Interchange supremum and expectation

Let $B_n:=\{f\in L^\infty_+\mid f\le n \}$, where we consider $L^\infty$ with the weak$^*$ topology. I have the following sets

$$D(z):=\{h\in L^0_+(\mathcal{F}_T)\mid h\le Z_T \mbox{ for a }Z\in Z(z)\}$$
where $Z(z)$ is the set of positive (RCLL) supermartingales on $[0,T]$ with $E[Z_0]\le z$. Moreover $U$ is a function from $\mathbb{R}_+\to\mathbb{R}$, strictly increasing, strictly concave and $C^1$. For a $h\in D(z)$ I want to prove

$$\sup_{f\in B_n} E[U(f)-hf]=E[\sup_{f\in B_n}\{U(f)-fh\}]$$

One direction is clear: $E[U(f)-hf]\le E[\sup_{f\in B_n}\{U(f)-fh\}]$, hence

$$\sup_{f\in B_n} E[U(f)-hf]\le E[\sup_{f\in B_n}\{U(f)-fh\}]$$

the inequality $”\ge “$ is bothering me. How can I tackle this problem? I know that the equality is true. Thanks for your help

Solutions Collecting From Web of "Interchange supremum and expectation"

Fix $U:\mathbb R^+\to \mathbb R$ as above.

For every $y\in\mathbb R$ the function $g_y: [0,n]\to \mathbb R$ defined by
$$ g_y(x) = U(x) -xy $$ is continuous in the compact interval [0,n].

Therefore $g_y$ is bounded and attains its bounds and we may set
$$G(y) = \sup_{x\in[0,n]} g_y(x)$$
and define a function $F:\mathbb R\to [0,n]$ such that
$$g_y(F(y)) =G(y).$$

That is $F(y)$ is the maximizer of $g_y$.

Therefore we have
$$E(\sup_{f\in B_n} U(f) – fh) \leq E(\sup_{x\in[0,n]} U(x) – xh) = E(G(h)).$$

It’s easy to see from your conditions on $U$ that $F$ is continuous.

Thus the composition $X(\omega) = F(h(\omega))$ is measurable and we have $X\in B_n$.

So $$\sup_{f\in B_n}E( U(f) – fh) \geq E( U(X) – Xh) = E(g_h(X)) = E(G(h)).$$

Hence $$\sup_{f\in B_n}E( U(f) – fh) \geq E(\sup_{f\in B_n} U(f) – fh).$$