I am studying a proof, and I do not understand one of the steps.
Let
$$u(x,y)=\int_{\mathbb R}f(s)\frac{y}{\pi(y^2+(x-s)^2)}\mathrm ds,$$
where $f\in L^1(\mathbb R)$ and $(x,y)\in \mathbb R\times (0,+\infty)$.
I have to show that:
$$\sup_{y\geqslant 0} \int_{\mathbb R} \vert u(x,y)\vert\mathrm d x<\infty.$$
I do not understand why this should be true.
May be a well-chosen change of variable? But I don't see which one...
Do you have any idea on how to prove this?
Note that
$$\int_{-\infty}^\infty |u(x,s)|\,dx\le \int_{-\infty}^\infty \int_{-\infty}^\infty |f(s)|\frac{|y|}{\pi(y^2+(x-s)^2)}\,ds\,dx$$
Then, we can use Fubini's Theorem to interchange the order of integration and write
$$\begin{align} \int_{-\infty}^\infty \int_{-\infty}^\infty |f(s)|\frac{|y|}{\pi(y^2+(x-s)^2)}\,ds\,dx&=\int_{-\infty}^\infty |f(s)|\underbrace{\left(\int_{-\infty}^\infty \frac{|y|}{\pi(y^2+(x-s)^2)}\,dx\right)}_{=1}\,ds\\\\ &=\int_{-\infty}^\infty |f(s)|\,ds\\\\ &<\infty \end{align}$$