Generalized Dobrushin ergodicity coefficient and uniform ergodicities of Markov operators

Abstract

In this paper the stability and the perturbation bounds of Markov operators acting on abstract state spaces are investigated. Here, an abstract state space is an ordered Banach space where the norm has an additivity property on the cone of positive elements. We basically study uniform ergodic properties of Markov operators by means of so-called a generalized Dobrushin’s ergodicity coefficient. This allows us to get several convergence results with rates. Some results on quasi-compactness of Markov operators are proved in terms of the ergodicity coefficient. Furthermore, a characterization of uniformly P-ergodic Markov operators is given which enable us to construct plenty examples of such types of operators. The uniform mean ergodicity of Markov operators is established in terms of the Dobrushin ergodicity coefficient. The obtained results are even new in the classical and quantum settings.

1 Introduction

It is known that Doeblin and Dobrushin [9, 21] characterized the contraction rate of Markov operators which act on a space of measures equipped with the total variation norm as follows: Let us consider a finite Markov chain with a transition (row stochastic) matrix $\mathbb {P}=(p_{ij})\in \mathbb {R}^{n\times n}$. It defines a Markov operator $P:\mathbb {R}^n\rightarrow \mathbb {R}^n$ such that $P\mathbf {x}=\mathbf {x}\mathbb {P}$, where the elements of $\mathbb {R}^n$ are row vectors. The set of probability measures can be identified with the standard simplex $\mathcal {K}=\{(x_i)\in \mathbb {R}^n: \ x_i\ge 0, \ \sum _{i=1}^n x_i=1\}$. The total variation norm is nothing but one half of the $\ell _1$ norm $\Vert \cdot \Vert _1$ on $\mathbb {R}^n$. One can introduce the following coefficient

$$\begin{aligned} \delta (P)=\sup _{\mu ,\nu \in \mathcal {K}, \mu \ne \nu }\frac{\Vert P\mu -P\nu \Vert _1}{\Vert \mu -\nu \Vert _1}. \end{aligned}$$

This coefficient is characterized by Doeblin and Dobrushin [9] as follows:

$$\begin{aligned} \delta (P)= & {} \frac{1}{2}\max _{i<j}\sum _{k=1}^n|p_{ik}-p_{jk}| \end{aligned}$$

(1)

$$\begin{aligned}= & {} 1-\min _{i<j}\sum _{k=1}^n\min \{p_{ik},p_{jk}\}. \end{aligned}$$

(2)

It is known that if $\delta (P)<1$ (this condition is often called Dobrushin condition) then $P^n$ converges to its invariant distribution with exponential rate [9, 42]. Moreover, this condition also gives the spectral gap of the operator P (see [42]). The Dobrushin condition played a major role as a source of inspiration for many mathematicians to do interesting work on the theory of Markov processes (see for example [21, 31, 42]).

Let us consider the following example: Let $T: \mathbb {R}^3 \rightarrow \mathbb {R}^3$ be the Markov operator which is given by the matrix

$$\begin{aligned}\begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad \frac{1}{2} &{}\quad \frac{1}{2} \end{pmatrix}. \end{aligned}$$

It is clear that $T^n$ converges to P, where

$$\begin{aligned} P = \begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 1&{}\quad 0 \end{pmatrix}. \end{aligned}$$

One can calculate that $\delta (T)=1$. From this, we infer that $T^n$ converges, but $\delta (T)=1$. Hence, the investigation of the sequence $\{T^n\}$ in terms of $\delta (T)$ is not effective. Hartfiel et al. [18, 19] introduced a generalized coefficient which covers the mentioned type of convergence in the finite-dimensional setting. To the best knowledge of the authors, such coefficient is not studied even in the classical $L^1$-spaces. Therefore, the main aim of this paper is to define an analogue of the coefficient mentioned above in a more general setting, i.e. for ordered Banach spaces, such that it will cover all known classical spaces as particular cases. Moreover, we are going to investigate uniform asymptotic stabilities of Markov operators on ordered Banach spaces. We notice that the consideration of these types of Banach spaces is convenient and important for the study of several properties of physical and probabilistic processes in an abstract framework which covers the classical and quantum cases (see [2, 11]). In this setting, certain limiting behaviors of Markov operators were investigated in [3, 5, 12, 15, 41].

Our purpose is to investigate stability and perturbation bounds of Markov operators acting on abstract state spaces. More precisely, an abstract state space is an ordered Banach space where the norm has an additivity property on the cone of positive elements. Examples of these spaces include all classical $L^1$-spaces and the space of density operators acting on some Hilbert spaces [2, 24]. Moreover, any Banach space can be embedded into some abstract spaces (see Example 2.3(c)). There are a few results in the literature on uniform convergence of iterates of bounded linear operators on Banach spaces (see, e.g. [11, 23, 27, 29, 30, 40, 44]). In the present paper, we study the asymptotic stability (in the sense of uniform topology) of Markov operators based on the so-called generalized Dobrushin’s ergodicity coefficient. This allows us to get several convergence results with rates. We notice that the Dobrushin coefficient (which extends $\delta (P)$ to abstract state spaces) has been introduced and studied in [15, 36, 37], for Markov operators acting on abstract state spaces.

The paper is organized as follows. In Sect. 2, we provide preliminary definitions and results on properties of abstract state spaces. In Sect. 3, we define a generalized Dobrushin ergodicity coefficient $\delta _P(T)$ of Markov operators with respect to a projection P and study its properties. Some results on quasi-compactness of Markov operators are proved in terms of this coefficient. At the end of that section, we give some connection of $\delta _P(T)$ to the spectral gap of T. Furthermore, in Sect. 4, the uniform P-ergodicity of Markov operators is studied in terms of the generalized Dobrushin ergodicity coefficient. This allows us to establish certain category results for the set of uniformly P-ergodic Markov operators. An application of the main result of this section is to get results on uniform ergodicities of linear bounded operators on Banach spaces. In Sect. 5, we give a characterization of uniformly P-ergodic Markov operators which enables us to explicitly construct such operators. Finally, in Sect. 6, we establish perturbation bounds for the uniform P-ergodic Markov operators. It is noticed that perturbation bounds have important applications in the theory of probability and quantum information (see, [14, 32, 33, 43]). Moreover, the results are even new in the classical and quantum settings.

2 Preliminaries

In this section, we recall some necessary definitions and results about abstract state spaces.

Let X be an ordered vector space with a cone $X_+=\{x\in X{:} \ x\ge 0\}$. A subset $\mathcal {K}$ is called a base for X, if $\mathcal {K}=\{x\in X_+{:}\ f(x)=1\}$ for some strictly positive (i.e. $f(x)>0$ for $x>0$) linear functional f on X. An ordered vector space X with generating cone $X_+$ (i.e. $X=X_+-X_+$) and a fixed base $\mathcal {K}$, defined by a functional f, is called an ordered vector space with a base [2]. Let U be the convex hull of the set $\mathcal {K}\cup (-\mathcal {K})$, and let

$$\begin{aligned} \Vert x\Vert _{\mathcal {K}}=\inf \{\lambda \in \mathbb {R}_+{:}\ x\in \lambda U\}. \end{aligned}$$

Then one can see that $\Vert \cdot \Vert _{\mathcal {K}}$ is a seminorm on X. Moreover, one has $\mathcal {K}=\{x\in X_+{:} \ \Vert x\Vert _{\mathcal {K}}=1\}$, $f(x)=\Vert x\Vert _{\mathcal {K}}$ for $x\in X_+$. Assume that the seminorm becomes a norm, and X is complete space w.r.t. this norm and $X_+$ is closed subset, then $(X,X_+,\mathcal {K},f)$ is called abstract state space. In this case, $\mathcal {K}$ is a closed face of the unit ball of X, and U contains the open unit ball of X. If the set U is radially compact [2], i.e. $\ell \cap U$ is a closed and bounded segment for every line $\ell$ through the origin of X, then $\Vert \cdot \Vert _{\mathcal {K}}$ is a norm. The radial compactness is equivalent to the coincidence of U with the closed unit ball of X. In this case, X is called a strong abstract state space. In the sequel, for the sake of simplicity, instead of $\Vert \cdot \Vert _{\mathcal {K}}$, the standard notation $\Vert \cdot \Vert$ is used. To better understand the difference between a strong abstract state space and a more general class of base norm spaces, the reader is referred to [46].

A positive cone $X_+$ of an ordered Banach space X is said to be $\lambda$-generating if, given $x\in X$, we can find $y,z\in X_+$ such that $x = y- z$ and $\Vert y\Vert + \Vert z\Vert \le \lambda \Vert x\Vert$. The norm on X is called regular (respectively, strongly regular) if, given x in the open (respectively, closed) unit ball of X, y can be found in the closed unit ball with $y\ge x$ and $y\ge -x$. The norm is said to be additive on $X_+$ if $\Vert x + y\Vert = \Vert x\Vert + \Vert y\Vert$ for all $x, y\in X_+$. If $X_+$ is 1-generating, then X can be shown to be strongly regular. Similarly, if $X_+$ is $\lambda$-generating for all $\lambda > 1$, then X is regular [46]. The following results are well-known.

Theorem 2.1

[45, p. 90] Let X be an ordered Banach space with closed positive cone $X_+$. Then te following statements are equivalent:

• (i) X is an abstract state space;

• (ii) X is regular, and the norm is additive on $X_+$;

• (iii) $X_+$ is $\lambda$-generating for all $\lambda > 1$, and the norm is additive on $X_+$.

Theorem 2.2

[46] Let X be an ordered Banach space with closed positive cone $X_+$. Then the following statements are equivalent:

• (i) X is a strong abstract state space;

• (ii) X is strongly regular, and the norm is additive on $X_+$;

• (iii) $X_+$ is 1-generating and the norm is additive on $X_+$.

In this paper, we consider a general abstract state space for which the convex hull of the base $\mathcal {K}$ and $-\mathcal {K}$ is not assumed to be radially compact (in our previous papers [13, 36, 37] this condition was essential). This consideration has an important advantage: whenever X is an ordered Banach space with a generating cone $X_+$ whose norm is additive on $X_+$, then X admits an equivalent norm that coincides with the original norm on $X_+$ and renders X that base norm space. Hence, to apply the results of the paper one would then only have to check that if the norm is additive on $X_+$.

Example 2.3

Let us provide some examples of abstract state spaces.

• (a) Let M be a von Neumann algebra. Let $M_{h,*}$ be the Hermitian part of the predual space $M_*$ of M. As a base $\mathcal {K}$ we define the set of normal states of M. Then $(M_{h,*},M_{*,+},\mathcal {K},\mathbf{1}\!\!\mathrm{I})$ is a strong abstract state spaces, where $M_{*,+}$ is the set of all positive functionals taken from $M_*$, and $\mathbf{1}\!\!\mathrm{I}$ is the unit in M. In particular, if $M=L^\infty (E,\mu )$, then $M_{*}=L^1(E,\mu )$ is an abstract state space.

• (b) Let A be a real ordered linear space and, as before, let $A_+$ denote the set of positive elements of A. An element $e\in A_+$ is called order unit if for every $a\in A$ there exists a number $\lambda \in \mathbb {R}_+$ such that $-\lambda e\le a\le \lambda e$. If the order is Archimedean, then the mapping $a\rightarrow \Vert a\Vert _e=\inf \{\lambda > 0{:}\ {-}\lambda e\le a\le \lambda e\}$ is a norm. If A is a Banach space with respect to this norm, the pair (A, e) is called an order-unit space with the order unite. An element $\rho \in A^*$ is called positive if $\rho (x)\ge 0$ for all $a\in A_+$. By $A^*_+$ we denote the set of all positive functionals. A positive linear functional is called a state if $\rho (e)=1$. The set of all states is denoted by S(A). Then it is well-known that $(A^*,A^*_+,S(A),e)$ is a strong abstract state space [2]. In particular, if ${\mathfrak {A}}_{sa}$ is the self-adjoint part of an unital $C^*$-algebra, ${\mathfrak {A}}_{sa}$ becomes order-unit spaces, hence $({\mathfrak {A}}_{sa}^*,{\mathfrak {A}}_{sa,+}^*,S({\mathfrak {A}}_{sa}),\mathbf{1}\!\!\mathrm{I})$ is a strong abstract state space.

