Abstract

We prove strong and weak convergence theorems of modified hybrid proximal-point algorithms for finding a common element of the zero point of a maximal monotone operator, the set of solutions of equilibrium problems, and the set of solution of the variational inequality operators of an inverse strongly monotone in a Banach space under different conditions. Moreover, applications to complementarity problems are given. Our results modify and improve the recently announced ones by Li and Song (2008) and many authors.

1. Introduction

Let be a Banach space with norm , a nonempty closed convex subset of , let denote the dual of and is the pairing between and .

Consider the problem of finding where is an operator from into . Such is called a zero point of . When is a maximal monotone operator, a well-known method for solving (1.1) in a Hilbert space is the proximal point algorithm and where and , then Rockafellar [1] proved that the sequence converges weakly to an element of .

In 2000, Kamimura and Takahashi [2] proved the following strong convergence theorem in Hilbert spaces, by the following algorithm: where , then the sequence converges strongly to , where is the projection from onto . These results were extended to more general Banach spaces see [3, 4].

In 2004, Kohsaka and Takahashi [4] introduced the following iterative sequence for a maximal monotone operator in a smooth and uniformly convex Banach space: and where is the duality mapping from into and .

Recently, Li and Song [5] proved a strong convergence theorem in a Banach space, by the following algorithm: and with the coefficient sequences , and satisfying , , , and . Where is the duality mapping from into and . Then, they proved that the sequence converges strongly to , where is the generalized projection from onto .

Let be a nonempty closed convex subset of , and let be a monotone operator of into . The variational inequality problem is to find a point such that The set of solutions of the variational inequality problem is denoted by . Such a problem is connected with the convex minimization problem, the complementarity problem, the problem of finding a point satisfying , and so on. An operator of into is said to be inverse-strongly monotone if there exists a positive real number such that for all . In such a case, is said to be -inverse-strongly monotone. If an operator of into is -inverse-strongly monotone, then is Lipschitz continuous, that is, for all .

In a Hilbert space , Iiduka et al. [6] proved that the sequence defined by: and where is the metric projection of onto and is a sequence of positive real numbers, converges weakly to some element of .

In 2008, Iiduka and Takahashi [7] introduced the folowing iterative scheme for finding a solution of the variational inequality problem for an inverse-strongly monotone operator in a Banach space and for every , where is the generalized metric projection from onto , is the duality mapping from into and is a sequence of positive real numbers. They proved that the sequence generated by (1.9) converges weakly to some element of .

Let be a bifunction of into and a real-valued function. The mixed equilibrium problem, denoted by , is to find such that If , the problem (1.10) reduces into the equilibrium problem for , denoted by , is to find such that If , the problem (1.10) reduces into the minimize problem, denoted by , is to find such that The above formulation (1.11) was shown in [8] to cover monotone inclusion problems, saddle point problems, variational inequality problems, minimization problems, optimization problems, variational inequality problems, vector equilibrium problems, and Nash equilibria in noncooperative games. In addition, there are several other problems, for example, the complementarity problem, fixed point problem, and optimization problem, which can also be written in the form of an . In other words, the is an unifying model for several problems arising in physics, engineering, science, optimization, economics, and so forth. In the last two decades, many papers have appeared in the literature on the existence of solutions of ; see, for example, [811] and references therein. Some solution methods have been proposed to solve the ; see, for example, [9, 1121] and references therein. In 2005, Combettes and Hirstoaga [12] introduced an iterative scheme of finding the best approximation to the initial data when is nonempty and they also proved a strong convergence theorem.

Recall, a mapping is said to be nonexpansive if for all . We denote by the set of fixed points of . If is bounded closed convex and is a nonexpansive mapping of into itself, then is nonempty (see [22]). A mapping is said to be quasi-nonexpansive if and for all and . It is easy to see that if is nonexpansive with , then it is quasi-nonexpansive. We write (, resp.) if converges (weakly, resp.) to . Let be a real Banach space with norm and let be the normalized duality mapping from into given by for all , where denotes the dual space of and the generalized duality pairing between and . It is well known that if is uniformly convex, then is uniformly continuous on bounded subsets of .

Let be a closed convex subset of , and let be a mapping from into itself. A point in is said to be an asymptotic fixed point of [23] if contains a sequence which converges weakly to such that . The set of asymptotic fixed points of will be denoted by . A mapping from into itself is said to be relatively nonexpansive [2426] if and for all and . The asymptotic behavior of a relatively nonexpansive mapping was studied in [27, 28]. is said to be -nonexpansive, if for . is said to be relatively quasi-nonexpansive if and for and .

