Fall 2010 - Problem 10

Solution.

Suppose otherwise, and let $\func{f}{\D \setminus \set{0}}{A}$ be conformal, where $A$ is the annulus in the question. In particular, $f$ is bounded and holomorphic near $0$, so it extends to a holomorphic function $\func{g}{\D}{\C}$.

We claim that $g$ is injective. Suppose otherwise, so that $w = g\p{0} = g\p{z}$ for some $z \in \D \setminus \set{0}$. Let $U$ be an open ball around $0$ and $V$ an open ball around $z$ such that $U \cap V = \emptyset$ and $U, V \subseteq \D$. By the open mapping theorem, $g\p{U}$ and $g\p{V}$ are open. By assumption, $w \in g\p{U} \cap g\p{V}$, so their intersection is also open and non-empty. In particular, $\p{g\p{U} \cap g\p{V}} \setminus \set{w}$ is non-empty, so there exist $z_1 \in U \setminus \set{0}$ and $z_2 \in V \setminus \set{z}$ such that $g\p{z_1} = g\p{z_2}$. By construction of $U$ and $V$, $z_1 \neq z_2$ and $z_1, z_2 \in \D \setminus \set{0}$, but this implies that $f\p{z_1} = f\p{z_2}$, contradicting injectivity of $f$. Thus, $g$ must have been injective to begin with.

By continuity, $g\p{0} \in \cl{A}$, but by injectivity, $g\p{0} \notin A$, so $g\p{0}$ is a boundary point of $A$. Now let $U$ be an open neighborhood of $0$ in $\D$, so that $g\p{U}$ is an open neighborhood of $g\p{0}$ by the open mapping theorem. $g\p{0}$ is a boundary point of $A$, so $g\p{U} \cap \p{\cl{A}}^\comp \neq \emptyset$, but by continuity, $g\p{U} \subseteq \cl{A}$, a contradiction. Thus, no such $g$ could have existed to begin with, so there is no conformal mapping from $\D \setminus \set{0}$ to $A$, which completes the proof.