Fixed Points for Multivalued Mappings and the Metric Completeness

Abstract

We consider the equivalence of the existence of fixed points of single-valued mappings and multivalued mappings for some classes of mappings by proving some equivalence theorems for the completeness of metric spaces.

1. Introduction

The Banach contraction principle [1] states that for a complete metric space , every contraction on , that is, for some , for all , has a (unique) fixed point.

Connell [2] gave an example of a noncomplete metric space on which every contraction on has a fixed point. Thus contractions cannot characterize the metric completeness of

Theorem 1.1 (see [3, Kannan]).

Let be a complete metric space. Let be a Kannan mapping on , that is, for some , for all Then has a (unique) fixed point.

Subrahmanyam [4] proved that Kannan mappings can be used to characterize the completeness of the metric. That is, a metric space is complete if and only if every Kannan mapping on has a fixed point.

In 2008 Suzuki [5] introduced a new type of mappings and presented a generalization of the Banach contraction principle in which the completeness can also be characterized by the existence of fixed points of these mappings. Define a nonincreasing function from onto by

(1.1)

Theorem 1.2 (see [5]).

For a metric space , the following are equivalent:

(i) is complete;

(ii)every mapping on such that there exists , implies for all has a fixed point.

In 2008, Kikkawa and Suzuki [6] partially extended Theorem 1.2 to multivalued mappings.

Theorem 1.3 (see [6]).

Define a strictly decreasing function from onto by . Let be a complete metric space and let be a multivalued mapping with bounded and closed values. Assume that there exists such that

(1.2)

for all , then there exists such that

Obviously, the converse of Theorem 1.3 is valid since for all

Moţ and Petruşel [7] proved the following theorem which is a generalization of Kikkawa and Suzuki Theorem.

Theorem 1.4 (see [7]).

Let be a complete metric space and let be a multivalued mapping with closed values and satisfies the following: if for nonnegative numbers with and for each , one has

(1.3)

Then has a fixed point.

In this paper, we will characterize the completeness of a metric space by the existence of fixed points for both single-valued and multivalued mappings. We first aim to extend, in Section 3, the Suzuki's result (Theorem 1.2) to more general classes of mappings. We then consider multivalued mappings in Section 4. We also show in this section that the converse of Theorem 1.4 is true.

2. Preliminaries

Let be a complete metric space and let be a mapping. We say that is a Caristi mapping if there exists a lower semicontinuous function such that is bounded below and

(2.1)

Recall that a mapping is lower semicontinuous if for each and for every , there exists a neighborhood of such that for all

For a metric space , let and denote, respectively, a collection of all nonempty closed subsets of and a collection of all nonempty bounded closed subsets of . Let be the Hausdorff metric on . That is, for ,

(2.2)

where is the distance from a point in to a subset of .

The next theorem plays important roles in this paper.

Theorem 2.1 (see cf. [8]).

If is a mapping of a complete metric space into the family of all nonempty closed subsets of and is a lower semicontinuous function such that the following condition holds:

(2.3)

then has at least one fixed point.

3. Completeness and Single-valued Mappings

In 2008, Kikkawa and Suzuki [9] proved fixed point theorems for some generalized Kannan mappings. Let be a nonincreasing function defined from [0,1) onto (1/2,1] by

(3.1)

Theorem 3.1 (see [9]).

Let be a complete metric space and let be a mapping on . Let and put Assume that

(3.2)

for all , then has a unique fixed point and holds for every

Theorem 3.2 (see [9]).

Let be a complete metric space and let be a mapping on . Suppose that there exists such that

(3.3)

for all Then has a unique fixed point and holds for every

The above theorems inspire us to present another version of Theorem 1.2. Before doing that we present first the following theorem. The proof of which is a mild modification of the proofs in [5, 9].

Theorem 3.3.

Let be a complete metric space and let be a mapping on such that there exists , implies for all , then has a fixed point.

Proof.

Since , holds for every , and thus

(3.4)

If for some , then and we get a fixed point of

Suppose now that

(3.5)

We fix and define a sequence in by .

Then , and so . Thus is a Cauchy sequence. Since is complete, converges to some point

We show that

(3.6)

Suppose Since as , there exists such that for each Observe that

(3.7)

Hence for each Letting we get for all and we obtain (3.6).

As in the proof of [5, Theorem  1.2], we show that for some from which it is proved that is a fixed point of For this purpose, we assume for all and find a contradiction. We show, by induction, that

(3.8)

From (3.6) we have

(3.9)

Suppose . Thus

(3.10)

Thus (3.8) holds and now we find a contradiction in each of the following cases.

Case 1 ().

We have .

Assume then

(3.11)

which is a contradiction. So

(3.12)

Hence , which is a contradiction.

Case 2 ().

We have .

We show, by induction, that

(*)

If then

(3.13)

which is a contradiction. Therefore

Suppose . Thus

(3.14)

If then

(3.15)

which is a contradiction. Hence and thus (*) holds.

For , We have , which is a contradiction.

Case 3 ().

We claim that or Suppose not,

(3.16)

which is a contradiction. So there exists a subsequence of such that :

(3.17)

Thus , which is a contradiction.

In fact the following theorem shows that the converse of Theorem 3.3 is valid.

Theorem 3.4.

