Weyl’s Equidistribution Theorem

Definition 1.1

Let $x\in\mathbb{R}$ be a real number. Then

(1) Let $[x]$, the integer part of $x$, denote the greatest integer less than or equal to $x$.

(2) Let $\langle x\rangle := x -[x]$ denote the fractional part of $x$.

(3) Given $x,y\in\mathbb{R}$, if $x-y\in\mathbb{Z}$ we say

\[\begin{align*} x \equiv y \text{ mod } \mathbb{Z} \text{ or } x\equiv y \text{ mod} 1 \end{align*}\]

Of course $x\equiv y$ if and only if $\langle x \rangle = \langle y \rangle$.

Problem

Consider the collection $\{ \langle n\gamma\rangle : n\in\mathbb{N}\}$: is it dense in $[0,1)$? Kronecker’s theorem: yes if $\gamma\notin\mathbb{Q}$.

Definition 1.2 (Definition of equidistribued sequence)

For every interval $(a,b)\subset [0,1)$, if

\[\begin{align*} \lim_{N\to\infty} \frac{\lvert \{ n\in \{1,2,\ldots, N\}: \xi_n\in (a,b)\}\rvert }{N}=b-a, \end{align*}\]

then we call $\{ \xi_n\}$ an equidistributed sequence.

Theorem 1.3 (Weyl’s Equidistribution theorem)

If $\gamma\notin \mathbb{Q}$, then $\{\langle n\gamma \rangle\}$ is equidistributed in $[0,1)$.

Observation

Let $\chi_{(a,b)}$ denote the characteristic function of $(a,b)\subset [0,1)$, extended as a 1-periodic function. Then we observe that

\[\begin{align*} \lvert \{ 1\leq n \leq N: \langle n\gamma \rangle \in (a,b)\}\rvert = \sum_{n=1}^N \chi_{(a,b)}(n\gamma) \end{align*}\]

Then the theorem 1.3 can be rephrased as follows:

Theorem 1.5

Given any $\gamma \notin \mathbb{Q}$, and any $(a,b)\in [0,1)$,

\[\begin{align*} \lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N \chi_{(a,b)}(n\gamma) = \int_0^1 \chi_{(a,b)}(x) dx \end{align*}\]

Lemma

If $f$ is continuous and 1-periodic on $[0,1)$, and $\gamma \notin \mathbb{Q}$, then

\[\begin{align*} \frac{1}{N}\sum_{n=1}^N f(n\gamma) \rightarrow \int_0^1 f(x) dx. \end{align*}\]

<Proof>

1. Case: $f\in \{1, e^{2\pi ix}, \ldots, e^{2\pi ikx},\ldots \}$. For $f\equiv 1$, it is obvious. Otherwise,

\[\begin{align*} \int_0^1 e^{2\pi ikx}dx &= \int_0^1 \cos(2\pi kx) + i\sin(2\pi kx) dx \\ &=[\sin (2\pi kx) 2\pi k -i \cos(2 \pi kx) 2\pi k]_0^1 \\ &= i(- \cos(2\pi k) + \cos(0))2\pi k \\ &=0. \end{align*}\]

Since $e^{2\pi ikx}\neq 1$ and

\[\begin{align*} \frac{1}{N} \sum_{n=1}^N f(n\gamma) &= \frac{1}{N} \sum_{n=1}^N e^{2\pi kn\gamma} \\ &=\frac{e^{2\pi k\gamma}}{N} \frac{1-e^{2\pi ikN\gamma}}{1-e^{2\pi ik\gamma}} \end{align*}\]

$\frac{1-e^{2\pi ikN\gamma}}{1-e^{2\pi ik\gamma}}$ is bounded, it goes to $0$ as $N\to \infty$.

1. Case: triogonometric polynomials. The problem is linear, so the lemma holds for linear combinations of exponentials.

2. Case: $f$ is a continuous periodic function. Given $\epsilon>0$, choose trigonometric polynomial $P$ such that

\[\begin{align*} \sup_{x\in [0,1)} \lvert P(x) - f(x)\rvert < \frac{\epsilon}{3}. \end{align*}\]

By case 2,

\[\begin{align*} \left\lvert \sum_{n=1}^N P(n\gamma) - \int_0^1 P(x)dx \right\rvert < \frac{\epsilon}{3}. \end{align*}\]

Then we get the conclusion as follows:

\[\begin{align*} \left\lvert \frac{1}{N}\sum_{n=1}^N f(n\gamma) - \int_0^1 f(x)dx \right\rvert &\leq \left\lvert \frac{1}{N} \sum_{n=1}^N f(n\gamma) - \frac{1}{N}\sum_{n=1}^N P(n\gamma)\right\rvert \\ &+ \left\lvert \frac{1}{N}\sum_{n=1}^N P(n\gamma) - \int_0^1 P(x)dx \right\rvert \\ &+ \left\lvert\int_0^1 P(x)dx - \int_0^1 f(x)dx \right\rvert \\ & < \epsilon \end{align*}\] \[\tag*{$\square$}\]

Given $\epsilon>0$, take

\[\begin{align*} f^{+}_{\epsilon} \text{ and } f^{-}_{\epsilon} \end{align*}\]

to be continuous 1-periodic function such that

1) approximate $\chi_{(a,b)}$ from above and below

2) bounded by 1

3) agree on $\chi_{(a,b)}$ except in intervals of total length $2\epsilon$.

