Application to Lipschitzian and Integral Systems via a Quadruple Coincidence Point in Fuzzy Metric Spaces

Citation: Hammad, H.A.; De la Sen, M. Application to Lipschitzian and Integral Systems via a Quadruple Coincidence Point in Fuzzy Metric Spaces. Mathematics 2022,10, 1905. https://doi.org/10.3390/ math10111905 Academic Editors: Salvador Romaguera and Manuel Sanchis Received: 5 May 2022 Accepted: 31 May 2022 Published: 2 June 2022 Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. Copyright: © 2022 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/). mathematics Article Application to Lipschitzian and Integral Systems via a Quadruple Coincidence Point in Fuzzy Metric Spaces Hasanen A. Hammad 1,2,* and Manuel De la Sen 3 1Department of Mathematics, Unaizah College of Sciences and Arts, Qassim University, Buraydah 52571, Saudi Arabia 2Department of Mathematics, Faculty of Science, Sohag University, Sohag 82524, Egypt 3Institute of Research and Development of Processes, Department of Electricity and Electronics, Faculty of Science and Technology, University of the Basque Country, 48940 Leioa, Bizkaia, Spain; [email protected] *Correspondence: [email protected] Abstract: In this paper, the results of a quadruple coincidence point (QCP) are established for commuting mapping in the setting of fuzzy metric spaces (FMSs) without using a partially ordered set. In addition, several related results are presented in order to generalize some of the prior findings in this area. Finally, to support and enhance our theoretical ideas, non-trivial examples and applications for finding a unique solution for Lipschitzian and integral quadruple systems are discussed. Keywords: quadruple coincidence point; commuting mapping; Lipschitzian mappings; an integral equation; fuzzy metric spaces MSC: 47H10; 54H25 1. Introduction Fixed-point (FP) theory has many applications, not only in nonlinear analysis and its trends—including solutions of differential and integral equations, functional equations arising from dynamic programming, topologies, and dynamic systems—but also in economics, game theory, biological sciences, computer sciences, chemistry, etc. [1–4]. There is no doubt that the study of fuzzy sets is extremely important for their many applications, such as in the control of ill-defined, complex, and non-linear systems. It is more common to find solutions for control problems that are difficult to solve with the classical control theory. Fuzzy set theory is becoming an increasingly important tool, especially in