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 : C → H be a nonlinear operator. It is well known that the variational inequality problem VI(C, A) is to find u ∈ C such that

$$⟨Au,v-u⟩\ge 0,\forall v\in C.$$

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, 1–25] and the references therein. Let us start with Korpelevich's extragradient method which was introduced by Korpelevich [6] in 1976 and which generates a sequence {x _n } via the recursion:

$$\left\{\begin{array}{c}{y}_n=P_C\left[x_n-\lambda A{x}_n\right],\\ x_{n+1}=P_C\left[x_n-\lambda A{y}_n\right],n\ge 0,\end{array}\right.$$

(1.1)

where P _C is the metric projection from Rⁿ onto C, A : C → H is a monotone operator and λ is a constant. Korpelevich [6] 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., [16–22, 26] and references therein. Especially, Nadezhkina and Takahashi [23] introduced the following iterative method which combines Korpelevich's extragradient method and a CQ method:

$$\begin{array}{c}{x}_0=x\in C,\\ y_n=P_C\left[x_n-{\lambda }_{n}A{x}_n\right],\\ z_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)S{P}_C\left[x_n-{\lambda }_{n}A{y}_n\right],\\ C_n=\left\{z\in C:\parallel z_n-z\parallel \le \parallel x_n-z\parallel \right\},\\ Q_n=\left\{z\in C:⟨x_n-z,x-x_n⟩\ge 0\right\},\\ x_{n+1}=P_{C_n\cap Q_n}x,n\ge 0,n\ge 0,\end{array}$$

where P _C is the metric projection from H onto C, A : C → H is a monotone k-Lipschitz-continuous mapping, S : C → C 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. [25] suggested a new iterative method as follows:

$$\begin{array}{c}{y}_n=P_C\left[x_n-{\lambda }_{n}A{x}_n\right],\\ z_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)S_n{P}_C\left[x_n-{\lambda }_{n}A{y}_n\right],\\ C_n=\left\{z\in C:\parallel z_n-z\parallel \le \parallel x_n-z\parallel \right\},\\ \mathrm{fi}nd{x}_{n+1}\in C_{n}suchthat\\ ⟨x_n-x_{n+1}+e_n-{\sigma }_{n}A{x}_{n+1},x_{n+1}-x⟩\ge -{\epsilon }_n,\forall x\in C_n,\end{array}$$

where A : C → H is a pseudomonotone, k-lipschitz-continuous and (w, s)-sequentially-continuous mapping, ${\left\{S_i\right\}}_{i=1}^N:C\to C$ 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 ${\bigcap }_{i=1}^{N}Fix\left(S_i\right)\cap \Omega$ 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

$$\begin{array}{c}{x}_0\in C,\\ y_n=\left(1-{\gamma }_n\right)x_n+{\gamma }_n{P}_C\left[x_n-{\lambda }_{n}A{x}_n\right],\\ z_n=\left(1-{\alpha }_n-{\beta }_n\right)x_n+{\alpha }_n{y}_n+{\beta }_{n}S{P}_C\left[x_n-{\lambda }_{n}A{y}_n\right],\\ C_n=\left\{z\in C:\parallel z_n-z{\parallel }^2\le \parallel x_n-z{\parallel }^2+\left(3-3{\gamma }_n+{\alpha }_n\right)b^2\parallel A{x}_n{\parallel }^2\right\},\\ Q_n=\left\{z\in C:⟨x_n-z,x_0-x_n⟩\ge 0\right\},\\ x_{n+1}=P_{C_n\cap Q_n}{x}_0,\end{array}$$

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 P_F(S)∩Ω.

Motivated and inspired by the works of Nadezhkina and Takahashi [23], Ceng et al. [25], and Ceng et al. [27], 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 : C → H is called monotone if

$$⟨Au-Av,u-v⟩\ge 0,\forall u,v\in C.$$

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

$$\parallel Sx-Sy\parallel \le \parallel x-y\parallel ,\forall x,y\in C.$$

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

$$\parallel u-u_0\parallel =inf\left\{\parallel u-x\parallel :x\in C\right\}.$$

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:

• (i) ||P _C [x] - P _C [y] || ≤ ||x - y|| for all x, y ∈ H.

