Perturbation Theory for Second Order Elliptic Operators with BMO Antisymmetric Part

In the present paper we study perturbation theory for the $L^p$ Dirichlet problem on bounded chord arc domains for elliptic operators in divergence form with potentially unbounded antisymmetric part in BMO. Specifically, given elliptic operators $L_0 = \mbox{div}(A_0\nabla)$ and $L_1 = \mbox{div}(A_1\nabla)$ such that the $L^p$ Dirichlet problem for $L_0$ is solvable for some $p>1$; we show that if $A_0 - A_1$ satisfies certain Carleson condition, then the $ L^q$ Dirichlet problem for $L_1$ is solvable for some $q \geq p$. Moreover if the Carleson norm is small then we may take $q=p$. We use the approach first introduced in Fefferman-Kenig-Pipher '91 on the unit ball, and build on Milakis-Pipher-Toro '11 where the large norm case was shown for symmetric matrices on bounded chord arc domains. We then apply this to solve the $L^p$ Dirichlet problem on a bounded Lipschitz domain for an operator $L = \mbox{div}(A\nabla)$, where $A$ satisfies a Carleson condition similar to the one assumed in Kenig-Pipher '01 and Dindo\v{s}-Petermichl-Pipher '07 but with unbounded antisymmetric part.


Introduction
The study of perturbations of elliptic operators in divergence form L := div(A∇·) goes back to a result from Dahlberg [Dah86]. Specifically, given elliptic operators L 0 = div(A 0 ∇) and L 1 = div(A 1 ∇), where we know that the L p Dirichlet problem for L 0 is solvable, he considered the discrepancy function ε(X) := |A 0 (X) − A 1 (X)|, and showed that if the measure dµ(Z) = sup X∈B(Z,δ(Z)/2)) ε(X) 2 δ(X) dZ, with δ(Z) := dist(Z, ∂Ω), is a Carleson measure with vanishing Carleson norm, then the solvability of the L p Dirichlet problem is transferred to L 1 := div(A 1 ∇·) with the same exponent p.
Actually this was formulated in terms of properties of the corresponding elliptic measures ω 0 and ω 1 . We know that the L p Dirichlet problem for L = div(A∇·) is solvable iff the elliptic measure ω associated with L belongs to the reverse Hölder space B p (σ), where σ denotes the surface measure on ∂Ω. In this language Dahlberg showed that if the Carleson norm of µ is small, then ω 0 ∈ B p (σ) implies ω 1 ∈ B p (σ). One natural question that arose was if the condition on µ could be relaxed to draw the weaker conclusion that ω 0 ∈ A ∞ (σ) implies ω 1 ∈ A ∞ (σ), where A ∞ (σ) = q>1 B q (σ); i.e. transferring solvability to L 1 but not necessarily with the same exponent. After some progress was made in [Fef89] it was finally shown in [FKP91] that if the Carleson norm of µ is bounded, then ω 0 ∈ A ∞ (σ) implies ω 1 ∈ A ∞ (σ). To summarize two different types of results were established (L) If the Carleson norm of µ is bounded then ω 0 ∈ A ∞ (σ) implies ω 1 ∈ A ∞ (σ).
In [Dah86] and [FKP91] the results were only proved for symmetric matrices in the case Ω = B n ⊂ R n . Since then there has been some work to extend this result to more general domains. In [MPT11] the authors extend (L) to the case where Ω is a bounded chord arc domain (see Definition 2.1). In the case of the unbounded domain of the upper halfspace R n+1 + , progress has recently been made in [CHM19] where the authors show both type (L) and (S) results. In the second part [CHMT20] they also prove a type (L) results for non-symmetric bounded matrices. However, they mention that their approach, which uses Carleson measure estimates, cannot be improved to obtain type (S) results.
