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Fall 2015 - Problem 7

Schwarz reflection principle

Assume that $f\p{z}$ is analytic in $\set{z \mid \abs{z} < 1}$ and continuous on $\set{z \mid \abs{z} \leq 1}$ . If $f\p{z} = f\p{1/z}$ when $\abs{z} = 1$ , prove that $f\p{z}$ is constant.

Solution.

Define

g\p{z} = \begin{cases} f\p{z} & \text{if } \abs{z} \leq 1 \\ f\p{1/z} & \text{if } \abs{z} \geq 1, \end{cases}

which is well-defined by assumption. $g$ is certainly holomorphic when $\abs{z} < 1$ and when $\abs{z} > 1$ (as a composition of two holomorphic functions). It is continuous near $\abs{z} = 1$ by assumption as well, so by the Schwarz reflection principle, $g$ is entire. But $f$ is bounded by some $M > 0$ on $\abs{z} \leq 1$ by continuity, so if $\abs{z} > 1$ , then $1/z \in \D$ , so $g$ obeys the same bound on all of $\C$ . By Liouville's theorem, $g$ is constant, and hence so is $f$ .