This seems plausible, given the properties of the unit ball of $c_0$.

I have a compact set in a complex Banach space $X$ whose closed convex hull has uncountably many extreme points. It would be nice to deduce from this that $X$ contains no copy of $c_0$. I have been searching, but could find no proof either way, and I cannot see how to prove it myself---well, not yet... But in the mean time, perhaps this is already known.

  • 10
    $\begingroup$ Consider the set $\{(x_1,x_2,x_3,\ldots):x_1^2+x_2^2\le1\text{ and }x_i=0\text{ for }i\ge3\}$. $\endgroup$ – Yoav Kallus Sep 25 '15 at 0:34

As stated your question admits an immediate answer because the extreme point structure of finite dimensional convex sets in infinite-dimensional Banach spaces is not related to the structure of the Banach space: for any such set we can find an affine (and thus, preserving extreme structure) map into any other infinite-dimensional Banach space.

Comment of Yoav Kallus is an illustration of this.

On the other hand, there is a very interesting theory of extreme points of unit balls of Banach spaces, in which $c_0$ plays an important role. See, for example, the paper of Fonf on Polyhedral Banach spaces.

Possibly it is worthwhile to redesign your question.

| cite | improve this answer | |

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.