I would like to bound an exponential sum of the shape $$ S = \sum_{ab \leq N \\ a, b > U} \Lambda(a) \omega(b) e^{2 \pi i F(ab)}, $$ where $\Lambda$ and $\omega$ are some weights (actually $\Lambda$ is von Mangoldt function), and $F$ is a polynomial in one variable, and $U$ some positive quantity. Here I obtained this from a use of Vaughn's identity.

I would like to obtain an upper bound of this sum $S$ by removing the weights. I know I could consider dyadic intervals for $a$ say, and then apply Cauchy-Schwarz twice to obtain an exponential sum without the weights. Here it seems necessary to consider dyadic intervals as the estimate gets very bad otherwise.

I was wondering if there are any other ways to remove the weights? In particular, I wanted to do it without having to have to break up the sum, for example into dyadic intervals as in this case. Any comments/suggestions would be appreciated.

PS What I was really interested in was an exponential sum estimate of the shape $$ T = \sum_{a_1b_1 \leq N \\ a_1, b_1 > U} \sum_{a_2b_2 \leq N \\ a_2, b_2 > U}\Lambda(a_1) \omega(b_1) \Lambda(a_2) \omega(b_2) e^{2 \pi i F(a_1b_1, a_2b_2)}, $$ here $U = N^{\varepsilon}$ and $F$ is a polynomial in two variables. The issue I have when I consider dyadic intervals for $a_1, a_2$ in this case is I will have to deal with a sum that looks like, for example $$ T_1 = \sum_{a_1b_1 \leq N \\ N^{\varepsilon} < a_1 \leq 2 N^{\varepsilon} \\ b_1 > U} \sum_{a_2b_2 \leq N \\ N/2 <a_2 \leq N \\ b_2 > U} \Lambda(a_1) \omega(b_1) \Lambda(a_2) \omega(b_2) e^{2 \pi i F(a_1b_1, a_2b_2)}. $$ Here I can use Cauchy-Schwarz to remove the weights but the resulting exponential sum without the weights has a very different ranges of summation for $a_1$ and $a_2$. At the moment I only know how to handle the resulting sum when the range of summation for $a_1$ and $a_2$ are of the same magnitude (or very close, preferably at most a log power factor).

  • $\begingroup$ Why do you want to avoid dyadic intervals? $\endgroup$ – Will Sawin Sep 27 '17 at 6:48
  • $\begingroup$ @Will Sawin: probably because it would introduce logarithmic losses in the estimate. It is however a bit surprising that these actually matter in Johny T.'s application. $\endgroup$ – js21 Sep 27 '17 at 7:24
  • $\begingroup$ @WillSawin I explained in PS the reason why I wanted to avoid dyadic intervals. Thanks! $\endgroup$ – Johnny T. Sep 27 '17 at 9:41
  • $\begingroup$ @JohnnyT. : then I do not understand why you expect to be able to handle the full sum better than the one restricted to dyadic intervals. $\endgroup$ – js21 Sep 27 '17 at 12:00
  • $\begingroup$ Can't you look for cancellation only in the sum over $a_2,b_2$, or only in the sum over $a_1,b_1$, and bound the other trivially, reducing to the first case? Is that not enough cancellation for your purposes? $\endgroup$ – Will Sawin Sep 27 '17 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.