Viscosity iterative method for a new general system of variational inequalities in Banach spaces

Abstract

In this paper, we study a new iterative method for finding a common element of the set of solutions of a new general system of variational inequalities for two different relaxed cocoercive mappings and the set of fixed points of a nonexpansive mapping in real 2-uniformly smooth and uniformly convex Banach spaces. We prove the strong convergence of the proposed iterative method without the condition of weakly sequentially continuous duality mapping. Our result improves and extends the corresponding results announced by many others.

MSC:46B10, 46B20, 47H10, 49J40.

1 Introduction

Let X be a real Banach space and $X^{\ast }$ be its dual space. Let C be a subset of X and let T be a self-mapping of C. We use $F(T)$ to denote the set of fixed points of T. The duality mapping $J:X\to 2^{X^{\ast }}$ is defined by $J(x)=\{x^{\ast }\in X^{\ast }|〈x,x^{\ast }〉={\parallel x\parallel }^2,\parallel x^{\ast }\parallel =\parallel x\parallel \}$, $\mathrm{\forall }x\in X$. If X is a Hilbert space, then $J=I$, where I is the identity mapping. It is well-known that if X is smooth, then J is single-valued, which is denoted by j.

Recall that a mapping $f:C\to C$ is a contraction on C, if there exists a constant $\alpha \in (0,1)$ such that $\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. We use ${\mathrm{\Pi }}_C$ to denote the collection of all contractions on C. This is ${\mathrm{\Pi }}_C=\{f|f:C\to C\text{ a contraction}\}$. A mapping $T:C\to C$ is said to be nonexpansive if $\parallel T(x)-T(y)\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$. Let $A:C\to X$ be a nonlinear mapping. Then A is called

• (i) L-Lipschitz continuous (or Lipschitzian) if there exists a constant $L\ge 0$ such that

  $$\parallel Ax-Ay\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in C;$$

• (ii) accretive if there exists $j(x-y)\in J(x-y)$ such that

  $$〈Ax-Ay,j(x-y)〉\ge 0,\mathrm{\forall }x,y\in C;$$

• (iii) α-inverse strongly accretive if there exist $j(x-y)\in J(x-y)$ and $\alpha >0$ such that

  $$〈Ax-Ay,j(x-y)〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C;$$

• (iv) relaxed $(c,d)$-cocoercive if there exist $j(x-y)\in J(x-y)$ and two constants $c,d\ge 0$ such that

  $$〈Ax-Ay,j(x-y)〉\ge (-c){\parallel Ax-Ay\parallel }^2+d{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in C.$$

Let C be a nonempty closed convex subset of a real Hilbert space H. Recall that the classical variational inequality is to find $x^{\ast }\in C$ such that

$$〈A{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,$$

(1.1)

where $A:C\to H$ is a nonlinear mapping. Variational inequality theory has emerged as an important tool in studying a wide class of obstacle, unilateral, free, moving, equilibrium problems arising in several branches of pure and applied sciences in a unified and general framework. The variational inequality problem has been extensively studied in the literature; see [1–8] and the references cited therein.

In 2006, Aoyama et al. [9] first considered the following generalized variational inequality problem in Banach spaces. Let $A:C\to X$ be an accretive operator. Find a point $x^{\ast }\in C$ such that

$$〈A{x}^{\ast },j(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C.$$

(1.2)

Problem (1.2) is very interesting as it is connected with the fixed point problem for a nonlinear mapping and the problem of finding a zero point of an accretive operator in Banach spaces; see [10–13] and the references cited therein.

In 2010, Yao et al. [14] introduced the following system of general variational inequalities in Banach spaces. For given two operators $A,B:C\to X$, they considered the problem of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{cc}〈A{y}^{\ast }+x^{\ast }-y^{\ast },j(x-x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈B{x}^{\ast }+y^{\ast }-x^{\ast },j(x-y^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \end{array}$$

(1.3)

which is called the system of general variational inequalities in a real Banach space and the set of solutions of problem (1.3) denoted by ${\mathrm{\Omega }}_1$. Yao et al. proved the following strong convergence theorem.

Theorem YNNLY Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X which admits a weakly sequentially continuous duality mapping. Let $Q_C$ be the sunny nonexpansive retraction from X onto C. Let the mappings $A,B:C\to X$ be α-inverse-strongly accretive with $\alpha \ge K^2$ and β-inverse-strongly accretive with $\beta \ge K^2$, respectively, with ${\mathrm{\Omega }}_1\ne \mathrm{\varnothing }$. For a given $x_0\in C$, let the sequence $\{x_n\}$ be generated iteratively by

$$\{\begin{array}{c}{y}_n=Q_C(x_n-B{x}_n),\hfill \\ x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{Q}_C(y_n-A{y}_n),n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are three sequences in $(0,1)$. Suppose that the sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfy the following conditions:

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $\mathrm{\forall }n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

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

Then $\{x_n\}$ converges strongly to $Q^{\mathrm{\prime }}u$ where $Q^{\mathrm{\prime }}$ is the sunny nonexpansive retraction of C onto ${\mathrm{\Omega }}_1$.

In 2011, Katchang and Kumam [15] introduced the following system of general variational inequalities in Banach spaces. For given two operators $A,B:C\to X$, they considered the problem of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{cc}〈\lambda A{y}^{\ast }+x^{\ast }-y^{\ast },j(x-x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈\mu B{x}^{\ast }+y^{\ast }-x^{\ast },j(x-y^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \end{array}$$

(1.4)

which is called the system of general variational inequalities in a real Banach space and the set of solutions of problem (1.4) denoted by ${\mathrm{\Omega }}_2$. Katchang and Kumam proved the following strong convergence theorem.

Theorem KK Let C be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space X which admits a weakly sequentially continuous duality mapping. Let $S:C\to C$ be a nonexpansive mapping and $Q_C$ be a sunny nonexpansive retraction from X onto C. Let the mappings $A,B:C\to X$ be β-inverse-strongly accretive with $\beta \ge \lambda K^2$ and γ-inverse-strongly accretive with $\gamma \ge \mu K^2$, respectively, and let K be the 2-uniformly smooth constant of X. Let f be a contraction of C into itself with coefficient $\alpha \in [0,1)$. Suppose that $F:={\mathrm{\Omega }}_2\cap F(S)\ne \mathrm{\varnothing }$. For a given $x_0=x\in C$, let the sequence $\{x_n\}$ be generated iteratively by

$$\{\begin{array}{c}{y}_n=Q_C(x_n-\mu B{x}_n),\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}S{Q}_C(y_n-\lambda A{y}_n),n\ge 0,\hfill \end{array}$$

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are three sequences in $(0,1)$. Suppose that the sequences $\{{\alpha }_n\}$, $\{{\beta }_n\}$ and $\{{\gamma }_n\}$ satisfy the following conditions:

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $\mathrm{\forall }n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

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

Then $\{x_n\}$ converges strongly to $\overline{x}=Q_{F}f(\overline{x})$ and $(\overline{x},\overline{y})$ is a solution of problem (1.4), where $\overline{y}=Q_C(\overline{x}-\mu B\overline{x})$ and $Q_F$ is the sunny nonexpansive retraction of C onto F.

