Is $B_{\ell_1}$ weak-metrizable?

Question

I know that for a Banach space $X$, the unit ball $B_{X}$ is weak metrizable if and only if $X^*$ is separable. My question is that

Is $B_{\ell_1}$ weak-metrizable?

Answer 1

As Norbert said, this is false. When considering sequences, we don't get the accurate picture of weak topology on the unit ball of $\ell^1$: in fact, we can't even tell it from the norm topology.

Theorem (Schur) A sequence in $\ell^1$ converges weakly if and only if it converges in the norm.

An accessible proof of this theorem is easy to find online: see this blog post or these notes; there is also a Wikipedia page (though without a proof).