Hyers-Ulam Stability of Iterative Equation in the Class of Lipschitz Functions

Abstract

Hyers-Ulam stability is a basic sense of stability for functional equations. In the present paper we discuss the Hyers-Ulam stability of a kind of iterative equations in the class of Lipschitz functions. By the construction of a uniformly convergent sequence of functions we prove that, for every approximate solution of such an equation, there exists an exact solution near it.

1. Introduction

It is pointed out that the theory of Ulam’s type stability (also quite often connected, e.g., with the names of Bourgin, Găvruţa, Aoki, Hyers, and Rassias) is a very popular subject of investigations at the moment in [1]. For more details of Ulam’s type stability, we refer the audience to [1–3]. In [1], the authors present a survey of some selected recent developments (results and methods) in the theory of Ulam’s type stability, such as the Forti method and the methods of fixed points and stability in non-Archimedean spaces. These results and methods have not been treated at all or have been treated only marginally. The very book of Jung [2] covers and offers almost all classic results on the Hyers-Ulam-Rassias stability in an integrated and self-contained fashion. And the authors of [3] discussed Hyers-Ulam stability for functional equations in single variable, including the forms of linear functional equations, nonlinear functional equations, and iterative equations. And they also clarified the relation between Hyers-Ulam stability and other senses of stability which are used for functional equations.

Let 𝐶(𝑋,𝑋) be the set of all continuous self-mappings on a topological space𝑋. For any 𝑓∈𝐶(𝑋,𝑋), let 𝑓^𝑚 denote the𝑚th iteration of 𝑓; that is, 𝑓^𝑚=𝑓∘𝑓^𝑚−1, 𝑓⁰=id, 𝑚=1,2,…. Equations having iterations as their main operation, that is, including iterations of the unknown mapping, are called iterative equations. It is one of the most interesting classes of functional equations [4–7]. As a natural generalization of the problem of iterative roots, iterative equations of the following form: 𝜆₁𝑓(𝑥)+𝜆₂𝑓²(𝑥)+⋯+𝜆_𝑛𝑓^𝑛(𝑥)(𝑥),[],=𝐹𝑥∈𝐼=𝑎,𝑏 (1) are known as polynomial-like iterative equations. Here 𝑛≥2 is an integer, 𝜆_𝑖∈ℝ(𝑖=1,2,…,𝑛), 𝐹∶𝐼→ℝ is a given mapping, and 𝑓∶𝐼→𝐼 is unknown. As mentioned in [8, 9], polynomial-like iterative equations are important not only in the study of functional equations but also in the study of dynamical systems.

In 1986, Zhang [10] constructed an interesting operator called “structural operator” for (1) and used the fixed point theory in Banach space to get the solutions of (1). From then on (1) and other types of equations were discussed extensively by employing this idea (see [8, 9, 11–16] and references therein). In 2002, by means of a modification of Zhang’s method applied in [10], Kulczycki and Tabor [11] investigated the existence of Lipschitzian solutions of the iterative equation ^∞_𝑖=1𝑎_𝑖𝑓^𝑖(𝑥)=𝐹(𝑥),𝑥∈𝔹, (2) where 𝔹 is a compact convex subset of ℝ^𝑁 and 𝐹∶𝔹→𝔹 is a given Lipschitz function. They generalized the results of Zhang [10] to a more general case.

In 2002, Xu and Zhang [17] discussed the Hyers-Ulam stability for a nonlinear iterative equation which includes the polynomial-like iterative equation. In 2003, Agarwal et al. [3] investigated the Hyers-Ulam stability for a general form of iterative equations which include the polynomial-like iterative equation with variable coefficients. Motivated by the above results, in this paper, we will discuss the Hyers-Ulam stability of (2). As in [11], we give a result on the Hyers-Ulam stability of a functional equation with a more general form firstly. And, by means of this result, the Hyers-Ulam stability of (2) is investigated. In fact, we want to generalize the result of Xu and Zhang [17] to high-dimensional case.

