# Common Fixed Points for Weakψ-Contractive Mappings in Ordered Metric Spaces with Applications

Authors: Sumit Chandok; Simona Dinu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We obtain some new common fixed point theorems satisfying a weak contractive condition in the framework of partially ordered metric spaces. The main result generalizes and extends some known results given by some authors in the literature.


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## Body

## 1. Introduction and Preliminaries

Fixed point and common fixed point theorems for different types of nonlinear contractive mappings have been investigated extensively by various researchers (see [1–41]). Fixed point problems involving weak contractions and the mappings satisfying weak contractive type inequalities have been studied by many authors (see [10–20] and references cited therein).Recently, many researchers have obtained fixed point, common fixed point, coupled fixed point, and coupled common fixed point results in partially ordered metric spaces (see [3, 6–8, 10–12, 29, 30, 32, 36]) and other spaces (see [5, 15, 31, 35, 38, 40, 41]).Let(X,≤) be a partially ordered set and f,g two self-mappings on X. A pair (f,g) of self-mappings of X is said to be weakly increasing [4] if fx≤gfx and gx≤fgx for any x∈X. An ordered pair (f,g) is said to be partially weakly increasing if fx≤gfx for all x∈X.Note that a pair(f,g) is weakly increasing if and only if the ordered pairs (f,g) and (g,f) are partially weakly increasing.Example 1 (see [3]).
LetX=[0,1] be endowed with usual ordering and f,g:X→X two mappings given by fx=x2 and gx=x. Clearly, the pair (f,g) is partially weakly increasing. But gx=x≰x=fgx for any x∈(0,1) implies that the pair (g,f) is not partially weakly increasing.
Let(X,≤) be a partially ordered set. A mapping f:X→X is called a weak annihilator of a mapping g:X→X if fgx≤x for all x∈X.Example 2 (see [3]).
LetX=[0,1] be endowed with usual ordering and f,g:X→X be two mappings given by fx=x2 and gx=x3. It is clear that fgx=x6≤x for x∈X implies that f is a weak annihilator of g.Let(X,≤) be a partially ordered set. A mapping f is called a dominating if x≤fx for any x∈X.Example 3 (see [3]).
LetX=[0,1] be endowed with usual ordering and f:X→X a mapping defined by fx=x1/3, since x≤x1/3=fx for x∈X implies that f is a dominating mapping.A subsetW of a partially ordered set X is said to be well ordered if every two elements of W are comparable.LetM be a nonempty subset of a metric space (X,d). Let S and T be mappings from a metric space (X,d) into itself. A point x∈M is a common fixed (resp., coincidence) point of S and T if x=Sx=Tx(resp.,Sx=Tx). The set of fixed points (resp., coincidence points) of S and T is denoted by F(S,T) (resp., C(S,T)).In 1986, Jungck [24] introduced the more generalized commuting mappings in metric spaces, called compatible mappings, which also are more general than the concept of weakly commuting mappings (that is, the mappings S,T:X→X are said to be weakly commuting if d(STx,TSx)≤d(Sx,Tx) for all x∈X) introduced by Sessa [34] as follows.Definition 4.
LetS and T be mappings from a metric space (X,d) into itself. The mappings S and T are said to be compatible if
(1)limn→∞d(STxn,TSxn)=0,
whenever {xn} is a sequence in X such that limn→∞Sxn=limn→∞Txn=z for some z∈X.In general, commuting mappings are weakly commuting and weakly commuting mappings are compatible, but the converses are not necessarily true and some examples can be found in [24–26].In [27], Jungck and Rhoades introduced the concept of weakly compatible mappings and proved some common fixed point theorems for these mappings.Definition 5.
The mappingsS and T are said to be weakly compatible if they commute at coincidence points of S and T.In Djoudi and Nisse [21], we can find an example to show that there exists weakly compatible mappings which are not compatible mappings in metric spaces.LetΨ denote the set of all functions ψ:[0,∞)5→[0,∞) such that(a)
ψis continuous;(b)
ψ is strictly increasing in all the variables;(c)
for allt∈[0,∞)∖{0},
(2)ψ(t,t,t,0,2t)<t,ψ(t,t,t,2t,0)<t,ψ(0,0,t,t,0)<t,ψ(0,t,0,0,t)<t,ψ(t,0,0,t,t)<t.It is easy to verify that the following functions are from the class Ψ, see [18]:
(3)ψ(t1,t2,t3,t4,t5)=kmax{t1,t2,t3,t42,t52},2222222222fork∈(0,1);ψ(t1,t2,t3,t4,t5)=kmax{t1,t2,t3,t4+t52},2222222222fork∈(0,1).Definition 6 (see [18]).
Let(X,≤) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a metric space. The mapping f:X→X is said to be a ψ- contractive mapping, if
(4)d(fx,fy)≤ψ(d(x,y),d(x,fx),d(y,fy),222222d(x,fy),d(y,fx)),
for x≥y.Recently, Chen introducedψ-contractive mappings. The purpose of this paper is to extend the results of Chen for four mappings, in the framework of ordered metric spaces.


