Weak-* Closed Linear Subspace of Normed Dual Space is Isometrically Isomorphic to a Normed Dual Space

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 $Y$ be a linear subspace of $X^\ast$ closed in $\struct {X^\ast, w^\ast}$.

Then there exists a normed vector space $Z$ such that:
 * $Y$ is isometrically isomorphic to $Z^\ast$.

Proof
From Set is Closed iff Equals Topological Closure, we have:
 * $\map {\cl_{w^\ast} } Y = Y$

From Closure in Weak-* Topology in terms of Annihilators, we have:
 * $Y = \paren { {}^\bot Y}^\bot$

where:
 * ${}^\bot Y$ denotes the annihilator of $Y \subseteq X^\ast$
 * $\paren { {}^\bot Y}^\bot$ denotes the annihilator of ${}^\bot Y \subseteq X$.

From Normed Dual Space of Normed Quotient Vector Space is Isometrically Isomorphic to Annihilator, we have that:
 * $Y = \paren { {}^\bot Y}^\bot$ is isometrically isomorphic to $\paren {X/{}^\bot Y}^\ast$.

Setting $W = {}^\bot Y$ and:
 * $Z = X/W$

we get the result.