The problem of finding solutions of (1.4) by using iterative methods has been studied by many others; see [16–19] and the references cited therein.

In this paper, we focus on the problem of finding $(x^{\ast },y^{\ast },z^{\ast })\in C\times C\times C$ such that

$$\{\begin{array}{cc}〈{\lambda }_1{A}_1{y}^{\ast }+x^{\ast }-y^{\ast },j(x-x^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈{\lambda }_2{A}_2{z}^{\ast }+y^{\ast }-z^{\ast },j(x-y^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \\ 〈{\lambda }_3{A}_3{x}^{\ast }+z^{\ast }-x^{\ast },j(x-z^{\ast })〉\ge 0,\hfill & \mathrm{\forall }x\in C,\hfill \end{array}$$

(1.5)

which is called a new general system of variational inequalities in Banach spaces, where $A_i:C\to X$ are three mappings, ${\lambda }_i>0$ for all $i=1,2,3$. In particular, if $A_3=0$ and $z^{\ast }=x^{\ast }$, then problem (1.5) reduces to problem (1.4). If we add up the requirement that ${\lambda }_i=1$ for $i=1,2$, then problem (1.5) reduces to problem (1.3).

In this paper, motivated and inspired by the idea of Katchang and Kumam [15] and Yao et al. [14], we introduce a new iterative method for finding a common element of the set of solutions of a new general system of variational inequalities in Banach spaces for two different relaxed cocoercive mappings and the set of fixed points of a nonexpansive mapping in real 2-uniformly smooth and uniformly convex Banach spaces. We prove the strong convergence of the proposed iterative algorithm without the condition of weakly sequentially continuous duality mapping. Our result improves and extends the corresponding results announced by many others.

2 Preliminaries

In this section, we recall the well-known results and give some useful lemmas that are used in the next section.

Let X be a Banach space and let $U=\{x\in X:\parallel x\parallel =1\}$ be a unit sphere of X. X is said to be uniformly convex if for each $ϵ\in (0,2]$, there exists a constant $\delta >0$ such that for any $x,y\in U$,

$$\parallel x-y\parallel \ge ϵ\text{implies}\parallel \frac{x+y}{2}\parallel \le 1-\delta .$$

The norm on X is said to be Gâteaux differentiable if the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

(2.1)

exists for each $x,y\in U$ and in this case X is said to be smooth. X is said to have a uniformly Frechet differentiable norm if the limit (2.1) is attained uniformly for $x,y\in U$ and in this case X is said to be uniformly smooth. We define a function $\rho :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, called the modulus of smoothness of X, as follows:

$$\rho (\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:x,y\in X,\parallel x\parallel =1,\parallel y\parallel =\tau \}.$$

It is known that X is uniformly smooth if and only if ${lim}_{\tau \to 0}\rho (\tau )/\tau =0$. Let q be a fixed real number with $1<q\le 2$. Then a Banach space X is said to be q-uniformly smooth if there exists a constant $c>0$ such that $\rho (\tau )\le c{\tau }^q$ for all $\tau >0$. For $q>1$, the generalized duality mapping $J_q:X\to 2^{X^{\ast }}$ is defined by

$$J_q(x)=\{f\in X^{\ast }:〈x,f〉={\parallel x\parallel }^q,\parallel f\parallel ={\parallel x\parallel }^{q-1}\},\mathrm{\forall }x\in X.$$

In particular, if $q=2$, the mapping $J_2$ is called the normalized duality mapping (or duality mapping), and usually we write $J_2=J$. If X is a Hilbert space, then $J=I$. Further, we have the following properties of the generalized duality mapping $J_q$.

• (1) $J_q(x)={\parallel x\parallel }^{q-2}{J}_2(x)$ for all $x\in X$ with $x\ne 0$.

• (2) $J_q(tx)=t^{q-1}{J}_q(x)$ for all $x\in X$ and $t\in [0,\mathrm{\infty })$.

• (3) $J_q(-x)=-J_q(x)$ for all $x\in X$.

It is known that if X is smooth, then J is a single-valued function, which is denoted by j. Recall that the duality mapping j is said to be weakly sequentially continuous if for each $\{x_n\}\subset X$ with $x_n\to x$, we have $j(x_n)\to j(x)$ weakly-∗. We know that if X admits a weakly sequentially continuous duality mapping, then X is smooth. For details, see [20].

Lemma 2.1 [21]

Let X be a q-uniformly smooth Banach space with $1\le q\le 2$. Then

$${\parallel x+y\parallel }^q\le {\parallel x\parallel }^q+q〈y,J_q(x)〉+2{\parallel Ky\parallel }^q$$

for all $x,y\in X$, where K is the q-uniformly smooth constant of X.

Lemma 2.2 [22]

In a Banach space X, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,j(x+y)〉,\mathrm{\forall }x,y\in X,$$

where $j(x+y)\in J(x+y)$.

Lemma 2.3 [23]

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

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n,n\ge 1,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

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

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

Let C be a nonempty closed convex subset of a smooth Banach space X and let D be a nonempty subset of C. A mapping $Q:C\to D$ is said to be sunny if

$$Q(Qx+t(x-Qx))=Qx,$$

whenever $Qx+t(x-Qx)\in C$ for $x\in C$ and $t\ge 0$. A mapping $Q:C\to D$ is called a retraction if $Qx=x$ for all $x\in D$. Furthermore, Q is a sunny nonexpansive retraction from C onto D if Q is a retraction from C onto D, which is also sunny and nonexpansive. A subset D of C is called a sunny nonexpansive retraction of C if there exists a sunny nonexpansive retraction from C onto D.

It is well known that if X is a Hilbert space, then a sunny nonexpansive retraction $Q_C$ is coincident with the metric projection from X onto C.

Lemma 2.4 [24]

Let C be a closed convex subset of a smooth Banach space X. Let D be a nonempty subset of C and $Q:C\to D$ be a retraction. Then the following are equivalent:

• (a) Q is sunny and nonexpansive.

• (b) ${\parallel Qx-Qy\parallel }^2\le 〈x-y,j(Qx-Qy)〉$, $\mathrm{\forall }x,y\in C$.

• (c) $〈x-Qx,j(y-Qx)〉\le 0$, $\mathrm{\forall }x\in C$, $y\in D$.

