1 EA (x) := √ H(t)(4πt)− 2 e− 4t detA

for (x, t) ∈ Rn+1 .

(8.3.13)

Moreover, from (8.3.12), Theorem 8.1, and Proposition 4.41, it follows that EA belongs to S (Rn+1 ) ∩ L1loc (Rn+1 ) ∩ C ∞ (Rn+1 \ {0}). The case when A has complex entries, is symmetric, and satisfies (8.3.2). As observed in Remark 7.50, under the current assumptions, A continues to be invertible. Also, (7.8.25) holds. In addition, under the current assumptions 1 (detA) 2 is unambiguously defined (see in Remark 7.52).

8.3. FUNDAMENTAL SOLUTIONS FOR GENERAL . . .

285

These comments show that the function EA from (8.3.13) continues to be well-defined under the current assumption on A if ln is replaced by the principal branch of the complex log (defined for points z ∈ C\(−∞, 0] so that z a = ea log z for each a ∈ R). In addition, EA continues to belong to L1loc (Rn+1 ) (this can be seen by a computation similar to that in (8.1.6), keeping in mind (7.8.23)), and EA ∈ C ∞ (Rn+1 \ {0}). Moreover, from (7.8.23) and Exercise 4.5 it follows that EA belongs to S (Rn+1 ). The goal is to prove that this expression is a fundamental solution for LA in the current case. Making use of (7.8.45) for every (x, t) ∈ Rn+1 with t = 0 differentiating pointwise we obtain (A−1 x)·x (A−1 x) · x (A−1 x)·x n ajk ∂j ∂k e− 4t − = e− 4t , 4t2 2t

(8.3.14)

(A−1 x) · x (A−1 x)·x (A−1 x)·x n = t−n/2 e− 4t . − ∂t t−n/2 e− 4t 2 4t 2t

(8.3.15)

n j,k=1

while

From (8.3.14), (8.3.15), and the expression for EA we may conclude that ∀ x ∈ Rn , ∀ t ∈ R \ {0}.

LA EA (x, t) = 0

(8.3.16)

In addition, for each ϕ ∈ C0∞ (Rn+1 ) we may compute n ! " LA EA , ϕ = − EA , ∂t ϕ + ajk ∂j ∂k ϕ j,k=1

⎡ = − lim ⎣ ε→0+

∞

Rn

ε

⎤ n EA (x, t) ∂t ϕ(x, t) + ajk ∂j ∂k ϕ(x, t) dx dt⎦ j,k=1

& = lim+ ε→0

Rn

EA (x, ε)ϕ(x, ε) dx + ε

1 = lim+ √ ε→0 detA n

π− 2 = lim √ ε→0+ detA n

∞

π− 2 = ϕ(0) √ detA

n

(4πε)− 2 e−

(A−1 x)·x 4ε

'

Rn

(LA EA )(x, t)ϕ(x, t) dx dt

ϕ(x, ε) dx

Rn

e−(A

−1

y)·y

√ ϕ(2 εy, ε) dy

Rn

e−(A

−1

y)·y

dy.

(8.3.17)

Rn

For the fourth equality √ in (8.3.17) we have used (8.3.16), for the fifth the change of variables x = 2 εy, while for the sixth equality we applied Lebesgue’s

286


dominated convergence theorem. From (8.3.17) we then conclude that EA is a fundamental solution for LA in Rn+1 if and only if

e−(A

−1

y)·y

n

dy = π 2

√ detA.

(8.3.18)

Rn

The fact that we already know that EA is a fundamental solution for LA in Rn+1 in the case when A ∈ Mn×n (R) satisfies A = A and condition (8.3.2), implies that formula (8.3.18) holds for this class of matrices. By using the same circle of ideas as the ones employed in proving (7.8.50) (based on Lemma 7.53), we conclude that (8.3.18) holds for the larger class of matrices A ∈ Mn×n (C) satisfying A = A and condition (8.3.2). Hence, EA is indeed a fundamental solution for LA under the current assumptions on A. Next, we claim that the hypotheses of Proposition 5.7 hold in the case of the operator P (D) := LA . To justify this, note that if ξ = (ξ1 , . . . , ξn+1 ) ∈ Rn+1 is such that P (ξ) = 0 then iξn+1 +

n

ajk ξj ξk = 0.

(8.3.19)

j,k=1

Taking reals parts, condition (8.3.2) implies ξ1 = · · · = ξn = 0 which, in combination with (8.3.19), also forces ξn+1 = 0. Hence, ξ = 0 as wanted. Applying now Proposition 5.7 gives that if u ∈ S (Rn+1 ) is an arbitrary fundamental solution for LA in Rn+1 , then u = EA + P in S (Rn+1 ), for some polynomial P in Rn+1 satisfying LA P = 0 pointwise in Rn+1 . In summary, the above analysis proves the following result. Theorem 8.8. Suppose A = ajk 1≤j,k≤n ∈ Mn×n (C) satisfies (8.3.2) and consider the operator LA associated to A as in (8.3.1). Then the function defined by 1 n − H(t)(4πt)− 2 e detAsym

EA (x) :=

(A−1 sym x) · x 4t

for all (x, t) ∈ Rn+1 , (8.3.20)

)∩L1loc (Rn+1 )∩C ∞ (Rn+1 \{0}) and is a fundamental solution Above, Asym := 12 (A + A ), detAsym is defined as in

belongs to S (R for LA in Rn+1 . Remark 7.50, and log denotes the principal branch of the complex logarithm (defined for points z ∈ C \ (−∞, 0] so that z a = ea log z for each a ∈ R). Moreover,

n+1

(8.3.21) u ∈ S (Rn+1 ) : LA u = δ in S (Rn+1 ) = EA + P : P polynomial in Rn+1 satisfying LA P = 0 .

¨ 8.4. FUNDAMENTAL SOLUTION FOR THE SCHRODINGER ...

8.4

287

Fundamental Solution for the Schr¨ odinger Operator

Let x ∈ Rn and t ∈ R. The operator 1i ∂t − Δx is called the (time-dependent) Schr¨ odinger operator in Rn+1 (with zero potential). In this section we determine a fundamental solution for this operator. By Theorem 5.13, there exists E ∈ S (Rn+1 ) such that 1 i

∂t − Δx E = δ(x) ⊗ δ(t) in S (Rn+1 ).

(8.4.1)

Fix such a distribution E and take the partial Fourier transform Fx of (8.4.1) (recall the discussion of the partial Fourier transform at the end of Sect. 4.2) to obtain 1 3 3x = 1(ξ) ⊗ δ(t) in S (Rn+1 ). ∂t Ex + |ξ|2 E (8.4.2) i Fix ξ ∈ Rn and consider the equation 1 ∂t u + |ξ|2 u = δ i

in D (R).

(8.4.3)

2

2

Using Example 5.11, we obtain that iH(t)e−i|ξ| t and −iH(−t)e−i|ξ| t are solu2 tion of this equation. This suggests considering F := iH(t)e−i|ξ| t that belongs to S (Rn+1 ) and satisfies (based on a computation similar to that from (8.1.4)) 1 ∂t F + |ξ|2 F = 1(ξ) ⊗ δ(t) i

in S (Rn+1 ).

(8.4.4)

Then Fx−1 (F ) ∈ S (Rn+1 ) and 1 i

∂t − Δx Fx−1 (F ) = δ(x) ⊗ δ(t) in S (Rn+1 ).

(8.4.5)

To compute Fx−1 (F ), pick ϕ ∈ S(Rn+1 ) and, based on the properties of Fx as well as the expression for F , write $

$ % $ % % Fx−1 (F ), ϕ = (2π)−n Fx F ∨ , ϕ = (2π)−n F ∨ , Fx ϕ 2 H(t)e−i|ξ| t (Fx ϕ)(ξ, t) dξ dt = (2π)−n i Rn+1

−n

= (2π)

i

∞

&

H(t) 0

e Rn

−i|ξ|2 t

' (Fx ϕ)(ξ, t) dξ dt.

(8.4.6)

$ % 2 For each t > 0, the integral over Rn in (8.4.6) equals Fx (e−it|x| ), ϕ(·, t) , which π n2

|ξ|2 by Example 4.24 (with a = t) is equal to it e− 4it ϕ(ξ, t) dξ. Hence, for Rn each ϕ ∈ S(Rn+1 ) we have

288

CHAPTER 8. THE HEAT OPERATOR AND RELATED VERSIONS $ −1 % Fx (F ), ϕ = (2π)−n i

∞

H(t)

π n2 & it

0 n

= i1− 2

∞

n

H(t)(4πt)− 2

&

e

−|ξ|2 4it

' ϕ(ξ, t) dξ dt

Rn

' |ξ|2 e− 4it ϕ(ξ, t) dξ dt.

(8.4.7)

Rn

0 |x|2

n

We remark here that H(t)(4πt)− 2 e− 4it ∈ L1loc (Rn+1 ) only if n = 1. Hence Fx−1 (F ) is of function type only if n = 1. In general, this distribution belongs to S (Rn+1 ) and its action on ϕ ∈ S(Rn+1 ) is given as in (8.4.7). In summary we proved the following result. Theorem 8.9. The distribution E ∈ S (Rn+1 ) defined by ' & ∞ 2 1− n −n − |ξ| 2 2 4it H(t)(4πt) e ϕ(ξ, t) dξ dt, E, ϕ := i 0

Rn

∀ ϕ ∈ S(Rn+1 ), (8.4.8)

1 i ∂t −Δx

is a fundamental solution for the Schr¨ odinger operator in R . In particular, if n = 1 then E is of function type and is given by the L1loc (R2 ) 1

E(x, t) = H(t)(4πt)− 2 e−

|x|2 4it

,

x ∈ R, t ∈ R.

n+1

(8.4.9)

Further Notes for Chap. 8. The heat equation is one example of what is commonly referred to as linear evolution equations. Originally derived in physics from Fourier’s law and conservation of energy (see, e.g., [71] for details), the heat equation has come to play a role of fundamental importance in mathematics and applied sciences. In mathematics, the heat operator is the prototype for a larger class, called parabolic partial differential operators, that includes the operators studied in Sect. 8.3. The Schr¨ odinger operator is named after the Austrian physicist Erwin Schr¨ odinger who first introduced it in 1926. It plays a fundamental role in quantum mechanics, where it describes how the quantum state of certain physical systems changes over time.

Chapter 9

The Wave Operator The operator := ∂t2 −Δx , x ∈ Rn , t ∈ R, is called the wave operator in Rn+1 . The wave operator arises from modeling vibrations in a string, membrane or elastic solid. The goal is to determine fundamental solutions for this operator that are tempered distributions. As an application we discuss the generalized Cauchy problem for the wave operator.

9.1

Fundamental Solution for the Wave Operator

By Theorem 5.13, we know that the wave operator admits a fundamental solution E ∈ S (Rn+1 ). Hence, by (4.1.25) and Exercise 2.81, we have ∂t2 E − Δx E = δ(x) ⊗ δ(t) in S (Rn+1 ).

(9.1.1)

In this section we determine an explicit expression for two such fundamental solutions. Fix E ∈ S (Rn+1 ) that satisfies (9.1.1) and apply the partial Fourier transform Fx to this equation (recall the discussion about partial Fourier transforms from the last part of Sect. 4.2) to obtain 3x + |ξ|2 E 3x = 1(ξ) ⊗ δ(t) in S (Rn+1 ), ∂t2 E

(9.1.2)

3x := Fx (E) ∈ S (Rn+1 ). For ξ ∈ Rn \ {0} fixed consider the initial where E value problem (in the variable t) ⎧ 2 ⎨ d 2 v + |ξ|2 v = 0 in R, dt (9.1.3) ⎩(∂ v)(0) = 1, v(0) = 0, t which admits the solution v(t) = sin(t|ξ|) for t ∈ R. By Example 5.11, it follows |ξ| that the distributions vH and −vH ∨ are fundamental solutions for the operator D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6 9, © Springer Science+Business Media New York 2013

289

CHAPTER 9. THE WAVE OPERATOR

290 d2 dt2

+ |ξ|2 in D (R). In addition, vH and −vH ∨ belong to S (R) (based on (b) in Theorem 4.13 and Exercise 4.115), thus 2 2 d d 2 2 + |ξ| + |ξ| (vH) = δ and (−vH ∨ ) = δ in S (R). dt2 dt2 (9.1.4) Moreover, there exists c ∈ (0, ∞) such that H(±t) sin(t|y|) ≤ c|t| for (y, t) ∈ B(0, 1) \ {0} ⊂ Rn+1 . |y|

(9.1.5)

In particular, if we define the functions f± (y, t) := H(±t)

sin(t|y|) |y|

for (y, t) ∈ Rn+1 with t = 0,

(9.1.6)

then (9.1.5) implies f± ∈ L1loc (Rn+1 ). Furthermore, (1 + |y|2 + t2 )−n−2 f± ∈ L1 (Rn+1 ),

(9.1.7)

∈ S (Rn+1 ). Based on all these thus by Example 4.3, we obtain H(±t) sin(t|y|) |y| facts, if we introduce F + := H(t)

sin(t|ξ|) , |ξ|

F − := −H(−t)

sin(t|ξ|) , |ξ|

(9.1.8)

then F + , F − ∈ S (Rn+1 ) and ∂t2 F ± + |ξ|2 F ± = 1(ξ) ⊗ δ(t)

(9.1.9) in S (Rn+1 ).

(9.1.10)

In particular, Fx−1 (F + ) and Fx−1 (F − ) are meaningfully defined in S (Rn+1 ). Thus, if we set sin(t|ξ|) E+ := Fx−1 H(t) , (9.1.11) |ξ| sin(t|ξ|) , E− := Fx−1 −H(−t) |ξ|

(9.1.12)

we have E± ∈ S (Rn+1 ),

supp E+ ⊆ R × [0, ∞), n

in S (Rn+1 ),

(9.1.13)

supp E− ⊆ R × (−∞, 0].

(9.1.14)

(∂t2 − Δx )E± = δ

n

The next task is to find explicit expressions for Fx−1 (F ± ). To this end, fix a function ϕ ∈ S(Rn+1 ) and, with the operation ·∨ considered only in the variable x, write $ $ % % E+ , ϕ = (2π)−n (Fx Fx (E+ ))∨ , ϕ = (2π)−n Fx (E+ ), (Fx ϕ)∨

9.1. FUNDAMENTAL SOLUTION FOR THE WAVE OPERATOR

291

sin t|ξ| ϕ 3x (−ξ, t) dξ dt. (9.1.15) |ξ| Rn+1

Note that if one replaces ϕ 3x (−ξ, t) above with Rn eix·ξ ϕ(x, t) dx then the order 1 of integration in the resulting iterated integral may not be switched since |ξ| is not integrable at infinity, thus Fubini’s theorem does not apply. This is why we should proceed with more care and, based on Lebesgue’s dominated convergence theorem, we introduce a convergence factor that enables us to eventually make the use of Fubini’s theorem. More precisely, we have E+ , ϕ = (2π)−n F + (ξ, t) ϕ 3x (−ξ, t) dξ dt −n

H(t)

= (2π)

Rn+1 −n

= lim (2π) ε→0+

Rn+1

−n

F + (ξ, t)e−ε|ξ| ϕ 3x (−ξ, t) dξ dt

= lim+ (2π) ε→0

ϕ(x, t)

e

Rn+1

ix·ξ−ε|ξ|

+

F (ξ, t) dξ

dx dt

Rn

= lim+ Eε , ϕ,

(9.1.16)

ε→0

where −n

Eε (x, t) := (2π)

eix·ξ−ε|ξ|

H(t) Rn

sin(t|ξ|) dξ, |ξ|

∀ x ∈ Rn , ∀ t ∈ R. (9.1.17)

To compute the last limit in (9.1.16), we separate our analysis into three cases: n = 1, n = 2p + 1 with p ≥ 1, and n = 2p with p ≥ 1. The Case n = 1 Fix ε > 0 and x ∈ R and define the function ∞ sin(t|ξ|) fε (t) := dξ eixξ−ε|ξ| |ξ| −∞ Then

∂t fε (t) =

∞

e

ixξ−ε|ξ|

−∞

1 = 2

&

∞

e 0

∞

cos(t|ξ|) dξ =

ξ(ix+it−ε)

eixξ−ε|ξ|

−∞

∞

dξ +

e

t ∈ R.

for

'

ξ(ix−it−ε)

(9.1.18)

eit|ξ| + e−it|ξ| dξ 2

dξ

0

' & 0 0 1 ξ(ix−it+ε) ξ(ix+it+ε) + e dξ + e dξ 2 −∞ −∞ & ' 1 1 1 1 1 = − − + + 2 ix + it − ε ix − it − ε ix − it + ε ix + it + ε =

ε ε + , 2 2 (x + t) + ε (x − t)2 + ε2

∀ t ∈ R.

(9.1.19)


292

Consequently, (9.1.19) and the fact that fε (0) = 0 imply x+t x−t fε (t) = arctan − arctan for t ∈ R. ε ε

(9.1.20)

Making use of (9.1.20) back in (9.1.17) (written for n = 1) then gives x+t x−t 1 H(t) arctan − arctan , ∀ x, t ∈ R. Eε (x, t) = 2π ε ε (9.1.21) Hence, for ϕ ∈ S(R2 ) fixed, we may write lim Eε (x, t), ϕ(x, t)

ε→0+

1 = lim + ε→0 2π

(9.1.22)

∞

&

arctan

R

0

x+t ε

− arctan

x−t ε

' ϕ(x, t) dt dx.

To continue with our calculation, we further decompose

∞

arctan

0

R

∞

x+t ε

ϕ(x, t) dx dt

−t

arctan

= 0

−∞

x+t ε

∞

∞

ϕ(x, t) dx dt +

arctan 0

−t

x+t ε

=: Iε + IIε .

ϕ(x, t) dx dt (9.1.23)

By Lebesgue’s dominated convergence theorem, π ∞ −t lim Iε = − ϕ(x, t) dx dt 2 0 ε→0+ −∞ and lim+ IIε =

ε→0

π 2

∞

∞

ϕ(x, t) dx dt. 0

(9.1.24)

(9.1.25)

−t

Similarly,

∞

arctan

lim+

ε→0

R

0

=−

π 2

∞

x−t ε

ϕ(x, t) dx dt

t

ϕ(x, t) dx dt + −∞

0

π 2

∞

∞

ϕ(x, t) dx dt. 0

(9.1.26)

t

Combining (9.1.22)–(9.1.26) permits us to write lim Eε (x, t), ϕ(x, t)

(9.1.27)

ε→0+

1 = 4

0

∞

& −

−t

−∞

∞

ϕ(x, t) dx + −t

' ϕ(x, t) dx dt

9.1. FUNDAMENTAL SOLUTION FOR THE WAVE OPERATOR +

=

1 2

1 4

∞

&

−∞

0 ∞

t

' ϕ(x, t) dx dt

t

t

ϕ(x, t) dx dt = −t

0

∞

ϕ(x, t) dx −

293

1 2

R2

H(t − |x|)ϕ(x, t) dx dt.

From (9.1.27) and (9.1.16) we then conclude that1 E+ (x, t) = H(t−|x|) for 2 (x, t) ∈ R2 . In summary, we proved that ⎧ ⎪ ⎨ E+ (x, t) = H(t − |x|) , ∀ (x, t) ∈ R2 , is a tempered distribution 2 ⎪ ⎩ 2 and satisfies (∂t − ∂x2 )E+ = δ in S (R2 ). (9.1.28) Remark 9.1. (1) The reasoning used to obtain E+ also yields that ⎧ ⎪ ⎨ E− (x, t) := H(−t − |x|) , ∀ (x, t) ∈ R2 , is a tempered distribution 2 ⎪ ⎩ 2 and satisfies (∂t − ∂x2 )E− = δ in S (R2 ). (9.1.29) (2) An inspection of (9.1.28) and (9.1.29) reveals that supp E+ = {(x, t) ∈ R2 : |x| ≤ t}

and

supp E− = {(x, t) ∈ R2 : |x| ≤ −t}.

(9.1.30)

The Case n = 2p + 1, p ≥ 1 It is immediate from (9.1.17) that Eε (x, t) is invariant under orthogonal transformations in the variable x. Hence, Eε (y, t) = Eε (|y|, 0, . . . , 0, t) for every y ∈ Rn and t ∈ R. As such, in what follows we may assume that x = (|x|, 0, . . . , 0). Fix such an x and set r := |x|. With t ∈ R fixed, using polar coordinates (see (13.8.1), (13.8.2), and (13.8.3)), rewrite the expression from (9.1.17) as Eε (x, t) = (2π)−n H(t)

0

∞

(0,π)n−2

2π

eiρr cos θ1 −ερ

0

sin tρ n−1 ρ ρ

× (sin θ1 )

· · · sin(θn−2 ) dθn−1 · · · dθ1 dρ ∞ π −n n−2 ρ sin(tρ) eiρr cos θ1 −ερ (sin θ1 )n−2 × = (2π) H(t) n−2

0 1 This

0

expression for a fundamental solution for the wave operator when n = 1 was first used by Jean d’Alembert in 1747 in connection with a vibrating string.


294 6 ×

7

2π

(sin θ2 )n−3 · · · sin(θn−2 ) dθn−1 · · · dθ2 dθ1 dρ

(0,π)n−3

0

= (2π)−n ωn−2 H(t)

∞

0

π

eiρr cos θ−ερ ρn−2 (sin θ)n−2 sin(tρ) dθ dρ 0

(9.1.31) where ωn−2 denotes the surface area of the unit ball in Rn−1 (see also (13.6.6) for more details on why the expression inside the right brackets in (9.1.31) is equal to ωn−2 ). To proceed with the computation of the integrals in the rightmost term in (9.1.31), recall that n = 2p + 1 and, for ρ > 0 fixed, set π eiρr cos θ (sin θ)2p−1 dθ, ∀ p ∈ N. (9.1.32) Ip := ρ2p−1 0

We claim that 2p ∂r Ip for all p ≥ 1. (9.1.33) r In order to prove (9.1.33) note that, for each p ≥ 1, integration by parts yields π 2p+1 Ip+1 = ρ eiρr cos θ (sin θ)2p+1 dθ Ip+1 = −

0

−1 iρr 6

π

= ρ2p+1

∂θ (eiρr cos θ )(sin θ)2p dθ 0

7 θ=π π i 2p iρr cos θ 2p iρr cos θ 2p−1 = ρ (sin θ) −2p e (sin θ) cos θ dθ e r 0 θ=0 2pi 2p =− ρ r =−

π

eiρr cos θ (sin θ)2p−1 cos θ dθ 0

2pi 2p 1 ρ ∂r r iρ

π

eiρr cos θ (sin θ)2p−1 dθ = − 0

2p (∂r Ip ), r

(9.1.34)

as wanted. By induction, from the recurrence relation in (9.1.33) it follows that (p−1) 1 p−1 Ip = (−2) (p − 1)! ∂r I1 , ∀ p ∈ N. (9.1.35) r As for I1 , we have π −1 iρr cos θ θ=π 2 sin(ρr) I1 = ρ e . eiρr cos θ sin θ dθ = = θ=0 ir r 0

(9.1.36)

Recalling (9.1.31) and the fact that n = 2p + 1 for some p ∈ N, formulas (9.1.35) and (9.1.36) yield ∞ Eε (x, t) = (2π)−n ωn−2 H(t) Ip e−ερ sin(tρ) dρ (9.1.37) 0


295

= (2π)−n ωn−2 H(t)(−2)p−1 (p − 1)! 2 ×

∞

×

e−ερ sin(tρ)

0

1 ∂r r

p−1

sin(ρr) r

dρ.

Furthermore, using (13.5.6) we have n−1

ωn−2 =

2π p 2π 2 2π p n−1 = = Γ(p) (p − 1)! Γ 2

(9.1.38)

which further simplifies the expression in (9.1.37) to ' p−1 & ∞ 1 1 Eε (x, t) = 2(−2π)−p−1 H(t) ∂r e−ερ sin(tρ) sin(ρr) dρ . r r 0 (9.1.39) Our next claim is that ' & ∞ ε 1 ε e−ερ sin(tρ) sin(ρr) dρ = − . 2 ε2 + (t − r)2 ε2 + (t + r)2 0 Indeed, ∞ e 0

−ερ

1 sin(tρ) sin(ρr) dρ = 2

∞

(9.1.40)

e−ερ [cos(t − r)ρ − cos(t + r)ρ] dρ

0

1 = Re 2

∞

[e−ερ+i(t−r)ρ − e−ερ+i(t+r)ρ ] dρ

0

& ' 1 1 1 = Re − + 2 −ε + i(t − r) −ε + i(t + r) =

& ' ε + i(t − r) 1 −ε − i(t + r) Re 2 + 2 2 ε + (t − r)2 ε + (t + r)2

=

' & 1 ε ε − , 2 ε2 + (t − r)2 ε2 + (t + r)2

(9.1.41)

proving (9.1.40). The identity resulting from (9.1.41) further simplifies the expression in (9.1.39) as (p−1) & ' 1 1 ε ε ∂r − Eε (x, t) = (−2π)−p−1 H(t) r r ε2 +(t−r)2 ε2 +(t+r)2 (9.1.42) for every x ∈ Rn and every t ∈ R, where r = |x|. Recall from (9.1.16) that in order to determine E+ we further need to compute lim+ Eε (x, t) in S (Rn+1 ). With this goal in mind, fix ϕ ∈ S(Rn+1 ) and ε→0

use (9.1.42) in concert with (13.8.8) to write


296 Eε , ϕ = (−2π)−p−1

∞ ∞ 0

0

∞

1 ∂B(0,r)

r

∂r

p−1 & ε r

1 1 − 2 ε2 + (t − r)2 ε + (t + r)2

' ×

× ϕ(ω, t) dσ(ω) dr dt = (−2π)−p−1 ε

0

∞

0

1 ∂r r

&

p−1 1 r

1 1 − 2 ε2 + (t − r)2 ε + (t + r)2

' ×

×

ϕ(ω, t) dσ(ω) dr dt. ∂B(0,r)

(9.1.43)

A natural question to ask is whether, if p ≥ 2, we may integrate by parts (p− 1) times in the r variable in the rightmost expression in (9.1.43). Observe that, at least formally, ' ' & ∞ & ∞ 1 d d 1 g(r) dr (9.1.44) f (r) f (r) g(r) dr = − r dr dr r 0 0 if f (r)g(r) f (r)g(r) = 0 = lim+ . r→∞ r r r→0 lim

(9.1.45)

In the setting of the last expression in (9.1.43), for each ε, t > 0 fixed, and assuming p ≥ 2, if we let f (r) := g(r) :=

1 ∂r r

p−2 & ' 1 1 1 − , r > 0, r ε2 + (t − r)2 ε2 + (t + r)2 ϕ(ω, t) dσ(ω) = rn−1 ϕ(rω, t) dσ(ω), r > 0,

(9.1.46)

(9.1.47)

S n−1

∂B(0,r)

then these functions satisfy (9.1.45) and (9.1.44). Proceeding by induction (with p ≥ 2), we apply (p − 1) times formula (9.1.44), pick up (−1)p−1 in the process (the last factor 1r bundled up with the derivative) and write ∞ ∞ ε ε Eε , ϕ = (2π)−p−1 − × ε2 + (t − r)2 ε2 + (t + r)2 0 0 7 6 p−1 1 ∂r ϕ(ω, t) dσ(ω) dr dt. × r r ∂B(0,r) (9.1.48) 1

Note that (9.1.48) is also valid if p = 1 without any need of integration by parts.


297

We are left with taking the limit as ε → 0+ in (9.1.48) a task we complete by using Lemma 9.3 (which is stated and proved at the end of this subsection). Specifically, we apply Lemma 9.3 with 7 p−1 6 1 1 ∂r ϕ(ω, t) dσ(ω) . (9.1.49) h(r) := r r ∂B(0,r) Note that the second equality in (9.1.47) and the fact that ϕ ∈ S(Rn+1 ) guarantee that h in (9.1.49) satisfies the hypothesis of Lemma 9.3. These facts combined with Lebesgue’s dominated convergence theorem yield 7 p−1 ∞ 6 1 1 ∂r lim Eε , ϕ = (2π)−p−1 π ϕ(ω, t) dσ(ω) dt. r r ∂B(0,r) ε→0+ 0 r=t

(9.1.50) In summary, we proved the following result: If n = 2p + 1, for some p ∈ N, then E+ ∈ S (Rn+1 ) defined by −p−1

E+ , ϕ = (2π)

∞

6

π 0

1 ∂r r

7 p−1 1 ϕ(ω, t) dσ(ω) dt r ∂B(0,r) r=t

(9.1.51) for ϕ ∈ S(Rn+1 ) is a fundamental solution for the wave operator in Rn+1 . The reasoning used to obtain (9.1.51) also yields an expression for E− . More precisely, similar to (9.1.16), we arrive at E− , ϕ = lim+ Eε , ϕ for ϕ ∈ S(Rn+1 ) where, this time, 0 ∞ Eε , ϕ = −(2π)−p−1 −∞

0

ε→0

ε ε − 2 2 2 ε + (t − r) ε + (t + r)2

×

7 6 p−1 1 ∂r ϕ(ω, t) dσ(ω) dr dt × r r ∂B(0,r) 1

−p−1

∞

= (2π)

0

0

∞

ε ε − 2 ε2 + (t − r)2 ε + (t + r)2

×

(9.1.52)

7 6 p−1 1 ∂r ϕ(ω, −t) dσ(ω) dr dt. × r r ∂B(0,r) 1


298

Then applying Lemma 9.3, we obtain: If n = 2p + 1, for some p ∈ N, then E− ∈ S (Rn+1 ), defined by E− , ϕ = (2π)−p−1 π

∞ 1 0

r ∂r

p−1 1

r

ϕ(ω, −t) dσ(ω) dt, (9.1.53) ∂B(0,r) r=t

for ϕ ∈ S(Rn+1 ), is a fundamental solution for the wave operator in Rn+1 . Remark 9.2. (1) The distributions E+ and E− from (9.1.51) and (9.1.53), respectively, satisfy supp E+ = {(x, t) ∈ R2p+1 × [0, ∞) : |x| = t},

(9.1.54)

supp E+ = {(x, t) ∈ R2p+1 × (−∞, 0] : |x| = −t}.

