# Some Valid Generalizations of Boyd and Wong Inequality andψ,ϕ-Weak Contraction in Partially Ordered b−Metric Spaces

Authors: Noor Jamal; Thabet Abdeljawad; Muhammad Sarwar; Nabil Mlaiki; Panda Sumati Kumari
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this manuscript, we useψ,ϕ-weak contraction to generalize coincidence point results which are established in the context of partially ordered b-metric spaces. The presented work explicitly generalized some recent results from the existing literature. Examples are also provided to show the authenticity of the established work.


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## Body

## 1. Introduction

Banach contraction principle [1] is one of the most important result in analysis. It is widely considered as the source of fixed point theory. Banach principle has been modified by many researchers by changing either the contractive condition or the underlying spaces. Boyd and Wong [2] modified the contraction condition in metric spaces X,d with control function(1)dft,fb≤ψdt,b,for allt,binX.Theϕ-weak contraction condition was initiated by Alber and Guerre-Delabriere in the Hilbert spaces, see [3]. Rhoudes, in [4], showed that every ϕ-weak contractive mapping in metric spaces has a unique fixed point. The generalized ϕ-weak contractive condition in complete metric spaces was introduced by Zhang and Song in [5]. In [6], Dutta and Choudhury studied unique fixed point results with the help of ψ,ϕ-weak contraction in metric spaces. Doric, in [7], studied the result of [6] with generalized ψ,ϕ-weak contraction in metric spaces. The concept of fixed point results for two pairs of mappings was introduced by Jungck in [8], for commuted mappings. Jungck in [8, 9] states this new assumption, with compatibility and weak compatibility of mappings. Some further work for generalized ψ,ϕ-weak contraction in metric spaces can be found in [6, 10, 11]. Ran and Reurring, in [12], introduced partially ordered metric spaces. A b-metric space with partial ordering is called partially ordered b-metric space. Recently, ψ,ϕ-weak contraction in partially ordered b-metric spaces gain the attention of many researcher. In this direction, fixed point and coincidence point results are discussed by many authors. For more details of fixed point and coincidence point results and their applications, comparison of different contraction conditions, and related results in b-metric spaces, we refer the reader to [13–40] along with the references mentioned therein. To show that a sequence is Cauchy in metric-type spaces, Jovanovic et al. [41] proved Lemma 1. As b-metric spaces have discontinuous structure, therefore, in this manuscript, we give a generalization of Boyd and Wong [2] inequality and ψ,ϕ-weak contraction, to establish coincidence and fixed point results.


## 2. Preliminaries

Definition 1.
(see [42]). Let X,≺¯ be a partially ordered set, and assume that the map d:X×X⟶R+ satisfies the following conditions, For all b,t,r∈X and s≥1:b1db,t=0 if and only if b=tb2db,t=dt,bb3db,r≤sdb,t+dt,r
Then,X,d,≺¯ is called partially ordered b-metric space.Lemma 1.
(see [41]). “Let X,d be b-metric space and s≥1. When a sequence am satisfies the following condition,(2)dam,am+1≤Kdam−1,am,for some, 0<K<1/s and m=1,2,….
Then,am is b-Cauchy sequence in X,d.Definition 2.
(see [9]). The pair of mappings f,g is compatible in the metric space X,d if and only if(3)limm⟶∞dfgnm,gfnm=0,whenever nm in sequence such that(4)limm⟶∞fnm=limm⟶∞gnm=r,withr∈X.Definition 3.
(see [8]). In the metric space X,d, the pair of mappings f, g:X⟶X is weakly compatible if f,g is commutative at the point of coincidence (i.e., gfn=fgn whenever fn=gn).Definition 4.
(see [43]). The pair of mappings f, g:X⟶X defined on a partially ordered set X is called(a)
Weakly increasing iffn≺¯gfn and gn≺¯fgn, for all n∈X(b)
Partially weakly increasing if∀n∈X, fn≺¯gfnDefinition 5.
(see [44, 45]). Let f, g, h:X⟶X be three mappings on partially ordered set X,≺¯ such that fn≺¯hn and gn≺¯hn. Then, pair f,g is said to be(i)
Weakly increasing with respect toh if and only if for all n∈X, fn≺¯gw, for all w∈h−1fn and gn≺¯fw, and for all w∈h−1gn(ii)
Partially weakly increasing w.r.th if and only if fn≺¯gw for all w∈h−1fn.Theorem 1.
(see [45]). Let self-mappings f,g be continuous mappings in partially ordered complete metric space X,≺¯,d with fX⊆gX. Assume that the pair f,g is weakly compatible, whereas f is partially weakly increasing with respect to g such that gb and gt are comparable elements and satisfying the following condition:(5)ψdfb,ft≤ψdgb,gt−ϕdgb,gt,for allb,tinX,where ϕ,ψ:0,∞⟶0,∞ are altering distance functions. Then, mappings f, g have coincidence point.Theorem 2.
