# Common Coupled Fixed Point Theorem for Geraghty-Type Contraction in Partially Ordered Metric Spaces

Authors: Xiao-lan Liu; Mi Zhou; Boško Damjanović
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We study the existence and uniqueness of common coupled fixed point of four self-mappings for Geraghty-type contraction using weakly compatible mappings in partially ordered metric spaces with common limit range property (denoted by(CLRST)), the property of E.A, and so on. It is noted that the continuity of mappings and completeness of spaces can be removed. Our results improve, extend, complement, and generalize several existing results in the literature. Also, some examples are provided to illustrate the usability of our results.


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## Body

## 1. Introduction and Preliminaries

In 1922, Banach proposed the contraction mapping theorem which is famously known as the Banach Contraction Principle in metric spaces. It is also noted that Banach Contraction Principle is one of the pivotal results of fixed point theory in metric spaces. From then on, many researchers worked on this. They improved, extended, and generalized this theorem in different ways in the setting of different metric spaces.In 1982, Sessa first studied common fixed point theorems for weakly commuting pair of mappings [1]. Afterward, in 1986, Jungck [2] weakened weakly commuting mappings to compatible mappings in metric spaces and proved that compatible pair of mappings commute on the sets of coincidence points of the involved mappings. In 1998, Jungck and Rhoades [3] proposed the notion of weak compatibility if they commute at their coincidence points and proved that compatible mappings are weakly compatible but the reverse does not hold. In 2002, Aamri and Moutawakil [4] introduced the property (E.A) of one pair and the common property (E.A) of two pairs and obtained common fixed point theorems in metric spaces. In 2005, Liu et al. [5] used common property (E.A) to obtain the corresponding fixed point theorems. Later, in 2008, Kubiaczyk and Sharma [6] introduced the property (E.A) in PM spaces and got some fixed point theorems. The property (E.A) always requires the completeness (or closedness) of underlying subspaces for the existence of common fixed point. Hence they coined the idea of common limit in the range property (called CLR) which relaxes the requirement of completeness (or closedness) of the underlying subspace. In 2011, Sintunavarat and Kumam [7] introduced (CLRS) property and got the fixed point theorem in fuzzy metric spaces. Soon, Imdad et al. [8] introduced (CLRST) property and obtained some fixed point theorems in Menger spaces. In 2012, Jain et al. [9] extended the concept of CLR property in the coupled case and established the fixed point theorems. In 2013, Karapinar et al. [10] utilized (CLRST) property in symmetric spaces to obtain some common fixed point theorems. In 2014, Imdad et al. [11] proved some common fixed point theorems for two pairs of non-self-weakly compatible mappings by means of (CLRST) property in symmetric spaces. In sequel, Roldán-López-De-Hierro and Sintunavarat [12] generalized some results of Jain et al. [9] by using the generalized contractive conditions and the (CLR) property in fuzzy metric spaces.Inspired by the above works, this paper utilizes (CLRST) property to obtain the common fixed point theorems for Geragthy-type contraction in partially ordered metric spaces. For more details about fixed point theorems under Geragthy-type contraction in some generalized metric spaces, see [13–18]. At the same time, uniqueness of common fixed point is obtained. Finally, we illustrate some examples to support our results.To begin with, we give some basic notions with relevant to common fixed point of mappingsA:X×X→X and S:X→X.Definition 1.
Let(X,d,⪯) be a partially ordered metric space. An element (x,y)∈X×X is called a coupled fixed point of the mapping A:X×X→X if (1)Ax,y=x,Ay,x=y.Definition 2.
An element(x,y)∈X×X is called
(i) a coupled coincidence point of the mappingA:X×X→X and S:X→X if (2)Ax,y=Sx,Ay,x=Sy.
(ii) a common coupled fixed point of the mappingA:X×X→X and S:X→X if (3)Ax,y=Sx=x,Ay,x=Sy=y.
(iii) a common fixed point of the mappingA:X×X→X and S:X→X if (4)Ax,x=Sx=x.Definition 3.
Let(X,d,⪯) be a partially ordered metric space and A:X×X→X and S:X→X be two mappings. Then
(i)(A,S) is said to be weakly compatible if (5)SAx,y=ASx,Sy,SAy,x=ASy,Sx,for all x, y∈X, such that (6)Ax,y=Sx,Ay,x=Sy.
(ii)(A,S) is said to be compatible if (7)limn→∞⁡dASxn,Syn,SAxn,yn=0,limn→∞⁡dASyn,Sxn,SAyn,xn=0for sequences {xn}, {yn} in X such that (8)limn→∞⁡Axn,yn=limn→∞⁡Sxn,limn→∞⁡Ayn,xn=limn→∞⁡Syn.Proposition 4.
Let(X,d,⪯) be a partially ordered metric space and A:X×X→X and S:X→X be two mappings. If (A,S) is compatible, then (A,S) is weakly compatible.Proof.