the rapidly evolving discipline of artificial intelligence, such as in expert systems and neural networks. It creates completely new opportunities for the application of fuzzy sets in chemical engineering [5–8]. The concept of fuzzy sets was initiated by Zadeh [ 9 ] in 1965. Many mathematicians used these sets to introduce interesting concepts into the field of mathematics, such as fuzzy logic, fuzzy differential equations, and fuzzy metric spaces. It is known that an FMS is an important generalization of an ordinary metric space where the topological definitions are extended, and there are possible applications in several areas. Many mathematicians have considered this problem in many ways. For example, the authors of [ 10 ] modified the concept of an FMS that was initiated by Kramosil and Michalek [ 11 ] and defined the Hausdorff topology of an FMS. For more details about this idea, we advise the reader to see [12–17]. In 2011, the coupled fixed-point (FP) [ 18 ] result was extended to a tripled FP in partially ordered metric spaces by Berinde and Borcut [ 19 ]. Using these spaces, they introduced exciting results of tripled FP theorems. For more details, see [20–24]. Mathematics 2022,10, 1905. https://doi.org/10.3390/math10111905 https://www.mdpi.com/journal/mathematics Mathematics 2022,10, 1905 2 of 16 In the setting of FMSs, coupled FP results were presented and some important theorems were given by Zhu and Xiao [ 25 ] and Hu [ 26 ]. Elagan et al. [ 27 ] studied the existence of an FP in a locally convex topology generated by fuzzy n−normed spaces. Motivated by the results of the notions of coupled and tripled FPs in partially ordered metric spaces, Karapinar [ 28 ] suggested the concept of a quadruple FP and proved some related consequences of FPs in the same spaces. Based on the last two paragraphs, in this publication, a QCP is considered, and some new and relevant FP results in FMSs are reported. Our paper’s strength is determined by two factors. First, we can adapt it to complete metric spaces (CMSs) so as to achieve Karapinar’s results [ 28 ] (in non-fuzzy sets). So, our paper covers and unifies a large number of outcomes in the same direction. Secondly, we can apply the theoretical conclusions to Lipschitzian and integral quadruple systems in order to discover a unique solution. Finally, non-trivial examples are mentioned and discussed. 