# How do you prove that $p(n \xi)$ for $\xi$ irrational and $p$ a polynomial is uniformly distributed modulo 1?

The Weyl equidistribution theorem states that the sequence of fractional parts ${n \xi}$, $n = 0, 1, 2, \dots$ is uniformly distributed for $\xi$ irrational.

This can be proved using a bit of ergodic theory, specifically the fact that an irrational rotation is uniquely ergodic with respect to Lebesgue measure. It can also be proved by simply playing with trigonometric polynomials (i.e., polynomials in $e^{2\pi i k x}$ for $k$ an integer) and using the fact they are dense in the space of all continuous functions with period 1. In particular, one shows that if $f(x)$ is a continuous function with period 1, then for any $t$, $\int_0^1 f(x) dx = \lim \frac{1}{N} \sum_{i=0}^{N-1} f(t+i \xi)$. One shows this by checking this (directly) for trigonometric polynomials via the geometric series. This is a very elementary and nice proof.

The general form of Weyl's theorem states that if $p$ is a monic integer-valued polynomial, then the sequence ${p(n \xi)}$ for $\xi$ irrational is uniformly distributed modulo 1. I believe this can be proved using extensions of these ergodic theory techniques -- it's an exercise in Katok and Hasselblatt. I'd like to see an elementary proof.

Can the general form of Weyl's theorem be proved using the same elementary techniques as in the basic version?

• Correct me if I'm wrong, but it seems like this would be better suited for MathOverflow.net? – Nick Jul 20 '10 at 21:47
• Nick, I don't think it's research-level. Since many people committing to this site expressed a desire for an MO-like site that would admit questions in pure math at, say, an undergraduate level, I think it should stay here. – Jamie Banks Jul 20 '10 at 22:17
• It's definitely not research-level, but I actually doubt it'd get closed on MO, since it's a fact that most books mention without proof; I don't really see why it would be terribly out of place on either site. – Akhil Mathew Jul 20 '10 at 23:16
• fair, Akhil; I'm glad to have it here, though, so that we indicate the acceptability of higher-level math questions (non-grade school) – Jamie Banks Jul 21 '10 at 0:08
• (I also think these questions will get slightly easier once we have jsMath or similar LaTeX support on math.SE) – Jamie Banks Jul 22 '10 at 6:14

## 1 Answer

There is a fairly good exposition in Terry Tao's post, see Corollaries 4-6. Here is a sketch:

We prove the more general statement: Let $p(n)= \chi n^d + a_{d-1} n^{d-1} + \cdots + a_1 n + a_0$ be any polynomial, with $\chi$ irrational. Then $p(n) \mod 1$ is equidistributed. Our proof is by induction on $d$; the base case $d=1$ is standard.

Set $e(x) = e^{2 \pi i x}$. By the standard trickery with exponential polynomials, it is enough to show $$\sum_{n=0}^{N-1} e(p(n)) = o(N).$$

Choose a positive integer $h$. With a small error, we can replace the sum by $$\sum_{n=0}^{N-1} (1/h) \left( e(p(n)) + e(p(n+1)) + \cdots + e(p(n+h-1)) \right).$$ By Cauchy-Schwarz, this is bounded by $$\frac{\sqrt{N}}{h} \left[ \sum_{n=0}^{N-1} \left( e(p(n)) + \cdots + e(p(n+h-1)) \right) \overline{ \left( e(p(n)) + \cdots + e(p(n+h-1)) \right)} \right]^{1/2}.$$

Expanding the inner sum, we get $h^2$ terms of the form $e(p(n) - p(n+k))$. There are $h$ terms where $k=0$; these each sum up to $N$. For the other $h^2-h$ terms, the sum is of the form $\sum_{n=0}^{N-1} e(q(n))$, where $q$ has leading term $\chi d n^{d-1}$. By induction, each of these sums is $o(N)$.

So the quantity in the square root is $$hN+o(N)$$ where the constant in the $o$ depends on $h$ and $\chi$. Putting it all together, we get a bound of $$N/\sqrt{h} + o(N).$$

Since $h$ was arbitrary, this proves the result.