Weak limit of iterates of some random-valued functions and its application

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Introduction
Fix a probability space (Ω, A, P ) and a complete and separable metric space (X, ρ).
Let B denote the σ-algebra of all Borel subsets of X. We say that f : X × Ω → X is a random-valued function (shortly: an rv-function) if it is measurable with respect to the product σ-algebra B ⊗ A. The iterates of such an rv-function are given by f 0 (x, ω 1 , ω 2 , . . .) = x, f n (x, ω 1 , ω 2 , . . .) = f f n−1 (x, ω 1 , ω 2 , . . .), ω n 416 K. Baron AEM for n ∈ N, x ∈ X and (ω 1 , ω 2 , . . .) from Ω ∞ defined as Ω N . Note that f n : X × Ω ∞ → X is an rv-function on the product probability space (Ω ∞ , A ∞ , P ∞ ). More exactly, for n ∈ N the nth iterate f n is B ⊗ A n -measurable, where A n denotes the σ-algebra of all sets of the form {(ω 1 , ω 2 , . . .) ∈ Ω ∞ : (ω 1 , . . . , ω n ) ∈ A} with A from the product σ-algebra A n . (See [8,Sec. 1.4], [6].) Let f : X × Ω → X be an rv-function. A result on the a.s. convergence of (f n (x, ·)) n∈N for X being the unit interval may be found in [8,Sec. 1.4B]. The paper [6] by Rafa l Kapica 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 Banach lattice. A simple criterion for the convergence in law of (f n (x, ·)) n∈N to a random variable independent of x ∈ X was proved in [1] and applied to the equation with ϕ as the unknown function. This criterion reads as follows.
(H) There exists a λ ∈ (0, 1) such that and Thus, denoting by π f n (x, ·) the distribution of f n (x, ·), i.e., π f n (x, B) = P ∞ (f n (x, ·) ∈ B) for n ∈ N ∪ {0}, x ∈ X and B ∈ B, hypothesis (H) guarantees the existence of a probability Borel measure π f on X such that holds for x ∈ X and for any continuous and bounded u : X → R; more exactly, cf. also [2, Theorem 3.1], and for x ∈ X, n ∈ N and a non-expansive u mapping X into [−1, 1]. Rafa l Kapica strengthened this estimation showing, see [7, Corollary 5.6 and Lemma 3.1], that (5) holds for x ∈ X, n ∈ N and a non-expansive u : X → R. Since it is explicitly stated there only for a non-expansive and bounded Vol. 94 (2020) Weak limit of iterates of some random-valued functions 417 u : X → R we prove it for a non-expansive u mapping X into a separable Banach space making use of (5) for non-expansive and bounded u : X → R only. Having done that we characterize the solvability of (1) in the class of Lipschitz functions with the aid of this limit distribution π f . Moreover, we propose a characterization of π f for some special rv-functions in Hilbert spaces.

Solvability of the equation
Following [3] given an rv-function f : X × Ω → X such that (H) holds and a Lipschitz F mapping X into a separable Banach space Y define for x ∈ X and n ∈ N, and note that according to [3, Theorem 2.1] there exists a y 0 ∈ Y such that for every x ∈ X the sequence (F n (x)) n∈N converges to y 0 , and any Lipschitz solution ϕ : X → Y of (1) has the form where c is a constant from Y . With the aid of the limit distribution π f we characterize this pointwise limit of (F n ) n∈N and, making use of [3, Theorem 2.1(ii)], the solvability of (1) as follows (cf. [1, Corollary 4.1]).
and Eq. (1) has a Lipschitz solution ϕ : As announced above, we start with the following lemma.
for x ∈ X, n ∈ N and for any non-expansive u mapping X into a separable Banach space.

418
K. Baron AEM Proof. First of all let us observe that for every x ∈ X, n ∈ N and (ω 1 , where the value f n (x, ω 1 , ω 2 , . . .) depends only on x and (ω 1 , . . . , ω n ), and by (2) we have Hence, applying the Fubini theorem, for x ∈ X and n ∈ N we get Consequently, for any non-expansive u mapping X into a separable Banach space and for every x ∈ X, n ∈ N we obtain moreover, Let u be a non-expansive mapping of X into a separable Banach space Y . To show that (8) holds for x ∈ X and n ∈ N we may assume that Y is a real space.
Fix x ∈ X, n ∈ N and then a y * ∈ Y * such that y * ≤ 1 and ∞) is non-expansive and |τ k (t)| ≤ |t| for t ∈ R. Consequently, since (5) holds for every non-expansive and bounded u : Vol. 94 (2020) Weak limit of iterates of some random-valued functions 419 and, by (9) and (10), Hence, taking (3) and (4) into account, applying the Lebesgue dominated convergence theorem and passing with k to the limit in (12) we get and (8) follows now from (11).
Proof of the theorem. An easy induction shows that i.e., Let L be a Lipschitz constant for F . Putting u = 1 L F we have (8), whence for x ∈ X and n ∈ N. It proves (6)

A characterization of the limit distribution
Obviously the problem of characterization of the limit distribution π f arises.
The following theorem provides a characterization via functional equations for some special rv-functions in Hilbert spaces. More exactly, we characterize π f via a functional equation for its characteristic function ϕ f : X → C, then the characteristic function of π f is the only solution ϕ : X → C of the equation which is continuous at zero and fulfils ϕ(0) = 1, where γ stands for the characteristic function of ξ.

Examples
1. Fix an integer n ≥ 2 and a Lipschitz mapping F of R into a separable Banach space Y and consider the equation In this case A simple calculation shows that the characteristic function of the uniform distribution U (0, 1), i.e. the function φ : R → C given by Hence and from Theorem 3.1 we infer that π f = U (0, 1), i.e., π f (B) = λ 1 (B ∩[0 assuming that ξ : Ω → R is a random variable with the Gaussian law N (m, σ 2 ), i.e. the equation where m is a real number and σ is a positive real number. In this case and, as a simple calculation shows, the function φ : R → C given by Hence and from Theorem 3.1 we infer that Applying now Theorem 2.1 we see that Eq. (15) has a Lipschitz solution ϕ : In particular (cf., e.g., [5,  2.2. If α ∈ (−1, 1), ξ : Ω → R is a random variable with the standard Gaussian law N (0, 1), n ∈ N and α 0 , α 1 , . . . , α n ∈ R, then the equation 3. Let X be a real separable Hilbert space. Following [4] denote by L + 1 (X) the set of all linear, symmetric and positive self-mappings of X with finite trace.
Given a linear and symmetric Λ : X → X with Λ < 1 and a Lipschitz mapping F of X into a separable Banach space Y consider the equation assuming now that ξ : Ω → X is a random variable with the Gaussian law N (m, Q), where m ∈ X and Q ∈ L + 1 (X), and ΛQ = QΛ.
In this case Consequently A is symmetric and positive. As Q has finite trace, so has A and A ∈ L + 1 (X). Moreover, the function φ : X → C given by In particular, if Λ : X → X is linear and symmetric with Λ < 1 and ξ : Ω → X is a random variable with the Gaussian law N (m, Q), where m ∈ X, Q ∈ L + 1 (X) and (17)