# Coincidence Point Results on Relation TheoreticFw,Rg-Contractions and Applications

Authors: Muhammad Aslam; Hassen Aydi; Samina Batul; Amna Naz
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Motivated by the ideas ofF-weak contractions and F,Rg-contractions, the notion of Fw,Rg-contractions is introduced and studied in the present paper. The idea is to establish some interesting results for the existence and uniqueness of a coincidence point for these contractions. Further, using an additional condition of weakly compatible mappings, a common fixed-point theorem and a fixed-point result are proved for Fw,Rg-contractions in metric spaces equipped with a transitive binary relation. The results are elaborated by illustrative examples. Some consequences of these results are also deduced in ordered metric spaces and metric spaces endowed with graph. Finally, as an application, the existence of the solution of certain Voltera type integral equations is investigated.


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## Body

## 1. Introduction and Preliminaries

In the development of the metric fixed-point theory, one of the main pillar is the Banach contraction principle [1], which states that every contraction on a complete metric space has a unique fixed point. Due to its extensive application potential, this concept has been observed in various forms over the years (see [2–9]).The concept ofF-contractions was introduced by Wardowski [10]. He proved some new fixed-point results for such kind of contractions. He built these results in a different way rather than traditional ways as done by many authors. Later on, fixed points for F-contractions were proved by Secelean [11] using an iterated function. Abbas et al. [12] extended the work of Wardowski and established various results of fixed points using F-contraction mappings. For further related works on F-contractions, see [13–16].The idea ofF,R-contractions was established by Sawangsup et al. [17]. They used this idea to demonstrate some fixed-point consequences using a binary relation. It is further investigated by Imdad et al. [18]. In present paper, we study the results presented by Alfaqih et al. [19] and we define Fw,Rg-contractions. We also prove similar results for Fw,Rg-contractions.Recall that a binary relationR on nonempty set X is said to be a partial order if it is reflexive, antisymmetric, and transitive. Moreover, the inverse or transpose or dual relation of R, denoted by R−1, is defined by
(1)R−1=x,y∈X2:y,x∈R.The symmetric closure ofR, denoted by Rs, is defined as the set R∪R−1, that is, Rs≔R∪R−1. In fact, Rs is the smallest symmetric relation on X containing R.Notice that there is another binary relationR∦⊆R on X, which is defined by kR∦ℓ, whenever kRℓ and k≠ℓ.Definition 1 [10].
LetF be the set of functions F:0,∞⟶ℝ such that
(F1) F is strictly increasing;
(F2) For every sequence βn⊂0,∞, limn⟶∞βn=0ifflimn⟶∞Fβn=−∞;
(F3) There is k∈0,1 so that limβ⟶0+βkFβ=0.
The following functions are inF:
(2)Fβ=lnβ,Fβ=lnβ+β,Fβ=lnβ2+β,Fβ=−1β.Many papers in literature deal with the concept ofF-contractions (see [20–22]). Throughout this work, the set of all continuous functions verifying F2 is denoted by F.Definition 2.
LetX≠ϕ and R be a binary relation on X. A sequence ςn⊆X is such that ςnRςn+1foralln∈ℕ0, then it is called an R preserving sequence.Definition 3.
Consider a metric spaceX,d with a binary relation R. Then, X is called R complete if each R preserving Cauchy sequence is convergent in X.Definition 4 [23].
LetX,d be a metric space and R be a binary relation on X, T:X⟶X and x∈X. We say that T is R-continuous at x if for each R-preserving sequence ςn⊆X so that ςn⟶x, we have Tςn⟶Tx. Also, T is named to be R-continuous if it is R-continuous at any element of X.Definition 5 [23].
LetX,d be a metric space and R be a binary relation on X and T,g:X⟶X and x∈X. We say that T is g,R-continuous at x if for each sequence ςn⊆X so that gςn is R-preserving and gςn⟶gx, we have Tςn⟶Tx. Also, T is named to be g,R-continuous if it is g,R-continuous at any element of X.Definition 6 [24].
