A simple proof of Dubinin's theorem
Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 2 (1995) no. 3, pp. 347-355
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Let $\Omega$ be a domain formed by removing $n$ radial segments connecting the circles $\{z:| z |=r_0\}$ and $\{z:|z|=1\}$ from the unit disk $\mathbf D$. Let $\Omega_0$ be a domain of the same type which is invariant with respect to rotation by the angle $2\pi/n$. If $\omega(z)$ and $\omega_0(z)$ are the harmonic measures of the unit circle with respect to these domains, then the inequality $$\omega_0\geq\omega_0(0),$$ holds, and the equality is possible only if the domain $\Omega$ coincides with $\Omega_0$ up to rotation. This proposition is known as the Gonchar problem which has been proved by Dubinin. The aim of this paper is to give a more simple proof of this theorem.
@article{JMAG_1995_2_3_a8,
author = {A. E. Fryntov},
title = {A simple proof of {Dubinin's} theorem},
journal = {\v{Z}urnal matemati\v{c}eskoj fiziki, analiza, geometrii},
pages = {347--355},
year = {1995},
volume = {2},
number = {3},
language = {en},
url = {http://geodesic.mathdoc.fr/item/JMAG_1995_2_3_a8/}
}
A. E. Fryntov. A simple proof of Dubinin's theorem. Žurnal matematičeskoj fiziki, analiza, geometrii, Tome 2 (1995) no. 3, pp. 347-355. http://geodesic.mathdoc.fr/item/JMAG_1995_2_3_a8/