Measure-theoretic equicontinuity and rigidity

TL;DR

This paper characterizes measure-theoretic rigidity in topological dynamical systems through the lens of measure-$A$-equicontinuity and mean-equicontinuity, linking these properties to subsequences and IP-sets.

Contribution

It establishes new equivalences between rigidity and measure-$A$-equicontinuity, including for IP-sets and positive density subsequences, and extends results to functions.

Findings

Rigidity is equivalent to existence of certain measure-$A$-equicontinuity subsequences.

Positive density measure-$A$-mean-equicontinuity implies rigidity.

Results extend to functions within the dynamical systems context.

Abstract

Let $(X,T)$ be a topological dynamical system and $\mu$ be a invariant measure, we show that $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some subsequence $A$ of $\mathbb N$ such that $(X,T)$ is $\mu$-$A$-equicontinuous if and only if there exists some IP-set $A$ such that $(X,T)$ is $\mu$-$A$-equicontinuous. We show that if there exists a subsequence $A$ of $\mathbb N$ with positive upper density such that $(X,T)$ is $\mu$-$A$-mean-equicontinuous, then $(X,\mathcal{B},\mu,T)$ is rigid. We also give results with respect to functions.

Equations

$$\underline{D}(A)=\underset{n\to \infty }{\mathrm{liminf}}\frac{|A\cap [0,n-1]|}{n},$$

$$\bar{D}(A)=\underset{n\to \infty }{\mathrm{limsup}}\frac{|A\cap [0,n-1]|}{n}.$$

$$FS\{b_i\}=\{b_{i_1}+b_{i_2}+\cdots +b_{i_k}:i_1<i_2<\cdots <i_k,k\in \mathbb{N}\}.$$

$$x,y\in K,d(x,y)<\delta \Rightarrow d(T^{a_n}x,T^{a_n}y)<ϵ,\forall n\in \mathbb{N}.$$

$$H_{\mu }(\xi )=-\sum _{A\in \xi }\mu (A)\log \mu (A),$$

$$H_{\mu }(\xi |\eta )=H_{\mu }(\xi \vee \eta )-H_{\mu }(\eta )=-\sum _{A\in \xi }\sum _{B\in \eta }\mu (A\cap B)\log \frac{\mu (A\cap B)}{\mu (B)}.$$

$$h_{\mu }^S(T,\xi )=\underset{n\to \infty }{\mathrm{limsup}}\frac{1}{n}{H}_{\mu }(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\xi ).$$

$$h_{\mu }^S(T)=\underset{\alpha }{\sup }{h}_{\mu }^S(T,\alpha ),$$

$$\prod _{j\in \mathbb{N}}\overline{\{U_T^n{f}_j:n\in A\}}\text{ is compact in }(L^2(\mu ){)}^{\mathbb{N}}.$$

$$\{(U_T^n{f}_1,\cdots ,U_T^n{f}_j,\cdots ):n\in A\}\subset \prod _{j\in \mathbb{N}}\overline{\{U_T^n{f}_j:n\in A\}}$$

$$\{(U_T^n{f}_1,\cdots ,U_T^n{f}_j,\cdots ):n\in A\}\subset \bigcup _{i=1}^{N}B((U_T^{n_i}{f}_1,\cdots ,U_T^{n_i}{f}_j,\cdots ),\frac{1}{k}).$$

$$\begin{array}{rl}d((U_T^n{f}_1,\cdots ,U_T^n{f}_j,\cdots ),(U_T^{n_1}{f}_1,\cdots ,U_T^{n_1}{f}_j,\cdots ))& =\\ \sum _{j\in \mathbb{N}}\frac{\Vert U_T^n{f}_j-U_T^{n_1}{f}_j{\Vert }_{L^2}}{2^j}& =\sum _{j\in \mathbb{N}}\frac{\Vert U_T^{n-n_1}{f}_j-f_j{\Vert }_{L^2}}{2^j}<\frac{1}{k}.\end{array}$$

$$U_T^{t_k}{f}_j\stackrel{L^2}{⟶}{f}_j\text{ as }k\to \infty \text{ for each }{f}_j.$$

$$x,y\in K,d(x,y)<\delta \Rightarrow d(T^{a}x,T^{a}y)<ϵ,\forall a\in A.$$

$$K=\bigcup _{i=1}^L(B(x_i,\frac{\delta }{2})\cap K).$$

$$diam(T^a(B(x_i,\frac{\delta }{2})\cap K))\le ϵ.$$

$$T^a(B(x_i,\frac{\delta }{2})\cap K)\subset U_{a,i}.$$

$$B(x_i,\frac{\delta }{2})\cap K\subset \bigcap _{a\in A}{T}^{-a}{U}_{a,i},\forall 1\le i\le L.$$

$${\mathcal{E}}_n^1=\{U_i{\}}_{i=1}^{m(n,1)},{\mathcal{E}}_n^2=\{V_i{\}}_{i=1}^{m(n,2)}.$$

$${\beta }_1=\{K\cap (U_i-\bigcup _{k=1}^{i-1}{U}_k):i=1,\dots ,m(n,1)\},$$

$${\beta }_2=\{K^c\cap (V_i-\bigcup _{k=1}^{i-1}{V}_k):i=1,\dots ,m(n,2)\}.$$

$$\begin{array}{rl}{H}_{\mu }(\beta )& =-\sum _{A\in {\beta }_1}\mu (A)\text{log}\mu (A)-\sum _{B\in {\beta }_2}\mu (B)\text{log}\mu (B)\\ & \le (\mu (K)\text{log}|{\beta }_1|-\mu (K)\text{log}\mu (K))+(\mu (K^c)\text{log}|{\beta }_2|-\mu (K^c)\text{log}\mu (K^c))\\ & \le \mu (K)\text{log}N(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\mathcal{U}|K)+\mu (K^c)\text{log}N(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\mathcal{U})+\text{log}2\\ & \le \mu (K)\text{log}L+n\mu (K^c)\text{log}N(\mathcal{U})+\text{log}2.\end{array}$$

$$H_{\mu }(T^{-s_i}\alpha |\beta )\le {ϵ}_0,\forall 1\le i\le n.$$

$$\begin{array}{rl}& H_{\mu }(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\alpha )\le H_{\mu }((\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\alpha )\vee \beta )\le H_{\mu }(\beta )+H_{\mu }(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\alpha |\beta )\\ & \le \mu (K)\text{log}L+n\mu (K^c)\text{log}N(\mathcal{U})+\text{log}2+\sum _{i=1}^n{H}_{\mu }(T^{-s_i}\alpha |\beta )\\ & \le \mu (K)\text{log}L+n\mu (K^c)\text{log}N(\mathcal{U})+\text{log}2+n{ϵ}_0.\end{array}$$

$$h_{\mu }^S(T,\alpha )=\underset{n\to \infty }{\mathrm{limsup}}\frac{1}{n}{H}_{\mu }(\underset{i=1}{\overset{n}{\bigvee }}{T}^{-s_i}\alpha )\le \mu (K^c)\text{log}N(\mathcal{U})+{ϵ}_0\le \tau \text{log}N(\mathcal{U})+{ϵ}_0.$$