2. Basic Assumptions, Definitions, and Lemmas

Let 𝔹 be a compact convex subset of ℝ^𝑁, 𝑁∈ℤ⁺, with nonempty interior. Let 𝐶𝔹,ℝ^𝑁=𝑓∶𝔹⟶ℝ^𝑁∣𝑓iscontinuous. (3) In 𝐶(𝔹,ℝ^𝑁), we use the supremum norm 𝑓_𝔹=sup_𝑥∈𝔹𝑓(𝑥),for∀𝑓∈𝐶𝔹,ℝ^𝑁, (4) where ‖⋅‖ denotes the usual metric of ℝ^𝑁. Obviously, 𝐶(𝔹,ℝ^𝑁) is a complete metric space.

Definition 1 (see [11]). Let 𝔹 be a convex compact subset of ℝ^𝑁 with nonempty interior. For 𝑚∈[0,1], and 𝑀∈[1,∞], one defines Lip_𝔹(𝑚,𝑀)∶=𝑓∶𝔹⟶𝔹∣𝑓iscontinuous,𝑓_𝜕𝔹=id|_𝜕𝔹,𝑚≤𝑓(𝑥)𝑦,∀𝑥,𝑦∈𝔹,𝑥−𝑦−𝑓≤𝑀𝑥−𝑦 (5) where 𝜕𝔹 denotes the boundary of 𝔹.

For a given function 𝑓∈𝐶(𝔹,ℝ^𝑁), we define its Lipschitz constant by 𝑓𝑓(𝑥)𝑦.lip∶=inf𝑀≥0∣−𝑓≤𝑀𝑥−𝑦∀𝑥,𝑦∈𝔹 (6)

Lemma 2 (see [11]). Let 𝑘∈[0,1], 𝐾∈[1,∞], and 𝑓∈Lip_𝔹(𝑘,𝐾) be arbitrary. One assumes additionally that 𝑓 is invertible. Then 𝑓⁻¹∈Lip_𝔹1𝐾,1𝑘. (7)

Lemma 3 (see [11]). For every 𝑚∈(0,1], the mapping ℐ∶Lip_𝔹(𝑚,∞)∋𝑓⟶𝑓⁻¹∈Lip_𝔹10,𝑚 (8) is well defined with Lipschitz constant1/𝑚.

Lemma 4 (see [11]). For 𝐾, 𝑀<∞, 𝐹∈Lip_𝔹(0,𝐾), the mapping defined by 𝒮_𝐹∶Lip_𝔹(0,𝑀)∋𝑓⟶𝑓∘𝐹∈Lip_𝔹(0,𝐾𝑀) (9) is Lipschitz with constant 1.

Lemma 5. Let 𝑚∈[0,1], and 𝑀∈[1,∞] be given. For any 𝑓∈Lip_𝔹(𝑚,𝑀), 𝑓∶𝔹→𝔹 is surjective.

Proof. Note that 𝔹 is homeomorphic to𝐷^𝑁, where 𝐷^𝑁={𝑥∈ℝ^𝑁∶‖𝑥‖≤1} is the unit ball of ℝ^𝑁. Without any loss of generality, let 𝔹=𝐷^𝑁. Since 𝑓|_𝜕𝔹=id|_𝜕𝔹, set 𝑓|_𝜕𝔹=𝑓₀; we have deg(𝑓₀)=1, where deg(𝑓₀) denotes the degree of 𝑓₀. Suppose that 𝑓 is not surjective. Let 𝑥₀∈𝔹⧵𝑓(𝔹). Since 𝑓|_𝜕𝔹=id|_𝜕𝔹, we know that 𝑥₀∉𝜕𝔹; then there exists a retraction mapping 𝑟∶𝔹⧵{𝑥₀}→𝜕𝔹. Thus 𝑟∘𝑓∶𝔹→𝜕𝔹 is a continuous mapping and 𝑓₀=(𝑟∘𝑓)|_𝜕𝔹. This means that deg(𝑓₀)=0, a contradiction. So 𝑓 is surjective.

