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# Strong convergence of a hybrid method for monotone variational inequalities and fixed point problems

## Abstract

In this paper, we suggest a hybrid method for finding a common element of the set of solution of a monotone, Lipschitz-continuous variational inequality problem and the set of common fixed points of an infinite family of nonexpansive mappings. The proposed iterative method combines two well-known methods: extragradient method and CQ method. Under some mild conditions, we prove the strong convergence of the sequences generated by the proposed method.

Mathematics Subject Classification (2000): 47H05; 47H09; 47H10; 47J05; 47J25.

## 1 Introduction

Let H be a real Hilbert space with inner product 〈· , ·〉 and induced norm || · ||. Let C be a nonempty closed convex subset of H. Let A : CH be a nonlinear operator. It is well known that the variational inequality problem VI(C, A) is to find u C such that

$⟨ A u , v - u ⟩ ≥ 0 , ∀ v ∈ C .$

The set of solutions of the variational inequality is denoted by Ω.

Variational inequality theory has emerged as an important tool in studying a wide class of obstacle, unilateral and equilibrium problems, which arise in several branches of pure and applied sciences in a unified and general framework. Several numerical methods have been developed for solving variational inequalities and related optimization problems, see [1, 125] and the references therein. Let us start with Korpelevich's extragradient method which was introduced by Korpelevich  in 1976 and which generates a sequence {x n } via the recursion:

$y n = P C [ x n - λ A x n ] , x n + 1 = P C [ x n - λ A y n ] , n ≥ 0 ,$
(1.1)

where P C is the metric projection from Rn onto C, A : CH is a monotone operator and λ is a constant. Korpelevich  proved that the sequence {x n } converges strongly to a solution of V I(C, A). Note that the setting of the space is Euclid space Rn .

Korpelevich's extragradient method has extensively been studied in the literature for solving a more general problem that consists of finding a common point that lies in the solution set of a variational inequality and the set of fixed points of a nonexpansive mapping. This type of problem aries in various theoretical and modeling contexts, see e.g., [1622, 26] and references therein. Especially, Nadezhkina and Takahashi  introduced the following iterative method which combines Korpelevich's extragradient method and a CQ method:

$x 0 = x ∈ C , y n = P C [ x n - λ n A x n ] , z n = α n x n + ( 1 - α n ) S P C [ x n - λ n A y n ] , C n = { z ∈ C : ∥ z n - z ∥ ≤ ∥ x n - z ∥ } , Q n = { z ∈ C : ⟨ x n - z , x - x n ⟩ ≥ 0 } , x n + 1 = P C n ∩ Q n x , n ≥ 0 , n ≥ 0 ,$

where P C is the metric projection from H onto C, A : CH is a monotone k-Lipschitz-continuous mapping, S : CC is a nonexpansive mapping, {λ n } and {α n } are two real number sequences. They proved the strong convergence of the sequences {x n }, {y n } and {z n } to the same element in Fix(S) ∩ Ω. Ceng et al.  suggested a new iterative method as follows:

$y n = P C [ x n - λ n A x n ] , z n = α n x n + ( 1 - α n ) S n P C [ x n - λ n A y n ] , C n = { z ∈ C : ∥ z n - z ∥ ≤ ∥ x n - z ∥ } , fi n d x n + 1 ∈ C n s u c h t h a t ⟨ x n - x n + 1 + e n - σ n A x n + 1 , x n + 1 - x ⟩ ≥ - ε n , ∀ x ∈ C n ,$

where A : CH is a pseudomonotone, k-lipschitz-continuous and (w, s)-sequentially-continuous mapping, ${ S i } i = 1 N :C→C$ are N nonexpansive mappings. Under some mild conditions, they proved that the sequences {x n }, {y n } and {z n } converge weakly to the same element of $⋂ i = 1 N Fix ( S i ) ∩Ω$ if and only if lim inf n Ax n , x - x n 〉 ≥ 0, x C. Note that Ceng, Teboulle and Yao's method has only weak convergence. Very recently, Ceng, Hadjisavvas and Wong further introduced the following hybrid extragradient-like approximation method

