Continuous dependence of the weak limit of iterates of some random-valued vector functions

Abstract

Given a probability space $(\Omega ,\mathcal {A},\mathbb {P})$, a complete separable Banach space X with the $\sigma$-algebra $\mathcal B(X)$ of all its Borel subsets, an operator $\Lambda :\Omega \rightarrow L(X,X)$ and $\xi :\Omega \rightarrow X$ we consider the $\mathcal {B}(X)\otimes \mathcal A$-measurable function $f:X\times \Omega \rightarrow X$ given by $f(x,\omega )=\Lambda (\omega )x+\xi (\omega )$ and investigate the continuous dependence of a weak limit $\pi ^f$ of the sequence of iterates $(f^n(x,\cdot ))_{n\in \mathbb {N}}$ of f, defined by $f^0(x,\omega )=x, f^{n+1}(x,\omega )=f(f^n(x,\omega ),\omega _{n+1})$ for $x\in X$ and $\omega =(\omega _1,\omega _2,\dots )$. Moreover for X taken as a Hilbert space we characterize $\pi ^f$ via the functional equation

$$\begin{aligned} \varphi ^f(u)=\int _{\Omega }\varphi ^f(\Lambda (\omega )u)\varphi ^{\xi }(u)\mathbb {P}(d\omega ) \end{aligned}$$

with the aid of its characteristic function $\varphi ^f$. We also indicate the continuous dependence of a solution of that equation.

1 Introduction

Fix a probability space $(\Omega ,\mathcal {A},\mathbb {P})$ and a separable Banach space X. By $\mathcal {B}(X)$ we denote the family of all Borel subsets of X. A map $f:X\times \Omega \rightarrow X$ measurable with respect to the product algebra $\mathcal {B}(X)\otimes \mathcal {A}$ (shortly: $\mathcal {B}(X)\otimes \mathcal {A}$-measurable) is called a random-valued function or an rv-function. By $f^n$ we denote the n-th iterate of f, given by

$$\begin{aligned} f^0(x,\omega )=x\ \ \text {and} \ \ f^n(x,\omega _1,\ldots \omega _n)=f(f^{n-1}(x,\omega _1,\ldots ,\omega _{n-1}),\omega _n) \end{aligned}$$

for $n\in \mathbb N, x\in X$ and $\omega =(\omega _1,\omega _2,\dots )$ from $\Omega ^\infty$ defined as $\Omega ^\mathbb N$. Note that the map $f^n:X\times \Omega ^{\infty }\rightarrow X$ is $\mathcal {B}(X)\otimes \mathcal {A}_n$-measurable, where $\mathcal {A}_n$ denotes the $\sigma$-algebra of all the sets $\left\{ (\omega _1,\omega _2\ldots ):(\omega _1,\ldots ,\omega _n)\in A\right\}$ and A belongs to the product $\sigma$-algebra $\mathcal {A}^n$. Since $f^n$ depends only on the first n coordinates of $\omega$, we can identify $f^n(\cdot ,\omega _1,\dots ,\omega _n)$ with $f^n(\cdot ,\omega )$. So $f^n$ is an rv-function on the probability space $(\Omega ^{\infty },\mathcal {A}_n,\mathbb {P}^{\infty })$ and also on $(\Omega ^{n},\mathcal {A}^n,\mathbb {P}^{n})$. These iterates were defined by K. Baron and M. Kuczma [2], and independently by Ph. Diamond [7] to solve iterative functional equations. In particular they form forward type iterations and are the prototype of random dynamical systems. A result on almost sure (a.s., for short) convergence of $(f^n(x,\cdot ))_{n\in \mathbb N}$ for $X=[0,1]$ can be found in [17, Sec. 1.4 B]. A simple and useful criterion for a weak convergence of distributions of $f^n(x,\cdot ), n\in \mathbb N$ to a probability Borel measure $\pi ^f$ independent of $x\in X$ for X being a Polish space was proved in [1] and applied to some linear inhomogeneous functional equation.

One of the most important cases of rv-functions is the so called random affine map (see e.g. [11]), which is given by

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

(1.1)

where $\eta :\Omega \rightarrow \mathbb R, \xi :\Omega \rightarrow X$ are $\mathcal {A}$-measurable. These maps are related to perpetuities, see for instance [11, 12, 16]); they are also applied to refinement type equations [15]. Substituting a random vector $\eta$ into a random operator, we will consider rv-functions of the form

$$\begin{aligned} (x,\omega ) \longmapsto \Lambda (\omega ) x+\xi (\omega ), \end{aligned}$$

(1.2)

where $\Lambda (\omega ):X\rightarrow X$ is a continuous and bounded operator for $\omega \in \Omega$. A function (1.2) will be called a generalized random affine map or GRAM, for short. However, the main motivation to study such rv-functions is the work of K. Baron [5], where a special case of map (1.2) with the same operator $\Lambda (\omega )$ for any $\omega$ was examined.

The first aim of the present paper is to give some natural conditions under which the sequence of iterates of GRAM’s f converges in law to $\pi ^f$, and to establish the continuity of the operator $f\longmapsto \pi ^f$ by showing how $\pi ^f$ change if $\Lambda$ and $\xi$ do. This extends the main result of [3] as well as [4, Theorem 1] and [14, Theorem 5.2].

In the case when X is a real Hilbert space a characterization of a limit distribution $\pi ^f$ by its characteristic function $\varphi ^f$ via the linear functional equation $\varphi ^f(u)=\varphi ^f(\Lambda ^*(u))\cdot \varphi ^{\xi }(u)$ was established in [5]. Referring to that paper we will show that the function $\varphi ^f$ for GRAM’s f is only one solution of the equation

$$\begin{aligned} \varphi ^f(u)= \varphi ^{\xi }(u)\int _{\Omega }\varphi ^f(\Lambda ^*(\omega )u)\mathbb {P}(d\omega ) \end{aligned}$$

in a class of characteristic functions. Moreover, we will indicate continuous dependence in such a characterisation of the limit distribution.

2 Notions and basic facts

Throughout the paper $(X,\Vert \cdot \Vert )$ is a separable Banach space and $(\Omega , \mathcal {A},\mathbb {P})$ is a given probability space. We write B(X) for a space of all Borel and bounded functions endowed with the supremum norm $\Vert \cdot \Vert _{\infty }$ and C(X) for its subspace containing all continuous (and bounded) functions. A space of all linear and continuous operators $\Lambda :X\rightarrow X$ will be denoted by L(X, X). We use the symbol $\mathcal {M}_1(X)$ to denote the space of all probability measures defined on $\mathcal {B}(X)$. For short, we will write $\int \varphi d\mu$ instead of $\int _X \varphi (x) \mu (dx)$ for Bochner integrable $\varphi$ and $\mu \in \mathcal M_1(X)$ if there is no confusion. We also consider a family of all measures with finite first moment given by

$$\begin{aligned} \mathcal {M}_1^1(X)=\left\{ \mu \in \mathcal {M}_1(X):\int \rho (x,x_0)\mu (dx)<\infty \right\} \end{aligned}$$

for some $x_0\in X$. (Clearly $\mathcal {M}_1^1(X)$ does not depend on $x_0$.) Recall that a measure $\mu *\nu$ is a convolution of measures $\mu$ and $\nu$ if

$$\begin{aligned} \mu *\nu (B)=\int \mu (B-x)\nu (dx)\qquad \text {for every} \ B\in \mathcal {B}(X). \end{aligned}$$

We write $\mu _{\chi }$ to denote a probability distribution of the random variable $\chi$. Random variables $\chi :\Omega \rightarrow X$, $\zeta :\Omega \rightarrow Y$ are called independent if

$$\begin{aligned} \mu _{(\chi ,\zeta )}=\mu _{\chi }\otimes \mu _{\zeta }, \end{aligned}$$

where $\mu _{(\chi ,\zeta )}$ is their joint probability distribution. We say that a sequence $(\mu _n)$ of measures from $\mathcal {M}_1(X)$ converges weakly to $\mu$ if $\int fd\mu _n\xrightarrow [n\rightarrow \infty ]{} \int f d\mu$ for every $f\in C(X)$. We introduce the symbol $d_{FM}$ to denote the Fortet–Mourier metric (also known as the bounded Lipschitz distance) given by

$$\begin{aligned} d_{FM}(\mu ,\nu )=\sup \left\{ \left| \int fd\mu -\int fd\nu \right| :f\in Lip_1(X), \Vert f\Vert _{\infty }\le 1 \right\} , \end{aligned}$$

and additionally $d_H$ to denote the Huthinson metric given by

$$\begin{aligned} d_{H}(\mu ,\nu )=\sup \left\{ \left| \int fd\mu -\int fd\nu \right| :f\in Lip_1(X) \right\} , \end{aligned}$$

where

$$\begin{aligned} Lip_1(X)=\left\{ f\in B(X):\ |f(x)-f(y)|\le \Vert x-y\Vert \ \text {for} \ x,y\in X\right\} . \end{aligned}$$

Note that the distance between some measures in the Huthinson metric may be infinite. It is known (see [9, Theorem 11.3.3]) that weak convergence is metrizable by the Fortet–Mourier norm.

With an rv-function $f:X\times \Omega \rightarrow X$ we may associate a linear operator $P:\mathcal {M}_1(X)\rightarrow \mathcal {M}_1(X)$ by the formula

$$\begin{aligned} P\mu (A)=\int _{X}\int _{\Omega }\mathbbm {1}_A(f(x,\omega ))\mathbb {P}(d\omega )\mu (dx), \end{aligned}$$

(2.1)

which will be used in this paper. It can be shown that P is the Markovian transition operator for the distribution $\pi _n$ of $f^n$ given by

$$\begin{aligned} \pi _n(x,A)=\mathbb {P}^{\infty }(\left\{ \omega \in \Omega ^{\infty }:f^n(x,\omega )\in A\right\} ), \end{aligned}$$

i.e.

$$\begin{aligned} P\pi _n(x,A)=\pi _{n+1}(x,A)\quad {\text {for}}\; x\in X, \ A\in \mathcal {B}(X). \end{aligned}$$

By the convergence in distribution or in law of the sequence of iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}$ we mean that the sequence $(\pi _n(x,\cdot ))_{n\in \mathbb {N}}$ converges weakly to a probability distribution.

