Dimension drop for harmonic measure on Ahlfors regular boundaries

We show that given a domain $\Omega\subseteq \mathbb{R}^{d+1}$ with uniformly non-flat Ahlfors $s$-regular boundary and $s\geq d$, the dimension of its harmonic measure is strictly less than $s$.

two-sided uniform domains. We say a domain Ω is C-uniform if for all x, y ∈ Ω there is a curve γ ⊆ Ω so that dist(z, Ω c ) ≥ C −1 min{ℓ(x, z), ℓ(y, z)} where ℓ(a, b) denotes the length of the subarc of γ between a and b. A domain is two-sided uniform if both Ω and Ω c are C-uniform domains for some C.
Since two-sided uniform domains have boundaries with dimension at least d and a Wolff snowflake can have dimension less than d, this example also shows that a dimension drop for harmonic measure can occur for a domain that is quite nice in terms of its connectivity and boundary properties. It is an interesting problem to identify some general criteria for when a whole class of domains Ω satisfy dim ω Ω < dim ∂Ω.
A dimension drop for harmonic measure occurs for some domains whose boundaries have some self-similar structure. This phenomenon was first observed by Carleson [Car85] for complements of planar Cantor sets whose boundaries have dimension at least 1 (rather, he showed for a particular class of Cantor sets C, dim ω Ω < 1). Later, Jones and Wolff showed the same result but for uniformly perfect sets satisfying a certain uniform disconnectedness property (see [JW88] or [GM08, Section X.I.2]). Makarov and Volberg showed dim ω Ω < dim ∂Ω when ∂Ω belongs to a more general class of Cantor sets (with dim ∂Ω possibly below 1) [MV86] and then to Cantor repellers of any dimension [Vol92], that is, sets K for which there are smooth disjoint domains U i ⊆ C compactly contained in a domain U ⊆ C and univalent maps f i : U i → U for which K is the unique compact set such that K = f i (U) (and in Volberg's result, two of the maps need to be linear). Urba'nski and Zdunik have also shown that the attractors of conformal iterated function systems (IFS) have a dimension drop when either the limit set is contained in a real-analytic curve, if the IFS consists of similarities only, or if the IFS is irregular (see [UZ02]). See also [Mey09,Pop98].
A common thread to many of these results is etiher some uniform disconnectivity property (see equations (XI.2.1)-(XI.2.3) in [GM08], [JW88], [Bat96,Lemma 2.5], and [Car85, Lemma 5]), or some self-similar or "dynamically defined" structure [MV86,Vol92,Vol93], and usually in order to exploit some ergodic theory. Notable exceptions are Jones and Wolff, and Batakis gave a non-ergodic proof for a wide class of Cantor sets [Bat96] that also works in higher dimensions, and later studied how the dimension is continuous with respect to the parameters defining the Cantor set [Bat00,Bat06].
In the present paper, we develop a different criterion for when the harmonic measure is strictly less than the dimension of the boundary. Though it assumes some strong conditions on the Hausdorff measure on the boundary, it requires no self-similar structure or uniform-disconnectedness. Instead, it assumes some uniform non-flatness condition (which will hold for any self-similar set in R d+1 of dimension at least d that isn't a d-dimensional plane), and also holds in higher dimensions.
Main Theorem. Given d ∈ N, C 1 > 0, and β > 0, there are constants s 0 < d and κ ∈ (0, 1) so that the following holds. Let s 0 < s ≤ d + 1 and Ω ⊆ R d+1 be a connected domain whose boundary is C 1 -Ahlfors s-regular, where the infimum is over all d-dimensional planes V ⊆ R d+1 . Then dim ω Ω < κs, meaning there is a set K with dim K < κs so that ω Ω (K c ) = 0.
Some remarks are in order. Firstly, this theorem does not cover all the fractals considered by Batakis, which may or may not satisfy (1.1). Secondly, nor does it tackle sets all sets with dimension s < d. However, the domains Batakis and others considered need their boundaries to be totally disconnected or be defined in some recursive way, whereas the domains we consider can be connected and quite random. Also notice that if s > d, then (1.1) implies there is β > 0 depending on s so that (1.2) holds, so (1.2) is not needed in this case to guarantee a dimension drop; however, if (1.2) holds for some β and s > d is close enough to d depending on β, we can have that dim ω Ω < d.