## 2. Main Results

Now, we give the main results in this paper.Theorem 7.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T, f, g, and S are self-mappings on X, the pairs (T,f) and (S,g) are partially weakly increasing with f(X)⊆T(X) and g(X)⊆S(X), and the dominating mappings f and g are weak annihilators of T and S, respectively. Further, suppose that for any two comparable elements x,y∈X, and ψ∈Ψ,
(5)d(fx,gy)≤ψ(d(Sx,Ty),d(Sx,fx),d(Ty,gy),22222d(Sx,gy),d(Ty,fx)),
holds. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and either(a)
f and S are compatible, f or S is continuous, and g,T are weakly compatible or(b)
g and T are compatible, g or T is continuous, and f,S are weakly compatible, then f,g,S, and T have a common fixed point in X. Moreover, the set of common fixed points of f,g,S, and T is well ordered if and only if f,g,S, and T have one and only one common fixed point in X.Proof.
Letx0∈X be an arbitrary point. Since f(X)⊆T(X) and g(X)⊆S(X), we can construct the sequences {xn} and {yn} in X such that
(6)y2n-1=fx2n-2=Tx2n-1,y2n=gx2n-1=Sx2n,
for each n≥1. By assumptions, we have
(7)x2n-2≤fx2n-2=Tx2n-1≤fTx2n-1≤x2n-1,x2n-1≤gx2n-1=Sx2n≤gSx2n≤x2n,
for each n≥1. Thus, for each n≥1, we have xn≤xn+1. Without loss of generality, we assume that y2n≠y2n+1 for each n≥1.
Now, we claim that for alln∈ℕ, we have
(8)d(yn+1,yn+2)<d(yn,yn+1).
Suppose to the contrary thatd(y2n,y2n+1)≤d(y2n+1,y2n+2) for some n∈ℕ. Since y2n and y2n+1 are comparable, from (5), we have
(9)d(y2n+1,y2n+2)=d(fx2n,gx2n+1)≤ψ(d(Sx2n,Tx2n+1),d(Sx2n,fx2n),d(Tx2n+1,gx2n+1),22222d(Sx2n,gx2n+1),d(Tx2n+1,fx2n))=ψ(d(y2n,y2n+1),d(y2n,y2n+1),d(y2n+1,y2n+2),22222d(y2n,y2n+2),d(y2n+1,y2n+1))=ψ(d(y2n,y2n+1),d(y2n,y2n+1),22222d(y2n+1,y2n+2),d(y2n,y2n+2),0)≤ψ(d(y2n,y2n+1),d(y2n,y2n+1),d(y2n+1,y2n+2),22222d(y2n,y2n+1)+d(y2n+1,y2n+2),0)≤ψ(d(y2n+1,y2n+2),d(y2n+1,y2n+2),22222d(y2n+1,y2n+2),2d(y2n+1,y2n+2),0)<d(y2n+1,y2n+2),
which is a contradiction. Hence d(y2n+1,y2n+2)≤d(y2n,y2n+1) for each n≥1.
Similarly, we can prove thatd(y2n+1,y2n)≤d(y2n,y2n-1) for each n≥1.
Therefore, we can conclude that (8) holds.
Let us denotecn=d(yn+1,yn). Then, from (8), cn is a nonincreasing sequence and bounded below. Thus, it must converge to some c≥0. If c>0, then by the above inequalities, we have c≤cn+1≤ψ(cn,cn,cn,2cn,0). Taking the limit, as n→∞, we have c≤c≤ψ(c,c,c,2c,0)<c, which is a contradiction. Hence,
(10)d(yn+1,yn)⟶0.
Now, we show that{yn} is a Cauchy sequence.
Suppose that{yn} is not a Cauchy sequence. Then, there exists ϵ>0 for which we can find two sequences of natural numbers {m(k)} and {n(k)} with n(k)>m(k)>k such that
(11)d(ym(k),yn(k))≥ϵ,d(ym(k),yn(k)-1)<ϵ.
From (11), it follows that
(12)ϵ≤d(ym(k),yn(k))≤d(ym(k),yn(k)-1)+d(yn(k)-1,yn(k))<ϵ+d(yn(k)-1,yn(k)).
Letting k→∞ and using (10), we have
(13)limk→∞d(ym(k),yn(k))=ϵ.
Again,(14)d(ym(k)-1,yn(k)-1)≤d(ym(k)-1,ym(k))+d(ym(k),yn(k))+d(yn(k),yn(k)-1),d(ym(k),yn(k))≤d(ym(k),ym(k)-1)+d(ym(k)-1,yn(k)-1)+d(yn(k)-1,yn(k)).
