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ISRN Applied Mathematics
Volume 2012 (2012), Article ID 539125, 16 pages
http://dx.doi.org/10.5402/2012/539125
Research Article

Quadruple Fixed Point Theorems in Partially Ordered Metric Spaces Depending on Another Function

1Université de Sousse Institut Supérieur d'Informatique et des Technologies de Communication de Hammam Sousse, Route GP1, H. Sousse 4011, Tunisia
2Department of Mathematics, Atılım University, 06836 İncek, Ankara, Turkey

Received 27 February 2012; Accepted 30 April 2012

Academic Editor: G. Wang

Copyright © 2012 Hassen Aydi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We prove quadruple fixed point theorems in partially ordered metric spaces depending on another function. Also, we state some examples showing that our results are real generalization of known ones in quadruple fixed point theory.

1. Introduction and Preliminaries

Basic topological properties of an ordered set like convergence were introduced by Wolk [1]. In 1981, Monjardet [2] considered metrics on partially ordered sets. Ran and Reurings [3] proved an analog of Banach contraction mapping principle in partially ordered metric spaces. In their pioneering work, they also provide applications to matrix equations. As an extension, Nieto and Rodríguez-López [4] discovered further fixed point theorems in partially ordered metric spaces. For some other related results in ordered metric spaces, see, for example, [57].

Bhaskar and Lakshmikantham in [8] introduced the concept of coupled fixed point of a mapping 𝐹𝑋×𝑋𝑋 and investigated the existence and uniqueness of a coupled fixed point theorem in partially ordered complete metric spaces. Lakshmikantham and Ćirić in [9] defined mixed 𝑔-monotone property and coupled coincidence point in partially ordered metric spaces. They also proved related fixed point theorems. Later, various results on coupled fixed point have been obtained, see, for example, [920].

Following this trend, Berinde and Borcut [21] introduced the concept of tripled fixed point in ordered sets. The following two definitions are from [21].

Definition 1.1. Let (𝑋,) be a partially ordered set and 𝐹𝑋×𝑋×𝑋𝑋. The mapping 𝐹 is said to have the mixed monotone property if, for any 𝑥,𝑦,𝑧𝑋, 𝑥1,𝑥2𝑋,𝑥1𝑥2𝑥𝐹1𝑥,𝑦,𝑧𝐹2,𝑦,𝑦,𝑧1,𝑦2𝑋,𝑦1𝑦2𝐹𝑥,𝑦1,𝑧𝐹𝑥,𝑦2,𝑧,𝑧1,𝑧2𝑋,𝑧1𝑧2𝐹𝑥,𝑦,𝑧1𝐹𝑥,𝑦,𝑧2.(1.1)

Definition 1.2. Let 𝐹𝑋×𝑋×𝑋𝑋. An element (𝑥,𝑦,𝑧) is called a tripled fixed point of 𝐹 if 𝐹(𝑥,𝑦,𝑧)=𝑥,𝐹(𝑦,𝑥,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥)=𝑧.(1.2)

Also, Berinde and Borcut [21] proved the following theorem.

Theorem 1.3. Let (𝑋,,𝑑) be a partially ordered set and suppose that there is a metric 𝑑 on 𝑋 such that (𝑋,𝑑) is a complete metric space. Suppose also that 𝐹𝑋×𝑋×𝑋𝑋 be a mapping such that it has the mixed monotone property and there exist 𝑗,𝑟,𝑙0 with 𝑗+𝑟+𝑙<1 such that 𝑑(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑤))𝑗𝑑(𝑥,𝑢)+𝑟𝑑(𝑦,𝑣)+𝑙𝑑(𝑧,𝑤),(1.3)
for any 𝑥,𝑦,𝑧𝑋 for which 𝑥𝑢, 𝑣𝑦, and 𝑧𝑤. Additionally suppose that either 𝐹 is continuous or 𝑋 has the following properties:(1)if a nondecreasing sequence 𝑥𝑛𝑥, then 𝑥𝑛𝑥 for all 𝑛,(2)if a nonincreasing sequence 𝑦𝑛𝑦, then 𝑦𝑦𝑛 for all 𝑛.
If there exist 𝑥0,𝑦0,𝑧0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0), 𝑦0𝐹(𝑦0,𝑥0,𝑧0), and 𝑧0𝐹(𝑧0,𝑦0,𝑥0), then there exist 𝑥,𝑦,𝑧𝑋 such that 𝐹(𝑥,𝑦,𝑧)=𝑥,𝐹(𝑦,𝑥,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥)=𝑧,(1.4)
that is, 𝐹 has a tripled fixed point.

The notion of fixed point of order 𝑁3 was first introduced by Samet and Vetro [22]. Very recently, Karapınar used the notion of quadruple fixed point and obtained some quadruple fixed point theorems [23] in partially ordered metric spaces. This work motivated the following studies [2427] which provide further fixed point theorems on quadruple fixed points.

From now on, we denote 𝑋4=𝑋×𝑋×𝑋×𝑋.

Definition 1.4 (see [24]). Let 𝑋 be a nonempty set and let 𝐹𝑋4𝑋 be a given mapping. An element (𝑥,𝑦,𝑧,𝑤)𝑋×𝑋3 is called a quadruple fixed point of 𝐹 if 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤.(1.5)
Let (𝑋,𝑑) be a metric space. The mapping 𝑑𝑋4𝑋, given by 𝑑((𝑥,𝑦,𝑧,𝑤),(𝑢,𝑣,,𝑙))=𝑑(𝑥,𝑦)+𝑑(𝑦,𝑣)+𝑑(𝑧,)+𝑑(𝑤,𝑙),(1.6) defines a metric on 𝑋4, which will be denoted for convenience by 𝑑.

