Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable

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Theorem

Let $X$ be a separable normed vector space.

Let $X^\ast$ be the normed dual space of $X$.

Let $B^-_{X^\ast}$ be the closed unit ball of $X^\ast$.

Let $w^\ast$ be the weak-$\ast$ topology on $B^-_{X^\ast}$.


Then $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable.


Proof

Let $\sequence {x_n}_{n \in \N}$ be a countable dense subset of $X$.

Let $\sequence {x_n^\wedge }_{n \in \N}$ be the evaluation maps associated with $\sequence {x_n}_{n \in \N}$.

We show that $\sequence {x_n^\wedge }_{n \in \N}$ separates points.

Suppose that:

$\map {x_n^\wedge } f = \map {x_n^\wedge } g$

for each $f, g \in X^\ast$.

Then:

$\map f {x_n} = \map g {x_n}$

for each $n \in \N$.

Since $f$ and $g$ are continuous and $\sequence {x_n}_{n \in \N}$ is dense in $X$, we have that $\sequence {x_n^\wedge }_{n \in \N}$ separates points as desired.

Now let $\sigma$ be the initial topology generated by $\sequence {x_n^\wedge \restriction_{B^-_{X^\ast} } }_{n \in \N}$

From Initial Topology Generated by Countable Family of Functions Separating Points is Metrizable, $\sigma$ is metrizable.

Note that the weak-$\ast$ topology on $B^-_{X^\ast}$ is generated by $\set {y^\wedge \restriction_{B^-_{X^\ast} } : y \in X}$ from Subspace Topology on Initial Topology is Initial Topology on Restrictions.

So $\sigma \subseteq w^\ast$.

Hence the identity mapping $\iota : \struct {B^-_{X^\ast}, w^\ast} \to \struct {B^-_{X^\ast}, \sigma}$ is continuous.

From the Banach-Alaoglu Theorem, $\struct {B^-_{X^\ast}, w^\ast}$ is compact.

Since $\sigma$ is metrizable, it is Hausdorff from Metric Space is Hausdorff.

From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $\iota$ is a homeomorphism.

Hence $\sigma = w^\ast$, and $\struct {B^-_{X^\ast}, w^\ast}$ is metrizable.

$\blacksquare$


Also see


Sources

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