$x 0 ∈ C , y n = ( 1 - γ n ) x n + γ n P C [ x n - λ n A x n ] , z n = ( 1 - α n - β n ) x n + α n y n + β n S P C [ x n - λ n A y n ] , C n = { z ∈ C : ∥ z n - z ∥ 2 ≤ ∥ x n - z ∥ 2 + ( 3 - 3 γ n + α n ) b 2 ∥ A x n ∥ 2 } , Q n = { z ∈ C : ⟨ x n - z , x 0 - x n ⟩ ≥ 0 } , x n + 1 = P C n ∩ Q n x 0 ,$

for all n ≥ 0. It is shown that the sequences {x n }, {y n }, {z n } generated by the above hybrid extragradient-like approximation method are well defined and converge strongly to PF(S)∩Ω.

Motivated and inspired by the works of Nadezhkina and Takahashi , Ceng et al. , and Ceng et al. , in this paper we suggest a hybrid method for finding a common element of the set of solution of a monotone, Lipschitz-continuous variational inequality problem and the set of common fixed points of an infinite family of nonexpansive mappings. The proposed iterative method combines two well-known methods: extragradient method and CQ method. Under some mild conditions, we prove the strong convergence of the sequences generated by the proposed method.

## 2 Preliminaries

In this section, we will recall some basic notations and collect some conclusions that will be used in the next section.

Let C be a nonempty closed convex subset of a real Hilbert space H. A mapping A : CH is called monotone if

$⟨ A u - A v , u - v ⟩ ≥ 0 , ∀ u , v ∈ C .$

Recall that a mapping S : CC is said to be nonexpansive if

$∥ S x - S y ∥ ≤ ∥ x - y ∥ , ∀ x , y ∈ C .$

Denote by Fix(S) the set of fixed points of S; that is, Fix(S) = {x C : Sx = x}.

It is well known that, for any u H, there exists a unique u0 C such that

$∥ u - u 0 ∥ = inf { ∥ u - x ∥ : x ∈ C } .$

We denote u0 by P C [u], where P C is called the metric projection of H onto C. The metric projection P C of H onto C has the following basic properties:

1. (i)

||P C [x] - P C [y] || ||x - y|| for all x, y H.

2. (ii)

x - P C [x], y - P C [x]〉 ≤ 0 for all x H, y C.

3. (iii)

The property (ii) is equivalent to

$∥ x - P C [ x ] ∥ 2 + ∥ y - P C [ x ] ∥ 2 ≤ ∥ x - y ∥ , ∀ x ∈ H , y ∈ C .$
4. (iv)

In the context of the variational inequality problem, the characterization of the projection implies that

$u ∈ Ω ⇔ u = P C [ u - λ A u ] , ∀ λ > 0 .$

Recall that H satisfies the Opial's condition ; i.e., for any sequence {x n } with x n converges weakly to x, the inequality

$lim i n f n → ∞ ∥ x n − x ∥ < lim i n f n → ∞ ∥ x n − y ∥$

holds for every y H with yx.

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${ S i } i = 1 ∞$ be infinite family of nonexpansive mappings of C into itself and let ${ ξ i } i = 1 ∞$ be real number sequences such that 0 ≤ ξ i ≤ 1 for every i N. For any n N, define a mapping W n of C into itself as follows:

$U n , n + 1 = I , U n , n = ξ n S n U n , n + 1 + ( 1 - ξ n ) I , U n , n - 1 = ξ n - 1 S n - 1 U n , n + ( 1 - ξ n - 1 ) I , ⋮ U n , k = ξ k S k U n , k + 1 + ( 1 - ξ k ) I , U n , k - 1 = ξ k - 1 S k - 1 U n , k + ( 1 - ξ k - 1 ) I , ⋮ U n , 2 = ξ 2 S 2 U n , 3 + ( 1 - ξ 2 ) I , W n = U n , 1 = ξ 1 S 1 U n , 2 + ( 1 - ξ 1 ) I .$
(2.1)

Such W n is called the W -mapping generated by ${ S i } i = 1 ∞$ and ${ ξ i } i = 1 ∞$.

We have the following crucial Lemmas 3.1 and 3.2 concerning W n which can be found in . Now we only need the following similar version in Hilbert spaces.

