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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 538912, 29 pages
doi:10.1155/2012/538912
Research Article

Iterative Algorithms for Solving the System of Mixed Equilibrium Problems, Fixed-Point Problems, and Variational Inclusions with Application to Minimization Problem

Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangmod, Thrungkru, Bangkok 10140, Thailand

Received 7 October 2011; Accepted 1 November 2011

Academic Editor: Yeong-Cheng Liou

Copyright © 2012 Tanom Chamnarnpan and Poom Kumam. 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 a new iterative algorithm for solving a common solution of the set of solutions of fixed point for an infinite family of nonexpansive mappings, the set of solution of a system of mixed equilibrium problems, and the set of solutions of the variational inclusion for a 𝛽 -inverse-strongly monotone mapping in a real Hilbert space. We prove that the sequence converges strongly to a common element of the above three sets under some mild conditions. Furthermore, we give a numerical example which supports our main theorem in the last part.

1. Introduction

Let 𝐶 be a closed convex subset of a real Hilbert space 𝐻 with the inner product , and the norm . Let 𝐹 be a bifunction of 𝐶 × 𝐶 into , where is the set of real numbers, 𝜑 𝐶 be a real-valued function. Let Λ be arbitrary index set. The system of mixed equilibrium problem is for finding 𝑥 𝐶 such that 𝐹 𝑘 ( 𝑥 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑥 ) 0 , 𝑘 Λ , 𝑦 𝐶 . ( 1 . 1 ) The set of solutions of (1.1) is denoted by S M E P ( 𝐹 𝑘 ) , that is, 𝐹 S M E P 𝑘 = 𝑥 𝐶 = 𝐹 𝑘 ( 𝑥 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑥 ) 0 , 𝑘 Λ , 𝑦 𝐶 . ( 1 . 2 ) If Λ is a singleton, then problem (1.1) becomes the following mixed equilibrium problem: finding 𝑥 𝐶 such that 𝐹 ( 𝑥 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑥 ) 0 , 𝑦 𝐶 . ( 1 . 3 ) The set of solutions of (1.3) is denoted by M E P ( 𝐹 ) .

If 𝜑 0 , the problem (1.3) is reduced into the equilibrium problem [1] for finding 𝑥 𝐶 such that 𝐹 ( 𝑥 , 𝑦 ) 0 , 𝑦 𝐶 . ( 1 . 4 ) The set of solutions of (1.4) is denoted by E P ( 𝐹 ) . This problem contains fixed-point problems, includes as special cases numerous problems in physics, optimization, and economics. Some methods have been proposed to solve the system of mixed equilibrium problem and the equilibrium problem, please consult [219].

Recall that, a mapping 𝑆 𝐶 𝐶 is said to be nonexpansive if 𝑆 𝑥 𝑆 𝑦 𝑥 𝑦 , ( 1 . 5 ) for all 𝑥 , 𝑦 𝐶 . If 𝐶 is a bounded closed convex and 𝑆 is a nonexpansive mapping of 𝐶 into itself, then 𝐹 ( 𝑆 ) is nonempty [20]. Let 𝐴 𝐶 𝐻 be a mapping, the Hartmann-Stampacchia variational inequality for finding 𝑥 𝐶 such that 𝐴 𝑥 , 𝑦 𝑥 0 , 𝑦 𝐶 . ( 1 . 6 ) The set of solutions of (1.6) is denoted by V I ( 𝐶 , 𝐴 ) . The variational inequality has been extensively studied in the literature [2128].

Iterative methods for nonexpansive mappings have recently been applied to solve convex minimization problems. Convex minimization problems have a great impact and influence on the development of almost all branches of pure and applied sciences. A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space 𝐻 : 1 𝜃 ( 𝑥 ) = 2 𝐴 𝑥 , 𝑥 𝑥 , 𝑦 , 𝑥 𝐹 ( 𝑆 ) , ( 1 . 7 ) where 𝐴 is a linear bounded operator, 𝐹 ( 𝑆 ) is the fixed point set of a nonexpansive mapping 𝑆 , and 𝑦 is a given point in 𝐻 [29].

We denote weak convergence and strong convergence by notations and , respectively. A mapping 𝐴 of 𝐶 into 𝐻 is called monotone if 𝐴 𝑥 𝐴 𝑦 , 𝑥 𝑦 0 , ( 1 . 8 ) for all 𝑥 , 𝑦 𝐶 . A mapping 𝐴 of 𝐶 into 𝐻 is called 𝛼 -inverse-strongly monotone if there exists a positive real number 𝛼 such that 𝐴 𝑥 𝐴 𝑦 , 𝑥 𝑦 𝛼 𝐴 𝑥 𝐴 𝑦 2 , ( 1 . 9 ) for all 𝑥 , 𝑦 𝐶 . It is obvious that any 𝛼 -inverse-strongly monotone mappings 𝐴 are monotone and Lipschitz continuous mapping. A linear bounded operator 𝐴 is strongly positive if there exists a constant 𝛾 > 0 with the property 𝐴 𝑥 , 𝑥 𝛾 𝑥 2 , ( 1 . 1 0 ) for all 𝑥 𝐻 . A self-mapping 𝑓 𝐶 𝐶 is a contraction on 𝐶 if there exists a constant 𝛼 ( 0 , 1 ) such that 𝑓 ( 𝑥 ) 𝑓 ( 𝑦 ) 𝛼 𝑥 𝑦 , ( 1 . 1 1 ) for all 𝑥 , 𝑦 𝐶 . We use Π 𝐶 to denote the collection of all contraction on 𝐶 . Note that each 𝑓 Π 𝐶 has a unique fixed point in 𝐶 .

Let 𝐵 𝐻 𝐻 be a single-valued nonlinear mapping and 𝑀 𝐻 2 𝐻 be a set-valued mapping. The variational inclusion problem is to find 𝑥 𝐻 such that 𝜃 𝐵 ( 𝑥 ) + 𝑀 ( 𝑥 ) , ( 1 . 1 2 ) where 𝜃 is the zero vector in 𝐻 . The set of solutions of problem (1.12) is denoted by 𝐼 ( 𝐵 , 𝑀 ) . The variational inclusion has been extensively studied in the literature, see, for example, [3032] and the reference therein.

A set-valued mapping 𝑀 𝐻 2 𝐻 is called monotone if for all 𝑥 , 𝑦 𝐻 , 𝑓 𝑀 ( 𝑥 ) , and 𝑔 𝑀 ( 𝑦 ) impling 𝑥 𝑦 , 𝑓 𝑔 0 . A monotone mapping 𝑀 is maximal if its graph 𝐺 ( 𝑀 ) = { ( 𝑓 , 𝑥 ) 𝐻 × 𝐻 𝑓 𝑀 ( 𝑥 ) } of 𝑀 is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping 𝑀 is maximal if and only if for ( 𝑥 , 𝑓 ) 𝐻 × 𝐻 , 𝑥 𝑦 , 𝑓 𝑔 0 for all ( 𝑦 , 𝑔 ) 𝐺 ( 𝑀 ) impling 𝑓 𝑀 ( 𝑥 ) .

Let 𝐵 be an inverse-strongly monotone mapping of 𝐶 into 𝐻 , and let 𝑁 𝐶 𝑣 be normal cone to 𝐶 at 𝑣 𝐶 , that is, 𝑁 𝐶 𝑣 = { 𝑤 𝐻 𝑣 𝑢 , 𝑤 0 , f o r a l l 𝑢 𝐶 } , and define 𝑇 𝑣 = 𝐵 𝑣 + 𝑁 𝐶 𝑣 , i f 𝑣 𝐶 , , i f 𝑣 𝐶 . ( 1 . 1 3 ) Then, 𝑇 is a maximal monotone and 𝜃 𝑇 𝑣 if and only if 𝑣 V I ( 𝐶 , 𝐵 ) (see [33]).

