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Let $E$ be a locally compact space, $X \subseteq E$ be a convex compact set. The Milman theorem states that if $Z \subseteq X$ is such that $X = \overline{ \mathop{\mathrm{conv}}Z}$, then $\mathop{\mathrm{ex}}{X} \subseteq \overline{Z}$. Is it true that if $Z$ is closed and no three points in $Z$ are in the same line then $\mathop{\mathrm{ex}}{X} = Z$? Here $\mathop{\mathrm{ex}}{X}$ is a set of extreme points of $X$.

The answer is negative in general. Now I'm looking for conditions that waranty $\mathop{\mathrm{ex}} \overline{\mathop{\mathrm{conv}}Z} = Z$ in multidimensional case.

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Let $X$ be a triangle in $\mathbb{R}^2 = C(\{0,1\})$ and let $Z$ be the set of its vertices plus one point in the interior of $X$... –  t.b. Apr 29 '12 at 15:26
@t.b. Great thanks! Maybe there are some general additional conditions on $Z$ to satisfy $\mathop{\mathrm{ex}}{X} = Z$? –  Nimza Apr 29 '12 at 15:49
In finite dimensions a sufficient condition is that $Z$ be affinely independent in the sense that for each $z \in Z$ we have that $z$ is not contained in the affine subspace spanned by $Z \smallsetminus \{z\}$. In infinite dimensions you want $Z$ to be closed and a Choquet simplex (see chapter 10 of Phelps's book in the references there for a full discussion). –  t.b. Apr 29 '12 at 16:19
@t.b. I found the definition in Phelps' book at page 52. Here author says that simplex must be convex. But I think that the set $\mathop{\mathrm{ex}}{X}$ can't be convex. –  Nimza Apr 29 '12 at 18:53

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