Partial differential equations and complex analysis


Published on

  • Be the first to comment

No Downloads
Total views
On SlideShare
From Embeds
Number of Embeds
Embeds 0
No embeds

No notes for slide

Partial differential equations and complex analysis

  1. 1. STEVEN G. KRANTZWashington University in St. Louis, Department of MathematicsPartial Differential Equationsand Complex AnalysisBased on notes by Estela A. Gavosto and Marco M. PelosoCRC PRESSBoca Raton Ann Arbor London Tokyo
  2. 2. Library of Congress Cataloging-in-Publication DataKrantz, Steven G., 1951- Partial differential equations and complex analysis / Steven G. Krantz. p. cm. Includes bibliographical references (p. ) and index. ISBN 0-8493-7155-4 1. Differential equations. Partial. 2. Functions of a complex variable.3. Mathematical analysis. 4. Functions of several complex variables. I. Title. QA374.K9 1992 515 .35-dc20 92-11422 CIPThis book represents information obtained from authentic and highly regarded sources.Reprinted material is quoted with permission, and sources are indicated. A wide varietyof references are listed. Every reasonable effort has been made to give reliable dataand information, but the author and the publisher cannot assume responsibility for thevalidity of all materials or for the consequences of their use.All rights reserved. This book, or any parts thereof, may not be reproduced in any formwithout written consent from the publisher.This book was formatted with ItTJ¥( by Archetype Publishing Inc., P.O. Box 6567,Champaign, IL 61826.Direct all inquiries to CRC Press, Inc., 2000 Corporate Blvd., N.W., Boca Raton, Horida,33431. © 1992 by CRC Press, Inc. International Standard Book Number 0-8493-7155-4 Library of Congress Card Number 92-11422 Printed in the United States of America 12 34567 890 Printed on acid-free paper
  3. 3. To the memory of my grandmother,Eda Crisafulli.
  4. 4. Contents Preface xi1 The Dirichlet Problem in the Complex Plane 11.1 A Little Notation 11.2 The Dirichlet Problem 21.3 Lipschitz Spaces 71.4 Boundary Regularity for the Dirichlet Problem for the Laplacian on the Unit Disc 101.5 Regularity of the Dirichlet Problem on a Smoothly Bounded Domain and Conformal Mapping 202 Review of Fourier Analysis 262.1 The Fourier Transform 262.2 Schwartz Distributions 362.3 Convolution and Friedrichs Mollifiers 432.4 The Paley-Wiener Theorem 483 PseudodifferentialOperators 523.1 Introduction to Pseudodifferential Operators 523.2 A Formal Treatment of Pseudodifferential Operators 563.3 The Calculus of Pseudodifferential Operators 654 Elliptic Operators 764.1 Some Fundamental Properties of Partial Differential Operators 76
  5. 5. viii4.2 Regularity for Elliptic Operators 814.3 Change of Coordinates 874.4 Restriction Theorems for Sobolev Spaces 895 Elliptic Boundary Value Problems 955.1 The Constant Coefficient Case 955.2 Well-Posedness 995.3 Remarks on the Solution of the Boundary Value Problem in the Constant Coefficient Case 1085.4 Solution of the Boundary Value Problem in the Variable Coefficient Case 1095.5 Solution of the Boundary Value Problem Using Pseudodifferential Operators 1155.6 Remarks on the Dirichlet Problem on an Arbitrary Domain, and a Return to Conformal Mapping 1235.7 A Coda on the Neumann Problem 1266 A Degenerate Elliptic Boundary Value Problem 1286.1 Introductory Remarks 1286.2 The Bergman Kernel 1316.3 The Szego and Poisson-Szego Kernels 1386.4 The Bergman Metric 1436.5 The Dirichlet Problem for the Invariant Laplacian on the Ball 1486.6 Spherical Harmonics 1546.7 Advanced Topics in the Theory of Spherical Hannonics: the Zonal Harmonics 1606.8 Spherical Harmonics in the Complex Domain and Applications 1727 The a-Neumann Problem 1847.1 Introduction to Hermitian Analysis 1857.2 a The Formalism of the Problem 1897.3 Formulation of the a-Neumann Problem 1967.4 The Main Estimate 2017.5 Special Boundary Charts, Finite Differences, and Other Technical Matters 2087.6 First Steps in the Proof of the Main Estimate 2187.7 Estimates in the Sobolev -1/2 Norm 2247.8 Conclusion of the Proof of the Main Estimate 2347.9 The Solution of the a-Neumann Problem 242
  6. 6. ix8 Applications of the a-Neumann Problem 2528.1 An Application to the Bergman Projection 2528.2 Smoothness to the Boundary of Biholomorphic Mappings 2568.3 Other Applications of [) Techniques 2639 The Local Solvability Issue and a Look Back 2699.1 Some Remarks about Local Solvabilitiy 2699.2 The Szego Projection and Local Solvability 2709.3 The Hodge Theory for the Tangential Cauchy-Riemann Complex 2749.4 Commutators, Integrability, and Local Solvability 277 Table of Notation 283 Bibliography 287 Index 295
  7. 7. PrefaceThe subject of partial differential equations is perhaps the broadest and deepestin all of mathematics. It is difficult for the novice to gain a foothold in thesubject at any level beyond the most basic. At the same time partial differentialequations are playing an ever more vital role in other branches of mathematics.This assertion is particularly true in the subject of complex analysis. It is my experience that a new subject is most readily learned when presentedin vitro. Thus this book proposes to present many of the basic elements of linearpartial differential equations in the context of how they are applied to the study ofcomplex analysis. We shall treat the Dirichlet and Neumann problems for ellipticequations and the related Schauder regularity theory. Both the classical point ofview and the pseudodifferential operators approach will be covered. Then weshall see what these results say about the boundary regularity of biholomorphicmappings. We shall study the a-Neumann problem, then consider applications tothe complex function theory of several variables and to the Bergman projection.The book culminates with applications of the a-Neumann problem, includinga proof of Feffermans theorem on the boundary behavior of biholomorphicmappings. There is also a treatment of the Lewy unsolvable equation fromseveral different points of view. We shall explore some partial differential equations that are of current interestand that exhibit some surprises. These include the Laplace-Beltrami operatorfor the Bergman metric on the ball. Along the way, we shall give a detailedtreatment of the Bergman kernel and associated metric, the Szego kernel, andthe Poisson-Szego kernel. Some of this material, particularly that in Chapter 6,may be considered ancillary and may be skipped on a first reading of this book. Complete and self-contained proofs of all results are provided. Some of theseappear in book form for the first time. Our treatrlent of the a-Neumann problemparallels some classic treatments, but since we present the problem in a concretesetting we are able to provide more detail and a more leisurely pace. Background required to read this book is a basic grounding in real and com-plex analysis. The book Function Theory of Several Complex Variables by thisauthor will also provide useful background for many of the ideas seen here.Acquaintance with measure theory will prove helpful. For motivation, exposure
  8. 8. xii Prefaceto the basic ideas of differential equations (such as one would encounter in asophomore differential equations course) is useful. All other needed ideas aredeveloped here. A word of warning to the reader unversed in reading tracts on partial differ-ential equations: the metier of this subject is estimates. To keep track of theconstants in these estimates would be both wasteful and confusing. (Although incertain aspects of stability and control theory it is essential to name and catalogall constants, that is not the case here.) Thus we denote most constants by Gor G; the values of these constants may change from line to line, even thoughthey are denoted with the same letter. In some contexts we shall use the nowpopular symbol ;S to mean "is less than or equal to a constant times ...." This book is based on a year-long course given at Washington University inthe academic year 1987-88. Some of the ideas have been presented in earliercourses at UCLA and Penn State. It is a pleasure to thank Estela Gavosto andMarco Peloso who wrote up the notes from my lectures. They put in a lot ofextra effort to correct my omissions and clean up my proofs and presentations.I also thank the other students who listened to my thoughts and provided usefulremarks. -S.G.K.
