# Prove Sequences are Uniformly Distributed Modulo 1

Can anybody show me the following Proof?

Prove that if $X_n$ for $n=1,2,\ldots,\infty$ is a real sequence that is uniformly distributed modulo $1$, and if $Y_n$, $n=1,2,\ldots,\infty$ is a real sequence such that $\lim_{n\to\infty} X_n – Y_n = \alpha$ then $Y_n$ is uniformly distributed modulo $1$.

Remark: This includes the situation where $X_n – Y_n = \varepsilon$ tends to zero as $n$ tends to infinity.

Hint: Prove for zero case first. Show that $X_n$ and $Y_n$ are close.

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Welcome to math.SE community! Please, acquaintance yourself with the FAQ, it is really helpful (among others there are some tips for how to make your questions/answers readable). Also, from the wording of your question I suspect that it might be a homework problem. Please note, that there is a (homework) tag, it will cause that you will get not only solution, but probably also a lot of useful hints how to work it out yourself. On the other hand, homework assignments without the (homework) tag are treated really badly. Good luck! –  dtldarek Apr 18 '12 at 19:58
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## 2 Answers

Hint: for any $b \in [0,1]$, consider how many $Y_n$ for $1 \le n \le N$ are in $[0,b]$ modulo $1$. If $|Y_n - X_n| < \epsilon$, what interval for $X_n$ would guarantee $Y_n$ is in this interval? What interval must $X_n$ be in if $Y_n$ is in this interval?

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@ Robert. The way I was trying to prove this was using the Weyl's criterion. I thought I need to consider a counting function. Could you please expand upon your hint? Thanks Ric –  Ric Apr 18 '12 at 21:56
For large $N$, estimate the fraction of $n \in [1, N]$ such that $Y_n$ is in the interval $[0,b]$ modulo $1$, using what you know about the fractions where $X_n$ is in certain other intervals. –  Robert Israel Apr 18 '12 at 22:03
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As an alternative to Robert's answer, it can also be done with the Weyl criterion. Let $X_n-Y_n=\alpha+b_n$ with $b_n\to0$. Then $$e^{2\pi ihY_n}=e^{2\pi ihX_n}e^{-2\pi ih\alpha}e^{-2\pi ihb_n}$$ The second term on the right is a constant, so it can be pulled out of any sum. The third term, you should show it can be written as $1+c_n$ with $c_n\to0$.

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This is what I have constructed for the zero case: I was wondering if you could show me for the case where alpha is an irrational number? Proof: Zero case: lim┬(n→∞)⁡〖(x_n-y_n )=0〗 From hypothesis and the continuity of the exponential function it follows that lim┬(n→∞)⁡〖(e^(2πivx_n )-e^(2πivy_n ) )=0〗 and if 〈y_n 〉 a sequence of real numbers is converging to a finite limit L, then lim┬(N→∞)⁡〖1/N〗 ∑_(n=1)^N▒〖y_n=L〗. Taking y_n=e^(2πivx_n )-e^(2πivy_n ) we have lim┬(N→∞)⁡〖1/N〗 ∑_(n=1)^N▒〖(e^(2πivx_n )-e^(2πivy_n ) )=0〗 –  Ric Apr 19 '12 at 13:56
and since by assumption lim┬(N→∞)⁡〖1/N ∑_(n=1)^N▒〖e^(2πivx_n )=0〗〗 then by Weyl’s criterion we also have lim┬(n→∞)⁡〖1/N ∑_(n=1)^N▒〖e^(2πivy_n )=0〗〗 and the sufficiency of Weyl’s criterion proves the sequence 〈y_n 〉 to be uniformly distributed modulo 1. Thanks Ric –  Ric Apr 19 '12 at 13:56
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