In 2009, Takahashi and Zembayashi [29] introduced the following shrinking projection method of closed relatively nonexpansive mappings as follows: for every , where is the duality mapping on , satisfies and for some . Then, they proved that the sequence converges strongly to .

In 2009, Qin et al. [30] modified the Halpern-type iteration algorithm for closed quasi--nonexpansive mappings (or relatively quasi-nonexpansive) defined by Then, they proved that under appropriate control conditions the sequence converges strongly to .

Recently, Ceng et al. [31] proved the following strong convergence theorem for finding a common element of the set of solutions for an equilibrium and the set of a zero point for a maximal monotone operator in a Banach space Then, the sequence converges strongly to , where is the generalized projection of onto .

In this paper, motivated and inspired by Li and Song [5], Iiduka and Takahashi [7], Takahashi and Zembayashi [29], Ceng et al. [31] and Qin et al. [30], we introduce the following new hybrid proximal-point algorithms defined by : and Under appropriate conditions, we will prove that the sequence generated by algorithms (1.18) and (1.19) converges strongly to the point and converges weakly to the point , respectively. The results presented in this paper extend and improve the corresponding ones announced by Li and Song [5] and many authors in the literature.

2. Preliminaries

A Banach space is said to be strictly convex if for all with and . Let be the unit sphere of . Then, the Banach space is said to be smooth provided exists for each . It is also said to be uniformly smooth if the limit is attained uniformly for . The modulus of convexity of is the function defined by A Banach space is uniformly convex if and only if for all . Let be a fixed real number with . A Banach space is said to be -uniformly convex if there exists a constant such that for all ; see [32, 33] for more details. Observe that every -uniform convex is uniformly convex. One should note that no Banach space is -uniform convex for . It is well known that a Hilbert space is 2-uniformly convex and uniformly smooth. For each , the generalized duality mapping is defined by for all . In particular, is called the normalized duality mapping. If is a Hilbert space, then , where is the identity mapping. It is also known that if is uniformly smooth, then is uniformly norm-to-norm continuous on each bounded subset of .

We know the following (see [34]):(1)if is smooth, then is single-valued,(2)if is strictly convex, then is one-to-one and holds for all with ,(3)if is reflexive, then is surjective,(4)if is uniformly convex, then it is reflexive,(5)if is uniformly convex, then is uniformly norm-to-norm continuous on each bounded subset of .

The duality from a smooth Banach space into is said to be weakly sequentially continuous [35] if implies , where implies the weak convergence.

Lemma 2.1 (see [36, 37]). If be a 2-uniformly convex Banach space. Then, for all one has where is the normalized duality mapping of and .

The best constant in Lemma is called the 2-uniformly convex constant of ; see [32].

Lemma 2.2 (see [36, 38]). If a p-uniformly convex Banach space and let be a given real number with . Then, for all , and where is the generalized duality mapping of and is the p-uniformly convexity constant of .

Lemma 2.3 (see Xu [37]). Let be a uniformly convex Banach space. Then, for each , there exists a strictly increasing, continuous, and convex function such that and for all and .

Let be a smooth, strictly convex, and reflexive Banach space and let be a nonempty closed convex subset of . Throughout this paper, we denote by the function defined by Following Alber [39], the generalized projection is a map that assigns to an arbitrary point the minimum point of the functional , that is, , where is the solution to the minimization problem existence and uniqueness of the operator follows from the properties of the functional and strict monotonicity of the mapping . It is obvious from the definition of function that (see [39]) If is a Hilbert space, then .

If is a reflexive, strictly convex and smooth Banach space, then for , if and only if . It is sufficient to show that if , then . From (2.9), we have . This implies that . From the definition of , one has . Therefore, we have ; see [34, 40] for more details.

Lemma 2.4 (see Kamimura and Takahashi [3]). Let be a uniformly convex and smooth real Banach space and let be two sequences of . If and either or is bounded, then .

Lemma 2.5 (see Alber [39]). Let be a nonempty, closed, convex subset of a smooth Banach space and . Then, if and only if

Lemma 2.6 (see Alber [39]). Let be a reflexive, strictly convex, and smooth Banach space, let be a nonempty closed convex subset of and let . Then,

Let be a strictly convex, smooth, and reflexive Banach space, let be the duality mapping from into . Then, is also single-valued, one-to-one, and surjective, and it is the duality mapping from into . Define a function as follows (see [4]): for all and . Then, it is obvious that and .

Lemma 2.7 (see Kohsaka and Takahashi [4, Lemma 3.2]). Let be a strictly convex, smooth, and reflexive Banach space, and let be as in (2.12). Then, for all and .

