# Convexity of Solutions for an Iterative Equation in Banach Spaces

Authors: Xiaobing Gong
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

By applying Schauder's fixed point theorem we investigate the existence
of increasing (decreasing) solutions of the iterative equationℋf∘f=F and further give conditions under which those solutions are convex or concave. As corollaries we obtain results on iterative equation
Gfx,fn1,…,fnkx=F(x)
in Banach spaces, where n1,n2,…,nk≥2.


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## Body

## 1. Introduction

Iterative root problem [1, 2], being a weak version of the problem of embedding flows, plays an important role in the theory of dynamical systems. As a natural generalization of the iterative root problem, the polynomial-like iterative equation
(1)λ1f(x)+λ2f2(x)+⋯+λmfm(x)=F(x),x∈S,
where S is a subset of a linear space over ℝ, F:S→S is a given function, λis (i=1,…,m) are real constants, f:S→S is the unknown function, and fi is the ith iterate of f, that is, fi(x)=f(fi-1(x)) and f0(x)=x for all x∈S, is one of the important iterative functional equations [3, 4] and was studied in many papers. For S⊂ℝ, while some works (e.g., [5–11]) are contributed to the case of linear F, there are many results given to the case of nonlinear F, for example, [12, 13] for m=2, [14] for general m, [15] for smoothness, and [16] for analyticity. Some efforts were also devoted to (1) in high-dimensional spaces such as in [17, 18]; radially monotonic solutions were discussed in high-dimensional Euclidean spaces by properties of orthogonal group in [18], and the existence of convex solutions was proved by introducing a partial order in Banach spaces in [17]. A general iterative equation can be presented as
(2)G(f(x),fn1(x),…,fnk(x))=F(x),x∈S,
where k>0 and n1,n2,…,nk≥2. In 1995, C0 solutions of (2) were discussed in [19], and as continuations of   [19], C1 and C2 solutions were studied in [20, 21], respectively, for S⊂ℝ. In 2007, by lifting maps on the unit circle 𝕋1 and maps on the torus 𝕋n, the existence, uniqueness, and stability of continuous solutions for (2) were proved on the unit circle in [22]. A more general iterative functional equation
(3)ℋ(f)∘f=F
was studied in [23, 24] in high-dimensional spaces, where ℋ is a operator. Equation (3) is a generalization of iterative equation (2). In fact, if ℋ(f)=G(f0,fn1-1,…,fnk-1), then (3) becomes (2). In [23], the existence of Lipschitzian solutions for (3) was proved on a compact convex subset of ℝN, and by using this result, the existence of Lipschitzian solutions for equation
(4)∑n=1+∞‍λnfn(x)=F(x)
was investigated on a compact interval of ℝ and a compact convex subset of ℝN,N>1. Later, the results were partially generalized to an arbitrary closed (not necessarily convex) subset of a Banach space and the existence of solutions for iterative functional equations
(5)∑n=-∞+∞‍Anfn(x)=F(x),A0f(x)+∑n=1+∞‍Anfn(x)=F(x),
was proved in [24], where Ans are bounded linear operators on the Banach space.Convexity is an important property of functions and the study of convexity for iterative equations can be traced to 1968 when Kuczma and Smajdor [25] investigated the convexity of iterative roots. Some recent results can be found from [17, 26–28]. In [27, 28], convexity of solutions for (1) was discussed on a compact interval, and in [26], nondecreasing convex solutions for (1) on open intervals were discussed. In [17], convexity of solutions for (1) was studied in Banach spaces. Up to now, there are no further results on monotonicity and convexity of solutions for (2) and (3) in Banach spaces. In fact, there are much more difficulties on monotonicity and convexity of solutions for these two equations in Banach spaces.In this paper we study monotonicity and convexity of solutions for (2) and (3) in Banach spaces and generalize the results in [17]. Using Schauder’s fixed point theorem, we discuss increasing (decreasing) solutions for (3) and further give conditions under which those solutions are convex or concave. As corollaries, we obtain results on (2). The uniqueness and continuous dependence of those solutions are also discussed.