2. Preliminaries Hereafter, we will refer to ζ as a non-empty set, Ω(ρ , σ , τ , υ) as Ωρστυ , Ψ(ρ , σ , κ) as Ψρσ(κ), and ω(ρ,σ)as ωρσ. The usual metric space is a non-empty set ζ equipped with a function ω:ζ×ζ→R+ such that for all ρ,σ,τ∈ζ, the following conditions are true: •ωρσ ≥0, •ωρσ =0 if ρ=σ, •ωρσ ≤ωρτ +ωτσ. The pair (ζ,ω)is called an MS. A mapping k:ζ→ζ on an MS (ζ,ω) is called Lipschitzian if there is v≥ 0 such that ωkρkσ≤vωρσ,∀ρ,σ∈ζ. The smallest constant v —denoted by vk —that satisfies the above inequality is called the Lipschitz constant for k . It is clear that a Lipschitzian mapping (LM) is a contraction with vk<1. Theorem 1 ([ 29 ]) . Let (ζ,ω) be a complete MS and let Q:ζ→ζ be a contraction mapping, that is, the following inequality is true: ω(Qx,Qy)≤kω(Qx,Qy),for all x,y∈ζ, where k∈[ 0, 1 ) . Then, Q has a unique FP x∗ in ζ . Moreover, for x0∈ζ , the sequence (Qnx0)n∈N converges to x∗. For examples on LMs, let ζ=R and let ki:ζ→ζ be defined by k1(ρ) = Λ , k2(ρ) = µρ , k3(ρ) = cos ρ , k4(ρ) = 1 1+ρ , k5(ρ) = 1 (1+ρ)2 , and k6(ρ) = arcsin ρ , where Λ,µ∈R. Definition 1 ([ 30 ]) . A mapping ?:[ 0, 1 ]2→[ 0, 1 ] is called a κ -norm if it is nondecreasing in both arguments, associative, commutative, and has 1as identity. For all `∈[ 0, 1 ] , the sequence {?m`}∞ m=1 is inductively defined by ?1`=` , ?m`=?m−1`? ` . A triangular norm ? is of Υ -type if {?m`}∞ m=1 is equicontinuous at `= 1, that is, for each e∈( 0, 1 ) , there is κ∈( 0, 1 ) such that if `∈(1−κ, 1], then ?m` > 1−efor each m ∈N. The most famous continuous κ -norm of the Υ -type is ?=min , which satisfies min(`1 , `2)≥ `1`2for all `1,`2∈[0, 1]. The results below include a wide range of κ-norms of the Υ-type. Mathematics 2022,10, 1905 3 of 16 Lemma 1 ([ 30 ]) . Assume that ? is a κ -norm and $∈( 0, 1 ] is a real number. Define ?$ by ρ?$σ=ρ?σ if max{ρ , σ} ≤ 1 −$ , and ρ?$σ=min{ρ , σ} if max{ρ , σ}> 1 −$ . Then, ?$ is a κ-norm of the Υ-type. Definition 2 ([ 11 ]) . Let ζ6=∅ be an arbitrary set, let ? be a continuous κ -norm, and let Ψ: ζ×ζ×[ 0, ∞)→[ 0, 1 ] be a fuzzy set. We say that (ζ,Ψ,?) is an FMS if the function Ψ satisfies the hypotheses below for each ρ,σ,τ∈ζ, and κ,µ>0 : (fms 1) Ψρσ(0) = 0; (fms 2) Ψρσ(κ) = 1⇔ρ=σ; (fms 3) Ψρσ(κ) = Ψσρ(κ); (fms 4) Ψρσ(.):[0, ∞)→[0, 1]is left continuous; (fms 5) Ψρσ(κ)?Ψστ(µ)≤Ψρτ(κ+µ). Here, we also consider (ζ,Ψ) an FMS under ? , and we will only consider the FMS that verifies: (D) lim κ→∞Ψρσ(κ) = 1,∀ρ,σ∈ζ. Lemma 2 ([12]).On the infinite set [0, ∞),Ψρσ(.)is a non-decreasing function. Definition 3 ([ 10 ]) . Assume that (ζ,Ψ) is an FMS under some κ -norm; a sequence {ρm} ⊂ ζ is called: • Convergent to ρ∈ζ , and we write lim m→∞ρm=ρ if, for every e> 0, κ> 