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UNIFORM RECTIFIABILITY AND HARMONIC MEASURE II:POISSON KERNELS IN Lp IMPLY UNIFORM RECTIFIABILITY

STEVE HOFMANN, JOSE MARIA MARTELL, AND IGNACIO URIARTE-TUERO

Abstract. We present the converse to a higher dimensional, scale-invariant ver-sion of a classical theorem of F. and M. Riesz [RR]. More precisely, forn ≥ 2, foran ADR domainΩ ⊂ Rn+1 which satisfies the Harnack Chain condition plus aninterior (but not exterior) Corkscrew condition, we show that absolute continuityof harmonic measure with respect to surface measure on∂Ω, with scale invari-ant higher integrability of the Poisson kernel, is sufficient to imply uniformlyrectifiable of∂Ω.

Contents

1. Introduction 2

1.1. Notation and Definitions 3

1.2. Statement of the Main Result 7

2. Proof of Theorem1.22 8

2.1. Preliminaries 8

2.2. Step 1: Passing to the approximating domains 13

2.3. Step 2: LocalTb theorem for square functions 21

2.4. Step 3: Good-λ inequality for the square function and thenon-tangential maximal function 27

2.5. Step 4: Proof of (2.61) 36

2.6. UR forΩ 37

References 40

Date: September 25, 2012.2010Mathematics Subject Classification.31B05, 35J08, 35J25, 42B99, 42B25, 42B37.Key words and phrases.Harmonic measure, Poisson kernel, uniform rectifiability,Carleson mea-

sures,A∞ Muckenhoupt weights.The first author was supported by NSF grant DMS-0801079. The second author was supported

by MINECO Grant MTM2010-16518 and ICMAT Severo Ochoa project SEV-2011-0087.This work has been possible thanks to the support and hospitality of the University of Missouri-

Columbia (USA), the Consejo Superior de Investigaciones Cientıficas(Spain), theUniversidadAutonoma de Madrid(Spain), and theAustralian National University(Canberra, Australia). Thefirst two authors would like to express their gratitude to these institutions.

1

http://arxiv.org/abs/1202.3860v2

2 STEVE HOFMANN, JOSE MARIA MARTELL, AND IGNACIO URIARTE-TUERO

1. Introduction

This paper is a sequel to the work of the first two named authors[HM], in whichwe have presented a higher dimensional, scale invariant version of the classicaltheorem of F. and M. Riesz [RR]. The F. and M. Riesz Theorem states that fora simply connected domainΩ in the complex plane, with a rectifiable boundary,one has that harmonic measure is absolutely continuous withrespect to arclengthmeasure on the boundary. In [HM], we proved a scale invariant version of the latterresult in higher dimensions. That is, we showed that forΩ ⊂ Rn+1, n ≥ 2, satisfy-ing certain quantitative topological properties, whose boundary is rectifiable in anappropriate quantitative sense, one has that harmonic measure satisfies a scale in-variant version of absolute continuity with respect to surface measure (the so-called“weak-A∞” property; cf. Definition1.15below). To be more precise, assuming thatΩ satisfies interior “Corkscrew” and “Harnack Chain” conditions (these are scaleinvariant versions of the topological properties of openness and path connected-ness; cf. Definitions1.1and1.3below), and that∂Ω is “Uniformly Rectifiable” (aquantitative, scale invariant version of rectifiability; cf. Definition 1.6), we showedthat harmonic measure belongs to weak-A∞ with respect to surface measure on∂Ω. Let us note that the weak-A∞ property implies that the Poisson kernel (i.e., theRadon-Nikodym derivative of harmonic measure with respectto surface measure),satisfies a scale-invariant higher integrability condition (cf. (1.20).)

In the present paper, we obtain a converse to the main result of [HM], that is, weshow that ifΩ ⊂ Rn+1 satisfies interior Corkscrew and Harnack Chain conditions,if∂Ω is n-dimensional “Ahlfors-David Regular” (cf. Definition1.4), and if harmonicmeasure is absolutely continuous with respect to surface measure, with Poissonkernel satisfying the scale invariant higher integrability condition (1.20), then∂Ωis Uniformly Rectifiable (cf. Theorem1.22). We observe that this result is in thespirit of the solution of the Painleve problem ([To], but see also [Ch], [MMV ], [Da]and [Vo]), in which analytic information is used to establish rectifiability propertiesof a set, via the use ofTbtheory. In our case, we use a so called “localTb” theorem,related to the technology of the solution of the Kato square root problem [HMc],[HLMc], [AHLMcT], and proved in [GM].

We observe that our work here and in [HM] may be viewed as a “large constant”analogue of the series of papers by Kenig and Toro [KT1, KT2, KT3]. Thesepapers say that in the presence of a Reifenberg flatness condition and Ahlfors-David regularity, logk ∈ VMO iff ν ∈ VMO, wherek is the Poisson kernelwith pole at some fixed point, andν is the unit normal to the boundary. More-over, given the same background hypotheses, the condition that ν ∈ VMO isequivalent to a uniform rectifiability (UR) condition with vanishing trace, thuslogk ∈ VMO ⇐⇒ vanishing UR. On the other hand, our large constant ver-sion “almost” says “ logk ∈ BMO ⇐⇒ UR”, given interior Corkscrews andHarnack Chains. Indeed, it is well known that theA∞ condition (i.e., weak-A∞plus the doubling property) implies that logk ∈ BMO, while if log k ∈ BMOwithsmall norm, thenk ∈ A∞.

We also point out that another antecedent of our main result here has appeared inthe work of Lewis and Vogel [LV]. The main emphasis of the latter pair of authors

UNIFORM RECTIFIABILITY AND HARMONIC MEASURE II 3

concerns so called “p-harmonic measure” (the analogue of harmonic measure cor-responding to thep-Laplacian), but specializing their results to the classical Lapla-cian, they show that for a domainΩ, for which harmonic measureω is an Ahlfors-David regular measure on∂Ω (thus, the Poisson kernelk = dω/dσ is boundedbetween two positive constants), then∂Ω is uniformly rectifiable. The assumptionthatk ≈ 1 is of course the strongest form of the higher integrability/(weak) reverseHolder conditions that we consider here.

We refer the reader to the introduction of [HM], for a detailed historical surveyof related work.

1.1. Notation and Definitions.

• We use the lettersc,C to denote harmless positive constants, not necessarily thesame at each occurrence, which depend only on dimension and the constants ap-pearing in the hypotheses of the theorems (which we refer to as the “allowableparameters”). We shall also sometimes writea . b anda ≈ b to mean, respec-tively, thata ≤ Cb and 0< c ≤ a/b ≤ C, where the constantsc andC are asabove, unless explicitly noted to the contrary. At times, weshall designate byMa particular constant whose value will remain unchanged throughout the proofof a given lemma or proposition, but which may have a different value duringthe proof of a different lemma or proposition.

• Given a domainΩ ⊂ Rn+1, we shall use lower case lettersx, y, z, etc., to denotepoints on∂Ω, and capital lettersX,Y,Z, etc., to denote generic points inRn+1

(especially those inRn+1 \ ∂Ω).

• The open (n+ 1)-dimensional Euclidean ball of radiusr will be denotedB(x, r)when the centerx lies on∂Ω, or B(X, r) when the centerX ∈ Rn+1 \ ∂Ω. A“surface ball” is denoted∆(x, r) := B(x, r) ∩ ∂Ω.• Given a Euclidean ballB or surface ball∆, its radius will be denotedrB or r∆,

respectively.

• Given a Euclidean or surface ballB = B(X, r) or∆ = ∆(x, r), its concentric dilateby a factor ofκ > 0 will be denoted byκB := B(X, κr) or κ∆ := ∆(x, κr).

• For X ∈ Rn+1, we setδ(X) := dist(X, ∂Ω).

• We letHn denoten-dimensional Hausdorff measure, and letσ := Hn∣∣∂Ω

denotethe “surface measure” on∂Ω.

• For a Borel setA ⊂ Rn+1, we let 1A denote the usual indicator function ofA, i.e.1A(x) = 1 if x ∈ A, and 1A(x) = 0 if x < A.

• For a Borel setA ⊂ Rn+1, we let int(A) denote the interior ofA. If A ⊂ ∂Ω,then int(A) will denote the relative interior, i.e., the largest relatively open setin ∂Ω contained inA. Thus, forA ⊂ ∂Ω, the boundary is then well defined by∂A := A \ int(A).

• For a Borel setA, we denote byC(A) the space of continuous functions onA, byCc(A) the subspace ofC(A) with compact support inA, and byCb(A) the space


of bounded continuous functions onA. If A is unbounded, we denote byC0(A)the space of continuous functions onA converging to 0 at infinity.

• For a Borel subsetA ⊂ ∂Ω, we set>

Af dσ := σ(A)−1

∫A f dσ.

• We shall use the letterI (and sometimesJ) to denote a closed (n+1)-dimensionalEuclidean cube with sides parallel to the co-ordinate axes,and we letℓ(I ) denotethe side length ofI . We useQ to denote a dyadic “cube” on∂Ω. The latter exist,given that∂Ω is ADR (cf. [DS1], [Ch]), and enjoy certain properties which weenumerate in Lemma1.11below.

Definition 1.1. (Corkscrew condition). Following [JK], we say that a domainΩ ⊂ Rn+1 satisfies the “Corkscrew condition” if for some uniform constant c > 0and for every surface ball∆ := ∆(x, r), with x ∈ ∂Ω and 0< r < diam(∂Ω), thereis a ballB(X∆, cr) ⊂ B(x, r) ∩ Ω. The pointX∆ ⊂ Ω is called a “Corkscrew point”relative to∆. We note that we may allowr < C diam(∂Ω) for any fixedC, simplyby adjusting the constantc.

Remark1.2. We note that, on the other hand, everyX ∈ Ω, with δ(X) < diam(∂Ω),may be viewed as a Corkscrew point, relative to some surface ball ∆ ⊂ ∂Ω. Indeed,setr = Kδ(X), with K > 1, fix x ∈ ∂Ω such that|X− x| = δ(X), and let∆ := ∆(x, r).

Definition 1.3. (Harnack Chain condition). Again following [JK], we say thatΩ satisfies the Harnack Chain condition if there is a uniform constantC such thatfor everyρ > 0, Λ ≥ 1, and every pair of pointsX,X′ ∈ Ω with δ(X), δ(X′) ≥ ρand|X − X′| < Λ ρ, there is a chain of open ballsB1, ..., BN ⊂ Ω, N ≤ C(Λ), withX ∈ B1, X′ ∈ BN, Bk ∩ Bk+1 , Ø andC−1 diam(Bk) ≤ dist(Bk, ∂Ω) ≤ C diam(Bk).The chain of balls is called a “Harnack Chain”.

We remark that the Corkscrew condition is a quantitative, scale invariant versionof the fact thatΩ is open, and the Harnack Chain condition is a scale invariantversion of path connectedness.

Definition 1.4. (Ahlfors-David regular ). We say that a closed setE ⊂ Rn+1 isn-dimensional ADR (or simply ADR) if there is some uniform constantC suchthat

(1.5)1C

rn ≤ Hn(E ∩ B(x, r)) ≤ C rn, ∀r ∈ (0,R0), x ∈ E,

whereR0 is the diameter ofE (which may be infinite). WhenE = ∂Ω, the boundaryof a domainΩ, we shall sometimes for convenience simply say that “Ω has theADR property” to mean that∂Ω is ADR.

Definition 1.6. (Uniform Rectifiability ). Following David and Semmes [DS1,DS2], we say that a closed setE ⊂ Rn+1 is n-dimensional UR (or simply UR)(“Uniformly Rectifiable”), if it satisfies the ADR condition(1.5), and if for someuniform constantC and for every Euclidean ballB := B(x0, r), r ≤ diam(E), cen-tered at any pointx0 ∈ E, we have the Carleson measure estimate

(1.7)∫∫

B|∇2S1(X)|2 dist(X,E) dX ≤ Crn,


whereS f is the single layer potential off , i.e.,

(1.8) S f (X) := cn

∫

E|X − y|1−n f (y) dHn(y).

Here, the normalizing constantcn is chosen so thatE(X) := cn|X|1−n is the usualfundamental solution for the Laplacian inRn+1. WhenE = ∂Ω, the boundary ofa domainΩ, we shall sometimes for convenience simply say that “Ω has the URproperty” to mean that∂Ω is UR.

We note that there are numerous characterizations of uniform rectifiability givenin [DS1, DS2]. Let us note that, by “T1 reasoning”, the Carleson measure condition(1.7) is equivalent to the globalL2 bound

(1.9)∫∫

Rn+1|∇2S f (X)|2 δ(X) dX ≤ C ‖ f ‖2L2(∂Ω).

The condition (1.7) will be most useful for our purposes, and appears in [DS2,Chapter 3, Part III]. We remark that the UR sets are preciselythose for which all“sufficiently nice” singular integrals are bounded onL2 (see [DS1]).

Definition 1.10. (“Big Pieces”). Given a closed setE ⊂ Rn+1 such thatE is n-dimensional ADR, and a collectionS of domains inRn+1, we say thatE has “bigpieces of boundaries ofS ” (denotedE ∈ BP(∂S)) if there is a constant 0< α ≤ 1such that for everyx ∈ E, and 0< r < diam(E), there is a domainΩ′ ∈ S such that

Hn(∂Ω′ ∩ B(x, r) ∩ E) ≥ αHn(B(x, r) ∩ E) ≈ αrn.

Lemma 1.11. (Existence and properties of the “dyadic grid”) [DS1, DS2], [Ch].Suppose that E⊂ Rn+1 satisfies the ADR condition(1.5). Then there exist constantsa0 > 0, η > 0 and C1 < ∞, depending only on dimension and the ADR constants,such that for each k∈ Z, there is a collection of Borel sets (“cubes”)

Dk := Qkj ⊂ E : j ∈ Ik,

whereIk denotes some (possibly finite) index set depending on k, satisfying

(i) E = ∪ jQkj for each k∈ Z

(ii ) If m ≥ k then either Qmi ⊂ Qkj or Qm

i ∩ Qkj = Ø.

(iii ) For each( j, k) and each m< k, there is a unique m such that Qkj ⊂ Qm

i .

(iv) Diameter(

Qkj

)≤ C12−k.

(v) Each Qkj contains some “surface ball”∆

(xk

j , a02−k)

:= B(xk

j , a02−k)∩ E.

(vi) Hn(

x ∈ Qkj : dist(x,E \ Qk

j ) ≤ τ2−k)≤ C1 τ

η Hn(

Qkj

), for all k, j and

for all τ ∈ (0, a0).

A few remarks are in order concerning this lemma.

• In the setting of a general space of homogeneous type, this lemma has beenproved by Christ [Ch]. In that setting, the dyadic parameter 1/2 should be re-placed by some constantδ ∈ (0, 1). It is a routine matter to verify that one may


takeδ = 1/2 in the presence of the Ahlfors-David property (1.5) (in this morerestrictive context, the result already appears in [DS1, DS2]).

• For our purposes, we may ignore thosek ∈ Z such that 2−k& diam(E), in the

case that the latter is finite.

• We shall denote byD = D(E) the collection of all relevantQkj , i.e.,

D := ∪kDk,

where, if diam(E) is finite, the union runs over thosek such that 2−k. diam(E).

• Properties (iv) and (v) imply that for each cubeQ ∈ Dk, there is a pointxQ ∈ E,a Euclidean ballB(xQ, r) and a surface ball∆(xQ, r) := B(xQ, r) ∩ E such thatr ≈ 2−k ≈ diam(Q) and

(1.12) ∆(xQ, r) ⊂ Q ⊂ ∆(xQ,Cr),

for some uniform constantC. We shall denote this ball and surface ball by

(1.13) BQ := B(xQ, r) , ∆Q := ∆(xQ, r),

and we shall refer to the pointxQ as the “center” ofQ.

