Weak-* Closed Linear Subspace of Normed Dual Space is Isometrically Isomorphic to a Normed Dual Space
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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.
$\blacksquare$