# Fixed Point Theory of Weak Contractions in Partially Ordered Metric Spaces

Authors: Chi-Ming Chen; Jin-Chirng Lee; Chao-Hung Chen
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.


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## Body

## 1. Introduction and Preliminaries

Throughout this paper, byℝ+, we denote the set of all nonnegative real numbers, while ℕ is the set of all natural numbers. Let (X,d) be a metric space, D a subset of X, and f:D→X a map. We say f is contractive if there exists α∈[0,1) such that, for all x,y∈D,(1)d(fx,fy)≤α·d(x,y).
The well-known Banach's fixed point theorem asserts that if D=X, f is contractive and (X,d) is complete, then f has a unique fixed point in X. In nonlinear analysis, the study of fixed points of given mappings satisfying certain contractive conditions in various abstract spaces has been investigated deeply. The Banach contraction principle [1] is one of the initial and crucial results in this direction. Also, this principle has many generalizations. For instance, Alber and Guerre-Delabriere in [2] suggested a generalization of the Banach contraction mapping principle by introducing the concept of weak contraction in Hilbert spaces. In [2], the authors also proved that the result of Eslamian and Abkar [3] is equivalent to the result of Dutta and Choudhury [4]. Later, weakly contractive mappings and mappings satisfying other weak contractive inequalities have been discussed in several works, some of which are noted in [4–16].In 2008, Dutta and Choudhury proved the following theorem.Theorem 1 (see [4]).
Let(X,d) be a complete metric space, and let f:X→X be such that
(2)ψ(d(fx,fy))≤ψ(d(x,y))-ϕ(d(x,y)),foreachx,y∈X,
where ψ,ϕ:ℝ+→ℝ+ are continuous and nondecreasing, and ψ(t)=ϕ(t)=0 if and only if t=0. Then f has a fixed point in X.Recently, Eslamian and Abkar [3] proved the following theorem.Theorem 2 (see [3]).
Let(X,d) be a complete metric space, and let f:X→X be such that
(3)ψ(d(fx,fy))≤α(d(x,y))-β(d(x,y)),foreachx,y∈X,
where ψ,α,β:ℝ+→ℝ+ are such that ψ is continuous and nondecreasing, α is continuous, β is lower semicontinuous, and
(4)ψ(t)-α(t)+β(t)>0∀t>0,ψ(t)=0ifandonlyift=0,α(0)=β(0)=0.
Then f has a fixed point in X.In the recent, fixed point theory has developed rapidly in partially ordered metric spaces (e.g., [17–22]).In 2012, Choudhury and Kundu [23] proved the following fixed point theorem as a generalization of Theorem 2.Theorem 3 (see [23]).
Let(X,⊑) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space and let f:X→X be a nondecreasing mapping such that
(5)ψ(d(fx,fy))≤α(d(x,y))-β(d(x,y)),foreachx,y∈Xsuchthatx⊑y,
where ψ,α,β:ℝ+→ℝ+ are such that ψ is continuous and nondecreasing, α is continuous, β is lower semicontinuous, and
(6)ψ(t)-α(t)+β(t)>0∀t>0,ψ(t)=0ifandonlyift=0,α(0)=β(0)=0.
Also, if any nondecreasing sequence {xn} in X converges to ν, then one assumes that
(7)xn⊑ν∀n∈ℕ.
If there exists x0∈X with x0⊑fx0, then f and g have a coincidence point in X.In this paper, we prove two new fixed point theorems in the framework of partially ordered metric spaces. Our results generalize and improve many recent fixed point theorems in the literature.


