Weak-* Metrizability of Closed Unit Ball in Normed Dual of Normed Vector Space implies Original Space is Separable
Theorem
Let $X$ be a normed vector space such that:
- $\struct {B_{X^\ast}^-, w^\ast}$ is metrizable
where $X^\ast$ is the normed dual of $X$ and $B_{X^\ast}^-$ is the closed unit ball of $X^\ast$.
Then $X$ is separable.
Proof
Let:
- $K = \struct {B_{X^\ast}^-, w^\ast}$
From the Banach-Alaoglu Theorem, $K$ is compact.
From Weak-* Topology is Hausdorff, $K$ is Hausdorff.
Since $K$ is metrizable, $\map \CC K$ is separable from Space of Complex-Valued Continuous Functions on Compact Hausdorff Space is Separable iff Space is Metrizable.
Define $T : X \to \map \CC K$ by:
- $T x = x^\wedge \restriction_{B_{X^\ast}^-}$
Clearly $T$ is a linear transformation.
Further, we have for each $f \in X^\ast$ with $\norm f_{X^\ast} = 1$:
- $\ds \cmod {\map {\paren {T x} } f} = \cmod {\map f x} \le \norm x$
from Fundamental Property of Norm on Bounded Linear Functional.
From Existence of Support Functional, for each $x \in X$ there exists $f \in X^\ast$ with $\norm f_{X^\ast} = 1$ and $\map f x = \norm x$.
Hence we have:
- $\norm {T x}_{\map \CC K} = \norm x$
So $T$ is a linear isometry, hence so is $T^{-1} : T \sqbrk X \to X$.
Since $\map \CC K$ is separable, $T \sqbrk X$ is separable.
From Separability of Normed Vector Space preserved under Isometric Isomorphism, $X$ is separable.
$\blacksquare$
See also
- Closed Unit Ball in Normed Dual Space of Separable Normed Vector Space is Weak-* Metrizable shows that the converse of this result is also true.
Sources
- 2001: Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos Santalucía, Jan Pelant and Václav Zizler: Functional Analysis and Infinite-Dimensional Geometry ... (previous) ... (next): Lemma $3.23$