Generalized convex combinations and infinite dimensional Jensen inequality In Cedric Villani's Topics in Optimal Transportation, a passage in the proof of Theorem 1.27 goes like this:
Let $X,Y$ be polish spaces, $\mu\in\Delta(X),\nu\in\Delta(Y)$ $C\subset X\times Y$ open.
Let $\tilde{\Phi}_C=\{(\varphi,\psi)\in L^1(\mu)\times L^1(\nu)\ |\ \varphi\ \text{upp. semi-cont}; \ 0\leq\varphi\leq 1,\ -1\leq\psi\leq 0;\ \varphi(x)+\psi(y)\leq\mathbf{1}_C(x,y)\}$.
Observe that  $\tilde{\Phi}_C\subset L^1(\mu)\times L^1(\nu)$ is convex.
Consider the linear functional:
$$J: (\varphi,\psi)\in\tilde{\Phi}_C\mapsto \int_X\varphi d\mu+\int_Y\psi d\nu\in\mathbb{R}$$
1. One can show that any $(\varphi,\psi)\in\tilde{\Phi}_C$ can be rewrittens as
$$(\varphi,\psi)=\int_0^1 (\mathbf{1}_{A(s)},-\mathbf{1}_{B(s)})ds$$
where $\forall s\in [0,1]\ A(s)=\{\varphi\geq s\}$ (closed thanks to upper continuity) and $B(s)=\{\psi\leq -s\}$.
Question 1: Villani refers to this representation saying that the pair $(\varphi,\psi)$ is a convex combination of pairs $(\mathbf{1}_{A},-\mathbf{1}_{B})$  with $A$ closed. This is intuitively clear, but what is the precise definition of "generalised" convex combination villani is employing?
2. It is claimed that since $J$ is linear, for any $(\varphi,\psi)$ there is $A,B$ (depending on $(\varphi,\psi)$) such that:
$$J((\varphi,\psi))=J\left(\int_0^1 (\mathbf{1}_{A(s)},-\mathbf{1}_{B(s)})ds\right)\leq J((1_A,-1_B))$$
Question 2: again, intuitively, this is clear: if $J$ was a function defined over a finite dimensional domain, clearly the value on a convex combination is no larger than the value at one of the extremes. But how can we be sure of it in infinite dimension? Is some kind of infinite dimensional Jensen inequality (linear implying convex) applied here ? Or is there a direct proof?
Along the second line, I was thinking of:
$$\int_X\varphi(x)d\mu(x)=\int_X\int_0^1\mathbf{1}_{A(s)}(x)dsd\mu(x)=\int_0^1\mu(\{\varphi\geq s\} ds$$
Now, if $s\in[0,1]\to\mu(\{\varphi\geq s\})\in\mathbb{R}$ was continuous, then we would be done I think, via Weierstrass. But I don't think this need to be the case in general ($\mu$ may have atoms). Similarly for $\int_Y\psi(y)d\nu(y)$.\
Any help would be precious.
P.S. In the case where we deal with a linear functional defined over a compact convex subset of a Banach space, I think a similar problem could be adressed using the Choquet representation theorem (https://en.wikipedia.org/wiki/Choquet_theory). But here I don't know whether $\tilde{\Phi}_C$ is compact (I don't remember compactness criteria in $L^1$, my functional analysis is suer-rusty) and more fundamentally, I don't think $\mathsf{Ext}(\tilde{\Phi}_C)\subseteq\{(1_A,-1_B)\ |\ A\ \text{closed}\}$. We are only proving that the latter is a "convex basis" for $\tilde{\Phi}_C$
 A: To solve point 2 we don't need an infinite dimensional Jensen's inequality.It would be an overkill here.
Indeed, it is sufficient to keep on reasoning along the lines described in the question. Using Fubini and linearity:
$$J(\varphi,\psi)=\int_X\varphi(x)d\mu(x)+\int_Y\psi d\nu(y)=$$
$$\int_X\int_0^1\mathbf{1}_{A(s)}(x)dsd\mu(x)+\int_Y\int_0^1-\mathbf{1}_{B(s)}(y)dsd\nu(y)$$
$$=\int_0^1\underbrace{\left[\int_X\mathbf{1}_{A(s)}(x)d\mu(x)-\int_Y\mathbf{1}_{B(s)}(y)d\nu(y)\right]}_{f(s)}ds$$
$$=\mathbb{E}[f(U)]$$
where $U$ is uniformly distributed over $[0,1]$.
Then, by contrapositive $\exists s^*$ such that:
$$f(s^*)=\int_X\mathbf{1}_{A(s^*)}(x)-\int_Y\mathbf{1}_{B(s^*)}(y)\geq \mathbb{E}[f(U)]$$
And we conclude. Continuity of $f$ is not needed, as we do not need to bound $\mathbb{E}[f(U)]$ by  $\max_{s\in [0,1]} f(s)$ to the ends of the proof.
Still I would be curious about Jensen's inequality in infinite dimension: it would allow to tackle similar problems where this kind of "one-dimensional reduction" is not available.
