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

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, [5–7].

Bhaskar and Lakshmikantham in [8] introduced the concept of coupled fixed point of a mapping 𝐹∶𝑋×𝑋→𝑋 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, [9–20].

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 𝐹∶𝑋×𝑋×𝑋→𝑋. The mapping 𝐹 is said to have the mixed monotone property if, for any 𝑥,𝑦,𝑧∈𝑋, 𝑥𝑦₁,𝑥₂∈𝑋,𝑥₁≤𝑥₂⟹𝐹𝑥₁,𝑦,𝑧≤𝐹𝑥₂,𝑦,𝑧₁,𝑧,𝑦₂∈𝑋,𝑦₁≤𝑦₂⟹𝐹𝑥,𝑦₁,𝑧≥𝐹𝑥,𝑦₂,𝑧₁,,𝑧₂∈𝑋,𝑧₁≤𝑧₂⟹𝐹𝑥,𝑦,𝑧₁≤𝐹𝑥,𝑦,𝑧₂.(1.1)

Definition 1.2. Let 𝐹∶𝑋×𝑋×𝑋→𝑋. 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 𝐹∶𝑋×𝑋×𝑋→𝑋 be a mapping such that it has the mixed monotone property and there exist 𝑗,𝑟,𝑙≥0 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,𝑧∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, and 𝑧≤𝑤. Additionally suppose that either 𝐹 is continuous or 𝑋 has the following properties:(1)if a nondecreasing sequence 𝑥_𝑛→𝑥, then 𝑥_𝑛≤𝑥 for all 𝑛,(2)if a nonincreasing sequence 𝑦_𝑛→𝑦, then 𝑦≤𝑦_𝑛 for all 𝑛.
If there exist 𝑥₀,𝑦₀,𝑧₀∈𝑋 such that 𝑥₀≤𝐹(𝑥₀,𝑦₀,𝑧₀), 𝑦₀≥𝐹(𝑦₀,𝑥₀,𝑧₀), and 𝑧₀≤𝐹(𝑧₀,𝑦₀,𝑥₀), then there exist 𝑥,𝑦,𝑧∈𝑋 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 𝑁≥3 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 [24–27] 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 𝐹∶𝑋⁴→𝑋 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 𝑑∶𝑋⁴→𝑋, given by 𝑑((𝑥,𝑦,𝑧,𝑤),(𝑢,𝑣,ℎ,𝑙))=𝑑(𝑥,𝑦)+𝑑(𝑦,𝑣)+𝑑(𝑧,ℎ)+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 𝐹∶𝑋⁴→𝑋 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,𝑤∈𝑋, 𝑥𝑦₁,𝑥₂∈𝑋,𝑥₁≤𝑥₂implies𝐹𝑥₁,𝑦,𝑧,𝑤≤𝐹𝑥₂,𝑦,𝑧,𝑤₁,𝑧,𝑦₂∈𝑋,𝑦₁≤𝑦₂implies𝐹𝑥,𝑦₂,𝑧,𝑤≤𝐹𝑥,𝑦₁,𝑧,𝑤₁,𝑤,𝑧₂∈𝑋,𝑧₁≤𝑧₂implies𝐹𝑥,𝑦,𝑧₁,𝑤≤𝐹𝑥,𝑦,𝑧₂,𝑤₁,,𝑤₂∈𝑋,𝑤₁≤𝑤₂implies𝐹𝑥,𝑦,𝑧,𝑤₂≤𝐹𝑥,𝑦,𝑧,𝑤₁.(1.7)

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

2. Main Results

We start with the following definition (see, e.g., [28–31]).

Definition 2.1. Let (𝑋,𝑑) be a metric space. A mapping 𝑇∶𝑋→𝑋 is said to be ICS if 𝑇 is injective, continuous, and it has the property: for every sequence {𝑥_𝑛} in 𝑋, if {𝑇𝑥_𝑛} is convergent then, {𝑥_𝑛} 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)lim_𝑟→𝑡+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 𝑇∶𝑋→𝑋 is an ICS mapping and 𝐹∶𝑋⁴→𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝜙∈Φ 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,𝑠∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, 𝑧≤𝑟, and 𝑠≤𝑤. Additionally assume that either(a)𝐹 is continuous, or(b)𝑋 has the following properties:(i) if nondecreasing sequence 𝑥_𝑛→𝑥 (respectively, 𝑧_𝑛→𝑧), then 𝑥_𝑛≤𝑥 (respectively, 𝑧_𝑛≤𝑧) for all 𝑛,(ii) if nonincreasing sequence 𝑦_𝑛→𝑦 (respectively, 𝑤_𝑛→𝑤), then 𝑦_𝑛≥𝑦 (respectively, 𝑤_𝑛≥𝑤) for all 𝑛.
