Annihilator of Subspace of Banach Space is Subspace of Normed Dual

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ be a Banach space over $\GF$.

Let $M$ be a vector subspace of $X$.

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

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

Let $M^\bot$ be the annihilator of $M$.

Then $M^\bot$ is a linear subspace of $X^\ast$.

Proof
For each $x \in X$, we have:
 * $\map { {\mathbf 0}_{X^\ast} } x = 0$ for each $x \in X$.

In particular:
 * $\map { {\mathbf 0}_{X^\ast} } x = 0$ for each $x \in M$

so that:
 * ${\mathbf 0}_{X^\ast} \in M^\bot$

So we have $M^\bot \ne \O$.

From One-Step Vector Subspace Test, it is enough to show that for each $f, g \in M^\bot$ and $\lambda \in \GF$ we have:
 * $f + \lambda g \in M^\bot$

Let $f, g \in M^\bot$ and $\lambda \in \GF$.

Then $\map f x = 0$ and $\map g x = 0$.

So $0 = \map f x + \lambda \map g x = \map {\paren {f + \lambda g} } x$.

So we have $f + \lambda g \in M^\bot$.

From One-Step Vector Subspace Test, $M^\bot$ is a linear subspace of $X^\ast$.