THE DIRICHLET PROBLEM WITH BM O BOUNDARY DATA AND ALMOST-REAL COEFFICIENTS

Transcription

1 THE DIRICHLET PROBLEM WITH BM O BOUNDARY DATA AND ALMOST-REAL COEFFICIENTS ARIEL BARTON Abstract. It is known that a function, harmonic in a Lipschitz domain, is the Poisson extension of a BMO function if and only if its gradient satisfies a Carleson-measure condition. We show that the same is true of functions that satisfy elliptic equations in two-dimensional Lipschitz domains, provided the coefficients are independent of one coordinate and have small imaginary part. Contents. Introduction 2. Preliminaries Definitions Bounds on solutions to elliptic equations Layer potentials 7 3. Existence of solutions A L 2 estimate in special Lipschitz domains A L 2 estimate in general Lipschitz domains BM O estimates in Lipschitz domains Converses and uniqueness 27 References 35. Introduction Consider the Dirichlet problem { div A u = 0 in V,.) u = f on where A is a uniformly elliptic matrix, that is, where there exist constants Λ > λ > 0 such that.2) λ η 2 Re η AX)η, ξ AX)η Λ ξ η for all X R 2 and all η, ξ C 2. In this paper we study the Dirichlet problem.) in the case where the boundary data f lies in the space BMO of functions of bounded mean oscillation, where the 200 Mathematics Subject Classification. Primary 35J25, Secondary 3A25. Key words and phrases. Elliptic equations, Dirichlet problem, Lipschitz domains, Carleson measures, bounded mean oscillation, layer potentials.

2 2 ARIEL BARTON domain V is a two-dimensional Lipschitz domain, and where the coefficients A are complex but have small imaginary part, and are t-independent in the sense that.3) Ax, t) = Ax, s) for all x, s, t R. This paper is an extension of the author s earlier work [8], which instead considered boundary data f in the Lebesgue space L p ). The study of elliptic boundaryvalue problems has a long and rich history; we refer the reader to Section. of [8] for a more detailed history of the Dirichlet problem with boundary data in L p ). In this paper we will prove the following theorem. Theorem.4. Suppose that A is t-independent and elliptic, and that V R 2 is a two-dimensional Lipschitz domain with connected boundary. Then there exist constants ε > 0 and C > 0, depending only on the constants λ, Λ in the bound.2) and the Lipschitz character of V, such that if Im A L < ε, then the following holds. For all f BMO ) there exists a function u that satisfies the conditions.) and such that, for all X 0 and all r > 0,.5) σ BX 0, r)) V BX 0,r) ux) 2 distx, ) dx C f 2 BMO. Furthermore, u is unique among solutions to the problem.) for which the supremum in the estimate.6) is finite. Conversely, if div A u = 0 in V and if.6) sup sup X 0 r>0 σ BX 0, r)) then f = u V BX 0,r) ux) 2 distx, ) dx C 2, exists and lies in BMO ), with f BMO C C. We will also establish the following maximum principle for such coefficients. Theorem.7. Let A and V be as in Theorem.4. Then there is a constant ε > 0 such that, if Im A L < ε, then for each f L ) there exists a unique function u that solves the problem.) and such that u L V ). Furthermore, there is a constant C such that u L V ) C f L ). In Theorems.4 and.7 the coefficients A are taken to be independent of the second coordinate. We would like to emphasize that the domain V in these theorems may be bounded, and thus the distinguished direction in which A is constant) need not be transverse to the boundary. This is the setting of the papers [24, 34, 8], but is distinct from the setting of many other papers such as [23, 26, 6, 2, 3,, 33, 8, 2, 20, 5]), where the coefficients A are taken to be constant specifically in a direction transverse to the boundary. We now review the history of the BM O-Dirichlet problem. In [5], Fefferman proved that the Poisson extension of a BM O function in the upper half-space satisfies the bound.5). In [3], Fabes, Johnson and Neri proved that a function u harmonic in the upper half-space that satisfies the condition.6) was necessarily the Poisson extension of a BMO function. Thus, Theorem.4 is valid for harmonic functions that is, for A I) in the upper half-space. Later in [4], Fabes and Neri extended these results to harmonic functions in arbitrary starlike Lipschitz domains.

3 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 3 Many of the known results concerning the Dirichlet problem.) for more general coefficients involve boundary data in Lebesgue spaces L p ), rather than in BMO ). We say that the L p -Dirichlet problem is well-posed if, for each f L p ), there exists a unique function u that solves the problem.) and such that the nontangential maximal function Nu lies in L p ), where NuX) = sup{ uy ) : X Y < + a) disty, )} for some constant a > 0. The following result suggests a connection between the L p -Dirichlet problem and the BMO-Dirichlet problem of Theorem.4. Suppose that V is a bounded Lipschitz domain of arbitrary dimension), that X 0 V, and that the L p -Dirichlet problem is well-posed in V for some real coefficient matrix A. Then the inequalities.8) Nu 2 dσ ux) 2 distx, ) dx C Nu 2 dσ C V are valid for any u that satisfies div A u = 0 in V and ux 0 ) = 0, where C depends on A, V and X 0. This was established by Dahlberg in [9] for harmonic functions) and by Dahlberg, Jerison and Kenig in [0] for more general real coefficients). In [24], Kenig, Koch, Pipher and Toro proved the converse: that if the estimate.8) holds in all bounded Lipschitz domains V R n, for a given real elliptic coefficient matrix A, then for each such V there is some p < such that the L p -Dirichlet problem is well-posed in V. We observe that under these conditions, the BMO-Dirichlet problem is well-posed in all such domains. In the later part of [24], the authors showed that the bound.8) holds in two-dimensional Lipschitz domains for real t-independent coefficients. In [2], the bound.8) was shown to hold for real t-independent coefficients in higher dimensions, at least in the domain above a Lipschitz graph.) Thus, Theorem.4 follows from [24] under the additional assumption that A is real-valued. We will use these results to generalize to A with small imaginary part. In [2], Dindos, Kenig, and Pipher found another relationship between the BM O- Dirichlet problem and the L p -Dirichlet problem for real coefficients. Specifically, they showed that if A is an arbitrary coefficient matrix, elliptic and real-valued but not necessarily t-independent, and if V is a Lipschitz domain, then the BM O- Dirichlet problem is well-posed in V if and only if there is some p, with < p <, such that the L p -Dirichlet problem is well-posed in V. The outline of this paper is as follows. In Section 2 we will define our terminology and some important tools, and will review some known results. In Section 3, we will prove the existence result of Theorem.4; that is, we will show that for each f BMO ), there exists a function u that solves the problem.) and satisfies the bound.5). In Section 4, we will study functions u that satisfy the bound.6); this will allow us to prove uniqueness of solutions and the converse result of Theorem.4, and will also allow us to prove the maximum principle Theorem.7). The results in this paper were part of my dissertation work at the University of Chicago. I would like to thank my advisor, Carlos Kenig, for his guidance and advice. I would also like to thank Svitlana Mayboroda and the anonymous reviewer for their comments, which have greatly improved the exposition of this paper.

