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(In another question Nate Eldredge said I should ask this.)

Let $X$ be a Banach space, $X^\ast$ the dual space, and $B_{X^\ast}$ the unit ball of $X^\ast$. In the weak* topology for $X^\ast$, does one of these imply the other?

(a) $X^\ast$ is weak* separable

(b) $B_{X^\ast}$ is weak* separable

Martin Sleziak
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GEdgar
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  • If $\overline{D} = B_{X^\ast}$ then $\overline{\bigcup_{n=1}^\infty nD} = \bigcup_{n=1}^\infty nB_{X^\ast} = X^\ast$, so (b) implies (a). – Martin Jun 01 '13 at 17:30
  • @Martin: Indeed. Just a nitpick: Write that as $\overline{\bigcup_{n=1}^\infty nD} \supseteq \bigcup_{n=1}^\infty nB_{X^\ast} = X^\ast$, since the closure of a union does not necessarily equal the union of the closures. The conclusion is unchanged by this, of course. – Harald Hanche-Olsen Jun 01 '13 at 17:40
  • @HaraldHanche-Olsen Right, that was a bit of a sloppy way of writing it. Thanks for the correction. – Martin Jun 01 '13 at 17:57
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    In light of Martin's answer I've added the [tag:set-theory] tag. – Nate Eldredge Jun 01 '13 at 20:47
  • I know an answer, and (if necessary) will post it eventually. – GEdgar Jun 02 '13 at 13:47
  • @NateEldredge I am not sure why the tag (set-theory) is suitable here. Is it because of using: "union of countably many countable sets is countable"? Then probably (elementary-set-theory) would be better. (I have asked this also in the tagging chat room.) – Martin Sleziak Apr 24 '17 at 04:28
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    @MartinSleziak: The paper cited by Martin (Avilés, Plebanek, Rodríguez) uses some pretty heavy set theory. Though GEdgar gives a solution that doesn't. – Nate Eldredge Apr 24 '17 at 05:52
  • @NateEldredge Thanks for the reply. – Martin Sleziak Apr 24 '17 at 07:59

2 Answers2

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(b) implies (a):

Let $D$ be a countable dense set in $B_{X^\ast}$. Then $\overline{\bigcup_{n=1}^\infty nD} \supseteq \bigcup_{n=1}^\infty nB_{X^\ast} = X^\ast$ and $\bigcup_{n=1}^\infty nD$ is countable.

(a) does not imply (b):

provided the preprint Antonio Avilés, Grzegorz Plebanek, José Rodríguez, A weak* separable $C(K)^\ast$ space whose unit ball is not weak* separable (2011) is correct. I don't know if there are easier examples among less special spaces than C(K)*-spaces. If I interpret the introduction correctly, reference [14] could contain further examples.

Martin
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  • Very nice reference! It appears the paper of Avilés, Plebanek and Rodríguez has been accepted by Transactions of the AMS (see here) so that is some evidence in favor of its correctness. – Nate Eldredge Jun 01 '13 at 20:52
  • But earlier examples exist where the space is not required to be of the form $C(K)$ ... – GEdgar Jun 02 '13 at 00:38
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    @GEdgar: yes, I believe you. As I said, I don't know. If you know, then why don't you indicate what they are instead of just pointing out their existence? – Martin Jun 02 '13 at 02:23
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OK, here is the answer I know. This explanation is essentially copied from my paper ("Measurability in a Banach space, II." Indiana Univ. Math. J. 28 (1979) 559--579 LINK) Example 5.6. But the space comes from a paper of W. B. Johnson & J. Lindenstrauss ("Some remarks on weakly compactly generated Banach spaces." Israel J. Math. 17 (1974), 219–230. LINK)

. . . . . . . . . . . . . . . .

In $l^\infty$, let $e_n$ be the unit vector $e_n(k) = \delta_{kn}$. Let $\{A_\gamma : \gamma \in \Gamma\}$ be an uncountable collection of infinite subsets of $\mathbb N$, such that the intersection of any two is finite.

aside. For example, consider the infinite binary tree. It has countably many nodes, but uncountably many branches. Let the nodes be indexed by $\mathbb N$ and let the branches be the sets $A_\gamma$.

Let $\varphi_\gamma \in l^\infty$ be the characteristic function of $A_\gamma$. For finite linear combinations, define a norm $$ \left\|\sum c_\gamma \varphi_\gamma + \sum b_n e_n\right\| = \max\left( \left\|\sum c_\gamma \varphi_\gamma + \sum b_n e_n\right\|_\infty, \left(\sum|c_\gamma|^2\right)^{1/2}\right) . $$
Let $X$ be the completion of these finite linear combinations under this norm. The closed linear span $X_0$ of $\{e_n: n \in \mathbb N\}$ is a subspace isomorphic to $c_0$, and $X/X_0$ is isomorphic to $l^2(\Gamma)$. The dual of $X$ is isomorphic to $l^1 \oplus l^2(\Gamma)$; the pairing identifies the unit vectors of $l^1$ with the linear functionals $e_n^\ast(f) = f(n)$, and the unit vectors of $l^2(\Gamma)$ with the linear functionals $e_\gamma^\ast(f) = \lim_{n \in A_\gamma} f(n)$.

The countable set $\{e_n^\ast : n \in \mathbb N\} \subset X^\ast$ separates points of $X$, so its rational linear span is a countable set that is weak* dense in $X^\ast$.

$X$ admits no countable norming set of functionals, since any countable subset of $X^*$ is contained in $l^1 \oplus l^2(\Gamma_0)$ for some countable $\Gamma_0 \subset \Gamma$.

definition. A norming set for $X$ is a set $R \subseteq B_{X^\ast}$ such that for all $x \in X$,$$\|x\| = \sup\{|u(x)|: u \in R\} .$$

To show that $B_{X^\ast}$ is not weak* separable, it suffices to show: any weak* dense subset of $B_{X^\ast}$ is a norming set. Indeed, let $R$ be weak* dense in $B_{X^\ast}$. Let $x_0 \in X$. Write $\|x_0\| = \alpha$. Let $\epsilon > 0$ be arbitrary. Since $R$ is a norming set, there exists $u_0 \in R$ with $|u_0(x_0)| \gt \alpha - \epsilon$. Replacing $x_0$ by $-x_0$ if necessary, we may assume $u_0(x_0)$ is positive. The set $$ T_\epsilon := \{ u \in B_{X^\ast}: u(x_0)>\alpha-\epsilon\} $$ is a nonempty weak* open set in $B_{X^\ast}$. Since $R$ is weak* dense, we have $R \cap T_\epsilon \ne \varnothing$. So $$ \sup\{ |u(x_0)|: u \in R\} \gt \alpha - \epsilon . $$ This is true for any $\epsilon > 0$, so $$ \sup\{ |u(x_0)|: u \in R\} \ge \alpha = \|x_0\|, $$ as required.

GEdgar
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  • Thanks for adding this example. Of course this is much simpler than the one I found. – Martin Jun 15 '13 at 17:33
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    There is also "Weak star separability" by E.N. Dancer and B. Sims, BULL.AUSTRAL.MATH.SOC.(1979). The authors use the Johnson-Lindenstrauss example as well. – Ramiro de la Vega Sep 17 '13 at 19:20