Boundary distortions and change of module under extension of a doubly connected domain

A. Yu. Solynin

Zapiski Nauchnykh Seminarov POMI, Investigations on linear operators and function theory. Part 20, Tome 201 (1992), pp. 157-163

Résumé

Let $\mathcal{F}(p,r)$ denote the class of univalent analytic functions $f(z)$ in the domain $\mathcal{K}(\rho)=\{z: \rho|z|1\}$, satisfying $|f(z)|=1$ for $|z|=1$ and $r|f(z)|1$ for $z\in\mathcal{K}(\rho)$. Let $f(z;\rho,r)$ map $\mathcal{K}(\rho)$ onto the domain $\mathcal{K}(r)\setminus[r,s]$ and let $f(z;\rho,r)\in\mathcal{F}(\rho,r)$. THEOREM 2. Let $f(z)\in\mathcal{F}(\rho,r)$, $f(z)\ne e^{i\alpha}f(z;\rho,r)$, $\alpha\in\mathbb{R}$, and $\Phi(t)$ be a strictly convex monotone function of $t>0$. Then $$ \int_0^{2\pi}\Phi(|f'(e^{i\theta})|)d\theta\int_0^{2\pi}\Phi(|f'(e^{i\theta};\rho,r)|)d\,\theta. $$ The proof of this theorem is based on the Golusin–Komatu equation. If $E$ is a continuum in the disk $U_R=\{z: |z|$, then $M(R,E)$ denotes the conformal module of the doubly connected component of $U_R\setminus E$; let $\varepsilon(m)=\{E: \overline{U}_r\subset E\subset U_1,\ M(1, E)=m^{-1}\}$. PROBLEM. Find the maximum of $M(R, E)$, $R>1$, and the minimum of cap $E$ over all $E$ in $\varepsilon(m)$. This problem was posed by V. V. Koževnikov in a lecture to the seminare on geometric function theory at the Kuban university in 1980, and by D. Gaier (see [2]). The solution of this problem is given by the following theorem. THEOREM 3. Let $E^*=\overline{U}_m\cup[m, s]$. If $R>1; E, E^*\in\varepsilon(m)$ and $E\ne e^{i\alpha}E^*$, $\alpha\in\mathbb{R}$, then $$ M(R, E)(R,E^*),\quad \mathrm{cap}\,E^*\mathrm{cap}\,E. $$ A similar statement is also proved for continue lying in the half-plane. ADDENDUM. When the paper was ready for publication, the author obtained a letter from R. Laugesen with the information that he had also proved Theorem 3 by a different method based on results of A. Baernstein. II on potential theory.