Abstract

We prove a number of quadruple fixed point theorems under 𝜙-contractive conditions for a mapping 𝐹𝑋4𝑋 in ordered metric spaces. Also, we introduce an example to illustrate the effectiveness of our results.

1. Introduction and Preliminaries

The notion of coupled fixed point was initiated by Gnana Bhaskar and Lakshmikantham [1] in 2006. In this paper, they proved some fixed point theorems under a set of conditions and utilized their theorems to prove the existence of solutions to some ordinary differential equations. Recently, Berinde and Borcut [2] introduced the notion of tripled fixed point and extended the results of Gnana Bhaskar and Lakshmikantham [1] to the case of contractive operator 𝐹𝑋×𝑋×𝑋𝑋, where 𝑋 is a complete ordered metric space. For some related works in coupled and tripled fixed point, we refer readers to [332].

For simplicity we will denote the cross product of 𝑘 copies of the space 𝑋 by 𝑋𝑘.

Definition 1.1 (see [2]). Let 𝑋 be a nonempty set and 𝐹𝑋3𝑋 a given mapping. An element (𝑥,𝑦,𝑧)𝑋3 is called a tripled fixed point of 𝐹 if 𝐹(𝑥,𝑦,𝑧)=𝑥,𝐹(𝑦,𝑥,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥)=𝑧.(1.1) Let (𝑋,𝑑) be a metric space. The mapping 𝑑𝑋3𝑋, given by 𝑑((𝑥,𝑦,𝑧),(𝑢,𝑣,𝑤))=𝑑(𝑥,𝑦)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑤),(1.2) defines a metric on 𝑋3, which will be denoted for convenience by 𝑑.

Definition 1.2 (see [2]). Let (𝑋,) be a partially ordered set and 𝐹𝑋3𝑋 a mapping. One says that 𝐹 has the mixed monotone property if 𝐹(𝑥,𝑦,𝑧) is monotone nondecreasing in 𝑥 and 𝑧 and is monotone nonincreasing in 𝑦; that is, for any 𝑥,𝑦,𝑧𝑋, 𝑥1,𝑥2𝑋,𝑥1𝑥2𝑥,implies𝐹1𝑥,𝑦,𝑧𝐹2,𝑦,𝑦,𝑧1,𝑦2𝑋,𝑦1𝑦2,implies𝐹𝑥,𝑦2,𝑧𝐹𝑥,𝑦1,𝑧,𝑧1,𝑧2𝑋,𝑧1𝑧2,implies𝐹𝑥,𝑦,𝑧1𝐹𝑥,𝑦,𝑧2.(1.3)

Let us recall the main results of [2] to understand our motivation toward our results in this paper.

Theorem 1.3 (see [2]). Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋3𝑋 be a continuous mapping such that 𝐹 has the mixed monotone property. Assume that there exist 𝑗,𝑘,𝑙[0,1) with 𝑗+𝑘+𝑙<1 such that 𝑑(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑤))𝑗𝑑(𝑥,𝑢)+𝑘𝑑(𝑦,𝑣)+𝑙𝑑(𝑧,𝑤)(1.4) for all 𝑥,𝑦,𝑧,𝑢,𝑣,𝑤𝑋 with 𝑥𝑢, 𝑦𝑣, and 𝑧𝑤. If there exist 𝑥0,𝑦0,𝑧0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0), 𝑦0𝐹(𝑦0,𝑥0,𝑦0), and 𝑧0𝐹(𝑧0,𝑦0,𝑥0), then 𝐹 has a tripled fixed point.

Theorem 1.4 (see [2]). Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋3𝑋 be a mapping having the mixed monotone property. Assume that there exist 𝑗,𝑘,𝑙[0,1) with 𝑗+𝑘+𝑙<1 such that 𝑑(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑤))𝑗𝑑(𝑥,𝑢)+𝑘𝑑(𝑦,𝑣)+𝑙𝑑(𝑧,𝑤)(1.5) for all 𝑥,𝑦,𝑧,𝑢,𝑣,𝑤𝑋 with 𝑥𝑢, 𝑦𝑣, and 𝑧𝑤. Assume that 𝑋 has the following properties: (i) if a nondecreasing sequence 𝑥𝑛𝑥, then 𝑥𝑛𝑥 for all 𝑛𝐍, (ii) if a nonincreasing sequence 𝑦𝑛𝑦, then 𝑦𝑛𝑦 for all 𝑛𝐍. If there exist 𝑥0,𝑦0,𝑧0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0), 𝑦0𝐹(𝑦0,𝑥0,𝑦0), and 𝑧0𝐹(𝑧0,𝑦0,𝑥0), then 𝐹 has a tripled fixed point.

