The L 2 regularity problem for elliptic equations satisfying a Carleson measure condition

Transcription

1 The L 2 regularity problem for elliptic equations satisfying a Carleson measure condition Martin Dindoš, Jill Pipher & David J. Rule 28th October 29 Abstract We prove that the L 2 regularity problem is solvable for the elliptic equation n j,k=1 j(a jk k u) = when i,j,k ia jk (x) 2 x n dx is a Carleson measure with a sufficiently small constant, i,j,k ia jk (x) C/x n and the bottom row of the coefficient matrix has the particular form (,,...,, 1). This is done in any dimension n. This was proved in the case n = 2 earlier in [9] without the assumption on the bottom row of (a jk ). 1 Introduction Consider the equation L(u) := div(a u) = n j (a jk k u) =, (1) j,k=1 in a domain Ω R n which is either a bounded Lipschitz domain or the domain above the graph of a Lipschitz function. The matrix A is assumed to be uniformly elliptic, that is, there exist constants λ and Λ such that λ ξ 2 Aξ ξ Λ ξ 2. (2) There have been many works, going back to the work of Dahlberg [3, 6], dedicated to proving estimates such as N(u) L p ( Ω) C u L p ( Ω), (3a) N( u) L p ( Ω) C ν A u L p ( Ω), (3b) or N( u) L p ( Ω) C τ u L p ( Ω), (3c) 1

2 where ν is the outward unit normal to Ω and τ is the projection of the gradient onto the tangent plane of Ω. The function N(u) is the nontangential maximal function of u and is defined by N(u)(q) = sup u, Γ(q) where Γ(q) is a non-tangential cone at the point q Ω. When the estimate (3a) exists we say the L p Dirichlet problem is solvable. Respectively, when estimates (3b) or (3c) exist we say the L p Neumann problem or the L p regularity problem is solvable. In practice it is often necessary to replace the non-tangential maximal function N in estimates (3b) and (3c) with a variant denoted by Ñ which takes the supremum of averages over balls in the cone Γ(q) with radii comparable to the distance to the boundary. This is because the gradient of weak solutions to (1) need not be locally finite so N( u) may not be well-defined. The results presented here will be proved in the particular case that the data and coefficients are smooth, although the estimates obtained will be independent of this smoothness. As a consequence, it makes sense to consider N( u). We refer the reader to the beginning of Section 4 for the precise assumptions we make and the method by which we can then pass to the general case. Such estimates allow one to formulate boundary value problems for solutions to (1) with data specifying either u (Dirichlet data) or ν A u (Neumann data) on Ω merely in L p. They also allow one to prove non-tangential convergence of the solution to the data. For arbitrary measurable matrices A satisfying (2), however, it is not always possible to prove these estimates [1]. Nevertheless, under certain conditions on A, they can be proved. The new result we wish to present here is that the L 2 regularity problem is solvable when i a jk (x) 2 x n dx (4) i,j,k is a Carleson measure with small constant (see Definition 2.1), i,j,k ia jk (x) C/x n and the last row of the coefficient matrix has the special form a nk = for k < n and a nn = 1. We will do this for Ω =, the upper half-plane, although the case of a domain above the graph of a Lipschitz function may be easily reduced to this case. This is formulated as Theorem 3.1. We do not expect the condition on the bottom row of A to be necessary, although we require it in our proof. We note here that we eventually want to replace the condition (4) by an averaging condition on coefficients of A as was done in [8] for the Dirichlet problem. In this paper, a solvability result was first established with a condition similar to (4). Then this condition was replaced 2

3 by a weaker averaging condition by the use of perturbation theory that is know for the Dirichlet problem. What we present here should therefore be considered as a first step on the road that will bring our knowledge of the Neumann problem to the same level as the Dirichlet problem. The paper is organised as follows. In Section 2 we fix some notation, state various definitions and recall some well-known results that we will use later. In Section 3 we motivate the hypotheses and state the main result, Theorem 3.1. In the Section 4 we prove Theorem 3.1, using three different tools. The first tool is that of distributional inequalities, which have been used before in this context (for example, [14]). This is combined with the use of an auxilary inhomogeneous equation, which was introduced in [9], although the idea is applied in a different manner here. Finally, we adapt the use of Rellich identities (see for example, [12, 13, 17]) to complete the proof. 2 Preliminaries For a point x R n we will write x = (x, x n ), where x = (x 1,..., x n 1 ) is the projection onto the first n 1 components of x. Given a Lipshtiz function φ, we define the domain Ω φ = {x x n > φ(x )}, so Ω = is the upper half-plane. The non-tangential maximal function (of a function u: Ω φ R) adapted to this domain with aperture a is N φ,a (u)(q) = sup u Γ a(q) where Γ a (q) = {x x q < a(x n q n )} is a cone with vertex q Ω φ and aperture a. Of course, it makes sense to require a < φ 1 L (R), so that Γ a (q) Ω φ for each q Ω φ. In a similar fashion we can adapt the square function to such a domain by defining ( S φ,a (u)(q) = u(x) 2 Γ a(q) ) 1 dx 2. (x n φ(q )) n 2 This notation differs slightly from common usage, in that we have written u on the right-hand side, rather than u. However, our calculations will be clearer if we use this convention. With this definition the usual square function is S,a ( u). When u is vector-valued, N φ,a (u) and S φ,a (u) are defined in the same way, but with u meaning the usual vector norm. We will also need truncated versions of the non-tangential maximal function and square function. These will be denoted by Sφ,a d (u) and N φ,a d (u) respectively, where Γ a (q) is replaced with Γ d a(q) = {x x q < a(x n q n ), < x n q n < d}. 3

