The Krein-Milman theorem shows that a compact convex set in a Hausdorff locally convex topological vector space is the convex hull of its extreme points.
It seems this implies that a compact convex set in such a space must have an extreme point.
I am interested in whether there is a very simple elementary argument that shows that a compact convex set must have an extreme point.
I have such an argument, but since it uses compactness of the unit ball, it is not so good if the space is infinite dimensional.
In point of fact, I am using this in R^n, but if there is a way to put it that can generalize to infinite dimensions then that would seem preferable for the students.
Best Answer
The main question is what generality you want. As soon as you have just one strictly convex function in the space, any point where it attains its maximum on $K$ is an extreme point. If we were talking about separable normed spaces, the construction of such function would be trivial: $F(x)=\sum_j 2^{-j}(1+\|x_j\|)^{-1}\|x-x_j\|$ where $x_j$ is any countable dense set would work just fine. In a strictly convex normed space $F(x)=\|x\|$ would be an even simpler example. The problem is that you want it in an abstract locally convex topological linear space, so some form of AC seems, indeed, inavoidable.