One version of the equidistribution property raised 2 questions that I am trying to answer. Let a sequence $\{s_1,s_2,s_3,... \}$be equidistributed (ED) on an interval $[a,b]$ if for any subinterval $[c,d]$

$$\lim_{n\to\infty}\frac{\left|\{s_1,s_2,s_3,...\}\cap [c,d] \right| }{n} = \frac{c-d}{b-a}.$$

So for a sequence which is not equidistributed, there is at least one interval $[x,y]$ on which the property ED does not hold.

Let us say a sequence has the midpoint property (MP) if for any subinterval $[c,d]$ of $[a,b]$ the mean value of those elements of $s_n$ contained in $[c,d]$ as $n\to \infty$ is $(c+d)/2.$

Suppose a sequence does not have the midpoint property so that for some subinterval $[c,d]$ the limit of the mean of $s_n$ is $ \neq (c+d)/2. $ By expanding $[c,d]$ by equal lengths on either side we can say that any interval (created this way) containing $[c,d]$ also does not have the MP property. In fact $[a,b]$ cannot have the MP property since it contains an interval $[c,d]$ which does not have the property. (This is obviously not a proof.)

So I have two questions.

  1. Can we show: ED $\leftrightarrow$ MP?

  2. Is there an argument that if a sequence is not ED then it is not ED on any subinterval of $[a,b]?$

  1. I'm pretty sure ED and MP are equivalent. ED is equivalent to investigating $\sum_{k\le n} \chi_{[c,d]}(s_k)$, where $\chi_{[c,d]}$ is the indicator function of the interval $[c,d]$. On the other hand, MP is equivalent to investigating $\sum_{k\le n} s_k\cdot \chi_{[c,d]}(s_k)$ (think about why this is true). And both sums can be attacked using Fourier analysis (indeed, the classical Weyl's criterion shows this explicitly for ED).

  2. That proposed implication is false. Let $\{s_k\}$ be any sequence that is ED on $[0,1]$, and define a new sequence $\{t_k\}$ by $$ t_k = \begin{cases} s_{k/2}, &\text{if $k$ is even}, \\ 3/2, &\text{if $k$ is odd}. \end{cases} $$ Then $\{t_k\}$ is not ED on $[0,2]$, but it does have the correct equidistribution on all subintervals of $[0,1]$.

(That's one interpretation of your question 2. The other interpretation is clearly false: the $\{s_k\}$ from above is not ED on $[0,2]$ but it is ED on the subinterval $[0,1]$ of $[0,2]$.)

  • $\begingroup$ This is really a complete answer, thank you. I was looking for useful ways to exploit that the fractional part of log[n] is not equidistributed. Not much seems to follow from non-ED. There is an interval on which not ED... $\endgroup$ – daniel May 11 '17 at 19:13

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