# Quadruple Best Proximity Points with Applications to Functional and Integral Equations

Authors: Hasanen A. Hammad; Rashwan A. Rashwan; A. Nafea; Fahd Jarad
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

This manuscript is devoted to obtaining a quadruple best proximity point for a cyclic contraction mapping in the setting of ordinary metric spaces. The validity of the theoretical results is also discussed in uniformly convex Banach spaces. Furthermore, some examples are given to strengthen our study. Also, under suitable conditions, some quadruple fixed point results are presented. Finally, as applications, the existence and uniqueness of a solution to a system of functional and integral equations are obtained to promote our paper.


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## Body

## 1. Introduction and Preliminaries

Fixed point (FP) theory has many applications not only in the nonlinear analysis and its trends, from solutions of differential and integral equations, functional equations arising from dynamical programming, topology, and a dynamical system, but also in economics, game theory, biological sciences, computer sciences, and chemistry, etc. [1–4].The FP technique became more attractive and elegant when Banach [5] introduced his principle, which is stated as follows: a mapping Γ:Y⟶Y defined on a complete metric space (MS) Y,d has a unique FP if Γ is a contraction, i.e., dΓσ1,Γσ2≤kdσ1,σ2,k∈0,1. He used this method for studying the existence of solutions for some integral equations.The FP technique was extended to a coupled and tripled FP by Bhaskar and Lakshmikantham [6] and Berinde and Borcut [7], respectively. Many researchers have worked in these directions and obtained exciting results and life applications that serve the scientific communities, which in turn has led to the fixed points being brilliant and pioneering in the field of functional analysis until the present time. For more details, see [8–20].Not only did the matter stop here, but Karapinar and Sadarangani [21] were able to generalize the triple point to the quadruple and generalized the previous results on this scale in terms of theories and applications. After that, this trend spreads to others; see, for instance, [22–25].In 1978, Pathak and B. Fisher [26] were able to merge the state and decision space to clarify the importance of the FP methodology in finding the solution to the following functional system:
(1)ησ1=supσ2∈Dℵσ1,σ2,ηIσ1,σ2,where S and D are the state space and the decision space, respectively; σ1∈S,I denotes the transformation of the process; and ησ1 refers to the optimal return function with the initial state σ1. The above system is called a functional equation arising from dynamical programming which is commonly used in modeling and optimization problems. To clarify the participation of fixed and coincidence points and to delve deeper into this trend, we guide the reader to read these papers [27–30].On the other hand, the importance of the FP technique lies in the fact that it presents a unified process and an important tool in solving equations that do not have to be linear. In the case ofdz,Γz≠0, that is, a contraction mapping Γ does not possess a FP; it became necessary to search a pointzthat makesdz,Γzthe minimum, meaning the pointzis in close proximity toΓz.Assume thatZ,O are nonempty closed subsets of a complete MS Y and Γ:Z⟶O is a given mapping. A point z∈Z is called the best proximity point (BPP, for short) if dz,Γz=dZ,O, where dZ,O is described as
(2)dZ,O=infdz,o:z∈Z,o∈O.It should be noted that ifZ∩O≠∅, then a BPP reduces to a usual FP.The initial paper concerned with the BPP was presented by Fan [31]. This direction is very interesting in optimization, so many researchers have discussed and developed this direction in several applications; see, [32–36].Now, we need the definitions below.Definition 1 (see [37]).
A Banach spaceY is said to be
(i)
strictly convex, if(3)σ1=σ2=1withσ1≠σ2impliesσ1+σ22<1,∀σ1,σ2∈Y,(ii)
uniformly convex, if for anyε∈0,2, there is δ>0 so that(4)σ1≤1,σ2≤1withσ1−σ2≥εimpliesσ1+σ22<1−δ,∀σ1,σ2∈Y.Clearly, a uniformly convex Banach space (UCBS) is strictly convex, but the opposite does not hold.Definition 2 (see [21]).
Assume thatY,d is a MS and Z≠∅ is a subset of Y. A point σ1,σ2,σ3,σ4∈Z4 is called a quadruple fixed point (QFP, for short) of the map Θ:Z4⟶Z if
(5)σ1=Θσ1,σ2,σ3,σ4,σ2=Θσ2,σ3,σ4,σ1,σ3=Θσ3,σ4,σ1,σ2,σ4=Θσ4,σ1,σ2,σ3.Our paper is arranged as follows: In Section2, some new definitions and supporting examples are presented. Also, the convergence of quadruple best proximity (QBP, for short) points for a pair of cyclic contraction mappings without and with property UC∗ are obtained in the context of metric spaces (MSs). Moreover, quadruple fixed point (QFP) results for cyclic contraction mappings are established, and an example for supporting the above results is discussed in Section 3. In Section 4, the existence of a solution for quadruple functional equations arising in dynamical programming is discussed, and an example for supporting the results is presented. Ultimately, in Section 5, the existence of solutions for a system of quadruple integral equations is given, and an example is obtained to strengthen this contribution.


