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. 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 0-8493-7155-41. 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-dc2092-11422CIPThis 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 890Printed on acid-free paper 3. To the memory of my grandmother,Eda Crisafulli. 4. ContentsPreface xi1 The Dirichlet Problem in the Complex Plane 11.1 A Little Notation11.2 The Dirichlet Problem21.3 Lipschitz Spaces 71.4 Boundary Regularity for the Dirichlet Problem for theLaplacian on the Unit Disc101.5 Regularity of the Dirichlet Problem on a SmoothlyBounded Domain and Conformal Mapping202 Review of Fourier Analysis262.1 The Fourier Transform 262.2 Schwartz Distributions362.3 Convolution and Friedrichs Mollifiers 432.4 The Paley-Wiener Theorem483 PseudodifferentialOperators 523.1 Introduction to Pseudodifferential Operators523.2 A Formal Treatment of Pseudodifferential Operators563.3 The Calculus of Pseudodifferential Operators654 Elliptic Operators764.1 Some Fundamental Properties of Partial DifferentialOperators 76 5. viii4.2Regularity for Elliptic Operators 814.3Change of Coordinates 874.4Restriction Theorems for Sobolev Spaces 895Elliptic Boundary Value Problems 955.1The Constant Coefficient Case955.2Well-Posedness 995.3Remarks on the Solution of the Boundary Value Problem in the Constant Coefficient Case 1085.4Solution of the Boundary Value Problem in the Variable Coefficient Case 1095.5Solution of the Boundary Value Problem Using Pseudodifferential Operators 1155.6Remarks on the Dirichlet Problem on an Arbitrary Domain, and a Return to Conformal Mapping1235.7A Coda on the Neumann Problem1266A Degenerate Elliptic Boundary Value Problem 1286.1Introductory Remarks 1286.2The Bergman Kernel 1316.3The Szego and Poisson-Szego Kernels1386.4The Bergman Metric 1436.5The Dirichlet Problem for the Invariant Laplacian on the Ball1486.6Spherical Harmonics1546.7Advanced Topics in the Theory of Spherical Hannonics: the Zonal Harmonics1606.8Spherical Harmonics in the Complex Domain and Applications 1727The a-Neumann Problem1847.1Introduction to Hermitian Analysis 1857.2a The Formalism of the Problem 1897.3Formulation of the a-Neumann Problem 1967.4The Main Estimate2017.5Special Boundary Charts, Finite Differences, and Other Technical Matters2087.6First Steps in the Proof of the Main Estimate2187.7Estimates in the Sobolev -1/2 Norm 2247.8Conclusion of the Proof of the Main Estimate 2347.9The Solution of the a-Neumann Problem242 6. ix8 Applications of the a-Neumann Problem2528.1 An Application to the Bergman Projection 2528.2 Smoothness to the Boundary of Biholomorphic Mappings 2568.3 Other Applications of [) Techniques2639 The Local Solvability Issue and a Look Back2699.1 Some Remarks about Local Solvabilitiy2699.2 The Szego Projection and Local Solvability 2709.3 The Hodge Theory for the Tangential Cauchy-RiemannComplex2749.4 Commutators, Integrability, and Local Solvability277Table of Notation283Bibliography 287Index295 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. xiiPrefaceto 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. 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 byz == 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 operatora2a2~ = ox2 + oy2 . When the Euclidean plane is studied as a real analytic object, it is convenientto study differential equations using the partial differential operatorsa aax 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 axayCheck 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. 2The 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 satisfiesaf / 0 Z == 0on a planar open set may depend on z but cannot depend on z.Observe thato- z == 1 o -z==Ooz oz oo- z ==0oz 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 U1such 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. 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 Uu(z) == l (z) if z E S n Uau (z) == lav 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. 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,znif n>0 un(z) == { Z-n1 fn - 0. 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 )/ = 11 271: l(e i(II-t)Pr(t) dt - l(eill)l.(1.2.3.1)Observe, using the sum from which we obtained the Poisson kernel, thatf27rf27r 1 IPr(t)1 dt = 1Pr(t) dt= 1. 00Thus we may rewrite (1.2.3.1) as rio271: [I (e i(lI-t) - 1 (e ill )] Pr (t) dt =ir1 tl Q then Aexl ~ A ex . Also prove that theWeierstrass nowhere-differentiable function 00 F(O) ==L 2- ej i2iej=Ois in Al (0, 27r) but not in LiPl (0, 21r). Construct an analogous example, foreach positive integer k, of a function in A kLip 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)? ILet 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 COC 1Examples 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 (~::) IIlal::;k Loo(U)then elementary arguments show that C k + 1 (U) is contained in, but is nowheredense in, Ck(U). 17. Lipschitz Spaces 9It 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 oforder not exceeding k extend continuously to (j.II. We say that a function f is in C k (0) if there is an open neighborhoodW 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 Eauand each multiindex Q such that IQ I ::; k the limitao: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. 10 The Dirichlet Problem in the Complex Plane1.4 Boundary Regularity for the Dirichlet Problem for the Laplacianon 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 - 7ry 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).PROOFSincehIPy(x - t)1 dt =h Py(x - t) dt = 1,it follows that u is bounded by"f" 00 . 19. Boundary Regularity11B .AX+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 thatlu(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 satisfies1:;2U (X,y)1 ~ Cy 2a-for (x, y) E U. Fact 2: The function u satisfiesfor (x, y) E U. 20. 12The 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 ddt (u()(t))) dtII (IHllaul I :S Joay -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 that2a2u I == I a u IIay2 ax2Then2a2u I == I a u Iay2 ax2 i:I= I ::2 Py(x - t)f(t) dtl= Ii: ::2Py(x - t)f(t) dtl= Ii: ::2Py(x - t)f(t) dtl(1.4.1.1)Now, from the Fundamental Theorem of Calculus, we know that 21. Boundary Regularity13As a result, line (1.4.1.1) equalsIi:::2 Py (X - t) [J(t) - f(x)] dtl < Ci: ::2I Py (X - t)llx - til> dtcjOO12Y (3(x - tf - ~2) IIX _ tl a dt - 00 [ (x - t) 2 + y2]2C JOO Iy(3t - y2) Iit ia dt -00 (t 2 + y2)3 OO 1C a-2 yJ -00(t 2 + 1)2-a/2 dtC ya-2 .This completes the proof of Fact 1.PROOF OF FACT 2 First notice that, for any x E IR,:; ~17r -0000 [(x - t)2 + y2] 2I(x - tf + y - 22y21If(t)1 dtl y=2 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 + 2EE < - + - ==E.22This proves the result.IPROPOSITION 2.1.5Let fELl (I~N). Thenj 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(~).PROOFWe calculate that J(~) =Jj(t)eite dt = J f( _t)eite dt = Jf(t)e-ite dt = 1.IPROPOSITION 2.1.7If p is a rotation of}RN then we define pf (x) == f (p( x)). ThenPi == p(j). 37. The Fourier Transform29PROOF 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)) Jf(s)e i f,.p-(s) ds =Jf(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. IIf I, g are L 1 functions, then we define their convolution to be the functionf * g(x) =Jf(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. 30Review of Fourier AnalysisPROPOSITION 2.1.10If 1,9 E 1, thenr;-9(~) == j(~) . g(~).PROOF We calculate thatf7g(0 = J * g)(tk~t = JJ(fdt 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 , thenJj(~)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 thatf(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 *fand then pass to the limit asE ~ 0+.LEMMA 2.1.12We have 39. The Fourier Transform31PROOF It is enough to do the case N== 1. Set I == J~oo e- t2 dt. Then I "I= I: I:e- s2dse- t2dt =IIIR2e-!(s,t)!2 dsdt {21rroo_r 2 = Jo Joerdrd()=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].INow 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=le(f,/2+ix)2 e-e/ 4dx=e-e /4l e(f,/2+ix)2 dx== e-~2 /4 { e(~/2+ix/2)2 ~ dxJIR 2==~e-~2 /41 e(Z/2)2 dz. (2.1.12.1)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. SinceIr== 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. 32 Review of Fourier Analysis O-iN-----~ EfFIGURE 2.1Thus lim 1== - lim1 .