1. Introduction

Let be a Banach space with the dual and let denote the dual space of . If , then is called reflexive. We denote by and the sets of positive integers and real numbers, respectively. Also, we denote by the normalized duality mapping from to defined by where denotes the generalized duality pairing. Recall that if is smooth, then is single-valued, and if is uniformly smooth, then is uniformly norm-to-norm continuous on bounded subsets of . We shall still denote by the single-valued duality mapping.

Let be a nonempty subset of be a mapping and let a function with for all and , where is a constant. A mapping is said to be relaxed semimonotone [1] if the following two conditions hold:(i)for each fixed is relaxed monotone; that is,(ii)for each fixed , is completely continuous; that is, for any net in , in weak topology of , then has a subsequence in norm topology of .

In case for all and , is called semi-monotone [2]. The following is an example of - semi-monotone mapping.

Example 1.1. Let , and where is a constant. Then, is relaxed - semi-monotone with
Let be a bifunction, a mapping, and , two real-valued functions, and let be a - semi-monotone mapping. We consider the problem of finding such that which is called the generalized mixed equilibrium problem with respect to relaxed semi-monotone mapping (GMEP()). The set of such is denoted by , that is,

Now, let us consider some special cases of the problem (1.5).(a) In the case of , (1.5) is deduced to the following variational-like inequality problem: The problem (1.7) was studied by Fang and Huang [1]. Using the KKM technique and - monotonicity of the mapping , they [1] obtained the existence of solutions of the variational-like inequality problem (1.7) in a real Banach space.(b) In the case of and for all , the problem (1.5) is deduced to the following variational inequality problem: The problem (1.8) was studied by Chen [2]. They obtained the existence results of solutions in a real Banach space.

When is a reflexive Banach space, we know , where is the duality mapping defined by , for all , which is an isometric mapping, so we may regard under an isometry. The following problems can be derived as special cases of the problem (1.5).(c) In case is reflexive (i.e., ), and for all , the problem (1.5) is deduced to the following variational inequality problem: The problem (1.9) was studied by Chen [2].(d) If is reflexive (i.e., ) and , (1.5) is deduced to the mixed equilibrium problem: The problem (1.10) was considered and studied by Ceng and Yao [3]; Cholamjiak and Suantai [4].(e)In the case of and , (1.5) is deduced to the following classical equilibrium problem: The set of all solution of (1.11) is denoted by , that is, Numerous problems in physics, optimization, and economics can be reduced to find a solution of the equilibrium problem, variational inequality problem, and related optimization problems; see, for instance, [5–11]. Some methods have been proposed to solve the equilibrium problem in a Hilbert space; see, for instance, Blum and Oettli [12]; Combettes and Hirstoaga [13]; Moudafi [14].

Let be a nonempty, closed convex subset of . A mapping is called nonexpansive if for all . Also a mapping is called asymptotically nonexpansive if there exists a sequence with as such that for all and for each . The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk [15] as an important generalization of nonexpansive mappings. Denote by the set of fixed points of , that is, . There are several methods for approximating fixed points of a nonexpansive mapping; see, for instance, [16–21]. Furthermore, since 1972, a host of authors have studied weak and strong convergence problems of the iterative processes for the class of asymptotically nonexpansive mappings; see, for instance, [22–25]. In 1953, Mann [16] introduced the following iterative procedure to approximate a fixed point of a nonexpansive mapping in a Hilbert space : where the initial point is taken in arbitrarily and is a sequence in . However, we note that Mann's iteration process (1.13) has only weak convergence, in general; for instance, see [26–28]. In 2003, Nakajo and Takahashi [29] introduced the following iterative algorithm for the nonexpansive mapping in the framework of Hilbert spaces: where , and is the metric projection from a Hilbert space onto . They proved that generated by (1.14) converges strongly to a fixed point of . Xu [30] extended Nakajo and Takahashi's theorem to Banach spaces by using the generalized projection.

Matsushita and Takahashi [17] introduced the following iterative algorithm in the framework of Banach spaces: where denoted the convex closure of the set is a sequence in with , and is the metric projection from onto .

Very recently, Dehghan [24] introduced the following iterative algorithm for finding fixed points of an asymptotically nonexpansive mapping in a uniformly convex and smooth Banach space: where denotes the convex closure of the set is the normalized duality mapping, is a sequence in with , and is the metric projection from onto . The strong convergence theorem of the iterative sequence defined by (1.16) is obtained in a uniformly convex and smooth Banach space.

In this paper, motivated and inspired by the above results, we first suggest and analyze the new generalized mixed equilibrium problem with respect to relaxed - semi-monotone mapping. Using the KKM technique, we obtain the existence of solutions for such problem in a Banach space. Next, we also introduce a hybrid projection algorithm for finding a common element in the solution set of a generalized mixed equilibrium problem and the fixed point set of an asymptotically nonexpansive mapping. The strong convergence theorem of the proposed sequence is obtained in a Banach space setting. The main results extend various results existing in the current literature.

