# holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set.

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.

• Surely, this isn't what it is meant by the author: $f_n(z)=z^n$ converges to $0$ uniformly on compact sets of the open unit disc, but surely it doesn't converge uniformly on the whole disc. Probably, the idea is that if you have the theorem for every $U\Subset\Omega$, then you have it for $\Omega$ as well. – wisefool Aug 17 '13 at 15:58
• @wisefool How does this imply the theorem? – Alex Aug 17 '13 at 17:41

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:

Theorem 1. If statement $A$ holds on every compact subset of domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.

Theorem 2. If statement $A$ holds on domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.

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.

• Oh, I see. Yeah, thank you very much, this is very helpful. – Alex Aug 17 '13 at 23:36

Let us assume the following result: if $\{f_n\}$ is a sequence of functions on some open set $U$ which converge uniformly on $U$ to $f$ (i.e. $\sup_{U}|f_n-f|\to 0$ for $n\to\infty$), then the sequence $\{f'_n\}$ converges as well uniformly on the compact sets of $U$ to a function $g$ and we have $f'=g$. This is the theorem proved in the book by Stein, if I understand well.

Now, let $\{f_n:\Omega\to\mathbb{C}\}_n$ be sequence of holomorphic functions converging uniformly on compact subsets of $\Omega$ to $f$. For any compact set $K\subset\Omega$, we can find an open set $U$ such that $K\subset U\Subset \Omega$. As $U\Subset\Omega$, $f_n\vert_{U}\to f\vert_{U}$ uniformly: $F=\overline{U}$ is compact, hence $$\sup_{U}|f_n\vert_U-f\vert_U|=\sup_{U}|f_n-f|=\sup_F|f_n-f|\to 0$$ when $n\to \infty$. Therefore, by our assumption $(f_n\vert_U)'$ converges uniformly on compact sets of $U$ to $g$, such that $g=f'$.

Obviousy, $(f_n\vert_U)'=f_n'\vert_U$ and then $$\sup_{K}|f'_n-f'|=\sup_K|f'_n\vert_U-f'\vert_U|\to 0$$ when $n\to\infty$. And this is the thesis.

• What Stein proved was if $f_n$ converges uniformly to $f$ on $\Omega$, then for any compact set $K\subset \Omega$ , $f'_n$ converges uniformly to $f'$on $K$. – Alex Aug 17 '13 at 21:38
• And in your answer, are you reproving what Stein did or just showing that how we got the WLOG condition from the condition of the theorem? – Alex Aug 17 '13 at 21:40
• Stein proved the theorem with the hypotesis The functions converge uniformly on the open set. Therefore he proves that, whenever you have a sequence converging uniformly on an open set, the derivatives converge to the derivative of the limit. Given this result, I show that the general theorem (where the convergence is on the compacts of the open set and not on the whole open set) follows (rather easily). – wisefool Aug 17 '13 at 21:44
• I am afraid that Stein proved that the derivatives converge to the derivative of the limit on any compact subset of the open set, not on the whole set in Thm 5.3. – Alex Aug 17 '13 at 22:59
• Fortunately for me, someone else explained it... let me just note that if Stein had proved (meaning that, he had given proof) of the fact you say in your first comment to this answer, you would have asked a very silly question.. – wisefool Aug 18 '13 at 0:08