To be a bit more precise and fix notations, let $X$ be a Banach space (over $\mathbb{R}$ or $\mathbb{C}$), $X^{\ast\ast}$ its second dual (as a Banach space). Here and in the following we identify $X$ as a (norm) closed subspace of $X^{\ast\ast}$ via the canonical embedding $J : X \hookrightarrow X^{\ast\ast}$. Now let $S\_X$ (resp. $S\_{X^{\ast\ast}}$) denote the unit sphere (the set of vectors with norm $1$) of $X$ (resp. $X^{\ast\ast}$).


**Question 1** Is it true that $S\_X$ is dense in $S\_{X^{\ast\ast}}$ with respect to the $\sigma(X^{\ast\ast}, X^\ast)$ topology?


**Question 2** Do the above have an affirmative answer in the special case where $X = A$ is a $C^\ast$-algebra, hence $X^{\ast\ast}=A^{\ast\ast}$ is its enveloping von Neumann algebra?


**Question 3** This is a little more general than Question 2. Let $M$ be a von Neumann algebra acting on some Hilbert space $H$, $A$ a $\ast$-subalgebra of $M$ (not necessarily norm closed) such that $A$ is non-degenerate as an algebra of operators on $H$ and such that the double commutant of $A$ is $M$. Is it true that $S\_A := \{a \in A \mid \|a\| = 1\}$ is dense in $S\_M := \{x \in M \mid \|x\| = 1\}$ with respect to the weak operator topology?


If we replace unit spheres by closed unit balls, all of the above questions have an affirmative answer (Question 1 is Goldstine's theorem, and Question 2 and 3 part of Kaplansky's density theorem). I was wondering whether the above finer statements still hold. Do we have (counter-)examples?



Is the unit sphere of a Banach space dense in the unit sphere of its second dual with respect to the weak-$\ast$ topology
I think a simple rescaling argument works. I do Question 1, Q3 being similar. Given $F\in S\_{X^{\*\*}}$, by Goldstine there is a net $(x\_i)$ in $B\_X$ converging weak$^\*$ to $F$. Given $\epsilon>0$ there is $f\in S\_{X^\*}$ with $|F(f)|>1-\epsilon$ and so
$$ 1-\epsilon < |F(f)| = \lim\_i |f(x\_i)| \leq \|f\|\liminf\_i\|x\_i\| \leq \liminf\|x\_i\| \leq 1, $$
from which it follows that $\liminf \|x\_i\|=1$. So for $\epsilon>0$ there is $i\_0$ with $\|x\_i\| > 1-\epsilon$ for $i\geq i\_0$, but as $\|x\_i\|\leq 1$, we conclude that actually $\lim\_i \|x\_i\| = 1$.


Thus, wlog $x\_i\not=0$ for all $i$, and so we can define $y\_i = \|x\_i\|^{-1} x\_i \in S\_X$ for each $i$. Then $\lim\_i \|x\_i - y\_i\| = \lim\_i \|x\_i\| |1-\|x\_i\|^{-1}| = 0$, and so also $(y\_i)\rightarrow F$ weak$^\*$, as required.


