A Krasnoselskii–Ishikawa Iterative Algorithm for Monotone Reich Contractions in Partially Ordered Banach Spaces with an Application

Abstract

Iterative algorithms have been utilized for the computation of approximate solutions of stationary and evolutionary problems associated with differential equations. The aim of this article is to introduce concepts of monotone Reich and Chatterjea nonexpansive mappings on partially ordered Banach spaces. We describe sufficient conditions for the existence of an approximate fixed-point sequence (AFPS) and prove certain fixed-point results using the Krasnoselskii–Ishikawa iterative algorithm. Moreover, we present some interesting examples to highlight the superiority of our results. Lastly, we provide both weak and strong convergence results for such mappings and consider an application of our results to prove the existence of a solution to an initial value problem.

1. Introduction

The study of the existence of fixed points for monotone nonexpansive and asymptotically nonexpansive mappings in the setting of a metric space have been widely investigated. Let U be a normed linear space. A self mapping $\mathcal{S}:U\to U$ is said to be nonexpansive if $\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel$ for all $\mu ,\omega \in U$. The set of fixed points of $\mathcal{S}$ is denoted by $\mathcal{F}\left(\mathcal{S}\right)$, that is, $\mathcal{F}\left(\mathcal{S}\right)=\{\mu \in U:\mathcal{S}\mu =\mu \}$. A mapping $\mathcal{S}$ is called quasi-nonexpansive if $\parallel \mathcal{S}\mu -\nu \parallel \le \parallel \mu -\nu \parallel$ for all $\mu \in U$, and $\nu \in \mathcal{F}\left(\mathcal{S}\right)$. A mapping $\mathcal{S}:U\to U$ is said to be asymptotically nonexpansive if there exists a sequence $\left\{a_n\right\}$ with $a_n\ge 1$ and ${lim}_{n\to \infty }{a}_n=1$ such that the condition $\parallel {\mathcal{S}}^n\mu -{\mathcal{S}}^n\omega \parallel \le a_n\parallel \mu -\omega \parallel$ holds for all $n\in \mathbb{N}$. A sequence $\left\{{\mu }_n\right\}$ in U is asymptotically regular if ${lim}_{n\to \infty }\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$ for all $n\in \mathbb{N}$. We know that every nonexpansive mapping or asymptotically nonexpansive mapping on a non-empty, closed, bounded, convex subset of a uniformly convex Banach space has at least one fixed point [1,2,3]. Debnath and La Sen [4] studied the fixed points of a broad category of set-valued maps that may include discontinuous maps extending the notions of orbitally continuous and asymptotically regular mappings. In [5], the authors established fixed-point results for some asymptotically regular multivalued mappings satisfying the Kannan type contractive condition without assuming compactness of the underlying metric space or continuity of underlying mapping.

Many authors have studied an approximate fixed-point sequence (AFPS) and approximate fixed-point property (AFPP) for different types of mappings [6,7,8,9]. A sequence $\left\{{\mu }_n\right\}$ in a normed space U is said to be an (AFPS) for a self mapping $\mathcal{S}:U\to U$ if ${lim}_{n\to \infty }\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0$. Matouškovà and Reich [10] found that every infinite dimensional Banach space contained an unbounded closed convex set with the approximate point fixed property for nonexpansive mappings. These results have become an important tool in solving a variety of problems, such as integral equations, partial differential equations, optimization problems and boundary value problems; see [10,11,12,13,14,15].

Recently, Som et al. in [16] introduced two types of mappings, Reic-type nonexpansive and Chatterjea-type nonexpansive mappings, and discussed that these classes of mappings possess an APFS under some conditions using the Krasnoselskii iteration method in Banach spaces. They explored some properties of the fixed-point sets of these mappings like closedness, convexity, and remotality, and obtained sufficient conditions under which a Reich-type nonexpansive mapping reduces to that of a nonexpansive one. For further considerations on Reich and Chatterjea contractions, see [17,18,19,20]. Let ${\mu }_1\in U$ be arbitrary. Then, consider the Krasnoselskii-Ishikawa iteration defined by the sequence $\left\{{\mu }_n\right\}\subset U$ as follows:

$${\mu }_{n+1}=\lambda \mathcal{S}{\mu }_n+(1-\lambda ){\mu }_n,\left(KIS\right)$$

where $\lambda \in (0,1),$$n\in \mathbb{N}$ and $\mathcal{S}$ is a self map of U. Very recently, Popescu and Stan [21] have obtained certain interesting results related to Reich- and Chatterjea-type nonexpansive mappings. In ref. [22], Debnath et al. studied the existence and uniqueness of fixed points of Reich-type G-contraction on closed and bounded subsets of a metric space endowed with a graph. Khamsi and Khan [23] discussed the behavior of the Krasnoselskii–Ishikawa iteration process for monotone nonexpansive mappings in $L_1\left([0,1]\right)$; see also [12,24].

The fixed-point theory of monotone contractions defined on partially ordered metric spaces has been researched by a number of scholars. The contractivity criterion on the nonlinear map is only considered to hold elements that are comparable in the partial order in such findings. The reader is directed to the ground-breaking work on the subject, which has applications in integral, differential, and nonlinear fractional evolution equations and equilibrium problems; see [2,25,26,27,28,29,30,31,32,33,34].

Iterative algorithms have been utilized for the computation of approximate solutions of stationary and evolutionary problems associated with differential equations. This paper is organized as follows: In Section 2, some basic definitions and propositions are stated. In Section 3, we develop sufficient conditions for the existence of an approximate fixed point sequence (AFPS) with certain fixed-point results using the Krasnoselskii–Ishikawa iteration algorithm. A number of examples are presented to illustrate the results. In Section 4, we provide some weak and strong convergence results for these mappings. In Section 5, we study the existence of a solution for a nonlinear differential equation. Section 6 is concerned with conclusion.

2. Preliminaries

Let $(U,\parallel .\parallel )$ be a Banach space endowed with a partial order ⪯. Throughout, it is assumed that order intervals are convex and closed. Recall that an order interval is any of the subsets

$$[\mu ,\to )=\{\nu \in U;\mu ⪯\nu \}and(\leftarrow ,\omega ]=\{\nu \in U;\nu ⪯\omega \},$$

for any $\mu ,\omega \in U$. Therefore, the order interval $[\mu ,\omega ]$ for all $\mu ,\omega \in U$ is given by

$$[\mu ,\omega ]=\{\nu \in U;\mu ⪯\nu ⪯\omega \}=[\mu ,\to )\cap (\leftarrow ,\omega ],$$

is also closed and convex for any $\mu ,\omega \in U$.

Definition 1

([24]). Let E be a nonempty subset of the ordered Banach space U. A map $\mathcal{S}:E\to E$ is said to be

(i) 

monotone if $\mathcal{S}\mu ⪯\mathcal{S}\omega$ for all $\mu ,\omega \in E$ with $\mu ⪯\omega$;

(ii) 

monotone nonexpansive if $\mathcal{S}$ is monotone and $\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel$ for any $\mu ,\omega \in E$ with $\mu ⪯\omega$.

The sequence $\left\{{\mu }_n\right\}$ is said to be bounded in a partially ordered set $(U,⪯)$ if there exists a point $\nu \in U$ such that ${\mu }_n⪯\nu$ for all $n\in \mathbb{N}$.

Now, we give the definition of ordered Reich and Chatterjea nonexpansive mappings.

Definition 2.

Let U be an ordered normed space, wherein E is a non-empty closed convex subset of U. A mapping $\mathcal{S}:E\to E$ is said to be an ordered Reich nonexpansive mapping if there exist non-negative real numbers $\alpha ,\beta ,\gamma$ with $\alpha +\beta +\gamma =1$ such that

$$\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\mu \parallel +\gamma \parallel \omega -\mathcal{S}\omega \parallel ,$$

(1)

for all $\mu ,\omega \in E$ with $\mu ⪯\omega$.

The mapping $\mathcal{S}$ is said to be an ordered Chatterjea nonexpansive mapping if there exist non-negative real numbers $\alpha ,\beta ,\gamma$ with $\alpha +\beta +\gamma =1$ such that the condition

$$\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\omega \parallel +\gamma \parallel \omega -\mathcal{S}\mu \parallel ,$$

(2)

holds, for all $\mu ,\omega \in E$ such that $\mu ⪯\omega$.

The following proposition is proved in [35,36].

Proposition 1.

Let $(U,d,⪯)$ be a partially ordered metric space. Assume that the sequences $\left\{{\mu }_n\right\}$ and $\left\{{\omega }_n\right\}$ in U with $t_n\in [0,1)$ satisfy following conditions for all $n\in \mathbb{N}$:

(i) 

${\mu }_{n+1}\in [{\mu }_n,{\omega }_n]$ with $d({\mu }_n,{\omega }_{n+1})=t_{n}d({\mu }_n,{\omega }_n)$;

(ii) 

$d({\omega }_{n+1},{\omega }_n)\le d({\mu }_{n+1},{\mu }_n).$

Then,

$$\left(1+\sum _{s=i}^{i+n-1}{t}_s\right)d({\omega }_i,{\mu }_i)\le d({\omega }_{i+n},{\mu }_i)+\prod _{s=i}^{i+n-1}{(1-t_s)}^{-1}[d({\omega }_i,{\mu }_i)-d({\omega }_{i+n},{\mu }_{i+n})]$$

for all $n\ge 1$ and $i\ge 0$.

