Is $X^\ast$ is weak* separable equivalent to $B_{X^\ast}$ is weak* separable?

(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

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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.

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)

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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.