3

1

Is anything similar to the following inequality true,

$\displaystyle P\{\max_{n \leq k \leq m} |A_k f - A_n f| > \epsilon\} \leq C \frac{||A_m f - A_n f||_1}{\epsilon}$

where $A_n f = \frac{\sum_{i=0}^{n-1} T^i \circ f}{n}$, and $T$ is a measure-preserving transformation?

My motivation for thinking it might be true is that something similar is true for martingales, namely

$\displaystyle P\{ \max_{n \leq k \leq m} |M_k - M_n| > \epsilon\} \leq \frac{||M_m - M_n||_1}{\epsilon}$,

by Doob's Submartingale Inequality (Ville's Inequality?), and I know there are many similarities between backward martingales and ergodic averages. However, I can't seem to deduce this from the Maximal Ergodic Theorem in the same way I can for the martingale case.

Any references or counterexamples would be helpful. Ergodic theory is not my specialty. Thank you!

flag

1 Answer

5

I think the following should provide a counter example.

Take $T: [0,1] \to [0,1]$ given by $T(x) = x + \frac{p}{q} + \delta \mod(1)$, where $\delta > 0$ is chosen small and so that $T$ is ergodic with respect to the Lebesgue measure.

Then $\|A_q f - A_{2 q} f\|_1 = O(\delta)$. Furthermore $$ \|A_q f - A_{q + 1} f\|_1 \geq \frac{1}{q+1} \|f\| - O(\frac{1}{q^2} ) - O(\delta). $$ This means choosing $q$ large enough and $\delta$ small enough, one obtains the counter example.

link|flag
Indeed this does seem to be a counterexample. Thank you! – Jason Rute May 3 2011 at 14:16

Your Answer

Get an OpenID
or

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