# Common Fixed Points of Four Maps SatisfyingF-Contraction on b-Metric Spaces

Authors: Muhammad Nazam; Ma Zhenhua; Sami Ullah Khan; Muhammad Arshad
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We manifest some common fixed point theorems for four maps satisfying Círíc typeF-contraction and Hardy-Rogers type F-contraction defined on complete b-metric spaces. We apply these results to infer several new and old corresponding results. These results also generalize some results obtained previously. We dispense examples to validate our results.


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## Body

## 1. Introduction and Preliminaries

After the famous Banach Contraction Principle, a large number of researchers revealed many fruitful generalizations of Banach’s fixed point theorem. One of these generalizations is known asF-contraction presented by Wardowski [1]. Wardowski [1] evinced the fact that every F-contraction defined on complete b-metric space has a unique fixed point. The concept of F-contraction proved to be another milestone in fixed point theory and numerous research papers on F-contraction have been published (see, e.g., [2–8]). Recently, Cosentino and Vetro [9] established a fixed point result for Hardy-Rogers type F-contraction and Mınak et al. [10] presented a fixed point result for Círíc type generalized F-contraction.In 1989, Bakhtin [11] investigated the concept of b-metric spaces; however, Czerwik [12] initiated study of fixed point of self-mappings in b-metric spaces and proved an analogue of Banach’s fixed point theorem. Since then, numerous research articles have been published comprising fixed point theorems for various classes of single-valued and multivalued operators in b-metric spaces (see, e.g., [13–19]).We shall bring into use the idea of Círíc typeF-contraction and Hardy-Rogers type F-contraction comprising four self-mappings defined on b-metric space. We present common fixed point results for four self-maps satisfying Círíc type and Hardy-Rogers type F-contraction on b-metric space. We apply our results to infer several new and old results.We denote0,∞ by R+, 0,∞ by R0+, -∞,+∞ by R, and set of natural numbers by N.We bring back into reader’s mind some definitions and properties ofb-metric.Definition 1 (see [12]).
LetM be a nonempty set and s≥1 be a real number. A function db:M×M→[0,∞) is said to be a b-metric if, for all θ,ρ,σ∈M, one has (db1)
θ=ρ if and only if db(θ,ρ)=0,(db2)
db(θ,ρ)=db(ρ,θ),(db3)
db(θ,ρ)≤sdb(θ,σ)+db(σ,ρ).
In this case, the pair(M,db,s) is called a b-metric space (with coefficient s).Definition1 allows us to remark that the class of b-metric spaces is effectually more general than that of metric spaces because a b-metric is a metric when s=1. The following example describes the significance of a b-metric.Example 2.
Let(M,d) be a metric space and db(θ,ρ)=(d(θ,ρ))r, r>1 is a real number. Then db is b-metric space with s=2r-1. Apparently, (db1) and (db2) of Definition 1 are satisfied. If 1<r<∞, then the convexity of the function f(θ)=θr(θ>0) implies (1)j+l2r≤12jr+lrthat gives (j+l)r≤2r-1(jr+lr). Thus, for all θ,ρ,σ∈M, one has (2)dbθ,σ=dθ,σr≤dθ,ρ+dρ,σr≤2r-1dθ,ρr+dρ,σr=2r-1dbθ,ρ+dbρ,σ.Therefore, db(θ,ρ)≤sdb(θ,ρ)+db(ρ,σ), where s=2r-1, which shows that (M,db,s) is a b-metric space. Nevertheless, if (M,d) is a metric space, then (M,db,s) may not be a metric space. Indeed, if M=R and d(θ,ρ)=|θ-ρ| (a usual metric), then db(θ,ρ)=[d(θ,ρ)]2 does not define a metric on M.For the notions like convergence, completeness, and Cauchy sequence in the setting ofb-metric spaces, the reader is referred to Aghajani et al. [20], Czerwik [12], Amini-Harandi [21], Huang et al. [13], Hussain et al. [14], Radenović et al. [15], Khamsi and Hussain [16], Latif et al. [22], Parvaneh et al. [17], Roshan et al. [18], and Zabihi and Razani [23].In line with Wardowski [1], Cosentino et al. [24] investigated a nonlinear function F:R+→R complying with the following axioms:(F1)
F is strictly increasing.(F2)
For each sequencern of positive numbers, the following equation holds: limn→∞rn=0 if and only if limn→∞F(rn)=-∞.(F3)
There existsθ∈(0,1) such that limrn→0+(rn)θF(rn)=0.(F4)
τ+F(srn)≤F(rn-1) implies τ+F(snrn)≤F(sn-1rn-1), for each n∈N and some τ>0.We denote byΔFs the set of all functions satisfying the conditions (F1)–(F4).Example 3.