Lemma 6. If 𝑓, 𝑔 are homeomorphisms of 𝔹 with Lipschitz constant 𝑀, then 𝑓−𝑔_𝔹≤𝑀𝑓⁻¹−𝑔⁻¹_𝔹. (10)

Proof. Note that 0=𝑓⁻¹∘𝑓−𝑔⁻¹∘𝑔_𝔹=𝑓⁻¹∘𝑓−𝑔⁻¹∘𝑓+𝑔⁻¹∘𝑓−𝑔⁻¹∘𝑔_𝔹. (11) So we have 𝑓⁻¹∘𝑓−𝑔⁻¹∘𝑓=−𝑔⁻¹∘𝑓+𝑔⁻¹∘𝑔. (12) Furthermore 𝑓⁻¹−𝑔⁻¹_𝔹=𝑓⁻¹∘𝑓−𝑔⁻¹∘𝑓_𝔹=𝑔⁻¹∘𝑓−𝑔⁻¹∘𝑔_𝔹≥1𝑀𝑓−𝑔_𝔹. (13) This means that ‖𝑓−𝑔‖_𝔹≤𝑀‖𝑓⁻¹−𝑔⁻¹‖_𝔹.

Lemma 7. Suppose that 𝑓, 𝑔∶𝔹→𝔹 are continuous mappings and lip(𝑓)=𝑀, where𝑀is a positive constant. Then 𝑓^𝑘−𝑔^𝑘_𝔹≤^𝑘−1_𝑗=0𝑀^𝑗𝑓−𝑔_𝔹,∀𝑘=1,2,…. (14)

Proof. We use induction to prove this lemma. For 𝑘=1, ‖𝑓−𝑔‖_𝔹≤‖𝑓−𝑔‖_𝔹. Suppose that this is true, for 𝑘; then 𝑓^𝑘+1−𝑔^𝑘+1_𝔹≤𝑓∘𝑓^𝑘−𝑓∘𝑔^𝑘_𝔹+𝑓∘𝑔^𝑘−𝑔∘𝑔^𝑘_𝔹≤𝑀𝑓^𝑘−𝑔^𝑘_𝔹+𝑓−𝑔_𝔹≤𝑀^𝑘−1_𝑗=0𝑀^𝑗𝑓−𝑔_𝔹+𝑓−𝑔_𝔹=^𝑘_𝑗=0𝑀^𝑗𝑓−𝑔_𝔹. (15)

Let 𝐴 be a subset of Lip_𝔹(0,∞) and satisfy(A1):for all 𝑓, 𝑔∈𝐴, 𝑓∘𝑔∈𝐴;(A2): if 𝑓∈𝐴 is invertible, then 𝑓⁻¹∈𝐴;(A3): 𝐴 is closed and convex.

And let 𝔄=𝐴⊂Lip_𝔹(0,∞)∣(A1),(A2),and(A3)hold. (16) It is easy to see that Lip_𝔹(0,∞)∈𝔄.

Let 𝐴∈𝔄 and let 𝒫∶𝐴→𝐴 be a map. For any 𝐹∈𝐴 we will consider a functional equation of the following form: 𝒫𝑓∘𝑓=𝐹, (17) where 𝑓∈𝐴 is unknown.

Definition 8. If, for every 𝑔∈Lip_𝔹(0,∞) such that ‖𝐹−𝒫(𝑔)∘𝑔‖_𝔹≤𝛿, there exists a solution 𝑓 of (17) such that ‖𝑓−𝑔‖_𝔹≤𝐾𝛿, where 𝐾>0 is a constant and 𝐾 does not depend on the choice of 𝑔, then (17) is said to have Hyers-Ulam stability.

