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Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 Research Article Strong convergence theorems for maximal monotone operators and continuous pseudocontractive mappings Jong Soo Jung Department of Mathematics, Dong-A University, Busan 49315, Korea. Communicated by Y. J. Cho Abstract We introduce a new iterative algorithm for finding a common element of the solution set of the variational inequality problem for a continuous monotone mapping, the zero point set of a maximal monotone operator, and the fixed point set of a continuous pseudocontractive mapping in a Hilbert space. Then we establish strong convergence of the sequence generated by the proposed algorithm to a common point of three sets, which is a solution of a certain variational inequality. Further, we find the minimum-norm element in common set of three sets. As applications, we consider iterative algorithms for the equilibrium problem coupled with fixed point problem. ©2016 All rights reserved. Keywords: Maximal monotone operator, continuous monotone mapping, continuous pseudocontractive mapping, fixed points, variational inequality, fixed points, zeros, minimum-norm point. 2010 MSC: 47H05, 47H09, 47H10, 47J05, 47J20, 47J25, 49M05. 1. Introduction In the real world, many nonlinear problems arising in applied areas are mathematically modeled as nonlinear operator equations and the operator is decomposed as the sum of two nonlinear operators. The nonlinear operator equations can be reduced to the monotone inclusion problems or fixed point problems for nonlinear operators. As the most popular techniques for solving the nonlinear operator equations, many authors formulated the nonlinear operator equations as finding a zero of the sum of two nonlinear operators or as finding a fixed point of a nonlinear mapping. Let H be a real Hilbert space with the inner product , ·i, and let C be a nonempty closed convex subset of H . For the mapping T : C C , we denote the fixed point set of T by F ix(T ), that is, Email address: [email protected] (Jong Soo Jung) Received 2016-03-03

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Page 1: Strong convergence theorems for maximal monotone operators ... · Strong convergence theorems for maximal monotone operators and continuous pseudocontractive mappings Jong Soo Jung

Available online at www.tjnsa.comJ. Nonlinear Sci. Appl. 9 (2016), 4409–4426

Research Article

Strong convergence theorems for maximal monotoneoperators and continuous pseudocontractivemappings

Jong Soo Jung

Department of Mathematics, Dong-A University, Busan 49315, Korea.

Communicated by Y. J. Cho

Abstract

We introduce a new iterative algorithm for finding a common element of the solution set of the variationalinequality problem for a continuous monotone mapping, the zero point set of a maximal monotone operator,and the fixed point set of a continuous pseudocontractive mapping in a Hilbert space. Then we establishstrong convergence of the sequence generated by the proposed algorithm to a common point of three sets,which is a solution of a certain variational inequality. Further, we find the minimum-norm element incommon set of three sets. As applications, we consider iterative algorithms for the equilibrium problemcoupled with fixed point problem. ©2016 All rights reserved.

Keywords: Maximal monotone operator, continuous monotone mapping, continuous pseudocontractivemapping, fixed points, variational inequality, fixed points, zeros, minimum-norm point.2010 MSC: 47H05, 47H09, 47H10, 47J05, 47J20, 47J25, 49M05.

1. Introduction

In the real world, many nonlinear problems arising in applied areas are mathematically modeled asnonlinear operator equations and the operator is decomposed as the sum of two nonlinear operators. Thenonlinear operator equations can be reduced to the monotone inclusion problems or fixed point problemsfor nonlinear operators. As the most popular techniques for solving the nonlinear operator equations, manyauthors formulated the nonlinear operator equations as finding a zero of the sum of two nonlinear operatorsor as finding a fixed point of a nonlinear mapping.

Let H be a real Hilbert space with the inner product 〈·, ·〉, and let C be a nonempty closed convexsubset of H. For the mapping T : C → C, we denote the fixed point set of T by Fix(T ), that is,

Email address: [email protected] (Jong Soo Jung)

Received 2016-03-03

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4410

Fix(T ) = x ∈ C : Tx = x.Let F : C → 2H be a maximal monotone operator. Many problems can be formulated as finding a zero

of a maximal monotone operator F in a Hilbert space H, that is, a solution of the inclusion problem 0 ∈ Fx.(A typical example is to find a minimizer of a convex functional.) A classical method for solving the problemis proximal point algorithm, proposed by Martinet [20, 21] and generalized by Rockafellar [27, 28]. In thecase of F = A+B, where A and B are monotone operators, the problem is reduced to as follows:

find z ∈ C such that 0 ∈ (A+B)z. (1.1)

The solution set of the problem (1.1) is denoted by (A+B)−10. As we know, the problem (1.1) is very generalin the sense that it includes, as special cases, convexly constrained linear inverse problem, split feasibilityproblem, convexly constrained minimization problem, fixed point problems, variational inequalities, Nashequilibrium problem in noncooperative games, and others; see, for instance, [2, 8, 12, 17, 22, 24, 25] and thereferences therein.

Let A be a nonlinear mapping of C into H. The variational inequality problem is to find a u ∈ C suchthat

〈v − u,Au〉 ≥ 0, ∀v ∈ C. (1.2)

This problem is called Hartmann-Stampacchia variational inequality (see [13, 31]). We denote the setof solutions of the variational inequality problem (1.2) by V I(C,A). Also variational inequality theoryhas emerged as an important tool in studying a wide class of numerous problem in physics, optimization,variational inequalities, minimax problem, Nash equilibrium problem in noncooperative games and others;see, for instance, [4, 6, 18, 19, 40] and the references therein.

Recently, in order to study the monotone inclusion problem (1.1) coupled with fixed point problem forthe nonlinear mapping T , many authors have introduced some iterative methods for finding an element ofFix(T )∩(A+B)−10, where A is an α-inverse-strongly monotone mapping of C into H, and B is a set-valuedmaximal monotone operator on H. For instance, in case that T is a nonexpansive mapping of C into itself,see [35, 37, 42, 45] and the references therein, and in case that T is a k-strictly pseudocontractive mappingof C into itself, see [16]. For a Lipschitzian pseudocontractive mapping T of C into itself, refer to [30].

Many researchers have also invented some iterative methods for finding an element of V I(C,A)∩Fix(T ),where A and T are nonlinear mappings. For instance, in case that A is an α-inverse-strongly monotonemapping of C into H and T is a nonexpansive mapping of C into itself, see [9, 14, 15, 23, 32, 36] and thereferences therein, and in case that A is a continuous monotone mapping of C into H and T is a continuouspseudocontractive mapping of C into itself, see [7, 38, 44].

