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.

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: 𝑑(𝑇𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦)𝑑(𝑥,𝑦)+𝛽⋅𝑑(𝑥,𝑦),(1.1) 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→𝑋. We say that 𝑇 is a nondecreasing mapping if for 𝑥,𝑦∈𝑋, 𝑥≤𝑦⇒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→𝑋 be a continuous and nondecreasing mapping such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝛼⋅𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦)𝑑(𝑥,𝑦)+𝛽⋅𝑑(𝑥,𝑦),for𝑥,𝑦∈𝑋,𝑥≥𝑦,𝑥≠𝑦,(2.1) with 𝛼+𝛽<1. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. If 𝑇𝑥₀=𝑥₀, then the proof is finished. Suppose that 𝑥₀<𝑇𝑥₀. Since 𝑇 is a nondecreasing mapping, we obtain by induction that 𝑥₀<𝑇𝑥₀≤𝑇²𝑥₀≤⋯≤𝑇^𝑛𝑥₀≤𝑇^𝑛+1𝑥₀≤⋯(2.2) Put 𝑥_𝑛+1=𝑇𝑥_𝑛. If there exists 𝑛≥1 such that 𝑥_𝑛+1=𝑥_𝑛, then from 𝑥_𝑛+1=𝑇𝑥_𝑛=𝑥_𝑛, 𝑥_𝑛 is a fixed point and the proof is finished. Suppose that 𝑥_𝑛+1≠𝑥_𝑛 for 𝑛≥1.
Then, from (2.1) and as the elements 𝑥_𝑛 and 𝑥_𝑛−1 are comparable, we get, for 𝑛≥1, 𝑑(𝑥_𝑛+1,𝑥_𝑛)=𝑑(𝑇𝑥_𝑛,𝑇𝑥_𝑛−1)≤𝛼𝑑(𝑥_𝑛,𝑇𝑥_𝑛)⋅𝑑(𝑥_𝑛−1,𝑇𝑥_𝑛−1)𝑑(𝑥_𝑛,𝑥_𝑛−1)+𝛽𝑑(𝑥_𝑛,𝑥_𝑛−1)=𝛼𝑑(𝑥_𝑛,𝑥_𝑛+1)⋅𝑑(𝑥_𝑛−1,𝑥_𝑛)𝑑(𝑥_𝑛,𝑥_𝑛−1)+𝛽𝑑(𝑥_𝑛,𝑥_𝑛−1)=𝛼⋅𝑑(𝑥_𝑛,𝑥_𝑛+1)+𝛽⋅𝑑(𝑥_𝑛,𝑥_𝑛−1).(2.3) The last inequality gives us 𝑑(𝑥_𝑛+1,𝑥_𝑛𝛽)≤1−𝛼𝑑(𝑥_𝑛,𝑥_𝑛−1).(2.4) Again, using induction 𝑑(𝑥_𝑛+1,𝑥_𝑛𝛽)≤()1−𝛼^𝑛𝑑(𝑥₁,𝑥₀),(2.5) Put 𝑘=𝛽/(1−𝛼)<1.
Moreover, by the triangular inequality, we have, for 𝑚≥𝑛, 𝑑(𝑥_𝑚,𝑥_𝑛)≤𝑑(𝑥_𝑚,𝑥_𝑚−1)+𝑑(𝑥_𝑚−1,𝑥_𝑚−2)+⋯+𝑑(𝑥_𝑛+1,𝑥_𝑛)≤(𝑘^𝑚−1+𝑘^𝑚−2+⋯+𝑘^𝑛)𝑑(𝑥₁,𝑥₀𝑘)≤(^𝑛1−𝑘)𝑑(𝑥₁,𝑥₀),(2.6) and this proves that 𝑑(𝑥_𝑚,𝑥_𝑛)→0 as 𝑚,𝑛→∞.
So, {𝑥_𝑛} is a Cauchy sequence and, since 𝑋 is a complete metric space, there exists 𝑧∈𝑋 such that lim_𝑛→∞𝑥_𝑛=𝑧.
Further, the continuity of 𝑇 implies 𝑇𝑧=𝑇(lim_𝑛→∞𝑥_𝑛)=lim_𝑛→∞𝑇𝑥_𝑛=lim_𝑛→∞𝑥_𝑛+1=𝑧,(2.7) 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:if(𝑥_𝑛)isanondecreasingsequencein𝑋suchthat𝑥_𝑛→𝑥,then𝑥=sup{𝑥_𝑛}.(2.8)