(2.1.12.2) N~oo JrNN~oo Jf N Now we combine (2.1.12.1) and (2.1.12.2) to see thatWe conclude that, in jRI ,and in }RN we have 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 == GEELI. 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. 34Review 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, thenf(O) = (27r)-N Jj(~) d~.(2.1.14.1 ) 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 (2.1.14.1) 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 andj - g == 0 E 1, this is immediate from eitherthe theorem or (2.1.13). ISince the Fourier transform is univalent, it is natural to ask whether it issurjective. We havePROPOSITION 2.1.17The operatoris not onto. 43. The Fourier Transform35PROOF 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)NIf(xW dx.PROOF Define g( x) == I * 1E c~ (IRN ). Then ,,~ ,,~ ,,~ .... 29 == I . I == I . I == I . f == III . (2.1.18.1)Now g(O) = f * 1(0) = Jf( -t)!( -t) dt= Jf(t)/(t) dt = J2If(t)1 dt.By Fourier inversion and formula (2.1.18.1) we may now conclude thatJ 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. 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 concludethatF : 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. Defineif x ~ 0if x < o. 45. Schwartz Distributions 37Then one checks, using lHopitals Rule, that A E Coo (IR). Seth(x) == A( -x - 1) . A(X + 1)E C~(I~).I:Moreover, if we defineg(x)= h(t) dt,then the functionf(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 setsNE,f,m == {: L lal:SPa,(3() f >fdx E C. We see that this functional is continuous by noticing that If >(x)f(x) dxl ::; sup 1>1 Ilflll = C . Po,o(. 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 have5 ::1 ~ (O) satisfiesI (0)1 ::; sup I (x) I ==PO.I(). xElR 3. If f E LP (RN ),1 :::; p ::; 00, thenf induces a distribution: Tf : S 3 f--> JdxE 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 CI(x) I ::; 1 + Ixl N +1 (Po,o( )+ PN+I,O()) . Finally, IILp f ::; c[J(1 + I~IN+I ) P , dx] lip . [Po,o() + PN+I,O()] . 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 E5 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 greatinterest, especially to mathematical physicists, to have such an operation. 47. Schwartz Distributions 39Colombeau [CMB] has developed a substitute operation. We shall say nomore 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, byObserve that in case the distribution J-l is induced by integration against aC~ function f, then the definition is compatible with what integration byparts 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= -100f(x)(x) dx -[00 f(x)(x) dx= _roo x(x) dx + fOx(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: Letn~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. 40 Review of Fourier AnalysisLEMMA 2.2.3If f E S then j E S.PROOFThis 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 ofu makes good sense. Moreover, by 2.2.5below, lu()1 == lu()1 ~LPn,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 sumCLPn,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 by00 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. Schwartz Distributions41in 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 wehaveIL()I ~ C L L Pn,f3() (2.2.5.1)Inl~f 1f3I~mSKETCH OF PROOF If an inequality like (2.2.5.1) 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:Llal:SPa,f3() < E} .1.6I:SmSince L is continuous, the inverse image of an open set under L is open.ConsiderThere existE, f, m such thatThus Llal:SfPn,f3() 0 such thatIJ-l() I :S CL PK,n() Inl:::;Nfor every testing function .2.3 Convolution and Friedrichs MollifiersRecall that two integrable functionsfand 9 are convolved as follows:f *g = Jf(x - t)g(t) dt = Jg(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 iscompactly 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 2.3.1We 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). 52. 44 Review of Fourier AnalysisPROOFWe calculate that (a * g)(cP) = a( * cP) = aX[1 (x - t)cP(t) dt] .Here the superscript on Q denotes the variable in which Q is acting. This last Next we introduce the concept of Friedrichs mollifiers. Let E C~ besupported in the ball B (0, 1). For convenience we assume that 2: 0, althoughthis is not crucial to the theory. Assume that J ( x) dx == 1. Set E (x) == NE- (x/ E). The family {E} will be called a family of Friedrichs mollifiers in honor ofK. O. Friedrichs. The use of such families to approximate a given functionby smooth functions has become a pervasive technique in modem analysis. Infunctional analysis, such a family is sometimes called a weak approximation tothe identity (for reasons that we are about to see). Observe that J E (x) dx == 1for every E > O.LEMMA 2.3.3If f E LP(RN ), 1~p ~ 00,thenPROOF The case p == 00 is obvious, so we shall assume that 1 :::; p < 00.Then we may apply Jensens inequality, with the unit mass measure E(X) dx,to see that Ilf*cP,II1p= 111 Pf(x-t)cP,(t)dtI dx:=; 11 If(x - t)IPlcP,(t)1 dt dx= 11If(x - t)IP dx cP,(t) dt= 1IlflltpcP,(t) dt== 11111~pIREMARK The function IE == f * E is certainly Coo Gust differentiate underthe integral sign) but it is generally not compactly supported unless I is.I 53. Convolution and Friedrichs Mollijiers 45LEMMA 2.3.4For 1 :::; p < 00 we havePROOF We will use the following claim: For 1 :::; p < 00 we havewhere 7ft f (x)== ! (x - h). Assume the claim for the moment.NowII!< - !11~p 111 !(x - t)t) dt - 1!(x)t) dtl[ 111[!(x - t) - f(x)]t) dtl[ 111[Ttf(x) - f(x)]t) dtl[ < 1IITt! - fll~pt) dt (t=J-U)== 1II -! liP () TJi,E! LP J-l dJ-l.In the inequality here we have used Jensens inequality. Now the claim and theLebesgue dominated convergence theorem yield that lifE - fllLP -+ O. To prove the claim, first observe that if 1/J E GYe then I Th1/J - 1/J I sup -+ 0by uniform continuity. It follows that II Th 1/J - 1/J II Lp -+ O. Now if ! E LP isarbitrary and E > 0, then choose 1/J E Ce such that II! - 1/J II Lp < E/2. Thenlim supII Thf - !IILP :::; lim sup II Th(! -1/J)IILP + lim sup IITh1/J -1/JII :::; E. h---+Oh---+O h---+OSince E > 0 was arbitrary, the claim follows. ILEMMA 2.3.5If ! E C e thenfE -+ ! uniformly.PROOF Let 1] > 0 and choose E > 0 such that if Ix-yl < E then If(x)- f(y)1 0 such that (2.4.1.2)PROOF First let us assume that Q E V with support in B(O, A); we shall thenprove that U == & satisfies (2.4.1.1). Let h E C~ (I~N) satisfy h( t) == 1 whenIt I :::; 1/2 and h(t) == 0 when Itl2 1. For (E eN fixed and nonzero, we setOne checks that ( E V and ((x) == eix ( when Ixl :::; A+ 1/(21(1). Moreover,((x) == 0 when I(/(Ixl - A) 2 1.Now for any E V we know that 58. leVleW oJ ~"ourier Analysisfor some C, K. As a result,I&(()I == IQX(eix()1 ==IQx (((x))1:S CL sup IDJ3(1 1J3I~K B(O,A):S C(I(I + I)Ksuple ix .( I xEB(O,A)Next let us assume thatQis a C~ function supported in B(O, A). We shallprove (2.4.1.2). NowI&(()I= If a(x)e ix -E f(x - t)K(t) dtconverges for almost every x when f E LP and that T is a bounded operatorfrom LP to LP, 1 < p < 00. The second type of operator is called a Riesz potential. The Riesz potentialof order ex has kernel 0< ex < N,where CN,Q is a positive constant that will be of no interest here. The Rieszpotentials are sometimes called fractional integration operators because theFourier multiplier corresponding to k Qis c~ Q I~I-Q. If we think about the factthat multiplication on the Fourier transform side by ( -I ~ 1) ex > 0, corresponds2Q ,to applying a power of the Laplacian-that is, it corresponds to differentiation- 64. 