0
$\begingroup$

Let $E$ be a locally convex Hausdorff space and $X$ a compact convex set. Is it true that $\overline{conv}(\overline{ex(X)})=\overline{conv}(ex(X))$? Here $conv(X)$ is the convex hull of the subset $X$ and $ex(X)$ is the extremal points set. I think that it's true, but I can't find a proof. I tried to first picking a convergent net $(x_\alpha)_{\alpha \in A}$ in the first set and then showing that there is another net $(y_\beta)_{\beta \in B}$ in the second set converging to the same limit, but I couldn't construct that second net.

Edit: I was read Lectures on Choquet's Theorem by R.Phelps, and I stumbled upon this question when he was proving that the Krein-Milman theorem is equivalent to an integral representation theorem. This question arises when you are trying to prove that this integral representation of the points implies Krein-Milman, so I can't use it to prove this!

$\endgroup$
1
$\begingroup$

Krein Milman says that $X=\overline{conv}(ex(X))$ since $X$ is compact thus closed, $\overline{ex(X)}\subset X$, you deduce that $conv(\overline{ex (X))}\subset conv(X)=\overline{conv}(e(X))=X$. Since $e(X)\subset\overline{e(X)}$, we deduce that $\overline{conv}(e(X))\subset\overline{conv}(\overline{e(X)})$ and henceforth $\overline{conv}(e(X))=\overline{conv}(\overline{e(X)})$.

https://en.wikipedia.org/wiki/Krein%E2%80%93Milman_theorem

$\endgroup$
  • $\begingroup$ The extremal points of a convex set need not to be closed.An example can of that case can be found in Barry Simon's book about Convexity! $\endgroup$ – Affonso Jul 3 '17 at 4:11
  • $\begingroup$ Also, I was reading Lectures on Choquet Theorem by Robert Phelps exactly on the part that he proves that the Krein Milman theorem is equivalent to an integral representation theorem, so I couldn't use it! I will edit the question $\endgroup$ – Affonso Jul 3 '17 at 4:13
0
$\begingroup$

Now I figure it out. I was looking in this site and found this Convex hull of extreme points . Just note that $ex(X) \subset conv(ex(X))$, the details are an exercise!

$\endgroup$
  • $\begingroup$ So you don't need $X$ be convex and compact . $\endgroup$ – Red shoes Jul 3 '17 at 3:59
  • $\begingroup$ Yeah, you are right, I put these hypothesis because this is the case I'm interested. $\endgroup$ – Affonso Jul 3 '17 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.