Kakutani's Theorem
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Theorem
Let $X$ be a normed vector 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 if and only if $\struct {B_X^-, w}$ 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:
- $\iota : \struct {X, w} \to \struct {\iota X, w^\ast}$ 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.
$\Box$
Sufficient Condition
Suppose that $\struct {B_X^-, w}$ is compact.
From Evaluation Linear Transformation on Normed Vector Space is Weak to Weak-* Continuous Embedding into Second Normed Dual, $\iota : \struct {X, w} \to \struct {X^{\ast \ast}, w^\ast}$ is continuous.
From Continuous Image of Compact Space is Compact, $\iota B_X^-$ is a compact subset of $\struct {X^{\ast \ast}, w^\ast}$.
In particular, $\iota B_X^-$ is closed in $\struct {X^{\ast \ast}, w^\ast}$.
So:
- $\map {\cl_{w^\ast} } {\iota B_X^-} = \iota B_X^-$
from Set is Closed iff Equals Topological Closure.
From Goldstine's Theorem, we have that:
- $\map {\cl_{w^\ast} } {\iota B_X^-} = B_{X^{\ast \ast} }^-$
So, we have:
- $\iota B_X^- = B_{X^{\ast \ast} }^-$
From Normed Vector Space is Reflexive iff Closed Unit Ball in Original Space is Mapped to Closed Unit Ball in Second Dual, $X$ is reflexive.
$\blacksquare$
Source of Name
This entry was named for Shizuo Kakutani.