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EDIT: I also asked this question on mathoverflow, since it might be too specialized for

I've been wondering about the following, I don't know if anyone knows the answer :

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 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 continuous from above : 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 $E \subseteq F$.

My question is the following :

Is analytic capacity continuous from below? 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

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I have tried my level in posting the answer, I don't know whether its upto mark or not. – Iyengar Jan 23 '12 at 17:52

Your question is rather a good one. Let me try to say what I know about it.

Its a simple consequence of the well-known fact that " Analytic capacity has following property "

If $K_a\subseteq K_b$ then its well known fact that $\gamma(K_a)\le\gamma(K_b)$.

So by using that what we can conlude is that the sequence {$\gamma(K_n)$ } is decreasing ( if you assume that {$K_n$} is a decreasing sequence of compact subsets ) , and you can also notice that its bounded below by $\gamma(K)$. So it now implies the following $$\gamma(K)\le\displaystyle\lim_{n\to\infty}\gamma(K_n)=\lim_{n\to\infty} \rm{Inf} \ \gamma(K_n).$$

Using your notation , let $F_n$ be the Ahlfors function of $K_n$ then its a known thing that they always form a normal family on every $\mathbb{C}\backslash K_n$ and also we have :

For every $K\subset\Omega$ there corresponds a number $M(K)$ ( $M(K)\lt\infty$ ) such that $|f(z)|\le M(K)$ for all $f\in X$ and $z\in K$ where $X\subset H(\Omega)$ for some region $\Omega$.

So I think one can prove the above statement by using " Cauchy's Formula " and knowing some basic rules of Convergence and some background in Analysis. ( If you are still looking a proof, I can add it, but its quite large, and I need to refer some old Analysis books for proof. Anyway if you are looking for a proof of above statement let me know ) .

Coming back , by combining the above statement and Famous Cantor's Diagonal argument we can conclude that sub-sequence of {$F_n$} uniformly converge ( on compact subsets of $\mathbb{C}\backslash K$ ) to a function $F$ analytic and bounded by one on $\mathbb{C}\backslash K$ . So one can clearly imply that $$\displaystyle\lim_{n\to\infty} \rm{Inf} \ \gamma(K_n) = \lim_{n\to\infty}\rm{Inf}\ F^{\prime}_{n}(\infty)\le F^{\prime}_{n}(\infty) = | F^{\prime}_{n}(\infty)|\le\gamma(K)$$

So now we reached our destination to say that , so it implies from above statements that $$\displaystyle\lim_{n\to\infty} \ \gamma(K_n)=\gamma(K)$$

So to add something to this, in a more broader manner, you are asking for increasing sequence, so let me show you a clear idea how can you achieve it.

I think that Analytic capacity is some sort of measure , measures the size of a set as a non-removable singularity . So it may be compared and taken as a normal measure which satisfies the property $$\mu(A_n)\to \mu(A)$$ as $n\to \infty$.

So I hope that is what you are looking for. Where $A=\bigcup^{\infty}_{n=1} A_n$ , and its nothing but increasing sequence $$A_1\subset A_2\subset A_3......$$

But I am not sure that one can consider the Analytic capacity to be a measure in a strict manner. But it can be extended to that notion.

And also something to add, the upper continuity you are talking about is perfectly defined for Hausdorff measure. There is a standard theorem which says " Let $E$ be a union of increasing sequence {$E_n$} of subsets of $\mathbb{C}$ then for any $S\in[0,2]$ , it follows that $$\displaystyle \lim_{n\to \infty} H^S(E_n)= H^S(E)$$. But there are some theorems that establish the relationship between analytic capacity and Hausdorff Measure, but I don't know how far it can be used to answer your theorem.

Thank you.

( Please feel free to write comments and valuable feed-back )

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I think that a blind man should not lead people, I think that what the answer I have given is correct, but if there are any mistakes, or if the entire answer is a mistake please let me know, please don't down-vote, just comment to delete, so that I will delete the answer . Thank you. – Iyengar Jan 23 '12 at 17:05
Thank you for taking the time to think about my question, but this does not answer it. If I understand correctly, in the first part, you prove that analytic capacity is continuous from above, and I'm already aware of that. I'm asking whether it is continuous from below. As for the second part of your answer, note that analytic capacity is far from being a measure, and it cannot be "extended to that notion". This is mainly because analytic capacity only depends on the outer boundary of the compact. – Kalim Jan 23 '12 at 18:25
@MalikYounsi : Thanks a lot for your response, I can say that we can take the analytic capacity to be a measure, I have seen some papers where the author himself stated that analytic capacity can be compared to a measure. Let me say how it can be done, let us see the similarities between the Analytic capacity and measure. If {$E_n$} are the sequence of sets, then $\displaystyle\mu(\bigcup^{\infty}_{i=1} E_i) \le \sum^{\infty}_{i=1}\mu (E_i)$ , so the same analogousness can be seen on the side of Analytic capacity ( continuous ) , it can be written as Contd... – Iyengar Jan 24 '12 at 6:24
Contd : .. In the same way the continuous analytic capacity can be written as $\displaystyle \alpha(\bigcup^{\infty}_{i=1} E_i)\le C\ \sum^{\infty}_{i=1} \alpha(E_i)$ where $C$ is an absolute constant , so I think we can compare the analytic capacity to measure. And once if we can do it, we can prove the above thing easily. I don't know how far I am correct. Its just an attempt, I am extremely sorry if the things I have said doesn't make any sense, Thank you @MalikYounsi – Iyengar Jan 24 '12 at 6:28
Maybe this answer is not the OP was looking for, but definitely does not deserve a downvote. – timur Aug 24 '12 at 2:36

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