4 4 ARIEL BARTON 2. Preliminaries In this section, we will begin by defining our terminology. We will then remind the reader of some known results concerning solutions to elliptic equations; we will conclude this section by discussing layer potentials, an important tool in the construction of solutions to boundary-value problems. 2.. Definitions. We say that div A u = 0 in a domain V in the weak sense if 2.) η A u = 0 for all η C0 V ). Thus, solutions u to the differential equation div A u = 0 need not be smooth; they are only assumed to lie in the local Sobolev space W,loc 2 V ) of functions with one weak derivative. In this paper, we will work only in two-dimensional Lipschitz domains, defined as follows. Definition 2.2. We say that the domain Ω R 2 is a special Lipschitz domain if, for some Lipschitz function ϕ and unit vector e, Ω = Ω e, ϕ) = { x e + t e : t > ϕx) } ), where e 0 = e. 0 We refer to M = ϕ L R) as the Lipschitz constant of Ω. We say that the domain V is a Lipschitz domain if either V is a special Lipschitz domain, or if there is some positive scale r, some constants M > 0 and c 0, and some finite set {X j } N j= of points with X j, such that the following conditions hold. First, N BX j, r j ) for some r j with r < r j < c 0 r. c 0 j= Second, for each X j, there is some special Lipschitz domain Ω j = Ω e j, ϕ j ) with X j Ω j and with Lipschitz constant at most M, such that Z j Ω j = Z j V, where Z j = {X j + x e j + t e j : x < 2r j, t < 4 + 4M)r j }. If V is a special Lipschitz domain let N = c 0 = ; otherwise let M, N, c 0 be as above. We refer to the triple M, N, c 0 ) as the Lipschitz character of V. We will let V refer to an arbitrary Lipschitz domain and reserve Ω for special Lipschitz domains. If V is a Lipschitz domain, we let σ denote the surface measure on and let ν denote the unit outward normal to V. We let ffl denote the average integral ffl E f dµ = µe) E f dµ. If V R2 is a Lipschitz domain and f is defined on, then the BMO-norm of f is given by 2.3) f BMO ) = sup inf F C fy ) F 2 dσy ) ) /2 where the supremum is taken over all sets that are both bounded and connected. Observe that if f is constant on each connected component of, then f BMO ) = 0; we let BMO ) be the the space of all equivalence classes of

5 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 5 functions f with f BMO ) <. We remind the reader that, if, and if, are bounded connected sets, then for some absolute constant C, 2.4) f ffl /2 f ) dσ) dσ 2 f BMO ) C + C logσ )/σ)). If X, we define the nontangential cone γx) by 2.5) γx) = γ a X) = {Y V : X Y < + a) disty, )} where a is a fixed positive number. The exact value of a is usually irrelevant provided a > 0.) The nontangential maximal function is thus given by 2.6) N ux) = sup uy ) = sup{ uy ) : X Y < + a) disty, )}. Y γx) By u = f on, or u = f, we mean that f is the nontangential limit of f on the boundary. That is, there is some a > 0 such that, for almost every dσ) X, we have that 2.7) lim uy ) = fx). Y X, Y γx) Definition 2.8. We define a convenient set of tents QX, R) as follows. If X, then X BX j, r j ) for one of the points X j of Definition 2.2. Let e j, ϕ j be as in Definition 2.2. Then X = x e j + ϕ jx) e j for some x R. If 0 < r < r j, we define QX, r) = {y e j We let X, r) = QX, r). + s e j : x y < r, ϕ j y) < s < ϕ j y) M)r.} Observe that QX, r) is a simply connected, bounded Lipschitz domain whose Lipschitz character depends only on M, and that V BX, r) QX, r) V BX, Cr). Furthermore, QX, r) is starlike with respect to the point X + + 2M)r e j. It should be noted that QX, r) depends on our choice of Ω j and e j, and also that if V is not a special Lipschitz domain, then QX, r) is defined only for r/σ ) sufficiently small. Finally, we will let C and ε denote positive constants whose value may change from line to line, but that in general depend only on the ellipticity constants λ, Λ in formula.2) and on the Lipschitz character of any relevant domains V ; any other dependency will be indicated explicitly. By a b, we mean that for some such constant C, the inequalities C a b C a are valid Bounds on solutions to elliptic equations. In this section we discuss some well-known results concerning elliptic partial differential equations, as well as some results valid in the special case of two dimensions or of t-independent coefficients. To prove Corollary 2.34 below, we will need to consider weak solutions to the inhomogeneous equation div A u = div f, where u W,loc 2 V ) is a weak solution in V if η A u = η f for all η C0 V ).

6 6 ARIEL BARTON The following lemmas are thus stated in that generality. However, we will most commonly use these lemmas in the case where f 0, that is, where div A u = 0. Lemma 2.9 The Caccioppoli inequality). Let A be an elliptic matrix, and suppose that div A u = div f in BX, r) for some f L 2 BX, r)). Then there exists a constant C depending only on the constants λ, Λ in formula.2) such that u 2 C r 2 u 2 + C f 2. BX,r/2) BX,r)\BX,r/2) BX,r) Remark 2.0. More generally, if V is a Lipschitz domain, then for any u with u L 2 BX, r) V ), div A u = div f in V BX, r) and u = 0 on B0, r), we have that BX,r/2) V u 2 C r 2 where C depends only on λ and Λ. V BX,r)\BX,r/2) u 2 + C V BX,r) Lemma 2. Theorem 2 in [29]). Let A be elliptic, and suppose that div A u = div f in BX, r). Then there exists a p 0 > 2, depending only on the constants λ, Λ in formula.2), such that if f L p BX, r)) for some 2 < p < p 0, then u L p loc BX, r)), and ) /p u p C BX,r/2) ) /2 u 2 + C BX,r) f 2 ) /p f p. BX,r) In two dimensions, Lemmas 2.9 and 2. together with Morrey s inequality have an immediate corollary. Lemma 2.2. Let A be elliptic, and suppose that div A u = div f in BX 0, r) for some f L p BX 0, r)), p > 2. Assume that the ambient dimension is 2. For some C, α > 0, depending only on λ, Λ and p, 2.3) ux) uy ) C for all X, Y BX 0, r/2). Furthermore, X Y α r α 2.4) sup ux) C BX 0,r/2) { ) /2 u 2 + r BX 0,r) ) /2 u 2 + Cr BX 0,r) ) /p } f p BX 0,r) ) /p f p. BX 0,r) Lemma 2.2 is valid in higher dimensions for real coefficients; this was first proven in [], [32] and [3] for A symmetric, and extended to nonsymmetric real equations in [30]. It is, however, not valid in higher dimensions for general complex coefficients; see [28] and [7] for specific counterexamples. Now, suppose that Ax, t) = Ax) is a 2 2 t-independent matrix. Let the matrix B 0 X) be given by 2.5) B 0 X) = B A 0 X) = ) a X) a 2 X) 0 We have the following lemma from [6]. ) a X) a for AX) = 2 X). a 2 X) a 22 X)