Let be a reflexive, strictly convex, and smooth Banach space. Let be a closed convex subset of . Because is strictly convex and coercive in the first variable, we know that the minimization problem has a unique solution. The operator is said to be the generalized projection of on .

A set-valued mapping with domain and range is said to be monotone if for all . We denote the set by . is maximal monotone if its graph is not properly contained in the graph of any other monotone operator. If is maximal monotone, then the solution set is closed and convex.

Let be a reflexive, strictly convex, and smooth Banach space, it is known that is a maximal monotone if and only if for all .

Define the resolvent of by . In other words, for all . is a single-valued mapping from to . Also, for all , where is the set of all fixed points of . Define, for , the Yosida approximation of by . We know that for all and .

Lemma 2.8 (see Kohsaka and Takahashi [4, Lemma 3.1]). Let be a smooth, strictly convex, and reflexive Banach space, a maximal monotone operator with , and . Then, for all and .

Let be an inverse-strongly monotone mapping of into which is said to be hemicontinuous if for all , the mapping of into , defined by , is continuous with respect to the weak topology of . We define by the normal cone for at a point , that is,

Theorem 2.9 (see Rockafellar [1]). Let be a nonempty, closed, convex subset of a Banach space and a monotone, hemicontinuous operator of into . Let be an operator defined as follows: Then, is maximal monotone and .

Lemma 2.10 (see Tan and Xu [41]). Let and be two sequence of nonnegative real numbers satisfying the inequality If , then exists.

For solving the mixed equilibrium problem, let us assume that the bifunction and is convex and lower semicontinuous satisfies the following conditions: (A1) for all ,(A2) is monotone, that is, for all , (A3) for each , (A4) for each , is convex and lower semicontinuous.

Motivated by Blum and Oettli [8], Takahashi and Zembayashi [29, Lemma 2.7] obtained the following lemmas.

Lemma 2.11 (see [29, Lemma 2.7]). Let be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space , let be a bifunction from to satisfying (A1)–(A4), let , and let . Then, there exists such that

Lemma 2.12 (see Takahashi and Zembayashi [29]). Let be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space and let be a bifunction from to satisfying (A1)–(A4). For all and , define a mapping as follows: for all . Then, the followings hold: (1) is single-valued, (2) is a firmly nonexpansive-type mapping, that is, for all , (3), (4) is closed and convex.

Lemma 2.13 (see Takahashi and Zembayashi [29]). Let be a closed, convex subset of a smooth, strictly convex, and reflexive Banach space , let a bifunction from to satisfying (A1)–(A4) and let . Then, for and ,

Lemma 2.14. Let be a closed convex subset of a smooth, strictly convex and reflexive Banach space . Let is convex and lower semicontinuous and be a bifunction from to satisfying (A1)–(A4). For and , then there exists such that Define a mapping as follows: for all . Then, the followings hold: (1) is single-valued,(2) is firmly nonexpansive, that is, for all , ,(3), (4) is closed and convex.

Proof. Define a bifunction as follows:
It is easily seen that satisfies (A1)–(A4). Therefore, in Lemma 2.14 can be obtained from Lemma 2.12 immediately.

3. Strong Convergence Theorem

In this section, we prove a strong convergence theorem for finding a common element of the zero point of a maximal monotone operator, the set of solutions of equilibrium problems, and the set of solution of the variational inequality operators of an inverse strongly monotone in a Banach space by using the shrinking hybrid projection method.

Theorem 3.1. Let be a 2-uniformly convex and uniformly smooth Banach space and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4) let be a lower semicontinuous and convex function, and let be a maximal monotone operator. Let for and let be an -inverse-strongly monotone operator of into with and for all and . Let be a sequence generated by with and , for , where is the generalized projection from onto , is the duality mapping on . The coefficient sequence , satisfying , , and for some with , is the 2-uniformly convexity constant of . Then, the sequence converges strongly to .