56PseudodifferentiDl Operatorsthen it is reasonable that a Fourier multiplier 1~1J3 with (3 < 0 should correspondto integration of some order. Now the classical idea of creating a calculus is to consider the smallest algebragenerated by the singular integral operators and the Riesz potentials. Unfortu-nately, it is not the case that the composition of two singular integrals is asingular integral, nor is it the case that the composition of a singular integraland a fractional integral is (in any simple fashion) an operator of one of the com-ponent types. Thus, while this calculus could be used to solve some problems,it is rather clumsy.Here is a second, and rather old, attempt at a calculus of pseudodifferentialoperators:DEFINITION 3.1.1 A function p(x,~) is said to be a symbol of order m if pis Coo, has compact support in the x variable, and is homogeneous of degreem in ~ when ~ is large. That is, we assume that there is an M > 0 such thatif I~I > ]v! and A > 1 then It is possible to show that symbols so defined, and the corresponding operatorsform an algebra in a suitable sense. These may be used to study elliptic operatorseffectively. But the definition of symbol that we have just given is needlessly restrictive.For instance, the symbol of even a constant coefficient partial differential oper-ator is not generally homogeneous and we would have to deal with only the toporder terms. It was realized in the mid-1960s that homogeneity was superfluousto the intended applications. The correct point of view is to control the decayof derivatives of the symbol at infinity. In the next section we shall introducethe Kohn-Nirenberg approach to pseudodifferential operators.3.2 A Formal Treatment of Pseudodifferential OperatorsNow we give a careful treatment of an algebra of pseudodifferential operators.We begin with the definition of the symbol classes.DEFINITION 3.2.1 KOHN-NIRENBERG [KONlj Let m E}R. We say that asmoothfunction a(x,~) on }RN x}RN is a symbol of order m if there is a compactset K ~ }RN such that supp a ~ K x}RN and, for any pair of multiindices Q, (3, 65. A Formal Treatment of Pseudodifferential Operators57there is a constant Cn ,/3 such that(3.2.1.1)We write a E sm.As a simple example, if E C~ (IR N ), == 1 near the origin, defineThen a is a symbol of order m. We leave it as an exercise for the reader toverify condition (3.2.1.1). For our purposes, namely the local boundary regularity of the Dirichlet prob-lem, the Kohn-Nirenberg calculus will be sufficient. We shall study this calculusin detail. However, we should mention that there are several more general cal-culi that have become important. Perhaps the most commonly used calculus isthe Hormander calculus [HOR2]. Its symbols are defined as follows:DEFINITION 3.2.2 Let m E IR and 0 ~ p, 8 ~ 1. We say that a smoothfunction a( x,~) lies in the symbol class S;::c5 if The Kohn-Nirenberg symbols are special cases of the Hormander symbolswith p == 1 and 8 == 0 and with the added convenience of restricting the xsupport to be compact. Hormanders calculus is important for the study of thea-Neumann problem (treated in our Chapter 8). In that context symbols of classS:/2,1/2 arise naturally. Even more general classes of operators, which are spatially inhomogenousand nonisotropic in the phase variable ~, have been developed. Basic referencesare [BEF2], [BEA 1], and [HOR5]. Pseudodifferential operators with "roughsymbols" have been studied by Meyer [MEY] and others. The significance of the index m in the notation smis that it tells us how thecorresponding pseudodifferential operator acts on certain function spaces. Whileone may formulate results for C k spaces, Lipschitz spaces, and other classes offunctions, we find it most convenient at first to work with the Sobolev spaces.DEFINITION 3.2.3 If E V then we define the norm IIIIHS = 11lls == (/ 1(~)12 (1 + 1~12r d~) 1/2 .We let HSC~N) be the closure of V with respect toII I/s. 66. 58 Pseudodifferential Operators In the case that s is a nonnegative integer,for 1~llarge. Therefore E HSif and only if. (L I~IQ) IQI~S 2 E L .This last condition means that ~Q E L 2 for all multiindicesQ with lad ~ s.That is, (:x)" E L 2 Va such that lal~ s.Thus we havePROPOSITION 3.2.4If s is a nonnegative integer thenHere derivatives are interpreted in the sense of distributions. Notice in passing that if s>r then HS ~ Hr because The Sobolev spaces tum out to be easy to work with because they are modeledon L 2-indeed, each HS is canonically isomorphic as a Hilbert space to 2(exercise). But they are important because they can be related to more classicalspaces of smooth functions. That is the content of the Sobolev imbeddingtheorem:THEOREM 3.2.5 SOBOLEVLet s > N/2. If f E HS(IR N ), then f can be corrected on a set of measurezero to be continuous. More generally, if k E {O, 1,2, ...} and if f E HS, s > N/2 + k, then f canbe corrected on a set of measure zero to be C k .PROOF For the first part of the theorem, let fE H s. By definition, there existj E V such that lIj - jllHs -+ O. Then(3.2.5.1) 67. A Formal Treatment of Pseudodifferential Operators 59Our plan is to show that {j} is an equibounded, equicontinuous family offunctions. Then the Ascoli-Arzehi theorem [RUD1] will imply that there is asubsequence converging uniformly on compact sets to a (continuous) functiong. But (3.2.5.1) guarantees that a subsequence of this subsequence convergespointwise to the function f. So f == 9 almost everywhere and the requiredassertion follows.To see that {j} is equibounded, we calculate thatIcPj(x)1 = c11 e-ix.F,Jj(~) d~1 ~ c1IJj(~)1(1 + 1~12)s/2(1 + 1~12)-s/2 d~ ~ c 1IJj(~)12(1 + 1~12)s d~ . 1 + 1~12)-s d~)( )1/2 ((1 1/2 .Using polar coordinates, we may see easily that, for s > N /2,Thereforeand { j} is equibounded.To see that {j} is equicontinuous, we writeObserve that le-ix,~ - e-iy~ I ::; 2 and, by the mean value theorem,Then, for any 0 < E < 1, le-ix,~_ e-iY~1 == le-ix,~ _ e-iY~ll-t:le-ix,~ _ e-iY~It: ::; 21-lx _ ylt:I~It:.Therefore IcPj(x) - cPj(y)1~c1IJj(~)lIx ylI~I d~-~ C1x - 1IJj(~)1 + 1~12)/2 d~ yl (1~ C1x - yl IlcPj IIHB( J(1 + It;Y)-s+, d~ ) 1/2 . 68. 60 Pseudodifferential OperatorsIf we select 0 < E < 1 such that -s + E < -N/2, then we find that J(1 +1~12)-s+E d~ is finite. It follows that the sequence {j} is equicontinuous andwe are done. The second assertion may be derived from the first by a simple inductiveargument. We leave the details as an exercise.IREMARKS 1. If s == N /2, then the first part of the theorem is false (exercise). 2. The theorem may be interpreted as saying that HS ~ C k for s > k + N /2.In other words, the identity provides a continuous imbedding of H S into C k .A converse is also true. Namely, if HS ~ C k for some nonnegative integer kthen s > k + N /2.To see this, notice that the hypotheses Uj ---+ U in HS and Uj ---+ v in C kimply that U == v. Therefore the inclusion of HS into C k is a closed map. It istherefore continuous by the closed graph theorem. Thus there is a constant Csuch thatFor x E }RN fixed and a a multiindex withlal ::;k, the tempered distributione~ defined bye~ () = (:xcxcx )( x)is bounded in (C k )* with bound independent of x and a (but depending on k).Hence {e~} form a bounded set in (HS)* == H-s (this point is discussed indetail in Lemma 3.2.9 below). As a result, for lal ::; k we have thatis finite, independent of x and a. But this can only happen if 2(k - s)< -N,that is, if s > k + N /2. IExercise: Imitate the proof of the Sobolev theorem to prove Rellichs lemma:If s > r, then the inclusion map i : H S ---+ HT is a compact operator. 69. A Formal Treatment of Pseudodifferential Operators 61THEOREM 3.2.6Let p E 8 m and define the associated pseudodifferential operator P == Op(p) ==Tp byThencontinuously.REMARKSNotice that if m > 0, then we lose smoothness under P. Like-wise, if m < 0 then P is essentially a fractional integration operator and wegain smoothness. We say that the pseudodifferential operator Tp has order mprecisely when its symbol is of order m. Observe also that in the constant coefficient case (which is misleadingly sim-ple), we would have p(x,~) == p(~) and the proof of the theorem would be asfollows:IIP(;)II;-m = Jl(ffl))(~)12 + 1~12)s-m d~ (1 = JIp(~)(~W(1 + 1~12y-m d~ ~ J1(~)12(1 + 1~12)s d~ c == cllII; I To prove the theorem in full generality is more difficult. We shall break it upinto several lemmas.LEMMA 3.2.7For any complex numbers a, b we have1 + lalT+lbf ~ 1 + la - blPROOF We have1 + lal ~ 1 +la - bl + Ibl ~ 1+la - bl + Ibl + Iblla - bl == (1 + la - bl)(l + Ibl).I 70. 