3. Main Results

The following technical lemma will be useful to develop further results.

Lemma 1.

Let $(U,d,⪯)$ be a partially ordered Banach space and E be a nonempty convex subset of U. Let $\mathcal{S}:E\to E$ be a monotone mapping satisfying the following condition:

$$\xi d(\mu ,\mathcal{S}\mu )\le d(\mu ,\omega )\Rightarrow d(\mathcal{S}\mu ,\mathcal{S}\omega )\le d(\mu ,\omega ),$$

(3)

where $\xi \in (0,1)$. Assume that there exists ${\mu }_1\in E$ such that ${\mu }_1$ and $\mathcal{S}{\mu }_1$ are comparable. Consider the sequence $\left\{{\mu }_n\right\}$ in E as defined by $KIS$, such that $\xi <\lambda$ and $\lambda \in (0,1)$. Then, for any $n\ge 1$ and $i\ge 0$, we have

$$(1+n\lambda )d(\mathcal{S}{\mu }_i,{\mu }_i)\le d(\mathcal{S}{\mu }_{i+n},{\mu }_i)+{(1-\lambda )}^{-n}[d(\mathcal{S}{\mu }_i,{\mu }_i)-d(\mathcal{S}{\mu }_{i+n},{\mu }_{i+n})].$$

Moreover, if $\left\{{\mu }_n\right\}$ has two subsequences which converge to ω and ν, then $\omega =\nu$.

Proof. 

The induction method is employed to demonstrate ${\mu }_n⪯{\mu }_{n+1}⪯\mathcal{S}{\mu }_n$ for any $n\ge 1$. By assumption, there is ${\mu }_1⪯\mathcal{S}{\mu }_1$. Using the convexity of the order interval $[{\mu }_1,\mathcal{S}{\mu }_1]$ and KIS, we obtain

$${\mu }_1⪯\lambda \mathcal{S}{\mu }_1+(1-\lambda ){\mu }_1={\mu }_2⪯\mathcal{S}{\mu }_1.$$

As $\mathcal{S}$ is monotone, $\mathcal{S}{\mu }_1⪯\mathcal{S}{\mu }_2$, which implies

$${\mu }_1⪯{\mu }_2⪯\mathcal{S}{\mu }_1⪯\mathcal{S}{\mu }_2.$$

Thus, by induction, we have

$${\mu }_n⪯{\mu }_{n+1}⪯\mathcal{S}{\mu }_n⪯\mathcal{S}{\mu }_{n+1},$$

for all $n\in \mathbb{N}$. This implies that

$${\mu }_{n+1}\in [{\mu }_n,\mathcal{S}{\mu }_n].$$

(4)

Thus, by KIS, we have

$$d({\mu }_{n+1},\mathcal{S}{\mu }_n)=(1-\lambda )d({\mu }_n,\mathcal{S}{\mu }_n).$$

Furthermore,

$$d({\mu }_{n+1},{\mu }_n)=\lambda d({\mu }_n,\mathcal{S}{\mu }_n).$$

(5)

Since $\xi <\lambda$, for all $\lambda \in (0,1)$, there is

$$\xi d({\mu }_n,\mathcal{S}{\mu }_n)\le \lambda d({\mu }_n,\mathcal{S}{\mu }_n)=d({\mu }_{n+1},{\mu }_n).$$

Now, (3) this implies that

$$d(\mathcal{S}{\mu }_{n+1},\mathcal{S}{\mu }_n)\le d({\mu }_{n+1},{\mu }_n)$$

(6)

for all $n\in \mathbb{N}$. All of the assumptions of Proposition 1 hold, with ${\omega }_n=\mathcal{S}{\mu }_n$. On the other hand

$$\left\{\begin{array}{c}(1+n\lambda )\le 1+{\sum }_{s=i}^{i+n-1}{t}_s,\hfill \\ {\prod }_{s=i}^{i+n-1}{(1-t_s)}^{-1}\le {(1-\lambda )}^{-n},\hfill \\ d({\mu }_i,\mathcal{S}{\mu }_{i+n})\le \delta \left(E\right)=sup\{d(\mu ,\omega ):\mu ,\omega \in E\},\hfill \end{array}\right.$$

(7)

for all $n\in \mathbb{N}$ and $\lambda \in (0,1)$, which implies that

$$(1+n\lambda )d(\mathcal{S}{\mu }_i,{\mu }_i)\le d(\mathcal{S}{\mu }_{i+n},{\mu }_i)+{(1-\lambda )}^{-n}[d(\mathcal{S}{\mu }_i,{\mu }_i)-d(\mathcal{S}{\mu }_{i+n},{\mu }_{i+n})]$$

for any $i\ge 0$ and $n\ge 1$ as required.

Next, let $\left\{{\mu }_{n_k}\right\}$ be a subsequence of $\left\{{\mu }_n\right\}$ that converges to $\omega$. Fix $m\ge 1$. Since the order interval $[{\mu }_k,\to )$ is closed and convex and $\left\{{\mu }_n\right\}$ is monotone increasing, then $\omega \in [{\mu }_m,\to )$. Hence ${\mu }_m⪯\omega$ for any $m\ge 1$. Consequently, if $\left\{{\mu }_n\right\}$ has another subsequence which converges to $\nu$, it is necessary to have $\omega =\nu$. Indeed, since ${\mu }_n⪯\omega ,$ for any $n\ge 1,$ we get $\omega ⪯\nu$. Similarly, $\nu ⪯\omega$, which implies that $\omega =\nu$. □

Remark 1.

Under the assumptions of Lemma 1, if we assume $\mathcal{S}{\mu }_1⪯{\mu }_1$, then we have

$$\mathcal{S}{\mu }_n⪯{\mu }_{n+1}⪯{\mu }_n$$

for any $n\ge 1.$ Moreover, the conclusion on equality of limits of subsequences of ${\mu }_n$ holds.

Theorem 1.

Let $(U,\parallel .\parallel ,⪯)$ be a partially ordered Banach space and E be the nonempty closed convex and bounded subset of U. Let a self-mapping $\mathcal{S}$ of E be a monotone and an ordered Reich nonexpansive mapping with coefficients $\alpha ,\beta ,\gamma$ such that $\gamma <1$. Assume that there exists ${\mu }_1\in E$ such that ${\mu }_1$ and $\mathcal{S}{\mu }_1$ are comparable. Furthermore, suppose that for any $\mu ,\omega$ in E with $\mu ⪯\omega$, we have

$$\frac{(1-\gamma )}{6}\parallel \mu -\mathcal{S}\mu \parallel \le \parallel \mu -\omega \parallel \Rightarrow \parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel .$$

(8)

Then, $\mathcal{S}$ has an AFPS in E, which is asymptotically regular.

Proof. 

Let ${\mu }_1\in E$ such that ${\mu }_1⪯\mathcal{S}{\mu }_1.$ Considering the sequence $\left\{{\mu }_n\right\}$ in U defined by KIS, the following is obtained:

$$\parallel {\mu }_{n+1}-{\mu }_n\parallel =\lambda \parallel \mathcal{S}{\mu }_n-{\mu }_n\parallel .$$

(9)

In addition, we have

$$\parallel {\mu }_{n+2}-{\mu }_{n+1}\parallel =\lambda \parallel \mathcal{S}{\mu }_{n+1}-{\mu }_{n+1}\parallel$$

(10)

for all $n\ge 1$ where $\lambda \in [\frac{1}{2},1)$.

Now, for any $\mu \in U$, with $\mu ={\mu }_n,$ and $\omega ={\mu }_{n+1}$, using (1), we get

$$\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \alpha \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\gamma \parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel ,$$

(11)

where ${\mu }_n⪯{\mu }_{n+1}.$

From (11), we get

$$\lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \lambda \alpha \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \lambda \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\gamma \lambda \parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel .$$

(12)

Using (9) and (10) in (12), we get

$$\lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \alpha \lambda \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-{\mu }_n\parallel +\gamma \parallel {\mu }_{n+2}-{\mu }_{n+1}\parallel .$$

(13)

Again, by KIS, we obtain,

$$\parallel {\mu }_{n+2}-{\mu }_{n+1}\parallel \le \lambda \parallel \mathcal{S}{\mu }_{n+1}-\mathcal{S}{\mu }_n\parallel +(1-\lambda )\parallel {\mu }_{n+1}-{\mu }_n\parallel .$$

(14)

Then, using (14) in (13), we have

$$\begin{array}{cc}\hfill \lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & \le \alpha \lambda \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-{\mu }_n\parallel +\gamma \lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & +\gamma (1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

Thus,

$$(\lambda -\gamma \lambda )\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \alpha \lambda \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-{\mu }_n\parallel +\gamma (1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

Since $\lambda <1$, then $(1-\lambda )<1$. We thus obtain

$$\lambda (1-\gamma )\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\alpha \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-{\mu }_n\parallel +\gamma \parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

We get that

$$\lambda (1-\gamma )\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel ,$$

where $\alpha +\beta +\gamma =1$. On the other hand,

$$\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel =\parallel {\mu }_n-\mathcal{S}{\mu }_n+\mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel .$$

Then, using (9), we have

$$\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \frac{1}{\lambda }\parallel {\mu }_n-{\mu }_{n+1}\parallel +\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel .$$

(15)

