A Generalization of Kannan's Fixed Point Theorem

Abstract

In order to observe the condition of Kannan mappings, we prove a generalization of Kannan's fixed point theorem. Our theorem involves constants and we obtain the best constants to ensure a fixed point.

1. Introduction

A mapping on a metric space is called Kannan if there exists such that

(1.1)

for all . Kannan [1] proved that if is complete, then every Kannan mapping has a fixed point. It is interesting that Kannan's theorem is independent of the Banach contraction principle [2]. Also, Kannan's fixed point theorem is very important because Subrahmanyam [3] proved that Kannan's theorem characterizes the metric completeness. That is, a metric space is complete if and only if every Kannan mapping on has a fixed point. Recently, Kikkawa and Suzuki proved a generalization of Kannan's fixed point theorem. See also [4–8].

Theorem 1.1 (see [9]).

Define a nonincreasing function from into by

(1.2)

Let be a mapping on a complete metric space . Assume that there exists such that

(1.3)

for all . Then has a unique fixed point . Moreover holds for every .

Remark 1.2.

is the best constant for every .

From this theorem, we can tell that a Kannan mapping with is much stronger than a Kannan mapping with .

While and play the same role in (1.1), and do not play the same role in (1.3). So we can consider "'' instead of ".'' And it is a quite natural question of what is the best constant for each pair . In this paper, we give the complete answer to this question.

2. Preliminaries

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

We use two lemmas. The first lemma is essentially proved in [5].

Lemma 2.1 (see [5, 9]).

Let be a metric space and let be a mapping on . Let satisfy for some . Then for , either

(2.1)

holds.

The second lemma is obvious. We use this lemma several times in the proof of Theorem 4.1.

Lemma 2.2.

Let , , , and be four real numbers such that and . Then holds.

3. Fixed Point Theorem

In this section, we prove a fixed point theorem.

We first put and () by

(3.1)

See Figure 1.

Figure 1

()

Theorem 3.1.

Define a nonincreasing function from into by

(3.2)

Let be a mapping on a complete metric space . Assume that there exists such that

(3.3)

for all . Then has a unique fixed point . Moreover holds for every .

Proof.

We put

(3.4)

Since , holds. From the assumption, we have

(3.5)

and hence

(3.6)

for all . Since

(3.7)

we have

(3.8)

and hence

(3.9)

for all .

Fix and put for . From (3.6), we have

(3.10)

So is a Cauchy sequence in . Since is complete, converges to some point .

We next show

(3.11)

Since converges, for sufficiently large , we have

(3.12)

and hence

(3.13)

Therefore we obtain

(3.14)

for all . By (3.11), we have

(3.15)

and hence

(3.16)

Let us prove that is a fixed point of . In the case where , arguing by contradiction, we assume . Then we have

(3.17)

So for sufficiently large ,

(3.18)

holds and hence

(3.19)

Thus we obtain

(3.20)

which is a contradiction. Therefore we obtain .

In the case where , if we assume , then we have

(3.21)

which is a contradiction. Therefore holds.

In the case where , we consider the following two cases.

(i)There exist at least two natural numbers satisfying

(ii) for sufficiently large .

In the first case, if we assume , then cannot be Cauchy. Therefore . In the second case, we have by (3.16), for sufficiently large . From the assumption,

(3.22)

Since , we obtain .

In the case where , we note that . By Lemma 2.1, either

(3.23)

holds for every . Thus there exists a subsequence of such that

(3.24)

for . From the assumption, we have

(3.25)

Since , we obtain . Therefore we have shown in all cases.

From (3.11), the fixed point is unique.

Remark 3.2.

We have shown , dividing four cases. It is interesting that the four methods are all different. We can rewrite by

(3.26)

4. The Best Constants

In this section, we prove the following theorem, which informs that is the best constant for every .

Theorem 4.1.

Define a function as in Theorem 3.1. For every , there exist a complete metric space and a mapping on such that has no fixed points and

(4.1)

for all .

Proof.

We put and by (3.4).

In the case where , define a complete subset of the Euclidean space by . We also define a mapping on by for . Then does not have any fixed points and

(4.2)

for all .

In the case where , we put

(4.3)

We note that . Define a complete subset of the Euclidean space by

(4.4)

where for . Define a mapping on by , and for . Then we have

(4.5)

for . Since

(4.6)

we have

(4.7)

for . For with , since

(4.8)

we have

(4.9)

In the case where , we note that . We also note that . Let be the Banach space consisting of all functions from into (i.e., is a real sequence) such that . Let be the canonical basis of . Define a complete subset of by

(4.10)

where

(4.11)

for . We note that

(4.12)

for with . Define a mapping on by , , and for . Then we have

(4.13)

for . Since

(4.14)

we have

(4.15)

Since , we have

(4.16)

for . We have

(4.17)

for . For with , we have

(4.18)

This completes the proof.