In this paper, as a continuation of study in this direction, we introduce a new iterative algorithm forfinding a common element of the set Fix(T ) of fixed points of a continuous pseudocontractive mapping T ,the solution set V I(C,A) of the variational inequality problem (1.2), where A is a continuous monotonemapping, and the set B−10 of zero points of B, where B is a multi-valued maximal monotone operatoron H. Then we establish strong convergence of the sequence generated by the proposed algorithm to acommon point of three sets, which is a solution of a certain variational inequality, where the constrained setis Fix(T )∩V I(C,A)∩B−10. As a direct consequence, we find the unique minimum-norm element of Fix(T )∩V I(C,A) ∩ B−10. Moreover, as applications, we consider iterative algorithms for the equilibrium problemcoupled with fixed point problem of continuous pseudocontractive mappings. Our results extend, improveand unify most of the results that have been proven for these important classes of nonlinear mappings.

2. Preliminaries and lemmas

In the following, we write xn x to indicate that the sequence xn converges weakly to x. xn → ximplies that xn converges strongly to x.

Let H be a real Hilbert space with the inner product 〈·, ·〉 and the induced norm ‖ · ‖, and let C be anonempty closed convex subset of H. A mapping A of C into H is called monotone if

〈x− y,Ax−Ay〉 ≥ 0, ∀x, y ∈ C.

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4411

A mapping A of C into H is called α-inverse-strongly monotone (see [14, 19]) if there exists a positive realnumber α such that

〈x− y,Ax−Ay〉 ≥ α‖Ax−Ay‖2, ∀x, y ∈ C.Clearly, the class of monotone mappings includes the class of α-inverse-strongly monotone mappings.

A mapping T of C into H is said to be pseudocontractive if

‖Tx− Ty‖2 ≤ ‖x− y‖2 + ‖(I − T )x− (I − T )y‖2, ∀x, y ∈ C,

and T is said to be k-strictly pseudocontractive (see [5]) if there exists a constant k ∈ [0, 1) such that

‖Tx− Ty‖2 ≤ ‖x− y‖2 + k‖(I − T )x− (I − T )y‖2, ∀x, y ∈ C,

where I is the identity mapping. Note that the class of k-strictly pseudocontractive mappings includes theclass of nonexpansive mappings as a subclass. That is, T is nonexpansive (i.e., ‖Tx − Ty‖ ≤ ‖x − y‖,∀x, y ∈ C) if and only if T is 0-strictly pseudocontractive. Clearly, the class of pseudocontractive mappingsincludes the class of strictly pseudocontractive mappings and the class of nonexpansive mappings as asubclass. Moreover, this inclusion is strict due to an example in [10] (see, also Example 5.7.1 and Example5.7.2 in [1]).

A mapping G : C → C is said to be κ-Lipschitzian and η-strongly monotone with constants κ > 0 andη > 0 if

‖Gx−Gy‖ ≤ κ‖x− y‖ and 〈Gx−Gy, x− y〉 ≥ η‖x− y‖2, ∀x, y ∈ C,respectively. A mapping V : C → C is said to be l-Lipschitzian with a constant l ≥ 0 if

‖V x− V y‖ ≤ l‖x− y‖, ∀x, y ∈ C.

Let B be a mapping of H into 2H . The effective domain of B is denoted by dom(B), that is, dom(B) =x ∈ H : Bx 6= ∅. A multi-valued mapping B is said to be a monotone operator on H if 〈x− y, u− v〉 ≥ 0for all x, y ∈ dom(B), u ∈ Bx, and v ∈ By. A monotone operator B on H is said to be maximal if its graphis not properly contained in the graph of any other monotone operator on H. For a maximal monotoneoperator B on H and r > 0, we may define a single-valued operator JBr = (I+ rB)−1 : H → dom(B), whichis called the resolvent of B. Let B be a maximal monotone operator on H and let B−10 = x ∈ H : 0 ∈ Bx.It is well-known that B−10 = Fix(JBr ) for all r > 0 is closed and convex ([3]), and the resolvent JBr is firmlynonexpansive, that is,

‖JBr x− JBr y‖2 ≤ 〈x− y, JBr x− JBr y〉, ∀x, y ∈ H, (2.1)

and that the resolvent identity

JBλ x = JBµ

λx+

(1− µ

λ

)JBλ x

)(2.2)

holds for all λ, µ > 0 and x ∈ H.In a real Hilbert space H, the following hold:

‖x− y‖2 = ‖x‖2 + ‖y‖2 − 2〈x, y〉, (2.3)

and‖αx+ βy‖2 = α‖x‖2 + β‖y‖2 − αβ‖x− y‖2 ≤ α‖x‖2 + β‖y‖2, (2.4)

for all x, y ∈ H and α, β ∈ (0, 1) with α + β = 1. For every point x ∈ H, there exists a unique nearestpoint in C, denoted by PCx, such that

‖x− PCx‖ = inf‖x− y‖ : y ∈ C.

PC is called the metric projection of H onto C. It is well known that PC is nonexpansive and PC ischaracterized by the property

u = PCx⇐⇒ 〈x− u, u− y〉 ≥ 0, ∀x ∈ H, y ∈ C. (2.5)

We need the following lemmas for the proof of our main results.

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4412

Lemma 2.1 ([1]). In a real Hilbert space H, the following inequality holds:

‖x+ y‖2 ≤ ‖x‖2 + 2〈y, x+ y〉, ∀x, y ∈ H.

Lemma 2.2 ([33]). Let xn and zn be bounded sequences in a real Banach space E, and let γn be asequence in [0, 1] which satisfies the following condition:

0 < lim infn→∞

γn ≤ lim supn→∞

γn < 1.

Suppose that xn+1 = γnxn + (1− γn)zn for all n ≥ 1 and

lim supn→∞

(‖zn+1 − zn‖ − ‖xn+1 − xn‖) ≤ 0.

Then limn→∞ ‖zn − xn‖ = 0.

Lemma 2.3 ([39]). Let sn be a sequence of nonnegative real numbers satisfying

sn+1 ≤ (1− ξn)sn + ξnδn, ∀n ≥ 1,

where ξ and δn satisfy the following conditions:

(i) ξn ⊂ [0, 1] and∑∞

n=1 ξn =∞;

(ii) lim supn→∞ δn ≤ 0 or∑∞

n=1 ξn|δn| <∞.

Then limn→∞ sn = 0.

The following lemmas are Lemma 2.3 and Lemma 2.4 of Zegeye [43], respectively.

Lemma 2.4 ([43]). Let C be a closed convex subset of a real Hilbert space H. Let A : C → H be a continuousmonotone mapping. Then, for r > 0 and x ∈ H, there exists z ∈ C such that

〈y − z,Az〉+1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C.

For r > 0 and x ∈ H, define Ar : H → C by

Arx =

z ∈ C : 〈y − z,Az〉+

1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C

.

Then the following hold:

(i) Ar is single-valued;

(ii) Ar is firmly nonexpansive, that is,

‖Arx−Ary‖2 ≤ 〈x− y,Arx−Ary〉, ∀x, y ∈ H;

(iii) Fix(Ar) = V I(C,A);

(iv) V I(C,A) is a closed convex subset of C.

