In complex analysis, the analytic capacity of a compact subset K of the complex plane is a number that denotes "how big" a bounded analytic function from <math>\mathbb{C}\setminus E </math> can become. Roughly speaking, <math>\gamma(E)</math> measures the size of the unit ball of the space of bounded analytic functions outside E. It was first introduced by Ahlfors in the 1940s while studying the removability of singularities of bounded analytic functions.
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Definition
Let <math>K\subset\mathbb{C}</math> be compact. Then its analytic capacity is defined to be
- <math>\gamma(K) = \sup \{|f'(\infty)|;\ f\in\mathcal{H}^\infty(\mathbb{C}\setminus K),\ \|f\|_\infty\leq 1,\ f(\infty)=0\}</math>
Here, <math> \mathcal{H}^\infty (U) </math> denotes the set of bounded analytic functions <math> U\to\mathbb{C} </math>, whenever <math>U</math> is an open subset of the complex plane. Further,
- <math> f'(\infty):= \lim_{z\to\infty}z\left(f(z)-f(\infty)\right) </math>
- <math> f(\infty):= \lim_{z\to\infty}f(z) </math>
(note that usually <math> f'(\infty)\neq \lim_{z\to\infty} f'(z) </math>)
Ahlfors function
For each compact <math>E\subset\mathbb{C}</math>, there exists a unique extremal function, i.e. <math>f\in\mathcal{H}^\infty(K)</math> such that <math>\|f\|\leq 1</math>, <math>f(\infty)=0</math> and <math>f'(\infty)=\gamma(K)</math>. This function is called the Ahlfors function of K This can be proved by using a normal family argument involving Montel's Theorem.
Proof of existence for a continuum
There is a relatively simple proof of the existence of an Ahlfors function, based on the Riemann mapping theorem, if we assume additionally that K is connected. We define an open subset of <math>\mathbb{C}</math> to be simply connected if it is connected and its complement in the Riemann sphere <math> S^2 </math> is connected. If K is compact and connected, we can assume <math>E\ne\varnothing</math> (otherwise <math>\mathcal{H}^\infty(\mathbb{C}\setminus K)=\mathcal{H}^\infty(\mathbb{C})=\{\text{constant functions}\}</math> by Liouville's theorem and hence <math>\gamma(K)=0</math>). Then there exists a unique connected component U of <math>\mathbb{C}\setminus K</math> that is unbounded. The claim is that U is simply connected. To see this, note that <math>\infty\in U </math> and therefore <math> S^2\setminus U=\mathbb{C}\setminus U </math>. So let <math> f:\mathbb{C}\setminus U\to\{-1,\,1\} </math> be continuous with respect to the discrete topology on <math>\{-1,\,1\}</math>. Write
- <math> \mathbb{C}\setminus U = K\cup\bigcup_{j\in J} V_j </math>
where the <math> V_j </math> are the bounded connected components of <math>\mathbb{C}\setminus K </math>. Then, since K and each <math> V_j</math> are connected, <math> f_K,\ f_{V_j} </math> are constant. But <math>U</math> is open in <math>\mathbb{C}\setminus K</math>, hence in <math>\mathbb{C}</math>. So <math>\mathbb{C}\setminus U</math> is closed and
- <math> \overline{V_j}\subseteq \mathbb{C}\setminus K\quad\forall j\in J.</math>
But each <math> V_j</math> is open in <math>\mathbb{C}\setminus K</math>, hence in <math>\mathbb{C}</math>, so that <math>\overline{V_j}\cap E\ne\varnothing </math>. Hence f must be constant, whence <math> \mathbb{C}\setminus U</math> is connected, so U is simply connected. The Riemann mapping theorem now yields a biholomorphism <math>g\ :\ U\to B(0,1)</math> such that <math>g(\infty)=0</math> and <math>g'(\infty)>0</math>. (Here, <math>B(0,1)</math> denotes the unit disk in <math>\mathbb{C}</math>.) Defining <math> g(z)=0 </math> for each <math>z\in\mathbb{C}\setminus(U\cup K)</math>, this defines a holomorphic map <math>g\ :\ \mathbb{C}\setminus K\to B(0,1)</math>. In particular, <math> g\in\mathcal{H}^\infty(\mathbb{C}\setminus E)</math>, so that <math>\gamma(E)\geq g'(\infty)</math>. To prove the reverse inequality, let <math>f\in\mathcal{H}^\infty</math> with <math>\|f\|_\infty\leq 1,\ f(\infty)=0</math> and put <math>F:= f\circ g^{-1}</math>. Then <math>F\ :\ B(0,1)\to \overline{B(0,1)} </math> is analytic (since f and g are),
- <math> F(0)= f(g^{-1}(0)) = f(\infty)=0 </math>
and so we may apply the Schwarz lemma to F. Hence, <math>F'(0)\leq 1</math>. Thus,
- <math> 1 \geq F'(0) = \frac{f'(g^{-1}(0))}{g'(g^{-1}(0))} = \frac{f'(\infty)}{g'(\infty)}</math>
which gives us <math>f'(\infty)\leq g'(\infty)</math>. Taking the supremum over all such f, we get <math>\gamma(E)\leq g'(\infty)</math>. This concludes the proof.
