THE PLANAR CANTOR SETS OF ZERO ANALYTIC CAPACITY

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1 JOURNAL OF THE AERICAN ATHEATICAL SOCIETY Volume 6, Number, Pages 9 28 S (02) Article electronically published on July 0, 2002 THE PLANAR CANTOR SETS OF ZERO ANALYTIC CAPACITY AND THE LOCAL T (b)-theore JOAN ATEU, XAVIER TOLSA, AND JOAN VERDERA. Introduction In this paper we characterize the planar Cantor sets of zero analytic capacity. Our main result answers a question of P. attila [a] and completes the solution of a long-standing open problem with a curious history, which goes back to 972. We refer the reader to [I2, p. 53] and [a] for more details. oreover, we confirm a conjecture of Eiderman [E] concerning the analytic capacity of the N-th approximation of a Cantor set. Before formulating our main results, we recall the definition of the basic objects involved. The analytic capacity of a compact subset E of the complex plane C is () γ(e) =sup f ( ), where the supremum is taken over all analytic functions f on C\E such that f on C\E. Although there has recently been important progress on our understanding of analytic capacity (see the survey papers [D], [V3] and the references given there), many basic questions about γ remain unanswered. One of the oldest is the semiadditivity problem, that is, the problem of showing the existence of an absolute constant C such that (2) γ(e F ) C {γ(e)+γ(f )}, for all compact sets E and F. If (2) were true, then one would have powerful geometric criteria for rational approximation, which are otherwise missing (see [V2] and [Vi2]). On the other hand, it has recently been established that a close variant of γ, called positive analytic capacity, is indeed semi-additive. The positive analytic capacity of a compact set E is γ + (E) =supµ(e), Received by the editors August 7, athematics Subject Classification. Primary 30C85; Secondary 42B20, 30E20. Key words and phrases. Analytic capacity, Cauchy integral, Cantor sets, T (b)-theorem, positive analytic capacity. The authors were partially supported by the grants BF , HPRN and 200- SGR The second author was supported by a arie Curie Fellowship of the European Union under contract HPFCT c 2002 American athematical Society

2 20 JOAN ATEU, XAVIER TOLSA, AND JOAN VERDERA where the supremum is taken over the positive Borel measures µ supported on E such that the Cauchy potential f = z µ is a function in L (C) with f. Since ( z µ) ( ) =µ(e), we clearly have γ + (E) γ(e). Since γ + is semi-additive, as shown in [T], it is clear that (2) follows from the inequality (3) γ(e) Cγ + (E), E compact C, where the positive constant C does not depend on E. Wewillseebelowthatour main result provides a proof of (3) for a particular (but significant) class of sets E. To the best of our knowledge, the first mention of (3) that can be found in the literature is in [DO]. An equivalent form of (3), which involves enger curvature, has recently been conjectured by elnikov (see [D]). NowweturnourattentiontoCantorsets. Givenasequenceλ = (λ n ) n=, 0 λ n /3, we construct a Cantor set by the following algorithm. Consider the unit square Q 0 =[0, ] [0, ]. At the first step we take four closed squares inside Q 0, of side-length λ, with sides parallel to the coordinate axes, such that each square contains a vertex of Q 0. At step 2 we apply the preceding procedure to each of the four squares produced at step, but now using the proportion factor λ 2. Then we obtain 6 squares of side-length σ 2 = λ λ 2. Proceeding inductively, we have at the n-thstep4 n squares Q n j, j 4n, of side-length σ n = n j= λ j. Write E n = E(λ,...,λ n )= and define the Cantor set associated with the sequence λ =(λ n ) n= by the identity E = E(λ) = E n. Our main result reads as follows. Theorem. The Cantor set E(λ) has zero analytic capacity if and only if (4 n σ n ) 2 =. n= The assumption λ n /3 for the Cantor sets E(λ) is purely technical. Actually Theorem (as well as Theorem 2 below) holds for any sequence (λ n ) n with 0 < λ n < /2. See Remark 2 at the end of the paper for more details. attila showed in [a] that the above condition is necessary and our contribution in this paper is to prove the sufficiency. The special case λ n =/4, n, was obtained independently by Garnett [G] and Ivanov [I] in the 970 s and, since then, the corner quarters Cantor set has become the favorite example of a set of zero analytic capacity and positive length. P. Jones gave in [J] an alternative proof of Garnett s result, based on harmonic measure. Recently Jones approach has been used to establish the vanishing of the analytic capacity of E(λ) forsome special classes of sequences λ =(λ n ) n= with 4n σ n tending to infinity [GY]. Theorem follows from a more precise result on the analytic capacity of the set E N = E(λ,...,λ N ). The asymptotic behaviour of γ + (E N ) is completely n= 4 n j= Q n j,

