Let $f_n$ be a sequence of holomorphic functions on an open, connected set $D \in \mathbb{C}$ with $|f_n(z)| \leq 1$ for all $z \in D$ and all $n \geq 1$. Let $A \in D$ be the set of all $z \in D$ for which $lim_{n \to \infty} f_n(z)$ exists. Show that if $A$ has an accumulation point, then there is a holomorphic function $f$ on $D$ with $f_n$ converging uniformly on compact subsets of $D$.
Thoughts so far: It's given that the sequence is locally bounded, hence normal by Montel's Theorem. Thus there is a subsequence of $f_n$ that converges to a holomorphic function on compact subsets of $D$. Now, we know that at each point, each subsequence has a further convergent subsequence, and hence the original sequence converges point-wise.
Could someone give me a hint as to how we can go from the fact that on every sequence in $f_n$ has subsequence to converging to a holomorphic function on compact subsets to showing that the original sequence converges on compact subsets to a holomorphic function? Also, it doesn't seem that we need the fact that $A$ has an accumulation point, is this so? Was this assumption unneeded?
Context: I'm studying for a qual, so just a hint would be most helpful for now. Thank you.
Best Answer
Try this: without the accumulation point assumption, it is possible that subsequences of $f_n$ can converge to different limit functions. However, if these limit functions are holomorphic and agree on a set with an accumulation point, then what do you know about them?