0, there is m0∈N such that Ψρmρ(κ)>1−efor all m ≥m0. • A Cauchy sequence if, for every e> 0, κ> 0, there is m0∈N such that Ψρmρj(κ)> 1 −e for all m,j≥m0. •An FMS is called complete if every Cauchy sequence is convergent. Definition 4 ([ 11 ]) . We say that a function k:ζ→ζ defined on an FMS is continuous at ρ0∈ζ if lim m→∞kρm=kρ0 for any {ρm} ∈ ζ such that lim m→∞ρm=ρ0 . As is familiar, for ρ0∈ζ , we will denote k−1(ρ0) = {ρ∈ζ:kρ=ρ0}. Remark 1 ([ 11 ]) . If `1≤`2 , then ρ`1≥ρ`2 provided that ρ∈[ 0, 1 ] and `1 , `2∈( 0, ∞) . This fact will be expressed here as follows: 0< `1≤`2≤1implies that Ψρσ(κ)`1≥Ψρσ(κ)`2≥Ψρσ(κ). For any κ -norm ? , it is obvious that ?≤min . So, if (ζ,Ψ) is an FMS via min , then (ζ,Ψ)is an FMS under any κ-norm. In the examples below, we only define Ψρσ(κ)for κ>0 and ρ6=σ. Example 1 ([ 10 ]) . For κ> 0and ρ6=σ , we define an FMS in different ways from an MS (ζ,ω) as follows: •Ψω ρσ(κ) = κ κ+ωρσ •Ψe ρσ(κ) = e−ωρσ κ•Ψo ρσ(κ) = 0, if κ≤ωρσ, 1, if κ>ωρσ. It is obvious that, under the product ?= ., (ζ,Ψω) is an FMS, which is called the standard FMS on (ζ,ω) . In addition, (ζ,Ψω) , (ζ,Ψe) , and (ζ,Ψo) are FMSs under min . This is a standard method for seeing the MS (ζ,ω)as an FMS, though it is not as well known. Moreover, (ζ,ω)is a CMS iff (ζ,Ψω),(ζ,Ψe), or (ζ,Ψo)is a complete FMS. 3. Main Results We begin this section with the following simple definition. Definition 5. Assume that Ω:ζ4→ζand k:ζ→ζare two mappings. •We say that Ωand kare commuting if kΩρστυ =Ωkρkσkτkυ,∀`,σ,ρ,υ∈ζ. Mathematics 2022,10, 1905 4 of 16 •We say that (ρ,σ,τ,υ)∈ζ4is a QCP of Ωand kif Ωρστυ =kρ,Ωστυρ =kσ,Ωτυρσ =kτand Ωυρστ =kυ. Theorem 2. Assume that ? is a κ -norm of the Υ -type such that µ?κ≥µκ for all µ , κ∈[ 0, 1 ] . Suppose that (ζ,Ψ,?) is a complete FMS and Ω:ζ4→ζ , k:ζ→ζ are two mappings such that (a) Ωζ4⊆k(ζ), (b) kis continuous, (c) kis commuting with Ω, (d) for all ρ,σ,τ,υ,b ρ,bσ,b τ,b υ∈ζ, ΨΩρστυΩb ρbσb τb υ(κv)≥Ψkρkb ρ(κ)`1?Ψkσkbσ(κ)`2?Ψkτkb τ(κ)`3?Ψkυkb υ(κ)`4, (1) where v∈( 0, 1 ) and `1 , `2 , `3 , `4 are real numbers in [ 0, 1 ] such that `1+`2+`3+`4≤ 1. Then, the following conclusions hold. (1) There is a unique ρ∈ζsuch that ρ=kρ=Ωρρρρ. In particular, (2) There is at least a QCP for the mappings k and Ω ; moreover, in the case of Ω=ρ0 , there is a constant on ζ4 . This holds only if the inverse of the mapping k exists and it satisfies .k−1(ρ0) = {ρ0}; then, we have (3) (ρ,ρ,ρ,ρ)is a unique QCP of kand Ω. Note that, to avoid the unidentified quantity 0 0 , we consider here Ψkρkb ρ(κ)0= 1 for all κ>0 