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 :


 * $\iota$ is surjective.

That is:


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

Proof
From the definition of a reflexive space, we have that $X$ is reflexive :


 * $\iota$ 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 $\iota$ is surjective.

That is:


 * $X$ is reflexive $\iota$ is surjective.