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I am reading a paper and there is such description as title. Why?

I have an example: $(0,1)$. This is a convex set but not closed, so I cannot find an extreme point. However if convex and compact,

I read some related problems:

  1. Exposed point of a compact convex set
    There must be at least one exposed point. But an extreme point is not necessary equal to an exposed point.
  2. Convex hull of extreme points
    A convex hull $P$ of finite points. Then $P$ is the convex hull of its extreme points.

It seems there is a requirement "finite points" to guarantee the topic?

glS
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Denny
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  • Duplicate? http://math.stackexchange.com/questions/1384579/prove-that-the-set-of-extreme-points-of-a-compact-convex-set-is-not-empty – A.Γ. Jul 02 '16 at 22:10
  • Yes, that is what I want. – Denny Jul 02 '16 at 22:13
  • What is the setting here, are we talking about Euclidean space $\mathbb{R}^n$ or a normed space, or a locally convex topological space? –  Jul 03 '16 at 02:53
  • Actually the paper is about positive semidefinite matrices with unity trace and rank one, which form extreme points – Denny Jul 03 '16 at 21:33

1 Answers1

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In a finite dimensional space (which is the case here, according to a comment), the existence of an extreme point of a compact convex set $K$ is easy to prove. Take any point $x\in K$ at which the norm $\|x\|$ is maximized. If there is $y\ne 0$ such that $x\pm y \in K$, then $$ 2\|x\|^2 \ge \|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2+2\|y\|^2> 2\|x\|^2 $$ a contradiction.

  • What is this contradicting? That $x$ is not and extreme point? How do I know there is a $y$ such that $x \pm y \in K$? – INQUISITOR Mar 02 '20 at 19:40