Convergence to common solutions of various problems for nonexpansive mappings in Hilbert spaces

Abstract

In this paper, motivated and inspired by Ceng and Yao (J. Comput. Appl. Math. 214(1):186-201, 2008), Iiduka and Takahashi (Nonlinear Anal. 61(3):341-350, 2005), Jaiboon and Kumam (Nonlinear Anal. 73(5):1180-1202, 2010), Kim (Nonlinear Anal. 73:3413-3419, 2010), Marino and Xu (J. Math. Anal. Appl. 318:43-52, 2006) and Saeidi (Nonlinear Anal. 70:4195-4208, 2009), we introduce a new iterative scheme for finding a common element of the set of solutions of a mixed equilibrium problem for an equilibrium bifunction, the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of some variational inequality problem, and the set of fixed points of a left amenable semigroup $\{T_t:t\in S\}$ of nonexpansive mappings with respect to W-mappings and a left regular sequence $\{{\mu }_n\}$ of means defined on an appropriate space of bounded real-valued functions of the semigroup S. Furthermore, we prove that the iterative scheme converges strongly to a common element of the above four sets. Our results extend and improve the corresponding results of many others.

MSC:43A65, 47H05, 47H09, 47H10, 47J20, 47J25, 74G40.

1 Introduction

Let H be a real Hilbert space, let C be a nonempty closed convex subset of H, and let $P_C$ be the metric projection of H onto C. Let $\phi :C\to \mathbb{R}$ be a real-valued function and $\theta :C\times C\to \mathbb{R}$ be an equilibrium bifunction with $\theta (u,u)=0$ for each $u\in C$. We consider the mixed equilibrium problem (for short, MEP) is to find $x^{\ast }\in C$ such that

$$\mathit{MEP}:\theta (x^{\ast },y)+\phi (y)-\phi \left(x^{\ast }\right)\ge 0,\mathrm{\forall }y\in C.$$

In particular, if $\phi \equiv 0$, this problem reduces to the equilibrium problem (for short, EP), which is to find $x^{\ast }\in C$ such that

$$\mathit{EP}:\theta (x^{\ast },y)\ge 0,\mathrm{\forall }y\in C.$$

Denote the set of solutions of MEP by Ω. The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems and the equilibrium problems as special cases.

A mapping T of C into itself is called nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,$$

for all $x,y\in C$. We denote by $F(T)$ the set of fixed points of T. It is well known that $F(T)$ is closed convex. Recall that a mapping $f:C\to C$ is called contractive if there exists a constant $\alpha \in (0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel ,$$

for all $x,y\in C$.

In 2000, Moudafi [1] introduced the viscosity approximation method for nonexpansive mappings (see [2] for further developments in both Hilbert and Banach spaces).

Starting with an arbitrary initial $x_0\in H$, define a sequence $\{x_n\}$ recursively by

$$x_{n+1}=(1-{\alpha }_n)T{x}_n+{\alpha }_{n}f(x_n),n\ge 0,$$

(1.1)

where ${\alpha }_n$ is a sequence in $(0,1)$. It is proved that under certain appropriate conditions imposed on $\{{\alpha }_n\}$, the sequence $\{x_n\}$ generated by (1.1) strongly converges to the unique solution $x^{\ast }$ in $F(T)$ of the variational inequality

$$〈(f-I)x^{\ast },x-x^{\ast }〉\le 0,\mathrm{\forall }x\in F(T)$$

(see [1, 2]).

Let A be a strongly positive bounded linear operator on H, that is, there exists a constant $\overline{\gamma }>0$ such that

$$〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2,$$

for all $x\in H$.

In 2006, Marino and Xu [3] considered the following iterative method:

$$x_{n+1}=(I-{\alpha }_{n}A)T{x}_n+{\alpha }_n\gamma f(x_n),n\ge 0,$$

(1.2)

where $0<\gamma <\frac{\overline{\gamma }}{\alpha }$, α is a contraction coefficient of f. They proved that if the sequence $\{{\alpha }_n\}$ satisfies appropriate conditions, then the sequence $\{x_n\}$ generated by (1.2) converges strongly to the unique solution of the variational inequality

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,x\in F(T),$$

which is the optimality condition for the minimization problem

$$\underset{x\in F(T)}{min}\frac{1}{2}〈Ax,x〉-h(x),$$

where h is a potential function for γf (i.e., $h^{\prime }(x)=\gamma f(x)$, for $x\in H$).

A set-valued mapping $T:H\to 2^H$ is called monotone if for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply $〈x-y,f-g〉\ge 0$. A monotone mapping $T:H\to 2^H$ is maximal if its graph $G(T)$ is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for every $(y,g)\in G(T)$ implies $f\in Tx$. Let A be a monotone mapping of C into H, and let $N_{C}v$ be the normal cone to C at $v\in C$, i.e.,

$$N_{C}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in C\}$$

and define

$$Tv=\{\begin{array}{cc}Av+N_{C}v,\hfill & \text{if }v\in C,\hfill \\ \mathrm{\varnothing },\hfill & \text{if }v\notin C.\hfill \end{array}$$

Then T is maximal monotone, and $0\in Tv$ if and only if $v\in \mathit{VI}(C,A)$; see [4].

In 2005, for finding an element of $F(T)\cap \mathit{VI}(C,A)$, Iiduka and Takahashi [5] proposed a new iterative sequence: $x_1=x\in C$ and

$$x_{n+1}={\alpha }_{n}x+(1-{\alpha }_n)T{P}_C(x_n-{\lambda }_{n}A{x}_n),n\ge 1$$

(1.3)

and obtained a strong convergence theorem in a Hilbert space.

Let $\{T_n\}$ be a sequence of nonexpansive mappings of C into itself, and let $\{{\lambda }_n\}$ be a sequence of nonnegative numbers in $[0,1]$. For each $n\ge 1$, define a mapping $W_n$ of C into itself as follows:

$$\begin{array}{r}{U}_{n,n+1}=I,\\ U_{n,n}={\lambda }_n{T}_n{U}_{n,n+1}+(1-{\lambda }_n)I,\\ U_{n,n-1}={\lambda }_{n-1}{T}_{n-1}{U}_{n,n}+(1-{\lambda }_{n-1})I,\\ ⋮\\ U_{n,k}={\lambda }_k{T}_k{U}_{n,k+1}+(1-{\lambda }_k)I,\\ U_{n,k-1}={\lambda }_{k-1}{T}_{k-1}{U}_{n,k}+(1-{\lambda }_{k-1})I,\\ ⋮\\ U_{n,2}={\lambda }_2{T}_2{U}_{n,3}+(1-{\lambda }_2)I,\\ W_n=U_{n,1}={\lambda }_1{T}_1{U}_{n,2}+(1-{\lambda }_1)I.\end{array}$$

(1.4)

Such a mapping $W_n$ is called the W-mapping generated by $T_1,T_2,\dots ,T_n$ and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$. The concept of W-mapping was introduced in [6, 7] and [8].

In 2008, Ceng and Yao [9] introduced the hybrid iterative scheme

$$\{\begin{array}{c}{x}_0\in C\text{arbitrary},\hfill \\ \theta (y_n,x)+\phi (x)-\phi (y_n)+\frac{1}{r}〈K^{\prime }(y_n)-K^{\prime }(x_n),\eta (x,y_n)〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ x_{n+1}={\alpha }_{n}f(W_n{x}_n)+{\beta }_n{x}_n+{\gamma }_n{W}_n{y}_n,\hfill \end{array}$$

(1.5)

where $K^{\prime }(x)$ is the Fréchet derivative of K at x. They proved the sequences $\{x_n\}$ and $\{y_n\}$ generated by the hybrid iterative scheme (1.5) converge strongly to a common element of the set of solutions of MEP and the set of common fixed points of finitely many nonexpansive mappings.

Recall the mapping B is said to be relaxed $(\xi ,\nu )$-cocoercive, if there exist two constants $\xi ,\nu >0$ such that

$$〈Bx-By,x-y〉\ge -\xi {\parallel Bx-By\parallel }^2+\nu {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

This class of mappings has been considered by many authors; for example, [10, 11].

In this paper, motivated and inspired by Ceng and Yao [9], Iiduka and Takahashi [5], Jaiboon and Kumam [12], Kim [13], Marino and Xu [3] and Saeidi [14], we introduce a new iterative scheme:

$$\{\begin{array}{c}{x}_0\in C\text{arbitrary},\hfill \\ \theta (z_n,x)+\phi (x)-\phi (z_n)+\frac{1}{r_n}〈K^{\prime }(z_n)-K^{\prime }(x_n),\eta (x,z_n)〉\ge 0,\hfill \\ y_n=(1-{\gamma }_n)x_n+{\gamma }_n{T}_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n,\hfill \\ x_{n+1}={\alpha }_n\gamma f(W_n{x}_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}A)T_{{\mu }_n}{W}_n{y}_n,\hfill \end{array}$$

(1.6)

for all $x\in C$, $n\ge 0$, for finding a common element of the set of solutions of a mixed equilibrium problem for an equilibrium bifunction, the set of fixed points of an infinite family of nonexpansive mappings, the set of solutions of some variational inequality problem and the set of fixed points of a left amenable semigroup $\{T_t:t\in S\}$ of nonexpansive mappings with respect to W-mappings and a left regular sequence $\{{\mu }_n\}$ of means defined on an appropriate space of bounded real-valued functions of the semigroup S. Furthermore, we prove that the proposed iterative scheme (1.6) converges strongly to a common element of the above four sets. Our result extends and improves the corresponding results of many others.