## 2. Fixed Point Results (I)

We start with the following definition.Definition 4.
Let(X,⊑) be a partially ordered set and f:X→X. Then f is said to be monotone nondecreasing if, for x,y∈X,
(8)x⊑y⟹fx⊑fy.Let(X,⊑) be a partially ordered set. x,y∈X are said to be comparable if either x⊑y or y⊑x holds.In the section, we denote byΨ the class of functions ψ:ℝ+3→ℝ+ satisfying the following conditions:  (ψ1)
ψ is an increasing, continuous function in each coordinate;(ψ2)
for allt∈ℝ+, ψ(t,t,t)≤t, ψ(0,0,t)≤t and ψ(t,0,0)≤t.Next, we denote byΦ the class of functions ϕ:ℝ+→ℝ+ satisfying the following conditions:  (ϕ1)
ϕ is a continuous, nondecreasing function;(ϕ2)
ϕ(t)>0 for t>0 and ϕ(0)=0;(ϕ3)
ϕ is subadditive; that is, ϕ(t1+t2)≤ϕ(t1)+ϕ(t2) for all t1,t2>0.And we denote the following sets of functions:(9)Θ={φ:ℝ+→ℝ+suchthatφiscontinuous},Ξ={ξ:ℝ+→ℝ+suchthatξislowercontinuous}.LetX be a nonempty set, and let (X,⊑) be a partially ordered set endowed with a metric d. Then, the triple (X,⊑,d) is called a partially ordered complete metric space.We now state the main fixed point theorem for(φ,ψ,ϕ,ξ)-contractions in partially ordered metric spaces, as follows.Theorem 5.
Let(X,⊑,d) be a partially ordered complete metric space. Let f:X→X be monotone nondecreasing, and
(10)φ(d(fx,fy))≤ψ(ϕ(d(x,y)),ϕ(d(x,fx)),ϕ(d(y,fy)))-ξ(max{d(x,y),d(x,fx),d(y,fy)}),
for all comparable x,y∈X, where φ∈Θ, ψ∈Ψ, ϕ∈Φ, and ξ∈Ξ, and
(11)φ(t)-ϕ(t)+ξ(t)>0∀t>0,φ(t)=0ifandonlyift=0,ϕ(0)=ξ(0)=0.
Suppose that either  (a)
f is continuous or(b)
if any nondecreasing sequence{xn} in X converges to ν, then one assumes that
(12)xn⊑ν∀n∈ℕ.If there exists x0∈X with x0⊑fx0, then f has a fixed point in X.Proof.
Sincef is nondecreasing, by induction, we construct the sequence {xn} recursively as
(13)xn=fnx0=fxn-1∀n∈ℕ.
Thus, we also conclude that
(14)x0⊏x1=fx0⊑x2=fx1⊑⋯⊑xn=fxn-1⊑⋯.
If any two consecutive terms in (14) are equal, then the f has a fixed point, and hence the proof is completed. So we may assume that
(15)d(xn-1,xn)≠0,∀n∈ℕ.
Now, we claim that d(xn,xn+1)≤d(xn-1,xn) for all n∈ℕ. If not, we assume that d(xn-1,xn)<d(xn,xn+1) for some n∈ℕ; substituting x=xn and y=xn+1 in (10) and using the definition of the function ψ, we have
(16)ψ(ϕ(d(xn,xn+1)),ϕ(d(xn,fxn)),ϕ(d(xn+1,fxn+1)))=ψ(ϕ(d(xn,xn+1)),ϕ(d(xn,xn+1)),ϕ(d(xn+1,xn+2)))≤ϕ(d(xn+1,xn+2)),ξ(max{d(xn,xn+1),d(xn,fxn),d(xn+1,fxn+1)})=ξ(max{d(xn,xn+1),d(xn,xn+1),d(xn+1,xn+2)})=ξ(d(xn+1,xn+2)),
and hence
(17)φ(d(xn+1,xn+2))=φ(d(fxn,fxn+1))≤ϕ(d(xn+1,xn+2))-ξ(d(xn+1,xn+2)).
Since φ(t)-ϕ(t)+ξ(t)>0 for all t>0, we have that d(xn+1,xn+2)=0, which contradicts (15). Therefore, we conclude that
(18)d(xn,xn+1)≤d(xn-1,xn)∀n∈ℕ.
From the previous argument, we also have that for each n∈ℕ(19)φ(d(xn,xn+1))≤ϕ(d(xn-1,xn))-ξ(d(xn-1,xn)).
