# Nonunique Coincidence Point Results via Admissible Mappings

Authors: Erdal Karapınar; Chi-Ming Chen; Andreea Fulga
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper is aimed at presenting some coincidence point results using admissible mapping in the framework of the partialb-metric spaces. Observed results of the article cover a number of existing works on the topic of “investigation of nonunique fixed points.” We express an example to indicate the validity of the observed outcomes.


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## Body

## 1. Introduction and Preliminaries

In 1974, Ćirić [1] published the first paper on nonunique fixed point theory. Despite Banach’s theorem, Ćirić [1] focused only on the existence of a fixed point, but not the uniqueness. The motivation of Ćirić [1] was inspired by Banach’s motivation. As it is known, Banach’s fixed point theorem is abstracted from Picard’s paper, in which Picard [2] analyzed both the existence and uniqueness of the solution of the certain differential equation (see [3–5]). On the other hand, not all differential or integral equations have a unique solution. In the differential/integral equations, nonunique solutions are also crucial, for example, periodic solutions. Consequently, Ćirić [1] investigated the corresponding fixed point theorems that would be a tool in finding periodic solutions of the differential/integral equations. In the last five decades, a number of nonunique fixed point results have been reported in two ways: either proposing a new contraction type or changing the structure. The first example for the changing the contraction inequality, in the standard set-up, was given by Achari [6] in 1976 and Pachpatte [7] in 1973. Fifteen years later, Ćirić and Jotić [8] proposed a new type of contraction inequalities in the context of complete metric space. This trend was followed by the attractive results [9–13]. On the other hand side, in [14–17], the authors observed several characterizations of the unique fixed point results in the setting of complete b-metric spaces. Indeed, among the several extensions of metric structure, the true extension is the b-metric space. For this reason, observed nonunique fixed theorems in the context of b-metric space is very interesting and important, see also [18–20]. In addition, in [21–23], the characterization of fixed point theorems in partial metric spaces is crucial due to the potential application in the domain theory of computer science. Regarding the applied mathematics, nonunique fixed point results in cone metric spaces have taken attention [24].In this paper, we consider a nonunique fixed point theorem in the context of the very general frame, partialb-metric spaces. An illustrative example is a set-up to indicate the validity of the main theorem.LetM be a nonempty set, a real number s≥1, and ℕ=1,2,3,⋯.In this case, the triplet M,pb,s forms a partial b-metric space, on short pb-ms.Undoubtedly, b-metric spaces (and ordinary metric spaces) are closely related to partial b-metric spaces. Definitely, a b-metric space (s≥1) is a partial b-metric space with zero self-distance and a partial metric space is a partial b-metric space with s=1. Moreover, a partial b-metric can define a b-metric space. Indeed, for example, let pb be a partial b-metric on M. Then, the functions b′p,bp,bp,m:M⟶M, where
(1)bpu,v=pbu,v,ifu≠v,0,ifu=v,(2)b′pu,y=2pbu,y−pbu,u−pby,y,(3)bp,mu,y=pbu,y−minpbu,u,pby,yare b-metrics on M.Definition 1.
A functionpb:M×M⟶0,∞ is a partial b-metric on M if for all u,y,w∈M, it satisfies the following conditions:
pb1u=y⟺pbu,u=pbu,y=pby,y
pb2pbu,u≤pbbu,y
pb3pbu,y≤py,u
pb4pbu,y≤spbu,w+pbw,u−pbw,wExample 1.
(see [25]). Let pb be a partial metric on the set M. Then, the functions pb:M×M⟶0,∞ are given for all u,y∈M by
(1)
pbu,y=pu,y+bu,y is a partial b-metric on M (where b is a b-metric (s>1) on M)(2)
pbu,y=pu,yr for r≥1, define a partial b-metrics on M with coefficient s=2r−1Remark 2.
Frompb1 and pb2, it follows that if u,y∈M are such that pbu,y=0, then u=y.Definition 3.