Let 𝑀 𝐻 2 𝐻 be a set-valued maximal monotone mapping, then the single-valued mapping 𝐽 𝑀 , 𝜆 𝐻 𝐻 defined by 𝐽 𝑀 , 𝜆 ( 𝑥 ) = ( 𝐼 + 𝜆 𝑀 ) 1 ( 𝑥 ) , 𝑥 𝐻 , ( 1 . 1 4 ) is called the resolvent operator associated with 𝑀 , where 𝜆 is any positive number and 𝐼 is the identity mapping. It is worth mentioning that the resolvent operator is nonexpansive, 1-inverse-strongly monotone, and that a solution of problem (1.12) is a fixed point of the operator 𝐽 𝑀 , 𝜆 ( 𝐼 𝜆 𝐵 ) for all 𝜆 > 0 , (for more details see [34]).

In 2000, Moudafi [35] introduced the viscosity approximation method for nonexpansive mappings and proved that if 𝐻 is a real Hilbert space, the sequence { 𝑥 𝑛 } defined by the iterative method below, with the initial guess 𝑥 0 𝐶 is chosen arbitrarily, 𝑥 𝑛 + 1 = 𝛼 𝑛 𝑓 𝑥 𝑛 + 1 𝛼 𝑛 𝑆 𝑥 𝑛 , 𝑛 0 , ( 1 . 1 5 ) where { 𝛼 𝑛 } ( 0 , 1 ) satisfies certain conditions and converges strongly to a fixed point of 𝑆 (say 𝑥 𝐶 ), which is then a unique solution of the following variational inequality: ( 𝐼 𝑓 ) 𝑥 , 𝑥 𝑥 0 , 𝑥 𝐹 ( 𝑆 ) . ( 1 . 1 6 )

In 2006, Marino and Xu [29] introduced a general iterative method for nonexpansive mapping. They defined the sequence { 𝑥 𝑛 } generated by the algorithm 𝑥 0 𝐶 , 𝑥 𝑛 + 1 = 𝛼 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝛼 𝑛 𝐴 𝑆 𝑥 𝑛 , 𝑛 0 , ( 1 . 1 7 ) where { 𝛼 𝑛 } ( 0 , 1 ) , and 𝐴 is a strongly positive linear bounded operator. They proved that if 𝐶 = 𝐻 , and the sequence { 𝛼 𝑛 } satisfies appropriate conditions, then the sequence { 𝑥 𝑛 } generated by (1.17) converges strongly to a fixed point of 𝑆 (say 𝑥 𝐻 ) which is the unique solution of the following variational inequality: ( 𝐴 𝛾 𝑓 ) 𝑥 , 𝑥 𝑥 0 , 𝑥 𝐹 ( 𝑆 ) , ( 1 . 1 8 ) which is the optimality condition for the minimization problem m i n 𝑥 𝐹 ( 𝑆 ) E P ( 𝐹 ) 1 2 𝐴 𝑥 , 𝑥 ( 𝑥 ) , ( 1 . 1 9 ) where is a potential function for 𝛾 𝑓 (i.e., ( 𝑥 ) = 𝛾 𝑓 ( 𝑥 ) for 𝑥 𝐻 ).

For finding a common element of the set of fixed points of nonexpansive mappings and the set of solution of the variational inequalities. Let 𝑃 𝐶 be the projection of 𝐻 onto 𝐶 . In 2005, Iiduka and Takahashi [36] introduced the following iterative process for 𝑥 0 𝐶 , 𝑥 𝑛 + 1 = 𝛼 𝑛 𝑢 + 1 𝛼 𝑛 𝑆 𝑃 𝐶 𝑥 𝑛 𝜆 𝑛 𝐴 𝑥 𝑛 , 𝑛 0 , ( 1 . 2 0 ) where 𝑢 𝐶 , { 𝛼 𝑛 } ( 0 , 1 ) , and { 𝜆 𝑛 } [ 𝑎 , 𝑏 ] for some 𝑎 , 𝑏 with 0 < 𝑎 < 𝑏 < 2 𝛽 . They proved that under certain appropriate conditions imposed on { 𝛼 𝑛 } and { 𝜆 𝑛 } , the sequence { 𝑥 𝑛 } generated by (1.20) converges strongly to a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the variational inequality for an inverse-strongly monotone mapping (say 𝑥 𝐶 ) which solve some variational inequality 𝑥 𝑢 , 𝑥 𝑥 0 , 𝑥 𝐹 ( 𝑆 ) V I ( 𝐶 , A ) . ( 1 . 2 1 )

In 2008, Su et al. [37] introduced the following iterative scheme by the viscosity approximation method in a real Hilbert space: 𝑥 1 , 𝑢 𝑛 𝐻 𝐹 𝑢 𝑛 + 1 , 𝑦 𝑟 𝑛 𝑦 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 𝑥 0 , 𝑦 𝐶 , 𝑛 + 1 = 𝛼 𝑛 𝑓 𝑥 𝑛 + 1 𝛼 𝑛 𝑆 𝑃 𝐶 𝑢 𝑛 𝜆 𝑛 𝐴 𝑢 𝑛 , ( 1 . 2 2 ) for all 𝑛 , where { 𝛼 𝑛 } [ 0 , 1 ) and { 𝑟 𝑛 } ( 0 , ) satisfing some appropriate conditions. Furthermore, they proved that { 𝑥 𝑛 } and { 𝑢 𝑛 } converge strongly to the same point 𝑧 𝐹 ( 𝑆 ) V I ( 𝐶 , 𝐴 ) E P ( 𝐹 ) , where 𝑧 = 𝑃 𝐹 ( 𝑆 ) V I ( 𝐶 , 𝐴 ) E P ( 𝐹 ) 𝑓 ( 𝑧 ) .

Let { 𝑇 𝑖 } be an infinite family of nonexpansive mappings of 𝐻 into itself, and let { 𝜆 𝑖 } be a real sequence such that 0 𝜆 𝑖 1 for every 𝑖 𝑁 . For 𝑛 1 , we defined a mapping 𝑊 𝑛 of 𝐻 into itself as follows: 𝑈 𝑛 , 𝑛 + 1 𝑈 = 𝐼 , 𝑛 , 𝑛 = 𝜆 𝑛 𝑇 𝑛 𝑈 𝑛 , 𝑛 + 1 + 1 𝜆 𝑛 𝑈 𝐼 , 𝑛 , 𝑘 = 𝜆 𝑘 𝑇 𝑘 𝑈 𝑛 , 𝑘 + 1 + 1 𝜆 𝑘 𝑈 𝐼 , 𝑛 , 2 = 𝜆 2 𝑇 2 𝑈 𝑛 , 3 + 1 𝜆 2 𝑊 𝐼 , 𝑛 = 𝑈 𝑛 , 1 = 𝜆 1 𝑇 1 𝑈 𝑛 , 2 + 1 𝜆 1 𝐼 . ( 1 . 2 3 )

In 2011, He et al. [38] introduced the following iterative process for { 𝑇 𝑛 𝐶 𝐶 } which is a sequence of nonexpansive mappings. Let { 𝑧 𝑛 } be the sequence defined by 𝑧 𝑛 + 1 = 𝜖 𝑛 𝑧 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝑊 𝑛 𝐾 1 𝑟 1 , 𝑛 𝐾 2 𝑟 2 , 𝑛 𝐾 𝐾 𝑟 𝐾 , 𝑛 𝑧 𝑛 , 𝑛 𝑁 . ( 1 . 2 4 ) The sequence { 𝑧 𝑛 } defined by (1.24) converges strongly to a common element of the set of fixed points of nonexpansive mappings, the set of solutions of the variational inequality, and the generalized equilibrium problem. Recently, Jitpeera and Kumam [39] introduced the following new general iterative method for finding a common element of the set of solutions of fixed point for nonexpansive mappings, the set of solution of generalized mixed equilibrium problems, and the set of solutions of the variational inclusion for a 𝛽 -inverse-strongly monotone mapping in a real Hilbert space.