Remark 1.5. In [23, 24, 27], the notion of quadruple fixed point is called quartet fixed point.

Definition 1.6 (see [24]). Let (𝑋,) be a partially ordered set and 𝐹𝑋4𝑋 be a mapping. We say that 𝐹 has the mixed monotone property if 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nondecreasing in 𝑥 and 𝑧 and is monotone nonincreasing in 𝑦 and 𝑤; that is, for any 𝑥,𝑦,𝑧,𝑤𝑋, 𝑥1,𝑥2𝑋,𝑥1𝑥2𝑥implies𝐹1𝑥,𝑦,𝑧,𝑤𝐹2,𝑦,𝑦,𝑧,𝑤1,𝑦2𝑋,𝑦1𝑦2implies𝐹𝑥,𝑦2,𝑧,𝑤𝐹𝑥,𝑦1,𝑧,𝑧,𝑤1,𝑧2𝑋,𝑧1𝑧2implies𝐹𝑥,𝑦,𝑧1,𝑤𝐹𝑥,𝑦,𝑧2,𝑤,𝑤1,𝑤2𝑋,𝑤1𝑤2implies𝐹𝑥,𝑦,𝑧,𝑤2𝐹𝑥,𝑦,𝑧,𝑤1.(1.7)

In this paper, we prove some quadruple fixed point theorems in partially ordered metric spaces depended on another function 𝑇𝑋𝑋.

2. Main Results

We start with the following definition (see, e.g., [2831]).

Definition 2.1. Let (𝑋,𝑑) be a metric space. A mapping 𝑇𝑋𝑋 is said to be ICS if 𝑇 is injective, continuous, and it has the property: for every sequence {𝑥𝑛} in 𝑋, if {𝑇𝑥𝑛} is convergent then, {𝑥𝑛} is also convergent.

Let Φ be the set of all functions 𝜙[0,)[0,) such that(1)𝜙(𝑡)<𝑡 for all 𝑡(0,+),(2)lim𝑟𝑡+𝜙(𝑟)<𝑡 for all 𝑡(0,+).

Our first result is given by the following theorem.

Theorem 2.2. Let (𝑋,) be a partially ordered set and suppose that there is a metric 𝑑 on 𝑋 such that (𝑋,𝑑) is a complete metric space. Suppose also that 𝑇𝑋𝑋 is an ICS mapping and 𝐹𝑋4𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝜙Φ such that 𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑟,𝑠))𝜙(max{𝑑(𝑇𝑥,𝑇𝑢),𝑑(𝑇𝑦,𝑇𝑣),𝑑(𝑇𝑧,𝑇𝑟),𝑑(𝑇𝑤,𝑇𝑠)})(2.1)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑟,𝑠𝑋 for which 𝑥𝑢, 𝑣𝑦, 𝑧𝑟, and 𝑠𝑤. Additionally assume that either(a)𝐹 is continuous, or(b)𝑋 has the following properties:(i) if nondecreasing sequence 𝑥𝑛𝑥 (respectively, 𝑧𝑛𝑧), then 𝑥𝑛𝑥 (respectively, 𝑧𝑛𝑧) for all 𝑛,(ii) if nonincreasing sequence 𝑦𝑛𝑦 (respectively, 𝑤𝑛𝑤), then 𝑦𝑛𝑦 (respectively, 𝑤𝑛𝑤) for all 𝑛.
If there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0,𝑤0), 𝑦0𝐹(𝑦0,𝑧0,𝑤0,𝑥0), 𝑧0𝐹(𝑧0,𝑤0,𝑥0,𝑦0), and 𝑤0𝐹(𝑤0,𝑥0,𝑦0,𝑧0), then there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤.(2.2)
that is, 𝐹 has a quadruple fixed point.