## 2. Main Results

This part is devoted to present the convergence of QBP points for a pair of cyclic contraction mappings in the setting of ordinary MSs.We begin this part with the definitions below.Definition 3.
LetZ and O be two nonempty closed subsets of a MS Y,d. The pair Z,O is called to satisfy the property UC, if there exist σ1n,σ4n⊂Z and σ2n,σ3n⊂O such that
(6)dσ1n,σ2n⟶dZ,O,dσ1n,σ3n⟶dZ,O,dσ4n,σ2n⟶dZ,O,dσ4n,σ3n⟶dZ,O,as n⟶∞, then
(7)dσ1n,σ4n⟶0,asn⟶∞.Example 1.
Suppose thatZ and O are two nonempty subsets of a MS Y,d with dZ,O=0. Then, Z,O satisfies the property UC.Example 2.
Assume thatZ,Z′,O, and O′ are nonempty subsets of a MS Y,d with Z⊂Z′,O⊂O′, and dZ,O=dZ′,O′. If the pair Z,O satisfies the property UC, then the pair Z′,O′ satisfies also the property UC.Example 3.
Suppose thatZ and O are nonempty subsets of a UCBS, then the pair Z,O satisfies the property UC, if one of the hypotheses below holds:
(i)
Z is convex(ii)
Z is convex relatively compactDefinition 4.
Assume thatY,d is a MS and Z, O are two nonempty subsets of Y. We say that the pair Z,O has the property UC∗ if Z,O has the property UC and the stipulation below is fulfilled:
If the sequencesσ1n,σ4n in Z and σ2n,σ3n in O so that the following statements are fulfilled:
(†1) dσ4n,σ2n⟶dZ,O and dσ4n,σ3n⟶dZ,O
(†2) ∀ε>0,∃N∈ℕ so that
(8)dσ1m,σ2n≤dZ,O+ε,∀m>n≥N,dσ1m,σ3n≤dZ,O+ε,∀m>n≥N,then there is N1∈ℕ so that
(9)dσ1m,σ4n≤dZ,O+ε,∀m>n≥N1.Example 4.
For nonempty subsetsZ and O of a MS Y,d, assume that dZ,O=0. Then, the pair Z,O posses the property UC∗.Example 5.
Assume that all requirements of Example2 hold. If the pair Z,O has the property UC∗, then the pair Z′,O′ has the property UC∗ too.Example 6.
Assume thatZ,O are two nonempty subsets of a UCBS and Z is convex. Then, Z,O verifies the property UC∗.Definition 5.
Assume thatY,d is a MS and Z, O are nonempty closed subsets of Y. Also assume Θ:Z4⟶O is a given mapping. We say that a quadruple σ1,σ2,σ3,σ4∈Z4 is a QBP point of Θ if,
(10)dσ1,Θσ1,σ2,σ3,σ4=dσ2,Θσ2,σ3,σ4,σ1=dσ3,Θσ3,σ4,σ1,σ2=dσ4,Θσ4,σ1,σ2,σ3=dZ,O.
Clearly, whenZ=O in Definition 5, then a QBP point reduces to a QFP.Definition 6.
Assume thatY,d is a MS and Z, O are nonempty closed subsets of Y. We say that the mappings Θ:Z4⟶O and Ω:O4⟶Z are cyclic contractions if there is ℓ∈0,1 so that the inequality below holds:
(11)dΘσ1,σ2,σ3,σ4,Ωϑ1,ϑ2,ϑ3,ϑ4≤ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4+1−ℓdZ,O,for all σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4.Notice that, if the pairΘ,Ω is a cyclic contraction, then the pair Ω,Θ is a cyclic contraction too.Example 7.
ConsiderY=ℝ equipped with the distance dσ1,σ2=σ1−σ2. Let Z=3,9 and O=−9,−3. Obviously, dZ,O=6. Describe two mappings Θ:Z4⟶O and Ω:O4⟶Z as
(12)Θσ1,σ2,σ3,σ4=−σ1−σ2−σ3−σ4−128,Ωϑ1,ϑ2,ϑ3,ϑ4=−ϑ1−ϑ2−ϑ3−ϑ4+128,for all σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4, respectively. For each σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4 and let ℓ=1/2, we have
(13)dΘσ1,σ2,σ3,σ4,Ωϑ1,ϑ2,ϑ3,ϑ4=−σ1−σ2−σ3−σ4−128−−ϑ1−ϑ2−ϑ3−ϑ4+128≤σ1−ϑ1+σ2−ϑ2+σ3−ϑ3+σ4−ϑ48+3=ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4+1−ℓdZ,O.This leads to the pairΘ,Ω as a cyclic contraction with ℓ=1/2.Example 8.
ConsiderY=ℝ4 endowed with
(14)dσ1,σ2,σ3,σ4,ϑ1,ϑ2,ϑ3,ϑ4=maxσ1−ϑ1,σ2−ϑ2,σ3−ϑ3,σ4−ϑ4,for all σ1,σ2,σ3,σ4,ϑ1,ϑ2,ϑ3,ϑ4∈Y and suppose that
(15)Z=σ1,0,0,0∈ℝ4:0≤σ1≤1,O=ϑ1,1,1,1∈ℝ4:0≤ϑ1≤1.Clearly,dZ,O=1. Define Θ:Z4⟶O and Ω:O4⟶Z by
(16)Θσ1,0,0,0,σ2,0,0,0,σ3,0,0,0,σ4,0,0,0=σ1+σ2+σ3+σ44,1,1,1,Ωϑ1,1,1,1,ϑ2,1,1,1,ϑ3,1,1,1,ϑ4,1,1,1=ϑ1+ϑ2+ϑ3+ϑ44,0,0,0,respectively. Then, we obtain
(17)dΘσ1,0,0,0,σ2,0,0,0,σ3,0,0,0,σ4,0,0,0,Ωϑ1,1,1,1,ϑ2,1,1,1,ϑ3,1,1,1,ϑ4,1,1,1=dσ1+σ2+σ3+σ44,1,1,1,ϑ1+ϑ2+ϑ3+ϑ44,0,0,0=1.Also, if(18)σ1,0,0,0,σ2,0,0,0,σ3,0,0,0,σ4,0,0,0∈Z,ϑ1,1,1,1,ϑ2,1,1,1,ϑ3,1,1,1,ϑ4,1,1,1∈O,then one can write
(19)ℓ4dσ1,0,0,0,ϑ1,1,1,1+dσ2,0,0,0,ϑ2,1,1,1+dσ3,0,0,0,ϑ3,1,1,1+dσ4,0,0,0,ϑ4,1,1,1+1−ℓdZ,O=ℓ4maxσ1−ϑ1,1,1,1+maxσ2−ϑ2,1,1,1+maxσ3−ϑ3,1,1,1+maxσ4−ϑ4,1,1,1+1−ℓdZ,O=ℓ4×4+1−ℓ=1,for any ℓ<1. In addition, let
(20)σ1,σ2,σ3,σ4=σ1,0,0,0,σ2,0,0,0,σ3,0,0,0,σ4,0,0,0∈Z4,ϑ1,ϑ2,ϑ3,ϑ4=ϑ1,1,1,1,ϑ2,1,1,1,ϑ3,1,1,1,ϑ4,1,1,1∈O4,then it follows from (17) and (19) that
(21)dΘσ1,σ2,σ3,σ4,Ωϑ1,ϑ2,ϑ3,ϑ4≤ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4+1−ℓdZ,O.Thus, the pairΘ,Ω is a cyclic contraction.The lemma below is very important in the sequel.Lemma 7.
Assume thatY,d is a MS and Z,O are nonempty closed subsets of Y. Let Θ:Z4⟶O and Ω:O4⟶Z be two cyclic contraction mappings. If σ10,σ20,σ30,σ40∈Z4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are defined as follows:
(22)σ12n+1=Θσ12n,σ22n,σ32n,σ42n,σ12n+2=Ωσ12n+1,σ22n+1,σ32n+1,σ42n+1,σ22n+1=Θσ22n,σ32n,σ42n,σ12n,σ22n+2=Ωσ22n+1,σ32n+1,σ42n+1,σ12n+1,σ32n+1=Θσ32n,σ42n,σ12n,σ22n,σ32n+2=Ωσ32n+1,σ42n+1,σ12n+1,σ22n+1,σ42n+1=Θσ42n,σ12n,σ22n,σ32n,σ42n+2=Ωσ42n+1,σ12n+1,σ22n+1,σ32n+1,for all n≥0, then we get
(23)dσ12n,σ12n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O,dσ22n,σ22n+1⟶dZ,O,dσ22n+1,σ22n+2⟶dZ,O,dσ32n,σ32n+1⟶dZ,O,dσ32n+1,σ32n+2⟶dZ,O,dσ42n,σ42n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O.Proof.