3. Main Results

Lemma 9. Let 𝔹 be a convex compact subset of ℝ^𝑁 with nonempty interior. Let 𝐴∈𝔄 and 𝐹∈𝐴 with lip(𝐹)≤𝑀₀. Let 𝒫∶𝐴→𝐴 be given. If there exists a decreasing function 𝛼∶[1,∞]→[0,1] and a constant 𝑀∈[1,∞] such that (P1):for any 𝑓∈𝐴, 𝒫(𝑓)∈Lip_𝔹(𝛼(lip(𝑓)),∞);(P2): 𝑀⋅𝛼(𝑀)≥𝑀₀.
Then the following two facts hold:(c1):for any 𝑓∈𝐴∩Lip_𝔹(0,𝑀), 𝒫(𝑓) is a homeomorphism with lip((𝒫(𝑓))⁻¹)≤1/𝛼(𝑀);(c2): for any given 𝑓₀∈𝐴∩Lip_𝔹(0,𝑀), the sequence 𝑓_𝑘=𝒮_𝐹∘ℐ∘𝒫𝑓_𝑘−1,𝑘=1,2,…, (18) is well defined and each 𝑓_𝑘∈𝐴∩Lip_𝔹(0,𝑀) is surjective.

Proof. (c1): since 𝐹|_𝜕𝔹=id|_𝜕𝔹, we get that 𝑀₀≥1. By inequality (P2) and the range of the function 𝛼, we know that both 𝛼(𝑀)∈(0,1] and 𝑀≥𝑀₀ hold. Therefore 𝐹∈𝐴∩Lip_𝔹(0,𝑀), which implies that 𝐴∩Lip_𝔹(0,𝑀) is nonempty. For any 𝑥₁≠𝑥₂∈𝔹 and any 𝑓∈𝐴∩Lip_𝔹(0,𝑀), since lip(𝑓)≤𝑀 and 𝛼 is decreasing, we have 𝒫𝑓𝑥₁−𝒫𝑓𝑥₂≥𝛼lip𝑓𝑥₁−𝑥₂≥𝛼(𝑀)𝑥₁−𝑥₂>0. (19) This means that 𝒫(𝑓) is injective. Since 𝒫(𝑓)∈𝐴, by Lemma 5, we obtain that 𝒫(𝑓) is surjective. Thus 𝒫(𝑓) is a homeomorphism. And lip((𝒫(𝑓))⁻¹)≤1/𝛼(𝑀) follows from Lemma 2.
(c2): by Lemma 3, the mapping ℐ∶Lip_𝔹(𝛼(𝑀),∞)∋𝑔⟶𝑔⁻¹∈Lip_𝔹10,𝛼(𝑀) (20) is continuous. By Lemma 4 and assumptions of 𝐹, the mapping 𝒮_𝐹∶Lip_𝔹10,𝛼(𝑀)∋𝑔⟶𝑔∘𝐹∈Lip_𝔹𝑀0,⁰𝛼(𝑀) (21) is well defined and continuous.
Now, by the assumption (P1) and the fact 𝒫∶𝐴→𝐴, it is easy to see that, for every 𝑔∈Lip_𝔹(0,𝑀)∩𝐴, 𝒫𝑔∈Lip_𝔹(𝛼(𝑀),∞)∩𝐴. (22) Since (A2) holds and by Lemma 2, we know that all elements of Lip_𝔹(𝛼(𝑀),∞) are invertible. Moreover we can get that ℐ∘𝒫𝑔∈Lip_𝔹10,𝛼(𝑀)∩𝐴, (23) and by means of (A1) we have 𝒮_𝐹∘ℐ∘𝒫𝑔∈Lip_𝔹𝑀0,⁰𝛼(𝑀)∩𝐴. (24) Finally, by inequality (P2), we obtain that the mapping 𝒮_𝐹∘ℐ∘𝒫∶Lip_𝔹(0,𝑀)∩𝐴⟶Lip_𝔹(0,𝑀)∩𝐴 (25) is well defined.
Take any 𝑓₀∈𝐴∩Lip_𝔹(0,𝑀). By means of the mapping (25), it is easy to see that the sequence 𝑓_𝑘=𝒮_𝐹∘ℐ∘𝒫𝑓_𝑘−1,𝑘=1,2,…, (26)
is well defined and each 𝑓_𝑘∈𝐴∩Lip_𝔹(0,𝑀). By virtue of Lemma 5, it is easy to see that all 𝑓_𝑘 are surjective.

