1
$\begingroup$

Let $\mathbf X := (X, \mathcal F, \mu)$ be a standard probability space. For an ergodic measure preserving transformation $T$, we define the ergodic robustness $\mathcal R(T)$ of $T$ as follows:

For $0 \leq r \leq 1$, let $C_r \subset \mathbb N^{\mathbb N}$ be the subset of monotonically increasing sequences whose natural density exists and is greater than of equal to $r$.

Define the quantity $E(T)$ by

$$E(T) := \inf \Big\{r \in [0, 1]\;\big| \text{ For all } f \in L^1 (X), (n_k) \in C_r, \lim_{K \to \infty} \frac{1}{K} \sum_{k = 0}^{K-1} f(T^{n_k} (x)) = \int f d\mu \text{ for a.e } x \in X\Big\}.$$

Finally, define $\mathcal R(T) = 1 - E(T)$.

Question: Do there exist ergodic measure preserving transformations $T$ with $\mathcal R(T)$ arbitrarily close to $1$? That is, for every $\varepsilon > 0$, does there exist an ergodic transformation $T$ with $\mathcal R(T) > 1 - \varepsilon$?

Remark: The choice to use $\mathcal R(T)$ instead of $E(T)$ is purely aesthetic to fit the terminology.

$\endgroup$
1
$\begingroup$

Yes. Let $\nu$ be a probability measure on $[0,1]$. Let $T$ be the left shift on a sequence space $X:=[0,1]^{\mathbb N}$ equipped with the product $\sigma$-field and the product measure $\mu:=\nu^{\mathbb N}$. Then for every strictly increasing sequence $\{n_k\}$ of positive lower density and $f \in L^1 (X)$ we have $$(*) \quad \lim_{K \to \infty} \frac{1}{K} \sum_{k = 0}^{K-1} f(T^{n_k} (x)) = \int f d\mu \text{ for a.e } x \in X \, ,$$ so $E(T)=0$. To verify $(*)$, it suffices to check it for a dense collection of functions in $L^1$. This reduction follows from a Theorem of Garsia (a version of the Banach principle) as stated e.g. in Theorem 4.2 of [1], see also its application in Theorem 4.3.

If $f$ depends only on the first $q$ coordinates, then $(*)$ follows (for all sequences $n_k$) from the law of large numbers if you separate $n_k$ into $q$ subsequences where each subsequence has gaps at least $q$. Since such functions $f$ are dense, this completes the proof.

[1] Reich, Jakob I. "On the individual ergodic theorem for subsequences." The Annals of Probability 5, no. 6 (1977): 1039-1046. https://projecteuclid.org/euclid.aop/1176995673

$\endgroup$
3
  • $\begingroup$ This is a great solution.. $\endgroup$
    – Nate River
    Jun 22 at 3:30
  • $\begingroup$ This system is strong mixing right? I would think that this (or weak mixing maybe) is a necessary condition for the robustness to be $1$. Hmm.. $\endgroup$
    – Nate River
    Jun 22 at 3:35
  • $\begingroup$ Oh I mean, a necessary condition for an ergodic system to satisfy that. $\endgroup$
    – Nate River
    Jun 22 at 3:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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