# Coincidence Point Theorems for(α,β,γ)-Contraction Mappings in Generalized Metric Spaces

Authors: Chaiporn Thangthong; Anchalee Khemphet
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

The result of our study is that a coincidence point of two mappingsP and Q can be achieved when the ordered pair (P,Q) is an (α,β,γ)-contraction with respect to a generalized metric space. Moreover, with some additional condition, a common fixed point can be obtained as a consequence of our main theorems. Further, we apply our findings to some examples and integral equation problems.


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## Body

## 1. Introduction

There has been a wide range of research in discovering fixed points, or the only fixed point, of certain types of mappings that are contractions in the past. Many aspects have been used to accomplish the result. At the very beginning, Geraghty [1] generally developed the Banach contraction principle by considering the class Θ whose elements are functions θ:[0,∞)→[0,1) such that (1)θtn⟶1⟹tn→0.In 2012, Samet et al. [2] studied the existing results for α-ψ-contractions. His concept was given in the following definition. Suppose that X≠∅ and α is a real-valued function on X×X.Definition 1 (see [2]).
LetA be a self-mapping on X and u,v∈X. If α(Au,Av)≥1 whenever α(u,v)≥1, then we say that A is α-admissible.Later, Karapinar [3] added more conditions to Definition 1.Definition 2 (see [3]).
LetA be an α-admissible self-mapping on X and u,v,w∈X. If α(u,w)≥1 and α(w,v)≥1 imply α(u,v)≥1, then we say that A is triangular α-admissible.Furthermore, another essential part in this topic is a metric space. There were a large number of literatures that worked not only on a metric space, but also on other topological spaces; for examples, see [4–6]. Three years ago, Jleli and Samet [7] defined a generalized metric, known as a JS-metric. The advantage of their idea is that many topological spaces are covered by the JS-metric space. With this reason, results of fixed point theorems on JS-metric spaces have been recently interesting (e.g., see [8]).LetD:X×X→[0,∞] be a function and, for u∈X, denote a set of sequences {un} in X such that limn→∞D(un,u)=0 by C(D,X,u).Definition 3 (see [7]).
Suppose that, for anyu,v∈X, (D1)
ifD(u,v)=0, then u=v;(D2)
D(u,v)=D(v,u);(D3)
there is aK>0 so that, for each w,z∈X, if {un}∈C(D,X,w), then D(w,z)≤Klimsupn→∞D(un,z).
Then, we callD a generalized metric, or a JS-metric on X, and also (X,D) a generalized metric space, or a JS-metric space.Definition 4 (see [7]).
Suppose that(X,D) is a JS-metric space and {un}⊆X. (1)
If{un}∈C(D,X,u) for some u∈X, then we say that {un}D-converges to u.(2)
Iflimm,n→∞D(un,um)=0, then we say that {un} is D-Cauchy.(3)
If anyD-Cauchy sequence {un} in XD-converges to some u in X, then we say that (X,D) is D-complete.Proposition 5 (see [7]).
Suppose that(X,D) is a JS-metric space. Let {un}⊆X and u,v∈X. If {un}∈C(D,X,u) and {un}∈C(D,X,v), then u=v.Definition 6 (see [7]).
Suppose that(X,D) is a JS-metric space. Let P be a self-mapping on X and {un}⊆X. If {un}∈C(D,X,u0) implies {Pun}∈C(D,X,Pu0) for some u0∈X, then we say that P is continuous at u0. Moreover, if P is continuous at every u∈X, then we say P is continuous.Next, Martínez-Moreno et al. [9] had a new perspective to obtain common fixed points of particular contractive mappings on a space with two metrics.Inspired by the above, we consider some existence results for a coincidence point of two functions when the ordered pair of these functions is an(α,β,γ)-contraction on a space with two JS-metrics. In addition, some examples and an application of an integral equation are presented.