Lemma 2.5 ([43]). Let C be a closed convex subset of a real Hilbert space H. Let T : C → H be a continuouspseudocontractive mapping. Then, for r > 0 and x ∈ H, there exists z ∈ C such that

〈y − z, Tz〉 − 1

r〈y − z, (1 + r)z − x〉 ≤ 0, ∀y ∈ C.

For r > 0 and x ∈ H, define Tr : H → C by

Trx =

z ∈ C : 〈y − z, Tz〉 − 1

r〈y − z, (1 + r)z − x〉 ≤ 0, ∀y ∈ C

.

Then the following hold:

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4413

(i) Tr is single-valued;

(ii) Tr is firmly nonexpansive, that is,

‖Trx− Try‖2 ≤ 〈x− y, Trx− Try〉, ∀x, y ∈ H;

(iii) Fix(Tr) = Fix(T );

(iv) Fix(T ) is a closed convex subset of C.

The following lemmas can be easily proven (see [40]), and therefore, we omit their proof.

Lemma 2.6. Let H be a real Hilbert space. Let V : H → H be an l-Lipschitzian mapping with a constantl ≥ 0, and let G : H → H be a κ-Lipschitzian and η-strongly monotone mapping with constants κ, η > 0.Then for 0 ≤ γl < µη,

〈(µG− γV )x− (µG− γV )y, x− y〉 ≥ (µη − γl)‖x− y‖2, ∀x, y ∈ C.

That is, µG− γV is strongly monotone with constant µη − γl.

Lemma 2.7. Let H be a real Hilbert space H. Let G : H → H be a κ-Lipschitzian and η-strongly monotoneoperator with constants κ > 0 and η > 0. Let 0 < µ < 2η

κ2 and 0 < t < ξ ≤ 1. Then ξI − tµG : H → H is a

contractive mapping with a constant ξ − tτ , where τ = 1−√

1− µ(2η − µκ2).

Lemma 2.8. Let C be a closed convex subset of a real Hilbert space H. Let A : C → H be a nonlinearmapping, and let B : dom(B) ⊂ C → 2H be a maximal monotone operator. Then V I(C,A) ∩ B−10 is asubset of (A+B)−10

Proof. Let z ∈ V I(C,A) ∩B−10. Then we have, for v ∈ Bu,

〈u− z,Az〉 ≥ 0 and 〈z − u,−v〉 ≥ 0.

Thus, we derive〈z − u,−Az − v〉 = 〈u− z,Az〉+ 〈z − u,−v〉 ≥ 0.

Since B is maximal monotone, −Az ∈ Bz, that is, z ∈ (A+B)−10.

3. Iterative algorithms

Throughout the rest of this paper, we always assume the following:

• H is a real Hilbert space with the inner product 〈·, ·〉 and the induced norm ‖ · ‖;• C is a nonempty closed subspace of H;

• B : H → 2H is a maximal monotone operator with dom(B) ⊂ C;

• B−10 is the set of zero points of B, that is, B−10 = z ∈ H : 0 ∈ Bz;• JBrn : H → dom(B) is the resolvent of B for rn ∈ (0,∞);

• G : C → C is a κ-Lipschitzian and η-strongly monotone mapping with constants κ, η > 0;

• V : C → C is a l-Lipschitzian mapping with constant l > 0;

• Constants µ > 0 and γ ≥ 0 satisfy 0 < µ < 2ηκ2 and 0 ≤ γl < τ , where τ = 1−

√1− µ(2η − µκ2);

• A : C → H is a continuous monotone mapping;

• V I(C,A) is the solution set of the variational inequality problem (1.2) for A;

• T : C → C is a continuous pseudocontractive mapping with Fix(T ) 6= ∅;• Arn : H → C is a mapping defined by

Arnx =

z ∈ C : 〈y − z,Az〉+

1

rn〈y − z, z − x〉 ≥ 0, ∀y ∈ C

for x ∈ H and rn ∈ (0,∞);

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4414

• Trn : H → C is a mapping defined by

Trnx =

z ∈ C : 〈Tz, y − z〉 − 1

rn〈y − z, (1 + rn)z − x〉 ≤ 0, ∀y ∈ C

for x ∈ H and rn ∈ (0,∞);

• Fix(T ) ∩ V I(C,A) ∩B−10 6= ∅.

By Lemma 2.4 and Lemma 2.5, we note that Arn and Trn are nonexpansive, V I(C,A) = Fix(Arn) andFix(Trn) = Fix(T ).

Now, we propose a new iterative algorithm for finding a common element of Fix(T )∩V I(C,A)∩B−10,where T is a continuous pseudocontractive mapping, A is a continuous monotone mapping, and B is amulti-valued maximal monotone operator on H.

Algorithm 3.1. For an arbitrarily chosen x1 ∈ C, let the iterative sequence xn be generated byyn = αnγV xn + (1− αnµG)xn,

xn+1 = βnxn + (1− βn)TrnJBrnArnyn, ∀n ≥ 1,

(3.1)

where αn and βn are two sequences in (0, 1), and rn ⊂ (0,∞).

Theorem 3.2. Suppose that Fix(T )∩V I(C,A)∩B−10 6= ∅. Let the sequence xn be generated iterativelyby algorithm (3.1). Let αn, βn ⊂ (0, 1) and rn ⊂ (0,∞) satisfy the following conditions:

(C1) limn→∞ αn = 0;

(C2)∑∞

n=1 αn =∞;

(C3) 0 < lim infn→∞ βn ≤ lim supn→∞ βn < 1;

(C4) 0 < a ≤ rn <∞ and limn→∞ |rn+1 − rn| = 0.

Then xn converge strongly to a point q ∈ Fix(T ) ∩ V I(C,A) ∩B−10, which is the unique solution of thefollowing variational inequality:

〈(γV − µG)q, q − p〉 ≥ 0, ∀p ∈ Fix(T ) ∩ V I(C,A) ∩B−10. (3.2)

Proof. First, let Q = PΩ, where Ω := Fix(T ) ∩ V I(C,A) ∩ B−10. Then, by Lemma 2.4 (iv), Lemma 2.5(iv), PΩ is well-defined. Also, it is easy to show that Q(I − µG + γV ) : C → C is a contractive mappingwith a constant 1− (τ − γl). In fact, from Lemma 2.7 we have

‖Q(I − µG+ γV )x−Q(I − µG+ γV )y‖ ≤ ‖(I − µG+ γV )x− (I − µG+ γV )y‖≤ ‖(I − µG)x− (I − µG)y‖+ γ‖V x− V y‖≤ (1− τ)‖x− y‖+ γl‖x− y‖= (1− (τ − γl))‖x− y‖

for any x, y ∈ C. So, Q(I − µG + γV ) is a contractive mapping with a constant 1 − (τ − γl) < 1. Thus,by Banach contraction principle, there exists a unique element q ∈ C such that q = PΩ(I − µG + γV )q.Equivalently, q is a solution of the variational inequality (3.2) (see (2.5)). We can show easily the uniquenessof a solution of the variational inequality (3.2). Indeed, noting that 0 ≤ γl < τ and µη ≥ τ ⇐⇒ κ ≥ η, itfollows from Lemma 2.6 that

〈(µG− γV )x− (µG− γV )y, x− y〉 ≥ (µη − γl)‖x− y‖2.