• (ii) 〈x - P _C [x], y - P _C [x]〉 ≤ 0 for all x ∈ H, y ∈ C.

• (iii) The property (ii) is equivalent to

  $$\parallel x-P_C\left[x\right]{\parallel }^2+\parallel y-P_C\left[x\right]{\parallel }^2\le \parallel x-y\parallel ,\forall x\in H,y\in C.$$

• (iv) In the context of the variational inequality problem, the characterization of the projection implies that

  $$u\in \Omega \iff u=P_C\left[u-\lambda Au\right],\forall \lambda >0.$$

Recall that H satisfies the Opial's condition [28]; i.e., for any sequence {x _n } with x _n converges weakly to x, the inequality

$$\underset{n\to \infty }{{\displaystyle \lim inf}}\parallel x_n-x\parallel <\underset{n\to \infty }{{\displaystyle \lim inf}}\parallel x_n-y\parallel$$

holds for every y ∈ H with y ≠ x.

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\left\{S_i\right\}}_{i=1}^{\infty }$ be infinite family of nonexpansive mappings of C into itself and let ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }$ 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:

$$\begin{array}{c}{U}_{n,n+1}=I,\\ U_{n,n}={\xi }_n{S}_n{U}_{n,n+1}+\left(1-{\xi }_n\right)I,\\ U_{n,n-1}={\xi }_{n-1}{S}_{n-1}{U}_{n,n}+\left(1-{\xi }_{n-1}\right)I,\\ ⋮\\ U_{n,k}={\xi }_k{S}_k{U}_{n,k+1}+\left(1-{\xi }_k\right)I,\\ U_{n,k-1}={\xi }_{k-1}{S}_{k-1}{U}_{n,k}+\left(1-{\xi }_{k-1}\right)I,\\ ⋮\\ U_{n,2}={\xi }_2{S}_2{U}_{n,3}+\left(1-{\xi }_2\right)I,\\ W_n=U_{n,1}={\xi }_1{S}_1{U}_{n,2}+\left(1-{\xi }_1\right)I.\end{array}$$

(2.1)

Such W _n is called the W -mapping generated by ${\left\{S_i\right\}}_{i=1}^{\infty }$ and ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }$.

We have the following crucial Lemmas 3.1 and 3.2 concerning W _n which can be found in [29]. 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${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)$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 lim_n→∞U_n,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${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)$is nonempty, and let ξ₁, ξ₂, ⋯ be real numbers such that 0 < ξ _i ≤ b < 1 for any i ∈ N. Then, $Fix\left(W\right)={\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)$.

Lemma 2.3. (see [30]) Using Lemmas 2.1 and 2.2, one can define a mapping W of C into itself as: Wx = lim_n→∞W _n x = lim_n→∞U_n,1x, for every x ∈ C. If {x _n } is a bounded sequence in C, then we have

$$\underset{n\to \infty }{lim}\parallel W{x}_n-W_n{x}_n\parallel =0.$$

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

Lemma 2.4. ([31]) Let C be a nonempty closed convex subset of a real Hilbert space H. Let S : C → C 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. ([32]) 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

$$\parallel x_n-u\parallel \le \parallel u-q\parallel foralln.$$

Then x _n → q.

We adopt the following notation:

• For a given sequence {x _n } ⊂ H, ω _w (x _n ) denotes the weak ω-limit set of {x _n }; that is, ${\omega }_w(x_n):=\{x\in H:\{x_{n_j}\}$ 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 : C → H be a monotone, k-Lipschitz-continuous mapping and let${\left\{S_n\right\}}_{n=1}^{\infty }$be an infinite family of nonexpansive mappings of C into itself such that${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega \ne \varnothing$. Let x₁ = x₀ ∈ C. For C₁ = C, let {x _n }, {y _n } and {z _n } be sequences generated by

$$\begin{array}{c}{y}_n=P_{C_n}\left[x_n-{\lambda }_{n}A{x}_n\right],\\ z_n={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)W_n{P}_{C_n}\left[x_n-{\lambda }_{n}A{y}_n\right],\\ C_{n+1}=\left\{z\in C_n:\parallel z_n-z\parallel \le \parallel x_n-z\parallel \right\},\\ x_{n+1}=P_{C_{n+1}}\left[x_0\right],n\ge 1,\end{array}$$

(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$P_{{\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega }\left[x_0\right]$.