2. Preliminaries

Let be a real Banach space, and let be the unit sphere of . A Banach space is said to be strictly convex if for any ,

It is also said to be uniformly convex if for each , there exists such that for any ,

It is known that a uniformly convex Banach space is reflexive and strictly convex. Define a function called the modulus of convexity of as follows:

Then is uniformly convex if and only if for all . A Banach space is said to be smooth if the limit exists for all . Let be a nonempty, closed, and convex subset of a reflexive, strictly convex, and smooth Banach space . Then for any , there exists a unique point such that

The mapping defined by is called the metric projection from onto . The following theorem is wellknown.

Theorem 2.1 (see [31]). Let be a nonempty, closed convex subset of a smooth Banach space and let , and . Then the following are equivalent:(a) is a best approximation to .(b) is a solution of the variational inequality: where is a duality mapping and is the metric projection from onto .

It is wellknown that if is a metric projection from a real Hilbert space onto a nonempty, closed, and convex subset , then is nonexpansive. But, in a general Banach space, this fact is not true.

In the sequel, we will need the following lemmas.

Lemma 2.2 (see [32]). Let be a uniformly convex Banach space, let be a sequence of real numbers such that for all , and let and be sequences in such that , and . Then .

Theorem 2.3 (see [33]). Let be a bounded, closed, and convex subset of a uniformly convex Banach space . Then there exists a strictly increasing, convex, and continuous function such that and for all , , with and nonexpansive mapping of into .

Theorem 2.4 (see [24]). Let be a bounded, closed, and convex subset of a uniformly convex Banach space . Then there exists a strictly increasing, convex, and continuous function such that and for all , ; with and an asymptotically nonexpansive mapping of into with the sequence .

Now, let us recall the following well-known concepts and results.

Definition 2.5. Let be a subset of topological vector space . A mapping is called a KKM mapping if for and , where co denotes the convex hull of the set .

Lemma 2.6 (see [34]). Let be a nonempty subset of a Hausdorff topological vector space , and let be a KKM mapping. If is closed for all and is compact for at least one , then .

Theorem 2.7 (see [35] (Kakutani-Fan-Glicksberg Fixed Point Theorem)). Let E be a locally convex Hausdorff topological vector space and a nonempty, convex, and compact subset of . Suppose is a upper semi-continuous mapping with nonempty, closed, and convex values. Then has a fixed point in .

Definition 2.8 (see [36]). Let be a nonempty, closed convex of a Banach space . Let and let be two mappings. is said to be -hemicontinuous if, for any fixed , the mapping defined by is continuous at .
For solving the mixed equilibrium problem, let us assume the following conditions for a bifunction :
(A1) for all ;(A2) is monotone, that is, for all ;(A3)for all , is weakly upper semicontinuous;(A4)for all , is convex. The following lemmas can be found in [37].

Lemma 2.9 (see [37]). Let be a nonempty, bounded, closed, and convex subset of a smooth, strictly convex and reflexive Banach space , let be an -hemicontinuous and relaxed monotone mapping. Let be a bifunction from to satisfying (A1) and (A4), and let be a lower semicontinuous and convex function from to . Let and . Assume that(i);(ii) for any fixed , the mapping is convex. Then the following problems (2.9) and (2.10) are equivalent. Find such that: Find such that

Lemma 2.10 (see [37]). Let be a nonempty, bounded, closed, and convex subset of a smooth, strictly convex, and reflexive Banach space , let be an -hemicontinuous and relaxed monotone mapping. Let be a bifunction from to satisfying (A1), (A3), and (A4), and let be a lower semicontinuous and convex function from to . Let and . Assume that(i) for all ;(ii)for any fixed , the mapping is convex and lower semicontinuous;(iii) is weakly lower semicontinuous; that is,  for any net converges to in implies that . Then, the solution set of the problem (2.9) is nonempty, that is, there exists such that

3. Existence Results of Generalized Mixed Equilibrium Problem

In this section, we prove the following crucial lemma concerning the generalized mixed equilibrium problem with respect to relaxed - semi-monotone mapping (GMEP()) in a real Banach space with the smooth and strictly convex second dual space.

