Suppose we have a dynamical system and a sequence of functions $0=f_0\leq f_1\leq\cdots\leq f_k$. Define $J_{r,\lambda}$ to be the set of points $x$ such that there are $j_0<j_1<\cdots<j_r$ so that, for each $i<r$, $\sup_n A_n(f_{j_{i+1}}-f_{j_i})(x)>\lambda$. Using the maximal ergodic theorem, we see that $\mu(J_{1,\lambda})\leq ||f_k||_1/\lambda$. Is there a similar bound on $\mu(J_{r,\lambda})$ (maybe even $||f_k||_1/r\lambda$)?

Note that this doesn't seem to trivially reduce to multiple applications of the maximal ergodic theorem, because the sequences of $j$'s might overlap between different points, and one point might see $r$ jumps from $0$ to $j_r$, while another point doesn't see any jumps until some $j'>j_r$.

(We have some ideas that might show this, at least for a slightly more restrictive definition of $J_{r,\lambda}$---that there are $n_0<\cdots<n_{r-1}$ so that, for each $i<r$, $A_{n_i}(f_{j_{i+1}}-f_{j_i})(x)>\lambda$---but they're messy, and it seems like a clever use of the maximal ergodic theorem should give a slick proof, but we don't see how to do it.)

  • $\begingroup$ When you say a similar bound, there is the identical bound $\mu(J_{r,\lambda})\le \|f_k\|_1/\lambda$, right? (just since $J_{r,\lambda}\subset J_{1,\lambda}$ unless I'm mistaken) - maybe you're looking for a bound that is of strictly smaller order than this (as a function of $r$)? Or is your question "what is the sharp bound"? $\endgroup$ – Anthony Quas Dec 12 '16 at 4:01
  • $\begingroup$ @AnthonyQuas I was hoping for a bound that goes to 0 as r goes to infinity $\endgroup$ – Henry Towsner Dec 12 '16 at 4:59

You can't do substantially better than $\|f_k\|_\lambda$. Here's a simple example: Consider $X=\{0,1,\ldots,2^{N}-1\}$, equipped with normalized counting measure. The transformation is $T(x)=x+1\bmod 2^{N}$. The functions are $$ f_i(x)=\begin{cases} 2&\text{if $x<i$;}\\ 0&\text{otherwise.} \end{cases} $$ for $i=0,\ldots,2^{N}$.

Let $\lambda=1$. Now if $j<2^N-2^{N/2}$ and $1\le n\le N/2$, consider $f_{j+2^{n}}-f_{j+2^{n-1}}$. This takes values 2 on the range $j+2^{n-1}$ to $j+2^n-1$ and 0's elsewhere. The maximum of the Birkhoff averages evaluated at $j$ is therefore 1. Hence for each $j\in \{0,\ldots,2^N-2^{N/2}-1\}$, $j\in J_{N/2,1}$. That is, $\mu(J_{N/2,1})\ge 1-2^{-N/2}$, but $\|f_{2^N}\|=2$. You cannot do better than $\mu(J_{r,\lambda})\le \|f_k\|/(2\lambda)$.

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    $\begingroup$ It occurs to me that if you play with the construction a little (using $M^n$ instead of $2^N$, you cannot actually obtain a better bound than you get from the trivial inclusion $J_{r,\lambda}\subset J_{1,\lambda}$. That is: $\mu(J_{r,\lambda})\le \|f_k\|/\lambda$ and this is sharp. (This feels like it would be a nice exercise on maximal ergodic theorems somewhere). $\endgroup$ – Anthony Quas Dec 12 '16 at 19:50

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