Proof. Let 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝑥𝐹0,𝑦0,𝑧0,𝑤0,𝑦0𝑦𝐹0,𝑧0,𝑤0,𝑥0𝑧,(2.3)0𝑧𝐹0,𝑤0,𝑥0,𝑦0,𝑤0𝑤𝐹0,𝑥0,𝑦0,𝑧0.(2.4)
Set 𝑥1𝑥=𝐹0,𝑦0,𝑧0,𝑤0,𝑦1𝑦=𝐹0,𝑧0,𝑤0,𝑥0,𝑧1𝑧=𝐹0,𝑤0,𝑥0,𝑦0,𝑤1𝑤=𝐹0,𝑥0,𝑦0,𝑧0.(2.5)
Then, 𝑥0𝑥1, 𝑦0𝑦1, 𝑧0𝑧1, and 𝑤0𝑤1. Again, define 𝑥2=𝐹(𝑥1,𝑦1,𝑧1,𝑤1), 𝑦2=𝐹(𝑦1,𝑧1,𝑤1,𝑥1), 𝑧2=𝐹(𝑧1,𝑤1,𝑥1,𝑦1), and 𝑤2=𝐹(𝑤1,𝑥1,𝑦1,𝑧1). Since 𝐹 has the mixed monotone property, we have 𝑥0𝑥1𝑥2, 𝑦2𝑦1𝑦0, 𝑧0𝑧1𝑧2, and 𝑤2𝑤1𝑤0. By continuing this process, we can construct four sequences {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛}, and {𝑤𝑛} in 𝑋 such that 𝑥𝑛+1𝑥=𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛,𝑦𝑛+1𝑦=𝐹𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛,𝑧𝑛+1𝑧=𝐹1,𝑤1,𝑥1,𝑦1,𝑤𝑛+1𝑤=𝐹𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛.(2.6)
Since 𝐹 has the mixed monotone property, by using a mathematical induction it is easy to see that 𝑥𝑛𝑥𝑛+1,𝑦𝑛+1𝑦𝑛,𝑧𝑛𝑧𝑛+1,𝑤𝑛+1𝑤𝑛,for𝑛=0,1,2,,(2.7)
Assume that, for some 𝑛, 𝑥𝑛=𝑥𝑛+1,𝑦𝑛=𝑦𝑛+1,𝑧𝑛=𝑧𝑛+1,𝑤𝑛=𝑤𝑛+1.(2.8) Then, by (2.6), (𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛) is a quadruple fixed point of 𝐹. Therefore, in the rest of the proof, for any 𝑛 we will assume that 𝑥𝑛𝑥𝑛+1or𝑦𝑛𝑦𝑛+1or𝑧𝑛𝑧𝑛+1or𝑤𝑛𝑤𝑛+1.(2.9)
Since 𝑇 is injective, for any 𝑛, 𝑑0<max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1.(2.10)
Due to (2.1), (2.6), and (2.7), we have 𝑑𝑇𝑥𝑛,𝑇𝑥𝑛+1𝑥=𝑑𝑇𝐹𝑛1,𝑦𝑛1,𝑧𝑛1,𝑤𝑛1𝑥,𝑇𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑑𝜙max𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛,𝑑𝑇𝑦𝑛+1,𝑇𝑦𝑛𝑦=𝑑𝑇𝐹𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑦,𝑇𝐹𝑛1,𝑧𝑛1,𝑤𝑛1,𝑥𝑛1𝑑𝜙max𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛,𝑑𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1𝑧=𝑑𝑇𝐹𝑛1,𝑤𝑛1,𝑥𝑛1,𝑦𝑛1𝑧,𝑇𝐹𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑑𝜙max𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛,𝑑𝑇𝑥𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑤𝑛+1,𝑇𝑤𝑛𝑤=𝑑𝑇𝐹𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑤,𝑇𝐹𝑛1,𝑥𝑛1,𝑦𝑛1,𝑧𝑛1𝑑𝜙max𝑇𝑤𝑛1,𝑇𝑤𝑛,𝑑𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛.(2.11)
Using the fact that 𝜙(𝑡)<𝑡 for all 𝑡>0 together with (2.11), we obtain that 𝑑0<max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1𝑑𝜙max𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛𝑑<max𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛.(2.12)
It follows that 𝑑max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1𝑑<max𝑇𝑥𝑛1,𝑇𝑥𝑛,𝑑𝑇𝑦𝑛1,𝑇𝑦𝑛,𝑑𝑇𝑧𝑛1,𝑇𝑧𝑛,𝑑𝑇𝑤𝑛1,𝑇𝑤𝑛.(2.13)
Thus, max{𝑑(𝑇𝑥𝑛,𝑇𝑥𝑛+1),𝑑(𝑇𝑦𝑛,𝑇𝑦𝑛+1),𝑑(𝑇𝑧𝑛,𝑇𝑧𝑛+1),𝑑(𝑇𝑤𝑛,𝑇𝑤𝑛+1)} is a positive decreasing sequence. Hence, there exists 𝑟0 such that lim𝑛+𝑑max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1=𝑟.(2.14)
Suppose that 𝑟>0. Letting 𝑛+ in (2.12), we obtain that 0<𝑟lim𝑛+𝜙𝑑max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1=lim𝑡𝑟+𝜙(𝑡)<𝑟,(2.15)
which is a contradiction. Therefore, we deduce that lim𝑛+𝑑max𝑇𝑥𝑛,𝑇𝑥𝑛+1,𝑑𝑇𝑦𝑛,𝑇𝑦𝑛+1,𝑑𝑇𝑧𝑛,𝑇𝑧𝑛+1,𝑑𝑇𝑤𝑛,𝑇𝑤𝑛+1=0.(2.16)
We will show that {𝑇𝑥𝑛}, {𝑇𝑦𝑛}, {𝑇𝑧𝑛}, and {𝑇𝑤𝑛} are Cauchy sequences. Assume the contrary, that is, either {𝑇𝑥𝑛} or {𝑇𝑦𝑛} or {𝑇𝑧𝑛} or {𝑇𝑤𝑛} is not a Cauchy sequence, consequently, lim𝑛,𝑚+𝑑𝑇𝑥𝑚,𝑇𝑥𝑛0orlim𝑛,𝑚+𝑑𝑇𝑦𝑚,𝑇𝑦𝑛0,(2.17)
or lim𝑛,𝑚+𝑑(𝑇𝑧𝑚,𝑇𝑧𝑛)0 or lim𝑛,𝑚+𝑑(𝑇𝑤𝑚,𝑇𝑤𝑛)0. This means that there exists 𝜀>0 for which we can find subsequences of integers (𝑚𝑘) and (𝑛𝑘) with 𝑛𝑘>𝑚𝑘>𝑘 such that 𝑑max𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘,𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘,𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘𝜀.(2.18)
Furthermore, corresponding to 𝑚𝑘, we can choose 𝑛𝑘 in such a way that it is the smallest integer with 𝑛𝑘>𝑚𝑘 and satisfying (2.18). Then, 𝑑max𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘1<𝜀.(2.19)
By the triangle inequality and (2.19), we have 𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘1+𝑑𝑇𝑥𝑛𝑘1,𝑇𝑥𝑛𝑘<𝜖+𝑑𝑇𝑥𝑛𝑘1,𝑇𝑥𝑛𝑘.(2.20)
Thus, by (2.16), we obtain lim𝑘+𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘lim𝑘+𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘1𝜀.(2.21)
Similarly, we have lim𝑘+𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘lim𝑘+𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘1𝜀,lim𝑘+𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘lim𝑘+𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘1𝜀,lim𝑘+𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘lim𝑘+𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘1𝜀.(2.22)