Hence,

$$\lambda (1-\gamma )\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \frac{\lambda (1-\gamma )}{\lambda }\parallel {\mu }_n-{\mu }_{n+1}\parallel +\lambda (1-\gamma )\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel .$$

(16)

Using (15) into (16), we obtain

$$\begin{array}{cc}\hfill \lambda (1-\gamma )\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & <(1-\gamma )\parallel {\mu }_n-{\mu }_{n+1}\parallel +\frac{\lambda (1-\gamma )}{\lambda (1-\gamma )}\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & =(2-\gamma )\parallel {\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

Which implies

$$\frac{\lambda (1-\gamma )}{(2-\gamma )}\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

(17)

Now, since $1/2\le \lambda <1$ and $0\le \gamma <1$, we get

$$\frac{1-\gamma }{4}<\frac{\lambda (1-\gamma )}{2-\gamma }.$$

Therefore, (17) implies

$$\frac{(1-\gamma )}{4}\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

The following is obtained:

$$\begin{array}{cc}\hfill \frac{(1-\gamma )}{4}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel & \le \frac{1-\gamma }{4}\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel +\frac{1-\gamma }{4}\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & <\parallel {\mu }_n-{\mu }_{n+1}\parallel +\frac{1-\gamma }{4\lambda (1-\gamma )}\parallel {\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \parallel {\mu }_n-{\mu }_{n+1}\parallel +\frac{1}{4\lambda }\parallel {\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \frac{3}{2}\parallel {\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

For any $1/2\le \lambda <1$, we have that

$$\frac{(1-\gamma )}{6}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel ,$$

which implies that

$$\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

(18)

Note that $\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel$ is decreasing if, and only if, $\parallel \mathcal{S}{\mu }_{n+1}-{\mu }_{n+1}\parallel$ is decreasing, which follows from the triangle inequality, (18) and (9),

$$\begin{array}{cc}\hfill \parallel \mathcal{S}{\mu }_{n+1}-{\mu }_{n+1}\parallel & \le \parallel \mathcal{S}{\mu }_{n+1}-\mathcal{S}{\mu }_n\parallel +\parallel \mathcal{S}{\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \parallel {\mu }_n-{\mu }_{n+1}\parallel +\parallel \mathcal{S}{\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \lambda \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\parallel \mathcal{S}{\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

(19)

From the KIS definition, we obtain

$$\parallel \mathcal{S}{\mu }_n-{\mu }_{n+1}\parallel =(1-\lambda )\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel .$$

(20)

From (19) and (20), the implication is that

$$\begin{array}{cc}\hfill \parallel \mathcal{S}{\mu }_{n+1}-{\mu }_{n+1}\parallel & \le \lambda \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +(1-\lambda )\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \\ \hfill & \le \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \end{array}$$

for all $n\in \mathbb{N}$. Consequently, $\parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel$ is a decreasing and bounded sequence. Thus, there exists $R\ge 0$ such that

$$\underset{n\to \infty }{lim}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =R\ge 0.$$

Since $\xi =(1-\gamma )/6<\lambda <1$, the inequality obtained in Lemma 1 implies that

$$(1+n\xi )\parallel {\mu }_i-\mathcal{S}{\mu }_i\parallel \le (1+n\lambda )\parallel {\mu }_i-\mathcal{S}{\mu }_i{\parallel \le \delta \left(E\right)+(1-\lambda )}^{-n}\left[\parallel {\mu }_i-\mathcal{S}{\mu }_i\parallel -\parallel {\mu }_{i+n}-\mathcal{S}{\mu }_{i+n}\parallel \right],$$

for all $i\ge 0,$ and $n\ge 1$, with $\delta \left(E\right)=sup\{\parallel \mu -\omega \parallel ;\mu ,\omega \in E\}<+\infty .$ If we let $i\to +\infty$, we get

$$(1+n\lambda )R\le \delta \left(E\right)+{(1-\lambda )}^{-n}[R-R].$$

Then,

$$R\le \frac{\delta \left(E\right)}{1+n\lambda }$$

holds for all $n\ge 1$. Clearly this is possible only if $R=0,$ which implies ${lim}_{n\to \infty }\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0.$ Thus, the sequence $\left\{{\mu }_n\right\}$ is an AFPS for $\mathcal{S}$. Further, we have

$$\parallel {\mu }_n-{\mu }_{n+1}\parallel =\lambda \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \to 0,asn\to \infty .$$

Therefore, the AFPS $\left\{{\mu }_n\right\}$ is asymptotically regular. □

Theorem 2.

Under the assumptions of Theorem 1, $\mathcal{S}$ has a fixed point, provided $\alpha <1$.

Proof. 

By Theorem 1, $\mathcal{S}$ has an AFPS $\left({\mu }_n\right)$ which is an asymptotically regular sequence.

By triangle inequality and asymptotic regularity, for $m\ge n\ge 1,$ we obtain

$$\begin{array}{cc}\hfill \parallel {\mu }_m-{\mu }_n\parallel & \le \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\parallel \mathcal{S}{\mu }_m-\mathcal{S}{\mu }_n\parallel +\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \\ \hfill & \le \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\alpha \parallel {\mu }_m-{\mu }_n\parallel +\beta \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\gamma \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \\ \hfill & +\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel ,\hfill \end{array}$$

which implies that

$$(1-\alpha )\parallel {\mu }_m-{\mu }_n\parallel \le \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\beta \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\gamma \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \to 0,$$

as $m\ge n\to \infty$, $\alpha <1$. Therefore,$\left\{{\mu }_n\right\}$ is Cauchy sequence in E and there exists $\mu$ in E such that ${\mu }_n\to \mu .$ Then,

$$\parallel \mathcal{S}{\mu }_m-\mathcal{S}{\mu }_n\parallel \le \alpha \parallel {\mu }_m-{\mu }_n\parallel +\beta \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\gamma \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \to 0,$$

as $n,m\to \infty$. This implies that $\mathcal{S}{\mu }_n$ is a Cauchy sequence in E. Therefore, the following is obtained:

$$\parallel \mathcal{S}{\mu }_n-\mu \parallel \le \parallel \mathcal{S}{\mu }_n-{\mu }_n\parallel +\parallel {\mu }_n-\mu \parallel \to 0$$

as $n\to \infty$. Again, there is

$$\parallel \mathcal{S}{\mu }_n-\mathcal{S}\mu \parallel \le \alpha \parallel {\mu }_n-\mu \parallel +\beta \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\gamma \parallel \mu -\mathcal{S}\mu \parallel .$$

As $n\to \infty$, we get

$$\parallel \mu -\mathcal{S}\mu \parallel \le \gamma \parallel \mu -\mathcal{S}\mu \parallel <\parallel \mu -\mathcal{S}\mu \parallel ,$$

which implies that $\mu =\mathcal{S}\mu ,$ i.e., $\mu$ is a fixed point of $\mathcal{S}$. □

The next theorem provides sufficient conditions for the existence of AFPS in case S is an ordered Chatterjea nonexpansive mapping.

Theorem 3.

Let $(U,\parallel .\parallel ,⪯)$ be a partially ordered Banach space and E be a nonempty closed convex and bounded subset of U. Let $\mathcal{S}:E\to E$ be a monotone and an ordered Chatterjea nonexpansive mapping with coefficients $\alpha ,\beta ,\gamma$ such that $\alpha +\beta +\gamma =1$ and $\beta <1$. Assume that there exists ${\mu }_1\in E$ such that ${\mu }_1$ and $\mathcal{S}{\mu }_1$ are comparable. Assume that for any $\mu ,\omega$ in E with $\mu ⪯\omega$, we have

$$\frac{(1-\beta )}{7}\parallel \mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel \Rightarrow \parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel .$$

Then, $\mathcal{S}$ has an AFPS in E, which is asymptotically regular.

Proof. 

Fix ${\mu }_1\in E$. It is assumed that ${\mu }_1⪯\mathcal{S}{\mu }_1.$ Consider the sequence $\left\{{\mu }_n\right\}$, defined by KIS in E, for all $n\ge 1$ where $\lambda \in [\frac{1}{2},1)$.

By our hypothesis, we obtain

$$\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\omega \parallel +\gamma \parallel \omega -\mathcal{S}\mu \parallel .$$

(21)

Putting $\mu ={\mu }_n$, and $\omega ={\mu }_{n+1}$ in (21), we get

$$\begin{array}{cc}\hfill \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & \le \alpha \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel +\gamma \parallel {\mu }_{n+1}+\mathcal{S}{\mu }_n\parallel \hfill \\ \hfill & \le \alpha \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \{\parallel {\mu }_n-{\mu }_{n+1}\parallel +\parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel \}\hfill \\ \hfill & +\gamma \{\parallel {\mu }_{n+1}-{\mu }_n\parallel +\parallel {\mu }_n+\mathcal{S}{\mu }_n\parallel \}\hfill \\ \hfill & =(\alpha +\beta +\gamma )\parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel +\gamma \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \\ \hfill & =\parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-\mathcal{S}{\mu }_{n+1}\parallel +\gamma \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel \hfill \end{array}$$

(22)

for all $n\ge 1$. Using (9) and (10) in (22), we obtain

$$\lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \lambda \parallel {\mu }_n-{\mu }_{n+1}\parallel +\beta \parallel {\mu }_{n+1}-{\mu }_{n+2}\parallel +\gamma \parallel {\mu }_n-{\mu }_{n+1}\parallel ,$$

(23)

where ${\mu }_n⪯{\mu }_{n+1}$.