Theorem 10. Let 𝔹 be a convex compact subset of ℝ^𝑁 with nonempty interior. Let 𝐴∈𝔄 and 𝐹∈𝐴 with lip(𝐹)≤𝑀₀. Let 𝒫∶𝐴→𝐴 be given and let there exist a decreasing function 𝛼∶[1,∞]→[0,1] and a constant 𝑀∈[1,∞] such that (P1) and (P2) of Lemma 9 hold. Moreover, if there is a constant 0<𝜂<𝛼(𝑀) such that 𝒫𝑔₁−𝒫𝑔_2𝔹≤𝜂𝑔₁−𝑔_2𝔹,∀𝑔₁,𝑔₂∈𝐴∩Lip_𝔹(0,𝑀), (27) then, for any 𝑔∈𝐴∩Lip_𝔹(0,𝑀) with 𝐹−𝒫𝑔∘𝑔_𝔹≤𝛿, (28) where 𝛿>0 is a constant, there exists a unique continuous solution 𝑓∈𝐴∩Lip_𝔹(0,𝑀) of (17) such that 𝑓−𝑔_𝔹≤𝐾𝛿, (29) where 1𝐾=𝛼(𝑀).−𝜂 (30)

Proof. Since all the assumptions of Lemma 9 hold, so we can construct a sequence {𝑓_𝑘} of functions as follows. We take 𝑓₀=𝑔 (31) first and then by means of equality (18) repeatedly, define 𝑓_𝑘=𝒮_𝐹∘ℐ∘𝒫𝑓_𝑘,𝑘=1,2,…. (32) By the conclusions (c1) and (c2) of Lemma 9, both 𝒫(𝑓_𝑘−1) and 𝑓_𝑘 are well defined for all 𝑘>1. And each 𝒫(𝑓_𝑘) is a homeomorphism of 𝔹 onto itself with lip((𝒫(𝑓_𝑘))⁻¹)≤1/𝛼(𝑀).
Now we claim that both 𝑓_𝑘−𝑓_𝑘−1𝔹≤1𝛼(𝑀)𝜂𝛼(𝑀)^𝑘−1𝑓𝛿,𝐹−𝒫_𝑘∘𝑓_𝑘𝔹≤𝜂𝛼(𝑀)^𝑘𝛿 (33) hold for 𝑘=1,2,…. We prove these two inequalities inductively.
For 𝑘=1, by virtue of the definition of 𝑓₁, we have 𝑓₁−𝑓_0𝔹=𝒮_𝐹∘ℐ∘𝒫𝑓₀−𝑓_0𝔹=𝒫𝑓₀⁻¹∘𝐹−𝒫𝑓₀⁻¹∘𝒫𝑓₀∘𝑓_0𝔹≤1𝛼(𝑀)𝐹−𝒫𝑓₀∘𝑓_0𝔹=1𝛼(𝑀)𝛿. (34) And noting that 𝐹=𝒫(𝑓_𝑘−1)∘𝑓_𝑘 and 𝑓_𝑘∶𝔹→𝔹 is surjective for 𝑘=1,2,…, we have 𝐹−𝒫𝑓₁∘𝑓_1𝔹=𝒫𝑓₀∘𝑓₁−𝒫𝑓₁∘𝑓_1𝔹=𝒫𝑓₀−𝒫𝑓_1𝔹≤𝜂𝑓₁−𝑓_0𝔹=𝜂𝛼(𝑀)𝛿. (35)
