Failure of rational approximation on some Cantor type sets

Let $A(K)$ be the algebra of continuous functions on a compact set $K\subset\mathbb C$ which are analytic on the interior of $K$, and $R(K)$ the closure (with the uniform convergence on $K$) of the functions that are analytic on a neighborhood of $K$. A counterexample of a question made by A. O'Farrell about the equality of the algebras $R(K)$ and $A(K)$ when $K=(K_{1}\times[0,1])\cup([0,1]\times K_{2})\subseteq\mathbb C$, with $K_{1}$ and $K_{2}$ compact subsets of $[0,1]$, is given. Also, the equality is proved with the assumption that $K_{1}$ has no interior.


Introduction
Consider a compact set K of the complex plane. Let A(K) be the algebra of continuous functions on K which are analytic on the interior of K, and R(K) the closure (with the uniform convergence on K) of the functions that are analytic on a neighborhood of K. Obviously, R(K) ⊆ A(K).
In the 60's, Vitushkin gave a description in analytic terms of the compact sets K for which R(K) = A(K) (see [Vi]), but there is still no characterization of those compact sets in a geometric way. Nevertheless, there have been important advances in this area recently, as can be seen in the articles of Xavier Tolsa [To1] and [To2] and the one of Guy David [Da]. In this direction, Anthony G. O'Farrell raised the following question (private communication): Question 1.1. Let K 1 and K 2 be two compact subsets of [0, 1] and define It is known that the identity holds if one of the compact sets K 1 or K 2 has no interior. For completeness, we include a proof of that fact at the end of the paper. However, it was not known whether the identity holds or not in general. In this paper we provide an example of a compact set K which gives a negative answer to the question. The set K is constructed as follows: Let C(1/3) be the ternary Cantor set on the interval [0, 1], i.e., where I 1 0 = [0, 1] and each I j n is an interval of length 3 −n obtained by dividing the intervals of length 3 −n+1 in three equal parts and excluding the central part. Call z j n the center of I j n . Consider a sequence δ n > 0 such that δ n < 3 −n−1 and define J j n = (z j n − δ n /2, z j n + δ n /2), where z j n is the center of I j n . Let With this construction of K we will prove the main result of the paper: Theorem 1.2. For a suitable choice of the sequence δ n , R(K) = A(K).
In the whole paper M 1 stands for the 1-dimensional Hausdorff content and α denotes the continuous analytic capacity (see [Vi]). Remember that, given a compact set F ⊆ C, where the supremum is taken over all continuous functions f : C −→ C which are analytic on C \ F , and uniformly bounded by 1 on C. If f satisfies all these properties, we say that f is admissible for α and F . By definition,

Proof of the main result
In the two following lemmas, we shall obtain some estimates of the Hausdorff content of [0, 1] 2 \K that will be useful to show that the algebras R(K) and A(K) are not equal for a suitable choice of the sequence δ n . Then Proof. Since R is the union of the squares R j n for 0 ≤ n ≤ n 0 and 1 ≤ j ≤ 2 n and each square has side length δ, we have The inequality δ < 3 −n 0 +2 is equivalent to n 0 < 2 − log 3 δ. Then, using that log 3 δ = log 2 δ/ log 2 3, we can deduce that As we will see in the proof of the following lemma, the important fact of the preceding one is that M 1 (R) is bounded by something that tends to zero as δ decreases, rather than the exact value of the bound.

It is clear that
4 n k=1 P k n . By construction, we also have M 1 (P 1 n ∩ G) = M 1 (P k n ∩ G) for all k = 1, . . . , 4 n . Therefore, Call X n the horitzontal strip of the cross P 1 n and Y n the vertical one. Because of the symmetry of the compact set K and the subadditivity of M 1 , Observe that G is a countable union of rectangles, and on X n all those rectangles have the sides of length less or equal than δ n . So, the set 3 n (X n ∩ G) := {3 n x : x ∈ X n ∩ G} can be included by a translation in a set R := n 0 n=0 2 n j=1 R j n like the one of the preceding lemma, if we take δ = 3 n δ n and n 0 ∈ N such that 3 −n 0 +1 ≤ δ < 3 −n 0 +2 . Applying the lemma we obtain, M 1 (X n ∩ G) < 3 −n 8(3 n δ n ) η = 3 n(η−1) 8δ η n with η = 1 − 1 log 2 3 , and then, Given ε > 0, it is easy to find a decreasing sequence δ n that makes the last sum less than ε, because η > 0.
Proof of theorem 1.2. As Vitushkin proved in [Vi] (see also [Ga], theorem VIII.8. If C = C(1/3) × C(1/3), we known that α(C) > 0 because dim(C) > 1, where dim(·) denotes the Hausdorff dimension. Observe that C ⊆ ∂K and it does not depend on the chosen sequence δ n . This implies that α(∂K) ≥ α(C), so it is guarantied a minimum of continuous analytic capacity on the boundary of K for any sequence δ n .

A(K) = R(K) when K 1 has no interior
Now, as we said at the beginning of the paper, we proceed to give an affirmative answer to the question 1.1 with the assumption that K 1 has no interior. We need an auxiliary lemma that we guess is already known, so we only sketch the proof.
Lemma 3.1. Fix δ > 0 and n ∈ N. Let R be a rectangle with sides of length δ and nδ and put R = n j=1 Q j , where Q j squares of side length δ with pairwise disjoint interiors. Let E j ⊆ Q j and suppose there exists C 0 > 0 such that α(E j ) ≥ C 0 δ for all j. Then, there exists a constant C 1 > 0 depending only on C 0 such that Hint of the proof. Given admissible functions f j for α and E j , one can find a function f admissible for α and n j=1 E j such that j |f j (∞)| = C 1 |f (∞)| using Vitushkin's localization scheme with a modified triple zero lemma (see [Ve] or [Vi]), where one uses the fact that the sets E j are aligned. Then, one can prove the lemma by taking supremums.
From now on, we shall denote by C an absolute constant that may change its value at different occurrences.
Proof. By Vitushkin's theorem, it is known that R(K) = A(K) if and only if there exists an absolute constant C > 0 such that α(Q \ intK) ≤ Cα(Q \ K) for all open squares Q.
Fix a square Q of side length l > 0. We can suppose that Q \ K is not empty, so there exists a square F ⊆ Q \ K. Let π x and π y be the projections onto the horitzontal and vertical coordinate axis respectively. Then, π y (F ) ⊆ π y (Q) \ K 2 and we can find an interval F y ⊆ π y (F ) of length l/n for n big enough.
On the other hand, if we split π x (Q) into intervals I j for j = 1, . . . , n with pairwise disjoint interiors and length l/n, we can also find intervals F j x ⊆ (π x (Q) \ K 1 ) ∩ I j for j = 1, . . . , n, because K 1 has no interior. Therefore, n j=1 (F j x × F y ) ⊆ Q \ K and α(F j x × F y ) ≥ C 0 l/n. Now we are ready to use the preceding lemma with the squares Q j = F y × I j , the subsets E j = F j x × F y and δ = l/n, and we obtain We can finally deduce that for every open square Q, so R(K) = A(K).
We are grateful to Anthony O'Farrell for the communication of another proof of theorem 3.2 which uses annihilating measures instead of Vitushkin's theorem.