A Proposal for Revisiting Ćirić and Caristi Type Theorems in Metric Spaces

Abstract

In this paper, we revisit the renowned fixed point theorems of Ćirić and Caristi. We propose some new fixed point theorems in a metric space with partial order. To make our results effective, several examples are presented.

1. Introduction and Preliminary

This work is motivated by some recent works on the extension of Banach Contraction Principle to metric spaces with a partial order [1]. Caristi’s fixed point theorem is maybe one of the most useful extension of Banach Contraction Principle [2–4]. It has been successfully applied in many topics such as differential equations, convex minimization, operator theory, variational inequalities, and control theory. For known Caristi-type fixed point results in the literature, see [5–13]. Recall that this theorem states that any map has a fixed point provided that is complete and there exists a lower semicontinuous map such that , for every . The proofs given to Caristi’s result vary and use different techniques (see [14, 15]).

Using the combined Ćirić–Caristi condition, we introduce new fixed point theorems under hypotheses of the formwhere the “Dominated Function” can be chosen to befor certain functions and . Other corresponding forms under some advanced settings such as “partial order” are also discussed. To the best of our knowledge, we provide all the possible conditions to make the Caristi-type fixed point theorems appropriate and applicable in most situations.

More precisely, the renowned results [14, 16, 17] for a single-valued map are the following.

Theorem 1 (see Theorem 1 in [16]). Let be a complete metric space and be a mapping. Suppose that there exists such thatfor all , where

Then, has a unique fixed point in .

Definition 1. Let be a partially ordered complete metric space. We say that verifies the condition (OSC) if for any decreasing sequence in such that exists, then there exists and .

Theorem 2 (see Theorem 5 in [14]). Let be a partially ordered set and suppose that there exists a distance in such that is a complete metric space satisfying the (OSC) property. Let be a monotone increasing mapping. Assume there exists a lower semicontinuous function such that

Then, has a fixed point if and only there exists such that .

In this article, we prove new fixed point theorems of Caristi type and Ćirić type. All optional conditions for dominated functions are presented and discussed.

2. Main Results

Theorem 3. Let be a partially ordered complete metric space satisfying the (OSC) property and be a monotonically increasing map such that there exists a function satisfying

Then, for any such that , the sequence defined by converges to a fixed point of .

Proof. On the one hand, the case, where there exists , such that , gives as a fixed point of .
On the other hand, if for every , then, by induction, the sequence is monotonically decreasing, and taking and in (6), one obtains, for every ,Since the sequence is necessarily positive and monotonically decreasing, the series is convergent and, by the comparison principle, is also convergent. Since, for every , one hasSo, for every , one obtains the inequalityconcluding the convergence of the series . Hence, is a Cauchy sequence. Set . Since has the (OSC) property, we obtain for all . By (6), one hasand so . Therefore, .

Example 1. Let be endowed with the usual distance and the order “” be defined byLet and be two functions defined on bywhere is monotonically increasing on .
We need only to consider two cases to check the hypothesis of Theorem 3. Case 1: let with . We have So, Case 2: and . We have

Remark 1. (i)In Example 1, does not satisfy the Banach Contraction Principle, for which we take and , and we have .(ii)In Example 1, does not satisfy the inequality of Caristi, for which we take , and we have and . So, .

Corollary 1. Let be a partially ordered complete metric space satisfying the (OSC) property and be a monotonically increasing map such that there exists a function satisfyingfor some . Then, for any such that , the sequence , defined by , converges to a fixed point of .

Proof. If we consider the function defined on by , we obtain . By applying Theorem 3, we get that admits a fixed point.

Example 2. Let endowed with the usual distance and the order “” defined byLet and be two functions defined on bywhere has the condition (OSC) and is monotonically increasing on .
We need only to consider five cases to check the hypothesis of Corollary 1: Case 1: with . We have and So, Case 2: and . We have and , so Case 3: and . We have and so Hence, Case 4: and . We have , then . So, Case 5: with . We have , then . So,This shows that inequality (16) is satisfied and all the hypotheses of Corollary 1 are verified. In addition, has the fixed points and every .

Remark 2. (i)Example 2 does not satisfy the inequality for . If we take the third case and , we have .(ii)Example 2 does not verify the inequality of Theorem 4 in [11]. If we take and , we have but .(iii)Note that , so the mapping is not nonexpansive.

Definition 2. Let be a partially ordered complete metric space and be a mapping. We say that satisfies the condition if, for any monotone decreasing sequence in such that there exists and for all , we have .

Theorem 4. Let be a partially ordered complete metric space and a monotonically increasing map such that there exists a function satisfyingwhere

Assume that there exists a point satisfying . If we further add one of the following hypotheses, (C1) is continuous (C2) The map is lower semicontinuous (C3) has the condition , has the condition , and is lower semicontinuous then has at least one fixed point.

