Positive integral Let $\alpha \geq0.$ Let $f:\mathbb{R^q} \to \mathbb{R}$ be a bounded measurable function such that
$$\int_{\mathbb{R}^q\times\mathbb{R}^q}\frac{|f(x)f(y)|}{|x-y|^{\alpha}}dxdy<\infty.$$
Prove that
$$\int_{\mathbb{R}^q\times\mathbb{R}^q}\frac{f(x)f(y)}{|x-y|^{\alpha}}dxdy\geq 0.$$
I tried to use Fubini's theorem and a change of variable $u=x-y,$ but this didn't work.
Any ideas? Does it help to find $c,\beta>0$ such that $|x|^{-\alpha}=|\,\unicode{x2219}\,|^{-\beta}*|\,\unicode{x2219}\,|^{-\beta}(x)$ ? How?
 A: A not so easy way to prove this is to observe that
\begin{align}
\frac{1}{|x-y|^\alpha} = C_{\alpha, q}\int^\infty_0 \int_{\mathbb{R}^q} s^{\frac{\alpha+q}{2}-1}e^{-\pi |x-z|^2s}e^{-\pi |y-z|^2s}dzds 
\end{align}
where $C_{\alpha, q}$ is some positive constant. Then, we have that
\begin{align}
\int_{\mathbb{R}^q\times \mathbb{R}^q}\frac{f(x)f(y)}{|x-y|^\alpha}d xd y =&\,  C_{\alpha, q}\int^\infty_0 s^{\frac{\alpha+q}{2}-1}\int_{\mathbb{R}^q} \int_{\mathbb{R}^q\times \mathbb{R}^q}f(x)e^{-\pi |x-z|^2s} f(y) e^{-\pi |y-z|^2s}d xd ydzds \\
=&\, C_{\alpha, q}\int^\infty_0 s^{\frac{\alpha+q}{2}-1} \int_{\mathbb{R}^q}\left( \int_{\mathbb{R}^q}f(x)e^{-\pi |x-z|^2s} dx\right)^2dz ds\ge 0.
\end{align}
The above formula is called the Fefferman-de la Llave representation of the singular kernel $|x-y|^{-\alpha}$.
Alternative: An easier way to prove the claim is to observe that
\begin{align}
I=\int_{\mathbb{R}^q\times \mathbb{R}^q}\frac{f(x)f(y)}{|x-y|^\alpha}d xd y = \int_{\mathbb{R}^q} \overline{f(x)}\ (u_{\alpha}\ast f)(x)\, dx
\end{align}
since $f(x) = \overline{f(x)}$, where $u_\alpha(x)  = |x|^{-\alpha}$.
Then by Plancherel's formula, we have that
\begin{align}
I= \int_{\mathbb{R}^q} \overline{\widehat{f}(\xi)}\ \widehat{(u_\alpha\ast f)}(\xi)\ d\xi = \int_{\mathbb{R}^q} \ \widehat{u_\alpha}(\xi)\, |\widehat{f}(\xi)|^2\ d\xi.
\end{align}
Since
\begin{align}
\widehat{u_\alpha}(\xi) = c_{\alpha, q} |\xi|^{q-\alpha}
\end{align}
where $c_{\alpha, q}>0$, then we arrived at the desired result.
