# 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.

$\blacksquare$