Beijing lectures in harmonic analysis

Annals of Mathematics Studies Number 112 BEIJING LECTURES IN HARMONIC ANALYSIS EDITED BY E. M. STEIN PRINCETON UNI...

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Annals of Mathematics Studies Number 112

BEIJING LECTURES IN HARMONIC ANALYSIS

EDITED BY

E. M. STEIN

PRINCETON UNIVERSITY PRESS

PRINCETON, NEW JERSEY 1986

Copyright C 1986 by Princeton University Press ALL RIGHTS RESERVED

The Annals of Mathematics Studies are edited by

William Browder, Robert P. Langlands, John Milnor, and Elias M. Stein Corresponding editors:

Stefan Hildebrandt, H. Blaine Lawson, Louis Nirenberg, and David Vogan

Clothbound editions of Princeton University Press

books are printed on acid-free paper, and binding materials are chosen for strength and durability. Pa. perbacks, while satisfactory for personal collections,

are not usually suitable for library rebinding ISBN 0-691-08418-1 (cloth)

ISBN 0-691-08419-X (paper)

Printed in the United States of America by Princeton University Press, 41 William Street Princeton, New Jersey

Library of Congress Cataloging in Publication data will

be found on the last printed page of this book

TABLE OF CONTENTS

vii

PREFACE NON-LINEAR HARMONIC ANALYSIS, OPERATOR THEORY AND P.D.D. by R.R. Coif man and Yves Meyer

3

MULTIPARAMETER FOURIER ANALYSIS

by Robert Fefferman

47

ELLIPTIC BOUNDARY VALUE PROBLEMS ON LIPSCHITZ DOMA INS

by Carlos E. Kenig

131

INTEGRAL FORMULAS IN COMPLEX ANALYSIS

by Steven G. Krantz

185

VECTOR FIELDS AND NONISOTROPIC METRICS 241

by Alexander Nagel

OSCILLATORY INTEGRALS IN FOURIER ANALYSIS by E. M. Stein

307

AVERAGES AND SINGULAR INTEGRALS OVER LOWER DIMENSIONAL SETS 357

by Stephen Wainger

INDEX

423

v

PREFACE In September 1984 a summer school in analysis was held at Peking University. The subjects dealt with were topics of current interest in the closely interrelated areas of Fourier analysis, pseudo-differential and singular integral operators, partial differential equations, real-variable theory, and several complex variables. Entitled the "Summer Symposium of Analysis in China," the conference was organized around seven series of expository lectures whose purpose was to give both an introduction of the basic material as well as a description of the most recent results in these areas. Our objective was to facilitate further scientific exchanges between the mathematicians of our two countries and to bring the students of the summer school to the level of current research in those important fields. On behalf of all the visiting lecturers I would like to acknowledge our great appreciation to the organizing committee of the conference: Professors M. T. Cheng and D. G. Deng of Peking University, S. Kung of the University of Science and Technology of China, S. L. Wang of Hangzhou University, and R. Long of the Institute of Mathematics of the Academia Sinica. Their efforts helped to make this a most fruitful and enjoyable meeting. E. M. STEIN

vii

Beijing Lectures in Harmonic Analysis

NON-LINEAR HARMONIC ANALYSIS,

OPERATOR THEORY AND P.D.E.

R. R. Coifman and Yves Meyer

Our purpose is to describe a certain number of results involving the study of non-linear analytic dependence of some functionals arising naturally in P.D.E. or operator theory. To be more specific we will consider functionals i.e., functions defined on a Banach space of functions (usually on Rn ) with values in another Banach space of functions or operators. Such a functional F:B1 -1 B2 is said to be real analytic around 0 in

B1 if we can expand it in a power series around 0 i.e. 00

F(f) = I Ak(f ) k=0

where Ak(f) is a "homogeneous polynomial" of degree k in f. This means that there is a k multilinear function

Ak(fI...fk):B1 xB1...xB1 B2 (linear in each argument) such that Ak(f) = Ak(f, f, f) and k

(1)

IIAk(fI...fk)IIB2 < Ck

11

IIfjIIB1

j=1

for some constant C . (This last estimate guarantees the convergence of the series in the ball IIf1I < C .)

3

4

R. R. COIFMAN AND YVES MEYER

Certain facts can be easily verified. In particular if F is analytic it can be extended to a ball in Bi (the complexification of B1 ) and the extension is holomorphic from B c to B2 i.e., F(f +zg) is a holomorphic (vector valued) function of z e C, JzJ < 1 , Yf, g sufficiently small. The converse is also true. Any such holomorphic function can be exx the kth Frechet differenpanded in a power series, (where Ak is k!

tial at 0 ). We will concentrate our attention on very concrete functionals arising in connection with differential equations or complex analysis, and would

like to prove that they depend analytically on certain functional parameters. As you know there are two ways to proceed. 1. Expand in a power series and show that one has estimates (1). 2. Extend the functional to the complexification as "formally holomorphic" and prove some boundedness estimates. Let L denote a differential operator like a(x) dx

xfR,

a(z)az

zEC

= div A(x) grad, A _ (aij)

aij(x) ai j(x)

x f Rn

x e Rn

the coefficients a(x) (or aij(x) ) will be assumed to belong to some Banach space B1 of functions (for example L°° ). It is natural to ask when such objects as: L-1

,

fL sgnL, e-tL , e-t\/L ,

or more generally, c(L) (where (k: C C ), can be defined as a bounded operator (say on L2 or some Soboleff space), and a functional calculus developed i.e., (k1(L)02(L) _ 0102(L).

NON-LINEAR HARMONIC ANALYSIS

5

Many questions arise:

a) Does F(a) = c(L) viewed as an operator valued function of a depend analytically on a ? This is equivalent to asking whether we can consider complex valued

coefficients in L and still have estimates on (A(L). b) What is the largest domain of coefficients a for which we have estimates for c(L) ? This question is the same as asking what is the largest BI for which (1) holds, and what is the domain of holomorphy of

F(a) in this space. The answer to question a) will require first that we understand methods for expanding functionals in a power series, and second, that the nature of the multilinear operators Ak be sufficiently well understood to provide estimates (1). As for question b) we will see that the largest spaces possible for the coefficients involve rough coefficients and leads us to work with coefficients in L°°, B.M.O. and other "exotic spaces." We now start with a fundamental example related to the Cauchy integral. We let La

l+a -dx

with hall < 1 a(x), real valued.

If we define h(x) = x+A(x), A'(x) = a. We then have

Laf=\_d f°h-110h=i

UhdxUhlf

where

Uhf =f°h. Of course, in this case, if we use the Fourier transform we can define

(I_ d)f idx

r eix&O(e)f(e)de

R. R. COIF MAN AND YVES MEYER

This gives, for example 00

sgn

d

elx6sgn 6 (e)de = n dx) f = J

f

P. V.

f t dt = H(f

x-t

-00

Thus we can define

n sgn(La )f = nUhsgn

r li dx a Ulf h =pv J

00

f(t)(l+a(t) x-t+A(x)-A(t)

dt

-00

(where we used the observation that 95(ULU-1) = U¢(L)U-1 ). We view

F(a) = sgn L. as an operator on L2(R) and wish to know whether it is analytic on L°° or if we can replace a by complex a and still have a bounded operator. If we do this, writing a = a + ip jfaIl0 < 1 , we find

F(a) f =

f

f(t) (1+ia+i$)

x-t+A(x) + iB(x) - A(t) - iB(t)

f t)[(l+a)/(1+a)](1+a)

x+A(x)-t-A(t)+i(B x)-B(t))

dt

dt

= UhCUhif1

where

h = x+A(x) Cf =

('

J

f(t) (1+Bi(t)) x-t+iB1(x)-iB1(t)dt

-00

f1(t) = f(t) i+a Bi(t)

B1 = B°h-1


7

Since Uh is bounded on L2 it would suffice to prove that C is bounded on L2 for all B such that B' is small. We could also try to prove this by expanding -irr sgn(La)f = J (xf t)+A(x)-A(t) dt = C' (-1)k

r

(A(x)_A(t)\kf(1+a)dtJ

-

x-t

/

x-t

Observe that the operators are of the form

T(f) = ft

(Ax_ A t) f(t) dt = fk(xt) f(t)dt

We will prove Theorem I: Let T a Cc(C) and A(x) such that

IA xx_A t

I

< M and T(f) = p.v. J

(`A(xx-y (Y) f(y) dy

Then the operator T is bounded on L2(R) (and LP 1 < p < oe ). This result will then be extended to Rn and other settings. We now return to the interpretation of C as the Cauchy integral for the curve z(t) = t + iA(t) where A is Lipschitz

as we can see its boundedness in L2 is equivalent to the analytic dependence of C(a)f on the curve a. This now is related to the lectures by C. Kenig (to which we shall return later).

8

R. R. COWMAN AND YVES MEYER

Let us consider a more general version of the Cauchy integral.

Let I' be a rectifiable curve through 0, s be the arc length parameter

s

z'(s) = ea(s) i.e., z(s) = r eia(t)dt 0

The Cauchy integral on r is given as:

f

cr(f) = p.v.

00

f t z' t dt

z(s)- z(t)

-00

r

J

00

00

1

z s- Z (t s-t

f t z' t dt s-t

rJ 0rz` ss-t-z t/I fs-tt)

dt

-00

if we assume 01 2Ix-xI I

I

E

for some E>0. 3°

II + II < t


13

(This last condition followed from k(x,y) = -k(x,y) and 1°, 2°) then the conditions * are verified. It can be shown that if * is valid then T can be represented as a limit of integral operators whose kernels satisfy the conditions 1°, 2°, 3° We will refer to 1°, 2° as standard estimates, and to 3° as the weak cancellation (or boundedness) property. This last condition is independent of 1°, 2°, and can be proved in a variety of ways, as we shall see. To see how the theorem can be applied to reduce the degree of nonlinearity a of a "polynomial" and to obtain estimates consider

f(y) C(aIf) n = ft!J,A(xX_A(Y)ln Y J Y !

dY

We check by a simple integration by parts that Cn(a,l)(x) = Cn-I(a,a)(x)

If we make the induction hypothesis that Cn_I(a,f) is bounded on L2 it would follow by the preceding remarks that Cn_I(a,f) maps L°° to B.M.O. from which we deduce that Cn(a,l) a B.M.O. and that 3C > 0 such that IIC5(a,f )IIL2 2 it isn't as simple. Consider for a moment a finite dimensional vector space V and a bilinear transformation AN xV - V. We can ask when is it true that

20


there exists a basis in V el

eN a linear operator A defined on

V®V diagonalized by ei®ej A (ei®ej) = Ai,j ei®ej such that

A(ei,ej) = Ai j e,(i,j) for some map n(i,j) of

(1,..,N)2 -(1,...,N) i.e. we want the diagram to commute

ti A V®V

where

v(ei, ej) = ei®ej n(ei®ej) = en(i,j)

This is clearly the situation for periodic functions. It would be interesting to understand the bilinear operators admitting such a diagonizable lift to the tensor product. This observation indicates that the hypothesis concerning commutation with translations, imposes severe restrictions on the nature of a multilinear operation. We now return to the line on which we realize multilinear operations as

Ak(fI...fk)(x)

ffeix(e1+...ek)VeI...ek)(eI)f(e2)...f((k)deidek =

Rk and

21


nk(eieix, ...,

elekx) (x) =

eix(e1+...ek)Ak(eI

... Xk) .

(Note that this realization is only valid for multilinear operations verifying some mild continuity conditions.) This realization permits us for example to show that the study of fo lAt * b (kt * fat can easily be converted to the study of

J to * (iAt*b 95t* fdtt 0

which we considered previously in the proof of Theorem [II]. In fact

0

(' t/it*b(kt*f 0

if we take we have

dt

_

('eix(eI

J

(f(tei) 3(te2)dt

2)

2

(ow

with support in 1
0, C>0. In fact, this will imply that for ¢(x) such that

+IeIN)d5

III

4 and

CI-3M M.

36


In that case consider the smallest function AI > A with A'(y) > 23 .

Ik

then have AI(y) = A(y) except for disjoint intervals Ik on which AI(y) = M . But

< 2M II-UlkI +

+

fiA(Y)dY = "_.UIk

'fUlk

=2MIII-3MIUlkI 3MIUIkI < (2M

- m1(Ai)) II I < (2M - M) III =III

i.e.,

IuIkI < 4 III

.