7 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 7 Lemma 2.6 Lemme II.3 in [6]). If div A u = 0 in BX, r) R 2, and Ax, t) = Ax) is t-independent, then div A u t = 0 in BX, r). Furthermore, if B 0 is as in formula 2.5), then ) ũt B0 A X) ux) = u t for some ũ t that satisfies div à ũ t = 0, for à = / det A)AT. Here A T denotes the matrix transpose of A. Thus, we may apply Lemmas 2.9, 2. and 2.2 to the components of B0 A u. In particular, by the bound 2.4), Lemma 2.9 and the Poincaré inequality, if div A u = 0 in BX, r) then 2.7) ux) C ) /2 u 2. BX,r)\BX,r/2) 2.3. Layer potentials. We will construct solutions to Theorem.4 using layer potentials, defined as follows. If A is an elliptic 2 2 matrix defined on R 2, then for each X R 2, there exists a function Γ X = Γ A X such that 2.8) AY ) Γ X Y ) ηy ) dy = ηx) for all η C0 R 2 ). R 2 This function Γ X is called the fundamental solution for div A with pole at X. It was constructed by Kenig and Ni in the appendix to [25] in the case of real coefficients, and by Auscher, McIntosh and Tchamitchian in Theorem 3.6 of [4] in the case of complex coefficients. In dimensions higher than two, a construction of the fundamental solution may be found in [22].) We will need some properties of the fundamental solution in two dimensions; see Theorem 2.6, Lemma 2.7 and Lemma 4. in [27] for the case where A is real, and Chapter 4 of [8] for the case where A is complex. If A is elliptic and t-independent, then there is a fundamental solution Γ A X that satisfies 2.9) Γ A C XY ) X Y. If A is merely elliptic then 2.20) sup r>0 r X Y 2r Γ A XY ) 2 dy C. If we require Γ X to satisfy the bound 2.9) or 2.20), then Γ X is unique up to an additive constant. We may choose additive constants such that, if X, Y R 2 and X Y, then 2.2) Γ AT X Y ) = Γ A Y X). Let V be a Lipschitz domain. If f : C is a function, and X R 2 \, the classical double layer potential Df is given by 2.22) DfX) = DV A fx) = νy ) A T Y ) Γ AT X Y )fy ) dσy ). If A is discontinuous, then the conormal derivative is defined weakly, that is, by the formula 2.23) νy ) A T Y ) Γ AT X Y )fy ) dσy ) = A T Y ) Γ AT X Y ) F Y ) dy V

8 8 ARIEL BARTON for any F W 2 V ) with F = f on and with F 0 near X. This defines DfX) for all smooth, compactly supported functions f. We let 2.24) KfX) = K A V fx) = lim Z X, Z γx) DA V fz) so Kf = Df in the sense of formula 2.7). We will show Corollary 2.34) that if f BMO ) or f L p ) for some < p <, then KfX) exists for almost every X. We will occasionally need a somewhat more general potential. If f : C 2 is a vector-valued function, define 2.25) T fx) = TV A fx) = B0 AT Y ) Γ AT X Y ) fy ) dσy ) where B 0 is as in formula 2.5). If Y Y < 2 X Y, then by Lemma 2.6 and the bounds 2.3) and 2.9), 2.26) B AT 0 Y ) Γ AT X Y ) B AT 0 Y ) Γ AT X Y ) C Y Y α X Y +α and so the kernel B0 AT Y ) Γ AT X Y ) of T is locally Hölder continuous in Y. By formula 2.2) and the bound 2.3), it is locally Hölder continuous in X as well. Observe that 2.27) Df = T B AT 0 ) T ) Aν f). We state without proof some simple results concerning layer potentials; these results are well known in the case of harmonic functions and are straightforward to establish for the more general case. First, suppose X / and that p <. Then T fx) converges for all f L p ), and the mapping f DfX) is well-defined and extends to a unique bounded functional on L p ). Also, by formula 2.2), div A T f) = 0 and div A Df) = 0 in V and in V C. If V is bounded, then by formulas 2.8) and 2.23), D in V and D 0 in V C. More generally, if V is a Lipschitz domain with compact boundary, and if f is constant on each connected component of, then Df is constant in each connected component of R 2 \. If V = Ω is a special Lipschitz domain, then D is still constant in Ω and in Ω C in the sense that, if X, X Ω or if X, X Ω C, then for all X 0 Ω, lim D Ω BX r 0,r)X) D Ω BX0,r)X ) = 0. Thus, by the bounds 2.9) and 2.4), if f BMO ) then Df is well-defined up to an additive constant. This paper builds on the results of [8]. Specifically, we will need the following lemmas. Lemma 2.28 Theorem 6. in [8]). Let V R 2 be a Lipschitz domain and let A be t-independent and elliptic. Suppose in addition that A is smooth. Then there is some ε 0 > 0, depending only on the constants λ, Λ in formula.2), such that if Im A L R) < ε 0, then for any < p < we have that 2.29) NT A V f) L p ) Cp) f L p )

9 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 9 where N is as in formula 2.6) and where Cp) depends only on p, λ, Λ and the Lipschitz character of V. Lemma 2.30 Section 2.4 in [8]). Let A and V be as in Lemma 2.28, and suppose in addition that is connected. Then there is some ε > 0 and some p 0 <, depending only on λ, Λ and the Lipschitz character of V, such that if Im A L R) < ε, then the operator K is invertible on L p ) for any p 0 < p <. If K is invertible on L p ), then by formula 2.27) and the bound 2.29), for every f L p ) the function u = DK f) exists and satisfies div A u = 0 in V, 2.3) u = f on, Nu L p ) C f L p ). This is the classic method of layer potentials for constructing solutions to the Dirichlet problem. [8] used the following uniqueness result to complete the proof that the L p -Dirichlet problem is well-posed. Lemma 2.32 Theorem 8.3 in [8]). Let V R 2 be a Lipschitz domain with connected boundary and let A be t-independent and elliptic. Then there is some ε > 0 and some p 0 <, depending only on λ, Λ and the Lipschitz character of V, such that if Im A L R) < ε, and if there is some p with p 0 < p < such that div A u = 0 in V, u = 0 on, Nu L p ), then u 0 in V. We will also need one result concerning the behavior of K on BMO ). Lemma 2.33 Corollary 9.3 in [8]). Let A and V be as in Lemma Then there is some ε > 0 such that if Im A L R) < ε, then the operator K is invertible on H ). Here H ) denotes the Hardy space dual to BMO ) and K represents the operator adjoint; thus, K is invertible on BMO ) provided Im A L < ε and is connected. These lemmas imply that the L p -Dirichlet problem is well-posed for smooth coefficients A. In [8], standard approximation techniques were used to pass to rough coefficients. In Corollary 2.34, we will use such approximation techniques to show that layer potentials are bounded, even for rough coefficients; this is somewhat more involved than the argument treating only the L p -Dirichlet problem. Corollary Lemmas 2.28, 2.30 and 2.33 are valid even if A is not smooth. Furthermore, if A, V are as in Lemma 2.28, except that A need not be smooth, then T f exists almost everywhere in provided f L p ) for some < p <, and KfX) exists for a.e. X provided f L p ) or f BMO ). Proof. We wish to exploit the fact that the above lemmas are valid for smooth coefficients. Let A j x, t) = A j x) = A ϕ j x), where ϕ j x) = jϕjx) for some smooth, nonnegative, compactly supported function ϕ with ϕ =. Observe that R A j is smooth, t-independent, Im A j L Im A L, and that A j satisfies the

10 0 ARIEL BARTON ellipticity condition.2) with the same constants λ, Λ as A. Furthermore, if U is a bounded set and q <, then A j A in L q U) as j. Observe that Lemma 2.28 is valid for A j ; to show that it is valid for A as well, we must control T A T Aj. We begin by controlling the kernels of these operators. Specifically, we will begin by bounding the L 2 norm of Γ A X ΓAj X. Fix some positive number r. Let f be a smooth vector field supported in B0, 3r). Define ux) = u j X) = B0,3r) B0,3r) fy ) Γ AT X Y ) dy = B0,3r) fy ) Γ AT j X Y ) dy. div fy ) Γ AT X Y ) dy, We begin with some preliminary bounds on u and u. By the bound 2.9), we have that if p > 2, then ux) is uniformly bounded and ux) Cp ) min r 2/p, r 2 2/p distx, supp f) ) f L p B0,r)). Similarly, we may bound u in terms of div f; thus u is locally in L 2 and we may apply the lemmas in Section 2.2. By the definition 2.8) of Γ A Y and by formula 2.2), we have that ηx) AX) ux) dx = div fy ) ηx) AX) Γ A Y X) dx dy R 2 R 2 R 2 = div fy ) ηy ) dy = fy ) ηy ) dy R 2 R 2 and so div A u = div f in R 2. Thus by Lemmas 2.9 and 2., we have that if R > 0 and if 2 < p < p 0, where p 0 is as in Lemma 2., then ) /p u p Cp ) B0,R) R 2/p We may take the limit as R ; this yields that ) /2 ) u 2 + Cp ) /p f p. B0,2R) B0,R) u L p R 2 ) C f L p B0,3r)). Now we wish to consider u u j. Observe that div A j u u j ) = div A j u div f = diva j A) u. By the bound 2.4), we then have that if X < R/2 and 2 < p 2 < p, then ) /2 ux) u j X) C u u j 2 B0,R) ) /p2 + CR 2/p2 A A j ) u p2. B0,R) Fix some δ > 0. By our bounds on u above, if R is large enough, depending only on r and δ, then the first term is at most δ f L p B0,3r)) /2. We may assume

