Abstract

We introduce a new iterative scheme for finding a common element of the set of solutions of a general system of variational inequalities, the set of solutions of a mixed equilibrium problem, and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Using the demiclosedness principle for nonexpansive mappings, we prove that the iterative sequence converges strongly to a common element of the above three sets under some control conditions, and we also give some examples for mappings which satisfy conditions of the main result.

1. Introduction

Let 𝐻 be a real Hilbert space with inner product , and 𝐶 a nonempty closed convex subset of 𝐻. A mapping 𝑇𝐶𝐶 is said to be nonexpansive mapping if 𝑇𝑥𝑇𝑦𝑥𝑦 for all 𝑥,𝑦𝐶. The fixed point set of 𝑇 is denoted by 𝐹(𝑇)={𝑥𝐶𝑇𝑥=𝑥}.

Halpern [1] studied the following iteration formula for approximating a fixed point of 𝑇. For an arbitrary 𝑣𝐶, let the sequence {𝑥𝑛} be defined by 𝑥1𝐶,𝑥𝑛+1=𝛼𝑛𝑣+1𝛼𝑛𝑇𝑥𝑛,𝑛1,(1.1) where {𝛼𝑛} is a sequence of real numbers in [0,1] that satisfies the following conditions: 𝐶1lim𝑛𝛼𝑛=0 and 𝐶2𝑛=1𝛼𝑛=.

Actually, Halpern studied the special case of (1.1) in which 𝛼𝑛=𝑛𝜎, 𝜎(0,1), and 𝑣=0 and proved that {𝑥𝑛} converges strongly to a fixed point of 𝑇. Under a different restriction on the parameter {𝛼𝑛}, Lions [2] improved the result of Halpern, still in Hilbert spaces. He proved strong convergence of {𝑥𝑛} to a fixed point of 𝑇, where the real sequence {𝛼𝑛} satisfies the following conditions: 𝐶1lim𝑛𝛼𝑛=0,𝐶2𝑛=1𝛼𝑛=,𝐶3lim𝑛||𝛼𝑛+1𝛼𝑛||𝛼2𝑛+1=0.(1.2) Reich [3] proved that the result of Halpern remains true when 𝐻 is uniformly smooth. It was observed that both Halpern's and Lion's conditions on the real sequence {𝛼𝑛} excluded the canonical choice 𝛼𝑛=1/(𝑛+1). This was overcome by Wittmann [4] who proved, still in Hilbert spaces, the strong convergence of {𝑥𝑛} to a fixed point of 𝑇 if {𝛼𝑛} satisfies the following conditions: 𝐶1lim𝑛𝛼𝑛=0,𝐶2𝑛=1𝛼𝑛=,𝐶3𝑛=1||𝛼𝑛+1𝛼𝑛||<.(1.3)

Recall that the classical variational inequality, denoted by VI(𝐶,𝐴), is to find an 𝑥𝐶 such that 𝐴𝑥,𝑦𝑥0,𝑦𝐶.(1.4) The variational inequality, nonconvex variational inequality, and mixed variational inequality have been widely studied in the literature; see, for example, Ceng et al. [514], Chang et al. [15], Noor [1618], Peng and Yao [1921], Plubtieng and Punpaeng [22], Yao et al. [23], Zeng and Yao [24], Zhao and He [25], and the references therein.

For solving the variational inequality problem in the finite-dimensional Euclidean space 𝑛 under the assumption that a set 𝐶𝑛 is closed and convex, a mapping 𝐴 of 𝐶 into 𝑛 is monotone and 𝑘-Lipschitz continuous, and VI(𝐶,𝐴) is nonempty, Korpelevich [26] introduced the following so-called extragradient method: 𝑥0𝑦=𝑥𝐶,𝑛=𝑃𝐶𝑥𝑛𝜆𝐴𝑥𝑛,𝑥𝑛+1=𝑃𝐶𝑥𝑛𝜆𝐴𝑦𝑛,(1.5) for every 𝑛=0,1,2,, where 𝜆(0,1/𝑘) and 𝑃𝐶 is the projection of 𝑛 onto 𝐶. He showed that the sequences {𝑥𝑛} and {𝑦𝑛} generated by this iterative process converge to the same point 𝑧VI(𝐶,𝐴). In 2007, Y. Yao and J. C. Yao [27] introduced a new iterative scheme for finding the common element of the set of fixed points of nonexpansive mapping and the set of solutions of the variational inequality for 𝛼-inverse-strongly monotone mappings in a real Hilbert space.

Let C be a nonempty closed convex subset of a real Hilbert space 𝐻.

Let 𝐴,𝐵𝐶𝐻 be two mappings. In 2008, Ceng et al. [12] considered the following problem of finding (𝑥,𝑦)𝐶×𝐶 such that𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐶,𝜇𝐵𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐶,(1.6) which is called a general system of variational inequalities, where 𝜆>0 and 𝜇>0 are two constants. In particular, if 𝐴=𝐵, then the problem (1.6) reduces to finding (𝑥,𝑦)𝐶×𝐶 such that𝜆𝐴𝑦+𝑥𝑦,𝑥𝑥0,𝑥𝐶,𝜇𝐴𝑥+𝑦𝑥,𝑥𝑦0,𝑥𝐶,(1.7) which was defined by Verma [28] and is called the new system of variational inequalities. Further, if we add up the requirement that 𝑥=𝑦, then the problem (1.7) reduces to the classical variational inequality VI(𝐶,𝐴). Ceng et al. [12] introduced and studied a relaxed extragradient method for finding a common element of the set of solutions of the problem (1.6) for the 𝛼- and 𝛽-inverse-strongly monotone mappings and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Let 𝑥1=𝑣𝐶, and {𝑥𝑛}, {𝑦𝑛} are given by 𝑦𝑛=𝑃𝐶𝑥𝑛𝜇𝐵𝑥𝑛,𝑥𝑛+1=𝑎𝑛𝑣+𝑏𝑛𝑥𝑛+1𝑎𝑛𝑏𝑛𝑆𝑃𝐶𝑦𝑛𝜆𝐴𝑦𝑛,𝑛1,(1.8) where 𝜆(0,2𝛼), 𝜇(0,2𝛽), and {𝑎𝑛},{𝑏𝑛}[0,1]. Then, they proved that the sequence {𝑥𝑛} converges strongly to a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of the problem (1.6) under some control conditions.

Let 𝜑𝐶{+} be a proper extended real-valued function and 𝐹 a bifunction from 𝐶×𝐶 to , where is the set of real numbers. Ceng and Yao [13] considered the following mixed equilibrium problem:nd𝑥𝐶suchthat𝐹(𝑥,𝑦)+𝜑(𝑦)𝜑(𝑥),𝑦𝐶.(1.9)

The set of solutions of (1.9) is denoted by MEP(𝐹,𝜑). It is easy to see that 𝑥 is a solution of the problem (1.9) impling that 𝑥dom𝜑={𝑥𝐶𝜑(𝑥)<+}.

If 𝜑=0, then the mixed equilibrium problem (1.9) becomes the following equilibrium problem:nd𝑥𝐶suchthat𝐹(𝑥,𝑦)0,𝑦𝐶.(1.10) If 𝐹=0, then the mixed equilibrium problem (1.9) reduces to the convex minimization problemnd𝑥𝐶suchthat𝜑(𝑦)𝜑(𝑥),𝑦𝐶.(1.11)

The set of solutions of (1.10) is denoted by EP(𝐹).