In this paper we consider non-symmetric elliptic matrices A with BMO antisymmetric part on bounded chord-arc domains. Recall that a matrix A is elliptic that there exists λ 0 such that λ 0 |ξ| 2 ≤ ξ T A(X)ξ ≤ λ −1 0 |ξ| 2 , ∀ξ ∈ R n , a.e. X ∈ Ω. (1.1) Note that even in the case where the matrix A is not symmetric, ellipticity is only a condition on the symmetric part of the matrix A s . For the antisymmetric part A a we request that A a BMO(Ω) ≤ Λ 0 , i.e.
These operators where first studied in the elliptic case by Li and Pipher in [Li19], where they showed that the usual elliptic theory holds. This gives one hope that type (L) and (S) results can be generalized to these operators and indeed we obtain such results in Theorem 1.4 and Theorem 1.5 respectively. Instead of using the Carleson measure µ from before, we use a slightly more generalised version that respects the unboundedness of A, namely for a large fix 1 ≤ r < ∞ depending on n, λ 0 and Λ 0 . We can easily see that if we restrict to a bounded matrix A this new Carleson measure µ ′ has smaller Carleson norm than the old Carleson measure µ which means that if the Carleson norm of µ exists and the pertubation result from [MPT11] applies then the Carleson norm of µ ′ exists and Theorem 1.4 applies. Hence, the perturbation result [MPT11,Theorem 8.1] is a special case of Theorem 1.4.
(The Carleson norm · C is defined in Definition 2.2.) These theorems can then be used to extend the L p solvability result from [KP01] and [DPP07]. In [DPP07]  Inspired by this approach, we show that Theorem 1.6. Let Ω be a bounded Lipschitz domain with Lipschitz constant K > 0 and L 0 = div(A∇·) an elliptic operator satisfying (1.1) and (1.2) and α r defined like in (1.7). For every 1 < p < ∞ there exists r = r(n, λ 0 , Λ 0 ) > 1 and ε = ε(p) > 0 such that if α r (Z) 2 δ(Z) −1 dZ C < ε and K < ε, then ω ∈ B p (σ), i.e. the L p Dirichlet problem is solvable, Similarly Theorem 1.4 is used to show the generalization of [KP01] in Theorem 1.8. Let Ω be a bounded Lipschitz domain and L 0 = div(A∇·) an elliptic operator satisfying (1.1) and (1.2) and α r defined like in (1.7). Then there exists r = r(n, λ 0 , Λ 0 ) > 1 such that if then the corresponding elliptic measure satisfies ω ∈ A ∞ (σ), i.e. the L p Dirichlet problem is solvable for some 1 < p < ∞, These two theorems give us a larger class of operators that solve the L p Dirichlet problem on bounded Lipschitz domains than was previously known, replacing the oscillation of A in the Carleson condition with an L p mean oscillation for a large p > 2. Finally we note that in the t-independent case -that is when Ω = R n+1 + and the matrix is independent of the direction transverse to the boundary -solvability of the Dirichlet problem for matrices with BMO antisymmetric part was shown in [HLMP22], specifically that it is solvable for some 1 < p < ∞.
The paper is organized as follows: We start with Section 2 on preliminaries. Then, in Section 3, we outline the proof of Theorem 1.4, which closely follows that of [MPT11] and [FKP91]; this contains a subsection with results needed to prove the key identity The meat of the proof of Theorem 1.4 is proving Lemma 3.3 and Lemma 5.3, which is done in Sections 4 and 5 respectively. With these results established, Theorem 1.5 follows quickly in Section 6. Finally, in Section 7 we prove Theorem 1.8 and Theorem 1.6.
We note that [FKP91] contains some gaps that were unfortunately carried over to [MPT11], but these have been rectified in this paper. For more details see remarks 4.1 and 4.10.

Preliminaries
Here and in the following sections we shall allow all our constants to depend on n, λ 0 and Λ 0 .
Here B(Q, r) denotes the n-dimensional ball with radius r and center Q and σ denotes the surface measure. The best constant C in above condition is called the Ahlfors regularity constant.
