On asymptotic periodicity of kernel double Markovian operators

Abstract

It is proved that a kernel, doubly Markovian operator T is asymptotically periodic if and only if its deterministic $\sigma$-field $\varSigma _d(T)$ (equivalently $\varSigma _d(T^*)$) is finite. It follows that kernel doubly Markovian operator T is asymptotically periodic if and only if $T^*$ is asymptotically periodic.

1 Introduction

Stochastic Markov processes may be described, defined and studied with the use of different mathematical theories. Depending on needs and context we may introduce them as a family of random variables (elements) $\{ \xi _t \}_{t\in \varTheta }$, defined on a fixed probability space and satisfying the so-called Markov property, or by the system of transition probabilities $\{ P(x,A) : x\in S, \ A \in {\mathcal {G}} \}$, where $(S, {\mathcal {G}})$ is a fixed measurable space (called a phase space). In another approach, one can entirely use functional analysis language and say that a Markov process is exactly a linear, positive operator T (called Markov or stochastic) defined on some Banach function space ${\mathfrak {X}} \ ( = L^p$ or C(S)), which satisfies other additional properties (for instance $T\mathbf{1} = \mathbf{1}$ if $\mathbf{1} \in {\mathfrak {X}}$). The methods and techniques of functional analysis are commonly accepted in a noncommutative Markov processes theory.

The evolution and asymptotic properties of a Markov process $\{ \xi _t \}_{t\in \varTheta }$ (once it is carried over to the operator theory framework) is reflected as an asymptotic behaviour (limit) of iterates $T^n$. There are many monographs on Markov processes, where probability theory and functional analysis intertwine. We shall borrow ideas, notions and inspiration from [8], which still may serve as a very good source of knowledge on limits of Markov processes (mainly for discrete time). Some other monographs [4, 6, 18, 19, 21, 23] were used once working on this project.

We revisit the notion of smoothness (smoothing) defined as some property of trajectories $T^nf$, where $f \in L^1(\mu )$ are densities (i.e. $f\ge 0$ and $\int f d\mu = 1$). The general idea of smoothing says that as long as integrals $\int _E T^nf d\mu$ do not go uncontrolled on small sets E, the asymptotic behavior of $T^n$ becomes periodic. The notion of smoothness may be attributed to A. Lasota (cf. [14, 16, 17]), who jointly with collaborators established many interesting results concerning limit properties of Markov processes (Markov, stochastic operators) under smoothing assumptions.

This paper has been ignited by Y. Iwata (see [10]), who considered constrictivity and smoothness for the class of Markovian operators possessing a strictly positive stationary density. His paper is focused entirely on kernel Markov operators. Here we give some commentaries on Iwata work and generalize his main result. Finishing this section it is perhaps worth to mention that besides Lasota’s seminal papers [16, 17] on constrictivity, essential contribution to this topic was added by Komornik, Li, Yorke, Sine, Emel’yanov, Wolff, Miklavčič and Bartoszek (cf. [1, 2, 5, 7, 11,12,13,14, 17, 20, 22]).

2 Preliminaries and notation

Let $({\mathfrak {X}}, \Vert \cdot \Vert )$ be a (real) Banach space. A linear operator $T : {\mathfrak {X}} \rightarrow {\mathfrak {X}}$ (in this paper we deal with contractions $\Vert T \Vert \le 1$) is called constrictive if there exists a norm compact set ${\mathcal {K}} \subset {\mathfrak {X}}$ such that for every $\Vert x\Vert \le 1$ the trajectory $T^nx$ is attracted to ${\mathcal {K}}$ (i.e. $\lim _{n\rightarrow \infty }$ dist$(T^nx, {\mathcal {K}}) = 0$). If ${\mathcal {K}}$ is weakly compact and $\lim _{n\rightarrow \infty }$ dist$(T^nx, {\mathcal {K}}) = 0$, for all x from the unit ball of ${\mathfrak {X}}$, then T is said to be weakly constrictive. After Lasota, ${\mathcal {K}}$ is called constrictor (or weak constrictor, respectively).