(9.1.55)

(2) In the case when n = 3 (thus, for p = 1), formulas (9.1.51) and (9.1.53) become ∞ 1 1 E+ , ϕ = ϕ(ω, t) dσ(ω) dt 4π 0 t ∂B(0,t) 9 8 H(t) δ∂B(0,t) , ϕ , (9.1.56) = 4πt ∞ 1 1 E− , ϕ = ϕ(ω, −t) dσ(ω) dt 4π 0 t ∂B(0,t) 9 8 H(−t) δ∂B(0,−t) , ϕ , = − 4πt

(9.1.57)

for every ϕ ∈ S(R4 ) where, for each R ∈ (0, ∞), the symbol δ∂B(0,R) stands for the distribution defined as in Exercise 2.128 corresponding to Σ := ∂B(0, R). (3) If n = 2p, p ∈ N, the approach used to obtain (9.1.51) works up to the point where the general formula for Ip was obtained. More precisely, with ρ > 0 fixed, if we define π n eiρr cos θ (sin θ)n dθ, ∀ n ≥ 2, (9.1.58) Jn := ρ 0

then the recurrence formula Jn = − n−1 r ∂r Jn−2 is valid for all n ≥ 2 (observe π that Ip = J2p−1 ). However, the integral J0 = 0 eiρr cos θ dθ cannot be computed explicitly (as opposed to the computation of I1 ), hence the recurrence formula for Jn may not be used inductively to obtain an explicit expression for Jn . This is why, in order to obtain explicit expressions for E± when n is even we resort to a proof different than the one used when n is odd.


299

Lemma 9.3. If h : [0, ∞) → R is continuous and bounded, then for each t ∈ (0, ∞) we have ∞ ε ε − lim h(r) dr = πh(t). (9.1.59) ε2 + (t − r)2 ε2 + (t + r)2 ε→0+ 0 Proof. If t ≥ 0 is fixed, then via suitable changes of variables we obtain ∞ ε ε − lim+ h(r) dr ε2 + (t − r)2 ε2 + (t + r)2 ε→0 0 6 = lim+ ε→0

∞

= −∞

∞

− εt

h(t + ελ) dλ − 1 + λ2

∞ t ε

7 h(−t + ελ) dλ 1 + λ2

(9.1.60)

h(t) dλ = πh(t), 1 + λ2

where for the second to the last equality in (9.1.60) we applied Lebesgue’s dominated convergence theorem. The Method of Descent In order to treat the case when n is even, we use a procedure called the method of descent. The ultimate goal is to use the method of descent to deduce from a fundamental solution for the wave operator in dimension n + 1 with n even, a fundamental solution for the wave operator in dimension n. To set the stage we make the following definition. Definition 9.4. A sequence {ψj }j∈N ⊂ C0∞ (R) is said to converge in a dominated fashion to 1 if the following two conditions are satisfied: (i) for each compact subset K of R there exists j0 = j0 (K) ∈ N such that ψj (x) = 1 for all x ∈ K if j ≥ j0 ; (ii) for each q ∈ N0 , the sequence {ψj }∞ j∈N is uniformly bounded on R. (q)

An example of a sequence {ψj }j∈N converging in a dominated fashion to 1 is given by x ψj (x) := ψ , ∀ x ∈ R, ∀ j ∈ N, (9.1.61) j where ψ ∈ C0∞ (R)

satisfied ψ(x) = 1 whenever |x| ≤ 1.

(9.1.62)

In what follows we will use the notation x = (x , xn ) ∈ Rn , x ∈ Rn−1 , xn ∈ R,

∂ = (∂ , ∂n ), ∂ = (∂1 , . . . , ∂n−1 ). (9.1.63)


300

Definition 9.5. A distribution u ∈ D (Rn ) is called integrable with respect ∞ {ψj }j∈N to xn if for any ϕ ∈ C0∞ (Rn−1 ) and any sequence ⊂ C0 (R) converging in a dominated fashion to 1, the sequence u, ϕ ⊗ ψj j∈N is convergent and its limit does not depend on the selection of the sequence {ψj }j∈N . Suppose u ∈ D (Rn ) is integrable with respect to xn and consider a sequence {ψj }j∈N ⊂ C0∞ (R) that converges in a dominated fashion to 1. For each j ∈ N define the linear mapping uj : D(Rn−1 ) → C,

uj (ϕ) := u, ϕ ⊗ ψj ,

Then uj ∈ D (Rn−1 ) for every j ∈ N and

∀ ϕ ∈ C0∞ (Rn−1 ).

(9.1.64)

lim uj − uk , ϕ = 0 for every

j,k→∞

function ϕ ∈ C0∞ (Rn−1 ), thus the sequence {uj }j∈N is Cauchy in D (Rn−1 ) (see the definition after (13.1.27)). Since D (Rn−1 ) is sequentially complete (recall Fact 2.18), it follows that there exists some u0 ∈ D (Rn−1 ) with the property that D (Rn−1 )

and u0 , ϕ = lim uj , ϕ ∀ ϕ ∈ C0∞ (Rn−1 ).

uj −−−−−−→ u0 j→∞

(9.1.65)

j→∞

Moreover, u0 is independent of the choice of the sequence {ψj }j∈N converging in a dominated fashion to 1 (prove this as an exercise). The distribution u0 will

∞ be called the integral of u with respect to xn and will be denoted by −∞ u dxn . The reason for using this terminology and notation for u0 is evident from the next proposition. We denote by Ln−1 the Lebesgue measure in Rn−1 . Proposition 9.6. If f ∈ L1loc (Rn ) is a function with the property that R

|f (x , xn )| dxn < ∞ f ∈ L (K × R) 1

and

for Ln−1 -almost every x ∈ Rn−1 for every compact set K ⊂ R

n−1

(9.1.66)

,

then the distribution uf ∈ D (Rn ) determined by f (recall

∞ (2.1.6)) is integrable with respect to xn . In addition, the distribution u0 := −∞ uf dxn is of function type and is given by the function g(x ) :=

R

f (x , xn ) dxn

defined for Ln−1 -almost every x ∈ Rn−1 . (9.1.67)

Proof. Since f ∈ L1loc (Rn ) we have that uf ∈ D (Rn ) as in (2.1.6) is welldefined. Fix ϕ ∈ C0∞ (Rn−1 ). Since f is absolutely integrable on supp ϕ × R, if {ψj }j∈N ⊂ C0∞ (R) is a sequence converging in a dominated fashion to 1, we may apply Fubini’s theorem and then Lebesgue’s dominated convergence theorem to write

9.1. FUNDAMENTAL SOLUTION FOR THE WAVE OPERATOR $ % lim uf , ϕ ⊗ ψj = lim

j→∞

ϕ(x )

j→∞

Rn−1

ϕ(x )

= Rn−1

R

R

301

f (x , xn )ψj (xn ) dxn dx

f (x , xn ) dxn dx ,

(9.1.68)

and the desired conclusion follows. The proposition that is the engine in the method of descent is proved next. Proposition 9.7. Let m ∈ N0 and let P (∂) = P (∂ , ∂n ) be a constant coefficient, linear operator of order m in Rn . Define the differential operator P0 (∂ ) := P (∂ , 0) in Rn−1 . If f ∈ D (Rn−1 ) and u ∈ D (Rn ) are such that (9.1.69) P (∂)u = f (x ) ⊗ δ(xn ) in D (Rn )

∞ and u is integrable with respect to xn , then u0 := −∞ u dxn is a solution of the equation P0 (∂ )u0 = f in D (Rn−1 ). Proof. Fix a sequence {ψj }j∈N ⊂ C0∞ (R) that converges in a dominated fashion to 1 and let ϕ ∈ C0∞ (Rn−1 ). Using the definition of u0 we may write $ % $ $ % % P0 (∂ )u0 , ϕ = u0 , P0 (−∂ )ϕ = lim

j→∞

! = lim

j→∞

u, (P0 (−∂ )ϕ) ⊗ ψj

" !

(9.1.70)

u, P (−∂)(ϕ ⊗ ψj ) + u, (P0 (−∂ )ϕ) ⊗ ψj −P (−∂)(ϕ ⊗ ψj )

" .

We claim that ! lim

j→∞

" u , (P0 (−∂ )ϕ) ⊗ ψj − P (−∂)(ϕ ⊗ ψj ) = 0.

(9.1.71)

Assume the claim for now. Then returning to (9.1.70) we have $ $ $ % % % P0 (∂ )u0 , ϕ = lim u, P (−∂)(ϕ ⊗ ψj ) = lim P (∂)u , ϕ ⊗ ψj j→∞

j→∞

$ % $ % = lim f ⊗ δ, ϕ ⊗ ψj = lim f, ϕ ψj (0) = f, ϕ, j→∞

j→∞

(9.1.72)

where for the last equality in (9.1.72) we used property (i) in Definition 9.4 with K = {0}. Hence, the desired conclusion follows. To prove (9.1.71), observe that m (q) P0 (−∂ )ϕ ⊗ ψj − P (−∂)(ϕ ⊗ ψj ) = Pq (∂ )ϕ ⊗ ψj q=1

where Pq is a differential operator of order ≤ m−q. Then for each q ∈ {1, . . . m}, (q) the sequence ψj + ψj j∈N also converges in a dominated fashion to 1, which combined with the fact that u is integrable with respect to xn , further yields


302 $ (q) % lim u, Pq (∂ )ϕ ⊗ ψj

j→∞

$ $ % (q) % = lim u, Pq (∂ )ϕ ⊗ (ψj + ψj ) − lim u, Pq (∂ )ϕ ⊗ ψj j→∞

$ % $ % = u0 , Pq (∂ )ϕ − u0 , Pq (∂ )ϕ = 0,

j→∞

(9.1.73)

proving (9.1.71). The proof of the proposition is now complete.

The Case n = 2p, p ≥ 1 We are now ready to proceed with determining a fundamental solution for the wave operator in the case when n = 2p, p ≥ 1. The main idea is to 2p+1 2 ∂j being the wave operator in Rn+2 , use Proposition 9.7 with P := ∂t2 − j=1

f := δ(x1 , . . . , x2p ) ⊗ δ(t), and u equal to E2p+1 , the fundamental solution from 2p (9.1.51). Note that under these assumptions we have P0 := ∂t2 − ∂j2 which is j=1

integrable with respect to x2p+1 , the wave operator in Rn+1 . Thus, if E2p+1 is

∞ then by Proposition 9.7 it follows that u := −∞ E2p+1 dx2p+1 satisfies ⎛ ⎝∂t2 −

2p

⎞ ∂j2 ⎠ u = δ(x1 , . . . , x2p ) ⊗ δ(t)

in D (R2p+1 ).

(9.1.74)

j=1

Therefore, u is a fundamental solution for the wave operator corresponding to n = 2p. Let us first show that the distribution E2p+1 given by the formula in (9.1.51) is integrable with respect to x2p+1 . Fix an arbitrary function ϕ ∈ C0∞ (R2p+1 ) and let {ψj }j∈N ⊂ C0∞ (R) be a sequence that converges in a dominated fashion to 1. Then, using the notation x = (x , x2p+1 ) ∈ R2p × R, we have lim

j→∞

E2p+1 , ϕ ⊗ ψj

= lim 2−(p+1) π −p j→∞

0

∞

⎡ ⎛ p−1 ⎢ 1 ⎜1 ∂r ⎣ ⎝ r r

⎡ ⎛ p−1 ∞ ⎢ 1 ⎜1 −(p+1) −p =2 π ∂r ⎣ ⎝ r r 0

⎞⎤

⎟⎥ ϕ(x , t)ψj (x2p+1 ) dσ(x)⎠⎦

x∈R2p+1 , |x|=r

x∈R2p+1 , |x|=r

r=t

⎞⎤ ⎟⎥ ϕ(x , t) dσ(x)⎠ ⎦

dt

dt,

(9.1.75)

r=t

where for the last equality in (9.1.75) we used Lebesgue’s dominated convergence theorem (here we note that the second equality in (9.1.47) and the properties satisfied by {ψj }j∈N play an important role). With (9.1.75) in hand, we may conclude that E2p+1 is integrable with respect to x2p+1 . Consequently, if one

9.1. FUNDAMENTAL SOLUTION FOR THE WAVE OPERATOR sets E2p :=

∞ −∞

303

E2p+1 dx2p+1 , then E2p ∈ D (R2p+1 ) and

E2p , ϕ = 2−(p+1) π −p

∞ 0

⎡ ⎛ p−1 ⎢ 1 ⎜1 ∂r ⎣ ⎝ r

⎞⎤

⎟⎥

ϕ(x , t) dσ(x)⎠⎦

r

x∈R2p+1 , |(x ,x

2p+1 )|=r

dt

r=t

(9.1.76)

for every ϕ ∈ C0∞ (R2p+1 ). In addition, by Proposition 9.7, it follows that E2p is a fundamental solution for the wave operator in Rn+1 . Note that the function ϕ appearing under the second integral in (9.1.76) does not depend on the variable x2p+1 . Hence, it is natural to proceed further in order to rewrite (9.1.76) in a form that does not involve the variable x2p+1 . Fix r > 0 and denote by B(0, r) the ball in Rn of radius r and centered at 0 ∈ Rn . Define the mappings P± : B(0, r) → R2p+1 by setting P± (x ) := (x1 , x2 , . . . , x2p , ±

r2 − |x |2 ), ∀ x = (x1 , . . . , x2p ) ∈ B(0, r). (9.1.77)

Then P+ and P− are parametrizations of the (open) upper and lower, respectively, hemispheres of the surface ball in R2p+1 of radius r and centered at 0. Hence (keeping in mind Definition 13.38 and Definition 13.39), we may write ϕ(x , t) dσ(x) = ϕ(x , t) |∂1 P+ × · · · × ∂2p P+ | dx B(0,r)

x∈R2p+1 , |(x ,x2p+1 )|=r

ϕ(x , t) |∂1 P− × · · · × ∂2p P− | dx .

+ B(0,r)

(9.1.78) A direct computation based on (9.1.77) and (13.6.4) further yields 1 0 . . . 0 ∓ √ x1 r 2 −|x |2 0 1 . . . 0 ∓ √ 2x2 2 r −|x | ∂1 P± × · · · × ∂2p P± = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . =: det A± , 0 0 . . . 1 ∓ √ x2p 2 −|x |2 r e1 e2 . . . e2p e2p+1

(9.1.79)

where ej is the unit vector in R2p+1 with 1 on the jth position, for each j ∈ {1, . . . , 2p + 1}. Hence, the components of the vector ∂1 P± × · · · × ∂2p P± are (−1)k+1 det A± k ∈ {1, . . . , 2p, 2p + 1}, (9.1.80) k, ± where A± by deleting column k and k is the 2p × 2p matrix obtained from A row 2p + 1. It is easy to see from (9.1.79) that

det A± 2p+1 = 1,

k det A± k = (−1)

∓xk r2

− |x |2

,

∀ k ∈ {1, . . . , 2p}.

(9.1.81)


304 Consequently,

r

|∂1 P± × · · · × ∂2p P± | =

r2 − |x |2

.

(9.1.82)

Formula (9.1.82) combined with (9.1.78) and (9.1.76) gives ∞ 6 p−1 1 r 1 −p E2p , ϕ = (2π)

r

0

= (2π)

−p

∞

6

1 r

0

∂r

r

p−1

ϕ(x , r)

x ∈B(0,r)

B(0,r)

r 2 − |x |2

dx

r 2 − |x |2

dx

dt r=t

7

ϕ(x , t)

∂r

7

dt.

(9.1.83)

r=t

In summary, we proved: If n = 2p, for some p ∈ N, then E+ ∈ S (Rn+1 ), defined by −p

E+ , ϕ = (2π)

0

∞

6

1 ∂r r

p−1 x∈Rn , |x| 2m, and are homogeneous polynomials of degree 2m − n when n ≤ 2m. Moreover, there exists a function Φ ∈ MM×M C ∞ (Rn \ {0}) that is positive homogeneous of degree 2m − n such that (11.3.24) E(x) = Φ(x) + ln |x| P(x), ∀ x ∈ Rn \ {0}. 4. For each β ∈ Nn0 with |β| ≥ 2m − 1, the restriction to Rn \ {0} of the matrix distribution ∂ β E is of class C ∞ and positive homogeneous of degree 2m − n − |β|. 5. For each β ∈ Nn0 there exists Cβ ∈ (0, ∞) such that the estimate

β

|∂ E(x)| ≤

⎧ Cβ ⎪ ⎪ ⎪ ⎨ |x|n−2m+|β|

if either n is odd, or n > 2m, or if |β| > 2m−n,

⎪ ⎪ ⎪ ⎩ Cβ (1 + | ln |x||) |x|n−2m+|β|

if 0 ≤ |β| ≤ 2m − n, (11.3.25)

holds for each x ∈ Rn \ {0}.

348

CHAPTER 11. MORE ON FUNDAMENTAL SOLUTIONS . . .

3 (with the “hat” denoting the 6. When restricted to Rn \ {0}, the entries of E ∞ Fourier transform) are C functions and, moreover, −1 3 E(ξ) = (−1)m L(ξ) for each ξ ∈ Rn \ {0}. (11.3.26) 7. Writing EL in place of E to emphasize the dependence on L, the fundamental solution EL with entries as in (11.3.19)–(11.3.20) satisfies ∗ EL = EL∗ , EL = EL , EL = EL , (11.3.27) and EλL = λ−1 EL ∀ λ ∈ C \ {0}. 8. Any fundamental solution U ∈ MM×M S (Rn ) of the system L in Rn is of the form U = E + Q where E is as in (11.3.19)–(11.3.20) and Q is an M × M matrix with entries that are polynomials in Rn and with columns, Qk , k ∈ {1, . . . , M }, that satisfy the pointwise equations LQk = 0 ∈ CM in Rn for each k ∈ {1, . . . , M }. Proof. To facilitate the subsequent discussion, denote by P jk (ξ) 1≤j,k≤M the inverse of the characteristic matrix L(ξ), that is, −1 P jk (ξ) := L(ξ) , ∀ ξ ∈ Rn \ {0}. (11.3.28) 1≤j,k≤M

Then the entries Ejk 1≤j,k≤M of the matrix E are given at each point x in Rn \ {0} by 1 (n−1)/2 Ejk (x) = Δ (x·ξ)2m−1 sgn (x·ξ)P jk (ξ) dσ(ξ) 4(2π i)n−1 (2m − 1)! x S n−1 (11.3.29) if n is odd, and −1 n/2 Δ (x · ξ)2m ln |x · ξ|P jk (ξ) dσ(ξ) (11.3.30) Ejk (x) = (2π i)n (2m)! x S n−1 if n is even. The proof of the theorem is completed in eight steps. Step I. We claim that if we set q := 0 if n is even and q := 1 if n is odd, then x · ξ −1 (n+q)/2 2m+q Ejk (x) = Δ P jk (ξ) dσ(ξ) (x·ξ) log (2π i)n (2m+q)! x i S n−1 (11.3.31) for x ∈ Rn \ {0} and j, k ∈ {1, . . . , M }, where log denotes the principal branch of the complex logarithm defined for points z ∈ C \ {x : x ∈ R, x ≤ 0}. Suppose first q = 0, and start from (11.3.30), then use that n is even, πthus i the formula log x·ξ sgn (x·ξ) for the term under the integral sign = ln |x·ξ|− i 2

11.3. FUNDAMENTAL SOLUTIONS FOR HIGHER-ORDER...

349

and the fact that the integral over S n−1 of the function (x · ξ)2m sgn (x · ξ)P jk (ξ) (which is odd in ξ) is zero, to obtain that (11.3.31) holds. Moving on to the case when n is odd, consider the function t2m+q log (t/i) if t ∈ R \ {0}, F (t) := (11.3.32) 0 if t = 0. It is not difficult to see that F ∈ C ∞ (R \ {0}) ∩ C 2m+q−1 (R),

if k = 1, . . . , 2m + q − 1, (11.3.33) and that for each k ∈ {1, . . . , 2m} there exists a constant Cm,q,k such that d k dt

F (t) =

F (k) (0) = 0

(2m + q)! 2m+q−k t t log + Cm,q,k t2m+q−k (2m + q − k)! i

in R \ {0}.

(11.3.34) In particular, by the chain rule, if β ∈ Nn0 is such that |β| ≤ 2m + q − 1, then ∂xβ [F (x · ξ)] = F (|β|) (x · ξ)ξ β , ∀ x ∈ Rn , ∀ ξ ∈ S n−1 . (11.3.35) Based on (11.3.35) and (11.3.34) for each x ∈ Rn \ {0} we may write x · ξ Δx P jk (ξ) dσ(ξ) (x · ξ)2m+1 log i S n−1 jk F (x · ξ)P (ξ) dσ(ξ) = F (x · ξ)P jk (ξ) dσ(ξ) = Δx S n−1

S n−1

= 2m(2m + 1) S n−1

+ Cm,q

S n−1

(x · ξ)2m−1 log

x · ξ i

P jk (ξ) dσ(ξ)

(x · ξ)2m−1 P jk (ξ) dσ(ξ)

= 2m(2m+1) S n−1

(x · ξ)2m−1 ln |x · ξ|− πi sgn(x · ξ) P jk (ξ) dσ(ξ) 2

= −πim(2m + 1)

S n−1

(x · ξ)2m−1 sgn(x · ξ)P jk (ξ) dσ(ξ). (11.3.36)

Hence, if one starts with the expression in the right-hand side of (11.3.31), then one transfers Δ under the integral sign using from (11.3.36), one arrives at the expression of Ejk from (11.3.29). This completes the proof of Step I. Step II. Proof of the fact that the entries of E are C ∞ and even in Rn \ {0}. That the functions in (11.3.29) and (11.3.30) are even is immediate from their respective expressions. To show that the components of E belong to C ∞ (Rn \ {0}), for each ∈ {1, . . . , n} let e be the unit vector in Rn with one on the th component, and consider the open set O := Rn \ {λ e : λ ≤ 0} = x = (x1 , . . . , xn ) ∈ Rn : x > 0 . (11.3.37)


350

Then for each given x ∈ O define the linear map R,x : Rn → Rn by R,x (ξ) := ξ −

(ξ · x)(|x| + 2x ) − ξ |x|2 ξ |x| + ξ · x + e , |x|(|x| + x ) |x|(|x| + x )

By Exercise 4.125 (with ζ = transformation and

x |x|

x · R,x (ξ) = |x|ξ ,

ξ ∈ Rn .

(11.3.38) and η := e ) we have that this is a unitary ∀ x ∈ O ,

∀ ξ ∈ Rn .

(11.3.39)

and λ > 0

(11.3.40)

is of class C ∞ .

(11.3.41)

Also, R,λ x = R,x

whenever x ∈ O

and the joint application O × Rn (x, ξ) → R,x (ξ) ∈ Rn

Fix j, k ∈ {1, . . . , M }. Using the invariance under unitary transformations of the operation of integration over S n−1 , for each x ∈ O we may then write x · ξ (x · ξ)2m+q · log P jk (ξ) dσ(ξ) i S n−1 x · R (ξ) ,x = (x · R,x (ξ))2m+q · log P jk (R,x (ξ)) dσ(ξ) i S n−1 |x|ξ (|x|ξ )2m+q · log P jk (R,x (ξ)) dσ(ξ) = i n−1 S ξ B 2m+q 2m+q = |x| ξ ln |x| + log P jk (R,x (ξ)) dσ(ξ). i n−1 S (11.3.42) From this representation, (11.3.41), and (11.3.31), it is clear that we have Ejk ∈ C ∞ (O ). To complete the proof of Step II, there remains to observe n O . that Rn \ {0} = =1

Step III. Proof of part (3) in the statement of the theorem. To facilitate the discussion, fix j, k ∈ {1, . . . , M }, introduce Qjk (x) := (x · ξ)2m+q P jk (ξ) dσ(ξ), ∀ x ∈ Rn ,

(11.3.43)

S n−1

and define Ψjk : Rn \ {0} → C by setting x · ξ (x · ξ)2m+q · log P jk (ξ) dσ(ξ) − ln |x| Qjk (x) Ψjk (x) := i S n−1 (11.3.44)


351

for each Rn \ {0}. Observe that Qjk is a polynomial of degree 2m + q that vanishes when n is odd [since in that case the integrand in (11.3.43) is odd]. Also, from our earlier analysis in (11.3.42), we know that the integral in the right-hand side of (11.3.44) depends in a C ∞ fashion on the variable x ∈ Rn \ {0}. These comments and (11.3.44) imply that Ψjk is of class C ∞ in Rn \ {0}. Our next goal is to prove that Ψjk

is positive homogeneous of degree 2m + q in Rn \ {0}. (11.3.45)

To this end, first note that for each ∈ {1, . . . , n} we may write 2m+q Qjk (x) = |x| ξ2m+q P jk (R,x (ξ)) dσ(ξ), ∀ x ∈ O , (11.3.46) S n−1

by (11.3.43) and (11.3.39). Consequently, from (11.3.46), (11.3.42), and (11.3.44), we deduce that for each ∈ {1, . . . , n}, ξ 2m+q Ψjk (x) := |x| P jk (R,x (ξ)) dσ(ξ), ∀ x ∈ O . ξ2m+q log i n−1 S In turn, this and (11.3.40) readily show that Ψjk is positive homogeneous of degree 2m + q when restricted to the cone-like region O . Since Rn \ n {0} = O , the claim in (11.3.45) follows. =1

Next, an induction argument shows that for each ξ ∈ Rn \{0} and k, N ∈ N satisfying N ≥ 2k, the following formulas hold: Δkx (x · ξ)N =

N! (x · ξ)N −2k |ξ|2k (N − 2k)!

(11.3.47)

and Δkx (x · ξ)N ln |x| = (ln |x|)Δkx (x · ξ)N +

k

(11.3.48)

c(r, k, N, n)|x|−2r (x · ξ)N −2k+2r .

r=1

We now observe that if Pjk is the (j, k)-entry in the matrix P from (11.3.23), then −1 (n+q)/2 Δ (x · ξ)2m+q P jk (ξ) dσ(ξ), Pjk (x) = (2π i)n (2m + q)! x S n−1 (11.3.49) as seen from identity (11.3.47) with N := 2m + q and k := (n + q)/2. It is also immediate from (11.3.49) that Pjk is identically zero when either n is odd (due to parity considerations) or n > 2m (due to degree considerations). In addition, formula (11.3.49) shows that Pjk is a homogeneous


352

polynomial of degree 2m − n whenever n ≤ 2m. Finally, we note that combining (11.3.43), (11.3.49), and (11.3.48) used with N := 2m + q and k := (n + q)/2, yields −1 Δ(n+q)/2 (ln |x|)Qjk (x) = (ln |x|)Pjk (x) (11.3.50) (2π i)n (2m + q)! x (n+q)/2 −2r + Cr |x| (x · ξ)2m−n+2r P jk (ξ) dσ(ξ), ∀ x ∈ Rn \ {0}, S n−1

r=1

for some constants Cr depending only on r, n, q, and m. It is easy to see that the sum in the right-hand side of (11.3.50) gives rise to a function that belongs to C ∞ (Rn \ {0}) and is positive homogeneous of degree 2m − n. At this point, if we define Φjk (x) :=

−1 Δ(n+q)/2 Ψjk (x) (2π i)n (2m + q)! x

(n+q)/2

+

Cr |x|−2r

(11.3.51)

r=1

S n−1

(x · ξ)2m−n+2r P jk (ξ) dσ(ξ),

∀ x ∈ Rn \ {0},

then (11.3.24) follows from (11.3.31), (11.3.44), (11.3.50), and (11.3.51). Moreover, from (11.3.45) and (11.3.51) it is clear that Φjk is positive homogeneous of degree 2m − n, while the regularity of Ψjk established earlier entails Φjk ∈ C ∞ (Rn \ {0}). Step IV. Proof of the fact that E ∈ MM×M (S (Rn )). Fix j, k ∈ {1, . . . , M } and recall (11.3.24). By what we proved in Step III, Φjk is positive homogeneous of degree 2m − n in Rn \ {0}, and since m ≥ 1 we may invoke Exercise 4.51 to conclude that Φjk ∈ L1loc (Rn ). Now the estimate from Exercise 4.51 and Example 4.3 give Φjk ∈ S (Rn ). In addition, by Exercise 4.104, it follows that (ln |x|)Pjk ∈ S (Rn ). In summary, we have Ejk ∈ S (Rn ). Step V. Proof of (11.3.22). Componentwise, (11.3.22) reads as follows: for each j, k ∈ {1, . . . , M }, M

Lxjr

Erk (x − y) =

r=1

0

if j = k,

δy (x)

if j = k,

in S (Rn ),

(11.3.52)

where the superscript x denotes the fact that the operator Ljr (defined as in (11.3.1)) is applied in the variable x. To justify this, fix some numbers j, k, r ∈ {1, . . . , M }. By (11.3.34) and (11.3.35) we have Lxjr F (x · ξ) =

(2m + q)! q!

