Are the weak* and the sequential weak* closures the same? 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. 
 A: 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 $X$.
A: 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 $\{ \sqrt n  e_n~ | \quad n \in N  \}.$   
