Viscosity Approximation Methods for Two Accretive Operators in Banach Spaces
We introduced a viscosity iterative scheme for approximating the common zero of two accretive operators in a strictly convex Banach space which has a uniformly Gâteaux differentiable norm. Some strong convergence theorems are proved, which improve and extend the results of Ceng et al. (2009) and some others.
1. Introduction and Preliminaries
Let be a real Banach space, a nonempty closed convex subset of , and a mapping. Recall that is nonexpansive if , for all . A point is a fixed point of provided that . Denote by the set of fixed points of ; that is, . Throughout this paper, we assume that is a nonexpansive mapping such that . Recall that a self-mapping is a contraction on if there exists a constant such that . Let is a contraction with constant . The normalized duality mapping from into is given by , , where denotes the dual space of and denotes the generalized duality pairing.
A Banach space is said to be strictly convex if , for all with . It is said to be uniformly convex if , for any two sequences , in such that and .
The norm of is said to be Gâteaux differentiable if exists for each in its unit sphere . Such an is called a smooth Banach space. The norm is said to be uniformly Gâteaux differentiable if, for each , the limit is attained uniformly for . It is well known that is smooth if and only if the duality mapping is single valued and that, if has a uniformly Gâteaux differentiable norm, is uniformly norm to weak* continuous on each bounded subset of (cf. ).
Let be a subset of . Then is called a retraction from onto if for all . A retraction is said to be sunny if for all and whenever . A subset of is said to be a sunny nonexpansive retract of if there exists a sunny nonexpansive retraction of onto . In a smooth Banach space , it is known that is a sunny nonexpansive retraction if and only if the following condition holds (cf. [2, page 48]):
Recall that an operator with and in is said to be accretive if, for each and , , there exists a such that . An accretive operator is -accretive if , for all . Denote by the set of zeros of ; that is, .
Denote by the resolvent of ; that is, . It is well known that , for all . And if is convex, then is a nonexpansive mapping from to . If is a Hilbert space, then is a maximal monotone operator if and only if is an -accretive operator.
Recently, the approximation of zeros of accretive operators has been studied extensively (see, e.g., [3–9]). Specially, Ceng et al.  studied the following composite iterative scheme in uniformly smooth Banach spaces: where is an arbitrary (but fixed) element. They proved that generated by (3) converges strongly to a zero of -accretive operator under certain appropriate conditions.
Very recently, Chen et al.  considered the following viscosity iterative scheme in a reflexive Banach space having a weakly sequentially continuous duality mapping: where . Under some conditions, they showed that generated by (4) converges strongly to a zero of -accretive operator .
In this paper, motivated by [10–14], we will consider the following so-called composite viscosity iterative scheme for finding a common zero of two -accretive operators: where and are -accretive operators, such that , , , and . Under some conditions, we will prove that generated by (5) converges strongly to a common zero of and in a strictly convex and reflexive Banach space having a uniformly Gâteaux differentiable norm, which improve the corresponding results in [10–13].
Lemma 1 (see ). In a Banach space , the following inequality holds: where .
Lemma 3 (the resolvent identity ). For , and ,
Lemma 4 (see [3, Theorem 4.1, page 287]). Let be a uniformly smooth Banach space, be a closed convex subset of , a nonexpansive mapping with , and . Then defined by the following converges strongly to a point in . If, moreover, one defines by then solves the variational inequality
Recall that a mapping is said to be weakly contractive [15, 16] if where is a continuous and strictly increasing function such that is positive on and . As a special case, if for , where , then the weakly contractive mapping is a contraction with constant . Rhodes  obtained the following result for weakly contractive mapping (see also ).
Lemma 5 (see [17, Theorem 2]). Let be a complete metric space and a weakly contractive mapping on . Then has a unique fixed point in .
Lemma 6. Let and be two sequences of nonnegative real numbers and a sequence of positive numbers satisfying the conditions(i),(ii).
Let the recursive inequality be given where is a continuous and strict increasing function on with . Then .
2. Main Results
Throughout this section, we assume the following:
(i) is a strictly convex Banach space which has a uniformly Gâteaux differentiable norm, and is a nonempty closed convex subset of .
(ii) Take , . Obviously is nonexpansive mapping and , if is a strictly convex Banach space. Indeed, it is easy to see that . Let , ; then From the above formula, we obtain , so . Similarly, . But Then the strict convexity of implies that , that is, , or .
Theorem 7. Let be a strictly convex Banach space which has a uniformly Gâteaux differentiable norm, two m-accretive maps in such that is convex and , and a fixed contraction mapping with contract constant . Suppose that , , and satisfy the following conditions:(i), , as ;(ii), and as ;(iii), , and .
Let be the composite viscosity process defined by Then converges strongly to , where is the unique solution of the following variational inequality:
Proof. First, by using Lemma 4, we know that there exists the unique solution of a variational inequality
where and is defined by for each and .
Next, we will divide our discussion into the following steps.
Step 1. We will show that is bounded.
In fact, take . Then Therefore, Using the induction method, we have which implies that , , , and are all bounded. Since , then is bounded. Following the conditions of (i) and (ii), we obtain that
Step 2. We show that .
