Fixed point solutions of generalized mixed equilibrium problems and variational inclusion problems for nonexpansive semigroups

Abstract

In this paper, we introduce a composite iterative method for solving a common element of the set of solutions of fixed points for nonexpansive semigroups, the set of solutions of generalized mixed equilibrium problems and the set of solutions of the variational inclusion for a β-inverse strongly monotone mapping in a real Hilbert space. We prove that the sequence converges strongly to a common element of the above sets under some mild conditions. Our results improve and extend the corresponding results of Kumam and Wattanawitoon (Math. Comput. Model. 53:998-1006, 2011), Shehu (Math. Comput. Model. 55:1301-1314, 2012), Plubtieng and Punpaeng (Math. Comput. Model. 48:279-286, 2008), Li et al. (Nonlinear Anal. 70:3065-3071, 2009), Plubtieng and Wangkeeree (Bull. Korean Math. Soc. 45:717-728, 2008) and some authors.

MSC:46C05, 47H09, 47H10.

1 Introduction

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. Recall that a mapping $T:C\to C$ is said to be nonexpansive if

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

We denote by $F(T)$ the set of fixed points of T. Let $P_C$ be the projection of H onto the convex subset C. Moreover, we also denote by ℝ the set of all real numbers.

A one-parameter family $\mathrm{\Im }=\{T(t):0\le t<\mathrm{\infty }\}$ is said to be a nonexpansive semigroup on C if it satisfies the following conditions:

• (1) $T(0)x=x$ for all $x\in C$;

• (2) $T(s+t)=T(s)T(t)$ for all $s,t\ge 0$;

• (3) $\parallel T(t)x-T(t)y\parallel \le \parallel x-y\parallel$ for all $x,y\in C$ and $t\ge 0$;

• (4) for all $x\in C$, $t\mapsto T(t)x$ is continuous.

We denote by $F(\mathrm{\Im })$ the set of all common fixed points of ℑ, that is, $F(\mathrm{\Im })=\{x\in C:T(t)x=x,0\le t<\mathrm{\infty }\}$. It is well known that $F(\mathrm{\Im })$ is closed and convex.

A mapping A of C into H is called monotone if $〈Ax-Ay,x-y〉\ge 0$ for all $x,y\in C$. A mapping A is called α-inverse strongly monotone if there exists a positive real number α such that $〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2$ for all $x,y\in C$. A mapping A is called α-strongly monotone if there exists a positive real number α such that $〈Ax-Ay,x-y〉\ge \alpha {\parallel x-y\parallel }^2$ for all $x,y\in C$. It is obvious that any α-inverse strongly monotone mappings A is a monotone and $\frac{1}{\alpha }$-Lipschitz continuous mapping. A linear bounded operator A is called strongly positive if there exists a constant $\overline{\gamma }>0$ with the property $〈Ax,x〉\ge \overline{\gamma }{\parallel x\parallel }^2$ for all $x\in H$.

Let $B:H\to H$ be a single-valued nonlinear mapping and $M:H\to 2^H$ be a set-valued mapping. The variational inclusion problem is to find $x\in H$ such that

$$0\in B(x)+M(x),$$

(1.1)

where 0 is the zero vector in H.

The set of solutions of (1.1) is denoted by $I(B,M)$ (see [1–3] and the reference therein).

A set-valued mapping $M:H\to 2^H$ is called monotone if for all $x,y\in H$, $f\in M(x)$ and $g\in M(y)$ imply $〈x-y,f-g〉\ge 0$. A monotone mapping M is maximal if its graph $G(M)=\{(f,x)\in H\times H:f\in M(x)\}$ of M is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping M is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for all $(y,g)\in G(M)$ imply $f\in M(x)$.

Let $M:H\to 2^H$ be a set-valued maximal monotone mapping. Then the single-valued mapping $J_{M,\lambda }:H\to H$ defined by

$$J_{M,\lambda }(x)={(I+\lambda M)}^{-1}(x),x\in H,$$

is called the resolvent operator associated with M, where λ is any positive number and I is the identity mapping. It is well known that the resolvent operator $J_{M,\lambda }$ is nonexpansive, 1-inverse strongly monotone and that a solution of problem (1.1) is a fixed point of the operator $J_{M,\lambda }(I-\lambda B)$ for all $\lambda >0$, where I denotes the identity operator on H (see [4]).

Peng and Yao [5] considered the following generalized mixed equilibrium problem of finding $x\in C$ such that

$$\theta (x,y)+\phi (y)-\phi (x)+〈Fx,y-x〉\ge 0,\mathrm{\forall }y\in C,$$

(1.2)

where $F:C\to H$ is a nonlinear mapping, $\phi :C\to \mathbb{R}$ is a function and $\theta :C\times C\to \mathbb{R}$ is a bifunction. The set of solutions of problem (1.2) is denoted by $GMEP$.

In the case of $F=0$, problem (1.2) reduces to the following mixed equilibrium problem of finding $x\in C$ such that

$$\theta (x,y)+\phi (y)-\phi (x)\ge 0,\mathrm{\forall }y\in C,$$

which was considered by Ceng and Yao [6]. $GMEP$ is denoted by $MEP$.

In the case of $\phi =0$, problem (1.2) reduces to the following generalized equilibrium problem of finding $x\in C$ such that

$$\theta (x,y)+〈Fx,y-x〉\ge 0,\mathrm{\forall }y\in C,$$

which was studied by Takahashi and Takahashi [7].

In the case of $\phi =0$ and $F=0$, problem (1.2) reduces to the equilibrium problem of finding $x\in C$ such that

$$\theta (x,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.3)

The set of solution of (1.3) is denoted by $EP(\theta )$.

In the case $\theta =0$ and $\phi =0$, problem (1.2) reduces to the classical variational inequality of finding $x\in C$ such that

$$〈Fx,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(1.4)

The set of solutions of problem (1.4) is denoted by $VI(F,C)$.

The problem (1.2) is very general in the sense that it includes, as special cases, optimization problem, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games and others (see [6, 8–10]).

Peng and Yao [5] considered iterative methods for finding a common element of the set of solutions of problem (1.2), the set of solutions of problem (1.4), and the set of fixed points of a nonexpansive mapping.

Let $G_1,G_2:C\times C\to \mathbb{R}$ be two bifunctions, $\varphi ,\psi :C\to \mathbb{R}$ be two functions and $B_1,B_2:C\to H$ be two nonlinear mappings. We consider the generalized mixed equilibria problem of finding $(\overline{x},\overline{y})\in C\times C$ such that

$$\{\begin{array}{c}{G}_1(\overline{x},x)+\varphi (x)-\varphi (\overline{x})+〈B_1\overline{y},x-\overline{x}〉+\frac{1}{{\mu }_1}〈\overline{x}-\overline{y},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ G_2(\overline{y},y)+\psi (y)-\psi (\overline{y})+〈B_2\overline{x},y-\overline{y}〉+\frac{1}{{\mu }_2}〈\overline{y}-\overline{x},y-\overline{y}〉\ge 0,\mathrm{\forall }y\in C,\hfill \end{array}$$

(1.5)

where ${\mu }_1>0$ and ${\mu }_2>0$ are two constants.

In the case $\varphi =\psi =0$, problem (1.5) reduces to the following problem of the general system of generalized equilibria of finding $(\overline{x},\overline{y})\in C\times C$ such that

$$\{\begin{array}{c}{G}_1(\overline{x},x)+〈B_1\overline{y},x-\overline{x}〉+\frac{1}{{\mu }_1}〈\overline{x}-\overline{y},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ G_2(\overline{y},y)+〈B_2\overline{x},y-\overline{y}〉+\frac{1}{{\mu }_2}〈\overline{y}-\overline{x},y-\overline{y}〉\ge 0,\mathrm{\forall }y\in C,\hfill \end{array}$$

(1.6)

which was considered by Ceng and Yao [9].

In the case $G_1=G_2=0$, problem (1.6) reduces to the following problem of the general system of variational inequalities of finding $(\overline{x},\overline{y})\in C\times C$ such that

$$\{\begin{array}{c}〈{\mu }_1{B}_1\overline{y}+\overline{x}-\overline{y},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\mu }_2{B}_2\overline{x}+\overline{y}-\overline{x},y-\overline{y}〉\ge 0,\mathrm{\forall }y\in C,\hfill \end{array}$$

(1.7)

which was considered by Ceng, Wang and Yao [11].

In particular, if $B_1=B_2=D:C\to H$ is a nonlinear mapping, then problem (1.7) reduces to the following problem of the system of variational inequalities of finding $(\overline{x},\overline{y})\in C\times C$ such that

$$\{\begin{array}{c}〈{\mu }_{1}D\overline{y}+\overline{x}-\overline{y},x-\overline{x}〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\mu }_{2}D\overline{x}+\overline{y}-\overline{x},y-\overline{y}〉\ge 0,\mathrm{\forall }y\in C,\hfill \end{array}$$

(1.8)

which was studied by Verma [12].

If $\overline{x}=\overline{y}$ in (1.8), then (1.8) reduces to the classical variational inequality (1.4).

For solving the mixed equilibrium problem, let us give the following assumptions for the bifunction θ, φ and the set C:

(H1) $\theta (x,y)=0$ for all $x\in C$;

(H2) θ is monotone, i.e., $\theta (x,y)+\theta (y,x)\le 0$ for all $x,y\in C$;

(H3) for each $x,y,z\in C$

$$\underset{t\to 0}{lim\hspace{0.17em}sup}\theta (tz+(1-t)x,y)\le \theta (x,y);$$

(H4) for each $x\in C$, $y\mapsto \theta (x,y)$ is convex and lower semicontinuous;

(A1) for each $x\in H$ and $r>0$, there exist a bounded subset $D\subset C$ and $y\in C$ such that for any $z\in C\mathrm{\setminus }D$

$$\theta (z,y)+\phi (y)-\phi (z)+\frac{1}{r}〈y-z,z-x〉<0;$$

(A2) C is a bounded set.