Very recently, Karapınar introduced the notion of quadruple fixed point and obtained some fixed point theorems on the topic [33]. Extending this work, quadruple fixed point is developed and related fixed point theorems are proved in [3439].

Definition 1.5 (see [34]). Let 𝑋 be a nonempty set and 𝐹𝑋4𝑋 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.(1.6) 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),(1.7) defines a metric on 𝑋4, which will be denoted for convenience by 𝑑.

Remark 1.6. In [33, 34, 38], the notion of quadruple fixed point is called quartet fixed point.

Definition 1.7 (see [34]). Let (𝑋,) be a partially ordered set and 𝐹𝑋4𝑋 a mapping. One says 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,𝑥1𝑥2𝑥,implies𝐹1𝑥,𝑦,𝑧,𝑤𝐹2,𝑦,𝑦,𝑧,𝑤1,𝑦2𝑋,𝑦1𝑦2,implies𝐹𝑥,𝑦2,𝑧,𝑤𝐹𝑥,𝑦1,𝑧,𝑧,𝑤1,𝑧2𝑋,𝑧1𝑧2,implies𝐹𝑥,𝑦,𝑧1,𝑤𝐹𝑥,𝑦,𝑧2,𝑤,𝑤1,𝑤2𝑋,𝑤1𝑤2,implies𝐹𝑥,𝑦,𝑧,𝑤2𝐹𝑥,𝑦,𝑧,𝑤1.(1.8)

By following Matkowski [40], we let Φ be the set of all nondecreasing functions 𝜙[0,+)[0,+) such that lim𝑛+𝜙𝑛(𝑡)=0 for all 𝑡>0. Then, it is an easy matter to show that(1)𝜙(𝑡)<𝑡 for all 𝑡>0,(2)𝜙(0)=0.

In this paper, we prove some quadruple fixed point theorems for a mapping 𝐹𝑋4𝑋 satisfying a contractive condition based on some 𝜙Φ.

2. Main Results

Our first result is the following.

Theorem 2.1. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a continuous mapping such that 𝐹 has the mixed monotone property. Assume that there exists 𝜙Φ such that 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))𝜙(max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)})(2.1) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. 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 𝐹 has a quadruple fixed point.