Since I< Ai(y) < M2, we have 3M-3M M and we construct A 11 as above. A11 = 2M(x-a)-A(x) except on a set of meas
6111 such that Hx rE , Vy eE ,

K1(x,y) = K(x,y)

.

38


Then T maps L°° to BMO with IITIIL°°,BMO

C77 C0 . LP(X) defined by

A(A,F) = J

frk(t1,t2. tk,s)[JAi(Ut .x)F(Usx)dt ds 1

satisfies

I}A(A,F)IILP(X)` 11

IIA'IIL°°(X)IIFIILP(X)

(note that the constants are the same).

The examples we have in mind are the following. X = Rn, dx Lebesque measure, e a unit vector Utx = x-te

If we take Hf

fR

.

f x t dt, A f = f x -te dt . Or consider

f

Ut(ei9) = ei(e-t)

X = T1

4f - c rf(0-t)

J

1t

dt

ctg t

If we take k

Ak(a,f) =

f(A(x)_A(y)dy

x-y

f

SE--y

k

f(A(x)_A(x_t))

A? t dt


41

then Nk,e(A' f) =

ftf

/

k

t'

A(x-te),j f(x-te) dt t

Multiplying by i2(e) and integrating on IeI = 1 in Rn we get

S2(e)

A(x)-A(x-te)

k

t

f(x-te) dt t

f

(A(x)-A(y)lk x-y I K(x-y)f(y)dy where

k(Y) _

WYIYI) IYIn

is odd for K even

even for K odd.

§6.

We now are in a position to recall the various ingredients which we discussed previously and recast them in a general setting. We considered

operators L which are "small" perturbations of an operator L0 for example, we took L0 = -A and L = -A+Y- blJ

ax-E

e0 and

then proceeded to expand functions of L, F(L) as a power series in terms of the coefficients of L-L0 i.e., we wrote 00

F(L) = F(L0) + I Ak(b) k=1

where the Ak(b) was an operator valued homogeneous polynomial of dAk in b. Of course such perturbations could be shown to converge only in a

little ball around 0 (in the space of coefficients). It is then appropriate to ask how far can one extend the results and what is the natural domain of analyticity or holomorphy of the function F(L). This question is

42


meaningless if the Banach space in which we prove analyticity (in a neighborhood of 0) is not the largest possible space for which such estimates can be proved. A first task is to identify this largest space, which we will call the space of holomorphy. Once this space has been found it is natural to ask for the domain of holomorphy of the corresponding functional. Let us return flow to our first example where F(a) = sgn(1i dx)

where a is a function on R, and F(a) is an operator on L2(R). The series expansion was f(y) (7 A (X

F(a)f=1: J

(1-a(y))dy,

y

0

A'=a

Let us consider for simplicity the commutator series Fo(a) f > ` I (A(xX y (Y))k X_ y fly

J

=

(f )

0

We wish to find the smallest norm

III

III

(i.e., the largest Banach space)

for which *

IIAk(a)IIL2 L2 0, f t L1(Rn), and a > 0.

Then there exist disjoint cubes Qk such that

(1) a
a . For these Q' we stop the bisection process. For the

10

rest, we continue until we first arrive at a cube Q' such that 1 f ,f > a at which point we stop. 1

I

Q

The cubes Q' at which we stop are then our Qk. By construction f > a. Let Qk be the cube containing Qk which was bisected

IQkI fQk

Qk. Then

1IkIf1k f < a.

It follows that

and since we did not stop at

k,


f 1Q

k

1

k1 1 f

49

f< 2na ,

f < 1Q k

Qk

proving (1). Notice that (3) follows from (1) because 1Qk' < a fQk f so summing on k, we have

('

IUQk1o IB(x;r)1

B(x,r)

Going along with this we also define MSf(x) =

sup

1

xEQ dyadic cube IQI

JI If(t)I dt Q

(Recall that a dyadic interval of RI is one of the form [j2k,(j+1)2k] j,k E Z and a dyadic cube is a product of dyadic intervals of equal length recall also the basic property of dyadic cubes-if Q1,Q2 are dyadic either Q1 f1Q2 = 0, Q1 C Q2 or Q2 C QI .) Then the following simple lemma sheds some light on the relationship of the Calderon-Zygmund lemma to the Maximal Operator.

ROBERT FEFFERMAN

50

LEMMA. Let f > 0 c L1(Rn) and a > 0. Let Qk be Calderon-Zygmund cubes as above, and let Qk denote the double of Qk. Then there exist positive numbers c and C such that (1) UQkC 1Mf > ca(

(2) UQk ) (Mf > Cal

(3) Furthermpre, if Qk are dyadic, then UQk )1MSf > Cal.

Proof. (1) Let x c Qk. Then there exists a ball B(x;r) such that x c Qk C B(x;r) and B(xx;rl < Cn. Then k Mf(x) >_

Ifl >

1

B(x;r)

,/

B(x;r)

IQkl

1

(IX)l) IV

IfI > 1 a Cn

Qk

(2) Let x / UQk. Let r > 0. Then we estimate

ff

I

B(x;r)

f+ I

f

B(x,r)fld[UQk]

Qi

< ajB(x;r)I }

Qi

< alB(x;r)I ± 2na

Key point : if Qj fl B(x ;r)

0 then Q) C B(x;l Or) so that

I IQiI 1

IITfII

all
0 and f c L1(X). Set 1

f dt

if x cQk,

IQ kI

Qk

g(x) = f(x)

if x / Qk .

and b(x) = f (x) - g(x) . Then Il ITg(x)IN > a1I
1 and weak 1-1. The Estimates for Pointwise Convergence of Singular Integrals on L1(Rn) . Suppose that K(x) is a classical Calder 6n-Zygmund kernel and let KE(x) = K(x) )(IXI>E(x), for e > 0. We are interested in the 4.

existence a.e. of lim f * KE(x) for f e Ll(Rn). In order to know this, it E-40

55


is enough to show that T*f(x) = sup If * KE(x)I satisfies the weak type E>0

estimate IIT*f(x) > all < a f

If I .

R

It turns out that by using the Hardy

Littlewood maximal operator it is not difficult to prove T*f(x) < CIM(Tf)(x) + Mf(x)l which immediately gives the boundedness of T* on Lp(Rn) for p > 1 . However, it fails to give the weak type inequality for functions on

LI(Rn). This inequality follows easily from the observation that T* is a singular integral. Let 1

if IxI < 1

O(x) c C (Rn), O(x) =

0 if lxI >2 and set K,(x) = K(x)[1 -0 QL)]

Then

IK,(x) -K(x)I

I In XIxI0

sup If *kI is weak type (1,1). In order to do this let H:Rn , L((0,-);de) E>0

be given by H(x)(E) = II

(x). Then IH(x)-H(x+h)IL
0 for all x. Use the Calderon-Zygmund decomposition with a = Ci , j c Z for some C > 0 sufficiently large, to get (dyadic)

cubes Qk where 1 f f - C). Define f .l decomposition,

as in the Calder6n-Zygmund

IQ)kI Q)k

1

f

if x c Q

IQ,

k

Qjk f(x)

and A f =fj+1-fj, then observe that: i

if x

U Qk

56

ROBERT FEFFERMAN

(1) jf lives on k Qk and has mean value 0 on each Q. (2) Aif is constant on every Qk for i < j.

(3) fJ - 0 as j,--o and f.J

+Do

f

as j -.+00 so f =

From (1) and (2) it is clear that the A

i

f

are orthogonal so that

1/2

If IIL2(Rn)

F Af. j=-

= (Y IlojfIIL2)= II (Y lojf(x)12)

1/2

I1L2

Finally, observe that the square function (I IA f(x)12)1 /2 is j

essentially just the dyadic maximal function. In fact, if CJ all < C

J lf(x)l>a/2

and so

f(x)dx

57

MULTIPARAMETER FOURIER ANALYSIS IIMf II L(log

f(log a) k-1 a

J

L)k-1
0. We have

J

f(x)dx < J f(x)dx< Y J

M&f(x)>Ca

k

UQk

f < Ca j IQkl < CaI{Mf > call .

Qk

k

This yields 00

ra

5 If(x)I(log+Msf(x))kdx
Cna

00

Cna

1

0. The nature of the real variable theory involved does not seem to depend at all on the dimension n. In marked contrast, it turns out that a study of the analogous operators commuting with a multiparameter family of dilations reveals that the number of parameters is enormously important, and changes in the number of parameters drastically change the results. Let us begin by giving the most basic example, which dates back to Jessen, Marcinkiewicz, and Zygmund. We are referring to a maximal operator on Rn which commutes with the full n-parameter group of dilations

(xl,x2, ,xn) - (Six1,82x2, ,Snxn), where Si > 0 is arbitrary. This is the "strong maximal operator," M(n), defined by

M(n)f(x) = sup 1 xER IRI

JR

If(t)I dt

where R is a rectangle in Rn whose sides are parallel to the axes. Unlike the case of the Hardy-Littlewood operator, M(n) does not satisfy IIxIM(n)(f )(x) > al I < a Ilf II I(Rn L

)

For instance when n = 2 and when f8 = 8-2X(Ixl Iail a llfll

L(Iog

L)n_t

(Q0)

The proof is strikingly simple. Define Mi to be the 1-dimensional maximal function in the ith coordinate direction. Consider the case

n = 2, which is already entirely typical. Let R be a rectangle containing the point (x t , 3E2)1 say R = I x J . Then

ffif(xt.x2)idxidx2 = III f(th ff(xi.x2)dx2)dxi

R

I

R

(2.1)

J

lf(xl,x2)ldx2 < Mx2f(xt,-2)

111

J so (2.1) is

< lIl

M2f(x1,x2)dxl <Mx1(Mx2f)(x1 ,x2)

60

ROBERT FEFFERMAN

Thus, for all (x ,x2) 'r QO' M(2)f(x1,x2)

aII < a'IIMX2fIIL1 (Q

cnIURkI 1

n-1

/

(2)

IleXp(I X'lk)

II'LI(B)
0 where Rs,t

r t t >' L c(s,t) i(s,t) E flx l -2'2J =[-2'2] -2 L2

where 0 is a function increasing in each variable separately, fixing the other variable. In other words, Zygmund next conjectured that since is a 2-parameter family of rectangles in R3, the corresponding maximal operator, which we shall call MZ should behave like the model 2-parameter 1

operator M(2) in R2: 11Mzf(x) > a, JxJ < 1 #{ < a 1{"L(1og L)(Ixl c}URkf

(2) 111

kllexp(L) < C

69


To prove this, order the Rk so that the z side lengths are decreasing. With no loss of generality, we may assume that IRkf1 ['Uk Rj11 < 2 (Rkl

,

that there are finitely many Rk and that the Rk are dyadic. (In fact, we may assume this because if IRI fR IfI > a for some R e 91 containing x ,

then there exists a dyadic R1 whose R1 (double) contains x such that IR' 1R1 +fj> C .) Now let R1 =R and, given R1, ,Rk, select Rk+l

as follows: Let Rk+1 be the first R on the list of Rk so that

1 fexP(x)dx 0 separately. Define a k-parameter family of rectangles

2,

0(tl,...,tk) +S6(t1,...,tk) Rt1 ,t2,...,tk i=1

and a maximal operator on Rn by

MJ)(f)(x) = t1 t 2

supsk >o

Rt1,...,tk 1

f R

lf(x+y)ldy t1,

,tk

Then lUxJlxI < 1,M,(D(f)(x) > all -q for some rl > 0. Let us list some properties of AP classes: (a) If p > 0 and w e AP then pw a AP with the same norm as w. (/3) If w e AP and S > 0 then w(3x) a AP with the same norm as w. AP, where 1 1, = 1. (y) If w e AP then w-1 /(P-1) a P

P

(8) If w e AP then w e A°°. In fact, if w e AP by (a) and (13) it is

enough to show that if IQI = 1 , and fQ w = 1 then IEI > 1/2 implies w(E) > r).

(For, in general if Q is arbitrary of side 8 and I E I > 1/2IQI , consider w(8x) on Q/8 and multiply w(8x) by the right constant p to have fl pw(Sx) = 1 . Then pw(Sx) on E/8 would have /8

[pw(Sx)](E/8) > [pw(bx)](Q/8)

w (E) >

rl

w(Q)

But by the AP condition, p-1

r)_1

< (f W-1/(P-1) JE

rw-1/(P-1)

C.

