I need some help to approach the following exercise.
Let $B$ be a separable Banach space and let $\gamma : [0,1] \rightarrow B$ be a continuous curve with respect to the weak topology of $B$. Prove that $\gamma$ is a Borel function with respect to the Borel sets of the strong topology of $B$. Is this still true if $B$ is not separable?
The key is that in any separable Banach space, the weak and strong topologies generate the same Borel $\sigma$-algebra. Recall that a strongly closed and convex set is weakly closed, so in particular every closed ball is weakly closed. And every strongly open set is a countable union of closed balls (by second countability), so every strongly open set is weakly Borel.
I am not sure about the non-separable case, but I would guess this is false. I would look for a counterexample in a non-separable Hilbert space like $\ell^2(\mathbb{R})$ or $\ell^2(\omega_1)$.
