Banach–Mazur theorem

Remarks

The idea of the proof: let $B$ be the closed unit ball of the dual space $X^*$ with the weak-* topology. Since $X$ is separable, $B$ is a compact metrizable space, so it is a continuous image of the Cantor set $\Delta$ under some map $\varphi$. The map $x \mapsto (t \mapsto \varphi(t)(x))$ is a linear isometry of $X$ into $C(\Delta)$, by the Hahn–Banach theorem. Each function in $C(\Delta)$ extends to $[0,1]$ by linear interpolation on the open intervals removed from $[0,1]$, and this extension is a linear isometry from $C(\Delta)$ into $C([0,1])$.

For example, the Hilbert space $\ell^2$ and every space $L^p([0,1])$ with $1 \le p < \infty$ are isometric to subspaces of $C([0,1])$. Separability is necessary: $C([0,1])$ is separable, so each of its subspaces is separable, and the non-separable space $\ell^\infty$ does not embed in it.

Going deeper

For a student

The theorem says that $C([0,1])$ is universal for the class of separable Banach spaces. It is the linear analogue of the fact that every compact metric space is a continuous image of the Cantor set, and the proof uses exactly that fact. The weak-* compactness of the dual ball comes from the Banach–Alaoglu theorem. Universality makes $C([0,1])$ itself very complicated: every property that fails in some separable space fails in some subspace of $C([0,1])$.