Definition 1

A linear operator $T: {\mathfrak {X}} \rightarrow {\mathfrak {X}}$ is called asymptotically periodic if there exist a system of vectors $g_1, \ldots , g_r \in {\mathfrak {X}}$, $\varLambda _1, \ldots ,\varLambda _r \in {\mathfrak {X}}^*$ and a permutation $\alpha$ of the set of indices $\{ 1, \ldots ,r \} ,$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| T^nx - \sum _{j=1}^r \varLambda _j(x) g_{\alpha ^n (j)} \right\| = 0 \ \end{aligned}$$

for all $x\in {\mathfrak {X}}$. If $r=1$, i.e. $\lim _{n\rightarrow \infty } \Vert T^n x - \varLambda (x) g \Vert = 0$ for all $x\in {\mathfrak {X}}$, then we say that T is asymptotically stable.

Let $(X, {\mathcal {F}}, \mu )$ be a $\sigma$-finite measure space and ${\mathfrak {X}} = L^p(\mu )$ stand for the classical Banach lattice of real valued measurable functions satisfying $\int _X |f |^p d\mu < \infty$ with the ordinary norm $\Vert \cdot \Vert _p$ (functions equal $\mu$ almost everywhere are identified). A linear operator $T : L^1(\mu ) \rightarrow L^1(\mu )$ is called Markov if $Tf\ge 0$ and $\int _X Tf d\mu = \int _X f d\mu$ for all nonnegative $f\in L^1_+(\mu )$. In other words, T is Markov if $T\ge 0$ and $T^* \mathbf{1} = \mathbf{1}$, where $T^* : L^{\infty }(\mu ) \rightarrow L^{\infty }(\mu )$. Clearly, the operator norm $\Vert T \Vert = 1$, so Markov operators are contractions. $f\in L^1(\mu )$ is called a density if $f\ge 0$ and $\int _X f d\mu = 1$. The closed convex set of all densities in $L^1(\mu )$ is denoted by ${\mathcal {D}}$.

It was proved in [17] (Theorem 1.1) that, if T is a constrictive Markov operator on $L^1(\mu )$, then T is asymptotically periodic. This result was subsequently extended on $L^p(\mu )$ spaces, general Banach lattices and recently on ordered Banach spaces with bases (cf. [1, 5, 20, 22]). We do not aim to dive into abstract regions but focus on classical spaces ${\mathfrak {X}} = L^1(\mu )$. We notice that if there exists an invariant (stationary) density $Tf_* = f_*$, which is strictly positive $\mu$ a.e., then we may introduce ${\widetilde{T}}f = T(ff_*)/f_*$, which is defined on $L^1(f_*\mu )$. Clearly for all $f \in L^1(f_*\mu )$ we have ${\widetilde{T}}^nf = T^n(ff_*)/f_*$, so the dynamics of $T^n$ may be deduced from ${\widetilde{T}}^n$. The Markov operator ${\widetilde{T}}$ satisfies ${\widetilde{T}}\mathbf{1} = \mathbf{1}$, where $\mathbf{1}$ is simply a density function on the probability measure space $(X, {\mathcal {F}}, f_*\mu )$.

Therefore, without loss of generality, we shall assume throughout this paper that $(X, {\mathcal {F}}, \mu )$ is a probability measure space and $\mathbf{1} \in L^1(\mu )$ is a stationary density of a Markov operator $T : L^1(\mu ) \rightarrow L^1(\mu )$. Such Markov operators T are called double Markovian (double stochastic), as $T^* : L^{\infty }(\mu ) \rightarrow L^{\infty }(\mu )$ may be extended to $L^1(\mu ) \supseteq L^{\infty }(\mu )$ and $T^*$ becomes again a Markov operator. Simply both T and $T^*$ are Markovian linear operators acting simultaneously on $L^1(\mu )$ and $L^{\infty }(\mu )$ as positive linear contractions. Notice that $T^{**} = T$ (see [4], Proposition 1.1).

We recall an introduced by Lasota and Komornik notion of smoothness.

