Analytic capacity

Analytic capacity

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 mathbb{C}setminus E can become. Roughly speaking, gamma(E) 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.

Definition

Let Ksubsetmathbb{C} be compact. Then its analytic capacity is defined to be

:gamma(K) = sup {|f'(infty)|; finmathcal{H}^infty(mathbb{C}setminus K), |f|_inftyleq 1, f(infty)=0}

Here, mathcal{H}^infty (U) denotes the set of bounded analytic functions U omathbb{C} , whenever U is an open subset of the complex plane. Further,

: f'(infty):= lim_{z oinfty}zleft(f(z)-f(infty) ight)

: f(infty):= lim_{z oinfty}f(z)

(note that usually f'(infty) eq lim_{z oinfty} f'(z) )

Ahlfors function

For each compact Esubsetmathbb{C}, there exists a unique extremal function, i.e. finmathcal{H}^infty(mathbb Csetminus K) such that |f|leq 1, f(infty)=0, and f'(infty)=gamma(K),. This function is called the Ahlfors function of "K". Its existence can be proved by using a normal family argument involving Montel's theorem.

Analytic capacity in terms of Hausdorff dimension

Let ext{dim}_H denote Hausdorff dimension and H^1 denote 1-dimensional Hausdorff measure. Then H^1(E)=0 implies gamma(E)=0 while ext{dim}_H(E)>1 guarantees gamma(E)>0.However, the case when ext{dim}_H(E)=1 and H^1(E)in(0,infty] is more difficult.

Positive length but zero analytic capacity

Given the partial correspondence between the 1-dimensional Hausdorff measure of a compact subset of mathbb{C} and its analytic capacity, it might be conjectured that gamma(E)=0 Leftrightarrow H^1(E)=0. 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 E_0:= [0,1] imes [0,1] be the unit square. Then, E_1 is the union of 4 squares of side length frac{1}{4} and these squares are located in the corners of E_1. In general, E_n is the union of 4^n squares (denoted by Q_n^j) of side length 4^{-n}, each Q_n^j being in the corner of some Q_{n-1}^k. Put E:=igcap E_n

Then H^1(E)=frac{1}{sqrt{2 but gamma(E)=0

Vitushkin's Conjecture

Suppose ext{dim}_H E=1 and H^1(E)>0. Vitushkin's conjecture states that

: gamma(E)=0 Leftrightarrow E ext{ is purely unrectifiable}

In this setting, "E" is (purely) unrectifiable if and only if H^1(EcapGamma)=0 for all rectifiable curves (or equivalently, C^1 -curves or (rotated) Lipschitz graphs) Gamma.

Guy David published a proof in 1998 for the case when, in addition to the hypothesis above, H^1(E). Until now, very little is known about the case when H^1(E) 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 mathbb{C} are removable?"It is easy to see that "K" is removable if and only if gamma(K)=0 . 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

*
*
* 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


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