Weyl’s Equidistribution Theorem
Definition 1.1
Let $x\in\mathbb{R}$.
- The integer part of $x$, denoted by $[x]$, is the greatest integer less than or equal to $x$.
- The fractional part of $x$ is $\langle x\rangle:=x-[x]$.
- For $x,y\in\mathbb{R}$, we say that $x$ and $y$ are congruent modulo 1 if $x-y\in\mathbb{Z}$. We write
Thus, $x\equiv y\pmod{1}$ if and only if $\langle x\rangle=\langle y\rangle$.
Problem
Is the set ${\langle n\gamma\rangle:n\in\mathbb{N}}$ dense in $[0,1)$? Kronecker’s theorem tells us that the answer is yes when $\gamma\notin\mathbb{Q}$.
Weyl’s equidistribution theorem gives a stronger conclusion: the proportion of terms in any interval converges to the length of that interval.
Definition 1.2 (Equidistributed sequence)
A sequence $(\xi_n)_{n\geq 1}$ in $[0,1)$ is equidistributed if, for every $0\leq a<b\leq 1$,
\[\lim_{N\to\infty} \frac{\bigl|\{n\in\{1,\ldots,N\}:\xi_n\in(a,b)\}\bigr|}{N} =b-a.\]Theorem 1.3 (Weyl’s equidistribution theorem)
If $\gamma\notin\mathbb{Q}$, then $(\langle n\gamma\rangle)_{n\geq 1}$ is equidistributed in $[0,1)$.
In particular, ${\langle n\gamma\rangle:n\in\mathbb{N}}$ is dense in $[0,1)$, since every interval of positive length contains terms of the sequence.
Observation 1.4
For $0\leq a<b\leq 1$, let $\chi_{(a,b)}$ be the characteristic function of $(a,b)$ on $[0,1)$, extended to $\mathbb{R}$ as a 1-periodic function. Then
\[\bigl|\{1\leq n\leq N:\langle n\gamma\rangle\in(a,b)\}\bigr| =\sum_{n=1}^N\chi_{(a,b)}(n\gamma).\]Thus, Theorem 1.3 can be restated as follows.
Theorem 1.5
If $\gamma\notin\mathbb{Q}$, then, for every $0\leq a<b\leq 1$,
\[\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\chi_{(a,b)}(n\gamma) =\int_0^1\chi_{(a,b)}(x)\,dx =b-a.\]Lemma
If $f:\mathbb{R}\to\mathbb{C}$ is continuous and 1-periodic, and $\gamma\notin\mathbb{Q}$, then
\[\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma) =\int_0^1 f(x)\,dx.\]Proof. We consider three cases.
1. Exponential functions. Let $f(x)=e^{2\pi ikx}$, where $k\in\mathbb{Z}$. If $k=0$, both the average and the integral equal 1. If $k\neq 0$, then
\[\int_0^1 e^{2\pi ikx}\,dx =\left[\frac{e^{2\pi ikx}}{2\pi ik}\right]_0^1 =0.\]Since $\gamma$ is irrational and $k\neq 0$, we have $e^{2\pi ik\gamma}\neq 1$. The geometric-series formula gives
\[\begin{aligned} \frac{1}{N}\sum_{n=1}^N e^{2\pi ikn\gamma} &=\frac{e^{2\pi ik\gamma}}{N} \frac{1-e^{2\pi ikN\gamma}}{1-e^{2\pi ik\gamma}}. \end{aligned}\]Consequently,
\[\left|\frac{1}{N}\sum_{n=1}^N e^{2\pi ikn\gamma}\right| \leq\frac{2}{N\lvert 1-e^{2\pi ik\gamma}\rvert} \longrightarrow 0.\]2. Trigonometric polynomials. Every trigonometric polynomial is a finite linear combination of the functions $e^{2\pi ikx}$, with $k\in\mathbb{Z}$. The conclusion therefore follows from Case 1 by linearity.
3. Continuous periodic functions. Let $\epsilon>0$. By the uniform approximation theorem for continuous periodic functions (for example, Fejér’s theorem), there is a trigonometric polynomial $P$ such that
\[\sup_{x\in[0,1]}\lvert P(x)-f(x)\rvert<\frac{\epsilon}{3}.\]By Case 2, for all sufficiently large $N$,
\[\left|\frac{1}{N}\sum_{n=1}^N P(n\gamma)-\int_0^1 P(x)\,dx\right| <\frac{\epsilon}{3}.\]For these $N$, periodicity and the triangle inequality give
\[\begin{aligned} \left|\frac{1}{N}\sum_{n=1}^N f(n\gamma)-\int_0^1 f(x)\,dx\right| &\leq\left|\frac{1}{N}\sum_{n=1}^N\bigl(f(n\gamma)-P(n\gamma)\bigr)\right|\\ &\quad+\left|\frac{1}{N}\sum_{n=1}^N P(n\gamma)-\int_0^1 P(x)\,dx\right|\\ &\quad+\left|\int_0^1\bigl(P(x)-f(x)\bigr)\,dx\right|\\ &<\epsilon. \end{aligned}\]This proves the lemma. $\square$
Proof of Theorem 1.5
Let $\epsilon>0$. Choose continuous 1-periodic functions $f^-\epsilon$ and $f^+\epsilon$ such that
\[0\leq f^-_\epsilon(x)\leq\chi_{(a,b)}(x)\leq f^+_\epsilon(x)\leq 1,\]and both functions agree with $\chi_{(a,b)}$ except in small neighborhoods of the endpoints whose total length in $[0,1)$ is at most $2\epsilon$. These functions can be constructed using linear ramps near the endpoints, with periodicity taken into account at 0 and 1.
