6
$\begingroup$

Let $c>1$, and let $A$ denote the set $$ \Big\{ \lfloor n^c \rfloor, \quad 1 \leq n \leq N \Big\}. $$ Thus $A$ consists of the first $N$ elements of a so-called Piatetski-Shapiro sequence.

The additive energy $E(A)$ of $A$ is defined as the number of solutions $(a_1,a_2,a_3,a_4) \in A^4$ of the equation $a_1 - a_2 = a_3 -a_4$.

Question: Is there an upper bound for $E(A)$ known? In particular, is it true that $$ E(A) \ll N^{3 - \varepsilon} $$ for some (small) $\varepsilon>0$, as $N \to \infty$?

(The case $c \geq 2$ is quite easy, but how about the other values of $c$ in the range $(1,2)$? Note that for $c < 2$ the set $A$ is not necessarily convex.)

Improve this question
$\endgroup$
10
  • $\begingroup$ See Corollary 3.2 in ams.org/mathscinet-getitem?mr=1772612 for set version of the statement that you want. You can then use Balog-Szemeredi-Gowers to deduce the energy version. $\endgroup$
    – Boris Bukh
    Commented Oct 20, 2015 at 15:46
  • $\begingroup$ To apply Corollary 3.2 from this paper it is necessary to have strict convexity (noted before the statement of Theorem 1) - which we do not have in the present case, since convexity is ruined by the floor-function. $\endgroup$
    – Kurisuto Asutora
    Commented Oct 20, 2015 at 17:27
  • 1
    $\begingroup$ ... See the introduction of this paper "Exceptional set of a representation with fractional powers" by Balanzario, Garaev, and Zuazua for a discussion of this result. This isn't quite a proof since one needs to consider all restricted sumsets of positive density within $A + A$ and not just the full set, however one might be able to adapt the ideas from that proof or use the "restriction theorem" for the P-S sequence of Mirek to pass to the general case (arxiv.org/abs/1305.0043). $\endgroup$
    – Mark Lewko
    Commented Oct 20, 2015 at 19:07
  • 1
    $\begingroup$ I haven't checked this carefully but here is an alternate approach that seems simpler. By BSG / "routine" additive combinatorics it suffices to show that $|A+A| > |A|^{1+c}$ for some fixed $c>0$ and $A$ a large subset of $[N]$ (that is $|A|>N^{1-a}$ for some small $a$). It thus should suffice to show that $\max(|A+A|, |f(A)+f(A)|) > |A|^{1+c}$ for $c>0$ where $f(x) = \lfloor x^{c} \rfloor$. $\endgroup$
    – Mark Lewko
    Commented Oct 20, 2015 at 21:16
  • 1
    $\begingroup$ Using Elekes' relation between sum-product and incidence theorems (see the sketch in my answer here mathoverflow.net/questions/217557/…), this should follow from a Szemeredi-Trotter theorem for translates of the curve $(x, \lfloor x^{c} \rfloor )$. However the crossing number proof of Szemeredi-Trotter only uses that translates of this curve intersects in $O(1)$ places and should apply here. $\endgroup$
    – Mark Lewko
    Commented Oct 20, 2015 at 21:18

2 Answers 2

Reset to default
0
$\begingroup$

In "$L_1$-Norms of Exponential Sums and the Corresponding Additive Problem" Garaev and Kueh prove that if $F$ is in $C^3([1,N])$ with $F'(x)>0$, $F''(x)>0$ and $F'''(x)<0$, then the additive energy of $A=\{\lfloor F(1)\rfloor,\ldots,\lfloor F(N)\rfloor\}$ satisfies $$ \frac{N^4}{F(N)+1}\ll E(A)\ll (F'(1)+1)N^{5/2} +\frac{N^2\log N}{F''(N)}.$$

In particular, if $F(x)=Ax^c$, with $A>0$ and $1< c\leq 3/2$, then $$ N^{4-c}\ll_{c,A} E(A)\ll_{c,A} N^{4-c}\log N.$$

This paper builds on earlier work of Garaev, which presumably uses indicator trick mentioned in the first answer to bound the exponential sums, however the exponential sum estimates do not appear explicitly in the Garaev-Kueh paper.

PS: The ISSN for the Garaev-Kueh paper is 0232-2064.

PPS: I think the method mentioned by Eric might work for $1<c<2$, but I didn't write down all the details. You should be able to figure it out by staring at Heath-Brown's paper for a while (you should be able to avoid eq (5) if you're not summing over primes; proving (5) occupies the next 20 pages; in particular, you can re-apply Lemma 1 to the exponential sum at the end of the chain of inequalities on p. 247; overall I think the error term is $O(N^{1/2}(\log N)^b)$ for some power $b$.

Improve this answer
$\endgroup$
1
$\begingroup$

For $1<c<c_0$, for some constant $c_0$ the circle method will yield an asymptotic of size $$E(A)\sim \mathfrak{S}N^{4/c-1.}$$ where $\mathfrak{S}$ is a constant. In fact, when $c$ is not too large, one can even obtain an asymptotic for the additive energy of the Piatetski-Shapiro primes (see Balog and Friedlander). Letting $S_{A}(\theta)=\mathbb{E}_{n\leq N}1_{A}(n)e(n\theta)$ we may write $$E(A)=N^{3}\int_{0}^{1}|S_{A}(\theta)|^{4}d\theta.$$ The sum $S_A(\theta)$ can be handled by noting that $[(n+1)^{1/c}]-[n^{1/c}]$ is the indicator function for the Piatetski-Shapiro sequence, and then using the truncated Fourier series for the sawtooth function.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I believe what you say is true however this really isn't an answer unless you can fill in the details or provide a reference. We know things must be somewhat subtle given that such statements aren't known for the full range of $c$. $\endgroup$
    – Mark Lewko
    Commented Oct 21, 2015 at 23:50
  • $\begingroup$ @Mark These exponential sums are not difficult to deal with - I added some explanation of what to do. Also, I provided Friedlander and Balog as a reference. $\endgroup$
    – Eric Naslund
    Commented Oct 22, 2015 at 0:11
  • $\begingroup$ Heath-Brown's paper "The Pjateckii-Sapiro Prime Number Theorem" has a nice exposition of the indicator trick + sawtooth method. $\endgroup$
    – Brendan Murphy
    Commented Oct 22, 2015 at 1:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .