Strong Law of Large Numbers for Iterates of Some Random-Valued Functions

Abstract

Assume $(\Omega , {\mathscr {A}}, P)$ is a probability space, X is a compact metric space with the $\sigma$-algebra ${\mathscr {B}}$ of all its Borel subsets and $f: X \times \Omega \rightarrow X$ is ${\mathscr {B}} \otimes {\mathscr {A}}$-measurable and contractive in mean. We consider the sequence of iterates of f defined on $X \times \Omega ^{{\mathbb {N}}}$ by $f^0(x, \omega ) = x$ and $f^n(x, \omega ) = f\big (f^{n-1}(x, \omega ), \omega _n\big )$ for $n \in {\mathbb {N}}$, and its weak limit $\pi$. We show that if $\psi :X \rightarrow {\mathbb {R}}$ is continuous, then for every $x \in X$ the sequence $\left( \frac{1}{n}\sum _{k=1}^n \psi \big (f^k(x,\cdot )\big )\right) _{n \in {\mathbb {N}}}$ converges almost surely to $\int _X\psi d\pi$. In fact, we are focusing on the case where the metric space is complete and separable.

1 Introduction

Fix a probability space $(\Omega ,{{\mathscr {A}}},P)$ and a metric space X.

Let ${\mathscr {B}}$ denote the $\sigma$-algebra of all Borel subsets of X. We say that $f: X \times \Omega \rightarrow X$ is a random-valued function (shortly: an rv-function) if it is measurable with respect to the product $\sigma$-algebra ${\mathscr {B}} \otimes {\mathscr {A}}$. The iterates of such an rv-function are given by

$$\begin{aligned} f^0 (x, \omega _1, \omega _2, \ldots ) = x, \quad f^n(x, \omega _1, \omega _2, \ldots ) = f\big (f^{n-1} (x, \omega _1, \omega _2, \ldots ), \omega _n\big ) \end{aligned}$$

for $n \in {\mathbb {N}}$, $x \in X$ and $(\omega _1, \omega _2, \ldots )$ from $\Omega ^{\infty }$ defined as $\Omega ^{{\mathbb {N}}}$. Note that $f^n : X \times \Omega ^{\infty } \rightarrow X$ is an rv-function on the product probability space $(\Omega ^{\infty }, {\mathscr {A}}^{\infty }, P^{\infty } )$. More exactly, for $n \in {\mathbb {N}}$ the n-th iterate $f^n$ is ${\mathscr {B}} \otimes {\mathscr {A}}_n$-measurable, where ${\mathscr {A}}_n$ denotes the $\sigma$-algebra of all sets of the form

$$\begin{aligned} \{ (\omega _1, \omega _2, \ldots ) \in \Omega ^{\infty }: \ (\omega _1, \ldots , \omega _n) \in A \} \end{aligned}$$

with A from the product $\sigma$-algebra ${\mathscr {A}}^n$. See [10, Sec. 1.4], [8].

A result on a.s. convergence of $\big (f^n(x, \cdot )\big )_{n \in {\mathbb {N}}}$ for X being the unit interval can be found in [10, Sec. 1.4B]. The paper [7] brings theorems on the convergence a.s. and in $L^1$ of those sequences of iterates in the case where X is a closed subset of a separable Banach lattice. A simple criterion for the convergence in law of $\big (f^n (x, \cdot )\big )_{n \in {\mathbb {N}}}$ to a random variable independent of $x \in X$ was proved in [1], assuming that X is complete and separable. In [2] it has been strengthened and applied to obtain a weak law of large numbers for iterates of random-valued functions. In the present paper we are interested in a strong law of large numbers. We will be based on the following Brunk-Prokhorov-type theorem, see [11, Theorem 3.3.1] and [6, Corollary 3.1].

(C) Let $({\mathscr {F}} _n)_{n \in {\mathbb {N}}}$ be an increasing sequence of sub-$\sigma$-algebras of ${\mathscr {A}}$ and $(\xi _n)_{n\in {\mathbb {N}}}$ a sequence of random variables such that $\xi _n$ is ${\mathscr {F}} _n$-measurable and ${\mathbb {E}}(\xi _{n+1}|{\mathscr {F}} _n)=0$ for each $\ n \in {\mathbb {N}}$. If $(a_n)_{n \in {\mathbb {N}}}$ is an increasing and unbounded sequence of positive reals and

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{{\mathbb {E}}\big (|\xi _n|^2\big )}{a_n^2}<\infty , \end{aligned}$$

then

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{a_n}\sum _{k=1}^{n}\xi _k=0 \text { a.s.} \end{aligned}$$

2 A Scheme

Assume X is a metric space and $f: X \times \Omega \rightarrow X$ an rv-function.