Throughout this paper Ω will denote a bounded CAD.
Proposition 2.3. Let b be a constant anti-symmetric matrix and let u ∈ W 1,2 (E) and v ∈ W 1,2 0 (E), with E ⊂ Ω measurable. Then Proof. Note that if b is a constant anti-symmetric matrix and E ⊂ Ω, then for Denoting by (A a i ) E the matrix with component wise mean values it follows that (2.4)
• w ∈ B ∞ (µ) iff there exists c > 0 such that for a.e. x ∈ R n and balls wdµ.
It is easy to see that the following hold: For more properties of these spaces we refer the reader to [Gra09].
Suppose now that ν is another doubling measure on ∂Ω. We say that ν ∈

The L p Dirichlet boundary value problem
is the cone of aperture α and δ denotes the distance to the boundary ∂Ω.
Let L = div(A∇·), where A(X) ∈ R n×n is a matrix, satisfying (1.1) and (1.2). We say that u ∈ W 1,2 loc (Ω) is weak solution to the equation where C ∞ 0 (Ω) denotes the space of all smooth functions with compact support. We know (see e.g. [Li19]) that if f ∈ C 0 (∂Ω) then there exists a u ∈ W 1,2 (Ω) ∩ C 0 (Ω) such that Definition 2.9. Let α > 0. We say the L p Dirichlet problem for the operator L is solvable, if for all boundary data f ∈ L p (∂Ω) ∩ C(∂Ω) the solution u above satisfies the estimate Recall there exists a measure ω X such that that for u above This is called the elliptic measure with pole at X. As noted in the introduction the introduction the L p Dirichlet problem is solvable iff ω ∈ B p ′ (σ). For a proof see e.g. [Ken94] and the references therein.

Properties of solutions
In this sections we include some important results from Li's thesis [Li19] that will be used later. These results hold for solutions in NTA domains. First we have a reverse Holder inequality for the gradient.
Proposition 2.10 (Lemma 3.1.2). Let u ∈ W 1,2 loc (Ω) be a weak solution. Let X ∈ Ω and let B R = B R (X) be such that B R ⊂ Ω and let 0 < σ < 1. Then there exists a p > 2 such that Furthermore we have Caccioppoli's inequality.
Proposition 2.11. For a C = C(n, λ, Λ 0 ) < ∞ it holds for a solution u and We have a Harnack inequality.
We have boundary Hölder estimates.
An important corollary of the above result is the following lemma This result will be of central importance later.

Properties of the Green's function
Next for properties of the Greens function, also taken from [Li19].
and for 0 < θ < 1 we have Proposition 2.21. Let L * be the adjoint operator to L and let G * be the corresponding greens functions. Then Finally we note the relation between the Greens function and the elliptic measure ω X which implies that it is doubling. .

Nontangential behaviour and the square function in chord arc domains
Recall that the nontangential maximal function is given by  where Ω is a NTA domain. Let v : Ω → R and let 0 < p < ∞, α, β > 0, 2 > η > 0. Then , where the implied constant depends on the character of Ω, the doubling constant of µ and β/α. Now let L i = div(A i ), i = 0, 1, where A i satisfies (1.1) and (1.2), and let u i be the solution to the Dirichlet problem for L i with boundary data f ∈ L p (∂Ω) ∩ C(∂Ω). Set F := u 1 − u 0 and note in particular that F = 0 on ∂Ω.
Lemma 2.30. Let 0 < η < 2. For a solution u we have (2.32) Proof. First, note that by Proposition 2.12 we have, for a solution u . Using this and the triangle inequality we then have We also have an Almost Caccioppoli inequality for F .