(see [26, 27]). Let un be a sequence on the pb‐msM,pb,s≥1(1)
un is pb-convergent to u∈M if limn→∞pbu,un=pbu,u(2)
un is pb-Cauchy if limn,q→∞pbun,uq exists and is finite(3)
un is 0-pb-Cauchy if limn,q→∞pbun,uq=0(4)
M,pb,s≥1 is pb-complete if every pb-Cauchy sequence in M is pb-convergent(4)limn,q→∞pbun,uq=limn→∞pbun,u=pbu,u(5)
M,pb,s≥1 is 0-pb-complete if every 0-pb-Cauchy sequence we can find u∈M such that(5)limn,q→∞pbun,uq=limn→∞pbun,u=pbu,u=0
Moreover, in [26], the following interesting results were proved.Lemma 4.
(see [26]). Every pb-complete pb-ms M,pb,s≥1 is 0-pb-complete.Lemma 5.
(see [26]). The pb-ms M,pb,s≥1 is 0-pb-complete if and only if the b-metric space M,bp,s≥1 is complete, where the b-metric bp was defined in (3).
They also showed that the converse affirmation does not hold.
LetR,S to self-mappings on the set M. We say that
(i)
S commutes with R on M if RSu=SRu for all u∈M(ii)
a pointz∈M is a point of coincidence of R and S if we can find u∗∈M such that z=Ru∗=Su∗(iii)
a pointu∗∈M is a common fixed point of R and S if Ru∗=u∗=Su∗
We will use the following notations:(6)CcR,SM=u∈M∣Ru=SuM∗=M\CcR,SM.In [28], the notion of R-β-admissible mapping was introduced as follows:
(i)
Let the functionβ:M×M⟶0,∞ and R,S:M⟶M. The mapping S is said to be R-β-admissible if(7)βRu,Ry≥1impliesβSu,Sy≥1,for all u,y∈M.
In case thatR=IM, the mapping S is said to be β-admissible.
LetM,pb,s≥1 be a pb-ms and β;M×M⟶0,+∞. The space M is β-regular if for every sequence zn in M such that zn⟶z and βzn,zn+1≥1, there exists a subsequence znl of zn such that
(8)βznl,z∗≥1,for all l∈ℕ.Lemma 6.
LetR,S:M⟶M such that S is a R-β-admissible. If there exists u0∈M such that βRu0,Su0≥1, then
(9)βRun,Run+1≥1,where the sequence un in M is defined by Sun=Run+1, for each n∈ℕ∪0.Proof.
By the assumptionβRu0,Su0≥1, since the mapping S is R-β-admissible, we get
(10)βRu0,Ru1=βRu0,Su0≥1impliesβRu1,Ru2=βSu0,Su1≥1,and by induction, it follows that
(11)βRun,Run+1≥1,for n∈ℕ∪0.


## 2. Main Results

Following the idea in [29], we state the following results useful in the sequel.Lemma 7.
LetM,pb,s≥1 be a pb-ms. If un is a sequence in M such that there exists zn in M, satisfying the inequality
(12)pbun,un+1≤cpbun−1,un,for any n∈ℕ, then the sequence is un and is 0-pb-Cauchy.Proof.
First of all, by (12), we get
(13)pbun,un+1≤cnpbu0,u1,for all n∈ℕ. On the other hand, by using pb4, we can derive that
(14)pbun,un+q≤spbun,un+1+pbun+1,un+q−pbun+1,un+1≤spbun,un+1+s2pbun+1,un+2+pbun+2,un+2,un+q−−pbun+1,un+1−pbun+2,un+2⋯≤spbun,un+1+s2pbun+1,un+2+⋯+sqpbun+q−1,un+q−−∑l=1q−1pbun+l,un+l≤sqpbun,un+1+pbun+1,un+2+⋯+pbun+q−1,un+q−−∑l=1q−1pbun+l,un+l.(1)
Ifc∈0,1/s, by (13) and (14), we get(15)pbun,un+q≤∑l=0q−1sl+1cn+lpbu0,u1−∑l=1q−1pbun+l,un+l≤scn∑l=0q−1sclpbu0,u1=scn1−scq1−sc⟶0asn,q⟶∞.Therefore, un is a 0-pb-Cauchy sequence.