4 Definition 2.1 Given a function u: Ω φ R, the measure x (x n φ(x )) u(x) 2 dx is called a Carleson measure on Ω φ if ( ) 1 1 q C φ,a (u)(q) := sup (x n φ(x )) u(x) 2 2 dx Q q Q t φ a(q) is in L ( Ω φ ), where the supremum is taken over all cubes Q in R n 1 containing q and t φ a(q) = {x x Q and x n (φ(x ), φ(x ) + diam(q)/a)} is a tent over Q. The number C (u) L ( Ω φ ) is the Carleson measure constant. We will also need a localised dyadic version of C φ,a (u). For a cube R, we will write Cφ,a R (u)(q) to mean C R φ,a(u)(q) := ( ) 1 1 sup (x n φ(x )) u(x) 2 2 dx, q Q R 16Q t φ a(16q) where the supremum is taken over dyadic cubes Q which contain q and are contained in R. The cube 16Q is the concentric enlargement of Q by a factor of 16 (and similarly for factors different from 16). Obviously, if R R, then Cφ,a R (u)(q) C φ,a R (u)(q). It will often be convenient to write S φ,a (u)(q ) := S φ,a (u)(q), N φ,a (u)(q ) := N φ,a (u)(q), Cφ,a R (u)(q ) := Cφ,a R (u)(q), etc., but this need not cause confusion. We now recall a fact proved by Coifman, Meyer and Stein [2, (4.1)] that we will use later. We have u(x)v(x) (x n φ(x ))dx C(a, b) C φ,b (u)(x )S,a (v)(x )dx. (5) Ω φ R n 1 Definition 2.2 Given a non-empty open proper subset D R n a Whitney decomposition of the set D is a family of closed dyadic cubes {Q j } j such that: (a) j Q j = D and the Q j have disjoint interiors; (b) diam(q j ) dist(q j, D c ) 4diam(Q j ); and (c) If the boundaries of Q j and Q k touch then 1 4 diam(q j) diam(q k ) 4. 4

5 The existence of Whitney decompositions is proved in, for example, [11, pa-34]. We remark that an examination of the proof therein reveals if {Q j} j is the Whitney decomposition of the set D D, then when Q j Q k we have Q j Q k. Throughout the paper we will denote the dilation of a cube Q by a factor 5 by Q and Q = (Q ) = 25Q. 3 Motivation and the Main Result The motivation for the conditions placed upon the matrix A come from the following example due to Kenig and Pipher [15]. Consider the Dirichlet problem for the Laplacian = n i=1 2 i : { u =, in Ωφ ; u = f, on Ω φ. It is well-known [4] that the L 2 Dirichlet problem is solvable when φ is Lipschitz, that is, we have the estimate N φ,a (u) L 2 ( Ω) C f L 2 ( Ω φ ). (6) We now consider the transformation Φ: Ω φ, first introduced by Dahlberg, Kenig and Stein (see [5] and [7]), defined as Φ(x) = (x, c x n + (θ xn φ)(x )), (7) where {θ t } t> is a smooth compactly supported approximate identity and c can be chosen large enough, depending only on φ L (R n 1 ), so that Φ is one-to-one. One may compute that the function Φ enjoys the properties Φ(x) C, 2 Φ(x) C/x n, and x x n 2 Φ(x) 2 dx is a Carleson measure. Moreover, it is straightforward to see that the composition w = u Φ is such that div(a w) = in, where A = (det Φ )((Φ ) 1 ) t (Φ ) 1. Therefore A inherits from Φ the properties (i) A(x) C /x n and (ii) x n A(x) 2 dx is a Carleson measure with constant C, 5

6 for some constant C, and from (6) we can see the corresponding non-tangential estimate N,a (w) L 2 (R n 1 ) C f (, φ( )) L 2 (R n 1 ) (8) holds for some a < a. So the natural question is: Are the conditions (i) and (ii) sufficient to conclude estimate (8) for an arbitrary solution to a divergence form equation? Kenig and Pipher [15] prove an L p version of such an estimate holds under closely related conditions, where p > 1 may possibly be large and Dindoš, Petermichl and Pipher [8] show that the L p version holds for any given 1 < p < under the same assumptions, but when C is sufficiently small (depending on p). In fact, both of these results are proved for bounded Lipschitz domains for an elliptic equation which has lower-order drift terms satisfying a similar Carleson measure condition. Given [13], the same motivation can be used to justify posing the same question regarding the regularity and Neumann problems. This was studied for domains in the plane in [9] and is also the subject of this paper. We now state our main result. Theorem 3.1 Suppose u solves L(u) = in, with A satisfying (i) and (ii) above, and a nk = for k < n and a nn = 1. If C is sufficiently small (determined in terms of n, λ, Λ and a), then there exists a constant C, which depends only on n, C, λ, Λ and a, such that We prove this in the following section. N,a ( u) L 2 ( ) C τ u L p ( ). (9) 4 The Proof of Theorem 3.1 First, we need to define precisely how we obtain solutions to (1). We produce a unique weak solution u to (1), whose gradient is in L 2 (), via the method presented in [16, Lem 1.1] (or equally, [18, Lem 1.1]). The proof there is carried out when n = 2, but the same method yields an analogous result for any dimension n. All our results will be proved under the a priori assumptions that the coefficients are smooth, A is the identity matrix outside a ball centred at the origin, and the Dirichlet data u R n + is smooth and compactly supported. Under the a priori smoothness assumptions, the solution u is smooth and u(x) = O(x δ n+1 ) as x for each δ > (see [18, Thm B.1]). From this we can further conclude α u(x) = O(x δ α n+1 ) as x. Consequently all the integrals we consider converge. Once our results are proved, we may then drop the smoothness assumptions, passing 6

