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# Convergence of the Sequence of Successive Approximations to a Fixed Point

*Fixed Point Theory and Applications*
**volume 2010**, Article number: 716971 (2010)

## Abstract

If is a complete metric space and is a contraction on , then the conclusion of the Banach-Caccioppoli contraction principle is that the sequence of successive approximations of starting from any point converges to a unique fixed point. In this paper, using the concept of -distance, we obtain simple, sufficient, and necessary conditions of the above conclusion.

## 1. Introduction

The following famous theorem is referred to as the *Banach-Caccioppoli contraction principle*. This theorem is very forceful and simple, and it became a classical tool in nonlinear analysis.

Theorem 1.1 (see Banach [1] and Caccioppoli [2]).

Let be a complete metric space and let be a self contraction on , that is, there exists such that for all . Then the following holds.

(A) has a unique fixed point , and converges to for any .

We note that the conclusion of Kannan's fixed point theorem [3] is also (A). See Kirk's survey [4]. Recently, we obtained that (A) holds if and only if is a strong Leader mapping [5, 6].

Theorem 1.2 (see [6]).

Let be a mapping on a complete metric space . Then the following are equivalent.

(i)(A) holds.

(ii) is a strong Leader mapping, that is, the following hold.

(a)For and , there exist and such that

for all , where is the identity mapping on .

(b)For , there exist and a sequence in such that

for all and .

The following theorem is proved in [7, 8].

Theorem 1.3 (see Rus [7] and Subrahmanyam [8]).

Let be a complete metric space and let be a continuous mapping on . Assume that there exists satisfying for all . Then the following holds.

(B) converges to a fixed point for every .

We obtained a condition equivalent to (B) in [9].

Theorem 1.4 (see [9]).

Let be a mapping on a complete metric space . Then the following are equivalent.

(i)(B) holds.

(ii)The following hold.

(a)For and , there exist and such that

for all .

(b)For , there exist and a sequence in such that

for all and .

We sometimes call a mapping satisfying (A) a *Picard operator* [10]. We also call a mapping satisfying (B) a *weakly Picard operator* [11–13].

We cannot tell that the conditions (ii) of Theorems 1.2 and 1.4 are simple. Motivated by this, we obtain simpler conditions which are equivalent to Conditions (A) and (B).

## 2. Preliminaries

Throughout this paper, we denote by , , and the sets of positive integers, integers and real numbers, respectively.

In 2001, Suzuki introduced the concept of -distance in order to improve results in Tataru [14], Zhong [15, 16], and others. See also [17].

Definition 2.1 (see [18]).

Let be a metric space. Then a function from into is called a *-distance* on if there exists a function from into and the following are satisfied:

() for all ,

() and for all and , and is concave and continuous in its second variable,

() and imply for all ,

() and imply that ,

() and imply that .

The metric is a -distance on . Many useful examples and propositions are stated in [9, 18–23] and references therein. The following fixed point theorems are proved in [18].

Theorem 2.2 (see [18]).

Let be a complete metric space and let be a mapping on . Assume that there exist a -distance and such that for all . Assume the following.

(i)If , , and , then .

Then (B) holds. Moreover, if , then .

Theorem 2.3 (see[18]).

Let be a complete metric space and let be a mapping on . Assume that is a contraction with respect to some -distance , that is, there exist a -distance and such that

for all . Then (A) and hold.

The following lemmas are useful in our proofs.

Lemma 2.4 (see [18]).

Let be a metric space and let be a -distance on . If sequences and in satisfy and for some , then . In particular for and imply that .

Lemma 2.5 (see [18]).

Let be a metric space and let be a -distance on . If a sequence in satisfies , then is a Cauchy sequence. Moreover if a sequence in satisfies , then .

The following lemmas are easily deduced from Lemmas 2.4 and 2.5.

Lemma 2.6.

Let be a metric space and let be a -distance on . Then for every and , there exists such that and imply that .

Lemma 2.7.

Let be a metric space and let be a -distance on . Assume that a sequence in satisfies , , and . Then .