• (c) Let X be a Banach space over $\mathbb {R}$. Consider a new Banach space $\mathcal {X}=\mathbb {R}\oplus X$ with a norm $\Vert (\alpha ,x)\Vert =\max \{|\alpha |,\Vert x\Vert \}$. Define a cone $\mathcal {X}_+=\{(\alpha ,x){:} \ \Vert x\Vert \le \alpha , \ \alpha \in \mathbb {R}_+\}$ and a positive functional $f(\alpha ,x)=\alpha$. Then one can define a base $\mathcal {K}=\{(\alpha ,x)\in \mathcal {X}{:}\ f(\alpha ,x)=1\}$. Clearly, we have $\mathcal {K}=\{(1,x){:}\ \Vert x\Vert \le 1\}$. Then $(\mathcal {X},\mathcal {X}_+,\mathcal {K},f)$ is an abstract state space [24]. Moreover, X can be isometrically embedded into $\mathcal {X}$. Using this construction one can study several interesting examples of abstract state spaces.

• (d) Let A be the disc algebra, i.e. the sup-normed space of complex-valued functions which are continuous on the closed unit disc, and analytic on the open unit disc. Let $X =\{f\in A{:}\ f(1)\in \mathbb {R}\}$. Then X is a real Banach space with the following positive cone $X_+=\{f\in X{:} f(1)=\Vert f\Vert \}=\{f\in X{:} \ f(1)\ge \Vert f\Vert \}$. The space X is an abstract state space, but not strong one (see [46] for details).

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space. A linear operator $T{:} X\rightarrow X$ is called positive, if $Tx\ge 0$ whenever $x\ge 0$. A positive linear operator $T{:} X\rightarrow X$ is said to be Markov, if $T(\mathcal {K})\subset \mathcal {K}$. It is clear that $\Vert T\Vert =1$, and its adjoint operator $T^*{:} X^*\rightarrow X^*$ acts an ordered Banach space $X^*$ with unit f, and moreover, $T^*f=f$. Now for each $y\in X$ we define a linear operator $T_y{:} X\rightarrow X$ by $T_y(x)=f(x)y$.

From the definition of Markov operator, one can prove the following auxiliary fact.

Lemma 2.4

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space and let T be a Markov operator on X. Then for any $x\in X$, we have $f(Tx)=f(x)$.

Example 2.5

Let us consider several examples of Markov operators.

• 1. Let $X=L^1(E,\mu )$ be the classical $L^1$-space. Then any transition probability P(x, A) defines a Markov operator T on X, whose dual $T^*$ acts on $L^\infty (E,\mu )$ as follows

  $$\begin{aligned} (T^*f)(x)=\int f(y)P(x,dy), \ \ f\in L^\infty . \end{aligned}$$

• 2. Let M be a von Neumann algebra, and consider $(M_{h,*},M_{*,+},\mathcal {K},\mathbf{1}\!\!\mathrm{I})$ as in (a) Example 2.3. Let $\Phi {:} M\rightarrow M$ be a positive, unital ($\Phi (\mathbf{1}\!\!\mathrm{I})=\mathbf{1}\!\!\mathrm{I}$) linear mapping. Then the operator given by $(Tf)(x)=f(\Phi (x))$, where $f\in M_{h,*}, x\in M$, is a Markov operator.

• 3. Let $X=C[0,1]$ be the space of real-valued continuous functions on [0, 1]. Denote

  $$\begin{aligned} X_+=\big \{x\in X{:} \ \max _{0\le t\le 1}|x(t)-x(1)|\le 2 x(1)\big \}. \end{aligned}$$

  Then $X_+$ is a generating cone for X, and $f(x)=x(1)$ is a strictly positive linear functional. Then $\mathcal {K}=\{x\in X_+{:} \ f(x)=1\}$ is a base corresponding to f. One can check that the base norm $\Vert x\Vert$ is equivalent to the usual one $\Vert x\Vert _{\infty }=\max \limits _{0\le t\le 1}|x(t)|$. Due to closedness of $X_+$ we conclude that $(X,X_+,\mathcal {K},f)$ is an abstract state space. Let us define a mapping T on X as follows:

  $$\begin{aligned} (Tx)(t)=tx(t). \end{aligned}$$

  It is clear that T is a Markov operator on X.

• 4. Let X be a Banach space over $\mathbb {R}$. Consider the abstract state space $(\mathcal {X},\mathcal {X}_+,\tilde{\mathcal {K}},f)$ constructed in (c) Example 2.3. Let $T{:}X\rightarrow X$ be a linear bounded operator with $\Vert T\Vert \le 1$. Then the operator $\mathcal {T}{:} \mathcal {X}\rightarrow \mathcal {X}$ defined by $\mathcal {T}(\alpha ,x)=(\alpha ,Tx)$ is a Markov operator.

• 5. Let A be the disc algebra, and let X be the abstract state space as in (d) Example 2.3. A mapping T given by $Tf(z)=zf(z)$ is clearly a Markov operator on X.

Definition 2.6

[36] Let $(X,X_+,\mathcal {K},f)$ be an abstract state space, and let $T{:}X\rightarrow X$ be a Markov operator. Then the Dobrushin’s ergodicity coefficient of T is given by

$$\begin{aligned} \delta (T)=\sup _{x\in N,\ x\ne 0}\frac{\Vert Tx\Vert }{\Vert x\Vert }, \end{aligned}$$

(3)

where

$$\begin{aligned} N=\{x\in X{:} \ f(x)=0\}. \end{aligned}$$

(4)

Remark 2.7

We note that if $X=L^1(E,\mu )$, the notion of the Dobrushin ergodicity coefficient was studied in [7] and [9]. In a non-commutative setting, i.e. when $X^*$ is a von Neumann algebra, such a notion was introduced in [34]. We should stress that this coefficient has been independently defined in [15].

3 Generalized Dobrushin ergodicity coefficient

In this section, we introduce a generalized notion of the Dobrushin’s ergodicity coefficient (3), and investigate its properties.

Definition 3.1

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space and let $T{:}X\rightarrow X$ be a linear bounded operator. Consider a non-trivial projection operator $P{:} X\rightarrow X$ (i.e. $P^2=P$). Then we define

$$\begin{aligned} \delta _P(T)=\sup _{x\in N_P,\ x\ne 0}\frac{\Vert Tx\Vert }{\Vert x\Vert }, \end{aligned}$$

(5)

where

$$\begin{aligned} N_P=\{x\in X{:} \ Px=0\}. \end{aligned}$$

(6)

If $P=I$, we put $\delta _P(T)=1$. The quantity $\delta _P(T)$ is called the generalized Dobrushin ergodicity coefficient ofTwith respect toP.

We notice that if $X=\mathbb {R}^n$, then there are some formulas to calculate this coefficient (see [18, 19]).

In the following remarks, let us have a brief comparison between the coefficients $\delta _P(T)$ and $\delta (T)$.

Remark 3.2

Let $y_0\in \mathcal {K}$ and consider the projection $Px=f(x)y_0$. Then one can see that $N_P$ coincides with

$$\begin{aligned} N= \{ x\in X;\ f(x)=0\}, \end{aligned}$$

and in this case $\delta _P(T)=\delta (T)$. Hence, $\delta _P(T)$ indeed is a generalization of $\delta (T)$.

Remark 3.3

Let P be a Markov projection on X. Then, for any Markov operator $T{:}X\rightarrow X$

$$\begin{aligned} \delta _P(T) \le \delta (T). \end{aligned}$$

Indeed, it is enough to show that $N_P\subseteq N$. Let $x\in N_P$, so $Px=0$. Due to Lemma 2.4, we have

$$\begin{aligned} N=\{x\in X;\ f(Px)=0\}, \end{aligned}$$

which yields $x\in N$, so $N_P\subseteq N$.

In what follows, we examine main properties of $\delta _P(T)$.

Proposition 3.4

Let $T{:}X\rightarrow X$ be a linear bounded operator. If P and Q are two projections on X such that $Q\le P$ (i.e. $QP=PQ=Q$), then $\delta _P(T)\le \delta _Q(T)$.

Proof

Assume that $Q\le P$. Then for every $x\in N_P$ we get $Qx=QPx=0$, therefore $N_P\subseteq N_Q$. Hence, we get the desired inequality. $\square$

Corollary 3.5

If P and Q are orthogonal projections on X, then $\delta _{P+Q}(T)\le \delta _P(T)$.

Proof

As P and Q are orthogonal projections, $P+Q$ is a projection which dominates P, hence the corollary follows directly from the previous proposition. $\square$

Before establishing our main result of this section, we need the following auxiliary fact.

Lemma 3.6

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space and let P be a Markov projection. Then for every $x\in N_P$ there exist $u,v\in \mathcal {K}$ with $u-v\in N_P$ such that

$$\begin{aligned} x=\alpha (x)(u-v), \end{aligned}$$

where $\alpha (x)\in \mathbb {R}_+$ and $\alpha (x)\le \frac{\lambda }{2}\Vert x\Vert$.

Proof

Given any $x\in N_P$, we have $Px=0$. As $X_+$ is $\lambda$-generating of X, there exist $x_+,x_-\in X_+$ such that $x=x_+-x_-$ with $\Vert x_+\Vert +\Vert x_-\Vert \le \lambda \Vert x\Vert$. Clearly $Px_+=Px_-$. As P a Markov projection

$$\begin{aligned} \Vert Px_+\Vert =f(Px_+)=f(x_+)=\Vert x_+\Vert , \end{aligned}$$

which yields $\Vert x_+\Vert =\Vert x_-\Vert$. Therefore,

$$\begin{aligned} x= & {} \frac{x_+}{\Vert x_+\Vert }\Vert x_+\Vert - \frac{x_-}{\Vert x_-\Vert }\Vert x_+\Vert \\= & {} \Vert x_+\Vert \left( \frac{x_+}{\Vert x_+\Vert }- \frac{x_-}{\Vert x_-\Vert } \right) . \end{aligned}$$

letting $u=\frac{x_+}{\Vert x_+\Vert }$ and $v=\frac{x_-}{\Vert x_-\Vert }$, so $u,v\in \mathcal {K}$. Moreover, $Pu=Pv$, then $u-v\in N_P$, and letting $\alpha (x):=\Vert x_+\Vert \le \frac{\lambda }{2}\Vert x\Vert$, hence the lemma is proved. $\square$

Let us denote by $\Sigma (X)$ the set of all Markov operators defined on X, and by $\Sigma _P(X)$ we denote the set of all Markov operators T on X with $PT=TP$.

Now, we prove the following essential result about main properties of $\delta _P$.

Theorem 3.7

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space, P be a projection on X and let $T,S\in \Sigma (X)$. Then:

• (i) $0\le \delta _P(T)\le 1$;

• (ii) $|\delta _P(T)-\delta _P(S)|\le \delta _P(T-S)\le \Vert T-S\Vert$;

• (iii) if $P\in \Sigma (X)$, one has

  $$\begin{aligned} \delta _P(T)\le \frac{\lambda }{2}\sup \{\Vert Tu-Tv\Vert ;\ u,v\in \mathcal {K}\ \text {with}\ u-v\in N_P \}. \end{aligned}$$

  (7)

• (iv) if $H{:} X\rightarrow X$ is a bounded linear operator such that $HP=PH$, then

  $$\begin{aligned} \delta _P(TH)\le \delta _P(T)\Vert H\Vert ; \end{aligned}$$

• (v) if $H{:} X\rightarrow X$ is a bounded linear operator such that $PH=0$, then

  $$\begin{aligned} \Vert TH\Vert \le \delta _P(T)\Vert H\Vert ; \end{aligned}$$

• (vi) if $S\in \Sigma _P(X)$, then

  $$\begin{aligned} \delta _P(TS)\le \delta _P(T)\delta _P(S). \end{aligned}$$

Proof

• (i) As T is a Markov operator and by the definition of $\delta _P$ one gets $0\le \delta _P(T)\le \Vert T\Vert = 1$. (ii) The second inequality is immediately obtained from (5). To establish the first one, take any $\epsilon >0$. Then there exists an $x_\epsilon \in N_P$, with $\Vert x_\epsilon \Vert =1$ such that $\delta _P(T)\le \Vert Tx_\epsilon \Vert +\epsilon$. Hence,

  $$\begin{aligned} \delta _P(T)-\delta _P(S)\le & {} \Vert Tx_\varepsilon \Vert +\varepsilon -\sup _{x\in N_P,\Vert x\Vert =1}\Vert Sx\Vert \\\le & {} \Vert Tx_\varepsilon \Vert -\Vert S x_\varepsilon \Vert +\varepsilon \\\le & {} \Vert (T-S)x_\varepsilon \Vert +\varepsilon \\\le & {} \sup _{x\in N_P{:} \ \Vert x\Vert =1}\Vert (T-S)x\Vert +\varepsilon \\= & {} \delta _P(T-S)+\varepsilon \end{aligned}$$

  which implies the assertion.