  9. 9. 1The Dirichlet Problem in the Complex Plane1.1 A Little NotationLet IR denote the real number line and C the complex plane. The complex andreal coordinates are related in the usual fashion by z == x + iy.We will spend some time studying the unit disc {z E C : Izi < I}, and wedenote it by the symbol D. The Laplace operator (or Laplacian) is the partialdifferential operator a2 a2 ~ = ox2 + oy2 . When the Euclidean plane is studied as a real analytic object, it is convenientto study differential equations using the partial differential operators a a ax and ay .This is so at least in part because each of these operators has a null space(namely the functions depending only on y and the functions depending only onx, respectively) that plays a significant role in our analysis (think of the methodof guessing solutions to a linear differential equation having the form u(x )v(y)). In complex analysis it is more convenient to express matters in terms of thepartial differential operators and ~ 8z == ~ (~+i~). 2 ax ayCheck that a continuously differentiable function f(z) = u(z) + iv(z) thatsatisfies 8 f / 8 z == 0 on a planar open set U is in fact holomorphic (use the
  10. 10. 2 The Dirichlet Problem in the Complex PlaneCauchy-Riemann equations). In other words, a function satisfying of/oz == 0may depend on z but not on z. Likewise, a function that satisfies af / 0 Z == 0on a planar open set may depend on z but cannot depend on z. Observe that o - z == 1 o -z==O oz oz o o - z ==0 oz ozz == 1.Finally, the Laplacian is written in complex notation as1.2 The Dirichlet ProblemIntroductory RemarksThroughout this book we use the notation C k (X) to denote the space of func-tions that are k-times continuously differentiable on X-that is, functions thatpossess all derivatives up to and including order k and such that all thosederivatives are continuous on X. When X is an open set, this notion is self-explanatory. When X is an arbitrary set, it is rather complicated, but possible,to obtain a complete understanding (see [STSI]). For the purposes of this book, we need to understand the case when X isa closed set in Euclidean space. In this circumstance we say that f is C kon X if there is an open neighborhood U of X and a C k function j on U 1such that the restriction of to X equals f. We write f E Ck(X). In casek == 0, we write either CO(X) or C(X). This definition is equivalent to allother reasonable definitions of C k for a non-open set. We shall present a moredetailed discussion of this matter in Section 3. Now let us formulate the Dirichlet problem on the disc D. Let ¢ E C(oD).The Dirichlet problem is to find a function U E C(D) n C 2 (D) such that ~U(z) == 0 if zED U(z) == ¢ ( z) if z E aD.REMARK Contrast the Dirichlet problem with the classical Cauchy problemfor the Lapiacian: Let S ~ lR2 == C be a smooth, non-self-intersecting curve
  11. 11. The Dirichlet Problem 3 s uFIGURE 1.1(part of the boundary of a smoothly bounded domain, for instance). Let U bean open set with nontrivial intersection with S (see Figure 1.1). Finally, let ¢oand ¢l be given continuous functions on S. The Cauchy problem is then ~u(z) == 0 if z E U u(z) == ¢l (z) if z E S n U au (z) == ¢l av if z E S n U.Here v denotes the unit normal direction at z E S. Notice that the solution to the Dirichlet problem posed above is unique: ifUl and U2 both solve the problem, then Ul - U2 is a hannonic function havingzero boundary values on D. The maximum principle thea implies that Ul == U2.In particular, in the Dirichlet problem the specifying of boundary values alsouniquely determines the normal derivative of the solution function u. However, in order to obtain uniqueness in the Cauchy problem, we mustspecify both the value of U on S and the normal derivative of u on S. How canthis be? The reason is that the Dirichlet problem is posed with a simple closedboundary curve; the Cauchy problem is instead a local one. Questions of whenfunction theory reflects (algebraic) topology are treated, for instance, by the deRham theorem and the Atiyah-Singer index theorem. We shall not treat themin this book, but refer the reader to [GIL], [KRl], [DER]. I
  12. 12. 4 The Dirichlet Problem in the Complex PlaneThe Solution of the Dirichlet Problem in L2Define functions ¢n on 8 D by n E Z.Notice that the solution of the Dirichlet problem with data ¢n is u n (re i8 ) ==r lnl ein8 . That is, zn if n >0 un(z) == { Z -n 1 ·f n - 0. <The functions {¢n}~=-oo form a basis for L 2 (8D) That is, if f E L 2 (8D) thenwe define ~ ir 27r an = f(t)e- int dt. 27f oIt follows from elementary Riesz-Fischer theory that the partial sums N SN == L a n ein8 ~ f (1.2.1) n=-Nin the L 2 topology. If 0 :::; r < 1 then observe that L 00 anrlnlein8 n=-oois an Abel sum for the Fourier series L:~oo a n ein8 of f. It follows from (1.2.1)that 00 S(r,O) == L anrlnlein8 ~ f(e i8 ) (1.2.2) n=-ooin the L 2 topology as r ~ 1- . In fact, the sum in (1.2.2) converges uniformly on compact subsets of thedisc. The computation that we are now about to do will prove this statement:We have 00 S(r,O) == L anrlnlein8 n=-oo
  13. 13. The Dirichlet Problem 5If we sum the two geometric series and do the necessary algebra then we findthat 27r 1 2 S 0 - -1 (r, ) - 21r 0 f it 1 - r d ( e ) 1 - 2r cos (0 - t) + r 2 t.This last formula allows one to do the estimates to check for uniform conver-gence, and thus to justify the change of order of the sum and the integral. We set =~ 2 Pr ( lj;) 1- r 27r 1 - 2r cos (l/J) + r2 and we call this function the Poisson kernel. Since the function u(re iO ) == S(r, 0)is the limit of the partial sums TN(r, 0) == E:=-N anrlnleinO, and since eachof the partial sums is hannonic, u is hannonic. Moreover, the partial sumTN is the harmonic function that solves the Dirichlet problem for the dataf N(e iO ) == E:=_ N an einO . We might hope that u is then the solution of theDirichlet problem with data f. This is in fact true:THEOREM 1.2.3Let f(e it ) be a continuous function on aD. Then the function u(reiO ) == { J~7r.f(ei(O-t))Pr(t) dt if 0 :::; r < 1 f(e~O) ifr == 1solves the Dirichlet problem on the disc with data f.PROOF Pick E > O. Choose 8 > 0 such that if Is - tl < 8, then If( eis ) -f (e it )I < Eo Fix a point eiO E aD. We will first show that limr-t 1- u(re iO ) ==f(e iO ) == u(e iO ). Now, for 0 < r < 1, lu(re ill ) - I(e ill )/ = 11 2 71: l(e i(II-t»)Pr(t) dt - l(eill)l. (, using the sum from which we obtained the Poisson kernel, that f27r f27r 1 IPr(t)1 dt = 1 Pr(t) dt = 1. 0 0Thus we may rewrite ( as rio 2 71: [I (e i(lI-t») - 1 (e ill )] Pr (t) dt = i r 1 tl<8 [I (e i(lI-t») - f( eill )] Pr (t) dt +l<::,t <::27r-JI( ei(lI-t») - I( eill )] Pr(t) dt == I + II.
