Let $a,q,N$ be integers such that $N/2 \leq q \leq N$ and $a/q \notin \mathbb{Z}$.

Is the following estimate true, and, if so, how can it be proved? \[\left|\sum_{1 \leq p \leq N} \exp(2\pi i p a/q) \right|\leq |a|^{o(1)} N^{o(1)},\] where $f(x)=x^{o(1)}$ means $\lim_{x \rightarrow \infty}\frac{\log f(x)}{\log x}=0$, or equivalently $f(x) = O_{\epsilon} (x^{\epsilon})$ for all $\epsilon > 0$. Can it be obtained, for example, from Vinogradov-type estimates?

It would even be useful to know whether the estimate holds in the following average sense: $$\sum_{\substack{N/2 \leq q \leq N \\ a/q \notin \mathbb{Z}}} \left|\sum_{1 \leq p \leq N} \exp(2\pi i p a/q) \right|\leq |a|^{o(1)} N^{1+o(1)}.$$

$N$ may be assumed to be as large as required.

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    $\begingroup$ A bound this strong seems really unlikely. I don't think we can have more than logarithmic cancellation, because by the prime number theorem there are significantly more primes closer to 0. $\endgroup$ – Will Sawin Jan 27 '17 at 5:15

No, this is definitely not true. First of all, there's no way we should expect better than square-root cancellation in an exponential sum without an incredible amount of structure, which the primes do not possess. Second, it's known to be false for certain rational numbers, and conditionally known to be false for all rational numbers, because then the exponential sum essentially counts primes in arithmetic progressions.

For example, take $\frac aq=\frac14$; then $$ \sum_{p\le x} e^{2\pi i p a/q} = \sum_{p\le x} e^{\pi i p/2} = -1 + i \big(\pi(x;4,1) - \pi(x;4,3) \big), $$ where $\pi(x;4,b)$ denotes the number of primes up to $x$ that are congruent to $b$ (mod $4$). Littlewood proved that the quantity in parentheses is $\Omega(\sqrt x \log\log\log x/\log x)$, which is far larger than $x^{o(1)}$.

  • $\begingroup$ I understand the first estimate is false for some specific pairs $(a,q)$. Is it possible that the first estimate or the average estimate holds for sufficiently large $N$ or for an increasing sequence $(N_k)_{k=1}^{\infty}$? When you say "conditionally" you mean "conditionally on GRH"? $\endgroup$ – Linden Jan 27 '17 at 18:38
  • $\begingroup$ What type of structure is needed to obtain sufficient cancellation? $\endgroup$ – Linden Jan 27 '17 at 18:43
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    $\begingroup$ Asking the estimate to hold for an increasing sequence $\{N_k\}$ is probably true, but really not the right question to ask. A good analogy is a random walk on the integers (each step is $+1$ or $-1$ each with probability $1/2$): it's not the case that the position of the random walk at time $N$ is $N^{o(1)}$—it's more like $N^{1/2}$. But of course a random walk will cross from positive to negative and back, and while it's crossing, it's $O(1)$ even. So occasionally it's small, but not for any real reason. $\endgroup$ – Greg Martin Jan 27 '17 at 19:19
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    $\begingroup$ The average estimate is even worse, because you're forcing $p$ and $q$ to be the same size, which (for small $a$, say) will completely skew the distribution to one side. There are known exponential-sum techniques for dealing with sums with a smoothly varying component like $1/q$: I recommend reading Montgomery's "Ten Lectures" for example. $\endgroup$ – Greg Martin Jan 27 '17 at 19:21
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    $\begingroup$ Yes, "conditionally" means "conditionally on GRH". As for structure, something as strong as an arithmetic progression (so that the sum is a geometric series) is basically needed. $\endgroup$ – Greg Martin Jan 27 '17 at 19:22

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