If there exist 𝑥₀,𝑦₀,𝑧₀,𝑤₀∈𝑋 such that 𝑥₀≤𝐹(𝑥₀,𝑦₀,𝑧₀,𝑤₀), 𝑦₀≥𝐹(𝑦₀,𝑧₀,𝑤₀,𝑥₀), 𝑧₀≤𝐹(𝑧₀,𝑤₀,𝑥₀,𝑦₀), and 𝑤₀≥𝐹(𝑤₀,𝑥₀,𝑦₀,𝑧₀), then there exist 𝑥,𝑦,𝑧,𝑤∈𝑋 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,𝑤₀∈𝑋 such that 𝑧𝑥₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑤₀,𝑦₀≥𝐹𝑦₀,𝑧₀,𝑤₀,𝑥₀,(2.3)₀≤𝐹𝑧₀,𝑤₀,𝑥₀,𝑦₀,𝑤₀≥𝐹𝑤₀,𝑥₀,𝑦₀,𝑧₀.(2.4)
Set 𝑧𝑥₁₁=𝐹𝑥₀,𝑦₀,𝑧₀,𝑤₀,𝑦₁=𝐹𝑦₀,𝑧₀,𝑤₀,𝑥₀,=𝐹𝑧₀,𝑤₀,𝑥₀,𝑦₀,𝑤₁=𝐹𝑤₀,𝑥₀,𝑦₀,𝑧₀.(2.5)
Then, 𝑥₀≤𝑥₁, 𝑦₀≥𝑦₁, 𝑧₀≤𝑧₁, and 𝑤₀≥𝑤₁. 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 𝑥₀≤𝑥₁≤𝑥₂, 𝑦₂≤𝑦₁≤𝑦₀, 𝑧₀≤𝑧₁≤𝑧₂, and 𝑤₂≤𝑤₁≤𝑤₀. By continuing this process, we can construct four sequences {𝑥_𝑛}, {𝑦_𝑛}, {𝑧_𝑛}, and {𝑤_𝑛} in 𝑋 such that 𝑧𝑥_𝑛+1𝑛+1=𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝑤_𝑛,𝑦_𝑛+1=𝐹𝑦_𝑛,𝑧_𝑛,𝑤_𝑛,𝑥_𝑛,=𝐹𝑧₁,𝑤₁,𝑥₁,𝑦₁,𝑤_𝑛+1=𝐹𝑤_𝑛,𝑥_𝑛,𝑦_𝑛,𝑧_𝑛.(2.6)
Since 𝐹 has the mixed monotone property, by using a mathematical induction it is easy to see that 𝑥_𝑛≤𝑥_𝑛+1,𝑦_𝑛+1≤𝑦_𝑛,𝑧_𝑛≤𝑧_𝑛+1,𝑤_𝑛+1≤𝑤_𝑛,for𝑛=0,1,2,…,(2.7)
Assume that, for some 𝑛∈ℕ, 𝑥_𝑛=𝑥_𝑛+1,𝑦_𝑛=𝑦_𝑛+1,𝑧_𝑛=𝑧_𝑛+1,𝑤_𝑛=𝑤_𝑛+1.(2.8) Then, by (2.6), (𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝑤_𝑛) is a quadruple fixed point of 𝐹. Therefore, in the rest of the proof, for any 𝑛∈ℕ we will assume that 𝑥_𝑛≠𝑥_𝑛+1or𝑦_𝑛≠𝑦_𝑛+1or𝑧_𝑛≠𝑧_𝑛+1or𝑤_𝑛≠𝑤_𝑛+1.(2.9)
Since 𝑇 is injective, for any 𝑛∈ℕ, 0<max𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1,𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1.(2.10)
Due to (2.1), (2.6), and (2.7), we have 𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1≤=𝜙𝑑𝑇m𝐹ax𝑥𝑑_𝑛−1,𝑇𝑦𝑥_{𝑛−1𝑛−1},𝑧_𝑛−1,𝑤_𝑛−1,𝑇𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝑤_𝑛𝑑𝑇𝑦_𝑛+1,𝑇𝑥_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛,,𝑇𝑦_𝑛≤=𝜙𝑑𝑇m𝐹ax𝑦_𝑛𝑑,𝑧_𝑛𝑇,𝑦𝑤_𝑛,_𝑛−1𝑥_𝑛,𝑇𝐹𝑦_𝑛−1,𝑧_𝑛−1,𝑤_𝑛−1,𝑥_𝑛−1𝑑𝑇𝑧_𝑛,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛,𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛,,𝑇𝑧_𝑛+1≤=𝑑𝜙𝑇m𝐹ax𝑧𝑑_𝑛−1,𝑇𝑤𝑧_{𝑛−1𝑛−1},𝑥_𝑛−1,𝑦_𝑛−1,𝑇𝐹𝑧_𝑛,𝑤_𝑛,𝑥_𝑛,𝑦_𝑛𝑑𝑇𝑤_𝑛+1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛,𝑑𝑇𝑥_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,,𝑇𝑤_𝑛≤=𝜙𝑑𝑇m𝐹ax𝑤𝑑_𝑛,𝑥𝑇_𝑛𝑤,𝑦_𝑛,_𝑛−1𝑧_𝑛,𝑇𝐹𝑤_𝑛−1,𝑥_𝑛−1,𝑦_𝑛−1,𝑧_𝑛−1,𝑇𝑤_𝑛,𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛.(2.11)
