Partial differential equations and complex analysis

STEVEN G. KRANTZ
Washington University in St. Louis, Department of Mathematics

Partial Differential Equations
and Complex Analysis

Based on notes by Estela A. Gavosto and Marco M. Peloso

CRC PRESS
Boca Raton Ann Arbor London Tokyo

Library of Congress Cataloging-in-Publication Data

Krantz, 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
CIP

This book represents information obtained from authentic and highly regarded sources.
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Direct all inquiries to CRC Press, Inc., 2000 Corporate Blvd., N.W., Boca Raton, Horida,
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© 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

To the memory of my grandmother,
Eda Crisafulli.

Contents

Preface xi

1 The Dirichlet Problem in the Complex Plane 1
1.1 A Little Notation 1
1.2 The Dirichlet Problem 2
1.3 Lipschitz Spaces 7
1.4 Boundary Regularity for the Dirichlet Problem for the
Laplacian on the Unit Disc 10
1.5 Regularity of the Dirichlet Problem on a Smoothly
Bounded Domain and Conformal Mapping 20

2 Review of Fourier Analysis 26
2.1 The Fourier Transform 26
2.2 Schwartz Distributions 36
2.3 Convolution and Friedrichs Mollifiers 43
2.4 The Paley-Wiener Theorem 48

3 PseudodifferentialOperators 52
3.1 Introduction to Pseudodifferential Operators 52
3.2 A Formal Treatment of Pseudodifferential Operators 56
3.3 The Calculus of Pseudodifferential Operators 65

4 Elliptic Operators 76
4.1 Some Fundamental Properties of Partial Differential
Operators 76

viii

4.2 Regularity for Elliptic Operators 81
4.3 Change of Coordinates 87
4.4 Restriction Theorems for Sobolev Spaces 89

5 Elliptic Boundary Value Problems 95
5.1 The Constant Coefficient Case 95
5.2 Well-Posedness 99
5.3 Remarks on the Solution of the Boundary Value Problem
in the Constant Coefficient Case 108
5.4 Solution of the Boundary Value Problem in the Variable
Coefficient Case 109
5.5 Solution of the Boundary Value Problem Using
Pseudodifferential Operators 115
5.6 Remarks on the Dirichlet Problem on an Arbitrary
Domain, and a Return to Conformal Mapping 123
5.7 A Coda on the Neumann Problem 126

6 A Degenerate Elliptic Boundary Value Problem 128
6.1 Introductory Remarks 128
6.2 The Bergman Kernel 131
6.3 The Szego and Poisson-Szego Kernels 138
6.4 The Bergman Metric 143
6.5 The Dirichlet Problem for the Invariant Laplacian
on the Ball 148
6.6 Spherical Harmonics 154
6.7 Advanced Topics in the Theory of Spherical Hannonics:
the Zonal Harmonics 160
6.8 Spherical Harmonics in the Complex Domain and
Applications 172

7 The a-Neumann Problem 184
7.1 Introduction to Hermitian Analysis 185
7.2 a
The Formalism of the Problem 189
7.3 Formulation of the a-Neumann Problem 196
7.4 The Main Estimate 201
7.5 Special Boundary Charts, Finite Differences, and Other
Technical Matters 208
7.6 First Steps in the Proof of the Main Estimate 218
7.7 Estimates in the Sobolev -1/2 Norm 224
7.8 Conclusion of the Proof of the Main Estimate 234
7.9 The Solution of the a-Neumann Problem 242

ix

8 Applications of the a-Neumann Problem 252
8.1 An Application to the Bergman Projection 252
8.2 Smoothness to the Boundary of Biholomorphic Mappings 256
8.3 Other Applications of [) Techniques 263

9 The Local Solvability Issue and a Look Back 269
9.1 Some Remarks about Local Solvabilitiy 269
9.2 The Szego Projection and Local Solvability 270
9.3 The Hodge Theory for the Tangential Cauchy-Riemann
Complex 274
9.4 Commutators, Integrability, and Local Solvability 277