(2)
Ifc∈1/s,1, thus cn⟶0 (as n⟶∞). Moreover, there exits l∈ℕ such that cl<1/s. This means l>−logs/logc. Again, by (13) together with (14), we have(16)pbunl,un+1l≤slpbunl,unl+1+⋯+pbunl+l−1,un+1l−−∑j=1l−1pbunl+j,unl+j≤sl∑j=0l−1cnl+jpbu0,u1−∑j=1l−1pbunl+j,unl+j≤slcnl∑j=0l−1pbu0,u1≤cnlsl·pbu0,u11−c⟶0asn⟶∞.Thereby, letting λ=cl<1/s by Case (i), we get that the sequence unl is 0-pb-Cauchy sequence, which means that
(17)limn,q→∞pbunl,uql=0.On the other hand,
(18)pbuln/l,un≤spbuln/l,uln/l+1+pbuln/l+1,un−pbuln/l+1,uln/l+1≤slpbuln/l,uln/l+1+⋯+pbun−1,un−−pbuln/l+1,uln/l+1+⋯+pbun−1,un−1,and using (13), we have
(19)pbuln/l,un≤slcln/l+⋯+cn−1pbu0,u1≤slcln/lpbu0,u11−c⟶0asn⟶∞.
Finally, combining relations (19) and (17) and keeping in mind pb4, we have
(20)pbun,uq≤spbun,uln/l+pbuln/l,uq−pbuln/l,uln/l≤spbun,uln/l+s2pbuln/l,ulq/l+s2pbulq/l,uq−−pbuln/l,uln/l−pbulq/l,ulq/l≤spbun,uln/l+s2pbuln/l,ulq/l+s2pbulq/l,uq⟶0asn,q⟶∞.Thereupon, the sequence un is 0-pb-Cauchy.Theorem 8.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exists κ∈0,1 such that
(21)βRu,RyminpbSu,Sy,pbSy,Ry−−minbpSu,Ry,bpSy,Ru≤κmaxpbRu,Ry,pbSu,Ru,for all u,y∈M, such that u≠y when u,y∈CcR,SM. Suppose also that
(a)
SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms(b)
S is R-β-admissible, and there exists u0∈M such that βRu0,Su0≥1(c)
M is β-regular
Then, the mappingsS and R have a point of coincidence.Proof.
Letu0 be an arbitrary point in M, such that βRu0,Su0≥1. Thus, since SM⊂RM, there exists u1∈M such that Su0=Ru1. Thereupon, Su1∈SM⊂RM and we can find u2∈M such that Su1=Ru2. In this way, we can build a sequence un⊆M as follows:
(22)havingdefinedun∈M,weletun+1∈MsuchthatSun=Run+1,for all n∈ℕ∪0. Letting u=un and y=un+1 in (ref1T1) and taking into account Lemma 6, we have
(23)minpbSun,Sun+1,pbSun+1,Run+1−−minbp,mSun,Run+1,bp,mSun+1,Run≤βRun,Run+1minpbSun,Sun+1,pbSun+1,Run+1−−minbpSun+1,Run≤κmaxpbRun,Run+1,pbSun,Run. Keeping in mind (22), we get
(24)minpbRun+1,Run+2,pbRun+1,Run,pbRun+2,Run+1−−minbpRun+1,Run+1,bpRun+2,Run≤κmaxpbRun,Run+1,pbRun+1,Run=κpbRun,Run+1,which is equivalent with
(25)minpbRun+1,Run+2,pbRun+1,Run−−minbpRun+1,Run+1,bpRun+2,Run≤κpbRun,Run+1.Therefore, we get
(26)pbRun+1,Run+2≤κpbRun,Run+1,for any n∈ℕ∪0.Let now zn be a sequence in M, with zn=Run+1=Sun, n∈ℕ∪0. First of all, we mention that zn≠zn+1 for every n∈ℕ. Indeed, if we suppose that there exists m0∈ℕ∪0 such that zm0=zm0+1, thus by (22), we have
(27)Rum0+1=Sum0=zm0=zm0+1=Sum0+1,so that zm0+1 is a point of coincidence. Thus, zn≠zn+1 for every ℕ∪0 and (28) can be rewritten as
(28)pbzn,zn+1≤κpbzn−1,zn.Therefore, according to Lemma 7, the sequence zn is 0-pb-Cauchy. Since the space is 0-pb-complete, it follows that there is z∈M such that
