Research Article | Open Access

# Weak and Strong Convergence Theorems for Zeroes of Accretive Operators in Banach Spaces

**Academic Editor:**Alberto Cabada

#### Abstract

The purpose of this paper is to present two new forward-backward splitting schemes with relaxations and errors for finding a common element of the set of solutions to the variational inclusion problem with two accretive operators and the set of fixed points of nonexpansive mappings in infinite-dimensional Banach spaces. Under mild conditions, some weak and strong convergence theorems for approximating this common elements are proved. The methods in the paper are novel and different from those in the early and recent literature. Our results can be viewed as the improvement, supplementation, development, and extension of the corresponding results in the very recent literature.

#### 1. Introduction

Let be a nonempty closed convex subset of a real Hilbert space . Let be a single-valued nonlinear mapping and let be a multivalued mapping. The so-called quasivariational inclusion problem is to find a such that The set of solutions of (1) is denoted by . A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see, for instance, [1â€“4]. The problem (1) includes many problems as special cases as follows.(1)If , where is a proper convex lower semicontinuous function and is the subdifferential of , then the variational inclusion problem (1) is equivalent to find such that which is called the mixed quasivariational inequality (see, Noor [5]).(2)If , where is a nonempty closed convex subset of and is the indicator function of , that is, then the variational inclusion problem (1) is equivalent to find such that This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).

In [7], Zhang et al. investigated the problem of finding a common element of the set of solutions to the inclusion problem and the set of fixed points of nonexpansive mappings by considering the following iterative algorithm: where is an -cocoercive mapping, is a maximal monotone mapping, is a nonexpansive mapping, and is a sequence in . Under mild conditions, they obtained a strong convergence theorem.

In [8], Manaka and Takahashi introduced the following iteration: where is a sequence in (0, 1), is a positive sequence, is a nonexpansive mapping, is an inversely strongly monotone mapping, is a maximal monotone operator, and is the resolvent of . They showed that the sequence generated in (6) converges weakly to some provided that the control sequence satisfies some restrictions.

It is well known that, the quasivariational inclusion problem in the setting of Hilbert spaces has been extensively studied in the literature; see, for instance, [3â€“14]. However, there is little work in the existing literature on this problem in the setting of Banach spaces (though there were some work on finding a common zero of a finite family of accretive operators [15â€“17]). The main difficulties are due to the fact that the inner product structure of a Hilbert space fails to be true in a Banach space. To overcome these difficulties, LĂ³pez et al. [18] use the new technique to carry out certain initiative investigations on splitting methods for accretive operators in Banach spaces. They considered the following algorithms with errors in Banach spaces: where , and are the resolvent of . Then they studied the weak and strong convergence of the algorithms (7) and (8), respectively.

Motivated and inspired by Zhang et al. [7], Manaka and Takahashi [8], LĂ³pez et al. [18], and Cho et al. [19], the purpose of this paper is to introduce two iterative forward-backward splitting methods for finding a common element of the set of solutions of the variational inclusion problem (1) with -accretive operators and inverse strongly accretive operators and the set of fixed points of nonexpansive mappings in the setting of Banach space. Under suitable conditions, some weak and strong convergence theorems for approximating to this common elements are proved. The results presented in the paper not only improve and extend the main result in Zhang et al. [7], but also replenish and extend the corresponding results in Manaka and Takahashi [8], LĂ³pez et al. [18], and Cho et al. [19].

#### 2. Preliminaries

Throughout this paper, we denote by and a real Banach space and the dual space of , respectively. Let be a subset of and be a mapping on . We use to denote the set of fixed points of . Let be a real number. The generalized duality mapping is defined by for all , where denotes the generalized duality pairing between and . In particular, is called the normalized duality mapping and for . If is a Hilbert space, then , where is the identity mapping. It is well known that if is smooth, then is single-valued, which is denoted by .

The norm of a Banach space is said to be GĂ¢teaux differentiable if the limit exists for all on the unit sphere . If, for each , the limit (10) is uniformly attained for , then the norm of is said to be uniformly GĂ¢teaux differentiable. The norm of is said to be FrĂ©chet differentiable if, for each , the limit (10) is attained uniformly for .

