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For a compact set $K$ in the complex plane, define the analytic capacity of $K$ by $$\gamma(K) := \sup |f'(\infty)|$$ where the supremum is taken over all functions $f$ holomorphic and bounded by $1$ in the complement of $K$ : $f \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|f\|_{\infty} \leq 1$. Here

$$f'(\infty) = \lim_{z \rightarrow \infty} z(f(z)-f(\infty)).$$

A theorem due to Ahlfors states that for each compact $K$, there always exists a unique (in the unbounded component of the complement of $K$) function $F$, called the Ahlfors function of $K$, such that $F \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|F\|_{\infty} \leq 1$, and $F'(\infty)=\gamma(K)$.

It's not hard to show that $\gamma$ is outer-regular, in the sense that if $(K_n)$ is a decreasing sequence of compact sets, then $$\gamma(\cap_n K_n) = \lim_{n\rightarrow \infty} \gamma(K_n).$$ This essentially follows from Montel's theorem and the fact that $\gamma(E) \subseteq \gamma(F)$ whenever $E \subseteq F$.

Question:

Is analytic capacity inner regular? More precisely, if $(K_n)$ is a sequence of compact sets such that $$K_1 \subseteq K_2 \subseteq K_3 \subseteq \dots$$ and such that $K:=\cup_n K_n$ is compact, then is it true that $\gamma(K) = \lim_{n \rightarrow \infty} \gamma(K_n)?$

I could not find anything in the litterature.

Thank you, Malik

EDIT

  1. As pointed out by Fedja in the comments, analytic capacity is comparable to a quantity which is continuous from below, see the article "Painleve's problem and the semiadditivity of analytic capacity" by Xavier Tolsa.

  2. The answer is yes if the compact sets $K_n$ and $K$ are connected. Indeed, for connected compact sets, analytic capacity is equal to logarithmic capacity, and logarithmic capacity is inner regular.

  3. The answer is yes if $K$ is a compact set whose boundary consists of a finite number of analytic and pairwise disjoint Jordan curves, provided we replace the condition $K:=\cup_n K_n$ by the condition that each compact subset of the interior of $K$ is eventually contained in some $K_n$. This easily follows from the fact that in this case, the Ahlfors function of $K$ extends analytically across the boundary of $K$.

EDIT I contacted Xavier Tolsa, and according to him, it's an open problem, related to the so called capacitability problem. It's not known if the Borel sets are capacitable.

I'll leave the question open though, because I'd be very interested to hear about sufficient conditions or similar results.

Malik Younsi
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    Slight quibble: the Ahlfors function may be unique in the unbounded component of the complement of $K$, but can't be uniquely determined in any bounded component. – Robert Israel Jan 20 '12 at 17:53
  • Yes, of course you're right. It is unique in the unbounded component of the complement of $K$. Thank you for the comment, I'll edit accordingly. – Malik Younsi Jan 20 '12 at 20:56
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    The best I can say is that it is comparable up to a constant factor to a quantity continuous from below in your sense. – fedja Jan 23 '12 at 11:26
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    @Fedja : Thank you for taking the time to comment on this. Are you referring to the quantity $\gamma^{+}$ and the following article from Tolsa on the comparability of $\gamma^{+}$ and analytic capacity ?

    Tolsa, Xavier "Painlevé's problem and the semiadditivity of analytic capacity."

     – Malik Younsi Jan 23 '12 at 14:15
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    Yes, that's it. It is not clear to me, however, if we can mix that argument with the classical definition somehow and get what you want. We can ask Xavier directly though. If he doesn't know, you are out of luck :). – fedja Jan 24 '12 at 03:10
  • @Malik: Using the results in the paper mentioned, I think that you get a positive result whenever the set $E=\bigcap_n\overline{\partial K\setminus K_n}$ has zero analytic capacity. $E$ is just the set of points on the boundary of $K$ where $K_n$ are not dense, so it is a nowhere dense closed subset of $\partial K$. For any nowhere dense closed subset of $\partial K$, you can always find $K_n\subseteq\partial K$ with $E=\bigcap_n\overline{\partial K\setminus K_n}$, and $\lim\gamma(K_n)$ is independent of the choice of $K_n$. For a counterexample, you need to find some sufficiently nasty $K,E$. – George Lowther Jan 25 '12 at 00:53
  • ...although, on second thoughts, I'm not quite sure that $\gamma(E)=0$ is really a sufficient condition. – George Lowther Jan 25 '12 at 01:00
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    @fedja : It is not clear to me at all either. Thank you for the suggestion, I'm going to contact X. Tolsa. Hopefully, I'll be lucky!

    @George : I don't really see why $\gamma(E)=0$ is a sufficient condition, but I'd be happy to hear more about it. I agree though that for a counterexample, we probably need to think about complicated compact sets. The problem though is that it's very difficult to determine the analytic capacity of even slightly complicated sets!

     – Malik Younsi Jan 25 '12 at 14:37
  • @Malik: no, I'm not sure that $\gamma(E)=0$ is sufficient anymore. – George Lowther Jan 25 '12 at 23:09

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