Remarks Connected with the Weak Limit of Iterates of Some Random-Valued Functions and Iterative Functional Equations

Abstract The paper consists of two parts. At first, assuming that (Ω, A, P) is a probability space and (X, ϱ) is a complete and separable metric space with the σ-algebra 𝒝 of all its Borel subsets we consider the set 𝒭c of all 𝒝 ⊗ 𝒜-measurable and contractive in mean functions f : X × Ω → X with finite integral ∫ Ω ϱ (f(x, ω), x) P (dω) for x ∈ X, the weak limit π f of the sequence of iterates of f ∈ 𝒭c, and investigate continuity-like property of the function f ↦ π f, f ∈ 𝒭c, and Lipschitz solutions φ that take values in a separable Banach space of the equation φ(x)=∫Ωφ(f(x,ω))P(dω)+F(x). \varphi \left( x \right) = \int_\Omega {\varphi \left( {f\left( {x,\omega } \right)} \right)P\left( {d\omega } \right)} + F\left( x \right). Next, assuming that X is a real separable Hilbert space, Λ: X → X is linear and continuous with ||Λ || < 1, and µ is a probability Borel measure on X with finite first moment we examine continuous at zero solutions φ : X → 𝔺 of the equation φ(x)=μ⌢(x)φ(Λx) \varphi \left( x \right) = \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\frown$}}\over \mu } \left( x \right)\varphi \left( {\Lambda x} \right) which characterizes the limit distribution π f for some special f ∈ 𝒭c.

Fix a probability space (Ω, A, P ) and a complete and separable metric space (X, ). Let B denote the σ-algebra of all Borel subsets of X.
(H) There exists a λ ∈ (0, 1) such that Thus, denoting by π f n (x, ·) the distribution of f n (x, ·), i.e., hypothesis (H) guarantees the existence of a probability Borel measure π f on X such that for x ∈ X and for any continuous and bounded u : X → R; moreover, as observed in [3] (see also [6]), In [2] we considered continuity-like property of the function f → π f . In [4] we characterized the limit distribution π f via a functional equation for its characteristic function for some special rv-functions in Hilbert spaces. In the present paper we are strengthening the result of [2], apply it to equation (1) and consider also the equation used for the above mentioned characterization of the limit distribution.
1. Assuming that (Ω, A, P ) is a probability space and (X, ) is a complete and separable metric space, consider the set R c of all rv-functions f : with a λ f ∈ [0, 1) and (2) holds. Put also The theorem in [2] says that if f, g ∈ R c , then for every non-expansive u : X → [−1, 1]. In fact the above inequality was proved there for every non-expansive and bounded u : X → R. But if f ∈ R c , then (3) holds and so every Lipschitz function mapping X into a separable Banach space is Bochner integrable with respect to π f . Therefore we can ask whether (4) holds also for such a function. The theorem reads as follows.
for every non-expansive u mapping X into a separable Banach space.
Proof. Let u be a non-expansive mapping of X into a separable Banach space Y . To show that (5) holds we may assume that Y is a real space.
Fix y * ∈ Y * such that y * ≤ 1 and Consequently, since (4) holds for every non-expansive and bounded u : X → R, for every k ∈ N we have and Hence, applying the Lebesgue dominated convergence theorem and passing with k to the limit in (7) we get and (5) follows now from (6).
The following example shows that both sides of (5) can be equal and nonzero.
Denoting by F L the smallest Lipschitz constant for a Lipschitz function F we have the following corollary concerning Lipschitz solutions ϕ of (1).
Corollary 2. Assume n ≥ 2 is an integer and F is a Lipschitz mapping of R into a separable Banach space Y with 1 0 F (x)dx = 0. If reals a 0 , . . . , a n−1 satisfy then equation (9) has no Lipschitz solution ϕ : R → Y .
2. Assuming now that X is a real separable Hilbert space, X = {0}, Λ : X → X is linear and continuous with Λ < 1, and µ is a probability Borel measure on X, consider the equation whereμ denotes the Fourier transform of µ, Theorem 2. If µ has a finite first moment, then there exists a probability Borel measure ν on X with a finite first moment such thatν solves (10), and for any continuous at zero solution ϕ : X → C of (10) we have ϕ = ϕ(0)ν; in particular, every continuous at zero solution ϕ : X → C of (10) is of class C 1 and Lipschitz.
Remind that a probability Borel measure ν on X has a finite first moment provided the integral X x ν(dx) is finite. We shall prove Theorem 2 later on, together with the next one and with the following remark.
Remark. If µ has a finite first moment and Λ is injective, then for every c ∈ C the set of all discontinuous at zero solutions ϕ : X → C of (10) such that ϕ(0) = c and ϕ | X\{0} is of class C 1 and Lipschitz has the cardinality of the continuum.
Theorem 2 implies that for every Borel and integrable with respect to µ function ξ : X → X the equation has exactly one continuous at zero solution ϕ ξ : X → C such that ϕ ξ (0) = 1, and it is of class C 1 and Lipschitz. Consequently, we have the operator ξ → ϕ ξ , ξ ∈ L 1 (µ, X), and a kind of its continuity gives the following theorem.
Theorem 3. If ξ, η : X → X are Borel and integrable with respect to µ, then Proofs. Consider the probability space (X, B, µ) and, given Borel ξ : X → X integrable with respect to µ, the rv-function f on it defined by as well as the limit distribution π f . Put π ξ = π f . According to [4,Theorem 3.1]π ξ solves (11). Since the first moment of π ξ is finite,π ξ is of class C 1 and Lipschitz.
To prove Theorem 2 put ν = π id X and let ϕ : X → C be a continuous at zero solution of (10). Then and lim n→∞ Λ n x = 0 for x ∈ X. Since X e i(Λ k x|z) µ(dz) ≤ X e i(Λ k x|z) µ(dz) = 1 for k ∈ N ∪ {0} and x ∈ X, it shows that if ϕ(0) = 0, then ϕ = 0, and if ϕ(0) = 0, then Consequently, for every c ∈ C equation (10) has at most one continuous at zero solution ϕ : X → C satisfying ϕ(0) = c and by the first part of the proof cν is such a solution.