Cauchy-Schwartz inequality in a convolution

Question

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!

Answer 1

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 $$ (g,h):=\int g\overline h $$

Answer 2

I think the comment of Mike Hawk pretty much answers the question. The remark about the Fourier transform of $|x - y|^{-\lambda}$ being positive can be used since, by the Plancherel formula $$ \langle u, v \rangle_{\lambda} = \int \int \frac{\bar{u}(x) v(y)}{|x-y|^\lambda} = \int \int \psi_\lambda(\xi, \eta) \widehat u(\xi) \widehat v(\eta), $$ where $\psi_\lambda(\xi,\eta) \geq 0$ is the Fourier transform of $|x - y|^{-\lambda}$. The identity you want to prove is just the Cauchy-Schwartz inequality for the inner product $\langle \cdot, \cdot \rangle_\lambda$.