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.

  • $\begingroup$ 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

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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.