$$x,y\in K,d(x,y)<\delta \Rightarrow d(T^{a_n}x,T^{a_n}y)<ϵ,\forall n\in \mathbb{N}.$$

$$x,y\in K,d(x,y)<\delta \Rightarrow |f(T^{a_n}x)-f(T^{a_n}y)|<ϵ,\forall n\in \mathbb{N}.$$

$$x,y\in K,d(x,y)<\delta \Rightarrow |U_T^{a_n}f(x)-U_T^{a_n}f(y)|<ϵ,\forall n\in \mathbb{N}.$$

$$||U_T^{a_{n_i}}f-U_T^{a_{n_j}}f|{|}_{L^2}^2>9{ϵ}^2+4ϵ,\forall i\ne j\in \mathbb{N}.$$

$$\mu (E)<\eta \Rightarrow {\int }_E|f{|}^{2}d\mu <ϵ,\forall E\subset X.$$

Full text

1. introduction

Throughout this paper, a topological dynamical system(t.d.s. for short) is a pair $(X,T)$, where $X$ is an non-empty compact metric space with a metric $d$ and $T$ is a homeomorphism from $X$ to itself. For a t.d.s. $(X,T)$, denote by $M(X,T)$ the set of $T$-invariant regular Borel probability measures on $X$. It is well known that there exists some $\mu\in M(X,T)$, thus $(X,\mathcal{B},\mu,T)$ can be viewed as a measure preserving system(m.p.s. for short), where $\mathcal{B}$ is the Borel $\sigma$-algebra.

A t.d.s. $(X,T)$ is called equicontinuous if $\{T^{n}:n\in\mathbb{N}\}$ is uniformly equicontinuous. For a t.d.s., equicontinuity represents predictability, one may see that the dynamical behaviour of such a system is very “rigid”. For example, it is well known that a transitive t.d.s. $(X,T)$ is equicontinuous if and only if it is topologically conjugate to a minimal rotation on a compact abelian metric group, if and only if it has topological discrete spectrum(see [18]).

The analogous concept of equicontinuity for a m.p.s. was also introduced. While studying cellular automata(a subclass of t.d.s.), Gilman [7, 8] introduced a notion of $\mu$-equicontinuity. Later Huang $et\ al$ [11] introduced a different definition of $\mu$-equicontinuity(which under some conditions are equivalent [4]) and showed that $\mu$-equicontinuous systems have discrete spectrum. García-Ramos [5] introduced a weakening of $\mu$-equicontinuity called $\mu$-mean-equicontinuity and showed that if a m.p.s. $(X,\mathcal{B},\mu,T)$ is an ergodic system, then it has discrete spectrum if and only if it is $\mu$-mean-equicontinuous. Recently, Huang $et\ al$ [10] proved that for a general m.p.s. $(X,\mathcal{B},\mu,T)$, it has discrete spectrum if and only if it is $\mu$-mean-equicontinuous.

The notion of rigidity in ergodic theory was first introduced by Furstenberg and Weiss in [3]. A m.p.s. $(X,\mathcal{B},\mu,T)$ is called rigid if there exists a subsequence $\{t_{k}\}$ of $\mathbb{N}$ such that $T^{t_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$ for all $f\in L^{2}(\mu)$. So rigidity is a spectral property. A function $f\in L^{2}(\mu)$ is called rigid if there exists a subsequence $\{t_{k}\}$ of $\mathbb{N}$ such that $T^{t_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$. Note that the sequence $\{t_{k}\}$ may depend on $f$. It is clear that if $(X,\mathcal{B},\mu,T)$ is rigid, then every function in $L^{2}(\mu)$ is rigid. An interesting result showed that the converse is true (see [2, Corollary 2.6]).

The notion of uniformly rigidity was introduced by Glasner and Maon in [9] as a topological analogue of rigidity in ergodic theory. A t.d.s. $(X,T)$ is uniformly rigid if there exists a subsequence $\{t_{k}\}$ of $\mathbb{N}$ such that $T^{t_{k}}\rightarrow id$ uniformly on $X$, where $id$ is the identity mapping.

The concept of rigidity is close to equicontinuity. In [13], the authors proved the following theorem:

Theorem 1.1**.**

[13, Lemma 4.1]** Let $(X,T)$ be a t.d.s.. If $(X,T)$ is uniformly rigid, then there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous. If in addition $(X,T)$ is an E-system, the converse holds.

In fact, in this paper we prove that:

Theorem 1.2**.**

Let $(X,T)$ be a t.d.s.. Then $(X,T)$ is uniformly rigid if and only if there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous, if and only if there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $A$-equicontinuous.

It is natural to consider the corresponding relation between measure-theoretic rigidity and measure-theoretic equicontinuity. Following this idea, in our paper we prove that:

Theorem 1.3**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous if and only if there exists some IP-set $A$ such that $(X,T)$ is $\mu$-$A$-equicontinuous.

Inspired by the work in [10], we also show that:

Theorem 1.4**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $(X,T)$ is $\mu$-$A$-mean-equicontinuous, then $(X,\mathcal{B},\mu,T)$ is rigid.

Since rigidity implies zero entropy, we have following corollary:

Corollary 1.5**.**

Let $(X,T)$ be a t.d.s.. If there exists a subsequence $A$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $(X,T)$ is $A$-mean-equicontinuous, then the topological entropy of $(X,T)$ is zero.

2. preliminaries

In this section we recall some notions and aspects of dynamical systems.

2.1. Subsets of non-negative integers

In the article, integers and natural numbers are denoted by $\mathbb{Z}$ and $\mathbb{N}$ respectively. Let $F$ be a set, denote by $|F|$ the number of elements of $F$. Let $A$ be a subset of $\mathbb{N}$. Define the lower density $\underline{D}(A)$ and the upper density $\bar{D}(A)$ of $A$ by:

[TABLE]

[TABLE]

If $\underline{D}(A)=\bar{D}(A)$, denote by $D(A)=\underline{D}(A)=\bar{D}(A)$ the density of $A$.

Let $\{b_{i}:i\in\mathbb{N}\}\subset\mathbb{N}$. Define

[TABLE]

We say $F\subset\mathbb{N}$ is an $IP$-set if there exists a subset $\{b_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $F=FS\{b_{i}\}$.

2.2. Discrete spectrum in measurable dynamics

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s., one can define the m.p.s. $(X\times X,\mathcal{B}\times\mathcal{B},\mu\times\mu,T)$ by $T(x,y)=(Tx,Ty)$ for $(x,y)\in X\times X$. One can define the Koopman operator $U_{T}$ as an isometry of $L^{p}(\mu)$ into $L^{p}(\mu)$ by $U_{T}f=f\circ T$, where $1\leq p<\infty$. We say a function $f\in L^{2}(\mu)$ is almost periodic if $\{U_{T}^{n}f:n\in\mathbb{Z}\}$ is pre-compact in $L^{2}(\mu)$, that is the closure of $\{U_{T}^{n}f:n\in\mathbb{Z}\}$ is compact in $L^{2}(\mu)$. We say that $(X,\mathcal{B},\mu,T)$ has discrete spectrum if for all $f\in L^{2}(\mu)$, $f$ is almost periodic.

