I have seen the following lemma in my Complex Analysis class:
Let $D \subset \mathbb{C}$ be open and connected, let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions holomorphic in $D$. Assume:
$(f_n)_{n \in \mathbb{N}}$ is local bounded
and there is a dense set $\mathcal{D} \subset D$ such that the sequence $(f_n(z))_{n \in \mathbb{N}}$ converges $\forall z \in \mathcal{D}$. Then $(f_n)_{n \in \mathbb{N}}$ converges locally uniformly on $D$ to a holomorphic function $f$.
What I am not sure about is the first step of the proof. It says that it suffices to prove the following: let$z_0 \in D$, pick $\delta > 0$ such that $\overline{B(z_0,2\delta)} \subset D$. Then $\forall \varepsilon >0, \exists N>0$ such that $\forall n,m \geq N$ and $\forall Z \in B(z_0,\delta)$, we have $|f_n(z)-f_m(z)| \leq \varepsilon$ $(1)$
I understand the proof of $(1)$, but I am not sure how it implies the lemma. My attempt so far:
Set $B = \overline{B(z_0,\delta)}$, $\mathcal{F} = \{ f_n|_{B} : n \in \mathbb{N}\}$. I have managed to show that $\mathcal{F}$ is bounded and equicontinuous using $(1)$ and uniform boundedness, and so by Arzela-Ascoli it must be compact. This implies that there is a subsequence of $(f_n|_B)_{n \in \mathbb{N}}$ that converges uniformly on $B$ to some $f_{z_0,\delta} : B \to \mathbb{C}$. But I am not where to go from there. I want to end up with a function $f$ defined on the whole of $D$, and such that $(f_n)_{n \in \mathbb{N}}$ converges to it locally uniformly (that is $\forall K \subset D$ compact, $f_n|_K \to f|_K$ uniformly). Thank you for your help!
Best Answer
(1) above implies $f_n$ is uniformly Cauchy sequence in $\overline{B(z_0,\frac{\delta)}{2}}$, so in particular they converge uniformly to a continuous function $f$ there by basic convergence of functions theory, nothing to do with analyticity; Morera then shows $f$ analytic there since uniform convergence means you can integrate on small triangles inside; since $z_0$ is arbitrary you are done at least as small discs around any point go, so you get analytic $f$ and uniform convergence on such discs, and then any compact is a finite union of such so you are done; local boundness is crucial in proving (1) from the hypothesis