Lemma 2.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let S1, S2, be nonexpansive mappings of C into itself such that$⋂ n = 1 ∞ Fix ( S n )$is nonempty, and let ξ1, ξ2, be real numbers such that 0 < ξ i b < 1 for any i N. Then, for every x C and k N, the limit limn→∞Un,kx exists.

Lemma 2.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let S1, S2, be nonexpansive mappings of C into itself such that$⋂ n = 1 ∞ Fix ( S n )$is nonempty, and let ξ1, ξ2, be real numbers such that 0 < ξ i b < 1 for any i N. Then, $Fix ( W ) = ⋂ n = 1 ∞ Fix ( S n )$.

Lemma 2.3. (see ) Using Lemmas 2.1 and 2.2, one can define a mapping W of C into itself as: Wx = limn→∞W n x = limn→∞Un,1x, for every x C. If {x n } is a bounded sequence in C, then we have

$lim n → ∞ ∥ W x n - W n x n ∥ = 0 .$

We also need the following well-known lemmas for proving our main results.

Lemma 2.4. () Let C be a nonempty closed convex subset of a real Hilbert space H. Let S : CC be a nonexpansive mapping with Fix(S) ≠ . Then S is demiclosed on C, i.e., if y n z C weakly and y n - Sy n y strongly, then (I - S)z = y.

Lemma 2.5. () Let C be a closed convex subset of H. Let {x n } be a sequence in H and u H. Let q = P C [u]. If {x n } is such that ω w (x n ) C and satisfies the condition

$∥ x n - u ∥ ≤ ∥ u - q ∥ f o r a l l n .$

Then x n q.

• For a given sequence {x n } H, ω w (x n ) denotes the weak ω-limit set of {x n }; that is, $ω w ( x n ) : = { x ∈ H : { x n j }$ converges weakly to x for some subsequence {n j } of {n}}.

• x n x stands for the weak convergence of (x n ) to x;

• x n x stands for the strong convergence of (x n ) to x.

## 3 Main results

In this section we will state and prove our main results.

Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let A : CH be a monotone, k-Lipschitz-continuous mapping and let${ S n } n = 1 ∞$be an infinite family of nonexpansive mappings of C into itself such that$⋂ n = 1 ∞ Fix ( S n ) ∩Ω≠∅$. Let x1 = x0 C. For C1 = C, let {x n }, {y n } and {z n } be sequences generated by

$y n = P C n [ x n - λ n A x n ] , z n = α n x n + ( 1 - α n ) W n P C n [ x n - λ n A y n ] , C n + 1 = { z ∈ C n : ∥ z n - z ∥ ≤ ∥ x n - z ∥ } , x n + 1 = P C n + 1 [ x 0 ] , n ≥ 1 ,$
(3.1)

where W n is W -mapping defined by (2.1). Assume the following conditions hold:

(i) {λ n } [a, b] for some a, b (0, 1/k);

(ii) {α n } [0, c] for some c [0, 1).

Then the sequences {x n }, {y n } and {z n } generated by (3.1) converge strongly to the same point$P ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω [ x 0 ]$.

Next, we will divide our detail proofs into several conclusions. In the sequel, we assume that all assumptions of Theorem 3.1 are satisfied.

Conclusion 3.2. (1) Every C n is closed and convex, n ≥ 1;

1. (2)

$⋂ n = 1 ∞ Fix ( S n ) ∩Ω⊂ C n + 1 , ∀ n ≥1$,

2. (3)

{x n+1} is well defined.

Proof. First we note that C1 = C is closed and convex. Assume that C k is closed and convex. From (3.1), we can rewrite Ck+1as

$C k + 1 = { z ∈ C k : ⟨ z - x k + z k 2 , z k - x k ⟩ ≥ 0 } .$

It is clear that Ck+1is a half space. Hence, Ck+1is closed and convex. By induction, we deduce that C n is closed and convex for all n ≥ 1. Next we show that $⋂ n = 1 ∞ Fix ( S n ) ∩Ω⊂ C n + 1 , ∀ n ≥1$.

