# Elementary proof that every compact convex subset of a normed vector space has an extreme point

I've shown that the following result is valid for $V=\Bbb R^v$:

If $K\subset V$ is compact and convex and $V$ is a $v$-dimensional Banach space, then $K$ has at least one extreme points.

The proof I know uses the parallelogram law. Can we do something similar for general finite dimensional Banach spaces?

Obs.: The proof of Krein-Milman Theorem (for locally convex topological vector spaces) needs the existence of extreme points in general, and the proof of this uses Zorn's Lemma. I was wondering if there is a proof using something different, maybe similar to the parallelogram law...

## 1 Answer

A point being extreme (or not) has nothing to do with the norm or topology of the space; it's just a concept of linear algebra. Topology enters the picture only when we talk about $K$ being compact. But then again, on a finite-dimensional space all norms induce the same topology, so they choice of a norm does not matter.

Upshot: if $V$ is a finite-dimensional normed vector space and $K$ is its compact subset, consider the Euclidean norm on $V$ instead and take the point of $K$ with maximal norm. That's an extreme point, by the parallelogram law.

• So, even when the finite dimensional space is not isometric to some euclidean, if we equip it with the euclidean norm, the parallelogram law is satisfied? – Filburt Aug 12 '17 at 20:07
• Yes, that is the case. Although they are not isometric, they are isomorphic, so we can have an inner product (= parallelogram law) – Filburt Aug 12 '17 at 20:16
• Simpler way to put it: forget the original norm and use Euclidean one instead. – user357151 Aug 12 '17 at 20:25