A new metric for statistical properties of long time behaviors

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ScienceDirect J. Differential Equations ••• (••••) •••–••• www.elsevier.com/locate/jde

A new metric for statistical properties of long time behaviors Liqi Zheng a,b , Zuohuan Zheng c,a,b,∗,1 a Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China b University of Chinese Academy of Sciences, Beijing 100049, China c College of Mathematics and Statistics, Hainan Normal University, Haikou, Hainan 571158, China

Received 17 September 2019; revised 31 December 2019; accepted 9 February 2020

Abstract Let (X, T ) be a topological dynamical system with metric d. We define a new function F (x, y) = n  d(T k x, T σ (k) y) by using permutation group Sn . It’s shown F (x, y) = lim sup inf n1

n→+∞ σ ∈Sn

k=1 n  1 lim inf n d(T k x, T σ (k) y) exists when x, y ∈ X are generic points. Applying this function, we n→+∞ σ ∈Sn k=1

prove (X, T ) is uniquely ergodic if and only if F (x, y) = 0 for any x, y ∈ X. The characterizations of ergodic measures and physical measures by F (x, y) are given. We introduce the notion of weak mean equicontinuity and prove that (X, T ) is weak mean equicontinuous if and only if the time averages n  f ∗ (x) = lim n1 f (T k x) are continuous for all f ∈ C(X). n→+∞ k=1 © 2020 Elsevier Inc. All rights reserved.

MSC: primary 54H20; secondary 37A20, 37B05, 37B45

* Corresponding author at: Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China. E-mail addresses: [email protected] (L. Zheng), [email protected] (Z. Zheng). 1 The second author is supported by the NSF of China (No. 11671382), CAS Key Project of Frontier Sciences (No. QYZDJ-SSW-JSC003), the Key Lab. of Random Complex Structures and Data Sciences CAS and National Center for Mathematics and Interdisplinary Sciences CAS.

https://doi.org/10.1016/j.jde.2020.02.010 0022-0396/© 2020 Elsevier Inc. All rights reserved.

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Keywords: Generic point; Unique ergodicity; Time average; Weak mean equicontinuity

1. Introduction Throughout this paper, a topological dynamical system (for short t.d.s.) is a pair (X, T ), where X is a non-empty compact metric space with a metric d and T is a continuous map from X to itself. When studying long time behaviors, people firstly focused on equicontinuous systems, because they have simple dynamical behaviors [1,2]. But only the cumulative effect of points in orbits can influence statistical properties of long time behaviors, so it is reasonable to ignore where positions are for some points in orbits when studying statistical properties of long time behaviors. For this purpose, mean-L-stable systems were introduced [3–5]. We call a dynamical system (X, T ) mean-L-stable if for any ε > 0, there is a δ > 0 such that d(x, y) < δ implies d(T n x, T n y) < ε for all n ∈ N except a set of upper density (see Section 2 for definition) less than ε. Recently, Li, Tu and Ye [11] introduced mean equicontinuous systems. A dynamic system is called mean equicontinuous if for any ε > 0, there exists a δ > 0 such that whenever x, y ∈ X with d(x, y) < δ, 1 d(T k x, T k y) < ε. n n−1

lim sup n→+∞

k=0

In their paper, they proved that a dynamic system is mean equicontinuous if and only if it is mean-L-stable. We refer to [6–10] for further study on mean equicontinuity and related subjects. The highlight in this paper is to ignore the order of points in orbits, for the order makes no sense when studying statistical properties of long time behaviors. In order to state our idea precisely, we introduce some new notations. For any x, y ∈ X, we define 1 F (x, y) = lim sup inf d(T k x, T σ (k) y), σ ∈S n n n→+∞ n

k=1

F (x, y) = lim inf inf

n→+∞ σ ∈Sn

n 1

n

d(T k x, T σ (k) y),

k=1

N (F ) = {(x, y) ∈ X × X|F (x, y) = 0}, N (F ) = {(x, y) ∈ X × X|F (x, y) = 0}, where Sn is the n-order permutation group. If F (x, y) = F (x, y), we say F (x, y) exists, and define 1 F (x, y) = lim inf d(T k x, T σ (k) y), n→+∞ σ ∈Sn n n

k=1

N (F ) = {(x, y) ∈ X × X|F (x, y) = 0}. It is easy to obtain that

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N (F ) = N (F ) ⊂ N (F ). F (x, y) and F (x, y) are functions which can measure the difference between distributions of Orb(x) and Orb(y), where Orb(x) = {x, T x, T 2 x, · · · } and Orb(y) = {y, T y, T 2 y, · · · } are the orbits of x and y respectively. When x and y are generic points (see Section 2 for definition), we can deduce that F (x, y) = F (x, y). Theorem 1.1. Let (X, T ) be a t.d.s. If x, y ∈ X are generic points, then F (x, y) exists. In [3], Fomin proved that a minimal mean-L-stable system is uniquely ergodic. And then in [5], Oxtoby proved a more general result that each transitive mean-L-stable system is uniquely ergodic. In our paper, we shall give new characterizations of unique ergodicity by F (x, y) and F (x, y). Theorem 1.2. Let (X, T ) be a t.d.s. Then the following statements are equivalent: (1) (X, T ) is uniquely ergodic; (2) N (F ) = X × X; (3) N (F ) = X × X. In the study of invariant measures, the set N (F )can play  an important role. We derive that a invariant measure μ is ergodic if and only if (μ ×μ) N (F ) = 1. In the last few decades, physical measures is a hot topic of invariant measures. We find out (X, T ) has physical  measures (see Section 2 for definition) with respect to m ∈ M(X) if and only if (m × m) N (F ) (Q × Q) > 0, where M(X) is the set of all regular Borel probability measures of X and Q is the set of all generic points. With respect to N (F ), we obtain the same results. In order to make clear statement of our results, we introduce the following two notions. Definition 1.1. Let (X, T ) be a t.d.s. We say (X, T ) is F -continuous at x ∈ X if for any ε > 0, there is a δ > 0 such that whenever d(x, y) < δ, we have F (x, y) < ε. Denote by C(F ) all F continuous points. If C(F ) = X, we say (X, T ) is F -continuous. In this case, we also call (X, T ) weak mean equicontinuous. Definition 1.2. Let (X, T ) be a t.d.s. We say (X, T ) is F-continuous at x ∈ X if for any ε > 0, there is a δ > 0 such that whenever d(x, y) < δ, F (x, y) exists and F (x, y) < ε. Denote by C(F ) all F-continuous points. If C(F ) = X, we say (X, T ) is F -continuous. Since X is compact, it is easy to deduce that (X, T ) is F -continuous if and only if for any ε > 0, there is a δ > 0 such that whenever x, y ∈ X with d(x, y) < δ, we have F (x, y) < ε. Similarly, we can derive that (X, T ) is F -continuous if and only if for any ε > 0, there is a δ > 0 such that whenever x, y ∈ X with d(x, y) < δ, F (x, y) exists and F (x, y) < ε. Obviously, mean equicontinuity implies weak mean equicontinuity. But in general, weak mean equicontinuity does not imply mean equicontinuity. The following example is a weak mean equicontinuous but not mean equicontinuous system. T : S 1 −→ S 1

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 T (x) =

1 − 2(x − 12 )2 , x ∈ [0, 12 ) 1 2

− 2(x − 1)2 , x ∈ [ 12 , 1)

.

