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Fall 2011 - Problem 10

Riemann mapping theorem, Schwarz lemma

Let $\Omega \subseteq \C$ be a simply connected domain with $\Omega \neq \C$ , and $\func{f}{\Omega}{\Omega}$ be a holomorphic mapping. Suppose that there exist points $z_1, z_2 \in \Omega$ , $z_1 \neq z_2$ , with $f\p{z_1} = z_1$ and $f\p{z_2} = z_2$ . Show that $f$ is the identity on $\Omega$ , i.e., $f\p{z} = z$ for all $z \in \Omega$ .

Solution.

By the Riemann mapping theorem, there exists a conformal map $\func{\phi}{\Omega}{\D}$ . Hence, $\func{g = \phi \circ f \circ \phi^{-1}}{\D}{\D}$ is a holomorphic map. Thus,

g\p{\phi\p{z_j}} = \phi\p{f\p{z_j}} = \phi\p{z_j}

for $j = 1, 2$ . For $a \in \D$ , consider the Blaschke factor $\func{\psi_a}{\D}{\D}$ given by

\psi_a\p{z} = \frac{z - a}{1 - \conj{a}z},

which is an automorphism of the disk that maps $a$ to $0$ . Consider $\func{h = \psi_{\phi\p{z_1}} \circ g \circ \psi_{\phi\p{z_1}}^{-1}}{\D}{\D}$ , and notice that

h\p{0} = \p{\psi_{\phi\p{z_1}} \circ g}\p{\phi\p{z_1}} = \psi_{\phi\p{z_1}}\p{\phi\p{z_1}} = 0.

Moreover,

h\p{\psi_{\phi\p{z_1}}\p{\phi\p{z_2}}} = \p{\psi_{\phi\p{z_1}} \circ g}\p{\phi\p{z_2}} = \psi_{\phi\p{z_1}}\p{\phi\p{z_2}},

so $\psi_{\phi\p{z_1}}\p{\phi\p{z_2}}$ is a fixed point of $h$ . By the Schwarz lemma, this implies that $h$ is the identity, and so

\id = h = \psi_{\phi\p{z_1}} \circ g \circ \psi_{\phi\p{z_1}}^{-1} \implies g = \id.

Similarly,

\id = g = \phi \circ f \circ \phi^{-1} \implies f = \id,

which completes the proof.