## 2. Main Results

First, we assume throughout this section that all functionsP and Q are self-mappings defined on X.Definition 7.
Foru,v,w∈X, if (1)
α(Qu,Qv)≥1 implies α(Pu,Pv)≥1,(2)
α(u,w)≥1 and α(w,v)≥1 imply α(u,v)≥1,
then we say thatP is triangular α-admissible with respect to Q.Next, we defineB as the class of mappings β:[0,∞]→[0,1) such that (2)βtn⟶1⟹tn→0.Note that β(t) is defined at t=∞. Also, let γ:[0,∞]→[0,∞] be a nondecreasing continuous function satisfying (3)γt=0⟺t=0.Denote the class of all such functions γ by Υ.Definition 8.
Suppose that(X,D) is a JS-metric space. If (1)
P is triangular α-admissible with respect to Q,(2)
there existβ∈B and γ∈Υ,(4)αQu,QvγDPu,Pv≤βDQu,QvγDQu,Qvfor allu,v∈X,
then we say that the pair(P,Q) is an (α,β,γ)-contraction with respect to D.Here, we are interested in the existence of a coincidence point ofP and Q, where (P,Q) is an (α,β,γ)-contraction with respect to some generalized metric on X. This can be done under suitable relations between P and Q.Definition 9.
Suppose that(X,D) and (Y,D′) are two JS-metric spaces and {un}⊆X. If A:X→Y and B:X→X are functions such that {Bun} being D-Cauchy in (X,D) implies {Aun} is D′-Cauchy in (Y,D′), then we say that A is B-Cauchy on X.Last but not least, we need the comparison notations for any two generalized metrics. IfD and D′ are two generalized metrics on X, the notation D≥D′ represents D(u,v)≥D′(u,v) for every u,v∈X. If the inequality fails for some u,v∈X, we use the notation D≱D′. All other inequality signs can be defined in the same fashion.Theorem 10.
Suppose that(X,D′) is a D′-complete JS-metric space and D is a generalized metric on X. If (i)
(P,Q) is an (α,β,γ)-contraction with respect to D,(ii)
P(X) is a subspace of Q(X),(iii)
there is au0∈X, α(Qu0,Pu0)≥1, and sup{D(Qu0,Pv):v∈X}<∞,(iv)
P:X,D→(X,D′) is Q-Cauchy on X whenever D≱D′,(v)
P and Q commute,(vi)
P,Q:(X,D′)→(X,D′) are continuous,
then there exists a coincidence point ofP and Q.Proof.
From assumption(iii), let u0∈X such that α(Qu0,Pu0)≥1 and sup{D(Qu0,Pv):v∈X}<∞. Since P(X)⊆Q(X) and Pu0∈X, a sequence {un} in X can be constructed such that (5)Qun=Pun-1for all n∈N. Observe that if Qun0=Qun0-1 for some n0∈N, then Qun0-1=Pun0-1, and so we are done. Assume that Qun≠Qun-1 for each n∈N.
Sinceα(Qu0,Qu1)=α(Qu0,Pu0)≥1 and P is triangular α-admissible with respect to Q, α(Qu1,Qu2)=α(Pu0,Pu1)≥1. Repeating this process inductively, we obtain that(6)αQun,Qun+1≥1for each n∈N.
Our task is now to prove that{Qun} is D-Cauchy.
Assume that this is not true. Equivalently, there is anϵ>0 so that, for each k∈N,(7)DQunk,Qumk≥ϵfor some mk≥nk≥k. By inequality (6), together with the assumption that P is triangular α-admissible with respect to Q, we have that (8)αQunk,Qumk≥1for all k∈N. Then, by assumption(i), (9)γDQunk+1,Qumk+1=γDPunk,Pumk≤αQunk,QumkγDPunk,Pumk≤βDQunk,QumkγDQunk,Qumk.Continuing to apply this concept totally nk+1 times, we finally get the equality (10)γDQunk+1,Qumk+1≤∏i=0nkβDQunk-i,Qumk-iγDQu0,Qumk-nk.Let 0≤ik≤nk such that (11)βDQunk-ik,Qumk-ik=max⁡βDQunk-i,Qumk-i:0≤i≤nk.Denote δ=limsupk→∞β(D(Qunk-ik,Qumk-ik)).
Notice that ifδ<1, then limk→∞γ(D(Qunk+1,Qumk+1))=0, and so (12)limk→∞DQunk+1,Qumk+1=0which contradicts inequality (7).
