Exchanging the expectation and supremum

Question

Suppose $\epsilon$ is a Rademacher variable such that $\mathrm{P} (\epsilon = + 1) = \mathrm{P} (\epsilon = - 1) = \frac{1}{2}$. And suppose $u \in V \subseteq \mathbb{R}$, I see from some articles that \begin{equation*} \mathrm{E} \sup_{u \in V} \epsilon f (u) = \frac{1}{2} \sup_{u, v \in V} \big( f(u) - f(v)\big) \end{equation*} for any mapping $f : \mathbb{R} \rightarrow \mathbb{R}$. But I don't know how to prove this.

Could anyone help me？ Thanks so much in advance.

Answer 1

We can write the left side as $\frac 1 2\sup_u f(u)+\frac1 2 \sup_u [-f(u)]$.
The stated equality follows from the following:

$\sup_u f(u)-\inf_u f(u)=\sup_{u,v} (f(u)-f(v)$

and

$\sup_u (-1)f(u)=-\inf_u f(u)$.