• (iii) For all $x\in N_P$, by Lemma 3.6 there exist $u,v\in \mathcal {K}$ with $u-v\in N_P$ such that

  $$\begin{aligned} x=\alpha (x)(u-v),\ \text {where}\ \alpha (x)\in \mathbb {R}_+ \ \text {with}\ \alpha (x)\le \frac{\lambda }{2}\Vert x\Vert . \end{aligned}$$

  Therefore,

  $$\begin{aligned} \frac{\Vert T(x)\Vert }{\Vert x\Vert }= & {} \frac{\alpha (x)}{\Vert x\Vert }\Vert T(u)-T(v)\Vert \\\le & {} \frac{\lambda }{2}\Vert T(u)-T(v)\Vert . \end{aligned}$$

  Hence, by the definition of $\delta _p$ and the previous inequality, we obtain (7).

• (iv) Suppose that H is a bounded linear operator on X which commutes with P. For all $x\in N_P$, we have

  $$\begin{aligned} PHx=HPx=0, \end{aligned}$$

  then $Hx\in N_P$. Therefore,

  $$\begin{aligned} \Vert THx\Vert\le & {} \delta _P(T)\Vert Hx\Vert \\\le & {} \delta _P(T) \Vert H\Vert \Vert x\Vert , \end{aligned}$$

  which implies that

  $$\begin{aligned} \frac{\Vert THx\Vert }{\Vert x\Vert }\le \delta _P(T)\Vert H\Vert ,\ \forall \ x\in N_P \end{aligned}$$

  and hence we have $\delta _P(TH)\le \delta _P(T)\Vert H\Vert$.

• (v) if H is a bounded linear operator on X with $PH=0$, then for all $x\in X$, $Hx \in N_P$. Therefore,

  $$\begin{aligned} \Vert THx\Vert\le & {} \delta _P(T)\Vert Hx\Vert \\\le & {} \delta _P(T) \Vert H\Vert \Vert x\Vert , \end{aligned}$$

  which yields

  $$\begin{aligned} \frac{\Vert THx\Vert }{\Vert x\Vert }\le \delta _P(T)\Vert H\Vert ,\ \forall \ x\in X. \end{aligned}$$

• (vi) As $S\in \Sigma _P(X)$, we have $Sx\in N_P$, for all $x\in N_P$. Then

  $$\begin{aligned} \Vert T(Sx)\Vert\le & {} \delta _P(T)\Vert Sx\Vert \\\le & {} \delta _P(T) \delta _P(S)\Vert x\Vert , \end{aligned}$$

  which implies

  $$\begin{aligned} \frac{\Vert TSx\Vert }{\Vert x\Vert }\le \delta _P(T)\delta _P(S),\ \forall \ x\in N_P, \end{aligned}$$

  then we get

  $$\begin{aligned} \delta _P(TS)\le \delta _P(T)\delta _P(S), \end{aligned}$$

  and hence the theorem is proved. $\square$

Now, let us consider the case of strong abstract state spaces. In this setting, by Theorem (2.2), $X_+$ is 1-generating and the norm is additive on $X_+$. Following the arguments of the proof of Lemma 3.6, one can prove the next result.

Lemma 3.8

Let $(X,X_+,\mathcal {K},f)$ be a strong abstract state space and let P be a Markov projection. Then for every $x,y\in X$ with $x-y\in N_P$ there exist $u,v\in \mathcal {K}$ with $u-v\in N_P$ such that

$$\begin{aligned} x-y=\frac{\Vert x-y\Vert }{2}(u-v). \end{aligned}$$

Consequently, (7) can be modified as follows:

Proposition 3.9

Let $(X,X_+,\mathcal {K},f)$ be a strong abstract state space, P be a Markov projection on X and let $T\in \Sigma (X)$. Then:

$$\begin{aligned} \delta _P(T)= \frac{1}{2}\sup \{\Vert Tu-Tv\Vert ;\ u,v\in \mathcal {K}\ \text {with}\ u-v\in N_P \}. \end{aligned}$$

(8)

Hence, we have the following result.

Corollary 3.10

Let $(X,X_+,\mathcal {K},f)$ be a strong abstract state space, P be a Markov projection on X and $T\in \Sigma (X)$. If $\delta _P(T)=0$, then $T=TP$.

Proof

If $\delta _P(T)=0$, then by (8) we have $Tu=Tv$, for all $u,v\in \mathcal {K}$ with $u-v\in N_P$. As P is a Markov projection, we have $Pu-u\in N_P$. Then

$$\begin{aligned} Tu=TPu,\ \forall u\in \mathcal {K}. \end{aligned}$$

If $x\in X_+$, then

$$\begin{aligned} Tx=\Vert x\Vert T\left( \frac{x}{\Vert x\Vert }\right) =\Vert x\Vert TP\left( \frac{x}{\Vert x\Vert }\right) =TPx. \end{aligned}$$

Now, for all $x\in X$, $x=x_+-x_-$, ($x_+,x_-\in X_+$). Therefore,

$$\begin{aligned} Tx=TPx_+-TPx_-=TPx, \end{aligned}$$

which proves the assertion. $\square$

From now, we consider general abstract state spaces. The following proposition is crucial in our investigations.

Proposition 3.11

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space, and let P be a projection on X. If $T\in \Sigma _P(X)$ and $\delta _P(T^{n_0})<1$ for some $n_0\in \mathbb {N}$, then $\Vert T^n(I-P)\Vert \rightarrow 0$.

Proof

Given such $n_0\in \mathbb {N}$ and let $\rho =\delta _P(T^{n_0})$. Then for a large $n\in \mathbb {N}$, we write $n=kn_0+r\ (k,r\in \mathbb {N}\ \text {and}\ r<n_0)$ and by (vi) of Theorem 3.7

$$\begin{aligned} \delta _P(T^n)= \delta _P(T^{kn_0}T^r) \le \rho ^k \delta _P(T^r). \end{aligned}$$

Again using (v) of the same theorem, we have

$$\begin{aligned} \Vert T^n(I-P)\Vert \le \delta _P(T^n)\Vert I-P\Vert \le 2 \rho ^k \delta _P(T^r)\le 2\rho ^{\lfloor \frac{n}{n_0}\rfloor } \rightarrow 0\ (\text {as}\ n\rightarrow \infty ), \end{aligned}$$

which proves the assertion. $\square$

It is clear that if $T\in \Sigma _P(X)$, then $T\in \Sigma _{I-P}(X)$. Therefore, it would be interesting to know a relation between $\delta _P(T)$ and $\delta _{I-P}(T)$. Next result clarifies this question.

Proposition 3.12

Let $T\in \Sigma _P(X)$. Then at most one of the following statements is valid:

• (i) there exists $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$;

• (ii) there exists $n_0\in \mathbb {N}$ such that $\delta _{I-P}(T^{n_0})<1$.

Proof

Suppose that there exist $n_0,m_0\in \mathbb {N}$ such that

$$\begin{aligned} \delta _P(T^{n_0})<1\ \ \ \text {and}\ \ \ d_{I-P}(T^{m_0})<1. \end{aligned}$$

Then by Proposition 3.11

$$\begin{aligned} \Vert T^{n}(I-P)\Vert \rightarrow 0. \end{aligned}$$

As $T\in \Sigma _{I-P(X)}$ and using the same argument

$$\begin{aligned} \Vert T^{n}P\Vert \rightarrow 0. \end{aligned}$$

Then

$$\begin{aligned} \Vert T^{n}\Vert = \Vert T^{n}(P+(I-P))\Vert \le \Vert T^{n}(P)| + \Vert T^n(I-P)\Vert \rightarrow 0, \end{aligned}$$

which contradicts the Markovianity of T. $\square$

Corollary 3.13

If $T\in \Sigma _P(X)$ and $\delta _P(T^{n_0})<1$ for some $n_0\in \mathbb {N}$, then $\delta _{I-P}(T^{n})=1$, for all $n\in \mathbb {N}$.

Let us recall that a bounded linear operator T on a Banach space X is called quasi-compact if there exists an $n_0\in \mathbb {N}$ such that $\Vert T^{n_0}-K\Vert <1$, for some compact operator K on X. Quasi-compact operators have been extensively studied in [20, 28].

It is natural to ask: whether T would be a quasi-compact in terms of $\delta _P$? Next result sheds some light on this question.

Theorem 3.14

Let $T\in \Sigma _P(X)$ and TP be quasi-compact on X. If there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$, then T is quasi-compact.

Proof

The quasi-compactness of TP yields the existence of $m_0\in \mathbb {N}$ and a compact operator K such that

$$\begin{aligned} \Vert (TP)^{m_0}-K\Vert <1. \end{aligned}$$

On the other hand, the existence of $n_0\in \mathbb {N}$ with $\delta _P(T^{n_0})<1$, due to Proposition 3.11 implies

$$\begin{aligned} \Vert T^n(I-P)\Vert =\Vert T^n-T^nP\Vert \rightarrow 0. \end{aligned}$$

(9)

Then, for any positive $\varepsilon$ with $0<\varepsilon <1-\Vert (TP)^{m_0}-K\Vert$, by (9) one finds $n_1\in \mathbb {N}$ (we may assume that $n_1>m_0$) such that

$$\begin{aligned} \Vert T^{n_1}-T^{n_1}P\Vert <\varepsilon . \end{aligned}$$

Let $K_1=T^{n_1-m_0}K$, which is clearly compact. Then

$$\begin{aligned} \Vert T^{n_1}-K_1\Vert\le & {} \Vert T^{n_1}-T^{n_1}P\Vert +\Vert T^{n_1}P-K_1\Vert \\< & {} \varepsilon +\Vert T^{n_1-m_0}(T^{m_0}-K)\Vert \\\le & {} \varepsilon +\Vert T^{m_0}-K\Vert <1, \end{aligned}$$

which means that T is quasi-compact. $\square$

From this proposition we immediately get the following one.

Corollary 3.15

Let $T\in \Sigma _P(X)$ and P be compact on X. If there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$, then T is quasi-compact.

Let X be an abstract state space. Its complexification $\tilde{X}$ is defined by $\tilde{X}=X+iX$ with a reasonable norm $\Vert \cdot \Vert _\mathbb {C}$ (see [38] for details). In this setting, X is called the real part of $\tilde{X}$. The positive cone of $\tilde{X}$ is defined as $X_+$. A vector $f \in \tilde{X}$ is called positive, which we denote by $f \ge 0$, if $f \in X_+$. For two elements $f,g\in \tilde{X}$ we write, as usual, $f \le g$ if $g-f \ge 0$. In the dual space $\tilde{X}^*$ of $\tilde{X}$, one can introduce an order as follows: a functional $\varphi \in \tilde{X}^*$ fulfils $\varphi \ge 0$ if and only if $\langle \varphi , x\rangle \ge 0$ for all $x \in X_+$; we denote the positive cone in $\tilde{X}^*$ by $\tilde{X}^*_+:= (\tilde{X}^*)_+$. In what follows, we assume that the norm $\Vert \cdot \Vert _\mathbb {C}$ is taken as

$$\begin{aligned} \Vert x+iy\Vert _\infty = \sup _{0\le t\le 2\pi } \Vert x\cos t-y\sin t\Vert . \end{aligned}$$

We note that all other complexification norms on $\tilde{X}$ are equivalent to $\Vert \cdot \Vert _\infty$, and moreover, $\Vert \cdot \Vert _\infty$ is the smallest one among all reasonable norms.

A linear mapping $T{:}X\rightarrow X$ can be uniquely extended to $\tilde{T}{:}\tilde{X} \rightarrow \tilde{X}$ by $\tilde{T}(x+iy)=Tx+iTy$. The operator $\tilde{T}$ is called the extension of T and it is well-known that $\Vert \tilde{T}\Vert =\Vert T\Vert$. In what follows, a mapping $\tilde{T}{:}\tilde{X}\rightarrow \tilde{X}$ is called Markov if it is the extension of a Markov operator T. Let $\tilde{P}$ be the extension of a projection $P{:}X\rightarrow X$, and define

$$\begin{aligned} \tilde{\delta }_{\tilde{P}}(\tilde{T})=\sup _{x\in N_{\tilde{P}}}\frac{\Vert \tilde{T}x\Vert _\infty }{\Vert x\Vert _\infty }, \end{aligned}$$

where $N_{\tilde{P}}=\{x\in \tilde{X};\ \tilde{P}x=0 \}$.