(x · ξ)q log

x·ξ + Cm,q (x · ξ)q Ljr (ξ) (11.3.53) i


353

for every x ∈ Rn \ {0} and every ξ ∈ S n−1 such that x·ξ = 0. To continue, fix ϕ ∈ C0∞ (Rn ) and write !

Lxjr

S n−1

" F (x · ξ)P rk (ξ) dσ(ξ) , ϕ(x)

!

= S n−1

" F (x · ξ)P rk (ξ) dσ(ξ), Ljr ϕ(x)

F (x · ξ)P rk (ξ) dσ(ξ)Ljr ϕ(x) dx

= Rn

S n−1

=

P

rk

(ξ)

S n−1

x∈Rn , x·ξ>0

F (x · ξ)Ljr ϕ(x) dx dσ(ξ)

P rk (ξ)

+ S n−1

x∈Rn , x·ξ 0 we have ψj (x) = −1 θ(y) dy = 1 if j ≥ j0 , where j0 ∈ N is such that j0 x−j02 ≤ −1 and j0 x−1 ≥ 1. This shows that

1 ψj converges pointwise to H as j → ∞. In addition, |ψj (x)| ≤ −1 |θ(y)| dy < ∞ for every j ∈ N. Hence, by applying Lebesgue’s dominated convergence theorem, lim ψj (x)ϕ(x) dx = H(x)ϕ(x) dx for each ϕ ∈ C0∞ (R). j→∞

R

R

D (R)

This proves that ψj −−−− → H as wanted. j→∞

Exercise 2.113 If ϕ ∈ C0∞ (R) then the support condition for ϕ guarantees ∞ that only finitely many terms in the sum ϕ(j) (j) are nonzero, hence u is j=1

well-defined. Clearly u is linear. If K is a compact set in R and k ∈ N is such that K ⊆ [−k, k], then |u(ϕ)| ≤

k (j) ϕ (j) ≤ k j=1

sup

x∈K, j≤k

(j) ϕ (x)

for ϕ ∈ C0∞ (R) with supp ϕ ⊆ K.

This proves that u ∈ D (R). Suppose the distribution u has finite order. Then there exists a nonnegative integer k0 with the property that for each compact set K ⊂ R there is a finite constant CK ≥ 0 such that |u, ϕ| ≤ CK sup |ϕ(j) (x)| for every ϕ ∈ C0∞ (R) x∈K, j≤k0

with supp ϕ ⊆ K. In particular, from this and the definition of u it follows that there exists that whenever ϕ ∈ C0∞ (R) satisfies the condition C ∈1 [0, ∞) 3such supp ϕ ⊆ k0 + 2 , k0 + 2 we have |ϕ(k0 +1) (k0 + 1)| = |u, ϕ| ≤ C

sup

|ϕ() (x)|.

(12.2.8)

x∈ k0 + 12 ,k0 + 13 , ≤k0

Now consider θ ∈ C0∞ (−1/2, 1/2) satisfying θ(0) = 1 and construct the sequence of smooth functions ϕj (x) := θ(jx − j(k0 + 1)) for x ∈ R and j ∈ N. Then, for each j ∈ N, we have 1 3 supp ϕj ⊆ k0 + , k0 + , 2 2 ()

ϕj = j θ() (j · −j(k0 + 1)), (k0 +1)

and ϕj

(k0 + 1) = j k0 +1 .

∀ ∈ N0 ,

12.2. SOLUTIONS TO EXERCISES FROM SECT. 2.10

385

Combining all these facts with (12.2.8) it follows that for each j ∈ N, j k0 +1 ≤ C

j |θ() (x)|

sup

x∈ − 12 , 12 , ≤k0

≤ C max θ() L∞ ([−1/2,1/2]) : ≤ k0 j k0 .

(12.2.9)

Choosing now j sufficiently large in (12.2.9) leads to a contradiction. Hence u does not have finite order. Exercise 2.114 Note that for each j ∈ N we have fj ∈ L1loc (Rn ). Take some ϕ ∈ C0∞ (Rn ) and suppose R > 0 is such that supp ϕ ⊂ B(0, R). Then ϕ(x) − ϕ(0) 1 1 ϕ(0) dx + (12.2.10) fj , ϕ = 1 1 dx. n− n− j B(0,R) |x| j j j B(0,R) |x| Using the mean value theorem and then (13.8.6) we may further write 1 ϕ(x) − ϕ(0) ∇ϕL∞ (Rn ) 1 dx ≤ dx n−1− 1j j B(0,R) |x|n− 1j j B(0,R) |x| =

ωn−1 ∇ϕL∞ (Rn ) j

1

R

1

ρ j dρ = 0

ωn−1 ∇ϕL∞ (Rn ) R j +1 −−−→ 0. j→∞ j+1 (12.2.11)

One more use of (13.8.6) implies 1 j

B(0,R)

1 |x|

n− 1j

dx =

ωn−1 j

R

1

ρ j −1 dρ

0 1

= ωn−1 R j −−−→ ωn−1 . j→∞

(12.2.12)

By combining (12.2.10)–(12.2.12) we obtain lim fj , ϕ = ωn−1 ϕ(0) = ωn−1 δ, ϕ . j→∞

The desired conclusion now follows. Exercise 2.115 Note that fε ∈ L1loc (R) for every ε > 0, hence fε ∈ D (R). Also, lim fε (x) = 0 for every x ∈ R \ {0}. For a given function ϕ ∈ C0∞ (R) let ε→0+

R > 0 be such that supp ϕ ⊂ (−R, R) and for ε > 0 write

R

ϕ(0)ε xε ϕ(x) − ϕ(0) dx + · x2 + ε2 x π

R

dx =: I + II. 2 + ε2 x −R −R (12.2.13) xε Then x2 +ε2 ≤ 1/2 for each x ∈ R if ε > 0, so we may apply Lebesgue’s dominated convergence theorem to conclude that lim I = 0. Also, integrating the 1 fε , ϕ = π

ε→0+

CHAPTER 12. SOLUTIONS TO SELECTED EXERCISES

386

second term and then taking the limit yields lim+ II = ε→0

2ϕ(0) π

ϕ(0). In conclusion, lim+ fε , ϕ = ϕ(0) = δ, ϕ as desired.

lim arctan(R/ε) =

ε→0+

ε→0

∞ n Exercise √ 2.116 Let ϕ ∈ C0 (R ) and ε > 0. Using first the change of variables x = 2 εy and then Lebesgue’s dominated convergence theorem, we have |x|2 n fε , ϕ = (4πε)− 2 e− 4ε ϕ(x) dx (12.2.14)

=π

−n 2

Rn

e

−|y|2

√ n ϕ(2 εy) dy −−−−→ π− 2 +

ε→0

Rn

2

e−|y| ϕ(0) dy = ϕ(0). Rn

Exercise 2.117 Note that for every ε > 0 we have |fε± | ≤ 1ε , hence these are functions in L1loc (R) ⊂ D (R). Let ϕ ∈ C0∞ (R) and let R > 0 be such that supp ϕ ⊂ (−R, R). Then fε± (x), ϕ Since

x x2 +ε2

R

= ϕ(0) −R

x ∓ iε dx + x2 + ε2

R

−R

x ∓ iε [ϕ(x) − ϕ(0)] dx =: I + II. x2 + ε2 (12.2.15)

is odd, we further obtain

I = ∓2iϕ(0) arctan (R/ε) −−−−→ ∓iπϕ(0) = ∓iπδ, ϕ. ε→0+

x∓iε 2 2 ε→0+ x +ε

As for II, since lim

=

1 x

(12.2.16)

for every x = 0 and

x ∓ iε |ϕ(x) − ϕ(0)| ∈ L1 (−R, R) , x2 + ε2 ϕ(x) − ϕ(0) ≤ |x| we may apply Lebesgue’s dominated convergence theorem to obtain R ! 1 " ϕ(x) − ϕ(0) dx = P.V. , ϕ . (12.2.17) lim II = x x ε→0+ −R Exercise 2.118 You may use the following outline:

(a) Show that for f ∈ L1 (R) one has lim R f (x) sin(jx) dx = 0 by reducing j→∞

to the case f ∈ C0∞ (R) based on density arguments. $ % (b) For ϕ ∈ C0∞ (R) and R > 0, write the expression π1 sinxjx , ϕ(x) as the sum of two integrals, one over the region {x ∈ R : |x| ≤ R}, the other over {x ∈ R : |x| > R}, and use (a) to obtain the desired conclusion. Recall that R sinx x dx = π.


387

Exercise 2.119 Let ϕ ∈ C0∞ (R).

√ (a) In the expression for fj , ϕ use the change of variables x = y/ j and D (R)

then Lebesgue’s dominated convergence theorem to conclude fj −−−−→ δ. j→∞

(b) Integrate by parts m + 1 times to conclude that |fj , ϕ| = j m cos(jx)ϕ(x) dx ≤ j −1 |ϕ(m+1) (x)| dx −−−→ 0, j→∞ R

R

D (R)

hence fj −−−−→ 0. j→∞

(c) Use a reasoning similar to the one in the proof of (a) to conclude that D (R)

fj −−−− → δ, this time via the change of variables x = y/j. j→∞

(d) Not convergent since if the function ϕ ∈ C0∞ (R) is such that ϕ(0) = 0, then uj , ϕ = (−1)j ϕ(1/j) and the sequence {(−1)j ϕ(1/j)}j∈N is not convergent. (e) Use the mean value theorem to obtain that uj , ϕ −−−→ ϕ (0) = −δ , ϕ. j→∞

D (R)

→ P.V. x1 . (f) fj −−−− j→∞

(g) Use a reasoning similar to the one in the proof of (a) to conclude that D (R)

fj −−−− → δ, this time via the change of variables x = y/j and recalling j→∞

(sin y)2 that R y2 dy = π. (h) Integrate by parts m + 1 times and then use Lebesgue’s dominated conD (R)

→ 0. vergence theorem to conclude that fj −−−− j→∞

(j) Integrate by parts twice to obtain ∞ uj , ϕ = iϕ(0) + i eijx ϕ (x) dx 0

1 1 = iϕ(0) − ϕ (0) − j j

∞

eijx ϕ (x) dx −−−→ iϕ(0) = iδ, ϕ.

0

j→∞

(12.2.18)

Exercise 2.120 (H(· − a)) = δa in D (R). Exercise 2.121 (uf ) = aδa + H(· − a) in D (R).


388

Exercise 2.122 Let ϕ ∈ C0∞ (R). Then integration by parts yields −

f (x)ϕ (x) dx =

0

−∞

R

=−

sin(x)ϕ (x) dx −

∞

sin(x)ϕ (x) dx

0

0

∞

cos(x)ϕ(x) dx + −∞

cos(x)ϕ(x) dx,

(12.2.19)

0

hence (uf ) = cos(x)H(x) − cos(x)H(−x) in D (R). Similarly,

f (x)ϕ (x) dx = 2ϕ(0) +

0

−∞

R

sin(x)ϕ(x) dx −

∞

sin(x)ϕ(x) dx, (12.2.20) 0

hence (uf ) = 2δ − sin(x)H(x) + sin(x)H(−x) in D (R). Exercise 2.123 Let a, b ∈ I be such that a < x0 < b. Since for each x ∈ [a, x0 ) x we have f (x) = f (a) + a f (t) dt, Lebesgue’s dominated convergence theorem

x gives that lim f (x) exists and equals f (a) + a 0 f (t) dt. A similar argument − x→x0

b proves that lim f (x) = f (b) − x0 f (t) dt. Note that f ∈ L1 [a, b] . Suppose x→x+ 0

now that ϕ ∈ C0∞ (I) satisfies supp ϕ ⊂ [c, d] for some c < x0 < d. Then for ε > 0 small enough we have (uf ) , ϕ = −uf , ϕ = −

d

f (t)ϕ (t) dt

c

x0 −ε

= −f (x0 − ε)ϕ(x0 − ε) +

f (t)ϕ(t) dt −

c

d

+ f (x0 + ε)ϕ(x0 + ε) +

f (t)ϕ(t) dt.

x0 +ε

f (t)ϕ (t) dt

x0 −ε

(12.2.21)

x0 +ε

Send ε → 0+ in (12.2.21) and observe that lim+ ε→0

x0 +ε x0 −ε

f (t)ϕ (t) dt = 0 by

Lebesgue’s dominated convergence theorem. The case when x0 is not in the interior of the support of ϕ is simpler, and can be handled via a direct integration by parts. Exercise 2.125 Use Exercise 2.123 and induction. Exercise 2.126 Use Exercise 2.123 and the fact that since {xk }k∈N has no accumulation point in I, for each R > 0 only finitely many terms of the sequence {xk }k∈N are contained in (−R, R). Exercise 2.127 Use Exercise 2.126.


389

Exercise 2.128 Clearly δΣ is well-defined and linear. Also, for each compact set K ⊂ Rn and each ϕ ∈ C0∞ (Rn ) satisfying supp ϕ ⊆ K we have |δΣ (ϕ)| ≤ σ(Σ ∩ K) sup |ϕ(x)|.

(12.2.22)

x∈K

This shows that δΣ ∈ D (Rn ) and has order zero. Also δΣ , ϕ = 0 if supp ϕ ∩ Σ = ∅, thus supp δΣ ⊆ Σ. To prove that Σ ⊆ supp δΣ , note that if x∗ ∈ Σ, then there exists a neighborhood U (x∗ ) of x∗ and a local parametrization P of class C 1 near x∗ as in (13.6.2)–(13.6.3). In particular, if u0 ∈ O is such that P (u0 ) = x∗ , then the vectors ∂1 P (u0 ), . . . , ∂n−1 P (u0 ), are linearly independent. Upon recalling the cross product from (13.6.4), this ensures that c0 := ∂1 P (u0 ) × · · · × ∂n−1 P (u0 ) = 0.

(12.2.23)

Since P is of class C 1 , it follows that ∂1 P (u) × · · · × ∂n−1 P (u) ≥ c0 /2 for ⊆ O of u0 in Rn−1 . Using each u belonging to some small open neighborhood O the fact that P : O → P (O) is a homeomorphism (see Proposition 13.37), it follows that there exists some open set V (x∗ ) in Rn contained in U (x∗ ) and = V (x∗ ) ∩ Σ. By further shrinking containing x∗ with the property that P (O) O if necessary, there is no loss of generality in assuming that V (x∗ ) is bounded. Now consider 0 < r1 < r2 < ∞ such that V (x∗ ) ⊆ B(x∗ , r1 ) ⊆ B(x∗ , r2 ). Pick a function ϕ ∈ C0∞ B(x∗ , r2 ) with ϕ ≥ 0, ϕ = 1 in a neighborhood of B(x∗ , r1 ) (see Proposition 13.26). Then ϕ(x) dσ(x) ≥ ϕ(x) dσ(x) (12.2.24) δΣ , ϕ = Σ

=

O

Σ∩V (x∗ )

> 0. ∂1 P (u) × · · · × ∂n−1 P (u) du ≥ c0 |O|

(12.2.25)

In a similar manner, for each function g satisfying g ∈ L∞ (K ∩ Σ) for any compact set K ⊆ Rn , we have gδΣ ∈ D (Rn ) and for each compact K one has |(gδΣ )(ϕ)| ≤ gL∞ (K) σ(Σ ∩ K) sup |ϕ(x)|, x∈K

∀ ϕ ∈ C0∞ (Rn ), supp ϕ ⊆ K.

Exercise 2.129 Use integration by parts (see Theorem 13.41) and Exercise 2.128. Exercise 2.130 Use the definition of distributional derivative, integration by parts (cf. Theorem 13.41), and Exercise 2.128. Exercise 2.131 Let ϕ ∈ C0∞ (Rn ) be such that supp ϕ ∩ ∂B(0, R) = ∅. In this scenario, we have |x|2ϕ−R2 ∈ C0∞ (Rn ), hence we may write ! u, ϕ = (|x|2 − R2 )u,

" ϕ = 0. |x|2 − R2


390

This proves that supp u ⊆ ∂B(0, R), thus u is compactly supported. Examples of distributions satisfying the given equation include δ∂B(0,R) and δx0 , for any x0 ∈ Rn with |x0 | = R. Exercise 2.132 Use Example 2.50. Exercise 2.133 The derivative of order m is equal to: (a) sgn(x) if m = 1 and 2δ (m−2) if m ≥ 2; (b) 2δ (m−1) ; (c)

n

δ (2j) − sin x H if m = 2n + 1, n ∈ N, and

j=0

n

δ (2j−1) + cos x H if m = 2n,

j=1

n ∈ N;

(d) (sin x H) = cos x H and use (c); (e) δ1 − δ−1 − 2xχ[−1,1] if m = 1, (δ1 ) − (δ−1 ) + 2δ1 + 2δ−1 + 2χ[−1,1] if m = 2, (m−1)

and δ1

(m−1)

− δ−1

(m−2)

+ 2δ1

(m−2)

+ 2δ−1

(m−3)

− 2δ1

(m−3)

+ 2δ−1

if m ≥ 3.

Exercise 2.134 For ϕ ∈ C0∞ (R2 ) use the change of variables u = x+y, v = x−y u−v and the reasoning in (1.1.6)–(1.1.8) with ψ(u, v) := ϕ u+v , for u ∈ [2, 4], 2 2 v ∈ [0, 2], to write % $ 2 [(∂12 ϕ)(x, y) − (∂22 ϕ)(x, y)] dx dy (∂1 − ∂22 )χA , ϕ = A

4

2

=2

∂u ∂v ψ(u, v) dv du 2

0

= 2[ψ(4, 2) − ψ(2, 2) − ψ(4, 0) + ψ(2, 0)] = 2[ϕ(3, 1) − ϕ(2, 0) − ϕ(2, 2) + ϕ(1, 1)] $ % = 2[δ(3,1) − δ(2,0) − δ(2,2) + δ(1,1) ], ϕ .

(12.2.26)

Exercise 2.135 Fix ϕ ∈ C0∞ (R2 ). Then, using the change of variables x = t+ y we obtain $ % ∂1 (uf ), ϕ = − χ[0,1] (x − y)∂1 ϕ(x, y) dx dy R2

=− = R

R2

χ[0,1] (t)(∂1 ϕ)(t + y, y) dt dy

[ϕ(y, y) − ϕ(1 + y, y)] dy.

(12.2.27)


391

Similarly,

$ % ∂2 (uf ), ϕ =

R

[ϕ(x, x − 1) − ϕ(x, x)] dx.

(12.2.28)

Combining (12.2.27)–(12.2.28) it follows that ∂1 (uf ) − ∂2 (uf ) = 0 in D (R2 ) and, in turn, that ∂12 (uf ) − ∂22 (uf ) = (∂1 + ∂2 ) ∂1 (uf ) − ∂2 (uf ) = 0 in D (R2 ). Exercise 2.138 The uniqueness statement is a consequence of Proposition 2.45. Let K be a compact set in Rn such that K ⊂ Ω. Refine {Ωj }j∈I to a finite subcover {Ωk }N k=1 , with k ∈ I for k = 1, . . . , N , of K. Consider a partition of unity {ϕk }1≤k≤N subordinate to the cover {Ωk }N k=1 of K, as given by Theorem 13.29. Hence, for each k ∈ N we have ϕk ∈ C0∞ (Ω) satisfies supp ϕk ⊂ Ωk and N ϕk = 1 in a neighborhood of K. Next, for each function ϕ ∈ C0∞ (Ω) with k=1

supp ϕ ⊆ K define uK (ϕ) :=

N

uk , ϕk ϕ. Show that

k=1

N

uk , ϕk ϕ is indepen-

k=1

dent of the cover of K chosen and of the partition of unity, thus uK : DK (Ω) → C is well-defined. The map uK is clearly linear. Now set u : D(Ω) → C to be the map given by u(ϕ) := uK (ϕ) for each ϕ ∈ C0∞ (Ω) such that supp ϕ ⊆ K. Show that this map is well-defined, satisfies u ∈ D (Ω), and uΩj = uj in D (Ωj ) for every j ∈ I.

Exercise 2.139 Fix ϕ0 ∈ C0∞ (Rn ) with the property that Rn ϕ0 (x) dx = 1.

Next, let ϕ ∈ C0∞ (Rn ) be arbitrary and

set λ := Rn ϕ(x) dx. Hence, if we set ψ := ϕ − λϕ0 , then ψ ∈ C0∞ (Rn ) and Rn ψ(x) dx = 0, so 0 = u, ψ = u, ϕ − λu, ϕ0 .

(12.2.29)

Consequently, u, ϕ = λu, ϕ0 = c, ϕ, where c := u, ϕ0 ∈ C. Exercise 2.140 You may proceed by completing the following steps. Fix some arbitrary ϕ ∈ C0∞ (Rn ).

Step I. Prove that Rn ϕ(x) dx = 0 if and only if there exist ϕ1 , . . . , ϕn ∈ C0∞ (Rn ) n ∂j ϕj . Do so by induction over n. Corresponding to n = 1 such that ϕ = j=1 show that if a, b ∈ R satisfy a < b and for ϕ ∈ C0∞ (a, b) one defines φ1 (x) :=

b

x ∞ a ϕ(t) dt, x ∈ (a, b), then φ1 ∈ C0 (a, b) if and only if a ϕ(x) dx = 0. Step II. Show that the statement from Step I continues to hold if Rn is replaced by (a1 , b1 ) × · · · × (an , bn ), where aj , bj ∈ R, aj < bj for each j = 1, . . . , n. Step III. Fix aj , bj ∈ R, aj < bj for j = 1, . . . , n and consider the n-dimensional rectangle Q := (a1 , b1 ) × · · · × (an , bn ). Let ϕ0 ∈ C0∞ (Q) be such that


392

ϕ0 dx = 1. Then, if ϕ ∈ C0∞ (Q) is arbitrary, the function defined by

ψ := ϕ − Q ϕ dx ϕ0 belongs to C0∞ (Q) and satisfies Q ψ dx = 0. As such, Step II applies and shows that there exist ϕ1 , . . . , ϕn ∈ C0∞ (Q) such that n

ϕ = Q ϕ dx ϕ0 + ∂j ϕj . In turn, this permits us to write Q

j=1

u, ϕ = u, ϕ0

ϕ dx − Q

n

$ % ∂j u, ϕj = u, ϕ0 , ϕ ,

(12.2.30)

j=1

which shows if cQ := u, ϕ0 ∈ C, then uQ = cQ in D (Q). Step IV. Since Ω is connected and open, it is path connected. Now combine this with the fact that u is locally constant (as proved in Step III) to finish the proof. Exercise 2.141 By Proposition 2.73, it suffices to prove that there exists some v ∈ D (Rn−1 ) such that u, ϕ ⊗ ψ = v, ϕδ, ψ for every ϕ ∈ C0∞ (Rn−1 ) and every ψ ∈ C0∞ (R). Fix ϕ ∈ C0∞ (Rn−1 ), ψ ∈ C0∞ (R), and consider ψ0 ∈ C0∞ (R) with the property that ψ0 (0) = 1. Then there exists some h ∈ C0∞ (R) satisfying ψ(xn )−ψ(0)ψ0 (xn ) = xn h(xn ) for every xn ∈ R. This and the fact that xn u = 0 allows us to write u, ϕψ = u, ϕ(ψ − ψ(0)ψ0 ) + u, ϕψ0 ψ(0) = xn u, ϕh + u, ϕψ0 ψ(0) = u, ϕψ0 δ, ψ.

(12.2.31)

Now define v : D(Rn−1 ) → C by v(ϕ) := u, ϕψ0 for ϕ ∈ C0∞ (Rn−1 ), and show that v ∈ D(Rn−1 ). By (12.2.31) this v does the job. Exercise 2.142 Fix ψ ∈ C0∞ (R) with the property that ψ(0) = 1. Use Exercise 2.141 and induction to show that u = c δ(x1 ) ⊗ · · · ⊗ δ(xn ), where c := u, ψ ⊗ · · · ⊗ ψ ∈ C.

Exercise 2.143 Fix ψ ∈ C0∞ (R) with the property that R ψ(s) ds = 1. Given any function ϕ ∈ C0∞ (Rn ), at each point x = (x , xn ) ∈ Rn−1 × R we may write ϕ(x) = ϕ(x) − ψ(xn ) = ∂xn

xn

−∞

R

ϕ(x , s) ds + ψ(xn )

ϕ(x , t) − ψ(t)

R

ϕ(x , s) ds

R

ϕ(x , s) ds dt + ψ(xn )

ϕ(x , s) ds.

R

Since ∂n u = 0 in D (Rn ), this yields ! " u, ϕ = u , ψ(xn ) ϕ(x , s) ds . R

(12.2.32)


393

In particular, if ϕ = ϕ1 ⊗ ϕ2 for some ϕ1 ∈ C0∞ (Rn−1 ) and some ϕ2 ∈ C0∞ (R), then " ! ϕ2 (s) ds = u, ϕ1 ⊗ ψ1, ϕ2 . (12.2.33) u, ϕ = u , ϕ1 ⊗ ψ R

: C0∞ (Rn−1 ) → n−1

Define v C by v(θ) := u, θ ⊗ ψ for every θ ∈ C0∞ (Rn−1 ). Then v ∈ D (R ) and u(x , xn ) = v(x ) ⊗ 1 when restricted to C0∞ (Rn−1 ) ⊗ C0∞ (R). The desired conclusion follows by recalling Proposition 2.73. Exercise 2.146 Fix j ∈ N and note that for each x = (x1 , . . . , xn ) ∈ Rn we may write j j 1 1 eix1 ξ1 dξ1 ⊗ · · · ⊗ eixn ξn dξn . (12.2.34) fj (x) = 2π −j 2π −j Also, for each j ∈ N and each k ∈ {1, . . . , n}, the fundamental theorem of calculus gives j j sin(jxk ) ixk ξk e dξk = cos(xk ξk ) dξk = 2 , assuming xk = 0. xk −j −j (12.2.35) Now use (12.2.34), (12.2.35), Exercise 2.118, and part (d) in Theorem 2.80 to finish the proof. Exercise 2.147 Note that u = −δ is a solution for the equation in (1). Hence, if u is any other solution of the equation (x − 1)u = δ, then setting v := u + δ it follows that (x − 1)v = 0 in D (R). Next use this and the reasoning from Example 2.68 to conclude that the general solution for the equation in (1) is −δ + c δ1 , with c ∈ C. Fix ψ ∈ C0∞ (R) with the property that ψ(0) = 1. Show that any solution u of the equation in (2) satisfies $ % ϕ(x) − ϕ(0)ψ(x) dx+ u, ψδ, ϕ , ∀ ϕ ∈ C0∞ (R) (12.2.36) a(x) u, ϕ = x R and use this to obtain that the general solution of the equation in (2) is va + c δ, c ∈ C, where va is the distribution given by ϕ(x) − ϕ(0)ψ(x) dx, ∀ ϕ ∈ C0∞ (R). a(x) va , ϕ := x R Similarly, any solution u of the equation in (3) satisfies ! ϕ − ϕ(0)ψ " $ % + u, ψδ, ϕ , ∀ ϕ ∈ C0∞ (R), (12.2.37) u, ϕ = v, x so the general solution for (3) is w + c δ, where c ∈ C and w is the distribution given by ! ϕ − ϕ(0)ψ " w, ϕ := v, , ∀ ϕ ∈ C0∞ (R). x


394

Exercise 2.148 (a) Since H ∈ L1loc (R) and MR := (x, y) ∈ [0, ∞) × [0, ∞) : |x + y| ≤ R is a compact set in R2 for each R > 0, by Remark 2.84 and Theorem 2.85, it follows that H ∗ H is well-defined and belongs to D (R). Fix a compact set K in R and let R > 0 be such that K ⊂ (−R, R). Pick now ϕ ∈ C0∞ (R) with supp ϕ ⊆ K, and suppose that ψ ∈ C0∞ (R2 ) satisfies ψ = 1 on MK := (x, y) ∈ [0, ∞) × [0, ∞) : x + y ∈ K . Then H ∗ H, ϕ =

H(x)H(y)ϕ(x + y)ψ(x, y) dy dx R

∞

R

∞

ϕ(x + y)ψ(x, y) dy dx

= 0

(12.2.38)

0

Note that

(x, y) ∈ [0, ∞) × [0, ∞) : |x + y| ≤ R = (x, y) : 0 ≤ x ≤ R, 0 ≤ y ≤ R − x ,

hence

R

R−x

H ∗ H, ϕ =

R

R−x

ϕ(x + y)ψ(x, y) dy dx = 0

0

R

R

=

R

ϕ(z) dz dx = 0

x

ϕ(x + y) dy dx 0

0

z

ϕ(z) dx dz = 0

0

R

zϕ(z) dz 0

= xH, ϕ.

(12.2.39)

Alternatively, one may use Remark 2.86, to observe that H ∗ H in the distributional sense is the distribution given by the function obtained by taking the convolution, in the sense of (2.8.2), of the function H with itself. Hence, (H ∗ H)(x) =

∞

χ[0,∞) (x − y) dy = xH(x)

for every x ∈ R.

