Abstract

The notion of coupled fixed point is introduced in by Gnana Bhaskar and Lakshmikantham (2006). Very recently, the concept of tripled fixed point is introduced by Berinde and Borcut (2011). In this paper, quadruple fixed point is introduced, and some new fixed point theorems are obtained.

1. Introduction and Preliminaries

Very recently, Berinde and Borcut [1] introduced the concept of triple fixed point. Their contributions are inspired from the remarkable paper of Gnana Bhaskar and Lakshmikantham [2] in which they introduced the notion of coupled fixed point and proved some fixed point theorems under certain condition. Later, Lakshmikantham and Ljubomir Ćirić in [3] extended these results by defining of 𝑔-monotone property. Many authors focused on coupled fixed point theory and proved remarkable results (see, e.g., [49]).

Here we recall the basic definitions and results from which triple and quadruple fixed point [10] notions are inspired. Let (𝑋,𝑑) be a metric space and 𝑋2=𝑋×𝑋. Then the mapping 𝜌=𝑋2×𝑋2𝐸 such that 𝜌((𝑥1,𝑦1),(𝑥2,𝑦2))=𝑑(𝑥1,𝑥2)+𝑑(𝑦1,𝑦2) forms a metric on 𝑋2. A sequence ({𝑥𝑛},{𝑦𝑛})𝑋2 is said to be a double sequence of 𝑋.

Definition 1.1 (see [2]). Let (𝑋,) be partially ordered set and 𝐹𝑋×𝑋𝑋. 𝐹 is said to have mixed monotone property if 𝐹(𝑥,𝑦) is monotone nondecreasing in 𝑥 and is monotone nonincreasing in 𝑦, that is, for any 𝑥,𝑦𝑋, 𝑥1𝑥2𝑥𝐹1𝑥,𝑦𝐹2,𝑦,for𝑥1,𝑥2𝑦𝑋,1𝑦2𝐹𝑥,𝑦2𝐹𝑥,𝑦1,for𝑦1,𝑦2𝑋.(1.1)

Definition 1.2 (see [2]). An element (𝑥,𝑦)𝑋×𝑋 is said to be a couple fixed point of the mapping 𝐹𝑋×𝑋𝑋 if 𝐹(𝑥,𝑦)=𝑥,𝐹(𝑦,𝑥)=𝑦.(1.2)

Throughout this paper, let (𝑋,) be partially ordered set and 𝑑 a metric on 𝑋 such that (𝑋,𝑑) is a complete metric space. Further, the product spaces 𝑋×𝑋 satisfy the following:(𝑢,𝑣)(𝑥,𝑦)𝑢𝑥,𝑦𝑣;(𝑥,𝑦),(𝑢,𝑣)𝑋×𝑋.(1.3)

The following two results of Gnana Bhaskar and Lakshmikantham in [2] were extended to class of cone metric spaces in [7].

Theorem 1.3. Let 𝐹𝑋×𝑋𝑋 be a continuous mapping having the mixed monotone property on 𝑋. Assume that there exists a 𝑘[0,1) with 𝑘𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))2[]𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣),𝑢𝑥,𝑦𝑣.(1.4) If there exists 𝑥0,𝑦0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0) and 𝐹(𝑦0,𝑥0)𝑦0, then, there exists 𝑥,𝑦𝑋 such that 𝑥=𝐹(𝑥,𝑦) and 𝑦=𝐹(𝑦,𝑥).

Theorem 1.4. Let 𝐹𝑋×𝑋𝑋 be a mapping having the mixed monotone property on 𝑋. Suppose that 𝑋 has the following properties: (i)if a nondecreasing sequence {𝑥𝑛}𝑥, then 𝑥𝑛𝑥, for all n; (ii)if a nonincreasing sequence {𝑦𝑛}𝑦, then 𝑦𝑦𝑛, for all n. Assume that there exists a 𝑘[0,1) with 𝑘𝑑(𝐹(𝑥,𝑦),𝐹(𝑢,𝑣))2[]𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣),𝑢𝑥,𝑦𝑣.(1.5) If there exists 𝑥0,𝑦0𝑋 such that 𝑥0𝐹(𝑥0,𝑦0) and 𝐹(𝑦0,𝑥0)𝑦0, then, there exists 𝑥,𝑦𝑋 such that 𝑥=𝐹(𝑥,𝑦) and 𝑦=𝐹(𝑦,𝑥).

Inspired by Definition 1.1, Berinde and Borcut [1] introduced the following partial order on the product space 𝑋3=𝑋×𝑋×𝑋: (𝑢,𝑣,𝑤)(𝑥,𝑦,𝑧)i𝑥𝑢,𝑦𝑣,𝑧𝑤,(1.6) where (𝑢,𝑣,𝑤),(𝑥,𝑦,𝑧)𝑋3. Regarding this partial order, we state the definition of the following mapping.

