A recurrence-type strong Borel–Cantelli lemma for Axiom A diffeomorphisms

Pub Date : 2024-09-18 DOI:10.1017/etds.2024.64

ALEJANDRO RODRIGUEZ SPONHEIMER

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引用次数: 0

Abstract

Let

$(X,\mu ,T,d)$

be a metric measure-preserving dynamical system such that three-fold correlations decay exponentially for Lipschitz continuous observables. Given a sequence

$(M_k)$

that converges to

$0$

slowly enough, we obtain a strong dynamical Borel–Cantelli result for recurrence, that is, for

$\mu $

-almost every

$x\in X$

$$ \begin{align*} \lim_{n \to \infty}\frac{\sum_{k=1}^{n} \mathbf{1}_{B_k(x)}(T^{k}x)} {\sum_{k=1}^{n} \mu(B_k(x))} = 1, \end{align*} $$

where

$\mu (B_k(x)) = M_k$

. In particular, we show that this result holds for Axiom A diffeomorphisms and equilibrium states under certain assumptions.

查看原文

公理 A 差分的递推型强博雷尔-康特利定理

让$(X,\mu ,T,d)$是一个度量保全的动力系统，对于利普齐兹连续观测值，三折相关性呈指数衰减。给定一个足够慢地收敛到 $0$ 的序列 $(M_k)$，我们会得到一个强动力学的 Borel-Cantelli 递归结果，即对于 $\mu $ - 几乎每一个 $x\in X$ ， $$ \begin{align*}\limit_{n \to \infty}\frac{sum_{k=1}^{n}\mathbf{1}_{B_k(x)}(T^{k}x)} {\sum_{k=1}^{n}\mu(B_k(x))} = 1, \end{align*}$$ 其中 $\mu (B_k(x)) = M_k$ 。我们特别指出，在某些假设条件下，这一结果对于公理 A 差分和平衡态都是成立的。

本文章由计算机程序翻译，如有差异，请以英文原文为准。

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