Lemma 1

If $\varphi :X \rightarrow {\mathbb {R}}$ is Borel and $\varphi \circ f^n(x,\cdot )$ is integrable for $P^\infty$ for each $x \in X$ and $n \in {\mathbb {N}}$, then the function $\alpha :X \rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} \alpha (x)=\int _\Omega \varphi \big (f(x,\omega )\big )P(d\omega ) \end{aligned}$$

(1)

is Borel and

$$\begin{aligned} {\mathbb {E}}\big (\varphi \circ f^{n+1}(x,\cdot )|{\mathscr {A}}_n \big )=\alpha \circ f^n(x,\cdot ) \quad \text {for } x \in X \text { and } n \in {\mathbb {N}}. \end{aligned}$$

Proof

Since $\varphi \circ f$ is ${\mathscr {B}} \otimes {\mathscr {A}}$-measurable, by Fubini’s theorem $\alpha$ is Borel. Consequently, for every $x \in X$ and $n \in {\mathbb {N}}$ the function $\alpha \circ f^n(x,\cdot )$ is ${\mathscr {A}}_n$-measurable and for each $A \in {\mathscr {A}}^n$ we have

$$\begin{aligned}&\int _{\{\omega \in \Omega ^{\infty }: \ (\omega _1, \ldots , \omega _n) \in A \}}\varphi \big ( f^{n+1}(x,\omega )\big )P^\infty (d\omega ) \\&\quad =\int _{\{ \omega \in \Omega ^{\infty }: \ (\omega _1, \ldots , \omega _n) \in A \}}\varphi \left( f\big (f^n(x,\omega ),\omega _{n+1}\big )\right) P^\infty (d\omega ) \\&\quad =\int _{\{ \omega \in \Omega ^{\infty }: \ (\omega _1, \ldots , \omega _n) \in A \}}\left( \int _\Omega \varphi \left( f\big (f^n(x,\omega ),\omega _{n+1}\big )\right) P(d\omega _{n+1})\right) P^\infty (d\omega ) \\&\quad =\int _{\{ \omega \in \Omega ^{\infty }: \ (\omega _1, \ldots , \omega _n) \in A \}}\alpha \big (f^n(x,\omega )\big )P^\infty (d\omega ). \end{aligned}$$

$\square$

The following theorem is in fact a scheme of proving a strong law of large numbers for iterates of random-valued functions.

Proposition 1

Let $\psi :X \rightarrow {\mathbb {R}}$ and assume that there exists a Borel and bounded $\varphi :X \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \varphi (x)=\int _{\Omega }\varphi \big (f(x,\omega )\big )P(d\omega )+\psi (x) \quad \text {for } x \in X. \end{aligned}$$

(2)

If $(a_n)_{n \in {\mathbb {N}}}$ is an increasing and unbounded sequence of positive reals such that

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{1}{a_n^2}<\infty , \end{aligned}$$

then, for every $x \in X$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{a_n}\sum _{k=1}^n\psi \circ f^k(x,\cdot )=0 \quad a.e. \ for \ P^\infty . \end{aligned}$$

(3)

Proof

Define $\alpha :X \rightarrow {\mathbb {R}}$ by (1). Since $\varphi$ is bounded, $|\varphi (x)|\le M$ for every $x \in X$ with an $M \in (0,\infty )$. Obviously also $|\alpha (x)|\le M$ for every $x \in X$. Fix $x \in X$ and put

$$\begin{aligned} \xi _n=\varphi \circ f^n(x,\cdot )-\alpha \circ f^{n-1}(x,\cdot ) \quad \text {for } n \in {\mathbb {N}}. \end{aligned}$$

(4)

Then $|\xi _n|\le 2M$ and by Lemma 1, ${\mathbb {E}}(\xi _{n+1}|{\mathscr {A}} _n)=0$ for each $n \in {\mathbb {N}}$. It now follows from Brunk-Prokhorov-type theorem (C) that

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{a_n}\sum _{k=1}^{n}\big (\varphi \circ f^k(x,\cdot )-\alpha \circ f^{k-1}(x,\cdot )\big )=0 \quad \text {a.e. for } P^\infty . \end{aligned}$$

(5)

Since $\psi =\varphi -\alpha$, for every $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned} \sum _{k=1}^n\psi \circ f^k(x,\cdot )&=\sum _{k=1}^n\big (\varphi \circ f^k(x,\cdot )-\alpha \circ f^{k-1}(x,\cdot )\big )\\&\quad +\sum _{k=1}^n\big (\alpha \circ f^{k-1}(x,\cdot )-\alpha \circ f^k(x,\cdot )\big ), \end{aligned} \end{aligned}$$

i.e.,

$$\begin{aligned} \sum _{k=1}^n\psi \circ f^k(x,\cdot ) =\sum _{k=1}^n\big (\varphi \circ f^k(x,\cdot )-\alpha \circ f^{k-1}(x,\cdot )\big ) +\alpha (x) -\alpha \circ f^n(x,\cdot )\nonumber \\ \end{aligned}$$

(6)

for every $n \in {\mathbb {N}}$. Moreover, $|\alpha \circ f^n(x,\cdot )|\le M$. Consequently (3) holds. $\square$

3 The Weak Limit

Assume now the following hypothesis (H).