In particular,
\[b-a-2\epsilon\leq\int_0^1 f^-_\epsilon(x)\,dx, \qquad \int_0^1 f^+_\epsilon(x)\,dx\leq b-a+2\epsilon.\]The pointwise inequalities imply
\[\frac{1}{N}\sum_{n=1}^N f^-_\epsilon(n\gamma) \leq\frac{1}{N}\sum_{n=1}^N\chi_{(a,b)}(n\gamma) \leq\frac{1}{N}\sum_{n=1}^N f^+_\epsilon(n\gamma).\]Applying the lemma yields
\[\begin{aligned} b-a-2\epsilon &\leq\int_0^1 f^-_\epsilon(x)\,dx\\ &=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f^-_\epsilon(n\gamma)\\ &\leq\liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\chi_{(a,b)}(n\gamma)\\ &\leq\limsup_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\chi_{(a,b)}(n\gamma)\\ &\leq\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f^+_\epsilon(n\gamma)\\ &=\int_0^1 f^+_\epsilon(x)\,dx\\ &\leq b-a+2\epsilon. \end{aligned}\]Since $\epsilon>0$ is arbitrary, the limit exists and equals $b-a$. This proves Theorem 1.5 and hence Theorem 1.3. $\square$
Corollary
The conclusion of the lemma also holds for every 1-periodic function $f:\mathbb{R}\to\mathbb{C}$ that is Riemann integrable on $[0,1]$.
Proof. First, observe that Theorem 1.5 also applies to half-open intervals $[a,b)$. Indeed, since $\gamma$ is irrational, the fractional parts $\langle n\gamma\rangle$ are all distinct: equality for two indices would imply that $(n-m)\gamma\in\mathbb{Z}$. Thus, changing whether an endpoint is included changes the number of terms counted by at most one, which does not affect the limiting proportion.
It follows by linearity that the conclusion holds for any step function that is a finite linear combination of periodically extended characteristic functions of half-open intervals.
Now suppose that $f$ is real-valued. Given $\epsilon>0$, choose a partition
\[\mathscr{P}=\{0=x_0<x_1<\cdots<x_m=1\}\]such that the upper and lower Darboux sums satisfy
\[\mathscr{U}(f,\mathscr{P})-\mathscr{L}(f,\mathscr{P})<\epsilon.\]For $j=1,\ldots,m$, define
\[m_j:=\inf_{y\in[x_{j-1},x_j]}f(y), \qquad M_j:=\sup_{y\in[x_{j-1},x_j]}f(y).\]Define the lower and upper step functions on $[0,1)$ by
\[f_L(x):=\sum_{j=1}^m m_j\chi_{[x_{j-1},x_j)}(x), \qquad f_U(x):=\sum_{j=1}^m M_j\chi_{[x_{j-1},x_j)}(x),\]and extend both functions periodically to $\mathbb{R}$. Using half-open intervals makes their values at partition points unambiguous. We have
\[f_L(x)\leq f(x)\leq f_U(x) \qquad\text{for every }x\in\mathbb{R},\]and
\[\int_0^1 f_L(x)\,dx =\mathscr{L}(f,\mathscr{P}) \leq\int_0^1 f(x)\,dx \leq\mathscr{U}(f,\mathscr{P}) =\int_0^1 f_U(x)\,dx.\]Since $f_L$ and $f_U$ are finite linear combinations of characteristic functions,
\[\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f_L(n\gamma) =\mathscr{L}(f,\mathscr{P}),\]and
\[\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f_U(n\gamma) =\mathscr{U}(f,\mathscr{P}).\]Writing $I:=\int_0^1 f(x)\,dx$, we obtain
\[\begin{aligned} I-\epsilon &<\mathscr{L}(f,\mathscr{P})\\ &\leq\liminf_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma)\\ &\leq\limsup_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma)\\ &\leq\mathscr{U}(f,\mathscr{P})\\ &<I+\epsilon. \end{aligned}\]Because $\epsilon>0$ is arbitrary, the limit exists and equals $I$:
\[\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma) =\int_0^1 f(x)\,dx.\]For complex-valued $f$, apply the real-valued result separately to its real and imaginary parts. $\square$
References
- Elias M. Stein and Rami Shakarchi, Fourier Analysis: An Introduction.
- Math 139 Fourier Analysis Notes.