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I am working on the decoupling inequality developed by Bourgain and Demeter: https://arxiv.org/abs/1604.06032.

Is there an example where we have strict inequality in Theorem 1.1, say in the case $n=2$ with $\delta^{-\alpha}$ losses in the power? Here by stricty inequality I mean $\alpha>0$ is a fixed positive constant depending on $p$ only.

This question may seem stupid, but I have not found a reference for that.

Edited: precisely, I am asking the following question.

For $g\in L^1([0,1])$ and $I\subseteq [0,1]$, define $E_I g(x,y)=\int_I g(s)e^{2\pi i (s x+s^2 y)}ds$. Is it true that for every $2\leq p\leq 6$ and every $\epsilon>0$, there is a constant $c_\epsilon>0$, depending on $p$ and $\epsilon$ only, such that for any $g\in L^1([0,1])$, any$\delta\in 2^{-2\mathbb N}$ and any ball $B$ of radius $\delta^{-1}$, we have $$ \left\| Eg\right\|_{L^p(w_B)}\geq c_\epsilon \delta^{\epsilon}\left(\sum_{j=1}^{\delta^{-1/2}}\left\|E_{[(j-1)\delta^{1/2},j\delta^{1/2}]}g\right\|^2_{L^p(w_B)}\right)^{1/2}? $$

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    $\begingroup$ The answer to your edited question is "no", as follows by applying the Bourgain-Demeter inequality with $\epsilon$ replaced by (say) $\epsilon/2$. $\endgroup$
    – Terry Tao
    Sep 21, 2020 at 15:37
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    $\begingroup$ (Assuming of course that you meant to write $\delta^{-\epsilon}$ instead of $\delta^\epsilon$.) $\endgroup$
    – Terry Tao
    Sep 21, 2020 at 15:48
  • $\begingroup$ @TerryTao Thanks for the answer! I actually kind of figured out the answer in some other way. We may choose $g$ such that $Eg_j$ has very sparse physical support. Then the LHS becomes essentially $\left\|\left\|E_j\right\|_{L^p}\right\|_{l^p(j)}$, and this can be arbitrarily smaller than the RHS for $p>2$. Is that correct? $\endgroup$ Sep 22, 2020 at 1:15
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    $\begingroup$ If you are quantifying over all $g$ in your question, then yes. If you are asking whether the inequality stated holds for at least one $g$ (which is the usual interpretation of what it means for the opposing inequality valid for all $g$ to be "sharp"), then no. $\endgroup$
    – Terry Tao
    Sep 22, 2020 at 1:45
  • $\begingroup$ @TerryTao Sorry for my unclear presentation. I actually meant that $c_\epsilon$ is independent of $g$; I re-edited my question. Also, I do think that I meant $\delta^{\epsilon}$ instead of $\delta^{-\epsilon}$ since I would allow an $\epsilon$-loss in the reverse direction. $\endgroup$ Sep 22, 2020 at 3:09

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Certainly for $n=2$ the $\delta^{-\epsilon}$ factor can not be dispensed with entirely. It is known that

$$ || \sum_{n\leq N} e(nx + n^2 t) ||_{L^4(dxdt)} \approx N^{1/2} \log^{1/6} N $$

This is essentially the two dimensional case of Vinogradov's mean value theorem, and is discussed in Bourgain's original 1991 discrete restriction papers on the periodic Schrodinger equation. A precise asymptotic for this quantity can be found in: https://mathscinet.ams.org/mathscinet-getitem?mr=2661311.

It is unclear what one should expect in the higher dimensional cases. It is also unclear, for instance, if the $N^{\epsilon}$ factor can be removed at the endpoint in the multi-linear restriction theorem.

Even reducing the $\delta^{-\epsilon}$ factor to a logarithmic factor for $n=2$ (for decoupling or the discrete parabolic restriction theorem) is open, interesting, and seems to require new ideas.

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  • $\begingroup$ Thanks for your answer. However, I guess you are not exactly answering my question. I am considering an example which gives strict inequality, not that $\epsilon$ can be removed or not. $\endgroup$ Apr 24, 2020 at 3:44
  • $\begingroup$ I don't understand what you mean by "gives strict inequality", then. $\endgroup$
    – Mark Lewko
    Apr 24, 2020 at 3:46
  • $\begingroup$ I still don't understand what you are looking for (following your revision to the question). The claim in Theorem 1.1 is that $\epsilon$ can be taken to be any real number greater than $0$ provided, of course, the implied constant is taken sufficiently large depending on $\epsilon$. $\endgroup$
    – Mark Lewko
    Apr 24, 2020 at 3:51
  • $\begingroup$ Basically I am asking whether or not the reverse inequality is true for decoupling. $\endgroup$ Apr 24, 2020 at 4:40
  • $\begingroup$ Thomas: I suggest you write down the precise inequality you would like to see disproved. If there was an extermal or almost extremal for some fixed $\epsilon$, then the result wouldn't hold for all $\epsilon>0$, contradicting the statement of the theorem. $\endgroup$
    – Mark Lewko
    Apr 24, 2020 at 5:14

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