# Fixed Point Theorems forψ-Contractive Mappings in Ordered Metric Spaces

Authors: Chi-Ming Chen
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We obtain some new fixed point theorems forψ-contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., Harjani et al., 2011 and 2010).


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## Body

## 1. Introduction and Preliminaries

Throughout this paper, byℝ+, we denote the set of all real nonnegative 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,(1.1)d(fx,fy)≤α⋅d(x,y).The well-known Banach fixed point theorem asserts that ifD=X, f is contractive and (X,d) is complete, then f has a unique fixed point in X. It is well known that the Banach contraction principle [1] is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, a mapping f:X→X is called a quasicontraction if there exists k<1 such that(1.2)d(fx,fy)≤k⋅max⁡{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)},for anyx,y∈X. In 1974, Ćirić [2] introduced these maps and proved an existence and uniqueness fixed point theorem.In 1972, Chatterjea [3] introduced the following definition.Definition 1.1.
Let(X,d) be a metric space. A mapping f:X→X is said to be a 𝒞-contraction if there exists α∈(0,1/2) such that for all x,y∈X, the following inequality holds:
(1.3)d(fx,fy)≤α⋅(d(x,fy)+d(y,fx)).Choudhury [4] introduced a generalization of 𝒞-contraction as follows.Definition 1.2.
Let(X,d) be a metric space. A mapping f:X→X is said to be a weakly 𝒞-contraction if for all x,y∈X,
(1.4)d(fx,fy)≤12(d(x,fy)+d(y,fx)-ϕ(d(x,fy),d(y,fx))),
where ϕ:ℝ+2→ℝ+ is a continuous function such that ϕ(x,y)=0 if and only if x=y=0.In [3, 4], the authors proved some fixed point results for the 𝒞-contractions. In [5], Harjani et al. proved some fixed point results for weakly 𝒞-contractive mappings in a complete metric space endowed with a partial order.In the following, we assume that the functionψ:ℝ+5→ℝ+ satisfies the following conditions:(C1)
ψ is a strictly increasing and continuous function in each coordinate, and(C2)
for allt∈ℝ+∖{0}, ψ(t,t,t,0,2t)<t, ψ(t,t,t,2t,0)<t, ψ(0,0,t,t,0)<t, and ψ(t,0,0,t,t)<t.Example 1.3.
Letψ:ℝ+5→ℝ+ denote
(1.5)ψ(t1,t2,t3,t4,t5)=k⋅max⁡{t1,t2,t3,t42,t52},fork∈(0,1).
Then, ψ satisfies the above conditions (C1) and (C2).Now, we define the following notion of aψ-contractive mapping in metric spaces.Definition 1.4.
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
(*)d(fx,fy)≤ψ(d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)),
for x≥y.Using Example1.3, it is easy to get the following examples of ψ-contractive mappings.Example 1.5.
LetX=ℝ+ endowed with usual ordering and with the metric d:X×X→ℝ+ given by
(1.6)d(x,y)=|x-y|,forx,y∈X.
Let ψ:ℝ+5→ℝ+ denote
(1.7)ψ(t1,t2,t3,t4,t5)=34⋅max⁡{t1,t2,t3,t42,t52},
where t1=d(x,y), t2=d(x,fx), t3=d(y,fy), t4=d(x,fy),and t5=d(y,fx), for all x,y∈X. Let f:X→X denote
(1.8)f(x)=13x.
Then, f is a ψ-contractive mapping.Example 1.6.
LetX=ℝ+×ℝ+ endowed with the coordinate ordering (i.e., (x,y)≤(z,w)⇔x≤z and y≤w) and with the metric d:X×X→ℝ+ given by
(1.9)d(x,y)=|x1-y1|+|x2-y2|,forx=(x1,x2),y=(y1,y2)∈X.
Let ψ:ℝ+5→ℝ+ denote
(1.10)ψ(t1,t2,t3,t4,t5)=34⋅max⁡{t1,t2,t3,t42,t52},
where t1=d(x,y), t2=d(x,fx), t3=d(y,fy), t4=d(x,fy), and t5=d(y,fx), for all x,y∈X. Let f:X→X denote
(1.11)f(x)=13x.
Then, f is a ψ-contractive mapping.In this paper, we obtain some new fixed point theorems forψ-contractive mappings in ordered metric spaces. Our results generalize or improve many recent fixed point theorems in the literature (e.g., [5, 6]).