(H) $(X,\rho )$ is a complete and separable metric space and $f: X \times \Omega \rightarrow X$ is an rv-function such that

$$\begin{aligned} \int _{\Omega } \rho \big (f(x, \omega ), f(z, \omega )\big ) P(d\omega ) \le \lambda \rho (x, z) \quad \text {for } x, z \in X \end{aligned}$$

(7)

with a $\lambda \in (0,1)$, and

$$\begin{aligned} \int _{\Omega } \rho \big (f(x, \omega ), x\big ) P(d \omega ) <\infty \quad \text {for } x \in X. \end{aligned}$$

(8)

Then (see [1, Theorem 3.1]) there exists a probability Borel measure $\pi ^f$ on X such that for every $x \in X$ the sequence of distributions of $f^n(x,\cdot ), \ n \in {\mathbb {N}}$, converges weakly to $\pi ^f$. See also [3, Lemma 2.2] and [9, Corollary 5.6 and Lemma 3.1].

This limit distribution $\pi ^f$ plays an important role in solving functional equations, in particular in the class of Hölder continuous functions. We call a function $\psi :X \rightarrow {\mathbb {R}}$ Hölder continuous with exponent $\delta \in (0,1]$ if there is a constant $L \in [0,\infty )$ such that

$$\begin{aligned} |\psi (x)-\psi (z)| \le L\rho (x,z)^\delta \quad \text {for } x, z \in X. \end{aligned}$$

Moreover we call a function Hölder continuous if it is Hölder continuous with an exponent $\delta \in (0,1]$. The following theorem (see [3, Theorem 2.1] and [4, Corollary 2.6]) will be useful to us.

(B) Assume (H). If $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with exponent $\delta \in (0,1]$, then it is integrable for $\pi ^f$ and if additionally

$$\begin{aligned} \int _X\psi (x)\pi ^f(dx)=0, \end{aligned}$$

(9)

then there exists a Hölder continuous with exponent $\delta$ function $\varphi :X \rightarrow {\mathbb {R}}$ such that (2) holds.

4 Main Results

In what follows $(X,\rho )$ is a metric space and $f: X \times \Omega \rightarrow X$ is an rv-function.

We start with a simple consequence of Proposition 1 and (B). It is a special case of Theorem 2 given below, but shows our approach without technical details.

Theorem 1

If $(X,\rho )$ is complete and separable with finite diameter and (7) holds with a $\lambda \in (0,1)$, then for every Hölder continuous $\psi :X \rightarrow {\mathbb {R}}$ and for each $x \in X$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{n}\sum _{k=1}^n\psi \circ f^k(x,\cdot )=\int _X\psi d\pi ^f \quad a.e. \ for \ P^\infty . \end{aligned}$$

(10)

Proof

Fix a Hölder continuous $\psi :X \rightarrow {\mathbb {R}}$. Replacing $\psi$ by $\psi -\int _X\psi d\pi ^f$ we may assume that (9) holds. By (B) there is a Hölder continuous $\varphi :X \rightarrow {\mathbb {R}}$ satisfying (2). Since X is bounded, so is $\varphi$. Applying now Proposition 1 with $a_n=n$ for $n \in {\mathbb {N}}$ we obtain (3) which ends the proof. $\square$

Since continuous real functions defined on a compact metric space can be uniformly approximated by Lipschitz functions (see [5, 11.2.4]), Theorem 1 implies the following corollary.

Corollary 1

If $(X,\rho )$ is compact and (7) holds with a $\lambda \in (0,1)$, then we have (10) for every continuous $\psi :X \rightarrow {\mathbb {R}}$ and for each $x \in X$.

Theorem 2

Assume (H). Let $x \in X$ and

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{\int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^{2\delta } P^\infty (d \omega )}{a_n^2}<\infty \end{aligned}$$

with a $\delta \in (0,1]$ and an increasing and unbounded sequence $(a_n)_{n \in {\mathbb {N}}}$ of positive reals. If $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with exponent $\delta$, then

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{a_n}\sum _{k=1}^n\big (\psi \circ f^k(x,\cdot )-\int _X\psi d\pi ^f\big )=0 \quad a.e. \ for \ P^\infty . \end{aligned}$$

(11)

The proof will be based on three lemmas.

Assume that $(X,\rho )$ is separable, (7) holds with a $\lambda \in (0,1)$, (8) is satisfied and $\varphi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with exponent $\delta \in (0,1]$, i.e.,

$$\begin{aligned} |\varphi (x)-\varphi (z)| \le L\rho (x,z)^\delta \quad \text {for } x, z \in X \end{aligned}$$

(12)

with an $L \in [0,\infty )$.

Lemma 2

For every $x \in X$ and $n \in {\mathbb {N}}$ we have

$$\begin{aligned}&\int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )P^\infty (d \omega ) \le \frac{1}{1-\lambda }\int _\Omega \varrho \big (f(x,\omega ),x\big )P(d\omega ), \\&\quad \int _{\Omega ^\infty } |\varphi \big (f^n(x,\omega )\big )|P^\infty (d \omega ) \le L\left( \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )P^\infty (d \omega )\right) ^\delta +|\varphi (x)|. \end{aligned}$$

Proof

Fix $x \in X, \ n \in {\mathbb {N}}$ and assume for the inductive proof that

$$\begin{aligned} \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )P^\infty (d \omega ) \le \sum _{k=0}^{n-1}\lambda ^k\int _\Omega \varrho \big (f(x,\omega ),x\big )P(d\omega ). \end{aligned}$$