Some domains with 2-dimensional boundaries in R 3 not covered by previous results that are admissible for the above theorem are the complements of the tetrahedral Sierpinski Gasket and C × [0, 1] where C is the 1-dimensional 4-corner Cantor set. Note their boundaries are either connected or have large connected components. Some boundaries with no selfsimilar structure include bi-Lipschitz images of these sets, or a snowflaked image of R 2 in R 3 (that is, the image of a map f : R 2 → R 3 satisfying |f (x) − f (y)| ∼ |x − y| 1/s with s > 1).
The proof of the Main Theorem is modelled after that of [Bat96] and relies on a trick introduced by Bourgain in [Bou87]. Similar to those papers, the first step we need to take is to show that inside any "cube" on the boundary of our domain, the s-dimensional density of harmonic measure dips (or increases) inside a sub-cube of comparable size (compare the bottom of page 480 in [Bou87] or [Bat96, Lemma 2.7]). After showing this, the proof is similar to those above: this dip causes harmonic measure to concentrate elsewhere in the cube (see [Bou87,Lemma 2] or [Bat96, Lemma 2.8]) and one can iterate this to show that harmonic measure is in fact supported on a set of dimension less than s.
This density drop is easier to show when s > d using a touching-point argument, and is more quantitative. To prove the density drop allowing for s = d (or slightly smaller than d) assuming non-flatness, we use a compactness argument to show that, if this weren't true, then we could find a domain with d-regular boundary such that the density of its harmonic measure was uniformly bounded over all small balls on the boundary, but then harmonic measure would be absolutely continuous with respect to ddimensional Hausdorff measure. This gives us a lot of structural information by the following result: Thus, the boundary of our domain has tangents, but this violates (1.2).
Using compactness arguments for harmonic measure is quite common, and the author first learned of it from the work of Kenig and Toro [KT99]. For more recent applications, see [HMM + 17, Section 3], [BE17], and [AM15], and the references therein, which are primarily concerned with uniform domains. See also [AMT17] for an example of compactness arguments used in non-uniform domains.
We don't know whether our result holds for all s ∈ (d − 1, d + 1]. If s < d, the answer to this question relies on knowing whether harmonic measure is always singular with respect to H s -measure if the boundary is Ahlfors s-regular. As far as the author knows ,this is an open question. An answer in the affirmative would mean that the arguments here could be used again to obtain the whole range (d − 1, d + 1].
We would like to thank Mihalis Mourgoglou and Xavier Tolsa for their comments on the manuscript.

PRELIMINARIES
We will let B(x, r) = {y : |x − y| ≤ r} and B = B(0, 1). If B = B(x, r), we let λB = B(x, λr), x B = x, and r B = r. We will denote by H s and H s ∞ the s-dimensional Hausdorff measure and Hausdorff content respectively. For a reference on geometric measure theory and Hausdorff measure, see [Mat95].
We will write a b if there is a constant C > 0 so that a ≤ Cb and a t b if the constant depends on the parameter t. As usual we write a ∼ b and a ∼ t b to mean a b a and a t b t a respectively. We will assume all implied constants depend on d and hence write ∼ instead of ∼ d . Whenever For a domain Ω and x ∈ Ω, we let ω x Ω denote the harmonic measure for Ω with pole at x and G Ω (·, ·) the associated Green function. For a reference on harmonic measure and the Green function, see [AG01].
Given a domain Ω and a ball B centered on ∂Ω, we say x is a c-corkscrew We will say a domain Ω has lower s-content regular complement with constant c 1 if for all B centered on ∂Ω and 0 < r B < diam ∂Ω, The following lemma is due to Bourgain for R 3 in [Bou87, Lemma 1]. The proof in R d+1 is identical and shown in [AHM + 16, Lemma 3.4].
