Suppose $f$ is a continuous real-valued function on $\mathbb{R}$ such that $f(x+1)=f(x)$ for every $x$, Let $\gamma$ be an irrational number.

Question

Suppose $f$ is a continuous real-valued function on $\mathbb{R}$ such that $f(x+1)=f(x)$ for every $x$, Let $\gamma$ be an irrational number. Prove that $$ \lim_{n\to \infty}\frac{1}{n}\sum_{j=1}^{n}f(j\gamma)=\int_{0}^{1}f(x)dx. $$

Answer 1

Let $e_k(x) = e^{2 \pi i k x}$, it is straightforward to verify that the result holds for $e_k$.

Since $f$ is $1$-periodic and continuous, for any $\epsilon>0$ there is some function $p$ of the form $p(x) = \sum_{k=0}^n f_n e_k(x)$ such that $|f(x) -p(x)| < \epsilon$ for all $x$ (see, for example, Rudin, "Real & Complex Analysis", Theorem 4.25).

Note that the theorem holds for $p$ using linearity of summation and integration.

Choose $p$ such that $|f(x) -p(x)| < {1 \over 3}\epsilon$ for all $x$. Now choose $N$ such that if $n \ge N$, we have $|{1 \over n} \sum_{k=1}^n p(k \gamma) - \int_0^1 p(x) dx | < {1 \over 3}\epsilon$.

Then $|{1 \over n} \sum_{k=1}^n (f(k \gamma)-p(k \gamma)) - \int_0^1 (f(x)-p(x)) dx | < {2 \over 3} \epsilon$. Since \begin{eqnarray} {1 \over n} \sum_{k=1}^n (f(k \gamma)-p(k \gamma)) - \int_0^1 (f(x)-p(x)) dx &=& {1 \over n} \sum_{k=1}^n f(k \gamma) - \int_0^1 f(x) dx \\ & & \ \ - ({1 \over n} \sum_{k=1}^n p(k \gamma) - \int_0^1 p(x) dx) \end{eqnarray} we see that $|{1 \over n} \sum_{k=1}^n f(k \gamma) - \int_0^1 f(x) dx| < \epsilon$, for all $n \ge N$.