  14. 14. 6 The Dirichlet Problem in the Complex PlaneNow the term I does not exceed 21r J r1 tl<8 IPr(t) IE dt ::; E1 0 IPr(t) Idt = E.For the second, notice that 8 < t < 27r - 8 implies that 1 1 - r2 IPr(t)1 == Pr(t) == - - - - - 27r 1 - 2r cos t + r 2 1 1 - r2 27r (1 - 2r cos t + r 2 cos 2 t) + r 2 (1 - cos2 t) 1 1 - r2 <-----2 - 27r (1 - r cos t) 1 1 - r2 < 27r 84 /8 .Thus 1r 1 1 - r2 II < 2 - l 8 2 sup III . - - - dt ~ 0 4 27r 8 /8as r -+ 1-. In fact, the proof shows that the convergence is uniform in (). Putting together our estimates on I and I I, we find that iO lim sup lu(re ) - I( e ) I iO < E. r----+l-Since E >0 was arbitrary, we see that lim sup Iu(re ) - iO I (e iO ) I == o. r----+l- The proof is nearly complete. For () E aD and E > 0 fixed, choose 8 > 0such that (i) lu(e iO ) - u(ei1/J) I < E/2 when le iO - ei1/J1 < 8;(ii) lu( re i1/J) - u(ei1/J) I < E/2 when r > 1 - 8,0 :S tt/J ~ 27r.Let zED satisfy Iz - eiol < 8. Then lu(z) - u(eio)1 :S lu(z) - u(z/lzl)l + lu(z/Izl) - u(eio)1 E E < 2 + 2where we have applied (ii) and (i) respectively. This shows that u is continuousat the boundary (it is obviously continuous elsewhere) and completes the proof. I
  15. 15. Lipschitz Spaces 71.3 Lipschitz SpacesOur first aim in this book is to study the boundary regularity for the Dirich-let problem: if the data f is "smooth," then will -the solution of the Dirichletproblem be smooth up to the boundary? This is a venerable question in thetheory of partial differential equations and will be a recurring theme throughoutthis book. In order to formulate the question precisely and give it a carefulanswer, we need suitable function spaces. The most naive function spaces for studying the question formulated in thelast paragraph are the C k spaces, mentioned earlier. However, these spacesare not the most convenient for our study. The reason, which is of centralimportance, is as follows: We shall learn later, by a method of Hormander[H03], that the boundary regularity of the Dirichlet problem is equivalent to theboundedness of certain singular integral operators (see [STSI]) on the boundary.Singular integral operators, central to the understanding of many problems inanalysis, are not bounded on the C k spaces. (This fact explains the mysteriouslyimprecise formulation of regularity results in many books on partial differentialequations. It also means that we shall have to work harder to get exact regularityresults.) Because of the remarks in the preceding paragraph, we now introduce the scaleof Lipschitz spaces. They will be somewhat familiar, but there will be somenew twists to which the reader should pay careful attention. A comprehensivestudy of these spaces appears in [KR2]. Now let U ~ ~N be an open set. Let 0 < Q < 1. A function f on U is saidto be Lipschitz of order Q, and we write f E An, if If(x + h) - f(x)1 sup h:;tO Ihl a + Ilfllux,(u) == IlfIIA,,(U) < 00. x,x+hEUWe include the term IIfIILoo(u) in this definition in order to guarantee that theLipschitz norm is a true norm (without this term, constant functions would have"norm" zero and we would only have a semi-norm). In other contexts it isuseful to use IlfIlLP(u) rather than IlfIILOO(U). See [KR3] for a discussion ofthese matters. When Q == 1 the "first difference" definition of the space An makes sense,and it describes an important class of functions. However, singular integraloperators are not bounded on this space. We set this space of functions apartby denoting it differently: sup If(x + h) - f(x)1 Ihl + IIIII Loo(U) < 00. h=¢.O x,x+hEU
  16. 16. The Dirichlet Problem in the Complex PlaneThe space Lipl is important in geometric applications (see [FED]), but less soin the context of integral operators. Therefore we define IIfIIA == If(x + h) + I(x - h) - 2/(x)1 IIIII 1 (U) sup h:;iO Ihl + LCXJ(U) < 00. x,x+h,x-hEU Inductively, if 0 < k E Z and k < a ::; k + 1, then we define a function fon U to be in Aex if I is bounded, I E C1(U), and any first derivative Djllies in Aex - l . Equivalently, I E A ex if and only if f is bounded and, for everynonnegative integer f < Q and multiindex {3 of total order not exceeding f wehave (8 j 8x)(3 I exists, is continuous, and lies in Aex - f . The space Lip k 1 < k E Z, is defined by induction in a similar fashion.REMARK As an illustration of these ideas, observe that a function 9 is inAS / 2(U) if 9 is bounded and the derivatives 8gj8xj, 82gj8xj8xk exist and liein A1/ 2 • Prove as an exercise that if Q > Q then Aexl ~ A ex . Also prove that theWeierstrass nowhere-differentiable function 00 F(O) == L 2- e j i2ie j=Ois in Al (0, 27r) but not in LiPl (0, 21r). Construct an analogous example, foreach positive integer k, of a function in A k Lip k If U is a bounded open set with smooth boundary and if 9 E Aex(U) thendoes it follow that 9 extends to be in A ex (U)? I Let us now discuss the definition of C k spaces in some detail. A functionf on an open set U ~ }RN is said to be k-times continuously differentiable,written lEek (U), if all partial derivatives of I up to and including order kexist on U and are continuous. On }Rl , the function I (x) == Ixllies in CO C 1•Examples to show that the higher order Ok spaces are distinct may be obtainedby anti-differentiation. In fact, if we equip Ok (U) with the norm IlfIICk(U) == IlfIIL(X)(U) + L II (~::) II lal::;k Loo(U) then elementary arguments show that C k + 1 (U) is contained in, but is nowheredense in, Ck(U).
  17. 17. Lipschitz Spaces 9 It is natural to suspect that if all the k th order pure derivatives (a/ax j )f f existand are bounded, 0 ::; /!, ::; k, then the function has all derivatives (includingmixed ones) of order not exceeding k and they are bounded. In fact Mityaginand Semenov [MIS] showed this to be false in the strongest possible sense.However, the analogous statement for Lipschitz spaces is true-see [KR2]. Now suppose that U is a bounded open set in ~N with smooth boundary. Wewould like to talk about functions that are C k on U == U u au. There are threeways to define this notion: I. We say that a function f is in C k (0) if f and all its derivatives on U of order not exceeding k extend continuously to (j. II. We say that a function f is in C k (0) if there is an open neighborhood W of 0 and a C k function F on W such that Flo == f. III. We say that a function f is in Ck(U) if f E Ck(U) and for each Xo E au and each multiindex Q such that IQ I ::; k the limit ao: lim U3x----+xo -a f(x) xO: exists. We leave as an exercise for the reader to prove the equivalence of thesedefinitions. Begin by using the implicit function theorem to map U locally to aboundary neighborhood of an upper half-space. See [HIR] for some help.REMARK A basic regularity question for partial differential equations is asfollows: consider the Laplace equation .6u == f.If f E ;0: (~N ), then where (Le. in what smoothness class) does the function ulive (at least locally)? In many texts on partial differential equations, the question is posed as "Iff E C k(~N) then where does u live?" The answer is generally given as"u E Cl~:2-f for any E > 0." Whenever a result in analysis is formulated inthis fashion, it is safe to assume that either the most powerful techniques are notbeing used or (more typically) the results are being formulated in the languageof the incorrect spaces. In fact, the latter situation obtains here. If one usesthe Lipschitz spaces, then there is no t-order loss of regularity: f E Ao: (~N)implies that u is locally in AO:+ 2 (IR N ). Sharp results may also be obtained byusing Sobolev spaces. We shall explore this matter in further detail as the bookdevelops. I
  18. 18. 10 The Dirichlet Problem in the Complex Plane1.4 Boundary Regularity for the Dirichlet Problem for the Laplacian on the Unit DiscWe begin this discussion by posing a question: Question: If we are given a "smooth" function f on the boundary of the unit disc D, then is the solution u to the Dirichlet problem for the Laplacian, with boundary data f, smooth up to closure of D? That is, if f E C k (8D), then is u E Ck(D)?It turns out that the answer to this question is "no." But the reason is that weare using the wrong spaces. We can only get a sharp result if we use Lipschitzspaces. Thus we have: Revised Question: If we are given a "smooth" function f on the boundary of the unit disc D, then is the solution u to the Dirichlet problem for the Laplacian, with boundary data f, smooth up to the closure of D? That is, if f E Ao:(8D), then is u E Ao:(D)? We still restrict our detailed considerations to ~2 for the moment. Also, it isconvenient to work on the upper half-space U == {(x, y) E ~2 : y > O}. Wethink of the real line as the boundary of U. By conformally mapping the disc tothe upper half-space with the Cayley transformation ¢(z) == i(l - z)/(l + z),we may calculate that the Poisson kernel for U is the function P (x) y -! x 2 + y2 - 7r y For simplicity, we shall study Ao: for 0 < Q < 2 only. We shall see later thatthere are simple techniques for extending results from this range of Q to all Q.Now we have the following theorem.THEOREM 1.4.1Fix 0 < Q < 1. If f E Ao:(~) then u(x, y) = Pyf(x, y) == l Py(x - t)f(t) dtlies in Ao: (11).PROOF Since h IPy(x - t)1 dt = h Py(x - t) dt = 1,it follows that u is bounded by "f" £00 .
  19. 19. Boundary Regularity 11 B .A X+H . XFIGURE 1.2 Now fix X == (Xl, X2) E U. Fix also an H == (hI, h2) such that X +H E U.We wish to show that lu(X + H) - u(X)1 ~ CIHla.Set A == (XI,X2 + IHI),B == (Xl + h l ,X2 + h2 + IHI). Clearly A,B lie in Ubecause X, X + H do. Refer to Figure 1.2. Thenlu(X + H) - u(X)1 ~ lu(X) - u(A)1 + lu(A) - u(B)1 + lu(B) - u(X + H)I -=1+11+111. For the estimate of 1 we will use the following two facts: Fact 1: The function u satisfies 1:;2 U (X,y)1 ~ Cy 2 a -for (x, y) E U. Fact 2: The function u satisfiesfor (x, y) E U.
