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 $0$.)


  • $\begingroup$ For your second sum, try making the substitution $n = a t$ and try to get a sum where the summand is of the form $\{ \alpha i \}$ $\endgroup$
    – user899
    Feb 22, 2011 at 14:58

1 Answer 1


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 $\{ b t \}^2 / 2$.)

  • $\begingroup$ Thanks for the post. Does it follow that Weyl's Theorem doesn't necessarily play a role here? Or is it just lurking behind the Riemann summation? $\endgroup$
    – user02138
    Feb 21, 2011 at 8:29
  • 1
    $\begingroup$ I'm not sure. I have always thought that what you called Weyl's theorem to be just a simple characterization of equidistributed sequences. I think Weyl's theorem refers to en.wikipedia.org/wiki/Weyl's_criterion_for_equidistribution . Anyhow, to use the sum-integral equivalence, you need to know your sequence is equidstributed which I think is doable with what I call Weyl's theorem. $\endgroup$
    – user899
    Feb 21, 2011 at 9:10
  • $\begingroup$ The "reason" that your problem has nothing to do with Weyl's theorem is that the sequence $u_k$ does not depend on $n$, whereas your summands do depend on $a = n/t$. $\endgroup$ Feb 22, 2011 at 15:24
  • $\begingroup$ Thanks for the comment. Yes, I finally realized this point after I worked through yjj's answer. $\endgroup$
    – user02138
    Feb 22, 2011 at 23:09

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