Proof. Suppose 𝑥0,𝑦0,𝑧0,𝑤0𝑋 are such that 𝑥0𝐹(𝑥0,𝑦0,𝑧0,𝑤0), 𝑦0𝐹(𝑦0,𝑧0,𝑤0,𝑥0), 𝑧0𝐹(𝑧0,𝑤0,𝑥0,𝑦0), and 𝑤0𝐹(𝑤0,𝑥0,𝑦0,𝑧0). Define 𝑥1𝑥=𝐹0,𝑦0,𝑧0,𝑤0,𝑦1𝑦=𝐹0,𝑧0,𝑤0,𝑥0,𝑧1𝑧=𝐹0,𝑤0,𝑥0,𝑦0,𝑤1𝑤=𝐹0,𝑥0,𝑦0,𝑧0.(2.2) 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. Continuing this process, we can construct four sequences (𝑥𝑛), (𝑦𝑛), (𝑧𝑛), and (𝑤𝑛) in 𝑋 such that 𝑥𝑛𝑥=𝐹𝑛1,𝑦𝑛1,𝑧𝑛1,𝑤𝑛1𝑥𝑛+1𝑥=𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛,𝑦𝑛+1𝑦=𝐹𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑦𝑛𝑦=𝐹𝑛1,𝑧𝑛1,𝑤𝑛1,𝑥𝑛1,𝑧𝑛𝑧=𝐹𝑛1,𝑤𝑛1,𝑥𝑛1,𝑦𝑛1𝑧𝑛+1𝑧=𝐹𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛,𝑤𝑛+1𝑤=𝐹𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑤𝑛𝑤=𝐹𝑛1,𝑥𝑛1,𝑦𝑛1,𝑧𝑛1.(2.3) If, for some integer 𝑛, we have (𝑥𝑛+1,𝑦𝑛+1,𝑧𝑛+1,𝑤𝑛+1)=(𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛), then 𝐹(𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛)=𝑥𝑛, 𝐹(𝑦𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛)=𝑦𝑛, 𝐹(𝑧𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛)=𝑧𝑛, and 𝐹(𝑤𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛)=𝑤𝑛; that is, (𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛) is a quadruple fixed point of 𝐹. Thus, we will assume that (𝑥𝑛+1,𝑦𝑛+1,𝑧𝑛+1,𝑤𝑛+1)(𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛) for all 𝑛; that is, we assume that 𝑥𝑛+1𝑥𝑛,𝑦𝑛+1𝑦𝑛, or 𝑧𝑛+1𝑧𝑛 or 𝑤𝑛+1𝑤𝑛. For any 𝑛, we have 𝑑𝑥𝑛+1,𝑥𝑛𝐹𝑥=𝑑𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑥,𝐹𝑛1,𝑦𝑛1,𝑧𝑛1,𝑤𝑛1𝑑𝑥𝜙max𝑛,𝑥𝑛1𝑦,𝑑𝑛,𝑦𝑛1𝑧,𝑑𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1,𝑑𝑦𝑛,𝑦𝑛+1𝐹𝑦=𝑑𝑛1,𝑧𝑛1,𝑤𝑛1,𝑥𝑛1𝑦,𝐹𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑑𝑦𝜙max𝑛1,𝑦𝑛𝑧,𝑑𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1𝑥,𝑑𝑛1,𝑥𝑛,𝑑𝑧𝑛+1,𝑧𝑛𝐹𝑧=𝑑𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑧,𝐹𝑛1,𝑤𝑛1,𝑥𝑛1,𝑦𝑛1𝑑𝑧𝜙max𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1𝑥,𝑑𝑛,𝑥𝑛1𝑦,𝑑𝑛,𝑦𝑛1,𝑑𝑤𝑛,𝑤𝑛+1𝐹𝑤=𝑑𝑛1,𝑥𝑛1,𝑦𝑛1,𝑧𝑛1𝑤,𝐹𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑑𝑦𝜙max𝑛1,𝑦𝑛𝑧,𝑑𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1𝑥,𝑑𝑛1,𝑥𝑛.