For any x, y ∈ S 1 , F (x, y) = 0, so (S 1 , T ) is weak mean equicontinuous. But 0 and 12 are not mean equicontinuous points, which shows (S 1 , T ) is not mean equicontinuous. On the one hand, F -continuity implies F -continuity. On the other hand, we can prove in an F -continuous system (X, T ), all the points are generic points. Combining this with Theorem 1.1, we deduce that an F -continuous t.d.s. is F -continuous. Hence, F -continuity is equivalent to F continuity. We conclude the relations as follows: equicontinuity ⇒ mean equicontinuity ⇒ weak mean equicontinuity ⇔ F -continuity. We say a system is chaotic if the positions of points in orbits are sensitive to initial values. While weak mean equicontinuous systems may be chaotic, but in the view of measure theory, they are stable, for the distributions of points in orbits are not sensitive to initial values. Birkhoff Ergodic Theorem shows that for any integrable function f , the time average f ∗ is also integrable. It is natural to ask in which case the time average operator can preserve continuity of observe functions? In [4], Auslander shows in a mean-L-stable system, the time average operator maps continuous functions to continuous ones. In our paper, we will show that weak mean equicontinuous systems are exact the systems in which the time average operator can preserve continuity of observe functions. Theorem 1.3. Let (X, T ) be a t.d.s. Then (X, T ) is weak mean equicontinuous if and only if the time averages f ∗ are continuous for all f ∈ C(X). We organize this paper as follows. In Section 2 we introduce some basic notions and results needed in the paper. In Section 3 we show some propositions of F (x, y) and N (F ) which are useful in the sequel. In Section 4 we prove Theorem 1.1. In Section 5 we study invariant measures by F (x, y) and F (x, y), and prove Theorem 1.2. Meanwhile new characterizations of ergodic measures and physical measures are given. In Section 6 we introduce weak mean equicontinuous systems and provide the proof of Theorem 1.3. Acknowledgments. We would like to express our deep gratitude to Professor Wen Huang for his valuable comments and suggestions. We also thank Weisheng Wu and Qianying Xiao very much for their valuable advice. 2. Preliminaries In this section we recall some notions and results of topological dynamical systems which are needed in our paper. Note that N denotes the set of all non-negative integers and N + denotes the set of all positive integers in this paper. 2.1. Density Let F ⊂ N, we define the upper density D(F ) of F by D(F ) = lim sup n→+∞

#(F ∩ [0, n − 1]) , n

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where #(·) is the number of elements of a set. Similarly, the lower density D(F ) of F is defined by D(F ) = lim inf

n→+∞

#(F ∩ [0, n − 1]) . n

One may say F has density D(F ) if D(F ) = D(F ), in which case D(F ) is equal to this common value. 2.2. Invariant measures Suppose (X, T ) is a t.d.s. The σ -algebra of Borel subsets of X will be denoted by B(X). Let M(X)  be thecollection of all regular Borel probability measures defined on the measurable space X, B(X) . In the weak∗ topology, M(X) is a nonempty compact set (see for example [12], Theorem 6.5).   We say μ ∈ M(X) is T -invariant if μ T −1 (A) = μ(A) holds for any A ∈ B(X). Denote by M(X, T ) the collection of all T -invariant regular Borel probability measures defined on the  measurable space X, B(X) . In the weak∗ topology, M(X, T ) is a nonempty compact convex set (see for example [12], Corollary 6.9.1, Theorem 6.10). We say μ ∈ M(X, T ) is ergodic if for any A ∈ B(X) with T −1 A = A, μ(A) = 0 or μ(A) = 1 holds. Denote by M e (X, T ) the collection of all ergodic measures on (X, T ). It is well known that M e (X, T ) is the collection of all extreme points of M(X, T ) (see for example [12], Theorem 6.10). Thus, M e (X, T ) is nonempty. We say (X, T ) is uniquely ergodic if M e (X, T ) is singleton. Since M e (X, T ) is the set of extreme points of M(X, T ), then (X, T ) is unique ergodic if and only if M(X, T ) is singleton. n  Given x ∈ X, we have { n1 δT k x } ∞ n=1 ⊂ M(X), where δy is the Dirac measure supported on k=1

y. Denote by Mx the collection of all limit points of { n1

n 

δT k x } ∞ n=1 . Since M(X) is compact,

k=1 Mx the

measure set generated by x. we have Mx = ∅. Moreover, Mx ⊂ M(X, T ). We call A point x ∈ X is called generic point if for any f ∈ C(X), the time average 1 f (T k x) n→+∞ n

f ∗ (x) = lim

n

k=1

exists. It is easy to derive that x is a generic point if and only if Mx consists of a single measure. We call μ ∈ Mx is generated by x if x is a generic point. It is well known that when μ is an ergodic measure, there is a generic point x ∈ X such that μ is generated by x (see for example [12], Lemma 6.13). We call a generic point x is an ergodic point if the invariant measure generated by x is ergodic. A Borel subset E ⊂ X is said to have invariant measure one if μ(E) = 1 for all μ ∈ M(X, T ). Let Q denote the set of all generic points, and QT denote the set of all ergodic points. In [5], Oxtoby proved that Q and QT are both Borel sets of invariant measure one. Next, we define physical measures in a general way. Definition 2.1. Let (X, T ) be a t.d.s. and m ∈ M(X). We say μ ∈ M(X, T ) is a physical measure with respect to m if m B(μ) > 0, where

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1 δT k x = μ}. n→+∞ n n

B(μ) = {x ∈ X| lim

k=1

For any A ⊂ X, let  χA (x) =

1, x ∈ A

0, x ∈ /A

.

The following Lemma is well known (see for example [12], Remarks on page:149). Lemma 2.1. Let (X, T ) be a t.d.s. If x ∈ X is a generic point and μ is generated by x, then for any open set U ⊂ X and any closed set V ⊂ X, we have 1 1 χU (T k x) ≥ μ(U ) and lim sup χV (T k x) ≤ μ(V ). n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

Given μ ∈ M(X). Since X is compact, there are finite mutually disjoint subsets of X such that the diameter of each subset is small enough and the sum of their measures are closed enough to one. Hence we have the following result. Lemma 2.2. Let μ ∈ M(X). Then for any ε > 0, there are finite mutually disjoint closed sets 0 {k }kk=1 such that μ(

k0 

k ) ≥ 1 − ε and diam(k ) ≤ ε, ∀k = 1, 2, · · · , k0 .

k=1 s

0 Similarly, there are finite mutually disjoint open sets {Vs }s=1 such that

μ(

s0 

Vs ) ≥ 1 − ε and diam(Vs ) ≤ ε, ∀s = 1, 2, · · · , s0 .

s=1

3. Some properties of F (x, y) and N(F ) In this section, we will show some properties of F (x, y) and N (F ), which play important roles in the next sections. Proposition 3.1. Let (X, T ) be a t.d.s. Then (1) For any sequences {xk }nk=1 and {yk }nk=1 of X, we have inf

σ ∈Sn

n 

d(xk , yσ (k) ) = inf

k=1

In particular, for any x, y ∈ X, we have

σ ∈Sn

n  k=1

d(yk , xσ (k) ).

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inf

σ ∈Sn

7

1 1 d(T k x, T σ (k) y) = inf d(T k y, T σ (k) x). σ ∈Sn n n n

n

k=1

k=1

(2) For any sequences {xk }nk=1 , {yk }nk=1 and {zk }nk=1 of X, we have inf

σ ∈Sn

n 

d(xk , zσ (k) ) ≤ inf

n 

σ ∈Sn

k=1

d(xk , yσ (k) ) + inf

σ ∈Sn

k=1

n 

d(yk , zσ (k) ).

k=1

In particular, for any x, y, z ∈ X, we have inf

σ ∈Sn

1 1 1 d(T k x, T σ (k) z) ≤ inf d(T k x, T σ (k) y) + inf d(T k y, T σ (k) z). σ ∈Sn n σ ∈Sn n n n

n

n

k=1

k=1

k=1

(3) For any x, y ∈ X, we have F (x, y) = F (y, x) and F (x, y) = F (y, x). (4) For any x, y, z ∈ X, we have F (x, z) ≤ F (x, y) + F (y, z) and F (x, z) ≤ F (x, y) + F (y, z). Proof. (1) There exists a σ1 ∈ Sn such that n 

d(xk , yσ1 (k) ) = inf

k=1

σ ∈Sn

n 

d(xk , yσ (k) ).

k=1

Let σ2 ∈ Sn such that σ1 σ2 = σ2 σ1 be the indentity element of Sn . Then we have n  k=1

d(xk , yσ1 (k) ) =

n 

d(xσ2 σ1 (k) , yσ1 (k) ) =

k=1

n 

d(xσ2 (k) , yk ) ≥ inf

σ ∈Sn

k=1

Thus, inf

σ ∈Sn

n  k=1

With the same reason, we can get

d(xk , yσ (k) ) ≥ inf

σ ∈Sn

n  k=1

d(yk , xσ (k) ).

n  k=1

d(yk , xσ (k) ).