11 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA R > 4r, and so the bound ux) u j X) δ 2 f L p B0,3r)) + CR 2/p2 B0,R) A A j ) u p2 ) /p2 is valid for all X < 2r and all f smooth and supported in B0, 3r). By Hölder s inequality, we have that ) /p2 ) /q ) /p A A j ) u p2 A A j q u p B0,R) B0,R) B0,R) for some q <. Recall that A j A in L q B0, R)); thus, there is some j depending on R and δ such that ux) u j X) δ f L p B0,r)) for all X < 2r and all f smooth and supported in B0, 3r). But recall that ux) u j X) = f Γ AT j X ) ΓAT X. B0,3r) Thus, our pointwise bound on u u j implies that, if j is large enough depending only on r and δ), then B0,3r)\BX,η) Γ A X Γ Aj X L p B0,3r)) δ for all X < 2r. Fix some η > 0. By Hölder s inequality Γ AT X Γ AT j 2 X B0,3r)\BX,η) Γ AT B0,3r)\BX,η) X Γ AT j X Γ A T X p ) /p Γ AT j X ) /p p. By the bound 2.9) the first term is at most Cr, η), and by the above arguments, if j is large enough then the second term is at most δ. Thus, Recall that Γ AT X Γ AT j X T A fx) = L 2 B0,3r)\BX,η)) δ Cr, η). B AT 0 Y ) Γ AT X Y ) fy ) dσy ). We wish to bound the difference B0 AT Γ AT X and T Aj. Observe that B AT BAT j 0 Γ AT j X A in the kernels of T 0 Y ) Γ AT X Y ) B AT j 0 Y ) ΓAT j X Y ) = B0 AT Y ) B AT j 0 Y )) Γ AT X Y ) + B AT j 0 Y ) Γ AT X Y ) Γ AT j X Y )) and that B AT j 0 B0 AT in L q U) for any q < and any bounded set U. Thus, again using Hölder s inequality and the bound 2.9) on Γ AT X, we have that for any ε > 0 and any r > η > 0, B0 AT Γ AT X B AT j 0 Γ AT j X L 2 B0,3r)\BX,η)) ε

12 2 ARIEL BARTON for all j sufficiently large. We now improve to a pointwise bound. Choose some r > η > 0 and some X and Y with X < 2r, Y < 2r and X Y > η. Define v ) w = B AT 0 Γ AT X, vj w j ) = B AT j 0 Γ AT j X. By Lemma 2.6, div ÃT v = 0 and div ÃT j v j = 0 away from X. As before, div ÃT j v v j) = div ÃT j v = divãt j ÃT ) v. Applying the bound 2.4) to the function v v j, we see that if X Y > 2η and if p 2 > 2, then ) /2 vy ) v j Y ) C v v j 2 BY,η/2) ) /p2 + Cη ÃT ÃT j p2 v p2. BY,η) The first term is at most C B AT 0 Γ AT X BAT j 0 Γ AT j X L 2 B0,3r)\BX,η)). By the bound 2.7), Lemma 2.9 and the bound 2.9), we have that v C/η 2 in BY, η). Thus, since A j A in L p2 B0, 3r)), we have that v j v uniformly in X and Y. A similar argument is valid for w. Thus, for any ζ > 0 and any r > η > 0, we have that there is some j such that B0 AT Y ) Γ AT X Y ) B AT j 0 Y ) ΓAT j X Y ) ζ for all X, Y B0, 2r) with X Y > 2η. Define N η,r F Z) = sup{ F X) : X γz) \ BZ, 2η), X < r}. We use the above remarks to bound N η,r T A B0,2r) f)). Specifically, if j is large enough, then we have that N η,r T A B0,2r) f))x) Nη,r T Aj B0,2r) f))x) + ζ f dσ and so the bound N η,r T A B0,2r) f)) L p ) Cp) f L p ) B0,2r) follows from the uniform bound on NT Aj f) Lp ). Bounding N η,r T A \B0,2r) f)) is much more straightforward. If is compact then we consider only r large enough that B0, 2r). Otherwise, V is a special Lipschitz domain. By the bound 2.9) and by the definition 2.25) of T, T A \B0,2r) f)x) Cr /p f L p ) for all X < r and so N η,r T A \B0,2r) f)) L p ) C f L p ). Combining these results gives the estimate N η,r T A f) Lp ) Cp) f B0,2r). This estimate is uniform in η and r; letting η 0 and r, we recover the same estimate on NT A f).

13 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 3 We thus have that the bound 2.29) is valid, and so Lemma 2.28 is valid even if A is not smooth. [8] used the assumption that A was smooth only to prove boundedness of layer potentials; thus, Lemma 2.28 implies Lemmas 2.30 and 2.33 even if A is not smooth. Finally, we establish that T f, Df have nontangential limits for f L p ), < p <, and for f BMO ). There is a bounded invertible matrix B such that, if g is smooth and compactly supported, then T B g) has a nontangential limit at all points in ; see Lemma 5.7 in [8]. By standard techniques, if the bound 2.29) is valid, then we may extend the condition that T f has a nontangential limit at almost every point in to all f L p ), not only f of the form f = B g for g C0. By formula 2.27), if f L p ) then Df has a nontangential limit at X and so KfX) exists for almost every X. If f BMO ) and is compact, then Kf exists because f L 2 ). If V = Ω is a special Lipschitz domain and f BMO Ω), then let be any connected bounded subset of Ω. We may take f dσ = 0. Let with dist, Ω \ ) > 0. Write f = f + f 2, where f = f in and f 2 = f in Ω \. Then Kf X) exists for almost every X Ω because f L 2 Ω). By the bound 2.9), formula 2.2) and the bound 2.3), if X, Z are far from supp f 2 then Df 2 X) Df 2 Z) Ω X Z α X Y +α f 2Y ) dσy ) and so by the bound 2.4), Kf 2 X) exists for all X. proof. 3. Existence of solutions This completes the In this section, we establish the existence result in Theorem.4 by proving the following theorem. We will complete the proof of Theorem.4 in Section 4. Theorem 3.. Suppose that A is t-independent and elliptic, and that V R 2 is a Lipschitz domain. Assume that the boundary layer potential K is bounded and invertible on BMO ). Then for every g BMO ), there exists a function u with div A u = 0 in V, u = g, and such that σ BX 0, r)) V BX 0,r) ux) 2 distx, ) dx C g 2 BMO where C depends only on λ, Λ, the Lipschitz character of V, and the BMO bound of K and K. By Lemma 2.33 and Corollary 2.34, if A is elliptic and t-independent and if Im A L is small enough, and if V is a Lipschitz domain with connected boundary, then the conditions of Theorem 3. hold. The solution u is constructed by the classic method of layer potentials, that is, by letting u = DK g). If we let f = K g, then because K is invertible on BMO ), there is some C such that f BMO ) C g BMO ). Thus, to prove Theorem 3., we need only show that the inequality 3.2) DfX) 2 distx, ) dx C f 2 BMO ) σ BX 0, r)) V BX 0,r)