Table of Notation 283

Bibliography 287

Index 295

Preface

The subject of partial differential equations is perhaps the broadest and deepest
in all of mathematics. It is difficult for the novice to gain a foothold in the
subject at any level beyond the most basic. At the same time partial differential
equations 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 presented
in vitro. Thus this book proposes to present many of the basic elements of linear
partial differential equations in the context of how they are applied to the study of
complex analysis. We shall treat the Dirichlet and Neumann problems for elliptic
equations and the related Schauder regularity theory. Both the classical point of
view and the pseudodifferential operators approach will be covered. Then we
shall see what these results say about the boundary regularity of biholomorphic
mappings. We shall study the a-Neumann problem, then consider applications to
the complex function theory of several variables and to the Bergman projection.
The book culminates with applications of the a-Neumann problem, including
a proof of Fefferman's theorem on the boundary behavior of biholomorphic
mappings. There is also a treatment of the Lewy unsolvable equation from
several different points of view.
We shall explore some partial differential equations that are of current interest
and that exhibit some surprises. These include the Laplace-Beltrami operator
for the Bergman metric on the ball. Along the way, we shall give a detailed
treatment of the Bergman kernel and associated metric, the Szego kernel, and
the 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 these
appear in book form for the first time. Our treatrlent of the a-Neumann problem
parallels some classic treatments, but since we present the problem in a concrete
setting 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 this
author will also provide useful background for many of the ideas seen here.
Acquaintance with measure theory will prove helpful. For motivation, exposure

xii Preface

to the basic ideas of differential equations (such as one would encounter in a
sophomore differential equations course) is useful. All other needed ideas are
developed 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 the
constants in these estimates would be both wasteful and confusing. (Although in
certain aspects of stability and control theory it is essential to name and catalog
all constants, that is not the case here.) Thus we denote most constants by G
or G'; the values of these constants may change from line to line, even though
they are denoted with the same letter. In some contexts we shall use the now
popular 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 in
the academic year 1987-88. Some of the ideas have been presented in earlier
courses at UCLA and Penn State. It is a pleasure to thank Estela Gavosto and
Marco Peloso who wrote up the notes from my lectures. They put in a lot of
extra 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 useful
remarks.

-S.G.K.

1
The Dirichlet Problem in the Complex Plane

1.1 A Little Notation
Let IR denote the real number line and C the complex plane. The complex and
real 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 we
denote it by the symbol D. The Laplace operator (or Laplacian) is the partial
differential operator

a2 a2
~ = ox2 + oy2 .

When the Euclidean plane is studied as a real analytic object, it is convenient
to 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 on
x, respectively) that plays a significant role in our analysis (think of the method
of 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 the
partial differential operators

and ~
8z == ~ (~+i~).
2 ax ay
Check that a continuously differentiable function f(z) = u(z) + iv(z) that
satisfies 8 f / 8 z == 0 on a planar open set U is in fact holomorphic (use the

2 The Dirichlet Problem in the Complex Plane

Cauchy-Riemann equations). In other words, a function satisfying of/oz == 0
may depend on z but not on z. Likewise, a function that satisfies af / 0 Z == 0
on 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 as

1.2 The Dirichlet Problem
Introductory Remarks
Throughout 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 that
possess all derivatives up to and including order k and such that all those
derivatives 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 is
a closed set in Euclidean space. In this circumstance we say that f is C k
on X if there is an open neighborhood U of X and a C k function j on U
1
such that the restriction of to X equals f. We write f E Ck(X). In case
k == 0, we write either CO(X) or C(X). This definition is equivalent to all
other reasonable definitions of C k for a non-open set. We shall present a more
detailed 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 problem
for the Lapiacian: Let S ~ lR2 == C be a smooth, non-self-intersecting curve

The Dirichlet Problem 3

s

u

FIGURE 1.1

(part of the boundary of a smoothly bounded domain, for instance). Let U be
an open set with nontrivial intersection with S (see Figure 1.1). Finally, let ¢o
and ¢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: if
Ul and U2 both solve the problem, then Ul - U2 is a hannonic function having
zero boundary values on D. The maximum principle thea implies that Ul == U2.
In particular, in the Dirichlet problem the specifying of boundary values also
uniquely determines the normal derivative of the solution function u.
However, in order to obtain uniqueness in the Cauchy problem, we must
specify both the value of U on S and the normal derivative of u on S. How can
this be? The reason is that the Dirichlet problem is posed with a simple closed
boundary curve; the Cauchy problem is instead a local one. Questions of when
function theory reflects (algebraic) topology are treated, for instance, by the de
Rham theorem and the Atiyah-Singer index theorem. We shall not treat them
in this book, but refer the reader to [GIL], [KRl], [DER]. I


The Solution of the Dirichlet Problem in L2
Define 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) then
we define

~ ir
27r
an = f(t)e- int dt.
27f o
It follows from elementary Riesz-Fischer theory that the partial sums
N
SN == L a n ein8 ~ f (1.2.1)
n=-N

in the L 2 topology.
If 0 :::; r < 1 then observe that

L
00

anrlnlein8
n=-oo

is 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=-oo

in the L 2 topology as r ~ 1- .
In fact, the sum in (1.2.2) converges uniformly on compact subsets of the
disc. The computation that we are now about to do will prove this statement:
We have
00

