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Let $\gamma(x,y)$ be some complex valued function in $L^{2}(\mathbb{R}^{2})$ such that $$ \gamma(x,y)=\overline{\gamma(y,x)},\forall x,y\in \mathbb{R} $$ Let $S=(1-\Delta)^{1/2}$ acting on $\gamma$. If we know that $$ \forall f\in L^{2}(\mathbb{R}), \int f(z)(\int \gamma(x,z)f(x)dx)dz>0 $$ Then can we conclude that $$ |S(\gamma)(x,y)|^{2}\le S(\gamma)(x,x)S(\gamma)(y,y),\forall x,y\in \mathbb{R} $$ from some spectral type argument? If $\gamma(x,y)$ is a positive semi-definite Hermitian matrix, then this should be easy ($l^{2}$ instead of $L^{2}$ case). But for $L^{2}$ I do not know how to prove it.

My thought is we have $$ |S\gamma(x,y)|^{2}=\int S(\gamma)(x,y)S(\gamma)(y,x)dxdy=\int \gamma(x,y)(1-\Delta)\gamma(y,x)dxdy $$ but this does not seem to be helpful at all. The main difficulty I have is $S\gamma$ should be a positive, Hermitian and self-adjoint integral operator. But here we are not integrating it against anything, and I do not know how to make use of its properties.

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  • $\begingroup$ Awesome profile page there, Bombyx. $\endgroup$ – Robert Lewis Jun 19 '14 at 7:03
  • $\begingroup$ I asked this on MO yesterday, hopefully the problem is resolved now. $\endgroup$ – Bombyx mori Jun 20 '14 at 19:59
  • $\begingroup$ Shouldn't that be a "Cauchy-Schwarz type formula"? Or "Cauchy-Bunyakovsky-Schwarz type inequality"? The "Schwarz" doesn't stand for Laurent Schwarz, but Hermann Schwarz en.wikipedia.org/wiki/Hermann_Amandus_Schwarz ... $\endgroup$ – UwF Jun 21 '14 at 13:13
  • $\begingroup$ Btw, to see the MO post mathoverflow.net/questions/172266/… $\endgroup$ – UwF Jun 21 '14 at 13:21

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