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