Harmonic Analysis and Partial Differential Equations

HARMONIC ANALYSIS AND PARTIAL DIFFERENTIAL EQUATIONS BJÖRN E. J. DAHLBERG CARLOS E. KENIG ISSN 0347-2809 DEPARTMENT OF ...

Author: Dahlberg B.E.J. | Kenig C.E.

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HARMONIC ANALYSIS AND PARTIAL DIFFERENTIAL EQUATIONS BJÖRN E. J. DAHLBERG CARLOS E. KENIG ISSN 0347-2809

DEPARTMENT OF MATHEMATICS CHALMERS UNIVERSITY OF TECHNOLOGY AND THE UNIVERSITY OF GÖTEBORG GÖTEBORG 1985/1996

FOREWORD These lecture notes are based on a course I gave rst at University of Texas, Austin during the academic year 1983 - 1984 and at University of Göteborg in the fall of 1984. My purpose in those lectures was to present some of the required background in order to present the recent results on the solvability of boundary value problems in domains with bad boundaries. These notes concentrate on the boundary value problems for the Laplace operator; for a complete survey of results, we refer to the survey article by Carlos Kenig; I am very grateful for this kind permission to include it here. It is also my pleasure to acknowledge my gratitude to Peter Kumlin for excellent work in preparing these notes for publication. January 1985 Björn E. J. Dahlberg

i

ii

Contents 0 Introduction

1

1 Dirichlet Problem for Lipschitz Domain. The Setup

11

2 Proofs of Theorem 1.1 and Theorem 1.2

19

3 Proof of Theorem 1.6

25


37


47

6 Dirichlet Problem for Lipschitz domains. The nal arguments for the L2-theory 51 7 Existence of solutions to Dirichlet and Neumann problems for Lipschitz domains. The optimal Lp-results 57 Index : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 64 Appendix 1 C. E. Kenig: Recent Progress on Boundary Value Problems on Lipschitz Domains : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 67 Appendix 2 B. E. Dahlberg/C. E. Kenig: Hardy spaces and the Neumann Problem in Lp for Laplace's equation in Lipschitz domains : : : : : : : : : 107

iii

iv

Chapter 0 Introduction In this course we will study boundary value problems (BVP:s) for linear elliptic PDE:s with constant coecients in Lipschitz-domains , i.e., domains where the boundary @

locally is given by the graph of Lipschitz function. We recall that a function ' is Lipschitz if there exists a constant M < 1 such that j'(x) ? '(z)j M jx ? zj for all x and z. y = '(x)

x

To solve the BVP:s we will reformulate the problems in terms of integral equations. It therefore becomes necessary to study singular integral operators of Calderón-Zygmund type, which we prove to be Lp-bounded for 1 < p < 1 and invertible. The Lp-boundedness is a consequence of the Lp-boundedness of the Cauchy integral (Coifman, McIntosh and Meyer) Z Tf (z) = wf (?w)z dw ? where ? is a Lipschitz-curve (method of rotation). The invertability will be proved by a new set of ideas recently developed by Dahlberg, Kenig and Verchota. Among the BVP:s which can be solved by this technique are the 1

Dirichlet problem Neumann problem the clamped plate problem

u = 0 in

u = f on @

u = 0 in

@u @n = f on @

8
0

Thus with X = Lp(Rn) and Y = fu : u harmonic in Rn++1 and u satises ()g we have the implication

f 2 X ) u 2 Y: However, we can also reverse the implication since a harmonic function u which satises () has non-tangential limits a.e. on @ Rn++1, the limit-function u0 = u(; 0) 2 Lp(Rn) and u(x; y) = py u0(x). Sketch of a proof. Assume u harmonic function in Rn++1 that satises (). The semigroup properties of fpy gy0 implies

u(x; y + ) = py u(x); > 0; y > 0 where u(x) = u(x; ). () ) un * v in Lp(Rn) ) py un (x) ! py v(x)

as n # 0 as n # 0; y > 0:

But py un (x) = u(x; y + n ) and thus

u(x; y) = py v(x) where v 2 Lp(Rn): For the proof of the existence of non-tangential limits of py u0 we refer to e.g. Stein/Weiss [2]. The notion of solution of the Dirichlet problem and any other problem, is sound only if we have such a matching between the boundary value f of u and the solution u itself, i.e., we should not accept concepts of solution which are so weak such that the reversed implication is impossible. Now assume that is a bounded (connected) domain in Rn; n 3 with C 2 boundary. (To avoid technicalities, we have assumed n 6= 2). Consider the Dirichlet problem (D)

u = 0 in

uj@ = f 2 C (@ )

Let r denote (?1) (the fundamental solution) of the Laplace operator in Rn, that is, (n=2) r(x) = cn jxj1n?2 ; cn = ? (2 ?1n)! ? 2 ?1 n ?2n=2 n

3

and set

R(x; y) = r(x ? y): For f 2 C (@ ) we dene

Z

@ R(P; Q)f (Q)d(Q) @nQ Z@

Sf (P ) = R(P; Q)f (Q)d(Q)

Df (P ) =

@

P 2= @

P 2= @

Thus Df and Sf denote the double layer potential and single layer potential resp. Here d is the surface measure on @ and @n@Q is the directional derivative along the unit outward normal for @ at Q. It is immediate that Df (P ) = 0; P 2 Rn n @

and Df will be our candidate for solution of (D). It remains to study the behaviour of Df at @ . Part of that story is Lemma 1. If f 2 C (@ ), then 1) Df 2 C ( ) 2) Df 2 C ({ ). More precisely: Df can be extended as a continuous function from inside to and from outside to { . Let D+ f and D? f denote the restrictions of these functions to @ resp. Set K (P; Q) = @n@Q R(P; Q) for P 6= Q; P; Q 2 @ . We note that i) K 2 C (@ @ n f(P; P ) : P 2 @ g) ii) jK (P; Q)j jP ?QCjn?2 for P; Q 2 @ and some C < 1. ii) is a consequence of the regularity of the boundary and can be seen as follows: Assume @ given by the graph of the C 2-function '. Set P = (x; '(x) and Q = (y; '(y)). Then K (P; Q) = !1 hPjP??Q;QnjnQi where h; i is the inner product in `2(Rn) and n nQ = p(r'(y); ?2 1) : jr'(y)j + 1 4

Since ' is a C 2 function, we have that

'(x) = '(y) + hx ? y; r'(y)i + e(x; y) where je(x; y)j = O(jx ? yj2): Hence

'(y); ?1)ij C je(x; y)j C : jK (P; Q)j C jhP ? Q;jP(r n n ? Qj jP ? Qj jP ? Qjn?2

This estimate is uniform in P and Q since @ compact. For f 2 C (@ ) dene

Tf (P ) =

Z

@

K (P; Q)f (Q)d(Q); P 2 @ :

We can now formulate

Lemma 2 (jump relation for D). 1) D+ = 21 I + T 2) D? = ? 12 I + T and

Lemma 3. T : C (@ ) ! C (@ ) is compact. Sketch of proof of Lemma 3. Dene the operators Tn by

Tnf (P ) = for f 2 C (@ ), where

Z

@

Kn (P; Q)f (Q)d(Q); P 2 @

Kn (P; Q) = sign (K (P; Q)) min(n; jK (P; Q)j); n 2 Z+ : Thus Kn is continuous on @ @ and Arzela-Ascoli's theorem implies that Tn is a compact operator on C (@ ). Furthermore since kTnk supQ2@ kKn (; Q)k1 C < 1, where C is independent of n we see that

Tn ! T in the space B = fbounded linear operators on C (@ )g. But the compact operators in B form a closed subspace in B , and hence T is compact. 5

Proof of Lemma 1 and 2. Some basic facts: Z

@ R(P; Q)d(Q) = 1, if P 2

@ @nQ Proof: Apply Green's formula to the harmonic function R(?; Q) in n B (P ) for > 0 small, where B (P ) = fx 2 Rn : jP ? xj g. Z @ R(P; Q)d(Q) = 0, if P 2= . 2) @ @nQ Proof: Exercise. Z 3) K (P; Q)d(Q) = 12 , if P 2 @

@

Proof: Exercise. Let P 2 @ . We want to show that Df (Q) ! 21 f (P ) + Tf (P ) as 3 Q ! P A: Assume P 2= supp f : Easy. B: Assume f (P ) = 0: We need. Z 4) 9C > 0 : @ R(P; Q) d(Q) < C for all P 2= @

@ @nQ Proof: Exercise. 4) implies the estimate 1)

kDf kL1 (Rn n@ ) C kf kL1 (@ ): Choose ffk g C (@ ) with P 2= supp fk such that kf ? fk kL1(@ ) ! 0 as k ! 1: T bounded operator implies Tfk(P ) ! Tf (P ) as k ! 1. Hence jDf (Q) ? Tf (P )j C k(f ? fk )kL1 (Rnn@ ) + jDfk (Q) ? Tfk (P )j + + jTfk(P ) ? Tf (P )j ! 0 as k ! 1 and 3 Q ! P: C: Enough to check f 1. The result follows from basic facts 1) and 3). Hence we have proved Lemma 1 and 2 part 1). Part 2) follows analogously. 6

We now return to the single layer potential and observe that Sf is harmonic in Rn n @

and continuous in Rn if f 2 C (@ ). Next we want to compare the normal derivative of Sf with Df at @ . Since @ is C 2 we have following result: For " > 0 small enough ] ? "; "[@ 3 (t; P ) ! P + tnp 2 V is a dieomorphism, where np is the outward unit normal of @ at P , and V is a neighborhood of @ . For P 2 @ and t 2] ? "; "[ set Z DSf (P + tnp) = @n@ R(P + tnp; Q)f (Q)d(Q): @ p The close relations between Df and DSf is formulated in Lemma 4. If f 2 C (@ ) then 1) DSf 2 C (V \ ) 2) DSf 2 C (V \ { ) (Compare Lemma 1). Let D+ Sf be the restriction to @ of the function DSf extended to V \ from inside and D? Sf the restriction to @ of the function DSf extended to V \ { from outside. But R(P; Q) = R(Q; P ) so with Rn (P; Q) = K (Q; P ), which is the real-valued kernel in

T f (P ) =

Z

@

K (P; Q)f (Q)d(Q) P 2 @ ;

we have that T is the adjoint operator of T . Lemma 5 (jump relations for DS ). 1) D+ S = ? 12 I + T 2) D? S = 12 I + T . Proof of Lemma 4 and 5. Let f 2 C (@ ) and dene

wf (P ) =

Df (P ) + DSf (P ) P 2 V n @

Tf (P ) + T f (P ) P 2 @ :

Claim: wf 2 C (V ). Proof: wf continuous on V n @ and on @ . Hence it is enough to show taht

wf (P + tnp) ! wf (P ) uniformly for P 2 @ as t ! 0: 7

Assume 0 2 C (@ ) such that 0 0 1; 0 = 1 in a neighborhood of P and supp 0 B (P ): Decompose f as

f = f1 + f2 0f + (1 ? 0)f: A: wf2 (P + tnp) ! wf2 (P ) as t ! 0. Easy B: Assume t 6= 0

11111111111111 00000000000000 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 0 00000000000000 11111111111111 00000000000000 11111111111111 = + 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111

00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 t
0, but not for domains with boundaries with less regularity. Remark: We observe that the method is non-constructive as a consequence of the soft arguments (i.e., compactness arguments) we have used. Hence it is not possible to solve the Dirichlet problem for, say Lipschitz-domains by approximating with C 2 domains

k , solve some Dirichlet problems for these and obtain an approximation of a solution for

, since we do not have any estimates of the inverses of the D+ :s. 9

References [1] Folland, G: Introduction to partial dierential equations. Math. Notes 17, Princeton U.P. [2] Stein, E.M./ Weiss, G: Introduction to Fourier Analysis on Euclidean Spaces. Princeton U.P.

10

Chapter 1 Dirichlet Problem for Lipschitz Domain. The Setup A function ' : Rn ! R such that

j'(x) ? '(y)j M jx ? yj for all x; y 2 Rn is called Lipschitz function. A bounded domain Rn+1 is called Lipschitz domain if @

can be covered by nitely many right circular cylinders L whose bases are at a positive distance from @ such that to each cylider L there is a Lipschitz function ' : Rn ! R and a coordinate system (x; y); x 2 Rn; y 2 R such that the y-axis is parallel to the axis of symmetry of L and L \ = L \ f(x; y) : y > '(x)g and L \ @ = L \ f(x; y) : y = '(x)g. A domain D Rn+1 is called special Lipschitz domain if there is a Lipschitz function ' : Rn ! R such that D = f(x; y) : y > '(x)g and @D = f(x; y) : y = '(x)g. In this and all proceeding chapters we reserve the notation for bounded Lipschitz domains and D for special Lipschitz domains respectively. With a cone ? we mean a circular cone which is open. A cone ? with vertex at a point P 2 @C , where C Rn+1 is a domain, is called a nontangential cone if there is a cone ?0 and a > 0 such that ; 6= (? \ B (P )) n fP g ?0 \ B (P ) C:

Br (Q) is our standard notation for the ball fx 2 Rn : jx ? Qj rg. We say that a function u dened in a domain C has nontangential limit L at a point P 2 @C if u(Q) ! L as Q ! P; Q 2 ? for all nontangential cones ? with vertices at P . Finally we dene the nontangential maximal function M u for > 1 and function u dened in Lipschitz domain by

M u(P ) = supfju(Q)j : jP ? Qj < dist (Q; @ ); Q 2 g; P 2 @ : 11

One of the main results in this course will be the existence of a solution to the Dirichlet problem

u = 0 in Rn+1 uj@ = f 2 L2(@ )

where is a bounded Lipschitz domain. By this we mean that there exists a harmonic function u in which converges nontangentially to f almost everywhere with respect to the surface measure d(@ ) and that the maximal function M u 2 L2(@ ) for > 1. The starting point for our enterprise of proving the existence of a solution to Dirichlet problem for the Lipschitz domain is the double layer potential

Dg(P ) =

Z

@ R(P; Q)g(Q)d(Q) P 2 ; @ @nQ

where R(P; Q) is the fundamental solution for Laplace equation in Rn+1 (multiplied with ?1) and g 2 L2(@ ). Since Dg is harmonic in , we are done if we can show that for some choice of g we have the right behaviour of Dg at @ . However, this is not easy since for K (P; Q) = @n@Q R(P; Q) P; Q 2 @ P 6= Q, we only have the estimate jK (P; Q)j jP ?QCjn?1 which cannot be improved in general. Thus we have to rely on the cancellation properties of K (P; Q), and the operator T which appeared in Chapter 0 can only be dened as a principal value operator. Before we study the case with a general bounded Lipschitz domain we treat the case with a special Lipschitz domain D. From this we obtain the result for using standard patching techniques (see Appendix 2). Consider Z Dg(P ) = @n@ R(P; Q)g(Q)d(Q) P 2 D @D Q where D = f(x; y) : y > '(x)g for a Lipschitz function ' : Rn ! R. We remark that ' Lipschitz function implies that '0 exists a.e. so the denition of Dg makes sense and @ R(P; Q) = C hnQ ; P ? Qi , with n = p(r'(x); ?1) for Q = (x; '(x)), exists a.e. Q n @nQ jP ? Qjn+1 jr'(x)j2 + 1 d(@D). To state the rst proposition, we need some more notation: For every measure R g and measurable set A with (A) 6= 0 we let A gd, and each -measurable function Z denote the mean value (1A) gd. Furthermore for g 2 L1loc(@D) we dene the maximal A function Mg by Z

Mg(P ) = sup

r>0 @D\Br (P )

jg(Q)jd(Q); P 2 @D:

The following result is crucial. 12

Proposition 1.1. Let D = f(x; y) : y > '(x)g where ' : Rn ! R is a Lipschitz function with k'0 k1 = A. Let P = (x; y) 2 D and P 2 (x; '(x)) 2 @D and set = y ? '(x). Assume g 2 Lp(@D) for some p where 1 < p < 1. Then jDg(P ) ? Tg(P )j CMg(P ) where

T

g(P ) =

Z

@DnB(P )

K (P ; Q)g(Q)d(Q):

The constant C depends only on the dimension a. Before we prove this proposition, we make some remarks on the maximal function M .

1) Another maximal function, which is quite similar to the Hardy-Littlewood maximal function M is M dened by

M g(P ) =

Z

sup

Br (Q)3P @D\Br (Q)

jg(Q0)jd(Q0); P 2 @D

for g 2 L1loc(@D). We immediately observe that Mg M g CMg for some dimensional constant C , i.e., M and M are equivalent. 2) Let denote the projection : @D ! Rn where (x; '(x)) 7! x and dene the maximal function M~ by Z

~ (x) = sup Mg

r>0 Br (x)

jg ?1(y)jdy; x 2 Rn;

for g 2 L1loc(@D). Since ' is a Lipschitz function, we see that M and M~ are equivalent. 3) M is bounded in L1 with norm 1, i.e., kMgk1 kgk1 for all g 2 L1 4) M is a weak (1; 1) operator, i.e., there exists a C > 0 such that jfx : Mg(x) > gj C kgk1 for all g 2 L1: 5) M is bounded in Lp; 1 0 such that kMgkp Cpkgkp for all g 2 Lp: Here 3) is trivial, 4) can be proven by a covering lemma argument and 5) follows from 3), 4) and Marcinkiewicz' interpolation theorem (see Stein [1]). For later reference we state 13

Marcinkiewicz' interpolation theorem. Let 1 p < q 1 and let T be a subad-

ditive operator dened on Lp + Lq . Assume T is a weak (p; p) operator and a weak (q; q) operator. Then T is bunded on Lr where p < r < q. An operator T is a weak (p; p) operator if there exists a constant C > 0 such that ? jfx : Tg(x)j > gj C kgkp p for all g 2 Lp and > 0: Hence, if T is bounded on Lp, then T is a weak (p; p) operator, but the converse is not true in general.

11111111111111111111111111 00000000000000000000000000 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 Proof of Proposition 1.1.11111111111111111111111111 00000000000000000000000000 D

P

P

jDg(P ) ? T +C

C

Z

Z@DnfjP ?Qj>g

Z

hnQ ; P ? Qi ? hnQ ; P ? Qi jg(Q)jd jP ? Qjn+1 @D\fjP ?Qj>g jP ? Qjn+1 hnQ ; P ? Qi jP ? Qjn+1 jg(Q)jd(Q)

g(P )j C

( + jQ ? P j)n+1 jg(Q)jd(Q) + 1 jg(Q)jd(Q) +C @DnfjP ?Qj>g n where we have applied the mean value theorem to the rst integral. The second integral is CMg(P ) and the rst integral can also be estimated from above with the same bound according to Lemma 1.1. Let 0Rbe a radial decreasing function dened in Rn. Assume f 2 L1 +L1 and set mf (Rx) = supr>0 Br (x) jf (x)jdx for x 2 Rn. Then f (x) Bmf (x) for all x 2 Rn where B = (x)dx. Z @D\fjP ?Qj>g

14

If we take this lemma for granted for a moment and set (x) = (jxj +)n+1 ;

the rst integral above is bounded from above by CMg(P ) and we are done.

Proof of Lemma 1.1. It is enough to prove the lemma for 0 f 2 C01; 2 C01 and x = 0. R Set S n = @B1(0) and A(r) = Br (0) f (x)dx. We obtain

f (0) = =

Z

n ZR1

0

(jxj)f (x)dx =

Z 1

(r)A0(r)dr = ?

0Z 1 0

(r)rn?1

Z

Sn

f (rw)d(w)dr =

0(r)A(r)dr ?

Set f 1 in the calculations above and we get ? proven.

Z 1

0

R1 0 (r)B

0

0(r)jB

r (0)dr

r (0)jdr mf (0):

= B . The lemma is

If we dene the operator T by Tg(P ) = sup jTg(P )j P 2 @D >0

for g 2 Lp(@D), then

jDg(P )j C (Tg(P ) + Mg(P )) for all P = (x; y) 2 D and P = (x; '(x)) 2 @D. Thus if we can prove that T is bounded on Lp (@D), then jDg(P )j < 1 for a.e. P 2 @D. We remark that with some additional

considerations one can prove that jDg(Q)j C (Tg(P ) + Mg(P )) for all Q in a nontangential cone ? with vertex at P 2 @D, and thus supQ2? jg(Q)j in non-tangential cones ? with vertices at P 2 @D for almost every P 2 @D. It then follows that Dg has nite nontangential limit a.e. d(@D) (see Dahlberg [2]). The limitfunction belongs to Lp(@D). If we also can prove that the limitfunction is equal to f for some choice of g, we are done. To be successful in our approach, we have to study the operators T for > 0 and T. This calls for some denitions. Let S (Rn) denote the Schwartz class (i.e., the space of all C 1-functions in Rn which together with all their derivatives die out faster than any power of x at innity) with the usual topology. T is called a singular integral operator (SIO) if T : S (Rn) ?! S (Rn) is linear and continuous and there exists a kernel K such that for all ', 2 C01(Rn) with supp ' \ supp =

hT'; i =

Z Z

K (x; y)'(y) (x)dydx; 15

where h; i is the usual S ? S paring. We observe that K does not determine T uniquely. Consider for instance Tf = f 0 for which K = 0 is a kernel. We say that a kernel K is of Calderón-Zygmund type (CZ-type) if 1) jK (x; y)j jx ?C yjn 2) jrxK (x; y)j + jry K (x; y)j jx ?Cyjn+1 3) K (x; y) = ?K (y; x). The operator-kernels, we will study, will be of the form x)) ? (y; '(y)))i i = 1; 2; : : : ; n + 1 Ki (x; y) = j(((x;x;''(x()) ? (y; '(y))jn+1 where (a)i denotes the i-th component of a 2 Rn+1. We observe that these kernels are of CZ -type and adopt the convention that whenever we discuss kernels K , they are assumed to be of CZ -type unless we explicitly state the converse. Starting with a kernel K , we can form a well-dened SIO with K as the kernel namely the principal value operator (PVO) T . Note that for '; 2 S (Rn)) Z Z Z Z 1 K (x; y)'(y) (x)dydx = 2 K (x; y)('(y) (x) ? '(x) (y))dydx jx?yj>" jx?yj>" since K (x; y) = ?K (y; x) and thus lim

"!0

Z Z

jx?yj>"

K (x; y)'(y) (x)dydx

exists since j'(y) (x) ? '(x) (y)j = O(jx ? yj) and '; 2 S (Rn) decay fast enough at innity. Hence T : S (Rn) 3 ' 7! T' 2 S (Rn) where

hT'; i = "lim !0

Z Z

jx?yj>"

K (x; y)'(y) (x)dydx

is a SIO. We leave the proof of continuity of T as an exercise. From now on we assume that all operators T are PVO with kernels K of CZ -type. To achieve our goal to establish the existence of a solution to the Dirichlet problem for Lipschitz domains, we will prove the following sequence of theorems Theorem 1.1. If T bounded on L2, then T is a weak (1; 1) operator. This implies 16

Theorem 1.1'. If T bounded on L2, then T bounded on Lp for 1 0 j jx?yj>" K (x; y)g(y)dyj.