The definition of KIS and (23) imply that

$$\begin{array}{cc}\hfill \parallel {\mu }_{n+1}-{\mu }_{n+2}\parallel & \le \lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel +(1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \lambda \parallel x_n-x_{n+1}\parallel +b\parallel x_{n+1}-x_{n+2}\parallel +c\parallel x_n-x_{n+1}\parallel \hfill \\ \hfill & +(1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel ,\hfill \end{array}$$

and

$$\parallel {\mu }_{n+1}-{\mu }_{n+2}\parallel \le \frac{(1+\gamma )}{(1-\beta )}\parallel {\mu }_n-{\mu }_{n+1}\parallel ,$$

(24)

then

$$\lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \le \parallel {\mu }_{n+1}-{\mu }_{n+2}\parallel +(1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

(25)

Now using (9), (24) and (25) we get

$$\begin{array}{cc}\hfill \parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & \le \parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \frac{1}{\lambda }\parallel {\mu }_n-{\mu }_{n+1}\parallel +\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill \Rightarrow \lambda \parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & \le \parallel {\mu }_n-{\mu }_{n+1}\parallel +\lambda \parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & \le \parallel {\mu }_n-{\mu }_{n+1}\parallel +\parallel {\mu }_{n+1}-{\mu }_{n+2}\parallel +(1-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & \le (2-\lambda )\parallel {\mu }_n-{\mu }_{n+1}\parallel +\frac{(1+\gamma )}{(1-\beta )}\parallel {\mu }_n-{\mu }_{n+1}\parallel \hfill \\ \hfill & <\left(\frac{3}{2}+\frac{2}{1-\beta }\right)\parallel {\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

This implies that

$$\begin{array}{cc}\hfill \frac{1}{2}\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel & \le \lambda \parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel \hfill \\ \hfill & <\left(\frac{3}{2}+\frac{2}{1-\beta }\right)\parallel {\mu }_n-{\mu }_{n+1}\parallel .\hfill \end{array}$$

Therefore,

$$\frac{(1-\beta )}{7}\parallel {\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

By our hypothesis, we obtain

$$\parallel \mathcal{S}{\mu }_n-\mathcal{S}{\mu }_{n+1}\parallel <\parallel {\mu }_n-{\mu }_{n+1}\parallel .$$

According to Theorem 1, the sequence $\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel$ is decreasing for any $n\ge 1.$ Set ${lim}_{n\to \infty }\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =R.$ Since $\xi =(1-\beta )/7<\lambda <1$, the inequality is obtained in Lemma 1, which in turn implies that

$$(1+n\xi )\parallel {\mu }_i-\mathcal{S}{\mu }_i\parallel \le (1+n\lambda )\parallel {\mu }_i-\mathcal{S}{\mu }_i{\parallel \le \delta \left(E\right)+(1-\lambda )}^{-n}[\parallel {\mu }_i-\mathcal{S}{\mu }_i\parallel -\parallel {\mu }_{i+n}-\mathcal{S}{\mu }_{i+n}\parallel ]$$

for all $i\ge 0$ and $n\ge 1$. If we let $i\to +\infty$, we get

$$(1+n\lambda )R\le \delta \left(E\right).$$

Then,

$$R\le \frac{\delta \left(E\right)}{1+n\lambda }$$

holds for all $n\ge 1$. Clearly this is possible only if $R=0,$ and hence ${lim}_{n\to \infty }\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0.$

Thus the sequence $\left\{{\mu }_n\right\}$ is an AFPS for $\mathcal{S}$ which is asymptotically regular. □

Theorem 4.

Suppose that all the conditions of Theorem 3 are satisfied. Further, assume that, for any $ϵ>0$, there exists $\delta >0$ such that

$$\parallel \mu -\omega \parallel +\parallel \mu -\mathcal{S}\omega \parallel +\parallel \omega -\mathcal{S}\mu \parallel <3ϵ+\delta \Rightarrow \parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \frac{ϵ}{2}.$$

(26)

Then, $\mathcal{S}$ has a fixed point in E.

Proof. 

According to Theorem 3, $\mathcal{S}$ has an AFPS $\left\{{\mu }_n\right\}$, such that ${\mu }_n⪯{\mu }_{n+1}$ for all $n\ge 1$, $1/2\le \lambda <1$ and satisfies the following relation:

$$\underset{n\to \infty }{lim}\parallel {\mu }_{n+1}-{\mu }_n\parallel =\lambda \underset{n\to \infty }{lim}\parallel \mathcal{S}{\mu }_n-{\mu }_n\parallel =0.$$

(27)

Suppose that $\left\{{\mu }_n\right\}$ is not a Cauchy sequence. Then, there exists $ϵ>0$ such that

$$\underset{n,m\to \infty }{lim\; sup}\parallel {\mu }_n-{\mu }_m\parallel \ge 2ϵ$$

(28)

for any $n\ge m$ and $n\ge 1$. By hypothesis, there exists $\delta >0$ such that (26) holds. Without the loss of generality, we take $\delta <ϵ$. Since $\left\{{\mu }_n\right\}$ is asymptotically regular from (27), there exists $N\in \mathbb{N}$ such that

$$\parallel {\mu }_{n+1}-{\mu }_n\parallel \le \parallel \mathcal{S}{\mu }_n-{\mu }_n\parallel <\frac{\delta }{12},$$

for all $n\ge N$. Pick $m,n>N$ and $m\le n$, so that $\parallel {\mu }_n-{\mu }_m\parallel >2ϵ$. For i in $[m,n]$, there is

$$\left|\parallel {\mu }_m-{\mu }_i\parallel -\parallel {\mu }_m-{\mu }_{i+1}\parallel \right|\le \parallel {\mu }_i-{\mu }_{i+1}\parallel <\frac{\delta }{12}.$$

Since $\parallel {\mu }_m-{\mu }_{m+1}\parallel <ϵ$, $\parallel {\mu }_n-{\mu }_m\parallel >ϵ+\delta$, there exists $i\in [m,n]$. This implies that

$$ϵ+\frac{\delta }{6}\le \parallel {\mu }_m-{\mu }_i\parallel \le ϵ+\frac{\delta }{4}.$$

(29)

We obtain

$$\begin{array}{cc}\hfill & \parallel {\mu }_m-{\mu }_i\parallel +\parallel {\mu }_m-\mathcal{S}{\mu }_i\parallel +\parallel {\mu }_i-\mathcal{S}{\mu }_m\parallel \hfill \\ \hfill & \le \parallel {\mu }_m-{\mu }_i\parallel +\parallel {\mu }_m-{\mu }_i\parallel +\parallel {\mu }_i-\mathcal{S}{\mu }_i\parallel +\parallel {\mu }_i-{\mu }_m\parallel +\parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel \hfill \\ \hfill & <3\left(ϵ+\frac{\delta }{4}\right)+2\frac{\delta }{12}=3ϵ+\frac{11\delta }{12}<3ϵ+\delta .\hfill \end{array}$$

Hence, we get

$$\parallel \mathcal{S}{\mu }_m-\mathcal{S}{\mu }_i\parallel \le ϵ.$$

It follows that

$$\begin{array}{cc}\hfill \parallel {\mu }_m-{\mu }_i\parallel & \le \parallel {\mu }_m-\mathcal{S}{\mu }_m\parallel +\parallel \mathcal{S}{\mu }_m-\mathcal{S}{\mu }_i\parallel +\parallel \mathcal{S}{\mu }_i-{\mu }_i\parallel \hfill \\ \hfill & <ϵ+2\frac{\delta }{12}=ϵ+\frac{\delta }{6},\hfill \end{array}$$

which is a contradiction. Thus $\left\{{\mu }_n\right\}$ is a Cauchy sequence and hence is convergent to some $\mu \in E$. Clearly, $\mathcal{S}{\mu }_n\to \mu$ as $n\to \infty .$

Since $\left\{{\mu }_n\right\}$ is increasing in bounded set E, then ${\mu }_n⪯\mu$ for all $n\ge 1$, where $\mu \in E$. Consider

$$\begin{array}{cc}\hfill \parallel \mu -\mathcal{S}\mu \parallel & \le \parallel \mu -{\mu }_n\parallel +\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\parallel \mathcal{S}{\mu }_n-\mathcal{S}\mu \parallel \hfill \\ \hfill & \le \parallel \mu -{\mu }_n\parallel +\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\alpha \parallel {\mu }_n-\mu \parallel +\beta \parallel {\mu }_n-\mathcal{S}\mu \parallel +\gamma \parallel \mu -\mathcal{S}{\mu }_n\parallel .\hfill \end{array}$$

Taking the limit as $n\to \infty$, we obtain

$$\parallel \mu -\mathcal{S}\mu \parallel \le \beta \parallel \mu -\mathcal{S}\mu \parallel ,$$

which gives $\parallel \mu -\mathcal{S}\mu \parallel =0;$ thus, $\mu$ is a fixed point of $\mathcal{S}$. □

Example 1.