Additional properties assuming finite connectivity
Let <math>U:= \mathbb{C}\setminus E</math>. If <math>n\in\mathbb{N}</math> and E has n components, then the Ahlfors function is analytic across <math>\partial U</math>. Moreover, if <math>\partial U</math> is smooth, then <math>f(\partial U)=\{1\}</math>.
Analytic capacity in terms of Hausdorff dimension
Let <math>\text{dim}_H</math> denote Hausdorff dimension and <math> H^1 </math> denote 1-dimensional Hausdorff measure. Then <math> H^1(E)=0</math> implies <math>\gamma(E)=0</math> while <math> \text{dim}_H(E)>1</math> guarantees <math>\gamma(E)>0</math>. However, the case when <math> \text{dim}_H(E)=1</math> and <math> H^1(E)\in(0,\infty] </math> is more difficult.
Positive length but zero analytic capacity
Given the partial correspondence between the 1-dimensional Hausdorff measure of a compact subset of <math>\mathbb{C}</math> and its analytic capacity, it might be conjectured that <math>\gamma(E)=0\ \Leftrightarrow\ H^1(E)=0</math>. However, this conjecture is false. A counterexample was first given by A. G. Vitushkin, and a much simpler one by J. Garnett in his 1970 paper. This latter example is the linear four corners Cantor set, constructed as follows: Let <math>E_0:= [0,1]\times[0,1]</math> be the unit square. Then, <math>E_1</math> is the union of 4 squares of side length <math>\frac{1}{4}</math> and these squares are located in the corners of <math>E_1</math>. In general, <math>E_n</math> is the union of <math>4^n</math> squares (denoted by <math>Q_n^j</math>) of side length <math>4^{-n}</math>, each <math>Q_n^j</math> being in the corner of some <math>Q_{n-1}^k</math>. Put <math>E:=\bigcap E_n</math> Then <math>H^1(E)=\frac{1}{\sqrt{2}}</math> but <math>\gamma(E)=0</math>
Vitushkin's Conjecture
Suppose <math>\text{dim}_H E=1 </math> and <math>H^1(E)>0</math>. Vitushkin's conjecture states that
- <math> \gamma(E)=0\ \Leftrightarrow\ E \ \text{ is purely unrectifiable} </math>
In this setting, E is (purely) unrectifiable if and only if <math>H^1(E\cap\Gamma)=0</math> for all rectifiable curves (or equivalently, <math> C^1 </math>-curves or (rotated) Lipschitz graphs) <math>\Gamma</math>. Guy David published a proof in 1998 for the case when, in addition to the hypothesis above, <math> H^1(E)<\infty</math>. Until now, very little is known about the case when <math> H^1(E) </math> is infinite (even sigma-finite).
Removable sets and Painlevé's problem
The compact set K is called removable if, whenever Ω is an open set containing K, every function which is bounded and holomorphic on the set Ω\K has an analytic extension to all of Ω. By Riemann's principle for removable singularities (see "Riemann's theorem" in Removable singularity), every singleton is removable. This motivated Painlevé to pose a more general question in 1880: "Which subsets of <math>\mathbb{C}</math> are removable?" It is easy to see that K is removable if and only if <math>\gamma(K)=0 </math>. However, analytic capacity is a purely complex-analytic concept, and much more work needs to be done in order to obtain a more geometric characterization.
References
- Mattila, Pertti (1995). Geometry of sets and measures in Euclidean spaces. Cambridge University Press. ISBN 0-521-65595-1.
- Pajot, Hervé (2002). Analytic Capacity, Rectifiability, Menger Curvature and the Cauchy Integral, Lecture Notes in Mathematics. Springer-Verlag.
- J. Garnett, Positive length but zero analytic capacity, Proc. Amer. Math. Soc. 21 (1970), 696-699
- G. David, Unrectifiable 1-sets have vanishing analytic capacity, Rev. Math. Iberoam. 14 (1998) 269-479
