Copyright © 2008 Misako Kikkawa and Tomonari Suzuki. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Contractions are always continuous and Kannan mappings are not necessarily continuous. This is a very big difference between both mappings. However, we know that relaxed both mappings are quite similar. In this paper, we discuss both mappings from a new point of view.

1. Introduction

Let be a metric space and let be a mapping on . Then is called a contraction if there exists such that (1.1) for all . is called Kannan if there exists such that (1.2) for all . We know that if is complete, then every contraction and every Kannan mapping have a unique fixed point, see [1, 2]. We know that both conditions are independent, that is, there exist a contraction, which is not Kannan, and a Kannan mapping, which is not a contraction. Thus we cannot compare both conditions directly. So we compare both indirectly.

Fact 1

Banach fixed-point theorem, which is often called the Banach contraction principle, is very important because it is a very forceful tool in nonlinear analysis. We think that Kannan fixed-point theorem is also very important because Subrahmanyam [3] proved that Kannan theorem characterizes the metric completeness of underlying spaces, that is, a metric space is complete if and only if every Kannan mapping on has a fixed point. On the other hand, Connell [4] gave an example of a metric space such that is not complete and every contraction on has a fixed point. Thus the Banach theorem cannot characterize the metric completeness of . Therefore, we consider that the notion of contractions is stronger from this point of view.

Fact 2

Using the notion of -distances, Suzuki [5] considered some weaker contractions and Kannan mappings and proved the following.(i) If is a contraction with respect to a -distance, then is Kannan with respect to another -distance. (ii) If is Kannan with respect to a -distance, then is a contraction with respect to another -distance. That is, both conditions are completely the same.

Recently, Suzuki [6] proved the following theorem, see also [7].

Theorem 1.1 (see [6]). Define a nonincreasing function from onto by (1.3) Then for a metric space , the following are equivalent:

(i) is complete, (ii) every mapping on , satisfying the following, has a fixed point: there exists such that implies for all .

Remark 1.2. is the best constant for every .

The purpose of this paper is to prove a Kannan version of Theorem 1.1. Then we compare the theorem (Theorem 2.2) with Theorem 1.1 and attempt to judge which is stronger from our new point of view.

2. Kannan Mappings

Throughout this paper we denote by the set of all positive integers and by the set of all real numbers.

In this section, we prove our main result. We begin with the following lemma.

Lemma 2.1. Let be a metric space and let be a mapping on Let satisfy for some Then for either (2.1) holds.

Proof. We assume (2.2) Then we have (2.3) This is a contradiction.

The following theorem is a Kannan version of Theorem 1.1.

Theorem 2.2. Define a nonincreasing function from into by (2.4) Let be a complete metric space and let be a mapping on . Let and put . Assume that (2.5) for all , then has a unique fixed point and holds for every .

Proof. Since , holds. From the assumption, we have (2.6) and hence (2.7) for . Let . Put and for all . From (2.7), we have (2.8) So is a Cauchy sequence in and by the completeness of , there exists a point such that .

We next show (2.9) Since , there exists such that for all with . Then we have (2.10) and hence (2.11) Therefore, we obtain (2.12) for with .

Let us prove that is a fixed point of . In the case where , arguing by contradiction, we assume that . Then we have, from (2.9), (2.13) and hence (2.14) This is a contradiction. Therefore, we obtain . In the case where , from Lemma 2.1, either (2.15) holds for . Thus there exists a subsequence of such that (2.16) for . From the assumption, we have (2.17) Since , we have . Therefore, we have shown in both cases.

From (2.9), we obtain that the fixed point is unique.

Remark 2.3. Since for every , we can consider that Kannan is stronger from our new point of view. Though and are different, we remark that the graphs of and are quite similar.

The following theorem shows that is the best constant for every .

Theorem 2.4. Define a function as in Theorem 2.2. For every putting , there exist a complete metric space and a mapping on such that has no fixed points and (2.18) for all .

Proof. In the case where , define a complete subset of the Euclidean space by . We also define a mapping on by for . Then dose not have a fixed point and (2.19) for all . In the case where , define a complete subset of the Euclidean space by (2.20) where for all . Define a mapping on by , , and for . Then the following are obvious: (i), (ii) for . Also, we have (2.21) for .

3. Generalized Kannan Mappings

It is a very natural question of whether or not another fixed-point theorem with exists. In this section, we give a positive answer to this problem.

Theorem 3.1. Define a nonincreasing function as in Theorem 1.1. Let be a complete metric space and let be a mapping on . Suppose that there exists such that (3.1) for all . Then has a unique fixed point and holds for every .

Proof. Since , we have, from the assumption, (3.2) and hence (3.3) for . Let . Put and for all . As in the proof of Theorem 2.2, we can prove that converges to some .

We next show (3.4) Since , we have for sufficiently large . Hence we obtain, from the assumption, (3.5) for with .

Let us prove that is a fixed point of . In the case where , we note (3.6) We will show, by induction, (3.7) for with . When , (3.7) becomes (3.3), thus (3.7) holds. We assume for some with . Since (3.8) we have , and hence (3.9) Therefore, by the assumption, we have (3.10) By induction, (3.7) holds for with . Arguing, by contradiction, we assume . Then from (3.7), holds for all . Then by (3.4), we have (3.11) This implies , which contradicts (3.7). Therefore, we obtain . In the case where , as in the proof of Theorem 2.2, we can show that there exists a subsequence of such that for . From the assumption, we have (3.12) Since , the above inequality implies that . Therefore, we have shown that in both cases.

From (3.4), we obtain that the fixed point is unique.

Remark 3.2. When the second author was proving Theorem 1.1, he did not feel that was natural. However, since the above proof is easier to understand how works, the authors can faintly feel that is natural.

The following theorem shows that is the best constant for every .

Theorem 3.3. Define a function as in Theorem 1.1. Then for any there exist a complete metric space and a mapping on such that has no fixed points and (3.13) for all .

Proof. We have already shown the conclusion in the case where or because holds. So let us consider the case where . Define a complete subset of the Euclidean space by where , , and for . Define a mapping on by for . Then the following are obvious:

(i), (ii), (iii) for , (iv). Since (3.14) we have the following:(i) for , (ii) for , (iii) for . This completes the proof.

Figure 1

Acknowledgment

The second author is supported in part by Grants-in-Aid for Scientific Research from the Japanese Ministry of Education, Culture, Sports, Science, and Technology.

References

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6. T. Suzuki, “A generalized Banach contraction principle that characterizes metric completeness,” Proceedings of the American Mathematical Society, vol. 136, pp. 1861–1869, 2008.
7. M. Kikkawa and T. Suzuki, “Three fixed point theorems for generalized contractions with constants in complete metric spaces,” 2007, to appear in Nonlinear Analysis: Theory, Methods & Applications.