Then, applying Fubini’s theorem, (7) and the above inequality, we obtain

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty }&\rho \big (f^{n+1}(x, \omega ), x\big )P^\infty (d \omega ) \\&\le \int _{\Omega ^\infty } \rho \big (f\big (f^n (x, \omega _1, \omega _2, \ldots ), \omega _{n+1}\big ),f(x, \omega _{n+1})\big )P^\infty \big (d (\omega _1, \omega _2, \ldots )\big ) \\&\quad + \int _\Omega \rho (f(x,\omega _{n+1}),x) P(d\omega _{n+1}) \\&\le \lambda \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )P^\infty (d \omega ) + \int _\Omega \varrho \big (f(x,\omega ),x\big )P(d\omega ) \\&\le \sum _{k=0}^n\lambda ^k\int _\Omega \varrho \big (f(x,\omega ),x\big )P(d\omega ) \end{aligned} \end{aligned}$$

which ends the proof of the first part. To get the second one observe that by (12) and Jensen’s inequality for every $x \in X$ and $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty } |\varphi \big (f^n(x,\omega )\big )|P^\infty (d \omega ) \le L\int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^\delta P^\infty (d \omega )+|\varphi (x)| \\ \le L\left( \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big ) P^\infty (d \omega )\right) ^\delta +|\varphi (x)|. \end{aligned} \end{aligned}$$

$\square$

Lemma 2 makes sense to define a Borel function $\alpha :X \rightarrow {\mathbb {R}}$ by (1).

Lemma 3

For every $x \in X$ and $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty }|\varphi \big (f^n(x,\omega )\big )-\alpha \big (f^{n-1}(x,\omega )\big )|^2 P^\infty (d \omega ) \\ \le 8 L^2\int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^{2\delta }P^\infty (d \omega ). \end{aligned} \end{aligned}$$

Proof

Since, for every $\omega \in \Omega ^\infty$ and $\omega ' \in \Omega$,

$$\begin{aligned} \begin{aligned} |\varphi \big (f^n(x,\omega )\big )-&\varphi \big (f\big (f^{n-1}(x,\omega ),\omega '\big )\big )| \le L\rho \big (f^n(x,\omega ),f\big (f^{n-1}(x,\omega ),\omega '\big )\big )^\delta \\&\le L \left( \rho \big (f^n(x,\omega ),x\big )^\delta +\rho \big (f\big (f^{n-1}(x,\omega ),\omega '\big ),x\big )^\delta \right) , \end{aligned} \end{aligned}$$

for every $\omega \in \Omega$ we have

$$\begin{aligned} \begin{aligned} |\varphi \big (&f^n(x,\omega )\big )-\alpha \big (f^{n-1}(x,\omega )\big )|^2 \\&=\left| \int _\Omega \left( \varphi \big (f^n(x,\omega )\big )-\varphi \big (f\big (f^{n-1}(x,\omega ),\omega '\big )\big )\right) P(d\omega ')\right| ^2 \\&\le L^2\left( \rho \big (f^n(x,\omega ),x\big )^\delta + \int _\Omega \rho \big (f\big (f^{n-1}(x,\omega ),\omega '\big ),x\big )^\delta P(d\omega ')\right) ^2 \\&\le 4 L^2\left( \rho \big (f^n(x,\omega ),x\big )^{2\delta } + \left( \int _\Omega \rho \big (f\big (f^{n-1}(x,\omega ),\omega '\big ),x\big )^\delta P(d\omega ')\right) ^2\right) . \end{aligned} \end{aligned}$$

Hence, applying Jensen’s inequality and Fubini’s theorem,

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty }&|\varphi \big (f^n(x,\omega )\big )-\alpha \big (f^{n-1}(x,\omega )\big )|^2 P^\infty (d \omega ) \\&\le 4 L^2\Big (\int _{\Omega ^\infty }\rho \big (f^n(x,\omega ),x\big )^{2\delta }P^\infty (d\omega ) \\&\quad +\int _{\Omega ^\infty }\left( \int _\Omega \rho \big (f\big (f^{n-1}(x,\omega ),\omega '\big ),x\big )^{2\delta } P(d\omega ')\right) P^\infty (d\omega )\Big ) \\&= 8 L^2\int _{\Omega ^\infty }\rho \big (f^n(x,\omega ),x\big )^{2\delta }P^\infty (d\omega ). \end{aligned} \end{aligned}$$

$\square$

Lemma 4

Let $(b_n)_{n \in {\mathbb {N}}}$ be a converging to zero sequence of positive reals. If $x \in X$ and there is a $p \in (0,\infty )$ such that

$$\begin{aligned} \sum _{n=1}^{\infty }b_n^p\int _{\Omega ^\infty }\rho \big (f^n(x,\omega ),x\big )^{p\delta }P^\infty (d\omega )< \infty , \end{aligned}$$

then

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }b_n\,\alpha \circ f^n(x,\cdot )=0 \quad a.e. \ for \ P^\infty . \end{aligned}$$

