Fall 2017 - Problem 11

Solution.

Suppose $\inf\,\set{\abs{w} \mid w \notin f\p{\D}} = 1 + \epsilon > 1$ for some $\epsilon > 0$. Equivalently, $f\p{\D}^\comp \subseteq \set{\abs{w} \geq 1 + \epsilon}$ so taking complements, $f\p{\D} \supseteq \set{\abs{w} < 1 + \epsilon} = B\p{0, 1 + \epsilon}$.

Let $U = f^{-1}\p{B\p{0, 1 + \epsilon}} \subseteq \D$, which is open since $f$ is continuous. Because $f$ is injective, $f$ restricts to a biholomorphism $g^{-1} = \func{\res{f}{U}}{U}{B\p{0, 1 + \epsilon}}$, so $\func{g}{B\p{0, 1 + \epsilon}}{\D}$ with $g\p{0} = 0$. Applying the Schwarz lemma to $g\p{\p{1 + \epsilon}z}$, we get $\abs{g\p{z}} \leq \frac{\abs{z}}{1 + \epsilon}$ for $z \in B\p{0, 1 + \epsilon}$. Thus, for $z \in U$, $f\p{z} \in B\p{0, 1 + \epsilon}$ and so we may apply the inequality:

But $0 \in U$, so for $z$ close enough to $0$,

a contradiction. Thus, $\inf\,\set{\abs{w} \mid w \notin f\p{\D}} \leq 1$ to begin with.

To prove the second claim, first, if $f\p{z} = z$, then $f\p{\D} = \D$, so $d\p{f\p{\D}^\comp, 0} = 1$ and we have equality. Conversely, suppose we had equality, which means

As before, let $U = f^{-1}\p{\D}$ and let $\func{g}{\D}{U \subseteq \D}$ be the holomorphic inverse of $f$ restricted to $U$.

so by the Schwarz lemma, $g\p{z} = z$ and so $f\p{z} = z$ as well, which completes the proof.