## 2. Preliminaries

As in [17], in order to discuss monotonicity and convexity of solutions in Banach spaces, we need to introduce a partial order. For convenience, we use the conventions of [17]. As in [29], a nonempty subset K of a real vector space X is called a cone if x∈K implies that ax∈K for all a>0. A nonempty and nontrivial (≠{θ}, where θ denotes the zero element of X) subset K⊂X is called an order cone in X if K is a convex cone and satisfies K∩(-K)={θ}. Having chosen such an order cone K in X, we can define a partial order x≤Ky in X, simply called the K-order, if
(6)y-x∈K.
A real vector space X equipped with a K-order is called an ordered vector space, abbreviated by OVS and denoted by (X,K). A real Banach space (X,∥·∥) associated with a K-order is called an ordered real Banach space, abbreviated by OBS and denoted by (X,K,∥·∥), if K is closed. One can define increasing (decreasing) operators as in [30] in an ordered real vector space (X,K). An operator f:D⊂X→X is said to be increasing(resp.,decreasing) in the sense of the K-order if x≤Ky implies f(x)≤Kf(y) (resp., f(x)≥Kf(y)). An operator f:D→X, where D⊂X is a convex subset, is said to be convex(resp.,concave) in the sense of the K-order if f(λx+(1-λ)y)≤Kλf(x)+(1-λ)f(y) (resp., f(λx+(1-λ)y)≥Kλf(x)+(1-λ)f(y)) for all λ∈[0,1] and for every pair of distinct comparable points x,y∈D (i.e., either x≤Ky or x≥Ky).LetΩ be a compact convex subset of an ordered real Banach space (X,K,∥·∥) with nonempty interior, and let C(Ω,X) consist of all continuous functions f:Ω→X. C(Ω,X) is a Banach space equipped with the norm ∥f∥C(Ω,X)≔supx∈Ω∥f(x)∥. For 0≤m≤M<+∞, define
(7)E+(Ω,m,M)≔{f∈C(Ω,X):m(y-x)≤Kf(y)-f(x)≤KM(y-x)ifx≤Ky,∥f(y)-f(x)∥≤M∥y-x∥ifxandyarenotcomparable},E-(Ω,m,M)≔{f∈C(Ω,X):m(y-x)≤Kf(x)-f(y)≤KM(y-x)ifx≤Ky,∥f(y)-f(x)∥≤M∥y-x∥ifxandyarenotcomparable},Ecv+(Ω,m,M)≔{f∈E+(Ω,m,M):fisconvexonΩinK-order},Ecc+(Ω,m,M)≔{f∈E+(Ω,m,M):fisconcaveonΩinK-order},C+(Ω,m,M)≔{f∈E+(Ω,m,M):f(Ω)⊂Ω},C-(Ω,m,M)≔{f∈E-(Ω,m,M):f(Ω)⊂Ω},Ccv+(Ω,m,M)≔{f∈Ecv+(Ω,m,M):f(Ω)⊂Ω},Ccc+(Ω,m,M)≔{f∈Ecc+(Ω,m,M):f(Ω)⊂Ω},I(f)≔inf{M:∀x,y∈Ω,θ≤Kf(y)-f(x)≤KM(y-x)ifx≤Ky,∥f(y)-f(x)∥≤M∥y-x∥ifxandyarenotcomparable},D(f)≔inf{M:∀x,y∈Ω,θ≤Kf(x)-f(y)≤KM(y-x)ifx≤Ky,∥f(y)-f(x)∥≤M∥y-x∥ifxandyarenotcomparable}.
Similar to Lemma  2.2 in [17], C+(Ω,m,M), C-(Ω,m,M), Ccv+(Ω,m,M), and Ccc+(Ω,m,M) are compact convex subsets of C(Ω,X).As shown in [29, 30], an order cone K in an ordered real Banach space (X,∥·∥) is said to be normal if there exists a constant N>0 such that ∥x∥≤N∥y∥ if θ≤Kx≤Ky in X. The smallest constant N, denoted by N(K), is called the normal constant of K.


## 3. Main Result

We first discuss monotonicity and convexity of solutions for iterative functional equation (3) in the ordered real Banach space (X,K,∥·∥) such that K is normal and N(K)≤1. Consider (3) with the following hypothesis: (H1)
ℋ(f)(x)=x-𝒫(f)(x).


### 3.1. Increasing and Decreasing Solutions

Theorem 1.
Suppose that (H1) holds andF∈E+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:C+(Ω,0,∞)→E+(Ω,0,∞) such that
(8)𝒫(f)∈E+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(9)p1≤p(1-α(p))
and 𝒫|C+(Ω,0,p) is continuous, then (3) has a solution f∈C+(Ω,0,p).Proof.
Define a mappingL:C+(Ω,0,p)→C(Ω,X) by
(10)Lf(x)=F(x)+𝒫(f)∘f(x).
We first prove that L is a self-mapping on C+(Ω,0,p). Obviously, 𝒫(f)∘f is well defined and F+𝒫(f)∘f∈C(Ω,X). By F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω and the definition of C+(Ω,0,p), (F+𝒫(f)∘f)(Ω)⊂Ω. Further, when x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K, by (8) and F∈E+(Ω,0,p1), we have
(11)∥Lf(x)-Lf(y)∥=∥(F(x)-F(y))+(𝒫(f)∘f(x)-𝒫(f)∘f(y))∥≤∥F(x)-F(y)∥+∥𝒫(f)∘f(x)-𝒫(f)∘f(y)∥≤p1∥x-y∥+α(I(f))∥f(x)-f(y)∥≤(p1+pα(p))∥x-y∥,
where the monotonicity of the function α is employed, which implies that
(12)∥Lf(x)-Lf(y)∥≤p∥x-y∥
because of (9). When x,y∈Ω are comparable, suppose that x≤Ky. By the definition of C+(Ω,0,p), θ≤Kf(y)-f(x)≤Kp(y-x); thus f(x)≤Kf(y). Hence, by (8), we get
(13)θ≤K𝒫(f)∘f(y)-𝒫(f)∘f(x)≤Kα(I(f))(f(y)-f(x)).
Consequently, we have
(14)θ≤KLf(y)-Lf(x)=(F(y)-F(x))+(𝒫(f)∘f(y)-𝒫(f)∘f(x))≤p1(y-x)+α(I(f))(f(y)-f(x))≤(p1+pα(p))(y-x),
where the monotonicity of the function α is employed, which implies that
(15)θ≤KLf(y)-Lf(x)≤Kp(y-x)
because of (9). Thus, (12) and (15) imply that L is a self-mapping on C+(Ω,0,p). The continuity of 𝒫|C+(Ω,0,p) implies that L is continuous on C+(Ω,0,p). Since C+(Ω,0,p) is a compact convex subset, by Schauder’s fixed point theorem, we see that L has a fixed point f∈C+(Ω,0,p). Thus, f is an increasing solution of (3). The proof is completed.The following is devoted to decreasing solutions.Theorem 2.
Suppose that (H1) holds andF∈E-(Ω,0,p1). Let 𝒫:C-(Ω,0,∞)→E+(Ω,0,∞) such that
(16)𝒫(f)∈E+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If condition (9) holds for a constant p∈(0,∞) and 𝒫|C-(Ω,0,p) is continuous, then (3) has a solution f∈C-(Ω,0,p).The proof is almost the same as the proof of Theorem1; we omit it here.


