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Let $\Omega$ be a convex compact set in $\mathbb{R}^n$, $f\colon \Omega \to \mathbb{R}$ be a convex function. Consider an optimization problem $$ \int\limits_{\Omega}f(x)\,\mu(dx) \to \max\limits_{\mu \in M}, $$ where $M$ is a set of Borel probability measures on $\Omega$ such that $$ \int\limits_{\Omega} x \, \mu (dx) = x^{\ast} \in \mathop{\mathrm{int}}\Omega \;\;\; \text{($x^{\ast}$ is fixed)} $$ How to show that if the maximization problem is solvable then there exists an optimal measure $\mu^{\ast}$ such that $$ \mu^{\ast} = p_{1}\delta(x-x_{1})+\ldots+p_{n+1}\delta(x-x_{n+1}), $$ where $p_{i} \geqslant 0$, $\sum_{i=1}^{n+1}p_{i}=1$ and $x_{1},\ldots,x_{n+1}$ are extreme points of $\Omega$? I tried to combine Krein-Milman and Caratheodory theorems, but without success.

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I might not understand something correctly, but if $f$ has a unique maximzer on the boundary of $\omega$, wouldn't any measure satisfying the second condition not be a maximizer? –  Michael Greinecker Dec 27 '12 at 22:53
    
@MichaelGreinecker You're right, but there are no maximizers at all in this case? In other words, can we say that if maximizer in class $M$ exists then there exists a maximizer of specified type? –  Nimza Dec 27 '12 at 23:05
    
This may be related to @Michael's question: is $x^*$ fixed, or can it vary with $\mu$? –  user53153 Dec 28 '12 at 5:50
    
@PavelM $x^{\ast}$ is fixed. –  Nimza Dec 28 '12 at 6:30
    
Good. Then the existence is not hard. The set of all admissible measures is compact in weak* topology (obtained from the dual of $C(\Omega)$). Take a sequence $\mu_n$ such that $\int f\,d\mu_n$ tends to the supremum of $\int f\,d\mu$ over all admissible measures. Any weak* cluster point will realize the supremum... But I don't see how to use Caratheodory's theorem to reduce the support to $n+1$ points... By the way, Chapter 10 of Simon's book on convexity covers Choquet theory, which concerns such extremal problems. –  user53153 Dec 28 '12 at 7:16
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