Using the fact that 𝜙(𝑡)<𝑡 for all 𝑡>0 together with (2.11), we obtain that 0<≤𝜙maxmax𝑑𝑇𝑑𝑥_𝑛𝑇,𝑥𝑇𝑥_{𝑛+1𝑛−1},𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1<max𝑑𝑇𝑥_𝑛−1,𝑇𝑥_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛,𝑇𝑥_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛.(2.12)
It follows that max<𝑑m𝑇ax𝑥_𝑛,𝑑𝑇𝑥𝑇𝑥_𝑛+1,_𝑛−1𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1,𝑇𝑥_𝑛,𝑑𝑇𝑦_𝑛−1,𝑇𝑦_𝑛,𝑑𝑇𝑧_𝑛−1,𝑇𝑧_𝑛,𝑑𝑇𝑤_𝑛−1,𝑇𝑤_𝑛.(2.13)
Thus, max{𝑑(𝑇𝑥_𝑛,𝑇𝑥_𝑛+1),𝑑(𝑇𝑦_𝑛,𝑇𝑦_𝑛+1),𝑑(𝑇𝑧_𝑛,𝑇𝑧_𝑛+1),𝑑(𝑇𝑤_𝑛,𝑇𝑤_𝑛+1)} is a positive decreasing sequence. Hence, there exists 𝑟≥0 such that lim_𝑛→+∞max𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1,𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1=𝑟.(2.14)
Suppose that 𝑟>0. Letting 𝑛→+∞ in (2.12), we obtain that 0<=𝑟li≤m_𝑡→𝑟+lim_𝑛→+∞𝜙max𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1,𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1𝜙(𝑡)<𝑟,(2.15)
which is a contradiction. Therefore, we deduce that lim_𝑛→+∞max𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛+1,𝑑𝑇𝑦_𝑛,𝑇𝑦_𝑛+1,𝑑𝑇𝑧_𝑛,𝑇𝑧_𝑛+1,𝑑𝑇𝑤_𝑛,𝑇𝑤_𝑛+1=0.(2.16)
We will show that {𝑇𝑥_𝑛}, {𝑇𝑦_𝑛}, {𝑇𝑧_𝑛}, and {𝑇𝑤_𝑛} are Cauchy sequences. Assume the contrary, that is, either {𝑇𝑥_𝑛} or {𝑇𝑦_𝑛} or {𝑇𝑧_𝑛} or {𝑇𝑤_𝑛} is not a Cauchy sequence, consequently, lim_{𝑛,𝑚→+∞}𝑑𝑇𝑥_𝑚,𝑇𝑥_𝑛≠0orlim_{𝑛,𝑚→+∞}𝑑𝑇𝑦_𝑚,𝑇𝑦_𝑛≠0,(2.17)
or lim_{𝑛,𝑚→+∞}𝑑(𝑇𝑧_𝑚,𝑇𝑧_𝑛)≠0 or lim_{𝑛,𝑚→+∞}𝑑(𝑇𝑤_𝑚,𝑇𝑤_𝑛)≠0. This means that there exists 𝜀>0 for which we can find subsequences of integers (𝑚_𝑘) and (𝑛_𝑘) with 𝑛_𝑘>𝑚_𝑘>𝑘 such that max𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘,𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘,𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘≥𝜀.(2.18)
Furthermore, corresponding to 𝑚_𝑘, we can choose 𝑛_𝑘 in such a way that it is the smallest integer with 𝑛_𝑘>𝑚_𝑘 and satisfying (2.18). Then, max𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘−1<𝜀.(2.19)
By the triangle inequality and (2.19), we have 𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤<𝜖𝑑+𝑇𝑑𝑥_𝑚𝑘𝑇,𝑥𝑇𝑥_{𝑛𝑘−1𝑛𝑘−1}+𝑑𝑇𝑥_𝑛𝑘−1,𝑇𝑥_𝑛𝑘,𝑇𝑥_𝑛𝑘.(2.20)
Thus, by (2.16), we obtain lim_𝑘→+∞𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘−1≤𝜀.(2.21)
Similarly, we have lliimm_{𝑘→+∞𝑘→+∞}𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘−1≤𝜀,lim_𝑘→+∞𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘−1≤𝜀,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘−1≤𝜀.(2.22)
