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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 264909, 9 pages

http://dx.doi.org/10.1155/2014/264909

## A New Iterative Scheme of Modified Mann Iteration in Banach Space

College of Applied Mathematics, Chengdu University of Information Technology, No. 24, Block 1, Xuefu Road, Chengdu 610225, China

Received 5 October 2013; Revised 28 December 2013; Accepted 29 December 2013; Published 11 February 2014

Academic Editor: Sehie Park

Copyright © 2014 Jinzuo Chen et al. 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

We introduce the modified iterations of Mann's type for nonexpansive mappings and asymptotically nonexpansive mappings to have the strong convergence in a uniformly convex Banach space. We study approximation of common fixed point of asymptotically nonexpansive mappings in Banach space by using a new iterative scheme. Applications to the accretive operators are also included.

#### 1. Introduction

Let be a real Banach space, a nonempty closed convex subset of , and a mapping. Recall that is a nonexpansive mapping [1] if for all , and is asymptotically nonexpansive [2] if there exists a sequence with for all and such that for all integers and . A point is a fixed point of provided . Denote by the set of fixed points of ; that is, .

Iterative methods are often used to solve the fixed point equation . One classical iteration process is introduced in 1953 by Mann [3] which is well known as Mann iteration process and is defined as follows: where the sequence is chosen in and the initial guess is arbitrarily chosen.

There exists rich literature on the convergence of Mann iteration for different classes of operators considered on various spaces. Mann’s iteration method (1) has been proved to be a powerful method for solving nonlinear operator equations involving nonexpansive mapping, asymptotically nonexpansive mapping, and other kinds of nonlinear mapping; see [3–11] and the references therein.

It is known that Mann’s iteration method (1) is in general not strongly convergent for nonexpansive mappings. So to get strong convergence, one has to modify the iteration method (1). In this regard, we will show the modified iteration in Section 3.

Motivated and inspired by the research going on in these fields, we suggest and analyze now new modified Mann’s iteration for finding the common fixed point of the nonexpansive mappings and asymptotically nonexpansive mappings in Banach space. We propose the modified Mann’s iteration and consider the strong convergence of the approximate solutions for nonexpansive and asymptotically nonexpansive in Banach space.

We suggest and analyze the following iterative: and if there exists two sequences and generated by

Our second modification of Mann’s iteration method (1) is adaption to (2) for finding a zero of an -accretive operator , for which we assume that the zero set . Our iterations process is given by (4), and sequences and are as follows (6), where, for each , is the resolvent of . We prove that not only defined by (4) but also and generated by (6) converge strongly to a zero of under certain assumptions in a uniformly Banach space.

We write to indicate that the sequence converges strongly to . Using is to denote the set of common fixed point of the mappings , , and , and using is to denote the set of common fixed points of the mappings and .

#### 2. Preliminaries

This section collects some lemmas, which will be used in the proofs for the main results in the next section.

Lemma 1 (see [8]). *Let , , and be sequences of nonnegative real numbers satisfying the inequality
**
If and , then*(1)* exists;*(2)* whenever .*

Lemma 2 (see [12]). *Suppose that is a uniformly convex Banach space and for all . Let and be two sequences of such that , and hold for some ; then .*

Lemma 3 (see [10]). *A mapping : with a nonempty fixed point set in will be said to satisfy Condition .**If there is a nondecreasing function with for all such that for all , where .*

Lemma 4 (see [13]). *Given a number , let be a uniformly convex Banach space; then there exists a continuous strictly increasing function with , such that
**
for all and such that , and .*

Lemma 5 (see [14]). *Let , be sequences of nonnegative real numbers such that . If , then .*

Lemma 6 (see [15]). *For , , and , the following identity holds
*

#### 3. Convergence to a Common Fixed Point of Nonexpansive Mappings

In this part, we prove our main theorem for finding a common fixed point of nonexpansive mappings in Banach space.

Theorem 7. *Let be a nonempty closed convex subset of a uniformly convex Banach space , and let , , and be three nonexpansive commuting mappings of satisfying Condition and . Given that , , and are sequences in () such that , , for all .**Define a sequence in by algorithm (2); then strongly converges to a common fixed point of , , and .*

*Proof. *First, we observe that is bounded; if we take an arbitrary fixed point of , noting that and , we have
By Lemma 1 and , exists. Denote
Hence, is bounded, so are and . Now
Since , this implies that
Moreover, implies that
Thus,
given by Lemma 2 that
By (10) and , then we have
That is,
where for all , for all and for .

Next, we prove that is a Cauchy sequence.

Since arbitrarily and exists, consequently, exists by Lemma 3. From Lemma 3 and (16), we get
Since is a nondecreasing function satisfying , for all , therefore, we have .

Let ; since , therefore, there exists a constant such that, for all , we have . There must exist , such that
From (18), it can be obtained that, when and ,

This implies that is a Cauchy sequence in a closed convex subset of a Banach space . Thus, it must converge to a point in ; let .

For all , as , thus, there exists a number such that, when ,

In fact, implies that using number above, when , we have . In particular, . Thus, there must exist , such that
From (22) and (23), we get
As is an arbitrary positive number, thus, .

Let
Then we have
hence,
for . Summing from to , we have
where ; since , we get
Since , from Lemma 5, we get . Hence,
Since and are nonexpansive mappings, we have
Since , , it follows from Lemma 1 that exists. Therefore, from (30), we get
Let
Using the same argument we can get
Since is a nonexpansive mapping, we have
Since , , it follows from Lemma 1 that exists. Therefore, from (34), we get

Then using the same argument, we can show that converges strongly to a common fixed point of , , and .