Lemma 2.5 [25]

If X is strictly convex and uniformly smooth and if $T:C\to C$ is a nonexpansive mapping having a nonempty fixed point set $F(T)$, then the set $F(T)$ is a sunny nonexpansive retraction of C.

Lemma 2.6 [26]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X and let $\{b_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n<1$. Suppose that $x_{n+1}=(1-b_n)y_n+b_n{x}_n$ for all integers $n\ge 1$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.7 [27]

Let C be a closed convex subset of a strictly convex Banach space X. Let $T_1$ and $T_2$ be two nonexpansive mappings from C into itself with $F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Define a mapping S by

$$Sx=\lambda T_{1}x+(1-\lambda )T_{2}x,\mathrm{\forall }x\in C,$$

where λ is a constant in $(0,1)$. Then S is nonexpansive and $F(S)=F(T_1)\cap F(T_2)$.

Lemma 2.8 [28]

Let X be a real smooth and uniformly convex Banach space and let $r>0$. Then there exists a strictly increasing, continuous and convex function $g:[0,2r]\to \mathbb{R}$ such that $g(0)=0$ and $g(\parallel x-y\parallel )\le {\parallel x\parallel }^2-2〈x,j(y)〉+{\parallel y\parallel }^2$ for all $x,y\in B_r$.

Lemma 2.9 [23]

Let X be a uniformly smooth Banach space, let C be a closed convex subset of X, let $T:C\to C$ be a nonexpansive mapping with $F(T)\ne \mathrm{\varnothing }$ and let $f\in {\mathrm{\Pi }}_C$. Then the sequence $\{x_t\}$ defined by $x_t=tf(x_t)+(1-t)T{x}_t$ converges strongly to a point in F(T) as $t\to 0$. If we define a mapping $Q:{\mathrm{\Pi }}_C\to F(T)$ by $Q(f):={lim}_{t\to 0}{x}_t$, $\mathrm{\forall }f\in {\mathrm{\Pi }}_C$, then $Q(f)$ solves the following variational inequality:

$$〈(I-f)Q(f),j(Q(f)-p)〉\le 0,\mathrm{\forall }f\in {\mathrm{\Pi }}_C,p\in F(T).$$

Lemma 2.10 [17]

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let the mapping $A:C\to X$ be relaxed $(c,d)$-cocoercive and $L_A$-Lipschitzian. Then we have

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+2(\lambda c{L}_A^2-\lambda d+K^2{\lambda }^2{L}_A^2){\parallel x-y\parallel }^2,$$

where $\lambda >0$ and K is the 2-uniformly smooth constant of X. In particular, if $0<\lambda \le \frac{d-c{L}_A^2}{K^2{L}_A^2}$, then $I-\lambda A$ is a nonexpansive mapping.

In order to prove our main result, the next lemma is crucial for proving the main theorem.

Lemma 2.11 Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X with the 2-uniformly smooth constant K. Let $Q_C$ be the sunny nonexpansive retraction from X onto C and let $A_i:C\to X$ be a relaxed $(c_i,d_i)$-cocoercive and $L_i$-Lipschitzian mapping for $i=1,2,3$. Let $G:C\to C$ be a mapping defined by

$$\begin{array}{rl}G(x)=& Q_C[Q_C(Q_C(x-{\lambda }_3{A}_{3}x)-{\lambda }_2{A}_2{Q}_C(x-{\lambda }_3{A}_{3}x))\\ -{\lambda }_1{A}_1{Q}_C(Q_C(x-{\lambda }_3{A}_{3}x)-{\lambda }_2{A}_2{Q}_C(x-{\lambda }_3{A}_{3}x))],\mathrm{\forall }x\in C.\end{array}$$

If $0<{\lambda }_i\le \frac{d_i-c_i{L_i}^2}{K^2{L_i}^2}$ for all $i=1,2,3$, then $G:C\to C$ is nonexpansive.

Proof For all $x,y\in C$, by Lemma 2.10, we have

$$\begin{array}{rl}\parallel G(x)-G(y)\parallel =& \parallel Q_C[Q_C(Q_C(I-{\lambda }_3{A}_3)x-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)x)\\ -{\lambda }_1{A}_1{Q}_C(Q_C(I-{\lambda }_3{A}_3)x-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)x)]\\ -Q_C[Q_C(Q_C(I-{\lambda }_3{A}_3)y-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)y)\\ -{\lambda }_1{A}_1{Q}_C(Q_C(I-{\lambda }_3{A}_3)y-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)y)]\parallel \\ \le & \parallel Q_C(Q_C(I-{\lambda }_3{A}_3)x-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)x)\\ -{\lambda }_1{A}_1{Q}_C(Q_C(I-{\lambda }_3{A}_3)x-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)x)\\ -[Q_C(Q_C(I-{\lambda }_3{A}_3)y-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)y)\\ -{\lambda }_1{A}_1{Q}_C(Q_C(I-{\lambda }_3{A}_3)y-{\lambda }_2{A}_2{Q}_C(I-{\lambda }_3{A}_3)y)]\parallel \\ =& \parallel (I-{\lambda }_1{A}_1)Q_C(I-{\lambda }_2{A}_2)Q_C(I-{\lambda }_3{A}_3)x\\ -(I-{\lambda }_1{A}_1)Q_C(I-{\lambda }_2{A}_2)Q_C(I-{\lambda }_3{A}_3)y\parallel \\ \le & \parallel x-y\parallel ,\end{array}$$

which implies that G is nonexpansive. □

Lemma 2.12 [29]

Let C be a nonempty closed convex subset of a real smooth Banach space X. Let $Q_C$ be the sunny nonexpansive retraction from X onto C. Let $A_i:C\to X$ be three possibly nonlinear mappings. For given $x^{\ast },y^{\ast },z^{\ast }\in C$, $(x^{\ast },y^{\ast },z^{\ast })$ is a solution of problem (1.5) if and only if $x^{\ast }\in F(G)$, $y^{\ast }=Q_C(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast })$ and $z^{\ast }=Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })$, where G is the mapping defined as in Lemma  2.11.

3 Main results

We are now in a position to state and prove our main result.