0. (C) From (E) it is immediate (see also (y)) that w c AP implies w c Aq for some q < p. (r)) If f is a locally integrable function in some LP space and 0 < a < 1 then (Mf )a c A 1 , i.e., M((Mf )a)(x) < C(Mf )a(x) (for w c A implies w c AP for all p > 1 ). To prove this let f e Lp(Rn) be given and a c (0,1). Let Q be a

cube centered at z, and Q its double. Then write f = XQf + XcQf = f1 + f0. We must show that

I M(f 1)a dx < CM(f )a(z )

IQI

Q

and

M(fo )adx

to .

We need to estimate A * q5t

(x) for

1t

2

10

x / CO. To do this let us make the following definitions. If R is a rectangle then R1, R2 will be its sides, so that R = R1 xR2 . Let

AI(x)

= I fR(x),

A?(x)

= L f R(x) . 91k. IR2l=2J

JR11=23

Then to estimate A * 6t1,t2(x) , since supp(q5t( -x)) C R(x; t) = S, in the definition of A we need only consider those fR for which R f1S For any such rectangle R c Rk , since R C (0, minimumr (sill 1`

where w denotes again

,

R il

1010

Then

=P

0.

125


IA

A * ctlt2x) =

I A * q5t1t2(x) 2i/Is2I 0, then u(X) has the limit f(Q) for almost every Q . The theorem is most easily proved by writing down a formula for the 2

solution, u(X) = fas P(X,Q) f(Q) do(Q) , where P(X,Q) = n Iin IX-Q The estimate now follows as an easy consequence of the Hardy-Littlewood maximal theorem. An analogous formula holds for the Neumann problem.

ELLIPTIC BOUNDARY VALUE PROBLEMS

135

This time, it is more difficult to obtain the estimates. One needs to use the Calderon-Zygmund theory of singular integrals, and the HardyLittlewood maximal theorem.

The case of the Laplacian in the ball is relatively easy because of the existence of explicit formulas for the solution. What should we do in the case of a general domain, where explicit formulas are not available? What should we do to study systems of equations? What happens to our solutions as the domains become less smooth? We hope to give a systematic answer to these questions in the rest of this paper. §1. Historical comments and preliminaries

(a) The method of layer' potentials for Laplace's equation on smooth domains.

DEFINITION. A bounded domain D is called a Lipschitz domain with Lipschitz constant less than or equal to M if for any Q e aD there is a

ball B with center at Q, a coordinate system (isometric to the usual

coordinate system) x'= (xi, , xn_I), xn, with origin at Q and a function (h: Rn-I R such that 4(O)=O, IOW)--0(0I <Mlx'-y'I and Df1B = (X =(x,xn):xn>o(x'){(1B .

If for each Q the function 0 can be chosen in CI(Rn-I), then D is called a C1 domain. If in addition, 74 satisfies a Holder condition of order a, IV(*')-V (y')I M1x'I ! satisfies ri n B CD. Thus, Lipschitz domains satisfy the interior and exterior cone condition.

The function 0 satisfying the Lipschitz condition 10(x')--*(y')I < Mlx'- y'l is differentiable almost everywhere and 170 e L00(Rn-I), 111741I,, M.

136

CARLOS E. KENIG

Surface measure a is defined for each Borel subset E C d D fl B by

f

a(E) =

(1 + IO95(x')12)I

/2 dx

E

where E* = ix': (x', O(x')) eE 1.

The unit outer normal to dD given in the coordinate system by /2

exists for almost every x'. The unit normal at Q will be denoted by NQ. It exists for almost every Q e dD , with respect to da. (NO(x'), -1)/(1 + lvq,(X,)I 2)1

In order to motivate the use of the method of layer potentials, we need to recall some formulas from advanced calculus, and some definitions. We will start with the derivation corresponding to the Dirichlet problem. We first recall the fundamental solution F(X) to Laplace's equation in Rn : AF = 8, where

n>2

1 (n-2)wnjX1n-2

F(X) =

1-

TIT

log jXJ

n

2

where wn is the surface area of the unit sphere in Rn. F(X) is the electrical potential in free space induced by a unit charge at the origin. with It provides a formula for a solution w to the equation Aa

( eCp(Rn),

w(X) = F *#(X) = fF(xYY)dv. Rn

It will be convenient to put F(X,Y) = F(X-Y). Notice that tYF(X,Y) _ 5(X-Y). The fundamental solution in a bounded domain is known as the Green function G(X,Y). It is the function on D xD continuous for X j Y and satisfying A G(X,Y) = 5(X-Y), X e D ; G(X,Y) = 0, X e D, Y e dD.


137

G(X,Y) as a function of Y is the potential induced by a unit charge at X that is grounded to zero potential on aD . The Green function can be obtained if one knows how to solve the Dirichlet problem. In fact, let uX(Y) be the harmonic function with boundary values uX(Y)laD = F(X,Y)IcD. Then, G(X,Y) = F(X,Y)-uX(Y) .

(1)

On the other hand, if we know G(X,Y), we can formally write down the solution to the Dirichlet problem. In fact, u(X) =

5u(Y)(XY)dY

r u(Y)AYG(X,Y)dY = JD

D

=

=

5Eu(Y)AyG(xv) - Au(Y) G(X,Y)ldY = D

f[u(Q)

(X,Q) -

(Q)G(X,Q)J da(Q) =

Q

Q

aD

=

Ju(Q) aN G(X,Q)da(Q) 4

,

aD

where the fourth equality follows from Green's formula. Thus, we have derived the formula

(2)

5f(Q) aN (X,Q)da(Q)

u(X) =

Q

aD

for the harmonic function u with boundary values f. The problem with

CARLOS E. KENIG

138

formula (2) is of course that we don't know G(X,Q). Because of formulas (1) and (2), C. Neumann proposed the formula

5f(Q) aN (X,Q)da(Q) _

w(X) =

Q

0 1 «n

IX--Q InI<X_Q,NQ> da(Q)

aD

as a first approximation to the solution of the Dirichlet problem, Au = 0

in D, uI

D=f.

w(X) is known as the double layer potential of f. First of all w is harmonic in D. Also, one can show that as X - Q e W, w(X) 2 f(Q) + Kf(Q), where K is the operator on aD given by

Kf(Q) = wn

J

0, a e (0,0.1), if Proposition

(M,e) holds, then Proposition ((1-a)M,1.1e) holds.

We postpone the proof of Proposition 3.1.13, and show first how Proposition 3.1.12 and Proposition 3.1.13 yield Lemma 3.1.11. Proof of Lemma 3.1.11. We will show that Proposition (M,a) holds for any M,e. Fix M,a, and choose R so large that if e(10M) is as in Proposition R

3.1.12, then (1.1)Re(10M) > E. Pick now aj > 0 so that

II (1-aj)=1/10

j=1

Then, since Proposition (10M, e(10M)) holds, by Proposition 3.1.12, applying Proposition 3.1.13 R times we see that Proposition (M,a) holds. We will not sketch the proof of Proposition 3.1.13. We first note that it

suffices to show that IINO(Du)IIL2(3D,da)

vector q(X) = q'(X)), where q'(X) =

X1

Our solution of (3.2.2) will

4nIXI3

be given in the form of a double layer potential, u(X) = Kg(X) _ faD (H'(Q)F(X-Q){g(Q)do(Q), where (H'(Q) r(X-Q))if = Sijge(X-Q) nj(Q) +

178

CARLOS E. KENIG

arig

(X-Q)nj(Q). We will also use the single layer potential u(X) _

Sg(X) = faD r(X-Q)g(Q)do(Q).

In the same way as one establishes 3.1.4, one has: LEMMA 3.2.3. Let Kg, Sg be defined as above, with g c L2(cD, do,).

Then, they both solve Au = Vp in D, and D-, div u = 0 in D and

D-. Also (a)

IIN(Kg)IIL2((?D,da)

(b)

(Kg)±(P)

(c)

IIN(VS0IIL2((3D,da)

=

5

(P) == +-.)

CII9IIL2(OD,da)

nl(P)g3(P)

nl(P)n)(P)

2

2

+ P.V. faD

(e)

(HSg)±(P)

'

± 2 g(P) - p.v. fdD IH (Q)F(P-Q)Ij(Q)da(Q)

+

aXl (Sg )j

(d)

III g IIL2(aD,aa)

aPi

F(P - Q) g (Q) d a (Q)

2 g(P) + P.V. faD {H(P)r(P-Q))g(Q)da(Q) ,

where (H(X)r(X-Q))ig = ni(x) aXIP (X-Q) - Sil qf(X-Q)nj(X). J

For the proof of this lemma in the case of smooth domains, see [25].

The proof of Theorem 3.2.2 (at least the existence part of it), reduces to the invertibility in L20D, da) of the operator 2 I+K, where Kg(P) - P.V. faD {H'(Q)T'(P-Q)jg(Q)da(Q). As in previous cases, it is enough

to show (3.2.4)

I2

I-K*)2L (aD,da)

+I(2

I+K*) 9

L2OD,da) 1

179


This is shown by using the following two integral identities.

LEMMA 3.2.5. Let h be a constant vector in Rn, and suppose that Du = Vp, div u = 0 in D, and that u,p and their derivatives are suitably small at co. Then,

f

hFnPaS

1

a

Sdo=2 J

hFaX F

1

aD

da-2

f

PnS

h

fa

da. F

3D

aD

LEMMA 3.2.6. Let h,p and u be as in 3.2.5. Then,

fhr

J hFnFp2da=2 aD

aNpda-2

3D

+2

fhrA i

aNda+

3D

ass fhns_.da. our

3D

The proofs of 3.2.5 and 3.2.6 are simple applications of the properties of u,p, and the divergence theorem. Choosing h = e3, we see that, from 3.2.6 we obtain

COROLLARY 3.2.7. Let u,p be as in 3.2.5. Then,

fP2do = 0. az

f

is

189


An arbitrary 1-form is written

u(z) = a(z)dz + b(z)dz

(1.8)

and we define the exterior differentials au = ab dz A dz , X = . dz A dz

(1.9)

.

Clearly du = au + au. Recall STOKES' THEOREM. If ci CC Rn is a bounded domain with smooth

boundary and u is a smooth form on ci then

fu = an

fdu. sZ

In our new notation, if ci C CI ti R2 and u is a 1-form as in (1.8), (1.9), then Stokes' Theorem becomes

fu= fau+u ale

11

(1.10)

(t

as

j)dzA.

a Now we can prove

THEOREM. If SZ C C is smoothly bounded and f is holomorphic on a neighborhood of ci then

f(z)

=

2ni

f.!2dC, all zci -z aci

.

190

STEVEN G. KRANTZ

Proof. Fix z e it. Let E < distance (z, aft). Define D(z,E) = 1C E C : IC-zI < El and it = it\D(z,E). We apply Stokes' Theorem to the 1-form u(C) _

f(t -z

dC on the domain it (note that u has smooth coefficients on a

ti

neighborhood of the closure of it, but not on all of it ). Thus, by (1.10),

fu(i= fdu=_j(2)dAd. ti

ti

ti ti

This last is 0 by (1.7). Thus, since aft = dl2 U 3D(z,e) (with suitable orientations), we have

fu() = aft

J

u(O

3D(z,E)

Parametrizing 3D(z,E) by C = z + Eeie, 0 < 0 < 21r, we obtain

f-!d= J

21r

('

f(z+Eei0)id9 2aif(z)

o

as E -. 0+. This completes the proof. 0 REMARK 1. In the proof of the theorem, if we assume that

f is smooth

but not necessarily holomorphic, then the right side of (1.11) does not

dC A dC. Thus the proof of the

vanish; instead it equals theorem yields

f(z)

2ai

ff

C -z

2ni

J

f(/a -z

A dJ.

This formula, valid for all f c C'(KI), will be valuable later on.

(1.12)

191


REMARK 2. As it stands, the method of Stokes' theorem will not generalis holomorphic with an ize to Cn. Namely, we used the fact that isolated singular point at z . In Cn , n > 2 , holomorphic functions never have isolated singularities. A more sophisticated approach will therefore be needed.