LetF:R+→R be defined by (a)
F(r)=ln⁡(r),(b)
F(r)=r+ln⁡(r).
It is easy to check that (a) and (b) are members ofΔFs.Definition 4 (see [25]).
Let(M,db,s) be a b-metric space. The pair {f,g} is said to be compatible if and only if (3)limn→∞dbfgrn,gfrn=0,wheneverrnisasequenceinMsothatlimn→∞frn=limn→∞grn=tfor  somet∈M.Lemma 5.
Let(M,db,s) be a b-metric space. If there exist two sequences {rn}, {sn} such that (4)limn→∞dbrn,sn=0,limn→∞rn=tforsomet∈M,Then limn→∞sn=t.Proof.
Due to the triangular inequality we have(5)dbsn,t≤sdbsn,rn+dbrn,t,and the result follows after applying limit as n→∞.


## 2. Círíc Type Fixed Point Theorem

In this section we set up a fixed point theorem for four self-maps satisfying Círíc typeF-contraction. We explain this theorem through an example and discuss its consequences.Theorem 6.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFs and τ>0, the inequality (6)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FMr1,r2,holds, where (7)Mr1,r2=max⁡dbSr1,Tr2,dbfr1,Sr1,dbgr2,Tr2,dbSr1,gr2+dbfr1,Tr22s,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Proof.
Letr0∈M. As f(M)⊆T(M), there exists r1∈M such that f(r0)=T(r1). Since g(r1)∈S(M), we can choose r2∈M such that g(r1)=S(r2). In general, r2n+1 and r2n+2 are chosen in M such that f(r2n)=T(r2n+1) and g(r2n+1)=S(r2n+2). Define a sequence {jn} in M such that j2n=f(r2n)=T(r2n+1) and j2n+1=g(r2n+1)=S(r2n+2) for all n≥0; we show that {jn} is a Cauchy sequence. Assume that db(f(r2n),g(r2n+1))>0; then from contractive condition (6), we get (8)Fsdbj2n,j2n+1=Fsdbfr2n,gr2n+1≤FMr2n,r2n+1-τ,for all n∈N∪{0}, where (9)Mr2n,r2n+1=max⁡dbSr2n,Tr2n+1,dbfr2n,Sr2n,dbgr2n+1,Tr2n+1,dbSr2n,gr2n+1+dbfr2n,Tr2n+12s=max⁡dbj2n-1,j2n,dbj2n,j2n-1,dbj2n+1,j2n,dbj2n-1,j2n+1+dbj2n,j2n2s=max⁡dbj2n-1,j2n,dbj2n,j2n+1.If M(r2n,r2n+1)=db(j2n,j2n+1), then (10)Fsdbj2n,j2n+1≤Fdbj2n,j2n+1-τ,which is a contradiction due to F1. Therefore, (11)Fsdbj2n,j2n+1≤Fdbj2n-1,j2n-τ,for all n∈N. Similarly, (12)Fsdbj2n-1,j2n≤Fdbj2n-2,j2n-1-τ,for all n∈N. Hence, from (11) and (12), we have (13)Fsdbjn,jn+1≤Fdbjn-1,jn-τ,for all n∈N. Let bn=db(jn,jn+1) for each n∈N, from (13) and axiom (F4) we have (14)τ+Fsnbn≤Fsn-1bn-1,n∈N.Repeating the process, we obtain (15)Fsnbn≤Fb0-nτ,n∈N.On taking limit n→∞ in (15), we have limn→∞Fsnbn=-∞; due to axiom (F2) we get limn→∞snbn=0 and (F3) implies that there exists κ∈0,1 such that (16)limn→∞snbnκFsnbn=0.From (15), for all n∈N, we obtain (17)snbnκFsnbn-snbnκFb0≤-snbnκnτ≤0.On taking limit n→∞ in (17), we have (18)limn→∞nsnbnκ=0.This implies there exists n1∈N, such that nsnbnκ≤1 for all n≥n1 or (19)snbn≤1n1/κ∀n≥n1.To