62Pseudodifferential OperatorsLEMMA 3.2.8If P E sm then, for any multiindex aand integer k > 0, we haveHere F x denotes the Fourier transform in the x variable.PROOFIf a is any multiindex andr is any multiindex such that I,I == k, then 1YIIFx 17(D~p(x,~)) 1== IFx (D1D~p(x,~)) (1])1 ::; IID~+Yp(x,~)IILI(x) ~ Ck,o: (1 + 1~I)m.As a result,This is what we wished to prove.ILEMMA 3.2.9We have thatPROOF Observe thatBut then HS and H-s are clearly dual to each other by way of the pairing(I, g) = Jj(~)g(~) dfI The upshot of the last lemma is that, in order to estimate the H S norm of afunction (or Schwartz distribution) cP, it is enough to prove an inequality of theformfor every 1/J E V. 71. A Formal Treatment of Pseudodifferential Operators 63PROOF OF THEOREM 3.2.6Fix E V. Let p E Smand let P = Op(p). Then PcjJ(x) = 1 O(~) d~.e-ix.p(x,DefineSX(A,~) = 1eiX)o.p(x,~)dx.This function is well defined since p is compactly supported in x. Then P(1]) = 11 ~)(~) d~eiTJx e-ix.p(x, dx= 1 ~)(~)eix(TJ-O d~ 1 p(x,dx= 1Sx(1]-~,~)(~)dfWe want to estimate IIPIs-m. By the remarks following Lemma 3.2.9, it isenough to show that, for l/J E V,We have 11 PcjJ(x)ijj(x) dxl = 11 P(~)~(~) d~1=11 (1 Sx(~- 1], 1])(1]) d1]) ~(~) d~1= 1 Sx(~1 - 1],1])(1+ 11]I)-s(1 + Iw s- m x ~ ( ~) (1 + 1 1) m - s (17) (1 + ~ 1171) s d 17 d~ .DefineK(~,1]) = ISx(~ -1],1])(1 + 11]I)-s(1 + Iws-mlWe claim thatJIK(~, d~ 1])1 :::; CandJIK(~,1])1 d1]:::; C. 72. 64Pseudodifferential Operators Assume the claim for the moment. Then x ( / 1(17)1 2 (1 + 1171 2 )S d17 ) 1/2 == CII7/JIIHm-s . 1IIIHsThat is the desired estimate. It remains to prove the claim.By Lemma 3.2.8 we know thatBut now, by Lemma 3.2.7, we have IK(~,17)1 == ISx(~ -17,17)(1 + 117I)-s(1 + Iws-ml::; Ck(l + 11JI)m(l + I~ -1JI)-k(l + 11J1)-S(l + 1~l)s-m 1 + 11J1) m-s-k= Ck ( 1 + I~I . (1 + I~ - 171)::; Ck(l + I~ _1Jl)m-s(l + I~ - 1J1)-k.We may specify k as we please, so we choose it so large that m-s-k ~ -N-1.Then the claim is obvious and the theorem is proved. I 73. The Calculus of Pseudodifferential Operators653.3 The Calculus of Pseudodifferential OperatorsThe three central facts about our pseudodifferential operators are these: 1. If p E sm then Tp: HS -+ Hs-m. 2. If p E sm then (Tp )* is "essentially" Tp . In particular, the symbol of(Tp )* lies in sm. 3. If p E sm, q E sn, then T p 0 Tq is "essentially" Tpq In particular, thesymbol of T p 0 T q lies in sm+n.We have already proved (1); in this section we shall give precise formulationsto (2) and (3) and we shall prove them. We begin with (2), and for motivation consider a simple example. Let A ==a(x)(ajaXl). Let us calculate A*. If , ljJ E V then (A * , ljJ) 2 == (, AljJ) 2 = J (a(x):~(x)(X)) dx = - JO~ 1 (a( x) ( X ) ) if;(x) dx = J(aaft) --a(x) OXI - OXI (X) (x) Ij;(x) dx.ThenA* == -a ( x ) - a - -aft (x ) - aXlaXl = Op ( -a(x)( -i6) - OXI (X) )aft = Op (i 6 a(x) -::1 (X)) .Thus we see in this example that the "principal part" of the adjoint operator(that is, the term with the highest degree monomial in ~ of the symbol of A *)is i~ 1ft (x), and this is just the conjugate of the symbol of A. In general it turns out that the symbol of A * for a general pseudodifferentialoperator A is given by the asymptotic expansiona)ex- 1 L DC; (of, a(A) a! . ex 74. 66 Pseudodifferential OperatorsHere D~ == (ia j ax)Q. We shall learn more about asymptotic expansions later.The basic idea of an asymptotic expansion is that, in a given application, theasymptotic expansion may be written in more precise form asOne selects k so large that the error term k is negligible. If we apply this asymptotic expansion to the operator a(X)ajaxl that wasjust considered, it yields that o-(A*) =i~la(x) - ~a(x), UXIwhich is just what we calculated by hand. Now let us look at an example to motivate how compositions of pseudodif-ferential operators will behave. Let the dimension N be 1 and letd and B == b(x) dx .Then a(A) == a(x)( -i~)and a(B)== b(x)( -i~). Moreover, if E D then(A 0 B)() = (a(x) d~) (b(X) ~~)Thus we see that 0-( A 0 B) = a(x) ~~ (x) (-i~)+ a(x )b(x)( -i~f.Notice that the principal part of the symbol of A0 B isa(x)b(x)( -i~)2 == a(A) . a(B). In general, the Kohn-Nirenberg formula says (in }RN) that o-(A 0 B) = L Q od (a)Q (o-(A)) . D~(o-(B)).1 fJ~(3.3.1)Recall that the commutator, or bracket, of two operators is [A,B]== AB - BA. 75. The Calculus of Pseudodifferential Operators67Here juxtaposition of operators denotes composition. A corollary of the Kohn-Nirenberg formula is that a([A, B]) = L (8/8~Y>a(A)D~a(B) -, (8/80aa(B)D~a(A) Q. Inl>O(notice here that the Q == 0 term cancels out) so that a([A, B]) has order strictlyless than (order(A) + order(B)). This phenomenon is illustrated concretely inJRl by the operators A == a(x)d/dx, B == b(x)d/dx. One calculates that dbda) d AB - BA == ( a(x) dx (x) - b(x) dx (x) dxwhich has order one instead of two.Our final key result in the development of pseudodifferential operators is theasymptotic expansion for a symbol. We shall first have to digress a bit on thesubject of asymptotic expansions.Let f be a Coo function defined in a neighborhood of 0 in R Then 00 1 dn ff(x) r-..J "-(0) x n .-(3.3.2) L..-i n! dx noWe are certainly not asserting that the Taylor expansion of an arbitrary Coo func-tion converges back to the function, or even that it converges at all (genericallyjust the opposite is true). This formal expression (3.3.2) means instead the following: Given an N > 0there exists an M > 0 such that whenever m > M and x is small then thepartial sum Sm satisfiesNow we present a notion of asymptotic expansion that is related to this one,but is specially adapted to the theory of pseudodifferential operators:DEFINITION 3.3.3Let {aj} be symbols in UmS m . We say that another symbola satisfiesif for every L E jR+ there is an M E Z+ such that M a - LajE S-L. j=l 76. 68Pseudodifferential OperatorsDEFINITION 3.3.4Let K CC }RN be a fixed compact set. Let WK be theset of symbols with x-support in K. If pEW K, then we will think of thecorresponding pseudodifferential operator P asP : CC:(K)~ C~(K).(This makes sense because P(x)== J e-ix,~p(x, ~)(~) d~.) Now our main result isTHEOREM 3.3.5Fix a compact set K and pick p E 8 m n WK. Let P == Op(p). Then P* hassymbol in 8 m n WK given by We will prove this theorem in stages. There is a technical difficulty that arisesalmost immediately: Recall that if an operator T is given by integration againsta kernel K (x, y), then the roles of x and yare essentially symmetric. If weattempt to calculate the adjoint of T by formal reasoning, there is no difficultyin seeing that T* is given by integration against the kernel K (y, x). However,at the symbol level matters are different. Namely, in our symbols p(x, ~), therole of x and ~ is not symmetric. If we attempt to calculate the symbol of Op(p)by a formal calculation, then this lack of symmetry serves as an obstruction. It was Hormander who determined a device for dealing with the problem justdescribed. We shall now describe his method. We introduce a new class ofsymbols r(x,~, y). Such a smooth function on }RN x }RN X }RN is said to be inthe symbol class Tm if there is a compact set K such thatsupp r(x,~, y) ~ K xandsupp r(x,~,y) ~ K yand, for any multiindices 0:, (3, "f, there is a constant Cn ,f3,-y such thatThe corresponding operator R is defined by R(x)= JJei(y-xHr(x,~,y)(y) dyd~. (3.3.6) 77. The Calculus of Pseudodifferential Operators 69Notice that the integral is not absolutely convergent and must therefore be in-terpreted as an iterated integral.PROPOSITION 3.3.7Let r E T m have x- and y-supports contained in a compact set K. Thenthe operator R defined as in (3.3.6) defines a pseudodifferential operator ofKohn-Nirenberg type with symbol pEW K having an asymptotic expansionp(x,~) " L ~!alD~r(x,~,y)ly=xnPROOFWe calculate that JeiY~r(x, ~,y )>(y) dy = (r(x, ~, .)>(-))= (i3 (x, y, .) * J(-)) (~).Here f3 indicates that we have taken the Fourier transform of r in the thirdvariable. By the definition of R we haveR>(x) =JJei(-x+Y)~r(x,~,y)>(y)dyd~J = e-ix.~ [i3(x,~,.) * J(-)] (~) d~ = JJi3(x,~, ~ - TJ)J(TJ) dTJe-ix,~ d~ = JJi3(x,~, ~ - TJ)e-ix,(~-f}) d~J(TJ)e-ix.f} dT] == J p(x, y)J(TJ)e-ix.f} dTJHereJi3(x,~, ~ TJ)e-ix(~-f}) ~p(x, TJ) == - =Je-ix~i3(x, ~ + TJ,~) d~.Now if we expand the function f3 (x, TJ + .,~) in a Taylor expansion in powersof ~, it is immediate that p has the claimed asymptotic expansion. In particular,one sees that p E sm. In detail, we have i3(x, 1] +~,~) = L a;i3(x, TJ,~) ~~ + R.Inl(y)dyp(x,~)d~"j;(x)dx. 1Let us suppose for the moment that p is compactly supported in~. With thisextra hypothesis the integral is absolutely convergent and we may write (4), P*7/J) = 14>(Y) [11ei(x-y) p(x, ~)7/J(x) d~ dx ] dy. (3.3.5.1)Thus we have P*7/J(y) = 1 ei(x-YHp(x,~)7/J(x)d~dx.1 Now let p E C~ be a real-valued function such that p == 1 on K. Setr(x,~,y) == p(x) p(y,~).Then P*7/J(y) =11 ~)p(Y)7/J(x) d~ ei(x-YH p(x,dx=1ei(x-YHr(y,~,x)7/J(x)d~dx== R1/J(y),where we define R by means of the multiple symbol r. (Note that the roles ofx and y here have unfortunately been reversed.) 