• Let us now specialize to the case thatE = ∂Ω, with Ω satisfying the Corkscrewcondition. GivenQ ∈ D(∂Ω), we shall sometimes refer to a “Corkscrew pointrelative toQ”, which we denote byXQ, and which we define to be the corkscrewpoint X∆ relative to the ball∆ := ∆Q (cf. (1.12), (1.13) and Definition1.1). Wenote that

(1.14) δ(XQ) ≈ dist(XQ,Q) ≈ diam(Q).

• For a dyadic cubeQ ∈ Dk, we shall setℓ(Q) = 2−k, and we shall refer to thisquantity as the “length” ofQ. Evidently,ℓ(Q) ≈ diam(Q).

• For a dyadic cubeQ ∈ D, we letk(Q) denote the “dyadic generation” to whichQ belongs, i.e., we setk = k(Q) if Q ∈ Dk; thus,ℓ(Q) = 2−k(Q).

Definition 1.15. (A∞, Adyadic∞ and weak-A∞). Given a surface ball∆ = B ∩ ∂Ω,

a Borel measureω defined on∂Ω is said to belong to the classA∞(∆) if there arepositive constantsC andθ such that for every∆′ = B′ ∩ ∂Ω with B′ ⊆ B, and everyBorel setF ⊂ ∆′, we have

(1.16) ω(F) ≤ C

(σ(F)σ(∆′)

)θω(∆′).

If we replace the surface balls∆ and∆′ by a dyadic cubeQ and its dyadic subcubesQ′, with F ⊂ Q′, then we say thatω ∈ Adyadic

∞ (Q):

(1.17) ω(F) ≤ C

(σ(F)σ(Q′)

)θω(Q′).

Similarly, ω ∈ weak-A∞(∆), with ∆ = B ∩ ∂Ω, if for every ∆′ = B′ ∩ ∂Ω with2B′ ⊆ B, we have

(1.18) ω(F) ≤ C

(σ(F)σ(∆′)

)θω(2∆′)


As is well known [CF], [GR], [Sa], the A∞ (resp. weak-A∞) condition is equiv-alent to the property that the measureω is absolutely continuous with respect toσ,and that its density satisfies a reverse Holder (resp. weak reverse Holder) condi-tion. In this paper, we are interested in the case thatω = ωX, the harmonic measurewith pole atX. In that setting, we letkX := dωX/dσ denote the Poisson kernel, sothat (1.16) is equivalent to the reverse Holder estimate

(1.19)

(?∆′

(kX)q

dσ

)1/q

≤ C?∆′

kX dσ ,

for someq > 1 and for some uniform constantC. In particular, when∆′ = ∆, andX = X∆, a Corkscrew point relative to∆, the latter estimate reduces to

(1.20)∫

∆

(kX∆)q

dσ ≤ Cσ(∆)1−q.

Similarly, (1.18) is equivalent to

(1.21)

(?∆′

(kX)q

dσ

)1/q

≤ C?

2∆′kX dσ .

Assuming that the latter bound holds with∆′ = ∆, and withX = X∆, then oneagain obtains (1.20).

1.2. Statement of the Main Result. Our main result is as follows. We shall usethe terminology that a connected open setΩ ⊂ Rn+1 is a 1-sided NTA domainif it satisfiesinterior (but not necessarily exterior) Corkscrew and Harnack Chainconditions.

Theorem 1.22.LetΩ ⊂ Rn+1, n ≥ 2, be a 1-sided NTA domain, whose boundaryis n-dimensional ADR. Suppose also that harmonic measureω is absolutely con-tinuous with respect to surface measure and that the Poissonkernel k= dω/dσsatisfies the scale invariant estimate

(1.23)∫

∆

(kX∆)p

dσ ≤ Cσ(∆)1−p,

for some1 < p < ∞ and for all surface balls∆. Then∂Ω is UR.

Remark1.24. As mentioned above (1.23) is (apparently) weaker thanωX∆ beingin weak-A∞(∆). However,a posteriori, we obtain that these two conditions areequivalent. Namely, Theorem1.22 shows that∂Ω is UR and therefore we canapply [HM, Theorem 1.26] to obtain thatωX∆ belongs to weak-A∞(∆).

Theorem1.22leads to an immediate “self-improvement” of itself, in which thehypotheses are assumed to hold only in an appropriate “big pieces” sense.

Theorem 1.25. Let E ⊂ Rn+1 be a closed set and assume that E is n-dimensionalADR. Suppose further that E∈ BP(∂S) (cf. Definition1.10), whereS is a collec-tion of 1-sided NTA domains with n-dimensional ADR boundaries, with uniformcontrol of all of the relevant Corkscrew, Harnack Chain and ADR constants. As-sume also that there exist1 < p < ∞ and C ≥ 1 such that for everyΩ′ ∈ S wehave that harmonic measureωΩ′ is absolutely continuous with respect to surface


measure and that the Poisson kernel kΩ′ = dωΩ′/dσΩ′ satisfies the scale invariantestimate

(1.26)∫

∆′

(kX∆′Ω′

)pdσΩ′ ≤ CσΩ′(∆

′)1−p,

and for all surface balls∆′ onΩ′. Then E is UR.

The proof of this result is as follows. Letx ∈ E and 0< r < diam(E). Underthe hypotheses of Theorem1.25, there isΩ′ satisfying the hypotheses of Theorem1.22, with the property that for some 0< α ≤ 1, we have

(1.27) σ(∂Ω′ ∩ B(x, r) ∩ E

)≥ ασ(B(x, r) ∩ E).

By Theorem1.22, (1.26) implies that∂Ω′ is UR with uniform control of the con-stants (since the ADR and 1-sided NTA constants are uniformly controlled, andalso p andC in (1.26) are independent ofΩ′). This and (1.27) implies thatE hasbig pieces of UR sets and thereforeE is UR, see [DS2].

2. Proof of Theorem 1.22

2.1. Preliminaries. We collect some of the definition and auxiliary results from[HM] that will be used later. In the sequel,Ω ⊂ Rn+1, n ≥ 2, will be a connected,open set,ωX will denote harmonic measure forΩ, with pole atX, andG(X,Y) willbe the Green function. At least in the case thatΩ is bounded, we may, as usual,defineωX via the maximum principle and the Riesz representation theorem, afterfirst using the method of Perron (see, e.g., [GT, pp. 24–25]) to construct a harmonicfunction “associated” to arbitrary continuous boundary data.1 For unboundedΩ,we may still define harmonic measure via a standard approximation scheme, see[HM, Section 3] for more details. We note for future reference thatωX is a non-negative, finite, outer regular Borel measure.

The Green function may now be constructed by setting

(2.1) G(X,Y) := E (X − Y) −∫

∂Ω

E (X − z) dωY(z),

whereE (X) := cn|X|1−n is the usual fundamental solution for the Laplacian inRn+1.We choose the normalization that makesE positive.

Lemma 2.2(Bourgain [Bo]). Suppose that∂Ω is n-dimensional ADR. Then thereare uniform constants c∈ (0, 1) and C ∈ (1,∞), such that for every x∈ ∂Ω, andevery r∈ (0, diam(∂Ω)), if Y ∈ Ω ∩ B(x, cr), then

(2.3) ωY(∆(x, r)) ≥ 1/C > 0 .

In particular, if Ω satisfies the Corkscrew and Harnack Chain conditions, then forevery surface ball∆, we have

(2.4) ωX∆(∆) ≥ 1/C > 0 .

We next introduce some terminology.

1Since we have made no assumption as regards Wiener’s regularity criterion, our harmonic func-tion is a generalized solution, which may not be continuous up to the boundary.


Definition 2.5. A domainΩ satisfies thequalitative exterior Corkscrew condi-tion if there existsN ≫ 1 such thatΩ has exterior corkscrew points at all scalessmaller than 2−N. That is, there exists a constantcN such that for every surface ball∆ = ∆(x, r), with x ∈ ∂Ω andr ≤ 2−N, there is a ballB(Xext

∆ , cN r) ⊂ B(x, r) ∩Ωext.

Let us observe that ifΩ satisfies the qualitative exterior Corkscrew condition,then every point in∂Ω is regular in the sense of Wiener. Moreover, for 1-sided NTAdomains, the qualitative exterior Corkscrew points allow local Holder continuity atthe boundary (albeit with bounds which may depend badly onN).

Lemma 2.6([HM, Lemma 3.11]). There are positive, finite constants C, depend-ing only on dimension, and c(n, θ), depending on dimension andθ ∈ (0, 1), suchthat the Green function satisfies

G(X,Y) ≤ C |X − Y|1−n(2.7)

c(n, θ) |X − Y|1−n ≤ G(X,Y) , if |X − Y| ≤ θ δ(X) , θ ∈ (0, 1) .(2.8)

Moreover, if every point on∂Ω is regular in the sense of Wiener, then

(2.9) G(X,Y) ≥ 0

(2.10) G(X,Y) = G(Y,X) , ∀X,Y ∈ Ω , X , Y .

Lemma 2.11 ([HM, Lemma 3.30]). Let Ω be a 1-sided NTA domain with n-dimensional ADR boundary, and suppose that every x∈ ∂Ω is regular in the senseof Wiener. Fix B0 := B(x0, r0) with x0 ∈ ∂Ω, and∆0 := B0 ∩ ∂Ω. Let B := B(x, r),x ∈ ∂Ω, and∆ := B∩ ∂Ω, and suppose that2B ⊂ B0. Then for X∈ Ω \ B0 we have

(2.12) rn−1G(X∆,X) ≤ CωX(∆).

If, in addition,Ω satisfies the qualitative exterior corkscrew condition, then

(2.13) ωX(∆) ≤ Crn−1G(X∆,X).

The constants in(2.12) and (2.13) dependonly on dimension and on the constantsin the ADR and 1-sided NTA conditions.

Corollary 2.14 ([HM, Corollary 3.36]). Suppose thatΩ is a 1-sided NTA domainwith n-dimensional ADR boundary and that it also satisfies the qualitative exteriorCorkscrew condition. Let B:= B(x, r), x ∈ ∂Ω, ∆ := B∩ ∂Ω and X∈ Ω \ 4B. Thenthere is a uniform constant C such that

ωX(2∆) ≤ CωX(∆).

We next introduce some “discretized” and “geometric” sawtooth and Carlesonregions from [HM, Section 3]. Given a “dyadic cube”Q ∈ D(∂Ω), thediscretizedCarleson regionDQ is defined to be

(2.15) DQ :=

Q′ ∈ D : Q′ ⊆ Q.

Given a familyF of disjoint cubesQ j ⊂ D, we define theglobal discretizedsawtoothrelative toF by

(2.16) DF := D \⋃

FDQ j ,


i.e.,DF is the collection of allQ ∈ D that are not contained in anyQ j ∈ F . Givensome fixed cubeQ, thelocal discretized sawtoothrelative toF by

(2.17) DF ,Q := DQ \⋃

FDQ j = DF ∩ DQ.

We also introduce the “geometric” Carleson regions and sawtooths. Let us firstrecall that we writek = k(Q) if Q ∈ Dk (cf. Lemma1.11), and in that case the“length” of Q is denoted byℓ(Q) = 2−k(Q). We also recall that there is a Corkscrewpoint XQ, relative to eachQ ∈ D (in fact, there are many such, but we just pickone). LetW = W(Ω) denote a collection of (closed) dyadic Whitney cubes ofΩ, so that the cubes inW form a pairwise non-overlapping covering ofΩ, whichsatisfy

(2.18) 4 diam(I ) ≤ dist(4I , ∂Ω) ≤ dist(I , ∂Ω) ≤ 40 diam(I ) , ∀ I ∈ Wand also

(1/4) diam(I1) ≤ diam(I2) ≤ 4 diam(I1) ,

wheneverI1 and I2 touch. Letℓ(I ) denote the side length ofI , and writek = kI ifℓ(I ) = 2−k. There areC0 ≥ 1000

√n andm0 ≥ 0 large enough (depending only on

the constants in the Corkscrew condition and in the dyadic cube construction) sothat for every cubeQ ∈ D(2.19)WQ :=

I ∈ W : k(Q) −m0 ≤ kI ≤ k(Q) + 1 , and dist(I ,Q) ≤ C0 2−k(Q) ,

satisfies thatXQ ∈ I for someI ∈ WQ, and for each dyadic childQ j of Q, therespective Corkscrew pointsXQ j ∈ I j for someI j ∈ WQ. Moreover, we mayalways find anI ∈ WQ with the slightly more precise property thatk(Q)−1 ≤ kI ≤k(Q) and(2.20)

WQ1 ∩WQ2 , Ø , whenever 1≤ ℓ(Q2)ℓ(Q1)

≤ 2 , and dist(Q1,Q2) ≤ 1000ℓ(Q2) .

We introduce some notation: given a subsetA ⊂ Ω, we write X →A Y if theinterior of A contains all the balls in a Harnack Chain (inΩ), connectingX to Y,and if, moreover, for any pointZ contained in any ball in the Harnack Chain, wehave dist(Z, ∂Ω) ≈ dist(Z,Ω \ A) , with uniform control of the implicit constants.We denote byX(I ) the center of a cubeI ∈ Rn+1, and we recall thatXQ denotes adesignated Corkscrew point relative toQ that can be assumed to be the center ofsome Whitney cubeI such thatI ⊂ BQ ∩ Ω andℓ(I ) ≈ ℓ(Q) ≈ dist(I ,Q).

For eachI ∈ WQ, we form a Harnack Chain, call itH(I ), from the centerX(I )to the Corkscrew pointXQ. We now denote byW(I ) the collection of all Whitneycubes which meet at least one ball in the chainH(I ), and we set

W∗Q :=

⋃

I∈WQ

W(I ).

We also define, forλ ∈ (0, 1) to be chosen momentarily,

(2.21) UQ :=⋃

W∗Q

(1+ λ)I =:⋃

I∈W∗Q

I ∗ .


By construction, we then have that

(2.22) WQ ⊂ W∗Q ⊂ W and XQ ∈ UQ , XQ j ∈ UQ ,

for each childQ j of Q. It is also clear that there are uniform constantsk∗ andK0

such that

k(Q) − k∗ ≤ kI ≤ k(Q) + k∗ , ∀I ∈ W∗Q(2.23)

X(I )→UQ XQ , ∀I ∈ W∗Q

dist(I ,Q) ≤ K0 2−k(Q) , ∀I ∈ W∗Q ,

wherek∗, K0 and the implicit constants in the conditionX(I )→UQ XQ, depend onlyon the “allowable parameters” (sincem0 andC0 also have such dependence) andonλ. Thus, by the addition of a few nearby Whitney cubes of diameter also com-parable to that ofQ, we can “augment”WQ so that the Harnack Chain conditionholds inUQ.

We fix the parameterλ so that for anyI , J ∈ W,

dist(I ∗, J∗) ≈ dist(I , J)

int(I ∗) ∩ int(J∗) , Ø ⇐⇒ ∂I ∩ ∂J , Ø(2.24)

(the fattening thus ensures overlap ofI ∗ andJ∗ for any pairI , J ∈ W whose bound-aries touch, so that the Harnack Chain property then holds locally, with constantsdepending uponλ, in I ∗ ∪ J∗). By choosingλ sufficiently small, we may alsosuppose that there is aτ ∈ (1/2, 1) such that for distinctI , J ∈ W,

(2.25) τJ ∩ I ∗ = Ø .