Theorem 3.1 Let X be a uniformly convex and 2-uniformly smooth Banach space with the 2-uniformly smooth constant K, let C be a nonempty closed convex subset of X and $Q_C$ be a sunny nonexpansive retraction from X onto C. Let the mappings $A_i:C\to X$ be relaxed $(c_i,d_i)$-cocoercive and $L_i$-Lipschitzian with $0<{\lambda }_i<\frac{d_i-c_i{L_i}^2}{K^2{L_i}^2}$ for all $i=1,2,3$. Let f be a contractive mapping with the constant $\alpha \in (0,1)$ and let $S:C\to C$ be a nonexpansive mapping such that $\mathrm{\Omega }=F(S)\cap F(G)\ne \mathrm{\varnothing }$, where G is the mapping defined as in Lemma  2.11. For a given $x_1\in C$, let $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ be the sequences generated by

$$\{\begin{array}{c}{z}_n=Q_C(x_n-{\lambda }_3{A}_3{x}_n),\hfill \\ y_n=Q_C(z_n-{\lambda }_2{A}_2{z}_n),\hfill \\ x_{n+1}=a_{n}f(x_n)+b_n{x}_n+(1-a_n-b_n)S{Q}_C(y_n-{\lambda }_1{A}_1{y}_n),n\ge 1,\hfill \end{array}$$

(3.1)

where $\{a_n\}$ and $\{b_n\}$ are two sequences in $(0,1)$ such that

(C1) ${lim}_{n\to \mathrm{\infty }}{a}_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

(C2) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n<1$.

Then $\{x_n\}$ converges strongly to $q\in \mathrm{\Omega }$, which solves the following variational inequality:

$$〈q-f(q),j(q-p)〉\le 0,\mathrm{\forall }f\in {\mathrm{\Pi }}_C,p\in \mathrm{\Omega }.$$

Proof Step 1. We show that $\{x_n\}$ is bounded.

Let $x^{\ast }\in \mathrm{\Omega }$ and $t_n=Q_C(y_n-{\lambda }_1{A}_1{y}_n)$. It follows from Lemma 2.12 that

$$\begin{array}{rl}{x}^{\ast }=& Q_C[Q_C(Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })-{\lambda }_2{A}_2{Q}_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast }))\\ -{\lambda }_1{A}_1{Q}_C(Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })-{\lambda }_2{A}_2{Q}_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast }))].\end{array}$$

Put $y^{\ast }=Q_C(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast })$ and $z^{\ast }=Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })$. Then $x^{\ast }=Q_C(y^{\ast }-{\lambda }_1{A}_1{y}^{\ast })$ and

$$x_{n+1}=a_{n}f(x_n)+b_n{x}_n+(1-a_n-b_n)S{t}_n.$$

From Lemma 2.10, we have $I-{\lambda }_i{A}_i$ ($i=1,2,3$) is nonexpansive. Therefore

$$\begin{array}{rl}\parallel t_n-x^{\ast }\parallel & =\parallel Q_C(y_n-{\lambda }_1{A}_1{y}_n)-Q_C(y^{\ast }-{\lambda }_1{A}_1{y}^{\ast })\parallel \le \parallel y_n-y^{\ast }\parallel \\ =\parallel Q_C(z_n-{\lambda }_2{A}_2{z}_n)-Q_C(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast })\parallel \le \parallel z_n-z^{\ast }\parallel \\ =\parallel Q_C(x_n-{\lambda }_3{A}_3{x}_n)-Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })\parallel \le \parallel x_n-x^{\ast }\parallel \end{array}$$

(3.2)

and $\parallel S{t}_n-x^{\ast }\parallel \le \parallel t_n-x^{\ast }\parallel$. It follows from (3.2) that

$$\begin{array}{rl}\parallel x_{n+1}-x^{\ast }\parallel & =\parallel a_{n}f(x_n)+b_n{x}_n+(1-a_n-b_n)S{t}_n-x^{\ast }\parallel \\ \le a_n\parallel f(x_n)-x^{\ast }\parallel +b_n\parallel x_n-x^{\ast }\parallel +(1-a_n-b_n)\parallel t_n-x^{\ast }\parallel \\ \le a_n\parallel f(x_n)-x^{\ast }\parallel +(1-a_n)\parallel x_n-x^{\ast }\parallel \\ \le a_n\alpha \parallel x_n-x^{\ast }\parallel +a_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +(1-a_n)\parallel x_n-x^{\ast }\parallel \\ =a_n\parallel f\left(x^{\ast }\right)-x^{\ast }\parallel +(1-a_n(1-\alpha ))\parallel x_n-x^{\ast }\parallel .\end{array}$$

By induction, we have

$$\parallel x_{n+1}-x^{\ast }\parallel \le max\{\frac{\parallel f(x^{\ast })-x^{\ast }\parallel }{1-\alpha },\parallel x_1-x^{\ast }\parallel \}.$$

Therefore, $\{x_n\}$ is bounded. Hence $\{y_n\}$, $\{z_n\}$, $\{t_n\}$, $\{A_1{y}_n\}$, $\{A_2{z}_n\}$, $\{S{t}_n\}$, $\{f(x_n)\}$ and $\{A_3{x}_n\}$ are also bounded.

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

By nonexpansiveness of $Q_C$ and $I-{\lambda }_i{A}_i$ ($i=1,2,3$), we have

$$\begin{array}{rcl}\parallel t_{n+1}-t_n\parallel & =& \parallel Q_C(y_{n+1}-{\lambda }_1{A}_1{y}_{n+1})-Q_C(y_n-{\lambda }_1{A}_1{y}_n)\parallel \\ \le & \parallel y_{n+1}-y_n\parallel \\ =& \parallel Q_C(z_{n+1}-{\lambda }_2{A}_2{z}_{n+1})-Q_C(z_n-{\lambda }_2{A}_2{z}_n)\parallel \\ \le & \parallel z_{n+1}-z_n\parallel \\ =& \parallel Q_C(x_{n+1}-{\lambda }_3{A}_3{x}_{n+1})-Q_C(x_n-{\lambda }_3{A}_3{x}_n)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel .\end{array}$$

(3.3)

Let $w_n=\frac{x_{n+1}-b_n{x}_n}{1-b_n}$, $n\in \mathbb{N}$. Then $x_{n+1}=b_n{x}_n+(1-b_n)w_n$ for all $n\in \mathbb{N}$ and

$$\begin{array}{rl}{w}_{n+1}-w_n& =\frac{x_{n+2}-b_{n+1}{x}_{n+1}}{1-b_{n+1}}-\frac{x_{n+1}-b_n{x}_n}{1-b_n}\\ =\frac{a_{n+1}f(x_{n+1})+(1-a_{n+1}-b_{n+1})S{t}_{n+1}}{1-b_{n+1}}-\frac{a_{n}f(x_n)+(1-a_n-b_n)S{t}_n}{1-b_n}\\ =\frac{a_{n+1}}{1-b_{n+1}}(f(x_{n+1})-S{t}_{n+1})+\frac{a_n}{1-b_n}(S{t}_n-f(x_n))+S{t}_{n+1}-S{t}_n.\end{array}$$

(3.4)

By (3.3), (3.4) and nonexpansiveness of S, we have

$$\parallel w_{n+1}-w_n\parallel -\parallel x_{n+1}-x_n\parallel \le \frac{a_{n+1}}{1-b_{n+1}}\parallel f(x_{n+1})-S{t}_{n+1}\parallel +\frac{a_n}{1-b_n}\parallel S{t}_n-f(x_n)\parallel .$$

By this together with (C1) and (C2), we obtain that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel w_{n+1}-w_n\parallel -\parallel x_{n+1}-x_n\parallel \le 0.$$

Hence, by Lemma 2.6, we get $\parallel x_n-w_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Consequently,

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

(3.5)

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$.

