A Fixed Point Theorem for Mappings Satisfying a Contractive Condition of Rational Type on a Partially Ordered Metric Space

Abstract

The purpose of this paper is to present a fixed point theorem using a contractive condition of rational type in the context of partially ordered metric spaces.

(ProQuest: ... denotes non-US-ASCII text omitted.)

Recommended by A. Zafer

Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain

Received 23 June 2010; Accepted 14 September 2010

1. Introduction

In [1], Jaggi proved the following fixed point theorem.

Theorem 1.1.

Let T be a continuous selfmap defined on a complete metric space (X,d) . Suppose that T satisfies the following contractive condition: [figure omitted; refer to PDF] for all x,y∈X , x≠y , and for some α,β∈[0,1) with α+β<1 , then T has a unique fixed point in X .

The aim of this paper is to give a version of Theorem 1.1 in partially ordered metric spaces.

Existence of fixed point in partially ordered sets has been considered recently in [2-15]. Tarski's theorem is used in [7] to show the existence of solutions for fuzzy equations and in [9] to prove existence theorems for fuzzy differential equations. In [5, 6, 8, 11, 14], some applications to matrix equations and to ordinary differential equations are presented. In [3, 6, 16], it is proved that some fixed theorems for a mixed monotone mapping in a metric space endowed with a partial order and the authors apply their results to problems of existence and uniqueness of solutions for some boundary value problems.

In the context of partially ordered metric spaces, the usual contractive condition is weakened but at the expense that the operator is monotone. The main idea in [8, 14] involves combining the ideas in the contraction principle with those in the monotone iterative technique [16].

2. Main Result

Definition 2.1.

Let (X,≤) be a partially ordered set and T:X[arrow right]X . We say that T is a nondecreasing mapping if for x,y∈X , x≤y[implies]Tx≤Ty .

In the sequel, we prove the following theorem which is a version of Theorem 1.1 in the context of partially ordered metric spaces.

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. Let T:X[arrow right]X be a continuous and nondecreasing mapping such that [figure omitted; refer to PDF] with α+β<1 . If there exists _x0 ∈X with _x0 ≤T_x0 , then T has a fixed point.

Proof.

If T_x0 =_x0 , then the proof is finished. Suppose that _x0 <T_x0 . Since T is a nondecreasing mapping, we obtain by induction that [figure omitted; refer to PDF] Put _xn+1 =T_xn . If there exists n≥1 such that _xn+1 =_xn , then from _xn+1 =T_xn =_xn , _xn is a fixed point and the proof is finished. Suppose that _xn+1 ≠_xn for n≥1 .

Then, from (2.1) and as the elements _xn and _xn-1 are comparable, we get, for n≥1 , [figure omitted; refer to PDF] The last inequality gives us [figure omitted; refer to PDF] Again, using induction [figure omitted; refer to PDF] Put k=β/(1-α)<1 .

Moreover, by the triangular inequality, we have, for m≥n , [figure omitted; refer to PDF] and this proves that d(_xm ,_xn )[arrow right]0 as m,n[arrow right]∞ .

So, {_xn } is a Cauchy sequence and, since X is a complete metric space, there exists z∈X such that _{lim n[arrow right]∞xn} =z .

Further, the continuity of T implies [figure omitted; refer to PDF] and this proves that z is a fixed point.

This finishes the proof.

In what follows, we prove that Theorem 2.2 is still valid for T , not necessarily continuous, assuming the following hypothesis in X : [figure omitted; refer to PDF]

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 (2.8). Let T:X[arrow right]X be a nondecreasing mapping such that [figure omitted; refer to PDF] with α+β<1 . If there exists _x0 ∈X with _x0 ≤T_x0 , then T has a fixed point.

Proof.

Following the proof of Theorem 2.2, we only have to check that Tz=z .

As {_xn } is a nondecreasing sequence in X and _xn [arrow right]z , then, by (2.8), z=sup {_xn } . Particularly, _xn ≤z for all n∈... .

Since T is a nondecreasing mapping, then T_xn ≤Tz , for all n∈... or, equivalently, _xn+1 ≤Tz for all n∈... . Moreover, as _x0 <_x1 ≤Tz and z=sup {_xn } , we get z≤Tz .

Suppose that z<Tz . Using a similar argument that in the proof of Theorem 2.2 for _x0 ≤T_x0 , we obtain that {^Tn z} is a nondecreasing sequence and _{lim n[arrow right]∞}^Tn z=y for certain y∈X .

Again, using (2.8), we have that y=sup {^Tn z} .

Moreover, from _x0 ≤z , we get _xn =^Tn_x0 ≤^Tn z for n≥1 and _xn <^Tn z for n≥1 because _xn ≤z<Tz≤^Tn z for n≥1 .

As _xn and ^Tn z are comparable and distinct for n≥1 , applying the contractive condition we get [figure omitted; refer to PDF] Making n[arrow right]∞ in the last inequality, we obtain [figure omitted; refer to PDF] As β<1 , d(z,y)=0 , thus, z=y .

