Let $C$ be a nonempty compact convex subset of ${\mathbb R}^d$ such that every extreme point of $C$ is an exposed point of $C$. Does it follow from this that every face of $C$ is an exposed face?

3$\begingroup$ Glossary: face= subset $F$ of $C$ such that every segment $S$ in $C$ whose interior meets $F$ is contained in $F$. (Singletons that are faces are extremal points). Exposed face= subset of $C$ that occurs as intersection of a closed halfspace with $C$ (this is necessarily a face). Exposed point = singleton that is an exposed face. (If $C$ is considered as face one should exclude exclude it in the question to avoid trivialities, since it's not exposed as soon as $C$ has nonempty interior.) $\endgroup$ – YCor Nov 30 '20 at 11:01

$\begingroup$ @YCor, "with C" or "with the boundary of C"? $\endgroup$ – Wlod AA Nov 30 '20 at 15:46
I don't think so. In dimension 3, first let $S$ be a "stadium" in the plane xy, namely: the segment joining $(1,1,0)$ to $(1,1,0)$, the halfcircle centered at $(1,0,0)$, radius $1$, with middle point $(2,0)$, and its opposite. Also consider the cube with vertices $(\pm 1,\pm 1,\pm 1)$. Consider the convex hull $K$ of this whole set of points.
The extremal points then are: the 8 vertices of the cube, and the points in the open half circles. They are all exposed. (Notably, the boundary points $(\pm 1,\pm 1,0)$ of half circles, which are extremal and nonexposed within the stadium, are not extremal in $K$ as they're not vertices in the cube.) For instance, the vertex $(1,1,1)$ is exposed, using the halfspace $\{x+y+z\le 3\}$. For points in open half circles, vertical halfspaces do the job.
The vertical edges of the cube, say the one joining $(1,1,1)$ and $(1,1,1)$, are faces. But are not exposed: indeed, looking around $(1,1,0)$, the only closed halfspace containing $K$ with $(1,1,0)$ as boundary point is the halfspace $\{y\le 1\}$. But its intersection with $K$ is a whole 2dimensional cube face.

$\begingroup$ Thank you for your answer! It is very nice and simple. $\endgroup$ – Janko Bracic Nov 30 '20 at 12:19

$\begingroup$ I am still thinking about conditions which ensure that every face is an exposed face. Are there some results related to this problem? Thinking about given counterexamples I wonder if the following is true. Let $C$ be as in the question. If, for every affine subspace $A\subseteq {\mathbb R}^n$, all extreme points of $A\cap C$ are exposed (or $A\cap C=\emptyset$), then every face of $C$ is exposed. $\endgroup$ – Janko Bracic Dec 1 '20 at 7:06
I believe I found a counterexample.
We first set $d=3$ and $$ C_1:=B_1(0)\cup [0,1]\times [1,1]\subset \mathbb R^2 \\ C_2:=C_1\times [1,1]\subset \mathbb R^3, $$ where $B_1(0)$ denotes the closed unit ball in $\mathbb R^2$.
Now, the set $$ F_1 := \{0\}\times \{1\}\times [1,1] \subset C_2 $$ is a face of $C_2$ which is not an exposed face.
However, the points $(0,\pm 1,\pm 1)$ are extreme points of $C_2$ which are not exposed points, and thus $C_2$ is not a counterexample. This can be solved by defining $$ C:= \{(x,y,z)\in C_2 \mid x+z \geq 1,\, xz \geq 1\}. $$ With this modification, the points $(0,\pm 1,\pm 1)$ are still extreme points of $C$ are also exposed points. It can also be checked that all other extreme points are exposed points.
However, the face $F_1$ is still a face of $C$ which is not an exposed face, so $C$ constitutes a counterexample.

$\begingroup$ Thank you for the answer! It is very nice. It is interesting that you and YCor have built your examples from the "stadium". $\endgroup$ – Janko Bracic Nov 30 '20 at 12:23