<Home>Qualifying Exams><Analysis>Fall 2012 - Problem 10

Fall 2012 - Problem 10

normal families, Urysohn subsequence principle

Let $\set{f_n}_{n\in\N}$ be a sequence of holomorphic functions on $\D$ satisfying $\abs{f_n\p{z}} \leq 1$ for all $z \in \D$ and $n \in \N$ . Let $A \subseteq \D$ be the set of all $z \in \D$ for which the limit $\lim_{n\to\infty} f_n\p{z}$ exists. Show that if $A$ has an accumulation point in $\D$ , then there exists a holomorphic function $f$ on $\D$ such that $f_n \to f$ locally uniformly on $\D$ as $n \to \infty$ .

Solution.

Let $K \subseteq \D$ be a compact set, and let $\delta = \frac{d\p{K, \D^\comp}}{2} > 0$ . By the Cauchy integral formula, for any $z \in K$ , we have $\cl{B\p{0, \delta}} \subseteq \D$ and so

\abs{f_n'\p{z}} \leq \frac{1}{2\pi} \int_{\partial B\p{0,\delta}} \frac{\abs{f_n\p{\zeta}}}{\abs{\zeta - z}^2} \,\diff\abs{\zeta} \leq \frac{1}{2\pi} \frac{\norm{f_n}_{L^\infty}}{\delta^2} 2\pi \delta = \frac{1}{\delta} \eqqcolon M_K.

Thus, by the fundamental theorem of calculus, given any $z, w \in K$ , we have

\abs{f\p{z} - f\p{w}} \leq \int_z^w \abs{f'\p{\zeta}} \,\diff\abs\zeta \leq M_K\abs{z - w},

i.e., $\set{f_n}_n$ is equicontinuous on $K$ and uniformly bounded on $M_K$ , so we may apply Arzelà-Ascoli on any compact subset of $\D$ . In particular, by taking $K = \cl{B\p{0, 1 - \frac{1}{n}}}$ and using a diagonal argument, we find a subsequence $\set{f_{n_k}}_k$ which converges locally uniformly to some $f$ on all of $\D$ . Since each $f_n$ is holomorphic, it follows that $f$ is itself holomorphic (e.g., by Morera's theorem). We will show that $f_n \to f$ locally uniformly itself by showing that any subsequence has a further subsequence which converges locally uniformly to $f$ .

Let $\set{f_{n_k}}_k$ be subsequence which converges locally uniformly to some $g$ and let $K \subseteq \D$ be compact. Notice that by pointwise convergence on $A$ , we see that

g\p{z} = \lim_{n\to\infty} f_n\p{z} = f\p{z},

i.e., $f$ and $g$ are two holomorphic functions which agree on a set containing a limit point, so $f = g$ on all of $\D$ , i.e., $\set{f_{n_k}}_k$ converges locally uniformly to $f$ .

Now let $\set{f_{n_k}}_k$ be any subsequence. By Arzelà-Ascoli, it admits a further subsequence which converges locally uniformly, and by the previous paragraph, this subsequence converges to $f$ . Thus, every subsequence of $\set{f_n}_n$ admits a further subsequence which converges locally uniformly to $f$ , and so $f_n \to f$ locally uniformly.

Indeed, if this were not the case, then there exist $K \subseteq \D$ compact and $\epsilon > 0$ such that for each $n \geq 1$ , we obtain $z_k \in K$ and $n_k$ so that $\set{f_{n_k}}_k$ is a subsequence and $\abs{f_{n_k}\p{z_k} - f\p{z_k}} \geq \epsilon$ . But clearly any further subsequence cannot converge uniformly to $f$ on $K$ , which is a contradiction. This completes the proof.