Proof

If $n \in {\mathbb {N}}$ and $\omega \in \Omega$, then by (1), (12), Jensen’s inequality and (7) we have

$$\begin{aligned} \begin{aligned} |\alpha \big (f^n(x,\omega )\big )|&\le \int _\Omega |\varphi \left( f\big (f^n(x,\omega ),\omega '\big )\right) |P(d\omega ') \\&\le L\int _\Omega \rho \left( f\big (f^n(x,\omega ),\omega '\big ),f(x,\omega ')\right) ^\delta P(d\omega ') \\&\quad +L\int _\Omega \rho \big (f(x,\omega '),x\big )^\delta P(d\omega ') +|\varphi (x)| \\&\le L\lambda ^\delta \rho \big (f^n(x,\omega ),x\big )^\delta +L\left( \int _\Omega \rho \big (f(x,\omega ),x\big ) P(d\omega )\right) ^\delta +|\varphi (x)|. \end{aligned} \end{aligned}$$

Now to finish the proof it is enough to show that $\lim _{n \rightarrow \infty }b_n\xi _ n=0$ a.e. for $P^\infty$, where $\xi _n=\rho \big (f^n(x,\cdot ),x\big )^\delta$ for $n \in {\mathbb {N}}$. To this end observe that by Markov’s inequality for every $n \in {\mathbb {N}}$ and $\varepsilon > 0$ we have

$$\begin{aligned} P^\infty (b_n\xi _n\ge \varepsilon ) \le \frac{{\mathbb {E}}(\xi _n^p)}{(\frac{\varepsilon }{b_n})^p}=\frac{1}{\varepsilon ^p}b_n^p{\mathbb {E}}(\xi _n^p). \end{aligned}$$

Hence it follows from the assumption of the lemma that for every $\varepsilon > 0$ the series $\sum _{n=1}^\infty P^\infty (b_n\xi _n\ge \varepsilon )$ converges. Consequently, $\lim _{n \rightarrow \infty }b_n\xi _ n=0$ a.e. for $P^\infty$. $\square$

Proof of Theorem 2. Fix a Hölder continuous with exponent $\delta$ function $\psi :X \rightarrow {\mathbb {R}}$. Replacing $\psi$ by $\psi -\int _X\psi d\pi ^f$ we may assume that (9) holds. By (B) there is a Hölder continuous with exponent $\delta$ function $\varphi :X \rightarrow {\mathbb {R}}$ satisfying (2). Now using Lemma 2 define a Borel function $\alpha :X \rightarrow {\mathbb {R}}$ by (1). Since $\psi =\varphi -\alpha$, (6) follows. Applying Lemmas 1 and 3, and the Brunk-Prokhorov-type theorem (C) to the sequence of random variables $(\xi _n)_{n \in {\mathbb {N}}}$ defined by (4), we have (5). Finally, by Lemma 4 with $b_n=\frac{1}{a_n}, \ n \in {\mathbb {N}}$, and $p=2$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{a_n}\alpha \circ f^n(x,\cdot )=0 \quad \text {a.e. for } P^\infty . \end{aligned}$$

This, (5), (6) and (9) give (11). $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \Box$

Corollary 2

Assume (H). If $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with an exponent $\delta \le \frac{1}{2}$, then we have (10) for each $x \in X$.

Proof

It is enough to observe that by Jensen’s inequality and Lemma 2 for every $x \in X$ we have

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^{2\delta } P^\infty (d \omega )&\le \left( \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big ) P^\infty (d \omega )\right) ^{2\delta } \\&\le \left( \frac{1}{1-\lambda }\int _\Omega \varrho \big (f(x,\omega ),x\big )P(d\omega )\right) ^{2\delta }, \end{aligned} \end{aligned}$$

and then to apply Theorem 2 with $a_n=n, \ n \in {\mathbb {N}}$. $\square$

To get a result for exponents $\delta > \frac{1}{2}$ we accept the following hypothesis (H$_\delta$) with parameter $\delta \in (0,\infty )$.

(H$_\delta$) $(X,\rho )$ is a complete and separable metric space, $f: X \times \Omega \rightarrow X$ is an rv-function such that

$$\begin{aligned} \rho \big (f(x,\omega ),f(z,\omega )\big ) \le \xi (\omega )\rho (x,z) \quad \text {for } \omega \in \Omega \ \text {and } x, z \in X, \end{aligned}$$

(13)

where $\xi :\Omega \rightarrow [0,\infty )$ is a random variable for which ${\mathbb {E}}(\xi ^{2\delta })<1$, and

$$\begin{aligned} \int _{\Omega } \rho \big (f(x_0, \omega ), x_0\big )^{2\delta } P(d \omega ) < \infty \end{aligned}$$

with an $x_0 \in X$.

Remark 1

If $\delta \ge \frac{1}{2}$, then (H$_\delta$) implies (H).