prove {jn} as a Cauchy sequence, we use (19) and, for m>n≥n1, we consider (20)dbjn,jm≤∑i=nm-1sibi≤∑i=n∞sibi≤∑i=n∞1n1/κ.The convergence of the series ∑i=n∞‍1/i1/κ entails limn,m→∞db(jn,jm)=0. Hence {jn} is a Cauchy sequence in M,db,s. Since M,db,s is a complete b-metric space, there exists υ∈M such that limn→∞db(jn,υ)=0. Also we have(21)limn→∞fr2n=limn→∞Tr2n+1=limn→∞gr2n+1=limn→∞Sr2n+2=υ.We show that υ is a common fixed point of f, g, S, and T. Since S is continuous, therefore, (22)limn→∞Sfr2n=Sυ=limn→∞S2r2n+2.Since, the pair {f,S} is compatible, so, (23)limn→∞dbfSr2n,Sfr2n=0,byLemma5limn→∞fSr2n=Sυ.Now put r1=S(r2n) and r2=r2n+1 in (6) and suppose on the contrary that db(S(υ),υ)>0; we obtain (24)FdbfSr2n,gr2n+1≤FMSr2n,r2n+1-τ,where (25)MSr2n,r2n+1=max⁡dbS2r2n,Tr2n+1,dbfSr2n,S2r2n,dbgr2n+1,Tr2n+1,dbS2r2n,gr2n+1+dbfSr2n,Tr2n+12s.Taking the upper limit in (24), we have (26)FdbSυ,υ≤FdbSυ,υ-τ<FdbSυ,υ,a contradiction; hence, db(S(υ),υ)=0 implies S(υ)=υ. Again since T is continuous, therefore, (27)limn→∞Tgr2n+1=Tυ=limn→∞T2r2n+1.Since the pair {g,T} is compatible, so, (28)limn→∞dbgTr2n+1,Tgr2n+1=0,byLemma5limn→∞gTr2n+1=Tυ.Now put r2=T(r2n+1) and r1=r2n in (6) and suppose on the contrary that db(υ,T(υ))>0; we obtain (29)Fdbfr2n,gTr2n+1≤FMr2n,Tr2n+1-τ,where (30)Mr2n,Tr2n+1=max⁡dbSr2n,T2r2n+1,dbfr2n,Sr2n,dbgTr2n+1,T2r2n+1,dbSr2n,gTr2n+1+dbfr2n,T2r2n+12s.Taking the upper limit in (29), we have (31)Fdbυ,Tυ≤Fdbυ,Tυ-τ<Fdbυ,Tυ,a contradiction; hence, db(υ,T(υ))=0 which then implies T(υ)=υ. Now assuming db(f(υ),υ)>0 on the contrary, from (6) we get (32)Fdbfυ,gr2n+1≤FMυ,r2n+1-τ,where (33)Mυ,r2n+1=max⁡dbSυ,Tr2n+1,dbfυ,Sυ,dbgr2n+1,Tr2n+1,dbSυ,gr2n+1+dbfυ,Tr2n+12s.Taking the upper limit in (32) and using the fact that S(υ)=T(υ)=υ, we have (34)Fdbfυ,υ≤Fdbfυ,υ-τ<Fdbfυ,υ,a contradiction; hence, db(f(υ),υ)=0 which then implies f(υ)=υ. Finally, assume on the contrary db(υ,g(υ))>0. From (6) and the fact that S(υ)=T(υ)=f(υ)=υ, we obtain (35)Fdbυ,gυ=Fdbfυ,gυ≤FMυ,υ-τ,where (36)Mυ,υ=max⁡dbSυ,Tυ,dbfυ,Sυ,dbgυ,Tυ,dbSυ,gυ+dbfυ,Tυ2s.This implies that (37)Fdbυ,gυ≤Fdbυ,gυ-τ<Fdbυ,gυ,a contradiction; hence, db(g(υ),υ)=0 which then implies g(υ)=υ. Hence, υ proves to be a common fixed point of the four maps f, g, S, and T. Also υ is unique; indeed if ω is another fixed point of f, g, S, and T, then from (6), we have (38)Fdbυ,ω=FdbSυ,Tω≤FMυ,ω-τ,where (39)Mυ,ω=max⁡dbSυ,Tω,dbfυ,Sυ,dbgω,Tω,dbSυ,gω+dbfυ,Tω2s.Thus, from (38), we have (40)Fdbυ,ω<Fdbυ,ω,which is a contradiction. Hence, υ=ω and υ is a unique common fixed point of the four maps f, g, S, and T.The following example explains Theorem6.Example 7.