Set $t n = P C n [ x n - λ n A y n ]$ for all n ≥ 1. Pick up $u∈ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$. From property (iii) of P C , we have

(3.2)

Since u Ω and y n C n C, we get

$⟨ A u , y n - u ⟩ ≥ 0 .$

This together with the monotonicity of A imply that

$⟨ A y n , y n - u ⟩ ≥ 0 .$
(3.3)

Combine (3.2) with (3.3) to deduce

$∥ t n - u ∥ 2 ≤ ∥ x n - u ∥ 2 - ∥ x n - t n ∥ 2 + 2 λ n ⟨ A y n , y n - t n ⟩ (1) = ∥ x n - u ∥ 2 - ∥ x n - y n ∥ 2 - 2 ⟨ x n - y n , y n - t n ⟩ - ∥ y n - t n ∥ 2 (2) + 2 λ n ⟨ A y n , y n - t n ⟩ (3) = ∥ x n - u ∥ 2 - ∥ x n - y n ∥ 2 - ∥ y n - t n ∥ 2 (4) + 2 ⟨ x n - λ n A y n - y n , t n - y n ⟩ . (5) (6)$
(3.4)

Note that $y n = P C n [ x n - λ n A x n ]$ and t n C n . Then, using the property (ii) of P C , we have

$⟨ x n - λ n A x n - y n , t n - y n ⟩ ≤ 0 .$

Hence,

(3.5)

From (3.4) and (3.5), we get

(3.6)

Therefore, from (3.6), together with z n = α n x n + (1 α n )W n t n and u = W n u, we get

$∥ z n - u ∥ 2 = ∥ α n ( x n - u ) + ( 1 - α n ) ( W n t n - u ) ∥ 2 (1) ≤ α n ∥ x n - u ∥ 2 + ( 1 - α n ) ∥ W n t n - u ∥ 2 (2) ≤ α n ∥ x n - u ∥ 2 + ( 1 - α n ) ∥ t n - u ∥ 2 (3) ≤ ∥ x n - u ∥ 2 + ( 1 - α n ) ( λ n 2 k 2 - 1 ) ∥ x n - y n ∥ 2 (4) ≤ ∥ x n - u ∥ 2 , (5) (6)$
(3.7)

which implies that

$u ∈ C n + 1 .$

Therefore,

$⋂ n = 1 ∞ F i x ( S n ) ∩ Ω ⊂ C n + 1 , ∀ n ≥ 1 .$

This implies that {xn+1} is well defined. □

Conclusion 3.3. The sequences {x n }, {z n } and {t n } are all bounded and limn→∞|| x n - x0 || exists.

Proof. From $x n + 1 = P C n + 1 [ x 0 ]$, we have

$⟨ x 0 - x n + 1 , x n + 1 - y ⟩ ≥ 0 , ∀ y ∈ C n + 1 .$

Since $⋂ n = 1 ∞ Fix ( S n ) ∩Ω⊂ C n + 1$, we also have

$⟨ x 0 - x n + 1 , x n + 1 - u ⟩ ≥ 0 , ∀ u ∈ ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω .$

So, for $u∈ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$, we have

$0 ≤ ⟨ x 0 - x n + 1 , x n + 1 - u ⟩ (1) = ⟨ x 0 - x n + 1 , x n + 1 - x 0 + x 0 - u ⟩ (2) = - ∥ x 0 - x n + 1 ∥ 2 + ⟨ x 0 - x n + 1 , x 0 - u ⟩ (3) ≤ - ∥ x 0 - x n + 1 ∥ 2 + ∥ x 0 - x n + 1 ∥ ∥ x 0 - u ∥ . (4) (5)$

Hence,

$∥ x 0 - x n + 1 ∥ ≤ ∥ x 0 - u ∥ , ∀ u ∈ ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω ,$
(3.8)

which implies that {x n } is bounded. From (3.6) and (3.7), we can deduce that {z n } and {t n } are also bounded.