(see [46]). Let f, g:X⟶X be mappings on a partially ordered complete b-metric space X,≺¯,d. Assume that f, g is partially weakly increasing and satisfying the following condition:(6)ψdft,gb≤ψMt,b−ϕMt,b,∀t,b∈X,where ϕ,ψ:0,∞⟶0,∞ are altering distance and lower semicontinuous functions, respectively, with ϕ0=0 and(7)Mt,b=maxdt,b,dt,ft,db,gb,dt,gb+db,ft2s,where s≥1. Furthermore, assume that either f or g is continuous mapping or space X is regular. Then, f,g have a common fixed point.Theorem 3.
(see [19]). “Let X,≺¯,d be partially ordered complete b-metric space and f, g, h, and k:X⟶X be continuous mappings such that fX⊆hX and gX⊆kX. Suppose for comparable elements kb and ht, the following condition hold(8)ψs2dfb,gt≤ψMb,t−ϕMb,t,∀b,t∈X,Mb,t=maxdkb,ht,dkb,fb,dht,gt,dkb,gt+dht,fb2s,with altering distance functions, ϕ,ψ:0,∞⟶0,∞ and s≥1. Let the pairs f,k and g,h be compatible and the pairs f,g and g,f be partially weakly increasing w.r.t h and k, respectively. Then, f,g,h, and k have a coincidence point.”Theorem 4.
(see [18]). “Let f, g, h, k:X⟶X be continuous mappings on a partially ordered complete b-metric space X,≺¯,d with fX⊆hX and gX⊆kX. Assume that compatible pairs f,h and g,k and comparable elements kt and hb satisfy the following condition:(9)sϵdft,gb≤Mt,b,Mt,b=maxdkt,hb,dkt,ft,dhb,gb,dkt,gb+dhb,ft2s,where ϵ>1, while s≥1 and the pairs of mappings f,g and g,f are partially weakly increasing w.r.t k and h, respectively. When ha and ka are comparable, then fa=ga=ha=ka.”Theorem 5.
(see [20]). Let the mappings f, g:X⟶X be defined on a partially ordered complete b-metric space X,≺¯,d and f be g-weakly isotone increasing. Let ψ:0,∞⟶0,∞ be a function such that ψn<n for all n>0 and ψ0=0 satisfy the condition:(10)s4dft,gb≤Mt,b,∀t,b∈X,Mt,b=maxψdt,b,ψdt,ft,ψdb,gb,ψdt,gb+db,ft2s,where s≥1. If the mappings f or g are continuous in X. Then, f,g has a common fixed point.


## 3. Main Results

We begin with the following result.Theorem 6.
LetA, B, G, F:X⟶X be continuous mappings on a partially ordered complete b-metric space, X,≺¯,d with AX⊆GX and BX⊆FX. Assume that compatible pairs A,F and B,G and comparable elements Gb and Ft satisfy the following contractive condition:(11)sϵdAt,Bb≤ψMt,b,∀t,b∈X,where ϵ>1 and ψ:0,∞⟶0,∞ satisfy ψn≤n for all n>0, ψ0=0, and(12)Mt,b∈dFt,Gb,dGt,At,dFb,Bb,dFt,Bb+dGb,At2s,where s≥1 and the pairs A,B and B,A are partially weakly increasing w.r.t G,F, respectively. Then, A,F and B,G have a coincidence point. Furthermore, if b is a coincidence point for comparable elements Gb and Fb, then Ab=Bb=Fb=Gb.Proof.
Lett0∈X. Since AX⊆FX and BX⊆GX, then there exist t1,t2∈X such that At0=Ft1 and Bt1=Gt2. The sequence bm is constructed as follows:(13)b2m+1=At2m=Ft2m+1,b2m+2=Bt2m+1=Gt2m+2,wherem=0,1,2,….
The pairs of mappingsA,B and B,A are partially weakly increasing w.r.t F and G, respectively. Therefore,(14)Ft1=At0≺¯Bt1=Gt2≺¯At2=Ft3,for allt1∈F−1At0,t2∈G−1Bt1.
By repeating this process, we deduce that(15)b2m+1≺¯b2m+2,∀m∈N∪0.
Assume that(16)b2m≠b2m+1,∀m∈N.
We discuss three steps:Step I: at this step, we have to prove thatdb2m+1,b2m≤db2m,b2m−1.SinceFt2m=b2m and Gt2m+1=b2m+1 are comparable, therefore, from (11), we have(17)sϵdb2m+1,b2m=sϵdAt2m,Bt2m−1≤ψMt2m,t2m−1,with(18)Mt2m,t2m−1∈db2m,b2m−1,db2m,b2m+1,db2m−1,b2m,db2m,b2m+db2m−1,b2m+12s,Mt2m,t2m−1∈db2m,b2m−1,db2m,b2m+1,db2m+1,b2m−12s.To show thatdb2m,,b2m+1≤db2m−1,b2m, suppose that(19)db2m−1,b2m<db2m,b2m+1.Here, we discuss three possible cases ofMt2m,t2m−1.Case (1): ifMt2m,t2m−1=db2m,b2m−1, from (17), we have(20)sϵdb2m,b2m+1≤ψdb2m,b2m−1.Sinceψn≤n for n>0, therefore(21)sϵdb2m,b2m+1≤db2m,b2m−1,which contradict assumption (19).Case (2): ifMt2m,t2m−1=db2m,b2m+1, since ψn≤n for n>0, therefore, from (17), one can write(22)sϵdb2m,b2m+1≤ψdb2m,b2m+1≤db2m,b2m+1,which is held unlessdb2m,b2m+1=0, which implies that b2m=b2m+1, but it contradicts (16).Therefore,(23)db2m+1,b2m≤db2m,b2m−1.Case (3): ifMt2m,t2m−1=db2m−1,b2m+1/2s.Using triangle inequality and assumption (19), we have(24)Mt2m,t2m−1≤sdb2m−1,b2m+db2m,b2m+12s<db2m,b2m+1.Sinceψn≤n for all n>0, thus, from (17), one has(25)sϵdb2m,b2m+1≤ψdb2m,b2m+1<db2m,b2m+1.This is again contradiction. Hence, for all cases, we conclude that(26)db2m+1,b2m≤db2m,b2m−1.Step II: in this step, we have to prove that the sequencebm is b-Cauchy by using Lemma 1, for all three cases. Case (1): ifMt2m,t2m−1=db2m,b2m−1.From (17), one can write(27)sϵdb2m,b2m+1≤ψdb2m,b2m−1.Asψn≤n for n>0, so(28)sϵdb2m,b2m+1≤db2m,b2m−1,db2m,b2m+1≤1/sϵdb2m,b2m−1.TakingK=1/sϵ, we have(29)db2m,b2m+1≤Kdb2m,b2m−1,forK∈0,1s.Case (2): ifMt2m,t2m−1=db2m,b2m+1, from (26), we have db2m,b2m+1≤db2m,b2m−1. From (17), one has(30)sϵdb2m,b2m+1≤ψdb2m,b2m−1.Sinceψn≤n, for n>0, therefore(31)sϵdb2m,b2m+1≤db2m,b2m−1.Thus,(32)db2m,b2m+1≤1sϵdb2m,b2m−1.LetK=1/sϵ. Then,(33)db2m,b2m+1≤Kdb2m,b2m−1,forK∈0,1s.Case (3): ifMt2m,t2m−1=db2m−1,b2m+1/2s.Using triangle inequality and (26), it is reduced to case (2). Therefore, (2) holds for three cases.Similarly, ifFt2m+1=b2m+1 and Gt2m+2=b2m+2, one can easily show that(34)db2m+1,b2m+2≤Kdb2m,b2m+1.Hence,(35)db2m+1,b2m+2≤Kdb2m,b2m+1withb2m≠b2m+1∀m∈N.Definedk=dbk,bk+1. Suppose dk0=0, for some k0, then bk0=bk0+1.In casek0=2m, then b2m=b2m+1. We have(36)Mt2m,t2m+1∈0,db2m+1,b2m+2,0,0+db2m,b2m+22s=0,db2m+2,b2m+1,db2m+2,b2m2s.If(37)Mt2m,t2m+1=db2m+2,b2m+1.Then, from (17),(38)sϵdb2m+1,b2m+2≤ψdb2m+2,b2m+1.Sinceψn≤n for all n>0 and ψ0=0, therefore,(39)sϵdb2m+1,b2m+2≤db2m+1,b2m+2.This is only possible ifdb2m+1,b2m+2=0. Hence, b2m+1=b2m+2.Thus,b2m=b2m+1=b2m+2. Similarly, one can show for the remaining two cases.By the same process, ifk0=2m+1, then b2m+1=b2m+2, which gives b2m+2=b2m+3.Therefore,bk is a constant sequence with k≥k0, and (2) is also valid for constant sequence.Hence,bm is b-Cauchy sequence. The completeness of space X implies that there exist b∈X such that bm⟶b. Consequently, the subsequences will also converge to b∈X:(40)At2m⟶b,Bt2m+1⟶b,Gt2m+1⟶b,Ft2m⟶b.Step III: now, we show thatb is a coincidence point of A,F, since(41)limm⟶∞dAt2m,b=limm⟶∞dGt2m+1,b=limm⟶∞db2m+1,b=0,limm⟶∞dBt2m+1,b=limm⟶∞dFt2m+2,b=limm⟶∞db2m+2,b=0.
As from the compatibility ofA,F, we have(42)limm⟶∞dFAt2m,AFt2m=0.
