I have a question that might be easy for the experts.
Let $E$ be a Banach space (non-separable) and let $E'$ be its dual space. Suppose that $X\subset E'$ and assume that $X$ is separable with respect to the weak* topology.
My question is the following:
Are the sequential weak* closure and the weak* closure of $X$ equal?
Google was not able to help me on this.
On
Let $E=E'=L_{2}(R)$ then $ X=l_{2}(N) \subset E'= E $, this is well-defined by considering a trivial extension from functions in $l_{2}$ to $L_{2}(R)$ (define $f=0$ for all $x \in R\setminus N$ ). Then $X$ is a separable Hilbert space and weak-star and weak topology on $X$ coincides!
Obviously sequentially weakly closed does not imply weakly closedness in $X$. one counter example is the set $\{ \sqrt n e_n~ | \quad n \in N \}.$
The answer is no.
For a specific, non-separable counter-example take $E=\ell_\infty$. Then the set $$X=\{e_n^*\colon n\in \mathbb{N}\}\subseteq \ell_\infty^*$$ consisting of evaluation functionals is bounded but not weak* compact. (Here $\langle e_n^*, f\rangle = f(n)$ for $f\in \ell_\infty$.) However, by boundedness, the weak* closure of $X$ is compact in the weak* topology hence different from $X$. There are no non-eventually constant convergent sequences in $X$ so the cluster points of $X$ cannot be realised as limits of sequences.
To be more precise, the weak*-sequential closure of $X$ is $X$ itself but $X$ is not weak*-closed (because it is not weak*-compact) so the weak* closure of $X$ is strictly bigger than $X$.