Normed Vector Space is Reflexive iff Surjective Evaluation Linear Transformation

Theorem

Let $\struct {X, \norm \cdot_X}$ be a normed vector space.

Let $\struct {X^{\ast \ast}, \norm \cdot_{X^{\ast \ast} } }$ be the second normed dual of $\struct {X, \norm \cdot_X}$.

Let $\iota : X \to X^{\ast \ast}$ be the evaluation linear transformation.

Then $X$ is reflexive if and only if:

$\iota$ is surjective.

That is:

$\iota X = X^{\ast \ast}$.

Proof

From the definition of a reflexive space, we have that $X$ is reflexive if and only if:

$\iota$ is an isometric isomorphism.

From Evaluation Linear Transformation on Normed Vector Space is Linear Isometry, we have:

$\iota$ is a linear isometry.

From Linear Isometry is Injective: Corollary, we then have:

$\iota$ is an isometric isomorphism if and only if $\iota$ is surjective.

That is:

$X$ is reflexive if and only if $\iota$ is surjective.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $26.1$: The Second Dual