Definition 1.5 (see [1]). Let (𝑋,) be partially ordered set and 𝐹𝑋3𝑋. We say that 𝐹 has the mixed monotone property if 𝐹(𝑥,𝑦,𝑧) is monotone nondecreasing in 𝑥 and 𝑦, and it is monotone nonincreasing in 𝑦, that is, for any 𝑥,𝑦,𝑧𝑋𝑥1,𝑥2𝑋,𝑥1𝑥2𝑥𝐹1𝑥,𝑦,𝑧𝐹2,𝑦,𝑦,𝑧1,𝑦2𝑋,𝑦1𝑦2𝐹𝑥,𝑦1,𝑧𝐹𝑥,𝑦2,𝑧,𝑧1,𝑧2𝑋,𝑧1𝑧2𝐹𝑥,𝑦,𝑧1𝐹𝑥,𝑦,𝑧2.(1.7)

Definition 1.6 (see [1]). An element (𝑥,𝑦,𝑧,𝑤)𝑋3 is called a quadruple fixed point of 𝐹𝑋3𝑋 if 𝐹(𝑥,𝑦,𝑧)=𝑥,𝐹(𝑦,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥)=𝑧.(1.8)

For a metric space (𝑋,𝑑), the function 𝜌𝑋3[0,), given by 𝜌((𝑥,𝑦,𝑧),(𝑢,𝑣,𝑤))=𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑤),(1.9) forms a metric space on 𝑋3; that is, (𝑋3,𝜌) is a metric induced by (𝑋,𝑑).

Theorem 1.7. Let (𝑋,) be partially ordered set, and let (𝑋,𝑑) be a complete metric space. Let 𝐹𝑋×𝑋×𝑋𝑋 be a continuous mapping having the mixed monotone property on 𝑋. Assume that there exist constants 𝑎,𝑏,𝑐[0,1) such that 𝑎+𝑏+𝑐<1 for which 𝑑(𝐹(𝑥,𝑦,𝑧),𝐹(𝑢,𝑣,𝑤))𝑎𝑑(𝑥,𝑢)+𝑏𝑑(𝑦,𝑣)+𝑐𝑑(𝑧,𝑤)(1.10) for all 𝑥𝑢,𝑦𝑣,𝑧𝑤. If there exist 𝑥0,𝑦0,𝑧0𝑋 such that 𝑥0𝑥𝐹0,𝑦0,𝑧0,𝑦0𝑦𝐹0,𝑥0,𝑦0,𝑧0𝑥𝐹0,𝑦0,𝑧0,(1.11) then there exist 𝑥,𝑦,𝑧𝑋 such that 𝐹(𝑥,𝑦,𝑧)=𝑥,𝐹(𝑦,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥)=𝑧.(1.12)

The aim of this paper is to introduce the concept of quadruple fixed point and prove the related fixed point theorems.

2. Quadruple Fixed Point Theorems

Let (𝑋,) be partially ordered set and (𝑋,𝑑) a complete metric space. We consider the following partial order on the product space 𝑋4=𝑋×𝑋×𝑋×𝑋: (𝑢,𝑣,𝑟,𝑡)(𝑥,𝑦,𝑧,𝑤)i𝑥𝑢,𝑦𝑣,𝑧𝑟,𝑡𝑤,(2.1) where (𝑢,𝑣,𝑟,𝑡),(𝑥,𝑦,𝑧,𝑤)𝑋4. Regarding this partial order, we state the definition of the following mapping.

Definition 2.1. Let (𝑋,) be partially ordered set and 𝐹𝑋4𝑋. We say that 𝐹 has the mixed monotone property if 𝐹(𝑥,𝑦,𝑧,𝑤) is monotone nondecreasing in 𝑥 and 𝑧, and it is monotone nonincreasing in 𝑦 and 𝑤, that is, for any 𝑥,𝑦,𝑧,𝑤𝑋𝑥1,𝑥2𝑋,𝑥1𝑥2𝑥𝐹1𝑥,𝑦,𝑧,𝑤𝐹2,𝑦,𝑦,𝑧,𝑤1,𝑦2𝑋,𝑦1𝑦2𝐹𝑥,𝑦1,𝑧,𝑤𝐹𝑥,𝑦2,𝑧,𝑧,𝑤1,𝑧2𝑋,𝑧1𝑧2𝐹𝑥,𝑦,𝑧1,𝑤𝐹𝑥,𝑦,𝑧2,𝑤,𝑤1,𝑤2𝑋,𝑤1𝑤2𝐹𝑥,𝑦,𝑧,𝑤1𝐹𝑥,𝑦,𝑧,𝑤2.(2.2)

Definition 2.2. An element (𝑥,𝑦,𝑧)𝑋4 is called a triple fixed point of 𝐹𝑋4𝑋 if 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤.(2.3)

For a metric space (𝑋,𝑑), the function 𝜌𝑋4[0,), given by 𝜌((𝑥,𝑦,𝑧,𝑤),(𝑢,𝑣,𝑟,𝑡))=𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑟)+𝑑(𝑤,𝑡),(2.4) forms a metric space on 𝑋4; that is, (𝑋4,𝜌) is a metric induced by (𝑋,𝑑). Let Φ denote the all functions 𝜙[0,)[0,) which satisfies that lim𝑡𝑟𝜙(𝑡)>0 for all 𝑟>0 and lim𝑡0+𝜙(𝑡)=0.

The aim of this paper is to prove the following theorem.