Remark 2.2. We will state some lemmas below that assume the CDC, but keep in mind the CDC is implied by (2.1) for s > d − 1. This can be seen from [HKM06, Lemma 2.31]). Alternatively, Ancona showed in [Anc86, Lemma 3] that the CDC is equivalent to the property that, for some α > 0, for all balls B centered on ∂Ω with r B < diam ∂Ω. This property is implied by Bourgain's lemma (see for example [AM18, Lemma 2.3]. In particular, the maximum principle implies the following lemma. be an open set that satisfies the CDC and let x ∈ ∂Ω. Then there is α > 0 so that for all 0 < r < diam(Ω), where α and the implicit constant depend on n and the CDC constant.
We will typically use the above lemma when φ(x) = 0.
The above lemma has evolved over many years and this isn't the most general statement, but it will suit our purposes. It follows from the proof of Lemma 3.5 in [AH08]: it is assumed there that the domain is John (and so in particular bounded) but is not necessary for the above statement. A version of this is also shown in [AHM + 16] that works for general bounded domains without the CDC, but it is only for d > 1.
Recall that a Harnack chain between two points x, y ∈ Ω is a sequence of balls B 1 , ..., B n for which 2B i ⊆ Ω. By Harnack's inequality, there is M > 0 so that for any non-negative harmonic function u on Ω, Lemma 2.6. [AMT17, Lemma 2.9] Let Ω j ⊂ R d+1 be a sequence of domains with lower s-content regular complements, ∞ also has lower s-content regular complement with the same constant.
This is not how it is stated in [AMT17], but it follows from the proof. Indeed, the lemma is stated for CDC domains, but they use the fact that CDC domains have lower s-content regular complements for a particular dimension and constant, and then use that characterization to prove the lemma. For x ∈ R n , r > 0, µ a Radon measure, and an affine d-dimensional plane V , let

RECTIFIABILITY
where the infimum is over all d-dimensional affine planes V .
As a corollary, if we let α E := α H d | E , we get the following. This follows from Lemma 3.2 since F may be covered by countably many d-dimensional Lipschitz graphs and since α µ (x, r) ≤ (s/r) d+1 α µ (x, s) for r < s.
Remark 3.4. The proof of the corollary is much simpler than envoking Lemma 3.2, and is quite standard, but we couldn't find a short reference for it. It can actually be proven more simply by the techniques in Chapters 14-16 of [Mat95], although for the sake of brevity we didn't want to recall too much background in order to do this.

NON-FLATNESS OR BIG DIMENSION IMPLIES CHANGE IN DENSITY
For a Radon measure µ, s ≥ 0, and a ball B, we define Then there is δ = δ(d, c, M, C 1 , A 0 , A) > 0 and a ball B ′ ⊆ 5A −1 B so that G(x, y).
Thus, there is a universal constant λ > 0 so that Moreover, E λ is open and also contains x. By the maximum principle, it must also be connected, so there is a curve γ ⊆ E λ joining x to a point in 2A −1 B. Note that if y ∈ γ and δ Ω (y) < ε, and ξ ∈ ∂Ω is closest to y, then x ∈ B(ξ, c/2), and so and so y ∈ E λ for ε small enough (depending on A and d). Thus, dist(y, ∂Ω) ≥ ε > 0 for all y ∈ γ. Thus, we can find a Harnack chain from x to a point y ∈ γ ∩ 2A −1 B of uniformly bounded length (depending on ε, d, and A).
Again, let ξ be the closest point in ∂Ω to y. Let v = y−ξ |y−ξ| and for t > 0 let B t = B(ξ, tv). Then for t > 0 small, In particular, for s > d, Θ s ω x (2B t ) t d−s , and so for t small enough, Θ s ω x (2B t ) > M. Since y ∈ 2A −1 B, |ξ − y| ≤ 2A −1 , so for t small enough, we can also guarantee that 2B t ⊆ 5A −1 B, and so 2B t is our desired ball.
