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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.

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Suppose $z$ is not in the convex hull of $K$. You want to show it is not in the $A(\Omega)$-hull (I assume from your notation that you are working in one complex variable). Since $K$ is compact there is a line in the plane separating $z$ from $K$; I think this should now suggest how to use Hormander's hint/sketch. (To simplify things further, you could WLOG suppose that $z=0$ by translating the problem in the plane.) – Yemon Choi Mar 29 '11 at 5:09
Thanks. I seemed to only use compactness, that there is an $\epsilon$ disc around $z$ not touching $K$, so I was able to choose an $a$ that led to $1>\sup_K |e^{az}|$. I don't know if I'm missing something because I didn't touch any other properties of the convex hull. – John C Mar 29 '11 at 6:16
@John C:It is not * on top of K it is ^ on top of ^.He is saying that the holomorphic hull of the holomorphic hull of K is the holomorphic hull of K. – Mohan Ramachandran Mar 29 '11 at 18:25
up vote 2 down vote accepted

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 $\Omega$.

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Rather than Krein-Milman, I think 2-dimensional Hahn-Banach would suffice, perhaps? – Yemon Choi Mar 29 '11 at 6:07
Thanks. I do not see $K^*$ defined anywhere even in the list of notation used before the first chapter. I definitely got some intuitive sense, but I was hoping to find a simple elementary answer. – John C Mar 29 '11 at 6:11
Yes, Krein-Milman's theorem is not really what I was looking for, I just wanted some result that says that the convex compact set is the intersection of its 'supporting' half planes. – Olivier Bégassat Mar 29 '11 at 6:16
@John C You really only need to understand the sets $\lbrace z\in\mathbb{C}~\mathrm{s.t.}~|\exp (az)|\leq M \rbrace$ for given $a\in\mathbb{C}\setminus\lbrace 0\rbrace$ and $M>0$. Try to figure them out, they are closed half planes of which you can find the outwards normal vector (it should be $\overline{a}$) Then you should convince yourself intuitively that the convex hull of $K$ is equal to the intersection of all half planes containing $K$ and 'meeting its border'. The half planes I describe in my answer are precisely those. – Olivier Bégassat Mar 29 '11 at 6:24
@Olivier: about the paragraph "I don't know what K^* stands for [...]" and the example with $1/z$: the space $A(\Omega)$ that the OP says is just analytic functions, if I remember correctly, usually denotes the analytic functions continuous up to the boundary, i.e. the intersection $\mathcal{O}(\Omega)\cap\mathcal{C}^0(\bar{\Omega})$. – Qfwfq Mar 29 '11 at 17:09

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