S(r,O) == L anrlnlein8
n=-oo

The Dirichlet Problem 5

If we sum the two geometric series and do the necessary algebra then we find
that
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 each
of the partial sums is hannonic, u is hannonic. Moreover, the partial sum
TN is the harmonic function that solves the Dirichlet problem for the data
f N(e iO ) == E:=_
N an einO . We might hope that u is then the solution of the
Dirichlet problem with data f. This is in fact true:

THEOREM 1.2.3
Let 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 == 1
solves 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. (1.2.3.1)

Observe, using the sum from which we obtained the Poisson kernel, that
f27r f27r
1 IPr(t)1 dt = 1 Pr(t) dt = 1.
0 0

Thus we may rewrite (1.2.3.1) as

r
io
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.


Now 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 /8
as 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 > 0
such 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 + 2'
where we have applied (ii) and (i) respectively. This shows that u is continuous
at the boundary (it is obviously continuous elsewhere) and completes the proof.
I

Lipschitz Spaces 7

1.3 Lipschitz Spaces
Our 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 Dirichlet
problem be smooth up to the boundary? This is a venerable question in the
theory of partial differential equations and will be a recurring theme throughout
this book. In order to formulate the question precisely and give it a careful
answer, we need suitable function spaces.
The most naive function spaces for studying the question formulated in the
last paragraph are the C k spaces, mentioned earlier. However, these spaces
are not the most convenient for our study. The reason, which is of central
importance, is as follows: We shall learn later, by a method of Hormander
[H03], that the boundary regularity of the Dirichlet problem is equivalent to the
boundedness of certain singular integral operators (see [STSI]) on the boundary.
Singular integral operators, central to the understanding of many problems in
analysis, are not bounded on the C k spaces. (This fact explains the mysteriously
imprecise formulation of regularity results in many books on partial differential
equations. It also means that we shall have to work harder to get exact regularity
results.)
Because of the remarks in the preceding paragraph, we now introduce the scale
of Lipschitz spaces. They will be somewhat familiar, but there will be some
new twists to which the reader should pay careful attention. A comprehensive
study of these spaces appears in [KR2].
Now let U ~ ~N be an open set. Let 0 < Q < 1. A function f on U is said
to 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+hEU

We include the term IIfIILoo(u) in this definition in order to guarantee that the
Lipschitz 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 is
useful to use IlfIlLP(u) rather than IlfIILOO(U). See [KR3] for a discussion of
these matters.
When Q == 1 the "first difference" definition of the space An makes sense,
and it describes an important class of functions. However, singular integral
operators are not bounded on this space. We set this space of functions apart
by denoting it differently:

sup If(x + h) - f(x)1
Ihl +
IIIII Loo(U) < 00.
h=¢.O
x,x+hEU

The Dirichlet Problem in the Complex Plane

The space Lipl is important in geometric applications (see [FED]), but less so
in 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 f
on U to be in Aex if I is bounded, I E C1(U), and any first derivative Djl
lies in Aex - l . Equivalently, I E A ex if and only if f is bounded and, for every
nonnegative integer f < Q and multiindex {3 of total order not exceeding f we
have (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 in
AS / 2(U) if 9 is bounded and the derivatives 8gj8xj, 82gj8xj8xk exist and lie
in A1/ 2 •
Prove as an exercise that if Q' > Q then Aexl ~ A ex . Also prove that the
Weierstrass nowhere-differentiable function

00

F(O) == L 2- e j i2ie

j=O

is in Al (0, 27r) but not in LiPl (0, 21r). Construct an analogous example, for
each 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) then
does 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 function
f 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 k
exist 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 obtained
by 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 nowhere
dense in, Ck(U).

Lipschitz Spaces 9

It is natural to suspect that if all the k th order pure derivatives (a/ax j )f f exist
and are bounded, 0 ::; /!, ::; k, then the function has all derivatives (including
mixed ones) of order not exceeding k and they are bounded. In fact Mityagin
and 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. We
would like to talk about functions that are C k on U == U u au. There are three
ways 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 these
definitions. Begin by using the implicit function theorem to map U locally to a
boundary neighborhood of an upper half-space. See [HIR] for some help.

REMARK A basic regularity question for partial differential equations is as
follows: consider the Laplace equation

.6u == f.

