I am trying to adapt a result in Phelps : "Lectures on Choquet's Theorem" in chapter 15.

Let $X$ be a convex subset of a locally convex space $E$, let $P_1$ denote the set of all regular Borel probability measures on $X$. We define the following two partial ordering with the hope to show that, under some aditional constraints, they match.

Definition : For two non-negative measure $\mu$ and $\nu$ on $X$, we say that $\mu\preccurlyeq \nu$ if for all bounded convex continuous function $h:X\to\mathbb R$, $\mu(h)\leq \nu(h)$.

Definition : For two non-negative measure $\mu$ and $\nu$ on $X$, we say that $\mu\curlyeqprec \nu$ if there is a dilatation $T:x\to T_x$ such that $\nu=\mu T$ (defined on page 93 of the link above).

A theorem by Hardy-Littlewood-Polya-Blackwell-Stein-Sherman-Cartier (on page 94) states

If $X$ is compact and if $\mu$ and $\nu$ regular Borel probability measures on $X$ then $\mu\preccurlyeq\nu$ if and only if $\mu\curlyeqprec\nu$.

To me it feels like compactness is used only in the second part of the proof when they use a result of Bourbaki on page 96. I am wondering if the proof can be modified to avoid the use of compactness. Here is my proof and saddly I cannot be sure it is correct.

The outline of the proof is to show that $A\triangleq\{ (\epsilon_x,\lambda)\in P_1\times P_1 : \epsilon_x\preccurlyeq\lambda \} = \{ (\epsilon_x,\lambda)\in P_1\times P_1 : \epsilon_x\curlyeqprec\lambda \}\triangleq B$ where $\varepsilon_x(A)=\mathbf 1(x\in A)$ for all $x\in X$ and $A\subseteq X$ measurable. Then we show that $C\triangleq\{ (\mu,\nu):\mu\preccurlyeq \nu \}$ and $D\triangleq\{ (\mu,\nu):\mu\curlyeqprec \nu \}$ are respectively the closed convex hulls of $A$ and $B$ which means that $C=D$. The proof relies on Jensen inequality that gives $\mu\curlyeqprec\nu\Rightarrow\mu\preccurlyeq \nu$ indeed if we have a kernel $T$ such that $\nu=\mu T$ and a convex function $h$, then for any $x\in X$, $T_x(h)\geq h(x)$, averaging over $\mu$ gives that $\nu(h)\geq\mu(h)$. It also relies on the definition of the upper concave enveloppe of a bounded function $g$ defined for all $x\in X$ as $\bar g(x)=\sup\{ \lambda(g) : \lambda\sim \epsilon_x \}$ (where $\lambda\sim\epsilon_x$ means that $\lambda$ averages to $x$, i.e. for all affine function $h$ on $X$, $\lambda(h)=h(x)$). This function is concave and such that $g\leq \bar g$ and, as we will show, $\lambda\sim\epsilon_x\Leftrightarrow \epsilon_x\preccurlyeq \lambda\Leftrightarrow\epsilon_x\curlyeqprec \lambda$ which means that $\bar g(x) = \sup\{ \lambda(g) : \epsilon_x\preccurlyeq \lambda \} = \sup\{ \lambda(g) : \epsilon_x\curlyeqprec \lambda \}$.

Jensen inequality gives that $A\supseteq B$ so we prove that $A\subseteq B$. Suppose that $\epsilon_x\preccurlyeq \lambda$, then for all $f\in X^*$, both $f$ and $-f$ are convex hence $f(x)=\varepsilon_x(f)=\lambda(f)$ and so $\lambda$ averages to $x$ therefore $\lambda\sim\epsilon_x$. The statement $\lambda\sim\epsilon_x\Leftrightarrow \epsilon_x\curlyeqprec \lambda$ is a tautology, hence $A=B$.

The proof of $C$ and $D$ are the closed convex hulls of respectively $A$ and $B$ are very similar and differ only in few steps. in order to show that $C$ is the closed convex hull of $A$, it is enough to show that for all affine function $L$ on $P_1\times P_1$ such that $L\geq 0$ on $A$ we have $L\geq 0$ on $C$, this is because the closed onvex hull of a set is the intersection of all half spaces that contains it. without loss of generality we can write for all $(\alpha,\beta)\in P_1\times P_1$, $L(\alpha,\beta)=\alpha(f)-\beta(g)$ for some function affine functions $f$ and $g$ on $X$. Assuming that $L\geq 0$ on $A$ means that for all $\epsilon_x\preccurlyeq \lambda$, we have $f(x)=\epsilon_x(f)\geq \lambda(g)$. Assume that $\mu\preccurlyeq \nu$, then $\mu(-\bar g) \leq \nu(-\bar g)$ by concavity of $g$ and so $\mu(\bar g)\geq \nu(\bar g)$. From $g\leq \bar g$ we get that $\nu(g)\leq \nu(\bar g)$. Since for any $\epsilon_x\preccurlyeq \lambda$ we have $f(x)\geq \lambda(g)$, we have $f(x)\geq \sup\{ \lambda(g): \epsilon_x\preccurlyeq \lambda\}=\bar g(x)$ and so we have $\mu(f)\geq\mu(\bar g)\geq \nu(\bar g)\geq \nu(g)$. This means that $L\geq 0$ on $C$ and shows that the closed convex hull of $A$ is $C$.

To show the closed convex hull of $B$ is $D$ we take a similar approach. Again suppose that $L : (\alpha,\beta)\to \alpha(f)-\beta(g)$ is such that $L\geq 0$ on $B$, we show $L\geq 0$ on $D$. For all $\epsilon_x\curlyeqprec\lambda$, we have $f(x)\geq \lambda(g)$ and so $f(x)\geq \sup\{ \lambda(g) : \epsilon_x\curlyeqprec\lambda \}=\bar g(x)$ which yields $\mu(f)\geq \mu(\bar g)$. By Jensen inequality, $\mu\curlyeqprec \nu$ implies $\mu\preccurlyeq \nu$ which implies $\mu(\bar g)\geq \nu(\bar g)$ since $\bar g$ is concave. Finally $\nu(g)\leq \nu(\bar g)$ follows again from $g\leq \bar g$ and so $\mu(f)\geq\mu(\bar g)\geq \nu(\bar g)\geq \nu(g)$. From this we deduce that $D$ is the closed convex hull of $B$ and so $C=D$.

To me this proof feels correct but I can't know that for a fact so any review or comment would be welcome, I can add more detail if needed. It also seems that this proof does not rely on compactness of $X$ and so if this proof is correct the original result can be generalized. I wonder if the compactness we actually need is the weak one, and here we have it, indeed the set of probability measures on $X$ is bounded (with the $\sup$ norm, the norm of the difference between two probability measures is at most $2$) and we also have that this set is weakly closed, because it is the intersection of closed half spaces. This means that the set of probability measures on $X$ is weakly compact, this may be true only if $X$ is itself weakly compact, but I don't know where this would make the proof wrong otherwise.



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