Since

$$x_{n+1}-x_n=a_n(f(x_n)-x_n)+(1-a_n-b_n)(S{t}_n-x_n),$$

therefore

$$\parallel S{t}_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.6)

Next, we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-t_n\parallel =0$. From Lemma 2.1 and nonexpansiveness of $Q_C$, we have

$$\begin{array}{rcl}{\parallel z_n-z^{\ast }\parallel }^2& =& {\parallel Q_C(x_n-{\lambda }_3{A}_3{x}_n)-Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }-{\lambda }_3(A_3{x}_n-A_3{x}^{\ast })\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\lambda }_3〈A_3{x}_n-A_3{x}^{\ast },j(x_n-x^{\ast })〉\\ +2K^2{\lambda }_3^2{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\lambda }_3(-c_3{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2+d_3{\parallel x_n-x^{\ast }\parallel }^2)\\ +2K^2{\lambda }_3^2{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ \le & {\parallel x_n-x^{\ast }\parallel }^2+2{\lambda }_3{c}_3{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2-\frac{2{\lambda }_3{d}_3}{L_3^2}{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ +2K^2{\lambda }_3^2{\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ =& {\parallel x_n-x^{\ast }\parallel }^2-2{\lambda }_3(\frac{d_3}{L_3^2}-c_3-K^2{\lambda }_3){\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\end{array}$$

(3.7)

and

$$\begin{array}{rcl}{\parallel y_n-y^{\ast }\parallel }^2& =& {\parallel Q_C(z_n-{\lambda }_2{A}_2{z}_n)-Q_C(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast })\parallel }^2\\ \le & {\parallel z_n-z^{\ast }-{\lambda }_2(A_2{z}_n-A_2{z}^{\ast })\parallel }^2\\ \le & {\parallel z_n-z^{\ast }\parallel }^2-2{\lambda }_2〈A_2{z}_n-A_2{z}^{\ast },j(z_n-z^{\ast })〉\\ +2K^2{\lambda }_2^2{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ \le & {\parallel z_n-z^{\ast }\parallel }^2-2{\lambda }_2(-c_2{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2+d_2{\parallel z_n-z^{\ast }\parallel }^2)\\ +2K^2{\lambda }_2^2{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ \le & {\parallel z_n-z^{\ast }\parallel }^2+2{\lambda }_2{c}_2{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2-\frac{2{\lambda }_2{d}_2}{L_2^2}{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ +2K^2{\lambda }_2^2{\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ =& {\parallel z_n-z^{\ast }\parallel }^2-2{\lambda }_2(\frac{d_2}{L_2^2}-c_2-K^2{\lambda }_2){\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2.\end{array}$$

(3.8)

Similarly, we have

$$\begin{array}{rcl}{\parallel t_n-x^{\ast }\parallel }^2& =& {\parallel Q_C(y_n-{\lambda }_1{A}_1{y}_n)-Q_C(y^{\ast }-{\lambda }_1{A}_1{y}^{\ast })\parallel }^2\\ \le & {\parallel y_n-y^{\ast }-{\lambda }_1(A_1{y}_n-A_1{y}^{\ast })\parallel }^2\\ \le & {\parallel y_n-y^{\ast }\parallel }^2-2{\lambda }_1〈A_1{y}_n-A_1{y}^{\ast },j(y_n-y^{\ast })〉\\ +2K^2{\lambda }_1^2{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2\\ \le & {\parallel y_n-y^{\ast }\parallel }^2-2{\lambda }_1(-c_1{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2+d_1{\parallel y_n-y^{\ast }\parallel }^2)\\ +2K^2{\lambda }_1^2{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2\\ \le & {\parallel y_n-y^{\ast }\parallel }^2+2{\lambda }_1{c}_1{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2-\frac{2{\lambda }_1{d}_1}{L_1^2}{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2\\ +2K^2{\lambda }_1^2{\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2\\ =& {\parallel y_n-y^{\ast }\parallel }^2-2{\lambda }_1(\frac{d_1}{L_1^2}-c_1-K^2{\lambda }_1){\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2.\end{array}$$

(3.9)

Substituting (3.7) and (3.8) into (3.9), we have

$$\begin{array}{rcl}{\parallel t_n-x^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-2{\lambda }_3(\frac{d_3}{L_3^2}-c_3-K^2{\lambda }_3){\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ -2{\lambda }_2(\frac{d_2}{L_2^2}-c_2-K^2{\lambda }_2){\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ -2{\lambda }_1(\frac{d_1}{L_1^2}-c_1-K^2{\lambda }_1){\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2.\end{array}$$

(3.10)

By the convexity of ${\parallel \cdot \parallel }^2$, we obtain

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& ={\parallel a_{n}f(x_n)+b_n{x}_n+(1-a_n-b_n)S{t}_n-x^{\ast }\parallel }^2\\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2+(1-a_n-b_n){\parallel S{t}_n-x^{\ast }\parallel }^2\\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2+(1-a_n-b_n){\parallel t_n-x^{\ast }\parallel }^2.\end{array}$$

(3.11)

Substituting (3.10) into (3.11), we have

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2\le & a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)({\parallel x_n-x^{\ast }\parallel }^2-2{\lambda }_3(\frac{d_3}{L_3^2}-c_3-K^2{\lambda }_3){\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ -2{\lambda }_2(\frac{d_2}{L_2^2}-c_2-K^2{\lambda }_2){\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ -2{\lambda }_1(\frac{d_1}{L_1^2}-c_1-K^2{\lambda }_1){\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2)\\ =& a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-a_n){\parallel x_n-x^{\ast }\parallel }^2\\ -(1-a_n-b_n)2{\lambda }_3(\frac{d_3}{L_3^2}-c_3-K^2{\lambda }_3){\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\\ -(1-a_n-b_n)2{\lambda }_2(\frac{d_2}{L_2^2}-c_2-K^2{\lambda }_2){\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\\ -(1-a_n-b_n)2{\lambda }_1(\frac{d_1}{L_1^2}-c_1-K^2{\lambda }_1){\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2,\end{array}$$

which implies

$$\begin{array}{c}(1-a_n-b_n)2{\lambda }_3(\frac{d_3}{L_3^2}-c_3-K^2{\lambda }_3){\parallel A_3{x}_n-A_3{x}^{\ast }\parallel }^2\hfill \\ +(1-a_n-b_n)2{\lambda }_2(\frac{d_2}{L_2^2}-c_2-K^2{\lambda }_2){\parallel A_2{z}_n-A_2{z}^{\ast }\parallel }^2\hfill \\ +(1-a_n-b_n)2{\lambda }_1(\frac{d_1}{L_1^2}-c_1-K^2{\lambda }_1){\parallel A_1{y}_n-A_1{y}^{\ast }\parallel }^2\hfill \\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\hfill \\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+\parallel x_n-x_{n+1}\parallel (\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel ).\hfill \end{array}$$

