Submultiplicativity of convolution in the L1 norm

For $x\in G$, [F5] applied to the integrable function $y\mapsto f(y)g(y^{-1}x)$ gives $\mathrm{\mid }(f\ast g)(x)\mathrm{\mid }\le {\int }_G\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }d\mu (y)$.

The kernel $H(x,y):=\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }$ is real, nonnegative, continuous, and compactly supported: it is a product of continuous functions, and $H(x,y)\mathrm{\ne }0$ forces $y\in \mathrm{supp}f$ and $x\in \mathrm{supp}f\cdot \mathrm{supp}g$, a compact set.

For each $y\in G$ left invariance [F3] applied to $H_y(x):=\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }$ gives ${\int }_{G}H(x,y)d\mu (x)=\mathrm{\mid }f(y)\mathrm{\mid }{\int }_G\mathrm{\mid }g(y^{-1}x)\mathrm{\mid }d\mu (x)=\mathrm{\mid }f(y)\mathrm{\mid }{\int }_G\mathrm{\mid }g(w)\mathrm{\mid }d\mu (w)=\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_1$.

Integrating step 1.2 over $y$ for each $x$ and then over $x$, the iterated integral ${\int }_G{\int }_{G}H(x,y)d\mu (y)d\mu (x)$ exists and, by [F2] under [A1], equals the reverse iterated integral ${\int }_G{\int }_{G}H(x,y)d\mu (x)d\mu (y)$.

Chaining steps 1.1, 2.1 and 1.3, $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1={\int }_G\mathrm{\mid }(f\ast g)(x)\mathrm{\mid }d\mu (x)\le {\int }_G{\int }_{G}H(x,y)d\mu (y)d\mu (x)={\int }_G{\int }_{G}H(x,y)d\mu (x)d\mu (y)=\mathrm{\parallel }g{\mathrm{\parallel }}_1{\int }_G\mathrm{\mid }f(y)\mathrm{\mid }d\mu (y)=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$. ∎