Weighted inequalities for gradients on non-smooth domains ...

where ɛ>0 is a constant depending on the domain Ω. To see that this generalization is the naturalone, think of (0.5) as a function of x ∈ R d and replace ω with Lebesgue measure. It’s importantto note that, for any x, the sum in (0.5) contains essentially only one term.Finally there is the right-hand term l(Q) d+1 . We will replace this with l(Q b )ω(Q b ).With the precise definitions still to follow, the rephrased version of (0.2) is⎛∫µ(T (Q b )) 1/q ⎜⎝∂Ω⎡⎤∞∑ 2⎣ω(Q −jɛb )ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σ(x) dω(x) ⎠1/p ′ ≤ cl(Q b )ω(Q b ),and our main theorem (Theorem 3.1), which we prove in Section 3, isTheorem 3.1. Let Ω ⊂ R d+1 be a bounded Lipschitz domain, and let ω be harmonic measureon ∂Ω for some fixed point X 0 ∈ Ω. Suppose that v ∈ L 1 (∂Ω,dω) is a non-negative function andµ is a positive Borel measure on Ω. Define σ ≡ v 1−p′ , and suppose that σdω ∈ A ∞ (ω) on ∂Ω. If1

or, more succinctly:|φ (Q) (x)| ≤ √ ω(Q)∞∑( 2−jαj=0ω(2 j Q))χ Rj (Q)(x).2) Smoothness. For any x and y in R d ,|φ (Q) (x) − φ (Q) (y)| ≤( |x − y|l(Q)) β√ ∑∞ ( 2−jαω(Q)j=0ω(2 j Q))(χ Rj (Q)(x)+χ Rj (Q)(y)).Note that, given the size condition, the smoothness condition is only meaningful when |x−y| ≤l(Q).3) Cancellation. For every finite linear combination ∑ Q γ Qφ (Q) ,∫R d | ∑ Qγ Q φ (Q) | 2 dω ≤ ∑ Q|γ Q | 2 .All of our results depend on the next theorem.Theorem 1.1. Let {φ (Q) } Q∈D be a standard familyof functions, and let ν ∈ A ∞ (ω). If0

The key to our argument lies in defining the right maximal function. Let us assume that wehave a fixed finite linear combination f = ∑ Q λ Qφ (Q) .IfI ∈D,wedefineS(I) ≡{Q ∈D: Q ⊄ I}.It is useful to think of S(I) as the family of dyadic cubes that “surround” I. Ifx ∈ I, wedefineF (I,x) ≡∑Q:Q∈S(I)λ Q φ (Q) (x),and we do not define F (I,x)forx/∈ I. Ifx I is the center of I, then we set F (I) ≡ F (I,x I ). Theright maximal function for the Littlewood-Paley function g ∗ (f) turns out to beF ∗ (x) ≡ sup |F (I)|.I:x∈ICorresponding to F ∗ (x) is a “maximal square function” adapted to g ∗ (f). For x ∈ I, wedefine⎛G(I,x) ≡ ⎝ ∑Q∈S(I)⎡∞∑|λ Q | 2 ⎣j=02 −j(2α−τ)⎤⎞ω(2 j Q) χ R j (Q)(x) ⎦⎠and we do not define G(I,x)forx/∈ I. We similarly define G(I) ≡ G(I,x I )andG ∗ (x) ≡ sup G(I).I:x∈IIn order to prove Theorem 1.1, we shall prove seven fairly elementary lemmas, followed by adifficult lemma, which is really the heart of the proof of Theorem 1.1. These lemmas are directlyanalogous to, respectively, Lemmas 1–7 and Lemma 1.9 in [Wi]. Our more general formulation ofthe φ (Q) ’s requires us to surmount some non-trivial technical obstacles.Lemma 1.2. For ω-a. e. x, |f(x)| ≤F ∗ (x).Proof. The inequality is obviously true Lebesgue almost everywhere. However, the onlyexceptional points lie on the faces of dyadic cubes, and these have ω-measure 0, because ω isdoubling. QED.Lemma 1.3. There is a constant C such that G ∗ (x) ≤ Cg ∗ (f)(x) almost everywhere.Proof. Let I ∈Dand x ∈ I. We need to show that G(I) ≤ Cg ∗ (f)(x), for which it is clearlysufficient to show that⎡⎤(G(I)) 2 ≤ C∑∞∑|λ Q | 2 ⎣2 −j(2α−τ)ω(2 j Q) χ R j (Q)(x) ⎦ ,where (recall the definition above)(G(I)) 2 =Q:Q∈S(I)∑Q:Q∈S(I)j=0⎡∞∑|λ Q | 2 ⎣j=072 −j(2α−τ)1/2⎤ω(2 j Q) χ R j (Q)(x I ) ⎦ .,

Comparing the sums termwise, we see that our inequality amounts to having∞∑j=02 −j(2α−τ)ω(2 j Q) χ R j (Q)(x I ) ≤ C∞∑j=02 −j(2α−τ)ω(2 j Q) χ R j (Q)(x)for any x ∈ I and any Q ∈ S(I). Let us now consider two cases: l(Q) ≥ l(I) andl(Q)

By the Cauchy-Schwarz inequality and our size estimate 1),⎛⎜(I) ≤ ⎝∑Q:Q∈S(I)l(Q)≤l(I)⎛⎜= G(I,x) ⎝⎛⎜≤ CG(I) ⎝⎡∞∑|λ Q | 2 ⎣∑Q:Q∈S(I)l(Q)≤l(I)∑Q:Q∈S(I)l(Q)≤l(I)j=02 −j(2α−τ⎤⎞ω(2 j Q) χ R j (Q)(x) ⎦⎟⎠1/2 ⎛⎜⎝⎡⎤⎞∞∑ω(Q) ⎣2 −jτω(2 j Q) χ R j (Q)(x) ⎦⎟⎠j=0⎡⎤⎞∞∑ω(Q) ⎣2 −jτω(2 j Q) χ R j (Q)(x) ⎦⎟⎠j=0∑Q:Q∈S(I)l(Q)≤l(I)1/21/2,⎡⎤⎞∞∑ω(Q) ⎣2 −jτω(2 j Q) χ R j (Q)(x) ⎦⎟⎠where the last line follows from Lemma 1.4. We need to show that the second factor in the lastline is bounded by an absolute constant.Write R d \ I = ∪ l I l , where the I l ’s are congruent copies of I (thus, {I}∪{I l } l tiles R d ). Weset⎡⎤H(l) ≡∑∞∑⎣2 −jτω(2 j Q) χ R j (Q)(x) ⎦ .Q:Q⊂I lω(Q)j=0For k =0, 1, 2, ...,letS k denote the set of I l ’s such that I l ∩ R k (I) ≠ ∅. These cubes are at adistance of approximately 2 k l(I) fromx ∈ I. Indeed, there is a constant C, independent of x, suchthat x ∈ R k ′(I l ) for some k with |k − k ′ |≤C.Let us temporarily fix I l ∈ S k and consider Q ⊂ I l with l(Q) =2 −m l(I). If x ∈ I andd(x, ∂I) ≥ ηl(I), then x ∈ R j (Q) for some j which is approximately equal to m + k; i.e., there isC such that x ∈ R j (Q) for|m + k − j| ≤C. This is simply another way of saying that d(x, Q) isapproximately 2 m+k l(Q) =2 k l(I). Therefore, for such I l ,H(l) ≤ C∞∑m=0∑Q:Q⊂I ll(Q)=2 −m l(I)ω(Q) 2−(m+k)τω(2 m+k Q) .j=01/2Notice that, because of ω’s doubling property, all of the numbers ω(2 m+k Q) are comparable toω(2 k I), with comparability constants which only depend on d and ω’s doubling constant. Therefore,H(l) ≤ C[ω(2 k I)] −1∞ ∑m=0∑Q:Q⊂I ll(Q)=2 −m l(I)∑ ∞= C2 −kτ [ω(2 k I)] −1 2 −mτ ω(I l )m=0≤ C2 −kτ ω(I l )[ω(2 k I)] −1 .92 −(m+k)τ ω(Q)

