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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.

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  • $\begingroup$ I assume you mean $|f_n(z)| \le 1$? $\endgroup$ – Christopher A. Wong Aug 24 '15 at 1:45
  • $\begingroup$ Yes, thank you. $\endgroup$ – user19817 Aug 24 '15 at 3:50
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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?

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  • $\begingroup$ I must be missing something. If we look at the sequence point-wise, say consider $z_0 \in D$, we have that each subsequence of $f_n(z_0)$ has a further subsequence that converges (since $f_n$ is a normal family by Montel's theorem). So doesn't this give us that $f_n(z_0)$ converges? And thus the subsequences of $f_n$ must converge to the same limit functions? $\endgroup$ – user19817 Aug 24 '15 at 3:59
  • $\begingroup$ Your claim about pointwise convergence cannot be true. Consider the sequence of holomorphic functions $z, -z, z, -z, \ldots$. Then the sequence is bounded on the unit disc, but $A = \{0\}$ and of course the sequence does not converge. $\endgroup$ – Christopher A. Wong Aug 24 '15 at 6:55
  • $\begingroup$ I accept your counter-example, so what I said must be false. But what is wrong with my logic? (I'm sure I misunderstand what it means to be a normal family). $\endgroup$ – user19817 Aug 24 '15 at 23:29
  • $\begingroup$ It is false that if every subsequence has a further convergent subsequence, then the original sequence converges. $\endgroup$ – Christopher A. Wong Aug 25 '15 at 10:50

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