Again, by (2.19), we have 𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑚𝑘1+𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1+𝑑𝑇𝑥𝑛𝑘1,𝑇𝑥𝑛𝑘𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑚𝑘1+𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑚𝑘+𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘1+𝑑𝑇𝑥𝑛𝑘1,𝑇𝑥𝑛𝑘<𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑚𝑘1+𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑚𝑘+𝜀+𝑑𝑇𝑥𝑛𝑘1,𝑇𝑥𝑛𝑘.(2.23)
Letting 𝑘+ and using (2.16), we get lim𝑘+𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘lim𝑘+𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1𝜀,lim𝑘+𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘lim𝑘+𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1𝜀,lim𝑘+𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘lim𝑘+𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1𝜀,lim𝑘+𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘lim𝑘+𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1𝜀.(2.24)
Using (2.18) and (2.24), we have lim𝑘+𝑑max𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘,𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘,𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘=lim𝑘+𝑑max𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1=𝜀.(2.25)
Now, using inequality (2.1), we obtain 𝑑𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘𝑥=𝑑𝑇𝐹𝑚𝑘1,𝑦𝑚𝑘1,𝑧𝑚𝑘1,𝑤𝑚𝑘1𝑥,𝑇𝐹𝑛𝑘1,𝑦𝑛𝑘1,𝑧𝑛𝑘1,𝑤𝑛𝑘1𝑑𝜙max𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘𝑦=𝑑𝑇𝐹𝑚𝑘1,𝑧𝑚𝑘1,𝑤𝑚𝑘1,𝑥𝑚𝑘1𝑦,𝑇𝐹𝑛𝑘1,𝑧𝑛𝑘1,𝑤𝑛𝑘1,𝑥𝑛𝑘1𝑑𝜙max𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1,𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘𝑧=𝑑𝑇𝐹𝑚𝑘1,𝑤𝑚𝑘1,𝑥𝑚𝑘1,𝑦𝑚𝑘1𝑧,𝑇𝐹𝑛𝑘1,𝑤𝑛𝑘1,𝑥𝑛𝑘1,𝑦𝑛𝑘1𝑑𝜙max𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1,𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘𝑤=𝑑𝑇𝐹𝑚𝑘1,𝑥𝑚𝑘1,𝑦𝑚𝑘1,𝑧𝑚𝑘1𝑤,𝑇𝐹𝑛𝑘1,𝑥𝑛𝑘1,𝑦𝑛𝑘1,𝑧𝑛𝑘1𝑑𝜙max𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1,𝑑𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1(2.26)
From (2.26), we deduce that 𝑑max𝑇𝑥𝑚𝑘,𝑇𝑥𝑛𝑘,𝑑𝑇𝑦𝑚𝑘,𝑇𝑦𝑛𝑘,𝑑𝑇𝑧𝑚𝑘,𝑇𝑧𝑛𝑘,𝑑𝑇𝑤𝑚𝑘,𝑇𝑤𝑛𝑘𝑑𝜙max𝑇𝑥𝑚𝑘1,𝑇𝑥𝑛𝑘1,𝑑𝑇𝑦𝑚𝑘1,𝑇𝑦𝑛𝑘1,𝑑𝑇𝑧𝑚𝑘1,𝑇𝑧𝑛𝑘1,𝑑𝑇𝑤𝑚𝑘1,𝑇𝑤𝑛𝑘1.(2.27)
Letting 𝑘+ in (2.27) and by using (2.25), we get that 0<𝜀lim𝑡𝜀+𝜙(𝑡)<𝜀,(2.28)
which is a contradiction. Thus, {𝑇𝑥𝑛}, {𝑇𝑦𝑛}, {𝑇𝑧𝑛}, and {𝑇𝑤𝑛} are Cauchy sequences in (𝑋,𝑑). Since 𝑋 is a complete metric space, {𝑇𝑥𝑛}, {𝑇𝑦𝑛}, {𝑇𝑧𝑛}, and {𝑇𝑤𝑛} are convergent sequences.
Since 𝑇 is an ICS mapping, there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that lim𝑛+𝑥𝑛=𝑥,lim𝑛+𝑦𝑛=𝑦,lim𝑛+𝑧𝑛=𝑧,lim𝑛+𝑤𝑛=𝑤.(2.29)
Since 𝑇 is continuous, we have lim𝑛+𝑇𝑥𝑛=𝑇𝑥,lim𝑛+𝑇𝑦𝑛=𝑇𝑦,lim𝑛+𝑇𝑧𝑛=𝑇𝑧,lim𝑛+𝑇𝑤𝑛=𝑇𝑤.(2.30)
Suppose now the assumption (𝑎) holds, that is, 𝐹 is continuous. By (2.6), (2.29), and (2.30) we obtain 𝑥=lim𝑛+𝑥𝑛+1=lim𝑛+𝐹𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛=𝐹lim𝑛+𝑥𝑛,lim𝑛+𝑦𝑛,lim𝑛+𝑧𝑛,lim𝑛+𝑤𝑛=𝐹(𝑥,𝑦,𝑧,𝑤),𝑦=lim𝑛+𝑦𝑛+1=lim𝑛+𝐹𝑦𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛=𝐹lim𝑛+𝑦𝑛,lim𝑛+𝑧𝑛,lim𝑛+𝑤𝑛,lim𝑛+𝑥𝑛=𝐹(𝑦,𝑧,𝑤,𝑥),𝑧=lim𝑛+𝑧𝑛+1=lim𝑛+𝐹𝑧𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛=𝐹lim𝑛+𝑧𝑛,lim𝑛+𝑤𝑛,lim𝑛+𝑥𝑛,lim𝑛+𝑦𝑛=𝐹(𝑧,𝑤,𝑥,𝑦),𝑤=lim𝑛+𝑤𝑛+1=lim𝑛+𝐹𝑤𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛=𝐹lim𝑛+𝑤𝑛,lim𝑛+𝑥𝑛,lim𝑛+𝑦𝑛,lim𝑛+𝑧𝑛=𝐹(𝑤,𝑥,𝑦,𝑧).(2.31)
We have proved that 𝐹 has a quadruple fixed point.
Suppose now the assumption (𝑏) holds. Since {𝑥𝑛} and {𝑧𝑛} are nondecreasing with 𝑥𝑛𝑥 and 𝑧𝑛𝑧 and also {𝑦𝑛} and {𝑤𝑛} are nonincreasing, with 𝑦𝑛𝑦 and 𝑤𝑛𝑤, we have 𝑥𝑛𝑥,𝑦𝑛𝑦,𝑧𝑛𝑧,𝑤𝑛𝑤(2.32)
for all 𝑛. Consider now 𝑑(𝑇𝑥,𝑇𝐹(𝑥,𝑦,𝑧,𝑤))𝑑𝑇𝑥,𝑇𝑥𝑛+1+𝑑𝑇𝑥𝑛+1,𝑇𝐹(𝑥,𝑦,𝑧,𝑤)=𝑑𝑇𝑥,𝑇𝑥𝑛+1𝑥+𝑑𝑇𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛,𝑇𝐹(𝑥,𝑦,𝑧,𝑤)𝑑𝑇𝑥,𝑇𝑥𝑛+1𝑑+𝜙max𝑇𝑥𝑛,𝑇𝑥,𝑑𝑇𝑦𝑛,𝑇𝑦,𝑑𝑇𝑧𝑛,𝑑,𝑇𝑧𝑇𝑤𝑛.,𝑇𝑤(2.33)
Taking 𝑛 and using (2.30), the right-hand side of (2.33) tends to 0, so we get that 𝑑(𝑇𝑥,𝑇𝐹(𝑥,𝑦,𝑧,𝑤))=0. Thus, 𝑇𝑥=𝑇𝐹(𝑥,𝑦,𝑧,𝑤), and since 𝑇 is injective, we get that 𝑥=𝐹(𝑥,𝑦,𝑧,𝑤). Analogously, one finds that 𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤.(2.34)
Thus, we proved that 𝐹 has a quadrupled fixed point. This completes the proof of Theorem 2.2.