Thus,δ=1. Then, there is a subsequence of {β(D(Qunk-ik,Qumk-ik))} which converges to 1. Without loss of generality, assume that (13)limk→∞βDQunk-ik,Qumk-ik=1.By the definition of β, (14)limk→∞DQunk-ik,Qumk-ik=0.Therefore, there exists a k0∈N such that (15)DQunk0-ik0,Qumk0-ik0<ϵ2.It follows that, also from inequality (7), (16)γϵ≤γDQunk0,Qumk0≤∏j=1ik0βDQunk0-j,Qumk0-jγDQunk0-ik0,Qumk0-ik0<γDQunk0-ik0,Qumk0-ik0≤γϵ2.This is a contradiction since γ is nondecreasing. Thus, {Qun} is D-Cauchy.
The next goal is to show that{Qun} is also D′-Cauchy. It can be observed that if D≥D′, we are done. Assume that D≱D′. Since P is Q-Cauchy on X, {Pun} is D′-Cauchy. Consequently, (17)limn,m→∞D′Qun+1,Qum+1=limn,m→∞D′Pun,Pum=0,and so {Qun} is D′-Cauchy.
Since(X,D′) is D′-complete, one can find a c∈X satisfying (18)limn→∞D′Qun,c=limn→∞D′Pun,c=0.That is, (19)Qun,Pun∈CD′,X,c.By assumption(vi), (20)PQun∈CD′,X,PcandQPun∈CD′,X,Qc.Since P and Q commute, Pc=Qc. This completes the proof.As a consequence of Theorem10, if D=D′, then we have the following theorem. Besides, we can replace properties (v) and (vi) by other conditions as stated in the theorem below.Theorem 11.
Suppose that(X,D) is a D-complete JS-metric space. Assume that (i)
(P,Q) is an (α,β,γ)-contraction with respect to D;(ii)
P(X) is a subspace of Q(X);(iii)
there is au0∈X, α(Qu0,Pu0)≥1, and sup{D(Qu0,Pv):v∈X}<∞;(iv)
either (a) or (b) holds:
(a)
P and Q are continuous mappings that commute.(b)
Q(X) is D-complete and, for any {un}⊆X satisfying α(un,un+1)≥1 for each n∈N, if there is a c∈X such that {un}∈C(D,X,c), then α(un,c)≥1 for all n∈N.
Then, there exists a coincidence point ofP and Q.Proof.
It is easy to see that if condition(a) is true, then applying Theorem 10 to the case D=D′ yields the desired result. Assume that statement(a) does not hold. Thus,(b) must be valid. Let {un}⊆X and Qun=Pun-1 for each n∈N. An argument similar to the one used in Theorem 10 shows that {Qun} is D-Cauchy and α(Qun,Qun+1)≥1 for all n∈N. From(b), (21)limn→∞DQun,Qc=limn→∞DPun,Qc=0for some c∈X. That is, (22)Qun,Pun∈CD,X,Qc.Again, since condition(b) holds, for any n∈N, α(Qun,Qc)≥1. From (4) and assumption(i), it follows that (23)γDPun,Pc≤αQun,QcγDPun,Pc≤βDQun,QcγDQun,Qc<γDQun,Qc.Since γ is nondecreasing, (24)DPun,Pc)<DQun,Qc,and so (25)limn→∞DPun,Pc)=0.Consider (26)DQc,Pc≤Klimsupn→∞DPun,Pcfor some K>0. Therefore, D(Qc,Pc)=0. Hence, Qc=Pc, completing the proof.Adding some extra condition to Theorem10, the coincidence point is actually a common fixed point. This can be shown in the following theorem. Denote (27)CP,Q=u∈X:Pu=Qu.Theorem 12.
Suppose that(X,D′) is a D′-complete JS-metric space and D is a generalized metric on X. If all assumptions (i)-(vi) in Theorem 10 are satisfied and α(Qu,Qv)≥1 whenever Qu≠Qv, where u,v∈C(P,Q), then there exists a common fixed point of P and Q.Proof.
According to Theorem10, C(P,Q)≠∅. Then, we can let u,v∈X so that Pu=Qu and Pv=Qv.
Suppose thatQu≠Qv. By the assumption, α(Qu,Qv)≥1. From the fact that (P,Q) is an (α,β,γ)-contraction with respect to D, we have that (28)γDPu,Pv≤αQu,QvγDPu,Pv≤βDQu,QvγDQu,Qv<γDQu,Qv=γDPu,Pvwhich leads to a contradiction. Therefore, Qu=Qv.
Next, letc=Qu=Pu. Since P and Q commute, Qc=QPu=PQu=Pc. Thus, c∈C(P,Q). Referring to the proof above, we can conclude that Pc=Qc=Qu=c. Hence, the proof is complete.We give examples to illustrate Theorems10 and 11, respectively.Example 13.