If f E ;'0: (~N ), then where (Le. in what smoothness class) does the function u
live (at least locally)?
In many texts on partial differential equations, the question is posed as "If
f 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 in
this fashion, it is safe to assume that either the most powerful techniques are not
being used or (more typically) the results are being formulated in the language
of the incorrect spaces. In fact, the latter situation obtains here. If one uses
the 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 by
using Sobolev spaces. We shall explore this matter in further detail as the book
develops. I


1.4 Boundary Regularity for the Dirichlet Problem for the Laplacian
on the Unit Disc
We 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 we
are using the wrong spaces. We can only get a sharp result if we use Lipschitz
spaces. 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 is
convenient to work on the upper half-space U == {(x, y) E ~2 : y > O}. We
think of the real line as the boundary of U. By conformally mapping the disc to
the 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 that
there are simple techniques for extending results from this range of Q to all Q.
Now we have the following theorem.

THEOREM 1.4.1
Fix 0 < Q < 1. If f E Ao:(~) then

u(x, y) = Pyf(x, y) == l Py(x - t)f(t) dt

lies 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 .

Boundary Regularity 11

B

.A

X+H
.
X

FIGURE 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 U
because X, X + H do. Refer to Figure 1.2. Then

lu(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 satisfies

for (x, y) E U.


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
o

Thus, 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 ax2

Then

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· (1.4.1.1)

Now, from the Fundamental Theorem of Calculus, we know that


As a result, line (1.4.1.1) 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(~).


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-
I
when 0 ::; Yo < 2.
We have proved Facts I and 2 and therefore have completed our estimates of
term I.
The estimate of I I I is just the same as the estimate for I and we shall say
no 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) I

Assume 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, in
absolute value, does not exceed GIHla.
The tenn 112 is estimated in the same way that we estimated I. I


THEOREM 1.4.2
Fix I < 0: < 2. If f E An(I~) then

u(x, y) == 1. Py(x - t)f(t) dt

lies in An (71).

DISCUSSION OF THE PROOF It follows from the first theorem that u E Af3, all
o < 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 see
that
82u I C n-2
8 y 2 ::; y .
I
Likewise, because 8 f / 8x E An -I, we may prove as in Facts 1 and 2 in the
estimate of I in the proof of the last theorem that

a2u I ::; c y n-2 .
Iax8y
But then our usual arguments show that au/ay E An-I. We leave details to
the reader.
The proof is now complete. I


THEOREM 1.4.3
If 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 1.4.4
Fix y > O. Take (xo, YO) E U. Then

u(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. I

LEMMA 1.4.5
If f E Al (~), then

PROOF 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


(since (82 / 8t 2 ) P y (t) is even and has mean value zero). By hypothesis, the last
line does not exceed

C JI::2Py(t)!ltldt. (1.4.5.1 )

But recall, as in the proof of 1.4.1, that

Therefore we may estimate line (1.4.5.1) by

This is what we wanted to prove. I

COROLLARY 1.4.6
Our calculations have also proved that

REMARK 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, then
we define

~kv(X) == v(X + H) + v(X - H) - 2v(X).

LEMMA 1.4.7
If 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


PROOF 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 1V'2 v (X + tH)I.
ItI::; 1

That proves the lemma. I

FINAL 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(XO'Yo+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 no
sense. 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.


Finally, the same reasoning gives

11111::; IHI 2 . sup 1V'2 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. I

REMARK 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.
I

THEOREM 1.4.8
Let V ~ IRm, W ~ IRn be open with smooth boundary. Fix 0 < 0: < (3
and assume that T is a linear operator such that T : Aa(V) ~ Aa(W) and
T : 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, is
discussed in detail in [KR2]. The subject of interpolation is discussed in a
broader 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 from
functions on IR to functions on U. We know that

and

We may conclude from the theorem that

Thus we have a neater way of seeing that Poisson integration is well behaved
on AI.

REMARK Twenty years ago it was an open question whether, if T is a bounded
linear operator on CO and on C 2 , it follows that T is a bounded linear operator
on 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.


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, while
Al does. We shall not encounter LiPI any more in this book. The space Al was
invented 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 for
the disc that lies in An (aD), 0 < a < 2, then the solution u to the Dirichlet
problem with that data is An up to the boundary. We did this by transferring
the problem to the upper half-space by way of the Cayley transform and then
using explicit calculations with the Poisson kernel for the half-space.

1.5 Regularity of the Dirichlet Problem on a Smoothly Bounded
Domain and Conformal Mapping
We 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 has
smooth boundary if the boundary consists of finitely many curves and each of
these is locally the graph of a COO function.

In practice it is more convenient to have a different definition of domain with
smooth boundary. A function p is called a defining function for U if p is defined
in 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 function
p that is smooth (or C k ). Yet a third definition of smooth boundary is that the
boundary consists of finitely many curves rj, each of which is the trace of a
smooth curve r( t) with nonvanishing gradient. We invite the reader to verify
that these three definitions are equivalent.