2.3. Measure-theoretic equicontinuity

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$ and $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$.

We say that $K$ is $A$-equicontinuous if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say $(X,T)$ is $A$-equicontinuous if $X$ is $A$-equicontinuous.

We say $(X,T)$ is $\mu$-$A$-equicontinuous if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $A$-equicontinuous.

By the regularity of $\mu$, the compactness of $K$ in the definition above can be ignored. It is clear that the definition of $\mu$-$A$-equicontinuity is independent on the choice of metric $d$.

2.4. Sequence entropy for a measure

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s. and $S=\{s_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Suppose $\xi$ and $\eta$ are two finite measurable partitions of $X$. Define $\xi\vee\eta=\{A\cap B:A\in\xi,B\in\eta\}$. It is easy to see $\xi\vee\eta$ is also a finite partition of $X$. The entropy of $\xi$, written $H_{\mu}(\xi)$, is defined by

[TABLE]

and the entropy of $\xi$ given $\eta$, written $H_{\mu}(\xi|\eta)$, is defined by

[TABLE]

The sequence entropy of $\xi$ with respect to $(X,\mathcal{B},\mu,T)$ along $S$ is defined by

[TABLE]

And the sequence entropy of $(X,\mathcal{B},\mu,T)$ along $S$ is

[TABLE]

where supremum is taken over all finite measurable partitions. When $S=\mathbb{N}$, we simply write $h_{\mu}(T)$ and it is called the entropy of $(X,\mathcal{B},\mu,T)$.

3. measure-theoretic equicontinuity and rigidity

In this section, we study the relation between measure-theoretic equicontinuity and rigidity. We prove that $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous. In the rest we give an IP-version of our result.

3.1. Measure-theoretic equicontinuity, rigidity and sequence entropy

In this subsection, we prove that $\mu$-$A$-equicontinuity implies rigidity via sequence entropy.

First we give a characterization of measure-theoretic rigidity. It is worth pointing out the following theorem:

Theorem 3.1**.**

[12, Theorem 3.10]** Let $(X,\mathcal{B},\mu,T)$ be a m.p.s.. Then $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some IP-set $F$ such that $h_{\mu}^{A}(T)=0$ for any subsequence $A$ of $F$.

Lemma 3.2**.**

[12, Lemma 3.3]** Let $(X,\mathcal{B},\mu,T)$ be a m.p.s. and $B\in\mathcal{B}$. Let $A$ be a subsequence of $\mathbb{N}$. Then ${\{U_{T}^{n}1_{B}:n\in A\}}$ is pre-compact in $L^{2}(\mu)$ if and only if $h_{\mu}^{S}(T,\{B,B^{c}\})=0$ for any subsequence $S$ of $A$.

Now we can give a characterization of rigidity:

Theorem 3.3**.**

Let $(X,\mathcal{B},\mu,T)$ be a m.p.s.. Then $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$ such that $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, if and only if ${\{U_{T}^{n}f:n\in A\}}$ is pre-compact in $L^{2}(\mu)$ for all $f\in L^{2}(\mu)$.

Proof.

We only need to prove that: If there exists some subsequence $A$ of $\mathbb{N}$ such that $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, then $(X,\mathcal{B},\mu,T)$ is rigid. By Lemma 3.2, we have that for any $B\in\mathcal{B}$, $\overline{\{U_{T}^{n}1_{B}:n\in A\}}$ is compact in $L^{2}(\mu)$. Let $\{f_{j}:j\in\mathbb{N}\}$ be a dense subset of $L^{2}(\mu)$, then for any $j\in\mathbb{N}$, $\overline{\{U_{T}^{n}f_{j}:n\in A\}}$ is compact in $L^{2}(\mu)$. Hence

[TABLE]

It follows that

[TABLE]

is totally bounded. For each $k\in\mathbb{N}$, there exist $n_{1},\ldots,n_{N}\in A$ such that

[TABLE]

Without loss of generality, we assume $B((U_{T}^{n_{1}}f_{1},\cdots,U_{T}^{n_{1}}f_{j},\cdots),\frac{1}{k})$ contains infinite many elements of the left set. Hence there are infinite $n\in A$ such that

[TABLE]

Take $t_{k}=n-n_{1}$, since $n$ can be sufficiently large, we can choose $\{t_{k}\}$ increasing. It follows that $\|U_{T}^{t_{k}}f_{j}-f_{j}\|_{L^{2}}<\frac{2^{j}}{k}$ for all $j,k\in\mathbb{N}$. Hence

[TABLE]

It follows that $U_{T}^{t_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$ for each $f\in L^{2}(\mu)$. ∎

Now we study the relation between measure-theoretic equicontinuity and measure-theoretic sequence entropy.

Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Let $\mathcal{U}$ be a finite open cover of $X$ and $K$ be a subset of $X$. Denote by $N(\mathcal{U}|K)$ the minimum among the cardinalities of the subset of $\mathcal{U}$ which covers $K$, and set $C_{A}(\mathcal{U}|K)=\lim\limits_{n\rightarrow\infty}N(\bigvee\limits_{i=1}^{n}T^{-a_{i}}\mathcal{U}|K)$, where $\mathcal{U}\vee\mathcal{V}=\{U\cap V:U\in\mathcal{U},V\in\mathcal{V}\}$ for two finite open covers $\mathcal{U},\mathcal{V}$ of $X$. We recall that for two finite covers $\alpha,\beta$ of $X$, $\alpha\preceq\beta$ means that for each $B\in\beta$, there exists $A\in\alpha$ such that $B\subset A.$

The following lemma is useful(see [14, Lemma 2.3]):

Lemma 3.4**.**

Let $\alpha=\{A_{1},A_{2}\ldots,A_{k}\}$ be a Borel partition of $X$ and $\epsilon>0$, then there exists a finite open cover $\mathcal{U}$ with $k$ elements such that for any $j\in\mathbb{N}$ and any Borel partition $\beta$ satisfying $T^{-j}\mathcal{U}\preceq\beta$, $H_{\mu}(T^{-j}\alpha|\beta)\leq\epsilon$.

Now we prove that $\mu$-$A$-equicontinuity implies rigidity. The idea is from [11, Proposition 5.4].

Theorem 3.5**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous, then $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$, hence $(X,\mathcal{B},\mu,T)$ is rigid.

Proof.

Claim: For any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that for any finite open cover $\mathcal{U}$, $C_{A}(\mathcal{U}|K)<\infty$.

Proof of Claim: Given $\tau>0$, since $(X,T)$ is $\mu$-$A$-equicontinuous, there exists a compact $A$-equicontinuous subset $K$ of $X$ with $\mu(K)>1-\tau$. For any finite open cover $\mathcal{U}$, let $\epsilon>0$ be a Lebesgue number of $\mathcal{U}$. Since $K$ is $A$-equicontinuous, there exists $\delta>0$ such that

[TABLE]

As $K$ is compact, there exist $x_{1},\ldots,x_{L}\in K$ such that

[TABLE]

For any $a\in A$ and $1\leq i\leq L$, we have that

[TABLE]

Hence there exists $U_{a,i}\in\mathcal{U}$ such that

[TABLE]

Thus

[TABLE]

This implies $C_{A}(\mathcal{U}|K)\leq L<\infty$. The proof of Claim is complete.

