Vector Subspace of Normed Dual Space is Weak-* Dense iff 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 $F \subseteq X^\ast$ be a vector subspace of $X^\ast$.

Then $F$ is everywhere dense in $\struct {X^\ast, w^\ast}$ if and only if it separates points.

Proof

Necessary Condition

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

From Weak-* Dense Subset of Normed Dual Space Separates Points, $F$ separates points.

$\Box$

Sufficient Condition

Let $F$ be a vector subspace separating the points of $X^\ast$.

Suppose that $F$ is not everywhere dense in $\struct {X^\ast, w^\ast}$.

From Set is Closed iff Equals Topological Closure, we then have that $\map {\cl_{w^\ast} } F \ne X^\ast$.

Then $\map {\cl_{w^\ast} } F$ is a proper closed vector subspace in $\struct {X^\ast, w^\ast}$.

From Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Hausdorff Locally Convex Space, there exists $\Phi \in \struct {X^\ast, w^\ast}^\ast$ such that $\Phi \ne 0$ and:

$\map \Phi f = 0$ for $f \in \map {\cl_{w^\ast} } F$.

From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $x \in X$ such that:

$\Phi = x^\wedge$

We then have:

$\map {x^\wedge} f = 0$ for $f \in F$.

That is:

$\map f x = 0$ for all $f \in F$.

Since $F$ separates points, we obtain $x = 0$, contradicting that $\Phi \ne 0$.

Hence we have reached a contradiction.

So $F$ is everywhere dense in $\struct {X^\ast, w^\ast}$.

$\blacksquare$