Lemma 3.1. Let be a real Banach space with the smooth and strictly convex second dual space , let be a nonempty bounded closed convex subset of , let be a relaxed - semi-monotone mapping. Let be a bifunction satisfying (A1), (A3), and (A4), and let be a proper lower semicontinuous and convex function. Let and . Assume that(i) for all ;(ii)for any fixed , the mapping is convex and lower semicontinuous;(iii)for each is finite-dimensional continuous: that is, for any finite-dimensional subspace is continuous;(iv) is convex lower semicontinuous. Then there exists such that

Proof. Let be a finite-dimensional subspace with . For each , consider the following problem: find such that
Since is bounded closed and convex, is continuous on and relaxed monotone for each fixed , from Lemma 2.10, we know that problem (3.2) has a solution .
Now, define a set-valued mapping as follows: It follows from Lemma 2.9 that, for each fixed : Since every convex lower semicontinuous function in Banach spaces is weakly lower semicontinuous, the proper convex lower semicontinuity of and , condition (ii), (A3) and (A4) implies that has nonempty bounded closed and convex values. Using (A3) and the complete continuity of , we can conclude that is upper semicontinuous. It follows from Theorem 2.7 that has a fixed point , that is, Let and let By (3.5) and Lemma 2.9, we know that is nonempty and bounded. Denote by the -closure of in . Then, is compact in .
For any , , we know that , so has the finite intersection property. Therefore, it follows that Let . We claim that Indeed, for each , let be such that and . Then, there exists such that . The definition of implies that that is for all . Using the complete continuity of , (A3), (ii), the continuity of , the convex lower semicontinuity of , , and , and letting , we get From Lemma 2.9, we have Hence, we complete the proof.

Setting and in Lemma 3.1, we have the following result.

Corollary 3.2. Let be a real Banach space with the smooth and strictly convex second dual space , let be a nonempty bounded closed convex subset of . Let be a bifunction satisfying (A1), (A3), and (A4). Let and . Then there exists such that

If is reflexive (i.e., ) smooth and strictly convex real Banach space, then we have the following result.

Corollary 3.3. Let be a reflexive smooth and strictly convex Banach space, let be a nonempty bounded closed convex subset of , let be a relaxed semi-monotone mapping. Let be a bifunction satisfying (A1), (A3), and (A4), and let be a proper lower semicontinuous and convex function. Let and . Assume that(i) for all ;(ii)for any fixed , the mapping is convex and lower semicontinuous;(iii)for each is finite-dimensional continuous.(iv) is convex lower semicontinuous.Then, there exists such that

If is reflexive (i.e., ) smooth and strictly convex, is semi-monotone, then we obtain the following result.

Corollary 3.4. Let be a reflexive smooth and strictly convex Banach space, let be a nonempty bounded closed convex subset of , let be a semi-monotone mapping. Let be a bifunction satisfying (A1), (A3), and (A4), and let be a proper lower semicontinuous and convex function. Assume that, for any and ,(i)for any fixed , the mapping is convex and lower semicontinuous;(ii)for each is finite-dimensional continuous.Then, there exists such that

Theorem 3.5. Let be a real Banach space with the smooth and strictly convex second dual space , let be a nonempty, bounded, closed, and convex subset of , let be a relaxed - semi-monotone mapping. Let be a bifunction from to satisfying (A1)–(A4) and let be a lower semicontinuous and convex function from to . For any , define a mapping as follows: for all . Assume that(i) for all ;(ii)for any fixed , the mapping is convex and lower semicontinuous;(iii)for each is finite-dimensional continuous: that is, for any finite-dimensional subspace is continuous;(iv) is convex lower semicontinuous;(v) for any , ;(vi)for any , . Then, the following holds: is single-valued; for all ;; is nonempty, closed, and convex.

Proof. For each , by Lemma 2.10, we conclude that is nonempty.We prove that is single-valued. Indeed, for and , let . Then, Hence, Adding the two inequalities, from (i) we have From (A2), we have That is, Calculating the right-hand side of (3.22), we have and so, In (3.24) exchanging the position of and , we get Adding the inequalities (3.24) and (3.25) and using (v) and (vi), we have Hence, Since is monotone and is strictly convex, we obtain that and hence . Therefore, is single-valued. For , we have Adding the above two inequalities and by (i) and (A2), we get that is After calculating (3.30), we have In (3.30), exchanging the position of and , we get Adding the inequalities (3.31) and (3.32), use (i) and (vi), we have It follows from (iv) that Hence, Next, we show that . Indeed, we have the following: Hence, . Finally, we prove that is nonempty, closed, and convex. For each , we define the multivalued mapping by Since , we have . We prove that is a KKM mapping on . Suppose that there exists a finite subset of , and with such that for all . Then From (A1), (A4), (ii), and the convexity of , we have which is a contradiction. Thus, is a KKM mapping on .
Next, we prove that is closed for each . For any , let be any sequence in such that . We claim that . Then, for each , we have By monotonicity of , we obtain that By (A3), (i), (ii), (iv), lower semicontinuity of , and the complete continuity of , we obtain the following Hence, From Lemma 2.9, we have This shows that , and hence is closed for each . Thus, is also closed.
Next, we observe that is weakly compact. In fact, since is bounded, closed, and convex, we also have , which is weakly compact in the weak topology. By Lemma 2.6, we can conclude that .
Finally, we prove that is convex. In fact, let , and for . From (2), we know that This yields that Similarly, we also have It follows from (3.46) and (3.47) that Hence, and hence is convex. This completes the proof.