2 Preliminaries

Let S be a semigroup. We denote by $B(S)$ the space of all bounded real-valued functions defined on S with supremum norm. For each $s\in S$, we define the left and right translation operators $l_s$ and $r_s$ on $B(S)$ by

$$(l_{s}f)(t)=f(st)\text{and}(r_{s}f)(t)=f(ts)$$

for each $t\in S$ and $f\in B(S)$, respectively. Let X be a subspace of $B(S)$ containing 1. An element μ in the dual space $X^{\ast }$ of X is said to be a mean on X if $\parallel \mu \parallel =\mu (1)=1$. For $s\in S$, we can define a point evaluation ${\delta }_s$ by ${\delta }_s(f)=f(s)$ for each $f\in X$. It is well known that μ is a mean on X if and only if

$$\underset{s\in S}{inf}f(s)\le \mu (f)\le \underset{s\in S}{sup}f(s)$$

for each $f\in X$.

Let X be a translation invariant subspace of $B(S)$ (i.e., $l_{s}X\subset X$ and $r_{s}X\subset X$ for each $s\in S$) containing 1. Then a mean μ on X is said to be left invariant (resp. right invariant) if

$$\mu (l_{s}f)=\mu (f)(\text{resp. }\mu (r_{s}f)=\mu (f))$$

for each $s\in S$ and $f\in X$. A mean μ on X is said to be invariant if μ is both left and right invariant [15–17]. X is said to be left (resp. right) amenable if X has a left (resp. right) invariant mean. X is amenable if X is left and right amenable. Moreover, $B(S)$ is amenable when S is a commutative semigroup or a solvable group. However, the free group or semigroup of two generators is not left or right amenable. In this case, we say that the semigroup S is an amenable semigroup (see [18, 19]). A semigroup S is left reversible if S has the finite intersection property for right ideals. Every left reversible semigroup S, $\mathit{WAP}(S)$ the space of weakly almost period functions on S has a left invariant mean. If S is both left and right reversible, then $\mathit{WAP}(S)$ has an invariant mean. Each group or amenable semigroup is left and right reversible (see [20, 21]).

A net $\{{\mu }_{\alpha }\}$ of means on X is said to be asymptotically left (resp. right) invariant if

$$\underset{\alpha }{lim}({\mu }_{\alpha }(l_{s}f)-{\mu }_{\alpha }(f))=0(\text{resp. }\underset{\alpha }{lim}({\mu }_{\alpha }(r_{s}f)-{\mu }_{\alpha }(f))=0)$$

for each $f\in X$ and $s\in S$, and it is said to be left (resp. right) strongly asymptotically invariant (or strong regular) if

$$\underset{\alpha }{lim}\parallel l_s^{\ast }{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0(\text{resp. }\underset{\alpha }{lim}\parallel r_s^{\ast }{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0)$$

for each $s\in S$, where $l_s^{\ast }$ and $r_s^{\ast }$ are the adjoint operators of $l_s$ and $r_s$, respectively. Such nets were first studied by Day in [18] where they were called weak^∗ invariant and norm invariant, respectively.

It is easy to see that if a semigroup S is left (resp. right) amenable, then the semigroup $S^{\prime }=S\cup \{e\}$, where $e{s}^{\prime }=s^{\prime }e=s^{\prime }$ for all $s^{\prime }\in S$, is also left (resp. right) amenable and conversely.

Let S be a semigroup, and let C be a closed and convex subset of H. Let $F(T)$ denote the fixed point set of T. Then $\mathrm{\Im }=\{T_s:s\in S\}$ is called a representation of S as nonexpansive mappings on C if $T_s$ is nonexpansive with $T_e=I$ and $T_{st}=T_s{T}_t$ for each $s,t\in S$ (cf. [22–30]). We denote by $F(\mathrm{\Im })$ the set of common fixed points of $\{T_s:s\in S\}$, i.e.,

$$F(\mathrm{\Im })=\bigcap _{s\in S}F(T_s)=\bigcap _{s\in S}\{x\in C:T_{s}x=x\}.$$

Lemma 2.1 ([19, 31])

Let S be a semigroup and C be a closed convex subset of a Hilbert space H. Let $\mathrm{\Im }=\{T_s:s\in S\}$ be a nonexpansive semigroup on C such that $\{T_{t}u:t\in S\}$ is bounded for some $u\in C$, let X be a subspace of $B(S)$ such that $1\in X$ and the mapping $t\mapsto 〈T_{t}x,y〉$ is an element of X for each $x\in C$ and $y\in H$, and μ be a mean on X. If we write $T_{\mu }x$ instead of $\int T_{t}xd\mu (t)$, then the following hold:

• (i) $T_{\mu }$ is a nonexpansive mapping from C into itself,

• (ii) $T_{\mu }x=x$ for each $x\in F(\mathrm{\Im })$,

• (iii) $T_{\mu }x\in \overline{co}\{T_{t}x:t\in S\}$ for each $x\in C$, where $\overline{co}A$ is the closed convex hull of A.

Let C be a nonempty subset of a Hilbert space H and $T:C\to H$ be a mapping. Then T is said to be demiclosed at $v\in H$ if, for any sequence $\{x_n\}$ in C, the following implication holds:

$$x_n\rightharpoonup u\in C\text{and}T{x}_n\to v\text{imply}Tu=v,$$

where → (resp. ⇀) denotes strong (resp. weak) convergence.

Lemma 2.2 ([32])

Let C be a nonempty closed convex subset of a Hilbert space H and suppose that $T:C\to H$ is nonexpansive. Then, the mapping $I-T$ is demiclosed at zero.

Let C be a nonempty subset of a normed space E, and let $x\in E$. An element $y_0\in C$ is said to be the best approximation to x if

$$\parallel x-y_0\parallel =d(x,C),$$

where $d(x,C)={inf}_{y\in C}\parallel x-y\parallel$. The number $d(x,C)$ is called the distance from x to C. Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H. Then, for any $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

The mapping $P_C$ is called the metric projection of H onto C. It is well known that $P_C$ is a nonexpansive mapping of H onto C and satisfies

$$〈x-y,P_{C}x-P_{C}y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2$$

for every $x,y\in H$. Moreover, $P_{C}x$ is characterized by the following properties: $P_{C}x\in C$ and for all $x\in H$, $y\in C$,

$$〈x-P_{C}x,y-P_{C}x〉\le 0$$

(2.1)

and

$${\parallel x-y\parallel }^2\ge {\parallel x-P_{C}x\parallel }^2+{\parallel y-P_{C}x\parallel }^2.$$

It is easy to see that the following is true:

$$u\in \mathit{VI}(C,B)⟺u=P_C(u-\lambda Bu),\lambda >0.$$

(2.2)

In this paper, for solving the mixed equilibrium problems for an equilibrium bifunction $\theta :C\times C\to \mathbb{R}$, we assume that θ satisfies the following conditions:

(E1) $\theta (x,x)=0$ for all $x\in C$;

(E2) θ is monotone, i.e., $\theta (x,y)+\theta (y,x)\le 0$ for all $x,y\in C$;

(E3) for each $x,y,z\in C$,

$$\underset{t\downarrow 0}{lim}\theta (tz+(1-t)x,y)\le \theta (x,y);$$

(E4) for each $x\in C$, the function $y\mapsto \theta (x,y)$ is convex and lower semicontinuous.

Definition 2.1 (1) Let $F:C\to H$ and $\eta :C\times C\to H$ be two mappings. Then F is called:

• (i) η-monotone if

  $$〈F(x)-F(y),\eta (x,y)〉\ge 0,\mathrm{\forall }x,y\in C,$$

• (ii) η-strongly monotone with constant α if there exists a constant $\alpha >0$ such that

  $$〈F(x)-F(y),\eta (x,y)〉\ge \alpha {\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C,$$

• (iii) Lipschitz continuous with constant β if there exists a constant $\beta >0$ such that

  $$\parallel F(x)-F(y)\parallel \le \beta \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

If $\eta (x,y)=x-y$, for all $x,y\in C$, then the definitions (i) and (ii) reduce to the definition of monotonicity and strong monotonicity, respectively.

• (2) A mapping $\eta :C\times C\to H$ is called Lipschitz continuous with constant λ if there exists a constant $\lambda >0$ such that

  $$\parallel \eta (x,y)\parallel \le \lambda \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

• (3) A differentiable function $K:C\to \mathbb{R}$ on a convex set C is called:

• (i) η-convex [33] if

  $$K(y)-K(x)\ge 〈K^{\prime }(x),\eta (y,x)〉,\mathrm{\forall }x,y\in C,$$

where $K^{\prime }(x)$ is the Fréchet derivative of K at x,

• (ii) η-strongly convex with constant σ [34] if there exists a constant $\sigma >0$ such that

  $$K(y)-K(x)-〈K^{\prime }(x),\eta (y,x)〉\ge \frac{\sigma }{2}{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

• (4) A mapping $F:C\to \mathbb{R}$ is called sequentially continuous at $x_0$ [35], if $F(x_n)\to F(x_0)$ for each sequence $\{x_n\}$ satisfying $x_n\rightharpoonup x_0$. F is called sequentially continuous on C if it is sequentially continuous at each point of C.