14 4 ARIEL BARTON is true for all f BMO ). We remark that the bound 3.2) is valid without any assumptions on K or Im A L ; we will prove that the bound 3.2) holds for all two-dimensional Lipschitz domains V and all elliptic t-independent coefficient matrices A. 3.. A L 2 estimate in special Lipschitz domains. A major step in the proof of the estimate 3.2) is the following lemma. Lemma 3.3. Suppose that V R 2 is a Lipschitz domain and that f L 2 ). If A is elliptic and t-independent, then 3.4) DfX) 2 distx, ) dx C f 2 L 2 ). V In this section, we will prove that Lemma 3.3 is valid provided V = Ω is a special Lipschitz domain; we will move to general Lipschitz domains in Section 3.2. Our proof will be by means of T b) theorems. We remark that in the case where Ω is the domain above a Lipschitz graph that is, where e T = 0 ) ), Lemma 3.3 follows from the results of Rosén in [33], and was also established by means of T b) theorems by Grau de la Herran and Hofmann in [8]. We will use a square-function T b) theorem of Semmes; this result is a special case of the extension iii) on page 724 of [35]. Theorem 3.5. Suppose that Ψ t : R R C 2 2 is a matrix-valued function for each t > 0, and that for some constants α > 0 and C 0 <, Ψ t satisfies Ψ t x, y) C 0 t α t + x y ) +α, Ψ tx, y) Ψ t x, y ) C 0 y y α t + x y ) +α whenever y y < 2 x y or y y < 2 t. If f : R C 2 is a vector-valued function, or more generally if f : R C 2 m, let Θ tfx) = Ψ t x, y) fy) dy. R Suppose that there is a constant C <, and a matrix-valued function b : R C 2 2, such that for all y R and all intervals Q R, Q 2 dx dt by) C, Θ t bx) C Q, 0 Q t by) dy) C. Q Then there is a constant C depending only on C 0, C and α such that, if f L 2 R C 2 ), then Θ tfx) 2 dx dt C f t 2 L 2 R). R 2 + Although this theorem is sufficient for our purposes, we observe that more general results are known; see, for example, Hofmann s paper [9] for a nice survey of T b) theorems. We reformulate this theorem in terms of special Lipschitz domains. If Ω = {x e + t e : t > ϕx)}, we may prove the following theorem by letting and applying Theorem 3.5. Ψ t x, y) = tψx e + ϕx) + t) e, y e + ϕy) e)

15 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 5 Theorem 3.6. Let Ω be a special Lipschitz domain. Suppose that Ψ : Ω Ω C 2 2, and that for some constants α > 0 and C 0 <, Ψ satisfies 3.7) 3.8) ΨX, Y ) C 0 X Y 2, ΨX, Y ) ΨX, Y Y Y α ) C 0 X Y 2+α whenever Y Y < 2 X Y. Define ΘfX) = Ω ΨX, Y ) fy ) dσy ). Suppose that there is a constant C <, and a matrix-valued function b : Ω C 2 2, such that for all X 0, Y Ω, all r > 0 and all bounded connected sets Ω, 3.9) 3.0) 3.) BX 0,r) Ω by ) C, ΘbX) 2 distx, Ω) dx C r, by ) dσy )) C. Then there is a constant C, depending only on C 0, C, α and the Lipschitz constant of Ω, such that if f L 2 Ω C 2 ), then 3.2) Θfx) 2 distx, Ω) dx C f L2 R). Ω Let Ω be a special Lipschitz domain; we now prove that Lemma 3.3 is valid with V = Ω. By formula 2.27), we need only prove that 3.3) T fx) 2 distx, Ω) dx C f 2 L 2 ) for all f L 2 Ω C 2 ). Recall that T fx) = Ω Ω BAT 0 Γ AT f dσ. Let X ΨX, Y ) = X Γ AT X Y )) T B AT 0 Y ) T = X B AT 0 Y ) Γ AT X Y )) T so that T fx) = Ω ΨX, Y ) fy ) dσy ). Defining Θ as in Theorem 3.6, we have that Θ f = T f and so the bound 3.3) follows from the bound 3.2). We claim that Ψ satisfies the estimates 3.7) and 3.8). Let ux) = B AT 0 Y ) Γ AT X Y ). By the bound 2.9) and formula 2.5), ux) C/ X Y. By formula 2.2) and the definition 2.) of weak solution, u satisfies div A u = 0 away from Y. We use the bound 2.7) to bound ux) ; this yields that C 3.4) X B0 AT Y ) Γ AT X Y )) X Y 2 and so the estimate 3.7) holds. Similarly, the estimate 3.8) follows from the estimate 2.26). Let 3.5) by ) = B0 AT Y ) T ) AY )νy ) τy ) )

16 6 ARIEL BARTON where AνY ) τy ) ) is the 2 2 matrix whose columns are AY )νy ) and τy ), ν is the unit outward normal vector to Ω, and τ is the unit tangent vector given by ) 0 3.6) ν = τ. 0 We need only show that b satisfies the bounds 3.9), 3.0) and 3.) to show that the bound 3.3) is valid. The bound 3.9) follows from the ellipticity condition.2) and the definition 2.5) of B0 AT. To establish the bound 3.0), observe that ) ΘbX) = ΨX, Y )by ) dσy ) = X DX) τy ) Γ AT X Y ) dσy ). Ω Recall that DX) is constant in Ω and in Ω C. Let r be a large positive number. Then because B0, r) is bounded and Γ X is continuous away from X, we have that τy ) Γ AT X Y ) dσy ) = 0. Ω B0,r)) If X, X Ω, then for r X + X, we have that by the bounds 2.3) and 2.9), Γ AT X Y ) Γ AT X Y ) dσy ) Cr X X α Ω B0,r) r +α which approaches 0 as r ; thus τy ) ΓAT Ω X Y ) dσy ) is constant in Ω. Thus, ΘbX) = 0, uniformly in Ω, and so the bound 3.0) holds The bound 3.). To complete the proof of Lemma 3.3 for special Lipschitz domains, we need only show that the bound 3.) is valid. This derivation is somewhat tedious but involves no deep theorems. Recall that the condition 3.) states that ffl by ) dσy ) must be invertible for every Ω connected. Choose some such. Let the endpoints of the segment be given by x a, t a ) and x b, t b ); we order these two endpoints so that if ν is the unit outward normal to Ω and τ is the unit tangent vector given by formula 3.6), then τ points from x a, t a ) to x b, t b ). Let x = x b x a, t = t b t a. Thus 2 x + 2 t is the distance between the two endpoints. Observe that because Ω is a special Lipschitz domain, we have that σ) 2 x + 2 t. Example 3.7. As a motivating example, consider the special case where A = I is the 2 2 identity matrix. Notice that B0 I = I as well, and so by ) = νy ) τy ) ). We have that ) x τy ) dσy ) =, t a vector whose length is comparable to σ). Furthermore, ) 0 νy ) dσy ) = τy ) dσy ) = 0 Thus, by ) dσy ) = Ω t ). x ) ) t νy ) τy ) dσy ) = x. σ) x t The right-hand side is clearly invertible, and so the bound 3.) is valid.

