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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?

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    $\begingroup$ $S_X$ is weakly dense in $B_X$, hence it is weak$^*$ly dense in $B_{X^{**}$. $\endgroup$ May 28 at 14:04
  • $\begingroup$ @M.González $S_X$ being weakly dense in $B_X$ holds if and only if $X$ is infinite dimensional, and the density in $B_{X^{\ast\ast}}$ is not what I want... $\endgroup$ May 28 at 14:07
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    $\begingroup$ $X$ infinite dimensional. $S_X \subset S_{X^{**}} \subset B_{X^{**}}$ and $S_X$ is weak* dense in $B_{X^{**}}$ and therefore weak* dense in $S_{X^{**}}$. $\endgroup$ May 28 at 15:11
  • $\begingroup$ @Gerald Edgar, I see the point. Very nice argument! $\endgroup$ May 28 at 15:21
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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.

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