Recently, Shehu [13] studied the problem of finding a common element of the set of common fixed points of a one-parameter nonexpansive semigroup, the set of solutions to a variational inclusion, and the set of solutions to a generalized equilibrium problem. More precisely, the author proved the following theorem.

Theorem 1.1 Let C be a nonempty closed and convex subset of a real Hilbert space H. Let θ be a bifunction from $C\times C$ to ℝ satisfying (H1)-(H4), ψ a μ-inverse strongly monotone mapping of C into H, B an α-inverse strongly monotone mapping of C into H and $M:H\to 2^H$ a maximal monotone mapping. Let $\mathrm{\Im }=\{T(u):0\le u<\mathrm{\infty }\}$ be a one-parameter nonexpansive semigroup on H such that $\mathcal{F}=F(\mathrm{\Im })\cap I(B,M)\cap EP(\theta )\ne \varphi$ and suppose $f:H\to H$ is a contraction mapping with a constant $\gamma \in (0,1)$. Let $\{t_n\}\subset (0,\mathrm{\infty })$ be a real sequence such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$. Suppose ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{u_n\}}_{n=1}^{\mathrm{\infty }}$ are generated by $x_1\in H$,

$$\begin{array}{c}\theta (u_n,y)+〈\psi x_n,y-u_n〉+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)(\frac{1}{t_n}{\int }_0^{t_n}T(u)[{\alpha }_{n}f(x_n)+(1-{\alpha }_n)J_{M,\lambda }(u_n-\lambda A{u}_n)]du)\hfill \end{array}$$

for all $n\ge 1$, where ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ are sequences in $(0,1)$ and ${\{r_n\}}_{n=1}^{\mathrm{\infty }}\subset (0,\mathrm{\infty })$ satisfying:

• (i) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$,

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$,

• (iii) $\lambda \in (0,2\alpha ]$,

• (iv) $0<a\le r_n\le b<2\mu$, ${\sum }_{n=1}^{\mathrm{\infty }}|r_{n+1}-r_n|<\mathrm{\infty }$,

• (v) ${lim}_{n\to \mathrm{\infty }}\frac{|t_n-t_{n-1}|}{t_n}\frac{1}{{\alpha }_n(1-{\beta }_n)}=0$.

Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to z, where $z=P_{\mathcal{F}}f(z)$.

In this paper, motivated by Shehu [13], Kumam and Wattanawitoon [14], Li et al. [15], Plubtieng and Punpaeng [10], Plubtieng and Wangkeeree [16], we introduce the following general iterative scheme for finding a common element of the set of common fixed points of a one-parameter nonexpansive semigroup, the set of solutions of the generalized mixed equilibrium problem (1.2), the set of solutions to a variational inclusion (1.1), and the set of solutions of the generalized mixed equilibria problem (1.5), which solves the variational inequality

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

where $\mathcal{F}=F(\mathrm{\Im })\cap GMEP\cap \mathrm{\Omega }\cap I(B,M)$ and Ω is the set of solutions of the generalized equilibria problem (1.5).

The results obtained in this paper improve and extend the recent results announced by [10, 13–16] and many others.

2 Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. For every point $x\in H$ there exists a unique nearest point of C, denoted by $P_{C}x$ such that $\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$ for all $y\in C$. Such a $P_C$ is called the metric projection of H onto C. We know that $P_C$ is a firmly nonexpansive mapping of H onto C, i.e.,

$$〈x-y,P_{C}x-P_{C}y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2,\mathrm{\forall }x,y\in H.$$

Further, for any $x\in H$ and $z\in C$, $z=P_{C}x$ if and only if

$$〈x-z,z-y〉\ge 0,\mathrm{\forall }y\in C.$$

It is also known that H satisfies Opial’s condition [17], that is, if for each sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ in H which converges weakly to a point $x\in H$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel ,\mathrm{\forall }y\in H,y\ne x.$$

In order to prove our main results in the next section, we need the following lemmas.

Lemma 2.1 ([6])

Let C be a nonempty closed convex subset of H. Let $\theta :C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (H1)-(H4) and let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex function. For $r>0$ and $x\in H$ define a mapping

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

for all $x\in H$. Assume that either (A1) or (A2) holds.

Then the following results hold:

• (i) $T_r^{(\theta ,\phi )}(x)\ne \varphi$ for each $x\in H$ and $T_r^{(\theta ,\phi )}$ is single-valued;

• (ii) $T_r^{(\theta ,\phi )}$ is firmly nonexpansive, i.e., for any $x,y\in H$

  $${\parallel T_r^{(\theta ,\phi )}x-T_r^{(\theta ,\phi )}y\parallel }^2\le 〈T_r^{(\theta ,\phi )}x-T_r^{(\theta ,\phi )}y,x-y〉;$$

• (iii) $F(T_r^{(\theta ,\phi )})=MEP(\theta ,\phi )$;

• (iv) $MEP(\theta ,\phi )$ is closed and convex.

By similar argument as in the proof of Lemma 2.2 in [6], we have the following result.

Lemma 2.2 Let C be a nonempty closed convex subset of H. Let $G_1,G_2:C\times C\to \mathbb{R}$ be two bifunctions satisfying conditions (H1)-(H4). Let $\varphi ,\psi :C\to \mathbb{R}$ be two lower semicontinuous and convex functions with restriction (A1) or (A2) and let the mappings $B_1,B_2:C\to H$ be ${\beta }_1$-inverse strongly monotone and ${\beta }_2$-inverse strongly monotone, respectively. Let ${\mu }_1\in (0,2{\beta }_1)$ and ${\mu }_2\in (0,2{\beta }_2)$. Then for given $\overline{x},\overline{y}\in C$, $(\overline{x},\overline{y})$ is a solution of problem (1.5) if and only if $\overline{x}$ is a fixed point of the mapping $\mathrm{\Gamma }:C\to C$ defined by

$$\mathrm{\Gamma }(x)=T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)],\mathrm{\forall }x\in C,$$

where $\overline{y}=T_{{\mu }_2}^{(G_2,\psi )}(\overline{x}-{\mu }_2{B}_2\overline{x})$.

Remark 2.1 Under the conditions of Lemma 2.2, the set of fixed points of the mapping Γ is denoted by Ω.

Proposition 2.1 ([7])

Let C, H, θ, φ and $T_r^{(\theta ,\phi )}$ be as in Lemma 2.1. Then the following holds:

$${\parallel T_s^{(\theta ,\phi )}x-T_t^{(\theta ,\phi )}x\parallel }^2\le \frac{s-t}{s}〈T_s^{(\theta ,\phi )}x-T_t^{(\theta ,\phi )}x,T_s^{(\theta ,\phi )}x-x〉$$

for all $s,t>0$ and $x\in H$.

Lemma 2.3 ([18])

Assume that T is a nonexpansive self-mapping of a nonempty closed convex subset of C of a real Hilbert space H. If T has a fixed point, then $I-T$ is demiclosed, that is, when ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in C converging weakly to some $x\in C$ and the sequence ${\{(I-T)x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to some y, it follows that $(I-T)x=y$.

Lemma 2.4 ([4])

Let $M:H\to 2^H$ be a maximal monotone mapping and let $B:H\to H$ be a monotone and Lipschitz continuous mapping. Then the mapping $M+B:H\to 2^H$ is a maximal monotone mapping.

Lemma 2.5 ([19])

Let C be a nonempty bounded closed convex subset of a Hilbert space H and let $\mathrm{\Im }=\{T(s):0\le s<\mathrm{\infty }\}$ be a nonexpansive semigroup on C. Then for any $h\ge 0$,

$$\underset{t\to \mathrm{\infty }}{lim}\underset{x\in C}{sup}\parallel \frac{1}{t}{\int }_0^{t}T(s)xds-T(h)(\frac{1}{t}{\int }_0^{t}T(s)xds)\parallel =0.$$

Lemma 2.6 ([12])

Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\delta }_n)a_n+b_n+{\sigma }_n,n\in N,$$

where ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}\subset (0,1)$ and ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\sigma }_n\}}_{n=1}^{\mathrm{\infty }}$ are sequences in ℝ such that

• (i) ${lim}_{n\to \mathrm{\infty }}{\delta }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\delta }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{b_n}{{\delta }_n}\le 0$;

• (iii) ${\sigma }_n\ge 0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$.