Consider, for eachn∈ℕ,(24)dσ12n,σ12n+1=dσ12n,Θσ12n,σ22n,σ32n,σ42n=dΩσ12n−1,σ22n−1,σ32n−1,σ42n−1,ΘΩσ12n−1,σ22n−1,σ32n−1,σ42n−1,Ωσ22n−1,σ32n−1,σ42n−1,σ12n−1,Ωσ32n−1,σ42n−1,σ12n−1,σ22n−1,Ωσ42n−1,σ12n−1,σ22n−1,σ32n−1≤ℓ4dσ12n−1,Ωσ12n−1,σ22n−1,σ32n−1,σ42n−1+dσ22n−1,Ωσ22n−1,σ32n−1,σ42n−1,σ12n−1+dσ32n−1,Ωσ32n−1,σ42n−1,σ12n−1,σ22n−1+dσ42n−1,Ωσ42n−1,σ12n−1,σ22n−1,σ32n−1+1−ℓdZ,O=ℓ4dΘσ12n−2,σ22n−2,σ32n−2,σ42n−2,ΩΘσ12n−2,σ22n−2,σ32n−2,σ42n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+dΘσ22n−2,σ32n−2,σ42n−2,σ12n−2,ΩΘσ22n−2,σ32n−2,σ42n−2,σ12n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2Θσ42n−2,σ12n−2,σ22n−2,σ32n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+dΘσ32n−2,σ42n−2,σ12n−2,σ22n−2,ΩΘσ32n−2,σ42n−2,σ12n−2,σ22n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+dΘσ42n−2,σ12n−2,σ22n−2,σ32n−2,ΩΘσ42n−2,σ12n−2,σ22n−2,σ32n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2Θσ22n−2,σ32n−2,σ42n−2,σ12n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2+1−ℓdZ,O.
Using (11), we have
(25)dσ12n,σ12n+1≤ℓ4ℓ4dσ12n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+dσ22n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+dσ32n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2+dσ42n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+1−ℓdZ,O+ℓ4dσ22n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+dσ32n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2+dσ42n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+dσ12n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+1−ℓdZ,O+ℓ4dσ32n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2,dσ42n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+dσ12n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+dσ22n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+1−ℓdZ,O+ℓ4dσ42n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+dσ12n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+dσ22n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+dσ32n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2+1−ℓdZ,O+1−ℓdZ,O=ℓ24dσ12n−2,Θσ12n−2,σ22n−2,σ32n−2,σ42n−2+dσ22n−2,Θσ22n−2,σ32n−2,σ42n−2,σ12n−2+dσ32n−2,Θσ32n−2,σ42n−2,σ12n−2,σ22n−2+dσ42n−2,Θσ42n−2,σ12n−2,σ22n−2,σ32n−2+1−ℓ2dZ,O.
By mathematical induction, we obtain for eachn∈ℕ that
(26)dσ12n,σ12n+1≤ℓ2n4dσ10,Θσ10,σ20,σ30,σ40+dσ20,Θσ20,σ30,σ40,σ10+dσ30,Θσ30,σ40,σ10,σ20+dσ40,Θσ40,σ10,σ20,σ30+1−ℓ2dZ,O.
Passingn⟶∞, we find that
(27)dσ12n,σ12n+1⟶dZ,O.
Again, for eachn∈ℕ, by induction, one can write
(28)dσ12n+1,σ12n+2≤ℓ2n4dσ11,Θσ11,σ21,σ31,σ41+dσ21,Θσ21,σ31,σ41,σ11+dσ31,Θσ31,σ41,σ11,σ21+dσ41,Θσ41,σ11,σ21,σ31+1−ℓ2dZ,O,this yields after passing n⟶∞,(29)dσ12n+1,σ12n+2⟶dZ,O.
Analogously, we have(30)dσ22n,σ22n+1⟶dZ,O,dσ22n+1,σ22n+2⟶dZ,O,dσ32n,σ32n+1⟶dZ,O,dσ32n+1,σ32n+2⟶dZ,O,dσ42n,σ42n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O.
This finishes the required proof.Lemma 8.
Assume thatY,d is a MS and Z, O are nonempty closed subsets of Y so that Z,O and O,Z satisfy the property UC. Let the mappings Θ:Z4⟶O and Ω:O4⟶Z be cyclic contractions. If σ10,σ20,σ30,σ40∈Z4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are defined as (22), ∀n∈ℕ∪0, then for each ε>0, there is N0>0 so that
(31)14dσ12m,σ12n+1+dσ22m,σ22n+1+dσ32m,σ32n+1+dσ42m,σ42n+1<dZ,O+ε,∀m>n≥N0.Proof.
According to Lemma7, we get
(32)dσ12n,σ12n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O,dσ22n,σ22n+1⟶dZ,O,dσ22n+1,σ22n+2⟶dZ,O,dσ32n,σ32n+1⟶dZ,O,dσ32n+1,σ32n+2⟶dZ,O,dσ42n,σ42n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O.
BecauseZ,O fulfills the property UC, then we have
(33)dσ12n,σ12n+2⟶0,dσ22n,σ22n+2⟶0,dσ32n,σ32n+2⟶0,dσ42n,σ42n+2⟶0.
Also,O,Z verifies the property UC, we have
(34)dσ12n+1,σ12n+3⟶0,dσ22n+1,σ22n+3⟶0,dσ32n+1,σ32n+3⟶0,dσ42n+1,σ42n+3⟶0.
Assuming (31) is not true. Then, for each k∈ℕ with mk>nk≥k, there is ε′>0 so that
(35)14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1≥dZ,O+ε′.
Hence, we can select the smallest integermk with mk>nk fulfilling (35). Therefore,
(36)14dσ12mk−2,σ12nk+1+dσ22mk−2,σ22nk+1+dσ32mk−2,σ32nk+1+dσ42mk−2,σ42nk+1<dZ,O+ε′.
Thus, we obtain(37)dZ,O+ε′≤14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1≤14dσ12mk,σ12mk−2+dσ12mk−2,σ12nk+1+dσ22mk,σ22mk−2+dσ22mk−2,σ22nk+1+dσ32mk,σ32mk−2+dσ32mk−2,σ32nk+1+dσ42mk,σ42mk−2+dσ42mk−2,σ42nk+1<14dσ12mk,σ12mk−2+dσ22mk,σ22mk−2+dσ32mk,σ32mk−2+dσ42mk,σ42mk−2+dZ,O+ε′.
Ask⟶∞, one can write
(38)14dσ12mk,σ12mk−2+dσ22mk,σ22mk−2+dσ32mk,σ32mk−2+dσ42mk,σ42mk−2⟶dZ,O+ε′.
Applying the triangle inequality, we get(39)14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1≤14dσ12mk,σ12mk+2+dσ12mk+2,σ12nk+3+dσ12nk+3,σ12nk+1+dσ22mk,σ22mk+2+dσ22mk+2,σ22nk+3+dσ22nk+3,σ22nk+1+dσ32mk,σ32mk+2+dσ32mk+2,σ32nk+3+dσ32nk+3,σ32nk+1+dσ42mk,σ42mk+2+dσ42mk+2,σ42nk+3+dσ42nk+3,σ42nk+1=14dσ12mk,σ12mk+2+dΩσ12mk+1,σ22mk+1,σ32mk+1,σ42mk+1,Θσ12nk+2,σ22nk+2,σ32nk+2,σ42nk+2+dσ12nk+3,σ12nk+1+14dσ22mk,σ22mk+2+dΩσ22mk+1,σ32mk+1,σ42mk+1,σ12mk+1,Θσ22nk+2,σ32nk+2,σ42nk+2,σ12nk+2++dσ22nk+3,σ22nk+1+14dσ32mk,σ32mk+2+dΩσ32mk+1,σ42mk+1,σ12mk+1,σ22mk+1,Θσ32nk+2,σ42nk+2,σ12nk+2,σ22nk+2+dσ32nk+3,σ32nk+1+14dσ42mk,σ42mk+2+dΩσ42mk+1,σ12mk+1,σ22mk+1,σ32mk+1,Θσ42nk+2,σ12nk+2,σ22nk+2,σ32nk+2+dσ42nk+3,σ42nk+1.
Applying (11), we obtain that
