1
$\begingroup$

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 $$\left|\underset{\underset{n_{1}+n_{2}=k}{2\leq n_{1},\, n_{2}\leq N}}{\sum}\left(\log n_{1}\log n_{2}\, e\left(c\left(n_{1}+n_{2}\right)\right)\right)\right|\leq\left|\underset{2\leq n_{1},\, n_{2}\leq N}{\sum}\left(\log n_{1}\log n_{2}\,e\left(c\left(n_{1}+n_{2}\right)\right)\right)\right|?$$

$\endgroup$
1
  • $\begingroup$ Where do the parentheses go around the "log(n_1)" and "log(n_2)" terms? Thanks. $\endgroup$ Feb 24, 2014 at 19:51

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ It is not forbidden to take $c=1$. $\endgroup$ Feb 25, 2014 at 2:46
  • $\begingroup$ If c=1, then the inequality holds. It's if there's cancellation that it fails. $\endgroup$ Feb 25, 2014 at 3:50

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.