Kernel of Bounded Linear Transformation is Annihilator of Image of Dual Operator

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

Let $X$ and $Y$ be normed vector spaces over $\GF$.

Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.

Let $T : X \to Y$ be a Bounded linear transformation.

Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$.

Then we have:
 * $\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$

where ${}^\bot$ denotes the annihilator operation on subspaces of $X^\ast$.

Proof
Let $x \in \map \ker T$.

Let $\phi \in T^\ast \sqbrk {Y^\ast}$.

Then from the definition of the dual operator, there exists $f \in Y^\ast$ such that $f \circ T = \phi$.

Then we have:

Since this holds for all $\phi \in T^\ast \sqbrk {Y^\ast}$, we obtain $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.

So we have:
 * $\map \ker T \subseteq {}^\bot T^\ast \sqbrk {Y^\ast}$

Now let $x \in {}^\bot T^\ast \sqbrk {Y^\ast}$.

Then for all $\phi \in T^\ast \sqbrk {Y^\ast}$ we have $\map \phi x = 0$.

That is, for all $f \in Y^\ast$, we have:
 * $\map {\paren {T^\ast f} } x = \map f {T x} = 0$

from the definition of the dual operator.

From Normed Dual Space Separates Points, we obtain $T x = {\mathbf 0}_Y$ and hence $x \in \map \ker T$.

So we obtain:
 * ${}^\bot T^\ast \sqbrk {Y^\ast} \subseteq \map \ker T$

Hence we can conclude:
 * $\map \ker T = {}^\bot T^\ast \sqbrk {Y^\ast}$