Is analytic capacity continuous from below? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:44:13Z http://mathoverflow.net/feeds/question/86220 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86220/is-analytic-capacity-continuous-from-below Is analytic capacity continuous from below? Malik Younsi 2012-01-20T15:54:03Z 2012-01-31T21:36:29Z <p>I've been wondering about the following, I don't know if anyone knows the answer :</p> <p>For a compact set $K$ in the complex plane, define the <em>analytic capacity</em> 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</p> <p>$$f'(\infty) = \lim_{z \rightarrow \infty} z(f(z)-f(\infty)).$$</p> <p>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 <em>Ahlfors function</em> of $K$, such that $F \in H^{\infty}(\mathbb{C}_ {\infty} \setminus K)$, $\|F\|_{\infty} \leq 1$, and $F'(\infty)=\gamma(K)$.</p> <p>It's not hard to show that $\gamma$ is <em>continuous from above</em> : 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 $E \subseteq F$.</p> <p>My question is the following :</p> <p>Is analytic capacity <em>continuous from below</em>? 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 $\gamma(K) = \lim_{n \rightarrow \infty} \gamma(K_n)?$</p> <p>I could not find anything in the litterature.</p> <p>Thank you, Malik</p> <p><strong>EDIT ( 12-01-2012)</strong> I edit the question to add what I know so far about this question :</p> <ol> <li><p>As pointed out by Fedja in the comments, analytic capacity is <em>comparable</em> to a quantity which is continuous from below, see the article "Painleve's problem and the semiadditivity of analytic capacity" by Xavier Tolsa.</p></li> <li><p>The answer is yes if the compact sets $K_n$ and $K$ are connected. Indeed, for connected compact sets, analytic capacity is equal to <em>logarithmic capacity</em>, and logarithmic capacity is continuous from below.</p></li> <li><p>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 mainly follows from the fact that in this case, the Ahlfors function of $K$ extends analytically across the boundary of $K$. See for example the book "analytic capacity and measure" by Garnett, p. 18.</p></li> </ol> <p><strong>EDIT ( 31-01-2012)</strong> 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. </p> <p>I'll leave the question open though, because I'd be very interested to hear about sufficient conditions or similar results.</p>