The following is proved at Page 442 of [18]. However we give a proof because we use reductio ad absurdum in [18].

Lemma 2.8 (see [18]).

Let be a nondecreasing function from into itself satisfying . Define a function from into itself by

Then , for all ; and is concave and continuous.

Proof.

It is clear that , , and is concave. We shall prove that is continuous at . Fix . Then there exists such that . Choose with . Fix . Let and such that and . Since , we have

Since and are arbitrary, we obtain . Thus, .

The following is obvious.

Lemma 2.9.

Let be a mapping on a set . Let be a subset of such that . Define a sequence of subsets of by

Then the following hold.

(i)For every and , if and only if for and .

(ii) for with .

(iii) for every .

## 3. Condition (B)

In this section, we discuss Condition (B).

Theorem 3.1.

Let be a complete metric space and let be a mapping on . Assume that there exist a -distance , , and such that

for all . Then (B) holds. Moreover, if , then .

Proof.

Assume that , , and . Then we have

and hence, . By Lemma 2.7, we obtain . By Theorem 2.2, we obtain the desired result.

As a direct consequence of Theorem 3.1, we obtain the following.

Corollary 3.2.

Let be a complete metric space and let be a mapping on . Assume that there exist a -distance and such that

for all . Then (B) holds.

Corollary 3.2 characterizes Condition (B).

Theorem 3.3.

Let be a mapping on a metric space such that (B) holds. Then there exist a -distance and satisfying (3.3).

Proof.

Let be fixed. We note that every periodic point is a fixed point. That is, if satisfies for some , then . Define a mapping from onto by for , where is the set of all fixed points of . Define a mapping from into the set of subsets of by

Since is a fixed point of , we have

Next, we define a function from into satisfying

for all . We put for . It is obvious that for . Define a sequence of subsets of by

Then by Lemma 2.9,

for with . We put for . We note that

Put

It is obvious that , , and . So,

for and . Define an equivalence relation on as follows: if and only if there exist such that . By Axiom of Choice, there exists a mapping on such that

Let with . Then we put for . Define a sequence of subsets of by

Then we have for with ; and

We put for with and . We have defined . We note that implies that .

Next, we define a -distance by

where . We note that implies either of the following.

(i).

(ii)There exist , , and such that , , , and . (In this case, , , , and hold.)

We shall show that is a -distance. Let . If and , then . So we have

If or , then

These imply (1). We shall define a function from into . For , we put . For , we put . Since converges to , there exists a strictly increasing sequence in such that implies that for . Since , we can define a nondecreasing function from into such that . It is obvious that . Put

Then satisfies (2) and by Lemma 2.8. In order to show (3), we assume that and . Then without loss of generality, we may assume that . Thus for . It is obvious that for with . We consider the following two cases.

(i)There exists such that for .

(ii)There exists a subsequence of such that .

In the first case, since exactly consists of one element and for , holds for all . So . Thus, holds for every . In the second case, we note that for all . Hence . Put . Since , there exists a sequence in such that . Since for all , there exists a sequence in such that . Since , we have and . So we obtain . We note that for all . Let . In the case where , we have

In the other case, where , we have , and hence,

Therefore we have shown (3). Let us prove (4). We assume that and . Without loss of generality, we may assume that . We consider the following two cases.

(i)There exists such that for .

(ii)There exists a subsequence of such that .

In the first case, we have

In the second case, as in the proof of (3), there exist , a sequence in , and a sequence in such that , , and . We note that . If , then . If , then . Therefore

Let us prove (5). We assume that . We note that . In the case where , we have . In the other case, where there exist , , and such that , , , and , we have

Hence

Thus, we obtain and . So we have

Therefore

which imply (5). Therefore we have shown that is a -distance on .

We shall show (3.3). Let . Since , , , and , we have

If , then holds. So we have

If , then we have

Therefore (3.3) holds.

Remark 3.4.

We have proved that, for every , there exists a -distance satisfying (3.3).