Remark2.26. We note that any sufficiently small choice ofλ (say 0< λ ≤ λ0) willdo for our purposes.

Of course, there may be some flexibility in the choice of additional Whitneycubes which we add to form the augmented collectionW∗

Q, but having made sucha choice for eachQ ∈ D, we fix it for all time.

We may then define theCarleson boxassociated toQ by

(2.27) TQ := int

⋃

Q′∈DQ

UQ′

.

Similarly, we may define geometric sawtooth regions as follows. As above, give afamily F of disjoint cubesQ j ⊂ D, we define theglobal sawtoothrelative toFby

(2.28) ΩF := int

⋃

Q′∈DF

UQ′

,

and again given some fixedQ ∈ D, thelocal sawtoothrelative toF by

(2.29) ΩF ,Q := int

⋃

Q′∈DF ,Q

UQ′

.


We also define as follows the “Carleson box”T∆ associated to a surface ball∆ :=∆(x∆, r). Let k(∆) denote the uniquek ∈ Z such that 2−k−1 < 200r ≤ 2−k, and set

(2.30) D∆ := Q ∈ Dk(∆) : Q∩ 2∆ , Ø.

We then define

(2.31) T∆ := int

⋃

Q∈D∆TQ

.

Given a surface ball∆ := ∆(x, r), let B∆ := B(x, r), so that∆ = B∆ ∩ ∂Ω. Then,

(2.32)54

B∆ ∩Ω ⊂ T∆.

and there existsκ0 large enough such that

(2.33) T∆ ⊂ κ0B∆ ∩ Ω.

Lemma 2.34([HM, Lemma 3.61]). Suppose thatΩ is a 1-sided NTA domain withan ADR boundary. Then all of its Carleson boxes TQ and T∆, and sawtooth re-gionsΩF , andΩF ,Q are also 1-sided NTA domains with ADR boundaries. In allcases, the implicit constants are uniform, and depend only on dimension and onthe corresponding constants forΩ.

Lemma 2.35([HM, Lemma 3.62]). Suppose thatΩ is a 1-sided NTA domain withan ADR boundary and thatΩ also satisfies the qualitative exterior Corkscrew con-dition. Then all of its Carleson boxes TQ and T∆, and sawtooth regionsΩF , andΩF ,Q satisfy the qualitative exterior Corkscrew condition. In all cases, the implicitconstants are uniform, and depend only on dimension and on the correspondingconstants forΩ.

Corollary 2.36 ([HM, Corollary 3.69]). Suppose thatΩ is a 1-sided NTA domainwith n-dimensional ADR boundary and that it also satisfies the qualitative exteriorCorkscrew condition. There is a uniform constant C such thatfor every pair ofsurface balls∆ := B ∩ ∂Ω, and ∆′ := B′ ∩ ∂Ω, with B′ ⊆ B, and for everyX ∈ Ω \ 2κ0B, whereκ0 is the constant in(2.33) below, we have

1CωX∆(∆′) ≤ ω

X(∆′)ωX(∆)

≤ CωX∆(∆′).

Remark2.37. Let us recall that the dilation factorλ, defining the fattened WhitneyboxesI ∗ and Whitney regionsUQ, is allowed to be any fixed positive number nolarger than some smallλ0 (cf. (2.21) and Remark2.26). For the rest of the paperwe now fix 0 < λ ≤ λ0/2, so that, in particular, the previous results apply notonly to the Carleson boxes and sawtooths corresponding toUQ as defined above,but also to those corresponding to “fatter” Whitney regionsU∗Q = ∪W∗

QI ∗∗ with

I ∗∗ = (1 + 2λ) I . We will work with these fatter regions in Sections2.4 and2.5below.


2.2. Step 1: Passing to the approximating domains.We first observe that with-out loss of generality we can assume thatp ≤ 2: If p > 2, (1.23) and Holder’sinequality imply the same estimate withp = 2.

We define approximating domains as follows. For each large integer N, setFN := DN. We then letΩN := ΩFN denote the usual (global) sawtooth with respectto the familyFN (cf. (2.23), (2.21) and (2.28).) Thus,

(2.38) ΩN = int

⋃

Q∈D: ℓ(Q)≥2−N+1

UQ

,

so thatΩN is the union of fattened Whitney cubesI ∗ = (1+ λ)I , with ℓ(I ) & 2−N,and the boundary ofΩN consists of portions of faces ofI ∗ with ℓ(I ) ≈ 2−N. Byvirtue of Lemma2.34, eachΩN satisfies the ADR, Corkscrew and Harnack Chainproperties. We note that, for each of these properties, the constants are uniform inN, and depend only on dimension and on the corresponding constants forΩ.

By constructionΩN satisfies the qualitative exterior corkscrew condition (Def-inition 2.5) since it has exterior corkscrew points at all scales. 2−N. By Lemma2.35the same statement applies to the Carleson boxesTQ andT∆, and to the saw-tooth domainsΩF andΩF ,Q (all of them relative toΩN) and even to Carlesonboxes within sawtooths.

We write ωN for the corresponding harmonic measure andkN for the corre-sponding Poisson kernel (we know by [DJ] thatωN is absolutely continuous withrespect to surface measure, sinceΩN enjoys aqualitative2-sided Corkscrew con-dition, i.e., it has exterior Corkscrew points at all scales. 2−N). We are going toshow that the scale invariant estimate (1.23) passes uniformly to the approximat-ing domains. To be precise, for all surface balls∆N, defined with respect to theapproximating domainΩN (i.e.,∆N = B∩ ∂ΩN with B centered at∂ΩN), we have

(2.39)∫

∆N

(k

X∆N

N

)pdσN ≤ CσN(∆N)1−p,

whereC and 1< p ≤ p are independent ofN. To prove (2.39), we shall first needto establish several preliminary facts.

Given X ∈ ΩN ⊂ Ω we write x for a point in∂Ω such thatδ(X) = |X − x|,analogously ˆxN ∈ ∂ΩN with δN(X) = |X − xN| (hereδN stands for the distance tothe boundary ofΩN). Set∆X = B(x, δ(X)) ∩ ∂Ω and∆N

X = B(xN, δN(X)) ∩ ∂ΩN.

Let X ∈ ΩN ⊂ Ω with δN(X) ≈ 2−N, thusδ(X) ≈ 2−N. We claim that for everyY ∈ ΩN we have

(2.40)ωY

N(∆NX)

σN(∆NX)≤ CωY(∆X)σ(∆X)

,

whereC is independent ofN.

We first consider the caseΩN bounded. For fixedY ∈ ΩN we setu(Y) = ωYN(∆N

X)andv(Y) = ωY(∆X). Note thatu, v are harmonic inΩN. For everyY ∈ ∆N

X , we haveu(Y) = 1 . ωY(∆X) = v(Y) by Lemma2.2and the Harnack chain condition. Also,if Y ∈ ∂ΩN \ ∆N

X we haveu(Y) = 0 ≤ ωY(∆X) = v(Y). Thus, maximum principle


yields thatu(Y) . v(Y) for everyY ∈ ΩN and then we obtain as desired

ωYN(∆N

X)

σN(∆NX)≤ CωY(∆X)σ(∆X)

,

whereC is independent ofN (notice that we have used thatσN(∆NX) ≈ δN(X)n ≈

δ(X)n ≈ σ(∆X).)

Let us treat the caseΩN unbounded. LetM ≫ 1 and setBNM = B(xN,M δN(X))

and∆NM = BN

M ∩ ∂ΩN. TakeΩN,M = T∆NM

whereT∆NM⊂ ΩN denotes the Carleson

box corresponding to∆NM for the domainΩN. By (2.32) and (2.33) (withΩ replaced

byΩN) we have

54

BMN ∩ ΩN ⊂ ΩN,M, ΩN,M ⊂ κ0BM

N ∩ ΩN.

These and the fact thatM ≫ 1 allow us to obtain thatX ∈ ΩN,M, xN ∈ ∂ΩN∩∂ΩN,M,δΩN,M (X) = δN(X) ≈ 2−N, and

∆NX = B(xN, δN(X)) ∩ ∂ΩN = B(xN, δN(X)) ∩ ∂ΩN,M.

Then, we proceed as in the previous case and consideruM(Y) = ωYΩN,M

(∆NX) and

v(Y) = ωY(∆X) for Y ∈ ΩN,M. The same argument above and the maximum prin-ciple in the bounded domainΩN,M yields uM(Y) . v(Y) for everyY ∈ ΩN,M withconstant independent onN, M andX. On the other hand notice that the solutionsuM(Y) = ωY

ΩN,M(∆N

X) are monotone increasing on any fixedΩN,M0, asM0 ≤ M →∞, by the maximum principle. We then obtain thatuM(Y) → u(Y) := ωX

N(∆NX),

uniformly on compacta, by Harnack’s convergence theorem. Thus,u(Y) . v(Y) foreveryY ∈ ΩN and then the previous argument leads to the desired estimate.

Proposition 2.41. There exists1 < p ≤ p such that for all N≫ 1 if we let∆N = B∩ΩN be a surface ball with r(B) ≈ 2−N, then

(2.42)

(?∆N

(k

X∆N

N

)pdσN

) 1p

. σN(∆N)−1

where the constant is uniform in N. Furthermore, for every Y∈ ΩN \2κ0B we have

(2.43)

(?∆N

(kY

N

)pdσN

) 1p

.ωY

N(∆N)σN(∆N)

,

where the constant is uniform in N.

Proof. Let us momentarily assume (2.42) and we obtain (2.43). As observed aboveΩN is a 1-sided NTA domain with ADR boundary and satisfies the qualitative ex-terior corkscrew condition. Thus we can use Corollary2.36: we take∆N = B∩ΩN

with r(B) ≈ 2−N and for everyY ∈ ΩN \ 2κ0B we have?

(∆N)′kY

N dσN ≈ ωYN(∆N)

?(∆N)′

kX∆N

N dσN, B′ ⊂ B.


Then forσN-a.e. x ∈ ∆N we takeB′ = B(x, r) and letr → 0 to obtainkYN(x) ≈

ωYN(∆N) k

X∆N

N (x) with constants that do not depend onN. Consequently, we have(?∆N

(kY

N

)pdσN

) 1p

. ωYN(∆N)

(?∆N

(k

X∆N

N

)pdσN

) 1p

, Y ∈ ΩN \ 2κ0B.

This and (2.42) clearly give (2.43).

We next show (2.42). Write ∆N = ∆N(xN, r) = B(xN, r) ∩ ∂ΩN with xN ∈ ∂ΩN

and r ≈ 2−N. We takex ∈ ∂Ω such thatδ(xN) = |xN − x| ≈ 2−N and set∆ =∆(x,M 2−N) with M (independent ofN) large enough. Note that

(2.44) B(xN, 4r) ∩ ΩN ⊂ ∪I I ∗ ,

where the union runs over a collection of Whitney boxesI ∈ W(Ω), of uniformlybounded cardinality, all with int(I ∗) ⊂ ΩN, ℓ(I ) ≈ 2−N, andB(xN, 4r) ∩ I ∗ , Ø.Notice that we haveB(xN, 4r)∩∂ΩN ⊂ ∪I∂I ∗. For each suchI , there is aQI ∈ DFN

such thatℓ(I ) ≈ ℓ(QI ) ≈ 2−N ≈ dist(I ,QI ) ,

so by definitionI ∈ W∗QI

. Thus, for any suchI we have that

(2.45) |xQI − x| . ℓ(QI ) + d(QI , I ) + ℓ(I ) + 4r + |xN − x| ≤ M 2−N ,

for M chosen large enough, and thereforexQI ∈ ∆.

Let k(∆) denote the uniquek ∈ Z such that 2−k−1 < M 2−N ≤ 2−k (below wemay need to makeM larger). We define

P := int

⋃

Q∈D∆ΩFN ,Q

.

whereD∆ is defined in (2.30) (note thatk(∆) is slightly different.)

Let us make some observations about this setP. Note first thatDFN is the col-lection of all dyadic cubes whose side length are at least 2−N+1. Thus for everyQ ∈ D∆, we have thatℓ(Q) = 2−k(∆) andDFN,Q consists of all dyadic subcubes ofQwhose side length is at least 2−N+1, i.e., thoseQ′ ∈ DQ with 2−N+1 ≤ ℓ(Q) ≤ 2−k(∆)

and these are a finite number. Note also that the cardinality of D∆ is finite anddepends only onM and the ADR constants, and thereforeP is the union of a fi-nite number (the cardinality depends only onM and on the parameters that ap-pear in the definition ofW∗

Q) of fattened Whitney boxes that are “close” to∆and have size comparable to 2−N. Note also, thatP ⊂ ΩN. It is easy to see thatB(xN, 4r) ∩ ΩN ⊂ P. Namely, for everyY ∈ B(xN, 4r) ∩ ΩN we have shown aboveusing (2.44) that there existsI ∗ ∋ Y with ℓ(I ) ≈ 2−N, I ∈ W∗

QIand xQI ∈ ∆.

Therefore there isQ∗ ∈ D∆ such thatQI ⊂ Q∗. Thus,Y ∈ ΩFN,QI ⊂⋃

Q∈D∆ ΩFN ,Q.Notice thatB(xN, 4r) ∩ ΩN is open so there exists a ballBY centered aty that iscontained in this set. For everyZ ∈ BY we have just shown thatz ∈

⋃Q∈D∆ ΩFN ,Q,

this eventually leads toY ∈ P.

Lemma 2.46. P is an NTA domain with ADR boundary, that is, it satisfies theinterior and exterior corkscrew conditions, the (interior) Harnack chain condition,and ∂P is n-dimensional ADR; moreover, all of the corresponding constants are


independent of N. Consequently,ωP, the harmonic measure for the domainP, isabsolutely continuous with respect to surface measure and its Poisson kernel, kP,satisfies the scale invariant estimate

(2.47)∫

∆P

(k

X∆P

P

) pdσP ≤ CσP(∆

P)1−p,

for some1 < p < ∞ and for all surface balls∆P, where C andp are independentof N.

The proof of this result is given below.

Fix y ∈ ∆N = B(xN, r) ∩ ∂ΩN. Theny ∈ ∂I ∗ with I ∈ W(Ω), ℓ(I ) ≈ 2−N andint(I ∗) ⊂ ΩN. We recall thatP is a finite union of fattened Whitney boxesJ∗ ∈W(Ω), and we definePI∗ to be the interior of the union of those boxesJ∗ ⊂ P thatoverlapI ∗. It is easy to see thatI ⊂ PI∗ ⊂ P. We shall see from the proof of Lemma2.46 thatPI∗ is an NTA domain with constants independent ofN. Moreover, weclaim that takingM above larger we have thatPI∗ is strictly contained inP, andin fact P \ PI∗ contains some Whitney boxJ ∈ W(Ω). We may verify this claimas follows. We fixQI ∈ FN with ℓ(QI ) ≈ ℓ(I ) ≈ 2−N and dist(I ,QI ) ≈ 2−N.Notice that, forM large enough, there is someQ ∈ D∆ such thatQI ⊂ Q (cf.(2.44)-(2.45).) For thisQ, takeJ ∈ W∗

Q which is clearly inP. Note that

ℓ(J) ≈ ℓ(Q) ≈ M 2−N ≈ M ℓ(I )

and therefore forM large enough we have thatI ∗ ∩ J∗ = Ø, by properties ofWhitney cubes (since otherwiseℓ(I ) ≈ ℓ(J)). This shows thatPI∗ ( P.

