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Volume 2020 |Article ID 3527430 | https://doi.org/10.1155/2020/3527430

Chenguang Wang, Jinxiu Mao, Zengqin Zhao, "A Fixed-Point Theorem for Ordered Contraction-Type Decreasing Operators in Banach Space with Lattice Structure", Journal of Function Spaces, vol. 2020, Article ID 3527430, 7 pages, 2020. https://doi.org/10.1155/2020/3527430

A Fixed-Point Theorem for Ordered Contraction-Type Decreasing Operators in Banach Space with Lattice Structure

Academic Editor: Chuanjun Chen
Received31 Mar 2020
Accepted01 May 2020
Published05 Jun 2020

Abstract

In this work, we mainly improve the results in Amini-Harandi and Emami (2010). By introducing a new kind of ordered contraction-type decreasing operator in Banach space, we obtain a unique fixed point by using the iterative algorithm. An example is also presented to illustrate the theorem.

1. Introduction

In this work, we obtain a unique fixed point for a kind of ordered contraction-type decreasing operator in Banach space by using the iterative algorithm. The fixed-point study is mainly focused on two aspects. On the one hand, it is about the research of contraction-type mapping, for example, in [18]. On the other hand, it is about the study of monotone operators with concavity and convexity, for example, in [915]. There is little research on operators that only satisfy the partial-order constrictions. Applications of operator theory in fractional differential equations can be seen in [1643].

The following generalization of Banach’s contraction principle is due to Geraghty [44].

Lemma 1. Let be a complete metric space and let be a map. Suppose there exists such that for each , where denotes the class of those functions which satisfy the condition . Then has a unique fixed-point , and converges to , for each .

Very recently, Amini-Harandi and Emami [6] proved the following existence theorem which is a version of Lemma 1 in the context of partially ordered complete metric spaces:

Lemma 2. Let () be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be an increasing map such that there exists an element with . Suppose that there exists such that Assume that either is continuous or is such that if an increasing sequence in , then .
Besides, if for each , there exists , which is comparable to and .
Then has a unique fixed point.

We found that in (2), the contraction is concerning a metric. But in fact, the relation of partial order does not play any role in (2). That is to say, in (1), the constriction is not effective because . A question appears naturally in the authors’ minds: “Can the contraction condition be merely about partial order so that can be effective?” The authors have been haunted by this question since it was found. Driven by this idea, we introduce a new kind of ordered contraction-type decreasing operator in Banach space with lattice structure and obtain a unique fixed point of the operator. Our results are helpful and meaningful for studies of fixed point. Comparing to [6], our improvements are in three aspects.

First, the contraction is merely about partial order, and the relation of partial-order does play an important role in the contraction condition. This has never been seen. Second, we consider the situation when the operator is decreasing. Third, we only use the iterative algorithm, and we can start the iterative process with any initial point, i.e., we do not need any assumptions of the existence of upper or lower solutions. An example is also presented to illustrate the theorem.

The outline of this paper is as follows. In the remainder of this section, we will give some preliminaries. In Section 2 of this paper, we present the existence and uniqueness theorem. In Section 3, an example is illustrated.

Definition 3 (see [45]). Let be a real Banach space. A nonempty convex closed set is called a cone if (i)(ii); is the zero element in

In the case that is a given cone in a real Banach space , a partial order “” can be induced on by . The cone is called normal if there exists a constant , such that for all implies that . Details about cones and fixed point of operators can be found in [45, 46].

Definition 4 (see [47, 48]). We call a set a lattice under the partial ordering , if sup and inf exist for arbitrary .

Lemma 5 (see [45]). A cone is normal if and only if there exists a norm in which is equivalent to such that for any is monotone. The equivalence of means that there exist such that .

Lemma 6 (see [45]). Let be a normal cone in a real Banach space . Suppose that is a monotone sequence which has a subsequence converging to , then also converges to . Moreover, if is an increasing sequence, then ; if is a decreasing sequence, then .

2. The Main Results

We suppose that is a partially ordered Banach space. is a normal cone. The partial-order “” on is induced by the cone .