Now we prove $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$. Assume the contrary, there exists a subsequence $S$ of $A$ such that $h_{\mu}^{S}(T)>0$, then there exists a Borel partition $\alpha$ of $X$ and $\epsilon_{0}>0$ such that $h_{\mu}^{S}(T,\alpha)>\epsilon_{0}$. By Lemma 3.4, there exists a finite open cover $\mathcal{U}$ of $X$ such that for any $j\in\mathbb{N}$ and any Borel partition $\beta$ satisfying $T^{-j}\mathcal{U}\preceq\beta$, $H_{\mu}(T^{-j}\alpha|\beta)\leq\epsilon_{0}$.

For any $\tau>0$, by Claim, there exists a compact set $K$ with $\mu(K)>1-\tau$ and $C_{A}(\mathcal{U}|K)<\infty$, set $C_{A}(\mathcal{U}|K)=L$. For any $n\in\mathbb{N}$, we pick out two sub-collections $\mathcal{E}^{1}_{n},\mathcal{E}^{2}_{n}$ of $\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$ such that $\mathcal{E}^{1}_{n}$ is a cover of $K$ with $N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}|K)$ elements, and $\mathcal{E}^{2}_{n}$ is a cover of $X$ with $N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U})$ elements. Enumerate them by

[TABLE]

Set

[TABLE]

[TABLE]

Then $\beta_{1}$ is a partition of $K$ with $|\beta_{1}|\leq N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}|K)$ and $\beta_{2}$ is a partition of $K^{c}$ with $|\beta_{2}|\leq N(\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U})$. Put $\beta=\beta_{1}\cup\beta_{2}$, then $\beta$ is a partition of $X$ and $\beta\succeq\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$. Hence we have

[TABLE]

On the other hand, since $\beta\succeq\bigvee\limits_{i=1}^{n}T^{-s_{i}}\mathcal{U}$, by the construction of $\mathcal{U}$,

[TABLE]

It follows that

[TABLE]

Hence

[TABLE]

Let $\tau\rightarrow 0,$ we have $h_{\mu}^{S}(T,\alpha)\leq\epsilon_{0}$, it is a contradiction. Hence $h_{\mu}^{S}(T)=0$ for any subsequence $S$ of $A$. By Theorem 3.3, $(X,\mathcal{B},\mu,T)$ is rigid. ∎

3.2. Measure-theoretic rigidity and equicontinuity

In this subsection, we prove that if $(X,\mathcal{B},\mu,T)$ is rigid, then we can find some subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous.

First we give a useful characterization of measure-theoretic equicontinuity:

Proposition 3.6**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Then the following statements are equivalent:

1. (1)

$(X,T)$* is $\mu$-$A$-equicontinuous.* 2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ and $\delta>0$ such that

[TABLE]

Proof.

(1)$\Rightarrow$(2) is obvious.

(2)$\Rightarrow$(1): Given $\tau>0$, for any $l\in\mathbb{N}$, by (2) there exist a compact set $K_{l}$ with $\mu(K_{l})>1-\frac{\tau}{2^{l}}$ and $\delta_{l}>0$ such that for any $x,y\in K_{l}$ with $d(x,y)<\delta_{l}$, we have $d(T^{a_{n}}x,T^{a_{n}}y)<\frac{1}{l}$ for all $n\in\mathbb{N}.$ Let $K=\bigcap\limits_{l=1}^{\infty}K_{l}$, then $\mu(K)>1-\tau$ and $K$ is compact. It is easy to see that $K$ is $A$-equicontinuous, hence $(X,T)$ is $\mu$-$A$-equicontinuous. ∎

Now we prove that rigidity implies $\mu$-$A$-equicontinuity. Note that the following theorem is stronger.

Theorem 3.7**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $U_{T}^{a_{n}}d\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}d$, where $d$ is the metric on $X$, then $(X,T)$ is $\mu$-$A$-equicontinuous.

Proof.

It is similar to Theorem 4.10. ∎

Combining Theorem 3.5 we have:

Theorem 3.8**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous.

3.3. Results with respect to a function

In this subsection we give corresponding results with respect to a function.

Following the idea in [6, 19], we introduce the notion of $f$-$\mu$-$A$-equicontinuity. we prove that for a function $f\in L^{2}(\mu)$, $f$ is rigid if and only if there exists some subsequence $A$ of $\mathbb{N}$ such that $f$ is $\mu$-$A$-equicontinuous.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$, $f\in L^{2}(\mu)$ and $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$.

We say that $K$ is $f$-$A$-equicontinuous(or $f$* is $A$-equicontinuous on $K$*) if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say $f$ is $\mu$-$A$-equicontinuous if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $f$-$A$-equicontinuous.

Similar as Proposition 3.6, we give a characterization of $f$-$\mu$-$A$-equicontinuity:

Proposition 3.9**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Then the following statements are equivalent:

1. (1)

$f$* is $\mu$-$A$-equicontinuous.* 2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ and $\delta>0$ such that

[TABLE]

Now we prove that for a function $f\in L^{2}(\mu)$, $f$-$\mu$-$A$-equicontinuity implies rigidity.

Theorem 3.10**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $f$ is $\mu$-$A$-equicontinuous, then $\{U_{T}^{a_{n}}f:n\in\mathbb{N}\}$ is pre-compact in $L^{2}(\mu)$, hence $f$ is rigid.

Proof.

Assume that $\{U_{T}^{a_{n}}f:n\in\mathbb{N}\}$ is not pre-compact in $L^{2}(\mu)$, then there exists $\epsilon>0$ and a subsequence $\{n_{i}\}$ of $\mathbb{N}$ such that

[TABLE]

Since $f\in L^{2}(\mu)$, there exists $\eta>0$ such that

[TABLE]

Claim: For any subsequence $\{s_{i}\}$ of $\mathbb{N}$, we have

[TABLE]

Proof of Claim: Let $M=\{x:|f(T^{s_{i}}x)|\rightarrow+\infty\}$. It is easy to see that $M$ is measurable. By Fatou’s lemma

[TABLE]

Note that

[TABLE]

we have $+\infty\cdot\mu(M)<+\infty$, hence $\mu(M)=0$. The proof of Claim is complete.

