Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual

Theorem
Let $X$ be a normed vector space.

Let $X^\ast$ be the normed dual of $X$.

Let $X^{\ast \ast}$ be the second norm dual.

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

Let $B_X^-$ be the closed unit ball of $X$.

Let $B_{X^{\ast \ast} }^-$ be the closed unit ball of $X^{\ast \ast}$.

Then $X$ is reflexive $\iota B_X^- = B_{X^{\ast \ast} }^-$.

Necessary Condition
Suppose that $X$ is reflexive.

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


 * $\norm {\iota x}_{X^{\ast \ast} } = \norm x_X$

Hence if $\norm x_X \le 1$, we have $\norm {\iota x}_{X^{\ast \ast} } \le 1$.

So $\iota B_X^- \subseteq B_{X^{\ast \ast} }^-$.

Conversely, suppose that $\Phi \in B_{X^{\ast \ast} }$.

Then there exists $x \in X$ such that $\Phi = \iota x$.

Then we have:


 * $1 \ge \norm \Phi_{X^{\ast \ast} } = \norm {\iota x}_{X^{\ast \ast} } = \norm x_X$

So $x \in B_X^-$ and $\Phi \in \iota B_X^-$.

So we obtain $\iota B_X^- = B_{X^{\ast \ast} }^-$.

Sufficient Condition
Conversely suppose that $\iota B_X^- = B_{X^{\ast \ast} }^-$.

Then we have: