proof of $\lim_\limits{m\to\infty}\frac1m\sum_\limits{n=1}^m f(an)=\int_\limits{0}^1fdx$

Question

I have taken a compact seminar about Fourier analysis and someone mentioned the following: Let $f\in C(\mathbb R)$ with period 1. Let $a$ be an arbitrary irrational number. Then $$ \lim_{m\to\infty}\frac1m\sum_{n=1}^m f(an)=\int_\limits{0}^1fdx $$

Since it wasn't mentioned any proof, I've tried to verify the equation by showing it first for $f(x)=\exp(2\pi ikx)$ but didn't get far. Anybody could help using Fourier?

Answer 1

This result is known as Weyl's ergodic theorem.

Using Fejér's theorem, $f$ is a uniformly approximate by elements in $\textrm{Vect}(\{e^{2i\pi n\cdot};n\in\mathbb{Z}\})$. Hence, it suffices to prove that for all $n\in\mathbb{N}$, one has: $$\lim_{m\to+\infty}\frac{1}{m}\sum_{k=0}^{m-1}e^{2i\pi kn\theta}=0.$$ To see that, notice that $n\theta$ is not an integer, so that $e^{2i\pi n\theta}\neq 1$. Hence using a geometric summation, one has: $$\left|\sum_{k=0}^{m-1}e^{2i\pi k\theta}\right|\leqslant\frac{2}{|1-e^{2i\pi n\theta}|}.$$

Answer 2

The statement is called 'ergodic theorem'. Pointwisely the statement is true. Since the function is continuous, it is true for every given point. Prove first that $Tx=ax$ holds ergodicity. Then use ergodic theorem is one kind of proof.

Answer 3

Not a complete answer, hopefully it helps though: If $f$ is constant, then we are done. If $f$ is of the form above ($exp(2 \pi i k)$ for $k$ not equal to zero) then the sum divided by $m$ you have written should go to zero when $m$ gets large - the sum can be seen as a geometric series + there you see why $a$ needs to be irrational!. If you approximate $f$ by a nice function (I think $K_N*f$ should work) and consider its Fourier coefficients, the first coefficient will be, up to epsilon, $\int_0^1 f(x)$. This one you will recover, the rest will disappear as desired. So its an epsilon & triangle inequality business