(12.2.40)

0

(b) −xH(−x) (c) (x2 − 2 + 2 cos x)H(x) 2 2 (d) x2 H(x) − (x−1) H(x − 1) 2 (e) Exercise 2.128 ensures that δ∂B(0,r) is compactly supported, so the given convolution is well-defined. Also, by Exercise 2.94, |x|2 ∗ δ∂B(0,r) ∈ C ∞ (Rn ) and equals


δ∂B(0,r) , |x − y| =

|x − y| dσ(y) = r

2

2

∂B(0,r)

=r

395

n−1 S n−1

n−1 S n−1

|x − rω|2 dσ(ω)

[r2 + |x|2 − 2rx · ω] dσ(ω)

= (rn+1 + |x|2 rn−1 )ωn−1 ,

∀ x ∈ Rn .

(12.2.41)

For the second equality in (12.2.41) we used the change of variables y = rω, ω ∈ S n−1 , while for the last equality we used the fact that since x · ω as a function in ω is odd, its integral over S n−1 is zero. D (Rn )

Exercise 2.149 uj −−−−→ 0, j→∞

D (Rn )

vj −−−−→ 0. j→∞

Given that uj , vj ∈ E (Rn ), we D (Rn )

deduce that uj ∗ vj ∈ E (Rn ) for each j ∈ N. Also, uj ∗ vj −−−−→ δ. j→∞

Exercise 2.150 The limits in (a) and (b) do not exist. Since for each j ∈ N we have fj ∈ E (R) and gj ∈ C ∞ (R), Exercise 2.94 may be used to conclude that D (Rn )

fj ∗ gj ∈ C ∞ (R) and that fj ∗ gj = 1 for every j. Thus, fj ∗ gj −−−−→ 1. j→∞

Exercise 2.151 Note that Λ is well-defined based on Proposition 2.93. You may want to use Theorem 13.5 to prove the continuity of Λ. Exercise 2.152 For f : Rn → C set f ∨ (x) := f (−x), x ∈ Rn . If u ∈ D (Rn ) is such that u ∗ ϕ = 0 for every ϕ ∈ C0∞ (Rn ), then 0 = (u ∗ ϕ∨ )(0) = u, ϕ for every ϕ ∈ C0∞ (Rn ), thus u = 0. This proves the uniqueness part in the statement. As for existence, given Λ as specified, define u0 : D(Rn ) → C by u0 (ϕ) := δ, Λ(ϕ∨ ) for ϕ ∈ C0∞ (Rn ). Being a composition of linear and continuous maps, u0 is linear and continuous, thus u0 ∈ D (Rn ). Also, if ϕ ∈ C0∞ (Rn ) is fixed, we have $ % $ % (u0 ∗ ϕ)(x) = u0 , ϕ(x − ·) = u0 , (ϕ∨ )(· − x) = δ, Λ(ϕ(· + x) $ % = δ, (Λϕ)(· + x) = Λ(ϕ)(x), ∀ x ∈ Rn . (12.2.42) For the first equality in (12.2.42) we used Proposition 2.93, the third equality is based on the definition of u0 , while the forth equality uses the fact that Λ is commutes with translations. Exercise 2.153 From hypotheses we obtain u, P = 0 for every polynomial P in Rn . Now use Lemma 2.75 to conclude that u, ϕ = 0 for every ϕ ∈ C0∞ (Rn ). Let ψ ∈ C0∞ (Rn ) be such that ψ = 1 in a neighborhood of supp u. Then for every function ϕ ∈ C ∞ (Rn ) we have 0 = u, ψϕ = u, ϕ since the support condition on u implies u, (1 − ψ)ϕ = 0.


396

Exercise 2.154 Fix ϕ ∈ C ∞ (R) and write k ϕ 1j − kϕ(0) − ϕ (0) ln k j=1

⎡ ⎤ k k 1 1 ⎦ = ϕ j − ϕ(0) − 1j ϕ (0) + ϕ (0) ⎣ j − ln k . j=1

Since

(12.2.43)

j=1

1 ϕ j − ϕ(0) − 1j ϕ (0) ≤

1 ∞ j 2 ϕ L ([0,1])

∀ j ∈ N,

taking the limit as k → ∞ in (12.2.43) [also recall Euler’s constant γ from (4.6.23)] we obtain u(ϕ) =

∞ ϕ 1j − ϕ(0) − 1j ϕ (0) + γ ϕ (0). j=1

∞ Now apply Fact 2.55 with K := [0, 1], m := 2 and C := j=1 u ∈ E (R). Also, show that supp u = {0} ∪ 1j : j ∈ N .

1 j2

+ γ to conclude

Exercise 2.155 Note that fj ∈ L1comp (R) for each j ∈ N, hence {fj }j∈N ⊂ E (R). If ϕ ∈ C ∞ (R) then we may write 1/j j j 1/j fj , ϕ − δ, ϕ = ϕ(x) − ϕ(0) dx ≤ |ϕ(x) − ϕ(0)| dx 2 −1/j −1/j 2 1 ϕ L∞ ([−1,1]) → 0 as j → ∞. j

≤

(12.2.44)

Exercise 2.156 Since fj ∈ L1comp(R) for each j ∈ N, we have {fj }j∈N ⊂ E (R). Let ϕ ∈ C0∞ (R) and suppose R ∈ (0, ∞) is such that supp ϕ ⊂ (−R, R). Then for j ∈ N with j ≥ R we have R j 2Rϕ ∞ ϕ(x) ϕ(x) L (R) dx = dx ≤ (12.2.45) |fj , ϕ| = j j j −j −R D (Rn )

which proves that fj −−−−→ 0. Suppose next that there exists a distribution

u ∈ E (R ) such that n

for every ϕ ∈

j→∞ E (Rn ) fj −−−−→ j→∞

C0∞ (R).

u. In particular, we have lim fj , ϕ = u, ϕ j→∞

Together with what we have proved before, this forces


397

u = 0. However, fj , 1 = 2 for every j ∈ N, which contradicts the fact that u = 0. Thus, {fj }j∈N does not converge in E (Rn ). Exercise 2.157 Since fj ∈ C0∞ (R) for each j ∈ N, we have {fj }j∈N ⊂ E (R). Let ϕ ∈ C ∞ (Rn ). Then using the change of variables jx = y we may write j n ψ(jx)ϕ(x) dx − ϕ(0) ψ(x) dx |fj , ϕ − δ, ϕ| = Rn

Rn

= ψ(y)ϕ(y/j) dy − ϕ(0) ψ(x) dx Rn Rn ≤ |ϕ(y/j) − ϕ(0)||ψ(y)| dy → 0 as j → ∞, (12.2.46) supp ψ

where the convergence is based on Lebesgue’s dominated convergence theorem. Exercise 2.158 For each k ∈ {1, . . . , n} consider the sequence {δxj , ϕk }j∈N where the function ϕk is defined by ϕk (x) := xk for each x = (x1 . . . , xn ) ∈ Rn .

12.3

Solutions to Exercises from Sect. 3.3

Exercise 3.32 Let R > 0 be such that supp ϕ ⊂ B(0, R). Then ϕj ∈ C0∞ (Rn ) and supp ϕj ⊂ B(0, jR) for each j ∈ N. Since there is no compact K ⊂ Rn such that supp ϕj ⊂ K for all j ∈ N, the sequence {ϕj }j∈N does not converge 1 −j in D(Rn ). Also, for each α ∈ Nn0 we have ∂ α ϕj (x) = j |α| e (∂ α ϕ)(x/j). Hence n if we also take β ∈ N0 arbitrary, then sup |xβ ∂ α ϕj (x)| ≤ e−j j −|α|

x∈Rn

|xβ (∂ α ϕ)(x/j)|

sup x∈B(0,jR)

≤ e−j j −|α|

|xβ |∂ α ϕL∞ (B(0,R))

sup x∈B(0,jR)

≤ e−j j |β|−|α|R|β| ∂ α ϕL∞ (B(0,R)) −−−→ 0, j→∞

(12.3.1)

which implies that {ϕj }j∈N converges to zero in S(Rn ). 1 Exercise 3.33 For each α ∈ Nn0 we have ∂ α ϕj (x) = j |α|+1 (∂ α ϕ)(x/j), for each n n x ∈ R . Consequently, for every compact subset K of R we may write

sup |∂ α ϕj (x)| ≤ j −|α|−1 sup |(∂ α ϕ)(x/j)| x∈K

x∈K

≤ j −|α|−1 ∂ α ϕL∞ (Rn ) −−−→ 0, j→∞

E(Rn )

which proves that ϕj −−−−→ 0. Moreover, if j→∞ O := x = (x1 , . . . , xn ) ∈ Rn : xj = 0 for j = 1, . . . , n

(12.3.2)


398

we claim that there exists x∗ ∈ O with the property that ϕ(x∗ ) = 0. Indeed, if this were not to be the case, we would have ϕ = 0 in O, hence ϕ = 0 in Rn since ϕ is continuous and O is dense in Rn . Having fixed such a point x∗ we then proceed to estimate sup |xβ ϕj (x)| = j −1 sup |xβ ϕ(x/j)| = j |β|−1 sup |xβ ϕ(x)|

x∈Rn

x∈Rn

x∈Rn

≥ j |β|−1 |xβ∗ | |ϕ(x∗ )| −−−→ ∞,

(12.3.3)

j→∞

for β ∈ Nn0 satisfying |β| > 1. Thus, {ϕj }j∈N does not converge in S(Rn ). Exercise 3.34 Clearly {ϕj }j∈N ⊂ C ∞ (R). If m ∈ N then for each j ∈ N0 we have m! 1 1 ψ (k) (x)θ(m−k) (x/j) m−k , = j k!(m − k)! j m

(m) ϕj (x)

∀ x ∈ R.

(12.3.4)

k=0

Hence, if ∈ N0 , then using the properties of θ and ψ we may write sup |x

x∈R

(m) ϕj (x)|

m 1 x m! = sup ψ (k) (x)θ(m−k) (x/j) m−k j x∈R k!(m−k)! j k=0

≤

m m! 1 x 1 sup |ψ (m) (x)|+ sup e−x θ(m−k) (x/j) m−k j |x|≤1 j x>1 k!(m − k)! j k=0

≤

C C C + sup |e−x x | ≤ −−−→ 0. j j x>1 j j→∞

(12.3.5)

S(Rn )

In conclusion, ϕj −−−−→ 0. j→∞

Exercise 3.35 (a) Not in S(Rn ) since it is not bounded given that

lim

x1 →−∞

e−x1 = ∞.

2

(b) Since e−|x| ∈ S(Rn ) and |x|2n! ∈ L(Rn ), by (a) in Theorem 3.13, their product is in S(Rn ). n

n

(c) (1 + |x|2 )2 is a polynomial function and if (1 + |x|2 )−2 ∈ S(Rn ) then their product, which is equal to 1, would belong to S(Rn ), which is not true. 2

(d) Show first that sin(e−|x| ) ∈ S(Rn ). Then, given that 2

follows that

sin(e−|x| ) 1+|x|2

∈ S(Rn ).

1 1+|x|2

∈ L(Rn ), it


399

−|x|2

) (e) Not in S(Rn ) since lim (1 + |x|2 )n+1 cos(e (1+|x|2 )n = ∞. |x|→∞

2

2

(f) Set ϕ(x) := e−|x| sin(ex1 ), x ∈ Rn . If ϕ were to belong to S(Rn ) then ∂1 ϕ and x1 ∂ϕ would be bounded. However, for every x = (x1 , . . . , xn ) ∈ Rn , 2

2

2

2

2

2

∂1 ϕ(x) = 2e−x2 −x3 −···−xn x1 [cos(ex1 ) − e−x1 sin(ex1 )].

(12.3.6)

In particular, 2

(∂1 ϕ)(x1 , 0, . . . , 0) + 2x1 ϕ(x1 , 0, . . . , 0) = 2x1 cos(ex1 ) which is not bounded. (g) Since A is positive definite, there exists a real, symmetric, positive definite 2 n × n matrix B such that B 2 = A. Hence, ϕ(x) = e−|Bx| for every x ∈ Rn , which means that ϕ is the composition between the function 2 e−|x| ∈ S(Rn ) (recall Exercise 3.3) and the linear transformation B that maps S(Rn ) into itself. Recalling Exercise 3.16 we may conclude that ϕ ∈ S(Rn ). Exercise 3.36 From part (g) in Exercise 3.35 we have f ∈ S(Rn ). Let B ∈ 2 Mn×n (R) be such that B 2 = A, so that f (x) = e−|Bx| for every x ∈ Rn . Now use Exercise 3.23 and Example 3.21.

1 1/2 1/2 1 and Exercise 3.36 applies and yields Exercise 3.37 Let A :=

. Then f (x) = e−(Ax)·x for every x ∈ R2

2 2 2π f3(ξ) = √ e−(ξ1 −ξ1 ξ2 +ξ2 )/3 3

for ξ = (ξ1 , ξ2 ) ∈ R2 .

(12.3.7)

Exercise 3.38 Use (b) in Theorem 3.20 and Example 3.21. ix·x 1 Exercise 3.39 Since sin(x · x0 ) = 2i e 0 − e−ix·x0 matters reduced to the case n = 1 after which we may apply Exercise 3.22. Hence, if x = (x1 , . . . , xn ), x0 = (x01 , . . . , x0n ), and ξ = (ξ1 , . . . , ξn ), we may write 2 F e−a|x| sin(x · x0 ) (ξ) =

1 2i

n

n 2 2 F e−axj +ixj x0j − F e−axj −ixj x0j

j=1

=

1 2i

j=1

n π n2 a

e−

(ξj −x0j )2 4a

j=1

=

1 π n 2

2i

a

e

|ξ−x |2 − 4a0

−

n

e−

(ξj +x0j )2 4a

j=1

− e−

|ξ+x0 |2 4a

.

(12.3.8)


400

Exercise 3.40 Fix ϕ ∈ S(R). Suppose ψ ∈ S(R) is such that ψ = ϕ. Then

ϕ(x) dx = R

R2

lim

R1 ,R2 →∞

ψ (x) dx =

R1

lim

[ψ(R2 ) − ψ(R1 )] = 0.

R1 ,R2 →∞

x

Conversely, if R ϕ(x) dx = 0, then the function ψ(x) := 0 ϕ(t) dt, for x ∈ R, belongs to S(R) and ψ = ϕ. Exercise 3.41 Use Exercise 3.40.

12.4


Exercise 4.103 Using the change of variables y = −x on the region corresponding to x < 0, we have |x|≥ε

ϕ(x) dx = x

ε

∞

ϕ(x) − ϕ(−x) dx, x

∀ ϕ ∈ S(Rn ).

(12.4.1)

Moreover, for each ε > 0 and each ϕ ∈ S(R) we may write ∞ 1 ϕ(x) |ϕ(x) − ϕ(−x)| |ϕ(x) − ϕ(−x)|x dx ≤ dx + dx |x|≥ε x x x2 ε 1 ∞ dx sup |xϕ(x)| ≤ sup |ϕ (x)| + 2 (12.4.2) x2 x∈R x∈R 1 so that P.V. x1 is well-defined and ! dx 1 " sup |xk ϕ(j) (x)|, P.V. , ϕ ≤ 1 + 2 x x∈R, k=0,1, j=0,1 |x|>1 |x|

∀ ϕ ∈ S(R), (12.4.3)

hence (4.1.2) holds for m = k = 1. Since P.V. x1 is also linear, we necessarily have P.V. x1 ∈ S (Rn ). Exercise 4.104 Use Exercise 2.107, (2.1.9), and Exercise 4.5. Exercise 4.105 Use Exercise 2.109 and (4.1.25). Exercise 4.106 Use a reasoning similar to the proof of the fact that the function in (4.1.27) is not a tempered distribution to conclude that eax H(x) and e−ax H(−x) do not belong to S (R) while e−ax H(x), eax H(−x) ∈ S (R).


401

Exercise 4.107 Let ϕ ∈ S(R) and j ∈ N. (a) We may write (

) x ϕ(x) − ϕ(0) x x2 dx , ϕ = ϕ(x) dx+ · 2 −2 2 −2 2 −2 x +j x |x|>1 x + j |x|≤1 x + j x dx. (12.4.4) +ϕ(0) 2 + j −2 x |x|≤1

The last integral in (12.4.4) is equal to zero (the function integrated is odd) while for the other two integrals in (12.4.4) apply Lebesgue’s dominated convergence theorem to obtain that the first converges to

ϕ(x) ϕ(x)−ϕ(0) dx. In conclux |x|>1 x dx and the second converges to |x|≤1 sion,

x x2 +j −2

S (R)

−−−→ P.V. x1 . j→∞

(b) Making the change of variables x = y/j and then applying Lebesgue’s dominated convergence theorem write ! " 1 ϕ(x) ϕ(y/j) , ϕ dx = dy −−−→ πϕ(0), = 2 −2 2 −2 2 j→∞ j(x + j ) ) R j(x + j R y +1 (12.4.5) hence

1

j(x2 +j −2 )

S (R)

−−−→ πδ. j→∞

(c) See Exercise 2.118.

sin(jx) S (R) −−−→ πx j→∞

2

D (R)

δ. 2

(d) Prove first that ej δj −−−−→ 0. If {ej δj }j∈N would converge in S (R) it j→∞

would have to converge to 0. However, for the test function e− one has " ! 2 j2 x2 ej δj , e− 2 = e 2 −−−→ ∞.

x2 2

∈ S(R)

j→∞

Exercise 4.108 Use the change of variables jx = y and Lebesgue’s dominated S (R)

convergence theorem to show that j n θ(jx) −−−→ δ. j→∞

Exercise 4.109 Based on Lebesgue’s dominated convergence theorem, it D (R)

2

follows that fj −−−−→ ex . On the other hand, the sequence {fj }j∈N does not j→∞

convergence in S (R) since $ % 2 fj , e−x /2 =

j

e −j

x2 /2

dx −−−→ j→∞

ex R

2

/2

dx = ∞.


402

Exercise 4.110 If there were some f ∈ Lp (R) such that lim fj − f Lp(R) = 0, j→∞

then lim fj (x) = f (x) for almost every x ∈ R. Since lim fj (x) = 0 for j→∞

j→∞

every x ∈ R, we have that f = 0 almost everywhere on R. This leads to a

j contradiction given that we would have 0 = lim fj pLp (R) = j−1 1 dx = 1. j→∞

Hence, the sequence {fj }j∈N does not converge in Lp (R). If ϕ ∈ S(R) then j dx C 2 −−−→ 0, ≤ (12.4.6) |fj , ϕ| ≤ sup |x ϕ(x)| 2 x j(j − 1) j→∞ x∈R j−1 S (R)

which shows that fj −−−→ 0. j→∞

Exercise 4.111 Using Lebesgue’s dominated convergence theorem it is not difficult to check that uj , ϕ −−−→ ex sin(ex ), ϕ for every ϕ ∈ C0∞ (R). j→∞

If ϕ ∈ S(R), integration by parts yields j j x x x e sin(e )ϕ(x) dx = − cos(e )ϕ(x) −j + −j

j

cos(ex )ϕ (x) dx.

(12.4.7)

−j

Since ϕ ∈ S(R) one has cos(ex )ϕ ∈ L1 (R) and j C j cos(ex ) xϕ(x)−j ≤ −−−→ 0. cos(ex )ϕ(x)−j = x j j→∞

Thus, lim uj , ϕ = R cos(ex )ϕ (x) dx and if u : S(R) → C is such that j→∞

u, ϕ :=

cos(ex )ϕ (x) dx

R

for every ϕ ∈ S(R),

then u is a well-defined, linear mapping and dx sup |(1 + x2 )ϕ (x)| |u, ϕ| ≤ 2 1 + x x∈R R

for every ϕ ∈ S(R).

(12.4.8)

(12.4.9)

S (R)

Hence u ∈ S (R) and uj −−−→ u. j→∞

Exercise 4.113 Fix f ∈ S (R) and let ψ ∈ S(R) be such that ψ(0) = 1. Then, if u is a solution of the equation xu = f in S (R), for each ϕ ∈ S(R) we have ! ! ϕ − ϕ(0)ψ " ϕ − ϕ(0)ψ " u, ϕ = u, x + u, ϕ(0)ψ = f, + u, ψδ, ϕ. x x (12.4.10) Set

g(x) :=

ϕ(x)−ϕ(0)ψ(x) x

if x ∈ R \ {0},

ϕ (0) − ϕ(0)ψ (0)

if x = 0.

(12.4.11)


1

Note that g ∈ S(R), since g(x) =

403

ϕ (tx) − ϕ(0)ψ (tx) dt for x ∈ R. Thus,

0

if we define

ϕ(x) :=

ϕ(x)−ϕ(0)ψ(x) x

if |x| > 1,

g(x)

if |x| ≤ 1,

(12.4.12)

then ϕ ∈ S(R) and xϕ ) = ϕ. There remains to observe that the mapping wf : S(R) → C,

wf , ϕ := f, ϕ

∀ ϕ ∈ S(R),

(12.4.13)

where ϕ is as in (12.4.12) satisfies wf ∈ S (R), in order to conclude that u = wf + cδ, some c ∈ R. 2

Exercise 4.114 Suppose u ∈ S (Rn ) is a solution of the equation e−|x| u = 1 in S (Rn ). Then for ϕ ∈ C0∞ (Rn ) we have $ $ $ 2 % 2 2 % 2 % u, ϕ = u, e−|x| e|x| ϕ = 1, e|x| ϕ = e|x| , ϕ , (12.4.14) 2 which shows that uC ∞ (Rn ) = e|x| . Since C0∞ (Rn ) is sequentially dense in S(Rn ), 0

2

this would imply that u = e|x| . However, as proved following Remark 4.15, 2 e|x| ∈ S (Rn ). Exercise 4.115 Since H is Lebesgue measurable and (1 + x2 )−1 H ∈ L1 (R), by Exercise 4.5 it follows that H defines a tempered distribution. We have seen that H ∈ D (R) and H = δ in D (R) (recall Example 2.37). Since δ ∈ S (R), of by (4.1.25) it follows that H = δ in S (R). Taking the Fourier transform the 1 3 last equation we arrive at iξ H = 1 in S (R). On the other hand, ξ P.V. ξ = 1 3 − P.V. 1 = 0 in in S (R) by Exercise 4.103 and (2.3.6). Consequently, ξ iH ξ 3 − P.V. 1 = cδ, for S (R). Now Example 2.68 may be used to conclude that iH ξ some c ∈ C. Hence 3 = −iP.V. 1 − cδ H in S (R). (12.4.15) ξ 3 to the function ϕ(x) := e−x2 ∈ S(R). First, by ExamTo determine c apply H ple 3.21 write ∞ % $ % $ √ % √ $ 2 2 −x2 = H, πe−x /4 = 3 e−x2 = H, e5 π e−x /4 dx H, √

=

π 2

0 ∞

e−x

2

/4

−x2 /4 (0) = π. dx = e

(12.4.16)

−∞ −x2

Second, from (12.4.15) and the fact that e x is an odd function, we obtain 2 ! " 2 1 e−x − i P.V. − cδ, e−x = −i lim dx − c = −c. (12.4.17) ξ ε→0+ |x|>ε x 3 = −i P.V. 1 + πδ. Combining (12.4.15)–(12.4.17), we arrive at H ξ


404

Exercise 4.116 You may use Exercise 4.115 to show that 1 (ξ) = 2πiH(ξ) − iπ = iπ sgn (ξ) F P.V. x

in S (R).

Exercise 4.117 (a) sgn = H − H ∨ , hence using Exercise 4.115 one may write 1 1 1 ∨ = πδ − i P.V. 3 −H 5 − πδ − i P.V. = −2i P.V. s5 gn = H ξ ξ ξ

in S (R).

(12.4.18) Alternatively, you may use Exercise 4.116 and take another Fourier transform. 5 5k = x kδ 3 = (−D)k 3 (b) If k is even, then |x|k = xk , so that |x| δ3 = 2πik δ (k) in k k S (R). If k is odd, then |x| = x sgn x, thus 1 (k) 5k = x k sgn x = (−D)k sgn x = −2ik+1 P.V. |x| ξ

in

S (R). (12.4.19)

(c) You may use Example 4.34. Alternatively, take the Fourier transform of = sin(ax) in S (R), then use (c) in Theorem 4.25 the identity x · sin(ax) x and (4.2.42) to conclude that & ' sin(ax) = πδ−a − πδa F x

in S (R).

Now invoke Exercise 2.120 and Proposition 2.40 to arrive at F

sin(ax) x

= π sgn(a)χ[−|a|,|a|] + c

in S (R),

and then show that c = 0 by applying the Fourier transform to the last identity and recalling (4.2.39). (d) Take the Fourier transform of the identity x·

sin(ax) sin(bx) = sin(ax) sin(bx) x

in S (R),

then use (c) in Theorem 4.25 and (4.2.45) to conclude that & ' iπ sin(ax) sin(bx) [δa+b − δa−b − δb−a + δ−a−b ] = F x 2 Now use Exercise 2.120 and Proposition 2.40 to obtain

in S (R).

12.4. SOLUTIONS TO EXERCISES FROM SECT. 4.11 F

405

sin(ax) sin(bx) x iπ [H(x−a−b)−H(x−a+b)−H(x−b+a)+H(x+a+b)] +c0 2 iπ sgn(b) χ[−a−|b|,−a+|b|] −χ[a−|b|,a+|b|] +c0 = (12.4.20) 2 =

in S (R), and then show that c0 = 0 by applying the Fourier transform to the last identity. 2 2 1 eix − e−ix , (4.2.21) and (4.2.50), we (e) Using the identity sin(x2 ) = 2i may write 1 5 2) = −ix2 eix2 − e sin(x 2i ' √ & 2 π i π −i ξ2 −i π i ξ4 4 4 4 −e e = e e 2i ' √ & 2 π i π−ξ2 −i π−ξ 4 4 −e = . e 2i

(12.4.21)

(f) Use (4.6.25), (4.2.3), and Proposition 4.30 to show ln |x| = −2πγδ − πwχ(−1,1)

in S (R).

Exercise 4.118 From Exercise 2.128 we know that δ∂B(0,R) ∈ E (R3 ), thus by (b) in Theorem 4.33 it follows that % $ (ξ) = δ∂B(0,R) , e−ix·ξ = e−ix·ξ dσ(x), ∀ ξ ∈ R3 . δ∂B(0,R) ∂B(0,R)

(12.4.22) Check that δ∂B(0,R) is invariant under orthogonal transformations, and conclude is invariant under orthogonal transformations. Fix ξ ∈ R3 \ {0} that δ∂B(0,R) and show that there exists an orthogonal transformation A ∈ M3×3 (R) such

(ξ) = that Aξ = (|ξ|, 0, 0), and furthermore that δ∂B(0,R) e−ix1 |ξ| dσ(x). ∂B(0,R) Now use spherical coordinates to compute the latter integral and conclude that (ξ) = 4πR sin(R|ξ|) . Treat separately the case ξ = (0, 0, 0). δ∂B(0,R) |ξ| Exercise 4.119 Use Lemma 4.27. Exercise 4.120 Since χ[−1,1] ∈ L1 (R), from (4.1.8) it follows that χ[−1,1] ∈ S (R). Also, by Exercise 4.26, we have that χ [−1,1] in S (R) is the tempered distribution given by the function

406


1

e−ixξ dx =

−1

In particular, since

sin ξ ξ

2 sin ξ , ξ

ξ ∈ R \ {0}.

1 ∈ L1 (R) we have χ [−1,1] ∈ L (R).

Exercise 4.122 See Example 4.36 and deduce that g3 = πδa + πδ−a in S (R). Exercise 4.123 Use Exercise 4.119. Exercise 4.124 If P (x) = |α|≤m aα xα , then the condition P (tx) = t−k P (x) |α| α −k α n in Rn for each t > 0 implies |α|≤m t aα x = t |α|≤m aα x in R for each t > 0. Hence, for each α ∈ Nn0 we have t|α| aα = t−k aα for every t > 0, or equivalently that t|α|+k aα = aα for every t > 0. Now take t → 0+ in the last equality to obtain aα = 0. Exercise 4.125 Check, via a direct calculation, that R w := w −

w · (ζ + η) w · [ζ − (1 + 2η · ζ)η] ζ− η, 1+η·ζ 1+η·ζ

∀ w ∈ Rn , (12.4.23)

and then use (12.4.23) to prove that R R = RR = In×n . Now (4.11.4) is easy to verify. Exercise 4.126 Fix c ∈ R \ {0} and use the fact that cosine is an odd function, the fundamental theorem of calculus, and Fubini’s theorem to write R R cos(cρ) − cos ρ cos(|c|ρ) − cos ρ dρ = dρ ρ ρ ε ε R |c| =− sin(rρ) dr dρ ε

=− = 1

1

|c|

R

sin(rρ) dρ dr 1 |c|

ε

cos(Rr) cos(εr) − dr. r r

(12.4.24)

In particular, estimate (4.11.7) readily follows from this formula. Going further, we note that an integration by parts gives |c| sin(Rr) r=|c| cos(Rr) 1 |c| sin(Rr) dr = + dr. (12.4.25) r Rr R 1 r2 r=1 1 By combining (12.4.24) with (12.4.25) we arrive at


R

ε

sin(|c|R) sin R 1 cos(cρ) − cos ρ dρ = − + ρ |c|R R R |c| cos(εr) dr. − r 1

407

1

|c|

sin(Rr) dr r2 (12.4.26)

Passing to limit R → ∞ and ε → 0+ in (12.4.26) yields (4.11.5) after observing (e.g., by Lebesgue’s dominated convergence theorem) that |c| |c| cos(εr) dr dr = = ln |c|. (12.4.27) lim r r ε→0+ 1 1 Next, lim

ε→0+

R→∞

ε

R

sin(cρ) dρ = lim ρ ε→0+

R→∞

Rc εc

sin t dt t

= sgn c lim

ε→0+

R→∞

ε

R

π sin t dt = sgn c, t 2

(12.4.28)

by a suitable change of variables and the well-known fact (based on a residue

R calculation) that lim ε→0+ ε sint t dt = π2 . This proves (4.11.6). Finally, to R→∞

justify (4.11.8) use the fact that whenever 0 < a < b < ∞ an integration by parts gives b cos t t=b b cos t sin t dt = − − dt. (12.4.29) t t t2 t=a a a

12.5


d m Exercise 5.15 The ordinary differential equation dx v = 0, with initial conditions v(0) = v (0) = · · · = v (m−2) (0) = 0, and v (m−1) (0) = 1, has the 1 xm−1 for x ∈ R. Hence, by Example 5.11 the unique solution v(x) = (m−1)! d m 1 function u := (m−1)! xm−1 H is a fundamental solution for dx in R. In addition, by Exercise 4.115 we have H ∈ S (R) which, when combined with the 1 xm−1 ∈ L(R), implies (as a consequence of (b) in Theorem 4.13) fact that (m−1)! 1 xm−1 H ∈ S (R). If v ∈ S (R) is an arbitrary fundamental solution that (m−1)! d m for dx , then by Proposition 5.7 we have that v − u = P for some polynomial d m in R satisfying dx P = 0 in R. This forces P to be a polynomial of degree less than or equal to m − 1. Exercise 5.16 First prove that P1 (Dx ) ⊗ P2 (Dy ), ϕ1 (x) ⊗ ϕ2 (y) = δx ⊗ δy , ϕ1 (x) ⊗ ϕ2 (y) for every ϕ1 ∈ C0∞ (Rn ) and every ϕ2 ∈ C0∞ (Rm ), and then use Proposition 2.73.


