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:


1. If is reflexive, what is the weak-star sequential closure of $S^\*$?
2. If $X$ is non-reflexive, what are the weak-star and weak-star sequential closure of $S^\*$



Weak star closure of unit sphere in dual space
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](https://mathoverflow.net/users/127871/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](https://link.springer.com/chapter/10.1007/978-1-4612-5200-9_13)