Setu(Z) = GN(Z,X∆N) andv(Z) = GP(Z,X∆N) whereGN, GP are respectivelythe Green functions associated with the domainsΩN andP. Notice thatu andv arenon-negative harmonic functions inPI∗ (sincePI∗ ⊂ P ⊂ ΩN) and we can assumethatX∆N < PI∗: indeed, by the Harnack chain condition, and our observation abovethatPI∗ ( P, we may assume thatX∆N is the center of a Whitney boxJ ⊂ P \ PI∗.Notice thatu

∣∣2∆N = v

∣∣2∆N = 0 (since 2∆N ⊂ ∂ΩN and sinceB(xN, 4 r) ∩ ΩN ⊂ P

implies that 2∆N = B(xN, 2 r) ∩ ∂P). Thereforeu = v ≡ 0 in B(y, 2−N−N1) ∩ ∂PI∗

(hereN1 is a fixed big integer such thatB(y, 2−N−N1) ∩ ∂PI∗ ⊂ 2∆N. As mentionedbefore,PI∗ is an NTA domain with constants independent ofN and we have the(full) comparison principle (see [JK, Lemma 4.10]):

u(Z)v(Z)

≈ u(X(I ))v(X(I ))

,

for everyZ ∈ B(y, 2−N−N1/C) ∩ PI∗ with C large independent ofN. We remind thereader thatX(I ) stands for the center of the Whitney boxI . Notice that

|X(I ) − X∆N | ≈ dist(X(I ), ∂P) ≈ dist(X∆N , ∂P) ≈ 2−N .

Thus, the Harnack chain condition and the size estimates forthe Green functionsimply thatu(X(I ))/v(X(I )) ≈ 1. Henceu(Z) ≈ v(Z) for all Z ∈ B(y, 2−N−N1/C)∩PI∗ .

As observed aboveΩN is a 1-sided NTA with ADR boundary and also satisfiesthe qualitative exterior corkscrew condition, so that, in particular, we may apply


Lemma2.11. We takes≪ 2−N and write∆N(y, s) = B(y, s) ∩ΩN to obtain

u(X∆N(y,s))

s=

GN(X∆N(y,s),X∆N)

s≈ ω

X∆N

N (∆N(y, s))σN(∆N(y, s))

.

The same can be done withGP (indeedP is NTA) after observing that∆N(y, s)is also a surface ball forP and that after using Harnack if neededX∆N(y,s) is acorkscrew point with respect toP for that surface ball. Then,

v(X∆N(y,s))

s=

GP(X∆N(y,s),X∆N)

s≈ ω

X∆N

P (∆N(y, s))σP(∆N(y, s))

Note that ifs is small enough thenZ = X∆N(y,s) ∈ B(y, 2−N−N1/C)∩PI∗ . This allowsus to gather the previous estimates and conclude that

ωX∆N

N (∆N(y, s))σN(∆N(y, s))

≈ ωX∆N

P (∆N(y, s))σP(∆N(y, s))

.

Next we use the Lebesgue differentiation theorem and obtain thatkX∆N

N (y) ≈ kX∆N

P (y)for a.e.y ∈ ∆N. Then if we set ˜p = minp, p and use (2.47) we obtain

(?∆N

(k

X∆N

N

)pdσN

) 1p

.

(?∆N

(k

X∆N

P

)pdσP

) 1p

≤(?∆N

(k

X∆N

P

)pdσP

) 1p

. σP(∆N)−1 ≈ σN(∆N)−1.

This completes the proof of Proposition2.41.

Proof of Lemma2.46. The proof is very elementary. The corkscrew conditions areas follows. Fix∆P = ∆P(x, r) with 0 < r < diam(∂P) ≈ 2−N. Sincex ∈ ∂P, which isa finite union of fattened boxes of size comparable to 2−N, x ∈ ∂I ∗ with ℓ(I ) ≈ 2−N.The fact thatI ∗ is a cube yields easily that you can always find a corkscrew point inthe segment that jointsx andX(I ) (the center ofI ) with constants independent ofNand this is an interior corkscrew point forP. For the exterior corkscrew conditionwe notice that∂I ∗ can be covered by the union of the Whitney boxes that areneighbors ofI (i.e., whose boundaries touch). Thenx ∈ J with ∂I ∩ ∂J , Ø. Notethat J ⊂ int(J∗) and henceJ∗ cannot be one of the Whitney boxes that defineP.Take J′ a neighbor ofJ: if int(( J′)∗) ⊂ P, then (J′)∗ “bites” a small portion ofJ(since (J′)∗ is a small dilation ofJ′); otherwise (J′)∗ does not “bite”J. Eventually,we see that the part ofJ minus all these “bites” is contained inPc. Note thatx isin one of this portions and therefore we can find a corkscrew point in the segmentbetweenx and the center ofJ. This is possible sincer . 2−N.

We show the Harnack chain condition. LetX1,X2 ∈ P. Then, for eachi = 1, 2there existsQi ∈ D∆ andQ′i ∈ DFN ,Qi with Xi ∈ int(I ∗i ) whereI i ∈ W∗

Q′i. Note

that ℓ(Qi) ≈ ℓ(Q′i ) ≈ ℓ(I i ) ≈ 2−N. If I ∗1, I ∗2 overlap then the condition is clear.Otherwise, dist(I ∗1, I

∗2) ≈ dist(I1, I2) ≈ 2−N and|X1 − X2| ≈ 2−N. Moreover, we can

construct an appropriate chain in each fattened box betweenXi andX(I i). In fact, ifδP(Xi) ≈ 2−N the number of balls is bounded by a dimensional constant, whereas ifδP(Xi) ≪ 2−N then we simply use that each cubeI ∗i has the Harnack chain property


with uniform constants. Therefore, it suffices to assume thatXi is the center ofI ∗i .In such a caseδP(Xi) ≈ 2−N, i = 1, 2, |X1−X2| ≈ 2−N and therefore we need to finda Harnack chain whose cardinality is uniformly bounded (inN).

Therefore we fix two fattened Whitney boxesI ∗1, I ∗2 and we want to find aHarnack chain between the centers of such boxes. As just mentioned, if I ∗1, I ∗2overlap we can easily construct a finite (depending only onλ) chain betweenX(I1) and X(I2) (with respect to the domainI ∗1 ∪ I ∗2) such that diam(Bi) ≈ 2−N,dist(Bi , ∂(I ∗1 ∪ I ∗2)) ≈ 2−N. Let us observe that in such a case

dist(Bi , ∂P) ≤ diam(P) ≈ 2−N ≈ dist(Bi , ∂(I∗1 ∪ I ∗2)) ≤ dist(Bi , ∂P)

Thus the constructed Harnack chain is valid for the domainP. If I ∗1, I ∗2 do not over-lap it suffices to find a chain of fattened boxesJ∗1, . . . , J

∗K in P such thatJ∗1 = I ∗1,

J∗K = I ∗2 and with the property thatJ∗k and J∗k+1 overlap. Notice thatK is uni-formly bounded sinceP is a finite union of fattened Whitney boxes with cardinalitybounded uniformly inN. Notice that ifQ1, Q2 ∈ D∆ we haveℓ(Q1) = ℓ(Q2) anddist(Q1,Q2) ≤ 4ℓ(Q2). Therefore, the construction of the setsWQ guarantees thatWQ1 ∩ WQ2 is non-empty (cf. (2.20)) and there existsI ∈ WQ1 ∩ WQ2. Thisleads to show thatI ⊂ int(UQ1)∩ int(UQ2) which in turn givesI ⊂ ΩFN ,Q1∩ΩFN,Q2.To conclude we just observe that eachΩFN ,Qi enjoys the Harnack chain propertyby Lemma2.34and this means that we can connect any box inΩFN,Qi with I asbefore.

To show that∂P is ADR we proceed as follows. Notice that∂P ⊂ ∪Q∈D∆∂ΩFN,Q

where each∂ΩFN ,Q is ADR by Lemma2.34and the cardinality ofD∆ is finite anddepends only onM and the ADR constants ofΩ. Thus the upper bound of theADR condition for∂P follows at once with constants that are independent ofN.For the lower bound, letB = B(x, r) with x ∈ ∂P andr ≤ diamP ≈ 2−N. As ∂P iscomprised of partial faces of fattened Whitney cubesI ∗ with ℓ(I ) ≈ 2−N, thenx liesin a subsetF of a (closed) face ofI ∗, with F ⊂ B∩∂P, andHn(F∩B) = Hn(F) & rn,as desired. Again the constants are independent ofN.

Having shown thatP is an NTA domain with ADR boundary, with constantsindependent ofN, we may invoke [DJ] to deduce absolute continuity of harmonicmeasure, along with the desired scale invariant estimate (2.47) for the Poisson ker-nel, with constants independent ofN.

With these preliminaries in hand, we now turn to the proof of (2.39). To thisend, we fix∆N, and consider three cases. Ifr(∆N) ≈ 2−N then (2.42) gives thedesired estimate. Assume now thatr(∆N) ≫ 2−N. We cover∆N by those dyadiccubesQN ∈ DN(∂ΩN) that meet∆N (we remind the reader thatDN(∂ΩN) are thedyadic cubes associated to the ADR set∂ΩN with ℓ(QN) = 2−N). Note that foreach of those cubes we haveX∆N ∈ ΩN \ 2κ0B(xQN ,C 2−N) and then we can use(2.43) with Y = X∆N and the surface ball∆N(xQN ,C 2−N):

∫

∆N

(k

X∆N

N

)pdσN ≤

∑

QN

σN(QN)?

QN

(k

X∆N

N

)pdσN

.

∑

QN

σN(QN)?∆N(xQN ,C 2−N)

(k

X∆N

N

)pdσN


.

∑

QN

σN(QN)

(ω

X∆N

N

(∆N(xQN ,C 2−N)

)

σN(∆N(xQN ,C 2−N)

))p

.

∑

QN

σN(QN)

ω

X∆N

N (∆NXQN

)

σN(∆NXQN

)

p

,

where in the last estimate we have used thatωN is doubling: let us observe that∆N(xQN ,C 2−N) and∆N

XQNare balls whose radii are comparable to 2−N and the

distance of their centers is controlled by 2−N, and thatδN(X∆N) ≈ r(∆N) ≫ 2−N.Let us observe thatXQN ∈ ΩN andδ(XQN) ≈ δN(XQN) ≈ 2−N ≪ δN(X∆N) ≈ δ(X∆N).Then we use (2.40) to obtain

(2.48)∫

∆N

(k

X∆N

N

) pdσN .

∑

QN

σN(QN)

(ωX

∆N (∆XQN )

σ(∆XQN )

)p

Given QN, its centerxQN is in ∂ΩN ⊂ Ω and therefore there exist ˜xQN ∈ ∂Ωsuch thatδ(xQN) = |xQN − xQN | ≈ 2−N. Then using the dyadic cube structure in∂Ω, there exists a uniqueR(QN) ∈ DN(∂Ω) that contains ˜xQN . We note that∆XQN

and∆R(QN) are surface balls in∂Ω with radii comparable to 2−N and whose centersare separated at mostC 2−N. This means that∆XQN ⊂ C∆R(QN). On the other

hand, one can show thatC∆R(QN) ⊂ C′ ∆X∆N by using thatQN meets∆N and that

r(∆N) ≫ 2−N.

Next we want to control the overlapping of the familyC∆R(QN)QN . It is easyto see that ifC∆R(QN) ∩C∆R(QN

1 ) , Ø then|xQN − xQN1| . 2−N = ℓ(QN

1 ) = ℓ(QN).

Thus the fact that the cubesQN are all dyadic disjoint with side length 2−N yields

#QN : C∆R(QN) ∩C∆R(QN1 ) , Ø ≤ #QN : |xQN − xQN

1| . 2−N ≤ C.

Gathering the previous facts we obtain that∑

QN

χ∆XQN≤∑

QN

χC∆R(QN )≤ CχC′ ∆X

∆N.

Set∆0 = C′ ∆X∆N and observe that by the Harnack Chain propertyωX

∆N ≈ ωX∆0 .Thus plugging these estimates into (2.48) we conclude that

∫

∆N

(k

X∆N

N

)pdσN .

∑

QN

σN(QN)

(ωX∆0(∆XQN )

σ(∆XQN )

)p

.

∑

QN

∫

∆XQN

(kX∆0

)pdσ

.

∫

∆0

(kX∆0

)pdσ ≤

(∫

∆0

(kX∆0

)pdσ

) pp

σ(∆0)p− p

p . σ(∆0)1−p ≈ σN(∆N)1−p,

where we have used ˜p ≤ p and our hypothesis (1.23) and where the involvedconstants are independent ofN. This completes the caser(∆N)≫ 2−N.

To complete Step 1, we need to establish (2.39) for r := r(∆N)≪ 2−N. Set∆N =

B(xN, r)∩∂ΩN. Given∆N(xN,C 2−N) we consider the associatedP as before. For afixedy ∈ ∆N = B(xN, r) ∩ ∂ΩN, we have thaty ∈ ∂I ∗ with ℓ(I ) ≈ 2−N and int(I ∗) ⊂


ΩN. We definePI∗ as before and recall that it is an NTA domain with constantsindependent ofN. Let us note that 5∆N ⊂ ∂ΩN ∩ ∂PI∗, sincer ≪ 2−N. We letY∆N denote a Corkscrew point with respect to∆N, for the domainPI∗. Let c < 1 besmall enough such thatY∆N < TPI∗

∆N(y,2c r), whereTPI∗∆N(y,2c r) is the Carleson box with

respect to∆N(y, 2c r) relative to the domainPI∗ (note thatc depends on the variousgeometric constants ofPI∗ which are all independent ofN, see (2.33)). Notice thatby Lemma2.34, TPI∗

∆N(y,2c r) inherits the (interior) Corkscrew and Harnack Chain

conditions and also the ADR property fromPI∗. Moreover,TPI∗∆N(y,2c r) also inherits

the exterior corkscrew condition, with uniform bounds. Indeed, consider a surfaceball ∆⋆(z, ρ) ⊂ ∂TPI∗

∆N(y,2c r), with z ∈ ∂TPI∗∆N(y,2c r), andρ . r ≈ diam(TPI∗

∆N(y,2c r)).

If dist(z, ∂PI∗) ≤ ρ/100, thenB(z, ρ) ∩(Rn+1 \ TPI∗

∆N(y,2c r)

)contains an exterior

Corkscrew point inherited fromPI∗. On the other hand, if dist(z, ∂PI∗) > ρ/100,thenz lies on a face of some fattened Whitney cubeJ∗ ∈ W(PI∗ ), of side lengthℓ(J∗) & ρ, for which there is some adjoining Whitney cubeJ′ containing an exteriorCorkscrew point inB(z, ρ) ∩

(PI∗ \ TPI∗

∆N(y,2c r)

). ConsequentlyTPI∗

∆N(y,2c r) is an NTAdomain with constants independent ofN, r andy.

Setu(Z) = GN(Z,Y∆N) andv(Z) = GPI∗ (Z,Y∆N) whereGN, GPI∗ are respectivelythe Green functions associated with the domainsΩN andPI∗. Thenu andv arenon-negative harmonic functions inTPI∗

∆N(y,2c r) (sinceY∆N < TPI∗∆N(y,2c r)). Notice that

for c small,r ≪ 2−N, andy ∈ ∆N = ∆N(xN, r) we have that

∆N1 := ∆N(y, cr) :=

(B(y, c r) ∩ ∂ΩN

)⊂(∂ΩN ∩ ∂PI∗ ∩ ∂TPI∗

∆N(y,2c r)

).