Moreover,limm⟶∞dAt2m,b=0 and limm⟶∞dFt2m,b=0.
AsA and F are continuous mappings, therefore(43)limm⟶∞dFAt2m,Fb=limm⟶∞AFt2m,Ab=0.
Using triangle inequality twice, one can write(44)dFb,Ab≤sdFb,FAt2m+dFAt2m,Ab,dFb,Ab≤sdFb,FAt2m+s2dFAt2m,AFt2m+dAFt2m,Ab.
Taking limitm⟶∞ and using (41) and (42) in the above inequality, we obtain(45)limm⟶∞dFb,Ab≤0.
Thus,dFb,Ab=0. Thus, Ab=Fb. Therefore, b is coincidence at point of A,F.
Similarly,Gb=Bb.
From comparability of elementFb,Gb and (11), one can write(46)sϵdAb,Bb≤ψMb,b,where(47)Mb,b∈dFb,Gb,dFb,Ab,dGb,Bb,dFb,Bb+dGb,Ab2s=dAb,Bb,0,0,dAb,Bbs.
IfMb,b=dAb,Bb, then(48)sϵdAb,Bb≤ψdAb,Bb.
This is possible ifAb=Bb. Hence, Ab=Bb=Fb=Gb.
On the same process, from the other two cases, we can show thatAb=Bb=Fb=Gb.
In the following, the result condition of continuity and compatibility is relaxed for mappings.Theorem 7.
LetA, B, G, F:X⟶X be four mappings on regular partially ordered complete b-metric spaces X,≺¯,d such that AX⊆FX, BX⊆GX, and FX, GX are b-closed subsets of X. Suppose the pairs A,B and B,A are partially weakly increasing with respect to F, G, respectively, whereas pairs A,G,B,F are weak compatible and satisfy the following condition:(49)sϵdAt,Bb≤ψMt,b,∀t,b∈X,where ϵ>1 and ψ:0,∞⟶0,∞ with ψn≤n for n>0, where ψ0=0, and(50)Mt,b∈dGt,Fb,dGt,At,dFb,Bb,dGt,Bb+dFb,At2s,with s≥1. Then, A,B,F, and G have a coincidence point. Furthermore, if Gb,Fb are comparable, then Ab=Bb=Fb=Gb.Proof.
Following the lines of the proof of Theorem6, one can easily show that there exists a sequence bm which converges to some b∈X. Therefore,(51)limm⟶∞dbm,b=0.
Ifb2m+1⊆FX and b2m+2⊆GX but FX and GX are b-closed subsets of X, then there must be some x,k∈X, from which b=Fx and b=Gk. Hence, from the construction of the sequence given in Theorem 6, we have(52)limm⟶∞Ft2m+1=limm⟶∞b2m+1=b=Gk,limm⟶∞Bt2m+1=limm⟶∞b2m+2=b=Gk.
Now, we prove thatAb=Gb.
From regularity ofX, we have Ft2m+1≺¯Gk and Bt2m+1≺¯Gk.
Thus, from (49), one can write(53)sϵdAk,Bt2m+1≤ψMk,t2m+1,where(54)Mk,t2m+1∈dGk,Ft2m+1,dGk,Ak,dFt2m+1,Bt2m+1,dGk,Bt2m+1+dFt2m+1,Ak2s≤dGk,Gk,dGk,Ak,dGk,Gk,dGk,Gk+dGk,Ak2s=0,dGk,Ak,0,dGk,Ak2s.
Sinceψn≤n for all n>0 and ψ0=0 and using (54) in (53), we obtain(55)sϵdAk,Gk≤ψdGk,Ak2s≤dGk,Ak.
This is only possible ifdGk,Ak=0. It implies that Gk=Ak. Hence, b=Gk=Ak. Now, from weak coAb=Gbmpatibility of A and G, we can write Ab=AGk=GAk=Gb. Thus, . Similarly, we can show that Fb=Bb. Furthermore, following the last lines of the proof of Theorem 6, we have Ab=Gb=Fb=Bb.
By puttingG=F=I(identity mapping) in Theorems (6 and 7), then the following corollary is obtained with(56)Mt,b∈dt,b,dt,At,db,Bb,dt,Bb+db,At2s.Corollary 1.
LetA, B:X⟶X be two mappings on X,≺¯,d. Let A,B be partially weakly increasing, whereas ψ:0,∞⟶0,∞ with ψn≤n for n>0 and ψ0=0, satisfy the following condition with comparable elements t,b∈X(57)sϵdAt,Bb≤ψMt,b.Also,(i)
A or B is continuousOr(ii)
X is regularThen,b is a common fixed point of A, B.Example 1.
LetX=0,∞. Define dt,b=t−b2, for all t,binX, then clearly d is a partially ordered b-metric on X with s=2 and partially ordering ≺¯ is defined by(58)t≺¯b⇔b≤t,for allt,binX.
