Three-circles theorems (Q1082484)
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scientific article; zbMATH DE number 3973269
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Three-circles theorems |
scientific article; zbMATH DE number 3973269 |
Statements
Three-circles theorems (English)
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1986
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Results of the Three Circles Theorem type are obtained for bounded nonvanishing analytic functions with domain \(U=\{| z| <1\}\). The following theorem is established with the aid of Pick's lemma: Let f be an allowed function. Suppose that \(0<\rho <R<1\). Let \(m_ 1=\log \sup \{| f(z)|:\) \(z\in U\}\). Let \(M(r,f)=\max \{| f(z)|:| z| =r\}\). Suppose that \(M(\rho,f)=1\) and \(m_ 1>0\). Let \(\kappa =\log M(R,f)/m_ 1\). Then \(1-\kappa \geq (1+\rho)^{-1}(1-R)(1+R)^{- 1}.\) The inequality is sharp. Let \({\mathcal P}\) denote the set of analytic functions F with domain U which have positive real part and satisfy \(F(0)=1\). Let \({\mathcal P}_{{\mathbb{R}}}\) denote the subset of \({\mathcal P}\) whose members take real values at the points of (-1,1). The following extremal problems are treated: (i) Given m, \(1\leq m\leq (1+\rho)/(1-\rho)\), determine \[ M=\sup \{M(R,G): G\in {\mathcal P},\quad M(\rho,G)\leq m\}. \] (ii) Given \(m_ 0\) satisfying \((1- \rho)/(1+\rho)\leq m_ 0\leq (1+\rho)/(1-\rho)\), determine \(M_ 0=\sup \{g(R):\) \(g\in {\mathcal P}_{{\mathbb{R}}}\), \(g(\phi)\leq m_ 0\}\). It is shown that for (ii) there is a unique extremal function, \(g_{\alpha}\), given by \[ g_{\alpha}(z)=\alpha (1+z)(1-z)^{-1}+(1-\alpha)(1-z)(1+z)^{- 1} \] where \(\alpha\) is determined by \(g_{\alpha}(\rho)=m_ 0\). Use is made of the finite Pick-Nevanlinna interpolation theory. Results of the following kind are obtained for (i): The extremal is unique and M is a concave function of m. Under the assumption that \(1<m\leq (1+\rho)/(1- \rho)\) the extremals have Herglotz measure with finite support the number of elements of which tends to \(+\infty\) as \(m\downarrow 1\).
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Three Circles Theorem
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Pick-Nevanlinna interpolation
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Herglotz measure
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