That is, µG − γV is strongly monotone for 0 ≤ γl < τ < µη. Hence the variational inequality (3.2) hasonly one solution. Below we will use q ∈ Fix(T ) ∩ V I(C,A) ∩ B−10 to denote the unique solution of thevariational inequality (3.2).

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From now on, by conditions (C1) and (C3), without loss of generality, we assume that αn(1−βn)(τ−γl) <1 for n ≥ 1. And we put wn := Arnyn, un := JBrnwn (= JBrnArnyn), and zn := Trnun (= TrnJ

Brnwn).

We divide the proof into several steps.Step 1. We show that xn is bounded. To this end, let p ∈ Fix(T ) ∩ V I(C,A) ∩ B−10. It is obvious

that p = JBrnArnp, p = TrnJBrnArnp, and Trnp = p. From Lemma 2.7 we obtain

‖yn − p‖ = ‖αn(γV xn − µG)p+ (I − αnµG)xn − (I − αnµG)p‖≤ (1− αnτ)‖xn − p‖+ αnγ‖V xn − V p‖+ αn‖γV p− µGp‖≤ (1− αnτ)‖xn − p‖+ αnγl‖xn − p‖+ αn‖γV p− µGp‖

= (1− (τ − γl)αn)‖xn − p‖+ (τ − γl)‖γV p− µGp‖τ − γl

.

(3.3)

Thus, since TrnJBrnArn is nonexpansive (by Lemma 2.4 and Lemma 2.5), from (3.3) we deduce

‖xn+1 − p‖ ≤ βn‖xn − p‖+ (1− βn)‖TrnJBrnArnyn − p‖≤ βn‖xn − p‖+ (1− βn)‖yn − p‖

≤ βn‖xn − p‖+ (1− βn)

[(1− (τ − γl)αn)‖xn − p‖+ (τ − γl)‖γV p− µGp‖

τ − γl

]= (1− (1− βn)αn(τ − γl))‖xn − p‖+ (1− βn)αn(τ − γl)‖γV p− µGp‖

τ − γl

≤ max

‖xn − p‖,

‖γV p− µGp‖τ − γl

.

Using an induction, we have

‖xn − p‖ ≤ max

‖x1 − p‖,

‖γV p− µGp‖τ − γl

.

Hence, xn is bounded. Also, yn, V xn, Gxn, wn = Arnyn, un = JBrnwn and zn = Trnunare bounded. And, from (3.1) and condition (C1) it follows that

‖yn − xn‖ = αn‖γV xn − µGxn‖ → 0 as n→∞. (3.4)

Step 2. We show that limn→∞ ‖xn+1 − xn‖ = 0. For this purpose, first, we notice

‖yn − yn−1‖ = ‖αnγV xn − (I − αnµG)xn − αn−1γV xn−1 − (I − αn−1µG)xn−1‖≤ ‖(αn − αn−1)(γV xn−1 − µGxn−1)‖+ αnγ‖V xn − V xn−1‖

+ ‖(I − αnµG)xn − (I − αnµG)xn−1‖≤ |αn − αn−1|(γ‖V xn−1‖+ µ‖Gxn−1‖) + αnγl‖xn − xn−1‖

+ (1− ταn)‖xn − xn−1‖= (1− (τ − γl)αn)‖xn − xn−1‖+ |αn − αn−1|M1,

(3.5)

where M1 > 0 is an appropriate constant. Let wn = Arnyn and wn−1 = Arn−1yn−1 again. Then we get

〈y − wn, Awn〉+1

rn〈y − wn, wn − yn〉 ≥ 0, ∀y ∈ C (3.6)

and

〈y − wn−1, Awn−1〉+1

rn−1〈y − wn−1, wn−1 − yn−1〉 ≥ 0, ∀y ∈ C. (3.7)

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Putting y := wn−1 in (3.6) and y := wn in (3.7), we obtain

〈wn−1 − wn, Awn〉+1

rn〈wn−1 − wn, wn − yn〉 ≥ 0, (3.8)

and

〈wn − wn−1, Awn−1〉+1

rn−1〈wn − wn−1, wn−1 − yn−1〉 ≥ 0. (3.9)

Adding up (3.8) and (3.9), we deduce

−〈wn − wn−1, Awn −Awn−1〉+ 〈wn−1 − wn,wn − ynrn

− wn−1 − yn−1

rn−1〉 ≥ 0.

Since F is monotone, we get

〈wn−1 − wn,wn − ynrn

− wn−1 − yn−1

rn−1〉 ≥ 0,

and hence〈wn − wn−1, wn−1 − wn + wn − yn−1 −

rn−1

rn(wn − yn)〉 ≥ 0. (3.10)

From (3.10) we derive

‖wn − wn−1‖2 ≤ 〈wn − wn−1, wn − yn + yn − yn−1 −rn−1

rn(wn − yn)〉

= 〈wn − wn−1, yn − yn−1 +

(1− rn−1

rn

)(wn − yn)〉

≤ ‖wn − wn−1‖[‖yn − yn−1‖+

1

a|rn − rn−1|‖wn − yn‖

].

This implies that

‖wn − wn−1‖ ≤ ‖yn − yn−1‖+1

a|rn − rn−1|‖wn − yn‖. (3.11)

Moreover, from the resolvent identity (2.2) and (3.11) we induce

‖JBrnwn − JBrn−1

wn−1‖ = ‖JBrn−1

(rn−1

rnwn +

(1− rn−1

rn

)JBrnwn

)− JBrn−1

wn−1‖

≤ ‖rn−1

rn(wn − wn−1) +

(1− rn−1

rn

)(JBrnwn − wn−1)‖

≤ ‖wn − wn−1‖+|rn − rn−1|

a‖JBrnwn − wn‖

≤ ‖yn − yn−1‖+ |rn − rn−1|(‖wn − yn‖

a+‖JBrnwn − wn‖

a

).