In particular,

\[\begin{align*} f^{-}_{\epsilon}(x) \leq \chi_{(a,b)}(x)\leq f^{+}_{\epsilon}(x), \end{align*}\]

and

\[\begin{align*} b-a-2\epsilon \leq \int_0^1 f^{-}_\epsilon(x) dx\quad \text{ and } \quad \int_0^1 f^{+}_\epsilon(x) dx \leq b-a+2\epsilon. \end{align*}\]

Moreover,

\[\begin{align*} \frac{1}{N}\sum_{n=1}^N f^-_\epsilon(n\gamma) \leq \frac{1}{N} \sum_{n=1}^N \chi_{(a,b)}(n\gamma) \leq \frac{1}{N} \sum_{n=1}^N f^+_\epsilon(n\gamma). \end{align*}\]

By the lemma,

\[\begin{align*} b-a-2\epsilon &\leq \int_0^1f^-_\epsilon(x)dx = \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N f^-_\epsilon(n\gamma) \\ &\leq \liminf_{N\to\infty} \frac{1}{N} \chi_{(a,b)}(n\gamma) \\ &\leq \limsup_{N\to\infty} \frac{1}{N} \chi_{(a,b)}(n\gamma) \\ &\leq \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N f^+_\epsilon(n\gamma) =\int_0^1 f^+_\epsilon (x) dx \\ &\leq b-a +2\epsilon. \end{align*}\]

and the above is true for all $\epsilon$ we see the desired limit exists and equals the desired $b-a$.

\[\tag*{$\square$}\]

Corollary

The lemma holds even if $f$ is Riemann integrable.

<Proof>

Note that a step function is finite linear combination of characteristic function. Assume $f$ is real-valued function and consider a partition $\mathscr{P}=\{x_0=0, \ldots, x_N=1\}$ of $[0,1]$ with $x_0 < x_1 < \cdots < x_N$. Define

\[\begin{align*} f_U(x) = \sup_{x_{j-1} \leq y \leq x_j} f(y), \quad f_L(x) = \inf_{x_{j-1}\leq y \leq x_j} f(y), \text{ for } x\in [x_{j-1}, x_j]. \end{align*}\]

Then $f_L(x) \leq f(x) \leq f_U(x)$ for all $x\in [0,1]$ and

\[\begin{align*} \mathscr{L}(f, \mathscr{P})=\int_0^1 f_L(x)dx \leq \int_0^1f(x)dx\leq \int_0^1 f_U(x)dx=\mathscr{U}(f,\mathscr{P}). \end{align*}\]

Since $f$ is Riemann integrable function, for a given $\epsilon >0$, there is a partition $\mathscr{P}$ such that $\int_0^1 f_U - \int_0^1 f_L\leq \epsilon$.

By Theorem 1.5,

\[\begin{align*} \frac{1}{N} \sum_{n=1}^N f_U (n\gamma) \to \int_0^1 f_U(x)dx \text{ as } N\to\infty \end{align*}\]

since $f_U$ is finite linear combination of characteristic function. Similarly,

\[\begin{align*} \frac{1}{N} \sum_{n=1}^N f_L (n\gamma) \to \int_0^1 f_L(x)dx \text{ as } N\to\infty. \end{align*}\]

Combining these inequalities,

\[\begin{align*} \int_0^1 f_L=\lim_{N\to\infty}\frac{1}{N} \sum_{n=1}^N f_L (n\gamma) &\leq \liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f (n\gamma) \\ &\leq \limsup_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f (n\gamma) \\ &\leq\lim_{N\to\infty}\frac{1}{N} \sum_{n=1}^N f_L (n\gamma) \\ &=\int_0^1 f_U. \end{align*}\]

Then

\[\begin{align*} \limsup_{N\to\infty}\frac{1}{N} \sum_{n=1}^N f (n\gamma) -\liminf_{N\to\infty} \frac{1}{N} \sum_{n=1}^N f (n\gamma) \leq \int_0^1 f_U -\int_0^1 f_L \leq \epsilon. \end{align*}\]

Since the choice of $\epsilon>0$ is arbitrary, the limit of $\frac{1}{N} \sum_{n=1}^N f_L (n\gamma)$ exists and

\[\begin{align*} \mathscr{L}(f,\mathscr{P})\leq \frac{1}{N} \sum_{n=1}^N f (n\gamma) \leq \mathscr{U}(f, \mathscr{P}) \end{align*}\]

Since $f$ is integrable, for any $\epsilon >0$ there is a partition $\mathscr{P}$ such that $\mathscr{U}(f,\mathscr{P})- \mathscr{L}(f,\mathscr{P}) <\epsilon$. Then

\[\begin{align*} \mathscr{U}(f,\mathscr{P}) \leq \mathscr{L}(f,\mathscr{P}) +\epsilon \leq \int_0^1f _\epsilon. \end{align*}\]

Similarly,

\[\begin{align*} \mathscr{L}(f,\mathscr{P}) \leq \mathscr{U}(f,\mathscr{P}) -\epsilon \leq \int_0^1f -\epsilon. \end{align*}\]

Thus,

\[\begin{align*} \int_0^1 f -\epsilon \leq \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma) \leq \int_0^1 f +\epsilon. \end{align*}\]

Then we can conclude that

\[\begin{align*} \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n\gamma) =\int_0^1 f(x)dx \end{align*}\]

\(\tag*{$\square$}\)

Reference

• Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
• Math 139 Fourier Analysis Notes