Holomorphically Convex Hull a Subset of the convex hull of - MathOverflow

Holomorphically Convex Hull a Subset of the convex hull of

2011-03-29T04:48:52Z

This comes from Hörmander's "An Introduction to Complex Analysis in Several Variables".

We defined the $A(\Omega)$-hull (analytic functions in an open set $\Omega$). $\hat{K}$ of a compact set $K\subset\Omega$ by $\hat{K}=\{z;z\in\Omega, |f(z)|\leq\sup_K |f| \operatorname{for every } f\in A(\Omega) \}$.

The book says, if we consider $f(z)=e^{az}$ for every complex number $a$, we obtain $\hat{K}\subseteq \operatorname{convex hull of }K$.

I do not get how he concluded this result. I do not know how to turn a $\hat{z}\in\hat{K}$ into a linear combination of elements $z\in K$ using the exp function. Are there specific $a$ I need to choose? Can I construct this?

Also, he says "Furthermore, it is clear that $\overset{*}{K}=\hat{K}$ ". I'm assuming that the $K$ with the weird star mark on top represents the convex hull? Even then, I do not understand the reverse inclusion.

Answer by Olivier Bégassat for Holomorphically Convex Hull a Subset of the convex hull of

2011-03-29T05:18:19Z

The exponential function grows in module as the exponential of the real part. Therefore, the set of all $z$ such that $|exp(az)|\leq \sup_K |exp(a\times\cdot)|$ is a half space containing $K$, and meeting $K$. You get all such half spaces, if you vary $a$ in $\mathbb{C}$ or even on the unit circle. Their intersection is the convex hull of $K$ by some famous theorem on convex sets (Krein-Milman?).

So by restricting yourself to the exponential functions you get the convex hull of $K$. The hull you're interested in is a subset of that set.

I don't know what $K^*$ stands for, but it won't be the convex hull in general. for instance, if you take $\Omega=\mathbb{C}\setminus\lbrace 0\rbrace$ and $K=$ the unit circle, and $f(z)=z, ~g(z)=\frac{1}{z}$, you see that the hull you're interested in is just $K$ itself. The convex hull may not even be a subset of $\Omega$.