If we now sum over the I l ’s in a fixed S k ,weget∑H(l) ≤ C2 −kτ [ω(2 k I)] ∑ −1 ω(I l )l:I l ∈S k l:I l ∈S k≤ C2 −kτ [ω(2 k I)] −1 ω(2 k I)≤ C2 −kτ ,where the next-to-last line follows from ω’s doubling property. Summing over all k yields ∑ H(l) ≤C, and term (I) has been bounded.Handling (II). This one’s easier.(II)=|≤∑Q:Q∈S(I)l(Q)>l(I)∑Q:Q∈S(I)l(Q)>l(I)λ Q(φ(Q) (x I ) − φ (Q) (x) ) |( ) β |xI − x| √ ∑ ∞ ( 2−jα|λ Q |ω(Q)l(Q)ω(2 j Q)j=0)(χ Rj (Q)(x I )+χ Rj (Q)(x)).Since we are considering cubes Q that are larger than I, ifx I ∈ R j (Q) andx ∈ I, then x ∈ R j ′(Q)for some j ′ such that |j − j ′ |≤C. Therefore, our sum (II) is actually less than or equal toC∑Q:Q∈S(I)l(Q)>l(I)Continuing, this is bounded by:C=C∑Q:Q∈S(I)l(Q)>l(I)∞∑( ) β |xI − x| √ ∑∞ ( 2−jα|λ Q |ω(Q)l(Q)ω(2 j Q)( l(I)|λ Q |l(Q)∑k=1 Q:l(Q)=2 k l(I)⎛∞∑≤CG(I) ⎝k=1 Q:l(Q)=2 k l(I)j=0) β√ ∑ ∞ ( 2−jαω(Q)j=0|λ Q |2 −kβ√ ω(Q)∑ω(2 j Q)∞∑( 2−jαj=0j=0ω(2 j Q))χ Rj (Q)(x I ).)χ Rj (Q)(x I ))χ Rj (Q)(x I )⎞∞∑( ) 22 −2kβ −jτω(Q)ω(2 j χ Rj (Q)(x I ) ⎠Q)where the last line follows from the Cauchy-Schwartz inequality.We now need to show that the second factor is bounded by a constant.We temporarily fix k and consider∑Q:l(Q)=2 k l(I)ω(Q)∞∑( 2−jτj=010ω(2 j Q)1/2)χ Rj (Q)(x I ). (1.3),

If χ Rj (Q)(x I ) ≠ 0 then Q ⊂ 2 k+j′ I for some j ′ such that |j − j ′ |≤C. Also, for such Q, ω(2 j Q)will be comparable to ω(2 k+j′ I), because ω is doubling. Therefore, for each j,∑( ) 2−jτω(2 j ≤ C2 −jτ .Q)ω(Q)Q:l(Q)=2 k l(I)χ Rj (Q) (x I )≠0Summing on j, we get that (1.3) is bounded by a constant. If we now multiply this by 2 −2kβ andsum on k, we get our result. Lemma 1.5 is proved. QED.If I is a dyadic cube, we define N(I) ≡ {I ∗ ∈ D : I ∗ ⊂ I, l(I ∗ ) = l(I)/2}, the “nextgeneration” of cubes “begotten” by I.Lemma 1.6. If I ∗ ∈ N(I), then G(I) ≤ CG(I ∗ ).Proof. By Lemma 1.4, G(I) =G(I,x I ) ≤ CG(I,x I ∗). But G(I ∗ )=G(I ∗ ,x I ∗) ≥ G(I,x I ∗).QED.Lemma 1.7. There is a positive constant C such that, for all I ∗ ∈ N(I) and all x ∈ I, G ∗ (x) ≥CG(I ∗ ).Proof. If x ∈ I ∗ , there is nothing to prove. So, let x ∈ I \ I ∗ ,andletL ⊂ I \ I ∗ be a dyadiccube such that x ∈ L and L is smaller than any of the cubes Q in the sum ∑ λφ (Q) defining f.Then:⎡⎤G(L) 2 =∑∞∑|λ J | 2 ⎣2 −j(2α−τ)ω(2 j J) χ R j (J)(x L ) ⎦≥J:J∈S(L)∑J:J∈S(I ∗ )j=0⎡∞∑|λ J | 2 ⎣j=02 −j(2α−τ)⎤ω(2 j J) χ R j (J)(x L ) ⎦ ,because the second sum excludes the J’s contained in I ∗ . (Technically, the first sum excludes theJ’s contained in L, but, because L is so small, these contribute nothing to the sum.)The lemma will be proved once we show that, for J ∈ S(I ∗ ),∞∑j=02 −j(2α−τ)ω(2 j J) χ R j (J)(x L ) ≥ C∞∑j=02 −j(2α−τ)ω(2 j J) χ R j (J)(x I ∗).But this follows from what are (by now) “the usual reasons.” Simply note that, if J ∈ S(I ∗ ),then |x L − x J |≤C(d)|x I∗ − x J |. Thus, if 2−j(2α−τ)ω(2 j J)χ Rj (J)(x I ∗) ≠ 0, then 2−j′ (2α−τ)χω(2 j′ J) Rj ′ (J)(x L )will also be non-zero, for some j ′ such that |j − j ′ |≤C. The doubling property of ω ensures thatω(2 j′ J) ≤ Cω(2 j J), which finishes the proof of Lemma 1.7. QED.Lemma 1.8. There is a C such that, if I ∗ ∈ N(I), then |F (I) − F (I ∗ )|≤CG(I ∗ ).Proof.|F (I) − F (I ∗ )| = |F (I,x I ) − F (I ∗ ,x I ∗)|≤|F (I,x I ) − F (I,x I ∗)| + |F (I,x I ∗) − F (I ∗ ,x I ∗)|.11