Suppose thatA(x,y)=Sx, and A(y,x)=Sy for some x, y∈X. Consider the constant sequence xn=x, and yn=y for all n∈N. It is obvious that A(xn,yn)=Sxn→Sx as n→∞, and A(yn,xn)=Syn→Sy as n→∞. Since (A,S) is compatible, then d(A(Sxn,Syn),SA(xn,yn))→0, and d(A(Syn,Sxn),SA(yn,xn))→0. Thus, A(Sx,Sy)=SA(x,y), and A(Sy,Sx)=SA(y,x). So, (A,S) is weakly compatible.The following example shows that the pair of weakly compatible mappings is not necessarily compatible.Example 5.
Let(X,d) be a metric space where X=0,3. Define A:X×X→X and S:X→X as follows: (9)Ax,y=3-x-y,0≤x<1,1≤y≤2;2,0≤x<1,2<y<3;2,1≤x≤3,0≤y<1;2,1≤x≤3,1≤y<3,Sx=x,0≤x<12,1≤x≤3.Letting xn=1-1/n, yn=1+1/2n, A(xn,yn)=1+1/2n→1, Sxn=1-1/n→1, A(yn,xn)=2, Syn=2. But (10)limn→∞⁡dASxn,Syn,SAxn,yn=limn→∞⁡dA1-1n,2,S1+12n=limn→∞⁡d1n,2=2≠0,limn→∞⁡dASyn,Sxn,SAyn,xn=limn→∞⁡dA2,1-1n,S2=limn→∞⁡d2,2=0.Thus, the pair (A,S) is not compatible.
Also, the coupled coincidence point ofA and S is (1/2,2). It is, namely, that for x=1/2, y=2, A(x,y)=3-1/2-2=1/2=Sx, A(y,x)=2=Sy. Then we get (11)SAx,y=S12=12=A12,2=ASx,Sy.Meanwhile, (12)SAy,x=S2=2=A2,12=ASy,Sx.It implies that (A,S) is weakly compatible but not compatible.The above example shows that the notion of weakly compatibility is more general than compatibility.Definition 6.
Let(X,d,⪯) be a partially ordered metric space and A:X×X→X,B:X×X→X,S:X→X,T:X→X be four mappings.
(i) The pair(A,S) is said to satisfy the property (E.A) if there exists a sequence {xn} in X such that (13)limn→∞⁡Axn,xn=limn→∞⁡Sxn=z,for  somez∈X.
(ii) Two pairs(A,S) and (B,T) are said to satisfy the property (E.A) if there exist two sequences {xn}, {yn} in X such that (14)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Byn,yn=limn→∞⁡Tyn=z,for  somez∈X.
(iii) The pair(A,S) is said to have the common limit range property with respect to the mapping S (denoted by (CLRS)) if there exists a sequence {xn} in X such that (15)limn→∞⁡Axn,xn=limn→∞⁡Sxn=z,wherez∈SX.
(iv) Two pairs(A,S) and (B,T) are said to have the common limit range property with respect to mappings S and T (denoted by (CLRST)) if there exist two sequences {xn}, {yn} in X such that (16)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Byn,yn=limn→∞⁡Tyn=z,wherez∈SX∩TX.
LetΘ denote the class of all functions θ:0,∞×0,∞→0,1 which satisfies the following conditions:
(i)θ(s,t)=θ(t,s) for all s, t∈0,∞;
(ii) For any two sequences{sn} and {tn} of nonnegative real numbers (17)θsn,tn⟶1⟹sn,tn⟶0.Lemma 7.
Let(X,d,⪯) be a partially ordered metric space and A,B:X×X→X,S,T:X→X be four self-mappings satisfying the following conditions:
(i)A(X,X)⊆T(X) [resp., B(X,X)⊆S(X)];
(ii) The pair(A,S) satisfies the (CLRS) property [resp., the pair (B,T) satisfies the (CLRT) property];
(iii)T(X) is a closed subset of X [resp., S(X) is a closed subset of X];
(iv) There existsθ∈Θ such that, for all x, y, u, v∈X, (18)dAx,y,Bu,v≤θdSx,Tu,dSy,Tv×max⁡dSx,Tu,dSy,Tv.Then the pairs(A,S) and (B,T) share the (CLRST) property.Proof.
Since the pair(A,S) satisfies the (CLRS) property, there exists a sequence {xn} in X such that (19)limn→∞⁡Axn,xn=limn→∞⁡Sxn=z,wherez∈SX.In view of conditions (i) and (iii), for {xn}⊂X, there exists a sequence {yn}⊂X such that A(xn,xn)=Tyn. It follows that (20)limn→∞⁡Tyn=limn→∞⁡Axn,xn=z,wherez∈SX∩TX.Therefore, it suffices to prove that limn→∞⁡B(yn,yn)=z. In fact, by (iv), putting (x,y)=(xn,xn), (u,v)=(yn,yn), we can obtain that (21)dAxn,xn,Byn,yn≤θdSxn,Tyn,dSxn,Tyn×max⁡dSxn,Tyn,dSxn,Tyn<max⁡dSxn,Tyn,dSxn,Tyn=dSxn,Tyn.Letting n→∞, it follows that (22)limn→∞⁡dAxn,xn,Byn,yn<limn→∞⁡dSxn,Tyn=0.Thus, limn→∞⁡B(yn,yn)=z.
