JAMJournal of Applied Mathematics1687-00421110-757XHindawi Publishing Corporation75645310.1155/2012/756453756453Research ArticleFixed Point Theorems for ψ-Contractive Mappings in Ordered Metric SpacesChenChi-MingLiuYansheng1Department of Applied MathematicsNational Hsinchu University of Education, Hsinchu 30014Taiwannhcue.edu.tw20121712012201219092011251120112012Copyright © 2012 Chi-Ming Chen.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We obtain some new fixed point theorems for ψ-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).

1. Introduction and Preliminaries

Throughout this paper, by +, we denote the set of all real nonnegative numbers, while is the set of all natural numbers. Let (X,d) be a metric space, D a subset of X; and f:DX a map. We say f is contractive if there exists α[0,1) such that for all x,yD,d(fx,fy)αd(x,y).

The well-known Banach 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. It is well known that the Banach contraction principle  is a very useful and classical tool in nonlinear analysis. Also, this principle has many generalizations. For instance, a mapping f:XX is called a quasicontraction if there exists k<1 such thatd(fx,fy)kmax{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)},

for any x,yX. In 1974, Ćirić  introduced these maps and proved an existence and uniqueness fixed point theorem.

In 1972, Chatterjea  introduced the following definition.

Definition 1.1.

Let (X,d) be a metric space. A mapping f:XX is said to be a 𝒞-contraction if there exists α(0,1/2) such that for all x,yX, the following inequality holds: d(fx,fy)α(d(x,fy)+d(y,fx)).

Choudhury  introduced a generalization of 𝒞-contraction as follows.

Definition 1.2.

Let (X,d) be a metric space. A mapping f:XX is said to be a weakly 𝒞-contraction if for all x,yX, d(fx,fy)12(d(x,fy)+d(y,fx)-ϕ(d(x,fy),d(y,fx))), where ϕ:+2+ 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 , 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 function ψ:+5+ satisfies the following conditions:

ψ is a strictly increasing and continuous function in each coordinate, and

for all t+{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.

Let ψ:+5+ denote ψ(t1,t2,t3,t4,t5)=kmax{t1,t2,t3,t42,t52},  for  k(0,1). Then, ψ satisfies the above conditions (C1) and (C2).

Now, we define the following notion of a ψ-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:XX 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 xy.

Using Example 1.3, it is easy to get the following examples of ψ-contractive mappings.

Example 1.5.

Let X=+ endowed with usual ordering and with the metric d:X×X+ given by d(x,y)=|x-y|,  for  x,yX. Let ψ:+5+ denote ψ(t1,t2,t3,t4,t5)=34max{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,yX. Let f:XX denote f(x)=13x. Then, f is a ψ-contractive mapping.

Example 1.6.

Let X=+×+ endowed with the coordinate ordering (i.e., (x,y)(z,w)xz and yw) and with the metric d:X×X+ given by d(x,y)=|x1-y1|+|x2-y2|,for  x=(x1,x2),  y=(y1,y2)X. Let ψ:+5+ denote ψ(t1,t2,t3,t4,t5)=34max{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,yX. Let f:XX denote f(x)=13x. Then, f is a ψ-contractive mapping.

In this paper, we obtain some new fixed point theorems for ψ-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

Definition 2.1.

Let (X,) be a partially ordered set and f:XX. Then one says that f is monotone nondecreasing if, for x,yX, xyfxfy.

We now state the main fixed point theorem for ψ-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:XX be a continuous and nondecreasing ψ-contractive mapping. If there exists x0X with x0fx0, then f has a fixed point in X.

Proof.

If f(x0)=x0, then the proof is finished. Suppose that x0<f(x0). Since f is nondecreasing mapping, by induction, we obtain that x0<fx0f2x0f3x0fnx0fn+1x0. Put xn+1=fxn=fn+1x0 for n{0}. Then, for each n, from (*), and, as the elements xn and xn-1 are comparable, we get 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, 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 c0. If c>0, then by the above inequalities, we have ccn+1ψ(cn,cn,cn,0,2cn). Passing to the limit, as n, we have ccψ(c,c,c,0,2c)<c, which is a contradiction. So c=0.

We next claim that that the following result holds.

