- Research Article
- Open Access

# Fixed Points for Multivalued Mappings and the Metric Completeness

- S. Dhompongsa
^{1}Email author and - H. Yingtaweesittikul
^{1}

**2009**:972395

https://doi.org/10.1155/2009/972395

© S. Dhompongsa and H. Yingtaweesittikul. 2009

**Received:**24 December 2008**Accepted:**6 May 2009**Published:**14 May 2009

## 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.

## Keywords

- General Classis
- Differential Geometry
- Closed Subset
- Fixed Point Theorem
- Simple Proof

## 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.

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]).

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]).

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

*a Caristi mapping*if there exists a lower semicontinuous function such that is bounded below and

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

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]).

then has at least one fixed point.

## 3. Completeness and Single-valued Mappings

Theorem 3.1 (see [9]).

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

Theorem 3.2 (see [9]).

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.

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

We fix and define a sequence in by .

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

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

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

Case 1 ( ).

We have .

Hence , which is a contradiction.

Case 2 ( ).

We have .

which is a contradiction. Therefore

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

For , We have , which is a contradiction.

Case 3 ( ).

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;

for all has a fixed point;

for all has a fixed point;

for all has a fixed point;

for all has a fixed point;

for all has a fixed point;

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.

Let .

Case : .

Case : .

Case : which is impossible.

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

Case 1 ( ).

Case 2 ( ).

Therefore (4.6) holds.

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

Case 1 ( ).

Case 2 ( ).

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.

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

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]).

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

Theorem 5.2 ([10]).

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.

Hence has a fixed point by Theorem 2.1.

Proof of Theorem 5.2.

Thus has a fixed point by Theorem 2.1.

## Declarations

### Acknowledgment

The authors would like to thank the Thailand Research Fund (grant BRG4780016) for its support.

## Authors’ Affiliations

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