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;

• (2) ${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega \subset C_{n+1},{\forall }_n\ge 1$,

• (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 C_k+1as

$$C_{k+1}=\left\{z\in C_k:⟨z-\frac{x_k+z_k}{2},z_k-x_k⟩\ge 0\right\}.$$

It is clear that C_k+1is a half space. Hence, C_k+1is closed and convex. By induction, we deduce that C _n is closed and convex for all n ≥ 1. Next we show that ${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega \subset C_{n+1},{\forall }_n\ge 1$.

Set $t_n=P_{C_n}\left[x_n-{\lambda }_{n}A{y}_n\right]$ for all n ≥ 1. Pick up $u\in {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$. From property (iii) of P _C , we have

$$\begin{array}{lll}\hfill \parallel t_n-u{\parallel }^2& \le \parallel x_n-{\lambda }_{n}A{y}_n-u{\parallel }^2-\parallel x_n-{\lambda }_{n}A{y}_n-t_n{\parallel }^2& \hfill \text{(1)}\\ =\parallel x_n-u{\parallel }^2-\parallel x_n-t_n{\parallel }^2+2{\lambda }_n⟨A{y}_n,u-t_n⟩& \hfill \text{(2)}\\ =\parallel x_n-u{\parallel }^2-\parallel x_n-t_n{\parallel }^2+2{\lambda }_n⟨A{y}_n,u-y_n⟩+2{\lambda }_n⟨A{y}_n,y_n-t_n⟩.& \hfill \text{(3)}\\ \hfill \text{(4) }\end{array}$$

(3.2)

Since u ∈ Ω and y _n ∈ C _n ⊂ C, we get

$$⟨Au,y_n-u⟩\ge 0.$$

This together with the monotonicity of A imply that

$$⟨A{y}_n,y_n-u⟩\ge 0.$$

(3.3)

Combine (3.2) with (3.3) to deduce

$$\begin{array}{lll}\hfill \parallel t_n-u{\parallel }^2& \le \parallel x_n-u{\parallel }^2-\parallel x_n-t_n{\parallel }^2+2{\lambda }_n⟨A{y}_n,y_n-t_n⟩& \hfill \text{(1)}\\ =\parallel x_n-u{\parallel }^2-\parallel x_n-y_n{\parallel }^2-2⟨x_n-y_n,y_n-t_n⟩-\parallel y_n-t_n{\parallel }^2& \hfill \text{(2)}\\ +2{\lambda }_n⟨A{y}_n,y_n-t_n⟩& \hfill \text{(3)}\\ =\parallel x_n-u{\parallel }^2-\parallel x_n-y_n{\parallel }^2-\parallel y_n-t_n{\parallel }^2& \hfill \text{(4)}\\ +2⟨x_n-{\lambda }_{n}A{y}_n-y_n,t_n-y_n⟩.& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

(3.4)

Note that $y_n=P_{C_n}\left[x_n-{\lambda }_{n}A{x}_n\right]$ and t _n ∈ C _n . Then, using the property (ii) of P _C , we have

$$⟨x_n-{\lambda }_{n}A{x}_n-y_n,t_n-y_n⟩\le 0.$$

Hence,

$$\begin{array}{lll}\hfill ⟨x_n-{\lambda }_{n}A{y}_n-y_n,t_n-y_n⟩& =⟨x_n-{\lambda }_{n}A{x}_n-y_n,t_n-y_n⟩+⟨{\lambda }_{n}A{x}_n-{\lambda }_{n}A{y}_n,t_n-y_n⟩& \hfill \text{(1)}\\ \le ⟨{\lambda }_{n}A{x}_n-{\lambda }_{n}A{y}_n,t_n-y_n⟩& \hfill \text{(2)}\\ \le {\lambda }_{n}k\parallel x_n-y_n\parallel \parallel t_n-y_n\parallel .& \hfill \text{(3)}\\ \hfill \text{(4) }\end{array}$$