Forx,y∈X, a path of length pp∈ℕ in R from x to y is a finite sequence u0,u1,⋯⋯,up⊆X such that u0=x,up=y, and ui,ui+1∈R for every i∈0,1,..⋯,p−1. Also, a subset L⊆X is called R connected if for any two elements x,y∈L, there is a path from x to y in R.Definition 7 [23].
LetX,d be a metric space and R be a binary relation on X and T,g:X⟶X. The pair T,g is R-compatible if for each sequence ςn⊆X so that Tςn and gςn are R-preserving and limn⟶∞gςn=limn⟶∞Tςn=x∈X,
(3)limn⟶∞dgTςn,Tgςn=0.Definition 8.
Letf and g be self-maps of a set X. If x=fx=gx for some x∈X, then x is said to a common fixed point of f and g.Definition 9 [25].
Letf,g:X⟶X. If w=fx=gx for some x∈X, then x is said to be a coincidence point of f and g, and w is said to be a point of coincidence of f and g.
f and g are said to be weakly compatible if they commute at their coincidence point, i.e., if fx=gx for some x∈X, then fgx=gfx.Definition 10 [26].
LetM,d be a metric space endowed with a binary relation R. Such a R is named to be d-self closed if for each R-preserving sequence ςn⊆M so that ςn⟶x, there is ςnk of ςn so that ςnk,x∈R∀k∈ℕ0.Definition 11 [23].
LetM be a nonempty set and T,g:M⟶M. A binary relation R on M is called T,g closed if for any x,y∈M, gxRgy yields that TxRTy.Lemma 12 [27, 28].
Consider a metric spaceX,d and a sequence km in X. If km is not Cauchy in X, then are ε>0 and kmj and ktj of km so that
(4)j≤mj≤tj,dkmj,ktj−1≤ε<dkmj,ktj∀j∈N0.
Moreover, ifkm is so that limm⟶∞dkm,km+1=0, then
(5)limj⟶∞dkmj,ktj=limj⟶∞dkmj−1,ktj−1=ε.Lemma 13 [29].
LetX be a nonempty set and g:X⟶X. Then, there is a subset E⊆X so that gE=gX and g:E⟶E is one to one.


## 2. Main Results

We begin this section by introducing the idea ofFw,Rg-contractions as follows.Definition 14.
Consider a metric spaceX,d endowed with a transitive binary relation R on X and Q,g:X⟶X. Then, T is called an Fw,Rg-contractions if there exist F∈F and τ>0 such that
(6)τ+FdQk,Qℓ≤Fmaxdgk,gℓ,dgk,Qk,dgℓ,Qℓ,dgk,Qℓ+dgℓ,Qk2,for all k,l∈X with gkR∦gℓ and QkR∦Qℓ.Remark 15.
EveryF,Rg contraction is an Fw,Rg contraction, but the converse of statement is not true.
The following result is easy to prove. We omit it.Proposition 16.
LetX,d be a metric space endowed with a transitive binary relation R. Given Q,g:X⟶X. Then, for each F∈F, we have equivalence of the two following statements:
(a)
∀k,l∈X so that gk,gℓ∈R∦ and Qk,Qℓ∈R∦(7)τ+FdQk,Qℓ≤Fmaxdgk,gℓ,dgk,Qk,gℓ,Qℓ,dgk,Qℓ+dgℓ,Qk2.(b)
∀k,l∈X such that either gk,gℓ, Qk,Qℓ∈R∦ or gℓ,gk, Qℓ,Qk∈R∦(8)τ+FdQk,Qℓ≤Fmaxdgk,gℓ,dgk,Qk,gℓ,Qℓ,dgk,Qℓ+dgℓ,Qk2.Theorem 17.
Consider a metric spaceX,d equipped with R (where R is a transitive binary relation) and Q,g:X⟶X. Assume that:
(1)
there existsk0∈X such that gk0RQk0(2)
R is Q,g-closed(3)
Q is an Fw,Rg-contraction(4)
(a)
A subsetK of X exists such that QX⊆K⊆gX and K is R-complete(b)
One of the subsequent conditions is fulfilled:(i)
Q is g,R-continuous, or(ii)
Q and g are continuous, or(iii)
R∣K is d-self closed in condition that (6) holds for all k,l∈X with gkRgℓ and QkR∦Qℓor on the other hand:
(α) (α1) ∃ a subset L of X such that QX⊆gX⊆L and L is R- complete,
(α2) Q,g is an R-compatible pair,
(α3) Q and g are R-continuous.
