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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.

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  • $\begingroup$ With $dy$ the left Haar measure then $d(xy)=dy, d(yx)=\Delta(x) dy$, $\Delta(zx)d(yzx) = \Delta(x)\Delta(z)dy$, $\Delta(y^{-1}x^{-1}) d(yx)=\Delta(y^{-1}) dy$ which is a right Haar measure thus $= C dy^{-1}$ $\endgroup$ – reuns Jun 13 '19 at 18:23
  • $\begingroup$ I don't quite follow you. Does that mean that $\Delta(y^{-1}x) d\lambda(y) = d\lambda(yy^{-1}x) = d\lambda(x)$ ? $\endgroup$ – M.G Jun 13 '19 at 18:39
  • $\begingroup$ No, it means that $\Delta(xy^{-1}) dy = \Delta(x)C dy^{-1}$. Do you see that $d(yx)$ is a left Haar measure, thus $= \Delta(x)dy$ ? $\endgroup$ – reuns Jun 13 '19 at 18:41
  • $\begingroup$ I think so, yes. $\endgroup$ – M.G Jun 13 '19 at 18:45

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