Assume that they are true for integer 𝑘. Then, by the same arguments, we can get that 𝑓_𝑘+1−𝑓_𝑘𝔹=𝒫𝑓_𝑘⁻¹∘𝐹−𝒫𝑓_𝑘⁻¹∘𝒫𝑓_𝑘∘𝑓_𝑘𝔹≤1𝛼(𝑀)𝐹−𝒫𝑓_𝑘∘𝑓_𝑘𝔹≤1𝛼(𝑀)𝜂𝛼(𝑀)^𝑘𝑓𝛿,𝐹−𝒫_𝑘+1∘𝑓_𝑘+1𝔹=𝒫𝑓_𝑘∘𝑓_𝑘+1−𝒫𝑓_𝑘+1∘𝑓_𝑘+1𝔹=𝒫𝑓_𝑘−𝒫𝑓_𝑘+1𝔹≤𝜂𝑓_𝑘+1−𝑓_𝑘𝔹=𝜂𝛼(𝑀)^𝑘+1𝛿. (36) Thus (33) is proved by induction.
For any positive integers 𝑘 and 𝑠 with 𝑘>𝑠, we have 𝑓_𝑘−𝑓_𝑠𝔹≤𝑓_𝑘−𝑓_𝑘−1𝔹+𝑓_𝑘−1−𝑓_𝑘−2𝔹+⋯+𝑓_𝑠+1−𝑓_𝑠𝔹≤1𝛼(𝑀)𝜂𝛼(𝑀)^𝑘−11𝛿+𝛼(𝑀)𝜂𝛼(𝑀)^𝑘−2𝛿1+⋯+𝛼(𝑀)𝜂𝛼(𝑀)^𝑠𝛿=𝛿𝛼(𝑀)⋅𝜂/𝛼(𝑀)^𝑠−𝜂/𝛼(𝑀)^𝑘(𝑀).1−𝜂/𝛼 (37) Since 𝛼(𝑀)>𝜂>0, it follows from inequality (37) that 𝑓_𝑘−𝑓_𝑠𝔹⟶0,as𝑘>𝑠⟶∞. (38) As a Cauchy sequence, {𝑓_𝑘}^∞_𝑘=0 converges uniformly in the Banach space 𝐶(𝔹,ℝ^𝑁). Let lim_𝑘→∞𝑓_𝑘=𝑓. (39) Clearly, 𝑓∈𝐴∩Lip_𝔹(0,𝑀). From 𝐹−𝒫𝑓∘𝑓_𝔹=lim_𝑘→∞𝐹−𝒫𝑓_𝑘∘𝑓_𝑘𝔹=lim_𝑘→∞𝜂𝛼(𝑀)^𝑘𝛿=0, (40) we know that 𝑓 is a solution of (17). Furthermore, 𝑓−𝑔_𝔹=lim_𝑘→∞𝑓_𝑘−𝑓_0𝔹≤lim_𝑘→∞𝑓_𝑘−𝑓_𝑘−1𝔹+⋯+𝑓₁−𝑓_0𝔹≤lim_𝑘→∞1𝛼(𝑀)𝜂𝛼(𝑀)^𝑘−11𝛿+𝛼(𝑀)𝜂𝛼(𝑀)^𝑘−2𝛿1+⋯+𝛼(𝑀)𝛿=1𝛼(𝑀)−𝜂𝛿. (41) This proves (29).
Concerning uniqueness, we assume that there is another continuous solution 𝜙∈𝐴∩Lip_𝔹(0,𝑀) for (17), such that 𝜙−𝑔_𝔹≤𝜀, (42) where 𝜀>0 only depends on 𝛿. By Lemma 5, we know that 𝐹∶𝔹→𝔹 is also surjective. By means of Lemma 6 and the assumption that 𝒫 is Lipschitzian, it follows that 𝑓−𝜙_𝔹=𝒫𝑓⁻¹∘𝐹−𝒫𝜙⁻¹∘𝐹_𝔹=𝒫𝑓⁻¹−𝒫𝜙⁻¹_𝔹≤1𝛼(𝑀)𝒫𝑓−𝒫𝜙_𝔹≤𝜂𝛼(𝑀)𝑓−𝜙_𝔹. (43) This implies that 𝜂1−𝛼(𝑀)𝑓−𝜙_𝔹≤0. (44) However, 𝛼(𝑀)>𝜂>0; this implies that ‖𝑓−𝜙‖_𝔹=0; that is, 𝑓=𝜙.