If 𝜑=0 and 𝐹(𝑥,𝑦)=𝐴𝑥,𝑦𝑥 for all 𝑥,𝑦𝐶, where 𝐴 is a mapping from 𝐶 into 𝐻, then (1.9) reduces to the classical variational inequality and EP(𝐹)=VI(𝐶,𝐴).

The equilibrium problems and variational inequality problems can be applied for the problems in science and technology, economics, optimizations, and control theory. The problems of finding a solution of the systems of variational inequalities which is also a solution of the mixed equilibrium problems and the problem of finding a solution of the mixed equilibrium problems under the constraints that it is also a solution of the system of variational inequalities are very useful and important for studying problems in science and applied science.

Inspired and motivated by these facts, we introduce a new iteration process for finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of a general system of variational inequalities in a real Hilbert space. Start with an arbitrary 𝑣𝐶, and let 𝑥1𝐶, {𝑥𝑛},{𝑢𝑛}, and {𝑦𝑛} be the sequences generated by𝐹𝑢𝑛𝑢,𝑦+𝜑(𝑦)𝜑𝑛+1𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑦0,𝑦𝐶,𝑛=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛,𝑥𝑛+1=𝑎𝑛𝑣+1𝑎𝑛𝑇𝑃𝐶𝑦𝑛𝜆𝐴𝑦𝑛,𝑛1,(1.12) where 𝜆>0 and 𝜇>0 are two constants, {𝑟𝑛}(0,) and {𝑎𝑛}[0,1]. Using the demiclosedness principle for nonexpansive mappings, we will show that the sequence {𝑥𝑛} converges strongly to a common element of the above three sets under some control conditions.

2. Preliminaries

In this section, we recall the well-known results and give some useful lemmas that will be used in the next section.

Let 𝐻 be a real Hilbert space with inner product , and 𝐶 a nonempty closed convex subset of 𝐻.

A mapping 𝐴𝐶𝐻 is called monotone if 𝐴𝑥𝐴𝑦,𝑥𝑦0,𝑥,𝑦𝐶.(2.1)𝐴 is called 𝛼-strongly monotone if there exists a positive real number 𝛼>0 such that 𝐴𝑥𝐴𝑦,𝑥𝑦𝛼𝑥𝑦2,𝑥,𝑦𝐶.(2.2) This implies that 𝐴𝑥𝐴𝑦𝛼𝑥𝑦,(2.3) that is, 𝐴 is 𝛼-expansive and, when 𝛼=1, it is expansive.

We can see easily that the following implications in monotonicity, strong monotonicity, and expansiveness hold: strongmonotonicitymonotonicityexpansiveness.(2.4)

The mapping 𝐴 is called 𝐿-Lipschitz continuous (or Lipschitzian) if there exists a constant 𝐿0 such that 𝐴𝑥𝐴𝑦𝐿𝑥𝑦,𝑥,𝑦𝐶.(2.5)𝐴 is called 𝛼-inverse-strongly monotone (or 𝛼-cocoercive) if there exists a positive real number 𝛼>0 such that 𝐴𝑥𝐴𝑦,𝑥𝑦𝛼𝐴𝑥𝐴𝑦2,𝑥,𝑦𝐶.(2.6) It is obvious that every 𝛼-inverse-strongly monotone mapping 𝐴 is monotone and Lipschitz continuous. Also, if 𝐴 is 𝛼-strongly and 𝐿-Lipschitz continuous, then 𝐴 is (𝛼/𝐿2)-inverse-strongly monotone, but inverse-strongly monotone need not be strongly monotone.

A mapping 𝐴 is called relaxed c-cocoercive, if there exists a constant 𝑐>0 such that 𝐴𝑥𝐴𝑦,𝑥𝑦(𝑐)𝐴𝑥𝐴𝑦2,𝑥,𝑦𝐶.(2.7) A mapping 𝐴 is called relaxed (c,d)-cocoercive, if there exist two constants 𝑐,𝑑>0 such that 𝐴𝑥𝐴𝑦,𝑥𝑦(𝑐)𝐴𝑥𝐴𝑦2+𝑑𝑥𝑦2,𝑥,𝑦𝐶.(2.8) For 𝑐=0, 𝐴 is 𝑑-strongly monotone. This class of mappings is more general than the class of strongly monotone mappings. As a result, we have the following implication: 𝑑-strong monotonicity relaxed (𝑐,𝑑)-cocoercivity.

It is known that if the operator is Lipschitz continuous, then the relaxed cocoercivity is strongly monotone, but the strongly monotone does not imply the cocoercivity as shown in the following example.

Example 2.1. Let 𝐻=𝑅, 𝐶=[1,), and 𝐴𝐶𝐻 be defined by 𝐴𝑥=𝑥2, 𝑥𝐶. For 𝑥,𝑦𝐶, we have 𝑥𝐴𝑥𝐴𝑦,𝑥𝑦=2𝑦2||||(𝑥𝑦)=(𝑥+𝑦)𝑥𝑦2||||2𝑥𝑦2.(2.9) Hence, 𝐴 is 2-strongly monotone. If 𝐴 is 𝜇-cocoercive for some 𝜇>0, then 𝐴𝑥𝐴𝑦,𝑥𝑦=(𝑥+𝑦)|𝑥𝑦|2𝜇|𝑥2𝑦2|2. This implies 𝑥+𝑦1/𝜇 for all 𝑥,𝑦[1,) which is a contradiction. Hence, 𝐴 is not 𝜇-cocoercive for any 𝜇>0.

For every point 𝑥𝐻, there exists a unique nearest point in 𝐶, denoted by 𝑃𝐶𝑥, such that 𝑥𝑃𝐶𝑥𝑥𝑦,𝑦𝐶.(2.10)𝑃𝐶 is called the metric projection of 𝐻 onto 𝐶. It is well known that 𝑃𝐶 is a nonexpansive mapping of 𝐻 onto 𝐶 and satisfies𝑥𝑦,𝑃𝐶𝑥𝑃𝐶𝑃𝑦𝐶𝑥𝑃𝐶𝑦2,𝑥,𝑦𝐻.(2.11) Obviously, this immediately implies that𝑃(𝑥𝑦)𝐶𝑥𝑃𝐶𝑦2𝑥𝑦2𝑃𝐶𝑥𝑃𝐶𝑦2,𝑥,𝑦𝐻.(2.12) Recall that 𝑃𝐶𝑥 is characterized by the following properties: 𝑃𝐶𝑥𝐶 and𝑥𝑃𝐶𝑥,𝑦𝑃𝐶𝑥0,𝑥𝑦2𝑥𝑃𝐶𝑥2+𝑃𝐶𝑥𝑦2,(2.13) for all 𝑥𝐻 and 𝑦𝐶; see Goebel and Kirk [29] for more details.

