Analysis Problem of the Day 57

Today’s problem appeared as Problem 8 on the UCLA Spring 2022 Analysis Qual:

Problem 8. Let $V \subseteq U \subseteq \mathbb{C}$ be open, suppose $U$ is connected and $K \subseteq U$ is compact. Show that for every $\epsilon>0$ there exists $\delta>0$ such that for $f \in H(U)$ and $|f| \leq \delta$ on $V, |f| \leq 1$ on $U,$ one has $|f| \leq \epsilon$ on $K.$

Solution: Suppose not. Then, for some $\epsilon>0$ and $\delta_n=\frac{1}{n}>0,$ there exists $f_n$ holomorphic on $U$ such that $|f_n| \leq \frac{1}{n}$ on $V, |f_n| \leq 1$ on $U,$ but $|f_n(x_n)| > \epsilon$ for some $x_n \in K.$ Since $f_n$ is locally uniformly bounded, by Montel’s theorem it has a normally convergent subsequence $f_{n_k} \to f$ on $U.$ It follows that $|f| =0$ on $V,$ so $f=0$ on $U.$ But $\|f_n\|_{L^\infty(K)} \geq \epsilon$ for all $n,$ so $f_{n_k}$ cannot converge uniformly to $f$ on $K,$ which is a contradiction. Thus, the claim holds.

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