Again, by (2.19), we have 𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤≤𝑑𝑑𝑇𝑇𝑥𝑥_{𝑚𝑘𝑚𝑘},𝑇𝑥_𝑚𝑘−1+𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1+𝑑𝑇𝑥_𝑛𝑘−1,𝑇𝑥_𝑛𝑘+𝑑𝑇𝑥,_𝑚𝑘𝑇𝑥_𝑚𝑘−1+𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑚𝑘<𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘−1+𝑑𝑇𝑥_𝑛𝑘−1,𝑇𝑥_𝑛𝑘,𝑇𝑥_𝑚𝑘−1+𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑚𝑘+𝜀+𝑑𝑇𝑥_𝑛𝑘−1,𝑇𝑥_𝑛𝑘.(2.23)
Letting 𝑘→+∞ and using (2.16), we get lliimm_{𝑘→+∞𝑘→+∞}𝑑𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1≤𝜀,lim_𝑘→+∞𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1≤𝜀,lim_𝑘→+∞𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1≤𝜀,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘≤lim_𝑘→+∞𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1≤𝜀.(2.24)
Using (2.18) and (2.24), we have lim_𝑘→+∞m=axlim𝑑_𝑘→+∞𝑇𝑥_𝑚𝑘,𝑇𝑥_𝑛𝑘,𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘,𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘=𝜀.max𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1(2.25)
Now, using inequality (2.1), we obtain 𝑑𝑇𝑥=𝑑_𝑚𝑘,𝑇𝑇𝑥𝐹_𝑛𝑘𝑥_𝑚𝑘−1≤𝜙max𝑑𝑇,𝑥𝑦_{𝑚𝑘−1𝑚𝑘−1},𝑧_𝑚𝑘−1,𝑤_𝑚𝑘−1,𝑇𝐹𝑥_𝑛𝑘−1,𝑦_𝑛𝑘−1,𝑧_𝑛𝑘−1,𝑤_𝑛𝑘−1𝑑𝑇𝑦_𝑚𝑘,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1=𝑑,𝑇𝑇𝑦𝐹_𝑛𝑘𝑦_𝑚𝑘−1≤𝜙max𝑑𝑇,𝑦𝑧_{𝑚𝑘−1𝑚𝑘−1},𝑤_𝑚𝑘−1,𝑥_𝑚𝑘−1,𝑇𝐹𝑦_𝑛𝑘−1,𝑧_𝑛𝑘−1,𝑤_𝑛𝑘−1,𝑥_𝑛𝑘−1𝑑𝑇𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1,𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1,=𝑑,𝑇𝑇𝑦𝐹_𝑛𝑘𝑧_𝑚𝑘−1≤𝜙max𝑑𝑇,𝑧𝑤_{𝑚𝑘−1𝑚𝑘−1},𝑥_𝑚𝑘−1,𝑦_𝑚𝑘−1,𝑇𝐹𝑧_𝑛𝑘−1,𝑤_𝑛𝑘−1,𝑥_𝑛𝑘−1,𝑦_𝑛𝑘−1𝑑𝑇𝑦_𝑚𝑘,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1,𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1,=𝑑,𝑇𝑇𝑦𝐹_𝑛𝑘𝑤_𝑚𝑘−1≤𝜙max𝑑𝑇,𝑤𝑥_{𝑚𝑘−1𝑚𝑘−1},𝑦_𝑚𝑘−1,𝑧_𝑚𝑘−1,𝑇𝐹𝑤_𝑛𝑘−1,𝑥_𝑛𝑘−1,𝑦_𝑛𝑘−1,𝑧_𝑛𝑘−1,𝑇𝑤_𝑛𝑘−1,𝑑𝑇𝑥_𝑚𝑘−1,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1(2.26)
From (2.26), we deduce that max≤𝑑𝜙𝑇𝑥ma_𝑚𝑘x,𝑇𝑥𝑑_𝑛𝑘𝑇,𝑥𝑑𝑇_𝑚𝑘−1𝑦_𝑚𝑘,𝑇𝑦_𝑛𝑘,𝑑𝑇𝑧_𝑚𝑘,𝑇𝑧_𝑛𝑘,𝑑𝑇𝑤_𝑚𝑘,𝑇𝑤_𝑛𝑘,𝑇𝑥_𝑛𝑘−1,𝑑𝑇𝑦_𝑚𝑘−1,𝑇𝑦_𝑛𝑘−1,𝑑𝑇𝑧_𝑚𝑘−1,𝑇𝑧_𝑛𝑘−1,𝑑𝑇𝑤_𝑚𝑘−1,𝑇𝑤_𝑛𝑘−1.(2.27)
Letting 𝑘→+∞ in (2.27) and by using (2.25), we get that 0<𝜀≤lim_𝑡→𝜀+𝜙(𝑡)<𝜀,(2.28)
which is a contradiction. Thus, {𝑇𝑥_𝑛}, {𝑇𝑦_𝑛}, {𝑇𝑧_𝑛}, and {𝑇𝑤_𝑛} are Cauchy sequences in (𝑋,𝑑). Since 𝑋 is a complete metric space, {𝑇𝑥_𝑛}, {𝑇𝑦_𝑛}, {𝑇𝑧_𝑛}, and {𝑇𝑤_𝑛} are convergent sequences.
