Math-1983 · mathematics · Dynamical systems

Math-1983

Multiple ergodic averages — Problem 33

Status Open · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

Fix distinct exponents $a,b\in(0,1)$. Let $\xi_1,\eta_1,\xi_2,\eta_2,\ldots$ be mutually independent Bernoulli variables with $\mathbb P(\xi_k=1)=k^{-a}$ and $\mathbb P(\eta_k=1)=k^{-b}$. Write $u_n(\omega)$ and $v_n(\omega)$ for the increasing enumerations of the two selected sets. Both enumerations are infinite almost surely.

The target is one event of probability one on which the following holds for every probability space $(X,\mathcal B,\mu)$, commuting invertible measure-preserving $T,S$, and bounded functions $f,g$: \[ \frac1N\sum_{n=1}^N (f\circ T^{u_n(\omega)})(g\circ S^{v_n(\omega)}) \longrightarrow \mathbb E_\mu(f\mid\mathcal I_T)\cdot \mathbb E_\mu(g\mid\mathcal I_S). \] Here $\mathcal I_T$ is the completed $\sigma$-algebra of sets invariant modulo $\mu$ under $T$, and similarly for $S$. Prove convergence in $L^2(\mu)$ and, as the stronger version, pointwise $\mu$-almost everywhere.

The same index $n$ pairs the $n$th selected time of each process; the selection indicators are not paired by their original locations. The exponents are fixed in advance, and the full-probability event may depend on them but not on the dynamical system. For pointwise convergence, the dynamical null set may depend on the chosen observables. No ergodicity assumption is imposed, so the asserted limit is a product of conditional expectations rather than necessarily a constant.

Short English statement

Do independently sampled time sequences with different growth exponents force double ergodic averages to separate into the two invariant means?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.