Corollary 1   A Banach space is isometrically isomorphic to a closed subspace of $C(X)$ for a compact Hausdorff space $X$ .

Proof. Let $X$ be the unit ball of . By the Banach-Alaoglu theorem it is compact in the weak-$*$ topology. Define the map by $(\Phi f)(\varphi) = \varphi(f)$ . This is linear and we have for :
With the Hahn-Banach theorem it follows that there is a such that $\varphi(f) = \|f\|$ . Thus $\|\Phi(f)\|_{\infty} = \|f\|$ and $\Phi$ is an isometric isomorphism, as required.