Since $f$ is $\mu$-$A$-equicontinuous, there exists a $f$-$A$-equicontinuous set $K_{0}$ in $X$ with $\mu(K_{0})>1-\eta$, hence there exists $\delta>0$ such that

[TABLE]

By the compactness of $X$, there exist

[TABLE]

where $diamX_{i}<\delta,\ 1\leq i\leq m_{0}$. We assume $\mu(X_{i})>0,\ 1\leq i\leq m;\ \mu(X_{i})=0,\ m+1\leq i\leq m_{0}$. Let $K=X_{1}\cup X_{2}\cup\cdots\cup X_{m}$, then $K\subset K_{0}$ and $\mu(K)=\mu(K_{0})>1-\eta$. Since $\mu(X_{1})>0$ and

[TABLE]

we can find $y_{1}\in X_{1}$ and $|f(T^{a_{n_{i}}}y_{1})|$ not convergent to $+\infty$, hence there exists a subsequence $\{a_{n_{i}}^{1}\}\subset\{a_{n_{i}}\}$ such that $\{f(T^{a_{n_{i}}^{1}}y_{1})\}$ is a Cauchy sequence. Note that $\mu(X_{2})>0$ and

[TABLE]

we can find $y_{2}\in X_{2}$ and a subsequence $\{a_{n_{i}}^{2}\}\subset\{a_{n_{i}}^{1}\}$ such that $\{f(T^{a_{n_{i}}^{2}}y_{2})\}$ is a Cauchy sequence. Working inductively, we can find $y_{i}\in X_{i},1\leq i\leq m$ and a subsequence $\{s_{i}\}\subset\{a_{n_{i}}\}$ such that $\{f(T^{s_{i}}y)\}$ is a Cauchy sequence for all $y\in\{y_{1},y_{2}\ldots,y_{m}\}$. Hence there exists $N\in\mathbb{N}$ such that

[TABLE]

Choose $i_{0}\neq j_{0}>N$, now we estimate $||U_{T}^{s_{i_{0}}}f-U_{T}^{s_{j_{0}}}f||_{L^{2}}^{2}$.

[TABLE]

We first estimate $\int_{K}|f\circ T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu$. If $x\in K$, then there exists $y\in\{y_{1},y_{2}\ldots,y_{m}\}$ such that $d(x,y)<\delta$. Since $x,y\in K_{0}$, we have

[TABLE]

hence

[TABLE]

Since $|f(T^{s_{i_{0}}}y)-f(T^{s_{j_{0}}}y)|<\epsilon,$ it follows that

[TABLE]

Hence $\int_{K}|f\circ T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu<9\epsilon^{2}$.

Now we estimate $\int_{K^{c}}|f\circ T^{s_{i_{0}}}-f\circ T^{s_{j_{0}}}|^{2}d\mu$.

[TABLE]

Since

[TABLE]

the right side $<4\epsilon$. Hence

[TABLE]

It follows that

[TABLE]

Note that $\{s_{i}\}\subset\{a_{n_{i}}\}$ and

[TABLE]

it is a contradiction. So $\{U_{T}^{a_{n}}f:n\in\mathbb{N}\}$ is pre-compact in $L^{2}(\mu)$.

Now we proof $f$ is rigid. Since $\{U_{T}^{a_{n}}f:n\in\mathbb{N}\}$ is pre-compact in $L^{2}(\mu)$, for every $k\in\mathbb{N}$, there exists $N\in\mathbb{N}$ such that

[TABLE]

For $n\in\mathbb{N}$, there exists $1\leq i\leq N$ such that

[TABLE]

Let $n_{k}=a_{n}-a_{i}$, since $a_{n}\rightarrow+\infty$ we can choose $\{n_{k}:k\in\mathbb{N}\}$ increasing, hence we have $||U_{T}^{n_{k}}f-f||_{L^{2}}<\frac{1}{k},\ U_{T}^{n_{k}}f\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f$. ∎

The next theorem shows that for a function $f\in L^{2}(\mu)$, rigidity implies $f$-$\mu$-$A$-equicontinuity.

Theorem 3.11**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $U_{T}^{a_{n}}f\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f$, then $f$ is $\mu$-$A$-equicontinuous.

Proof.

$\forall\tau>0,\forall\epsilon>0$, let

[TABLE]

It is easy to see that $\{A_{N}\}$ is increasing and

[TABLE]

Hence $\mu(\bigcup\limits_{N=1}^{\infty}A_{N})=1$ so there exists $N\in\mathbb{N}$ such that $\mu(A_{N})>1-\dfrac{\tau}{2}$. Since $f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, by Lusin’s theorem, there exists a compact subset $K_{0}$ of $X$ with $\mu(K_{0})>1-\dfrac{\tau}{2}$ such that $f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are continuous hence uniformly continuous on $K_{0}$. Then there exists $\delta>0$ such that

[TABLE]

Choose a compact set $K\subset A_{N}\cap K_{0}$ with $\mu(K)>1-\tau$. For any $x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, then $|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon$; if $n>N$, then

[TABLE]

Hence $|U_{T}^{a_{n}}f(x)-U_{T}^{a_{n}}f(y)|<\epsilon$ for all $n\in\mathbb{N}$. By Proposition 3.9, $f$ is $\mu$-$A$-equicontinuous. ∎

Corollary 3.12**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Then $f$ is rigid if and only if there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $f$ is $\mu$-$A$-equicontinuous.

Using the notion of $f$-$\mu$-$A$-equicontinuity, we give a different proof of Theorem 3.8:

alternative proof of Theorem 3.8.

$(\Leftarrow)$: Assume $(X,T)$ is $\mu$-$A$-equicontinuous. For any $f\in C(X)$, it is clear that $f$ is $\mu$-$A$-equicontinuous. By Theorem 3.10,

[TABLE]

Since $C(X)$ is dense in $L^{2}(\mu)$, we have

[TABLE]

By Theorem 3.3, $(X,\mathcal{B},\mu,T)$ is rigid.

$(\Rightarrow)$: Assume $(X,\mathcal{B},\mu,T)$ is rigid. Then there exists a subsequence $\{n_{i}\}$ of $\mathbb{N}$ such that

[TABLE]

Let $\{f_{n}:n\in\mathbb{N}\}$ be a dense subset of $C(X)$. It is well known that

[TABLE]

is a compatible metric of $X$. Without loss of generality, assume $d$ is the metric on $X$. Since $U_{T}^{n_{i}}f_{1}\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f_{1}$, we can find a subsequence $\{n_{i}^{1}\}$ of $\{n_{i}\}$ such that $U_{T}^{n_{i}^{1}}f_{1}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f_{1}$. Note that $U_{T}^{n_{i}}f_{2}\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f_{2}$, we have $U_{T}^{n_{i}^{1}}f_{2}\stackrel{{\scriptstyle L^{2}}}{{\longrightarrow}}f_{2}$, so we can find a subsequence $\{n_{i}^{2}\}$ of $\{n_{i}^{1}\}$ such that $U_{T}^{n_{i}^{2}}f_{2}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f_{2}$. Working inductively and by diagonal procedure, we can find a subsequence $A=\{a_{i}\}$ of $\{n_{i}\}$ such that $U_{T}^{a_{i}}f_{n}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f_{n}$ for all $n\in\mathbb{N}$. By Theorem 3.11, $f_{n}$ is $\mu$-$A$-equicontinuous for all $n\in\mathbb{N}$.