Lemma 2.3 ([9])

Let $K:C\to \mathbb{R}$ be differentiable η-strongly convex with a constant $\sigma >0$, and let $\eta :C\times C\to H$ be a mapping such that $\eta (x,y)+\eta (y,x)=0$ for all $x,y\in C$. Then $K^{\prime }:C\to H$ is η-strongly monotone with constant $\sigma >0$.

Lemma 2.4 ([36])

A Hilbert space H is said to satisfy Opial’s condition if for each sequence $\{x_n\}$ in H, the condition $x_n\rightharpoonup x$ implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel$$

for all $y\in H$ with $y\ne x$.

Lemma 2.5 Let H be a real Hilbert space. Then

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉,$$

for all $x,y\in H$.

Let C be a nonempty closed convex subset of a real Hilbert space H, $\phi :C\to \mathbb{R}$ be a real-valued function and $\theta :C\times C\to \mathbb{R}$ be an equilibrium bifunction. Let r be a positive parameter. For a given point $x\in C$, consider the auxiliary problem for the mixed equilibrium problem (for short, $\mathit{MEP}(x,r)$) which consists of finding $y\in C$ such that

$$\theta (y,z)+\phi (z)-\phi (y)+\frac{1}{r}〈K^{\prime }(y)-K^{\prime }(x),\eta (z,y)〉\ge 0,\mathrm{\forall }z\in C,$$

where $\eta :C\times C\to H$ and $K^{\prime }(x)$ is the Fréchet derivative of a functional $K:C\to \mathbb{R}$ at x. Let $S_r:C\to C$ be the mapping such that for each $x\in C$, $S_r(x)$ is the solution set of $\mathit{MEP}(x,r)$, i.e.,

$$\begin{array}{rcl}{S}_r(x)& =& \{y\in C:\theta (y,z)+\phi (z)-\phi (y)\\ +\frac{1}{r}〈K^{\prime }(y)-K^{\prime }(x),\eta (z,y)〉\ge 0,\mathrm{\forall }z\in C\}\end{array}$$

(2.3)

for all $x\in C$.

We first need the following important and interesting result.

Lemma 2.6 ([9])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex functional. Let $\theta :C\times C\to \mathbb{R}$ be an equilibrium bifunction satisfying conditions (E1)-(E4). Assume that

• (i) $\eta :C\times C\to H$ is Lipschitz continuous with constant $\lambda >0$ such that

• (a) $\eta (x,y)+\eta (y,x)=0$, $\mathrm{\forall }x,y\in C$,

• (b) for each fixed $y\in C$, $x\mapsto \eta (y,x)$ is sequentially continuous from the weak topology to the weak topology,

• (ii) $K:C\to \mathbb{R}$ is η-strongly convex with constant $\sigma >0$ and its derivative $K^{\prime }$ is sequentially continuous from the weak topology to the strong topology,

• (iii) for each $x\in C$, there exist a bounded subset $D_x\subseteq C$ and $z_x\in C$ such that for any $y\in C\setminus D_x$,

  $$\theta (y,z_x)+\phi (z_x)-\phi (y)+\frac{1}{r}〈K^{\prime }(y)-K^{\prime }(x),\eta (z_x,y)〉<0.$$

Then the following hold:

• (1) $S_r$ is single-valued;

• (2) $S_r$ is a firmly nonexpansive-type mapping, i.e., for all $x,y\in H$,

  $${\parallel S_{r}x-S_{r}y\parallel }^2\le 〈S_{r}x-S_{r}y,x-y〉;$$

• (3) (j) $S_r$ is nonexpansive if $K^{\prime }$ is Lipschitz continuous with constant $\nu >0$ such that $\sigma \ge \lambda \nu$;

(jj) $〈K^{\prime }(x_1)-K^{\prime }(x_2),\eta (u_1,u_2)〉\ge 〈K^{\prime }(u_1)-K^{\prime }(u_2),\eta (u_1,u_2)〉$, $\mathrm{\forall }(x_1,x_2)\in C\times C$, where $u_i=S_r(x_i)$, $i=1,2$;

• (4) $F(S_r)=\mathrm{\Omega }$;

• (5) Ω is a closed and convex subset of C.

Remark 2.1 In particular, from Lemma 2.6, whenever $K(x)=\frac{{\parallel x\parallel }^2}{2}$ and $\eta (x,y)=x-y$ for each $(x,y)\in C\times C$, then $S_r$ is firmly nonexpansive, i.e.,

$$〈x_1-x_2,S_r(x_1)-S_r(x_2)〉\ge {\parallel S_r(x_1)-S_r(x_2)\parallel }^2,\mathrm{\forall }(x_1,x_2)\in C\times C.$$

We need the following results concerning the W-mapping $W_n$.

Lemma 2.7 ([37])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots$ be nonexpansive mappings of C into H such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$ is nonempty, and let ${\lambda }_1,{\lambda }_2,\dots$ be real numbers such that $0<{\lambda }_i\le b<1$ for any $i\in \mathbb{N}$. Then, for every $x\in C$ and $k\in \mathbb{N}$, the limit ${lim}_{n\to \mathrm{\infty }}{U}_{n,k}x$ exists.

Using Lemma 2.7, one can define a mapping W of C into H as

$$Wx=\underset{n\to \mathrm{\infty }}{lim}{W}_{n}x=\underset{n\to \mathrm{\infty }}{lim}{U}_{n,1}x,$$

for every $x\in C$.

Remark 2.2 ([37])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots$ be nonexpansive mappings of C into H such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$ is nonempty, and let ${\lambda }_1,{\lambda }_2,\dots$ be real numbers such that $0<{\lambda }_i\le b<1$ for any $i\in \mathbb{N}$. Then $F(W)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$.

Remark 2.3 ([38])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots$ be nonexpansive mappings of C into H such that ${\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$ is nonempty. If $\{x_n\}$ is an arbitrary bounded sequence in C, then we have

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

Lemma 2.8 ([39])

Let $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a Hilbert space H and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$$

for all integers $n\ge 0$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Then ${lim}_{n\to \mathrm{\infty }}\parallel z_n-x_n\parallel =0$.

Lemma 2.9 ([3])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with a coefficient $\overline{\gamma }>0$ and $0<\rho \le {\parallel A\parallel }^{-1}$. Then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$.

Lemma 2.10 ([2])

Assume $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-b_n)a_n+b_n{c}_n,$$

where $\{b_n\}$ is a sequence in $(0,1)$ and $\{c_n\}$ is a sequence in ℝ such that

• (1) ${\sum }_{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$,

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{c}_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|b_n{c}_n|<\mathrm{\infty }$.

Then

$$\underset{n\to \mathrm{\infty }}{lim}{a}_n=0.$$

3 Main result: strong convergence theorems

In this section, we deal with the strong convergence of hybrid viscosity approximation scheme (1.6) for finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of an infinite family of nonexpansive mappings, the set of fixed points of a left amenable semigroup of nonexpansive mappings and the set of solutions of variational inequality in a Hilbert space.

Theorem 3.1 Let S be a semigroup, $\mathrm{\Im }=\{T_t:t\in S\}$ be a nonexpansive semigroup on H such that $F(\mathrm{\Im })\ne \mathrm{\varnothing }$, X be a left invariant subspace of $B(S)$ such that $1\in X$, and the function $t\mapsto 〈T_{t}x,y〉$ is an element of X for each $x,y\in H$. Let $\{{\mu }_n\}$ be a left strong regular sequence of means on X such that ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$. Let C be a nonempty closed convex subset of a real Hilbert space H and $\{T_i\}$ be an infinite family of nonexpansive mappings from C into itself such that $T_i(F(\mathrm{\Im })\cap \mathrm{\Omega })\subset F(\mathrm{\Im })$ for each $i\in \mathbb{N}$. Let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex functional. Let $\theta :C\times C\to \mathbb{R}$ be an equilibrium bifunction satisfying conditions (E1)-(E4), and let $T_1,T_2,\dots$ be an infinite family of nonexpansive mappings of C into H. Let $r>0$, $\gamma >0$ be two constants. Let f be a contraction of C into itself with a coefficient $\alpha \in (0,1)$, and let A be a strongly positive bounded linear operator with a coefficient $\overline{\gamma }>0$ such that $0<\alpha \gamma <\overline{\gamma }<\alpha \gamma +1$. Let $B:C\to H$ be an L-Lipschitzian and relaxed $(\xi ,\nu )$-cocoercive mapping. Suppose that $\mathcal{F}={\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n)\cap F(\mathrm{\Im })\cap \mathrm{\Omega }\cap \mathit{VI}(C,B)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ be sequences in $[0,1]$ such that ${\alpha }_n+{\beta }_n\le 1$, and let the sequence $\{{\delta }_n\}\subset (0,\mathrm{\infty })$. Assume that:

(C1) $\eta :C\times C\to H$ is Lipschitz continuous with constant $\lambda >0$ such that

• (a) $\eta (x,y)+\eta (y,x)=0$, $\mathrm{\forall }x,y\in C$,

• (b) for each fixed $y\in C$, $x\mapsto \eta (y,x)$ is sequentially continuous from the weak topology to the weak topology,

(C2) $K:C\to \mathbb{R}$ is η-strongly convex with constant $\sigma >0$ and its derivative $K^{\prime }$ is not only sequentially continuous from the weak topology to the strong topology, but also Lipschitz continuous with constant $\nu >0$ such that $\sigma \ge \lambda \nu$,

(C3) for each $x\in C$, there exist a bounded subset $D_x\subset C$ and $z_x\in C$ such that for any $y\in C\setminus D_x$,

$$\theta (y,z_x)+\phi (z_x)-\phi (y)+\frac{1}{r}〈K^{\prime }(y)-K^{\prime }(x),\eta (z_x,y)〉<0,$$

(C4)

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

• (ii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$,

$0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\gamma }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\gamma }_n<1$,

• (iii) ${lim}_{n\to \mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|=0$, $a\le {\delta }_n\le b$ for some a, b with $0\le a\le b\le \frac{2(\nu -\xi L^2)}{L^2}$,

• (iv) ${lim}_{n\to \mathrm{\infty }}|{\gamma }_{n+1}-{\gamma }_n|=0$,

(C5) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$.

