# Is every face exposed if all extreme points are exposed?

Let $$C$$ be a non-empty 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?

• 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.) – YCor Nov 30 '20 at 11:01
• @YCor, "with C" or "with the boundary of C"? – 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 half-circle 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 non-exposed 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 2-dimensional cube face.

• Thank you for your answer! It is very nice and simple. – Janko Bracic Nov 30 '20 at 12:19
• 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. – 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 $$C:= \{(x,y,z)\in C_2 \mid x+z \geq -1,\, x-z \geq -1\}.$$ 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.

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