The mutual singularity of harmonic measure and Hausdorff measure of codimension smaller than one
aa r X i v : . [ m a t h . C A ] A p r THE MUTUAL SINGULARITY OF HARMONIC MEASURE AND HAUSDORFFMEASURE OF CODIMENSION SMALLER THAN ONE
XAVIER TOLSAA
BSTRACT . Let Ω ⊂ R n +1 be open and let E ⊂ ∂ Ω with < H s ( E ) < ∞ , for some s ∈ ( n, n + 1) ,satisfy a local capacity density condition. In this paper it is shown that the harmonic measure cannot bemutually absolutely continuous with the Hausdorff measure H s on E . This answers a question of Azzamand Mourgoglou, who had proved the same result under the additional assumption that Ω is a uniformdomain.
1. I
NTRODUCTION
In this paper we study the relationship between harmonic measure and Hausdorff measure of codi-mension smaller than in R n +1 . The importance of harmonic measure is mainly due to its connectionwith the Dirichlet problem for the Laplacian. Indeed, recall that given a domain Ω ⊂ R n +1 and a point p ∈ Ω , the harmonic measure with pole at p is the measure ω p satisfying the property that, for anyfunction f ∈ C ( ∂ Ω) ∩ L ∞ ( ∂ Ω) , the integral R f dω p equals the value at p of the harmonic extensionof f to Ω .The study of the metric and geometric properties of harmonic measure has been a classical subjectin mathematical analysis since the Riesz brothers theorem [RR] asserting that harmonic measure isabsolutely continuous with respect to arc length measure on simply connected planar domains withrectifiable boundary. In the plane, complex analysis plays a very important role in connection withharmonic measure, essentially because of the invariance of harmonic measure by conformal mappings.This fact makes the case of planar domains rather special.In the plane it is known that the dimension of harmonic measure is at most by a celebrated resultof Jones and Wolff [JW]. This means that there exists a set E ⊂ ∂ Ω of Hausdorff dimension at most which has full harmonic measure. Furthermore, such set E can be taken so that it has σ -finite length, asshown by Wolff [Wo1]. More precise results for simply connected planar domains had been obtainedpreviously by Makarov [Mak1], [Mak2].In higher dimensions one has to use real analysis techniques to study harmonic measure. The codi-mension is still quite special, mainly because of the relationship between harmonic measure andrectifiability. For instance, in [AHM TV] it was shown that the mutual absolute continuity between har-monic measure and n -dimensional Hausdorff measure on a subset E ⊂ ∂ Ω , Ω ⊂ R n +1 , implies the n -rectifiability of E . Also, under the assumption that ∂ Ω is n -AD-regular, that is H n ( B ( x, r ) ∩ ∂ Ω) ≈ r n for all x ∈ ∂ Ω , < r ≤ diam( ∂ Ω) , many recent works have been devoted to relate quantitative prop-erties of harmonic measure and other analytic or geometric properties of the domain. See for example[Az1], [GMT], [HLMN], [HM1], [HM2], [HMM], [HMU], [MT].One of the main differences between the planar case and the higher dimensional case is that in R n +1 ,with n ≥ , there exist domains where the dimension of harmonic measure is larger than n . Thiswas proved by Wolff in [Wo2]. An important open problem consists of finding the sharp value for the Partially supported by by 2017-SGR-0395 (Catalonia) and MTM-2016-77635-P, MDM-2014-044 (MINECO, Spain). upper bound of the dimension of harmonic measure in R n +1 , n ≥ . In [Bo], Bourgain showed thatthis sharp value is strictly smaller than n + 1 . In [Jo], Jones conjectured that the sharp bound shouldbe n + 1 − /n . However, for the moment there have been no significative advances on this openproblem. On the other hand, the techniques of Bourgain [Bo] play an important role in more recentresults asserting that in some classes of sets (for example, in some self-similar sets) the dimension ofharmonic measure is strictly smaller than the dimension of the set. See [Ba1], [Ba2], and [Az2].As mentioned above, the current paper deals with harmonic measure in the case of codimensionless than . Although the main result of the paper is not directly related to the above Jones’ conjec-ture, I think that this contributes to a better understanding of the behavior of harmonic measure in thiscodimension.To state precisely the main result, we need some additional notation. For n ≥ , let Ω ⊂ R n +1 beopen and let E ⊂ ∂ Ω be a non-empty set. We say that the local capacity density condition (or localCDC) holds in E is there exists constants r E > and c E > such that(1.1) Cap( B ( x, r ) ∩ Ω c ) ≥ c E r n − for all x ∈ E and < r ≤ r E ,where Cap stands for the Newtonian capacity (see Section 2.2 for the definition). We denote by ω theharmonic measure in Ω .The main result of this paper is the following. Theorem 1.1.
Given n ≥ and s > n , let Ω ⊂ R n +1 be open and let E ⊂ ∂ Ω be such H s ( E ) < ∞ . Suppose that the harmonic measure ω and the Hausdorff measure H s are mutually absolutelycontinuous in E and that the local CDC holds in E . Then H s ( E ) = ω ( E ) = 0 . In other words, harmonic measure cannot be mutually absolutely continuous with Hausdorff measureof codimension less than in any subset of positive harmonic measure, under the local CDC assump-tion. Recall that the same result was proved in [AM] by Azzam and Mourgoglou under the additionalassumption that Ω is a uniform domain. Recall also that, roughly speaking, a domain is called uni-form if it satisfies an interior porosity assumption (the so-called interior corkscrew condition), and aquantitative connectivity condition in terms of Harnack chains.The methods in the current paper are very different from the ones used in [AM]. The new maintool is an identity obtained by integration by parts (see Section 3.1), whose application requires latersome rather delicate stopping time arguments. On the other hand, the arguments in [AM] use blowupsand tangent measures, and it seems that the uniformity assumption is important. Indeed, in their work,Azzam and Mourgoglou leave as an open question the possibility of eliminating the uniformity as-sumption. They also ask the same questions about the CDC: can this be avoided? While Theorem 1.1confirms that uniformity is not necessary, it is still an open problem to know if the CDC is required.In fact, in [AM] a non-degeneracy condition weaker (at least, a priori) than the CDC is used. I thinkthat, quite likely, in the arguments below one may be able to replace the local CDC assumption by thenon-degeneracy condition of Azzam-Mourgoglou. However, I have preferred to state Theorem 1.1 interms of the local CDC, which is closer to the usual CDC. On the other hand, the techniques in thecurrent paper do not look very useful for codimensions larger than , unlike the arguments in [AM],which are applied by the authors to derive other related results.Nevertheless, it seems the statement in Theorem 1.1 does not hold for s < n . Actually, according toan example constructed by Alexander Volberg, there is a domain Ω ⊂ R satisfying the CDC such thatharmonic measure is mutually absolutely continuous with H s | ∂ Ω for some < s < (cf. [AM]).The aforementioned integration by parts formula (see (3.1)) required for the proof of Theorem 1.1is a generalization of a formula that has already been applied to some problems involving harmonic or ARMONIC MEASURE AND HAUSDORFF MEASURE 3 elliptic measure and rectifiability in works such as [HLMN] or [AGMT], and it goes back to some workof Lewis and Vogel [LV], at least. 2. P
RELIMINARIES
In the paper, constants denoted by C or c depend just on the dimension and perhaps other fixedparameters (such as the constant c E involved the local CDC, for example). We will write a . b if thereis C > such that a ≤ Cb . We write a ≈ b if a . b . a .2.1. Measures.