7 to the general case using the method used to prove Theorem B.2 in [18]. This is where we are forced to abandon an estimate on N( u) and introduce the averaged version Ñ( u) mentioned in Section 1. As this is a repeat of well-known arguments, we leave the details to the interested reader. We now introducing the auxilary inhomogeneous equation. Simply differentiating equation (1) in the ith direction we obtain div(a v i ) = div(( i A)v) for i = 1, 2,..., n, (1) where v i = i u and v = (v 1, v 2,..., v n ) t = u. The idea now is to view (1) as an equation in v i and use distributional inequality techniques to prove N,a (v) L 2 ( ) C v L 2 ( ) (11) We then have a separate argument, inspired by the Rellich identity techniques of [12, 13], but implimented in a different fashion, to prove u L 2 ( ) C τ u L 2 ( ). (12) Of course, these two estimates combine to prove (9). The more technically challenging estimate will be (11) and the proof of this will not be completed until after Theorem 4.8, once we have proved a string of estimates. The proof of (12) is carried out after this, at the end of the paper. We now begin the task of proving (11) with an important lemma. This is the essential tool to control the non-tangential maximal function by the square function. Lemma 4.1 Let u solve (1) and set v = u so v solves system (1). Let φ and φ + be two non-negative Lipschitz functions and let Q be a cube in R n 1 with r := diam(q). Suppose φ + φ and φ + (x ) φ(x ) 12diam(Q)/a for x Q. Then for sufficiently small a, which can be determined in terms φ + L (R n 1 ), there exists a constant C, depending only on λ, Λ, C, φ + L (R n 1 ) and a, such that v(, φ + ( )) 2 L 2 (Q) C(rn 1 min Q (C Q φ,a ( v))2 + C N φ,a (v) 2 L 2 (Q ) + N φ,a (v) L 2 (Q ) S 3r/a φ,a ( v) L 2 (Q ) + r n 1 v(x r ) 2 ), where x r is any point in {x x Q, φ + (x ) + r/2 x n φ + (x ) + 6r/a}. Proof. In order to facilitate the use of integration by parts, we will use the mapping Φ: R 2 + Ω φ+ defined by (7) with φ + replacing φ. This pulls back v in Ω φ+ to a solution w = v Φ in of the equation div(ã w i) = div(f i w) for i = 1, 2,..., n, (13) 7

8 where and à = (det Φ )(Φ 1 ) t (A Φ)(Φ 1 ) F i = (det Φ )(Φ 1 ) t ( i A Φ) (14) The coefficient matrix à satifies the ellipticity condition (2) with the constants λ and Λ being replaced by multiples of λ and Λ, respectively, which can be determined in terms of φ + L (R n 1 ). For each i, the matrix F i satisfies the properties (i) and (ii) with A replaced by F and C replaced by a multiple of C, which again can be determined in terms of φ + L (R n 1 ). We choose a smooth function ξ 1 : R n 1 R such that ξ 1 (x ) = 1 for x Q, ξ 1 16/r, with support contained in the concentric dilation (9/8)Q. Choose another function ξ 2 : [, ) R such that ξ 2 (x n ) = 1 for x n [, r], ξ 1 5/r and support contained in [, 2r]. Now define ξ(x, x n ) = ξ 1 (x )ξ 2 (x n ). We now calculate as in [14], [15], [17] and [9]. First of all, for each i = 1, 2,..., n, w i (x, ) 2 ξ 1 (x ) dx = n (wi 2 ξ)(x) dx = 2w i ( n w i )ξ w 2 i ξ 1 ξ 2 (15) The second term on the right-hand side of (15) is controlled by r 1 K w2 i where K = {x = (x, x n ) x Q, r/3 x n 7r/a}. Let x r be any point in K and choose K and K to be appropriate concentric enlargements of K. We set c = 1 w K K i. Using [1, Thm 8.17] and Poincaré s inequality, we may further estimate this term by r 1 (w i w i (x r )) 2 dx + r 1 wi 2 (x r ) dx K Cr n 1 osc K (w i ) 2 + Cr n 1 w i (x r ) 2 K Cr n 1 sup w i c 2 + Cr n 1 w i (x r ) 2 K Cr 1 w i c 2 dx + Cr n 1+2(1 n/q) F w 2 L q (K ) + Cr n 1 w i (x r ) 2 K Cr (w i ) 2 dx + CC N φ,a (w) 2 L 2 (Q ) + Crn 1 w i (x r ) 2 K C Q min C Q Q φ,a ( w i) 2 + CC N φ,a (w) 2 L 2 (Q ) + Crn 1 w i (x r ) 2. 8