Following [1] and [13] we consider a family of rv-functions $f:X\times \Omega \rightarrow X$ which satisfy:

(H$_f$):

There exists $\lambda _f\in (0,1)$ such that

$$\begin{aligned} \int _\Omega \Vert f(x,\omega )-f(y,\omega )\Vert \mathbb P(d\omega )\le \lambda _f \Vert x-y\Vert \quad \textrm{for}\quad x,y\in X \end{aligned}$$

and

$$\begin{aligned} \int _\Omega \Vert f(x,\omega )-x\Vert \mathbb P(d\omega )<\infty \quad \mathrm{for\; some \;(thus \;all)} \quad x\in X. \end{aligned}$$

A simple criterion [13, Corollary 5.6], cf. [1, Theorem 3.1], for the convergence in distribution of iterates of rv-functions reads as follows:

Proposition 2.1

Assume that an rv-function $f:X\times \Omega \rightarrow X$ satisfies (H$_f$). Then for every $x\in X$ the sequence of iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}$ converges in distribution and the limit $\pi ^f$ does not depend on x. Moreover $\pi ^f\in \mathcal M_1^1(X)$ and

$$\begin{aligned} d_H(\pi _n(x,\cdot ),\pi ^f) \le \frac{{\lambda _f}^n}{1-\lambda _f}\int _\Omega \Vert f(x,\omega ) -x\Vert \mathbb P(d\omega ) \end{aligned}$$

for $n\in \mathbb N$ and $x\in X$.

The geometric rate of convergence allows us to formulate a result concerning the continuity of $f\longmapsto \pi ^f$. We cite a part of [14, Theorem 4.1] that will be useful in the next section.

Proposition 2.2

Assume that rv-functions f, g satisfy (H$_f$) and (H$_g$), respectively. Then for limit distributions $\pi ^f$ and $\pi ^g$, occurring in Proposition 2.1, we have

$$\begin{aligned} d_H(\pi ^f,\pi ^g)\le \min \left\{ \frac{1}{1-\lambda _f}\inf _{x\in X}{\alpha _g(x)},\frac{1}{1-\lambda _g}\inf _{x\in X}{\alpha _f(x)}\right\} , \end{aligned}$$

(2.2)

where

$$\begin{aligned} \alpha _h(x)=\sup _{n\in \mathbb {N}_0}\int _{\Omega ^{\infty }}\int _{\Omega }\Vert f(h^n(x,\omega ),\varpi )-g(h^n(x,\omega ),\varpi )\Vert \mathbb {P}(d\varpi )\mathbb {P}^{\infty }(d\omega ) \end{aligned}$$

(2.3)

for $h\in \{f,g\}$.

Remark 2.3

By condition (H$_g$) we mean (H$_f$) in which all functions f’s are replaced by g’s. A similar convention will be used considering condition (U$_g$) in the next section.

3 Continuous dependence of the limit distribution of generalized random affine maps

Fix $\Lambda :\Omega \rightarrow L(X,X)$ and $\mathcal A$-measurable $\xi :\Omega \rightarrow X$. Since X is separable, we may consider equivalently the weak, strong (in Bochner’s sense), and Borel measurability of the random variable $\xi$. To get some results concerning the convergence in law of GRAM’s (1.2) we need to show that (1.2) is an rv-function. To do this we will introduce the following:

Definition 3.1

We call a map $\Lambda :\Omega \rightarrow L(X,X)$ a random operator, if it is $\mathcal {A}$-measurable, i.e. $\Lambda ^{-1}(B)\in \mathcal A$ for every Borel subset B of L(X, X).

Proposition 3.2

If $\Lambda :\Omega \rightarrow L(X,X)$ is a random operator, then a function $\Lambda (\cdot )x:\Omega \rightarrow X$ is $\mathcal {A}$-measurable for every $x\in X$.

Proof

Fix $x\in X$ and define $\varphi _x:L(X,X)\rightarrow X$ by $\varphi _x(T)=Tx$. It is obvious that $\varphi _x$ is linear, and since

$$\begin{aligned} \Vert \varphi _x(T)\Vert =\Vert Tx\Vert \le \Vert x \Vert \cdot \Vert T\Vert \end{aligned}$$

it is bounded (thus continuous). Now fix $B\in \mathcal {B}(X)$ then we have

$$\begin{aligned} \left\{ \omega \in \Omega :\Lambda (\omega )x\in B\right\}&=\left\{ \omega \in \Omega :\varphi _x(\Lambda (\omega ))\in B\right\} \\ {}&=\left\{ \omega \in \Omega :\Lambda (\omega )\in \varphi ^{-1}_x(B)\right\} \in \mathcal {A}. \end{aligned}$$

$\square$

Remark 3.3

One can show that for a separable space X if $\Lambda (\cdot )x:\Omega \rightarrow X$ is $\mathcal {A}$-measurable for every $x\in X$ and $\Lambda (\omega ):X\rightarrow X$ is continuous for every $\omega \in \Omega$ then a map $\Lambda :\Omega \times X\rightarrow X$ with $\Lambda (x,\omega )=\Lambda (\omega )x$ is $\mathcal {A}\otimes \mathcal {B}(X)$-measurable. Moreover, $\xi$ extended to $\xi :X\times \Omega \rightarrow X$ by $\xi (x,\omega )=\xi (\omega )$ is $\mathcal {A}\otimes \mathcal {B}(X)$-measurable. Since the sum of $\mathcal {A}\otimes \mathcal {B}(X)$-measurable functions on separable values is also $\mathcal {A}\otimes \mathcal {B}(X)$-measurable it follows that $f:\Omega \times X\rightarrow X$ given by (1.2) is an rv-function.

The main result of this section concerns the continuous dependence of the limit of iterates of GRAM’s. We will formulate it for a family of rv-functions $f:X\times \Omega \rightarrow X$ which satisfy:

(U$_f$):

The function $f:X\times \Omega \rightarrow X$ has the form $f(x,\omega )=\Lambda _f(\omega )x+\xi _f(\omega )$, where $\xi _f:\Omega \rightarrow X$ is $\mathcal A$-measurable,

$$\begin{aligned} \mathbb E\Vert \xi _f\Vert =\int _\Omega \Vert \xi _f(\omega )\Vert \mathbb P(d\omega )<\infty , \end{aligned}$$

and $\Lambda _f:\Omega \rightarrow L(X,X)$ is a random operator satisfying

$$\begin{aligned} \mathbb {E}\Vert \Lambda _f(\cdot )\Vert =\int _{\Omega }\Vert \Lambda _f(\omega )\Vert \mathbb {P}(d\omega )<1, \end{aligned}$$

where $\Vert \Lambda _f(\omega )\Vert$ is the operator norm of $\Lambda _f(\omega )$.

Theorem 3.4

Assume that rv-functions f, g satisfy (U$_f$) and (U$_g$), respectively. Then the sequences of iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}, (g^n(x, \cdot ))_{n\in \mathbb {N}}$ are convergent in law to the probability distributions $\pi ^f, \pi ^g\in \mathcal {M}_1^1(X)$, respectively, the limits do not depend on $x\in X$, and

$$\begin{aligned} d_H(\pi ^f,\pi ^g)\le \min&\left\{ \frac{1}{1-\mathbb {E}\Vert \Lambda _f(\cdot )\Vert }\left( \frac{\mathbb {E}\Vert \xi _g\Vert }{1-\mathbb {E}\Vert \Lambda _g(\cdot )\Vert }\alpha +\beta \right) ,\right. \\&\left. \frac{1}{1-\mathbb {E}\Vert \Lambda _g(\cdot )\Vert }\left( \frac{\mathbb {E}\Vert \xi _f\Vert }{1-\mathbb {E}\Vert \Lambda _f(\cdot )\Vert }\alpha +\beta \right) \right\} ,\end{aligned}$$

where $\alpha =\mathbb {E}\Vert \Lambda _f(\cdot )-\Lambda _g(\cdot )\Vert , \ \beta =\mathbb {E}\Vert \xi _f-\xi _g\Vert .$

Proof

At the beginning let us observe that (U$_f$) implies (H$_f$). Indeed,

$$\begin{aligned} \int _\Omega \Vert f(x,\omega )-f(y,\omega )\Vert \mathbb P(d\omega ) \le \Vert x-y\Vert \int _{\Omega }\Vert \Lambda _f(\omega )\Vert \mathbb {P}(d\omega ) \quad \textrm{for}\quad x,y\in X \end{aligned}$$

and

$$\begin{aligned} \int _\Omega \Vert f(0,\omega )\Vert \mathbb P(d\omega )= \int _\Omega \Vert \xi _f(\omega )\Vert \mathbb P(d\omega )<\infty . \end{aligned}$$

By Proposition 2.1 we infer that there exist probability distributions $\pi ^f, \ \pi ^g\in \mathcal {M}_1^1(X)$ such that for every $x\in X$ the sequences $(f^n(x, \cdot ))_{n\in \mathbb {N}}$, $(g^n(x, \cdot ))_{n\in \mathbb {N}}$ are convergent in law to $\pi ^f, \pi ^g$, respectively.

