## Uniform distribution of polynomials mod 1

Last week Lam Wai Kit posted an excellent introductory article of ergodic theory and its applications to number theory. As stated in the note, one consequence of Birhoff recurrence theorem is that for any non-constant polynomial ${p(x)}$, the fractional part of the sequence ${p(1), p(2), p(3), \ldots}$ can be arbitrarily close to ${0}$.

In fact, it is a special case of the following well-known result.

Theorem 1 Suppose ${p(x)}$ is a non-constant real polynomial with irrational leading coefficient, then ${p(1), p(2), p(3), \ldots}$ is uniformly distributed mod 1.

The theorem is a consequence of Weyl’s criterion on uniform distribution.

Proposition 2 (Weyl’s criterion) Let ${\{x_n\}}$ be a sequence of real numbers. Then ${\{x_n\}}$ is uniformly distributed mod 1 if and only if for any non-zero integer ${m}$,

$\displaystyle \lim_{N\rightarrow \infty} \frac 1N \sum_{n=1}^N e(mx_n) = 0,$

here ${e(x) = e^{2\pi i x}}$.

Proof: (Sketch). By Weierstrass’ theorem, any step function with support in ${[0,1]}$ can be approximated by linearly combinations of ${\{e(mx)\}}$ with ${L^{\infty}}$ norm. $\Box$

Weyl’s criterion leads us to the investigation of the following summation

$\displaystyle S(I,f) = \sum_{x \in I} e(f(x)).$

Here ${I=[A,B]}$ is an interval, ${A are integers. Here and what follows, the summation ${x \in I}$ runs through integers. Let ${|I|=B-A+1}$ be the number of integers in ${I}$. Throughout the rest of the note, ${f(x)}$ and ${p(x)}$ are polynomials of degree ${k}$ with irrational leading coefficient.

We first deal with special case ${f(x)}$ being linear.

Proposition 3 Let ${I=[A,B]}$, where ${A < B}$ are integers. Let ${f(x) = \alpha x + \beta}$ with ${\alpha}$ irrational. Then

$\displaystyle |S(I,f)| \le \frac{1}{\sin (\alpha \pi)}.$

Proof:

$\displaystyle |S(I,f)| = \left|e(\alpha A+\beta) \frac{e(\alpha(B-A+1)) - 1}{e(\alpha)-1}\right|$

$\displaystyle = \left|\frac{e(\alpha(B-A+1/2)) - e(-\alpha/2)}{e(\alpha/2)-e(-\alpha/2)} \right|\le \frac{1}{\sin (\alpha \pi)}$

$\Box$

Let

$\displaystyle \Delta f(x;h) = f(x+h)-f(x)$

and inductively

$\displaystyle \Delta^{r+1}(f) = \Delta(\Delta^r f(\,\cdot\,;h_1, \ldots, h_r))(x,h_{r+1})$

$\displaystyle = \Delta^rf(x+h_{r+1};h_1,\ldots, h_r) - \Delta^r f(x; h_1, \ldots, h_r).$

Proposition 4

$\displaystyle |S(I,f)|^{2^j} \le (2|I|)^{2^j-j-1} \sum_{|h_1| \le B-A} \cdots \sum_{|h_j| \le B-A} \left| \sum_{x \in I_{h_1, \ldots, h_j}} e(\Delta^{j}f(x; h_1, \ldots, h_j)) \right|,$

where ${I_\emptyset=I}$,

$\displaystyle I_{\emptyset} \subset I_{h_1} \subset I_{h_1,h_2} \ldots$

are intervals (can be empty). In fact

$\displaystyle I_{h_1, \ldots, h_{\ell+1}} = I_{h_1, \ldots, h_{\ell}} \bigcap (I_{h_1, \ldots, h_{\ell}}-h_{\ell+1})$

where for ${I=[A,B]}$, ${I-h=[A-h,B-h]}$.

Proof: By induction on ${j}$, for ${j=1}$,

$\displaystyle |S(I,f)|^2 = \sum_{y \in I} \sum_{x \in I} e(f(y)-f(x)).$

Let ${h=y-x}$, the above is

$\displaystyle \sum_{|h| \le B-A} \sum_{x \in I_h} e(\Delta f(x;h))$

where ${I_h = I \bigcap (I-h)}$.

Suppose the statement is true for ${j}$, then by the Cauchy-Schwarz inequality

$\displaystyle |S(I,f)|^{2^{j+1}} \le (2|I|)^{2^{j+1}-2j-2} (2|I|))^{j} \left|\sum_{x \in I_{h_1, \ldots, h_j}} e(\Delta^{j}f(x; h_1, \ldots, h_j))\right|^2$

As above, the innermost summand is

$\displaystyle \sum_{|h_{j+1}| \le B-A} \sum_{x \in I_{h_1, \ldots, h_{j+1} }} e(\Delta^{j+1}f(x; h_1, \ldots, h_j,h_{j+1}))$

This completes the proof. $\Box$

Proof of Theorem 1. Let ${j=k-1}$, then ${\Delta^{k-1}f}$ is linear with leading coefficient irrational. Let ${f(x)=mp(x)}$ for some non-zero integer ${m}$. Let ${I=[1,N]}$. By the previous proposition and proposition 3

$\displaystyle |S(I,f)|^{2^{k-1}} \le (2N)^{2^{k-1}-k} \sum_{|h_1|< N} \cdots \sum_{|h_{k-1}| < N} \sum_{x \in I_{h_1,h_2, \ldots, h_{k-1}}} \Delta^{k-1}f(x; h_1, \ldots, h_{k-1}) .$

$\displaystyle \ll N^{2^{k-1}-k} N^{k-1} = N^{2^{k-1}-1}.$

Thus

$\displaystyle N^{-1} |S(I,f)| \le N^{-1/2^{k-1}} \rightarrow 0, \text{ as } N \rightarrow \infty.$

The theorem then follows from Weyl’s criterion.