Consider the real sequence space $U={\ell }^p$; let $1<p<\infty$ be equipped with the usual ordering and standard norm

$${\parallel \mu \parallel }_p=(\sum _{i=1}^{\infty }|{\mu }_i{{|}^p)}^{\frac{1}{p}}.$$

Define $A=\left\{\mu =({\mu }_1,{\mu }_2,{\mu }_3,...)\in {\ell }^p:{\mu }_1\le 1,{\mu }_j=0\forall j\ne 1\right\}.$ A is a closed, bounded and convex subset of U. Consider that the usual order ⪯ on U and $\mathcal{S}:A\to A$ is defined by

$$\mathcal{S}\left(\mu \right)=\left\{\begin{array}{cc}(1-\frac{{\mu }_1}{2},0,0,...)\hfill & \mathrm{if}0\le {\mu }_1<\frac{1}{4},\hfill \\ (\frac{{\mu }_1}{4}+\frac{1}{2},0,0,...)\hfill & \mathrm{if}\frac{1}{4}\le {\mu }_1\le 1.\hfill \end{array}\right.$$

Consider $\mu =(1/8,0,0,...)$, $\omega =(1/4,0,0,...)$; then, $\mathcal{S}\left(\mu \right)=(15/16,0,...)$ and $\mathcal{S}\left(\omega \right)=(9/16,0,0,...)$, and we obtain

$$\begin{array}{cc}\hfill \left|\right|\mathcal{S}\left(\mu \right)-{\mathcal{S}\left(\omega \right)\left|\right|}_p& =\left|\frac{15}{16}-\frac{9}{16}\right|=\frac{6}{8}\hfill \\ \hfill & \ge \frac{1}{8}=\left|\frac{1}{8}-\frac{1}{4}\right|={\parallel \mu -\omega \parallel }_p,\hfill \end{array}$$

which implies that $\mathcal{S}$ is not a nonexpansive mapping.

Now, we will show that $\mathcal{S}$ is monotone and an ordered Reich nonexpansive mapping. Choosing $\alpha =\beta =\gamma =\frac{1}{3}$, we consider different cases as follows:

(i)

Let $0\le {\mu }_1\le {\omega }_1<\frac{1}{4}$; we have

$$\begin{array}{cc}\hfill {\alpha \parallel \mu -\omega \parallel }_p{+\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel }_p+\gamma {\parallel \omega -\mathcal{S}\left(\omega \right)\parallel }_p& =\frac{1}{3}|{\mu }_1-{\omega }_1|+\frac{1}{3}\left(\left|{\mu }_1-1+\frac{{\mu }_1}{2}\right|+\left|{\omega }_1-1+\frac{{\omega }_1}{2}\right|\right)\hfill \\ \hfill & =\frac{1}{3}|{\mu }_1-{\omega }_1|+\frac{1}{3}\left(\left|\frac{3{\mu }_1}{2}-1\right|+\left|\frac{3{\omega }_1}{2}-1\right|\right)\hfill \\ \hfill & \ge \frac{1}{3}\left|{\mu }_1-{\omega }_1\right|+\frac{1}{3}\left|\left(\frac{3{\mu }_1}{2}-1\right)-\left(\frac{3{\omega }_1}{2}-1\right)\right|\hfill \\ \hfill & =\frac{1}{3}\left|{\mu }_1-{\omega }_1\right|+\frac{1}{2}\left|{\mu }_1-{\omega }_1\right|\hfill \\ \hfill & \ge \frac{1}{2}\left|{\mu }_1-{\omega }_1\right|={\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel }_p.\hfill \end{array}$$

(ii)

Let $\frac{1}{4}\le {\mu }_1\le {\omega }_1\le 1$; we get

$$\begin{array}{cc}\hfill {\alpha \parallel \mu -\omega \parallel }_p{+\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel }_p+\gamma {\parallel \omega -\mathcal{S}\left(\omega \right)\parallel }_p& =\frac{1}{3}|{\mu }_1-{\omega }_1|+\frac{1}{3}\left(\left|{\mu }_1-\frac{{\mu }_1}{4}-\frac{1}{2}\right|+\left|{\omega }_1-\frac{{\omega }_1}{4}-\frac{1}{2}\right|\right)\hfill \\ \hfill & =\frac{1}{3}|{\mu }_1-{\omega }_1|+\frac{1}{3}\left(\left|\frac{3{\mu }_1}{4}-\frac{1}{2}\right|+\left|\frac{3{\omega }_1}{4}-\frac{1}{2}\right|\right)\hfill \\ \hfill & \ge \frac{1}{3}\left|{\mu }_1-{\omega }_1\right|+\frac{1}{3}\left|\left(\frac{3{\mu }_1}{4}-\frac{1}{2}\right)-\left(\frac{3{\omega }_1}{4}-\frac{1}{2}\right)\right|\hfill \\ \hfill & \ge \frac{7}{12}\left|{\mu }_1-{\omega }_1\right|\ge \frac{3}{12}\left|{\mu }_1-{\omega }_1\right|={\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel }_p.\hfill \end{array}$$

(iii)

Let $0\le {\mu }_1<\frac{1}{4}$, and $\frac{1}{4}\le {\omega }_1\le 1$; we obtain that

$$\begin{array}{cc}\hfill {\alpha \parallel \mu -\omega \parallel }_p{+\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel }_p+\gamma {\parallel \omega -\mathcal{S}\left(\omega \right)\parallel }_p& =\frac{1}{3}|{\mu }_1-{\omega }_1|+\frac{1}{3}\left(\left|\frac{3{\mu }_1}{2}-1\right|+\left|\frac{3{\omega }_1}{4}-\frac{1}{2}\right|\right)\hfill \\ \hfill & \ge \frac{1}{3}\left|{\mu }_1-y_1\right|+\frac{1}{3}\left|\left(\frac{3{\mu }_1}{2}-1\right)+\left(\frac{3{\omega }_1}{4}-\frac{1}{2}\right)\right|\hfill \\ \hfill & =\frac{1}{3}\left|{\mu }_1-{\omega }_1\right|+\frac{1}{4}\left|2-2{\mu }_1-{\omega }_1\right|\hfill \\ \hfill & \ge \frac{1}{4}\left|2-2{\mu }_1-{\omega }_1\right|={\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel }_p.\hfill \end{array}$$

Hence $\mathcal{S}$ is an ordered Rich nonexpansive mapping with coefficients $(1/3,1/3,1/3)$. The required AFPS is ${\mu }_n=(2n/(3n+1),0,0,...)$ for all $n\in \mathbb{N}$ with $\mu =(2/3,0,0,...)$ as a fixed point of $\mathcal{S}$. The mapping $\mathcal{S}$ satisfies the condition (8) of Theorem 1 for all $\mu ,\omega \in A_1\cup A_2$ such that $A_1=\{\mu =({\mu }_1,0,0,...),\omega =({\omega }_1,0,0,...):{\mu }_1\in [0,1/5],{\omega }_1\in [1/4,1]\}$ and $A_2=\{\mu =({\mu }_1,0,0,...),\omega =({\omega }_1,0,0,...):{\mu }_1\in [0,1/4),{\omega }_1\in [3/10,1]\}.$

Example 2.

Let $U={\mathbb{R}}^2$, and $A=\{\mu =({\mu }_1,{\mu }_2):({\mu }_1,{\mu }_2)\in [0,1]\times [0,1]\}$ be a subset of U with the norm $\parallel \mu \parallel =\parallel ({\mu }_1,{\mu }_2)\parallel =|{\mu }_1|+|{\mu }_2|$. Define the mapping $\mathcal{S}:A\to A$ by

$$\mathcal{S}({\mu }_1,{\mu }_2)=\left\{\begin{array}{cc}(1-{\mu }_1,1-{\mu }_2),\hfill & \mathrm{if}({\mu }_1,{\mu }_2)\in [0,\frac{1}{2}]\times [0,1],\hfill \\ \left(\frac{1+{\mu }_1}{3},\frac{1+{\mu }_2}{3}\right),\hfill & \mathrm{if}({\mu }_1,{\mu }_2)\in (\frac{1}{2},1]\times [0,1].\hfill \end{array}\right.$$

For $\mu =(0,0)$ and $\omega =(0.51,0.25)$, we have $\mathcal{S}\left(\mu \right)=(1,1)$, $\mathcal{S}\left(\omega \right)=(151/300,5/12)$,

$$\parallel \mathcal{S}\left(\mu \right)-\mathcal{S}\left(\omega \right)\parallel =1.08>0.76=\parallel \mu -\omega \parallel .$$

Therefore, $\mathcal{S}$ is not a nonexpansive mapping.