LetM=[0,1] and define db:M×M→R0+ by dbr1,r2=|r1-r2|2, so M,db,2 is a complete b-metric space. Define the mappings f,g,S,T:M→M, for all r∈M by (41)fr=r316,gr=r38,Sr=r38,Tr=r34.Clearly, f, g, S, and T are self-mappings complying with f(M)⊆T(M), g(M)⊆S(M). We note that the pair {f,S} is compatible. If {rn} is a sequence in M satisfying (42)limn→∞frn=limn→∞Srn=t,for  somet∈M,then (43)limn→∞frn-t2=limn→∞Srn-t2=0,and equivalently (44)limn→∞rn316-t2=limn→∞rn38-t2=0implies (45)limn→∞rn16-316t2=limn→∞rn8-38t2=0.Uniqueness of limit gives t1/16=t1/8 which implies t=0,1. Due to the continuity of f, S we have (46)limn→∞dbfSrn,Sfrn=limn→∞fSrn-Sfrn2=ft-St2=0-02=0fort=0∈M.Similarly, the pair {g,T} is compatible. Now for each r1,r2∈M, consider (47)dbfr1,gr2=fr1-gr22=r1316-r2382=r138+r2342r138-r2342≤16561+1812dbTr1,Sr2=672443046721dbTr1,Sr2≤672443046721Mr1,r2.The above inequality can be written as (48)ln⁡430467216724+ln⁡dbfr1,gr2≤ln⁡Mr1,r2.Define the function F:R+→R by F(r)=ln⁡(r), for all r∈R+>0. Hence, for all r1,r2∈M such that db(f(r1),g(r2))>0, τ=ln⁡43046721/6724 we obtain (49)τ+Fdbfr1,gr2≤FMr1,r2.Thus, contractive condition (6) is satisfied for all r1,r2∈M. Hence, all the hypotheses of Theorem 6 are satisfied; note that f,g,S,T have a unique common fixed point r=0.The following theorems can easily be obtained by chasing the proof of Theorem6. Theorem 8 generalizes the result presented by Bianchini [26].Theorem 8 (Bianchini type).
Let(M,db,s) be a complete b-metric space and f, g, S, T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFsτ>0, the inequality (50)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM3r1,r2holds, where (51)M3r1,r2=kmax⁡dbfr1,Sr1,dbgr2,Tr2,withk∈0,s,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem9 generalizes the result presented by Sehgal [27].Theorem 9 (Sehgal type).
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFsτ>0, the inequality (52)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM4r1,r2holds, where (53)M4r1,r2=kmax⁡dbSr1,Tr2,dbfr1,Sr1,dbgr2,Tr2,withk∈0,s,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem10 generalizes the result presented by Círíc [28].Theorem 10.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFsτ>0, the inequality (54)Fdbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM5r1,r2holds, where for all k∈[0,s)(55)M5r1,r2=kmax⁡dbSr1,Tr2,dbfr1,Sr1,dbgr2,Tr2,dbgr2,Sr1,dbfr1,Tr2,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem 11.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFsτ>0, the inequality (56)Fdbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM6r1,r2holds, where for all k∈[0,s)(57)M6r1,r2=kmax⁡dbSr1,Tr2,dbfr1,Sr1+dbgr2,Tr22,dbgr2,Sr1+dbfr1,Tr22s,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem 12.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If for all r1,r2∈M, k∈[0,s) the inequality (58)sdbfr1,gr2≤kMr1,r2holds, where (59)Mr1,r2=max⁡dbSr1,Tr2,dbfr1,Sr1,dbgr2,Tr2,dbSr1,gr2+dbfr1,Tr22s,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Proof.