From $x n = P C n [ x 0 ]$ and $x n + 1 = P C n + 1 [ x 0 ] ∈ C n + 1 ⊂ C n$, we have

$⟨ x 0 - x n , x n - x n + 1 ⟩ ≥ 0 .$
(3.9)

As above one can obtain that

$0 ≤ - ∥ x 0 - x n ∥ 2 + ∥ x 0 - x n ∥ ∥ x 0 - x n + 1 ∥ ,$

and therefore

$∥ x 0 - x n ∥ ≤ ∥ x 0 - x n + 1 ∥ .$

This together with the boundedness of the sequence {x n } imply that limn→∞|| x n - x0 || exists.

Conclusion 3.4. limn→∞||xn+1- x n || = limn→∞||x n - y n || = limn→∞||x n - z n || = limn→∞||x n - t n || = 0 and limn→∞||x n - W n x n || = limn→∞||x n - Wx n || = 0.

Proof. It is well known that in Hilbert spaces H, the following identity holds:

$∥ x - y ∥ 2 = ∥ x ∥ 2 - ∥ y ∥ 2 - 2 ⟨ x - y , y ⟩ , ∀ x , y ∈ H .$

Therefore,

$∥ x n + 1 - x n ∥ 2 = ∥ ( x n + 1 - x 0 ) - ( x n - x 0 ) ∥ 2 (1) = ∥ x n + 1 - x 0 ∥ 2 - ∥ x n - x 0 ∥ 2 - 2 ⟨ x n + 1 - x n , x n - x 0 ⟩ , (2) (3)$

and by (3.9)

$∥ x n + 1 - x n ∥ 2 ≤ ∥ x n + 1 - x 0 ∥ 2 - ∥ x n - x 0 ∥ 2 .$

Since limn→∞||x n - x0|| exists, we get ||xn+1- x0||2 - ||x n - x0||2 → 0. Therefore,

$lim n → ∞ ∥ x n + 1 - x n ∥ = 0 .$

Since xn+1 C n , we have

$∥ z n - x n + 1 ∥ ≤ ∥ x n - x n + 1 ∥ ,$

and hence

$∥ x n - z n ∥ ≤ ∥ x n - x n + 1 ∥ + ∥ x n + 1 - z n ∥ (1) ≤ 2 ∥ x n + 1 - x n ∥ (2) → 0 . (3) (4)$

For each $u∈ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$, from (3.7), we have

$∥ x n - y n ∥ 2 ≤ 1 ( 1 - α n ) ( 1 - λ n 2 k 2 ) ( ∥ x n - u ∥ 2 - ∥ z n - u ∥ 2 ) (1) ≤ 1 ( 1 - α n ) ( 1 - λ n 2 k 2 ) ( ∥ x n - u ∥ + ∥ z n - u ∥ ) ∥ x n - z n ∥ . (2) (3)$

Since ||x n - z n || → 0 and the sequences {x n } and {z n } are bounded, we obtain ||x n - y n || → 0.

We note that following the same idea as in (3.6) one obtains that

$∥ t n - u ∥ 2 ≤ ∥ x n - u ∥ 2 + ( λ n 2 k 2 - 1 ) ∥ y n - t n ∥ 2 .$

Hence,

It follows that

$∥ t n - y n ∥ 2 ≤ 1 ( 1 - α n ) ( 1 - λ n 2 k 2 ) ( ∥ x n - u ∥ 2 - ∥ z n - u ∥ 2 ) (1) ≤ 1 ( 1 - α n ) ( 1 - λ n 2 k 2 ) ( ∥ x n - u ∥ + ∥ z n - u ∥ ) ∥ x n - z n ∥ (2) → 0 . (3) (4)$

Since A is k-Lipschitz-continuous, we have ||Ay n - At n || → 0. From

$∥ x n - t n ∥ ≤ ∥ x n - y n ∥ + ∥ y n - t n ∥ ,$

we also have

$∥ x n - t n ∥ → 0 .$

Since z n = α n x n + (1 - α n )W n t n , we have

$( 1 - α n ) ( W n t n - t n ) = α n ( t n - x n ) + ( z n - t n ) .$

Then,

$( 1 - c ) ∥ W n t n - t n ∥ ≤ ( 1 - α n ) ∥ W n t n - t n ∥ (1) ≤ α n ∥ t n - x n ∥ + ∥ z n - t n ∥ (2) ≤ ( 1 + α n ) ∥ t n - x n ∥ + ∥ z n - x n ∥ (3) (4)$

and hence || t n - W n t n || → 0. To conclude,

$∥ x n - W n x n ∥ ≤ ∥ x n - t n ∥ + ∥ t n - W n t n ∥ + ∥ W n t n - W n x n ∥ (1) ≤ ∥ x n - t n ∥ + ∥ t n - W n t n ∥ + ∥ t n - x n ∥ (2) ≤ 2 ∥ x n - t n ∥ + ∥ t n - W n t n ∥ . (3) (4)$