Then, the pairQ,g admits a coincidence point.Proof.
In the above two cases (11) and α, note that QX⊆gX. Using assumption (6), we get gk0RQk0. If Qk0=gk0, then a coincidence point of Q,g is k0. This completes the proof. Suppose that Qk0≠gk0. Since QX⊆gX, there must exist k1∈X such that gk1=Qk0. Similarly, there is k2∈X such that gk2=Qk1. Proceeding in this way, we can construct a sequence km⊆X such that
(9)gkm+1=Qkm∀m∈ℕ0.
Now, we will provegkm is an R-preserving sequence, that is,
(10)gkmRgkm+1∀m∈ℕ0.
By using induction, we will prove this claim. If we putm=0 in (9) and use condition (6), we get gkoRgk1. This implies that the above statement holds for m=0. Suppose that (10) is accurate for m=j≥1, that is, gkjRgkj+1. Since R is Q,g-closed, we get QkjRQkj+1, and so gkj+1Rgkj+2.
Hence, our claim is true for allm∈ℕ0. By using (9) and (10), we can conclude that Qkm is also an R-preserving sequence, that is,
(11)QkmRQkm+1∀m∈ℕ0.
IfQkm0=Qkm0+1 for some m0∈ℕ0, then km0 is a coincidence point of Q,g.
Suppose on the contrary thatQkm≠Qkm+1 for al m∈ℕ0. With the help of (9), (10), (11), and condition (10), we can see that
(12)τ+FdQkm−1,Qkm≤Fmaxdgkm−1,gkm,dgkm−1,Qkm−1,dgkm,Qkm,dgkm−1,Qkm+dgkm,Qkm−12∀m∈ℕ0.=Fmaxdgkm−1,Qkm−1,dgkm,Qkm,dgkm−1,Qkm2.≤Fmaxdgkm−1,Qkm−1,dgkm,Qkm,dgkm−1,gkm+dgkm,Qkm2.=Fmaxdgkm−1,gkm,dgkm,Qkm.
Now,maxdgkm−1,gkm,dgkm,Qkm cannot be dgkm,Qkm. Otherwise,
(13)τ+Fdgkm,gkm+1≤dgkm,gkm+1,which is a contradiction. Hence, maxdgkm−1,gkm,dgkm,Qkm=dgkm−1,gkm. Therefore,
(14)τ+FdQkm−1,Qkm≤Fdgkm−1,gkm⇒FdQkm−1,Qkm≤Fdgkm−1,gkm−τ.
Takeγm=dgkm,gkm+1. With the help of above condition, we obtain
(15)Fγm≤Fγm−1−τ≤Fγm−2−2τ..⋯≤Fγ0−mτ∀m∈ℕ.
By usingF2 and taking m⟶∞ in above inequality, we obtain
(16)limm⟶∞Fγm=−∞.
This together withF2 imply that
(17)limm⟶∞γm=dgkm,gkm+1=0.
Now, we will show thatgkm is a Cauchy sequence. We argue by contradiction. In this case, Lemma 12 guarantees the existence of ε>0 and two subsequences gkmj and gktj of gkm such that
(18)dgkmj,gktj−1≤ε<dgkmj,gktj,with
(19)j≤mj≤tj,∀j∈ℕ0,(20)limj⟶∞dgkmj,gktj=dgkmj−1,gktj−1=ε.
This implies that there isj0∈ℕ0 so that dgkmj−1,gktj−1>0∀j≥j0.
SinceR is transitive, one writes
(21)gkmj−1R∦gktj−1andQkmj−1R∦Qktj−1∀j≥j0.