(3.5)

From (3.4) and (3.5), we get

$$\begin{array}{lll}\hfill \parallel t_n-u{\parallel }^2& \le \parallel x_n-u{\parallel }^2-\parallel x_n-y_n{\parallel }^2-\parallel y_n-t_n{\parallel }^2+2{\lambda }_{n}k\parallel x_n-y_n\parallel \parallel t_n-y_n\parallel & \hfill \text{(1)}\\ \le \parallel x_n-u{\parallel }^2-\parallel x_n-y_n{\parallel }^2-\parallel y_n-t_n{\parallel }^2+{\lambda }_n^2{k}^2\parallel x_n-y_n{\parallel }^2+\parallel y_n-t_n{\parallel }^2& \hfill \text{(2)}\\ =\parallel x_n-u{\parallel }^2+\left({\lambda }_n^2{k}^2-1\right)\parallel x_n-y_n{\parallel }^2& \hfill \text{(3)}\\ \le \parallel x_n-u{\parallel }^2.& \hfill \text{(4)}\\ \hfill \text{(5) }\end{array}$$

(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

$$\begin{array}{lll}\hfill \parallel z_n-u{\parallel }^2& =\parallel {\alpha }_n\left(x_n-u\right)+\left(1-{\alpha }_n\right)\left(W_n{t}_n-u\right){\parallel }^2& \hfill \text{(1)}\\ \le {\alpha }_n\parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\parallel W_n{t}_n-u{\parallel }^2& \hfill \text{(2)}\\ \le {\alpha }_n\parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\parallel t_n-u{\parallel }^2& \hfill \text{(3)}\\ \le \parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\left({\lambda }_n^2{k}^2-1\right)\parallel x_n-y_n{\parallel }^2& \hfill \text{(4)}\\ \le \parallel x_n-u{\parallel }^2,& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$$

(3.7)

which implies that

$$u\in C_{n+1}.$$

Therefore,

$$\bigcap _{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega \subset C_{n+1},\forall n\ge 1.$$

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

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

Proof. From $x_{n+1}=P_{C_{n+1}}\left[x_0\right]$, we have

$$⟨x_0-x_{n+1},x_{n+1}-y⟩\ge 0,\forall y\in C_{n+1}.$$

Since ${\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega \subset C_{n+1}$, we also have

$$⟨x_0-x_{n+1},x_{n+1}-u⟩\ge 0,\forall u\in \bigcap _{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega .$$

So, for $u\in {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$, we have

$$\begin{array}{lll}\hfill 0& \le ⟨x_0-x_{n+1},x_{n+1}-u⟩& \hfill \text{(1)}\\ =⟨x_0-x_{n+1},x_{n+1}-x_0+x_0-u⟩& \hfill \text{(2)}\\ =-\parallel x_0-x_{n+1}{\parallel }^2+⟨x_0-x_{n+1},x_0-u⟩& \hfill \text{(3)}\\ \le -\parallel x_0-x_{n+1}{\parallel }^2+\parallel x_0-x_{n+1}\parallel \parallel x_0-u\parallel .& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$$

Hence,

$$\parallel x_0-x_{n+1}\parallel \le \parallel x_0-u\parallel ,\forall u\in \bigcap _{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega ,$$

(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 $x_n=P_{C_n}\left[x_0\right]$ and $x_{n+1}=P_{C_{n+1}}\left[x_0\right]\in C_{n+1}\subset C_n$, we have

$$⟨x_0-x_n,x_n-x_{n+1}⟩\ge 0.$$

(3.9)

As above one can obtain that

$$0\le -\parallel x_0-x_n{\parallel }^2+\parallel x_0-x_n\parallel \parallel x_0-x_{n+1}\parallel ,$$

and therefore

$$\parallel x_0-x_n\parallel \le \parallel x_0-x_{n+1}\parallel .$$

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

Conclusion 3.4. lim_n→∞||x_n+1- x _n || = lim_n→∞||x _n - y _n || = lim_n→∞||x _n - z _n || = lim_n→∞||x _n - t _n || = 0 and lim_n→∞||x _n - W _n x _n || = lim_n→∞||x _n - Wx _n || = 0.