(2.4) From (2.4), it follows that 𝑑𝑥max𝑛+1,𝑥𝑛𝑦,𝑑𝑛,𝑦𝑛+1𝑧,𝑑𝑛+1,𝑧𝑛𝑤,𝑑𝑛,𝑤𝑛+1𝑑𝑥𝜙max𝑛,𝑥𝑛1𝑦,𝑑𝑛,𝑦𝑛1𝑧,𝑑𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1.(2.5) By repeating (2.5) 𝑛 times, we get that 𝑑𝑥max𝑛+1,𝑥𝑛𝑦,𝑑𝑛,𝑦𝑛+1𝑧,𝑑𝑛+1,𝑧𝑛𝑤,𝑑𝑛,𝑤𝑛+1𝑑𝑥𝜙max𝑛,𝑥𝑛1𝑦,𝑑𝑛,𝑦𝑛1𝑧,𝑑𝑛,𝑧𝑛1𝑤,𝑑𝑛,𝑤𝑛1𝜙2𝑑𝑥max𝑛1,𝑥𝑛2𝑦,𝑑𝑛1,𝑦𝑛2𝑧,𝑑𝑛1,𝑧𝑛2𝑤,𝑑𝑛1,𝑤𝑛1𝜙𝑛𝑑𝑥max1,𝑥0𝑦,𝑑1,𝑦0𝑧,𝑑1,𝑧0𝑤,𝑑1,𝑤0.(2.6) Now, we will show that (𝑥𝑛), (𝑦𝑛), (𝑧𝑛), and (𝑤𝑛) are Cauchy sequences in 𝑋. Let 𝜖>0. Since lim𝑛+𝜙𝑛𝑑𝑥max1,𝑥0𝑦,𝑑1,𝑦0z,𝑑1,𝑧0𝑤,𝑑1,𝑤0=0(2.7) and 𝜖>𝜙(𝜖), there exist 𝑛0 such that 𝜙𝑛𝑑𝑥max1,𝑥0𝑦,𝑑1,𝑦0𝑧,𝑑1,𝑧0𝑤,𝑑1,𝑤0<𝜖𝜙(𝜖)𝑛𝑛0.(2.8) This implies that 𝑑𝑥max𝑛+1,𝑥𝑛𝑦,𝑑𝑛,𝑦𝑛+1𝑧,𝑑𝑛+1,𝑧𝑛𝑤,𝑑𝑛,𝑤𝑛+1<𝜖𝜙(𝜖)𝑛𝑛0.(2.9) For 𝑚,𝑛, we will prove by induction on 𝑚 that 𝑑𝑥max𝑛,𝑥𝑚𝑦,𝑑𝑛,𝑦𝑚𝑧,𝑑𝑛,𝑧𝑚𝑤,𝑑𝑛,𝑤𝑚<𝜖𝑚𝑛𝑛0.(2.10) Since 𝜖𝜙(𝜖)<𝜖, then by using (2.9) we conclude that (2.10) holds when 𝑚=𝑛+1. Now suppose that (2.10) holds for 𝑚=𝑘. For 𝑚=𝑘+1, we have 𝑑𝑥𝑛,𝑥𝑘+1𝑥𝑑𝑛,𝑥𝑛+1𝑥+𝑑𝑛+1,𝑥𝑘+1𝐹𝑥𝜖𝜙(𝜖)+𝑑𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑥,𝐹𝑘,𝑦𝑘,𝑧𝑘,𝑤𝑘𝑑𝑥𝜖𝜙(𝜖)+𝜙max𝑛,𝑥𝑘𝑦,𝑑𝑛,𝑦𝑘𝑧,𝑑𝑛,𝑧𝑘𝑤,𝑑𝑛,𝑤𝑘<𝜖𝜙(𝜖)+𝜙(𝜖)=𝜖.(2.11) Similarly, we show that 𝑑𝑦𝑛,𝑦𝑘+1𝑑𝑧<𝜖,𝑛,𝑧𝑘+1𝑑𝑤<𝜖,𝑛,𝑤𝑘+1<𝜖.(2.12) Hence, we have 𝑑𝑥max𝑛,𝑥𝑘+1𝑦,𝑑𝑛,𝑦𝑘+1𝑧,𝑑𝑛,𝑧𝑘+1𝑤,𝑑𝑛,𝑤𝑘+1<𝜖.(2.13) Thus, (2.10) holds for all 𝑚𝑛𝑛0. Hence, (𝑥𝑛), (𝑦𝑛), (𝑧𝑛), and (𝑤𝑛) are Cauchy sequences in 𝑋.
Since 𝑋 is a complete metric space, there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that (𝑥𝑛), (𝑦𝑛), (𝑧𝑛) and (𝑤𝑛) converge to 𝑥, 𝑦, 𝑧, and 𝑤, respectively. Finally, we show that (𝑥,𝑦,𝑧,𝑤) is a quadruple fixed point of 𝐹. Since 𝐹 is continuous and (𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛)(𝑥,𝑦,𝑧,𝑤), we have 𝑥𝑛+1=𝐹(𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛)𝐹(𝑥,𝑦,𝑧,𝑤). By the uniqueness of limit, we get that 𝑥=𝐹(𝑥,𝑦,𝑧,𝑤). Similarly, we show that 𝑦=𝐹(𝑦,𝑧,𝑤,𝑥), 𝑧=𝐹(𝑧,𝑤,𝑥,𝑦), and 𝑤=𝐹(𝑤,𝑥,𝑦,𝑧). So, (𝑥,𝑦,𝑧,𝑤) is a quadruple fixed point of 𝐹.

