# 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:

1. If đť‘‹ 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 $$S^*$$

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
• By the Josefson-Nissenzweig theorem, $0$ is in the weak$^*$ sequential closure of $S^*$ if $X$ is infinite dimensional. Commented Jul 29, 2022 at 18:49