4. Iterative Equations in ℝ^𝑁

Theorem 11. Let 𝔹 be a convex compact subset of ℝ^𝑁 with nonempty interior. Let 𝐹∈Lip_𝔹(0,∞) be arbitrary with lip(𝐹)≤𝑀₀. Assume that 𝑎_𝑘∈[0,1],𝑘=0,1,2,…, and there exists a constant 𝑝∈[1,∞] such that(a1):∑^∞_𝑘=0𝑎_𝑘=1;(a2): 𝑎₀𝑝>𝑎₀𝑝−∑^∞_𝑘=1𝑎_𝑘𝑝^𝑘+1≥𝑀₀;(a3): ∑^∞_𝑘=1𝑎_𝑘∑^𝑘−1_𝑗=0𝑝^𝑗+∑^∞_𝑘=1𝑎_𝑘𝑝^𝑘<𝑎₀hold. Then the equation ^∞_𝑘=0𝑎_𝑘𝑓^𝑘+1(𝑥)=𝐹(𝑥),𝑥∈𝔹, (45) has Hyers-Ulam stability.

Proof. Let 𝐴=Lip_𝔹(0,∞). For ∀𝑓∈𝐴, define 𝒫𝑓=^∞_𝑘=0𝑎_𝑘𝑓^𝑘. (46) We claim that, for ∀𝑓∈𝐴, 𝒫(𝑓)∈𝐴. For ∀𝑥∈𝔹, by the compactness and convexity of 𝔹 and the assumption (a1), we have ^∞_𝑘=0𝑎_𝑘𝑓^𝑘(𝑥)∈co𝑥,𝑓(𝑥),𝑓²(𝑥),…⊂𝔹, (47) where co{𝑥,𝑓(𝑥),𝑓²(𝑥),…} is the convex hull of the set {𝑥,𝑓(𝑥),𝑓²(𝑥),…}.Moreover, by means of the assumption (a1) and the definition of the mapping 𝒫 and the set 𝐴, it is easy to see that 𝒫(𝑓)|_𝜕𝔹=id|_𝜕𝔹. Now we prove the continuity of 𝒫(𝑓). By the compactness of 𝔹, we first denote 𝐿=max_𝑥∈𝔹{‖𝑥‖}. (48) By the assumption (a1), for any 𝜖>0, there exists a positive integer 𝑙₀ such that ^∞_𝑘=𝑙0+1𝑎_𝑘≤𝜖.4𝐿 (49) Note that each 𝑓^𝑘, 𝑘=0,1,2,…, 𝑙₀, is continuous. Then, for any 𝑥∈𝔹, there is a 𝛿>0 such that when ‖𝑦−𝑥‖<𝛿 the fact 𝑓^𝑘𝑦−𝑓^𝑘(𝑥≤𝜖)2𝑙₀+1,𝑘=0,1,2,…,𝑙₀, (50) holds. Thus we can get that 𝑦(𝑥≤𝒫𝑓−𝒫𝑓)_𝑙0𝑘=0𝑎_𝑘𝑓^𝑘𝑦−𝑓^𝑘(𝑥)+^∞_𝑘=𝑙0+1𝑎_𝑘𝑓^𝑘𝑦−𝑓^𝑘(𝑥≤_𝑙0𝑘=0)𝜖2𝑙₀+1+2𝐿^∞_𝑘=𝑙0+1𝑎_𝑘=𝜖. (51) This means that the fact 𝒫∶𝐴→𝐴 holds.
By virtue of the mapping 𝒫, (45) can be rewritten as 𝒫𝑓∘𝑓=𝐹. (52)
Since 𝐹|_𝜕𝔹=id|_𝜕𝔹, this implies that 𝑀₀≥1.And by the assumption (a2), we see that 𝑎₀>0. For 𝑥∈[1,∞), define 𝛼(𝑥)=max𝑎₀−^∞_𝑘=1𝑎_𝑘𝑥^𝑘,0. (53) Since 𝛼(𝑥)≤𝑎₀≤1 we obtain that 𝛼∶[1,∞]→[0,1] and it is easy to see that 𝛼 is decreasing.
For any 𝑥∈[1,∞) and any 𝑔∈Lip_𝔹(0,∞) with lip(𝑔)=𝑥, we claim that 𝒫(𝑔)∈Lip_𝔹(𝛼(𝑥),∞). It is enough to show that, for ∀𝑏,𝑐∈𝔹, 𝛼(𝑥)‖𝑏−𝑐‖≤𝒫𝑔(𝑏)−𝒫𝑔(𝑐). (54) But 𝒫𝑔(𝑏)𝑔(𝑐=−𝒫)_∞𝑘=0𝑎_𝑘𝑔^𝑘(𝑏)−𝑔^𝑘(𝑐)≥𝑎₀‖𝑏−𝑐‖−^∞_𝑘=1𝑎_𝑘𝑔^𝑘(𝑏)−𝑔^𝑘(𝑐)≥𝑎₀‖𝑏−𝑐‖−^∞_𝑘=1𝑎_𝑘𝑥^𝑘‖‖=𝑎𝑏−𝑐₀−^∞_𝑘=1𝑎_𝑘𝑥^𝑘‖𝑏−𝑐‖=𝛼(𝑥)‖𝑏−𝑐‖. (55) It follows from the assumption (a2) that 𝑝⋅𝛼𝑝=𝑎₀𝑝−^∞_𝑘=1𝑎_𝑘𝑝^𝑘+1≥𝑀₀. (56) Thus (P1) and (P2) of Lemma 9 also hold. Moreover, we can get that 𝒫𝑓−𝒫𝑔_𝔹=^∞_𝑘=0𝑎_𝑘𝑓^𝑘−𝑔^𝑘_𝔹≤^∞_𝑘=0𝑎_𝑘𝑓^𝑘−𝑔^𝑘_𝔹≤^∞_𝑘=1𝑎_𝑘^𝑘−1_𝑗=0𝑝^𝑗𝑓−𝑔_𝔹 (57) hold. Let 𝜂=∑^∞_𝑘=1𝑎_𝑘∑^𝑘−1_𝑗=0𝑝^𝑗. By the assumption (a2), we know that there must be a positive integer 𝑘₀ such that 𝑎_𝑘0≠0. So we know that 𝜂>0. Furthermore, by the assumption (a3), we have 0<𝜂<𝛼(𝑝). Then, by Theorem 10, we get the conclusion.

