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 $0$.

  • $\begingroup$ Why do we have the other inclusion, $\overline{A(\Omega)} \subseteq A(K)$, though? Is $A(\Omega)$ a Banach space? $\endgroup$ – rosecabbagedragon Jan 1 at 13:25

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