Kakutani's Theorem

Theorem
Let $X$ be a Banach space.

Let $w$ be the weak topology on $X$.

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

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

Then $X$ is reflexive $\struct {B_X^-, w}$ is compact.

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

Let $w^\ast$ be the weak-* topology on $X^{\ast \ast}$.

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

Necessary Condition
Suppose that $X$ is reflexive.

By Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual, $\iota B_X^- = B_{X^{\ast \ast} }^-$.

From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Homeomorphism onto Image, we have that:


 * $\iota : \struct {X, w} \to \struct {X^\ast, w^\ast}$ is a homeomorphism.

From Restriction of Homeomorphism is Homeomorphism, we have that:


 * $\iota : \struct {B_X^-, w} \to \struct {\iota B_X^-, w^\ast} = \struct {B_{X^{\ast \ast} }^-, w^\ast}$ is a homeomorphism.

From the Banach-Alaoglu Theorem, $\struct {B_{X^{\ast \ast} }^-, w^\ast}$ is compact.

Since $\struct {B_X^-, w}$ is homeomorphic to $\struct {B_{X^{\ast \ast} }^-, w^\ast}$, we have that $\struct {B_X^-, w}$ is compact.