Since, for allr1,r2∈M, we have (60)dbfr1,gr2≤kMr1,r2,thus, if db(f(r1),g(r2))>0, we have (61)τ+ln⁡dbfr1,gr2≤ln⁡Mr1,r2,where τ=ln⁡(s/k)>0; thus, the contractive condition (58) reduces to (6) with F(r)=ln⁡(r). Hence, Theorem 6 is a generalization of [29, Theorem 3.1].


## 3. Hardy and Rogers Type Fixed Point Theorem

In this section, we set up a fixed point theorem for four self-maps satisfying Hardy and Rogers typeF-contraction. We give an example to validate this theorem and discuss its consequences.Theorem 13.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFsτ>0, the inequality (62)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM1r1,r2holds, where (63)M1r1,r2=a1dbSr1,Tr2+a2dbfr1,Sr1+a3dbgr2,Tr2+a4dbSr1,gr2+dbfr1,Tr2withai≥0i=1,2,3,4suchthata1+a2+a3+2sa4=1,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Proof.
Letr0∈M. As f(M)⊆T(M), there exists r1∈M such that f(r0)=T(r1). Since g(r1)∈S(M), we can choose r2∈M such that g(r1)=S(r2). In general r2n+1 and r2n+2 are chosen in M such that f(r2n)=T(r2n+1) and g(r2n+1)=S(r2n+2). Define a sequence {jn} in M such that j2n=f(r2n)=T(r2n+1) and j2n+1=g(r2n+1)=S(r2n+2) for all n≥0; we show that {jn} is a Cauchy sequence. Assume that db(f(r2n),g(r2n+1))>0; then from contractive condition (62), we get (64)Fsdbj2n,j2n+1=Fsdbfr2n,gr2n+1≤FM1r2n,r2n+1-τ,for all n∈N∪{0}, where (65)M1r2n,r2n+1=a1dbSr2n,Tr2n+1+a2dbfr2n,Sr2n+a3dbgr2n+1,Tr2n+1+a4dbSr2n,gr2n+1+dbfr2n,Tr2n+1=a1dbj2n-1,j2n+a2dbj2n,j2n-1+a3dbj2n+1,j2n+a4dbj2n-1,j2n+1+dbj2n,j2n=a1+a2+sa4dbj2n-1,j2n+a3+sa4dbj2n,j2n+1.Now from (64) we have (66)Fsdbj2n,j2n+1≤Fa1+a2+sa4dbj2n-1,j2n+a3+sa4dbj2n,j2n+1-τ.Since F is strictly increasing, (66) implies (67)sdbj2n,j2n+1≤a1+a2+sa4dbj2n-1,j2n+a3+sa4dbj2n,j2n+1,1-a3-sa4dbj2n,j2n+1<s-a3-sa4dbj2n,j2n+1≤a1+a2+sa4dbj2n-1,j2n,dbj2n,j2n+1≤a1+a2+sa41-a3-sa4dbj2n-1,j2n.Since a1+a2+a3+2sa4=1, therefore(68)dbj2n,j2n+1≤a1+a2+sa41-a3-sa4dbj2n-1,j2n=dbj2n-1,j2n.Thus, from (66) we obtain (69)Fsdbj2n,j2n+1≤Fdbj2n-1,j2n-τ,for all n∈N. Similarly, (70)Fsdbj2n-1,j2n≤Fdbj2n-2,j2n-1-τ,for all n∈N. Hence, from (69) and (70), we have (71)Fsdbjn,jn+1≤Fdbjn-1,jn-τ,for all n∈N. Let bn=db(jn,jn+1) for each n∈N; from (71) and axiom (F4) we have (72)τ+Fsnbn≤Fsn-1bn-1,n∈N.Repeating the process, we obtain (73)Fsnbn≤Fb0-nτ,n∈N.On taking limit n→∞ in (73), we have limn→∞Fsnbn=-∞; due to axiom (F2) we get limn→∞snbn=0 and (F3) implies that there exists κ∈0,1 