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Hoi, i'm struggling a little with something :p

Suppose we know $\frac{1}{n}\sum_{k=1}^n a_k-p \to 0$ and $\frac{1}{n}\sum_{k=1}^n|b_k-q|\to 0$.

Is there a way we can conclude $\frac{1}{n}\sum_{k=1}^na_kb_k-pq\to 0$

It seems like something obvious, but i can't prove it :/

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$a_k,b_k,p,q$ positive real numbers.. – DinkyDoe Nov 9 '12 at 1:49
up vote 1 down vote accepted

Actually you have no chance without making some additional assumptions. Let $p=q=0$. The Cesaro averaging allows you to have occasional terms of any size that is $o(n)$, which translates into $o(n^2)$ for the products. But that's too much! An explicit counterexample is $a_n=b_n=n^{2/3}$ when $n=2^k$ and $0$ otherwise.

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I forgot to mention that $a_k,b_k,p,q$ are all bounded between 1, and 0 :P – DinkyDoe Nov 9 '12 at 14:42
Well, how about $a_k=1-b_k=0$ for odd $k$ and $1$ for even $k$ then?. We still have $\frac 12\frac 12\ne 0$ – fedja Nov 9 '12 at 17:38

Thank you for the counterexample. Let me give a little background info, cause it is about weak-mixing and Ergodicity of a transformation T. Apparantly, if we assume T is weakly mixing, and S Ergodic, then $T\times S$ is Ergodic. This amounts to assuming:

$\frac{1}{n}\sum_{i=1}^n|a_k-p|\to 0$ (absolute convergence on one hand) $\frac{1}{n}\sum_{i=1}^n b_k-q\to 0$ (convergence on the other hand)

and showing this implies $\frac{1}{n}\sum_{i=1}^na_kb_k-pq \to 0$

\begin{align*} \frac{1}{n}\sum_{k=1}^n a_kb_k-pq & = \frac{1}{n} \sum_{k=1}^n (a_k-p)(b_k-q) + \frac{1}{n} \sum_{k=1}^np(b_k-q)+ \frac{1}{n} \sum_{k=1}^nq(a_k-p)\\ & \to \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n (a_k-p)(b_k-q)+0+0 \end{align*}

Maybe I'm missing something, but what extra conditions do we need to make sure this limit is 0. ?

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