## 2. Main Results

We start with the following definition.Definition 2.1.
Let(X,≤) be a partially ordered set and f:X→X. Then one says that f is monotone nondecreasing if, for x,y∈X,
(2.1)x≤y⟹fx≤fy.We now state the main fixed point theorem forψ-contractive mappings in ordered metric spaces when the operator is nondecreasing, as follows.Theorem 2.2.
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 continuous and nondecreasing ψ-contractive mapping. If there exists x0∈X with x0≤fx0, then f has a fixed point in X.Proof.
Iff(x0)=x0, then the proof is finished. Suppose that x0<f(x0). Since f is nondecreasing  mapping, by induction, we obtain that
(2.2)x0<fx0≤f2x0≤f3x0≤⋯≤fnx0≤fn+1x0≤⋯.
Put xn+1=fxn=fn+1x0 for n∈ℕ∪{0}. Then, for each n∈ℕ, from (*), and, as the elements xn and xn-1 are comparable, we get
(2.3)d(xn+1,xn)≤ψ(d(xn,xn-1),d(xn,fxn),d(xn-1,fxn-1),d(xn,fxn-1),d(xn-1,fxn))≤ψ(d(xn,xn-1),d(xn,xn+1),d(xn-1,xn),d(xn,xn),d(xn-1,xn+1))≤ψ(d(xn,xn-1),d(xn,xn+1),d(xn-1,xn),0,d(xn-1,xn)+d(xn,xn+1)),
and so we can deduce that, for each n∈ℕ,
(2.4)d(xn+1,xn)≤d(xn,xn-1).
Let we denote cm=d(xm+1,xm). Then, cm 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
(2.5)c≤cn+1≤ψ(cn,cn,cn,0,2cn).
Passing to the limit, as n→∞, we have
(2.6)c≤c≤ψ(c,c,c,0,2c)<c,
which is a contradiction. So c=0.
We next claim that that the following result holds.
For eachγ>0, there is n0(γ)∈ℕ such that for all m>n>n0(γ),(*)d(xm,xn)<γ.
We will prove (*) by contradiction. Suppose that (*) is false. Then, there exists some γ>0 such that for all k∈ℕ, there exist mk and nk with mk>nk>k such that
(2.7)d(xmk,xnk)≥γ,d(xmk-1,xnk)<γ.
Using the triangular inequality:
(2.8)γ≤d(xmk,xnk)≤d(xmk,xmk-1)+d(xmk-1,xnk)<γ+d(xmk,xmk-1),
and letting k→∞, we have
(2.9)lim⁡k→∞d(xmk,xnk)=γ.
Sincef is a ψ-contractive mapping, we also have(2.10)γ≤d(xmk,xnk)=d(fxmk-1,fxnk-1)≤ψ(d(xmk-1,xnk-1),d(xmk-1,xmk),d(xnk-1,xnk),d(xmk-1,xnk),d(xnk-1,xmk))≤ψ(cmk-1+d(xmk,xnk)+cnk-1,cmk-1,cnk-1,cmk-1+d(xmk,xnk),d(xmk,xnk)+cnk-1).
Letting k→∞. Then, we get
(2.11)γ≤ψ(γ,0,0,γ,γ)<γ,
a contradiction. It follows from (*) that the sequence {xn} must be a Cauchy sequence.
Similary, we also conclude that for eachn∈ℕ,
(2.12)d(xn,xn+1)≤ψ(d(xn-1,xn),d(xn-1,fxn-1),d(xn,fxn),d(xn-1,fxn),d(xn,fxn-1))≤ψ(d(xn-1,xn),d(xn-1,xn),d(xn,xn+1),d(xn-1,xn+1),d(xn,xn))≤ψ(d(xn-1,xn),d(xn-1,xn),d(xn,xn+1),d(xn-1,xn)+d(xn,xn+1),0),
and so we have that for each n∈ℕ,
(2.13)d(xn,xn+1)≤d(xn-1,xn).
Let us denotebm=d(xm,xm+1). Then, bm is a nonincreasing sequence and bounded below. Thus, it must converge to some b≥0. If b>0, then by the above inequalities, we have
(2.14)b≤bn+1≤ψ(bn,bn,bn,2bn,0).Passing to the limit, as n→∞, we have
(2.15)b≤b≤ψ(b,b,b,2b,0)<b,which is a contradiction. So b=0. By the above argument, we also conclude that {xn} is a Cauchy sequence.