such that (74)limn→∞snbnκFsnbn=0.From (73), for all n∈N, we obtain (75)snbnκFsnbn-snbnκFb0≤-snbnκnτ≤0.On taking limit n→∞ in (75), we have (76)limn→∞nsnbnκ=0.This implies there exists n1∈N, such that nsnbnκ≤1 for all n≥n1 or (77)snbn≤1n1/κ∀n≥n1.To prove {jn} as a Cauchy sequence, we use (77) and for m>n≥n1; we consider (78)dbjn,jm≤∑i=nm-1sibi≤∑i=n∞sibi≤∑i=n∞1n1/κ.The convergence of the series ∑i=n∞‍1/i1/κ entails limn,m→∞db(jn,jm)=0. Hence {jn} is a Cauchy sequence in M,db,s. Since M,db,s is a complete b-metric space, there exists υ∈M such that limn→∞db(jn,υ)=0. Also we have(79)limn→∞fr2n=limn→∞Tr2n+1=limn→∞gr2n+1=limn→∞Sr2n+2=υ.We show that υ is a common fixed point of f, g, S, and T. Since S is continuous, therefore, (80)limn→∞Sfr2n=Sυ=limn→∞S2r2n+2.Since, the pair {f,S} is compatible, so, (81)limn→∞dbfSr2n,Sfr2n=0,byLemma5limn→∞fSr2n=Sυ.Now put r1=S(r2n) and r2=r2n+1 in (62) and suppose on the contrary that db(S(υ),υ)>0; we obtain (82)FdbfSr2n,gr2n+1≤FM1Sr2n,r2n+1-τ,where (83)M1Sr2n,r2n+1=a1dbS2r2n,Tr2n+1+a2dbfSr2n,S2r2n+a3dbgr2n+1,Tr2n+1+a4dbS2r2n,gr2n+1+dbfSr2n,Tr2n+1.Taking the upper limit in (82) and considering the fact that a1+2a4<1, we have (84)FdbSυ,υ≤FdbSυ,υ-τ<FdbSυ,υ,a contradiction; hence, db(S(υ),υ)=0 implies S(υ)=υ. Again since T is continuous, therefore, (85)limn→∞Tgr2n+1=Tυ=limn→∞T2r2n+1.Since, the pair {g,T} is compatible, (86)limn→∞dbgTr2n+1,Tgr2n+1=0,byLemma5limn→∞gTr2n+1=Tυ.Now put r2=T(r2n+1) and r1=r2n in (62) and suppose on the contrary that db(υ,T(υ))>0; we obtain (87)Fdbfr2n,gTr2n+1≤FM1r2n,Tr2n+1-τ,where (88)M1r2n,Tr2n+1=a1dbSr2n,T2r2n+1+a2dbfr2n,Sr2n+a3dbgTr2n+1,T2r2n+1+a4dbSr2n,gTr2n+1+dbfr2n,T2r2n+1.Taking the upper limit in (87), we have (89)Fdbυ,Tυ≤Fdbυ,Tυ-τ<Fdbυ,Tυ,a contradiction; hence, db(υ,T(υ))=0 which then implies T(υ)=υ. Assume that db(f(υ),υ)>0 and from (62) we get (90)Fdbfυ,gr2n+1≤FM1υ,r2n+1-τ,where (91)M1υ,r2n+1=a1dbSυ,Tr2n+1+a2dbfυ,Sυ+a3dbgr2n+1,Tr2n+1+a4dbSυ,gr2n+1+dbfυ,Tr2n+1.Taking the upper limit in (90) and using the fact that S(υ)=T(υ)=υ, we have (92)Fdbfυ,υ≤Fdbfυ,υ-τ<Fdbfυ,υ,a contradiction; hence, db(f(υ),υ)=0 which then implies f(υ)=υ. Finally, assume that db(υ,g(υ))>0. From (62) and the equation S(υ)=T(υ)=f(υ)=υ, we obtain (93)Fdbυ,gυ=Fdbfυ,gυ≤FM1υ,υ-τ,where (94)M1υ,υ=a1dbSυ,Tυ+a2dbfυ,Sυ+a3dbgυ,Tυ+a4dbSυ,gυ+dbfυ,Tυ.This implies that (95)Fdbυ,gυ≤Fdbυ,gυ-τ<Fdbυ,gυ,a contradiction; hence, db(g(υ),υ)=0 which then implies