(40)14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1≤14dσ12mk,σ12mk+2+ℓ4dσ12mk+1,σ12nk+2+dσ22mk+1,σ22nk+2+dσ32mk+1,σ32nk+2+dσ42mk+1,σ42nk+2+1−ℓdZ,O+dσ12nk+3,σ12nk+1+dσ22mk,σ22mk+2+ℓ4dσ22mk+1,σ22nk+2+dσ32mk+1,σ32nk+2+dσ42mk+1,σ42nk+2+dσ12mk+1,σ12nk+2+1−ℓdZ,O+dσ22nk+3,σ22nk+1+dσ32mk,σ32mk+2+ℓ4dσ32mk+1,σ32nk+2+dσ42mk+1,σ42nk+2+dσ12mk+1,σ12nk+2+dσ22mk+1,σ22nk+2+1−ℓdZ,O+dσ32nk+3,σ32nk+1+dσ42mk,σ42mk+2+ℓ4dσ42mk+1,σ42nk+2+dσ12mk+1,σ12nk+2+dσ22mk+1,σ22nk+2+dσ32mk+1,σ32nk+2+1−ℓdZ,O+dσ42nk+3,σ42nk+1=14dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1+dσ22mk,σ22mk+2+dσ22nk+3,σ22nk+1+dσ32mk,σ32mk+2+dσ32nk+3,σ32nk+1+dσ42mk,σ42mk+2+dσ42nk+3,σ42nk+1+ℓ4dσ12mk+1,σ12nk+2+dσ22mk+1,σ22nk+2+dσ32mk+1,σ32nk+2+dσ42mk+1,σ42nk+2+1−ℓdZ,O.
It follows that(41)14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1≤14dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1+dσ22mk,σ22mk+2+dσ22nk+3,σ22nk+1+dσ32mk,σ32mk+2+dσ32nk+3,σ32nk+1+dσ42mk,σ42mk+2+dσ42nk+3,σ42nk+1+ℓ4dΘσ12mk,σ22mk,σ32mk,σ42mk,Ωσ12nk+1,σ22nk+1,σ32nk+1,σ42nk+1+dΘσ22mk,σ32mk,σ42mk,σ12mk,Ωσ22nk+1,σ32nk+1,σ42nk+1,σ12nk+1+dΘσ32mk,σ42mk,σ12mk,σ22mk,Ωσ32nk+1,σ42nk+1,σ12nk+1,σ22nk+1+dΘσ42mk,σ12mk,σ22mk,σ32mk,Ωσ42nk+1,σ12nk+1,σ22nk+1,σ32nk+1+1−ℓdZ,O≤14dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1+dσ22mk,σ22mk+2+dσ22nk+3,σ22nk+1+dσ32mk,σ32mk+2+dσ32nk+3,σ32nk+1+dσ42mk,σ42mk+2+dσ42nk+3,σ42nk+1+ℓ24dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1+1−ℓdZ,O.
Lettingn⟶∞, we have
(42)dZ,O+ε′≤ℓ2dZ,O+ε′+1−ℓ2dZ,O=dZ,O+ℓ2ε′,which is a contradiction since ℓ<1. This implies that (31) is fulfilled, and this finishes the proof.Lemma 9.
Assume thatY,d is a MS and Z, O are nonempty closed subsets of Y so that Z,O and O,Z satisfy the property UC∗. Let the mappings Θ:Z4⟶O and Ω:Z4⟶O be cyclic contractions. If σ10,σ20,σ30,σ40∈Z4 and for all n∈ℕ∪0, the sequences σ1n,σ2n,σ3n,σ4n in Y are defined by (22). Then, σ12n,σ22n,σ32n, and σ42n are Cauchy sequences.Proof.
Based on Lemma7, one can get
(43)dσ12n,σ12n+1⟶dZ,O,dσ12n+1,σ12n+2⟶dZ,O.
AsZ,O satisfies the property UC, then dσ12n,σ12n+2⟶0. Similarly, since O,Z verifies the property UC, then dσ12n+1,σ12n+3⟶0.
Now, we claim that,∀ε>0, ∃N∈ℕ so that
(44)dσ12mk,σ12nk+1≤dZ,O+ε,∀m>n≥N.
Assume that (44) is not true. Then, ∀k∈ℕ, ∃ε>0, and mk>nk≥k such that
(45)dσ12mk,σ12nk+1>dZ,O+ε.
Therefore, we can select a smallest integermk with mk>nk fulfilling (45). Hence, one can get
(46)dZ,O+ε<dσ12mk,σ12nk+1≤dσ12mk,σ12mk−2+dσ12mk−2,σ12nk+1≤dσ12mk,σ12mk−2+dZ,O+ε.
Settingk⟶∞, we have
(47)dσ12mk,σ12nk+1⟶dZ,O+ε.
Using Lemma8, we can write
(48)14dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1<dZ,O+ε,for all mk>nk≥k. Applying the triangle inequality, we get
(49)dZ,O+ε<dσ12mk,σ12nk+1≤dσ12mk,σ12mk+2+dσ12mk+2,σ12nk+3+dσ12nk+3,σ12nk+1=dσ12mk,σ12mk+2+dΩσ12mk+1,σ22mk+1,σ32mk+1,σ42mk+1,Θσ12nk+2,σ22nk+2,σ32nk+2,σ42nk+2+dσ12nk+3,σ12nk+1≤dσ12mk,σ12mk+2+ℓ4dσ12mk+1,σ12mk+2+dσ22mk+1,σ22mk+2+dσ32mk+1,σ32mk+2+dσ42mk+1,σ42mk+2+1−ℓdZ,O+dσ12nk+3,σ12nk+1=ℓ4dΘσ12mk,σ22mk,σ32mk,σ42mk,Ωσ12nk+1,σ22nk+1,σ32nk+1,σ42nk+1+dΘσ22mk,σ32mk,σ42mk,σ12mk,Ωσ22nk+1,σ32nk+1,σ42nk+1,σ12nk+1+dΘσ32mk,σ42mk,σ12mk,σ22mk,Ωσ32nk+1,σ42nk+1,σ12nk+1,σ22nk+1+dΘσ42mk,σ12mk,σ22mk,σ32mk,Ωσ42nk+1,σ12nk+1,σ22nk+1,σ32nk+1+dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1+1−ℓdZ,O≤ℓ4ℓ4dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1+1−ℓdZ,O+ℓ4dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1+dσ12mk,σ12nk+1+1−ℓdZ,Oℓ4dσ32mk,σ32nk+1+dσ42mk,σ42nk+1+dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+1−ℓdZ,O+ℓ4dσ42mk,σ42nk+1+dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+1−ℓdZ,O+dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1+1−ℓdZ,O=ℓ24dσ12mk,σ12nk+1+dσ22mk,σ22nk+1+dσ32mk,σ32nk+1+dσ42mk,σ42nk+1+1−ℓ2dZ,O+dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1<ℓ2dZ,O+ε+1−ℓ2dZ,O+dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1=ℓ2ε+dZ,O+dσ12mk,σ12mk+2+dσ12nk+3,σ12nk+1.
Lettingn⟶∞, we obtain that
(50)dZ,O+ε≤dZ,O+ℓ2ε.This is a contradiction. This achieves the inequality (44). It follows from (44), dσ12n,σ12n+1⟶dZ,O and the property UC∗ of dZ,O that σ12n is a Cauchy sequence. By the same manner, we can show that σ22n,σ32n, and σ42n are Cauchy sequences. This finishes the proof.Now, via the propertyUC∗, we shall discuss the existence and convergence of QBP points.Theorem 10.
Assume thatZ and O are nonempty closed subsets of Y so that the property UC∗ are satisfied on Z,O and O,Z. Let the mappings Θ:Z4⟶O and Ω:O4⟶Z be cyclic contractions. If σ10,σ20,σ30,σ40∈Z4 and for all n∈ℕ∪0, the sequences σ1n,σ2n,σ3n,σ4n in Y are described as (22). Then, Θ has a QBP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4 and Ω has a QBP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈O4. Moreover, we have
(51)σ12n⟶ℸ1,σ22n⟶ℸ2,σ32n⟶ℸ3,σ42n⟶ℸ4,σ12n+1⟶ℸ1′,σ22n+1⟶ℸ2′,σ32n+1⟶ℸ3′,σ42n+1⟶ℸ4′.
In addition, ifℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then
(52)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.Proof.
Based on Lemma7, we conclude that dσ12n,σ12n+1⟶dZ,O. From Lemma 9, we find that σ12n,σ22n,σ32n, and σ42n are Cauchy sequences. Thus, there are ℸ1,ℸ2,ℸ3,ℸ4∈Z so that σ12n⟶ℸ1, σ22n⟶ℸ2, σ32n⟶ℸ3, and σ42n⟶ℸ4. Hence, we have
(53)dZ,O≤dℸ1,σ12n−1≤dℸ1,σ12n+dσ12n,σ12n−1.
Passingn⟶∞ in (53), we find that
(54)dℸ1,σ12n−1⟶dZ,O.
By the same method, we have(55)dℸ2,σ12n−1⟶dZ,O,dℸ3,σ32n−1⟶dZ,O,dℸ4,σ42n−1⟶dZ,O.
Now, consider(56)dσ12n,Θℸ1,ℸ2,ℸ3,ℸ4=dΩσ12n−1,σ22n−1,σ32n−1,σ42n−1,Θℸ1,ℸ2,ℸ3,ℸ4≤ℓ4dσ12n−1,ℸ1,dσ22n−1,ℸ2,dσ32n−1,ℸ3,dσ42n−1,ℸ4+1−ℓdZ,O.
Passingn⟶∞, we obtain
(57)dℸ1,Θℸ1,ℸ2,ℸ3,ℸ4=dZ,O.
Analogously, we can obtain(58)dℸ2,Θℸ2,ℸ3,ℸ4,ℸ1=dZ,O,dℸ3,Θℸ3,ℸ4,ℸ1,ℸ2=dZ,O,dℸ4,Θℸ4,ℸ1,ℸ2,ℸ3=dZ,O.
