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.



A quadratic estimate for the Schrödinger equation
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


$$ \|u\_0\|\_{H^{(d-2)/4}}^2 \lesssim \|u\_0\|\_{H^{-1/2}} \|u\_0\|\_{H^{(d-1)/2}}$$