Definition 2

We say that a Markovian operator $T : L^1(\mu ) \rightarrow L^1(\mu )$ is smoothing if there exist constants $0< \eta < 1$ and $\delta >0$ such that for every density $f\in {\mathcal {D}}$ there exists a natural $n_f$, such that for all $n \ge n_f$ we have

$$\begin{aligned} \int _E T^nf d\mu \le \eta \ \ , \ \mathrm{for \ all \ } E \in {\mathcal {F}} \ \mathrm{satisfying } \ \mu (E) \le \delta . \end{aligned}$$

Asymptotic periodicity, smoothness and constrictivity of Markovian operators on $L^1(\mu )$ appear to be closely related through the Komornik and Lasota works (see [11,12,13,14]). Gathering their results we arrive to the following characterization.

Theorem 1

Let $(X, {\mathcal {F}}, \mu )$ be a probability space and $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a Markov operator. Then the following are equivalent:

• (1) T is asymptotically periodic,

• (2) T is constrictive,

• (3) T is weakly constrictive,

• (4) T is smoothing.

In [10] the author studied conditions for smoothness (constrictivity) of doubly Markovian and kernel operators. He borrowed most of his ideas and techniques from the Foguel’s monograph (chapters V and VIII). However the power of [8] has not been exploited to its limits and potential. Let us start recalling necessary notation from [8].

A Markov operator $T:L^1(\mu ) \rightarrow L^1(\mu )$ is said to be conservative if $\sum \nolimits _{n=0}^{\infty } T^nf(x) = +\infty \ \ \mu \ a.e.$ for some (equivalently for all) strictly positive $f\in L^1(\mu )$. If T is conservative and $T^*h\le h$ for some $h\in L^{\infty }(\mu )$ then $T^*h = h$. Clearly, each Markov operator with strictly positive stationary density is conservative. In particular, doubly Markovian operators are conservative. A conservative operator T is called ergodic (totally ergodic), if $T^*h = h$ ($T^{*k} h= h$ for some $k \ge 1$) implies that $h =$ const $\mu \ a.e.$ where $h\in L^{\infty }(\mu )$. Ergodicity means that $T^*\mathbf{1}_A = \mathbf{1}_A$ holds only when $\mu (A) = 0$ or $\mu (A) = 1$. Let us recall that conservative Markov operators T (in particular all doubly Markovian operators) are non-disappearing, i.e. if $T^*f = 0$ for some $f \ge 0$, then $f=0$. Hence, (see Lemma 0 in [15] for all details) if $T^*g = \mathbf{1}_A$ with $0 \le g \le 1$, then there exists a unique $E\in {\mathcal {F}}$ such that $g = \mathbf{1}_E$. The family of all $A\in {\mathcal {F}}$, such that for every n there exists $A_n\in {\mathcal {F}}$, such that $T^{*n}{} \mathbf{1}_A = \mathbf{1}_{A_n}$, is denoted by $\varSigma _{\text {d}}(T)$. $\varSigma _{\text {d}}(T)$ is a sub $\sigma$-algebra if T is double Markovian, and it is called a deterministic $\sigma$-algebra. By $\varSigma _1 (T)$ we denote a sub $\sigma$-algebra of $\varSigma _{\text {d}}(T)$ of all those sets $A \in {\mathcal {F}}$ such that for every natural n we have $T^{*n}T^n \mathbf{1}_A = T^nT^{*n}{} \mathbf{1}_A = \mathbf{1}_A$. By the symmetry $\varSigma _1(T)= \varSigma _1(T^*)$.

Given a doubly Markovian operator T consider supports $B_f = \{ x\in X : f(x) > 0 \}$ of a nonnegative $f\in L^1(\mu )$ and $B_{Tf} = \{ x\in X : Tf(x) > 0 \}$ corresponding to its image Tf. We notice that

$$\begin{aligned} \Vert f \Vert _1&= \Vert Tf \Vert _1 = \int _{B_{Tf}} Tf d\mu = \int _X f \cdot (T^*\mathbf{1}_{B_{Tf}}) d\mu \\&= \int _{B_f} f\cdot (T^*\mathbf{1}_{B_{Tf}}) \le \int _{B_f} f d\mu = \Vert f \Vert _1. \end{aligned}$$

Hence $T^*\mathbf{1}_{B_{Tf}} \ge \mathbf{1}_{B_f}$. We obtain $\mu (B_{Tf}) \ge \mu (B_f)$ as $T^*$ is also Markovian.