In this paper, we modify the iterative methods (1.17), (1.22), and (1.24) by purposing the following new general viscosity iterative method: 𝑥 0 , 𝑢 𝑛 𝐶 , 𝑢 𝑛 = 𝐾 𝐹 𝑁 𝑟 𝑛 , 𝑛 𝐾 𝐹 𝑁 1 𝑟 𝑛 1 , 𝑛 𝐾 𝐹 𝑁 2 𝑟 𝑛 2 , 𝑛 𝐾 𝐹 2 𝑟 2 , 𝑛 𝐾 𝐹 1 𝑟 1 , 𝑛 𝑥 𝑛 𝑥 , 𝑛 𝑁 𝑛 + 1 = 𝑃 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝐽 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 , ( 1 . 2 5 ) for all 𝑛 , where { 𝛼 𝑛 } ( 0 , 1 ) , { 𝑟 𝑛 } ( 0 , 2 𝜎 ) , and 𝜆 ( 0 , 2 𝛽 ) satisfy some appropriate conditions. The purpose of this paper shows that under some control conditions the sequence { 𝑥 𝑛 } converges strongly to a common element of the set of common fixed points of nonexpansive mappings, the solution of the system of mixed equilibrium problems, and the set of solutions of the variational inclusion in a real Hilbert space. Moreover, we apply our results to the class of strictly pseudocontractive mappings. Finally, we give a numerical example which supports our main theorem in the last part. Our results improve and extend the corresponding results of Marino and Xu [29], Su et al. [37], He et al. [38], and some authors.

2. Preliminaries

Let 𝐻 be a real Hilbert space and 𝐶 be a nonempty closed and convex subset of 𝐻 . Recall that the (nearest point) projection 𝑃 𝐶 from 𝐻 onto 𝐶 assigns to each 𝑥 𝐻 and the unique point in 𝑃 𝐶 𝑥 𝐶 satisfies the property 𝑥 𝑃 𝐶 𝑥 = m i n 𝑦 𝐶 𝑥 𝑦 , ( 2 . 1 ) which is equivalent to the following inequality 𝑥 𝑃 𝐶 𝑥 , 𝑃 𝐶 𝑥 𝑦 0 , 𝑦 𝐶 . ( 2 . 2 ) The following characterizes the projection 𝑃 𝐶 . We recall some lemmas which will be needed in the rest of this paper.

Lemma 2.1. The function 𝑢 𝐶 is a solution of the variational inequality if and only if 𝑢 𝐶 satisfies the relation 𝑢 = 𝑃 𝐶 ( 𝑢 𝜆 𝐵 𝑢 ) for all 𝜆 > 0 .

Lemma 2.2. For a given 𝑧 𝐻 , 𝑢 𝐶 , 𝑢 = 𝑃 𝐶 𝑧 𝑢 𝑧 , 𝑣 𝑢 0 , 𝑣 𝐶 .
It is well known that 𝑃 𝐶 is a firmly nonexpansive mapping of 𝐻 onto 𝐶 and satisfies 𝑃 𝐶 𝑥 𝑃 𝐶 𝑦 2 𝑃 𝐶 𝑥 𝑃 𝐶 𝑦 , 𝑥 𝑦 , 𝑥 , 𝑦 𝐻 . ( 2 . 3 ) Moreover, 𝑃 𝐶 𝑥 is characterized by the following properties: 𝑃 𝐶 𝑥 𝐶 and for all 𝑥 𝐻 , 𝑦 𝐶 , 𝑥 𝑃 𝐶 𝑥 , 𝑦 𝑃 𝐶 𝑥 0 . ( 2 . 4 )

Lemma 2.3 (see [40]). Let 𝑀 𝐻 2 𝐻 be a maximal monotone mapping, and let 𝐵 𝐻 𝐻 be a monotone and Lipshitz continuous mapping. Then the mapping 𝐿 = 𝑀 + 𝐵 𝐻 2 𝐻 is a maximal monotone mapping.

Lemma 2.4 (see [41]). Each Hilbert space 𝐻 satisfies Opial's condition, that is, for any sequence { 𝑥 𝑛 } 𝐻 with 𝑥 𝑛 𝑥 , the inequality l i m i n f 𝑛 𝑥 𝑛 𝑥 < l i m i n f 𝑛 𝑥 𝑛 𝑦 , hold for each 𝑦 𝐻 with 𝑦 𝑥 .

Lemma 2.5 (see [42]). Assume { 𝑎 𝑛 } is a sequence of nonnegative real numbers such that 𝑎 𝑛 + 1 1 𝛾 𝑛 𝑎 𝑛 + 𝛿 𝑛 , 𝑛 0 , ( 2 . 5 ) where { 𝛾 𝑛 } ( 0 , 1 ) and { 𝛿 𝑛 } is a sequence in such that(i) 𝑛 = 1 𝛾 𝑛 = , (ii) l i m s u p 𝑛 𝛿 𝑛 / 𝛾 𝑛 0 or 𝑛 = 1 | 𝛿 𝑛 | < .Then l i m 𝑛 𝑎 𝑛 = 0 .

Lemma 2.6 (see [43]). Let 𝐶 be a closed convex subset of a real Hilbert space 𝐻 , and let 𝑇 𝐶 𝐶 be a nonexpansive mapping. Then 𝐼 𝑇 is demiclosed at zero, that is, 𝑥 𝑛 𝑥 , 𝑥 𝑛 𝑇 𝑥 𝑛 0 , ( 2 . 6 ) implying 𝑥 = 𝑇 𝑥 .

For solving the mixed equilibrium problem, let us assume that the bifunction 𝐹 𝐶 × 𝐶 and the nonlinear mapping 𝜑 𝐶 satisfy the following conditions: (A1) 𝐹 ( 𝑥 , 𝑥 ) = 0 for all 𝑥 𝐶 ; (A2) 𝐹 is monotone, that is, 𝐹 ( 𝑥 , 𝑦 ) + 𝐹 ( 𝑦 , 𝑥 ) 0 for any 𝑥 , 𝑦 𝐶 ; (A3)for each fixed 𝑦 𝐶 , 𝑥 𝐹 ( 𝑥 , 𝑦 ) is weakly upper semicontinuous; (A4)for each fixed 𝑥 𝐶 , 𝑦 𝐹 ( 𝑥 , 𝑦 ) is convex and lower semicontinuous; (B1)for each 𝑥 𝐶 and 𝑟 > 0 , there exist a bounded subset 𝐷 𝑥 𝐶 and 𝑦 𝑥 𝐶 such that for any 𝑧 𝐶 𝐷 𝑥 , 𝐹 𝑧 , 𝑦 𝑥 𝑦 + 𝜑 𝑥 1 𝜑 ( 𝑧 ) + 𝑟 𝑦 𝑥 𝑧 , 𝑧 𝑥 < 0 , ( 2 . 7 ) (B2) 𝐶 is a bounded set.

Lemma 2.7 (see [44]). Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻 . Let 𝐹 𝐶 × 𝐶 be a bifunction mapping satisfying (A1)–(A4), and let 𝜑 𝐶 be a convex and lower semicontinuous function such that 𝐶 d o m 𝜑 . Assume that either (B1) or (B2) holds. For 𝑟 > 0 and 𝑥 𝐻 , then there exists 𝑢 𝐶 such that 1 𝐹 ( 𝑢 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑢 ) + 𝑟 𝑦 𝑢 , 𝑢 𝑥 0 . ( 2 . 8 ) Define a mapping 𝐾 𝑟 𝐻 𝐶 as follows: 𝐾 𝑟 1 ( 𝑥 ) = 𝑢 𝐶 𝐹 ( 𝑢 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑢 ) + 𝑟 , 𝑦 𝑢 , 𝑢 𝑥 0 , 𝑦 𝐶 ( 2 . 9 ) for all 𝑥 𝐻 . Then, the following hold: (i) 𝐾 𝑟 is single-valued; (ii) 𝐾 𝑟 is firmly nonexpansive, that is, for any 𝑥 , 𝑦 𝐻 , 𝐾 𝑟 𝑥 𝐾 𝑟 𝑦 2 𝐾 𝑟 𝑥 𝐾 𝑟 𝑦 , 𝑥 𝑦 ;(iii) 𝐹 ( 𝐾 𝑟 ) = M E P ( 𝐹 ) ; (iv) M E P ( 𝐹 ) is closed and convex.