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→𝑋 be a nondecreasing mapping such that 𝑑(𝑇𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦)𝑑(𝑥,𝑦)+𝛽𝑑(𝑥,𝑦),for𝑥,𝑦∈𝑋,𝑥≥𝑦,𝑥≠𝑦,(2.9) with 𝛼+𝛽<1. If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. Following the proof of Theorem 2.2, we only have to check that 𝑇𝑧=𝑧.
As {𝑥_𝑛} is a nondecreasing sequence in 𝑋 and 𝑥_𝑛→𝑧, then, by (2.8), 𝑧=sup{𝑥_𝑛}. Particularly, 𝑥_𝑛≤𝑧 for all 𝑛∈ℕ.
Since 𝑇 is a nondecreasing mapping, then 𝑇𝑥_𝑛≤𝑇𝑧, for all 𝑛∈ℕ or, equivalently, 𝑥_𝑛+1≤𝑇𝑧 for all 𝑛∈ℕ. Moreover, as 𝑥₀<𝑥₁≤𝑇𝑧 and 𝑧=sup{𝑥_𝑛}, we get 𝑧≤𝑇𝑧.
Suppose that 𝑧<𝑇𝑧. Using a similar argument that in the proof of Theorem 2.2 for 𝑥₀≤𝑇𝑥₀, we obtain that {𝑇^𝑛𝑧} is a nondecreasing sequence and lim_𝑛→∞𝑇^𝑛𝑧=𝑦 for certain 𝑦∈𝑋.
Again, using (2.8), we have that 𝑦=sup{𝑇^𝑛𝑧}.
Moreover, from 𝑥₀≤𝑧, we get 𝑥_𝑛=𝑇^𝑛𝑥₀≤𝑇^𝑛𝑧 for 𝑛≥1 and 𝑥_𝑛<𝑇^𝑛𝑧 for 𝑛≥1 because 𝑥_𝑛≤𝑧<𝑇𝑧≤𝑇^𝑛𝑧 for 𝑛≥1.
As 𝑥_𝑛 and 𝑇^𝑛𝑧 are comparable and distinct for 𝑛≥1, applying the contractive condition we get 𝑑(𝑥_𝑛+1,𝑇^𝑛+1𝑧)=𝑑(𝑇𝑥_𝑛,𝑇(𝑇^𝑛𝑧))≤𝛼𝑑(𝑥_𝑛,𝑇𝑥_𝑛)⋅𝑑(𝑇^𝑛𝑧,𝑇^𝑛+1𝑧)𝑑(𝑥_𝑛,𝑇^𝑛𝑧)+𝛽𝑑(𝑥_𝑛,𝑇^𝑛𝑧)=𝛼𝑑(𝑥_𝑛,𝑥_𝑛+1)⋅𝑑(𝑇^𝑛𝑧,𝑇^𝑛+1𝑧)𝑑(𝑥_𝑛,𝑇^𝑛𝑧)+𝛽𝑑(𝑥_𝑛,𝑇^𝑛𝑧).(2.10) Making 𝑛→∞ in the last inequality, we obtain 𝑑(𝑧,𝑦)≤𝛽𝑑(𝑧,𝑦).(2.11) As 𝛽<1, 𝑑(𝑧,𝑦)=0, thus, 𝑧=𝑦.
Particularly, 𝑧=𝑦=sup{𝑇^𝑛𝑧} 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)}⊂ℝ² and consider the usual order (𝑥,𝑦)≤(𝑧,𝑡)⇔𝑥≤𝑧,𝑦≤𝑡.(2.12)

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 for𝑥,𝑦∈𝑋,thereexistsalowerboundoranupperbound.(2.13)

In [8], it is proved that the above-mentioned condition is equivalent,for𝑥,𝑦∈𝑋,thereexists𝑧∈𝑋whichiscomparableto𝑥and𝑦.(2.14)