Therefore,ℸ1,ℸ2,ℸ3,ℸ4 is a QBP point of Θ.
Analogously, we can prove that there areℸ1′,ℸ2′,ℸ3′,ℸ4′∈O so that σ12n+1⟶ℸ1′, σ22n+1⟶ℸ2′, σ32n+1⟶ℸ3′, and σ42n+1⟶ℸ4′. Moreover, we get
(59)dℸ1′,Θℸ1′,ℸ2′,ℸ3′,ℸ4′=dZ,O,dℸ2′,Θℸ2′,ℸ3′,ℸ4′,ℸ1′=dZ,O,dℸ3′,Θℸ3′,ℸ4′,ℸ1′,ℸ2′=dZ,O,dℸ4′,Θℸ4′,ℸ1′,ℸ2′,ℸ3′=dZ,O.
Hence,ℸ1′,ℸ2′,ℸ3′,ℸ4′ is a QBP point of Ω.
Ultimately, letℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then we claim that (52) holds. For each n∈ℕ, one can write
(60)dσ12n,σ12n+1=dΩσ12n−1,σ22n−1,σ32n−1,σ42n−1,Θσ12n,σ22n,σ32n,σ42n≤ℓ4dσ12n−1,σ12n+dσ22n−1,σ22n+dσ32n−1,σ32n+dσ42n−1,σ42n+1−ℓdZ,O.
Lettingn⟶∞, we get
(61)dℸ1,ℸ1′≤ℓ4dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′+1−ℓdZ,O.
Also,∀n∈ℕ, we get
(62)dσ22n,σ22n+1=dΩσ22n−1,σ32n−1,σ42n−1,σ12n−1,Θσ22n,σ32n,σ42n,σ12n≤ℓ4dσ22n−1,σ22n+dσ32n−1,σ32n+dσ42n−1,σ42n+dσ12n−1,σ12n+1−ℓdZ,O.
Passingn⟶∞, one can obtain
(63)dℸ2,ℸ2′≤ℓ4dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′+dℸ1,ℸ1′+1−ℓdZ,O.
Similarly, we obtain(64)dℸ3,ℸ3′≤ℓ4dℸ3,ℸ3′+dℸ4,ℸ4′+dℸ1,ℸ1′+dℸ2,ℸ2′+1−ℓdZ,O,dℸ4,ℸ4′≤ℓ4dℸ4,ℸ4′+dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+1−ℓdZ,O.
It follows from (61), (63), (64), and (66) that
(65)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′≤ℓdℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′+41−ℓdZ,O.This leads to
(66)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′≤4dZ,O.
Since(67)dZ,O≤dℸ1,ℸ1′,dZ,O≤dℸ2,ℸ2′,dZ,O≤dℸ3,ℸ3′,dZ,O≤dℸ4,ℸ4′,then we have
(68)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′≥4dZ,O.
According to (66) and (68), we have
(69)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.
This finishes the proof.It should be noted that every pair of nonempty closed subsetsZ and O of a UCBS so that Z is convex fulfills the property UC∗, then we can state the result below.Corollary 11.
Assume thatY is a UCBS and Z, O are nonempty closed subsets of Y. Assume also Θ:Z4⟶O and Ω:O4⟶Z are cyclic contraction mappings. If σ10,σ20,σ30,σ40∈Z4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are defined as (22), for each n∈ℕ∪0, then Θ has a QBP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4 and Ω has a QBP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈Z4. Moreover, we get
(70)σ12n⟶ℸ1,σ22n⟶ℸ2,σ32n⟶ℸ3,σ42n⟶ℸ4,σ12n+1⟶ℸ1′,σ22n+1⟶ℸ2′,σ32n+1⟶ℸ3′,σ42n+1⟶ℸ4′.
In addition, ifℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then
(71)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.The following example supports Corollary11.Example 9.
LetY=ℝ be a UCBS equipped with the usual norm. Take Z=3,5 and O=−5,−3. Obviously, dZ,O=8. Describe two mappings Θ:Z4⟶O and Ω:O4⟶Z as
(72)Θσ1,σ2,σ3,σ4=−σ1−σ2−σ3−σ4−168,Ωϑ1,ϑ2,ϑ3,ϑ4=−ϑ1−ϑ2−ϑ3−ϑ4+168,for each σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4, respectively. For all σ1,σ2,σ3,σ4∈Z4, ϑ1,ϑ2,ϑ3,ϑ4∈O4, and fixed ℓ=1/2, we have
(73)dΘσ1,σ2,σ3,σ4,Ωϑ1,ϑ2,ϑ3,ϑ4=−σ1−σ2−σ3−σ4−168−−ϑ1−ϑ2−ϑ3−ϑ4+168≤σ1−ϑ1+σ2−ϑ2+σ3−ϑ3+σ4−ϑ48+4=ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4+1−ℓdZ,O.
Thus, the mappingsΘ and Ω are cyclic contractions with ℓ=1/2. Because Z and O are closed convex, the pairs Z,O and O,Z justify the property UC∗. Therefore, all requirements of Corollary 11 are fulfilled. Thus, Θ has a QBP point and Ω has a QBP point. We note that a point 4,4,4,4∈Z4 is a unique QBP point of Θ and a point −4,−4,−4,−4∈V4 is a unique QBP point of Ω. Therefore, we obtain
(74)d4,−4+d4,−4+d4,−4+d4,−4=32=4dZ,O.In a compact subset of a MS, we can obtain the QBP point result as follows.Theorem 12.
Assume thatY,d is a MS and Z, O are nonempty compact subsets of Y. Assume also Θ:Z4⟶O and Ω:O4⟶Z are cyclic mappings. If σ10,σ20,σ30,σ40∈Z4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are defined in (22), for each n≥0. Then Θ has a QBP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4 and Ω has a QBP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈Z4. Moreover, we get
(75)σ12n⟶ℸ1,σ22n⟶ℸ2,σ32n⟶ℸ3,σ42n⟶ℸ4,σ12n+1⟶ℸ1′,σ22n+1⟶ℸ2′,σ32n+1⟶ℸ3′,σ42n+1⟶ℸ4′.
In addition, ifℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then
(76)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.Proof.
Sinceσ10,σ20,σ30,σ40∈Z and (22) holds for each n∈ℕ∪0, we get
(77)σ12n,σ22n,σ32n,σ42n∈Z,σ12n+1,σ22n+1,σ32n+1,σ42n+1∈O.
The compactness ofZ illustrates that the sequences σ12n,σ22n,σ32n, and σ42n have the convergent subsequences σ12nk,σ22nk,σ32nk, and σ42nk, respectively, so that
(78)σ12nk⟶ℸ1∈Z,σ22nk⟶ℸ2∈Z,σ32nk⟶ℸ3∈Z,σ42nk⟶ℸ4∈Z.
Now, we have(79)dZ,O≤dℸ1,σ12nk−1≤dℸ1,σ12nk+dσ12nk,σ12nk−1.
Applying Lemma7, we find that
(80)dσ12nk,σ12nk−1⟶dZ,O.
Takingn⟶∞ in (79), we get
(81)dℸ1,σ12nk−1⟶dZ,O.
By the same manner, one can obtain(82)dℸ2,σ22nk−1⟶dZ,O,dℸ3,σ32nk−1⟶dZ,O,dℸ4,σ42nk−1⟶dZ,O.
Notice that(83)dZ,O=dσ12nk,Θℸ1,ℸ2,ℸ3,ℸ4=dΩσ12nk−1,σ22nk−1,σ32nk−1,σ42nk−1,Θℸ1,ℸ2,ℸ3,ℸ4≤ℓ4dσ12nk−1,ℸ1,dσ22n−1,ℸ2,dσ32n−1,ℸ3,dσ42n−1,ℸ4+1−ℓdZ,O.
Asn⟶∞, we have
(84)dℸ1,ℸ1,ℸ2,ℸ3,ℸ4=dZ,O.
Analogously, we have(85)dℸ2,ℸ2,ℸ3,ℸ4,ℸ1=dZ,O,dℸ3,ℸ3,ℸ4,ℸ1,ℸ2=dZ,O,dℸ4,ℸ4,ℸ1,ℸ2,ℸ3=dZ,O.
Thus,Θ has a QBP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4. By the same argument, since O is compact, we can also claim that Ω has a QBP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈O4. To prove
(86)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.we can follow the same approach used in the proof of Theorem 10.


