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In a problem I'm working on I found myself with a point $y\in \mathbb{R}^m$ lying at the boundary of a non-closed convex set $K$. I'd like to express it as as "infinite convex combination" $$y=\sum_{i=1}^\infty \lambda_i y_i,$$ where $\lambda_i \ge 0$, $\sum_i \lambda_i=1$ and $y_i \in K$ for all indices $i$. May I do so?

Thank you.

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$y$ doesn't belong to $K$? So $K$ is not closed? –  littleO Nov 11 '12 at 3:21
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@littleO: Yes, exactly. That's the problem. –  Giuseppe Negro Nov 11 '12 at 3:23
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I don't know whether this is helpful. But in Proposition 2.6 of mth.msu.edu/~shapiro/pubvit/downloads/numrangenotes/…, the author proved on $\mathbb{C}$, a finite convex combination is the same as an infinite convex combination. –  Hui Yu Nov 11 '12 at 3:37
    
@HuiYu: This means that the answer to the present question is negative, as $\text{conv}_\infty(K)=\text{conv}(K)=K$, if I'm not mistaken. –  Giuseppe Negro Nov 11 '12 at 3:40
    
Yes. That's what I mean, at least in $\mathbb{R}^2=\mathbb{C}$. –  Hui Yu Nov 11 '12 at 3:41

1 Answer 1

up vote 1 down vote accepted

No. The following is from Convex series (G.J.O. Jameson).

Definition: A set is CS-closed if it contains the sum of every convergent convex series of its elements.

Example (i): An open convex set, $A$, is CS-closed, since $\lambda A+(1-\lambda)\overline{A}\subseteq A$ for $0<\lambda <1$.

This holds for any Hausdoff topological linear space.

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