3 ANALYTIC CAPACITY OF CANTOR SETS 2 understood: for some constant C>andallN =, 2,... one has N C ( ) /2 ( N γ + ) /2 (4) (4 n σ n= n ) 2 (E N ) C. (4 n σ n= n ) 2 The upper estimate is due to Eiderman [E] and a different proof has been given in [T2]. The lower estimate was proved by attila in [a]. However, the result was not explicitly stated in [a], presumably because at that time the main object of interest was γ rather than γ +. An indication of how one proves the lower estimate in (4) will be provided at the end of Section 2. See also [E, p. 82]. Theorem follows from the upper estimate in (4) and the next result. Theorem 2. There exists a positive constant C 0 such that (5) γ(e N ) C 0 γ + (E N ),N=, 2,... If λ n =/4, n, then combining Theorem 2 with (4), we get γ(e N ) C N,N=, 2,..., which improves considerably urai s inequality [u] γ(e N ) C,N=2, 3,..., log N the best estimate known up to now. The main tool used in our proof of Theorem 2 is the local T (b)-theorem of. Christ [CH2], a particular version of which will be discussed and stated in Section 2. Section 2 also contains some basic facts on the Cauchy transform and the Plemelj formulae. The proof of Theorem 2 is presented in Section 3. Our notation and terminology are standard. For example D(z,r) istheopen disk centered at z and of radius r, ds is the arclength measure on a rectifiable arc and P Q means that C Q P CQ for some absolute constant C>. The symbols C, C,C,C 0,C,... stand for absolute constants with a definite value. We will also use the symbol A to denote an absolute constant that may vary at different occurrences. Remark. The second author [T3] has recently proved that Theorem 2 also holds for a general compact set. In particular, this implies the semi-additivity of analytic capacity. The proof in [T3] also involves an induction argument and an appropriate T (b)-theorem as in the present paper. 2. Background results 2.. Cauchy integrals. Fix an integer >0andletE = E(λ,...,λ )be the -th approximation of the Cantor set associated with the sequence (λ n ) n=. Then E is the union of 4 closed squares Q j, j 4,and E is the union of the 4 closed piecewise linear curves Q j. For a Borel measure µ supported on E set dµ(ζ) C(µ)(z) = ζ z, z / E, and dµ(ζ) C(µ)(z) = lim ε 0 ζ z >ε ζ z, z E,