Given $x_0\in C$ is arbitrary, then the sequences $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ generated iteratively by (1.6) converge strongly to $x^{\ast }\in \mathcal{F}$, where $x^{\ast }=P_{\mathcal{F}}(\gamma f+(I-A))x^{\ast }$, which solves the following variational inequality:

$$〈(\gamma f-A)x^{\ast },x-x^{\ast }〉\le 0,\mathrm{\forall }x\in \mathcal{F}.$$

Lemma 3.1 $\parallel (1-{\beta }_n)I-{\alpha }_{n}A\parallel \le 1-{\beta }_n-{\alpha }_n\overline{\gamma }$.

Proof Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, we may assume, without loss of generality, that ${\alpha }_n\le (1-{\beta }_n){\parallel A\parallel }^{-1}$. Since A is a linear bounded self-adjoint operator on H, we have

$$\parallel A\parallel =sup\left\{\right|〈Ax,x〉|:x\in H,\parallel x\parallel =1\}.$$

Observe that

$$\begin{array}{rcl}〈((1-{\beta }_n)I-{\alpha }_{n}A)x,x〉& =& 1-{\beta }_n-{\alpha }_n〈Ax,x〉\\ \ge & 1-{\beta }_n-{\alpha }_n\parallel A\parallel \\ \ge & 0,\end{array}$$

which shows that $(1-{\beta }_n)I-{\alpha }_{n}A$ is positive. By Lemma 2.9, we have

$$\parallel (1-{\beta }_n)I-{\alpha }_{n}A\parallel \le 1-{\beta }_n-{\alpha }_n\overline{\gamma }.$$

 □

Lemma 3.2 Let B be an L-Lipschitzian and relaxed $(\xi ,\nu )$-cocoercive mapping and ${\delta }_n\le \frac{2(\nu -\xi L^2)}{L^2}$, then

$$\parallel (I-{\delta }_{n}B)x-(I-{\delta }_{n}B)y\parallel \le \parallel x-y\parallel ,$$

for all $x,y\in C$.

Proof Since B is an L-Lipschitzian and relaxed $(\xi ,\nu )$-cocoercive mapping and ${\delta }_n\le \frac{2(\nu -\xi L^2)}{L^2}$, we have

$$\begin{array}{c}{\parallel (I-{\delta }_{n}B)x-(I-{\delta }_{n}B)y\parallel }^2\hfill \\ ={\parallel x-y\parallel }^2-2{\delta }_n〈Bx-By,x-y〉+{\delta }_n^2{\parallel Bx-By\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2-2{\delta }_n(-\xi {\parallel Bx-By\parallel }^2+\nu {\parallel x-y\parallel }^2)+{\delta }_n^2{L}^2{\parallel x-y\parallel }^2\hfill \\ \le {\parallel x-y\parallel }^2+2{\delta }_n\xi L^2{\parallel x-y\parallel }^2-2{\delta }_n\nu {\parallel x-y\parallel }^2+{\delta }_n^2{L}^2{\parallel x-y\parallel }^2\hfill \\ =(1+2{\delta }_n\xi L^2-2{\delta }_n\nu +{\delta }_n^2{L}^2){\parallel x-y\parallel }^2\le {\parallel x-y\parallel }^2,\hfill \end{array}$$

for all $x,y\in C$. Thus,

$$\parallel (I-{\delta }_{n}B)x-(I-{\delta }_{n}B)y\parallel \le \parallel x-y\parallel ,$$

for all $x,y\in C$. □

Lemma 3.3 $\parallel z_n-p\parallel \le \parallel x_n-p\parallel$, $\mathrm{\forall }p\in \mathcal{F}$.

Proof From (2.3), we note that $z_n=S_{r_n}{x}_n$. From Lemma 2.6, we get

$$\parallel z_n-p\parallel \le \parallel S_{r_n}{x}_n-S_{r_n}p\parallel \le \parallel x_n-p\parallel$$

for all $p\in \mathcal{F}$. □

Lemma 3.4 $\{x_n\}$, $\{y_n\}$, $\{z_n\}$, $\{W_n{y}_n\}$, $\{W_n{x}_n\}$ and $\{f(W_n{x}_n)\}$ are all bounded.

Proof Let $p\in \mathcal{F}$. Since $p\in \mathit{VI}(C,B)$, from (2.2), we get $p=P_C(I-{\delta }_{n}B)p$. From Lemma 2.1, Lemma 3.3 and $W_n$, $P_C$ being nonexpansive, we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & =& \parallel (1-{\gamma }_n)x_n+{\gamma }_n{T}_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n-p\parallel \\ \le & (1-{\gamma }_n)\parallel x_n-p\parallel +{\gamma }_n\parallel T_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n-p\parallel \\ \le & (1-{\gamma }_n)\parallel x_n-p\parallel +{\gamma }_n\parallel (I-{\delta }_{n}B)z_n-(I-{\delta }_{n}B)p\parallel \\ \le & (1-{\gamma }_n)\parallel x_n-p\parallel +{\gamma }_n\parallel z_n-p\parallel \\ \le & \parallel x_n-p\parallel .\end{array}$$

From (1.6) and Lemma 3.1, we obtain

(3.1)

for all $n\ge 0$. It follows by mathematical induction that

$$\parallel x_{n+1}-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel \gamma f(p)-Ap\parallel }{\overline{\gamma }-\gamma \alpha }\},n\ge 0.$$

Therefore, $\{x_n\}$ is bounded. We also deduce that $\{y_n\}$, $\{z_n\}$, $\{W_n{y}_n\}$, $\{W_n{x}_n\}$ and $\{f(W_n{x}_n)\}$ are all bounded. □

Lemma 3.5 Let the mapping $W_n$ be generated iteratively by (1.5). If $\{{\omega }_n\}$ is a bounded sequence in H, then

• (1) ${lim}_{n\to \mathrm{\infty }}\parallel W_{n+1}{\omega }_n-W_n{\omega }_n\parallel =0$.

• (2) ${lim}_{n\to \mathrm{\infty }}\parallel T_{{\mu }_{n+1}}{\omega }_n-T_{{\mu }_n}{\omega }_n\parallel =0$.

Proof (1) We shall use M to denote the possible different constants appearing in the following argument. From (1.5), since $T_i$ and $U_{n,i}$ are nonexpansive, we have

$$\begin{array}{rcl}\parallel W_{n+1}{\omega }_n-W_n{\omega }_n\parallel & =& \parallel {\lambda }_1{T}_1{U}_{n+1,2}{\omega }_n+(1-{\lambda }_1){\omega }_n-{\lambda }_1{T}_1{U}_{n,2}{\omega }_n-(1-{\lambda }_1){\omega }_n\parallel \\ \le & {\lambda }_1\parallel U_{n+1,2}{\omega }_n-U_{n,2}{\omega }_n\parallel \\ =& {\lambda }_1\parallel {\lambda }_2{T}_2{U}_{n+1,3}{\omega }_n+(1-{\lambda }_2){\omega }_n-{\lambda }_2{T}_2{U}_{n,3}{\omega }_n-(1-{\lambda }_2){\omega }_n\parallel \\ \le & {\lambda }_1{\lambda }_2\parallel U_{n+1,3}{\omega }_n-U_{n,3}{\omega }_n\parallel \\ \le & {\lambda }_1{\lambda }_2\cdots {\lambda }_n\parallel U_{n+1,n+1}{\omega }_n-U_{n,n+1}{\omega }_n\parallel \\ \le & M\prod _{i=1}^n{\lambda }_i,\end{array}$$

which implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel W_{n+1}{\omega }_n-W_n{\omega }_n\parallel =0.$$

• (2) Let $q\in F(\mathrm{\Im })$. Then $\parallel T_s{\omega }_n-q\parallel \le \parallel {\omega }_n-q\parallel$. Also, we have

  $$\parallel T_s{\omega }_n\parallel \le \parallel {\omega }_n-q\parallel +\parallel q\parallel$$

for all $s\in S$ and $n\ge 0$. Since $\{{\omega }_n\}$ is bounded and ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$, we get

$$\begin{array}{rcl}\parallel T_{{\mu }_{n+1}}{\omega }_n-T_{{\mu }_n}{\omega }_n\parallel & =& sup\left\{\right|〈T_{{\mu }_{n+1}}{\omega }_n-T_{{\mu }_n}{\omega }_n,z〉|:\parallel z\parallel =1\}\\ =& sup\left\{\right|{\mu }_{n+1}(s)〈T_s{\omega }_n,z〉-{\mu }_n(s)〈T_s{\omega }_n,z〉|:\parallel z\parallel =1\}\\ \le & \parallel {\mu }_{n+1}-{\mu }_n\parallel \cdot \underset{s\in S}{sup}\parallel T_s{\omega }_n\parallel \\ \le & \parallel {\mu }_{n+1}-{\mu }_n\parallel (\parallel {\omega }_n-q\parallel +\parallel q\parallel )\\ ⟶& 0.\end{array}$$

 □

Lemma 3.6 ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel z_{n+1}-z_n\parallel =0$.