The set of (positive) Radon measure in R n +1 is denoted by M + ( R n +1 ) . The Hausdorff s -dimensional measure and Hausdorff s -dimensional content are denoted ty H s and H s ∞ , respectively.Given µ ∈ M + ( R n +1 ) , its supper s -dimensional density at x is defined by Θ s, ∗ ( x, µ ) = lim sup r → µ ( B ( x, r ))(2 r ) s . Recall that, given an H s -measurable set E ⊂ R n +1 with < H s ( E ) < ∞ , we have(2.1) − s ≤ Θ s, ∗ ( x, H s | E ) ≤ for H s -a.e. x ∈ E .See [Mat, Chapter 6], for example.2.2. Newtonian capacity and harmonic measure.
The fundamental solution of the negative Lapla-cian in R n +1 , n ≥ , equals E ( x ) = c n | x | n − , where c n = ( n − H n ( S n ) , with S n being the unit hypersphere in R n +1 .The Newtonian potential of a measure µ ∈ M + ( R n +1 ) is defined by U µ ( x ) = E ∗ µ ( x ) , and the Newtonian capacity of a compact set F ⊂ R n +1 equals Cap( F ) = sup (cid:8) µ ( F ) : µ ∈ M + ( R n +1 ) , supp µ ⊂ F, k U µ k ∞ ≤ (cid:9) . It is well known that k U µ k ∞ = k U µ k ∞ ,F , and that there exist a unique measure that attains the supremum in the definition of Cap( F ) . If µ attainsthat supremum, then it turns out that U µ ( x ) = 1 for quasievery x ∈ F (denoted also q.e. in F ), i.e., forall x ∈ F with the possible exception of a set of zero Newtonian capacity. The probability measure µ F = 1Cap( F ) µ is called equilibrium measure (of F ), and it holds that U µ F ( x ) = 1Cap( F ) for q.e. x ∈ F .Recall that we denote by ω the harmonic measure on an open set Ω . The associated Green functionis denoted by g ( · , · ) . The following result is quite well known, but we prove it here for the reader’sconvenience. XAVIER TOLSA
Lemma 2.1.
Given n ≥ , let Ω ⊂ R n +1 be open and let B be a closed ball centered at ∂ Ω . Then ω x ( B ) ≥ c ( n ) Cap( B ∩ ∂ Ω) r ( B ) n − for all x ∈ B ∩ Ω , with c ( n ) > .Proof. Let µ B ∩ ∂ Ω be the equilibrium measure for B ∩ ∂ Ω , and let µ = Cap( B ∩ ∂ Ω) µ B ∩ ∂ Ω , sothat k U µ k ∞ ≤ and k µ k = Cap( B ∩ ∂ Ω) . Notice that, for all x ∈ B c , U µ ( x ) = Z c n | x − y | n − dµ ( y ) ≤ c n k µ k ( r ( B )) n − . Consider the function f ( x ) = U µ ( x ) − c n k µ k ( ( B )) n − . Using that f ( x ) ≤ in B c , k f k ∞ ≤ , and that f is harmonic in Ω , by the maximum principle we deduce that, for all x ∈ Ω , ω x ( B ) ≥ f ( x ) . In particular, for x ∈ B we have ω x ( B ) ≥ Z c n | x − y | n − dµ ( y ) − c n k µ k ( r ( B )) n − ≥ c n k µ k ( r ( B )) n − − c n k µ k ( r ( B )) n − = c n (cid:0) n − − ( ) n − (cid:1) Cap( B ∩ ∂ Ω) r ( B ) n − , which proves the lemma. (cid:3) We recall also the following lemma, whose proof can be found in [AHM TV].
Lemma 2.2.
Let n ≥ and Ω ⊂ R n +1 be open. Let B be a closed ball centered at ∂ Ω . Then, for all a > , (2.2) ω x ( aB ) & inf z ∈ B ∩ Ω ω z ( aB ) r ( B ) n − g ( x, y ) for all x ∈ Ω \ B and y ∈ B ∩ Ω ,with the implicit constant independent of a . Combining the two preceding lemmas, choosing a = 8 , we obtain: Lemma 2.3.
Let n ≥ , s > n − , and Ω ⊂ R n +1 be open. Let B be a closed ball centered at ∂ Ω .Then, (2.3) ω x (8 B ) & n Cap(2 B ∩ ∂ Ω) r ( B ) n − g ( x, y ) for all x ∈ Ω \ B and y ∈ B ∩ Ω .
3. T
HE KEY IDENTITY AND THE MAIN IDEA
The key identity.Lemma 3.1 (Key identity) . Let Ω ⊂ R n +1 be open, let ψ ∈ C ∞ c (Ω) , and let u : Ω → R be harmonicand positive in supp ψ . Then, for any α ∈ R , Z |∇ u | u α ψ dx = 12 α ( α − Z |∇ u | u α − ψ dx (3.1) − Z ∇ ( |∇ u | ) · ∇ ψ u α dx + 12 Z |∇ u | ∇ ( u α ) · ∇ ψ dx. ARMONIC MEASURE AND HAUSDORFF MEASURE 5
In the lemma we denoted |∇ u | = X i,j ( ∂ i,j u ) . The identity (3.1), in the particular case α = 1 , was already used in connection with harmonic measurein [LV] and [HLMN]. Proof.