Now, we will show that $\mathcal{S}$ is an ordered Reich nonexpansive mapping. Choosing $\alpha =\beta =\gamma =1/3$, we consider different cases as follows:

(i)

Let $\mu =({\mu }_1,{\mu }_2),\omega =({\omega }_1,{\omega }_2)\in [0,\frac{1}{2}]\times [0,1]$; we then have

$$\begin{array}{cc}\hfill \parallel \mathcal{S}\left(\mu \right)-\mathcal{S}\left(\omega \right)\parallel & =\parallel (1-{\mu }_1,1-{\mu }_2)-(1-{\omega }_1,1-{\omega }_2)\parallel \hfill \\ \hfill & =\parallel ({\omega }_1-{\mu }_1,{\omega }_2-{\mu }_2)\parallel =|{\omega }_1-{\mu }_1|+|{\omega }_2-{\mu }_2|\hfill \\ \hfill & =|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2|=\parallel \mu -\omega \parallel \hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel +\gamma \parallel \omega -\mathcal{S}\left(\omega \right)\parallel & =\frac{1}{3}\parallel \mu -\omega \parallel +\frac{1}{3}\left[\right(|2{\mu }_1-1|\hfill \\ \hfill & +|2{\mu }_2-1\left|\right)]+\frac{1}{3}\left[\right(|2{\omega }_1-1|+|2{\omega }_2-1\left|\right)]\hfill \\ \hfill & \ge \frac{1}{3}\parallel \mu -\omega \parallel +\frac{1}{3}\left[\right(|2{\mu }_1-1|+|1-2{\omega }_1\left|\right)]\hfill \\ \hfill & +\frac{1}{3}\left[\right(|2{\mu }_2-1|+|1-2{\omega }_2\left|\right)]\hfill \\ \hfill & \ge \frac{1}{3}\parallel \mu -\omega \parallel +\frac{2}{3}\left(\right|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2\left|\right)\hfill \\ \hfill & =\parallel \mu -\omega \parallel =\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel .\hfill \end{array}$$

(ii)

Let $\mu =({\mu }_1,{\mu }_2),\omega =({\omega }_1,{\omega }_2)\in (\frac{1}{2},1]\times [0,1]$; we then get

$$\begin{array}{cc}\hfill \parallel \mathcal{S}\left(\mu \right)-\mathcal{S}\left(\omega \right)\parallel & =\parallel (\frac{1+{\mu }_1}{3},\frac{1+{\mu }_2}{3})-(\frac{1+{\omega }_1}{3},\frac{1+{\omega }_2}{3})\parallel \hfill \\ \hfill & =\frac{1}{3}\parallel ({\mu }_1-{\omega }_1,{\mu }_2-{\omega }_2)\parallel \hfill \\ \hfill & =\frac{1}{3}\left(\right|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2\left|\right)=\frac{1}{3}\parallel \mu -\omega \parallel .\hfill \end{array}$$

In addition,

$$\begin{array}{cc}\hfill \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel +\gamma \parallel \omega -\mathcal{S}\left(\omega \right)\parallel & =\frac{1}{3}\parallel \mu -\omega \parallel +\frac{1}{3}\left(\right|2{\mu }_1-1|+|2{\mu }_2-1\left|\right)\hfill \\ \hfill & +\frac{1}{3}\left(\right|2{\omega }_1-1|+|2{\omega }_2-1\left|\right)\hfill \\ \hfill & \ge \frac{1}{3}\parallel \mu -\omega \parallel +\frac{1}{3}\left(\right|2{\mu }_1-1|+|1-2{\omega }_1\left|\right)\hfill \\ \hfill & +\frac{1}{3}\left(\right|2{\mu }_2-1|+|1-2{\omega }_2\left|\right)\hfill \\ \hfill & \ge \frac{1}{3}\parallel \mu -\omega \parallel +\frac{4}{3}\left(\right|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2\left|\right)\hfill \\ \hfill & =\frac{5}{3}\parallel \mu -\omega \parallel =5\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \ge \parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel .\hfill \end{array}$$

(iii)

Let $\mu =({\mu }_1,{\mu }_2)\in [0,\frac{1}{2}]\times [0,1]$, and $\omega =({\omega }_1,{\omega }_2)\in (\frac{1}{2},1]\times [0,1]$; then, we have

$$\begin{array}{cc}\hfill \parallel \mathcal{S}\left(\mu \right)-\mathcal{S}\left(\omega \right)\parallel & =\parallel (1-{\mu }_1,1-{\mu }_2)-(\frac{1+{\omega }_1}{3},\frac{1+{\omega }_2}{3})\parallel \hfill \\ \hfill & =\frac{1}{3}\parallel (2-3{\mu }_1-{\omega }_1,2-3{\mu }_1-{\omega }_2)\parallel \hfill \\ \hfill & =\frac{1}{3}\left[\right|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2|+2|1-2{\mu }_1|+2|1-2{\mu }_2\left|\right]\hfill \\ \hfill & =\frac{1}{3}\parallel \mu -\omega \parallel +\frac{2}{3}\left[\right|1-2{\mu }_1|+|1-2{\mu }_2\left|\right]\le \parallel \mu -\omega \parallel +2,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\left(\mu \right)\parallel +\gamma \parallel \omega -\mathcal{S}\left(\omega \right)\parallel & =\frac{1}{3}\parallel \mu -\omega \parallel +\frac{1}{3}\left(\right|2{\mu }_1-1|+|2{\mu }_2-1\left|\right)\hfill \\ \hfill & +\frac{1}{3}\left(\right|2{\omega }_1-1|+|2{\omega }_2-1\left|\right)\hfill \\ \hfill & \ge \frac{1}{3}\parallel \mu -\omega \parallel +\frac{2}{3}\left(\right|{\mu }_1-{\omega }_1|+|{\mu }_2-{\omega }_2\left|\right)\hfill \\ \hfill & \ge \parallel \mu -\omega \parallel +2\ge \parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel .\hfill \end{array}$$

Then, $\mathcal{S}$ is monotone and an ordered Reich nonexpansive mapping with coefficients $(1/3,1/3,1/3)$. The AFPS is $\left({\mu }_n\right)$ where ${\mu }_n=((n+1)/2n,(n+1)/2n)$ for all $n\in \mathbb{N}$. Furthermore, $\mathcal{S}$ has a fixed point $\mu =(0.5,0.5)$. Since $\alpha <1$, the mapping $\mathcal{S}$ satisfies the condition (8) of Theorem 1 for all $\mu \prec \omega$ in $A_1$, where $A_1=\{\mu =({\mu }_1,{\mu }_2),\omega =({\omega }_1,{\omega }_2):({\mu }_1,{\mu }_2)\in [0,\frac{1}{2}]\times \left\{1\right\};({\omega }_1,{\omega }_2)\in (\frac{1}{2},1]\times [0,1]\}$.

4. Convergence Results

In this section we establish certain strong and weak convergence results in a partially ordered Banach space.

Definition 3.

Let $(U,\parallel .\parallel ,⪯)$ be a partially ordered Banach space.

(i) 

According to [37], a space U satisfies the weak-Opial property if for any sequence $\left\{{\mu }_n\right\}$ in U which converges weakly to ν; thus, we have

$$\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel <\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\omega \parallel$$

for all $\omega \ne \nu$ in U;

(ii) 

According to [17], a space U satisfies the monotone weak-Opial property if for any monotone sequence $\left\{{\mu }_n\right\}$ in U which converges weakly to ν; thus, we have

$$\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel <\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\omega \parallel$$

for all $\omega \ne \nu$ in U, and ω is greater or less than all the elements of the sequence $\left\{{\mu }_n\right\}$.

Proposition 2.

Let $(U,\parallel .\parallel ,⪯)$ be a partially ordered Banach space and E be a nonempty closed and convex subset of U. Let $\mathcal{S}:E\to E$ be a monotone and an ordered Reich nonexpansive mapping satisfying the conditions of Theorem 2. Then, $\mathcal{S}$ is quasi-nonexpansive, provided $\gamma \le \beta <1$.

Proof. 

By Theorem 2, $\mathcal{F}\left(\mathcal{S}\right)$ is nonempty. Suppose $\nu \in \mathcal{F}\left(\mathcal{S}\right)$ and $\mu \in E$ are such that $\mu ⪯\nu$:

$$\begin{array}{cc}\hfill \parallel \mathcal{S}\nu -\mathcal{S}\mu \parallel =\parallel \nu -\mathcal{S}\mu \parallel & \le \alpha \parallel \nu -\mu \parallel +\beta \parallel \nu -\mathcal{S}\nu \parallel +\gamma \parallel \mu -\mathcal{S}\mu \parallel \hfill \\ \hfill & \le \alpha \parallel \nu -\mu \parallel +\gamma \parallel \mu -\nu \parallel +\gamma \parallel \nu -\mathcal{S}\mu \parallel \hfill \\ \hfill \Rightarrow (1-\gamma )\parallel \nu -\mathcal{S}\mu \parallel & \le (\alpha +\gamma )\parallel \mu -\nu \parallel ,\hfill \end{array}$$

which implies that

$$\parallel \nu -\mathcal{S}\mu \parallel \le \frac{\alpha +\gamma }{1-\gamma }\parallel \nu -\mu \parallel .$$

Since $1-\gamma >0$, and $\frac{\alpha +\gamma }{1-\gamma }=\frac{1-\beta }{1-\gamma }\le 1$, we obtain

$$\parallel \nu -\mathcal{S}\mu \parallel \le \parallel \nu -\mu \parallel$$

as required. □

Lemma 2.

Let E be a nonempty subset of an ordered Banach space $(U,\parallel .\parallel ,⪯)$ and $\mathcal{S}:E\to E$ be a monotone and an ordered Reich nonexpansive mapping satisfying the conditions of Theorem 1. If for each $\mu ,\omega \in E$ with $\mu ⪯\omega$, and $\gamma \le \beta <1$, then the following estimates hold true:

(i) 

$\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel \le \parallel \mu -\mathcal{S}\mu \parallel$;

(ii) 

At least one of the following ((a) and (b)) holds:

(a) 

$\frac{1-\gamma }{6}\parallel \mu -\mathcal{S}\mu \parallel \le \parallel \mu -\omega \parallel$;

(b) 

$\frac{1-\gamma }{6}\parallel {\mathcal{S}}^2\mu -\mathcal{S}\mu \parallel \le \parallel \mathcal{S}\mu -\omega \parallel$;

The condition (a) implies $\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \le \parallel \mu -\omega \parallel$,and the condition (b) implies $\parallel {\mathcal{S}}^2\mu -\mathcal{S}\omega \parallel \le \parallel \mathcal{S}\mu -\omega \parallel$;

(iii)

$\parallel \mu -\mathcal{S}\omega \parallel \le \frac{2+\alpha +\beta }{1-\gamma }\parallel \mu -\mathcal{S}\mu \parallel +\frac{1-\beta }{1-\gamma }\parallel \mu -\omega \parallel .$

Proof. 