Theorem 2.3. Let (𝑋,) be partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a mapping having the mixed monotone property on 𝑋. Assume that for all 𝑥𝑢,𝑦𝑣,𝑧𝑤, 1𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑟,𝑡))4[]1𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑟)+𝑑(𝑤,𝑡)𝜙4[],𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑟)+𝑑(𝑤,𝑡)(2.5) where 𝜙Φ. Suppose there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝑥𝐹0,𝑦0,𝑧0,𝑤0,𝑦0𝑥𝐹0,𝑤0,𝑧0,𝑦0,𝑧0𝑧𝐹0,𝑦0,𝑥0,𝑤0,𝑤0𝑧𝐹0,𝑤0,𝑥0,𝑦0.(2.6) Suppose 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. Then there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤.(2.7)

Proof. We construct a sequence {(𝑥𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛)} in the following way. Set𝑥1𝑥=𝐹0,𝑦0,𝑧0,𝑤0𝑥0,𝑦1𝑥=𝐹0,𝑤0,𝑧0,𝑦0𝑦0,𝑧1𝑧=𝐹0,𝑦0,𝑥0,𝑤0𝑧0,𝑤1𝑧=𝐹0,𝑤0,𝑥0,𝑦0𝑤0,(2.8) and by the mixed monotone property of 𝐹, for 𝑛1, inductively we get 𝑥𝑛𝑥=𝐹𝑛1,𝑦𝑛1,𝑧𝑛1,𝑤𝑛1𝑥𝑛1𝑥0,𝑦𝑛𝑥=𝐹𝑛1,𝑤𝑛1,𝑧𝑛1,𝑦𝑛1𝑦𝑛1𝑦0,𝑧𝑛𝑧=𝐹𝑛1,𝑦𝑛1,𝑥𝑛1,𝑤𝑛1𝑧𝑛1𝑧0,𝑤𝑛𝑧=𝐹𝑛1,𝑤𝑛1,𝑥𝑛1,𝑦𝑛1𝑤𝑛1𝑤0.(2.9) Due to (2.5) and (2.9), we have 𝑑𝑥1,𝑥2𝐹𝑥=𝑑0,𝑦0,𝑧0,𝑤0𝑥,𝐹1,𝑦1,𝑧1,𝑤114𝑑𝑥0,𝑥1𝑦+𝑑0,𝑦1𝑧+𝑑0,𝑧1𝑤+𝑑0,𝑤11𝜙4𝑑𝑥0,𝑥1𝑦+𝑑0,𝑦1𝑧+𝑑0,𝑧1𝑤+𝑑0,𝑤114𝑑𝑥0,𝑥1𝑦+𝑑0,𝑦1𝑧+𝑑0,𝑧1𝑤+𝑑0,𝑤1,𝑑𝑦(2.10)1,𝑦2𝐹𝑥=𝑑0,𝑤0,𝑧0,𝑦0𝑥,𝐹1,𝑤1,𝑧1,𝑦114𝑑𝑥0,𝑥1𝑤+𝑑0,𝑤1𝑧+𝑑0,𝑧1𝑦+𝑑0,𝑦11𝜙4𝑑𝑥0,𝑥1𝑤+𝑑0,𝑤1𝑧+𝑑0,𝑧1𝑦+𝑑0,𝑦114𝑑𝑥0,𝑥1𝑤+𝑑0,𝑤1𝑧+𝑑0,𝑧1𝑦+𝑑0,𝑦1,𝑑𝑧(2.11)1,𝑧2𝐹𝑧=𝑑0,𝑦0,𝑥0,𝑤0𝑧,𝐹1,𝑦1,𝑥1,𝑤114𝑑𝑧0,𝑧1𝑦+𝑑0,𝑦1𝑥+𝑑0,𝑥1𝑤+𝑑0,𝑤11𝜙4𝑑𝑧0,𝑧1𝑦+𝑑0,𝑦1𝑥+𝑑0,𝑥1𝑤+𝑑0,𝑤114𝑑𝑧0,𝑧1𝑦+𝑑0,𝑦1𝑥+𝑑0,𝑥1𝑤+𝑑0,𝑤1,𝑑𝑤(2.12)1,𝑤2𝐹𝑧=𝑑0,𝑤0,𝑥0,𝑦0𝑧,𝐹1,𝑤1,𝑥1,𝑦114𝑑𝑧0,𝑧1𝑤+𝑑0,𝑤1𝑥+𝑑0,𝑥1𝑦+𝑑0,𝑦11𝜙4𝑑𝑧0,𝑧1𝑤+𝑑0,𝑤1𝑥+𝑑0,𝑥1𝑦+𝑑0,𝑦114𝑑𝑧0,𝑧1𝑤+𝑑0,𝑤1𝑥+𝑑0,𝑥1𝑦+𝑑0,𝑦1.(2.13) Regarding (2.5) together with (2.14) we have 𝑑𝑥2,𝑥3𝐹𝑥=𝑑1,𝑦1,𝑧1,𝑤1𝑥,𝐹2,𝑦2,𝑧2,𝑤214𝑑𝑥1,𝑥2𝑦+𝑑1,𝑦2𝑧+𝑑1,𝑧2𝑤+d1,𝑤21𝜙4𝑑𝑥1,𝑥2𝑦+𝑑1,𝑦2𝑧+𝑑1,𝑧2𝑤+𝑑1,𝑤214𝑑𝑥1,𝑥2𝑦+𝑑1,𝑦2𝑧+𝑑1,𝑧2𝑤+𝑑1,𝑤2,𝑑𝑦2,𝑦3𝐹𝑥=𝑑1,𝑤1,𝑧1,𝑦1𝑥,𝐹2,𝑤2,𝑧2,𝑦214𝑑𝑥1,𝑥2𝑤+𝑑1,𝑤2𝑧+𝑑1,𝑧2𝑦+𝑑1,𝑦21𝜙4𝑑𝑥1,𝑥2𝑤+𝑑1,𝑤2𝑧+𝑑1,𝑧2𝑦+𝑑1,𝑦214𝑑𝑥1,𝑥2𝑤+𝑑1,𝑤2𝑧+𝑑1,𝑧2𝑦+𝑑1,𝑦2,𝑑𝑧2,𝑧3𝐹𝑧=𝑑1,𝑦1,𝑥1,𝑤1𝑧,𝐹2,𝑦2,𝑥2,𝑤214𝑑𝑧1,𝑧2𝑦+𝑑1,𝑦2𝑥+𝑑1,𝑥2𝑤+𝑑1,𝑤21𝜙4𝑑𝑧1,𝑧2𝑦+𝑑1,𝑦2𝑥+𝑑1,𝑥2𝑤+𝑑1,𝑤214𝑑𝑧1,𝑧2𝑦+𝑑1,𝑦2𝑥+𝑑1,𝑥2𝑤+𝑑1,𝑤2,𝑑𝑤2,𝑤3𝐹𝑧=𝑑1,𝑤1,𝑥1,𝑦2𝑧,𝐹2,𝑤2,𝑥2,𝑦214𝑑𝑧1,𝑧2𝑤+𝑑1,𝑤2𝑥+𝑑1,𝑥2𝑦+𝑑1,𝑦21𝜙4𝑑𝑧1,𝑧2𝑤+𝑑1,𝑤2𝑥+𝑑1,𝑥2𝑦+𝑑1,𝑦214𝑑𝑧1,𝑧2𝑤+𝑑1,𝑤2𝑥+𝑑1,𝑥2𝑦+𝑑1,𝑦2.(2.14)
