Is analytic capacity continuous from below?

Solution 1:

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 )