For solving the mixed equilibrium problem, let us assume the following assumptions for the bifunction 𝐹,𝜑 and the set 𝐶:(A1)𝐹(𝑥,𝑥)=0 for all 𝑥𝐶;(A2)𝐹 is monotone, that is, 𝐹(𝑥,𝑦)+𝐹(𝑦,𝑥)0 for all 𝑥,𝑦𝐶;(A3) for each 𝑦𝐶, 𝑥𝐹(𝑥,𝑦) is weakly upper semicontinuous;(A4) for each 𝑥𝐶, 𝑦𝐹(𝑥,𝑦) is convex;(A5) for each 𝑥𝐶, 𝑦𝐹(𝑥,𝑦) is lower semicontinuous;(B1) for each 𝑥𝐻 and 𝑟>0, there exist a bounded subset 𝐷𝑥𝐶 and 𝑦𝑥𝐶 such that for any 𝑧𝐶𝐷𝑥𝐹𝑧,𝑦𝑥𝑦+𝜑𝑥+1𝑟𝑦𝑥𝑧,𝑧𝑥<𝜑(𝑧);(2.14)(B2)𝐶 is a bounded set.

In the sequel we will need to use the following lemma.

Lemma 2.2 (see [30]). Let 𝐶 be a nonempty closed convex subset of 𝐻. Let 𝐹 be a bifunction from 𝐶×𝐶 to satisfying (A1)–(A5), and let 𝜑𝐶{+} be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For 𝑟>0 and 𝑥𝐻, define a mapping 𝑇𝑟𝐻𝐶 as follows 𝑇𝑟1(𝑥)=𝑧𝐶𝐹(𝑧,𝑦)+𝜑(𝑦)+𝑟𝑦𝑧,𝑧𝑥𝜑(𝑧),𝑦𝐶(2.15) for all 𝑥𝐻. Then the following conclusions hold: (1)for each 𝑥𝐻, 𝑇𝑟(𝑥);(2)𝑇𝑟 is single valued;(3)𝑇𝑟 is firmly nonexpansive, that is, for any 𝑥,𝑦𝐻, 𝑇𝑟(𝑥)𝑇𝑟(𝑦)2𝑇𝑟𝑥𝑇𝑟𝑦,𝑥𝑦;(2.16)(4)𝐹(𝑇𝑟)=MEP(𝐹,𝜑);(5)MEP(𝐹,𝜑) is closed and convex.

We also need the following lemmas.

Lemma 2.3 (see [31]). Let {𝑎𝑛} be a sequence of nonnegative real numbers satisfying the property 𝑎𝑛+11𝑡𝑛𝑎𝑛+𝑏𝑛+𝑡𝑛𝑐𝑛,𝑛1,(2.17) where {𝑡𝑛}, {𝑏𝑛}, and {𝑐𝑛} satisfy the restrictions (i)𝑛=1𝑡𝑛=,(ii)𝑛=1𝑏𝑛<,(iii)limsup𝑛𝑐𝑛0.(2.18) Then lim𝑛𝑎𝑛=0.

Lemma 2.4 (see [32]). Let (𝐻,,) be an inner product space. Then, for all 𝑥,𝑦,𝑧𝐻 and 𝛼,𝛽,𝛾[0,1] with 𝛼+𝛽+𝛾=1, one has 𝛼𝑥+𝛽𝑦+𝛾𝑧2=𝛼𝑥2+𝛽𝑦2+𝛾𝑧2𝛼𝛽𝑥𝑦2𝛼𝛾𝑥𝑧2𝛽𝛾𝑦𝑧2.(2.19)

Lemma 2.5 (see [29] (demiclosedness principle)). Assume that 𝑇 is a nonexpansive self-mapping of a nonempty closed convex subset 𝐶 of a real Hilbert space 𝐻. If 𝑇 has a fixed point, then 𝐼𝑇 is demiclosed: that is, whenever {𝑥𝑛} is a sequence in 𝐶 converging weakly to some 𝑥𝐶 (for short, 𝑥𝑛𝑥𝐶) and the sequence {(𝐼𝑇)𝑥𝑛} converges strongly to some 𝑦 (for short, (𝐼𝑇)𝑥𝑛𝑦), it follows that (𝐼𝑇)𝑥=𝑦. Here 𝐼 is the identity operator of 𝐻.

The following lemma is an immediate consequence of an inner product.

Lemma 2.6. In a real Hilbert space 𝐻, there holds the inequality 𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦,𝑥,𝑦𝐻.(2.20)

In 2009, Kangtunyakarn and Suantai [33] introduced a new mapping called the 𝑆-mapping as follows.

Let {𝑇𝑖}𝑁𝑖=1 be a finite family of nonexpansive mappings of 𝐶 into itself. For each 𝑛, and 𝑗=1,2,,𝑁, let 𝛼𝑗(𝑛)=(𝛼1𝑛,𝑗,𝛼2𝑛,𝑗,𝛼3𝑛,𝑗) be such that 𝛼1𝑛,𝑗,𝛼2𝑛,𝑗,𝛼3𝑛,𝑗[0,1] with 𝛼1𝑛,𝑗+𝛼2𝑛,𝑗+𝛼3𝑛,𝑗=1. Very recently, in 2009, Kangtunyakarn and Suantai [33] introduced the new mapping 𝑆𝑛𝐶𝐶 as follows: 𝑈𝑛,0𝑈=𝐼,𝑛,1=𝛼1𝑛,1𝑇1𝑈𝑛,0+𝛼2𝑛,1𝑈𝑛,0+𝛼3𝑛,1𝑈𝐼,𝑛,2=𝛼1𝑛,2𝑇2𝑈𝑛,1+𝛼2𝑛,2𝑈𝑛,1+𝛼3𝑛,2𝑈𝐼,𝑛,3=𝛼1𝑛,3𝑇3𝑈𝑛,2+𝛼2𝑛,3𝑈𝑛,2+𝛼3𝑛,3𝑈𝐼,𝑛,𝑁1=𝛼1𝑛,𝑁1𝑇𝑁1𝑈𝑛,𝑁2+𝛼2𝑛,𝑁1𝑈𝑛,𝑁2+𝛼3𝑛,𝑁1𝑆𝐼,𝑛=𝑈𝑛,𝑁=𝛼1𝑛,𝑁𝑇𝑁𝑈𝑛,𝑁1+𝛼2𝑛,𝑁𝑈𝑛,𝑁1+𝛼3𝑛,𝑁𝐼.(2.21) The mapping 𝑆𝑛 is called the 𝑆-mapping generated by 𝑇1,𝑇2,,𝑇𝑁 and 𝛼1(𝑛),𝛼2(𝑛),,𝛼𝑁(𝑛). Nonexpansivity of each 𝑇𝑖 ensures the nonexpansivity of 𝑆𝑛. They also showed the following useful fact.

Lemma 2.7 (see [33]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝑋. Let {𝑇𝑖}𝑁𝑖=1 be a finite family of nonexpansive mappings of 𝐶 into itself with 𝑁𝑖=1𝐹(T𝑖), and let 𝛼𝑗=(𝛼𝑗1,𝛼𝑗2,𝛼𝑗3), 𝑗=1,2,,𝑁, where 𝛼𝑗1,𝛼𝑗2,𝛼𝑗3[0,1], 𝛼𝑗1+𝛼𝑗2+𝛼𝑗3=1, 𝛼𝑗1(0,1) for all 𝑗=1,2,,𝑁1, 𝛼𝑁1(0,1] and 𝛼𝑗2,𝛼𝑗3[0,1) for all 𝑗=1,2,,𝑁. Let 𝑆 be the 𝑆-mapping generated by 𝑇1,𝑇2,,𝑇𝑁 and 𝛼1,𝛼2,,𝛼𝑁. Then 𝐹(𝑆)=𝑁𝑖=1𝐹(𝑇𝑖).