Since 𝑇 is an ICS mapping, there exist 𝑥,𝑦,𝑧,𝑤∈𝑋 such that lim_𝑛→+∞𝑥_𝑛=𝑥,lim_𝑛→+∞𝑦_𝑛=𝑦,lim_𝑛→+∞𝑧_𝑛=𝑧,lim_𝑛→+∞𝑤_𝑛=𝑤.(2.29)
Since 𝑇 is continuous, we have lim_𝑛→+∞𝑇𝑥_𝑛=𝑇𝑥,lim_𝑛→+∞𝑇𝑦_𝑛=𝑇𝑦,lim_𝑛→+∞𝑇𝑧_𝑛=𝑇𝑧,lim_𝑛→+∞𝑇𝑤_𝑛=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==𝐹lim_𝑛→+∞l𝑥im_{𝑛+1𝑛→+∞}=lim_𝑛→+∞𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝑤_𝑛𝑥_𝑛,lim_𝑛→+∞𝑦_𝑛,lim_𝑛→+∞𝑧_𝑛,lim_𝑛→+∞𝑤_𝑛𝑦=lim_𝑛→+∞=𝐹(𝑥,𝑦,𝑧,𝑤),=𝐹l𝑦im_{𝑛+1𝑛→+∞}=lim_𝑛→+∞𝐹𝑦_𝑛,𝑧_𝑛,𝑤_𝑛,𝑥_𝑛𝑦_𝑛,lim_𝑛→+∞𝑧_𝑛,lim_𝑛→+∞𝑤_𝑛,lim_𝑛→+∞𝑥_𝑛𝑧=lim_𝑛→+∞=𝐹(𝑦,𝑧,𝑤,𝑥),=𝐹l𝑧im_{𝑛+1𝑛→+∞}=lim_𝑛→+∞𝐹𝑧_𝑛,𝑤_𝑛,𝑥_𝑛,𝑦_𝑛𝑧_𝑛,lim_𝑛→+∞𝑤_𝑛,lim_𝑛→+∞𝑥_𝑛,lim_𝑛→+∞𝑦_𝑛𝑤=lim_𝑛→+∞=𝐹(𝑧,𝑤,𝑥,𝑦),=𝐹l𝑤im_{𝑛+1𝑛→+∞}=lim_𝑛→+∞𝐹𝑤_𝑛,𝑥_𝑛,𝑦_𝑛,𝑧_𝑛𝑤_𝑛,lim_𝑛→+∞𝑥_𝑛,lim_𝑛→+∞𝑦_𝑛,lim_𝑛→+∞𝑧_𝑛=𝐹(𝑤,𝑥,𝑦,𝑧).(2.31)
We have proved that 𝐹 has a quadruple fixed point.
Suppose now the assumption (𝑏) holds. Since {𝑥_𝑛} and {𝑧_𝑛} are nondecreasing with 𝑥_𝑛→𝑥 and 𝑧_𝑛→𝑧 and also {𝑦_𝑛} and {𝑤_𝑛} are nonincreasing, with 𝑦_𝑛→𝑦 and 𝑤_𝑛→𝑤, we have 𝑥_𝑛≤𝑥,𝑦_𝑛≥𝑦,𝑧_𝑛≤𝑧,𝑤_𝑛≥𝑤(2.32)
for all 𝑛. Consider now 𝑑(𝑇𝑥,𝑇𝐹(𝑥,𝑦,𝑧,𝑤))≤=𝑑𝑑𝑇𝑇𝑥𝑥,,𝑇𝑇𝑥𝑥_𝑛+1𝑛+1+𝑑𝑇𝑥_𝑛+1,𝑇𝐹(𝑥,𝑦,𝑧,𝑤)≤𝑑𝑇𝑥,𝑇𝑥_𝑛+1+𝑑𝑇𝐹𝑥_𝑛,𝑦_𝑛,𝑧_𝑛,𝑤_𝑛,𝑇𝐹(𝑥,𝑦,𝑧,𝑤)+𝜙m𝑑ax𝑇𝑤𝑑_𝑛𝑇𝑥_𝑛,𝑇𝑥,𝑑𝑇𝑦_𝑛,𝑇𝑦,𝑑𝑇𝑧_𝑛,𝑇𝑧,,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 𝑇∶𝑋→𝑋 is an ICS mapping and 𝐹∶𝑋⁴→𝑋 is such that 𝐹 has the mixed monotone property. Assume that there exists 𝜙∈Φ such that 𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑡,𝑠))≤𝜙𝑑(𝑇𝑥,𝑇𝑢)+𝑑(𝑇𝑦,𝑇𝑣)+𝑑(𝑇𝑧,𝑇𝑡)+𝑑(𝑇𝑤,𝑇𝑠)4(2.35)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑡,𝑠∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, 𝑧≤𝑡, and 𝑠≤𝑤. Additionally suppose that either(a)𝐹 is continuous, or(b)𝑋 has the following property:(i) if non-decreasing sequence 𝑥_𝑛→𝑥 (resp., 𝑧_𝑛→𝑧), then 𝑥_𝑛≤𝑥 (resp., 𝑧_𝑛≤𝑧) for all 𝑛,(ii) if non-increasing sequence 𝑦_𝑛→𝑦 (resp., 𝑤_𝑛→𝑤), then 𝑦_𝑛≥𝑦 (resp., 𝑤_𝑛≥𝑤) for all 𝑛.