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

Lemma 2.7 ([19])

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

3 Main results

Now we state and prove our main results.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $\theta ,G_1,G_2:C\times C\to \mathbb{R}$ be three bifunctions which satisfy assumptions (H1)-(H4) and $\phi ,\varphi ,\psi :C\to \mathbb{R}$ be three lower semicontinuous and convex functions with restriction (A1) or (A2). Let $F,B,B_1,B_2:C\to H$ be ζ-inverse strongly monotone, β-inverse strongly monotone, ${\beta }_1$-inverse strongly monotone and ${\beta }_2$-inverse strongly monotone, respectively and $M:H\to 2^H$ be a maximal monotone mapping. Let $\mathrm{\Im }=\{T(u):0\le u<\mathrm{\infty }\}$ be a one-parameter nonexpansive semigroup on H such that $\mathcal{F}=F(\mathrm{\Im })\cap GMEP\cap \mathrm{\Omega }\cap I(B,M)\ne \varphi$. Let $\{t_n\}\subset (0,\mathrm{\infty })$ be a real sequence such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$. Let f be a contraction from C into itself with a constant α ($0<\alpha <1$) and let A be a strongly positive linear bounded operator with coefficient $\overline{\gamma }>0$ such that $\parallel A\parallel \le 1$. Assume that $0<\gamma <\frac{\overline{\gamma }}{\alpha (1-{\alpha }_n\overline{\gamma })}$. Let $x_1\in C$ and let $\{x_n\}$ be a sequence defined by

$$\{\begin{array}{c}{z}_n=T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n),\hfill \\ y_n=T_{{\mu }_1}^{(G_1,{\varphi }_1)}[T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n\hfill \\ \phantom{x_{n+1}=}+((1-{\beta }_n)I-{\alpha }_{n}A)[\frac{1}{t_n}{\int }_0^{t_n}T(u)(J_{M,\lambda }(y_n-\lambda B{y}_n))du],n\ge 1,\hfill \end{array}$$

(3.1)

where ${\alpha }_n\in (0,1)$, ${\mu }_1\in (0,2{\beta }_1]$, ${\mu }_2\in (0,2{\beta }_2]$, $\lambda \in (0,2\beta ]$, ${\delta }_n\in (0,2\zeta ]$ satisfy the following conditions:

(C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$;

(C2) ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

(C3) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$;

(C4) ${lim}_{n\to \mathrm{\infty }}\frac{|t_n-t_{n-1}|}{t_n}\frac{1}{{\alpha }_n}=0$.

Then $\{x_n\}$ converges strongly to $x^{\ast }=P_{\mathcal{F}}(\gamma f+(I-A))(x^{\ast })$, which solves the following variational inequality:

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

and $(x^{\ast },y^{\ast })$ is a solution of problem (1.5), where $y^{\ast }=T_{{\mu }_2}^{(G_2,\psi )}(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$.

Proof Since F is a ζ-inverse strongly monotone mapping, we have

$$\begin{array}{rl}{\parallel (I-{\delta }_{n}F)x-(I-{\delta }_{n}F)y\parallel }^2& ={\parallel x-y\parallel }^2-2{\delta }_n〈x-y,Fx-Fy〉+{\delta }_n^2{\parallel Fx-Fy\parallel }^2\\ \le {\parallel x-y\parallel }^2+{\delta }_n({\delta }_n-2\zeta ){\parallel Fx-Fy\parallel }^2.\end{array}$$

(3.2)

In similar way, we can obtain

$$\begin{array}{r}\parallel (I-\lambda B)x-(I-\lambda B)y\parallel \le {\parallel x-y\parallel }^2+\lambda (\lambda -2\beta ){\parallel Bx-By\parallel }^2,\\ \parallel (I-{\mu }_1{B}_1)x-(I-{\mu }_1{B}_1)y\parallel \le {\parallel x-y\parallel }^2+{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_{1}x-B_{1}y\parallel }^2,\\ \parallel (I-{\mu }_2{B}_2)x-(I-{\mu }_2{B}_2)y\parallel \le {\parallel x-y\parallel }^2+{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_{2}x-B_{2}y\parallel }^2.\end{array}$$

(3.3)

Noticing that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0={lim}_{n\to \mathrm{\infty }}{\beta }_n$, we may assume, with no loss of generality, that $\frac{{\alpha }_n}{1-{\beta }_n}<{\parallel A\parallel }^{-1}$ for all $n\ge 1$. From Lemma 2.7 we know that if $0\le \rho \le {\parallel A\parallel }^{-1}$, then $\parallel I-\rho A\parallel \le 1-\rho \overline{\gamma }$. Since A is a strongly positive bounded linear operator on H, we have

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

Observe that

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

This shows that $(1-{\beta }_n)I-{\alpha }_{n}A$ is positive. It follows that

$$\begin{array}{rl}\parallel (1-{\beta }_n)I-{\alpha }_{n}A\parallel & =sup\{〈((1-{\beta }_n)I-{\alpha }_{n}A)x,x〉:x\in H,\parallel x\parallel =1\}\\ =sup\{1-{\beta }_n-{\alpha }_n〈Ax,x〉:x\in H,\parallel x\parallel =1\}\\ \le 1-{\beta }_n-{\alpha }_n\overline{\gamma }.\end{array}$$

We divide the proof into several steps.

Step 1. $\{x_n\}$ is bounded.

Indeed, take $p\in \mathcal{F}$ arbitrarily. Since $p=T_{{\delta }_n}^{(\theta ,\phi )}(p-{\delta }_{n}Fp)$, F is ζ-inverse strongly monotone and $0<{\delta }_n\le 2\zeta$, we obtain for any $n\ge 1$

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2& ={\parallel T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(p-{\delta }_{n}Fp)\parallel }^2\\ \le {\parallel (x_n-p)-{\delta }_n(F{x}_n-Fp)\parallel }^2\\ ={\parallel x_n-p\parallel }^p-2{\delta }_n〈x_n-p,F{x}_n-Fp〉+{\delta }_n^2{\parallel F{x}_n-Fp\parallel }^2\\ \le {\parallel x_n-p\parallel }^2+{\delta }_n({\delta }_n-2\zeta ){\parallel F{x}_n-Fp\parallel }^2\\ \le {\parallel x_n-p\parallel }^2.\end{array}$$

(3.4)

Putting $u_n=T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)$, $w_n=J_{M,\lambda }(y_n-\lambda B{y}_n)$, and $u=T_{{\mu }_2}^{(G_2,\psi )}(p-{\mu }_2{B}_{2}p)$, we have

$$\begin{array}{rl}{\parallel u_n-u\parallel }^2& ={\parallel T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-T_{{\mu }_2}^{(G_2,\mu )}(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le {\parallel (z_n-p)-{\mu }_2(B_2{z}_n-B_{2}p)\parallel }^2\\ ={\parallel z_n-p\parallel }^2-2{\mu }_2〈z_n-p,B_2{z}_n-B_{2}p〉+{\mu }_2^2{\parallel B_2{z}_n-B_{2}p\parallel }^2\\ \le {\parallel z_n-p\parallel }^2+{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_2{z}_n-B_{2}p\parallel }^2\\ \le {\parallel z_n-p\parallel }^2\end{array}$$

(3.5)

and

$$\begin{array}{rl}{\parallel w_n-p\parallel }^2& ={\parallel J_{M,\lambda }(y_n-\lambda B{y}_n)-J_{M,\lambda }(p-\lambda Bp)\parallel }^2\\ \le {\parallel (y_n-p)-\lambda (B{y}_n-Bp)\parallel }^2\\ ={\parallel y_n-p\parallel }^2-2\lambda 〈y_n-p,B{y}_n-Bp〉+{\lambda }^2{\parallel B{y}_n-Bp\parallel }^2\\ \le {\parallel y_n-p\parallel }^2+\lambda (\lambda -2\beta ){\parallel B{y}_n-Bp\parallel }^2\\ \le {\parallel y_n-p\parallel }^2.\end{array}$$

(3.6)

And since

$$\begin{array}{rl}p& =T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(p-{\mu }_2{B}_{2}p)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(p-{\mu }_2{B}_{2}p)]\\ =T_{{\mu }_1}^{(G_1,\varphi )}(u-{\mu }_1{B}_{1}u),\end{array}$$

we know that for any $n\ge 1$

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel T_{{\mu }_1}^{(G_1,\varphi )}(u_n-{\mu }_1{B}_1{u}_n)-T_{{\mu }_1}^{(G_1,\varphi )}(u-{\mu }_1{B}_{1}u)\parallel }^2\\ \le {\parallel (u_n-u)-{\mu }_1(B_1{u}_n-B_{1}u)\parallel }^2\\ ={\parallel u_n-u\parallel }^2-2{\mu }_1〈u_n-u,B_1{u}_n-B_{1}u〉+{\mu }_1^2{\parallel B_1{u}_n-B_{1}u\parallel }^2\\ \le {\parallel u_n-u\parallel }^2+{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{u}_n-B_{1}u\parallel }^2\\ \le {\parallel u_n-u\parallel }^2\\ \le {\parallel z_n-p\parallel }^2.\end{array}$$

(3.7)

Since A is a strongly positive linear bounded operator with coefficient $\overline{\gamma }>0$, we have

$$\begin{array}{r}\parallel x_{n+1}-p\parallel \\ =\parallel {\alpha }_n(\gamma f(x_n)-Ap)+{\beta }_n(x_n-p)+((1-{\beta }_n)I-{\alpha }_{n}A)\frac{1}{t_n}{\int }_0^{t_n}[T(u)w_n-T(u)p]du\parallel \\ \le {\alpha }_n\parallel \gamma f(x_n)-Ap\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel w_n-p\parallel \\ \le {\alpha }_n\gamma \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \gamma f(p)-Ap\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel x_n-p\parallel \\ \le [1-{\alpha }_n(\overline{\gamma }-\gamma \alpha )]\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f(p)-Ap\parallel .\end{array}$$

By induction, we obtain for all $n\ge 1$

$$\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{1}{\overline{\gamma }-\gamma \alpha }\parallel \gamma f(p)-Ap\parallel \}.$$

Hence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is bounded. So are ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{z_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{w_n\}}_{n=1}^{\mathrm{\infty }}$.

