## equidistribution on the unit circle of particular sequences of finite subsets

Given a strictly convex function $g : [0, 1] \to \mathbb{R}$, I'm curious about the asymptotic distribution of the points $\exp{(2 \pi i N g(n / N))}$ for $n = 1, 2, \dots, N$, counted with multiplicity, as $N$ gets large. To be precise, I'm curious what sorts of bounds one can expect for the "discrepancy" $N^{-1} \#\{n : a < Ng(n/N) \pmod{1} < b\} - (b - a),$ and in particular if these points become equidistributed as $N$ goes to infinity.

For what it's worth, I believe the answer is "yes." I have a guess about how to prove it, but I am not an analytic number theorist, and, even if my proof is right, I'm very curious if people know 1) better ways to prove it, or 2) preexisting results that imply it.

As for my humble stab at a sketch of a proof: First, I believe I have a proof in the case $g(x) = x^2$ using the Erd\"os-Tur\'an Inequality (here). (In this inequality, one has a bound on the above discrepancy which can shown to go to zero using what I understand are well-known formulas for quadratic Gauss sums.) Perhaps this is also true for arbitrary quadratic polynomials, and perhaps one can get good bounds on the discrepancy. For a general strictly convex function, $g$, I think it might be possible to approximate $g$ on small intervals (w.r.t. $N$) by a quadratic polynomial, and use the discrepancies computed in the polynomial cases to bound the discrepancy for $g$.

My proof for $g(x) = x^2$ will be provided only upon request: I am ashamed of it's ugliness. Thanks in advance.

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By the Erdos-Turan inequality, this would follow if you could show that $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} = o(N)$ uniformly in $m$. I'm not sure how one would go about showing this though. – Peter Humphries Jul 4 at 8:50
Is there a reason to think much can be done for general g? – Charles Matthews Jul 4 at 9:19
The quadratic polynomial approach sounds rather like "stationary phase" theory. For exponential sums this is supposed to register with the work of Van der Corput. – Charles Matthews Jul 4 at 9:27
Have you checked to see if there's anything like this in the book by Kuipers and Niederreiter? – Gerry Myerson Jul 4 at 12:47

Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then $$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$ So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that $$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$
So now let $\mu_N$ be the probability measure on $[0,1]$ given by $$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : N g(n/N) \in B \pmod{1}\right\}$$ for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency $$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$ satisfies $$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$ for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ and using the earlier bound on $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that $$D(N) = O\left(N^{-1/3}\right)$$ and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.
It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.