Let be a normed space, be a proper closed convex cone with nonempty interior. Define the positive polar cone by . Let be a weak*-compact convex base of , i.e, is weak*-compact convex, and for any nonzero there exists unique and such that .

Given , , and define . By the weak*-compactness of , the infimum is attained.

Question: Show that it is attained at an extreme point of , i.e. is an extreme point if there does not exist two distinct , and such that .

It seems this can be done by applying the Krein Milman theorem.

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I don’t know anything about optimization and Krein Milman, but I’d like to try ^_^.

Let . (Btw, your does not depend on your “given ” right? )

Then is closed and convex in in weak* topology, thus by Krein Milman theorem is the closed convex hull of its extreme points. Take an extreme point of . We claim that it is also an extreme point of . To see this, let such that , . Then forces , i.e. and so , as is an extreme point of . Therefore is an extreme point of .

Yes, given is a typo.

The proof is nearly the same as what I think last night, except the topology on should be the weak* topology. Since your is weak*-compact and convex, the Krein Milman theorem can still be applied.

[Corrected, thanks! -KKK]Let me state the general Krein Milman theorem:

Let be a nonempty compact convex set of a Hausdorff LCS (locally convex space, in our case, with weak* topology is an example). Then is the closed convex hull of its extreme points.

Perhaps I can add some small remarks.

If the weak topology and the weak* topology are concerned, we usually add weakly and weak* (not weak*ly, don’t care about grammar) to the related property, for example, weakly closed, weak* open etc.

In fact, the above problem is just a small part of a proof, which concerns characterizing a certain concept of minimality to the solution of some other optimization problem that the objective function is an extreme point of . This result at least guarantees there exists some such point.