# If int_(RR) |f(x)|dx and int_(RR) |g(x)|dx are bounded, show that int_(RR) |(f ***g)(x)|dx <= int_(RR) |f(x)|dx * int_(RR) |g(x)|dx.

If ${\int }_{\mathbb{R}}|f\left(x\right)|dx$ and ${\int }_{\mathbb{R}}|g\left(x\right)|dx$ are bounded, show that
${\int }_{\mathbb{R}}|\left(f\star g\right)\left(x\right)|dx\le {\int }_{\mathbb{R}}|f\left(x\right)|dx\cdot {\int }_{\mathbb{R}}|g\left(x\right)|dx$
From the definition we have $\left(f\star g\right)\left(x\right)={\int }_{\mathbb{R}}f\left(x-\tau \right)g\left(\tau \right)\phantom{\rule{thinmathspace}{0ex}}d\tau$. I guess I need to do work on this right side, but I don't see how to simplify to get the desired inequality
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vyhlodatis
Applying the integral to the convolution:
$|{\int }_{\mathbb{R}}\left(f\star g\right)\left(x\right)dx|=|{\int }_{\mathbb{R}}{\int }_{\mathbb{R}}f\left(x-\tau \right)g\left(\tau \right)d\tau dx|=|{\int }_{\mathbb{R}}{\int }_{\mathbb{R}}f\left(x-\tau \right)g\left(\tau \right)dxd\tau |$
$\le {\int }_{\mathbb{R}}{\int }_{\mathbb{R}}|f\left(x-\tau \right)||g\left(\tau \right)|dxd\tau ={\int }_{\mathbb{R}}|f\left(x\right)|dx\cdot {\int }_{\mathbb{R}}|g\left(x\right)|dx$