Banach–Stone theorem

Remarks

The idea of the proof: the extreme points of the unit ball of the dual of $C(X)$ are the functionals $\pm \delta_x$, with $\delta_x(f) = f(x)$. The adjoint of $T$ is an isometry of the duals and maps extreme points to extreme points. So it sends each $\delta_y$ to $h(y) \delta_{\varphi(y)}$, and the weak-* topology makes $\varphi$ and $h$ continuous.

Example: on $X = Y = [0, 1]$, the map $(T f)(t) = -f(1 - t)$ is a surjective linear isometry. Here $h = -1$ and $\varphi(t) = 1 - t$.

For complex scalars the same holds with $h$ a continuous function of absolute value $1$. As a consequence, $X$ and $Y$ are homeomorphic if and only if $C(X)$ and $C(Y)$ are isometrically isomorphic. Isometry is needed. Let $S = \set{0} \cup \set{1/n : n \ge 1}$, a convergent sequence with its limit. Then $C(S)$ and $C(S \times \set{1, 2})$ are linearly isomorphic, but $S$ has one limit point and $S \times \set{1, 2}$ has two.

Going deeper

For a student

The theorem shows that the Banach space $C(X)$, with its norm, determines the compact space $X$ up to homeomorphism. Gelfand–Naimark theory gives a similar result for $C(X)$ as an algebra, and the Banach–Stone theorem gets it from the metric structure alone. Without the isometry condition the result fails badly: by a theorem of Milutin, $C(X)$ is isomorphic to $C([0, 1])$ for every uncountable compact metric space $X$.