If a uniformly equicontinuous family of functions is analytic on an open disk in the complex plane, it has compact closure by Montel's theorem (and Arzela-Ascoli). Is it possible that this set is compact by showing it to be complete and totally bounded? Do uniformly convergent sequences of uniformly equicontinuous functions in the complex plane (on an open disk) converge to a uniformly continuous, analytic function?


I think I can answer my own question.

Let $\mathcal F$ be a uniformly equiconinuous set of functions, analytic on some $G\subseteq \mathbb C$. By Montel's theorem, we have that $\overline{\mathcal F}$ is compact. We show that $\mathcal F=\overline{\mathcal F}$ by demonstrating that $\mathcal F$ contains its limit points. Let $\{f_n\}\subseteq\mathcal F$ be a sequence of functions such that $f_n\rightarrow f$ pointwise, and $f_n\neq f$ for all $n$. Consider any closed loop $D\in G$. Since $D$ is compact, the convergence on $D$ is uniform. Therefore $f$ is continuous. Furthermore, we may apply the Cauchy-Goursat Theorem,

$$\int_D f=\lim_{n\to\infty}\int_D f_n=0$$

Since this is true for every closed path $D$, $f$ is analytic on $G$ by Morera's Theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.