By taking 𝜙(𝑡)=𝑘𝑡, where 𝑘[0,1), in Theorem 2.1, we have the following.

Corollary 2.2. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a continuous mapping such that 𝐹 has the mixed monotone property. Assume that there exists 𝑘[0,1) such that 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))𝑘max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}(2.14) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. 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 𝐹 has a quadruple fixed point.

As a consequence of Corollary 2.2, we have the following.

Corollary 2.3. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a continuous mapping such that 𝐹 has the mixed monotone property. Assume that there exist 𝑎1,𝑎2,𝑎3,𝑎4[0,1) with 𝑎1+𝑎2+𝑎3+𝑎4<1 such that 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))𝑎1𝑑(𝑥,𝑢)+𝑎2𝑑(𝑦,𝑣)+𝑎3𝑑(𝑧,)+𝑎4𝑑(𝑤,𝑙)(2.15) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. 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 𝐹 has a quadruple fixed point.

By adding an additional hypothesis, the continuity of 𝐹 in Theorem 2.1 can be dropped.

Theorem 2.4. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a mapping having the mixed monotone property. Assume that there exists 𝜙Φ such that 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))𝜙(max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)})(2.16) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. Assume also that 𝑋 has the following properties: (i) if a nondecreasing sequence 𝑥𝑛𝑥, then 𝑥𝑛𝑥 for all 𝑛𝐍, (ii) if a nonincreasing sequence 𝑦𝑛𝑦, then 𝑦𝑛𝑦 for all 𝑛𝐍.
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 𝐹 has a quadruple fixed point.

