Show that for $f,g \in L^{1}(G)$ we have $f*g \in L^{1}(G)$.

Question

Let G be a locally compact group and consider ($L^{1}(G), \|\cdot\|_{1}$) where $\|f\|_{1}=\int_{G}|f(x)|dx$ for $f \in L^{1}(G)$.

Define convolution $*$ by $(f*g)(y)=\int_{G}f(x)g(x^{-1}y)dx$ for $f,g \in L^{1}(G)$ and $y \in G$

Show that for $f,g \in L^{1}(G)$ we have $f*g \in L^{1}(G)$.

I know that this amounts to showing that $\int_{G}|(f*g)(y)|dy=\int_{G}|\int_{G}f(x)g(x^{-1}y)dx|dy<\infty$,but I really don't know how to show that this integral is finite.

Answer 1

$\int |g(x^{-1}y)|dy=\int |g(y)|dy$ since you are integrating w.r.t. Haar measure. Hence, By Fubini/Tonelli Theorem we get $\int \int |g(x^{-1}y)|dy|f(x)|dx=\|f\|_1\|g\|_1$.

Answer 2

$$\int \int |f(x)g(x^{-1})|dxdy=^{TONELLI}\int|f(x)|(\int|g(x^{-1}y|dy)dx=||f||_1||g||_1$$

by a simple change of variables(exploiting the fact the $G$ is a group with a Haar measure)