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 $$ \left\| Eg\right\|_{L^p(w_B)}\geq c_\epsilon \delta^{\epsilon}\left(\sum_{j=1}^{\delta^{-1/2}}\left\|E_{[(j-1)\delta^{1/2},j\delta^{1/2}]}g\right\|^2_{L^p(w_B)}\right)^{1/2}? $$

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ (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
  • 1
    $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.