Linear Subspace is Subset of Double Annihilator

Theorem

Let $X$ be a Banach space.

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

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

Then:

$N \subseteq \paren { {}^\bot N}^\bot$

where:

${}^\bot N$ denotes the annihilator of $N \subseteq X^\ast$

$\paren { {}^\bot N}^\bot$ denotes the annihilator of ${}^\bot N \subseteq X$.

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Proof

From Set Complement inverts Subsets, we can equivalently show:

$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$

Let $f \in X^\ast \setminus \paren { {}^\bot N}^\bot$, then:

$\map f x \ne 0$ for some $x \in {}^\bot N$.

By the definition of the annihilator of a subspace of $X$, we have:

$\map g x = 0$ for all $g \in N$.

In particular, $g \in X^\ast \setminus N$.

So we obtain:

$X^\ast \setminus \paren { {}^\bot N}^\bot \subseteq X^\ast \setminus N$

and hence:

$N \subseteq \paren { {}^\bot N}^\bot$

from Set Complement inverts Subsets.

$\blacksquare$