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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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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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DOI: https://doi.org/10.1155/2010/716971