### 3.2. Convex and Concave Solutions

On the basis of the last subsection, we can discuss convexity of solutions for (3).Theorem 3.
Suppose that (H1) holds andF∈Ecv+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:Ccv+(Ω,0,∞)→Ecv+(Ω,0,∞) such that
(17)𝒫(f)∈Ecv+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(18)p1≤p(1-α(p)),
and 𝒫|Ccv+(Ω,0,p) is continuous, then (3) has a solution f∈Ccv+(Ω,0,p).In order to prove Theorem3, we need the following lemma.Lemma 4 (see [17, Lemma 3.1]).
Let(X,K,∥·∥) be an ordered real Banach space. Then composition f∘g is convex (resp., concave) if both f and g are convex (resp., concave) and increasing. In particular, for increasing convex (resp., concave) operator f, the iterate fk is also convex (resp., concave).Proof of Theorem3.
Define a mappingL:Ccv+(Ω,0,p)→C(Ω,X) as in Theorem 1. In order to prove that L is a self-mapping on Ccv+(Ω,0,p), it suffices to prove that F+𝒫(f)∘f is convex in the sense of K-order on Ω. In fact, by (17), we know 𝒫(f)(x) is increasing and convex on Ω. Hence, by Lemma 4 and f∈Ccv+(Ω,0,p), 𝒫(f)∘f is convex in the sense of K-order on Ω. Consequently, by F∈Ecv+(Ω,0,p1),
(19)F(tx+(1-t)y)+𝒫(f)∘f(tx+(1-t)y)≤tF(x)+(1-t)F(y)+t𝒫(f)∘f(x)+(1-t)𝒫(f)∘f(y)=t(F(x)+𝒫(f)∘f(x))+(1-t)(F(y)+𝒫(f)∘f(y))
for every pair of distinct comparable points x,y∈Ω and t∈[0,1]. So L is an self-mapping on Ccv+(Ω,0,p). The remaining part of the proof is the same as the proof of Theorem 1. This completes the proof.Similarly, we can prove the following results for concavity of solutions.Theorem 5.
Suppose that (H1) holds andF∈Ecc+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:Ccc+(Ω,0,∞)→Ecc+(Ω,0,∞) such that
(20)𝒫(f)∈Ecc+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is a increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(21)p1≤p(1-α(p)),
and 𝒫|Ccc+(Ω,0,p) is continuous, then (3) has a solution f∈Ccc+(Ω,0,p).


## 3.1. Increasing and Decreasing Solutions

Theorem 1.
Suppose that (H1) holds andF∈E+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:C+(Ω,0,∞)→E+(Ω,0,∞) such that
(8)𝒫(f)∈E+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(9)p1≤p(1-α(p))
and 𝒫|C+(Ω,0,p) is continuous, then (3) has a solution f∈C+(Ω,0,p).Proof.
Define a mappingL:C+(Ω,0,p)→C(Ω,X) by
(10)Lf(x)=F(x)+𝒫(f)∘f(x).
We first prove that L is a self-mapping on C+(Ω,0,p). Obviously, 𝒫(f)∘f is well defined and F+𝒫(f)∘f∈C(Ω,X). By F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω and the definition of C+(Ω,0,p), (F+𝒫(f)∘f)(Ω)⊂Ω. Further, when x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K, by (8) and F∈E+(Ω,0,p1), we have
(11)∥Lf(x)-Lf(y)∥=∥(F(x)-F(y))+(𝒫(f)∘f(x)-𝒫(f)∘f(y))∥≤∥F(x)-F(y)∥+∥𝒫(f)∘f(x)-𝒫(f)∘f(y)∥≤p1∥x-y∥+α(I(f))∥f(x)-f(y)∥≤(p1+pα(p))∥x-y∥,
where the monotonicity of the function α is employed, which implies that
(12)∥Lf(x)-Lf(y)∥≤p∥x-y∥
because of (9). When x,y∈Ω are comparable, suppose that x≤Ky. By the definition of C+(Ω,0,p), θ≤Kf(y)-f(x)≤Kp(y-x); thus f(x)≤Kf(y). Hence, by (8), we get
(13)θ≤K𝒫(f)∘f(y)-𝒫(f)∘f(x)≤Kα(I(f))(f(y)-f(x)).
Consequently, we have
(14)θ≤KLf(y)-Lf(x)=(F(y)-F(x))+(𝒫(f)∘f(y)-𝒫(f)∘f(x))≤p1(y-x)+α(I(f))(f(y)-f(x))≤(p1+pα(p))(y-x),
where the monotonicity of the function α is employed, which implies that
(15)θ≤KLf(y)-Lf(x)≤Kp(y-x)
because of (9). Thus, (12) and (15) imply that L is a self-mapping on C+(Ω,0,p). The continuity of 𝒫|C+(Ω,0,p) implies that L is continuous on C+(Ω,0,p). Since C+(Ω,0,p) is a compact convex subset, by Schauder’s fixed point theorem, we see that L has a fixed point f∈C+(Ω,0,p). Thus, f is an increasing solution of (3). The proof is completed.The following is devoted to decreasing solutions.Theorem 2.
Suppose that (H1) holds andF∈E-(Ω,0,p1). Let 𝒫:C-(Ω,0,∞)→E+(Ω,0,∞) such that
(16)𝒫(f)∈E+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If condition (9) holds for a constant p∈(0,∞) and 𝒫|C-(Ω,0,p) is continuous, then (3) has a solution f∈C-(Ω,0,p).The proof is almost the same as the proof of Theorem1; we omit it here.