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

We estimate $\parallel y_{n+1}-y_n\parallel$. From (3.1), we have

$$\begin{array}{c}\begin{array}{rl}{\parallel y_{n+1}-y_n\parallel }^2& ={\parallel T_{{\mu }_1}^{(G_1,\varphi )}(u_{n+1}-{\mu }_1{B}_1{u}_{n+1})-T_{{\mu }_1}^{(G_1,\varphi )}(u_n-{\mu }_1{B}_1{u}_n)\parallel }^2\\ \le {\parallel (u_{n+1}-u_n)-{\mu }_1(B_1{u}_{n+1}-B_1{u}_n)\parallel }^2\\ \le {\parallel u_{n+1}-u_n\parallel }^2+{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{u}_{n+1}-B_1{u}_n\parallel }^2\\ \le {\parallel u_{n+1}-u_n\parallel }^2\\ ={\parallel T_{{\mu }_2}^{(G_2,\psi )}(z_{n+1}-{\mu }_2{B}_2{z}_{n+1})-T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)\parallel }^2\\ \le {\parallel (z_{n+1}-z_n)-{\mu }_2(B_2{z}_{n+1}-B_2{z}_n)\parallel }^2\\ \le {\parallel z_{n+1}-z_n\parallel }^2+{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_2{z}_{n+1}-B_2{z}_n\parallel }^2\\ \le {\parallel z_{n+1}-z_n\parallel }^2,\end{array}\hfill \\ \begin{array}{r}\parallel (x_{n+1}-{\delta }_{n+1}F{x}_{n+1})-(x_n-{\delta }_{n}F{x}_n)\parallel \\ =\parallel x_{n+1}-x_n-{\delta }_{n+1}(F{x}_{n+1}-F{x}_n)+({\delta }_n-{\delta }_{n+1})F{x}_n\parallel \\ \le \parallel x_{n+1}-x_n-{\delta }_{n+1}(F{x}_{n+1}-F{x}_n)\parallel +|{\delta }_n-{\delta }_{n+1}|\parallel F{x}_n\parallel \\ \le \parallel x_{n+1}-x_n\parallel +|{\delta }_n-{\delta }_{n+1}|\parallel F{x}_n\parallel \end{array}\hfill \end{array}$$

(3.8)

and

$$\begin{array}{rl}\parallel z_{n+1}-z_n\parallel =& \parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_{n+1}-{\delta }_{n+1}F{x}_{n+1})-T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel \\ \le & \parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_{n+1}-{\delta }_{n+1}F{x}_{n+1})-T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel \\ +\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel \\ \le & \parallel (x_{n+1}-{\delta }_{n+1}F{x}_{n+1})-(x_n-{\delta }_{n}F{x}_n)\parallel \\ +\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel F{x}_n\parallel \\ +\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel .\end{array}$$

(3.9)

Without loss of generality, let us assume that there exists a real number a such that ${\delta }_n>a>0$ for all n. Utilizing Proposition 2.1 we have

$$\begin{array}{r}\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)\parallel \\ \le \frac{|{\delta }_{n+1}-{\delta }_n|}{{\delta }_{n+1}}\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(I-{\delta }_{n}F)x_n-x_n\parallel \\ \le \frac{|{\delta }_{n+1}-{\delta }_n|}{a}\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(I-{\delta }_{n}F)x_n-x_n\parallel .\end{array}$$

(3.10)

It follows from (3.8)-(3.10) that

$$\begin{array}{rl}\parallel w_{n+1}-w_n\parallel & =\parallel J_{M,\lambda }(y_{n+1}-\lambda B{u}_{n+1})-J_{M,\lambda }(y_n-\lambda B{y}_n)\parallel \\ \le \parallel (y_{n+1}-\lambda B{y}_{n+1})-(y_n-\lambda B{y}_n)\parallel \\ \le \parallel y_{n+1}-y_n\parallel \end{array}$$

(3.11)

and

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel \le & \parallel z_{n+1}-z_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel F{x}_n\parallel \\ +\frac{|{\delta }_{n+1}-{\delta }_n|}{a}\parallel T_{{\delta }_{n+1}}^{(\theta ,\phi )}(I-{\delta }_{n}F)x_n-x_n\parallel .\end{array}$$

(3.12)

Put $h_n=\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du$ for all $n\ge 1$. We note that

$$\begin{array}{rl}\parallel h_n-h_{n-1}\parallel =& \parallel \frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-\frac{1}{t_{n-1}}{\int }_0^{t_{n-1}}T(u)w_{n-1}du\parallel \\ \le & \parallel \frac{1}{t_n}{\int }_0^{t_n}[T(u)w_n-T(u)w_{n-1}]du\parallel \\ +|\frac{1}{t_n}-\frac{1}{t_{n-1}}|\parallel {\int }_0^{t_{n-1}}[T(u)w_{n-1}-T(u)p]du\parallel \\ +\frac{1}{t_n}\parallel {\int }_{t_{n-1}}^{t_n}[T(u)w_{n-1}-T(u)p]du\parallel \\ \le & \parallel y_n-y_{n-1}\parallel +\frac{2|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel .\end{array}$$

(3.13)

Using (3.12) and (3.13) we get

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ =\parallel [{\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}A)h_n]\\ -[{\alpha }_{n-1}\gamma f(x_{n-1})+{\beta }_{n-1}{x}_{n-1}+((1-{\beta }_{n-1})I-{\alpha }_{n-1}A)h_{n-1}]\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\gamma |{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})\parallel +{\beta }_n\parallel x_n-x_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel +(1-{\alpha }_n\overline{\gamma })\parallel h_n-h_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel A{h}_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|\parallel h_n-h_{n-1}\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\gamma |{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})\parallel +{\beta }_n\parallel x_n-x_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel +(1-{\alpha }_n\overline{\gamma })[\parallel y_n-y_{n-1}\parallel +\frac{2|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel ]\\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel A{h}_{n-1}\parallel +|{\beta }_n-{\beta }_{n-1}|[\parallel y_n-y_{n-1}\parallel +\frac{2|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel ]\\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\gamma |{\alpha }_n-{\alpha }_{n-1}|\parallel f(x_{n-1})\parallel +{\beta }_n\parallel x_n-x_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel +(1-{\alpha }_n\overline{\gamma })[\parallel x_n-x_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel F{x}_{n-1}\parallel \\ +\frac{|{\delta }_n-{\delta }_{n-1}|}{a}\parallel T_{{\delta }_n}^{(\theta ,\phi )}(I-{\delta }_{n-1}F)x_{n-1}-x_{n-1}\parallel ]\\ +(1-{\alpha }_n\overline{\gamma })\frac{2|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel A{h}_{n-1}\parallel \\ +|{\beta }_n-{\beta }_{n-1}|[\parallel x_n-x_{n-1}\parallel +|{\delta }_n-{\delta }_{n-1}|\parallel F{x}_{n-1}\parallel \\ +\frac{|{\delta }_n-{\delta }_{n-1}|}{a}\parallel T_{{\delta }_n}^{(\theta ,\phi )}(I-{\delta }_{n-1}F)x_{n-1}-x_{n-1}\parallel ]\\ +|{\beta }_n-{\beta }_{n-1}|\frac{2|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel \\ \le [1-(\overline{\gamma }-\gamma \alpha ){\alpha }_n]\parallel x_n-x_{n-1}\parallel +\frac{4|t_n-t_{n-1}|}{t_n}\parallel y_{n-1}-p\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|(\gamma \parallel f(x_{n-1})\parallel +\parallel A{h}_{n-1}\parallel )+{\beta }_n[\parallel x_n\parallel +\parallel x_{n-1}\parallel ]\\ +|{\beta }_n-{\beta }_{n-1}|[\parallel x_n\parallel +2\parallel x_{n-1}\parallel ]\\ +2|{\delta }_n-{\delta }_{n-1}|[\parallel F{x}_{n-1}\parallel +\frac{1}{a}\parallel T_{{\delta }_n}^{(\theta ,\phi )}(I-{\delta }_{n-1}F)x_{n-1}-x_{n-1}\parallel ]\\ =[1-(\overline{\gamma }-\gamma \alpha ){\alpha }_n]\parallel x_n-x_{n-1}\parallel +D[\frac{4|t_n-t_{n-1}|}{t_n}+|{\alpha }_n-{\alpha }_{n-1}|+{\beta }_n\\ +|{\beta }_n-{\beta }_{n-1}|+2|{\delta }_n-{\delta }_{n-1}|],\end{array}$$

where D = max{$3{sup}_{n\ge 1}\parallel x_n\parallel$, ${sup}_{n\ge 1}(\gamma \parallel f(x_n)\parallel +\parallel A{h}_n\parallel )$, ${sup}_{n\ge 1}(\parallel F{x}_n\parallel +\frac{1}{a}\parallel T_{{\delta }_n}^{(\theta ,\phi )}(I-{\delta }_{n}F)x_n-x_n\parallel )$, ${sup}_{n\ge 1}\parallel y_n-p\parallel$}. From Lemma 2.6, taking ${\delta }_n=(\overline{\gamma }-\gamma \alpha ){\alpha }_n$, $b_n=\frac{4|t_n-t_{n-1}|}{t_n}D$, ${\sigma }_n=D(|{\alpha }_n-{\alpha }_{n-1}|+{\beta }_n+|{\beta }_n-{\beta }_{n-1}|+2|{\delta }_n-{\delta }_{n-1}|)$, it follows that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

Step 3. ${lim}_{n\to \mathrm{\infty }}\parallel F{x}_n-Fp\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_1{u}_n-B_{1}u\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_2{z}_n-B_{2}p\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel B{y}_n-Bp\parallel =0$.

