# Uniform convergence of the derivative function sequence on compact subsets

Let me make it short here: I believe the following proposition is true, as it is given a "proof" on page 54-55 in Elias Stein and Rami Shakarchi's Complex Analysis (Princeton University Press), which to me, however, is not convincing.

Anyway, the proposition in question says that

Given $\Omega\subset \Bbb C$ an arbitrary open subset, if $\{f_n\}$, a sequence of holomorphic functions on $\Omega$, converges uniformly to some function $f$ on all compact subsets of $\Omega$, then $f$ is also holomorphic on $\Omega$ and $\{f'_n\}$ converges uniformly to $f'$ on all compact subsets of $\Omega$.

I already understand how the first part of the assertion is proved, but I am uncertain about the second part (namely, the part in italics). Can anyone refer me to a proof or disproof? Beware, though, that there is no restriction put on the open set $\Omega$! Thanks in advance.

For those who are interested, I have doubts about Stein's proof because it wants to prove by showing that $f_n-f$ converges uniformly to $0$ on any $$\Omega_{\delta}:=\{z\in\Omega\mid d(z,\partial \Omega)<\delta\}.$$ This I think is only true when we can cover each $\Omega_\delta$ with a compact subset of $\Omega$, which seems hopeless when $\Omega$ is, say, an unbounded region.

• It is the other way around: You can cover each compact set $K \subset \Omega$ by some $\Omega_\delta$. – Martin R Apr 12 '16 at 14:28
• @MartinR This I understand. But now, following Steins' thread, what we need to show is that $f_n-f$ converges uniformly to $0$ on $\Omega_\delta$, while what we have proved is that this holds on all compact subsets of $\Omega$. – Vim Apr 12 '16 at 14:38

The method of making the $\Omega_\delta$ work (which does not appear explicitly in this proof) is that we only need the result on compacta. Any compact subset of $U$ is covered by suitable $\Omega_\delta$, so no compact subset of $U$ can be a witness to failure of the theorem.