It yields that(23)limn→∞⁡Sxn=limn→∞⁡Axn,xn=limn→∞⁡Byn,yn=limn→∞⁡Tyn=z,wherez∈SX∩TX;that is, the pairs (A,S) and (B,T) share the (CLRST) property.Remark 8.
It can be pointed that Lemma7 generalizes Lemma 3.2 in [8], though still from four self-mappings to four self-mappings, but here A and B are defined in product space X×X. Simultaneously, it is straight forward to notice that Lemma 7 improves Lemma 1 of [11] from symmetric spaces to general partially ordered metric spaces.


## 2. Main Results

Now, we state and prove our main results.Theorem 9.
Let(X,d,⪯) be a complete partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose the following assumptions hold:
(i)A(X,X)⊆T(X), B(X,X)⊆S(X);
(ii)A and S are continuous;
(iii)(A,S) and (B,T) are both compatible;
(iv) There existsθ∈Θ such that for all x, y, u, v∈X, the inequality (18) holds.
ThenA, B, S and T have a unique common fixed point in X.Proof.
Fix arbitraryx0, y0∈X. From condition (i), A(X,X)⊆T(X), there exist x1, y1∈X such that (24)Tx1=Ax0,y0,Ty1=Ay0,x0.Since B(X,X)⊆S(X), there exist x2, y2∈X such that (25)Sx2=Bx1,y1,Sy2=By1,x1.Continuing in this way, we can construct four sequences {xn}, {yn}, {zn}, and {zn′} in X such that (26)z2n+1=Tx2n+1=Ax2n,y2n,z2n+1′=Ty2n+1=Ay2n,x2n,z2n+2=Sx2n+2=Bx2n+1,y2n+1,z2n+2′=Sy2n+2=By2n+1,x2n+1.
Step  1. We will show that {zn}, {zn′} in X are Cauchy sequences. Putting (x,y)=(x2n,y2n), (u,v)=(x2n+1,y2n+1) in the equality (18), we get (27)dz2n+1,z2n+2=dAx2n,y2n,Bx2n+1,y2n+1≤θdSx2n,Tx2n+1,dSy2n,Ty2n+1×max⁡dSx2n,Tx2n+1,dSy2n,Ty2n+1<max⁡dSx2n,Tx2n+1,dSy2n,Ty2n+1=max⁡dz2n,z2n+1,dz2n′,z2n+1′.Likewise, (x,y)=(y2n,x2n),(u,v)=(y2n+1,x2n+1) in the equality (18); we get: (28)dz2n+1′,z2n+2′=dAy2n,x2n,By2n+1,x2n+1≤θdSy2n,Ty2n+1,dSx2n,Tx2n+1×max⁡dSy2n,Ty2n+1,dSx2n,Tx2n+1<max⁡dSy2n,Ty2n+1,dSx2n,Tx2n+1=max⁡dz2n′,z2n+1′,dz2n,z2n+1.By the mentioned above, we can obtain that (29)max⁡dz2n+1,z2n+2,dz2n+1′,z2n+2′max⁡dz2n,z2n+1,dz2n′,z2n+1′≤θdz2n,z2n+1,dz2n′,z2n+1′<1.Letting n→∞, it follows that (30)θdz2n,z2n+1,dz2n′,z2n+1′⟶1.By the property of θ∈Θ, we get (31)dz2n,z2n+1⟶0,dz2n′,z2n+1′⟶0asn⟶∞.Analogously, we can prove that (32)dz2n+1,z2n+2⟶0,dz2n+1′,z2n+2′⟶0asn⟶∞.In conclusion, we obtain that (33)dzn,zn+1⟶0,dzn′,zn+1′⟶0asn⟶∞.Therefore, (34)dn=max⁡dzn,zn+1,dzn′,zn+1′⟶0asn⟶∞.
Now, suppose to the contrary that,{zn}, {zn′} are not Cauchy sequences. Then there exists ϵ>0, for which we can find subsequences {znk}, {zmk} of {zn} and {znk′}, {zmk′} of {zn′} with nk>mk≥k such that (35)rk=maxdznk,zmk,dznk′,zmk′≥ϵ,max⁡dznk-1,zmk,dznk-1′,zmk′<ϵ.