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inf

σ ∈Sn

n 

d(yk , xσ (k) ) ≥ inf

σ ∈Sn

k=1

n 

d(xk , yσ (k) ).

k=1

Hence, inf

σ ∈Sn

n 

d(xk , yσ (k) ) = inf

σ ∈Sn

k=1

n 

d(yk , xσ (k) ).

k=1

(2) There are σ1 , σ2 ∈ Sn such that n 

d(xk , yσ1 (k) ) = inf

σ ∈Sn

k=1

n 

d(xk , yσ (k) )

k=1

and n 

d(yk , zσ2 (k) ) = inf

σ ∈Sn

k=1

n 

d(yk , zσ (k) ).

k=1

Let σ3 = σ2 σ1 , then we have n 

d(xk , zσ3 (k) ) ≤

k=1

n 

d(xk , yσ1 (k) ) +

k=1

=

n 

d(yσ1 (k) , zσ3 (k) )

k=1

d(xk , yσ1 (k) ) +

k=1

= inf

n 

σ ∈Sn

n 

d(yk , zσ2 (k) )

k=1 n 

d(xk , yσ (k) ) + inf

σ ∈Sn

k=1

n 

d(yk , zσ (k) ).

k=1

Thus, inf

σ ∈Sn

n  k=1

d(xk , zσ (k) ) ≤ inf

σ ∈Sn

n 

d(xk , yσ (k) ) + inf

k=1

σ ∈Sn

n 

d(yk , zσ (k) ).

k=1

By (1) and (2), we can easily deduce (3) and (4).  F (x, y) and F (x, y) are functions which can measure the difference between distributions of Orb(x) and Orb(y). When x and y are in the same orbit, the distributions of Orb(x) and Orb(y) are same. Thus, F (x, y) = 0. Next, we shall show that F |Orb(x)×Orb(y) and F |Orb(x)×Orb(y) are constants for any x, y ∈ X. Proposition 3.2. Let (X, T ) be a t.d.s. For any x, y ∈ X and r, s ∈ N, we have F (T r x, T s y) = F (x, y)

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and F (T r x, T s y) = F (x, y). If F (x, y) exists, we also have F (T r x, T s y) = F (x, y). Proof. By Proposition 3.1, we have F (T r x, y) ≤ F (x, y) + F (T r x, x) = F (x, y). On the other hand, we have F (x, y) ≤ F (T r x, y) + F (x, T r x) = F (T r x, y). Thus, F (T r x, y) = F (x, y). Similarly, we can deduce that F (T r x, T s y) = F (T r x, y). Hence, we have F (T r x, T s y) = F (x, y). With the same reason, we can deduce the last two equalities.  By Proposition 3.2, we can deduce N (F ) and N (F ) are both invariant sets with respect to T r × T s for any r, s ∈ N. Given x ∈ X, let N (F, x) = {y ∈ X|F (x, y) = 0} and N (F , x) = {y ∈ X|F (x, y) = 0}. Then we have the following proposition. Proposition 3.3. Let (X, T ) be a t.d.s. Then for any x ∈ X, N (F, x) and N (F , x) are both Borel invariant sets.

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Proof. For any m, n ∈ N + and any σ ∈ Sn , let O(x, Fn,σ ,

1 1 ) = {y ∈ X|Fn,σ (x, y) < }, m m

where 1 d(T k x, T σ (k) y). n n

Fn,σ (x, y) =

k=1

Then O(x, Fn,σ , m1 ) is an open set. Since N (F, x) =

∞ ∞  ∞ 

O(x, Fn,σ ,

1 ) m

O(x, Fn,σ ,

1 ), m

m=1 r=1 n=r σ ∈Sn

and N (F , x) =

∞  ∞  ∞ m=1 r=1 n=r σ ∈Sn

we derive that N(F, x) and N (F , x) are both Borel sets. On the other hand, by Proposition 3.2 we derive that N (F, x) and N (F , x) are both invariant sets.  The following proposition provides a way to estimate the upper bound of F (x, y). s

0 Proposition 3.4. Let (X, T ) be a t.d.s. and {Us }s=1 be mutually disjoint subsets of X. Given x, y ∈ X. If the following two conditions hold:

(1) There is ε > 0 such that diam(Us ) ≤ ε holds for any s ∈ {1, 2, · · · , s0 }; (2) For any s ∈ {1, 2, · · · , s0 }, there is as ≥ 0 such that 1 χUs (T k x) ≥ as n→+∞ n n

lim inf

k=1

and 1 χUs (T k y) ≥ as . n→+∞ n n

lim inf

k=1

Then we have F (x, y) ≤ ε

s0  s=1

where M = diam(X).

as + M(1 −

s0  s=1

as ),

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Proof. Given δ > 0, there is an n1 > 0 such that for any n > n1 and any s ∈ {1, 2, · · · , s0 }, we have 1 1 χUs (T k x) ≥ lim inf χUs (T k x) − δ ≥ as − δ. n→+∞ n n n

n

k=1

k=1

(3.1)

Similarly, there is an n2 > 0 such that for any n > n2 and any s ∈ {1, 2, · · · , s0 }, we have 1 χUs (T k y) ≥ as − δ. n n

(3.2)

k=1

Given n > max{n1 , n2 }. For any s ∈ {1, 2, · · · , s0 }, let Nn (Us , x) = {k ∈ N + |T k x ∈ Us , k ≤ n} and Nn (Us , y) = {k ∈ N + |T k y ∈ Us , k ≤ n}. By (3.1) and (3.2), there exists rs,n ∈ N with as n + 1 ≥ rs,n ≥ (as − δ)n

(3.3)

such that     # Nn (Us , x) ≥ rs,n , # Nn (Us , y) ≥ rs,n . r

r

s,n s,n Thus there are subsets {kx,s,r }r=1 and {ky,s,r }r=1 of N + such that

{kx,s,1 , kx,s,2 , · · · , kx,s,rs,n } ⊂ Nn (Us , x) and {ky,s,1 , ky,s,2 , · · · , ky,s,rs,n } ⊂ Nn (Us , y).   Since Nn (Us1 , x) Nn (Us2 , x) = ∅ and Nn (Us1 , y) Nn (Us2 , y) = ∅ hold for any s1 , s2 ∈ {1, 2, · · · , s0 } with s1 = s2 , there is σn ∈ Sn such that σn (kx,s,r ) = ky,s,r

(3.4)

holds for any s ∈ {1, 2, · · · , s0 } and any r ∈ {1, 2, · · · , rs,n }. Let A=

s0 

r

s,n {kx,s,r }r=1 and B = {1, 2, · · · , n}\A.

s=1

Then by (3.3), we have #(A) =

s0  s=1

rs,n ≤

s0  (as n + 1) s=1

(3.5)

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12

and #(B) = n −

s0 

rs,n ≤ n −

s=1

s0  (as − δ)n.

(3.6)

s=1

By (3.4), we deduce that d(T k x, T σn (k) y) ≤ ε

(3.7)

holds for any k ∈ A. On the other hand, for any k ∈ B we have d(T k x, T σn (k) y) ≤ diam(X) = M.

(3.8)

Hence, we derive that 1 1 d(T k x, T σ (k) y) ≤ d(T k x, T σn (k) y) n n n

inf

σ ∈Sn

n

k=1

k=1

=

1 n

d(T k x, T σn (k) y) +

k∈A

1 d(T k x, T σn (k) y) n k∈B

ε M ≤ #(A) + #(B) n n s0  ε (as n + 1) + ≤ n s=1



by (3.7) , (3.8)



0   M (as − δ)n , n− n

s

s=1

where the last inequality comes from (3.5) and (3.6). Let n → +∞, then we have s0 s0 n     1 k σ (k) F (x, y) = lim sup inf d(T x, T y) ≤ ε as + M 1 − (as − δ) . σ ∈S n n n→+∞ k=1

s=1

s=1

Let δ → 0, and then we deduce that F (x, y) ≤ ε

s0 

as + M(1 −

s=1

This finishes the proof of Proposition 3.4.

s0 

as ).

s=1



With respect to F (x, y), we have the similar proposition. 0 Proposition 3.5. Let (X, T ) be a t.d.s. and {Us }ss=1 be mutually disjoint subsets of X. Given x, y ∈ X. If the following two conditions hold:

(1) There is ε > 0 such that diam(Us ) ≤ ε holds for any s ∈ {1, 2, · · · , s0 };