Discussion of III. If Q C C is a smoothly bounded domain and if E > 0, define S1E _ {z a iZ : dist(z, 00) > E 1. If E is sufficiently small, say

0 < E < E0, then 0E will also be smoothly bounded. Define

H2(fl) = f holomorphic on 0:

f If(C)I2ds(C) < 4

sup O<E<E

0

.

do E

(Here ds is the element of arc length.) That this definition is unambiguous is a technical matter (see (31, Ch. 81). We shall need PROPOSITION ([311). Each f e H2(fl) has associated to it a unique f e L2((3Q). The Poisson integral of f is f. Also

sup 0

f

If(()I2da( )I/2

I

II f

L2

(f)

E

If ITE: 09 E

i3f is normal projection then (f I

au

E)

=

(17 E)-I' f in L2 (do)

ffda an

for f,g a H2(SZ) .

(1.13)

192

STEVEN G. KRANTZ

BASIC LEMMA. if K C Sl is compact then there is a constant C = C(K)

such that

all f E H2(0) .

sup If(z)I < C1101 H 2s

(1.14)

zEK

Proof. Fix z E K. Let Et < U (distance (K, 311) ). Then

1f(z)1 °

1

z dC

2m

I

3D(z,el)

I

(Stokes) 1

dC

2771

C -z a1Z

el

< --L-

ffIds

- 2nE1

aloe 1

(Schwartz) 1/+

< C(E1)

J

If(y S)l2ds(OI/2

< C(K) IIf1IH2 .

Now (1.13) and (1.14) imply that H2(fl) is a Hilbert space. Fix

z c fl and define the functional Oz :H2(11), C

fF-f(z).

193


Then the lemma, with K = {z#, shows that ¢z is continuous. The Riesz Representation Theorem yields a unique kz f H2(il) such that

95z(f)=, all f eH2. In other words,

f(z)

ff())dsQ) ,

all f fH2(Sl), z eSl.

(1.15)

ail

Formula (1.12) is called the Szego formula and kz(c) = S(z, C) the Szego kernel (see [31, Ch. 11 for further details). REMARK 1. Formula (1.15) has the advantage of working on any smoothly

bounded domain (even in Cn ), and the disadvantage of being non-

explicit. As an exercise, check that when Sl = A C C and z = 0 then S(z,0 _= 21r. Then use Mobius transformations to calculate

for

any z f A. You will rediscover the Cauchy formula! §2. The Cauchy -Fanta ppie Formula

Now we begin to consider integral formulas in Cn. For purposes of differential analysis we introduce the notation

1 !a

a

=2

'

aa

1

2

(1)"ij

dzj = dxj +idyj , dzj = dxj - idyj,

j

It is easily checked that

T =1,

a_ zj =

= = 1 ,

j

and all other pairings are 0.

j = 1,...,n ,

194

STEVEN G. KRANTZ

If a =

are tuples of non-negative

integers (mufti-indices) then we write

dza = dza, A ... A dzak ,

di- 13

= dz16A ... A d Q

1

A differential form is written

u=Iaa,6 dzaACZ /3

(2.1)

a,f3

with smooth coefficients a

a/.

(If 0 < p,q e Z and the sum in (2.1)

ranges over jai = p, 1131 = q only, then u is called a form of type (p,q) .) We then define

au =

t dzj n dza F. dz

ru =

a# dzj A dza Adz c3i

a calculation (or functoriality), du = au + au. A C1 function f is called holomorphic if Ju = 0. (Note: this means that df = 0, (9zj

so f is holomorphic in the one variable sense in each

j

variable separately.) Finally, we introduce two special forms: if w = is an n-tuple of smooth functions then we define the Leray form to be 77(w) = I (-1)l+lwj A dw1 A

A dwj-1 A dwj+1 A - A dwn .

j=1

Likewise m(w) =dw1 A... Adwn.


195

We define a constant

W(n)= J

W(C)AW(C)

B(0,1)

Here B(z,r) = IC a Cn : IC -z I < r 1.

Now we may formulate a generalization of approach II in Section 1.

THEOREM (The Cauchy-Fantappi(; formula). Let Sl C Cn be a smoothly

bounded domain. Assume that w = wj

cC°°(SZ x SZ\A),

and n

I wj(z,4)'(Cj-zj)-1 on iZxSZ\A.

(2.2)

jj=1

If f e Cl(1) is holomorphic on 0, then for any z e 11 we have

f(z)

_W (n)

f

1(i) i w) A

W (C) .

do

Before proving this result, we make some detailed remarks. REMARK 1.

In case n = 1, then w = wl =

1

4-z 1

rl(w)_---,

nW(n)

C-z

The Cauchy-Fantappie formula becomes

f(z)

m

f do

which is just Cauchy's formula.

f(C) dC, z

(of necessity). So

_2ni 1

(2 . 3)

196

STEVEN G. KRANTZ

REMARK 2. As soon as n > 2 , the condition (2.2) no longer uniquely determines w. However an interesting example is given by

1-z1

(wi(z,0,...,wn(z,0)

w(z,0 _

.,

_

Cn-n

.

I.

Let us calculate what the theorem says for this w in case n = 2. Now r!(w) A c(C) _ (w 1dw2-w2dw 1) A dC1 Ad C2

-

2dc1 + w I

w1

aa1

- w2

1

dC2

aC2

w2

a 1

1A

1

dC, A dC2

(X2

which by direct calculation

- 2z IC -z

q dc-1 + IC _z q

A dC1 A dC2

Thus, by the theorem, we get a form of the Bochner-Martinelli formula:

f(z) = z(2)

4=z2)

5 f(O

IC-zI

A dC1 Ad

2

for f c C 1(Sl), holomorphic on Q.

Now we turn to the proof of the Cauchy -Fa ntappie formula. For sim-

plicity, we restrict attention to n = 2 . Let 2

a=(a1,a2) cC00(SlxSZ\O): 1 aj(z,c) (cj-zj) =1 j=1

.


197

If al,a2 E `.f then we define

B(a',a2) =det(a',dal)

_I

E(a)ao(I) A

aES2

We claim that B has three key properties:

(1) B(a', al) does not depend on aI ; (2) c3CB(a1, a2) = O ; (3)

If a I, al, RI, R2 E f then B(a1, al) - B((31, R2) is J exact.

Assuming (1), (2) and (3) for the moment, let us complete the proof (note

that (1) is used only to prove (3)). Letting wI

aI =a2

c EJ

=

(C2-z2)/K zI2

we have (observe that B(at,a2) = q(w) )

J f(())7(w)AW(C)=

ff()B(ata2)

A w(C)

do

ag

Note that, by (2),

dC(f(() B(at , a2) A W(C)) = a.f(C) A B(a1, a2)

=0+0=0.

A Ev(C) + f(C) dCB(a1, a2) A W(C)

(2.5)

198

STEVEN G. KRANTZ

Hence, letting 0 < e < dist (z, SO), we have by Stokes' Theorem that (2.5)

=

f(C)B(a1,a2) A 0)(C)

J

aB(z,E)

f f(C)IB(al,a2)-B(j61, 62)1 Atd(C)

(2.6)

a6(z,E)

+

B(j6 1,62)AW(C).

J aB(z,E)

But the first integral

J

=

f(()dA A W(C)

aB(z,e)

(for some A, by (3))

= f d(f(() A A A W(C)) 3B(z,e)

(2.7)

=0,

by Stokes' Theorem, since a9B(z,E) = 0. Thus (2.5)-(2.7) give

f Al which, by (2.4),

f f(C)B(j61,,62) A W(C) 3B(z,E)


=

A W(C)

J

I C-z l4

3B(z,E)

=

1 E4

f (C) rl(- z) A W(C) .

J 3B(z,E)

=13B(z,E) J f (z) rK -) A 0)(t) + ((E) 4

(Stokes)

4f(z)

= 64

2((-z)Aw(C)+O(E)

J

B(z,E)

= 4f(z)E4

2(a(J)Au(C)+0(E)

J B(0,1)

= f(z) 2W(2) +

O(E)

Letting E - 0 yields the desired result. 0

We conclude this section by proving (1)-(3). For (1), we have

B(a1, a2) = det

1

(C1-z1)(C2-z2)

det

199

STEVEN G. KRANTZ

200

which, by adding row 2 to row 1,

det

1

(C1-zl)(C2-z2)

aC ((C2-z2)a2)

((C22)a2

_

(Cl

z1) ? a2

This calculation is correct for C1

z2. The full result follows

z 1 , C2

by continuity.

For (2), imitate the proof of (1). For (3), use (1) to write B(a1, a2)

- B((31, p2) = B(a1, a2-P2)

ai (ai-bi) = det a21 'j (a 2-b2

Now an easy calculation, as in (1), shows that this last equals aA where

A = det a2

a2

2

This completes our discussion of the Cauchy-Fantappie formula.

§3. Introduction to the a problem One of the principal problems in complex function theory is the construction of holomorphic functions with specified properties. In one dimension, there are a number of highly developed techniques: Runge and Mergelyan theorems, power series, infinite products, integral formulas, and so on. In several variables, these techniques are either unavailable, much less useful, or much less accessible.


201

The two most prevalent techniques for constructing functions in several complex variables are sheaf theory and the inhomogeneous Cauchy-Riemann equations. The latter interact strongly with the subject of integral formulas, and in any case are a more flexible technique than the former. To these we now turn our attention. n The setup for our study is that, for a given form f = ,SI f j(z) dz j , we

seek a function u such that au = f . Notice that since 0 = d2 = a2 + ;72, linear independence considerations yield that '2 = 0. + Hence au = f necessitates of = 0. A simple calculation shows that, for n > 2 , this compatibility condiaft

n=1

af k

all j, k. Notice, however, that when

vkj the condition df = 0 is always vacuously satisfied. This differ-

tion is equivalent to

=

,

ence can be explained in part by the fact that the equation au = f is

really n equations

f

one unknown (namely u ). For n > 1

1 the system is then "over-determined" and a compatibility condition is necessary. For n = 1 the system is not over-determined. The three basic considerations about a PDE are existence, uniqueness and regularity. It is easy to check that a is elliptic on functions in the interior of a given domain; hence, if u exists, it will be smooth whenever f is (we will see this in a more elementary fashion later). So interior regularity is not a problem. Also, since the kernel of 3 consists of all holomorphic functions, uniqueness is out of the question. So, for us,

existence is the main issue. The following example shows that the compatibility condition of = 0 does not by itself guarantee existence of u. EXAMPLE. Let S1 C C2 be given by 11 = (B(0,4)\B(0,2)) U B ((2,0), 2/

202

STEVEN G. KRANTZ

Let U = B ((1,0), 2) and V = B ((1,0), 4) as shown. Let 0 e C° (U) satisfy

--1 on V. Finally, let

Then f is smooth and j 'closed on Q since

zl 1

is well-defined and

holomorphic on supp (drl) fl 11. If there existed a u satisfying Al = f

then the function h =_ u

zt-1

would be holomorphic (Jh=0) on

cl\IB (1,0),41 n 1zt=11 . But u would necessarily be smooth near (1,0) (since f is) hence h has a singularity at, for instance, (1,0). Thus we have created a function holomorphic on B(0,4)\B(0,2) which does not continue analytically to (1,0). This contradicts the Haitogs extension phenomenon (an independent proof of this phenomenon will be given momentarily).

C1

Now that we know that du = f is not always solvable, let us turn to an example where it is useful to be able to solve the a equation.


203

EXAMPLE. Consider the following question for an open domain it CC Cn :

(3.1)

If w = fl fl 1zn = OI' 0 and if g is holomorphic on co (in an obvious sense), can we find G holomorphic on ft such that GIC0 = g ?

1 is the unit ball, then the trivial extension g(zi,,zn_1,0) will suffice. However if 0 = B(0,2)\B(0,1) C C2 then g(z1,0) = 1/z, is holomorphic on a) but could not have an extension G (else the Hartogs extension phenomenon would be contradicted). o If

THEOREM. Suppose that w C Cn is a connected open set such that whenever f is a smooth a-closed (0.1) form on fZ then the equation au = f has a smooth solution. Then the answer to (3.1) is "yes."

Proof. Let u: Cn - Cn be given by (z1,..,zn-1,0). Let B = ;z f l : nz / col. Then B, w are disjoint relatively closed subsets of fZ, so there is a C°° function 4 on fZ such that 0 = 1 on a relative neighborhood of co and = 0 on a relative neighborhood of B. ti Define F on fZ by ti

c(z) f(n(z))

if

0

else.

z e supp q

F(z) _

ti Then F gives a C°° (but certainly not holomorphic) extension of f to Q.