408

12.6


Exercise 6.27 (a) The general solution for the ordinary differential equation v − a2 v = 0 in R is v(x) := c1 eax + c2 e−ax for x ∈ R, where c1 , c2 ∈ C. Hence, if we 1 further impose the conditions v(0) = 0 and v (0) = 1 then c1 = 2a = −c2 , 1 ax 1 −ax H satisfies and by Example 5.11, we have that u := 2a e H − 2a e u − a2 u = δ in D (R). d2 2 (b) By Theorem 5.13 there exists E ∈ S (R) such that dx E = δ 2 − a in S (R). Fix such an E and apply the Fourier transform to the last 3 = 1 in S (R). Since 2 1 2 ∈ L(R) we may equation. Then −(ξ 2 + a2 )E ξ +a 3 = 2 1 2 = f , and, multiply the last equality by 2 1 2 to conclude that −E ξ +a

ξ +a

furthermore, that f3 = −2πE ∨ . Therefore we can determine f3 as soon as d2 2 we find a fundamental solution E ∈ S (R) for the operator dx 2 − a . 2

d 2 By Theorem 6.15, dx is a hypoelliptic operator in R. As a conse2 − a quence of Remark 6.7 we have [with u as in (a)] that u − E ∈ C ∞ (R) and is a classical solution of the ordinary differential equation v − a2 v = 0 in R. Thus,

E = u + c1 eax + c2 e−ax 1 ax 1 e H − e−ax H + c1 eax H + c1 eax H ∨ + c2 e−ax H + c2 e−ax H ∨ 2a 2a & & ' ' 1 1 ax + c1 e H + − + c2 e−ax H + c1 eax H ∨ + c2 e−ax H ∨ . = 2a 2a (12.6.1) =

1 The condition E ∈ S (R) implies (in view of Exercise 4.106) that 2a + 1 −a|x| . c1 = 0 and c2 = 0, which when used in (12.6.1) give E = − 2a e Consequently, f3(x) = πa e−a|x| for x ∈ R.

Exercise 6.29 Recall Exercise 6.4. Consider u := 1 + δ in D (Rn ). It follows that sing supp u = {0} and supp u = Rn , thus sing supp u ⊂ supp u. Exercise 6.30 Since ∂ α δx0 Rn \{x } = 0 we have sing supp ∂ α δx0 ⊆ {x0 }. To see 0 that the latter inclusion is in fact equality, use Example 2.13 if α = (0, . . . , 0) and if |α| > 0 use the fact that the order of the distribution ∂ α δx0 equals |α| while any distribution of function type is of order zero. Exercise 6.31 there exists an open set ω ⊆ Ω with If x ∈ Ω is such that x ∈ ω and uω ∈ C ∞ (ω), then (au)ω ∈ C ∞ (ω), which gives Ω \ sing supp u ⊆ Ω \ sing supp (au).


409

Exercise 6.32 Since P.V. x1 = x1 and x1 belongs to C ∞ (Rn \ {0}), it n \{0} R follows that sing supp P.V. x1 ⊆ {0}. Now using Example 2.11 we have that the distribution P.V. x1 is not of function type. 1 Exercise 6.33 First prove that P (x)· P (x) = 1 in S (Rn ). Then take the Fourier transform to arrive at 1 (12.6.2) P (−D)F (ξ) = (2π)n δ in S (Rn ). P

Use (12.6.2), (6.1.3), and Example 6.4, to show that it is not possible to have 1 sing supp F = ∅. P Exercise 6.34 sing supp u = R × {0}. To prove this you may want to use the fact that " ! ϕ(x, 0) ϕ(x, 0) − ϕ(0, 0) 1 dx + dx u, ϕ = P.V. , ϕ(x, 0) = x x x |x|≥1 |x|≤1 (12.6.3) for every ϕ ∈ C0∞ (R2 ).

12.7


Exercise 7.58 If the equation has two solutions E1 , E2 ∈ S (Rn ), set E := E1 − E2 . Then ΔE − E = 0 in S (Rn ) implies (after an application of the 3 = 0 in S (Rn ). Since 1 2 ∈ L(Rn ) (recall Fourier transform) −(|ξ|2 + 1)E 1+|ξ| 1 n Exercise 3.12), by part (a) in Theorem 3.13 we have 1+|ξ| 2 ϕ ∈ S(R ) for every n ϕ ∈ S(R ). Thus, ! 3 ϕ = (1 + |ξ|2 )E, 3 E,

" 1 ϕ = 0, 2 1 + |ξ|

∀ ϕ ∈ S(Rn ).

3 = 0 in S (Rn ); thus, after applying the Fourier transform This proves that E we obtain E = 0 in S (Rn ). In turn, the latter implies E1 = E2 in S (Rn ) and the proof of the uniqueness statement is complete. Regarding the existence statement, suppose E ∈ S (Rn ) satisfies ΔE − E = δ in S (Rn ). This equation, 3 = 1 in S (Rn ). Since via the Fourier transform, is equivalent to −(|ξ|2 + 1)E 1 1 n 2 Exercise 4.6) and (|ξ| + 1) 1+|ξ|2 = 1 in S (Rn ), it follows 1+|ξ|2 ∈ S (R ) (use 1 is a solution of ΔE − E = δ in S (Rn ), which must that E := −F −1 1+|ξ| 2 be unique based on the earlier reasoning.


410

3= Exercise 7.59 Suppose there exists E ∈ L1 (Rn ) with ΔE = δ. Then −|ξ|2 E 0 n 2 4 1 3 3 ∈ L ⊂ C (R ) [recall (3.1.3)], hence [|ξ| E] 1. We know that E = 0. ξ=(0,...,0) This leads to a contradiction. Exercise 7.62 Since EΔ ∈ S (Rn ) is a fundamental solution for Δ, we have ΔEΔ = δ in S (Rn ). Fix j ∈ {1, . . . , n}. Then, Δ(∂j EΔ ) = ∂j δ in S (Rn ). Take the Fourier transform of the last equation to arrive at in S (Rn ).

|ξ|2 F (∂j EΔ ) = −iξj Since n ≥ 2, by Example 4.3 we have |ξ|2 ·

ξj |ξ|2 2

ξj |ξ|2

(12.7.1)

∈ S (Rn ). Also, |ξ|2 ∈ L(Rn ), thus

∈ S (Rn ) (recall (b) in Theorem 4.13), and it is not difficult to check

that |ξ| ·

ξj |ξ|2

= ξj in S (Rn ). These facts combined with (12.7.1) imply

ξj (12.7.2) |ξ|2 F (∂j EΔ ) + i 2 = 0 in S (Rn ). |ξ| ξ Thus, supp F (∂j EΔ ) + i |ξ|j2 ⊆ {0} and by Exercise 4.35, it follows that F (∂j EΔ ) = −i

ξj 4j (ξ) +P |ξ|2

in S (Rn ),

(12.7.3)

where Pj is a polynomial in Rn . Now a direct computation gives that if n ≥ 2, then 1 xj ∂j EΔ = · in S (Rn ). (12.7.4) ωn−1 |x|n Hence, ∂j EΔ is positive homogeneous of degree 1 − n, which in turn when combined with Proposition 4.57 implies that F (∂j EΔ ) is positive homogeneous of degree −1. Thus, the term in the right-hand side of (12.7.3) is positive ξ homogeneous of degree −1. Since |ξ|j2 is positive homogeneous of degree −1, we 4j (ξ) is positive homogeneous of degree −1, and furthermore, by conclude that P Proposition 4.57 and Exercise 4.55, that the polynomial Pj is positive homogeneous of degree 1 − n ≤ −1. Now invoking Exercise 4.124 we obtain Pj ≡ 0. The latter when used in (12.7.3) proves (7.10.3). Identities (7.10.4) and (7.10.5) follow from (7.10.3) and (12.7.4). Exercise 7.63 Since EΔ2 ∈ S (Rn ) is a fundamental solution for Δ2 , we have Δ2 EΔ2 = δ in S (Rn ). Fix j, k ∈ {1, . . . , n}. Then, Δ2 (∂j ∂k EΔ2 ) = ∂j ∂k δ in S (Rn ). Take the Fourier transform of the last equation to arrive at |ξ|4 F (∂j ∂k EΔ2 ) = −ξj ξk

in S (Rn ).

Under the current assumption on n, by Exercise 4.4 we have ξj ξk |ξ|4

(12.7.5) 1 |ξ|2

∈ S (Rn ). Also,

∈ L1loc (Rn ) and in view of Example 4.3 one may infer that

ξj ξk |ξ|4

∈ S (Rn ).

12.7. SOLUTIONS TO EXERCISES FROM SECT. 7.10 In addition, |ξ|4 ∈ L(Rn ), thus |ξ|4 · ξj ξk |ξ|4

ξj ξk |ξ|4 , 4

Theorem 4.13), and |ξ|4 · = ξj ξk , |ξ| · combined with (12.7.5) imply

|ξ|4 · 1 |ξ|2

ξj ξk |ξ|4 F (∂j ∂k EΔ2 ) + =0 |ξ|4 Thus, supp F (∂j ∂k EΔ2 ) +

ξj ξk |ξ|4

1 |ξ|2

411

∈ S (Rn ) (recall (b) in

= |ξ|2 in S (Rn ). These facts

in S (Rn ).

(12.7.6)

⊆ {0} and by Exercise 4.35, it follows that

F (∂j ∂k EΔ2 ) = −

ξj ξk 5 +R jk (ξ) |ξ|4

in S (Rn ),

(12.7.7)

where Rjk is a polynomial in Rn . It is easy to check that ∂j ∂k EΔ2 = −

1 δjk 1 xj xk · n−2 + · 2(n − 2)ωn−1 |x| 2ωn−1 |x|n

in S (Rn ).

(12.7.8)

Hence, ∂j ∂k EΔ2 is positive homogeneous of degree 2 − n which, in turn, when combined with Proposition 4.57, implies that F (∂j ∂k EΔ2 ) is positive homogeneous of degree −2. Thus, the term in the right-hand side of (12.7.7) is positive ξj ξk homogeneous of degree −2. Since |ξ| 4 is positive homogeneous of degree −2, 5 we may conclude that Rjk (ξ) is positive homogeneous of degree −2 and furthermore (by Proposition 4.57 and Exercise 4.55) that the polynomial Rjk is positive homogeneous of degree 2 − n ≤ −1. Now invoking Exercise 4.124 we obtain Rjk ≡ 0. The latter when used in (12.7.7) proves (7.10.6). Identity (7.10.7) follows from (7.10.6) and (12.7.8). As for identity (7.10.8), apply the Fourier transform to (7.10.7) and then use Proposition 4.61 with λ = n − 2. Exercise 7.64 If P (D) is elliptic, then first show that there exists C ∈ (0, ∞) such that Pm (ξ) ≥ C for ξ ∈ S n−1 ; thus, conclude Pm (ξ) ≥ C|ξ|m for every ξ ∈ Rn \ {0}. Use the latter and the fact that |P (ξ)| ≥ |Pm (ξ)| −

aα ξ α

|α|≤m−1

to obtain the desired conclusion. Conversely, suppose that there exist C, R ∈ (0, ∞) such that |P (ξ)| ≥ C|ξ|m for every ξ ∈ Rn \ B(0, R) and that Pm (ξ∗ ) = 0 for some ξ∗ ∈ Rn \ {0}. Then for every λ > R/|ξ∗ | we have 0 < Cλ |ξ∗ | m

m

m−1 |α| α ≤ |P (λξ∗ )| = aα λ ξ∗ ≤ cj λj |α|<m

j=1

and obtain a contradiction by dividing with λm and then letting λ → ∞.


412

Exercise 7.65 Fix ϕ ∈ C0∞ (Rn ). Using the fact that ϕ is compactly supported, Lebesgue’s dominated convergence theorem, formula (13.7.4) twice, and the fact that A is symmetric, we may write EA (x)LA ϕ(x) dx = lim EA (x)LA ϕ(x) dx ε→0+

Rn

6 = lim

ε→0+

Rn \B(0,ε)

1 + ε

Rn \B(0,ε)

(LA EA )(x)ϕ(x) dx −

EA (x)(A∇ϕ(x)) · x dσ(x) ∂B(0,ε)

[ϕ(x) − ϕ(0)] A∇EA (x) · x dσ(x)

∂B(0,ε)

ϕ(0) + ε

1 ε

7 A∇EA (x) · x dσ(x) .

(12.7.9)

∂B(0,ε)

For the second equality in (12.7.9) we also used the fact that the outward unit normal to ∂B(0, ε) is ν(x) = 1ε x, for x ∈ ∂B(0, ε). Analyze each of the integrals in the rightmost side of (12.7.9). First, a direct computation shows that LEA = 0 pointwise in Rn \ {0}. Second, (7.8.48), (7.8.25), and the mean value theorem for ϕ, imply 1 [ϕ(x) − ϕ(0)] A∇EA (x) · x dσ(x) ≤ Cε −−−−→ 0, (12.7.10) ε ∂B(0,ε) ε→0+ where the constant C in (12.7.10) depends on A, ∇ϕL∞ (Rn ) and n but not on ε. Third, thanks to (7.8.25) we have 1 EA (x)(A∇ϕ(x)) · x dσ(x) ε ∂B(0,ε) Cε if n ≥ 3 ≤ −−−−→ 0. (12.7.11) C(| ln ε| + 1)ε if n = 2 ε→0+ Fourth, (7.8.48) and a natural change of variables allow us to write ε 1 dσ(x) √ A∇EA (x) · x dσ(x) = n ε ∂B(0,ε) ωn−1 detA ∂B(0,ε) [(A−1 x) · x] 2 dσ(x) 1 √ = . (12.7.12) −1 x) · x] n 2 n−1 [(A ωn−1 detA S Combining (12.7.9)–(12.7.12) and (7.8.47), we conclude that, under the current assumptions on A, for every ϕ ∈ C0∞ (Rn ) we have dσ(x) ϕ(0) √ EA (x)LA ϕ(x) dx = (12.7.13) n . −1 ωn−1 detA S n−1 [(A x) · x] 2 Rn


413

In particular, (12.7.13) holds if A has real entries, is symmetric, and satisfies (7.8.8). In this latter situation, we already proved that EA is a fundamental solution for LA , thus the left-hand side in (12.7.13) is equal to EA , LA ϕ = LA EA , ϕ = δ, ϕ = ϕ(0), which means that (7.8.50) holds in the case when A has real entries, is symmetric and satisfies (7.8.8). As in the proof of Theorem 7.54, the identity (7.8.50) may be extended to the class of complex symmetric matrices satisfying condition (7.8.8). Based on what we have just proved and (12.7.13) we deduce that EA (x)LA ϕ(x) dx = ϕ(0) (12.7.14) Rn

C0∞ (Rn )

and every matrix A ∈ Mn×n (Cn ) satisfying A = holds for every ϕ ∈ A as well as condition (7.8.8). The latter, combined with the fact that EA ∈ L1loc (Rn ) further implies that EA is a fundamental solution for LA whenever A ∈ Mn×n (C) is symmetric and satisfies (7.8.8).

12.8


Exercise 9.12 Let E be the fundamental solution for the heat operator ∂t − Δx in Rn+1 as in (8.1.7). Then E ∈ L1loc (Rn+1 ) and we claim that for each x ∈ Rn \ {0}, the function E(x, t) is absolutely integrable with respect to t ∈ R. n |x|2 Indeed, n using n the change of variables τ = 4t , (13.5.1), the fact that Γ 2 = 2 − 1 Γ 2 − 1 , and (13.5.6), we obtain ∞ ∞ n 1 Γ(n/2 − 1) E(x, t) dt = − n/2 n−2 e−τ τ 2 −2 dτ = − n/2 n−2 4π |x| 4π |x| −∞ 0 =−

1 1 · (n − 2)ωn−1 |x|n−2

if x ∈ Rn \ {0}.

(12.8.1)

Hence, we may apply Proposition 9.6 and Proposition 9.7, to conclude that the distribution ∞ 1 1 − E(x, t) dt = · n−2 , x ∈ Rn \ {0}, (n − 2)ω |x| n−1 −∞ is a fundamental solution for Δ in Rn , when n ≥ 3.

12.9


n Exercise 10.31 We have Δu + ∇p = 0 in D (Ω) and div u = 0 in D (Ω). Apply div to the first equation and conclude that Δp = 0 in D (Ω). Next apply Δ to the first equation. Finally, use Theorem 7.10 and Theorem 7.23. Exercise 10.32 Follow the outline from Exercise 10.16 with the adjustment that the integrals in (10.3.24) this time will be bounded by C∇ϕL∞ (Rn ) ε(1+| ln ε|).


414

Exercise 10.33 Follow the outline from Exercise 10.30 with the adjustment that the integrals in (10.6.22) this time will be bounded by C∇ϕL∞ (Rn ) ε(1+| ln ε|). Exercise 10.34 Note that u is a solution for (10.7.3) if and only if u − Ax − b is a solution for (10.5.3). Exercise 10.35 Show that Theorem 10.3 applies.

12.10


Exercise 11.19 Follow the outline of the proof of Theorem 11.1. Exercise 11.20 Compute L2 EL1 L2 using the analogue of (11.3.31) written for the operator L := L1 L2 , and make use of a suitable version of (11.3.53). Deduce that (n+q)/2

P (x) = cm1 ,m2 ,n,q Δ

S n−1

−1 (x · ξ)2m1 +q L2 (ξ) dσ(ξ)

(12.10.1)

for every x ∈ Rn , from which all the desired properties of P follow. Exercise 11.21 Start with (11.3.31) written for L = Δm and use (7.5.24), (7.5.25), and (7.5.13), in order to deduce that (11.7.3) holds with x ∈ Rn , P (x) = Pm+ n+q , n+q (x) + C(n, q)Δ(n+q)/2 |x|q , (12.10.2) 2

2

where C(n, q) is as in (7.5.26).

Chapter 13

Appendix 13.1

Summary of Topological and Functional Analytic

A topology on a set X is a family τ of subsets of X satisfying the following properties: 1. X, ∅ ∈ τ ; 2. If {Aα }α∈I is a family of sets contained in τ , then

α∈I

Aα ∈ τ ;

3. If A1 , A2 ∈ τ , then A1 ∩ A2 ∈ τ . If X and τ are as above, the pair (X, τ ) is called a topological space and the elements of τ are called open sets. If x is an element of X (sometimes referred to as point), then any open set containing x is called an open neighborhood of x. Any set that contains an open neighborhood of x is called a neighborhood of x. A sequence {xj }j∈N of elements of a topological space (X, τ ) are said to converge to x ∈ X, and we write lim xj = x, if every neighborhood of x contains all but j→∞

a finite number of the elements of the sequence. A family B of subsets of X is a base for a topology τ on X if B is a subfamily of τ and for each x ∈ X, and each neighborhood U of x, there is V ∈ B such that x ∈ V ⊂ U . We say that the base B generates the topology τ . An equivalent characterization of bases that is useful in applications is as follows. A base for a topological space (X, τ ) is a collection B of open sets in τ such that every open set in τ can be written as a union of elements of B. It is important to note that not any family of subsets of a given set is a base for some topology on the set. Given a set X and a family B of subsets of X, this family B is a base for some topology on X if and only if ⎧ B = X, ⎨ ⎩

B∈B

∀ B1 , B2 ∈ B and ∀ x ∈ B1 ∩ B2 , ∃ B ∈ B such that x ∈ B ⊆ B1 ∩ B2 . (13.1.1)


415

CHAPTER 13. APPENDIX

416

If τ1 and τ2 are two topologies on a set X, such that every member of τ2 is also a member of τ1 , then we say that τ1 is finer (or, larger) than τ2 , and that τ2 is coarser (or, smaller) than τ1 . If (X, τ ) is a topological space and E ⊆ X, then the topology induced by τ on E is the topology for which all open sets are intersections of open sets in τ with E. Let (X, τ ) and (Y, τ ) be two topological spaces. The topology on X × Y with base the collection of products of open sets in X and open sets in Y [which satisfies (13.1.1)] is called the product topology. A function f : X → Y is called continuous at x ∈ X if the inverse image under f of every open neighborhood of f (x) is an open neighborhood of x. It is called continuous on X if it is continuous at every x ∈ X. In particular, it is easy to see that if f : X → Y is continuous then lim f (xj ) = f (x) for every convergent sequence j→∞

{xj }j∈N of X converging to x ∈ X, that is, f is sequentially continuous. While in general the converse is false, in the case when X is in fact a metric space (see below), if f : X → Y is sequential continuous, then f is continuous. A topological space (X, τ ) is said to be separated, or Hausdorff, if for any two distinct elements x and y of X, there exist U , a neighborhood of x and V , a neighborhood of y, such that U ∩ V = ∅. In a Hausdorff space the limit of a convergent sequence is unique. A metric space is a set X equipped with a distance function (also called metric). This is a function d : X × X → [0, ∞) satisfying the following properties: 1. If x, y ∈ X, then d(x, y) = 0 if and only if x = y (nondegeneracy); 2. d(x, y) = d(y, x) for every x, y ∈ X (symmetry); 3. d(x, y) ≤ d(x, z) + d(z, y) for every x, y, z ∈ X (triangle inequality). Let d be a metric on X. The open balls B(x, r) := {y ∈ X : d(x, y) < r}, x ∈ X, r ∈ (0, ∞), are then the base of a Hausdorff topology on X (since it satisfies (13.1.1)), denoted τd . For a sequence {xj }j∈N of points in X and x ∈ X one has lim xj = x in the topology τd , if and only if the sequence of numbers j→∞

{d(xj , x)}j∈N converges to 0 as j → ∞. A sequence {xj }j∈N of points in (X, τd ) is called Cauchy if d(xj , xk ) → 0 as j, k → ∞. A metric space is called complete if every Cauchy sequence is convergent. A topological space X is called metrizable if there exists a distance function for which the open balls form a base for the topology (i.e., the topology on X coincides with τd ). A vector space X over C becomes a topological vector space if equipped with a topology that is compatible with its vector structure, that is, the operation of vector addition, and also the operation of multiplication by a complex number, are continuous maps X × X → X and C × X → X, respectively (where X × X and C × X are endowed each with the corresponding product topology). Observe that in order to specify a topology on a vector space, it suffices to give the system of neighborhoods of the zero element 0 ∈ X, since the system of

13.1. SUMMARY OF TOPOLOGICAL AND FUNCTIONAL . . .

417

neighborhoods of any other element of X is obtained from this via translation. In fact, it suffices to give a base of neighborhoods of 0. This is a family B of neighborhoods of 0 such that every neighborhood of 0 contains a member of B. A set E ⊆ X is then open if and only if, for every x ∈ E, there exists U ∈ B such that x + U ⊆ E, where x + U := {x + y : y ∈ U }. A topological vector space is called locally convex if it has a base of neighborhoods of 0 consisting of sets that are convex, balanced, and absorbing. A set U is called convex provided tx + (1 − t)y ∈ U whenever x, y ∈ U and t ∈ [0, 1]. A set U is called balanced if cx ∈ U whenever x ∈ U , c ∈ C and |c| ≤ 1. A set U is called absorbing provided for each x ∈ X there exists t > 0 such that x ∈ tU := {ty : y ∈ U }. A seminorm on a vector space X is a function p : X → R satisfying the properties 1. p(cx) = |c|p(x), for every c ∈ C and every x ∈ X (positively homogeneous); 2. p(x + y) ≤ p(x) + p(y), for every x, y ∈ X (sub-additive). In particular, a seminorm p on X satisfies p(0) = 0 and p(x) ≥ 0 for every x ∈ X. A family P of seminorms on X is called separating if, for each x ∈ X, x = 0, there exists p ∈ P such that p(x) = 0. Given a separating family of seminorms P on X, let B be the collection of sets of the form (13.1.2) x : p(x) < ε ∀ p ∈ P0 , P0 ⊆ P, P0 finite, ε > 0. Then B is a base of neighborhoods of 0 of a locally convex vector space topology τP on X called the topology generated by the family of seminorms P. In this context, it may be readily verified that if Y is a linear subspace of X and P|Y := {p|Y : p ∈ P}, then the topology induced on Y by τP coincides with τP|Y .

(13.1.3)

Conversely, if X is a locally convex topological vector space, for each U convex, balanced, and absorbing neighborhood of 0, the mapping pU : X → R defined by pU (x) := inf{t > 0 : t−1 x ∈ U }, x ∈ X (called the Minkowski functional associated with U ) is a seminorm. It is then not hard to see that the topology on X is generated by this family of seminorms (in the manner described above). Let P = {pj }j∈N be a countable family of seminorms that is also separating (thus, if x ∈ X and pj (x) = 0 for all j ∈ N, then x = 0). The topology generated by this family is metrizable. Indeed, the function d : X × X → R defined by d(x, y) :=

∞ j=1

2−j

pj (x − y) , 1 + pj (x − y)

for each x, y ∈ X,

(13.1.4)

is a distance on X and the topology τd induced by the metric d coincides with the topology generated by P. In the converse direction, it can be shown that the


418

topology of a locally convex space that is metrizable, endowed with a translation invariant metric, can be generated by a countable family of seminorms. A locally convex topological vector space that is metrizable and complete is called a Fr´ echet space. Thus, if a family P = {pj }j∈N of seminorms generates the topology of a Fréchet space X, then whenever pj (xk − xl ) → 0 as k, l → ∞ for every j ∈ N, there exists x ∈ X such that pj (xk − x) → 0 as k → ∞ for every j ∈ N. Let X bea vector space, {Xj }j∈J be a family of vector subspaces of X such Xj and if j1 , j2 ∈ J then there exists j3 ∈ J with Xj1 ⊂ Xj3 that X = j∈J

and Xj2 ⊂ Xj3 . Also assume that there exist topologies τj on Xj such that if Xj1 ⊂ Xj2 ; then the topology τj1 is finer than the topology τj2 j1 induced on Xj1 by Xj2 . Let W := W ⊂ X : W balanced, convex and such that W ∩ Xj is a neighborhood of 0 in Xj , ∀ j ∈ N . (13.1.5) Then W is a base of neighborhoods of 0 in a locally convex topology τ on X. Call this topology the inductive limit topology on X. If in addition whenever Xj1 ⊂ Xj2 we also have τj2 j1 = τj1 , we call the inductive limit strict. If the topology τ on X is the strict inductive limit of the topologies of an increasing sequence {Xn }n∈N , then the topology induced on Xn by the topology τ on X coincides with the initial topology τn on Xn , for every n ∈ N. In general, if X is a topological vector space, its dual space is the collection of all linear mappings f : X → C (also referred to as functionals) that are continuous with respect to the topology on X. In the case when X is a locally convex topological vector space and P is a family of seminorms generating the topology of X, then a seminorm q on X is continuous if and only if there exist N ∈ N, seminorms p1 , . . . , pN ∈ P, and a constant C ∈ (0, ∞), such that |q(x)| ≤ C max {p1 (x), . . . , pn (x)},

∀ x ∈ X.

(13.1.6)

This fact then gives a criterion for continuity for functionals on X, since if f : X → C is a linear mapping, then q(x) := |f (x)|, for x ∈ X, is a seminorm on X. In addition, if the family of seminorms P has the property that for any p1 , p2 ∈ P there exists p3 ∈ P with the property that max{p1 (x), p2 (x)} ≤ p3 (x) for every x ∈ X, then a linear functional f : X → C is continuous if and only if there exist a seminorm p ∈ P and a constant C ∈ (0, ∞), such that |f (x)| ≤ Cp(x),

∀ x ∈ X.

(13.1.7)

Given a nonempty set X and a family F of mappings f : X → C, denote by τF the collection of all unions of finite intersections of sets f −1 (V ), for f ∈ F and V open set in C. Then τF is a topology on X, and is the weakest topology on X that makes every f ∈ F continuous. We will refer to it as the F -topology on X.