Letting k→∞ in the above inequalities and using (10) and (13), we have
(15)limk→∞d(ym(k)-1,yn(k)-1)=ϵ.
Again,
(16)d(yn(k)-1,ym(k))≤d(ym(k)-1,yn(k)-1)+d(ym(k)-1,ym(k)),d(ym(k)-1,yn(k)-1)≤d(ym(k)-1,ym(k))+d(yn(k)-1,ym(k)).
Letting k→∞ in the above inequalities and using (10) and (15), we have
(17)limk→∞d(yn(k)-1,ym(k))=ϵ.
Similarly, we have
(18)limk→∞d(ym(k)-1,yn(k))=ϵ.
Also, again from (10), (15), and the inequality
(19)d(ym(k)-1,yn(k)+1)-d(ym(k)-1,yn(k))≤d(yn(k),yn(k)+1),
it follows that
(20)limk→∞d(ym(k)-1,yn(k)+1)=ϵ.
Now, we have
(21)d(ym(k),yn(k)+1)≤ψ(d(ym(k)-1,yn(k)),d(ym(k)-1,ym(k)),22222222222d(yn(k),yn(k)+1),d(ym(k)-1,yn(k)+1),222222222222d(yn(k),ym(k))).
Letting k→∞, we get
(22)ϵ≤ψ(ϵ,0,0,ϵ,ϵ)<ϵ,
which is a contradiction. Thus {yn} is a Cauchy sequence. Since X is a complete metric space, there exists z∈X such that yn→z. Therefore, we have
(23)limn→∞y2n+1=limn→∞Tx2n+1=limn→∞fx2n=z,limn→∞y2n+2=limn→∞Sx2n+2=limn→∞gx2n+1=z.
Assume that S is continuous. Since f and S are compatible, we have
(24)limn→∞fSx2n+2=limn→∞Sfx2n+2=Sz.
Also, x2n+1≤gx2n+1=Sx2n+2. Now, we have
(25)d(fSx2n,gx2n+1)≤ψ(d(S2x2n,Tx2n+1),d(S2x2n,fSx2n),222222222d(Tx2n+1,gx2n+1),d(S2x2n,gx2n+1),222222222222d(Tx2n+1,fSx2n)).
Letting n→∞, we get
(26)d(Sz,z)≤ψ(d(Sz,z),0,0,d(Sz,z),d(z,Sz))<d(Sz,z),
which implies that Sz=z.
Now, it follows thatx2n+1≤gx2n+1 and gx2n+1→z, x2n+1≤z. From (5), we have
(27)d(fz,gx2n+1)≤ψ(d(Sz,Tx2n+1),d(Sz,fz),d(Tx2n+1,gx2n+1),22222222d(Sz,gx2n+1),d(Tx2n+1,fz)).
Letting n→∞, we get
(28)d(fz,z)≤ψ(0,d(fz,z),0,0,d(z,fz))<d(fz,z),
which implies that fz=z. Since f(X)⊆T(X), there exists w∈X such that Sz=z=fz=Tw. Suppose that gw≠Tw. Since z≤fz=Tw≤fTw≤w implies z≤w, from (5), we obtain
(29)d(Tw,gw)=d(fz,gw)≤ψ(d(Sz,Tw),d(Sz,fz),d(Tw,gw),22222d(Sz,gw),d(Tw,fz))=ψ2(0,0,d(Tw,gw),d(Tw,gw),0)<d(Tw,gw),
which implies that Tw=gw. Since g and T are weakly compatible, gz=gfz=gTw=Tgw=Tfz=Tz. Thus, z is a coincidence point of T and g.
Now,x2n≤fx2n and fx2n→z implies x2n≤z. Thus, from (5), we obtain
(30)d(fx2n,gz)≤ψ(d(Sx2n,Tz),d(Sx2n,fx2n),22222d(Tz,gz),d(Sx2n,gz),d(Tz,fx2n)).
Letting n→∞, we get
(31)d(z,gz)≤ψ(d(z,gz),0,0,d(z,gz),d(z,gz))<d(z,gz),
which implies that gz=z. Therefore, we have fz=gz=Sz=Tz=z.
Iff is continuous, then, following the similar arguments, also we get the result.
Similarly, the result follows when (b) holds.
Now, suppose that the set of common fixed points ofT, S, f, and g is well ordered.
We claim that common fixed points ofT, S, f, and g are unique.
Assume thatTu=Su=fu=gu=u and Tv=Sv=fv=gv=v, but u≠v. Then, from (5), we have
(32)d(u,v)=d(fu,gv)≤ψ(d(Su,Tv),d(Su,fu),d(Tv,gv),111111d(Su,gv),d(Tv,fu))=ψ(d(u,v),0,0,d(u,v),d(v,u))<d(u,v).
This implies that d(u,v)=0, and hence u=v.
Conversely, ifT, S, f, and g have only one common fixed point, then the set of common fixed point of f, g, S, and T being singleton is well ordered. This completes the proof.Example 8.
