The convolution is in $L^1$

Question

According to my notes:

The convolution is in $L^1$. Indeed

$$\left| \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} f(y) g(x-y) dy\right) dx\right| \leq \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y) g(x-y)| dx dy= ||f||_{L^1(\mathbb{R}^n)} ||g||_{L^1(\mathbb{R}^n)}$$

First of all, the formula of the convolution is this:

$$(f \ast g) (x)=\int_{\mathbb{R}^n} f(y) g(x-y) dy$$

So in order to show that the convolution is in $L^1$ why do we take a double integral?

Also why does it hold that $\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y) g(x-y)| dx dy= ||f||_{L^1(\mathbb{R}^n)} ||g||_{L^1(\mathbb{R}^n)}$ ?

Don't we have that $||f||_{L^1(\mathbb{R}^n)}=\int_{\mathbb{R}^n} |f(x)| dx$ and $||g||_{L^1(\mathbb{R}^n)}=\int_{\mathbb{R}^n} |g(x)| dx$ ?

EDIT: Could you explain to me why we can use Fubini?

Answer 1

Note that in order to show that the function $f \ast g$ is in $L^1$ we just have to integrate its absolute value over the variable, in this case over $x$. For the detailed calculation consider \begin{equation*} \begin{split} \int_{\mathbb{R}^n}| (f \ast g)(x)| dx=&\int_{\mathbb{R}^n}\left| \int_{\mathbb{R}^n} f(y) g(x-y) dy\right| dx \\ \leq& \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y) g(x-y)| \;dy\; dx\\ =& \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y)| | g(x-y)| \;dy\; dx\\ =&\int_{\mathbb{R}^n} |f(y)|\left(\int_{\mathbb{R}^n} | g(x-y)|\; dx\right)\; dy \\ =&\int_{\mathbb{R}^n} |f(y)|\left(\int_{\mathbb{R}^n} | g(v)| \;dv\right)\; dy\\ =&\int_{\mathbb{R}^n} |f(y)|\|g\|_{L^1(\mathbb{R}^n)} \;dy \\ =& ||f||_{L^1(\mathbb{R}^n)} ||g||_{L^1(\mathbb{R}^n)} \end{split} \end{equation*}

where we used Fubini to interchange $dx,dy$ and we substituted $x-y$ with $v$.