  20. 20. 12 The Dirichlet Problem in the Complex Plane Fact 2 follows from Fact 1, as we shall see below. Once we have Fact 2,the estimation of I proceeds as follows: Let 1(t) == (Xl, X2 + t), 0 ::; t ::; IHI.Then, noting that 11(t) I ~ t, we have lu(X) - u(A)1 = Jo (IHI d dt (u()(t))) dt I I (IHllaul I :S Jo ay -y(t) dt {IHI :S C Jo h(t)I"-1 dt (IHI :S C J t,,-l dt oThus, to complete the estimate on I, it remains to prove our two facts.PROOF OF FACT 1 First we exploit the harmonicity of u to observe that 2 a2u I == I a u I· I ay2 ax2Then 2 a2u I == I a u I ay2 ax 2 i: I = I ::2 Py(x - t)f(t) dtl = Ii: ::2 Py(x - t)f(t) dtl = Ii: ::2 Py(x - t)f(t) dtl· (, from the Fundamental Theorem of Calculus, we know that
  21. 21. Boundary Regularity 13As a result, line ( equals Ii: ::2 Py (X - t) [J(t) - f(x)] dtl < C i: ::2 I Py (X - t)llx - til> dt cjOO 1 2Y (3(x - tf - ~2) IIX _ tl a dt - 00 [ (x - t) 2 + y2] 2 C JOO Iy(3t - y2) Iit ia dt -00 (t 2 + y2)3 OO 1 C a-2 y J -00 (t 2 + 1)2-a/2 dt C ya-2 .This completes the proof of Fact 1.PROOF OF FACT 2 First notice that, for any x E IR, :; ~1 7r -00 00 [(x - t)2 + y2] 2 I(x - tf + y - 22y21If(t)1 dtl y=2 <~ - 7r 1 -00 00 If(t)1 (x - t)2 + y2 dtl y=2 ::; C . IIfIlLOO(~).
  22. 22. 14 The Dirichlet Problem in the Complex Plane Now, if Yo ~ 2, then from Fact 1 we may calculate that 1~~(xO,Yo)1 ~ 11 ~:~(XO,Y)dY+ ~~(XO,2)1 YO ~ l YO y"-2dy+ 1~~(XO,2)1 < Ca [ya-l - 0 + 2a - l ] + C2 A nearly identical argument shows that au( xo,yO) I ~ ay c a 1 Yo- Iwhen 0 ::; Yo < 2. We have proved Facts I and 2 and therefore have completed our estimates ofterm I. The estimate of I I I is just the same as the estimate for I and we shall sayno more about it. For the estimate of II, we write A == (aI, a2) and B == (b l , b2). Then lu(A) - u(B)1 == lu(al a2) - u(b l , b2)1 ::; lu( aI, a2) - u(b l , a2) I + lu(b l , a2) - u(b l , b2) IAssume for simplicity that al < bl as shown in Figure 1.2. Now set 7](t) =(al + t, a2),O < t < bl - al. Then bl - al d I I I == l o - (u 0 7]) (t) dt dt = Jo rbl-al d dt 1 00 -oc Pa2 (al + t - 8)f(8) d8 dt I I bl-all°O -d Pa2(al + t - d == l o -00 t s) [f(s) - f(t)] ds dt. We leave it as an exercise for the reader to see that this last expression, inabsolute value, does not exceed GIHla. The tenn 112 is estimated in the same way that we estimated I. I
  23. 23. Boundary Regularity 15THEOREM 1.4.2Fix I < 0: < 2. If f E An(I~) then u(x, y) == 1. Py(x - t)f(t) dtlies in An (71).DISCUSSION OF THE PROOF It follows from the first theorem that u E Af3, allo < 13 < 1. In particular, u is bounded. It remains to see that au au ax E An - I and ay E An-I.But ~~ = :x JPy(t)f(x - t) dt = JPy(t)j(x - t) dt.Now f E An - I by definition and we have already considered the case 0 <0: - 1 < 1. Thus au/ax lies in An - I (71). To estimate au/ay, we can instead estimate 82u 8y2 J == - ax 2 8 2 Py(t)f(x - t) dt = - :x J Py(t)j(x - t) dt = - J :x Py(x - t)f(t) dt =- J :x Py(x - t)f(t) dt.Using the ideas from the estimates on I in the proof of the last theorem, we seethat 82u I C n-2 8 y 2 ::; y . ILikewise, because 8 f / 8x E An -I, we may prove as in Facts 1 and 2 in theestimate of I in the proof of the last theorem that a2u I ::; c y n-2 . Iax8yBut then our usual arguments show that au/ay E An-I. We leave details tothe reader. The proof is now complete. I
  24. 24. 16 The Dirichlet Problem in the Complex PlaneTHEOREM 1.4.3If f E AI(~), then u(x, y) == l Py(x - t)f(t) dt We break the proof up into a sequence of lemmas.LEMMA 1.4.4Fix y > O. Take (xo, YO) E U. Thenu(xo, Yo) = l Y y ()~~2 u(xo, Yo + y) dy - y :y u(xo, Yo + y) + u(xo, y + Yo)·PROOF Note that when y == 0, then the right-hand side equals O-O+u(xo, Yo).Also, the partial derivative with respect to y of the right side is identically zero.That completes the proof. ILEMMA 1.4.5If f E Al (~), thenPROOF Notice that, because u is harmonic, I::2 ul = I::2 ul =I ::2 f Py(x - t)f(t) dtl = If (::2 Py(x - t)) f(t) dtl = If (::2 Py(x - t)) f(t) dtl = IJ (::2 Py(t)) f(x - t) dtl = ~ IJ (:t: Py(t)) [I(x + t) + I(x - t) - 2f(x)] dtl
  25. 25. Boundary Regularity 17(since (82 / 8t 2 ) P y (t) is even and has mean value zero). By hypothesis, the lastline does not exceed C JI::2Py(t)!ltldt. ( ) But recall, as in the proof of 1.4.1, that Therefore we may estimate line ( byThis is what we wanted to prove. ICOROLLARY 1.4.6Our calculations have also proved thatREMARK Similar calculations prove that !a~3ay u(x, Y)I ::; C· y- 2, Ia:;y2 (X,y)1 ::; C·y-2, U 1:;3 U (X,y)!::; C·y- 2, !a:~y u(x, y) I ::; C . y-l. I If v is a function, H == (h l ,h2 ),X == (XI,X2) and if X,X +H E U, thenwe define ~kv(X) == v(X + H) + v(X - H) - 2v(X).LEMMA 1.4.7If all second derivatives of the function v exist, then we have l~kv(X)1 ::; C· IHI 2 . sup 1~2v(X + tH)I. Itl~l
  26. 26. 18 The Dirichlet Problem in the Complex PlanePROOF We apply the mean value theorem twice to the function ¢J(t) == v(X +tH). Thus l~kv(X)1 == 1¢(I) + ¢( -1) - 2¢(O)1 == I(¢(l) - ¢(O)) - (¢(O) - ¢J(-1))1 == 11 . ¢ (~ l) - 1 . ¢ (~2) I == I¢" (~3) . (~l - ~2) I ~ 21¢"(~3)1 ~ CIHI 2 sup 1V2 v (X + tH)I. ItI::; 1That proves the lemma. IFINAL ARGUMENT IN THE PROOF OF THEOREM 1.4.3 Fix X = (xo, Yo), H ==(h l ,h2 ). Then, by Lemma 1.4.4, ~ku(xo, Yo) = l Y y ~k (a~~2 u(xo, Yo + y)) dy -y~H 2 ({) ayU(XOYo+y) ) +~HU(XO,yo+Y) 2 =:1+11+111,any y > O. Now we must have that h 2 < Yo or else ~1Iu(xo, Yo) makes nosense. Thus Yo + tH + y 2 y, all -1 ~ t ~ 1. Applying Lemma 1.4.5, we see that III:::; (Y y ·4 sup I !~2 u(xo + th), Yo + th 2 + y) I dy Jo Itl::;l uy :::; c l Y y . (y)-l dy ~ C·y.Next, Lemma 1.4.7 and the remark preceding it yield 1111 ~ C·y·IHI 2 sup Itl::;l 2 ! IV uy U(Xo+th ,Yo+y+th2 1 )1 ~ CylHI2 . (y)-2 == CIHI2 . y-l.
