Kakutani's Theorem

# Kakutani's Theorem

Proposition 1: Let $X$ be a normed linear space and let $J : X \to X^{**}$ be the natural embedding. Then $J$ is a topological homeomorphism between $X$ with the weak topology and $J(X)$ with the weak-* topology. |

Theorem 2 (Kakutani's Theorem): Let $X$ be a Banach space. Then $X$ is reflexive if and only if the closed unit ball of $X$ is weakly compact. |

**Proof:**Let $B_X$ denote the closed unit ball of $X$ and let $B_{X^{**}}$ denote the closed unit ball of $X^{**}$. That is:

\begin{align} \quad B_X = \{ x \in X : \| x \| \leq 1 \} \end{align}

(2)
\begin{align} \quad B_{X^{**}} = \{ \varphi \in X^{**} : \| \varphi \| \leq 1 \} \end{align}

- $\Rightarrow$ Suppose that $X$ is reflexive. Then the natural embedding $J : X \to X^{**}$ defined for each $x \in X$ by $J(x) = \hat{x}$ (where $\hat{x} : X^* \to \mathbb{R}$ is defined for all $f \in X^*$ by $\hat{x} (f) = f(x)$) is onto. Actually, $J$ is an isomorphism from $X$ to $X^{**}$ and so $J$ is an isomorphism from $B_{X}$ to $B_{X^{**}}$.

- By proposition 1 above, $J$ is a topological homeomorphism from $B_X$ with the weak topology to $B_{X^{**}}$ with the weak-* topology.

- Since $X^*$ is a normed linear space, by Alaoglu's theorem we have that the closed unit ball of $X^{**}$, $B_{X^{**}}$ is weak-* compact. Since $J$ is a homeomorphism, this means that $B_X$ is weakly compact.

- $\Leftarrow$ Suppose that $B_X$ is weakly compact. Since $J : X \to X^{**}$ is homeomorphism from $X$ with the weak topology to $X^{**}$ with the weak-* topology, we have that $J(B_X)$ is weak-* compact (since $J$ is continuous) Furthermore, since $B_X$ is convex and $J$ is linear, we have that $J(B_X)$ is convex.

- Suppose that $X$ is not reflexive. Let $\varphi \in B_{X^{**}} \setminus J(B_X)$. By The Separation Theorems there exists a linear functional $f \in X^*$ with $\| f \| = 1$ such that:

\begin{align} \quad \varphi(f) < \inf_{\psi \in J(B_X)} \psi(f) = \inf_{x \in B_X} f(x) \end{align}

- Since $\| f \| = 1$ we have that $|f(x)| \leq \| x \|$. Since $x \in B_X$ implies that $\| x \| \leq 1$ we see that $\inf_{x \in B_X} f(x) = -1$, and hence:

\begin{align} \quad \varphi(f) < -1 \quad (*) \end{align}

- However, since $\varphi \in B_{X^{**}}$ we have that $\| \varphi \| \leq 1$ and since $\| f \| = 1$, we have that:

\begin{align} \quad |\varphi(f)| \leq \| \varphi \| \| f \| \leq 1 \end{align}

- That is:

\begin{align} \quad -1 \leq \varphi (f) \quad (**) \end{align}

- So from $(*)$ and $(**)$ we have that $-1 < -1$, a contradiction. Therefore the assumption that $B_{X^{**}} \neq J(B_X)$ is false. Hence $J(B_X) = B_{X^{**}}$. So $X = X^{**}$, that is, $X$ is reflexive. $\blacksquare$

Recall from the Closed Unit Ball of X Being Weakly Compact Implies X is a Banach Space page that if $X$ is a normed space and the closed unit ball $B_X$ is weakly compact, then $X$ is a Banach space.

Observe that if $X$ is reflexive, then $X$ is isomorphic to $X^{**}$ (via the canonical embedding $J : X \to X^{**}$) and that $X^{**} = B(X^*, \mathbb{C})$ is always a Banach space since $\mathbb{C}$ is. Thus $X$ is a Banach space.

So Kakutani's Theorem does NOT require the assumption that $X$ is a Banach space since both parts of the theorem imply $X$ is a Banach space. We state this modified version of Kakutani's theorem below.

Corollary 2: Let $X$ be a normed space. Then $X$ is reflexive if and only if the closed unit ball of $X$ is weakly compact. |