419

Let X be a vector space, F be a separating vector space of linear functionals on X and τF be the F -topology on X. Then (X, τF ) is a locally convex topological space and its dual is F . In particular, a sequence {xj }j in X satisfies xj → 0 in τF as j → ∞ if and only if f (xj ) → 0 as j → ∞ for every f ∈ F . If X is a topological vector space and X denotes its dual space, then every x ∈ X induces a linear functional Fx on X defined by Fx (Λ) := Λ(x) for every Λ ∈ X and if we set F := {Fx : x ∈ X}, then F separates points in X . In particular, the F -topology on X , called the weak∗-topology on X , is locally convex and every linear functional on X that is continuous with respect to the weak∗-topology is precisely of the form Fx , for some x ∈ X. Moreover, a sequence {Λj }j ⊂ X converges to some Λ ∈ X , weak∗-topology on X , if and only if Λj (x) → Λ(x) as j → ∞ for every x ∈ X. An inspection of the definition of the weak∗-topology yields a description of the open sets in this topology. More precisely, if X is a topological vector space and for A ⊆ X and ε ∈ (0, ∞) we set (13.1.8) OA,ε := f ∈ X : |f (x)| < ε, ∀ x ∈ A , then the following equivalence holds: O ⊆ X is a weak∗-open neighborhood of 0 ∈ X ⇐⇒ there exist a set I, OAj ,εj . Aj ⊆ X finite and εj > 0 for each j ∈ I, such that O = j∈I

(13.1.9) The transpose of any linear and continuous operator between two topological vector spaces is always continuous at the level of dual spaces equipped with weak∗-topologies. Proposition 13.1. Assume X and Y are two given topological vector spaces, and denote by X , Y their duals, each endowed with the corresponding weak∗topology. Also, suppose T : X → Y is a linear and continuous operator, and define its transpose T t as the mapping T t : Y → X ,

T t (y ) := y ◦ T

for each

y ∈ Y .

(13.1.10)

Then T t is well-defined, linear, and continuous. Proof. For each y ∈ Y it follows that y ◦ T is a composition of two linear and continuous mappings. Hence, y ◦ T ∈ X which proves that T t : Y → X is well-defined. It is also clear from (13.1.10) that T t is linear. There remains to prove that T t is continuous. By linearity it suffices to check that T t is continuous at 0. With this goal in mind, fix an arbitrary finite subset A of X along with an arbitrary number ε ∈ (0, ∞), and define (13.1.11) OA,ε := x ∈ X : |x (x)| < ε, ∀ x ∈ A . := {T x : x ∈ A} and set Furthermore, introduce A := y ∈ Y : |y (y)| < ε, ∀ y ∈ A . O A,ε

(13.1.12)


420

) ⊆ OA,ε . Invoking the description from (13.1.9) we may conclude Then T t (O A,ε t that T is continuous at 0, and this finishes the proof of the proposition. Proposition 13.2. Suppose that X, Y , Z are topological vector spaces, and denote by X , Y , Z their duals, each endowed with the corresponding weak∗topology. In addition, assume that T : X → Y and R : Y → Z are two linear and continuous operators. Then (R ◦ T )t = T t ◦ Rt .

(13.1.13)

In particular, if T : X → Y is a linear, continuous, bijective map, with continuous inverse T −1 : Y → X, then T t : Y → X is also bijective and has a continuous inverse (T t )−1 : X → Y that satisfies (T t )−1 = (T −1 )t . Proof. Formula (13.1.13) is immediate from definitions, while the claims in the last part of the statement are direct consequences of (13.1.13) and the fact that the transpose of the identity is also the identity. We also state and prove an embedding result at the level of dual spaces endowed with the weak∗-topology. Proposition 13.3. Suppose X and Y are topological vector spaces such that X ⊆ Y densely and the inclusion map ι : X → Y , ι(x) := x for each x ∈ X, is continuous. Then Y endowed with the weak∗-topology embeds continuously in the space X endowed with the weak∗-topology, in the sense that the mapping ιt : Y −→ X

(13.1.14)

is well-defined, linear, injective and continuous. Under the identification of Y with ιt (Y ) ⊆ X , we therefore have Y → X .

(13.1.15)

Proof. The fact that ιt is well-defined, linear, and continuous follows directly from Proposition 13.1 and assumptions. Assume now that y ∈ Y is such that ιt (y ) = 0. Then from the fact that y ◦ ι : X → C is zero we deduce that y X = 0. Since y : Y → C is continuous and X is dense in Y , we necessarily have that y vanishes on Y , forcing y = 0 in Y . Keeping in mind that ιt is linear, this implies that ιt is injective. Theorem 13.4 (Hahn–Banach Theorem). Let X be a vector space (over complex numbers) and suppose p : X → [0, ∞) is a seminorm on X. Also assume that Y is a linear subspace of X and that ϕ : Y → C is a linear functional dominated by p on Y , that is, |φ(y)| ≤ p(y),

∀ y ∈ Y.

(13.1.16)

Then there exists a linear functional Φ : X → C satisfying Φ(y) = φ(y)

∀ y ∈ Y,

|Φ(x)| ≤ p(x)

∀ x ∈ X.

(13.1.17)


421

In the next installment we shall specialize the above considerations to various specific settings used in the book. The reader is reminded that Ω ⊆ Rn denotes a fixed, nonempty, arbitrary open set. The Topological Vector Space E(Ω) By τ we denote the topology on C ∞ (Ω) generated by the following family of seminorms: ⎧ ∞ ⎪ ⎨ pK,m : C (Ω) → R, ⎪ ⎩ pK,m (ϕ) :=

sup

x∈K, α∈Nn 0 , |α|≤m

|∂ α ϕ(x)|,

∀ ϕ ∈ C ∞ (Ω),

(13.1.18)

where K ⊂ Ω is a compact set and m ∈ N0 . Consequently, a sequence ϕj ∈ C ∞ (Ω), j ∈ N, converges in τ to a function ϕ ∈ C ∞ (Ω) as j → ∞, if and only if for any compact set K ⊂ Ω and any m ∈ N0 one has lim

sup

sup |∂ α (ϕj − ϕ)(x)| = 0.

j→∞ α∈Nn , |α|≤m x∈K 0

We will use the notation

E(Ω) = C ∞ (Ω), τ .

(13.1.19)

(13.1.20)

The space E(Ω) is locally convex and metrizable since its topology is defined by the family of countable seminorms {pKm ,m }m∈N0 where ∞ ( ˚m+1 ⊂ Ω, Km is compact for each m ∈ N0 and Ω = Km ⊂ K Km . m=0

(13.1.21) In addition, τ is independent of the family {Km }m∈N0 with the above properties and E(Ω) is complete. Thus, E(Ω) is a Frechét space. The Topological Vector Space E (Ω) Based on the discussion on dual spaces for locally convex topological vector spaces (see (13.1.7) and the remarks preceding it), it follows that the dual space of E(Ω) is the collection of all linear functionals u : E(Ω) → C for which there exist m ∈ N, a compact set K ⊂ Rn with K ⊂ Ω, and a constant C > 0 such that |u(ϕ)| ≤ C

sup

sup |∂ α (ϕ)(x)| ,

x∈K α∈Nn 0 , |α|≤m

∀ ϕ ∈ C ∞ (Ω).

(13.1.22)

This dual space will be endowed with the weak∗-topology induced by E(Ω) and we denote this topological space by E (Ω). Hence, if for each ϕ ∈ C ∞ (Ω) we consider the evaluation mapping Fϕ taking any functional u from the dual of E(Ω) into the number Fϕ (u) := u(ϕ) ∈ C, then the family F := {Fϕ : ϕ ∈ E(Ω)} separates points in the dual of E(Ω), and the weak∗-topology on this dual is the F -topology on it. In particular, if {uj }j∈N is a sequence in E (Ω) and u ∈ E (Ω), then uj → u in E (Ω) as j → ∞ ⇐⇒ (13.1.23) uj (ϕ) → u(ϕ) as j → ∞, ∀ ϕ ∈ C ∞ (Ω).


422

Moreover, a sequence {uj }j∈N in E (Ω) is Cauchy provided lim (uj −uk )(ϕ) = 0 j,k→∞

for every ϕ ∈ C ∞ (Ω) and the weak∗-topology on the dual of E(Ω) is locally convex and complete. The Topological Vector Space DK (Ω) Let K ⊆ Ω be a compact set in Rn . Denote by DK (Ω) the topological vector space of functions {f ∈ C ∞ (Ω) : supp f ⊆ K} with the topology induced by τ , the topology in E(Ω). Then the topology on DK (Ω) is generated by the family of seminorms ⎧ ⎨ pm : DK (Ω) → R, ⎩ pm (ϕ) :=

sup

x∈K, α∈Nn 0 , |α|≤m

|∂ α ϕ(x)|,

∀ ϕ ∈ C ∞ (Ω), supp ϕ ⊆ K, (13.1.24)

where m ∈ N0 . Hence, DK (Ω) is a Fréchet space. In addition, a linear mapping u : DK (Ω) → C is continuous if and only if there exist m ∈ N and a constant C > 0, both depending on K, such that |u(ϕ)| ≤ C

sup

x∈K, α∈Nn 0 , |α|≤m

|∂ α (ϕ)(x)| ,

∀ ϕ ∈ C ∞ (Ω), supp ϕ ⊆ K.

(13.1.25) The Topological Vector Space D(Ω) The topological vector space on C0∞ (Ω) endowed with the inductive limit topology of the Frechét spaces DK (Ω) will be denoted by D(Ω). In this setting, we have ϕj → ϕ in D(Ω) as j → ∞ ∃ K ⊆ Ω compact set such that ϕj ∈ DK (Ω), ∀ j, and ⇐⇒ → ϕ in DK (Ω) as j → ∞. ϕj − (13.1.26) The topology D(Ω) is locally convex and complete but not metrizable (thus not Fréchet) and is the strict inductive limit of the topologies {DKj (Ω)}j∈N , where the family {Kj }j∈N is as in (13.1.21). We also record an important result that is proved in [59, Theorem 6.6, p. 155]. Theorem 13.5. Let X be a locally convex topological vector space and suppose the map Λ : D(Ω) → X is linear. Then Λ is continuous if and only if for every D(Ω)

sequence {ϕj }j∈N in C0∞ (Ω) satisfying ϕj −−−→ 0 we have lim Λ(ϕj ) = 0 in X. j→∞

j→∞

The Topological Vector Space D (Ω) The dual space of D(Ω) endowed with the weak∗-topology is denoted by D (Ω). Hence, if {uj }j∈N is a sequence in D (Ω) and u ∈ D (Ω), then uj → u in D (Ω) as j → ∞ ⇐⇒ uj (ϕ) → u(ϕ) as j → ∞, ∀ ϕ ∈ C0∞ (Ω).

(13.1.27)


423

In addition, a sequence {uj }j∈N in D (Ω) is called Cauchy provided lim (uj − uk )(ϕ) = 0

j,k→∞

for every ϕ ∈ C0∞ (Ω). The weak∗-topology on the dual of D(Ω) is locally convex and complete and an inspection of this topology reveals that it coincides with the topology defined by the family of seminorms D (Ω) u → max |u(ϕj )|, 1≤j≤m

m ∈ N, ϕ1 , . . . , ϕm ∈ D(Ω).

(13.1.28)

The Topological Vector Space S(Rn ) The Schwartz class of rapidly decreasing functions is the vector space S(Rn ) := ϕ ∈ C ∞ (Rn ) : ∀ α, β ∈ Nn0 , sup |xβ ∂ α ϕ(x)| < ∞ , (13.1.29) x∈Rn

endowed with the topology generated by the family of seminorms {pk,m }k,m∈N0 defined by ⎧ n ⎪ ⎨ pk,m : S(R ) → R, (13.1.30) sup |xβ ∂ α ϕ(x)|, ∀ ϕ ∈ S(Rn ). ⎪ ⎩ pk,m (ϕ) := n x∈R , |α|≤m, |β|≤k

Hence, the topology generated by the family of seminorms {pk,m }k,m∈N0 on S(Rn ) is locally convex, metrizable, and since it is also complete, the space S(Rn ) is Frechét. Moreover, a sequence ϕj ∈ S(Rn ), j ∈ N, converges in S(Rn ) to a function ϕ ∈ S(Rn ) as j → ∞, if and only if for every m, k ∈ N0 one has β α x ∂ (ϕj − ϕ)(x) = 0. (13.1.31) lim sup j→∞ x∈Rn , |α|≤m, |β|≤k

The Topological Vector Space S (Rn ) By the discussion about dual spaces for locally convex topological vector spaces, it follows that the dual space of S(Rn ) is the collection of all linear functions u : S(Rn ) → C for which there exist m, k ∈ N0 , and a finite constant C > 0, such that |u(ϕ)| ≤ C sup sup xβ ∂ α ϕ(x) , ∀ ϕ ∈ S(Rn ). (13.1.32) x∈Rn α,β∈Nn 0 , |α|≤m, |β|≤k

We endow the dual of S(Rn ) with the weak∗-topology and denote the resulting locally convex topological vector space by S (Rn ). Hence, if {uj }j∈N is a sequence in S (Rn ) and u ∈ S (Rn ), then uj → u in S (Rn ) as j → ∞ ⇐⇒ uj (ϕ) → u(ϕ) as j → ∞, ∀ ϕ ∈ S(Rn ). Also, a sequence {uj }j∈N in S (Rn ) is called Cauchy provided lim (uj − uk )(ϕ) = 0 for every ϕ ∈ S(Rn ).

j,k→∞

(13.1.33)


424

13.2

Summary of Basic Results from Calculus, Measure Theory, and Topology

Proposition 13.6 (Multinomial theorem). If x = (x1 , . . . , xn ) ∈ Rn and N ∈ N are arbitrary, then n N N! xα . xj = (13.2.1) α! j=1 |α|=N

Theorem 13.7 (Binomial theorem). For any x, y ∈ Cn and any γ ∈ Nn0 we have γ! xα y β , (13.2.2) (x + y)γ = α!β! α+β=γ

(with the convention that z 0 := 1 for each z ∈ C). In the particular case when x = (1, . . . , 1) ∈ Cn and y = (−1, . . . , −1) ∈ Cn , formula (13.2.2) yields 0=

α+β=γ

γ! (−1)|β|, α!β!

∀ γ ∈ Nn0 with |γ| > 0.

(13.2.3)

Proposition 13.8 (Leibniz’s formula). Suppose that U ⊆ Rn is an open set, N ∈ N, and f, g : U → C are two functions of class C N in U . Then ∂ α (f g) =

β≤α

α! (∂ β f )(∂ α−β g) β!(α − β)!

in U,

(13.2.4)

for every multi-index α ∈ Nn0 of length ≤ N . It is useful to note that for each α, β ∈ Nn0 and x ∈ Rn , β

α

∂ (x ) =

⎧ ⎨ ⎩

α! xα−β (α − β)!

if β ≤ α,

0

otherwise.

(13.2.5)

Theorem 13.9 (Taylor’s formula). Assume U ⊆ Rn is an open convex set, and that N ∈ N. Also, suppose that f : U → C is a function of class C N +1 on U . Then for every x, y ∈ U one has f (x) =

1 (x − y)α (∂ α f )(y) α!

(13.2.6)

|α|≤N

+

|α|=N +1

N +1 α!

1

(1 − t)N (x − y)α (∂ α f )(tx + (1 − t)y) dt. 0

13.2. SUMMARY OF BASIC RESULTS FROM CALCULUS...

425

In particular, for each x, y ∈ U there exists θ ∈ (0, 1) with the property that f (x) =

1 (x − y)α (∂ α f )(y) α!

(13.2.7)

|α|≤N

+

|α|=N +1

1 (x − y)α (∂ α f )(θx + (1 − θ)y). α!

Theorem 13.10 (Rademacher’s theorem). If f : Rn → R is a Lipschitz function with Lipschitz constant less than or equal to M , for some M ∈ (0, ∞), then f is differentiable almost everywhere and ∂k f L∞ (Rn ) ≤ M for each k = 1, . . . , n. Theorem 13.11 (Lebesgue’s differentiation theorem). If f ∈ L1loc (Rn ), then 1 |f (y) − f (x)| dy = 0 for almost every x ∈ Rn . (13.2.8) lim+ ε→0 |B(x, ε)| B(x,ε) In particular, lim+

ε→0

1 |B(x, ε)|

f (y) dy = f (x)

for almost every x ∈ Rn .

(13.2.9)

B(x,ε)

Theorem 13.12 (Lebesgue’s dominated convergence theorem). Let (X, μ) be a positive measure space and assume that g ∈ L1 (X, μ) is a nonnegative function. If {fj }j∈N is a sequence of μ-measurable, complex valued functions on X, such that |fj (x)| ≤ g(x) for μ-almost every x ∈ X and f (x) := lim fj (x) exists (in j→∞

C) for μ-almost every x ∈ X, then f ∈ L1 (X, μ) and lim X |fj − f | dμ = 0. j→∞

In particular, lim X fj dμ = X f dμ. j→∞

Theorem 13.13 (Young’s inequality). Assume that 1 ≤ p, q, r ≤ ∞ are such that p1 + 1q = 1r + 1. Then for every f ∈ Lp (Rn ), g ∈ Lq (Rn ) it follows that f ∗ g is well-defined almost everywhere in Rn , f ∗ g ∈ Lr (Rn ), and f ∗ gLr (Rn ) ≤ f Lp(Rn ) gLq (Rn ) . Theorem 13.14. Let X and Y be two Hausdorff topological spaces. (a) If Λ : X → Y is continuous, then Λ is sequentially continuous. (b) If Λ : X → Y is sequentially continuous and X is metrizable, then Λ is continuous. For a proof of Theorem 13.14 see [59, Theorem, p. 395]. Definition 13.15. A rigid transformation, or isometry, of the Euclidean space Rn is any distance preserving mapping of Rn , that is, any function T : Rn → Rn satisfying |T (x) − T (y)| = |x − y|,

∀ x, y ∈ Rn .

(13.2.10)


426

A rigid transformation of Rn is any distance preserving mapping of Rn , i.e., any function T : Rn → Rn satisfying |T x−T y| = |x−y| for every x, y ∈ Rn . The rigid transformations of the Euclidean space Rn are precisely those obtained by composing a translation with a mapping in Rn given by an orthogonal matrix. In other words, a mapping T : Rn → Rn is a rigid transformation of Rn if and only if there exist x0 ∈ Rn and an orthogonal matrix A : Rn → Rn with the property that (13.2.11) T (x) = x0 + Ax, ∀ x ∈ Rn . For a proof of the following version of the Arzelà–Ascoli theorem see [57, Corollary 34, p. 179]. Theorem 13.16 (Arzel` a–Ascoli theorem). Let F be an equicontinuous family of real-valued functions on a separable space X. Then each sequence {fj }j∈N in F which is bounded at each point has a subsequence {fjk }k∈N that converges pointwise to a continuous function, the converges being uniform on each compact subset of X. Theorem 13.17 (Riesz’s representation theorem for positive functionals). Let X be a locally compact Hausdorff topological space and Λ a positive linear functional on the space of continuous, compactly supported functions on X (denoted by C00 (X)). Then there exists a unique σ-algebra M on X, which contains all Borel sets on X, and a unique measure μ : M → [0, ∞] that represents Λ, that is, the following hold:

(i) Λf = X f dμ for every continuous, compactly supported function f on X; (ii) μ(K) < ∞ for every compact K ⊂ X; (iii) For every E ∈ M we have μ(E) = inf{μ(V ) : E ⊂ V, V open}; (iv) μ(E) = sup {μ(K) : K ⊂ E, K compact} for every open set E and every E ∈ M with μ(E) < ∞; (v) If E ∈ M, A ⊂ E, and μ(E) = 0, then μ(A) = 0. Theorem 13.18 (Riesz’s representation theorem for complex functionals). Let X be a locally compact Hausdorff topological space and consider the space of continuous functions on X vanishing at infinity, that is, Coo (X) := {f ∈ C 0 (X) : ∀ ε > 0, ∃ compact K ⊂ X such that |f (x)| < ε for x ∈ X \ K}.

(13.2.12)

Then Coo (X) is the closure in the uniform norm of C00 (X) and for every bounded linear functional Λ : Coo (X) → C there exists a unique regular complex Borel measure μ on X such that Λf = X f dμ for every f ∈ Coo (X) and Λ = |μ|(X).

13.2. SUMMARY OF BASIC RESULTS FROM CALCULUS...

427

Theorem 13.19 (Riesz’s representation theorem for locally bounded functionals). Let X be a locally compact Hausdorff topological space and assume that Λ : C00 (X) → R is a linear functional that is locally bounded, in the sense that for each compact set K ⊂ X there exists a constant CK ∈ (0, ∞) such that |Λf | ≤ CK sup |f (x)|, x∈K

∀ f ∈ C00 (X) with supp f ⊆ K.

(13.2.13)

Then there exist two measures μ1 , μ2 , taking Borel sets from X into [0, ∞], and satisfying properties (ii)–(iv) in Theorem 13.17, such that Λf = f dμ1 − f dμ2 for every f ∈ C00 (X). (13.2.14) X

X

The reader is warned that since both μ1 and μ2 are allowed to take the value ∞, their difference μ1 − μ2 is not well-defined in general. This being said, μ1 − μ2 is a well-defined finite signed measure on each compact subset of X. Proposition 13.20 (Urysohn’s lemma). If X is a locally compact Hausdorff space and K ⊂ U ⊂ X are such that K is compact and U is open, then there exists a function f ∈ C00 (U ) that satisfies f = 1 on K and 0 ≤ f ≤ 1. Theorem 13.21 (Vitali’s convergence theorem). Let (X, μ) be a positive measure space with μ(X) < ∞. Suppose {fk }k∈N is a sequence of functions in L1 (X, μ) and that f is a function on X (all complex-valued) satisfying: (i) fk (x) → f (x) for μ-almost every x ∈ X as k → ∞; (ii) |f | < ∞ μ-almost everywhere in X; (iii) {fk }k∈N is uniformly integrable, in the sense that for every ε > 0 there exists δ > 0 such that for every k ∈ N we have E fk dμ < ε whenever E ⊆ X is a μ-measurable set with μ(E) < δ. Then f ∈ L1 (X, μ) and

fk − f dμ = 0.

lim

k→∞

In particular, lim

k→∞ X

fk dμ =

X

(13.2.15)

X

f dμ.

See, for example, [58, p. 133]. Proposition 13.22. Let (X, μ) be a positive measure space and suppose that f ∈ L1 (X, μ). Then for every ε > 0 there exists δ > 0 such that for every μ-measurable set A ⊆ X satisfying μ(A) < δ we have A |f | dμ < ε. Proof. Consider the measure λ := |f |μ on X. Then λ is absolutely continuous with respect to μ and the (ε, δ) characterization of absolute continuity of measures (see, e.g., [58, Theorem 6.11, p. 124]) yields the desired conclusion.


428

13.3

Custom-Designing Smooth Cut-Off Functions

Lemma 13.23. Let f : R → R be the function defined by e−1/x , if x > 0, f (x) := ∀ x ∈ R. 0, if x ≤ 0,

(13.3.1)

Then f is of class C ∞ on R. Proof. Denote by C the collection of functions g : R → R for which there exists a polynomial P such that e−1/x P (1/x), if x > 0, ∀ x ∈ R. (13.3.2) g(x) := 0, if x ≤ 0, Recall that if h : R → R is a continuous function that is differentiable on R \ {0} and for which there exists L ∈ R such that lim− h (x) = L = lim+ h (x), then x→0

x→0

h is also differentiable at the origin and h (0) = L. An immediate consequence of this fact is that any g ∈ C is differentiable and g ∈ C. In turn, this readily gives that any g ∈ C is of class C ∞ on R. Since f defined in (13.3.1) clearly belongs to C, it follows that f is of class C ∞ on R.

Lemma 13.24. The function φ : Rn → R defined by 1 Ce |x|2 −1 if x ∈ B(0, 1), φ(x) := (13.3.3) 0 if x ∈ Rn \ B(0, 1), −1

∞ 2 where C := ωn−1 0 e1/(ρ −1) ρn−1 dρ ∈ (0, ∞), satisfies the following properties: φ(x) dx = 1. (13.3.4) φ ∈ C ∞ (Rn ), φ ≥ 0, supp φ ⊆ B(0, 1), and Rn

Proof. That φ ≥ 0 and supp φ ⊆ B(0, 1) is immediate from its definition. Also, since φ(x) = f (1 − |x|2 ) for x ∈ Rn where f is as in (13.3.1), invoking

Lemma 13.23 it follows that φ ∈ C ∞ (Rn ). Finally, the condition Rn φ(x) dx = 1 1

follows upon observing that based on (13.8.9) we have Rn e |x|2 −1 dx = 1/C. Proposition 13.25. Let F0 , F1 ⊂ Rn be two nonempty sets with the property that dist(F0 , F1 ) > 0. Then there exists a function ψ : Rn → R with the following properties: ψ ∈ C ∞ (Rn ),

0 ≤ ψ ≤ 1,

∀ α ∈ Nn0 ∃ Cα ∈ (0, ∞)

such that

ψ = 0 on F0 ,

ψ = 1 on F1 ,

|∂ α ψ(x)| ≤

Cα dist (F0 , F1 )|α|

and ∀ x ∈ Rn . (13.3.5)

13.4. PARTITION OF UNITY

429

)1 := x ∈ Rn : dist (x, F1 ) ≤ r/4 . Proof. Let r := dist(F0 , F1 ) > 0 and set F 4 n Also, with φ as in Lemma 13.24, define the function θ(x) := r φ(4x/r) for x ∈ Rn . Then θ ∈ C ∞ (Rn ), θ ≥ 0, supp θ ⊆ B(0, r/4), and θ(x) dx = 1. (13.3.6) Rn

We claim that the function ψ : Rn → R defined by ψ := χF1 ∗ θ has the desired properties. To see why this is true, note that since θ(x − y) dy ∀ x ∈ Rn , (13.3.7) ψ(x) = 1 F

from the properties of θ it is immediate that ψ ∈ C ∞ (Rn ) and 0 ≤ ψ ≤ 1. Furthermore, for α ∈ Nn0 and x ∈ Rn we may write |α| n α 4 4 (∂ φ) 4(x − y)/r dy r r F 1 α |α| Cα ∂ φ(y) dy = Cα r−|α| = , (13.3.8) ≤ 4r dist (F0 , F1 )|α| Rn

where Cα := 4|α| Rn ∂ α φ(y) dy is a positive, finite number independent of r. We are left with checking the fact that ψ = j on Fj , j = 0, 1. First, if x ∈ F0 , )1 , )1 , hence θ(x − y) = 0 for every y ∈ F then |x − y| ≥ 3r/4 for every y ∈ F which when combined with (13.3.7) implies ψ(x) = 0. Second, if x ∈ F1 , then ) B(x, r/4)

⊆ F1 . Since the support of θ(x − ·) ⊂ B(x, r/4) the latter implies ψ(x) = Rn θ(x − y) dy = 1. The proof of the proposition is complete. |∂ α ψ(x)| ≤

A trivial yet useful consequence of the above result is as follows. Proposition 13.26. If U ⊆ Rn is open and K ⊂ U is compact, then there exists a function ψ : Rn → R that is of class C ∞ , satisfies 0 ≤ ψ(x) ≤ 1 for every x ∈ Rn and ψ(x) = 1 for every x ∈ K, and which has compact support, contained in U . Proof. Since K is a compact contained in U we have r := dist (K, Rn \ U ) is contained in (0, ∞). Proposition 13.25 applied with F1 := K and with set F0 := {x ∈ Rn : dist (x, K) > r/2} then yields the desired function ψ.

13.4

Partition of Unity

Lemma 13.27. If C ⊂ Rn is compact, and U ⊆ Rn is an open set such that ˚ ⊂ D ⊂ U. C ⊂ U , then there exists a compact set D ⊆ Rn such that C ⊂ D Proof. Let V = U c ∩ B(0, R), where R > 0 is large enough so that U ⊂ B(0, R). Then V is compact and disjoint from C so r := dist(V, C) = inf x − y > 0. x∈C

y∈V


430 Hence, if we set D :=

(

B(x, r/4),

(13.4.1)

x∈C

˚ ⊂ D ⊂ U. then D is compact and C ⊂ D Lemma 13.28. Suppose that K ⊆ Rn is a compact set and that {Oj }1≤j≤k is a finite open cover of K. Then there are compact sets Dj ⊂ Oj , 1 ≤ j ≤ k, with the property that k ( ˚j . D K⊂ (13.4.2) j=1

k Proof. Set C1 := K \ j=2 Oj ⊆ O1 . Since C1 is compact, Lemma 13.27 shows ˚1 ⊂ D1 ⊂ O1 . that there exists a compact set D1 with the property that C1 ⊂ D Next, proceeding inductively, suppose that m ∈ N is such that 1 ≤ m < k and that m sets Dk 1 , . . . , Dm have been constructed with the property that compact m ˚ D ∪ O K⊂ j=1 j j=m+1 j . Introduce ⎛ Cm+1 := K \ ⎝

m ( j=1

˚j ∪ D

k (

⎞ Oj ⎠ .

