Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense

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

Then:

$M^\bot = \set { {\mathbf 0}_{X^\ast} }$

if and only if:

$M$ is everywhere dense in $X$.

Proof

Necessary Condition

Suppose that:

$M$ is not everywhere dense in $X$.

From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $\map \cl M$ is a closed linear subspace of $X$.

From Set is Closed iff Equals Topological Closure, we have $X \ne \map \cl M$.

So $\map \cl M$ is a proper closed linear subspace of $X$.

From Existence of Distance Functional, there exists $f \in X^\ast \setminus \set { {\mathbf 0}_{X^\ast} }$ such that:

$\map f x = 0$ for all $x \in M$

so that $f \in M^\bot$.

So $M^\bot \ne \set { {\mathbf 0}_{X^\ast} }$

From Proof by Contraposition, we have:

if $M^\bot = \set { {\mathbf 0}_{X^\ast} }$ then $M$ is everywhere dense in $X$.

$\Box$

Sufficient Condition

Suppose that:

$M$ is everywhere dense in $X$.

Let $f \in M^\bot$.

Then for all $x \in M$ we have $\map f x = 0$.

From Metric Space is Hausdorff, $\GF$ is Hausdorff.

Since $M$ is everywhere dense, we have:

$\map f x = 0$ for all $x \in X$.

So if $f \in M^\bot$, then $f = {\mathbf 0}_{X^\ast}$.

Further, we have ${\mathbf 0}_{X^\ast} \in M^\bot$.

Hence we obtain:

$M^\bot = \set { {\mathbf 0}_{X^\ast} }$

$\blacksquare$