Now we proof $(X,T)$ is $\mu$-$A$-equicontinuous. For any $\tau>0$ and $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $\sum\limits_{n=N+1}^{\infty}\dfrac{1}{2^{n}}<\dfrac{\epsilon}{2}$. Since $f_{1},\ldots,f_{N}$ are $\mu$-$A$-equicontinuous, there exists a compact set $K$ with $\mu(K)>1-\tau$ such that $K$ is $f_{i}$-$A$-equicontinuous, $1\leq i\leq N$. Hence there exists $\delta>0$ such that for any $x,y\in K$ and $d(x,y)<\delta$, we have

[TABLE]

For any $m\in\mathbb{N}$, we have

[TABLE]

By Proposition 3.6, $(X,T)$ is $\mu$-$A$-equicontinuous. ∎

Now we give the IP-version of our main result:

Theorem 3.13**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Then $(X,\mathcal{B},\mu,T)$ is rigid if and only if there exists an IP-set $F$ such that $(X,T)$ is $\mu$-$F$-equicontinuous.

Proof.

By Theorem 3.8, we only need to prove $(\Rightarrow)$. From the proof above, there exists a subsequence $A=\{a_{i}\}$ of $\mathbb{N}$ such that $U_{T}^{a_{i}}f_{n}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}f_{n}$ for all $n\in\mathbb{N}$. Hence

[TABLE]

Since $T^{a_{n}}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}id$, for each $i\in\mathbb{N}$, we can find $n_{i}\in\mathbb{N}$ such that

[TABLE]

Let $F=FS\{a_{n_{i}}\}$, we prove that $(X,T)$ is $\mu$-$F$-equicontinuous. For any $\tau>0$ and $\epsilon>0$, there exists $N\in\mathbb{N}$ such that $\dfrac{1}{2^{N-1}}<\min\{\tau,\dfrac{\epsilon}{3}\}$. Note that

[TABLE]

we have

[TABLE]

Let $K=\bigcap\limits_{i=N}^{\infty}\bigcap\limits_{1\leq i_{1}<\cdots<i_{k}<i}T^{-(a_{n_{i_{1}}}+\cdots a_{n_{i_{k}}})}A_{i}$, then $\mu(K)>1-\tau$. Since

[TABLE]

are uniformly equicontinuous, there exists $\delta>0$ such that

[TABLE]

For $x,y\in K,d(x,y)<\delta$, we prove that $d(T^{a}x,T^{a}y)<\epsilon$ for all $a\in F$. Let

[TABLE]

where $1\leq i_{1}<\cdots<i_{k}<N\leq i_{k+1}<\cdots<i_{t}$. Since $i_{t}\geq N$, we have

[TABLE]

hence $T^{a_{n_{i_{1}}}+\cdots a_{n_{i_{t-1}}}}x\in A_{i_{t}}$. It follows that

[TABLE]

Similarly

[TABLE]

[TABLE]

[TABLE]

Hence

[TABLE]

Similarly

[TABLE]

Then

[TABLE]

By Proposition 3.6, $(X,T)$ is $\mu$-$F$-equicontinuous. ∎

Similarly we have:

Theorem 3.14**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Then $f$ is rigid if and only if there exists an IP-set $F$ such that $f$ is $\mu$-$F$-equicontinuous.

3.4. The topological case

In [13], the authors proved the following theorem:

Theorem 3.15**.**

[13, Lemma 4.1]** Let $(X,T)$ be a t.d.s.. If $(X,T)$ is uniformly rigid, then there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous. If in addition $(X,T)$ is an E-system, the converse holds.

In fact we have:

Theorem 3.16**.**

Let $(X,T)$ be a t.d.s.. Then $(X,T)$ is uniformly rigid if and only if there exists an IP-set $A$ such that $(X,T)$ is $A$-equicontinuous, if and only if there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $A$-equicontinuous.

Proof.

We only need to prove that: If there exists a subsequence $A$ of $\mathbb{N}$ such that $(X,T)$ is $A$-equicontinuous, then $(X,T)$ is uniformly rigid.

Assume there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $(X,T)$ is $\mu$-$A$-equicontinuous. By Ascoli’s Theorem, $\{T^{a_{n}}:n\in\mathbb{N}\}$ is pre-compact in $C(X,X)$ with uniform topology. For every $k\in\mathbb{N}$, there exists $N\in\mathbb{N}$ such that

[TABLE]

For $n\in\mathbb{N}$, there exists $1\leq i\leq N$ such that

[TABLE]

Let $n_{k}=a_{n}-a_{i}$, since $a_{n}\rightarrow+\infty$ we can choose $\{n_{k}:k\in\mathbb{N}\}$ increasing, hence we have $||T^{n_{k}}-id||_{\infty}<\frac{1}{k},\ T^{n_{k}}\rightarrow id$ uniformly. ∎

4. measure-theoretic mean equicontinuity and rigidity

In [5], the author introduced the concept of measure-theoretic mean equicontinuity and defined a notion called $\mu$-mean-equicontinuity. In [10], the authors introduced a similar notion called $\mu$-equicontinuity in the mean and showed that they are equivalent. In fact, it is shown in [10] that they are equivalent to discrete spectrum for a general $\mu\in M(X,T)$.

Following the idea, in this section we consider the measure-theoretic mean equicontinuity along some subsequence $A$ of $\mathbb{N}$. We study the relation between measure-theoretic mean equicontinuity and rigidity.

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$, let $K$ be a subset of $X$ and $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$.

For $x,y\in X$, define

[TABLE]

We say that $K$ is $A$-equicontinuous in the mean if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say $(X,T)$ is $A$-equicontinuous in the mean if $X$ is $A$-equicontinuous in the mean.

We say $(X,T)$ is $\mu$-$A$-equicontinuous in the mean if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $A$-equicontinuous in the mean.

Similar as Propositin 3.6, we have:

Proposition 4.1**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Then the following statements are equivalent:

1. (1)

$(X,T)$* is $\mu$-$A$-equicontinuous in the mean.* 2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ and $\delta>0$ such that

[TABLE]

We say that $K$ is $A$-mean-equicontinuous if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say $(X,T)$ is $A$-mean-equicontinuous if $X$ is $A$-mean-equicontinuous.

We say $(X,T)$ is $\mu$*-$A$-mean-equicontinuous * if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $A$-mean-equicontinuous.

The notions of $\mu$-$A$-equicontinuous in the mean and $\mu$-$A$-mean-equicontinuous are equivalent:

Theorem 4.2**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$, then the following statements are equivalent:

1. (1)

$(X,T)$* is $\mu$-$A$-equicontinuous in the mean.* 2. (2)

$(X,T)$* is $\mu$-$A$-mean-equicontinuous.*

Proof.

The proof is similar to Theorem 4.8. ∎

Remark 4.3*.*

As the same in [15], it is easy to see that the definition of $\mu$-$A$-mean-equicontinuity is independent on the choice of metric $d$.

Remark 4.4*.*

The property of mean-equicontinuity in a topological dynamical system has also been studied, we refer readers to [1, 15, 16].

Now we prove the main theorem of this section. The idea is from [17, Theorem 2.21].

Theorem 4.5**.**

Let $(X,T)$ be a t.d.s. and $\mu\in M(X,T)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $(X,T)$ is $\mu$-$A$-mean-equicontinuous, then $(X,\mathcal{B},\mu,T)$ is rigid.

