Weak-* Dense Subset of Normed Dual Space Separates Points

Theorem

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

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

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

Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.

Let $D$ be everywhere dense in $\struct {X^\ast, w^\ast}$.

Then $D$ separates points.

Proof

Suppose that $D$ is everywhere dense in $\struct {X^\ast, w^\ast}$.

Let $x, y \in X$ be such that:

$\map f x = \map g x$ for each $f, g \in D$.

Then:

$\map {x^\wedge} f = \map {x^\wedge} g$ for each $f, g \in D$.

From Characterization of Continuity of Linear Functional in Weak-* Topology, $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.

From Metric Space is Hausdorff, $\GF$ is Hausdorff.

By hypothesis, $D$ is everywhere dense in $\struct {X^\ast, w^\ast}$.

Hence from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide, we obtain:

$\map {x^\wedge} f = \map {x^\wedge} g$ for each $f, g \in X^\ast$.

Then:

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

From Normed Dual Space Separates Points, we obtain $x = y$.

$\blacksquare$