To see this one simply uses Caccioppoli for u 0 and u 1 , and the triangle inequalitŷ We can define the square function of a function u ∈ W 1,2 (Ω) by . By setting f := |∇u|δ we can consider the square function operator as restriction of a more general operator More results of this operator can be found in [MPT11], but we are just going to recall Proposition 4.5 Proposition 2.34. We have for 0 < p < ∞ and two apertures α, β ≥ 1 In fact this holds for any doubling measure µ. For our purposes this means that the L p norms of Square functions with different apertures are comparable.
By [Chr90] there exists a family of "cubes" {Q k α ⊂ ∂Ω; k ∈ Z, α ∈ I k ⊂ N} where each scale k decomposes ∂Ω in a way that for every k ∈ Z Furthermore, the following properties hold: The last listed property implies together with the Ahlfors regularity of the surface measure that σ(Q k α ) ≈ 8 −k(n−1) . Similarly, the doubling property Proposition 2.23 of the elliptic measure guarantees us Now we can define a decomposition of (∂Ω, 4R 0 ). For k ∈ Z, α ∈ I k , set where λ is chosen so small that the {I k α } α∈I k have finite overlaps and The scale k 0 is chosen such that k 0 is the largest integer with 4R 0 < λ8 −k0+1 . Additionally, for ε > 0 we set the scale k ε as the smallest integer such that I k α ⊂ (∂Ω, ε) for all k ≥ k ε . The choices of k 0 and k ε guarantee that Furthermore, we define the enlarged decomposition and it is clear that that I k α ⊂Î k α and that theÎ k α have finite overlap. Observe that we can cover I k α by balls and On the other hand for Z ∈ (∂Ω, 4R 0 ) there exists Z * ∈ ∂Ω with |Z * − Z| = δ(Z) and an I k α such that Z * ∈ Q k α and Thus, for every X ∈ B(Z, δ(Z)/4) Lastly, we also observe that if P ∈ Q k α and Z ∈Î k α then, for M = 8 4 , We are also going to need an intermediate decomposition in cubesĨ k α defined byĨ The enlargement compared to I k α here is chosen such that (2.44) Furthermore, there is the following important Proposition.
Proposition 2.45. It holds Proof: Using the ball covering we get 3 Proof of Theorem 1.4 We use the method presented in [MPT11]. Recall that A 0 and A 1 satisfy (1.1) and (1.2), and that ω 0 and ω 1 denote their elliptic measures. Furthermore, F = u 1 − u 0 and β r is defined in (1.3). First, we need to prove This theorem corresponds to Theorem 2.9 in [MPT11] with the discrepency function α instead of β r . Using this the authors of [MPT11] first prove Theorem 3.2 (Theorem 8.2). Let Ω be a bounded CAD and let After that, they conclude Theorem 1.4 (which is Theorem 8.1 in their notation). If we replace α by β r , we can conclude Theorem 1.4 from Theorem 3.2 and Theorem 3.2 from Theorem 3.1 in the exact same way. The only modification needed is the substitution of every discrepancy function α by β r .
Hence we are left with proving Theorem 3.1. For this purpose we introduce the following two lemmas.
where the apertureM = 2 8 is at least twice as large as α.
Lemma 3.4. Under the assumptions of Theorem 3.1 it holds for any aperture Morally these lemmas correspond to Lemma 2.9 and Lemma 2.10 in [FKP91] or Lemma 7.7 and Lemma 7.8 in [MPT11].
Proof of Theorem 3.1: We assume Lemma 3.3 and Lemma 3.4 and recall that Combining these facts we have that Thus, with ε 0 small enough we can push the termÑ α [F ] 2 back to the left hand side to get By Lemma 2.30 we obtain

The difference function F
Our first proposition is a generalization of Lemma 3.12 in [CHM19] which uses the same proof strategy.