## 3. Quadruple Fixed Point Technique

This part is devoted to present new QFP consequences in the sense of cyclic contraction mappings.Theorem 13.
Assume thatY,d is a MS and Z, O are nonempty closed subsets of Y. Let the mappings Θ:Z4⟶O and Ω:O4⟶Z be cyclic contractions. If σ10,σ20,σ30,σ40∈U4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are described as (22), for each n≥0. If dZ,O=0, then Θ has a QFP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4 and Ω has a QFP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈O4. Moreover, we get
(87)σ12n⟶ℸ1,σ22n⟶ℸ2,σ32n⟶ℸ3,σ42n⟶ℸ4,σ12n+1⟶ℸ1′,σ22n+1⟶ℸ2′,σ32n+1⟶ℸ3′,σ42n+1⟶ℸ4′.
In addition, ifℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then Θ and Ω have a common QFP in Z∩O4.Proof.
BecausedZ,O=0, we find that the pairs Z,O and O,Z justify the property UC∗. Using Theorem 10, we see that Θ has a QBP point ℸ1,ℸ2,ℸ3,ℸ4∈Z4, that is,
(88)dℸ1,Θℸ1,ℸ2,ℸ3,ℸ4=dℸ2,Θℸ2,ℸ3,ℸ4,ℸ1=dℸ3,Θℸ3,ℸ4,ℸ1,ℸ2=dℸ4,Θℸ4,ℸ1,ℸ2,ℸ3=dZ,O,and Ω has a QBP point ℸ1′,ℸ2′,ℸ3′,ℸ4′∈O4, that is,
(89)dℸ1′,ℸ1′,ℸ2′,ℸ3′,ℸ4′=dℸ2′,ℸ2′,ℸ3′,ℸ4′,ℸ1′=dℸ3′,Ωℸ3′,ℸ4′,ℸ1′,ℸ2′=dℸ4′,Ωℸ4′,ℸ1′,ℸ2′,ℸ3′=dZ,O.
From (88) and since dZ,O=0, we obtain
(90)ℸ1=Θℸ1,ℸ2,ℸ3,ℸ4,ℸ2=Θℸ2,ℸ3,ℸ4,ℸ1,ℸ3=Θℸ3,ℸ4,ℸ1,ℸ2,ℸ4=Θℸ4,ℸ1,ℸ2,ℸ3.This means that ℸ1,ℸ2,ℸ3,ℸ4 is a QFP of Θ.
Again, from (89) and since dZ,O=0, we get
(91)ℸ1′=Ωℸ1′,ℸ2′,ℸ3′,ℸ4′,ℸ2′=Ωℸ2′,ℸ3′,ℸ4′,ℸ1′,ℸ3′=Ωℸ3′,ℸ4′,ℸ1′,ℸ2′,ℸ4′=Ωℸ4′,ℸ1′,ℸ2′,ℸ3′.This means that ℸ1′,ℸ2′,ℸ3′,ℸ4′ is a QFP of Ω.
Now, letℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′. From Theorem 10, one can write
(92)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=4dZ,O.
SincedZ,O=0, we have
(93)dℸ1,ℸ1′+dℸ2,ℸ2′+dℸ3,ℸ3′+dℸ4,ℸ4′=0.It follows that
(94)ℸ1=ℸ1′=ℸ2=ℸ2′,ℸ3=ℸ3′=ℸ4=ℸ4′.
Therefore, the quadrupleℸ1,ℸ2,ℸ3,ℸ4∈Z∩O4 is a common QFP of Θ and Ω. This is enough to end the proof.Example 10.
ConsiderY=ℝ equipped with the usual norm. Take Z=−4,0 and O=0,4. Describe two mappings Θ:Z4⟶O and Ω:O4⟶Z as
(95)Θσ1,σ2,σ3,σ4=−σ1+σ2+σ3+σ48,Ωϑ1,ϑ2,ϑ3,ϑ4=−ϑ1+ϑ2+ϑ3+ϑ48,for each σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4, respectively. Then, dZ,O=0 and Θ,Ω are cyclic contractions with ℓ=1/2. Moreover, for each σ1,σ2,σ3,σ4∈Z4 and ϑ1,ϑ2,ϑ3,ϑ4∈O4, we get
(96)dΘσ1,σ2,σ3,σ4,Ωϑ1,ϑ2,ϑ3,ϑ4=−σ1+σ2+σ3+σ48+ϑ1+ϑ2+ϑ3+ϑ48≤18σ1−ϑ1+σ2−ϑ2+σ3−ϑ3+σ4−ϑ4=ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4+1−ℓdZ,O.
Therefore, all postulates of Theorem13 are justified. Then, Θ and Ω have a common QFP 0,0,0,0∈Z∩O4.PuttingZ=O in the above theorem, we have the result below.Corollary 14.
Assume thatY,d is a complete MS and Z≠∅ is a closed subset of Y. Assume also that Θ:Z4⟶Z and Ω:Z4⟶Z are cyclic contraction mappings so that ifσ10,σ20,σ30,σ40∈Z4 and the sequences σ1n,σ2n,σ3n,σ4n in Y are defined as (22), for each n≥0, then Θ has a QFP ℸ1,ℸ2,ℸ3,ℸ4∈Z4 and Ω has a QFP ℸ1′,ℸ2′,ℸ3′,ℸ4′∈Z4. Also, we obtain
(97)σ12n⟶ℸ1,σ22n⟶ℸ2,σ32n⟶ℸ3,σ42n⟶ℸ4,σ12n+1⟶ℸ1′,σ22n+1⟶ℸ2′,σ32n+1⟶ℸ3′,σ42n+1⟶ℸ4′.
Moreover, ifℸ2=ℸ3=ℸ4 and ℸ2′=ℸ3′=ℸ4′, then Θ and Ω have a common QFP in Z4.The following corollary is very important in the application part. We get this result by placingΘ=Ω in Corollary 14.Corollary 15.
LetY,d be a complete MS and Z be a nonempty closed subset of Y. Assume also the mapping Θ:Z4⟶Z verifying
(98)dΘσ1,σ2,σ3,σ4,Θϑ1,ϑ2,ϑ3,ϑ4≤ℓ4dσ1,ϑ1+dσ2,ϑ2+dσ3,ϑ3+dσ4,ϑ4,for all σ1,σ2,σ3,σ4,ϑ1,ϑ2,ϑ3,ϑ4∈Z4, and ℓ∈0,1. Then, Θ has a unique QFP ℸ1,ℸ2,ℸ3,ℸ4∈Z4.