9 The first term on the right-hand side of (15) is 2w i ( n w i )ξ dx = = w i ( n w i )ξ[ n (x n )] dx = 2 ( n w i ) 2 ξx n dx + 2 =: I + II + III. w i ( n w i )ξ 1 ξ 2x n dx w( 2 nw i )ξx n dx [ n (w i ( n w i )ξ)]x n dx Observe that we are free to assume à is upper triangular, provided we introduce lower order terms. Indeed, writing à = (ã jk) jk, div(ã w i) = j,k j (ã jk k w i ) = j=k = j=k j (ã jk k w i ) + j<k( j (ã jk k w i ) + k (ã kj j w i )) j (ã jk k w i ) + j<k j ((ã jk + ã kj ) k w i ) + j<k(( k ã kj ) j w i ( j ã kj ) k w i ) = div(ā w i) + B w i, Using the fact that w i solves (13) and writing Ā = {ā jk} jk, we see that II is equal to 2 w i div(f i w)ξx n dx 2 w i j (ā jk k w i )ξx n dx j k, j<n 2 w i B w i ξx n dx =: II 1 + II 2 + II 3. Integrating by parts, we have II 2 = 2 ( j w i )ā jk ( k w i )ξx n dx + 2 j k, j<n j k, j<n w i ā jk ( k w i )( j ξ)x n dx. (16) 9

10 The first integral may be combined with I to produce w i Ā w iξ2x n dx C Q min C Q Q φ,a ( w i) 2, using the ellipicity condition (2), which Ā also satisfies, and the assumption φ + φ 11diam(Q)/a. We now use (5) to control the second integral in (16) by C w( w)( ξ) x n dx v( v)( ξ) Φ 1 y n dy Ω φ CC ( ξ ) S 3r/a φ,a (v( v))(y )dy Q C N φ,a (v) L 2 (Q ) S 3r/a φ,a (χ Q,a v) L 2 (Q ), where we have extended ξ to R n by zero and used that C ( ξ ) C, which may be checked directly, and the simple observation Thus we obtain S 3r/a φ,a (v( v))(x ) N φ,a (v)(x )S 3r/a φ,a ( v)(x ). II 2 C( Q min C Q Q φ,a ( w i) 2 + N φ,a (v) L 2 (Q ) S 3r/a φ,a ( v) L 2 (Q )). Again, integrating by parts we see that II 1 = 2 w i F i wξx n dx w i ξf i w e n dx, w i F i w ξx n dx where e n = (,,...,, 1) t. Now we claim that F i w e n =. To see this, first observe that suffices to show the n-th entry of the vector F i w is zero, and consequently, it is enough to show that the bottom row of the matrix F i is zero. Now, the bottom row of i A is identically zero as we are assuming that a nk = for k < n and a nn = 1. Moreover, when j < n, n Φ j =, as may be seen from (7), so the jn-th entries of the matrix Φ 1 are zero when j < n. Therefore, using formula (14), we see that the bottom row of the matrix F i is indeed zero. 1

11 Consequently the third term of II 1 is zero. The remaining terms can be controlled using (5), as above. Indeed, w i F i wξx n dx R C C S 3r/a φ,a ( v)n φ,a(v) dx, and n + Q w i F i w ξx n dx C C S φ,a ( ξ Φ 1 )N φ,a (v) 2 dx. Q Thus, we find II 1 CC ( N φ,a (v) 2 L 2 (Q ) + N φ,a(v) L 2 (Q ) S 3r/a φ,a ( v) L 2 (Q )). The terms II 3 and III can also be dealt with in this way. So, combining all of these estimates and summing in i, we obtain the lemma. Our second lemma is very similar to Lemma 4.1, but it controls the square function by the boundary value of the gradient u = v. Lemma 4.2 Let u solve (1) and set v = u so v solves system (1). Let φ be a non-negative Lipschitz function. Then for each b >, there exists a constant C, depending only on λ, Λ, C, φ + L (R n 1 ) and b, such that S φ,b (v) 2 L 2 ( Ω φ ) C( v 2 L 2 ( Ω φ ) + C N φ,b (v) 2 L 2 ( Ω φ ) + C N φ,b (v) L 2 ( Ω φ ) S φ,b ( v) L 2 ( Ω φ )). Proof. We will repeat the proof of Lemma 4.1 and so obtain an estimate localised to a cube Q. The cut-off function ξ and the dilation factor of Q may need to be modified when b is large, however, otherwise the result remains the same. We then will pass to the limit r := diam(q) to obtain the lemma. First we obtain (15) and observe that the second term on the righthand side will tend to zero as r by our a priori smoothness assumptions, which mean w has sufficient decay, and the dominated convergence theorem, because ξ C/r. The same is true of III. The term I is again combined with the first term on the right-hand side of (16), but this time bounded below to obtain w i Ā w iξ2x n dx C S r,b( w i ) 2 L 2 (Q). Once again, the dominated convergence theorem can be used to see that the last integral in (16) will tend to zero as r. Terms II 1 and II 3 can all be dealt with as before and controlled by CC ( N φ,b (v) 2 L 2 (Q ) + N φ,b(v) L 2 (Q ) S φ,b ( v) L 2 (Q )). 11