204

STEVEN G. KRANTZ

We now seek a v such that F + v is holomorphic and F + vIw = f. With this in mind, we take v of the form zn u and we want d(F+v) = 0 or

=0. Now f o n is holomorphic on supp (b and zn is holomorphic so all that remains is

(f°n) - zn' 3u=0 or

n) zn

(3.2)

The critical fact is that, by construction, 4 = 0 in a neighborhood of n f1

{zn=0# so the right-hand side of (3.2) is smooth on Q. Also it is

easily checked to be 3 closed. Thus our hypothesis is satisfied and a ti u satisfying (3.2) exists. Therefore F - F +v has all the desired properties.

Our two examples show that solving the 3 equation is (i) subtle and (ii) useful. Thus we have ample motivation to prove our next result. LEMMA. Let (he CS(C) , k > 1 , and define f = (z)dz. Then u(z)

=

2i

dC A dC

J J ; -z C

satisfies u e Ck(C) and 3u = f . Proof. We have


0 "U

az

205

a (-L ff 2ni

)g-

C

2ni

f(a/)(+z) do A dC

f C

1

2ni

II

w d6 (o d A dC

D (0, R)

where D(0,R) is a disc which contains supp 0. We apply Remark 1 of II in Section 1 to 0 on D(0,R) to obtain that the last line (C)dC 2ni

J

z

aD(0,R)

The integral vanishes since 0 = 0 on (91)(0,R), hence au = f . Observe finally that u e Ck by differentiation under the integral sign. o REMARK. In general, the u given by the lemma will not be compactly supported. Indeed if f f 0 0 and if u were compactly supported (say u C D(0,R) ) then a contradiction arises as follows:

0=

J aD(0,R)

udC=

J D(O,R)

The supports of solutions to the a problem explain many phenomena in one and several complex variables. We explore this theme later. Meanwhile, contrast the Lemma and Remark with the following result.

STEVEN G. KRANTZ

206

THEOREM. Let n > 1 and let D(z) _ q5 1dz 1 +-.. + (kndz 1 be Xclosed on Cn and suppose each (kj CC(Cn). Then for any 1 < j < n the of

function 1 uj(z) _ - 217i

ff

dC A dC

(kl(z

C_zj

C

satisfies uj a Ck(Cn) and aut = 40. Moreover

uj = of for all j,

Proof. Fix 1 < m < n. We need to check that auj fim m = j then the result follows from the lemma. d0j

compatibility condition

=

Sam

-m azj

= (kj

uj(z)

rn

µ0 > 0.

VECTOR FIELDS AND NONISOTROPIC METRICS

285

We are now in a position to define balls in terms of a single exponen-

tial map. For

E Al set 2n

ajXj+yT (c), where

77co-QIrl=exp 1

IajJ<S, 1<j 0

X2+Y2 2 there is a constant Am so that if J q(z) is a real polynomial of degree at most m with a

(0) = 0,

az

0 < j < m , and if AO(z)>O for JzI 0. The general result now follows by dividing

a general polynomial 4 by I laza(90 ( 0)1

.

Finally, in order to show sup IQ.0g(ei9)J < Ct(S)

when 0 is a polynomial of degree < m, it again suffices to check this for S = 1

.

But sup IQ.08(e1O)I = 111.0111

0

is a norm on this space of polynomials. Hence III(bPHH < CA(0,1) < Ct(1).

As an easy corollary of the theorem we obtain an estimate for the Szego kernel S(z,C) for the domain 1 on the diagonal z = C. Recall

that the orthogonal projection S of L2(af) onto H2(1Z) is called the Szego projection, and formally, this projection is given by integrating against a kernel: Sf(z) =

ff()S(zs)da() 0 il

,

z e iZ .


297

In fact S(z, J) = E 0j(z)0j(C) where (0j! is a complete orthonormal basis for H2(H), and this series converges uniformly on compact subsets of SZ x Q. (See Krantz [l0a], Chapter 1, for further details.) Now it is easy to check that for z f SZ S(z,z)1 12 = sup IF(z)I

where the supremum is taken over all F e H2(1) with by our theorem, if F E H2(SZ),

IF(z)I
0, 1 = {(71,r)lr),r) a 2, the space


L2(e4rr(rlx+rb(x))dx)

301

contains the constants. Let P. be the projection

of L2(e4n(r)x+rb(x))dx) onto the constants, if (,l,r) a E, and let otherwise. Let

P'7,7

=0

P: L2(e4n(rlx+rb(x))dxdrldr) -, L2(e4n(rlx+rb(x))dxdrldr)

be defined by Pg(x,rl,r) =

where g,7 r(x) = g(x,rl,r). It is then easy to check that P is an orthogonal

projection whose range is precisely the null space of dx on L2(e4n(rlx+rb (x)) dx drl dr) . Thus if we set P

.` -IM

PMtA

then P is the orthogonal projection of L2(dxdydt) onto the null space

of L. Now if (rl, r) e

1 Pr1,r g = < 1,1 > 00

r g(r)e4n(rlr+rb(r))dr 00 00

r e41r(Tlr+rb(r))dr

Thus 00

P77,rg(x) = J

g(y) It 7,r(Y)dy

ALEXANDER NAGEL

302

where

Kll,r(x,Y) =

Thus if f f L2(R3, dx dy dt)

Pf(x,y,t)

If

f(r,s,u)S((x,y,t); (r,s,u))drds dy

where

S((x,Y,t); (r,s,u)) =

ff K,?,r(x,r) dry dr

00

fe

00

e21rq((x+r) + i(y-s))

27rrt(b(x)+b(r))+i(t-u)]

foe

0

00

f

e41r(nr-rb (r)) dr

-00

This is the kernel we have to estimate. We begin by estimating the inner integral. For r > 0, set 00

e21r-q(,k+it)

F A+it r -00

r e4n[r7r-rb(r)]dr -00

d

di

dr.

303


and 77 by r) + rb'(2) it follows that

Then replacing r by r + 2

F(A+it, r) _ I_

2rrr 2b e

(2)

(`]

+itb'

00

e2ni77td77

f+

/1

00

-00

J

2(2)1J

2(r+)+rbe 47rl77r+rrb ' (-rb LLL

-00

Let G(r) = rrb'(2) - rb (r+2) + rb(2)

.

Then

G'(r) = rb'C2) - rb'(r+2)

G"(r)= -rb"(r+2) Since G(0) = G'(0) = 0, we have m

G(r)

1 b())

r j=2

1

( -) r)

Hence

[2b(3) + itb'(2)1

277r

F(a+it,r) = e

L

JJ

00

e2rrjr7td,7

f J

-00

00

4n nr-r

m

f

1 b (j) ( ) r7

Now choose p = µ(A, r) so that

I In

j=2

1 b(j)(a>rµj12 2

=1

1!

and in the last integral, make the change of variables r - gr,

77

77.

ALEXANDER NAGEL

304

Then

F(a+it, r) =

e2ar [2b()

+

e2ai7,(

itb "%]

rl

A-2

M

477 r)r-

.j

e

-00

j

where ai = 1

dr

00

Ia.2 = 1 . and hence Ym 2 7

2

We now make two observations. First, in terms of size, m µ(X, r)

1 ti Y z

Ib(j) ( l \2/ I rl

/i

This is clear from the definition of µ . Second, the collection of functions f4a r - a i m 2 e

6a (n)

dr

m

where a =

am), Y-IajI2 = 1

,

and

r

'I' air) is convex, is a com2

pact set of functions in the Schwartz class S(R). Thus e2rrr L2b

F(A+it, r) =

rj'l

+ itb (2/] t1o, r)-1

\/

.

ea 1

From this, one can make estimates on the size of F(A+it,r) and its derivatives. Finally we have 00

S((x,y,t); (r,s,u)) 5 0

e-2ar[b(x)+b(r)+i(t-u)]F(x+r+i(y-s),r)dr


305

and we can use the estimates on F to estimate S. A consequence is, for example: IS((x,y,t); (r,s,u))l
1 for some fixed k, dxk

and we wish to obtain an estimate for f b a

which is independent

of a and b. Then a simple scaling argument shows that the only possible estimate for the integral is 0(A-Ihk). That this is indeed the case goes back to van der Corput.

PROPOSITION 2. Suppose 0 is real-valued and smooth in [a,b]. If I0tkl(x)I > 1 , then

b

f a

holds when

ea(bi"ldx

2

(ii) or k = 1 , if in addition it is assumed that 0'(x) is monotonic. Proof. Let us show (ii) first. We have b

b

e'4dx

=f

a

b

r e1X4t-D(1)dx + i1

D(eiAO)dx = -

b e

ix 4v a

a

a

The boundary terms are majorized by 2/A, while b

r ei

b

r eat i-j1 x (-1) dx

t_D(1)dx

'0,

aJ

a

a

b

0, (1-ik)-y'4/2

where we have fixed the z-'/r4/2 in the plane slit along the negative halfprincipal branch of axis. The power series expansion of (w-i)-`/'`e/2 (which holds for lwl < 1 ), then gives the desired asymptotic expansion (1.4). Step 2. Observe next that if -q e Co and a is a non-negative integer,

then 00

f

(1.5)

eax2xe77(x)dx < AX-1/2

/2

CIO

To prove this let a be a C°° function with the property that a(x) = 1 for lxl < 1 , and a(x) = 0 when lxl > 2 , and write

J

eikx2xe>)(x)dx = feiXx2xfrl(x)a(x/E)dx + J eikx2xeri(x)(1-a(x/E))dx

The first integral is dominated by CEe+1 . The second integral can be written as

5e2(tr)N [xf-q(x) [1-a(x/E)]]dx with tDf =

1

dx

f

A simple computation then shows that this term

.

is majorized by CN ,\N

f

lxle-2N dx

= C' \-NEe-2N-1 N

lxl>E

if

e - 2N < - 1 . Altogether then the integral in (1.5) is bounded by

.

E. M. STEIN

314 CN#E/+1 +A-NCQ-2N+1 I

(with N > e 21) A similar

,

(/but

and we need only take e = .1-112

to get the conclusion (1.5). simpler) argument of integration by parts also shows

that

fe2ex)dx =

(1.6)

0(A-N), every N > 0

whenever e c 8, and 6 vanishes near the origin. Step 3. We prove the proposition first in the case q(x) = x2 . To do this write

fe2r(x) dx = feiAXe x2(ex2 is a Co function which is each N, write the taylor expansion where

1

on the support of Vi. Now for

N

eX2c

(x) _ Y` bjxj + xN+IRN(x) = P(x) + XN+1RN(x) jj=0+

Substituting in the above gives three terms 00

f

(2)

eiAX2 e_

C2

xi dx

00

00

(b)

(C)

J00

fe2P(x)e2(1_(x))dx.

OSCILLATORY INTEGRALS IN FOURIER ANALYSIS

315

For (a) we use (1.4); for (b) we use (1.5); and for (c) we use (1.6). It is then easy to see that their combination gives the desired asymptotic exeiAx2

c/i(x)dx . pansion for f Let us now consider the general case when k = 2. We can then write q(x) = c(x-x0)2 + 0(x-x0)3 with c 4 0 and set q(x) = c(x-x0)2[1+e(x)] , 1 where a is a smooth function which is 0(x-x0), and hence when x is sufficiently close to x0. Moreover, q'(x) 0, when x x0 but x lies sufficiently close to x0. Let us now fix such a neighborhood of x0, and let y = (x-x0)(1+e(x))I12. Then the mapping x y is a

diffeomorphism of that neighborhood of x0 to a neighborhood of y = 0,

and of course cy2 = 4(x). Thus

fei'(o)clr(x)dx = reilcy2

(y)dy

e Co if the support of 0 lies in our fixed neighborhood of x0. The expansion (1.3) (for k=2 ), is then proved as a consequence of the special case treated before. with

t

REMARKS:

(1) The proof for higher k is similar and is based on the fact that

f

eixxke xkxedx = ck,Q(1- i'\)-R+I)/k

.

