Caristi Fixed Point Theorem in Metric Spaces with a Graph

Abstract

We discuss Caristi’s fixed point theorem for mappings defined on a metric space endowed with a graph. This work should be seen as a generalization of the classical Caristi’s fixed point theorem. It extends some recent works on the extension of Banach contraction principle to metric spaces with graph.

Dedicated to Rashed Saleh Alfuraidan and Prof. Miodrag Mateljevi’c for his 65th birthday

1. Introduction

This work was motivated by some recent works on the extension of Banach contraction principle to metric spaces with a partial order [1] or a graph [2]. Caristi’s fixed point theorem is maybe one of the most beautiful extensions of Banach contraction principle [3, 4]. Recall that this theorem states the fact that any map has a fixed point provided that is a complete metric space and there exists a lower semicontinuous map such that for every . Recall that is called a fixed point of if . This general fixed point theorem has found many applications in nonlinear analysis. It is shown, for example, that this theorem yields essentially all the known inwardness results [5] of geometric fixed point theory in Banach spaces. Recall that inwardness conditions are the ones which assert that, in some sense, points from the domain are mapped toward the domain. Possibly, the weakest of the inwardness conditions, the Leray-Schauder boundary condition, is the assumption that a map points of anywhere except to the outward part of the ray originating at some interior point of and passing through .

The proofs given to Caristi’s result vary and use different techniques (see [3, 6–8]). It is worth to mention that because of Caristi’s result of close connection to the Ekeland’s [9] variational principle, many authors refer to it as Caristi-Ekeland fixed point result. For more on Ekeland’s variational principle and the equivalence between Caristi-Ekeland fixed point result and the completeness of metric spaces, the reader is advised to read [10].

2. Main Results

Maybe one of the most interesting examples of the use of metric fixed point theorems is the proof of the existence of solutions to differential equations. The general approach is to convert such equations to integral equations which describes exactly a fixed point of a mapping. The metric spaces in which such mapping acts are usually a function space. Putting a norm (in the case of a vector space) or a distance gives us a metric structure rich enough to use the Banach contraction principle or other known fixed point theorems. But one structure naturally enjoyed by such function spaces is rarely used. Indeed we have an order on the functions inherited from the order of . In the classical use of Banach contraction principle, the focus is on the metric behavior of the mapping. The connection with the natural order is usually ignored. In [1, 11], the authors gave interesting examples where the order is used combined with the metric conditions.

Example 1 (see [1]). Consider the periodic boundary value problem where and is a continuous function. Clearly any solution to this problem must be continuously differentiable on . So the space to be considered for this problem is . The above problem is equivalent to the integral problem where and the Green function is given by Define the mapping by Note that if is a fixed point of , then is a solution to the original boundary value problem. Under suitable assumptions, the mapping satisfies the following property: (i)if , then we have ;(ii)if , then  for a constant independent of and .The contractive condition is only valid for comparable functions. It does not hold on the entire space . This condition led the authors in [1] to use a weaker version of the Banach contraction principle to prove the existence of the solution to the original boundary value problem, a result which was already known [12] using different techniques.

Let be a partially ordered set and . We will say that is monotone increasing if The main result of [1] is the following theorem.

Theorem 2 (see [1]). Let be a partially ordered set and suppose that there exists a distance in such that is a complete metric space. Let be a continuous and monotone increasing mapping such that there exists with If there exists , with , then has a fixed point.

Clearly, from this theorem, one may see that the contractive nature of the mapping is restricted to the comparable elements of not to the entire set . The detailed investigation of the example above shows that such mappings may exist which are not contractive on the entire set . Therefore the classical Banach contraction principle will not work in this situation. The analogue to Caristi’s fixed theorem in this setting is the following result.

Theorem 3. Let be a partially ordered set and suppose that there exists a distance in such that is a complete metric space. Let be a continuous and monotone increasing mapping. Assume that there exists a lower semicontinuous function such that Then has a fixed point if and only if there exists , with .

Proof. Clearly, if is a fixed point of , that is, , then we have . Assume that there exists such that . Since is monotone increasing, we have , for any . Hence Hence is a decreasing sequence of positive numbers. Let . For any , we have Therefore is a Cauchy sequence in . Since is complete, there exists such that . Since is continuous, we conclude that ; that is, is a fixed point of .

The continuity assumption of may be relaxed if we assume that satisfies the property (OSC).

Definition 4. Let be a partially ordered set. Let be a distance defined on . One says that satisfies the property (OSC) if and only if for any convergent decreasing sequence in , that is, , for any , one has .

One has the following improvement to Theorem 3.

Theorem 5. Let be a partially ordered set and suppose that there exists a distance in such that is a complete metric space. Assume that satisfies the property (OSC). Let be a monotone increasing mapping. Assume that there exists a lower semicontinuous function such that Then has a fixed point if and only if there exists , with .