The rest of the proof runs similarly to the proof of [14, Theorem 5.2] which concerns (1.1). For the convenience of the reader we repeat the relevant computations after appropriate changes for the case of GRAM’s, thus making our exposition self-contained. So fix $k\in \mathbb {N}$ and let us define $\Lambda _k:\Omega ^{\infty }\rightarrow L(X,X)$ and $\xi _k:\Omega ^{\infty }\rightarrow X$ by $\Lambda _k (\omega )=\Lambda _f(\omega _{k}), \ \xi _k (\omega )=\xi _f(\omega _{k})$, where $\omega =(\omega _1,\omega _2\ldots )\in \Omega ^{\infty }$, and observe that for $\omega \in \Omega ^{\infty }$ and $x\in X$

$$\begin{aligned} f^n(x,\omega )&=\bigodot _{i=0}^{n-1}\Lambda _{n-i}(\omega )x+\bigodot _{i=0}^{n-2}\Lambda _{n-i}(\omega )\xi _{1}(\omega )+\\&\quad +\bigodot _{i=0}^{n-3}\Lambda _{n-i}(\omega )\xi _{2}(\omega )+\ldots +\Lambda _{n}(\omega )\xi _{n-1}(\omega )+\xi _{n}(\omega ),\end{aligned}$$

where

$$\begin{aligned} \bigodot _{i=0}^{n-k}\Lambda _{n-i}(\omega )=\Lambda _{n}(\omega )\circ \Lambda _{n-1}(\omega )\circ \Lambda _{n-2}(\omega )\circ \ldots \circ \Lambda _{k}(\omega ) \end{aligned}$$

and $\circ$ is a composition. From that

$$\begin{aligned} f^n(0,\omega )=\sum _{k=2}^{n}\bigodot _{i=0}^{n-k}\Lambda _{n-i}(\omega )\xi _{k-1}(\omega )+\xi _n(\omega ). \end{aligned}$$

Then

$$\begin{aligned}&\Vert g(f^n(0,\omega ),\overline{\omega })-f(f^n(0,\omega ),\overline{\omega })\Vert \\&\le \Vert \Lambda _g(\overline{\omega })-\Lambda _f(\overline{\omega })\Vert \times \left( \sum _{k=2}^{n}\left\| \bigodot _{i=0}^{n-k}\Lambda _{n-i}(\omega )\xi _{k-1}(\omega ) \right\| +\Vert \xi _n(\omega )\Vert \right) \\&\quad +\Vert \xi _g(\overline{\omega })-\xi _f(\overline{\omega })\Vert \end{aligned}$$

and from the inequality

$$\begin{aligned} \left\| \bigodot _{i=0}^{n-k}\Lambda _{n-i}(\omega )\xi _{k-1}(\omega ) \right\| \le \Vert \xi _{k-1}(\omega )\Vert \prod _{i=k}^{n}\Vert \Lambda _i(\omega )\Vert \end{aligned}$$

we have

$$\begin{aligned} \Vert g(f^n(0,\omega ),\overline{\omega })&-f(f^n(0,\omega ),\overline{\omega })\Vert \le \Vert \Lambda _g(\overline{\omega })-\Lambda _f(\overline{\omega })\Vert \\&\times \left( \sum _{k=2}^{n}\Vert \xi _{k-1}(\omega )\Vert \prod _{i=k}^{n}\Vert \Lambda _i(\omega )\Vert +\Vert \xi _n(\omega )\Vert \right) +\Vert \xi _g(\overline{\omega })-\xi _f(\overline{\omega })\Vert .\end{aligned}$$

Since $\Vert \xi _{k-1}\Vert , \Vert \Lambda _k(\cdot )\Vert , \ldots , \Vert \Lambda _n(\cdot )\Vert$ are independent it follows that

$$\begin{aligned}\int _{\Omega ^{\infty }}\int _{\Omega }&\Vert g(f^n(0,\omega ),\overline{\omega })-f(f^n(0,\omega ),\overline{\omega })\Vert \mathbb {P}^{\infty }(d\omega )\mathbb {P}(d\overline{\omega }) \\&\le \int _{\Omega }\Vert \Lambda _g(\overline{\omega })-\Lambda _f(\overline{\omega })\Vert \mathbb {P}(d\overline{\omega })\int _{\Omega ^{\infty }}\Bigg (\sum _{k=2}^{n}\Vert \xi _{k-1}(\omega )\Vert \\&\quad \times \prod _{i=k}^{n}\Vert \Lambda _i(\omega )\Vert +\Vert \xi _n(\omega )\Vert \Bigg )\mathbb {P}^{\infty }(d\omega )+\int _{\Omega }\Vert \xi _g(\overline{\omega })-\xi _f(\overline{\omega })\Vert \mathbb {P}(d\overline{\omega })\\&=\alpha \sum _{k=2}^{n}\mathbb {E}\Vert \xi _{k-1}\Vert \prod _{i=k}^{n}\mathbb {E}\Vert \Lambda _{i}(\cdot )\Vert +\mathbb {E}\Vert \xi _n\Vert +\beta \\&=\alpha \sum _{k=2}^{n+1}\mathbb {E}\Vert \xi _{f}\Vert \cdot ( \mathbb {E}\Vert \Lambda _{f}(\cdot )\Vert )^{n-k+1}+\beta \\&=\alpha \mathbb {E}\Vert \xi _{f}\Vert \frac{1-\left( \mathbb {E}\Vert \Lambda _{f}(\cdot )\Vert \right) ^{n}}{1- \mathbb {E}\Vert \Lambda _{f}(\cdot )\Vert }+\beta .\end{aligned}$$

Therefore for the function $\alpha _f(x)$ given by (2.3) we obtain

$$\begin{aligned} \inf _{x\in X}\alpha _f(x)\le \alpha _f(0)\le \alpha \frac{\mathbb {E}\Vert \xi _{f}\Vert }{1- \mathbb {E}\Vert \Lambda _{f}(\cdot )\Vert }+\beta \qquad \text {for} \ x\in X. \end{aligned}$$

A similar inequality holds for $\alpha _g(x)$. Taking $\lambda _f=\mathbb {E}\Vert \Lambda _{f}(\cdot )\Vert , \lambda _g=\mathbb {E}\Vert \Lambda _{g}(\cdot )\Vert$ and applying Proposition 2.2 we finish the proof. $\square$

Corollary 3.5

Assume that rv-functions f, g have the form

$$\begin{aligned} f(x,\omega )=\Lambda _f x+\xi _f(\omega ), \qquad g(x,\omega )=\Lambda _g x+\xi _g(\omega ) \end{aligned}$$

with $\Lambda _f,\Lambda _g\in L(X,X)$ such that $\Vert \Lambda _f\Vert <1$, $\Vert \Lambda _g\Vert <1$ and $\xi _f, \xi _g:\Omega \rightarrow X$ such that $\mathbb {E}\Vert \xi _f\Vert <\infty$, $\mathbb {E}\Vert \xi _g\Vert <\infty$. Then the sequences of iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}$, $(g^n(x, \cdot ))_{n\in \mathbb {N}}$ are convergent in law to the probability distributions $\pi ^f$, $\pi ^g\in \mathcal {M}_1^1(X)$, respectively, the limits do not depend on $x\in X$, and

$$\begin{aligned} d_H(\pi ^f,\pi ^g)\le \min&\left\{ \frac{1}{1-\Vert \Lambda _f\Vert }\left( \frac{\mathbb {E}\Vert \xi _g\Vert }{1-\Vert \Lambda _g\Vert }\alpha +\beta \right) ,\right. \\&\left. \frac{1}{1-\Vert \Lambda _g\Vert }\left( \frac{\mathbb {E}\Vert \xi _f\Vert }{1-\Vert \Lambda _f\Vert }\alpha +\beta \right) \right\} ,\end{aligned}$$

where $\alpha =\Vert \Lambda _f-\Lambda _g\Vert , \ \beta =\mathbb {E}\Vert \xi _f-\xi _g\Vert .$

Corollary 3.5 given above extends the main result of [3] as well as [4, Theorem 1]. Due to this result we can generalize [4, Theorem 3] and [5, Theorem 3.1]; see Theorems 4.10, 4.22.

4 Characterisation of the limit distribution

Let $(\Omega ,\mathcal {A},\mathbb {P})$ be a probability space. In this section X is a separable real Hilbert space with the inner product $(\cdot |\cdot )$. However in cases when it is not needed we will emphasize it. We define a characteristic function $\varphi ^f$ of the rv-function f, assuming that the iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}$ converge in law and the limit does not depend on x; in such a case we denote by $\pi ^f$ the distribution of the limit, i.e.

$$\begin{aligned} \pi ^f_n(x,\cdot )\xrightarrow [n\rightarrow \infty ]{w}\pi ^f. \end{aligned}$$

Definition 4.1

A function $\varphi ^{\chi }:X\rightarrow \mathbb {C}$ given by

$$\begin{aligned} \varphi ^{\chi }(u)=\int _Xe^{i(u|z)}\mu _{\chi }(dz) \end{aligned}$$

is called a characteristic function of the X-valued random variable $\chi$ with distribution $\mu _{\chi }$.

Definition 4.2

A function $\varphi ^{f}:X\rightarrow \mathbb {C}$ given by

$$\begin{aligned} \varphi ^{f}(u)=\int _Xe^{i(u|z)}\pi ^f(dz) \end{aligned}$$

is called a characteristic function of the rv-function f.

The problem of characterization of the limit distribution $\pi ^f$ via a functional equation for its characteristic function $\varphi ^f$ was considered in [5]. The author showed that for the rv-function f given by

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

with $\Lambda \in L(X,X)$ such that $\Vert \Lambda \Vert <1$ and a random variable $\xi :\Omega \rightarrow X$ such that $\mathbb {E}\Vert \xi \Vert <\infty$ its characteristic function $\varphi ^f$ is the only solution of the equation

$$\begin{aligned} \varphi ^f(u)=\varphi ^f(\Lambda ^*(u))\cdot \varphi ^{\xi }(u), \end{aligned}$$

where $\Lambda ^*$ stand for the adjoint operator to $\Lambda$, which satisfies $(\Lambda ^*u|z)=(u|\Lambda z)$ for every $u,z\in X$. Our goal is to generalize this result to GRAM’s. First we give some preceding facts, which will be needed in the general setting.

Lemma 4.3

Let X be a Banach space. Assume that a random operator $\Lambda :\Omega \rightarrow L(X,X)$ and a random variable $\xi :\Omega \rightarrow X$ are independent. If $x\in X$, then $\Lambda (\cdot ) x:\Omega \rightarrow X$ and $\xi :\Omega \rightarrow X$ are independent.