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 smoothly
to the boundary. It is a classical result of Caratheodory [CAR] that if a simply
connected domain U has boundary consisting of a Jordan curve, then any con-
formal map of the domain to the disc extends univalently and bicontinuously to
the boundary. It is less well known that Painleve, in his thesis [PAl], proved that
when U has smooth boundary then the conformal mapping extends smoothly to
the boundary. In fact, Painleve's result long precedes that of Caratheodory.
We shall present here a modem approach to smoothness to the boundary
for conformal mappings. These ideas come from [KERl]. See also [BKR]
for a self-contained approach to these matters. Our purpose here is to tie the
smoothness-to-the-boundary issue for mappings directly to the regularity theory
of 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


and 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. By
Caratheodory's 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 smoothly
bounded 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 following
lemma.

LEMMA 1.5.3 HOPF'S LEMMA
Let 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. Then
the one-sided normal derivative satisfies
au
av (P) < O.
PROOF Suppose without loss of generality that u > 0 on n near P and that
n
u(P) == O. Let B R be a ball that is internally tangent to at P. We may assume
that the center of this ball is at the origin and that P has coordinates (R, 0, ... ,0).
Then, by Harnack's 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


Now let us return to the u from the Dirichlet problem that we considered
prior to line (1.5.2). Hopf's 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 and
72 a generic second derivative. We conclude that

Hence (again using Hopf's lemma),

17 2 "'1 <
f' -
~
l7ul
< C".
-

In the same fashion, we may prove that l7 k ¢1 ~ Ck , any k E {1,2, ...}. This
means (use the fundamental theorem of calculus) that ¢ E Coo (0).
We have amved at the following situation: Smoothness to the boundary
of confonnal maps implies regularity of the Dirichlet problem on a smoothly
bounded domain. Conversely, regularity of the Dirichlet problem can be used,
together with Hopf's 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 the
Dirichlet problem for a more general class of operators that is invariant under
smooth changes of coordinates. We will study these operators by (i) localizing
the problem and (ii) mapping the smooth domain under a diffeomorphism to
an upper half-space. It will tum out that elliptic operators are invariant under
these operations. We shall then use the calculus of pseudodifferential operators
to prove local boundary regularity for elliptic operators.
There is an important point implicit in our discussion that deserves to be
brought into the foreground. The Laplacian is invariant under conformal trans-
formations (exercise). This observation was useful in setting up the discussion
in 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 expand
to 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 under
a diffeomorphic change of coordinates. For simplicity, we restrict attention to


]R2 with coordinates (x, y). Let

¢ ( x, y) == (¢ 1 (x, y), cP2 (x, y)) == (x', y')
be a diffeomorphism of }R2. Let

In (x', y') coordinates, the operator ~ becomes

In an effort to see what the new operator has in common with the old one, we
introduce the notation

where
a a a a
axO: axr 1 ax~2 ... ax~n

is a differential monomial. Its "symbol" (for more on this, see the next two
chapters) is defined to be

The 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 the
coefficient of ~i~j in the symbol. Thus

The 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).


Then

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 positive
definiteness property of the symbol that will be crucial to the success of our
attack on elliptic operators.

2
Review of Fourier Analysis

2.1 The Fourier Transform
A 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 this
matter. We have opted for this definition because of its simplicity. We note that
the significance of the exponentials eit.~ is that the only continuous multiplicative
homomorphisms of}RN into the circle group are the functions ¢~ (t) == eit.~.
(We leave this as an exercise for the reader. A thorough discussion appears
in [KAT] or [BAC].) These functions are called the characters of the additive
group }RN.

Basic Properties of the Fourier Transform
PROPOSITION 2.1.1
If f E £1 (}RN) then

PROOF Observe that

Ij(~) I ::; J If(t)1 dt. I

The Fourier Transform 27

PROPOSITION 2.1.2
If f E L 1 (IR N ), f is differentiable, and af/axj ELI, then

PROOF 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. I

PROPOSITION 2.1.3
If f E Ll(I~N) and iXjf E L 1 (IR N ), then

-- a A

(iXjf) = 8~j f·

PROOF Integrate by parts. I

PROPOSITION 2.1.4 THE RIEMANN-LEBESGUE LEMMA
If f E L 1(IR N ), then

lim Ij(~)1 == o.
~-+oo

PROOF First assume that f E C~ (IR N ). We know that

and

28 Review of Fourier Analysis

Then (1 + lel 2 )j is bounded. Therefore

III :::; C le~oo O.
1 + 1~12

This proves the result for f E C~.
Now let 9 E L 1 be arbitrary. By elementary measure theory, there is a
function ¢ E Cc(I~N) such that IIg - ¢IILI < E/4. It is then straightforward to
construct 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 2
This proves the result. I

PROPOSITION 2.1.5
Let 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. Equip
this space with the supremum norm. Then the Fourier transform maps L 1 to Co
continuously, 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. I

PROPOSITION 2.1.7
If p is a rotation of}RN then we define pf (x) == f (p( x)). Then Pi == p(j).


