This is part of the definition of constructing the reduced $C$-norm.
Let $G$ be a locally compact hausdorff group, $\nu$ a Haar measure that is both left and right invariant, $\xi\in B(L^2(G))$, Then we define $\pi(f):C_c(G) \rightarrow L^2(G)$ by $$ \pi(f)(\xi) (\gamma) = \int_{\eta \in G} f(\gamma \eta^{-1}) \xi (\eta) \, d\nu (\eta) $$
It is claimed that
$$||\pi(f) \xi||_2 \le ||f||_1 ||\xi||_2$$ $\pi(f)$ can now be regarded as an operator $L^1(G) \rightarrow L^2(G)$.
How is this so? Moreover, it is claimed that
$$||f||_r = || \pi(f)||$$
in fact defines a $C^*$ norm (rather than a seminorm) - is this true, and if so, how?
You have, using that convolution is commutative, Minkowski's Inequality, and the invariance of $\nu$, \begin{align} \|\pi(f)\xi\|_2 &=\left(\int_G\left|\int_G g(\gamma\eta^{-1})f(\eta)\,d\nu(\eta) \right|^2\,d\nu(\gamma)\right)^{1/2} \\ &\leq\int_G\left(\int_G |g(\gamma\eta^{-1})|^2\,|f(\eta)|^2\,d\nu(\gamma) \right)^{1/2}\,d\nu(\eta)\\ &\leq\int_G|f(\eta)|^2\,\left(\int_G |g(\gamma\eta^{-1})|^2\,d\nu(\gamma) \right)^{1/2}\,d\nu(\eta)\\ &=\|f\|_1\,\|g\|_2. \end{align} If $\pi(f)=0$, then $f*\xi=0$ for all $\xi\in L^2(G)$ which implies that $f=0$, $\|\cdot\|_r$ is a norm.
As for C$^*$, $$ \pi(f*g)\xi=(f*g)*\xi=f*(g*\xi)=\pi(f)\pi(g)\xi, $$ so $\pi$ is multiplicative. After you check that $\pi(f^*)=\pi(f)^*$, you have that $\pi$ is a $*$-homomorphism. Then $$ \|f\|_r^2=\|\pi(f)\|^2=\|\pi(f)^*\pi(f)\|=\|\pi(f^*f)\|=\|f^*f\|_r. $$