(i)

By the definition of an ordered Reich nonexpansive mapping, we obtain

$$\begin{array}{cc}\hfill \parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel & \le \alpha \parallel \mu -\mathcal{S}\mu \parallel +\beta \parallel \mu -\mathcal{S}\mu \parallel +\gamma \parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel \hfill \\ \hfill \Rightarrow (1-\gamma )\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel & \le (\alpha +\beta )\parallel \mu -S\mu \parallel ,\hfill \end{array}$$

which implies

$$\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel \le \frac{1-\gamma }{1-\gamma }\parallel \mu -\mathcal{S}\mu \parallel =\parallel \mu -\mathcal{S}\mu \parallel ;$$

(ii)

Assume on the contrary that $(1-\gamma )/6\parallel \mu -\mathcal{S}\mu \parallel >\parallel \mu -\omega \parallel$ and $(1-\gamma )/6\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel >\parallel \mathcal{S}\mu -\omega \parallel$. Then, by triangle inequality with the assumption (i), we find that

$$\begin{array}{cc}\hfill \parallel \mu -\mathcal{S}\mu \parallel & \le \parallel \mu -\omega \parallel +\parallel \omega -\mathcal{S}\mu \parallel \hfill \\ \hfill & <\frac{1-\gamma }{6}\parallel \mu -\mathcal{S}\mu \parallel +\frac{1-\gamma }{6}\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel ,\hfill \\ \hfill \Rightarrow \parallel \mu -\mathcal{S}\mu \parallel & <\frac{1-\gamma }{3}\parallel \mu -\mathcal{S}\mu \parallel .\hfill \end{array}$$

Since $(1-\gamma )/3<1$, we obtain $\parallel \mu -\mathcal{S}\mu \parallel <\parallel \mu -\mathcal{S}\mu \parallel$, which is a contradiction. Therefore, at least one of (a) and (b) holds. Hence, (ii) holds;

(iii)

For the first case, by our hypotheses, we have

$$\begin{array}{cc}\hfill \parallel \mu -\mathcal{S}\omega \parallel & \le \parallel \mu -\mathcal{S}\mu \parallel +\parallel \mathcal{S}\mu -\mathcal{S}\omega \parallel \hfill \\ \hfill & \le \parallel \mu -\mathcal{S}\mu \parallel +\alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\mu \parallel +\gamma \parallel \omega -\mathcal{S}\omega \parallel \hfill \\ \hfill & \le (1+\beta )\parallel \mu -\mathcal{S}\mu \parallel +(\alpha +\gamma )\parallel \mu -\omega \parallel +\gamma \parallel \mu -\mathcal{S}\omega \parallel \hfill \\ \hfill \Rightarrow (1-\gamma )\parallel \mu -\mathcal{S}\omega \parallel & \le (1+\beta )\parallel \mu -\mathcal{S}\mu \parallel +(\alpha +\gamma )\parallel \mu -\omega \parallel .\hfill \end{array}$$

Since $\alpha +\beta +\gamma =1$, and $1-\gamma >0$, then

$$\parallel \mu -\mathcal{S}\omega \parallel \le \frac{1+\beta }{1-\gamma }\parallel \mu -\mathcal{S}\mu \parallel +\frac{1-\beta }{1-\gamma }\parallel \mu -\omega \parallel .$$

For the second case, using (i), we get

$$\begin{array}{cc}\hfill \parallel \mu -\mathcal{S}\omega \parallel & \le \parallel \mu -\mathcal{S}\mu \parallel +\parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel +\parallel {\mathcal{S}}^2\mu -\mathcal{S}\omega \parallel \hfill \\ \hfill & \le \parallel \mu -\mathcal{S}\mu \parallel +\parallel \mu -\mathcal{S}\mu \parallel +\alpha \parallel \mathcal{S}\mu -\omega \parallel +\beta \parallel \mathcal{S}\mu -{\mathcal{S}}^2\mu \parallel +\gamma \parallel \omega -\mathcal{S}\omega \parallel \hfill \\ \hfill & \le (2+\alpha )\parallel \mu -S\mu \parallel +\alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\mu \parallel +\gamma \parallel \mu -\omega \parallel +\gamma \parallel \mu -\mathcal{S}\omega \parallel \hfill \\ \hfill \Rightarrow (1-\gamma )\parallel \mu -\mathcal{S}\omega \parallel & \le (2+\alpha +\beta )\parallel \mu -\mathcal{S}\mu \parallel +(\alpha +\gamma )\parallel \mu -\omega \parallel ,\hfill \end{array}$$

where $1-\gamma >0$; thus, we have

$$\parallel \mu -\mathcal{S}\omega \parallel \le \frac{2+\alpha +\beta }{1-\gamma }\parallel \mu -\mathcal{S}\mu \parallel +\frac{1-\beta }{1-\gamma }\parallel \mu -\omega \parallel ,$$

where $\gamma \le \beta <1$, and $\alpha +\beta +\gamma =1$. □

As an application of the above Lemma, we obtain the following result:

Theorem 5.

Let E be a nonempty convex and compact subset of an ordered Banach space $(U,\parallel .\parallel ,⪯)$ and $\mathcal{S}:E\to E$ be a monotone and an ordered Reich nonexpansive mapping satisfying the conditions of Theorem 2. Suppose that $\left\{{\mu }_n\right\}$ is the sequence defined by KIS. Then, $\left\{{\mu }_n\right\}$ converges strongly to a fixed point of $\mathcal{S}$, provided $\gamma \le \beta <1.$

Proof. 

From Theorem 1, we have

$$\underset{n\to \infty }{lim}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0.$$

Since E is compact, there exists a subsequence $\left\{{\mu }_{n_j}\right\}$ of $\left\{{\mu }_n\right\}$ such that $\left\{{\mu }_{n_j}\right\}\to \nu$ as $j\to \infty$ for $\nu \in E$. and $\nu \in E$. By Lemma 2, we obtain

$$\parallel {\mu }_{n_j}-\mathcal{S}\nu \parallel \le \frac{2+\alpha +\beta }{1-\gamma }\parallel {\mu }_{n_j}-\mathcal{S}{\mu }_{n_j}\parallel +\frac{1-\beta }{1-\gamma }\parallel {\mu }_{n_j}-\nu \parallel$$

for all $j\in \mathbb{N}$. Taking the limit as $j\to \infty$, we obtain that $\left\{{\mu }_{n_j}\right\}$ converges to $\mathcal{S}\nu$, which implies $\nu =\mathcal{S}\nu$. By Proposition 2, and the sequence $\left\{{\mu }_n\right\}$, which is defined by KIS, we infer that

$$\parallel {\mu }_{n+1}-\nu \parallel \le \lambda \parallel \mathcal{S}{\mu }_n-\nu \parallel +(1-\lambda )\parallel {\mu }_n-\nu \parallel \le \parallel {\mu }_n-\nu \parallel$$

for all $n\ge 1$. ${lim}_{n\to \infty }\parallel x_n-\nu \parallel$ exists for every $\nu \in \mathcal{F}\left(\mathcal{S}\right)$, so $\left\{{\mu }_n\right\}$ converges strongly to a fixed point of $\mathcal{S}$. □

Proposition 3.

Let $(U,\parallel .\parallel ,⪯)$ be a partially ordered Banach space and E a nonempty convex, closed and bounded subset of U with a monotone weak-Opial property. Let $\mathcal{S}:E\to E$ be a monotone Reich nonexpansive mapping satisfying the conditions of Theorem 2. If $\left\{{\mu }_n\right\}$ converges weakly to ν, then, $\mathcal{S}\nu =\nu ,$ provided $\gamma \le \beta <1$.

Proof. 

From Theorem 1, we obtain

$$\underset{n\to \infty }{lim}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0.$$

(30)

Lemma 2 verifies that

$$\parallel {\mu }_n-\mathcal{S}\nu \parallel \le \frac{2+\alpha +\beta }{1-\gamma }\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel +\frac{1-\beta }{1-\gamma }\parallel {\mu }_n-\nu \parallel$$

for all $n\in \mathbb{N}$. As $\frac{1-\beta }{1-\gamma }<1$, so

$$\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\mathcal{S}\nu \parallel \le \underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel .$$

We show that $\mathcal{S}\nu =\nu$. Suppose that $\mathcal{S}\nu \ne \nu$; by the monotone weak-Opial property, we obtain

$$\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel <\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\mathcal{S}\nu \parallel ,$$

which is a contradiction. Thus, $\mathcal{S}\nu =\nu$. □

Theorem 6.