SinceX is complete, there exists μ∈X such that limn→∞xn=μ. Moreover, the continuity of f implies that
(2.16)μ=lim⁡n→∞xn+1=lim⁡n→∞f(xn)=f(μ).
So we complete the proof.In what follows, we prove that Theorem2.2 is still valid for f not necessarily continuous, assuming the following hypothesis in X(which appears in Theorem 1 of [7]).If{xn} is a nondecreasing sequence in X, such that
(**)xn⟶x,thenxn≤x∀n∈N.Theorem 2.3.
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. Assume that X satisfies (**), and let f:X→X be a nondecreasing ψ-contractive mapping. If there exists x0∈X with x0≤f(x0), then f has a fixed point in X.Proof.
Following the proof of Theorem2.2, we only have to check that f(μ)=μ. As {xn} is a nondecreasing sequence in X and xn→μ, then the condition (**) gives us that xn≤μ for every n∈ℕ. Since f:X→X is a nondecreasing ψ-contractive mapping, we have
(2.17)d(xn+1,fμ)=d(fxn,fμ)≤ψ(d(xn,μ),d(xn,fxn),d(μ,fμ),d(xn,fμ),d(μ,fxn))≤ψ(d(xn,μ),d(xn,xn+1),d(μ,fμ),d(xn,fμ),d(μ,xn+1)).
Letting n→∞ and using the continuity of ψ, we have
(2.18)d(μ,fμ)≤ψ(0,0,d(μ,fμ),d(μ,fμ),0)<d(μ,fμ),
and this is a contraction unless d(μ,fμ)=0, or equivalently, μ=fμ.In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems2.2 and 2.3. This condition is the following and it appears in [8]:(2.19)forx,y∈X,thereexistsalowerboundoranupperbound.
In [7], it is proved that the above-mentioned condition is equivalent to the following:
(***)forx,y∈X,thereexistsz∈Xwhichiscomparabletoxandy.Theorem 2.4.
Adding condition (***) to the hypothesis of Theorem 2.2 (or Theorem 2.3) and the condition for all t∈ℝ+, ψ(t,0,2t,t,t)<t(or, ψ(t,2t,0,0,t)<t) to the function ψ, one obtains the uniqueness of the fixed point of f.Proof.
Suppose that there existμ,ν∈X which are fixed points of f. We distinguish two cases.
Case 1.
Ifμ and ν are comparable and μ≠ν, then fnμ=μ is comparable to fnν=ν for all n∈ℕ, and
(2.20)d(μ,ν)=d(fnμ,fnν)≤ψ(d(fn-1μ,fn-1ν),d(fn-1μ,fnμ),d(fn-1ν,fnν),d(fn-1μ,fnν),d(fn-1ν,fnμ))≤ψ(d(μ,ν),d(μ,μ),d(ν,ν),d(μ,ν),d(ν,μ))=ψ(d(μ,ν),0,0,d(μ,ν),d(ν,μ))<d(μ,ν),
and this is a contradiction unless d(μ,ν)=0, that is, μ=ν.Case 2.
Ifμ and ν are not comparable, then there exists x∈X comparable to μ and ν. Monotonicity of f implies that fnx is comparable to fnμ and fnν for all n∈ℕ. We also distinguish two cases.Subcase 2.1.
If there existsn0∈ℕ with fn0x=μ, then we have
(2.21)d(μ,ν)=d(fμ,fν)=d(fn0+1x,fn0+1ν)≤ψ(d(fn0x,fn0ν),d(fn0x,fn0+1x),d(fn0ν,fn0+1ν),d(fn0x,fn0+1ν),d(fn0ν,fn0+1x))=ψ(d(μ,ν),d(μ,fμ),d(ν,ν),d(μ,ν),d(ν,fμ))=ψ(d(μ,ν),0,0,d(μ,ν),d(ν,μ))<d(μ,ν),
and this is a contradiction unless d(μ,ν)=0, that is, μ=ν.Subcase 2.2.
For alln∈ℕ with fnx≠μ, since f is a nondecreasing ψ-contractive mapping, we have
(2.22)d(μ,fnx)=d(fnμ,fnx)≤ψ(d(fn-1μ,fn-1x),d(fn-1μ,fnμ),d(fn-1x,fnx),d(fn-1μ,fnx),d(fn-1x,fnμ))≤ψ(d(μ,fn-1x),d(μ,μ),d(fn-1x,fnx),d(μ,fnx),d(fn-1x,μ))≤ψ(d(μ,fn-1x),0,d(fn-1x,μ)+d(μ,fnx),d(μ,fnx),d(fn-1x,μ)).