Proposition 3.5. Suppose that Ω ⊂ R n is a bounded CAD and let L 0 , L 1 be two elliptic operators. Let u 0 ∈ W 1,2 (Ω) be the weak solution of L 0 u 0 = 0 in Ω, and let G 1 be the Green's function of L 1 . Then Proof. We begin by fixing X 0 ∈ Ω and taking a cut-off function Due to Proposition 2.18 we have that G 1 (·, X 0 )ψ ε ∈ W 1,2 0 (Ω), which giveŝ Thus if we can show that I ε (X 0 ) + II k ε (X 0 ) → 0 as ε → 0 for a.e. X 0 ∈ Ω, then we are done. We start by considering the first term. Note that where r > 2 will be fixed later. Notice that the first term is bounded since A ∈ L r loc (Ω). To deal with the second term we decompose the ball Using Proposition 2.10, Caccioppoli and Proposition 2.20 we get for r sufficiently large Choosing r > 2n we get Now, for the second term we note that ∇ϕ ε ∞ ≈ ε −1 and obtain with Proposition 2.20 and Hölder's inequality Combining both integrals we havê on Ω. Thus letting ε go to zero finishes the proof.
Analogous to the proof of Proposition 3.5 we get for large k Thus sup k M [g k ] 1/2 < ∞ a.e. whence letting ε → 0 in (3.9) yields (3.8) as desired.
Let Q ∈ ∂Ω and X ∈ Γ α (Q). Let G 0 be the Green's function corresponding to L 0 and G * 0 its adjoint. As before let F = u 1 − u 0 and note that by Proposition 3.6, Proposition 3.5 and Proposition 2.21 Remark 4.1. Note that in [MPT11] and [FKP91] the authors claim that the above holds simply by the Green's function property and integration by parts. However, the Green's function property (2.19) only holds for ϕ ∈ W 1,p 0 (Ω) with p > n ≥ 2 and so cannot be applied directly. Thus proving propositions like Proposition 3.5 and Proposition 3.6 are necessary even when A 0 , A 1 are bounded and symmetric.
We will start by splitting F we then split F 1 further Here B(X) := B(X, δ(X)/4) andG 0 denotes the Green function for L 0 on 2B(X). We also set K(Z, Y ) := G 0 (Z, Y ) −G 0 (Z, Y ). Since µ is a doubling measure we have by Lemma 2.29 that Hence it is enough to show the pointwise bound B(X) in almost every Q ∈ ∂Ω and X ∈ Γ α (Q) to conclude Lemma 3.3 . We will consider each of the terms separately.
Thus ∇F ρ m L 2 (2B(X)) ρ,X 1 and we conclude thatF ρ m ∈ W 1,2 0 (2B(X)). Now we can hide this term on the left side and arrive at Next note that We know that if δ(X) ≥ 4R 0 , then ε = 0 on B(X). Hence assume now that δ(X) ≤ 4R 0 . Then we can use observation (2.41) and Proposition 2.45, to obtain For the remaining term using Proposition 2.10 yields Using the Poincare inequality forF ρ m yields whence putting everything together (4.6) We will now show that and then so that 4.5 and 4.6 combine to B(X) as desired.

this means that
i.e.F ρ m (Z) →F m (Z). Thus by (4.4) we may apply the dominated convergence theorem to conclude as desired.
We know that ∇ψ s ∞ Since ∇u 1 , ∇û m ∈ L 2r r−2 ( 3 2 B(X)), using Hölder and the fact that M L p →L p < ∞ for p > 1 we obtain and similarly Because the implicit constant in this inequality depends on neither m nor s the statement (4.8) is proved. Thus B(X) as desired.
Remark 4.10. This complicated approximation argument is not present in neither [MPT11] nor [FKP91]. Instead it follows from the equation fromF 1 satisfying the equation There are two problems with this. The first one is that it is not clear if the weak derivative ∇F 1 even exists in L 1 loc (2B(X)) due to the low regularity of the Green's function. But even if we were to haveF 1 ∈ W 1,2 0 (2B(X)), the Green's function property (2.19) only holds for ϕ ∈ W 1,p 0 (2B(X)) with p > n ≥ 2 and so cannot be used directly. Thus even when A 0 , A 1 are bounded and symmetric an approximation argument like the one carried out above is necessary.