79. The Calculus of Pseudodifferential Operators71By Proposition 3.3.7, P* is then a classical pseudodifferential operator withsymbol p* whose asymptotic expansion isp*(x,~) " ~ ~! of D; [p(x)p(y, ~)] Iy=x" L J,af D;p(x, ~). Q. aWe have used here the fact that p == 1 on K. The theorem is thus proved withthe extra hypothesis of compact support of the symbol in ~. To remove the extra hypothesis, let E ergo satisfy == 1 if I~ I ~ 1 and == 0 if I~I 2: 2. LetObserve that Pj ~ P in the e k topology on compact sets for any k. Also, bythe special case of the theorem already proved,The proof is completed now by letting j ~00.ITHEOREM 3.3.8 KOHN-NIRENBERGLet P E l1 K n sm, q E l1 K n sn. Let P, Q denote the pseudodifferentialoperators associated with p, q respectively. Then Po Q == Op(a) where 1. a E l1 K n sm+n; 2. a ~ La ~8rp(x, ~)D~q(x, ~).PROOFWe may shorten the proof by using the following trick: write Q =(Q*) * and recall that Q* is defined byQ*J(y) = / / ei(x-YHJ(x)q(x,~)dxd(,= (/ eix~J(x)q(x,~) dx ) ---- (y).Here we have used (3.3.5.1).ThenQJ(x) = (J eiY~J(y)q.(y,~) dy) ---- (x),(3.3.8.1 ) 80. 72 Pseudodifferential Operatorswhere q* is the symbol of Q*. (Note that q* is not if; however, we do knowthat if is the principal part of q*.) Then, using (3.3.8.1), we may calculate that (P 0 Q)((x) = J e-ixep(x, ~)(Q)(O d~ = JJe-ixep(x,Oeiyeq*(y,~)>(y)dyd~ = JJe-i(x-y)e [p(x,~)q*(y,~)] >(y)dyd~.Set q == q*. Define r(x,~,y) == p(x,~) . ij(y, ~).One verifies directly that r E Tn+m. We leave this as an exercise. Thus R,the associated operator, equals P 0 Q. By Proposition 3.3.7 there is a classicalsymbol a such that R == Op( a) anda(x,O "-J L ~! of D~r(x,~, y)ly=x QDeveloping this last line we obtain a(x,O "-J L ~!OfD~ (p(x,Oq(y,~))ly=xQ "-J L ~!Of [p(x,OD~q(y,O] Iy=xQ "-J L~or [p(x,OD~q(x,O]o.Q "-J L all! [or1p(x,o] a;! [otD~2D~lq(X,~)] QI,Q2QI """ oI! 8 ~ p ( x, ~ ) D x ~ 1 ~ 0 1 ! 8Q2 D x q x, ~ ) ] 2 ~ QI ["""Q2 _( f"V QI Q2 "-J L~o{p(X,~)D~lq(X,~).0. QI 81. The Calculus of Pseudodifferential Operators 73Here we have used the fact that the expression inside the brackets is just theasymptotic expansion for the symbol of (Q*) *. That completes the proof. I The next proposition is a useful device for building pseudodifferential oper-ators. Before we can state it we need a piece of terminology: we say that twopseudodifferential operators P and Q are equal up to a smoothing operator ifP - Q E Sk for all k < O. In this circumstance we write P rv Q.PROPOSITION 3.3.9Let Pj, j == 0, 1,2, ..., be symbols of order mj, mj ~ -00. Then there is asymbol P Esmo, unique modulo smoothing operators, such thatPROOF Let l/J : ~n ~ [0, 1] be a Coo function such that l/J == 0 when Ixl ::; 1and l/J == 1 when Ixl 2: 2. Let 1 < t l < t2 < ... be a sequence of positivenumbers that increases to infinity. We will specify these numbers later. Define00 p(x,~) == L l/J(~/tj )Pj (x, ~). j=ONote that for every fixed x, ~ the sum is finite, for l/J (~/ tj) == 0 as soon astj> I~I. Thus P is a well-defined Coo function.Our goal is to choose the tjS so that P has the correct asymptotic expansion.We claim that there exist {t j} such that Assume the claim for the moment. Then for any multiindicesQ, {3 we have 00ID~Drp(x,~)1 ::; L ID~Dr (l/J(~/tj)Pj(x,~))1 j=O 00 ::; L2- (1 + 1~l)mJ-IQI j j=O 82. 74 Pseudodifferential OperatorsIt follows that p Esmo Now we want to show that p has the right asymptotic expansion. Let < k E Z be fixed. We will show thatk-lp- LPj j=Olives in smk. We havek-l- L (l-7/J(~/tj))Pj(x,~)j=O == q(x,~)+ s(x, ~).It follows directly from our construction that q(x,~) E k sm Since [l-ljJ(~/tj)]has compact support in B(O, 2t l ) for every j, it follows that s(x,~) E S-oo.Thenk-lP - LPj E sm kj=Oas we asserted. We wish to see that P is unique modulo smoothing terms. Suppose thatqE smo and q L~OPj. Then f"V P- q== (p - LPj) - (q - LPj)J fj-l and 1~12: fj implies Cj(l + 1~l)mJ-mJ-l :::; 2- j .Then it follows thatwhich establishes the claim and finishes the proof of the proposition. I 84. 4Elliptic Operators4.1 Some Fundamental Properties of Partial Differential OperatorsWe begin this chapter by discussing some general properties that it is desirablefor a partial differential operator to have. We will consider why these proper-ties are desirable and illustrate with examples. We follow this discussion byintroducing an important, and easily recognizable, class of partial differentialoperators that enjoy these desirable properties: the elliptic operators. Our first topic of discussion is locality and pseudolocality. Let T : C~ ~Coo. We say that T is local if whenever a testing function vanishes on anopen set U then T also vanishes on U. The most important examples of localoperators are differential operators. In fact, the converse is true as well:THEOREM 4.1.1 PEETREIfT: Cr: ~ COOis a linear operator that is local then T is a R.artial differential operator.PROOF See [HEL].I The calculation of a derivative at a point involves only the values of thefunction at points nearby. Thus the notion of locality is well suited to differ-entiation. In particular, it means that if T is local and == l/J on an open setthen T == Tl/J on that open set. For the purposes of studying regularity fordifferential operators, literal equality is too restrictive and not actually necessary.Therefore we make the following definitions:DEFINITION 4.1.2 Let ex E V and U ~ IR N an open set. If there is a Coo 85. Properties of Differential Operators77function f on U such that for all E V that are supported in U we have n( =1j(x)>(x) dx,then we say that 0 is Coo on u. The singular support of a distribution 0 is defined to be the complement ofthe union of all the open sets on which 0 is Coo.DEFINITION 4.1.3A linear operator T : V ~ Vis said to be pseudolocalif whenever U is an open set and 0 E Vis Coo on U then To is Coo on U.Now we haveTHEOREM 4.1.4If T is a pseudodifferential operator, then T is pseudoloeal.The theorem may be restated as "the singular support of Tu is contained in thesingular support of u for every distribution u." A sort of converse to this theorem was proved by R. Beals in [BEA2-4]. Thatis, in some sense the only pseudolocal operators are pseudodifferential. We shallnot treat that result in detail here. The proof of the theorem will proceed in stages. First, we need to define howa pseudodifferential operator operates on a distribution. Let P be a pseudodif-ferential operator and let u E V have compact support. We want Pu to be adistribution. For any testing function , we set (Pu, ) == (u, t P).Here t P is the transpose of P which we define byfor ,l/J E V. To illustrate the definition, suppose that u is given by integration against anLl function f. Suppose also that the pseudodifferential operator P is given byintegration against the kernel K(x, x - y) with K(, y) E L 1 and K(x, .) ELI.Then P has symbol K (x, ~). We see that 2 (Pu)( =1[1 K(x,x-y)j(Y)dY] >(x)dx =J~ K(x, x - y)>(x) dX] j(y) dy =Jt P>(y)f(y) dy. 86. 