(3.12)

Substituting (3.5) into (3.12), we derive

‖JBrnwn − JBrn−1

wn−1‖ ≤ (1− (τ − γl)αn)‖xn − xn−1‖+ |αn − αn−1|M1 + |rn − rn1 |M2, (3.13)

where M2 > 0 is an appropriate constant.On the other hand, since zn = TrnJ

Brnwn and zrn−1 = Trn−1J

Brnwn−1, we have

〈y − zn, T zn〉 −1

rn〈y − zn, (1 + rn)zn − JBrnwn〉 ≤ 0, ∀y ∈ C, (3.14)

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and

〈y − zn−1, T zn−1〉 −1

rn−1〈y − zn−1, (1 + rn−1)zn−1 − JBrn−1

wn−1〉 ≤ 0, ∀y ∈ C. (3.15)

Putting y := zn−1 in (3.14) and y := zn in (3.15), we get

〈zn−1 − zn, T zn〉 −1

rn〈zn−1 − zn, (1 + rn)zn − JBrnwn〉 ≤ 0, (3.16)

and

〈zn − zn−1, T zn−1〉 −1

rn−1〈zn − zn−1, (1 + rn−1)zn−1 − JBrn−1

wn−1〉 ≤ 0. (3.17)

Adding up (3.16) and (3.17), we obtain

〈zn−1 − zn, T zn − Tzn−1〉 − 〈zn−1 − zn,(1 + rn)zn − JBrnwn

rn−

(1 + rn−1)zn−1 − JBrn−1wn−1

rn−1〉 ≤ 0. (3.18)

Using the fact that T is pseudocontractive, we have by (3.18)

〈zn−1 − zn,zn − JBrnwn

rn−zn−1 − JBrn−1

wn−1

rn−1〉 ≥ 0,

and hence〈zn−1 − zn, zn − zn−1 + zn−1 − JBrnwn −

rnrn−1

(zn−1 − JBrn−1wn−1)〉 ≥ 0. (3.19)

From (3.19) we deduce

‖zn − zn−1‖2 ≤ 〈zn−1 − zn, JBrn−1wn−1 − JBrnwn +

(1− rn

rn−1

)(zn−1 − JBrn−1

wn−1)〉

≤ ‖zn−1 − zn‖(‖JBrn−1

wn−1 − JBrnwn‖+|rn − rn−1|

a‖zn−1 − JBrn−1

wn−1‖).

Thus we obtain

‖zn − zn−1‖ ≤ ‖JBrn−1wn−1 − JBrnwn‖+

|rn − rn−1|a

‖zn−1 − JBrn−1wn−1‖. (3.20)

Substituting (3.13) into (3.20) yields

‖zn − zn−1‖ ≤ (1− (τ − γl)αn)‖xn − xn−1‖+ |αn − αn−1|M1 + |rn − rn−1|M2

+|rn − rn−1|

a‖zn−1 − JBrn−1

wn−1‖

≤ ‖xn − xn−1‖+ |αn − αn−1|M1 + |rn − rn−1|(M2 +M3),

(3.21)

where M3 > 0 is an appropriate constant. In view of conditions (C1) and (C4), we find from (3.21)

lim supn→∞

(‖zn − zn−1‖ − ‖xn − xn−1‖ ≤ 0.

Thus, by Lemma 2.2, we havelimn→∞

‖zn − xn‖ = 0. (3.22)

Since xn+1 − xn = (1− βn)(zn − xn), by (3.22) and condition (3), we conclude

limn→∞

‖xn+1 − xn‖ = 0.

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Step 3. We show that limn→∞ ‖yn−wn‖ = 0, where wn = Arnyn. To show this, let p ∈ Fix(T )∩V I(C,A)∩B−10. Then, since p = Arnp, we deduce

‖wn − p‖2 = ‖Arnyn −Arnp‖2

≤ 〈wn − p, yn − p〉

=1

2(‖yn − p‖2 + ‖wn − p‖2 − ‖yn − wn‖2),

and hence‖wn − p‖2 ≤ ‖yn − p‖2 − ‖yn − wn‖2.

Thus we have‖TrnJBrnwn − p‖

2 ≤ ‖wn − p‖2 ≤ ‖yn − p‖2 − ‖yn − wn‖2.

This implies

‖yn − wn‖2 ≤ ‖yn − p‖2 − ‖TrnJBrnwn − p‖2

≤ (‖yn − p‖+ ‖TrnJBrnwn − p‖)(‖yn − p‖ − ‖TrnJBrnwn − p‖)

≤ (‖yn − p‖+ ‖TrnJBrnwn − p‖)‖yn − TrnJBrnwn‖

≤ (‖yn − p‖+ ‖TrnJBrnwn − p‖)(‖yn − xn‖+ ‖xn − TrnJBrnwn‖)

= (‖yn − p‖+ ‖TrnJBrnwn − p‖)(‖yn − xn‖+

‖xn − xn+1‖1− βn

).

Hence, by (3.4), condition (C3) and Step 2, we obtain

limn→∞

‖yn − wn‖ = 0.

Step 4. We show that limn→∞ ‖JBrnwn − yn‖ = 0. To this end, let p ∈ Fix(T ) ∩ V I(C,A) ∩ B−10. First,by (3.3), we observe

‖yn − p‖ ≤ (1− (τ − γl)αn)‖xn − p‖+ αn‖γV p− µGp‖≤ ‖xn − p‖+ αn‖γV p− µGp‖.

(3.23)

Then, since JBrn is firmly nonexpansive (see (2.1)) and JBrnp = p, we derive from (2.3)

‖JBrnwn − p‖2 ≤ 〈JBrnwn − p, wn − p〉

≤ 1

2(‖JBrnwn − p‖

2 + ‖wn − p‖2 − ‖(JBrnwn − p)− (wn − p)‖2)

=1

2(‖JBrnwn − p‖

2 + ‖yn − p‖2 − ‖JBrnwn − yn + yn − wn‖2)

≤ 1

2(‖JBrnwn − p‖

2 + ‖wn − p‖2 − ‖JBrnwn − yn‖2 − ‖yn − wn‖2 + 2‖JBrnwn − yn‖‖yn − wn‖),

and so

‖JBrnwn − p‖2 ≤ ‖wn − p‖2 − ‖JBrnwn − yn‖

2 − ‖yn − wn‖2 + 2‖JBrnwn − yn‖‖yn − wn‖≤ ‖wn − p‖2 − ‖JBrnwn − yn‖

2 + 2‖JBrnwn − yn‖‖yn − wn‖≤ ‖yn − p‖2 − ‖JBrnwn − yn‖

2 + 2‖JBrnwn − yn‖‖yn − wn‖.(3.24)

Thus, by (3.1), (3.23) and (3.24), we obtain

‖xn+1 − p‖2 ≤ βn‖xn − p‖2 + (1− βn)‖TrnJBrnwn − p‖2

≤ βn‖xn − p‖2 + (1− βn)‖JBrnwn − p‖2

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≤ βn‖xn − p‖2 + (1− βn)(‖yn − p‖2 − ‖JBrnwn − yn‖2 + 2‖JBrnwn − yn‖‖yn − wn‖)

≤ βn‖xn − p‖2 + (1− βn)(‖xn − p‖2 + 2αn‖xn − p‖‖γV p− µGp‖+ α2n‖γV p− µGp‖2)

− (1− βn)‖JBrnwn − yn‖2 + 2‖JBrnwn − yn‖‖yn − wn‖

≤ ‖xn − p‖2 + αn(2‖xn − p‖‖γV p− µGp‖+ αn‖γV p− µGp‖2)

− (1− βn)‖JBrnwn − yn‖2 + 2‖JBrnwn − yn‖‖yn − wn‖.