Using condition (10), we have for all j≥j0,
(22)τ+FdQkmj−1,Qktj−1≤Fmaxdgkmj−1,gktj−1,dgkmj−1,Qkmj−1,dgktj−1,Qktj−1,dgkmj−1,Qktj−1+dgktj−1,Qkmj−12.
Denote(23)maxdgkmj−1,gktj−1,dgkmj−1,Qkmj−1,dgktj−1,Qktj−1,dgkmj−1,Qktj−1+dgktj−1,Qkmj−12=Dkmj−1,ktj−1.
IfDkmj−1,ktj−1=dgkmj−1,gktj−1 or it is equal to dgkmj−1,Qktj−1+dgktj−1,Qkmj−1/2 then taking j⟶∞ and using (20), we get
(24)limj⟶∞Dkmj−1,ktj−1=ε.
SinceF is continuous, letting m⟶∞ in (22) and using (20) and (24), we get
(25)τ+Fε≤Fε,which is a contradiction. On the other hand, if Dkmj−1,ktj−1=dgkmj−1,Qkmj−1 or it is equal to dgktj−1,Qktj−1 then letting m⟶∞ in (22), using continuity of F and (20) together with condition F2, we get τ+Fε≤−∞, which is again a contradiction. Thus, gkm is a Cauchy sequence.
Let the condition (11) hold. With the help of (9), we obtain gkm⊆QX. Therefore, gkm is R-preserving Cauchy in K. By utilizing R-completeness of K, there is l∈K so that gkm⟶l. As K⊆gX, there is v∈X so that l=gv. Hence, by using (2),
(26)limm⟶∞gkm=limm⟶∞Qkm=gv.
In order to prove thatv is coincidence point of Q,g, we will use three different cases of condition b. First of all, suppose that Q is g,R-continuous. By utilizing (10) and (26), we get
(27)limm⟶∞Qkm=Qv.
By utilizing (26) and (27), we get Qv=gv. This shows that v is a coincidence point of Q,g.
Now, suppose the second case ofb, that is, Q and g are continuous. Since X≠ϕ and g:X⟶X, by using Lemma 13, there is B⊆X so that gB=gX and g:B⟶B is one-one. Define a mapping f:gB⟶gX by
(28)fgb=Qb∀gb∈gBwhereb∈B.
Recall thatg is one-one and QX⊆gX, so f is well-defined mapping. As Q and g are continuous, f is also continuous. Now, utilizing the fact that gX=gB, we can rewrite condition a as QX⊆K⊆gB, so that, without loss of generality, we can select a sequence km in B and v∈B. By using (26), (28), and continuity of f, we have
(29)Qv=fgv=flimm⟶∞gkm=limm⟶∞fgkm=limm⟶∞Qkm=gv.
Finally, assume that condition (iii) of (b) holds, which implies thatR∣K is d-self closed and (2.1) detain ∀k,l∈X, with gkRgℓ and QkR∦Qℓ. As gkm⊆K, gkm is R∣K preserving due to (10) and with the help of (26) gkm⟶gv. So, there is a subsequence gkmj⊆gkm such that
(30)gkmj,gv∈R∣K⊆R∀j∈ℕ0.
Utilizing condition (b) and (30), one writes
(31)Qkmj,Qv∈R∣K⊆R∀j∈ℕ0.
Now, letq=j∈ℕ:Qkmj=Qv. If the set q is infinite, then Qkmj has a subsequence Qkmjp, such that Qkmjp=Qv. This implies that limp⟶∞Qkmjp=Qv∀p∈ℕ. By using (26), we have limm⟶∞Qkm=gv. So we obtain Qv=gv.
If the setq is finite, then Qkmj has a subsequence Qkmjp such that Qkmjp≠Qv∀p∈ℕ. Next, we will show that limp⟶∞Qkmjp=Qv. With the help of (30), (31) and kmjp≠Qv∀p∈ℕ, we have
(32)gkmjp,gv∈R∣K⊆R∀p∈ℕ0,(33)Qkmjp,Qv∈R∣K⊆RandQkmjp≠Qv∀p∈ℕ0.