Indeed, from (3.1), (3.4), (3.5), and (3.7) we get

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ =\parallel {\alpha }_n(\gamma f(x_n)-Ap)+{\beta }_n(x_n-p)+((1-{\beta }_n)I\\ -{\alpha }_{n}A)\frac{1}{t_n}{\int }_0^{t_n}[T(u)(J_{M,\lambda }(y_n-\lambda B{y}_n))-T(u)(J_{M,\lambda }(p-\lambda Bp))]du{\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-p\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2\\ +(1-{\beta }_n-{\alpha }_n\overline{\gamma })[{\parallel u_n-u\parallel }^2+{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{u}_n-B_{1}u\parallel }^2]\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma })[{\parallel z_n-p\parallel }^2\\ +{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_2{z}_n-B_{2}p\parallel }^2+{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{u}_n-B_{1}u\parallel }^2]\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma })[{\parallel x_n-p\parallel }^2\\ +{\delta }_n({\delta }_n-2\zeta ){\parallel F{x}_n-Fp\parallel }^2+{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_2{z}_n-B_{2}p\parallel }^2\\ +{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{u}_n-B_{1}u\parallel }^2].\end{array}$$

Therefore

$$\begin{array}{r}{\delta }_n(2\zeta -{\delta }_n){\parallel F{x}_n-Fp\parallel }^2+{\mu }_2(2{\beta }_2-{\mu }_2){\parallel B_2{z}_n-B_{2}p\parallel }^2\\ +{\mu }_1(2{\beta }_1-{\mu }_1){\parallel B_1{u}_n-B_{1}u\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_n-x_{n+1}\parallel \to 0$ as $n\to \mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}\parallel F{x}_n-Fp\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel B_2{z}_n-B_{2}p\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel B_1{u}_n-B_{1}u\parallel =0$. Similarly, from (3.4), (3.6), and (3.7) we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel w_n-p\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(1-{\alpha }_n\overline{\gamma }){\parallel x_n-p\parallel }^2+\lambda (\lambda -2\beta ){\parallel B{y}_n-Bp\parallel }^2,\end{array}$$

which implies that

$$\lambda (2\beta -\lambda ){\parallel B{y}_n-Bp\parallel }^2\le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .$$

We also have ${lim}_{n\to \mathrm{\infty }}\parallel B{y}_n-Bp\parallel =0$.

Step 4. We claim that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel w_n-y_n\parallel =0$.

Indeed, from Lemma 2.1, (3.4), (3.5), and (3.7) we have

$$\begin{array}{rl}{\parallel u_n-u\parallel }^2=& {\parallel T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-T_{{\mu }_2}^{(G_2,\psi )}(p-{\mu }_2{B}_{2}p)\parallel }^2\\ \le & 〈(z_n-{\mu }_2{B}_2{z}_n)-(p-{\mu }_2{B}_{2}p),u_n-u〉\\ =& \frac{1}{2}[{\parallel (z_n-{\mu }_2{B}_2{z}_n)-(p-{\mu }_2{B}_{2}p)\parallel }^2+{\parallel u_n-u\parallel }^2\\ -{\parallel (z_n-{\mu }_2{B}_2{z}_n)-(p-{\mu }_2{B}_2{z}_n)-(u_n-u)\parallel }^2]\\ \le & \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel u_n-u\parallel }^2-{\parallel (z_n-u_n)-(p-u)\parallel }^2\\ +2{\mu }_2〈(z_n-u_n)-(p-u),B_2{z}_n-B_{2}p〉-{\mu }_2^2{\parallel B_2{z}_n-B_{2}p\parallel }^2]\end{array}$$

and

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2=& \parallel T_{{\mu }_1}^{(G_1,\varphi )}(u_n-{\mu }_1{B}_1{u}_n)-T_{{\mu }_1}^{(G_1,\varphi )}(u-{\mu }_1{B}_{1}u)\parallel \\ \le & 〈(u_n-\mu B_1{u}_n)-(u-{\mu }_1{B}_{1}u),y_n-p〉\\ =& \frac{1}{2}[{\parallel (u_n-{\mu }_1{B}_1{u}_n)-(u-{\mu }_1{B}_{1}u)\parallel }^2+{\parallel y_n-p\parallel }^2\\ -{\parallel (u_n-{\mu }_1{B}_1{u}_n)-(u-{\mu }_1{B}_{1}u)-(y_n-p)\parallel }^2]\\ \le & \frac{1}{2}[{\parallel u_n-u\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1〈B_1{u}_n-B_{1}u,(u_n-y_n)+(p-u)〉-{\mu }_1^2{\parallel B_1{u}_n-B_{1}u\parallel }^2]\\ \le & \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1〈B_1{u}_n-B_{1}u,(u_n-y_n)+(p-u)〉],\end{array}$$

which imply that

$$\begin{array}{rl}{\parallel u_n-u\parallel }^2\le & {\parallel x_n-p\parallel }^2-{\parallel (z_n-u_n)-(p-u)\parallel }^2\\ +2{\mu }_2〈(z_n-u_n)-(p-u),B_2{z}_n-B_{2}p〉-{\mu }_1^2{\parallel B_2{z}_n-B_{2}p\parallel }^2\end{array}$$

(3.14)

and

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2\le & {\parallel x_n-p\parallel }^2-{\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel .\end{array}$$

(3.15)

It follows from (3.15) that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(1-{\alpha }_n\overline{\gamma }){\parallel x_n-p\parallel }^2\\ -(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel ,\end{array}$$

which gives

$$\begin{array}{r}(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel (u_n-y_n)+(p-u)\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel \\ +2{\mu }_1(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel B_1{u}_n-B_{1}u\parallel \to 0$ as $n\to \mathrm{\infty }$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (u_n-y_n)+(p-u)\parallel =0.$$

(3.16)

Also, from (3.14) we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel u_n-u\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma })[{\parallel x_n-p\parallel }^2\\ -{\parallel (z_n-u_n)-(p-u)\parallel }^2+2{\mu }_2〈(z_n-u_n)-(p-u),B_2{z}_n-B_{2}p〉].\end{array}$$

So, we have

$$\begin{array}{r}(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel (z_n-u_n)-(p-u)\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel \\ +2{\mu }_2\parallel (z_n-u_n)-(p-u)\parallel \parallel B_2{z}_n-B_{2}p\parallel .\end{array}$$

Note that $\parallel B_2{z}_n-B_{2}p\parallel \to 0$ as $n\to \mathrm{\infty }$. Then we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (z_n-u_n)-(p-u)\parallel =0.$$

(3.17)

In addition, from the firm nonexpansivity of $T_{{\delta }_n}^{(\theta ,\phi )}$, we obtain

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2=& {\parallel T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)-T_{{\delta }_n}^{(\theta ,\phi )}(p-{\delta }_{n}Fp)\parallel }^2\\ \le & 〈(x_n-{\delta }_{n}F{x}_n)-(p-{\delta }_{n}Fp),z_n-p〉\\ =& \frac{1}{2}[{\parallel (x_n-{\delta }_{n}F{x}_n)-(p-{\delta }_{n}Fp)\parallel }^2+{\parallel z_n-p\parallel }^2\\ -{\parallel (x_n-{\delta }_{n}F{x}_n)-(p-{\delta }_{n}Fp)-(z_n-p)\parallel }^2]\\ \le & \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel z_n-p\parallel }^2-{\parallel x_n-z_n-{\delta }_n(F{x}_n-Fp)\parallel }^2]\\ =& \frac{1}{2}[{\parallel x_n-p\parallel }^2+{\parallel z_n-p\parallel }^2-{\parallel x_n-z_n\parallel }^2\\ +2{\delta }_n〈F{x}_n-Fp,x_n-z_n〉-{\delta }_n^2{\parallel F{x}_n-Fp\parallel }^2],\end{array}$$

which implies that

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2& \le {\parallel x_n-p\parallel }^2-{\parallel x_n-z_n\parallel }^2+2{\delta }_n〈F{x}_n-Fp,x_n-z_n〉\\ \le {\parallel x_n-p\parallel }^2-{\parallel x_n-z_n\parallel }^2+2{\delta }_n\parallel F{x}_n-Fp\parallel \parallel x_n-z_n\parallel .\end{array}$$

(3.18)

From (3.7) and (3.18), we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel z_n-p\parallel }^2\\ \le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma })[{\parallel x_n-p\parallel }^2\\ -{\parallel x_n-z_n\parallel }^2+2{\delta }_n\parallel F{x}_n-Fp\parallel \parallel x_n-z_n\parallel ].\end{array}$$

It follows that

$$\begin{array}{rl}(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel x_n-z_n\parallel }^2\le & {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel \\ +2(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\delta }_n\parallel F{x}_n-Fp\parallel \parallel x_n-z_n\parallel .\end{array}$$

Since $\parallel F{x}_n-Fp\parallel \to 0$ as $n\to \mathrm{\infty }$, we obtain

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

(3.19)

Thus, from (3.16), (3.17), and (3.19) we obtain

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-y_n\parallel & =\underset{n\to \mathrm{\infty }}{lim}\parallel (z_n-u_n)-(p-u)+(u_n-y_n)+(p-u)\parallel \\ \le \underset{n\to \mathrm{\infty }}{lim}\parallel (z_n-u_n)-(p-u)\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel (u_n-y_n)+(p-u)\parallel \\ =0\end{array}$$

and

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-y_n\parallel & \le \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-z_n\parallel +\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-y_n\parallel \\ =0.\end{array}$$