The first difference is ≤ CG(I) ≤ G(I ∗ ), by Lemmas 1.5 and 1.6. The second is less than or equalto∑|λ Q ||φ (Q) (x I ∗)|;Q:Q⊂IQ⊄I ∗which, by the Cauchy-Schwartz inequality (see the proof of Lemma 1.5) is less than or equal to⎛G(I ∗ ⎜∑) ⎝Q:Q⊂IQ⊄I ∗⎡⎤⎞∞∑ω(Q) ⎣2 −jτω(2 j Q) χ R j (Q)(x I ∗) ⎦⎟⎠j=0But the first part of the proof of Lemma 1.5 shows that the second factor is bounded by a constant.QED.We are now ready to prove Lemma 1.9, from which Theorem 1.1 will follow as a corollary.Lemma 1.9. Let Q 0 ∈Dbe the unit dyadic cube, and let {φ (Q) } Q be a familyof functionssatisfying 1), 2), and 3). Let f = ∑ Q λ Qφ (Q) be a finite linear combination such that λ Q =0forall Q not contained in Q 0 . Then: For all δ>0, there is a γ>0 such thatω ({x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}) ≤ δω(Q 0 ). (1.4)Remark .Ifν ∈ A ∞ (ω), then Lemma 1.9 immediately implies, as a corollary, the same conclusionfor ν; i.e., for all δ>0 there is a γ>0 such thatν ({x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}) ≤ δν(Q 0 ).We will use this corollary to obtain Theorem 1.1.Proof of Lemma 1.9. Let A>1 be a large number, to be chosen presently. Let {I k } k bethe family of maximal dyadic subcubes of Q 0 having the property that, for some I ∗ ∈ N(I k ),G(I ∗ ) >Aγ.By Lemmas 1.6 and 1.7, if A is chosen large enough, the set {x ∈ Q 0 : G ∗ (x) >γ} will contain∪ k I k . We henceforth assume that A has been chosen ‘large enough.’ On the other hand, noticethat, if x/∈∪ k I k , then G ∗ (x) ≤ Aγ:this will be important.With {I k } k now fixed, let {J l } l be the maximal subcubes of Q 0 such that, first, no J l iscontainedinanyI k , and second, |F (J l )| > 1. Denote the union {I k }∪{J l } by P, andlet{P i } i bethe family of maximal cubes from P.We claim that{x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ} ⊂∪ i P i . (1.5)To see this, suppose that x belongs to the left-hand side of (1.5). Since F ∗ (x) > 1, x must belongto some cube J such that |F (J)| > 1. If this cube J were contained in some I k , then we would haveG ∗ (x) >γ, a contradiction. Therefore, x belongs to one of the special cubes J l .But∪ l J l ⊂∪ i P i .121/2.

Thus, our problem has now reduced to controlling the size of ∑ i: |F (P i )|>1 ω(P i).The reader may wonder why we throw the cubes I k into P at all, since only the cubes J l areneeded to cover {x ∈ Q 0 : F ∗ (x) > 1, G ∗ (x) ≤ γ}. The reason will soon become apparent. But,essentially:we use the family {I k } k to control the size of G ∗ (x) globally on Q 0 (i.e., even at pointswhere F ∗ (x) ≤ 1). This is very much in the spirit of the proof of the classical good-λ inequalityfor the dyadic square function.Define F 1 = {Q ⊂ Q 0 : ∀i(Q ⊄ P i )} and F 2 = {Q ⊂ Q 0 : Q ⊂ P i for some i}; andsetf i = ∑ Q∈F iλ Q φ (Q) for i =1, 2. It is obvious that f = f 1 + f 2 . Corresponding to f 1 and f 2 ,wedefineF i (I,x)=∑λ Q φ (Q) (x)Q:Q∈S(I)Q∈F iF i (I) =F i (I,x I )Fi ∗ (x) = sup |F i (I)|I:x∈I⎛⎡⎜∑∞∑G i (I,x)= ⎝ |λ Q | 2 ⎣Q∈S(I)Q∈F ij=02 −j(2α−τ)⎤⎞ω(2 j Q) χ R j (Q)(x) ⎦⎟⎠G i (I) =G i (I,x I )G ∗ i (x) = sup G i (I);I:x∈Iwhere, as before, we do not define F i (I,x)orG i (I,x)forx/∈ I.Before going on, let us note—what is easy to see—that F (I,x) =F 1 (I,x)+F 2 (I,x) andG(I,x) ≤ G 1 (I,x)+G 2 (I,x). It is also easy to see that that each G i (I,x) ≤ G(I,x).For any cube Q, defineC(Q) ≡{x ∈ Q : |x − x Q | 1ω(C(P i )).1/2Clearly,∑∑∑ω(C(P i )) ≤ω(C(P i )) +ω(C(P i ))i: |F (P i )|>1i: |F 1 (P i )|>1/2i: |F 2 (P i )|>1/2≡ (I)+(II).Let us consider (I) first. EachP i satisfies G(P i ) ≤ γ. Therefore, if x ∈ C(P i ), we have (byLemma 1.5) |F 1 (P i ) − F 1 (P i ,x)| ≤Cγ. If we take γ small enough, then this difference will be lessthan 1/4, and having |F 1 (P i )| > 1/2 will force |F 1 (P i ,x)| > 1/4 onallofC(P i ). Let us assumethat γ is so chosen. We get:∑ω(C(P i )) ≤ ∑ ω({x ∈ C(P i ): |F 1 (P i ,x)| > 1/4}).i: |F 1 (P i )|>1/2 i13

It is this last sum which we will now control. Recall that to this point we have used the decayand smoothness properties of the functions φ (Q) , but not their almost-orthogonality. Now is thetime to apply 3).Our argument relies on a TRICK: If x ∈ P i , then F 1 (P i ,x)=f 1 (x). This is true because (seealso page 41 in [Wi]):F 1 (P i ,x)=∑λ Q φ (Q) (x)Q: Q∈F 1Q⊄P i= ∑λ Q φ (Q) (x)Q: Q∈F 1= f 1 (x),where the second line follows because having Q ∈F 1 automatically implies Q ⊄ P i .Because of property 3),∫|f 1 | 2 dω ≤ ∑Q: Q∈F 1|λ Q | 2 .We rewrite and bound the second sum as∑∫ ⎛ ⎞|λ Q | 2 = ⎝∑ |λ Q | 2⎠ dω.ω(Q)Q: Q∈F 1 Q: x∈Q∈F 1We claim that∑ |λ Q | 2ω(Q) ≤ Cγ2Q: x∈Q∈F 1ω-a.e. Proofofclaim.Ifx ∈ P i , then∑ |λ Q | 2ω(Q) =∑ |λ Q | 2ω(Q)Q: x∈Q∈F 1 Q: P i ⊂Q∈F 1∑ ∑ ∞ [ ]2≤|λ Q | 2 −j(2α−τ)ω(2 j Q) χ R j (Q)(x)Q: P i ⊂Q∈F 1 j=0≤ C(G(P i )) 2≤ Cγ 2 .If x/∈∪ i P i , then∑ |λ Q | 2ω(Q) ≤ ∑ ∑∞ [ ]2|λ Q | 2 −j(2α−τ)ω(2 j Q) χ R j (Q)(x)Q: x∈Q∈F 1 Q j=0≤ (G ∗ (x)) 2≤ (Aγ) 2 ,as noted above.14