Proof Define a sequence $\{u_n\}$ by

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)u_n$$

for all $n\ge 0$. Observe that from the definition of $u_n$, we get

$$\begin{array}{rcl}{u}_{n+1}-u_n& =& \frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}\gamma f(W_{n+1}{x}_{n+1})+((1-{\beta }_{n+1})I-{\alpha }_{n+1}A)T_{{\mu }_{n+1}}{W}_{n+1}{y}_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_n\gamma f(W_n{x}_n)+((1-{\beta }_n)I-{\alpha }_{n}A)T_{{\mu }_n}{W}_n{y}_n}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\gamma f(W_{n+1}{x}_{n+1})-\frac{{\alpha }_n}{1-{\beta }_n}\gamma f(W_n{x}_n)+T_{{\mu }_{n+1}}{W}_{n+1}{y}_{n+1}\\ -T_{{\mu }_n}{W}_n{y}_n+\frac{{\alpha }_n}{1-{\beta }_n}A{T}_{{\mu }_n}{W}_n{y}_n-\frac{{\alpha }_{n+1}}{1-{\beta }_n}A{T}_{{\mu }_{n+1}}{W}_{n+1}{y}_{n+1}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\gamma f(W_{n+1}{x}_{n+1})-A{T}_{{\mu }_{n+1}}{W}_{n+1}{y}_{n+1})\\ +\frac{{\alpha }_n}{1-{\beta }_n}(A{T}_{{\mu }_n}{W}_n{y}_n-\gamma f(W_n{x}_n))+T_{{\mu }_{n+1}}{W}_{n+1}{y}_{n+1}\\ -T_{{\mu }_{n+1}}{W}_{n+1}{y}_n+T_{{\mu }_{n+1}}{W}_{n+1}{y}_n-T_{{\mu }_n}{W}_n{y}_n.\end{array}$$

(3.2)

From (2.3), we note that $z_n=S_{r_n}{x}_n$ and $z_{n+1}=S_{r_{n+1}}{x}_{n+1}$, we have

(3.3)

(3.4)

for all $x\in C$. Putting $x=z_{n+1}$ in (3.3) and $x=z_n$ in (3.4), we have

(3.5)

(3.6)

After multiplying (3.5) and (3.6) by $r_n$ and adding them together, we obtain

$$〈\eta (z_{n+1},z_n),K^{\prime }(z_n)-K^{\prime }(x_n)-\frac{r_n}{r_{n+1}}(K^{\prime }(z_{n+1})-K^{\prime }(x_{n+1}))〉\ge 0.$$

Hence,

$$\begin{array}{c}〈\eta (z_{n+1},z_n),K^{\prime }(z_n)-K^{\prime }(z_{n+1})+K^{\prime }(x_{n+1})-K^{\prime }(x_n)\hfill \\ +(1-\frac{r_n}{r_{n+1}})(K^{\prime }(z_{n+1})-K^{\prime }(x_{n+1}))〉\ge 0.\hfill \end{array}$$

Then, by Lemma 2.3, we have

$$\begin{array}{c}〈\eta (z_{n+1},z_n),K^{\prime }(x_{n+1})-K^{\prime }(x_n)+(1-\frac{r_n}{r_{n+1}})(K^{\prime }(z_{n+1})-K^{\prime }(x_{n+1}))〉\hfill \\ \ge 〈\eta (z_n,z_{n+1}),K^{\prime }(z_n)-K^{\prime }(z_{n+1})〉\ge \sigma {\parallel z_n-z_{n+1}\parallel }^2,\hfill \end{array}$$

and hence

$$\begin{array}{c}\sigma {\parallel z_{n+1}-z_n\parallel }^2\hfill \\ \le \parallel \eta (z_{n+1},z_n)\parallel (\parallel K^{\prime }(x_{n+1})-K^{\prime }(x_n)\parallel +(1-\frac{r_n}{r_{n+1}})\parallel K^{\prime }(z_{n+1})-K^{\prime }(x_{n+1})\parallel )\hfill \\ \le \lambda \parallel z_{n+1}-z_n\parallel (\nu \parallel x_{n+1}-x_n\parallel +(1-\frac{r_n}{r_{n+1}})\nu \parallel z_{n+1}-x_{n+1}\parallel ).\hfill \end{array}$$

Without loss of generality, we assume that there exists a real number k such that $r_n>k>0$ for all $n\ge 0$, we have

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \frac{\lambda \nu }{\sigma }\parallel x_{n+1}-x_n\parallel +\frac{\nu }{\sigma }\frac{1}{k}|r_{n+1}-r_n|\parallel z_{n+1}-x_{n+1}\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\frac{\nu }{\sigma k}|r_{n+1}-r_n|M^{\prime },\end{array}$$

(3.7)

where $M^{\prime }=sup\{\parallel z_n-x_n\parallel :n\ge 0\}$.

Setting $v_n=P_C(I-{\delta }_{n}B)z_n$ for all $n\ge 0$, from Lemma 3.2, we have

$$\begin{array}{rcl}\parallel v_{n+1}-v_n\parallel & \le & \parallel P_C(I-{\delta }_{n+1}B)z_{n+1}-P_C(I-{\delta }_{n}B)z_n\parallel \\ \le & \parallel (I-{\delta }_{n+1}B)z_{n+1}-(I-{\delta }_{n}B)z_n\parallel \\ \le & \parallel (I-{\delta }_{n+1}B)z_{n+1}-(I-{\delta }_{n+1}B)z_n\parallel +\parallel ({\delta }_{n+1}-{\delta }_n)B{z}_n\parallel \\ \le & \parallel z_{n+1}-z_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel B{z}_n\parallel .\end{array}$$

From (3.7), we get

(3.8)

Also, we have

(3.9)

From (3.2), we obtain

(3.10)

Combining (3.8), (3.9) and (3.10), we obtain

(3.11)

Thus, it follows from (3.11), Lemma 3.4, Lemma 3.5 and condition (C4) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel u_{n+1}-u_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

By Lemma 2.8, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x_n\parallel =0.$$

Consequently, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel u_n-x_n\parallel =0.$$

(3.12)

From (3.7), we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_{n+1}-z_n\parallel =0.$$

 □

Lemma 3.7 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_t{x}_n\parallel =0$ for all $t\in S$.

Proof Let $p\in \mathcal{F}$ and put

$$M_0=max\{\parallel x_0-p\parallel ,\frac{\parallel \gamma f(p)-Ap\parallel }{\overline{\gamma }-\gamma \alpha }\}.$$

Set $D=\{y\in H:\parallel y-p\parallel \le M_0\}$. We remark that D is a bounded closed convex set, $\{x_n\},\{y_n\},\{z_n\}\subset D$ and it is invariant under ℑ and $W_n$ for all $n\in \mathbb{N}$. We will show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{y\in D}{sup}\parallel T_{{\mu }_n}y-T_t{T}_{{\mu }_n}y\parallel =0,$$

(3.13)

for all $t\in S$. Let $\epsilon >0$, by [40] (Theorem 1.2), there exists $\delta >0$ such that

$$\overline{co}{F}_{\delta }(T_t;D)+B_{\delta }\subset F_{\epsilon }(T_t;D)$$

(3.14)

for all $t\in S$. By [40] (Corollary 1.1), there exists a natural number N such that

$$\parallel \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y-T_t\left(\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\right)\parallel \le \delta ,$$

(3.15)

for all $t,s\in S$ and $y\in D$. Since $\{{\mu }_n\}$ is left strong regular, there exists $n_0\in \mathbb{N}$ such that $\parallel {\mu }_n-l_{t^i}^{\ast }{\mu }_n\parallel \le \frac{\delta }{M_0+\parallel p\parallel }$ for $n\ge n_0$ and $i=1,2,\dots ,N$. Then we have

(3.16)

for all $n\ge n_0$. By Lemma 2.1, we have

$$\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}yd{\mu }_n(s)\in \overline{co}\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^i}(T_{s}y):s\in S\}.$$

(3.17)

It follows from (3.14)-(3.17) that

$$\begin{array}{rcl}{T}_{{\mu }_n}y& =& \int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}yd{\mu }_n(s)+(T_{{\mu }_n}y-\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}yd{\mu }_n(s))\\ \in & \overline{co}\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y:s\in S\}+B_{\delta }\\ \subset & \overline{co}{F}_{\delta }(T_t;D)+B_{\delta }\\ \subset & T_{\epsilon }(T_t;D),\end{array}$$

for all $y\in D$ and $n\ge n_0$. Therefore,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{y\in D}{sup}\parallel T_t{T}_{{\mu }_n}y-T_{{\mu }_n}y\parallel \le \epsilon .$$