Since $J_{M,\lambda }$ is 1-inverse strongly monotone, we have

$$\begin{array}{rl}{\parallel w_n-p\parallel }^2=& {\parallel J_{M,\lambda }(y_n-\lambda B{y}_n)-J_{M,\lambda }(p-\lambda Bp)\parallel }^2\\ \le & 〈(y_n-\lambda B{y}_n)-(p-\lambda Bp),w_n-p〉\\ =& \frac{1}{2}[{\parallel (y_n-\lambda B{y}_n)-(p-\lambda Bp)\parallel }^2+{\parallel w_n-p\parallel }^2\\ -{\parallel (y_n-\lambda B{y}_n)-(p-\lambda Bp)-(w_n-p)\parallel }^2]\\ \le & \frac{1}{2}({\parallel y_n-p\parallel }^2+{\parallel w_n-p\parallel }^2-{\parallel (y_n-w_n)-\lambda (B{y}_n-Bp)\parallel }^2)\\ \le & \frac{1}{2}({\parallel y_n-p\parallel }^2+{\parallel w_n-p\parallel }^2-{\parallel y_n-w_n\parallel }^2+2\lambda 〈y_n-w_n,B{y}_n-Bp〉\\ -{\lambda }^2{\parallel B{y}_n-Bp\parallel }^2),\end{array}$$

which implies that

$${\parallel w_n-p\parallel }^2\le {\parallel y_n-p\parallel }^2-{\parallel y_n-w_n\parallel }^2+2\lambda \parallel y_n-w_n\parallel \parallel B{y}_n-Bp\parallel .$$

(3.20)

Substituting (3.15) into (3.20), we have

$$\begin{array}{rl}{\parallel w_n-p\parallel }^2\le & {\parallel x_n-p\parallel }^2-{\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel \\ -{\parallel y_n-w_n\parallel }^2+2\lambda \parallel y_n-w_n\parallel \parallel B{y}_n-Bp\parallel .\end{array}$$

(3.21)

It follows from (3.21) that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel w_n-p\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+(1-{\alpha }_n\overline{\gamma }){\parallel x_n-p\parallel }^2\\ -(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel \\ -(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-w_n\parallel }^2+2\lambda (1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel y_n-w_n\parallel \parallel B{y}_n-Bp\parallel ,\end{array}$$

which gives

$$\begin{array}{r}(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel y_n-w_n\parallel }^2\\ \le {\alpha }_n{\parallel \gamma f(x_n)-Ap\parallel }^2+\parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\\ -(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel (u_n-y_n)+(p-u)\parallel }^2\\ +2{\mu }_1(1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel B_1{u}_n-B_{1}u\parallel \parallel (u_n-y_n)+(p-u)\parallel \\ +2\lambda (1-{\beta }_n-{\alpha }_n\overline{\gamma })\parallel y_n-w_n\parallel \parallel B{y}_n-Bp\parallel .\end{array}$$

Since ${\alpha }_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$, $\parallel (u_n-y_n)+(p-u)\parallel \to 0$, $\parallel B{y}_n-Bp\parallel \to 0$ as $n\to \mathrm{\infty }$, we have ${lim}_{n\to \mathrm{\infty }}\parallel y_n-w_n\parallel =0$.

Step 5. We show ${lim}_{n\to \mathrm{\infty }}\parallel T(u)y_n-y_n\parallel =0$.

Denote $h_n=\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du$. From (3.1), ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$ we have

$$\begin{array}{rl}\parallel x_{n+1}-h_n\parallel & =\parallel {\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_{n}A)h_n-h_n\parallel \\ \le {\alpha }_n\parallel \gamma f(x_n)-A{h}_n\parallel +{\beta }_n\parallel x_n-h_n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.22)

Let $K=\{w\in C:\parallel w-p\parallel \le \frac{1}{\overline{\gamma }-\gamma \alpha }\parallel \gamma f(p)-Ap\parallel \}$. Then K is a nonempty bounded closed convex subset of C which is $T(u)$-invariant for each $u\in [0,\mathrm{\infty })$ and contains $\{x_n\}$. It follows from Lemma 2.5 that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel h_n-T(u)h_n\parallel =0,u\ge 0$$

(3.23)

and from (3.22) and (3.23), we have

$$\begin{array}{rl}\parallel x_{n+1}-T(u)x_{n+1}\parallel & \le \parallel x_{n+1}-h_n\parallel +\parallel h_n-T(u)h_n\parallel +\parallel T(u)h_n-T(u)x_{n+1}\parallel \\ \le 2\parallel x_{n+1}-h_n\parallel +\parallel h_n-T(u)h_n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Hence

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T(u)x_n\parallel =0.$$

(3.24)

Furthermore, from Step 4 we have for every $u\in [0,\mathrm{\infty })$ that

$$\parallel T(u)y_n-T(u)x_n\parallel \le \parallel y_n-x_n\parallel \to 0\text{as }n\to 0.$$

(3.25)

So, we obtain from (3.24)

$$\begin{array}{rl}\parallel T(u)y_n-x_n\parallel & \le \parallel T(u)y_n-T(u)x_n\parallel +\parallel T(u)x_n-x_n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.26)

Hence, we have for every $u\in [0,\mathrm{\infty })$ that

$$\begin{array}{rl}\parallel T(u)y_n-y_n\parallel & \le \parallel T(u)y_n-x_n\parallel +\parallel x_n-y_n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

Step 6. We show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈\gamma f(x^{\ast })-A{x}^{\ast },x_n-x^{\ast }〉\le 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈\gamma f(x^{\ast })-A{x}^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\le 0$, where $x^{\ast }=P_{\mathcal{F}}(\gamma f(x^{\ast })+(I-A)(x^{\ast }))$.

Indeed, take a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

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

(3.27)

Since $\{y_{n_i}\}$ is bounded, we can assume that $y_{n_i}\rightharpoonup w$. First, we prove that $w\in F(\mathrm{\Im })$.

Assume the contrary that $w\ne T(u)w$ for some $u\in [0,\mathrm{\infty })$. Then by Opial’s condition, we obtain from Step 5 that

$$\begin{array}{rl}\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_j}-w\parallel & <\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_j}-T(u)w\parallel \\ \le \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\parallel y_{n_j}-T(u)y_{n_j}\parallel +\parallel T(u)y_{n_j}-T(u)w\parallel )\\ \le \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_j}-w\parallel .\end{array}$$

This is a contraction. Hence, $w\in F(\mathrm{\Im })$.

Next, let us show that $w\in GMEP$.

From $z_n=T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)$, we obtain

$$\theta (z_n,y)+\phi (y)-\phi (z_n)+\frac{1}{{\delta }_n}〈y-z_n,z_n-(x_n-{\delta }_{n}F{x}_n)〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from (H2) that

$$\phi (y)-\phi (z_n)+\frac{1}{{\delta }_n}〈y-z_n,z_n-(x_n-{\delta }_{n}F{x}_n)〉\ge \theta (y,z_n),\mathrm{\forall }y\in C.$$

Replacing n by $n_i$, we have

$$\phi (y)-\phi (z_{n_i})+〈y-z_{n_i},F{x}_{n_i}〉+〈y-z_{n_i},\frac{z_{n_i}-x_{n_i}}{{\delta }_{n_i}}〉\ge \theta (y,z_{n_i}),\mathrm{\forall }y\in C.$$

(3.28)

Let $z_t=ty+(1-t)w$ for all $t\in [0,1]$ and $y\in C$. Then we have $z_t\in C$. It follows from (3.28) that

$$\begin{array}{rl}〈z_t-z_{n_i},F{z}_t〉\ge & 〈z_t-z_{n_i},F{z}_t〉-\phi (z_t)+\phi (z_{n_i})-〈z_t-z_{n_i},F{x}_{n_i}〉\\ -〈z_t-z_{n_i},\frac{z_{n_i}-x_{n_i}}{{\delta }_{n_i}}〉+\theta (z_t,z_{n_i})\\ \le & 〈z_t-z_{n_i},F{z}_t-F{z}_{n_i}〉+〈z_t-z_{n_i},F{z}_{n_i}-F{x}_{n_i}〉\\ -\phi (z_t)+\phi (z_{n_i})-〈z_t-z_{n_i},\frac{z_{n_i}-x_{n_i}}{{\delta }_{n_i}}〉+\theta (z_t,z_{n_i}).\end{array}$$

Since $\parallel z_{n_i}-x_{n_i}\parallel \to 0$, we have $\parallel F{z}_{n_i}-F{x}_{n_i}\parallel \to 0$ as $n_i\to \mathrm{\infty }$. From the monotonity of F, we have

$$〈F{z}_t-F{z}_{n_i},z_t-z_{n_i}〉\ge 0.$$

From (H4), $\frac{z_{n_i}-x_{n_i}}{{\delta }_{n_i}}\to 0$ and $z_{n_i}\rightharpoonup w$, we have

$$〈z_t-w,F{z}_t〉\ge -\phi (z_t)+\phi (w)+\theta (z_t,w)$$

(3.29)

as $n_i\to \mathrm{\infty }$. By (H1), (H4), and (3.29), we obtain

$$\begin{array}{rl}0& =\theta (z_t,z_t)+\phi (z_t)-\phi (z_t)\\ \le t\theta (z_t,y)+(1-t)\theta (z_t,w)+t\phi (y)+(1-t)\phi (w)-\phi (z_t)\\ \le t[\theta (z_t,y)+\phi (y)-\phi (z_t)]+(1-t)〈z_t-w,F{z}_t〉\\ =t[\theta (z_t,y)+\phi (y)-\phi (z_t)]+(1-t)t〈y-w,F{z}_t〉.\end{array}$$

Hence we obtain

$$0\le \theta (z_t,y)+\phi (y)-\phi (z_t)+(1-t)〈y-w,F{z}_t〉.$$

Putting $t\to 0$, we have

$$0\le \theta (w,y)+\phi (y)-\phi (w)+〈y-w,Fw〉,\mathrm{\forall }y\in C.$$

This implies that $w\in GMEP$.

Next, we prove that $w\in \mathrm{\Omega }$.

