Banach Space is Reflexive iff Normed Dual is Reflexive
Theorem
Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a Banach space over $\Bbb F$.
Let $X^\ast$ be the normed dual space of $X$.
Then:
- $X$ is reflexive if and only if $X^\ast$ is reflexive.
Proof
Let $X^{\ast \ast}$ be the second normed dual of $X$.
Let $X^{\ast \ast \ast}$ be the normed dual of $X^{\ast \ast}$.
Necessary Condition
Suppose that $X$ is reflexive.
We want to show that $X^\ast$ is reflexive.
From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, for each $\Phi \in X^{\ast \ast \ast}$ we aim to find $\phi \in X^\ast$ such that:
- $\map \Phi F = \map {\phi^\wedge} F = \map {\map {J_{X^\ast} } f} F$ for each $F \in X^{\ast \ast}$
where $J_{X^\ast}$ is the evaluation linear transformation $X^\ast \to X^{\ast \ast \ast}$.
Since $X$ is reflexive, for each $F \in X^{\ast \ast}$ there exists $x \in X$ such that:
- $F = x^\wedge = \map {J_X} x$
where $J_X$ is the evaluation linear transformation $X \to X^{\ast \ast}$.
So, it suffices to find $\phi$ such that for each $x \in X$ we have:
\(\ds \map \Phi {x^\wedge}\) | \(=\) | \(\ds \map {\phi^\wedge} {x^\wedge}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} \phi\) | Definition of Evaluation Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x\) | Definition of Evaluation Linear Transformation |
for each $x \in X$.
We show that this actually defines an element $\phi \in X^\ast$.
From Evaluation Linear Transformation on Normed Vector Space is Linear Transformation from Space to Second Normed Dual, we have that:
- $x \mapsto x^\wedge$ is a linear transformation from $X$ to $X^{\ast \ast}$.
Further, from Evaluation Linear Transformation on Normed Vector Space is Linear Isometry:
- $x \mapsto x^\wedge$ is a bounded linear transformation from $X$ to $X^{\ast \ast}$.
Since we also have that:
- $\Phi$ is a bounded linear functional $X^{\ast \ast} \to \Bbb F$
we have:
- $\phi$ is a bounded linear functional $X \to \Bbb F$
from Composition of Bounded Linear Transformations is Bounded Linear Transformation.
That is:
- $\phi \in X^\ast$
as required.
$\Box$
Sufficient Condition
Suppose that $X^\ast$ is reflexive.
Aiming for a contradiction, suppose that $X$ is not reflexive.
From Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation, this means that:
- for all $\Phi \in X^{\ast \ast \ast}$ there exists $\phi \in X^\ast$ such that $\Phi = \phi^\wedge$
but:
- for some $F \in X^{\ast \ast}$ there does not exist $x \in X$ such that $F = x^\wedge$.
From Image of Evaluation Linear Transformation on Banach Space is Closed Linear Subspace of Second Dual, we have that:
- $\map J X = \set {x^\wedge \in X^{\ast \ast} : x \in X}$ is a closed linear subspace of $X^{\ast \ast}$.
From Existence of Distance Functional, there exists $\Phi \in X^{\ast \ast \ast}$ such that:
- $\norm {\Phi}_{X^{\ast \ast \ast} } = 1$
and:
- $\map \Phi F = 0$ for all $F \in \map J X$.
That is:
- $\map \Phi {x^\wedge} = 0$ for all $x \in X$.
Since $X^\ast$ is reflexive, there exists $\phi \in X^\ast$ such that:
- $\Phi = \phi^\wedge$
Then for each $x \in X$ we have:
\(\ds 0\) | \(=\) | \(\ds \map \Phi {x^\wedge}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\phi^\wedge} {x^\wedge}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {x^\wedge} \phi\) | Definition of Evaluation Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x\) | Definition of Evaluation Linear Transformation |
Then we have:
- $\Phi \equiv 0$
so:
- $\norm \Phi_{X^{\ast \ast \ast} } = 0 \ne 1$
a contradiction.
So we have that $X$ is reflexive.
$\blacksquare$