Lemma 3.16

Let X be a normed space, $T{:}X\rightarrow X$ be an operator and let $\tilde{T}$ be its extension. Then

$$\begin{aligned} \tilde{\delta }_{\tilde{P}}(\tilde{T})= \delta _P(T). \end{aligned}$$

Proof

As $\tilde{T}$ is the extension of T, $\tilde{\delta }_{\tilde{P}}(\tilde{T})\ge \delta _P(T)$. On the other hand, if $\tilde{x}\in N_{\tilde{P}}\ (\tilde{x}=x+iy)$, then $Px=Py=0$, i.e. both x and y belong to $N_P$. Therefore,

$$\begin{aligned} \Vert \tilde{T}(x+iy)\Vert _\infty= & {} \Vert Tx+iTy\Vert _\infty \\= & {} \sup _{0\le t\le 2\pi }\Vert T(x)\cos t -T(y)\sin t \Vert \\= & {} \sup _{0\le t\le 2\pi }\Vert T(x\cos t -y\sin t) \Vert \\\le & {} \delta _P(T)\sup _{0\le t\le 2\pi }\Vert x\cos t - y \sin t\Vert \ (\text {by } (v) \text { of Theorem } 3.7)\\= & {} \delta _P(T)\Vert x+iy\Vert _\infty , \end{aligned}$$

hence $\tilde{\delta }_{\tilde{P}}(\tilde{T})\le \delta _P(T)$, which completes the proof. $\square$

Now, let $S\in \Sigma (X)$ and let P be a projection on X. Recall that $X=PX\oplus (I-P)X$ and so the dual $X^*=(PX)^*\oplus ((I-P)X)^*$. Assume that $\lambda$ is an eigenvalue of S, in the following we discuss the comparison between $|\lambda |$ and $\delta _P(T)$.

Theorem 3.17

Let P be a Markov projection on a complex space X and let $S \in \Sigma _P(X)$. If one of the following conditions is satisfied:

• (i) $\lambda \ne 1$ is an eigenvalue of S in $(I-\tilde{P})\tilde{X}$; or

• (ii) $\lambda \ne 1$ is an eigenvalue of $S^*$ in $((I-\tilde{P})\tilde{X})^*$,

then $|\lambda |\le \delta _P(S)$.

Proof

If (i) is satisfied and $x\in (1-\tilde{P})\tilde{X}$ is a corresponding eigenvector to $\lambda$ with $\Vert x\Vert _\infty =1$, then $x\in N_{\tilde{P}}$ and

$$\begin{aligned} |\lambda | =\Vert \lambda x\Vert _\infty =\Vert \tilde{S}x\Vert _\infty \le \sup _{x\in N_{\tilde{P}}}\Vert \tilde{S}x\Vert _\infty \le \tilde{\delta }_{\tilde{P}}(\tilde{S})= \delta _P(S). \end{aligned}$$

Assume that (ii) is satisfied. Notice that for $y\in \tilde{X}^*$, the set

$$\begin{aligned} \{|y(\tilde{x})|;\ \tilde{x}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\} \end{aligned}$$

is bounded by $\Vert y\Vert$. Let $G{:} \tilde{X}^*\rightarrow \mathbb {R}$ be defined as follows:

$$\begin{aligned} G(y) =\sup \{|y(\tilde{x})|;\ \tilde{x}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\},\ y\in \tilde{X}^*. \end{aligned}$$

Now, $\tilde{S}^*y\in \tilde{X}^*$ and

$$\begin{aligned} G(\tilde{S}^*y)= & {} \sup \{|\tilde{S}^*y(\tilde{x})|;\ \tilde{x}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\}\\= & {} \sup \{|y(\tilde{S}(\tilde{x}))|;\ \ \tilde{x}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\}\\= & {} \sup \left\{ \left| \Vert \tilde{S}(\tilde{x})\Vert _\infty y\left( \frac{\tilde{S}(\tilde{x})}{\Vert \tilde{S}(\tilde{x})\Vert _\infty }\right) \right| ;\ \ \tilde{x}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\right\} \\\le & {} \tilde{\delta }_{\tilde{P}}(\tilde{S}) \sup \left\{ \left| y\left( \frac{\tilde{S}(\tilde{x})}{\Vert \tilde{S}(\tilde{x})\Vert _\infty }\right) \right| ;\ \ \tilde{x}\in N_P\ \text {and}\ \Vert \tilde{x}\Vert _\infty \le 1\right\} \\\le & {} \delta _P(S) \sup \left\{ |y(\tilde{v})|;\ \tilde{v}\in N_{\tilde{P}}\ \text {and}\ \Vert \tilde{v}\Vert _\infty \le 1\right\} \ (\text {since}\ \tilde{S}(N_{\tilde{P}})\subseteq N_{\tilde{P}}) )\\= & {} \delta _P(S)G(y). \end{aligned}$$

If $\lambda$ is an eigenvalue of $\tilde{S}^*$ in $((I-\tilde{P})\tilde{X})^*$, then for a corresponding eigenvector $\tilde{y}\in ((I-\tilde{P})\tilde{X})^*$ we have

$$\begin{aligned} |\lambda | G(\tilde{y}) =G(\lambda \tilde{y})= G(\tilde{S}^*\tilde{y})\le \delta _P(S)G(\tilde{y}). \end{aligned}$$

As $\tilde{y}$ is a non-zero eigenvector of $\tilde{S}^*$ which belongs to $((I-\tilde{P})\tilde{X})^*$, there exists $x_0\in (1-\tilde{P})\tilde{X}$ (consequently $x_0\in N_{\tilde{P}}$) such that $\tilde{y}(x_0)\ne 0$. Then we get $G(\tilde{y}) \ne 0$ and hence the proof is completed. $\square$

Remark 3.18

We notice that there are many works devoted to the spectral properties of Markov operators (see for example, [1, 16]). One of them is its spectral gap. Namely, we say that a Markov operator T on X (here X is a complex abstract state space) has a spectral gap, if one has $\Vert T(I-P)\Vert <1$, where P is a Markov projection such that $PT=TP=P$. This is clearly equivalent to $\delta _P(T)<1$. When X is taken as a non-commutative $L_p$-spaces, the spectral gap of Markov operator has been recently studied in [8]. In the classical setting, this gap has been extensively investigated by many authors (see for example, [25]).

We can stress that if T has a spectral gap, then 1 has to be an isolated point of the spectrum. Indeed, choose an arbitrary $\varepsilon >0$ with $\varepsilon <1-\delta _P(T)$. Assume that $\lambda$ is an element of the spectrum of T such that $|1-\lambda |<\varepsilon$ with corresponding eigenvector x. Then, it is clear that $y=x-Px$ belongs to $N_P$, therefore, one gets

$$\begin{aligned} Ty=Tx-TPx=Tx-PTx=\lambda (x-Px)=\lambda y \end{aligned}$$

hence, y is an eigenvector with eigenvalue of $\lambda$, and we have

$$\begin{aligned} \Vert Ty\Vert =|\lambda |\Vert y\Vert >\delta _p(T)\Vert y\Vert , \end{aligned}$$

which contradicts to $\delta _P(T)<1$.

Going further, we just emphasize that if T has a spectral gap, then one has $\Vert T^n-P\Vert \rightarrow 0$, which is called as a uniformP-ergodicity. Next sections will be devoted to this notion.

4 Uniformly P-ergodic operators

In this section, we study uniform P-ergodicities of Markov operators on abstract state spaces.

Definition 4.1

Let P be a projection on X. A bounded operator $T{:}X\rightarrow X$ is called uniformly P-ergodic if $\Vert T^n-P\Vert \rightarrow 0$, as $n\rightarrow \infty$.

Let us prove the following results for uniform P-ergodicity.

Proposition 4.2

Let P and Q be two projection operators on X with $Q\le P$ and let $T\in \Sigma _Q(X)$. If T is uniformly P-ergodic, then TQ is uniformly Q-ergodic.

Proof

Suppose that T is uniformly P-ergodic. Then $T^n\rightarrow P$ as $n\rightarrow \infty$, therefore we have $(TQ)^n=QT^n\rightarrow QP=Q$, which proves the statement. $\square$

Proposition 4.3

If T is uniformly P-ergodic operator on X, then $TP=PT=P$, and in addition, if $T\in \Sigma (X)$, then $P\in \Sigma (X)$.

Proof

Assume that T is uniformly P-ergodic. Then

$$\begin{aligned} T^{n+1}= TT^n\rightarrow TP, \end{aligned}$$

similarly

$$\begin{aligned} T^{n+1}= T^nT\rightarrow PT, \end{aligned}$$

so $PT=TP=P$.

As $T\in \Sigma (X)$, $T^n\in \Sigma (X)$, for all $n\in \mathbb {N}$. Therefore, for every $x\in \mathcal {K}$, one has $f(Px)=\lim \limits _{n\rightarrow \infty }f(T^nx)=1$, hence $P\in \Sigma (X)$. $\square$

Consequently, in the case of strong abstract state spaces, we deduce the following result.

Corollary 4.4

Let $(X,X_+,\mathcal {K},f)$ be a strong abstract state space, P be a projection on X and let $T\in \Sigma (X)$. If T is uniformly P-ergodic and $\delta _P(T)=0$, then $T=P$.

Proof

Directly follows by combining the previous proposition and Theorem 3.10. $\square$

Proposition 4.5

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space (i.e. $\lambda$-generating). If T is uniformly P-ergodic, then there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$.

Proof

The uniformly P-ergodicity of T implies the existence of an $n_0\in \mathbb {N}$ such that

$$\begin{aligned} \Vert T^{n_0}-P\Vert <\frac{1}{2\lambda }. \end{aligned}$$

By (iii) of Theorem 3.7, we have

$$\begin{aligned} \delta _P(T^{n_0})\le & {} \frac{\lambda }{2}\sup \Vert T^{n_0}u-T^{n_0}v\Vert \ \ \ (\ u,v\in \mathcal {K},\ \text {and}\ Pu=Pv)\\= & {} \frac{\lambda }{2}\sup \Vert T^{n_0}u-Pu+Pv - T^{n_0}v\Vert \\\le & {} \frac{\lambda }{2}( \sup \Vert T^{n_0}u-Pu\Vert + \sup \Vert T^{n_0}v-Pv \Vert )\\\le & {} \frac{\lambda }{2}( \Vert T^{n_0}-P\Vert + \Vert T^{n_0}-P \Vert )\\< & {} 1, \end{aligned}$$

which is the desired assertion. $\square$

Conversely, we have the following theorem:

Theorem 4.6

Let $T\in \Sigma _P(X)$ be such that $TP=P$. If there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$, then T is uniformly P-ergodic.

Proof

Assume that there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$. By Proposition 3.11

$$\begin{aligned} \Vert T^{n}(I-P)\Vert \rightarrow 0,\ \text {as}\ n\rightarrow \infty . \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert T^n-P\Vert =\Vert T^n-T^nP\Vert = \Vert T^n(I-P)\Vert \rightarrow 0, \end{aligned}$$

hence T is uniformly P-ergodic. $\square$

Corollary 4.7

Let $T\in \Sigma _P(X)$. Then T is uniformly P-ergodic if and only if

$$\begin{aligned} TP=P\ \text {and} \ \exists n_0\in \mathbb {N}\ \text {such that} \ \delta _P(T^{n_0})<1. \end{aligned}$$

Moreover, there are constants $C, \alpha \in \mathbb {R}_+$ and $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \left\| T^n - P \right\| \le C e^{-\alpha n}, \,\,\,\,\, \forall n\ge n_0. \end{aligned}$$

Now, we would like to provide an application of the deduced results above to the case of linear operators which are defined on arbitrary Banach spaces.

Theorem 4.8

Let X be any Banach space over $\mathbb {R}$. Assume that $T{:} X\rightarrow X$ is a linear bounded operator with $\Vert T\Vert \le 1$ and $P{:} X\rightarrow X$ is a projection operator with $T P = P T =P$. Then the following statements are equivalent:

• (i) T is uniformly P-ergodic;

• (ii) there is an $n_0\in \mathbb {N}$ such that $\Vert T^{n_0}_{|_{I-P}}\Vert <1$, where $T_{|_{I-P}}$ denotes the restriction of T to the subspace $(I-P)(X)$.