and all ρ,b ρ∈ζ. Proof. We divide the proof into two cases: Case 1. When Ω⊆ζ is constant, that is, there is ρ0∈ζ such that, for all ρ,σ,τ,υ∈ζ, Ωρστυ =ρ0 . Since Ω and k are commuting, one can write kρ0=kΩρστυ =Ωkρkσkτkυ=ρ0 . Therefore, ρ0=kρ0=Ωρ0ρ0ρ0ρ0 and (ρ0 , ρ0 , ρ0 , ρ0) is a QCP of Ω and k . On the other hand, assume that k−1(ρ0) = {ρ0} and (ρ,σ,τ,υ)∈ζ4 is another QCP of Ω and k . Then, kρ=Ωρστυ =ρ0 , so ρ∈k−1(ρ0) = {ρ0} . In the same manner, we can write ρ=σ=τ=υ=ρ0; hence, (ρ0,ρ0,ρ0,ρ0)is a unique QCP of Ωand k. Case 2. Assume that Ω∈ζ is not constant; for this, let (`1,`2,`3,`4)6= ( 0, 0, 0, 0 ) . In this case, we consider j and m to be non-negative integers and κ∈[ 0, ∞) . This case is divided into five steps. St1 . Deriving four sequences {ρm} , {σm} , {τm} , and {υm} : Suppose that ρ0 , σ0 , τ0 , υ0 are arbitrary points in ζ . As Ωζ4⊆k(ζ) , we can select ρ1 , σ1 , τ1 , υ1∈ζ so that kρ1=Ωρ0σ0τ0υ0,kσ1=Ωσ0τ0υ0ρ0 , kτ1=Ωτ0υ0ρ0σ0 and kυ1=Ωυ0ρ0σ0τ0 . Again, with Ωζ4⊆k(ζ) , we can select ρ2 , σ2 , τ2 , υ2∈ζ so that kρ2=Ωρ1σ1τ1υ1 , kσ2=Ωσ1τ1υ1ρ1 , kτ2=Ωτ1υ1ρ1σ1 , and kυ2=Ωυ1ρ1σ1τ1 . Continuing with the same scenario, we can construct {ρm} , {σm} , {τm} , and {υm} so that for m≥ 0, kρm+1=Ωρmσmτmυm , kσm+1=Ωσmτmυmρm , kτm+1=Ωτmυmρmσm, and kυm+1=Ωυmρmσmτm. St2 . {ρm} , {σm} , {τm} , and {υm} are Cauchy sequences. For m≥ 0 and all κ> 0, we define Ξm(κ) = Ψkρmkρm+1(κ)?Ψkσmkσm+1?Ψkτmkτm+1?Ψkυmkυm+1. Ξmis a non-decreasing function and κ−κv ≤κ≤κ v, so we get Ξm(κ−κv)≤Ξm(κ)≤Ξmκ v, for all κ>0 and m≥0. (2) It follows from (1) that, for all m∈Nand all κ≥0, Ψkρmkρm+1(κ)=ΨΩρm−1σm−1τm−1υm−1Ωρmσmτmυm(κ) ≥Ψkρm−1kρmκ v`1?Ψkσm−1kσmκ v`2 ?Ψkτm−1kτmκ v`3?Ψkυm−1kυmκ v`4; (3) Mathematics 2022,10, 1905 5 of 16 Ψkσmkσm+1(κ)=ΨΩσm−1τm−1υm−1ρm−1Ωσmτmυmρm(κ) ≥Ψkσm−1kσmκ v`1?Ψkτm−1kτmκ v`2 ?Ψkυm−1kυmκ v`3?Ψkρm−1kρmκ v`4; (4) Ψkτmkτm+1(κ)=ΨΩτm−1υm−1ρm−1σm−1Ωτmυmρmσm(κ) ≥Ψkτm−1kτmκ v`1?Ψkυm−1kυmκ v`2 ?Ψkρm−1kρmκ v`3?Ψkσm−1kσmκ v`4; (5) Ψυmkυm+1(κ)=ΨΩυm−1ρm−1σm−1τm−1Ωυmρmσmτm(κ) ≥Ψkυm−1kυmκ v`1?Ψkρm−1kρmκ v`2 ?Ψkσm−1kσmκ v`3?Ψkτm−1kτmκ v`4; (6) It follows from (3)–(6) and Remark 1that Ψkρmkρm+1(κ) ≥Ψkρm−1kρmκ v`1?Ψkσm−1kσmκ v`2?Ψkρm−1kρmκ v`3?Ψkυm−1kυmκ v`4 ≥Ψkρm−1kρmκ v?Ψkσm−1kσmκ v?Ψkρm−1kρmκ v?Ψkυm−1kυmκ v =Ξm−1(κ v); Ψkσmkσm+1(κ) ≥Ψkσm−1kσmκ v`1?Ψkτm−1kτmκ v`2?Ψkυm−1kυmκ v`3?Ψkρm−1kρmκ v`4 ≥Ψkσm−1kσmκ v?Ψkτm−1kτmκ v?Ψkυm−1kυmκ v?Ψkρm−1kρmκ v =Ξm−1(κ v); Ψkτmkτm+1(κ) ≥Ψkτm−1kτmκ v`1?Ψkυm−1kυmκ v`2?Ψkρm−1kρmκ v`3?Ψkσm−1kσmκ v`4 ≥Ψkτm−1kτmκ v?Ψkυm−1kυmκ v?Ψkρm−1kρmκ v?Ψkσm−1kσmκ v =Ξm−1(κ v); and Ψυτmkυm+1(κ) ≥Ψkυm−1kυmκ v`1?Ψkρm−1kρmκ v`2?