Corollaries to Kakutani's Theorem

Corollaries to Kakutani's Theorem

Recall from the Kakutani's Theorem page that Kakutani's theorem states that if $X$ is a Banach space then $X$ is reflexive if and only if the closed unit ball $B_X$ of $X$ is weakly compact.

We will now look at some important corollaries to Kakutani's theorem.

Corollary 1: Let $X$ be a Banach space. If $X$ is reflexive then every norm closed, bounded, and convex subset $K$ of $X$ is weakly compact.
  • Proof: By Kakutani's theorem we have that the closed unit ball $B_X$ of $X$ is weakly compact. Therefore any norm closed ball of $X$ is weakly compact.
  • Let $K \subseteq X$ be norm closed, bounded, and convex. By Mazur's Theorem, since $K$ is convex and norm closed, it is weakly closed. Since $K$ is bounded, there exists a norm closed ball $B$ such that $K \subseteq B$.
  • We have established that $B$ must be weakly compact. Therefore $K$ is a weakly closed subset of a weakly compact set $B$ and thus $K$ is weakly compact. $\blacksquare$
Corollary 2: Let $X$ be a Banach space. If $X$ is reflexive then the closed unit ball $B_{X^{*}}$ of $X^*$ is weak-* sequentially compact.
  • Proof: Since $X$ is reflexive we have that the weak topology on $X^*$ is the same as the weak-* topology on $X^*$.
  • By Alaoglu's Theorem we have that the closed unit ball $B_{X^{*}}$ of $X^*$ is weak-* compact, and so it is weakly compact.
  • Since $X^*$ is a Banach space and $B_{X^{*}}$ is weakly compact, we have by Kakutani's theorem that $X^*$ is reflexive.
  • From the theorem on the Every Bounded Sequence in a Reflexive Space X has a Weakly Convergent Subsequence page, we have that every bounded sequence in $X^*$ has a weak-* convergent subsequence. Since $B_{X^*}$ is bounded, we have that every sequence in $B_{X^*}$ has a weak-* convergent subsequence. Moreover, since $B_{X^{*}}$ is weak-* closed, we have that every sequence in $B_{X^*}$ has a weak-* convergent subsequence that converges IN $B_{X^*}$.
  • Hence $B_{X^*}$ is weak-* sequentially compact. $\blacksquare$
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