If there exist 𝑥₀,𝑦₀,𝑧₀,𝑤₀∈𝑋 such that 𝑥₀≤𝐹(𝑥₀,𝑦₀,𝑧₀,𝑤₀), 𝑦₀≥𝐹(𝑦₀,𝑧₀,𝑤₀,𝑥₀), 𝑧₀≤𝐹(𝑧₀,𝑤₀,𝑥₀,𝑦₀) and 𝑤₀≥𝐹(𝑤₀,𝑤₀,𝑦₀,𝑧₀), then there exist 𝑥,𝑦,𝑧,𝑤∈𝑋 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 𝑇∶𝑋→𝑋 is an ICS mapping and 𝐹∶𝑋⁴→𝑋 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,𝑠∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, 𝑧≤𝑡, and 𝑠≤𝑤. Suppose that either (a)𝐹 is continuous, or(b)𝑋 has the following property:(i)if nondecreasing sequence 𝑥_𝑛→𝑥 (resp., 𝑧_𝑛→𝑧), then 𝑥_𝑛≤𝑥 (resp., 𝑧_𝑛≤𝑧) for all 𝑛,(ii)if nonincreasing sequence 𝑦_𝑛→𝑦 (resp., 𝑤_𝑛→𝑤), then 𝑦_𝑛≥𝑦 (resp., 𝑤_𝑛≥𝑤) for all 𝑛.
If there exist 𝑥₀,𝑦₀,𝑧₀,𝑤₀∈𝑋 such that 𝑥₀≤𝐹(𝑥₀,𝑦₀,𝑤₀,𝑧₀), 𝑦₀≥𝐹(𝑦₀,𝑧₀,𝑤₀,𝑥₀), 𝑧₀≤𝐹(𝑧₀,𝑤₀,𝑥₀,𝑦₀), and 𝑤₀≥𝐹(𝑤₀,𝑤₀,𝑦₀,𝑧₀) then, there exist 𝑥,𝑦,𝑧,𝑤∈𝑋 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 𝑇∶𝑋→𝑋 is an ICS mapping and 𝐹∶𝑋⁴→𝑋 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))≤𝑘4(𝑑(𝑇𝑥,𝑇𝑢),𝑑(𝑇𝑦,𝑇𝑣),𝑑(𝑇𝑧,𝑇t),𝑑(𝑇𝑤,𝑇𝑠))(2.39)
for any 𝑥,𝑦,𝑧,𝑤,𝑢,𝑣,𝑡,𝑠∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, 𝑧≤𝑡 and 𝑠≤𝑤. Suppose that either(a)𝐹 is continuous, or(b)𝑋 has the following property:(i)if nondecreasing sequence 𝑥_𝑛→𝑥 (resp., 𝑧_𝑛→𝑧), then 𝑥_𝑛≤𝑥 (resp., 𝑧_𝑛≤𝑧) for all 𝑛,(ii)if nonincreasing sequence 𝑦_𝑛→𝑦 (resp., 𝑤_𝑛→𝑤), then 𝑦_𝑛≥𝑦 (resp., 𝑤_𝑛≥𝑤) for all 𝑛.
