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Dudziak - Vitushkins Conjecture for Removable Sets

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Dudziak Vitushkins Conjecture for Removable Sets
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Vitushkins conjecture, a special case of Painleves problem, states that a compact subset of the complex plane with finite linear Hausdorff measure is removable for bounded analytic functions if and only if it intersects every rectifiable curve in a set of zero arclength measure. Chapters 1-5 of the book provide important background material on removability, analytic capacity, Hausdorff measure, arclength measure, and Garabedian duality that will appeal to many analysts with interests independent of Vitushkins conjecture. The fourth chapter contains a proof of Denjoys conjecture that employ.;Preface: Painlevs Problem; 1 Removable Sets and Analytic Capacity; 1.1 Removable Sets; 1.2 Analytic Capacity; 2 Removable Sets and Hausdorff Measure; 2.1 Hausdorff Measure and Dimension; 2.2 Painlevs Theorem; 2.3 Frostmans Lemma; 2.4 Conjecture and Refutation: The Planar Cantor Quarter Set; 3 Garabedian Duality for Hole-Punch Domains; 3.1 Statement of the Result and an Initial Reduction; 3.2 Interlude: Boundary Correspondence for H(U); 3.3 Interlude: An F. & M. Riesz Theorem; 3.4 Construction of the Boundary Garabedian Function.

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James J. Dudziak Universitext Vitushkins Conjecture for Removable Sets 10.1007/978-1-4419-6709-1_1 Springer Science+Business Media, LLC 2010
1. Removable Sets and Analytic Capacity
James J. Dudziak 1
(1)
Lyman Briggs College, Michigan State University, East Lansing, MI 48825, USA
James J. Dudziak
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Abstract
For now and forevermore, let K be a compact subset of the complex plane .This will be restated for emphasis many times in what follows but just as often will be tacitly assumed and not mentioned.
1.1 Removable Sets
For now and forevermore, let K be a compact subset of the complex plane Picture 1 .This will be restated for emphasis many times in what follows but just as often will be tacitly assumed and not mentioned. For the sake of those readers who skip prefaces, we repeat a definition: K is removable for bounded analytic functions , or more concisely removable , if for each open superset U of K in Picture 2 , each function that is bounded and analytic on Picture 3 extends across K to be analytic on the whole of U . These analytic extensions must be bounded on the whole of U since they are continuous and so also bounded on K . Thus the definition may be equivalently restated as follows: K is removable if for each open superset U of K in Vitushkins Conjecture for Removable Sets - image 4 , each element of Vitushkins Conjecture for Removable Sets - image 5 extends to an element of Picture 6 . Of course, for any open set V of Picture 7 , Picture 8 denotes the Banach algebra of all functions bounded and analytic on V . What can we say about a removable K ?
First, a removable K must have no interior. For if there were a point z 0 in the interior of K , then the function Vitushkins Conjecture for Removable Sets - image 9 would be a function nonconstant, bounded, and analytic on Picture 10 which, by Liouvilles Theorem [, 10.23], would not extend analytically to all of Vitushkins Conjecture for Removable Sets - image 11 .
An immediate consequence of this first observation is that the analytic extension of any element of Vitushkins Conjecture for Removable Sets - image 12 to an element of Vitushkins Conjecture for Removable Sets - image 13 is unique and of the same supremum norm, i.e., Vitushkins Conjecture for Removable Sets - image 14 and Picture 15 are isometrically isometric. This explains the terminology: removing K from U has made no difference to Picture 16 .
Second, a removable K must have connected complement. For if Picture 17 had more than one component, then the function which is one on the unbounded component and zero on all the bounded components would be a function nonconstant, bounded, and analytic on Picture 18 which, by Liouvilles Theorem [, 10.23], would not extend analytically to all of Picture 19 .
Third, a removable K must be totally disconnected , i.e., a removable K can contain no nontrivial connected subset. To see this we suppose otherwise and deduce a contradiction. So let C be a nontrivial connected subset of a removable K . Replacing C with its closure, we may assume that C is closed. Let U be the component of Vitushkins Conjecture for Removable Sets - image 20 containing . Of course, Vitushkins Conjecture for Removable Sets - image 21 denotes the extended complex plane (also known as the Riemann sphere). By our second observation, U contains all of Picture 22 . It is an exercise in point-set topology, which we leave to the reader, to show that Vitushkins Conjecture for Removable Sets - image 23 is a nontrivial connected subset of Vitushkins Conjecture for Removable Sets - image 24 . Fix Vitushkins Conjecture for Removable Sets - image 25 and set Vitushkins Conjecture for Removable Sets - image 26 . Then is a proper subregion of for which is connected Hence by the many - photo 27 is a proper subregion of for which is connected Hence by the many equivalences to simple connectivity - photo 28 for which is connected Hence by the many equivalences to simple connectivity and the - photo 29 is connected. Hence, by the many equivalences to simple connectivity and the Riemann Mapping Theorem [, 10.18]. Clearly then, f does not map Picture 30 onto the open unit disc at the origin. With this contradiction we are done.
As an aside, we note that the third observation subsumes the first two since any totally disconnected K must have no interior and a connected complement. While the first part of this assertion is trivial, the reader may find verifying the second part one of those exercises in mere point-set topology that is a wee bit frustrating!
Turning to concrete examples, any single point is removable since bounded analytic functions can be analytically continued across isolated singularities. This assertion is simple [, 10.16]:
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