Application of Peter-Weyl theorem: groups with no small-subgroups admit a faithful finite-dimensional representation

Question

My version of the Peter-Weyl theorem says that if $G$ is a compact group, then the matrix coefficients of $G$ are uniformly dense in $C(G)$. Consequently, the matrix coefficients are also dense in $L^2(G).$

Consider the following fragment from Bump's book "Lie groups":

I have two questions:

(1) Why does the Peter-Weyl theorem imply that such a matrix coefficient $f$ exists?

I tried to construct some continuous function with Urysohn's lemma and then use the Peter-Weyl theorem to approximate this but I do not succeed in forcing $f(1)=0$. Maybe I should use a translation? How should I construct $f$?

(2) Why if $f$ is constant on the kernel of $\pi$, we have $\ker \pi \subseteq U$?

Maybe we can show that $f < 1$ on $\ker \pi$?

Answer 1

Building upon the OP's idea of using Urysohn's lemma the argument can be completed as follows.

1. Because $G$ is normal as a topological space, Urysohn's lemma implies the existence of a continuous function $g:G\to\Bbb{R}$ such that $g(x)=2$ for all $x\notin U$ and $g(1)=0$.
2. By Peter-Weyl theorem, there exists a matrix coefficient $f$ of $G$ such that the sup-norm $||f-g||<1/2$.
3. So for the matrix coefficient $f$ from step 2 we have $f(1)<1/2$ and $f(x)>3/2$ for all $x\in U$.
4. Consider the function $h:G\to\Bbb{C}$ defined by $h(g)=f(g)-f(1)$. The space of matrix coefficients of $G$ is closed under linear combinations, and contains the constant functions. Hence $H$ is a matrix coefficient of some representation $\pi$ of $G$.
5. By definitiion $h(1)=f(1)-f(1)=0$ and for all $x\notin U$ we have $h(x)=f(x)-f(1)>1$.
6. If $g\in G$ is in the kernel of the representation $\pi$, then $\pi(g)=\pi(1)$. Because $h$ is a matrix coefficient of $\pi$, this implies that $h(g)=h(1)=0$. Therefore $g\in U$, and thus $\operatorname{ker}(\pi)\subseteq U$.
7. As $U$ contains no non-trivial subgroups, we can conclude that $\operatorname{ker}(\pi)$ is trivial.