Lemma 4.2. Given d ∈ N, M, c, C 1 > 0, and β > 0, there is s 0 < d so that the following holds. Let s 0 < s ≤ d + 1 and Ω ⊆ R d+1 be a connected domain so that (1.1) and (1.2) hold. Let A > 4, A 0 > 0 and suppose Proof. Again, we can assume B = B. Suppose instead that for all j ∈ N we could find a domain Ω j with C 1 -Ahlfors s j -regular boundary with d − 1 j < s j ≤ d + 1 containing 0 and x j a c-corkscrew point in B\2A −1 B so that for all j, ω By 2.6, we may pass to a subsequence so that In particular, one can show that (4.1) for all balls B centered on ∂Ω 0 contained in 5A −1 B, and Note that by the previous lemma, (4.1) is impossible if s > d, thus we must have s = d. In particular, ω x 0 Ω 0 ≪ H d in 5A −1 B. Since the ∂Ω j are uniformly Ahlfors d-regular, we can also pass to a subsequence so that ∂Ω j ∩ 2B converges in the Hausdorff metric to a set Σ so that if σ = H d | Σ , then σ(B) ∼ r d for all B ⊆ 2B centered on Σ. It is also not hard to show that for this new set we have bβ Σ (x, r) ≥ β > 0 for all x ∈ Σ ∩ B, 0 < r < 1.
By (4.1) and (4.2) and ω x 0 Ω 0 (5A −1 B) > 0 and ω x 0 for some x ∈ E. In particular, if ε > 0, there is r > 0 small enough and a plane V so that Without loss of generality, we can assume . and so V ∩ B(x, r) = ∅ for ε > 0 small. But by (4.3), for all r > 0 so that B(x, r) ⊆ B (since x ∈ 5A −1 B), there is either y ∈ B(x, r)∩Σ so that dist(y, V ) ≥ δr, or there is y ∈ V ∩B(x, r) so that dist(y, Σ) ≥ δr. In the former case, if we let 0 ≤ φ ≤ 1 be a 1 2r -Lipschitz function equal to 1 on B(x, 2r) and zero outside B(x, 4r), then ψ(z) = φ(z) dist(z, V ) is a 6-Lipschitz function (recall dist(z, V ) ≤ 8r for all z ∈ B(x, 4r) since V ∩ B(x, 4r) = ∅) and ψ(z) ≥ δr/2 on B(y, δr/2). Hence, which is a contradiction for ε small enough. The case that there is y ∈ V ∩ B(x, r) so that dist(y, Σ) ≥ δr has a similar proof and we omit it. Proof. Without loss of generality, B 0 = B. Letω = ω Ω\B . We can also assume that (4.6) ω p 1 10 By the Strong Markov property 1 and for ε > 0 small enough, By Lemma 2.1, ω x (aB) 1 on ∂B ∩ Ω, so by the maximum principle, So for ε > 0 small enough depending on M 1 and a, Let B j be a covering of E by boundedly many balls centered on E of radius ε/4 (whose total number depends only on ε and d), so 2B j ⊆ Ω. We claim we there are t > 0 (depending on ε and M 1 ) and j so that if B ′ = B j , theñ If not, then for each j eitherω p (B j ) < tω p ( 1 10 B) (let J 1 denote the set of these j) or ω x B j ( 1 10 B) < t (let J 2 denote the set of these j). For j ∈ J 2 , Harnack's principle implies ω x ( 1 10 B) t for all x ∈ B j . These alternatives and the fact that harmonic measure is at most 1 imply ω p 1 10 B (4.8) which contradicts (4.7) for t small enough. 1 This follows from the Brownian motion definition of harmonic measure, but for a direct proof, see the appendix in [AAM16].
and ω x B ′ ( 1 10 B) ≥ t. Let M > 0. Now x B ′ is a ε 2a -corkscrew point for aB, and so Lemma 4.2 (applied with aB in place of B, A = 10, and A 0 = t) implies there is δ > 0 depending on M, d, C 1 , and β and B ⊆ 1 ε,M 1 ,a ω p (aB)Mr s B . Thus, for M large enough (depending on ε and M 1 , and recall a is a universal constant), we have ω p (B)r −s B > M 1 a −s ω p (aB), which proves the lemma.

PROOF OF THE MAIN THEOREM
Let C 1 , β, s, d, β, and Ω ⊆ R d+1 be as in the Main Theorem and set ω = ω p Ω for some p ∈ Ω.
We recall the following version of "dyadic cubes" for metric spaces, first introduced by David [Dav88] but generalized in [Chr90] and [HM12].
