Let $G$ be a locally compact group with unimodular Haar measure $\mu$. We consider the Hilbert space $\mathscr{H}:= L\_{\mu}^2(G)$ together with the unitary representation $\pi : G \to U(\mathscr{H})$ given by $\pi(g): \mathscr{H} \to \mathscr{H}, f \mapsto f \circ r\_g$, where $r\_g$ is right multiplication by $g$. Hence $(\pi(g)f)(x) = f(xg)$.


We studies the following: For a continuous function $\varphi \in C\_c(G)$ with compact support we define $$\pi(\varphi):= \int\_G \varphi(g)\pi(g) \, d\mu(g).$$ More explicitly, for a measurable function $f: G \to \mathbb{C}$ we define $$(\pi(\varphi)f)(x) :=\int\_G \varphi(g)f(xg) \, d\mu(g).$$


My question in now: Does this define a well defined representation of $C\_c(G)$ on $\mathscr{H}$, where the problem is, that if $f \in L\_{\mu}^2(G)$, then we need to show that $\pi(\varphi)f$ is in $L\_{\mu}^2(G)$.


I doubt that this is true for a general $G$. However, I can prove the statement if $G$ is compact as follows: Note that if $G$ is compact, then $L\_{\mu}^2(G) \subset L\_{\mu}^1(G)$. Hence for $f \in L\_{\mu}^2(G)$,
\begin{align\*}
||\pi(\varphi)f||\_2^2 &= \int\_G|(\pi(\varphi)f)(x)|^2 \, d\mu(x) \\
&= \int\_G \bigg| \int\_G \varphi(g)f(xg) \, d\mu (g) \bigg|^2 d\mu(x) \\
& \leq ||\varphi||\_{\infty}^2 \int\_G \bigg| \int\_G f(xg) \, d\mu (g) \bigg|^2 d\mu(x) \\
&\leq ||\varphi||\_{\infty}^2 \int\_G ||f||\_1^2 \,d \mu(x) \\
&= ||\varphi||\_{\infty}^2 \cdot ||f||\_1^2 \cdot \mu(G) < \infty.
\end{align\*}
Where we used the invariance of the Haar measure in line 4. And the last term is less than infinity, since $\varphi$ is continuous of compact support and hence bounded, $L\_{\mu}^2(G) \subset L\_{\mu}^1(G)$ and hence for $f \in L\_{\mu}^2(G)$ we have $||f||\_1 < \infty$ and finally since $G$ is compact, the Haar measure is finite.


So the problem is can, one do a similar calculation if $G$ is not compact, or if one can find a counterexample. 



Convergence of some object depending on functions with compact support
It's true: $\pi(\varphi)f\in L^2(G,\mu)$.


Let $M$ be the support of $\varphi$, so 
$$(\pi(\varphi)f)(x) :=\int\_M \varphi(g)f(xg) \, d\mu(g).$$
By the Cauchy-Schwarz inequality, we have 
$$|(\pi(\varphi)f)(x)|^2\le \|\varphi\|^2\_2\int\_M|f(xg)|^2d\mu(g)=\|\varphi\|^2\_2\int\_{xM}|f(g)|^2d\mu(g),$$
so
$$\int\_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2\_2\int\_G\int\_{G}\mathbb{1}\_{xM}(g)|f(g)|^2d\mu(g)d\mu(x).$$
Since $g\in xM$ $\Leftrightarrow$ $x\in gM^{-1}$, this yields (I should assume $G$ $\sigma$-compact to apply Fubini, but otherwise since $f$ has $\sigma$-compact support I apply it inside a suitable $\sigma$-compact open subgroup)
$$\int\_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2\_2\int\_G\int\_{gM^{-1}}d\mu(x)|f(g)|^2d\mu(g).$$
So
$$\int\_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2\_2\mu(M^{-1})\int\_G|f(g)|^2d\mu(g)=\|\varphi\|^2\_2\mu(M^{-1})\|f\|\_2^2<\infty.$$


Note that I only used (twice) left-invariance of $\mu$ (in the non-unimodular case, the right-regular representation $\pi$ is not unitary, however).


