I'm looking for a precise statement of a theorem from convex analysis. It concerns a convex set A of real functions that are concave on compact convex set K(maybe in some locally convex space?). For each f in A there is k in K so that f(k) is nonnegative. The conclusion is that there is k' in K so that f(k') is nonnegative for all f in A.
Let K be a compact convex subset of a normed space E and
A(K) = { f:K -> R / f concave on K}.
If for each f in A(K) there is kf in K such that f(kf) is nonnegative, then there is z in K such that f(z) is nonnegative for all f in A(K).
REMARK: If E is only a locally convex space, then we must add the condition " E is Hausdorff". In all such situation K is closed, being compact and the closedness is necessary......z is on the boundary(border) of K.
I have recommended this answer just because in the given formulation it is [trivially] true. Indeed, it is impossible that every f of A(K) possesses a nonnegative value at some point. In other words, since A(K) contains all negative constant functions.
Dear Dinu, I hope you didnt get angry by the above pseudo-joke since it should be obvious to everybody that your formulation possesses just a forgotten assumption that f-s are taken from A which is a convex subset of A(K).
BUT there is an objection about the position of z on the boundary of K. Here is an example with K:=[-1,1], A:={ (u x; x in K) : -1
No problem, dear Joachim! It's ok!
I was in a hurry and wrote some nonsense! I thought I remembered a theorem from convex optimization regarding proximal point, but I was wrong! Anyway, it's clear z is not necessary on the boundary and, it's possible the closedness of K be not necessary!
Sorry for mistakes!
Don't be sorry, be happy dear Dinu!
PS I am still trying to prove the correct version. Perhaps I will find it :-)
I'm pretty sure K in Hausdorff locally-convex is enough. I found a statement for the weaker version for a convex set of affine maps but I don't think the proof would be much different than the concave function case.
However, the statement is in French and when it refers to the convex set of affine functions it states the set is "s.c.s." and that no function has value of +infinity. Not sure what "s.c.s" means in this context. This is Lemma 3 from Maurey's "Theoremes de factorization pour les operators lineaires a valeurs dans les spaces Lp".
Geoff Diestel I think "s.c.s." is the abreviation from "semi-continu supérieur" which means "upper semi-continuous"
Geoff Diestel Joachim Domsta But I can't understand what signifies the property "s.c.s" for a set ! At the same time I think every function from A=A(K) must be strictly concave !?
If it works for (strictly) concave functions, then it works for (strictly) convex functions too.....with some change.
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