ConsiderX=[0,1]∪{2,3,4,…} with usual ordering and
(33)d(x,y)={|x-y|ifx,y∈[0,1],x≠y;x+yifatleastoneofxory∉[0,1],x≠y;0ifx=y.
Then (X,≤,d) is a complete partially ordered metric space.
Letf,g,S, and T be self-mappings on X defined as
(34)f(x)={0ifx=0;12ifx∈(0,12];1ifx∈(12,1];xifx∈{2,3,4,…};g(x)={0ifx=0;12ifx∈(0,12];xifx∈(12,1]∪{2,3,4,…};T(x)={0ifx≤12;12ifx∈(12,1];x-1ifx∈{2,3,4,…};S(x)={0ifx≤12;2x-1ifx∈(12,1];xifx∈{2,3,4,…}.
Define function ψ:[0,∞)5→[0,∞) by the formula
(35)ψ(t1,t2,t3,t4,t5)=67max{t1,t2,t3,t4+t52}.
Note that f,g,S, and T satisfy all the conditions given in Theorem 7. Moreover, 0 is a common fixed point of f,g,S, and T.Iff=g, then we have the following result.Corollary 9.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T, f, and S are self-mappings on X, the pairs (T,f) and (S,f) are partially weakly increasing with f(X)⊆T(X) and f(X)⊆S(X), and the dominating mapping f is a weak annihilator of T and S. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(36)d(fx,fy)≤ψ(d(Sx,Ty),d(Sx,fx),d(Ty,fy),22222d(Sx,fy),d(Ty,fx))
holds. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and either(a)
f,S are compatible, f or S is continuous, and f,T are weakly compatible or(b)
f,T are compatible, f or T is continuous, and f,S are weakly compatible, then f,S, and T have a common fixed point in X. Moreover, the set of common fixed points of f,S, and T is well ordered if and only if f,S, and T have one and only one common fixed point in X.Corollary 10.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T, f, and g are self-mappings on X, the pairs (T,f) and (T,g) are partially weakly increasing with f(X)⊆T(X) and g(X)⊆T(X), and the dominating mappings f and g are weak annihilators of T. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(37)d(fx,gy)≤ψ(d(Tx,Ty),d(Tx,fx),d(Ty,gy),22222d(Tx,gy),d(Ty,fx))
holds. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and either(a)
f,T are compatible, f or T is continuous, and g,T are weakly compatible or(b)
g,T are compatible, g or T is continuous, and f,T are weakly compatible, then f,g, and T have a common fixed point in X. Moreover, the set of common fixed points of f,g, and T is well ordered if and only if f,g, and T have one and only one common fixed point in X.Corollary 11.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T and f are self-mappings on X, the pair (T,f) is partially weakly increasing with f(X)⊆T(X), and the dominating mapping f is a weak annihilator of T. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(38)d(fx,fy)≤ψ(d(Tx,Ty),d(Tx,fx),d(Ty,fy),22222d(Tx,fy),d(Ty,fx))
holds. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and, further, f,T are compatible, f or T is continuous, and f,T are weakly compatible, then f and T have a common fixed point in X. Moreover, the set of common fixed points of f and T is well ordered if and only if f and T have one and only one common fixed point in X.