Now, with the help of (32), (33), Proposition 16 and the fact that (2.1) is satisfied, we get
(34)FdQkmjp,Qv≤Fmaxdgkmjp,gv,dgkmjp,Qkmjp,dgv,Qv,dgkmjp,Qv+dgv,Qkmjp2−τ.
Denote(35)maxdgkmjp,gv,dgkmjp,Qkmjp,dgv,Qv,dgkmjp,Qv+dgv,Qkmjp2=Dkmjp,v.
IfDkmjp,v=dgkmjp,gvthen, we have
(36)τ+FdQkmjp,Qv≤Fdgkmjp,gv⇒FdQkmjp,Qv≤Fdgkmjp,gv−τ.
By using (26), (F2) and taking p⟶∞, we get limp⟶∞Qkmjp=Qv. If Dkmjp,v=dgkmjp,gv then, we have
(37)τ+FdQkmjp,Qv≤Fdgkmjp,gv⇒FdQkmjp,Qv≤Fdgkmjp,gv+Ldgkmjp,gkmjp−1−τ.
By using (26), (F2) and taking p⟶∞, we get limp⟶∞Qkmjp=Qv. Now, if Dkmjp,v=dgv,Qv, then
(38)FdQkmjp,Qv≤Fdgkmjp,Qkmjp−τ.
By using (26), (F2) and taking p⟶∞, we get limp⟶∞Qkmjp=Qv. If Dkmjp,v=dgkmjp,Qv+dgv,Qkmjp/2, then, we have
(39)τ+FdQkmjp,Qv≤Fdgkmjp,Qv+dgv,Qkmjp2⇒limp⟶∞FdQkmjp,Qv≤limp⟶∞Fdgkmjp,Qkmjp+dgkmjp,Qkmjp2−τ=limp⟶∞Fdgkmjp,gkmjp−1+dgkmjp,gkmjp−12−τ.
By using (26), (F2) and taking p⟶∞, we get
(40)limp⟶∞Qkmjp=Qv.
From (26) and (40), we obtain Qv=gv. Hence, when the set q is finite or infinite, v is a coincidence point of Q and g. Now, if α holds, then gkm⊆L, and hence gkm is an R-preserving Cauchy sequence in L. Since L is R-complete, there is u∈L so that
(41)limm⟶∞gkm=u.
Using Equations (9) and (41), one gets
(42)limm⟶∞Qkm=u.
Now, with the help of (10), (41), and continuity of g, we have
(43)limm⟶∞ggkm=glimm⟶∞gkm=gu.
Utilizing (11), (42) and continuity of g to find
(44)limm⟶∞gQkm=glimm⟶∞Qkm=gu.
AsQkm and gkm are R-preserving due to (10), (11) and
(45)limm⟶∞Qkm=limm⟶∞gkm=u.
Now, using (41), (42), and condition (α2),
(46)limm⟶∞dgQkm,Qgkm=0.
Next, we will demonstrate thatu is a coincidence point of Q,g. Making use of (10), (41) and the R-continuity of Q, we get
(47)limm⟶∞Qgkm=Qlimm⟶∞gkm=Qu.
With the use of (44), (46), and (47), we get
(48)dgu,Qu=dlimm⟶∞gQkm,limm⟶∞Qgkm=limm⟶∞dgQkm,Qgkm=0⇒Qu=gu.
This implies thatu is a coincidence point of Q,g.☐Theorem17 does not guarantee the uniqueness of a coincidence point. The following theorem guarantees that coincidence point is unique.Theorem 18.
Suppose all hypothesis of Theorem17 are true except α and assume that gu and gv are R-comparable for all u≠v∈ coin Q,g, and one of Q or g is one-one, then there is a unique coincidence point of Q,g.Proof.
The setcoinQ,g is nonempty, because of Theorem 17. Consider two elements u,v∈coinQ,g, then by definition of coinQ,g, we have gv,gu∈R and Qu=gu, Qv=gv. This implies Qu,Qv∈R.
Now, ifgu=gv, we obtain Qv=gv=gu=Qu, and hence, v=u, because one of Q and g is one-one.