Thus it is crucial for us to be able to prove L2-boundedness of T . This is done in two steps. Theorem 1.3. If ' : R ?! R Lipschitz function and K (x; y) = x?y+i('(1x)?'(y)) , then the corresponding operator T is L2 bounded. Theorem 1.4. If ' : Rn ?! R Lipschitz function and y)))i i = 1; 2; : : : ; n + 1; Ki (x; y) = j(((x;x;''(x(x))))??(y;(y;''(y()) jn+1 then the corresponding operators Ti are L2 bounded.

Finally we prove

Theorem 1.5. Dj@D is invertible. To prove Theorem 1.3, we will characterize those kernels K of CZ -type which correspond to L2 bounded operators T . This is done by a theorem of Daivd and Journé [3]. Theorem 1.6. T bounded on L2 i T 1 2 BMO. The denition of BMO and the theorem and its proof will be discussed in Chapter 3.

References [1] E. M. Stein: Singular integral and dierentiability properties of functions, Princeton University Press 1970. [2] B.E.J. Dahlberg: Harmonic functions in Lipschitz domains, Proceedings of Symposia in Pure Mathematics Vol XXXV, Part 1 (1979) pp. 313-322. [3] G. David and J.-L. Journé: A boundedness criterion for generalized Calderón-Zygmund operators, Preprint.

17

18

Chapter 2 Proofs of Theorem 1.1 and Theorem 1.2 We recall that T is a PVO with kernel K of CZ-type. In this chapter we give a proof of the following result of Calderón-Zygmund [1].

Theorem 1.1: If T bounded on L2, then T is a weak (1; 1) operator. The following bound on T is due to Cotlar [3].

Theorem 1.2: RIf T bounded on L2, then T is bounded on Lp for 1 0 j jx?yj>" K (x; y)f (y)dyj, Proof of Theorem 1.1. The idea of the proof is the same as when T is a translationinvariant L2-bounded operator (See Stein [2]). Thus we show that there exists C > 0 such that

jfx 2 Rn : jTf (x)j > gj C kfk1 for all f 2 L1 and > 0

by splitting f in a good part g, which is a L2-function and a bad part b. This is done with the following lemma

Lemma (Calderón-Zygmund decomposition). Let f 2 L1(Rn) and > 0. Then there exist cubes Qj ; j = 1; 2; : : : such that 1) jQj \ Qk j = 0 for j 6= k 2) jf (x)j a.e. for x 2 Rn n [1 j =1 Qj

19

R

3) Qj jf (x)jdx < 2n . The proof is based on a recursive stop-time argument and can be found e.g. Stein [2]. Now set

g(x) =

x 2 Rn n [1j=i Qj j = 1; 2; : : : Qj jf (x)jdx x 2 Qj

Rf (x)

and b(x) = f (x) ? g(x). We immediately observe that j 1) supp b [1 j =1 Q R 2) Qj b(x)dx = 0 j = 1; 2; : : : and 3) kgk2 2n kf k1. Furthermore, jfx 2 Rn : jTf (x)j > gj jfx 2 Rn : jTg(x)j > 2 gj + jfx 2 Rn : jTb(x)j > 2 gj jfx 2 Rn : jTg(x)j > 2 gj + j [1j=1 2Qj j + jfx 2 Rn n [1j=1 2Qj : jTb(x)j > 2 gj

where 2Qj is the cube with the same center as Qj , which we denote yj , with sides parallel with Qj and with doubled sidelengths compared with Qj . Here jfx 2 Rn : jTg(x)j > 2 gj 42 kTgk2 C2 kgk2 C 1 kf k1 and 1 n X 1 n j [j=1 2Qj j 2 jQj j 2 kf k1 j =1 so it remains to estimate jfx 2 Rn n [1j=1 2Qj : jTb(x)j > 2 gj and this is the point where we use the properties of the kernel K . Set

bj (x) = For, x 2= 2Qj we have

Tbj (x) =

Z

Qj

b(x) x 2 Qj 0 otherwise

(K (x; y) ? K (x; yj ))bj (y)dy 20

and thus

jTbj (x)j C where we used

Z

Qj

Z

jy ? yj j jb (y)jdy j Qj jx ? y jn+1

bj (y)dy = 0. Integrating these inequalities gives Z

Rn n[1 j=1 2Qj

C

jTb(x)jdx

1 Z X

j =1 Qj

1 Z X j =1 Rn n2Qj

jTbj (x)jdx

jbj (y)jdy = C kbk1 C kf k1

and consequently

jfx 2 Rn n [1j=1 2Qj : jTb(x)j > 2 gj 2 kTbkL1(Rnn[1j=12Qj) C kfk1 :

The proof is done.

Corollary: If T is bounded on L2, then T is bounded on Lp for 1 < p < 1. Proof. Marcinkiewicz' interpolation theorem and Theorem 1.1 implies that T is bounded on Lp for 1 < p 2. But the adjoint operator T of T is a PVO with CZ-kernel K (x; y) = K (y; x) and T is bounded on L2. An application of Marcinkiewicz' interpolation theorem and Theorem 1.1 to T gives T bounded on Lp for 1 < p 2. Hence T = (T ) is bounded on Lp for 2 p < 1 by duality and we are done. Proof of Theorem 1.2. The proof is an easy consequence of Cotlar's inequality, i.e., if T bounded on L2 then

(2.1) Tf C (T )(Mf + MTF ) where M is the Harcy-Littlewood maximal function; since M is bounded on Lp and T is bounded on Lp according to the corollary above. Remains to show Cotlar's inequality: It is enough to prove (2.1) for x = 0. Fix an " > 0. We will show that (2.2) T"f (0) C (Mf (0) + MTF (0)) where C is independent of ". Set " f1(x) = f0 (x) jjxxjj < " 21

and f2(x) = f (x) ? f1(x). Thus T"f (0) = Tf2(0). The strategy is to prove that for jxj < 2" we have (2.3) jTf2(x) ? Tf2(0)j CMf (0) where the constant C is independent of x and ". Assume (2.3) to be true for a moment and argue as follows. ~ (0) jTf (x)j + jTf1(x)j + CMf ~ (0): jT"f (0)j = jTf2(0)j jTf2(x)j + CMf Consider the two cases ~ (0). 1) 13 jT"f (0)j CMf ~ (0). 2) 13 jT"f (0)j > CMf For case 1) inequality (2.2) is trivial. For case 2) set = jT"f (0)j; B = B "2 (0) and dene E1 = fx 2 B : jTf (x)j > 3 g E2 = fx 2 B : jTf1(x)j > 3 g: Here B = E1 [ E2 and thus

We see that

1 jjEB1jj + jjEB2jj :

Z 3 jE1j jTf (x)jdx 3jB j MTf (0) B

and

jE2j C kf1k1 C jB j Mf (0)

where we have used that T is a weak (1; 1) operator. Thus C (Mf (0) + MTf (0)) and it only remains to show (2.3), which is a straightforward calculation:

jTf2(x) ? Tf2(0)j

Z

jyj>"

jK (x; y) ? K (0; y)jjf (y)jdy

Z ? " " ?n jf (y )jdy CMf (0) j f ( y ) j dy C min ; " C jyjn+1 jyj>" jy jn+1 Z

for jxj < 2" where we have applied Lemma 1.2. This completes the proof. 22

References [1] A. P. Calderon/A. Zygmund: On the existence of certain singular integrals, Acta Math. 88 (1952) pp. 85-139. [2] E. M. Stein: Singular integrals and dierentiability properties of functions. Princeton University Press 1970. [3] M. Cotlar: Some generalizations of the Hardy-Littlewood maximal theorem Rev. Mat. Cuyana 1(1955) pp. 85-104.

23

24

Chapter 3 Proof of Theorem 1.6 We begin this section by introducing some of the tools we need to prove the L2-boundedness of T . BMOp(Rn): For f 2 L1loc(Rn) we set Z ?

1 kf kp; = sup jf (x) ? fQjpdx p ; 1 p < 1

Q cube

Q

R

where fQ = Q f (x)dx and dene the space BMOp to consist of those functions f such that kf kp; < 1. Thus (BMOp; k kp;) becomes a semi-normed vectorspace with semi-norm vanishing on the constant functions. The letters BMO stand for bounded mean oscillation. Examples of BMO-functions: 1) log jxj 2 BMOp(Rn) 2) L1(Rn) BMOp(Rn) Z 3) log jx ? yjd(y) 2 BMOp(Rn) for nite measures . To be able to work with this space, we only need to know three basic facts. R

Fact 1: Let f 2 L1loc. If for all cubes Q, there exist constants CQ such that (Q jf (x) ? CQjpdx)1=p `, then f 2 BMOp and kf kp; 2`. Proof. Exercise.

This fact can be used to prove following proposition. Proposition 3.1. If T bounded on L2, then T : L1 ! BMO. Proof. The rst step in the proof is to give a denition of the function Tf where f 2 L1 . We therefore introduce fQj g = the set of all cubes Q with centers with rational coordinates

25

and with rational sidelengths. Set E = [j @Qj . For each pair (x1; x2) 2 (Rn n E ) (Rn n E ) choose a cube Q 2 fQj g such that x1; x2 2 Q. Set f1 = f 2Q and f2 = f ? f1. Dene

F (x1; x2) = Tf1(x1) ? Tf1(x2) +

Z

Rn

(K (x1; y) ? K (x2; y))f2(y)dy:

We note that F is dened a.e. and that F is independent of Q (as long as x1; x2 2 Q). Check it! Furthermore, for a.e. x1 2 Rn and x2 2 Rn; F (x; x1) ? F (x; x2) is a constant (regarded as a function of x). We now dene Tf as the class x ! F (x; x1) for a.e. x1 2 Rn. It remains to show that T : L1 ! BMO is bounded. It is enough to show that Z

jF (x; xQ) ? Tf1(xQ)jdx C kf kL1 ; f 2 L1 (Rn ) Q

for all cubes Q 2 fQj g. But Z

Z

Z

jTf1(x)jdx ( jTf1(x)j2dx)1=2 C ( jf1 (x)j2dx)1=2 C kf k1 Q Q Q

since

j

Z

Z

(K (x; y) ? K (xQ; y))f2(y)dyj j n jK (x; y) ? K (xQ; y)jjf (y)jdy R n2Q j x ? x Qj C n jy ? x jn+1 dykf k1 C kf k1 for x 2 Q: Q R n2Q Rn Z

The proposition follows.

Fact 2: John-Nirenberg inequality Theorem: Let ' 2 BMO (Rn). Then there exists constants, C > 0; > 0, depending only on n, such that

jfx 2 Q : j'(x) ? 'Qj > gj C jQj exp ? k 'k ?

for all > 0 and cubes Q. Sketch of a proof. It is enough to show that Z

? sup exp k'k j'(x) ? 'Qj dx C < 1: Q cube Q

26

Assume k'k = 1 and ' 2 L1 . Since the constants C and will be independent of k'k1, the result follows for a general '. Fix a cube Q. Consider all cubes Qj in the dyadic mesh of Q and choose a t > 1. Let Q~ j denote those dyadic cubes which are maximal with respect to inclusion satisfying Z

j'(x) ? 'Q jdx > t Q~ j

and Clearly Q~ j Q and

j'(x) ? 'Qj t a.e. for x 2 Q n [1j=1 Q~ j : j [1j=1 Q~ j j 1t k' ? 'QkL1(Q) 1t jQj:

The maximality of Q~ j implies that

Z

j'(x) ? 'Q jdx t Q j

where Q j is the minimal cube in the dyadic mesh of Q with respect to inclusion for which Q~ j 6 Q j . Furthermore

Q~ j Q j

j'Q~Zj ? 'Qj j'Q~j ? 'Qj j + j'Qj ? 'Qj ~ j'(x) ? 'Qj jdx + t QZj n 2 j'(x) ? 'Qj jdx + t Qj n (2 + 1)t:

Q R

Set X (; Q) = supQj 2dyadicmeshofQ Qj exp(j'(x) ? 'Qj j)dx which is < 1 since ' 2 L1 . From the properties of Q~ j it follows that Z Z 1 exp(j'(x) ? 'Q j)dx etdx + 1 j Q j Qn[j=1 Q~ j Q Z 1 X + jQ1 j jQ~ j j exp(j'(x) ? 'Q~j j)dx exp(t(2n + 1)) Q~ j j =1 et + 1t exp(t(2n + 1))X (; Q) 27

Take supremum over all cubes Q. Thus sup X (; Q)[1 ? 1t exp(t(2n + 1))] et

Q cube

which implies supQ cube X (; Q) C if > 0 small enough. The proof is done.

Remark: It is an easy consequence of John-Nirenberg's inequality that the norms k kp; and k k k k1; are equivalent for 1 0 there exists a C > 0 such that xp C exp( x) for x > 0. Choose = 2 and apply the inequality above. Hence Z ? ' k k'k kp; C sup exp 2 j'(xk)'?k 'Qj dx = Q ZQ cube 1 ? ? 1 exp 2 t d jfx 2 Q : j'(xk)'?k 'Qj > tgj = C sup jQj Q cube Z0 1 ? 1 C sup jQj exp 2 t C jQj exp(?t) (?)dt = C Q cube 0

and k'kp; C k'k. The inequality k'k k'kp; follows from Hölder's inequality.

Fact 3: Connection between BMO and Carleson measures. Carleson measures originally appeared as answers to the following question.

Question: Which positive measures on Rn++1 have the property Z Z

Rn+1 +

jPy f (x)j2d(x; y) C ()kf k22 for all f 2 L2(Rn)

where Py f (x) = py f (x) with the Poisson kernel py (x) = cn (jxj2+yy2 ) n+1 ? 2

To obtain a necessary condition on consider f = Q, i.e., f is the characteristic function for a cube Q Rn. We immediately observe that Py f (x) C > 0 for f(x; y) : x 2 12 Q; 0 < y < `(Q)g where `(Q) = side length of Q. Set Q~ = f(; ) 2 Rn++1 : 2 Q; 0 < < `(Q)g. Hence (C ) (Q~ ) C jQj for all cubes Q Rn is a necessary condition on . We call a positive measure a Carleson measure if satises (C) and inf fC : (Q~ ) C jQj for all cubes Qg is called the Carleson norm for . 28

Lemma 3.1. Let be a continuous function in Rn++1 and set u(x) = supfju(; )j : jx ? j < g; x 2 Rn:

Let be a Carleson measure. Then (f(x; y) 2 Rn++1 : ju(x; y)j > g) C jfx 2 R : u(x) > gj for all > 0, where C only depends on n and the Carleson norm of .

Proof. The lemma is a consequence of following geometric fact: For every covergin fQj g of countably many cubes there is a subcovering fQ0j g such that [Qj = [Q0j and each x 2 [Qj belongs to at most 2n of the Q0j 's. We leave the proof of this fact as an exercise. For > 0 set E = f(x; y) 2 Rn++1 : ju(x; y)j > g and for each (x; y) 2 E dene Q~ (x; y) = f(; ) 2 Rn++1 : k ? xk < y; 0 < < yg Q(x; y) = f 2 Rn : k ? xk < yg where kxk = maxi=1;:::;n jxij. Select a covering of E consisting of countably many cubes Q~ (x; y) by a compactness argument. It is obvious that u() > for all 2 Q(x; y) and hence (E) X ([Q~ (x; y)) = ([Q~X (x; y)0) (Q~ (x; y)0) C jQ(x; y)0j 2nC jf 2 Rn : u() > gj:

We can now answer the question posed above by Theorem 3.1. If is a Carleson measure on Rn++1, then Z Z

Rn+1 +

jPy f (x)jpd(x; y) C (p; ; n)kf kpp; 1 < p 1:

Proof. Let py (x) denote the Poisson kernel. If jx ? xj < y, then py (x ? t) Cpy (x ? t) for all t 2 Rn where C is independent of x; x and y. Furthermore Py f (x) py f (x) CMf (x) and thus supfjPy f (x)j : jx ? xj < yg CMf (x): Lemma 3.1 implies Theorem 3.1 since M is bounded on Lp for 1 < p 1.

29

Remark: Theorem 3.1 is also valid for all operators of the form Pt f (x) = 't f (x)

where ' is a smooth function which decays at innity and such that j'(x)j (x) for some ? radial function 2 L1(Rn). 't(x) denotes t1n ' xt . We leave the proof of this remark as an exercise. We now introduce two families of operators denoted Pt and Qt of which the rst is an approximation of the identity and the second is an approximationR of the zero operator. Let '; be smooth functions that decay at innity such that Rn '(x)dx = 1 and R ( Rn x)dx = 0. Dene Pt and Qt by Pd t f ( ) = '^(t )f^( ) Qd t f ( ) = ^(t )f^( ) for nice functions f in Rn . We immediately observe Lemma 3.2. If f 2 L2(Rn), then Z 1 kQtf k22 dtt C ( )kf k22: 0 Proof. Apply Plancherel's formula. Theorem 3.2. If f 2 BMO (Rn), then

d(x; t) = jQtf (x)j2 dxdt t

is a Carleson measure with Carleson norm C ( )kf k2.

To carry through the argument in the proof of this theorem, we need a lemma. Lemma 3.3. If f 2 BMO (Rn) and Q0 is the unit cube (centered at 0), then Z jf (x) ? fQ0 j dx C kf k Rn 1 + jxjn+1 where C only depends on n. Proof. For a > 0 let aQ denote the cube with sides parallel with the sides of Q and of lengths a times the sidelengths of Q and with the same center as Q. We observe that for every cube Q Rn Z n Z 2 1 jfQ ? f2Qj jQj jf (x) ? f2Qjdx j2Qj jf (x) ? f2Qjdx 2n kf k: Q 2Q

30

Set Qj = 2j Q0 for j 2 N and assume kf k = 1. Here jfQj+1 ? fQj j 2n for j 2 N which implies jfQj+1 ? fQ0 j (j + 1)2n for j 2 N: Hence Z 1 Z jf (x) ? fQ0 j dx + jf (x) ? fQ0 j dx = X n +1 n+1 Rn 1 + jxj j =0 Qj+1 nQj 1 + jxj Z 1 ?Z X jf (x) ? fQj+1 j dx + j f ( x ) ? f Q0 j + dx n +1 2j(n+1) Q0 1 + jxj j =0 Qj+1 Z 1 X j fQj+1 ? fQ0 j (2n?j + (j + 1)22n?j ) + 1 = C + j (n+1) dx + 1 2 Qj+1 j =0 which completes the proof. Proof of Theorem 3.2. Qtf (x) is a well-dened function in Rn++1 since Qt1 = 0. We want to prove that for each cube Q Rn

()

Z Z

2 j Qtf (x)j2 dxdt t C ( )kf kjQj: ~

Q

It is enough to consider Q = unit cube Q0 since BMO is scale- and translation invariant, i.e., kf k = kftsk where fts(x) = f (t(x ? s)) and dtt is scale invariant. Furthermore we may assume f2Q0 = 0 since Qt1 = 0. Thus we have to prove () for Q = Q0 and all f 2 BMO with f2Q0 = 0. Set f1 = f2Q0 and f2 = f ? f1. Then Qtf = Qtf1 + Qtf2 and we obtain Z Z Z Z dxdt 2 jQtf1j t jQtf1j2 dxdt n +1 t Q0 R+ C ( )kf1k22 C ( )kf k2 from lemma 3.2 and for (x; t) 2 Q~ 0 we obtain Z ? jQtf2(x)j n t1n j x ?t z jjf2(z)jdz R n2Q0 Z C ( ) n tn+1 + jxt ? zjn+1 jf2(z)jdz ZR n2Q0 C ( )t n 1 +jf j(zzj)nj+1 dz C ( )tkf k R according to lemma 3.3. This completes the proof. We have now prepared the tools we need to prove the theorem of David and Journé. 31

Theorem 1.6: If T is a PVO with CZ type kernel K , then T is bounded on L2 i T 1 2 BMO: Here the only if-part follows from Proposition 3.1. The if-part is the hard part. The proof we present is due to Coifman/Meyer [1]. Choose ' 2 C01(Rn) such that ' radial with support in the unit ball B1(0) and such that

'^(jj) = 1 + O(jj4) as jj ! 0: d Dene Pt as above by Pd t f ( ) = '^(tj j)f^( ). Analogously dene Qt by Q tf ( ) = (tj j)2'(tj j)f ( ) and Rt by Rd t f ( ) = (j j)?1 '^0(tj j)f^( ). Hence Pt ; Qt and Rt commutes and d P 2 = 2R Q : dt t t t t This implies 1 d P 2TP 2 = 1 (R Q TP 2 + P 2TR Q ): 2 dt t t t t t t t t t The idea is as follows: We want to show jh1; T2ij C k1k2k2k2 for all 1; 2 2 S (Rn): We note that h1; Pt2TPt22i ! 0 as t ! 1 and hence it is enough to prove Z 1 j dtd h1; Pt2TPt22idtj C k1k2k2k2 for all 1; 2 2 S (Rn) 0 since P0 is the identity operator. Furthermore, it is enough to prove Z 1 j h1; RtQtTPt22i dtt j C k1k2k2k2 for all 1; 2 2 S (Rn) 0 since Pt ; Qt; Rt are selfadjoint operators and T = ?T . We need the following estimate. Lemma 3.4. Let '; 2 C01(Rn) with support in the unit ball B1(0) and assume Z

Rn

(x)dx = 0:

Then

jh tx; T'yt ij Cpt(x ? y)

where

1 ? z?x ; 'y (z ) = 1 '? z?y x t t (z ) = tn t tn t

and pt is the Poisson kernel.

32

Proof. The argument consists of a straightforward calculation where we use the CZ type properties of the kernel K .