Proof.

Without loss of generality, we assume the metric $d\leq 1$ and $\bar{D}(A)>\frac{1}{M},\ M\in\mathbb{N}$. By Theorem 3.7 and Theorem 3.8, we only need to prove ${\{U_{T}^{a_{n}}d:n\in\mathbb{N}\}}$ is pre-compact in $L^{1}(\mu\times\mu)$. Assume the contrary, then there exists $\epsilon>0$ and a subsequence $\{n_{i}\}$ of $\mathbb{N}$ such that

[TABLE]

Since $(X,T)$ is $\mu$-$A$-equicontinuous in the mean, there exist a compact set $K$ with $\mu(K)>1-\epsilon$ and $\delta>0$ such that

[TABLE]

By the compactness of $X$, there exist $K=\bigcup\limits_{i=1}^{N}X_{i}$, where $diamX_{i}<\delta,1\leq i\leq N$, $X_{i}\cap X_{j}=\emptyset,1\leq i\neq j\leq N$. Without loss of generality, we assume $X_{i}\neq\emptyset$ for all $1\leq i\leq N$ and choose $x_{i}\in X_{i},1\leq i\leq N$.

Let $C=(\frac{1}{[\epsilon]}+1)^{N^{2}}$. Choose $n_{0}\in\mathbb{N}$ sufficiently large such that

[TABLE]

Note that $diam_{\bar{d}_{n_{0}}^{A}}X_{i}\leq\epsilon,1\leq i\leq N$. Let $X_{0}=K^{c},$ then $\{X_{0},X_{1},\ldots,X_{N}\}$ is a partition of $X$ with

[TABLE]

For any $1\leq s\leq n_{0}$, define

[TABLE]

For any $(x,y)\in X_{i}\times X_{j}$, we have

[TABLE]

Sum $s$ from $1$ to $n_{0}$, we have

[TABLE]

Hence

[TABLE]

since $diam_{\bar{d}_{n_{0}}^{A}}X_{i}\leq\epsilon,1\leq i\leq N.$ Integrate on $X_{i}\times X_{j}$, we have

[TABLE]

Note that

[TABLE]

we have

[TABLE]

It follows that

[TABLE]

Note that

[TABLE]

we have

[TABLE]

Denote the left set by $S$. Divide $[0,1]$ into $\frac{1}{[\epsilon]}+1$ intervals averagely, then $[0,1]^{N^{2}}$ is divided into $(\frac{1}{[\epsilon]}+1)^{N^{2}}=C$ cubes. Since

[TABLE]

$\{b_{ij}\}_{1\leq i,j\leq N}$ must be in one of the $C$ cubes. By pigeonhole principle, there exists $S_{0}\subset S$ with $|S_{0}|\geq[\frac{[\frac{n_{0}}{2M}]}{C}]\geq\frac{n_{0}}{4MC}$ such that

[TABLE]

It follows that for any $s\neq s^{\prime}\in S_{0}$,

[TABLE]

Enumerate $S_{0}$ as $1\leq s_{1}<s_{2}<\cdots<s_{k}\leq n_{0},\ k\geq\frac{n_{0}}{4MC}.$ Consider the set

[TABLE]

it contains $10kMC>2n_{0}$ elements. On the other hand, it has an upper bound of $2n_{0}$, hence there must be some $i\neq j,\ p\neq q$ with $a_{s_{p}}+a_{n_{i}}=a_{s_{q}}+a_{n_{j}}$. It follows that

[TABLE]

A contradiction. ∎

Since rigidity implies zero entropy, we have following corollary:

Corollary 4.6**.**

Let $(X,T)$ be a t.d.s.. If there exists a subsequence $A$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $(X,T)$ is $A$-mean-equicontinuous, then the topological entropy of $(X,T)$ is zero.

4.1. Results with respect to a function

Let $(X,T)$ be a t.d.s., let $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$, let $K$ be a subset of $X$ and $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$.

For $x,y\in X$, define

[TABLE]

and

[TABLE]

Then $d_{f}$ and $\bar{d}_{n,f}^{A}$ are pseudo-metrics.

We say that $K$ is $f$-$A$-equicontinuous in the mean if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say $f$ is $A$-equicontinuous in the mean if $X$ is $f$-$A$-equicontinuous in the mean.

We say $f$ is $\mu$-$A$-equicontinuous in the mean if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $f$-$A$-equicontinuous in the mean.

Proposition 4.7**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$. Then the following statements are equivalent:

1. (1)

$f$* is $\mu$-$A$-equicontinuous in the mean.* 2. (2)

For any $\tau>0$ and $\epsilon>0$, there exist a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ and $\delta>0$ such that

[TABLE]

We say that $K$ is $f$-$A$-mean-equicontinuous if for any $\epsilon>0$, there exists $\delta>0$ such that

[TABLE]

We say that $f$ is $A$-mean-equicontinuous if $X$ is $f$-$A$-mean-equicontinuous.

We say that $f$ is $\mu$*-$A$-mean-equicontinuous * if for any $\tau>0$, there exists a compact subset $K$ of $X$ with $\mu(K)>1-\tau$ such that $K$ is $f$-$A$-mean-equicontinuous.

The notions of $f$-$\mu$-$A$-equicontinuous in the mean and $f$-$\mu$-$A$-mean-equicontinuous are equivalent. Note that the proof of Theorem 4.8 is similar to [10, Theorem 4.3].

Theorem 4.8**.**

Let $(X,T)$ be a t.d.s., $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. Let $A=\{a_{i}:i\in\mathbb{N}\}$ be a subsequence of $\mathbb{N}$, then the following statements are equivalent:

1. (1)

$f$* is $\mu$-$A$-equicontinuous in the mean.* 2. (2)

$f$* is $\mu$-$A$-mean-equicontinuous.*

Proof.

(1)$\Rightarrow$(2) is obvious.

(2)$\Rightarrow$(1): Fix $\tau>0$ and $\epsilon>0$. There exists a compact set $K_{0}$ with $\mu(K_{0})>1-\dfrac{\tau}{2}$ such that $K_{0}$ is $f$-$A$-mean-equicontinuous. Then there exists $\delta_{0}>0$ such that for all $x,y\in K_{0}$ with $d(x,y)<\delta_{0}$, we have

[TABLE]

By the compactness of $X$, there exist $K_{0}=\bigcup\limits_{i=1}^{m}X_{i}$, where $diamX_{i}<\delta_{0},1\leq i\leq m,\ m\in\mathbb{N}.$ Without loss of generality, assume $X_{i}\neq\emptyset$ and choose $x_{i}\in X_{i},1\leq i\leq m.$ For $1\leq j\leq m$ and $N\in\mathbb{N}$, let

[TABLE]

It is easy to see that for each $1\leq j\leq m$, $\{A_{N}(x_{j})\}_{N=1}^{\infty}$ is increasing and

[TABLE]

Choose $N\in\mathbb{N}$ such that

[TABLE]

Since $f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, there exists a compact set $K_{1}$ with $\mu(K_{1})>1-\dfrac{\tau}{2}$ such that $f,U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are uniformly continuous on $K_{1}$. Then there exists $\delta_{1}>0$ such that for any $x,y\in K_{1}$ and $d(x,y)<\delta_{1}$, we have

