Prove that every exposed point is a extreme point

Question

Let $C$ be a non-empty convex subset of $\mathbb{R}^n$.

We say that $x\in C$ is a extreme point of $C$ if for every $z,y\in C$ and $t\in [0,1]$ such that $x=ty+(1-t)z$ we have $x=z$ or $x=y$. Or equivalently, if for every $z,y\in C$ and $t\in (0,1)$ such that $x=ty+(1-t)z$ we have $x=z=y$.

In other hand, we say that $x\in C$ is a exposed point of $C$ if there exists a supporting hyperplane $H$ such that $H\cap C = \{x \}$.

How can I prove that every exposed point of $C$ is a extreme point of $C$?

Your definition of exposed point is not quite right. You want $H$ to be a supporting hyperplane, not just any hyperplane. — Robert Israel, May 05 '13 at 21:23

score 2 · Answer 1 · answered May 06 '13 at 01:21

2

Hint: Suppose that the exposed point $x$ is not an extreme point. That is $x$ is an interior point of a segment $I\subset C$. How $I$ and $H$ are disposed?

answered May 06 '13 at 01:21

Alex Ravsky

90,434

score 2 · Answer 2 · answered Feb 19 '14 at 13:29

Let $c$ is a exposed point of $C$ and $u,v\in C$ and $0<\alpha<1$. Let $c=\alpha u+(1-\alpha)v$ and $c$ is not extreme point of $C$. Then $c\neq u$ and $c\neq v$.However there exist $x^{*}\in X^{*}$ such that for every $x\neq c$ in $C$, $x^{*}(c)>x^{*}(x)$. In paricular $$x^{*}(c)>x^{*}(u) , x^{*}(c)>x^{*}(v)$$ Then we have $$x^{*}(c)=x^{*}(\alpha u+(1-\alpha)v)=\alpha x^{*}(u)+(1-\alpha)x^{*}(v)<x^{*}(c)$$ that it is contradiction.

Prove that every exposed point is a extreme point

2 Answers2