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I've been reading Barvinok's A Course in Convexity and I come across with the following problem that I don't know how to tackle. I would like to take advice for how to start.

Let $A\subset\mathbb{R}^d$ be a closed convex set. Prove that each extreme point of $A$ is a limit of exposed points of $A$.

For context, given $K\subset\mathbb{R}^d$ a closed and convex set, we say that $F\subset K$ is a face of $K$ if there exists an affine hyperplane $H$ which isolates $K$ and such that $F=K\cap H$. If $F$ is a point, we say that that point is an exposed point.

Let $A\subset\mathbb{R}^d$ be a convex set. We say that $a\in A$ is an extreme point of $A$ if for each point $b,c\in A$ such that $a=\frac{b+c}{2}$, one must have $b=c=a$.

I don't have any major idea, I've tried to argument by contradiction, saying: Let $x\in A$ be an extreme point.

If $x$ is already an exposed point, it is done.

If $x$ is not an exposed point and it is not the limit of a sequence of them, then there must be an $\varepsilon>0$ such that $B(x,\varepsilon)$ does not contain any exposed point. I have not been able to "milk" this statement.

Any help would be appreciated!

Floromidante
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  • Do you have any tools under your belt? Or should it be shown just from these definitions? – While I Am Apr 23 '21 at 18:29
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    At the time that the exercise appears, one should know Caratheodory's Theorem, Helly's Theorem and some separation theorems. But the exercise is flagged as a difficult one, so is more than possible that its answer is more than just a mere combination of this theorem. – Floromidante Apr 25 '21 at 04:44

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This is a result known as Straszewicz's theorem. You can find its proof in Rockafeller's Convex Analysis (Theorem 18.6). Also, see Proof of Straszewicz's theorem

Xiangliang Li
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