  27. 27. Boundary Regularity 19Finally, the same reasoning gives 11111::; IHI 2 . sup 1V2 u (xo+th l ,yo+th2 +y)1 It I:::; I ::; C . IHI 2 . y-I.Now we take y == 21HI. The estimates on 1,11,111 then combine to give 1~~u(X)1 ::; CIHI·This is the desired estimate on u. IREMARK We have proved that if f E Al (IR) then u == Pyf E Al (U) us-ing the method of direct estimation. An often more convenient, and natural,methodology is to use interpolation of operators, as seen in the next theorem. ITHEOREM 1.4.8Let V ~ IRm, W ~ IRn be open with smooth boundary. Fix 0 < 0: < (3and assume that T is a linear operator such that T : Aa(V) ~ Aa(W) andT : A{J(V) ~ A{J(W). Then, for all 0: < 1 < (3 we have T : A,(V) ~ A,(W). Interpolation of operators, presented in the context of Lipschitz spaces, isdiscussed in detail in [KR2]. The subject of interpolation is discussed in abroader context in [STW] and [BOL]. Here is an application of the theorem: Let 0: == 1/2, {3 == 3/2, and let T be the Poisson integral operator fromfunctions on IR to functions on U. We know thatandWe may conclude from the theorem thatThus we have a neater way of seeing that Poisson integration is well behavedon AI.REMARK Twenty years ago it was an open question whether, if T is a boundedlinear operator on CO and on C 2 , it follows that T is a bounded linear operatoron C l . The answer to this question is negative; details may be found in [MIS].In fact, AI is the appropriate space that is intennediate to CO and C2.
  28. 28. 20 The Dirichlet Problem in the Complex Plane One might ask how Poisson integration is behaved on Lip I (IR). Set f(x) ==Ixl· ¢(x) where ¢ E Cgo(IR), ¢ == 1 near O. One may calculate that (x, Y ) == P yf (x ) == ~ In [ (1 - 2x) + y2 . (1 +2x) + y2] 2 2 U 2- 2 2 X +y x +y +x 1- x 1+ x] [ arctan -y- - arctan -y- + (smooth error).Set x == O. Then, for y small, Again, we see that the classical space Lipi does not suit our purposes, whileAl does. We shall not encounter LiPI any more in this book. The space Al wasinvented by Zygmund (see [ZYG]), who called it the space of smooth functions.He denoted it by A*. Here is what we have proved so far: if ¢ is a piece of Dirichlet data forthe disc that lies in An (aD), 0 < a < 2, then the solution u to the Dirichletproblem with that data is An up to the boundary. We did this by transferringthe problem to the upper half-space by way of the Cayley transform and thenusing explicit calculations with the Poisson kernel for the half-space.1.5 Regularity of the Dirichlet Problem on a Smoothly Bounded Domain and Conformal MappingWe begin by giving a precise definition of a domain "with smooth boundary":DEFINITION 1.5.1 Let U ~ C be a bounded domain. We say that U hassmooth boundary if the boundary consists of finitely many curves and each ofthese is locally the graph of a COO function. In practice it is more convenient to have a different definition of domain withsmooth boundary. A function p is called a defining function for U if p is definedin a neighborhood W of au, l p i= 0 on au, and wnu == {z E W : p(z) < O}.Now we say that U has smooth (or C k ) boundary if U has a defining functionp that is smooth (or C k ). Yet a third definition of smooth boundary is that theboundary consists of finitely many curves rj, each of which is the trace of asmooth curve r( t) with nonvanishing gradient. We invite the reader to verifythat these three definitions are equivalent.
  29. 29. Regularity of the Dirichlet Problem 21 "",---- ........... , / ", " /_-........" U ", /-- ........ "" I / " / / ~ laD Iw / I / ~ I I I I aD / I I I / / I I / I ( / / I " au / / / ",/ "-..... -. - _ / "B / "........ " ........ _--_/ ", /FIGURE 1.3 Our motivating question for the present section is as follows: Let n ~ C be a bounded domain with smooth boundary. Assume that! E An (an). If u E C(n) satisfies (i) u is harmonic on nand (ii) ulan == !, then does it follow that u E An(n)? Here is a scheme for answering this question:Step 1: Suppose at first that U is bounded and simply connected.Step 2: By the Riemann mapping theorem, there is a conformal mapping ¢ : U ~ D. Here D is the unit disc. We would like to reduce our problem to the Dirichlet problem on D for the data ! 0 ¢ - 1• In order to carry out this program, we need to know that ¢ extends smoothlyto the boundary. It is a classical result of Caratheodory [CAR] that if a simplyconnected domain U has boundary consisting of a Jordan curve, then any con-formal map of the domain to the disc extends univalently and bicontinuously tothe boundary. It is less well known that Painleve, in his thesis [PAl], proved thatwhen U has smooth boundary then the conformal mapping extends smoothly tothe boundary. In fact, Painleves result long precedes that of Caratheodory. We shall present here a modem approach to smoothness to the boundaryfor conformal mappings. These ideas come from [KERl]. See also [BKR]for a self-contained approach to these matters. Our purpose here is to tie thesmoothness-to-the-boundary issue for mappings directly to the regularity theoryof the Dirichlet problem for the Laplacian. Refer to Figure 1.3. Let W be a collared neighborhood of au. Set au ==aw n U and let aD == ¢(aU). Define B to be the region bounded by aD
  30. 30. 22 The Dirichlet Problem in the Complex Planeand aD. We solve the Dirichlet problem on B with boundary data I if (E aD f(() ={ 0 if (E aD.Call the solution u. Consider v == u 0 ¢ : U ~ nt Then, of course, v is still hannonic. ByCaratheodorys theorem, v extends to au, au, and I if (E au v == { 0 if (E au.Suppose that we knew that solutions of the Dirichlet problem on a smoothlybounded domain with Coo data are in fact Coo on the closure of the domain.Then, if we consider a first-order derivative V of v, we obtain IVvl == IV(u 0 ¢)I == l7ull7¢1 ~ C.It follows that (1.5.2)This will prove to be a useful estimate once we take advantage of the followinglemma.LEMMA 1.5.3 HOPFS LEMMALet n cc }RN have C 2 boundary. Let u E C(n) with u harmonic and noncon-stant on n. Let PEn and assume that u takes a local minimum at P. Thenthe one-sided normal derivative satisfies au av (P) < O.PROOF Suppose without loss of generality that u > 0 on n near P and that nu(P) == O. Let B R be a ball that is internally tangent to at P. We may assumethat the center of this ball is at the origin and that P has coordinates (R, 0, ... ,0).Then, by Harnacks inequality (see [KR1]), we have for 0 < r < R that R2 - r 2 u(r, 0, ... , 0) ~ c· R2 + r 2 hence u(r,O, ... ,O)-u(R,O, ... ,O) , 0 - - - - - - - - - - - < -c < . r-R -Therefore ou() , ov p ~ -c < o.This is the desired result. I
  31. 31. Regularity of the Dirichlet Problem 23 Now let us return to the u from the Dirichlet problem that we consideredprior to line (1.5.2). Hopfs lemma tells us that l7ul 2 c > 0 near aD. Thus,from (1.5.2), we conclude that 17¢1 ~ C. (1.5.4)Thus we have bounds on the first derivatives of ¢. To control the second derivatives, we calculate that C 2 17 2 vl == 17(7v) 1== 17(7(u 0 ¢))I == 1 (7 u ( ¢) . 7¢)1== 1(7 2U . [7¢] 2 ) + (7 u . 7 2 ¢) 7 I·Here the reader should think of 7 as representing a generic first derivative and72 a generic second derivative. We conclude thatHence (again using Hopfs lemma), 17 2 "1 < f - ~ l7ul < C". -In the same fashion, we may prove that l7 k ¢1 ~ Ck , any k E {1,2, ...}. Thismeans (use the fundamental theorem of calculus) that ¢ E Coo (0). We have amved at the following situation: Smoothness to the boundaryof confonnal maps implies regularity of the Dirichlet problem on a smoothlybounded domain. Conversely, regularity of the Dirichlet problem can be used,together with Hopfs lemma, to prove the smoothness to the boundary of con-formal mappings. We must find a way out of this impasse. Our solution to the problem posed in the last paragraph will be to study theDirichlet problem for a more general class of operators that is invariant undersmooth changes of coordinates. We will study these operators by (i) localizingthe problem and (ii) mapping the smooth domain under a diffeomorphism toan upper half-space. It will tum out that elliptic operators are invariant underthese operations. We shall then use the calculus of pseudodifferential operatorsto prove local boundary regularity for elliptic operators. There is an important point implicit in our discussion that deserves to bebrought into the foreground. The Laplacian is invariant under conformal trans-formations (exercise). This observation was useful in setting up the discussionin the present section. But it turned out to be a point of view that is too narrow:we found ourselves in a situation of circular reasoning. We shall thus expandto a wider universe in which our operators are invariant under diffeomorphisms.This type of invariance will give us more flexibility and more power. Let us conclude this section by exploring how the Laplacian behaves undera diffeomorphic change of coordinates. For simplicity, we restrict attention to
  32. 32. 24 The Dirichlet Problem in the Complex Plane]R2 with coordinates (x, y). Let ¢ ( x, y) == (¢ 1 (x, y), cP2 (x, y)) == (x, y)be a diffeomorphism of }R2. LetIn (x, y) coordinates, the operator ~ becomesIn an effort to see what the new operator has in common with the old one, weintroduce the notationwhere a a a a axO: axr 1 ax~2 ... ax~nis a differential monomial. Its "symbol" (for more on this, see the next twochapters) is defined to beThe symbol of the Laplacian ~ == (a 2 / ax 2 ) + (a 2 / ay2) is (J(~) == ~f + ~i·Now associate to (J(~) a matrix ..4~ == (aij)1:S;i,j:S;2, where aij = aij(x) is thecoefficient of ~i~j in the symbol. ThusThe symbol of the transfonned Laplacian (in the new coordinates) is (J(¢*(~)) == 17¢112~f + 17¢212~i ax ay ax ay,] +2 [ - - + - - ~le2 ax ay By By + (lower order tenns).
