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?

  • 8
    $\begingroup$ $S_X$ is weakly dense in $B_X$, hence it is weak$^*$ly dense in $B_{X^{**}$. $\endgroup$ May 28, 2021 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, 2021 at 14:07
  • 6
    $\begingroup$ $X$ 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, 2021 at 15:11
  • $\begingroup$ @Gerald Edgar, I see the point. Very nice argument! $\endgroup$ May 28, 2021 at 15:21

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.