I am trying to prove a certain family of functions $\mathcal{F}$ is normal, and my proof got very stuck. I am trying to show that the family of analytic functions on $\mathbb{D}$, continuous on the boundary, satisfying $\int_{\partial\mathbb{D}}|f(z)|dz \leq M$ for a fixed $M > 0$ is normal. Using the mean value property one can show that $|f(0)| \leq \frac{M}{2\pi}$, so I have been trying to bootstrap this into local boundedness.
If we let $B_n$ be the closed ball about $0$ with radius $1 - \frac{1}{n}$ then these cover $\mathbb{D}$ so I am trying to show that the family $\mathcal{F}$ is bounded on an arbitrary $B_n$. If we let the boundary of $B_n$ be $C_n$, then for $z \in B_n$ by Cauchy's Integral Formula we have
$|f(z)| \leq \frac{1}{2\pi}\int_{C_{n+1}}|\frac{f(\zeta)}{\zeta - z}|dz \leq \frac{1}{2\pi}\int_{C_{n+1}}|\frac{f(\zeta)}{\zeta - z}|dz \leq \frac{1}{2\pi}\int_{C_{n+1}}|\frac{f(\zeta)}{\frac{1}{n}-\frac{1}{n+1}}|dz$,
so if $m = \frac{1}{n}-\frac{1}{n+1}$, then $|f(z)| \leq \frac{m}{2\pi}\int_{C_{n+1}}|f(z)|dz$. I don't know where to go from here. Ideally, it would be the case that $\int_{C_{n+1}}|f(z)|dz$ is eventually bounded (as it approaches $\partial\mathbb{D}$ in compact $\bar{\mathbb{D}}$) but I can't formalize this argument; I think this is the final step of the proof unless I am in a totally wrong direction.
Could somebody give me a hint on how to proceed? I would be really appreciative.
Thanks a lot!
Let $\mathcal F$ be the family. It suffices to show that, for every compact $K\subset\mathbb D$, there exists a $C_K>0$, such that $$ \lvert f(z)\rvert \le C_K\quad\text{for all}\,\,f\in\mathcal F\,\,\text{and}\,\, z\in K. $$ If $K\subset \mathbb D$ is compact, then there exists an $r\in (0,1)$, such that $$ K\subset\mathbb D_r=\{z\in\mathbb C: \lvert z\rvert<1\}. $$ For example $r=1-\frac{1}{2}\mathrm{dist}(K,\partial\mathbb D)$.
Next, observe that, if $f\in\mathcal F$ and $z\in\mathbb D_r$, then for every $z\in K$, $$ f(z)=\frac{1}{2\pi i}\int_{\lvert\zeta\rvert=1}\frac{f(\zeta)\,d\zeta}{\zeta-z} $$ and hence $$ \lvert\, f(z)\rvert\le \frac{1}{2\pi} \left| \int_{\lvert\zeta\rvert=1}\frac{f(\zeta)\,d\zeta}{\zeta-z}\right|\le \lvert f(z)\rvert\le \frac{1}{2\pi(1-r)}\int_{\partial\mathbb D}\lvert \,f\rvert=\frac{M}{2\pi(1-r)}. $$