Complex Analysis Prelim Question

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.

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.