## 3.2. Convex and Concave Solutions

On the basis of the last subsection, we can discuss convexity of solutions for (3).Theorem 3.
Suppose that (H1) holds andF∈Ecv+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:Ccv+(Ω,0,∞)→Ecv+(Ω,0,∞) such that
(17)𝒫(f)∈Ecv+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is an increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(18)p1≤p(1-α(p)),
and 𝒫|Ccv+(Ω,0,p) is continuous, then (3) has a solution f∈Ccv+(Ω,0,p).In order to prove Theorem3, we need the following lemma.Lemma 4 (see [17, Lemma 3.1]).
Let(X,K,∥·∥) be an ordered real Banach space. Then composition f∘g is convex (resp., concave) if both f and g are convex (resp., concave) and increasing. In particular, for increasing convex (resp., concave) operator f, the iterate fk is also convex (resp., concave).Proof of Theorem3.
Define a mappingL:Ccv+(Ω,0,p)→C(Ω,X) as in Theorem 1. In order to prove that L is a self-mapping on Ccv+(Ω,0,p), it suffices to prove that F+𝒫(f)∘f is convex in the sense of K-order on Ω. In fact, by (17), we know 𝒫(f)(x) is increasing and convex on Ω. Hence, by Lemma 4 and f∈Ccv+(Ω,0,p), 𝒫(f)∘f is convex in the sense of K-order on Ω. Consequently, by F∈Ecv+(Ω,0,p1),
(19)F(tx+(1-t)y)+𝒫(f)∘f(tx+(1-t)y)≤tF(x)+(1-t)F(y)+t𝒫(f)∘f(x)+(1-t)𝒫(f)∘f(y)=t(F(x)+𝒫(f)∘f(x))+(1-t)(F(y)+𝒫(f)∘f(y))
for every pair of distinct comparable points x,y∈Ω and t∈[0,1]. So L is an self-mapping on Ccv+(Ω,0,p). The remaining part of the proof is the same as the proof of Theorem 1. This completes the proof.Similarly, we can prove the following results for concavity of solutions.Theorem 5.
Suppose that (H1) holds andF∈Ecc+(Ω,0,p1), where p1∈(0,+∞) is a constant. Let 𝒫:Ccc+(Ω,0,∞)→Ecc+(Ω,0,∞) such that
(20)𝒫(f)∈Ecc+(Ω,0,α(I(f))),
where α:(0,+∞)→(0,+∞) is a increasing function and F(x)+𝒫(f)(y)∈Ω for any x,y∈Ω. If there exists p∈(0,∞) such that
(21)p1≤p(1-α(p)),
and 𝒫|Ccc+(Ω,0,p) is continuous, then (3) has a solution f∈Ccc+(Ω,0,p).


## 4. Iterative Equation in Banach Spaces

In this section, we are going to discuss monotonicity and convexity of solutions for (2) in the ordered real Banach space (X,K,∥·∥) such that K is normal and N(K)≤1. Consider (2) with the following hypotheses:(H2)
G(y0,y1,…,yk)=y0+g(y1,…,yk),
g(y1,…,yk)∈C(Ωk,X).Before discussing convexity, we prove the existence of increasing and decreasing solutions of (2).