Ψkσm−1kσmκ v`3?Ψkτm−1kτmκ v`4 ≥Ψkυm−1kυmκ v?Ψkρm−1kρmκ v?Ψkσm−1kσmκ v?Ψkτm−1kτmκ v =Ξm−1(κ v). This proves that, for all κ>0 and all m≥0, Ψkρmkρm+1(κ),Ψkσmkσm+1(κ),Ψkτmkτm+1(κ),Ψυτmkυm+1(κ)≥Ξm−1(κ v)≥Ξm−1(κ). (7) Putting κ−vκ instead of κ, we obtain, for all κ>0 and all m≥0, that Mathematics 2022,10, 1905 6 of 16 Ψkρmkρm+1(κ−vκ),Ψkσmkσm+1(κ−vκ),Ψkτmkτm+1(κ−vκ),Ψυτmkυm+1(κ−vκ) ≥Ξm−1(κ−vκ).(8) Since ?is commutative and ?≥., using (3)–(6), we deduce that Ξm(κ) = Ψkρmkρm+1(κ)?Ψkσmkσm+1?Ψkτmkτm+1?Ψkυmkυm+1 ≥Ψkρm−1kρmκ v`1?Ψkσm−1kσmκ v`2?Ψkτm−1kτmκ v`3?Ψkυm−1kυmκ v`4 ?Ψkρm−1kρmκ v`2?Ψkσm−1kσmκ v`3?Ψkτm−1kτmκ v`4?Ψkυm−1kυmκ v`1 ?Ψkρm−1kρmκ v`3?Ψkσm−1kσmκ v`4?Ψkτm−1kτmκ v`1?Ψkυm−1kυmκ v`2 ?Ψkρm−1kρmκ v`4?Ψkσm−1kσmκ v`1?Ψkτm−1kτmκ v`2?Ψkυm−1kυmκ v`3 =Ψkρm−1kρmκ v`1?Ψkρm−1kρmκ v`2?Ψkρm−1kρmκ v`3?Ψkρm−1kρmκ v`4 ?Ψkσm−1kσmκ v`2?Ψkσm−1kσmκ v`3?Ψkσm−1kσmκ v`4?Ψkρm−1kρmκ v`2 ?Ψkτm−1kτmκ v`3?Ψkτm−1kτmκ v`4?Ψkτm−1kτmκ v`1?Ψkτm−1kτmκ v`2 ?Ψkυm−1kυmκ v`4?Ψkυm−1kυmκ v`1?Ψkυm−1kυmκ v`2?Ψkυm−1kυmκ v`3. It follows that Ξm(κ)≥Ψkρm−1kρmκ v`1.Ψkρm−1kρmκ v`2.Ψkρm−1kρmκ v`3.Ψkρm−1kρmκ v`4 ?Ψkσm−1kσmκ v`2.Ψkσm−1kσmκ v`3.Ψkσm−1kσmκ v`4.Ψkρm−1kρmκ v`2 ?Ψkτm−1kτmκ v`3.Ψkτm−1kτmκ v`4.Ψkτm−1kτmκ v`1.Ψkτm−1kτmκ v`2 ?Ψkυm−1kυmκ v`4.Ψkυm−1kυmκ v`1.Ψkυm−1kυmκ v`2.Ψkυm−1kυmκ v`3 =Ψkρm−1kρmκ v`1+`2+`3+`4?Ψkσm−1kσmκ v`1+`2+`3+`4 ?Ψkτm−1kτmκ v`1+`2+`3+`4?Ψkυm−1kυmκ v`1+`2+`3+`4 ≥Ψkρm−1kρmκ v?Ψkσm−1kσmκ v?Ψkτm−1kτmκ v?Ψkυm−1kυmκ v =Ξm−1(κ v) By using (2), one can write Ξm(κ)≥Ξm−1(κ v)≥Ξm−1(κ)≥Ξm−1(κ−κv),∀κ>0, and m≥1. (9) By continuing in the same manner, we have Ξm(κ)≥Ξm−1(κ v)≥Ξm−2(κ v2)≥... ≥Ξ0(κ vm),∀κ>0, and m≥1, which leads to find that for all κ>0 lim m→∞Ξm(κ)≥lim m→∞Ξ0(κ vm) = 1⇒lim m→∞Ξm(κ) = 1. (10) Mathematics 2022,10, 1905 7 of 16 From (7) and (9), we have Ψkρmkρm+1(κ),Ψkσmkσm+1(κ),Ψkτmkτm+1(κ),Ψυτmkυm+1(κ)≥Ξm(κ)≥Ξm−1(κ−κv). (11) After that, we will prove that, for all κ>0 and all m,r≥1, Ψkρmkρm+r(κ),Ψkσmkσm+r(κ),Ψkτmkτm+r(κ),Ψυτmkυm+r(κ)≥?rΞm−1(κ−κv). (12) We can show this by induction in r≥ 1 as follows: Inequality (12) holds if r= 1 for all m≥ 1 and all κ> 0 by (11). Assume that (12) is true for all m≥ 1 and all κ> 0 for some r . Now, we prove the relation for r+ 1. It follows from (1), the induction assumption, and ?≥. that Ψkρm+1kρm+r+1(vκ) =ΨΩρmσmτmυmΩρm+rσm+rτm+rυm+r(vκ) ≥Ψkρmkρm+r(κ)`1?Ψkσmkσm+r(κ)`2?Ψkτmkτm+r(κ)`3?Ψkυmkυm+r(κ)`4 ≥(?rΞm−1(κ−κv))`1?(?rΞm−1(κ−κv))`2?(?rΞm−1(κ−κv))`3?(?rΞm−1(κ−κv))`4 ≥(?rΞm−1(κ−κv))`1.(?rΞm−1(κ−κv))`2.(?rΞm−1(κ−κv))`3.