[Math] For Banach space there is a compact topological space so that the Banach space is isometrically isomorphic with a closed subspace of $C(X)$.

Question

I want to prove that for Banach space V there is a compact topological space $X$ so that $V$ is isometrically isomorphic to a closed subspace of $C(X)$-continuous function on a (compact) topological space X, equipped with the supremum norm $\|f\|_\infty$ = $\sup_{x \in X} |f(x)|$.

Answer 1

Consider the canonical isometric embedding $\operatorname{ev}: V \to V^{\ast\ast}$.

Let $X$ be the unit ball of $V^\ast$, which is compact in the weak$^\ast$-topology by Alaoglu's theorem. By definition we have $\operatorname{ev}_v(\varphi) = \varphi(v)$ for a linear functional $\varphi \in X$ and $\operatorname{ev}_v$ is a continuous function on $X$. Finally observe that the sup-norm of $\operatorname{ev}_v$ on $X$ is the same as its norm as element of $V^{\ast\ast}$ and thus $v \mapsto \operatorname{ev}_v \in C(X)$ is isometric.