0

(2) Each constant aj that appears in the asymptotic expansion (1.3) depends on only finitely many derivatives of and Vi at x0. Note \r1-T(-i0'(x0))-I"2Vi(x0). Similarly e.g. that when k = 2, we have a0 = the bounds occurring in (1.3') depend on upper bounds of finitely many

derivatives of 0 and Vi in the support of t,, and a lower bound for 0(k)(x0).

sli ,

the size of the support of

316

E. M. STEIN

References : The reader may consult Erde1yi [8], Chapter II, where further

citations of the classical literature may be found. 2. Oscillatory integrals of the first kind, n > 2

Only some of the above results have analogues when n > 2, but the extension of Proposition 1 is simple. Continuing a terminology used above we say that a phase function (k defined in a neighborhood of a point xa

in R° has x0 as a critical point, if (Vg)(xo) = 0. PROPOSITION 4. Suppose cu c Co(Rn), and 0 is a smooth real-valued

function which has no critical points in the support of Vi. Then

I(A) =

r

O(A-N), as A

for every N > 0 .

Rn

Proof. For each xo in the support of ci, , there is a unit vector 6 and a small ball B(x0), centered at x0, so that (e,vx)o(x)> c > 0, for x c B(xp). Decompose the integral fek'0(x)t4(x)dx as a finite sum

I fe(')clFk(x)dx n

where each c/ik is C°° and has compact support in one of these balls. It then suffices to prove the corresponding estimate for each of these

integrals. Now choose a coordinate system xt,x2, ,xn so that xI lies along 6. Then

fe('c)ci/k(x)dx

'...,xn)Vk(xt,...,xn)dxI

=f(fe"(x

I

dx2,...,dxn .

But the inner integral is 0(A-N) by Proposition 1, and so our desired conclusion follows.


317

We can only state a weak analogue for the scaling principle, Proposition 2; it, however, will be useful in what follows.

PROPOSITION S. Suppose 0 t Co , q is real-valued, and for some multi-index a, at > 0,

(d)a

(x),>1

throughout the support of t# . Then

(2.1)

f eikO(x)tp(x)dx < Ck('A)' A-I /k (IIqj .L00+ II174IIL1) Rn

with k = ial , and the constant ck(¢) is independent of A and t/i and remains bounded as long as the Ck+1 norm of 0 remains bounded. Proof. Consider the real linear space of homogeneous polynomials of

degree k in Rn. Let d(k,n) denote its dimension. Of course {xa}1a`=k is a basis for this space. However it is not difficult to see that there are d(k,n) unit vectors 1` (1), (2), , e (d(k,n)) so that the homogeneous polynomials ( (J) x)k , j - 1, , d(k,n) , give another basis. This means that if

0(x0I axa

>

for some lal = k , there is a unit vector e = ;(x°), so that I(e, px)kO(xO)l > ak, with ak> 0 . is bounded we Moreover since we can assume that the Ck+1 norm of can also conclude that I(e, p,)(k(x)l > ak/2 whenever x c B(xo), where

B is the ball centered at x of fixed radius. (The radius of B can be taken to be a small multiple of the Ck+1 norm of (b.) Next choose an

318

E. M. STEIN

appropriate covering of Rn by such balls of fixed radius, and a corresponding partition of unity, 1 = F 173(x), with 0 < ijj < 1 , {pnj{ < bk , and each rlj supported in one of our balls. So k

fe'Acj'dx=

feuic

feuicci7j dx

lij dx.

To estimate Jei-k'ipj dx , with a determined as above, choose a coordinate system so that x1 lies along C. Then

fe'1c1rjdx

xn)dx1) dx2,...,dxn

=J W

For the inner integral we invoke (1.2) giving us an estimate of the form -1 /kA-1 / ckak

a

f'

(x1,...,xn) dx1 L

J I1

f

(x1,...,xn) dx1 I

A final integration in the other variables then leads to (2.1).

REMARK. Let us note that in R2 if q(x) = x 1 x2 , the above proposition gives no better than a decrease of order 'k-1 /2 , while the asymptotics of the proposition below shows that the true order is 0 . Let us go back for the moment to the case of one dimension. If

95

has a critical point at x0, and q' does not vanish of infinite order at x0, then after a smooth change of variables (b can be transformed to a simple canonical form , with '(x) = ±xk (for x near 0 ). There is no analogue of this in higher dimensions, except for k = 1 and in a special case corresponding to k = 2. To the asymptotics of the latter situation we now turn.

319


Suppose 4 has a critical point at xO. If the symmetric n x n matrix is invertible, then the critical point is said to be nonr7x (((

degenerate. It is an easy matter to see by the use of Taylor's expansion

that if xO is a non-degenerate critical point, then in fact it is an isolated critical point. PROPOSITION 6. Suppose q(xo) = 0, and (A has a non-generate critical point at x0. If i/r a Co and the support of q1 is a sufficiently small neighborhood of x°, then

e"(x)0(x)dx

(2.2)

A-n/2

ai X-j , as

A

j=o Rn

where the asymptotics hold in the same sense as (1.3), (1S).

Note. Again each of the constants aj appearing in the asymptotic expansion depends on only finite many values of derivatives of (A and 0 n / (-iµj)-1/21 O(x0), where at x0. Thus e.g. ao = (an/2 R i=I

1A1' µ2' "' µn are the eigenvalues of the matrix

a2(A(x0)

Similarly

2 axk 1

each of the bounds occurring in the error terms depend only on upper

bounds for finitely many derivatives of 0 and 0 in the support of Vi , the size of the support of

and a lower bound for det

a2(b(xo)

xk

The proof of the proposition follows closely the same pattern as that of Proposition 3. First, let Q(x) denote the unit quadratic form given by xn, where 0 < m < n , with m fixed. Q(x) The analogue of (1.4) is 00

(2.3)

r eikQ(x)e Ix12xedx s%

Rn

ti

A-n/2-JeJ/2 5" c)(m,e)),-i

j=0

E. M. STEIN

320

with P =

a multi-index, Pl =

El

and

P

xn ; also note that if one Pi is odd then (2.3) is identically zero. To prove (2.3) write it as a product xP =

n

(f°°

('

eax2x_x2 J Pdx

J j=1

n

and expand the function

11

j=1

e

2

P x J dx

(1

-oo

-lh-P./2

(1/X+i)

(for large A ) in a power

series in 1/X . The analogue of (1.5) is the statement that

f

(2.4)

ei,\Q(x)xPrl(x) dx
tsince Then

i

j=1

2n

Ix12 y, and the smaller cases hI _ {xi Ix)I2 >

1

J

= Rn we can find functions

n

Ix121

i = 1, , n, each

lZ , J

homogeneous of degree 0, and Co away from the origin, so that n

1=

fI.(x), x = 0, with ft. supported in F.. Then we can write

j=1

fe V Q(x)xPn(x)dx

fe'AQ(x)xP,r(x)SZ(x)dx

= i

In the cone C'i one uses integration by parts via pie"\Q(x) = eixQ(x) with Di(f) _- I Of NT T 1

7

This, together with the fact jxii > 1 Ixl in F., and I (tD.)Nj .(x)1 J2n

.


CN Ixl-2N

321

allows one to conclude the proof of (2.4) in analogy with that

of (1.5). A similar argument also show that whenever

e 8 and e vanishes

near the origin, then

(2.5)

1

e 4Q(x)e(x)dx

=

O(A-N), for every N > 0

.

We then combine (2.3), (2.4) and (2.5) as before to obtain the asymptotic formula (2.2) in the special case when c (x) = Q(x).

To pass to the general case one is then fortunate to be able to appeal to the change of variables guaranteed by Morse's lemma: Since cb(xo) = 0, (V )(x°) = 0, and the critical point is assumed to be non-degenerate, there exists a diffeomorphism of a small neighborhood of x0 in the ym x-space to the y-space, under which cb is transformed 2 Y2. ym+1 yn' Observe that the index m is the same as that of the form corresponding to

1 .92`b(xO) 2 oPx j(?xk

-

References. For a proof of Morse's lemma see Milnor [201, §2.

D. Fourier transforms of surface-carried measures Let S denote a smooth m-dimensional sub-manifold of Rn (not

necessarily closed). We let do denote the measure on S induced by the Lebesgue measure on Rn, and fix a function 0 in Co (Rn) . Consider now the finite Bore] measure dµ = O(x)do on Rn, which is of course carried on S . The problem we wish to deal with is that of finding estimates at infinity of the Fourier transform of µ, i.e. dµ(e). We shall consider two cases of this problem.

(1) Suppose first dim S = n -1 , and S has non-zero Gaussian curvature at each point. By this we mean the following: Let x0 be any point of S,

E. M. STEIN

322

and consider a rotation and translation of the underlying Rn so that the point xO is moved to the origin, and the tangent plane of S at xD becomes the hyperplane xn = 0. Then near the origin (i.e. near x0 ) the ,Xn_1) with ¢ e C*, surface S can be given as a graph xn = and 0(0) = 0, (%)(0) = 0. Now consider the (n-1) x (n-1) matrix 1

a2d?(o)

xk

_

are called the princi-

Its eigenvalues

/

1 0, if the support of 0- is a a sufficiently small neighborhood of the origin. Hence for the n in region 3° we may use Proposition 4 to conclude that the left-side of (3.2) is actually 0(11-N) for every N . The proof of Theorem 1 is therefore concluded.

i=1

l

4k(x)r/n+1 ,

REMARK. We have used only a special consequence of the asymptotic formula (2.2), namely the "remainder estimate" analogous to (1.3') when

N = r = 0. Had we used the full formula we can get an asymptotic expansion for dµ(e); its main term is explicitly expressible in terms of the Gaussian curvature at those points x c S , for which the normal is in the

direction 6 or -6. (2) We shall now consider the problem in a wider setting. Here S will be a smooth m-dimensional sub-manifold, with 1 < m < n-1 , and our assumptions on the non-vanishing curvature will be replaced by the more

E. M. STEIN

324

general assumption that at each point S has at most a finite order contact with any hyperplane. We shall call such sub-manifolds of finite type. (These have some analogy with the finite-type domains in several complex variables, which are also discussed in Nagel's lectures [21].) The precise definitions required for our considerations are as follows. We shall assume that we are considering S in a sufficiently small neighborhood of a given point, and then write S as the image of mapping Rn, defined in a neighborhood U of the origin in R n. (To get a smoothly embedded S we should also suppose that-the vectors , ,

'

axI ax2

, NM

are linearly independent for each x, but we shall not need that assumpin Rn . tion.) Now fix any point x0 c U C Rm , and any unit vector does not vanish of We shall assume that the function infinite order as x --)x0. Put another way, for each x0 E U and each unit vector 17, there is a multi-index a , with 1 < lal , so that 71

(3)a (fi(x)

0. Notice that if (x, ii') are sufficiently close to

(x 0, rj), then also

(d)a(x.).,).I x=x

,

'0.

The smallest k so that for

each unit vector q then 3a , jal < k, with UJL

a

(O(x)

0

77)1

0 will

x=x

be called the type of (A at x0. Also if UI is a compact set in U,

the type of 0 in UI will be the least upper bound of the types for x0 in Ut . THEOREM 2. Suppose S is a smooth m-dimensional manifold in Rn of finite type. Let dµ = Vida, with t/i E Co(Rm). Then

du(e)l0,

and in fact we can take e = 1/k, where k is the type of S inside the support of

t/i .

Proof. By a suitable partition of unity we can reduce the problem to showing that


J

ei (x)*

(x)dx =

325

0(IeI-I /k)

R`"

ti with (A as described above, and the support of 0- sufficiently small. Now we can write l; = AY), with Jill = , and X > 0 . Then we know that 11

there is an a , with tat < k, so that ()t2qs(x).?1

0, whenever x is

in the support of t (once the size of the support has been chosen small enough). Thus the conclusion (3.3) follows from (2.1) of Proposition 5.

