[Math] Uniform convergence of the derivative function sequence on compact subsets

complex-analysisreference-request

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.

Best Answer

Perhaps the proof of corollary 3.5.2 (p. 89) in Greene and Krantz (Function Theory of One Complex Variable, the link is to the main theorem from which the corollary takes its notation) is convincing?

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.

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