2h

y x t ; T't i = lim "#0

= lim "#0

Z Z

j?j>"

Z Z

j?j>"

K (; )( tx()'yt () ? tx()'yt ())dd

K (t + y; t + y)

? x?y

1

t

()'() ?

x?y t

1

()'())dd:

Hence, it is enough to prove the lemma for y = 0. Assume jxj < 10t: Then we obtain jh tx; T'ij t1n ? ? jCxj 2 n+12 = Cpt(x): 1+ t Assume jxj 10t: Then we obtain x t ; T'ij = j

Z Z

jh (K (; ) ? K (x; )) tx()'t()dd Z Z x jK (t; t) ? K (x; t)j 1t ()'()dd C jxjtn+1 C 2 t 2 n+12 = Cpt(x) (t + jxj ) which concludes the proof of the lemma. Now set Lt = QtTPt where

Ltf (x) =

Z

Rn

1t(x; y)f (y)dy

with j1t(x; y)j Cpt(x ? y) according to Lemma 3.4. We recall that it is enough to show that Z 1 j h1; RtQtTPt22i dtt j C k1k2k2k2 for all 1; 2 2 S (Rn): 0 Hence it is enough to show Z 1 j h1; RtQtTPt22i dtt j C (k1k22 + k2k22) for all 1; 2 2 S (Rn): 0 But Z 1 Z 1 dt 2 j h1; RtQtTPt 2i t j jhRt1; QtTPt22ij dtt 0Z 1 0 Z 1 dt 2 krt1k2 t + kQtTPt22k22 dtt I + II: 0

0

33

Here

I=

Z 1

0

k '^ t(jtjj j) ^1()k22 dtt C k^1k22 = C k1k22: 0

To cope with II, we rewirte QtTPt22 = LtPt2 as ((LtPt)2)(x) = Lt[Pt2 ? Pt2(x)](x) + Pt2(x)Lt1(x) = = Lt[Pt2 ? Pt 2(x)](x) + Pt 2(x)QtT 1(x) But T 1 2 BMO implies jQtT 1j2 dxdt t is a Carleson measure according to Theorem 3.2 and hence Z Z Z 1 dt 2jQ T 1j2 dxdt C k k2 2 j P ( x ) j kPt2(x)QtT 1k2 t = t 2 t 2 2 n+1 t R+

0

where we have used the remark to Theorem 3.1. Furthermore using Jensen's inequality A(x;Zt) jLt(Pt 2 ? Pt 2(x))(x)j2 = Z 2 = j 1t(x; y)(Pt2(y) ? P2 2(x))dyj C pt (x ? y)jPt2(y) ? Pt2(x)j2dy and thus

But Thus

Rn

Rn

Z 1

Z 1Z

A(x; t) dxdt t 0 Rn 0 Z 1Z Z pt(x ? y)jPt2(x) ? Pt2(y)j2dydx dtt = C Z0 1 ZRn ZRn =C pt (x)jPt2(x + y) ? Pt 2(y)j2dxdy dtt Z0 1 ZRn ZRn x) ? Pt2())()j2 ddx dtt = =C pt (x)j(Pt2( +\ Z0 1 ZRn ZRn 2 ihx;ij2 j^ ( )j2 ddx dt : =C p t (x)j'^(tj j)j j1 ? e 2 t n n

kLt[Pt2 ? Pt 2(x)](x)k22 dtt

0

R

R

Z

Rn Z 1

0

=

pt(x)j1 ? eihx;ij2dx = 2 ? 2e?jjt:

kLt[Pt2 ? Pt 2(x)](x)k22 dtt Z

Z 1

2(1 ? e?jjt)j'^(tjj)j2 dtt j^2()j2d n Z 1 ZR 0Z 1 ? jj dt ?j j t C n (1 ? e ) t + 1 j'^(tjj)j2 dtt j^2()j2d R 0 jj 2 C k2k2: This completes the proof.

C

34

References [1] R. R. Coifman/ Y. Meyer: personal communication.

35

36

Chapter 4 Proof of Theorem 1.3 In this section we prove

Theorem 1.3: If ' : R ! R Lipschitz function and K' (x; y) = x ? y + i('1(x) ? '(y)) , then the corresponding operator T' is bounded on L2 and kT'k C (k'0k1). We immediately observe that the kernel K' is of CZ type and thus the L2 boundedness of the PVO T' can be proved using Theorem 1.6, i.e., it is enough to prove T'1 2 BMO. However, this is not easy. We give the proof in two steps. A: There exists an "0 > 0 such that if k'0k1 "0 then kT'k C (k'0k1). B: Removal of the constraint k'0k1 "0. Part A was proved by Calderón [1]. Part B was proved by Coifman/McIntosh/Meyer [2]. The proof we present is due to David [3]. Proof of A: Assume ' 2 C01(Rn) and dene N KN (x; y) = ('((xx)??y')N(y+1)) N = 0; 1; 2; : : : : TN corresponding PVO with kernel KN . We remark that the TN 's are called commutators and arise naturally when one tries to construct a calculus of singular integral operators to handle dierential equations with nonsmooth coecients. WePrefer to Calderón [4] for an extensive discussion of commutators N and PDE's. Since T = 1 N =0 (?i) TN , it is enough to prove that

kTN k C N +1k'0kN1 N = 0; 1; 2; : : : : 37

We also note that it is enough to prove that there exists an "1 > 0 such that if k'0k1 "1, then

kTN k C N N = 1; 2; : : : : since T0 is the Hilbert transform and this operator is L2 bounded. There are many proofs of this fact and one proof is supplied by Theorem 1.6. To prove that TN N = 1; 2; : : : are L2 bounded we make the following observation. Lemma 4.1. If ' 2 C01, then

TN +1(1) = TN ('0); N = 0; 1; : : : : Proof. The lemma is a consequence of the identity

d ? '(x) ? '(y) N +1 = (N + 1) ('(x) ? '(y))N +1 ? (N + 1) ('(x) ? '(y))N '0(y) dy x ? y (x ? y)N +2 (x ? y)N +1 and

Z

('(x) ? '(y))N +1 dy = TN +11(x) = lim "#0 jx?yj>" (x ? y)N +2 1 ? '(x) ? '(x ? ") ? '(x) ? '(x + ") + T '0(x) = T '0(x): = lim N N "#0 N + 1 " ?"

A recursion argument using Proposition 3.1, Lemma 4.1 and the fact that ' Lipschitz function implies '0 2 L1 shows that TN ; N = 0; 1; 2; : : : are L2 bounded. What remains to be done is to show that

kTN k C N N = 1; 2; : : : ; for some choice of C > 0. Here, of course, k k denotes the norm k kL2 !L2 . To conclude the proof of A we note that

1: If K is a kernel of CZ type such that

jK (x; y)j + (jrxK (x; y)j + jry K (x; y)j)jx ? yj jx ?C1yjn

and if kT 1k C2, then

kT k Dn (C1 + C2) for some constant Dn which only depends on dimension n. 38

2: Under the same assumptions as in 1

kT kL1!BMO Dn (kT k + C1) This follows from the proofs of Proposition 3.1 and Theorem 1.6. Hence

kTN +1k C1 + C2kTN k N = 0; 1; 2; : : : for some constants C1; C2 independent of N and

kTN k C N N = 1; 2; : : : for some constant C > 0 follows. The proof of A is completed. Proof of B: We start with three lemmas. Lemma 4.2. There exists an "0 > 0 such that if

'(x) = Ax + (x); x 2 R where A 2 R and : R ! R Lipschitz function with k 0k1 "0 then

kT'k C0 for some constant C0 > 0 which is independent of A. Proof. Repeating the argument above, we see that if h : R ! C Lipschitz function with kh0k < 1 then kThk C where the constant C is independent of h. Furthermore

consider T' with kernel

K' (x; y) = x ? y + i(Ax + 1(x) ? Ay ? (y)) = 1 +1 iA x ? y + i(h1(x) ? h(y)) where h(x) = 1+(xiA) . Then the rst observation gives the desired result. Lemma 4.3 (David [3]). Assume ' : R ! R Lipschitz function and L 2 R such that

j'(x) + Lx ? ('(y) + Ly)j M jx ? yj x; y 2 R: Let I R be an interval. Then there exists a Lipschtz function '~ : R ! R and a L~ 2 R such that (i) jfx 2 I : '(x) = '~(x)gj 83 jI j (ii) j'~(x) + L~ x ? ('~(y) + L~ y)j 109 M jx ? yj x; y 2 R.

39

Remark: L~ 2 [L ? M; L + M ]. Proof. Without loss of generality we can assume that I = [0; 1]; M = 1; L = ? 54 ; U fx 2 I : '0(x) + L 0g has measure 21 . Check this! Hence ? 51 '0 59 . Dene '~ : R ! R by 8 x 15. Then there exists points zk 2 ; k = 1; 2; : : : ; 6 such that zk 2 and mink=6 1 jzk ? z1j > 2. Set Mk = fx 2 J : jf (x) ? zkj < 1g. The sets P6 Mk are mutually disjoint. Furthermore the points zk can be chosen such that jJ j > k=1 jMk j. Since C : ! R is continuous, there exist intervals Ik ; k = 1; 2; : : : ; 6 such that I Ik J and C (Ik) = zk . Hence

jJ j >

6 X

k=1

jMk j

6 X

k=1

jMk \ Ik j 6 31 jI j = 2jI j:

This contradicts jJ j = 2jI j and claim is proven. We now show that under the hypothesis in the lemma with = 1 Z

I

jf (x) ? C (I )jdx C jI j

for each interval I R. Observe that we do not know whether f 2 L1loc or not. Since the assumptions on f are scale- and translationinvariant we can assume I = [0; 1]. We can also assume C (I ) = 0. Now, set k = 100k k = 1; 2; : : : and let k = [j Ijk be the union of those intervals in the dyadic mesh of I which are maximal with respect to inclusion with the property jC (Ijk)j > k . We see that I 2= 1 and 1 2 3 : : : . For each Ijk in k there exists 41

an interval I~jk in the dyadic mesh of I which is minimal with respect to inclusion and with the property Ijk * I~jk . Hence jC (I~jk)j k and the claim above implies

k < jC (Ijk)j < k + 15: We claim that if we can prove that we are done.

j k+6 j 12 j k j

Proof. Consider Ak = fx 2 I : jf (x)j > k + 2g which is a measurable set and take a point of density x0 2 Ak . Then there exists a suciently small interval J in the dyadic mesh of I such that x0 2 J and 99 jJ j: jfx 2 J : jf (x)j > k + 2gj 100 This implies C (J ) > k and thus x0 2 J k . Hence Ak k except for a set of measure 0 and

jAk j j k j: But j k+6j 12 j k j implies that there exist constants B; C > 0 such that

jAk j Be?C(k+2) k = 1; 2; : : : : Choosing a slightly larger constant C we get

jfx 2 I : jf (x)j > gj Be?C: The remaining argument is the same as in the proof of the remark on page 29. Finally we give the argument for j k+6j 12 j k j. Proof. Set Ek = fx 2 I : jf (x) ? k j < 16g k = 1; 2; : : : . Hence Ek are mutually disjoint and jEk \ Ijk j jfx 2 Ijk : jf (x) ? C (Ijk )j < 1gj 31 jIjkj: Furthermore, for k k0 we obtain jIjk00 \ Ek j 31 jIjk00 \ k j

42

by taking the union overall Ijk Ijk00 and using jEk \ Ijk j 13 jIjkj. Hence

jIjk00 j

X

k0 1 and k'0k1. Theorem 6.2. If f 2 L2(@D), then there exists a u such that

u = 0 in D @u @n j@D = f on @D where the boundary values are taken in the sense np ru(Q) ! f (P ) as Q ! P nontangentialy a.e. @D and M (ru) 2 L2 (@D) with kM (ru)k2 C kf k2 where C only depends on > 1 and k'0k1 . We recall that M u(P ) = supfju(Q)j : jP ? Qj < dist (Q; @D)g; P 2 @D and that np denotes the outward unit normal at P 2 @D which is dened a.e. on @D. Corollary 6.1. Theorem 6.1 is valid if L2 replaced by Lp for 2 p 1. 51

This is a straightforward consequence of the maximum principle and interpolation between L2 and L1 , but apart from this result we discuss the Lp-theory for the Dirichlet problem and Neumann problem in Chapter 7. The proofs of Theorems 6.1 and 6.2 involves the layer potentials Z Df (P ) = @n@ r(P; Q)f (Q)d(Q); P 2 D Q Z@D S f (P ) = r(P; Q)f (Q)d(Q); P 2 D D

where r(P; Q) = cn jP ? Qj1?n. With Q = (x; '(x)) 2 @D and P = (z; y) 2 D Z

y ? '(x) ? (z ? x) r'(x) f (x)dx n+1 Rn (jx ? z j2 + ('(x) ? y )2) 2 p Z 1 + jr'(x)j2 S f (z; y) = cn n n?1 f (x)dx: R (jx ? z j2 + ('(x) ? y )2 ) 2

Df (z; y) = cn

From Proposition 1.1, the remark on page 17 and Theorems 1.2 and 1.4 it follows that

kM (Df )kp C ( ; k'0k1)kf kp;

f 2 Lp

for 1 yg is also a special Lipschitz domain and we have the following jump relations at the interface between D and D? .

Lemma 6.1. Let f 2 L2(@D) and let Tf (P ) = lim "#0

Z

@ r(P; Q)f (Q)d(Q); P 2 @D jP ?Qj>" @nQ

Then

Dj@D f (p) = 21 f (P ) + Tf (P ) Dj@D? f (P ) = ? 21 f (P ) + Tf (P )

@ Sj f (P ) = ? 1 f (P ) + T f (P ) @np @D 2 @ Sj f (P ) = 1 f (P ) + T f (P ) @nP @D? 2 where T is the adjoint operator of T . 52

a.e. @D a.e. @D a.e. @D a.e. @D

We observe that it is equivalent to solve the Dirichlet problem in D and the Neumann problem in D? . The jumprelations are similar to those in the case of regular boundary in chapter 0 with the dierence that here the operator T has to be interpreted as a PVO. The main ingredient in the proof is Proposition 1.1. The idea is to approximate @D by a C 2-boundary and use the result from Chapter 0. We leave the proof as an exercise. The nal step in the proof of Theorem 6.1 and 6.2 is to prove that 12 I + T and 21 I + T are invertible on L2. The proof of this is due to Verchota [1], who used an indentity due to Rellich [2]. See also Jerison/Kenig [3].

Lemma 6.2. Let f 2 L2(@D) and u = S f jD . Then there exists C1; C2 > 0 such that C1

Z

@D

? @u 2

@n d

Z

@D

jrt

uj2d

C2

Z

where rt denotes the tangential derivative.

? @u 2

@D

@n d

Remark: Let : @D ! Rn denote the projection mapping (x; '(x)) 7! x. Then rtu is r(u ?1) lifted with ?1 to @D. Proof. Assume f has compact support. Set e = (0; 1). Since u = 0 in D we get

div (jruj2e ? 2 @u ru) = 0 @y called Rellich identity. Apply the divergence theorem. Hence Z Z @u d: 2 () jruj he; nid = 2 @u @D @D @y @n Since 0 < c0 he; ni 1 for Lipschitz domain, we obtain

c0

Z

@D

jruj2d

2

Z

and Schwartz inequality implies Z

@D

jruj2d C

@u d jruj @n

@D Z

? @u 2

@D

Remains to prove the reversed inequality. @u = hru; ei @y 53

@n d

where e = he; nin + et on @D. Hence @u = @u he; ni + hru; e i t @y @n

? @u 2 + jrtuj2 and which we introduce in formula () above together with jruj2 = @n jhru; etij jrtuj. Then we obtain Z Z Z ? @u 2 @u hru; e id: 2 jrtuj he; nid = h e; n i d + 2 t @D @D @n @D @n Thus Z Z Z Z ? @u 2 1 ? ? ? ? @u 2 2 2 1 2 jr d C d jr tuj d + tuj d 2 @D @n @D @D @D @n and Z Z ? @u 2 d C jrtuj2d @n @D @D

To prove that 21 I + T and 21 I + T are invertible on L2, it is enough to prove that 12 I + T are invertible. We rst claim that there exists a C > 0 such that k( 21 I + T )f k2 C jjf k2 f 2 L2(@D) Proof. Assume k( 12 I + T )f k2 = "kf k2 for some f 2 L2 (@D). But @ Sj f k kr Sj f k = k 21 f + T f k2 =k @n t @D? 2 @D? 2 = krtSj@Df k2 k @ Sj@Df k2 = k ? 1 f + T f k2 @n 2 since rtS f is continuous across the boundary. Now f = ( 12 f + T f ) ? (? 21 f + T f ) and k 21 f + T f k2 k ? 21 f + T f k2 implies that " above cannot be too small. The proof is completed.

54

The nal step for proving the invertability of 12 I + T is done with a method of continuity argument. Let U s donote the operator 21 I + T where ' is replaced by s'. We note that U s : L2 ! L2 bounded such that for 0 s 1

kU sf k2 C kf k2; C independent of s kU sf ? U t f k2 C jt ? sjkf k2; C independent of s and t

(6.1) (6.2) (6.3)

U 0 = 21 I invertible

The invertability follows easily.

Proof. Set S = fs 2 [0; 1] : U s invertibleg. Then that S 6= follows from (6.3), that S is open which follows from (6.2) and that S is closed is a consequence of (6.1) and (6.2). We only indicate the closedness of S . Assume sj ! s and U (sj ) invertible. Take g 2 L2 and fj 2 L2 such that U (sj )fj = g. (6.1) implies fj * f in L2 for a subsequence. We can also assume U (s)fj * U (s)f . Claim: U (s)f = g. Take an h 2 L2. Then

jhU (s)f ? g; hij jhU (s)f ? U (s)fj ; hij + jh(U (s) ? U (sj ))fj ; hij ! 0 as j ! 1. Hence U (s)f = g. Remark on uniqueness of solutions in Theorem 6.1 and 6.2: The u appearing in Theorem 6.1 is unique while the u appearing in Theorem 6.2 is uniquely dened up to an additive constant.

References [1] G. C. Verchota: Layer potentials and boundary value problems for Laplace's equation in Lispchitz domain. (to appear in J. Funct. Anal.) [2] F. Rellich: Darstellung der eigenwerte von ru + u durch ein Randintegral Math. Z. 46 (1940) pp. 635 - 646. [3] D. S. Jerison/C. E. Kenig: The Neumann problem on Lipschitz domains. Bull AMS vol. 4 (1981) pp. 203 - 207.

55

56

Chapter 7 Existence of solutions to Dirichlet and Neumann problems for Lipschitz domains. The optimal Lp-results In this chapter we give parts of the proofs of Theorem 7.1. For every Lipschitz domain D = f(x; y) 2 Rn+1 : '(x) < yg for ' : Rn ! R Lipschitz function there exists an " = "(D) > 0 such that for all 2 ? " < p 1 and all f 2 Lp(@D) there exists an u such that u = 0 in D uj@D = f on @D where the boundary values are taken non-tangentially a.e. @D and such that M u 2 Lp (@D) with kM ukp C kf kp where C only depends on > 1 and k'0 k1 .

Theorem 7.2. For every Lipschiatz domain D as above there exists an " = "(D) > 0 such that for all 1 < p < 2 + " and all f 2 Lp (@D) there exists an u such that u = 0 in D @u j = f on @D @n @D

where the boundary values are taken in the sense np ru(Q) ! f (P ) as Q ! P nontangentialy a.e. @D and such that M ru 2 Lp (@D) with kM rukp C kf kp where C only depends on > 1 and k'0k1.

Remark: The ranges 2 ? " 2 it is possible to construct a Lipschitz domain such that kM rukp C kf kp fails. The situation is analogous for the Dirichlet problem. See Dahlberg [1]. The extension 2 ? " < p < 2 for the Dirichlet problem and 2 < p < 2 + " for the Neumann problem is done by a real variable argument using the result for p = 2 an a good inequality. The extension to 1 < p < 2 in the Neumann problem is shown by establishing that for f 2 Hat1 , the atomic H 1 on @D, the solution of the Neumann problem with data f satises kM ruk1 C kf kHat1 . This is done by estimating the maximal function of gradient of the L2 solutions of atoms using the regularity theory for uniformly elliptic equations in self-adjoint form. The full result then follows by interpolation. The extension to 2 < p 1 in the Dirichlet problem is a consequence of the maximumprinciple and interpolation. We begin to discuss the regularity theory for uniformly elliptic equations in self-adjoint form. Let A(x) = (aij (x)) be a n n dimensional symmetric matrix valued function in D where the entries aij (x) are bounded real-valued measurable functions. We assume that A(x) is uniformly elliptic on D, i.e., there exists a 1 such that n 1 jj2 X aij (x)ij jj2 for all 2 Rn i;j =1 n X

@ a (x) @ . We call u a (weak) solution of Lu = 0 in @xj ij @xj i;j =1 Z 2 D if u 2 L1; loc(D) and hAru; r'idx = 0 for all ' 2 C01(D). Here L21; loc denotes D the space of functions in L2loc(D) with distributional derivatives of rst order in L2loc(D). We say that u is a subsolution (supersolution) of Lu = 0 in D if u 2 L21; loc(D) and Z Z hAru; r'idx 0 ( hAru; r'idx 0) for all 0 ' 2 C01(D). The main result is Let L denote the operator

D

D

Theorem 7.3 (DeGiorgi [2], Nash [3]). If u is a solution of Lu = 0 in D, then u is

Hölder continuous.

This follows from

Theorem 7.4 (Harnack's inequality). If u 0 and Lu = 0 in D and if K D is a compact set, then

ess supK u C ess infK u where C = C (n; ; K; D).

Remark: Harnack's inequality is a quantitative version of the maximum principle. Remark: For notational convenience we let min and max denote ess inf and ess sup, resp. 58

Proof of Theorem 7.3. This is done using Harnack's inequality. Assume Lu = 0 in D = fx 2 Rn : jxj < 2g. Set

M (r) = maxjxjr u m(r) = minjxjr u

r < 1:

Then M (r) ? u and u ? m(r) are solutions and 0 in fx 2 Rn : jxj rg. Hence ? ? M (r) ? m 2r C (M (r) ? M r2 ? ? ? M 2r ? m(r) C m 2r ? m(r)

Add these inequalities and set (r) = M (r) ? m(r). We obtain ? 1 (r) 2r CC ? +1 and hence

(r) r (1) for some > 0: Note that we have used the same constant C in the repeated uses of Harnack's inequality. This is justied by the scale invariance properties of C . We leave it as an exercise to check this. Proof of Theorem 7.4. (The proof is due to Moser [4]). Set Q(h) = fx 2 Rn : jxj j < h2 g. By a covering argument it is enough to prove the theorem for K = Q(1) and D = Q(4). Thus we assume u 0 and Lu = 0 in Q(4). Set Z Z 1=p ? 1 p u dx updx 1=p '(p; h) = jQ(h)j Q(h) Q(h) ?

for 0 < h < 4 and ?1 0 small and then let " tend to zero in the estimates. Since max u = plim '(p; h) !1 Q(h) min u = p!?1 lim '(p; h); Q(h) the theorem is equivalent to show that for some C = C (; n) we have

'(1; 1) C'(?1; 1): The proof is based on three general inequalties relating integrals of functions v = v(x) to integrals of the gradient of v. We assume n 3. 59

Inequality A (Poincare's inequality) Z

Z

jv ? vQ(h)j2dx Ch2 jrvj2dx Q(h) Q(h)

R

where vQ(h) = Q(h) vdx.

