Let $X$ be a reflexive Banach Space and let $X^{*}$ be it’s dual, which is assumed to be seperable. Is the weak topology then metrizable on $X$?
My thoughts: X beeing reflexive implies (Eberlein-Smuljan Theorem) that every bounded sequence has a convergent subsequence. Thus, because the dual space is seperable, the closed unit ball is metrizable w.r.t. the weak topology. Hence every closed ball is metrizable w.r.t. the weak topology, hence the weak topology on X is metrizable.
Is this correct?