Lemma 2.8 (see [12]). For given 𝑥,𝑦𝐶, (𝑥,𝑦) is a solution of the problem (1.6) if and only if 𝑥 is a fixed point of the mapping 𝐺𝐶𝐶 defined by 𝐺(𝑥)=𝑃𝐶𝑃𝐶(x𝜇𝐵𝑥)𝜆𝐴𝑃𝐶(𝑥𝜇𝐵𝑥),𝑥𝐶,(2.22) where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Throughout this paper, the set of fixed points of the mapping 𝐺 is denoted by GVI(𝐶,𝐴,𝐵).

Next, we prove a lemma which is very useful for our consideration.

Lemma 2.9. Let 𝐺𝐶𝐶 be defined by 𝐺(𝑥)=𝑃𝐶𝑃𝐶(𝑥𝜇𝐵𝑥)𝜆𝐴𝑃𝐶(𝑥𝜇𝐵𝑥),𝑥𝐶,(2.23) where 𝜆,𝜇>0 and 𝐴,𝐵𝐶𝐻 are two mappings. If 𝐼𝜆𝐴 and 𝐼𝜇𝐵 are nonexpansive mappings, then 𝐺 is nonexpansive.

Proof. For any 𝑥,𝑦𝐶, we have 𝑃𝐺(𝑥)𝐺(𝑦)=𝐶𝑃𝐶(𝑥𝜇𝐵𝑥)𝜆𝐴𝑃𝐶(𝑥𝜇𝐵𝑥)𝑃𝐶𝑃𝐶(𝑦𝜇𝐵𝑦)𝜆𝐴𝑃𝐶(𝑦𝜇𝐵𝑦)2𝑃𝐶(𝑥𝜇𝐵𝑥)𝜆𝐴𝑃𝐶𝑃(𝑥𝜇𝐵𝑥)𝐶(𝑦𝜇𝐵𝑦)𝜆𝐴𝑃𝐶=(𝑦𝜇𝐵𝑦)(𝐼𝜆A)𝑃𝐶(𝐼𝜇𝐵)𝑥(𝐼𝜆𝐴)𝑃𝐶(𝐼𝜇𝐵)𝑦𝑥𝑦.(2.24) This shows that 𝐺 is a nonexpansive mapping.

3. Main Results

In this section, we prove strong convergence theorems of the iterative scheme (1.12) to a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of a general system of variational inequality in a real Hilbert space.