Inequality B (Sobolev's inequality). Set K = n ?n 2 . Z ?

Q(h)

1 jvj2Kdx K

Z

Z

C (h2 jrvj2dx + jvj2dx Q(h) Q(h)

Inequality C (John-Nirenberg's inequality). If kvk 1, then there exists > 0 and C only depeding on n such that Z

Q(2)

evdx

Z

Q(2)

e?v dx C:

The proof will be given in two parts viz.

Proposition 7.1. If u 0, subsolution in Q(4), then ?

p

max uC p?1 Q(1)

Z 2 ?

Q(2)

updx

1=p

for p > 1

Proposition 7.2. If u > 0, supersolution in Q(4), then Z ?

1 2 min u for 0 1, the theorem follows from the propositions. The proof of the propositions will be done by estimating '(p; h)='(p0; h0) for p > p0 and derive the desired estimates by iteration. We need the following lemma.

Lemma 7.1. If u > 0 is a subsolution in D and v = u , then for any function 2 C01(D) one has

Z

D

? 2jrvj2dx C

2 ? 1

2

Z

jrj2v2dx if > 21 : D

The same assertion is true for supersolutions if < 21 .

60

R

Proof. u subsolution in D implies that D hAru; r'idx 0 for all 0 ' 2 C01 (D). Choose 2 C01(D) and > 0 and set ' = u2. (This ' does not belong to C01(D), but we have the approximation business as an exercise to the reader). This implies Z

D

hAru; ruiu?12dx +

Z

D

hAru; ri2udx 0:

The uniform ellipticity of A implies 1 Z jruj2u?12dx + Z hAru; ri2udx 0: D D 1 . Then we obtain Introduce v = u where = + 2 Z ?1 jrvj22dx C Z jhru; rij jj ju +1 2 jju 2 jdx 2 D D

Thus

Z 1 C (jrjv)(jrvj jj)dx D Z

? jrvj22dx c

2

Z

jrj2v2dx; > 12 : D

2 ? 1 The argument for u supersolution and < 12 is similar. D

Now let 0 < h0 < h < 2h0 < 4 and choose 2 C01(Q(h)) such that 0 1; = 1 on Q(h0) and jrj C h ?1 h0 . Since u solution to Lu = 0 in Q(h), u is both a subsolution and a supersolution and Lemma 7.1 implies that for 6= 12 Z Z ? 2 1 2 dx: 2 jru j dx C u 0 2 (2 ? 1) (h ? h ) Q(h) Q(h0 ) Set p = 2 . Inequality B gives us Z Z ? upK dx) K1 C ? p 2 ? 1 2 up dx; p ? 1 hh0 ? 1 Q(h) Q(h0 ) i.e, ? 2 ? 2 '(Kp; h0) C p ?p 1 p hh0 ? 1 ? p '(p; h) if p > 1 and ? 2 '(Kp; h0) C hh0 ? 1 ? p '(p; h) if p < 0: 61

Now for p > 1 let

p = K p h = 1 + 2? h0 = h+1 We nd since

? Y i=1

= 0; 1; 2; : : :

Kip C that Kip ? 1 '(p+1 ; h+1 ) C =K '(p ; h ) for = 1; 2; : : :

and iteration yields

? ? '(p+1 ; h+1) C p ?p 1 2'(p; h) = C p ?p 1 2'(p; 2): But lim sup '(p ; h ) '(+1; 1) and the proof of Proposition 7.1 is completed.

!1

To prove Proposition 7.2, we rst note that

'(?1; 1) C '(?q; 2) q > 0 follows if we apply the same iteration technique as above to p < 0 and especially to ?q < 0 close to 0. What remains to be shown is that

'(?q; 2) C'(q; 2) and

'(q; 2) C'(p; 3) where 0 < p < K is the parameter that appears in the proposition. The rst inequality follows from Inequality C if q . Proof. It is enough to show that v = log u 2 BMO (Q(2)). Take any cube Q Q(2) and choose 2 C0(Q(3)) such that = 1 on Q. Since u is a supersolution Z

D

hAru; r'idx 0 for all 0 ' 2 C01(D):

Choose ' = 2 u1 . We get Z

Q(3)

hAru; rui2

1 dx C Z hAru; ri 1 dx u2 u Q(3) 62

and with Schwarz' inequality and the uniform ellipticity Z Z 1 2 2 jruj u2 dx C jrj jruj u1 dx: Hence

Q(3)

and Inequality A implies

Q(3)

Z

Q

jrvj2dx C Z

Q

Z

Q(3)

jrjdx

jv ? vQj2dx C

where C independent of Q Q(2). The inequality '(?q; 2) '(q; 2) follows from Inequality C . Finally, for 0 0 such that qK = p for some 2 N and q in Ineauality C . Finitely many applicaitons of Lemma 7.1 and Inequality B regarding u as a positive supersolution gives '(q; 2) C'(p; 3): This concludes the proof of Proposition 7.2. The rest of the proof the Theorem 7.1 and Theorem 7.2 can be found in Dahlberg/Kenig [5]. See Appendix. There can also be found the corresponding results for bounded Lipschitz domains using a patching technique. We nally remark that the solution of the Dirichlet problem is unique and the solution of the Neumann problem is unique to an additive constant.

References [1] Dahlberg, B. E. J.: Estimates of harmonic measure. Arch. Rat. Mech. Anal. 65 (1977) pp. 278 - 288. [2] DeGiori, E.: Sulle dierenziabilata e analiticita della estremali degli integrali multipli regotari. Mem. Acad. Sci. Torino 3 (1957) pp. 25 - 43. [3] Nash, J.: Continuity of the solutions of parabolic and elliptic equations. Amer. J. Math. 80 (1957) pp. 931 - 954. [4] Moser, J.: On Harnack's theorem for elliptic dierential operator Comm. P.A.M. 14 (1961) pp. 577 - 591. [5] Dahlberg, B. E. J. /Kenig, C. E.: Hardy spaces and the Lp Neumann problem for Laplace's equation in a Lipschitz domain. Preprint. 63

Index page

BMO : : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : 26 Calderón-Zygmund decomposition : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : : 21 Calderón-Zygmund type (CZ) kernel : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : 18 Carleson measure : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : 30 Carleson norm : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : 30 Cauchy integral : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : 1 characteristic function : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : 30 commutators : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : 39 cone : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : : 12 Cotlar inequality : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : 23 Hardy-Littlewood maximal function : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : 14 John-Nirenberg's inequality :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : 27 Lipschitz domain : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : 12 Lipschitz function : : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : : 12 Marcinkiewicz' interpolation theorem : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : 15 Method of rotations : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : 51 Nontangential cone : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: 12 Nontangential limit : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : 12 Nontangential maximal function :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : 12 Poisson kernel : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : : 29 Principal value operator (PVO) : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : 18 Rellich identity : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : 55 Schwartz class S : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : 17 Singular integral operators (SIO) : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : 17 64

Special Lipschitz domain : : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : 12 Solution to the Dirichlet Problem : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : : 12 subsolution : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : : 60 supersolution : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : :: : : : : : :: : : : : :: : : : : : :: : : : : : 60

65

66

Appendix 1

University of Minnesota Mathematics Report 84-133

Recent Progress on Boundary Value Problems on Lipschitz Domains by Carlos E. Kenig School of Mathematics University of Minnesota Minneapolis, MN 55455 U.S.A.

Supported in part by the NSF, and the Alfred P. Sloan Foundation

Introduction In this note we will describe, and sketch the proofs of some recent developments on boundary value problems on Lipschitz domains. In 1977, B. E. J. Dahlberg was able to show the solvability of the Dirichlet problem for Laplace's equation on a Lipschitz domain D, and with L2(@D; d) data and optimal estimates. In fact, he proved that given a Lipschitz domain D, there exists " = "(D) such that this can be done for data in Lp(@D; d); 2 ? " p 1. (See [6], [7] and [8]). Also, simple examples show that given p < 2, there exists a Lipschitz domain D where this fails in Lp(@D; d). Dahlberg's method consisted of a careful analysis of the harmonic measure. His techniques relied on positivity, Harnack's inequality and the maximum principle, and thus, they were not applicable to the Neumann problem, to systems of equations, or to higher order equations. In 1978, E. Fabes, M. Jodeit, Jr. and N.Riviere ([15]) were able to utilize A. P. Calderon's theorem ([1]) on the boundedness of the Cauchy integral on C 1 curves, to extend the classical method of layer potentials to C 1 domains. They were thus able to resolve the Dirichlet and Neumann problem for Laplace's equation, with Lp(@D; d) data, and optimal estimates, for C 1 domains. They relied on the Fredholm theory, exploiting the compactness of the layer potentials in the C 1 case. In 1979, D. Jerison and C. Kenig [20], [21] were able to give a simplied proof of Dahlberg's results, using an integral identity that goes back to Rellich ([33]). However, the method still relied on positivity. Shortly afterwards, D. Jerison and C. Kenig, ([22]) were also able to treat the Neumann problem on Lipschitz domains, with L2(@D; d) data and optimal estimates. To do so, they combined the Rellich type formulas with Dahlberg's results on the Dirichlet problem. This still relied on positivity, and dealt only with the L2 case, leaving the corresponding Lp theory open. In 1981, R. Coifman, A. McIntosh and Y. Meyer [3] established the boundedness of the Cauchy integral on any Lipschitz curve, opening the door to the applicability of the method of layer potentials to Lipschitz domains. This method is very exible, does not rely on positivity, and does not in principle dierentiate between a single equation or a system of equations. The diculty then becomes the solvability of the integral equations, since unlike in the C 1 case, the Fredholm theory is not applicable, because on a Lipschitz domain operators like the double layer potential are not compact. For the case of the Laplace equation, with L2(@D; @) data, this diculty was overcome by G. C. Verchota ([36]) in 1982, in his doctoral dissertation. He made the key observation that the Rellich identities mentioned before are the appropriate substitutes to compactness, in the case of Lipschitz domains. Thus, Verchota was able to recover the L2 results of Dahlberg [7] and of Jerison and Kenig [22], for Laplace's equation on a Lipschitz domain, but using the method of layer potentials. This paper is divided into two sections. The rst section which consists of two parts, deals with Laplace's equation on Lipschitz domains. The rst part explains the L2 results of 70

Verchota mentioned above. The second part deals with a sketch of recent joint work of B. Dahlberg and C. Kenig (1984) ([9]). We were able to show that given a Lipschitz domain D Rn, there exists " = "(D) such that one can solve the Neumann problem for Laplace's equation with data in Lp(@D; @); 1 < p 2 + ". Easy examples show that this range of p's is optimal. Moreover, we showed that the solution can be obtained by the method of layer potentials, and that Dahlbergs solution of the Lp Dirichlet problem can also be obtained by the method of layer potentials. We also obtained endpoint estimates for the Hardy space H 1(@D; d), which generalize the results for n = 2 in [25] and [26], and for C 1 domains in [16]. The key idea in this work is that one can estimate the regularity of the so-called Neumann function for D, by using the De Giorgi-Nash regularity theory for elliptic equations with bounded measurable coecients. This, combined with the use of the so-called 'atoms' yields the desired results. The second section, which consists of three parts, deals with higher order problems. In parts 1 and 2, we treat L2 boundary value problems for systems of equations. Part 1 deals with the systems of elastostatics, whicle part 2 deals with the Stokes system of hydrostatics. The results in part 1 are joint work of B. Dahlberg, C. Kenig and G. Verchota (see [12]), while the results in part 2 are joint work of E. Fabes, C. Kenig and G. Verchota (see [17]). The results obtained had not been previously available for general Lipschitz domains, although a lot of work has been devoted to the case of piecewise linear domains. (See [27], [28] and their bibliographies). For the case of C 1 domains, our results for the systems of elastostatics had been previously obtained by A. Gutierrez ([19]), using compactness and the Fredholm theory. This is, of course, not available for the case of Lipschitz domains. We are able to use once more the method of layer potentials. Invertability is shown again by means of Rellich type formulas. This works very well in the Dirichlet problem for the Stokes system (see part 2), but serious diculties occur for the systems of elastostatics (see part 1). These diculties are overcome by proving a Korn type inequality at the boundary. The proof of this inequality proceeds in three steps. One rst establishes it for the case of small Lipschitz constant. One then proves an analogous inequality for non-tangential maximal functions on any Lipschitz domain, by using the ideas of G. David ([13]), on icreasing the Lipschitz constant. Finally, one can remove the non-tangential maximal function, using the results on the Dirichlet problem for the Stokes system, which are established in part 2. See parts 1 and 2 for the details. Some partial results in this direction were previously announced in [26]. The third part of Section 2 deals with the Dirichlet problem for the biharmonic equation 2 ( a fourth order elliptic equation), on an arbitrary Lipschitz domain in Rn. This sketches joint work of B. Dahlberg, C. Kenig and G. Verchota ([11]). The case of C 1 domains in the plane was previsouly treated by J. Cohen and J. Gosselin [2], using layer potentials and compactness. We are able to reduce the problem, for an arbitrary Lipschitz domain in Rn, to a bilinear estimate for harmonic functions. This is a Lipschitz domain version of the paraproduct of J. M. Bony. See part 3 of Section 2 for further details. Compete proofs of the results explained in Section 1, part 2, and Section 2, will appear in future publicaitons. 71

Acknowledgements: As was mentioned before, the results in Section 1, part 2 are joint

work with B. Dahlberg, the results in Section 2, part 1, joint work with B. Dahlberg and G. Verchota, the results in Section 2, part 2, joint work with E. Fabes and G. Verchota, and the results in Seciton 2, part 3, joint work with B. Dahlberg and G. Verchota. It is a great pleasure to express my gratitutde to B. Dahlberg, E. Fabes and G. Verchota, for their contributions to our joint work. I would also like to thank A. McIntosh for pointing out the applicability of the continuity method in Seciton 1, part 1, and for pointing out to us the work of Necas ([30]). I also would like to thank G. David for making his unpublished result (Lemma 2.1.10 in Section 2, part 1) available to us. Part of this research was carried out while I was visiting the Center for Mathematical Analysis at the Australian National University, Princeton University and the University of Paris, Orsay. I would like to thank these institutions for their kind hospitality.

72

Section 1: Laplace's equation Part 1: The theory on a Lipschitz domain, for Laplace's equation, by the method of layer potentials A bounded Lipschitz domain D Rn is one which is locally given by the domain above the graph of a Lipschitz function. For such a domain D, the non-tangential region of opening at a point Q 2 @D is ?(Q) = fX 2 D : jX ? Qj < (1 + )dist (X; @D)g. All the results in this paer are valid, when suitably interpreted for all bounded Lipschitz domains in Rn; n 2, with the non-tangential approach regions dened above. For simplicity, in this exposition we will restrict ourselves to the case n 3 (and sometimes even to the case n = 3), and to domains D Rn; D = f(x; y) : y > '(x)g, where ' : Rn?1 ! R is a Lipschitz function, with Lipschitz constant M , i.e., j'(x) ? '(x0)j M jx ? x0j. D? = f(x; y) : y < '(x)g. For xed M 0 < M; ?e (x) = f(z; y) : (y ? '(x)) > ?M 0jz ? xjg D? , and ?i(x) = f(z; y) : (y ? '(x)) > M 0jz ? xjg D. Points in D will usually be denoted by X , while points on @D by Q = (x; '(x)) or simply by x. Nx or NQ will denote the unit normal to @D = at Q = (x; '(x)). If u is a function dened on Rn n , and Q 2 @D; u(Q) will denote lim X !Q u(X ) or lim X !Q u(X ), respectively. If u is a X 2?e(Q) X 2?i(Q) function dened on D; N (u)(Q) = supX 2?i(Q) ju(X )j. We wish to solve the problems u = 0 in D (D) uj = f 2 L2(@D; d) ; (N) @uuj = =0 f in2 LD2(@D; d) @D @N @D The results here are

Theorem 1.1.1: There exists a unique u such that N (u) 2 L2(@D; d), solving (D), where the boundary values are taken non-tangentially a.e. Moreover, the solution u has the form Z hX ? Q; NQi g(Q)d(Q); 1 u(X ) = ! n @D jQ ? X jn for some g 2 L2(@D; d).

Theorem 1.1.2. There exists a unique u tending to 0 at 1, such that N (ru) 2 L2(@D; d), solving (N) in the sense that NQ ru(X ) ! f (Q) as X ! Q non-tantentially a.e. Moreover, the solution u has the form Z ? 1 u(X ) = ! (n ? 2) jX ?1Qjn?2 g(Q)d(Q); n @D 73

for some g 2 L2(@D; d). In order to prove the above theorems, we introduce Z hX ? Q; NQi g(Q)d(Q) 1 Kg(x) = ! n @D jX ? Qjn and Z ? 1 Sg(X ) = ! (n ? 2) jX ?1Qjn?2 g(Q)d(Q): n @D If Q = (X; '(x)); X = (z; y), then Z y ? '(x) ? (z ? x) r'(x) g(x)dx 1 Kg(z; y) = ! n n Rn?1 [jx ? z j2 + ['(x) ? '(z )]2] 2 p

Z 1 + jr'(x)j2 ? 1 Sg(z; y) = ! (n ? 2) n?1 n?2 g (x)dx: n R [jx ? z j2 + ['(x) ? y ]2] 2

Theorem 1.1.3. a) If g 2 Lp(@D; d); 1 " [jx ? z j2 + ['(x) ? '(z )]2]n=2 exists a.e. and

kKgkLp(@D;d) C kgkLp(@D;d); 1 " [j2 ? xj2 + ['(z) ? '(x)]2]n=2 g(x)dx exists a.e. and in Lp(@D; d), and its Lp norm is bounded by C kgkLp(@D;d); 1 "

Corollary 1.1.4. (Nz rSg)(z) = 21 g(z) ? K g(z), where K is the L2(@D; d) adjoint of K . The proof of Theorem 1.1.3 is an easy consequence of the deep results of Coifman-McIntoshMeyer ([3]). 74

It is easy to see that (at least the existence part) of Theorems 1.1. and 1.1.2 will follow immediately if we can show that ( 12 I + K ) and 21 I + K ) are invertible on L2(@D; d). This is the result of G. Verchota ([36]).

Theorem 1.1.5. ( 12 I + K ); ( 21 I + K ) are invertible on L2(@D; d). In order to do so, we show that if f 2 L2(@D; d); k( 21 I + K )f kL2(@D;d) k( 21 I ? K )f kL2(@D;d), where the constants of equivalence depend only on the Lipschitz constant

M . Let us take this for granted, and show, for example, that 12 I + K is invertible. To do this, note rst that if T = 12 I + K ; kTf kL2 C kf kL2 , where C depends only on the Lipschitz constant M . For 0 t 1, consider the operator Tt = 21 I + Kt, where Kt is the operator corresponding to the domain dened by t'. Then, T0 = 21 I; T1 = T , and @ T : Lp(Rn?1) ! Lp(Rn?1); 1 < p < 1 with bound independent of t, by the theorem of @t t Coifman-McIntosh-Meyer. Moreover, for each t, kTtf kL2 C kf kL2 , C independent of t. The invertibility of T now follows from the continuity mehtod:

Lemma 1.1.6. Suppose that Tt : L2(Rn?1 ) ! L2(Rn?1) satisfy (a) kTtf kL2 C1kf kL2 (b) kTtf ? Tsf kL2 C2jt ? sjkf kL2 ; 0 t; s 1 (c) T0 : L2(Rn?1) ! L2(Rn?1) is invertible.

Then, T1 is invertible. The proof of 1.1.6 is very simple. We are thus reduced to proving (1.1.7) k( 21 I + K )f kL2 (@D;d) k( 12 I ? K )f kL2(@D;d). In order to prove (1.1.7), we will use the following formula, which goes back to Rellich [33] (see also [31], [30], [22]).

Lemma 1.1.8. Assume that u 2 Lip (D ); u = 0 in D, and u and its derivatives are suitablly small at 1. Then if en is the unit vector in the direction of the y-axis, Z

@D

hNQ; en

ijruj2d

=2

75

Z

@u @u d: @D @y @N

@ jruj2 = 2 @ ru ru, while div @u ru = Proof. Observe that div (enjruj2) = @y @y @y @ ru ru + @u div ru = @ ruru. Stokes' theorem now gives the lemma. @y @y @y ?

We will now deduce a few consequences of the Rellich identity. Recall that p 1 Nx = (?r'(x); 1)= 1 + jr'(x)j2, so that (1 + M 2 )1=2 hNx ; en i 1.

Corollary 1.1.9. Let u be as in 1.1.8, and let T1(x); T2(x); : : : ; Tn?1(x) be an orthogonal n?1 X basis for the tangent plane to @D at (x; '(x)). Let jrtu(x)j2 = jhru(x); Tj (x)ij2. Then, j =1

Z

Z

? @u 2

@D

@N d C

@D

jrtuj2d:

Proof. Let = en?hNx; eniNx , so that is a linear combination of T1(x); T2(x); : : : ; Tn?1(x). Then,

@u = hN ; e i @u + h; rui: x n @y @N Also, ? @u 2 + jrtuj2; jruj2 = @N

and so,

Z

? @u 2

Z

hNx; eni @N d + hNx ; enijrtuj2d = @DZ ? @u ? @u 2 d: = 2 hNx; eni @N + 2 h; rui @N @D Z

@D

@D

Hence,

Z

? @u 2

hNx ; eni @N d = @D

So,

Z

? @u 2

d C @D @N and the corollary follows.

Z

@D

Z

@D

hNx; enijrtuj2d ? 2

jrtuj2d + C (

Z

76

@D

Z

@D

@u d: h; rui @N

Z

jrtuj2d)1=2(

@D

? @u 2

@N d

1=2

;

Corollary 1.1.10. Let u be as in 1.1.8. Then, Z

@D

Proof.

Z

@D

jruj2d

2

?

Z

@D

jrtuj2d

c

? jruj2d 1=2

Z

? @u 2

@D

Z

@D

@u j2d j @N

Corollary 1.1.11. Let u be as in 1.1.8. Then Z

@D

jrtuj2d

@N d:

Z

@D

1=2

, and the corollary follows.