12 Summing in i and taking the limit r gives the lemma. For any continuous function v : Ω φ R n and µ R, define h φ,µ,a (v)(x ) = sup{x n x n φ(x ) and sup v > µ}. Γ a((x,x n)) Lemma 4.3 If v is such that h φ,µ,a (v) <, then h φ,µ,a (v) is Lipschitz with constant 1/a. Proof. See, for example, [15, Lem 3.5]. Lemma 4.4 Let u solve (1) and set v = u so v solves system (1) and let {Q j } j be a Whitney decomposition of {x N φ,a (v)(x ) > µ/24}. Given a > and φ, a Lipshitz function such that a φ L (R n 1 ) < 1, let Eµ,ρ j be the intersection of a cube Q j with {x N φ,a/12 (v)(x ) > µ, and C N φ,a (v)(x ) + C (Q j) ( v)(x ) ρµ}. There exists a sufficiently small choice of ρ, independent of Q j, so that, for each x E j µ,ρ, there is a cube R with x 6R and R Q j and for which for all z R. v(z, h φ,µ,a/12 (v)(z )) > µ/2 Proof. See [14, 3.14]. Let x E j µ,ρ and so, by definition, h φ,µ,a/12 (v)(x ) > φ(x) and so there exists an y on Γ a/12 (x, h φ,µ,a/12 (v)(x )) such that v(y) = µ and h φ,µ,a/12 (v)(y ) = y n. Set r = y n φ(x ) > and K = Γ a/12 ((x, φ(x ))) {z z n y n r /6}. Since Q j is a Whitney cube, r (12/11)diam(Q j )/a and so 3K t φ a(16q j ) Now by [1, Thm 8.17] we have that osc K (v) C(r n/2 v c L 2 (2K) + r 1 n/q ( A)v L q (2K)), for any constant c. But, by property (i), ( A)v (z) Cr 1 C N φ,a (v)(x ) for z 2K, so and so, using Poincaré s inequality, φ,a r 1 n/q ( A)v L q (2K) CC N φ,a (v)(x ) v(z) v(y) osc K (v) C(r 1 n/2 v L 2 (3K) + C N φ,a (v)(x )) C(C (Q j) φ,a ( v)(x ) + C N φ,a (v)(x )) Cρµ, 12

13 for any z K. Thus, we may choose ρ sufficiently small so that v(z) v(y) µ/2. Then, clearly, v(z, h φ,µ,a/12 (v)(z )) µ/2 for z y ar /72 and the lemma is proved. Theorem 4.5 Let u solve (1) and set v = u so v solves system (1). Let {Q j } j be a Whitney decomposition of {x N φ,a (v)(x ) > µ/24}. Fix a cube R and set F j equal to the intersection of Q j with {x N φ,a/12 (v)(x ) > µ, [ (C M R (N φ,a (v) 2 ) + M R (C R φ,a ( v) 2 ))(x ) ] 1 2 {x [ M R (N φ,a (v) 2 )(x ) ] 1 2 [ M R (S 3r/a φ,a ( v)2 )(x ) ] 1 2 ρµ}, ρµ} where r = diam(r ) and M R is the Hardy-Littlewood maximal function applied to functions restricted to R, that is M R (f) = M(χ R f) where χ R is the characteristic function of R. Given sufficiently small a > and a Lipschitz function φ such that a φ L (R n ) < 1, there exist constants c and c(ρ), independent of j, such that for all µ > F j c(ρ) Q j, provided Q j R. Moreover, c(ρ) tends to zero as ρ. Proof. Fix j, set Q := Q j and F := F j. Let Q h = {(x, h φ,µ,a/12 (v)(x )) x Q }. By Lemma 4.4, for each x F and sufficiently small ρ, we have that M h (vχ Q h )(x ) > µ/12, where M h is the Hardy-Littlewood maximal function on the graph of the function h φ,µ,a/12 (v). By property (b) of Definition 2.2, there exists a point q such that dist(q, q ) 4diam(Q) and N φ,a (v)(q ) µ/24. Therefore we can choose y Q such that y n φ(y ) 11diam(Q)/a such that v(y) µ/24. By Lemma 4.4, we have M h ((v v(y))χ Q h)(x ) M h (vχ Q h)(x ) v(y) > µ 24 for all x Q, and applying the weak-type estimate for the maximal function, we obtain F C (v v(y)) 2. µ 2 Q h Now we can apply Lemma 4.1 with x r = y, Q replaced with Q, φ + = h φ,µ,a/12 (v) and v i replaced with v i v i (y), provided a is sufficiently small. We may do this as φ + φ 12diam(Q )/a on Q by (b) of Definition 2.2, as 13

