Elementary proof of the equidistribution theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:18:14Z http://mathoverflow.net/feeds/question/75777 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75777/elementary-proof-of-the-equidistribution-theorem Elementary proof of the equidistribution theorem user8761468 2011-09-18T18:43:19Z 2012-10-09T13:24:54Z <p>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 <a href="http://mathworld.wolfram.com/WeylsCriterion.html" rel="nofollow">Weyl's criterion</a>.</p> <p>Thank you very much.</p> http://mathoverflow.net/questions/75777/elementary-proof-of-the-equidistribution-theorem/75819#75819 Answer by Denis Chaperon de LauziÃ¨res for Elementary proof of the equidistribution theorem Denis Chaperon de LauziÃ¨res 2011-09-19T05:22:58Z 2011-09-19T05:22:58Z <p>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). </p> <p>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".</p> http://mathoverflow.net/questions/75777/elementary-proof-of-the-equidistribution-theorem/75822#75822 Answer by Tony Huynh for Elementary proof of the equidistribution theorem Tony Huynh 2011-09-19T06:32:39Z 2011-09-19T06:32:39Z <p>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 <em>Fourier Analysis</em> book (there are 110 chapters in the book, so this really is one of the first things that is covered). </p> http://mathoverflow.net/questions/75777/elementary-proof-of-the-equidistribution-theorem/109158#109158 Answer by David Speyer for Elementary proof of the equidistribution theorem David Speyer 2012-10-08T16:32:43Z 2012-10-09T13:24:54Z <p>There is a really easy prof that just uses the Pigeonhole principle. Let $\alpha$ be irrational.</p> <p><b>Lemma:</b> For any $\delta >0$, there is an $n>0$ such that <code>$(n \alpha) \in (- \delta, \delta) \setminus \{ 0 \}$</code>.</p> <p><b>Proof:</b> Choose $N$ large enough that $1/N &lt; \epsilon$. 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, $((j-k) \alpha) \neq 0$. $\square$</p> <p>Now, fix $0 \leq p &lt; q \leq 1$ and $\epsilon>0$. Our goal is to show that <code>$$q-p-\epsilon \leq \lim_{N \to \infty} inf \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq \lim_{N \to \infty} sup \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq q-p+\epsilon.$$</code></p> <p>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 $|\rho| &lt; \delta$.</p> <p>Choose $n$ such that <code>$n \alpha \in (-\delta, \delta) \setminus \{ 0 \}$</code>. 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.</p> <p>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.</p>