Notice that |∇ u | = X i |∇ ∂ i u | . So (3.1) follows by summing from i = 1 to n + 1 the following identity: Z |∇ ∂ i u | u α ψ dx = 12 α ( α − Z | ∂ i u | |∇ u | u α − ψ dx (3.2) − Z ∇ ( | ∂ i u | ) · ∇ ψ u α dx + 12 Z | ∂ i u | ∇ ( u α ) · ∇ ψ dx. To prove this, we integrate by parts: Z |∇ ∂ i u | u α ψ dx = Z ∇ ∂ i u · ∇ ∂ i u u α ψ dx = Z ∇ ∂ i u · ∇ (cid:0) ∂ i u u α ψ (cid:1) dx − Z ∇ ∂ i u · ∇ (cid:0) u α ψ (cid:1) ∂ i u dx. The first integral on the right hand side vanishes because u is harmonic: Z ∇ ∂ i u · ∇ (cid:0) ∂ i u u α ψ (cid:1) dx = − Z ∆( ∂ i u ) (cid:0) ∂ i u u α ψ (cid:1) dx = 0 . Using also ∂ i u ∇ ∂ i u = ∇ ( | ∂ i u | ) , we get Z |∇ ∂ i u | u α ψ dx = − Z ∇ ( | ∂ i u | ) · ∇ (cid:0) u α ψ (cid:1) dx (3.3) = − Z ∇ ( | ∂ i u | ) · ∇ (cid:0) u α (cid:1) ψ dx − Z ∇ ( | ∂ i u | ) · ∇ ψ u α dx = − Z ∇ ( | ∂ i u | ψ ) · ∇ (cid:0) u α (cid:1) dx + 12 Z | ∂ i u | ∇ ψ · ∇ (cid:0) u α (cid:1) dx − Z ∇ ( | ∂ i u | ) · ∇ ψ u α dx. Finally, integrating by parts and taking into account that ∆ (cid:0) u α (cid:1) = α ( α − |∇ u | u α − , we deducethat the first term on the right hand side satisfies − Z ∇ ( | ∂ i u | ψ ) · ∇ (cid:0) u α (cid:1) dx = 12 Z | ∂ i u | ψ ∆ (cid:0) u α (cid:1) dx = 12 α ( α − Z | ∂ i u | ψ |∇ u | u α − dx. Plugging this into (3.3), we get (3.2). (cid:3)
XAVIER TOLSA
The strategy of the proof.
Let s > n be as in Theorem 1.1. By Bourgain’s theorem [Bo], it isclear that we can assume s ∈ ( n, n + 1) . Let a ∈ (0 , be such that s = n + a , and let α = 1 − a a , so that α ∈ (0 , too. We will apply the identity (3.1) with u equal to the Green function g ( · , p ) anda suitable function ψ . The choice of the preceding value of α is motivated by the fact that then theintegrals that appear in (3.1) scale like ℓ (cid:18) ω ( · ) ℓ n − (cid:19) α +2 ℓ n +1 = ω ( · ) (cid:18) ω ( · ) ℓ s (cid:19) α +1 , under the assumption that that u = g ( · , p ) scales like ω ( · ) ℓ − n .A key fact in our arguments is that the first term on the right hand side of (3.1) is negative (because α ( α − < ), while the left hand side is positive. These two terms should be considered as themain ones in (3.1), and the last two integrals should be considered as “boundary terms” because of thepresence of ∇ ψ in their integrands.Writing g ( x ) = g ( x, p ) , from (3.1) we get | α ( α − | Z |∇ g | g α − ψ dx (3.4) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z ∇ ( |∇ g | ) · ∇ ψ g α dx (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z |∇ g | ∇ ( g α ) · ∇ ψ dx (cid:12)(cid:12)(cid:12)(cid:12) − Z |∇ g | g α ψ dx. Using this inequality and assuming the existence of a set E ⊂ ∂ Ω with ω ( E ) > such that the harmonicmeasure and the Hausdorff measure H s are absolutely continuous on E , we will get a contradiction. Tothis end, we will construct an appropriate function ψ by some stopping time arguments involving theset E , and with this choice we will show that the integral on left hand side of (3.4) is much larger thanthe right hand side.To illustrate how we will apply the inequality (3.4) we consider a simple example. Suppose that E ⊂ B (0 , is compact and s -AD-regular (with s ∈ ( n, n + 1) ), that is, H s ( E ∩ B ( x, r )) ≈ r s forall x ∈ E, ≤ r ≤ diam E . Let Ω = R n +1 \ E and suppose that diam E ≈ . Observe that the s -AD-regularity of E ensures that the CDC holds. We will sketch how one can check that the harmonicmeasure ω for Ω is not comparable to H s | E , that is, ω cannot be of the form ω = h H s | E , for somedensity function h such that h ≈ . For the sake of contradiction, assume ω = h H s | E , with h ≈ (of course, this condition is much stronger than the mutual absolute continuity of ω and H s | E , but thissuffices for the example).Given a small parameter r ∈ (0 , / , consider a C ∞ function ψ satisfying χ B (0 , \ U r ( E ) ≤ ψ ≤ χ B (0 , \ U r / ( E ) (where U ρ ( E ) is the ρ -neighborhood of E ), with |∇ ψ | . /r in U r ( E ) , and |∇ ψ | . in B (0 , \ B (0 , . Suppose that the pole p for harmonic measure is far from E , say p B (0 , . Consider the second integral on the right hand side of (3.4). Notice that(3.5) (cid:12)(cid:12)(cid:12)(cid:12)Z |∇ g | ∇ ( g α ) · ∇ ψ dx (cid:12)(cid:12)(cid:12)(cid:12) . r Z U r ( E ) \ U r / ( E ) |∇ g | g α − dx + Z B (0 , \ B (0 , |∇ g | g α − dx. To estimate the first integral, consider a finite family of balls B h , h ∈ H , with radii r , centered at E ,which cover U r ( E ) and have bounded overlap. Then, for each ball B h , using the harmonicity of g and ARMONIC MEASURE AND HAUSDORFF MEASURE 7
Lemma 2.3, r Z B h \ U r / ( E ) |∇ g | g α − dx . r Z B h g α +2 dx . r n − (cid:18) ω (10 B h ) r n − (cid:19) α +2 = ω (10 B h ) (cid:18) ω (10 B h ) r s (cid:19) α +1 ≈ H s ( B h ∩ E ) . Thus, r Z U r ( E ) \ U r / ( E ) |∇ g | g α − dx ≤ X h ∈ H r Z B h \ U r / ( E ) |∇ g | g α − dx . X h ∈ H H s ( B h ∩ E ) . H s ( E ) . By analogous arguments, one can show that the last integral on the right hand side is at most C H s ( E ) ,and also the first integral on the right hand side of (3.4). These estimates together with (3.4) imply that(3.6) Z |∇ g | g α − ψ dx . H s ( E ) . To reach the desired contradiction it is reasonable to try to estimate the integral on the left hand sideabove from below. To this end, consider a ball B centered at E with radius r ( B ) ∈ (Λ r , , for someconstant Λ > . One can show that, for Λ big enough, the following holds: Z B \ U Λ − r ( B ) ( E ) |∇ g | g α − ψ dx = Z B \ U Λ − r ( B ) ( E ) |∇ g | g α − dx (3.7) & ω ( B ) (cid:18) ω ( B ) r ( B ) s (cid:19) α +1 ≈ H s ( B ∩ E ) . The detailed proof of this estimate is not too difficult but it would lead us too far. So we will justmention that this follows using Lemma 2.3 and other rather standard arguments (see Lemma 5.2 belowfor more details). Given r ∈ (Λ r , / , by covering U r ( E ) by a family of balls centered at E of radius r with bounded overlap and applying (3.7) to each ball, we infer that Z U r ( E ) \ U (2Λ) − r ( E ) |∇ g | g α − ψ dx & H s ( E ) . So assuming that r is of the form r = (2Λ) − N , N > , we get Z |∇ g | g α − ψ dx ≥ N − X k =1 Z U (2Λ) − k ( E ) \ U (2Λ) − k − ( E ) |∇ g | g α − ψ dx & ( N − H s ( E ) , which contradicts (3.6) if N is big enough, or equivalently, r is small enough.The proof of Theorem 1.1 will involve similar ideas to the ones above, but with additional technicalcomplications which will require the use of stopping time arguments. XAVIER TOLSA
4. T
HE BALL B , THE STOPPING CONSTRUCTION , AND THE FUNCTION ψ The ball B . From now we assume that we are under the assumptions of Theorem 1.1. Weconsider a point p ∈ Ω and we denote by ω the harmonic measure for Ω with respect to the pole p . Wealso denote µ = H s | E and we assume that < µ ( E ) < ∞ and that µ is absolutely continuous withrespect to ω . Our objective is to find a contradiction.By replacing E by a suitable subset if necessary, by standard methods (taking into account the upperbound for the upper density of µ in (2.1)) we may assume that there exists some δ > such that µ ( B ( x, r )) ≤ s r s for all x ∈ E and < r ≤ δ . Since µ ≪ ω , there exists some non-negative function h ∈ L ( ω ) such that µ = h ω . We consider apoint x ∈ E satisfying the following: x is a Lebesgue point for h with h ( x ) > and a density pointof E (both with respect to ω ), and there exists a sequence of radii r k → such that(4.1) ω ( B ( x , r k )) ≤ (200) n +2 ω ( B ( x , r k )) . For this last property, see, for example, Lemma 2.8 in [To]. Now, given some κ ∈ (0 , / , let δ ∈ (0 , δ ] be such that(4.2) ω ( B ( x, r )) Z B ( x,r ) | h − h ( x ) | dω ≤ κ h ( x ) for all r ∈ (0 , δ ] and(4.3) ω ( B ( x, r ) \ E ) ≤ κ ω ( B ( x, r )) for all r ∈ (0 , δ ] .The parameter κ will be fixed below, and depends only on n . We take now a radius e r ∈ (cid:0) , min( r E , δ , | p − x | ) (cid:1) such that (4.1) holds for e r = r k (recall that r E is defined by local CDC in (1.1)), and we denote B = B ( x , e r ) . From (4.2) we deduce that, for all r ∈ (0 , r ( B )] , µ ( B ( x , r )) = Z B ( x ,r ) h dω ≤ h ( x ) ω ( B ( x , r )) + Z B ( x ,r ) | h − h ( x ) | dω ≤ h ( x ) ω ( B ( x , r )) . Analogously, µ ( B ( x , r )) ≥ h ( x ) ω ( B ( x , r )) − Z B ( x ,r ) | h − h ( x ) | dω ≥ h ( x ) ω ( B ( x , r )) . We collect some of the properties about B in the next lemma. Lemma 4.1.
For all r ∈ (0 , r ( B )] , we have (4.4) h ( x ) ω ( B ( x , r )) ≤ µ ( B ( x , r )) ≤ h ( x ) ω ( B ( x , r )) . We also have (4.5) ω (100 B ) ≤ (200) n +1 ω ( B ) and µ (100 B ) ≤ n +1 µ ( B ) . Proof.
The estimates in (4.4) have been shown above, as well as the first inequality in (4.5). The secondinequality follows from the preceding estimates: µ (100 B ) ≤ h ( x ) ω (100 B ) ≤ h ( x ) (200) n +1 ω ( B ) ≤ n +1 µ ( B ) . (cid:3) ARMONIC MEASURE AND HAUSDORFF MEASURE 9
The bad balls and the function d ( · ) . We consider the constant(4.6) A = 4 ω (5 B ) µ ( B ) . Notice that, by Lemma 4.1, A ≈ h ( x ) − . For each x ∈ ∂ Ω ∩ B and r ∈ (0 , r ( B )] , we say that the ball B ( x, r ) is bad (and we write B ( x, r ) ∈ Bad ) if ω ( B ( x, r )) > A µ ( B ( x, r )) . Given some fixed parameter ρ ∈ (0 , r ( B )] , if there exists some r ∈ ( ρ , r ( B )] such that B ( x, r ) is bad, we denote(4.7) r ( x ) = sup (cid:8) r ∈ ( ρ , r ( B )] : B ( x, r ) is bad (cid:9) . Otherwise, we set r ( x ) = ρ . Using the openness of the balls in the definition of r ( x ) , it is easy to check that the supremum in (4.7)is attained and thus the ball B ( x, r ( x )) is bad if r ( x ) > ρ .Next we define the following regularized version of r ( · ) : d ( x ) = inf y ∈ B ∩ ∂ Ω ( r ( y ) + | x − y | ) , for x ∈ R n +1 . It is immediate to check that d ( · ) is a -Lipschitz function. Further, since r ( x ) ≤ r ( B ) for any x ∈ B ∩ ∂ Ω , we infer that d ( x ) ≤ r ( B ) for any x ∈ B ∩ ∂ Ω too.We need the following auxiliary result. Lemma 4.2.
Let x ∈ B ∩ ∂ Ω . For all r ∈ [ d ( x ) , r ( B )] , (4.8) ω ( B ( x, r )) ≤ A µ ( B ( x, r )) . Proof.
Suppose first that r ≥ r ( B ) / . In this case, using just that B ( x, r ) ⊂ B , B ⊂ B ( x, r ( B )) ,and the choice of A in (4.6), we infer that ω ( B ( x, r )) ≤ ω (3 B ) ≤ A µ ( B ) ≤ A µ ( B ( x, r ( B ))) ≤ A µ ( B ( x, r )) . Assume now that r < r ( B ) / . Let y ∈ B ∩ ∂ Ω be such that d ( x ) ≥ r ( y ) + | x − y | . Using that B ( x, r ) ⊂ B ( y, | x − y | + r ) ⊂ B ( y, r ) (because | x − y | ≤ d ( x ) ≤ r ) and that r ≥ d ( x ) ≥ r ( y ) and r ≤ r ( B ) , we get ω ( B ( x, r )) ≤ ω ( B ( y, r )) ≤ A µ ( B ( y, r )) . Now we take into account that B ( y, r ) ⊂ B ( x, | x − y | + 30 r ) ⊂ B ( x, r ) (again because | x − y | ≤ r ), and we derive ω ( B ( x, r )) ≤ A µ ( B ( y, r )) ≤ A µ ( B ( x, r )) . (cid:3) Now we apply Vitali’s r -covering theorem to get a finite subfamily of balls(4.9) { B i } i ∈ I ⊂ { B ( x, d ( x )) } x ∈ B ∩ ∂ Ω such that • the balls B i , i ∈ I , are pairwise disjoint, and • S x ∈ B ∩ ∂ Ω B ( x, d ( x )) ⊂ S i ∈ I B i . In the next lemma we show some elementary properties of the family { B i } i ∈ I . Lemma 4.3.