By view of (35) and triangle inequality, we get (36)ϵ≤rk=max⁡dznk,zmk,dznk′,zmk′<max⁡dznk,znk-1,dznk′,znk-1′+max⁡dznk-1,zmk,dznk-1′,zmk′<max⁡dznk,znk-1,dznk′,znk-1′+ϵ.Letting k→∞, we obtain: (37)rk=max⁡dznk,zmk,dznk′,zmk′⟶ϵ.Again by means of triangle inequality, we have (38)rk=max⁡dznk,zmk,dznk′,zmk′<max⁡dznk,znk+1,dznk′,znk+1′+max⁡dznk+1,zmk+1,dznk+1′,zmk+1′+max⁡dzmk+1,zmk,dzmk+1′,zmk′=max⁡dznk,znk+1,dznk′,znk+1′+max⁡dzmk+1,zmk,dzmk+1′,zmk′+max⁡dznk+1,zmk+1,dznk+1′,zmk+1′≤max⁡dznk,znk+1,dznk′,znk+1′+max⁡dzmk+1,zmk,dzmk+1′,zmk′+θdznk,zmk,dznk′,zmk′·max⁡dznk,zmk,dznk′,zmk′=max⁡dznk,znk+1,dznk′,znk+1′+max⁡dzmk+1,zmk,dzmk+1′,zmk′+θdznk,zmk,dznk′,zmk′·rk=dnk+dmk+θdznk,zmk,dznk′,zmk′·rk<dnk+dmk+rk.Thus,(39)rk≤dnk+dmk+θdznk,zmk,dznk′,zmk′·rk≤dnk+dmk+rk.Letting n→∞, it yields that θ(d(znk,zmk),d(znk′,zmk′))→1.
Now, by means of property ofθ, it follows that (40)dznk,zmk⟶0,dznk′,zmk′⟶0ask⟶∞,which implies that (41)limn→∞⁡rk=limn→∞⁡max⁡dznk,zmk,dznk′,zmk′=0,which contracts with (37). Thus, {zn}, {zn′} in X are Cauchy sequences.
Step  2. We will show that Sa=Ta and Sb=Tb. Suppose to the contrary that Sa≠Ta and Sb≠Tb. Since X is complete, there exist a, b∈X such that (42)limn→∞⁡zn=a,limn→∞⁡zn′=b.Also its subsequences converge as follows: (43)limn→∞⁡z2n+1=limn→∞⁡Tx2n+1=limn→∞⁡Ax2n,y2n=a,limn→∞⁡z2n+1′=limn→∞⁡Ty2n+1=limn→∞⁡Ay2n,x2n=b,limn→∞⁡z2n+2=limn→∞⁡Sx2n+2=limn→∞⁡Bx2n+1,y2n+1=a,limn→∞⁡z2n+2′=limn→∞⁡Sy2n+2=limn→∞⁡By2n+1,x2n+1=b.
SinceS and T are continuous, S2x2n→Sa, S2y2n→Sb and SA(x2n,y2n)→Sa, SA(y2n,x2n)→Sb, also T2x2n→Ta, T2y2n→Tb and TB(x2n,y2n)→Ta, TB(y2n,x2n)→Tb. Since A,S and (B,T) are compatible, we have A(Sx2n,Sy2n)→Sa, A(Sy2n,Sx2n)→Sb, and B(Tx2n,Ty2n)→Ta,B(Ty2n,Tx2n)→Tb.
Putting(x,y)=(Sx2n,Sy2n), (u,v)=(Tx2n,Ty2n) in the equality (18), we have (44)dASx2n,Sy2n,BTx2n,Ty2n≤θdSSx2n,TTx2n,dSSy2n,TTy2n×max⁡dSSx2n,TTx2n,dSSy2n,TTy2n<max⁡dSSx2n,TTx2n,dSSy2n,TTy2n.Letting n→∞, we obtain (45)dSa,Ta<max⁡dSa,Ta,dSb,Tb.
Putting(x,y)=(Sy2n,Sx2n), (u,v)=(Ty2n,Tx2n) in the equality (18), we have: (46)dASy2n,Sx2n,BTy2n,Tx2n≤θdSSy2n,TTy2n,dSSx2n,TTx2n×max⁡dSSy2n,TTy2n,dSSx2n,TTx2n<max⁡dSSy2n,TTy2n,dSSx2n,TTx2n.Letting n→∞, we obtain (47)dSb,Tb<max⁡dSa,Ta,dSb,Tb.
Combined with inequalities (45) and (47), it is a contraction. Therefore, (48)Sa=Ta,Sb=Tb.
Step  3. We will show that Sa=B(a,b). Putting (x,y)=(Sx2n,Sy2n) and (u,v)=(a,b) in the equality (18), we get (49)dASx2n,Sy2n,Ba,b≤θdSSx2n,Ta,dSSy2n,Tb×max⁡dSSx2n,Ta,dSSy2n,Tb<max⁡dSSx2n,Ta,dSSy2n,Tb.Letting n→∞, it yields that (50)dSa,Ba,b≤max⁡dSa,Ta,dSb,Tb=0.Hence (51)Sa=Ba,b=Ta.It suffices to show that B(a,b)=A(a,b). Putting (x,y)=(a,b) and (u,v)=(a,b) in the equality (18), we get (52)dAa,b,Ba,b≤θdSa,Ta,dSb,Tb×max⁡dSa,Ta,dSb,Tb.It implies that d(A(a,b),B(a,b))≤0. Thus, A(a,b)=B(a,b). Until now, we show that A(a,b)=B(a,b)=Sa=Ta.
Analogously, we can also obtain thatA(b,a)=B(b,a)=Sb=Tb.