References. Theorem 1 in its more precise form alluded to in the remark goes back to Hlawka [14]. See also Herz [13], Littman [18], Randol [25], and Hormander [16]. When S is a real-analytic sub-manifold not contained in any affine hyper-plane, then it is of finite type as defined above. For such real-analytic S estimates of the type (3.3) were proved by Bjork [2]. 4. Restriction theorems for the Fourier transform

The Fourier transform of a function in Lp(Rn) , 1 < p < 2 is most naturally thought of as an LP function (via the Hausdorff-Young Theorem) and so at first sight it is viewed as defined only almost-everywhere. This impression is further supported by the case p = 2 , when clearly the Fourier transform can be completely arbitrary on any given set of zero Lebesgue measure. It is therefore a noteworthy fact that whenever n > 2 and S is a sub-manifold of Rn (with some appropriate "curvature")

then there exists a p0 = p(S), p0 > 1 , so that every function in LP, 1 < p < p0 has a Fourier transform restricting to S (i.e. with respect to the induced measure on S ). Let us make this precise. Suppose that S is a given smooth sub-manifold in Rn, with da its induced. Lebesgue measure. We shall say that the LP restriction property holds for S , if there exists a q = q(p), so that the inequality

E. M. STEIN

326

1 /q

(4.1)

JSo

If(e)lgda(e)

< Ap,q(So) IIf tIp

holds for each f c 8, whenever So is an open subset of S with compact closure in S.

THEOREM 3. Suppose S is a smooth hypersurface in Rn with non-zero Gaussian curvature. Then the restriction property (4.1) holds for

1 0 and inequality (4.2)

e Co C. It will suffice to prove the

f

1 /2

o

AIIfII p

Rn

for PO = 2n + 32, and f e 5; the case 1 < p < p0 will then follow by

interpolation.* By covering the support of V1 by sufficiently many small open sets, it will be enough to prove (4.2) when (after a suitable rotation and translation of coordinates) the surface S can be represented (in the Now with dµ = V'da we support o f Vi) as a graph: en have that

J )?(e)I2dµ = J f(e)f(()d, = J T(f)(x)13dx where (Tf) (x) _ (f * K) (x), with

K(x) =

fe

*In fact the interpolation argument shows that we can take q so that (4.1) holds with q = (n+1} p'' which is the optimal relation between p and q.

327


Thus (4.2) follows from Holder's inequality if we can show that (4.3)

IIT(f )Ilpo < AIIfIIpo

where po is the dual exponent to p0. To prove (4.3) we consider the function Ks (initially defined for Re(s) > 0 ) by

(4.4) Ks(x) =

e s2 1'(s,'2)

fe

1+s

2nix

)I

7

W(e')de

n(

Rn

by e'; we have set (C') =

Here we have abbreviated

+Iph(e')I2)I/2 , so that (C')dC'= dµ; also function which equals

ri

is a Co(R)

near the origin. Now the change of variables en -+ en + 0(') in the above integral 1

shows that it equals

('

e 2ni(x'-e1+xn0(e'))Yf(e')de'

s(xn)

= Cs(xn)K(x)

Rn-I with 00

2

Cs(xn)

1'(s/2) ./

e

2nix nenlfnl-l+s n(fn)den

-00

Now it is well known that 4s has an analytic continuation in s which is an entire function; also Co = 1 ; and I4 (xn)I < clxnl-Re(s), where Ixnl > 1 , and the real part of s remains bounded. From these facts it follows that Ks has an analytic continuation to an entire function s (whose values are smooth functions of x 1'..., xn of at most polynomial growth). One can conclude as well that

E. M. STEIN

328

(a) KO(x) = K(x) ,

(b) IK-n/2+it(x)I < A, all x e R, all real (c) IK1+it(e)I < A, all

t

£ Rn, all real t

In fact (c) is immediate from our initial definition (4.4), and (b) follows from Theorem 1.

Now consider the analytic family Ts of operators defined by T5(f) _ f * Ks' From (b) one has IIT_n/2+it(f )IIL00 < AIIf1ILI , all real t

(4.5)

and from (c) and Plancherel's theorem one gets (4.6)

IITI+it(f )IIL2 < AIIfIIL2

,

all real t

,

An application of a known convexity property of operators (see [281) then shows that IIT0(f )II LP°, < AIIfJJ PO , with PO0 = 2n + 2 , and the proof of n+3

Theorem 3 is complete. REMARKS:

(i)

For hypersurfaces with non-zero Gaussian curvature this theorem is the best possible, only insofar as it is of the form (4.1) with q > 2 . If q is not required to be 2 or greater, then it may be

conjectured that a restriction theorem holds for such hypersurfaces in the wider range 1 < p < 2n/(n+1). This is known to be true when n = 2 (see also §7 below). (ii) For hypersurfaces for which only k principal curvatures are nonvanishing, Greenleaf [121 has shown that then the corresponding results hold with 1 < p < 2k 2 , giving an extension of +

Theorem 3.

(iii) In the case of dim(S) = 1 (i.e. in the case of a curve) there are a series of results extending our knowledge of the case n = 2 alluded to above. For further details one should consult the references cited below.


329

It would of course be of interest to know what are the exponents p and q (if any) for which the restriction holds if we are dealing with a given sub-manifold S. This problem is highlighted by the fact quite general sub-manifolds S (those which are of finite type in the sense described in §3) have the restriction property: THEOREM 4. Suppose S is a smooth m-dimensional sub-manifold of Rn of finite type. Then there exists a po = po(S), 1 < po, so that S has the LP restriction property (4.1) with q = 2 , and 1 < p < po. (In fact if

the type of S is k, we can take po = 2nk/(2nk-1 .) COROLLARY. Suppose S is real analytic and does not lie in any affine hyperplane. Then S has the LP restriction property for 1 < p < po, for some po > 1 .

Proof. As we saw above, it suffices to prove (4.3). However Tf = f *K, and K(x) = dµ(-x), Theorem 2 tells us that IK(x)l < AIxI-11k, So according to the theorem of fractional integration, (see [26], Chapter V), where a=n-1/k, and this we therefore get (4.3) with __ = P -n,

0

0

relation among exponents is the same as PO = 2nknk1

'

Q.E.D.

Further bibliographic remarks. The initial restriction theorem dates from 1967 but was unpublished. The sharp result for n = 2 was observed by C. Fefferman and the author and can be found essentially in [9]; see also Zygmund [33]. Further results are in Thomas [30], [31], Strichartz [29], Prestini [24], Christ [4], and Drury [7].

Oscillatory integrals of the first kind related to singular integrals A key oscillatory integral used in the theory of Hilbert transforms along curves is the following: 5.

00

(5.1)

P.V.

Jr

elpa(t) dt t,

E. M. STEIN

330

d

where Pa(t) is a real polynomial in t of degree d, Pa(t) = F a.O. j=o

It

was proved by Wainger and the author in [27], that the integral is bounded with a bound depending only on the degree d and independent of the

coefficients ao,al,...,ad. The relevance of such integrals can be better understood by consulting Wainger's lectures [32]. We shall be interested here in giving an n=dimensional generalization of this result. We formulate it as follows. Let K(x) be a homogeneous function of degree -n; suppose also that IK(x)I < AIxI-n (i.e. K is bounded on the unit sphere); moreover, we assume the usual cancellation property: f x,l-1 K(x') d a (x') = 0. We let P(x) _

I aaxa be any real polynomial of degree d. Ial 1 , we have only f"0

and that integral is estimate like

f00

.) Let us consider the second

integral. It equals 00

00

f eiPa(t) dt _ (' eiPb(t) dt J t t

f

Now since bd = 1, we see that (d/dt)d Pb(t) = d!, and hence


333

(' eipb(t) dt 0, and the metric could be defined in terms of the

usual distance. The second are the dilations (z,t) -+ (pz,p2t), and the appropriate quasi-distance (from the origin) is then (Izl4+t2)1 /4. The latter dilations and metric are closely tied with the realization of the Heisenberg group as the boundary of the generalized upper half-space holomorphically equivalent with the unit ball in Cn+1 . This point of view, as well as related generalizations, is elaborated in Nagel's lectures [211.

336

E. M. STEIN

In the present context the first type of dilations and corresponding metric would be appropriate if one considered expressions related to ordinary potential theory in Hm viewed as R2m+1 . However the two conflicting types of dilations (and related metrics) occur in e.g. the solutions of Ju = f . (One sees this for example in Krantz's lectures [17], where in the formula of Henkin we have a kernel made of products of functions each belonging to one of the two above homogeneities.)* Other expressions of this type occur in the explicit formulae for the solutions of the a-Neumann problem (see [1], Chapter 7). Let us now consider the simplest operator on the Heisenberg group displaying simultaneously these two homogeneities. The prime example is given by (6.1)

Tf =f*K

where convolution is with respect to the Heisenberg group, and the kernel K is a distribution of the form (6.2)

K(z,t) = L(z)S(t) .

L(z) is a standard Calder6n-Zygmund kernel in Cm = R2m, i.e. L(pz) _ p-2m L(z), L is smooth away from the origin, and L has vanishing mean value on the unit sphere. Here 6(t) is the Dirac delta function in the t-variable, and in an obvious sense is homogeneous 3(pt) = p-1 S(t) . Thus K is homogeneous at degree -2m - 1 with respect to the standard dilations, and at the same time homogeneous of degree -2m - 2 with respect to the other dilations; in both instances the degrees are the critical ones. We turn next to the question of proving that the operator (6.1) is bounded on L2(Hm). The most efficient way is to proceed via the Fourier transform in the t-variable. This leads to the problem of showing that the family of operators TA defined by

*In particular the terms AI and A2 that appear in §6 of [17].


(6.3)

337

(TA)(F)(z) = i L(z-w)e1AF(w)dw Cm

(with the anti-symmetric form which occurs in the multiplication law for the Heisenberg group) is bounded on L2(Cm) to itself, uniformly

in A, -DO«t
_

B( , ) , and F = f. Then the operators TA have the form

(6.4)

(Tf)(x) =

K(x-y)eiB(x,Y)f(y)dy

I

Rn

We shall suppose B is a real bilinear form, but we shall not suppose that B is necessarily anti-symmetric nor that K is homogeneous of

degree -n. THEOREM 7. Suppose K is homogeneous of degree -µ, 0 < g < n, smooth away from the origin, and with vanishing mean-value when p = n. (a) If B is non-degenerate, then the operator T given by (6.4) is

bounded on L2(Rn) to itself, for 0 < p < n ; when 1 < p < oo , the operator is bounded on Lp(Rn) to itself if -11 I < 2n I1

.

(b) If we drop the assumption that B is non-degenerate but require that

g = n, then T is bounded on LP(Rn) to itself for 1 < p < -o. The bound of T can then be taken to be independent of B. We shall give only the highlights of the proof, leaving the details, further variants, and applications to the papers cited below. Let us con-

sider first the L2 part of assertion (a) when n/2 < g < n. Suppose

ri

*For further details see Mauceri, Picardello and Ricci [19] and Geller and Stein [10].

E. M. STEIN

338

is a Co function, with 77(x)=l for IxI < 1/2, and ii(x) = 0, for 1x1 > 1 .

To is defined as in (6.4), but

We write T =

with K replaced by Ko = riK, and T with K replaced by K = (1-77)K. Observe first that since Ko(x-y) is supported where Ix-yl < 1 , estimating T0(f)(x) in the ball IxI < 1 involves only f(y) in the ball IYI < 2. We claim

5Ixl 0, and hence (6.7)

IL(x,y)I < AN(1 + Ix-yI)-N

,

N>0.

This shows that L is the kernel of a bounded operator on L2 proving the boundedness of T ,T and thus of T.. The proofs of the L2 boundedness when 0 < µ < n/2 (in part (a) of the theorem), and the L2 boundedness when µ = n but when B is not assumed to be nondegenerate, are refinements of the above argument.

E. M. STEIN

340

Let us now describe the main idea in proving the LP inequalities stated in (a) and (b) above. We shall need a generalization of BMO (and of H1 ) which may be of interest in its own right. Suppose E = leQI is a mapping from the collection of cubes Q in Rn to complex-valued functions on Rn so that IeQ(x)I =

,

Q(x),

all x

where XQ denotes the characteristic function of the cube Q. Let us define on "E-atom" to be a function a so that for some cube Q (i) a is supported in Q (ii) Ia(x)I < 1/IQI (iii) ,f a(x) Q(x) dx = 0

The space HE is then given by If if = I Ajaj , with each aj an E atom, and I IAil < ool. In a similar vein the function fE will be defined as

(6.8)

fE(x)

su IQI

I

!f -fQldx

,

Q

where

fQ

IQI

f

f U_Q

Q

and we take BMOE = IfIfE E L°°I .

Some of the basic facts about the standard H1 and BMO spaces* go through for HE and BMOE, and sometimes these come free of charge.