Let { B i } i ∈ I be the family of balls defined above. The following holds: (a) For each i ∈ I , r ( B i ) ≤ r ( B ) and B i ⊂ B . (b) For all x ∈ B i , with i ∈ I , r ( B i ) ≤ d ( x ) ≤ r ( B i ) . (c) If B i ∩ B j = ∅ , for i, j ∈ I , then r ( B i ) ≤ r ( B j ) ≤ r ( B i ) . (d) The balls B i , i ∈ I , have finite superposition. That is, X i ∈ I χ B i ≤ C , for some absolute constant C .Proof. Denote by x i the center of B i , so that B i = B ( x i , d ( x i )) . The statement in (a) is due to thefact that, for each i ∈ I , we have r ( B i ) = 12000 d ( x i ) ≤ r ( x i ) ≤ r ( B ) , with x i ∈ B .On the other hand, notice that, for all x ∈ B i , | d ( x ) − d ( x i ) | ≤ | x − x i | ≤ d ( x i ) , and thus d ( x i ) ≤ d ( x ) ≤ d ( x i ) , which gives (b).Concerning (c), given B i and B j with non-empty intersection, we consider x ∈ B i ∩ B j and we deduce that d ( x i ) ≤ d ( x ) ≤ d ( x j ) . Together with the converse estimate, this shows that r ( B i ) ≤ r ( B j ) ≤ r ( B i ) . To prove (d), let B i ,. . . , B i m be such that m \ j =1 B i j = ∅ . Suppose that B i has maximal radius among the balls B i ,. . . , B i m , so that m [ j =1 B i j ⊂ B i . ARMONIC MEASURE AND HAUSDORFF MEASURE 11
Since the balls B i ,. . . , B i m are pairwise disjoint, by the properties (c), (b) and the usual volumeconsiderations we deduce that m n +1 r ( B i ) n +1 ≤ m X j =1 r ( B i j ) n +1 ≤ (3000 r ( B i )) n +1 , and thus m . . (cid:3) The function ψ . Let ϕ be a radial C ∞ function such that χ B (0 , . ≤ ϕ ≤ χ B (0 , . , and let(4.10) ϕ i ( x ) = ϕ (cid:16) x − x i r i (cid:17) , where x i is the center of B i and r i its radius. Notice that ϕ ≡ on . B i and vanishes out of B i . Nextwe need to define some auxiliary functions θ j . First, by applying the r -covering theorem, we considera covering of B \ S i ∈ I . B i with balls of the form B ( z j , − d ( z j )) , with z j ∈ B \ S i ∈ I . B i ,so that the balls B ( z j , − d ( z j )) are disjoint. This implies that the dilated balls . B ( z j , − d ( z j )) have finite superposition, by arguments analogous to the ones in Lemma 4.3. For each j ∈ J , we define θ j ( x ) = ϕ (cid:16) x − z j − d ( z j ) (cid:17) . In this way, using the property (b) in the preceding lemma, for any j ∈ J ,(4.11) supp θ j ∩ [ i ∈ I B i = ∅ . We consider the functions e ϕ i = ϕ i P j ∈ I ϕ j + P j ∈ j θ j , i ∈ I. Notice that the denominator above is bounded away from in supp ϕ i , and thus e ϕ i ∈ C ∞ , with k∇ e ϕ i k ∞ . r − i . Further, by construction ≤ X i ∈ I e ϕ i ≤ in R n +1 . Also, taking into account (4.11), X i ∈ I e ϕ i ≡ in S i ∈ I B i and, since supp e ϕ i ⊂ B i , X i ∈ I e ϕ i ≡ in R n +1 \ S i ∈ I B i .We also denote ψ = (cid:18) − X i ∈ I e ϕ i (cid:19) ϕ (cid:18) x − x r ( B ) (cid:19) (recall that x is the center of B ). Finally, we let ψ = ψ . Lemma 4.4.
The following holds: (a) supp ψ ⊂ B \ S i ∈ I B i and ψ ≡ in B \ S i ∈ I B i . (b) supp ∇ ψ ⊂ S i ∈ I A ( x i , r i , r i ) ∪ A ( x , r ( B ) , r ( B )) . (c) |∇ ψ ( x ) | . r ( B i ) for all x ∈ B i . (d) |∇ ψ ( x ) | . r ( B ) for all x ∈ B \ S i ∈ I B i .The same properties hold for ψ . The proof of the lemma follows easily from the construction above, and we leave it for the reader.4.4.
The sets V , e V , and F . By Vitali’s r -covering theorem, there exists a subfamily Bad V ⊂ Bad such that • the balls from Bad V are pairwise disjoint, and • any ball from Bad is contained in some ball B , with B ∈ Bad V .We denote V = [ B ∈ Bad V B, e V = [ B ∈ Bad V B. Notice that V ⊂ e V and that all the bad balls are contained in V (not only the ones with radius largerthan ρ ).In the next lemma we show that e V is rather small, because of our choice of A above. Lemma 4.5.
We have µ (cid:0) e V (cid:1) ≤ X B ∈ Bad V µ (10 B ) ≤ µ ( B ) . Proof.
By the definition of bad balls and the disjointness of the family
Bad V , we get µ ( e V ) ≤ X B ∈ Bad V µ (10 B ) ≤ A X B ∈ Bad V ω ( B ) ≤ A ω (5 B ) , where, in the last inequality, we took into account that the bad balls are centered at B and have radiusat most r ( B ) . By the choice of A in (4.6), we are done. (cid:3) Next we need to consider another kind of bad set. We let F be the subset of the points x ∈ E ∩ B for which there exists some r ∈ (0 , r ( B )] such that ω ( B ( x, r )) ≤ κ h ( x ) − µ ( B ( x, r )) (recall that κ ∈ (0 , / will be fixed below). Lemma 4.6.
We have µ ( F ) ≤ Cκ µ ( B ) . Proof.
