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It is known that the bilinear Strichartz estimate $$\Vert e^{it\Delta}u_0\cdot \overline{e^{it\Delta}v_0}\Vert_{L_t^2L^2(\mathbb{R}^d)}\lesssim_{\delta} \Vert u_0\Vert_{H^{-1/2+\delta}(\mathbb{R}^d)}\Vert v_0\Vert_{H^{(d-1)/{2}-\delta}(\mathbb{R}^d)},\quad\delta>0$$ for the Schrodinger equation in dimension $d\geq 2$, fails for $\delta=0$ (it holds true with $\delta=0$ for dyadically localised data, but the dyadic contributions do not lead to a finite sum in general).

I'm wondering what happens in the case $u_0=v_0$, as in principle one may expect some cancellations when summing up the dyadic contributions.

More precisely, one can consider the following inequality: $$ \Vert e^{it\Delta}u_0\Vert_{L_t^4L^4(\mathbb{R}^d)}^2\lesssim\Vert u_0\Vert_{H^{-1/2}(\mathbb{R}^d)}\Vert u_0\Vert_{H^{(d-1)/{2}}(\mathbb{R}^d)}\qquad(*)$$ Is estimate (*) actually true? Thank you for any suggestion.

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In general spatial dimensions $d\geq 2$, (non-endpoint) Strichartz for Schrodinger holds with

$$ \| e^{it \Delta} u_0 \|_{L^{q}_t L^p_x} \lesssim \| u_0\|_{L^2}$$

where $p\in (2, \frac{2d}{d-2})$ and $q = \frac{4p}{d(p-2)}$. To get $L^4$ in time you want

$$ q = 4 \iff p = \frac{2d}{d-1} \tag{1}$$

So in $d = 2$ you have the $L^4_t L^4_x$ estimate based on initial date in $L^2$.

Notice next that we have the Sobolev inequality

$$ \|f\|_{L^4_x} \lesssim \|f\|_{W^{\frac{d-2}{4},p}} $$

where $p$ is as in (1). So this means we have the Strichartz inequality

$$ \|e^{it\Delta} u_0\|_{L^4_t L^4_x} \lesssim \| u_0\|_{H^{(d-2)/4}} $$

using that the linear Schrodinger equation commutes with differentiation.

A simple interpolation gives you finally that

$$ \|u_0\|_{H^{(d-2)/4}}^2 \lesssim \|u_0\|_{H^{-1/2}} \|u_0\|_{H^{(d-1)/2}}$$

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  • $\begingroup$ Thank you! Incidentally, your argument is essentially the same used in the proof of bilinear estimates when the data have comparable frequencies. This actually puzzles me a bit: in principle, setting e.g. $u:=e^{it\Delta}u_0$, in the bilinear expression we have terms of the form $Re(u_j\overline{u_k})$, with (say) $j>>k$. I would like to understand, intuitively, why the contributions of these terms can be summed. $\endgroup$
    – Capublanca
    Commented Nov 17, 2020 at 17:42
  • $\begingroup$ Do you have a handy reference for the proof of the bilinear estimates you are looking at? I can take a look and see if I can see what's going on. $\endgroup$ Commented Nov 17, 2020 at 18:00
  • $\begingroup$ Sure, for example you can refer to Lemma 3.4 in "Global well-posedness and scattering for the energy critical non-linear Schrodinger equation in $\mathbb{R}^3$", by the I- team. Thank you. $\endgroup$
    – Capublanca
    Commented Nov 17, 2020 at 18:13
  • $\begingroup$ I don't think it is so much that they can be summed, but that in the symmetric/quadratic case, when summing, you can choose to put (when $j\gg k$) $\| u_j u_k\|_{L^2}$ into $\|u_k\|_{H^{-1/2}} \|u_j\|_{H^{(d-1)/2}}$ instead of the other way around which is what the Lemma is trying to accomplish. $\endgroup$ Commented Nov 17, 2020 at 19:18

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