Let $X$ be a separable Banach space, the associated dual space is denoted by $X^*$ and the usual duality between $X$ and $X^*$ by $\langle , \rangle$. Let $B^*$ the closed unit ball of $X^*$.
Take $C$ nonempty weakly compact convex subsets of $X$. Can we say that:
$$ \sup_{x\in C}\|x\|=\sup_{x^*\in B^*}\sup_{x\in C}\langle x^*,x\rangle $$
It is obvious that $$ \sup_{x^*\in B^*}\sup_{x\in C}\langle x^*,x\rangle\leq \sup_{x\in C}\|x\| $$ But for inverse inequality. $C$ should it be bounded?
The reverse inequality holds without any assumptions on $C$.
Let $x \in C$. By the Hahn-Banach theorem, there exists $x^* \in B^*$ such that $\lVert x \rVert = \langle x^*,x\rangle$. Thus $$\lVert x \rVert \leq \sup_{y \in C} \,\langle x^*,y\rangle \leq \sup_{y^* \in B^*}\sup_{y\in C} \,\langle y^*,y\rangle.$$ Taking the supremum over $x \in C$ yields the result.