TITLE: Prove that $L^1$ is a Banach algebra with multiplication defined by convolution
QUESTION [5 upvotes]: To be more specific, prove that $L^1(\mathbb{R}^n)$ with multiplication defined by convolution:
$$
(f\cdot g)(x)=\int_\mathbb{R^n}f(x-y)g(y)dy
$$
is a Banach algebra. All the properties of Banach algebra are easily proved except the last one:
$$
\|f\cdot g\|\le \|f\|\|g\|.
$$
Could anyone give me a hint?
Actually I found a proof on the web, but I don't see why the last equality holds.
$$\|f* g\|_1  = \int |f*g(x)|dx \le \iint |f(x-y) g(y)|dydx = \|f\|_1 \|g\|_1$$

REPLY [2 votes]: \begin{align}
\iint |f(x-y)g(y)|\;dx\;dy &=
\int\left[\int|f(x-y)|\;dx\right]|g(y)|\;dy \\
&= \int\left[\int|f(x)|\;dx\right]|g(y)|\;dy \\
&= \left[\int|f(x)|\;dx\right]\;\left[\int |g(y)| \;dy\right]\\
& = \|f\|_1\;\|g\|_1
\end{align}