We say that a Markov operator T is regular, if there exists a family$\{ P(x, \cdot ) \}_{x\in X}$ of probability measures $P(x, \cdot )$ on $(X, {\mathcal {F}})$, such that for every $A\in {\mathcal {F}}$ the mapping $X \ni x \rightarrow P(x,A)$ is ${\mathcal {F}}$ measurable, and

$$\begin{aligned} T^*h(x) = \int _X h(y) P(x, dy), \ \ \ \int _A Tf(x) d\mu (x) = \int _X f(x) P(x, A) d\mu (x) \end{aligned}$$

for all $f\in L^1(\mu ),$ $A\in {\mathcal {F}}$ and $h\in L^{\infty }(\mu )$.

A double Markovian operator T is called integral (or kernel) if it has an integral representation $Tf(y) = \int _X k(x,y) f(x) d\mu (x) ,$ where $k : X \times X \rightarrow R_+$ is a jointly measurable function satisfying $\int _X k(x,y) d\mu (y) = 1$ for $\mu$ almost all x and $\int _X k(x,y) d\mu (x) = 1$ for $\mu$ almost all y. Clearly, all kernel Markov operators (see [9, 23]) are regular and $P(x,A) = \int _X k(x,y)\mathbf{1}_A(y) d\mu (y)$.

It follows that kernel, double Markovian operators are conservative and therefore Harris operators. In this situation $\varSigma _d(T)$ is atomic (the geometric structure of Harris operators and their analytical properties are well known, and are comprehensively presented in [8], Theorem D, p. 58).

3 Result

The main goal of this note is to upgrade results of [10] (see below).

Theorem 2

([10]) Let $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a kernel double Markovian operator. Then the following are equivalent:

• (1) T is smoothing (asymptotically periodic),

• (2) the deterministic sub $\sigma$-algebra $\varSigma _d(T^*)$ has at most finitely many atoms and if q denotes the least common multiple of orders of atoms in $\varSigma _d(T^*)$ then for every atom $W\in \varSigma _d(T^*)$ one has

  $$\begin{aligned} \lim _{n\rightarrow \infty } \mu (A\setminus \ \mathrm{supp}(T^{nq}{} \mathbf{1}_{B})) = 0 , \end{aligned}$$

  for all measurable $A, B \subseteq W.$

We will generalize the above mentioned theorem and shall prove that for kernel double Markovian operator T, in condition (2), finiteness of $\varSigma _d(T^*)$ is sufficient and no extra verifications like $\lim _{n\rightarrow \infty } \mu (A\setminus \mathrm{supp}(T^{nq}{} \mathbf{1}_{B})) = 0$ for all measurable $A, B \subseteq W$, need to be done. Definitely, this will bring the Iwata result to its refined and final shape. Let us mention that, looking from the point of view of structure of $\varSigma _d(T^*)$, asymptotic behaviour of iterates $T^nf$ was studied in [3]. Finiteness of $\varSigma _d(T^*)$ was guaranteed by assuming that T almost overlaps supports (see [3], Proposition 1), which was additionally supported with another assumption that trajectories $T^nf$ have strong limit points. However, doubly Markovian operators T were not necessarily kernel. Here we show that for kernel and doubly Markovian T, we can (if we wish) replace $\varSigma _d(T^*)$ for a (classical) deterministic $\sigma$-field $\varSigma _d(T)$. For this let us start with the following lemma, which is proved with all details, even though a specialist on ergodic theory of Markov operators may relatively easily extract its proof from [8].

Lemma 1

Let $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a double Markovian operator. If $\varSigma _d(T)$ is finite then $\varSigma _d(T) = \varSigma _1(T) \subseteq \varSigma _d(T^*)$.

Proof

We start from a general observation: if $T^*\mathbf{1}_E = \mathbf{1}_F$ then $\mu (E) = \mu (F)$, and $\mu (E) \ge \int _X \mathbf{1}_E T\mathbf{1}_F d\mu = \int _X T^*\mathbf{1}_E \mathbf{1}_F d\mu = \int _X \mathbf{1}_F \mathbf{1}_F d\mu = \mu (F) = \mu (E)$. Hence, $T\mathbf{1}_F = \mathbf{1}_E$. It follows from ([15], Lemma 0) that if $T^*\mathbf{1}_{E_1} = T^*\mathbf{1}_{E_2} = \mathbf{1}_F$, then $E_1 = E_2$ . Thus, $T^*\mathbf{1}_E = \mathbf{1}_F$ if and only $T\mathbf{1}_F = \mathbf{1}_E$.