g(υ)=υ. Hence, υ proves to be a common fixed point of the four maps f, g, S, and T. Also υ is unique. Indeed if ω is another fixed point of f, g, S, and T, then, from (62), we have (96)Fdbυ,ω=FdbSυ,Tω≤FM1υ,ω-τ,where (97)M1υ,ω=a1dbSυ,Tω+a2dbfυ,Sυ+a3dbgω,Tω+a4dbSυ,gω+dbfυ,Tω.Thus, from (96), we have (98)Fdbυ,ω<Fdbυ,ω,which is a contradiction. Hence, υ=ω and υ is a unique common fixed point of the four maps f, g, S, and T.The following example explains Theorem13.Example 14.
LetM=[0,∞) and define db:M×M→R0+ by dbr1,r2=|r1-r2|2, so, M,db,2 is a complete b-metric space. Define the mappings f,g,S,T:M→M, for all r∈M, by (99)fr=ln⁡1+r6,gr=ln⁡1+r7,Sr=e7r-1,Tr=e6r-1.Clearly, f, g, S, and T are continuous self-mappings complying with f(M)=T(M)=g(M)=S(M). We note that the pair {f,S} is compatible. Indeed, let {rn} be a sequence in M satisfying (100)limn→∞frn=limn→∞Srn=t,for  somet∈M.Then (101)limn→∞frn-t2=limn→∞Srn-t2=0,and equivalently (102)limn→∞ln⁡1+rn6-t2=limn→∞e7rn-1-t2=0implies (103)limn→∞rn-6et-62=limn→∞rn-ln⁡t+172=0.The uniqueness of limit gives 6et-6=ln⁡(t+1)/7; thus, t=0 is only possible solution. Due to the continuity of f, S we have (104)limn→∞dbfSrn,Sfrn=limn→∞fSrn-Sfrn2=ft-St2=0-02=0fort=0∈M.Similarly, the pair {g,T} is compatible. Now for each r1,r2∈M, consider (105)dbfr1,gr2=fr1-gr22=ln⁡1+r6-ln⁡1+r72≤r6-r72≤14227r-6r2≤1422e7r-e6r2=1422dbTr1,Sr2≤1422M1r1,r2.The above inequality can be written as (106)2ln⁡42+ln⁡dbfr1,gr2≤ln⁡M1r1,r2.Define the function F:R+→R by F(r)=ln⁡(r), for all r∈R+>0. Hence, for all r1,r2∈M, such that db(f(r1),g(r2))>0, τ=2ln⁡(42) we obtain (107)τ+Fdbfr1,gr2≤FMr1,r2.Thus, contractive condition (62) is satisfied for all r1,r2∈M. Hence, all the hypotheses of Theorem 13 are satisfied. Note that f, g, S, and T have a unique common fixed point r=0.Theorem15 generalizes the result presented by Reich [30] and can be proved by following the proof of Theorem 13.Theorem 15 (Reich type).
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If, for all r1,r2∈M, for some F∈ΔFs, τ>0, the inequality (108)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM2r1,r2holds, where for all ai≥0(i=1,2,3) such that a1+a2+a3=1(109)M2r1,r2=a1dbSr1,Tr2+a2dbfr1,Sr1+a3dbgr2,Tr2,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem16 generalizes the result presented by Chatterjea [31] and can be proved by following the proof of Theorem 13.Theorem 16 (Chatterjea type).