[TABLE]

Note that

[TABLE]

by the regularity of $\mu$, we can find pairwise disjoint compact sets

[TABLE]

such that

[TABLE]

Let $K=\bigcup\limits_{j=1}^{m}E_{j}$, then $K$ is compact and $\mu(K)>1-\tau$. Choose $\delta$ with

[TABLE]

For any $x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, since $x,y\in K_{1}$, we have $\dfrac{1}{n}\sum\limits_{i=0}^{n-1}|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon$; if $n>N$, since $d(x,y)<\delta$, there exists $1\leq j_{0}\leq m$ such that $x,y\in E_{j_{0}}$. Then

[TABLE]

Hence $\dfrac{1}{n}\sum\limits_{i=0}^{n-1}|f(T^{a_{i}}x)-f(T^{a_{i}}y)|<\epsilon.$ By Proposition 4.7, $f$ is $\mu$-$A$-equicontinuous in the mean. ∎

The main theorem of this subsection is:

Theorem 4.9**.**

Let $(X,T)$ be a t.d.s. $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ with $\bar{D}(A)>0$ such that $f$ is $\mu$-$A$-mean-equicontinuous, then $f$ is rigid.

In order to prove Theorem 4.9, we need the following theorem. The idea is from [10, Theorem 4.4].

Theorem 4.10**.**

Let $(X,T)$ be a t.d.s. $\mu\in M(X,T)$ and $f\in L^{2}(\mu)$. If there exists a subsequence $A=\{a_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $U_{T}^{a_{n}}d_{f}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}d_{f}$, then $f$ is $\mu$-$A$-equicontinuous.

Proof.

Given $\tau>0$ and $\epsilon>0$. Since $f\in L^{2}(\mu)$, there exists a compact set $K_{0}$ with $\mu(K_{0})>1-\dfrac{\tau}{4}$ such that $f$ is uniformly continuous on $K_{0}$. Then there exists $\delta_{0}>0$ such that for any $x,y\in K_{0}$ and $d(x,y)<\delta_{0}$, we have $d_{f}(x,y)<\dfrac{\epsilon}{8}$. By the compactness of $X$, there exist $K_{0}=\bigcup\limits_{i=1}^{C}X_{i}$, where $diamX_{i}<\delta_{0},1\leq i\leq C,\ C\in\mathbb{N}$. Without loss of generality, assume $X_{i}\neq\emptyset$ and choose $x_{i}\in X_{i},\ 1\leq i\leq C.$ It is easy to see that $K_{0}\subset\bigcup\limits_{i=1}^{C}B_{d_{f}}(x_{i},\dfrac{\epsilon}{8}).$ It follows that

[TABLE]

For any $N\in\mathbb{N}$, let

[TABLE]

It is easy to see that

[TABLE]

Note that $\{A_{N}\}$ is increasing, so there exists $N\in\mathbb{N}$ such that $\mu\times\mu(A_{N})>1-(\dfrac{\tau}{4C})^{2}$. By Fubini’s theorem there exists a subset $D$ of $X$ with $\mu(D)>1-\dfrac{\tau}{4C}$ such that $\forall x\in D,\mu((A_{N})_{x})>1-\dfrac{\tau}{4C},$ where $(A_{N})_{x}=\{y\in X:(x,y)\in A_{N}\}$.

Note that

[TABLE]

without loss of generality, assume $B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\cap D\neq\emptyset$ and choose

[TABLE]

Then

[TABLE]

Since $B_{d_{f}}(x_{i},\dfrac{\epsilon}{8})\subset B_{d_{f}}(y_{i},\dfrac{\epsilon}{4})$, we have

[TABLE]

Since $U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f\in L^{2}(\mu)$, there exists a compact set $K_{1}$ with $\mu(K_{1})>1-\dfrac{\tau}{4}$ such that $U_{T}^{a_{1}}f,\ldots,U_{T}^{a_{N}}f$ are uniformly continuous on $K_{1}$. Then there exists $\delta_{1}>0$ such that for any $x,y\in K_{1}$ and $d(x,y)<\delta_{1}$, we have

[TABLE]

Note that

[TABLE]

by the regularity of $\mu$, we can find pairwise disjoint compact sets

[TABLE]

such that

[TABLE]

and $\mu(\bigcup\limits_{i=1}^{C}E_{i})>1-\tau.$ Let $K=\bigcup\limits_{i=1}^{C}E_{i}$, then $K$ is compact and $\mu(K)>1-\tau$. Choose $\delta$ with

[TABLE]

For any $x,y\in K$ and $d(x,y)<\delta$, if $1\leq n\leq N$, since $x,y\in K_{1}$, $U_{T}^{a_{n}}d_{f}(x,y)<\epsilon$; if $n>N$, since $d(x,y)<\delta$, there exists $1\leq i_{0}\leq C$ such that $x,y\in E_{i_{0}}$. Then

[TABLE]

Note that $x,y\in(A_{N})_{y_{i_{0}}}$, we have $(y_{i_{0}},x),(y_{i_{0}},y)\in A_{N}$, hence

[TABLE]

Since $E_{i_{0}}\subset B_{d_{f}}(y_{i_{0}},\dfrac{\epsilon}{4}),$ we have $d_{f}(y_{i_{0}},x)<\dfrac{\epsilon}{4},$ it follows that $U_{T}^{a_{n}}d_{f}(y_{i_{0}},x)<\dfrac{\epsilon}{2}$. Similarly $U_{T}^{a_{n}}d_{f}(y_{i_{0}},y)<\dfrac{\epsilon}{2}$, so $U_{T}^{a_{n}}d_{f}(x,y)<\epsilon$. By Proposition 3.9, $f$ is $\mu$-$A$-equicontinuous. ∎

Now we prove Theorem 4.9:

proof of Theorem 4.9.

Replacing $U_{T}^{a_{n}}d$ by $U_{T}^{a_{n}}d_{f}$, the proof of Theorem 4.5 still works, so we have ${\{U_{T}^{a_{n}}d_{f}:n\in\mathbb{N}\}}$ is pre-compact in $L^{1}(\mu\times\mu)$. Then there exists a subsequence $S=\{s_{i}:i\in\mathbb{N}\}$ of $\mathbb{N}$ such that $U_{T}^{s_{n}}d_{f}\stackrel{{\scriptstyle a.e.}}{{\longrightarrow}}s_{f}$. By Theorem 4.10, $f$ is $\mu$-$S$-equicontinuous. By Theorem 3.10, $f$ is rigid. ∎

Remark 4.11*.*

When considering the measure-theoretic mean equicontinuity along some subsequence $A$ of $\mathbb{N}$, things may get relatively complicated, since some useful tools and methods in ergodic theory do not work. But there are still some interesting questions to consider.

Question 4.12**.**

Is the condition $\bar{D}(A)>0$ in Theorem 4.5 necessary?

Question 4.13**.**

If we take $A$ as $\mathbb{P}$, the set of all prime numbers, can we obtain some interesting results?