For this, we estimate first. From (16), we know that Then simple calculations show that It follows from (25) that In view of Lemma 3, we have If , then Similarly, Thus, let ; we have Substituting (30) into (26) we get where is a constant such that On the other hand, we have Simple calculations show that It follows that Substituting (31) into (35) we get where is a constant such that From conditions (i)–(iii), we have that Hence, noticing (36) and applying Lemma 2, we obtain . Then by (22) we obtain
Step 3. We prove that , .
In fact, since from (22) and (23), we have . Then Hence, we have
Step 4. We show that , .
To prove this, let be a subsequence of such that
By Lemma 4, , where . Then For each integer , let such that Using Lemma 1, we get and hence Since , , and are bounded, then . Therefore, We also know that Notice that , , , and is norm to weak* uniformly continuous on bounded subset of ; then we obtain From (48), (49), and the two results mentioned above, we have Using (22) and the property of , we obtain the result that
Step 5. .
Using (16), we have Applying Lemma 1, we obtain where . Put From the conditions (i) and (ii), the result of Step 4, and the facts that and , we know that , and . In view of Lemma 2, (54) reduces to then we know that This completes the proof.
Remark 8. If we modify (16) as follows: or Then, imitating the proof of Theorem 7, we can also get the result of Theorem 7. Therefore, from the compare of iterative scheme, the conclusions of [10, 11] are special cases of Theorem 7.
Example 9. Next we study the following optimization problem: where is an interior nonempty closed convex subset of a Hilbert space and are two proper convex and lower semicontinuous functionals. To solve optimization problem (61), we will list the following well known results.
Proposition 10 (see ). Let be a proper convex and lower semicontinuous functional. Then(i) ( denotes the subdifferential in the sense of convex analysis) is a maximal monotone mapping;(ii) if and only if .
In Hilbert space is a -accretive mapping. Thus , are two -accretive mappings. Solving optimization problem (61) is equivalent to finding a common zero of and .
Let Then the conditions (i), (ii), and (iii) of Theorem 7 are satisfied. For arbitrary the sequence generated by (16) converges strongly to a common zero of and , which is also the solution of the optimization problem (60).
Theorem 11. Let be two accretive maps in with and satisfying the following range conditions: , which are convex. Let , , , , and , be the same as those in Theorem 7. Let be a sequence generated by (16). Then converges strongly to , where is the unique solution of the following variational inequality:
Theorem 12. Let be two accretive maps in with and satisfying the following range conditions: , which are convex. Let , , , and be the same as those in Theorem 7. Let be a weakly contractive mapping with the function . Let be a sequence generated by Then converges strongly to , where is a sunny nonexpansive retraction from onto .
Proof. Since is a uniformly smooth Banach space, then there is a sunny nonexpansive retraction from onto . Then is a weakly contractive mapping of into itself. Indeed, for all , Lemma 5 assures that there exists a unique element such that . Such is an element of . Now we define a iterative scheme as follows: Let be the sequence generated by (66). Then Remark 8 (59) assures that converges strongly to as . For any , we have Thus, we obtain for the following recursive inequality: Since , it follows from Lemma 6 that . Hence This completes the proof.
In virtue of the weakly contractive mapping being a contraction, using Theorem 12 we may obtain the following.
Corollary 13. Let be two accretive maps in with and satisfying the following range conditions: , which are convex. Let , , , and be the same as those in Theorem 7. Let be a sequence generated by Then converges strongly to , where is a sunny nonexpansive retraction from onto .
Example 14. Next we give an essential example.
Let be an bounded domain in a Euclidean space with Lipschitz boundary . Let be a given function shch that for each (i) is a proper, convex, lower semicontinuous function with .(ii) (subdifferential of ) is the maximal monotone mapping on with and for each , the function is measurable for .
Let be a continuous, monotone function such that there exist constants satisfying (i) and (ii) for each .
Definition 15 (see ). One first defines a mapping ( is a sobolev space) by
for . Clearly is an everywhere defined, monotone, demicontinuous operator from into . Second one defines an operator for as follows.
(i) For one defines the domain of by Here is the proper, convex, l.s.c. function (see [19, Lemma 3.1]). For we set
(ii) For , one defines as the -closure of defined in (i) above.
For the operator one has following results.
Lemma 16 (see [19, Lemma 3.4]). is -accretive operator ().
Lemma 17 (see [19, Proposition 3.2]). Let such that . Then(i) , a.e. on and(ii) for a.e. .
Lemma 18 (see [19, Proposition 3.3]). Let for . Then .
Clearly for different , are two m-accretive operators. The above results show that For the sake of finding a common zero, Theorems 7, 11, and 12 provided three different iterative algorithms. Therefore the study of a common zero of two accretive operators makes sense.
Remark 19. The results presented in this paper substantially improve and extend the results of Ceng et al.  from the following aspects.(1)Theorems 7 and 12 extend the result on the iterative construction of the zero for a single accretive operator to the case of that for common zeros of two accretive operators. If we modify two accretive operators as finite accretive operators, then, imitating the proof of Theorem 7, we can also get the result of Theorem 7.(2)Our results include one or two different viscosity items. Remark 8 shows that the conclusions of Ceng et al. are special cases of this paper.(3)The viscosity item is changed from a contractive mapping to weakly contractive mapping in Theorem 12.
This work was supported by the National Natural Science Foundation of China (Grants no. 11101115, 11271106) and the Natural Science Foundation of Hebei Province (A2011201053).
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