Ifgu≠gv, then by utilizing condition (10) and Proposition 16, we get
(49)τ+FdQu,Qv≤Fdgu,gv,dgu,Qv,dgv,Qv,dgu,Qv+dgv,Qv2=FdQu,Qv.
Sinceτ>0, our assumption is false. Therefore, a unique coincidence point of Q,g exists.☐Theorem 19.
Consider above theorem and add a condition thatQ,g is a weakly compatible pair, then a unique common fixed point of Q,g exists.Proof.
Above theorem assures that the pairQ,g has a unique coincidence point. Let v be the common coincidence point and suppose z∈X be such that
(50)z=Qv=gv.
The weak compatibility ofQ and g leads to Qz=Qgv=gQv=gz. That is, z is a coincidence point of Q and g. Since v is unique, one writes z=v. That is, the uniqueness of a common fixed point. Since all the assumptions of Theorem 18 are true, the set coinQ,g is nonempty.☐Example 1.
LetX=0,∞ and define d:X×X⟶ℝbydx,y=x−y. Then, X,d is a complete metric space.
Consider the sequenceςn⊆X which is defined by ςn=nn+14n−1/3,n≥1.
Define the binary relationR on X by
(51)R=ςi,ςi,ςi,ςi+1suchthatςi≤ςi+1.DefineQ,g:X⟶X by
(52)Qx=x,if0≤x≤ς1,ς2,ifς1<x≤ς2,ςi+ςi+1−ςiςi+2−ςi+1x−ςi+1,ifςi+1≤x≤ςi+2,and
(53)gx=ςi+ςi+1−ςiςi+2−ςi+1x−ςi,ifςi≤x≤ςi+1,i=1,2,⋯.Observe that ifgxR∦gy and QxR∦Qy, then x=ςi and y=ςi+1 for i∈ℕ−1. Further, by choosing Fα=lnα and α∈0,+∞, we have
(54)FdQςi,Qςi+1=Fςi−1−ςi=Fςi−ςi−1=lnςi−ςi−1,Fmaxdgςi,gςi+1,dgςi,Qςi,dgςi+1,Qςi+1,dgςi,Qςi+1+dQςi,gςi+12=Fmaxςi+1−ςi,ςi−ςi−1,ςi+1−ςi,ςi+1−ςi−12=Fςi+1−ςi=lndgςi+1,Qxi+1.Now, forn=2,3,⋯ and for τ=ln3, we have
(55)τ+lnςi−ςi−1≤ςi+1−ςi.Therefore,(56)ln3+FdQςi,Qςi+1≤Fdgςi+1,Qxi+1,∀x,y∈XsuchthatgxR∦gyandQxR∦Qy.Moreover, all the assumptions of Theorem19 are true, and ς1 is the unique common fixed point of Q,g.On settingg=I in Theorem 19, we obtain the following result.Theorem 20.
Consider a self-mappingQ:X⟶X and let X,d be a metric space with a transitive binary relation R. Assume that:
(1)
∃k0∈X such that k0RQk0(2)
R is Q-closed(3)
Q is an Fw,R-contraction(4)
(α) ∃ a subset K of X such that QX⊆K and K is R-complete,
(η) one of these conditions hold:
(i)
Q is R-continuous, or(ii)
R∣K is d-self closed on condition that (1.1) with binary relation holds ∀k,l∈X with kRl and QkR∦Qℓ
Then, a fixed point ofQ exists. Furthermore, if
(e) u,v∈FixQ⇒u,v∈R.
Then, such fixed point ofQ is unique.Theorem 21.
Replace condition (e) of above theorem by:
(e∗) FixQ is Rs-connected,
thenQ has a unique fixed point.Proof.
Assume on contrary thatQ has more than one fixed point, say u and v with u≠v. Then, there exists a path Rs⊆FixQ. As it is from v to u of length q, let us denote the path by v0,⋯vq such that vp≠vp+1 for each p where 0≤p≤j−1. If v=u, it is a contradiction. Hence,
(57)v0=v,vq=uandvp,vp+1∈Rforeachp0≤p≤q−1.