### 4.1. Increasing and Decreasing Solutions

We first study increasing solutions. Consider (2) with the following hypothesis:(H3)
∃βi>0 such that
(22)θ≤Kg(y-1,…,y-k)-g(y1,…,yk)≤K∑i=1k‍βi(yi-y-i)ify-i≤Kyi,∥g(y1,…,yk)-g(y-1,…,y-k)∥≤∑i=1k‍βi∥yi-y-i∥foranyyi,y-i∈Ω.Theorem 6.
Suppose that (H2) and (H3) hold andF∈E+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(23)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈C+(Ω,0,M). Additionally, if
(24)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in C+(Ω,0,M) and depends continuously on F.In order to prove Theorem6, we need the following lemmas.Lemma 7 (see [17, Lemma 3.2]).
Let(X,K,∥·∥) be an ordered real Banach space such that K is normal, and let f,g∈C+(Ω,m,M) (resp., C-(Ω,m,M), Ccv+(Ω,m,M), and Ccc+(Ω,m,M)), where 0≤m≤M≤+∞. Then
(25)∥fk-gk∥C(Ω,X)≤∑j=0k-1‍M0j∥f-g∥C(Ω,X),∀k=1,2,….Lemma 8 (see [17, Lemma 3.3]).
Let(X,K,∥·∥) be an ordered real Banach space, and let f∈C-(Ω,m,M), where 0≤m≤M<+∞. Then
(26)-M2n-1(y-x)≤Kf2n-1(y)-f2n-1(x)≤K-m2n-1(y-x),n=1,2,…,m2n(y-x)≤Kf2n(y)-f2n(x)≤KM2n(y-x),n=1,2,…,
for all x≤Ky in Ω.Proof of Theorem6.
We apply Theorem1. Define
(27)𝒫(f)(x)=-g(fn1-1(x),…,fnk-1(x)).
By (H2), 𝒫(f)∈C(Ω,X). Next we prove that 𝒫(f) is an operator from C+(Ω,0,∞) to E+(Ω,0,∞) such that
(28)𝒫(f)∈E+(Ω,0,α(I(f))).
By the definition of C+(Ω,0,∞), for f∈C+(Ω,0,∞), we have
(29)θ≤Kf(y)-f(x)≤KI(f)(y-x)
if x≤Ky, and
(30)∥f(x)-f(y)∥≤I(f)∥x-y∥
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K. Note that
(31)θ≤Kfni-1(y)-fni-1(x)≤KI(f)ni-1(y-x),i=1,2,…,k
if x≤Ky, and
(32)∥fni-1(x)-fni-1(y)∥≤I(f)ni-1∥x-y∥,i=1,2,…,k
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K. By (H3), we get
(33)θ≤K𝒫(f)(y)-𝒫(f)(x)=g(fn1-1(x),…,fnk-1(x))-g(fn1-1(y),…,fnk-1(y))≤K∑i=1k‍βi(fni-1(y)-fni-1(x))≤K∑i=1k‍βiI(f)ni-1(y-x)
if x≤Ky, and
(34)∥𝒫(f)(y)-𝒫(f)(x)∥=∥g(fn1-1(x),…,fnk-1(x))-g(fn1-1(y),…,fnk-1(y))∥≤∑i=1k‍βi∥fni-1(y)-fni-1(x)∥≤∑i=1k‍βiI(f)ni-1∥y-x∥≤∑i=1k‍βiI(f)ni-1∥y-x∥
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K, where N(K)≤1 is employed. Hence, 𝒫(f)∈E+(Ω,0,∑i=1kβiI(f)ni-1). Let α(I(f))=∑i=1kβiI(f)ni-1. Obviously, function α is increasing on (0,+∞). By the definition of C+(Ω,0,+∞), we have fni-1(x)∈Ω,i=1,2,…,k for any f∈C+(Ω,0,+∞). Hence, F(x)+𝒫(f)(y)∈Ω,forallx,y∈Ω.
Letp=M, and let p1=M1; by (23),
(35)p(1-α(p))=M(1-∑i=1k‍βiMni-1)=M-∑i=1k‍βiMni≥M1.
Next, we prove that𝒫(f)|C+(Ω,0,M) is continuous. By (H3), for any f1,f2∈C+(Ω,0,M), by Lemma 7, we have
(36)∥𝒫(f1)-𝒫(f2)∥C+(Ω,X)=supx∈Ω∥g(f1n1-1(x),f1n2-1(x),…,f1nk-1(x))-g(f2n1-1(x),f2n2-1(x),…,f2nk-1(x))∥≤∑i=1k‍βi∥f1ni-1(x)-f2ni-1(x)∥C+(Ω,X)≤∑i=1k‍βi∑j=0ni-2‍Mj∥f1-f2∥C(Ω,X).
By Theorem 1, there exists an f∈C+(Ω,0,M) such that
(37)ℋ(f)∘f=F.
By (H2), we have
(38)G(f,fn1,…,fnk)=F.
This completes the proof of existence. Under the additional hypothesis (24), we see from (36) that the mapping L defined as (10) is a contraction mapping on the closed subset C+(Ω,0,M). The uniqueness of solution follows. Let f1,f2∈C+(Ω,0,M) be solutions of (2) with the given functions F1,F2 respectively. By the uniqueness, f1=F1+𝒫(f1)∘f1 and f2=F2+𝒫(f2)∘f2. Hence,
(39)∥f1-f2∥C(Ω,X)=∥(F1-F2)+(𝒫(f1)∘f1-𝒫(f2)∘f2)∥C(Ω,X)≤∥F1-F2∥C(Ω,X)+∥𝒫(f1)∘f1-𝒫(f2)∘f2∥C(Ω,X)≤∑i=1k‍βi∑j=0ni-1‍Mj∥f1-f2∥C(Ω,X)+∥F1-F2∥C(Ω,X).
By (24), we have ∥f1-f2∥C(Ω,X)≤(1/(1-∑i=1kβi∑j=0ni-1Mj))∥F1-F2∥C(Ω,X), implying that the solution of (2) depends continuously on F. The proof is completed.Similarly, we can prove the following result for decreasing solutions. We need the following hypothesis:(H4)
∃βi>0 such that
(40)θ≤Kg(y1,…,yk)-g(y-1,…,y-k)≤K∑i=1k‍βi(yi-y-i)ify-i≤Kyi,∥g(y1,…,yk)-g(y-1,…,y-k)∥≤∑i=1k‍βi∥yi-y-i∥foranyyi,y-i∈Ω.Theorem 9.
Suppose that (H2) and (H4) hold and all even order iterates in (2) vanish. Let F∈E-(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω, for all x,yi∈Ω and
(41)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈C-(Ω,0,M). Additionally, if
(42)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in C-(Ω,0,M) and depends continuously on F.By using Lemma8, the proof is almost the same as the proof of Theorem 6. We omit it here.