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13

+ (2) There is a subsequence {nr }∞ r=1 of N such that for any s ∈ {1, 2, · · · , s0 }, the following inequalities nr 1  χUs (T k x) ≥ as r→+∞ nr

lim

k=1

and nr 1  χUs (T k y) ≥ as . r→+∞ nr

lim

k=1

hold for some as ≥ 0. Then we have F (x, y) ≤ ε

s0 

as + M(1 −

s=1

s0 

as ),

s=1

where M = diam(X). 4. Proof of Theorem 1.1 In this section, we will prove Theorem 1.1. Assume the contrary that there are generic points x, y ∈ X such that F (x, y) does not exist, this is α = F (x, y) − F (x, y) > 0. Then we estimate the upper bound of F (x, y) and the lower bound of F (x, y), from which we deduce that F (x, y) − F (x, y) ≤ α2 . This contradicts with the assumption. So F (x, y) exists when x, y ∈ X are generic points. In the Proof of Theorem 1.1, we need the following lemma which is a direct corollary of Birkhoff-Von Neumann Theorem [13]. mn Lemma 4.1. Let X be a metric space and m, n ∈ N + . If {xi }ni=1 , {yi }ni=1 , {x j }mn j =1 and {y j }j =1 are subsequences of X such that

    # {j : x j = xi } = # {j : y j = yi } = m, ∀i = 1, 2, · · · n. Then we have inf

σ ∈Smn

mn 

d(x j , y σ (j ) ) = m · inf

σ ∈Sn

j =1

n 

d(xi , yσ (i) ).

i=1

Next, we provide the detailed proof of Theorem 1.1. Proof of Theorem 1.1. Let x, y ∈ X be generic points of (X, T ). Put α = F (x, y) − F (x, y). We assume that F (x, y) does not exist, then α > 0.

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14

Since x, y are generic points, there are μx , μy ∈ M(X, T ) such that 1 1 δT k x and μy = lim δT k y . n→+∞ n n→+∞ n n

n

k=1

k=1

μx = lim

α Let ε = 8+16M , where M = diam(X). By Lemma 2.2, there exist finite mutually disjoint open r0 sets {r }r=1 of X such that r0 

μx (r ) ≥ 1 − ε, diam(r ) ≤ ε, ∀r = 1, 2, · · · , r0 .

(4.1)

r=1 s

0 Similarly, there exist finite mutually disjoint open sets {Vs }s=1 of X such that

s0 

μy (Vs ) ≥ 1 − ε, diam(Vs ) ≤ ε, ∀s = 1, 2, · · · , s0 .

(4.2)

s=1

Without loss of generality, we can assume that μx (r ) > 0 and μy (Vs ) > 0 for any r ∈ r0 s0



{1, 2, · · · , r0 }, s ∈ {1, 2, · · · , s0 }. Let r0 +1 = X\ r and Vs0 +1 = X\ Vs . We select ser +1

r=1

s +1

s=1

0 0 and {ys }s=1 of X such that xr ∈ r and ys ∈ Vs for any r ∈ {1, 2, · · · , r0 + quences {xr }r=1 1}, s ∈ {1, 2, · · · , s0 + 1}. Let {xˆk }nk=1 and {yˆk }nk=1 be sequences such that xˆk = xr if T k x ∈ r and yˆk = ys if T k y ∈ Vs . By Proposition 3.1, we have

1 d(T k x, T σ (k) y) n n

inf

σ ∈Sn

≤ inf

σ ∈Sn

= inf

σ ∈Sn

k=1

1 1 1 d(T k x, xˆσ (k) ) + inf d(xˆσ (k) , yˆk ) + inf d(yˆk , T σ (k) y) σ ∈Sn n σ ∈Sn n n n

n

n

k=1

k=1

k=1

1 1 1 d(xˆk , T σ (k) x) + inf d(xˆk , yˆσ (k) ) + inf d(yˆk , T σ (k) y). σ ∈Sn n σ ∈Sn n n n

n

n

k=1

k=1

k=1

Similarly, we have 1 d(xˆk , yˆσ (k) ) n n

inf

σ ∈Sn

≤ inf

σ ∈Sn

= inf

σ ∈Sn

k=1

n 1

n 1 n

k=1 n  k=1

d(xˆk , T σ (k) x) + inf

σ ∈Sn

d(xˆk , T σ (k) x) + inf

σ ∈Sn

1 1 d(T σ (k) x, T k y) + inf d(T k y, yˆσ (k) ) σ ∈Sn n n 1 n

n

n

k=1

k=1

n  k=1

1 d(yˆk , T σ (k) y). n n

d(T k x, T σ (k) y) + inf

σ ∈Sn

k=1

(4.3)

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15

Thus, we deduce that 1 d(T k x, T σ (k) y) n n

inf

σ ∈Sn

k=1

1 1 1 d(xˆk , yˆσ (k) ) − inf d(xˆk , T σ (k) x) − inf d(yˆk , T σ (k) y). ≥ inf σ ∈Sn n σ ∈Sn n σ ∈Sn n n

n

n

k=1

k=1

k=1

n 1  d(T k x, T σ (k) y). n σ ∈Sn k=1

Our aim is to estimate the bounds of inf

(4.4)

And inequalities (4.3) and n 1  d(xˆk , yˆσ (k) ), n σ ∈Sn k=1

(4.4) show that to achieve this aim it suffices to estimate the bounds of inf n 1  d(xˆk , T σ (k) x) n σ ∈Sn k=1

inf

n 1  d(xˆk , T σ (k) x) σ ∈Sn n k=1 n  of inf n1 d(xˆk , yˆσ (k) ) σ ∈Sn k=1

upper bounds of inf lower bound

n 1  d(yˆk , T σ (k) y). n σ ∈Sn k=1

and inf

In the following, Lemma 4.2 shows the

n 1  d(yˆk , T σ (k) y), σ ∈Sn n k=1

and inf

Lemma 4.3 shows the

and Lemma 4.4 shows the upper bound of

n 1  d(xˆk , yˆσ (k) ). σ ∈Sn n k=1

inf

Given β > 0. By Lemma 2.1, for all sufficiently large n ∈ N + we have 1 1 χr (T k x) ≥ μx (r ) − β and χVs (T k y) ≥ μy (Vs ) − β n n n

n

k=1

k=1

(4.5)

hold for any r ∈ {1, 2, · · · , r0 } and s ∈ {1, 2, · · · , s0 }. Lemma 4.2. For all sufficiently large n ∈ N + , we have inf

σ ∈Sn

1 1 d(xˆk , T σ (k) x) ≤ ε + Mε + Mr0 β and inf d(yˆk , T σ (k) y) ≤ ε + Mε + Ms0 β. σ ∈Sn n n n

n

k=1

k=1

r0 Proof of Lemma 4.2. By (4.1), we have d(xˆk , T k x) ≤ ε if T k x ∈ r=1 r . On the other hand, we can estimate d(xˆk , T k x) ≤ M for diam(X) = M. Then we have n n 1 1 d(xˆk , T σ (k) x) ≤ d(xˆk , T k x) σ ∈Sn n n k=1 k=1

r0

r0 + k / r=1 r , k ≤ n}) (4.6) #({k ∈ N : T x ∈ r=1 r , k ≤ n}) #({k ∈ N + : T k x ∈ ≤ε· +M · n n

0  r , k ≤ n})  #({k ∈ N + : T k x ∈ rr=1 . ≤ε+M 1− n

inf

0 Since {r }r=1 are mutually disjoint, for sufficiently large n ∈ N + we have

r

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16

#({k ∈ N + : T k x ∈ n

r0

r=1 r , k

≤ n})

=

r0  #({k ∈ N + : T k x ∈ r , k ≤ n})

n

r=1

=

r0 n  1

n

r=1

≥

r0 

χr (T k x)

k=1

(4.7)



   μx (r ) − β by (4.5)

r=1

=

r0 

μx (r ) − r0 β.

r=1

Combining (4.6) with (4.7), we derive that  #({k ∈ N + : T k x ∈ 1 inf d(xˆk , T σ (k) x) ≤ ε + M 1 − σ ∈Sn n n n

r0

r=1 r , k

≤ n}) 

k=1

r0    ≤ε+M 1− μx (r ) + r0 β

(4.8)

r=1

≤ ε + Mε + Mr0 β, where the last inequality comes from (4.1). Similarly, we have 1 d(yˆk , T σ (k) y) ≤ ε + Mε + Ms0 β. n n

inf

σ ∈Sn

(4.9)

k=1

This finishes the proof of Lemma 4.2.



n 1  d(xˆk , yˆσ (k) ), n σ ∈Sn k=1 , r0 }, a2 = min{μy (Vs ), s =

To estimate the bounds of inf

we introduce some notations. Let a1 =

ε 1, 2, · · · , s0 } and a = min{ r0 +s , a1 , a2 }. min{μx (r ), r = 1, 2, · · · 0 + Then for any r ∈ {1, 2, · · · , r0 } and any s ∈ {1, 2, · · · , s0 }, there are nr , ms ∈ N such that

anr ≤ μx (r ) < a(nr + 1)

(4.10)

ams ≤ μy (Vs ) < a(ms + 1).

