Given a compact metric topological group $G$ and an irreducible unitary representation $\pi$ of $G$, I would like to show that $\pi$ is contained in the left-regular representation $\lambda$ of $G$.
One of my reference books is 'A Course in Abstract Harmonic Analysis' by Folland. In chapter 5. the Peter-Weyl Theorem is exposed. However, I content myself with the weaker statement above. I was hoping that you could explain how to prove the result above (maybe by using matrix coefficients or otherwise). Thank you very much in advance.
Let $u$ be any representation of the compact group $G$ on the Hilbert space $H$.
Fixing a unit vector $\xi $ in $H$, define the linear map $$ F:H\to L^2(G), $$ by $$ F(\eta )|_g = \langle \eta , u_g(\xi )\rangle, \quad\forall \eta\in H, \quad\forall g\in G. $$ For $h$ in $G$, notice that $$ \lambda _h(F(\eta ))|_g = F(\eta )|_{h^{-1}g} = \langle \eta , u_{h^{-1}g}(\xi )\rangle = $$$$ = \langle u_h(\eta ), u_g(\xi )\rangle = F(u_h(\eta ))|_g, $$ so we see that $$ \lambda _h\circ F= F\circ u_h, \tag 1 $$ meaning that $F$ is a covariant map.
Assuming from now on that $u$ is irreducible, notice that (1) implies that $F^*F$ commutes with $u$, so the Schur Lemma implies that $F=cI$, for some constant $c$. Defining $V=c^{-1/2}F$, we have that $V$ is an isometric operator and clearly $\lambda _h\circ V= V\circ u_h$.
The conclusion is that $V$ establishes a unitary equivalence between $u$ and the restriction of $\lambda $ to the range of $V$.