Lemma 2.8 (see [29]). Assume 𝐴 is a strongly positive linear bounded operator on a Hilbert space 𝐻 with coefficient 𝛾 > 0 and 0 < 𝜌 𝐴 1 , then 𝐼 𝜌 𝐴 1 𝜌 𝛾 .

Lemma 2.9 (see [38]). Let C be a nonempty closed and convex subset of a strictly convex Banach space. Let { 𝑇 𝑖 } 𝑖 𝑁 be an infinite family of nonexpansive mappings of C into itself such that 𝑖 𝑁 𝐹 ( 𝑇 𝑖 ) , and let { 𝜆 𝑖 } be a real sequence such that 0 𝜆 𝑖 b < 1 for every 𝑖 𝑁 . Then F ( W ) = 𝑖 𝑁 𝐹 ( 𝑇 𝑖 ) .

Lemma 2.10 (see [38]). Let C be a nonempty closed and convex subset of a strictly convex Banach space. Let { 𝑇 𝑖 } be an infinite family of nonexpansive mappings of C into itself, and let { 𝜆 𝑖 } be a real sequence such that 0 𝜆 𝑖 b < 1 for every 𝑖 𝑁 . Then, for every 𝑥 𝐶 and 𝑘 𝑁 , the limit l i m 𝑛 𝑈 𝑛 , 𝑘 exist.
In view of the previous lemma, we define 𝑊 𝑥 = l i m 𝑛 𝑈 𝑛 , 1 𝑥 = l i m 𝑛 𝑊 𝑛 𝑥 . ( 2 . 1 0 )

3. Strong Convergence Theorems

In this section, we show a strong convergence theorem which solves the problem of finding a common element of the common fixed points, the common solution of a system of mixed equilibrium problems and variational inclusion of inverse-strongly monotone mappings in a Hilbert space.

Theorem 3.1. Let 𝐻 be a real Hilbert space and 𝐶 a nonempty close and convex subset of 𝐻 , and let 𝐵 be a 𝛽 -inverse-strongly monotone mapping. Let 𝜑 𝐶 𝑅 be a convex and lower semicontinuous function, 𝑓 𝐶 𝐶 a contraction mapping with coefficient 𝛼 ( 0 < 𝛼 < 1), and 𝑀 𝐻 2 𝐻 a maximal monotone mapping. Let 𝐴 be a strongly positive linear bounded operator of 𝐻 into itself with coefficient 𝛾 > 0 . Assume that 0 < 𝛾 < 𝛾 / 𝛼 and 𝜆 ( 0 , 2 𝛽 ) . Let { 𝑇 𝑛 } be a family of nonexpansive mappings of 𝐻 into itself such that 𝜃 = 𝑛 = 1 𝐹 𝑇 𝑛 𝑁 𝑘 = 1 𝐹 S M E P 𝑘 𝐼 ( 𝐵 , 𝑀 ) . ( 3 . 1 ) Suppose that { 𝑥 𝑛 } is a sequence generated by the following algorithm for 𝑥 0 𝐶 arbitrarily and 𝑢 𝑛 = 𝐾 𝐹 𝑁 𝑟 𝑛 , 𝑛 𝐾 𝐹 𝑁 1 𝑟 𝑛 1 , 𝑛 𝐾 𝐹 𝑁 2 𝑟 𝑛 2 , 𝑛 𝐾 𝐹 2 𝑟 2 , 𝑛 𝐾 𝐹 1 𝑟 1 , 𝑛 𝑥 𝑛 𝑥 , 𝑛 𝑁 𝑛 + 1 = 𝑃 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝐽 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 , ( 3 . 2 ) for all 𝑛 = 1 , 2 , 3 , , where K 𝐹 𝑖 𝑟 𝑖 , 𝑛 𝑢 ( 𝑥 ) = 𝑛 𝐶 𝐹 𝑖 𝑢 𝑛 𝑢 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑛 + 1 𝑟 𝑖 , 𝑛 𝑦 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 , 0 , 𝑦 𝐶 𝑖 = 1 , 2 , 3 , , 𝑁 , ( 3 . 3 ) and the following conditions are satisfied(C1): { 𝜖 𝑛 } ( 0 , 1 ) , l i m 𝑛 0 𝜖 𝑛 = 0 , 𝑛 = 1 𝜖 𝑛 = , 𝑛 = 1 | 𝜖 𝑛 + 1 𝜖 𝑛 | < ; (C2): { 𝑟 𝑛 } [ 𝑐 , 𝑑 ] with 𝑐 , 𝑑 ( 0 , 2 𝜎 ) and 𝑛 = 1 | 𝑟 𝑛 + 1 𝑟 𝑛 | < .
Then, the sequence { 𝑥 𝑛 } converges strongly to 𝑞 𝜃 , where 𝑞 = 𝑃 𝜃 ( 𝛾 𝑓 + 𝐼 𝐴 ) ( 𝑞 ) which solves the following variational inequality: ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑝 𝑞 0 , 𝑝 𝜃 , ( 3 . 4 ) which is the optimality condition for the minimization problem m i n 𝑞 𝜃 1 2 𝐴 𝑞 , 𝑞 ( 𝑞 ) , ( 3 . 5 ) where h is a potential function for 𝛾 𝑓 (i.e., ( 𝑞 ) = 𝛾 𝑓 ( 𝑞 ) for 𝑞 𝐻 ).