One such case is the following assertion: Suppose f c Lpo , 1 < p < p0 < oo, and fE f LP. Then f c LP and *The standard situation arises of course when eQ = yQ, all Q.


Lp

OilLp

(6.9)

341

To prove this we need only observe that (if I)# < 2f0 , and use the result (see [10]) for the standard # function. The point of all of this is that for operators of the form (6.4), there is a naturally associated HE and BMOE theory, and it is given by choosing eQ(x) = e

(6.10)

-iB(x,cQ) ,

where cQ is the center of the cube Q. The basic step in the LP theory (besides an appropriate interpolation which goes via (6.9)), is the proof that when u = n our operator T maps L°° to BMOE . Let us give the proof in the case (a). We may assume that ilf 11 °° < 1 , and

suppose first that Q is a cube centered at the origin. Then we have to show that there exists a constant yQ , so that

(6.11)

1

Q,

f

JTf -yQ dx < A

Q

The corresponding inequality for a cube centered at another point, say CQ , then follows from the translation formula (6.6), (and this is the reason for defining eQ as we do). Turning to (6.11), the argument is not exactly the same as in the standard case (see e.g. Coifman's lectures [5] or [91), since we must split f into three parts to take into account the oscillations of e1B(x,y) . Suppose Q = QS , has side-lenghts S , then write

f = fl+f2+f3, where

Q28' fl = 0 elsewhere, CQ25 n QS-1

,

f2 = 0 elsewhere,

SQ2S n cQ 5-1 , f3 = 0 elsewhere.

E. M. STEIN

342

(Note that f2 occurs only when S < x/2/2 .) We have F = T(f) =

F1+F2+ F3, where FJ =T(fj). For F1 we make the usual estimate, using the fact that T is bounded on L2. Next observe that IK(x-y)eis(x,Y)- K(-Y)I < cS

if

1

1

+

IYIn+1

IYIn-1

'

x cQs and y e'-Q2s. Thus if yQ = f K(-y)f2(y)dy, we get that for

xCQ3

IF2(x)-yQI a1I < C(a) I E I

360

STEPHEN WAINGER

where C(a) may depend on A but not on E, then 1) holds for every f in L°°. If 12)

I1xI911BXE(x)>a1I < C(A)JEJ

then 2) holds for every f in

,

A discussion of this can be found in

[BF].

Let us try to see if 9) or 10) could be true in some simple cases. We consider for example the one-dimensional case. Here Br = Qr = lxl-r<x 1

for all x. We could still ask if 1) and 2) hold in some interesting class even though 9), 19), 11), and 12) fail. However an important idea of Stein shows that the failure [SI] of 9), 10), 11), and 12) implies that 1) and 2) fail even in the class of locally bounded functions. The statement of the main theorem of [SI] requires that the underlying space be compact. But if 1) or 2) were true for an LP class on Rn, it would also hold for the corresponding LP class on the torus. Furthermore, the theorem of Stein requires the hypothesis that 1 < p < 2. However due to the positive nature of the averages under consideration, his ideas can be modified to show that 1) fails for at least some L°° functions. See [SW]. Thus we obtain negative results in one dimension. Similar reasoning gives the same negative conclusion for question IA in any number of dimensions.

361

AVERAGES AND SINGULAR INTEGRALS

One need only consider

f

is as

above and h is a nice function. So there are no interesting positive results in problem IA. However as we shall see later there are positive results for problem IB in 3 or more dimensions. One might ask if there is a simple geometric reason why there should be positive answers for the sphere and only negative answers for the boundaries of squares. It turns out that the underlying basic reason that we have positive results for the boundary of balls and negative results for the boundary of squares is that spheres are round and boundaries of squares are flat. In other words an important word for us will be CURVATURE.

We will come back to the role of curvature in our problem in a little while, but first we shall discuss the other problems that we will consider. Problem II: Let y(t) be a curve passing through the origin in Rn . Is it

true that lim h J f(x-y(t))dt = f(x) a.e., for

f in

L'° or L2 or LI ?

Problem III: Let v(x) be a smooth vector field in Rn. Does h

lim

1

h-'0 h _

1

f(x-tv(x))dt = f(x)

a.e.

0

for f in

L°° or L2 or LI ?

Corresponding to problems II and III there are interesting singular

integrals. We let y(t) be a curve and v(x) be a smooth vector field as in problems II and III. We set a

13)

Hyf(x) =

J-a

f (x-y(t)) Lt .

(where sometimes we wish to think of a as finite and sometimes as and

00 ),

362

STEPHEN WAINGER

1

14)

Hvf(x)

f(x-tv(x)) dt

-1

We call Hy the Hilbert transform along the curve y and Hv the Hilbert transform along the vector field v(x). We then have the following two problems:

Problem II': Can we have an estimate 15)

IIHyf 11

LP

< CPIIIIILP

for some p's ? Problem III': Can we have an estimate 16)

IIHyfIILp 0


363

(K(x) is a function on Rn ) to the one-dimensional Hilbert transform

ff(x-t) dt

Hf(x) =

(f a function on R1 ). We will explain how the method of rotations can lead to problem II'.

Let K(x,y) be a function of two variables x and y which is odd, K(-x,-y) = -K(x,y)

and which has a "parabolic homogeneity," that is K(Ax.X2y) = 3 K(x,y)

18)

X3

We wish to consider the LP boundedness of the transformation 00

00

Tf(u,v) =

19)

f(u-x,v-y)K(x,y)dx dy .

I

We now introduce parabolic polar coordinates into 19) x = rcos 0 y = r2sinO and find 00

20)

277

Tf(u,v) = J

J

0

0

f(u-rcos 0, v-resin 0)

K(rcos 0, resin 0)r2N(0)drd0

where r2N(0) is the Jacobian factor in the change of variables. N(0) is smooth and N(0+n) = N(0). By 18 we see that

364

STEPHEN WAINGER o0

27r

r

21) Tf(u,v) =

N(O)K(cos d,sin O)dO

0

r C f(u-rcos 0,v-rsin 0) dr 0

2rr

00

-J

N(O)K(cos(O+n),sin(O+n))dd f

0

0

=

i f(u-rcos 0, v-rsin(O)) dr

since K is odd. Thus 217

r

fTf(u,v)

i f(u+rcos O,v+rsin 0) dr

N(O)K(cos 6,s in 0) dd

0

0

since

N(6+n) = N(6) .

Finally 277

r

21A) Tf(u,v) =

00

N(O)K(cost,sinO)dO

r r f(u-rcos6;v-r2sinO)dr

Now adding 21) and 21A) we find that 2n

f

Tf(u,v) = 2

ao

N(O)K(cos O,sin 0)dd f -00

0

i (u-rcos6,v-r2sin6)dr f

If K(cos O,sin 0) is in LT of [0,2n] , we can apply Minkowski's inequality n

IITfIILp < C f de IIHIILp 0

where


Hef =

f

365

00

f(x-yg(r))

dt

,

-00

with

ye(r) = (rcos O,r2sin O) .

Now we prove IITfIILp 0 where the difference K0(r, 0) - Ko(r , 0)

offers no cancellation, we find there is only one bad 0, 0 = 0. But still


383

n/4

f

JK0(r, 6)- K0(r', O')j dO = 1

.

J0

Let us consider now what happens with

We should expect no

help from the difference K(x,y) - K(x,y-1) when y > 0, if y-1 < 0. But this can only happen if y < 1 or 0 < 1/r . But over this set KE(x,y) is integrable at infinity 00

1 /r

KE(r, 0)d0rdr

1 50

5

1/r

00

Irfderdr

el-F

2

0

5

f1 00

0 an operator of CalderonZygmund type. This proved that Hy was bounded in LP 1 < p < oo if y = (t,t2). However it would be extremely difficult to carry over this proof to a three dimensional curve. For example it would be hard to derive an analogue of 48) for the curve (t,t2,t3). Essentially one needed a way to define a suitable analytic family Tz without using the asymptotic formula 48). Recall that Hyf = DP * f where

56)

Dp(e,rl) =

reieteint2 dt

384

STEPHEN WAINGER

So one might be tempted to define Hy,f = DP * f

57) w here

e

a irlt2 dt (1+t2)z/2

Dz(e,rl) p

58)

It turns out that 58) is not a good idea for a very important reason. By changing variables in formula 56) we see that Dp(Ac, A271) = Dp(c,rl) , for

any A > 0. Note that also the function mz(e,r/) defined in 52) also has this type of homogeneity, namely mz(Ac,A2r7) = mz((,77), for A > 0.

Now experience has shown that homobeneity is a powerful friend not

to be tossed away lightly. However Dp does not have this homogeneity. This situation can be remedied by defining

Hyf=Dzxf

59)

where 00

60)

Dp(c,rl) =

I

(1

+r72t4)-z /4 Xteir7t2 dt

-00

Note that for A > 0 61)

Dp(Ac, A2rl) = DP(c ,77) .

Let us see how formula 61) can help us. We would like to show 62)

C(z)

if Re z > -2 . By formula 61) we may assume r/ = ±1 , let us say Then by Van Der Carput's lemma with } = 2 , we see that

77

=1.


t

eXsel1

2

ds

1

ft d ese's2 dsI

nn1

is necessary in order that 77) hold. The situation for n = 2 , p > 2 is unknown at this time. I would like to present here Stein's original argument which proved

77) for p = 2 and n = 4 . We define

g(f) (x) _

78)

f

1 /2

00

tI dt Mtf(x)I2 dt

Assume that we could prove IIg(f )IIL2 < C(n)IIf1IL2 '

79)

and let us see how 77) would follow. Now

rnMrf(x)

CT-

snMsf(x)ds

0

=n

r sn-1 Msf(x)ds +

r sn ds Msf(x)ds


395

Thus r

Mrf(x) < n

r

r sn'1 Msf(x) ds + 0

n

r

sn d Msf(x)ds

0

= I(r) + 11(r)

.

Now l(r) is dominated by the Hardy-Littlewood Maximal function and

II(r) < n

f

sn-1 /2 s1 /2 Msf(x)ds

rJ

0

r

fr -rn J
0

0

sup

1f

E

Idµh *f(x,Y)-oh *f(x,Y)I2dh

E>0 E1/2

0

o 0

0

401

STEPHEN WAINGER

402

A classical argument (see (RI) shows sup Oh *fp < C tIfIILp Lp h>0

Thus by 95), we see E

sup I e fd1zh*f(xY)dhl

< CIIf11L2

E>0

L2

0

If f>0, E

(' J dµh*f(x,Y)dh 0

2h

E

f

f(x-t,y-t2)dtdh h

0

E

E

>

f r f(x-t,y-t2)

f

dh

/2

0

('E

E >

f

f(x-t,y-t2)dt

J

.

0

So from 96) we infer E

sup If

E>0

eJ

0

It remains to prove 96).

0 we may write the expression in 98) as

0

J0

µ(Xhe,)12h271)-0(Ahe,42h277)I2

By choosing A so that \2e2+X4772 = 1 , we see that it suffices to estimate 99) when e 2 +772 = 1 . In this case we see 1

1

(' 99)

JJ

dh dµ(he,h27)-cb(he,h27)I2 < C

0

since dµ(0) = 0(0) = 1 .

r h2 dh < C 0

404

STEPHEN WAINGER

Then from 27) Idµ(he,h27l)I < Ch-3

for some 6>0. So ('00

100)

('00

ld26 < C

T Idµ(he,h2rl)I < J

J

J

h

1

1

CN

Also k1;'"?)
h>o

y

.

We are now ready for the statement of problem IV and W. Problem IV: When do we have UHyf(x)11Lp 0. If r(t) is even IIHyfIIL2 0

for some C>0. If y(t) is odd IIHyfIIL2 2h(t) ,

t>0

for some C>0. There are generalizations of Theorem 8 to higher dimensions, and an investigation of the LP theory has begun. We know that

412

STEPHEN WAINGER

II

yfliL2 0, however the maximal functions can be

bounded on L2 (and in fact in LP for any p > 1) for some convex curves even if * fails. Recently Phong and Stein [PS] introduced a general problem of which

our problems are special cases. They consider at each point P in Rn a submanifold Mp of dimension say a and an a dimensional CalderonZygmund kernel K(P,Q). Then they consider

Tf(P) T (Q) K(P,Q)dm(Q)

where dm(Q) is a measure on Mp M. They show that if n > 3 and

k = n-1 IITfIILp

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