Repeating the same proof of Theorem 2.2, we may state the following corollary.

Corollary 2.3. Let (𝑋,) be a partially ordered set and suppose that there is a metric 𝑑 on 𝑋 such that (𝑋,𝑑) is a complete metric space. Suppose also 𝑇𝑋𝑋 is an ICS mapping and 𝐹𝑋4𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝜙Φ such that 𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑡,𝑠))𝜙𝑑(𝑇𝑥,𝑇𝑢)+𝑑(𝑇𝑦,𝑇𝑣)+𝑑(𝑇𝑧,𝑇𝑡)+𝑑(𝑇𝑤,𝑇𝑠)4(2.35)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑡,𝑠𝑋 for which 𝑥𝑢, 𝑣𝑦, 𝑧𝑡, and 𝑠𝑤. Additionally suppose that either(a)𝐹 is continuous, or(b)𝑋 has the following property:(i) if non-decreasing sequence 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧), then 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧) for all 𝑛,(ii) if non-increasing sequence 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤), then 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤) for all 𝑛.
If there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0,𝑤0), 𝑦0𝐹(𝑦0,𝑧0,𝑤0,𝑥0), 𝑧0𝐹(𝑧0,𝑤0,𝑥0,𝑦0) and 𝑤0𝐹(𝑤0,𝑤0,𝑦0,𝑧0), then there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤,(2.36)
that is, 𝐹 has a quadruple fixed point.

Corollary 2.4. Let (𝑋,) be a partially ordered set and suppose that there is a metric 𝑑 on 𝑋 such that (𝑋,𝑑) is a complete metric space. Suppose also that 𝑇𝑋𝑋 is an ICS mapping and 𝐹𝑋4𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝑘[0,1) such that 𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑡,𝑠))𝑘max{𝑑(𝑇𝑥,𝑇𝑢),𝑑(𝑇𝑦,𝑇𝑣),𝑑(𝑇𝑧,𝑇t),𝑑(𝑇𝑤,𝑇𝑠)}(2.37)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑡,𝑠𝑋 for which 𝑥𝑢, 𝑣𝑦, 𝑧𝑡, and 𝑠𝑤. Suppose that either (a)𝐹 is continuous, or(b)𝑋 has the following property:(i)if nondecreasing sequence 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧), then 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧) for all 𝑛,(ii)if nonincreasing sequence 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤), then 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤) for all 𝑛.
If there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑤0,𝑧0), 𝑦0𝐹(𝑦0,𝑧0,𝑤0,𝑥0), 𝑧0𝐹(𝑧0,𝑤0,𝑥0,𝑦0), and 𝑤0𝐹(𝑤0,𝑤0,𝑦0,𝑧0) then, there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤,(2.38)
that is, 𝐹 has a quadruple fixed point.

