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 $A,B$ are two nonempty, convex, weakly compacts subsets.
Let $x_1^*,...,x_n^*\in X^*$ and $q_1,...,q_n\in \mathbb{Q}$, such that: $$ \sup_{x\in A}{\langle x^*,x\rangle}>\sup_{x\in B}{\langle x^*,x\rangle} $$ with $x\sum_{i=1}^{n}{q_ix_1^*}$
Can we say that: there exists $i\in \{1,..,.n\}$, such that: $$ \sup_{x\in A}{\langle x_i^*,x\rangle}>\sup_{x\in B}{\langle x_i^*,x\rangle} $$
In general, no. For instance take $X=\mathbb C$, $A=\{1\}$, $B=\{2\}$, $n=1$, $x_1^*=1$, $q_1=-1$. Then $$ \sup_{x\in A}\langle x^*,x\rangle=-1>-2=\sup_{x\in B}\langle x^*,x\rangle $$ while $$\sup_{x\in A}\langle x_1^*,x\rangle=1<2=\sup_{x\in B}\langle x_1^*,x\rangle. $$
If you require $q_i\geq0$ for all $i$, then yes: because if $$ \sup_{x\in A}{\langle x_i^*,x\rangle}\leq\sup_{x\in B}{\langle x_i^*,x\rangle} $$ then $$ \sup_{x\in A}{\langle x^*,x\rangle}\leq\sup_{x\in B}{\langle x^*,x\rangle}. $$