(13.4.3)

j=m+2

Clearly, Cm+1 is a compact subset of Om+1 . By once again invoking Lemma 13.27, there exists a compact set Dm+1 ⊂ Om+1 such that Cm+1 ⊂ ˚m+1 . After k iterations, this procedure yields a family of sets C1 , . . . , Ck D which have all the desired properties. Theorem 13.29 (Partition of unity for compact sets). Let K ⊂ Rn be a compact set, and let {Oj }1≤j≤N be a finite open cover of K. Then there exists a finite collection of C ∞ functions ϕj : Rn → R, 1 ≤ j ≤ N , satisfying the following properties: (i) For every 1 ≤ j ≤ N , the set supp (ϕj ) is compact and contained in Oj ; (ii) For every 1 ≤ j ≤ N , one has 0 ≤ ϕj ≤ 1; (iii)

N

ϕj (x) = 1 for every x ∈ K.

j=1

The family {ϕj : 1 ≤ j ≤ N } is called a partition of unity subordinate to the cover {Oj }1≤j≤N of K. Proof. Let {Oj }1≤j≤N be any finite open cover for K. From Lemma 13.28 we N ˚j . D know that there exist compact sets Dj ⊆ Oj , 1 ≤ j ≤ N , such that K ⊂ j=1


431

By Proposition 13.26, for each 1 ≤ j ≤ N , choose a C ∞ function ηj : Rn → [0, ∞) that is positive on Dj and has compact support in Oj . It follows that N N ˚j , so we can define for each j ∈ {1, . . . , N } the D ηj (x) > 0 for all x ∈ j=1

j=1

function ψj :

N (

˚j → R, D

ψj (x) :=

i=1

ηj (x) , N ηk (x)

∀x ∈

N (

˚j . D

(13.4.4)

i=1

k=1 N ˚j . ˚⊆U ⊆ D By Lemma 13.28, there exists a compact set U ⊆ Rn with K ⊆ U j=1

We apply Proposition 13.26 to obtain a C ∞ function f : Rn → [0, 1] that satisfies ˚. Then for each f (x) = 1 for x ∈ K and having compact support contained in U j ∈ {1, . . . , N } we define the function ϕj := f ψj acting from Rn into R. It is ∞ n not hard to see that each ϕj is C in R , has compact support, contained in Oj , 0 ≤ ϕj ≤ 1, and ϕj (x) = 1 for all x ∈ K. 1≤j≤N

Definition 13.30. (i) A family {Fj }j∈I of subsets of Rn is said to be locally finite in E ⊆ Rn provided every x ∈ E has a neighborhood O ⊆ Rn with the property that the set {j ∈ I : Fj ∩ O = ∅} is finite. (ii) Given a collection of functions fj : Ω → R, j ∈ I, defined in some fixed fj is called locally finite in E ⊆ Rn subset Ω of Rn , the sum j∈I

provided the family of sets {x ∈ Ω : fj (x) = 0}, indexed by j ∈ I, is locally finite in E. Exercise 13.31. Show that a family {Fj }j∈I of subsets of Rn is locally finite in the open set E ⊆ Rn if and only if for every compact K ⊆ E the collection {j ∈ I : Fj ∩ K = ∅} is finite. Exercise 13.32. Show that if the family {Fj }j∈I of subsets of Rn is locally finite in E ⊆ Rn , then {Fj }j∈I is also locally finite in E. Exercise 13.33. Show that ifthe family {Fj }j∈I of closed subsets of Rn is locally finite in E ⊆ Rn , then (E ∩ Fj ) is a relatively closed subset of E. j∈I

Proof. Let x∗ ∈ E be such that x∗ ∈ / Fj for every j ∈ I. Then there exists r > 0 with the property that B(x∗ , r) ∩ Fj = ∅ for every j ∈ I \ I∗ , where I∗ is a finite subset of I. Hence, by eventually further decreasing r, it can be assumed that B(x∗ , r) ∩ Fj = ∅ for every I∗ as well. Thus, B(x∗ , r) ∩ E is a Fj . This proves that relative neighborhood of x∗ in E that is disjoint from j∈I E\ Fj is relatively open in E hence, (E ∩ Fj ) is a relatively closed j∈I

subset of E.

j∈I


432

Theorem 13.34 (Partition of unity for arbitrary opencovers). Let {Ok }k∈I be Ok . Then there exists an arbitrary family of open sets in Rn and set Ω := k∈I

an at most countable collection {ϕj }j∈J of C ∞ functions ϕj : Ω → R satisfying the following properties: (i) For every j ∈ J there exists k ∈ I such that ϕj is compactly supported in Ok ; (ii) For every j ∈ J, one has 0 ≤ ϕj ≤ 1 in Ω; (iii) The family of sets {x ∈ Ω : ϕj (x) = 0}, indexed by j ∈ J, is locally finite in Ω; (iv) ϕj (x) = 1 for every x ∈ Ω. j∈J

The family {ϕj }j∈J is called a partition of unity subordinate to the family {Ok }k∈I . Proof. Start by defining Ωj := x ∈ Ω : |x| ≤ j and dist (x, ∂Ω) ≥ 1j , Then Ω =

∞

j ∈ N,

(13.4.5)

Ωj and

j=1

Ωj ⊆ Rn is compact,

Ωj ⊂ ˚ Ωj+1

for every j ∈ N.

(13.4.6)

Proceed now to define the compact sets Kj := Ωj \ ˚ Ωj−1

for every j ≥ 3.

(13.4.7)

As such, from (13.4.7) and (13.4.6) we have c K j = Ωj ∩ ˚ Ωj−1 ⊆ ˚ Ωj+1 ∩ (Ωj−2 )c

for every j ≥ 3.

(13.4.8)

K2 := Ω2 ,

Finally, we define the following families of open sets O2 := Ok ∩ ˚ Ω3 : k ∈ I , Oj := Ok ∩ Ωj+1 \ ˚ Ωj−2 : k ∈ I Making use of (13.4.6), for every j ≥ 3 we have that ◦ ◦ ◦ Ωj+1 ∩ (Ωj−2 )c Ωj+1 \ Ωj−2 = Ωj+1 ∩ (Ωj−2 )c = ˚ c =˚ Ωj+1 ∩ Ωj−2 = ˚ Ωj+1 ∩ (Ωj−2 )c .

∀ j ≥ 3. (13.4.9)

(13.4.10)

Hence, (13.4.7), (13.4.9), and (13.4.10), imply that Oj is an open cover for Kj for every j ≥ 3. Also, from the definitions of K2 and O2 and (13.4.6), we obtain that O2 is an open cover for K2 . Since the Kj ’s are compact, these open covers can be refined to finite subcovers in each case. As such, for each


433

j = 2, 3, . . . , we can apply Theorem 13.29 to obtain a finite partition of unity {ϕ : ϕ ∈ Φj } for Kj subordinate to Oj . Also note that due to (13.4.10) and (13.4.9), we necessarily have that, for each j ∈ {2, 3, . . . }, Ωj ∩ O = ∅ for every O ∈ Ok , for every k ∈ N satisfying k ≥ j + 2. This ensures that the family {supp ϕ : ϕ ∈ Φj , j ≥ 2} is locally finite in Ω, so we can define s(x) :=

ϕ(x),

for every x ∈ Ω.

(13.4.11)

j≥2 ϕ∈Φj

Given that differentiability is a local property, it follows that s is of class C ∞ in Ω. Moreover, note that s(x) > 0 for every x ∈ Ω, since 0 ≤ ϕ ≤ 1 for all ϕ(x) = 1. ϕ ∈ Φj , j = 2, 3, . . . , and if x ∈ Ωj , some j ∈ {2, 3, . . . }, then ϕ∈Φj

Consequently, 1/s is also a C ∞ function in Ω. It is then clear that the collection Φ := ϕ/s : ϕ ∈ Φj , j = 2, 3, . . . is a partition of unity subordinate to the family of open sets {Ok }k∈I (in the sense described in the statement of the theorem). Theorem 13.35 (Partition of unity with preservation of indexes). Let {Ok }k∈I be an arbitrary family of open sets in Rn and set Ω := Ok . Then there exists k∈I

a collection {ψk }k∈I of C ∞ functions ψk : Ω → R satisfying the following properties: (i) For every k ∈ I the function ψk vanishes outside of a relatively closed subset of Ok ; (ii) For every k ∈ I, one has 0 ≤ ψk ≤ 1 in Ω; (iii) The family of sets {x ∈ Ω : ψk (x) = 0}, indexed by k ∈ I, is locally finite in Ω; (iv)

ψk (x) = 1 for every x ∈ Ω.

k∈I

Proof. Let {ϕj }j∈J be a partition of unity subordinate to the family {Ok }k∈I , and denote by f : J → I a function with the property that, for every j ∈ J, the function ϕj is compactly supported in Of (j) . That this exists is guaranteed by Theorem 13.34. For every k ∈ I then define ψk (x) :=

ϕj (x),

∀ x ∈ Ω.

(13.4.12)

j∈f −1 ({k})

Note that the sum is locally finite in Ω, hence ψk : Ω → R is a well-defined nonnegative function, of class C ∞ in Ω, for every k ∈ I. In addition, the result from Exercise 13.33 shows that, for every k ∈ I, ϕk vanishes outside a relatively


434

closed subset of Ok . Furthermore, since {f −1 ({k})}k∈I is a partition of J into mutually disjoint subsets, we may compute

ψk (x) =

ϕj (x) =

k∈I j∈f −1 ({k})

k∈I

ϕj (x) = 1 ∀ x ∈ Ω.

(13.4.13)

j∈J

Incidentally, this also shows that, necessarily, 0 ≤ ψk (x) ≤ 1 for every k ∈ I and x ∈ Ω. Finally, the fact that the family of sets {x ∈ Ω : ψk (x) = 0}, k ∈ I, is locally finite in Ω is inherited from the corresponding property of the ϕj ’s.

13.5

The Gamma and Beta Functions

The gamma function is defined as

∞

Γ(z) :=

tz−1 e−t dt,

for z ∈ C, Re z > 0.

(13.5.1)

0

√ It is easy to check that Γ(1) = 1, Γ(1/2) = π and via integration by parts that Γ(z + 1) = zΓ(z) for z ∈ C, Re z > 0. By analytic continuation, the function Γ(z) is extended to a meromorphic function defined for all complex numbers z except for z = −n, n ∈ N0 , where the extended function has simple poles with residue (−1)n /n! and this extension satisfies Γ(z + 1) = z Γ(z) for z ∈ C \ {−n : n ∈ N0 }.

(13.5.2)

By induction it follows that for every n ∈ N we have Γ(n) = (n − 1)!, Γ Γ

1 2 1 2

(13.5.3)

1 · 3 · 5 · · · (2n − 1) √ (2n)! √ +n = π = 2n π, n 2 2 n! √ (−1)n 2n (−1)n 22n n! √ −n = π= π. 1 · 3 · 5 · · · (2n − 1) (2n)!

(13.5.4) (13.5.5)

The volume of the unit ball B(0, 1) in Rn , which we denote by |B(0, 1)|, and the surface area of the unit sphere in Rn , denoted here by ωn−1 , have the following formulas: n

|B(0, 1)| =

π2 n , Γ 2 +1

n

2π 2 . Γ n2

(13.5.6)

Re z, Re w > 0.

(13.5.7)

ωn−1 = n|B(0, 1)| =

Next, consider the so-called beta function

1

tz−1 (1 − t)w−1 dt,

B(z, w) := 0

13.6. SURFACES IN RN AND SURFACE INTEGRALS

435

Clearly B(z, w) = B(w, z). Making the change of variables t = u/(u + 1), u ∈ (0, ∞), it follows that ∞ 1 z+w B(z, w) = uw−1 du (13.5.8) u+1 0 whenever Re z, Re w > 0. The basic identity relating the gamma and beta functions reads B(z, w) =

Γ(z)Γ(w) , Γ(z + w)

Re z, Re w > 0.

(13.5.9)

∞ This is easily proved starting with (13.5.8), writing Γ(z + w) = 0 tz+w−1 e−t dt and expressing B(z, w)Γ(z + w) as a double integral, then making the change of variables s := t/(u + 1). A useful consequence of identity (13.5.9) is the following formula π/2 b+1 a+1 b+1 1 1 Γ( a+1 a b 2 )Γ( 2 ) , if a, b > −1. (sin θ) (cos θ) dθ = B = 2 2 2 2 Γ( a+b+2 ) 0 2 (13.5.10) Indeed, making the change of variables u := (sin θ)2 , the integral in the leftmost

1 side of (13.5.10) becomes 12 0 u(a−1)/2 (1 − u)(b−1)/2 du. For further reference, let us also note here that π a+1 b+1 1 + (−1)b ·B , (sin θ)a (cos θ)b dθ = 2 2 2 0 =

b+1 1 + (−1)b Γ( a+1 2 )Γ( 2 ) · , a+b+2 2 Γ( 2 )

(13.5.11)

whenever a, b > −1. This is proved by splitting the domain of integration into (0, π/2) ∪ (π/2, π), making a change of variables θ → θ − π/2 in the second integral, and invoking (13.5.10).

13.6

Surfaces in Rn and Surface Integrals

Definition 13.36. Given n ≥ 2 and k ∈ N ∪ {∞}, a C k surface (or, surface of class C k ) in Rn is a subset Σ of Rn with the property that for every x∗ ∈ Σ there exists an open neighborhood U (x∗ ) such that Σ ∩ U (x∗ ) = P (O)

(13.6.1)

where O is an open subset of Rn−1 and P : O −→ Rn

is an injective function of class C k satisfying

rank [DP (u)] = n − 1, for all u = (u1 , . . . , un−1 ) ∈ O,

(13.6.2)


436

where DP is the Jacobian matrix of P , that is, D(P1 , . . . , Pn ) DP (u) = (u), D(u1 , . . . , un−1 )

u ∈ O.

(13.6.3)

The function P in (13.6.2) is called a local parametrization of class C k near x∗ and Σ ∩ U (x∗ ) a parametrizable patch. In the case when (13.6.1) holds when we formally take r = +∞, that is, in the case when Σ = P (O), we call P a global parametrization of the surface Σ. Proposition 13.37. If O is an open subset of Rn−1 and P : O → Rn satisfies (13.6.2)–(13.6.3) for k = 1 , then P : O → P (O) is a homeomorphism. Definition 13.38. Assume n ≥ 3. If v1 = (v11 , . . . , v1n ), . . . , vn−1 = (vn−1 1 , . . . , vn−1 n ) are n − 1 vectors in Rn , their cross product is defined as ⎛ v1 × v2 × · · · × vn−1

v11 v21 .. .

⎜ ⎜ ⎜ := det ⎜ ⎜ ⎝ vn−1 1 e1

v12 v22 .. .

... ... ... ... ...

vn−1 2 e2

v1n v2n .. . vn−1 n en

⎞ ⎟ ⎟ ⎟ ⎟, ⎟ ⎠

(13.6.4)

where the determinant is understood as computed by formally expanding it with respect to the last row, the result being a vector in Rn . More precisely, v1 × · · · × vn−1 :=

n

j+1

(−1)

⎛ ⎜ det ⎝

j=1

v11 .. .

... . . . . ..

vj−1 .. .

vj+1

vn−1 1

...

vn−1 j−1

vn−1 j+1

...

... .. .

v1n

(13.6.5) ⎞ ⎟ ⎠ ej .

. . . vn−1 n

Definition 13.39. Let Σ ⊂ Rn , n ≥ 2, be a C 1 surface and assume that P : O → Rn , with O open subset of Rn−1 , is a local parametrization of Σ of class C 1 near some point X ∗ ∈ Σ. Also, suppose that f : Σ → R is a continuous function on Σ that vanishes outside of a compact subset of P (O). We then define f (x) dσ(x) := (f ◦ P )(u) |∂1 P (u) × · · · × ∂n−1 P (u)| du1 . . . dun−1 O

Σ

if n ≥ 3, and

(13.6.6)

f (x) dσ(x) := Σ

O

(f ◦ P )(u)|P (u)| du

if n = 2.

(13.6.7)

In (13.6.6), dσ stands for the surface measure (or, surface area element), whereas in (13.6.7), dσ stands for the arc-length measure.

13.8. POLAR COORDINATES AND INTEGRALS ON SPHERES

13.7

437

Integration by Parts and Green’s Formula

Definition 13.40. We say that a nonempty open set Ω ⊆ Rn , where n ≥ 2, is a C k domain (or, a domain of class C k ), for some k ∈ N0 ∪ {∞}, provided the following holds. For every point x∗ ∈ ∂Ω there exist R > 0, an open interval I ⊂ R with 0 ∈ I, a rigid transformation T : Rn → Rn with T (x∗ ) = 0, along with a function φ of class C k that maps B(0, R) ⊆ Rn−1 into I with the property that φ(0) = 0 and, if C denotes the (open) cylinder B(0, R)×I ⊆ Rn−1 ×R = Rn , then C ∩ T (Ω) = {x = (x , xn ) ∈ C : xn > φ(x )},

(13.7.1)

C ∩ ∂T (Ω) = {x = (x , xn ) ∈ C : xn = φ(x )},

(13.7.2)

C ∩ (T (Ω))c = {x = (x , xn ) ∈ C : xn < φ(x )}.

(13.7.3)

If Ω ⊆ Rn is a C k domain for some k ∈ N0 ∪ {∞}, then it may be easily verified that ∂Ω is a C k surface. Theorem 13.41 (Integration by parts formula). Suppose Ω ⊂ Rn is a domain of class C 1 and ν = (ν1 , . . . , νn ) denotes its outward unit normal. Let k ∈ {1, . . . , n} and assume f, g ∈ C 1 (Ω) ∩ C 0 (Ω ) are such that ∂k f , ∂k g ∈ L1 (Ω) and there exists a compact set K ⊂ Rn with the property that f = 0 on Ω \ K. Then, (∂k f )g dx = − f (∂k g) dx + f gνk dσ, (13.7.4) Ω

Ω

∂Ω

where σ is the surface measure on ∂Ω. For the sense in which the last integral in (13.7.4) should be understood see Definition 13.39. An immediate corollary of Theorem 13.41 is Green’s formula that is stated next (recall (7.1.14)). Theorem 13.42 (Green’s formula). Suppose Ω ⊂ Rn is a bounded domain of class C 1 and ν denotes its outward unit normal. If f, g ∈ C 2 (Ω), then ∂f ∂g −g f Δg dx = gΔf dx + f dσ. (13.7.5) ∂ν ∂ν Ω Ω ∂Ω

13.8

Polar Coordinates and Integrals on Spheres

Assume that n ≥ 3 and R > 0 are fixed. For ρ ∈ (0, R), θj ∈ (0, π), j ∈ {1, . . . , n − 2}, and θn−1 ∈ (0, 2π), set x1 := ρ cos θ1 , x2 := ρ sin θ1 cos θ2 , x3 := ρ sin θ1 sin θ2 cos θ3 ,

(13.8.1)


438 .. .

xn−1 := ρ sin θ1 sin θ2 . . . sin θn−2 cos θn−1 , xn := ρ sin θ1 sin θ2 . . . sin θn−2 sin θn−1 . The variables θ1 , . . . , θn−1 , ρ are called polar coordinates. Definition 13.43. Assume that x∗ ∈ Rn , n ≥ 3, is a fixed point, and R > 0 is arbitrary. The standard parametrization of the ball B(x∗ , R) is defined as P : (0, π)n−2 × (0, 2π) × (0, R) −→ Rn , P(θ1 , θ2 , . . . , θn−1 , ρ) := x∗ + (x1 , x2 , . . . , xn ),

(13.8.2)

where x1 , . . . , xn are as in (13.8.1). The function P in (13.8.2) is injective, of class C ∞ , takes values in B(x∗ , R), its image differs from B(x∗ , R) by a subset of measure zero and det (DP)(θ1 , θ2 , · · · , θn−1 , ρ)=ρn−1 (sin θ1 )n−2 (sin θ2 )n−3 · · · (sin θn−2 ), (13.8.3) at every point in its domain, where DP denotes the Jacobian of P. Using this standard parametrization for the unit sphere in Rn , we see that

ωn−1 =

π

π

π

2π

... 0

0

0

0

(sin ϕ1 )n−2 (sin ϕ2 )n−3 . . . (sin ϕn−2 ) dϕn−1 dϕn−2 . . . dϕ1 . (13.8.4)

This parametrization of the sphere B(x∗ , R) may also be used to prove the following theorem. Theorem 13.44 (Spherical Fubini and polar coordinates). Let f ∈ L1loc (Rn ), n ≥ 2. Then for each x∗ ∈ Rn and each R > 0 the following formulas hold: R f dx = f dσ dρ, (13.8.5) 0

B(x∗ ,R)

∂B(x∗ ,ρ) R

f dx = S n−1

0

B(x∗ ,R)

R

f (x∗ + ρω)ρn−1 dσ(ω) dρ

f (ρω)ρn−1 dσ(ω) dρ.

= 0

0

0

f dσ dρ,

(13.8.8)

∂B(0,ρ) ∞

f (ρω)ρn−1 dσ(ω) dρ.

f dx = Rn

(13.8.7)

∂B(x∗ ,1)

Moreover, if f ∈ L1 (Rn ), then ∞ f dx = Rn

(13.8.6)

S n−1

(13.8.9)


439

Note that if P : O → S n−1 is a parametrization of the unit sphere in Rn and if R is a unitary transformation in Rn , then R ◦ P : O → S n−1 is also a parametrization of the unit sphere in Rn . Indeed, this function is injective, has C 1 components, its image is S n−1 (up to a negligible set) and Rank D(R ◦ P ) = dim Im [R(DP )] = dim Im (DP ) = Rank (DP ) = n − 1. (13.8.10) Hence, S n−1

f ◦ R dσ =

O

f (R ◦ P )|∂u1 P × · · · × ∂un−1 P | du1 . . . dun−1

=

O

f (R ◦ P )|∂u1 (R ◦ P ) × · · · × ∂un−1 (R ◦ P )| du1 · · · dun−1

=

f dσ.

(13.8.11)

S n−1

The same type of reasoning also yields the following result. Proposition 13.45. For each x = (x1 , . . . , xn ) ∈ Rn , define Rj (x) := (x1 , . . . , xj−1 , −xj , xj+1 , . . . , xn ),

1 ≤ j ≤ n,

Rjk (x) := (x1 , . . . , xj−1 , xk , xj+1 , . . . , xk−1 , xj , xk+1 , . . . , xn ), Then

1 ≤ j ≤ k ≤ n. (13.8.12)

S n−1

S n−1

f ◦ Rj dσ =

f dσ,

j = 1, 2, . . . , n,

(13.8.13)

f dσ,

1 ≤ j ≤ k ≤ n.

(13.8.14)

S n−1

f ◦ Rjk dσ =

S n−1

Proposition 13.46. Let v ∈ Rn \ {0}, n ≥ 2, be fixed. Then for any real-valued function f defined on the real line, there holds 1 f (v · θ) dσ(θ) = ωn−2 f (s|v|)( 1 − s2 )n−3 ds. (13.8.15) S n−1

−1

Proof. Since integrals over the unit sphere are invariant under orthogonal transformations, we may assume that v/|v| = e1 and, hence, using polar coordinates and (13.8.3), we have

S n−1

f (v · θ) dσ(θ) =

S n−1

f (|v|θ1 ) dσ(θ)

2π π

=

π

... 0

0

= ωn−2

π 0

0

f (|v| cos ϕ1 )

n−2 '

(sin ϕj )n−1−j dϕ1 · · · dϕn−2 dϕn−1

j=1

f (|v| cos ϕ1 )(sin ϕ1 )n−2 dϕ1 .

(13.8.16)


440

Making the change of variables s := cos ϕ1 in the last integral above shows that this matches the right-hand side of (13.8.15). Proposition 13.47. Let f ∈ C 0 (R) be positive homogeneous of degree m ∈ R and fix η = (η1 , . . . , ηn ) ∈ Rn . Then if n ∈ N with n ≥ 2, for j, k ∈ {1, . . . , n} one has f (η · ξ)ξj ξk dσ(ξ) = α|η|m δjk + β|η|m−2 ηj ηk (13.8.17) S n−1

where

1 f (ξ1 )(1 − ξ12 ) dσ(ξ), n − 1 S n−1 1 β= f (ξ1 )(nξ12 − 1) dσ(ξ). n − 1 S n−1 α=

Proof. For j, k ∈ {1, . . . , n} set qjk (η) :=

S n−1

f (η · ξ)ξj ξk dσ(ξ),

η ∈ Rn ,

(13.8.18)

(13.8.19)

and define the quadratic form Q(ζ, η) :=

n

qjk (η)ζj ζk ,

∀ ζ, η ∈ Rn .

(13.8.20)

j,k=1

Observe that we can write Q(ζ, η) = S n−1 f (η·ξ)(ζ ·ξ)2 dσ(ξ). By the invariance under rotations of integrals over S n−1 (see (13.8.11)), we have that for any rotation R in Rn Q(ζ, η) = f (η·R ξ)(ζ·R ξ)2 dσ(ξ) = Q(Rζ, Rη), ζ, η ∈ Rn . (13.8.21) S n−1

A direct computation also gives that Q(λ1 ζ, λ2 η) = λ21 λm 2 Q(ζ, η)

for all λ1 , λ2 > 0, ζ, η ∈ Rn .

(13.8.22)

Next we claim that Q(ζ, η) = α + β(η · ζ)2 ,

∀ ζ, η ∈ S n−1 .

(13.8.23)

To show that (13.8.23) holds, we first observe that it suffices to prove (13.8.23) when η = e1 . Indeed, if we assume that (13.8.23) is true for η = e1 , then for an arbitrary η let R be the rotation such that Rη = e1 . Then if we also take into account (13.8.20), we have Q(ζ, η) = Q(Rζ, Rη) = Q(Rζ, e1 ) = α + β(Rζ · e1 )2 = α + β(ζ · R e1 )2 = α + β(η · ζ)2 .

(13.8.24)


441

Hence, (13.8.23) will follow if we prove that S n−1

f (ξ1 )(ζ · ξ)2 dσ(ξ) = α + β ξ12 ,

∀ ζ ∈ S n−1 .

(13.8.25)

To see the later we let Ajk := S n−1 f (ξ1 )ξj ξk dσ(ξ) for j, k ∈ {1, . . . , n}. Assume j = k. Then either j = 1 or k = 1. If, say, j = 1 we use (13.8.13) to conclude that Ajk = 0 in this case. A similar reasoning applies if k = 1. Clearly, A11 = S n−1 f (ξ1 )ξ12 dσ(ξ). As for the case j = k = 1, we first observe that A22 = · · · = Ann since Ajj is independent of j ∈ {2, . . . , n} due to n

Ajj = S n−1 f (ξ1 ) dσ(ξ), which in turn implies that (13.8.14). Moreover, j=1

1 Ajj = n−1 S n−1 f (ξ1 ) dσ(ξ) − A11 = α for each j = 2, . . . , n. Combining all these we have that for ζ ∈ S n−1 , S n−1

f (ξ1 )(ζ · ξ)2 dσ(ξ) =

n

ζj ζk Ajk

j,k=1

=

ζ12

S n−1

f (ξ1 )ξ12

dσ(ξ) + α

n

ζj2

j=2

= α + βζ12 .

(13.8.26)

This concludes the proof of (13.8.25) which, in turn, implies (13.8.23). Now if ζ, η ∈ Rn \ {0}, we make use of (13.8.22) and (13.8.23) to write Q(ζ, η) = |ζ|2 |η|m Q =

ζ η , = α|ζ|2 |η|m + β|η|m−2 (ζ · η)2 |ζ| |η|

(13.8.27)

n α|η|m δjk + β|η|m−2 ηj ηk ζj ζk j,k=1

which proves (13.8.17). Proposition 13.48. Consider f (t) := |t| for t ∈ R, and let α, β be as in (13.8.17) for this choice of f . Then α=β=

2ωn−2 , n2 − 1

(13.8.28)

where ωn−2 denotes the surface measure of the unit ball in Rn−1 . Proof. Using the standard parametrization of S n−1 (see (13.8.1) with R = 1) we have


442 1 n−1

α=

ωn−2 n−1

=

π

π

π

2π

| cos ϕ1 |(1 − (cos ϕ1 )2 )(sin ϕ1 )n−2 (sin ϕ2 )n−3 ×

... 0

0

0

0

· · · × (sin ϕn−2 ) dϕn−1 dϕn−2 · · · dϕ1 π

| cos ϕ1 |(sin ϕ1 )n dϕ1 .

(13.8.29)

0

The change of variables θ = π−y yields − Using this back in (13.8.29) then gives 2 ωn−2 α= n−1

0

π 2

π π 2

cos θ(sin θ)n dθ =

π 2

0

cos y(sin y)n dy.

π 2 ωn−2 (sin θ)n+1 2 2 ωn−2 . (13.8.30) cos θ (sin θ) dθ = = 2 n−1 n + 1 0 n −1 n

Similar arguments in computing β give that ωn−2 π | cos θ|[n(cos θ)2 − 1](sin θ)n−2 dθ (13.8.31) n−1 0 π n ωn−2 π = ωn−2 | cos θ|(sin θ)n−2 dθ − | cos θ|(sin θ)n dθ n − 1 0 0 π2 π2 2 ωn−2 = cos θ (sin θ)n−2 dθ − n cos θ (sin θ)n dθ (n − 1) n−1 0 0

β=

=

π2 2 ωn−2 (sin θ)n+1 π2 2 ωn−2 . (sin θ)n−1 − n = 2 n−1 n+1 0 n −1 0

This finishes the proof. Recall from 0.0.7 thatz α = z1α1 z2α2 · · · znαn whenever z = (z1 , . . . , zn ) ∈ Rn and α = (α1 , α2 , . . . , αn ) ∈ Nn0 . Let us also introduce 2Nn0 := (2α1 , 2α2 , . . . , 2αn ) : α = (α1 , α2 , . . . , αn ) ∈ Nn0 .

