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YCor
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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 $K_2$?

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.

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.

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.

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Janko Bracic
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Faces of the intersection of convex sets

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.