Putting it all together, we get∫|f 1 | 2 dω ≤ Cγ 2 ω(Q 0 ),implying ω({x ∈ Q 0 : |f 1 (x)| > 1/4}) ≤ Cγ 2 . Therefore,(I) ≤ Cγ 2 ω(Q 0 ) ≤ (δ/2)ω(Q 0 ),if γ is taken sufficiently small.Now we look at (II). Reasoning precisely as we did for (I), we only need to control∑ω({x ∈ C(P i ): |F 2 (P i ,x)| > 1/4}).iWe will handle this last sum via a pure brute-force argument, using only the size condition 1).An estimate reminiscent of Carleson measures comes in at the end.Let κ be a number greater than 1, and chosen so that, for any cube Q, ω(κQ\Q) < (δ/4)ω(Q).Such a κ exists because ω is doubling. We defineD ≡∪ i (κP i \ P i ).We call this the “zone of death.” It consists of a union of thin bands (or shells) around the cubes P i ,inside which we may encounter bad edge effects when estimating f 2 and its associated functionals.Byourchoiceofκ, ω(D) ≤ (δ/4)ω(Q 0 ). Thus, it is sufficient to bound∑ω({x ∈ C(P i ) \ D : |F 2 (P i ,x)| > 1/4})by Cδ, where C depends on ω’s doubling constant.Fix P i .Ifx ∈ C(P i ) \ D, then|F 2 (P i ,x)| = | ∑ ∑λ Q φ (Q) (x)|Q: Q⊂P jij: j≠i⎛≤ G 2 (P i ,x) ⎝ ∑j: j≠i⎛≤ CG(P i ) ⎝ ∑⎛≤ Cγ ⎝ ∑j: j≠i≡ CγH i (x).j: j≠i∑∑Q: Q⊂P jω(Q)∑Q: Q⊂P jω(Q)Q: Q⊂P jω(Q)15[ ∞∑k=0[ ∞∑k=0[ ∞∑k=0] ⎞2⎠1/2−kτω(2 k Q) χ R k (Q)(x)] ⎞2⎠1/2−kτω(2 k Q) χ R k (Q)(x)] ⎞2⎠1/2−kτω(2 k Q) χ R k (Q)(x)

We claim that⎛H i (x) ≤ C ⎝ ∑ jω(P j )[ ∞∑k=0] ⎞2⎠1/2−kτω(2 k P j ) χ R k (P j )(x) . (1.6)Note that the right-hand side of (1.6) does not depend on i. Oncewehave(1.6), we will obtain∑∫iC(P i )\D⎛∫ ∑|F 2 (P i ,x)| 2 dω(x) ≤ Cγ 2 ⎝j≤ Cγ 2 ∑ jω(P j )[ ∞∑k=0] ⎞2 −kτω(2 k P j ) χ R k (P j )(x) ⎠ dω(x)∫ [ ∑ ∞2 −kτω(P j )ω(2 k P j ) χ R k (P j )(x)k=0]dω(x)≤ Cγ 2 ∑ jω(P j )≤ Cγ 2 ω(Q 0 ),and the bound for (II) will follow from Chebyshev’s inequality.Let us now fix a j ≠ i and consider the sum∑Q: Q⊂P jω(Q)[ ∞∑k=0]2 −kτω(2 k Q) χ R k (Q)(x)for x ∈ C(P i ) \ D. We rewrite the sum as:∞∑l=0∑Q: Q⊂P jl(Q)=2 −l l(P j )ω(Q)[ ∞∑k=0]2 −kτω(2 k Q) χ R k (Q)(x) .Now let’s fix l. Let k ′ be such that x Pi ∈ R k ′(P j ). If Q ⊂ P j , l(Q) = 2 −l l(P j ), andx ∈ C(P i ) \ D, then x ∈ R k (Q) for some k such that |k ′ + l − k| ≤C, where C only depends onthe dimension d. Conversely, if x ∈ C(P i ) \ D and belongs to R k (Q), then x ∈ R˜k(P j ) for some ˜ksatisfying |˜k + l − k| ≤C ′ . The reason for these inequalities is that the distance between x Q andx is comparable to the distance between x Pi and x Pj , with “comparability constants” dependingonly on d and κ. For all such Q, the ω-measures of 2 k Q will be comparable to ω(2 k−l P j ), andthus comparable to ω(2 k′ P j ), because ω is doubling. Therefore, for each fixed l ≥ 0, and all16

x ∈ C(P i ) \ D,∑Q: Q⊂P jl(Q)=2 −l l(P j )ω(Q)[ ∞∑k=0]2 −kτω(2 k Q) χ R k (Q)(x) ≤ C⎡⎡⎤τ˜k=k ′ ∑+C+C ′≤ C ⎣2−k′ χω(2 k′ P j )R˜k(P j )(x) ⎦ ⎢˜k=k ′ −C−C ′⎡⎣ 2−lτ= C2 −lτ ω(P j ) ⎣χω(2 k′ P j )R˜k(P j )(x) ⎦˜k=k ′ −C−C ′≤ C2 −lτ ω(P j )∞∑˜k=02−k′ τ˜k=k ′ ∑+C+C ′2 −k′ τω(2 k′ P j ) χ R˜k(P j )(x).k=0∑Q: l(Q)=2 −l l(P j )∑Q: Q⊂P jl(Q)=2 −l l(P j )⎤ω(Q) ⎥⎦˜k=0ω(Q)ω(2 k Q)⎤k ′ ∑+l+Ck=k ′ +l−CThis holds for every l ≥ 0. Summing over l we get:[∑∞]∑ 2 −kτ∞∑ω(Q)ω(2 k Q) χ 2 −k′ τR k (Q)(x) ≤ Cω(P j )ω(2 k′ P j ) χR˜k(P j )(x).Q: Q⊂P jWhen we sum this over j ≠ i, weget(1.6). Lemma 1.9 is proved.2 −kτ χ Rk (Q)(x)ProofofTheorem1.1. Recall that f = ∑ λ Q φ (Q) is a finite sum, from which it followsthat, for large x, |f(x)| ≤Cg ∗ (f)(x). This is because, if we take a single term in the sum definingf, weget√ω(Q)2−jα|λ Q φ (Q) |≤|λ Q |ω(2 k Q)when x ∈ R k (Q). However, for x ∈ R k (Q),g ∗ (f)(x) ≥|λ Q | 2−j(α−τ/2)√ω(2k Q)2 −jα≥|λ Q | √ω(2k Q) .After some canceling on both sides, our desired inequality,|λ Q φ (Q) |≤g ∗ (f)reduces to √ ω(Q) ≤ √ ω(2 k Q); which is true for k ≥ 0. (We note that the C in our inequality‘|f(x)| ≤g ∗ (f)(x)’ depends strongly on f, and on the fact that f is a finite sum.)Therefore, without loss of generality, we may assume that g ∗ (f) belongs to L p (dv), and thusthat G ∗ does too (because G ∗ ≤ cg ∗ (f)). This said, the Theorem 1.1 will follow from a good-λinequality: For every ɛ>0 there is a γ>0 so that, for all λ>0,ν ({x : F ∗ (x) > 2λ, G ∗ (x) ≤ γλ}) ≤ ɛν ({x : F ∗ (x) >λ}) .17