#### 4. Convergence to a Common Fixed Point of Asymptotically Nonexpansive Mappings

##### 4.1. There Exists One Sequence

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of one sequence.

Theorem 8. *Let be a nonempty closed convex subset of a uniformly convex Banach space , and let and be two asymptotically nonexpansive mappings of satisfying Condition (I) and . Given , , and are sequences in () and with such that , and for all .**Define a sequence in by algorithm (3); then strongly converges to a common fixed point of and .*

*Proof. *First, we observe that is bounded; if we take an arbitrary fixed point of , noting that and , we have
By Lemma 1 and , thus, exists. Denote , and put . Hence, is bounded, so are and . Now
Since , this implies that
Moreover, implies that
Thus,
given by Lemma 2 that
Now,
Hence, by (42),
Also note that
so that condition and (44) give
Next, we show .

We have
Hence, by (42) and (46), we get
By (37) we assume that , so for , and now,
That is,
where , for all , for all and for .

Next, we prove that is a Cauchy sequence.

As proved in Theorem 7, it is easy to see that is a Cauchy sequence in a closed convex subset of a Banach space . Thus, it must converge to a point in ; let .

For all , as , thus, there exists a number such that, when ,

In fact, implies that using number above, when , we get . In particular, . Thus, there must exist , such that
From (51) and (52), using the same argument in Theorem 7, we get .

Now, we return to prove .

It is easy to see that , and then combined with (44), we have ; thus, , and then we get
Moreover, implies that
thus,
Therefore, we have
Thus, we can show that converges strongly to a common fixed point of and immediately.

##### 4.2. There Exist Two Sequences and

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of two sequences.

Theorem 9. *Let be a nonempty bounded closed convex subset of a uniformly convex Banach space , and let and be two asymptotically nonexpansive mappings of satisfying Condition and . Given , , and are sequences in (0,1) and with such that , , and for all .**Define two sequences and in by the algorithm (5); then and strongly converge to a common fixed point of and .*

*Proof. *By the boundedness of , we obverse that both and are bounded; if we take an arbitrary fixed point of , noting that and , we have
By Lemma 1 and , , thus, exists. Denote
Similarly, we have
Since both and are bounded, put . We get that and are bounded. Now
Since , this implies that
Moreover, implies that
Thus,
given by Lemma 2 that
Also note that
so that condition and (64) give
We have
Hence, by (64) and (66), we get
and then we assume that , so for ; now by (57), we obtain that
where .

By and the convergence of , that is,
where , , for all , and for .

Next, we prove that is a Cauchy sequence.

Since arbitrarily and exists, consequently, exists by Lemma 3. From Lemma 3 and (68), we get
Since is a nondecreasing function satisfying , for all , therefore, we have
Let , since and , therefore, there exists a constant such that, for all , we have
in particular,
There must exist such that
From (70), it can be obtained that, when ,
This implies that is a Cauchy sequence in a closed convex subset of a Banach space . Thus, it must converge to a point in ; let .

For all , on lines similar to Theorem 8, from , it can be proved that . As is an arbitrary positive number, thus, . Similarly, we have , and then ( as ).

Finally, we prove .

Let , and put , for all . Then,
We have . Now,
Since and the boundedness of and , we get
Moreover, and imply that
Thus,
given by Lemma 2 that
So
so we obtain that for (64) and (82), and it means . This completes the proof.

#### 5. Application: Convergence to a Zero of Accretive Operators

Let be a real Banach space. Recall that an operator (possibly multivalued) with domain and range in is said to be accretive if, for each and , there exists a such that where is the normalized duality map from to the dual space given by An accretive operator is -accretive if for all . Denote the zero set of by

For an -accretive operator with and convex, the problem of finding a zero of , that is, has extensively been investigated due to its applications in related problems such as minimization problems, variational inequality problems, and nonlinear evolution equations.

It is known that the *resolvent* of , defined by
for , is a nonexpansive mapping from to and it is straightforward to see that coincides with the fixed point set of for any . Therefore, (87) is equivalent to the fixed point problem . Then an interesting approach to solving this problem is via iterative methods for nonexpansive mappings. We need the resolvent identity [15].

Theorem 10. *Let be a uniformly convex Banach space, and let be an m-accretive operator in such that , is nonexpansive commuting mappings for all () satisfying Condition . Given , , and are sequences in (0,1), such that , , and for some for all .**Define a sequence by algorithm (4); then strongly converges to a zero of .*

*Proof. *Take any arbitrary ; it follows from Lemma 1 that exists. From Lemma 2, it can be shown that . Since is nonexpansive for all , it follows from Lemma 6 that . Therefore, all the conditions in Theorem 7 are satisfied. The conclusion of Theorem 10 can be obtained from Theorem 7 immediately.

Theorem 11. *Let be a nonempty bounded closed convex subset of a uniformly convex Banach space , and let be an m-accretive operator in such that , is nonexpansive for all () satisfying Condition (I). Given , , and are sequences in (0,1) such that , , and for some for all .**Define two sequences and in by algorithm (6); then and strongly converge to a zero of .*

*Proof. *Only a sketch of the proof is given here.

Take any arbitrary ; it follows from Lemma 1 that and exist. From Lemma 2, it can be shown that and . Since is nonexpansive for all (), it follows from lemma 2.6 that and . Therefore, all the conditions in Theorems 7, 8, and 9, are satisfied and using the same argument in those theorems, the conclusion of Theorem 11 can be obtained immediately.

#### Authors’ Contribution

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 11171046) and supported by the Scientific Research Foundation of CUIT (no. KYTZ201004).

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