Cauchy-Schwartz inequality in a convolution I stumbled over this Cauchy-Schwartz inequality while reading (https://link.springer.com/article/10.1007%2Fs00208-014-1046-2#Equ12 , last page before acknowledgements). Here is the situation:
$$\int_{\mathbb{R}^d\times\mathbb{R}^d}\frac{g(x)h(y)}{|x-y|^{\lambda}}dxdy \leq \left(\int_{\mathbb{R}^d\times\mathbb{R}^d}\frac{g(x)g(y)}{|x-y|^{\lambda}}dxdy\right)^{\frac{1}{2}}\left(\int_{\mathbb{R}^d\times\mathbb{R}^d}\frac{h(x)h(y)}{|x-y|^{\lambda}}dxdy\right)^{\frac{1}{2}}$$
$\lambda < d$. I do not see how to handle this kind of inequality. They claim this holds since the Fouriertransform of $|x-y|^{\lambda}$ is positive. I am very interested in the argument leading to this since I do not know how to handle these inequalities (any reference is very appreciated, I did not find any). 
The motivation for this comes from the fact that I am stuck with showing that
$$\left|\left(|\int_{\mathbb{R}^d\times\mathbb{R}^d}u_n(x)u_n(y)f(x-y)dxdy|\right)^{\frac{1}{2}} - \left(|\int_{\mathbb{R}^d\times\mathbb{R}^d}u(x)u(y)f(x-y)dxdy|\right)^{\frac{1}{2}}\right| \leq \\\left(\left|\int_{\mathbb{R}^d\times\mathbb{R}^d}\{u_n(x)-u(x)\}\{u_n(y)-u(y)\}f(x-y)dxdy\right|\right)^{\frac{1}{2}}$$
where $u$, $u_n$ and $f$ all positive functions. $u$ lies in some $L^p$ space and $f$ in some weak $L^p$ space (such that the whole thing is well defined). I am very interested in the arguments!
Thank you!

It may have to do with the fact that $(u*v)^\wedge(\xi)=\hat{u}(\xi)\hat{v}(\xi)$,
for $u,v$ Schwartz-class functions, that is.
Edit: As Mike Hawk points out, it has to do with
$$ |(g,h)|\le \|g\|_2\,\|h\|_2, $$
where $$ (g,h):=\int g\overline h $$
