Image of Bounded Linear Transformation is Everywhere Dense iff Dual Operator is Injective
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 1
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 $Y$.
$\blacksquare$
Proof 2
From Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator, we have:
- $T \sqbrk X^\bot = \map \ker {T^\ast}$
where $T \sqbrk X^\bot$ denotes the annihilator of $T \sqbrk X$.
From Linear Transformation is Injective iff Kernel Contains Only Zero, we then have that $T^\ast$ is injective if and only if:
- $T \sqbrk X^\bot = \set { {\mathbf 0}_{Y^\ast} }$
From Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense, this is equivalent to:
- $T \sqbrk X$ is everywhere dense in $Y$.
This was the demand.
$\blacksquare$