  33. 33. Regularity of the Dirichlet Problem 25Then The matrix A(J(¢*(~)) is positive definite provided that the change of coordi-nates ¢ is a diffeomorphism (i.e., has nondegenerate Jacobian). It is this positivedefiniteness property of the symbol that will be crucial to the success of ourattack on elliptic operators.
  34. 34. 2Review of Fourier Analysis2.1 The Fourier TransformA thorough treatment of Fourier analysis in Euclidean space may be found in[STW] or [HOR4]. Here we give a sketch of the theory. If t, ~ E }RN then we let t· ~ == tl~1 + ... + tN~N.We define the Fourier transform of an f E £1 (I~N) by j(~) = Jf(t)eit.~ dt.Many references will insert a factor of 271" in the exponential or in the measure.Others will insert a minus sign in the exponent. There is no agreement on thismatter. We have opted for this definition because of its simplicity. We note thatthe significance of the exponentials eit.~ is that the only continuous multiplicativehomomorphisms of}RN into the circle group are the functions ¢~ (t) == eit.~.(We leave this as an exercise for the reader. A thorough discussion appearsin [KAT] or [BAC].) These functions are called the characters of the additivegroup }RN.Basic Properties of the Fourier TransformPROPOSITION 2.1.1If f E £1 (}RN) thenPROOF Observe that Ij(~) I ::; J If(t)1 dt. I
  35. 35. The Fourier Transform 27PROPOSITION 2.1.2If f E L 1 (IR N ), f is differentiable, and af/axj ELI, thenPROOF Integrate by parts: if f E C~ then ( ~) aXj == J af eit.~ dt atj - f··· 1f atj eit.~ dt· dtl d-t· dtN - af J... J ••• = - f··· Jf(t) C)~/it.t,) dtj dt) ... dij ... dtN = -i~j J..J f(t)eit.t, dt.The general case follows from a limiting argument. IPROPOSITION 2.1.3If f E Ll(I~N) and iXjf E L 1 (IR N ), then -- a A (iXjf) = 8~j f·PROOF Integrate by parts. IPROPOSITION 2.1.4 THE RIEMANN-LEBESGUE LEMMAIf f E L 1(IR N ), then lim Ij(~)1 == o. ~-+ooPROOF First assume that f E C~ (IR N ). We know thatand
  36. 36. 28 Review of Fourier AnalysisThen (1 + lel 2 )j is bounded. Therefore III :::; C le~oo O. 1 + 1~12This proves the result for f E C~. Now let 9 E L 1 be arbitrary. By elementary measure theory, there is afunction ¢ E Cc(I~N) such that IIg - ¢IILI < E/4. It is then straightforward toconstruct a ljJ E C~ such that 1I¢-ljJIILI < E/4. It follows that IIg-ljJIILI < E/2. Choose M so large that when I~I > M then 1"z,(~)1 < E/2. Then, for I~I > M,we have Ig(~)1 == l(g--=-ljJ)(~) + "z,(~)1 :S I(g --=-ljJ)(~)1 + 1"z,(~)1 E :S IIg - ljJIILI + 2 E E < - + - == E. 2 2This proves the result. IPROPOSITION 2.1.5Let fELl (I~N). Then j is uniformly continuous.PROOF Apply the Lebesgue dominated convergence theorem and Proposi-tion 2.1.4. I Let Co (}RN) denote the continuous functions on }RN that vanish at 00. Equipthis space with the supremum norm. Then the Fourier transform maps L 1 to Cocontinuously, with operator norm 1.PROPOSITION 2.1.6 A _If f E L 1 (}RN), we let j(x) == f(-x). Then j(~) == j(~).PROOF We calculate that J(~) = J j(t)eit·e dt = J f( _t)eit·e dt = J f(t)e-it·e dt = 1. IPROPOSITION 2.1.7If p is a rotation of}RN then we define pf (x) == f (p( x)). Then Pi == p(j).
  37. 37. The Fourier Transform 29PROOF Remembering that p is orthogonal, we calculate that p](E,) J (pf)(t)eif,.t dt = J f(p(t))eif,.t dt (s=g,(t)) J f(s)e i f,.p-(s) ds =J f(s)eip(f,).s ds (p(j)) (~). IPROPOSITION 2.1.8We havePROOF We calculate thatPROPOSITION 2.1.9If 8 > 0 and f E L 1(IR N ), then we set 08f(x)8- N f(x/8). Then (alif) = ali (I) 0 8f == 08j.PROOF We calculate that (alif) = J(ali/) (t)eit.f, dt = J f(8t)e it .f, dt = J f(t) ei(t/li)-f, 8- N dt = 8- N j(E,/8) = ali(j).That proves the first assertion. The proof of the second is similar. I If I, g are L 1 functions, then we define their convolution to be the function f * g(x) = J f(x - t)g(t) dt = J g(x - t)f(t) dt.It is a standard result of measure theory (see [RUD3]) that I *g so defined isan L 1 function and III * gllL1 ::; IIfllL1 IIgllLI.
  38. 38. 30 Review of Fourier AnalysisPROPOSITION 2.1.10If 1,9 E £1, then r;-9(~) == j(~) . g(~).PROOF We calculate that f7g(0 = J * g)(tk~·t = JJ (f dt f(t - s)g(s) ds ei~.t dt = JJf(t - s)ei~.(t-s) dt g(s)ei~s ds = j(~) . g(~).The reader may justify the change in the order of integration. IPROPOSITION 2.1.11If I, 9 E £1 , then Jj(~)g(~) d~ Jf(Og(~) d~. =PROOF This is a straightforward change in the order of integration. IThe Inverse Fourier TransformOur goal is to be able to recover 1 from j. This program entails several technicaldifficulties. First, we need to know that the Fourier transform is univalent inorder to have any hope of success. Second, we would like to say that f(t) = c· Jj(~)e-it.~ dt.In general, however, the Fourier transform of an £1 function is not integrable.Thus we need a family of summability kernels G E satisfying the following prop-erties: 1. G E * f ~ 1 as E ~ 0; 2. c:(~) == e-EI~12; 3. G E * 1 and G--;; 1 are both integrable.It will be useful to prove formulas about G E *f and then pass to the limit asE ~ 0+.LEMMA 2.1.12We have
  39. 39. The Fourier Transform 31PROOF It is enough to do the case N == 1. Set I == J~oo e- t2 dt. Then I "I = I: I: e- s2 ds e- t2 dt = II IR2 e-!(s,t)!2 dsdt {21r roo _r 2 = Jo Jo e rdrd()=7LThus I == ~, as desired. IREMARK Although this is the most common method for evaluating J e- 1x !2 dx,several other approaches are provided in [HEI]. I Now let us calculate the Fourier transform of e- 1xI2 • It suffices to treat theone-dimensional case because (e- 1xI2 f = IN e-lx!2eixof, dx = l e-x~eixlf,1 dX"""l e-x~eixNf,N dXN"Now when N == 1 we have l e- x2 ix + f, dx =l e(f,/2+ix)2 e-e/ 4 dx = e-e /4l e(f,/2+ix)2 dx == e-~2 /4 { e(~/2+ix/2)2 ~ dx J IR 2 == ~e-~2 /4 1 e(Z/2)2 dz. ( 2 JrHere, for ~ E R fixed, r == r ~ is the curve t ~ ~ + it. Let r N be the part ofthe curve between t == - Nand t == N. Since Ir == limN ---l>OO Jr N it is enough Jfor us to understand r N Refer to Figure 2.1. NowThereforeBut, as N -+ 00, 1 +1 -+ O. JEfl JEr
  40. 40. 32 Review of Fourier Analysis O-iN -----~­ EfFIGURE 2.1Thus lim 1 == - lim 1 . ( N~oo JrN N~oo Jf N Now we combine ( and ( to see thatWe conclude that, in jRI ,and in }RN we have
  41. 41. The Fourier Transform 33It is often convenient to scale this formula and write The function G(x) == (27r)-N/2 e- 1 I2 /2 is called the Gauss-Weierstrass ker- xnel. It is a summability kernel (see [KAT]) for the Fourier transform. Observethat G(~) == e-I~12 /2. On R N we defineThen -- -- ~(~) = (e-1~12 /2) = (avee-I~12 /2) = ave [(27r)N/2e-I~12/2] == E-N/2(27r)N/2e-I~12 /(2E). Now assume that !, j are in L 1 and are continuous. We apply Proposi-tion 2.1.11 with 9 == GE ELI. We obtain Jf~(x) Jj(~)C.(~) d~. dx =In other words, (2.1.13)Now e-EI~12 /2 - t 1 uniformly on compact sets. Thus J j(~)e-EI~12 d~ -tJ j(~) d~. That takes care of the right-hand side of (2.1.13). Next observe thatThus the left side of (2.1.13) equals (27r)N Jf(x)G,(x) dx = (27r)N J f(O)G,(x) dx +(27r)N j[f(x) - f(O)JG,(x) dx -t (27r)N ·1(0).