Proof. By following the same process in Theorem 2.1, we construct four Cauchy sequences (𝑥𝑛), (𝑦𝑛), (𝑧𝑛), and (𝑤𝑛) in 𝑋 with 𝑥1𝑥2𝑥𝑛𝑦,1𝑦2𝑦𝑛z,1𝑧2𝑧𝑛𝑤,1𝑤2𝑤𝑛,(2.17) such that 𝑥𝑛𝑥𝑋, 𝑦𝑛𝑦𝑋, 𝑧𝑛𝑧𝑋, and 𝑤𝑛𝑤𝑋. By the hypotheses on 𝑋, we have 𝑥𝑛𝑥, 𝑦𝑛𝑦, 𝑧𝑛𝑧, and 𝑤𝑛𝑤 for all 𝑛𝐍. From (2.16), we have 𝑑𝐹(𝑥,𝑦,𝑧,𝑤),𝑥𝑛+1𝐹𝑥=𝑑(𝑥,𝑦,𝑧,𝑤),𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑑𝜙max𝑥,𝑥𝑛,𝑑𝑦,𝑦𝑛,𝑑𝑧,𝑧𝑛,𝑑𝑤,𝑤𝑛,𝑑𝑦𝑛+1𝐹𝑦,𝐹(𝑦,𝑧,𝑤,𝑥)=𝑑𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑑𝑦,𝐹(𝑦,𝑧,𝑤,𝑥)𝜙max𝑛𝑧,𝑦,𝑑𝑛𝑤,𝑧,𝑑𝑛𝑥,𝑤,𝑑𝑛,𝑑𝐹,𝑥(𝑧,𝑤,𝑥,𝑦),𝑧𝑛+1𝐹𝑧=𝑑(𝑧,𝑤,𝑥,𝑦),𝐹𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑑𝜙max𝑥,𝑥𝑛,𝑑𝑦,𝑦𝑛,𝑑𝑧,𝑧𝑛,𝑑𝑤,𝑤𝑛,𝑑𝑤𝑛+1𝐹𝑤,𝐹(𝑤,𝑥,𝑦,𝑧)=𝑑𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑑𝑦,𝐹(𝑤,𝑥,𝑦,𝑧)𝜙max𝑛𝑧,𝑦,𝑑𝑛𝑤,𝑧,𝑑𝑛𝑥,𝑤,𝑑𝑛.,𝑥(2.18) From (2.18), we have 𝑑max𝐹(𝑥,𝑦,𝑧,𝑤),𝑥𝑛+1,𝑑𝑦𝑛+1,𝑑,𝐹(𝑦,𝑧,𝑤,𝑥)𝐹(𝑧,𝑤,𝑥,𝑦),𝑧𝑛+1,𝑑𝑤𝑛+1𝑑,𝐹(𝑤,𝑥,𝑦,𝑧)𝜙max𝑥,𝑥𝑛,𝑑𝑦,𝑦𝑛,𝑑𝑧,𝑧𝑛,𝑑𝑤,𝑤𝑛.(2.19) Letting 𝑛+ in (2.19), it follows that 𝑥=𝐹(𝑥,𝑦,𝑧,𝑤), 𝑦=𝐹(𝑦,𝑧,𝑤,𝑥), 𝑧=𝐹(𝑧,𝑤,𝑥,𝑦), and 𝑤=𝐹(𝑤,𝑥,𝑦,𝑧). Hence, (𝑥,𝑦,𝑧,𝑤) is a quadruple fixed point of 𝐹.

By taking 𝜙(𝑡)=𝑘𝑡, where 𝑘[0,1), in Theorem 2.4, we have the following result.

Corollary 2.5. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a mapping having the mixed monotone property. Assume that there exists 𝑘[0,1) such that 𝑑(𝐹(𝑥,𝑦,z,𝑤),𝐹(𝑢,𝑣,,𝑙))𝑘max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}(2.20) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. Assume also that 𝑋 has the following properties: (i) if a nondecreasing sequence 𝑥𝑛𝑥, then 𝑥𝑛𝑥 for all 𝑛𝐍, (ii) if a nonincreasing sequence 𝑦𝑛𝑦, then 𝑦𝑛𝑦 for all 𝑛𝐍. 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 𝐹 has a quadruple fixed point.

As a consequence of Corollary 2.5, we have the following.

Corollary 2.6. Let (𝑋,) be a partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a mapping having the mixed monotone property. Assume that there exist 𝑎1,𝑎2,𝑎3,𝑎4[0,1) with 𝑎1+𝑎2+𝑎3+𝑎4<1 such that 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))𝑎1𝑑(𝑥,𝑢)+𝑎2𝑑(𝑦,𝑣)+𝑎3𝑑(𝑧,)+𝑎4𝑑(𝑤,𝑙)(2.21) for all 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙. Assume that 𝑋 has the following properties: (i) if a nondecreasing sequence 𝑥𝑛𝑥, then 𝑥𝑛𝑥 for all 𝑛𝐍, (ii) if a nonincreasing sequence 𝑦𝑛𝑦, then 𝑦𝑛𝑦 for all 𝑛𝐍. 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 𝐹 has a quadruple fixed point.