By the Besicovitch covering theorem, there exists a covering of F by a family of balls B ( z i , s i ) ,with z i ∈ F , < s i ≤ r ( B ) / , such that ω ( B ( z i , s i )) ≤ κ h ( x ) − µ ( B ( z i , r i )) , and having finitesuperposition. That is, P i χ B ( z i ,s i ) . . Then we have: ω ( F ) ≤ X i ω ( B ( z i , s i )) ≤ κ h ( x ) − X i µ ( B ( z i , s i )) ≤ C κ h ( x ) − µ ( B ) ≤ C κ h ( x ) − µ ( B ) , ARMONIC MEASURE AND HAUSDORFF MEASURE 13 taking into account that all the balls B ( z i , s i ) are contained in B and the finite superposition of theballs in the before to last inequality, and the fact that B is doubling with respect to µ , by (4.5), in thelast inequality. As a consequence, µ ( F ) = Z F h dω ≤ h ( x ) ω ( F ) + Z B | h − h ( x ) | dω ≤ Cκ µ ( B ) + κ h ( x ) ω ( B ) ≤ C κ µ ( B ) . (cid:3)
5. P
ROOF OF T HEOREM s > n be as in Theorem 1.1. Recall that we assume s ∈ ( n, n + 1) and we denote a = s − n .Also, we take α = 1 − a a , so that both a, α ∈ (0 , . We will apply the identity (3.1) with u equal to the Green function g ( · , p ) ,the function ψ constructed in Section 4, and the preceding value of α . Recall that ψ supported in B and vanishes in a neighborghood of ∂ Ω ∩ B . Thus, g = g ( · , p ) is harmonic in supp ψ . Recall alsothat we have | α ( α − | Z |∇ g | g α − ψ dx (5.1) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z ∇ ( |∇ g | ) · ∇ ψ g α dx (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)Z |∇ g | ∇ ( g α ) · ∇ ψ dx (cid:12)(cid:12)(cid:12)(cid:12) − Z |∇ g | g α ψ dx. To achieve the desired contradiction to prove Theorem 1.1 we will show that the integral on the lefthand side tends to ∞ as ρ → , while the right hand side is much smaller than the left hand side.We denote I = Z |∇ g | g α − ψ dx,I = Z ∇ ( |∇ g | ) · ∇ ψ g α dx,I = Z |∇ g | ∇ ( g α ) · ∇ ψ dx,I = Z |∇ g | g α ψ dx. Estimate of I . Using the fact that |∇ ( |∇ g | ) | . |∇ g | |∇ g | and H ¨older’s inequality and recallingthat ψ = ψ , we get | I | . Z |∇ g | |∇ g | g α ψ |∇ ψ | dx ≤ (cid:18)Z |∇ g | g α ψ dx (cid:19) / (cid:18)Z |∇ g | g α ψ |∇ ψ | dx (cid:19) / . Observe that the first integral on the right hand side coincides with I . To deal with the last one, wesplit it as follows:(5.2) Z |∇ g | g α ψ |∇ ψ | dx ≤ X i ∈ I Z B i . . . + Z B \ S i ∈ I B i . . . . By Lemma 4.4 (c) and Caccioppoli’s inequality, for each i ∈ I we obtain Z B i |∇ g | g α |∇ ψ | dx . r i sup B i g ( x ) α Z B i |∇ g | dx (5.3) . r i sup B i g ( x ) α Z B i | g | dx . r n − i sup B i g ( x ) α +2 . By (2.3), we have(5.4) sup B i g ( x ) . ω (96 B i ) r n − i . Therefore, by the choice of α ,(5.5) Z B i |∇ g | g α |∇ ψ | dx . r n − i (cid:18) ω (96 B i ) r n − i (cid:19) α +2 = ω (96 B i ) (cid:18) ω (96 B i ) r si (cid:19) α +1 . By Lemma 4.2,(5.6) ω (96 B i ) ≤ ω ( B ( x i , d ( x i )) ≤ A µ ( B ( x i , d ( x i )) . Ad ( x i ) s ≈ Ar si . Thus,(5.7) Z B i |∇ g | g α |∇ ψ | dx . A α +1 ω (96 B i ) . Finally, Lemma 4.3 (a) and (d),(5.8) X i ∈ I Z B i |∇ g | g α |∇ ψ | dx . A α +1 X i ∈ I ω (96 B i ) . A α +1 ω (3 B ) . A α +2 µ ( B ) . Next we deal with the last integral on the right hand side of (5.2). We argue as in (5.2)-(5.8), but nowwe use the fact that |∇ ψ | . /r ( B ) in B \ S i ∈ I B i and we replace B i by B . Then, as in (5.5),we get(5.9) Z B \ S i ∈ I B i |∇ g | g α |∇ ψ | dx . ω (32 B ) (cid:18) ω (32 B ) r ( B ) s (cid:19) α +1 . Using now (4.5) and (4.6), we derive(5.10) Z B \ S i ∈ I B i |∇ g | g α |∇ ψ | dx . A α +2 µ ( B ) . Altogether, we obtain(5.11) | I | ≤ (cid:0) CA α +2 µ ( B ) (cid:1) / I / ≤ CA α +2 µ ( B ) + I . ARMONIC MEASURE AND HAUSDORFF MEASURE 15
Estimate of I . Using that ∇ ( g α ) = α g α − ∇ g , ∇ ψ = 4 ψ ∇ ψ , and H ¨older’s inequality, weget | I | . Z |∇ g | g α − ψ |∇ ψ | dx ≤ (cid:18)Z |∇ g | g α − ψ dx (cid:19) / (cid:18)Z g α +2 |∇ ψ | dx (cid:19) / . Observe that the first integral on the right hand side equals I . To estimate the second one we split it:(5.12) Z g α +2 |∇ ψ | dx ≤ X i ∈ I Z B i . . . + Z B \ S i ∈ I B i . . . . By Lemma 4.4 (c), for each i ∈ I , we obtain Z B i g α +2 |∇ ψ | dx . r n − i sup B i g ( x ) α +2 . As in (5.4), we have sup B i g ( x ) ≤ sup B i g ( x ) . ω (96 B i ) r n − i . Then, operating exactly as in (5.5)-(5.8), we derive X i ∈ I Z B i g α +2 |∇ ψ | dx . A α +2 µ ( B ) . To estimate the last integral on the right hand side of (5.12) we use the fact that |∇ ψ | . /r ( B ) in B \ S i ∈ I B i and we apply (2.3). Then we get Z B \ S i ∈ I B i g α +2 |∇ ψ | dx . r ( B ) n − sup B g ( x ) α +2 . ω (32 B ) (cid:18) ω (32 B ) r ( B ) s (cid:19) α +1 , which is the same estimate as in (5.9). Then, as in (5.10), we deduce Z B \ S i ∈ I B i g α +2 |∇ ψ | dx . A α +2 µ ( B ) . Therefore, | I | ≤ I / ( A α +2 µ ( B )) / ≤ | α (1 − α ) | I + C ( α ) A α +2 µ ( B ) . Lower estimate of I . From the identity (5.1) and the estimates for I and I we derive | α ( α − | I = | α ( α − | Z |∇ g | g α − ψ dx ≤ I + I − I ≤ (cid:18) CA α +2 µ ( B ) + I (cid:19) + (cid:18) | α (1 − α ) | I + C ( α ) A α +2 µ ( B ) (cid:19) − I . Hence,(5.13) I ≤ C ( α ) A α +2 µ ( B ) . In this section, by estimating I from below, we will contradict this inequality. To get a lower estimate for I we need to define some reasonably good set contained in B ∩ ∂ Ω . Tothis end, we need first to introduce another type of balls. Let I b ⊂ I be the subfamily of indices i suchthat r ( B i ) > ρ . Recall that I is the set of indices in (4.9) and r ( B i ) = d ( x i ) ≤ r ( x i ) .So if i ∈ I b , then r ( x i ) > ρ and thus B ( x i , r ( x i )) is a bad ball. We say that a ball B is useless (andwe write B ∈ Uss ) if it is centered at B ∩ ∂ Ω \ e V and(5.14) µ (cid:18) [ i ∈ I b :6 B i ∩ B = ∅ B i (cid:19) > ε µ ( B ) and µ ( B ) ≥ − s r ( B ) s , where ε ∈ (0 , / is a small parameter to be fixed below that will depend only on n .Recall now that, by Lemma 4.5, X B ∈ Bad V µ (10 B ) ≤ µ ( B ) . Hence there exists some ρ > such that(5.15) X B ∈ Bad V : r ( B ) ≤ ρ µ (10 B ) ≤ ε µ ( B ) . Notice that ρ may depend here on the particular measure µ , not only on n . We define U ( ρ ) = [ B ∈ Uss: r ( B ) ≤ ρ B. Lemma 5.1.
