8
$\begingroup$

Let $k\geq 2$. Consider the following norm of exponenetial sum: $$ I(N,p,k)=\int_0^1\int_0^1 \left|\sum_{n=0}^N e^{2\pi i (n x+n^k y)}\right|^p dxdy. $$ Bourgain mentioned on Page 118 of

https://math.mit.edu/classes/18.158/bourgain-restriction.pdf

that $I(N,6,2)\gtrsim N^3\log N$, where he referenced the following article: https://www.researchgate.net/publication/259308546_The_method_of_trigonometric_sums_in_number_theory.

But I did not find an explicit result in the above article that leads directly to the lower bound above.

So my questions are:

  1. What is the idea to prove the above lower bound? The famous Vinogradov's mean value theorem deals with upper bounds of $I(N,p,2)$, but not lower bounds.

  2. What is a reasonably sharp lower bound for $I(N,p,3)$, or particularly, $I(N,6,3)$? Note that this may not be in the direct form of Vinogradov's mean value theorem, as the $n^2$ term is missing here.

$\endgroup$
2
  • $\begingroup$ It seems to me that the result in Bourgain's paper is an application of Weyl's inequality in the form worked out by Vinogradov: if you want a brief introduction with a fairly complete list of references, you may want to have a look at this Q&A. By using that form of Weyl's estimate, you should be able to obtain Bourgain's lower bound. $\endgroup$ Apr 20, 2020 at 5:59
  • $\begingroup$ I wrote up a short note concerning (1) awhile back: gshakan.files.wordpress.com/2017/03/firstcaseofvmvt2.pdf. I don't get an asymptotic as is known but do get a decent lower bound pretty quickly. $\endgroup$ Apr 20, 2020 at 9:55

2 Answers 2

8
$\begingroup$

There are a few things to clear up.

The first is that, on the page in the Bourgain paper you mention, he actually proves the lower bound $I(N,6,2)\gg N^3\log N$ from the fact that

$$ \left\lvert\sum_{n=0}^N e(nx+n^2y)\right\rvert \gg N/q^{1/2}$$

whenever $\lvert x-b/q\rvert \ll 1/N$ and $\lvert y-a/q\rvert \ll 1/N^2$ for some fixed $1\leq a< q\leq N^{1/2}$ with $(a,q)=1$ and $1\leq b\leq q$. (The proof is simply summing the contribution from all such $a$ and $b$). It is this estimate which he invokes a reference for, rather than the lower bound for $I(N,6,2)$.

Secondly, the reference he gives is not to the paper you link to (which is a 1986 paper by Karatsuba-Vinogradov) but instead to Vinogradov's 1954 book with a similar title, usually translated to 'The Method of Trigonometric Sums in the Theory of Numbers'. I don't have a copy of this to hand to check the reference, but a quick search turned up a short note by Tamahiro Oh (https://www.maths.ed.ac.uk/~toh/Files/WeylSum.pdf) proving exactly this Weyl sum lower bound.

Finally, for $I(N,6,3)$, the situation is quite different, and here in fact an asymptotic formula is known: $$ I(N,6,3) = 6N^3 + O(N^2(\log N)^5).$$ This is a result of Vaughan and Wooley (On a certain nonary cubic form and related equations. Duke Mathematical Journal, 80(3), 669–735, 1995).

$\endgroup$
1
  • $\begingroup$ Thanks for pointing that out. $\endgroup$ Apr 24, 2020 at 0:46
7
$\begingroup$

The result for $I(N,6,2)$ was proved by Rogovskaya N. N. in the article An asymptotic formula for the number of solutions of a certain system of equations. The proof is elementary. Main idea is to replace the system $$x_ 1+x_ 2+x_ 3=y_ 1+y_ 2+y_ 3,\quad x^ 2_ 1+x^ 2_ 2+x^ 2_ 3=y^ 2_ 1+y^ 2_ 2+y^ 2_ 3 $$ by $$a_1+a_2+a_3=0,\quad a_1b_1+a_2b_2+a_3b_3=0,$$ where $a_i=x_i-y_i$ and $b_i=x_i+y_i$. Then you can count solutions of the last equation and sum the result over $a_i.$ The answer is nice $${\mathcal N}(P)=18\pi^{-2}P^ 3\log P+{\mathcal O}(P^ 3).$$ Probably this is the only case when trigonometric integral was calculated explicitely.

$\endgroup$
1
  • $\begingroup$ This proof is so nice. Thanks and I will take a look. $\endgroup$ Apr 24, 2020 at 0:46

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.