## Abstract

We present some fixed point results for nondecreasing and weakly increasing operators in a partially ordered metric space using implicit relations. Also we give an existence theorem for common solution of two integral equations.

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# Some Fixed Point Theorems on Ordered Metric Spaces and Application

## Abstract

## 1. Introduction

## 2. Implicit Relation

## 3. Fixed Point Theorem for Nondecreasing Mappings

## 4. Fixed Point Theorem for Weakly Increasing Mappings

## 5. Application

## References

## Acknowledgment

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*Fixed Point Theory and Applications*
**volume 2010**, Article number: 621469 (2010)

We present some fixed point results for nondecreasing and weakly increasing operators in a partially ordered metric space using implicit relations. Also we give an existence theorem for common solution of two integral equations.

Existence of fixed points in partially ordered sets has been considered recently in [1], and some generalizations of the result of [1] are given in [2–6]. Also, in [1] some applications to matrix equations are presented, in [3, 4] some applications to periodic boundary value problem and to some particular problems are, respectively, given. Later, in [6] O'Regan and Petruşel gave some existence results for Fredholm and Volterra type integral equations. In some of the above works, the fixed point results are given for nondecreasing mappings.

We can order the purposes of the paper as follows.

First, we give a slight generalization of some of the results of the above papers using an implicit relation in the following way.

In [1, 3], the authors used the following contractive condition in their result, there exists such that

(1.1)

Afterwards, in [2], the authors used the nonlinear contractive condition, that is,

(1.2)

where is anondecreasing function with for , instead of (1.1). Also in [2], the authors proved a fixed point theorem using generalized nonlinear contractive condition, that is,

(1.3)

for , where is as above. In the Section 3, we generalized the above contractive conditions using the implicit relation technique in such a way that

(1.4)

for , where is a function as given in Section 2. We can obtain various contractive conditions from (1.4). For example, if we choose

(1.5)

in (1.4), then, we have (1.3). Similarly we can have the contractive conditions in [7–9] from (1.4).

In some of the above mentioned theorems, the fixed point results are given for nondecreasing mappings. Also in these theorems the following condition is used:

(1.6)

In Section 4, we give some examples such that two weakly increasing mappings need not be nondecreasing. Therefore, we give a common fixed point theorem for two weakly increasing operators in partially ordered metric spaces using implicit relation technique. Also we did not use the condition (1.6) in this theorem. At the end, to see the applicability of our result, we give an existence theorem for common solution of two integral equations using a result of the Section 4.

Implicit relations on metric spaces have been used in many articles. See for examples, [10–15].

Let denote the nonnegative real numbers, and let be the set of all continuous functions satisfying the following conditions:

is nonincreasing in variables ;

there exists a right continuous function , for such that for ,

(2.1)

or

(2.2)

implies ;

, for all .

Example 2.1.

where , , .

Let and . If then which implies , a contradiction. Thus and . Similarly, let and then If , then Thus is satisfied with . Also , for all . Therefore, .

Example 2.2.

, where .

Let and . If then which is a contradiction. Thus and Similarly, let and then we have If then Thus is satisfied with Also for all Therefore, .

Example 2.3.

where is right continuous and , for

Let and If then which is a contradiction. Thus and Similarly, let and then we have If , then Thus is satisfied with Also , for all Therefore, .

Example 2.4.

, where , , and

Let and Then Similarly, let and then we have If then Thus is satisfied with . Also , for all . Therefore, .

We need the following lemma for the proof of our theorems.

Lemma 3.1 (See [16]).

Let be a right continuous function such that for every , then , where denotes the -times repeated composition of with itself.

Theorem 3.2.

Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Suppose is a nondecreasing mapping such that for all with ,

(3.1)

where . Also

(3.2)

or

(3.3)

hold. If there exists an with , then has a fixed point.

Proof.