@u j2d: j @N

In order to prove 1.1.7, let u = Sg. Because of 1.1.3c, rtu is continuous across the boundary, while by 1.1.4, ? @u 1 I ? K )g: = ( @N 2 We now apply 1.1.11 in D and D? , to obtain 1.1.7. This nishes the proof of 1.1.1 and 1.1.2. We now turn our attention to L2 regularity in the Dirichlet problem.

Denition 1.1.12. f 2 Lp1(); 1 < p < 1, if f (x; '(x)) has a distributional gradient in Lp(Rn?1). It is easy to check that if F is any extension to Rn of f , then rxF (x; '(x)) is well dened, and belongs to Lp(). We call this rtf . The norm in Lp1() will be krtf kLp (). Theorem 1.1.13. The single layer potential S maps L2() into L21() boundedly, and

has a bounded inverse.

Proof. The boundedness follows from 1.1.3 a). Because of the L2-Neumann theory, and @S (f )k 2 C kf k 2 . The argument used in the proof of 1.1.11, krtS (f )kL2 () C @N L () L () 1.1.5 now proves 1.1.13.

Theorem 1.1.14. Given f 2 L21(), there exists a harmonic function u, with kN (ru)kL2() C krtf kL2 (), and such that rtu = rtf (a.e.) non-tangentially on . u is unique (modulo constants), and we can choose u = S (g), where g 2 L2(). The existence part of 1.1.14 follows directly from 1.1.13. 77

Part 2: The Lp theory for Laplace's equation on a Lipschitz domain The main results in this section are:

Theorem 1.2.1. There exists " = "(M ) > 0 such that, given f 2 Lp(@D; d), 2 ? " p < 1, there exists a unique u harmonic in D, with N (u) 2 Lp(@D; d), such that u

converges non-tangentially almost everywhere to f . Moreover, the solution u has the form Z hX ? Q; NQi g(Q)d(Q); 1 u(x) = ! n @D jX ? Qjn

for some g 2 Lp(@D; d).

Theorem 1.2.2. The exists " = "(M ) > 0, such that, given f 2 Lp(@D; d); 1 < p 2+ ", there exists a unique u harmonic in D, tending to 0 at 1, with N (ru) 2 Lp(@D; d), such that NQ ru(X ) covnergens non-tangentially a.e. to f (Q). Moreover, u has the form Z ? 1 u(X ) = ! (n ? 2) jX ?1Qjn?2 g(Q)d(Q); n @D for some g 2 Lp(@D; d). Theorem 1.2.3. There exists " = "(M ) > 0 such that given f 2 Lp1(); 1 < p 2 + ", there exists a harmonic funciton u, with

kN (ru)kLp() C krtf kLp(); and such that rtu = rtf a.e. non-tangentially on , u is unique (modulo constants). Moreover, u has the form Z ? 1 u(x) = ! (n ? 2) jX ?1Qjn?2 g(Q)d(Q); n @D

for some g 2 Lp(@D; d). The case p = 2 of the above theorems was discussed in Part 1. The rst part of 1.2.1 (i.e., without the representation formula), is due to B. Dahlberg (1977) ([7]). Theorem 1.2.3 was rst proved by G. Verchota (1982) ([36]). The representation formula in 1.2.1, Theorem 1.2.2, and the proof that we are going to present of 1.2.3 are due to B. Dahlberg and C. Kenig (1984) ([9]). Just like in Seciton 1, 1.2.1, 1.2.2, and 1.2.3 follow from: 78

Theorem 1.2.4. There exists " = "(M ) > 0 such taht ( 12 I ? K ) is invertible in Lp(@D; d); 1 < p 2 + "; ( 12 I + K ) is invertible in Lp(@D; d); 2 ? " p < 1, and S : Lp(@D; @) ! Lp1(@D; d) is invertible, 1 < p 2 + ". In order to prove Theorem 1.2.4, just as in Part 1, it is enough to show that if u = Sf , f since, then, for 1 < p 2 + ",

@u

p krtukLp(@D;d) @N L (@D;d): This will be done by proving the following two theorems:

@u Theorem 1.2.5. Let u = 0 in D. Then kN (ru)kLp(@D;d) C @N Lp (@D;d); 1 < p

2 + ".

Theorem 1.2.6. Let u = 0 in D. Then kN (ru)kLp(@D;d) C krtukLp(@D;d); 1 < p 2 + ": We rst turn our attention to the case 1 < p < 2 of Theorem 1.2.5. In order to do so, we introduce some denitions. A surface ball B in is a set of the form (x; '(x)), where x belongs to a ball in Rn?1.

Denition 1.2.7. An atom R a on is a function supported in a surface ball B , with kakL1 1=(B ), and with ad = 0.

Notice that atoms are in particular L2 functions. The following interpolation theorem will be of importance to us.

Theorem 1.2.8. Let T be a linear operator such that kTf kL2() C kf kL2(), and such that for all atoms a, kTakL1() C . Then, for 1 < p < 2, kTf kLp() C kf kLp(). For a proof of this theorem, see [5]. Thus, in order to establish the case 1 < p < 1 of 1.2.5, @u is an atom, then kN (ru)k 1 C . By dilation and it suces to show that if a = @N L () translation invariance we can assume that '(0) = 0; supp a B1 = f(x; '(x)) : jxj < 1g. Let B be a large ball centred at (0; 0) in Rn, which contains (x; '(x)); jxj < 2. The diameter of B depends only on M . Since kakL2() (B1 )1=2 = C , by the L2-Neuman 1 theory, Z

@D\B

N (ru) C

Z

@D\B

79

N (ru)2d c:

R

Thus, we only have to estimate CB\@D(ru)d. We will do so by appealing to the regularity theory for divergence form elliptic equations. Consider the bi-Lipschitzian mapping : D ! D? given by (x; y) = (x; '(x) ? [y ? '(x)]). Dene u on D? by the formula u = u ?1 . A simple calculation shows that, in D? ; u veries (in the weak sense) the equation div (A(x; y)ru) = 0, where A(x; y) = J1(X ) (0)t(X ) (0)(X ), where X = ?1(x; y). It is easy to see that A 2 L1(D? ), and hA(x; y); i C jj2. Notice also @u B B \ @D. Dene now that supp @N 1 for (x; y) 2 D B (x; y) = IA(x; y) for (x; y) 2 D? ; and y) for (x; y) 2 D u~(x; y) = uu(x; (x; y) for (x; y) 2 D? : @u = 0 in @D n B , it is very easy to see that u~ is a (weak) solution in Rn n Because @N B of the divergence form elliptic equation with bounded measurable coecients, Lu~ = div B (x; y)ru~ = 0. In order to estimate u, (and hence ru) at 1, we use the following theorem of J. Serrin and H. Weinberger ([34]).

Theorem 1.2.9. Let u~ solve Lu~ = 0 in Rn n B , and suppose that ku~kL1(RnnB) < 1. Let g(X ) solve Lg = 0 in jX j > 1, with g(X ) jX j2?n . Then, u~(X ) = u~1 + g(X )+ v(X ), where Lv = 0 in Rn n B , and jv(X )j C ku~kL1(RnnB) jX j2?nR? , where > 0; C > 0 depend only on the ellipticity constants of L., Moreover, = c B (X )ru~(X ) r (X ), where 2 C 1(Rn); = 0 for X in 2B , and 1 for large X . Let us assume for the time being that u is bounded, and let us show that if is as in 1.2.9, then = 0. Pick a as in 1.2.9. In D; B (X ) = I , and so Z

D

B rur =

Z

D

ru r = "lim !0

where

Z

D"

ru r ;

D" = f(x; y) : j(x; y)j < ; y > '(x) + "g; and is large. The right-hand side equals Z Z @u @u ; lim = lim [ ? 1] "!0 @D" @N "!0 @D" @N since, by the harmonicity of u, Z @u = 0: @D" @N 80

Let

@D;" 1 = f(x; y) 2 @D" : y = '(x) + "g; and @D;" 2 = @D" n @D;" 1. Then, Z

Z Z Z @u @u @u [ ? 1] @N + "lim lim [ ? 1] @N = "lim [ ? 1] @N = [ ? 1]a = "!0 @D" !0 @D;" 1 !0 @D;" 2 @D Z Z Z = a? a = a = 0;

@D

@D

R

@D

R

since 0 on supp a. Moreover D? B ru~r = D ru r , where = , by our construction of B . The last term is also 0 by the same argument, and so = 0. We now show that u (and hence u~) is bounded. We Z will assume that n 4 for simplicity. ) d(Q), with kf k 2 C . Since kakL2() C , we know that u(X ) = Cn jX f?(Q L () Qjn?2 @D Now, for X 2 D1 = f(x; y) : y > '(x) + 1g; jX ?1Qjn?2 1 + jCQjn?2 2 L2() and so u 2 L1 (D1). Let now B be any ball in Rn so that 2B Rn n B , B is of unit size, and such thatR a xed fraction of B is contained in D1 . Since N (ru) 2 L2(), with norm uj2 C , and moreover on B \ D1; ju(x)j C . Therefore, by the less than C; 2B\D jr R ?R 2 2 1 = Poincare inequality 2B u~ C . But, since u~ solves Lu~ = 0; maxB u~ C 2B ju~j ) 2 C , ([29]). Therefore, u~ 2 L1 (Rn n B ); ku~kL1(Rn nB) C . Hence, since = 0R; ru = rv, and jv(x; y)j C=(jxj + jyj)n?2+ ; > 0. For R R0 = diam B , set b(R) = AR N (ru)2, where AR = f(x; '(x)) : R < jxj < 2Rg. For each xed R, let

N1(ru)(x) = supfjru(z; y)j : (z; y) 2 ?i (x); dis ((z; y); @D) Rg; N2(ru)(x) = supfjru(z; y)j : (z; y) 2 ?i (x); dist ((z; y); @D) Rg: In the set where the sup in N2 is taken, u is harmonic, and the distance of any point X to the boundary is comparable to jX j. Thus, using our bound on v, we see that R n ? 1+ n ? 1+ N2(ru)(x) C=jX j C=R , and so AR N2(ru)2 CR1?n?2 . Let now ?

= f(x; y) : '(x) < y < '(x) + CR; R < jX j < r?1Rg; 2 41 ; 12 : R R By the L2-Neumann theory in ; AR N1(ru)2d C @ jruj2d. Integrating in from 1=4 to 1=2 gives Z Z Z C C 2 2 u2 ; jruj dX R3 N1(ru) d R

1=4 n 1=2 C1 R<jX j 0 such that Z

m2+"dx c

Z

jf j2+" dx

@u . Let h = M (f 2 )1=2, where M denotes the Hardyfor all 0 < " "0, where f = @N Littlewood maximal operator. Let E = fx 2 Rn?1 : m(x) > g: We claim that Z

fm>;h]

C2jE

j + C

83

Z

fm >g

m2:

Let us assume the claim, and prove the desired estimate. First, note that Z

E

m2

Z

fm >;hg

m2 +

Z

fh>g

m2 C2jE

j + C

Z

fm>g

m2 +

Z

by the claim. Choose now and x so that C < 1=2. Then, Z

E

For " > 0,

Z

m2+"

="

Z 1

"?1

Z

Z

j + C

m2d "

m2:

fhg Z 1

"?1

Z

m2d

fm>g 0 Z 1 E ? Z 1+" jfm > gjd + C" "?1 m2 d: 0 h> 0

0Z

C"

m2 C2jE

fh>g

m2;

1

By a well-knwon inequality (see [18] for example), jEj Cjfm > gj. Thus, Z

m2+"

C"

Z 1 Z0

1+" jfm > gjd + C"

Z 1

Z

C" m2+" + C m2n" : If we now choose "0 so that

C"0 < 1=2; for " < "0;

Z

m2+"

If we now use Holder's inequality with exponents Z

m2+" C

?

Z

m2+"

2 ? 2+"

2+" 2

Z

0

? "?1

Z

h>

m2 d

Z

C m2n": and

2+" , "

M (f 2) 2+2 "

we see that

" 2+" ;

and the desired inequality follows from the Hardy-Littlewood maximal theorem. It remains to establish the claim. Let fQk g be a Whitney decomposition of the set E = fm > g, such that 3Qk E, and f3Qk g has bounded Fix k, we can assume R overlap. 2 that there exists x 2 Qk such that h(x) , and hence, 2Qk f C2jQk j. For 1 2, let Qk; = Qk , and Q~ k; = f(x; y) : x 2 Qk ; 0 < y < length (Qk )g: Q~ k; (and (Q~ k; )) is a Lipschitz domain, uniformly in k; . Also, by construction of Qk , there exists xk with dist (xk ; Qk ) length (Qk ) and such that m(xk ) . Let Ak; = @Qk; \ xk + Bk; = @Qk; \ Rn+ n Ak; ; 84

so that

@Qk; = Qk; [ Ak; [ Bk; : Note that the height of Bk; is dominated by C length (Qk ), and that jruj on Ak; . Let m1 be the maximal function of ru, corresponding to the domain Q~ k; (i.e., where the cones are truncated at height `(Qk )). Then, for x 2 Qk ; m(x) m1(x) + . Also, Z

Qk

C

m21

Z

Bk;

Z

m21 (using the L2-theory on Q~ k; )

@ Q~ k;

jruj2d + c

Z

Ak;

jruj2d + c

Z

2Qk

f2

C

Z

Bk;

jruj2d + C2jQk j:

Integrating in between 1 and 2, we see that Z

Qk

m21

C Z `(Qk ) Z jruj2 + C2jQ j C Z m2 + C2jQ j: k k `(Qk ) 0 2Qk 2Qk

Thus, Z

Qk

m2 C

Z

2Qk

m2 + C2jQk j:

Adding in k, we see that Z

fm >;hg

m2

C2jE

j + C

Z

fm >g

m2;

which is the claim. Note also that the same argument gives the estimate kN (ru)kp C krtukp; 2 < p < 2 + ", and the Lp theory is thus completed.

Section 2. Higher order boundary value problems Part 1: The systems of elastostatics In this part we will sketch the extension of the L2 results for the Laplace equation to the systems of linear elastostatics on Lipschitz domains. These results are joint work of B. Dahlberg, C. Kenig and G. Verchota, and will be discussed in detail in a forthcoming paper ([12]). Here we will describe some of the main ideas in that work. For simplicity, here we restrict our attention to domains D above the graph of a Lipschitz function ' : R2 ! R. 85

Let > 0; 0 be constants (Lame moduli). We will seek to solve the following boundary value problems, where ~u = (u1; u2; u3) (2.1.1)

~u + ( + )rdiv ~u = 0 in D ~uj@D = f~ 2 L2(@D; d)

(2.1.2)

~u + ( + )rdiv ~u = 0 in D (div ~u)N + fr~u + (r~u)tgN j@D = f~ 2 L2(@D; d):

(2.1.1) corresponds to knowing the displacement vector ~u on the boundary of D, while (2.1.2) corresponds to knowing the surface stresses on the boundary of D. We seek to solve (2.1.1) and (2.1.2) by the method of layer potentials. In order to do so, we introduce the Kelvin matrix of fundamental solutions (see [27] for example), ?(X ) = (?ij (X )); where ?ij (X ) = 4A jXijj + 4C XjXi Xj3j ; and A = 12 1 + 2 1+ ; C = 21 1 ? 2 1+ :

We will also introduce the stress operator T , where

T~u = (div ~u)N + fr~u + r~utgN: The double layer potential of a density ~g(Q) is then given by

~u(X ) = K~g(X ) =

Z

@D

fT (Q)?(X ? Q)gt~g(Q)d(Q);

where the operator T is applied to each column of the matrix ?. The single layer potential of a denisty ~g(Q) is

~u(X ) = S~g(X ) =

Z

@D

?(X ? Q) ~g(Q)d(Q):

Our main results here parallel those of Section 1, Part 1. They are 86

Theorem 2.1.3. (a) There exists a unique solution of problem 2.1.1 in D, with N (~u) 2 L2(@D; d). Moreover, the solution u has the form ~u(x) = K~g(x); ~g 2 L2(@D; d). (b) There exists a unique solution of (2.1.2) in D, which is 0 at innity, with N (r~u) 2 L2(@D; d). Moreover, the solution ~u has the form ~u(X ) = S~g(X ); ~g 2 L2(@D; d). (c) If the data f~ in 2.1.1 belongs to L21(@D; d), then we can solve (2.1.1), with N (r~u) 2 L2(@D; d). The proof of Theorem 2.1.3 starts out following the pattern we used to prove 1.1.1, 1.1.2 and 1.1.14. We rst show, as in Theorem 1.1.3, that the following lemma holds:

Lemma 2.1.4. Let K~g; S~g be dened as above, so that they both solve ~u + ( + )r div ~u = 0 in R3 n @D. Then, (a) kN (K~g)kLp(@D;d) C k~gkLp(@D;d); kN (rS~g)kLp(@D;d) C k~gkLp(@D;d); for 1 < p < 1: (b)

(K~g) (P ) = 21 ~g(P ) + K~g(P ) A + C ? @ n ( S~ g ) i (P )gj (P ) ? ni (P ) nj (P )hNp ; g (P )i + j ) (P ) = @Xi 2 Z ? @ ?(P ? Q)~g(Q)d(Q)) ; + p:v: @P j R

@D

i

where K~g(P ) = p:v: @DfT (Q)?(P ? Q)gt~g(Q)d(Q), and A; C are the constants in the denition of the fundamental solution. Thus, just as in Section 1, part 1 reduces to proving the invertibility on L2(@D; d) of 21 I + K; 12 I + K , and the invertibility from L2(@D; d) onto L21(@D; d) of S . Just as before, using the jump relations, it suces to show that if ~u(X ) = S~g(X ), then kT~ukL2(@D;d) krt~ukL2(@D;d): Before explaining the diculties in doing so, it is very useful to explain the stress operator T (and thus the boundary value problem 2.1.2), from the point of view of the theory of constant coecient second order elliptic systems. We go back to working on Rn, and use the summation convention. Let arsij; 1 r; s m; 1 i; j n be constants satisfying the ellipticity condition arsijij r s C jj2jj2 and the symmetry condition arsij = asrji. Consider vector valued functions ~u = (u1; : : : ; um) on Rn satisfying the divergence form system @ ars @ us = 0 in D: @Xi ij @Xj 87

From variational considerations, the most natural boundary conditions are Dirichlet cons @u @~ u ditions (~uj@D = f~) or Neumann type conditions, @ = ni arsij @X = fr . The interpretaj tion of problem 2.1.2 in this context is that we can nd constants arsij; 1 i; j 3; 1 r; s 3, which satisfy the ellipticity condition and the ssymmetry condition, and such that @ ars @u = 0 in D, and with T~u = @ ~u. In ~u +( + )rdiv ~u = 0 in D if and only if @X ij @ i @Xj order to obtain the equivalence between the tangential derivatives and the stress operator, we need an identity of the Rellich type. Such identities are available for general constant coecient systems (see [32], [30]).

Lemma 2.1.5 (The Rellich, Payne-Weinberger, Necas identities). Suppose that @ ars @ us = 0 in D, ars = asr ; ~h is a constant vector in Rn, and ~u and its derivatives ij ji @Xi ij @Xj are suitably small at 1. Then, Z

@ur @us d h`n` arsij @X i @Xj @D

=2

Z

@D

@ur n ars @us d: hi @X ` `j @X i

j

Proof. Apply the divergence theorem to the formula

@ ?h ars ? h ars ? h ars @ur @us = 0 @X` ` ij i `j j i` @Xi @Xj

Remark 1: Note that if we are dealing with the case m = 1; aij = I , and we choose ~h = en, we recover the identity we used before for Laplace's equation. Remark 2: Note that if we had the stronger ellipticityassumption that arsijir js C P`;t j`tj2, we would have, if @D = f(x; '(x)) : ' : Rn?1 ! R; kr'k1 M g, that krtukL2(@D;d) k @@u kL2(@D;d). In fact, if we take ~h = en, then we would have XZ

Z

@ur @us d = h` nàrsij @X i @Xj @D r @D Z Z Z r s ?X 1=2? @u @u 2 d 1=2: rs r 2 j = 2C hi @X nà`j @X d 2C j @u jru j d i j @D @D @ r @D

P R

jrurj2d

c

R

2 Thus, r @D jrurj2d C @D j @u @ j d . For the opposite inequality, observe that for each r; s; j xed, the vector hin` ars`j ? h` n` arsij is perpendicular to N . Because of Lemma 2.1.5 Z Z r @us @u @u` @us d: rs h` nàij @X @X d = 2 (h`n` arsij ? hinàrs`j) @X i j i @Xj @D @D

88

Hence, and so

Z

@D

? jruj2d C

Z

@D

? jrtuj2d 1=2

Z

@D

jruj2d

1=2

;

Z

Z Z @u 2 2 j @ j d c jruj d c jrtuj2: @D @D @D Remark 3: In the case in which we are interested, i.e., the case of the systems of elastostatics, @us @ur = (div ~u)2 + X ? @uj + @ui 2; arsij @X 2 i;j @Xi @Xj i @Xj which clearly does not satisfy X arsijir js C j`tj2;

`;t

since the quadratic form involves only the symmetric part of the matrix (ir ). In this case, u = T~u = (div ~u)N + fr~u + r~utgN . of course @~ @

Remark 4: The inequality

kr~ukL2(@D;d) C krt~ukL2(@D;d)

holds in the general case, directly from Lemma 2.1.5, by a more complicated algebraic argument. In fact, as in Remark 2, Z Z r @us @u` @us d; @u rs h` nàij @X @X d = 2 (h`n` arsij ? hinàrs`j) @X i j i @Xj @D @D and for xed r; s; j; (h`n` arsij ? hinàrs`j) is a tangential vector. Thus, Z Z Z r @us 1=2? ? @u rs 2 h`n` aij @X @X d C jr~uj2d 1=2: jrt~uj d i j @D @D @D rs ? 1 Consider now the matrix drs = (aij ni nj ) . This is a strictly positive matrix, since arsijij r s C jj2jj2. Moreover, s r ? @u ? @u rs @u @u = drs @ ? a ij @X @X r @ s i j t m r @us @u @u @u rt sm rs = drtniaij @X nà`k @X ? aij @X @X = j k i j t @u t @u @u @u = drsnk artk` @X nmasmv @X ? atv` @X @X = `t v tv @u` @u @u @u s = drsnk artkv atv` @X @X = @Xv nmam` @X` ? ` t @u v @u rt s t = fdrsnk akv nm am` ? av`g @X @X : v ` 89

Now, note that for t; ; ` xed fdrs nk artkv nm asm`?atv` g is perpendicular to N , by our denition of drs, and the symmetry of arsij:

drs nk artkv nmastm`nv ? atv`nv = artkv nk nv drsastm` nm ? atm`nm = = atrvk nv nk drs astm`nm ? atm` nm = tsastm` nm ? atm`nm = atm` nm ? atm`nm = 0: Therefore, Z

? @~ u ? @~u

h` n`drs @ @D

r

@ sd c

?

