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

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

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

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

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

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

Let $T \sqbrk X^\bot$ be the annihilator of $T \sqbrk X$.

Then:
 * $T \sqbrk X^\bot = \map \ker {T^\ast}$

Proof
We have:
 * $f \in \map \ker {T^\ast}$


 * $\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$.
 * $\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$.

That is, :
 * $\map f y = 0$ for all $y \in T \sqbrk X$

To conclude, we have $f \in \map \ker {T^\ast}$ :
 * $f \in T \sqbrk X^\bot$