# Runge Approximation Theorem in Hormanders text

NOTE : This excerpt comes form Hormander's text, An Introduction to COmplex Analysis of Several Variables.

STATEMENT: (Runge approximation theorem)Let $\Omega$ be an open set in $\mathbb{C}$ and $K$ a compact subset of $\Omega$. The following conditions on $\Omega$ and $K$ are equivalent:

(a) Every function which is analytic in a neighborhood of $K$ can be approximated uniformly on $K$ by functions in $A(\Omega)$.

(b)The open set $\Omega\backslash K=\Omega \cap K^c$ has no components which is relatively compact in $\Omega$.

(c)For every $z\in\Omega\backslash K$ there is a function $f\in A(\Omega)$ such that $$|f(z)|>\sup_k |f|$$

Proof:To prove that $(b)\rightarrow (a)$ it suffices to show that every measure which is orthogonal to $A(\Omega)$ is also orthogonal to every function which is analytic in a neighborhood of $K$, for the theorem is then a consequence of the Hahn-Banach theorem. Set $$\varphi(\zeta)=\int(z-\zeta)^{-1}d\mu(z),\;\;\;\;\; z\in K^c$$

By theorem 1.2.2, $\varphi$ is analytic in $K^c$, and when $\zeta\in \Omega^c$ we have $$\varphi^{(k)}(\zeta)=k!\int(z-\zeta)^{-k-1}d\mu(z)=0\;\;\;\;\;\;\text{for every}\;k$$

QUESTION: So I have two questions regarding Hormander's proof. The first one is how does the problem reduce to just using Hahn-Banach theorem. Secondly, why does $\varphi^{(k)}$ vanish for all $k$ when $\zeta\in\Omega^c$.

We are given that $$A(\Omega) \subseteq A(K)$$, and what we want to show is that $$\overline{A(\Omega)} = A(K)$$. One of the versions of the Hahn-Banach theorem states that if $$\overline{A(\Omega)} \subsetneq A(K)$$, then there's a bounded linear functional, i.e. a measure, that vanishes on $$A(\Omega)$$, but not on $$A(K)$$ (See this Wikipedia link). That shows how the result follows from Hahn-Banach theorem.
And as for why $$\int (z - \zeta)^{-1} d\mu(z)$$ is $$0$$, it's because we picked a measure $$\mu$$ which is orthogonal to $$A(\Omega)$$, which is why $$\int (z - \zeta)^{-1} d\mu(z)$$ is $$0$$.
• Why do we have the other inclusion, $\overline{A(\Omega)} \subseteq A(K)$, though? Is $A(\Omega)$ a Banach space? – rosecabbagedragon Jan 1 at 13:25