Proof

The implication (i)$\Rightarrow$(ii) is obvious. Let us prove (ii)$\Rightarrow$(i). First consider the abstract state space $(\mathcal {X},\mathcal {X}_+,\mathcal {K},f)$ which was introduced in Example 2.3-c. Define the operators $\mathcal {T},\mathcal {P}{:} \mathcal {X} \rightarrow \mathcal {X}$, respectively by

$$\begin{aligned} \mathcal {T}(\alpha , x)=(\alpha , Tx), \ \ \mathcal {P}(\alpha , x)=(\alpha , Px). \end{aligned}$$

It is clear that $\mathcal {T}$ and $\mathcal {P}$ are Markov operators. To prove that $\mathcal {T}$ is uniformly $\mathcal {P}$-ergodic, first we notice that

$$\begin{aligned} N_{\mathcal {P}}= \{(\alpha , x)\in \mathcal {X}{:} \ \mathcal {P}(\alpha , x)=0 \}=\{(0,x){:}\ x\in \text {ker}(P) \}. \end{aligned}$$

Therefore,

$$\begin{aligned} \delta _{\mathcal {P}}(\mathcal {T})= & {} \sup \{\Vert \mathcal {T}(\alpha , x)\Vert ;\ \Vert (\alpha , x)\Vert \le 1 \ \text {and}\ (\alpha , x) \in N_{\mathcal {P}} \}\\= & {} \sup \{\Vert (0, Tx)\Vert ;\ \Vert x\Vert \le 1 \ \text {and}\ x \in \text {ker}(P) \}\\= & {} \sup \{\Vert Tx\Vert ;\ \Vert x\Vert \le 1 \ \text {and}\ x \in (1-P)X\}\\= & {} \Vert T_{|_{I-P}}\Vert . \end{aligned}$$

Hence, from the condition we infer that $\delta _{\mathcal {P}}(\mathcal {T}^{n_0})<1$, then Theorem 4.6 implies $\mathcal {T}$ is uniformly $\mathcal {P}$-ergodic. Using the definition of the norm on $\mathcal {X}$, we obtain the required assertion. $\square$

Remark 4.9

A similar kind of result has been proved in [23]. An advantage of our approach is that we are working only with $\delta _P$, which will allow us to establish some category results for uniformly P-ergodic operators (see Theorem 4.12).

We now define a weaker condition than uniform P-ergodicity. Namely, a bounded linear operator $T{:}X\rightarrow X$ is called weaklyP-ergodic if

$$\begin{aligned} \delta _P(T^n)\rightarrow 0,\ \text {as}\ n\rightarrow \infty . \end{aligned}$$

The following result characterizes the concept of weak P-ergodicity of T.

Proposition 4.10

Let $T\in \Sigma _P(X)$. Then the following conditions are equivalent:

• (i) T is weakly P-ergodic;

• (ii) there exists an $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$.

Proof

• (i) $\Rightarrow$ (ii) If T is weakly P-ergodic, then it is obvious that there exists $n_0\in \mathbb {N}$ such that $\delta _P(T^{n_0})<1$.

• (ii) $\Rightarrow$ (i) Assume that such an $n_0\in \mathbb {N}$ exists and let $\rho =\delta _P(T^{n_0})$. Then for a large $n\in \mathbb {N}$, we write $n=kn_0+r\ (k,r\in \mathbb {N}\ \text {and}\ r<n_0)$ and by (vi) of Theorem 3.7, we have

  $$\begin{aligned} \delta _P(T^n)= \delta _P(T^{kn_0}T^r) \le \rho ^k \delta _P(T^r). \end{aligned}$$

  As n tends to 0, k also tends to 0, and hence the proof is completed. $\square$

Using Corollary 3.13, we immediately get the following fact.

Proposition 4.11

Let $T\in \Sigma _P(X)$. If T is weakly P-ergodic, then T is not weakly $(1-P)$-ergodic.

Let us now fix the following notations:

$$\begin{aligned} \Sigma _P^u(X)= & {} \{T\in \Sigma _P(X){:}\ T \ \text {is uniformly } P\text {-ergodic}\}, \\ \Sigma _P^w(X)= & {} \{T\in \Sigma _P(X){:}\ T \ \text {is weakly } P\text {-ergodic}\},\\ \Sigma _P^{inv}(X)= & {} \{T\in \Sigma _P(X){:}\ TP=P\}. \end{aligned}$$

Then, it is clear that

$$\begin{aligned} \Sigma _P^u(X)\subseteq \Sigma _P^w(X), \ \ \ \Sigma _P^u(X)\subseteq \Sigma _P^{inv}(X) \end{aligned}$$

Moreover,

$$\begin{aligned} \Sigma _P^u(X)= \Sigma _P^w(X)\cap \Sigma _P^{inv}(X). \end{aligned}$$

Theorem 4.12

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space and let P be a Markov projection on X. Then the set $\Sigma _P^{u}(X)$ is a norm dense and open subset of $\Sigma _P^{inv}(X)$.

Proof

Given any $T\in \Sigma _P^{inv}(X)$, $0<\varepsilon <2$, and let us denote

$$\begin{aligned} T^{(\varepsilon )}=\bigg (1-\frac{\varepsilon }{2}\bigg )T+\frac{\varepsilon }{2}P. \end{aligned}$$

It is clear that $T^{(\varepsilon )}\in \Sigma _P^{inv}(X)$ and

$$\begin{aligned} \Vert T-T^{(\varepsilon )}\Vert =\bigg \Vert \frac{\varepsilon }{2}P-\frac{\varepsilon }{2}T\bigg \Vert = \frac{\varepsilon }{2}\Vert P-T\Vert <\varepsilon . \end{aligned}$$

Now we show that $T^{(\varepsilon )}\in \Sigma _P^{u}(X)$. For all $x\in N_P$ by Lemma 3.6, $x=\alpha (x)(u-v),\ u,v\in \mathcal {K}$ with $u-v\in N_P$, and $0<\alpha (x)\le \frac{\lambda }{2}\Vert x\Vert$. Therefore,

$$\begin{aligned} \Vert T^{(\varepsilon )}(x)\Vert= & {} \alpha (x)\Vert T^{(\varepsilon )}(u-v)\Vert \\= & {} \alpha (x)\bigg \Vert \bigg (1-\frac{\varepsilon }{2}\bigg )T(u-v)+ \frac{\varepsilon }{2}P(u-v)\bigg \Vert \\= & {} \alpha (x)\bigg (1-\frac{\varepsilon }{2}\bigg )\bigg \Vert T(u-v)\bigg \Vert \\= & {} \bigg (1-\frac{\varepsilon }{2}\bigg )\Vert Tx\Vert \\\le & {} \bigg (1-\frac{\varepsilon }{2}\bigg )\Vert x\Vert , \end{aligned}$$

which implies $\delta _P(T^{(\varepsilon )})\le 1-\frac{\varepsilon }{2}$. Hence, by Theorem 4.6$T^{(\varepsilon )}\in \Sigma _P^u(X)$.

Now let us show that $\Sigma _P^{u}(X)$ is a norm open subset of $\Sigma _P^{inv}(X)$. First we establish that for every $n\in \mathbb {N}$, the set

$$\begin{aligned} \Sigma _{P,n}^{inv}(X)=\bigg \{T\in \Sigma _P^{inv}(X){:} \ \delta _P(T^n)< 1\bigg \} \end{aligned}$$

is an open subset of $\Sigma _P^{inv}(X)$. Indeed, take any $T\in \Sigma _{P,n}^{inv}(X)$ and letting $\alpha :=\delta _P(T^n)<1$, we choose $\beta$ such that $0<\beta <1$ and $\alpha +\beta <1$. Then, for any $H\in \Sigma _P^{inv}(X)$ with $\Vert H-T\Vert <\beta /n$ and using (ii) of Theorem 3.7, we obtain

$$\begin{aligned} |\delta _P(H^n)-\delta _P(T^n)|\le & {} \Vert H^n-T^n\Vert \\\le & {} \Vert H^{n-1}(H-T)\Vert +\Vert (H^{n-1}-T^{n-1})T\Vert \\\le & {} \Vert H-T\Vert +\Vert H^{n-1}-T^{n-1}\Vert \\&\vdots&\\\le & {} n\Vert H-T\Vert <\beta . \end{aligned}$$

Hence, the above inequality yields that $\delta _P(H^n)<\alpha +\beta <1$, i.e. $H\in \Sigma _{P,n}^{inv}(X)$. As

$$\begin{aligned} \Sigma _P^{u}(X)=\bigcup _{n\in \mathbb {N}}\Sigma _{P,n}^{inv}(X), \end{aligned}$$

we find that $\Sigma _P^{u}(X)$ is an open subset of $\Sigma _P^{inv}(X)$, which completes the proof. $\square$

Using the same arguments, one can prove the following theorem.

Theorem 4.13

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space and let P be a Markov projection on X. Then the set $\Sigma _P^{w}(X)$ is a norm dense and open subset of $\Sigma _P(X)$.

Remark 4.14

We notice that the Baire category theorem has a long history in ergodic theory [17], and it has many applications [4, 22]. Baire type considerations usually bring easy answers to existence problems. In [6] a particular case of Theorem 4.12 has been established for Markov operators, acting on the Schatten class $C_1$. We aim that our results in this direction will open new perspectives in the non-commutative ergodic theory.

5 Characterizations of uniformly P-ergodic Markov operators

In this section, we provide a large class of examples of uniformly P-ergodic operators on abstract state spaces. Precisely, we describe those uniformly P-ergodic operators in terms of the projection P. Afterwards, we use this characterization to deduce examples of uniformly P-ergodic on $\mathbb {R}^n$, on $\ell _1$ and on $L_1$- spaces.

Let X be an abstract state space. For an operator Q on X, let Rang(Q) and Fix(Q) denote the range and the fixed points of Q, respectively. We now prove the following auxiliary fact.

Lemma 5.1

Let X be a vector space, P be a projection operator on X and let Q be any operator on X. Then the following statements are equivalent:

• (i) $Rang(Q)\cap Fix(P)=\{0\}$ and $PQ=QP$;

• (ii) $PQ=QP=0$.

Proof

• $(i) \Rightarrow (ii)$ For every $x\in X$, $QPx\in Rang(Q)$. As

  $$\begin{aligned} P(QPx)= QP^2x=QPx, \end{aligned}$$

  we get $QPx \in Fix(P)$, then by the assumption $QPx=0$, and hence assertion (ii) follows.

• $(ii) \Rightarrow (i)$ Suppose that $PQ=QP=0$. If $x\in Rang(Q)\cap Fix(P)$, then, for some $s\in X$, one has

  $$\begin{aligned} x=Qs\ \ \text {and}\ \ Px=x. \end{aligned}$$

  Therefore,

  $$\begin{aligned} x=Px=P(Qs)=0, \end{aligned}$$

  which means the assertion (i). $\square$

Now, let us prove the following characterization result.

Theorem 5.2

Let P be a projection on X. Then T is uniformly P-ergodic if and only if T can be written as $T=P+Q$, where Q is an operator on X such that $PQ=QP=0$ and $\Vert Q^{n_0}\Vert <1$, for some $n_0\in \mathbb {N}$. Moreover, if $T\in \Sigma (X)$, then

$$\begin{aligned} \delta _P(T)\le \Vert Q\Vert \le 2\delta _P(T). \end{aligned}$$

Proof

Suppose that T is uniformly P-ergodic. Put $Q=T-P$, then Proposition 4.3 implies $PQ=QP=0$. Therefore, $T^n=P+Q^n$. Hence, the uniform P-ergodicity implies the existence of $n_0\in \mathbb {N}$ such that

$$\begin{aligned} \Vert Q^{n_0}\Vert =\Vert T^{n_0}-P\Vert <1. \end{aligned}$$

Conversely, suppose that $T=P+Q$ and Q satisfies the given hypotheses. Then for every $n\in \mathbb {N}$, we have

$$\begin{aligned} T^n=P+Q^n. \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert T^n-P\Vert =\Vert Q^n\Vert \le \Vert Q^{n_0}\Vert ^{[n/n_0]}\rightarrow 0\ \text {as}\ n\rightarrow \infty , \end{aligned}$$

so T is uniformly P-ergodic.

Now assume that T is a Markov operator. Then

$$\begin{aligned} \delta _P(T)= \sup _{x\in N_P,\ x\ne 0}\frac{\Vert Px+Qx\Vert }{\Vert x\Vert }= \sup _{x\in N_P,\ x\ne 0}\frac{\Vert Qx\Vert }{\Vert x\Vert } =\delta _P(Q)\le \Vert Q\Vert . \end{aligned}$$

Also, as $T\in \Sigma (X)$ we get $P\in \Sigma (X)$, Therefore, by Proposition 4.3

$$\begin{aligned} \Vert Q\Vert= & {} \Vert T-P\Vert \\= & {} \Vert T-TP\Vert \\= & {} \Vert T(I-P)\Vert \\\le & {} \delta _P(T)\Vert I-P\Vert \ \ \ (\text {using (v) of Theorem } 3.7)\\\le & {} 2\delta _P(T), \end{aligned}$$

This completes the proof. $\square$

From this theorem, we immediately get the following result.

