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Let $X$ be a Banach space. Assume $K$ is compact, and convex subset of $X$.

My first question: Do we have $K = \text{conv} \; (ext K)$? where $ext K$ denotes the set of all extrem points of $K$.

My second and main question: Assume $K \subset X^*$ which is convex, norm bounded and weak* compact. Do we have $K = \text{conv} \; (ext K)$?

I think answer of the first question is NO. Because by Krein-Miliman Theorem we only can say $K $ is the closure of $ \text{conv} \; (ext K)$ and I dont think being Banach is a bonus here.

Red shoes
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  • Re question1: You're right, see here for an explicit example showing why taking the closure is necessary – Alessandro Codenotti Aug 23 '19 at 19:46
  • Re question 2: Note that weak* compact implies norm bounded, so you have an unnecessary hypothesis – Alessandro Codenotti Aug 23 '19 at 19:53
  • Thanks, Do you have a proof for your second statement ? @AlessandroCodenotti – Red shoes Aug 23 '19 at 20:06
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    Yes, the argument used to prove that a weakly compact set is norm bounded can be used to show that the same holds for the weak* topology. In fact more can be said, in the weak* topology compact and norm bounded+weak* closed are equivalent, see here – Alessandro Codenotti Aug 23 '19 at 20:10
  • Also I think that counterexamples to question two can be constructed in the same way as for question one, since $X^\ast$ with the weak topology (so with the weak* topology for a reflexive Banach space $X$) is a locally convex topological vector space. – Alessandro Codenotti Aug 23 '19 at 20:13

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