Proof. It suffices to remark that 𝜙(𝑡)=𝑘𝑡 in Theorem 2.2.

Corollary 2.5. Let (𝑋,) be a partially ordered set and suppose that there is a metric 𝑑 on 𝑋 such that (𝑋,𝑑) is a complete metric space. Suppose also that 𝑇𝑋𝑋 is an ICS mapping and 𝐹𝑋4𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝑘[0,1) such that 𝑘𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑡,𝑠))4(𝑑(𝑇𝑥,𝑇𝑢),𝑑(𝑇𝑦,𝑇𝑣),𝑑(𝑇𝑧,𝑇t),𝑑(𝑇𝑤,𝑇𝑠))(2.39)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑡,𝑠𝑋 for which 𝑥𝑢, 𝑣𝑦, 𝑧𝑡 and 𝑠𝑤. Suppose that either(a)𝐹 is continuous, or(b)𝑋 has the following property:(i)if nondecreasing sequence 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧), then 𝑥𝑛𝑥 (resp., 𝑧𝑛𝑧) for all 𝑛,(ii)if nonincreasing sequence 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤), then 𝑦𝑛𝑦 (resp., 𝑤𝑛𝑤) for all 𝑛.
If there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑤0,𝑧0), 𝑦0𝐹(𝑦0,𝑧0,𝑤0,𝑥0), 𝑧0𝐹(𝑧0,𝑤0,𝑥0,𝑦0) and 𝑤0𝐹(𝑤0,𝑤0,𝑦0,𝑧0) then there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤(2.40)
that is, 𝐹 has a quadruple fixed point.

Proof. It suffices to take 𝜙(𝑡)=𝑘𝑡 in Corollary 2.3.

Now, we shall prove the existence and uniqueness of a quadruple fixed point. For a product 𝑋4 of a partial ordered set (𝑋,), we define a partial ordering in the following way: For all (𝑥,𝑦,𝑧,𝑤),(𝑢,𝑣,𝑡,𝑠)𝑋4, (𝑥,𝑦,𝑧,𝑤)(𝑢,𝑣,𝑡,𝑠)𝑥𝑢,𝑦𝑣,𝑧𝑡,𝑤𝑠.(2.41)

We say that (𝑥,𝑦,𝑧,𝑤) and (𝑢,𝑣,𝑡,𝑠) are comparable if (𝑥,𝑦,𝑧,𝑤)(𝑢,𝑣,𝑡,𝑠)or(𝑢,𝑣,𝑠,𝑡)(𝑥,𝑦,𝑧,𝑤).(2.42)

Also, we say that (𝑥,𝑦,𝑧,𝑤) is equal to (𝑢,𝑣,𝑡,𝑠) if and only if 𝑥=𝑢,𝑦=𝑣,𝑧=𝑡,𝑤=𝑠.

Theorem 2.6. In addition to hypotheses of Theorem 2.2, suppose that that for all (𝑥,𝑦,𝑧,𝑤),(𝑢,𝑣,𝑡,𝑠)𝑋4, there exists (𝑎,𝑏,𝑐,𝑑)𝑋4 such that (𝐹(𝑎,𝑏,𝑐,𝑑), 𝐹(𝑏,𝑐,𝑑,𝑎), 𝐹(𝑐,𝑑,𝑎,𝑏), 𝐹(𝑑,𝑎,𝑏,𝑐)) is comparable to (𝐹(𝑥,𝑦,𝑧,𝑤), 𝐹(𝑦,𝑧,𝑤,𝑥), 𝐹(𝑧,𝑤,𝑥,𝑦), 𝐹(𝑤,𝑥,𝑦,𝑧)) and (𝐹(𝑢,𝑣,𝑡,𝑠), 𝐹(𝑣,𝑡,𝑠,𝑢), 𝐹(𝑡,𝑠,𝑢,𝑣), 𝐹(𝑠,𝑢,𝑣,𝑡)). Then, 𝐹 has a unique quadruple fixed point (𝑥,𝑦,𝑧,𝑤).

