1
$\begingroup$

This question arises when I am reading Klainerman&Machedon's paper "On the Uniqueness of Solutions to the Gross-Pitaevskii Hierarchy". The author made a comment on page 3, which in effect is as follows:

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 we can conclude that $$ |S(\gamma)(x,y)|^{2}\le S(\gamma)(x,x)S(\gamma)(y,y),\forall x,y\in \mathbb{R} $$

I want to ask for a hint why this should be true, and how to prove it rigorously. I asked on math.SE site earlier but did not get an answer. I do not know if the question is too easy.

$\endgroup$

1 Answer 1

2
$\begingroup$

Let me introduce the notation $$ \langle \psi,\phi\rangle_\gamma = \int_{\mathbb{R}^2} \bar{\psi}(y) \gamma(x,y) \phi(x) ~\mathrm{d}x~\mathrm{d}y $$ so that $$ \langle \phi,\psi\rangle_\gamma = \overline{\langle \psi,\phi\rangle_\gamma}$$


Firstly, your definition of $S$ I think is not strictly what they have. Your $S$ should be $$S = S_x\cdot S_y = (1 - \partial_x^2)^\frac12 (1-\partial_y^2)^\frac12$$ This has the advantage that $$ \langle \psi,\phi\rangle_{S\gamma} = \langle \sqrt{1 - \partial^2} \psi,\sqrt{1 - \partial^2} \phi\rangle_{\gamma} $$ this comes from formally integrating by parts or using that $\sqrt{1-\partial^2}$ is self-adjoint on suitable spaces.


Now, fix $(x,y)\in \mathbb{R}^2$. If $x = y$ then the inequality you want to prove is trivial. So we look at $x \neq y$. Then we can choose disjoint neighbourhoods $U_x, U_y$. Let $f_x$ and $f_y$ be smooth, real valued functions supported in $U_x$ and $U_y$ respectively. Assume further that they are non-negative and not identically zero.

Define $f^{\lambda,\theta} = \frac1\lambda f_x + e^{i\theta} \lambda f_y$, where $\lambda\in \mathbb{R}_+$ and $\theta\in [0,2\pi]$.

Consider now $$ 0 < \langle \sqrt{1-\partial^2}f^{\lambda,\theta},\sqrt{1-\partial^2}f^{\lambda,\theta}\rangle = \langle f^{\lambda,\theta},f^{\lambda,\theta}\rangle_{S\gamma} $$ This implies $$ 0 < \frac1{\lambda^2} \langle f_x,f_x\rangle_{S\gamma} + \lambda^2 \langle f_y,f_y\rangle_{S\gamma} + 2\Re \langle f_x, e^{i\theta} f_y\rangle_{S\gamma} $$ Choose $$\lambda^2 = \sqrt{ \frac{\langle f_x,f_x\rangle_{S\gamma}}{\langle f_y,f_y\rangle_{S\gamma}}}$$ and $\theta$ such that $\langle f_x,e^{i\theta}f_y\rangle_{S\gamma}$ is purely real. Then we have $$ \left| \langle f_x,f_y\rangle_{S\gamma}\right|^2 \leq \langle f_x,f_x\rangle_{S\gamma} \langle f_y,f_y\rangle_{S\gamma} \tag{1}$$

Observe (1) holds for every $f_x\in C^\infty_0(U_x;\overline{\mathbb{R}_+})$ and $f_y\in C^\infty_0(U_y;\overline{\mathbb{R}_+})$. Now you just need to take approximations of identities to get the desired pointwise estimate from the integrated estimate (1).


Remark: the two parameter choice $\lambda,\theta$ is sometimes called the "amplification trick" which Terry Tao explained wonderfully on his blog.

$\endgroup$
4
  • $\begingroup$ I should probably add that the pointwise estimate only holds, through the approximations of identities, at Lebesgue points of $S\gamma$. If $S\gamma$ were continuous, then this is everywhere. If $S\gamma$ were $L^1$, then this would be almost everywhere. $\endgroup$ Jun 20, 2014 at 10:47
  • $\begingroup$ May I ask what is the reason you let $S_{\gamma}$ to be the product of $S_{x}$ and $S_{y}$? I thought $S_{\gamma}$ is the operator one get from inverting the $(1-\sum \xi_{i}^{2})^{1/2}$ using Fourier transform, and it is not obvious to me that it factorizes. $\endgroup$ Jun 20, 2014 at 23:19
  • $\begingroup$ I checked the notation again and you are right. Sorry! $\endgroup$ Jun 21, 2014 at 9:14
  • $\begingroup$ Good that it is resolved. I should remark that on the Fourier side you see clearly that the operator $(1 + |\xi|^2)^\frac12 (1+ |\eta|^2)^\frac12$ bounds $(1 + |\xi|^2 + |\eta|^2)^\frac12$. So their $S$ is a slightly stronger derivative. $\endgroup$ Jun 23, 2014 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.