(29)limn,q→∞pbzn,zq=limn→∞pbzn,z=pbz,z=0.But, on the other hand, since zn=Run+1 and the space RM,pb,s is 0-pb-complete, we can find u∗∈M, with z=Ru∗. Thus,
(30)limn→∞pbSun,Ru∗=limn→∞pbRun,Ru∗=pbRu∗,Ru∗=0.Supposing that Ru∗≠Su∗ for u=unl and y=u∗ and taking into account the β-regularity of the space M, we have
(31)minpbSunl,Su∗,pbSu∗,Ru∗−minbpSunl,Ru∗,bpSu∗,Runl≤≤βznl,zminpbSunl,Su∗,pbSu∗,Ru∗−minbpSunl,Ru∗,bpSu∗,Runl=βRunl,Ru∗minpbSunl,Su∗,pbSu∗,Ru∗−minbpSunl,Ru∗,bpSu∗,Runl≤κmaxpbRunl,Ru∗,pbSunl,Runl.If minpbSunl,Su∗,pbSu∗,Ru∗=pbSu∗,Ru∗, the above inequality becomes
(32)pbSu∗,Ru∗−minpbSunl,Su∗,pbSu∗,Ru∗−minbpSunl,Ru∗,bpSu∗,Runl≤κmaxpbRunl,Ru∗,pbSunl,Runl.Letting l⟶∞ and taking into account (28) and (30), we get
(33)pbSu∗,Ru∗=0,and by pb1, pb1, we have Su∗=Ru∗.If minpbSunl,Su∗,pbSu∗,Ru∗=pbSunl,Su∗, we find that liml→∞pbSunl,Su∗=0. On the other hand, by pb4,
(34)pbSu∗,Ru∗≤spbSu∗,Sunl+pbSunl,Ru∗−pbSunl,Sunl,and then, pbSu∗,Ru∗=0, as l⟶∞. This proves that z=Su∗=Ru∗, that is, z is a point of coincidence for S and R.Example 2.
LetM=0,∞ and pb:M×M⟶0,∞ be a partial b-metric, where pbu,y=maxu,y2. Let the mappings S,R:M⟶M,
(35)Su=u+12,ifu∈0,1,3,ifu>1,Ru=u+24,ifu∈0,1,u+510,ifu>1,
and the functionβ:M×M⟶0,∞,
(36)βx,v=2,forx=v=12,3,forx=v=3,1,forx,v≥4,0,otherwise.Obviously, since x=Ru≥4 for u≥35 we have
(i)
Foru,y≥35(37)βRu,Ry=1⟹βSu,Sv=β3,3=3>1,β12,12=βR0,R0=2⟹βS0,S0=β12,12=2,β3,3=βR25,R25=3⟹βS25,S25=β3,3=3.Moreover,
(38)βRu,RyminpbSu,Sy,pbbSy,Ry−−minbpSu,Ry,bpSy,Ru≤minpb3,3,pb3,Ry=9,≤κ·16≤κ·maxpbRu,Ry,pbSu,Ru,for any 9/16<κ<1.(ii)
All other cases are uninteresting due to the way the functionβ was defined
Consequently, by Theorem8, the mappings S,R have points of coincidence. These are 1/2=S0=R0, respectively, 3=S25=R25.Corollary 9.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exists κ∈0,1 such that
(39)minpbSu,Sy,pbSy,Ry−−minbpSu,Ry,bpSy,Ru≤κmaxpbRu,Rv,pbSu,Ru,for every u,y∈M, such that u≠y when u,y∈CcR,SM. If SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms, then the mappings S and R have a point of coincidence.Proof.
It is enough to chooseβu,y=1 in Theorem 8.Theorem 10.
LetM,pb,s≥1 be a complete pb-ms and a mapping S:M⟶M. Suppose that there exists κ∈0,1 such that
(40)βu,yminpbSu,Sy,pbSy,y−−minbpSu,y,bpSy,u≤κmaxpbu,y,pbSu,u,for every u,y∈M, such that u≠y. Suppose also that
(a)
S is β-admissible, and there exists u0∈M such that βu0,Su0≥1(b)
M is β-regular
Then, the mappingS has a fixed point.Proof.
PutR=IM in Theorem 8.Corollary 11.
LetM,pb,s≥1 be a complete pb-ms and a mapping S:M⟶M. Suppose that there exists κ∈0,1 such that
(41)minpbSu,Sy,pbSy,y−−minbpSu,y,bpSy,u≤κmaxpbu,y,pbSu,u,for every u,y∈M,u≠y. Then, the mapping S has a fixed point.Proof.