14 above. Observe that since we obviously have Q 1/2 v(x r ) C N φ,a (v) L 2 (Q ), we get, for any x F, µ 2 F C( Q min C Q Q φ,a ( v)2 + C N φ,a (v) 2 L 2 (Q ) + N φ,a (v) L 2 (Q ) S 3r/a φ,a ( v) L 2 (Q )) C Q ( M R (Cφ,a R ( v) 2 )(x ) + C M R (N φ,a (v) 2 )(x ) + M R (N φ,a (v) 2 )(x ) 1 2 MR (S 3r/a φ,a ( v)2 )(x ) 1 2 Cρ 2 µ 2 Q, since Q R and so the proof is complete. ) Corollary 4.6 Let u solve (1) and set v = u so v solves system (1). Let and {Q k } k be a Whitney decomposition of {x N φ,a (v)(x ) > µ /24} and let G k = {x N φ,a/12 (v)(x ) > µ } Q k. For each q > 2, Lipschitz function φ and sufficiently small a > such that a φ L (R) < 1, there exists constants C and c(ρ) such that c(ρ) as ρ and N φ,a/12 (v) L q (G k ) ) C ( C (Q k ) φ,a ( v) L q ((Q k ) ) + N 2φ,a (v) 1 2 L q ((Q k ) ) S3r/a φ,a ( v) 1 2 L q ((Q k ) ) + (c(ρ) + CC ) N φ,a (v) L q ((Q k ) ), where r = diam((q k ) ). Proof. Fix k. Let {Q j } j and {F j } j be as in Theorem 4.5 with R = Q k and let {F j } j (and {Q j} j ) be the same, but with µ replaced by µ and R = Q k. As we remarked below Definition 2.2, if µ µ and Q k Q j, then Q j Q k. This means, using Theorem 4.5, we have, for µ µ, G k ( j F j ) Q k ( j F j ) F j j:q k Q j c(ρ) j:q k Q j Q j = c(ρ) Q k ( j Q j ) (17) Now if µ < µ, then G k ( j F j ) F k 14

15 and Q k jq j. As a result of this, by Theorem 4.5, when µ < µ, G k ( j F j ) F k c(ρ) Q k = c(ρ) Q k ( j Q j ). (18) Now we can proceed by standard arguments. We have N φ,a/12 (v) L q (G k ) = 4 q qµ q 1 {x G k N φ,a/12 (v)(x) > µ/4} dµ 4 q qµ q 1 G k ( j F j ) dµ + CC (M (Q k ) (N φ,a(v) 2 )) 1 2 L q (R) + C (M (Q k ) (C (Q k ) φ,a ( v) 2 )) 1 2 L q (R) + C (M (Q k ) (N φ,a(v) 2 )) 1 2 (M(Q k ) (S3r/a(v)2 )) 1 2 L q (R) c(ρ) 4 q qµ q 1 Q k ( j Q j ) dµ + CC N φ,a (v) L p ((Q k )) + C C (Q k ) φ,a ( v) L p ((Q k ) ) + C N φ,a (v) 1 2 L q ((Q k ) ) S3r/a φ,a (v) 1 2 L q ((Q k ) ) c(ρ) N φ,a/12 (v) L q (Q k ) + CC N φ,a (v) L p ((Q k ) ) + C (Q k ) φ,a ( v) L p ((Q k ) ) + N φ,a (v) 1 2 L q ((Q k ) ) S3r/a φ,a (v) 1 2. L q ((Q k ) ) This completes the proof. We now need to turn the local L q inequality for q > 2 into a global (on ) L 2 estimate. To do this we will need a more local version of the notion of a global point of density from [2]. An x R n 1 is a point of dyadic regional γ-density with respect to F over R if, for each dyadic cube Q such that x 16Q 16R, F 16Q / 16Q γ. The set of such x will be denoted D R γ (F ). If R R, then a point of dyadic regional γ-density over R is a point of dyadic regional γ-density over R. Given a set S R n 1, we define a sawtooth domain over S to be φ,a Γ b (S) := x SΓ b ((x, )). Lemma 4.7 Let α < 2a < 2α < b, with a < 1/2, and Q R. Suppose that the graph of φ gives the boundary of Γ α (Dγ R (F )) and 32Q Dγ R (F ), then there exists a constant C(γ, a, α) such that ( ) ψ(y) y n dy C(γ, a, α) ψ(y) /yn n 2 dy dx, t φ a(16q) where Q = 16Q. F Q 15 Γ b ((x,))

