Corollaries to Alaoglu's Theorem

# Corollaries to Alaoglu's Theorem

Recall from the Alaoglu's Theorem page that Alaoglu's theorem states that if $X$ is a normed linear space then the closed unit ball of $X^*$ is weak-* compact. We will now look at some corollaries to Alaoglu's theorem.

Corollary 1: Let $X$ be a normed linear space. Then $X$ is isometrically isomorphic to a subspace of $C(K)$ where $K$ is a Hausdorff compact space. |

**Proof:**Let $K = B_{X^*}$ where $B_{X^*}$ is the closed unit ball of $X^*$. Since $X^*$ with the weak-* topology is Hausdorff and $B_{X^*}$ is a subspace of $X^*$ we have that $B_{X^*}$ is Hausdorff. Furthermore, Alaoglu's theorem tells us that $B_{X^*}$ is weak-* compact.

- Let $T : X \to C(K)$ be defined for each $x \in X$ by:

\begin{align} \quad T(x) = \hat{x} |_K \end{align}

- Clearly $T$ is linear. Furthermore, $T$ is bounded and further and isometry since:

\begin{align} \quad \| T(x) \| = \| \hat{x}_K \| = \sup_{f \in X^*, \| f \| \leq 1} |\hat{x}_K (f)| = \sup_{f \in X^*, \| f \| \leq 1} |\hat{x}(f)| = \| \hat{x} \| = \| x \| \end{align}

- So $T$ is an isometry and hence $T$ is bounded and more importantly, injective. Note that $T(X)$ is a subspace of $C(K)$. So $T : X \to T(X)$ is a bijective isometry. So $X$ is isometrically isomorphic to $\mathrm{range} (T) = T(X)$. $\blacksquare$

Corollary 2: Let $X$ be a normed linear space. Then the closed unit ball of $X^*$ has an extreme point. |

**Proof:**$X^*$ is a locally convex topological vector space with respect to the weak-* topology. If $B_{X^*}$ denotes the closed unit ball in $X^*$ then $B_{X^*}$ is nonempty and convex. By Alaoglu's theorem, $B_{X^*}$ is weak-* compact.

- So by The Krein-Milman Lemma we have that $B_{X^*}$ has an extreme point. $\blacksquare$