Step  4. Set (53)Aa,b=Ba,b=Sa=Ta=x,Ab,a=Bb,a=Sb=Tb=y.Since (A,S) and (B,T) are both compatible, we can get (54)Sx=SAa,b=ASa,Sb=Ax,y,Sy=SAb,a=ASb,Sa=Ay,x,Tx=TBa,b=BTa,Tb=Bx,y,Ty=TBb,a=BTb,Ta=By,x.We show that x=y. Suppose to the contrary that x≠y. Putting (x,y)=(a,b) and (u,v)=(b,a) in the equality (18), we get (55)dx,y=dAa,b,Bb,a≤θdSa,Tb,dSb,Ta×max⁡dSa,Tb,dSb,Ta<max⁡dSa,Tb,dSb,Ta.It implies that d(x,y)<max{d(x,y),d(x,y)}; it is a contradiction! So, we have x=y.
Step  5. We will show that Sx=Tx. Putting (x,y)=(x,y) and (u,v)=(x,y) in the equality (18), we get (56)dSx,Tx=dAx,y,Bx,y≤θdSx,Tx,dSy,Ty×max⁡dSx,Tx,dSy,Ty.It implies that Sx=Tx.
In sequel, we show thatSx=x. Putting (x,y)=(x,y) and (u,v)=(a,b) in the equality (18), we get (57)dSx,x=dAx,y,Ba,b≤θdSx,Ta,dSy,Tb×max⁡dSx,Ta,dSy,Tb<max⁡dSx,x,dSx,x.It implies that Sx=x. Therefore, we get x=Sx=Tx=A(x,x)=B(x,x). So, x is a unique common fixed point of A, B, S, and T in X.Sequentially, continuity ofA and S is removed, and the compatibility of (A,S) is relaxed to the weak compatibility of (A,S); we propose the following theorem. It is noted that the closedness of T(X) or S(X) is necessary.Theorem 10.
Let(X,d,⪯) be a complete partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose the following conditions hold:
(i)A(X,X)⊆T(X), B(X,X)⊆S(X);
(ii)(A,S) and (B,T) are both weakly compatible;
(iii) There existsθ∈Θ such that, for all x, y, u, v∈X, the inequality (18) holds;
(iv)T(X) or S(X) is closed in X.
ThenA, B, S and T have a unique common fixed point in X.Proof.
Letx0, y0∈X. Then followed by Theorem 9, we construct sequences {zn}, {zn′} in X. {zn}, {zn′} are two Cauchy sequences; since X is complete, {zn}, {zn′} are converge sequences and its subsequences converge as follows: (58)limn→∞⁡z2n+1=limn→∞⁡Tx2n+1=limn→∞⁡Ax2n,y2n=a,limn→∞⁡z2n+1′=limn→∞⁡Ty2n+1=limn→∞⁡Ay2n,x2n=b,limn→∞⁡z2n+2=limn→∞⁡Sx2n+2=limn→∞⁡Bx2n+1,y2n+1=a,limn→∞⁡z2n+2′=limn→∞⁡Sy2n+2=limn→∞⁡By2n+1,x2n+1=b.
Case  1 (T(X) is closed in X). Since T(X) is closed in X, there exists a1, b1∈X such that (59)a=Ta1=limn→∞⁡Tx2n+1,b=Tb1=limn→∞⁡Ty2n+1.
Step  1. Putting (x,y)=(x2n,y2n) and (u,v)=(a1,b1) in the equality (18), we get (60)dAx2n,y2n,Ba1,b1≤θdSx2n,Ta1,dSy2n,Tb1×max⁡dSx2n,Ta1,dSy2n,Tb1<max⁡dSx2n,Ta1,dSy2n,Tb1.Letting n→∞, it yields that (61)limn→∞⁡dAx2n,y2n,Ba1,b1≤limn→∞⁡max⁡dSx2n,Ta1,dSy2n,Tb1=0.It follows that B(a1,b1)=a=Ta1. Since (B,T) is weakly compatible, then TB(a1,b1)=B(Ta1,Tb1); that is, Ta=B(a,b). Similarly, we can obtain that Tb=B(b,a).
Step  2. If a≠Ta, putting (x,y)=(x2n,y2n) and (u,v)=(a,b) in the equality (18), we get (62)dAx2n,y2n,Ba,b≤θdSx2n,Ta,dSy2n,Tb×maxdSx2n,Ta,dSy2n,Tb<maxdSx2n,Ta,dSy2n,Tb.Letting n→∞, it yields that (63)limn→∞⁡dAx2n,y2n,Ba,b=da,Ta≤limn→∞⁡max⁡dSx2n,Ta,dSy2n,Tb=max⁡da,Ta,db,Tb.
Ifb≠Tb, putting (x,y)=(y2n,x2n) and (u,v)=(b,a) in the equality (18), we get (64)dAy2n,x2n,Bb,a≤θdSy2n,Tb,dSx2n,Ta×max⁡dSy2n,Tb,dSx2n,Ta<max⁡dSy2n,Tb,dSx2n,Ta.Letting n→∞, it yields that (65)limn→∞⁡dAy2n,x2n,Bb,a=db,Tb≤limn→∞⁡max⁡dSy2n,Tb,dSx2n,Ta=max⁡db,Tb,da,Ta,which is a contradiction! So, we have that (66)a=Ta,b=Tb.It follows that B(a,b)=a=Ta, B(b,a)=b=Tb.
