In Stein's complex analysis, there is one theorem 5.3 of Chapter 2 which says that if holomorphic functions $f_n$ converges uniformly to $f$ on any compact subset of an open set $\Omega$, then so does $f'_n$.
In his proof, he said "without loss of generality we can assume $f_n$ converges uniformly to $f$ on all of $\Omega$". So I think what he meant was that from the conditions of the theorem, we can get $f_n$ converges uniformly to $f$ on all of $\Omega$.
However, I don't know how to prove this assumption. Any help would be welcome.
Best Answer
Here's a useful topological fact.
Fact. If $K$ is a compact subset of a domain $\Omega$, then there exists a domain $U$ such that $K\subset U$ and $\overline{U}$ is a compact subset of $\Omega$.
Proof: Fix a point $z_0\in \Omega$. For $n=1,2,\dots$ let $G_n$ be the connected component of $z_0$ in the set $$\{z\in\Omega: |z|<n, \ \operatorname{dist}(z,\partial \Omega)>1/n\}$$ Using the connectedness of $\Omega$, one can show that $\bigcup_n G_n=\Omega$. Therefore, the sets $G_n$ form an open cover of $K$. There is a finite subcover; since the sets $G_n$ are nested, this means $K\subset G_n$ for some $n$. This $G_n$ is the desired $U$. $\quad\Box$
Back to the issue. It's not really about holomorphic functions and their derivatives. We are to prove the following:
But instead we prove
Indeed, suppose Theorem 2 is proved. Given $\Omega$ as in Theorem 1 and its compact set $K$, take $U$ from the topological fact. Since $\overline{U}$ is a compact subset of $\Omega$, property $A$ holds on $U$. Theorem 2 says that property $B$ holds on $K$, which was to be proved.