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 $X^\ast$ 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:

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

We have:

$f \in \map \ker {T^\ast}$

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

$\map f y = 0$ for all $y \in T \sqbrk X$

To conclude, we have $f \in \map \ker {T^\ast}$ if and only if:

$f \in T \sqbrk X^\bot$

$\blacksquare$