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 $X^\ast$ 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:

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.

Sufficient Condition
Suppose that $X^\ast$ is reflexive.

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:

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.