Suppose thatX=[0,1]. Given the generalized metrics D and D′ on X defined by (29)Du,v=u+v,u≠0andv≠0,u2,v=0,v2,u=0,and (30)D′u,v=Lu+v,u≠0andv≠0,Lu2,v=0,Lv2,u=0,where u,v∈X and L is a real number such that L>1, we have that (X,D′) is D′-complete. Let α be a function defined by (31)αu,v=1,u,v∈0,14withu≠0orv=0,0,otherwise.Given the self-mappings P and Q on X defined by (32)Pu=u4andQu=u2,some tedious manipulation yields assumptions (ii), (v), and (vi) in Theorem 10. Further, notice that 1/2∈X such that α(Q(1/2),P(1/2))=α(1/4,1/16)≥1 and sup{D(Q(1/2),Pv):v∈X}<∞.
Claim 1. P is triangular α-admissible with respect to Q.
Letu,v,w∈X. Assume that α(Qu,Qv)≥1. Then, u2,v2∈[0,1/4] and Qu=u2≠0 or Qv=v2=0. Accordingly, u,v∈[0,1/2], and u≠0 or v=0. It follows that u4,v4∈[0,1/16], and Pu=u4≠0 or Pv=v4=0. Therefore, α(Pu,Pv)≥1.
Next, assume thatα(u,w)≥1 and α(w,v)≥1. It can be observed that if w=0, then v=0, and if w≠0, then u≠0. That is, u≠0 or v=0. Therefore, α(u,v)≥1. Thus, we have Claim 1.
Claim 2. ( P , Q ) is an (α,β,γ)-contraction with respect to D, where β∈B and γ∈Υ are as follows: (33)γt=t2andβt=14,0≤t<1,1t2+2,t≥1for t∈[0,∞].
Givenu,v∈X, if α(Qu,Qv)=0, inequality (4) holds. Assume that α(Qu,Qv)≥1. Similar as above, u,v∈[0,1/2]. Consider the following cases.
Case 1. Q v = 0. We have that (34)αQu,QvγDPu,Pv=Pu/22=u44≤14u24=βDQu,QvγDQu,Qv.
Case 2. Q v ≠ 0. Then, Qu≠0. Consider (35)αQu,QvγDPu,Pv=Pu+Pv2=u4+v42≤14u2+v22=βDQu,QvγDQu,Qv.Therefore, (P,Q) is a (α,β,γ)-contraction with respect to D.
Claim 3. P : ( X , D ) → ( X , D ′ ) is Q-Cauchy on X.
Notice thatD≤D′. Suppose that {un}⊆X such that {Qun} is D-Cauchy. Let ϵ>0. There is a k∈N so that, for all m,n≥k, we have that (36)un2+um2=DQun,Qum<ϵL.With this k, (37)D′Pun,Pum≤Lun4+um4≤Lun2+um2<ϵfor any m,n≥k. This completes Claim 3.
Thus, by Theorem10, P and Q have a coincidence point, precisely, 0.Example 14.
LetX=[0,1]. Therefore, (X,D) is D-complete, where D is the generalized metric as defined in Example 13. Suppose that α is a function as follows: (38)αu,v=1,u≠0orv=0,0,otherwise.Define self-mappings P and Q on X by (39)Pu=uu+6andQu=u3.Note that P(X)⊆Q(X) and Q(X) is D-complete. Moreover, we have 1∈X such that α(Q1,P1)≥1 and sup{D(Q1,Pv):v∈X}<∞.
Claim 1. P is triangular α-admissible with respect to Q.
Letu,v,w∈X. Assume that α(Qu,Qv)≥1. Then, Qu≠0 or Qu=0. That is, u≠0 or v=0. Thus, Pu≠0 or Pv=0. Therefore, α(Pu,Pv)≥1.
Similar as in the proof of the previous example, ifα(u,w)≥1 and α(w,v)≥1, then α(u,v)≥1. Therefore, Claim 1 is obtained.
Claim 2. ( P , Q ) is an (α,β,γ)-contraction with respect to D, where β∈B and γ∈Υ defined by β(t)=1/2 and γ(t)=t/2 for t∈[0,∞], respectively.
Supposeu,v∈X. If α(Qu,Qv)=0, then inequality (4) holds. Assume that α(Qu,Qv)≥1. Consider the following cases.
Case 1. Q v = 0. We have that (40)αQu,QvγDPu,Pv=Pu/22=12u2u+6≤12u6=βDQu,QvγDQu,Qv.Case 2.Qv≠0. Then, Qu≠0. Consider (41)αQu,QvγDPu,Pv=Pu+Pv2=12uu+6+vv+6≤12u6+v6=βDQu,QvγDQu,Qv.Therefore, we have Claim 2.
Next, let{un}⊆X such that, for any n∈N, α(un,un+1)≥1. Assume that {un}∈C(D,X,c) for some c∈X. By the definition of α, for each n∈N, un≠0 or un+1=0. Fix n∈N. If un≠0, then α(un,c)≥1. Assume that un=0. Suppose that c≠0. Then, D(un,c)=D(0,c)=c/2≠0. This is a contradiction since {un}∈C(D,X,c). Thus, c=0. Therefore, we get that α(un,c)≥1. Since n is arbitrary, this is true for every n∈N. Hence, by Theorem 11, c=0 is a coincidence point of P and Q.