  42. 42. 34 Review of Fourier AnalysisThus we have evaluated the limits of the left- and right-hand sides of (2.1.13).We have proved the following theorem.THEOREM 2.1.14 THE FOURIER INVERSION FORMULAIf !, j E £1 and both are continuous, then f(O) = (27r)-N Jj(~) d~. ( ) Of course there is nothing special about the point 0 E R N . We now exploitthe compatibility of the Fourier transform with translations to obtain a moregeneral formula. First, we define (Th!)(x) == I(x - h)for any function I on RN and any h E RN . Then, by a change of variable inthe integral, T;] == eih.~ j(~).Now we apply formula ( in our theorem to T-h/: The result isorTHEOREM 2.1.15If I, j E £1 then for any h E R N we have f(h) = (2JT)-N Jj(~)e-ih.f, d~.COROLLARY 2.1.16The Fourier transform is univalent. That is, if I, 9 E £1 and j == 9 then f = 9almost everywhere.PROOF Since I - 9 E £1 and j - g == 0 E £1, this is immediate from eitherthe theorem or (2.1.13). I Since the Fourier transform is univalent, it is natural to ask whether it issurjective. We havePROPOSITION 2.1.17The operatoris not onto.
  43. 43. The Fourier Transform 35PROOF Seeking a contradiction, we suppose that the operator is in fact surjec-tive. Then the open mapping principle guarantees that there is a constant C > 0such that IIIIILI :S cllj"L~On IR , let g(~) be the characteristic function of the interval [-1, 1]. The in- Iverse Fourier transform of g is a nonintegrable function. But then {G 1/ j * g}forms a sequence that is bounded in supremum norm but whose inverse Fouriertransforms are unbounded in £1 norm. That gives the desired contradiction. IPlancherels FormulaPROPOSITION 2.1.18 PLANCHERELIf I E C~ (IR N ) then JIj(~)12 d~ = J (21T)N If(xW dx.PROOF Define g( x) == I * 1E c~ (IRN ). Then ,,~ ,,~ ,,~ .... 2 9 == I . I == I . I == I . f == III . ( g(O) = f * 1(0) = J f( -t)!( -t) dt = J f(t)/(t) dt = J 2 If(t)1 dt.By Fourier inversion and formula ( we may now conclude that J If(t)1 2 dt = g(O) = (21T)-N Jg(~) d~ = (27T)-N JIj(~)12 dfThat is the desired formula. ICOROLLARY 2.1.19If f E £2 (IR N ) then the Fourier transform of I can be defined in the followingfashion: Let Ij E C~ satisfy fj -+ I in the £2 topology. It follows fromthe proposition that {.fj } is Cauchy in £2. Let g be the £2 limit of this lattersequence. We set j == g. It is easy to check that the definition of j given in the corollary is independentof the choice of sequence Ij E C~ and that J11(01 d~ 2 = (21T)N J If(xW dx.
  44. 44. 36 Review of Fourier Analysis We now know that the Fourier transform F has the following mapping prop-erties: F: £1 ~ LX F: L 2 ~ £2.The Riesz-Thorin interpolation theorem (see [STW]) now allows us to concludethat F : £P ~ LP , 1 ~ p ~ 2,where p == p/ (p - 1). If p > 2 then F does not map LP into any nice functionspace. The precise norm of F on LP has been computed by Beckner [BEC].Exercises: Restrict attention to dimension 1. Consider the Fourier transform Fas a bounded linear operator on the Hilbert space £2 (IR N ). Prove that the fourroots of unity (suitably scaled) are eigenvalues of F. Prove that if p(x) is a Hermite polynomial (see [STW], [WHW]), then thefunction p(x)e-lxI2/2 is an eigenfunction of F. (Hint: (ix!) == (j) and == l-i~j.)2.2 Schwartz DistributionsThorough treatments of distribution theory may be found in [SCH], [HOR4],[TRE2]. Here we give a quick review. We define the space of Schwartz functions: s = {¢ E c oo (IRN ) : Pa,(3(¢) == x~~ Ix a (~) (3 ¢(x) I < 00, 0: = (0:1, .. " O:N), {3 = ({31,"" (3N) }.Observe that e- 1x12 E Sand p(x) . e- 1x12 E S for any polynomial p. Anyderivative of a Schwartz function is still a Schwartz function. The Schwartzspace is obviously a linear space. It is worth noting that the space of Coo functions with compact support (whichwe have been denoting by C~) forms a proper subspace of S. Since as recentlyas 1930 there was some doubt as to whether C~ functions are genuine functions(see [OSG]), it may be worth seeing how to construct elements of this space. Let the dimension N equal 1. Define if x ~ 0 if x < o.
  45. 45. Schwartz Distributions 37Then one checks, using lHopitals Rule, that A E Coo (IR). Set h(x) == A( -x - 1) . A(X + 1) E C~(I~). I:Moreover, if we define g(x) = h(t) dt,then the function f(x) == g(x + 2) . g( -x - 2)lies in C~ and is identically equal to a constant on (-1, 1). Thus we haveconstructed a standard "cutoff function" on ~l. On IR N , the functionplays a similar role. Exercise: [The Coo Urysohn lemma] Let K and L be disjoint closed setsin IR N . Prove that there is a Coo function ¢ on IR N such that ¢ == 0 on K and¢ == 1 on L. (Details of this sort of construction may be found in [HIR].)The Topology of the Space 5The functions Pn,(3 are seminorms on 5. A neighborhood basis of 0 for thecorresponding topology on 5 is given by the sets NE,f,m == {¢: L lal:S€ Pa,(3(¢) < E} . 1.6I:SmExercise: The space 5 cannot be normed.DEFINITION 2.2.1 A Schwartz distribution Q is a continuous linear functionalon 5. We write Q E 5.Examples: 1. If f EL I , then f induces a Schwartz distribution as follows: S:3 qH--> f </>fdx E C. We see that this functional is continuous by noticing that If </>(x)f(x) dxl ::; sup 1</>1· Ilfll£l = C . Po,o(</».
  46. 46. 38 Review of Fourier Analysis A similar argument shows that any finite Borel measure induces a distri- bution. 2. Differentiation is a distribution: On R I , for example, we have 5 ::1 ¢ ~ ¢(O) satisfies I¢ (0) 1 ::; sup I¢ (x) I == PO.I (¢). xElR 3. If f E LP (RN ), 1 :::; p ::; 00, then f induces a distribution: Tf : S 3 </J f--> J </Jf dx E c. To see that this functional is bounded, we first notice that (2.2.2) where 1/ p + 1/ p == 1. Now notice that (1 + Ix IN +I) I¢(x )I ::; C (po,o (¢) + PN + 1,0 ( ¢)) , hence C I</J(x) I ::; 1 + Ixl N +1 (Po,o( </J) + PN+I,O(</J)) . Finally, II</JII Lp f ::; c· [ J(1 + I~IN+I ) P , dx ] lip . [Po,o(</J) + PN+I,O(</J)] . As a result, (2.2.2) tells us that T f ( ¢) ::; ell f II Lp (Po ,0 ( ¢) + PN + I ,0 ( ¢)) .Algebraic Properties of Distributions (i) If Q, (3 E 5 then Q+ (3 is defined by (Q + (3) (¢) == Q(¢) + (3( ¢). Clearly Q + (3 so defined is a Schwartz distribution. (ii) If Q E 5 and C E C then CQ is defined by (CQ)(¢) == c[Q(¢)]. We see that CQ E 5.(iii) If 1/J E 5 and Q E 5 then define (1/JQ) (¢) == Q(1/J¢). It follows that 1/JQ is a distribution. (iv) It is a theorem of Laurent Schwartz (see [SCH]) that there is no contin- uous operation of multiplication on S. However, it is a matter of great interest, especially to mathematical physicists, to have such an operation.