Proof. For condition (C1), we may assume without loss of generality, and 𝜖 𝑛 ( 0 , 𝐴 1 ) for all 𝑛 . By Lemma 2.8, we have 𝐼 𝜖 𝑛 𝐴 1 𝜖 𝑛 𝛾 . Next, we will assume that 𝐼 𝐴 1 𝛾 .
Next, we will divide the proof into six steps.
Step 1. First, we will show that { 𝑥 𝑛 } and { 𝑢 𝑛 } are bounded. Since 𝐵 is 𝛽 -inverse-strongly monotone mappings, we have ( 𝐼 𝜆 𝐵 ) 𝑥 ( 𝐼 𝜆 𝐵 ) 𝑦 2 = 𝐼 𝑥 𝜆 𝐵 𝑥 𝐼 𝑦 + 𝜆 𝐵 𝑦 2 = 𝑥 𝑦 𝜆 𝐵 𝑥 + 𝜆 𝐵 𝑦 2 = ( 𝑥 𝑦 ) 𝜆 ( 𝐵 𝑥 + 𝐵 𝑦 ) 2 𝑥 𝑦 2 2 𝜆 𝑥 𝑦 𝐵 𝑥 + 𝐵 𝑦 + 𝜆 2 𝐵 𝑥 𝐵 𝑦 2 𝑥 𝑦 2 2 𝜆 𝛽 𝐵 𝑥 + 𝐵 𝑦 2 + 𝜆 2 𝐵 𝑥 𝐵 𝑦 2 𝑥 𝑦 2 + 𝜆 ( 𝜆 2 𝛽 ) 𝐵 𝑥 + 𝐵 𝑦 2 , ( 3 . 6 ) if 0 < 𝜆 < 2 𝛽 , then 𝐼 𝜆 𝐵 is nonexpansive.
Put 𝑦 𝑛 = 𝐽 𝑀 , 𝜆 ( 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 ) , 𝑛 0 . Since 𝐽 𝑀 , 𝜆 and 𝐼 𝜆 𝐵 are nonexpansive mapping, it follows that 𝑦 𝑛 = 𝐽 𝑞 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝐽 𝑀 , 𝜆 𝑢 ( 𝑞 𝜆 𝐵 𝑞 ) 𝑛 𝜆 𝐵 𝑢 𝑛 𝑢 ( 𝑞 𝜆 𝐵 𝑞 ) 𝑛 . 𝑞 ( 3 . 7 ) By Lemma 2.7, we have 𝑢 𝑛 = 𝐾 𝐹 𝑁 𝑟 𝑛 , 𝑛 𝐾 𝐹 𝑁 1 𝑟 𝑛 1 , 𝑛 𝐾 𝐹 𝑁 2 𝑟 𝑛 2 , 𝑛 𝐾 𝐹 2 𝑟 2 , 𝑛 𝐾 𝐹 1 𝑟 1 , 𝑛 𝑥 𝑛 𝜏 , f o r 𝑛 0 𝑘 𝑛 = 𝐾 𝐹 𝑘 𝑟 𝑘 , 𝑛 𝐾 𝐹 𝑘 1 𝑟 𝑘 1 , 𝑛 𝐾 𝐹 2 𝑟 2 , 𝑛 𝐾 𝐹 1 𝑟 1 , 𝑛 , f o r 𝑘 { 0 , 1 , 2 , , 𝑁 } , ( 3 . 8 ) and 𝜏 0 𝑛 = 𝐼 f o r a l l 𝑛 𝑁 , 𝑞 = 𝜏 𝐹 𝑘 𝑟 𝑘 , 𝑛 𝑞 , 𝑢 𝑛 = 𝜏 𝑁 𝑟 𝑘 , 𝑁 𝑥 𝑛 Then, we have 𝑢 𝑛 𝑞 2 = 𝜏 𝑁 𝑟 𝑘 , 𝑛 𝑥 𝑛 𝜏 𝐹 𝑘 𝑟 𝑘 , 𝑛 𝑞 2 = 𝑥 𝑛 𝑞 2 . ( 3 . 9 ) Hence, we get 𝑦 𝑛 𝑥 𝑞 𝑛 . 𝑞 ( 3 . 1 0 ) From (3.2), we deduce that 𝑥 𝑛 + 1 = 𝑃 𝑞 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑃 𝐶 𝑞 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐴 𝑞 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑞 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐴 𝑞 1 𝜖 𝑛 𝛾 𝑦 𝑛 𝑞 𝜖 𝛾 𝜖 𝑛 𝑥 𝑛 𝑞 + 𝜖 𝑛 ( + 𝛾 𝑓 𝑔 ) 𝐴 𝑞 1 𝜖 𝑛 𝛾 𝑥 𝑛 = 𝑞 1 𝜖 𝛾 𝛾 𝜖 𝑛 𝑥 𝑛 𝑞 𝜖 𝑛 = 𝛾 𝑓 ( 𝑞 ) 𝐴 𝑞 1 𝜖 𝛾 𝛾 𝜖 𝑛 𝑥 𝑛 + 𝑞 𝜖 𝛾 𝛾 𝜖 𝑛 𝛾 𝑓 ( 𝑞 ) 𝐴 𝑞 𝑥 𝛾 𝛾 𝜖 m a x 𝑛 , 𝑞 𝛾 𝑓 ( 𝑞 ) 𝐴 𝑞 . 𝛾 𝛾 𝜖 ( 3 . 1 1 ) It follows by induction that 𝑥 𝑛 𝑥 𝑞 m a x 0 , ( 𝑞 𝛾 𝑓 𝑞 ) 𝐴 𝑞 𝛾 𝛾 𝜖 , 𝑛 0 . ( 3 . 1 2 ) Therefore { 𝑥 𝑛 } is bounded, so are { 𝑦 𝑛 } , { 𝐵 𝑢 𝑛 } , { 𝑓 ( 𝑥 𝑛 ) } , and { 𝐴 𝑊 𝑛 𝑦 𝑛 } .
Step 2. We claim that l i m 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 = 0 and l i m 𝑛 𝑦 𝑛 + 1 𝑦 𝑛 = 0 . From (3.2), we have 𝑥 𝑛 + 1 𝑥 𝑛 = 𝑃 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑃 𝐶 𝜖 𝑛 1 𝑥 𝛾 𝑓 𝑛 1 + 𝐼 𝜖 𝑛 1 𝐴 𝑊 𝑛 𝑦 𝑛 1 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑊 𝑛 𝑦 𝑛 1 𝜖 𝑛 𝜖 𝑛 1 𝐴 𝑊 𝑛 𝑦 𝑛 1 + 𝛾 𝜖 𝑛 𝑓 𝑥 𝑛 𝑥 𝑓 𝑛 1 𝜖 + 𝛾 𝑛 𝜖 𝑛 1 𝑓 𝑥 𝑛 1 1 𝜖 𝑛 𝛾 𝑦 𝑛 𝑦 𝑛 1 + | | 𝜖 𝑛 𝜖 𝑛 1 | | 𝐴 𝑊 𝑛 𝑦 𝑛 + 𝛾 𝜖 𝜖 𝑛 𝑥 𝑛 𝑥 𝑛 1 | | 𝜖 + 𝛾 𝑛 𝜖 𝑛 1 | | 𝑓 𝑥 𝑛 1 . ( 3 . 1 3 ) Since 𝐽 𝑀 , 𝜆 and 𝐼 𝜆 𝐵 are nonexpansive, we also have 𝑦 𝑛 𝑦 𝑛 1 = 𝐽 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝐽 𝑀 , 𝜆 𝑢 𝑛 1 𝜆 𝐵 𝑢 𝑛 1 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝑢 𝑛 1 𝜆 𝐵 𝑢 𝑛 1 𝑢 𝑛 𝑢 𝑛 1 . ( 3 . 1 4 ) On the other hand, from 𝑢 𝑛 1 = 𝜏 𝑁 𝑟 𝑘 , 𝑛 1 𝑥 𝑛 1 and 𝑢 𝑛 = 𝜏 𝑁 𝑟 𝑘 , 𝑛 𝑥 𝑛 , it follows that 𝐹 𝑢 𝑛 1 𝑢 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑛 1 + 1 𝑟 𝑛 1 𝑦 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑥 𝑛 1 𝐹 𝑢 0 , 𝑦 𝐶 , ( 3 . 1 5 ) 𝑛 𝑢 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑛 + 1 𝑟 𝑛 𝑦 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 0 , 𝑦 𝐶 . ( 3 . 1 6 ) Substituting 𝑦 = 𝑢 𝑛 into (3.15) and 𝑦 = 𝑢 𝑛 1 into (3.16), we get 𝐹 𝑢 𝑛 1 , 𝑢 𝑛 𝑢 + 𝜑 𝑛 𝑢 𝜑 𝑛 1 + 1 𝑟 𝑛 1 𝑢 𝑛 𝑢 𝑛 1 , 𝑢 n 1 𝑥 𝑛 1 𝐹 𝑢 0 , 𝑛 , 𝑢 𝑛 + 1 𝑢 + 𝜑 𝑛 + 1 𝑢 𝜑 𝑛 + 1 𝑟 𝑛 𝑢 𝑛 + 1 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 0 . ( 3 . 