Corollary 5.3

Let X be a normed space and let P be a projection on X. If Q is an operator on X such that $PQ=QP=0$, then $T=P+\frac{r}{\Vert Q\Vert }Q$ is uniformly P-ergodic, for all $r\in (-1,1)$.

The deduced results above enable us to produce several examples of uniformly P-ergodic operators.

Example 5.4

Let us consider $\mathbb {R}^n$ and we denote by $E_i\ (1\le i\le n)$ the diagonal matrix units in $\mathbb {M}_n(\mathbb {R})$. Then the operator

$$\begin{aligned} T= \sum _{i=1}^mE_i + \sum _{k=m+1}^{n}r_kE_k,\ \ r_k\in \mathbb {R}\ \text {and}\ |r_k|<1, \end{aligned}$$

is uniformly P-ergodic, where $P= \sum _{i=1}^mE_i$. As in Theorem 5.2, we have $Q=\sum _{k=m+1}^{n}r_kE_k$. Indeed, $PQ=QP=0$ and $\Vert Q\Vert <1$.

Next example shows that the commutativity of P and Q in Theorem 5.2 is a necessary condition.

Example 5.5

Let us consider the following operators

$$\begin{aligned} Q= \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \frac{1}{2} &{}\quad \frac{1}{4} \end{pmatrix} \ \text {and} \ P=\begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}. \end{aligned}$$

Then P is a projection, $\Vert Q\Vert <1$, $PQ=0$ but $QP\ne 0$. Letting $T=P+Q$, we get that

$$\begin{aligned} T^n=\begin{pmatrix} 1 &{} \quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad \frac{4^{n+1}-1}{6\cdot 4^n} &{}\quad \frac{1}{4^n} \end{pmatrix} \end{aligned}$$

converges to

$$\begin{aligned} \tilde{P}=\begin{pmatrix} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad \frac{2}{3} &{}\quad 0 \end{pmatrix}. \end{aligned}$$

Hence, T is uniformly $\tilde{P}$-ergodic, but not uniformly P-ergodic. Indeed, $T=\tilde{P}+\tilde{Q}$, where

$$\begin{aligned} \tilde{Q}= \begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad -\frac{1}{6} &{}\quad \frac{1}{4} \end{pmatrix}. \end{aligned}$$

Next example shows that uniform P-ergodicity does not imply quasi-compactness.

Example 5.6

Consider the space $\ell _1$, the subspaces $\mathcal {A}=\{x\in \ell _1;\ x_{2n}=0\}$ and the operator $P{:}\ell _1 \rightarrow \mathcal {A}$ defined by

$$\begin{aligned} P(x)= (x_1+x_2, 0, x_3+x_4, 0, \ldots ). \end{aligned}$$

Then P is a projection on $\mathcal {A}$. We construct a class of uniformly P-ergodic operators on $\ell _1$ as follows:

Let $\mathcal {Q}{:}\ell _1 \rightarrow \ell _1$ be the operator defined by

$$\begin{aligned} x \mapsto \left( \frac{-x_2}{2},\frac{x_2}{2}, \frac{-x_4}{2}, \frac{x_4}{2}, \ldots \right) . \end{aligned}$$

It is clear that $\mathcal {Q}^{n} \rightarrow 0$, so for some $n_0\in \mathbb {N}$, we have $\Vert \mathcal {Q}^{n_0}\Vert <1$. Also, $P\mathcal {Q}=\mathcal {Q}P=0$. Then by Theorem 5.2, we have that the operator $T=P+\mathcal {Q}$ is uniformly P-ergodic, but one can see that T is not quasi-compact [20].

Now in the following example we construct uniformly P-ergodic operators on $L_1$-space:

Example 5.7

Let $(S, \mathcal {B}, \mu )$ be a probability measure space and consider the space $X=L^1(S, \mathcal {B}, \mu )$. We construct a class of uniformly P-ergodic operators on X as follows:

Let $f_i(t)\in L^\infty (\mu )$, for $1\le i\le n$, and let $E_1$ denote the subspace generated by $span\{f_i\}$. If P is a projection operator from X onto $E_1$, then the operator P can be written as follows

$$\begin{aligned} (Pf)(t):=\sum _{i=1}^n \Gamma _i(f)f_i(t), \end{aligned}$$

where $\Gamma _i$ are linear functionals on X, which can be represented as

$$\begin{aligned} \Gamma _i(f)= \int _S f(t)\gamma _i(t)d\mu ,\ \forall f\in X \end{aligned}$$

with

$$\begin{aligned} \gamma _i\in L^\infty (\mu ),\ \text {such that}\ \int _S \gamma _i(t)f_j(t) d\mu =\delta _{i,j}. \end{aligned}$$

Similarly, let us construct another projection Q on X: Let $g_i(t)\in L^\infty (\mu )$, for $1\le i\le m$, and let $E_2$ denote the subspace generated by $span\{g_i\}$. Let Q be a projection operator from X onto $E_2$ which is defined by

$$\begin{aligned} (Qf)(t):=\sum _{i=1}^m \Lambda _i(f)g_i(t), \end{aligned}$$

where $\Lambda _i$ are linear functionals on X, which can be represented as

$$\begin{aligned} \Lambda _i(f)= \int _S f(t)\lambda _i(t)d\mu , \ \forall f\in X \end{aligned}$$

with

$$\begin{aligned} \lambda _i\in L^\infty (\mu ),\ \text {such that}\ \int _S \lambda _i(t)g_j(t) d\mu =\delta _{i,j}. \end{aligned}$$

In addition, we assume that the choice of $\lambda _j(t)$ and $\gamma _i(t)$ satisfying

$$\begin{aligned} \lambda _j(t)f_i(t)=0\ \mu \ \text {a.e. and} \ \gamma _j(t)g_i(t)=0 \ \mu \ \text {a.e.} \end{aligned}$$

(10)

Then P and Q are projections from X onto $E_1$ and $E_2$, respectively. To show that $QP=0$, let $f\in X$ then we have

$$\begin{aligned} QP(f)= & {} Q\left( \sum _{i=1}^n \Gamma _i(f)f_i(t)\right) \\= & {} \sum _{i=1}^n \Gamma _i(f)Q(f_i(t))\\= & {} \sum _{i=1}^n \Gamma _i(f)\sum _{j=1}^m \Lambda _j(f_i)g_j(t)\\= & {} \sum _{i=1}^n\sum _{j=1}^m \Gamma _i(f) \Lambda _j(f_i)g_j(t)\\= & {} 0, \end{aligned}$$

since $\Lambda _j(f_i) = 0$ (see, (10)). Similarly, by the second part of (10) we get $PQ(f)=0$, for all $f\in X$. Therefore, Corollary 5.3 implies that $T=P+rQ$ is a uniformly P-ergodic operator on X, for all $r\in (-1,1)$.

6 On uniform and weak mean ergodicities

In this section, we are going to investigate uniform mean ergodicities of Markov operator.

Given a bounded linear operator $T{:}X\rightarrow X$, we set

$$\begin{aligned} A_n(T) =\frac{1}{n}\sum _{k=1}^{n}T^k. \end{aligned}$$

Recall that $T{:}X\rightarrow X$ is said to be

• (a) mean ergodic if for every $x\in X$

  $$\begin{aligned} \lim _{n\rightarrow \infty }A_n(T)x=Qx; \end{aligned}$$

• (b) uniformly mean ergodic if

  $$\begin{aligned} \lim _{n\rightarrow \infty }\Vert A_n(T)-Q\Vert =0; \end{aligned}$$

for some operator Q on X.

In this setting, it is well-known that Q is a projection [26], which is called the limiting projection ofT, and denoted by $Q_T$. Moreover, if $T\in \Sigma (X)$, then $Q_T$ is also Markov.

By analogy with the weak P-ergodicity, one may introduce the following notion. A linear operator T is called weaklyP-mean ergodic if

$$\begin{aligned} \lim _{n\rightarrow \infty }\delta _P(A_n(T))=0. \end{aligned}$$

It is clear that any uniformly mean ergodic operator is weakly $Q_T$-mean ergodic.

By Theorem 4.6, we obtain the following result.

Corollary 6.1

Assume that $T\in \Sigma (X)$ and T is mean ergodic with its limiting projection $Q_T$. If there exists an $n_0\in \mathbb {N}$ such that $\delta _{Q_T}(T^{n_0})<1$, then T is uniformly $Q_T$-ergodic.

Theorem 6.2

Assume that $T\in \Sigma (X)$ and T is mean ergodic with its limiting projection $Q_T$. If $T\in \Sigma _P^w(X)$, for some P, then $Q_T\le P$.

Proof

Suppose that $T\in \Sigma _P^w(X)$, so $\delta _P(T^{n_0})<1$ for some $n_0\in \mathbb {N}$. Then by Proposition 3.11, we have

$$\begin{aligned} \Vert T^{n}(I-P)\Vert \rightarrow 0. \end{aligned}$$

As $TQ_T=Q_T$, $A_n(T)Q_T=Q_TA_n(T)=Q_T$. Then

$$\begin{aligned} \Vert Q_T(I-P)\Vert= & {} \Vert Q_TA_{n}(T)(I-P)\Vert \\\le & {} \Vert A_{n}(T)(I-P)\Vert \\\le & {} \frac{1}{n}\sum _{k=1}^n\Vert T^k(I-P)\Vert \rightarrow 0, \end{aligned}$$

so $Q_T(I-P)=0$ which implies $Q_T=Q_TP$.

On the other hand,

$$\begin{aligned} \Vert (I-P)Q_T\Vert= & {} \Vert (I-P)A_n(T)Q_T\Vert \\\le & {} \Vert (I-P)A_n(T)\Vert \Vert Q_T\Vert \\\le & {} \Vert A_n(T)(I-P)\Vert \rightarrow 0, \end{aligned}$$

so $(I-P)Q_T=0$ which implies $Q_T=Q_TP$, and hence $Q_T\le P$. $\square$

It is natural to ask: when mean ergodic operator would be uniformly mean ergodic? Next result clarifies this question in terms of $\delta _P$.

Theorem 6.3

Assume that $T\in \Sigma (X)$ and T is mean ergodic with its limiting projection $Q_T$. Then the following statements are equivalent:

• (i) T is uniformly mean ergodic;

• (ii) there exists an $n_0\in \mathbb {N}$ such that $\delta _{Q_T}(A_{n_0}(T))<1$. Moreover,

  $$\begin{aligned} \Vert A_n(T)-Q_T\Vert \le \frac{2(n_0+1)}{1-\delta _{Q_T}(A_{n_0}(T))}\cdot \frac{1}{n}. \end{aligned}$$

Proof

We note that if $T=I$, then $Q_T=I$ and according to the definition $\delta _{Q_T}(T)=1$, hence the statement of the theorem follows. Therefore, in what follows it is always assumed $T\ne I$. The implications (i) $\Rightarrow$ (ii) directly follows using the same arguments as in the proof of Proposition 4.5, replacing $T^n$ by $A_n(T)$ and P by $Q_T$.