Theorem 3.1. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐹 be a function from 𝐶×𝐶 to satisfying (A1)–(A5) and 𝜑𝐶{+} a proper lower semicontinuous and convex function. Let 𝜆,𝜇>0, and let 𝐴,𝐵𝐶𝐻 be such that 𝐼𝜆𝐴 and 𝐼𝜇𝐵 are nonexpansive mappings. Let 𝑇 be a nonexpansive self-mapping of 𝐶 such that Ω=𝐹(𝑇)GVI(𝐶,𝐴,𝐵)MEP(𝐹,𝜑). Assume that either (B1) or (B2) holds and that 𝑣 is an arbitrary point in 𝐶. Let the sequences {𝑥𝑛}, {𝑦𝑛}, and {𝑢𝑛} be defined by (1.12), {𝑟𝑛}(0,) and {𝑎𝑛}[0,1] such that (C1)lim𝑛𝑎𝑛=0, lim𝑛(𝑎𝑛+1/𝑎𝑛)=1, and 𝑛=1𝑎𝑛=, (C2)liminf𝑛𝑟𝑛>0 and 𝑛=1|𝑟𝑛+1𝑟𝑛|<.If lim𝑛𝐴𝑦𝑛𝐴𝑦=0 and lim𝑛𝐵𝑢𝑛𝐵𝑥=0 for all 𝑥Ω and 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥), then {𝑥𝑛} converges strongly to 𝑥=𝑃Ω𝑣 and (𝑥,𝑦) is a solution of the problem (1.6), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Proof. Let 𝑥Ω and {𝑇𝑟𝑛} a sequence of mappings defined as in Lemma 2.2. It follows from Lemma 2.8 that 𝑥=𝑃𝐶𝑃𝐶𝑥𝜇𝐵𝑥𝜆𝐴𝑃𝐶𝑥𝜇𝐵𝑥.(3.1) Put 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥) and 𝑡𝑛=𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛), then 𝑥=𝑃𝐶(𝑦𝜆𝐴𝑦) and 𝑥𝑛+1=𝑎𝑛𝑣+1𝑎𝑛𝑇𝑡𝑛.(3.2) Since 𝐼𝜆𝐴, 𝐼𝜇𝐵, 𝑇𝑟𝑛, and 𝑃𝐶 are nonexpansive mappings, we have 𝑡𝑛𝑥2=𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛)𝑃𝐶(𝑦𝜆𝐴𝑦)2𝑦𝑛𝑦2=𝑃𝐶(𝑢𝑛𝜇𝐵𝑢𝑛)𝑃𝐶(𝑥𝜇𝐵𝑥)2𝑢𝑛𝑥2=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥2𝑥𝑛𝑥2,(3.3) which implies that 𝑥𝑛+1𝑥=𝑎𝑛𝑣+1𝑎𝑛𝑇𝑡𝑛𝑥𝑎𝑛𝑣𝑥+1𝑎𝑛𝑡𝑛𝑥𝑎𝑛𝑣𝑥+1𝑎𝑛𝑥𝑛𝑥max𝑣𝑥𝑥,1𝑥.(3.4) Thus, {𝑥𝑛} is bounded. Consequently, the sequences {𝑢𝑛}, {𝑦𝑛}, {𝑡𝑛}, {𝐴𝑦𝑛}, {𝐵𝑢𝑛}, and {𝑇𝑡𝑛} are also bounded. Also, observe that 𝑡𝑛+1𝑡𝑛=𝑃𝐶𝑦𝑛+1𝜆𝐴𝑦𝑛+1𝑃𝐶𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝑛+1𝑦𝑛=𝑃𝐶𝑢𝑛+1𝜇𝐵𝑢𝑛+1𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛𝑢𝑛+1𝑢𝑛.(3.5) On the other hand, from 𝑢𝑛=𝑇𝑟𝑛𝑥𝑛dom𝜑 and 𝑢𝑛+1=𝑇𝑟𝑛+1𝑥𝑛+1dom𝜑, we have 𝐹𝑢𝑛𝑢,𝑦+𝜑(𝑦)𝜑𝑛+1𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝐹𝑢0,𝑦𝐶,(3.6)𝑛+1𝑢,𝑦+𝜑(𝑦)𝜑𝑛+1+1𝑟𝑛+1𝑦𝑢𝑛+1,𝑢𝑛+1𝑥𝑛+10,𝑦𝐶.(3.7) Putting 𝑦=𝑢𝑛+1 in (3.6) and 𝑦=𝑢𝑛 in (3.7), we have 𝐹𝑢𝑛,𝑢𝑛+1𝑢+𝜑𝑛+1𝑢𝜑𝑛+1𝑟𝑛𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑥𝑛𝐹𝑢0,𝑛+1,𝑢𝑛𝑢+𝜑𝑛𝑢𝜑𝑛+1+1𝑟𝑛+1𝑢𝑛𝑢𝑛+1,𝑢𝑛+1𝑥𝑛+10.(3.8) From the monotonicity of 𝐹, we obtain that 𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑥𝑛𝑟𝑛𝑢𝑛+1𝑥𝑛+1𝑟𝑛+10,(3.9) and hence 𝑢𝑛+1𝑢𝑛,𝑢𝑛𝑢𝑛+1+𝑢𝑛+1𝑥𝑛𝑟𝑛𝑟𝑛+1𝑢𝑛+1𝑥𝑛+10.(3.10) Without loss of generality, let us assume that there exists a real number 𝑐 such that 𝑟𝑛>𝑐>0 for all 𝑛. Then, we have 𝑢𝑛+1𝑢𝑛2𝑢𝑛+1𝑢𝑛,𝑥𝑛+1𝑥𝑛+𝑟1𝑛𝑟𝑛+1𝑢𝑛+1𝑥𝑛+1𝑢𝑛+1𝑢𝑛𝑥𝑛+1𝑥𝑛+||||𝑟1𝑛𝑟𝑛+1||||𝑢𝑛+1𝑥𝑛+1,(3.11) and hence 𝑢𝑛+1𝑢𝑛𝑥𝑛+1𝑥𝑛+1𝑟𝑛+1||𝑟𝑛+1𝑟𝑛||𝑢𝑛+1𝑥𝑛+1𝑥𝑛+1𝑥𝑛+1𝑐||𝑟𝑛+1𝑟𝑛||𝑢𝑛+1𝑥𝑛+1.(3.12) It follows from (3.5) and (3.12) that 𝑡𝑛+1𝑡𝑛𝑥𝑛+1𝑥𝑛+1𝑐||𝑟𝑛+1𝑟𝑛||𝑢𝑛+1𝑥𝑛+1.(3.13) Next, we show that 𝑥𝑛+1𝑥𝑛0 as 𝑛. From (3.13), we have 𝑥𝑛+2𝑥𝑛+1=𝑎𝑛+1𝑣+1𝑎𝑛+1𝑇𝑡𝑛+1𝑎𝑛𝑣1𝑎𝑛𝑇𝑡𝑛=𝑎𝑛+1𝑎n𝑎𝑣+𝑛𝑎𝑛+1𝑇𝑡𝑛+1+1𝑎𝑛𝑇𝑡𝑛+1𝑇𝑡𝑛||𝑎𝑛+1𝑎𝑛||𝑣+𝑇𝑡𝑛+1+1𝑎𝑛𝑡𝑛+1𝑡𝑛||𝑎𝑛+1𝑎𝑛||𝑣+𝑇𝑡𝑛+1+1𝑎𝑛𝑥𝑛+1𝑥𝑛+1𝑐||𝑟𝑛+1𝑟𝑛||𝑢𝑛+1𝑥𝑛+11𝑎𝑛𝑥𝑛+1𝑥𝑛+1𝑐||𝑟𝑛+1𝑟𝑛||||𝑎𝑀+2𝑀𝑛+1𝑎𝑛||,(3.14) where 𝑀=max{𝑣,sup𝑛𝑇𝑡𝑛+1,sup𝑛𝑢𝑛+1𝑥𝑛+1}. By the assumptions on {𝑎𝑛} and {𝑟𝑛} and Lemma 2.3, we conclude that lim𝑛𝑥𝑛+1𝑥𝑛=0.(3.15) From (C1), (C2), (3.5), (3.12), and (3.15), we also have 𝑢𝑛+1𝑢𝑛0, 𝑡𝑛+1𝑡𝑛0, and 𝑦𝑛+1𝑦𝑛0, as 𝑛.