Theorem 5.1. Let X be a doubling metric space. Let X k be a nested sequence of maximal ρ k -nets for X where ρ < 1/1000 and let c 0 = 1/500. For each n ∈ Z there is a collection D k of "cubes," which are Borel subsets of X such that the following hold.
(1) For every integer k, X = Q∈D k Q.
(3) For Q ∈ D, let k(Q) be the unique integer so that Q ∈ D k and set ℓ(Q) = 5ρ k(Q) . Then there is ζ Q ∈ X k so that Let D be the Christ-David cubes for ∂Ω. Fix n 0 so that x 0 ∈ aB Q for all Q ∈ D n 0 . By rescaling, we can assume without loss of generality that n 0 = 0.
Lemma 5.2. Let n ≥ 0, M 2 > 0, and Q ∈ D n . There is N Q ∈ N so that N Q M 2 ,C 1 ,β,d 1 and such that there is Proof. Fix N ∈ N large enough so that ifQ ∈ D n+N is the cube with same center as Q, then where a is as in Lemma 4.3. Clearly we can assume Then we take Q ′ to be the largest cube containing the center of B that is also contained in B, so and the lemma follows in this case by picking M 1 ≫ M 2 2 . Now suppose that Θ s ω (B) ≥ M 1 Θ s ω (aBQ). Let N ′ be the largest integer for which 5ρ N ′ < ℓ(Q)/4. Then there are at most boundedly many cubes in D N ′ which cover B ∩ ∂Ω, and one of them, call it Q ′ , must have ω(Q ′ ) ω(B). Since Q ′ ∩ B = ∅ and ℓ(Q ′ ) = 5ρ N ′ < ℓ(Q)/4, and B ⊆ 1 2 BQ, we have Also, we have ℓ(Q ′ ) ∼ δ ℓ(Q) ∼ N ℓ(Q) and Again, the lemma follows by picking M 1 ≫ M 2 2 . In either case, since δ always depends on d, β, C 1 and M 2 , we have ℓ(Q ′ ) d,β,C 1 ,M 2 ℓ(Q), so if N Q is such that Q ′ ∈ D n+N Q , then N Q d,β,C 1 ,M 2 1, and we're done.
We now let ε > 0 be small and let N Q denote the integer from the previous lemma applied when M 2 = ε −1 . Fix a cube Q 0 ∈ D 0 and define families of sub-cubes of Q 0 in D ′ n inductively as follows. First let D ′ 0 = {Q 0 }, then if D ′ n has been defined and R ∈ D ′ n , let n R be so that R ∈ D n R , set Proof. Let R ′ be as in Lemma 5.2 with M 2 = ε −1 . Suppose first that By the Cauchy-Schwartz inequality, Also, by (5.7), Now (5.6) follows since, if R ∈ D ′ n , and since N R ′ β,C 1 ,dε 1, Hence, there is t = t(C 1 , β, d, ε) > 0 so that and the lemma follows in this case. Now suppose that By the Cauchy-Schwartz inequality, Now we use the fact that ω(R ′ ) ω(R) Thus, there is t = t(ε, C 1 , β, d) > 0 so that ω(R ′ ) ω(R) > t. Hence, we again have and again the lemma follows.
Let τ ∈ (0, 1) be small, we will fix its value later. Then Pick τ > 0 small enough so we still have that γ := c − τ 2 λ < 1. Let E n = {Q ∈ D ′ n : ω(Q) ≤ σ(Q) 1−τ }. Then Let ℓ > 0. Since the Christ cubes partition Q 0 , for each x ∈ F Q 0 we may find Q x ∋ x contained in Q 0 with ℓ(Q x ) < ℓ so that ω(Qx) σ(Qx) 1−τ > 1/2. Let Q j be the collection of maximal cubes from {Q x : x ∈ F Q 0 }. Then because the Q j are disjoint and diam Q j ≤ diam B Q j = 2ℓ(Q j ) < 2ℓ, Since our choice of Q 0 ∈ D 0 was arbitrary and D 0 partitions ∂Ω, this implies dim ω ≤ s(1 − τ ). This finishes the proof of the Main Theorem.