Let $C$ be a convex set in a normed space and let $F$ be a face of $C.$ By a "face", it is understood a set such that if the segment $u + (0,1)(v-u)$ with two end points in $C$ intersects $F$ then $u,v \in F$ (it can be shown, in fact, that $[u, v] \subset F$). Note, a face is not required to be convex as is often done.
I know that faces are closed in the induced topology ($F$ is a closed subset of $C$) and I know that closures preserve convexity ($\bar C$ is both closed and convex). Naturally, I wonder if $\bar F$ is a face of $\bar C.$
Prove or provide a counterexample: $\bar F$ a face of $\bar C.$
The reciprocal conclusion is easy: if $G$ is a face of $\bar C$ then $G \cap C$ is easily shown to be a face of $C$ (by hypothesis, the defining property of face is satisfied for $u,v \in \bar C,$ so it is also satisfied for $u,v \in C,$ from which it follows that $G \cap C$ is a face of $C$).
Also, since faces are closed in the induced topology, it is also true that $F = \bar F \cap C.$
Finally, I'd be very happy if the result were true in general and will be happy if the result is true for convex faces (i.e. assuming $F$ is also convex).
Here is a counter-example: take $$ C = (-1,+1)^2 \cup \{ (0,1) \} $$ with face $F$ given by $$ F=\{(0,1)\}. $$ Then $F$ is closed and convex, but not a face of $\bar C$.