SinceB(X,X)⊆S(X), then there exists a2, b2∈X such that B(a,b)=Sa2=a, B(b,a)=Sb2=b.
Step  3. Putting (x,y)=(a2,b2) and (u,v)=(x2n,y2n) in the equality (18), we get (67)dAa2,b2,Bx2n,y2n≤θdSa2,Tx2n,dSb2,Ty2n×max⁡dSa2,Tx2n,dSb2,Ty2n<max⁡dSa2,Tx2n,dSb2,Ty2n.Letting n→∞, it yields that (68)limn→∞⁡dAa2,b2,Bx2n,y2n=dAa2,b2,a≤limn→∞⁡max⁡dSa2,a,dSb2,b=max⁡da,a,db,b=0.It yields that A(a2,b2)=a=Sa2. Similarly, we can show that A(b2,a2)=b=Sb2. Since (A,S) is weakly compatible, then SA(a2,b2)=A(Sa2,Sb2); that is, Sa=A(a,b). Similarly, we can obtain that Sb=A(b,a).
Analogous to Step 2, we can obtain thatA(a,b)=a=Sa, A(b,a)=b=Sb. To sum up, we have that (69)Aa,b=Sa=Ba,b=Ta=a,Ab,a=Sb=Bb,a=Tb=b.
Step  4. Putting (x,y)=(a,b) and (u,v)=(b,a) in the equality (18), we get (70)da,b=dAa,b,Bb,a≤θdSa,Tb,dSb,Ta×max⁡dSa,Tb,dSb,Ta=θda,b,db,a×max⁡da,b,db,a. Since the property of θ∈Θ, d(A(a,b),B(b,a))=d(a,b)=0. Then, a=b.
Until now, we have proved thatA(a,a)=Sa=B(a,a)=Ta=a. Thus, a is a unique common fixed point of A, B, S and T.
Case  2 (S(X) is closed in X). This proof of Case 2 is similar as the Case 1 provided above.In the following, compared to Theorem10, completeness of (X,d,≤) is removed. And instead, we assume that (A,S) or (B,T) has property (E.A).Theorem 11.
Let(X,d,⪯) be a partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose the following conditions hold:
(i)A(X,X)⊆T(X), B(X,X)⊆S(X);
(ii)(A,S) and (B,T) are both weakly compatible;
(iii) There existsθ∈Θ such that for all x, y, u, v∈X, the inequality (18) holds;
(iv)T(X) is closed in X and (A,S) satisfies the property (E.A) or S(X) is closed in X and (B,T) satisfies the property (E.A).
ThenA, B, S and T have a unique common fixed point in X.Proof.
⁡
Case  1. T(X) is closed in X and (A,S) satisfies the property (E.A).
Step  1. Since (A,S) satisfies the property (E.A), then there exists a sequence {xn} in X such that (71)limn→∞⁡Axn,xn=limn→∞⁡Sxn=z,for  somez∈X.Since A(X,X)⊆T(X), there exists a sequence {yn} in X such that A(xn,xn)=Tyn. Hence limn→∞⁡Tyn=z. Let us show that limn→∞⁡B(yn,yn)=z. In fact, by means of the equality (18), putting (x,y)=(xn,xn), (u,v)=(yn,yn), we get (72)dAxn,xn,Byn,yn≤θdSxn,Tyn,dSxn,Tyn×max⁡dSxn,Tyn,dSxn,Tyn<max⁡dSxn,Tyn,dSxn,Tyn=dSxn,Tyn.Letting n→∞, it follows that (73)limn→∞⁡dAxn,xn,Byn,yn≤limn→∞⁡dSxn,Tyn=0.It yields that limn→∞⁡B(yn,yn)=z. Since T(X) is closed in X, then limn→∞⁡Tyn=Tu=z for some u∈X. Thus, Tu=z.
Step  2. Putting (x,y)=(xn,xn), (u,v)=(u,u) in the equality (18), we get (74)dAxn,xn,Bu,u≤θdSxn,Tu,dSxn,Tu×max⁡dSxn,Tu,dSxn,Tu<max⁡dSxn,Tu,dSxn,Tu=dSxn,Tu. Letting n→∞, it yields that limn→∞⁡d(A(xn,xn),B(u,u))≤limn→∞⁡d(Sxn,Tu)=0. Thus, B(u,u)=z. It follows that Tu=B(u,u)=z. Since (B,T) is weakly compatible, then B(Tu,Tu)=TB(u,u); that is, B(z,z)=Tz.
Step  3. Since B(X,X)⊆S(X), then there exists v∈X such that B(u,u)=Sv=z. We claim that A(v,v)=Sv=z. Putting (x,y)=(v,v), (u,v)=(u,u) in the equality (18), we get (75)dAv,v,Sv=dAv,v,Bu,u≤θdSv,Tu,dSv,Tu×max⁡dSv,Tu,dSv,Tu.It yields that d(A(v,v),Sv)≤0. Thus, A(v,v)=Sv=z. Since (A,S) is weakly compatible, then SA(v,v)=A(Sv,Sv); that is, A(z,z)=Sz.