Let be a metric space. Then the following are equivalent:

(i) is complete;

(ii)for each , every mapping on such that

(3.18)

for all has a fixed point;

(iii)for each , every mapping on such that

(3.19)

for all has a fixed point;

(iv)for each , every mapping on such that

(3.20)

for all has a fixed point;

(v)For nonnegative numbers with , every mapping on such that

(3.21)

for all has a fixed point;

(vi)for each , every mapping on such that

(3.22)

for all has a fixed point;

(vii)for each , every mapping on such that

(3.23)

for all has a fixed point.

Proof.

The implication (i)(vii) is exactly Theorem 3.3.

(vii)(vi). Let satisfy (3.22). We show that satisfies (3.23) to obtain a fixed point for . Let , . Thus , and (3.23) holds.

(vi)(v). Let satisfy (3.21). To show satisfies (3.22), let , , and . Notice that Thus So we get , and (3.22) holds.

(v)(ii). Let satisfy (3.18). To show satisfies (3.21), let Thus , and so and (3.21) holds.

(ii)(i). Follows the same proof of Theorem 1.2. Notice that, for

(vii)(iv). Let satisfy (3.20). To show satisfies (3.23), let . Thus .

(iv)(iii). Let satisfy (3.19). We show satisfies (3.20). Let , . Thus

(iii)(i). We know that every Kannan mapping belongs to the class of mappings in (iii). Thus is complete by Subrahmanyam [4].

4. Completeness and Multivalued Mappings

Inspired by Theorem 1.2 and Theorem 1.3, we prove the following theorem for a larger class of mappings under some certain assumptions.

Theorem 4.1.

Let be a metric space. Then the following are equivalent:

(i) is complete;

(ii)for each , every mapping such that implies , and the function is lower semicontinuous has a fixed point.

Observe that Theorem 4.1 is not covered by Theorem 3.4 when considering as single-valued mappings.

Proof of Theorem 4.1.

(i)(ii). Let be small enough so that and define . For any , we can find some satisfying . To apply Theorem 2.1, it remains to show that . We have Thus Note that

(4.1)

Let .

Case : .

Case : .

Case : which is impossible.

Hence

(4.2)

Thus has a fixed point by Theorem 2.1.

(ii)(i). Suppose is not complete.

Define a function as in the proof of Theorem 1.2 and a mapping as follows:

for each since and , there exists satisfying

We put and write

It is obvious that for all Since for all , for all , thus That is, does not have a fixed point. Note that

(4.3)

We have

(4.4)

(4.5)

Fix with To show that the mapping satisfies the condition in (ii), that is, for all ,

(4.6)

Observe that

(4.7)

Case 1 ().

(4.8)

Case 2 ().

(4.9)

Therefore (4.6) holds.

It remains to show that the mapping is lower semicontinuous, that is,

(4.10)

Suppose not, then there exists such that for all for each . Since such that We have , for all large . Thus for each , , for all large So for those . Consequently,

(4.11)

which impliest that

(4.12)

a contradiction. Thus the mapping is lower semicontinuous.

The converse of Theorem 1.4 is also valid by following the same proof of Theorem 1.2. Assuming that is not complete, we find a fixed point free mapping satisfying the condition in Theorem 1.4. Following the same proof of Theorem 1.2 by replacing by where , we obtain for all and is fixed point free. We now verify the condition in Theorem 1.4 for

Fix with We show that

(4.13)

Observe that

(4.14)

Case 1 ().

(4.15)

Case 2 ().

(4.16)

Therefore (4.13) holds, and the proof of the converse of Theorem 1.4 is complete.

Moreover, by following the proof of Theorem 1.4, we can partially extend the class of mappings and still obtain their fixed points. Notice that

Theorem 4.2.

Let be a metric space. Then the following are equivalent:

(i) is complete.

(ii)every mapping such that for each with and for each ,

(4.17)

has a fixed point.

Proof.

(i)(ii). Following the same proof of Theorem 1.4 by replacing in its proof by Thus we obtain a sequence such that

(1), for each and;

(2) for

Choose so that and therefore . We see that the sequence is Cauchy in , and so converges to some We show for each

Suppose Since as , there exists such that for each We have Hence for each Letting , we get for all as desired.

Next, we show for all For we obtain for each such that Clearly , for all Hence, as we get and so implying that for

Finally, we obtain

(4.18)

Thus and has a fixed point.

(ii)(i). Let and , we have implying . Hence is complete by the converse of Theorem 1.4.

5. Caristi Set-Valued Mappings

In 2008, Ćirić [10] proved the following fixed point theorems.

Theorem 5.1 ([10]).

Let be a complete metric space and let If there exist constants such that for any there is satisfying the following two conditions:

(5.1)

Then has a fixed point in provided a function is lower semicontinuous.

Theorem 5.2 ([10]).

Let be a complete metric space and If there exists a function satisfying

(5.2)

and such that for any there is satisfying the following two conditions:

(5.3)

Then has a fixed point in provided a function is lower semicontinuous.

We give a simple proof of each of these theorems.

Proof of Theorem 5.1.

Define a lower semi-continuous function by For any we can find some satisfying

(5.4)

We show that . Let . Clearly,

(5.5)

Hence has a fixed point by Theorem 2.1.

Proof of Theorem 5.2.

Let and . For each there exists such that

(5.6)

Furthermore, . Indeed,

(5.7)

Thus has a fixed point by Theorem 2.1.