Let denote the class of those functionals which satisfy the condition

Theorem 7 (main theorem). Suppose that is a closed subset, . is a lattice. is a decreasing operator and satisfies the following ordered contraction condition:
(H) Suppose that there exists such that Then has unique fixed-point . Moreover, constructing successively sequence for any initial , we have

Remark 8. Here, we study the decreasing operator while most of the contractions are about increasing operators. And the contraction condition (4) is merely about the partial order, while most of the contractions are about metric.

Remark 9. Two elements and in an ordered set are said to be comparable if either , and we denote it as .

Proof. Let , we have . So we have the following two cases.

Case 1. When is comparable with . Firstly, without loss of generality, we suppose that If , then the proof is finished. Suppose that . Since is decreasing, we obtain , and it is easy to prove that is increasing. Using the contractive condition (4), we have hence,

From (4), we have

Let

From (9) and (10), we have the following two conclusions: (a)There exists a functional such that for with (b)There exists such that We assert that the operator has unique fixed point in . And the unique fixed point of is also the unique fixed point of . In order to be clear, we divide the process of proof into three steps.

Step 1. We will use the method of iteration to construct a fixed point of . In fact, consider the iterative sequence

Since and the operator is increasing, we have

This means that is an increasing sequence. It follows from (12) that

Since is normal, from the equivalence of in Lemma 5 we have

Then, is a decreasing sequence and bounded as follows. So

Assume . Then, from (16), we have

The above inequality yields

And implies . Then, and .

Now we show that is a Cauchy sequence. On the contrary, assume that

For any fixed natural number . From (16), by the triangle inequality

Hence, we have .

Since and , then from which we obtain

But since , we get

This contradicts (21) and shows that is a Cauchy sequence in .

Since is closed, we can suppose that there exists a such that

Since is normal, (14) together with Lemma 5 implies that (26), together with (12) and the equivalence of and , implies that

So

Let , we obtain . So

This proves that is a fixed point of in and

Step 2. We will obtain the uniqueness of the fixed point of . On the contrary, if is another fixed point of , we will get .

In fact, the first case, whenis comparable with, without loss of generality, we suppose that. Since is increasing,

Moreover, by (12) and so

Consequently, the sequence is nonnegative and decreasing, and so . Now we show that

On the contrary, assume that . By passing to subsequences, if necessary, we may assume that exists. From (31), we obtain , and so . Since , then , and .

This contradiction proves .

It can be obtained that

The second case, when cannot compare with . Fromwhich is a lattice, we obtain satisfying i.e., is comparable with and is comparable with . Since is increasing, we know Moreover, by (12) So we have Similar to the process of (31), (32), and (33), From (36), we have So from (39) and (40), we get (33) together with (41) implies that is unique fixed point of .

Step 3. We will point that the unique fixed point of is also the unique fixed point of . Since

Thus, i.e., . From the uniqueness of the fixed point of , we know

So is the unique fixed point of in .

Case 2. Another case, when is not comparable to . From which is a lattice, we know there exists such that . That is, . Since is a decreasing operator, we have This shows that For any initial , constructing successively sequence From , we can get From (28), we know Since and from the arbitrary of in (28), we obtain (49) and (50) imply that holds. Similarly to the proof of Step 2 and Step 3 in case 1, we get that is the unique fixed point of .

3. An Example

Let , equipped with usual normal and usual partial order . . Then, . is a normal cone in . is a partially ordered Banach space. And is a lattice under the partial order induced by the cone .

Then, satisfying the assumptions of Theorem 7. Define the mapping by where is a fixed real number. Then, is nonincreasing. Define by , then . Now, for all with , we have so that and satisfy the assumption of Theorem 7. Observing that all the other conditions of Theorem 7 are also satisfied, has a unique fixed-point . Moreover, constructing successively sequence for any initial , we have

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

Supported by the Natural Science Foundation of China (11571197), Shandong Province Higher Educational Science and Technology Program (J17KB143), Shandong Education Science 13th Five-Year Plan Project (BYK2017003), and Jining University Youth Research Foundation (2015QNK102).

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