17 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 7 We must now show that the condition 3.) is valid even if A and thus B0 AT ) is not the identity matrix. Since b is bounded, it suffices to show that det by ) dσy ) σ)2 /C for some constant C > 0. The proof will take several steps. Our first step is to write by ) dσy ) in terms of the endpoints x a, t a ) and x b, t b ) of and in terms of the matrix A and its components a jk. We will see that this integral is independent of the particular path taken from x a, t a ) to x b, t b ). Lemma 3.8. We have that ) α β γ by ) dσy ) = δ ᾱ + β where, in the notation given above, xb a2 x) α = t x a 2a x) + ā2x) ) dx, γ = 2ā x) xb a2 x) β = x a 2a x) ā2x) ) dx, δ = 2ā x) ) a x) a and where Ax, t) = Ax) = 2 x). a 2 x) a 22 x) xb x a xb x a a x) dx, det Ax) a x) Notice that if A = I then α = t, γ = δ = x, and β = 0, and so this formula is in agreement with Example 3.7. Proof of Lemma 3.8. By assumption, Ω is a special Lipschitz domain. Recall that this means that there is some vector e and some Lipschitz function ϕ such that Ω = { x e + t e : t > ϕx) } ), e 0 = e. 0 Let ψx) = x e + ϕx) e; then ψ = ψ, ψ 2 ) is a parameterization of Ω. We have that x a, t a ) = ψa) and x b, t b ) = ψb) for some a, b R. Recall that by ) = B AT 0 Y ) T ) AY )νy ) τy ) ). We will compute explicit formulas for B0 AT, ν and τ. First, notice that the unit tangent vector τ satisfies τ ψy)) = ± ψ y)/ ψ y), and that the choice of signs is determined by the requirement that ν be the outward and not the inward normal vector to Ω. If we choose τ ψy)) = + ψ y)/ ψ y), then e, τ > 0. By formula 3.6) and the relation above between e and e, this implies that e, ν < 0, as desired. Writing explicit formulas for τ and ν in terms of the components ψ and ψ 2 of ψ, we see that ) ) τ ψy)) ψ = y) ψ y) ψ 2y), ν ψy)) ψ = 2 y) ψ y) ψ y). Next, recall from formula 2.5) that B AT 0 Y ) = a Y ) a 2 Y ) 0 ). dx.

18 8 ARIEL BARTON This means that and B AT 0 Y ) T ) = B AT 0 Y ) T ) AY ) = = a Y ) a Y ) a Y ) 0 a 2 Y ) a Y ) ) ) 0 a Y ) a 2 Y ) a 2 Y ) a Y ) a 2 Y ) a 22 Y ) ) a Y ) a 2 Y ). 0 det AY ) We want to evaluate b ψy)). We adopt the shorthand that a jk = a jk ψy)), det A = det A ψy)), and ψ j = ψ j y). Then ) ) B0 AT ψy)) T ) τ ψy)) 0 ψ = a ψ y) a 2 a ψ 2 ) ψ = a ψ y) a 2 ψ + a ψ 2 and Thus, ) ) B0 AT ψy)) T ) A ψy)) ν ψy)) a a = 2 ψ 2 a ψ y) 0 det A ψ ) a ψ = a ψ 2 a 2 ψ y) det A ψ. b ψy)) = = B0 AT ψy)) T ) A ψy)) ν ψy)) B0 AT ψy)) T ) τ ψy)) ) ) a ψ a ψ 2 a 2 ψ ψ y) det A ψ a ψ 2 a 2 ψ. Integrating, we see that b by ) dσy ) = b ψy)) ψ y) dy = = a b a a b a ) a ψ 2 a 2 ψ ψ det A ψ a ψ 2 a 2 ψ dy ) b ) a2 dy + ψ det A a 2 y) dy. ψ 2 y) 0 0 ψ 2y) The first integral is equal to t I. According to our shorthand, A and a jk are to be evaluated at ψy). But because A is t-independent, A ψy)) = Aψ y)). Therefore, the second integral is equal to b a a ψ y)) a a ) a2 ψ y)) ψ det Aψ y)) a 2 ψ y)) y) dy. We make the change of variables x = ψ y); thus, ) ψb) ) t 0 a2 x) by ) dσy ) = + dx. 0 t a x) det Ax) a 2 x) ψ a) )

19 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 9 Recall that ψ a) = x a and that ψ b) = x b. It is then elementary to compute that, if ) α β γ by ) dσy ) = δ ᾱ + β then xb α = t β = xb x a x a a2 x) 2a x) + ā2x) ) dx, 2ā x) ) γ = dx, δ = a2 x) 2a x) ā2x) 2ā x) xb x a xb x a a x) dx, det Ax) a x) This completes the proof. Our goal is to control det by ) dσy ). We will only be able to do so in the case where A is elliptic. We will need to rewrite the ellipticity condition.2) as a series of inequalities concerning the components a jk of A; these inequalities are the subject of the next lemma. ) a a Lemma 3.9. Let A = 2. The ellipticity condition a 2 a 22 λ η 2 Re η Aη for all η C 2 is true if and only if the three inequalities Re a λ, Re a 22 λ, a 2 + ā 2 2 Re a λ)re a 22 λ) are valid. Proof. We begin by making the following computation. Choose some η C 2. If we let η T = η η 2 ), then λ η 2 = λ η 2 + λ η 2 2 and and so Re η T Aη = Rea η η + a 22 η 2 η 2 + a 2 η η 2 + a 2 η 2 η ) = η 2 Re a + η 2 2 Re a 22 + Rea 2 + ā 2 ) η η 2 ) Re η T Aη λ η 2 = η 2 Re a λ) + η 2 2 Re a 22 λ) + Rea 2 + ā 2 ) η η 2 ). Suppose that the three inequalities Re a λ, Re a 22 λ, a 2 + ā 2 2 Re a λ)re a 22 λ) are valid. We then have that Re η T Aη λ η 2 η 2 Re a λ) + η 2 2 Re a 22 λ) a 2 + ā 2 η η 2 η 2 Re a λ) + η 2 2 Re a 22 λ) 2 Re a λ) η Re a 22 λ) η 2 = Re a λ) η Re a 22 λ) η 2 ) 2. The right-hand side is nonnegative for all η, and so A must be elliptic. Conversely, suppose that A is elliptic, and so Re η T Aη λ η 2 0 for all η. This means that 3.20) η 2 Re a λ) + η 2 2 Re a 22 λ) + Rea 2 + ā 2 ) η η 2 ) 0 dx.

20 20 ARIEL BARTON for any choice of η and η 2. If we choose η = and η 2 = 0, this implies that Re a λ. Similarly, if we choose η = 0 and η 2 = then we have that Re a 22 λ. Finally, we consider the third inequality. Choose η 2 = Re a 22 λ + ε and η 2 2 = Re a λ + ε for some ε > 0, and choose the moduli of η and η 2 such that is a negative real number. Then a 2 + ā 2 ) η η 2 0 η 2 Re a λ) + η 2 2 Re a 22 λ) + Rea 2 + ā 2 ) η η 2 ) η 2 Re a λ + ε) + η 2 2 Re a 22 λ + ε) + Rea 2 + ā 2 ) η η 2 ) = 2Re a λ + ε)re a 22 λ + ε) a 2 + ā 2 Re a λ + ε)re a 22 λ + ε). Solving, and using the fact that Re a λ + ε)re a 22 λ + ε) > 0, we see that 2 Re a λ + ε)re a 22 λ + ε) a 2 + ā 2 for any ε > 0; taking the limit as ε 0 + completes the proof. We now use our bounds on the components a jk of A to establish some upper and lower bounds on the quantities α, β, γ and δ. Lemma 3.2. Let α, β, γ and δ be as in Lemma 3.8, and suppose that A satisfies the ellipticity condition.2). We have the lower bounds and the upper bounds t x Λ λ Re α, x λ Re γ, Λ2 λ x Re δ α t + x Λ λ, γ λ x, β 2 Re δ λ x )Re γ), δ Λ2 λ x. Also, the real numbers x, Re γ, Re δ and Re δ λ x are either all nonpositive or all nonnegative. Proof. By Lemma 3.9, we have that Re a > λ and Re a 22 > λ; by the ellipticity bound.2), we have that a jk Λ and that det A Λ 2. This gives us in particular that Re = Re ā a a 2 λ Λ 2, a λ. Applying these inequalities gives us the two bounds on α, the two bounds on γ and the upper bound on δ. Notice that if x = 0, then x a = x b, and so β = γ = δ = 0; thus we are done. For the remainder of the proof we will assume that x 0. The integrand in the definition of γ is positive; therefore, Re γ and x are either both positive or both negative.

