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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 '12 at 8:50
Is there a reason to think much can be done for general g? –  Charles Matthews Jul 4 '12 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 '12 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 '12 at 12:47

1 Answer 1

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

$$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$


$$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$

for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ and using the earlier bound on $\displaystyle\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ 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.

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