PROOF 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)) (~). I
PROPOSITION 2.1.8
We have


PROPOSITION 2.1.9
If 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.


(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 is
an L 1 function and III * gllL1 ::; IIfllL1 IIgllLI.


PROPOSITION 2.1.10
If 1,9 E £1, then
r;-9(~) == j(~) . g(~).


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. I

PROPOSITION 2.1.11
If I, 9 E £1 , then

Jj(~)g(~) d~ Jf(Og(~) d~.
=

PROOF This is a straightforward change in the order of integration. I

The Inverse Fourier Transform
Our goal is to be able to recover 1 from j. This program entails several technical
difficulties. First, we need to know that the Fourier transform is univalent in
order 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 as
E ~ 0+.

LEMMA 2.1.12
We have


PROOF 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()=7L

Thus I == ~, as desired. I

REMARK 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 the
one-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. (2.1.12.1)
2 Jr
Here, for ~ E R fixed, r == r ~ is the curve t ~ ~ + it. Let r N be the part of
the curve between t == - Nand t == N. Since Ir
== limN ---l>OO Jr N' it is enough
J
for us to understand r N' Refer to Figure 2.1.
Now

Therefore

But, as N -+ 00,

1 +1 -+ O.
JEfl JEr


O-iN -----~
Ef

FIGURE 2.1

Thus

lim 1 == - lim 1 . (2.1.12.2)
N~oo JrN N~oo Jf N
Now we combine (2.1.12.1) and (2.1.12.2) to see that

We conclude that, in jRI ,

and in }RN we have


It 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-
x

nel. It is a summability kernel (see [KAT]) for the Fourier transform. Observe
that G(~) == e-I~12 /2.
On R N we define

Then

-- --
~(~) = (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~ -t

J j(~) d~. That takes care of the right-hand side of (2.1.13).
Next observe that

Thus 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).


Thus 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 FORMULA
If !, j E £1 and both are continuous, then

f(O) = (27r)-N Jj(~) d~. (2.1.14.1 )

Of course there is nothing special about the point 0 E R N . We now exploit
the compatibility of the Fourier transform with translations to obtain a more
general 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 in
the integral,
T;] == eih.~ j(~).
Now we apply formula (2.1.14.1) in our theorem to T-h/: The result is

or

THEOREM 2.1.15
If I, j E £1 then for any h E R N we have

f(h) = (2JT)-N Jj(~)e-ih.f, d~.
COROLLARY 2.1.16
The Fourier transform is univalent. That is, if I, 9 E £1 and j == 9 then f = 9
almost everywhere.

PROOF Since I - 9 E £1 and j - g == 0 E £1, this is immediate from either
the theorem or (2.1.13). I

Since the Fourier transform is univalent, it is natural to ask whether it is
surjective. We have

PROPOSITION 2.1.17
The operator

is not onto.


PROOF 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 > 0
such that

IIIIILI :S cllj"L~'
On IR , let g(~) be the characteristic function of the interval [-1, 1]. The in-
I

verse 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 Fourier
transforms are unbounded in £1 norm. That gives the desired contradiction.
I

Plancherel's Formula
PROPOSITION 2.1.18 PLANCHEREL
If 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 . (2.1.18.1)

Now

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 (2.1.18.1) we may now conclude that

J If(t)1 2 dt = g(O) = (21T)-N Jg(~) d~ = (27T)-N JIj(~)12 df
That is the desired formula. I

COROLLARY 2.1.19
If f E £2 (IR N ) then the Fourier transform of I can be defined in the following
fashion: Let Ij E C~ satisfy fj -+ I in the £2 topology. It follows from
the proposition that {.fj } is Cauchy in £2. Let g be the £2 limit of this latter
sequence. We set j == g.

It is easy to check that the definition of j given in the corollary is independent
of the choice of sequence Ij E C~ and that

J11(01 d~ 2
= (21T)N J If(xW dx.


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 conclude
that
F : £P ~ LP' , 1 ~ p ~ 2,
where p' == p/ (p - 1). If p > 2 then F does not map LP into any nice function
space. The precise norm of F on LP has been computed by Beckner [BEC].