This implies

(1− βn)‖JBrnwn − yn‖2 ≤ ‖xn − p‖2 − ‖xn+1 − p‖2 + αn(2‖xn − p‖‖γV p− µGp‖+ αn‖γV p− µGp‖2)

+ 2‖yn − wn‖‖JBrnwn − yn‖≤ (‖xn − p‖+ ‖xn+1 − p‖)‖xn − xn+1‖+ αnM5 + ‖yn − wn‖M6,

where M5 > 0 and M6 > 0 are appropriate constants. Thus, by conditions (C1), (C3), Step 2 and Step 3,we have

limn→∞

‖JBrnwn − yn‖ = 0.

Step 5. We show thatlim supn→∞

〈(γV − µG)q, yn − q〉 ≤ 0,

where q ∈ Fix(T )∩V I(C,A)∩B−10 is the unique solution of the variational inequality (3.2). To show this,we can choose a subsequence yni of yn such that

limi→∞〈(γV − µG)q, yni − q〉 = lim sup

n→∞〈(γV − µG)q, yn − q〉.

Since yni is bounded, there exists a subsequence ynij of yni which converges weakly to some point z.

Without loss of generality, we can assume that yni z.Now, we prove z ∈ Fix(T ) ∩ V I(C,A) ∩ B−10. First, we show that z ∈ Fix(T ). Put zn = TrnJ

Brnwn

again. Then, by Lemma 2.5, we have

〈y − zn, T zn〉 −1

rn〈y − zn, (1 + rn)zn − JBrnwn〉 ≤ 0, ∀y ∈ C. (3.25)

Put wt = tv+ (1− t)z for t ∈ (0, 1] and v ∈ C. Then wt ∈ C, and from (3.25) and pseudocontractivity of Tit follows that

〈zn − wt, Twt〉 ≥ 〈zn − wt, Twt〉+ 〈wt − zn, T zn〉 −1

rn〈wt − zn, (1 + rn)zn − JBrnwn〉

= − 〈wt − zn, Twt − Tzn〉 −1

rn〈wt − zn, zn − J − rnBwn〉 − 〈wt − zn, zn〉

≥ − ‖wt − zn‖2 −1

rn〈wt − zn, zn − JBrnwn〉 − 〈wt − zn, zn〉

= − 〈wt − zn, wt〉 − 〈wt − zn,zn − JBrnwn

rn〉.

(3.26)

Since ‖yn − zn‖ ≤ ‖yn − xn‖ + ‖xn − zn‖ → 0 as n → ∞ by (3.4) and (3.22), and ‖JBrnwn − yn‖ → 0 asn → ∞ by Step 4, it follows that zni z and JBrni

wni z as i → ∞. So, replacing n by ni and letting

i→∞, we derive from (3.26)〈z − wt, Twt〉 ≥ 〈z − wt, wt〉

and−〈v − z, Twt〉 ≥ −〈v − z, wt〉, ∀v ∈ C.

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Letting t→ 0 and using the fact T is continuous, we obtain

− 〈v − z, Tz〉 ≥ −〈v − z, z〉. (3.27)

Let v = Tz in (3.27). Then we have z = Tz, that is, z ∈ Fix(T ).Next, we prove that z ∈ V I(C,A). In fact, from the definition of Arnyn = wn we have

〈y − wn, Awn〉+ 〈y − wn,wn − ynrn

〉 ≥ 0, ∀y ∈ C. (3.28)

Set wt = tv + (1− t)z for all t ∈ (0, 1] and v ∈ C. Then, wt ∈ C, and from (3.28) it follows that

〈wt − wn, Awt〉 ≥ 〈wt − wn, Awt〉 − 〈wt − wn, Awn〉 − 〈wt − wn,wn − ynrn

= 〈wt − wn, Awt −Awn〉 − 〈wt − wn,wn − ynrn

〉.(3.29)

By Step 3, we have wn−ynrn

→ 0 as n→∞, and since yni z, wni z as i→∞. From monotonicity of Ait also follows that 〈wt − wn, Awt −Awn〉 ≥ 0. Thus, replacing n by ni , from (3.29) we derive

0 ≤ limi→∞〈wt − wni , Awt〉 = 〈wt − z, Fwt〉,

and hence〈v − z,Awt〉 ≥ 0, ∀v ∈ C.

If t→ 0, the continuity of A yields that

〈v − z,Az〉 ≥ 0, ∀v ∈ C.

This means that z ∈ V I(C,A).Finally, we prove that z ∈ B−10. To this end, recall un = JBrnwn again. Then, it follows that

wn ∈ (I + rnB)un.

That is, wn−unrn

∈ Bun. Since B is monotone, we know that for any (u, v) ∈ B,

〈un − u,wn − unrn

− v〉 ≥ 0. (3.30)

Since ‖wn − un‖ ≤ ‖wn − yn‖+ ‖yn − un‖ → 0 as n→∞ by Step 3 and Step 4, and yni z as i→∞, weobtain uni z as i→∞. By replacing n by ni in (3.30) and letting i→∞, we have

〈z − u,−v〉 ≥ 0.

Since B is maximal monotone, 0 ∈ Bz, that is, z ∈ B−10. Therefore, z ∈ Fix(T ) ∩ V I(C,A) ∩B−10.Now, since q is the unique solution of the variational inequality (3.2), we conclude

lim supn→∞

〈(γV − µG)q, yn − q〉 = limi→∞〈(γV − µG)q, yni − q〉

= 〈(γV − µG)q, z − q〉 ≤ 0.

Step 6. We show that limn→∞ ‖xn − q‖ = 0, where q ∈ Fix(T ) ∩ V I(C,A) ∩ B−10 is the unique solutionof the variational inequality (3.2). Indeed, from (3.1), Lemma 2.1 and Lemma 2.7 we derive

‖yn − q‖2 =‖αn(γV xn − µGq) + (I − αnµG)xn − (I − αnµG)q‖2

≤ ‖(I − αnµG)xn − (I − αnµG)q‖2 + 2αn〈γV xn − µGq, yn − q〉≤ (1− ταn)2‖xn − q‖2 + 2αnγ〈V xn − V q, yn − q〉+ 2αn〈(γV − µG)q, yn − q〉≤ (1− ταn)2‖xn − q‖2 + 2αnγl‖xn − q‖‖yn − q‖+ 2αn〈(γV − µG)q, yn − q〉≤ (1− ταn)2‖xn − q‖2 + 2αnγl‖xn − q‖(‖yn − xn‖+ ‖xn − q‖)

+ 2αn〈(γV − µG)q, yn − q〉= (1− 2(τ − γl)αn)‖xn − q‖2 + α2

nτ2‖xn − q‖2 + 2αnγl‖xn − q‖‖yn − xn‖

+ 2αn〈(γV − µG)q, yn − q〉.

