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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!

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    $\begingroup$ The Cauchy-Schwarz inequality holds in any inner product space. So consider the left hand side of the inequality as defining an inner product. The bit about the Fourier transform being positive corresponds to this inner product satisfying the axiom about positive definiteness. $\endgroup$ – Mike Hawk Jan 5 '18 at 18:07
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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 $$ (g,h):=\int g\overline h $$

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  • $\begingroup$ yes, I tried that, but the $|x-y|$ term does not fit in the picture (or I do not see why) $\endgroup$ – mvenzin Jan 5 '18 at 18:05
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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 $\langle \cdot, \cdot \rangle_\lambda$.

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  • $\begingroup$ Are you proposing to prove the Cauchy-Schwartz inequality on the RHS (on the Fourier side) or is this just to prove the homogeneity?. I do not see how on the RHS it is easier than in the middle.... $\endgroup$ – mvenzin Jan 8 '18 at 17:25

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