Inductively we have 𝑑𝑥𝑛+1,𝑥𝑛+2𝐹𝑥=𝑑𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛𝑥,𝐹𝑛+1,𝑦𝑛+1,𝑧𝑛+1,𝑤𝑛+114𝑑𝑥𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑧+𝑑𝑛,𝑧n+1𝑤+𝑑𝑛,𝑤𝑛+11𝜙4𝑑𝑥𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑧+𝑑𝑛,𝑧𝑛+1𝑤+𝑑𝑛,𝑤𝑛+114𝑑𝑥𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑧+𝑑𝑛,𝑧𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1,𝑑𝑦𝑛+1,𝑦𝑛+2𝐹𝑥=𝑑𝑛,𝑤𝑛,𝑧𝑛,𝑦𝑛𝑥,𝐹𝑛+1,𝑤𝑛+1,𝑧𝑛+1,𝑦𝑛+114𝑑𝑥𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑧+𝑑𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+11𝜙4𝑑𝑥𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑧+𝑑𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+114𝑑𝑥𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑧+𝑑𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1,𝑑𝑧𝑛+1,𝑧𝑛+2𝐹𝑧=𝑑𝑛,𝑦𝑛,𝑥𝑛,𝑤𝑛𝑧,𝐹𝑛+1,𝑦𝑛+1,𝑥𝑛+1,𝑤𝑛+114𝑑𝑧𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+11𝜙4𝑑𝑧𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+114𝑑𝑧𝑛,𝑧𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1,𝑑𝑤𝑛+1,𝑤𝑛+2𝐹𝑧=𝑑𝑛,𝑤𝑛,𝑥𝑛,𝑦𝑛𝑧,𝐹𝑛+1,𝑤𝑛+1,𝑥𝑛+1,𝑦𝑛+114𝑑𝑧𝑛,𝑧𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+11𝜙4𝑑𝑧𝑛,𝑧𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+114𝑑𝑧𝑛,𝑧𝑛+1𝑤+𝑑𝑛,𝑤𝑛+1𝑥+𝑑𝑛,𝑥𝑛+1𝑦+𝑑𝑛,𝑦𝑛+1.(2.15) Set 𝛿𝑛=𝑑(𝑥𝑛,𝑥𝑛1)+𝑑(𝑦𝑛,𝑦𝑛1)+𝑑(𝑧𝑛,𝑧𝑛1)+𝑑(𝑤𝑛,𝑤𝑛1). Due to (2.15), we conclude that {𝛿𝑛} is a nonincreasing sequence. Since it is bounded below, there is some 𝛿0 such that lim𝑛𝛿𝑛=𝛿.(2.16) We shall show that 𝛿=0. Suppose, to the contrary, that 𝛿>0.
Again by (2.15) and (2.9) together with (2.5), we have 𝛿𝑛𝛿𝑛14𝜙4𝛿𝑛.(2.17)
Letting 𝑛 in (2.17) and having in mind that we suppose lim𝑡𝑟𝜙(𝑡)>0 for all 𝑟>0 and lim𝑡0+𝜙(𝑡)=0, we have 1𝛿𝛿4𝜙4𝛿,(2.18) which is a contradiction. Thus, 𝛿=0, that is, lim𝑛𝛿𝑛=lim𝑛𝑑𝑥𝑛,𝑥𝑛1𝑦+𝑑𝑛,𝑦𝑛1𝑧+𝑑𝑛,𝑧𝑛1𝑤+𝑑𝑛,𝑤𝑛1=0.(2.19)
