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Problem: In Appendix (A.6) of Main paper is written

$$\lVert K(x; t_0, t_1, t_2, \frac{1}{2\pi}q_1, \frac{1}{2\pi}q_2)\rVert_3 \leq \prod_{\nu=1}^{d} \lVert p_{R^{\nu}}^{(d=1)}\rVert_3 \leq C \prod_{\nu=1}^{d}\frac{1}{\langle R^{\nu}\rangle^{\frac{1}{7}} }$$ where we have applied a known bound for the $\ell^3$-norm of the one dimensional propagator, following [11, 13].

Compacted definitions:

$$ \begin{align*} &K(x; t_0, t_1, t_2, \frac{1}{2\pi}q_1, \frac{1}{2\pi}q_2) =\\ &\quad\quad\quad e^{-ic(t_0 + t_1 + t_2)}\times \prod\limits_{\substack{\nu = 1}}^{d}\int_{0}^{2\pi}\frac{dp}{2\pi}e^{ipx^{\nu}}e^{i(t_0\cos(p) + t_1\cos(p + q_1^{\nu}) + t_2 \cos(p + q_2^{\nu}))} \end{align*} $$

and $t_0,t_1,t_2 \in \mathbb{R},\; u_1, u_2 \in \mathbb{T}^d = [0,1]^d, x \in \mathbb{Z}^d$. $R^{\nu} = |t_0 + t_1 e^{iq_1^\nu} + t_2 e^{iq_2^\nu}|$, $\langle x \rangle = \sqrt{1+x^2}$ for all $x \in \mathbb{R}$, and the free propagator is $p_t(x) = \int_{\mathbb{T}^d}dke^{i2\pi x\cdot k}e^{-i t\omega(k)}$ where $\omega(k)\colon \mathbb{T}^d \to \mathbb{R}$ is the dispersion relation satisfying $(DR1)$ through $(DR4)$ on page $91$ of the main paper.

Question: What is the referenced known bound and corresponding proof?

Investigation: See $(21)$ of $[11]$, but note that the paper has a typo on the RHS: $e^{it}$ should be $e^{-it}$. The bound they reference on the following page in Claim $4$ Proposition $5$ I suspect is the referenced bound. However, I am stuck on proving this claim and using it to obtain the above inequalities. In their listed reference is discussed integral limits which may be relevant. Regarding $[13]$, $(2)$ seems most relevant but I have yet to see how this is applied.

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    $\begingroup$ The question is well-posed, and I do not have a problem with its content. Perhaps it is however a good idea not to link directly to SciHub (as you do). While it may be argued that the latter provides an excellent and unique service to academia, SciHub is at least "grey zone" in many geographical areas. $\endgroup$ Commented Apr 6, 2021 at 7:35
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    $\begingroup$ To add to @SamSanders: while SciHub is currently up, it may one day go down (as it has a few times in the past). It would be better if you can replace/amend all SciHub links with actual bibliographic data, so your question can still make sense without dependence on a third party website's availability. $\endgroup$ Commented Apr 6, 2021 at 15:51
  • $\begingroup$ The links have been updated to not use scihub. $\endgroup$
    – hirotaFan
    Commented Apr 6, 2021 at 17:17

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