(?rΞm−1(κ−κv))`4 =(?rΞm−1(κ−κv))`1+`2+`3+`4≥?rΞm−1(κ−κv). Similarly, we arrive at Ψkρm+1kρm+r+1(vκ),Ψkσm+1kσm+r+1(vκ),Ψkτm+1kτm+r+1(vκ),Ψkυm+1kυm+r+1(vκ)≥?rΞm−1(κ−κv). From Definition 2(fms 5), (8), and the induction assumption, we get Ψkρm+1kρm+r+1(κ)=Ψkρm+1kρm+r+1(κ−κv +κv) ≥Ψkρmkρm+1(κ−κv)?Ψkρm+1kρm+r+1(κv) ≥Ξm−1(κ−κv)?(?rΞm−1(κ−κv)) =?r+1Ξm−1(κ−κv). In addition, the same result holds if we consider Ψkσm+1kσm+r+1(κ) , Ψkτm+1kτm+r+1(κ) , and Ψkυm+1kυm+r+1(κ) . This leads to (12) being true. This allows us to prove that {kρm} is Cauchy. Assume that κ> 0 and ε∈( 0, 1 ) are given. From this assumption, as ? is a κ -norm of the Υ -type, there is ϕ∈( 0, 1 ) such that ?r`1> 1 −ε for all `1∈( 1 −ϕ , 1 ] and for all r≥1. From (10), limm→∞Ξm(κ) = 1, so there is m0∈Nsuch that Ξm(κ−κv)>1−ϕ,∀m≥m0. Hence, by (12), we have Ψkρmkρm+r(κ),Ψkσmkσm+r(κ),Ψkτmkτm+r(κ),Ψυτmkυm+r(κ)>1−ε,∀m≥m0and r≥1. Thus, {kρm} is a Cauchy sequence. Similarly, {kσm} , {kτm} , and {kυm} are also Cauchy sequences. St3 . Proving that Ω and k have a QCP: As ζ is complete, there are ρ , σ , τ , υ∈ζ such that lim m→∞kρm=ρ, lim m→∞kσm=σ, lim m→∞kτm=τand lim m→∞kυm=υ. The continuity of kimplies that lim m→∞kkρm=kρ, lim m→∞kkσm=kσ, lim m→∞kkτm=kτ, and lim m→∞kkυm=kυ. Mathematics 2022,10, 1905 8 of 16 The commutativity of Ωand kleads to kkρm+1=kΩ(ρm,σm,τm,υm)=Ω(kρm,kσm,kτm,kυm). By (1), we get Ψkkρm+1Ωρστυ (κv)=ΨΩkρmσmτmυmΩρστυ (κv) ≥Ψkkρmkρ(κ)`1?Ψkkσmkσ(κ)`2?Ψkkτmkτ(κ)`3?Ψkkυmkυ(κ)`4 ≥Ψkkρmkρ(κ)?Ψkkσmkσ(κ)?Ψkkτmkτ(κ)?Ψkkυmkυ(κ). (13) As m→∞, in (13), we find that lim m→∞kkρm+1=Ωρστυ =kρ. Similarly, we deduce that Ωστυρ =kσ , Ωτυρσ =kτ , Ωυρστ =kυ . This shows that (ρ,σ,τ,υ)is a QCP of Ωand k. kρ=Ωρστυ,kσ=Ωστυρ,kτ=Ωτυρσ,kυ=Ωυρστ. (14) St4 . Showing that ρ=Ωρστυ , σ=Ωστυρ , τ=Ωτυρσ , and υ=Ωυρστ : From Stipulation (1), we get Ψkρkσm+1(κv) =ΨΩρστυΩσmτmυmρm(κv) ≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4; (15) Ψkσkτm+1(κv) =ΨΩστυρΩτmυmρmσm(κv) ≥Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4; (16) Ψkτkυm+1(κv) =ΨΩτυρσΩυmρmσmτm(κv) ≥Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4; (17) Ψkυkρm+1(κv) =ΨΩυρστ Ωρmσmτmυm(κv) ≥Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4. (18) We set ∇m(κv) = Ψkρkσm(κv)?Ψkσkτm(κv)?Ψkτkυm(κv)?Ψkυkρm(κv) for all κ>0 and m≥0. It follows from (15)–(18) that ∇m+1(κv) = Ψkρkσm+1(κv)?Ψkσkτm+1(κv)?Ψkτkυm+1(κv)?Ψkτkυm+1(κv) ≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4 ?Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4 ?Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4 ?Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4 =Ψkρkσm(κ)`1?Ψkρkσm(κ)`4?