Now, we shall prove that {𝑥𝑛},{𝑦𝑛},{𝑧𝑛}, and {𝑤𝑛} are Cauchy sequences. Suppose, to the contrary, that at least one of {𝑥𝑛},{𝑦𝑛},{𝑧𝑛}, and {𝑤𝑛} is not Cauchy. So, there exists an 𝜀>0 for which we can find subsequences {𝑥𝑛(𝑘)}, {𝑥𝑛(𝑘)} of {𝑥𝑛} and {𝑦𝑛(𝑘)}, {𝑦𝑛(𝑘)} of {𝑦𝑛} and {𝑧𝑛(𝑘)}, {𝑧𝑛(𝑘)} of {𝑧𝑛} and {𝑤𝑛(𝑘)}, {𝑤𝑛(𝑘)} of {𝑤𝑛} with 𝑛(𝑘)>𝑚(𝑘)𝑘 such that 𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝜀.(2.20) Additionally, corresponding to 𝑚(𝑘), we may choose 𝑛(𝑘) such that it is the smallest integer satisfying (2.20) and 𝑛(𝑘)>𝑚(𝑘)𝑘. Thus, 𝑑𝑥𝑛(𝑘)1,𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘)1,𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘)1,𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘)1,𝑤𝑚(𝑘)<𝜀.(2.21) By using triangle inequality and having (2.20), (2.21) in mind 𝜀𝑡𝑘𝑥=𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑥𝑑𝑛(𝑘),𝑥𝑛(𝑘)1𝑥+𝑑𝑛(𝑘)1,𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑛(𝑘)1𝑦+𝑑𝑛(𝑘)1,𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑛(𝑘)1𝑧+𝑑𝑛(𝑘)1,𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑛(𝑘)1𝑤+𝑑𝑛(𝑘)1,𝑤𝑚(𝑘)𝑥<𝑑𝑛(𝑘),𝑥𝑛(𝑘)1𝑦+𝑑𝑛(𝑘),𝑦𝑛(𝑘)1𝑧+𝑑𝑛(𝑘),𝑧𝑛(𝑘)1𝑤+𝑑𝑛(𝑘),𝑤𝑛(𝑘)1+𝜀.(2.22) Letting 𝑘 in (2.22) and using (2.16) lim𝑘𝑡𝑘=lim𝑘𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)=𝜀.(2.23) Again by triangle inequality, 𝑡𝑘𝑥=𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑥𝑑𝑛(𝑘),𝑥𝑛(𝑘)+1𝑥+𝑑𝑛(𝑘)+1,𝑥𝑚(𝑘)+1𝑥+𝑑𝑚(𝑘)+1,𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑛(𝑘)+1𝑦+𝑑𝑛(𝑘)+1,𝑦𝑚(𝑘)+1𝑦+𝑑𝑚(𝑘)+1,𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑛(𝑘)+1𝑧+𝑑𝑛(𝑘)+1,𝑧𝑚(𝑘)+1𝑧+𝑑𝑚(𝑘)+1,𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑛(𝑘)+1𝑤+𝑑𝑛(𝑘)+1,𝑤𝑚(𝑘)+1𝑤+𝑑𝑚(𝑘)+1,𝑤𝑚(𝑘)𝛿𝑛(𝑘)+1+𝛿𝑚(𝑘)+1𝑥+𝑑𝑛(𝑘)+1,𝑥𝑚(𝑘)+1𝑦+𝑑𝑛(𝑘)+1,𝑦𝑚(𝑘)+1𝑧+𝑑𝑛(𝑘)+1,𝑧𝑚(𝑘)+1𝑤+𝑑𝑛(𝑘)+1,𝑤𝑚(𝑘)+1.(2.24) Since 𝑛(𝑘)>𝑚(𝑘), then 𝑥𝑛(𝑘)𝑥𝑚(𝑘),𝑦𝑛(𝑘)𝑦𝑚(𝑘),𝑧𝑛(𝑘)𝑧𝑚(𝑘),𝑤𝑛(𝑘)𝑤𝑚(𝑘).(2.25) Hence from (2.25), (2.9) and (2.5), we have 𝑑𝑥𝑛(𝑘)+1,𝑥𝑚(𝑘)+1𝐹𝑥=𝑑𝑛(𝑘),𝑦𝑛(𝑘),𝑧𝑛(𝑘),𝑤𝑛(𝑘)𝑥,𝐹𝑚(𝑘),𝑦𝑚(𝑘),𝑧𝑚(𝑘),𝑤𝑚(𝑘)14𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)1𝜙4𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘),𝑑𝑦𝑛(𝑘)+1,𝑦𝑚(𝑘)+1𝐹𝑥=𝑑𝑛(𝑘),𝑤𝑛(𝑘),𝑧𝑛(𝑘),𝑦𝑛(𝑘)𝑥,𝐹𝑚(𝑘),𝑤𝑚(𝑘),𝑧𝑚(𝑘),𝑦𝑚(𝑘)14𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)1𝜙4𝑑𝑥𝑛(𝑘),𝑥𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑧+𝑑𝑛(𝑘),𝑧𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘),𝑑𝑧𝑛(𝑘)+1,𝑧𝑚(𝑘)+1𝐹𝑧=𝑑𝑛(𝑘),𝑦𝑛(𝑘),𝑥𝑛(𝑘),𝑤𝑛(𝑘)𝑧,𝐹𝑚(𝑘),𝑦𝑚(𝑘),𝑥𝑚(𝑘),𝑤𝑚(𝑘)14𝑑𝑧𝑛(𝑘),𝑧𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑥+𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)1𝜙4𝑑𝑧𝑛(𝑘),𝑧𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)𝑥+𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘),𝑑𝑤𝑛(𝑘)+1,𝑤𝑚(𝑘)+1𝐹𝑧=𝑑𝑛(𝑘),𝑤𝑛(𝑘),𝑥𝑛(𝑘),𝑦𝑛(𝑘)𝑧,𝐹𝑚(𝑘),𝑤𝑚(𝑘),𝑥𝑚(𝑘),𝑦𝑚(𝑘)14𝑑𝑧𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑥+𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘)1𝜙4𝑑𝑧𝑛(𝑘),𝑧𝑚(𝑘)𝑤+𝑑𝑛(𝑘),𝑤𝑚(𝑘)𝑥+𝑑𝑛(𝑘),𝑥𝑚(𝑘)𝑦+𝑑𝑛(𝑘),𝑦𝑚(𝑘).(2.26) Combining (2.24) with (2.26), we obtain that 𝑡𝑘𝛿𝑛(𝑘)+1+𝛿𝑚(𝑘)+1𝑥+𝑑𝑛(𝑘)+1,𝑥𝑚(𝑘)+1𝑦+𝑑𝑛(𝑘)+1,𝑦𝑚(𝑘)+1𝑧+𝑑𝑛(𝑘)+1,𝑧𝑚(𝑘)+1𝑤+𝑑𝑛(𝑘)+1,𝑤𝑚(𝑘)+1𝛿𝑛(𝑘)+1+𝛿𝑚(𝑘)+1+𝑡𝑘14𝜙4𝑡𝑘.(2.27) Letting 𝑘 and having in mind (2.19) we get a contradiction. This shows that {𝑥𝑛}, {𝑦𝑛}, {𝑧𝑛}, and {𝑤𝑛} are Cauchy sequences. Since 𝑋 is complete metric space, there exists 𝑥,𝑦,𝑧,𝑤𝑋 such that lim𝑛𝑥𝑛=𝑥,lim𝑛𝑦𝑛=𝑦,lim𝑛𝑧𝑛=𝑧,lim𝑛𝑤𝑛=𝑤.(2.28)
Suppose now the assumption (a) holds. Then by (2.9) and (2.28), we have 𝑥=lim𝑛𝑥𝑛=lim𝑛𝐹𝑥𝑛1,𝑦𝑛1,𝑧𝑛1,𝑤𝑛1=𝐹lim𝑛𝑥𝑛1,lim𝑛𝑦𝑛1,lim𝑛𝑧𝑛1,lim𝑛𝑤𝑛1=𝐹(𝑥,𝑦,𝑧,𝑤).(2.29) Analogously, we also observe that 𝑦=lim𝑛𝑦𝑛=lim𝑛𝐹𝑥𝑛1,𝑤𝑛1,𝑧𝑛1,𝑦𝑛1=𝐹(𝑥,𝑤,𝑧,𝑦),𝑧=lim𝑛𝑧𝑛=lim𝑛𝐹𝑧𝑛1,𝑦𝑛1,𝑥𝑛1,𝑤𝑛1=𝐹(𝑧,𝑦,𝑥,𝑤),𝑤=lim𝑛𝑤𝑛=lim𝑛𝐹𝑧𝑛1,𝑤𝑛1,𝑥𝑛1,𝑦𝑛1=𝐹(𝑧,𝑤,𝑥,𝑦).(2.30) Thus, we have 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤.(2.31)
Suppose now the assumption (b) holds. Since {𝑥𝑛},{𝑧𝑛} are nondecreasing and 𝑥𝑛𝑥, 𝑧𝑛𝑧, and also {𝑦𝑛},{𝑤𝑛} are nonincreasing and 𝑦𝑛𝑦,𝑤𝑛𝑤, then by assumption (b) we have 𝑥𝑛𝑥,𝑦𝑛𝑦,𝑧𝑛𝑧,𝑤𝑛𝑤(2.32) for all 𝑛. Consider now 𝑑(𝑥,𝐹(𝑥,𝑦,𝑧,𝑤))𝑑𝑥,𝑥𝑛+1𝑥+𝑑𝑛+1,𝐹(𝑥,𝑦,𝑧,𝑤)=𝑑𝑥,𝑥𝑛+1𝐹𝑥+𝑑𝑛,𝑦𝑛,𝑧𝑛,𝑤𝑛1,𝐹(𝑥,𝑦,𝑧,𝑤)4𝑑𝑥𝑛𝑦,𝑥+𝑑𝑛𝑧,𝑦+𝑑𝑛𝑤,𝑧+𝑑𝑛1,𝑤𝜙4𝑑𝑥𝑛𝑦,𝑥+𝑑𝑛𝑧,𝑦+𝑑𝑛𝑤,𝑧+𝑑𝑛.,𝑤(2.33) Taking 𝑛 in (2.33) and using (2.28), we get that 𝑑(𝑥,𝐹(𝑥,𝑦,𝑧,𝑤))=0. Thus, 𝑥=𝐹(𝑥,𝑦,𝑧,𝑤). Analogously, we get that 𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤.(2.34) Thus, we proved that 𝐹 has a quadruple fixed point.

