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I have been thinking about the following exercise from some complex analysis lecture notes I found online for a while but I can't seem to be able to conclude the desired result. The exercise says that

If $A$ is an open subset of the complex plane, and if $g: A \rightarrow \mathbb{C}$ is analytic, then there is a sequence of rational functions $(f_n)$ such that no pole of the $f_n$'s lies in $A$ (that is, that the poles are in $\overline{\mathbb{C}} \setminus A$) and such that $f_n \rightarrow g$ uniformly on compact subsets of $A$.

I know that I have to apply Runge's theorem here, but the problem is that Runge's theorem tells me that the sequence of functions $f_n$ converges uniformly to $g$ on a compact set $K \subseteq A$ and not on the whole $A$.

I thought that maybe I can approximate the open set $A$ by compact sets $K_n$ and take elements from a sequence $f_{n,k}$ that converges uniformly on each of the compacts, and then any given compact $K\subset A$ would have to be contained in one of the compacts $K_n$ approximating $A$ but I'm not sure if this is possible of if it this works.

I would appreciate any help with this exercise. Thanks.

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2 Answers

See Theorem 13.9 in Rudin's real and complex analysis, 3rd edition.

Your idea is the good one : what you need is the notion of compact exhaustion of $A$, that is a sequence of compact sets $K_n$ such that $A=\cup K_n$, each $K_n$ lies in the interior of $K_{n+1}$, every compact subset of $A$ lies in some $K_n$, and every component of $\overline{\mathbb{C}} \setminus K_n$ contains a component of $\overline{\mathbb{C}} \setminus A$.

The construction of such $K_n$'s can also be found in Rudin, Thm 13.3.

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Wow that's really helpful. Thank you. –  Student Apr 26 '12 at 17:04
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Let K be a compact subset of A and let $\epsilon>0$ we need to show that there exist f(a rational function) such that $|g-f|<\epsilon$ for all $z\in K$, as Malik says in the refernce by that there is a compact set $K_1$ such that $K\subseteq K_1\subseteq A$ and each component of $\bar{\mathbb{C}}\setminus K_1$ contains a component of $\bar{\mathbb{C}}\setminus A$. Hence $E\subseteq\bar{\mathbb{C}}\setminus K_1$ meets each component of $\bar{\mathbb{C}}\setminus K_1$, Now apply Runge's Theorem and conclude. you can conclude "Let A be an open subset of the plane and let E be a subset of $\bar{\mathbb{C}}\setminus A$ such that E meets every component of $\bar{\mathbb{C}}\setminus A$. Let $R(A,E)$ be the set of all rational functions with poles in E and consider $R(A,E)$ as a subspace of $H(A)$. If $g\in H(A)$ then there is a sequence ${f_n}\in R(A,E)$ such that $g=limf_n$", $H(A)$ is set of all analytic functions on A

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But it has to be a sequence that works for every possible compact set $K \subset A$ so I think that we basically need to do this to get a sequence for each $K_n$ coming from the compact exhaustion, and then that will produce a sequence $f_{n, k}$. Then maybe we can take the diagonal to get a sequence that works for all compacts at once? –  Student Apr 26 '12 at 17:19
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