Show that if $F,G \subseteq E$ are compact convex sets such that $\sigma_F=\sigma_G$ then $F=G$ (this requires a separation argument) where$$\sigma _F (x) := \max\{\langle x, u\rangle : u ∈ F\}.$$
Proof
Let $F$ and $G$ be non-empty compact convex sets such that $\sigma_F=\sigma_G$. We want to use Proper Separation that is, $rai(F) \cap rai(E)= \emptyset$
I am trying to prove this statement by using a separation argument. Can anyone help
Suppose to the contrary that $F \ne G$, wlog (due to symmetry of the problem), $F \not\subseteq G$. Then there is $u \in F$ with $u \not\in G$. By the seperation theorem, there is $x \in E^*$, such that $$ \sup_{v \in G} \def\<#1>{\left<#1\right>}\<x,v> < \<x,u> $$ But then $$ \sigma_G(x) < \<x,u> \le \sigma_F(x) $$ Contradiction.