Evaluation functionals and point masses

Evaluation is complex-linear and $\mathrm{\mid }f(t)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. The continuous constant function $1$ has norm one since $K\mathrm{\ne }\mathrm{\varnothing }$, and ${\delta }_t(1)=1$, so $\mathrm{\parallel }{\delta }_t\mathrm{\parallel }=1$.

Define the Borel measure ${\mu }_t(A)=1$ when $t\in A$ and $0$ otherwise. In a disjoint countable union at most one member contains $t$, proving countable additivity. Its mass is one. For inner regularity, a set containing $t$ contains the compact singleton $\{t\}$, and a set not containing $t$ has measure zero. For outer regularity, a set missing $t$ lies in the open set $K\setminus \{t\}$ of measure zero; for a set containing $t$, every open superset has measure one.

Integration of a simple Borel function against ${\mu }_t$ equals its value at $t$. Uniform simple approximation of a bounded complex Borel function extends this identity, since the integral error is bounded by the uniform error times ${\mu }_t(K)=1$. In particular $\int fd{\mu }_t=f(t)$ for $f\in C(K;\mathbb{C})$. Compact Hausdorff $K$ is LCH and $C_0(K)=C(K)$, so RMK uniqueness identifies this regular point mass as the representing measure.

Composition with a homeomorphism preserves continuity and the supremum norm. For every $f$, $(P^{\ast }{\delta }_t)(f)={\delta }_t(f\circ \varphi )=f(\varphi (t))={\delta }_{\varphi (t)}(f)$, which proves the transpose identity.