# Intuition behind Weyl's equidistribution theorem

Recently I've been studying Fourier analysis through Stein's book, and there is a section there dedicated to Weyl's Equidistribution Theorem, specifically,

A sequence $$(\xi_n)_n$$ in $$[0, 1)$$ is equidistributed if, and only if, for each $$k\in\mathbb{Z}_{\neq 0}$$ we have $$\lim_{N}\frac{1}{N}\sum_{n=1}^N e^{2\pi i k \xi_n}=0.$$

Now, I'm pretty confident that I understand Stein's proof, but I don't have any intuition for this result, and it uses a deep fact of Fourier analysis, that continuous functions can be approximated by trigonometric polynomials.

I think I got a nice intuition for the "only if" part: it is pretty straightforward to show (and intuitive to see) that if $$(\xi_n)_n$$ is equidistributed, so is $$(\{k\xi_n\})_n$$, and then for each $$N$$ the sequence $$e^{2\pi i k \xi_1}, \dots, e^{2\pi i k \xi_N}$$ behaves like the vertices of a $$N$$-sided regular polygon, and then its barycenter must be close to $$0$$, and the approximations get better as $$N\to\infty$$.

The "if" part is what I'm struggling with. I've tried the following. Suppose that $$\xi_n$$ is not equidistributed. Then there exists a interval $$[a, b)$$ and some $$\varepsilon>0$$ such that $$\left|\frac{\#\{1\le n\le N: \xi_n\in[a, b)\}}{N}-(b-a)\right|>\varepsilon$$ infinitely often. With some transformations, we can assume that $$a=0$$ (take $$\xi_n'=\{\xi_n-a\}$$). Moreover, we can assume that $$\frac{\#\{1\le n\le N: \xi_n\in[0, b)\}}{N}-b>\varepsilon$$ infinitely often (just take the interval [b, 1) otherwise and redo the previous step). Now, I hoped that there would be some $$k$$ such that, infinitely often, the points $$e^{2\pi ik\xi_n}$$ would cluster too much in some arc (because of the interval $$[0, b)$$), and this would prevent the mean from being too close from $$0$$. See the image below. But when I sat down to try to make sense of this argument, I couldn't get much further, as it is kinda hard to control the behavior of $$\{k\xi_n\}$$.

Does this idea have any truth to it, or is it a lost cause? Also, if you guys have other ideas, I'd like to read them.

I remind you thatt I'm just looking for some intuition, and the idea doesn't have to be super rigorous.

• I thunk the ituitio comes from probability. Equidistribution of a sequence $\xi_n$ is "similar" drawing i.i.d samples $(U_n:n\in\mathbb{N})$ from the uniform distribution in $[0,1]$. The law of large numbers number will give you the "expected" result $\frac1n\sum^n_{k=1}e^{2\pi I k U_n}\xrightarrow{n\rightarrow\infty}0$ for $k\in\mathbb{Z}\setminus\{0\}$. Jan 15 at 15:46
• you may want to hear directly from the Lion's mouth Jan 15 at 19:21
• Typo: you presumably meant "for $k\not0$, rather than $k>0$... Jan 15 at 21:15

If you want to prove the "if" part by contraposition you could go on as follows. As you've noted there is some interval $$[a,b)$$ such that
$$\left|\frac{\#\{1\le n\le N: \xi_n\in[a, b)\}}{N}-(b-a)\right|>\varepsilon$$
for infinitely many $$N$$'s, which is equivalent with
$$\left\vert\frac{1}{N}\sum_{n\leq N}\mathbb{1}_{[a,b)}(\xi_n)-(b-a)\right\vert>\varepsilon$$ but then also $$\left\vert\frac{1}{N}\sum_{n\leq N}\sum_k c_k e(k\xi_n)-(b-a)\right\vert>\varepsilon/2$$ for infinitely many $$N's$$ where the inner sum a sufficiently good approximation of $$\mathbb{1}_{[a,b)}$$ so $$\left\vert\sum_k c_k \frac{1}{N}\sum_{n\leq N}e(k\xi_n)-(b-a)\right\vert>\varepsilon/2$$ for infinitely many $$N$$'s, hence there exists a $$k_0$$ s.t. $$\lim_{N\to\infty}\frac{1}{N}\sum_{n\leq N}e(k_0\xi_n)\neq0$$ (with $$\neq0$$ I mean that the limit doesn't exist or it exists and is not equal to zero) (note that the sum over $$k$$ is a finite sum).
Although it may be a little fancier than one might like, depending on tastes, the hypothesis is that the average (divide-by-$$N$$) of a sum (over $$N$$) of (periodic) Dirac deltas at the $$\xi_k$$ has Fourier series converging (weakly!) to $$1$$ (all other Fourier coefficients go to zero). This has seemed a "meaningful", as opposed to "technical", explanation of the hypothesis.