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If two sequences $\{a_k\}$ and $\{b_k\}$ are such that

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k-b_k)=0$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k^2-b_k^2)=0$$

does it mean that the two sequences are asymptotically equally distributed? i.e., you can't distinguish one sequence from the other?

EDIT

Assume that two sequences are such that

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}(a_k^q-b_k^q)=0;\ \forall\ q\in\mathbb{N}$$

I'm asking this because a sentence in a paper claims that for two asymptotically equally distributed sequences, the above holds. I'm wondering if the converse is true, i.e., if the above hold for any sequence, are they asymptotically equally distributed?

EDIT 2

Definition: Asymptotically equally distributed sets

Two sets $\{a_k\}$ and $\{b_k\}$ are said to be asymptotically equally distributed if

$$-\infty<\alpha_1\leq \{a_k,b_k\}\leq\alpha_2<\infty$$

and

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}\left[f(a_k)-f(b_k)\right]=0$$

where $f(x)$ is any function continuous on $[\alpha_1,\alpha_2]$. This is called Weyl's theorem or Weyl's definition, but I don't remember the original paper this was discussed.

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2  
What do you mean by an "asyptotically equally distributed" pair of sequences? And that you can't "distinguish" one from the other? Please clarify your question. –  cardinal Mar 12 '11 at 17:31
    
Whatever "asymptotically equally distributed" means, it should imply that the corresponding limit for third moments is zero, and it should be easy to find a counterexample where this fails. –  Qiaochu Yuan Mar 12 '11 at 17:38
    
@Qiaochu Yuan: I've edited my question –  user7815 Mar 12 '11 at 17:50
    
Why not cite the paper? –  Raskolnikov Mar 12 '11 at 17:52
    
You still have not provided a definition of "asymptotically equally distributed." –  Qiaochu Yuan Mar 12 '11 at 18:06

1 Answer 1

up vote 2 down vote accepted

The answer to the question as it stands is no. Let $b_n = 0$ identically and let $a_n$ be a sparse enough sequence such that all of its asymptotic moments are zero but such that it is unbounded (e.g. let $a_n = k$ if $n = 2^k$ and $0$ otherwise). Then the first condition is not satisfied.

If you add the boundedness condition, then the answer is yes. Like yoyo says in the comments, by Stone-Weierstrass we can uniformly approximate any continuous function $f$ on an interval by polynomials, and it is not hard to see that the desired expression in $f$ respects uniform limits for bounded sequences.

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