Since 𝑥𝑛+1𝑥𝑛=𝑎𝑛𝑣𝑥𝑛+1𝑎𝑛𝑇𝑡𝑛𝑥𝑛,(3.16) we have that 𝑇𝑡𝑛𝑥𝑛0as𝑛.(3.17) Next, we prove that lim𝑛𝑥𝑛𝑢𝑛=0. From Lemma 2.2(3), we have 𝑢𝑛𝑥2=𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥2𝑇𝑟𝑛𝑥𝑛𝑇𝑟𝑛𝑥,𝑥𝑛𝑥=𝑢𝑛𝑥,𝑥𝑛𝑥1=2𝑢𝑛𝑥2+𝑥𝑛𝑥2𝑥𝑛𝑢𝑛2.(3.18) Hence, 𝑢𝑛𝑥2𝑥𝑛𝑥2𝑥𝑛𝑢𝑛2.(3.19) From Lemma 2.4, (3.3), and (3.19), we have 𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑡𝑛𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑢𝑛𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑥𝑛𝑥2𝑥𝑛𝑢𝑛2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥21𝑎𝑛𝑥𝑛𝑢𝑛2.(3.20) It follows that 1𝑎𝑛𝑥𝑛𝑢𝑛2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥2𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛.(3.21) From conditions (C1) and (3.15), we obtain lim𝑛𝑥𝑛𝑢𝑛=0.(3.22) By (3.17) and (3.22), we have 𝑇𝑡𝑛𝑢𝑛𝑇𝑡𝑛𝑥𝑛+𝑥𝑛𝑢𝑛0,as𝑛.(3.23) Next, we prove that 𝑇𝑡𝑛𝑡𝑛0 as 𝑛. From (2.11) and nonexpansiveness of 𝐼𝜇𝐵, we get 𝑦𝑛𝑦2=𝑃𝐶(𝑢𝑛𝜇𝐵𝑢𝑛)𝑃𝐶(𝑥𝜇𝐵𝑥)2𝑢𝑛𝜇𝐵𝑢𝑛𝑥𝜇𝐵𝑥,𝑦𝑛𝑦=12(𝑢𝑛𝜇𝐵𝑢𝑛)(𝑥𝜇𝐵𝑥)2+𝑦𝑛𝑦2(𝑢𝑛𝜇𝐵𝑢𝑛)(𝑥𝜇𝐵𝑥)(𝑦𝑛𝑦)212𝑢𝑛𝑥2+𝑦𝑛𝑦2(𝑢𝑛𝑥)(𝑦𝑛𝑦)2𝑢+2𝜇𝑛𝑥𝑦𝑛𝑦,𝐵𝑢𝑛𝐵𝑥𝜇2𝐵𝑢𝑛𝐵𝑥2.(3.24) By (3.3), we obtain 𝑦𝑛𝑦2𝑢𝑛𝑥2(𝑢𝑛𝑥)(𝑦𝑛𝑦)2𝑢+2𝜇𝑛𝑥𝑦𝑛𝑦,𝐵𝑢𝑛𝐵𝑥𝜇2𝐵𝑢𝑛𝐵𝑥2𝑥𝑛𝑥2(𝑢𝑛𝑥)(𝑦𝑛𝑦)2𝑢+2𝜇𝑛𝑥𝑦𝑛𝑦,𝐵𝑢𝑛𝐵𝑥𝜇2𝐵𝑢𝑛𝐵𝑥2.(3.25) Hence, 𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑦𝑛𝑦2𝑎𝑛𝑣𝑥2+1𝑎𝑛×𝑥𝑛𝑥2(𝑢𝑛𝑥)(𝑦𝑛𝑦)2𝑢+2𝜇𝑛𝑥𝑦𝑛𝑦,𝐵𝑢𝑛𝐵𝑥𝜇2𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥21𝑎𝑛(𝑢𝑛𝑥)(𝑦𝑛𝑦)2+1𝑎𝑛𝑢2𝜇𝑛𝑥𝑦𝑛𝑦𝐵𝑢𝑛𝐵𝑥,(3.26) which implies that 1𝑎𝑛(𝑢𝑛𝑥𝑦)𝑛𝑦2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑢2𝜇𝑛𝑥𝑦𝑛𝑦𝐵𝑢𝑛𝐵𝑥+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛.(3.27) By this together with (C1), (3.15), and lim𝑛𝐵𝑢𝑛𝐵𝑥=0, we obtain 𝑢𝑛𝑥𝑦𝑛𝑦0as𝑛.(3.28) From Lemma 2.6 and (2.12), it follows that (𝑦𝑛𝑡𝑛)+(𝑥𝑦)2=(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦𝑃)𝐶𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐶𝑦𝜆𝐴𝑦+𝜆(𝐴𝑦𝑛𝐴𝑦)2(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦𝑃)𝐶𝑦𝑛𝜆𝐴𝑦𝑛𝑃𝐶𝑦𝜆𝐴𝑦2+2𝜆𝐴𝑦𝑛𝐴𝑦,𝑦𝑛𝑡𝑛+𝑥𝑦(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦)2𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛)𝑃𝐶(𝑦𝜆𝐴𝑦)2+2𝜆𝐴𝑦𝑛𝐴𝑦𝑦𝑛𝑡𝑛+𝑥𝑦(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦)2𝑇𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛)𝑇𝑃𝐶(𝑦𝜆𝐴𝑦)2+2𝜆𝐴𝑦𝑛𝐴𝑦𝑦𝑛𝑡𝑛+𝑥𝑦𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝜆𝐴𝑦𝑇𝑡𝑛𝑥×𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝜆𝐴𝑦+𝑇𝑡𝑛𝑥+2𝜆𝐴𝑦𝑛𝐴𝑦𝑦𝑛𝑡𝑛+𝑥𝑦=𝑢𝑛𝑇𝑡𝑛+𝑥𝑦𝑢𝑛𝑦𝑛𝜆𝐴𝑦𝑛𝐴𝑦×𝑦𝑛𝜆𝐴𝑦𝑛𝑦𝜆𝐴𝑦+𝑇𝑡𝑛𝑥+2𝜆𝐴𝑦𝑛𝐴𝑦𝑦𝑛𝑡𝑛+𝑥𝑦.(3.29) By this together with (3.23), (3.28), and lim𝑛𝐴𝑦𝑛𝐴𝑦=0, we obtain (𝑦𝑛𝑡𝑛)+(𝑥𝑦)0 as 𝑛. This together with (3.17), (3.22), and (3.28), we obtain that 𝑇𝑡𝑛𝑡𝑛𝑇𝑡𝑛𝑥𝑛+𝑥𝑛𝑢𝑛+𝑢𝑛𝑦𝑛𝑥𝑦+𝑦𝑛𝑡𝑛+𝑥𝑦0,as𝑛.(3.30) Next, we show that limsup𝑛𝑣𝑥,𝑥𝑛𝑥0,(3.31) where 𝑥=𝑃Ω𝑣.
Indeed, since {𝑡𝑛} and {𝑇𝑡𝑛} are two bounded sequences in 𝐶, we can choose a subsequence {𝑡𝑛𝑖} of {𝑡𝑛} such that 𝑡𝑛𝑖𝑧𝐶 and limsup𝑛𝑣𝑥,𝑇𝑡𝑛𝑥=lim𝑖𝑣𝑥,𝑇𝑡𝑛𝑖𝑥.(3.32) Since lim𝑛𝑇𝑡𝑛𝑡𝑛=0, we obtain that 𝑇𝑡𝑛𝑖𝑧 as 𝑖.
Next, we show that 𝑧Ω.
(a) We first show 𝑧𝐹(𝑇).
Since 𝑡𝑛𝑖𝑧 and 𝑇𝑡𝑛𝑡𝑛0, we obtain by Lemma 2.5 that 𝑧𝐹(𝑇).
(b) Now, we show that 𝑧GVI(𝐶,𝐴,𝐵).
From (3.30) and (3.17), we have 𝑡𝑛𝑥𝑛𝑇𝑡𝑛𝑡𝑛+𝑇𝑡𝑛𝑥𝑛0as𝑛.(3.33) Furthermore, by Lemma 2.9, we have that 𝐺𝐶𝐶 is nonexpansive. Then, we have 𝑡𝑛𝑡𝐺𝑛=𝑃𝐶𝑦𝑛𝜆𝐴𝑦𝑛𝑡𝐺𝑛=𝑃𝐶𝑃𝑢𝑛𝜇𝐵𝑢𝑛𝑢𝜆𝐴𝑃𝑛𝜇𝐵𝑢𝑛𝑡𝐺𝑛=𝐺𝑢𝑛𝑡𝐺𝑛𝑢n𝑡𝑛𝑢𝑛𝑥𝑛+𝑥𝑛𝑡𝑛0as𝑛.(3.34) Again by Lemma 2.5, we have 𝑧GVI(𝐶,𝐴,𝐵).
(c) We show that 𝑧MEP(𝐹,𝜑). Since 𝑡𝑛𝑖𝑧 and 𝑥𝑛𝑡𝑛0, we obtain that 𝑥𝑛𝑖𝑧. From 𝑢𝑛𝑥𝑛0, we also obtain that 𝑢𝑛𝑖𝑧. By using the same argument as that in the proof of [30, Theorem  3.1, page 1825], we can show that 𝑧MEP(𝐹,𝜑). Therefore, there holds 𝑧Ω.
On the other hand, it follows from (2.13), (3.17), and 𝑇𝑡𝑛𝑖𝑧 as 𝑖 that limsup𝑛𝑣𝑥,x𝑛𝑥=limsup𝑛𝑣𝑥,𝑇𝑡𝑛𝑥=lim𝑖𝑣𝑥,𝑇𝑡𝑛𝑖𝑥=𝑣𝑥,𝑧𝑥0.(3.35) Hence, we have 𝑥𝑛+1𝑥2=𝑎𝑛𝑣+1𝑎𝑛𝑇𝑡𝑛𝑥,𝑥𝑛+1𝑥=𝑎𝑛𝑣𝑥,𝑥𝑛+1𝑥+1𝑎𝑛𝑇𝑡𝑛𝑥,𝑥𝑛+1𝑥𝑎𝑛𝑣𝑥,𝑥𝑛+1𝑥+121𝑎𝑛𝑡𝑛𝑥2+𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥,𝑥𝑛+1𝑥+121𝑎𝑛𝑥𝑛𝑥2+𝑥𝑛+1𝑥2,(3.36) which implies that 𝑥𝑛+1𝑥21𝑎𝑛𝑥𝑛𝑥2+2𝑎𝑛𝑣𝑥,𝑥𝑛+1𝑥.(3.37) By this together with (C1) and (3.35), we have by Lemma 2.3 that {𝑥𝑛} converges strongly to 𝑥. This completes the proof.

