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.
Best Answer
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).