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Spring 2021 - Problem 9

Schwarz lemma

Let $\Omega_1 \subseteq \Omega_2$ be bounded Jordan domains in $\C$ . We also assume that $0 \in \Omega_1$ . Now suppose $\func{f_1}{\D}{\Omega_1}$ and $\func{f_2}{\D}{\Omega_2}$ are Riemann mappings, satisfying $f_1\p{0} = f_2\p{0} = 0$ . Show that

\abs{f_1'\p{0}} \leq \abs{f_2'\p{0}}.

Solution.

First, notice that $\func{f_2^{-1} \circ f_1}{\D}{\D}$ is holomorphic, since we had Riemann mappings. Observe also that

\begin{aligned} f_2^{-1} \circ f_2 = \id &\implies \p{f_2^{-1}}'\p{f_2\p{z}} f_2'\p{z} = 1 \\ &\implies \p{f_2^{-1}}'\p{f_2\p{z}} = \frac{1}{f_2'\p{z}}. \end{aligned}

In particular, when $z = 0$ , we have

f_2^{-1}\p{0} = \frac{1}{f_2'\p{0}},

since $f_2\p{0} = 0$ . Thus, by the Schwarz lemma and the fact that $f_1\p{0} = 0$ , we see

\abs{\p{f_2^{-1} \circ f}'\p{0}} \leq 1 \implies \abs{f_2^{-1}\p{f_1\p{0}}f'\p{0}} \leq 1 \implies \abs{f_1'\p{0}} \leq \abs{f_2'\p{0}},

which was what we wanted to show.