## 4. Solving a System of Functional Equations

Here, we apply Corollary15 to discuss the existence of the solution for the following quadruple functional equations:
(99)η1σ1=supσ2∈Dgσ1,σ2+ℵσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2,η2σ1=supσ2∈Dgσ1,σ2+ℵσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2,η1Iσ1,σ2,η3σ1=supσ2∈Dgσ1,σ2+ℵσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η4σ1=supσ2∈Dgσ1,σ2+ℵσ1,σ2,η4Iσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,where S and D are a state and decision space, respectively; σ1∈S,g:S×D⟶ℝ; ℵ:S×D×ℝ4⟶ℝ; and I:S×D⟶Swhereℝ4=ℝ×ℝ×ℝ×ℝ.Assume thatRS is the set of all bounded real-valued functions on S. Consider
(100)υ=supσ1∈Sυσ1,∀υ∈RS.Moreover, define a distance onRS in the form of
(101)℧μ,ν=supσ1∈Sμσ1−νσ1,forallμ,ν∈RS.Obviously, the pairRS,℧ is a complete MS.In the theorem below, we will discuss the existence of the solution for the system (99).Theorem 16.
Assume that the following postulates are fulfilled:
(a1) the functions g:S×D⟶ℝ and I:S×D⟶S are bounded
(a2) for each σ1∈S,σ2∈D, and η1,η2,η3,η4,η1′,η2′,η3′,η4′∈ℝ, we have
(102)ℵσ1,σ2,η1,η2,η3,η4−ℵσ1,σ2,η1′,η2′,η3′,η4′≤18η1−η1′+η2−η2′+η3−η3′+η4−η4′.
Then, system (99) has a unique bounded solution (UBS).Proof.
Describe an operatorΨ on the space RS by
(103)Ψη1,η2,η3,η4σ1=supσ2∈Dgσ1,σ2+ℵσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2,for each η1,η2,η3,η4∈RS and σ1∈S,σ2∈D. The existence solution for system (99) is equivalent to find a QFP of the operator Ψ.
Clearly, the mappingΨ is well-defined (because the functions g and ℵ are bounded).
Hence, by the postulatea2, we obtain
(104)℧Ψη1,η2,η3,η4,Ψη1′,η2′,η3′,η4′=supσ1∈SΨη1,η2,η3,η4σ1−Ψη1′,η2′,η3′,η4′σ1=supσ1∈Ssupσ2∈Dgσ1,σ2+ℵσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2−supσ2∈Dgσ1,σ2+ℵσ1,σ2,η1′Iσ1,σ2,η2′Iσ1,σ2,η3′Iσ1,σ2,η4′Iσ1,σ2=supσ1∈Ssupσ2∈Dℵσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2−ℵσ1,σ2,η1′Iσ1,σ2,η2′Iσ1,σ2,η3′Iσ1,σ2,η4′Iσ1,σ2≤supσ1∈S18supσ2∈Dη1−η1′+η2−η2′+η3−η3′+η4−η4′≤18supσ1∈Sη1−η1′+η2−η2′+η3−η3′+η4−η4′=ℓ4℧η1,η1′+℧η2,η2′+℧η3,η3′+℧η4,η4′.
This implies that the contractive stipulation of Corollary15 holds with ℓ=1/2. Then, the mapping Ψ has a unique QFP, which is a UBS of problem (99).The example below justifies Theorem16.Example 11.
Consider a quadruple system of functional equations below:(105)η1σ1=supσ2∈ℝarctanσ1+4σ2+12+σ12+13+e3σ2+18η1I1+η1I+18η2I1+eη2I+18η3I1+lnη3I+18η4I1+η4I,η2σ1=supσ2∈ℝarctanσ1+4σ2+12+σ12+13+e3σ2+18η2I1+η2I+18η3I1+eη3I+18η4I1+lnη4I+18η1I1+η1I,η3σ1=supσ2∈ℝarctanσ1+4σ2+12+σ12+13+e3σ2+18η3I1+η3I+18η4I1+eη4I+18η1I1+lnη1I+18η2I1+η2I,η4σ1=supσ2∈ℝarctanσ1+4σ2+12+σ12+13+e3σ2+18η4I1+η4I+18η1I1+eη1I+18η2I1+lnη2I+18η3I1+η3I,for all σ1∈0,1.Clearly, system (105) is comparable to system (99) with S=0,1 and D=ℝ. Clearly, the postulate a1 of Theorem 16 is fulfilled. To achieve the postulate a2, we have
(106)ℵσ1,σ2,η1Iσ1,σ2,η2Iσ1,σ2,η3Iσ1,σ2,η4Iσ1,σ2−ℵσ1,σ2,η1′Iσ1,σ2,η2′Iσ1,σ2,η3′Iσ1,σ2,η4′Iσ1,σ2≤18η1I1+η1I−η1′I1+η1′I+18η2I1+eη2I−η2′I1+eη2′I+18η3′I1+lnη3′I−η3′I1+lnη3′I+18η4I1+η4I−η4′I1+η4′I≤18η1I−η1′I+η2I−η2′I+η3I−η3′I+η4I−η4′I≤18η1−η1′+η2−η2′+η3−η3′+η4−η4′.Therefore, the postulatea2 of Theorem 16 is fulfilled. Thus, problem (105) has a UBS in RS.


