Weyl's Equidistribution Theorem
Weyl's criterion lets us test uniform distribution by checking exponential sums. We will prove the equivalence between this test, interval counts, and averages of Riemann-integrable functions. 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
where \(\langle x \rangle\) denotes the fractional part of \(x\). Weyl's equidistribution criterion states that the following statements are equivalent:
- \(x_1, x_2, \cdots\) are equidistributed (mod 1).
-
For each non-zero integer \(k\), we have
\[ \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N e^{2\pi i k x_n}=0 \] -
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
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,
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 \(N\geq N_0\),
In a similar way, we can prove that
Therefore, we have
To see that (3) holds when \(f\) is complex valued, we need only consider the real and imaginary parts separately.
(2) ⇒ (3)
First let \(f\) be continuous and \(1\)-periodic. Trigonometric polynomials uniformly approximate such functions on the circle. For any \(\epsilon>0\), choose
By (2), the average of \(p(x_n)\) tends to its constant coefficient \(c_0=\int_0^1p(x)\,dx\). Thus
Letting \(\epsilon\to0\) proves the assertion for continuous periodic \(f\).
To pass to interval indicators, fix an arc \(I\) of the circle. For any \(\epsilon>0\), choose continuous periodic functions \(g_-\) and \(g_+\), with short linear ramps near the endpoints, such that
Their integrals bracket the length of \(I\). Applying the result just proved to \(g_-\) and \(g_+\) squeezes the limiting interval frequency to that length. The same construction with a small neighborhood of a single point shows that endpoint visits have density zero, so the choice of open, closed, or half-open endpoints does not affect the result. This proves (1). The argument in (1) ⇒ (3) now extends the conclusion to all Riemann-integrable functions.
Periodicity is essential in the first approximation step: a continuous function on \([0,1]\) whose endpoint values differ cannot be uniformly approximated there by trigonometric polynomials.
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