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 𝑑(𝑦,𝑧)=𝑑(𝑇𝑦,𝑇𝑧)≤𝛼𝑑(𝑦,𝑇𝑦)⋅𝑑(𝑧,𝑇𝑧)𝑑(𝑦,𝑧)+𝛽𝑑(𝑦,𝑧)=𝛼𝑑(𝑦,𝑦)⋅𝑑(𝑧,𝑧)𝑑(𝑦,𝑧)+𝛽𝑑(𝑦,𝑧)=𝛽𝑑(𝑦,𝑧).(2.15) 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 𝑇^𝑛𝑥 is comparable to 𝑇^𝑛𝑦=𝑦 and 𝑇^𝑛𝑧=𝑧 for 𝑛=0,1,2,….
If there exists 𝑛₀≥1 such that 𝑇^𝑛0𝑥=𝑦, then as 𝑦 is a fixed point, the sequence {𝑇^𝑛𝑥∶𝑛≥𝑛₀} is constant, and, consequently, lim_𝑛→∞𝑇^𝑛𝑥=𝑦.
On the other hand, if 𝑇^𝑛𝑥≠𝑦 for 𝑛≥1, using the contractive condition, we obtain, for 𝑛≥2, 𝑑(𝑇^𝑛𝑥,𝑦)=𝑑(𝑇^𝑛𝑥,𝑇^𝑛𝑦)≤𝛼𝑑(𝑇^𝑛−1𝑥,𝑇^𝑛𝑥)⋅𝑑(𝑇^𝑛−1𝑦,𝑇^𝑛𝑦)𝑑(𝑇^𝑛−1𝑥,𝑇^𝑛−1𝑦)+𝛽𝑑(𝑇^𝑛−1𝑥,𝑇^𝑛−1𝑦)≤𝛼𝑑(𝑇^𝑛−1𝑥,𝑇^𝑛𝑥)⋅𝑑(𝑦,𝑦)𝑑(𝑇^𝑛−1𝑥,𝑦)+𝛽𝑑(𝑇^𝑛−1𝑥,𝑦)=𝛽𝑑(𝑇^𝑛−1𝑥,𝑦).(2.16) Using induction, 𝑑(𝑇^𝑛𝑥,𝑦)≤𝛽^𝑛⋅𝑑(𝑥,𝑦),for𝑛≥2,(2.17) and as 𝛽<1, the last inequality gives us lim_𝑛→∞𝑇^𝑛𝑥=𝑦.
Hence, we conclude that lim_𝑛→∞𝑇^𝑛𝑥=𝑦.
Using a similar argument, we can prove that lim_𝑛→∞𝑇^𝑛𝑥=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]→ℝ,continuous} with the partial order given by 𝑥≤𝑦⇔𝑥(𝑡)≤𝑦(𝑡),for𝑡∈[0,1],(2.18) and the metric given by 𝑑(𝑥,𝑦)=sup{|𝑥(𝑡)−𝑦(𝑡)|∶𝑡∈[0,1]}(2.19) 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: if(𝑥_𝑛)isanondecreasingsequencein𝑋suchthat𝑥_𝑛→𝑥,then𝑥_𝑛≤𝑥∀𝑛∈ℕ.(3.1) 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→𝑋 be a nondecreasing mapping such that there exists 𝛼∈[0,1) satisfying 𝑑(𝑇𝑥,𝑇𝑦)≤𝛼𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦)𝑑(𝑥,𝑦),for𝑥,𝑦∈𝑋with𝑥≥𝑦,𝑥≠𝑦.(3.2) Suppose also that either 𝑇 is continuous or 𝑋 satisfies condition (2.8). If there exists 𝑥₀∈𝑋 with 𝑥₀≤𝑇𝑥₀, 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→𝑋 be defined by 𝑇(0,1)=(1,0),𝑇(1,0)=(0,1),𝑇(1,1)=(1,1).(3.3)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 𝑑(𝑇𝑥,𝑇𝑦)=√2,𝑑(𝑥,𝑇𝑥)=√2,𝑑(𝑦,𝑇𝑦)=√2,𝑑(𝑥,𝑦)=√2,(3.4) and the contractive condition of Theorem 1.1 is not satisfied because 𝑑(𝑇𝑥,𝑇𝑦)=√2≤𝛼𝑑(𝑥,𝑇𝑥)⋅𝑑(𝑦,𝑇𝑦)√𝑑(𝑥,𝑦)+𝛽𝑑(𝑥,𝑦)=𝛼2⋅√2√2+𝛽√2=(𝛼+𝛽)√2,(3.5) 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.