= Solution
Suppose first that $X$ is separable, and choose a norm-dense sequence $(x_n)$ in its unit ball. On the dual unit ball define
$$
d(f,g)=\sum_{n=1}^\infty2^{-n}
\frac{|(f-g)(x_n)|}{1+|(f-g)(x_n)|}.
$$
Uniform boundedness on the unit ball and density of the $x_n$ show that this metric induces the weak-star topology. Conversely, if $B_{X^*}$ is weak-star metrizable, the <Banach-Alaoglu theorem> makes it a compact metric space. Hence $C(B_{X^*})$ is separable. The evaluation map
$$
X\longrightarrow C(B_{X^*}),\qquad
x\longmapsto(f\mapsto f(x))
$$
is an isometry by the <Hahn-Banach theorem>. A subspace of a separable metric space is separable, so $X$ is separable. This proves the <weak-star metrizability criterion for a dual ball>.
If $X$ has a countable weakly dense subset $D$, the rational linear span of $D$ is weakly dense. Its norm closure is a convex set, so <Mazur theorem> says that its weak and norm closures agree. Thus $X$ is norm separable. The weak-star compact metric ball $B_{X^*}$ consequently has a countable weak-star dense subset, and the union of its integer dilates is weak-star dense in $X^*$. Therefore $X^*$ is weak-star separable.
It need not be weakly separable. Take $X=\ell^1$, whose dual is $\ell^\infty$. A weakly separable normed space is norm separable by the preceding convex-closure argument, whereas $\ell^\infty$ is not norm separable.
If the Banach space $X$ is reflexive, its closed unit ball identifies with the weak-star compact ball of $X^{**}$, hence is weakly compact. Conversely, if $B_X$ is weakly compact, its canonical image $J(B_X)$ is weak-star compact and therefore weak-star closed in $X^{**}$. <Goldstine theorem> says it is weak-star dense in $B_{X^{**}}$, so
$$
J(B_X)=B_{X^{**}}.
$$
Scaling proves that $J$ is surjective and $X$ is reflexive. This is the <weak compactness characterization of reflexivity>.
The <Krein-Milman theorem> says that a nonempty compact convex subset of a locally convex space is the closed convex hull of its extreme points. For reflexive $X$, the ball $B_X$ is weakly compact, so
$$
B_X=\overline{\operatorname{conv}}\operatorname{Ext}(B_X),
$$
where weak and norm closure agree for the convex hull by Mazur's theorem.
For the final claim, let $\mathcal H$ be the set of functions $\mathbb Z^2\to[0,1]$ with the mean-value property. It is a compact convex subset of the product $[0,1]^{\mathbb Z^2}$. If $f$ is extreme, its four unit translates also lie in $\mathcal H$, and the mean-value identity writes $f$ as their average. Extremality forces every translate to equal $f$, so $f$ is constant. Every extreme point is therefore constant. Krein--Milman now makes every member of $\mathcal H$ a limit of convex combinations of constant functions, and hence constant. This is the <bounded harmonic function theorem on the integer lattice>.
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