Proof. We first show that is bounded. Put , let , and let be a sequence of mapping define as Lemma 2.14 and . By (3.1) and Lemma 2.7, the convexity of the function in the second variable, we have Since and is -inverse-strongly monotone, we have and by Lemma 2.1, we obtain Substituting (3.3) and (3.4) into (3.2), we get By Lemmas 2.7, 2.8 and (3.5), we have It follows that From (3.1) and (3.7), we obtain So, we have . This implies that , for all .
From Lemma 2.5 and , we have From Lemma 2.6, one has for all and . Then, the sequence is bounded. Since and , we have Therefore, is nondecreasing. Hence, the limit of exists. By the construction of , one has that and for any positive integer . It follows that Letting in (3.12), we get . It follows from Lemma 2.4, that as , that is, is a Cauchy sequence. Since is a Banach space and is closed and convex, we can assume that , as . Since for all , we also have . From Lemma 2.4, we get . Since and by definition of , we have Noticing the conditions and , we obtain From again Lemma 2.4, So, by the triangle inequality, we get Since is uniformly norm-to-norm continuous on bounded sets, we have On the other hand, we observe that It follows that From (3.1), (3.5), (3.6), (3.7), and (3.8), we have and then From conditions , and (3.20), we obtain By again Lemma 2.4, we have .
Since is uniformly norm-to-norm continuous on bounded sets, we obtain Applying (3.5) and (3.6), we observe that and, hence, for all . Since , , and (3.20), we have From Lemmas 2.6, 2.7, and (3.4), we get From Lemma 2.4 and (3.27), we have Since is uniformly norm-to-norm continuous on bounded sets, we obtain Since is bounded, there exists a subsequence of such that . Since , then we get as .
Now, we claim that . First, we show that . Indeed, since , it follows from (3.24) that If , then it holds from the monotonicity of that for all . Letting , we get . Then, the maximality of implies .
Next, we show that . Let be an operator as follows: By Theorem 2.9, is maximal monotone and . Let . Since , we get . From , we have On the other hand, since , then by Lemma 2.5, we have Thus, It follows from (3.34) and (3.36) that where . From (3.29) and (3.30), we obtain . By the maximality of , we have and, hence, .
Next, we show that . Since . From Lemmas 2.13 and 2.14, we have Similarly by (3.20), and so Since is uniformly norm-to-norm continuous on bounded sets, we obtain From (3.1) and (A2), we also have Hence, From , , we get . Since , it follows by (A4) and the weak, lower semicontinuous of that For with and , let . Since and , we have and hence . So, from (A1), (A4), and the convexity of , we have Dividing by , we get . From (A3) and the weakly lower semicontinuity of , we have for all implies . Hence, .
Finally, we show that . Indeed, from and Lemma 2.5, we have Since , we also have Taking limit , we have By again Lemma 2.5, we can conclude that . This completes the proof.

Corollary 3.2. Let be a 2-uniformly convex and uniformly smooth Banach space, let be a nonempty, closed, convex subset of . Let be a bifunction from to satisfying (A1)–(A4) let be a lower semicontinuous and convex function, and let be a maximal monotone operator. Let for with . Let be a sequence generated by with and , for , where is the generalized projection from onto , is the duality mapping on . The coefficient sequence , satisfying , and . Then, the sequence converges strongly to .

Proof. In Theorem 3.1 if , then (3.1) reduced to (3.49).

4. Weak Convergence Theorem

In this section, we first prove the following strong convergence theorem by using the idea of Plubtieng and Sriprad [42].

Theorem 4.1. Let be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping is weak sequentially continuous. Let be a maximal monotone operator and let for . Let be a nonempty, closed, convex subset of such that , let be a bifunction from to satisfying (A1)–(A4), let be a lower semicontinuous and convex function, and let be an -inverse-strongly monotone operator of into with and for all and . Let be a sequence generated by and for , where is the generalized projection from onto , is the duality mapping on . The coefficient sequence , satisfying , and for some with , is the 2-uniformly convexity constant of . Then, the sequence converges strongly to an element of , which is a unique element such that where is the generalized projection from onto .

Proof. Put . Let , by Lemma 2.14 and nonexpansiveness of , we have By (4.1) and Lemma 2.7, the convexity of the function in the second variable, we obtain Since and is -inverse-strongly monotone, we also have Substituting (4.5) and (4.6) into (4.4) and (4.3), we get By Lemmas 2.7, 2.8, (4.7), and using the same argument in Theorem 3.1, (3.6), we obtain and hence by Lemma 2.6 and (4.7), we note that for all . So, from and Lemma 2.10, we deduce that exists. This implies that is bounded. It implies that , , , and are bounded. Define a function as follows: Then, by the same argument as in proof of [43, Theorem 3.1], we obtain is a continuous convex function and if , then . Hence, by [34, Theorem 1.3.11], there exists a point such that Put for all . We next prove that as . Suppose on the contrary that there exists such that, for each , there is satisfying . Since , we have for all . This implies that Since for all and is bounded, is bounded. By Lemma 2.3, there exists a stricly increasing, continuous, and convex function such that and for all . Now, choose satisfying . Hence, there exists such that for all . Thus, there exists satisfying the following: From (4.9), (4.14), and (4.16), we obtain for all . Hence, This is a contradiction. So, converges strongly to . Consequently, is the unique element of such that This completes the proof.