78Elliptic OperatorsThis calculation is consistent with the last definition.PROOF OF THEOREM 4.1.4Let U be an open set on which the distribution uis Coo. Fix x E U. Let E C~ (U) satisfy == 1 in a neighborhood of x.Finally, let ljJ E C~ (U) satisfy ljJ == 1 on the support of . Then we haveljJu E V, hence PljJu E C~. (Note here that we are using implicitly thefact that if a(P) E sm, then P maps HS to Hs-m, hence, by the Sobolevimbedding theorem, P maps C~ to Coo.) We wish to find the symbol of PljJ. This is where the calculus of pseudod-ifferential operators will come in handy. Let us write our operator aswhere the symbol M denotes a multiplication operator. Of course a(M ==(x) and a(M1jJ) == ljJ(x). Hence a(T) == (x)a(PoM1jJ)" J(x) (~~!8fPD~1/J ) .In the last equality we have used the Kohn-Nirenberg formula. But, on thesupport of , any derivative of ljJ of order at least 1 vanishes. As a result,a(T) ~ (x)a(P)(x,~).Therefore a(T) - (x)a(P) E S-kVk ~ o.In other words,T - Met> 0 P : H S ~ H s +kfor all k 2: 0 and every s. By Sobolevs theorem,But V == UsHs so our distribution u lies in some HS.We conclude thatSince PljJu is Coo, we may conclude that Pu E Coo. But == 1 in aneighborhood of x. Therefore Pu is Coo near x. Since x was an arbitrary pointof U, we are done. IDEFINITION 4.1.5 A linear operator T : V ~ Vis called hypoelliptic ifsing supp u ~ sing supp Tu. 87. Properties of Differential Operators 79Notice that the containment defining hypoellipticity is just the opposite from thatdefining pseudolocality. A good, but simple, example of a hypoelliptic operatoron the real line is djdx. For ifd - u == f dxand f is smooth on an open set U, then u is also smooth on U. The reason,of course, is that we may recover u from f on this open set by integration.A more interesting example of a hypoelliptic operator is the Laplacian ~ on~N, N 2: 2. Thus the equation ~u == f entails u being smooth wherever fis. This is proved, in analogy with the one-variable case, by constructing aright inverse (or at least a parametrix) for the operator~. If the right inverseor parametrix is a pseudodifferential operator, then the last theorem will tell usthat ~ is hypoelliptic. We now introduce an important class of pseudodifferential operators whichare hypoelliptic and enjoy several other appealing regularity properties. Theseare the elliptic operators:DEFINITION 4.1.6 We say that a symbol p Esm is elliptic on an open setU ~ ~N if there exists a continuous function c(x) > 0 on U such thatfor ~ large. A partial differential operator or, more generally, a pseudodiffer-ential operator L is elliptic precisely when its symbol a( L) is elliptic.Example 1Let ~ be the Laplacian. Then a(~) == Lf=1(-i~j)2follows thatla(~)1 == 1~12 2: 1 1~12on all of space. Hence ~ is elliptic. 0Example 2Now let1 if i == jo if i =I j.Then 88. 80Elliptic OperatorsIf Eij, i, j 1, ... ,N are Coo functions having small CO nonns, then theoperatoris elliptic. 0Example 3Let (aij) be a positive definite N x N matrix of constants. Let b1 , , bN bescalars. Then the partial differential operator 2La j -.8x . + Lbj -88x . . . Z,J88iXz J J. Jis elliptic. 0 If P is a partial differential operator that is elliptic of order m (usually m ispositive, but it is not necessary to assume this), and if Q ==L an (x) 8 n is apartial differential operator with continuous coefficients of strictly lower order,then P + Q is still elliptic, no matter what Q is. To see this, let a(P) == p(x,~)with Ip(x,~)I2: c(x)I~lm for 1~llarge. Then la(P + Q) I 2: la(P) I- la( Q) I 2: c(x)I~lm - Lan(x)(-i~)nInl 0 a continuous function. Select arelatively compact open set L ~ supp p(.,~) so that there is a constant CO > 0with a(x) > Co on L. We also select Ko > 0 such that, for I~I 2: K o and x E L,it holds that Ip(x,~)1 2: col~lm. Letl/J be a C~ function with support in the(x variable) domain of a(P) such that l/J == 1 on L. Also choose a function E Coo such that (~) == 1 for I~I 2: K o + 2 and (~) == 0 for I~I :::; K o + 1.Then setandQo == Op( qo).Observe that == 0 on a neighborhood of the zeroes of p(x, ~); hence qo is Coo.Furthermore, qo has compact support in x. Finally, qo E s-m.To see this, firstnotice thatMoreover, la~j qo(X,~)1 = 1/J(X)lp(x,~)(ac/>/a~j)~l(:,~~~)(ap(x,o/a~j)I < C. [(1 + IW 1(a>/a~j)(~)1 + (1 + IW mm 1- ] - (1+ 1~1)2m (1 + 1~1)2mfor lei large. But this isHowever, B/ Bej is compactly supported, so it follows thatArguing in a similar fashion, one can show thatSince x derivatives are harmless, we conclude that qo E s-m. 91. Regularity for Elliptic Operators83Now consider Qo 0 P. By the Kohn-Nirenberg fonnula,That is,a(Qo0 P) == qo(x, ~)p(x,~) + 1-1 (x, ~).(4.2.3.1)Notice that qo(x, ~)p(x,~) == l/J(x )(~) and 1-1 (x,~) E 5- 1 Now set a(Qo 0P) == ljJ(x) + r-l(x,~), where r-l is defined by this equation and (4.2.3.1).The equation that defines r -1 shows that we may suppose that r -1 has compactsupport in the x variable. Observe also that r -1 E 5- 1 .Definewhere is a C~ function that is identically equal to 1 on the x-support of r -1.Consider Qo + Ql == Op(qO + ql). We calculate (Qo + Ql) 0 P. By theKohn-Nirenberg fonnula, a((Qo + Ql) 0 p) == a(Qo 0 P) + a(Ql 0 P)== ljJ(x) + ,(x)r -1 - (x)(~)r -1 + f -2.Notice that 1-2 E 5- 2 Since (x)(l - (~)) is compactly supported in bothx and~, it follows that ,(x)(l- (~))r-l is smoothing. Therefore we maywritewith r -2 E 5- 2 . Now suppose inductively that we have constructed qo, ... ,qk-l such that,setting Qj == Op(qj) for j :::: 0, 1, ... , k - 1, we havewith r -k E 5- k . DefineLet Q be the pseudodifferential operator having symbol q, where 92. 84Elliptic Operators(here we are using 3.3.9). Thena(Q a P) == 1/;(x) + s(x, ~),with s(x,~) E S-oo, Le., QaP-1/;/ is smoothing. We also will write QaP==1/;/ + 5. Thus Q is a left parametrix for P. A similar construction yields a right parametrix Q for P. We write P a Q ==1/;/ +5. We will now show that a( Q - Q) E S-oc on a slightly smaller open setW cc L. By adjusting notation, one can of course arrange (after the proof)for the equations Q a P == 1/;/ + 5, P a Q == 1/;/ + 5, and a(Q - Q) E 5- 00to all be valid on the original open set L. We now interpolate an open setW cc V cc L. Let p E C~(V) satisfy p == 1 on W. Let J1 E C~(L)satisfy J1 == 1 on V. In what follows we will use continually the fact that thecomposition of any pseudodifferential operator with a smoothing operator is stillsmoothing. The identity of our smoothing operators may change from line toline, but we will denote them all by 5 or 5. NowpQJ1 == pQ(P a Q - 5)J1== pQ a P(Q - 5)J1== p(/ + S) 0 (Q - 5)J1== p(Q - 5 + 5Q - S5)J1== pQJ1 + SjL.Thus Q - Q is smoothing on W when applied to functions in C~(V). Weconclude that Q - Q is smoothing and we are done. I Now we can present our basic interior regularity result for elliptic partialdifferential (in fact, even pseudodifferential) operators.THEOREM 4.2.4Let U ~ ~N be an open set. Let P be a pseudodifferential operator that iselliptic of order m > 0 on U. If I E HI~c and u is a solution of the equationPu == I , then u E H lac m .s+PROOF The hard work has already been done. Since the theorem is local, wecan suppose that u and I have compact support. (It is important to developsome intuition about this: the point is to see that we can consider u ratherthan u and pi rather I, where , p are C~ cutoff functions. This amounts tocommuting past P and p past the parametrix, noticing that this process givesrise to error tenns of lower order, and then thinking about how the error terms 93. Regularity for Elliptic Operators85would affect the parametrix.) Let l/J E C~ satisfyl/J == 1 on suppu. Let Q bea left parametrix for P:S == QoP - l/JIis smoothing. Thenu == l/Ju == (Q0 P - S)u == Qf - SuoSince f E H S and Q is of order -m, it follows that Qf E H s +m ; also, Su issmooth. Then Qf - Su E Hs+m, that is, u E Hs+m.I Now that we have our basic regularity result in place we will use it, and somefunctional analysis, to prove our basic existence result. We would be remissnot to begin by mentioning the paramount discovery of Hans Lewy in 1956[LEW2]: not all partial differential operators are locally solvable. In fact, weshall discuss Lewys example in detail in Chapter 9. Although we do not attempt to fonnulate the most general local solvabilityresult, we present a result that will suffice in our applications. The most impor-tant ideas connected with local solvability of partial differential operators appearin [NTR], [BEFI], and the references therein.THEOREM 4.2.5Let P be an elliptic partial differential operator on a domain o. If f E V(O)and Xo E 0, then there exists a u E D (0) such that Pu == f in a neighborhoodofxo.PROOFWe may assume, by the usual arguments, that f has compact support-that is, f E [. Then f E H S for some s. Thus we may take the partialdifferential operator P to have compact support in a neighborhood K of Xo.Let l/J E C~ (0) satisfy l/J == 1 on K. Let Q be a right parametrix for P suchthatPo Q - l/JI== S (4.2.5.1)and S is smoothing. By looking at the left side of this equation we see thata(S) has x-support also lying in suppl/J. ThusS: H S~ C~(suppl/J).Then S : HS ~ HS is compact by Rellichs lemma. Therefore the equation (S + I)u == f (4.2.5.2) 94. 