## 5. Solving a System of Integral Equations

The existence of solutions for a system of quadruple integral equations is presented here by using the results of Corollary15.Consider the following problem:(107)σ1β=∫0⅁ℏβ,ζJ^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζdζ,σ2β=∫0⅁ℏβ,ζJ^β,ζ,σ2ζ,σ3ζ,σ4ζ,σ1ζdζ,σ3β=∫0⅁ℏβ,ζJ^β,ζ,σ3ζ,σ4ζ,σ1ζ,σ2ζdζ,σ4β=∫0⅁ℏβ,ζJ^β,ζ,σ4ζ,σ1ζ,σ2ζ,σ3ζdζ,where β,ζ∈0,⅁ with ⅁>0.Suppose thatℶ=C0,⅁,ℝn is endowed with
(108)σ1=supβ∈0,⅁σ1β,forallσ1∈ℶ.Moreover, define a distance onℶ in the form of
(109)℘r˘,u~=supβ∈0,⅁r˘β−u~β,forallr˘,u~∈ℶ.Hence,ℶ,⅁ is a complete MS.Theorem 17.
Suppose that the following hypotheses hold:
▲1 The functions J^:0,⅁×0,⅁×ℝn4⟶ℝn and ℏ:0,⅁×0,⅁⟶ℝn are continuous so that
(110)∫0⅁ℏβ,ζdζ≤⅁16,forallβ,ζ∈0,⅁.
▲2 For all σ1, σ2, σ3, σ4, σ4′, σ2′, σ3′, σ4′∈ℝn, we get
(111)J^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζ−J^β,ζ,σ1′ζ,σ2′ζ,σ3′ζ,σ4′ζ≤1⅁σ1−σ1′+σ2−σ2′+σ3−σ3′+σ4−σ4′,forallβ,ζ∈0,⅁.Then, problem (107) has a unique solution on C0,⅁,ℝn.Proof.
Define the mappingΛ:ℶ4⟶ℶ by
(112)Λσ1,σ2,σ3,σ4β=∫0⅁ℏβ,ζJ^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζdζ,β∈0,⅁.
The existence solution of (107) corresponds to finding a QFP of Λ.
Assume thatσ1, σ2, σ3, σ4, σ1′, σ2′, σ3′, σ4′∈ℝn, we get
(113)℘Λσ1,σ2,σ3,σ4,Λσ1′,σ2′,σ3′,σ4′=supβ∈0,⅁Λσ1,σ2,σ3,σ4β−Λσ1′,σ2′,σ3′,σ4′β=supβ∈0,⅁∫0⅁ℏβ,ζJ^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζdζ−∫0⅁ℏβ,ζJ^β,ζ,σ1′ζ,σ2′ζ,σ3′ζ,σ4′ζdζ≤supβ∈0,⅁∫0⅁ℏβ,ζJ^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζ−J^β,ζ,σ1′ζ,σ2′ζ,σ3′ζ,σ4′ζdζ≤supβ∈0,⅁∫0⅁ℏβ,ζdζ1⅁σ1−σ1′+σ2−σ2′+σ3−σ3′+σ4−σ4′≤116supβ∈0,⅁σ1−σ1′+σ2−σ2′+σ3−σ3′+σ4−σ4′=ℓ4℘σ1,σ1′+℘σ2,σ2′+℘σ3,σ3′+℘σ4,σ4′.
Hence, the stipulation of Corollary15 holds with ℓ=1/4. Therefore, Λ has a QFP, which in turn is considered the unique solution to (107).Example 12.
Consider a system of quadruple integral equations below:(114)σ1β=∫016β3+2ζ2+116σ1ζ1+sinσ1ζ+116σ2ζ2+cosσ2ζ+116σ3ζ1+σ3ζ+116σ4ζ3+σ4ζ2dζ,σ2β=∫016β3+2ζ2+116σ2ζ1+sinσ2ζ+116σ3ζ2+cosσ3ζ+116σ4ζ1+σ4ζ+116σ1ζ3+σ1ζ2dζ,σ3β=∫016β3+2ζ2+116σ3ζ1+sinσ3ζ+116σ4ζ2+cosσ4ζ+116σ1ζ1+σ1ζ+116σ2ζ3+σ2ζ2dζ,σ4β=∫016β3+2ζ2+116σ4ζ1+sinσ4ζ+116σ1ζ2+cosσ1ζ+116σ2ζ1+σ2ζ+116σ3ζ3+σ3ζ2dζ,for all β∈0,16.Problem (114) is another shape of problem (107) with ⅁=1,ℏβ,ζ=β3+2ζ2 and
(115)J^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζ=116σ1ζ1+sinσ1ζ+116σ2ζ2+cosσ2ζ+116σ3ζ1+σ3ζ+116σ4ζ3+σ4ζ2.Obviously, the hypothesis▲1 of Theorem 17 holds. For the hypothesis ▲2, one can write
(116)J^β,ζ,σ1ζ,σ2ζ,σ3ζ,σ4ζ−J^β,ζ,σ1′ζ,σ2′ζ,σ3′ζ,σ4′ζ≤116σ1ζ1+sinσ1ζ−σ1′ζ1+sinσ1′ζ+116σ2ζ2+cosσ2ζ−σ2′ζ2+cosσ2′ζ+116σ3ζ1+σ3ζ−σ3′ζ1+σ3′ζ+116σ4ζ3+σ4ζ2−σ4′ζ3+σ4′ζ2≤116σ1ζ−σ1′ζ+σ2ζ−σ2′ζ+σ3ζ−σ3′ζ+σ4ζ−σ4′ζ≤116σ1−σ1′+σ2−σ2′+σ3−σ3′+σ4−σ4′.Hence, the hypothesis▲2 of Theorem 17 is justified with ⅁=16. Therefore, the mapping J^ has a unique QFP which in turn is considered the unique solution to (114).


## 6. Conclusion

One of the central problems in approximation theory is to determine points that minimize the distance to a given point or subset. The best approximation has always attracted analysts because it carries enough potential to be extended especially with the functional analytic approach in nonlinear analysis. The best proximity point has many applications such as obtaining the existence of a unique solution for a variational inequality problem, integral and differential equations, and many other directions. The fixed point method is considered one of the distinguished methods for obtaining these points under cyclic contraction mappings, due to its smoothness and clarity. So, in this paper, the existence of a quadruple best proximity point for a cyclic contraction mapping is introduced in ordinary metric space. The validity of theoretical results in a uniformly convex Banach space was also discussed. Moreover, several examples are given to strengthen the theoretical results. Finally, our paper has been provided with applications on the existence and uniqueness of the solution to a system of functional and integral equations.



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