I'm trying to show that given a compact convex set $K$ in $R^d$, there must be at least one exposed point (where $v$ is exposed if there exists a hyperplane H such that $H \cap K = \{v\}$ . This is a homework problem, but I'm totally stuck and looking for a hint.
I know that convex compact sets are convex hulls of their extreme points, if that's at all useful. Thank you!
On
Let me do the proof in $d=2$ to make it easy for me but the same idea works in general.
We will assume that we are so unlucky that whenever we choose a direction there are not exposed points in that direction.
Take a direction $d_1$ (a functional) and consider the support set on $K$ in that direction. It is not a single point. It is a compact interval. Take another direction $d_2$ and the support set is not a single point. We consider now directions in between these two. The support set of any direction in between these two is contained in the triangle formed by the two support lines and the to closest end-points of the support sets.
So, whenever we consider a new direction in between the two last already considered we find a smaller triangle inside the previous triangle. This nested sequence of triangles converges to a point.
We take directions in the following way $d_1,d_2$, then $d_3$ between $d_1$ and $d_2$, then $d_4$ between $d_2,d_3$, then $d_5$ between $d_3,d_4$, .... etc, so that we are not moving towards an endpoint of the support sets already considered.
The limiting point is the support set of the limiting direction, and it is exposed because the support set in that direction can only be that point (because it is nested in between arbitrarily close support sets of other directions).
For higher dimension instead of a triangle you work with a simplex bounded by the necessary number of directions.
Hint: Pick a point $p$ and look for a point of maximum of the function $f(k)=||p-k||, k\in K$.