If there exist 𝑥₀,𝑦₀,𝑧₀,𝑤₀∈𝑋 such that 𝑥₀≤𝐹(𝑥₀,𝑦₀,𝑤₀,𝑧₀), 𝑦₀≥𝐹(𝑦₀,𝑧₀,𝑤₀,𝑥₀), 𝑧₀≤𝐹(𝑧₀,𝑤₀,𝑥₀,𝑦₀) and 𝑤₀≥𝐹(𝑤₀,𝑤₀,𝑦₀,𝑧₀) then there exist 𝑥,𝑦,𝑧,𝑤∈𝑋 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)⟺𝑥≤𝑢,𝑦≥𝑣,𝑧≤𝑡,𝑤≥𝑠.(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, 𝐹𝐹𝐹𝐹((((𝑥𝑤𝑦𝑧,,,,𝑦𝑧𝑤𝑥,,,𝑧,𝑤𝑦,𝑥,,𝑤,𝑧𝑥𝑦))))====𝑥𝑤𝑦𝑧,,,,𝐹𝐹𝐹𝐹((((𝑣𝑡𝑠𝑢,,,,𝑠𝑡𝑢𝑣,,,,𝑠𝑢𝑣𝑡,,,,𝑢𝑣𝑠𝑡))))====𝑣𝑡𝑠𝑢,,.,(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 {𝑎_𝑛},{𝑏_𝑛}, {𝑐_𝑛}, and {𝑑_𝑛} such that 𝑎₀=𝑎,𝑏₀𝑎=𝑏,𝑐₀=𝑐,𝑑₀=𝑑,forany𝑛≥1,_𝑛𝑏_𝑛=𝐹𝑎_𝑛−1,𝑏_𝑛−1,𝑐_𝑛−1,𝑑_𝑛−1,𝑐_𝑛=𝐹𝑏_𝑛−1,𝑐_𝑛−1,𝑑_𝑛−1,𝑎_𝑛−1,𝑑_𝑛=𝐹𝑐_𝑛−1,𝑑_𝑛−1,𝑎_𝑛−1,𝑏_𝑛−1,=𝐹𝑑_𝑛−1,𝑎_𝑛−1,𝑏_𝑛−1,𝑐_𝑛−1,(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 {𝑥_𝑛},{𝑦_𝑛}, {𝑧_𝑛}, and {𝑤_𝑛} and {𝑢_𝑛},{𝑣_𝑛}, {𝑡_𝑛}, and {𝑠_𝑛}. Then, it is easy that 𝑦𝑥_𝑛=𝐹(𝑥,𝑦,𝑧,𝑤),_𝑛𝑧=𝐹(𝑦,𝑧,𝑤,𝑥,),_𝑛𝑤=𝐹(𝑧,𝑤,𝑥,𝑦),_𝑛=𝐹(𝑤,𝑥,𝑦,𝑧),𝑢^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 𝑛≥1. 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 (𝑥,𝑦,𝑧,𝑤)≥𝑎_𝑛,𝑏_𝑛,𝑐_𝑛,𝑑_𝑛∀𝑛.(2.46)
By (2.46) and (2.1), we have 𝑑𝑇𝑥,𝑇𝑎_𝑛+1=≤𝑑𝜙𝑇𝐹m(a𝑥x,𝑦,𝑑𝑧,𝑤𝑇𝑥),,𝑇𝑇𝑎𝐹_𝑛𝑎_𝑛,𝑏_𝑛,𝑐_𝑛,𝑑_𝑛𝑑𝑇𝑏_𝑛+1,𝑑𝑇𝑦,𝑇𝑏_𝑛,𝑑𝑇z,𝑇𝑐_𝑛,𝑑𝑇𝑤,𝑇𝑑_𝑛,,𝑇𝑦≤=𝑑𝜙𝑇m𝐹ax𝑏_𝑛𝑑,𝑐_𝑛𝑇,𝑑𝑦,_𝑛,𝑇𝑎𝑏_𝑛𝑛,𝑇𝐹(𝑦,𝑧,𝑤,𝑥)𝑑𝑇𝑧,𝑇𝑐_𝑛+1,𝑑𝑇𝑧,𝑇𝑐_𝑛,𝑑𝑇𝑤,𝑇𝑑_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,=≤𝑑𝜙𝑇m𝐹ax(𝑧,𝑑𝑤,𝑇𝑥𝑧,𝑦,𝑇),𝑐𝑇_𝑛𝐹𝑐_𝑛,𝑑_𝑛,𝑎_𝑛,𝑏_𝑛𝑑𝑇𝑤,𝑇𝑑_𝑛+1,𝑑𝑇𝑤,𝑇𝑑_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,𝑑𝑇𝑦,𝑇𝑏_𝑛,=≤𝑑𝜙𝑇𝐹m(a𝑤x,𝑥𝑑,𝑦,𝑇𝑧𝑤),,𝑇𝑇𝐹𝑑_𝑛𝑑_𝑛,𝑎_𝑛,𝑏_𝑛,𝑐_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,𝑑𝑇𝑦,𝑇𝑏_𝑛,𝑑𝑇𝑧,𝑇𝑐_𝑛.(2.47)
It follows from (2.47) that max≤𝑑𝜙𝑇𝑧m,a𝑇x𝑐𝑑_𝑛+1,𝑇𝑑𝑧,𝑇𝑇𝑐𝑤_𝑛,𝑇𝑑_𝑛+1,𝑑𝑇𝑥,𝑇𝑎_𝑛+1,𝑑𝑇𝑦,𝑇𝑏_𝑛+1,𝑑𝑇𝑤,𝑇𝑑_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,𝑑𝑇𝑦,𝑇𝑏_𝑛.(2.48)