(4.11)

and

Let K = min{

r0  r=1

nr ,

s0  s=1

ms }. Without loss of generality, we can assume K = s0∗

r0  r=1

nr ≤

s0 

ms .

s=1

Then there is s0∗ ≤ s0 and integer sequence {m∗s }s=1 such that 1 ≤ m∗s ≤ ms , ∀s = 1, 2, · · · , s0∗

(4.12)

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17

and ∗

K=

s0 

m∗s .

(4.13)

s=1

By (4.1) we derive that 1 − Ka ≤ ε +

r0 

μx (r ) − a

r=1

=ε+

r0 

nr

r=1

r0  

μx (r ) − anr



(4.14)

r=1

  < ε + r0 a by (4.10) ≤ 2ε, ε where the last inequality comes from a ≤ r0 +s . 0 K K Construct the sequences {x i }i=1 and {y i }i=1 as follows:

xi =

yi =

⎧ ⎪ ⎨x1 , ⎪ ⎩xr+1 ,

⎧ ⎪ ⎨y1 , ⎪ ⎩ys+1 ,

i ≤ n1 r  j =1

nj < i ≤

r+1  j =1

nj , r = 1, 2, · · · , r0 − 1,

i ≤ m∗1 s  j =1

m∗j < i ≤

s+1  j =1

m∗j , s = 1, 2, · · · , s0∗ − 1.

K In the following we will use the distance between sequences {x i }K i=1 and {y i }i=1 to estimate the n  bounds of inf n1 d(xˆk , yˆσ (k) ). σ ∈Sn

k=1

Lemma 4.3. For all sufficiently large n ∈ N + , we have inf

σ ∈Sn

 1 MK d(xˆk , yˆσ (k) ) ≥ (a − β) · inf d(x k , y σ (k) ) − − 2Mε − MKβ. σ ∈SK n n n

K

k=1

k=1

Proof of Lemma 4.3. Let ln be the minimal integer such that ln ≥ (a − β)n. Then by (4.14) we have n − Kln ≤ 1 − K(a − β) ≤ 2ε + Kβ. n Claim 1. For all sufficiently large n ∈ N + , we have #({k|xˆk = xr , k ≤ n}) ≥ ln nr and #({k|yˆk = ys , k ≤ n}) ≥ ln m∗s

(4.15)

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18

hold for any r ∈ {1, 2, · · · , r0 } and any s ∈ {1, 2, · · · , s0∗ }. Proof of Claim 1. Combining (4.5) with (4.10), for sufficiently large n ∈ N + we have 1 χr (T k x) ≥ μx (r ) − β ≥ anr − β n n

k=1

holds for any r ∈ {1, 2, · · · , r0 }. If nr = 1, we have n 

χr (T k x) ≥ annr − βn = (a − β)n.

k=1

Since

n 

χr (T k x) is integer and ln is the minimal integer such that ln ≥ (a − β)n, we derive

k=1

that #({k|xˆk = xr , k ≤ n}) =

n 

χr (T k x) ≥ ln = ln nr .

k=1

If nr > 1, for sufficiently large n ∈ N + we have βn(nr − 1) − nr > 0. Then we deduce that n 

χr (T k x) ≥ annr − βn

k=1

= ln nr − ln nr + annr − βnnr + βn(nr − 1)   = ln nr + βn(nr − 1) − ln − (a − β)n nr ≥ ln nr + βn(nr − 1) − nr > ln n r , which is #({k|xˆk = xr , k ≤ n}) =

n 

χr (T k x) > ln nr .

k=1

Similarly, we can prove that for sufficiently large n ∈ N + , we have #({k|yˆk = ys , k ≤ n}) ≥ ln m∗s holds for any s ∈ {1, 2, · · · , s0∗ }.



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19

By Claim 1, for sufficiently large n ∈ N + there is a set A ⊂ {1, 2, · · · , n} such that #(A) = Kln

(4.16)

#({k|xˆk = xr , k ∈ A}) = ln nr

(4.17)

and

holds for any r ∈ {1, 2, · · · , r0 }. Let σ1 ∈ Sn such that 1 1 d(xˆk , yˆσ1 (k) ) = inf d(xˆk , yˆσ (k) ). σ ∈Sn n n n

n

k=1

k=1

For any s ∈ {1, 2, · · · , s0∗ }, we denote bs = #({k ≤ n|yˆσ1 (k) = ys , k ∈ / A}). Then by (4.16), we have ∗

s0 

bs ≤ n − #(A) = n − Kln .

(4.18)

s=1

For any s ∈ {1, 2, · · · , s0∗ } and sufficiently large n ∈ N + , we have #({k ∈ N + |yˆσ1 (k) = ys , k ∈ A}) + bs #({k ∈ N + |yˆσ1 (k) = ys , k ≤ n}) = n n n + #({k ∈ N |yˆk = ys , k ≤ n}) 1  χVs (T k y) = = n n k=1   ≥ μy (Vs ) − β by (4.5)   ≥ ms a − β by (4.11) ≥ m∗s (a − β), where the last inequality follows from (4.12). Therefore, #({k ∈ N + |yˆσ1 (k) = ys , k ∈ A}) bs ≥ m∗s (a − β) − . n n

(4.19)

By Claim 1 and (4.19), for sufficiently large n ∈ N + there is a σ2 ∈ Sn such that #({k|yˆσ2 (k) = ys , k ∈ A}) = ln m∗s and

(4.20)

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20

#({k ∈ N + |σ2 (k) = σ1 (k), yˆσ1 (k) = ys , k ∈ A}) bs ≥ m∗s (a − β) − n n hold for any s ∈ {1, 2, · · · , s0∗ }. Since #(A) = Kln , we deduce that #({k ∈ N + |σ2 (k) = σ1 (k), k ∈ A}) = #(A) − #({k ∈ N + |σ2 (k) = σ1 (k), k ∈ A}) ∗

≤ Kln −

s0 

#({k ∈ N + |σ2 (k) = σ1 (k), yˆσ1 (k) = ys , k ∈ A})

s=1 ∗

≤ Kln − n

s0  

m∗s (a − β) −

s=1 s0∗

= Kln −



bs  n ∗

m∗s (a

− β)n +

s=1



s0 

bs

s=1



∗

= K ln − (a − β)n +

s0 

bs (by (4.13))

s=1

≤ K + n − Kln , where the last inequality comes from (4.18) and the fact that ln is the minimal integer such that ln ≥ (a − β)n. Then we derive that 



d(xˆk , yˆσ2 (k) ) =

k∈A

d(xˆk , yˆσ2 (k) ) +

k∈A,σ1 (k)=σ2 (k)

≤





d(xˆk , yˆσ2 (k) )

k∈A,σ1 (k) =σ2 (k)

d(xˆk , yˆσ1 (k) ) + M · #({k ∈ N + |σ2 (k) = σ1 (k), k ∈ A})

k∈A

≤



d(xˆk , yˆσ1 (k) ) + M(K + n − Kln ).

k∈A

This shows 1 1 MK M(n − Kln ) d(xˆk , yˆσ2 (k) ) ≤ d(xˆk , yˆσ1 (k) ) + + n n n n k∈A

k∈A

≤

1 n

k∈A

d(xˆk , yˆσ1 (k) ) +

MK + 2Mε + MKβ, n

where the last inequality comes from (4.15). Let B = {σ2 (k)|k ∈ A} and H (A, B) = {σ ∈ Sn |σ (k) ∈ B, ∀k ∈ A}. Then σ2 ∈ H (A, B) and we have

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L. Zheng, Z. Zheng / J. Differential Equations ••• (••••) •••–•••

1 1 1 d(xˆk , yˆσ (k) ) = d(xˆk , yˆσ1 (k) ) ≥ d(xˆk , yˆσ1 (k) ) n n n n

inf

σ ∈Sn

21

n

k=1

k∈A

k=1

1 MK ≥ d(xˆk , yˆσ2 (k) ) − − 2Mε − MKβ n n

(4.21)

k∈A

≥

1 MK d(xˆk , yˆσ (k) ) − − 2Mε − MKβ. σ ∈H (A,B) n n inf

k∈A

By (4.17), (4.20) and Lemma 4.1, we can derive that 

inf

σ ∈H (A,B)

d(xˆk , yˆσ (k) ) = ln · inf

σ ∈SK

k∈A

K 

n 1  d(xˆk , yˆσ (k) ) n σ ∈Sn k=1

Combining (4.21) with (4.22), we estimate the lower bound of inf

inf

σ ∈Sn

(4.22)

d(x k , y σ (k) ).

k=1

as follows:

 1 ln MK d(xˆk , yˆσ (k) ) ≥ · inf d(x k , y σ (k) ) − − 2Mε − MKβ n n σ ∈SK n n

K

k=1

k=1

≥ (a − β) · inf

σ ∈SK

This finishes the proof of Lemma 4.3.