By the conditions (C1), (C2), (3.5) and $0<{\lambda }_i<\frac{d_i-c_i{L_i}^2}{K^2{L_i}^2}$ for each $i=1,2,3$, we obtain

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel A_3{x}_n-A_3{x}^{\ast }\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel A_2{z}_n-A_2{z}^{\ast }\parallel =0\text{and}\\ \underset{n\to \mathrm{\infty }}{lim}\parallel A_1{y}_n-A_1{y}^{\ast }\parallel =0.\end{array}$$

(3.12)

Let $r={sup}_{n\ge 1}\{\parallel x_n-x^{\ast }\parallel ,\parallel z_n-z^{\ast }\parallel ,\parallel y_n-y^{\ast }\parallel ,\parallel t_n-x^{\ast }\parallel \}$. By Lemma 2.4(b) and Lemma 2.8, we obtain

$$\begin{array}{rl}{\parallel t_n-x^{\ast }\parallel }^2=& {\parallel Q_C(y_n-{\lambda }_1{A}_1{y}_n)-Q_C(y^{\ast }-{\lambda }_1{A}_1{y}^{\ast })\parallel }^2\\ \le & 〈y_n-{\lambda }_1{A}_1{y}_n-(y^{\ast }-{\lambda }_1{A}_1{y}^{\ast }),j(t_n-x^{\ast })〉\\ =& 〈y_n-y^{\ast },j(t_n-x^{\ast })〉-{\lambda }_1〈A_1{y}_n-A_1{y}^{\ast },j(t_n-x^{\ast })〉\\ \le & \frac{1}{2}[{\parallel y_n-y^{\ast }\parallel }^2+{\parallel t_n-x^{\ast }\parallel }^2-g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)]\\ +{\lambda }_1〈A_1{y}^{\ast }-A_1{y}_n,j(t_n-x^{\ast })〉,\end{array}$$

which implies

$$\begin{array}{rcl}{\parallel t_n-x^{\ast }\parallel }^2& \le & {\parallel y_n-y^{\ast }\parallel }^2-g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ +2{\lambda }_1〈A_1{y}^{\ast }-A_1{y}_n,j(t_n-x^{\ast })〉\\ \le & {\parallel y_n-y^{\ast }\parallel }^2-g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ +2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel .\end{array}$$

(3.13)

And

$$\begin{array}{rl}{\parallel y_n-y^{\ast }\parallel }^2=& {\parallel Q_C(z_n-{\lambda }_2{A}_2{z}_n)-Q_C(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast })\parallel }^2\\ \le & 〈z_n-{\lambda }_2{A}_2{z}_n-(z^{\ast }-{\lambda }_2{A}_2{z}^{\ast }),j(y_n-y^{\ast })〉\\ =& 〈z_n-z^{\ast },j(y_n-y^{\ast })〉-{\lambda }_2〈A_2{z}_n-A_2{z}^{\ast },j(y_n-y^{\ast })〉\\ \le & \frac{1}{2}[{\parallel z_n-z^{\ast }\parallel }^2+{\parallel y_n-y^{\ast }\parallel }^2-g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)]\\ +{\lambda }_2〈A_2{z}^{\ast }-A_2{z}_n,j(y_n-y^{\ast })〉,\end{array}$$

which implies

$$\begin{array}{rcl}{\parallel y_n-y^{\ast }\parallel }^2& \le & {\parallel z_n-z^{\ast }\parallel }^2-g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ +2{\lambda }_2〈A_2{z}^{\ast }-A_2{z}_n,j(y_n-y^{\ast })〉\\ \le & {\parallel z_n-z^{\ast }\parallel }^2-g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ +2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel .\end{array}$$

(3.14)

Similarly, we have

$$\begin{array}{rl}{\parallel z_n-z^{\ast }\parallel }^2=& {\parallel Q_C(x_n-{\lambda }_3{A}_3{x}_n)-Q_C(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast })\parallel }^2\\ \le & 〈x_n-{\lambda }_3{A}_3{x}_n-(x^{\ast }-{\lambda }_3{A}_3{x}^{\ast }),j(z_n-z^{\ast })〉\\ =& 〈x_n-x^{\ast },j(z_n-z^{\ast })〉-{\lambda }_3〈A_3{x}_n-A_3{x}^{\ast },j(z_n-z^{\ast })〉\\ \le & \frac{1}{2}[{\parallel x_n-x^{\ast }\parallel }^2+{\parallel z_n-z^{\ast }\parallel }^2-g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)]\\ +{\lambda }_3〈A_3{x}^{\ast }-A_3{x}_n,j(z_n-z^{\ast })〉,\end{array}$$

which implies

$$\begin{array}{rcl}{\parallel z_n-z^{\ast }\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)\\ +2{\lambda }_3〈A_3{x}^{\ast }-A_3{x}_n,j(z_n-z^{\ast })〉\\ \le & {\parallel x_n-x^{\ast }\parallel }^2-g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)\\ +2{\lambda }_3\parallel A_3{x}^{\ast }-A_3{x}_n\parallel \parallel z_n-z^{\ast }\parallel .\end{array}$$

(3.15)

From (3.11), (3.13), (3.14) and (3.15), we have

$$\begin{array}{rl}{\parallel x_{n+1}-x^{\ast }\parallel }^2\le & a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2+(1-a_n-b_n){\parallel t_n-x^{\ast }\parallel }^2\\ \le & a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)[{\parallel y_n-y^{\ast }\parallel }^2-g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ +2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel ]\\ \le & a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)[{\parallel z_n-z^{\ast }\parallel }^2-g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ +2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel -g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ +2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel ]\\ \le & a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+b_n{\parallel x_n-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)[{\parallel x_n-x^{\ast }\parallel }^2-g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)\\ +2{\lambda }_3\parallel A_3{x}^{\ast }-A_3{x}_n\parallel \parallel z_n-z^{\ast }\parallel -g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ +2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel -g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ +2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel ]\\ =& a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+(1-a_n){\parallel x_n-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)\left(2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_3\parallel A_3{x}^{\ast }-A_3{x}_n\parallel \parallel z_n-z^{\ast }\parallel \right)\\ -(1-a_n-b_n)g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)\\ -(1-a_n-b_n)g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ -(1-a_n-b_n)g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right),\end{array}$$