If , then the proof is finished; so suppose . Now let for . Notice that, since and is nondecreasing, we have

(3.4)

Now since , we can use the inequality (3.1) for these points, then we have

(3.5)

and so

(3.6)

Now using , we have

(3.7)

and from there exists a right continuous function, , , for , such that for all ,

(3.8)

If we continue this procedure, we can have

(3.9)

and so from Lemma 3.1,

(3.10)

Next we show that is a Cauchy sequence. Suppose it is not true. Then we can find a and two sequence of integers with

(3.11)

We may also assume

(3.12)

by choosing to be the smallest number exceeding for which (3.11) holds. Now (3.9), (3.11), and (3.12) imply

(3.13)

and so

(3.14)

Also since

(3.15)

we have from (3.9) that

(3.16)

On the other hand, since , we can use the condition (3.1) for these points. Therefore, we have

(3.17)

and so

(3.18)

Now letting and using (3.14), we have, by continuity of that

(3.19)

From , we have . Therefore, letting in (3.16), we have This is a contradiction since for Thus is a Cauchy sequence in so there exists an with .

If (3.2) holds, then clearly . Now suppose (3.3) holds. Suppose . Now since , then from (3.3), for all . Using the inequality (3.1), we have

(3.20)

so letting from the last inequality, we have

(3.21)

which is a contradiction to . Thus and so .

Remark 3.3.

Note that if we take that

there exists a nondecreasing function with for each such that for ,

(3.22)

or

(3.23)

implies ,

Instead of in Theorem 3.2, again we can have the same result.

If we combine Theorem 3.2 with Example 2.1, we obtain the following result.

Corollary 3.4.

Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Suppose is a nondecreasing mapping such that for all with ,

(3.24)

where , , . Also

(3.25)

or

(3.26)

hold. If there exists an with , then has a fixed point.

Remark 3.5.

Theorem of [2] follows from Example 2.3, Remark 3.3, and Theorem 3.2.

Remark 3.6.

We can have some new results from other examples and Theorem 3.2.

Remark 3.7.

In Theorem [1], it is proved that if

(3.27)

then for every ,

(3.28)

where is the fixed point of such that

(3.29)

and hence has a unique fixed point. If condition (3.27) fails, it is possible to find examples of functions with more than one fixed point. There exist some examples to illustrate this fact in [3].

Now we give a fixed point theorem for two weakly increasing mappings in ordered metric spaces using an implicit relation. Before this, we will define an implicit relation for the contractive condition of the theorem.

Let be the set of all continuous functions satisfying and the following conditions:

there exists a right continuous function , for such that for ,

(4.1)

or

(4.2)

or

(4.3)

implies ;

and , for all .

We can easily show that, all functions in the Examples in Section 2 are in .

Definition 4.1 (See [17, 18]).

Let be a partially ordered set. Two mappings are said to be weakly increasing if and for all .

Note that, two weakly increasing mappings need not be nondecreasing.

Example 4.2.

Let endowed with usual ordering. Let defined by

(4.4)

then it is obvious that and for all . Thus and are weakly increasing mappings. Note that both and are not nondecreasing.

Example 4.3.

Let be endowed with the coordinate ordering, that is, and . Let be defined by and , then and . Thus and are weakly increasing mappings.

Example 4.4.

Let be endowed with the lexicographical ordering, that is, or if then . Let be defined by

(4.5)

then

(4.6)

Thus and are weakly increasing mappings. Note that but , then is not nondecreasing. Similarly, is not nondecreasing.

Theorem 4.5.

Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Suppose are two weakly increasing mappings such that for all comparable ,

(4.7)

where . Also

(4.8)

or

(4.9)

or

(4.10)

hold, then and have a common fixed point.

Remark 4.6.

Note that, in this theorem we remove the condition "there exists an with '' of Theorem 3.2. Again we can consider the result of Remark 3.7 for this theorem.

Proof of Theorem 4.5.