The following examples provide mappings 𝐴 and 𝐵 which satisfy those conditions in Theorem 3.1.

Example 3.2. Let 𝐴,𝐵𝐶𝐻 be 𝛼-inverse-strongly monotone and 𝛽-inverse-strongly monotone, respectively. If 𝜆(0,2𝛼) and 𝜇(0,2𝛽), then we have(1)𝐼𝜆𝐴 and 𝐼𝜇𝐵 are nonexpansive,(2)𝐴𝑦𝑛𝐴𝑦0 and 𝐵𝑢𝑛𝐵𝑥0 as 𝑛 for all 𝑥Ω and 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥), where {𝑦𝑛} and {𝑢𝑛} are sequences defined as in Theorem 3.1.

Proof. (1) For any 𝑥,𝑦𝐶, we have (𝐼𝜆𝐴)𝑥(𝐼𝜆𝐴)𝑦2=(𝑥𝑦)𝜆(𝐴𝑥𝐴𝑦)2=𝑥𝑦2+𝜆2𝐴𝑥𝐴𝑦22𝜆𝑥𝑦,𝐴𝑥𝐴𝑦𝑥𝑦2+𝜆2𝐴𝑥𝐴𝑦22𝜆𝛼𝐴𝑥𝐴𝑦2=𝑥𝑦2+𝜆(𝜆2𝛼)𝐴𝑥𝐴𝑦2𝑥𝑦2,(3.38) hence, 𝐼𝜆𝐴 is nonexpansive. Similarly, we can show that 𝐼𝜇𝐵 is nonexpansive.
(2) Let {𝑥𝑛}, {𝑦𝑛}, and {𝑢𝑛} be the sequences defined as in Theorem 3.1. From (3.3), we have 𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑡𝑛𝑥2=𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛)𝑃𝐶(𝑦𝜆𝐴𝑦)2𝑎𝑛𝑣𝑥2+1𝑎𝑛(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦)2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑦𝑛𝑦2+𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥2+1𝑎𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2,𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑡𝑛𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑦𝑛𝑦2𝑎𝑛𝑣𝑥2+1𝑎𝑛(𝑢𝑛𝜇𝐵𝑢𝑛)(𝑥𝜇𝐵𝑥)2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑢𝑛𝑥2+𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥2+1𝑎𝑛𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2.(3.39) Therefore, we have 1𝑎𝑛𝜆(𝜆2𝛼)𝐴𝑦𝑛𝐴𝑦2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛,1𝑎𝑛𝜇(𝜇2𝛽)𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛.(3.40) From (3.15) and (C1), we obtain 𝐴𝑦𝑛𝐴𝑦0,𝐵𝑢𝑛𝐵𝑥0as𝑛.(3.41)

Example 3.3. Let 𝐴𝐶𝐻 be an 𝐿𝐴-Lipschitzian and relaxed (𝑐,𝑑)-cocoercive mapping and 𝐵𝐶𝐻 an 𝐿𝐵-Lipschitzian and relaxed (𝑐,𝑑)-cocoercive mapping. If 0<𝜆<(2(𝑑𝑐𝐿2𝐴))/𝐿2𝐴 and 0<𝜇<(2(𝑑𝑐𝐿2𝐵))/𝐿2𝐵, then we have(1)𝐼𝜆𝐴 and 𝐼𝜇𝐵 are nonexpansive,(2)𝐴𝑦𝑛𝐴𝑦0 and 𝐵𝑢𝑛𝐵𝑥0 as 𝑛 for all 𝑥Ω and 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥), where {𝑦𝑛} and {𝑢𝑛} are sequences defined as in Theorem 3.1.

Proof. (1) For any 𝑥,𝑦𝐶, we have (𝐼𝜇𝐵)𝑥(𝐼𝜇𝐵)𝑦2=(𝑥𝑦)𝜇(𝐵𝑥𝐵𝑦)2=𝑥𝑦2+𝜇2𝐵𝑥𝐵𝑦22𝜇𝑥𝑦,𝐵𝑥𝐵𝑦𝑥𝑦2+𝜇2𝐵𝑥𝐵𝑦22𝜇𝑐𝐵𝑥𝐵𝑦2+𝑑𝑥𝑦2=𝑥𝑦2+𝜇2𝐵𝑥𝐵𝑦2+2𝜇𝑐𝐵𝑥𝐵𝑦22𝜇𝑑𝑥𝑦2𝑥𝑦2+𝜇2𝐿2𝐵𝑥𝑦2+2𝜇𝑐𝐿2𝐵𝑥𝑦22𝜇𝑑𝑥𝑦2=1+2𝜇𝑐𝐿2𝐵2𝜇𝑑+𝜇2𝐿2𝐵𝑥𝑦2𝑥𝑦2,(3.42) hence, 𝐼𝜇𝐵 is nonexpansive. Similarly, we can show that 𝐼𝜆𝐴 is nonexpansive.
(2) Let {𝑥𝑛}, {𝑦𝑛}, and {𝑢𝑛} be the sequences defined as in Theorem 3.1. From (3.3), we have 𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑡𝑛𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑦𝑛𝑦2𝑎𝑛𝑣𝑥2+1𝑎𝑛(𝑢𝑛𝜇𝐵𝑢𝑛)(𝑥𝜇𝐵𝑥)2𝑎𝑛𝑣𝑥2+1𝑎𝑛×𝑢𝑛𝑥22𝜇𝑢𝑛𝑥,𝐵𝑢𝑛𝐵𝑥+𝜇2𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛×𝑢𝑛𝑥2+2𝜇𝑐𝐵𝑢𝑛𝐵𝑥22𝜇𝑑𝑢𝑛𝑥2+𝜇2𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑥𝑛𝑥2+2𝜇𝑐+𝜇22𝜇𝑑𝐿2𝐵𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥2+1𝑎𝑛2𝜇𝑐+𝜇22𝜇𝑑𝐿2𝐵𝐵𝑢𝑛𝐵𝑥2,𝑥𝑛+1𝑥2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑡𝑛𝑥2=𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑃𝐶(𝑦𝑛𝜆𝐴𝑦𝑛)𝑃𝐶(𝑦𝜆𝐴𝑦)2𝑎𝑛𝑣𝑥2+1𝑎𝑛(𝑦𝑛𝜆𝐴𝑦𝑛)(𝑦𝜆𝐴𝑦)2𝑎𝑛𝑣𝑥2+1𝑎𝑛𝑥𝑛𝑥2+2𝜆𝑐+𝜆22𝜆𝑑𝐿2𝐴𝐴𝑦𝑛𝐴𝑦2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥2+1𝑎𝑛2𝜆𝑐+𝜆22𝜆𝑑𝐿2𝐴𝐴𝑦𝑛𝐴𝑦2.(3.43) Therefore, we have 1𝑎𝑛2𝜇𝑐+𝜇22𝜇𝑑𝐿2𝐵𝐵𝑢𝑛𝐵𝑥2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛,1𝑎𝑛2𝜆𝑐+𝜆22𝜆𝑑𝐿2𝐴𝐴𝑦𝑛𝐴𝑦2𝑎𝑛𝑣𝑥2+𝑥𝑛𝑥+𝑥𝑛+1𝑥𝑥𝑛+1𝑥𝑛.(3.44) From (3.15) and (C1), we obtain 𝐴𝑦𝑛𝐴𝑦0,𝐵𝑢𝑛𝐵𝑥0as𝑛.(3.45)

By using the same proof as in Examples 3.2 and 3.3, we can obtain the following example.