For each γ>0, there is n0(γ) 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, there exist mk and nk with mk>nk>k such that d(xmk,xnk)γ,d(xmk-1,xnk)<γ. Using the triangular inequality: γd(xmk,xnk)d(xmk,xmk-1)+d(xmk-1,xnk)<γ+d(xmk,xmk-1), and letting k, we have limkd(xmk,xnk)=γ.

Since f is a ψ-contractive mapping, we also haveγd(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 γψ(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 each n, 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, d(xn,xn+1)d(xn-1,xn).

Let us denote bm=d(xm,xm+1). Then, bm is a nonincreasing sequence and bounded below. Thus, it must converge to some b0. If b>0, then by the above inequalities, we have bbn+1ψ(bn,bn,bn,2bn,0).Passing to the limit, as n, we have bbψ(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.

Since X is complete, there exists μX such that limnxn=μ. Moreover, the continuity of f implies that μ=limnxn+1=limnf(xn)=f(μ).

So we complete the proof.

In what follows, we prove that Theorem 2.2 is still valid for f not necessarily continuous, assuming the following hypothesis in X  (which appears in Theorem 1 of ).

If {xn} is a nondecreasing sequence in X, such that   xnx,  then  xnxnN.

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:XX be a nondecreasing ψ-contractive mapping. If there exists x0X with x0f(x0), then f has a fixed point in X.

Proof.

Following the proof of Theorem 2.2, we only have to check that f(μ)=μ. As {xn} is a nondecreasing sequence in X and xnμ, then the condition (**) gives us that xnμ for every n. Since f:XX is a nondecreasing ψ-contractive mapping, we have 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 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 Theorems 2.2 and 2.3. This condition is the following and it appears in :for  x,yX,  there  exists  a  lower  bound  or  an  upper  bound. In , it is proved that the above-mentioned condition is equivalent to the following: for  x,yX,  there  exists  zX  which  is  comparable  to  x  and  y.

Theorem 2.4.

Adding condition (***) to the hypothesis of Theorem 2.2 (or Theorem 2.3) and the condition for all t+, 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 exist μ,νX which are fixed points of f. We distinguish two cases.

Case 1.

If μ and ν are comparable and μν, then fnμ=μ is comparable to fnν=ν for all n, and 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.

If μ and ν are not comparable, then there exists xX comparable to μ and ν. Monotonicity of f implies that fnx is comparable to fnμ and fnν for all n. We also distinguish two cases.

Subcase 2.1.

If there exists n0 with fn0x=μ, then we have 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 all n with fnxμ, since f is a nondecreasing ψ-contractive mapping, we have 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, 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 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 limnd(μ,fnx)=η, η0. We now claim that η=0. If η>0, then making n, we getη=limnd(μ,fnx)ψ(η,0,2η,η,η)<η,this is a contradiction. So η=0, that is, limnd(μ,fnx)=0.

Analogously, it can be proved that limnd(ν,fnx)=0.

Finally, the uniqueness of the limit gives us μ=ν.

This finishes the proof.

In the following, we present a fixed point theorem for a ψ-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:XX. Then one says that f is monotone nonincreasing if, for x,yX, xyfxfy.

Using a similar argument to that in the proof of Theorem 3.1 of , 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 x0X with x0fx0 or x0fx0, then inf{d(x,fx):xX}=0. Moreover, if in addition, X is compact and f is continuous, then f has a unique fixed point in X.

Proof.

If fx0=x0, then it is obvious that inf{d(x,fx):xX}=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 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, 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 η0. We claim that η=0. If η>0, then by the above inequalities and the continuity of ψ, letting n, we have η=limnd(fn+1x0,fnx0)ψ(η,η,η,2η,0)<η, which is a contradiction. So η=0, that is, limnd(fn+1x0,fnx0)=0. Consequently, inf{d(x,fx):xX}=0.

Further, since f is continuous and X is compact, we can find μX such that d(μ,fμ)=inf{d(x,fx):xX}=0,

and, therefore, μ is a fixed point of f.

The uniqueness of the fixed point is proved as in Theorem 2.4.

Acknowledgment

The authors would like to thank the referees whose helpful comments and suggestions led to much improvement in this paper.

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