K 

d(x k , y σ (k) ) −

k=1

MK − 2Mε − MKβ. n



Lemma 4.4. For all sufficiently large n ∈ N + , we have  1 MK inf d(xˆk , yˆσ (k) ) ≤ (a − β) · inf d(x k , y σ (k) ) + + 2Mε + MKβ. σ ∈Sn n σ ∈SK n n

K

k=1

k=1

Proof of Lemma 4.4. Let σK ∈ SK such that K  k=1

d(x k , y σK (k) ) = inf

σ ∈SK

K 

d(x k , y σ (k) ).

k=1

For sufficiently large n ∈ N + , Claim 1 shows that there is a partition {Ai }K+1 i=1 of {1, 2, · · · , n} such that #(Ai ) = ln and {xˆk |k ∈ Ai } = {x i }

(4.23)

hold for any i ∈ {1, 2, · · · , K}. Similarly, there is a partition {Bi }K+1 i=1 of {1, 2, · · · , n} such that #(Bi ) = ln and {yˆk |k ∈ Bi } = {y σK (i) } hold for any i ∈ {1, 2, · · · , K}.

(4.24)

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22

By (4.23) and (4.24), there exists σn ∈ Sn such that Bi = {σn (k)|k ∈ Ai }, ∀i = 1, 2, · · · , K. Thus for any i ∈ {1, 2, · · · , K}, we have 

d(xˆk , yˆσn (k) ) = ln · d(x i , y σK (i) ).

k∈Ai

Hence, we deduce that n 

d(xˆk , yˆσn (k) ) =

K+1 



d(xˆk , yˆσn (k) )

i=1 k∈Ai

k=1

=

K 

ln · d(x i , y σK (i) ) +

d(xˆk , yˆσn (k) )

k∈AK+1

i=1

≤ ln ·



K 

d(x i , y σK (i) ) + M(n − Kln )

i=1

≤ (ln − 1) ·

K 

d(x k , y σK (k) ) + MK + M(n − Kln ).

k=1

Since ln − 1 < (a − β)n, we have  1 MK M(n − Kln ) d(xˆk , yˆσn (k) ) ≤ (a − β) · d(x k , y σK (k) ) + + n n n n

K

k=1

k=1

≤ (a − β) ·

K 

d(x k , y σK (k) ) +

k=1

MK + 2Mε + MKβ, n

where the last inequality comes from (4.15). Hence, we can estimate the upper bound of n  inf n1 d(xˆk , yˆσ (k) ) as follows: σ ∈Sn

k=1

inf

σ ∈Sn

1 1 d(xˆk , yˆσ (k) ) ≤ d(xˆk , yˆσn (k) ) n n n

n

k=1

k=1

≤ (a − β) · inf

σ ∈SK

This finishes the proof of Lemma 4.4.



K  k=1

d(x k , y σ (k) ) +

MK + 2Mε + MKβ. n

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Given sufficiently large n ∈ N + . By (4.3), Lemma 4.2 and Lemma 4.4, we deduce that 1 inf d(T k x, T σ (k) y) σ ∈Sn n n

k=1

≤ (a − β) · inf

σ ∈SK

K 

d(x k , y σ (k) ) +

k=1

MK + 2Mε + MKβ + 2ε + 2Mε + M(r0 + s0 )β. n (4.25)

On the other hand, by (4.4), Lemma 4.2 and Lemma 4.3, we derive that 1 d(T k x, T σ (k) y) n n

inf

σ ∈Sn

k=1

≥ (a − β) · inf

σ ∈SK

K 

d(x k , y σ (k) ) −

k=1

MK − 2Mε − MKβ − 2ε − 2Mε − M(r0 + s0 )β. n (4.26)

Let n → +∞, and from (4.25) and (4.26) we deduce that F (x, y) ≤ (a − β) · inf

σ ∈SK

K 

d(x k , y σ (k) ) + 2Mε + MKβ + 2ε + 2Mε + M(r0 + s0 )β

k=1

and F (x, y) ≥ (a − β) · inf

σ ∈SK

K 

d(x k , y σ (k) ) − 2Mε − MKβ − 2ε − 2Mε − M(r0 + s0 )β.

k=1

This shows F (x, y) − F (x, y) ≤ 8Mε + 2MKβ + 4ε + 2M(r0 + s0 )β. Let β → 0, then we derive that F (x, y) − F (x, y) ≤ 8Mε + 4ε =

α , 2

which conflicts with the assumption. This shows that F (x, y) exists.



5. Invariant measures In this section, we study invariant measures by functions F (x, y) and F (x, y). And then we prove Theorem 1.2. Theorem 1.2 is a direct corollary of Theorem 5.1 and Theorem 5.3. The following proposition shows that F (x, y) = 0 implies the measure sets generated by x and y are the same.

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Proposition 5.1. Let (X, T ) be a t.d.s. Then Mx = My holds for any x, y ∈ X with F (x, y) = 0. Proof. Given x, y ∈ X with F (x, y) = 0. To prove Mx = My , it suffices to show Mx ⊂ My and My ⊂ Mx . In the following, we will prove that Mx ⊂ My . With the same reason, My ⊂ Mx holds. + Given μ ∈ Mx , there is a subsequence {nr }∞ r=1 of positive integers N such that for any f ∈ C(X), we have  nr 1  f (T k x) = f dμ. r→+∞ nr lim

k=1

(5.1)

X

Given f ∈ C(X) and ε > 0. There is δ = δ(ε, f ) > 0 such that whenever x1 , x2 ∈ X with d(x1 , x2 ) < δ, we have |f (x1 ) − f (x2 )| < ε.

(5.2)

Let L = max{|f (z)|}. Given σ ∈ Snr , we have z∈X

|

nr nr 1  1  f (T k x) − f (T k y)| nr nr k=1

k=1

nr  1  f (T k x) − f (T σ (k) y) | ≤ |f (T k x) − f (T σ (k) y)| nr k=1 k=1   + k σ (k)  # {k ∈ N |d(T x, T y) < δ, k ≤ nr }  ≤ε× by (5.2) nr   + # {k ∈ N |d(T k x, T σ (k) y) ≥ δ, k ≤ nr } + 2L × nr   # {k ∈ N + |d(T k x, T σ (k) y) ≥ δ, k ≤ nr } ≤ ε + 2L × . nr

=

1 | nr

nr 



Since   nr # {k ∈ N + |d(T k x, T σ (k) y) ≥ δ, k ≤ nr } 1  k σ (k) d(T x, T y) ≥ δ × , nr nr k=1

we deduce that |

nr nr nr 1  1  2L 1  f (T k x) − f (T k y)| ≤ ε + d(T k x, T σ (k) y). × nr nr δ nr k=1

k=1

k=1

Thus we have |

nr nr nr 1  1  1  2L f (T k x) − f (T k y)| ≤ ε + d(T k x, T σ (k) y). × inf σ ∈Snr nr nr nr δ k=1

k=1

k=1

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Let r → +∞, then we have lim sup | r→+∞

nr nr nr 1  1  1  2L f (T k x) − f (T k y)| ≤ ε + d(T k x, T σ (k) y) × lim sup inf nr nr δ r→+∞ σ ∈Snr nr k=1

k=1

k=1

=ε+

2L × F (x, y) = ε. δ

Let ε → 0, then we deduce that lim sup | r→+∞

nr nr 1  1  f (T k x) − f (T k y)| = 0. nr nr k=1

k=1

Combining this with (5.1), we have  nr nr 1  1  f (T k y) = lim f (T k x) = f dμ, r→+∞ nr r→+∞ nr lim

k=1

k=1

X

which implies μ ∈ My . Therefore, Mx ⊂ My . This finishes the proof of Proposition 5.1.



