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I am interested in studying the following problem: \begin{align} \sup_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x) \end{align} where $\mu$ is a probability measure. Assume that $\mathcal{D}$ is closed, covex and compact (in weak^{*} topology).

We know that if $f(x)$ is continuous and bounded then $\mu \to \sup_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x)$ is a continous functional and \begin{align} \sup_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x)&=\max_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x), \text{ this steps is due to contintinuity,}\\ &=\max_{\mu \in \text{Extrem Points of} \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x), \text{ this step is due to linearity of the functional,}\\ \end{align}

My question: Now, assumes that $f(x)$ is continuous and positive but not bounded from above. So, we can not claim that $\mu \to \sup_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x)$ is a continous functional.

However, can we still say that

\begin{align} \sup_{\mu \in \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x)=\sup_{\mu \in \text{Extrem Points of} \mathcal{D} } \int_{\mathbb{R}} f(x) d\mu(x), \end{align}

from the linearity?

The example I have in mind is $f(x)=x^2$.

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The answer is yes. Call ${\mathcal E}$ the set of extreme points of ${\mathcal D}$. It is a consequence of Choquet's theorem that for every $\mu\in{\mathcal D}$ there exists a probability measure $\nu=\nu_\mu$ on ${\mathcal E}$ such that $\mu=\int_{\mathcal E}\rho d\nu(\rho)$ and in particular $$\int_{\mathbb R}f(x)d\mu(x)=\int_{\mathcal E}\left(\int_{\mathbb R}f(x)d\rho(x)\right)d\nu(\rho)\leq\sup_{\rho\in{\mathcal E}}\int_{\mathbb R}f(x)d\rho(x)$$

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  • $\begingroup$ The Choquet's theorem states if $D$ is compact and convex and if $f$ is an affine function on $D$, then there exists a probability measure $\nu$ supported on $\mathcal{E}$ such that $f(c)=\int_{\mathcal{E}} f(e) d\rho(e)$. I have a question. How are you getting $\mu = \int_{\mathcal{E}} e d\nu(e)$. I don't think $\mu$ is an affince function. Can you explain this point a bit. $\endgroup$
    – Boby
    Commented Aug 4, 2017 at 1:57
  • $\begingroup$ I was thinking of the version of Choquet's theorem which says that every point of a compact convex set ${\mathcal D}$ is the barycenter of a probability measure on the extreme points of ${\mathcal D}$, although I realize now that is not exactly the version on wikipedia. But the version on wikipedia is good enough to answer your question: Let $F:{\mathcal D}\to{\mathbb R}$ be the affine function $F:\mu\mapsto\int_{\mathbb R}f(x)d\mu(x)$. Then we have $F(\mu)=\int_{\mathcal E}F(\rho)d\nu(\rho)\leq\sup_{\rho\in{\mathcal E}}F(\rho)$. $\endgroup$
    – Joel Moreira
    Commented Aug 4, 2017 at 2:13

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