Z

@D

? jrt~uj2d 1=2

Z

@D

jr~uj2d

1=2

Now, rs n n @u = n ars @u ? ars n n n @u = ? a i ij kj k j i @X kj k j @N r @ @X j i s s @u @us = @u rs rs rs rs = niaij @X ? aki nk nj ni @X = fniaij ? aki nk ninj g @X

? @~ u

s

s

s

j j s @u rs rs = fniaij ? aik nk ninj g @X : j

j

But, for i; r; s xed, arsij ? arsiknk nj is perpendicular to N , and so Z

@D

@ut gfas n n @u gd h` n` drsfartkj nk nj @N i` i ` @N

C

?

Z

@D

? jrt~uj2@ 1=2

Z

@D

jr~uj2d 1=2 +

Z

@D

jrt~uj2d :

We now choose ~h = en , so that h`n` C , and recall that (drs ) and (artkj nk nj ) are strictly positive denite matrices. We then see that Z Z Z Z 1=2 1=2? ? @~ u 2 2 2 jr~uj d + jrt~uj2d : jrt~uj d j @N j d C @D @D @D @D @~u j2 , the remark follows. Now, as jr~uj2 = jrt~uj2 + j @N

Remark 5: In order to show that R@D jrt~uj2d C R@D jT~uj2d, it suces to show that Z

@D

jr~uj2d

c

Z

@D

j(div ~u)I + fr~u + r~utgj2d:

In fact, if this inequality holds, we would clearly have that Z

@D

jr~uj2d C

Z

@D

90

jr~u + r~utj2d

(Korn type inequality at the boundary). The Rellich-Payne/Weinberger-Necas identity is, in this case (with ~h = en), Z nnf 2 jr~u + r~utj2 + (div ~u)2gd = @D Z u f(div ~u)N + fr~u + r~utgN gd: = 2 @~ @y @D

But then, Z

@D

jr~uj2d

C

?

Z

@D

? jr~uj2d 1=2

Z

@D

j(div ~u)N + fr~u + r~utgN j2d

1=2

:

The rest of part 1 is devoted to sketching the proof of the above inequality.

Theorem 2.1.6. Let ~u solve ~u + ( + )r div ~u = 0 in D; u~ = S (~g), where ~g is nice. Then, there exists a constant C , which depends only on the Lipschitz constant of ' so that Z

@D

jr~uj2d C

Z

@D


The proof of the above theorem proceeds in two steps. They are:

Lemma 2.1.7. Let ~u be as in Theorem 2.1.6. Then, Z

@D

N (r~u)2d

c

Z

@D

N ((div ~u)I + fr~u + r~utg)2d:

Lemma 2.1.8. Let ~u be as in Theorem 2.1.6. Then, Z

@D

N ((div ~u)I + fr~u + r~utg)2d C

Z

@D


Lemma 2.1.7 is proved by rst doing so in the case when the Lipschitz constant is small, and then passing to the general case by using the ideas of G. David ([13]). Lemma 2.1.8 is proved by observing that if ~v is any row of the matrix (div ~u)I + fr~u + r~utg, then ~v is a solution of the Stokes system 8 < ~v = rp in D div ~v = 0 in D (S) : ~vj@D = f~ 2 L2(@D; d) This is checked directly by using the system of equations ~u + ( + )rdiv ~u = 0. One then invokes the following Theorem of E. Fabes, C. Kenig and G. Verchota, whose proof will be presented in the next section. 91

Theorem 2.1.9. Given f~ 2 L2(@D; d), there exists a unique solution (~v; p) to system (S) with p tending to 0 at 1, and N (~v) 2 L2(@D; d). Moreover, kN (~v)kL2(@D;d) C kf~kL2(@D;d): We now turn to a sketch of the proof of Lemma 2.1.7. We will need the following unpublished real variable lemma of G. David ([14]).

Lemma 2.1.10. Let F : R Rn ! R be a function of two variables t 2 R; x = (x1; : : : ; xn) 2 Rn. Assume that for each x, the function t ! F (t; x) is Lipschitz, with Lipschitz constant less than or equal to M , and for each i; 1 i n, the function xi ! F (t; x) is Lipschitz, with Lipschitz constant less than or equal to Mi, for any choice of the other variables. Given an interval I J = I J1 : : : Jn, where the Ji's and I are 1-dimensional compact intervals, there exists a function G(t; x) : R Rn ! R with

the following properties:

(a) G(t; x) F (t; x) on I J (b) If E = f(t; x) 2 I J : F (t; x) = G(t; x)g, then jE j 83 jI jjJ j. (c) For each i, the function G(t; x1; x2; : : : ; xi?1; ?; xi+1; : : : ; xn) is Lipschitz, with Lipschitz constant less than or equal to Mi, and one of the following statements is true: 4M , or for each x, ?4M @G (t; x) M . Either for each x, ?M @G ( t; x ) @t 5 5 @t The proof of this lemma is the same as in the 1-dimensional case, treating x as a parameter (see [13]). Before we procedd with the proof of Lemma 2.1.7, we would like to point out that in the analogue of Lemma 2.1.7 for bounded domains, a normalization is necessary since if ~u(x) solves the systems of elastostatics so does ~u(x) + ~a + BX , where ~a is a constant vector, while B is any antisymmetric 3 3 matrix. The right-hand side of the inequality in the Lemma of course remains unchanged, while the left-hand side increases if B 'increases'. The most convenient normalization is that for some xed point X in the domain r~u(X )? r~u(X )t = 0. This also gives uniqueness modulo constants to problem 2.1.2 in bounded domains. We now need to introduce some denitions. Let D0 Rn+ be a xed, C 1 domain with f(x; 0) : jjxjj = max jxij 1g @D0, f(x; y) : 0 < y < 1; jjxjj 1g D0 f(x; y) : 0 < y < 2; jjxjj < 2g: If ' : Rn?1 ! R is Lipschitz, with jjr'jj M , we construct theRmapping T' : Rn+ ! Rn by T'(x; y) = (x; cy + y '(x)) where 2 C01(Rn?1) is radial, = 1, and c = c(M ) is 92

chosen so that T'(Rn+) f(x; y) : y > '(x)g, and so that T' is a bi-Lipschitzian mapping. Also, it is clear that T' is smooth for (x; y) with y > 0, and T'(x; 0) = (x; '(x)). We will denote by A' the point T'(0; 1). Lemma 2.1.7 is an easy consequence of the following result.

Lemma 2.1.11. Given M > 0 and ' with jjr'jj M , there exists a constant C = C (M )

such that for all functions ~u in D' , which are Lipschitz in D ', which satisfy ~u + ( + )rdiv ~u = 0 in D' and r~u(A') = r~u(A')t, we have kN' (r~u)kL2(@D;d) C kN'((div ~u)I + fr~u + r~utg)kL2(@D;d): Here N' is the non-tangential maximal operator corresponding to the domain D'. This lemma will be proved by a series of propositions. Before we proceed, we need to introduce one more denition. We say that proposition (M; ") holds if whenever ' is such that jjr'jj M , and there exists a constant vector ~a with jj~ajj M so that jjr' ?~ajj ", then for all Lipschitz functions ~u on D ' , with ~u + ( + )rdiv ~u = 0 in D' , with r~u(A') = r~ut(A') we have kN'(r~u)kL2(@D;d) C kN'((div ~u)I + fr~u + r~utgkL2(@D;d); where C = C (M; "). Note that if proposition (M; ") holds, then the corresponding estimates automatically hold for all translates, rotates or dilates of the domains D' when ' satises the conditions in proposition (M; "). In the rest of this section, a coordinate chart will be a translate, rotate or dilate of a domain D'. The bottom B' of @D' will be T'(@D0 \ (x; 0) : x 2 Rn?1).

Proposition 2.1.12. Given M > 0, there exists " = "(M ) so that propostion (M; ")

holds. We will not give the proof of Proposition 2.1.12 here. We will just make a few remarks about its proof. First, in this case the stronger estimate kN'(r~u)kL2(@D;d) C k(div ~u)N + fr~u + r~utgN kL2 (@D;d) holds. This is because in this case, the domain D' is a small perturbation of the smooth domain Dax . For the smooth domain Dax, we can solve problem 2.1.2 by the method of layer potentials (see [27], for example). If " is small, a perturbation analysis based on the theorem of Coifman-McIntosh-Meyer ([3]) shows that this is still the case. This easily gives the estimate claimed above.

Proposition 2.1.13. For all M > 0; " > 0; 2 (0; 0:1), if proposition (M; ") holds, then propostion (1 ? M; 1:1") holds. We postpone the proof of Proposition 2.1.13, and show rst how Proposition 2.1.12 and Proposition 2.1.13 yield Lemma 2.1.11. 93

Proof of Lemma 2.1.11. We will show that proposition (M; ") holds for any M; ". Fix M; ", and choose N so large that if "(10M ) is as in Proposition 2.1.12, then (1:1)N "(10M ) N Y ". Pick now j > 0 so that (1 ? j ) = 1=10. Then, since proposition (10M; "(10M )) j =1

holds, by Proposition 2.1.12, applying Proposition 2.1.13 N times we see that proposition (M; ") holds. We will now sketch the proof of Proposition 2.1.13. We rst note that it suces to show that

kN'(r~u)kL2(@D;d) C kN~'((div ~u)I + fr~u + r~vtgkL2(@D;d) where N~' is the non-tangential maximal operator with a wider opening of the non-tangential region. This follows because of classical arguments relating non-tangential maximal functions with dierent openings (see [18]) for example). Pick now ' with jjr'jj (1 ? )M , and such that there exists ~a with jjr' ? ~ajj 1:1"; jj~ajj (1 ? )M . We will choose N~' as follows: Since @D' n B' is smooth, it is easy to see that we can nd a nite number of coordinate charts (i.e., rotates, translates and dilates of D ), which are entirely contained in D', such that their bottoms B are contained in @D', such that T ((x; 0) : jjxjj < 1=2) cover @D', and such that the 's involved satisfy jjr jj (1 ? 2 )M and there exist ~a such that jj~a jj (1 ? 2 )M , and jjr ? ~a jj 1:11". The non-tantential region dening N~', on T ((x; 0) : jjxjj < 21 ) is dened as follows: let F f(x; 0) : jjxjj < 1=2g be a closed set. Consider the cone on Rn+; = f(x; y) 2 Rn+ : bjxj < yg, where b is a small constant. Consider now the domain DF on Rn+, given by DF = [x2F ((x; 0)+ ). Then DF is the domain above the graph of a Lipschitz function , for which jjrjj cb, for some absolute constant c (independent of F ). It is also easy to see that we can take now b so small, depending only on M and " such that T (DF ) is the domain above the graph of a Lipschitz function ~, with ~ , and which statises jjr ~jj (1 ? 10 )M; jjr ~ ? ~a jj 1:111": The non-tangential region dening N~', for Q 2 T ((x; 0) : jjxjj < 1=2) is then the image under T of (x; 0)+ , with b chosen as above, suitably truncated, and where Q = T ((x; 0). Let now, to lighten notation, m = N'(r~u); m = N~'((div ~u)I + fr~u + r~utg). For t > 0, consider the open-set Et = fm > tg. We now produce a Whitney type decomposition of Et into a family of disjoint sets fUj g with the property that each Uj is contained in T ((x; 0) : jjxjj < 1:2) for a coordinate chart D , each Uj containes T (Ij ), where Ij is a cube in jjxjj < 1=2, and is contained in T (Ij ), where Ij is a xed multiple of Ij . Finally, we can also assume that there exists a constant 0 such that if diam (Uj ) 0, there exists a point Qj in @D', with dist (Qj ; Uj ) diam Uj , such that m(Qj ) t. Let now 94

> 1 be given. We claim that there exists > 0 so small that if Ej = Uj \fm > t; m tg then (Ej ) (1 ? M )(Uj ), where M > 0. Assume the claim for the time being. Then, Z

@D'

m2d

=2

Z 1

0

X

j

t(Et)dt = 2 2

2 2

Z 1

0

Z 1

0

t(E t)dt =

t(Ej )dt + 2 2

Z 1

0

X

j

2 2

Z 1

0

t(Qj \ E t)dt

t(m > t)dt

2Z 1 tfm > tgdt = t(Qj )dt + 2 2 M) 0 0 j Z 2Z 2 2 m 2d: m d + 2 = (1 ? M ) @D' @D' X

2 2(1 ?

Z 1

Thus, if we choose > 1, but so that 2 (1 ? M ) < 1, the desired result follows. It remains to establish the claim. We argue by contradiction. Suppose not, then (Ej ) > (1 ? M )(Uj ). Let E~j = T ?1(Ej ). If M is chosen suciently small, we can gurarantee that jE~j \ Ij j :99jIj j. Let now Fj = E~j \ Ij , and construct now the Lipschitz function ~ )M; jjr ~ ? corresponding to it, as in the denition of N~'. Thus, ~ ; jjr ~jj (1 ? 10 ~a jj 1:111". We now apply Lemma 2.1.10 to ~, one variable at a time, to nd a Lipschitz function f , with f ~ on Ij , such that if Fj = fx 2 Ij : f = ~g, then jFj \ Fj j c(Uj ), )M , and such that there exists ~a , with jj~a jj (1 ? )M so that with jjrf jj (1 ? 10 f f 10 4 jjrf ? ~af jj 5 1:111" < ". We can also arrange the truncation of our non-tangential regions in such a way that on the appropriate rotate, translate and dilate of Df (which of course is contained in the corresponding coordinate chart associated to D , which is contained in D'),

j(div ~u)I + fr~u + r~utgj t: To lighten the exposition, we will still denote by Df the translate, rotate and dilate of Df . Note that proposition (M; ") applies to it it. We divide the sets Uj into two types. Type I are those with diam Uj 0, and type II those for which diam Uj 0. We rst deal with the Uj of type I. In this case, Df has diameter of the order of 1. Because of the solvability of problem 2.1.2 for balls, and our normalization, we see that on a ball B D' ; diam B 1; A' 2 B , we have Z

B

jr~uj2

Z

c jdiv ~uI + fr~u + r~utgj2: B

Joining Af to A' by a nite number of balls, and using interior regularity results for the system ~u +( + )r div ~u = 0, we see that jr~u(Af )j Ct, for some absolute constant 95

C . Then

Z

m2d

c

Z

Nf2(ru)d @Df T (FZj \Fj ) ? [ r ~ u ( A f ) ? r~ut(Af )] 2 2 2 C(Uj ) t + C Nf r~u ? d 2 Z@Df C(Uj )2t2 + C Nf2((div ~u)I + fr~u + r~utg)2d;

C(Uj

) 2t2

@Df

by (M; "). The last quantity is also bounded by C(Uj )2t2, which is a contradiction for small . Now, assume that Uj is of type II. Note that in this case there exists Qj 2 @D', with dist (Qj ; Uj ) diam Uj , and such that jr~u(x)j t for all x in the non-tantential region associated to Qj . Because of this, it is easy to see, using the arguments we used to bound jr~u(Af )j in case I, that for all X in a neighborhood of Af and also on the top part of Df , we have that jr~u(X )j t + Ct. Since for Q 2 T (Fj \ Fj ); m(Q) t, and > 1, if is small enough, we see that we must have Nf (r~u)(Q) m(Q). Hence, t ? Nf r~u ? r~u(Af ) ?2 r~u (Af ) (Q) ( ? 1 ? C)t ( ?2 1) t

if is small and Q 2 T (Fj \ Fj ). Thus, applying (M; ") to Df , we see that Z

t Nf (r~u ? r~u(Af ) ?2 r~u (Af ) 2d T (Fj \Fj ) Z t ? r~ Nf r~u ? u(Af ) ?2 r~u (Af ) 2d C(Uj )2t2; @Df

C ( ? 1)2t2(U

j)

a contradiction if is small. This nishes the proof of Proposition 2.1.13, and hence of Lemma 2.1.11.

Part 2: The Stokes system of linear hydrostatics In this part I will sketch the proof of the L2 results for the Stokes system of hydrostatics. These results are joint work of E. Fabes, C. Kenig and G. Verchota ([17]). We will keep using the notation introduced in Part 1. We seek a vector valued function ~u = (u1; u2; u3) and a scalar valued function p satisfying (2.2.1)

8
'(x); x 2 Rn?1g; where ' : Rn?1 ! R is Lipschitz continous; i.e., j'(x) ? '(x0)j mjx ? x0j. We start out by reviewing the Neumann problem with data f 2 L2(), = @D. Let us a priori assume that f is bounded and has compact support. Let Z 1 u(X ) = S (f )(X ) = ! (n ? 2) f (Q)jX ? Qj2?nd(Q); n 111

be the single layer potential of f . Since u is harmonic in D, we have the identity ? (2.1) r u = 0; div jruj2e ? 2 @u @y where e = (0; 1). The results in [2] show that kM (ru)kL2() C kf kL2(): Since ru(X ) = O(jX j1?n ) as X ! 1, it follows from (2.1), the estimate above, and the divergence theorem applied in the domains D;" = f(x; y) : y > '(x) + "; jxj2 + y2 < 2g, that Z ? @u @u d = 0; jruj2he; ni ? 2 @n (2.2) @y where the derivatives on are taken as the non-tangential limits from D, of the corresponding expressions in D. This is the Rellich ([21]) identity on . To exploit (2.2), let T1(x); T2(x); : : : ; Tn?1(x) be an orthonormal basis for the tangent plane to at (x; '(x)). The Ti(x) exist for a.e. x. Let

jrT u(x; '(x))j2 =

n?1 X j =1

jhru(x; '(x)); Ti(x)ij2:

A well-known argument (see [15], Corollary 2.1.11 for example) shows that there are constants c1; c2, that depend only on the Lipschitz constant m of ' such that Z Z Z @u 2 @u 2 2 d: C1 @n d jrT uj d C2 @n (2.3) Note that in the above estimate the values of the derivatives of u are taken as the limits from above the graph, but the same estimate can be obtained by taking limits from below the graph. Let now Tf and T?f denote the normal derivatives of Sf as a function in D and Rn n D respectivly. We then have (see [15]) (2.4) kTf kL2() ckf kL2(); where c = c(m). To establish (2.4), we recall the classical jump relation T + T? = I , where I is the identity operator (again, see [15] for a proof of this in our case). To prove (2.4) we only need to remark that jrT uj2 is continuous a.e. across (see [15]). Thus, (2.4) follows by application of (2.3) in D and Rn n D , together with the jump relation. The boundedness of T ([2]) together with (2.4) immediately show that T is one-to-one, with closed range. 112

To see that the range of T is all of L2(), we let U : L2(Rn?1) ! L2(Rn?1) be the pullback of T , i.e., Ug = T (g F ?1) F , where F : Rn?1 ! is given by F (x) = (x; '(x)). Letting Us denote the operator corresponding to the graph of x ! s'(x); 0 s 1, we see easily, using the results in [2], that kUs ? Us1 k C js ? s1j; 0 s; s1 1. (See [15] for the details.) Since U0 = 21 I , the continuity method shows that U1, and hence T , is onto (see [15], Lemma 2.1.6).

