Let $G$ be a compact group equipped with Haar measure $\lambda$. Let $\pi:G\to U(H)$ be a continuous unitary representation. Suppose $f\in L^1(G)$. It follows from the Riesz representation theorem that there exists a bounded linear map $T:H\to H$ such that $$\langle T\xi, \eta \rangle = \int_G f(x) \langle \pi(x)\xi, \eta \rangle d\lambda(x)$$ for all $\xi, \eta\in H$.
I am having trouble seeing why the existence of $T$ is given by the Riesz representation theorem. Can someone explain this or give some hints on how I might prove this fact? What I am stumped on is that this integral is seemingly dependent on two unknowns. I tried to view this integral as a function in the dual space of $L^1(G)$, where the coefficient functions $\pi_{\xi,\eta}$ are elements, but it didn't seem to help all that much.
Fix $\xi\in H$ and consider the map $$\tag1 \varphi_\xi:\eta\longmapsto \int_Gf(x)\,\langle\pi(x)\xi,\eta\rangle\,d\lambda(x). $$ Because $\pi$ is continuous and $G$ is compact, $\pi(G)$ is compact, hence bounded. So there exists $c>0$ with $\|\pi(x)\|\leq c$ for all $x\in G$. Then, using $|\langle\pi(x)\xi,\eta\rangle|\leq c\,\|\xi\|\,\|\eta\|$,
$$ |\varphi_\xi(\eta)|\leq\int_G|f(x)|\,|\langle\pi(x)\xi,\eta\rangle|\,d\lambda(x) \leq c\|\xi\|\,\|\eta\|\,\|f\|_1. $$ This shows that $\varphi_\xi$ is bounded, with $\|\varphi_\xi\|\leq c\|\xi\|\,\|f\|_1$. It is clear that $\varphi_\xi$ is linear. By the Riesz Representation Theorem there exists $T\xi\in H$ such that $$ \varphi_\xi(\eta)=\langle T\xi,\eta\rangle. $$ Because the expression in $(1)$ is also linear in $\varphi$ we have that $\varphi_{\xi_1+a\xi_2}=\varphi_{\xi_1}+a\varphi_{\xi_2}$. Then the uniqueness in the Riesz Representation Theorem gives you that $T$ is linear. Finally, we have $$ |\langle T\xi,\eta\rangle|\leq c\,\|f\|_1\,\|\xi\|\,\|\eta\| $$ for all $\xi,\eta$, which shows first that $\|T\xi\|\leq c\|f\|_1\,\|\xi\|$, and this in turn is $\|T\|\leq c\|f\|_1$ and $T$ is bounded.