Exercises: Restrict attention to dimension 1. Consider the Fourier transform F
as a bounded linear operator on the Hilbert space £2 (IR N ). Prove that the four
roots of unity (suitably scaled) are eigenvalues of F.
Prove that if p(x) is a Hermite polynomial (see [STW], [WHW]), then the
function p(x)e-lxI2/2 is an eigenfunction of F. (Hint: (ix!) == (j)' and == l'
-i~j.)

2.2 Schwartz Distributions
Thorough 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. Any
derivative of a Schwartz function is still a Schwartz function. The Schwartz
space is obviously a linear space.
It is worth noting that the space of Coo functions with compact support (which
we have been denoting by C~) forms a proper subspace of S. Since as recently
as 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.

Schwartz Distributions 37

Then one checks, using l'Hopital's 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 have
constructed a standard "cutoff function" on ~l. On IR N , the function

plays a similar role. '

Exercise: [The Coo Urysohn lemma] Let K and L be disjoint closed sets
in 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 5
The functions Pn,(3 are seminorms on 5. A neighborhood basis of 0 for the
corresponding topology on 5 is given by the sets

NE,f,m == {¢: L lal:S€
Pa,(3(¢) < E} .
1.6I:Sm

Exercise: The space 5 cannot be normed.

DEFINITION 2.2.1 A Schwartz distribution Q is a continuous linear functional
on 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(</».


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.


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 function
f(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 -00

Thus f' consists of integration against b( x) == -x (-00,0] + X[O,oo). This function
is often called the Heaviside function.

Exercise: Let n ~
R N be a smoothly bounded domain. Let v be the unit
outward normal vector field to 8n. Prove that -VXn E S'. (Hint: Use Green's
theorem. It will tum out that (- VXn) (¢) == Jan ¢ da, where da is area measure
on the boundary.)

The Fourier Transform
The principal importance of the Schwartz distributions as opposed to other dis-
tribution theories (more on those below) is that they are well behaved under the
Fourier transform. First we need a lemma:


LEMMA 2.2.3
If f E S then j E S.

PROOF This is just an exercise with Propositions 2.1.2 and 2.1.3: the Fourier
transform converts multiplication by monomials into differentiation and vice
versa. I

DEFINITION 2.2.4 If u is a Schwartz distribution, then we define a Schwartz
distribution u by
u(¢) == u(¢).

By the lemma, the definition of u makes good sense. Moreover, by 2.2.5
below,

lu(¢)1 == lu(¢)1 ~ L Pn,f3(¢)
lal+If3I~M

for some M > 0 (by the definition of the topology on S). It is a straightforward
exercise with 2.1.2 and 2.1.3 to see that the sum on the right is majorized by
the sum

C· L Pn,f3(¢)'
lal+It3I~M

In conclusion, the Fourier transform of a Schwartz distribution is also a Schwartz
distribution.

Other Spaces of Distributions
Let V == C~ and £ == Coo. Clearly V ~ S ~ £. On each of the spaces V and
£ we use the semi-norms

where K ~ R N is a compact set and Q == (Ql,' .. , QN) is a multiindex. These
induce 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 V
and £ respectively. Trivially, £' ~ V'. The functional in R l given by
00
j
J-l==L2 8j ,
j=l

where 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 union
of all open sets U such that J1 (¢) == 0 for all elements of C~ that are supported


in U. As an example, the support of the Dirac mass Do is the origin: when Do
is applied to any testing function ¢ with support disjoint from 0 then the result
is O.

Exercise: Let J.L E V'. Then J.L E £' if and only if J.L has compact support. The
elements of £' are sometimes referred to as the "compactly supported distribu-
tions."

PROPOSITION 2.2.5
A linear functional L on S is a Schwartz distribution (tempered distribution) if
and only if there is a C > 0 and integers m and f such that for all ¢ E S we
have

IL(¢)I ~ C· L L Pn,f3(¢)' (2.2.5.1)
Inl~f 1f3I~m

SKETCH OF PROOF If an inequality like (2.2.5.1) holds, then clearly L is
continuous.
For the converse, assume that L is continuous. Recall that a neighborhood
basis of 0 in S is given by sets of the form

NE,f,m == {¢ E S: L
lal:S€
Pa,f3(¢) < E} .
1.6I:Sm

Since L is continuous, the inverse image of an open set under L is open.
Consider

There exist E, f, m such that

Thus

L
lal:Sf
Pn,f3(¢) < E

1.6I:Sm

implies that

/L(¢)/ < 1.
That is the required result, with C == 1/ Eo I
Exercise: A similar result holds for V' and for £'.