Assume thatA,B are continuous mappings defined as follows:(59)At=ln1+t4,Bt=ln1+t5.
Since1+t/5≤et/5. Therefore, BAt=ln1+At/5≤At. Therefore, At≺¯BAt. Thus, the pair A,B is partially weakly increasing. Now, we show that(60)ψsϵdAt,Bb≤ψMt,b,∀t,b∈X.
Defineψm=bv, where 1≤b≤16 and take 1<ϵ≤4. Here, we discuss the following cases.Case (1): ift/4<b/5, then(61)1≤1+b/51+t/4≤1+b/41+t/4⟹0≤ln1+b/51+t/4≤ln1+b/41+t/4.Using mean value theorem forln1+x, where x∈t/4,b/5, we obtain(62)2ϵdAt,Bb=2ϵBb−At2=2ϵln1+b5−ln1+t42=2ϵln1+b/51+t/42≤2ϵln1+b/41+t/42≤2ϵln1+b4−ln1+t42≤2ϵb4−t42≤2ϵ−4dt,b=ψMt,b.Case (2): ifb/5<t/4, then 0<t/4−b/5≤t/4−b/4. Therefore, 2ϵdAt,Bb=2ϵAt−Bb2=2ϵln1+t/4−ln1+b/52=2ϵt/4−b/52≤2ϵt/4−b/42=2ϵ−4dt,b=ψMt,b.Hence, condition of Corollary1 holds, and 0 is coincidence point.Remark 1.
If we putψn=n in Theorems 6 and 7, we can obtain Theorems (2.1, 2.2) of [18], respectively.Remark 2.
AsA,B are partially weakly increasing, therefore A is B-weakly isotone increasing. Thus, by substitution ϵ=4, in Corollary 1, we get Theorem 2.1 of [20]. Moreover, if A=B and ϵ=4 in Corollary 1, then Corollary 2.4 of [20] is obtained.Remark 3.
Clearly, the conditions of our Corollary1 holds for Example 2.8 of [20] and the corresponding conclusion holds. By substituting ϵ=4 in Example 2, then condition of Theorem 2.1 of [20] does not hold but our condition (57) of Corollary 1 holds.
In the remaining part of this manuscript, we discuss coincidence point of two compatible pairs of mappings with generalizedψ,ϕ-weak contractive condition.
Throughout the rest of this paper, we consider the following,ϵ>1, and for all t,binX, we define(63)Mt,b∈dGt,Fb,dGt,At,dFb,Bb,dGt,Bb+dFb,At2s,and ϕ,ψ:0,∞⟶0,∞ are altering distance functions and X,≺¯,d is partially ordered complete b-metric space.Theorem 8.
LetA, B, G, F:X⟶X be continuous self-mappings on X,≺¯,d, where AX⊆FX and BX⊆GX. Suppose that the pairs A,F and B,G are compatible, whereas the pairs A,B and B,A are partially weakly increasing with respect to F and G, respectively, and satisfy the following condition:(64)ψsϵdAt,Bb≤ψMt,b−ϕMt,b,∀t,b∈X.ThenA,B,F, and G have a coincidence point. Moreover, if Gb,Fb are comparable then Ab=Bb=Fb=Gb.Proof.
Lett0∈X. Since AX⊆FX and BX⊆GX, so there exist t1,t2∈X such that At0=Ft1 and Bt1=Gt2. Construct the sequence bm as follows:(65)b2m+1=At2m=Ft2m+1,b2m+2=Bt2m+1=Gt2m+2,form=0,1,2,….
Since the pairsA,B and B,A are partially weakly increasing with respect to F and G, therefore,(66)Ft1=At0≺¯Bt1=Gt2≺¯At2=Ft3,wheret1∈F−1At0,t2∈G−1Bt1.
Repeating the above process, we can write(67)b2m+1≺¯b2m+2,∀m∈N∪0.
Assume that(68)b2m≠b2m+1,∀m∈N.