In particular,u∣∣∆N

1= v∣∣∆N

1= 0. As noted above,TPI∗

∆N(y,2c r) is an NTA domain with

constants independent ofN and thus by [JK, Lemma 4.10]) we have the comparisonprinciple:

u(Z)v(Z)

≈ u(X0)v(X0)

, ∀Z ∈ B(y, cr/C) ∩ TPI∗∆N(y,2c r) ,

with C sufficiently large but independent ofN, and whereX0 ∈ TPI∗∆N(y,2c r) is a

corkscrew point associated to the surface ball∆N1 , relative to each of the domains

ΩN, PI∗ andTPI∗∆N(y,2c r). By the Harnack chain condition inPI∗ and since

|X0 − Y∆N | ≈ dist(X0, ∂PI∗) ≈ dist(Y∆N , ∂PI∗) ≈ r ,

one can show thatu(X0)/v(X0) ≈ 1 which eventually leads tou(Z) ≈ v(Z) for allZ ∈ B(y, cr/C) ∩ TPI∗

∆N(y,2c r).

We can now apply Lemma2.11(since, as observed above,ΩN satisfies the re-quired “qualitative assumption”), so that if we takes ≪ r and write∆N(y, s) =B(y, s) ∩ΩN, we have

u(X∆N(y,s))

s=

GN(X∆N(y,s),X∆N)

s≈ ω

X∆N

N (∆N(y, s))σN(∆N(y, s))

.

The same can be done withGPI∗ (indeedPI∗ is NTA) after observing that∆N(y, s)is also a surface ball forPI∗ and that, after using Harnack if needed,X∆N(y,s) may


be taken to be a corkscrew point with respect toPI∗ for that surface ball. Then,

v(X∆N(y,s))

s=

GPI∗ (X∆N(y,s),X∆N)

s≈ω

X∆N

PI∗(∆N(y, s))

σPI∗ (∆N(y, s))

Note that if s is small enough thenZ = X∆N(y,s) ∈ B(y, cr/C) ∩ TPI∗∆N(y,2c r). This

allows us to gather the previous estimates and conclude that

ωX∆N

N (∆N(y, s))σN(∆N(y, s))

≈ω

X∆N

PI∗(∆N(y, s))

σPI∗ (∆N(y, s))

.

Next we use the Lebesgue differentiation theorem and obtain thatkX∆N

N (y) ≈ kX∆N

PI∗(y)

for a.e.y ∈ ∆N. Then

(?∆N

(k

X∆N

N

)pdσN

) 1p

.

(?∆N

(k

X∆N

PI∗

)pdσPI∗

) 1p

≤(?∆N

(k

X∆N

PI∗

) pdσPI∗

) 1p

. σPI∗ (∆N)−1 ≈ σN(∆N)−1

where we have used that ˜p ≤ p and (2.47). This completes the proof of (2.39)

Remark. We notice that to obtain (2.47), in place of using [DJ], we could haveinvoked [VV]: P is a polyhedral domain (its boundary consists of a finite numberof flat “faces”), is NTA and has the ADR property, thuskP is aRH2 weight. Thusp = 2 andp = minp, 2.

2.3. Step 2: LocalTb theorem for square functions. Having already establishedStep 1, we want to show thatΩN has the UR property with uniform bounds. Inthe last step we shall show that this ultimately implies thatΩ inherits this prop-erty. Therefore, in all the remaining steps, but the last one, we drop the alreadyfixed subindex/superindexN everywhere and writeΩ to denote the correspondingapproximating domainΩN. Our main assumption is that (2.39) holds, and thisrewrites as

(2.49)∫

∆

(kX∆)p

dσ ≤ Cσ(∆)1−p,

wherep andC are independent ofN.

We introduce some notation. We have already definedW =W(Ω) the collec-tion of Whitney boxes ofΩ. We also defineWext = W(Ωext) the collection ofWhitney boxes ofΩext = R

n+1 \ Ω. SetW0 = W ∪Wext —notice that we canget directly this collection by taking the Whitney decomposition of Rn+1 \ ∂Ω. Wecan then consider as beforeWQ with the same fixedC0 as before (this guaranteesamong other things thatWQ , Ø). With that fixedC0 we define analogouslyWext

Q(using only I ’s in Wext). Notice that we have not assumed exterior corkscrewpoints forΩext and thereforeWext

Q might be the null set. We then define

VQ =⋃

I∈WQ

I , Λ(x) =⋃

x∈Q∈DVQ ΛQ0(x) =

⋃

x∈Q∈DQ0

VQ


and analogously

VextQ =

⋃

I∈WextQ

I , Λext(x) =⋃

x∈Q∈DVext

Q ΛextQ0

(x) =⋃

x∈Q∈DQ0

VextQ ,

where x ∈ ∂Ω and Q0 ∈ D. We also consider the “two-sided” conesΛ(x) =Λ(x) ∪ Λext(x) andΛQ0(x) = ΛQ0(x) ∪Λext

Q0(x).

In this step we are going to apply the following localTb theorem for squarefunctions in [GM]:

Theorem 2.50(Local Tb theorem for square functions, [GM]). Let Ω ⊂ Rn+1,n ≥ 2, be a connected open set whose boundary∂Ω is ADR. We assume that thereis an exponent q∈ (1, 2], and a finite constantA0 > 1 such that for every Q∈ Dthere exists a function bQ satisfying

(2.51)∫

∂Ω

|bQ|q dσ ≤ A0σ(Q)

(2.52)

∣∣∣∣∫

QbQ dσ

∣∣∣∣ ≥1A0σ(Q)

(2.53)∫

Q

("ΛQ(x)

|∇2SbQ(Y)|2 dYδ(Y)n−1

) q2

dσ(x) ≤ A0σ(Q)

Then,

(2.54)"Rn+1|∇2S f (Y)|2 δ(Y) dY ≤ C ‖ f ‖2L2(∂Ω)

We observe that we can easily show∫

Rn+1|∇2S f (Y)|2 δ(Y) dY ≈

∫

∂Ω

∫

Λ(x)|∇2S f (Y)|2 dY

δ(Y)n−1 dσ(x)

which is nothing but the comparability of the “vertical” and“conical” square func-tions. In this way, we see (2.53) as anLq-testing condition for the local (conical)square function and the conclusion states theL2 boundedness of the (conical orvertical) square function.

In order to apply this result and define the functionsbQ we shall require somegeometric preliminaries. GivenQ ∈ D, we recall that there exists a surface ball∆Q = ∆(xQ, rQ) such that∆Q ⊂ Q ⊂ C∆Q = B(xQ,C rQ)∩∂Ω with rQ ≈ ℓ(Q). Letκ1 be large enough such thatC rQ < κ1ℓ(Q) and also withκ1 > κ0 whereκ0 is givenin (2.62) below. SetBQ = B(xQ, κ1 ℓ(Q)). We notice that ifx ∈ Q andY ∈ ΛQ(x)thenY ∈ VQ′ with x ∈ Q′ ∈ DQ andY ∈ I ∈ WQ′ . Thus,

|Y − xQ| ≤ ℓ(I ) + d(I ,Q′) + ℓ(Q′) + |x− xQ| . C0ℓ(Q′) + ℓ(Q) . C0ℓ(Q).

The same can be done forY ∈ ΛextQ (x) and therefore by takingκ1 sufficiently large

(depending onK0) we have

(2.55)⋃

x∈QΛ(x) ⊂ BQ.


Next we setBQ = κ2 BQ, with κ2 large enough so thatXQ, the corkscrew pointrelative to∆Q = BQ∩Ω, satisfiesXQ < 6 BQ (it suffices to takeκ2 = 6/c with c theconstant that appears in the corkscrew condition).

Bearing in mind the previous considerations, we are ready todefine our func-tionsbQ as follows: we setbQ = σ(Q) ηQ kXQ whereηQ is a smooth cut-off (definedin Rn+1) with 0 ≤ ηQ ≤ 1, suppηQ ⊂ 5 BQ, ηQ ≡ 1 in 4BQ, and‖∇ηQ‖∞ . ℓ(Q)−1.We also takeq = p and we recall that 1< p ≤ p ≤ 2.

Using Harnack’s inequality, thatq = p and (2.49) we obtain (2.51):∫

∂Ω

|bQ|q dσ ≤ σ(Q)q∫

5 ∆Q

(kXQ

)qdσ . σ(Q)q

∫

5 ∆Q

(k

X5∆Q

)qdσ

. Cσ(Q)qσ(5 ∆Q)1−q. σ(Q).

Regarding, (2.52) we have by Lemma2.2and the Harnack chain condition that∣∣∣∣∫

QbQ dσ

∣∣∣∣ = σ(Q)∫

QηQ kXQ dσ & σ(Q)ωXQ(∆Q) ≥ σ(Q)/C.

Next we show (2.53). Let X ∈ BQ ∩ Ωext. Then

∇2XSbQ(X) =

∫

∂Ω

∇2XE(X − y) bQ(y) dσ(y)

= σ(Q)∫

∂Ω

∇2XE(X − y) ηQ(y) dωXQ(y)

= σ(Q)∫

∂Ω

∇2XE(X − y) (ηQ(y) − 1)dωXQ(y)

+ σ(Q)∫

∂Ω

∇2XE(X − y) dωXQ(y)

= I1(X) + I2(X).

For I2 we observe thatu(Z) = ∇2XE(X − Z) is harmonic inΩ and C2(Ω) since

X ∈ Ωext. Thus, for everyZ ∈ Ω we have

(2.56) ∇2XE(X − Z) = u(Z) =

∫

∂Ω

u(y) dωZ(y) =∫

∂Ω

∇2XE(X − y) dωZ(y).

We would like to point out that we have implicitly used uniqueness of the solutionwhich follows from the maximum principle even for an unbounded domain since∇2

XE(X − Z) → 0 asZ → ∞. We apply (2.56) with Z = XQ and note that since|X − XQ| ≈ ℓ(Q) for everyX ∈ BQ ∩ Ωext, we therefore obtain

|I2(X)| = σ(Q)|∇2XE(X − XQ)| . σ(Q) |X − XQ|−(n+1) ≈ ℓ(Q)−1.

For I1, we observe thatΩ is an approximating domain whose boundary consistsof portions of faces of fattened Whitney cubes of size comparable to 2−N. Thus, its(outward) unit normalν is well defined a.e. on∂Ω, and we can apply the divergencetheorem to obtain

I1(X) = σ(Q)∇2X

∫

∂Ω

E(X − y) (ηQ(y) − 1)dωXQ(y)


= σ(Q)∇2X

∫

∂Ω

E(X − y) (ηQ(y) − 1)∇YG(XQ, y) · ν(y) dσ(y)

= σ(Q)∇2X

"Ω

divY(E(X − Y) (ηQ(Y) − 1)∇YG(XQ,Y)

)dY

= σ(Q)∇2X

"Ω

∇Y(E(X − Y)

)(ηQ(Y) − 1)∇YG(XQ,Y) dY

+ σ(Q)∇2X

"Ω

E(X − Y)∇ηQ(Y)∇YG(XQ,Y) dY

+ σ(Q)∇2X

"Ω

E(X − Y)(ηQ(Y) − 1) divY(∇YG(XQ,Y)) dY

=: I11(X) + I12(X) + 0 ,

where we have used that the term in the next-to-last line vanishes, sinceηQ − 1is supported inRn+1 \ 4 BQ, and sinceXQ ∈ BQ implies thatG(XQ, ·) is harmonicin Ω \ BQ. Notice that the integration by parts can be justified even whenΩ isunbounded, sinceE andG have sufficient decay at infinity.

We estimate the termsI11 and I12 in turn. First, notice that ifX ∈ BQ andY ∈ Ω \ 4 BQ we haver(BQ) . |Y − xQ| ≈ |Y − XQ| ≈ |Y − X|. Then, writingSk(Q) = 2k+1 BQ \ 2k BQ, k ≥ 2, and using thatηQ − 1 is supported inRn+1 \ 4 BQ,we have that for everyX ∈ BQ,

|I11(X)| ≤ σ(Q)"Ω

|∇2X∇YE(X − Y)| |ηQ(Y) − 1| |∇YG(XQ,Y)|dY(2.57)

. σ(Q)"Ω\4 BQ

|Y − X|−(n+2) |∇YG(XQ,Y)|dY

. σ(Q)∞∑

k=2

"Ω∩Sk(Q)

|Y − xQ|−(n+2) |∇YG(XQ,Y)|dY

. ℓ(Q)−2∞∑

k=2

2−(n+2)k"Ω∩Sk(Q)

|∇YG(XQ,Y)|dY

=: ℓ(Q)−2∞∑

k=2

2−(n+2)kIk.

To estimateIk we coverSk(Q) by a purely dimensional number of balls meetingSk(Q) and whose radii are 2k−5 r(BQ). Then, it suffices to get an estimate with theintegral restricted to such a ballBk. We may assume without loss of generality that2Bk 1 Ωext for otherwise we have thatΩ ∩ Sk(Q) ∩ Bk = Ø. We then have twocases: 2Bk ⊂ Ω and 2Bk 1 Ω. In the second case, since 2Bk is neither containedin Ω nor in Ωext, there existsyk ∈ ∂Ω ∩ 2Bk. We then setBk = B(yk, 3 r(Bk)),and observe that this ball is centered on∂Ω and containsBk. SinceG(XQ, ·) isharmonic in 2Bk∩Ω and vanishes on∂Ω, and sinceΩ is an approximating domainin which the Gauss/Green theorem holds, we can use Caccioppoli’s inequality atthe boundary to write

Ik,Bk :="Ω∩Sk(Q)∩Bk

|∇YG(XQ,Y)|dY


≤"Ω∩Bk

|∇YG(XQ,Y)|dY

. |Bk|12

("Ω∩Bk

|∇YG(XQ,Y)|2 dY

)1/2

. (2k ℓ(Q))n+1

2 r(Bk)−1("

Ω∩2 Bk

|G(XQ,Y)|2 dY

)1/2

. (2k ℓ(Q))n+1

2 −1(2k ℓ(Q))1−n |Bk|12

= 2k ℓ(Q) ,

by the size estimates for the Green function plus the fact that |XQ−Y| ≥ 2k−2 r(BQ)for Y ∈ Ω ∩ 2 Bk.

In the other case, i.e., when 2Bk ⊂ Ω, the situation is simpler: we just usethe (interior) Caccioppoli inequality and repeat the same computations withBk inplace ofBk. Summing over the collection ofBk’s coveringSk(Q), which for fixedk is a family of uniformly bounded cardinality, we conclude that

(2.58) Ik . 2k ℓ(Q).

We plug this estimate into (2.57) to obtain

|I11(X)| . ℓ(Q)−2∞∑

k=2

2−(n+2)kIk . ℓ(Q)−1.

Let us estimateI12(X). We notice that∇ηQ is supported in 5BQ \ 4 BQ. TakeY ∈ 5 BQ \ 4 BQ. If X ∈ BQ then |X − Y| ≈ |XQ − Y| ≈ ℓ(Q) which yields|∇2

XE(X −Y)| . |X −Y|−(n+1) ≈ ℓ(Q)−(n+1). Then using (2.58) with k = 2 we obtain

|I12(X)| . ℓ(Q)−2"Ω∩5 BQ\4 BQ

|∇YG(XQ,Y)|dY. ℓ(Q)−2I2 . ℓ(Q)−1.

Collecting our estimates forI11 and I12 we have shown that|I1(X)| . ℓ(Q)−1.Thus, for allX ∈ BQ ∩ Ωext we have

|∇2XSbQ(X)| ≤ C ℓ(Q)−1.

Remark2.59. In the previous argument, for the estimate ofI1 we have only usedthatX ∈ BQ and we have not used thatX ∈ Ωext.