Using the above inequality, we claim that for each n∈ℕ,
(2.23)d(μ,fnx)<d(μ,fn-1x).
If not, we assume that d(μ,fn-1x)≤d(μ,fnx), then by the definition of ψ and ψ(t,0,2t,t,t)<t, we have
(2.24)d(μ,fnx)≤ψ(d(μ,fn-1x),0,d(fn-1x,μ)+d(μ,fnx),d(μ,fnx),d(fn-1x,μ))≤ψ(d(μ,fnx),0,2d(fnx,μ),d(μ,fnx),d(fnx,μ))<d(μ,fnx),
which implies a contradiction. Therefore, our claim is proved.
This proves that the nonnegative decreasing sequence{d(μ,fnx)} is convergent. Put limn→∞d(μ,fnx)=η, η≥0. We now claim that η=0. If η>0, then making n→∞, we get(2.25)η=lim⁡n→∞d(μ,fnx)≤ψ(η,0,2η,η,η)<η,this is a contradiction. So η=0, that is, limn→∞d(μ,fnx)=0.
Analogously, it can be proved thatlimn→∞d(ν,fnx)=0.
Finally, the uniqueness of the limit gives usμ=ν.
This finishes the proof.In the following, we present a fixed point theorem for aψ-contractive mapping when the operator f is nonincreasing. We start with the following definition.Definition 2.5.
Let(X,≤) be a partially ordered set and f:X→X. Then one says that f is monotone nonincreasing if, for x,y∈X,
(2.26)x≤y⟹fx≥fy.
Using a similar argument to that in the proof of Theorem 3.1 of [5], we get the following point results.Theorem 2.6.
Let(X,≤) be a partially ordered set satisfying condition (***) and suppose that there exists a metric d in X such that (X,d) is a complete metric space, and let f be a nonincreasing ψ-contractive mapping. If there exists x0∈X with x0≤fx0 or x0≥fx0, then inf{d(x,fx):x∈X}=0. Moreover, if in addition, X is compact and f is continuous, then f has a unique fixed point in X.Proof.
Iffx0=x0, then it is obvious that inf{d(x,fx):x∈X}=0. Suppose that x0<fx0 (the same argument serves for x0<fx0). Since f is nonincreasing the consecutive terms of the sequence {fnx0} are comparable, we have
(2.27)d(fn+1x0,fnx0)≤ψ(d(fnx0,fn-1x0),d(fnx0,fn+1x0),d(fn-1x0,fnx0),d(fn-1x0,fn+1x0),d(fnx0,fnx0))≤ψ(d(fnx0,fn-1x0),d(fnx0,fn+1x0),d(fn-1x0,fnx0),d(fn-1x0,fn+1x0),0)≤ψ(d(fnx0,fn-1x0),d(fnx0,fn+1x0),d(fn-1x0,fnx0),d(fn-1x0,fnx0)+d(fnx0,fn+1x0)),
and so we conclude that for each n∈ℕ,
(2.28)d(fn+1x0,fnx0)<d(fnx0,fn-1x0).
Thus,{d(fn+1x0,fnx0)} is a decreasing sequence and bounded below, and it must converge to η≥0. We claim that η=0. If η>0, then by the above inequalities and the continuity of ψ, letting n→∞, we have
(2.29)η=lim⁡n→∞d(fn+1x0,fnx0)≤ψ(η,η,η,2η,0)<η,
which is a contradiction. So η=0, that is, limn→∞d(fn+1x0,fnx0)=0. Consequently, inf{d(x,fx):x∈X}=0.
Further, sincef is continuous and X is compact, we can find μ∈X such that
(2.30)d(μ,fμ)=inf⁡{d(x,fx):x∈X}=0,
and, therefore,μ is a fixed point of f.
The uniqueness of the fixed point is proved as in Theorem2.4.



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*Source: 756453-2012-01-18.xml*