Proof

Fix $x\in X$. Let us define $\tau _x:L(X,X)\times X\rightarrow X^2$ by

$$\begin{aligned} \tau _x(T,y)=(Tx,y). \end{aligned}$$

Observe that $\tau _x$ is well defined, continuous in product topology (by the continuity of T) and thus $\mathcal {B}(L(X,X))\otimes \mathcal {B}(X)$-measurable. Denote the distribution of $\Lambda (\cdot )x$ by $\mu _{\Lambda x}$. We claim that $\mu _{(\Lambda x,\xi )}(B)=\mu _{(\Lambda ,\xi )}(\tau _x^{-1}(B))$ for $B\in \mathcal {B}({X^2})$. Indeed we have

$$\begin{aligned}\mu _{(\Lambda x,\xi )}(B)&=\mathbb {P}(\left\{ \omega : (\Lambda (\omega )x,\xi (\omega ))\in B\right\} )= \mathbb {P}(\left\{ \omega : \tau _x(\Lambda (\omega ),\xi (\omega ))\in B\right\} )\\&=\mathbb {P}(\left\{ \omega : (\Lambda (\omega ),\xi (\omega ))\in \tau _x^{-1}(B)\right\} )= \mu _{(\Lambda ,\xi )}(\tau _x^{-1}(B)).\end{aligned}$$

It remains to show that $\mu _{\Lambda x}\otimes \mu _{\xi }(B)=\mu _{\Lambda }\otimes \mu _{\xi }(\tau _x^{-1}(B))$ for $B\in \mathcal {B}({X^2})$. Define $B_y=\left\{ \overline{x}:(\overline{x},y)\in B\right\}$ and now we have the following

$$\begin{aligned}\mu _{\Lambda x}\otimes \mu _{\xi }(B)&=\int _X\mu _{\Lambda x}(B_y)\mu _{\xi }(dy)=\int _X\mu _{\Lambda x}(\left\{ \overline{x}: (\overline{x},y)\in B\right\} )\mu _{\xi }(dy)\\ {}&=\int _X\mathbb {P}(\left\{ \omega :(\Lambda (\omega ) x,y)\in B\right\} )\mu _{\xi }(dy)\\ {}&=\int _X\mathbb {P}(\left\{ \omega :(\Lambda (\omega ),y)\in \tau _x^{-1}(B)\right\} )\mu _{\xi }(dy)=\mu _{\Lambda }\otimes \mu _{\xi }(\tau _x^{-1}(B)).\end{aligned}$$

Finally by the assumption of independence we obtain

$$\begin{aligned} \mu _{(\Lambda x,\xi )}(B)=\mu _{(\Lambda ,\xi )}(\tau _x^{-1}(B))=\mu _{\Lambda }\otimes \mu _{\xi }(\tau _x^{-1}(B))=\mu _{\Lambda x}\otimes \mu _{\xi }(B), \end{aligned}$$

which ends the proof. $\square$

Lemma 4.4

Let X be a Banach space and $n\in \mathbb {N}$. Assume that $\Lambda :\Omega \rightarrow L(X,X)$ is a random operator and $\psi :\Omega ^{n}\rightarrow X$, $\xi :\Omega \rightarrow X$ are random variables. Define $\psi _n:\Omega ^{\infty }\rightarrow X$, $\Lambda _{n+1}:\Omega ^{\infty }\rightarrow L(X,X)$, $\xi _{n+1}:\Omega ^{\infty }\rightarrow X$ by

$$\begin{aligned} \psi _n(\omega )=\psi (\omega _1,\ldots ,\omega _n), \qquad \Lambda _{n+1}(\omega )=\Lambda (\omega _{n+1}), \qquad \xi _{n+1}(\omega )=\xi (\omega _{n+1}) \end{aligned}$$

and $\Lambda \psi _{n+1}:\Omega ^{\infty }\rightarrow X$ by

$$\begin{aligned} \Lambda \psi _{n+1}(\omega )=\Lambda _{n+1}(\omega )\psi _n(\omega )=\Lambda (\omega _{n+1})\psi (\omega _1,\ldots ,\omega _n), \end{aligned}$$

where $\omega =(\omega _1,\omega _2,\ldots )\in \Omega ^{\infty }.$ If $\Lambda _{n+1}$ and $\xi _{n+1}$ are independent, then $\Lambda \psi _{n+1}$ and $\xi _{n+1}$ are also independent.

Proof

Fix $B\in \mathcal {B}(X^2)$. Put

$$\begin{aligned} \eta (\omega _1,\ldots ,\omega _{n+1})= \Lambda (\omega _{n+1})\psi (\omega _1,\ldots ,\omega _n) \end{aligned}$$

and

$$\begin{aligned} \zeta (\omega _1,\ldots ,\omega _{n+1})= (\eta (\omega _1,\ldots ,\omega _{n+1}),\xi (\omega _{n+1})) \end{aligned}$$

for $\omega _1,\dots ,\omega _{n+1}\in \Omega$. Then

$$\begin{aligned} \mu _{(\Lambda \psi _{n+1},\xi _{n+1})}(B)&=\mathbb {P}^{\infty }\Big (\Big \{(\omega _1, \omega _2\ldots ): \zeta (\omega _1,\ldots ,\omega _{n+1})\in B)\Big \}\Big )\\&=\mathbb {P}^{n+1}\Big (\Big \{(\omega _1,\ldots ,\omega _{n+1}): \zeta (\omega _1,\ldots ,\omega _{n+1})\in B)\Big \}\Big )\\&=\mathbb {P}^{n}\otimes \mathbb {P}\Big (\Big \{(\omega _1,\ldots ,\omega _{n+1}): \zeta (\omega _1,\ldots ,\omega _{n+1})\in B\Big \}\Big )\\&=\int _{\Omega ^n}\mathbb {P}\Big (\Big \{\omega _{n+1}: \zeta (\omega _1,\ldots ,\omega _{n+1})\in B)\Big \}\Big )d\mathbb {P}^n(d(\omega _1,\ldots ,\omega _n))\\&=\int _{\Omega ^n}\mu _{(\Lambda \psi (\omega _1,\ldots ,\omega _n),\xi )}(B)\mathbb {P}^n(d(\omega _1,\ldots ,\omega _n))\\&=\int _{\Omega ^n}\mu _{\Lambda \psi (\omega _1,\ldots ,\omega _n)}\otimes \mu _{\xi }(B)\mathbb {P}^n(d(\omega _1,\ldots ,\omega _n)), \end{aligned}$$

when the last equality holds due to Lemma 4.3. Therefore

$$\begin{aligned} \mu _{(\Lambda \psi _{n+1},\xi _{n+1})}(B)&=\int _{\Omega ^n}\int _X\mu _{\Lambda \psi (\omega _1,\ldots ,\omega _n)}(B_y) \mu _{\xi }(dy)\mathbb {P}^n(\omega _1,\ldots ,\omega _n)\\&=\int _X\int _{\Omega ^n}\mu _{\Lambda \psi (\omega _1,\ldots ,\omega _n)}(B_y) \mathbb {P}^n(d(\omega _1,\ldots ,\omega _n))\mu _{\xi }(dy)\\&=\int _X \mathbb {P}^n\otimes \mathbb {P}\Big (\Big \{(\omega _1,\ldots ,\omega _{n+1}):\eta (\omega _1,\ldots ,\omega _{n+1})\in B_y\Big \}\Big )\mu _{\xi }(dy)\\&=\int _X\mathbb {P}^{\infty }\Big (\Big \{(\omega _1,\omega _2\ldots ): \eta (\omega _1,\ldots ,\omega _{n+1})\in B_y\Big \}\Big )\mu _{\xi }(dy)\\&=\int _X\mathbb {P}^{\infty }\Big (\Big \{\omega : \Lambda \psi _{n+1}(\omega )\in B_y\Big \}\Big )\mu _{\xi }(dy)=\mu _{\Lambda \psi _{n+1}}\otimes \mu _{\xi _{n+1}}(B), \end{aligned}$$

which ends the proof. $\square$

Corollary 4.5

Let X be a separable Banach space. Assume that an rv-function $f:X\times \Omega \rightarrow X$ is given by (1.2), where $\Lambda :\Omega \rightarrow L(X,X)$ is a random operator and $\xi :\Omega \rightarrow X$ is a random variable. If $\Lambda$ and $\xi$ are independent, $x\in X$ and $n\in \mathbb {N}$, then $\Lambda _{n+1}(\cdot )f^n(x,\cdot ):\Omega ^{\infty }\rightarrow X$ with

$$\begin{aligned} \Lambda _{n+1}(\omega )f^n(x,\omega )=\Lambda (\omega _{n+1})f^n(x,\omega _1,\ldots ,\omega _n) \end{aligned}$$

and $\xi _{n+1}:\Omega ^\infty \rightarrow X$ with $\xi _{n+1}(\omega )=\xi (\omega _{n+1})$ are independent.

Having proved independence we also have to characterise the probability distribution of the sum of independent random variables. It is well known that such a distribution can be described as the convolution of each random variable distributions. More precisely, we have:

Theorem 4.6

Let X be a separable Banach space. If $\eta :\Omega \rightarrow X$, $\xi :\Omega \rightarrow X$ are independent random variables, then

$$\begin{aligned} \mu _{\eta +\xi }=\mu _{\eta }*\mu _{\xi }. \end{aligned}$$

Definition 4.7

If $\Lambda :\Omega \rightarrow L(X,X)$ is a random operator, then a map $\Lambda ^*:\Omega \rightarrow L(X,X)$ satisfying

$$\begin{aligned} (\Lambda ^*(\omega )x|y)=(x|\Lambda (\omega )y)\qquad \text {for every} \ \omega \in \Omega , x,y\in X \end{aligned}$$

is called an adjoint random operator to $\Lambda$.

Lemma 4.8

A function $\Lambda ^*:X\times \Omega \rightarrow X$ given by $\Lambda ^*(x,\omega )=\Lambda ^*(\omega )x$ is $\mathcal {B}(X)\otimes \mathcal {A}$-measurable.

Proof

According to Remark 3.3 it is enough to show that $\Lambda ^*(\cdot )x:\Omega \rightarrow X$ is $\mathcal {A}$-measurable for every $x\in X$. Fix $x\in X$ and observe that $(x|\Lambda (\omega )y):\Omega \rightarrow \mathbb {R}$ is $\mathcal {A}$-measurable for every $y\in X$. By the Riesz Representation Theorem for every linear functional $y^*:X\rightarrow \mathbb {R}$ there exists y such that

$$\begin{aligned} y^*\big (\Lambda ^*(\omega )x\big )=(\Lambda ^*(\omega )x|y) \qquad \text {for every} \ \omega \in \Omega . \end{aligned}$$

Therefore from the $\mathcal {A}$-measurability of $(x|\Lambda (\cdot )y):\Omega \rightarrow X$ we conclude that $\Lambda ^*(\cdot )x$ is weak measurable. Since X is separable, we may conclude that $\Lambda ^*(\cdot )x$ is strong measurable and consequently $\mathcal {A}$-measurable. $\square$

Remark 4.9

Note that $\Vert \Lambda ^*(\cdot )\Vert :\Omega \rightarrow [0,\infty )$ is $\mathcal {A}$-measurable due to the equality

$$\begin{aligned} \Vert \Lambda (\omega )\Vert =\Vert \Lambda ^*(\omega )\Vert \ \ \text {for every} \ \omega \in \Omega . \end{aligned}$$

The following theorem characterizes the limit distribution of GRAM’s and it generalizes [5, Theorem 3.1] (see Remark 4.12).