(ii) $\Rightarrow$ (i). Assume that $\rho = \delta _{Q_T}(A_{n_0}(T))<1$, for some $n_0\in \mathbb {N}$. Then

$$\begin{aligned} A_n(T)(I-T)= & {} A_n(T)-A_n(T)T \\= & {} \frac{1}{n}\sum _{k=0}^{n-1}T^k- \frac{1}{n}\sum _{k=0}^{n-1}T^{k+1}\\= & {} \frac{1}{n}(I-T^n), \end{aligned}$$

so, $\Vert A_n(T)(I-T)\Vert \le \frac{2}{n}$, and then

$$\begin{aligned} \Vert A_n(T)(I-T^k)\Vert \le \frac{2k}{n},\ k\in \mathbb {N}\end{aligned}$$

which implies

$$\begin{aligned} \Vert A_n(T)(I-A_{n_0}(T))\Vert= & {} \bigg \Vert A_n(T)\bigg (\frac{1}{n_0}\sum _{k=1}^{n_0}(I-T^k)\bigg )\bigg \Vert \\\le & {} \frac{1}{n_0}\sum _{k=1}^{n_0}\Vert A_n(T)(I-T^k)\Vert \\\le & {} \frac{n_0+1}{n}. \end{aligned}$$

Therefore,

$$\begin{aligned} \delta _{Q_T}(A_n(T)(I-A_{n_0}(T))) \le \frac{n_0+1}{n}. \end{aligned}$$

(11)

Using Properties (ii) and (vi) of Theorem 3.7, we have

$$\begin{aligned} \delta _{Q_T}(A_n(T)(I-A_{n_0}(T)))\ge & {} \delta _{Q_T}(A_n(T)) - \delta _{Q_T}(A_n(T)A_{n_0}(T))\\\ge & {} \delta _{Q_T}(A_n(T)) - \delta _{Q_T}(A_n(T))\delta _{Q_T}(A_{n_0}(T))\\= & {} \delta _{Q_T}(A_n(T))(1-\rho ). \end{aligned}$$

By (11) and as $\rho <1$, we have

$$\begin{aligned} \delta _{Q_T}(A_n(T))\le \frac{n_0+1}{1-\rho }\cdot \frac{1}{n}. \end{aligned}$$

(12)

Now,

$$\begin{aligned} \delta _{Q_T}(A_n(T))= & {} \sup _{y\in N_{Q_T}}\frac{\Vert A_n(T)y\Vert }{\Vert y\Vert }\\\ge & {} \sup _{x\in X} \frac{\Vert A_n(T)x-A_n(T)Q_Tx\Vert }{\Vert x-Q_Tx\Vert }\ \ \ \ \ (\text {for}\ y=x-Q_Tx)\\= & {} \sup _{x\in X} \frac{\Vert A_n(T)x-Q_Tx\Vert }{\Vert x-Q_Tx\Vert }\ \ \ (\text {since}\ \ A_n(T)Q_T=Q_T)\\\ge & {} \frac{1}{2}\sup _{x\in X} \frac{\Vert A_n(T)x-Q_Tx\Vert }{\Vert x\Vert }\\= & {} \frac{1}{2} \Vert A_n(T)-Q_T\Vert . \end{aligned}$$

Then by (12)

$$\begin{aligned} \Vert A_n(T)-Q_T\Vert \le \frac{2(n_0+1)}{1-\rho }\cdot \frac{1}{n} \end{aligned}$$

which yields the desired assertion. $\square$

Now, we are going to introduce an abstract analogue of the well-known Doeblin’s Condition [39].

Definition 6.4

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space, whose cone $X_+$ is $\lambda$-generating, let P be a Markov projection on X, and let $T\in \Sigma _P (X)$. We say that T satisfies condition$\mathfrak {D}_m$ if there exists a constant $\tau \in (0,1]$ and an integer $n_0\in \mathbb {N}$ and for every $x,y\in \mathcal {K}$ with $x-y\in N_P$, there exists $z_{xy}\in \mathcal {K}$ and $\varphi _{xy}\in X_{+}$ with

$$\begin{aligned} \sup \limits _{xy}\Vert \varphi _{xy}\Vert \le \eta , \end{aligned}$$

where

$$\begin{aligned} 0\le \eta <\tau +\frac{1}{\lambda }-1, \end{aligned}$$

(13)

such that

$$\begin{aligned} A_{n_0}(T)x+\varphi _{xy}\ge \tau z_{xy}, \ \ A_{n_0}(T)y+\varphi _{xy}\ge \tau z_{xy}. \end{aligned}$$

(14)

The next result characterize the weakly P-mean ergodic Markov operators in terms of the above condition $\mathfrak {D}_m$.

Theorem 6.5

Let $(X,X_+,\mathcal {K},f)$ be an abstract state space whose cone $X_+$ is $\lambda$-generating, and let P be a Markov projection on X. Assume that $T\in \Sigma _P (X)$. Then the following conditions are equivalent:

• (i) T satisfies condition $\mathfrak {D}_m$;

• (ii) there is an $n_0\in \mathbb {N}$ such that $\delta _P(A_{n_0}(T))<1$;

• (iii) T is weakly P-mean ergodic.

Proof

(i) $\Rightarrow$ (ii). By condition $\mathfrak {D}_m$, there is a $\tau \in (0,1]$, $n_0\in \mathbb {N}$ and for any two elements $x,y\in \mathcal {K}$ with $x-y\in N_P$, there exist $z_{xy}\in \mathcal {K}$, $\varphi _{xy}\in X_{+}$ with

$$\begin{aligned} \sup \limits _{xy}\Vert \varphi _{xy}\Vert \le \eta \end{aligned}$$

(15)

such that

$$\begin{aligned} A_{n_0}(T)x+\varphi _{xy}\ge \tau z_{xy}, \ \ A_{n_0}(T)y+\varphi _{xy}\ge \tau z_{xy}. \end{aligned}$$

(16)

Using the Markovianity of T, and the inequalities (16) with (15), we obtain

$$\begin{aligned} \Vert A_{n_0}(T)x+\varphi _{xy}-\tau z_{xy}\Vert= & {} f(A_{n_0}(T)x+\varphi _{xy}-\tau z_k)\\= & {} 1-(\underbrace{\tau -f(\varphi _{xy})}_{c})\\= & {} 1-c\le 1-(\tau -\eta ). \end{aligned}$$

By the same argument, one finds

$$\begin{aligned} \Vert A_{n_0}(T)y+\varphi _{xy}-\tau z_{xy}\Vert =1-c\le 1-(\tau -\eta ) \end{aligned}$$

Let us denote

$$\begin{aligned} x_1= & {} \frac{1}{1-c}(A_{n_0}(T)x+\varphi _{xy}-\tau z_{xy}),\\ y_1= & {} \frac{1}{1-c}(A_{n_0}(T)y+\varphi _{xy}-\tau z_{xy}). \end{aligned}$$

It is clear that both $x_1,y_1\in \mathcal {K}$.

So,

$$\begin{aligned} \Vert A_{n_0}(T)x-A_{n_0}(T)y\Vert =(1-c)\Vert x_1-y_1\Vert \le 2\bigg (1-(\tau -\eta )\bigg ). \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\lambda }{2}\Vert A_{n_0}(T)x-A_{n_0}(T)y\Vert \le \lambda \bigg (1-(\tau -\eta )\bigg ). \end{aligned}$$

(17)

By (13) and (iii) of Theorem 3.7, and using (17) we obtain,

$$\begin{aligned} \delta _P(A_{n_0}(T))\le \mu <1, \end{aligned}$$

where $\mu =\lambda (1-\tau +\eta )$, hence (ii) follows.

The implication (ii) $\Rightarrow$ (iii) immediately follows from the proof of the implication (ii) $\Rightarrow$ (i) of Theorem 6.3. Therefore, it is enough to establish (iii) $\Rightarrow$ (i). Assume that T is weakly P-mean ergodic. Then

$$\begin{aligned} \sup _{x,y\in \mathcal {K}, x-y\in N_P}\Vert A_n(T)x-A_n(T)y\Vert \rightarrow 0\ \ \ \text {as} \ \ n\rightarrow \infty . \end{aligned}$$

Therefore, one can find $n_0\in \mathbb {N}$ such that

$$\begin{aligned} \Vert A_{n_0}(T)x-A_{n_0}(T)y\Vert \le \frac{1}{4\lambda ^2}, \ \ \text {for all} \ \ x,y\in \mathcal {K}, x-y\in N_P. \end{aligned}$$

(18)

Now pick any $y_0\in \mathcal {K}$ with $x-y_0\in N_p$ and $y-y_0\in N_P$. Due to Lemma 3.6 we decompose

$$\begin{aligned} A_{n_0}(T)x-A_{n_0}(T)y_0= & {} (A_{n_0}(T)x-A_{n_0}(T)y_0)_+-(A_{n_0}(T)x-A_{n_0}(T)y_0)_-\nonumber \\ A_{n_0}(T)y-A_{n_0}(T)y_0= & {} (A_{n_0}(T)y-A_{n_0}(T)y_0)_+-(A_{n_0}(T)y-A_{n_0}(T)y_0)_-.\nonumber \\ \end{aligned}$$

(19)

Denote

$$\begin{aligned} \varphi _{x}=(A_{n_0}(T)x-A_{n_0}(T)y_0)_-, \ \ \varphi _{y}=(A_{n_0}(T)y-A_{n_0}(T)y_0)_- \end{aligned}$$

and define

$$\begin{aligned} \varphi _{xy}=\varphi _{x}+\varphi _{y}. \end{aligned}$$

It is clear that $\varphi _{xy}\in X_+$ and from (18) with Lemma 3.6, one gets

$$\begin{aligned} \sup _{x,y\in \mathcal {K}, x-y\in N_P}\Vert \varphi _{xy}\Vert \le \frac{1}{4\lambda }. \end{aligned}$$

Moreover, by (19) we obtain

$$\begin{aligned} A_{n_0}(T)x+\varphi _{xy}\ge & {} A_{n_0}(T)x+\varphi _{x}\\= & {} A_{n_0}(T)y_0+A_{n_0}(T)x-A_{n_0}(T)y_0+\varphi _{x}\\= & {} A_{n_0}(T)y_0+(A_{n_0}(T)x-A_{n_0}(T)y_0)_+\\\ge & {} A_{n_0}(T)y_0. \end{aligned}$$

Similarly, one gets

$$\begin{aligned} A_{n_0}(T)x+\varphi _{xy}\ge A_{n_0}(T)y_0. \end{aligned}$$

Now, by denoting $\tau =1$, $\eta =\frac{1}{4\lambda }$ and $z_{xy}=A_{n_0}(T)y_0$, we infer that the operator T satisfies the condition $\mathfrak {D}_m$. This completes the proof. $\square$

Remark 6.6

We notice that if in the condition $\mathfrak {D}_m$ one replaces $A_n(T)$ with some power of T, then we obtain the Deoblin’s condition for T which has been investigated in [10, 35, 37, 41]. We think that such type of result is even a new in the classical, i.e. X is taken as an $L^1$-space.

In the next example by means of Theorem 6.5, we show that weakly P-mean ergodic operator is not necessary to be uniformly mean ergodic.

Example 6.7

Recall Example 2.5(3). Namely, $X=C[0,1]$ with the cone

$$\begin{aligned} X_+=\big \{x\in X{:} \ \max _{0\le t\le 1}|x(t)-x(1)|\le 2 x(1)\big \}. \end{aligned}$$

Consider the Markov operator $T{:}X\rightarrow X$ given by $(Tx)(t)=tx(t).$

Let us establish that T satisfies the condition $\mathfrak {D}_m$. First, it is noted that

$$\begin{aligned} (A_n(T)x)(t)=\frac{1}{n}\frac{t-t^{n+1}}{1-t}x(t). \end{aligned}$$

We assume that $Px=x(1)$. Now take $x,y\in \mathcal {K}$. Put $\varphi _{xy}\equiv 0$, $\tau =1$ and $z_{xy}=c$, $c\in (0,1/2)$. Then the inequalities $A_{n_0}x\ge \tau z_{xy}$, $A_{n_0}y\ge \tau z_{xy}$ are equivalent to $A_{n_0}x-\tau z_{xy},A_{n_0}y-\tau z_{xy}\in X_+$, which is equivalent to

$$\begin{aligned} \max _{0\le t\le 1}|(A_{n_0}x)(t)-(A_{n_0}x)(1)|\le & {} 2\big ((A_{n_0}x)(1)-z_{xy}\big ), \\ \max _{0\le t\le 1}|(A_{n_0}y)(t)-(A_{n_0}y)(1)|\le & {} 2\big ((A_{n_0}y)(1)-z_k\big ). \end{aligned}$$

The last one can be rewritten as follows:

$$\begin{aligned} \max _{0\le t\le 1}\bigg |\frac{1}{n_0}\frac{t-t^{n_0+1}}{1-t}x(t)-x(1)\bigg |\le & {} 2(x(1)-c),\\ \max _{0\le t\le 1}\bigg |\frac{1}{n_0}\frac{t-t^{n_0+1}}{1-t}y(t)-y(1)\bigg |\le & {} 2(y(1)-c). \end{aligned}$$

Taking into account $x,y\in \mathcal {K}$, from the last ones, we have

$$\begin{aligned} \max _{0\le t\le 1}\bigg |\frac{1}{n_0}\frac{t-t^{n_0+1}}{1-t}x(t)-1)\bigg |\le & {} 2(1-c),\end{aligned}$$

(20)

$$\begin{aligned} \max _{0\le t\le 1}\bigg |\frac{1}{n_0}\frac{t-t^{n_0+1}}{1-t}y(t)-1\bigg |\le & {} 2(1-c). \end{aligned}$$

(21)

From the last expressions, we infer the existence of $n_0$ such that inequalities (20) and (21) are satisfied. This, due to Theorem 6.5, yields that T satisfies the condition $\mathfrak {D}_m$. Hence, T is weakly P-mean ergodic. However, one can see that T is not uniformly means ergodic.

Now, we give an application of Theorem 6.3.