Next we considerF 1 (Z). For large r and Z ∈ B(X) The first two terms are handled like forF 1 . Note that L 0 K(Z, ·) = 0 in 2B(X) and K(Z, ·) ≥ 0, so we may use Caccioppoli (Proposition 2.11) and then Harnack (Proposition 2.13) to deduce Next note that we can apply the bounds in Proposition 2.20 on the sum of two Green's functions, and hencê

The part away from the pole F 2
We consider a fixed Z ∈ B(X, δ(X)/8) so that the pole of the Green's function is uniformly away from Ω \ B(X), i.e. |Z − Y | ≥ δ(X)/8 for all Y ∈ Ω \ B(X), and we show the pointwise bound Let X * ∈ ∂Ω with |X * − X| = δ(X) and set ∆ j := ∆(X * , 2 j−1 δ(X)), Ω j := Ω ∩ B(X * , δ(X)2 j−1 ), R j := Ω j \ (Ω j−1 ∪ B(X)), Then we have We have Since we can cover (∂Ω, 4R 0 ) \ (∂Ω, ε) by the decomposition introduced in Subsection 2.6, by (2.36), we can writê (4.12) Further, we can use the ball cover introduced in Subsection 2.6 and its properties (2.39) and (2.38) together with Proposition 2.10 to conclude Now we can use Caccioppoli to obtain For the last term we apply the comparison principle and the doubling property of the elliptic measure and get Next, we can cover 5∆ 0 with N balls B(Q i , δ(X)/4) such that |Q i − Q j | < δ(X)/4, Q i ∈ 5∆ 0 . Hence Ω 0 ∩ (∂Ω, δ(X)/8) ⊂ i B(Q i , δ(X)/4). Note that N is independent of X and δ(X). Then we setÃ i = A(Q i , δ(X)/4) and the comparison principle gives for Y ∈ B(Q i , δ(X)/4) Furthermore we have with Harnack's inequality for all Y ∈ Ω 0 \ (∂Ω, δ(X)/16) Using the comparison principle for the Green's function gives then , (4.14) and hence in total . (4.15) Using Proposition 2.45 we get , and hence in total T u 1 (Z)dZ We say a dyadic boundary cube Q k α , k R0 ≤ k ≤ k ε belongs to J j , if and it belongs to J ∞ , if Furthermore, we denote by M ω0 the uncentered Hardy-Littlewood maximal function and we setÕ and hence Q k α ⊂Õ j . Thus, we also have The weak L 1 boundedness of the maximal function implies With Cauchy-Schwarz inequality we get Since for every two cubes Q k α , Q l β , l ≤ k either Q l β ⊂ Q k α or Q l β ∩ Q k α = ∅ holds, there is a disjoint collection of cubes in J j such that their union covers all the other cubes and we call them top cubes. We observe that for such a top cube Q k α and a sub cube Q l β ⊂ Q k α we haveÎ l β ⊂ T (∆(Z k α , (C 0 + 16λ)8 −k )) and the overlap of these Carleson regions of different top cubes Q k α is finite. We also know that the overlap of theÎ l β is finite. Hence Here we used Proposition 2.45 in the second last step and the property of cubes in J j in the last step. We denote S ε T ε u 1 (P ) 2 dω 0 (P ).
First for Y ∈ R j , we note that |A −1 − Y | ≥ δ(X)/4 because A −1 ∈ Ω −1 . Now, we can observe that Harnack's inequality implies G 0 (Z, Y ) ≈ G 0 (A −1 , Y ) where the implicit constants from Harnack's inequality are independent of Z and X.
