Consider $$S\_N:=\sum\_{n=1}^N \exp\left(2\pi i\left(\frac12n^2x+\alpha n\right)\right)$$


where $\alpha$ is irrational. For certain $x$ (say integer) we can get that this is bounded for all $N$. I am curious, for what $x\in \mathbb R$ is this sum "small" - that is $S\_N=o(\sqrt N)$?


Is the Lebesgue measure of such $x$ positive?



Is the Lebesgue measure of the $x$ so that this exponential sum is $o(\sqrt{N})$ positive?
Ok, apologies if this is overkill, but [this paper](https://arxiv.org/abs/2011.09306) shows that for almost every (irrational) $\alpha \in \mathbb{R}$, only a measure $0$ set of $x \in \mathbb{R}$ satisfy $S\_{N,\alpha}(x) = o(\sqrt{N})$ as $N \to \infty$.


Indeed, that paper,


Metric theory of Weyl sums  

Changhao Chen, Bryce Kerr, James Maynard, Igor Shparlinski


shows that there is an absolute constant $c > 0$ such that almost every pair $(x,\alpha) \in \mathbb{R}^2$ has $|S\_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$. It follows from basic measure theory that for almost every $\alpha \in \mathbb{R}$, it holds for almost every $x \in \mathbb{R}$ that $|S\_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$.