Theorem 6.8

Let X be a Banach space, $T{:}X\rightarrow X$ be a mean ergodic operator on X with $\Vert T\Vert \le 1$ and let P be a Markov projection on X. Then the following statements are equivalent:

• (i) there exists an $n_0\in \mathbb {N}$ such that $\Vert A_{n_0}(T)_{|_{I-P}}\Vert <1$;

• (ii) T is uniformly mean ergodic.

Proof

(i)$\Rightarrow$(ii). Now consider the abstract state space $(\mathcal {X}, \mathcal {X}_+,\mathcal {K},f)$ and the linear operator $\mathcal {T}(\alpha , x)=(\alpha , T(x))$. Due to Theorem 4.8, the operator $\mathcal {T}$ is Markov. Moreover, for every $(\alpha , x)\in \mathcal {X}$, one has

$$\begin{aligned} A_n(\mathcal {T})(\alpha , x)= & {} \frac{1}{n}\sum _{k=1}^n \mathcal {T}^k(\alpha , x)\\= & {} \frac{1}{n}\sum _{k=1}^n (\alpha , T^k(x))\\= & {} (\alpha , A_n(T)(x)). \end{aligned}$$

Hence, the mean ergodicity of T implies the convergence of $\{A_n(\mathcal {T})(\alpha , x)\}$, which shows that $\mathcal {T}$ is mean ergodic with its limiting projection $\mathcal {P}$. By the proof of the implication (ii)$\Rightarrow$(i) in Theorem 4.8, we have

$$\begin{aligned} \delta _{\mathcal {P}}(A_n(\mathcal {T}))= \Vert A_n(T)_{|_{I-P}}\Vert , \end{aligned}$$

hence, from the hypothesis of the theorem, for some $n_0\in \mathbb {N}$, one has

$$\begin{aligned} \delta _{\mathcal {P}}(A_{n_0}(\mathcal {T}))<1. \end{aligned}$$

So, Theorem 6.3 yields that $\mathcal {T}$ is uniformly mean ergodic, which implies the uniform mean ergodicity of T.

The implication (ii)$\Rightarrow$(i) can be proved in the reverse order. $\square$

Remark 6.9

We notice that in [29] relations between the uniform mean ergodicity and uniform convergence of the Abel averages have been studied.

7 Perturbation bounds and uniform P-ergodicity of Markov operators

This section is devoted to perturbation bounds for uniformly P-ergodic Markov operators. The case when P is a one-dimensional projection, this type of questions have been studied in [14, 32, 43]. For general projections, these kinds of bounds have not been investigated. Therefore, results of this section are new even in the classical case as well.

Recall that if T is uniformly P-ergodic, then by Corollary 4.7 there are constants $C, \alpha \in \mathbb {R}_+, n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \left\| T^n - P \right\| \le C e^{-\alpha n}, \,\,\,\,\, \forall n\ge n_0. \end{aligned}$$

In this section, we prove perturbation bounds in terms of C and $e^{\alpha }$. Moreover, we also give several bounds in terms of the Dobrushin’s ergodicity coefficient.

Theorem 7.1

Let $(X, X_+, \mathcal {K}, f)$ be an abstract state space (i.e. $\lambda$-generating), P be a projection on X and let $T,S\in \Sigma _P^{inv}(X)$. If $T\in \Sigma _P^u(X)$, then

$$\begin{aligned} \displaystyle&\left\| T^n x- S^n z \right\| \le {\left\{ \begin{array}{ll} \left\| x - z \right\| + n \left\| T-S\right\| , &{}\forall n \le \tilde{n},\\ \lambda C e^{-\alpha n} \left\| x- z \right\| + \big ( \tilde{n}+ \lambda C \frac{e^{-\alpha \tilde{n}} - e^{-\alpha n}}{1 - e^{-\alpha }}\big ) \left\| T-S\right\| , &{} \forall n \ge \tilde{n}+1 \end{array}\right. } \end{aligned}$$

where $\displaystyle \tilde{n}:= \bigg [\frac{\log (1/C)}{\log e^{-\alpha }}\bigg ]$, $C, \alpha \in \mathbb {R}_+$, $x,z\in \mathcal {K}$ and $x-z\in N_P$.

Proof

For every $n\in \mathbb {N}$, by induction, we have

$$\begin{aligned} S^n = T^n + \sum _{i=0}^{n-1} T^{n-i-1} \circ (S-T) \circ S^i. \end{aligned}$$

(22)

Let $x,z\in \mathcal {K}$ and $\ x-z\in N_P$. Then it follows from (22) that

$$\begin{aligned} T^n x - S^n z= & {} T^n x - T^n z - \sum _{i=0}^{n-1} T^{n-i-1} \circ (S-T) \circ S^iz \\= & {} T^n (x - z) - \sum _{i=0}^{n-1} T^{n-i-1} \circ (S-T) (z_i), \end{aligned}$$

where $z_i=S^iz$. Hence,

$$\begin{aligned} \left\| T^n x - S^n z \right\| \le \left\| T^n (x - z) \right\| + \sum _{i=0}^{n-1} \left\| T^{n-i-1} \circ (S-T) (z_i) \right\| . \end{aligned}$$

As $T,S\in \Sigma _p^{inv}(X)$, we have $P(S-T)=0$ and due to (v) of Theorem (3.7), one finds

$$\begin{aligned} \left\| T^{n-i-1} (S-T) (z_i) \right\| \le \delta _P(T^{n-i-1}) \left\| S-T\right\| , \end{aligned}$$

and

$$\begin{aligned} \left\| T^n (x - z) \right\| \le \delta _P(T^n) \left\| x -z\right\| . \end{aligned}$$

Hence,

$$\begin{aligned} \left\| T^n x - S^n z \right\|\le & {} \delta _P(T^n) \left\| x - z\right\| + \sum _{i=0}^{n-1} \delta _P(T^{n-i-1}) \left\| S-T \right\| \nonumber \\= & {} \delta _P(T^n) \left\| x - z\right\| + \left\| S-T \right\| \sum _{i=0}^{n-1} \delta _P (T^{i}). \end{aligned}$$

(23)

By

$$\begin{aligned} \left\| T^i u- T^i v \right\| \le \left\| T^i u- Pu\right\| + \left\| Pv- T^i v \right\| , \end{aligned}$$

with the fact $Pu=Pv$, and due to (iii) of Theorem (3.7), one gets

$$\begin{aligned} \displaystyle \delta _P(T^i ) \le \frac{\lambda }{2} \sup _{u,v\in \mathcal {K},u-v\in N_P} \left\| T^i u- T^i v \right\| \le \lambda \sup _{u\in \mathcal {K}} \left\| T^i u- P u\right\| . \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle \delta _P(T^n) \le {\left\{ \begin{array}{ll} 1, &{}\forall n \le \tilde{n},\\ \lambda C e^{-\alpha n}, &{} \forall n \ge \tilde{n}+1 \end{array}\right. } \end{aligned}$$

(24)

where $\tilde{n} = \bigg [\frac{\log (1/C)}{\log e^{-\alpha }}\bigg ]$.

From (24) we obtain

$$\begin{aligned} \displaystyle \sum _{i=0}^{n-1} \delta _P(T^i )= & {} \sum _{i=0}^{\tilde{n}-1} \delta _P(T^i) + \sum _{i=\tilde{n}}^{n-1} \delta _P(T^i) \nonumber \\\le & {} \tilde{n} + \sum _{i=\tilde{n}}^{n-1} \lambda C e^{-\alpha i} \nonumber \\= & {} \tilde{n} + \lambda C e^{-\alpha \tilde{n}} \frac{1-e^{-\alpha (n- \tilde{n})}}{1- e^{-\alpha }}, \, \, \forall n \ge \tilde{n}+1. \end{aligned}$$

(25)

Hence, we get the required assertion. $\square$

Corollary 7.2

Assume that the same hypotheses of Theorem 7.1 are satisfied. Then, for all $x \in \mathcal {K}$

$$\begin{aligned} \displaystyle&\left\| T^n x- S^n x \right\| \le {\left\{ \begin{array}{ll} n \left\| T-S\right\| , &{}\forall n \le \tilde{n},\\ \big ( \tilde{n}+ \lambda C \frac{e^{-\alpha \tilde{n}} - e^{-\alpha n}}{1 - e^{-\alpha }}\big ) \left\| T-S\right\| , &{} \forall n\ge \tilde{n}+1 \end{array}\right. } \end{aligned}$$

here as before, $\displaystyle \tilde{n}:= \bigg [\frac{\log (1/C)}{ \log e^{-\alpha }}\bigg ]$, $C,\alpha \in \mathbb {R}_+$.

The following theorem gives an alternative method of obtaining perturbation bounds in terms of $\delta _p(T^m)$.

Theorem 7.3

Let $(X, X_+, \mathcal {K}, f)$ be an abstract state space, P be a projection on X and let $S,T\in \Sigma ^{inv}_P(X)$. If $\delta _P (T^m) < 1$ holds for some $m\in \mathbb {N}$, then for every $x,z\in \mathcal {K}$ with $x-z\in N_P$ one has

$$\begin{aligned} \left\| T^n x - S^n z \right\|\le & {} \delta _P(T^m)^{\lfloor n/m \rfloor } \big (\left\| x - z\right\| + \max _{0< i< m} \left\| T^i - S^i \right\| \big ) \nonumber \\&+ \frac{1 - \delta _P(T^m)^{\lfloor n/m\rfloor }}{1 - \delta _P(T^m)} \left\| T^m - S^m\right\| , \,\,\,\, n \in \mathbb {N}. \end{aligned}$$

(26)

Proof

For any $n\le m$, due to $T^n x - S^n z = S^n (x - z) + (T^n - S^n) x$, we get

$$\begin{aligned} \left\| T^n x- S^n z \right\|\le & {} \left\| x -z\right\| + \left\| T^n - S^n \right\| \nonumber \\\le & {} \left\| x -z\right\| +\max \limits _{0<i<m} \left\| T^i -S^i\right\| . \end{aligned}$$

(27)

If $n < m$, then Eq. (26) reduces to (27). If $n\ge m$, we obtain

$$\begin{aligned} T^n x - S^n z= & {} T^m (T^{n-m} x) - S^m (S^{n-m} z) \\= & {} T^m (T^{n-m} x - S^{n-m} z) + (T^m - S^m) S^{n-m} z. \end{aligned}$$

Therefore, keeping in mind $S,T\in \Sigma ^{inv}_P(X)$ one finds

$$\begin{aligned} \left\| T^n x - S^n z \right\| \le \delta _P(T^m)\left\| T^{n-m} x - S^{n-m} z\right\| + \left\| T^m - S^m \right\| . \end{aligned}$$

Applying this relation to

$$\begin{aligned} \left\| T^{n-m} x - S^{n-m} z\right\| , \ldots , \left\| T^{n- m(\lfloor n/m\rfloor - 1)} x - S^{n- m(\lfloor n/m\rfloor - 1)} z \right\| \end{aligned}$$

and using (27) to bound $\left\| T^{n- m \lfloor n/m\rfloor } x- S^{n- m \lfloor n/m\rfloor } z \right\|$, we obtain

$$\begin{aligned} \left\| T^n x - S^n z \right\|\le & {} \delta _P(T^m)^{\lfloor n/m \rfloor } (\left\| x - z\right\| + \max _{0< i< m} \left\| T^i - S^i \right\| )\\&+ \bigg (\delta _P(T^m)^{\lfloor n/m \rfloor - 1} + \delta _P (T^m)^{\lfloor n/m \rfloor - 2} + \cdots + 1\bigg ) \left\| T^m - S^m \right\| , \\= & {} \delta _P(T^m)^{\lfloor n/m \rfloor } (\left\| x - z\right\| + \max _{0< i< m} \left\| T^i - S^i \right\| )\\&+ \frac{ 1- \delta _P(T^m)^{\lfloor n/m \rfloor }}{1-\delta _P(T^m)} \left\| T^m - S^m \right\| . \end{aligned}$$

The proof is completed. $\square$

Consequently, we get the following corollary which allows to estimate the dynamics of S to its fixed points set.

Corollary 7.4

Assume that the same hypotheses of Theorem 7.3 are satisfied. Then, for every $x \in \mathcal {K}$

$$\begin{aligned} \left\| S^n x-Px \right\|\le & {} \delta _P(T^m)^{\lfloor n/m \rfloor } \big (\left\| x - Px\right\| + \max _{0< i< m} \left\| T^i - S^i \right\| \big ) \\&+ \,\frac{1 - \delta _P(T^m)^{\lfloor n/m\rfloor }}{1 - \delta _P(T^m)} \left\| T^m - S^m\right\| , \,\,\,\, n \in \mathbb {N}. \end{aligned}$$