Let $X$ be a Banach space and : $$ \mathcal{P}_{wkc}(X)=\{A\subset X\, |\, A\,\text{is nonempty, weakly compact, convex\}}. $$ Let $A\in \mathcal{P}_{wkc}(X)$, we define the radius of $A$ by: $$ \|A\|:=\sup_{x\in A}{\|x\|} $$ Show that : $$ \|A\|=\sup_{x^*\in B^*}{s(x^*|A)} $$ with : ${s(x^*|A)}:=\sup_{x\in A}{\langle x^*,x\rangle}$ and $B^*$ is the unit ball of $X^*$.
My effort:
according to the definition of $ s (x^* | A) $, we have $$ s (x^* | A) \leq \|x^*\|\|A\| $$ Then : $s(x^*|A)\leq \|A\|$ for all $x^*\in B^*$, which implies : $\sup_{x^*\in B^*}{s(x^*|A)}\leq \|A\|$.
My problem is to show other inequality. An idea please.
Take a sequence $(x_n)$ of elements of $A$ such that for all $n$, $\lVert x_n\rVert\geqslant \lVert A\rVert-n^{-1}$. Then for each $x^*\in B^*$ and each $n\geqslant 1$, since $x_n\in A$, $$ s\left(x^*\mid A\right)\geqslant \langle x^*,x_n\rangle, $$ hence $$ \sup_{x^*\in B^*}s\left(x^*\mid A\right)\geqslant \sup_{x^*\in B^*}\langle x^*,x_n\rangle. $$ Now, you probably know that as a corollary of Hahn-Banach theorem, $\sup_{x^*\in B^*}\langle x^*,x_n\rangle=\lVert x_n\rVert$ hence you can conclude.