Next we letX ∈ BQ∩Ω, and suppose first thatδ(X) & ℓ(Q). If y ∈ 5∆Q we have|∇2

XE(X − y)| . |X − y|−(n+1) ≤ δ(X)−(n+1). ℓ(Q)−(n+1). Consequently,

|∇2XSbQ(X)| =

∣∣∣∣∫

∂Ω

∇2XE(X − y) bQ(y) dσ(y)

∣∣∣∣ . ℓ(Q)−1ωXQ(5 ∆Q) ≤ ℓ(Q)−1.

It remains to treat the caseX ∈ BQ ∩ Ω, with δ(X) ≤ cℓ(Q), wherec is to bechosen. Notice that

2ℓ(Q)/C ≤ δ(XQ) ≤ |XQ − X| + δ(X) ≤ |XQ − X| + cℓ(Q).


Then we pickc < C−1 we obtain|XQ − X| > ℓ(Q)/C. Next, we write as before∇2

XSbQ(X) = I1(X) + I2(X). For I1, sinceX ∈ BQ, by Remark2.59we concludethat |I1(X)| . ℓ(Q)−1. For I2 we observe that (2.1) gives

|I2(X)| = σ(Q)

∣∣∣∣∫

∂Ω

∇2XE(X − y) dωXQ(y)

∣∣∣∣ = σ(Q)∣∣∇2

X

(E(X − XQ) −G(X, XQ)

)∣∣

. σ(Q) |X − XQ|−(n+1) + σ(Q) |∇2XG(X, XQ)| . ℓ(Q)−1 + σ(Q) |∇2

XG(X, XQ)|.

Therefore, collecting all of our estimates, we have shown that for everyX ∈ BQ,

(2.60) |∇2XSbQ(X)| ≤ C ℓ(Q)−1 + σ(Q) |∇2

XG(X, XQ)|χBQ∩Ω∩δ(X)≤cℓ(Q)(X).

Notice that ifx ∈ ∂Ω then"ΛQ(x)

dYδ(Y)n−1 ≤

∑

x∈Q′∈DQ

∑

I∈WQ′∪WextQ′

"I

dYδ(Y)n−1

≈∑

x∈Q′∈DQ

∑

I∈WQ′∪WextQ′

ℓ(I )n+1

ℓ(I )n−1 .∑

x∈Q′∈DQ

ℓ(Q′)2. ℓ(Q)2

where we have used that #(WQ′∪WextQ′ ) . CC0. Therefore, using (2.55) and (2.60)

we have

∫

Q

("ΛQ(x)


) q2

dσ(x)

. ℓ(Q)−q∫

Q

("ΛQ(x)

dYδ(Y)n−1

) q2

dσ(x)

+ σ(Q)q∫

Q

("ΛQ(x)∩δ(Y)≤cℓ(Q)

|∇2YG(Y, XQ)|2 dY

δ(Y)n−1

) q2

dσ(x)

. σ(Q) + σ(Q)q∫

QSQu(x)q dσ(x)

where we setu(Y) = ∇YG(Y, XQ) and

SQu(x) =

("ΓQ(x)|∇u(Y)|2 dY

δ(Y)n−1

) 12

, ΓQ(x) :=⋃

x∈Q′∈DQ

UQ′ .

We claim that

(2.61)∫

QSQu(x)q dσ(x) . σ(Q)1−q.

Assuming this momentarily we obtain as desired (2.53):

∫

Q

("ΛQ(x)


) q2

dσ(x) . σ(Q).


Modulo the claim (2.61), we have now verified all the hypotheses of Theorem2.50and we therefore conclude that (2.54) holds. The latter in turn implies that∂Ω isUR (see [DS2, p. 44]).

To complete this stage of our proof, it remains to establish (2.61). We do this inSection2.5, after first proving a bound for the square function in terms of the non-tangential maximal function in Section2.4. As mentioned before, at present,Ωis actually an approximating domain (i.e.,Ω stands forΩN with N large enough).Thus, the conclusion that we have obtained in the current step is that for everyN ≫ 1, we have thatΩN has the UR property with uniform constants. In Section2.6 we show that this property may be transmitted toΩ, thereby completing theproof of Theorem1.22.

2.4. Step 3: Good-λ inequality for the square function and the non-tangentialmaximal function. We recall thatUQ =

⋃W∗

QI ∗ with I ∗ = (1 + λ)I , where 0<

λ ≤ λ0/2 was fixed above. As mentioned in Remark2.37, since 0< 2λ ≤ λ0,the fattened Whitney boxesI ∗∗ = (1 + 2λ)I , and corresponding Whitney regionsU∗Q :=

⋃W∗

QI ∗∗ enjoy the same properties as doI andUQ.

Let us set some notation: for fixedQ0 ∈ D, and forx ∈ Q0,

SQ0u(x) =

("ΓQ0(x)

|∇u(Y)|2 dYδ(Y)n−1

) 12

, NQ0,∗u(x) = supY∈ΓQ0(x)

|u(Y)|

whereΓQ0(x) :=

⋃

x∈Q∈DQ0

UQ, ΓQ0(x) :=⋃

x∈Q∈DQ0

U∗Q.

We also define a localized dyadic maximal operator

MQ0 f (x) = supx∈Q∈DQ0

?Q| f (y)|dσ(y).

Given anyQ ∈ DQ0, for any familyF ⊂ DQ0 of disjoint dyadic cubes we definethe fattened versions of the Carleson boxTQ and the local sawtoothΩF ,Q0 by

TQ0 = int

⋃

Q∈DQ0

U∗Q

, ΩF ,Q0 = int

⋃

Q∈DF ,Q0

U∗Q

.

It is straightforward to show that there existsκ0 (depending onK0) large enoughsuch that

(2.62) TQ0 ⊂ B(xQ0, κ0 ℓ(Q0)) ∩Ω.Let us writeBκ0Q0

= B(xQ0, κ0 ℓ(Q0)) and note that in particular, for everyx ∈ Q0

and every pairwise disjoint familyF ⊂ DQ0, we have

(2.63) ΓQ0(x) ⊂ ΓQ0(x) ⊂ TQ0 ⊂ Bκ0Q0∩Ω , ΩF ,Q0 ⊂ TQ0 ⊂ Bκ0Q0

∩ Ω.

Proposition 2.64. Given Q0 ∈ D, let u be harmonic in2Bκ0Q0∩ Ω. Then, for every

1 < q < ∞ we have

(2.65) ‖SQ0u‖Lq(Q0) ≤ C ‖NQ0,∗u‖Lq(Q0),


provided the left hand side is finite.

Remark.Let us emphasize that in this part of the argument,u is allowed to be anyharmonic function in 2Bκ0Q0

∩Ω, not only∇G(·, XQ) as above.

Proof. The proof is based on the standard “good-λ” argument of [DJK] (see also[Br]), but adjusted to our setting. Fix 1< q < ∞ and assume (qualitatively) thatthe left hand side of (2.65) is finite. In particular we have that

?Q0

SQ0u dσ ≤(?

Q0

(SQ0u)q dσ

) 1q

< ∞.

We consider first the case thatλ >>

Q0SQ0u dσ. Set

Eλ = x ∈ Q0 :MQ0(SQ0u)(x) > λ .Then by the usual Calderon-Zygmund decomposition argument, there exists a pair-wise disjoint family of cubesP j j ⊂ DQ0 \ Q0, which are maximal with respectto the property that

>P j

SQ0u dσ > λ, such thatEλ = ∪ jP j. Let P j denote the

dyadic parent ofP j . Then P j must contain a subset of positiveσ-measure onwhich SQ0u(x) ≤ λ, by maximality ofP j. Fix a cube inP j j , sayP0, and set

Fλ = x ∈ P0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λwhereβ > 1 is to be chosen and 0< γ < 1. We are going to show that

σ(Fλ) ≤ Cγθ σ(P0).

Assume thatσ(Fλ) > 0 otherwise there is nothing to prove. We claim that ifβ islarge enough, depending only on the ADR constants, thenσ(Fλ) < σ(P0). Other-wise, we would haveSQ0u(x) > βλ for a.e. x ∈ P0 and then by the maximality ofP0 and the fact that∂Ω is ADR,

β λ <

?P0

SQ0udσ ≤ C1

?P0

SQ0udσ ≤ C1 λ.

Choosingβ > C1, we obtain a contradiction. Thus,σ(Fλ) < σ(P0), so by the innerregularity ofσ, there exists a compact setF ⊂ Fλ ⊂ P0 such that

0 <12σ(Fλ) ≤ σ(F) ≤ σ(Fλ) < σ(P0).

Sinceσ(F) < σ(P0), it follows that int(P0) \ F is a non-empty open set. By astandard stopping time procedure, we may then subdivideP0 dyadically, to extracta pairwise disjoint family of cubesF = Q j j ⊂ DP0 \ P0, which are maximalwith respect to the property thatQ j ∩ F = Ø.

We see next thatF = P0 \ (∪FQ j). If x ∈ F ⊂ P0 andx ∈ Q j thenx ∈ Q j ∩ Fwhich contradicts the fact thatQ j ∩F = Ø. On the other hand, letx ∈ P0 \ (∪FQ j).Pick Qx

k ∈ DP0 the unique cube containingx with ℓ(Qxk) = 2−k ℓ(P0), k ≥ 1. Note

that Qxk ∩ F , Ø, otherwiseQx

k ⊂ Q j for some j and we would havex ∈ Q j. Letxk ∈ Qx

k ∩ F. Then|xk − x| . ℓ(Qxk) = 2−k ℓ(P0) and thereforexk → x ask → ∞.

SinceF is closed andxk ∈ F we conclude thatx ∈ F as desired.

Notice that for anyx ∈ F = P0 \ (∪FQ j) and for anyz ∈ P0 we have that


ΓQ0(x) =⋃

x∈Q∈DQ0

UQ =

( ⋃

x∈Q∈DQ0 ,Q⊂P0

UQ

)⋃( ⋃

x∈Q∈DQ0 ,Q)P0

UQ

)

= ΓP0(x)⋃( ⋃

Q∈DQ0 ,Q⊃P0

UQ

)⊂ ΓP0(x)

⋃( ⋃

z∈Q∈DQ0

UQ

)= ΓP0(x) ∪ ΓQ0(z).

Let us recall thatP0 is a Calderon-Zygmund cube inEλ and that we can pickz0 ∈ P0 with SQ0u(z0) ≤ λ (indeed we know that this happens in a set of positivemeasure inP0.) Then for anyx ∈ F ⊂ Fλ we have

β λ < SQ0u(x) =

("ΓQ0(x)

|∇u(Y)|2 dYδ(Y)n−1

) 12

(2.66)

≤("

ΓP0(x)|∇u(Y)|2 dY

δ(Y)n−1

) 12

+

("ΓQ0(z0)

|∇u(Y)|2 dYδ(Y)n−1

) 12

= SP0u(x) + SQ0u(z0)

≤ SP0u(x) + λ

and thereforeF ⊂ x ∈ P0 : SP0u(x) > (β − 1)λ.Next we claim that

(2.67)⋃

x∈FΓP0(x) ⊂

⋃

Q∈DF ,P0

UQ ⊂ ΩF ,P0

where ΩF ,P0 is the fattened version ofΩF ,P0 defined above. The second con-tainment in (2.67) is trivial (since UQ ⊂ U∗Q), so let us verify the first. Wetake Y ∈ ΓP0(x) with x ∈ F. Then, Y ∈ UQ where x ∈ Q ∈ DP0. Sincex ∈ F = P0 \ (∪FQ j) we must haveQ ∈ DF (otherwiseQ ⊂ Q j for someQ j ∈ Fand this would imply thatx ∈ Q j) and thereforeQ ∈ DF ,P0 which gives the firstinclusion.

Write ω⋆ = ωX0⋆ to denote the harmonic measure for the domainΩF ,P0 where

X0 = AP0 is given in [HM, Proposition 6.4] and [HM, Corollary 6.6] (which wemay apply toΩF ,P0 in place ofΩF ,P0 since 0< 2λ ≤ λ0, see Remark2.37). Letus also writeδ⋆(Y) to denote the distance fromY to ∂ΩF ,P0, andG⋆ to denote thecorresponding Green function. GivenY ∈ ΩF ,P0, let us chooseyY ∈ ∂ΩF ,P0 suchthat |Y − yY| = δ⋆(Y). By definition, forx ∈ F andY ∈ ΓP0(x), there is aQ ∈ DP0

such thatY ∈ UQ andx ∈ Q. Thus, by the triangle inequality, and the definition ofUQ, we have that forY ∈ ΓP0(x),

(2.68) |x− yY| ≤ |x− Y| + δ⋆(Y) ≈ δ(Y) + δ⋆(Y) ≈ δ⋆(Y) ,

where in the last step we have used that

(2.69) δ(Y) ≈ δ⋆(Y) for Y ∈⋃

Q∈DF ,P0

UQ .

Then, sinceF = P0 \ (∪FQ j) ⊂ ∂Ω ∩ ∂ΩF ,P0, see [HM, Proposition 6.1], we have

ω⋆(F) (λ (β − 1))2 =∫

F(λ (β − 1))2 dω⋆ ≤

∫

FSP0u(x)2 dω⋆(x)(2.70)


=

∫

F

"ΓP0(x)

|∇u(Y)|2 dYδ(Y)n−1 dω⋆(x)

.

"⋃

Q∈DF ,P0

UQ

|∇u(Y)|2 δ(Y)

(1δ(Y)n

∫

F∩B(yY,C δ⋆(Y))dω⋆(x)

)dY

.

"ΩF ,P0

|∇u(Y)|2 δ⋆(Y)ω

X0⋆ (∆⋆(yY,C δ⋆(Y))

(C δ⋆(Y))ndY,

where we have used (2.68), (2.67)and (2.69).

We now claim that forY ∈ ΩF ,P0, we have

(2.71)ω

X0⋆ (∆⋆(yY,C δ⋆(Y))

(C δ⋆(Y))n.

G⋆(X0,Y)

δ⋆(Y).

Indeed, if δ⋆(Y) < δ⋆(X0)/(2C), then (2.71) is immediate by Lemma2.11 and(2.10). Otherwise, we haveδ⋆(Y) ≈ δ⋆(X0) ≈ ℓ(P0) & |X0 − Y|, whence (2.71)follows directly from (2.8) and the Harnack Chain condition, and the fact that har-monic measure is a probability measure.

We recall that by hypothesis,u is harmonic in 2Bκ0Q0∩ Ω ⊃ ΩF ,P0 (cf. (2.63)).

LetL := ∇·∇ denote the usual Laplacian inRn+1, so thatL(u2) = 2 |∇u|2 in ΩF ,P0.Combining these observations with (2.70)-(2.71), we see that

ω⋆(F) (λ (β − 1))2 ."ΩF ,P0

|∇u(Y)|2 G⋆(X0,Y) dY(2.72)

=12

"ΩF ,P0

L(u2)(Y) G⋆(X0,Y) dY

= −12

u(X0)2 +12

∫

∂ΩF ,P0

u(y)2dωX0⋆ (y).

where the last step is a well known identity obtained using properties of the Greenfunction.

Let Y ∈ ΩF ,P0, so thatY ∈ U∗Q for someQ ∈ DF ,P0. By definition ofDF ,P0, thisQ cannot be contained in anyQ j ∈ F . Therefore,Q∩ F , Ø. Indeed, otherwiseQ∩ F = Ø, which by maximality of the cubes inF , would imply thatQ ⊂ Q j forsomeQ j ∈ F , a contradiction. Thus, there is somex ∈ Q∩ F which then satisfiesx ∈ P0 \ (∪FQ j) andx ∈ Q ∈ DP0. Hence,Y ∈ U∗Q ⊂ ΓP0(x) with x ∈ F. SinceF ⊂ Fλ we have that

|u(Y)| ≤ supZ∈ΓP0(x)

|u(Z)| ≤ NQ0,∗u(x) ≤ γ λ.

