On the problems of conformal maps with quasiconformal extension (Q1817247)

In this paper the following results are proved: Theorem 1. Let \(f(z)\) be analytic in \(D= \{z: |z|< 1\}\) with \(f(0)= 0\), \(f'(0)= 1\), \(f''(0)= 0\). If \(f(z)\) satisfies the condition. \[ \bigl|S_f(z) \bigr|\leq{2t \over \bigl(1- |z |^2\bigr)^2} \quad \text{for } z \in {\mathbf D} \] where \(S_f(z) = (f''/f')'- {1\over 2} (f''/f')^2\) is the well-known Schwarzian derivative and \(t\) is fixed \(0 \leq t<1\), then \(f(z)\) has a Hölder continuous extension to \(|z|\leq 1\) with \[ \bigl|f(z_1) - f(z_2) \bigr|\leq \left({4 \over 1- \sqrt{1-t}} \right)^{1- \sqrt {1-t}} {1+ \sqrt{1-t} (2^{1 - \sqrt {1-t}}-1) \over \sqrt{1-t}} |z_1- z_2|^{\sqrt {1-t}} \] for all \(z_1\), \(z_2\in \overline {\mathbf D} = \{z: |z|\leq 1\}\). The exponent \(\sqrt{1-t}\) is sharp. Theorem 2. Let \(f(z) = z/(1-a_2 z + \varphi (z))\) be analytic in \({\mathbf D}\), where \(\varphi (z)\) is analytic in \({\mathbf D}\) and such that \(\varphi (0)= \varphi'(0) =0\) and for some \(k<1\) \[ \left|{\varphi (z_1) \over z_1} - {\varphi (z_2) \over z_2} \right|\leq k |z_1 - z_2 |, \quad z_1 \in {\mathbf D},\;z_2 \in{\mathbf D}. \] Then the mapping \[ F(z)= \left\{\begin{matrix} f(z), \quad & |z|\leq 1 \\ z/ \bigl(1- a_2 z+z \overline z \varphi (1/ \overline z) \bigr),\quad & |z |\geq 1 \end{matrix} \right. \] is a quasiconformal extension of \(f(z)\) onto \(\overline\mathbb{C}\) and \(\mu_F(z) = |F_{\overline z}/F_z |\leq k\), \(z \in\mathbb{C}\).