Let ${\mathcal {A}} = \{ A_1, \ldots , A_a \}$ be the family of all atoms of $\varSigma _d(T)$. We partition ${\mathcal {A}}$ into subfamilies ${\mathcal {A}}_k$, such that $\mu (A) = \gamma _k$ for all $A\in {\mathcal {A}}_k$, where $\gamma _1< \gamma _2< \cdots < \gamma _m$. Let $A_l\in {\mathcal {A}}_1$ be arbitrary. We have $T^*\mathbf{1}_{A_l} = \mathbf{1}_{A_{\beta (l)}}$, where $A_{\beta (l)} \in \varSigma _d(T)$. Clearly, $A_{\beta (l)} \in {\mathcal {A}}_{1}$ is an atom. The transformation $A_l \rightarrow A_{\beta (l)}$ on the set ${\mathcal {A}}_1$ is a bijection (we do not claim that $\beta$ is one cycle).

Then considering $T^*\mathbf{1}_{A_l} = \mathbf{1 }_{A_{\beta (l)}}$ on ${\mathcal {A}}_2$ we obtain that again $A_l \rightarrow A_{\beta (l)}$, on the set ${\mathcal {A}}_2$, is a bijection. We continue with these arguments to exhaust the whole partition ${\mathcal {A}}$.

Because $\varSigma _d(T)$ is finite $T^* : \$span($\{ \mathbf{1}_{A_1}, \ldots , \mathbf{1}_{A_a} \}$) $\rightarrow$ span($\{ \mathbf{1}_{A_1}, \ldots , \mathbf{1}_{A_a} \})$ is linear, positive and invertible with $T^{*^{-1}} = T$. Hence, for any atom $A_l\in \varSigma _d(T)$, and any natural n, we have $T^n\mathbf{1}_{A_l} = \mathbf{1}_{A_{\alpha ^n (l) }}$, where $\alpha = \beta ^{-1}$ is a permutation of the set $\{ 1, 2, \ldots , a \}$. It follows that $T^{n}T^{*n}{} \mathbf{1}_{A} = T^{*n}T^{n}{} \mathbf{1}_{A} = \mathbf{1}_A$ for all $A\in {\mathcal {A}}$. By linearity $T^{n}T^{*n}{} \mathbf{1}_{A} = T^{*n}T^{n}{} \mathbf{1}_{A} = \mathbf{1}_A$ for all $A\in \varSigma _d(T)$, as $\mathbf{1}_A$ has a unique representation as a sum $\mathbf{1}_A = \sum _{j\in J}{} \mathbf{1}_{ A_j} ,$ where $J \subseteq \{ 1, 2, \ldots , a \}$ and $A_j \in {\mathcal {A}}$. We have obtained $\varSigma _d(T) = \varSigma _1(T) \subseteq \varSigma _d(T^*)$. $\square$

The following corollary is a direct consequence of the above Lemma 1.

Corollary 1

Let $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a double Markovian operator. If $\sigma$-algebras $\varSigma _d(T)$ and $\varSigma _d(T^*)$ are finite then $\varSigma _d(T) = \varSigma _1(T) = \varSigma _d(T^*)$.

Now we are ready to explain mutual relations between deterministic $\sigma$-fields $\varSigma _d(T)$ and $\sigma _d(T^*)$ if T is kernel.

Theorem 3

Let T be a kernel doubly Markovian operator. The deterministic $\sigma$-algebra $\varSigma _d(T)$ (equivalently $\varSigma _d(T^*)$) is finite if and only if $\varSigma _d(T^*)$ is finite (respectively $\varSigma _d(T)$ is finite) and then $\varSigma _d(T) = \varSigma _1(T) = \varSigma _d(T^*)$.