It follows in (18) that the sequence {d(xn,xn+1)} is monotone decreasing; it must converge to some η≥0. Taking limit as n→∞ in (19) and using the continuities of φ and ϕ and the lower semicontinuous of ξ, we get
(20)φ(η)≤ϕ(η)-ξ(η),
which implies that η=0. So we conclude that
(21)limn→∞d(xn,xn+1)=0.We next claim that{xn} is a Cauchy sequence; that is, for every ε>0, there exists n∈ℕ such that if p,q≥n, then d(xp,xq)<ε.Suppose, on the contrary, that there existsϵ>0 such that, for any n∈ℕ, there are pn,qn∈ℕ with pn>qn≥n satisfying
(22)d(xqn,xpn)≥ϵ.
Further, corresponding to qn≥n, we can choose pn in such a way that it the smallest integer with pn>qn≥n and d(xqn,xpn)≥ϵ. Therefore d(xqn,xpn-1)<ϵ. Now we have that for all n∈ℕ(23)ϵ≤d(xpn,xqn)≤d(xpn,xpn-1)+d(xpn-1,xqn)<d(xpn,xpn-1)+ϵ.By lettingn→∞. we get that(24)limn→∞d(xpn,xqn)=ϵ.
On the other hand, we have
(25)d(xpn,xqn)≤d(xpn,xpn-1)+d(xpn-1,xqn-1)+d(xqn-1,xqn),d(xpn-1,xqn-1)≤d(xpn-1,xpn)+d(xpn,xqn)+d(xqn,xqn-1).
Letting n→∞, then we get
(26)limn→∞d(xpn-1,xqn-1)=ϵ.
By (14), we have that the elements xpn and xqn are comparable. Substituting x=xpn-1 and y=xqn-1 in (10), we have that, for all n∈ℕ,
(27)ψ(ϕ(d(xpn-1,xqn-1)),ϕ(d(xpn-1,fxpn-1)),ϕ(d(xqn-1,fxqn-1)))≤ψ(ϕ(d(xpn-1,xqn-1)),ϕ(d(xpn-1,xpn)),ϕ(d(xqn-1,xqn))),M(xpn-1,xqn-1)=max{d(xpn-1,xqn-1),d(xpn-1,fxpn-1),d(xqn-1,fxqn-1)}=max{d(xpn-1,xqn-1),d(xpn-1,xpn),d(xqn-1,xqn)}.
By the previous argument and using inequality (10), we can conclude that
(28)φ(ϵ)≤ψ(ϕ(ϵ),0,0)-ξ(ϵ)≤ϕ(ϵ)-ξ(ϵ),
which implies that ϵ=0, a contradiction. Therefore, the sequence {xn} is a Cauchy sequence.SinceX is complete, there exists ν∈X such that
(29)limn→∞xn=ν.
Suppose that (a) holds. Then
(30)ν=limn→∞xn+1=limn→∞fxn=fν.Thus,ν is a fixed point in X.Suppose that(b) holds; that is, xn⊑ν for all n∈ℕ. Substituting x=xn and y=ν in (10), we have that
(31)φ(d(xn+1,fν))=φ(d(fxn,fν))≤ψ(ϕ(d(xn,ν)),ϕ(d(xn,fxn)),ϕ(d(ν,fν)))-ξ(max{d(xn,ν),d(xn,fxn),d(ν,fν)}).
Taking limit as n→∞ in equality (31), we have
(32)φ(d(ν,fν))≤ψ(ϕ(0),ϕ(o),ϕ(d(ν,fν)))-ξ(d(ν,fν))≤ϕ(d(ν,fν))-ξ(d(ν,fν)),
which implies that d(ν,fν)=0; that is ν=fν. So we complete the proof.If we let(33)ψ(ϕ(d(x,y)),ϕ(d(x,fx)),ϕ(d(y,fy)))=max{ϕ(d(x,y)),ϕ(d(x,fx)),ϕ(d(y,fy))},it is easy to get the following theorem.Theorem 6.
Let(X,⊑,d) be a partially ordered complete metric space. Let f:X→X be monotone nondecreasing, and
(34)φ(d(fx,fy))≤max{ϕ(d(x,y)),ϕ(d(x,fx)),ϕ(d(y,fy))}-ξ(max{d(x,y),d(x,fx),d(y,fy)}),
for all comparable x,y∈X, where φ∈Θ, ϕ∈Φ, and ξ∈Ξ, and
(35)φ(t)-ϕ(t)+ξ(t)>0∀t>0,φ(t)=0ifandonlyift=0,ϕ(0)=ξ(0)=0.
Suppose that either(a)
f is continuous or(b)
if any nondecreasing sequence{xn} in X converges to ν, then one assumes that
(36)xn⊑ν∀n∈ℕ. If there exists x0∈X with x0⊑fx0, then f has a fixed point in X.