Now, we prove a weak convergence theorem for the algorithm (4.20) below under different condition on data.

Theorem 4.2. Let be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping is weakly sequentially continuous. Let be a maximal monotone operator and let for . Let be a nonempty closed convex subset of such that , let be a bifunction from to satisfying (A1)–(A4), let be a lower semicontinuous and convex function, and let be an -inverse-strongly monotone operator of into with and for all and . Let be a sequence generated by and for , where is the generalized projection from onto , is the duality mapping on . The coefficient sequence , satisfying , and for some with , is the 2-uniformly convexity constant of . Then, the sequence converges weakly to an element of , where .

Proof. By Theorem 4.1, we have is bounded and so are , .
From (4.9), we obtain and then Since , and exists, then we have By again Lemma 2.4, we have . Since is uniformly norm-to-norm continuous on bounded sets, we obtain Apply (4.7), (4.8), and (4.9), we observe that and hence for all . Since , and , we have
From Lemmas 2.6, 2.7, and (4.7), we get From Lemma 2.4 and (4.27), we have Since is uniformly norm-to-norm continuous on bounded sets, we obtain
Since is bounded, there exists a subsequence of such that . It follows that and as .
Now, we claim that . First, we show that . Indeed, since , it follows that If , then it holds from the monotonicity of that for all . Letting , we get . Then, the maximality of implies .
Next, we show that . Let be an operator as follows: By Theorem 2.9, is maximal monotone and . Let . Since , we get . From , we have On the other hand, since . Then, by Lemma 2.5, we have Thus, It follows from (4.34) and (4.36) that where . From (4.29) and (4.30), we obtain . By the maximality of , we have and hence .
Next, we show . From . It follows from (4.7), (4.8), and (4.9) that or, equivalently, with and , yield that .
From Lemmas 2.13 and 2.14, for , This implies that . Noticing Lemma 2.4, we get Since is uniformly norm-to-norm continuous on bounded sets, we obtain
From (4.20) and (A2), we also have Hence, From , we get . Since , it follows by (A4) and the weakly lower semicontinuous of that For with and , let . Since and , we have and hence . So, from (A1), (A4), and the convexity of , we have Dividing by , we get . From (A3) and the weakly lower semicontinuity of , we have for all implies . Hence, .
By Theorem 4.1, the converges strongly to a point which is a unique element of such that By the uniform smoothness of , we also have .
Finally, we prove . From Lemma 2.5 and , we have Since is weakly sequentially continuous, and . Then, On the other hand, since is monotone, we have Hence, Since is strict convexity, it follows that . Therefore, the sequence converges weakly to . This completes the proof.

5. Application to Complementarity Problems

Let be a nonempty, closed convex cone in and A an operator of into . We define its polar in to be the set Then, the element is called a solution of the complementarity problem if The set of solutions of the complementarity problem is denoted by ; see [34], for more detial.

Theorem 5.1. Let be a 2-uniformly convex and uniformly smooth Banach space and let be a nonempty closed convex subset of . Let be a bifunction from to satisfying (A1)–(A4) let be a lower semicontinuous and convex function, and let be a maximal monotone operator. Let for and let be an -inverse-strongly monotone operator of into with and for all and . For an initial point with and , for , where is the generalized projection from onto and is the duality mapping on . The coefficient sequence , satisfying , , and for some with , is the 2-uniformly convexity constant of . Then, the sequence converges strongly to .

Proof. As in the proof Lemma 7.1.1 of Takahashi in [44], we have . So, we obtain the desired result.

Theorem 5.2. Let be a 2-uniformly convex and uniformly smooth Banach space whose duality mapping is weakly sequentially continuous. Let be a maximal monotone operator and let for . Let be a nonempty closed convex subset of such that , let be a bifunction from to satisfying (A1)–(A4), let be a proper lower semicontinuous and convex function, and let be an -inverse-strongly monotone operator of into with and for all and . Let be a sequence generated by and for , where is the generalized projection from onto , is the duality mapping on . The coefficient sequence , satisfying , and for some with , is the 2-uniformly convexity constant of . Then, the sequence converges weakly to an element of , where .

Proof. It follows by Lemma 7.1.1 of Takahashi in [44], we have . Hence, Theorem 4.2, converges weakly to an element of , where .

Acknowledgments

The authors would like to thank the referees for their careful readings and valuable suggestions to improve the writing of this paper. This research is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.