Faces of the intersection of convex sets

Question

Let $V$ be a normed real vector space and let $K_1, K_2\subseteq V$ be closed convex subsets such that the intersection $K_1\cap K_2$ is non-empty. Assume that $F_1$ is a face of $K_1$ and $F_2$ is a face of $K_2$ (face $F$ of a convex set $K$ is a convex subset such that $a=tb+(1-t)c$, for $a\in F$, $b, c\in K$, $0<t<1$, implies $b,c\in F$). If $F_1\cap F_2$ is non-empty, then it is a face of $K_1\cap K_2$.

Question: is every face $F$ of $K_1\cap K_2$ the intersection of $F_1, F_2$, where $F_1$ is a face of $K_1$ and $F_2$ is a face of $K_2$?

Probably the answer is known to experts working in convexity, however I am not able to find a reference in the literature. In my opinion the answer is affirmative, but maybe I am wrong and there are (even simple) counterexamples. It is possible that the answer is much simpler if one assumes that $V$ is finite dimensional. On the other hand it is also possible that the answer is known also for more general case when instead of the intersection of two convex sets we have the intersection of an arbitrary family of closed convex sets.

Answer 1

For finite-dimensional $V$, your definition of a face is equivalent to the definition of a poonem, according to part (i) of Exercise 7 on page 21 of the book Convex Polytopes by B. Gruenbaum; then the positive answer to your question is part (iii) of Exercise 9 on the same page.

Answer 2

It seems that the answer is Yes.

In the affine subspace $A$ spanned by $K:=K_1\cap K_2$, a face is the intersection of $K$ with a (closed) supporting hyperplane $\Pi$. By Hahn-Banach, there is an extension $\Pi_1$ of $\Pi$ as a closed hyperplane in $V$, so that $\Pi_1$ is a supporting hyperplane for $K_1$. Then $F_1=\Pi_1\cap K_1$ is a face of $K_1$. Likewise, $\Pi$ extends as a closed supporting hyperplane $\Pi_2$ of $K_2$, and $F_2=\Pi_2\cap K_2$ is a face of $K_2$. Eventually, $F=F_1\cap F_2$.

Answer 3

I found a sufficient condition that guarantees that the answer to your question is yes: If the relative interior of $F$ is not empty, then $F$ is the intersection of a face of $K_1$ and a face of $K_2$. A sufficient condition for $F$ to have nonempty relative interior would be that the dimension of $F$ be finite. Thereby, $K_1$ and $K_2$ can be any two convex sets in a real vector space (without assumptions on topology). A reference is Proposition 4.7 in the preprint Weis, S., A note on faces of convex sets (https://arxiv.org/abs/2404.00832). I don't know the answer to your question if the relative interior of $F$ is empty.

Also, a generalization of your question to infinite families fails. An example is the family of closed segments $[-\epsilon,1+\epsilon]$ for $\epsilon>0$. The intersection over this family is the unit interval $[0,1]$, whose two extreme points $0$ and $1$ cannot be written as intersections of faces of members of this family.