Thus,|u(Y)| ≤ γ λ and in particularu(Y) ≥ −γ λ, for all Y ∈ ΩF ,P0. Next, we apply[JK, Theorem 6.4] to the (qualitative) NTA domainΩF ,P0 (we recall that, in thepresent stage of the argument,Ω is actuallyΩN for some largeN, which satisfiesthe qualitative exterior Corkscrew condition (Definition2.5); thus, the boundeddomain ΩF ,P0 enjoys an exterior Corkscrew condition with constants thatmaydepend very badly onN.) Of course, the interior Corkscrew and Harnack chain


constants, as well as the ADR constants, are controlled uniformly in N). Conse-quently, we obtain thatu has non-tangential limit ˜ω⋆-a.e.. Since|u(Y)| ≤ γ λ foreveryY ∈ ΩF ,P0, its non-tangential limit therefore satisfies this same bound ω⋆-a.e., i.e., for ˜ω⋆-a.e.y ∈ ∂ΩF ,P0 we have|u(y)| ≤ γ λ. Consequently, by (2.72), weconclude that ˜ω⋆(F) (λ (β − 1))2 . (γ λ)2 and then ˜ω⋆(F) . γ2.

We use the notation of [HM, Lemma 6.15] (withΩF ,P0 and ωX0⋆ in place of

ΩF ,Q0 andωX0⋆ ). SinceP0 is not contained in anyQ j ∈ F ,

PF ν(P0) = ωX0⋆

(P0 \ (∪FQ j)

)+

∑

Q j∈F ,Q j(Q

σ(P0 ∩ Q j)σ(Q j)

ωX0⋆ (P j)

= ωX0⋆

(P0 \ (∪FQ j)

)+

∑

Q j∈F ,Q j(P0

ωX0⋆ (P j)

& ωX0⋆

(P0 \ (∪FQ j)

)+

∑

Q j∈F ,Q j(Q

ωX0⋆

(B(x⋆j , r j) ∩ ∂ΩF ,P0

)

≥ ωX0⋆ (∆P0

⋆ ),

where in the third line we have used the doubling property of ˜ωX0⋆ (plus a subdivi-

sion and Harnack Chain argument ifℓ(Q j) ≈ ℓ(P0)), and in the last line we haveused [HM, Proposition 6.12], along with [HM, Proposition 6.1] and [HM, Propo-sition 6.3] and the doubling property to ignore the difference betweenQ \ (∪FQ j)andQ∩∂ΩF ,Q0. Also, ∆P0

⋆ = ∆P0⋆ (x⋆P0

, tP0) is a surface ball such that ˜x⋆P0∈ ∂ΩF ,P0,

tP0 ≈ ℓ(P0), dist(P0, ∆P0⋆ ) . ℓ(P0), and the implicit constants may depend uponK0,

see [HM, Proposition 6.12]. Using Lemma2.2, Harnack chain and [HM, Corollary6.6] it is immediate to show that ˜ωX0

⋆ (∆P0⋆ ) ≥ C and thereforePF ν(P0) ≥ C. On

the other hand, sinceF = P0 \ (∪FQ j) we conclude that

PF ν(F)PF ν(P0)

. PF ν(F) = ωX0⋆ (F) . γ2.

Thus Harnack chain, [HM, Corollary 6.6] and [HM, Lemma 6.15] imply(2.73)

ωX∆P0 (F) ≤ ω

X∆P0 (F)

ωX∆P0 (P0)

≈ ωX0(F)ωX0(P0)

=PFωX0(F)PFωX0(P0)

.

(PF ν(F)PF ν(P0)

) 1θ

. γ2/θ,

where we recall that∆P0 ⊂ P0 ⊂ ∆P0.

We need the following auxiliary result, the proof is given below.

Lemma 2.74. Assume(2.49). Then, given a surface ball∆0 = B0 ∩ ∂Ω for every∆ = B∩ ∂Ω with B⊂ B0 we have

(2.75)

(?∆

(kX∆0

)pdσ

)1/p

≤ C?∆

kX∆0 dσ.

Notice that this result says thatkX∆0 ∈ RHp(∆0) ⊂ A∞(∆0) for all ∆0 and theconstants are uniform in∆0. SinceA∞(∆0) defines an equivalence relationship (onthe set of doubling measures on∆0) we obtain thatσ ∈ A∞(∆0, ω

X∆0) for all ∆0 andthe constants are uniform in∆0. Thus, there exist positive constantsC andϑ such


that for every∆0 = B0 ∩ ∂Ω, ∆ = B∩ ∂Ω with B ⊂ B0 and every Borel setE wehave

σ(E)σ(∆)

≤ C

(ωX∆0(E)

ωX∆0(∆)

)ϑ

We apply this with∆ = ∆0 = ∆P0 and withE = F ⊂ P0 ⊂ ∆P0. Then, Lemma2.2and (2.73) imply

σ(F)σ(P0)

≈ σ(F)

σ(∆P0).

(ω

X∆P0 (F)

ωX∆P0 (∆P0)

)ϑ. ω

X∆P0 (F)ϑ . γ2ϑ/θ.

Let us recall thatσ(Fλ) ≤ 2σ(F) and therefore we have obtained

(2.76) σ(x ∈ P0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

)≤ Cγ2ϑ/θσ(P0),

where all the constants are independent ofP0.

We recall thatP0 is an arbitrary cube inP j j , which is a family of Calderon-Zygmund cubes associated withEλ for λ >

>Q0

SQ0u dσ. In addition, forσ-a.e.

x ∈ Q0 such thatSQ0u(x) > βλ, we haveλ < β λ < SQ0u(x) ≤ MQ0

(SQ0u

)(x)

and thereforex ∈ Eλ. Using these observations and (2.76) in eachP j, we obtain

σ(x ∈ Q0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

)

= σ(x ∈ Q0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ ∩ Eλ

)

=∑

j

σ(x ∈ P j : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

)

≤ Cγ2ϑ/θ∑

j

σ(P j)

= Cγ2ϑ/θσ(Eλ)

= Cγ2ϑ/θσ(x ∈ Q0 :MQ0(SQ0u)(x) > λ

).

Thus, we have shown that for everyλ >>

Q0SQ0u dσ we have

(2.77) σ(x ∈ Q0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

)

≤ Cγθ′σ(x ∈ Q0 :MQ0(SQ0u)(x) > λ

).

Let us now consider the caseλ ≤>

Q0SQ0u dσ. We are going to show that

(2.78) σ(x ∈ Q0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

)≤ C γθ

′σ(Q0).

We repeat the previous computations withP0 = Q0 with the main differencethatP0 is no longer a maximal Calderon-Zygmund cube. We take the same setFλand we assume thatσ(Fλ) > 0, otherwise the desired estimate is trivial. Beforewe used the maximality ofP0 to show thatσ(Fλ) < σ(P0) for β large enough andthis was only used to obtain thatσ(F) < σ(P0). Here we cannot do that and weproceed as follows. By the inner regularity ofσ we can find a compact setF, suchthat Ø, F ⊂ Fλ ⊂ P0 and

0 <12σ(Fλ) ≤ σ(F) ≤ σ(Fλ) ≤ σ(P0).


If σ(Fλ) < σ(P0) or σ(F) < σ(Fλ) we setF = F and we haveσ(F) < σ(P0).Otherwise, i.e., ifσ(F) = σ(Fλ) = σ(P0), we use the following lemma:

Lemma 2.79. LetΩ have the ADR property with constant C1, i.e.,

C−11 rn ≤ σ(∆(x, r)) ≤ C1 rn

for all x ∈ ∂Ω and 0 < r < diam∂Ω. Let Q ∈ D and recall that there exists∆Q = ∆(xQ, rQ) with rQ ≈ ℓ(Q) with the property that∆Q ⊂ Q ⊂ ∆(xQ,C2 rQ) forsome uniform constant C2 ≥ 1. Takeτ0 = (2C2

1)−1/n and set∆τ0Q = ∆(xQ, τ0 rQ).Then,

∆τ0Q ⊂ ∆Q ⊂ Q and (2C4

1 Cn2)−1σ(Q) ≤ σ

(∆τ0Q

)≤ 3

4σ(Q).

Assuming this momentarily (the proof is given below), we apply this result toP0. SetF = F ∩ ∆τ0P0

⊂ F which is a compact set. As we haveF ⊂ P0 andσ(F) = σ(Fλ) = σ(P0) we obtain

(2C41 Cn

2)−1σ(Fλ) = (2C41 Cn

2)−1σ(P0) ≤ σ(∆τ0P0

)= σ

(∆τ0P0∩ F)+ σ(∆τ0P0\ F)

≤ σ(F) + σ(P0 \ F) = σ(F) ≤ σ(∆τ0P0

)≤ 3

4σ(P0) =

34σ(Fλ).

Thus, we have found a compact setF such that Ø, F ⊂ Fλ ⊂ P0 and

0 < (2C41 Cn

2)−1σ(Fλ) ≤ σ(F) < σ(Fλ) = σ(P0).

This allows us to run the stopping time argument and find the family F as before.We then continue the argument and notice that in (2.66), we used the maximalityof P0. Here, the analogous estimate is trivial: sinceP0 = Q0, it follows that forany x ∈ F ⊂ Fλ, we haveSP0u(x) = SQ0u(x) > βλ > (β − 1)λ and thereforeF ⊂ x ∈ P0 : SP0u(x) > (β − 1)λ. From this point the proof continues withoutchange (except for the fact that we haveσ(Fλ) . σ(F) in place ofσ(Fλ) ≤ 2σ(F)),so that (2.76) holds, and in the present case this is our desired estimate (2.78).

Let 1< q < ∞ and writeaQ0 =>

Q0SQ0u dσ < ∞. Then

‖SQ0u‖qLq(Q0) = β

q∫ ∞

0qλqσ

(x ∈ Q0 : SQ0u(x) > βλ

) dλλ

≤ βq∫ ∞

0qλqσ

(x ∈ Q0 : SQ0u(x) > βλ, NQ0,∗u(x) ≤ γ λ

) dλλ

+ (β/γ)q ‖NQ0,∗u‖qLq(Q0)

= βq∫ aQ0

0· · · dλλ+ βq

∫ ∞

aQ0

· · · dλλ+ (β/γ)q ‖NQ0,∗u‖

qLq(Q0)

= I + II + (β/γ)q ‖NQ0,∗u‖qLq(Q0).

For I we use (2.78) and Jensen’s inequality:

I ≤ C γθ′βqσ(Q0)

∫ aQ0

0qλq dλ

λ= C γθ

′βqσ(Q0) aq

Q0≤ Cγθ

′βq ‖SQ0u‖

qLq(Q0).

For II we use (2.77) and the fact thatMQ0 is bounded onLq(Q0) (notice thatMQ0

is the localized dyadic Hardy-Littlewood maximal functionand we can insert the


characteristic function ofQ0 in the argument for free):

II ≤ Cγθ′βq∫ ∞

aQ0

qλqσ(x ∈ Q0 :MQ0(SQ0u)(x) > λ

) dλλ

= Cγθ′βq ‖MQ0(SQ0u)‖qLq(Q0) ≤ C γθ

′βq ‖SQ0u‖

qLq(Q0).

Gathering the obtained estimates we conclude that

‖SQ0u‖qLq(Q0) ≤ C γθ

′βq ‖SQ0u‖

qLq(Q0) + (β/γ)q ‖NQ0,∗u‖

qLq(Q0).

Choosingγ small enough so thatC γθ′βq < 1

2 the first term in the right hand sidecan be hidden in the left hand side (since it is finite by assumption) and we concludeas desired (2.65).

Proof of Lemma2.74. Notice that in proving (2.75), we may suppose that the ballsB andB0 have respective radiirB ≪ rB0; otherwise, ifrB ≈ rB0, then (2.75) reducesimmediately to (2.49). We now may proceed as in the first part of the proof ofProposition2.41in order to use Corollary2.36: we take∆ = B∩ Ω and for everyY ∈ Ω \ 2κ0B we have?

∆′kY dσ ≈ ωY(∆)

?∆′

kX∆ dσ B′ ⊂ B.

By the Harnack chain condition this estimate holds forY = X∆0. Then forσ-a.e.x ∈ ∆ we takeB′ = B(x, r) and letr → 0 to obtainkX∆0 (x) ≈ ωX∆0 (∆) kX∆(x).Consequently, by (2.49) we have(?∆

(kX∆0

)pdσ

) 1p

. ωX∆0(∆)

(?∆

(kX∆)p

dσ

) 1p

.ωX∆0(∆)σ(∆)

=

?∆

kX∆0 dσ.

Proof of Lemma2.79. The proof is almost trivial. What∆τ0Q ⊂ ∆Q follows at

once from the fact thatτ0 < 1. On the other hand, let us notice that∆τ0Q ⊂∆(xQ, (3/2)1/n τ0 rQ) and therefore the ADR property gives

σ(∆τ0Q

)≤ σ

(∆(xQ, (3/2)1/n τ0 rQ)) ≤ C1

32τn0 rn

Q =34

C−11 rn

Q ≤34σ(∆Q)

and

σ(Q) ≤ σ(∆(xQ,C2 rQ)) ≤ C1 Cn2 rn ≤ C2

1 Cn2 τ−n0 σ(∆τ0Q ) ≤ 2C4

1 Cn2σ(∆τ0Q

).

2.4.1. Good-λ inequality for truncated cones.In order to apply Proposition2.64we need to know a priori that‖SQ0u‖Lq(Q0) < ∞ (qualitatively), which can be veri-fied as follows. SinceΩN satisfies the exterior corkscrew condition at small scales,we may invoke [JK] to control the square function by the non-tangential maximaloperator with constants that may depend onN (we recall that at this stageΩ = ΩN),but this gives, in particular, qualitative finiteness of thesquare function. The resultsof [JK] apply to square functions and non-tangential maximal functions defined viathe classical conesΓα(x) := Y ∈ Ω : |Y − x| < (1+ α) δ(Y), but our dyadic conesmay be compared to the classical cones by varying the aperture parameterα. Thus,


we may infer also that the our square function has (qualitatively) finite Lp norm,given finiteness of our non-tangential maximal function, sowe may then applyProposition2.64to obtain that (2.65) holds with uniform bounds.

There is, however, a different approach that consists of working with cones thatare “truncated” so that they stay away from the boundary. This in turns impliesthat the truncated square function is bounded and thereforethat the correspondingleft hand side is finite. Passing to the limit we conclude thatthe a priori finitenesshypothesis can be removed from Proposition2.64. We present this argument forthe sake of self-containment, and because the argument is ofindependent interest.

Before stating the precise result we introduce some notation. GivenQ0, we takek ≥ k(Q0) + 2 (recall thatk(Q0) is defined in such a wayℓ(Q0) = 2−k(Q0)) a largeenough integer (eventually,k ↑ ∞). We defineΓk

Q0(x) to be the truncated cone

where the cubes in the union satisfy additionallyℓ(Q) ≥ 2−k. The correspondingtruncated square function is written asSk

Q0.

Proposition 2.80. Given Q0 ∈ D, let u be harmonic in2Bκ0Q0∩ Ω. Then, for every

1 < q < ∞ and for every k≥ k(Q0) + 2 we have

(2.81) ‖SkQ0

u‖Lq(Q0) ≤ C ‖NQ0,∗u‖Lq(Q0),

where the constant C is independent of k and consequently(2.65) holds whether ornot the left hand side is finite.