Since $\epsilon >0$ is arbitrary, we obtain (3.13). Now, let $t\in S$ and $\epsilon >0$. Then there exists $\delta >0$, which satisfies (3.14). Take $L_0=(\gamma \alpha +\parallel A\parallel )M_0+\parallel \gamma f(p)-Ap\parallel$. From ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, (3.12) and (3.13), there exists $k\in \mathbb{N}$ such that ${\alpha }_n<\frac{\delta (1-{\beta }_n)}{2L_0}$, $T_{{\mu }_n}y\in F_{\delta }(T_t;D)$ for all $y\in D$ and $\parallel x_n-x_{n+1}\parallel <\frac{\delta (1-{\beta }_n)}{2{\beta }_n}$ for all $n\ge k$. We note that

$$\begin{array}{c}\parallel {\alpha }_n(\gamma f(W_n{x}_n)-A{T}_{{\mu }_n}{W}_n{y}_n)\parallel \hfill \\ \le {\alpha }_n(\gamma \parallel f(W_n{x}_n)-f(p)\parallel +\parallel \gamma f(p)-Ap\parallel +\parallel A{T}_{{\mu }_n}{W}_n{y}_n-Ap\parallel )\hfill \\ \le {\alpha }_n(\gamma \alpha \parallel W_n{x}_n-p\parallel +\parallel \gamma f(p)-Ap\parallel +\parallel A\parallel \parallel T_{{\mu }_n}{W}_n{y}_n-p\parallel )\hfill \\ \le {\alpha }_n(\gamma \alpha \parallel x_n-p\parallel +\parallel \gamma f(p)-Ap\parallel +\parallel A\parallel \parallel y_n-p\parallel )\hfill \\ \le {\alpha }_n((\gamma \alpha +\parallel A\parallel )\parallel x_n-p\parallel +\parallel \gamma f(p)-Ap\parallel )\hfill \\ \le {\alpha }_n((\gamma \alpha +\parallel A\parallel )M_0+\parallel \gamma f(p)-Ap\parallel )\hfill \\ \le \frac{\delta (1-{\beta }_n)}{2},\hfill \end{array}$$

for all $n\ge k$. Since

$$\begin{array}{c}{x}_{n+1}-{\beta }_n{x}_{n+1}\hfill \\ =(1-{\beta }_n)T_{{\mu }_n}{W}_n{y}_n+{\beta }_n{x}_n+{\alpha }_n(\gamma f(W_n{x}_n)-A{T}_{{\mu }_n}{W}_n{y}_n)-{\beta }_n{x}_{n+1},\hfill \end{array}$$

we get

$$\begin{array}{rcl}{x}_{n+1}& =& T_{{\mu }_n}{W}_n{y}_n+\frac{{\beta }_n}{1-{\beta }_n}(x_n-x_{n+1})\\ +\frac{{\alpha }_n}{1-{\beta }_n}(\gamma f(W_n{x}_n)-A{T}_{{\mu }_n}{W}_n{y}_n)\\ \in & F_{\delta }(T_t;D)+B_{\frac{\delta }{2}}+B_{\frac{\delta }{2}}\\ \subset & F_{\delta }(T_t;D)+B_{\delta }\\ \subset & F_{\epsilon }(T_t;D),\end{array}$$

for all $n\ge k$. This shows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-T_t{x}_n\parallel \le \epsilon .$$

Since $\epsilon >0$ is arbitrary, we get ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_t{x}_n\parallel =0$. □

Lemma 3.8 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel =0$.

Proof Since $x_{n+1}={\alpha }_n\gamma f(W_n{x}_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}A)T_{{\mu }_n}{W}_n{y}_n$, we have

$$\begin{array}{rcl}\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel & \le & \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T_{{\mu }_n}{W}_n{y}_n\parallel \\ =& \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \gamma f(W_n{x}_n)-A{T}_{{\mu }_n}{W}_n{y}_n\parallel +{\beta }_n\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel ,\end{array}$$

that is,

$$\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel \le \frac{1}{1-{\beta }_n}\parallel x_n-x_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \gamma f(W_n{x}_n)-A{T}_{{\mu }_n}{W}_n{y}_n\parallel .$$

It follows from condition (C4) and Lemma 3.6 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel =0.$$

 □

Lemma 3.9 ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$.

Proof For $p\in \mathcal{F}$, since $S_r$ is firmly nonexpansive, we have

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& =& {\parallel S_{r_n}{x}_n-S_{r_n}p\parallel }^2\\ \le & 〈S_{r_n}{x}_n-S_{r_n}p,x_n-p〉=〈z_n-p,x_n-p〉\\ =& \frac{1}{2}({\parallel z_n-p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel z_n-x_n\parallel }^2)\end{array}$$

and hence

$${\parallel z_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel z_n-x_n\parallel }^2.$$

(3.18)

Note that the following equality holds:

$${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$$

for all $t\in [0,1]$ and $x,y\in H$. So, from (1.6) and (3.18), we get

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel (1-{\gamma }_n)(x_n-p)+{\gamma }_n(T_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n-p)\parallel }^2\\ \le & {\gamma }_n{\parallel T_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n-p\parallel }^2+(1-{\gamma }_n){\parallel x_n-p\parallel }^2\\ -{\gamma }_n(1-{\gamma }_n){\parallel T_{{\mu }_n}{W}_n{P}_C(I-{\delta }_{n}B)z_n-x_n\parallel }^2\\ =& {\gamma }_n{\parallel z_n-p\parallel }^2+(1-{\gamma }_n){\parallel x_n-p\parallel }^2-(1-{\gamma }_n){\parallel y_n-x_n\parallel }^2\\ \le & {\gamma }_n({\parallel x_n-p\parallel }^2-{\parallel z_n-x_n\parallel }^2)+(1-{\gamma }_n){\parallel x_n-p\parallel }^2\\ =& {\parallel x_n-p\parallel }^2-{\gamma }_n{\parallel z_n-x_n\parallel }^2.\end{array}$$

(3.19)

Therefore, from Lemma 2.5, Lemma 3.1 and (3.19), we have

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel {\alpha }_n(\gamma f(W_n{x}_n)-Ap)+{\beta }_n(x_n-T_{{\mu }_n}{W}_n{y}_n)+(I-{\alpha }_{n}A)(T_{{\mu }_n}{W}_n{y}_n-p)\parallel }^2\hfill \\ \le {\parallel {\beta }_n(x_n-T_{{\mu }_n}{W}_n{y}_n)+(I-{\alpha }_{n}A)(T_{{\mu }_n}{W}_n{y}_n-p)\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma f(W_n{x}_n)-Ap,x_{n+1}-p〉\hfill \\ \le {(\parallel I-{\alpha }_{n}A\parallel \parallel y_n-p\parallel +{\beta }_n\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel )}^2\hfill \\ +2{\alpha }_n\parallel \gamma f(W_n{x}_n)-Ap\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {((1-{\alpha }_n\overline{\gamma })\parallel y_n-p\parallel +{\beta }_n\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel )}^2\hfill \\ +2{\alpha }_n\parallel \gamma f(W_n{x}_n)-Ap\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {(1-{\alpha }_n\overline{\gamma })}^2{\parallel y_n-p\parallel }^2+{\beta }_n^2{\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel }^2\hfill \\ +2(1-{\alpha }_n\overline{\gamma }){\beta }_n\parallel y_n-p\parallel \parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel \hfill \\ +2{\alpha }_n\parallel \gamma f(W_n{x}_n)-Ap\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {(1-{\alpha }_n\overline{\gamma })}^2({\parallel x_n-p\parallel }^2-\gamma {\parallel z_n-x_n\parallel }^2)+{\beta }_n^2{\parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel }^2\hfill \\ +2(1-{\alpha }_n\overline{\gamma }){\beta }_n\parallel y_n-p\parallel \parallel x_n-T_{{\mu }_n}{W}_n{y}_n\parallel \hfill \\ +2{\alpha }_n\parallel \gamma f(W_n{x}_n)-Ap\parallel \parallel x_{n+1}-p\parallel .\hfill \end{array}$$

Then we derive

(3.20)

So, from (C4), Lemma 3.6, Lemma 3.8 and (3.20), we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-z_n\parallel =0.$$

 □

Lemma 3.10 ${lim}_{n\to \mathrm{\infty }}\parallel y_n-v_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel z_n-v_n\parallel =0$, where $v_n=P_C(I-{\delta }_{n}B)z_n$ for all $n\ge 0$.