### 4.2. Convex and Concave Solutions

On the basis of the last subsection we can discuss convexity of solutions for (2). Consider (2) with the following hypothesis:(H5)
ifyi≤Ky-iory-i≤KyiinΩ, then
(43)tg(y1,…,yk)+(1-t)g(y-1,…,y-k)≤Kg(ty1+(1-t)y-1,…,tyk+(1-t)y-k).Theorem 10.
Suppose that (H2), (H3), and (H5) hold andF∈Ecv+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(44)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈Ccv+(Ω,0,M). Additionally, if
(45)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in Ccv+(Ω,0,M) and depends continuously on F.Proof.
Similar to Theorem6, by applying Theorem 3, it suffices to prove that 𝒫(f):=-g(fn1-1,…,fnk-1) is convex in the sense of K-order. In fact, each fni-1,i=1,2,…,k is convex in the sense of K-order because f is increasing and convex by Lemma 4. Hence, for every distinct comparable point x,y∈Ω, suppose x≤Ky; by (H3) and (H5), we have
(46)𝒫(f)(tx+(1-t)y)=-g(fn1-1(tx+(1-t)y),…,fnk-1(tx+(1-t)y))≤K-g(tfn1-1(x)+(1-t)fn1-1(y),…,tfnk-1(x)+(1-t)fnk-1(y))≤K-tg(fn1-1(x),…,fnk-1(x))-(1-t)g(fn1-1(y),…,fnk-1(y))=t𝒫(f)(x)+(1-t)𝒫(f)(y).
The proof is completed.Similarly, we can prove the following results for concavity of solutions. We need the following hypothesis:(H6)
ifyi≤Ky-iory-i≤KyiinΩ, then
(47)g(ty1+(1-t)y-1,…,tyk+(1-t)y-k)≤Ktg(y1,…,yk)+(1-t)g(y-1,…,y-k).Theorem 11.
Suppose that (H2), (H3), and (H6) hold andF∈Ecc+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(48)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈Ccv+(Ω,0,M). Additionally, if
(49)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in Ccv+(Ω,0,M) and depends continuously on F.Example 12.
LetX=C([0,1],ℝ) equipped with the norm ∥x∥=supt∈[0,1]|x(t)| for x∈X. Let
(50)Ω≔{x∈C([0,1],[0,1]):|x(t1)-x(t2)|≤|t1-t2|,t1,t2∈[0,1]},
a subset of X. Then, the equation
(51)G(f(x),f2(x),f3(x))=f(x)-132(f2(x))2-148ef3(x)=14x2,∀x∈Ω,
is an iterative equation of the form (2) in the infinite-dimensional setting, where
(52)G(y0,y1,y2)=y0-132y12-148ey2
and F(x)≔(1/4)x2. Note that
(53)K≔ℝ+[0,1]≔{x∈C([0,1],ℝ):x(t)≥0}
is a normal order cone in X and N(K)=1. Then, Ω is a compact convex subset of the ordered real Banach space (X,K,∥·∥). Clearly, (H2) is satisfied and
(54)g(y1,y2)=-132y12-148ey2.
If y-1≤Ky1,y-2≤Ky2,y-i,yi∈Ω,i=1,2, then
(55)0≤Kg(y-1,y-2)-g(y1,y2)=(132y12+148ey2)-(132y-12+148ey-2)≤K116(y1-y-1)+116(y2-y-2).
For any y-i,yi∈Ω,
(56)∥g(y1,y2)-g(y-1,y-)∥=∥(132y-12+148ey-2)-(132y12+148ey2)∥≤116∥y1-y-1∥+116∥y2-y-2∥,
where β1=1/16,β2=1/16. Hence, (H3) is satisfied.Ifyi≤Ky-i or y-i≤Kyi in Ω, then
(57)tg(y1,y2)+(1-t)g(y-1,y-2)=t(-132y12-148ey2)+(1-t)(-132y-12-148ey-2)=-132(ty12+(1-t)y-12)-148(tey2+(1-t)ey-2)≤K-132(ty1+(1-t)y-1)2-148ety2+(1-t)y-2≤Kg(ty1+(1-t)y-1,ty2+(1-t)y-2).
Hence, (H5) is satisfied. Similar to Example  4.2 in [17], F∈Ecv+(Ω,0,1/2). F(x)-g(y1,y2)=(1/4)x2+(1/32)y12+(1/48)ey2∈Ω for any x,y1,y2∈Ω. In this case,
(58)M-∑i=1k‍βiMni-M1=M-116M2-116M3-12≥0
for all M∈[1,3]; that is, inequality (44) holds. By Theorem 10, (51) has an increasing solution f∈Ccv+(Ω,0,M).Additionally, ifM=1, we have
(59)∑i=1K‍βi∑j=0nj-1‍Mj=116(1+M)+116(1+M+M2)=516<1.
Hence, the solution is unique and depends continuously on F.It is difficult to discuss the convexity of solutions for (2) without hypothesis (H2) and (3) without hypothesis (H1) because of the difficulties in discussing the inverse of a mapping in Banach spaces. The reason also leads to difficulties in discussing the convexity of solutions for (2) and (3) by applying the method in [23].