(13.8.32)

The next proposition deals with the issue of integrating arbitrary monomials on the unit sphere centered at the origin. Proposition 13.49. For each multi-index α = (α1 , . . . , αn ) ∈ Nn0 , ⎧ ⎪ ⎪0 ⎪ ⎨ |α| z α dσ(z) = n−1 1+|α| ⎪ 2 ! α! 2π 2 Γ( 2 ) S n−1 ⎪ ⎪ · ⎩ α |α|! Γ( |α|+n ) 2 !

2

where Γ is the gamma function introduced in (13.5.1).

if α ∈ 2Nn0 , (13.8.33) if α ∈ 2Nn0 ,

13.8. POLAR COORDINATES AND INTEGRALS ON SPHERES Proof. Fix an arbitrary k ∈ N and set qα := z α dσ(z), ∀ α ∈ Nn0 with |α| = k.

443

(13.8.34)

S n−1

Also, with a “dot” standing for the standard inner product in Rn , introduce Qk (x) :=

k! qα xα , α! n

x = (x1 , x2 , . . . , xn ) ∈ Rn .

(13.8.35)

α∈N0

|α|=k

Then (13.2.1) implies that Qk (x) =

S n−1

(z · x)k dσ(z),

∀ x ∈ Rn .

(13.8.36)

Let us also observe here that, if x ∈ S n−1 is arbitrary but fixed and if R is a rotation about the origin in Rn such that R−1 x = en := (0, . . . , 0, 1) ∈ Rn , then by (13.8.35) and the rotation invariance of integrals on S n−1 (cf. (13.8.11)), we have (Rz · x)k dσ(z) = Qk (en ). (13.8.37) Qk (x) = S n−1

By the homogeneity of Qk , (13.8.37) implies that Qk (x) = |x|k Qk (en ) for all x ∈ Rn and, hence, k! qα xα = |x|k Qk (en ) for all x ∈ Rn . α!

(13.8.38)

|α|=k

We now consider two cases: Case I: k is odd. In this the mapping S n−1 z → znk ∈ R is odd.

scenario, k In particular, Qk (x) = S n−1 zn dσ(z) = 0. This, in turn, along with (13.8.38) k! α n then force α! qα x = 0 for every x ∈ R . From (13.2.5) it is easy to deduce |α|=k

that for each β ∈ Nn0 we have ∂xγ [xα ]

⎧ ⎨0 =

if α = γ,

(13.8.39) α! if α = γ. 1 α = 0 for each γ ∈ Nn0 q x We may therefore conclude that qγ = ∂xγ α |α|=k α! with |γ| = k, in agreement with (13.8.33). m! 2β Case II: k is even. Suppose k = 2m for some m ∈ N. Then, |x|k = β! x x=0

⎩

|β|=m

and (13.8.38) becomes k! m! x2β Qk (en ) = qα xα β! α!

|β|=m

|α|=k

for all x ∈ Rn .

(13.8.40)


444

Fix γ ∈ Nn0 such that |γ| = k and observe that ⎧ ⎪ ' & ⎨ m! (2β)!Qk (en ) if γ = 2β, some β ∈ Nn , |β| = m, 0 γ m! 2β β! x Qk (en ) ∂x = ⎪ β! ⎩0 x=0 otherwise. (13.8.41) This, (13.8.40), and (13.8.39) then imply that ⎧ 0 ⎪ ⎪ ⎪ ⎨ qγ = |γ| ⎪ 2 ! γ! ⎪ ⎪ Q|γ|(en ) ⎩ γ · |γ|! ! 2

if γ ∈ 2Nn0 , (13.8.42) if γ ∈

2Nn0 .

We are now left with computing Q2m (en ) when m ∈ N. Using spherical coordinates, a direct computation gives that Q2m (en ) = (z · en )2m dσ(z) S n−1

2π π

= 0

0

π

(cos θ1 )2m

... 0

(sin θj )n−1−j dθ1 · · · dθn−2 dθn−1

n−2 j=1

π

(cos θ)2m (sin θ)n−2 dθ =

= ωn−2 0

2π

n−1 2

Γ( 12 + m) , Γ(m + n2 )

(13.8.43)

by (13.8.4) and (13.5.6) (considered with n − 1 in place of n), and (13.5.11). This once again agrees with (13.8.33), and the proof of Proposition 13.49 is finished. A simple useful consequence of the general formula (13.8.33) is the fact that ωn−1 δjk whenever 1 ≤ j, k ≤ n. zj zk dσ(z) = (13.8.44) n S n−1

13.9. TABLES OF FOURIER TRANSFORMS

13.9

445

Tables of Fourier Transforms

Fourier Transforms of Schwartz Functions

F(f )

f 2

e−λ|x| , λ ∈ C, Re(λ) > 0 e−ax

2

+ibx

, a > 0 and b ∈ R fixed

Location

πn

e−

π1

e−

λ

a

2

2

|ξ|2 4λ

Example 3.21

(ξ−b)2 4a

Exercise 3.22

n

e−(Ax)·x , A ∈ Mn×n (R), A = A ,

(A−1 ξ)·ξ π2 4 e− det A

Exercise 3.36

2 2 2π √ e−(ξ1 −ξ1 ξ2 +ξ2 )/3 3

Exercise 3.37

√

n

(Ax) · x > 0 for x ∈ R \ {0} 2

2

e−(x1 +x1 x2 +x2 ) for (x1 , x2 ) ∈ R2 2

e−a|x| sin(x · x0 ), for x ∈ Rn , n

if a > 0 and x0 ∈ R are fixed

1 2i

πn a

2

e−

|ξ−x0 |2 4a

− e−

|ξ+x0 |2 4a

Exercise 3.39


446

Fourier Transforms of Tempered Distributions in R Below x, ξ ∈ R.

u

F (u)

Location

1 , x2 +a2

a ∈ (0, ∞) given

π −a|ξ| e a

Example 4.23

a ∈ (0, ∞) given

−πi(sgn ξ)e−a|ξ|

Example 4.29

x , x2 +a2

χ[−a,a] , a ∈ (0, ∞) given

⎧ ⎨ 2

sin(aξ) ξ

for ξ ∈ R \ {0}

⎩ 2a

sin(b x), b ∈ R fixed

−iπδb + iπδ−b

cos(b x), b ∈ R

πδb + πδ−b

sin(b x) sin(c x), b, c ∈ R fixed

− π2

H P.V.

1 x

1 ξ

|x|k , k ∈ N even

2πik δ(k)

|x|k ,

−2ik+1

k ∈ N odd b ∈ R fixed

sin(b x) sin(c x) , x

ln |x|

Example 4.37

−cδ

Exercise 4.115

iπ sgn (ξ) −2i P.V.

sin(x2 )

Exercise 4.122

δb+c − δb−c − δc−b + δ−b−c

sgn x

sin(b x) , x

Example 4.36

−iP.V.

Example 4.34

for ξ = 0

b, c ∈ R fixed

Exercise 4.116 1 ξ

Exercise 4.117 Exercise 4.117

P.V.

1 (k) ξ

Exercise 4.117

π sgn(b)χ[−|b|,|b|] iπ 2 √

Exercise 4.117

sgn(c) χ[−b−|c|,−b+|c|] − χ[b−|c|,b+|c|]

π 2i

*

ei

π−ξ 2 4

− e−i

π−ξ 2

Exercise 4.117

+

4

−2πγδ − πwχ(−1,1) wχ(−1,1) from (4.6.21), γ from (4.6.23)

Exercise 4.117 Exercise 4.117

13.9. TABLES OF FOURIER TRANSFORMS

447

Fourier Transforms of Tempered Distributions Below x, ξ ∈ Rn and x , ξ ∈ Rn−1 . u

F (u)

Location

δ

1

Example 4.21

1

(2π)n δ

Exercise 4.31

2

e−ia|x| , a ∈ (0, ∞) given 2

e ia|x| , a ∈ (0, ∞) given

πn

e−i

2

a

πn 2

a

ei

nπ 4

Example 4.38

e

n−λ n Γ 2 2 Γ λ 2

2n−λ π

xj , |x|n

ξ −i ωn−1 |ξ|j2

xj , |x|λ+2

Example 4.24

|ξ|2 −i 4a

P.V. ∂j

xk |x|n

, j, k ∈ {1, . . . , n}

P.V. Θ, with Θ as in (4.4.1)

P.V.

xj |x|n+1

, j ∈ {1, . . . , n}

2 t ωn−1 (t2 +|x |2 ) n 2

δ

Γ n−λ 2

−

,

ξj ξk |ξ|2

S n−1

e−t|ξ

ξj ξk |ξ|4

ωn−1 δjk n

Θ(ω) log(i(ξ · ω)) dσ(ω)

ξj |ξ|

− iω2n

−

ξj Γ( λ +1) |ξ|n−λ 2

jk ωn−1 |ξ| 2 − 2ωn−1

ωn−1

Proposition 4.61 Corollary 4.62

n

n ≥ 3, j, k ∈ {1, . . . , n}

|ξ|λ−n

−i 2n−λ−1 π 2

n ≥ 2,

( )

λ ∈ [0, n − 1), j ∈ {1, . . . , n} xj xk , |x|n

ei

|x|−λ , λ ∈ (0, n) fixed n≥2

|ξ|2 4a

nπ 4

Corollary 4.62

Exercise 7.63 Proposition 4.70 Theorem 4.71 Proposition 4.81

|

Proposition 4.87

with t > 0 fixed xj 2 ωn−1 (t2 +|x |2 ) n 2

with t > 0 fixed, j ∈ {1, . . . , n − 1}

ξ

−i |ξj | e−t|ξ

|

Proposition 4.89

Bibliography [1] R. A. Adams and J. J. F. Fournier, Sobolev Spaces, Pure and Applied Mathematics, 140, 2nd edition, Academic Press, 2003. [2] D. R. Adams and L. I. Hedberg, Function Spaces and Potential Theory, Grundlehren der Mathematischen Wissenschaften, Vol. 314, Springer, Berlin, 1996. [3] S. Axler, P. Bourdon, and W. Ramey, Harmonic Function Theory, 2nd edition, Graduate Texts in Mathematics, 137, Springer-Verlag, New York, 2001. [4] F. Brackx, R. Delanghe, and F. Sommen, Clifford Analysis, Research Notes in Mathematics, 76, Pitman, Advanced Publishing Program, Boston, MA, 1982. [5] A. P. Calder´ on and A. Zygmund, On the existence of certain singular integrals, Acta Math., 88 (1952), 85–139. [6] M. Christ, Lectures on Singular Integral Operators, CBMS Regional Conference Series in Mathematics, No. 77, AMS, 1990. [7] G. David and S. Semmes, Singular Integrals and Rectifiable Sets in Rn : Beyond Lipschitz Graphs, Astérisque, No. 193, 1991. [8] J. J. Duistermaat and J. A. C. Kolk, Distributions. Theory and applications, Translated from the Dutch by J. P. van Braam Houckgeest, Cornerstones, Birkh¨ auser Boston, Inc., Boston, MA, 2010. [9] J. Duoandikoetxea. Fourier Analysis, Graduate Studies in Mathematics, AMS, 2000. [10] R. E. Edwards, Functional Analysis. Theory and Applications, Publications, Inc., New York, 1995.

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[11] L. Ehrenpreis, Solution of some problems of division. I. Division by a polynomial of derivation, Amer. J. Math. 76, (1954). [12] L. Evans, Partial Differential Equations, 2nd edition, Graduate Studies in Mathematics, 19, American Mathematical Society, Providence, RI, 2010. [13] L. C. Evans and R. F. Gariepy, Measure Theory and Fine Properties of Functions, Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, 1992. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6, © Springer Science+Business Media New York 2013

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[35] F. John, Plane Waves and Spherical Means Applied to Partial Differential Equations, Interscience Publishers, New York-London, 1955. [36] C. E. Kenig, Harmonic Analysis Techniques for Second Order Elliptic Boundary Value Problems, CBMS Regional Conference Series in Mathematics, 83, American Mathematical Society, Providence, RI, 1994. [37] S. G. Krantz and H. R. Parks, A Primer of Real Analytic Functions, 2-nd edition, Birkh¨ auser, Boston, 2002. [38] P. S. Laplace, Mémoire sur la théorie de l’anneau de Saturne, Mém. Acad. Roy. Sci. Paris (1787/1789), 201–234. ´ [39] P. S. Laplace, J. Ecole Polytéch. cah., 15 (1809), p. 240 (quoted in Enc. Math. Wiss. Band II, 1. Teil, 2. H¨ alfte, p. 1198). [40] G. Leoni, A First Course in Sobolev Spaces, Vol. 105, Graduate Studies in Mathematics, American Mathematical Soc., 2009. [41] S. Lojasiewicz, Division d’une distribution par une fonction analytique de variables réelles, C. R. Acad. Sci. Paris, 246 (1958), 683–686. [42] B. Malgrange, Existence et approximation des solutions des équations aux dérivées partielles et des équations de convolution, (French) Ann. Inst. Fourier, Grenoble 6 (1955–1956), p. 271–355. [43] V. G. Maz’ya, Sobolev Spaces, Springer-Verlag, Berlin-New York, 1985. [44] V. G. Maz’ya, Sobolev spaces with applications to elliptic partial differential equations, Second, revised and augmented edition, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 342, Springer, Heidelberg, 2011. [45] V. G. Maz’ya and T. Shaposhnikova, Theory of Sobolev multipliers. With applications to differential and integral operators, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 337, Springer-Verlag, Berlin, 2009. [46] W. McLean, Strongly Elliptic Systems and Boundary Integral Equations, Cambridge University Press, Cambridge, 2000. [47] Y. Meyer, Ondelettes et Opérateurs II. Opérateurs de Calder´ on-Zygmund, Actualités Mathématiques, Hermann, Paris, 1990. [48] Y. Meyer and R. R. Coifman, Ondelettes et Opérateurs III. Opérateurs Multilinéaires, Actualités Mathématiques, Hermann, Paris, 1991.

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[49] M. Mitrea, Clifford Wavelets, Singular Integrals, and Hardy Spaces, Lecture Notes in Mathematics, 1575, Springer-Verlag, Berlin, 1994. [50] I. Mitrea and M. Mitrea, Multi-Layer Potentials and Boundary Problems for Higher-Order Elliptic Systems in Lipschitz Domains, Lecture Notes in Mathematics 2063, Springer, 2012. [51] D. Mitrea and H. Rosenblatt, A General Converse Theorem for Mean Value Theorems in Linear Elasticity, Mathematical Methods for the Applied Sciences, Vol. 29, No. 12 (2006), 1349–1361. [52] D. Mitrea, M. Mitrea and M. Taylor, Layer potentials, the Hodge Laplacian and global boundary problems in nonsmooth Riemannian manifolds, Memoirs of American Mathematical Society, Vol. 150, No. 713, 2001. [53] D. Mitrea, I. Mitrea, M. Mitrea, and S. Monniaux, Groupoid Metrization Theory With Applications to Analysis on Quasi-Metric Spaces and Functional Analysis, Birkh¨ auser, Springer New York, Heidelberg, Dordrecht, London, 2013. [54] C. B. Morrey, Second order elliptic systems of differential equations. Contributions to the theory of partial differential equations, Ann. Math. Studies, 33 (1954), 101–159. [55] S. D. Poisson, Remarques sur une équation qui se présente dans la théorie de l’attraction des sphéroides, Bulletin de la Société Philomathique de Paris, 3 (1813), 388–392. [56] S. D. Poisson, Sur l’intégrale de léquation relative aux vibrations des surfaces élastiques et au mouvement des ondes, Bulletin de la Société Philomathique de Paris, (1818), 125–128. [57] H. L. Royden, Real Analysis, Macmillian Publishing Co., Inc., New York, 1968. [58] W. Rudin, Real and Complex Analysis, 2nd edition, International Series in Pure and Applied Mathematics, McGraw-Hill, Inc., 1986. [59] W. Rudin, Functional Analysis, 2nd edition, International Series in Pure and Applied Mathematics, McGraw-Hill, Inc., 1991. [60] L. Schwartz, Théorie des Distributions, I, II, Hermann, Paris, 1950–51. [61] Z. Shapiro, On elliptical systems of partial differential equations, C. R. (Doklady) Acad. Sci. URSS (N. S.), 46 (1945). [62] E.M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Mathematical Series, 30, Princeton University Press, Princeton, N.J., 1970. [63] E.M. Stein, Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals, Princeton Mathematical Series, 43, Monographs in Harmonic Analysis, III, Princeton University Press, Princeton, NJ, 1993. [64] E. M. Stein and G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, Princeton, New Jersey, 1971.

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[65] R. Strichartz, A guide to Distribution Theory and Fourier Transforms, World Scientific Publishing Co., Inc., River Edge, NJ, 2003. [66] H. Tanabe, Functional Analytic Methods for Partial Differential Equations, Monographs and Textbooks in Pure and Applied Mathematics, 204, Marcel Dekker, Inc., New York, 1997. [67] M. E. Taylor, Pseudodifferential Operators, Princeton Mathematical Series, 34, Princeton University Press, Princeton, N.J., 1981. [68] M. E. Taylor, Partial Differential Equations. I. Basic Theory, Applied Mathematical Sciences, 115, Springer-Verlag, New York, 1996. [69] M. E. Taylor, Tools for PDE. Pseudodifferential Operators, Paradifferential Operators, and Layer Potentials, Mathematical Surveys and Monographs, 81, American Mathematical Society, Providence, RI, 2000. [70] F. Tréves, Topological Vector Spaces, Distributions and Kernels, Dover Publications, 2006. [71] V. S. Vladimirov, Equations of Mathematical Physics, Marcel Dekker Inc., 1971. [72] V. S. Vladimirov, Methods of the Theory of Generalized Functions, Analytical Methods and Special Functions, 6, Taylor & Francis, London, 2002. [73] V. Voltera, Sur les vibrations des corps élastiques isotropes, Acta Math., 18 (1894), 161–232. [74] W. Wendland, Integral Equation Methods for Boundary Value Problems, SpringerVerlag, 2002. [75] J. T. Wloka, B. Rowley, and B. Lawruk, Boundary Value Problems for Elliptic Systems, Cambridge University Press, 1995. [76] W. P. Ziemer, Weakly Differentiable Functions: Sobolev Spaces and Functions of Bounded Variation, Graduate Text in Mathematics, 120, Springer-Verlag, New York, 1989.

Subject Index absorbing set, 417 Arzel` a-Ascoli’s Theorem, 426 balanced set, 417 base for a topology, 415 Beta function, 434 bi-harmonic, 237 Binomial Theorem, 424 Cauchy problem heat operator, 280 vibrating infinite string, 1 wave operator, 307 Cauchy–Clifford operator, 257 Cauchy operator, 252 Cauchy sequence, 416 Clifford algebra, 253 conjugate Poisson kernels, 165 conormal derivative, 271 continuous, 416 convergence in D(Ω), 11 E (Ω), 9 S(Rn ), 93 convex set, 417 convolution of distributions, 61 convolution of functions, 59 convolution, 115, 116 cross product in Rn , 436 cut-off functions, 428 differential operators bi-Laplacian, 207, 233 Cauchy–Riemann, 248 Dirac, 253, 255 heat, 277 higher order systems, 344 Lamé, 319 Laplacian, 207, 217, 220 poly-harmonic, 207, 240 scalar second order, 260 Schr¨ odinger, 287 second order parabolic, 282 Stokes, 334 wave, 289 dilations, 130 Dirac distribution δ, 21, 31, 36, 120 Dirichlet problem, 165, 370 distance, 416

distribution compactly supported, 40, 41 definition of, 17 of finite order, 19 of function type, 19 tempered, 109 distributional derivative, 29 elliptic operators, 206, 207 Euler’s constant, 150 Fourier transform, 89, 99, 105, 120, 123, 125, 142, 160 partial, 128 Fr´ echet space, 418 F-topology, 418 function spaces D(Ω), 11, 422 DK (Ω), 422 E (Ω), 9, 421 S(Rn ), 91, 423 fundamental solutions definition of, 191 for higher order systems, 348 for scalar second order operators, 261 for second order parabolic operators, 286 for strongly elliptic operators, 270 for the bi-Laplacian, 235 for the Cauchy operator, 249 for the Dirac operator, 256 for the heat operator, 279 for the Lamé operator, 323, 341 for the Laplacian, 220 for the poly-harmonic operator, 241 for the Schr¨ odinger operator, 288 for the Stokes operator, 336, 343 for the wave operator, 305 gamma function, 434 generalized volume potential, 179 Green’s formula, 437 Hahn-Banach Theorem, 420 harmonic function, 217, 225 Heaviside function definition of, 6 derivative, 31

D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6, © Springer Science+Business Media New York 2013

455

456 Hilbert transform, 167, 171, 253 hypoelliptic operators, 201, 203, 207, 279 inductive limit topology, 418 integral representation formula, 210, 357 integration by parts formula, 437 integration on surfaces, 436 interior estimates for elliptic homogeneous systems, 359 for hypoelliptic operators, 211 for the bi-Laplacian, 238 for the Lamé operator, 329, 331 for the Laplacian, 226 for the poly-harmonic operator, 245 invariant under orthogonal transformations, 131 inverse Fourier transform, 102 isometry, 425 layer potential operators double layer, 232, 367 harmonic double-layer, 162 single layer, 232, 366 layer potential representation formula, 229, 272 Lebesgue’s Differentiation Theorem, 425 Lebesgue’s Dominated Convergence Theorem, 425 Leibniz’s Formula, 424 Liouville’s theorem for bi-Laplacian, 238 for general operators, 191 for Lamé, 331 for Laplacian, 227 for systems, 312 for the heat operator, 280 Lipschitz function, 30, 77 locally finite, 431 Malgrange-Ehrenpreis theorem, 195 mean value formulas for bi-Laplacian, 237 for Lamé, 328 for Laplacian, 225 metric space complete, 416 definition of, 416

SUBJECT INDEX Multinomial Theorem, 424 multiplier, 168 Neumann problem, 372 Newtonian potential, 175, 177, 178, 232 open neighborhood, 415 open set, 415 order of a distribution, 19 orthogonal transformation, 131 parametrix, 205 parametrization, 436 Parseval’s identity, 104 partition of unity for arbitrary open covers, 432 for compact sets, 430 with preservation of indexes, 433 Planch´ erel’s identity, 106 Plemelj formula, 252 Poisson equation, 192, 223, 332 Poisson kernel, 162 Poisson problem for Lamé, 332 for the bi-Laplacian, 239 for the Laplacian, 227 polar coordinates, 438 positive definite matrix, 264 positive homogeneous distribution, 132 function, 131 principal symbol, 189 principal value distribution, 20, 136 projection, 253, 259 p-subaveraging, 361 Rademacher’s Theorem, 425 real-analytic function, 214, 216, 227, 238, 245, 331, 347, 359 restriction of a distribution, 34 reverse H¨ older estimate, 364 Riesz’s Representation Theorem for complex functionals, 426 for locally bounded functionals, 427 for positive functionals, 426 Riesz transforms, 167, 168, 221, 222, 228, 258 rigid transformation, 425 seminorm, 417 separating seminorms, 417 sequentially continuous, 416

SUBJECT INDEX sign function, 14 singular integral operator, 167, 172 singular support, 202 slowly increasing functions, 93 strongly elliptic matrix, 263 strongly elliptic operators, 263 p-subaveraging, 362, 364 support of an arbitrary function, 37 support of a distribution, 36 surface, 435 Taylor’s Formula, 424 tensor product of distributions, 55, 112 of functions, 48 the method of descent, 299 topological space definition of, 415 Hausdorff, 416 metrizable, 416 separated, 416

457 topological vector space definition, 416 dual, 418 locally convex, 417 topology coarser, 416 definition of, 415 finer, 416 induced, 416 product, 416 transpose of an operator, 7 unique continuation property, 215 Urysohn’s Lemma, 427 Vitali’s Convergencen Theorem, 427 weak derivative, 4 solution, 2 weak∗-topology, 419 Young’s Inequality, 425

Symbol Index α ! defined on page, xix |α| defined on page, xix C

Cauchy operator in C, 252 Cauchy-Clifford operator, 257 χE characteristic function of the set E, xx f ∨ , 104 u∨ , 124 C k (Ω), xx C k (Ω), xx C0k (Ω), xx C0∞ (Ω), xx Cn Clifford algebra with n generators, 253 Clifford algebra multiplication, 253 ∗ convolution of S(Rn ) with S (Rn ), 116 distributions, 61 functions, 59 v1 × v2 × · · · × vn−1 cross product in Rn , 436 D Dirac

operator,255 D = 1i ∂1 , . . . 1i ∂n , 91 δjk Kronecker symbol, xix δ Dirac distribution, 36 D(Ω) test functions, 11, 422 DK (Ω) test functions supported in K, 11, 422 D (Ω) general distributions, 26, 422 {ej }1≤j≤n orthonormal basis in Rn , xix E (Ω), 9, 421 E (Ω) compactly supported distributions, 41, 421 EA fundamental solution for LA elliptic, 270 parabolic, 286 γ Euler’s constant, 150 ! factorial, xix F Fourier transform, 89, 123 · Fourier transform, 89, 120 F −1 inverse Fourier transform, 102

Fx partial Fourier transform, 128 Fx−1 inverse partial Fourier transform, 129 Fm,n fundamental solution for Δm in Rn , 241 Γ gamma function, 434 H Heaviside function, 6 H Hilbert transform, 167 ,i complex imaginary unit, xix −A f dμ integral average, xx Δ Laplace operator, 165 Δ2 bi-Laplace operator, 207 Δm poly-harmonic operator, 207 ∂t − Δ heat operator, 277 ∂t2 − Δ wave operator, 289 Lpcomp (Ω), xx Lploc (Ω), xx LA elliptic, 262 parabolic, 282 Mn×m (R) matrices with entries from R, xxi, 314

the ring Mn×m D (Ω) , 309 Mn×m E (Ω) , 311 mΘ the Fourier transform of P.V. Θ, 142 N the Newtonian potential, 175 ∂f normal derivative of f , 220 ∂ν ∂νA conormal derivative, 271 ωn−1 surface measure of unit ball in Rn , xx, 434 ∂ α partial derivative of order α, xix ϕΔ defined on page, 60 ΠΦ generalized volume potential, 179 P harmonic Poisson kernel, 162 projection, 253, 259 pt defined on page, 163 P harmonic double layer, 162

D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis, Universitext, DOI 10.1007/978-1-4614-8208-6, © Springer Science+Business Media New York 2013

459

460 P (D) linear constant coefficient partial differential operator, 101, 189 P (x, ∂) linear partial differential operator, 7 P (ξ), 101, 189 Pm (ξ) principal symbol of P (D), 189 P.V. Θ principal value distribution associated with Θ, 136 P.V. x1 principal value distribution associated with x1 , 20

SYMBOL INDEX sgn x sign function, 14 sing supp u singular support of the distribution u, 202 L(Rn ) slowly increasing functions, 93 supp f support of an arbitrary function f , 37 supp u support of the distribution u, 36

Re real part of a complex number, xix Im imaginary part of a complex number, xix u|ω restriction of the distribution u to the open set ω , 34 Rj Riesz transform, 167

τt dilation, 130 ⊗ tensor product of distributions, 55 of functions, 48 of tempered distributions, 112 tx0 translation by x0 map of distributions, 66 of functions, 12 P transpose of the operator P , 7 TΘ defined on page, 167

S(Rn ) Schwartz functions, 91, 423 S (Rn ) tempered distributions, 111, 423

uf distribution associated with f , 19, 110 vertical limit of u to ∂Rn u|ver + , 165 ∂Rn +

Qj defined on page, 165 (qj )t defined on page, 166

Distributions, Partial Differential Equations, and Harmonic Analysis

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Harmonic Analysis and Partial Differential Equations

Harmonic Analysis and Partial Differential Equations

Problems in Distributions and Partial Differential Equations

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Fourier Analysis and Nonlinear Partial Differential Equations

Elements and analysis of partial differential equations

Applied Functional Analysis and Partial Differential Equations

Fourier analysis and nonlinear partial differential equations

Geometric partial differential equations and image analysis

Geometric Partial Differential Equations and Image Analysis

Partial Differential Equations. Nonlinear equations

Partial differential equations. Nonlinear equations

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Ordinary and Partial Differential Equations

Partial Differential Equations

Partial Differential Equations

Partial differential equations

Partial Differential Equations

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Elliptic partial differential equations

Nonlinear Partial Differential Equations

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Distributions, Partial Differential Equations, and Harmonic Analysis

Harmonic Analysis and Partial Differential Equations

Harmonic analysis and partial differential equations

Harmonic Analysis and Partial Differential Equations

Harmonic Analysis and Partial Differential Equations

Problems in Distributions and Partial Differential Equations

Distributions and Nonlinear Partial Differential Equations

Harmonic analysis and partial differential equations: in honor of Calderon

Fourier Analysis and Nonlinear Partial Differential Equations

Elements and analysis of partial differential equations

Applied Functional Analysis and Partial Differential Equations

Fourier analysis and nonlinear partial differential equations

Geometric partial differential equations and image analysis

Geometric Partial Differential Equations and Image Analysis

Partial Differential Equations. Nonlinear equations

Partial differential equations. Nonlinear equations

Partial Differential Equations

Partial Differential Equations

Partial Differential Equations

Applied partial differential equations

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