With respect to F (x, y), we have the following result similar to Proposition 5.1. Proposition 5.2. Let (X, T ) be a t.d.s. Then Mx ∩ My = ∅ for any x, y ∈ X with F (x, y) = 0. Proof. Given x, y ∈ X with F (x, y) = 0. There is a subsequence {nr }∞ r=1 of positive integers N + such that nr 1  lim inf d(T k x, T σ (k) y) = 0. r→+∞ σ ∈Snr nr k=1

Without loss of generality, we can assume that there exists μ ∈ Mx such that  nr 1  f (T k x) = f dμ r→+∞ nr lim

k=1

X

holds for any f ∈ C(X). With  the same reason in the proof of Proposition 5.1, we derive that μ ∈ My . Thus, Mx My = ∅.  When x ∈ X is a generic point of (X, T ), we can strengthen the Proposition 5.1 as follows. Proposition 5.3. Let (X, T ) be a t.d.s. and x, y ∈ X. If x is a generic point of (X, T ), then F (x, y) = 0 if and only if Mx = My .

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Proof. Proposition 5.1 shows that F (x, y) = 0 implies Mx = My . To finish the proof of Proposition 5.3, we need only to prove that Mx = My implies F (x, y) = 0. n  ε Let μ = lim n1 δT k x , then Mx = My = {μ}. Given ε > 0, and let η = 1+M , where M = n→+∞

k=1

0 such that diam(X). By Lemma 2.2, there are mutually disjoint open sets {Us }ss=1

μ(

s0 

Us ) ≥ 1 − η and diam(Us ) ≤ η, ∀s = 1, 2, · · · , s0 .

s=1

Combining Lemma 2.1 with Proposition 3.4, we have F (x, y) ≤ η

s0 

s0    μ(Us ) + M 1 − μ(Us ) ≤ η + Mη = ε.

s=1

s=1

Let ε → 0, then we deduce that F (x, y) = 0. This shows F (x, y) = 0.



Applying Proposition 5.3, we have the following theorem. Theorem 5.1. Let (X, T ) be a t.d.s. Then (X, T ) is uniquely ergodic if and only if N (F ) = X ×X. Proof. If (X, T ) is uniquely ergodic and μ is the unique ergodic measure. Then for any x, y ∈ X, we have Mx = My = {μ}, which implies that x and y are generic points. By Proposition 5.3, we derive that F (x, y) = 0. Thus (x, y) ∈ N (F ). Hence N (F ) = X × X. If N (F ) = X × X. Let μ1 and μ2 be ergodic measures on (X, T ). By Birkhoff pointwise ergodic theorem, there exist x, y ∈ X such that Mx = {μ1 } and My = {μ2 }. Since N (F ) = X × X, we have F (x, y) = 0. By Proposition 5.3, we deduce that Mx = My , which implies μ1 = μ2 . Thus, (X, T ) is uniquely ergodic.  When (X, T ) is a transitive weak mean equicontinuous system, we can deduce that N (F ) = X × X. Thus by Theorem 5.1, we have that a transitive weak mean equicontinuous system is uniquely ergodic. Corollary 5.2. Let (X, T ) be a transitive weak mean equicontinuous t.d.s. Then (X, T ) is uniquely ergodic. In particular, a transitive mean equicontinuous system is uniquely ergodic. Proof. Let x ∈ X be a transitive point of (X, T ). Then for any y ∈ X, there is a subsequence + mr x = y. Since (X, T ) is weak mean equicon{mr }+∞ r=1 of positive integers N such that lim T r→+∞

tinuous, we deduce that lim F (T mr x, y) = 0. By Proposition 3.2, we have that for any r ≥ 1, r→+∞

F (x, y) = F (T mr x, y). Thus, F (x, y) = 0. Given y1 , y2 ∈ X. By Proposition 3.1, we have F (y1 , y2 ) ≤ F (y1 , x) + F (x, y2 ) = 0, which shows (y1 , y2 ) ∈ N (F ). Thus N (F ) = X × X. By Theorem 5.1, we derive that (X, T ) is uniquely ergodic. 

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Combining Theorem 5.1 with Proposition 5.2, we can show a new characterization of unique ergodicity by N(F ). Theorem 5.3. Let (X, T ) be a t.d.s. Then (X, T ) is uniquely ergodic if and only if N (F ) = X ×X. Proof. Assume that (X, T ) is uniquely ergodic. By Theorem 5.1, we have N (F ) = X × X. Since N(F ) ⊂ N(F ), we derive that N (F ) = X × X. Conversely, we assume that N (F ) = X × X. Let μ1 and μ2 be ergodic measures of (X, T ). By Birkhoff pointwise ergodic theorem, there are x, y ∈ X such that Mx = {μ1 } and My = {μ2 }. Since N(F ) = X × X, we have F (x, y) = 0. By Proposition 5.2, we deduce that Mx ∩ My = ∅, which implies μ1 = μ2 . Thus, (X, T ) is uniquely ergodic.  Proposition 5.1 shows that N (F ) is a subset of all point pairs in X which can generate the same measure set. Combining Proposition 5.3 with the fact that Q is a set of invariant measure one, it is reasonable to regard N (F ) as the whole set of all point pairs in X which can generate the same measure set in the view of measure theory. Thus, there are close connections between N(F ) and invariant measures. Similarly, there are close connections between N (F ) and invariant measures. We state some of them as follows. Theorem 5.4. Let (X, T ) be a t.d.s. and μ ∈ M(X, T ). Then the following statements are equivalent: (1) μ is ergodic;   (2) (μ × μ)N(F ) = 1; (3) (μ × μ) N(F ) = 1. Proof. (3) ⇒  (2) Since Q = 1. Thus,  has invariant measure one, we have(μ × μ)(Q × Q)  (μ ×μ) N(F )  (Q ×Q) = 1. By Theorem 1.1, we have N (F ) (Q ×Q) = N (F ) (Q ×Q).     Hence, (μ × μ) N (F ) (Q × Q) = 1. This shows (μ × μ) N (F ) = 1. (2) ⇒ (1) Since   (μ × μ) N (F ) =



  μ N (F, x) dμ(x),

X

  there is x0 ∈ X such that μ N (F, x0 ) = 1. For QT is a Borel set of invariant measure one, there n   is y ∈ QT N (F, x0 ). Let μy = lim δT k y . Then by Proposition 5.3, we can derive that n→+∞ k=1

Mz = My = {μy } for any z ∈ N (F, x0 ). Thus given f ∈ C(X), we have 1 f (T k z) = n→+∞ n n



lim

k=1

holds for any z ∈ N (F, x0 ). Hence, we deduce that

f dμy X

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 f dμ = lim

n→+∞ N (F,x0 )

X



=

1 f (T k z)dμ(z) n n

k=1

1 f (T k z)dμ(z) n→+∞ n n

lim

k=1

N (F,x0 )



=

f dμy . X

This shows μ = μy . So μ is an ergodic measure. (1) ⇒ (3) By Birkhoff pointwise ergodic theorem and Proposition 5.3, there  exists  a measurable subset  of X such that μ() = 1 and  ×  ⊂ N (F ). Thus (μ × μ) N (F ) = 1. Since N(F ) ⊂ N(F ), we derive that (μ × μ) N (F ) = 1.  Theorem 5.5. Let (X, T ) be a t.d.s. and m ∈ M(X). Then the following statements are equivalent: (1) (X, T ) has  physical  with respect to m;  measures (2) (m × m)N(F ) (Q × Q) > 0; (3) (m × m) N(F ) (Q × Q) > 0. Proof. (2) ⇔ (3) By Theorem 1.1, we have N (F )



(Q × Q) = N (F )



(Q × Q).

  (1) ⇒ (2) Let μ be a physical measure of (X, T ) with respect to m. Then m B(μ) > 0. For any x, y ∈ B(μ), we have Mx = My= {μ}. Thus x, y ∈ Q. By Proposition 5.3, we have F (x, y) = 0, which shows (x, y) ∈ N (F ) (Q × Q). Thus, B(μ) × B(μ) ⊂ N (F )



(Q × Q).