4 22 JOAN ATEU, XAVIER TOLSA, AND JOAN VERDERA whenever the principal value integral exists. Let C + (µ)(z) (respectively, C (µ)(z)) stand for the non-tangential limit of C(µ)(w) asw tends to z from the interior of E (respectively, from the complement of E ). It follows from standard classical results that Cµ(z), C + µ(z) andc µ(z) exist for almost all z with respect to arclength measure ds on E. oreover one has the Plemelj formulae (see [V3]) { C + µ(z) =Cµ(z)+πif(z), (6) C µ(z) =Cµ(z) πif(z), where the identities hold for ds-almost all z E and µ = f(z)dz + µ s, f being integrable and µ s being singular with respect to ds. Assume that one has µ s =0and f(z) A, for ds-almost all z E, and that one wants to show (7) C(µ)(z) A, for ds-almost all z E. Then one only has to check that C(µ)(z) A, for z/ E, because then C (µ)(z) A, for ds-almost all z E, and thus the second identity in (6) gives (7) The local T (b)-theorem. The local T (b)-theorem is a criterion for the L 2 boundedness of a singular integral that was proved originally by. Christ in the setting of homogeneous spaces [CH2]. We state below a very particular version of Christ s result, which is adapted to the principal value Cauchy integral and to ameasureµ supported on E. The reader may think that µ is of the form µ = cds E for some (small) positive constant c. However, one should keep in mind that, for 4 n σ n, ds E does not satisfy condition (i) in the statement below with a constant independent of. As will become clear later, an appropriate choice of c is required to get (i) and (iii) with absolute constants. Theorem (Christ). Let µ be a positive Borel measure supported on E satisfying, for some absolute constant C, the following conditions: (i) µ(d(z,r)) Cr, z E, r>0. (ii) µ(d(z,2r)) Cµ(D(z,r)), z E, r>0. (iii) For each disc D centered at a point in E there exists a function b D in L (µ), b D supported on D, satisfying b D and C(b D µ) µ-almost everywhere on E,andµ(D) C b D dµ. Then (8) C(fµ) 2 dµ C f 2 dµ, f L 2 (µ), for some absolute constant C (depending only on C). The relevance of inequality (8) for our problem lies in the fact that it implies (9) µ(k) C γ + (K), K compact E, for some absolute constant C (depending only on C ).

5 ANALYTIC CAPACITY OF CANTOR SETS 23 The derivation of (9) from (8) goes through a well-known path: first, by classical Calderón-Zygmund theory one gets a weak (, ) inequality from (8); then, a surprisingly simple method to dualize a weak (, ) inequality leads immediately to (9). The original argument is in [DO]. Some years before [DO] Uy found a slightly different way of dualizing a weak (, ) inequality, which, however, does not yield (9) (see [Uy]). The interested reader will find additional information in [CH], [T] and [V]. Inequality (9) also explains why the lower estimate in (4) follows from attila s arguments in [a]; see Theorem 3.7 on p. 202 and the first paragraph after it. 3. Proof of Theorem 2 We first give a sketch of the argument. Assume that one can find a positive Borel measure µ supported on E N, E N = E(λ,...,λ N ), which satisfies (i) and (ii) with replaced by N, such that µ = γ(e N ) and the Cauchy integral is bounded on L 2 (µ). Then we get (9) with replaced by N, as we explained in Section 2. For K = E N (9) yields γ(e N )= µ Aγ + (E N ), as desired. In the actual argument we do not construct µ on E N. For reasons that will become clear later, we are forced to work in E with smaller than N. On the other hand, cannot be much smaller than N, because in the course of the subsequent reasoning one needs to have γ + (E ) Aγ + (E N ). Hence has to be chosen carefully, in such a way that the local T (b)-theorem can be applied to get the boundedness of the Cauchy integral on L 2 (µ), with an absolute constant. Now we start the proof of Theorem 2. Set a n =4 n σ n and S n = a 2 + a a 2. n We can assume, without loss of generality, that for each N>, there exists, <N, such that S S N (0) 2 <S +. Otherwise S N 2 <S and thus, by (4), γ + (E N ) A λ. On the other hand, taking into account the obvious estimate of analytic capacity by length, we clearly have γ(e N ) γ(e ) 2π length( E )= 8 π λ. Therefore (5) is trivial in the present case, provided C 0 is chosen to satisfy C 0 8 π A. Assume, then, that (0) holds and let us proceed to prove (5) by induction on N. ThecaseN = is obviously true. The induction hypothesis is γ(e n ) C 0 γ + (E n ), 0 <n<n, where the precise value of the absolute constant C 0 will be determined later. We distinguish two cases. Case : For some absolute constant C, to be determined later, () a γ(e(λ +,...,λ N )) C γ(e N ).