Proof

Assume (H$_\delta$) with a $\delta \ge \frac{1}{2}$. By Jensen’s inequality

$$\begin{aligned} {\mathbb {E}}\xi ={\mathbb {E}}\big ((\xi ^{2\delta })^{\frac{1}{2\delta }}\big ) \le \big ({\mathbb {E}}(\xi ^{2\delta })\big )^{\frac{1}{2\delta }} < 1 \end{aligned}$$

and

$$\begin{aligned} \int _{\Omega } \rho \big (f(x_0, \omega ), x_0\big ) P(d \omega ) \le \left( \int _{\Omega } \rho \big (f(x_0, \omega ), x_0\big )^{2\delta } P(d \omega )\right) ^{\frac{1}{2\delta }}. \end{aligned}$$

Moreover, for every $x \in X$,

$$\begin{aligned} \begin{aligned} \int _{\Omega } \rho \big (f(x, \omega ), x\big ) P(d \omega )&\le \int _{\Omega } \rho \big (f(x, \omega ), f(x_0,\omega ) \big ) P(d \omega ) \\&\quad +\int _{\Omega } \rho \big (f(x_0, \omega ), x_0)\big ) P(d \omega )+\rho (x_0,x) \\&\le \big ({\mathbb {E}}\xi +1\big )\rho (x,x_0)+\int _{\Omega } \rho \big (f(x_0, \omega ), x_0)\big ) P(d \omega ). \end{aligned} \end{aligned}$$

$\square$

Theorem 3

Assume (H$_\delta$) with a $\delta \in [\frac{1}{2},1]$. If $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with exponent $\delta$, then we have (10) for each $x \in X$.

Proof

By Remark 1 we have (H), and it follows from Theorem 2 that to finish the proof it is enough to show that for every $x \in X$ the sequence

$$\begin{aligned} \big (\int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^{2\delta } P^\infty (d \omega )\big )_{n \in {\mathbb {N}}} \end{aligned}$$

is bounded. This follows from the lemma that is stated below. $\square$

Let

$$\begin{aligned} \beta _p(x)=\int _\Omega \rho \big (f(x,\omega ),x\big )^p P(d\omega ) \quad \text {for } p \in (0,\infty ) \text { and } x \in X. \end{aligned}$$

Lemma 5

Assume (13) holds with a random variable $\xi :\Omega \rightarrow [0,\infty )$ and let p be a positive real. If ${\mathbb {E}}(\xi ^p)<1$ and $\beta _p (x_0) < \infty$ for an $x_0 \in X$, then $\beta _p (x) < \infty$ for every $x \in X$ and there exists a constant $c_p \in (0,\infty )$ such that

$$\begin{aligned} \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^p P^\infty (d \omega ) \le c_p\beta _p(x) \quad \text {for } x \in X \ \text {and } n \in {\mathbb {N}}. \end{aligned}$$

Proof

Fix $x \in X$. By (13) for every $\omega \in \Omega$ we have

$$\begin{aligned} \rho \big (f(x, \omega ), x\big )^p \le 3^p\left( \xi (\omega )^p \rho (x,x_0)^p+\rho \big (f(x_0, \omega ), x_0\big )^p+\rho (x_0,x)^p\right) , \end{aligned}$$

whence

$$\begin{aligned} \begin{aligned} \int _\Omega&\rho \big (f(x, \omega ), x\big )^p P(d\omega ) \\&\le 3^p\left( \big ({\mathbb {E}}(\xi ^p)+1\big )\rho (x,x_0)^p+\int _\Omega \rho \big (f(x_0, \omega ) x_0\big )^p P(d\omega )\right) <\infty . \end{aligned} \end{aligned}$$

Put now

$$\begin{aligned} \eta (\omega )=\rho \big (f(x,\omega ),x\big ) \quad \text {for } \omega \in \Omega , \end{aligned}$$

and

$$\begin{aligned} \xi _n(\omega _1, \omega _2, \ldots )=\xi (\omega _n), \quad \eta _n(\omega _1, \omega _2, \ldots )=\eta (\omega _n) \end{aligned}$$

for $n \in {\mathbb {N}}$ and $(\omega _1, \omega _2, \ldots ) \in \Omega ^\infty$. Then, by induction and (13),

$$\begin{aligned} \rho \big (f^n(x,\omega ), x\big ) \le \sum _{k=1}^{n}\eta _k (\omega )\xi _{k+1} (\omega )\cdot \ldots \cdot \xi _n (\omega ) \quad \text {for } \omega \in \Omega ^\infty \ \text {and } n \in {\mathbb {N}}, \end{aligned}$$

where $\prod _{j=n+1}^n\xi _j(\omega ):=1$. Consequently,

$$\begin{aligned} \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^p P^\infty (d \omega ) \le {\mathbb {E}}\big (\big (\sum _{k=1}^n\eta _k \prod _{j=k+1}^n\xi _j\big )^p\big ) \quad \text {for } n \in {\mathbb {N}}. \end{aligned}$$

Moreover, for every integer $n\ge 2$ and $k \in \{1,\ldots ,n-1\}$ the random variables $\eta _k$, $\xi _{k+1},\dots ,\xi _n$ are independent. Hence, if $p \in (0,1)$, then for every $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned} \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^p P^\infty (d \omega )&\le {\mathbb {E}}\big (\sum _{k=1}^n\eta _k^p \prod _{j=k+1}^n\xi _j^p\big ) =\sum _{k=1}^n{\mathbb {E}}(\eta _k^p)\prod _{j=k+1}^n{\mathbb {E}}(\xi _j^p) \\&=\sum _{k=1}^n{\mathbb {E}}(\eta ^p)\big ({\mathbb {E}}(\xi ^p)\big )^{n-k}= {\mathbb {E}}(\eta ^p)\frac{1-\big ({\mathbb {E}}(\xi ^p)\big )^n}{1-{\mathbb {E}}(\xi ^p)} \\&\le {\mathbb {E}}(\eta ^p)\frac{1}{1-{\mathbb {E}}(\xi ^p)}= \frac{1}{1-{\mathbb {E}}(\xi ^p)}\beta _p(x). \end{aligned} \end{aligned}$$