Example 12. For 𝐹(𝑥)=(𝑥₁,𝑥₂,𝑥₃,…,𝑥_𝑁) + dist(𝑥,𝑆^𝑁−1)(1,0,0,…,0) = (1+𝑥₁−𝑥²₁+𝑥²₂+⋯+𝑥²_𝑁,𝑥₂,…,𝑥_𝑁), where 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑁)∈𝐷^𝑁 and dist(𝑥,𝑆^𝑁−1) denotes the distance of the point 𝑥 from 𝑆^𝑁−1, obviously, 𝐹|_𝑆^𝑁−1=id|_𝑆^𝑁−1. By simple calculation, we get that, for any 𝑥, 𝑦∈𝐷^𝑁, the following two inequalities: 𝐹(𝑥)𝑦,‖𝐹(𝑥≤‖𝑥‖−𝐹≤2𝑥−𝑦)‖+dist𝑥,𝑆^𝑁−1=1 (58) hold. This implies that 𝐹∈Lip_𝐷𝑁(0,2). Then we consider the equation 254𝑓(𝑥)+255^∞_𝑖=11256^𝑖𝑓^𝑖+1(𝑥)=𝐹(𝑥),𝑥∈𝐷^𝑁,𝐹∶𝐷^𝑁⟶𝐷^𝑁,𝐹|_𝑆^𝑁−1=id|_𝑆^𝑁−1. (59) Denote 𝑎₀=254/255, 𝑎_𝑘=(1/256)^𝑖, 𝑘=1,2,…. Define 𝛼(𝑥)=max{(254/255) − ∑^∞_𝑖=1(1/256)^𝑖𝑥^𝑖,0}. Let 𝑀₀=2 and 𝑝=3. Then we have ^∞_𝑘=0𝑎_𝑘=1,254(3)=255×3>3𝛼190491(𝐹),64515>2≥lip254>255^∞_𝑘=11256^𝑘𝑘−1_𝑗=03^𝑗+^∞_𝑖=13256^𝑖=1021.64515 (60) By Theorem 11, (59) has Hyers-Ulam stability.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The research of the authors was supported by National Natural Science Foundation of China (Grant nos. 11101105 and 11001064).