Let {Ii λ} be the maximal dyadic cubes such that |F (Iλ i )| >λ. It is enough to prove that, for allɛ>0, there is a γ so that, for all i and all λ,ν ( {x ∈ I λ i : F ∗ (x) > 2λ, G ∗ (x) ≤ γλ} ) ≤ ɛν(I λ i ). (1.7)Since ν ∈ A ∞ (ω), it is enough to prove (1.7) with the measure ν replaced by ω; and that is whatwe shall do.From this point the proof is essentially identical to the (very short) proof of Theorem A in[Wi] (on page 45). Fix Ii λ. We can take G(Iλ i ) ≤ γλ, or there is nothing to prove. Let I be theunique dyadic cube such that Ii λ ∈ N(I). Because of Ii λ ’s maximality, |F (I)| ≤λ. By Lemma 1.8,we may therefore take |F (Ii λ )|≤1.1λ, if we choose γ small enough—and of course we do.Now let η>0besosmallthatω({x ∈ Iiλ : d(x, ∂Ii λ) .8λ, G ∗ (x) ≤ γλ} ) ≤ (ɛ/2)ω(I λ i ).But this is just a rescaled version of Lemma 1.9 (divide h by .8λ). The Theorem 1.1 is proved.QED.2. Lipschitz domains.We are going to apply the Theorem 1.1 to a family of functions {φ (Q) } Q defined on a part ofthe boundary of a bounded Lipschitz domain Ω ⊂ R d+1 . That is, ∂Ω is a finite union of translationsand rotations of graphs Γ i of functions ψ i : R d ↦→ R, and there is a fixed M such that each ψ isatisfies |ψ i (x) − ψ i (y)| ≤M|x − y| for all x and y.As we said in the introduction, our attention will be directed at points near ∂Ω, where we canpretend that we are working in a domain lying above a graph Γ i . But first we wish to say a fewwords about the points in Ω that are (relatively) far from the boundary.Recall our definition of Ω δ ≡{x ∈ Ω: d(x, ∂Ω) ≤ δ}. Given a measure µ defined on Ω, writeµ = µ 1 + µ 2 where µ 1 (E) =µ(E \ Ω δ )andµ 2 (E) =µ(E ∩ Ω δ ).We claim that an inequality of the form(∫ ) 1/q (∫1/p|∇u| q dµ 1 ≤ C |f| vdω) p (2.1)Ω∂Ω18

comes almost (but not quite) for free.We have, for any x/∈ Ω δ ,∫|∇u(x)| ≤C∂Ω|f| dω,where C depends on Ω, δ, and our choice of X 0 (which, without loss of generality, we may assumelies in Ω \ Ω δ ). This inequality comes from the fact thatlinked to the inequality:By Hölder’s Inequality,sup |∇u(x)| ≤Cδ −1 sup |u(x)|x/∈Ω δ x/∈Ω δ/2sup |u(x)| ≤C Ω,δ |f| dω.x/∈Ω δ/2∫∂Ω∫∂Ω(∫ ) 1/p (∫|f| dω ≤ |f| p vdω∂Ω∂Ω) 1/p′σdω .Thus, to get (2.1), it is sufficient to haveµ(Ω \ Ω δ ) 1/q (∫∂Ω) 1/p′σdω ≤ c ′ ,where c ′ is small positive constant depending on Ω, δ, andX 0 .A moment’s thought will show that this is just a “global” version of the sufficient conditionfrom Theorem 3.1, with ∂Ω (a bounded set, recall) playing the role of a cube Q b ,andΩ\ Ω δpretending to be T (Q b ). The bump function being integrated against σ is just χ ∂Ω . It has no “tail”because there is no room for one:the “cube” Q b fills up all of ∂Ω.To fit the pattern of Theorem 3.1, the constant c ′ should really be cdiam(Ω)ω(∂Ω) = cdiam(Ω).But, of course, it is—assuming we choose c properly!That is (almost) all we will say about the parts of Ω lying far from the boundary.Our problem now consists in finding an appropriate condition on the measure µ 2 which ensuresthat(∫(∫) 1/q|∇u| q dµ 2 ≤Ω|f| p vdω∂Ωholds for all f ∈ L ∞ , where we remember that µ 2 is supported in Ω δ .By cutting the integral up, taking δ small enough, rescaling, and doing a rotation, we canassume that the support of µ 2 —which we will henceforth call µ—lies in a set of the form) 1/pR≡{(x, y) : ψ(x)

Let F be the family of dyadic cubes Q ⊂ R d such that Q ⊂{x ∈ R d : |x| ≤3/2}. For eachcube Q ∈F,wedefineQ b = {(x, ψ(x)) : x ∈ Q}, the boundary cube corresponding to Q, andwedefineˆQ b = {(x, y) : x ∈ Q, ψ(x)

where, of course, R j (Q b )=2 j Q b \ 2 j−1 Q b when j>0 and equals Q b for j = 0 (see [JK][K]). And:|K(X Q i ,s) − K(XQ i ,s′ )|≤C( |s − s ′ |l(Q)) β ∑ ∞ ( 2−jαj=0ω(2 j Q b ))χ Rj (Q b )(s)(see [K]).All we need now is condition 3), the almost-orthogonality.Let G(X) be the Green’s function for Ω, with a pole at X 0 . By classical estimates for theGreen’s function and harmonic measure, if Q is one of our cubes, then ω(Q b ) is bounded above andbelow by positive constants times G(Z Q )l(Q) d−1 . We shall refer to this fact as ‘inequality (2.2).’In symbols:c 1 ω(Q b ) ≤ G(Z Q )l(Q) d−1 ≤ c 2 ω(Q b ). (2.2)(Note:the exponent on l(Q) isd − 1 and not the usual d − 2, because we are working in a subsetof R d+1 .)Let {λ Q } Q be an arbitary finite collection of real numbers indexed over F, andsetg(s) = ∑ λ Q φ (Q) (s).We wish to show that∫|g| 2 dω(s) ≤ C ∑ |λ Q | 2 .To this purpose, let f ∈ L 2 (∂Ω,ω) be continuous and satisfy ∫ fdω= 0. We consider the integral∂Ω∫gfdω. (2.3)Our job now is to show that this integral is less than or equal to a constant times⎛⎝ ∑ Q⎞ 1/2 (∫1/2|λ Q | 2 ⎠ |f| dω) 2 .∂ΩThe integral (2.3) equals∑λQ√ω(Qb )(u(X Q 1 ) − u(XQ 2 )).By Cauchy-Schwarz, this has magnitude less than or equal to(∑|λQ | 2) 1/2 (∑ω(Qb )|u(X Q 1 ) − u(XQ 2 )|2) 1/2,and so our problem reduces to showing∑∫ω(Qb )|u(X Q 1 ) − u(XQ 2 )|2 ≤ C|f| 2 dω. (2.4)21

