# About a result by Hardy-Littlewood-Polya-Blackwell-Stein-Sherman-Cartier on two order operators.

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.