## 4.1. Increasing and Decreasing Solutions

We first study increasing solutions. Consider (2) with the following hypothesis:(H3)
∃βi>0 such that
(22)θ≤Kg(y-1,…,y-k)-g(y1,…,yk)≤K∑i=1k‍βi(yi-y-i)ify-i≤Kyi,∥g(y1,…,yk)-g(y-1,…,y-k)∥≤∑i=1k‍βi∥yi-y-i∥foranyyi,y-i∈Ω.Theorem 6.
Suppose that (H2) and (H3) hold andF∈E+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(23)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈C+(Ω,0,M). Additionally, if
(24)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in C+(Ω,0,M) and depends continuously on F.In order to prove Theorem6, we need the following lemmas.Lemma 7 (see [17, Lemma 3.2]).
Let(X,K,∥·∥) be an ordered real Banach space such that K is normal, and let f,g∈C+(Ω,m,M) (resp., C-(Ω,m,M), Ccv+(Ω,m,M), and Ccc+(Ω,m,M)), where 0≤m≤M≤+∞. Then
(25)∥fk-gk∥C(Ω,X)≤∑j=0k-1‍M0j∥f-g∥C(Ω,X),∀k=1,2,….Lemma 8 (see [17, Lemma 3.3]).
Let(X,K,∥·∥) be an ordered real Banach space, and let f∈C-(Ω,m,M), where 0≤m≤M<+∞. Then
(26)-M2n-1(y-x)≤Kf2n-1(y)-f2n-1(x)≤K-m2n-1(y-x),n=1,2,…,m2n(y-x)≤Kf2n(y)-f2n(x)≤KM2n(y-x),n=1,2,…,
for all x≤Ky in Ω.Proof of Theorem6.
We apply Theorem1. Define
(27)𝒫(f)(x)=-g(fn1-1(x),…,fnk-1(x)).
By (H2), 𝒫(f)∈C(Ω,X). Next we prove that 𝒫(f) is an operator from C+(Ω,0,∞) to E+(Ω,0,∞) such that
(28)𝒫(f)∈E+(Ω,0,α(I(f))).
By the definition of C+(Ω,0,∞), for f∈C+(Ω,0,∞), we have
(29)θ≤Kf(y)-f(x)≤KI(f)(y-x)
if x≤Ky, and
(30)∥f(x)-f(y)∥≤I(f)∥x-y∥
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K. Note that
(31)θ≤Kfni-1(y)-fni-1(x)≤KI(f)ni-1(y-x),i=1,2,…,k
if x≤Ky, and
(32)∥fni-1(x)-fni-1(y)∥≤I(f)ni-1∥x-y∥,i=1,2,…,k
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K. By (H3), we get
(33)θ≤K𝒫(f)(y)-𝒫(f)(x)=g(fn1-1(x),…,fnk-1(x))-g(fn1-1(y),…,fnk-1(y))≤K∑i=1k‍βi(fni-1(y)-fni-1(x))≤K∑i=1k‍βiI(f)ni-1(y-x)
if x≤Ky, and
(34)∥𝒫(f)(y)-𝒫(f)(x)∥=∥g(fn1-1(x),…,fnk-1(x))-g(fn1-1(y),…,fnk-1(y))∥≤∑i=1k‍βi∥fni-1(y)-fni-1(x)∥≤∑i=1k‍βiI(f)ni-1∥y-x∥≤∑i=1k‍βiI(f)ni-1∥y-x∥
if x,y∈Ω are not comparable, that is, x-y∉K and y-x∉K, where N(K)≤1 is employed. Hence, 𝒫(f)∈E+(Ω,0,∑i=1kβiI(f)ni-1). Let α(I(f))=∑i=1kβiI(f)ni-1. Obviously, function α is increasing on (0,+∞). By the definition of C+(Ω,0,+∞), we have fni-1(x)∈Ω,i=1,2,…,k for any f∈C+(Ω,0,+∞). Hence, F(x)+𝒫(f)(y)∈Ω,forallx,y∈Ω.
Letp=M, and let p1=M1; by (23),
(35)p(1-α(p))=M(1-∑i=1k‍βiMni-1)=M-∑i=1k‍βiMni≥M1.
Next, we prove that𝒫(f)|C+(Ω,0,M) is continuous. By (H3), for any f1,f2∈C+(Ω,0,M), by Lemma 7, we have
(36)∥𝒫(f1)-𝒫(f2)∥C+(Ω,X)=supx∈Ω∥g(f1n1-1(x),f1n2-1(x),…,f1nk-1(x))-g(f2n1-1(x),f2n2-1(x),…,f2nk-1(x))∥≤∑i=1k‍βi∥f1ni-1(x)-f2ni-1(x)∥C+(Ω,X)≤∑i=1k‍βi∑j=0ni-2‍Mj∥f1-f2∥C(Ω,X).
By Theorem 1, there exists an f∈C+(Ω,0,M) such that
(37)ℋ(f)∘f=F.
By (H2), we have
(38)G(f,fn1,…,fnk)=F.
This completes the proof of existence. Under the additional hypothesis (24), we see from (36) that the mapping L defined as (10) is a contraction mapping on the closed subset C+(Ω,0,M). The uniqueness of solution follows. Let f1,f2∈C+(Ω,0,M) be solutions of (2) with the given functions F1,F2 respectively. By the uniqueness, f1=F1+𝒫(f1)∘f1 and f2=F2+𝒫(f2)∘f2. Hence,
(39)∥f1-f2∥C(Ω,X)=∥(F1-F2)+(𝒫(f1)∘f1-𝒫(f2)∘f2)∥C(Ω,X)≤∥F1-F2∥C(Ω,X)+∥𝒫(f1)∘f1-𝒫(f2)∘f2∥C(Ω,X)≤∑i=1k‍βi∑j=0ni-1‍Mj∥f1-f2∥C(Ω,X)+∥F1-F2∥C(Ω,X).
By (24), we have ∥f1-f2∥C(Ω,X)≤(1/(1-∑i=1kβi∑j=0ni-1Mj))∥F1-F2∥C(Ω,X), implying that the solution of (2) depends continuously on F. The proof is completed.Similarly, we can prove the following result for decreasing solutions. We need the following hypothesis:(H4)
∃βi>0 such that
(40)θ≤Kg(y1,…,yk)-g(y-1,…,y-k)≤K∑i=1k‍βi(yi-y-i)ify-i≤Kyi,∥g(y1,…,yk)-g(y-1,…,y-k)∥≤∑i=1k‍βi∥yi-y-i∥foranyyi,y-i∈Ω.Theorem 9.
Suppose that (H2) and (H4) hold and all even order iterates in (2) vanish. Let F∈E-(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω, for all x,yi∈Ω and
(41)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈C-(Ω,0,M). Additionally, if
(42)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in C-(Ω,0,M) and depends continuously on F.By using Lemma8, the proof is almost the same as the proof of Theorem 6. We omit it here.