Asvp∈FixQ, so Qvp=vp for each p∈0,1,.⋯,q. With the help of condition c, we obtain
(58)τ+Fdvp,vp+1≤Fmaxdvp,vp+1,dvp,vp+1,dvp+1,vp+1,dvp,vp+1+dvp+1,vp2.
That is,(59)τ+Fdvp,vp+1≤Fdvp,vp+1.
Sinceτ>0, our supposition is not true. Hence, Q has a unique fixed point.☐In the next section, we are presenting a significance of our results in ordered metric spaces.


## 3. Some Consequences in Ordered Metric Spaces

Definition 22.
LetX,d be a metric space and X,≼ be an ordered set, then the triplet X,d,≼ is known as an ordered metric space.Definition 23.
Consider self-mappingsQ,g:X⟶X and an ordered set X,≼. If, for any k,l∈X, gk≼gℓ implies that Qk≼Qℓ. Then, Q is g-increasing.Remark 24.
Notice that the notion ofQ is g-increasing is equal to say that ≼ is Q,g-closed.TakingR=≼ in Theorem 17 to 19 and with the help of Remark 24, we state the following result.Corollary 25.
Consider self-mappingsQ,g:X⟶X and an ordered metric space X,d,≼. Assume that:
(a)
∃k0∈X such that gk0≼Qk0(b)
Qisg-increasing(c)
There areτ>0 and F∈F so that
(60)τ+FdQk,Qℓ≤Fmaxdgk,gℓ,dgk,Qk,dgℓ,Qℓ,dgk,Qℓ+dgℓ,Qk2,(d)
∃ a subset K of X such that QX⊆K⊆gX and K is ≼ -complete(e)
EitherQ and g are continuous, or Q is g,≼-continuous. Then, a coincidence point of Q,g exists. Additionally, we suppose that(f)
Qu and gv are ≼-comparable for all distinct coincidence points u,v∈coinQ,g, then pair Q,g has a unique coincidence point
Furthermore, ifQ and g are weakly compatible, then Q,g has a unique common fixed point.
TakingR=≼ in Theorem 20 and with the help of Remark 24, we conclude the result given below.Corollary 26.
Consider an ordered metric spaceX,d,≼ and mapping Q:X⟶X. Suppose the that conditions given below are fulfilled:
(a)
∃k0∈X such that k0≼Qk0(b)
Q is ≼-increasing(c)
∃τ>0 and F∈F such that
(61)τ+FdQk,Qℓ≤Fmaxdk,l,dk,Qk,l,Qℓ,dk,Qℓ+dl,Qk2,(d)
A subsetK of X exists such that QX⊆K and K is ≼-complete(e)
Q is ≼-continuous. Then a fixed point of Q exists. Furthermore,(f)
if for any two fixed pointsu,v∈Q we have u,v∈≼, then Q has a unique fixed point


## 4. Applications to Metric Spaces Endowed with a Graph

Jachymski [30] in 2008 has instituted the idea of metric spaces endowed with a graph in order to generalize the idea of a partial ordering and specified the Banach contraction principle in metric spaces and partially ordered metric spaces. In this section, we are going to present an application of our results in the situating of complete metric spaces endowed with a graph.Corollary 27.
Consider self-mappingsQ,g:X⟶X on a metric space X,d endowed with a graph G=VG,EG. Define ≼ on X as u≼v if and only if there is an edge between u and v. Assume that all the conditions given in Corollary 25 are satisfied. Then a coincidence point of Q,g exists. Further, if we suppose that Qu and gv are comparable on edges for all distinct coincidence points u,v∈coinQ,g, then the pair Q,g has a unique coincidence point.Furthermore, a unique common fixed point ofQ,g exists if Q and g are weakly compatible.Corollary 28.
Consider a metric spaceX,d endowed with a graph G and a mapping Q:X⟶X. Define ≼ on X as u≼v if and only if there is an edge between u and v. Suppose that conditions given in Corollary 27 are fulfilled. Then, a fixed point of Q exists. Furthermore, if u,v∈FixQ are such that there is an edge between u and v, then a unique fixed point of Q exists.