Now we prove the following result.

Theorem 2.7. In addition to the hypotheses of Theorem 2.1 (resp., Theorem 2.4), suppose that 𝑥0𝑦0𝑧0𝑦0𝑥0𝑤0𝑧0𝑤0𝑦0𝑥0𝑦0𝑧0𝑤0𝑥0𝑤0𝑧0.(2.22) Then, 𝑥=𝑦=𝑧=𝑤.

Proof. Without loss of generality, we may assume that 𝑥0𝑦0, 𝑧0𝑦0, 𝑥0𝑤0, and 𝑧0𝑤0. By the mixed monotone property of 𝐹, we have 𝑥𝑛𝑦𝑛, 𝑧𝑛𝑦𝑛, 𝑥𝑛𝑤𝑛, and 𝑧𝑛𝑤𝑛 for all 𝑛𝐍. Thus, by (2.1), we have 𝑑𝑦𝑛+1,𝑥𝑛+1𝐹𝑦=𝑑𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑥,𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑑𝑦𝜙max𝑛,𝑥𝑛𝑧,𝑑𝑛,𝑦𝑛𝑤,𝑑𝑛,𝑧𝑛𝑥,𝑑𝑛,𝑤𝑛,𝑑𝑦(2.23)𝑛+1,𝑧𝑛+1𝐹𝑦=𝑑𝑛,𝑧𝑛,𝑤𝑛,𝑥𝑛𝑧,𝐹𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑑𝑦𝜙max𝑛,𝑧𝑛𝑧,𝑑𝑛,𝑤𝑛𝑤,𝑑𝑛,𝑥𝑛𝑥,𝑑𝑛,𝑦𝑛,𝑑𝑤(2.24)𝑛+1,𝑥𝑛+1𝐹𝑤=𝑑𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑥,𝐹𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑑𝑥𝜙max𝑛,𝑤𝑛𝑦,𝑑𝑛,𝑥𝑛𝑧,𝑑𝑛,𝑦𝑛𝑤,𝑑𝑛,𝑧𝑛,𝑑𝑤(2.25)𝑛+1,𝑧𝑛+1𝐹𝑤=𝑑𝑛,𝑥𝑛,𝑦𝑛,𝑧𝑛𝑧,𝐹𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑑𝑧𝜙max𝑛,𝑤𝑛𝑤,𝑑𝑛,𝑥𝑛𝑥,𝑑𝑛,𝑦𝑛𝑦,𝑑𝑛,𝑧𝑛.(2.26) By (2.23) and (2.26), we have 𝑑𝑦max𝑛+1,𝑥𝑛+1𝑦,𝑑𝑛+1,𝑧𝑛+1𝑤,𝑑𝑛+1,𝑥𝑛+1𝑤,𝑑𝑛+1,𝑧𝑛+1𝑑𝑦𝜙max𝑛,𝑥𝑛𝑦,𝑑𝑛,𝑧𝑛𝑤,𝑑𝑛,𝑥𝑛𝑤,𝑑𝑛,𝑧𝑛𝜙2𝑑𝑦max𝑛1,𝑥𝑛1𝑦,𝑑𝑛1,𝑧𝑛1𝑤,𝑑𝑛1,𝑥𝑛1𝑤,𝑑𝑛1,𝑧𝑛1𝜙𝑛+1𝑑𝑦max0,𝑥0𝑦,𝑑0,𝑧0𝑤,𝑑0,𝑥0𝑤,𝑑0,𝑧0.(2.27) By letting 𝑛+ in (2.27) and using the property of 𝜙 and the fact that 𝑑 is continuous on its variable, we get that max{𝑑(𝑦,𝑥),𝑑(𝑦,𝑧),𝑑(𝑤,𝑥),𝑑(𝑤,𝑧)}=0. Hence, 𝑦=𝑧=𝑥=𝑤.

Corollary 2.8. In addition to the hypotheses of Corollary 2.3 (resp., Corollary 2.5), suppose that 𝑥0𝑦0𝑧0𝑦0𝑥0𝑤0𝑧0𝑤0𝑦0𝑥0𝑦0𝑧0𝑤0𝑥0𝑤0𝑧0.(2.28) Then, 𝑥=𝑦=𝑧=𝑤.

