Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective/Proof 1

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 $T \sqbrk X$ is everywhere dense in $Y$ if and only if $T^\ast$ is injective.

Proof

Necessary Condition

Suppose that $T \sqbrk X$ is everywhere dense in $Y$.

Let $f \in Y^\ast$ be such that $T^\ast f = 0$.

That is:

$\map f {T x} = 0$

for all $x \in X$.

So:

$\map f y = 0$

for all $y \in T \sqbrk X$.

Since $T \sqbrk X$ is everywhere dense in $Y$, $f$ is continuous and $\GF$ is Hausdorff, we have that:

$\map f y = 0$ for all $y \in Y$

from Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide.

So $T^\ast f = 0$ implies $f = 0$.

So $T^\ast$ is injective from Linear Transformation is Injective iff Kernel Contains Only Zero.

$\Box$

Sufficient Condition

Suppose that $T \sqbrk X$ is not everywhere dense in $Y$.

Then $\map \cl {T \sqbrk X}$ is a proper closed linear subspace of $Y$.

From Existence of Distance Functional, there exists $f \in Y^\ast$ such that $f \ne 0$ and:

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

In particular:

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

so that:

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

Hence, we have:

$T^\ast f = 0$

while $f \ne 0$.

So from Linear Transformation is Injective iff Kernel Contains Only Zero, we conclude that $T^\ast$ is not injective.

Hence by Proof by Contraposition, if $T^\ast$ is injective then $T \sqbrk X$ is everywhere dense in $Y$.

$\blacksquare$