Utilizing Lemma 2.1, we have for all $x,y\in C$

$$\begin{array}{r}{\parallel \mathrm{\Gamma }(x)-\mathrm{\Gamma }(y)\parallel }^2\\ =\parallel T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)]\\ -T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)]{\parallel }^2\\ \le \parallel T_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-T_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)\\ -{\mu }_1(B_1{T}_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-B_1{T}_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)){\parallel }^2\\ \le {\parallel T_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-T_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)\parallel }^2\\ +{\mu }_1({\mu }_1-2{\beta }_1){\parallel B_1{T}_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-B_1{T}_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)\parallel }^2\\ \le {\parallel T_{{\mu }_2}^{(G_2,\psi )}(x-{\mu }_2{B}_{2}x)-T_{{\mu }_2}^{(G_2,\psi )}(y-{\mu }_2{B}_{2}y)\parallel }^2\\ \le {\parallel x-y-{\mu }_2(B_{2}x-B_{2}y)\parallel }^2\\ \le {\parallel x-y\parallel }^2+{\mu }_2({\mu }_2-2{\beta }_2){\parallel B_{2}x-B_{2}y\parallel }^2\\ \le {\parallel x-y\parallel }^2.\end{array}$$

This shows that $\mathrm{\Gamma }:C\to C$ is nonexpansive. Note that

$$\begin{array}{rl}\parallel y_n-\mathrm{\Gamma }(y_n)\parallel & =\parallel \mathrm{\Gamma }(z_n)-\mathrm{\Gamma }(y_n)\parallel \\ \le \parallel z_n-y_n\parallel \\ \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

According to Lemma 2.2 and Lemma 2.3, we obtain $w\in \mathrm{\Omega }$.

Lastly, we show that $w\in I(B,M)$.

In fact, since B is a β-inverse strongly monotone, B is monotone and Lipschitz continuous mapping. It follows from Lemma 2.4 that $M+B$ is a maximal monotone mapping. Let $(v,g)\in G(M+B)$. Then $g-B(v)\in M(v)$. Since $w_{n_i}=J_{M,\lambda }(y_{n_i}-\lambda B{y}_{n_i})$, we have $y_{n_i}-\lambda B{y}_{n_i}\in (I+\lambda M)(w_{n_i})$, i.e., $\frac{1}{\lambda }(y_{n_i}-w_{n_i}-\lambda A{y}_{n_i})\in M(w_{n_i})$. By virtue of the maximal monotonicity of $M+B$, we have

$$〈v-w_{n_i},g-B(v)-\frac{1}{\lambda }(y_{n_i}-w_{n_i}-\lambda B{y}_{n_i})〉\ge 0$$

and hence

$$\begin{array}{rl}〈v-w_{n_i},g〉\ge & 〈v-w_{n_i},B(v)+\frac{1}{\lambda }(y_{n_i}-w_{n_i}-\lambda B{y}_{n_i})〉\\ =& 〈v-w_{n_i},Bv-B{w}_{n_i}〉+〈v-w_{n_i},B{w}_{n_i}-B{y}_{n_i}〉\\ +〈v-w_{n_i},\frac{1}{\lambda }(y_{n_i}-w_{n_i})〉.\end{array}$$

It follows from ${lim}_{n\to \mathrm{\infty }}\parallel w_n-y_n\parallel =0$ that we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B{w}_n-B{y}_n\parallel =0$$

and

$$\begin{array}{c}{w}_{n_i}\rightharpoonup w,\hfill \\ \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈v-w_{n_i},g〉=〈v-w,g〉\ge 0.\hfill \end{array}$$

It follows from the maximal monotonicity of $M+B$ that $\theta \in (M+B)(w)$, that is, $w\in I(B,M)$. Therefore $\omega \in \mathcal{F}=F(\mathrm{\Im })\cap GMEP\cap \mathrm{\Omega }\cap I(B,M)$.

By $x^{\ast }=P_{\mathcal{F}}(\gamma f+(I-A))(x^{\ast })$, we obtain

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

and

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(\gamma f-A)x^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\\ =\underset{i\to \mathrm{\infty }}{lim}〈(\gamma f-A)x^{\ast },\frac{1}{t_{n_i}}{\int }_0^{t_{n_i}}T(u)w_{n_i}du-x^{\ast }〉\\ =〈(\gamma f-A)x^{\ast },w-x^{\ast }〉\\ \le 0,\end{array}$$

as required.

Step 7. We prove $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$.

By using (3.1), we have

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ =\parallel {\alpha }_n(\gamma f(x_n)-A{x}^{\ast })+{\beta }_n(x_n-x^{\ast })\\ +((1-{\beta }_n)I-{\alpha }_{n}A)(\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }){\parallel }^2\\ \le {\alpha }_n^2{\parallel \gamma f(x_n)-A{x}^{\ast }\parallel }^2\\ +{\parallel {\beta }_n(x_n-x^{\ast })+((1-{\beta }_n)I-{\alpha }_{n}A)(\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast })\parallel }^2\\ +2{\alpha }_n〈\gamma f(x_n)-A{x}^{\ast },{\beta }_n(x_n-x^{\ast })+((1-{\beta }_n)I-{\alpha }_{n}A)(\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast })〉\\ \le {\alpha }_n^2{\parallel \gamma f(x_n)-A{x}^{\ast }\parallel }^2+{\beta }_n^2{\parallel x_n-x^{\ast }\parallel }^2+{(1-{\beta }_n-{\alpha }_n\overline{\gamma })}^2{\parallel w_n-x^{\ast }\parallel }^2\\ +2{\beta }_n〈x_n-x^{\ast },((1-{\beta }_n)I-{\alpha }_{n}A)(\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast })〉\\ +2{\alpha }_n{\beta }_n〈\gamma f(x_n)-\gamma f\left(x^{\ast }\right),x_n-x^{\ast }〉+2{\alpha }_n{\beta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })〈\gamma f(x_n)-\gamma f\left(x^{\ast }\right),\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉.\end{array}$$

It follows from (3.4), (3.6), and (3.7) that

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ \le {\alpha }_n^2{\parallel \gamma f(x_n)-A{x}^{\ast }\parallel }^2+{\beta }_n^2{\parallel x_n-x^{\ast }\parallel }^2+{(1-{\beta }_n-{\alpha }_n\overline{\gamma })}^2{\parallel x_n-x^{\ast }\parallel }^2\\ +2{\beta }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma }){\parallel x_n-x^{\ast }\parallel }^2\\ +2{\alpha }_n{\beta }_n\gamma \alpha {\parallel x_n-x^{\ast }\parallel }^2+2{\alpha }_n{\beta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })\gamma \alpha {\parallel x_n-x^{\ast }\parallel }^2\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\\ \le [1-2{\alpha }_n(\overline{\gamma }-\gamma \alpha +{\alpha }_n\overline{\gamma }\gamma \alpha )]{\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n^2{\parallel \gamma f(x_n)-A{x}^{\ast }\parallel }^2\\ +2{\alpha }_n{\beta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\\ +{\alpha }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2\\ =(1-{\overline{\alpha }}_n){\parallel x_n-x^{\ast }\parallel }^2+{\overline{\beta }}_n,\end{array}$$

where

$$\begin{array}{c}{\overline{\alpha }}_n=2{\alpha }_n(\overline{\gamma }-\gamma \alpha +{\alpha }_n\overline{\gamma }\gamma \alpha ),\hfill \\ \begin{array}{rl}{\overline{\beta }}_n=& {\alpha }_n^2{\parallel \gamma f(x_n)-A{x}^{\ast }\parallel }^2+2{\alpha }_n{\beta }_n〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },x_n-x^{\ast }〉\\ +2{\alpha }_n(1-{\beta }_n-{\alpha }_n\overline{\gamma })〈\gamma f\left(x^{\ast }\right)-A{x}^{\ast },\frac{1}{t_n}{\int }_0^{t_n}T(u)w_{n}du-x^{\ast }〉\\ +{\alpha }_n^2{\overline{\gamma }}^2{\parallel x_n-x^{\ast }\parallel }^2.\end{array}\hfill \end{array}$$

It is easily seen that ${\overline{\alpha }}_n\to 0$ as $n\to \mathrm{\infty }$, ${\sum }_{n=1}^{\mathrm{\infty }}{\overline{\alpha }}_n=+\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\overline{\beta }}_n}{{\overline{\alpha }}_n}\le 0$. Hence, applying Lemma 2.6 we immediately obtain $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 3.1 Let us consider the following sequences:

$${\alpha }_n=\frac{1}{n^{\frac{1}{4}}},{\beta }_n=\frac{1}{n^2},{\delta }_n=\frac{1}{n^{\frac{1}{4}}}+\frac{1}{2}\text{and}{t}_n=n,\mathrm{\forall }n\ge 1.$$

It is easy to see that all hypotheses (C1)-(C4) of Theorem 3.1 are satisfied.

