Let $c>0$ a real number, let $N$ a large natural number and let $e\left(x\right):=e^{2\pi ix}$. Is it true that $\forall k\in\left[1,\,2N\right]$, $k$ natural number, that $$\left|\underset{\underset{n_{1}+n_{2}=k}{2\leq n_{1},\, n_{2}\leq N}}{\sum}\left(\log n_{1}\log n_{2}\, e\left(c\left(n_{1}+n_{2}\right)\right)\right)\right|\leq\left|\underset{2\leq n_{1},\, n_{2}\leq N}{\sum}\left(\log n_{1}\log n_{2}\,e\left(c\left(n_{1}+n_{2}\right)\right)\right)\right|?$$
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$\begingroup$ Where do the parentheses go around the "log(n_1)" and "log(n_2)" terms? Thanks. $\endgroup$– Bill BradleyCommented Feb 24, 2014 at 19:51
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1 Answer
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That's certainly not true as written. The right side is just $$ \left | \left( \sum_{2\le n\le N} e(cn)\log n \right)^2 \right| = \left|\sum_{2\le n\le N} e(cn)\log n \right|^2. $$
If you choose $c$ to be 1/2, this is just the square of an alternating series (and hence is bounded by $(\log N)^2$.
On the other hand, if you choose $k=N$ on the left (say $N$ is even), you've got approximately $N(\log N)^2$.
This can be done with irrational $c$ as well if you take a bit of care.
answered Feb 24, 2014 at 20:13
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$\begingroup$ It is not forbidden to take $c=1$. $\endgroup$ Commented Feb 25, 2014 at 2:46
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$\begingroup$ If c=1, then the inequality holds. It's if there's cancellation that it fails. $\endgroup$ Commented Feb 25, 2014 at 3:50