If $p \in [1,\infty )$, then by Minkowski’s inequality for every $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned}&\left( \int _{\Omega ^\infty } \rho \big (f^n(x, \omega ), x\big )^p P^\infty (d \omega )\right) ^{1/p} \le \sum _{k=1}^n\big ({\mathbb {E}}\big (\eta _k \prod _{j=k+1}^n\xi _j\big )^p\big )^{1/p} \\&\quad =\sum _{k=1}^n\big ({\mathbb {E}}(\eta _k^p)\prod _{j=k+1}^n{\mathbb {E}}(\xi _j^p)\big )^{1/p} \le \frac{1}{1-\big ({\mathbb {E}}(\xi ^p)\big )^{1/p}}\beta _p(x)^{1/p}. \end{aligned} \end{aligned}$$

$\square$

Corollary 3

Assume that either

• (i) (H$_\delta$) holds with a $\delta \in [\frac{1}{2},1]$ and $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with exponent $\delta$,

  or

• (ii) (H$_\frac{1}{2}$) is satisfied and $\psi :X \rightarrow {\mathbb {R}}$ is Hölder continuous with an exponent $\delta \le \frac{1}{2}$.

  Then for every bounded and nonempty $A \subset X$ and for almost all $\omega \in \Omega ^\infty$ with respect to $P^\infty$,

  $$\begin{aligned} \lim \limits _{n \rightarrow \infty }\sup \big \{\big |\frac{1}{n}\sum _{k=1}^n\psi \big (f^k(x,\omega )\big )-\int _X\psi d\pi ^f\big |: x \in A\big \}=0. \end{aligned}$$

Proof

It concerns both, (i) and (ii).

By induction,

$$\begin{aligned} \rho \big (f^n(x,\omega ),f^n(z,\omega )\big ) \le \left( \prod _{k=1}^n\xi _k(\omega )\right) \rho (x,z) \end{aligned}$$

for $x,z \in X, \ \omega \in \Omega ^\infty$ and $n \in {\mathbb {N}}$, with

$$\begin{aligned} \xi _n(\omega _1, \omega _2, \ldots )=\xi (\omega _n) \quad \text {for } (\omega _1, \omega _2, \ldots ) \in \Omega ^\infty \ \text {and } n \in {\mathbb {N}}. \end{aligned}$$

Hence

$$\begin{aligned} |\psi \big (f^n(x,\omega )\big )-\psi \big (f^n(z,\omega )\big )| \le L \left( \prod _{k=1}^n\xi _k(\omega )^\delta \right) \rho (x,z)^\delta \end{aligned}$$

for $x,z \in X, \ \omega \in \Omega ^\infty$ and $n \in {\mathbb {N}}$, with an $L \in (0,\infty )$.

Fix $z \in X$. Since, for every $x \in X, \ \omega \in \Omega ^\infty$ and $n \in {\mathbb {N}}$,

$$\begin{aligned} \begin{aligned}&\left| \frac{1}{n}\sum _{k=1}^n\psi \big ( f^k(x,\omega )\big )-\int _X\psi d\pi ^f\right| \le \frac{1}{n}\sum _{k=1}^n\big |\psi \big ( f^k(x,\omega )\big )-\psi \big ( f^k(z,\omega )\big )\big | \\&\quad \quad +\left| \frac{1}{n}\sum _{k=1}^n\psi \big ( f^k(z,\omega )\big )-\int _X\psi d\pi ^f\right| \\&\quad \le L \frac{1}{n}\sum _{k=1}^n\big (\prod _{j=1}^k\xi _j(\omega )^\delta \big ) \rho (x,z)^\delta +\left| \frac{1}{n}\sum _{k=1}^n\psi \big (f^k(z,\omega )\big )-\int _X\psi d\pi ^f\right| , \end{aligned} \end{aligned}$$

for every $r \in (0,\infty )$ and for every nonempty subset A of the ball with center at z and radius r, for every $\omega \in \Omega ^\infty$ and $n \in {\mathbb {N}}$ we have

$$\begin{aligned} \begin{aligned} \sup \big \{\big |\frac{1}{n}\sum _{k=1}^n&\psi \circ f^k(x,\omega )-\int _X\psi d\pi ^f\big |: x \in A\big \}\\&\le Lr^\delta \frac{1}{n}\sum _{k=1}^n\prod _{j=1}^k\xi _j(\omega )^\delta +\left| \frac{1}{n}\sum _{k=1}^n\psi \big (f^k(z,\omega )\big )-\int _X\psi d\pi ^f\right| . \end{aligned} \end{aligned}$$

In view of Theorem 3 and Corollary 2, to finish the proof it is enough to show that

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{n}\sum _{k=1}^n\prod _{j=1}^k \xi _j^\delta =0 \quad \text {a.e. for } P^\infty . \end{aligned}$$

(14)

