Normed Dual of Normed Vector Space is Separable iff Closed Unit Ball is Metrizable

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Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a normed vector space over $\GF$.

Let $w$ be the weak topology on $X$.

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

Let $B_X^-$ be the closed unit ball of $X$.


Then $X^\ast$ is separable if and only if $\struct {B_X^-, w}$ is metrizable.


Proof

Let $X^{\ast \ast}$ be the second normed dual.

Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.

Necessary Condition

Suppose that $X^\ast$ is separable.

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

$\struct {B_{X^{\ast \ast} }^-, w^\ast}$ is metrizable.

From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have:

$\iota B_X^- \subseteq B_{X^{\ast \ast} }^-$.

So:

$\struct {\iota B_X^-, w^\ast}$ is metrizable.

From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Homeomorphism onto Image, $\struct {X, w}$ is homeomorphic to $\struct {X^{\ast \ast}, w^\ast}$.

From Restriction of Homeomorphism is Homeomorphism, $\struct {B_X^-, w}$ is homeomorphic to $\struct {\iota B_X^-, w^\ast}$.

So $\struct {B_X^-, w}$ is metrizable.

$\Box$

Sufficient Condition

Suppose that $\struct {B_X^-, w}$ is metrizable.

From Metric Space is First-Countable, $\struct {B_X^-, w}$ is first-countable.

Let $\sequence {U_n}_{n \in \N}$ be a countable local basis for ${\mathbf 0}_X$.

Then there exists a finite subset $F_n \subseteq X^\ast$ and $\epsilon_n > 0$ such that:

$U_n = \set {x \in B_X^- : \cmod {\map g x} < \epsilon_n \text { for each } g \in F_n}$

Let:

$\ds Z = \map \cl {\map \span {\bigcup_{n \mathop = 1}^\infty F_n} }$

From Countable Union of Finite Sets is Countable, we have:

$\ds \bigcup_{n \mathop = 1}^\infty F_n$ is countable.

From Closed Linear Span of Countable Set in Topological Vector Space is Separable, we have that $Z$ is separable.

To finish, we show that $Z = X^\ast$.

Let $f \in X^\ast$ and $\epsilon > 0$.

From Open Sets in Weak Topology of Topological Vector Space:

$V = \set {x \in B_X^- : \cmod {\map f x} < \epsilon}$ is open in $\struct {B_X^-, w}$.

Then there exists $n \in \N$ such that $U_n \subseteq V$.

Let:

$\ds F = \bigcap_{g \in F_n} \ker g$

From Set of Linear Subspaces is Closed under Intersection, $F$ is a vector subspace of $X$.

Then we have:

$g \restriction_F = 0$ for each $g \in F_n$.

So since $U_n \subseteq V$, we have:

$\cmod {\map f x} < \epsilon$

for each $x \in F$.

So, we have:

$\norm {f \restriction_F}_{F^\ast} \le \epsilon$

Hence by:

Hahn-Banach Theorem: Real Vector Space if $\GF = \R$
Hahn-Banach Theorem: Complex Vector Space if $\GF = \C$

there exists $\tilde f \in X^\ast$ such that:

$\tilde f \restriction_F = \tilde f \restriction_F$

with:

$\norm {\tilde f}_{X^\ast} \le \epsilon$

So:

$\ds \bigcap_{g \in F_n} \ker g \subseteq \map \ker {f - \tilde f}$

From Condition for Linear Dependence of Linear Functionals in terms of Kernel, we have:

$f - \tilde f \in \map \span {F_n} \subseteq Z$

So, we have:

$\norm {f - \paren {f - \tilde f} }_{X^\ast} \le \epsilon$

with $f - \tilde f \in Z$.

So:

$\map d {f, Z} \le \epsilon$

This holds for all $\epsilon > 0$, so we have $\map d {f, Z} = 0$.

From Point at Distance Zero from Closed Set is Element, we have $f \in Z$.

So we indeed have $Z = X^\ast$.

$\blacksquare$


Sources

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