Closed unit ball of $X^*$ has extreme points.

banach-spacesfunctional-analysisweak-topology

Let $X$ be a Banach space. I have to show that the closed unit ball of the dual space $X^*$ has extreme points.

I just used the Banach-Alaoglu and Krein-Milman theorems to prove this but I'm not sure if this is correct because we haven't specified a topology on $X^*$. Because this argument obviously only holds when we consider the weak* topology so if anyone could clarify this that would be greatly appreciated.

Best Answer

The concept of extreme points doesn't depend on which topology you choose. So we could restate Krein-Milman as follows:

Let $V$ be a vector space, and $K \subset V$ a nonempty convex set. Suppose there exists a locally convex topology $\tau$ on $V$ such that $K$ is compact with respect to $\tau$. Then $K$ has extreme points (and indeed $K$ is equal to the $\tau$-closure of the convex hull of its extreme points).

Thus, taking $V = X^*$ and $K$ the closed unit ball, it suffices to come up with some locally convex topology $\tau$ in which $K$ is compact. Thanks to the Banach-Alaoglu theorem, an obvious choice is to take $\tau$ to be the weak-* topology, and then everything works.

So your proof is fine.

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