It is enough to chooseβu,y=1 in Theorem 10.Theorem 12.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exist κ∈0,1 and a>0 such that
(42)βRu,RyMS1u,y−a·NS,R1u,y≤κpbSu,RupbSy,Ry,where
(43)MS,R1u,y=minpbSu,Sy2,pbSy,Ry2,NS,R1u,y=minbpSu,RybpSy,Ru,pbSu,RypbSu,Sy,pbSy,RupbRu,Ry,for every u,y∈M, such that u≠y when u,y∈CcR,SM. Suppose also that:
(a)
SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms(b)
S is R-β-admissible, and there exists u0∈M such that βRu0,Su0≥1(c)
M is β-regular
Then, the mappingsS and R have a point of coincidence.Proof.
Starting with a pointu0∈M such that βRu0,Su0≥1, we build the sequences un, zn as in Theorem 8,
(44)zn=Run+1=Sun,foralln∈ℕ.Using the same arguments, we can assume that zn≠zn+1, also, for all n∈ℕ. Thus, for u=un, y=un+1,
(45)MS,R1un,un+1=minpbSun,Sun+12,pbSun+1,Run+12=minpbzn,zn+12,pbzn+1,zn2=pbzn+1,zn2,(46)NS,R1un,un+1=minbpSun,Run+1bpSun+1,Run,pbSun,Run+1pbSun,Sun+1,pbSun+1,RunpbRun,Run+1=minbpzn,znbpzn+1,zn−1,pbzn,zn+1pbzn,zn+1,pbzn+1,un−1pbun−1,zn=0,and taking into account Lemma 6, (42) becomes
(47)MS,R1un,un+1≤βRun,Run+1MS,R1un,un+1−a·NS,R1un,un+1≤κpbSun,Run·pbSun+1,Run+1.Taking into account (46), the above inequality turns into
(48)pbzn,zn+12≤κpbzn,zn−1pbzn+1,zn,or equivalent (since zn≠zn+1)
(49)pbzn,zn+1≤κpbzn,zn−1.Accordingly, from Lemma 7, it follows that the sequence zn is 0-pb-Cauchy and due to the completeness of the space, there exists z∈M such that limn⟶∞pbzn,z=pbz,z=0. Following the corresponding lines in Theorem 8, we can find u∗∈M such that Ru∗=z. Supposing that Ru∗≠Su∗ for u=unl and y=u∗ and taking into account the assumption (c),
(50)MS,R1unl,u∗≤βRunl,Ru∗MS,R1unl,u∗−a·NS,R1unl,u∗≤κpbSunl,Runl·pbSu∗,Ru∗,where
(51)MS,R1unl,u∗=minpbSunl,Su∗2,pbSu∗,Ru∗2,NS,R1unl,u∗=minbpSunl,Ru∗bpSu∗,Runl,pbSunl,Ru∗pbSu∗,Sunl,pbSu∗,RunlpbRu∗,Runl.Since liml→∞NS,R1unl,u∗=0 and liml→∞=pbSunl,Runl·pbSu∗,Ru∗=0 (by) letting l⟶∞ in (50), we have
(52)eitherpbSu∗,Ru∗2=0orliml→∞pbSunl,Su∗2=0.(1)
IfpbSu∗,Ru∗2=0, it follows that Su∗=Ru∗.(2)
Ifliml→∞pbSunl,Su∗2=0, by pb4(53)pbRu∗,Su∗≤spbRu∗,Sunl+pbSunl,Su∗−pbSunl,Sunl≤spbRu∗,Sunl+pbSunl,Su∗⟶0asl⟶∞,so pbRu∗,Su∗=0.
Thereupon,Ru∗=Su∗=z and z is a point of coincidence of R and S.Example 3.