Theorem 2.5. Given any f 2 L2() there is a harmnonic function u in D such that kM (ru)kL2() C (m)kf kL2(), and @u=@n = f a.e. on , in the sense of non-tangential convergence. Any two such harmonic functions dier by a constant. Let g = T ?1 f . If n 4 one such function is given by S (g). If n = 3 one such solution is given by u(X ) =

Z

@D

g(Q)fjX ? Qj2?n ? jX0 ? Qj2?ngd(Q);

where X0 is any xed point in Rn n D . Proof. The existence part and the representation formulas follow from the invertibility of T , and the results of Coifman-McIntosh-Meyer [2] mentioned before. It remains to establish uniqueness. We present here an argument which will be very useful for us later on in treating the Lp case 1 < p < 2. LetR ! be harmonic in D, with M (r!) 2 L2(), and @!=@n = 0 a.e. on . Note rst that DR jr!j2dX CR, where DR = D \ f(x; y) : jxj2 + y2 < R2g. Consider now the bi-Lipschitzian mapping : D ! Rn n D given by

(x; y) = (x; '(x) ? [y ? '(x)]) = (x; 2'(x) ? y): Dene ! on Rn n D by the formula ! = ! ?1. A simple calculation shows that in Rn n D , ! veries (in the weak sense) the equation div (A(x; y)r!) = 0, where 0 A(x; y) = 1=[J(X )]0(X ) t(X ), where X = ?1(x; y), 0 is the Jacobian matrix of , and J the Jacobian determinant of . It is easy to see that A 2 L1 (Rn n D ), and hA(x; y); i C jj2, where C = C (m). Let now B (x; y) =

I for (x; y) 2 D A(x; y) for (x; y) 2 Rn n D

and extend !(x; y) to all of Rn by setting it equal to !(x; y) in Rn n D . Since M (r!) 2 L2(), and @!=@n = 0 a.e. on , it is easy to see that the extended ! is a weak solution in all of Rn of the divergence form elliptic equation with bounded coecients R measurable 2 div B (x; y)r! = 0. The extended ! also satises the estimate jxj 0 such that for all n-dimensional balls B centered in Rn n f(x; y) : jxj < 2; jy ? '(x)j < 2g, of radius rR0 = r0(m), we have that jfX 2 B : !(X ) = 0gj d. Furthermore, on such balls B , B jr!j2 C , by the L2 estiamte for the non-tangential maximal R function of ru. Therefore, by a standard variant of the Poincaré inequality, we have B !2 C . The sub-mean value inequality for L-subsolutions ([19]) now establishes (2.8) in the caseR n 4. It remains to show (2.8) when n = 3. In this case, we let, for X 2 D; u(X ) = g(Q)fjX ? Qj2?n ?jX0 ? Qj2?ngd(Q). Schwarz's inequality now shows that ju(X )j C log(2+dist (X; )) whenever dist (X; ) 1. Let now X = (x; y) and set !(X ) = max(ju(X )j ? c; 0) for y > '(x) + 1, and 0 otherwise. If c is chosen large enough, then ! is zero in a neighborhood of f(x; y) : y = '(x) + 1g, and so ! is a subharmonic function in all of Rn. If ?0 = f(x; y) : y + y0 > ?2mjxjg, then ! is 0 on @ ?0 if y0 is large 115

enough, and ! has logarithmic growth at 1. By the Phragmen-Lindelöf Theorem (see [17]), ! is identically 0, which yields (2.8) as in the case n 4. Let g be the fundamental solution of L in Rn, with pole at 0. As is well-known, there are constants C1, and C2 which depend only on the ellipticity constants of L (and hence only on m), such that C1jX j2?n g(X ) C2jX j2?n (see [18]). By (2.8) and the asymptotic expansion of Serrin and Weinberger (Theorem 7 of [22]), there are constants ; ; ; and R0; 0 < ; R0, such that u(X ) = g(X ) + + v(X ); for X 2 Rn n f(x; '(x)) : jxj 1g where jv(X )j jX j2?n? for jX j R0, where ?1 can be bounded in terms of m, while ; ; and R0 can be bounded in terms of and kukL1(DnD). We next claim that R = 0. To show this, we recall that if L = div B r, Theorem 7 of [22] shows that = b hB ru; r i, where b is a constant that depends on g, and is any C 1 function on Rn, which is 0 for jxj R0 and equals 1 for jxj R1 > R0. Let R be large, and set, for 0 < , D( ) = f(x; y) : '(x) + < y < '(x) + R; jxj Rg: Clearly, for R large enough, 1 on A( ) = @D( ) ? B ( ) where B ( ) = f(x; '(x)+ ) : jxj Rg. Hence, Z

D

Z

hB ru; r i = hru; r i = lim !0 D Z

Z

DZ( )

hru; r i

@u @u = lim ( ? 1) ! 0 @n Z@D( ) @n Z@D( ) @u = lim ( ? 1) @n = ( ? 1)ad = 0: !0 = lim !0

B ( )

R

R

By our construction of B (see the proof of Theorem 2.5), Rn nD hB ru; r i = D hru; r ~i, where ~ = . This quanity is also zero. Let

a(R) =

Z

(R)

M (ru)2d;

where (R) = f(x; '(x)) : R jxj 2Rg; and let

a(1) =

Z

f(x;'(x)):jxj2g

116

M (ru)2d:

By Theorem 2.5, a(1) C kak2L2() C . For Q 2 , let (Q) = fX 2 D : jX ? Qj < 2 dist (X; )g and for Q 2 (R) set

1(Q) = fX 2 (Q) : jX ? Qj < Rg; 2(Q) = (Q) n 1(Q), and Mi(Q) = supfjru(X )j : X 2 i(Q)g; i = 1; 2. Observe that if X 2 2(Q), then u is harmonic in B = fY : jY ? X j Rg, where = (m) is small enough, and supY 2B ju(Y ) ? j CR2?n? . Since R jru(X )j CR?1?n B ju(y) ? jdY CR1?n? , it follows that Z

(2.9)

(R)

(M2(ru))2d CR1?n?2 :

For 2 I = [1=4; 1=2], set

= f(x; y) : '(x) < y < '(x) + R; R < jxj < ?1Rg; where is chosen so that for Q 2 (R); 1(Q) . From the L2 Neumann theory for bounded Lipschitz domains ([13]), it follows that Z Z Z ? @u 2 2 2d; M1(jruj) d C d C jr u j (R) @ @n @ r \D with C depending only on m, since @u=@n on @ \ @D. Integrating in on I yields Z

(R)

M1

(jruj2)d

CR?1 CR?3

Z

Z 1=4 n 1=2

jruj2dX

C1 RjxjC2 R

u2dX CR1?n?2 ;

where the next to the last inequality follows from the inequality of Caciopolli for solutions of Lu = 0 (see [12], for example). Putting this together with (2.9), we see that Z

(R)

M (ru)2d CR1?n?2 ;

which easily yields the lemma. We shall next study the boundary value properties of harmonic functions in D, with M (ru) 2 L1().

Lemma 2.10. Suppose that u is harmonic in D, and M (ru) 2 L1(). Then, ru has non-tangential limits a.e. on , and if @u=@n = hru(Q); n(Q)i, then @u=@n 2 Hat1 (),

with

(2.11)

@u

@n 1 Hat ()

C kM (ru)kL1(): 117

Proof. We rst remark that by the extension theorem of Varoupoulos ([24]), given g continuous and with compact support in , there is a continuous function G in D , which agrees with g on , which has bounded support, and such that jrGjdX is a Carleson measure, with Carleson norm bounded by C (m)kgkBMO(), i.e., for all Q 2 , and all r 0, we have Z

fX 2D:jX ?Qj 0 let u (X ) = u(X + (0; )). Then, Z Z @u @u g = lim g d d !0 @n @n Z Z hrG; ru idX C kgkBMO() M (ru)d; = lim !0 D

by the basic property of Carleson measures (see [11]). Recall now that VMO() is the closure in BMO () of the space of continuous functions with compact support, and that the dual space of VMO () is Hat1 () (see [3]). This concludes the proof of the lemma. We are now in position to solve the Neumann problem on a Lipschitz graph, with data in Hat1 ().

Theorem 2.12. Let f 2 Hat1 (). Then there exists a harmonic function u in D with M (ru) 2 L1(), and @u=@n = f non-tangentially a.e. on . The function u is unique

modulo constants. Furthermore, there are constants C1 = C1(m); C2 = C2 (m); C3 = C3 (m) such that C1kf kHat1 () kM (ru)kL1() C2kf kHat1 (); and

@u

@ T~ 1 j Hat ()

C3kf kHat1 ():

Proof. The exisatence of u follows directly from Lemma 2.7 (a). In order to show uniqueness, let us assume that ! is harmonic in D; M (r!) 2 L1(), and @!=@n = 0 nontangentialy a.e. on . We want to conclude that ! is a constant. From the sub-mean value property of jr!j it follows that jr!(X )j C fdist (X; )g1?n :

118

Therefore, adding a suitable constant to !, we have j!(X )j C fdist (X; )g2?n: For > 0, let ! (X ) = !(X + (0; )). By the estimates above, Sobolev's inequality and the assumption on M (r!), we have that Z

j! j(n?1)=(n?2)d C;

with C independent of . Let now L be the divergence form operator used in the proof of Lemma 2.7, and let G(X; Y ) be its fundamental solution in Rn . Fix X 2 D; jX j < R, and let 2 C01(Rn), be identically 1 for jX j < R, and 0 for jX j > 2R. We can also assume that Rjr j + R2j j C , where C is independent of R. Let G(Y ) = G(X; Y )+ G(X ; Y ), where X is the reection of X . Then, Z Z @! ! (X ) = G @n d + G! @@n d Z + G f2hr ; r! i + ! gdy D = I + II + III: Set F (R) = fX : R < jX j < 2Rg. Then

jII j CR1?n as R ! 1, and

jIII j C

Z

Z

\F (R)

j! jd

R1?n jr!

? C R1?n

jdY

+C

Z

\F (R) Z

j! j(n?1)=(n?2)d

(n?2)=(n?1)

!0

R?n j! j

D\F Z(R) DZ\F (R) ? ?n 2 ? n j! j(n?1)=(n?2)dY (n?2)=(n?1) M (r!)d + C R CR D\F (R)

Hence

! 0 as R ! 1:

! (X ) =

Z

G @! @n d:

Since G 2 L1 (), the dominated convergence theorem shows that !(X ) = lim !0 ! (X ) =

0. The estimate kM (ru)kL1() C2kf kHat1 () follows by construction, while the estimate C1kf kHat1 () kM (ru)kL1() follows by Lemma 2.10. Finally, it is enough to establish the estimate k@u=@ T~jkHat1 () C3kf kHat1 () when f is an atom. One rst shows that R (@u=@ T~j )d = 0. We see that this follows by considering the functions u ; > 0, and then passing to the limit. This fact, togther with (b) of Lemma 2.7 shows that @u=@ T~j is a molecule and the estimate follows by the general theory of [3]. We shall now treat the Neumann problem with Lp data. 119

Theorem 2.13. There exists a positive number " = "(n; m) such that for all f 2 Lp(); 1 < p < 2 + " there is a harmonic function u in D with kM (ru)kLp() C kf kLp() and @u=@n = f non-tangentially a.e. on . Furthermore, u is unique modulo constants.

Proof. We rst remark that uniqueness for 1 < p < (n ? 1) follows by the same argument as in the uniqueness part of Theorem 2.11. The case n = 3; p = 2 of uniqueness is proved in Theorem 2.5. The case n = 3; 2 2. We remark that this in fact follows from an abstract argument of A. P. Calderón ([1.B]). We present here an alternative proof, which also yields uniqueness. Let f 2 L1 () have compact support and let u be the L2 Neumann solution with data f , given by Theorem 2.5. Let H = f(x; y) 2 Rn; y > 0g and : H ! D be given by (x; y) = (x; y + '(x)) where is a bi-Lipschitzian mapping between H and D. Put v = ru , and for x 2 Rn?1, set m(x) = sup (x) jvj, where

(x) = f(x0; y) 2 H : y > jx0 ? xjg. Similarly, set (x) = f(x0; y) 2 H : y > jx0 ? xjg, and m(x) = sup (x) jvj, where 2 (0; 1) is a number to be chosen later. Our aim is to show that there exists " = "(n; m) > 0 such that, if 0 < ", then Z

m2+ dx C

Z

2+ d: f Rn?1 This clearly yields the desired existence results, in the range 2 < p 2 + ". In order to establish (2.14), we needR to introduce a bit more notation. Let g(x) = f (x; 0); x 2 Rn?1, and h(x) = supI (1=jI j I g2dx)1=2, where the sup is taken over all cubes I in Rn?1, that contain x. Finally, for > 0 let E = fx 2 Rn?1 : m(x) > g. We will show that, for suciently small, Z Z 2 (2.15) C jEj + C m2dx:

(2.14)

fm >; hg

fm >g

Let us assume (2.15) for the time being, and use it to establish (2.14). From (2.15) it immediately follows that, if is chosen suceintly small, Z

E

m2dx C2jE

j + C

Let now and N be positive numbers. Then, Z

Rn?1

[minfm; N g]2+dx =

Z N

0Z

C

0

? ?1

N

Z

fh>g\E

Z

fm>g

+1jE

m2dx:

m2dx d

jd + C

Z N

0

? ?1

By a classical argument (see [11]), jEj Cjfm > gj, and so Z

Rn?1

[minfm; N g]2+dx C=(2 + ) +C

Z

Rn?1

120

Z

Rn?1

Z

fh>g

[min(m; N )]2+dx

min(h; N ) m2dx:

m2dx d:

Thus, if we choose small enough, we have Z

Rn?1

[minfm; N g]2+dx C

Z

Rn?1

h m2dx:

The boundedness of f implies that h is also bounded, and hence the right-hand side is nite. By monotone convergence, we see that Z

Rn?1

Z

m2+ dx C

Rn?1

h m2dx < +1; R

R

(2.14) now follows from Hölder's inequality and the fact that Rn?1 h2+ C f 2+ d. It remains to establish (2.15). Let fIk g be a Whitney decomposition on E, such that the cubes 3Ik E, and f3Ik g has bounded overlap (see [25]). Here 3Ik denotes the cube with the same center as Ik , and sides 3 times those of Ik . We are only interested in Whitney cubes Ik such that Ik \ fh g 6= ;. For 2 < < 3, let Ik; = f(x; y) : x 2 Ik ; 0 < y < l(Ik)g, where l(Ik) denotes the side length of Ik . For a set F H , we let F~ = (F ). Clearly I~k; is a Lipschitz domain, with Lipschitz constants bounded independently of k; . Since Ik is a Whitney cube, there exists a point xk 2 Rn?1 n E with dist (xk ; Ik ) Cn l(Ik). Put Ak; = @Ik; \ (xk ); Bk; = (@Ik; \ H ) n Ak; . Since 2 (2; 3), the height of Bk; is bounded by Cl(Ik ), i.e., supfy : (x; y) 2 Bk; g Cl(Ik ). Also, jvj on Ak; . Since Ik \ fh g 6= ;, we have Z

3g I k

f 2d

c

Z

3Ik

g2dx C2jIkj:

From the L2 Neumann theory for bounded Lipschitz domains, applied to I~k; ([13]), we nd that Z Z ? @u 2 2 d m1dx C @n @ I Ik k; Z C jruj2d + C2jIk j;

] g

Bk;

where m1(x) = supfjv(x0; y)j : jx ? x0j < y < l(Ik)g, and > 0 is chosen so small that fX 2 H : jX ? (x; 0)j < l(Ik )g is contained in Ik; , for all x 2 Ik ; 2 (2; 3). If is chosen small enough, then, for all x 2 Ik , we have that

(X ) n f(x0; y) : jx ? x0j < y < l(Ik)g (xk ); and so, for all 2 (2; 3), we have Z

Ik

m2dx c

Z

jruj d + C jIk j: Bg 2

k;

121

2

Integration in from 2 to 3 gives Z

Ik

m2dx C

Z

m2dx + C2jIkj;

3Ik

which gives (2.14) by additon of k. For the uniqueness, in the case n = 3; 2 < p < 2 + ", note that, keeping the notation we used above, if M (ru) 2 Lp; 2 gjd C(2 + )

m2+dx;

which shows that m is 0 if is small enough.

3. Regularity properties for the Dirichlet problem on graphs We continue treating domains of the form D = f(x; y) 2 Rn : y > '(x); x 2 Rn?1g, where ' : Rn?1 ! R is Lipschitz continuous, i.e., j'(x) ? '(x0)j mjx ? x0j. We will say that f 2 Lp1(); 1 < p < 1, if g(x) = f (x; '(x)) has a gradient in Lp(Rn?1). It is easy to check that this is equivalent to the fact that if F is any extension of f to Rn; jrT F j (dened as in the remarks following (2.2)) belongs to Lp(). It is also easy to see that this is equivalent to the fact that, for any extension F of f; @F=@ T~j ; j = 1; : : : ; n ? 1, belong to Lp(), where T~1; : : : ; T~n?1 are the vector elds introduced in Lemma 2.6. Moreover, @F=@ T~j is independent of the particular choice of the extension F , and depends only on f . We put kf kLp1 () = krxgkLp(Rn?1). Clearly, Lp1() is a Banach space modulo constants. We start out by studying the properties of the single layer potential on the L21() spaces. The following lemma is the graph version of results of G. Verchota ([25]).

Lemma 3.1. The single layer potential S maps L2() onto L21() boundedly, and has a

bounded inverse.

122

Proof. The boundedness follows by the denition of L21 (), and the theorem of Coifman, McIntosh and Meyer ([2]). From (2.3) and (2.4), it follows that

krT S (f )kL2()

(3.2)

@

C1

@n S (f )

L2 ()

C kf kL2():

The lemma now follows as in the L2-Neumann case.

Theorem 3.3. For every f 2 L21() there is a harmonic function u in D with M (ru) 2 L2(), and such that @u=@ T~j = @f=@ T~j non-tangentially a.e. on ; 1 j n ? 1. Furthermore, u is unique modulo constants, and

kM (ru)kL2() ckf kL21 (); where C = C (n; m). Proof. The existence follows from Lemma 3.1. To show uniqueness, it is enough to show that if u is harmonic in D; M (ru) 2 L2(), and @u=@ T~j = 0; j = 1; : : : ; n ? 1 nontantentialy a.e. on , then u is a constant. By our assumption on u; jrT uj = 0 a.e. on . By the uniqueness in the Neumann problem (Theorem 2.5),

u(X ) = C +

Z

g(Q)fjX ? Qj2?n ? jX0 ? Qj2?n gd(Q);

where X0 2 Rn n D , and g 2 L2(). By (3.2), g 0, and so u is a constant. A vector-valued function A : ! RN is a vector-valued atom if A is supported on a ?1=2 surface R ball B = fP 2 : jP ? Qj < 1rg for some Q 2 , and r > 0; kAkL22 () f (B )g and Ad = 0. We say that f 2 H1;at() if there are functions fj 2 L1() with

@ f ; @ f ;::: ; @ f @ T~1 j @ T~2 j @ T~n?1 j

being vector-valued atoms and (3.3)

1 @ f =X j @~ fj ; k = 1; : : : ; n ? 1; ~ @ Tk j =1 @ Tk P

X

jj j < 1:

We also set kf kH11;at() = inf jj j, where the j 's are as in (3.4). Note that H11;at() is a Banach space modulo constants. 123

Lemma 3.5. The single layer potential S maps Hat1 () into H11;at() boundedly. Proof. The proof is standard and will be omitted.

Theorem 3.6. Given f 2 H11;at() there is a harmonic function u in D with M (ru) in L1(), and @u=@ T~j = @f=@ T~j ; j = 1; : : : ; n ? 1 non-tangentially a.e. on . Moreover u is

unique modulo constants, and

kM (ru)kL1() C kf kH11;at(): Proof. For the existence part of the theorem, it is enough to assume '(0) = 0, and to show that if ((@f=@ T~1); : : : ; (@f=@ T~n?1)) is a vector-valued atom supported in fQ 2 : jQj < 1g and u is the L21 solution of the Dirichlet problem with data f , given in Theorem 3.3, then

kM (ru)kL1() C where C depends only on the Lipschitz constant of . By adding a suitable constant to f , we may assume that f has support in B1 = fQ 2 ; jQj < R0g where R0 depends only on the Lipschitz constant of . Furthermore by Sobolev's inequality, kf kL2() C . By the L2 theory for the Dirichlet problem (see [5]) ju(X )j C = C (m) for X 2 D; jX j > 2R0 , and u(X ) takes the boundary value zero continuously on n B1. Let !(x) = 0 for Rn n D, and !(X ) = ju(X )j for X 2 D, so that ! is subharmonic in Rn n B1. By the Phragmen-Lindelöf theory (see [17]) we have j!(X )j C jX j2?n?, where C and only depend on m, and jxj > 2R0. Arguing as in the corresponding Neumann problem, we obtain the existence and the estiamte in Theorem 3.7. We remark that instead of the Phragmen-Lindelöf theory we could have used an odd reection of u to extend u as a solution of Lu = 0 in Rn n B1, and use the Serrin-Weinberger asymptotic expansion just as in the case of the Neumann problem. To show uniqueness, we assume that u is harmonic in D; M (ru) 2 L1(), and @u=@ T~j = 0; j = 1; : : : ; n ? 1, non-tngentially a.e. on . We must conclude that u is a constant in D. As in the corresponding uniqueness theorem for the Neumann problem, we have jru(X )j C fdist (X; )g1?n , and after Rwe add a suitable constant ju(X )j C fdist (x; )g2?n. Thus, by Sobolev's inequality, ju j(n?1)=(n?2)d c, and so u = 0 a.e. on . By the uniqueness in the Neumann problem, it is enough to show that f = 0 a.e., where f = @u=@n. Let b be a Lipschitz function on , with compact support. Let ! be the harmonic extension of b to D. By the Phragmen-Lindelöf principle, j!(X )j C jX j2?n? for X 2 D, X large, where C > 0; > 0. We will now show that for s; t > 0 we have Z Z @u t s !s @n d = ut @! d: @n 124

In fact, let R > 0 be large, and let

R = f(x; y) : jxj < R; '(x) < y < '(x) + Rg: We then have @ R = R [ SR [ TR, where R = f(x; '(x)) : jxj < Rg; SR = f(x; y) : jxj = R; '(x) < y < '(x) + Rg; and TR = f(x; y) : jxj < R; y = '(x) + Rg: Applying Green's theorem in R, we see that Z Z @u s t d: ut @! !s @n d = @ R @n @ R Since !s 2 L1 (D ), and N (ru) 2 L1(), Z

R

Also, while

!s (@ut=@n)d !

Z

Z

!s(@ut=@n)d:

t 2?n? R1?n Rn?1 ???! 0; !s @u d CR R!1 TR @n

Z

Z @u t 2 ? n ? M (ru)d ???! 0: !s d R R!1 nR SR @n Similarly, we know that, for s xed jr!s(X )j C jX j2?n? , and it isR locally bounded in . Thus, since ut 2 L(n?1)=(n?2)(); ut(@!s=@n) 2 L1(), and so ut(@!s=@n)d ! D R R u ( @! =@n ) d . Also, t s Z @! s ut @n d CR2?n R1?n? Rn?1 ! 0; TR while 1 Z 2R Z u @!s d C Z R R SR t @n R 2R n RZjutjjr!sjdX 2?n? jutjdX CR R

2R n R (n?2)=(n?1) 2?n? Z R ( n ? 1) = ( n ? 2) C R jutj dx Rn=n?1

2R 2?n? R C R R(n?2)=(n?1)Rn=n?1 = CR3?n? ???! 0: R!1 Thus, by a choice of an appropriate sequence of Rj 's tending to 1, theR claim follows. R Letting t # 0 we see that !sfd = 0, and then letting s # 0 we have bfd = 0 as desired.

125

Corollary 3.7. The single layer potential S is a bounded operator from Hat1 () onto

H11;at(). It has a bounded inverse, whose norm depends only on the Lipschitz constant of . We now turn to regularity results for the solution of the Dirichlet problem when the data are in Lp1(). This is a new proof of results of Verchota ([25]).

Theorem 3.8. There exists a positive number " = "(n; m) such that for all f 2 Lp1(); 1 < p < 2 + ", there is a harmonic function u in D with M (ru) in Lp(), and @u=@ T~j = @f=@ T~j ; j = 1; : : : ; n ? 1 non-tangentially a.e. on . Moreover, u is unique modulo constants and

kM (ru)kLp() C kf kLp1(); were C depends only on p; n and m. Proof. The case 2 < p < 2 + " follows in the same way as in the Neumann case. Since S is invertible from Hat1 () onto H11;at() and from L2() onto L21(), it follows by interpolation that S is invertible from Lp () onto Lp1(); 1 < p < 2, which gives existence for 1 0 such that S maps Lp () boundedly onto Lp1(), with a bounded inverse, for 1 < p < 2 + ". Furthermore S is a bounded invertible mapping from Hat1 () onto H11;at (). The operators @S=@n and D are bounded and invertible on Lp () for 1 < p < 2 + " and 2 ? " < p < 1 respectively. Furthermore @S=@n is a bounded invertible mapping on Hat1 (), and D is a bounded invertible mapping on BMO ().