By Theorem 3.1, we can obtain the following results immediately.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $\theta ,G_1,G_2:C\times C\to \mathbb{R}$ be three bifunctions which satisfy assumptions (H1)-(H4) and $\phi ,\varphi ,\psi :C\to \mathbb{R}$ be three lower semicontinuous and convex functions with restriction (A1) or (A2). Let $B,B_1,B_2:C\to H$ be β-inverse strongly monotone, ${\beta }_1$-inverse strongly monotone and ${\beta }_2$-inverse strongly monotone, respectively and $M:H\to 2^H$ be a maximal monotone mapping. Let $\mathrm{\Im }=\{T(u):0\le u<\mathrm{\infty }\}$ be a one-parameter nonexpansive semigroup on H such that ${\mathcal{F}}_1=F(\mathrm{\Im })\cap MEP\cap \mathrm{\Omega }\cap I(B,M)\ne \varphi$. Let $\{t_n\}\subset (0,\mathrm{\infty })$ be a real sequence such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$. Let f be a contraction from C into itself with a constant α ($0<\alpha <1$) and let A be a strongly positive linear bounded operator with coefficient $\overline{\gamma }>0$ such that $\parallel A\parallel \le 1$. Assume that $0<\gamma <\frac{\overline{\gamma }}{\alpha (1-{\alpha }_n\overline{\gamma })}$. Let $x_1\in C$ and let $\{x_n\}$ be a sequence defined by

$$\{\begin{array}{c}\theta (z_n,y)+\phi (y)-\phi (z_n)+\frac{1}{{\delta }_n}〈y-z_n,z_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ y_n=T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n\hfill \\ \phantom{x_{n+1}=}+((1-{\beta }_n)I-{\alpha }_{n}A)[\frac{1}{t_n}{\int }_0^{t_n}T(u)(J_{M,\lambda }(y_n-\lambda B{y}_n))]du,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where ${\alpha }_n\in (0,1)$, ${\mu }_1\in (0,2{\beta }_1]$, ${\mu }_2\in (0,2{\beta }_2]$, $\lambda \in (0,2\beta ]$, ${\delta }_n\in (0,2\zeta ]$ satisfy the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$;

• (iv) ${lim}_{n\to \mathrm{\infty }}\frac{|t_n-t_{n-1}|}{t_n}\frac{1}{{\alpha }_n}=0$.

Then $\{x_n\}$ converges strongly to $x^{\ast }=P_{{\mathcal{F}}_1}(\gamma f+(I-A))(x^{\ast })$, which solves the following variational inequality:

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in {\mathcal{F}}_1$$

and $(x^{\ast },y^{\ast })$ is a solution of problem (1.5), where $y^{\ast }=T_{{\mu }_2}^{(G_2,\psi )}(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$.

Proof In Theorem 3.1, for all $n\ge 0$, $z_n=T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n)$ is equivalent to

$$\theta (z_n,y)+\phi (y)-\phi (z_n)+〈F{x}_n,y-z_n〉+\frac{1}{{\delta }_n}〈y-z_n,z_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

(3.30)

Putting $F=0$, we obtain

$$\theta (z_n,y)+\phi (y)-\phi (z_n)+\frac{1}{{\delta }_n}〈y-z_n,z_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

By Theorem 3.1, we can easily get the desired conclusion. □

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $G_1,G_2:C\times C\to \mathbb{R}$ be two bifunctions which satisfy assumptions (H1)-(H4) and $\varphi ,\psi :C\to \mathbb{R}$ be two lower semicontinuous and convex functions with restriction (A1) or (A2). Let $F,B,B_1,B_2:C\to H$ be ζ-inverse strongly monotone, β-inverse strongly monotone, ${\beta }_1$-inverse strongly monotone and ${\beta }_2$-inverse strongly monotone, respectively, and let $M:H\to 2^H$ be a maximal monotone mapping. Let $\mathrm{\Im }=\{T(u):0\le u<\mathrm{\infty }\}$ be a one-parameter nonexpansive semigroup on H such that ${\mathcal{F}}_2=F(\mathrm{\Im })\cap VI(F,C)\cap \mathrm{\Omega }\cap I(B,M)\ne \varphi$. Let $\{t_n\}\subset (0,\mathrm{\infty })$ be a real sequence such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$. Let f be a contraction from C into itself with a constant α ($0<\alpha <1$) and let A be a strongly positive linear bounded operator with coefficient $\overline{\gamma }>0$ such that $\parallel A\parallel \le 1$. Assume that $0<\gamma <\frac{\overline{\gamma }}{\alpha (1-{\alpha }_n\overline{\gamma })}$. Let $x_1\in C$ and let $\{x_n\}$ be a sequence defined by

$$\{\begin{array}{c}{z}_n=P_C(x_n-{\delta }_{n}F{x}_n),\hfill \\ y_n=T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n\hfill \\ \phantom{x_{n+1}=}+((1-{\beta }_n)I-{\alpha }_{n}A)[\frac{1}{t_n}{\int }_0^{t_n}T(u)(J_{M,\lambda }(y_n-\lambda B{y}_n))du],\mathrm{\forall }n\ge 1,\hfill \end{array}$$

where ${\alpha }_n\in [0,1]$, ${\mu }_1\in (0,2{\beta }_1]$, ${\mu }_2\in (0,2{\beta }_2]$, $\lambda \in (0,2\beta ]$, ${\delta }_n\in (0,2\zeta ]$ satisfy the following conditions:

• (i) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\alpha }_{n+1}-{\alpha }_n|<\mathrm{\infty }$;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\beta }_{n+1}-{\beta }_n|<\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\delta }_n>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n+1}-{\delta }_n|<\mathrm{\infty }$;

• (iv) ${lim}_{n\to \mathrm{\infty }}\frac{|t_n-t_{n-1}|}{t_n}\frac{1}{{\alpha }_n}=0$.

Then $\{x_n\}$ converges strongly to $x^{\ast }=P_{{\mathcal{F}}_2}(\gamma f+(I-A))(x^{\ast })$, which solves the following variational inequality:

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in {\mathcal{F}}_2$$

and $(x^{\ast },y^{\ast })$ is a solution of problem (1.5), where $y^{\ast }=T_{{\mu }_2}^{(G_2,\psi )}(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$.

Proof Put $\theta =0$ and $\phi =0$ in Theorem 3.1. Then we have from (3.30)

$$〈F{x}_n,y-z_n〉+\frac{1}{{\delta }_n}〈y-z_n,z_n-x_n〉\ge 0,\mathrm{\forall }y\in C,n\ge 1.$$

That is,

$$〈y-z_n,x_n-{\delta }_{n}F{x}_n-z_n〉\le 0,\mathrm{\forall }y\in C.$$

It follows that $P_C(x_n-{\delta }_{n}F{x}_n)=z_n$ for all $n\ge 1$. We easily obtain the desired conclusion. □

Corollary 3.3 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $\theta ,G_1,G_2:C\times C\to \mathbb{R}$ be three bifunctions which satisfy assumptions (H1)-(H4) and $\phi ,\varphi ,\psi :C\to \mathbb{R}$ be three lower semicontinuous and convex functions with restriction (A1) or (A2). Let $F,B,B_1,B_2:C\to H$ be ζ-inverse strongly monotone, β-inverse strongly monotone, ${\beta }_1$-inverse strongly monotone, and ${\beta }_2$-inverse strongly monotone, respectively. Let $\mathrm{\Im }=\{T(u):0\le u<\mathrm{\infty }\}$ be a one-parameter nonexpansive semigroup on H such that ${\mathcal{F}}_3=F(\mathrm{\Im })\cap GMEP\cap \mathrm{\Omega }\cap VI(F,C)\ne \varphi$. Let $\{t_n\}\subset (0,\mathrm{\infty })$ be a real sequence such that ${lim}_{n\to \mathrm{\infty }}{t}_n=\mathrm{\infty }$. Let f be a contraction from C into itself with a constant α ($0<\alpha <1$) and let A be a strongly positive linear bounded operator with coefficient $\overline{\gamma }>0$ such that $\parallel A\parallel \le 1$. Assume that $0<\gamma <\frac{\overline{\gamma }}{\alpha (1-{\alpha }_n\overline{\gamma })}$. Let $x_1\in C$ and let $\{x_n\}$ be a sequence defined by

$$\{\begin{array}{c}{z}_n=T_{{\delta }_n}^{(\theta ,\phi )}(x_n-{\delta }_{n}F{x}_n),\hfill \\ y_n=T_{{\mu }_1}^{(G_1,\varphi )}[T_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{T}_{{\mu }_2}^{(G_2,\psi )}(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\beta }_n{x}_n\hfill \\ \phantom{x_{n+1}=}+((1-{\beta }_n)I-{\alpha }_{n}A)[\frac{1}{t_n}{\int }_0^{t_n}T(u)(P_C(y_n-\lambda B{y}_n))du],n\ge 1,\hfill \end{array}$$

where ${\alpha }_n\in (0,1)$, ${\mu }_1\in (0,2{\beta }_1]$, ${\mu }_2\in (0,2{\beta }_2]$, $\lambda \in (0,2\beta ]$, ${\delta }_n\in (0,2\zeta ]$ satisfy the conditions (C1)-(C4). Then $\{x_n\}$ converges strongly to $x^{\ast }=P_{{\mathcal{F}}_3}(\gamma f+(I-A))(x^{\ast })$, which solves the following variational inequality:

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in {\mathcal{F}}_3,$$

and $(x^{\ast },y^{\ast })$ is a solution of problem (1.5), where $y^{\ast }=T_{{\mu }_2}^{(G_2,\psi )}(x^{\ast }-{\mu }_2{B}_2{x}^{\ast })$.

Proof Taking $J_{M,\lambda }=P_C$ in Theorem 3.1, we can obtain desired conclusion immediately. □

Remark 3.2 Theorem 3.1 generalizes and improves Theorem 3.1 of Kumam and Wattanawitoon [14], Theorem 3.3 of Plubtieng and Punpaeng [10] and Theorem 3.1 of Shehu [13] in the following aspects:

• (1) Algorithm of Theorem 3.1 is different from algorithms in [10, 13, 14].

• (2) Theorem 3.1 includes Theorem 3.3 of Plubtieng and Punpaeng [10] as a special case.

• (3) Theorem 3.1 improves Theorem 3.1 of Kumam and Wattanawitoon [14] since the generalized equilibrium problem that is within [14] is extended to the generalized mixed equilibrium problem.