Norm triangle inequality for convolutions proof

Question

I'm trying to prove that $$\|f*g\|_{L_1}\le{\|f\|_{L_1}\|g\|_{L_1}}$$ with respect to a Haar measure over a group G. Using Fubini's theorem, I'm up to $$\|f*g\|_{L_1}\le{\int_G|f(y)|(\int_G|g(y^{-1}x)|dx)dy}.$$ I'm just a bit stuck regarding how to change variables to get the integrand in the second interval to a function of one variable. Does the Haar measure come in to play here? Many thanks in advance.

Answer 1

If $G$ is a locally compact group, then $G$ has a left invariant measure $$ \mu(aX)=\mu(X) $$ for a measurable set $X$ and $a\in G$ called the left Haar measure. Using this, we can conclude the following: $$ \int_G |f(y)|(\int_G |g(y^{-1}x)|dx)dy \\ =\int_G|f(y)|(\int_G|g(x)|dx)dy \\ =||f||_{L^1}||g||_{L^1}.~_{\square} $$