Particularly, z=y=sup {^Tn z} and, consequently, Tz≤z and this is a contradiction.

Hence, we conclude that z=Tz .

Now, we present an example where it can be appreciated that hypotheses in Theorem 2.2 do not guarantee uniqueness of the fixed point. This example appears in [8].

Let X={(1,0),(0,1)}⊂^...2 and consider the usual order [figure omitted; refer to PDF]

Thus, (X,≤) is a partially ordered set whose different elements are not comparable. Besides, (X,_d2 ) is a complete metric space considering, _d2 , the Euclidean distance. The identity map T(x,y)=(x,y) is trivially continuous and nondecreasing and assumption (2.1) of Theorem 2.2 is satisfied since elements in X are only comparable to themselves. Moreover, (1,0)≤T(1,0) and T has two fixed points in X .

In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition appears in [14] and [figure omitted; refer to PDF]

In [8], it is proved that the above-mentioned condition is equivalent, [figure omitted; refer to PDF]

Theorem 2.4.

Adding condition (2.14) to the hypotheses of Theorem 2.2 (or Theorem 2.3) one obtains uniqueness of the fixed point of T .

Proof.

Suppose that there exists z,y∈X which are fixed point.

We distinguish two cases.

Case 1.

If y and z are comparable and y≠z , then using the contractive condition we have [figure omitted; refer to PDF] As β<1 is the last inequality, it is a contradiction. Thus, y=z .

Case 2.

If y is not comparable to z , then by (2.14) there exists x∈X comparable to y and z . Monotonicity implies that ^Tn x is comparable to ^Tn y=y and ^Tn z=z for n=0,1,2,... .

If there exists _n0 ≥1 such that ^T_n0 x=y , then as y is a fixed point, the sequence {^Tn x:n≥_n0 } is constant, and, consequently, _{lim n[arrow right]∞}^Tn x=y .

On the other hand, if ^Tn x≠y for n≥1 , using the contractive condition, we obtain, for n≥2 , [figure omitted; refer to PDF] Using induction, [figure omitted; refer to PDF] and as β<1 , the last inequality gives us _{lim n[arrow right]∞}^Tn x=y .

Hence, we conclude that _{lim n[arrow right]∞}^Tn x=y .

Using a similar argument, we can prove that _{lim n[arrow right]∞}^Tn x=z .

Now, the uniqueness of the limit gives us y=z .

This finishes the proof.

Remark 2.5.

It is easily proved that the space C[0,1]={x:[0,1][arrow right]...,continuous} with the partial order given by [figure omitted; refer to PDF] and the metric given by [figure omitted; refer to PDF] satisfies condition (2.8). Moreover, as for x,y∈C[0,1] , the function max (x,y)(t)=max {x(t),y(t)} is continuous, (C[0,1],≤) satisfies also condition (2.14).

3. Some Remarks

In this section, we present some remarks.

Remark 3.1.

In [8], instead of condition (2.8), the authors use the following weaker condition: [figure omitted; refer to PDF] We have not been able to prove Theorem 2.3 using (3.1).

Remark 3.2.

If, in Theorems 2.2, 2.3, and 2.4, α=0 , then we obtain Theorems 2.1 , 2.2 , and 2.3 of [8].

If in the theorems of Section 2, β=0 , we obtain the following fixed point theorem in partially ordered complete metric spaces.

Theorem 3.3.

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. Let T:X[arrow right]X be a nondecreasing mapping such that there exists α∈[0,1) satisfying [figure omitted; refer to PDF] Suppose also that either T is continuous or X satisfies condition (2.8). If there exists _x0 ∈X with _x0 ≤T_x0 , then T has a fixed point.

Besides, if (X,≤) satisfies condition (2.14), then one obtains uniqueness of the fixed point.

Finally, we present an example where Theorem 2.2 can be applied and this example cannot be treated by Theorem 1.1.

Example 3.4.

Let X={(0,1),(1,0),(1,1)} and consider in X the partial order given by R={(x,x):x∈X} . Notice that elements in X are only comparable to themselves. Besides, (X,_d2 ) is a complete metric space considering _d2 the Euclidean distance. Let T:X[arrow right]X be defined by [figure omitted; refer to PDF] T is trivially continuous and nondecreasing, and assumption (2.1) of Theorem 2.2 is satisfied since elements in X are only comparable to themselves. Moreover, (1,1)≤T(1,1)=(1,1) and, by Theorem 2.2, T has a fixed point (obviously, this fixed point is (1,1) ).

On the other hand, for x=(0,1), y=(1,0)∈X , we have [figure omitted; refer to PDF] and the contractive condition of Theorem 1.1 is not satisfied because [figure omitted; refer to PDF] and thus α+β≥1 .

Consequently, this example cannot treated by Theorem 1.1.

Moreover, notice that in this example we have uniqueness of fixed point and (X,≤) does not satisfy condition (2.14). This proves that condition (2.14) is not a necessary condition for the uniqueness of the fixed point.

Acknowledgment

This research was partially supported by "Ministerio de Educación y Ciencia", Project MTM 2007/65706.

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