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!



Is this ergodic inequality true?
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.