4. Bounded Lipschitz domains In this seciton we will sketch the localization arguments which are necessary to extend the results in the last two sections to the case of general bounded Lipschitz domains in Rn. The L2 theory in the Neumann problem and the L2-regularity in the Dirichlet problem have been treated in [13] and [25]. The Lp regularity in the Dirichlet problem has been treated in [25]. From now on we will assume that D Rn; n 3, is a bounded Lipschitz domain such that D = Rn n D is connected. Atoms are dened as in the graph case, and the atomic 126

Hardy space Hat1 (@D) is also dened as in the graph case. We say that f 2 Lp1(@D) if f 2 Lp(@D; d) and for each coordinate chart (Z; '), there are Lp(Z \ @D) functions g1; : : : ; gn?1 so that Z Z @ h(x)f (x; '(x))dx h(x)gj (x; '(x))dx = Rn?1 Rn?1 @xj for all h 2 C01(Z \ Rn?1). It is easy to see that given f 2 Lp1(@D), it is possible to dene a unique vector rT f 2 Rn, at almost every Q 2 @D so that krT f kLp(@D;d) is equivalent to the sum over alla the coordinate cylinders in a given covering of @D of the Lp norms of the locally dened functions gj for f , occurring in the denition of Lp1(@D). The resulting vector eld, rT f , will be called the tangential gradient of f . If F is a function dened on Rn; rT F is orthogonal to the normal vector n, and rF = rT F + (@F=@n) n. In local coordinates, rT f may be realized as (g1(x; '(x)); g2(x; '(x)); : : : ; gn?1(x; '(x)); 0) ? h(g1(x; '(x)); : : : ; gn?1 (x; '(x)); 0); n(x;'(x))in(x;'(x)); Lp1(@D) may be normed by kf kLp1 (@D) = kf kLp(@D) + krT f kLp (@D). Before we proceed to dene the space H11;at(@D), we will make a few remarks about it in the graph case. We say that f is an H11;at() ? L2 atom if f is in L21(), it is supported in T~1)f; : : : ; (@=@ T~n?1)f ) (which automatically veries R a surface ball B , and A = ((@=@ ? () if f 2 L(n?1)=(n?2)(), kAkL2() (B ) 1=2. We say that f 2 H~ 11;atP Ad = 0) veries 1 2 and there exist H1;at() ? L atoms fj and numbers j with jj j < +1, such that P1 f = j=1 j fj , where the sum is taken in the sense of L(n?1)=(n?2)(). Moreover, if f 2 H11;at(), there exists a constant c such that f ? c 2 H~ 11;at(). Let R : Rn?1 ! be given by (x) = (x; '(x)). Then f 2 H~ 11;at(Rn?1) if and only if g(x) = Cn Rn?1 (h(y)=jx ? yjn?2)dy, where h 2 H11;at(Rn?1). In fact, such g(x) clearly belong to L(n?1)=(n?2)(Rn?1), and Lemma 3.5 shows that they are in fact in H~ 11;at(Rn?1). Conversely, if g 2 H~ 11;at(Rn?1), then R Pn n ? 2 g(x) = Cn Rn?1 (h(y)=jx ? yj )dy, where h(y) = j=1 Rj (@=@yj )g, where Rj are the classical Riesz transforms. Note that if we dene H~ 11;at(Rn?1 ) by using H11;at(Rn?1) ? Lp atoms, 1 < p 1, we obtain the same characterization of H~ 11;at(Rn?1), which shows that all these spaces coincide, and have comparable norms. The same fact of course remains true for H~ 11;at() This allows one to show in a very simple fashion that if is a Lipschitz function with compact support in , and f 2 H~ 11;at(), then f also belongs to H~ 11;at(). Our nal remark is that if f 2 H~ 11;at(), and u is the solution to the Dirichlet problem constructed in Theorem 3.7, then uj= f , in the sense of non-tangential convergence, R (n?1)=(n?2) d C , and ju(X )j C fdist (X; )g2?n . Moreover, the uniqueness (ju j) then follows without the addendum 'modulo constants'. We are now ready to dene H11;at(@D). We say that f is an H11;at(@D) ? L2 atom if f is supported in a coordinate cylinder (Z; '), and if is the graph of ', f is an H11;at() ? L2 atom. The space H11;at(@D) is then dened as the absolutely convergent sums of H11;at(@D)? 127

L2 atoms, where the convergence of the sum takes place in the L(n?1)=(n?2)(@D) norms. It is a Banach space, and if we replace L2 atoms by Lp atoms, 1 < p 1, we obtain the same space, with an equivalent norm. Also, if 2 Lip (@D), and f 2 H11;at(@D); f 2 H11;at(@D), and, if f 2 H11;at(@D), then f 2 L(n?1)=(n?2)(@D). Also, Lp1(@D) H11;at(@D), for any 1 < p 1. The non-tangential regions ?(Q); Q 2 @D, are dened as ?(Q) = fX 2 D : jX ? Qj < (1 + )dist (X; @D)g, while the non-tangential maximal function M (!)(Q) = supX 2?1(Q) j!(X )j. Finally, we recall that a bounded Lipschitz domain is called a starlike Lipschitz domain (with respect to the origin) if there exists ' : S n?1 ! R, where ' is strictly positive, and j'()?'(0)j mj ?0j; ; 0 2 S n?1 such that, in polar coordinates (r; ); = f(r; ) : 0 r < '()g. Note that if D is an arbitrary bounded Lipschitz domain, and (Z; ') is a coordinate chart, with kr'k1 m, then, for appropriate > 0; a > 0; b > 0 which depend only on m, the domain D \ U is a starlike Lipschitz domain with respect to X0 = (0; b), where U = f(x; y) : jxj < ; jtj < ag.

Lemma 4.1. Let be a starlike Lipschitz domain, and let u be the L2-solution of the Neumann problem with data an atom a, centred at Q0 2 @ . Then, there exists a constant C , which depends only on the Lipschitz constants of D such that

(a) (b) (c)

kM (ru)kL1(@ ) C; M (ru)2jQ ? Q0jn?1 d C; @

kukH11;at(@ ) C;

Z

if we subtract from u an appropriate constant. Proof. We may assume that the size of the support of a is small. We may also assume that

fy < '(x)g = D, where ' : Rn?1 ! R is Lipschitz with norm depending only on the Lipschitz character of , that @ \ @D fjX ? Q0j < r0g \ @ , where r0 depends only on the Lipschitz characater of @ , that Q0 is the origin and that supp a fjX ? Q0j < r0g. Let v be the solution of the Neuman problem in D, with data a, given by Lemma 2.7, and let ! be the L2-solution of the Neumann problem in , with data @!=@n = 0 on @ \ @D, and @!=@n = ?@v=@nj@ on @ n (@ \ @D). We clearly have u = v + !, and from this the lemma follows.

Lemma 4.2. Let be a bounded, starlike Lipschitz domain, and let u be harmonic in , with M (ru) 2 L1 (@ ) and either rT u = 0 or @u=@n = 0 non-tangentially a.e. on @ . Then, u is a constant.

128

Proof. Assume rst that @u=@n = 0 non-tangentially a.e. on @ . We can show that u is a constant using a variant of the uniqueness proof in Theorem 2.11, using a radial reection across our starlike surface. If rT u = 0, is constant on @ , we have, if b 2 Lip (@ ), and ! is its harmonic extension, that (with ur(x) = u(rx)) Z

Z @u r s !s @n d = ur @! d: @

@ @n R If we let r ! 1, the right-hand side tends to 0, while the left-hand side tends to @ b(@u=@n) d = 0, and so @u=@n = 0 a.e. on @ . Therefore u is constant by the previous result.

We are now in a position to give the solution of the Neumann problem with Hat1 (@D) data, for a general bounded Lipschitz domain D.

Theorem 4.3. Let D Rn be a bounded Lipschitz domain. If u is harmonic in D, with M (ru) 2 L1(@D), then @u=@n 2 Hat1 (@D) and (4.4)

@u

@n 1 Hat (@D)

C kM (ru)kL1(@D):

If f 2 Hat1 (@D), then there is a harmonic function u with M (ru) 2 L1(@D) and @u=@n = f non-tangentially a.e. on @D. Furthermore, u is unique modulo constants, and

(4.5)

kM (ru)kL1(@D) C kf kHat1 (@D);

u can be chosen so that (4.6)

kujjH11;at(@D) C kf kHat1 (@D):

Proof. As in the proof of Lemma 2.10, the estimate (4.4) follows from Green's formula, the extension theorem of Varopoulos ([24]) and the fact that the dual of VMO (@D) is Hat1 (@D). (See [8] for the exact form of the Varopoulos extension theorem that is needed here.) In the case when D is a bounded starlike Lipschitz domain, the rest of the theorem follows from Lemma 4.1 and Lemma 4.2.

We now pass to the general case. We rst establish uniqueness in the general case. Thus, M (ru) 2 L1(@D), and @u=@n = 0 a.e. on @D. We can cover a neighborhood of @D in D, 129

with nitely many bounded starlike Lipschitz domains i D, such that M i (ru), the non-tangential maximal function relative to the domain i, is in L1(@ i). Thus, if vi = uj i , we have @vi=@n 2 Hat1 (@ i). If also @ i B (Qi; 3r) \ @D, for some r > 0; Qi 2 @D, we can take the atoms in the atomic decomposition of @vi=@n to have supports that are so small that they are all contained in @ i=B (Qi; 2r) (since @vi=@n = 0 on B (Qi; 3r) \ @D). It then follows from (b) in Lemma 4.1, and the uniqueness for starlike Lipschitz domains, that M (ru) 2 L2(B (Qi; r) \ @D). Since [iB (Qi; r) \ @D can be taken to be @D, it follows that M (ru) 2 L2(@D), and hence u is a constant by the L2-theory (see [13] or [25]). To show (4.5), it is enough to show that if a is an atom with support contained in a ball of radius r, with r r0 = r0(D), then kM (ru)kL1(@D) C (D), where u is the solution of the L2-Neumann problem with data a. For 2 ( 41 ; 10) let D() be a domain of the form f(x; y) : '(x) < y < '(x) + 1; jxj < 2g, where ' is a Lispchitz function. We can choose numbers 1; 2 and coordinate systems so that the domains D() are starlike Lipschitz domains contained in D, for 14 < < 10. The number r0 is chosen in such a way that there are nitely many D (); 1 N = N (D) such that [@D (1=4) \ @D = @D, and such that, for any we have that either the support of a is contained in @D \ @D (4) or supp a \ @D (3) = ;. We rst claim that for each compact set K D, we have sup jruj C = C (K; D):

(4.7)

K

R

R

To see this, pick 2 L1(D); supp R K and (X )dX = 0. Letting !(X ) = Cn jX ? Y j2?n (Y )dY , we have that ! = ; @D (@!=@n)d = 0, and k@!=@nkL1(@D) C (K; D)kkL1(K). R Let h solve the Neumann problem in D with data @!=@n and D h(x)dx = 0. Then, Z

D

udX =

Z

D

u(! ? h) =

Z

@D

@u : (! ? h) @n

If we now note that the normal derivative of ! ? h is 0 on @D, and we use locally the graph reection argument that we used in the proof of Lemma 2.7, it follows that k! ? hkL1 (@D) C (K; D)kkL1(K) which yields (4.7). Let M ; be the non-tangential maximal operator associated to the domain D (). We can choose a suitable compact set K D so that, for all 2 (1=4; 10) we have (4.8)

Z

@D

M (ru)d

XZ

v

@D ()

jM ; (ru)jd + C sup jruj: K

In order to apply (4.8), we shall rst study the case when (supp a) \ D (3) = ;. From the 130

L2-Neumann theory, it follows that for 1=4 < < 3 we have Z

@D (1=4)

M ;1=4(ru)d

2

c

Z

M ; (ru)2d @D () 2 Z @u d C Z@D ()n@D @n C jru(X )jd: @D ()n@D

Integrating in from 1/2 to 1 now gives Z

@D (1=4)\@D

M ;1=4(ru)d

2

C C

Z

D (2) Z

jru(X )j2dX

D (3)

2

jru(X )jdX :

The last inequality follows from the graph reection and the reversed Hölder inequality for the gradient of the solution of a uniformly elliptic equation in divergence form (see [12]) together with the fact that one can lower the exponent on the right-hand side of such a reversed Hölder inequality. This last fact was proved by the present authors; see [10]. It is possible to use the R R graph reection because supp @u=@n \ D (3) = ;. Hence, ; 1 = 4 (ru)d C D (3) jru(X )jdX , and therefore, given " > 0 there is a compact D (1=4) M K" D such that (4.9)

Z

D (1=4)

M ;1=4(ru)d C"

Z

@D

M (ru)d + C (")

Z

K"

jruj:

If supp a D (4), we let v solve the Neumann problem in D (4), with data a on @D (4) \ @D, and 0 elsewhere on @D (4). Let ! = u ? v. Since @!=@n = 0 on @D (4) \ @D we have, from the argument leading to (4.9), that Z

@D (1=4)\@D

and therefore (4.10)

Z

@D (1=4)\@D

M ;1=4(r!)d C

M ;1=4(ru)d C"

Z

Z

+ C (")

@D Z

D (3)

jr!jdX

M (ru)d

K"

jru(X )jdX + C:

Using now (4.8), (4.9) and (4.10), and the weak estimate (4.7), we see that Z

@D

M (ru)d C"

Z

@D

M (ru)d + C (") + C;

131

and so, if we choose " small enough Z

@D

M (ru)d C = C (D);

which yields (4.5). Finally, note that because of (4.7), we can subtract a constant C from u so that supK ju ? C j CK for all K compact in D. Let v = u ? C . We claim that kvkRH11;at(@D (1)) C . In fact, we know that, because of the Poincaré inequality on @D (1); @D (1) jvj(n?1)=(n?2)d C . But, byR Lemma 4.1, there exists a constant C so that kv ? C kH11;at(@D (1)) C . But then, @D (1) jv ? C j(n?1)=(n?2)d C , and thus jC j C . Therefore kvkH11;at(@D (1)) C , and (4.6) follows for the case of atoms. The general case follows from this. We shall next study the regularity in the Dirichlet problem with H11;at(@D) data.

Lemma 4.11. Let f be an H11;at(@D) ? L2 atom. If u solves the Dirichlet problem with boundary values f , then

Z

(a) (b)

@D

(

Z

@D

? M (ru)2d)

Z

@D

M (ru)d C:

M (ru)2jQ ? Q0j("+1)(n?1)d

1="

C;

where Q0 is the center of the support of f .

(c)

Z

@D

M (ru)2jQ ? Q0jn?1d C:

Here C and " > 0 are independent of the H11;at(@D) ? L2 atom f . Proof. If we perform a change of scale so that the support of f is of size 1, we see that the arguments in the graph case (Theorem 3.7), yield the proof of Lemma 4.11.

Theorem 4.12. Let D Rn be a bounded Lipschitz domain. If u is harmonic in D, with M (ru) 2 L1 (@D), then u 2 H11;at(@D), and kukH11;at(@D) C kM (ru)kL1(@D): (4.13) If f 2 H11;at (@D), then there is a harmonic function u with M (ru) 2 L1 (@D) and u = f non-tangentially a.e. on @D. Furthermore, u is unique, kM (ru)kL1(@D) C kf kH11;at(@D):

132

Proof. Uniqueness follows from the uniqueness in Theorem 4.3. Next note that existence in the range 1 2 follows by a minor modication of the corresponding part of the proof of Theorem 2.12. In fact, the main dierence is that in the bounded case there are two kinds of Whitney cubes Ik , the small ones and the big ones. The small ones are treated just as in 2.12, while the big ones are of diameter comparable to that of D, and hence m1 is comparable to m on them. The rest of the proof is identical, and is therefore omitted.

Our next theorem deals with regularity in the Dirichlet problem with Lp1(@D) data. It was rst proved in [25].

Theorem 4.14. Let D Rn be a bounded Lipschitz domain. There exists a positive p number " = "(D) such that for all f 2 L1 (@D); 1 < p < 2+ ", there is a harmonic function u in D, with M (ru) in Lp(@D), and u = f non-tangentially a.e. on @D. Moreover, u is unique and

kM (ru)kLp(@D) C kf kLp1 (@D) where C depends only on p and D. Proof. Uniqueness follows from Theorem 4.12. Existence follows just as in Theorem 4.13 in the range 2 < p < 2 + ", while the case 1 < p < 2 follows by interpolation.

We will now study the Neumann problem and regularity in the Dirichlet problem for the domain D = Rn n D . The L2 theory for D can be found in [25]. We will let M be the non-tangential maximal operator associated to D , where the non-tangential regions are truncated. We let H~ at1 (@D) be dened as Hat1 (@D), but add the constant 1 to the atoms.

Theorem 4.15. Given f 2 H~ at1 (@D), there exists a harmonic function u in D with @u=@n = f non-tangentially a.e. on @D; u(X ) = o(1) at 1, kM (ru)kL1(@D) C kf kH~at1 (@D); and kukH11;at(@D) C kf kH~at1 (@D): Moreover, u is unique. There exists " = "(D) > 0 such that if f 2 Lp (@D); 1 < p < 2 + ", then kM (ru)kLp(@D) C kf kLp(@D) where C = C (p; D). Proof. The uniqueness reduces to the L2 -uniqueness just as in Theorem 4.3. For existence in the H~ at1 (@D) case, the atom 1 is taken care of by the L2-theory. The existence and the

133

estimate kM (ru)kL1(@D) C for the other atoms are the same as in the proof of Theorem 4.3, the only dierence being that the estiamte Z u D

C (K; D )kkL1(K)

is valid for all 2 L1 (D); supp K , since u(X ) = o(1) at 1. This fact also shows, by a small variation of the argument used in Theorem 4.3, that kukH11;at(@D) C . The case 1 < p < 2 + " of the theorem follows in the same way as in Theorem 4.13. Note also that if M (ru) 2 L1(@D), and u(X ) = o(1) at 1, then @u=@n 2 H~ at1 (@D). This is proved in a similar way to (4.4) in Theorem 4.3.

Theorem 4.16. Given f 2 H11;at(@D), there exists a harmonic function u in D with u = f on @D non-tangentially a.e., u(X ) = o(1) at 1, kM (ru)kL1(@D) C kf kH11;at(@D)

and

@u

@n 1 H~ at(@D)

C kf kH11;at(@D):

Moreover, u is unique. There exists " = "(D) > 0 such that if f 2 Lp1 (@D); 1 < p < 2 + ", then kM (ru)kLp(@D) C kf kLp(@D), where C = C (p; D). Proof. Uniqueness follows as in the proof of uniqueness in Theorem 4.12. Existence for atoms follows in the same way as in Lemma 4.11. The estimate k@u=@nkH~ at1 (@D) C kf kH11;at(@D) follows because of the remark before the statement of Theorem 4.16. The case 1 < p < 2 + " follows in the same way as in Theorem 4.14.

We are now ready to prove the sharp invertibility properties of the layer potentials. For P; Q 2 @D; P 6= Q, let K (P; Q) = !1 hQ ? P; n(Q)i; n where !n is the surface area of the unit sphere in Rn, and put Tf (P ) = p:v: K (P; Q)f (Q)d(Q). Also, let

R

@D

Z 1 Sf (P ) = ! (n ? 2) jP ? Qj2?nf (Q)d(Q): n @D

The boundedness properties of these operators are the same as for the corresponding operators in the graph case. 134

Theorem 4.17. There isp a number q0 p= q0(D); q0 2 (2; 1), such thatp 21 I ? T is an invertible Rmapping from L0 (@D) onto L0 (@D) for 1 < p < q0 , where L0 (@D) = ff 2 Lp(@D) : @D fd = 0g. Also, 21 I ? T is invertible from Hat1 (@D) onto Hat1 (@D). There is a number p0 = p0 (D); p0 2 (1; 2) such that 12 I + T is an invertible mapping of Lp (@D) onto Lp(@D) for p0 < p < 1. Also, 21 I + T is invertible from BMO (@D) onto BMO (@D). There is a number r0 = r0 (D); r0 2 (2; 1) such that S is an invertible mapping of Lp(@D) p onto L1 (@D). Also, S is an invertible mapping from Hat1 (@D) onto H11;at (@D).

Proof. The proof of this theorem is the same as the corresponding L2 case presented in [25], using the results of this ection. Finally, we give representation formulas for the solutions of the Dirichlet and Neumann problem, using layer potential.

Theorem 4.18. Let D Rn be a bounded Lipschitz domain, whose complement is connected. Let q0; p0 ; r0 be the numbers given in Theorem 4.17. Let f 2 Lp (@D); p0 < p < 1, and let Ru(X ) be the unique solution of the Dirichlet problem given in [5]. Then u(X ) = (1=!n ) @D(hX ? Q; n(Q)i=jX ? Qjn)( 12 I + T )?1(f )(Q)d(Q). The same holds when f 2 BMO (@D), and uRis the unique solution of the Dirichlet problem given in [9]. Let f 2 Lp (@D); 1 < p < q0; @D fd = 0, and let u(X ) be the unique (modulo constants) solution of the Neumann problem given in Theorem 4.13. Then, Z ? 1 u(X ) = ! (n ? 2) jX ? Qj2?n 21 I ? T )?1 (f )(Q)d(Q): n @D

The same holds when f 2 Hat1 (@D), and u is as in Theorem 4.3. Let f 2 Lp1(@D); 1 < p < r0 and let u(X ) be the unique solution of the Dirichlet problem given in Theorem 4.14. Then, Z 1 u(X ) = ! (n ? 2) jX ? Qj2?nS ?1(f )(Q)d(Q): n @D The same holds when f 2 H11;at(@D), and u is as in Theorem 4.12.

Proof. The proof follows from well-known properties of layer potentials (see [25], for example), the uniqueness in all the theorems mentioned and Theorem 4.17.

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[19] J. MOSER, On Harnack's theorem for elliptic dierential operators, Comm. Pure and App. Math. 14 (1961), 577-591. [20] J. NASH, Contunuity of the solutions of parabolic and elliptic equations, Amer. J. Math. 80 (1957), 931-954. [21] F. RELLICH, Darstellung der Eigenwerte von u + u durch ein Randintegral, Math. Z. 46 (1940), 635-646. [22] J. SERRIN and H. WEINBERGER, Isolated singularities of solutions of linear elliptic equations, Amer. J. of Math. 88 (1966), 258-272. [23] E. M. STEIN, Singular Integrals and Dierentiability Properties of Functions, Princeton University Press, Princeton New Jersey, 1970. [24] N. VAROPOULOS, A remark on BMO and bounded harmonic functions, Pacic J. Math. 73 (1977), 257-259. [25] G. VERCHOTA, Layer potentials and regularity for the Dirichlet problem for Laplace's equation in Lipschitz domains, J. of Funct. Anal. 59 (1984), 572-611. CHALMERS UNIVERSITY OF TECHNOLOGY AND GÖTEBOR G UNIVERSITY UNIVERSITY OF CHICAGO, ILLINOIS

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