Proof

The $\sigma$-algebras $\varSigma _d(T)$ and $\varSigma _d(T^*)$ are atomic as T and $T^*$ are kernel and conservative. By Lemma 1 we have $\varSigma _d(T) = \varSigma _1(T) \subseteq \varSigma _d(T^*)$. Denote $\{ W_1, \dots , W_a \}$ to be the family of all atoms of $\varSigma _d(T)$. We choose and fix for a while an atom $W_i = W$. Let $r_i = r$ be the period of W (i.e. r is the least $j\in {\mathbb {N}}_0$ such that $T^{*j}{} \mathbf{1}_W = \mathbf{1}_W$). We introduce locally $S : L^1(W, {\mathcal {F}}_W, \mu _W) \rightarrow L^1(W, {\mathcal {F}}_W, \mu _W)$ defined as $S = T^r$. Here ${\mathcal {F}}_W = \{E \cap W : E \in {\mathcal {F}} \}$ denotes the trace $\sigma$-algebra and $\mu _W(E) = \frac{\mu (E)}{\mu (W)}$ is the conditional probability measure. Clearly, S is a kernel doubly Markovian operator on $L^1(\mu _W)$. In particular, it is conservative and non-disappearing (and of course kernel).

Similarly if $E \in {\mathcal {F}}_W$ belongs to $\varSigma _d(S)$, then $S^{*k}{} \mathbf{1}_E = T^{*rk}{} \mathbf{1}_E = \mathbf{1}_{F_{rk}}$, for some $F_{rk} = E_k \in {\mathcal {F}}_W$. Again by [15] we have $T^{*j}{} \mathbf{1}_E = \mathbf{1}_{F_j}$ for all $0 \le j \le rk$. In other words $\varSigma _d(S) \subseteq \varSigma _d(T)$. But $W \in \varSigma _d(T)$ is taken to be an atom, so $E = W$. Thus, $\varSigma _d(S) = \{ \emptyset , W \}$ is trivial.

Let us suppose that $E \in \varSigma _d(S^*)$, for some $0< \mu _W(E) < 1$. Hence, $S^k\mathbf{1}_E = \mathbf{1}_{E_k}$ for all $k \in {\mathbb {N}}$, where $E_k \in {\mathcal {F}}_W$ satisfy $\mu _W(E_k) = \mu _W(E)$. It is well known that kernel doubly Markovian operators transform order intervals $[0, f] \subseteq L^1$ into norm $\Vert \cdot \Vert _1$ relatively compact subsets (see [21], Proposition IV.9.8). Thus we may find an increasing subsequence $k_j \rightarrow \infty$ such that $\lim _{j\rightarrow \infty }\Vert S^{k_j} \mathbf{1}_E - g \Vert _1 = 0$, where $0 \le g \le \mathbf{1}_W$. Actually $g = \mathbf{1}_F$ for some $F \in {\mathcal {F}}_W$ (g may be approximated by characteristic functions as close as we wish). Clearly $0< \mu _W(F) = \mu _W(E) < 1$. For an arbitrary $i\in {\mathbb {N}}$ let us consider $S^{*i}{} \mathbf{1}_F = \lim _{j\rightarrow \infty } S^{*i} S^{k_j} \mathbf{1}_{E}= \lim _{j\rightarrow \infty } S^{k_j - i}{} \mathbf{1}_E = \mathbf{1}_{F_i}$ for some $F_i \in {\mathcal {F}}_W$ (if necessary we may choose another subsequence from $k_j - i$ and again apply the compactness argument). Hence, $F \in \varSigma _d(S)$, a contradiction as W is an atom of $\varSigma _d(S)$. Thus, $\varSigma _d(S^*) = \{ \emptyset , W \}$ is trivial.

Applying these considerations to all (finitely many) atoms $W_j \in \varSigma _d(T)$ we obtain that each atom $F \in \varSigma _d(T^*)$ has a representation $F = \bigcup _{j=1}^a F\cap W_j \in \varSigma _{d}(T)$ as $F\cap W_j$ is $W_j \in \varSigma _d(T)$ or $\emptyset$. Hence, $\varSigma _d(T^*) \subseteq \varSigma _d(T) = \varSigma _1(T) \subseteq \varSigma _d(T^*)$ and all are finite. $\square$