21 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 2 We are left with the lower bound on δ and the upper bound on β. Consider the integrand in the integral defining β. We compute a 2 ā2 2 a = a 2ā 2 a 2 + a 2ā 2 a 2 a 2 a 2 a ) 2 ā2ā 2 ā ) 2 ā = a 2ā 2 + a 2 ā 2 a 2 2 Re a 2 a 2 a ) 2. In order to apply Lemma 3.9, we rewrite the first fraction in terms of the quantity a 2 + ā 2 2. Observe that a 2 + ā 2 2 = a 2 ā 2 + a 2 ā 2 + a 2 a 2 + ā 2 ā 2 = a 2 ā 2 + a 2 ā Rea 2 a 2 ) and so a 2 ā2 2 a = a 2 + ā 2 2 a 2 2 Re a 2 a 2 a 2 2 Re a 2 a 2 a ) 2. ā Combining the last two terms, we see that a 2 ā2 2 a ā = a 2 + ā 2 2 a 2 2 Re + ) )) a2 a 2 ā a a = a 2 + ā 2 2 a 2 4 Re Re a 2 a 2. a a We now apply Lemma 3.9 to see that a 2 ā2 2 a ā 4Re a λ)re a 22 λ) a 2 4 Re Re a 2 a 2 a a = 4 Re a Re a 22 a 2 4λ Re a a 2 4λRe a 22 λ) a 2 We have that and that Thus, a 2 ā2 a 4 Re a Re a 2 a 2 a. Re a Re a 22 a 2 = Re a a 2 Re a a 22 a Re a a 2 = Re a a 2 = Re a a ā = Re ā = Re a. ā 2 4 Re Re a a 22 4λ Re 4λRe a 22 λ) a a a a 2 4 Re a Re a 2 a 2 a = 4 Re a Re det A a 4λ Re a 4λRe a 22 λ) a 2. Because Re a 22 λ 0 we can disregard the final term. The left-hand side is nonnegative. Furthermore, Re/a ) > 0. Thus, 0 4 Re Re det A ) λ and so Re det A λ. a a a

22 22 ARIEL BARTON Since xb det A δ = x a a this implies that x and Re δ are either both positive or both negative, and moreover that Re δ λ x, as desired. Finally, we have that β 2 = xb x a 2 ) a2 ā2 2. a ā Recall that a 2 ā2 2 a ā 4 Re Re det A ) λ a a and so by Hölder s inequality, xb β 2 a 2 x a 4 ā2 2 ) xb a ā Re ) Re/a ) x a a xb Re det A ) xb λ Re ) = Re δ λ x )Re γ). x a a x a a This completes the proof. We can now establish the bound 3.). Lemma The bound det is valid, and thus the bound 3.) holds. by ) dσy ) 2 x + 2 t )/C Proof. Recall from Lemma 3.8 that ) α β γ b dσ = δ ᾱ + β and so det b dσ = αᾱ β β + 2 i Imα β) + γδ. We use Lemma 3.2. Applying our upper bound on β, we see that Re det b dσ αᾱ Re γ Re δ λ x Re γ) + Re γ Re δ Im γ Im δ) αᾱ + λ x Re γ Im γ Im δ. Applying our lower bound on Re γ, we see that 3.23) Re det b dσ αᾱ + λ2 Λ 2 2 x Im γ Im δ. We may also compute that 3.24) Im det b dσ = 2 Imα β) + Re γ Im δ + Re δ Im γ. We wish to show that at least one of these two quantities is bounded from below. We will consider the following four cases: x t λ/4λ), x > t λ/4λ), α 2 + λ 2 /2Λ 2 ) x 2 Im γ Im δ,

23 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 23 x > t λ/4λ), α 2 +λ 2 /2Λ 2 ) x 2 < Im γ Im δ, and α < x λ/4λ, x > t λ/4λ), α 2 +λ 2 /2Λ 2 ) x 2 < Im γ Im δ, and α x λ/4λ. Observe that these cases include all possible values of our parameters. Suppose that x t λ/4λ). Applying our lower bound on α and our upper bounds on γ and δ, we see that the bound 3.23) implies that Re det b dσ t 2 2 t x Λ λ + λ2 Λ 2 2 x and since x t λ/4λ), we have that Re det b dσ t 2 /2 2 x + 2 t )/C as desired. In the remaining three cases it suffices to show that det b dσ 2 x/c. Suppose that αᾱ + λ2 2Λ 2 x 2 Im γ Im δ 0. Then by formula 3.23), we have that Re det b dσ x 2 λ2 2Λ 2 2 x/c. Finally, suppose that x > t λ/4λ) and αᾱ + λ 2 /2Λ 2 ) x 2 < Im γ Im δ. In the remaining two cases we will bound Im det b dσ and not Re det b dσ from below. Observe that in particular, Im δ and Im γ are both positive or both negative. We already know that the same is true of Re γ and Re δ. So by formula 3.24), Im det b dσ Re γ Im δ + Re δ Im γ 2 α β 2 Re γ Re δ Im γ Im δ 2 α β. Using Lemma 3.2 to bound β, and using the bound α 2 + x 2 λ 2 /2Λ 2 ) < Im γ Im δ, we have that Im det b dσ > 2 α x λ 2 /2Λ 2 ) Re γ Re δ 2 α Re γre δ λ x ). If α < x λ/4λ, then Im det b dσ > 2 2 x λ 2 /2Λ 2 ) Re γ Re δ 2 x λ Re γ Re δ 4Λ 2 /2) x λ Λ Re γ Re δ and by the lower bounds on Re γ and Re δ in Lemma 3.2, the right-hand side is at least 2 x/c.

