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Krein-Milman theorem states that

If ${K}$ is a non-empty compact convex subset of a locally convex space $ {X}$, then ${\text{ext}\;K\neq\emptyset}$ and ${K=\overline{\text{co}}(\text{ext}\;K)}.$

Note that the closure is needed and here is a related question why closure is needed. Let's suppose the LCS $X$ (having predual) is equipped with weak* topology. My question is that under which condition on $K$ we will have $$ K=\mathrm{co}(\mathrm{ext}(K)).\quad(1)$$ Clearly when $K$ is a subset of finite dimensional space, then $(1)$ is true. But for the infinte case, is there any sufficient condition for $K$ to satify $(1)$? Or $(1)$ is true if and only if $K$ is a subset of finite dimensional space?

And again I am interested in the case of $X$ equipped with weak* topology. Thanks!

Lin2568
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  • It is possible to have $K$ satisfy $(1)$, but not have $K$ be contained in any finite-dimensional space. For example, take $K$ to be the unit ball in $\ell^2$, which is weak$^*$ compact (considering at as the dual of itself). Then $\operatorname{ext} K$ is the sphere, and any point in the ball lies in the line segment connecting some pair of diametrically opposed points. – Theo Bendit Jun 21 '22 at 16:00
  • In the above example, $K$ has a lot of extreme points, and I think the richness of $\operatorname{ext} K$ is key. If $\operatorname{ext} K$ were countable, for example, then a Baire Category Theorem argument shows $\operatorname{co} \operatorname{ext} K$ could not equal $K$. The (unclosed) convex hull would be the countable union of convex hulls of finite subsets of $\operatorname{ext} K$, so if they covered the norm-closed $K$, one would have to have to have an interior in $K$, and this ball would be compact, which would limit $K$ to being finite-dimensional. – Theo Bendit Jun 21 '22 at 16:14

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