(3.31)

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Thus, by (3.1) and (3.31), we have

‖xn+1 − q‖2 ≤ βn‖xn − q‖2 + (1− βn)‖TrnJBrnArnyn − q‖2

≤ βn‖xn − q‖2 + (1− βn)‖yn − q‖2

≤ βn‖xn − q‖2 + (1− βn)(1− 2(τ − γl)αn)‖xn − q‖2 + (1− βn)α2nτ

2M7

+ 2(1− βn)αnγl‖yn − xn‖M8 + 2(1− βn)αn〈(γV − µG)q, yn − q〉= (1− 2αn(1− βn)(τ − γl))‖xn − q‖2

+ 2αn(1− βn)(τ − γl)( 1

2αnτ2M7 + ‖yn − xn‖M8 + 〈(γV − µG)q, yn − q〉

τ − γl

)= (1− ξn)‖xn − q‖2 + ξnδn,

where M7 > 0 and M8 > 0 are appropriate constants, ξn = 2αn(1− βn)(τ − γl) and

δn =

( 12αnτ

2M7 + ‖yn − xn‖M8 + 〈(γV − µG)q, yn − q〉τ − γl

).

From conditions (C1), (C2), (C3), (3.4) and Step 5 it is easy to see that ξn → 0,∑∞

n=1 ξn = ∞ andlim supn→∞ δn ≤ 0. Hence, by Lemma 2.3, we obtain

limn→∞

‖xn − q‖ = 0.

This completes the proof.

From Theorem 3.2, we deduce immediately the following result.

Corollary 3.3. Suppose that Fix(T ) ∩ V I(C,A) ∩ B−10 6= ∅. Let the sequence αn, βn ⊂ (0, 1) andrn ⊂ (0,∞) satisfy the conditions (1) – (4) in Theorem 3.2. Let the sequence xn be generated iterativelyby

yn = (1− αn)xn

xn+1 = βnxn + (1− βn)TrnJBrnArnyn, ∀n ≥ 1,

(3.32)

where x1 ∈ C is an arbitrary initial guess. Then xn converge strongly to a point q in Fix(T )∩V I(C,A)∩B−10, which is the minimum-norm element in Fix(T ) ∩ V I(C,A) ∩B−10.

Proof. Take V ≡ 0, l = 0, G ≡ I, µ = 1, and τ = 1 in Theorem 3.2. Then the variational inequality (3.2) isreduced to the inequality

〈−q, q − p〉 ≥ 0, ∀p ∈ Fix(T ) ∩ V I(C,A) ∩B−10.

This is equivalent to ‖q‖2 ≤ 〈q, p〉 ≤ ‖q‖‖p‖ for all p ∈ Fix(T ) ∩ V I(C,A) ∩ B−10. It turns out that‖q‖ ≤ ‖p‖ for all p ∈ Fix(T ) ∩ V I(C,A) ∩ B−10. Therefore, q is the minimum-norm element in Fix(T ) ∩V I(C,A) ∩B−10.

Remark 3.4.

1) It is worth pointing out that our iterative algorithms (3.1) and (3.32) are new ones different from thosein the literature.

2) From Lemma 2.8, we know that Fix(T ) ∩ V I(C,A) ∩B−10 ⊂ Fix(T ) ∩ (A+B)−10. Thus, as resultsfor finding a common element of the fixed point set of continuous pseudocontractive mappings moregeneral than nonexpansive mappings and strictly pseudocontractive mappings and the zero point set ofsum of maximal monotone operators and continuous monotone mappings more general than α-inversestrongly monotone mappings, Theorem 3.2 and Corollary 3.3 extend, improve and unify most of theresults that have been proved for these important classes of nonlinear mappings; see for instance,[16, 30, 35, 37, 42, 45] and references therein.

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4. Applications

Let H be a real Hilbert space, and let g be a proper lower semicontinuous convex function of H into(−∞,∞]. Then the subdifferential ∂g of g is defined as follows:

∂g(x) = z ∈ H|g(x) + 〈z, y − x〉 ≤ g(y), y ∈ H

for all x ∈ H. From Rockafellar [26], we know that ∂g is maximal monotone. Let C be a closed convexsubset of H, and let iC be the indicator function of C, that is,

iC(x) =

0, x ∈ C,∞, x /∈ C.

(4.1)

Since iC is a proper lower semicontinuous convex function on H, the subdifferential ∂iC of iC is a maximalmonotone operator. It is well-known that if B = ∂iC , then to find a point u in (A+B)−10 is equivalent tofind a point u ∈ C such that

〈Au, v − u〉 ≥ 0, ∀v ∈ C. (4.2)

The following result is proved by Takahashi et al. [35].

Lemma 4.1 ([35]). Let C be a nonempty closed convex subset of a real Hilbert space H, let PC be the metricprojection from H onto C, let ∂iC be the subdifferential of iC , and let Jr be the resolvent of ∂iC for r > 0,where iC is defined by (4.1) and Jr = (I + r∂iC)−1. Then

u = Jrx⇐⇒ u = PCx, ∀x ∈ H, y ∈ C.

Applying Theorem 3.2, we can obtain a strong convergence theorem for finding a common element of theset of solutions to the variational inequality (4.2), the set of fixed points of a continuous pseudocontractivemapping T , and the set ∂i−1

C 0 of zero points of ∂iC .

Theorem 4.2. Suppose that Fix(T ) ∩ V I(C,A) ∩ ∂i−1C 0 6= ∅. Let αn, βn ⊂ (0, 1) and λn ⊂ (0, 2α)

satisfy the conditions (C1) – (C4) in Theorem 3.2. Let the sequence xn be generated iteratively byyn = αnγV xn + (1− αnµG)xn,

xn+1 = βnxn + (1− βn)TrnPCArnyn, ∀n ≥ 1,

where x1 ∈ C is an arbitrary initial guess. Then xn converge strongly to a point q in Fix(T )∩V I(C,A)∩∂i−1C 0, which is the unique solution of the following variational inequality:

〈(γV − µG)q, q − p〉 ≥ 0, ∀p ∈ Fix(T ) ∩ V I(C,A) ∩ ∂i−1C 0.

Proof. Put B = ∂iC . From Lemma 4.1 , we get JBrn = PC for all rn. Hence the desired result follows fromTheorem 3.2.

As in [34, 35], we consider the problem for finding a common element of the set of solutions of a mathe-matical model related to equilibrium problems and the set of fixed points of a continuous pseudocontractivemapping in a Hilbert space.