We discuss the proof in three steps.Step I: first, we prove thatdb2m+1,b2m≤db2m,b2m−1.SinceFt2m=b2m and Gt2m+1=b2m+1 are comparable, therefore (64) implies that(69)ψsϵdb2m+1,b2m=ψsϵdAt2m,Bt2m−1≤ψMt2m,t2m−1−ϕMt2m,t2m−1,where(70)Mt2m,t2m−1∈db2m,b2m−1,db2m,b2m+1,db2m−1,b2m,db2m,b2m+db2m−1,b2m+12sdb2m,b2m−1,db2m,b2m+1,db2m+1,b2m−12s.To show thatdb2m,,b2m+1≤db2m−1,b2m, suppose that(71)db2m−1,b2m<db2m,b2m+1.Here, we discuss three possible cases ofMt2m,t2m−1.Case (1): ifMt2m,t2m−1=db2m,b2m−1, then(72)ψsϵdb2m,b2m+1≤ψdb2m,b2m−1−ϕdb2m,b2m−1≤ψdb2m,b2m−1.Asψ is nondecreasing, therefore(73)sϵdb2m,b2m+1≤db2m,b2m−1,which contradicts assumption (69). Thus,(74)db2m+1,b2m≤db2m,b2m−1.Case (2): ifMt2m,t2m−1=db2m,b2m+1, then(75)ψsϵdb2m,b2m+1≤ψdb2m,b2m+1−ϕdb2m,b2m+1≤ψsϵdb2m,b2m+1−ϕdb2m,b2m+1.This is only possible ifϕdb2m,b2m+1=0. However, ϕn=0 if n=0. Hence, db2m,b2m+1=0, which implies that b2m=b2m+1 which contradicts (68). Hence,(76)db2m+1,b2m≤db2m,b2m−1.Case (3): ifMt2m,t2m−1=db2m−1,b2m+1/2s.Using triangle inequality and (69), we haveThen,(77)Mt2m,t2m−1≤sdb2m−1,b2m+db2m,b2m+12s≤db2m,b2m+1.(78)ψsϵdb2m,b2m+1≤ψdb2m,b2m+1−ϕdb2m−1,b2m+12s≤ψsϵdb2m,b2m+1−ϕdb2m−1,b2m+12s,which implies thatϕdb2m−1,b2m+1/2s=0, Therefore, db2m−1,b2m+1=0. So, b2m−1=b2m+1 contradicts (68).Therefore, in all three cases, we concluded that(79)db2m+1,b2m≤db2m,b2m−1.Step II: in this step, we will show that the sequencebm is b-Cauchy sequence by using Lemma 1 for all three cases.Case (1): ifMt2m,t2m−1=db2m,b2m−1.From (69), one can write(80)ψsϵdb2m,b2m+1≤ψdb2m,b2m−1−ϕdb2m,b2m−1.Sinceϕn>0 for n>0, therefore(81)ψsϵdb2m,b2m+1≤ψdb2m,b2m−1.Sinceψ is nondecreasing, therefore(82)sϵdb2m,b2m+1≤db2m,b2m−1,db2m,b2m+1≤1sϵdb2m,b2m−1,db2m,b2m+1≤Kdb2m,b2m−1,whereK=1/sϵ. Thus,(83)db2m,b2m+1≤Kdb2m,b2m−1.Case(2): whenMt2m,t2m−1=db2m,b2m+1, then from (79), one can write(84)Mt2m,t2m−1=db2m,b2m+1≤db2m,b2m−1.From (69), one has(85)ψsϵdb2m,b2m+1≤ψdb2m,b2m−1−ϕdb2m,b2m+1.Sinceϕn>0 for n>0, therefore(86)ψsϵdb2m,b2m+1≤ψdb2m,b2m−1.From nondecreasing property ofψ, we have(87)sϵdb2m,b2m+1≤db2m,b2m−1,db2m,b2m+1≤Kdb2m,b2m−1,whereK=1/sϵ with K∈0,1/s.Case (3): whenMt2m,t2m−1=db2m−1,b2m+1/2s.Then, by using triangle inequality and (79) it will be converted to case (2). Therefore, (2) holds for all cases.Similarly, ifFt2m+1=b2m+1 and Gt2m+2=b2m+2, again we have(88)db2m+1,b2m+2≤Kdb2m,b2m+1.Hence,(89)dbm+1,bm+2≤Kdbm,bm+1,forbm≠bm+1∀m∈N.Letdk=dbk,bk+1. Assume that dk0=0 for k0, then bk0=bk0+1.Whenk0=2m, it implies that b2m=b2m+1; therefore,(90)Mt2m,t2m+1∈0,db2m+1,b2m+2,0,0+db2m,b2m+22s=0,db2m+2,b2m+1,db2m+2,b2m2s.If(91)Mt2m,t2m+1=db2m+2,b2m+1.Then, from (64),(92)ψsϵdb2m+1,b2m+2≤ψdb2m+2,b2m+1−ϕdb2m+2,b2m+1.From the above, we can obtain(93)ψsϵdb2m+1,b2m+2≤ψdb2m+1,b2m+2.Sinceψ is nondecreasing, therefore(94)sϵdb2m+1,b2m+2≤db2m+1,b2m+2.It implies thatdb2m+1,b2m+2=0. Therefore, b2m+1=b2m+2.Hence,b2m=b2m+1=b2m+2.Therefore, the sequencebk is constant sequence for k≥k0. Therefore, (2) also holds for constant sequence bk. Thus, by Lemma 1, bm is a b-Cauchy sequence. From the completeness of X one can say that b-Cauchy sequence bm converges to some binX.Consequently,At2m⟶b,Bt2m+1⟶b,Ft2m+1⟶b,andGt2m⟶b.Step III: now, we show thatb is a coincidence point of A and F:(95)limm⟶∞dAt2m,b=limm⟶∞dGt2m+1,b=limm⟶∞db2m+1,b=0,limm⟶∞dBt2m+1,b=limm⟶∞dFt2m+2,b=limm⟶∞db2m+2,b=0.