which implies

$$\begin{array}{r}(1-a_n-b_n)g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)+(1-a_n-b_n)g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)\\ +(1-a_n-b_n)g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)\\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2\\ +(1-a_n-b_n)\left(2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_3\parallel A_3{x}^{\ast }-A_3{x}_n\parallel \parallel z_n-z^{\ast }\parallel \right)\\ \le a_n{\parallel f(x_n)-x^{\ast }\parallel }^2+\parallel x_n-x_{n+1}\parallel (\parallel x_n-x^{\ast }\parallel +\parallel x_{n+1}-x^{\ast }\parallel )\\ +(1-a_n-b_n)\left(2{\lambda }_1\parallel A_1{y}^{\ast }-A_1{y}_n\parallel \parallel t_n-x^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_2\parallel A_2{z}^{\ast }-A_2{z}_n\parallel \parallel y_n-y^{\ast }\parallel \right)\\ +(1-a_n-b_n)\left(2{\lambda }_3\parallel A_3{x}^{\ast }-A_3{x}_n\parallel \parallel z_n-z^{\ast }\parallel \right).\end{array}$$

By the conditions (C1), (C2), (3.5) and (3.12), we obtain

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}g\left(\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel \right)=0,\underset{n\to \mathrm{\infty }}{lim}g\left(\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel \right)=0\text{and}\hfill \\ \underset{n\to \mathrm{\infty }}{lim}g\left(\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel \right)=0.\hfill \end{array}$$

It follows from the properties of g that

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-y^{\ast }-(t_n-x^{\ast })\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-z^{\ast }-(y_n-y^{\ast })\parallel =0\text{and}\hfill \\ \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x^{\ast }-(z_n-z^{\ast })\parallel =0.\hfill \end{array}$$

Therefore

$$\begin{array}{rl}\parallel x_n-t_n\parallel \le & \parallel x_n-z_n-(x^{\ast }-z^{\ast })\parallel +\parallel z_n-y_n-(z^{\ast }-y^{\ast })\parallel \\ +\parallel y_n-t_n-(y^{\ast }-x^{\ast })\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.16)

By (3.6) and (3.16), we have

$$\begin{array}{rl}\parallel S{x}_n-x_n\parallel & \le \parallel S{x}_n-S{t}_n\parallel +\parallel S{t}_n-x_n\parallel \\ \le \parallel x_n-t_n\parallel +\parallel S{t}_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.17)

Define a mapping $W:C\to C$ as

$$Wx=\eta Sx+(1-\eta )Gx,\mathrm{\forall }x\in C,$$

where η is a constant in $(0,1)$. Then it follows from Lemma 2.7 that $F(W)=F(G)\cap F(S)$ and W is nonexpansive. From (3.16) and (3.17), we have

$$\begin{array}{rcl}\parallel x_n-W{x}_n\parallel & =& \parallel x_n-(\eta S{x}_n+(1-\eta )G{x}_n)\parallel \\ =& \parallel \eta (x_n-S{x}_n)+(1-\eta )(x_n-G{x}_n)\parallel \\ \le & \eta \parallel x_n-S{x}_n\parallel +(1-\eta )\parallel x_n-G{x}_n\parallel \\ =& \eta \parallel x_n-S{x}_n\parallel +(1-\eta )\parallel x_n-t_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.18)

Step 4. We claim that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉\le 0,$$

(3.19)

where $q={lim}_{t\to 0}{x}_t$ with $x_t$ being the fixed point of the contraction

$$x\mapsto tf(x)+(1-t)Wx.$$

From Lemma 2.9, we have $q\in F(W)=F(G)\cap F(S)=\mathrm{\Omega }$ and

$$〈(I-f)q,j(q-p)〉\le 0,\mathrm{\forall }f\in {\mathrm{\Pi }}_C,p\in \mathrm{\Omega }.$$

Since $x_t=tf(x_t)+(1-t)W{x}_t$, we have

$$\begin{array}{rl}\parallel x_t-x_n\parallel & =\parallel tf(x_t)+(1-t)W{x}_t-x_n\parallel \\ =\parallel (1-t)(W{x}_t-x_n)+t(f(x_t)-x_n)\parallel .\end{array}$$

It follows from (3.18) and Lemma 2.2 that

$$\begin{array}{rcl}{\parallel x_t-x_n\parallel }^2& =& {\parallel (1-t)(W{x}_t-x_n)+t(f(x_t)-x_n)\parallel }^2\\ \le & {(1-t)}^2{\parallel W{x}_t-x_n\parallel }^2+2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ \le & {(1-t)}^2{(\parallel W{x}_t-W{x}_n\parallel +\parallel W{x}_n-x_n\parallel )}^2\\ +2t〈f(x_t)-x_n,j(x_t-x_n)〉\\ =& {(1-t)}^2({\parallel W{x}_t-W{x}_n\parallel }^2+2\parallel W{x}_t-W{x}_n\parallel \parallel W{x}_n-x_n\parallel +{\parallel W{x}_n-x_n\parallel }^2)\\ +2t〈f(x_t)-x_t,j(x_t-x_n)〉+2t〈x_t-x_n,j(x_t-x_n)〉\\ \le & (1-2t+t^2){\parallel x_t-x_n\parallel }^2+{(1-t)}^2(2\parallel x_t-x_n\parallel \parallel W{x}_n-x_n\parallel +{\parallel W{x}_n-x_n\parallel }^2)\\ +2t〈f(x_t)-x_t,j(x_t-x_n)〉+2t{\parallel x_t-x_n\parallel }^2\\ =& (1+t^2){\parallel x_t-x_n\parallel }^2+f_n(t)+2t〈f(x_t)-x_t,j(x_t-x_n)〉,\end{array}$$

(3.20)

where $f_n(t)={(1-t)}^2(2\parallel x_t-x_n\parallel +\parallel W{x}_n-x_n\parallel )\parallel W{x}_n-x_n\parallel \to 0$ as $n\to \mathrm{\infty }$. It follows from (3.20) that

$$〈x_t-f(x_t),j(x_t-x_n)〉\le \frac{t}{2}{\parallel x_t-x_n\parallel }^2+\frac{f_n(t)}{2t}.$$

(3.21)

Let $n\to \mathrm{\infty }$ in (3.21), we obtain that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_t-f(x_t),j(x_t-x_n)〉\le \frac{t}{2}M,$$

(3.22)

where $M>0$ is a constant such that $M\ge {\parallel x_t-x_n\parallel }^2$ for all $t\in (0,1)$ and $n\ge 1$. Let $t\to 0$ in (3.22), we obtain

$$\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_t-f(x_t),j(x_t-x_n)〉\le 0.$$

(3.23)