Therefore, for each 𝑛≥1, max≤𝑑𝜙_𝑛𝑇𝑧,𝑇𝑐_𝑛,𝑑𝑇𝑤,𝑇𝑑_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,𝑑𝑇𝑦,𝑇𝑏_𝑛max𝑑𝑇𝑧,𝑇𝑐₀,𝑑𝑇𝑤,𝑇𝑑₀,𝑑𝑇𝑥,𝑇𝑎₀,𝑑𝑇𝑦,𝑇𝑏₀.(2.49)
It is known that 𝜙(𝑡)<𝑡 and lim_𝑟→𝑡+𝜙(𝑟)<𝑡 imply lim_𝑛→∞𝜙^𝑛(𝑡)=0 for each 𝑡>0. Thus, from (2.49), lim_𝑛→∞max𝑑𝑇𝑧,𝑇𝑐_𝑛,𝑑𝑇𝑤,𝑇𝑑_𝑛,𝑑𝑇𝑥,𝑇𝑎_𝑛,𝑑𝑇𝑦,𝑇𝑏_𝑛=0.(2.50)
This yields that lliimm_{𝑛→∞𝑛→∞}𝑑𝑇𝑥,𝑇𝑎_𝑛=0,lim_𝑛→∞𝑑𝑇𝑦,𝑇𝑏_𝑛=0,𝑑𝑇𝑧,𝑇𝑐_𝑛=0,lim_𝑛→∞𝑑𝑇𝑤,𝑇𝑑_𝑛=0.(2.51)
Analogously, we show that lliimm_{𝑛→∞𝑛→∞}𝑑𝑇𝑢,𝑇𝑎_𝑛=0,lim_𝑛→∞𝑑𝑇𝑣,𝑇𝑏_𝑛=0,𝑑𝑇𝑡,𝑇𝑐_𝑛=0,lim_𝑛→∞𝑑𝑇𝑠,𝑇𝑑_𝑛=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)=|𝑥−𝑦|, for all 𝑥,𝑦∈𝑋 and the usual ordering. Clearly, (𝑋,𝑑) is a complete metric space.
Let 𝑇∶𝑋→𝑋 and 𝐹∶𝑋⁴→𝑋 be defined by 𝑇𝑥=ln(𝑥)+1,𝐹(𝑥,𝑦,𝑧,𝑤)=8⎛⎜⎜⎝𝑥𝑦⎞⎟⎟⎠^1/3,∀𝑥,𝑦,𝑧,𝑤∈𝑋.(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,𝑡∈𝑋 for which 𝑥≤𝑢, 𝑣≤𝑦, 𝑧≤𝑠, and 𝑡≤𝑤, we have 𝑑(𝑇𝐹(𝑥,𝑦,𝑧,𝑤),𝑇𝐹(𝑢,𝑣,𝑠,𝑡))=1≤31||(ln𝑥−ln𝑦)−(ln𝑢−ln𝑣)||≤31||ln𝑥−ln𝑢||+13||ln𝑦−ln𝑣||=𝜙3(mmaaxx{||𝑑ln(𝑇𝑥𝑥−,𝑇ln𝑢)𝑢,||𝑑,||(l𝑇n𝑦𝑦,−𝑇𝑣l)n,𝑣𝑑||(𝑇𝑧,𝑇𝑠),𝑑(𝑇𝑤,𝑇𝑡)}),(2.54)which is the contractive condition (2.1). Moreover, taking 𝑥₀=𝑦₀=𝑧₀=8=𝑤₀, we have 𝑧𝑥₀₀≤𝐹𝑥₀,𝑦₀,𝑧₀,𝑤₀,𝑦₀≥𝐹𝑦₀,𝑧₀,𝑤₀,𝑥₀,≤𝐹𝑧₀,𝑤₀,𝑥₀,𝑦₀,𝑤₀≥𝐹𝑤₀,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, 1≤𝑧=𝑠 and 1≤𝑡=𝑤, we have 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑠,𝑡))=8||(2)^1/3−1||>1=𝑘4>𝑘4𝑑(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=ℝ with 𝑑(𝑥,𝑦)=|𝑥−𝑦| and natural ordering. Let 𝑇∶𝑋→𝑋 and 𝐹∶𝑋⁴→𝑋 be defined by 𝑇𝑥=𝑥/12 and 𝐹(𝑥,𝑦,𝑧,𝑤)=2/5(𝑥−𝑦+𝑧−𝑤). It is clear that 𝑇 is an ICS mapping and 𝐹 has the monotone property and continuous. Set 𝜙(𝑡)=2𝑡/3∈Φ. 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 𝑑(𝐹(𝑥,𝑦,𝑧,𝑤),𝐹(𝑢,𝑣,𝑠,𝑡))=25>𝑘4𝑑(𝑥,𝑢)=𝑘4[𝑑(𝑥,𝑢)+𝑑(𝑦,𝑣)+𝑑(𝑧,𝑠)+𝑑(𝑤,𝑡)],(2.57) for any 𝑘∈[0,1).