00Elliptic Operatorscan be solved if fEN 1-, whereN == {gE HS : (S* + 1)g == O};here the inner product is with respect to the Hilbert space HS. Notice that, sinceS* is also smoothing (why?) and 9 == -S* 9 for 9 E N, we know that N is afinite-dimensional space of Coo functions. Let gl , ... , 9 M be a basis of N. We claim that there exists a neighborhood Uof Xo such that gl , ... , 9 M are linearly independent as functionals on C~ (0U).To see this, suppose that U does not exist. Let Un be neighborhoods of Xo suchthat Un . xo. For each n, let C l, ... ,CnM be constants (not all zero) such that nM G n == LCnjgj == 0j=1as functionals on C~(OUn). By linearity, we may suppose that each G n hasnonn 1 in N for all n (note in passing that G n -# 0 for every n). Since N is finitedimensional, there is a convergent subsequence G nm such that G nm ~ G E Nand G has nonn 1. However, viewed as a functional on C~(O), G has support{xo}. Since G is Coo, we have a contradiction. Thus we have proved the existence of U. We may assume that U cc K.Now we can find a Coo function h with support in 0U such that f + h isperpendicular to N in the HS(O) topology. By (4.2.5.2), there is a distributionu such that (1 + S)u == h + f == fon U.By (4.2.5.1), we see that P (Qu) == f on U. Therefore Qu solves the differentialequation. IExercise: The proofs of the last two theorems, suitably adapted, show thatthe local solution exhibits a gain in smoothness, in the Sobolev topology, oforder m. It is also the case that elliptic partial differential equations exhibit a gain oforder m in the Lipschitz topology. We shall present no details for this assertion.Given the machinery that we have developed, it is only necessary to check thatif p Esmthen Op(p) : A~c ~ A~~m for any Q > O. This material is discussedin [TAY].Exercise: Prove that the elliptic operator P in Theorem 4.2.5 is surjective in asuitable sense by showing that its adjoint is injective. 95. Change of Coordinates874.3 Change of CoordinatesIt is a straightforward calculation to see, as we have indicated earlier, that anelliptic partial differential operator remains elliptic under a smooth change ofcoordinates. In fact pseudodifferential operators behave rather nicely underchange of coordinates, and that makes them a powerful tool. They are vital, forinstance, in the proof of the Atiyah-Singer index theorem because of this invari-ance and because they can be smoothly defonned more readily than classicalpartial differential operators. We shall treat coordinate changes in the presentsection. Let U, {; be open sets in ~N and lal + 1/2. Then nauhas trace in Hs-la l-I/2(lRN - 1 ). 100. 92Elliptic Operators Now we present a converse to the theorem. Again set S == {(x, 0) E IR N }.THEOREM 4.4.3Assume that cPo, ... , cPk are defined on S, each cPj E Hs-j-l/2CIR N - 1 ) withs > 1/2 + k. Then there exists a function f E HS (IR N ) such that D~ f hastrace cPj on S, j == 0, ... ,k. Moreover, k IIfIlHs(~N) ~ Cs,k L IlcPjllks-J-I/2(~N-I).j=OPROOF Let h E C~ (IR), h ==1 in a neighborhood of the origin, 0 ~ h ~ 1.We defineandThis is the function f that we seek. For if m is any integer, 0~ m ~ k, then ==J.e -LX I .~ I m ~ _1 l k ~., j=O J.(m) . J ".J. ( _ l)Jx8:-~j [h(xN(l + 1e1 2)1/2)] I . ;(~) d~8x NXN=O =J e- ix ,( ;;,(e)de == cPm(X).This verifies our first assertion. 101. Restriction Theorems for Sobolev Spaces93Now we need to check that f has the right Sobolev norm. We have IIfllk2 = { Ij(~W(1 + 1~12)s d~ J~N = k.N 1 ((,.))~ (~N )11 + 1~12)s d~ (uiNow we use the fact that I I:~=o (j 1 :::; C . (I:~=o I(j 1), where the constant C2 2depends only on k. The result is that the last line is 2t~ Ck.N (j~)218~~ h ((1 + f;1 2)1/2 ) 11 --- 2 28 X 1+ 1e1 2 . lj(~)1 (1 + I~I ) d~ (4.4.3.1)Now d~ == d~ d~N. We make the change of variablesThen 1 + 1~12 == 1 + 1~12 + I~NI2r---+ 1 + 1~12 + I~NI2(1 + 1~12)== (1 + 11 2 )(1 + IN2). 102. 94Elliptic OperatorsThen (4.4.3.1) is k::; Cf; (j~)2 kN-l k Ih(j)(~N)12 (1 + 1;1 2)1+ 1(1 + leI 2)1/2xl;(~/) 1 [( 1 + 1 ) (1 + 1N 1) ] s d~ d~ N 2~/12 ~ 2 k= CL: ~ ((J.)j=O JRN-Ilj(OI2(1 + leI 2)S-j-I/2deX kIh(j)(~N)1 (1 + I~NI2)s d~N2k~ Cs,k L: lIjll~s-j-I/2(~N-I). j=oThat completes the proof of our trace theorem. IREMARK The extension iii was constructed by a scheme based on ideas thatgo back at least to A. P. Calderon-see [STSI] and references therein. I 103. 5Elliptic Boundary Value Problems5.1 The Constant Coefficient CaseWe begin our study of boundary value problems by consideringn == }R~ , an =={(x , 0) : x E }RN -1 }. We will study the problem P(D)u ==f on }RN +{ Bj(D)u == gj on a}R~,j == 1, ... , k.Here P will be an elliptic operator. At first both P and the B j will have constantcoefficients. Our aim is to determine what conditions on the operators P, B jwill make this a well-posed and solvable boundary value problem. We shallassume that P is homogeneous of degree m > 0 and thatExamples: Leta aP(D) == 6. ==-a + -a 2Xl X 2.2Example 1First consider n == }R~. Let us discuss the boundary value problem ~u== 0ulan == 0 { g:: Ian ==1.This system has no solution because the second boundary condition is incon-sistent with the first. The issue here turns out to be one of transversality ofboundary conditions involving derivatives. See the next example. 0 104. 96 Elliptic Boundary Value ProblemsExample 2We repair the first example by making the second boundary condition transverse:the problem~u== 0{ ulan == 0g:2lan == 1has the unique solutionU(Xl X2) == X2.0Example 3Let n == IR~. Consider the boundary value problem~u== 0ulan == 91{ g:! Ian == 92In fact, take 91 (XI, X2) == Xl, 92(XI, X2) =1.Notice that any function of theform V(XI X2) == Xl + CX2 satisfies~v== 0{vlan == XIg:! Ian == 1.Hence the problem is sensible, but it has infinitely many solutions. 0Our goal is to be able to recognize problems that have one, and only one,solution.Necessary Conditions on the Operators B jFirst, the degree of each B j must be smaller than the degree of P. The reasonis that P is elliptic. According to elliptic regularity theory, all derivatives ofU of order m and above are controlled by f (the forcing term in the partialdifferential equation). Thus these derivatives are not free to be specified. Now we develop the Lopatinski condition. We shall assume for simplic-ity (and in the end see that this entails no loss of generality) that our partialdifferential operator P is homogeneous of degree m. Thus it has the form 105. The Constant Coefficient Case 97For each fixed ~ == (~1, ... ,~N -1), we considerP(~, D N). We will solveas an ordinary differential equation. Thus our problem has the fonnm Lap(t) (d dx) v = o. =0NThe solution ,v of such an equation will have the formr(~/) VJ-I V(XN)==L L Cj(~)(XN )eiXN~j(~/),j=1 =0where Al (~), A2 (~), ... , Ar (~) are the roots of P (~, . ) == 0 with multiplicitiesVI (~), V2(~), .. ,Vr(~). We restrict attention to those Aj with positive real part. Since these rootsalone will be enough to enable us to carry out our program, this choice isjustified in the end. However, an a priori theoretical justification for restrictingattention to these AS may be found in [HOR1]. After renumbering, let us say that we have retained the rootsAl (~ ), ... , Aro ( ~) , TO :::; T.Thenro v J - l V == L L Cj(~)(XN )eiXN~j(~/).j=1 =0 Fact: A fundamental observation for us will be that the number TO of roots(counting multiplicities) with positive imaginary part is independent of ~ E]RN -1{O}. This is proved by way of the following two observations:(i) Each Aj (~), j== 1, ... , TO, depends continuously on ~. (ii) There are no Aj with zero imaginary part (except possibly A == 0).We leave it to the reader to supply the details verifying that these two obser-vations imply that the number of roots is independent of ~ (use the ellipticityfor (ii. Now we summarize the situation: our problem is to solve, for fixed~, the systemP(~/, DN)v == 0 on 1R~Bj(~, DN)v== gj() on 8IR~ ,j = 1, ... , k. (5.1.1)From now on we take k to equalVo + ... + V ro 106. 