Proof. It is straightforward to see that the truncated cones are away from theboundary: the cubes that define the cones have side length at least 2−k and thus thecorresponding Whitney boxes have side length at lestC 2−k, thus dist(Γk

Q0, ∂Ω) &

2−k. Note thatu is harmonic in 2Bκ0Q0∩ Ω and therefore smooth in32 Bκ0Q0

∩ Y ∈Ω : δ(Y) & 2−k. Thus, (2.63) implies |∇u| ≤ CQ0,k,u in Γk

Q0(x), x ∈ Q0, and con-

sequentlySkQ0

u ∈ L∞(Q0). The bound may depend ofk, Q0 but we will use thisqualitatively and a not quantitatively. Note also, that ifQ ∈ DQ0 with ℓ(Q) = 2−k

thenΓkQ0

(x) does not depend onx ∈ Q (i.e. ΓkQ0

(x) = ΓkQ0

(x′) for everyx, x′ ∈ Q)and thusSk

Q0u is constant onQ.

Let us writeMkQ0

for the truncated dyadic maximal function where the cubesin the sup have side length at least 2−k. Associated toSk

Q0u, and for eachλ >>

Q0Sk

Q0u dσ we define the correspondingEλ; we clearly haveEλ = ∪ jP j with

P j being maximal cubes as before satisfying furtherℓ(P j) ≥ 2−k (note that as wehave observed thatSk

Q0u is constant on cubes of size 2−k we could have takenMQ0

obtaining the same familyP j with the same properties.) We also observe thatSk

Q0u ≤ Mk

Q0(Sk

Q0u) a.e. inQ0. Fix one of these cubesP0 and defineFλ with

the truncated square functionSkQ0

replacingSQ0 —we do not truncate the non-tangential maximal operator. Then we proceed as before, assume thatσ(Fλ) >0 and find the corresponding setF, and a familyF with the same properties asbefore. We can easily see thatΓk

Q0(x) ⊂ Γk

P0(x) ∪ Γk

Q0(z) for every x ∈ F and

z ∈ P0. Then, sinceP0 is a Calderon-Zygmund cube inEλ, we can pickz0 ∈ P0

with SkQ0

u(z) ≤ λ (indeed we know that this happens in a set of positive measurein


P0.) Then for anyx ∈ F ⊂ Fλ , we have

β λ < SkQ0

u(x) =

("Γk

Q0(x)|∇u(Y)|2 dY

δ(Y)n−1

) 12

(2.82)

≤("

ΓkP0

(x)|∇u(Y)|2 dY

δ(Y)n−1

) 12

+

("Γk

Q0(z0)|∇u(Y)|2 dY

δ(Y)n−1

) 12

= SkP0

u(x) + SkQ0

u(z0)

≤ SkP0

u(x) + λ

and thereforeF ⊂ x ∈ P0 : SkP0

u(x) > (β − 1)λ.Next, using thatSk

P0u(x) ≤ SP0u(x) we easily obtain (2.70) and from this point

the argument goes through without change. Thus, we obtain the following analogof (2.76)

(2.83) σ(x ∈ P0 : SkQ0

u(x) > βλ, NQ0,∗u(x) ≤ γ λ) ≤ C γ2ϑ/θσ(P0),

where all the constants are independent ofP0 andk.

Let us consider the caseλ ≤>

Q0Sk

Q0u dσ. The same argument as before works

in this case and we easily obtain

(2.84) σ(x ∈ Q0 : SkQ0

u(x) > βλ, NQ0,∗u(x) ≤ γ λ) ≤ C γθ′σ(Q0).

Gathering (2.83) and (2.84) and repeating the computations above we concludethat

‖SkQ0

u‖qLq(Q0) ≤ C γθ′βq ‖Sk

Q0u‖qLq(Q0) + (β/γ)q ‖NQ0,∗u‖

qLq(Q0),

where all the constants are independent ofk. As observed, the fact that we areworking with truncated square functions gives us thatSk

Q0u ∈ L∞(Q0) (we use this

in a qualitative way but not quantitatively). Thus, choosing γ small enough so thatC γθ

′βq < 1

2 the first term in the right hand side can be hidden in the left hand side.Hence, we obtain as desired (2.81)

‖SkQ0

u‖qLq(Q0) ≤ C ‖NQ0,∗u‖qLq(Q0)

whereC is independent ofk. Letting k ↑ ∞, the monotone convergence theoremgives (2.65) without assuming that the left hand side is finite. Thus we have proveda version of Proposition2.64where there is no need to assume that the left handside is finite.

2.5. Step 4: Proof of (2.61). In order to obtain (2.61) we shall use Proposition2.64, applied tou = ∇G, with our exponentq = p. Thus, we need to study thenon-tangential maximal function ofu, and it suffices to do this for each componentof u, i.e., for any given partial derivative ofG. More precisely, we setu(Y) =∂YjG(Y, XQ), whereXQ is the corkscrew point associated to∆Q defined in Section2.3. As mentioned there,XQ < 6 BQ with BQ = B(xQ, κ1 ℓ(Q)) andκ1 > κ0. InparticularXQ < 6Bκ0Q, so thatu is harmonic in 2Bκ0Q ∩ Ω.

Proposition 2.85. We have

‖NQ,∗u‖qLq(Q) ≤ Cσ(Q)1−q .


Proof. Let x ∈ Q andY ∈ ΓQ(x). By (2.63) applied toQ in place ofQ0 we haveΓQ(x) ⊂ Bκ0Q ∩ Ω and thereforeu is harmonic inB(Y, 3δ(Y)/4) ⊂ 7

4Bκ0Q. We thenobtain by the mean value property of harmonic functions, Caccioppoli’s inequality,the Harnack chain condition and Lemma2.11, that

|u(Y)| ≤?

B(Y,δ(Y)/2)|∇ZG(Z, XQ)|dZ . δ(Y)−1

(?B(Y, 34δ(Y))

G(Z, XQ)2 dZ

) 12

.G(Y, XQ)δ(Y)

≈ G(X∆Y, XQ)δ(Y)

≈ ωXQ(∆Y)δ(Y)n

where∆Y = ∆(y, δ(Y)), with y ∈ ∂Ω such that|Y − y| = δ(Y). It is easy to see thatwe can findC large enough so thatx ∈ ∆(y,C δ(Y)) and∆Y ⊂ C ∆Q. As usual, welet XC ∆Q

denote a Corkscrew point with respect toC ∆Q. Then,

|u(Y)| . 1δ(Y)n

∫

∆Y

kXQ dσ ≤ 1δ(Y)n

∫

∆(y,C δ(Y))kXQ χC ∆Q

dσ

.1δ(Y)n

∫

∆(y,C δ(Y))k

XC ∆Q χC ∆Qdσ . M(k

XC ∆Q χC ∆Q)(x) ,

where in the next-to-last inequality we have used the Harnack Chain condition.Therefore, using thatq = p and (2.49), we obtain

‖NQ,∗u‖qLq(Q) . ‖M(kXC ∆Q χC ∆Q

)‖qLq(Q) .

∫

C ∆Q

(k

XC ∆Q

)qdσ

. σ(C ∆Q)1−q. σ(Q)1−q.

Now we are ready to establish (2.61). Given Q ∈ D, we first apply Propo-sition 2.80 (or else Proposition2.64) with q, with Q in place of Q0, and withu = ∂YjG(·, XQ) as above, since the latter is harmonic in 2Bκ0Q ∩ Ω. Then, usingProposition2.85, we obtain as desired that

‖SQu‖qLq(Q) ≤ C ‖NQ,∗u‖qLq(Q) ≤ Cσ(Q)1−q .

2.6. UR for Ω. To conclude the proof of our main result we see that the UR prop-erty for the approximating domainsΩN with uniform bounds passes toΩ. As ob-served before, as a consequence of theTb theorem Theorem2.50we have obtainedthat (2.54) holds forΩN with uniform bounds. This in turn implies that∂ΩN is UR(see [DS2, p. 44]) with uniform bounds. To obtain that this property ispreservedwhen passing to the limit we use an argument, based on ideas ofGuy David, alongthe lines of [HM, Appendix C] (indeed the present situation is easier) wherewehave to switch the roles ofΩ andΩN.

To show thatΩ has the UR property we use the singular integral characteriza-tion. We recall that a closed,n-dimensional ADR setE ⊂ Rn+1 is UR if and only iffor all singular kernelsK as below, and corresponding truncated singular integrals


Tε, we have that

(2.86) supε>0

∫

E|Tε f |2 dHn ≤ CK

∫

E| f |2 dHn.

Here,

TE,ε f (x) = Tε f (x) :=∫

EKε(x− y) f (y) dHn(y) ,

whereKε(x) := K(x)Φ(|x|/ε), with 0 ≤ Φ ≤ 1, Φ(ρ) ≡ 1 if ρ ≥ 2, Φ(ρ) ≡ 0 ifρ ≤ 1, andΦ ∈ C∞(R), and where the singular kernelK is an odd function, smoothonRn+1 \ 0, and satisfying

|K(x)| ≤ C |x|−n(2.87)

|∇mK(x)| ≤ Cm |x|−n−m , ∀m= 1, 2, 3, . . . .(2.88)

We also introduce the following extension of these operators

(2.89) TE f (X) :=∫

EK(X − y) f (y) dHn(y) , X ∈ Rn+1 \ E.

We define non-tangential approach regionsΥEτ (x) as follows. LetWE denote the

collection of cubes in the Whitney decomposition ofRn+1 \ E, and setWτ(x) :=I ∈ WE : dist(I , x) < τℓ(I ). We then define

ΥEτ (x) :=

⋃

I∈Wτ(x)

I ∗

(thus, roughly speaking,τ is the “aperture” ofΥEτ (x)). ForF ∈ C(Rn+1\E) we may

then also define the non-tangential maximal function

NE∗,τ(F)(x) := sup

Y∈ΥEτ (x)|F(Y)|.

Let us recall [HM, Lemma C.5] which states that ifE ⊂ Rn+1 is n-dimensionalUR, we then have

(2.90)∫

E

(NE∗,τ (TE f )

)2dHn ≤ Cτ,K

∫

E| f |2dHn.

for every 0< τ < ∞ and withCτ,K depending only onn, τ, K and the UR constants.

After these preliminaries, in order to show that∂Ω is UR we take one of theprevious kernelsK and form the corresponding operatorsTε = T∂Ω,ε. Fix ε > 0and N ≥ 1 such thatε ≫ 2−N. We write∂Ω = ∪ jQ j with Q j ∈ DN(∂Ω). Forfixed j, if we write Q j for the dyadic parent ofQ j, we have by construction thatXQ j∈ int(UQ j

) ⊂ ΩN. Then in the segment joiningx j (the center ofQ j) andXQ j

there exists a point ˆx j ∈ ∂ΩN. Next we takeQ j(N) the unique cube inDN(∂ΩN)such that ˆx j ∈ Q j(N). Note that we have dist(Q j ,Q j(N)) ≈ 2−N. Since∂ΩN isADR with uniform bounds inN, any givenQ(N) ∈ D(∂ΩN) can serve in this wayfor at most a bounded number ofQ j ∈ D(∂Ω). Thus we have

(2.91)∑

Q j∈DN(∂Ω)

1Q j (N)(x) ≤ C , ∀x ∈ ∂ΩN.


As usual, we setσ := Hn|∂Ω, and we now also letσN := Hn|∂ΩN . For τ largeenough, we have that ifx′ ∈ Q j(N) andx ∈ Q j, thenx ∈ Υ ∂ΩN

τ (x′). Thus,

∫

∂Ω

|T∂ΩN f |2 dσ =∑

Q j∈DN(∂Ω)

∫

Q j

|T∂ΩN f |2 dσ

=∑

Q j∈DN(∂Ω)

1σN(Q j(N))

∫

Q j (N)

∫

Q j

|T∂ΩN f (x)|2 dσ(x) dσN(x′)

.

∑

Q j∈DN(∂Ω)

∫

Q j (N)

(N∂ΩN∗,τ

(T∂ΩN f

))2dσN ≤ Cτ,K

∫

∂ΩN

| f |2dσN ,

where in the last line we have used first the ADR properties of∂ΩN and∂Ω, andthen (2.91) and (2.90) with E = ∂ΩN (as we may do, since∂ΩN is UR with uniformbounds). In particular we observe thatCτ,K is independent ofN.

Thus we have shown thatT∂ΩN : L2(∂ΩN) → L2(∂Ω). Since the kernelK isodd, we therefore obtain by duality that

(2.92) T∂Ω : L2(∂Ω)→ L2(∂ΩN).

Let us now observe thatKε is odd, smooth away the origin and satisfies (2.87),(2.88) uniformly in ε. Thus (2.92) applies to the corresponding operatorT∂Ω,εdefined by means ofKε, that is, we haveT∂Ω,ε : L2(∂Ω) → L2(∂ΩN) with boundsthat are uniform onN andε. Then we proceed as in Case 2 of [HM, Appendix C]and write∫

∂Ω

|T∂Ω,ε f |2 dσ =∑

Q j∈DN(∂Ω)

∫

Q j

|T∂Ω,ε f |2 dσ

=∑

Q j∈DN(∂Ω)

1σN(Q j(N))

∫

Q j (N)

∫

Q j

|T∂Ω,ε f (x)|2 dσ(x) dσN(x′)

.

∑

Q j∈DN(∂Ω)

1σN(Q j(N))

∫

Q j (N)

∫

Q j

|T∂Ω,ε f (x) − T∂Ω,ε f (x′)|2 dσ(x) dσN(x′)

+∑

Q j∈DN(∂Ω)

∫

Q j (N)|T∂Ω,ε f (x′)|2 dσN(x′),

= I + II ,

where we have use the ADR property for both∂Ω and∂ΩN. To estimateII we use(2.91) and (2.92) applied toT∂Ω,ε:

II .∫

∂ΩN

|T∂Ω,ε f |2 dσN ≤ Cτ,K

∫

∂Ω

| f |2 dσ.

On the other hand, standard Calderon-Zygmund arguments that are left to the in-terested reader give that (2.87) and (2.88) imply

|T∂Ω,ε f (x) − T∂Ω,ε f (x′)| . M∂Ω f (x), x ∈ Q j , x′ ∈ Q j(N),


whereM∂Ω is the Hardy-Littlewood maximal operator on∂Ω and the constants areuniform inε andN, and where we have used that 2−N ≪ ε. Hence,

I .∑

Q j∈DN(∂Ω)

∫

Q j

M∂Ω f (x)2 dσ(x) =∫

∂Ω

M∂Ω f (x)2 dσ(x) .∫

∂Ω

| f (x)|2 dσ(x).

Gathering our estimates, we conclude that∫

∂Ω

|T∂Ω,ε f |2 dσ .∫

∂Ω

| f |2 dσ,

where the implicit constants are independent ofε andN, which gives at once that∂Ω is UR.

Remark2.93. Gathering together this section with [HM, Appendix C] we concludethat∂Ω is UR if and only ifΩN is UR with uniform bounds for allN ≫ 1: herewe have shown the right-to-left implication and the converse argument is given in[HM, Appendix C].

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Steve Hofmann, Department of Mathematics, University of Missouri, Columbia, MO 65211,USA

E-mail address: [email protected]

JoseMarıaMartell, Instituto deCienciasMatematicas CSIC-UAM-UC3M-UCM, ConsejoSu-perior de Investigaciones Cientıficas, C/ Nicolas Cabrera, 13-15, E-28049 Madrid, Spain


Ignacio Uriarte-Tuero, Department of Mathematics, Michigan State University, East Lansing,MI 48824, USA


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