Using the boundary Hölder continuity of the nonnegative solution G 0 (·, Y ) in Ω j−2 in Proposition 2.17 we can conclude that Now, let Y ∈ V j . We proceed similar to the case for Ω 0 . We have similar to (4.12) We replace (4.15) by the following similar argument. First, we can use Caccioppoli to obtain Note that N is independent of X and δ(X). Then we setÃ i = A(Q i , 2 j−5 δ(X)) and since dist( Since δ(Ã i ) = 2 j−5 δ(X) we can use Harnack to obtain With Proposition 2.22 we get then . (4.17) Combining this with (4.16) we get in total and hence Applying a stopping time argument like in the case of F 0 2 doing the exact same Next, consider Y ∈ W j . We cover W j by N balls B jl := B(X j l , 2 j−8 δ(X)) with X j l ∈ W j . Note here that N is independent of j. Using the comparison principle for the Green's function, the doubling property of the elliptic measure and G 0 (A j−2 ) ≈ G 0 (Y ) from Harnack we get Due to (4.16) we have in total Since for every Y ∈ B jl we have δ(Y ) ≥ (2 j−6 − 2 j−8 )δ(X) ≥ 2 j−7 δ(X) and |Y − X * | ≤ (2 j + 2 j−8 )δ(X) ≤ 2 j+1 δ(X) we have forM = 2 8 Since the finite ball cover has finite overlap we obtain similar to (4.12) From Proposition 2.10 we get analog to (4.13) Furthermore, we can observe that |B jl | ≈ δ(Y ) n ≈ (2 j δ(X)) n and B jl ⊂ B(Y, δ(Y )/2) for Y ∈ B jl . Then we have with Harnack that G 0 (Y ) ≈ G 0 (A j+1 ) and apply this together with the observations to obtain Using the comparison principle Proposition 2.22, Cauchy Schwarz inequality and the assumption of Theorem 3.1 yields Combining the estimates of V j and W j we get First, we observe that To see this take Y ∈ (∂Ω, 4R 0 ) \ (B(X) ∪ B(X * , 2 15 R 0 )) and since the {I k α } α,k cover (∂Ω, 4R 0 ) we have Y ∈ I k α . We know that that there exists P k α ∈ Q k α such and hence Q k α ⊂ ∂Ω \ ∆ 2 14 R0 . Analogously to (4.12), Since Y ∈ I k α is far away from Z and 0 we can use Harnack's inequality to conclude G 0 (Z, Y ) ≈ G 0 (Y ), where the implicit constants are independent of Y. Hence, with Cacciopolli, Proposition 2.22 and the doubling property of ω 0 we replace (4.15) From here on we can proceed with a stopping time argument like in the case F 0 2 doing the exact same with O j = {P ∈ ∂Ω \ ∆ 2 14 R0 ; T ε u 1 (P ) > 2 j } instead. This leads to To prove Lemma 3.4 we need the following "good-λ" inequality.

Proof of the "good-λ" inequality Lemma 5.1
To finish the proof of Lemma 3.4 we need to give the proof of Lemma 5.1. Consider a decomposition of {SM [F ] > λ} in Whitney balls ∆ j . We set and in the following drop the subscripts j. From Lemma 1 in [DJK84] we have that for every τ > 0 there exists a γ > 0 such that for the truncated square function holds for all points Q ∈ E, where Γ τ r M (Q) := ΓM (Q) ∩ B(Q, τ r). We setΩ = Q∈E Γ τ r M and want to define a partition of unity on Γ τ r M (Q). Therefore, recall from Subsection 2.6 the family of balls B(X l , λ8 −k−3 ) 1≤l≤N covering I k α and denote their center points by X k,l α . We claim the existence of a family (η k,l α ) k,l,α with the following properties Furthermore, we define D k (Q) := {I k α |I k α ∩ Γ τ r M (Q) = ∅}, D(Q) = k D k (Q), where we let k 0 be the scale from which on D k (Q) = ∅ for all k ≥ k 0 . We can observe that for the choiceM := 8M and for all I k α ∈ D(Q) we havê I k α ⊂ ΓM (Q). Thus In the last line we used the comparison principle Proposition 2.22 and write as before G 0 (X) = G 0 (0, X). Further we have In the second last line we used the observation that δ(X) ≈ diam(Q k α ). We now fix k, l, α.
