# Intersection of the boundary of convex set with probability simplex

The question I'm working on, (a version of) which I have been posting in a piecemeal manner over the last few days because I thought they would be of more general interest that way, is really the following:

Take a compact convex set $$A \subset \mathbb{R}^n$$. Take its boundary $$\partial A$$. Take its intersection with the $$n$$-dimensional probability simplex $$\Delta_n=\{x \in \mathbb{R}^n: x\geq 0, \sum\limits^n_ix_i\leq 1\}$$, i.e. take $$\partial A \cap \Delta_n = : B$$ (say). Assuming $$B \neq \emptyset$$ (not that this matters), can every point in the convex hull of $$B$$ be represented as a convex combination of at most two points in $$B$$?

As explained in a (now deleted) answer to this question of mine (and as is kind of obvious), the answer is yes if $$\partial A \cap \Delta_n = \partial A$$ or $$\partial \Delta_n$$. What about in general?

I'm totally open to the possibility that this may not be true, and a counterexample would also be most appreciated.

Thanks a lot for your help.

Both the simplex and A are convex so their intersection will be convex as well. If $$A\subset \Delta_n$$ then it is trivially true that any point in B is a convex combination of two points in $$\partial A$$.

If not, then there are some points that are in B but not A. These will be along a line joining two points in $$\partial A$$.

Therefore, either way—

• $B$ is a subset of $A$ ($A$ is cpt, $\therefore$ closed. So $\partial A \subseteq A$, and $B \subseteq \partial A$.) But the bigger problem is, I don't think we can say that any part of the boundary of the convex hull of a set which is not part of the boundary of the set is a line segment joining two points of the set. See this: math.stackexchange.com/questions/1324037/… Maybe you can suggest a correct version of the linked question (which is not correct as stated, as pointed out in comments)? Aug 3 at 7:13

OK I think I found an obvious counterexample. Let me know if this makes sense.

In $$\mathbb{R}^3$$, take the part of the unit sphere in the non-negative orthant (i.e. the "quarter"-sphere). Take the equilateral triangle which is the intersection of the plane $$x+y+z=1$$ with the non-negative orthant. This triangle lies in the convex hull of the surface of the "quarter"-sphere, i.e. in $$conv\{x^2+y^2+z^2=1,x,y,z\geq0\}$$, but none of its interior (when the triangle is viewed as a subset of the metric space constituted by the 2-D plane through it) points can be expressed as a convex combination of two points from $$\{x^2+y^2+z^2=1,x,y,z\geq0\}$$.

Now obviously the intersection of the boundary of the unit sphere with the 3-D unit simplex will not contain any points from the boundary of the sphere other than the three points $$(0,0,1),(0,1,0),(1,0,0)$$. But scale the above setup down (using $$x^2+y^2+z^2=\alpha^2$$ and $$x+y+z=\alpha$$ for small enough $$\alpha$$ will make the entire quarter sphere lie within the unit simplex), and we have the desired conterexample.

PS: Actually, while writing this I realized, the unit sphere itself is also a counterexample. In this case $$\partial A={x^2+y^2+z^2=1},B=\partial A \cap \Delta_3=\{(0,0,1),(0,1,0),(1,0,0)\}$$ and most points in $$convB$$ obviously require all three of the points in $$B$$ in their support!