## 3. Applications

The aim of the section is to apply our new results to mappings involving contractions of integral type. For this purpose, denote byΛ the set of functions μ:[0,∞)→[0,∞) satisfying the following hypotheses:(h1)
μ is a Lebesgue-integrable mapping on each compact of [0,∞);(h2)
for anyϵ>0, we have ∫0ϵ‍μ(t)>0.Corollary 12.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T, f, g, and S are self-mappings on X, the pairs (T,f) and (S,g) are partially weakly increasing with f(X)⊆T(X) and g(X)⊆S(X), and the dominating mappings f and g are weak annihilators of T and S, respectively. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(39)∫0d(fx,gy)‍α(s)ds≤∫0ψ(d(Sx,Ty),d(Sx,fx),d(Ty,gy),d(Sx,gy),d(Ty,fx))‍α(s)ds
holds, where α∈Λ. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and either(a)
f,S are compatible, f or S is continuous, and g,T are weakly compatible or(b)
g,T are compatible, g or T is continuous, and f,S are weakly compatible, then f,g,S, and T have a common fixed point in X. Moreover, the set of common fixed points of f,g,S, and T is well ordered if and only if f,g,S, and T have one and only one common fixed point in X.Corollary 13.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T, f, and g are self-mappings on X, the pairs (T,f) and (T,g) are partially weakly increasing with f(X)⊆T(X) and g(X)⊆T(X), and the dominating mappings f and g are weak annihilators of T. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(40)∫0d(fx,gy)‍α(s)ds≤∫0ψ(d(Tx,Ty),d(Tx,fx),d(Ty,gy),d(Tx,gy),d(Ty,fx))‍α(s)ds
holds, where α∈Λ. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and either(a)
f,T are compatible, f or T is continuous, and g,T are weakly compatible or(b)
g,T are compatible, g or T is continuous, and f,T are weakly compatible, then f,g, and T have a common fixed point in X. Moreover, the set of common fixed points of f,g, and T is well ordered if and only if f,g, and T have one and only one common fixed point in X.Corollary 14.
Let(X,≤) be a partially ordered set, and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Suppose that T and f are self-mappings on X, the pair (T,f) is a partially weakly increasing with f(X)⊆T(X), and the dominating mapping f is a weak annihilator of T. Further, suppose that there exists the function ψ∈Ψ such that, for any two comparable elements x,y∈X,
(41)∫0d(fx,fy)‍α(s)ds≤∫0ψ(d(Tx,Ty),d(Tx,fx),d(Ty,fy),d(Tx,fy),d(Ty,fx))‍α(s)ds
holds, where α∈Λ. If, for a nondecreasing sequence {xn} with xn≤yn for all n≥1, yn→u implies that xn≤u and, further, f,T are compatible, f or T is continuous, and f,T are weakly compatible, then f and T have a common fixed point in X. Moreover, the set of common fixed points of f and T is well ordered if and only if f and T have one and only one common fixed point in X.



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