Using that ∇η k,l α L ∞ ≈ 1 diam(Q k α ) , and applying Proposition 2.11 and Proposition 2.22 allows us to observe First, we consider II k,l,α

1
. For X ∈Ĩ k α applying Proposition 2.22 and Proposition 2.23 yields Hence, the use of (5.7) yields In the previous calculation we choose s > 1 independent of k and α to be a small constant such that the enlargement sĨ k α ofĨ k α is a sufficiently small enlargements, so that it fits betweenĨ k α andÎ k α , i.e.
Noting that fflĨ Putting everything together we get In the last line we used the defining properties of the set E. Continuing we use k ∈ B p (dσ), Hölder's inequality and boundedness of the maximal function to obtain Hence we have ω 0 (E) γ 2 ω 0 (∆) which finishes the proof of the "good λ-inequality".
6 Proof of Theorem 1.5 Most of the work for the proof of Theorem 1.5 has already been done in the form of the "good-λ" inequality. Recall that we want to show that ω 1 ∈ B p (dσ) which is equivalent to showing By assumption ω 0 ∈ B p (dσ) or equivalently σ ∈ A q (dω). Combining this observation with Lemma 3.3 and Lemma 5.3 we havê By Lemma 2.29, and with ε 0 small enough, we can push the first term back to the left hand side to get whence u 1 ∈ B p (σ) as desired.

Application
For this section let Ω be a bounded Lipschitz domain with Lipschitz constant K and consider the operator L = div(A∇·), where A is λ 0 -elliptic with A a BMO(Ω) ≤ Λ 0 . Set The aim of this section is to prove Theorem 1.8 and Theorem 1.6. But first we will prove a similar weaker result: |∇Â(X)|, 0 < η < 1/2.
It remains to show (7.3). Let ∆ ⊂ ∂Ω be a boundary ball with diam ∆ ≤ γ, where γ is taken small enough that T (∆) lies above a Lipschitz graph. Note that by [DP18, Cor 5.2] we havê (Again althoughÂ is assumed to be bounded in [DP18] this assumption is not necessary for Cor 5.2 to hold.) Thus assuming that f ∈ C 0 (∂Ω), the maximum principle implieŝ Hence dividing both sides with σ(∆) and taking supremum over our ∆ we obtain (7.3) as desired.
7.1 Proof of Theorem 1.8 and Theorem 1.6 In order to prove Theorem 1.8 and Theorem 1.6 we take inspiration from the approach for bounded matrices in [DPP07]. Our strategy is to construct a ma-trixÂ from A, apply Theorem 7.1 toÂ and then use our perturbation results to move fromÂ to A.
To begin with, let B(X) = B(X, δ(X)/2) and set Note that in order to apply Theorem 7.1 to our matrix we need it to be differentiable. Thus we will letÂ be the smoothing of A in the following sense: Take φ ∈ C ∞ c ( 1 5 B n ) be non-negative with´R n φ = 1, let φ t = t −n φ(X/t) and set A(X) := (φ δ(X) * A)(X).
Next we wish to apply our perturbation results with A 0 =Â and A 1 = A. ClearlyÂ λ 0 -elliptic. Moreover we can see that Â BMO(Ω) Λ 0 . To be more precise here, we distinguish two cases. First, we take a ball B ⊂ Ω and assume that for all X B ⊂B(X) . Then we can find a cover with balls (B(X i )) i such that the ballsB(X i ) have finite overlap, and | iB (X i )| |B|. Note here that the constants in the last inequality are independent of B. Due to Lemma 2.1 from [Jon80]