Proof. We define the sequence by for each . On the one hand, the case where there exists , such that , gives as a fixed point of . On the other hand, if for every , then, by induction, the sequence is monotonically decreasing, and taking and in (3), one obtains, for every ,where . So, the sequence is necessarily positive and monotonically decreasing, and therefore the series is convergent. Thus, is also convergent. For every , one hasTherefore, for every , one obtains the inequalityobtaining the convergence of the series . Hence, is a Cauchy sequence. Set .
(C1) and (C2) The case where is continuous is obvious. If the function is lower semicontinuous, we obtainThus, , i.e., .
(C3) Suppose has the condition , so there exists and is equal to . We have for all . Since is strictly increasing, , thus .
In what follows, we suppose that (because otherwise admits as a fixed point).
We put . The set is nonempty because . Let , so . Since is monotonically increasing, , thus .
Taking and in inequality (26), we haveLetting tend to , we obtainBy (33), we obtainWe define the partial order “” on bywhere “” is the usual order on . It is interesting to see that is monotonically increasing from to .
Assume that is a set of totally ordered subset of such that, for any , , or , where is the index set. Let . For any there exists such that and . We may assume that , so , that is, and are comparable. Hence, is totally ordered subset of , and by Zorn’s lemma, has a maximal totally ordered subset.
Let be a maximal totally ordered subset of . We consider and a sequence such that is decreasing and convergent to . By monotony of on , the sequence is also decreasing in . Therefore, for every integers , we have , which implies that is a Cauchy’s sequence and there exists a unique such that . So, , that is, for all . Moreover, by the lower semicontinuity of , we obtain .
Next, we show that for all . For that we distinguish two cases: Case 1: there exists satisfying , so we have , for any . It follows that and for all . Letting tend to infinity, we have , which means that for all . Case 2: let , , so there exists such that whenever . Hence, and for all . Letting tends to infinity, we obtain , and so .On the basis of the above discussion, we can claim that for all . Hence, is lower bound of in with respect to the order “.”
We have for every . Then, by inequality (26),and soSince has the condition , and when tends to infinity, we obtainWe claim that . Indeed, if , one hasSince is monotonically increasing, we have . Particularly, for all , which implies that . Thus, and, by the fact that , we get and for all . Since , we get that and is a totally ordered subset of . This contradicts the maximality of and finishes the proof.

Example 3. Let endowed with the usual distance and the usual order “.” We put, for all , and . Let and be the two functions defined on bywhere is a complete metric space and verifies the condition . is monotone increasing on and has the condition (actually, is continuous), is lower semicontinuous on , and .
We need only to consider two cases to check the hypothesis of Theorem 4: Case 1: let such that and . We have and , i.e., . So, Thus, Case 2: and , . We have . So,This shows that inequality (26) is satisfied and all the hypotheses hold (including condition (C3)). In addition, has the fixed point .
Note that , then does not satisfy the inequality of Caristi.

Corollary 2. Let be a partially ordered complete metric space and be a monotone increasing map such that there exists a function satisfyingwhere

Assume that there exists a point satisfying . If we further add one of the hypotheses , , or of Theorem 4, then has at least one fixed point.

Corollary 3. Let be a partially ordered complete metric space and let be a monotone increasing map such that there exists a function satisfyingfor some , where

Assume that there exists a point satisfying . If we further add one of the hypotheses (C1), (C2), or (C3) of Theorem 4, replacing the lower semicontinuity of in (C3) with the lower semicontinuity of , for each , then has at least one fixed point.

Proof. Let such that . By inequality (46), we haveIf we consider the function define on by , we obtain , and by hypothesis (C3), the function is lower semicontinuous. Applying Theorem 4, we deduce that admits a fixed point.
Our goal in the rest of this article is to generalize the recent Theorem 4 in [11] and to give a short proof of it.

Theorem 5. Let be a complete metric space and a map such that there exists a function satisfyingwhere

Then, has a fixed point.

Proof. Let and define the sequence by . On the one hand, the case where there exists , such that , gives as a fixed point of . On the other hand, if for every , then, taking and in (49), one obtains, for every ,where , which shows that the sequence is necessarily positive and monotonically decreasing. Therefore, the series is convergent, and by the comparison principle, is also convergent and .
Let , there exists such that, for all ,So,because . The sequence is then positive and bounded above byNext, we prove that is a Cauchy sequence. For this, we present two methods.

Method 1. Since is bounded, we get . We haveThus, .
Suppose that is not a Cauchy sequence. Then, there exists for which we can find subsequences and with , and is the smallest integer such thatSo,Taking and using , we obtainSimilarly, we haveLetting in the above two inequalities and using (56), we obtainNow, taking and in (49), one obtainsLetting , we obtainwhich is a contradiction. This shows that is a Cauchy sequence.

Method 2. We have for all . So, for all ,Since is convergent, the series is also convergent. Therefore, is a Cauchy sequence. Hence, the sequence is convergent. Let . Taking and in (49), one obtains, for every ,Letting , we have .

Example 4. Let with the usual distance . Let and be the functions defined on byWe need only to consider five cases to check the hypothesis of Theorem 5. Case 1: . Case 2: . Case 3: and . We have because . So, Case 4: and . We have and Case 5: and . We have andThis shows that inequality (49) is satisfied and all the hypotheses are verified. In addition, has the fixed point .
We notice that, in the third case, and , and we have .

Remark 3. If we take Example 1 of [11], satisfies condition (49) of our theorem and admits a fixed point. However, does not satisfy the Banach Contraction Principle nor the Ćirić contraction.
Theorem 4 in [11] treats the case, where in (49), and is replaced byBy triangular inequality, we obtainwhich gives rise to the following corollary.

Corollary 4. Let be a complete metric space and a map such that there exists a function satisfyingwhere

Then, has a fixed point.

Corollary 5. Let be a complete metric space and a map such that there exists a function satisfying, for all ,for some real . Then, has a fixed point.

Proof. If assumption (75) holds for some real then (49) holds, since for every , one haswhich allows to conclude that T admits a fixed point.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.