# Weyl’s Equidistribution Theorem

In this post, we will prove the Weyl’s Equidistribution theorem. A sequence of real numbers $x_1, x_2, \cdots$ is said to be equidistributed (mod 1) if for every sub-interval $(a,b)\subset [0,1]$, we have $$\lim_{N\to \infty}\frac{|\{1\leq n\leq N:\; \langle x_n \rangle\in (a,b)\}|}{N} = b-a$$ where $\langle x \rangle$ denotes the fractional part of $x$. Weyl’s equidistribution criteria states that the following statements are equivalent:

1. $x_1, x_2, \cdots$ are equidistributed (mod 1).
2. For each non-zero integer $k$, we have $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N e^{2\pi i k x_n}=0$$
3. For each Riemann integrable function $f:[0,1]\to\mathbb{C}$, we have $$\lim_{N\to \infty}\frac{1}{N}\sum_{n=1}^N f(\langle x_n \rangle) = \int_0^1 f(x) dx$$

Proof: (1) ⇒ (3)

Let $I=[a,b)\subseteq [0,1]$ and note that $$\frac{|\{1 \leq n \leq N: \langle x_n \rangle\in [a,b) \}|}{N}=\frac{1}{N}\sum_{n=1}^N \chi_{[a,b)}(\langle x_n \rangle)$$ where $\chi_{[a,b)}(x)$ equals 1 if $x\in [a,b)$ and 0 otherwise. This shows that (3) holds for the case when $f$ is a characteristic function. Now, let $\lambda_1, \lambda_2\in \mathbb{R}$ and $f_1$, $f_2$ be functions for which (3) holds. Then, \begin{aligned}\lim_{N\to \infty}\sum_{n=1}^N (\lambda_1 f_1 + \lambda_2 f_2)(\langle x_n \rangle) &= \lim_{N\to \infty} \frac{\lambda_1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) + \lim_{N\to \infty}\frac{\lambda_2}{N}\sum_{n=1}^Nf_2(\langle x_n\rangle) \\ &= \lambda_1\int_0^1 f_1(x) dx + \lambda_2 \int_0^1 f_2(x) dx \\ &= \int_0^1 (\lambda_1 f_1 + \lambda_2 f_2)(x) dx\end{aligned} Thus, (3) holds for all linear combinations of characteristic functions of subintervals of $[0,1]$.

Now, let $f:[0,1]\to \mathbb{R}$ be an integrable function, and let $\epsilon >0$. Choose step functions $f_1$ and $f_2$ such that:

• $f_1\leq f\leq f_2$ pointwise
• $\int_0^1 (f_2(x)-f_1(x))dx < \frac{\epsilon}{2}$
• There exists $N_0$ such that $\left|\int_0^1 f_1(x)dx - \frac{1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) \right| < \frac{\epsilon}{2}$ and $\left|\int_0^1 f_2(x)dx - \frac{1}{N}\sum_{n=1}^N f_2(\langle x_n\rangle) \right| < \frac{\epsilon}{2}$ for all $N\geq N_0$
It follows that for $N\geq N_0$, \begin{aligned} \int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) &\leq \int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f_1(\langle x_n\rangle) \\ &< \int_0^1 f(x) dx -\int_0^1 f_1(x) dx +\frac{\epsilon}{2} \\ &< \int_0^1 (f_2(x)-f_1(x)) dx + \frac{\epsilon}{2} \\ &< \epsilon \end{aligned} In a similar way, we can prove that $$\int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) > -\epsilon \quad \forall\; N\geq N_0$$ Therefore, we have $$\left|\int_0^1 f(x) dx - \frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) \right| < \epsilon \quad \forall\; N\geq N_0$$ To see that (3) holds when $f$ is complex valued, we need only consider the real and imaginary parts separately.

(2) ⇒ (3)

Let $f:[0,1]\to \mathbb{R}$ be continuous, and let $\epsilon > 0$. The Stone-Weierstrass Theorem allows us to choose a trigonometric polynomial $p$ such that: $$\sup_{x\in [0,1]} |f(x) - p(x)| < \frac{\epsilon}{3}$$ Also, (2) implies the existence of an $N_0$ such that for $N\geq N_0$, we have $$\left|\frac{1}{N}\sum_{n=1}^N p(\langle x_n \rangle)-\int_0^1 p(x) dx \right| < \frac{\epsilon}{3}$$ Now, \begin{aligned} &\; \left|\frac{1}{N}\sum_{n=1}^N f(\langle x_n\rangle) - \int_0^1 f(x) dx\right| \\ &= \left|\frac{1}{N}\sum_{n=1}^N (f(\langle x_n \rangle) - p(\langle x_n \rangle)) + \int_0^1 (p(x)-f(x))dx + \frac{1}{N}\sum_{n=1}^N p(\langle x_n \rangle) - \int_0^1 p(x) dx\right| \\ &< \frac{1}{N}\sum_{n=1}^N\left|f(\langle x_n \rangle) - p(\langle x_n \rangle) \right| + \int_0^1 \left|p(x)-f(x) \right| dx + \left|\frac{1}{N}\sum_{n=1}^N p(\langle x_n \rangle) - \int_0^1 p(x) dx \right| \\ &< \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} \\ &= \epsilon \end{aligned} for all $N\geq N_0$. Thus, (3) holds for continuous functions on $[0,1]$. By the proof of (1) ⇒ (3), it is sufficient to show that (3) holds for all step functions on $[0,1]$. If $g$ is a step function on $[0,1]$, we can find continuous functions $g_1, g_2$ such that $g_1\leq g\leq g_2$ and $\int_0^1 (g_1(x)-g_2(x))dx < \epsilon$. We again conclude that (3) holds for $g$.

The implications (3) ⇒ (1) and (3) ⇒ (2) are obvious.