Combining Theorem 6 in [9], we obtain the following.

Corollary 3.5.

Let be a mapping on a complete metric space . Then the following are equivalent.

(i)(B) holds.

(ii)There exists a -distance on satisfying the following.

(a)For and , there exist and such that

for all with .

(b)For , there exist and a sequence in such that

for all and with .

(iii)There exist a -distance and such that and for all .

## 4. Condition (A)

In this section, we discuss Condition (A).

Define a relation on as follows: if and only if either or holds.

Theorem 4.1.

Let be a complete metric space and let be a mapping on . Assume that there exist a -distance and such that

for all . Then (A) holds.

Proof.

By Theorem 3.1, (B) holds. Moreover, if , then . Let be fixed points of . Then

which implies that . Since , we obtain by Lemma 2.4. Thus the fixed point is unique.

Theorem 4.2.

Let be a complete metric space and let be a mapping on . Assume that there exist a -distance and such that

for all with . Then (A) holds.

Proof.

In the case where consists of one element, the conclusion obviously holds. So we consider the other case. Assume that , , and . We consider the following two cases:

(i) for sufficiently large ,

(ii)there exists a sequence of such that .

In the first case, we have

for sufficiently large , and hence, . By Lemma 2.7, we obtain . In the second case, we have

By Lemma 2.4, we obtain . By Theorem 2.2, (B) holds. Let be distinct fixed points of . Then

which implies a contradiction. Thus the fixed point is unique.

We shall show that Theorems 4.1 and 4.2 characterize Condition (A).

Theorem 4.3.

Let be a mapping on a metric space such that (A) holds. Then there exist a -distance and satisfying (4.1).

Proof.

Let , , , and be as in the proof of Theorem 3.3. Then holds. Fix . We consider the following two cases:

(i) and

(ii)either or

In the first case, holds by (A). Since

we obtain . Thus, . In the second case, we note that either or holds. Thus

If , then holds. So we have

If , then we have

Therefore (4.1) holds.

Theorem 4.4.

Let be a mapping on a metric space such that (A) holds. Then there exist a -distance and satisfying (4.3) for all with .

Proof.

The proof of Theorem 4.3 works.

Combining Theorem 7 in [9], we obtain the following.

Corollary 4.5.

Let be a mapping on a complete metric space . Then the following are equivalent.

(i)(A) holds.

(ii)There exists a -distance on satisfying the following.

(a)For and , there exist and such that

for all with .

(b)For , there exist and a sequence in such that

for all and with .

(iii)There exist a -distance and such that and for all .

(iv)There exist a -distance and such that and for all with .

## 5. Additional Result

Since Theorem 2.2 deduces Corollary 3.2, we can tell that Theorem 2.2 characterizes Condition (B). However, the following example tells that Theorem 2.3 does not characterize Condition (A).

Example 5.1.

Let be the set of all real sequences such that for , is strictly decreasing, and converges to . Let be a Hilbert space consisting of all the functions from into satisfying with inner product for all . Define a subset of by

where is defined by and for . Define a mapping on by

Then (A) holds. However, is not a contraction with respect to any -distance .

Proof.

It is obvious that (A) holds. Arguing by contradiction, we assume that is a contraction with respect to some -distance . That is, there exist a -distance and such that for all . Since

we have . By Lemma 2.6, there exists a strictly increasing sequence in such that

We choose such that . Fix with . Then we have

and hence,

This is a contradiction.

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## Acknowledgments

The author is supported in part by Grant-in-Aid for Scientific Research from Japan Society for the Promotion of Science. The author wishes to express his gratitude to the referees for careful reading and giving a historical comment.

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### Cite this article

Suzuki, T. Convergence of the Sequence of Successive Approximations to a Fixed Point.
*Fixed Point Theory Appl* **2010, **716971 (2010). https://doi.org/10.1155/2010/716971

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### Keywords

- Hilbert Space
- Equivalence Relation
- Point Theorem
- Nonlinear Analysis
- Successive Approximation