At the moment, I'm studying Ergodic Theory and I find myself a little stuck.
The Weyl Equidistribution Theorem states that the following are equivalent: 1. For any $f \in L^{1}([0,1])$ and sequence $\{ u_{k} \}_{k \geq 1}$ on $[0,1]$, \begin{eqnarray} \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^{n} f( u_{k} ) = \int_{0}^{1} f(x) \, dx. \end{eqnarray} 2. The sequence $\{ u_{k} \}_{k \geq 1}$ is uniformly distributed modulo 1, i.e., equidistributed on $[0,1]$.
For fixed real $b, t > 0$, how does one compute the following limit? \begin{eqnarray} \lim_{a \to \infty} \tfrac{1}{a t} \sum_{i = 1}^{at} \lbrace b ( t - \tfrac{i}{a} ) \rbrace \end{eqnarray} where $\lbrace \cdot \rbrace$ denotes the fractional part function. It is known that the sequence of fractional parts of the integral multiples of the irrational $\alpha$, i.e., $\{ \{ k \alpha \} \}_{k \geq 1}$, is equidistributed on $[0,1]$. I'm not sure how to work with the fractional part in the sum above.
Update: For fixed real $a, b > 0$, how does one compute the following limit? \begin{eqnarray} \lim_{t \to \infty} \tfrac{1}{a t} \sum_{i = 1}^{at} \lbrace b ( t - \tfrac{i}{a} ) \rbrace \end{eqnarray} Unfortunately, the Riemann summation trick doesn't work in this case. (I believe the answer is $\frac{1}{2}$ in the general case, and I think it follows from the above theorem and the average value of $\{ \cdot \}$ on $[0,1]$. However, if $a$ divides $b$, then the limit is $0$.)
Thanks!