## 5. Applications to Integral Equations

In this section, we present an application of Theorem21 by finding a solution of the integral equation of Volterra type given below:
(62)ut=∫0tKt,s,usds+ht,t∈0,1.Here,K:0,1×0,1×ℝ⟶ℝ and h:0,1⟶ℝ.LetX be the Banach space of all continuous functions u:0,1⟶ℝ. Define a norm on X as follows.u=maxt∈0,1ut. Then, the metric d on X is defined as du,v=u−v∀u,v∈X.Definition 29.
A functionα∈X such that
(63)αt≤∫0tKt,s,αsds+ht,t∈0,1,is called a lower solution for (62).Definition 30.
A functionβ∈X such that
(64)βt≥∫0tKt,s,βsds+ht,t∈0,1,is called an upper solution for (62).Now, we have enough material to prove the following results.Theorem 31.
Assume that in third variableK is nondecreasing and there is τ>0 such that
(65)Kt,s,u−Kt,s,v≤u−vτDu,v+1,for all t,s∈0,1 and u,v∈X, where Du,v=maxdu,v,du,Qu,v,Qv,du,Qv+dv,Qu/2. Then, the existence of a unique solution of the integral Equation (62) follows from the existence of lower solution of (62).Proof.
LetQut=∫0tKt,s,usds+ht for all u∈X, be a self operator on X. It is clear that u is a fixed point of the operator Q if and only if it is solution of the Equation (62).
LetR be the binary relation on X defined by
(66)R=u,v∈X×X:ut≤vtforallt∈0,1.
Now, for anyu,v∈R and for all t∈0,1(67)Qut=∫0tKt,s,usds+ht≤∫0tKt,s,vsds+ht=Qvt.
This implies thatQu,Qv∈R. That is, R is Q closed.
Now, letu,v∈R and consider
(68)Qut−Qvt=∫0tKt,s,us−Kt,s,vsds≤∫0tKt,s,us−Kt,s,vsds≤∫0tu−vτDu,v+1ds≤1τDu,v+1∫0tmaxt∈0,1ut−vtds=1τDu,v+1∫0tdu,vds≤1τDu,v+1∫0tDu,vds=Du,vτDu,v+1t,≤Du,vτDu,v+1sincet∈0,1.
Therefore, we have(69)Qut−Qvt≤Du,vτDu,v+1,∀t∈0,1.
On taking supremum on both sides of above inequality, we obtain(70)Qu−Qv≤Du,vτDu,v+1.
It yields that(71)τ−1Qu−Qv≤−1Du,v,or
(72)τ−1dQu,Qv≤−1Du,v.
By choosingFμ=−1/μ,μ>0, from the above inequality, we get
(73)τ+FdQu,Qv≤FDu,v.
Hence, inequality (6) is satisfied. We have defined binary relation R on X by uRv if and only if ut≤vt for all t∈0,1. Now, consider an R-preserving sequence un in C0,1 which converges to u∈X. Then, we have
(74)u0t≤u1t≤⋯unt≤ut+1s≤⋯,which gives us unt≤ut∀t∈0,1. Therefore, R is d self closed on X. To show that FixQ is Rs-connected, if u,v∈FixQ, then w=maxu,v∈C0,1. Since u≤w and v≤w, thus uRw and vRw. Therefore, all conditions of Theorem 21 are true. Hence, the conclusion holds.☐Now, in the situation where upper solution is presented, we have the following result.Theorem 32.
Consider that in third variableK is nonincreasing and there is τ>0 such that
(75)Kt,s,u−Kt,s,v≤ut−vtτDu,v+1,for all t,s∈0,1 and u,v∈X, where Du,v=maxdu,v,du,Qu,dv,Qv,du,Qv+dv,Qu/2. Then, the existence of a unique solution of the integral Equation (62) follows from the existence of an upper solution of (62).Proof.
Let the binary relation onX be defined by
(76)R=u,v∈X×X:us≥vsforallt∈0,1.
Now, proceeding as in Theorem31, we can conclude that all the assumptions of Theorem 21 are satisfied and it guarantees the existence of a unique solution of (62).☐



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*Source: 9937318-2021-06-11.xml*