In what follows, $G$ is a locally compact group with left Haar measure $\lambda$ and modular function $\Delta$.
I came across this statement :
It follows from a careful application of Hölder's inequality that $$ \|f \ast g\Delta^{1/q}\|_\infty \leq \|f\|_p \|g\|_q,$$ where as usual $\frac{1}{p} + \frac{1}{q}=1$.
Being unfamiliar with locally compact non-abelian groups, I need a bit of help with the proof.
Here is what I got so far :
\begin{equation}\begin{aligned} |f \ast g\Delta^{1/q}(x)| &\leq \int_G |f(x)| |g(y^{-1}x)|\Delta^{1/q}(y^{-1}x)\; d\lambda(y) \\&\leq~ \left(\int_G|f(x)|^p\; d\lambda(y)\right)^{1/p}\left(\int_G|g(y^{-1}x)|^q\Delta(y^{-1}x)\; d\lambda(y)\right)^{1/q}\\&\leq~ \|f\|_p \left(\int_G|g(y^{-1}x)|^q\Delta(y^{-1}x)\; d\lambda(y)\right)^{1/q}\end{aligned}\end{equation} Clearly, some change of variable is needed but I'm not quite sure how to proceed.