Let $X$ be a Banach space and for each $t \in [a,b]$ let $Y_t$ be a Banach space. Let $F_t:X \to Y_t$ be a bounded map for each $t$.
I know that for given $u \in X^*$ and for all $w \in X$, $$\frac{|\langle u, w \rangle_{X^*,X}|}{\lVert F_t w \rVert_{Y_t}}$$ is continuous with respect to $t.$ Is it true that $$\sup_{w \in X}\frac{|\langle u, w \rangle_{X^*,X}|}{\lVert F_t w \rVert_{Y_t}}$$ is measurable with respect to $t$?
Since $F_t$ is linear you have $$ \sup_{w \in X}\frac{|\langle u, w \rangle_{X^*,X}|}{\lVert F_t w \rVert_{Y_t}} =\sup_{w \in \operatorname{Sphere}_X}\frac{|\langle u, w \rangle_{X^*,X}|}{\lVert F_t w \rVert_{Y_t}} $$ Since $X$ is separable we have a countable dense subset $S:=\{w_n:n\in\mathbb{N}\}\subset \operatorname{Sphere}_X$. Since $S$ is dense in $\operatorname{Ball}_X$, then $$ \sup_{w \in \operatorname{Sphere}_X}\frac{|\langle u, w \rangle_{X^*,X}|}{\lVert F_t w \rVert_{Y_t}} =\sup_{n\in\mathbb{N}}\frac{|\langle u, w_n \rangle_{X^*,X}|}{\lVert F_t w_n \rVert_{Y_t}} $$ Hence the desired supremum is measurable as supremum of the sequence of measurable functions (note that continuous fnctions are measurable).