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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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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