1 7 ) From (A2), we obtain 𝑢 𝑛 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑥 𝑛 1 𝑟 𝑛 1 𝑢 𝑛 𝑥 𝑛 𝑟 𝑛 𝑢 0 , 𝑛 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑥 𝑛 1 𝑟 𝑛 1 𝑟 𝑛 𝑢 𝑛 𝑥 𝑛 0 , ( 3 . 1 8 ) so, 𝑢 𝑛 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑢 𝑛 + 𝑢 𝑛 𝑥 𝑛 1 𝑟 𝑛 1 𝑟 𝑛 𝑢 𝑛 𝑥 𝑛 0 . ( 3 . 1 9 ) It follows that 𝑢 𝑛 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑢 𝑛 + 𝑢 𝑛 𝑥 𝑛 𝑟 𝑛 1 𝑟 𝑛 𝑢 𝑛 𝑥 𝑛 0 , 𝑢 𝑛 𝑢 𝑛 1 , 𝑢 𝑛 1 𝑢 𝑛 𝑢 + 𝑛 𝑢 𝑛 1 , 𝑟 1 𝑛 1 𝑟 𝑛 𝑢 𝑛 𝑥 𝑛 0 . ( 3 . 2 0 ) Without loss of generality, let us assume that there exists a real number 𝑐 such that 𝑟 𝑛 1 > 𝑐 > 0 , for all 𝑛 . Then, we have 𝑢 𝑛 𝑢 𝑛 1 2 𝑢 𝑛 𝑢 𝑛 1 , 𝑟 1 𝑛 1 𝑟 𝑛 𝑢 𝑛 𝑥 𝑛 𝑢 𝑛 𝑢 𝑛 1 | | | | 𝑟 1 𝑛 1 𝑟 𝑛 | | | | 𝑢 𝑛 𝑥 𝑛 , ( 3 . 2 1 ) and hence 𝑢 𝑛 𝑢 𝑛 1 𝑥 𝑛 𝑥 𝑛 1 + 1 𝑟 𝑛 | | 𝑟 𝑛 𝑟 𝑛 1 | | 𝑢 𝑛 𝑥 𝑛 𝑥 𝑛 𝑥 𝑛 1 + 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | , ( 3 . 2 2 ) where 𝑀 1 = s u p { 𝑢 𝑛 𝑥 𝑛 𝑛 } . Substituting (3.22) into (3.14), we have 𝑦 𝑛 𝑦 𝑛 1 𝑥 𝑛 𝑥 𝑛 1 + 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | . ( 3 . 2 3 ) Substituting (3.23) into (3.13), we get 𝑥 𝑛 + 1 𝑥 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝑛 𝑥 𝑛 1 + 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | + | | 𝜖 𝑛 𝜖 𝑛 1 | | 𝐴 𝑊 𝑛 𝑦 𝑛 1 + 𝛾 𝜖 𝜖 𝑛 𝑥 𝑛 𝑥 𝑛 1 | | 𝜖 + 𝛾 𝑛 𝜖 𝑛 1 | | 𝑓 𝑥 𝑛 1 = 1 𝜖 𝑛 𝛾 𝑥 𝑛 𝑥 𝑛 1 + 1 𝜖 𝑛 𝛾 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | + | | 𝜖 𝑛 𝜖 𝑛 1 | | 𝐴 𝑊 𝑛 𝑦 𝑛 1 + 𝛾 𝜖 𝜖 𝑛 𝑥 𝑛 𝑥 𝑛 1 | | 𝜖 + 𝛾 𝑛 𝜖 𝑛 1 | | 𝑓 𝑥 n 1 1 𝜖 𝛾 𝛾 𝜖 𝑛 𝑥 𝑛 𝑥 𝑛 1 + 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | + | | 𝜖 𝑛 𝜖 𝑛 1 | | 𝐴 𝑊 𝑛 𝑦 𝑛 1 | | 𝜖 + 𝛾 𝑛 𝜖 𝑛 1 | | 𝑓 𝑥 𝑛 1 1 𝜖 𝛾 𝛾 𝜖 𝑛 𝑥 𝑛 𝑥 𝑛 1 + 𝑀 1 𝑐 | | 𝑟 𝑛 𝑟 𝑛 1 | | + 𝑀 2 | | 𝜖 𝑛 𝜖 𝑛 1 | | , ( 3 . 2 4 ) where 𝑀 2 = s u p { m a x { 𝐴 𝑊 𝑛 𝑦 𝑛 1 , 𝑓 ( 𝑥 𝑛 1 ) 𝑛 } } . Since conditions (C1)-(C2) and by Lemma 2.5, we have 𝑥 𝑛 + 1 𝑥 𝑛 0 as 𝑛 . From (3.23), we also have 𝑦 𝑛 + 1 𝑦 𝑛 0 as 𝑛 .Step 3. Next, we show that l i m 𝑛 𝐵 𝑢 𝑛 𝐵 𝑞 = 0 .
For 𝑞 𝜃 hence 𝑞 = 𝐽 𝑀 , 𝜆 ( 𝑞 𝜆 𝐵 𝑞 ) . By (3.6) and (3.9), we get 𝑦 𝑛 𝑞 2 = 𝐽 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝐽 𝑀 , 𝜆 ( 𝑞 𝜆 𝐵 𝑞 ) 2 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 ( 𝑞 𝜆 𝐵 𝑞 ) 2 𝑢 𝑛 𝑞 2 + 𝜆 ( 𝜆 2 𝛽 ) 𝐵 𝑢 𝑛 𝐵 𝑞 2 𝑥 𝑛 𝑞 2 + 𝜆 ( 𝜆 2 𝛽 ) 𝐵 𝑢 𝑛 𝐵 𝑞 2 . ( 3 . 2 5 ) It follows that 𝑥 𝑛 + 1 𝑞 2 = 𝑃 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑃 𝐶 ( 𝑞 ) 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐴 𝑞 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑞 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐴 𝑞 1 𝜖 𝑛 𝛾 𝑦 𝑛 𝑞 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 1 𝜖 𝑛 𝛾 𝑦 𝑛 𝑞 2 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 𝑞 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 + 𝑞 1 𝜖 𝑛 𝛾 𝑥 𝑛 𝑞 2 + 𝜆 ( 𝜆 2 𝛽 ) 𝐵 𝑢 𝑛 𝐵 𝑞 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 + 𝑥 𝑞 𝑛 𝑞 2 + 1 𝜖 𝑛 𝛾 𝜆 ( 𝜆 2 𝛽 ) 𝐵 𝑢 𝑛 𝐵 𝑞 2 . ( 3 . 2 6 ) So, we obtain 1 𝜖 𝑛 𝛾 𝜆 ( 2 𝛽 𝜆 ) 𝐵 𝑢 𝑛 𝐵 𝑞 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 𝑥 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 + 𝑥 𝑞 𝑛 + 1 𝑞 + 𝜉 𝑛 , ( 3 . 2 7 ) where 𝜉 𝑛 = 2 𝜖 𝑛 ( 1 𝜖 𝑛 𝛾 ) 𝛾 𝑓 ( 𝑥 𝑛 ) 𝐴 𝑞 𝑦 𝑛 𝑞 . By conditions (C1), (C3) and l i m 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 = 0 , then, we obtain that 𝐵 𝑢 𝑛 𝐵 𝑞 0 as 𝑛 .
Step 4. We show the following: (i) l i m 𝑛 𝑥 𝑛 𝑢 𝑛 = 0 ; (ii) l i m 𝑛 𝑢 𝑛 𝑦 𝑛 = 0 ; (iii) l i m 𝑛 𝑦 𝑛 𝑊 𝑛 𝑦 𝑛 = 0 . Since 𝐾 𝑟 𝑛 ( 𝑥 ) is firmly nonexpansive and (2.3), we observe that 𝑢 𝑛 𝑞 2 = 𝜏 𝑁 𝑟 𝑛 , 𝑛 𝑥 𝑛 𝜏 𝑁 𝑟 𝑛 , 𝑛 𝑞 2 𝑥 𝑛 𝑞 , 𝑢 𝑛 = 1 𝑞 2 𝑥 𝑛 𝑞 2 + 𝑢 𝑛 𝑞 2 𝑥 𝑛 𝑞 𝑢 𝑛 𝑞 2 1 2 𝑥 𝑛 𝑞 2 + 𝑢 𝑛 𝑞 2 𝑥 𝑛 𝑢 𝑛 2 , ( 3 . 2 8 ) it follows that 𝑢 𝑛 𝑞 2 𝑥 𝑛 𝑞 2 𝑥 𝑛 𝑢 𝑛 2 . ( 3 . 2 9 )