Example 3.4. Let 𝐴 be an 𝛼-inverse-strongly monotone mapping of 𝐶 into 𝐻 and 𝐵 be a L-Lipschitzian and relaxed (𝑐,𝑑)-cocoercive mapping of 𝐶 into 𝐻. If 𝜆(0,2𝛼) and 0<𝜇<(2(𝑑𝑐𝐿2))/𝐿2, then we have(1)𝐼𝜆𝐴 and 𝐼𝜇𝐵 are nonexpansive,(2)𝐴𝑦𝑛𝐴𝑦0 and 𝐵𝑢𝑛𝐵𝑥0 as 𝑛 for all 𝑥Ω and 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥), where {𝑦𝑛} and {𝑢𝑛} are sequences defined as in Theorem 3.1.

Let 𝒜 be the class of all 𝛼-inverse-strongly monotone mappings from 𝐶 into 𝐻, the class of all 𝛽-inverse-strongly monotone mappings from 𝐶 into 𝐻, 𝒞 the class of all 𝐿-Lipschitzian and relaxed (𝑐,𝑑)-cocoercive mappings from 𝐶 into 𝐻, and 𝒟 the class of all 𝐿1-Lipschitzian and relaxed (𝑐,𝑑)-cocoercive mappings from 𝐶 into 𝐻.

Theorem 3.5. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐹 be a function from 𝐶×𝐶 to satisfying (A1)–(A5) and 𝜑𝐶{+} a proper lower semicontinuous and convex function. Let {𝑇𝑖}𝑁𝑖=1 be a finite family of nonexpansive self-mappings of 𝐶 such that Ω=𝑁𝑖=1𝐹(𝑇𝑖)GVI(𝐶,𝐴,𝐵)MEP(𝐹,𝜑). Let 𝛼𝑗=(𝛼𝑗1,𝛼j2,𝛼𝑗3), 𝑗=1,2,,𝑁, where 𝛼𝑗1,𝛼𝑗2,𝛼𝑗3[0,1], 𝛼𝑗1+𝛼𝑗2+𝛼𝑗3=1, 𝛼𝑗1(0,1) for all 𝑗=1,2,,𝑁1, 𝛼𝑁1(0,1] and 𝛼𝑗2,𝛼𝑗3[0,1) for all 𝑗=1,2,,𝑁. Let 𝑆 be the 𝑆-mapping generated by 𝑇1,𝑇2,,𝑇𝑁 and 𝛼1,𝛼2,,𝛼𝑁. Assume that either (B1) or (B2) holds and that 𝑣 is an arbitrary point in 𝐶. Let 𝑥1𝐶 and {𝑥𝑛},{𝑢𝑛},{𝑦𝑛} the sequences generated by 𝐹𝑢𝑛𝑢,𝑦+𝜑(𝑦)𝜑𝑛+1𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑦0,𝑦𝐶,𝑛=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛,𝑥𝑛+1=𝑎𝑛𝑣+1𝑎𝑛𝑆𝑃C𝑦𝑛𝜆𝐴𝑦𝑛,𝑛1.(3.46) If one of the following conditions is satisfied: (1)𝐴𝒜, 𝐵, 𝜆(0,2𝛼), and 𝜇(0,2𝛽),(2)𝐴𝒞, 𝐵𝒟, 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2, and 0<𝜇<(2(𝑑𝑐𝐿21))/𝐿21,(3)𝐴𝒜, 𝐵𝒞, 𝜆(0,2𝛼), and 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2,and the sequences {𝑟𝑛} and {𝑎𝑛} are as in Theorem 3.1, then {𝑥𝑛} converges strongly to 𝑥=𝑃Ω𝑣 and (𝑥,𝑦) is a solution of the problem (1.6), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Proof. By Lemma 2.7, we obtain that 𝑆 is nonexpansive and 𝐹(𝑆)=𝑁𝑖=1𝐹(𝑇𝑖). Hence, the result is obtained directly from Theorem 3.1 and Examples 3.23.4.

From Theorem 3.1 and Examples 3.23.4, we obtain the following result.

Corollary 3.6. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐹 be a function from 𝐶×𝐶 to satisfying (A1)–(A5) and 𝜑𝐶{+} a proper lower semicontinuous and convex function. Let 𝑇 be a nonexpansive self-mapping of 𝐶 such that Ω=𝐹(𝑇)GVI(𝐶,𝐴,𝐵)MEP(𝐹,𝜑). Assume that either (B1) or (B2) holds and that 𝑣 is an arbitrary point in 𝐶. If one of the following conditions is satisfied: (1)𝐴𝒜, 𝐵, 𝜆(0,2𝛼), and 𝜇(0,2𝛽),(2)𝐴𝒞, 𝐵𝒟, 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2, and 0<𝜇<(2(𝑑𝑐𝐿21))/𝐿21,(3)𝐴𝒜, 𝐵𝒞, 𝜆(0,2𝛼), and 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2,and the sequences {𝑟𝑛}, {𝑎𝑛} are as in Theorem 3.1, then the sequence {𝑥𝑛} generated by (1.12) converges strongly to 𝑥=𝑃Ω𝑣 and (x,𝑦) is a solution of the problem (1.6), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Let 𝜑=0 in Theorem 3.1. From Theorem 3.1 and Examples 3.23.4, we obtain the following result.

Corollary 3.7. Let 𝐶 be a nonempty closed and convex subset of a real Hilbert space 𝐻. Let 𝐹 be a function from 𝐶×𝐶 to satisfying (A1)–(A5). Let 𝑇 be a nonexpansive self-mapping of 𝐶 such that Ω=𝐹(𝑇)GVI(𝐶,𝐴,𝐵)EP(𝐹). Let 𝑥1,𝑣𝐶 and {𝑥𝑛},{𝑢𝑛},{𝑦𝑛} be the sequences generated by 𝐹𝑢𝑛+1,𝑦𝑟𝑛𝑦𝑢𝑛,𝑢𝑛𝑥𝑛𝑦0,𝑦𝐶,𝑛=𝑃𝐶𝑢𝑛𝜇𝐵𝑢𝑛,𝑥𝑛+1=𝑎𝑛𝑣+1𝑎𝑛𝑇𝑃𝐶𝑦𝑛𝜆𝐴𝑦𝑛,𝑛1.(3.47) If one of the following conditions is satisfied: (1)𝐴𝒜, 𝐵, 𝜆(0,2𝛼), and 𝜇(0,2𝛽),(2)𝐴𝒞, 𝐵𝒟, 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2, and 0<𝜇<(2(𝑑𝑐𝐿21))/𝐿21,(3)𝐴𝒜, 𝐵𝒞, 𝜆(0,2𝛼), and 0<𝜆<(2(𝑑𝑐𝐿2))/𝐿2,and the sequences {𝑟𝑛}, {𝑎𝑛} are as in Theorem 3.1, then {𝑥𝑛} converges strongly to 𝑥=𝑃Ω𝑣 and (𝑥,𝑦) is a solution of the problem (1.6), where 𝑦=𝑃𝐶(𝑥𝜇𝐵𝑥).

Remark 3.8. In Theorem 3.5, if 𝐹0, then the sequence {𝑥𝑛} generated by (3.46) converges strongly to a solution of the minimization problem which is also a solution of a system of variational inequalities.

Acknowledgments

The authors would like to thank the referees for valuable suggestions on the paper and thank the National Research University Project under Thailand’s Office of the Higher Education Commission, the Centre of Excellence in Mathematics, the Thailand Research Fund and the Thaksin university for financial support.