Since the pairA,F is compatible, so we have(96)limm⟶∞dFAt2m,AFt2m=0.
Moreover,limm⟶∞dAt2m,b=0 and limm⟶∞dFt2m,b=0. Also, A and F are continuous mapping; therefore,(97)limm⟶∞dFAt2m,Fb=limm⟶∞dAFt2m,Ab=0.
Using triangle inequality, one can write(98)dFb,Ab≤sdFb,FAt2m+dFAt2m,Ab.
Again, applying triangle inequality on the second term, we have(99)dFb,Ab≤sdFb,FAt2m+s2dFAt2m,AFt2m+dAFt2m,Ab.
By taking limitm⟶∞ and using (96) and (97), we can write(100)limm⟶∞dFb,Ab≤0.
Therefore,dFb,Ab=0. Thus, Ab=Fb.
Similarly, we can show thatBb=Gb.
SinceGb and Fb are comparable, therefore from (64), we can obtain(101)ψsϵdAb,Bb≤ψMb,b−ϕMb,b,where(102)Mb,b∈dGb,Fb,dGb,Ab,dFb,Bb,dGb,Bb+dFb,Ab2s=dAb,Bb,dBb,Ab,dAb,Bb,0,=dAb,Bb.
Thus,(103)ψsϵdAb,Bb≤ψdAb,Bb−ϕdAb,Bb,which implies that Ab=Bb. Hence, Ab=Bb=Fb=Gb.
In the forthcoming, result condition of continuity and compatibility for mapping is relaxed.Theorem 9.
LetA, B, G, F:X⟶X, with AX⊆FX, BX⊆GX and FX, GX are b-closed subsets of regular partially ordered complete b-metric space X,≺¯,d. Assume that the pairs A,B and B,A are partially weakly increasing with respect to F and G, respectively, whereas A,G and B,F are weakly compatible and satisfy the following condition:(104)ψsϵdAt,Bb≤ψMt,b−ϕMt,b,∀t,b∈X.Then, these four mappings have a coincidence point. IfGb and Fb are comparable then Ab=Bb=Fb=Gb.Proof.
As Theorem8, one can easily construct a sequence bm which converges to some b∈X. Thus,(105)limm⟶∞dbm,b=0.
Ifb2m+1⊆FX, b2m+2⊆GX. Since FX and GX are b-closed subsets X, therefore, there exist some x,k∈X, such that b=Fx and b=Gk. Hence,(106)limm⟶∞Ft2m+1=limm⟶∞b2m+1=b=Gk,limm⟶∞Bt2m+1=limm⟶∞b2m+2=b=Gk.
Now, we prove thatAb=Gb.
SinceX,≺¯,d is regular, therefore(107)Ft2m+1≺¯Gk,Bt2m+1≺¯Gk.
Hence, (104) implies that(108)ψsϵdAk,Bt2m+1≤ψMk,t2m+1−ϕMk,t2m+1,where(109)Mk,t2m+1∈dGk,Ft2m+1,dGk,Ak,dFt2m+1,Bt2m+1,dGk,Bt2m+1+dFt2m+1,Ak2s≤dGk,Gk,dGk,Ak,dGk,Gk,dGk,Gk+dFt2m+1,Gk+dGk,Ak2s=0,dGk,Ak,0,dGk,Ak2s.
From (108) and (109), we have(110)ψsϵdAk,Gk≤ψdGk,Ak2s−ϕdGk,Ak2s≤ψsϵdGk,Ak2s−ϕdGk,Ak2s≤ψsϵdGk,Ak−ϕdGk,Ak2s,which implies that dGk,Ak=0. Hence, Gk=Ak. Thus, b=Gk=Ak. Now, from weak compatibility of Ab and G, one can write Ab=AGk=GAk=Gb. Thus, Ab=Gb. Similarly, we can show Fb=Bb. By following the last lines of the proof of Theorem 8 we have Ab=Bb=Fb=Gb.
By takingG=F, in Theorem 8, the following corollary is obtained with(111)Mt,b∈dFt,Fb,dFt,At,dFb,Bb,dFt,Bb+dFb,At2s.Corollary 2.
LetA, B, F:X⟶X be three continuous mappings on X,≺¯,d where AX∪BX⊆FX. Let pairs A,F and B,F be compatible, whereas A,B be partially weakly increasing with respect to F and satisfying the following condition:(112)ψsϵdAt,Bb≤ψMt,b−ϕMt,b,∀t,b∈X,then A,B, and F have a coincidence point.If we putA=B in Theorem 8, the following corollary is obtained with(113)Mt,b∈dGt,Fb,dGt,At,dFb,Ab,dGt,Ab+dFb,At2s.Corollary 3.