6 24 JOAN ATEU, XAVIER TOLSA, AND JOAN VERDERA Case 2: () does not hold. We deal first with Case 2. By the induction hypothesis applied to the sequence λ +,...,λ N and by (4) we have γ(e N ) C a γ(e(λ +,...,λ N )) C C 0a γ + (E(λ +,...,λ N )) C C a 0A ( N (4λ + 4λ n ) 2 Clearly, the inequality S S N 2 ( N and so, again by (4), n=+ = C C 0A( N ). /2 a 2 n=+ n a 2 n=+ n is equivalent to 2 ) /2 ( N ), /2 a 2 n= n ) /2 γ(e N ) C C 0Aγ + (E N ). If C = A, wherea is the constant in the preceding inequality, we get (5), as desired. Now let us now consider Case. Set γ(e N ) µ = length( E ) ds E, so that µ = γ(e N ). To check condition (i) in the local T (b)-theorem of Section 2, we consider two cases. If r σ,then γ(e N ) µ(d) 2r Cr, length( E ) because γ(e N ) γ(e ) length( E ). For r>σ we can replace arbitrary discs centered at points in E by the squares Q n j,0 n, 0 j 4n. In other words, it suffices to prove (2) µ(q n j ) Cl(Qn j ), 0 n, j 4n. To show this, given a disc D of radius r centered at z E, one considers a square Q n j D, wheren is chosen so that l(qn j ) is comparable to r. Then, µ(q n j )=γ(e N ) 4 n = 4γ(E N) length( E n ) l(qn j ) 4γ(E n) length( E n ) l(qn j ) 2 π l(qn j ), and so (2) is proved. It is also a simple matter to ascertain that (ii) holds.

7 ANALYTIC CAPACITY OF CANTOR SETS 25 Now our goal is to prove that hypothesis (iii) of the local T (b)-theorem is satisfied. Once this has been verified, (9) applied to K = E yields (3) γ(e N )= µ Aγ + (E ) A S /2 Assuming that (3) holds, to complete the proof we again distinguish two cases, according to whether /a 2 + is greater than S or not. If /a 2 + >S,then and so Hence γ + (E + ) S + = S + a 2 + S /2 + a 2, + a + = 4 length( E +). γ(e N ) γ(e + ) 2π length( E +) γ + (E + ) γ + (E N ), and thus (5) holds for a sufficiently big constant C 0. If /a 2 + S,thenS + S and so γ + (E N ) S /2 N S /2 which gives (5), with a big enough C 0,by(3). Summing up, we have reduced the proof of Theorem 2 to checking that hypothesis (iii) of the local T (b)-theorem is satisfied. As we already remarked when dealing with hypothesis (i), in proving (iii) we can replace discs centered at points in E by squares Q n j, j 4n,0 n. In other words, it is enough to show that, given a square Q n j,0 n, j 4n, there exists a function b n j in L (µ), supported on Q n j, satisfying bn j and C(bn j µ) dµ-almost everywhere on E and such that µ(q n j ) C b n j dµ. Let f be the Ahlfors function of E N. Then f is analytic on C\E N, f(z), z / E N, f( ) =0andf ( ) =γ(e N ). The non-tangential boundary value of f at ζ E N,whichexistsfords-almost all ζ E N, is denoted by f(ζ). Set ν = 2πi f(ζ)dζ E N,sothat f(z) = z ζ dν(ζ), z / E N. Fix a generation n, 0 n. Then, for some index k, k 4 n, γ(e N )= 4 n j= ν(q n j ) 4 n ν(q n k), or, equivalently, µ(q n k ) ν(qn k ). To define the function b n k associated with Qn k, we need to describe a simple preliminary construction.,.