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If for all r1,r2∈M for some F∈ΔFsτ>0 the inequality (110)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FadbSr1,gr2+dbfr1,Tr2holds for all a≥0 such that 2a=1, then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Similarly, Theorems17 and 18 can be proved by following the proof of Theorem 13.Theorem 17.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If for all r1,r2∈M for some F∈ΔFs, τ>0 the inequality (111)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM7r1,r2holds, where (112)M7r1,r2=a1r1,r2dbSr1,Tr2+a2r1,r2dbfr1,Sr1+a3r1,r2dbgr2,Tr2+a4r1,r2dbSr1,gr2+dbfr1,Tr2with ai(r1,r2)(i=1,2,3,4) are nonnegative functions such that(113)supr1,r2∈Ma1r1,r2+a2r1,r2+a3r1,r2+2sa4r1,r2=1,then f, g, S, and T have a unique common fixed point υ provided S, T are continuous and pairs {f,S}, {g,T} are compatible.Theorem 18.
Let(M,db,s) be a complete b-metric space and f, g, S, and T are self-mappings on (M,db,s) such that f(M)⊆T(M), g(M)⊆S(M). If for all r1,r2∈M for some F∈ΔFs, τ>0 the inequality (114)dbfr1,gr2>0impliesτ+Fsdbfr1,gr2≤FM8r1,r2holds, where (115)M8r1,r2=a1dbSr1,Tr2+a2+a32dbfr1,Sr1+dbgr2,Tr2+a4+a52sdbSr1,gr2+dbfr1,Tr2withai≥0,∑i=15ai=1,then f, g, S, and T have a unique common fixed point υ provided S,T are continuous and pairs {f,S}, {g,T} are compatible.


## 4. Application of Theorem6 to a System of Integral Equations

LetM=C([a,b],R) be the space of all continuous real valued functions defined on [a,b], a>0. Let the function db:M×M→[0,∞) be defined by (116)dbu,v=supt∈a,but-vt2,for all u,v∈C([a,b],R). Obviously, (M,db,2) is a complete b-metric space. We shall apply Theorem 6 to show the existence of a common solution of the system of Volterra type integral equations given by (117)ut=pt+∫atKt,r,Sutdr,wt=pt+∫atJt,r,Tvtdr,for all t, where p:M→R is a continuous function and K,J:[a,b]×[a,b]×M→R are lower semicontinuous operators. Now we prove the following theorem to ensure the existence of a solution of the system of integral equations (117).Theorem 19.
LetM=C([a,b],R) and define the mappings f,g:M→M by (118)fut=pt+∫atKt,r,Sutdr,gut=pt+∫atJt,r,Tvtdr,where p:M→R is a continuous function and K,J:[a,b]×[a,b]×M→R are lower semicontinuous operators. Assume the following conditions are satisfied: (i)
There exists a continuous functionH:M→[0,∞) such that (119)Kt,r,S-Jt,r,T≤HrSut-Tvtfor eacht,r∈[a,b] and S,T∈M.(ii)
There existsτ>0 and, for each r∈M, one has (120)∫atHrdr≤e-τs,t∈a,b.(iii)
There exists a sequence{rn} in M such that limn→∞db(fS(rn),Sf(rn))=0 and limn→∞db(gT(rn),Tg(rn))=0, whenever (121)limn→∞frn=limn→∞Srn=t,limn→∞grn=limn→∞Trn=tfor  somet∈M.
Then the system of integral equations given in (117) has a solution.Proof.
Following assumptions (i) and (ii), we have(122)dbfut,gvt=supt∈a,bfut-gvt2=supt∈a,b∫atKt,r,Sut-Jt,r,Tvtdr2≤supt∈a,b∫atHrSut-Tvtdr2≤supt∈a,bSut-Tvt2∫atHrdr2=dbSut,Tvt∫atHrdr2≤dbSut,Tvte-τs≤M1ut,vte-τs.Consequently, we have (123)sdbfut,gvt≤e-τM1ut,vt.As the natural logarithm belongs to ΔFs, applying it on above inequality and after some simplification, we get (124)τ+ln⁡sdbfut,gvt≤ln⁡M1ut,vt.So takingF(r)=ln⁡(r), all hypotheses of Theorem 6 are satisfied. Hence the system of integral equations given in (117) has a unique common solution.



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