## 3. Fixed Point Results (II)

In the section, we denote byΨ the class of functions ψ:ℝ+3→ℝ+ satisfying the following conditions:  (ψ1)
ψ is an increasing and continuous function in each coordinate;(ψ2)
fort∈ℝ+, ϕ(t,t,t)≤t, ϕ(t,0,0)≤t, and ϕ(0,0,t)≤t.Next, we denote byΘ the class of functions φ:ℝ+→ℝ+ satisfying the following conditions:  (φ1)
φ is continuous and nondecreasing;(φ2)
fort>0, φ(t)>0 and φ(0)=0.And we denote byΦ the class of functions ϕ:ℝ+→ℝ+ satisfying the following conditions.(ϕ1)
ϕ is continuous;(ϕ2)
fort>0, ϕ(t)>0 and ϕ(0)=0.We now state the main fixed point theorem for the(ψ,φ,ϕ)-contractions in partially ordered metric spaces, as follows.Theorem 7.
Let(X,⊑,d) be a partially ordered complete metric space, and let f:X→X be monotone nondecreasing, and
(37)φ(d(fx,fy))≤ψ(φ(d(x,y)),φ(d(x,fx)),φ(d(y,fy)))-ϕ(M(x,y))+L·m(x,y),
for all comparable x,y∈X and ψ∈Ψ, φ∈Θ, ϕ∈Φ, where L>0 and
(38)M(x,y)=max{d(x,y),d(x,fx),d(y,fy)},m(x,y)=min{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}.
Suppose that either(a)
f is continuous or(b)
if any nondecreasing sequence{xn} in X converges to ν, then one assumes that
(39)xn⊑ν∀n∈ℕ.If there exists x0∈X with x0⊑fx0, then f has a fixed point in X.Proof.
Iffx0=x0, then the proof is finished. Suppose that x0⊏fx0. Since f is nondecreasing, by induction, we construct the sequence {xn} recursively as
(40)xn=fnx0=fxn-1∀n∈ℕ.
Thus, we also conclude that
(41)x0⊏x1=fx0⊑x2=fx1⊑⋯⊑xn=fxn-1⊑⋯.
We now claim that
(42)limn→∞d(xn,xn+1)=0.
Put x=xn-1 and y=xn in (37). Note that
(43)m(xn-1,xn)=min{d(xn-1,xn),d(xn-1,fxn-1),d(xn,fxn),d(xn-1,fxn),d(xn,fxn-1)}=min{d(xn-1,xn),d(xn-1,xn),d(xn,xn+1),d(xn-1,xn+1),d(xn,xn)}=0.
So, we obtain that
(44)φ(d(xn,xn+1))=φ(d(fxn-1,fxn))≤ψ(φ(d(xn-1,xn)),φ(d(xn-1,fxn-1)),φ(d(xn,fxn)))-ϕ(M(xn-1,xn))≤ψ(φ(d(xn-1,xn)),φ(d(xn-1,xn)),φ(d(xn,xn+1)))-ϕ(M(xn-1,xn)),
where
(45)M(xn-1,xn)=max{d(xn-1,xn),d(xn-1,fxn-1),d(xn,fxn)}=max{d(xn-1,xn),d(xn-1,xn),d(xn,xn+1)}.
We now claim that
(46)d(xn,xn+1)<d(xn-1,xn),∀n∈ℕ.
If not, we assume that d(xn-1,xn)≤d(xn,xn+1); then φ(d(xn-1,xn))≤φ(d(xn,xn+1)), since φ is non-decreasing. Using inequality (44) and the conditions of the function ψ, we have that, for each n∈ℕ,
(47)φ(d(xn,xn+1))≤φ(d(xn,xn+1))-ϕ(d(xn,xn+1)),
which implies that ϕ(d(xn,xn+1))=0, and hence d(xn,xn+1)=0. This contradicts our initial assumption.
From the previous argument, we have that, for eachn∈ℕ,
(48)φ(d(xn,xn+1))≤φ(d(xn-1,xn))-ϕ(d(xn-1,xn)),d(xn,xn+1)<d(xn-1,xn).
And since the sequence {d(xn,xn+1)} is decreasing, it must converge to some η≥0. Taking limit as n→∞ in (48) and by the continuity of φ and ϕ, we get
(49)φ(η)≤φ(η)-ϕ(η),
and so we conclude that ϕ(η)=0 and η=0.
We next claim that{xn} is Cauchy; that is, for every ε>0, there exists n∈ℕ such that if p,q≥n, then d(xp,xq)<ε.