To this end observe that, by Jensen’s inequality, in the first case (i) we have

$$\begin{aligned} {\mathbb {E}}(\xi ^\delta )={\mathbb {E}}\big ((\xi ^{2\delta })^{\frac{1}{2}}\big ) \le \big ({\mathbb {E}}(\xi ^{2\delta })\big )^{\frac{1}{2}}<1, \end{aligned}$$

and in the second one

$$\begin{aligned} {\mathbb {E}}(\xi ^\delta )\le ({\mathbb {E}}\xi )^\delta <1. \end{aligned}$$

Therefore, applying the monotone convergence theorem and independence of $\xi _n, \ n \in {\mathbb {N}}$, we get

$$\begin{aligned} {\mathbb {E}}\left( \sum _{n=1}^\infty \prod _{k=1}^n \xi _k^\delta \right) =\sum _{n=1}^\infty {\mathbb {E}}\left( \prod _{k=1}^n \xi _k^\delta \right) =\sum _{n=1}^\infty \prod _{k=1}^n {\mathbb {E}}\left( \xi _k^\delta \right) =\sum _{n=1}^\infty \left( {\mathbb {E}}(\xi ^\delta )\right) ^n<\infty . \end{aligned}$$

Consequently, the series $\sum _{n=1}^\infty \prod _{k=1}^n \xi _k^\delta$ converges a.e. for $P^\infty$ and (14) follows.

$\square$

5 An Application to Random Affine Maps

Corollary 4

Assume X is a closed subset of a separable Banach space containing the origin, $\xi :\Omega \rightarrow {\mathbb {R}}$ and $\eta :\Omega \rightarrow X$ are random variables such that $\xi (\omega )X+\eta (\omega ) \subset X$ for $\omega \in \Omega$, and

$$\begin{aligned} \zeta _n(\omega _1,\omega _2,\ldots )=\sum _{k=1}^{n}\left( \prod _{j=k+1}^n\xi (\omega _j )\right) \eta (\omega _k) \quad \text {for } (\omega _1,\omega _2,\ldots ) \in \Omega ^\infty , \ n \in {\mathbb {N}}. \end{aligned}$$

If either $\delta \in (0,\frac{1}{2}]$ and

$$\begin{aligned} {\mathbb {E}}|\xi |< 1, \quad {\mathbb {E}}\Vert \eta \Vert <\infty , \end{aligned}$$

or $\delta \in [\frac{1}{2},1]$ and

$$\begin{aligned} {\mathbb {E}}(|\xi |^{2\delta })<1, \quad {\mathbb {E}}(\Vert \eta \Vert ^{2\delta })<\infty , \end{aligned}$$

then there exists a probability Borel measure $\mu$ on X such that

$$\begin{aligned} \int _X\Vert x\Vert \mu (dx) < \infty \end{aligned}$$

and for every Hölder continuous with exponent $\delta$ function $\psi :X \rightarrow {\mathbb {R}}$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{n}\sum _{k=1}^n\psi \circ \zeta _k=\int _X\psi d\mu \quad a.e. \ for \ P^\infty . \end{aligned}$$

Proof

The function $f: X \times \Omega \rightarrow X$ defined by

$$\begin{aligned} f(x,\omega )=\xi (\omega )x+\eta (\omega ) \end{aligned}$$

is an rv-function. It satisfies (H) in the first case, and (H$_\delta$) in the second one. By induction,

$$\begin{aligned} f^n(x,\omega _1,\omega _2,\ldots )=\left( \prod _{k=1}^n\xi (\omega _k)\right) x+\sum _{k=1}^{n}\left( \prod _{j=k+1}^n\xi (\omega _j )\right) \eta (\omega _k) \end{aligned}$$

for $x \in X, \ (\omega _1,\omega _2,\ldots ) \in \Omega ^\infty$ and $n \in {\mathbb {N}}$. Hence, $\zeta _n=f^n(0,\cdot )$ for $n \in {\mathbb {N}}$, so an application of Corollary 2 and Theorem 3 finishes the proof. $\square$

Remark 2

Let $\lambda \in (0,1)$ and let $\eta :\Omega \rightarrow [0,1-\lambda ]$ be a random variable. Put

$$\begin{aligned} \zeta _n(\omega _1,\omega _2,\ldots )=\sum _{k=1}^n\lambda ^{n-k}\eta (\omega _k) \end{aligned}$$

for $(\omega _1,\omega _2,\ldots ) \in \Omega ^\infty$ and $n \in {\mathbb {N}}$. By Corollary 4 there exists a probability Borel measure $\mu$ on [0, 1] such that for every Hölder continuous $\psi :[0,1] \rightarrow {\mathbb {R}}$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty }\frac{1}{n}\sum _{k=1}^n\psi \circ \zeta _k=\int _{[0,1]}\psi d\mu \quad \text {a.e. for } P^\infty . \end{aligned}$$

But, as observed in [2, Remark 4.3], if $(\psi \circ \zeta _n)_{n \in {\mathbb {N}}}$ converges in probability for a Borel $\psi : [0,1] \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} c|x-z| \le |\psi (x)-\psi (z)| \quad \text {for } x,z \in [0,1] \end{aligned}$$

with a constant $c \in (0,\infty )$, then $\eta$ is a.s. for P constant.