A quadratic estimate for the Schrödinger equation

Question

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.

Answer 1

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

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