Any reference on Jensen inequality for measurable convex functions on a Banach space?

The only proof of Jensen inequality (and most general version) that I know is a direct consequence of the Fenchel-Moreau Theorem : If $$X$$ is a locally convex Hausdorff topological space, let $$\mu$$ be a Borel probability measure and $$x\in X$$ be such that for all continuous linear functional $$x^\ast\in X^\ast$$, $$\int_X \langle y,x^\ast\rangle~d\mu=\langle x,x^\ast\rangle$$ then we say that $$\mu$$ averages to $$x$$, in symbol $$\mu\sim x$$. The Fenchel-Moreau theorem states that the bidual of a proper l.s.c. convex function $$f$$ is the function itself. Recall that $$f^\ast(x^\ast)=\sup_{x\in X} \langle x,x^\ast \rangle-f(x)$$ and $$f^{\ast\ast}(x)=\sup_{x^\ast\in X^\ast} \langle x,x^\ast\rangle-f^\ast(x^\ast)$$, the theorem states that on $$X$$, $$f=f^{\ast\ast}$$.

Suppose that $$\mu\sim x$$ and $$f$$ is a proper l.s.c. convex function, then by Fenchel's inequality, for any $$y\in X$$ and any $$x^\ast\in X^\ast$$, $$\langle y,x^\ast\rangle\leq f(y)+f^\ast(x^\ast)$$, taking the integral over $$\mu$$ we get \begin{align*} \langle x,x^\ast\rangle &= \int_X \langle y,x^\ast\rangle d\mu(y)\\ &\leq \int_X \left[f(y)+f^\ast(x^\ast)\right] d\mu(y)\\ &= \int_X f ~d\mu + f^\ast(x^\ast) \end{align*} and therefore $$\langle x,x^\ast\rangle-f^\ast(x^\ast)\leq \int_X f ~d\mu$$ for all $$x^\ast\in X^\ast$$, taking the supremum of the LHS over $$x^\ast\in X^\ast$$ we get $$f(x)=f^{\ast\ast}(x)\leq \int_X f~d\mu$$.

I am wondering if the result can be extended to any measurable convex function and if there is any literature on the subject. Is there something like

For any Borel probability measure $$\mu$$ such that $$\mu\sim x\in X$$ and any bounded measurable convex functional $$f:X\to\mathbb R$$, $$f(x)\leq \int_X f~d\mu$$.

Or is there any reason to believe that this would be false ? Also if true can we generalize to other measurable spaces $$X$$ where all points can be separated by measurable linear functional ?

I think that I have a hint of proof whenever $$X$$ is a convex open subset of a Banach space and $$f$$ is a bounded convex functional, I think this is along the lines of what @MaoWao suggests. I would like to generalize this proof to the case where $$X$$ is a closed and bounded subset of a Banach space. We prove that in this case $$f$$ is actually lower semi continuous, by way of contradiction. Suppose that there is $$x\in X$$ and $$x_n\to x$$ such that $$\liminf_n f(x_n) =f(x)-\delta$$ with $$\delta>0$$ (which we want to contradict). Since $$X$$ is an open set, there is $$\varepsilon>0$$ such that $$B_{2\varepsilon}(x)\subseteq X$$, for any $$n$$, define $$y_n=x-\varepsilon\frac{x_n-x}{\| x_n-x \|}\in B_{2\varepsilon}$$, in particular $$y_n\in X$$. Observe that for any $$n$$, $$x= (1-\alpha_n)x_n + \alpha_n y_n$$ with $$\alpha_n=\frac{\| x_n-x\|}{\|x_n-x\|+\varepsilon}$$ and therefore by convexity, \begin{align*} \Leftrightarrow&&(\| x_n-x\|+\varepsilon)f(x) &\leq \varepsilon f(x_n)+\| x_n-x\| f(y_n)\\ \Leftrightarrow&&\|x_n-x\| f(x) +\varepsilon (f(x)-f(x_n))\leq \|x_n-x\| f(y_n)\\ \Rightarrow && 0<\varepsilon \delta \leq \liminf_n\|x_n-x\| f(y_n)\\ \end{align*} But if $$f$$ is bounded then the RHS is $$0$$ which is a contradiction. Now since $$f$$ is l.s.c. then Jensen inequality applies and we are done.

I am much more interested in the case where $$X$$ is a closed, convex and bounded subset of Banach space, in this case it feels like a similar argument could be made by working in the largest relatively open subset of $$X$$ containing $$x$$, i.e. the largest set containing $$x$$ in it's relative interior, but there are many point I do not master here, any reference on that would be welcome, the only one I know is Rockafelar for finite dimension.

• It is likely that this question can be challenging. If there's no luck here, you could try on MathOverflow. Commented Oct 12, 2022 at 11:15
• Do you want any measurable convex function or any bounded measurable function? That makes a huge difference. If a convex function on an arbitrary TVS is only bounded on a non-empty open set, it is continuous everywhere. Commented Oct 12, 2022 at 14:47
• @MaoWao In my particular setting I am interested in a bounded measurable convex function on a bounded subset of Banach space. I think that your comment implies for this setting that (even without measurability) the function is lower semi continuous. Is that right ? Is there any reference I could read on the subject ? Commented Oct 12, 2022 at 19:02
• @P.Quinton The subset needs to be open (and convex of course), that's very important for this result. And yes, such a function is necessarily lower semicontinuous, but much more than that, even locally Lipschitz. You can find a proof here: users.mat.unimi.it/users/libor/AnConvessa/continuity_all.pdf. I don't have a good book reference for this kind of questions at hand. Commented Oct 13, 2022 at 15:52
• See also here: individual.utoronto.ca/jordanbell/notes/semicontinuous.pdf There are some references, which I cannot check right now. Commented Oct 13, 2022 at 15:55