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As the title, I think, in a closed, bounded, and convex set $C$ in $\mathbb{R}^n$, $C=\{\sum_{i=1}^{k}\lambda_k x_k:x_i\in C_i, \lambda_i\geq 0, \sum_{i=1}^{k}\lambda_k\leq 1\}$, where $C_i$ is a convex, bounded, and closed set for all $i\in\{1,2,3,...,k\}$, if a point $p$ satisfies the following conditions:

$\exists u,l\in C$ such that $l\leq p\leq u$, where $\leq$ is element-wise operator, that means at least one element in $p$, $l$, $r$, satisfying $l_i<p_i<u_i$ for some $i$.

then, $p$ is not an extreme point.

I want to prove that this statement is true. However, I only can show this may be true by drawing a graph. Can someone give me a hint about how to prove this statement? Thank you!

jerry
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  • Your question is unclear. What does it mean that a set has "finite dimensions"? – Bcpicao Dec 14 '23 at 15:40
  • If one takes $u=l=p$, then your claim would imply that every point is not an extreme point. Perhaps you need strict inequality? – kwerty Dec 14 '23 at 15:42
  • @Bcpicao Thank you for your notice. I change the statement to be a set in $\mathbb{R}^n$. That means each element in $C$ is an element in $\mathbb{R}^n$. But the cardinality of $C$ is infinite. – jerry Dec 14 '23 at 15:43
  • @JasonChoy Yes, I change the statement to be $\leq$ meaning that at least one element $i$ in $l,p,u$ such that $l_i<p_i<u_i$. – jerry Dec 14 '23 at 15:47
  • @Bcpicao Let say $C={\sum_{i=1}^{k}\lambda_k x_k:x_i\in C_i, \lambda_i\geq 0, \sum_{i=1}^{k}\lambda_k\leq 1}$, where $C_i$ is a convex, bounded, and closed set for all $i\in{1,2,3,...,k}$. – jerry Dec 14 '23 at 15:52

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How about this? $p$ is an extreme point.

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Robert Israel
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  • Yes... It seems like my statement is wrong. I think my statement should be if $p$ is an interior point of $C$, then $p$ is not an extreme point. Thank you a lot! – jerry Dec 14 '23 at 16:06
  • Hi Robert, given each $C_i$ is a hypercube, I think if $p$ is an extreme point of $C$, then for the other points (say $y$) in $C$, $p$ always has at least one row that is larger than or equal to the corresponding row of $y$. Is that right? – jerry Dec 27 '23 at 08:48
  • Add: there is no rotation on each $C_i$. – jerry Dec 27 '23 at 08:55