## 3. Application

We wish to apply our finding to the existence problem of a solution to the integral equation. This is one of the crucial uses of fixed point theorems that can be found in the literatures (see [10–13]).(42)ut=∫0Tpt,s,usds+btfor t∈[0,T], where T is a real number such that T>0.Suppose thatX=C([0,T],R) and (43)Du,v=maxt∈0,Tut+maxt∈0,Tvtfor u,v∈C([0,T],R). We have that (X,D) is a D-complete JS-metric space. The following theorem shows when the equation (42) has a solution if the integral equation is homogeneous.Theorem 15.
Consider equation (42). Suppose that (i)
p:[0,T]×[0,T]×R→R is continuous;(ii)
for anyu,v∈R, if u≤v, then p(t,s,u)≤p(t,s,v) and (44)pt,s,u+pt,s,v≤ln⁡1+u+vT,wheres,t∈[0,T];(iii)
there is au0∈X, u0(t)≤∫0Tp(t,s,u0(s))ds, where t∈[0,T];(iv)
sup{maxt∈0,T∫0Tpt,s,vsds:v∈X}<∞.
Then, the integral equation (42) has a solution.Proof.
Define the self-mappingsP and Q on X as follows: (45)Put=∫0Tpt,s,usds,and Qu(t)=u(t) for u∈X and t∈[0,T]. Suppose that α:X×X→R is a function defined by (46)αu,v=1,ut≤vtfor  anyt∈0,T,0,otherwise.Clearly, P(X)⊆Q(X), and P and Q are continuous mappings that commute. Moreover, by assumptions (iii) and (iv), it is straightforward to show that condition (iii) of Theorem 11 is satisfied. Our problem reduces to show that the pair (P,Q) is an (α,β,γ)-contraction with respect to D for some β∈B and γ∈Υ.
First, we establish thatP is triangular-α-admissible with respect to Q.
Assume thatα(Qu,Qv)≥1. Then, Qu(t)≤Qv(t); that is, u(t)≤v(t) for any t∈[0,T]. By assumption (ii), p(t,s,u)≤p(t,s,v). Therefore, (47)Put=∫0Tpt,s,usds≤∫0Tpt,s,vsds=Pvt.That is, α(Pu,Pv)≥1. It is simple to show that the second condition of Definition 7 holds. Thus, P is triangular-α-admissible with respect to Q.
Finally, it remains to prove inequality (4).
Ifu(t)>v(t) for some t∈[0,T], we are done. Suppose that u(t)≤v(t) for all t∈[0,T]. From assumption (ii), consider (48)Put+Pvt≤∫0Tpt,s,us+pt,s,vsds≤1T∫0Tln⁡(1+us+vs)ds≤ln⁡1+maxt∈0,TQut+maxt∈0,TQvt,and, then, (49)lnPut+Pvt+1≤ln∫0Tpt,s,us+pt,s,vsds+1≤ln1T∫0Tln⁡(1+us+vs)ds+1≤ln⁡ln⁡1+maxt∈0,TQut+maxt∈0,TQvt+1=ln⁡ln⁡1+maxt∈0,T⁡Qut+maxt∈0,T⁡Qvt+1ln1+maxt∈0,T⁡Qut+maxt∈0,T⁡Qvtln⁡1+maxt∈0,TQut+Qvt.This gives us the desired inequality for γ(t)=ln(t+1) and (50)βt=ln⁡ln⁡1+t+1ln⁡1+t,t>0,k,t=0for some k∈[0,1), where t∈[0,∞). Thus, (P,Q) is an (α,β,γ)-contraction with respect to D.
Hence, there exists a coincidence point ofP and Q which is a solution to the integral equation (42).



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