I met a problem that I cannot easily solve: $f: \mathbb{R} \to \mathbb{R}$ is a continuous 1-periodic function. $\alpha$ is irrational. We want to prove $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k\alpha) = \int_{0}^{1}f(x)dx$$ I first examined the orthonormal basis in $L^2([0,1])$, namely $\lbrace e^{2\pi i kx}, k\in \mathbb{Z}\rbrace$ and condition holds. Then by $L^2$ estimation, $\vert \int_{0}^{1}f(x)dx - \int_{0}^{1}f_m(x)dx\vert \leq \vert \vert f-f_m \vert \vert_2 \to 0 $ as $m \to \infty$, where $f_m$ is the linear combination of elements in $\lbrace e^{2\pi i kx}, k\in \mathbb{Z}\rbrace$. Now I only need to show we can also approximate left-hand side by such $f_m$, i.e., want to show:$$\lim_{m\to \infty}\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f_m(k\alpha) = \lim_{n\to \infty}\lim_{m \to \infty} \frac{1}{n}\sum_{k=1}^{n}f_m(k\alpha) = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k\alpha)$$ (because any $m$, $f_m$ is a finite combination of basis elements, thus $\lim\limits_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f_m(k\alpha) = \int_{0}^{1}f_m(x)dx$).But I'm stuck here now: I need to verify$\lim\limits_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f(k\alpha)$ exists and then to verify interchanging limit of $n$ and $m$ is legitimate (i.e.,$\lim\limits_{m\to \infty}\lim\limits_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}f_m(k\alpha) = \lim\limits_{n\to \infty}\lim\limits_{m \to \infty} \frac{1}{n}\sum_{k=1}^{n}f_m(k\alpha)$) .
Could anyone offer any hints/solutions? Any help would be thankful.
The crucial thing that I missed is a theorem in Fourier Analysis, which is called $\textbf{Fejer's Theorem}$: A continuous periodic function on $\mathbb{R}$ restricted on $[0,1]$ (a period) can be uniformly approximated by Cesaro mean of Fourier sum, i.e., $\displaystyle{\frac{\sum_\limits{N=0}^{n}S_N(f)}{n} \to f}$ uniformly. With this theorem, we simply apply the argument to function $$\displaystyle{f_n = \frac{\sum_\limits{N=0}^{n}S_N(f)}{n}}$$ and not hard to see that $f_n$ is a linear combination of $e^{2\pi i m x}$, $0 \leq m \leq n$. Also note that $\int_{0}^{1}f_n(x)dx = \int_{0}^{1}f(x)dx$, $\forall n$. Uniform convergence of $f_n$ also allows us to change the order of limit of the left-hand side.