Suppose, on the contrary, that there existsϵ>0 such that, for any n∈ℕ, there are pn,qn∈ℕ with pn>qn≥n satisfying
(50)d(xqn,xpn)≥ϵ.
Further, corresponding to qn≥n, we can choose pn in such a way that it the smallest integer with pn>qn≥n and d(xqn,xpn)≥ϵ. Therefore d(xqn,xpn-1)<ϵ. By the rectangular inequality, we have
(51)ϵ≤d(xpn,xqn)≤d(xpn,xpn-1)+d(xpn-1,xqn)<d(xpn,xpn-1)+ϵ.
Letting n→∞, then we get
(52)limn→∞d(xpn,xqn)=ϵ.
On the other hand, we have
(53)d(xpn,xqn)≤d(xpn,xpn-1)+d(xpn-1,xqn-1)+d(xqn-1,xqn),d(xpn-1,xqn-1)≤d(xpn-1,xpn)+d(xpn,xqn)+d(xqn,xqn-1).
By letting n→∞, we get that
(54)limn→∞d(xpn-1,xqn-1)=ϵ.
Using inequalities (37), (52), and (54) and putting x=xpn-1 and y=xqn-1, we have that
(55)φ(d(xpn,xqn))=φ(d(fxpn-1,fxqn-1))≤ψ(φ(d(xpn-1,xqn-1)),φ(d(xpn-1,fxpn-1)),φ(d(xqn-1,fxqn-1)))-ϕ(M(xpn-1,xqn-1))+L·m(xpn-1,xqn-1)=ψ(φ(d(xpn-1,xqn-1)),φ(d(xpn-1,xpn)),φ(d(xqn-1,xqn)))-ϕ(M(xpn-1,xqn-1))+L·m(xpn-1,xqn-1),
where
(56)M(xpn-1,xqn-1)=max{d(xpn-1,xqn-1),d(xpn-1,xpn),d(xqn-1,xqn)},m(xpn-1,xqn-1)=min{d(xpn-1,xqn-1),d(xpn-1,xpn),d(xqn-1,xqn),d(xpn-1,xqn),d(xqn-1,xpn)}.
Letting n→∞, then we obtain that
(57)limn→∞M(xpn-1,xqn-1)=ϵ,limn→∞m(xpn-1,xqn-1)=0,φ(ϵ)≤ψ(φ(ϵ),0,0)-ϕ(ϵ)≤φ(ϵ)-ϕ(ϵ).
This implies that ϕ(ϵ)=0, and hence ϵ=0. So we get a contraction. Therefore {xn} is a Cauchy sequence.
SinceX is complete, there exists ν∈X such that
(58)limn→∞xn=ν.
Suppose that(a) holds. Then
(59)ν=limn→∞xn+1=limn→∞fxn=fν.
Thus, ν is a fixed point in X
Suppose that(b) holds; that is, xn⊑ν for all n∈ℕ. Substituting x=xn and y=ν in (37), we have that
(60)φ(d(xn+1,fν))=φ(d(fxn,fν))≤ψ(φ(d(xn,ν)),φ(d(xn,fxn)),φ(d(ν,fν)))-ϕ(M(xn,ν))+L·m(xn,ν),
where
(61)M(xn,ν)=max{d(xn,ν),d(xn,fxn),d(ν,fν)},m(xn,ν)=min{d(xn,ν),d(xn,fxn),d(ν,fν),d(xn,fν)d(ν,fxn)}.
Letting n→∞, then we obtain that
(62)M(xn,ν)⟶d(ν,fν),m(xn,ν)⟶0,φ(d(ν,fν))≤ψ(φ(0),φ(0),φ(d(ν,fν)))-ϕ(d(ν,fν))≤φ(d(ν,fν))-ϕ(d(ν,fν)),
which implies that d(ν,fν)=0; that is, ν=fν. So we complete the proof.If we let(63)ψ(φ(d(x,y)),φ(d(x,fx)),φ(d(y,fy)))=max{φ(d(x,y)),φ(d(x,fx)),φ(d(y,fy))},it is easy to get the following theorem.Theorem 8.
Let(X,⊑,d) be a partially ordered complete metric space, and let f:X→X be monotone nondecreasing, and
(64)φ(d(fx,fy))≤max{φ(d(x,y)),φ(d(x,fx)),φ(d(y,fy))}-ϕ(M(x,y))+L·m(x,y),
for all comparable x,y∈X and φ∈Θ, ϕ∈Φ, where L>0 and
(65)M(x,y)=max{d(x,y),d(x,fx),d(y,fy)},m(x,y)=min{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}.
Suppose that either(a)
f is continuous or(b)
if any nondecreasing sequence{xn} in X converges to ν, then one assumes that
(66)xn⊑ν∀n∈ℕ.If there exists x0∈X with x0⊑fx0, then f has a fixed point in X.



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