Remark 1

In the above Theorem 3 the assumption that T is kernel is essential. In fact, consider the Frobenius–Perron operator associated with the transformation $\tau (x) = (2x) \ \mathrm{mod} \ 1$. It is commonly known that $\tau$ preserves the Lebesgue measure $\lambda$ on $X = [0,1)$. In particular, T is doubly Markovian. Clearly $\varSigma _d(T) = {\mathcal {B}}_{[0,1)}$ is the full $\sigma$-algebra (simply $T^*h(x) = h(\tau (x)),$ where $h\in L^{\infty }(\lambda ),$ is the composition operator). On the other hand $\tau$ is exact (T is asymptotically stable), so $\lim _{n\rightarrow \infty }T^n\mathbf{1}_A = \lambda (A)\mathbf{1}$ in the $L^1$ norm (cf. [18]). Hence,

$$\begin{aligned} \varSigma _d(T^*) = \{ A\in {\mathcal {B}}_{[0,1)} : T^j\mathbf{1}_A = \mathbf{1}_{A_j} \ \mathrm{for \ \ all \ } j = 0, 1, 2, \ldots \} = \{ \emptyset , [0,1) \} \end{aligned}$$

is the trivial $\sigma$-algebra, thus finite. Nevertheless $\varSigma _d(T) \ne \varSigma _d(T^*)$.

By [8] (Theorem A, p. 85) for any $f\in L^2(\mu )$ such that $\int _X f h d\mu = 0$ for all $h \in L^2(X, \varSigma _1(T), \mu )$ we have $\lim _{n\rightarrow \infty } T^n f = 0$ weakly. If T is kernel we have even strong convergence $\lim _{n\rightarrow \infty } \Vert T^n f\Vert _2 = 0$ (again apply [21], Proposition IV.9.8). We may easily modify it to $\lim _{n\rightarrow \infty } \Vert T^n f \Vert _1 = 0$. Therefore, if T is kernel, doubly Markovian with finite $\varSigma _d(T) = \varSigma _d(T^*) = \varSigma _1(T)$ and W is an atom of period r, then $L^2(W, \varSigma _1(S), \mu _W) = \{ t\mathbf{1}_W \ : t\in {\mathbb {R}} \}$. Thus, for all f satisfying $\int _W f d\mu _W = 0$ we have $\lim _{n\rightarrow \infty }\Vert S^n f \Vert _1 = 0$ and $\lim _{n\rightarrow \infty } \Vert S^{*n} f \Vert _1 = 0$. Notice that $\int _W [f - (\int _W fd\mu ) \mathbf{1}_W] d\mu _W = 0$ for all $f\in L^1(\mu )$. We have obtained:

Corollary 2

Let $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a kernel double Markovian operator. If $\varSigma _d(T)$ is finite (by Theorem 3 equivalently $\varSigma _d(T^*)$ is finite) then for each atom $W\in \varSigma _d(T)$ with period r we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert T^{rn}f - \frac{1}{\mu (W)}\left( \int _W fd\mu \right) \mathbf{1}_W \Vert _1 = 0, \end{aligned}$$

and by symmetry

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert T^{*rn}f - \frac{1}{\mu (W)}\left( \int _W fd\mu \right) \mathbf{1}_W \Vert _1 = 0, \end{aligned}$$

for all $f\in L^1(\mu )$ concentrated on W.

Moreover, denoting $\{ A_1, \ldots , A_a\}$ to be the family of all atoms of $\varSigma _d(T)$, there exists (see our Lemma 1) a permutation $\alpha : \{ 1, \ldots , a \} \rightarrow \{ 1, \ldots , a \}$ such that $T\mathbf{1}_{A_j} = \mathbf{1}_{A_{\alpha (j)}}$ and $T^*\mathbf{1}_{A_j} = \mathbf{1}_{A_{\alpha ^{-1} (j)}}$.

Now we are in a position to formulate the main result of the paper.