Theorem 4.10

Assume that an rv-function f has the form (1.2) with a random operator $\Lambda :\Omega \rightarrow L(X,X)$ and a random variable $\xi :\Omega \rightarrow X$ such that $\mathbb {E}\Vert \Lambda (\cdot )\Vert <1$, $\mathbb {E}\Vert \xi \Vert <\infty$. Moreover, assume that $\Lambda$ and $\xi$ are independent. Then the characteristic function $\varphi ^f$ of f is the only solution of the equation

$$\begin{aligned} \varphi ^f(u)= \varphi ^{\xi }(u)\int _{\Omega }\varphi ^f(\Lambda ^*(\omega )u)\mathbb {P}(d\omega ), \end{aligned}$$

(4.1)

which is continuous at zero, bounded and fulfills $\varphi ^f(0)=1$.

Lemma 4.11

Let $(\Omega ,\mathcal {A},\mathbb {P})$ be an arbitrary probability space. Suppose that the independent and identically distributed random variables $\zeta _i:\Omega \rightarrow \mathbb {R}$, $i\in \mathbb {N}$ fulfil the following properties

• 1. $\zeta _i \ge 0$

• 2. $0<\mathbb {E}\zeta _i <1$.

Then the sequence $(\prod _{i=1}^{n}\zeta _i)_{n\in \mathbb {N}}$ converges a.s. to zero.

Proof

To show convergence we will consider three cases:

I. If $\mathbb {E}\zeta _i=0=\int _{\Omega }\zeta _i(\omega )\mathbb {P}(d\omega )$, then $\zeta _i=0$ a.s., so is $\prod _{i=1}^{n}\zeta _i$.

II. Assume that $0<\mathbb {E}\zeta _i <1$ and $\mathbb {P}(\zeta _i=0)=p>0$. Then

$$\begin{aligned}\mathbb {P}\Bigg (\Bigg \{\omega \in \Omega : \ {}&\prod _{i=1}^{n}\zeta _i(\omega )\ne 0\Bigg \}\Bigg )\\ {}&=\mathbb {P}\Big (\Big \{\omega \in \Omega : \zeta _i(\omega )\ne 0, \ \text{ for } \text{ every } \ i\in \left\{ 1,\ldots , n\right\} \Big \}\Big ) \\ {}&= \prod _{i=1}^{n}\mathbb {P}\Big (\Big \{\omega \in \Omega : \zeta _i(\omega )\ne 0 \Big \}\Big )=(1-p)^n. \end{aligned}$$

Define a set $A_n=\left\{ \omega \in \Omega : \ \prod _{i=1}^{n}\zeta _i(\omega )\ne 0\right\}$ and observe that $A_{n+1}\subset A_n$, and

$$\begin{aligned} A=\bigcap _{n=1}^{\infty }A_n\supset \left\{ \omega \in \Omega : \ \prod _{i=1}^{\infty }\zeta _i(\omega )\ne 0\right\} . \end{aligned}$$

By the continuity of the measure it follows that

$$\begin{aligned} \mathbb {P}\left( \left\{ \omega \in \Omega : \ \prod _{i=1}^{\infty }\zeta _i(\omega )\ne 0\right\} \right) =0. \end{aligned}$$

III. Now assume that $0<\mathbb {E}\zeta _i <1$, and $\mathbb {P}(\zeta _i=0)=0$. From Jensen’s inequality we have $\mathbb {E}\log \zeta _i\le \log \mathbb {E}\zeta _i<0$. Observe that

$$\begin{aligned} \prod _{i=1}^{n}\zeta _i=e^{\log \prod _{i=1}^{n}\zeta _i}=\left( e^{\frac{1}{n}\sum _{i=1}^{n}\log \zeta _i}\right) ^n. \end{aligned}$$

If $-\infty <\mathbb {E}\log \zeta _1$ then by the independence of $\zeta _i's$ we can apply the Strong Law of Large Numbers, hence for $0<\epsilon <|\mathbb {E}\log \zeta _1|$ there exists $N_{\epsilon }\in \mathbb {N}$ such that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\log \zeta _i<\mathbb {E}\log \zeta _1+\epsilon \qquad \text {for every} \ n>N_{\epsilon }. \end{aligned}$$

Therefore for the same $n>N_{\epsilon }$ it holds that

$$\begin{aligned} \left( e^{\frac{1}{n}\sum _{i=1}^{n}\log \zeta _i}\right) ^n<e^{n(\mathbb {E}\log \zeta _1+\epsilon )}. \end{aligned}$$

Passing with n to the limit we obtain

$$\begin{aligned} \prod _{n=1}^{n}\zeta _i\xrightarrow {n\rightarrow \infty } 0\qquad \text {a.s.} \end{aligned}$$

(4.2)

If $\mathbb {E}\log \zeta _1=-\infty$, then we can apply theorem [10, Theorem 2.4.5], from which we conclude that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}\log \zeta _i\xrightarrow {n\rightarrow \infty }-\infty \qquad \text {a.s.} \end{aligned}$$

Hence

$$\begin{aligned} \left( \prod _{n=1}^{n}\zeta _i\right) ^{\frac{1}{n}}=e^{\frac{1}{n}\sum _{i=1}^n\log \zeta _i}\xrightarrow {n\rightarrow \infty } 0\qquad \text {a.s.} \end{aligned}$$

Summarizing we get convergence in all cases. $\square$

Proof of Theorem 4.10

A random operator $\Lambda :\Omega \rightarrow L(X,X)$ can be considered as an rv-function $\Lambda :X\times \Omega \rightarrow X$ due to its measurability (see Sect. 3) and consequently we can associate it with a linear operator Q given by

$$\begin{aligned} Q\mu (B)=\int _{X}\int _{\Omega }\mathbbm {1}_B(\Lambda (\omega )x)\mathbb {P}(d\omega )\mu (dx), \ \ \text {for} \ B\in \mathcal {B}(X). \end{aligned}$$

Now let us define $\pi ^{\Lambda f}_n:X\times \mathcal {B}(X)\rightarrow [0,1]$ by

$$\begin{aligned} \pi ^{\Lambda f}_n(x,B)=\mathbb {P}^{\infty }(\{(\omega _1,\omega _2\ldots ):\Lambda (\omega _{n+1})f^n(x,\omega _1,\ldots ,\omega _n)\in B\}) \end{aligned}$$

and observe that

$$\begin{aligned} \pi ^{\Lambda f}_n(x,\cdot )=Q\pi _{n}^f(x,\cdot ) \quad \text {for every} \ x\in X. \end{aligned}$$

Indeed, for fixed $x\in X$, $B\in \mathcal {B}(X)$ it holds that

$$\begin{aligned} \pi ^{\Lambda f}_n(x,B)&=\mathbb {P}^{\infty }(\{(\omega _1,\omega _2\ldots ):\Lambda (\omega _{n+1})f^n(x,\omega _1,\ldots ,\omega _n)\in B\})\\&=\int _{\Omega ^{\infty }}\mathbbm {1}_B(\Lambda (\omega _{n+1})f^n(x,\omega _1,\ldots ,\omega _n))\mathbb {P}^{\infty }(d(\omega _1,\omega _2\ldots ))\\&=\int _{\Omega }\int _{\Omega ^{\infty }}\mathbbm {1}_B(\Lambda (\overline{\omega })f^n(x,\omega _1,\ldots ,\omega _n))\mathbb {P}(d\overline{\omega })\mathbb {P}^{\infty }(d(\omega _1,\omega _2\ldots ))\\&=\int _{\Omega }\int _X\mathbbm {1}_B(\Lambda (\overline{\omega })y)\pi ^f_n(x,dy)\mathbb {P}(d\overline{\omega })=Q\pi _n^f(x,B). \end{aligned}$$

So now, by Corollary 4.5 and Theorem 4.6 we see that

$$\begin{aligned} \pi ^{ f}_{n+1}(x,\cdot )=\pi ^{\Lambda f}_{n}(x,\cdot )*\mu _{\xi } =Q\pi _{n}^f(x,\cdot )*\mu _{\xi }. \end{aligned}$$

It can be easily shown that the Markov operator Q has the Feller property. To do this let us see at first that

$$\begin{aligned} Q^*\psi (x)=\int _{\Omega }\psi (\Lambda (\omega )x)\mathbb {P}(d\omega ). \end{aligned}$$

For a fixed $\psi \in C(X)$ take an arbitrary $x_0\in X$ and note that for every $(x_n)_{n\in \mathbb {N}}$ such that $x_n\xrightarrow []{n\rightarrow \infty }x_0$ we have $\psi (\Lambda (\omega )x_n)\xrightarrow []{n\rightarrow \infty }\psi (\Lambda (\omega )x_0)$ for every $\omega \in \Omega$. Let us define $\varphi _n(\omega )=\psi (\Lambda (\omega )x_n)$ and $\varphi _0(\omega )=\psi (\Lambda (\omega )x_0)$. Since $|\varphi _n(\omega )|\le \Vert \psi \Vert _{\infty }$ for $\omega \in \Omega$, $n\in \mathbb {N}$ we can apply the Lebesgue Dominated Convergence theorem and hence

$$\begin{aligned} Q^*\psi (x_n)=\int _{\Omega } \varphi _n(\omega )\mathbb {P}(d\omega )\xrightarrow []{n\rightarrow \infty }\int _{\Omega } \varphi _0(\omega )\mathbb {P}(d\omega )=Q^*\psi (x_0). \end{aligned}$$

Because $x_0$, $(x_n)_{n\in \mathbb {N}}$ and $\psi$ are arbitrary, we have $Q^*(C(X))\subset C(X)$. From that and [18, Theorem 1.1, Ch. III] we can pass n to the limit and we obtain

$$\begin{aligned} \pi ^f=Q\pi ^f*\mu _{\xi }. \end{aligned}$$

Now from the definition of the characteristic function we make the following computations

$$\begin{aligned}\varphi ^f(u)&=\int _Xe^{i(u|z)}\pi ^f(dz) =\int _Xe^{i(u|z)}Q\pi ^f*\mu _{\xi }(dz)\\&=\int _X\int _Xe^{i(u|x+y)}Q\pi ^f(dx)\mu _{\xi } (dy)\\&=\int _X\int _Xe^{i(u|x)}\cdot e^{i(u|y)}Q\pi ^f(dx)\mu _{\xi } (dy)\\&=\int _X\int _XQ^*e^{i(u|x)}\cdot e^{i(u|y)}\pi ^f(dx)\mu _{\xi } (dy)\\&=\int _X\int _X\left[ \int _{\Omega }e^{i(u|\Lambda (\omega )x)}\mathbb {P}(d\omega )\right] \cdot e^{i(u|y)}\pi ^f(dx)\mu _{\xi } (dy)\\&=\int _X e^{i(u|y)}\mu _{\xi }(dy)\cdot \int _{\Omega }\int _X e^{i(u|\Lambda (\omega )x)}\pi ^f(dx)\mathbb {P}(d\omega )\\&=\varphi ^{\xi }(u)\int _{\Omega }\int _X e^{i(\Lambda ^*(\omega )u|x)}\pi ^f(dx)\mathbb {P}(d\omega ) =\varphi ^{\xi }(u)\int _{\Omega }\varphi ^f(\Lambda ^*(\omega )u)\mathbb {P}(d\omega ). \end{aligned}$$

This shows that $\varphi ^f$ satisfies (4.1).