24 24 ARIEL BARTON If α x λ/4λ, then Im det b dσ > 2 α Re γ Re δ 2 α Re γre δ λ x ) 2 α ) Re δ λx Re γ Re δ. Re δ But by Lemma 3.2, Re δ λx = λ x Re δ Re δ λ2 Λ 2 < and so by assumption on α and, again, the lower bounds on Re γ and Re δ in Lemma 3.2, ) Im det b dσ Re δ > 2 α λx Re γ Re δ 2 Re δ x/c. In all four cases, we have that det b dσ 2 x + 2 t )/C σ) 2 /C. Thus, the bound 3.) is valid. This completes the proof of Lemma 3.3 in the case of special Lipschitz domains A L 2 estimate in general Lipschitz domains. We now wish to move to Lipschitz domains with compact boundary. Again by formula 2.27), to complete the proof of Lemma 3.3, we need only show that, if V is a Lipschitz domain, then 3.25) T V fx) 2 distx, ) dx C f 2 L 2 ). V Choose some Lipschitz domain V with compact boundary. By Definition 2.2, there are N special Lipschitz domains Ω j, each with Lipschitz constant at most M, such that N Ω j BX j, r j ) j= where r j > σ )/C and X j, with Ω j BX j, 2r j ) = V BX j, 2r j ). So we may write f = N j= f j, where f j = 0 outside of BX j, r j ). Pick some j and note that T V fj T j fj, where T j = T Ωj. By Section 3., Ω j T j fj X) 2 distx, Ω j ) dx C f j 2 L 2 Ω j). If X BX j, 3 2 r j), then either distx, Ω j ) = distx, Ω j BX j, 2r j ) distx, ), or distx, Ω j ) = distx, Ω j \ BX j, 2r j )) > 2 r j > 3 X X j distx, ). 3 In either case distx, ) 3 distx, Ω j ) and so T V fj X) 2 distx, ) dx C f j 2 L 2 Ω. j) V BX j,3r j/2) Conversely, suppose X / BX j, 3 2 r j). By the bound 3.4), T V fj C X) X X j 2 f j Y ) dσy )

25 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA 25 and because σsupp f j ) Cr j, we have that T V fj X) C X X j 2 rj f j L2 ). We have that distx, ) X X j. Thus, we may readily compute that V \BX j,3r j/2) T V fj X) 2 distx, ) dx R 2 \BX j,3r j/2) C X X j 3 r j f j 2 L 2 ) dx C f j 2 L 2 ). Combining these estimates and summing over j completes the proof of the estimate 3.25) BM O estimates in Lipschitz domains. We have established that if V is a Lipschitz domain and if f L 2 ), then 3.26) DfX) 2 distx, ) dx C f 2 L 2 ). V In this section, we show that the estimate 3.2) holds, that is, that 3.27) DfX) 2 distx, ) dx C f 2 BMO ) σ BX 0, r)) V BX 0,r) for all f BMO ), all X 0 and all r > 0. This will complete the proof of Theorem 3.. The argument comes essentially from the proof of Theorem 3 in [6], where it was applied to the Poisson extension in the upper half-space. Choose some r > 0 and some X 0. Suppose that is compact and r > σ )/C. Then σ BX 0, r)) σ ). Recall that if F is constant on each connected component of, then DF is constant in V ; thus we may assume that ffl f dσ = 0 for each connected component ω of. Then, by the ω definition 2.3) of BMO ), f L 2 ) C σ ) f BMO ) and so the bound 3.27) follows immediately from the estimate 3.26). Otherwise, we may assume that r is small enough that QX 0, 2r) exists, where the tents Q are as in Definition 2.8. If V = Ω is a special Lipschitz domain then this is true for all r.) Then r σ BX 0, r)) σx 0, 2r)). We may assume that ffl X 0,2r) f dσ = 0. Let ω 0 be the connected component of containing X 0 ; we may assume that ffl ω f dσ = 0 for any connected component ω of with ω ω 0. We write f = f + f 2 + f 3 + f 4 ; we define the f i s as follows. If V = Ω is a special Lipschitz domain, then f = f f 2 = f f 3 = f 4 = 0. on X 0, 2r), on Ω \ X 0, 2r),

26 26 ARIEL BARTON Otherwise, is compact. Let k be the largest integer such that QX 0, 2 k r) exists; observe that σ )/C 2 k r σ ). Then let f = f f 2 = f f 3 = f on X 0, 2r), on X 0, 2 k r) \ X 0, 2r), on ω 0 \ X 0, 2 k r), f 4 = f on \ ω 0. By our definition 2.3) of BMO, we have that both f and f 4 are in L 2 ), with f L2 ) C r /2 f BMO ) and f 4 L2 ) Cσ ) /2 f BMO ). The bound 3.26) immediately yields that r Recall that V BX 0,r) Df X) 2 distx, ) dx C r f 2 L 2 ) C f 2 BMO ). DfX) = νy ) A T Y ) X Γ AT X Y )) fy ) dσy ). We will in fact be able to bound Df f )X) pointwise. By the bound 3.4) on X Γ AT X Y )), if X BX 0, r), then C C Df 4 X) σ ) 2 f 4 dσ σ ) f 4 3/2 L2 ) C σ ) f BMO ). By the bound 2.4), we have that f 3 L 2 ) σ ) Thus, recalling that 2 k r σ ), Finally, Df 3 X) Df 2 X) C + C log σω ) 0) f BMO ) σ) C k σ ) f BMO ). C 2 k r) f 3 3/2 L 2 ) Ck 2 k r) f BMO ) C r f BMO ). k j=2 k j=2 X 0,2 j r)\x 0,2 j r) C 2 j r) 3/2 f L 2 X 0,2 j r)) ν A T X Γ AT X ) f dσ and again using the bound 2.4), we have that k Cj Df 2 X) 2 j r) f BMO ) C r f BMO ). j=2 Thus, Df + f 2 + f 3 )X) C/r) f BMO ) for all X BX 0, r), and so Df + f 2 + f 3 )X) 2 distx, ) dx C f 2 BMO ) r V BX 0,r) as desired. Thus the bound 3.27) is valid and Theorem 3. is proven.

27 THE DIRICHLET PROBLEM WITH BMO BOUNDARY DATA Converses and uniqueness In this section, we will complete the proof of Theorem.4 by proving the following two theorems. Theorem 4.. Suppose that A is t-independent and elliptic, and that V R 2 is a Lipschitz domain. Then there is some ε > 0 and some C > 0, depending only on the Lipschitz character of V and the constants λ, Λ in the ellipticity condition.2), such that if Im A L < ε, then the following holds. Suppose that u satisfies div A u = 0 in V and 4.2) sup sup X 0 r>0 r Then f = u V BX 0,r) ux) 2 distx, ) dx C 2. exists and lies in BMO ), with f BMO C C. Theorem 4.3. Suppose u, A, and V are as in Theorem 4.. There is some ε > 0 such that, if Im A L < ε, if u satisfies 4.4) sup sup ux) 2 distx, ) dx X 0 r>0 σ BX 0, r)) C 2, and if u V BX 0,r) is a constant, then u is constant. We will also prove the maximum principle Theorem.7). Remark 4.5. We comment on the two conditions 4.2) and 4.4). First, observe that if V is a Lipschitz domain then σ BX 0, r)) Cr, and so the condition 4.4) implies that the bound 4.2) is valid at a cost of increasing C. Conversely, if V is a bounded or special Lipschitz domain, then the condition 4.2) implies that the bound 4.4) is valid at the same cost. We thus only need the more complicated formulation 4.4) if V C is bounded. By considering the function ux) = log X, which is harmonic in V = R 2 \ B0, ) and satisfies the bound 4.2) but not 4.4), we see that for such domains the condition 4.4) is necessary to ensure uniqueness. We begin by showing that the bound 4.2) is valid in all subdomains of V. Lemma 4.6. Let u be defined in some domain V. If sup sup ux) 2 distx, ) dx X 0 r>0 r C 2, and if U V, then sup sup X 0 U r>0 r V BX 0,r) U BX 0,r) ux) 2 distx, U) dx 3 C 2. Proof. Notice that if X 0 U, then the lemma follows immediately from the fact that U V and that distx, U) distx, ). Thus the only complication is the case where X 0 U \. Choose some X 0 U, and let r > 0. Let X 0 satisfy X 0 X 0 = distx 0, ). Let R = X 0 X 0 + r, so BX 0, r) BX 0, R). If R 3r, then because distx, U) distx, ) for all X U, we have that r distx, U) 3 distx, ). R