Proof Let $p\in \mathcal{F}$. Setting $v_n=P_C(I-{\delta }_{n}B)z_n$ for all $n\ge 0$, since $p\in \mathit{VI}(C,B)$, we have $p=P_C(I-{\delta }_{n}B)p$. From the L-Lipschitzian and relaxed $(\xi ,\nu )$-cocoercive mapping on B and Lemma 3.3, we have

(3.21)

From (1.6) and (3.21), we get

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel (1-{\gamma }_n)x_n+{\gamma }_n{T}_{{\mu }_n}{W}_n{v}_n-p\parallel }^2\\ =& {\parallel (1-{\gamma }_n)(x_n-p)+{\gamma }_n(T_{{\mu }_n}{W}_n{v}_n-p)\parallel }^2\\ \le & (1-{\gamma }_n){\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel v_n-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2+{\gamma }_n(2{\delta }_n\xi +{\delta }_n^2-\frac{2{\delta }_n\nu }{L^2}){\parallel B{z}_n-Bp\parallel }^2.\end{array}$$

(3.22)

From (1.6), Lemma 3.1 and (3.22), we have

(3.23)

It follows that

$$\begin{array}{rcl}0& \le & -{(1-{\alpha }_n\overline{\gamma })}^2{\gamma }_n(2{\delta }_n\xi +{\delta }_n^2-\frac{2{\delta }_n\nu }{L^2}){\parallel B{z}_n-Bp\parallel }^2\\ \le & {\alpha }_{n}M+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le & {\alpha }_{n}M+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_{n+1}-x_n\parallel \\ ⟶& 0,\end{array}$$

which implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B{z}_n-Bp\parallel =0.$$

(3.24)

On the other hand, since $P_C$ is firmly nonexpansive, by Lemma 3.2, we have

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& =& {\parallel P_C(I-{\delta }_{n}B)z_n-P_C(I-{\delta }_{n}B)p\parallel }^2\\ \le & 〈(I-{\delta }_{n}B)z_n-(I-{\delta }_{n}B)p,v_n-p〉\\ =& \frac{1}{2}\{{\parallel (I-{\delta }_{n}B)z_n-(I-{\delta }_{n}B)p\parallel }^2+{\parallel v_n-p\parallel }^2-{\parallel z_n-v_n-{\gamma }_n(B{z}_n-Bp)\parallel }^2\}\\ \le & \frac{1}{2}\{{\parallel z_n-p\parallel }^2+{\parallel v_n-p\parallel }^2-{\parallel z_n-v_n\parallel }^2\\ +2{\gamma }_n\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel -{\gamma }_n^2{\parallel B{z}_n-Bp\parallel }^2\},\end{array}$$

which yields that

$$\begin{array}{rcl}{\parallel v_n-p\parallel }^2& \le & {\parallel z_n-p\parallel }^2-{\parallel z_n-v_n\parallel }^2+2{\gamma }_n\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel \\ \le & {\parallel x_n-p\parallel }^2-{\parallel z_n-v_n\parallel }^2+2{\gamma }_n\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel .\end{array}$$

(3.25)

Combining (3.22) and (3.25), we obtain

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & (1-{\gamma }_n){\parallel x_n-p\parallel }^2+{\gamma }_n{\parallel v_n-p\parallel }^2\\ \le & (1-{\gamma }_n){\parallel x_n-p\parallel }^2+{\gamma }_n({\parallel x_n-p\parallel }^2-{\parallel z_n-v_n\parallel }^2\\ +2{\gamma }_n\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel )\\ =& {\parallel x_n-p\parallel }^2-{\gamma }_n{\parallel z_n-v_n\parallel }^2+2{\gamma }_n\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel .\end{array}$$

(3.26)

Therefore, from (3.23) and (3.26), we get

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & {\alpha }_{n}M+{\beta }_n(1-{\alpha }_n\overline{\gamma }){\parallel x_n-p\parallel }^2+(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-p\parallel }^2\\ \le & {\alpha }_{n}M+{\beta }_n(1-{\alpha }_n\overline{\gamma }){\parallel x_n-p\parallel }^2+(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma })\\ \times ({\parallel x_n-p\parallel }^2-\gamma {\parallel z_n-v_n\parallel }^2+2{\gamma }_n^2\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel )\\ \le & {\alpha }_{n}M+{(1-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-p\parallel }^2-(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\gamma }_n{\parallel z_n-v_n\parallel }^2\\ +2(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\gamma }_n^2\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel .\end{array}$$

Hence, we have

$$\begin{array}{c}(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\gamma }_n{\parallel z_n-v_n\parallel }^2\hfill \\ \le {\alpha }_{n}M+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_{n+1}-x_n\parallel \hfill \\ +2(1-{\alpha }_n\overline{\gamma })(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\gamma }_n^2\parallel z_n-v_n\parallel \parallel B{z}_n-Bp\parallel ,\hfill \end{array}$$

which implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-v_n\parallel =0.$$

(3.27)

Observe that

$$\begin{array}{rcl}\parallel y_n-v_n\parallel & \le & \parallel y_n-x_n\parallel +\parallel x_n-z_n\parallel +\parallel z_n-v_n\parallel \\ \le & {\gamma }_n\parallel T_{{\mu }_n}{W}_n{v}_n-x_n\parallel +\parallel x_n-z_n\parallel +\parallel z_n-v_n\parallel \\ \le & {\gamma }_n(\parallel T_{{\mu }_n}{W}_n{v}_n-T_{{\mu }_n}{W}_n{y}_n\parallel +\parallel T_{{\mu }_n}{W}_n{y}_n-x_n\parallel )\\ +\parallel x_n-z_n\parallel +\parallel z_n-v_n\parallel \\ \le & {\gamma }_n(\parallel v_n-y_n\parallel +\parallel T_{{\mu }_n}{W}_n{y}_n-x_n\parallel )+\parallel x_n-z_n\parallel +\parallel z_n-v_n\parallel ,\end{array}$$

and hence

$$(1-{\gamma }_n)\parallel y_n-v_n\parallel \le {\gamma }_n\parallel T_{{\mu }_n}{W}_n{y}_n-x_n\parallel +\parallel x_n-z_n\parallel +\parallel z_n-v_n\parallel .$$

Thus, from Lemma 3.8, Lemma 3.9, (3.27) and (C4), we derive

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-v_n\parallel =0.$$

 □

Lemma 3.11 $P_{\mathcal{F}}(\gamma f+(I-A))$ is a contraction of H into itself.

Proof From Lemma 3.1, we have

$$\begin{array}{c}\parallel P_{\mathcal{F}}(\gamma f+(I-A))x-P_{\mathcal{F}}(\gamma f+(I-A))y\parallel \hfill \\ \le \parallel \gamma f(x)+(I-A)x-\gamma f(y)-(I-A)y\parallel \hfill \\ \le \gamma \parallel f(x)-f(y)\parallel +\parallel (I-A)(x-y)\parallel \hfill \\ \le \gamma \alpha \parallel x-y\parallel +(1-\overline{\gamma })\parallel x-y\parallel \hfill \\ =(1-(\overline{\gamma }-\gamma \alpha ))\parallel x-y\parallel ,\hfill \end{array}$$

for all $x,y\in H$. From the condition $\overline{\gamma }$, $0<\alpha \gamma <\overline{\gamma }<\alpha \gamma +1$, we obtain $1-(\overline{\gamma }-\gamma \alpha )\in (0,1)$. Therefore, $P_{\mathcal{F}}(\gamma f+(I-A))$ is a contraction. □

Now, we prove Theorem 3.1.

Proof of Theorem 3.1 From Lemma 3.11 and the Banach contraction principle, $P_{\mathcal{F}}(\gamma f+(I-A))$ has a unique fixed point, say $x^{\ast }\in H$. That is, $x^{\ast }=P_{\mathcal{F}}(\gamma f+(I-A))x^{\ast }$. Then, using (2.1), $x^{\ast }$ is the unique solution of the variational inequality

$$〈(\gamma f-A)x^{\ast },x-x^{\ast }〉\le 0$$

(3.28)

for all $x\in \mathcal{F}$. Now, we show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },x_n-x^{\ast }〉\le 0.$$

(3.29)

To show this, we can choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },x_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈(\gamma f-A)x^{\ast },x_{n_i}-x^{\ast }〉.$$

(3.30)

Since $\{x_{n_i}\}$ is bounded, there exists a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$ which converges weakly to z. Without loss of generality, we can assume that $x_{n_i}\rightharpoonup z$. We need to show that $z\in \mathcal{F}=F(\mathrm{\Im })\cap \mathrm{\Omega }\cap ({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\cap \mathit{VI}(C,B)$.

• (I) Since $x_{n_i}\rightharpoonup z$, by Lemma 2.2 and Lemma 3.7, we get

  $$T_{t}z=z$$

for all $t\in S$. Therefore, $z\in F(\mathrm{\Im })$.

• (II) Now, we show that $z\in \mathrm{\Omega }$. Since $z_n=S_{r_n}{x}_n$, we derive

  $$\theta (z_n,x)+\phi (x)-\phi (z_n)+\frac{1}{r_n}〈K^{\prime }(z_n)-K^{\prime }(x_n),\eta (x,z_n)〉\ge 0,$$

for all $x\in C$. From the monotonicity of θ, we have

$$\theta (x,z_n)\le -\theta (z_n,x)\le \phi (x)-\phi (z_n)+\frac{1}{r_n}〈K^{\prime }(z_n)-K^{\prime }(x_n),\eta (x,z_n)〉,$$

and hence

$$\theta (x,z_{n_i})\le \phi (x)-\phi (z_{n_i})+〈\frac{K^{\prime }(z_{n_i})-K^{\prime }(x_{n_i})}{r_{n_i}},\eta (x,z_{n_i})〉.$$

Since $\frac{K^{\prime }(z_{n_i})-K^{\prime }(x_{n_i})}{r_{n_i}}\to 0$ and $z_{n_i}\rightharpoonup z$, from the lower semicontinuity of φ and (E4), we have

$$\theta (x,z)+\phi (z)-\phi (x)\le 0,$$

for all $x\in C$. For t with $0<t\le 1$ and $x\in C$, let $x_t=tx+(1-t)z$. Since $x\in C$ and $z\in C$, we have $x_t\in C$ and

$$\theta (x_t,z)+\phi (z)-\phi (x_t)\le 0.$$

From (E1), (E4) and the convexity of φ, we get

$$\begin{array}{rcl}0& =& \theta (x_t,x_t)+\phi (x_t)-\phi (x_t)\\ \le & t\theta (x_t,x)+(1-t)\theta (x_t,z)+t\phi (x)+(1-t)\phi (z)-\phi (x_t)\\ \le & t(\theta (x_t,x)+\phi (x)-\phi (x_t)).\end{array}$$

Hence,

$$\theta (x_t,x)+\phi (x)-\phi (x_t)\ge 0,$$

for all $x\in C$. From (E3) and the lower semicontinuity of φ, we have

$$\theta (z,x)+\phi (x)-\phi (z)\ge 0,$$

for all $x\in C$. Therefore, $z\in \mathrm{\Omega }$.