## 4.2. Convex and Concave Solutions

On the basis of the last subsection we can discuss convexity of solutions for (2). Consider (2) with the following hypothesis:(H5)
ifyi≤Ky-iory-i≤KyiinΩ, then
(43)tg(y1,…,yk)+(1-t)g(y-1,…,y-k)≤Kg(ty1+(1-t)y-1,…,tyk+(1-t)y-k).Theorem 10.
Suppose that (H2), (H3), and (H5) hold andF∈Ecv+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(44)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈Ccv+(Ω,0,M). Additionally, if
(45)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in Ccv+(Ω,0,M) and depends continuously on F.Proof.
Similar to Theorem6, by applying Theorem 3, it suffices to prove that 𝒫(f):=-g(fn1-1,…,fnk-1) is convex in the sense of K-order. In fact, each fni-1,i=1,2,…,k is convex in the sense of K-order because f is increasing and convex by Lemma 4. Hence, for every distinct comparable point x,y∈Ω, suppose x≤Ky; by (H3) and (H5), we have
(46)𝒫(f)(tx+(1-t)y)=-g(fn1-1(tx+(1-t)y),…,fnk-1(tx+(1-t)y))≤K-g(tfn1-1(x)+(1-t)fn1-1(y),…,tfnk-1(x)+(1-t)fnk-1(y))≤K-tg(fn1-1(x),…,fnk-1(x))-(1-t)g(fn1-1(y),…,fnk-1(y))=t𝒫(f)(x)+(1-t)𝒫(f)(y).
The proof is completed.Similarly, we can prove the following results for concavity of solutions. We need the following hypothesis:(H6)
ifyi≤Ky-iory-i≤KyiinΩ, then
(47)g(ty1+(1-t)y-1,…,tyk+(1-t)y-k)≤Ktg(y1,…,yk)+(1-t)g(y-1,…,y-k).Theorem 11.
Suppose that (H2), (H3), and (H6) hold andF∈Ecc+(Ω,0,M1), where M1∈(0,+∞) is a constant. If F(x)-g(y1,…,yk)∈Ω for any x,yi∈Ω and
(48)M1≤M-∑i=1k‍βiMni
for a constant M∈(0,+∞), then (2) has a solution f∈Ccv+(Ω,0,M). Additionally, if
(49)∑i=1k‍βi∑j=0ni-1‍Mj<1,
then the solution f is unique in Ccv+(Ω,0,M) and depends continuously on F.Example 12.
LetX=C([0,1],ℝ) equipped with the norm ∥x∥=supt∈[0,1]|x(t)| for x∈X. Let
(50)Ω≔{x∈C([0,1],[0,1]):|x(t1)-x(t2)|≤|t1-t2|,t1,t2∈[0,1]},
a subset of X. Then, the equation
(51)G(f(x),f2(x),f3(x))=f(x)-132(f2(x))2-148ef3(x)=14x2,∀x∈Ω,
is an iterative equation of the form (2) in the infinite-dimensional setting, where
(52)G(y0,y1,y2)=y0-132y12-148ey2
and F(x)≔(1/4)x2. Note that
(53)K≔ℝ+[0,1]≔{x∈C([0,1],ℝ):x(t)≥0}
is a normal order cone in X and N(K)=1. Then, Ω is a compact convex subset of the ordered real Banach space (X,K,∥·∥). Clearly, (H2) is satisfied and
(54)g(y1,y2)=-132y12-148ey2.
If y-1≤Ky1,y-2≤Ky2,y-i,yi∈Ω,i=1,2, then
(55)0≤Kg(y-1,y-2)-g(y1,y2)=(132y12+148ey2)-(132y-12+148ey-2)≤K116(y1-y-1)+116(y2-y-2).
For any y-i,yi∈Ω,
(56)∥g(y1,y2)-g(y-1,y-)∥=∥(132y-12+148ey-2)-(132y12+148ey2)∥≤116∥y1-y-1∥+116∥y2-y-2∥,
where β1=1/16,β2=1/16. Hence, (H3) is satisfied.Ifyi≤Ky-i or y-i≤Kyi in Ω, then
(57)tg(y1,y2)+(1-t)g(y-1,y-2)=t(-132y12-148ey2)+(1-t)(-132y-12-148ey-2)=-132(ty12+(1-t)y-12)-148(tey2+(1-t)ey-2)≤K-132(ty1+(1-t)y-1)2-148ety2+(1-t)y-2≤Kg(ty1+(1-t)y-1,ty2+(1-t)y-2).
Hence, (H5) is satisfied. Similar to Example  4.2 in [17], F∈Ecv+(Ω,0,1/2). F(x)-g(y1,y2)=(1/4)x2+(1/32)y12+(1/48)ey2∈Ω for any x,y1,y2∈Ω. In this case,
(58)M-∑i=1k‍βiMni-M1=M-116M2-116M3-12≥0
for all M∈[1,3]; that is, inequality (44) holds. By Theorem 10, (51) has an increasing solution f∈Ccv+(Ω,0,M).Additionally, ifM=1, we have
(59)∑i=1K‍βi∑j=0nj-1‍Mj=116(1+M)+116(1+M+M2)=516<1.
Hence, the solution is unique and depends continuously on F.It is difficult to discuss the convexity of solutions for (2) without hypothesis (H2) and (3) without hypothesis (H1) because of the difficulties in discussing the inverse of a mapping in Banach spaces. The reason also leads to difficulties in discussing the convexity of solutions for (2) and (3) by applying the method in [23].



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*Source: 164851-2013-09-15.xml*