Hence we have     (m × m) N (F ) (Q × Q) ≥ (m × m) B(μ) × B(μ) > 0.   (2) ⇒ (1)  There is x0 ∈ Q such that m N (F, x0 ) > 0. If not, for any x ∈ Q, we have m N(F, x) = 0. Then we derive that   (m × m) N (F ) (Q × Q) =



  m N (F, x) dm(x) = 0,

Q

which is a contradiction. Let μx0 be the invariant measure generated by x0 . Then by Proposition 5.3, we have B(μx0 ) = N(F, x0 ). Thus

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29

    m B(μx0 ) = m N (F, x0 ) > 0, which shows that μx0 is a physical measure with respect to m.  6. Weak mean equicontinuity In this section, we study F -continuity and F -continuity. Combining the following Proposition 6.1 with Theorem 1.1, we deduce that F -continuity is equivalent to F -continuity. Then we provide the proof of Theorem 1.3. Proposition 6.1. Let (X, T ) be an F -continuous t.d.s. Then all the points in X are generic points. Proof. Given x ∈ X. For any y ∈ N (F, x), there are {yn }∞ lim yn = y. n=1 ⊂ N (F, x) such that n→∞ By Proposition 3.1, we have F (x, y) ≤ F (x, yn ) + F (yn , y) = F (yn , y). Since (X, T ) is F -continuous, we have lim F (yn , y) = 0. Thus we deduce that n→+∞

F (x, y) = F (x, y) = 0, which implies y ∈ N (F, x). Hence N (F, x) is closed. By  Proposition  3.3, we know N (F, x) is an invariant set. Then by Theorem 1.2, we derive that N (F, x), T is uniquely ergodic, which implies that all the points in N (F, x) are generic. In particular, x is a generic point.  Since an F -continuous t.d.s. is F -continuous, the following is a direct corollary of Proposition 6.1 Proposition 6.2. Let (X, T ) be an F -continuous t.d.s. Then all the points in X are generic points. By Theorem 1.1 and Proposition 6.1, we can deduce the following theorem. Theorem 6.1. Let (X, T ) be a t.d.s. Then (X, T ) is F -continuous if and only if (X, T ) is F continuous. Proof. We need only to prove that F -continuity implies F -continuity. Suppose that (X, T ) is F -continuous. By Proposition 6.1, we derive that all the points in X are generic points. Then by Theorem 1.1, we have F (x, y) exists for any x, y ∈ X. Thus (X, T ) is F -continuous.  Next, we give the proof of Theorem 1.3. Proof of Theorem 1.3. Assume that (X, T ) is weak mean equicontinuous, then we will prove f ∗ is continuous for any f ∈ C(X). Given f ∈ C(X), by Proposition 6.1 we know f ∗ (x) exists for any x ∈ X. Fix υ > 0, then there is ε > 0 such that whenever x, y ∈ X with d(x, y) < ε, we have

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30

|f (x) − f (y)| <

υ , 2(2L + 1)

(6.1)

where L = max{|f (x)|}. x∈X

Given x, y ∈ X. For any σ ∈ Sn , we have |

1 1 f (T k x) − f (T k y)| n n n

n

k=1

k=1

n  1 f (T k x) − f (T σ (k) y) | ≤ |f (T k x − f (T σ (k) y)| n k=1 k=1   + k σ  # {k ∈ N |d(T x, T (k) y) < ε, k ≤ n}  υ ≤ × by (6.1) 2(2L + 1) n   + k σ (k) # {k ∈ N |d(T x, T y) ≥ ε, k ≤ n} + 2L × n   # {k ∈ N + |d(T k x, T σ (k) y) ≥ ε, k ≤ n} υ ≤ + 2L × . 2(2L + 1) n

1 = | n

n 



Since   n # {k ∈ N + |d(T k x, T σ (k) y) ≥ ε, k ≤ n} 1 k σ (k) d(T x, T y) ≥ ε × , n n k=1

we deduce that |

1 1 υ 2L 1  f (T k x) − f (T k y)| ≤ d(T k x, T σ (k) y). + × n n 2(2L + 1) ε n n

n

n

k=1

k=1

k=1

Thus we have |

1 1 1 υ 2L f (T k x) − f (T k y)| ≤ d(T k x, T σ (k) y). + × inf σ ∈Sn n n n 2(2L + 1) ε n

n

n

k=1

k=1

k=1

Let n → +∞, then we have lim sup | n→+∞

1 1 υ 2L f (T k x) − f (T k y)| ≤ + × F (x, y). n n 2(2L + 1) ε n

n

k=1

k=1

This implies |f ∗ (x) − f ∗ (y)| ≤

υ 2L + × F (x, y). 2(2L + 1) ε

(6.2)

Since (X, T ) is weak mean equicontinuous, there is δ > 0 such that whenever x, y ∈ X with d(x, y) < δ, we have

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F (x, y) <

31

ευ . 2(2L + 1)

Combining this with (6.2), we deduce that |f ∗ (x) − f ∗ (y)| ≤

υ 2L ευ υ + × = <υ 2(2L + 1) ε 2(2L + 1) 2

whenever x, y ∈ X with d(x, y) < δ, which implies f ∗ (x) ∈ C(X). Conversely, we will prove (X, T ) is weak mean equicontinuous with the assumption that f ∗ is continuous for any f ∈ C(X). If (X, T ) is not weak mean equicontinuous, there are x ∈ X, ε > 0 and {xm }∞ m=1 ⊂ X such that lim xm = x but F (x, xm ) ≥ ε. With the assumption, we know x and {xm }∞ m=1 are generic m→+∞

points.

n 1  δ k ,M n n→+∞ k=1 T x

Let μ = lim

= diam(X), η =

ε 2(M+5) . Then by Lemma 2.2, there exist finite

0 of X such that mutually disjoint closed subsets {s }ss=1

μ(

s0 

s )>1 − η and diam(s )<η, ∀s = 1, 2, · · · , s0 .

s=1

Take δ = min {d(s1 , s2 )}, r = min{ 5δ , η} and α = s1 =s2

ε 4Ms0 +1 .

For any s ∈ {1, 2, · · · , s0 }, let s

s

0 0 Us = {y ∈ X : d(y, s ) < r} and Vs = {y ∈ X : d(y, s ) < 2r}. Then {Us }s=1 and {Vs }s=1 are mutually disjoint open subsets of X and diam(Vs ) ≤ 5η for any s ∈ {1, 2, · · · , s0 }. We have the following claim:

Claim 2. For any xm ∈ X, there is sm ∈ {1, 2, · · · , s0 } such that 1 1 lim inf χUsm (T k x) > lim inf χVsm (T k xm ) + α. n→+∞ n n→+∞ n n

n

k=1

k=1

Proof of Claim 2. If not, there is xm ∈ X such that for any s ∈ {1, 2, · · · , s0 }, we have 1 1 χUs (T k x) ≤ lim inf χVs (T k xm ) + α. n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

(6.3)

Given s ∈ {1, 2, · · · , s0 }. By Lemma 2.1, we have 1 χUs (T k x) ≥ μ(Us ). n→+∞ n n

lim inf

k=1

Then combining (6.3) with (6.4), we deduce that

(6.4)

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32

1 1 χVs (T k xm ) ≥ lim inf χUs (T k x) − α n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

≥ μ(Us ) − α. Since Us ⊂ Vs , we have 1 1 χVs (T k x) ≥ lim inf χUs (T k x) ≥ μ(Us ). n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

Then by Proposition 3.4, we derive that F (x, xm ) ≤ 5η

s0      μ(Us ) − α + M 1 − μ(Us ) − α

s0   s=1

s=1

3ε ≤ 5η + Mη + Ms0 α ≤ , 4 which is a contradiction. This finishes the proof of Claim 2.



∞ By Claim 2, there is sm0 ∈ {1, 2, · · · , s0 } and a subsequence {xmp }∞ p=1 of {xm }m=1 such that for any p ∈ N + , we have

1 1 χUsm (T k x) > lim inf χVsm (T k xmp ) + α. 0 0 n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

(6.5)

Take f ∈ C(X) such that 0 ≤ f ≤ 1 and f |U s

m0

= 1, f |Vscm = 0. 0

Then we derive that 1 1 χUsm (T k x) ≤ lim inf f (T k x) = f ∗ (x) 0 n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

and 1 1 χVsm (T k xmp ) ≥ lim inf f (T k xmp ) = f ∗ (xmp ). 0 n→+∞ n n→+∞ n n

n

k=1

k=1

lim inf

Thus by (6.5), we deduce that f ∗ (x) ≥ f ∗ (xmp ) + α, which implies f ∗ (x) ∈ / C(X). This is a contradiction. Hence (X, T ) is weak mean equicontinuous. 

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