Continuous solutions to two iterative functional equations

Based on iteration of random-valued functions we study the problem of solvability in the class of continuous and Hölder continuous functions φ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi $$\end{document} of the equations φ(x)=F(x)-∫Ωφ(f(x,ω))P(dω),φ(x)=F(x)+∫Ωφ(f(x,ω))P(dω),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \varphi (x)=F(x)-\int _{\Omega }\varphi \big (f(x,\omega )\big )P(d\omega ),\\ \varphi (x)=F(x)+\int _{\Omega }\varphi \big (f(x,\omega )\big )P(d\omega ), \end{aligned}$$\end{document}where P is a probability measure on a σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sigma $$\end{document}-algebra of subsets of Ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Omega $$\end{document}.


Introduction
Fix a probability space (Ω, A, P ), a complete and separable metric space (X, ρ) with the σ-algebra B of all its Borel subsets, and a B ⊗ A-measurable function f : X × Ω → X.
This limit distribution π f plays an important role in solving (1) and (2), see [5,  (I) If F : X → R is continuous and bounded, then any continuous and bounded solution ϕ : X → R of (1) has the form if additionally F is Lipschitz, then (6) defines a Lipschitz solution ϕ : X → R of (1).
(II) If F : X → R is continuous and bounded and (2) has a continuous and bounded solution ϕ : X → R, then and any such solution has the form with a real constant c.
(III) If F : X → R is Lipschitz, then it is integrable for π f and (2) has a Lipschitz solution ϕ : X → R if and only if (7) holds.
The limit distribution π f and facts cited above will be used in the main part of the paper. A characterization of this limit for some special randomvalued functions in Hilbert spaces have been given by [ Actually we do not have a sufficiently satisfactory theorem to guarantee the existence of continuous solutions to the equations considered. An explanation of this situation is given in the paper [9] by Witold Jarczyk (see also [13,Note 3.8.4]). Namely, in the case where Ω is a singleton and X is a compact real interval, for the appropriate f the set of continuous F : X → R such that the equation has a continuous solution is small in the sense of Baire category. It is also small from the measure point of view (see [1]). We will go also in this direction but, above all, we are looking for conditions under which Eqs. (1) and (2)
We regard λ as fixed in (0, 1), and for any metric space X we define F(X) as the set of all continuous functions F : X → R such that there are a sequence (F n ) n∈N of real functions on X and constants ϑ ∈ (0, 1), L ∈ (0, 1 λ ) and α, β ∈ (0, ∞) such that Clearly any real Lipschitz function defined on X belongs to F(X).
The proof will be based on three lemmas. In each of them we assume (H).

Lemma 2.2. If F ∈ F(X), then the integrals
are finite, and the function is continuous.
Proof. Corresponding to F choose a sequence (F n ) n∈N of real functions on X and constants ϑ ∈ (0, 1), L ∈ (0, 1 λ ) and α, β ∈ (0, ∞) as in the definition of F(X). Then see also (III). Moreover, for every n ∈ N the function is Lipschitz: for x, z ∈ X, and therefore function (9), as their uniform limit, is continuous.
and for every n ∈ N the function is continuous.
Proof. By induction, (3) and (4), and an application of Lemma 2.2 with f replaced by f n , n ∈ N, finishes the proof.
Proof. Corresponding to F choose a sequence (F n ) n∈N of real functions on X and constants ϑ ∈ (0, 1), L ∈ (0, 1 λ ) and α, β ∈ (0, ∞) as in the definition of F(X), and put Then θ ∈ (0, 1), and by Lemmas 2.3 and 2.2, (5) with u = Fn βL n and (3) for every x, x 0 ∈ X and n ∈ N we have Proof of Theorem 2.1. It follows from Lemmas 2.2-2.4 that formula (6) defines a continuous function ϕ : X → R and arguing like in the proof of Theorem 3.1(ii) of [5] (see also the calculations below) we show that it solves (1). Assume now that also (7) holds. Then it follows from Lemmas 2.3 and 2.4 that formula (8) defines a continuous function ϕ 0 : X → R. Applying (11), Lemma 2.4, the Lebesgue dominated convergence theorem and the Fubini theorem we observe that for every x ∈ X the function ϕ 0 • f (x, ·) is integrable for P and
and for every n ∈ N let A n be a maximal for inclusion subset of X such that ρ(x, z) ≥ θ n for every pair of distinct points x, z of A n .
By the maximality, If n ∈ N and x, z are distinct points of A n , then by (12), It follows from this, using Kirszbraun-McShane extension theorem [7, Theorem 6.1.1], that for every n ∈ N there exists an F n : X → R such that If n ∈ N and x ∈ X, then there is a z ∈ A n such that ρ(x, z) < θ n , and ≤ βθ αn + βL n θ n = 2βϑ n .
Regarding the uniqueness of solutions, we have the following theorem.
We finish with a qualitative result. Following [6] by Jens Peter Reus Christensen we say that a Borel subset B of an abelian Polish group G is a Haar zero set if there is a probability Borel measure μ on G such that μ(B + x) = 0 for every x ∈ G. See also [8] where measurability in abelian Polish groups related to Christensen's Haar zero set is studied.
Assuming (H 0 ) we have in particular (4): Moreover one can consider the Banach space C(X) of all continuous real functions on X with the uniform norm and its subspace C f , Clearly C f is a closed linear subspace of C(X) and (see, e.g., [7,Corollary 11.2.5]) C(X) is a separable Banach space. We have also the following lemma.
Let (ii) F 2 is a Borel and dense subset of C f , and if F 2 = C f , then F 2 is of first category in C f and a Haar zero subset of C f . Proof. By Lemma 2.8 the formulas for ϕ ∈ C(X) and x ∈ X, define self-mappings T 1 , T 2 of C(X). Clearly, these operators are linear and continuous. Moreover, Furthermore, for every F ∈ T 2 (C(X)) Eq. (2) has a continuous solution ϕ : X → R. Hence (II) gives T 2 (C(X)) ⊂ C f , and T 2 (C(X)) = F 2 .
Applying now [1,Lemma] we see that F 1 is a Borel subset of C(X), and if F 1 = C(X), then F 1 is of first category in C(X) and a Haar zero subset of C(X), and F 2 is a Borel subset of C f , and if F 2 = C f , then F 2 is of first category in C f and a Haar zero subset of C f .
Since by (I) the set {F ∈ C(X) : F is Lipschitz} is contained in F 1 and (see [7,Theorem 11.2.4]) dense in C(X), the set F 1 is dense in C(X).