First of all we show that if or has a fixed point, then it is a common fixed point of and . Indeed, let be a fixed point of . Now assume . If we use the inequality (4.7), for , we have

(4.11)

which is a contradiction to . Thus and so is a common fixed point of and . Similarly, if is a fixed point of , then it is also a fixed point of . Now let be an arbitrary point of . If , the proof is finished, so assume . We can define a sequence in as follows:

(4.12)

Without loss of generality, we can suppose that the successive terms of are different. Otherwise, we are again finished. Note that since and are weakly increasing, we have

(4.13)

and continuing this process, we have

(4.14)

Now since and are comparable then, we can use the inequality (4.7) for these points then we have

(4.15)

and so

(4.16)

Now using , we have

(4.17)

and form there exists a right continuous function for , we have for all

(4.18)

Similarly, since and are comparable, then we can use the inequality (4.7) for these points then we have

(4.19)

and so

(4.20)

Now again using , we have

(4.21)

and form , we have for all ,

(4.22)

Therefore, from (4.18) and (4.22), we can have, for all

(4.23)

and so

(4.24)

Thus from Lemma 3.1, we have, since ,

(4.25)

Next we show that is a Cauchy sequence. For this it is sufficient to show that is a Cauchy sequence. Suppose it is not true. Then we can find an such that for each even integer , there exist even integers such that

(4.26)

We may also assumethat

(4.27)

by choosing to be the smallest number exceeding for which (4.26) holds. Now (4.24), (4.26), and (4.27) imply

(4.28)

and so

(4.29)

Also, by the triangular inequality,

(4.30)

Therefore, we get

(4.31)

Also we have

(4.32)

On the other hand, since and are comparable, we can use the condition (4.7) for these points. Therefore, we have

(4.33)

and so

(4.34)

Now, considering (4.29) and (4.31) and letting in the last inequality, we have, by continuity of , that

(4.35)

From , we have . Therefore, letting in (4.32), we have . This is a contradiction since for . Thus is a Cauchy sequence in , so is a Cauchy sequence. Therefore, there exists an with .

If (4.8) or (4.9) hold then clearly . Now suppose (4.10) holds. Suppose . Now since , then from (4.10), for all . Using the inequality (4.7), we have

(4.36)

So letting from the last inequality, we have

(4.37)

which is a contradiction to . Thus and so .

Remark 4.7.

We can have some new results from Theorem 4.5 with some examples for .

For example, we can have the following corollary.

Corollary 4.8.

Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Suppose are two weakly increasing mappings such that for all comparable ,

(4.38)

where is a right continuous function such that , for . Also

(4.39)

or

(4.40)

hold, then and have a common fixed point.

Proof.

Let , then it is obvious that . Therefore, the proof is complete from Theorem 4.5.

Consider the integral equations

(5.1)

The purpose of this section is to give an existence theorem for common solution of (5.1) using Corollary 4.8. This section is related to those [19–22].

Let be a partial order relation on .

Theorem 5.1.

Consider the integral equations (5.1).

(i) and are continuous;

(ii)for each ,

(5.2)

(iii)there exist a continuous function and a right continuous and nondecreasing function such that and for , such that

(5.3)

for each and comparable ;

(iv).

Then the integral equations (5.1) have a unique common solution in .

Proof.

Let with the usual supremum norm, that is, , for . Consider on the partial order defined by

(5.4)

Then is a partially ordered set. Also is a complete metric space. Moreover, for any increasing sequence in converging to , we have for any . Also for every , there exists which is comparable to and [6].

Define , by

(5.5)

Now from (ii), we have, for all

(5.6)

Thus, we have and for all . This shows that and are weakly increasing. Also for each comparable , we have

(5.7)

Hence for each comparable . Therefore, all conditions of Corollary 4.8 are satisfied. Thus the conclusion follows.

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The authors thank the referees for their appreciation, valuable comments, and suggestions.

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Altun, I., Simsek, H. Some Fixed Point Theorems on Ordered Metric Spaces and Application.
*Fixed Point Theory Appl* **2010**, 621469 (2010). https://doi.org/10.1155/2010/621469

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DOI: https://doi.org/10.1155/2010/621469

- Continuous Function
- Integral Equation
- Point Theorem
- Existence Theorem
- Contractive Condition