Bounded Linear Transformation is Into Linear Isomorphism iff Dual Operator is Surjective

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Theorem

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

Let $X$ and $Y$ be Banach spaces over $\GF$.

Let $X^\ast$ and $Y^\ast$ be the normed duals 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 $T$ is an into linear isomorphism if and only if $T^\ast$ is surjective.


Proof

Let $B_{X^\ast}^-$ and $B_{Y^\ast}^-$ be the closed unit balls of $X^\ast$ and $Y^\ast$ respectively.

Necessary Condition

We want to show that $T$ is a linear isomorphism considered as a map $X \to T \sqbrk X$.

Suppose that $T^\ast$ is surjective.

From Dual Operator is Bounded Linear Transformation, $T^\ast$ is a bounded linear transformation.

From Banach-Schauder Theorem, $T^\ast$ is an open mapping.

From Characterization of Open Linear Transformation between Normed Vector Spaces, there exists $\delta > 0$ such that $\delta B_{X^\ast}^- \subseteq T^\ast \sqbrk {B_{Y^\ast}^-}$.

Now, for each $x \in X$ we have:

\(\ds \norm {T x}_Y\) \(=\) \(\ds \sup_{f \in Y^\ast} \cmod {\map f {T x} }\) Norm in terms of Normed Dual Space
\(\ds \) \(=\) \(\ds \sup_{f \in Y^\ast} \cmod {\map {\paren {T^\ast f} } x}\) Definition of Dual Operator
\(\ds \) \(=\) \(\ds \sup_{g \in T^\ast \sqbrk {B_{Y^\ast}^-} } \cmod {\map g x}\)
\(\ds \) \(\ge\) \(\ds \sup_{g \in \delta B_Y^-} \cmod {\map g x}\) from Supremum of Subset since $\delta B_{X^\ast}^- \subseteq T^\ast \sqbrk {B_{Y^\ast}^-}$
\(\ds \) \(=\) \(\ds \delta \sup_{g \in B_Y^-} \cmod {\map g x}\) Multiple of Supremum
\(\ds \) \(=\) \(\ds \delta \norm x_X\)

Hence $\map \ker T = \set { {\mathbf 0}_X}$.

So from Linear Transformation is Injective iff Kernel Contains Only Zero, we have that $T$ is injective, hence bijective as a map $X \to T \sqbrk X$.

From Characterization of Invertible Bounded Linear Transformations, $T^{-1}$ is a bounded linear transformation.

Hence $T$ is a linear isomorphism as a map $X \to T \sqbrk X$, hence an into linear isomorphism.

$\Box$


Sufficient Condition

Suppose that $T$ is an into linear isomorphism.

Then $T^{-1} : T \sqbrk X \to X$ is a bounded linear transformation.

Let $f \in X^\ast$.

From Composite of Continuous Mappings is Continuous, $f \circ T^{-1} : T \sqbrk X \to \GF$ is continuous.

By Continuity of Linear Functionals, $f \circ T^{-1}$ is bounded.

From the Hahn-Banach Theorem, there exists $g \in Y^\ast$ such that:

$\map g y = \map f {T^{-1} y}$

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

Then for $x \in X$, we have:

\(\ds \map {\paren {T^\ast g} } x\) \(=\) \(\ds \map {\paren {g \circ T} } x\)
\(\ds \) \(=\) \(\ds \map {\paren {\paren {f \circ T^{-1} } \circ T} } x\)
\(\ds \) \(=\) \(\ds \map f x\) Composition of Mappings is Associative

So we have:

$T^\ast g = f$

So $T^\ast$ is surjective.

$\blacksquare$


Sources

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