Analysis Problem of the Day 18

Today’s problem is Problem 8 on the UCLA Fall 2024 Analysis Qual:

Problem 8: An analytic arc is the image $h(I)$ of an open subinterval $I \subseteq \mathbb{R}$ of a holomorphic injective function $h: U \to \mathbb{C}$ for an open neighborhood $U$ of $I.$ Let $\Omega \subset \mathbb{C}$ be a simply connected domain, $F: \Omega \to \mathbb{D}$ be a conformal mapping that extends to a homeomorphism $F: \overline{\Omega} \to \overline{\mathbb{D}},$ and $\gamma \subset \partial \Omega$ be an analytic arc. Show that there is an open set $W \supset \mathbb{D} \cup F^{-1}(\gamma)$ and a holomorphic function $G$ on $W$ such that $G=F$ on $\mathbb{D}.$

Solution: While the problem statement appears to be quite long and confusing, we are essentially being asked to construct an analytic continuation of the conformal map $F$ to some larger open set $\Omega.$ A very famous construction of an analytic extension of a holomorphic function on the unit disc is the Schwarz reflection principle. The idea in this problem will be that one can in fact apply the reflection principle across arbitrary analytic arcs.

Indeed, given an arbitrary analytic arc $\Gamma = h(I),$ the map $\sigma_\Gamma: z \to h(\overline{h^{-1}(z)})$ is an antiholomorphic map (that is, $\partial_z \sigma_\Gamma = 0$ ) which reflects points across $\Gamma.$ In our case, given the analytic arc $\Gamma := F^{-1}(\gamma) \subset \partial \mathbb{D} \subset U,$ one may define $W:= \mathbb{D} \cup U$ and define $G := \sigma_\gamma \circ F \circ \sigma_\Gamma$ on $U \setminus \mathbb{D}$ and set $G=F$ on $\mathbb{D} \cup \gamma.$ This is a holomorphic map on $U \setminus \mathbb{D},$ as it is the composition of two antiholomorphic maps and a holomorphic map. Moreover, it agrees with $F$ on $\Gamma,$ and so extends by a standard Morera’s theorem argument to a holomorphic map on $W$ such that $F=G$ on $\mathbb{D}.$

Remark: The above solution was inspired by the following StackExchange post.

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