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.



Faces of the intersection of convex sets
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](https://archive.org/download/springer_10.1007-978-1-4613-0019-9/10.1007-978-1-4613-0019-9.pdf); then the positive answer to your question is part (iii) of Exercise 9 on the same page. 


