Is this ergodic inequality true? 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!
 A: I guess you wanted to define $A_n f =\frac{1}{n}\sum_{k=0}^{n-1}f \circ T^k.$
Then you can prove that there exists constants $C,B>0$ such that:

$P\left\{\max_{n\leq k \leq m} |A_k f - A_n f|>\epsilon\right\} <Ce^{-B n \epsilon^2}(m-n).$

In ergodic theory by supposing that the invariant measure of $T$ is given by $P,$ you can get the inequality (under some extra hypotheses on the decay rate of mixing)
$P\{\max_{n\leq k \leq m} |A_k f - A_n f|>\epsilon\}\leq P\{\max_{n\leq k \leq m} |A_k f-\int  fdP |>\frac{\epsilon}{2}\}<(m-n)P\{|A_n f-\int  fdP |>\frac{\epsilon}{2}\}<Ce^{-B n \epsilon^2}(m-n)$
for $B,C>0$ constants.
On the other hand, I do not see the analog with Martingales. For example, the process
$\{\exp(\lambda \frac{1}{\sqrt{n}}\sum_{k=0}^{[tn]}f\circ T^k-\frac{1}{2}\lambda^2 t),t\geq 0\}$ looks like a Martingale when $n$ is big.
A: 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.