16 Proof. We follow [2, Lem 2]. After appyling Fubini s theorem, it clearly suffices to prove that if y t φ a(16q) Γ α (D R γ (F )), then F Q χ((x y )/(by n )) dx C(γ, a, α)y n 1 n, (19) where χ is the characteristic function of the unit ball. Since y t φ a(16q) Γ α (Dγ R (F )), there exists x Dγ R (F ) such that y x αy n and since 32Q Dγ R (F ), y n ( )16diam(Q) and α a x 64Q. Observe that since 2α < b, B( x, αy n ) B(y, by n ) c =, therefore F Q B( x, αy n ) F Q B( x, αy n ) B(y, by n ) + F Q B( x, αy n ) B(y, by n ) c F Q B(y, by n ) + B( x, αy n ) B(y, by n ) c = F Q B(y, by n ). Now, because y n ( 2 α + 1 a )16diam(Q) and x 64Q, there exists a dyadic cube Q with diameter comparable to y n such that x 16Q B( x, αy n ) Q Q 16R, so F Q B(y, b y n ) F Q B( x, αy n ) F 16Q γ 16Q. This proves (19) and with it the lemma. Theorem 4.8 Let u solve (1) and set v = u so v solves system (1). For a sufficiently small constant a, there exist constants C and b such that N,a/12 (v) L 2 (R n 1 ) C( S,b ( v) L 2 (R n 1 ) + C N,b (v) L 2 (R n 1 )) Proof. Choose a sufficiently small so that we may apply Corollary 4.6 with φ L (R n 1 ) 1/a and suppose b > 2a. Set E l,ρ µ = {x N,a/12 (v)(x ) > µ, S,b ( v)(x ) ρµ, N,b (v)(x ) lµ }. and F l,ρ µ = {x S,b ( v)(x ) ρµ, N,b (v)(x ) lµ }. 16

17 By standard arguments we have N,a/12 (v) 2 L 2 (R n 1 ) = 2µ {N,a/12 (v)(x ) > µ } dµ 2µ ( E l,ρ µ + {S,b ( v)(x ) > ρµ } + {N,c (v)(x ) > lµ } ) dµ 2µ E l,ρ µ dµ + 1 ρ 2 ( S,b( v) 2 L 2 (R) + 1 l 2 N,b(v) 2 L 2 (R n 1 ). Using [19, p62], we may pick l sufficiently large so that the last term may be hidden on the left-hand side. Now let {Q k } k be the Whitney decomposition of {x N φ,a (v)(x ) > µ /24}. We consider two cases. The first is k for which D (Q k ) γ (Q k F µ l,ρ ) is empty. Then, for each x 16(Q k ), there exists a dyadic cube R x, such that x 16R x 16(Q k ) and Fµ l,ρ 16R x γ 16R x. Obviously, x 16(Q k ) 16R x = 16(Q k). By the properties of dyadic cubes, we can pick out the maximal subcollection of {R x } x, which we label {R k } k. Now the {R x } x will be disjoint and have the property R k 16(Q k) C Q k, and so k Q k F l,ρ µ k (F l,ρ µ 16R k ) k l,ρ (Fµ 16R k 16R k ) 16R k Cγ16 n 1 k R k Cγ Q k. 17

18 Thus, if we fix γ sufficiently small, we have by [19, p62] that k k Cγ k 2µ Q k E l,ρ µ dµ 2µ Q k F l,ρ µ dµ 2µ Q k dµ (2) Cγ N,a (v) 2 L 2 (R) (1/4) N,a/12 (v) 2 L 2 (R), where the sum in k is only over those k for which D (Q k ) γ (Q l,ρ k Fµ ) is empty. We now consider those k for which D (Q k ) γ (Q k F µ l,ρ ) is non-empty. In this case, we may form the sawtooth domain Γ α (D (Q k ) γ (Q k F µ l,ρ )) for a < α < 2a. Let φ be the Lipschitz function which gives the boundary of Γ α (D (Q k ) γ (Q k F µ l,ρ )). Now, for q > 2, by Corollary 4.6 and Cauchy s inequality, µ q Q k Eµ l,ρ C N φ,a/12 (v) q G k ( ) C(ε) C (Q k ) φ,a ( v) q (Q k ) ( ) (21) + C(c(ρ) + C + ε) (N φ,b (v) q (Q k ( ) ) C(ε) C (Q k ) φ,a ( v) q + C(c(ρ) + C + ε)l q µ q Q k, (Q k ) since, for y (Q k ), N φ,b (v)(y ) N,b (v)(x ) for some x Q k F l,ρ µ. We now aim to show that, for y (Q k ), To see this take a dyadic cube Q (Q k ). If C (Q k ) φ,a ( v)(y ) Cρµ. (22) 32Q D (Q k ) γ (Q k Fµ l,ρ ) 18

19 then, by Lemma 4.7, (x n φ(x )) v 2 dx t φ a(16q) C(γ, a, α) C (Q l,ρ k Fµ ) 16Q Cρ 2 µ 2 Q. On the other hand, if (Q l,ρ k Fµ ) 16Q ( Γ b ((x,)) S,b ( v)(x ) 2 dx 32Q D (Q k ) γ (Q k Fµ l,ρ ) = ) v /yn n 2 dy dx then the distance from t φ a(16q) to is at least a fixed multiple of diam(q) =: r. Thus, for x t φ a(16q), x n Cr and so, for b sufficiently large, (x n φ(x )) v 2 dx r v 2 dx t φ a(16q) t φ a(16q) Cr n 1 v 2 dx t φ a(16q) x n 2 n C Q S,b ( v)(y ) C Q ρ 2 µ 2, for some y Fµ l,ρ and fixed b sufficiently large. Putting these two facts together proves (22). Substituting this in (21) we find Q k E l,ρ µ (C(ε)ρ q + C(c(ρ) + C + ε)) Q k. And so, choosing ε then ρ sufficiently small, we can argue as in (2) to complete the proof of the theorem. We can now prove (11). By Theorem 4.8 and Lemma 4.2 we have N,a/12 (v) 2 L 2 ( ) C( v 2 L 2 ( ) + C N φ,b (v) 2 L 2 ( ) + C N φ,b (v) L 2 ( ) S φ,b ( v) L 2 ( ) ). Therefore, (11) follows by [19, p62], Cauchy s inequality and taking C sufficiently small so that we may hide N φ,b (v) 2 L 2 ( ) on the left-hand side. Now we move on to prove (12). Let us assume A is symmetric, so that our equation becomes div(a u) + B u = and B satisfies conditions 19

