One can read in this document (Proposition 64, p. 60) that there exists a function $F \in L^2_tL^{1}_x (\mathbb{R}_t\times \mathbb{R}^3_x)$ which is supported on frequencies $|\xi| \sim 1$ and such that $\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt' \notin L^2_x$. Define $F(t,x)=\eta(t/T)\psi(x-te_3-\beta_te_1)$ where $\beta_t$ is a standard Brownian motion, $\psi$ is localized in frequency around $|\xi| \sim 2$ and $\eta$ a smooth cutoff.

Then it is not difficult to see that $\|F\|_{L^2_xL^1_x} \lesssim \sqrt T$ and $\|\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt'\|_{L^2_x}$ has infinite expectation (we take expectation with respect to the Brownian motion) hence the result. For the latter the proof uses Fourier-Plancherel and a lower bound on the frequency set $|\xi| \sim \xi_3+\xi_1^2$, which seems to work fine with the $\psi (\cdot -te_3-\beta_te_1)$.

However I was surprised by such a non-inuitive counterexample. If one takes $F(t,x)=\eta(t/T)\psi(x)$ then it is not a counter-example. Can someone explain to me the idea behind the clever substitution $\psi \to \psi (\cdot - te_3-\beta_te_1)$? I do not understand why this is a natural attempt? There might be some "singularity propagation on a cone" thing behind but it is not clear to me. For the use of the brownian motion I do not get the point of why it is usefull. Otherwise, is it possible to give a more intuitive counter-example?

Thank you very much,

One can read in Oh - Probabilistic perspectives in nonlinear dispersive PDEs (Proposition 64, p. 60) that there exists a function $F \in L^2_tL^{1}_x (\mathbb{R}_t\times \mathbb{R}^3_x)$ which is supported on frequencies $\lvert\xi\rvert \sim 1$ and such that $\int_{\mathbb{R}} e^{it\lvert\nabla\rvert}F(t')dt' \notin L^2_x$. Define $F(t,x)=\eta(t/T)\psi(x-te_3-\beta_te_1)$ where $\beta_t$ is a standard Brownian motion, $\psi$ is localized in frequency around $\lvert\xi\rvert \sim 2$ and $\eta$ is a smooth cutoff.

Then it is not difficult to see that $\|F\|_{L^2_xL^1_x} \lesssim \sqrt T$ and $\left\|\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt'\right\|_{L^2_x}$ has infinite expectation (we take expectation with respect to the Brownian motion), hence the result. For the latter the proof uses Fourier–Plancherel and a lower bound on the frequency set $\lvert\xi\rvert \sim \xi_3+\xi_1^2$, which seems to work fine with the $\psi (\cdot -te_3-\beta_te_1)$.

However I was surprised by such a non-inuitive counterexample. If one takes $F(t,x)=\eta(t/T)\psi(x)$ then it is not a counter-example. Can someone explain to me the idea behind the clever substitution $\psi \to \psi (\cdot - te_3-\beta_te_1)$? I do not understand why this is a natural attempt. There might be some "singularity propagation on a cone" thing behind but it is not clear to me. For the use of the Brownian motion I do not get the point of why it is useful. Otherwise, is it possible to give a more intuitive counter-example?

One can read in this document (Proposition 64, p. 60) that there exists a function $F \in L^2_tL^{1}_x (\mathbb{R}_t\times \mathbb{R}^3_x)$ which is supported on frequencies $|\xi| \sim 1$ and such that $\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt' \notin L^2_x$. Define $F(t,x)=\eta(t/T)\psi(x-te_3-\beta_te_1)$ where $\beta_t$ is a standard Brownian motion, $\psi$ is localized in frequency around $|\xi| \sim 2$ and $\eta$ a smooth cutoff.

Then it is not difficult to see that $\|F\|_{L^2_xL^1_x} \lesssim \sqrt T$ and $\|\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt'\|_{L^2_x}$ has infinite expectation (we take expectation with respect to the Brownian motion) hence the result. For the latter the proof uses Fourier-Plancherel and a lower bound on the frequency set $|\xi| \sim \xi_3+\xi_1^2$, which seems to work fine with the $\psi (\cdot -te_3-\beta_te_1)$.

However I was surprised by such a non-inuitive counterexample. If one takes $F(t,x)=\eta(t/T)\psi(x)$ then it is not a counter-example. Can someone explain to me the idea behind the clever substitution $\psi \to \psi (\cdot - te_3-\beta_te_1)$? Otherwise, is it possible to give a more intuitive counter-example?

Thank you very much,

One can read in this document (Proposition 64, p. 60) that there exists a function $F \in L^2_tL^{1}_x (\mathbb{R}_t\times \mathbb{R}^3_x)$ which is supported on frequencies $|\xi| \sim 1$ and such that $\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt' \notin L^2_x$. Define $F(t,x)=\eta(t/T)\psi(x-te_3-\beta_te_1)$ where $\beta_t$ is a standard Brownian motion, $\psi$ is localized in frequency around $|\xi| \sim 2$ and $\eta$ a smooth cutoff.

Then it is not difficult to see that $\|F\|_{L^2_xL^1_x} \lesssim \sqrt T$ and $\|\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt'\|_{L^2_x}$ has infinite expectation (we take expectation with respect to the Brownian motion) hence the result. For the latter the proof uses Fourier-Plancherel and a lower bound on the frequency set $|\xi| \sim \xi_3+\xi_1^2$, which seems to work fine with the $\psi (\cdot -te_3-\beta_te_1)$.

However I was surprised by such a non-inuitive counterexample. If one takes $F(t,x)=\eta(t/T)\psi(x)$ then it is not a counter-example. Can someone explain to me the idea behind the clever substitution $\psi \to \psi (\cdot - te_3-\beta_te_1)$? I do not understand why this is a natural attempt? There might be some "singularity propagation on a cone" thing behind but it is not clear to me. For the use of the brownian motion I do not get the point of why it is usefull. Otherwise, is it possible to give a more intuitive counter-example?

Thank you very much,