Proof. The set of quadruple fixed points of 𝐹 is not empty due to Theorem 2.2. Assume, now, (𝑥,𝑦,𝑧,𝑤) and (𝑢,𝑣,𝑡,𝑠) are two quadrupled fixed points of 𝐹, that is, 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑢,𝑣,𝑡,𝑠)=𝑢,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑦,𝐹(𝑣,𝑡,𝑠,𝑢)=𝑣,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑧,𝐹(𝑡,𝑠,𝑢,𝑣)=𝑡,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑤,𝐹(𝑠,𝑢,𝑣,𝑡)=𝑠.(2.43)
We shall show that (𝑥,𝑦,𝑧,𝑤) and (𝑢,𝑣,𝑡,𝑠) are equal. By assumption, there exists (𝑎,𝑏,𝑐,𝑑)𝑋4 such that (𝐹(𝑎,𝑏,𝑐,𝑑), 𝐹(𝑏,𝑐,𝑑,𝑎), 𝐹(𝑐,𝑑,𝑎,𝑏), 𝐹(𝑑,𝑎,𝑏,𝑐)) is comparable to (𝐹(𝑥,𝑦,𝑧,𝑤), 𝐹(𝑦,𝑧,𝑤,𝑥), 𝐹(𝑧,𝑤,𝑥,𝑦), 𝐹(𝑤,𝑥,𝑦,𝑧)) and (𝐹(𝑢,𝑣,𝑡,𝑠), 𝐹(𝑣,𝑡,𝑠,𝑢), 𝐹(𝑡,𝑠,𝑢,𝑣), 𝐹(𝑠,𝑢,𝑣,𝑡)).
Define sequences {𝑎𝑛},{𝑏𝑛}, {𝑐𝑛}, and {𝑑𝑛} such that 𝑎0=𝑎,𝑏0=𝑏,𝑐0=𝑐,𝑑0𝑎=𝑑,forany𝑛1,𝑛𝑎=𝐹𝑛1,𝑏𝑛1,𝑐𝑛1,𝑑𝑛1,𝑏𝑛𝑏=𝐹𝑛1,𝑐𝑛1,𝑑𝑛1,𝑎𝑛1,𝑐𝑛𝑐=𝐹𝑛1,𝑑𝑛1,𝑎𝑛1,𝑏𝑛1,𝑑𝑛𝑑=𝐹𝑛1,𝑎𝑛1,𝑏𝑛1,𝑐𝑛1,(2.44)
for all 𝑛. Further, set 𝑥0=𝑥, 𝑦0=𝑦, 𝑧0=𝑧, 𝑤0=𝑤 and 𝑢0=𝑢, 𝑣0=𝑣, 𝑡0=𝑡, 𝑠0=𝑠 and on the same way define the sequences {𝑥𝑛},{𝑦𝑛}, {𝑧𝑛}, and {𝑤𝑛} and {𝑢𝑛},{𝑣𝑛}, {𝑡𝑛}, and {𝑠𝑛}. Then, it is easy that 𝑥𝑛𝑦=𝐹(𝑥,𝑦,𝑧,𝑤),𝑛𝑧=𝐹(𝑦,𝑧,𝑤,𝑥,),𝑛𝑤=𝐹(𝑧,𝑤,𝑥,𝑦),𝑛𝑢=𝐹(𝑤,𝑥,𝑦,𝑧),𝑛𝑣=𝐹(𝑢,𝑣,𝑡,𝑠),𝑛𝑡=𝐹(𝑣,𝑡,𝑠,𝑢),𝑛𝑠=𝐹(𝑡,𝑠,𝑢,𝑣),𝑛=𝐹(𝑠,𝑢,𝑣,𝑡),(2.45)
for all 𝑛1. Since (𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑦,𝑧,𝑤,𝑥),𝐹(𝑧,𝑤,𝑥,𝑦),𝐹(𝑤,𝑥,𝑦,𝑧))=(𝑥1,𝑦1,𝑧1,𝑤1)=(𝑥,𝑦,𝑧,𝑤) is comparable to (𝐹(𝑎,𝑏,𝑐,𝑑),𝐹(𝑏,𝑐,𝑑,𝑎),𝐹(𝑐,𝑑,𝑎,𝑏))=(𝑎1,𝑏1,𝑐1,𝑑1), then it is easy to show (𝑥,𝑦,𝑧,𝑤)(𝑎1,𝑏1,𝑐1,𝑑1). Recursively, we get that 𝑎(𝑥,𝑦,𝑧,𝑤)𝑛,𝑏𝑛,𝑐𝑛,𝑑𝑛𝑛.(2.46)
By (2.46) and (2.1), we have 𝑑𝑇𝑥,𝑇𝑎𝑛+1𝑎=𝑑𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹𝑛,𝑏𝑛,𝑐𝑛,𝑑𝑛𝑑𝜙max𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛,𝑑𝑇z,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑏𝑛+1𝑏,𝑇𝑦=𝑑𝑇𝐹𝑛,𝑐𝑛,𝑑𝑛,𝑎𝑛𝑑,𝑇𝐹(𝑦,𝑧,𝑤,𝑥)𝜙max𝑇𝑦,𝑇𝑏𝑛,𝑑𝑇𝑧,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑧,𝑇𝑐𝑛+1𝑐=𝑑𝑇𝐹(𝑧,𝑤,𝑥,𝑦),𝑇𝐹𝑛,𝑑𝑛,𝑎𝑛,𝑏𝑛𝑑𝜙max𝑇𝑧,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛+1𝑑=𝑑𝑇𝐹(𝑤,𝑥,𝑦,𝑧),𝑇𝐹𝑛,𝑎𝑛,𝑏𝑛,𝑐𝑛𝑑𝜙max𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛,𝑑𝑇𝑧,𝑇𝑐𝑛.(2.47)
It follows from (2.47) that 𝑑max𝑇𝑧,𝑇𝑐𝑛+1,𝑑𝑇𝑤,𝑇𝑑𝑛+1,𝑑𝑇𝑥,𝑇𝑎𝑛+1,𝑑𝑇𝑦,𝑇𝑏𝑛+1𝑑𝜙max𝑇𝑧,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛.(2.48)
Therefore, for each 𝑛1, 𝑑max𝑇𝑧,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛𝜙𝑛𝑑max𝑇𝑧,𝑇𝑐0,𝑑𝑇𝑤,𝑇𝑑0,𝑑𝑇𝑥,𝑇𝑎0,𝑑𝑇𝑦,𝑇𝑏0.(2.49)
It is known that 𝜙(𝑡)<𝑡 and lim𝑟𝑡+𝜙(𝑟)<𝑡 imply lim𝑛𝜙𝑛(𝑡)=0 for each 𝑡>0. Thus, from (2.49), lim𝑛𝑑max𝑇𝑧,𝑇𝑐𝑛,𝑑𝑇𝑤,𝑇𝑑𝑛,𝑑𝑇𝑥,𝑇𝑎𝑛,𝑑𝑇𝑦,𝑇𝑏𝑛=0.(2.50)
This yields that lim𝑛𝑑𝑇𝑥,𝑇𝑎𝑛=0,lim𝑛𝑑𝑇𝑦,𝑇𝑏𝑛=0,lim𝑛𝑑𝑇𝑧,𝑇𝑐𝑛=0,lim𝑛𝑑𝑇𝑤,𝑇𝑑𝑛=0.(2.51)
Analogously, we show that lim𝑛𝑑𝑇𝑢,𝑇𝑎𝑛=0,lim𝑛𝑑𝑇𝑣,𝑇𝑏𝑛=0,lim𝑛𝑑𝑇𝑡,𝑇𝑐𝑛=0,lim𝑛𝑑𝑇𝑠,𝑇𝑑𝑛=0.(2.52)
Combining (2.51) to (2.52) yields that (𝑇𝑥,𝑇𝑦,𝑇𝑧,𝑇𝑤) and (𝑇𝑢,𝑇𝑣,𝑇𝑡,𝑇𝑠) are equal. The fact that 𝑇 is injective gives us 𝑥=𝑢, 𝑦=𝑣, 𝑧=𝑡, and 𝑤=𝑠.

