Question

I'm looking for references to (as many as possible) elementary proofs of the Weyl's equidistribution theorem, i.e., the statement that the sequence $\alpha, 2\alpha, 3\alpha, \ldots \mod 1$ is uniformly distributed on the unit interval. With "elementary" I mean that it does not make use of complex analysis in particular the Weyl's criterion.

Thank you very much.

Answer 1

Hardy and Wright give a proof based on continued fractions in Section 23.10 of "An introduction to the theory of numbers" (reference valid at least in the 4th edition).

Interestingly, their endnotes to Chapter 23 claim that the "equidistribution" form of the theorem is not due to Kronecker (the statement they call "Kronecker's theorem" is the density of $\{n\alpha\}$; they give a number of elementary proofs of that weaker result), but "independently to Bohl, Sierpinski and Weyl".

Answer 2

Not sure if this qualifies, but there is a short proof using Fourier Analysis. No hardcore stuff, just Fejér's Theorem. See Chapter 3 of Körner's Fourier Analysis book (there are 110 chapters in the book, so this really is one of the first things that is covered).

Answer 3

There is a really easy prof that just uses the Pigeonhole principle. Let $\alpha$ be irrational.

Lemma: For any $\delta >0$, there is an $n>0$ such that $(n \alpha) \in (- \delta, \delta) \setminus \{ 0 \}$.

Proof: Choose $N$ large enough that $1/N < \epsilon$. Divide $[0,1]$ up into $N$ segments of length $1/N$. By pigeonhole, there are $j$ and $k$ such that $(j \alpha)$ and $(k \alpha)$ land in the same segment. So $((j-k) \alpha) \in (-\delta, \delta)$. Since $\alpha$ is irrational, $((j-k) \alpha) \neq 0$. $\square$

Now, fix $0 \leq p < q \leq 1$ and $\epsilon>0$. Our goal is to show that $$q-p-\epsilon \leq \lim_{N \to \infty} inf \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq \lim_{N \to \infty} sup \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq q-p+\epsilon.$$

Choose $\delta >0$ such that $$\frac{(q-p)/\rho -1}{1/\rho+1} \geq q-p-\epsilon \quad \mbox{and} \quad \frac{(q-p)/\rho +1}{1/\rho-1} \leq q-p+\epsilon$$ for all $\rho$ with $|\rho| < \delta$.

Choose $n$ such that $n \alpha \in (-\delta, \delta) \setminus \{ 0 \}$. Write $\rho = (n \alpha)$. Break up the set of values $(k \alpha)$ up into arithmetic progressions based on $k$ modulo $n$. So each segment is of the form $(\beta + j \rho)$. It is enough to prove that the lim inf and lim sup contributed by each progression lie between $q-p-\epsilon $ and $q-p+\epsilon$, as the total contribution is a weighted average from the contributions from the progressions.

Break each progression up into segments according to $\lfloor \beta + j \rho \rfloor$. The initial segment and final segment each contain at most $1/ |\rho|$ terms. The segments in the middle contain between $1/\rho - 1$ and $1/\rho+1$ terms of which between $(q-p)/\rho -1$ and $(q-p)/\rho+1$ are between $p$ and $q$. Since $((q-p)/\rho-1)/(1/\rho+1)$ and $((q-p)/\rho+1)/(1/\rho-1)$ were chosen to lie in $(q-p-\epsilon, q-p+\epsilon)$, the average of all of these terms lies in the required interval. And the terms from the initial segments can only drag the average off by at most $(2/|\rho|)/(N/n)$, which goes to $0$. QED.