20 (i) and (ii) with B replacing A. We start in the spirit of [17, (3.1)] with e n = (,,...,, 1), x = (x, t) and χ = χ(t) a smooth function on [, ): div((χa u u)e n t) = t (χa u u)t + (χa u u) = 2(χA u t u)t + ( t χa u u)t + (χ t A u u)t + (χa u u) = 2div((χA u)( t u)t) 2χdiv(A u)( t ut) 2( t χ)(a u)( t ut) 2(χA u e n ) t u + ( t χa u u)t + (χ t A u u)t + (χa u u) = 2div((χA u)( t u)t) + 2χ(B u)( t ut) 2( t χ)(a u)( t ut) 2(χA u e n ) t u + (χ t A u u)t + (χ + t t χ)(a u u). Now, integrating in and using the divergence theorem, we obtain 2(χA u e n ) t u = + 2χ(B u)( t ut) 2( t χ)(a u)( t ut) + (χ t A u u)t (χ + t t χ)(a u u). Now, let us take χ to be a smooth function of t which is equal to n on [, 1/n) and vanishing on [2/n, ). Taking the limit as n and using the fact that (χ + t tχ)dt =, we obtain 2(A u e n ) t u 2C u t u + C u 2. Now by ellipticity, Cauchy s inequality and knowing C can be taken to be sufficiently small, we find t u 2 C T u t u 2. 2 This proves (12) and with it Theorem 3.1. References [1] L.A. Caffarelli, E.B. Fabes, and C.E. Kenig. Completely singular ellipticharmonic measures. Indiana Univ. Math., 3(6): ,

21 [2] R.R. Coifman, Y. Meyer, and E.M. Stein. Some new function spaces and their applications to harmonic analysis. J. Funct. Anal., 62(2):34 335, [3] B.E.J. Dahlberg. Estimates of harmonic measure. Arch. Rational Mech. Anal., 65(3): , [4] B.E.J. Dahlberg. On the Poisson integral for Lipschitz and C 1 -domains. Studia Math., 66:13 24, [5] B.E.J. Dahlberg. Approximation of harmonic functions. Ann. Inst. Fourier, 3(2):97 17, 198. [6] B.E.J. Dahlberg. Weighted norm inequalities for the Lusin area integral and the nontangential maximal functions for functions harmonic in a Lipschitz domain. Sudia Mathematica, 67: , 198. [7] B.E.J. Dahlberg, C.E. Kenig, J. Pipher, and G.C. Verchota. Area integral estimates for higher order elliptic equations and systems. Ann. Inst. Fourier, 47(5): , [8] M. Dindoš, S. Petermichl, and J. Pipher. The L p Dirichlet problem for second order elliptic operators and a p-adapted square function. Journal of Functional Analysis, 249: , 27. [9] M. Dindoš and D.J. Rule. Elliptic equations in the plane satisfying a carleson measure condition. Accepted by Revista Mathemática Iberoamericana. [1] D. Gilbarg and N.S. Trudinger. Elliptic Partial Differential Equations of Second Order. Springer-Verlag, Heidelberg, [11] L. Grafakos. Classical and Modern Fourier Analysis. Pearson Education Inc., Upper Saddle River, New Jersey, 24. [12] D.S. Jerison and C.E. Kenig. The Dirichlet problem in nonsmooth domains. Ann. of Math., 113(2): , [13] D.S. Jerison and C.E. Kenig. The Neumann problem on Lipschitz domains. Bull. Amer. Math. Soc., 4:23 27, [14] C.E. Kenig, H. Koch, J. Pipher, and T. Toro. A new approach to absolute continuity of elliptic measure, with applications to non-symmetric equations. Advances in Mathematics, 153: , 2. 21

22 [15] C.E. Kenig and J. Pipher. The Dirichlet problem for elliptic equations with drift terms. Publications in Mathematics, 45: , 21. [16] C.E. Kenig and D.J. Rule. The regularity and Neumann problem for non-symmetric elliptic operators. Trans. Amer. Math. Soc., 361(1):125 16, 29. [17] J. Pipher. Littlewood-paley estimates: Some applications to elliptic boundary value problems. In CRM Proceedings and Lecure Notes, volume 12, pages , Providence, Rhode Island, Amer. Math. Soc. [18] D.J. Rule. The Regularity and Neumann Problem for Non-Symmetric Elliptic Operators. PhD thesis, University of Chicago, 27. [19] E.M. Stein. Harmonic Analysis: Real-Variable Methods, Orthoganality and Oscillatory Integrals. Princeton University Press, Princeton, New Jersey,