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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.

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    $\begingroup$ this is Montel's theorem so you need just to prove local uniform boundness; if you want to prove from scratch as above, take an exhausting countable sequence of compacts ($K_n=\bar D(0, 1-1/n)$ will do) and use the diagonal method to finalize $\endgroup$
    – Conrad
    Commented Jun 30, 2020 at 22:50
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    $\begingroup$ @Conrad This is precisely the technique used in the proof of Montel's theorem, right? $\endgroup$ Commented Jun 30, 2020 at 22:56
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    $\begingroup$ @Just yes precisely as the OP seems to want to prove the thing from scratch - Montel obviously does it immediately as noted in your answer $\endgroup$
    – Conrad
    Commented Jun 30, 2020 at 22:57
  • $\begingroup$ Thanks! Really appreciate the responses. $\endgroup$
    – MSA2016
    Commented Jun 30, 2020 at 23:10

2 Answers 2

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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.

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The missing detail in this approach is to diagonalize over a compact exhaustion. This is an idea that's sometimes paired with Arzela-Ascoli arguments on qualifying exam problems, and worth knowing in qual prep. (Or, as in the other answer, one can use Montel's theorem, which is faster). What has already been shown is that for any sequence $(f_n)$ in $ \mathcal{F}$ and compact set $K \subset \mathbb{D}$, there is a subsequence $(f_{n_k})$ that converges uniformly on $K$ to a continuous limit $f$.

For all $k$, let $K_k = \overline{B_{1/k}(0)}$, $U_k = B_{1/k}(0)$. Define a sequence of subsequences $(f_{k, n})$ as follows ($f_{k, n}$ is the $n$-th term in the $k$-th subsequence). First, put $(f_{0, n}) = (f_n)$. Once $(f_{k, n})$ is defined, let $(f_{k+1, n})$ be a subsequence of $(f_{k, n})$ that converges uniformly on $K_{k+1}$, as we have shown must exist. Define $g_n = f_{n,n}$, so $(g_n)$ is a subsequence of $(f_n)$. Note that $g_n$ has a well-defined continuous limit $g$.

Now let $K$ be any compact subset of $\mathbb{D}$. $K$ is contained in $K_k$ for some $k$ so $g_n$ is eventually a subsequence of $f_{n, k}$, hence $g_n$ converges uniformly to $g$ on $K_k$, hence $K$. So $g_n$ is a sequence of holomorphic functions converging uniformly on compact subsets of $\mathbb{D}$ to $g$. By elementary complex analysis, $g$ is holomorphic and the result is shown.

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  • $\begingroup$ Awesome! Thanks so much. $\endgroup$
    – MSA2016
    Commented Jun 30, 2020 at 23:13

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