Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a measure-preserving (finite) system $(X,\mathcal{B},\mu,T)$, is it correct that the following are equivalent?

  1. For every $A,B\in\mathcal{B}$ , $\displaystyle\lim_{n\rightarrow\infty}\mu(A\cap T^{-n}B)=\mu(A)\mu(B)$.

  2. For every $A,B\in\mathcal{B}$ of positive measure, there is some $n_0\in\mathbb{N}$ such that for every $n>n_0$, $\mu(A\cap T^{-n}B)>0$.

Clearly 1 implies 2. Is the opposite direction also correct?

share|improve this question
    
Is every invariant measure a mixture of ergodix measures ? I think this is Choquet's thm but ..., are all ergodic measure mutually singular ? Can you use this to show the mixture must be trivial ? –  mike Dec 10 '12 at 16:06
2  
@mike: The ergodicity of $\mu$ follows from "2" immediately. –  23rd Dec 10 '12 at 17:12
4  
This exact question has already been asked in 1968 : it is proposed as a conjecture at the end of the paper “On weak mixing automorphisms” by James W. England and N.F.G. Martin, Bulletin of the AMS, 74, pp.505-507. –  Ewan Delanoy Dec 13 '12 at 13:44
    
Ergodicity implies convergence in the Cesàro sense, thus, should it exist, the limit in (1) is $\mu(A)\mu(B)$. –  Norbert Pintye Dec 19 '12 at 1:38
    
The answer seems to be No. The question is answered here: mathoverflow.net/questions/125245/silly-question-about-mixing –  Cantor Apr 11 '13 at 20:30

2 Answers 2

Yes, I'm sure. The definition of lightly mixing is (2.1) in the Rank-one lightly mixing paper: $\liminf_{n\to \infty} \mu(T^nA\cap B)>0$ for $A,B$ of positive measure. This definition implies (2) above. Lemma 2.4 of the paper shows the lightly mixing definition is equivalent to (2) where B=A (as you observed!). See the proof on page 2.

The Rank-one lightly mixing paper gives an example of a measure preserving transformation that satisfies (2) but is not strongly mixing. This shows that (1) and (2) are not equivalent.

As a side note, this rank-one transformation is attributed to Chacon from a paper in 1969 "Weakly mixing transformations which are not strongly mixing". Possibly, Chacon already knew lightly mixing was distinct from strongly mixing? Also, Peter Walters may have been the first to introduce lightly mixing in print, but he called it intermixing (circa 1970). I don't have the reference right now, but presumably Walters knew lightly/intermixing mixing is different than strongly mixing. Ciao

share|improve this answer
1  
Thanks. You don't have an account ? Any expertise in ergodic theory is welcomed here :) –  Stéphane Laurent Mar 6 '13 at 14:53

The answer is NO. The second condition is called lightly mixing, while the first condition is called strongly mixing. It was shown in a paper in 1991 that these conditions are not equivalent. There exists a lightly mixing measure preserving system which is not strongly mixing: http://link.springer.com/article/10.1007%2FBF02773841?LI=true.

share|improve this answer
    
Are you sure ? Lemma 2.4 of this paper says that lightly mixing is 2) with $B=A$. –  Stéphane Laurent Mar 5 '13 at 16:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.