LetA, F, G:X⟶X be three continuous mappings on X,≺¯,d such that AX⊆FX∪GX. Assume that pairs A,F and A,G are compatible, whereas A is partially weakly increasing with respect to G, F and satisfies the following condition:(114)ψsϵdAt,Ab≤ψMt,b−ϕMt,b,∀t,b∈X,then A,F, and G have a coincidence point.When we putA=B and G=F in Theorem 8, the following corollary is obtained with(115)Mt,b∈dFt,Fb,dFt,At,dFb,Ab,dFt,Ab+dFb,At2s.Corollary 4.
LetA, F:X⟶X be mappings on X,≺¯,d such that AX⊆FX. Assume that pair A,F is compatible, whereas A is partially weakly increasing with respect to F and satisfies the following condition:(116)ψsϵdAt,Ab≤ψMt,b−ϕMt,b,∀t,b∈X,eitherA and F are continuous or space X is regular.Then,A,F have a coincidence point.If we putF=G=I (identity mapping) in Theorem 8, then the following corollary is obtained with(117)Mt,b∈dt,b,dt,At,db,Bb,dt,Bb+db,At2s.Corollary 5.
LetA, B:X⟶X be mappings on X,≺¯,d such that the pair A,B is partially weakly increasing and satisfies the following condition:(118)ψsϵdAt,Bb≤ψMt,b−ϕMt,b,∀t,b∈X,eitherA and B are continuous or space X is regular. Then, A,B have a common fixed point.Example 2.
Lett≺¯b if and only if b≤t,for allt,binX, is partially ordering on X=0,∞. Define b-metric on X by dt,b=t−b2, with s=2. Suppose A,B,F, and G be continuous mappings defined by(119)At=sinh−1t,Bt=sinh−1t2,Gt=sinh4t,Ft=sinh8t.
First, we check the compatibility ofA,F. Let ta be sequence which converges to some a∈X. Then,(120)lima⟶∞da,Ata=lima⟶∞da,Fta=0,lima⟶∞sinh−1ta−a2=lima⟶∞sinh8ta−a2=0.
The continuity ofA, F implies that(121)lima⟶∞ta−sinha2=lima⟶∞ta−sinh−1a82=0.
Since limit of the sequence is unique, thereforesinha=sinh−1a/8, which implies that a=0. Also, A and F are continuous, so(122)lima⟶∞dAFta,FAta=lima⟶∞AFta−FAta2=0.
Hence, the pairA,F is compatible. Similarly, one can check the compatibility of the pair B,G.
Now, we have to prove thatA,B is partially weakly increases with respect to G. Assume that b∈G−1At, for t,b∈S,(123)At=Gb⟹sinh−1t=sinh4b.
Therefore,b=sinh−1sinh−1t/4. We know that(124)sinh8t≥sinh−1t⟹8t≥sinh−1sinh−1t.
t≥sinh−1sinh−1t/8. Hence,(125)At=sinh−1t≥sinh−1sinh−1sinh−1t8=sinh−1b2=Bb.
Therefore,(126)At≺¯Bb.
With the same process one can show thatB,A is partially weakly increases with respect to F.
Next, we check the validity of the following inequity:(127)ψsϵdAt,Bb≤ψMt,b−ϕMt,b,for allt,b∈X.
Letψm=bm and ϕm=b−1m, where 1<b<17 and 1<ϵ<2.
Apply mean value theorem tosinh−1 in intervals t,b/2 and sinh in intervals 8t,4b. Take(128)ψ2ϵdAt,Bb=2ϵbAt−Bb2≤2ϵbsinh−1t−sinh−1b22≤2ϵbt−b22≤2ϵb8t−4b282≤2ϵ−6bsinh8t−sinh4b2=2ϵ−6bFt−Gb2≤dFt,Gb=ψMt,b−ϕMt,b.
All conditions of Theorem8 are satisfied, and consequently 0 is a coincidence point.Remark 4.
Note that the conditions of our Theorem8 holds for example of [19] and the corresponding conclusion holds. However, when 1<ϵ<2, then inequality of Theorem 3 does not hold for our Example 2. Hence, Theorem 8 is a genuine generalization of Theorem 3.Remark 5.
By substitution,ϵ=2 in Theorem 8 and 9 and corollaries 3.7, 3.8, and 3.9, and we get the results of [19].Remark 6.
Theorems 2.4 and 2.6 of [45] are special case of Corollary 4. By putting G=I in Corollary 2, one can get Corollary 2.7 of [45].Remark 7.
Theorems 3.1 and 4.3 and Corollary 3.3 of [46] are special cases of Corollary 5.



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