Matrix coefficients are uniformly dense in C(G)

Since $f$ is continuous on the compact group, $f\in L^2(G)$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_2<+\mathrm{\infty }$; fix $\epsilon >0$ and choose $n$ with ${\sup }_x\mathrm{\mid }{T}_{k_n}f(x)-f(x)\mathrm{\mid }<\epsilon \mathrm{/}2$ by [L2], so that $\mathrm{\mid }f(x)-{\int }_G{k}_n(u)f(xu)du\mathrm{\mid }<\epsilon \mathrm{/}2$ for every $x$.

By [L1] applied to the $L^2$ class of $k_n$, there is a finite linear combination $p$ of matrix coefficients with $\mathrm{\parallel }{k}_n-p{\mathrm{\parallel }}_2<\epsilon \mathrm{/}(2\mathrm{\parallel }f{\mathrm{\parallel }}_2)$ when $f\mathrm{\ne }0$ (and the conclusion below is trivial when $f=0$).

For every $x$, the defining formula for the convolution operators gives $\mathrm{\mid }{T}_{p}f(x)-T_{k_n}f(x)\mathrm{\mid }\le {\int }_G\mathrm{\mid }p(x^{-1}y)-k_n(x^{-1}y)\mathrm{\mid }\mathrm{\mid }f(y)\mathrm{\mid }dy\le \mathrm{\parallel }p-k_n{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$ by Cauchy–Schwarz and Haar invariance. Hence $\mathrm{\parallel }{T}_{p}f-f{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon$ by step 1.1. The function $T_{p}f$ is a finite linear combination of matrix coefficients: if a summand of $p$ is ${\pi }_{ij}$, then ${\pi }_{ij}(x^{-1}y)={\sum }_k{\pi }_{ik}(x^{-1}){\pi }_{kj}(y),$ so its contribution to $T_{p}f(x)$ is a finite sum of the constants ${\int }_G{\pi }_{kj}(y)f(y)dy$ times the matrix coefficients $x\mapsto {\pi }_{ik}(x^{-1})$ of the contragredient representation. Thus a finite matrix-coefficient sum lies within $\epsilon$ of $f$ uniformly.

The zero function is approximated by the zero linear combination, and the same argument includes the one-element group. All noncanonical existence used above is contained in the Haar, Hilbert-space and representation-theoretic suppliers invoked in [L1] and [L2], under the Axiom of Choice assumed in [A1].