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Spring 2015 - Problem 7

normal families, Urysohn subsequence principle

Let $\D = \set{z \in \C \mid \abs{z} < 1}$ and $\C^+ = \set{z \in \C \mid \Im{z} > 0}$ . Suppose $\func{f_n}{\D}{\C^+}$ is a sequence of holomorphic functions and $f_n\p{0} \to 0$ as $n \to\infty$ . Show that $f_n\p{z} \to )$ uniformly on compact subsets of $\D$ .

Solution.

First, let $\func{\phi}{C^+}{\D}$ be a conformal mapping, e.g., the Cayley transform

\phi\p{z} = \frac{z - i}{z + i}.

Then $\set{\phi \circ f_n}_n$ is a uniformly bounded sequence of holomorphic functions, so they form a normal family. Thus, there exists a holomorphic $\func{g}{\D}{\D}$ and a subsequence such that $\phi \circ f_{n_k} \to g$ locally uniformly on $\D$ . By assumption, we get

g\p{0} = \lim_{k\to\infty} \p{\phi \circ f_{n_k}}\p{0} = \phi\p{0} = -1.

In other words, $g$ attains its maximum in the interior of $\D$ , so by the maximum principle, $g = -1$ . Thus, it follows that $f_{n_k} \to \phi^{-1}\p{-1} = 0$ locally uniformly on $\D$ . Indeed, on compact sets, $\phi$ is bounded, hence Lipschitz by the Cauchy integral formula, so it preserves locally uniform convergence.

Now suppose $\set{g_n}_n$ is a subsequence of $\set{f_n}_n$ . By the same argument but replacing $f_n$ with $g_n$ above, we see that $\set{g_n}_n$ admits a further subsequence which converges locally uniformly to $0$ as well. Hence, by the Urysohn subsequence principle, it follows that $\set{f_n}_n$ converges locally uniformly to $0$ on $\D$ , which completes the proof.