Corollary 2.4. Let (𝑋,) be partially ordered set and (𝑋,𝑑) a complete metric space. Let 𝐹𝑋4𝑋 be a mapping having the mixed monotone property on 𝑋. Assume that there exists 𝑘[0,1) such that 𝑘𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑟,𝑡))4[]𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑟)+𝑑(𝑤,𝑡)(2.35) for all 𝑥𝑢, 𝑦𝑣, 𝑧𝑤. Suppose there exist 𝑥0,𝑦0,𝑧0,𝑤0𝑋 such that 𝑥0𝑥𝐹0,𝑦0,𝑧0,𝑤0,𝑦0𝑥𝐹0,𝑤0,𝑧0,𝑦0,𝑧0𝑧𝐹0,𝑦0,𝑥0,𝑤0,𝑤0𝑧𝐹0,𝑤0,𝑥0,𝑦0.(2.36) Suppose 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, then there exist 𝑥,𝑦,𝑧,𝑤𝑋 such that 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤.(2.37)

Proof. It is sufficient to take 𝜙(𝑡)=((1𝑘)/2)𝑡 in previous theorem.

3. Uniqueness of Quadruple Fixed Point

In this section we shall prove the uniqueness of 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,𝑡),(𝑢,𝑣,𝑟,𝑡)𝑋×𝑋×𝑋×𝑋(𝑥,𝑦,𝑧,𝑤)(𝑢,𝑣,𝑟,𝑡)𝑥𝑢,𝑦𝑣,𝑧𝑟,𝑤𝑟.(3.1) We say that (𝑥,𝑦,𝑧,𝑤) is equal (𝑢,𝑣,𝑟,𝑡) if and only if 𝑥=𝑢,𝑦=𝑣,𝑧=𝑟, and 𝑤=𝑡.

Theorem 3.1. In addition to hypothesis of Theorem 2.3, suppose that for all (𝑥,𝑦,𝑧,𝑡),(𝑢,𝑣,𝑟,𝑡)X×𝑋×𝑋×𝑋; there exists (𝑎,𝑏,𝑐,𝑑)𝑋×𝑋×𝑋×𝑋 that is comparable to (𝑥,𝑦,𝑧,𝑡) and (𝑢,𝑣,𝑟,𝑡); then 𝐹 has a unique quadruple fixed point.