Since 𝐽 𝑀 , 𝜆 is 1-inverse-strongly monotone and by (2.3), we compute 𝑦 𝑛 𝑞 2 = 𝐽 𝑀 , 𝜆 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝐽 𝑀 , 𝜆 ( 𝑞 𝜆 𝐵 𝑞 ) 2 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 ( 𝑞 𝜆 𝐵 𝑞 ) , 𝑦 𝑛 = 1 𝑞 2 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 ( 𝑞 𝜆 𝐵 𝑞 ) 2 + 𝑦 𝑛 𝑞 2 𝑢 𝑛 𝜆 𝐵 𝑢 𝑛 𝑦 ( 𝑞 𝜆 𝐵 𝑞 ) 𝑛 𝑞 2 1 2 𝑢 𝑛 𝑞 2 + 𝑦 𝑛 𝑞 2 𝑢 𝑛 𝑦 𝑛 𝜆 𝐵 𝑢 𝑛 𝐵 𝑞 2 = 1 2 𝑢 𝑛 𝑞 2 + 𝑦 𝑛 𝑞 2 𝑢 𝑛 𝑦 𝑛 2 + 2 𝜆 𝑢 𝑛 𝑦 𝑛 , 𝐵 𝑢 𝑛 𝐵 𝑞 𝜆 2 𝐵 𝑢 𝑛 𝐵 𝑞 2 , ( 3 . 3 0 ) which implies that 𝑦 𝑛 𝑞 2 𝑢 𝑛 𝑞 2 𝑢 𝑛 𝑦 𝑛 2 𝑢 + 2 𝜆 𝑛 𝑦 𝑛 𝐵 𝑢 𝑛 . 𝐵 𝑞 ( 3 . 3 1 ) Substituting (3.31) into (3.26), we have 𝑥 𝑛 + 1 𝑞 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 𝑦 𝑛 𝑞 2 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 𝑞 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 𝑢 𝑛 𝑞 2 𝑢 𝑛 𝑦 𝑛 2 + 2 𝜆 𝑛 𝑢 𝑛 𝑦 𝑛 𝐵 𝑢 𝑛 𝐵 𝑞 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 . 𝑞 ( 3 . 3 2 ) Then, we derive 𝑥 𝑛 𝑢 𝑛 2 + 𝑢 𝑛 𝑦 𝑛 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 𝑥 𝑛 𝑞 2 𝑥 𝑛 + 1 𝑞 2 𝑢 + 2 𝜆 𝑛 𝑦 𝑛 𝐵 𝑢 𝑛 𝐵 𝑞 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 . 𝑞 = 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 𝑥 𝑛 𝑥 𝑛 + 1 𝑥 𝑛 + 𝑥 𝑞 𝑛 + 1 𝑢 𝑞 + 2 𝜆 𝑛 𝑦 𝑛 𝐵 𝑢 𝑛 𝐵 𝑞 + 2 𝜖 𝑛 1 𝜖 𝑛 𝛾 𝑥 𝛾 𝑓 𝑛 𝑦 𝐴 𝑞 𝑛 . 𝑞 ( 3 . 3 3 ) By condition (C1), l i m 𝑛 𝑥 𝑛 𝑥 𝑛 + 1 = 0 and l i m 𝑛 𝐵 𝑢 𝑛 𝐵 𝑞 = 0 .
So, we have 𝑥 𝑛 𝑢 𝑛 0 , 𝑢 𝑛 𝑦 𝑛 0 as 𝑛 . It follows that 𝑥 𝑛 𝑦 𝑛 𝑥 𝑛 𝑢 𝑛 + 𝑢 𝑛 𝑦 𝑛 0 , a s 𝑛 . ( 3 . 3 4 ) From (3.2), we have 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 1 + 𝑊 𝑛 𝑦 𝑛 1 𝑊 𝑛 𝑦 𝑛 𝑃 𝐶 𝜖 𝑛 1 𝑥 𝛾 𝑓 𝑛 1 + 𝐼 𝛼 𝑛 1 𝐴 𝑊 𝑛 𝑦 𝑛 1 𝑃 𝐶 𝑊 𝑛 𝑦 𝑛 1 + 𝑦 𝑛 1 𝑦 𝑛 𝜖 𝑛 1 𝛾 𝑓 𝑥 𝑛 1 𝐴 𝑊 𝑛 𝑦 𝑛 1 + 𝑦 𝑛 1 𝑦 𝑛 . ( 3 . 3 5 ) By condition (C1) and l i m 𝑛 𝑦 𝑛 1 𝑦 𝑛 = 0 , we obtain that 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 0 as 𝑛 .
Hence, we have 𝑥 𝑛 𝑊 𝑛 𝑥 𝑛 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 + 𝑊 𝑛 𝑦 𝑛 𝑊 𝑛 𝑥 𝑛 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 + 𝑦 𝑛 𝑥 𝑛 . ( 3 . 3 6 ) By (3.34) and l i m 𝑛 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 = 0 , we obtain 𝑥 𝑛 𝑊 𝑛 𝑥 𝑛 0 as 𝑛 .
Moreover, we also have 𝑦 𝑛 𝑊 𝑛 𝑦 𝑛 𝑦 𝑛 𝑥 𝑛 + 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 . ( 3 . 3 7 ) By (3.34) and l i m 𝑛 𝑥 𝑛 𝑊 𝑛 𝑦 𝑛 = 0 , we obtain 𝑦 𝑛 𝑊 𝑛 𝑦 𝑛 0 as 𝑛 .
Step 5. We show that 𝑞 𝜃 = 𝑛 = 1 𝐹 ( 𝑇 𝑛 ) ( 𝑁 𝑘 = 1 S M E P ( 𝐹 𝑘 ) ) 𝐼 ( 𝐵 , 𝑀 ) and l i m s u p 𝑛 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑞 0 . It is easy to see that 𝑃 𝜃 ( 𝛾 𝑓 + ( 𝐼 𝐴 ) ) is a contraction of 𝐻 into itself.
Indeed, since 0 < 𝛾 < 𝛾 / 𝜖 , we have 𝑃 𝜃 ( 𝛾 𝑓 + ( 𝐼 𝐴 ) ) 𝑥 𝑃 𝜃 ( 𝛾 𝑓 + ( 𝐼 𝐴 ) ) 𝑦 𝛾 𝑓 ( 𝑥 ) 𝑓 ( 𝑦 ) + 𝐼 𝐴 𝑥 y 𝛾 𝜖 𝑥 𝑦 + 1 𝛾 𝑥 𝑦 1 𝛾 + 𝛾 𝜖 𝑥 𝑦 . ( 3 . 3 8 ) Since 𝐻 is complete, then there exists a unique fixed point 𝑞 𝐻 such that 𝑞 = 𝑃 𝜃 ( 𝛾 𝑓 + ( 𝐼 𝐴 ) ) ( 𝑞 ) . By Lemma 2.2, we obtain that ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑤 𝑞 0 for all 𝑤 𝜃 .
Next, we show that l i m s u p 𝑛 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑞 0 , where 𝑞 = 𝑃 𝜃 ( 𝛾 𝑓 + 𝐼 𝐴 ) ( 𝑞 ) is the unique solution of the variational inequality ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑤 𝑞 0 for all 𝑤 𝜃 . We can choose a subsequence { 𝑦 𝑛 𝑖 } of { 𝑦 𝑛 } such that l i m s u p 𝑛 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑞 = l i m 𝑖 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑖 . 𝑞 ( 3 . 3 9 ) As { 𝑦 𝑛 𝑖 } is bounded, there exists a subsequence { 𝑦 𝑛 𝑖 𝑗 } of { 𝑦 𝑛 𝑖 } which converges weakly to 𝑤 . We may assume without loss of generality that 𝑦 𝑛 𝑖 𝑤 .