LetM=α1,α2,α3,α4,α5 and the partial b-metric pb:M×M⟶0,+∞ defined as follows (Table 1).Table 1pbu,yα1α2α3α4α5α10263042α22132131α36321422α430211457α542312276Let the functionβ:M×M⟶0,+∞, with
(54)βu,y=1,foru,y∈α5,α3,α3,α2,2,foru,y=α2,α2,0,otherwise,and two mappingsS,R:M⟶M (Table 2).Table 2α1α2α3α4α5Sα5α3α2α2α2Rα1α5α1α2α3First of all, we remark that(55)βα5,α3=βRα2,Rα5=1⟹βSα2,Sα5=βα3,α2=1,βα3,α2=βRα5,Rα4=1⟹βSα5,Sα4=βα2,α2=2,βα2,α2=βRα4,Rα4=2⟹βSα4,Sα4=βα2,α2=2,which shows as that (b) holds. Also, it is easy to see that (a) and (c) are satisfied, so it remains to be verified (42). We distinguish two cases as follows:
(1)
u,y=α2,α5(56)MS,R1α2,α5=minpbSα2,Sα52,pbSα5,Rα52=minpbα3,α22,pbbα2,α32=9,NS,R1α2,α5=minbpSα2,Rα5bpSα5,Rα2,⋯=minbpα3,α3bpα2,α5,⋯=0,pbSα2,Rα2pbSα5,Rα5=pbα3,α5pbα2,α3=22·3=66.(2)
u,y=α5,α4(57)MS,R1α5,α4=minpbSα5,Sα42,pbSα4,Rα42=minpbα2,α22,pbα2,α22=1,NS,R1α5,α4=minbpSα5,Rα4bpSα4,Rα5,⋯=minbpα2,α2bpα2,α3,⋯=0,pbSα5,Rα5pbSα4,Rα4=pbα2,α3pbα2,α2=3·1=3.So, for anyκ∈0,1, the inequality (42) holds. Therefore, the mappings S, R have a point of coincidence, which is z=α2.Corollary 13.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M→M. Suppose that there exist κ∈0,1 and a>0 such that
(58)MS,R1u,y−a·NS,R1u,y≤κpbSu,RupbSy,Ry,where
(59)MS,R1u,y=minpbSu,Sy2,pbSy,Ry22,NS,R1u,y=minbpSu,RybpSy,Ru,pbSu,RypbSu,Sy,pbSy,RupbRu,Ry,for every u,y∈M, such that u≠y when u,y∈CcR,SM. Then, the mappings S and R have a point of coincidence providing that SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms.Proof.
Putβu,y=1 in Theorem 12.Theorem 14.
LetM,pb,s≥1 be a complete pb-ms a mapping S:M⟶M. Suppose that there exists κ∈0,1 and a>0 such that
(60)βu,yMS1u,y−a·NS1u,y≤κpbSu,upbSy,y,where
(61)MS1u,y=minpbSu,Sy2,pbSy,y2,NS1u,y=minbpSu,ybpSy,u,pbSu,ypbSu,Sy,pbSy,upbu,y,for every u,y∈M,u≠y. Suppose also that
(a)
S is β-admissible, and there exists u0∈M such that βu0,Su0≥1(b)
M is β-regular
Then, the mappingS possesses a fixed point.Proof.
ChooseR=IM in Theorem 12.Corollary 15.
LetM,pb,s≥1 be a complete pb-ms a mapping S:M⟶M. Suppose that there exists κ∈0,1 and a>0 such that
(62)MS1u,y−a·NS1u,y≤κpbSu,upbSy,y,where
(63)MS1u,y=minpbSu,Sy2,pbSy,y2,NS1u,y=minbpSu,ybpSy,u,pbSu,ypbSu,Sy,pbSy,upbu,y,for every u,y∈M, u≠y. Then, the mapping S possesses a fixed point.Proof.
Putβu,y=1 in Theorem 14.Theorem 16.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exist κ∈0,1 and a>0 such that
(64)βRu,RyMS,R2u,y≤κ·NS,R2u,y,where
(65)MS,R2u,y=pbSu,SypbSy,Ry−a·minbpSu,Ry,bpSy,Ru,NS,R2u,y=pbRu,Ry·maxpbSu,Ru,pbSy,Ry,pbSu,Ry+pbSy,Ru2s,for every u,y∈M, such that u≠y when u,y∈CcR,SM. Suppose also that
(a)
SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms(b)
S is R-β-admissible and there exists u0∈M such that βRu0,Su0≥1(c)
M is β-regular
Then, the mappingsS and R have a point of coincidence.Proof.
