Weak star closure of unit sphere in dual space Let $X$ be a normed vector space with its dual $X^*$. Let $S^*$ be the unit sphere of $X^*$. We have known that if $X$ (or $X^*$ ) is reflexive then the weak-star and weak topology of $X^*$ coincide and thus the weak-star closure (or weak closure) of $S^*$ is the unit ball. I have the following questions:

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*If  is reflexive, what is the weak-star sequential closure of $S^*$?

*If $X$ is non-reflexive, what are the weak-star and weak-star sequential closure of $S^*$
 A: Suppose the weak$^*$ sequential closure of $S^*$ contains $0$.  So there is a sequence $(f_n) \subseteq X^*$ with $\|f_n\|=1$ for each $n$, and with $f_n\rightarrow 0$ weak$^*$.  For $f\in B^*$ the unit ball, we seek a bounded sequence $(t_n)$ of scalars such that $f+t_nf_n \in S^*$ for each $n$, as then $f+t_nf_n \rightarrow f$ weak$^*$.  We can find such $t_n$ from the triangle inequality,
$$ \big| \|f\| - |t_n| \big| \leq \|f + t_nf_n\| \leq \|f\| + |t_n|, $$
and using that $t\mapsto \|f+tf_n\|$ is continuous.  Conclude: the weak$^*$-sequential closure of $S^*$ is all of $B^*$.
So, when is $0$ in the weak$^*$ sequential closure of $S^*$?
Here is an easy argument if $X$ is separable.  Then $X$ hash a countably dense subset $\{x_k\}$, then for each $n$ by Hahn-Banach we can find $f_n\in S^*$ with $f_n(x_k)=0$ for $k\leq n$.  Given $x\in X$ and $\epsilon>0$ there is $k$ with $\|x-x_k\|<\epsilon$, and so for $n\geq k$ we have $|f_n(x)| = |f_n(x-x_k)| < \epsilon$.  Conclude that $f_n\rightarrow 0$ weak$^*$.
As Dirk Werner points out, in the general case, we can use the Josefson-Nissenzweig Theorem which exactly says that any (infinite-dimensional) Banach space has the property we need: $X^*$ has a weak$^*$-null sequence of norm one vectors.  Here is one source for this theorem: Extract from Diestel, Sequences and Series in Banach Spaces