Step  4. Putting (x,y)=(z,z), (u,v)=(u,u) in the equality (18), we get (76)dSz,z=dAz,z,Bu,u≤θdSz,Tu,dSz,Tu×max⁡dSz,Tu,dSz,Tu.Since the property of θ∈Θ, it yields that d(Sz,z)=0. Thus, Sz=A(z,z)=z.
Step  5. Putting (x,y)=(z,z), (u,v)=(z,z) in the equality (18), we get(77)dSz,Tz=dAz,z,Bz,z≤θdSz,Tz,dSz,Tz×max⁡dSz,Tz,dSz,Tz.Since the property of θ∈Θ, it yields that d(Sz,Tz)=0. Thus, Sz=Tz=z. Hence, we obtain that A(z,z)=Sz=B(z,z)=Tz=z. So, z is a common fixed point of A, B, S and T.
Step  6 (uniqueness). Suppose that there is another common fixed point t∈X of A, B, S and T. It is obviously that A(t,t)=St=B(t,t)=Tt=t. Putting (x,y)=(z,z), (u,v)=(t,t) in the equality (18), we get (78)dz,t=dAz,z,Bt,t≤θdSz,Tt,dSz,Tt×max⁡dSz,Tt,dSz,Tt.Since the property of θ∈Θ, it yields that d(z,t)≤0. Thus, z=t. So, z is unique common fixed point of A, B, S and T.
Case  2. S(X) is closed in X and (B,T) satisfies the property (E.A).
The proof can be followed by the line of process of Case 1. To avoid repetition, we omit it.The next theorem is that the property (E.A) of(A,S) or (B,T) can be replaced by the (CLRST) property of (A,S) and (B,T). At the same time, the containment relationships of A(X,X) (or B(X,X)) and TX (or S(X)) are removed. As we all know, under the (CLRST) property, it is not necessary to assume the completeness of the spaces, which is an important advantage compared with most of the theorems in fixed point theory.Theorem 12.
Let(X,d,⪯) be a partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose the following conditions hold:
(i)(A,S) and (B,T) share the (CLRST) property;
(ii)(A,S) and (B,T) are both weakly compatible;
(ii) There existsθ∈Θ such that for all x, y, u, v∈X, the inequality (18) holds.
ThenA, B, S and T have a unique common fixed point in X.Proof.
Since the pairs(A,S) and (B,T) share the (CLRST) property, there exist two sequences {xn}, {yn} in X such that (79)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Byn,yn=limn→∞⁡Tyn=z,wherez∈SX∩TX.Since z∈T(X), there exists a point u∈X such that Tu=z. Putting (x,y)=(xn,xn) and (u,v)=(u,u) in the equality (18), it yields that(80)dAxn,xn,Bu,u≤θdSxn,Tu,dSxn,Tu×max⁡dSxn,Tu,dSxn,Tu<max⁡dSxn,Tu,dSxn,Tu=dSxn,Tu. Letting n→∞, it follows that (81)limn→∞⁡dAxn,xn,Bu,u≤limn→∞⁡dSxn,Tu=0.Thus, B(u,u)=z. Hence, z=B(u,u)=Tu, which shows u is a coincidence point of the pair (B,T).
Asz∈S(X), there exists a point v∈X such that Sv=z; putting (x,y)=(v,v), (u,v)=(yn,yn) in the equality (18), we have (82)dAv,v,Byn,yn≤θdSv,Tyn,dSv,Tyn×max⁡dSv,Tyn,dSv,Tyn<max⁡dSv,Tyn,dSv,Tyn=dSv,Tyn.Letting n→∞, it follows that (83)limn→∞⁡dAv,v,Byn,yn≤limn→∞⁡dSv,Tyn=0.Thus, A(v,v)=z. Hence, z=A(v,v)=Sv, which shows v is a coincidence point of the pair (A,S).
Since the pair(B,T) is weakly compatible, and by the previous proof, z=B(u,u)=Tu; then B(Tu,Tu)=TB(u,u); it yields that B(z,z)=Tz. And since the pair (A,S) is weakly compatible, and by the previous proof, z=A(v,v)=Sv; then A(Sv,Sv)=SA(v,v); it yields that A(z,z)=Sz. Letting (x,y)=(z,z),(u,v)=(u,u) in the inequality (18), we obtain (84)dAz,z,Bu,u=dSz,z≤θdSz,Tu,dSz,Tu×max⁡dSz,Tu,dSz,Tu=θdSz,Tu,dSz,Tu×dSz,z.Since θ∈Θ, then θ(d(Sz,Tu),d(Sz,Tu))<1; it follows that (85)dSz,z=0.Thus,  Sz=z. Therefore, Sz=A(z,z)=z.
Sequentially, letting(x,y)=(z,z), (u,v)=(z,z) in the inequality (18), we obtain (86)dAz,z,Bz,z=dSz,Tz≤θdSz,Tz,dSz,Tz×max⁡dSz,Tz,dSz,Tz=θdSz,Tz,dSz,Tz×dSz,Tz.Since θ∈Θ, then θ(d(Sz,Tu),d(Sz,Tu))<1; it follows that (87)dSz,Tz=0.Thus, Sz=Tz. By the mentioned above, we have that A(z,z)=B(z,z)=Sz=Tz=z. Hence, A, S, B and T have a common fixed point z.
