Let $X$ be a separable Banach space, the associated dual space is denoted by $X^*$ and the usual duality between $X$ and $X^*$ by $\langle , \rangle$. For $C$ nonempty weakly compact convex subsets of $X$. we set $$ s(x^*, C) := \sup_ {x\in C} \langle x^* , x\rangle \qquad ;\qquad \|C\| := \sup_{x\in C}\|x\| $$ to define for the set C respectively its support function $s(., C)$ and radius $\|C\|$.
The support function satisfies the following simple properties that are stated for easy references. $$ s(x^*, C + C')= s(x^*, C) + s(x^*, C'), \qquad x^*\in X^*,~ C, C' \in 2^X \setminus \{\emptyset\}. $$ It is subadditive. $$ s(x^* + y^*,C)\leq s(x^*,C) + s(y^*,C), \qquad x^*,y^*\in X^*,~ C \in 2^X \setminus \{\emptyset\}. $$ Let $\{C_n\}$ are nonempty weakly compact convex subsets of $X$ such that : there exists $M>0$ $$ \|C_n\|\leq M,\qquad \forall n $$ Show that $\{s(. , C_n)\}$ is equicontinuous on $X^*$ (the strong topology is used on $X^*$).
My effort:
Let $x^*,y^*\in X^*$ and $n\in\mathbb{N}$ we have: $$ s(x^*,C_n)=s((x^*-y^*)+y^*,C_n)\leq s(x^*-y^*,C_n)+ s(y^*,C_n) $$ then $s(x^*,C_n)- s(y^*,C_n)\leq s(x^*-y^*,C_n)$. Without loss of generality we may suppose that $$|s(x^*,C_n)- s(y^*,C_n)|\leq s(x^*-y^*,C_n)$$ then $$|s(x^*,C_n)- s(y^*,C_n)|\leq \|x^*-y^*\|\|C_n\|\leq M\|x^*-y^*\|$$ thus $\{s(. , C_n)\}$ is equicontinuous on $X^*$.
Is what I wrote true? An idea please.
Your idea is right!
It is not clear to me that $$s(x^*-y^*,C)=s(y^*-x^*,C),$$ so one can not assume "without loss of generality that $|s(x^*,C)-s(y^*,C)|\leq s(x^*-y^*,C)$".
Fortunately, this can be adapted: we have $$s(x^*,C)-s(y^*,C)\leq s(x^*-y^*,C)\leq\|x^*-y^*\| \|C\|$$ and $$s(y^*,C)-s(x^*,C)\leq s(y^*-x^*,C)\leq\|y^*-x^*\| \|C\|$$ so that $$|s(x^*,C)-s(y^*,C)|\leq\|x^*-y^*\|\|C\| \, ,$$ and then you have the desired equicontinuity.