By the ordinary, differential mean value theorem, and the sub-mean-value property for harmonicfunctions,∫|u(X Q 1 ) − u(XQ 2 )|2 ≤ Cl(Q) −d−1 (l(Q)|∇u(X)|) 2 dX= Cl(Q) −d+1 ∫˜ ˜T (Q b )˜ ˜T (Q b )|∇u(X)| 2 dX, (2.5)where the constant C depends on the “usual” parameters.By successively applying the estimate from the last inequality, inequality (2.2), the Harnackproperty for G(X), and the bounded overlap property of the sets ˜T (Q˜b ), we obtain that∑ω(Qb )|u(X Q 1 ) − u(XQ 2 )|2 ≤ C ∑ Q∫≤ C∫≤ C∫G(Z Q )˜ ˜T (Q b )Ω˜ ˜T (Q b )|∇u(X)| 2 dXG(X) |∇u(X)| 2 dXG(X) |∇u(X)| 2 dX.But by Green’s Theorem and our normalization on f—i.e., u(X 0 )= ∫ fdω= 0—the last quantity∂Ωis less than or equal to C ∫ ∂Ω |f|2 dω. Therefore, modulo multiplication by a small positive constant,our family {φ (Q) } satisfies 1), 2), and 3) on the homogeneous space ∂Ω, with the Euclidean metricand measure ω.3. The weighted-norm theorem.Let’s briefly recap our situation. We have a bounded Lipschitz domain Ω, whose boundary canbe written as an overlapping union of (pieces of) graphs of Lipschitz functions ψ i (appropriatelyrotated, scaled, and translated). On each of these pieces we have a collection of dyadic boundarycubes that are near the origin. We can assume that we have enough pieces so that the union ofthese cubes covers all of ∂Ω. Let’s throw all of these cubes into a big family, which we will callG. For each one of these cubes Q b we can talk about ˆQ b , l(Q b ), and T (Q b ). It’s possible thatagivenQ b will have more than one definition of T (Q b )(orl(Q b ). This is okay. For a given Q b ,all of its possible values of l(Q b ) will be comparable (with comparability constants depending onour domain’s Lipschitz constant). There can be no more than C different T (Q b )’s, where C isthe number of pieces into which we have divided ∂Ω. Since we’re mainly interested in the size ofµ(T (Q b )), in our statement of Theorem 3.1 below, we can take µ(T (Q b )) to be the largest of thesenumbers.We can now state the precise form of Theorem 3.1, slightly rephrased from our original statementin the introduction:Theorem 3.1. Let Ω ⊂ R d+1 be a bounded Lipschitz domain, and let ω be harmonic measureon ∂Ω for some fixed point X 0 ∈ Ω. Suppose that v ∈ L 1 (∂Ω,dω) is a non-negative function and22

µ is a positive Borel measure on Ω. Define σ ≡ v 1−p′ , and suppose that σdω ∈ A ∞ (ω) on ∂Ω. If1

and chosen so that∑g(Q b )l(Q b ) −1 (u(X Q 1∣) − u(XQ 2 ))µ(T (Q b))∣Q b ∈G⎛≥ (1/2) ⎝ ∑ l(Q b ) −q |u(X Q 1 ) − u(XQ 2 )|q µ(T (Q b ) ⎠ . (3.2)Q b ∈G⎞1/qWe need to show that, for every such g, the left-hand side of (3.2) is not too big; i.e., that itis less than or equal to a constant timesWe define(∫1/p|f(s)| vdω(s)) p .∂ΩT (g)(s) ≡ ∑ Q b ∈Gg(Q b )µ(T (Q b ))l(Q b ) −1 (K(X Q 1 ,s) − K(XQ 2 ,s)),and notice that∑Q b ∈G∫g(Q b )l(Q b ) −1 (u(X Q 1 ) − u(XQ 2 ))µ(T (Q b)) =Recall that σ = v 1−p′ . The left-hand side of (3.2) will befor all g as we have defined, if(∫≤ C |f(s)| p vdω(s)∂Ω) 1/p∂Ωf(s) T (g)(s) dω(s).(∫|T (g)(s)| p′ σdω∂Ω) 1/p′⎛⎞≤ C ⎝ ∑ |g(Q)| q′ µ(T (Q b )) ⎠1/q ′ (3.3)Q∈Gfor all such g. It is this last inequality which we shall prove.WriteT (g)(s) = ∑ Q b ∈Gλ Q φ (Q) ,whereandφ (Q) (s) ≡ √ ω(Q b )(K(X Q 1 ,s) − K(XQ 2 ,s)),|λ Q |≤C |g(Q b)|µ(T (Q b ))l(Q b ) √ ω(Q b ) .24

The integral we need to estimate naturally breaks into two pieces. Let us recall the region wedenoted by R in Section 2:R≡{(x, y) : ψ(x) κ}. By our estimates onthe φ (Q) ’s,|Tg(x)| ≤C κ,Ω∑Q b ∈G|λ Q | √ ω(Q b )∞∑j=02 −jαω(2 j Q b ) χ R j (Q b )(x).2There is a C, independent of x ∈ℵ, such that−jαω(2 j Q b ) χ R j (Q b )(x) can be non-zero for at most Cmany j’s. For each of these j’s, 2 −j is essentially equal to l(Q b ), with the comparability constantsdepending on κ and Ω. Also, for such j, ω(2 j Q b ) is comparable to ω(∂Ω), which equals 1. Therefore,for x ∈ℵ,∑|Tg(x)| ≤C κ,Ω |λ Q | √ ω(Q b )l(Q b ) α≤ C ∑ Q b ∈GQ b ∈G|g(Q)|µ(T (Q b ))l(Q b ) −1 l(Q b ) α ;which looks funny—but we have a good reason for not combining the exponents in the l(Q b )’s.We assume that we have⎛∫µ(T (Q b )) 1/q ⎜⎝∂Ω⎡⎣ω(Q b )⎤∞∑ 2 −jɛω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σ(x) dω(x) ⎠1/p ′ ≤ cl(Q b )ω(Q b )for every Q b ∈G. We may replace the integral on the left-hand side of this inequality by an integralover the smaller region ℵ. Doing so, we may rewrite the inequality (after a change in c) as:where ɛ>0 is small.Thus, for x ∈ℵ,|Tg(x)| ≤c ∑ Q b ∈G(µ(T (Q b )) 1/q ≤ cl(Q b ) −ɛ/2 l(Q b ) √ )ω(Q b )(∫,ℵ σdω) 1/p ′|g(Q)|µ(T (Q b ))l(Q b ) −1 l(Q b ) α= c ∑ |g(Q)|µ(T (Q b )) 1/q′ µ(T (Q b )) 1/q l(Q b ) −1 l(Q b ) αQ b ∈G⎡⎤≤ c ⎣ ∑ |g(Q)|µ(T (Q b )) 1/q′ l(Q b ) α−ɛ/2√ (∫ω(Q b ) ⎦ σdωQ b ∈Gℵ25) −1/p′.