8 26 JOAN ATEU, XAVIER TOLSA, AND JOAN VERDERA Take a compactly supported C function ϕ on C, 0 ϕ, ϕds, Q 0 such that ϕ vanishes on 4 j= D(z j, /4), where the z j are the vertices of Q 0.Then C(ϕds Q0 ) A, as one checks easily. Set ( z v ) ϕ j (z) =ϕ j χ σ Q j, where vj is the left lower vertex of Q j. Hence C(ϕ j dµ) A and ϕ j dµ = length( Q j ) ϕ j ds, µ(q j ) 4 µ(q j ). Define b = b n k = ν(q j ) ϕ j ϕ j dµ. Q j Qn k For j k we construct b n j by simply translating bn k. We have Qn j = wn j + Qn k,for some complex number wj n.set b n j (z) =bn k (z wn j ), z C. Now we will prove that b n k satisfies condition (iii). Clearly, bdµ = ν(qn k ) µ(q n k). To show that b is bounded, it suffices to prove (4) ν(q j ) Aµ(Q j ), j 4, and for this we first remark that C(χ Q j ν)(z) A, z/ E N.Thisisprovedin[G2, Lemma 2.3 (a), p. 90]. Since C(χ Q j ν) is analytic outside Q j E N, we conclude that ν(q j ) Aγ(Q j E N ). Now notice that the set Q j E N can be obtained from E(λ +,...,λ N )bya dilation of factor σ and a translation. Hence, recalling (), γ(q j E N )=σ γ(e(λ +,...,λ N )) C 4 γ(e N )=C µ(q j ), which gives (4). It is worth noting at this point that the above inequality explains why cannot be taken to be N. Thanks to the discussion in Section 2 on Cauchy integrals and the Plemelj formulae, it becomes clear that we are only left with the task of proving C(bdµ)(z) A, z / E Q n k. Since C(χ Q n k ν)(z) A, z/ E N Q n k, we only need to estimate, for z/ E Q n k, the difference (5) C(bdµ)(z) C(χ Q n k ν)(z) = C(α j )(z), where Q j Qn k α j = ν(q j ) ϕ j dµ ϕ j dµ χ Q j ν.

9 ANALYTIC CAPACITY OF CANTOR SETS 27 We have dα j =0and C(α j )(z) A, z/ Q j, j 4, again using [G2, Lemma 2.3 (a), p. 90]. Thus, if z j is the center of Q j, (6) C(α j σ 2 )(z) A dist(z,q, z z j )2 j >σ. By the maximum principle, in estimating (5), we can assume that z zj for some j with Q j Q n k. Hence (5) is not greater than σ (7) A + A l j σ 2 dist(z,q l ) 2. For 0 n let Q n be the square in the n-th generation that contains Q j can estimate (7) by.we A + A n=0 Q l Q n \Q σ2 n+ dist(z,q l ) 2 A + A σ n n=0 σ 2 σn 2 4 n A + A ( 4 ) n A, 9 because σ = σ n λ n+ λ,0 n. 3 n This completes the construction of the function b n k associated with the square Q n k as required by hypothesis (iii) in the local T (b)-theorem. Now, by translation invariance it is clear that b n j for j k also satisfies (iii). This shows that the local T (b)-theorem can be applied to µ and thus completes the proof of Theorem 2. n=0 Remark 2. For simplicity, we assumed above that λ n /3 for all n. However, both Theorem and Theorem 2 hold for 0 <λ n < /2. Let us sketch the changes needed in the proof. First it should be noticed that the estimate (4) for γ + (E N )holdsfor0<λ n < /2. Indeed, the arguments for the left inequality in (4) in [a] are valid in this case. On the other hand, the right inequality in (4) is also true for 0 <λ n < /2. For example, arguing as in [T2], one can easily check that ( γ + (E N ) A + n N, λ n /3 (4 n σ n ) 2 ) /2 ( n N ) /2 (4 n σ n ) 2. The other places where the assumption λ n /3 has been used are (4) and (5). The inequality (4) also holds for 0 <λ n < /2. It follows from Vitushkin s estimates for the integral f(z) dz for piecewise Lyapunov curves Γ [Vi] (in our Γ case Γ = Q j ). To prove (5), one can use the sharper estimate C(α j )(z) Aσ γ(q j E N ) σ µ(q j dist(z,q AC ) j )2 dist(z,q, z z j )2 j >σ, instead of (6) (see [G2, pp. 2 3], for example). We leave the details for the reader.

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