It remains to show the uniqueness of the solution of (4.1). To do this, let us assume that $\varphi$ is a bounded, continuous at zero solution of (4.1) and $\varphi (0)=1$. Then observe that

$$\begin{aligned} \varphi (u)&=\int _{\Omega }\ldots \int _{\Omega }\varphi ^{\xi }(u)\prod _{i=2}^{n}\varphi ^{\xi }((\Lambda ^*)^{i-1}(\omega _1,\ldots ,\omega _{i-1})u)\times \\&\quad \times \varphi ((\Lambda ^*)^n(\omega _1,\ldots ,\omega _n )u)\mathbb {P}(d\omega _1)\ldots \mathbb {P}(d\omega _n), \end{aligned}$$

where

$$\begin{aligned} (\Lambda ^*)^i(\omega _1,\ldots ,\omega _i)u=\Lambda ^*(\omega _i)\circ \ldots \circ \Lambda ^*(\omega _1)u. \end{aligned}$$

It follows that for every $n\in \mathbb {N}$ we can write

$$\begin{aligned} \varphi (u) =\int _{\Omega ^{\infty }}\prod _{i=1}^{n}\varphi ^{\xi }((\Lambda ^*)^{i-1}(\omega )u)\varphi ((\Lambda ^*)^n(\omega )u)\mathbb {P}^{\infty }(d\omega ). \end{aligned}$$

(4.3)

Since $\Vert \Lambda ^*(\omega ) \Vert =\Vert \Lambda (\omega ) \Vert$ for every $\omega \in \Omega$, we have $\mathbb {E} \Vert \Lambda ^*(\cdot )\Vert =\mathbb {E} \Vert \Lambda (\cdot )\Vert <1$. Taking $\zeta _i(\omega )= \Vert \Lambda ^*(\omega _i)\Vert$ for $\omega =(\omega _1,\omega _2,\ldots )\in \Omega ^{\infty }$ we see that

$$\begin{aligned} \Vert (\Lambda ^*)^n(\omega )u\Vert \le \Vert u\Vert \prod _{i=1}^{n}\zeta _i(\omega ). \end{aligned}$$

By Lemma 4.11 we conclude that the sequence $\big (\Vert (\Lambda ^*)^n(\cdot )(u)\Vert \big )_{n\in \mathbb N}$ converges a.s. to zero.

Fix $n\in \mathbb {N}$ and let us define random variables $\eta _n, \theta _n:\Omega ^{\infty }\rightarrow \mathbb {C}$, respectively, by

$$\begin{aligned} \eta _n(\omega )=\prod _{i=1}^n\varphi ^{\xi }((\Lambda ^*)^{i-1}(\omega )u) \quad {\text{ a }nd}\quad \theta _n(\omega )=\varphi ((\Lambda ^*)^n(\omega )u). \end{aligned}$$

Hence we can rewrite (4.3) as

$$\begin{aligned} \varphi (u)=\int _{\Omega ^{\infty }}\theta _n(\omega )\eta _n(\omega )\mathbb {P}^{\infty }(d\omega ), \ \ n\in \mathbb {N}, u\in X \end{aligned}$$

and thus we obtain

$$\begin{aligned}\Bigg |\int _{\Omega ^{\infty }}\theta _n(\omega )\eta _n(\omega )\mathbb {P}^{\infty }(d\omega )&-\int _{\Omega ^{\infty }}\eta _n(\omega )\mathbb {P}^{\infty }(d\omega )\Bigg |\\&\le \int _{\Omega ^{\infty }}|\theta _n(\omega )-1|\cdot |\eta _n(\omega )|\mathbb {P}^{\infty }(d\omega )\\&\le \int _{\Omega ^{\infty }}|\theta _n(\omega )-1|\mathbb {P}^{\infty }(d\omega ). \end{aligned}$$

Observe that $|\theta _n(\omega )-1|\le \Vert \varphi \Vert _{\infty }+1$ and $(\theta _n)_{n\in \mathbb N}$ converges a.s. to 1, by the continuity of $\varphi$ at zero. Therefore, from the Lebesgue dominated convergence theorem it can be concluded that

$$\begin{aligned} \int _{\Omega ^{\infty }}|\theta _n(\omega )-1|\mathbb {P}^{\infty }(d\omega )\xrightarrow []{n\rightarrow \infty }0. \end{aligned}$$

Hence passing with n to the limit we obtain

$$\begin{aligned} \varphi (u)=\lim _{n\rightarrow \infty }\int _{\Omega ^{\infty }}\prod _{i=1}^n\varphi ^{\xi }((\Lambda ^*)^{i-1}(\omega )u)\mathbb {P}^{\infty }(d\omega ),\end{aligned}$$

(4.4)

which completes the proof. $\square$

Remark 4.12

Note that under the assumptions of Theorem 4.10 the following statements hold:

• (i) The characteristic function $\varphi ^f$ is the only solution of the equation (4.1), which is Lipschitz, continuous at zero and $\varphi (0)=1$.

• (ii) If $\Lambda$ does not depend on $\omega$, i.e. $\Lambda (\omega )$ is the same as $\omega$ changes, then $\varphi ^f$ is the only solution of the equation (4.1), which is continuous at zero and $\varphi (0)=1$.

To show assertion (i) observe that for a function $\varphi$ which is a solution of (4.1) and $M>0$, a Lipschitz constant of $\varphi$, the following inequalities hold,

$$\begin{aligned} \int _{\Omega ^{\infty }}|\varphi ((\Lambda ^*)^n(\omega )u)-1|\mathbb {P}^{\infty }(d\omega )&\le \int _{\Omega ^{\infty }}M\Vert (\Lambda ^*)^n(\omega )u)\Vert \mathbb {P}^{\infty }(d\omega ) \\&\le \Vert u\Vert M(\mathbb {E}\Vert \Lambda ^*(\cdot )\Vert )^n, \end{aligned}$$

which yields (4.4).

When (ii) holds, the formula (4.3) reduces to

$$\begin{aligned} \varphi (u)=\prod _{i=1}^{n}\varphi ^{\xi }((\Lambda ^*)^{i-1}u)\varphi ((\Lambda ^*)^nu) \end{aligned}$$

for any $n\in \mathbb N$. Passing with n to the limit we obtain

$$\begin{aligned} \varphi (u)=\prod _{i=1}^{\infty }\varphi ^{\xi }((\Lambda ^*)^{i-1}u)\varphi ((\Lambda ^*)^nu). \end{aligned}$$

(4.5)

$\square$

Remark 4.13

Note that the expression (4.4) is in fact the formula of the unique solution $\varphi$ of (4.1). In particular, when $\Lambda$ is independent of $\omega$, this solution takes the form (4.5) and it can also be found in [5, Theorem 3.1].

We now give an example of a GRAM which satisfies the assumptions of Theorem 4.10.

Example 4.14

Let us consider random variables $\xi :\Omega \rightarrow X$ and $\kappa :\Omega \rightarrow \mathbb {N}$. Take a countable family of linear bounded operators $T_i:X\rightarrow X$, $i\in \mathbb {N}$. We define $\Lambda :\Omega \rightarrow L(X,X)$ as

$$\begin{aligned} \Lambda (\omega )=T_{\kappa {(\omega )}}, \ \ \text {for} \ \omega \in \Omega . \end{aligned}$$

Then the following statements hold:

• (i) $\Lambda$ is a random operator.

• (ii) If $\xi$ and $\kappa$ are independent, then so are $\xi$ and $\Lambda$.

