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Spring 2014 - Problem 9

Hurwitz' theorem, Rouché's theorem

Prove Hurwitz' Theorem: Let $\Omega \subseteq \C$ be a connected open set and $\func{f_n,f}{\Omega}{\C}$ holomorphic functions. Assume that $f_n\p{z}$ converges uniformly to $f\p{z}$ on compact subsets of $\Omega$ . Prove that if $f_n\p{z} \neq 0$ , $\forall z \in \Omega$ , $\forall n$ , then either $f$ is identically equal to $0$ or $f\p{z} \neq 0$ , $\forall z \in \Omega$ .

Solution.

Suppose $f$ is not identically $0$ , but there exists $z_0 \in \Omega$ such that $f\p{z_0} = 0$ . Thus, $z_0$ is an isolated zero of $f$ , so there exists $R > 0$ such that $z_0$ is the only zero of $f$ in $\cl{B\p{z_0, R}} \subseteq \Omega$ . Hence, $\abs{f\p{z}} \geq \epsilon > 0$ on $\partial B\p{z_0, R}$ .

Since $f_n$ converges uniformly to $f$ on the compact set $\partial B\p{z_0, R}$ , there exists $N \in \N$ such that on $\partial B\p{z_0, R}$ ,

\abs{f_n\p{z} - f\p{z}} < \epsilon \leq \abs{f\p{z}}.

By Rouché's theorem, $f$ and $\p{f_n - f} + f = f_n$ have the same number of zeroes in $B\p{z_0, R}$ . But $f$ has one zero in this ball, and $f_n$ has no zeroes by assumption, a contradiction. Hence, no zero could have existed to begin with.