# Is analytic capacity continuous from below?

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

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

-
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.

@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