• (iii) The expected value of $\Lambda$ is equal to

  $$\begin{aligned} \mathbb {E}\Vert \Lambda (\cdot )\Vert =\sum _{i\in \mathbb {N}}\mu _{\kappa }(\left\{ i\right\} )\Vert T_i\Vert . \end{aligned}$$

• (iv) The adjoint random operator $\Lambda ^*$ has the form

  $$\begin{aligned} \Lambda ^*(\omega )=T^*_{\kappa (\omega )}. \end{aligned}$$

Assertion (i) follows from the fact that $\Lambda$ can be rewritten in the form

$$\begin{aligned} \Lambda (\omega )=\sum _{i\in \mathbb {N}}\mathbbm {1}_{\kappa ^{-1}(\left\{ i\right\} )}(\omega )T_i, \ \ \text {for} \ \omega \in \Omega . \end{aligned}$$

Hence it can be easily seen that $\Lambda$ is $\mathcal {A}$-measurable. To show statement (ii) assume that $\xi$ and $\kappa$ are independent and observe that $\mu _{\Lambda }$ has the form

$$\begin{aligned}\mu _{\Lambda }(A)&=\mathbb {P}\left( \bigcup _{i\in \mathbb {N}}\left\{ \omega :\kappa (\omega )=i\right\} \cap \left\{ \omega :T_i\in A\right\} \right) \\&=\sum _{i\in \mathbb {N}}\mathbb {P}(\left\{ \omega :\kappa (\omega )=i\right\} \cap \left\{ \omega :T_i\in A\right\} )\end{aligned}$$

and

$$\begin{aligned}\mathbb {P}(\left\{ \omega :\kappa (\omega )=i\right\} \cap \left\{ \omega :T_i\in A\right\} )&={\left\{ \begin{array}{ll} \mathbb {P}(\left\{ \omega :\kappa (\omega )=i\right\} ), &{} T_i\in A \\ 0, &{} T_i\notin A \end{array}\right. }\\&=\mu _{\kappa }(\left\{ i\right\} )\delta _{T_i}(A).\end{aligned}$$

From that

$$\begin{aligned}\mu _{\Lambda }(A)&=\sum _{i\in \mathbb {N}}\mu _{\kappa }(\left\{ i\right\} )\delta _{T_i}(A).\end{aligned}$$

Now fix $B\in \mathcal {B}(L(X,X))\otimes \mathcal {B}(X)$, define $B_T\in \mathcal {B}(\mathbb {N})\otimes \mathcal {B}(X)$ as

$$\begin{aligned} B_T=\left\{ (i,y)\in \mathbb {N}\times X:(T_i,y)\in B \right\} \end{aligned}$$

and observe that

$$\begin{aligned} B^{T_i}=\left\{ y\in X:(T_i,y)\in B \right\} =(B_T)^i, \end{aligned}$$

where $B^x=\left\{ y\in X:(x,y)\in B\right\}$, $x\in L(X,X)$. An easy computation shows that

$$\begin{aligned} \mu _{\Lambda }\otimes \mu _{\xi }(B)&=\int _{L(X,X)}\mu _{\xi }(B^x)\mu _{\Lambda }(dx)=\sum _{i\in \mathbb {N}}\mu _{\xi }(B^{T_i})\cdot \mu _{\kappa }(\left\{ i\right\} )\\&=\int _{\mathbb {N}}\mu _{\xi }((B_T)^i)\mu _{\kappa }(di)=\mu _{\kappa }\otimes \mu _{\xi }(B_T)=\mu _{(\kappa ,\xi )}(B_T)\\&=\mathbb {P}(\omega :(\kappa (\omega ),\xi (\omega ))\in B_T)=\mathbb {P}(\omega :(T_{\kappa (\omega )},\xi (\omega ))\in B))\\&=\mu _{(\Lambda ,\xi )}(B). \end{aligned}$$

Statement (iii) is obvious. Finally to show (iv) fix $i\in \mathbb {N}$ and observe that for $\omega \in \kappa ^{-1}(\left\{ i\right\} )$ we have

$$\begin{aligned} (\Lambda ^*(\omega )x|y)=(x|T_iy)=(T^*_ix|y) \ \ \text {for every} \ x,y\in X. \end{aligned}$$

Therefore $\Lambda ^*(\omega )=T_i^*$ for $\omega \in \kappa ^{-1}(\left\{ i\right\} )$. From that we obtain

$$\begin{aligned} \Lambda ^*(\omega )=\sum _{i\in \mathbb {N}}\mathbbm {1}_{\kappa ^{-1}(\left\{ i\right\} )}(\omega )T^*_i=T^*_{\kappa (\omega )}, \ \ \text {for} \ \omega \in \Omega . \end{aligned}$$

By statements (i)–(iv) we can consider an rv-function f of the form

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

and if we assume additionally that

$$\begin{aligned} \sum _{i\in \mathbb {N}}\mu _{\kappa }(\left\{ i\right\} )\Vert T_i\Vert<1\quad \textrm{and}\quad \mathbb {E}\Vert \xi \Vert <\infty , \end{aligned}$$

then Theorem 4.2 allows us to claim that (provided that $\kappa$ and $\xi$ are independent) the characteristic function $\varphi ^f$ is the only solution of the equation

$$\begin{aligned} \varphi (u)=\varphi ^{\xi }(u)\sum _{i\in \mathbb {N}}\mu _{\kappa }(\left\{ i\right\} )\varphi (T^*_iu), \ \ u\in X,\end{aligned}$$

(4.6)

which is bounded, continuous at zero and $\varphi (0)=1$. $\square$

It is worth pointing out that if we consider the class of solutions $\varphi$ of the equation (4.1) (or in particular of (4.6)) which do not have to be either bounded or Lipschitz, then such a class can contain more than one solution, which is shown in the example given below.

Example 4.15

Fix $a\in \mathbb {R}$ such that $|a|>1$ and $p\in \left( 0,\frac{1}{1+|a|}\right)$ and let $X=\mathbb {R}$. Let operators $T_i:\mathbb {R}\rightarrow \mathbb {R}, i\in \left\{ 1,2\right\}$ be given, respectively, by

$$\begin{aligned} T_1x=ax, \qquad T_2x=\frac{1}{a}x. \end{aligned}$$

Set a random variable $\kappa :\Omega \rightarrow \mathbb {N}$ with the following distribution

$$\begin{aligned} \mu _{\kappa }(\left\{ 1\right\} )=p, \qquad \mu _{\kappa }(\left\{ 2\right\} )=1-p. \end{aligned}$$

It can be easily seen that for a random operator $\Lambda$ given by

$$\begin{aligned} \Lambda (\omega )=T_{\kappa (\omega )}=\mathbbm {1}_{\kappa ^{-1}(\left\{ 1\right\} )}(\omega )T_1+\mathbbm {1}_{\kappa ^{-1}(\left\{ 2\right\} )}(\omega )T_2 \end{aligned}$$

we have

$$\begin{aligned} \mathbb {E}\Vert \Lambda (\cdot )\Vert =|a|\cdot p +\left| \frac{1}{a}\right| (1-p)<\frac{|a|^2-1}{|a|^2+|a|}+\frac{1}{|a|}=1. \end{aligned}$$

Observe furthermore that $\Lambda$ and $\Lambda ^*$ have the same distribution.

Now consider a random variable $\xi :\Omega \rightarrow \mathbb {R}$, independent of $\kappa$, with $\mu _{\xi }=\delta _0$. Then $\varphi ^\xi \equiv 1$. It is easy to check that $\varphi ^f\equiv 1$ and it is a solution of the equation

$$\begin{aligned} \varphi (u)=p\varphi (au)+(1-p)\varphi \left( \frac{u}{a}\right) .\end{aligned}$$

(4.7)

However it is not unique in a family of continuous at zero functions $\varphi$ which satisfy $\varphi (0)=1$. To this end, take a function $\varphi _0:\mathbb {R}\rightarrow \mathbb {R}$ with

$$\begin{aligned} \varphi _0(u)=|u|^{\log _{|a|}\left( \frac{1-p}{p}\right) }+1. \end{aligned}$$

Let us see that $\varphi _0$ is continuous on its domain, $\varphi _0(0)=1$ and

$$\begin{aligned}p\varphi _0(au)&+(1-p)\varphi _0\left( \frac{u}{a}\right) =p|u|^{\log _{|a|}\left( \frac{1-p}{p}\right) }\cdot |a|^{\log _{|a|}\left( \frac{1-p}{p}\right) }\\&+(1-p)|u|^{\log _{|a|}\left( \frac{1-p}{p}\right) }\cdot |a|^{-\log _{|a|}\left( \frac{1-p}{p}\right) }+1\\&=|u|^{\log _{|a|}\left( \frac{1-p}{p}\right) }+1=\varphi _0(u),\end{aligned}$$

so $\varphi ^f$ is not the unique continuous solution of the equation (4.7) having value 1 at zero. $\square$

For GRAM’s f given above, the natural question arises whether an operator $(\Lambda ,\xi ) \longmapsto \varphi ^f$ is continuous and what kind of continuity it has. Before we formulate an appropriate result, we present some additional facts in which $(X,\rho )$ is a metric space and

$$\begin{aligned} \ Lip_\alpha (X,Y)=\left\{ \varphi \in B(X,Y) :\Vert \varphi (x)-\varphi (y)\Vert \le \alpha \rho (x,y), \ x,y\in X \right\} \end{aligned}$$

for $\alpha \in (0,\infty )$, and B(X, Y) is a set of all bounded functions acting on X into Y.

Definition 4.16

Let $(X,\rho )$ be a separable and complete metric space and let $(Y,\Vert \cdot \Vert )$ be a Banach space. We denote a metric $d_H^{X,Y}$ on $\mathcal M_1(X)$ by the formula

$$\begin{aligned} d_H^{X,Y}(\mu ,\nu )=\sup {\left\{ \left\| \int _X\varphi (x)\mu (dx)-\int _X \varphi (x) \nu (dx)\right\| :\varphi \in Lip_1(X,Y) \right\} }. \end{aligned}$$

Proposition 4.17

Assume that spaces X and Y are nontrivial. Then the metric $d^{X,Y}_H$ is independent of the choise spaces X and Y, and moreover $d^{X,Y}_H(\mu ,\nu )=d_H(\mu ,\nu )$ for every $\mu ,\nu \in \mathcal {M}_1(X)$.

Proof

Fix $u\in Lip_1(X)$ and $x_0\in Y$ such that $\Vert x_0\Vert =1$. Put $\varphi _0 (x)=u(x)\cdot x_0$ for $x\in X$, then $\varphi _0\in Lip_1 (X,Y)$ and it is integrable in Bochner’s sense with respect to any probability measure, so we have

$$\begin{aligned}&\left| \int _X u(x)\mu (dx)-\int _X u(x)\nu (dx)\right| \\&\qquad \qquad \qquad \ =\frac{1}{\Vert x_0\Vert }\cdot \left\| x_0\left( \int _X u(x)\mu (dx)- \int _X u(x)\nu (dx) \right) \right\| \\&\qquad \qquad \qquad \ =\left\| \int _X\varphi _0 (x)\mu (dx)-\int _X \varphi _0 (x)\nu (dx)\right\| \le d^{X,Y}_H(\mu ,\nu ).\end{aligned}$$

Since u is arbitrary, we can take the supremum on the left hand side of the inequality and as a consequence we obtain $d_H\le d^{X,Y}_H.$

Now fix $\varphi \in Lip_1(X,Y)$ and $\mu , \nu \in \mathcal {M}_1(X)$. Then there exists $y^*\in Y^*$ such that $\Vert y^*\Vert =1$ and

$$\begin{aligned} \left\| \int _X\varphi (x) \mu (dx)-\int _X\varphi (x) \nu (dx)\right\| = \left| y^*\left( \int _X\varphi (x) \mu (dx)-\int _X\varphi (x) \nu (dx)\right) \right| \end{aligned}$$

by the Hahn–Banach theorem. Applying the Hille Theorem (see e.g. [8, Theorem 6 Ch. II]) we deduce that

$$\begin{aligned}&\left| y^*\left( \int _X\varphi (x) \mu (dx)-\int _X\varphi (x) \nu (dx)\right) \right| \\&\qquad \qquad \qquad \qquad \ =\left| \int _X y^*\circ \varphi (x) \mu (dx)-\int _X y^*\circ \varphi (x) \nu (dx)\right| \le d_H(\mu ,\nu ),\end{aligned}$$

and since $y^*\circ \varphi \in Lip_1(X)$ we finally obtain $d_H\ge d^{X,Y}_H.$ $\square$

Lemma 4.18

If $u\in X\setminus \{0\}$ and a function $\psi :X\rightarrow \mathbb {C}$ is given by $\psi (z)= e^{i(u|z)}$, then $\psi \in Lip_{\Vert u\Vert }(X,\mathbb {C})$.