98Elliptic Boundary Value ProblemsFor fixed ~ we will formulate a condition that guarantees that our system hasa unique solution: The Lopatinski Condition: For each fixed ~ =1= 0 and for each set of functions {gj}, the system (5.1.1) has a unique solution.Let us clarify what we are about to show: If, for each fixed ~/, the ordinarydifferential equation with boundary conditions (5.1.1) has a unique solution (nomatter what the data {gj}) then we will show that the full system described atthe beginning of the section has a unique solution. The condition amounts, aftersome calculation, to demanding the invertibility of a certain matrix. Let p(~) == Lj,k ajk~j~k with the matrix (ajk) positive definite. Then P(D) == ~ a j k - -aa a . ~ aXj XkObserve thatIn this example, for fixed ~/, the polynomial P(~/, TJ) is quadratic in TJ. Thepositive definiteness implies that it has just one root with positive imaginary part.So, if Lopatinskis condition is to be satisfied, we can have just one boundarycondition of degree 0 or 1. Let us consider two cases:(i) B is of degree zero (that is, B consists of multiplication by a function).Thus, by Lopatinski, we must be able to solveB(~/, D N)V == b(~/)V == g(~/)for every g. This is the same as being able to find, for every g(~), a functionc( ~/) such that b(~/)c(~/)eixN)q(() == g(~/)when XN == 0; in other words, we must be able to solve b(~/)C(~/)== g(~/). Wecan find such a c provided only that b(~/) does not vanish.(ii) B is of degree one. It is convenient to write B as B(D) ==a Lbo-. J aXj Jwhere D j == ia/axj. Assume for simplicity that the bj 8 are real. Accordingto the Lopatinksi condition, we must be able to solveon }RN -1 for any g. Recall that we have already ascertained that v == c(~/)eiXN)l((). 107. Well-Posedness99Because B has degree one we have a B(~/, D N ) == bli~l + b2i~2 + ... + bN-li~N-I + bNi aXNB(~/, D N)V== ib l ~l c(~/)eixN~1 (() + ib2~2c(~/)e~XN ~l (~/)+ ... + ibN_I~N_Ic(~/)eixN~I(() + ibNi)q (~)c(~)eiXN)q(() == g(~/).Setting x N == 0 givesSince the coefficients bj are real, the hypothesis that bN =1= 0 will guarantee thatwe can always solve this equation.5.2 Well-PosednessLet the system P(~/, DN)v ==f on R~ Bj(~/, DN)v == gj(~/) on aR~,j == 1, ... , k,(5.2.1)have the property that the operators P, B j have constant coefficients. The op-erator P is homogeneous of degree m. Assume that each operator B j is ho-mogeneous of order mj and that m > mj. The system is called well-posed ifConditions (A), (B), (C) below are met. (A) Regularity. The space of solutions of P(D)u == 0,XN >0 Bj(D)u == 0, XN == 0, 1 ~ j ~ k, in Hm (1R~) has finite dimension and there is a C>0such that IIvIlH" :::; c (IIP(D)VIIHO + ; I B j(D)vII H"-"J- / + IIV11HO)2 Iv E Hm(1R~). 108. .LUV Elliptic Boundary Value Problems (B) Existence. In the spacek C~(IR~) x IT C~(IRN-I)p=1 there is a subspace having finite codimension such that if (f, 91,92, ... , 9k) E , then the boundary value problem (5.2.1) has a solution u in Hm(IR~). (C) Let, be the operator of restriction to the hyperplane Then the set is closed in Now our theorem isTHEOREM 5.2.2The system (5.2.1) is well posed (that is, conditions (A), (B), and (C) are satisfied)if and only if the system satisfies Lopatinskis condition.The proof of this theorem will occupy the rest of the section. It will be brokenup into several parts.PART 1 OF THE PROOF It is plain that the failure of the existence part of theLopatinski condition implies that either B or C of well-posedness fails. We will thus show in this part that if the uniqueness portion of the Lopatin-ski condition fails then Condition A of well-posedness fails. The failure ofLopatinski uniqueness for some ~b =1= 0 means that the systemP(~b, DN)v== 0 Bj(~b, DN)v== 0has a nonzero solution. Call it v. Let cP1 E C~ (IR N-1 ), cP2 E C~ (IR) both beidentically equal to 1 near O. For T > 0 we defineWe will substitute UT into condition (A) of well-posedness and let T -+ +00to obtain a contradiction. 109. Well-Posedness 101First observe that lIuT I sup < 00. Now let Qmultiindex of order m. We calculate thatIIUTIIHm 2: IIDQuTII~o=JID~, [JI(X)J2(xN)eiT(X"~~)v(TXN)] 1 2 dxdxN~ JIJI(x)J2(xN)(~~)aTlaV(TxN)12C dxdxN2 - c J L C/3ID~,[JI(X)]J2(XN)(~~)Thlv(TxN)dxdxN13+,=0 1131>02-~ J L13+,=0C/3ID~,[JI(X)]J2(XN/T)(~~)Thlv(XN)dxdxN 1/31>02: cT T- 2m 1 - CT 2m - 2T- 1 2: cr2m - 1for T large. Therefore(5.2.2.1)for T large. On the other hand,P(D)UT = P(D) [JI(X)J2(xN)eiTX"~~v(TxN)]= JI (X)J2 (x N)P(D) [e iTX~~v(T xN)]+ (terms in which a derivative falls on a cutoff function) == 0 + (terms in which fewer than m derivativesfall on eiTxl~~v(TxN))(5.2.2.2)We have used here the fact that P(D) [eiT(X"~~)v(TxN)] = P(T~~, TDN)vI TxN == T m P(~b, D N )vl TxN == o. 110. 102Elliptic Boundary Value ProblemsAs a result, from (5.2.2.2), we obtain IIP(D)uTII~ ::; CT2m - 2. (5.2.2.3)Similarly,Bj(D) [eiTX"~~v(TxN)] = Bj(T~~, TD N )vl TxN == TmJ Bj(~b, D N )vI TxN ThereforeBj(D)[uyJ = Bj(D) [>I(X)>2(xN)eiYX"~~v(TxN)]== 0 + terms in which derivatives of total order not exceeding mj - 1 land on eiTxl~~v(TxN).As above, 2m 3IJ m-m J -1/2 < C [TmJ-1+m-mJ-1/2]2 == CT - I B(D)u T 112- (5.2.2.4)Therefore, substituting UT into condition (A) of the definition of well-posedness,we obtain (from (5.2.2.1), (5.2.2.3), and (5.2.2.4 that CT 2m - 1 ::; CT 2m - 2 + C".This inequality leads to a contradiction if we let T---+ +00.PART 2 OF THE PROOFAssume that the Lopatinski condition holds at all~ =1=O. We will prove that the system is well posed. This argument willproceed in several stages and will take the remainder of the section.Let u E Hm (IR~) be a solution of P(D)u == 0on IR~ { Bj(D)u == 0 on aIR~,j == 1, ... , k.LetThenP(~,DN)V==Oonn { Bj(~,DN)V == 0on an.The general solution of P(~, D N)V == 0 looks like ro v J -1 V(~,XN) == L L Cj(~)(XN)eiXN~j((). j=1 =0But the Lopatinski hypothesis then guarantees that such a v == 0 so that u == O. 111. WeU-Posedness 103To prove part (A) of well-posedness, it is therefore enough to show thatwhenever f E L2(1R~) and gj E Hm-mJ-I/2(IR.N-I), then the boundary valueproblemP(D)u== fBj(D)u == gj,j == 1, ... , khas a solution that satisfies the desired inequality. We shall need the followinglemma.LEMMA 5.2.3If P is a constant coefficient partial differential operator, then P always hasa fundamental solution. That is, if P is of order m, then there is a boundedoperatorfor every s E IR. such that P== P == 80 , If P is elliptic then is of order -mePROOFFirst consider the case N == 1. Select a number T E IR. such thatP(~+ iT) never vanishes. Then the fundamental solution operator is(-~- iT)(= JP(~ + iT) d~for 1J E V. We check thatP(D) == P(-D)1J ==J(P(-D)A(~~ - P(~ + IT)iT) d~ ==JP(~ + + ~~ dE" J( -~ -P(~iT)(IT)- iT)= iT) d~ = J - iT)d~ J(~ (Od~ = == (O).This proves the result in dimension 1. If N > 1, then we can reduce the problem to the one-dimensional case asfollows: By rotating coordinates, we may assume that the coefficient of ~1 inP is not zero. Fix ~ == (~2, ... , ~N). We can find a T, ITI ~ 1, such thatfor all ~1.Moreover, if we multiply P by a constant, we can assume that 112. 104 Elliptic Boundary Value Problemsfor all ~ 1. This inequality-that is, the choice of the constant to normalize theinequality-will depend on ~/. But the choice is uniform in a neighborhood ofthe fixed ~/. Thus to each fixed ~ we associate a neighborhood We and a realnumber T, ITIs1, such thaton We. Observe that }RN -1 is covered by these neighborhoods We. We may refinethis covering to a locally finite one WI, w2 , ... , with a 7j associated to eachW j . Now we replace the W j with their disjoint counterparts: defineW{WIW~W2W{WW 3W~W{These sets still cover }RN -1 and they are disjoint. Now we define the "Honnanderladder"Given E V, we defineNotice that, on H, we know thatIPI > 1 and (P(D))>=(P(-D)= { (P(-D)>)(-:6 ~iT,e) d~lH P(~I+lT,~) = i( -6 - iT, () d~ =L{ { ( -6 - i7j, 0 d6 df.1j ieEw; i IR = L f,{ ( -6,0 df.! d~j if, EW i IR J = { f(6,Od6d( l~N-I l~ == (O). 113. Well-Posedness105It is elementary to check that in case P is elliptic, then is an operator oforder -m. That completes the proof of the lemma.I We conclude this section by proving the inequality in part (A) of well-posedness. Notice that part (C) of well-posedness follows immediately fromthis inequality. Along the way, we prove the sufficiency of the Lopatinski con-dition for the existence of solutions to our system:PROPOSITION 5.2.4Let m > mj, j == 1, ... , k. Assume that our system satisfies the Lopatinskicondition. Then whenever f E L2(IR~) and 9j E Hm-m j -l/2(IRN -l), we mayconclude that the boundary value problemP(D)u ==f Bj(D)u == 9j, j == 1, ... , khas a unique solution u E H m (IR~ ). The solution satisfies the inequality inpart (A) of well-posedness.PROOF First we notice that the Lopatinski condition guarantees that the kernelof the system is zero. Since the system is linear, we conclude that solutions areunique once they exist. For existence, we begi