Let E be a nonempty convex and weakly compact subset of a partially ordered Banach space $(U,\parallel .\parallel ,⪯)$ with a monotone weak-Opial property. Let $\mathcal{S}:E\to E$ be a monotone and an ordered Reich nonexpansive mapping satisfying the condition of Theorem 2. Assume that there exists ${\mu }_1\in E$ such that ${\mu }_1⪯\mathcal{S}{\mu }_1$ and $\mathcal{F}\left(\mathcal{S}\right)$ is nonempty. Then, the sequence $\left\{{\mu }_n\right\}$ defined by KIS converges weakly to a fixed point of $\mathcal{S}$.

Proof. 

By Theorem 1, we have

$$\underset{n\to \infty }{lim}\parallel {\mu }_n-\mathcal{S}{\mu }_n\parallel =0.$$

Since E is weakly compact, there exists a subsequence $\left\{{\mu }_{n_j}\right\}$ of $\left\{{\mu }_n\right\}$ and $\nu \in E$ such that $\left\{{\mu }_{n_j}\right\}$ converges weakly to $\nu$. From Proposition 3, we infer that $\nu$ is a fixed point of $\mathcal{S}$. As in the proof of Theorem 1, we prove that $\{\parallel {\mu }_n-\nu \parallel \}$ is a decreasing sequence. Now, suppose that the sequence $\left\{{\mu }_n\right\}$ does not converge to $\nu$. Therefore, there exists a subsequence $\left\{{\mu }_{n_i}\right\}$ of $\left\{{\mu }_n\right\}$ which converges weakly to $\kappa$ such that $\nu \ne \kappa$ and $\mathcal{S}\kappa =\kappa$. Now from the monotone weak-Opial property,

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel & =\underset{j\to \infty }{lim\; inf}\parallel {\mu }_{n_j}-\nu \parallel <\underset{j\to \infty }{lim\; inf}\parallel {\mu }_{n_j}-\kappa \parallel =\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\kappa \parallel \hfill \\ \hfill & =\underset{i\to \infty }{lim\; inf}\parallel {\mu }_{n_i}-\kappa \parallel <\underset{i\to \infty }{lim\; inf}\parallel {\mu }_{n_i}-\nu \parallel =\underset{n\to \infty }{lim\; inf}\parallel {\mu }_n-\nu \parallel ,\hfill \end{array}$$

which is a contradiction. Thus, $\nu =\kappa$. Therefore, $\left\{{\mu }_n\right\}$ converges weakly to a fixed point of $\mathcal{S}$. This completes the proof. □

5. Application to Initial-Value Problems

Fixed-point theorems for monotone nonexpansive mappings in ordered metric spaces have been widely investigated and have found various applications in matrix equations and differential and integral equations [14,26,29,30,38,39,40]. In this section we discuss the existence of solutions of initial-value problems for nonlinear first-order ordinary differential equations.

Consider the differential equation

$$\left\{\begin{array}{c}{\mu }^{\prime }\left(t\right)=g(t,\mu \left(t\right)),t\in I,\hfill \\ \mu \left(0\right)={\nu }_0.\hfill \end{array}\right.$$

(31)

Theorem 7.

Let E be a compact subset of $U=C[0,1]$, with the space of continuous functions on $I=[0,1]$ with partial order ⪯ defined by $\mu ⪯\omega$ if, and only if, $\mu \left(t\right)\le \omega \left(t\right)$ for all $\mu ,\omega \in U$ and $t\in I$. Suppose U is equipped with a supremum norm defined by $\parallel \mu -\omega \parallel ={sup}_{t\in I}|\mu \left(t\right)-\omega \left(t\right)|$. Assume ${\nu }_0\in \mathbb{R}$, and the following conditions hold:

(a) 

$g:I\times E\to E$ is continuous;

(b) 

For any $t\in I$, the function $g(t,.)$ is nondecreasing;

(c) 

There exists a continuous function $f:I\to \mathbb{R}$ satisfying

$$\underset{t\in I}{sup}\{f\left(t\right)-{\int }_0^{t}g(s,f\left(s\right))ds\}\le {\nu }_0;$$

(d) 

There exist non-negative real numbers $\alpha ,\beta ,\gamma$ such that $\alpha +\beta +\gamma =1$ with $\alpha \le 1$, and for any continuous functions $\mu \left(t\right),\omega \left(t\right):I\to \mathbb{R}$ such that $\mu \left(t\right)⪯\omega \left(t\right)$ for all $t\in I$, we have

$$\begin{array}{cc}\hfill {\int }_0^t|g(s,\mu \left(s\right))-g(s,\omega \left(s\right))|ds\le & \alpha |\mu \left(t\right)-\omega \left(t\right)|+\beta |\mu \left(t\right)-{\nu }_0-{\int }_0^{t}g(s,\mu \left(s\right))ds|\hfill \\ \hfill & +\gamma |\omega \left(t\right)-{\nu }_0-{\int }_0^{t}g(s,\omega \left(s\right))ds|\hfill \end{array}$$

for all $t\in I$;

(e) 

$\mathcal{S}:E\to E$ is a mapping defined by

$$\mathcal{S}\mu \left(t\right)={\nu }_0+{\int }_0^{1}g(s,\mu \left(s\right))ds,t\in I,\lambda \ge 0.$$

Then, the nonlinear differential Equation (31) has a solution in $C[0,1]$, provided $\mathcal{S}$ admits an AFPS.

Proof. 

For $\mu ,\omega \in E$, such that $\mu ⪯\omega$, we have

$$\begin{array}{cc}\hfill \left|\mathcal{S}\mu \right(t)-\mathcal{S}\omega (t\left)\right|& =\left|{\nu }_0+{\int }_0^{t}g(s,\mu \left(s\right))ds-{\nu }_0-{\int }_0^{t}g(s,\omega \left(s\right))ds\right|\hfill \\ \hfill & \le {\int }_0^t|g(s,\mu \left(s\right))-g(s,\omega \left(s\right))|ds\hfill \\ \hfill & \le \alpha |\mu \left(t\right)-\omega \left(t\right)|+\beta |\mu \left(t\right)-{\nu }_0-{\int }_0^{t}g(s,\mu \left(s\right))|\hfill \\ \hfill & +\gamma |\omega \left(t\right)-{\nu }_0-{\int }_0^{t}g(s,\omega \left(s\right))|\hfill \\ \hfill & \le \alpha \left|\mu \right(t)-\omega (t\left)\right|+\beta \left|\mu \right(t)-\mathcal{S}\mu (t\left)\right|+\gamma \left|\omega \right(t)-\mathcal{S}\omega (t\left)\right|.\hfill \end{array}$$

Taking the supremum norm on both sides, we get

$$\parallel \mathcal{S}\mu \left(t\right)-\mathcal{S}\omega \left(t\right)\parallel \le \alpha \parallel \mu -\omega \parallel +\beta \parallel \mu -\mathcal{S}\mu \parallel +\gamma \parallel \omega -\mathcal{S}\omega \parallel .$$

Moreover, we show the mapping $\mathcal{S}$ is monotone. Let $\mu ,\omega \in E$ with $\mu ⪯\omega$. Then, $\mu \left(t\right)⪯\omega \left(t\right)$ for all $t\in I$. By (b), we obtain

$$\begin{array}{cc}\hfill \left|\mathcal{S}\mu \right(t)-\mathcal{S}\omega (t\left)\right|& =\left|{\nu }_0+{\int }_0^{t}g(s,\mu \left(s\right))ds-{\nu }_0-{\int }_0^{t}g(s,\omega \left(s\right))ds\right|\hfill \\ \hfill & \le {\int }_0^t|g(s,\mu \left(s\right))-g(s,\omega \left(s\right))|ds\ge 0,\mathrm{for}\mathrm{all}t\in I,\hfill \end{array}$$

which implies $\mathcal{S}\mu \le \mathcal{S}\omega$.

Hence, $\mathcal{S}$ is a monotone Reich nonexpansive map. Thus, all of the assumptions of Theorem 1 are satisfied; therefore, (31) has a solution in $E\subseteq C[0,1].$ □

Example 3.

Consider the following functional differential equation:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\mu }^{\prime }\left(t\right)=3e^{3t}fort\in [0,1],\hfill \\ \mu \left(0\right)=3.\hfill \end{array}\right.\end{array}$$

(32)

The problem has a solution as follows:

$$\mu \left(t\right)=e^{3t}+1,t\in [0,1].$$

Therefore, the continuous function $f:[0,1]\to \mathbb{R}$ defined by

$$f\left(t\right)=e^{3t},t\in [0,1]$$

satisfies the condition ${sup}_{t\in I}\{f\left(t\right)-{\int }_0^{t}g(s,f\left(s\right))ds\}\le 3$, where $g(t,f\left(t\right))=3e^{3t}$ for $t\in [0,1]$. Let ${\nu }_0=3.$ It is obvious that the function $g(t,\mu )=3e^{3t}$ satisfies hypotheses (a), (b) and (d) in Theorem 7. Hence, Theorem 7 guarantees the existence of solution of the problem (32).

6. Conclusions

The concepts of monotone Reich and Chatterjea nonexpansive mappings on partially ordered Banach spaces are introduced and studied. Certain sufficient conditions for the existence of approximate fixed-point sequences have been established to derive some fixed-point results by employing the Krasnoselskii–Ishikawa iteration method. Furthermore, we provided some weak and strong convergence results for such mappings together with an application to nonlinear differential equations. As a future research plan, we suggest to researchers to prove these results for partially ordered metric linear spaces, partially ordered convex metric spaces and partially ordered hyperbolic spaces.