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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?

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    $\begingroup$ @YCor is it better now? If there is need for more details about the proof in the document I can provide them. $\endgroup$
    – J.Mayol
    Commented Sep 18, 2020 at 15:40
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    $\begingroup$ The Brownian motion term is needed to destroy an "unwanted" cancellation in the Duhamel integral that would otherwise occur if this randomization was not present. I discuss this a little in my own unpublished notes on this example at math.ucla.edu/~tao/preprints/Expository/stein.dvi $\endgroup$
    – Terry Tao
    Commented Sep 18, 2020 at 18:30
  • $\begingroup$ @TerryTao thank you, your notes are nice, at least I understand a little more why the Brownian motion is needed. It seems like we can replace the Brownian motion with some $C^{1/2}$ function (with appropriate norm) and try to work with it, although this basically reduces to using the Brownian motion. One last thing: that the $-te_3$ is needed is because we want to "kill" the oscillatory factor $e^{it|\xi|}$ in order to force the integral to diverge? Am I right? $\endgroup$
    – J.Mayol
    Commented Sep 18, 2020 at 18:51
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    $\begingroup$ Yes. In physical space this corresponds to trying to create a "sonic boom" effect by having the light cones from the sources at different times all line up. $\endgroup$
    – Terry Tao
    Commented Sep 18, 2020 at 21:00

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