Proof. The set of quadruple fixed point of 𝐹 is not empty due to Theorem 2.3. Assume, now, that (𝑥,𝑦,𝑧,𝑡) and (𝑢,𝑣,𝑟,𝑡) are the quadruple fixed point of 𝐹, that is, 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥,𝐹(𝑢,𝑣,𝑟,𝑡)=𝑢,𝐹(𝑥,𝑤,𝑧,𝑦)=𝑦,𝐹(𝑢,𝑡,𝑟,𝑣)=𝑣,𝐹(𝑧,𝑦,𝑥,𝑤)=𝑧,𝐹(𝑟,𝑣,𝑢,𝑡)=𝑟,𝐹(𝑧,𝑤,𝑥,𝑦)=𝑤,𝐹(𝑟,𝑡,𝑢,𝑣)=𝑡.(3.2) We shall show that (𝑥,𝑦,𝑧,𝑤) and (𝑢,𝑣,𝑟,𝑡) are equal. By assumption, there exists (𝑎,𝑏,𝑐,𝑑)𝑋×𝑋×𝑋×𝑋 that is comparable to (𝑥,𝑦,𝑧,𝑡) and (𝑢,𝑣,𝑟,𝑡). Define sequences {𝑎𝑛},{𝑏𝑛},{𝑐𝑛}, and {𝑑𝑛} such that 𝑎=𝑎0,𝑏=𝑏0,𝑐=𝑐0,𝑑=𝑑0,𝑎𝑛𝑎=𝐹𝑛1,𝑏𝑛1,𝑧𝑛1,𝑑𝑛1,𝑏𝑛𝑎=𝐹𝑛1,𝑑𝑛1,𝑐𝑛1,𝑏𝑛1,𝑐𝑛𝑐=𝐹𝑛1,𝑏𝑛1,𝑎𝑛1,𝑑𝑛1,𝑑𝑛𝑐=𝐹𝑛1,𝑑𝑛1,𝑎𝑛1,𝑏𝑛1(3.3) for all 𝑛. Since (𝑥,𝑦,𝑧,𝑤) is comparable with (𝑎,𝑏,𝑐,𝑑), we may assume that (𝑥,𝑦,𝑧,𝑤)(𝑎,𝑏,𝑐,𝑑)=(𝑎0,𝑏0,𝑐0,𝑑0). Recursively, we get that 𝑎(𝑥,𝑦,𝑧,𝑤)𝑛,𝑏𝑛,𝑐𝑛,𝑑𝑛𝑛.(3.4) By (3.4) and (2.5), we have 𝑑𝑥,𝑎𝑛+1𝐹𝑎=𝑑(𝑥,𝑦,𝑧,𝑤),𝐹𝑛,𝑏𝑛,𝑐𝑛,𝑑𝑛14𝑑𝑥,𝑎𝑛+𝑑𝑦,𝑏𝑛+𝑑𝑧,𝑐𝑛+𝑑𝑤,𝑑𝑛1𝜙4𝑑𝑥,𝑎𝑛+𝑑𝑦,𝑏𝑛+𝑑𝑧,𝑐𝑛+𝑑𝑤,𝑑𝑛,𝑑𝑏𝑛+1𝐹𝑎,𝑦=𝑑𝑛,𝑑𝑛,𝑐𝑛,𝑏𝑛1,𝐹(𝑥,𝑤,𝑧,𝑦)4𝑑𝑎𝑛𝑑,𝑥+𝑑𝑛𝑐,𝑤+𝑑𝑛𝑏,𝑧+𝑑𝑛1,𝑦𝜙4𝑑𝑎𝑛𝑑,𝑥+𝑑𝑛𝑐,𝑤+𝑑𝑛𝑏,𝑧+𝑑𝑛,𝑑,𝑦𝑧,𝑐𝑛+1𝑐=𝑑𝐹(𝑧,𝑦,𝑥,𝑤),𝐹𝑛,𝑏𝑛,𝑎𝑛,𝑑𝑛14𝑑𝑧,𝑐𝑛+𝑑𝑦,𝑏𝑛+𝑑𝑥,𝑎𝑛+𝑑𝑤,𝑑𝑛1𝜙4𝑑𝑧,𝑐𝑛+𝑑𝑦,𝑏𝑛+𝑑𝑥,𝑎𝑛+𝑑𝑤,𝑑𝑛,𝑑𝑑𝑛+1𝐹𝑐,𝑤=𝑑𝑛,𝑑𝑛,𝑎𝑛,𝑏𝑛1,𝐹(𝑧,𝑤,𝑥,𝑦)4𝑑𝑐𝑛𝑑,𝑧+𝑑𝑛𝑎,𝑤+𝑑𝑛𝑏,𝑥+𝑑𝑛1,𝑦𝜙4𝑑𝑐𝑛𝑑,𝑧+𝑑𝑛𝑎,𝑤+𝑑𝑛𝑏,𝑥+𝑑𝑛.,𝑦(3.5) Set 𝛾𝑛=𝑑(𝑥,𝑎𝑛)+𝑑(𝑦,𝑏𝑛)+𝑑(𝑧,𝑐𝑛)+𝑑(𝑤,𝑑𝑛). Then, due to (3.5), we have 𝛾𝑛+1𝛾𝑛,𝑛.(3.6)

Hence, the sequence {𝛾𝑛} is decreasing and bounded below. Thus, there exists 𝛾0 such thatlim𝑛𝛾𝑛=𝛾.(3.7)

Now, we shall show that 𝛾=0. Suppose, to the contrary, that 𝛾>0. Again by (3.5), we have 𝛾𝑛+1𝛾𝑛144𝜙𝛾𝑛,𝑛.(3.8) Letting 𝑛 in (3.8), we obtain that 1𝛾𝛾44𝜙(𝛾),(3.9)

which is a contradiction. Therefore, 𝛾=0. That is, lim𝑛𝛾𝑛=0.(3.10)

Consequently, we havelim𝑛𝑑𝑥,𝑎𝑛=0,lim𝑛𝑑𝑦,𝑏𝑛=0,lim𝑛𝑑𝑧,𝑐𝑛=0,lim𝑛𝑑𝑤,𝑑𝑛=0.(3.11) Analogously, we show thatlim𝑛𝑑𝑢,𝑎𝑛=0,lim𝑛𝑑𝑣,𝑏𝑛=0,lim𝑛𝑑𝑟,𝑐𝑛=0,lim𝑛𝑑𝑠,𝑑𝑛=0.(3.12) Combining (3.11) and (3.12) yields that (𝑥,𝑦,𝑧,𝑤) and (𝑢,𝑣,𝑟,𝑡) are equal.

Example 3.2. Let 𝑋= with the metric 𝑑(𝑥,𝑦)=|𝑥𝑦|, for all 𝑥,𝑦𝑋 and the usual ordering.
Let 𝐹𝑋4𝑋 be given by 𝐹(𝑥,𝑦,𝑧,𝑤)=𝑥𝑦+𝑧𝑤16,𝑥,𝑦,𝑧,𝑤𝑋,(3.13) and let 𝜙[0,)[0,) be given by 𝜙(𝑡)=𝑡/4 for all 𝑡[0,).
It is easy to check that all the conditions of Theorem 2.3 are satisfied and (0,0,0,0) is the unique quadruple fixed point of 𝐹.