Lipschitzian solutions to linear iterative equations revisited

We study the problems of the existence, uniqueness and continuous dependence of Lipschitzian solutions 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 equations of the form φ(x)=∫Ωg(ω)φ(f(x,ω))μ(dω)+F(x),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \varphi (x)=\int _{\Omega }g(\omega )\varphi \big (f(x,\omega )\big )\mu (d\omega )+F(x), \end{aligned}$$\end{document}where μ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu $$\end{document} is a 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} and ∫Ωg(ω)μ(dω)=1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\int _{\Omega }g(\omega )\mu (d\omega )\!=\!1$$\end{document}.

Concerning the given functions f, g and F we assume the following hypotheses in which B stands for the σ-algebra of all Borel subsets of X and K ∈ {R, C}.
(H 1 ) The function f maps X × Ω into X and for every x ∈ X the function f (x, ·) is A-measurable, i.e.
{ω ∈ Ω : f (x, ω) ∈ B} ∈ A for all x ∈ X and B ∈ B. 162 K. Baron, J. Morawiec AEM (H 2 ) The function g : Ω → K is integrable, satisfies (2), As emphasized in [3, Section 0.3] iteration is the fundamental technique for solving functional equations in a single variable, and iterates usually appear in the formulae for solutions. We iterate, as in [2], the operator which transforms a Lipschitzian F : X → Y into Ω g(ω)F (f (x, ω))μ(dω), contrary to [3, Section 7.2D] where the Schauder fixed point theorem is used. The special case where g(ω) = 1 for all ω ∈ Ω and μ(Ω) = 1 was considered in [1, Section 4] on a base of iteration of random-valued functions.
For integrating vector functions we use the Bochner integral.

Existence and uniqueness
Assuming (H 1 )-(H 3 ) and making use of [2, Lemma 2.2] we define for all x ∈ X and n ∈ N, and note that for all x, z ∈ X and n ∈ N.
Our main result reads as follows.
where c is a constant from Y .
Proof. It follows from (3), (2) and (4) that for all x ∈ X and n ∈ N. Consequently, for every x ∈ X the series converges, i.e., (F n (x)) n∈N converges in Y . Hence and from (4) assertion (i) follows.
Passing to the proof of assertion (ii) assume that ϕ : X → Y is a Lipschitzian solution of (1) with a Lipschitz constant L ϕ and define for all x ∈ X and n ∈ N. Since and by (i) there is a c ∈ Y such that lim n→∞ ϕ n (x) = c for every x ∈ X.
Taking this and (8) into account we see that for every x ∈ X the series occurring in (5) converges and (5) holds. In particular, lim n→∞ F n (x) = 0 for every x ∈ X. Applying (5) and (4) we obtain (6). For the proof of the existence assume that lim n→∞ F n (x 0 ) = 0 for an x 0 ∈ X and making use of (4) define Φ : X → Y by Clearly, We shall show that Φ solves (1). To this end fix x ∈ X. According to [2, Lemma 2.2] the function is Bochner integrable and by (4) for all n ∈ N and ω ∈ Ω we have Hence, applying the dominated convergence theorem, (9), (3) and (2) we see that which ends the proof.
The following example shows that sometimes the limit of the sequence (F n ) n∈N can be easily calculated, but its value may be a surprise.

Continuous dependence
Assume (H 1 ) and (H 2 ), fix x 0 ∈ X and let (Y, · ) be a separable Banach space over K. In what follows we consider the linear space Lip(X, Y ) of all Lipschitzian functions mapping X into Y with the norm where F L stands for the smallest Lipschitz constant for F , and the linear subspace F of Lip(X, Y ) consists of all F ∈ Lip(X, Y ) such that for the sequence (F n ) n∈N defined by (3) we have lim n→∞ F n (x 0 ) = 0. It is clear that the norm · Lip depends on the fixed point x 0 , but for different points such norms are equivalent, and it follows from (4) that F does not depend on x 0 . Putting we shall prove the following theorem.
for every x ∈ X 166 K. Baron, J. Morawiec AEM defines the only Lipschitzian and vanishing at x 0 solution ϕ F : X → Y of (1), the operator is a linear homeomorphism of F onto F 0 and for every x ∈ X defines an F ∈ Lip(X, Y ). Since ψ is a Lipschitzian solution of (1), by Theorem 2.1 the function F is in F with Moreover, as ψ(x 0 ) = 0, ψ = ϕ F and Consequently, Lip Ω |g(ω)|ρ f (x 0 , ω), x 0 μ(dω) + 1 + λ , which ends the proof.
Remark 3.2. Since F 0 is closed in the Banach space (Lip(X, Y ), · Lip ), it follows from Theorem 3.1 that also F is a closed subspace of (Lip(X, Y ), · Lip ).