Let C be a nonempty closed convex subset of a Hilbert space H, and let us assume that a bifunctionΘ : C × C → R satisfies the following conditions:

(A1) Θ(x, x) = 0 for all x ∈ C;

(A2) Θ is monotone, that is, Θ(x, y) + Θ(y, x) ≤ 0 for all x, y ∈ C;

(A3) for each x, y, z ∈ C,limt↓0

Θ(tz + (1− t)x, y) ≤ Θ(x, y);

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(A4) for each x ∈ C, y 7→ Θ(x, y) is convex and lower semicontinuous.

Then the mathematical model related to the equilibrium problem (with respect to C) is to find x ∈ C suchthat

Θ(x, y) ≥ 0

for all y ∈ C. The set of such solutions x is denoted by EP (Θ). The following lemma was given in [2, 11].

Lemma 4.3 ([2, 11]). Let C be a nonempty closed convex subset of H, and let Θ be a bifunction of C × Cinto R satisfying (A1)–(A4). Then, for any r > 0 and x ∈ H, there exists z ∈ C such that

Θ(z, y) +1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C.

Moreover, if we define Kr : H → C as follows:

Krx =

z ∈ C : Θ(z, y) +

1

r〈y − z, z − x〉 ≥ 0, ∀y ∈ C

for all x ∈ H, then, the following hold:

(1) Kr is single-valued;

(2) Kr is firmly nonexpansive, that is, for any x, y ∈ H,

‖Krx−Kry‖2 ≤ 〈Krx−Kry, x− y〉;

(3) Fix(Kr) = EP (Θ);

(4) EP (Θ) is closed and convex.

We call such Kr the resolvent of Θ for r > 0. The following lemma was given in Takahashi et al. [35].

Lemma 4.4 ([35]). Let C be a nonempty closed convex subset of a real Hilbert space H, and let Θ be abifunction of C × C into R satisfying (A1)–(A4). Let AΘ be a multivalued mapping of H into itself defineby

AΘx =

z ∈ H : Θ(x, y) ≥ 〈y − x, z〉, x ∈ C,∅, x /∈ C.

Then, EP (Θ) = A−1Θ 0 and AΘ is a maximal monotone operator with dom(AΘ) ⊂ C. Moreover, for any

x ∈ H and r > 0, the resolvent KAΘr of Θ coincides with the resolvent of AΘ; that is,

KAΘr x = (I + rAΘ)−1x.

Applying Lemma 4.4 and Theorem 3.2, we can obtain the following results.

Theorem 4.5. Let Θ be a bifunction of C×C into R satisfying (A1)–(A4). Let AΘ be a maximal monotoneoperator with dom(AΘ) ⊂ C defined as in Lemma 4.4, and let KAΘ

r be the resolvent of Θ for r > 0. Supposethat Fix(T ) ∩ V I(C,A) ∩A−1

Θ 0 6= ∅. Let αn, βn ⊂ (0, 1) and rn ⊂ (0,∞) satisfy the conditions (C1)– (C4) in Theorem 3.2. Let xn be generated iteratively by

yn = αnγV xn + (1− αnµG)xn,

xn+1 = βnxn + (1− βn)TrnKAΘrn Arnyn, ∀n ≥ 1,

where x1 ∈ C is an arbitrary initial guess. Then the sequence xn converge strongly to a point q inFix(T ) ∩ V I(C,A) ∩A−1

Θ 0, which is the unique solution of the following variational inequality:

〈(γV − µG)q, q − p〉 ≥ 0, ∀p ∈ Fix(T ) ∩ V I(C,A) ∩A−1Θ 0.

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J. S. Jung, J. Nonlinear Sci. Appl. 9 (2016), 4409–4426 4424

Theorem 4.6. Let Θ be a bifunction of C×C into R satisfying (A1)–(A4). Let AΘ be a maximal monotoneoperator with dom(AΘ) ⊂ C defined as in Lemma 4.4, and let KAΘ

r be the resolvent of Θ for r > 0. Supposethat Fix(T ) ∩EP (Θ) 6= ∅. Let αn, βn ⊂ (0, 1) and rn ⊂ (0,∞) satisfy the conditions (C1) – (C4) inTheorem 3.2. Let the sequence xn be generated iteratively by

yn = αnγV xn + (1− αnµG)xn,

xn+1 = βnxn + (1− βn)TrnKAΘrn yn, ∀n ≥ 1,

where x1 ∈ C is an arbitrary initial guess. Then xn converge strongly to a point q in Fix(T ) ∩ EP (Θ),which is the unique solution of the following variational inequality:

〈(γV − µG)q, q − p〉 ≥ 0, ∀p ∈ Fix(T ) ∩ EP (Θ).

Proof. Take A ≡ 0 in Theorem 4.2. Then Arn in Lemma 2.4 is the identity mapping. From Lemma 4.4 wealso know that JAΘ

rn = KAΘrn for all n ≥ 1. Hence, the desired result follows from Theorem 4.2.

Remark 4.7.

1) As in Corollary 3.3, if we take V ≡ 0, l = 0, G ≡ I, µ = 1, and τ = 1 in Theorems 4.2, 4.5 and 4.6, thenwe can obtain the minimum-norm element in Fix(T ) ∩ V I(C,A) ∩ ∂i−1

C 0, Fix(T ) ∩ V I(C,A) ∩A−1Θ 0

and Fix(T ) ∩ EP (Θ), respectively.

2) From Lemma 2.8 it follows that Fix(T ) ∩ V I(C,A) ∩ ∂i−1C 0 ⊂ Fix(T ) ∩ (A + ∂iC)−10 = Fix(T ) ∩

V I(C,A) and Fix(T ) ∩ V I(C,A) ∩ A−1Θ 0 ⊂ Fix(T ) ∩ (A + AΘ)−10. So, Theorem 4.2, Theorem

4.5, and Theorem 4.6 also improve and unify the corresponding results for nonexpansive mappings,strictly pseudocontarctive mappings, Lipschitzian pseudocontractive mappings, and α-inverse stronglymonotone mappings; see, for instance, [16, 30, 35, 37, 42, 45], and the references therein.

3) For a certain iterative algorithm for finding a common element of the set (A + B)−10 of zero pointsof A + B for an α-inverse-strongly monotone mapping A on H and a set-valued maximal monotoneoperator B on H, the solution set of the mixed equilibrium problem and fixed point set for an infinitefamily of nonexpansive mappings, we can refer to [41]. For a certain hybrid projection method forfinding a common element of the set of zeros of a finite family maximal monotone operators and theset of common solutions of a system of generalized equilibrium problems in a certain Banach space,see [29].

Acknowledgment

1. This study was supported by research funds from Dong-A University.2. The author would like to thank the anonymous reviewers for their careful reading and valuable

suggests, which improved the presentation of this manuscript, and the editor for his valuable commentsalong with providing some recent related papers.

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