So, ||x n - W n x n || → 0 too. On the other hand, since {x n } is bounded, from Lemma 2.3, we have limn→∞||W n x n - Wx n || = 0. Therefore, we have

$lim n → ∞ ∥ x n - W x n ∥ = 0 .$

Finally, according to Conclusions 3.3-3.5, we prove the remainder of Theorem 3.1.

Proof. By Conclusions 3.3-3.5, we have proved that

$lim n → ∞ ∥ x n - W x n ∥ = 0 .$

Furthermore, since {x n } is bounded, it has a subsequence ${ x n j }$ which converges weakly to some $ũ∈C$; hence, we have $lim j → ∞ ∥ x n j -W x n j ∥=0$. Note that, from Lemma 2.4, it follows that I - W is demiclosed at zero. Thus $ũ∈Fix ( W )$. Since $t n = P C n [ x n - λ n A y n ]$, for every x C n we have

$⟨ x n - λ n A y n - t n , t n - x ⟩ ≥ 0$

hence,

$⟨ x - t n , A y n ⟩ ≥ ⟨ x - t n , x n - t n λ n ⟩ .$

Combining with monotonicity of A we obtain

$⟨ x - t n , A x ⟩ ≥ ⟨ x - t n , A t n ⟩ (1) = ⟨ x - t n , A t n - A y n ⟩ + ⟨ x - t n , A y n ⟩ (2) ≥ ⟨ x - t n , A t n - A y n ⟩ + ⟨ x - t n , x n - t n λ n ⟩ . (3) (4)$

Since limn→∞(x n - t n ) = limn→∞(y n - t n ) = 0, A is Lipschitz continuous and λ n a > 0, we deduce that

$⟨ x - ũ , A x ⟩ = lim n j → ∞ ⟨ x - t n j , A x ⟩ ≥ 0 .$

This implies that $ũ∈Ω$. Consequently, $ũ∈ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$ That is, $ω w ( x n ) ⊂ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$.

In (3.8), if we take $u= P ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω [ x 0 ]$, we get

$∥ x 0 - x n + 1 ∥ ≤ ∥ x 0 - P ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω [ x 0 ] ∥ .$
(3.10)

Notice that $ω w ( x n ) ⊂ ⋂ n = 1 ∞ Fix ( S n ) ∩Ω$. Then, (3.10) and Lemma 2.5 ensure the strong convergence of {xn+1} to $P ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω [ x 0 ]$. Consequently, {y n } and {z n } also converge strongly to $P ⋂ n = 1 ∞ F i x ( S n ) ∩ Ω [ x 0 ]$. This completes the proof.

Remark 3.5. Our algorithm (3.1) is simpler than the one in  and we extend the single mapping in  to an infinite family mappings. At the same time, the proofs are also simple.

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## Acknowledgements

The authors are extremely grateful to the referees for their useful comments and suggestions which helped to improve this paper. Yonghong Yao was supported in part by Colleges and Universities Science and Technology Development Foundation (20091003) of Tianjin, NSFC 11071279 and NSFC 71161001-G0105. Yeong-Cheng Liou was supported in part by NSC 100-2221-E-230-012. Jen-Chih Yao was partially supported by the Grant NSC 99-2115-M-037-002-MY3.

## Author information

Correspondence to Mu-Ming Wong.

### 4 Competing interests

The authors declare that they have no competing interests.

### 5 Authors' contributions

All authors participated in the design of the study and performed the converegnce analysis. All authors read and approved the final manuscript.

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• #### DOI

https://doi.org/10.1186/1687-1812-2011-53

### Keywords

• variational inequality problem
• fixed point problems; monotone mapping
• nonexpansive mapping 