Interior radii of symmetric not leaning domains (Q1603055)
From MaRDI portal
| This is the item page for this Wikibase entity, intended for internal use and editing purposes. Please use this page instead for the normal view: Interior radii of symmetric not leaning domains |
scientific article; zbMATH DE number 1758667
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Interior radii of symmetric not leaning domains |
scientific article; zbMATH DE number 1758667 |
Statements
Interior radii of symmetric not leaning domains (English)
0 references
13 November 2002
0 references
Let \(D\) be a domain of the complex sphere \(\overline{{\mathbb C}}\), \(D^*:=\{z:1/\overline{z}\in D\}\) \(g_D(z,z_0)\) denotes Green's function of the domain \(D\) with the pole \(z_0\). The interior radius of \(D\) with respect to \(z_0\) is given by \[ r(D,z_0)= \begin{cases} \exp(\lim_{z\to z_0}(g_D(z,z_0)+\log|z-z_0|)), & z_0\neq\infty;\\ \exp(\lim_{z\to z_0}(g_D(z,z_0)-\log|z|)), & z_0=\infty. \end{cases} \] As the author notes, many extremal problems for classes of analytic functions may be reduced to problems on not leaning domains. The present article is devoted to solving one extremal problem posed in the survey [\textit{V. N. Dubinin}, Russ. Math. Surv. 49, No. 1, 1-79 (1994); translation from Usp. Mat. Nauk 49, No. 1(295), 3-76 (1994; Zbl 0830.30014)]. The following result gives the solution: Theorem. Let \(B_0,\dots,B_n\) (\(n>2\)) be not leaning domains in \(\overline{{\mathbb C}}\); \(a_k\in B_k\), \(k=0,\dots,n\); \(a_0=0\), \(|a_k|=1\), \(k=1,\dots,n\); \(B_k=B_k^*\), \(k=1,\dots,n\). Then \[ \prod_{k=0}^n r(B_k,a_k)\leq {2^{2n+1/n}\over{(n^2-2)^{n/2+1/n}}} ({n-\sqrt{2}\over{n+\sqrt{2}}})^{\sqrt{2}}. \tag{1} \] If, in addition, the domains \(B_k\) possess classical Green's functions, then the equality in (1) is attained if and only if the points \(a_k\) and the domains \(B_k\) are poles and circular domains, respectively, of the quadratic differential \[ Q(z)dz^2=-{(\alpha z)^{2n}+(2n^2-2) (\alpha z)^n+1\over{z^2(( \alpha z)^n-1)^2}} dz^2,\quad |\alpha|=1. \] The article contains only a scheme of the proof. However, the precise references on the necessary results are presented.
0 references
analytic function
0 references
not leaning domains
0 references
Green's function
0 references
interior radius
0 references
quadratic differential
0 references