Closure in Weak-* Topology in terms of Annihilators

Theorem
Let $X$ be a Banach space.

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

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

Let $N$ be a linear subspace of $X^\ast$.

Then:
 * $\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$

where:
 * $\cl_{w^\ast}$ denotes closure in the weak-$\ast$ topology
 * ${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$
 * $\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$.

Proof
From Linear Subspace is Subset of Double Annihilator, we have:
 * $N \subseteq \paren { {}^\bot N}^\bot$

So, from Set Closure Preserves Set Inclusion:
 * $\map {\cl_{w^\ast} } N \subseteq \map {\cl_{w^\ast} } {\paren { {}^\bot N}^\bot}$

From Annihilator of Subspace of Banach Space is Weak-* Closed and Set is Closed iff Equals Topological Closure, we have:
 * $\map {\cl_{w^\ast} } N \subseteq \paren { {}^\bot N}^\bot$

It remains to show:
 * $\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$

From Set Complement inverts Subsets, we can equivalently show:
 * $X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$

Let:
 * $f \in X^\ast \setminus \map {\cl_{w^\ast} } N$

Applying Existence of Non-Zero Continuous Linear Functional vanishing on Proper Closed Subspace of Hausdorff Locally Convex Space to the locally convex space $\struct {X^\ast, w^\ast}$, there exists $\Phi \in \struct {X^\ast, w^\ast}^\ast$ such that $\Phi \ne {\mathbf 0}_{X^\ast}$ and:
 * $\map \Phi g = 0$ for each $g \in \map {\cl_{w^\ast} } N$

and:
 * $\map \Phi f \ne 0$

From Characterization of Continuity of Linear Functional in Weak-* Topology, there exists $x \in X$ such that $\Phi = x^\wedge$.

Then:
 * $\map {x^\wedge} g = \map g x = 0$ for all $g \in N$

So $x \in {}^\bot N$.

But we also have $0 \ne \map {x^\wedge} f = \map f x$, so $f$ does not vanish at some point in ${}^\bot N$.

So $f \not \in \paren { {}^\bot N}^\bot$ by the definition of the annihilator of a subspace of $X$.

We therefore have:
 * $X^\ast \setminus \map {\cl_{w^\ast} } N \subseteq X^\ast \setminus \paren { {}^\bot N}^\bot$

hence:
 * $\paren { {}^\bot N}^\bot \subseteq \map {\cl_{w^\ast} } N$

from Set Complement inverts Subsets.

We conclude:
 * $\map {\cl_{w^\ast} } N = \paren { {}^\bot N}^\bot$