# Is the unit sphere of a Banach space dense in the unit sphere of its second dual with respect to the weak-$\ast$ topology

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?

• $S_X$ is weakly dense in $B_X$, hence it is weak$^*$ly dense in $B_{X^{**}$. May 28 at 14:04
• @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... May 28 at 14:07
• $X$ infinite dimensional. $S_X \subset S_{X^{**}} \subset B_{X^{**}}$ and $S_X$ is weak* dense in $B_{X^{**}}$ and therefore weak* dense in $S_{X^{**}}$. May 28 at 15:11
• @Gerald Edgar, I see the point. Very nice argument! May 28 at 15:21

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.