We have µ ( U ( ρ )) . ε µ ( B ) . Proof.
Let B ∈ Uss with r ( B ) ≤ ρ and let B i , i ∈ I b , be such that B i ∩ B = ∅ . Notice that B i is contained in some bad ball (because d ( x i ) > ρ ), which in turn is contained in some ball B ′ , with B ′ ∈ Bad V . Thus, B i ⊂ B ′ . Now note that B is centered at ∂ Ω \ e V ⊂ (10 B ′ ) c , and observethat the condition B i ∩ B = ∅ implies that B intersects B ′ . These two facts ensure that(5.16) r ( B ) ≥ r (5 B ′ ) ≥ r (2000 B i ) . Then we deduce that B i ⊂ B i ⊂ B. The first inequality in (5.16) also implies that r ( B ′ ) ≤ ρ , which in turn gives that B i ⊂ [ B ′′ ∈ Bad V : r ( B ′′ ) ≤ ρ B ′′ ⊂ [ B ′′ ∈ Bad V : r ( B ′′ ) ≤ ρ B ′′ =: e V , with(5.17) µ ( e V ) ≤ ε µ ( B ) , by (5.15). From the first condition in (5.14), we deduce that µ (3 B ∩ e V ) ≥ µ (cid:18) [ i ∈ I b :6 B i ∩ B = ∅ B i (cid:19) > ε µ ( B ) . Using also the fact that µ (15 B ) . r ( B ) s . µ ( B ) (by the second condition in (5.14)), we get(5.18) µ (3 B ∩ e V ) ≥ c ε µ (15 B ) . ARMONIC MEASURE AND HAUSDORFF MEASURE 17
Now we apply the r -covering theorem to get a subfamily I U from the balls in Uss with radius notexceeding ρ such that • the balls B with B ∈ I U are pairwise disjoint, and • U ( ρ ) ⊂ S B ∈ I U B .From these properties and (5.18) and (5.17), we obtain µ ( U ( ρ )) ≤ X B ∈ I U µ (15 B ) . ε X B ∈ I U µ (3 B ∩ e V ) ≤ ε µ ( e V ) ≤ ε µ ( B ) . (cid:3) Now we are ready to define the aforementioned reasonably good set contained in B ∩ ∂ Ω . Firstdenote G = B ∩ ∂ Ω \ (cid:0) F ∪ e V ∪ U ( ρ ) (cid:1) , and recall that, by Lemmas 4.6, 4.5, and 5.1, µ ( G ) ≥ µ ( B ) − Cκ µ ( B ) − µ ( B ) − Cε µ ( B ) . We assume κ to be an absolute constant small enough so that Cκ ≤ / and also ε small enough sothat Cε ≤ / , and then we obtain µ ( G ) ≥ µ ( B ) . Next we need to define some families of balls centered at G inductively. Let G be the subset ofthose x ∈ G such that Θ s, ∗ ( x, µ ) ≥ − s . Note that, by (2.1), µ ( G \ G ) = 0 . By definition, for each η k ∈ (0 , r ( B ) / , for µ -a.e. x ∈ G there exists a ball B ix centered at x with radius r ( B ix ) ≤ η k suchthat(5.19) µ ( B ix ) ≥ − s r ( B ix ) s . Hence, by the r -covering theorem, we can extract a subfamily e F k ⊂ { B ix } x ∈ G such that(a) G ⊂ S B ∈ e F k B, and(b) the balls B , B ∈ e F k , are disjoint.Further, we can still extract a finite subfamily F k ⊂ e F k such that(5.20) µ (cid:18) [ B ∈F k B (cid:19) ≥ µ ( G ) ≥ µ ( B ) . Now we fix inductively the parameters η k as follows: first we take η = r ( B ) / , and next we set η k = ε min B ∈F k − r ( B ) , where ε ∈ (0 , / is some small constant to be chosen below. Notice that this choice ensures thatthe balls from the family F k are much smaller than the ones of the preceding families F , . . . , F k − .Remark that the balls B ( x, r ) centered at G (like the balls from the families F k ) satisfy a couple ofcrucial properties: • For r ∈ (0 , r ( B ) / ,(5.21) ω ( B ( x, r )) ≥ κ h ( x ) − µ ( B ( x, r )) , by the construction of F and G . • For r ∈ [ ρ , r ( B )] ,(5.22) ω ( B ( x, r )) ≤ A µ ( B ( x, r )) ≤ CA r s ≈ h ( x ) − r s , taking into account that B ( x, r ) does not belong to Bad and using (4.8) and the choice of A .The next lemma contains the key estimate that will allow to bound I from below. Lemma 5.2.
Let B ∈ F k \ Uss and suppose that ρ ≤ η k +1 . Denote (5.23) e B = B \ (cid:18) [ B ′ ∈F k +1 B ′ ∪ [ i ∈ I B i (cid:19) . Then, (5.24) Z e B |∇ g | g α − ψ dx & A α +2 µ ( B ) . Proof.