Next we claim that 𝑤 𝜃 . Since 𝑦 𝑛 𝑊 𝑛 𝑦 𝑛 0 , 𝑥 𝑛 𝑊 𝑛 𝑥 𝑛 0 , and 𝑥 𝑛 𝑦 𝑛 0 , and by Lemma 2.6, we have 𝑤 𝑛 = 1 𝐹 ( 𝑇 𝑛 ) .
Next, we show that 𝑤 𝑘 = 1 S M E P ( 𝐹 𝑘 ) . Since 𝑢 𝑛 = 𝜏 𝑁 𝑟 𝑘 , 𝑛 𝑥 𝑛 , for 𝑘 = 1 , 2 , 3 , , 𝑁 , we know that 𝐹 𝑘 𝑢 𝑛 𝑢 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑛 + 1 𝑟 𝑛 𝑦 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 0 , 𝑦 𝐶 . ( 3 . 4 0 ) It follows by (A2) that 𝑢 𝜑 ( 𝑦 ) 𝜑 𝑛 + 1 𝑟 𝑛 𝑦 𝑢 𝑛 , 𝑢 𝑛 𝑥 𝑛 𝐹 𝑘 𝑦 , 𝑢 𝑛 , 𝑦 𝐶 . ( 3 . 4 1 ) Hence, for 𝑘 = 1 , 2 , 3 , , 𝑁 , we get 𝑢 𝜑 ( 𝑦 ) 𝜑 𝑛 𝑖 + 1 𝑟 𝑛 𝑖 𝑦 𝑢 𝑛 𝑖 , 𝑢 𝑛 𝑖 𝑥 𝑛 𝑖 𝐹 𝑘 𝑦 , 𝑢 𝑛 𝑖 , 𝑦 𝐶 . ( 3 . 4 2 ) For 𝑡 ( 0 , 1 ] and 𝑦 𝐻 , let 𝑦 𝑡 = 𝑡 𝑦 + ( 1 𝑡 ) 𝑤 . From (3.42), we have 𝑦 0 𝜑 𝑡 𝑢 + 𝜑 𝑛 𝑖 1 𝑟 𝑛 𝑖 𝑦 𝑡 𝑢 𝑛 𝑖 , 𝑢 𝑛 𝑖 𝑥 𝑛 𝑖 + 𝐹 𝑘 𝑦 𝑡 , 𝑢 𝑛 𝑖 . ( 3 . 4 3 )
Since 𝑢 𝑛 𝑖 𝑥 𝑛 𝑖 0 , from (A4) and the weakly lower semicontinuity of 𝜑 , ( 𝑢 𝑛 𝑖 𝑥 𝑛 𝑖 ) / 𝑟 𝑛 𝑖 0 and 𝑢 𝑛 𝑖 𝑤 . From (A1) and (A4), we have 0 = 𝐹 𝑘 𝑦 𝑡 , 𝑦 𝑡 𝑦 𝜑 𝑡 𝑦 + 𝜑 𝑡 𝑡 𝐹 𝑘 𝑦 𝑡 , 𝑦 + ( 1 𝑡 ) 𝐹 𝑘 𝑦 𝑡 𝑦 , 𝑤 + 𝑡 𝜑 ( 𝑦 ) + ( 1 𝑡 ) 𝜑 ( 𝑤 ) 𝜑 𝑡 𝐹 𝑡 𝑘 𝑦 𝑡 𝑦 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑡 . ( 3 . 4 4 ) Dividing by 𝑡 , we get 𝐹 𝑘 𝑦 𝑡 𝑦 , 𝑦 + 𝜑 ( 𝑦 ) 𝜑 𝑡 0 . ( 3 . 4 5 ) The weakly lower semicontinuity of 𝜑 for 𝑘 = 1 , 2 , 3 , , 𝑁 , we get 𝐹 𝑘 ( 𝑤 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑤 ) . ( 3 . 4 6 ) So, we have 𝐹 𝑘 ( 𝑤 , 𝑦 ) + 𝜑 ( 𝑦 ) 𝜑 ( 𝑤 ) 0 , 𝑘 = 1 , 2 , 3 , , 𝑁 . ( 3 . 4 7 ) This implies that 𝑤 𝑁 𝑘 = 1 S M E P ( 𝐹 𝑘 ) .
Lastly, we show that 𝑤 𝐼 ( 𝐵 , 𝑀 ) . In fact, since B is 𝛽 -inverse strongly monotone, hence 𝐵 is a monotone and Lipschitz continuous mapping. It follows from Lemma 2.3 that 𝑀 + 𝐵 is a maximal monotone. Let ( 𝑣 , 𝑔 ) 𝐺 ( 𝑀 + 𝐵 ) , since 𝑔 𝐵 𝑣 𝑀 ( 𝑣 ) . Again since 𝑦 𝑛 𝑖 = 𝐽 𝑀 , 𝜆 ( 𝑢 𝑛 𝑖 𝜆 𝐵 𝑢 𝑛 𝑖 ) , we have 𝑢 𝑛 𝑖 𝜆 𝐵 𝑢 𝑛 𝑖 ( 𝐼 + 𝜆 𝑀 ) ( 𝑦 𝑛 𝑖 ) , that is, ( 1 / 𝜆 ) ( 𝑢 𝑛 𝑖 𝑦 𝑛 𝑖 𝜆 𝐵 𝑢 𝑛 𝑖 ) 𝑀 ( 𝑦 𝑛 𝑖 ) . By virtue of the maximal monotonicity of 𝑀 + 𝐵 , we have 𝑣 𝑦 𝑛 𝑖 1 , 𝑔 𝐵 𝑣 𝜆 𝑢 𝑛 𝑖 𝑦 𝑛 𝑖 𝜆 𝐵 𝑢 𝑛 𝑖 0 , ( 3 . 4 8 ) and hence 𝑣 𝑦 𝑛 𝑖 , 𝑔 𝑣 𝑦 𝑛 𝑖 1 , 𝐵 𝑣 + 𝜆 𝑢 𝑛 𝑖 𝑦 𝑛 𝑖 𝜆 𝐵 𝑢 𝑛 𝑖 = 𝑣 𝑦 𝑛 𝑖 , 𝐵 𝑣 𝐵 𝑦 𝑛 𝑖 + 𝑣 𝑦 𝑛 𝑖 , 𝐵 𝑦 𝑛 𝑖 𝐵 𝑢 𝑛 𝑖 + 𝑣 𝑦 𝑛 𝑖 , 1 𝜆 𝑢 𝑛 𝑖 𝑦 𝑛 𝑖 . ( 3 . 4 9 ) It follows from l i m 𝑛 𝑢 𝑛 𝑦 𝑛 = 0 , we have l i m 𝑛 𝐵 𝑢 𝑛 𝐵 𝑦 𝑛 = 0 and 𝑦 𝑛 𝑖 𝑤 , it follows that l i m s u p 𝑛 𝑣 𝑦 𝑛 , 𝑔 = 𝑣 𝑤 , 𝑔 0 . ( 3 . 5 0 ) It follows from the maximal monotonicity of 𝐵 + 𝑀 that 𝜃 ( 𝑀 + 𝐵 ) ( 𝑤 ) , that is, 𝑤 𝐼 ( 𝐵 , 𝑀 ) . Therefore, 𝑤 𝜃 . We observe that l i m s u p 𝑛 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑞 = l i m 𝑖 ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑊 𝑛 𝑦 𝑛 𝑖 𝑞 = ( 𝛾 𝑓 𝐴 ) 𝑞 , 𝑤 𝑞 0 . ( 3 . 5 1 )
Step 6. Finally, we prove 𝑥 𝑛 𝑞 . By using (3.2) and together with Schwarz inequality, we have 𝑥 𝑛 + 1 𝑞 2 = 𝑃 𝐶 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑃 𝐶 ( 𝑞 ) 2 𝜖 𝑛 𝑥 𝛾 𝑓 𝑛 + 𝐴 𝑞 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑞 2 𝐼 𝜖 𝑛 𝐴 2 𝑊 𝑛 𝑦 𝑛 𝑞 2 + 𝜖 2 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 2 𝜖 𝑛 𝐼 𝜖 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑥 𝑞 , 𝛾 𝑓 𝑛 𝐴 𝑞 1 𝜖 𝑛 𝛾 2 𝑦 𝑛 𝑞 2 + 𝜖 2 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 2 𝜖 𝑛 𝑊 𝑛 𝑦 𝑛 𝑥 𝑞 , 𝛾 𝑓 𝑛 𝐴 𝑞 2 𝜖 2 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑥 𝑞 , 𝛾 𝑓 𝑛 𝐴 𝑞 1 𝜖 𝑛 𝛾 2 𝑥 𝑛 𝑞 2 + 𝜖 2 𝑛 𝑥 𝛾 𝑓 𝑛 𝐴 𝑞 2 + 2 𝜖 𝑛 𝑊 𝑛 𝑦 𝑛 𝑥 𝑞 , 𝛾 𝑓 𝑛 𝛾 𝑓 ( 𝑞 ) + 2 𝜖 𝑛 𝑊 𝑛 𝑦 𝑛 𝑞 , 𝛾 𝑓 ( 𝑞 ) 𝐴 𝑞 2 𝜖 2 𝑛 𝐴 𝑊 𝑛 𝑦 𝑛 𝑥 𝑞 , 𝛾 𝑓 𝑛 𝐴 𝑞<