The expression in the brackets is bounded by⎛⎞1/q ′ ⎛⎝ ∑ |g(Q)| q′ µ(T (Q b )) ⎠ ⎝ ∑ ⎠ ,Q b ∈GQ b ∈Gl(Q b ) ɛ′ ω(Q b ) q/2 ⎞1/qwhere ɛ ′ > 0. By hypothesis, the first factor is ≤ 1. Since q ≥ 2, the second factor is no bigger than⎛⎞1/2⎝ ∑ l(Q b ) 2ɛ′ /q ω(Q b ) ⎠ ,Q b ∈Gwhich, since ɛ ′ > 0, is bounded by a constant.This means that, when x ∈ℵ, |Tg(x)| is no bigger than a constant timeswhich trivially implies that∫ℵ(∫σdωℵ) −1/p′|Tg(x)| p′ σdω≤ C.So, now we look at the x’s close to R.To keep ideas clear, let’s first consider the simple case p = q = 2. Having thrown out thepoints that are far from R, inequality (3.3) reduces to:∫|T (g)(s)| 2 σdω≤ C ∑ |g(Q)| 2 µ(T (Q b )), (3.4)∂Ω\ℵQ b ∈Gwhere all the cubes Q b are small and touch R. By Theorem 1.1, the left-hand side of (3.4) is lessthan or equal to a constant times⎛ ⎡⎤⎞∫⎝ ∑ ∞∑|λ Q | 2 ⎣2 −j(2α−τ)∂Ωω(2 j QQ b ∈G j=0 b ) χ R j (Q b )(x) ⎦⎠ σdω⎛≤ C ∑ ⎝g(Q)µ(T (Q b ))⎞ ⎡⎤2 ∫∣Q bl(Q∈G b ) √ ∞∑⎠ ⎣2 −j(2α−τ)ω(Q b ) ∣ ∂Ω ω(2 j Qj=0 b ) χ R j (Q b )(x) ⎦ σdω⎡⎤= C ∑ |g(Q)| 2 µ(T (Q b )) 2 ∫ ∞∑⎣2 −j(2α−τ)l(Q b ) 2 ω(Q b )Q b ∈G∂Ω ω(2 j Q) χ R j (Q b )(x) ⎦ σdω.j=0We want this last quantity to be less than or equal toC ∑ |g(Q)| 2 µ(T (Q b )).Q b ∈G,26

Comparing the sums term-by-term, we see that this will happen if, for every Q ∈F,⎡⎤∫ ∞∑µ(T (Q b )) ⎣2 −j(2α−τ)∂Ω ω(2 j Q b ) χ R j (Q b )(x) ⎦ σdω≤ Cl(Q b ) 2 ω(Q b );or, taking square roots:⎛∫µ(T (Q b )) 1/2 ⎝∂Ωj=0⎡⎤ ⎞∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦ σdω⎠j=01/2≤ Cl(Q b )ω(Q b ) 1/2 . (3.5)Thus, the appropriate sufficient condition, when p = q = 2, is the one given by Theorem 3.1.The proofs for more general p and q follow this same pattern. The main difficulty encounteredis that we can no longer freely exchange the order of integration and summation. Fortunately, thiscan be circumvented by some standard trickery.We shall consider the more difficult case first:1

We’ll save eyestrain and concentrate on bounding∫∂Ω|T (g)| 2 M r,ν (h) σdω.By Theorem 1.1, it is less than or equal to a constant times∫∂Ω⎛= ∑ Q b ∈G⎝ ∑ Q b ∈G|λ Q | 2 ∫⎡∞∑|λ Q | 2 ⎣∂Ωj=0j=02 −j(2α−τ)⎤⎞ω(2 j Q b ) χ R j (Q b )(x) ⎦⎠ M r,ν (h) σdω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦ M r,ν (h) σdω.But∫∂Ω⎛∫⎜≤ ⎝⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦ M r,ν (h) σdωj=0∂Ω⎛∫⎜≤ C ⎝⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦∂Ωj=0j=0p ′ /2⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦⎞⎟σdω⎠p ′ /2⎟σdω2/p ′ (∫⎞⎠2/p ′ ,(M r,ν (h)) s σdω∂Ω) 1/swhere the last line follows from (3.6) and our normalization on h.Thus,(∫|T (g)| p′ σdω∂Ω) 1/p′⎛ ⎛∑∫≤ C ⎜⎝ |λ Q | 2 ⎜⎝Q b ∈G∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠⎞2/p ′ 1/2⎟⎠.But, since q ′ ≤ 2, this last quantity is less than or equal to⎛ ⎛∑∫C ⎜ ⎜⎝ |λ Q | q′ ⎝Q b ∈G∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ⎞⎟⎠1/q ′ .Let’s recall that|λ Q |≤C |g(Q b)|µ(T (Q b ))l(Q b ) √ ω(Q b ) .28

Thus, the last quantity is bounded by⎛∑C ⎜⎝Q b ∈Gg(Q) q′ µ(T (Q b )) q′l(Q b ) q′ ω(Q b ) q′ /2⎛∫⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ⎞⎟⎠1/q ′ ,and this is supposed to be less than or equal to⎛⎞⎝ ∑ Q b ∈Gg(Q) q′ µ(T (Q b ))⎠1/q ′ .Comparing the sums termwise, we see that our inequality will follow if⎛∫µ(T (Q b )) q′⎜l(Q b ) q′ ω(Q b ) q′ /2⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ≤ cµ(T (Q b ))for all Q b ∈G; i.e., that⎛∫µ(T (Q b )) q′ −1 ⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ≤ cl(Q b ) q′ ω(Q b ) q′ /2 ,or, taking q ′ roots and simplifying,⎛∫µ(T (Q b )) 1/q ⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠1/p ′ ≤ cl(Q b ) √ ω(Q b ),which is the sufficient condition we stated in Theorem 3.1.Let’s now consider the easier case, 2

Using our bound on λ Q , the last quantity is less than or equal toC ∑ Q b ∈G(|g(Q b )|µ(T (Q b ))l(Q b ) √ ω(Q b )) p′⎛∫⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠ ;which, since p ′ ≥ q ′ , is bounded by⎛(∑C ⎜|g(Q b )|µ(T (Q b ))⎝l(Q b ) √ ω(Q b )Q b ∈G) q′⎛∫⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ⎞⎟⎠p ′ /q ′ .In order for our dual inequality to hold, the last quantity must be less than or equal to⎛⎞C ⎝ ∑ |g(Q b )| q′ µ(T (Q b ))Q b ∈G⎠p ′ /q ′ ,which will be true for all g’s if(µ(T (Q b ))l(Q b ) √ ω(Q b )) q′⎛∫⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q)(x) ⎦j=0p ′ /2⎞⎟σdω⎠q ′ /p ′ ≤ Cµ(T (Q b ))for all Q b ∈G; i.e., after some transposition,⎛∫µ(T (Q b )) 1/q ⎜⎝∂Ω⎡⎤∞∑⎣2 −j(2α−τ)ω(2 j Q b ) χ R j (Q b )(x) ⎦j=0p ′ /2⎞⎟σdω⎠1/p ′ ≤ Cl(Q b ) √ ω(Q b ),which is our condition from Theorem 3.1. QED.We shall now state the fairly straightforward corollary of Theorem 3.1 to solutions of uniformlyelliptic equations in divergence form.Recall that such an equation has the formLu ≡ div(A(x)∇u) =0, (3.7)where A(x) =[a i,j (x)] is a real, bounded, measurable, (d +1)× (d + 1) symmetric matrix satisfyingΛ −1 |ξ| 2 ≤ ∑ i,ja i,j ξ i ξ j ≤ Λ|ξ| 2for all x ∈ Ω and all ξ ∈ R d+1 , where Λ is a fixed positive constant. A u that satisfies (3.7) iscalled elliptic.30