  47. 47. Schwartz Distributions 39 Colombeau [CMB] has developed a substitute operation. We shall say no more about it here. (v) Schwartz distributions may be differentiated as follows: If J-l E S then (8/ 8x)(3 J-l E S is defined, for ¢ E S, by Observe that in case the distribution J-l is induced by integration against a C~ function f, then the definition is compatible with what integration by parts would yield. Let us differentiate the distribution induced by integration against the functionf(x) == Ixl on lIt Now, for ¢ E S, f(¢) == - f(¢) = - [ : f¢ dx = -100 f(x)¢(x) dx - [°00 f(x)¢(x) dx = _ roo x¢(x) dx + fO x¢(x) dx io -00 00 =- [x¢(x)]: + 1 ¢(x) dx + [x¢(x)]~oo - [°00 ¢(x) dx = roo ¢(x) dx _ fO ¢(x) dx. io -00Thus f consists of integration against b( x) == -x (-00,0] + X[O,oo). This functionis often called the Heaviside function.Exercise: Let n ~ R N be a smoothly bounded domain. Let v be the unitoutward normal vector field to 8n. Prove that -VXn E S. (Hint: Use Greenstheorem. It will tum out that (- VXn) (¢) == Jan ¢ da, where da is area measureon the boundary.)The Fourier TransformThe principal importance of the Schwartz distributions as opposed to other dis-tribution theories (more on those below) is that they are well behaved under theFourier transform. First we need a lemma:
  48. 48. 40 Review of Fourier AnalysisLEMMA 2.2.3If f E S then j E S.PROOF This is just an exercise with Propositions 2.1.2 and 2.1.3: the Fouriertransform converts multiplication by monomials into differentiation and viceversa. IDEFINITION 2.2.4 If u is a Schwartz distribution, then we define a Schwartzdistribution u by u(¢) == u(¢). By the lemma, the definition of u makes good sense. Moreover, by 2.2.5below, lu(¢)1 == lu(¢)1 ~ L Pn,f3(¢) lal+If3I~Mfor some M > 0 (by the definition of the topology on S). It is a straightforwardexercise with 2.1.2 and 2.1.3 to see that the sum on the right is majorized bythe sum C· L Pn,f3(¢) lal+It3I~MIn conclusion, the Fourier transform of a Schwartz distribution is also a Schwartzdistribution.Other Spaces of DistributionsLet V == C~ and £ == Coo. Clearly V ~ S ~ £. On each of the spaces V and£ we use the semi-normswhere K ~ R N is a compact set and Q == (Ql, .. , QN) is a multiindex. Theseinduce a topology on V and £ that turns them into topological vector spaces.The spaces V and £ are defined to be the continuous linear functionals on Vand £ respectively. Trivially, £ ~ V. The functional in R l given by 00 j J-l==L2 8j , j=lwhere 8j is the Dirac mass at j, is readily seen to be in V but not in £. The support of a distribution J1 is defined to be the complement of the unionof all open sets U such that J1 (¢) == 0 for all elements of C~ that are supported
  49. 49. Schwartz Distributions 41in U. As an example, the support of the Dirac mass Do is the origin: when Dois applied to any testing function ¢ with support disjoint from 0 then the resultis O.Exercise: Let J.L E V. Then J.L E £ if and only if J.L has compact support. Theelements of £ are sometimes referred to as the "compactly supported distribu-tions."PROPOSITION 2.2.5A linear functional L on S is a Schwartz distribution (tempered distribution) ifand only if there is a C > 0 and integers m and f such that for all ¢ E S wehave IL(¢)I ~ C· L L Pn,f3(¢) ( Inl~f 1f3I~mSKETCH OF PROOF If an inequality like ( holds, then clearly L iscontinuous. For the converse, assume that L is continuous. Recall that a neighborhoodbasis of 0 in S is given by sets of the form NE,f,m == {¢ E S: L lal:S€ Pa,f3(¢) < E} . 1.6I:SmSince L is continuous, the inverse image of an open set under L is open.ConsiderThere exist E, f, m such thatThus L lal:Sf Pn,f3(¢) < E 1.6I:Smimplies that /L(¢)/ < 1.That is the required result, with C == 1/ Eo IExercise: A similar result holds for V and for £.
  50. 50. 42 Review of Fourier AnalysisTHEOREM 2.2.6 STRUCTURE THEOREM FOR Vflu E V then k U == LDjJ-lj, j=1where J-lj is a finite Borel measure and each Dj is a differential monomial.IDEA OF PROOF For simplicity, restrict attention to R I . We know that thedual of the continuous functions with compact support is the space of finiteBorel measures. In a natural fashion, the space of C I functions with compactsupport can be identified with a subspace of the set of ordered pairs of Cefunctions: f +-+ (I, f). Then every functional on C~ extends, by the Hahn-Banach theorem, to a functional on C e x C e . But such a functional will begiven by a pair of measures. Combining this information with the definitionof derivative of a distribution gives that an element of the dual of C~ is of theform J-l1 + (J-l2). In a similar fashion, one can prove that an element of the dualof C~ must have the form J-li + (J-l2) + ... + (J-lk+I)(k). Finally, it is necessary to note that V is nothing other than the countableunion of the dual spaces (C~). I The theorem makes explicit the fact that an element of V can depend ononly finitely many derivatives of the testing function-that is, on finitely manyof the norms Pn,(3. We have already noted that the Schwartz distributions are the most convenientfor Fourier transform theory. But the space V is often more convenient in thetheory of partial differential equations (because of the control on the supportof testing functions). It will sometimes be necessary to pass back and forthbetween the two theories. In any given context, no confusion should result.Exercise: Use the Paley-Wiener theorem (discussed in Section 4) or some othertechnique to prove that if cP E V then ¢ (j. V. (This fact is often referred to asthe Heisenberg uncertainty principle. In fact, it has a number of qualitative andquantitative formulations that are useful in quantum mechanics. See [FEG] formore on these matters.)More on the Topology of V and VWe say that a sequence {cPj} ~ V converges to ¢ E V if 1. all the functions ¢j have compact support in a single compact set K o; 2. PK,n(¢j - ¢) -+ 0 for each compact set K and for every multiindex Q. I The enemy here is the example of the "gliding hump": On R , if ljJ is afixed Coo function and ¢j (x) == ljJ (x - j), then we do not want to say that thesequence {¢j} converges to O.
  51. 51. Convolution and Friedrichs Mollifiers 43 A functional J-l on V is continuous if J-l(¢j) -+ J-l( ¢) whenever ¢j -+ ¢. Thisis equivalent to the already noted characterization that there exist a compact Kand an N > 0 such that IJ-l(¢) I :S C L PK,n(¢) Inl:::;Nfor every testing function ¢.2.3 Convolution and Friedrichs MollifiersRecall that two integrable functions f and 9 are convolved as follows: f *g = Jf(x - t)g(t) dt = J g(x - t)f(t) dt.In general, it is not possible to convolve two elements of V. However, we maysuccessfully perform any of the following operations: 1. We may convolve an element J-l E V with an element 9 E D. 2. We may convolve two distributions J-l, v E V provided one of them is compactly supported. 3. We may convolve VI, ..• ,Vk E V provided that all except possibly one is compactly supported. We shall now learn how to make sense of convolution. This is one of thosetopics in analysis (of which there are many) where understanding is best achievedby remembering the proof rather than the statements of the results.DEFINITION 2.3.1 We define the following convolutions: 1. If J-l E V and 9 E V then we define (J-l * g)(¢) == J-l(g * ¢), all ¢ E V. 2. If J-l E S and 9 E S then we define (J-l * g)(¢) == J-l(g * ¢), all ¢ E S. 3. If J-l E £ and 9 E V then we define (J-l * g)(¢) == J-l(g * ¢), all ¢ E £. Recall here that g( x) == g( - x). Observe in part (1) of the definition that9*¢ E V, hence the definition makes sense. Similar remarks apply to parts (2)and (3) of the definition. In part (3), we must assume that 9 E V; otherwise9 * ¢ does not necessarily make sense.LEMMA 2.3.2If Q E V and 9 E V then Q * 9 is a function. What is more, If we let7h¢(x) == ¢(x - h) then (Q * 9 )(x) == a(Txg).