We state some examples showing that our results are effective.

Example 2.7. Let 𝑋=[1,64] with the metric 𝑑(𝑥,𝑦)=|𝑥𝑦|, for all 𝑥,𝑦𝑋 and the usual ordering. Clearly, (𝑋,𝑑) is a complete metric space.
Let 𝑇𝑋𝑋 and 𝐹𝑋4𝑋 be defined by 𝑥𝑦𝑇𝑥=ln(𝑥)+1,𝐹(𝑥,𝑦,𝑧,𝑤)=81/3,𝑥,𝑦,𝑧,𝑤𝑋.(2.53)
It is clear that 𝑇 is an ICS mapping, 𝐹 has the mixed monotone property and continuous.
Set 𝜙(𝑡)=2𝑡/3. Taking 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑠,𝑡𝑋 for which 𝑥𝑢, 𝑣𝑦, 𝑧𝑠, and 𝑡𝑤, we have 1𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑠,𝑡))=3||||1(ln𝑥ln𝑦)(ln𝑢ln𝑣)3||||+1ln𝑥ln𝑢3||||1ln𝑦ln𝑣3||||,||||maxln𝑥ln𝑢ln𝑦ln𝑣=𝜙(max{𝑑(𝑇𝑥,𝑇𝑢),𝑑(𝑇𝑦,𝑇𝑣),𝑑(𝑇𝑧,𝑇𝑠),𝑑(𝑇𝑤,𝑇𝑡)}),(2.54)which is the contractive condition (2.1). Moreover, taking 𝑥0=𝑦0=𝑧0=8=𝑤0, we have 𝑥0𝑥𝐹0,𝑦0,𝑧0,𝑤0,𝑦0𝑦𝐹0,𝑧0,𝑤0,𝑥0,𝑧0𝑧𝐹0,𝑤0,𝑥0,𝑦0,𝑤0𝑤𝐹0,𝑥0,𝑦0,𝑧0.(2.55)
Therefore, all the conditions of Theorem 2.2 hold and (8,8,8,8) is the unique quadruple fixed point of 𝐹, since also the hypotheses of Theorem 2.6 hold.
On the other hand, we can not apply Corollary 15 of Karapınar [27] to this example. Indeed, for 𝑥=1=𝑦=𝑣, 𝑢=2, 1𝑧=𝑠 and 1𝑡=𝑤, we have ||𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑠,𝑡))=8(2)1/3||>114>𝑘4=𝑘𝑑(𝑥,𝑢)4[],𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑠)+𝑑(𝑤,𝑡)(2.56)for any 𝑘[0,1).

Example 2.8. Let 𝑋= with 𝑑(𝑥,𝑦)=|𝑥𝑦| and natural ordering. Let 𝑇𝑋𝑋 and 𝐹𝑋4𝑋 be defined by 𝑇𝑥=𝑥/12 and 𝐹(𝑥,𝑦,𝑧,𝑤)=2/5(𝑥𝑦+𝑧𝑤). It is clear that 𝑇 is an ICS mapping and 𝐹 has the monotone property and continuous. Set 𝜙(𝑡)=2𝑡/3Φ. It is clear that all conditions of Theorem 2.2 are satisfied and (0,0,0,0) is the desired quadruple point.
Note that Corollary 15 of Karapınar [27] is not applicable. Indeed, for 𝑥=0, 𝑢=1 and 𝑦=𝑣=𝑧=𝑠=𝑡=𝑤=0, we have 2𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑠,𝑡))=5>𝑘4𝑘𝑑(𝑥,𝑢)=4[],𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑠)+𝑑(𝑤,𝑡)(2.57) for any 𝑘[0,1).

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