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

$$\parallel x-y{\parallel }^2=\parallel x{\parallel }^2-\parallel y{\parallel }^2-2⟨x-y,y⟩,\forall x,y\in H.$$

Therefore,

$$\begin{array}{lll}\hfill \parallel x_{n+1}-x_n{\parallel }^2& =\parallel \left(x_{n+1}-x_0\right)-\left(x_n-x_0\right){\parallel }^2& \hfill \text{(1)}\\ =\parallel x_{n+1}-x_0{\parallel }^2-\parallel x_n-x_0{\parallel }^2-2⟨x_{n+1}-x_n,x_n-x_0⟩,& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

and by (3.9)

$$\parallel x_{n+1}-x_n{\parallel }^2\le \parallel x_{n+1}-x_0{\parallel }^2-\parallel x_n-x_0{\parallel }^2.$$

Since lim_n→∞||x _n - x₀|| exists, we get ||x_n+1- x₀||² - ||x _n - x₀||² → 0. Therefore,

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =0.$$

Since x_n+1∈ C _n , we have

$$\parallel z_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel ,$$

and hence

$$\begin{array}{lll}\hfill \parallel x_n-z_n\parallel & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-z_n\parallel & \hfill \text{(1)}\\ \le 2\parallel x_{n+1}-x_n\parallel & \hfill \text{(2)}\\ \to 0.& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

For each $u\in {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$, from (3.7), we have

$$\begin{array}{lll}\hfill \parallel x_n-y_n{\parallel }^2& \le \frac{1}{\left(1-{\alpha }_n\right)\left(1-{\lambda }_n^2{k}^2\right)}\left(\parallel x_n-u{\parallel }^2-\parallel z_n-u{\parallel }^2\right)& \hfill \text{(1)}\\ \le \frac{1}{\left(1-{\alpha }_n\right)\left(1-{\lambda }_n^2{k}^2\right)}\left(\parallel x_n-u\parallel +\parallel z_n-u\parallel \right)\parallel x_n-z_n\parallel .& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$$

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

$$\parallel t_n-u{\parallel }^2\le \parallel x_n-u{\parallel }^2+\left({\lambda }_n^2{k}^2-1\right)\parallel y_n-t_n{\parallel }^2.$$

Hence,

$$\begin{array}{lll}\hfill \parallel z_n-u{\parallel }^2& \le {\alpha }_n\parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\parallel t_n-u{\parallel }^2& \hfill \text{(1) }\\ \le {\alpha }_n\parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\left(\parallel x_n-u{\parallel }^2+\left({\lambda }_n^2{k}^2-1\right)\parallel y_n-t_n{\parallel }^2\right)& \hfill \text{(2) }\\ =\parallel x_n-u{\parallel }^2+\left(1-{\alpha }_n\right)\left({\lambda }_n^2{k}^2-1\right)\parallel y_n-t_n{\parallel }^2.& \hfill \text{(3) }\\ \hfill \text{(4) }\end{array}$$

It follows that

$$\begin{array}{lll}\hfill \parallel t_n-y_n{\parallel }^2& \le \frac{1}{\left(1-{\alpha }_n\right)\left(1-{\lambda }_n^2{k}^2\right)}\left(\parallel x_n-u{\parallel }^2-\parallel z_n-u{\parallel }^2\right)& \hfill \text{(1)}\\ \le \frac{1}{\left(1-{\alpha }_n\right)\left(1-{\lambda }_n^2{k}^2\right)}\left(\parallel x_n-u\parallel +\parallel z_n-u\parallel \right)\parallel x_n-z_n\parallel & \hfill \text{(2)}\\ \to 0.& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

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

$$\parallel x_n-t_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-t_n\parallel ,$$

we also have

$$\parallel x_n-t_n\parallel \to 0.$$

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

$$\left(1-{\alpha }_n\right)\left(W_n{t}_n-t_n\right)={\alpha }_n\left(t_n-x_n\right)+\left(z_n-t_n\right).$$