Let be the modulus of smoothness of defined by

A Banach space is said to be uniformly smooth if as . Let . A Banach space is said to be -uniformly smooth, if there exists a fixed constant such that . It is well known that is uniformly smooth if and only if the norm of is uniformly FrĂ©chet differentiable. If is -uniformly smooth, then and is uniformly smooth, and hence the norm of is uniformly FrĂ©chet differentiable, in particular, the norm of is FrĂ©chet differentiable. Typical examples of both uniformly convex and uniformly smooth Banach spaces are , where . More precisely, is -uniformly smooth for every .

A Banach space is said to be uniformly convex if, for any , there exists such that, for any , implies . It is known that a uniformly convex Banach space is reflexive and strictly convex.

A Banach space is said to have the Kadec-Klee property if for every sequence in , and together imply .

A Banach space is said to satisfy Opialâ€™s condition if for any sequence in the condition that converges weakly to implies that the inequality holds for every with .

*Definition 1. *A mapping is said to be (1)nonexpansive if
(2)-contractive if for all , there exists such that
(3)accretive if for all , there exists such that
(4)-strongly accretive if for all , there exists and such that
(5)-inverse strongly accretive if for all , there exists and such that

*Definition 2. *A set-valued mapping is said to be(1)accretive if for any , there exists , such that for all and ,
(2)-accretive if is accretive and for every (equivalently, for some) , where is the identity mapping.

Let be -accretive. The mapping , defined by
is called the resolvent operator associated with , where is any positive number and is the identity mapping. It is well known that is single-valued and nonexpansive.

In order to prove our main results, we need the following lemmas.

Lemma 3 (see [20]). *Let be a Banach space and be a generalize duality mapping. Then for any given , the following inequality holds:
**
In particular, we have, for any given ,
*

Lemma 4 (see [21]). *Let be a sequence of nonnegative numbers satisfying the property:
**
where , , and satisfy the restrictions:*(i)*,*(ii)*, ,*(iii)*.**Then, .*

Lemma 5 (see [22]). *Let , , and be sequences of nonnegative real numbers satisfying the inequality
**
If and , then exists. In particular, whenever there exists a subsequence in which strongly converges to zero.*

Lemma 6 (see [20]). *Let , , be given.*(i)*If is uniformly convex, then there exists a continuous, strictly increasing, and convex function with such that
where , .*(ii)*If be a real -uniformly smooth Banach space, then there exists a constant such that
*

Lemma 7 (see [23]). *Let be a uniformly convex Banach space, a closed convex subset of , and a nonexpansive mapping. Then, is demiclosed at zero.*

Lemma 8 (see [24]). *If is a uniformly convex Banach space and is a closed convex bounded subset of , there is a continuous strictly increasing function with such that
**
for all and nonexpansive mappings .*

Lemma 9 (see [25]). *Let be a real reflexive Banach space such that its dual has the Kadec-Klee property. Let be a bounded sequence in and ; here denotes the weak -limit set of . Suppose exists for all . Then .*

Lemma 10. *Let be a nonempty closed convex subset of a real -uniformly smooth Banach space . Let the mapping be a -inverse strongly accretive operator. Then the following inequality holds:
**
In particular, if , then is nonexpansive.*

*Proof. *Indeed, for all , it follows from Lemma 6 that
It is clear that if , then is nonexpansive. This completes the proof.

Lemma 11. *Assume that is a nonempty closed subset of a real uniformly convex and -uniformly smooth Banach space . Suppose that is a single-valued and -inverse strongly accretive operator for some and is an -accretive operator in , with and . Moreover, denote by
**
and by
**
Then, it is holds for all that .*

*Proof. *From the definition of , it follows that

This lemma alludes to the fact that in order to solve the inclusion problem (1), it suffices to find a fixed point of . Since is already split, an iterative algorithm for corresponds to a splitting algorithm for (1). However, to guarantee convergence (weak or strong) of an iterative algorithm for , we need good metric properties of such as nonexpansivity. To this end, some relate geometric conditions on the underlying space are very necessary (see Lemmas 12 and 13 below).

Lemma 12 (see [18]). *Assume that is a nonempty closed subset of a real uniformly convex and -uniformly smooth Banach space . Suppose that is a single-valued and -inverse strongly accretive operator for some and is an -accretive operator in , with and . Then, the following relations hold.*(i)*Given and ,
*(ii)*Given , there exists a continuous, strictly increasing, and convex function with such that for all ,
*

Lemma 13. *Assume that is a nonempty closed subset of a real uniformly convex and -uniformly smooth Banach space . Suppose that is a nonexpansive mapping and is a single-valued and -inverse strongly accretive operator for some and is a -accretive operator in , with and . Assume . Then .*

*Proof. *Suppose that , it is sufficient to show that . Indeed, for , we have by Lemma 12 that
The property of and the condition together imply that
It turns out that
which imply
Noticing the assumption of , we can deduce . This means that .

