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Let $J^s = (\mathbb I - \Delta)^{s/2}$, and, $\hat{\mathcal{R}_x u}(\xi)=\frac{-i \xi_1}{|\xi|} \hat{u}(\xi)$, and $u \in H^\infty(\mathbb{T}^2)$, where $u:(\mathbb R, \mathbb{T}^2) \to \mathbb C.$

Using the following:

$$ \int u J^s (\nabla \bar{u}) = \int \hat{u} (1 + |\xi|^2)^{s/2} (-i) \xi \bar{\hat{u}} = - \int (1 + |\xi|^2)^{s/2} i\xi \hat{u} \bar{\hat{u}} = - \int J^s(\nabla u) \bar{u}.$$

I am trying to move $J^s$ from the Riesz transform to $u^2$, is it possible? Namely, I want to get

$$\int_{\mathbb{T}^2} J^s(\partial_x\mathcal{R}_x (\overline{u})) u^2 J^s(\overline{u}) \lesssim \int_{\mathbb{T}^2} \partial_x\mathcal{R}_x (\overline{u}) J^s(u^2) J^s(\overline{u}).$$

I tried so hard for three days, then I figured maybe it is impossible! However, I could not get disproof either!

Any help is appreciated, thanks in advance.

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    $\begingroup$ Take $s=2$ so that $J^s=(1-\Delta)$ and you will see the difficulty $\endgroup$ Commented Jun 11, 2022 at 18:39
  • $\begingroup$ @PieroD'Ancona I know it is difficult but my question is it passible? $\endgroup$
    – Mr. Proof
    Commented Jun 11, 2022 at 23:54

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