We will only sketch the proof, because, basically, we use the same technique that was used in the above theorems. Indeed, foru=un, y=un+1, where the sequences zn, un are defined in Theorem 8, we have
(66)MS,R2un,un+1=pbSun,Sun+1pbSun+1,Run+1−−a·minbpSun,Run+1,bpSun+1,Run=pbzn,zn+1pbzn+1,zn−a·minbpzn,zn,bpzn+1,zn−1=pbzn,zn+12,NS,R2un,un+1=pbRun,Run+1·maxpbSun,Run,pbSun+1,Run+1,pbSun,Run+1+pbSun+1,Run2s=maxpbzn−1,zn,pbzn,zn+1,pbzn,zn+pbzn+1,zn−12s≤pbzn−1,zn·mpbzn−1,zn·maxpbzn−1,zn,pbzn,zn+1,pbzn,zn+spbzn+1,zn+pbzn,zn+1−pbzn,zn2s=pbzn−1,zn·maxpbzn−1,zn,pbzn,zn+1,pbzn+1,zn+pbzn,zn+12=pbzn−1,zn·maxpbzn−1,zn,pbzn,zn+1.Thus, the inequality (64) becomes
(67)pbzn,zn+12≤κpbzn−1,zn·maxpbzn−1,zn,pbzn,zn+1.Since for the case maxpbzn−1,zn,pbzn,zn+1=pbzn,zn+1 we get pbzn,zn+12≤κpbzn−1,zn·pbzn,zn+1, or pbzn,zn+1≤κpbzn−1,zn<pbzn−1,zn, which is a contradiction, we conclude that maxpbzn−1,zn,pbzn,zn+1=pbzn−1,zn and then (67) becomes
(68)pbzn,zn+1≤κpbzn−1,zn,for any n∈ℕ. Therefore, by Lemma L2A and using similar arguments as in Theorems 8 and 12, there exists u∗∈M such that
(69)limn→∞pbSun,Ru∗=limn→∞pbRun,Ru∗=pbRu∗,Ru∗=0.Finally, we claim that Su∗=Ru∗. From the assumptions (c), there exists a subsequences unl of un such that βunl,u∗≥1. Thus, replacing u by unl and y by u∗, we get (as l⟶∞)
(70)limn→∞MS,R2unl,u∗=limn→∞pbSunl,Su∗pbSu∗,Ru∗−−a·minbpSunl,Ru∗,bpSu∗,Runl=pbSu∗,Ru∗·limn→∞pbSunl,Su∗,limn→∞NS,R2unl,u∗=limn→∞pbRunl,Ru∗·maxpbSunl,Runl,pbSu∗,Ru∗,pbSunl,Ru∗+pbSu∗,Runl2s=0.Consequently, (64) becomes pbSu∗,Ru∗·limn→∞pbSunl,Su∗=0 and the rest is just a verbatim repetition of the lines in the previous proofs.Corollary 17.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exist κ∈0,1 and a>0 such that
(71)MS,R2u,y≤κ·NS,R2u,y,where
(72)MS,R2u,y=pbSu,SypbSy,Ry−a·minbpSu,Ry,bpSy,Ru,NS,R2u,y=pbRu,Ry·maxpbSu,Ru,pbSy,Ry,pbSu,Ry+pbSy,Ru2s,for every u,y∈M, such that u≠y when u,y∈CcR,SM. If SM⊂RM and RM,pb,s is a 0-pb-complete pb-ms, then, the mappings S and R have a point of coincidence.Proof.
Letβu,y=1 in Theorem 16.Theorem 18.
LetM,pb,s≥1 be a complete pb-ms and a mapping S:M⟶M. Suppose that there exist κ∈0,1 and a>0 such that
(73)βu,yMS2u,y≤κ·NS2u,y,where
(74)MS2u,y=pbSu,SypbSy,y−a·minbpSu,y,bpSy,u,NS2u,y=pbu,y·maxpbSu,u,pbSy,y,pbSu,y+pbSy,u2s,for every u,y∈M. Suppose also that
(i)
S is β-admissible, and there exists u0∈M such that βRu0,Su0≥1(ii)
M is β-regular
Then, the mappingS admits a fixed point.Proof.
ChooseR=IM.Corollary 19.
LetM,pb,s≥1 be a complete pb-ms and two mappings R,S:M⟶M. Suppose that there exist κ∈0,1 and a>0 such that
(75)MS2u,y≤κ·NS2u,y,where
(76)MS2u,yMS2u,y=pbSu,SypbSy,y−a·minbpSu,y,bpSy,u,NS2u,y=pbu,y·maxpbSu,u,pbSy,y,pbSu,y+pbSy,u2s,for every u,y∈M. Then, the mapping S has a fixed point.



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*Source: 5538833-2021-03-29.xml*