Then,

$$\begin{array}{lll}\hfill \left(1-c\right)\parallel W_n{t}_n-t_n\parallel & \le \left(1-{\alpha }_n\right)\parallel W_n{t}_n-t_n\parallel & \hfill \text{(1)}\\ \le {\alpha }_n\parallel t_n-x_n\parallel +\parallel z_n-t_n\parallel & \hfill \text{(2)}\\ \le \left(1+{\alpha }_n\right)\parallel t_n-x_n\parallel +\parallel z_n-x_n\parallel & \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

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

$$\begin{array}{lll}\hfill \parallel x_n-W_n{x}_n\parallel & \le \parallel x_n-t_n\parallel +\parallel t_n-W_n{t}_n\parallel +\parallel W_n{t}_n-W_n{x}_n\parallel & \hfill \text{(1)}\\ \le \parallel x_n-t_n\parallel +\parallel t_n-W_n{t}_n\parallel +\parallel t_n-x_n\parallel & \hfill \text{(2)}\\ \le 2\parallel x_n-t_n\parallel +\parallel t_n-W_n{t}_n\parallel .& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

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

$$\underset{n\to \infty }{lim}\parallel x_n-W{x}_n\parallel =0.$$

□

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

$$\underset{n\to \infty }{lim}\parallel x_n-W{x}_n\parallel =0.$$

Furthermore, since {x _n } is bounded, it has a subsequence $\left\{x_{n_j}\right\}$ which converges weakly to some $ũ\in C$; hence, we have ${lim}_{j\to \infty }\parallel x_{n_j}-W{x}_{n_j}\parallel =0$. Note that, from Lemma 2.4, it follows that I - W is demiclosed at zero. Thus $ũ\in Fix\left(W\right)$. Since $t_n=P_{C_n}\left[x_n-{\lambda }_{n}A{y}_n\right]$, for every x ∈ C _n we have

$$⟨x_n-{\lambda }_{n}A{y}_n-t_n,t_n-x⟩\ge 0$$

hence,

$$⟨x-t_n,A{y}_n⟩\ge ⟨x-t_n,\frac{x_n-t_n}{{\lambda }_n}⟩.$$

Combining with monotonicity of A we obtain

$$\begin{array}{lll}\hfill ⟨x-t_n,Ax⟩& \ge ⟨x-t_n,A{t}_n⟩& \hfill \text{(1)}\\ =⟨x-t_n,A{t}_n-A{y}_n⟩+⟨x-t_n,A{y}_n⟩& \hfill \text{(2)}\\ \ge ⟨x-t_n,A{t}_n-A{y}_n⟩+⟨x-t_n,\frac{x_n-t_n}{{\lambda }_n}⟩.& \hfill \text{(3)}\\ \hfill \text{(4)}\end{array}$$

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

$$⟨x-ũ,Ax⟩=\underset{n_j\to \infty }{lim}⟨x-t_{n_j},Ax⟩\ge 0.$$

This implies that $ũ\in \Omega$. Consequently, $ũ\in {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$ That is, ${\omega }_w\left(x_n\right)\subset {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$.

In (3.8), if we take $u=P_{{\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega }\left[x_0\right]$, we get

$$\parallel x_0-x_{n+1}\parallel \le \parallel x_0-P_{{\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega }\left[x_0\right]\parallel .$$

(3.10)

Notice that ${\omega }_w\left(x_n\right)\subset {\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega$. Then, (3.10) and Lemma 2.5 ensure the strong convergence of {x_n+1} to $P_{{\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega }\left[x_0\right]$. Consequently, {y _n } and {z _n } also converge strongly to $P_{{\bigcap }_{n=1}^{\infty }Fix\left(S_n\right)\cap \Omega }\left[x_0\right]$. This completes the proof.

Remark 3.5. Our algorithm (3.1) is simpler than the one in [23] and we extend the single mapping in [23] to an infinite family mappings. At the same time, the proofs are also simple.