# Relation between runge domain and polynomial convexity

Are these concepts the same? Just to state the definitions

Definition 1 A domain $\Omega \in \mathbb{C}^n$ is a Runge domain if every function $f \in H(\Omega)$ can be approximated, uniformly on compact subsets of $\Omega$, by polynomials in $\mathbb{C}^n$.

Here $H$ stands for the space of holomorphic functions. While the definition of polynomial convexity is stated as

Definiton 2 Let $\mathcal{P}$ denote the set of holomorphic polynomials on $\mathbb{C}^n$. Let $K$ be a compact set in $\mathbb{C}^n$ and let $\|P\|_K = \sup\limits_{z\in K}|P(z)|$ be the sup-norm of $P \in \mathcal{P}$ on $K$. The set $$\hat{K} = \{ z \in \mathbb{C}^n : |P(z)| \leq \|P\|_K \ , \ P \in \mathcal{P} \}$$ is called the polynomially convex hull of $K$. If the compact subsets $K \subset \omega$ have compact polynomial hulls $\hat{K}$, then we have a polynomial convex domain.

I have a book that gives an example [Wermer] which shows a domain $\Omega\in\mathbb{C}^2$ which is biholomorphic to a bidisc, but which is not a runge domain.

However another paper says the following:

"On the other hand , biholomorphic images of polydiscs can fail to be polynomically convex, see [Wermer]"

Just for completeness sake, here is the theorem both papers refers to

Theorem 1 [Wermer]> There is a bounded domain in $\mathbb{C}^2$ which analytically (holomorphic) equivalent to the bidisk, but which is not a runge domain

So one paper refers says that the domain is not Runge, while the other says that it fails to be polynomically convex.

Hence my question is as follows: Is being a Runge domain equivalent to being polynomically convex?

• Your definition of polynomially convex is not correct: the strict inequality sign must be replaced by $\leq$ and you should require that $\hat K$ be compact but not that $\hat K=K$. – Georges Elencwajg May 22 '15 at 22:49
• @GeorgesElencwajg I agree with the strict equality sign, but what do you mean by your second comment? I took the definition straight from here encyclopediaofmath.org/index.php/Polynomial_convexity. – N3buchadnezzar May 22 '15 at 23:06
• I mean exactly what I wrote. The point is that neither you nor the linked article define what a polynomially convex domain is but only what a polynomially convex compact set is. For polynomially convex domains you should require that compact subsets $K\subset \Omega$ have compact polynomial hulls $\hat K$, but not that $K=\hat K$. By the way, with your definition no nonempty domain in the universe is polynomially convex: I challenge you to find just one (even in $\mathbb C$) if you cling to your false definition !! – Georges Elencwajg May 22 '15 at 23:59
• I agree with you, I am very new to this subject so sorry for coming of as arrogant. – N3buchadnezzar May 23 '15 at 0:01
• No, no you are not arrogant at all . You have adopted the right atitude: defend your position and if you see that something is indeed wrong, say so. By the way there are excellent books on the subject, written by real masters: Grauert-Fritzsche, L.Kaup-B.Kaup, Shabat . You would certainly profit from them. – Georges Elencwajg May 23 '15 at 0:12

## 1 Answer

The concepts are not equivalent:

The open subset $\Omega=\mathbb C^2\setminus\{0\}\subset \mathbb C^2$ is Runge but not polynomially convex: indeed for $K$ the unit sphere $||Z||=1$ centered at the origin $\hat K$ is the non compact set $0\lt||Z||\leq 1$.