• (III) We show that $z\in F(W)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$. Assume that $z\notin F(W)$, then $z\ne Wz$. Since $z\in F(\mathrm{\Im })\cap \mathrm{\Omega }$, by our assumption, we have $T_{i}z\in F(\mathrm{\Im })$, $\mathrm{\forall }i\in \mathbb{N}$ and then $W_{n}z\in F(\mathrm{\Im })$. From Lemma 2.1, we get

  $$T_{{\mu }_n}{W}_{n}z=W_{n}z,$$

  (3.31)

for all $n\in \mathbb{N}$. Since

$$\parallel y_n-x_n\parallel \le \parallel y_n-v_n\parallel +\parallel v_n-z_n\parallel +\parallel z_n-x_n\parallel ,$$

from Lemma 3.9 and Lemma 3.10, we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-x_n\parallel =0.$$

(3.32)

By Lemma 2.4, Lemma 3.8, (3.31) and (3.32), we obtain

$$\begin{array}{c}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-z\parallel \hfill \\ <\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-Wz\parallel \hfill \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\parallel x_{n_i}-T_{{\mu }_{n_i}}{W}_{n_i}{y}_{n_i}\parallel +\parallel T_{{\mu }_{n_i}}{W}_{n_i}{y}_{n_i}-T_{{\mu }_{n_i}}{W}_{n_i}{x}_{n_i}\parallel \hfill \\ +\parallel T_{{\mu }_{n_i}}{W}_{n_i}{x}_{n_i}-T_{{\mu }_{n_i}}{W}_{n_i}z\parallel +\parallel T_{{\mu }_{n_i}}{W}_{n_i}z-Wz\parallel )\hfill \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\parallel y_{n_i}-x_{n_i}\parallel +\parallel x_{n_i}-z\parallel )\hfill \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-z\parallel .\hfill \end{array}$$

This is a contraction. Therefore, $z\in F(W)={\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)$.

• (IV) We show that $z\in \mathit{VI}(C,B)$. Let $U:H\to 2^H$ be a set-valued mapping defined by

  $$Ux=\{\begin{array}{cc}Bx+N_{C}x,\hfill & \text{if }x\in C,\hfill \\ \mathrm{\varnothing },\hfill & \text{if }x\notin C,\hfill \end{array}$$

where $N_{C}x$ is the normal cone to C at $x\in C$. By assumption of B, we have

$$\begin{array}{rcl}〈Bx-By,x-y〉& \ge & -\xi {\parallel Bx-By\parallel }^2+\nu {\parallel x-y\parallel }^2\\ \ge & (\nu -\xi L^2){\parallel x-y\parallel }^2\\ \ge & 0,\end{array}$$

(3.33)

which implies that B is monotone. Thus, U is a maximal monotone. Let $(u,v)\in G(U)$. Since $v-Bu\in N_{C}u$ and $v_n=P_C(I-{\delta }_{n}B)z_n\in C$, we have

$$〈u-v_n,v-Bu〉\ge 0.$$

(3.34)

On the other hand, from (2.1), we have

$$〈u-v_n,v_n-(I-{\delta }_{n}B)z_n〉\ge 0$$

and hence

$$〈u-v_n,\frac{v_n-z_n}{{\delta }_n}+B{z}_n〉\ge 0.$$

It follows by (3.33) and (3.34) that

$$\begin{array}{rcl}〈u-v_{n_i},v〉& \ge & 〈u-v_{n_i},Bu〉\\ \ge & 〈u-v_{n_i},Bu〉-〈u-v_{n_i},\frac{v_{n_i}-z_{n_i}}{{\delta }_{n_i}}+B{z}_{n_i}〉\\ =& 〈u-v_{n_i},Bu-B{z}_{n_i}-\frac{v_{n_i}-z_{n_i}}{{\delta }_{n_i}}〉\\ =& 〈u-v_{n_i},Bu-B{v}_{n_i}〉+〈u-v_{n_i},B{v}_{n_i}-B{z}_{n_i}〉\\ -〈u-v_{n_i},\frac{v_{n_i}-z_{n_i}}{{\delta }_{n_i}}〉\\ \ge & 〈u-v_{n_i},B{v}_{n_i}-B{z}_{n_i}〉-〈u-v_{n_i},\frac{v_{n_i}-z_{n_i}}{{\delta }_{n_i}}〉.\end{array}$$

From Lemma 3.9 and Lemma 3.10, we obtain $〈u-z,v〉\ge 0$ as $i\to \mathrm{\infty }$. Since U is maximal monotone, we have $z\in U^{-1}(0)$. Therefore, $z\in \mathit{VI}(C,B)$. By (I)-(IV), $z\in \mathcal{F}=F(\mathrm{\Im })\cap \mathrm{\Omega }\cap ({\bigcap }_{n=1}^{\mathrm{\infty }}F(T_n))\cap \mathit{VI}(C,B)$. Since $x^{\ast }=P_{\mathcal{F}}(\gamma f+(I-A))x^{\ast }$, from (3.28), we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },x_n-x^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },x_{n_i}-x^{\ast }〉\\ =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },z_{n_i}-x^{\ast }〉\\ =& 〈(\gamma f-A)x^{\ast },z-x^{\ast }〉\\ \le & 0.\end{array}$$

• (V) Finally, we prove that $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ converge strongly to $x^{\ast }$. From (1.6), we obtain

  $$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& \parallel {\alpha }_n(\gamma f(W_n{x}_n)-A{x}^{\ast })+{\beta }_n(x_n-x^{\ast })\\ +((1-{\beta }_n)I-{\alpha }_{n}A)(T_{{\mu }_n}{W}_n{y}_n-x^{\ast }){\parallel }^2\\ \le & {\parallel {\beta }_n(x_n-x^{\ast })+((1-{\beta }_n)I-{\alpha }_{n}A)(T_{{\mu }_n}{W}_n{y}_n-x^{\ast })\parallel }^2\\ +2{\alpha }_n〈\gamma f(W_n{x}_n)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {({\beta }_n\parallel x_n-x^{\ast }\parallel +\parallel ((1-{\beta }_n)I-{\alpha }_{n}A)(T_{{\mu }_n}{W}_n{y}_n-x^{\ast })\parallel )}^2\\ +2{\alpha }_n\gamma 〈f(W_n{x}_n)-f\left(x^{\ast }\right),x_{n+1}-x^{\ast }〉\\ +2{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {({\beta }_n\parallel x_n-x^{\ast }\parallel +(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel y_n-x^{\ast }\parallel )}^2\\ +2{\alpha }_n\gamma \alpha \parallel x_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel +2{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\\ \le & {(1-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n\gamma \alpha ({\parallel x_n-x^{\ast }\parallel }^2+{\parallel x_{n+1}-x^{\ast }\parallel }^2)\\ +2{\alpha }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉,\end{array}$$

which implies that

$$\begin{array}{c}{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le \frac{1-2{\alpha }_n\overline{\gamma }+{({\alpha }_n\overline{\gamma })}^2+{\alpha }_n\gamma \alpha }{1-{\alpha }_n\gamma \alpha }{\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \alpha }〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ =(1-\frac{2(\overline{\gamma }-\gamma \alpha ){\alpha }_n}{1-{\alpha }_n\gamma \alpha }){\parallel x_n-x^{\ast }\parallel }^2+\frac{{({\alpha }_n\overline{\gamma })}^2}{1-{\alpha }_n\gamma \alpha }{\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \alpha }〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉\hfill \\ =(1-\frac{2(\overline{\gamma }-\gamma \alpha ){\alpha }_n}{1-{\alpha }_n\gamma \alpha }){\parallel x_n-x^{\ast }\parallel }^2\hfill \\ +\frac{2(\overline{\gamma }-\gamma \alpha ){\alpha }_n}{1-{\alpha }_n\gamma \alpha }(\frac{{\alpha }_n{\overline{\gamma }}^2}{2(\overline{\gamma }-\gamma \alpha )}{\parallel x_n-x^{\ast }\parallel }^2+\frac{1}{\overline{\gamma }-\gamma \alpha }〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_{n+1}-x^{\ast }〉).\hfill \end{array}$$

It follows that

$${\parallel x_{n+1}-x^{}}^{}$$