Let $\mathcal{F}$ be a family of complex functions analytic on the open unit disc $\mathbb{D}$ and continuous on the closed unit disc $\overline{\mathbb{D}}$ such that
$$
\int_0^{2\pi}|f(e^{i\theta})|d\theta \leq M < \infty.
$$
Show that $\mathcal{F}$ is a normal family.
It suffices to show that $\mathcal{F}$ is locally uniformly bounded (for all $w \in \mathbb{D}$ there exists $\delta > 0$ and $M < \infty$ such that $|z – w| < \delta \implies |f(z)| < M$ for all $f \in \mathcal{F}$). We believe that we should apply Cauchy's integral formula much like how this answer suggests, but I do not believe that we can apply Cauchy's integral formula to $\partial\mathbb{D}$ since it is not contained in $\mathbb{D}$, the region in which all the functions are analytic. Any assistance would be appreciated.
Best Answer
Fix $f\in \mathcal{F}$ and let $g(s):=\frac{1}{2\pi} \int_0^{2\pi} |f(se^{it})|dt$.
Since $f$ is uniformly continuous on the unit disk, for every $\varepsilon>0$ there exists $\delta>0$ such that if $1-\delta<s<1$, then $|f(z)-f(sz)|<\varepsilon$ for all $x\in \partial \mathbb D$. The reverse triangle inequality then gives $g(s)\to g(1)$ as $s\to 1^-$.
Fix two numbers $0<r<s<1$. By Cauchy's integral formula applied on the disk of radius $s$ we have, for any $|z|<r$, $$ |f(z)|\leq \dfrac{1}{2\pi} \int_0^{2\pi} \dfrac{|f(se^{it})|}{|z-se^{it}|}\, dt \leq \dfrac{g(s)}{(s-r)}. $$ Keeping $z,r$ fixed, and letting $s\to 1^-$ we arrive at $$ |f(z)|\leq \dfrac{M}{2\pi(1-r)}, \qquad |z|<r. $$