On P.23 of https://arxiv.org/pdf/1705.01703.pdf, they seemed to suggest that by the non-negativity of $\psi\big(\frac{k}{N}\big)$ for all $k$, $$ K_N(\xi_0 n)\left[1-\cos\bigg(\frac{2\pi\xi_0 n}{p}\bigg)\right]=\sum_{k\in \mathbb{Z}}e_p(kn)\left(\psi(k/N)-\frac{\psi\big(\frac{k-1}{N}\big)+\psi\big(\frac{k+1}{N}\big)}{2}\right)$$ and $$ \psi(k/N)-\frac{\psi\big(\frac{k-1}{N}\big)+\psi\big(\frac{k+1}{N}\big)}{2}\ll\frac{1}{N^2}\sum_{j=-8}^{8}\psi\left(\frac{k}{N}-\frac{j}{4}\right) $$ imply $$ K_N(\xi_0 n)\left[1-\cos\bigg(\frac{2\pi\xi_0 n}{p}\bigg)\right]\ll\frac{1}{N^2}K_N(\xi_0 n) $$ I fail to see why it is obvious, I would imagine that since $\cos\big(\frac{2\pi kn}{p}\big)$ can take both positive and negative value it would make the conclusion less obvious. Any hint of why their claim is true would be greatly appreciated.
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10$\begingroup$ We are not claiming this pointwise estimate; instead we are claiming the integrated estimate in which both sides are averaged against $\prod_{\xi \in S \backslash \xi_0} K_N(\xi n)$. This is accomplished by expanding both sides into Fourier coefficients and using the facts previously mentioned. $\endgroup$– Terry TaoCommented Oct 25, 2022 at 20:38
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$\begingroup$ Got it. I misunderstood. Thanks! $\endgroup$– Jonathan LamCommented Oct 27, 2022 at 2:15
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