On the other hand, we have

$$\begin{array}{rl}〈f(q)-q,j(x_n-q)〉=& 〈f(q)-q,j(x_n-q)〉-〈f(q)-q,j(x_n-x_t)〉\\ +〈f(q)-q,j(x_n-x_t)〉-〈f(q)-x_t,j(x_n-x_t)〉\\ +〈f(q)-x_t,j(x_n-x_t)〉-〈f(x_t)-x_t,j(x_n-x_t)〉\\ +〈f(x_t)-x_t,j(x_n-x_t)〉\\ =& 〈f(q)-q,j(x_n-q)-j(x_n-x_t)〉+〈x_t-q,j(x_n-x_t)〉\\ +〈f(q)-f(x_t),j(x_n-x_t)〉+〈f(x_t)-x_t,j(x_n-x_t)〉.\end{array}$$

It follows that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉\le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)-j(x_n-x_t)〉\\ +\parallel x_t-q\parallel \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x_t\parallel +\alpha \parallel x_t-q\parallel \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x_t\parallel \\ +\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(x_t)-x_t,j(x_n-x_t)〉.\end{array}$$

Noticing that j is norm-to-norm uniformly continuous on a bounded subset of C, it follows from (3.23) and ${lim}_{t\to 0}{x}_t=q$ that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉=\underset{t\to 0}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,j(x_n-q)〉\le 0.$$

Hence (3.19) holds.

Step 5. Finally, we show that $x_n\to q$ as $n\to \mathrm{\infty }$.

From (3.2), we have

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2=& 〈x_{n+1}-q,j(x_{n+1}-q)〉\\ =& 〈a_n(f(x_n)-q)+b_n(x_n-q)+(1-a_n-b_n)(S{t}_n-q),j(x_{n+1}-q)〉\\ =& a_n〈f(x_n)-f(q),j(x_{n+1}-q)〉+b_n〈x_n-q,j(x_{n+1}-q)〉\\ +(1-a_n-b_n)〈S{t}_n-q,j(x_{n+1}-q)〉+a_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & a_n\alpha \parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +b_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ +(1-a_n-b_n)\parallel S{t}_n-q\parallel \parallel x_{n+1}-q\parallel +a_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & a_n\alpha \parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +b_n\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel \\ +(1-a_n-b_n)\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +a_n〈f(q)-q,j(x_{n+1}-q)〉\\ =& (1-a_n(1-\alpha ))\parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +a_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & \frac{1-a_n(1-\alpha )}{2}({\parallel x_n-q\parallel }^2+{\parallel x_{n+1}-q\parallel }^2)+a_n〈f(q)-q,j(x_{n+1}-q)〉\\ \le & \frac{1-a_n(1-\alpha )}{2}{\parallel x_n-q\parallel }^2+\frac{1}{2}{\parallel x_{n+1}-q\parallel }^2+a_n〈f(q)-q,j(x_{n+1}-q)〉,\end{array}$$

which implies

$${\parallel x_{n+1}-q\parallel }^2\le (1-a_n(1-\alpha )){\parallel x_n-q\parallel }^2+a_n(1-\alpha )\frac{2〈f(q)-q,j(x_{n+1}-q)〉}{1-\alpha }.$$

It follows from Lemma 2.3, (3.19) and condition (C1) that $\{x_n\}$ converges strongly to q. This completes the proof. □

Example 3.2 Let $X=\mathbb{R}$ and $C=[0,1]$. Define the mappings $S,f:C\to C$ and $A_1,A_2,A_3:C\to X$ as follows:

$$S(x)=\frac{x}{3},f(x)=\frac{x}{2}+3,A_1(x)=x,A_2(x)=2x\text{and}{A}_3(x)=3x.$$

Then it is obvious that S is nonexpansive, f is contractive with a constant $\alpha =\frac{1}{2}$, $A_1$ is relaxed $(\frac{1}{2},1)$-cocoercive and 1-Lipschitzian, $A_2$ is relaxed $(\frac{1}{4},2)$-cocoercive and 2-Lipschitzian and $A_3$ is relaxed $(\frac{1}{9},3)$-cocoercive and 3-Lipschitzian. In this case, we have $\mathrm{\Omega }=F(S)\cap F(G)=\{0\}$. In the terms of Theorem 3.1, we choose the parameters ${\lambda }_1$, ${\lambda }_2$, ${\lambda }_3$. Then the sequence $\{x_n\}$ generated by (3.1) converges to $q=0\in \mathrm{\Omega }$, which solves the following variational inequality:

$$〈q-f(q),j(q-p)〉\le 0,\mathrm{\forall }p\in \mathrm{\Omega }.$$

Let $A_3=0$ in Theorem 3.1, we obtain the following result.

Corollary 3.3 Let X be a uniformly convex and 2-uniformly smooth Banach space with the 2-uniformly smooth constant K, let C be a nonempty closed convex subset of X and $Q_C$ a sunny nonexpansive retraction from X onto C. Let the mappings $A_i:C\to X$ be relaxed $(c_i,d_i)$-cocoercive and $L_i$-Lipschitzian with $0<{\lambda }_i<\frac{d_i-c_i{L}_i^2}{K^2{L}_i^2}$, for all $i=1,2$. Let f be a contractive mapping with the constant $\alpha \in (0,1)$ and let $S:C\to C$ be a nonexpansive mapping such that $F=F(S)\cap {\mathrm{\Omega }}_2\ne \mathrm{\varnothing }$, where ${\mathrm{\Omega }}_2$ is the set of solutions of problem (1.4). For a given $x_1\in C$, let $\{x_n\}$ and $\{y_n\}$ be the sequences generated by

$$\{\begin{array}{c}{y}_n=Q_C(x_n-{\lambda }_2{A}_2{x}_n),\hfill \\ x_{n+1}=a_{n}f(x_n)+b_n{x}_n+(1-a_n-b_n)S{Q}_C(y_n-{\lambda }_1{A}_1{y}_n),n\ge 1,\hfill \end{array}$$

where $\{a_n\}$ and $\{b_n\}$ are two sequences in $(0,1)$ such that

(C1) ${lim}_{n\to \mathrm{\infty }}{a}_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

(C2) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n<1$.

Then $\{x_n\}$ converges strongly to $q\in F$, which solves the following variational inequality:

$$〈q-f(q),j(q-p)〉\le 0,\mathrm{\forall }f\in {\mathrm{\Pi }}_C,p\in F.$$

Remark 3.4 (i) Since $L^p$ for all $p\ge 2$ is uniformly convex and 2-uniformly smooth, we see that Theorem 3.1 is applicable to $L^p$ for all $p\ge 2$.

• (ii) The problem of finding solutions for a finite number of variational inequalities can use the same idea of a new general system of variational inequalities in Banach spaces.