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.

References

• Hannigan-Daley, Brad. Equidistribution and Weyl’s criterion. Retrieved from http://individual.utoronto.ca/hannigandaley/equidistribution.pdf. Accessed 5 Feb. 2020.
• Stein, Elias M. and Shakarchi, Rami. Fourier Analysis: An Introduction. Princeton University Press, 2003