Proof

Since $(u|z)\in \mathbb {R}$ for every $u,z\in X$, it follows that

$$\begin{aligned}\left| \psi (z)-\psi (y)\right|&=\left| e^{i(u|z)}-e^{i(u|y)}\right| =\sqrt{2-2\cos {((u|z)-(u|y))}}\\&=2\left| \sin {\frac{(u|z-y)}{2}}\right| \le \left| 2\cdot \frac{(u|z-y)}{2}\right| \le \Vert u\Vert \cdot \Vert z-y\Vert .\end{aligned}$$

Then the proof is completed. $\square$

Proposition 4.19

Let $f,g:X\times \Omega \rightarrow X$ be rv-functions. Assume that the iterates $(f^n(x, \cdot ))_{n\in \mathbb {N}}$, $(f^n(x, \cdot ))_{n\in \mathbb {N}}$ converge in law to $\pi ^f$ and $\pi ^g$, respectively, and the limits $\pi ^f, \pi ^g$ do not depend on x. Then the following inequality for the characteristic functions $\varphi ^f$ and $\varphi ^g$ holds

$$\begin{aligned} \left| \varphi ^f(u)-\varphi ^g(u)\right| \le \Vert u\Vert \cdot d_H(\pi ^f,\pi ^g),\end{aligned}$$

(4.8)

for every $u\in X$.

Proof

Fix $u\in X\setminus \{0\}$ and define $\psi :X\rightarrow \mathbb {C}$ as $\psi (z)=e^{i(u|z)}$. Then $\frac{1}{\Vert u\Vert }\psi \in Lip_{1}(X,\mathbb {C})$, by Lemma 4.18. Using Proposition 4.17 we see that

$$\begin{aligned}\frac{1}{\Vert u\Vert }\left| \varphi ^f(u)-\varphi ^g(u)\right|&=\left| \int \frac{1}{\Vert u\Vert }e^{i(u|z)}\pi ^f(dz)-\int \frac{1}{\Vert u\Vert }e^{i(u|z)}\pi ^g(dz)\right| \\&\le d^{X,\mathbb {C}}_H(\pi ^f,\pi ^g)=d_H(\pi ^f,\pi ^g).\end{aligned}$$

This ends the proof. $\square$

Remark 4.20

Inequality (4.8) can not be strengthened by

$$\begin{aligned} \left\| \varphi ^f-\varphi ^g\right\| _{\infty }\le d_H(\pi ^f,\pi ^g),\end{aligned}$$

(4.9)

which is shown in the example given below.

Example 4.21

Fix $a\in \mathbb {R}$. For $n\in \mathbb {N}$ let $\xi _n:\Omega \rightarrow X$ be a random variable with uniform distribution on the interval $\left[ a,a+\frac{1}{n}\right]$. (Obviously, we assume such ${\xi _n}'s$ can be constructed. It is possible for instance on the space $(\Omega ,\mathcal A,\mathbb P)$ as a unit interval with Lebesgue measure.) Define rv-functions $f_n,g:X\times \Omega \rightarrow X$ by

$$\begin{aligned} f_n(x,\omega )=\xi _n(\omega ),\quad g(x,\omega )=a. \end{aligned}$$

Observe that the k-th iterate of $f_n$ satisfies $f_n^k(x,\omega _1,\ldots ,\omega _k)=\xi _n(\omega _k)$ and $g^k(x,\omega )=a$. So we can write

$$\begin{aligned}\pi _k^{f_n}(A)&=\mathbb {P}^{\infty }\left( \left\{ (\omega _1,\omega _2,\ldots )\in \Omega ^{\infty }:f_n^{k}(x,\omega _1,\ldots ,\omega _k)\in A \right\} \right) \\&=\mathbb {P}^{\infty }\left( \left\{ (\omega _1,\omega _2,\ldots )\in \Omega ^{\infty }:\xi _n(\omega _k)\in A\right\} \right) \\&=\mathbb {P}\left( \left\{ \omega _k\in \Omega :\xi _n(\omega _k)\in A\right\} \right) =\int _A n\mathbbm {1}_{[a,n+\frac{1}{n}]}dx =\pi ^{f_n}(A).\end{aligned}$$

Additionally let us see that

$$\begin{aligned} \pi _k^g(A)=\delta _a(A)=\pi ^g(A). \end{aligned}$$

The characteristic functions of the above distributions have the following forms

$$\begin{aligned}\varphi ^{f_n}(u)&=\int _{\mathbb {R}}e^{iux}\pi ^{f_n}(dx)=\frac{n}{iu}e^{iua}\left( e^{iu\frac{1}{n}}-1\right) ,\\ \varphi ^g(u)&=\int _{\mathbb {R}}e^{iux}\pi ^{g}(dx)=e^{iua}.\end{aligned}$$

For every $c\in Lip_1(\mathbb {R})$ we have the following computation

$$\begin{aligned}\Bigg |\int _{\mathbb {R}}c(x)\pi ^{f_n}(dx)&-\int _{\mathbb {R}}c(x)\pi ^g(dx)\Bigg |\\&=\Bigg |n\int _{\mathbb {R}}c(x)\cdot \mathbbm {1}_{[a,a+\frac{1}{n}]} dx-c(a)\Bigg |\\&=\Bigg |n\int _{\mathbb {R}}c(x)\cdot \mathbbm {1}_{[a,a+\frac{1}{n}]} dx-n\int _{\mathbb {R}}c(a)\cdot \mathbbm {1}_{[a,a+\frac{1}{n}]} dx\Bigg |\\&\le n\int _{\mathbb {R}}|x-a|\mathbbm {1}_{[a,a+\frac{1}{n}]}dx=\frac{1}{2n}.\end{aligned}$$

Taking supremum over all $c\in Lip_1(\mathbb {R})$ we obtain

$$\begin{aligned} d_H(\pi ^{f_n},\pi ^g)\le \frac{1}{2n}\xrightarrow {n\rightarrow \infty }0. \end{aligned}$$

It is easily seen that $\varphi ^{f_n}(u)\xrightarrow {n\rightarrow \infty }\varphi ^g(u)$ for every $u\in X$, but

$$\begin{aligned} \left| \varphi ^{f_n}(u)-\varphi ^g(u)\right| \xrightarrow {u\rightarrow +\infty }1 \end{aligned}$$

for every $n\in \mathbb {N}$. From that

$$\begin{aligned} \left\| \varphi ^{f_n}-\varphi ^g\right\| _{\infty }\ge 1 \ \ \text {for every} \ n\in \mathbb {N}. \end{aligned}$$

Therefore the sequence $\big (\varphi ^{f_n}\big )_{n\in \mathbb N}$ is not convergent to $\varphi ^g$ in the supremum norm $\Vert \cdot \Vert _{\infty }$. $\square$

Now we turn to formulating the second theorem of this section that extends [4, Theorem 3]. We note that in this theorem a real separable Hilbert space X is considered and $\varphi ^f,\varphi ^g$ denote the characteristic functions of $\pi ^f, \pi ^f$, which result from Theorem 3.4. The announced theorem is a straightforward consequence of Theorem 3.4 and Lemma 4.19, and reads as follows.

Theorem 4.22

Assume that rv-functions f, g satisfy (U$_f$) and (U$_g$), respectively. Then

$$\begin{aligned}\left| \varphi ^f(u)-\varphi ^g(u)\right| \le \Vert u\Vert \cdot \min&\left\{ \frac{1}{1-\mathbb {E}\Vert \Lambda _f(\cdot )\Vert }\left( \frac{\mathbb {E}\Vert \xi _g\Vert }{1-\mathbb {E}\Vert \Lambda _g(\cdot )\Vert }\alpha +\beta \right) ,\right. \\&\left. \frac{1}{1-\mathbb {E}\Vert \Lambda _g(\cdot )\Vert }\left( \frac{\mathbb {E}\Vert \xi _f\Vert }{1-\mathbb {E}\Vert \Lambda _f(\cdot )\Vert }\alpha +\beta \right) \right\} ,\end{aligned}$$

where $\alpha =\mathbb {E}\Vert \Lambda _f(\cdot )-\Lambda _g(\cdot )\Vert , \ \beta =\mathbb {E}\Vert \xi _f-\xi _g\Vert .$

Remark 4.23

The main results of [4, 5] concern rv-functions of the form $f(x,\omega )=\Lambda x+\xi _f(\omega )$ with $\Lambda \in L(X,X)$. In particular the author examines a kind of continuity of the operator $\xi _f\longmapsto \varphi ^f$. Note that this is one case in our results, when $\alpha =0$. Under appropriate assumptions we have

$$\begin{aligned} d_H(\pi ^f,\pi ^g)\le \frac{\mathbb E\Vert \xi _f-\xi _g\Vert }{1-\Vert \Lambda \Vert } \end{aligned}$$

as well as

$$\begin{aligned} \left| \varphi ^f(u)-\varphi ^g(u)\right| \le \frac{ \Vert u\Vert }{1-\Vert \Lambda \Vert }\mathbb E\Vert \xi _f-\xi _g\Vert . \end{aligned}$$