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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.

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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—

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  • $\begingroup$ $B$ 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)? $\endgroup$
    – Canine360
    Aug 3 at 7:13
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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!

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