Uniqueness. Assume that t is another common fixed point of A, B, S and T. It follows that t=A(t,t)=B(t,t)=St=Tt. Letting (x,y)=(z,z), (u,v)=(t,t) in the equality (18), we obtain (88)dAz,z,Bt,t=dz,t≤θdSz,Tt,dSz,Tt×max⁡dSz,Tt,dSz,Tt=θdSz,Tt,dSz,Tt×dz,t.Since θ∈Θ, then θ(d(Sz,Tt),d(Sz,Tt))<1; it follows that (89)dz,t=0,and we obtain that z=t. Thus, A, B, S and T have a unique common fixed point z.If we utilize Lemma7, then we have the following.Corollary 13.
Let(X,d,⪯) be a complete partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose the following conditions hold:
(i) All the conditions in Lemma7 hold;
(ii)(A,S) and (B,T) are both weakly compatible.
ThenA, B, S and T have a unique common fixed point in X.Proof.
By Lemma7, we can get that (A,S) and (B,T) share the (CLRST) property. Then by use of Theorem 12, we can obtain the conclusions.Instead of the(CLRST) property of (A,S) and (B,T) in Theorem 12, we utilize the common property (E.A) to obtain fixed point theorems. It is noticed that it should require the closedness of S(X) and T(X).Theorem 14.
Let(X,d,⪯) be a partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose that the inequality (18) of Lemma 7 and the following hypotheses hold:
(i) The pairs(A,S) and (B,T) have the common property (E.A);
(ii)S(X) and T(X) are closed subsets in X;
(iii) Both the pairs(A,S) and (B,T) are weakly compatible.
ThenA, B, S and T have a unique common fixed point.Proof.
If the pairs(A,S) and (B,T) have the common property (EA), then there exist two sequences {xn} and {yn} in X such that (90)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Byn,yn=limn→∞⁡Tyn=z,for  somez∈X.Since T(X) is closed, then limn→∞⁡Tyn=z=Tu for some u∈X. And S(X) is closed; then limn→∞⁡Sxn=z=Sv for some v∈X. The rest of the proof can run on the lines of Theorem 12.Corollary 15.
The result of Theorem14 holds if condition (ii) is replaced by condition (ii′):
(ii′)B(X,X)¯⊆S(X) and A(X,X)¯⊆T(X) where B(X,X)¯ denoted the closure of B(X,X),A(X,X)¯ denoted the closure of A(X,X).Corollary 16.
The result of Theorem14 holds if condition (ii) is replaced by condition (ii′′):
(ii′′)A(X,X) and B(X,X)are closed subsets in X×X and A(X,X)⊆T(X) and B(X,X)⊆S(X).In order to show that the common property(E.A) of two pairs (A,S) and (B,T) can be deduced from containment of A(X,X)⊆T(X) and property (E.A) of (A,S), we propose the following theorem.Theorem 17.
Let(X,d,⪯) be a partially ordered metric space and A,B:X×X→X,S,T:X→X be the mappings. Suppose that the inequality (18) of Lemma 7 and the following hypotheses hold:
(i)A(X,X)⊆T(X);
(ii)T(X) is closed in X and (A,S) satisfies the property (E.A).
Then(A,S) and (B,T) share the common property (E.A).Proof.
Since(A,S) satisfies the property (E.A), there exists a sequence {xn}⊂X such that(91)limn→∞⁡Axn,xn=limn→∞⁡Sxn=z,for  somez∈X.Since A(X,X)⊆T(X), for each xn, there exists a corresponding yn∈X such that A(xn,xn)=Tyn. Therefore, we have that (92)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Tyn=z,for  somez∈X.It suffices to show that limn→∞⁡B(yn,yn)=z.
Letting(x,y)=(xn,xn) and (u,v)=(yn,yn) in the inequality (18), we obtain (93)dAxn,xn,Byn,yn≤θdSxn,Tyn,dSxn,Tyn×max⁡dSxn,Tyn,dSxn,Tyn<max⁡dSxn,Tyn,dSxn,Tyn.Letting n→∞, we obtain that (94)limn→∞⁡dAxn,xn,Byn,yn≤limn→∞⁡max⁡dSxn,Tyn,dSxn,Tyn=0.It follows that limn→∞⁡B(yn,yn)=z.
Thus, we have that(95)limn→∞⁡Axn,xn=limn→∞⁡Sxn=limn→∞⁡Tyn=limn→∞⁡Byn,yn=z,for  somez∈X.Thus, (A,S) and (B,T) share the common property (E.A).Remark 18.
Theorem17 shows that our common property (E.A) of two pairs (A,S) and (B,T) is weaker than containment of A(X,X)⊆T(X) and property (E.A) of (A,S). It is, namely, that Theorem 17 is indeed a generalization of Theorem 3.4 in [14].



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*Source: 9063267-2018-03-18.xml*