Weak-* Metrizability of Closed Unit Ball in Normed Dual of Normed Vector Space implies Original Space is Separable

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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


Sources

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