This is a complex analysis qualifying examination question. I'm an incoming grad student and we get the opportunity to have a freebie attempt on one qualifying exam. I have $\textit{some}$ experience in complex analysis from undergrad, and some of the advanced topics that were not covered have been tricky to learn on my own. So here goes.
Let $\mathcal{F}$ be a family of analytic functions on $\mathbb{D}$ (the unit disk). Suppose that for all $0<r<1$, \begin{equation}M_r:=\sup_{f\in \mathcal{F}}\int_{\lvert z\rvert=r}\lvert f(z)\rvert \lvert dz\rvert<\infty.\end{equation}
Prove that $\mathcal{F}$ is a normal family.
From my understanding, this means that I must show that for any sequence $f_n$ in $\mathcal{F}$, there exists an analytic function $f:\mathbb{D}\rightarrow \mathbb{C}$ and a subsequence $(n_k)$ such that $f_{n_k}\to f$ uniformly on any compact $K\subset \mathbb{D}$.
Here is my proof attempt. Let $f_n$ be a given sequence in $\mathcal{F}$ and $K$ be a compact subset of $\mathbb{D}$. My goal is to use the Arzela-Ascoli Theorem on $\mathcal{F}$ to prove the existence of the required subsequence and limit function $f$. We must show that $\mathcal{F}$ is uniformly bounded and equicontinuous.
I will only show equicontinuity since uniform boundedness is the same technique. Let $f\in \mathcal{F}$ and $z_1,z_2\in K$ be given. Let also $\varepsilon>0$ be given. Since $K$ is compact, there exists an $r<1$ such that $K\subset \{\lvert z\rvert <r\}$. Since both $K$ and $\{\lvert z\rvert =r\}$ are compact and disjoint, $d:=\text{dist}(K,\{\lvert z\rvert =r\})>0$. Pick $\delta=\varepsilon\frac{2\pi d^2}{M_r}$ and suppose $\lvert z_1-z_2\rvert <\delta$. Then, by Cauchy's Integral Formula, we have
\begin{equation}
\begin{split}
\lvert f(z_1)-f(z_2)\rvert =& \frac{1}{2\pi}\lvert \int_{\lvert z\rvert =r} f(z)\left(\frac{1}{z-z_1}-\frac{1}{z-z_2}\right)dz\rvert\\
\leq & \frac{1}{2\pi}\int_{\lvert z\rvert =r}\lvert f(z)\frac{z_1-z_2}{(z-z_1)(z-z_2)}\rvert \lvert dz\rvert\\
\leq &\frac{\lvert z_1-z_2\rvert}{2\pi d^2}\int_{\lvert z\rvert =r}\lvert f(z)\rvert \lvert dz\rvert \\
\leq &\frac{M_r}{2\pi d^2}\lvert z_1-z_2\rvert \\
<& \varepsilon.
\end{split}
\end{equation}
Then, since $\mathcal{F}$ is equicontinuous and uniformly bounded, by Arzela-Ascoli's theorem, we should be done. However, it only tells me that there is a continuous limit function $f:K\rightarrow \mathbb{C}$ and a subsequence $f_{n_k}\rightarrow f$ uniformly on $K$. How can I show that this limit function can be extended to all of $\mathbb{D}$, that it is independent of $K$ and that it is also analytic? I think I'm not seeing whats going on with this whole normal family buisness. Any help is appreciated.
Best Answer
Why not use Montel's theorem? This is usually taught when the course gets to normal families, so you are not nuking the mosquito here.
This theorem states that a family is normal if and only if it is locally uniformly bounded. So, let's show that $\mathcal{F}$ is locally uniformly bounded. Let $K\subset\mathbb{D}$ be a compact subset of the disk. Then we can find $r\in(0,1)$ and such that $K\subset D(0,r)\subset D(0,r+\varepsilon)\subset\mathbb{D}$ for any sufficiently small $\varepsilon>0$. Since
$$M_r=\sup_{f\in\mathcal{F}}\bigg(r\cdot\int_0^{2\pi}|f(re^{i\theta})|d\theta\bigg)<\infty $$
we have that if $f\in\mathcal{F}$ and $z\in K$ it is by Cauchy's formula $$f(z)=\int_{|z|=r+\varepsilon}\frac{f(\zeta)}{\zeta-z}d\zeta,$$ so applying the triangular inequality we get $$|f(z)|\leq (r+\varepsilon)\int_0^{2\pi}\frac{|f((r+\varepsilon)e^{i\theta})|}{|(r+\varepsilon)e^{i\theta}-z|}d\theta\leq\frac{1}{\varepsilon}M_{r+\varepsilon}$$ and this is true for any $\varepsilon>0$ such that $r+\varepsilon<1$. Note that the bound $\frac{M_{r+\varepsilon}}{\varepsilon}$ does not depend on $f$ or $z$, so $\mathcal{F}$ is uniformly bounded on compact sets, which is exactly what we wanted to show.