By using Fourier Series Prove that $\frac{1}{n+1}\sum_{k=0}^n f(x+k\alpha)\to\int_0^1 f(y)dy$.

Question

Here we assume that $f$ is an $1$-periodic continuous function, $x\in[0,\,1]$ and $\alpha$ is an irrational number in $(0,\,1)$.

I couldn't recall which theorem in the Fourier series has a similar form to the limit I want to prove.

Answer 1

The OP is basically Weyl's theorem on equi-distribution.

This result can be proved by methods of Calculus (Fourier series as required by the OP) with a little help from the Stone-Weierstrass theorem about density of trigonometric polynomials on the space of continuous functions on the circle (or continuous periodic functions).

Consider the unit circle $\mathbb{S}^1=\{z\in\mathbb{C}:|z|=1\}$ equipped with the $\sigma$-algebra inherited as a subspace of $\mathbb{R}^2$ and the measure $\lambda_{\mathbb{S}^1}:=\frac{1}{2\pi}\lambda_1$, where $\lambda_1$ is the the arch-length measure. (Equivalently, this is the real line $\mod 1$ with the Borel $\sigma$-algebra and Lebesgue measure restricted to $[0,1]$.)

For any $\theta\in(0,1)$ define the rotation map $$ \begin{align} R_\theta&:\mathbb{S}^1\rightarrow\mathbb{S}^1\\ z&\mapsto z e^{2\pi i\theta} \end{align} $$

Or equivalently, if $z=e^{i2\pi x}$, $R_\theta(x)=x+\theta\mod 1$

Define $R^0_\theta(z)=z$, and or $k\geq1$, $R^k_\theta(z)=R_\theta(R^{k-1}_\theta(z))$. That is, $$R^k_\theta(z)=ze^{2\pi i k\theta}, \qquad z\in\mathbb{S}^1$$ or equivalently, in terms of sum $\mod 1$, if $z=e^{2\pi i x}$, $$ R^k_\theta(x)=x+k\theta\mod 1$$

Theorem (Weyl): Suppose $R_\theta$ is an irrational rotation. Then, for every bounded Riemann integrable function $f:\mathbb{S}^1:\rightarrow\mathbb{C}^1$, and any $z\in\mathbb{S}^1$, (or equivalently, any $1$-periodic continuous function on the real line Riemann integrable in $[0,1]$) $$ \lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f\circ R^k_\theta(z)=\int_{\mathbb{S}^1} f(w)\,\lambda_{\mathbb{S}^1}(dw) $$ or in terms of the real line $\mod 1$, if $z=e^{2\pi ix}$ $$ \lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f(x+k\theta)=\int^1_0 f(t)\,dt $$

Here is a sketch of the proof:

For any function $f$ on $\mathbb{S}^1$, let $S_nf(z)=\frac{1}{n}\sum^{n-1}_{j=0}f(R^j_\theta(z))$. Consider polynomials $f_k(z)=z^k$, $k\in\mathbb{\mathbb{Z}}$.

For $k=0$, $f_0\equiv 1$ and so, $$\begin{align}S_nf_0(x)\equiv1&=\int_{\mathbb{S}^1} f_0(z)\,dw\\ &=\frac{1}{2\pi}\int^{2\pi}_0f_0(e^{ikx})\,dx=\int^1_0f_0(e^{2\pi it})\,dt \end{align} $$

For $|k|\geq1$, $$\begin{align} S_nf_k(z)&=\frac{z^k}{n}\sum^{n-1}_{j=0}e^{i2\pi\theta k j}= \frac{z^k}{n}\frac{1-e^{in 2\pi k\theta}}{1-e^{i2\pi k\theta}}\xrightarrow{n\rightarrow\infty}0\\ &=\int_{\mathbb{S}^1}f_k\,\lambda_{\mathbb{S}^1}=\frac{1}{2\pi}\int^{2\pi}_0 e^{ikx}\,dx=\int^1_0 e^{2\pi ikt}\,dt \end{align}$$ since $\theta$ is irrational and so, $e^{i2\pi k\theta}\neq1$ for all $k\in\mathbb{Z}$. This means that the statement holds for all trigonometric polynomials. By the (complex) Stone-Weierstrass theorem, the result then holds for any $f\in\mathcal{C}(\mathbb{S}^1)$.

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To extend the result to any Riemann integrable function $f$, it is enough to assume that $f$ is real valued. Given $\varepsilon>0$, we can choose continuous functions $g_\varepsilon$ and $h_\varepsilon$ such that $ g_\varepsilon < f\leq h_\varepsilon$, and $$ \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} -\varepsilon <\int_{\mathbb{S}^1} g_\varepsilon\, \lambda_{\mathbb{S}^1}\leq \int_{\mathbb{S}^1}h_\varepsilon \,\lambda_{\mathbb{S}^1}< \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} +\varepsilon $$ Then $$ S_ng_\varepsilon-\int g_\varepsilon -\varepsilon \leq S_nf(z)-\int f \leq S_nh_\varepsilon(z)-\int h_\varepsilon+\varepsilon $$ whence we conclude that $$-\varepsilon\leq \liminf_{n\rightarrow\infty}S_nf(z)-\int f\leq\limsup_n S_nf-\int f\leq\varepsilon$$ for all $\varepsilon>0$ and $z\in\mathbb{S}^1$. This completes the proof.

Answer 2

Otherwise there is a solution using only Parseval's theorem.

For $f$ continuous $1$-periodic let $$F_n(x)= \frac1{n+1}\sum_{k=0}^n f(x+k\alpha)- \int_0^1 f(t)dt$$

Note that $\sup_x |F_n(x+h)-F_n(x)| \le \sup_x |f(x+h)-f(x)|$, ie. the family $(F_n)_{n\ge 1}$ is uniformly continuous. We find the Fourier coefficients $$c_0(F_n)=0,\qquad c_m(F_n)=\int_0^1 F_n(x)e^{-2i\pi mx}dx = c_m(f)\frac1{n+1}\frac{1-e^{2i\pi (n+1)m\alpha}}{1-e^{2i\pi m\alpha}}$$

So $|c_m(F_n)| \le |c_m(f)|$ and $\lim_{n\to \infty} c_m(F_n)=0$. This implies that $$\lim_{n\to \infty}\int_0^1 |F_n(x)|^2dx=\lim_{n\to \infty}\sum_m |c_m(F_n)|^2= 0$$ Which in turn, by uniform continuity of the $F_n$ family, implies that $$\lim_{n\to \infty}\sup_x |F_n(x)|= 0$$