THEOREM 2.2.6 STRUCTURE THEOREM FOR V'
flu E V' then
k
U == LDjJ-lj,
j=1
where 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 the
dual of the continuous functions with compact support is the space of finite
Borel measures. In a natural fashion, the space of C I functions with compact
support can be identified with a subspace of the set of ordered pairs of Ce
functions: 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 be
given by a pair of measures. Combining this information with the definition
of derivative of a distribution gives that an element of the dual of C~ is of the
form J-l1 + (J-l2)'. In a similar fashion, one can prove that an element of the dual
of 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 countable
union of the dual spaces (C~)'. I

The theorem makes explicit the fact that an element of V' can depend on
only finitely many derivatives of the testing function-that is, on finitely many
of the norms Pn,(3.
We have already noted that the Schwartz distributions are the most convenient
for Fourier transform theory. But the space V' is often more convenient in the
theory of partial differential equations (because of the control on the support
of testing functions). It will sometimes be necessary to pass back and forth
between the two theories. In any given context, no confusion should result.

Exercise: Use the Paley-Wiener theorem (discussed in Section 4) or some other
technique to prove that if cP E V then ¢ (j. V. (This fact is often referred to as
the Heisenberg uncertainty principle. In fact, it has a number of qualitative and
quantitative formulations that are useful in quantum mechanics. See [FEG] for
more on these matters.)

More on the Topology of V and V'
We 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 a
fixed Coo function and ¢j (x) == 'ljJ (x - j), then we do not want to say that the
sequence {¢j} converges to O.

Convolution and Friedrichs Mollifiers 43

A functional J-l on V is continuous if J-l(¢j) -+ J-l( ¢) whenever ¢j -+ ¢. This
is equivalent to the already noted characterization that there exist a compact K
and an N > 0 such that

IJ-l(¢) I :S C L PK,n(¢)
Inl:::;N

for every testing function ¢.

2.3 Convolution and Friedrichs Mollifiers
Recall 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 may
successfully 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 those
topics in analysis (of which there are many) where understanding is best achieved
by 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 that
9*¢ 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; otherwise
9 * ¢ does not necessarily make sense.

LEMMA 2.3.2
If Q E V' and 9 E V then Q * 9 is a function. What is more, If we let
7h¢(x) == ¢(x - h) then (Q * 9 )(x) == a(Txg).



(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~ be
supported in the ball B (0, 1). For convenience we assume that ¢ 2: 0, although
this is not crucial to the theory. Assume that J ¢( x) dx == 1. Set ¢E (x) ==
N
E- ¢(x/ E).
The family {¢E} will be called a family of Friedrichs mollifiers in honor of
K. O. Friedrichs. The use of such families to approximate a given function
by smooth functions has become a pervasive technique in modem analysis. In
functional analysis, such a family is sometimes called a weak approximation to
the identity (for reasons that we are about to see). Observe that J ¢E (x) dx == 1
for every E > O.

LEMMA 2.3.3
If f E LP(RN ), 1~p ~ 00, then

PROOF The case p == 00 is obvious, so we shall assume that 1 :::; p < 00.
Then we may apply Jensen's inequality, with the unit mass measure ¢E(X) dx,
to see that

Ilf*cP,II1p = 111 P
f(x-t)cP,(t)dtI dx

:=; 1
1 If(x - t)IPlcP,(t)1 dt dx

= 11 If(x - t)IP dx cP,(t) dt

= 1IlflltpcP,(t) dt

== 11111~p· I
REMARK The function IE == f * ¢E is certainly Coo Gust differentiate under
the integral sign) but it is generally not compactly supported unless I is. I

Convolution and Friedrichs Mollijiers 45

LEMMA 2.3.4
For 1 :::; p < 00 we have

PROOF We will use the following claim: For 1 :::; p < 00 we have

where 7ft f (x) == ! (x - h). Assume the claim for the moment.
Now

II!< - !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~p¢«t) dt

(t=J-U)
== 1II -! liP ()
TJi,E! LP¢ J-l dJ-l.

In the inequality here we have used Jensen's inequality. Now the claim and the
Lebesgue 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 -+ 0
by uniform continuity. It follows that II Th 1/J - 1/J II Lp -+ O. Now if ! E LP is
arbitrary and E > 0, then choose 1/J E Ce such that II! - 1/J II Lp < E/2. Then

lim sup II Thf - !IILP :::; lim sup II Th(! -1/J)IILP + lim sup IITh1/J -1/JII :::; E.
h---+O h---+O h---+O

Since E > 0 was arbitrary, the claim follows. I

LEMMA 2.3.5
If ! E C e then fE -+ ! uniformly.

PROOF Let 1] > 0 and choose E > 0 such that if Ix-yl < E then If(x)- f(y)1 <
1]. Then

If«x) - !(x)1 11 f(x - t)¢«t) dt - !(x)1
=

= I/[f(x - t) - f(x)J¢«t) dtl

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