Denote by x B the center of B and let ϕ B ( y ) = ϕ (cid:16) y − x B r ( B ) (cid:17) , where ϕ is the radial C ∞ function appearing in (4.10), so that supp ϕ B ⊂ B , ϕ ≡ in B , and k∇ ϕ B k ∞ . /r ( B ) . Then we have r ( B ) Z B |∇ g | dx & (cid:12)(cid:12)(cid:12)(cid:12)Z ∇ g · ∇ ϕ B dx (cid:12)(cid:12)(cid:12)(cid:12) = Z ϕ B dω ≥ ω ( B ) ≥ κ h ( x ) − µ ( B ) , taking into account (5.21) for the last inequality.Next we will show that r ( B ) R B \ e B |∇ g | dx is small if the parameters ε and ε above are small.To this end, given any ball B ′ intersecting B and centered at x B ′ ∈ B ∩ ∂ Ω with radius r ( B ′ ) ∈ [ d ( x B ′ ) , r ( B )] , we write: Z B ∩ B ′ |∇ g | dx . r ( B ′ ) n sup B ′ g ( x ) , applying H ¨older’s inequality and Caccioppoli’s inequality. By (2.3), the last supremum can be boundedabove by ω (16 B ′ ) r ( B ′ ) − n , and then we get(5.25) Z B ∩ B ′ |∇ g | dx . r ( B ′ ) ω (16 B ′ ) . We split(5.26) Z B \ e B |∇ g | dx ≤ Z B ∩ S B ′∈F k +1 B ′ |∇ g | dx + Z B ∩ S i ∈ I \ Ib B i |∇ g | dx + Z B ∩ S i ∈ Ib B i |∇ g | dx. To deal with the first integral on the right hand side, recall that the balls B ′ , with B ′ ∈ F k +1 , aredisjoint and their radius is at most ε r ( B ) , by construction. Thus, r ( B ) Z B ∩ S B ′∈F k +1 B ′ |∇ g | dx . r ( B ) X B ′ ∈F k +1 :16 B ′ ⊂ B r ( B ′ ) ω (16 B ′ ) . ε ω (2 B ) . ARMONIC MEASURE AND HAUSDORFF MEASURE 19
The second integral on the right hand side of (5.26) is estimated analogously. In this case we use thatall the balls B i with i ∈ I \ I b have radius equal to ρ / ≪ r ( B ) , and that the balls B i , i ∈ I \ I b ,have bounded overlap, by Lemma 4.3 (d). Then, by (5.25), we deduce r ( B ) Z B ∩ S i ∈ I \ Ib B i |∇ g | dx . r ( B ) X i ∈ I \ I b :96 B i ⊂ B r ( B i ) ω (96 B i ) . ε ω (2 B ) . Regarding the last integral on the right hand side of (5.26), we will use the fact that(5.27) µ (cid:18) [ i ∈ I b :6 B i ∩ B = ∅ B i (cid:19) ≤ ε µ ( B ) , because B is assumed to be not useless. Note that given i ∈ I b such that B ∩ B i = ∅ , we have r (6 B i ) ≤ r ( B ) . Otherwise, B ⊂ B i , which violates the condition (5.27). Then, applying (5.25)again, we deduce r ( B ) Z B ∩ S i ∈ Ib B i |∇ g | dx . r ( B ) X i ∈ I b :6 B i ∩ B = ∅ r ( B i ) ω (96 B i ) . X i ∈ I b :6 B i ∩ B = ∅ ω (96 B i ) . A X i ∈ I b :6 B i ∩ B = ∅ µ (960 B i ) . A µ (cid:18) [ i ∈ I b :6 B i ∩ B = ∅ B i (cid:19) . Aε µ (2 B ) , by the finite superposition of the balls B i and (5.27).We take into account now that, by (5.22), the s -growth of µ , and (5.19), ω (2 B ) . h ( x ) − r (2 B ) s ≈ h ( x ) − r ( B ) s . h ( x ) − µ ( B ) . Therefore, r ( B ) Z B ∩ S B ′∈F k +1 B ′ |∇ g | dx + 1 r ( B ) Z B ∩ S i ∈ I \ Ib B i |∇ g | dx . ε h ( x ) − µ ( B ) . Then, using also that µ ( B ) & r s & µ (2 B ) , h ( x ) − ≈ A , and recalling the splitting (5.26), wededuce r ( B ) Z e B |∇ g | dx ≥ ( κ − Cε − Cε ) h ( x ) − µ ( B ) ≥ κ h ( x ) − µ ( B ) ≈ κ A µ ( B ) ≈ κ A µ ( B ) , assuming ε and ε small enough.Next, applying H ¨older’s inequality, we get Z e B |∇ g | dx ≤ (cid:18)Z e B |∇ g | g α − dx (cid:19) / (cid:18)Z B g (2 − α ) / dx (cid:19) / . To estimate the last integral on the right hand side we take into account that, by (2.3) and (5.22), g ( x ) . ω (8 B ) r ( B ) n − . A µ (80 B ) r ( B ) n − ≈ A µ ( B ) r ( B ) n − for all x ∈ B . Hence, Z B g (2 − α ) / dx . A (2 − α ) / µ ( B ) (2 − α ) / r ( B ) n +1 − (2 − α )( n − / . Therefore, Z e B |∇ g | g α − dx & (cid:18) Z e B |∇ g | dx (cid:19) (cid:16) A (2 − α ) / µ ( B ) (2 − α ) / r ( B ) n +1 − (2 − α )( n − / (cid:17) − & (cid:0) κ A r ( B ) µ ( B ) (cid:1) (cid:16) A (2 − α ) / µ ( B ) (2 − α ) / r ( B ) n +1 − (2 − α )( n − / (cid:17) − = κ Aµ ( B ) (cid:18) Aµ ( B ) r ( B ) s (cid:19) α ≈ κ A α µ ( B ) . To finish the proof of the lemma it just remains to notice that κ is some absolute constant dependingjust on n and that ψ = 1 on e B . (cid:3) Now we are ready to obtain the lower estimate for I required to complete the proof of Theorem 1.1.Let N > be an arbitrarily large integer, and choose ρ ∈ (0 , η N +1 ] such that ρ ≪ ρ too. Let k bethe minimal integer such that η k ≤ ρ . If B ∈ F k with k ∈ [ k , N ] , then r ( B ) ≤ ρ by constructionand so B Uss (since B is centered in U ( ρ ) c ). Then we can apply Lemma 5.2 to deduce that Z e B |∇ g | g α − ψ dx & A α +2 µ ( B ) , with e B defined in (5.23). Then we get Z |∇ g | g α − ψ dx ≥ N X k = k X B ∈F k Z e B |∇ g | g α − ψ dx & N X k = k X B ∈F k A α +2 µ ( B ) , using the fact that the regions e B above do not overlap. Now, by the doubling property of the balls from F k and (5.20), for each k we have X B ∈F k µ ( B ) ≈ X B ∈F k µ (80 B ) ≥ µ ( G ) ≥ µ ( B ) . Thus, I = Z |∇ g | g α − ψ dx & ( N − k ) A α +2 µ ( B ) . Taking N big enough, we contradict (5.13), as wished.R EFERENCES [AHM TV] J. Azzam, S. Hofmann, J.M. Martell, S. Mayboroda, M. Mourgoglou, X. Tolsa, and A. Volberg.
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