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!
Best Answer
First make the substitution $n = a t$, then your sum becomes $$\frac{1}{n} \sum_{i=1}^n \{(b t) (1 - i/n) \}.$$ Now this is really just a Riemann sum which converges (as $n \rightarrow \infty$) to $$ \int_0^1 \{(bt) (1-i) \} di$$ Putting $u = (bt)(1-i)$ the integral becomes $$ \frac{1}{b t} \int_0^{bt} \{u\} du $$ and this is equal to $$\frac{1}{2 bt} ( \lfloor{b t\rfloor} + \{ b t \}^2 ).$$ (Because there are $\lfloor{b t\rfloor}$ triangles of area $1/2$ and one triangle of area $\{ b t \}^2 / 2$.)