Proof. We proceed as we did in the proof of Theorem 3. Let such that . Write , . Then we have the fact that   is decreasing and exists in . Since we did not assume continuous, then may not be a fixed point of . The idea is to use transfinite induction to build a transfinite orbit to help catch the fixed point. Note that since satisfies (OSC), then we have . Since is monotone increasing, then we will have . These basic facts so far will help us seek the transfinite orbit , where is the set of all ordinals. This transfinite orbit must satisfy the following properties: (1), for any ;(2), if is a limit ordinal;(3), whenever ;(4), whenever .Clearly the above properties are satisfied for any . Let be an ordinal number. Assume that the properties are satisfied by . We have two cases as follows. (i)If , then set .(ii)Assume that is a limit ordinal. Set . Then one can easily find an increasing sequence of ordinals , with , such that . Property will force to be Cauchy. Since is complete, then exists in . The property (OSC) will then imply . Let us show that . Let . If , for all , then we have But . Hence which implies that . Hence . Assume otherwise that there exists such that . Hence which implies . In any case, we have , for any . Therefore we have Hence . Set . Let us prove that satisfies all properties . Clearly and are satisfied. Let us focus on and . Let . We need to show that . If is a limit ordinal, this is obvious. Assume that exists. We have two cases; if exists, then we have . Since is monotone increasing, then ; that is, . Otherwise, if is an ordinal limit, then . Since is monotone increasing, then we have which implies , for any . Therefore we have , which completes the proof of . Let us prove . Let . First assume that exists. Then, in the proof of , we saw that . Our assumption on will then imply If , we are done. Otherwise, if , then we use the induction assumption to get The triangle inequality will then imply Next we assume that is a limit ordinal. Then there exists an increasing sequence of ordinals , with , such that . Given , assume that we have , for all . In this case, we have seen that . Otherwise, let us assume that there exists such that . In this case, from our induction assumption and the triangle inequality, we get Using the lower semicontinuity of , we conclude that which completes the proof of . By the transfinite induction we conclude that the transfinite orbit exists which satisfies the properties . Using Proposition ([13, page 284]), there exists an ordinal such that , for any . In particular, we have , for any . Property will then force ; that is, . Therefore has a fixed point.
One may wonder if Theorem 5 is truly an extension of the main results of [1, 2, 11]. The following example shows that it is the case.
Example  6. Let be the classical Banach space with the natural pointwise order generated by . Let be the positive cone of . Define by
First note that is a closed subset of . Hence is complete for the norm-1 distance. Also it is easy to check that the property (OSC) holds in this case. Note that, for any , we have . Also we have ; that is, is a fixed point of for any . Note that for any we have
Therefore all the assumptions of Theorem 5 are satisfied. But fails to satisfy the assumptions of [1, 2, 11]. Indeed if we take
then we have and , for any . Therefore is not continuous since converges uniformly (and in norm-1 as well) to . Note also that , for . So any Lipschitz condition on the partial order of will not be satisfied by in this case.

3. Caristi’s Theorem in Metric Spaces with Graph

It seems that the terminology of graph theory instead of partial ordering sets can give more clear pictures and yield to generalize the theorems above. In this section, we give the graph versions of our two main results.

Throughout this section we assume that is a metric space and is a directed graph (digraph) with set of vertices and set of edges containing all the loops; for any . We also assume that has no parallel edges (arcs) and so we can identify with the pair . Our graph theory notations and terminology are standard and can be found in all graph theory books, like [14, 15]. A digraph is called an oriented graph; if whenever , then .

Let be a partially ordered set. We define the oriented graph on as follows. The vertices of are the elements of , and two vertices are connected by a directed edge (arc) if . Therefore, has no parallel arcs as .

If are vertices of the digraph , then a directed path from to of length is a sequence of vertices such that

A closed directed path of length from to , that is, , is called a directed cycle. An acyclic digraph is a digraph that has no directed cycle.

Given an acyclic digraph, , we can always define a partially order on the set of vertices of by defining that whenever there is a directed path from to .

Definition 7. One says that a mapping is -edge preserving if is said to be a Caristi -mapping if there exists a lower semicontinuous function such that

One can now give the graph theory versions of our two mean Theorems 3 and 5 as follows.

Theorem 8. Let be an oriented graph on the set with containing all loops and suppose that there exists a distance in such that is a complete metric space. Let be continuous, -edge preserving, and a -Caristi mapping. Then has a fixed point if and only if there exists , with .

Proof. has all the loops. In particular, if is a fixed point of , , then we have . Assume that there exists such that . Since is -edge preserving, we have , for any . Hence Hence is a decreasing sequence of positive numbers. Let . For any , we have Therefore is a Cauchy sequence in . Since is complete, there exists such that . Since is continuous, we conclude that ; that is, is a fixed point of .

The following definition is needed to prove the analogue to Theorem 5.

Definition 9. Let be an acyclic oriented graph on the set with containing all loops. One says that satisfies the property (OSC) if and only if satisfies (OSC).

The analogue to Theorem 5 may be stated as follows.

Theorem 10. Let be an oriented graph on the set with containing all loops and suppose that there exists a metric in such that is a complete metric space. Assume that satisfies the property (OSC). Let be a -edge preserving and a Caristi -mapping. Then has a fixed point if and only if there exists , with .

The proof of Theorem 10 is similar to the proof of Theorem 5. In fact it is easy to check that these two theorems are equivalent.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.