Ψkρkσm(κ)`3?Ψkρkσm(κ)`2 ?Ψkσkτm(κ)`2?Ψkσkτm(κ)`1?Ψkσkτm(κ)`4?Ψkσkτm(κ)`3 ?Ψkτkυm(κ)`3?Ψkτkυm(κ)`2?Ψkτkυm(κ)`1?Ψkτkυm(κ)`4 ?Ψkυkρm(κ)`4?Ψkυkρm(κ)`3?Ψkυkρm(κ)`2?Ψkυkρm(κ)`1, Mathematics 2022,10, 1905 9 of 16 which implies that ∇m+1(κv)≥Ψkρkσm(κ)`1.Ψkρkσm(κ)`4.Ψkρkσm(κ)`3.Ψkρkσm(κ)`2 ?Ψkσkτm(κ)`2.Ψkσkτm(κ)`1.Ψkσkτm(κ)`4.Ψkσkτm(κ)`3 ?Ψkτkυm(κ)`3.Ψkτkυm(κ)`2.Ψkτkυm(κ)`1.Ψkτkυm(κ)`4 ?Ψkυkρm(κ)`4.Ψkυkρm(κ)`3.Ψkυkρm(κ)`2.Ψkυkρm(κ)`1 =Ψkρkσm(κ)`1+`2+`3+`4?Ψkσkτm(κ)`1+`2+`3+`4 ?Ψkτkυm(κ)`1+`2+`3+`4?Ψkυkρm(κ)`1+`2+`3+`4 ≥Ψkρkσm(κ)?Ψkσkτm(κ)?Ψkτkυm(κ)?Ψkυkρm(κ)=∇m(κ). This implies that ∇m+1(κv)≥ ∇m(κ) for all m≥ 0 and all κ> 0. Repeating this process, ∇m(κ)≥ ∇m−1(κ v)≥ ∇m−2(κ v2)≥... ≥ ∇0(κ vm),∀κ>0 and m≥1. (19) From (15)–(19), we conclude that Ψkρkσm+1(κv)≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4 ≥ ∇m(κ)≥ ∇0(κ vm);(20) Ψkσkτm+1(κv)≥Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4 ≥ ∇m(κ)≥ ∇0(κ vm);(21) Ψkτkυm+1(κv)≥Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4 ≥ ∇m(κ)≥ ∇0(κ vm);(22) Ψkυkρm+1(κv)≥Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4 ≥ ∇m(κ)≥ ∇0(κ vm).(23) Thus, Ψkρkσm+1(κv),Ψkσkτm+1(κv),Ψkτkυm+1(κv),Ψkυkρm+1(κv)≥ ∇0(κ vm),∀κ>0 and m≥1. Taking the limit as m→∞ in (20)–(23) and using limm→∞∇0(κ vm) = 1, for all κ> 0, we get limm→∞kρm=kυ , limm→∞kσm=kρ , limm→∞kτm=kσ , and limm→∞kυm=kτ . This shows, together with (14), that Ωρστυ =kρ=lim m→∞kσm=σ,Ωστυρ =kσ=lim m→∞kτm=τ, Ωτυρσ =kτ=lim m→∞kυm=υ,Ωυρστ =kυ=lim m→∞kρm=ρ. St5 . We shall prove that ρ=σ=τ=υ . We set Π(κ) = Ψρσ(κ)?Ψστ(κ)?Ψτυ(κ)? Ψυρ(κ)for all κ>0. Then, according to (1), we can write Ψρσ(κv)=ΨΩρστυΩστυρ (κv)≥Ψkρkσ(κ)`1?Ψkσkτ(κ)`2?Ψkτkυ(κ)`3?Ψkυkρ(κ)`4 =Ψστ(κ)`1?Ψτυ(κ)`2?Ψυρ(κ)`3?Ψρσ(κ)`4;(24) Ψστ(κv)=ΨΩστυρΩτυρσ (κv)≥Ψkσkτ(κ)`1?Ψkτkυ(κ)`2?Ψkυkρ(κ)`3?Ψkρkσ(κ)`4 =Ψτυ(κ)`1?Ψυρ(κ)`2?Ψρσ(κ)`3?Ψστ(κ)`4;(25) Ψτυ(κv)=ΨΩτυρσΩυρστ (κv)≥Ψkτkυ(κ)`1?Ψkυkρ(κ)`2?Ψkρkσ(κ)`3?Ψkσkτ(κ)`4 =Ψυρ(κ)`1?Ψρσ(κ)`2?Ψστ(κ)`3?Ψτυ(κ)`4;(26) Ψυρ(κv)=ΨΩυρστ Ωρστυ (κv)≥Ψkυkρ(κ)`1?Ψkρkσ(κ)`2?Ψkσkτ(κ)`3?Ψkτkυ(κ)`4 =Ψρσ(κ)`1?Ψστ(κ)`2?Ψτυ(κ)`3?Ψυρ(κ)`4.(27) Mathematics 2022,10, 1905 16 of 16 Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The authors are grateful to the Spanish Government and the European Commission for Grant IT1207-19. Conflicts of Interest: The authors declare no conflict of interest. References 1. Border, K.C. Fixed Point Theorems with Applications to Economics and Game Theory; Econometrics, Statistics and Mathematical Economics; Cambridge University Press: Cambridge, UK, 1985. 2. 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