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Spring 2020 - Problem 7

normal families

Let $\mathcal{F}_M$ be the set of functions holomorphic on $\D$ and continuous on $\cl{\D}$ that satisfy

\int_0^{2\pi} \abs{f\p{e^{it}}} \,\diff{t} \leq M < \infty.

Show that every sequence $\set{f_n}_n$ in $\mathcal{F}_M$ contains a subsequence that converges uniformly on compact subsets of $\D$ .

Solution.

It suffices to show that $\mathcal{F}_M$ is a normal family, i.e., uniformly bounded on each compact set. Let $0 < r < R < 1$ and $f \in \mathcal{F}_M$ . Observe that $f_R\p{z} = f\p{Rz}$ is holomorphic on a neighborhood of $\cl{\D}$ , so by the Cauchy integral formula, we have for any $z \in \cl{B\p{0, r}}$ that

\begin{gathered} f_R\p{z} = \frac{1}{2\pi i} \int_{\partial \D} \frac{f_R\p{\zeta}}{\zeta - z} \,\diff\zeta \\ \implies \abs{f_R\p{z}} \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{\abs{f_R\p{e^{i\theta}}}}{1 - r} \,\diff\theta. \end{gathered}

Notice that $f_R \to f$ everywhere on $\partial\D$ by continuity, so by uniform continuity on $\cl{\D}$ , we may simply send $R \to 1$ to get

\abs{f\p{z}} \leq \frac{1}{2\pi\p{1 - r}} \int_0^{2\pi} \abs{f\p{e^{i\theta}}} \,\diff\theta \leq \frac{M}{2\pi\p{1 - r}}.

Since $f$ was arbitrary and the constant depends only on $r$ , we see that $\mathcal{F}_M$ is a normal family, so any sequence has a subsequence which converges locally uniformly on $\D$ .