Next we give a weak convergence theorem in a Banach space .

#### 3. Main Results

Theorem 14. *Let be a uniformly convex and -uniformly smooth Banach space. Let be -inverse strongly accretive, let be -accretive, and let be nonexpansive. Assume that . We define a sequence by the perturbed iterative scheme:
**
where , , , and . Assume that*(i)* and ;*(ii)*;
*(iii)*.**Then converges weakly to some point .*

*Proof. *We divide the proof into several steps.

First we prove that exist for any point .

Putting , one has
where
Then the iterative formula (38) turns into the form
Thus, by virtue of nonexpansivity of and , it follows that
It follows from (42) and the condition (i) that
Since , we can deduce due to Lemma 13. And is nonexpansive due to Lemma 12 and condition (iii). Therefore, we can get from (41) that
In view of (43), (44), and Lemma 5, we can get that exists. Therefor is bounded.

Next, we show .

Let be such that , for all , and let . By (41), Lemmas 3, and 12, we have
Meanwhile, by the fact that , , and (45), we can get that
Thanks to the conditions of (ii),(iii), and (43) and existence of , one has
Consequently,
*Stepâ€‰â€‰1.* We prove .

Noticing (43) and Lemma 6, we have
which implies
where . From (50), assumptions (ii), (43), and existence of , it turns out that
It follows from the property of that
Noticing (43), we have
By (48) and (53), we get
*Stepâ€‰â€‰2.* We prove .

Since , there exists such that for all . Then, by Lemma 12, we have
It follows from (48), (53), and (54) that
By Lemmas 7 and 13, we get
Finally, we show that converges weakly to a fixed point of .

Indeed, it suffices to show that consists of exactly only one point. To this end, we suppose that two different points and are in . Then there exist two different subsequences and such that and as and . Define by
Then can be written
where
Thanks to the nonexpansivity of , we have
It follows from (43) that
Let
Apply Lemma 8 to the closed convex bounded subset to obtain
Since exists, (62) and (64) together imply that
Furthermore, we have
After taking first and then in (66) and using (62) and (65), we get
So that exists for all . It follows from Lemma 9 that . This completes the proof.

*Remark 15. *Compared with the known results in the literature, our results are very different from those in the following aspects. (i)Theorem 14 improves and extends Theorem 3.1 of Manaka and Takahashi [8] and Theorem 3 of Kamimura and Takahashi [9] from Hilbert spaces to uniformly convex and -uniformly smooth Banach spaces.(ii)Theorem 14 also improves and extends Theorem 3.6 of LĂ³pez et al. [18] from the problem of finding an element of to the problem of finding an element of .

Theorem 16. *Let be a uniformly convex and -uniformly smooth Banach space. Let be -inverse strongly accretive, let be -accretive, let be -contractive, and let be nonexpansive. Assume that . We define a sequence by the perturbed iterative scheme:
**
where . Assume that , , and satisfying the following conditions:*(i)* and ;*(ii)* and ;*(iii)* and .**Then converges strongly to some point , which is the unique solution of the variational inequality , .*

*Proof. *Let be a sequence generated by
where . Hence to show the desired result, it suffices to prove that . Indeed, since and are both nonexpansive under the condition of (iii), it follows that
By virtue of Lemma 4 and (70), one has .

We first prove that the sequences is bounded.

Thanks to (69) and Lemma 13, we have
By induction, we have
Hence, is bounded, so are and .

Next we prove that
Putting , it follows from Lemma 12 that
where , , and . Hence from (69) and (74) we have
where . It follows from Lemma 4, (ii), and (iii) that .

Again from Lemmas 3 and 12, we obtain
where . It follows immediately from (76) that
Hence we obtain that
By condition (iii), there exists such that for all . Then, by Lemma 12, we get
*Stepâ€‰â€‰1.* We show .

From (73), (78), (79), and (ii), we have
Lemmas 7 and 13 imply that
Next we prove that
Equivalently (should ), we need to prove that
To this end, let satisfy . By Xuâ€™s Theorem 4.1 [26], we get as , which is the unique solution of the variational inequality:
Using subdifferential inequality, we deduce that
which implies that