Example 2.9. Let 𝑋=[0,1] with usual order. Define 𝑑𝑋×𝑋𝑋 by 𝑑(𝑥,𝑦)=|𝑥𝑦|. Define 𝐹𝑋4𝑋 by 𝐹1(𝑥,𝑦,𝑧,𝑤)=0,max{𝑦,𝑤}min{𝑥,𝑧},4(min{𝑥,𝑧}max{𝑦,𝑤}),max{𝑦,𝑤}<min{𝑥,𝑧}.(2.29) Then, (a)(𝑋,𝑑,) is a complete ordered metric space, (b)for 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙, we have that 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)},(2.30)(c)holds for all 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙,(d)𝐹 has the mixed monotone property.

Proof. To prove (𝑏), given 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,,𝑙𝑋 with 𝑥𝑢, 𝑦𝑣, 𝑧, and 𝑤𝑙, we examine the following cases.
Case 1. If max{𝑦,𝑤}min{𝑥,𝑧}, and max{𝑣,𝑙}min{𝑢,𝑤}. Here, we have 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=02max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.31)
Case 2. If max{𝑦,𝑤}min{𝑥,𝑧} and max{𝑣,𝑙}<min{𝑢,}. This case is impossible since 𝑦𝑣<min{𝑢,}min{𝑥,𝑧},𝑤𝑙<min{𝑢,}min{𝑥,𝑧}.(2.32) So, max{𝑦,𝑤}<min{𝑥,𝑧}.(2.33)
Case 3. If max{𝑦,𝑤}<min{𝑥,𝑧} and max{𝑣,𝑙}min{𝑢,}.
This case will have different possibilities.
(i) Let max{𝑦,𝑤}=𝑦 and max{𝑣,𝑙}=𝑣. Suppose that 𝑣; then 𝑦𝑣𝑦 and hence min{𝑥,z}max{𝑦,𝑤}=min{𝑥,𝑧}𝑦𝑧𝑦=𝑧+𝑦𝑧+𝑣𝑦=𝑑(𝑧,)+𝑑(𝑦,𝑣)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.34) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.35) Suppose that 𝑢𝑣; then 𝑢𝑦𝑣𝑦 and hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑦𝑥𝑦=𝑥𝑢+𝑢𝑦(𝑥𝑢)+(𝑣𝑦)=𝑑(𝑥,𝑢)+𝑑(𝑣,𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.36) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.37)
(ii) Let max{𝑦,𝑤}=𝑦 and max{𝑣,𝑙}=𝑙. Suppose that 𝑙; then 𝑦𝑙𝑦 and (since 𝑤𝑦) hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑦𝑧𝑦=𝑧+𝑦𝑧+𝑙𝑦𝑧+𝑙𝑤=𝑑(𝑧,)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.38) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.39)
Suppose that 𝑢𝑙; then 𝑢𝑦𝑙𝑦 and (since 𝑤𝑦 ) hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑦𝑥𝑦=𝑥𝑢+𝑢𝑦(𝑥𝑢)+(𝑙𝑦)𝑥𝑢+𝑙𝑤=𝑑(𝑥,𝑢)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.40) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.41)
(iii) Let max{𝑦,𝑤}=𝑤 and max{𝑣,𝑙}=𝑣. Suppose that 𝑣; then 𝑤𝑣𝑤, but 𝑦𝑤, and hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑤𝑧𝑤=𝑧+𝑤𝑧+𝑣𝑤𝑧+𝑣𝑦=𝑑(𝑧,)+𝑑(𝑦,𝑣)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.42) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑤)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.43)
Suppose that 𝑢𝑣; then 𝑢𝑤𝑣𝑤 and hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑤𝑥𝑤=𝑥𝑢+𝑢𝑤(𝑥𝑢)+(𝑣𝑤)𝑥𝑢+𝑣𝑦=𝑑(𝑥,𝑢)+𝑑(𝑣,𝑦)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.44) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑤)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.45)
(iv) Let max{𝑦,𝑤}=𝑤 and max{𝑣,𝑙}=𝑙. Suppose that 𝑙; then 𝑤𝑙𝑤 and hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑤𝑧𝑤=𝑧+𝑤𝑧+𝑙𝑤=𝑑(𝑧,)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.46) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑤)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.47)
Suppose that 𝑢𝑙; then 𝑢𝑤𝑙𝑤 and hence min{𝑥,𝑧}max{𝑦,𝑤}=min{𝑥,𝑧}𝑤𝑥𝑤=𝑥𝑢+𝑢𝑤(𝑥𝑢)+(𝑙𝑤)=𝑑(𝑥,𝑢)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.48) Therefore, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))=𝑑4=1(min{𝑥,𝑧}max{𝑦,𝑤}),041(min{𝑥,𝑧}𝑤)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.49)
Case 4. (i) If max{𝑦,𝑤}<min{𝑥,𝑧} and max{𝑣,𝑙}<min{𝑢,}.
Since 𝑥𝑢 and 𝑧, then min{𝑥,𝑧}min{𝑢,}, and also since 𝑦𝑣 and 𝑤𝑙, then max{𝑣,𝑙}max{𝑦,𝑤}. Thus, 1𝑑(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑤))=𝑑41(min{𝑥,𝑧}max{𝑦,𝑤}),4=1(min{𝑢,}max{𝑣,𝑙})4||||.(min{𝑥,𝑧}min{𝑢,})+(max{𝑣,𝑙}max{𝑦,𝑤})(2.50)
(ii) If min{𝑢,}=𝑢 and max{𝑣,𝑙}=𝑣, then min{𝑥,𝑧}min{𝑢,}𝑥𝑢 and max{𝑣,𝑙}max{𝑦,𝑤}𝑣𝑦. Thus, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))4[]=1(𝑥𝑢)+(𝑣𝑦)4[]1𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.51)
(iii) If min{𝑢,}= and max{𝑣,𝑙}=𝑣, then min{𝑥,𝑧}min{𝑢,}𝑧 and max{𝑣,𝑙}max{𝑦,𝑤}𝑣𝑦, hence 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))4[]=1(𝑧)+(𝑣𝑦)4[]1𝑑(𝑧,)+𝑑(𝑦,𝑣)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.52)
(iv) If min{𝑢,}=𝑢 and max{𝑣,𝑙}=𝑙, then min{𝑥,𝑧}min{𝑢,}𝑥𝑢 and max{𝑣,𝑙}max{𝑦,𝑤}𝑙𝑤, and hence 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))4[]=1(𝑥𝑢)+(𝑙𝑤)4[]1𝑑(𝑥,𝑢)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.53)
(v) If min{𝑢,}= and max{𝑣,𝑙}=𝑙, then min{𝑥,𝑧}min{𝑢,}𝑧 and max{𝑣,𝑙}max{𝑦,𝑤}𝑙𝑤, and hence 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,,𝑙))4[]=1(𝑧)+(𝑙𝑤)4[]1𝑑(𝑧,)+𝑑(𝑤,𝑙)2max{𝑑(𝑥,𝑢),𝑑(𝑦,𝑣),𝑑(𝑧,),𝑑(𝑤,𝑙)}.(2.54)
To prove (c), let 𝑥,𝑦,𝑧,𝑤𝑋. To show that 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nondecreasing in 𝑥, let 𝑥1,𝑥2𝑋 with 𝑥1𝑥2.
If max{𝑦,𝑤}min{𝑥1,𝑧}, then 𝐹(𝑥1,𝑦,𝑧,𝑤)=0𝐹(𝑥2,𝑦,𝑧,𝑤). If max{𝑦,𝑤}<min{𝑥1,𝑧}, then 𝐹𝑥1=1,𝑦,𝑧,𝑤4𝑥min11,𝑧max{𝑦,𝑤}4𝑥min2𝑥,𝑧max{𝑦,𝑤}=𝐹2.,𝑦,𝑧,𝑤(2.55) Therefore, 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nondecreasing in 𝑥. Similarly, we may show that 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nondecreasing in 𝑧.
To show that 𝐹(𝑥,y,𝑧,𝑤) is monotone nonincreasing in 𝑦, let 𝑦1,𝑦2𝑋 with 𝑦1𝑦2. If max{𝑦2,𝑤}min{𝑥,𝑧}, then 𝐹(𝑥,𝑦2,𝑧,𝑤)=0𝐹(𝑥1,𝑦,𝑧,𝑤). If max{𝑦2,𝑤}<min{𝑥,𝑧}, then 𝐹𝑥,𝑦2=1,𝑧,𝑤4𝑦min{𝑥,𝑧}max21,𝑤4𝑦min{𝑥,𝑧}max1,𝑤=𝐹𝑥,𝑦2.,𝑧,𝑤(2.56) Therefore, 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nonincreasing in 𝑦. Similarly, we may show that 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nonincreasing in 𝑤.
Thus, by Theorem 2.1 (let 𝜙(𝑡)=(𝑡/2)), 𝐹 has a unique quadruple fixed point, namely, (0,0,0,0). Since the condition of Theorem 2.7 is satisfied, (0,0,0,0) is the unique quadruple fixed point of 𝐹.

Remark 2.10. We notice that for, 𝐹𝑋2𝑛𝑋,(𝑛), it is very natural to consider the analog of Theorem 2.1–Theorem 2.7 to get fixed points. Moreover, for 𝐹𝑋2𝑛+1𝑋(𝑛), the analog of Theorem 7–Theorem 11 of Berinde and Borcut [2] yields fixed points.