Theorem 4

Let $T : L^1(\mu ) \rightarrow L^1(\mu )$ be a kernel doubly Markovian operator. Then the following are equivalent

• (1) $\varSigma _d(T)$ (or $\varSigma _d(T^*)$) is finite with atoms $\{ A_1, A_2, \ldots , A_a \}$,

• (2) $\varSigma _d(T) = \varSigma _d(T^*) = \varSigma _1(T)$ are finite,

• (3) T and $T^*$ are simultaneously asymptotically periodic with

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left\| T^n f - \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_{{\alpha ^n(j)}}} \right\| _1 = \lim _{n\rightarrow \infty }\left\| T^{*n} f - \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_{\alpha ^{-n}(j)}} \right\| _1 = 0, \end{aligned}$$

  for all $f\in L^1(\mu )$, where the permutation $\alpha$ is described in our Lemma 1,

• (4) T is asymptotically periodic,

• (5) T is constrictive,

• (6) T is weakly constrictive,

• (7) T is smoothing.

Proof

The implication $(1) \Rightarrow (2)$ is proved in our Theorem 3, $(3) \Rightarrow (4)$ is obvious. The implications $(4) \Leftrightarrow (5) \Leftrightarrow \ldots \Leftrightarrow (7)$ may be recognized today as classical and hold for general (doubly) Markovian operators.

$(2) \Rightarrow (3)$ follows from Corollary 2. Let $A_1, A_2, \ldots$, $A_a$ be atoms of $\varSigma _d(T) = \varSigma _d(T^*)$. For a fixed $f\in L^1(\mu )$ consider its partition $f = \sum _{j=1}^a \mathbf{1}_{A_j} f$. Let q be the least common multiple of all (minimal) periods of atoms. Then

$$\begin{aligned} \lim _{n\rightarrow \infty }T^{qn}f= & {} \sum _{j=1}^a \lim _{n\rightarrow \infty } T^{qn}(\mathbf{1}_{A_j} f )\\= & {} \sum _{j=1}^a \frac{1}{\mu (A_j)}\left( \int _{A_j} fd\mu \right) \mathbf{1}_{A_j}, \ \end{aligned}$$

in the $L^1$ norm.

On limit functions the iterates behave periodically:

$$\begin{aligned} T^k \left( \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_j} \right) = \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_{\alpha ^k(j)}} \end{aligned}$$

and

$$\begin{aligned} T^{*k}\left( \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_j} \right) = \sum _{j=1}^a \left( \int _{A_j} f d\mu \right) \mathbf{1}_{A_{\alpha ^{-k}(j)}} . \end{aligned}$$

The operators T and $T^*$ are asymptotically periodic.

The proof of implication $(7) \Rightarrow (1)$ may be borrowed from [10]. For the sake of completeness of this paper we provide its proof (especially because we have replaced $\varSigma _d(T^*)$ for $\varSigma _d(T)$). Actually, we need only a fraction of Iwata’s argument.

Let $\delta> 0 , \eta > 0$ be as in Definition 2. Suppose that an atom $W\in \varSigma _d(T^*)$ satisfies $\mu (W) \le \delta$. For all $n\ge 1$ we have $\mu (W_n) = \mu (W)$, where $T^{n}{} \mathbf{1}_W = \mathbf{1}_{W_n}$. Then $T^{*n}{} \mathbf{1}_{W_n} = \mathbf{1}_W$. Now let us consider the density $f = \frac{1}{\mu (W)}{} \mathbf{1}_W$. We have

$$\begin{aligned} \int _{W_n} T^n f d\mu = \int _X f T^{*n}{} \mathbf{1}_{W_n} d\mu = \int _X f \mathbf{1}_W d\mu = 1 \end{aligned}$$

for all $n\in {\mathbb {N}}$, where $\mu (W_n ) \le \delta$. But the smoothness assumption requires $\int _{W_n} T^n f d\mu \le \eta < 1$ if n is large enough, a contradiction. Hence $\mu (W) \ge \delta$ are separated from 0 for all atoms $W\in \varSigma _d(T^*)$. The $\sigma$-algebra $\varSigma _d(T^*)$ is finite, so $\varSigma _d(T)$ is also finite. $\square$

Corollary 3

Let $T : L^1(\mu ) \mapsto L^1(\mu )$ be a kernel doubly Markovian operator. Then the following are equivalent

• (1) T is asymptotically stable,

• (2) $T^*$ is asymptotically stable (as a doubly Markovian operator on $L^1(\mu ))$,

• (3) $\varSigma _d(T) = \{ \emptyset , X \}$,

• (4) $\varSigma _d(T^*) = \{ \emptyset , X \}$.