The Dirichlet problem for the operator L is solvable on bounded Lipschitz domains Ω, andthe solution is written in terms of elliptic measure ω L . Unfortunately, this measure need not beA ∞ with respect to surface measure on ∂Ω—but it is doubling. It also has a Green’s function, G L ,which satisfies estimates (relative to ω L ) analogous to those that hold for G and ω. Indeed, theproofs we have given concerning weighted norm inequalities of the form(∫ ) 1/q (∫|∇u| q dµ ≤ |f| p vdωΩ δ ∂Ω(where u is f’s harmonic extension, and ω is harmonic measure for the Laplacian), will go throughalmost verbatim for weighted norm inequalities of the form) 1/p(∫ ) 1/q (∫) 1/p|∇u| q dµ ≤ |f| p vdω LΩ δ ∂Ωwhere u is now f’s elliptic extension. The one difficulty is that ∇u need not be defined pointwise.This requires us to rephrase our weighted norm inequality somewhat, and to look carefully at theproof of one of the auxiliary results used in the proof of Theorem 3.1.Let us assume that we have our family of boundary cubes G, as used in the proof of Theorem3.1. In that proof we used the fact that, for every Q b ∈G,sup T (Qb )) |∇u| was “morally equivalent”tosup l(Q b ) −1 |u(X) − u(Y )| (3.8)X,Y ∈T (Q b )when u was harmonic. In our rephrased form of Theorem 3.1, we will essentially replace |∇u| withthe right-hand side of (3.8).Let G be a family of boundary cubes as described above. Let {X Qb } and {Y Qb } be two familiesof points in Ω, both indexed over G, such that, for all Q b ∈G, X Qb ∈ T (Q b )andY Qb ∈ T (Q b ). Weshall call such a double sequence hyperbolically close.Theorem 3.2. Let Ω ⊂ R d+1 be a bounded Lipschitz domain, and let ω L be elliptic measure on∂Ω for some fixed point X 0 ∈ Ω. LetG be a familyof boundarycubes as described above. Supposethat v ∈ L 1 (∂Ω,dω L ) is a non-negative function and µ is a positive Borel measure on Ω. Defineσ ≡ v 1−p′ , and suppose that σdω L ∈ A ∞ (ω L ) on ∂Ω. If1

where c is a small positive constant depending on L, p, q, Ω, and the choice of the point X 0 ;andɛ>0 depends on the domain Ω and on L. (However, c does NOT depend on the particular familyG.)Sketch of Proof. The only troublesome part comes in verifying the properties 1), 2), and 3)of the family√φ (Qb )(s) ≡ ω L (Q b ))(K L (X Qb ,s) − K L (Y Qb ,s)),where we use K L to denote the elliptic kernel function analogous to K. The decay and smoothnessfollow from classical estimates just as before [K]. And, as in the first case, we prove the almostorthogonalityvia a duality argument. The only estimate that does not follow as in the harmoniccase is Poincaré:∫|u(X Qb ) − u(Y Qb )| 2 ≤ Cl(Q b ) −d+1 |∇u| 2 dX;˜T (Q b )but that also is classical ([DJK]).Appendix.Suppose that ν is a doubling measure on R d and h is a non-negative function that is locallyintegrable with respect to ν. Wedefine,forx ∈ R d ,M ν (h)(x) ≡where, as usual, Q denotes a cube in R d .supQ: x∈Q∫1hdν,ν(Q) QTheorem A1. Let 0 0. I claim that∫1hdν >R,ν(Q) Qwhich implies that M ν (h) ≡∞ν-a.e. Let L>0 be large and let {Q k } be the maximal dyadicsubcubes of Q 0 such that∫1hdν >L.ν(Q k ) Q kLet us note that, for any L, ∑ k ν(Q k) ≥ ν(E) > 0. Therefore,∫1hdν ≥ 1 ∑(Lν(Q k )) ≥ Lν(E)/ν(Q),ν(Q) Q ν(Q)which we can make >Rby taking L large enough.k32

So, if M ν (h) is not identically infinite, it is finite ν-a.e. Henceforth we will assume thatM ν (h) < ∞ ν-a.e.Let Q be a cube and let ˜Q be Q’s concentric triple. Following the standard procedure, writeh = hχ ˜Q+ hχ R d \ ˜Q ≡ h 1 + h 2 .Sinceν is doubling, standard arguments show thatM ν (h 2 )(x) ≤ C 1 M ν (h 2 )(y) ≤ C 2 M ν (h 2 )(x)for all x and y in Q, where C 1 and C 2 are finite positive constants depending only on d and on ν’sdoubling constant. Therefore, if E ⊂ Q and ν(E) ≤ δν(Q),∫∫∫(M ν (h 2 )) β dν ≤ Cδ (M ν (h 2 )) β dν ≤ Cδ (M ν (h)) β dν.EQQTo handle h 1 , we recall that M ν satisfies a weak-type (1, 1) inequality, relative to ν. This hasthe consequence that, for any 0

References.[DJK] B. Dahlberg, D. Jerison, C. Kenig, “Area integral estimates for elliptic differential operatorswith nonsmooth coefficients,” Arkiv Mat. 125 (1984), 97-108.[JK] D. Jerison, C. Kenig, “Boundary behavior of harmonic functions in non-tangentially accessibledomains,” Advances in Math. 46 (1982), 80-147.[K] C. Kenig, Harmonic Analysis Techniques for Second Order Elliptic Boundary Value Problems,American Mathematical Society (1994), Providence RI.[L1] D. Luecking, “Forward and reverse Carleson inequalities for functions in the Bergman spacesand their derivatives,” Amer. J. Math. 107 (1985), 85-111.[L2] —, “Embedding derivatives of Hardy spaces into Lebesgue spaces,” Proc. London Math. Soc.3 (1991), 595-619.[Sh1] N. A. Shirokov, “Some generailisations of the Littlewood-Paley Theorem,” J. Soviet Math. 8(1977), 119-129.[Sh2] —, “Some embedding theorems for spaces of harmonic functions,” J. Soviet Math. 14 (1980),1173-1176.[Ve] I. E. Verbitsky, “Imbedding theorems for the spaces of analytic functions with mixed norms,”Acad. Sci. Kishinev, Moldova (preprint) (1987).[Vi] I. V. Videnskii, “On an analogue of Carleson measures,” Soviet Math. Dokl. 37 (1988), 186-190.[WhWi] R. L. Wheeden and J. M. Wilson, “Weighted norm estimates for gradients of half-space extensions,”Indiana Univ. Math. J. 44 (1995), 917-969.[Wi] J. M. Wilson, “Global orthogonality implies local almost-orthogonality,” Revista MatemáticaIberoamericana 16 (2000), 29-48.Authors’ addresses:Caroline SweezyDepartment of Mathematical SciencesNew Mexico State UniversityLas Cruces NM 88003Email:csweezy@nmsu.edu.J. Michael WilsonDepartment of MathematicsUniversity of VermontBurlington VT 05405Email:wilson@emba.uvm.edu.34