Generalized Mann Viscosity Implicit Rules for Solving Systems of Variational Inequalities with Constraints of Variational Inclusions and Fixed Point Problems

Abstract

In this work, let X be Banach space with a uniformly convex and q-uniformly smooth structure, where $1<q\le 2$. We introduce and consider a generalized Mann-like viscosity implicit rule for treating a general optimization system of variational inequalities, a variational inclusion and a common fixed point problem of a countable family of nonexpansive mappings in X. The generalized Mann-like viscosity implicit rule investigated in this work is based on the Korpelevich’s extragradient technique, the implicit viscosity iterative method and the Mann’s iteration method. We show that the iterative sequences governed by our generalized Mann-like viscosity implicit rule converges strongly to a solution of the general optimization system.

1. Introduction

Throughout this work, we always suppose that C is a non-empty, convex and closed subset of a real Banach space X. $X^{*}$ will be used to present the dual space of space X. In this present work, we let the norms of X and $X^{*}$ be presented by the denotation $\parallel ·\parallel$. Let T be a nonlinear self mapping with fixed points defined on subset C.

One use $⟨·,·⟩$ to denote the duality pairing. The possible set-valued normalized duality mapping $J:X\to 2^{X^{*}}$ is defined by

$$J\left(x\right):=\{\varphi \in X^{*}:⟨\varphi ,x⟩=\parallel \varphi {\parallel }^2=\parallel x{\parallel }^2\},\forall x\in X.$$

Banach space X is said to be a smooth space (has a Gâteaux differentiable norm) if ${lim}_{t\to 0^{+}}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$ exists for all $\parallel x\parallel =\parallel y\parallel =1$. J is norm-to-weak${}^{*}$ continuous single-valued map in such a space. X is also said to be a uniformly smooth space (has a uniformly Fréchet differentiable norm) if the above limit is attained uniformly for $\parallel x\parallel =\parallel y\parallel =1$ and J is norm-to-norm uniformly continuous on bounded sets in such a space. X is said to be a strictly convex space if $\parallel (1-\lambda )x+\lambda y\parallel <1,\forall \lambda \in (0,1)$ for all $\parallel x\parallel =\parallel y\parallel =1$. Space X is said to be uniformly convex if, for each $\epsilon \in (0,2]$, we have a constant $\delta >0$ such that $\parallel \frac{x+y}{2}\parallel >1-\delta \Rightarrow \parallel x-y\parallel <\epsilon$ for all $\parallel x\parallel =\parallel y\parallel =1$. A uniformly convex Banach space yields a strictly convex Banach space. Under the reflexive framework, X is strictly convex if and only if $X^{*}$ is smooth.

Next, we suppose X is smooth, i.e., J is single-valued. Let $A_1,A_2:C\to X$ be two nonlinear single-valued mappings. One is concerned with the problem of approximating $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}⟨x^{*}+{\mu }_1{A}_1{y}^{*}-y^{*},J(x^{*}-x)⟩\le 0,\forall x\in C,\hfill \\ ⟨y^{*}+{\mu }_2{A}_2{x}^{*}-x^{*},J(y^{*}-x)⟩\le 0,\forall x\in C,\hfill \end{array}\right.$$

(1)

with two real positive constants ${\mu }_1$ and ${\mu }_2$. This optimization system is called a general system of variational inequalities (GSVI). In particular, in the case that $X=H$ is Hilbert, then GSVI (1) is reduced to the following GSVI of finding $(x^{*},y^{*})\in C\times C$ such that

$$\left\{\begin{array}{c}⟨x^{*}+{\mu }_1{A}_1{y}^{*}-y^{*},x^{*}-x⟩\ge 0,\forall x\in C,\hfill \\ ⟨y^{*}+{\mu }_2{A}_2{x}^{*}-x^{*},y^{*}-x⟩\ge 0,\forall x\in C,\hfill \end{array}\right.$$

with two real positive constants ${\mu }_1$ and ${\mu }_2$. This was introduced and studied in [1]. Additionally, if $A=A_1=A_2$ and $x^{*}=y^{*}$, then GSVI (1) becomes the variational problem of finding $x^{*}\in C$ such that $⟨A{x}^{*},J(x-x^{*})⟩\ge 0,\forall x\in C$. In 2006, Aoyama, Iiduka and Takahashi [2] proposed an iterative scheme of finding its approximate solutions and claimed the weak convergence of the iterative sequences governed by the proposed algorithm. Recently, many researchers investigated the variational inequality problem through gradient-based or splitting-based methods; see, e.g., [3,4,5,6,7,8,9,10,11,12,13]. Some stability results can be found at [14,15].

In 2013, Ceng, Latif and Yao [16] analyzed and introduced an implicit computing method by using a double-step relaxed gradient idea in the setting of 2-uniformly smooth and uniformly convex space X with 2-uniform smoothness coefficient ${\kappa }_2$. Let ${\Pi }_C:X\to C$ be a retraction, which is both sunny and nonexpansive. Let $f:C\to C$ be a contraction with constant $\delta \in (0,1)$. Let the mapping $A_i:C\to X$ be ${\alpha }_i$-inverse-strongly accretive for $i=1,2$. Let ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of nonexpansive self single-valued mappings on C such that $\Omega ={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\ne \varnothing$, where $\mathrm{GSVI}(C,A_1,A_2)$ stands for the set of fixed points the mapping $G:={\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(I-{\mu }_2{A}_2)$. For a arbitrary initial $x_0\in C$, let $\left\{x_n\right\}$ be the sequence generated by

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+(1-{\alpha }_n){\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(I-{\mu }_2{A}_2)x_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)S_n{y}_n,\forall n\ge 0,\hfill \end{array}\right.$$

with $0<{\mu }_i<\frac{2{\alpha }_i}{{\kappa }_2}$ for $i=1,2$, where $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ are sequences of real numbers in $(0,1)$ satisfying the restrictions: ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${lim}_{n\to \infty }{\alpha }_n=0$, ${lim\; sup}_{n\to \infty }{\beta }_n<1$ and ${lim\; inf}_{n\to \infty }{\beta }_n>o$. They got convergence analysis of $\left\{x_n\right\}$ to $x^{*}\in \Omega$, which treats the variational inequality: $⟨(I-f)x^{*},J(x^{*}-p)⟩\le 0,\forall p\in \Omega$. Recently, projection-like methods, including sunny nonexpansive retractions, have largely studied in Hilbert and Banach spaces; see, e.g., [17,18,19,20,21,22,23,24] and the references therein.

2. Preliminaries

Next, we let X be a space with uniformly convex and q-uniformly smooth structures. Then the following inequality holds:

$${\parallel x+y\parallel }^q\le {\kappa }_q{\parallel y\parallel }^q+{\parallel x\parallel }^q+q⟨y,{\mathfrak{J}}_q\left(x\right)⟩,\forall x,y\in E,$$

where ${\kappa }_q$ is the smoothness coefficient. Let ${\Pi }_C,A_1,A_2,G,{\left\{S_n\right\}}_{n=0}^{\infty }$ be the same mappings as above. Assume that $\Omega ={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\ne \varnothing$. Suppose that $F:C\to X$ is a $\eta$-strongly accretive operator with constants $k,\eta >0$ and k-Lipschitzian, and $f:C\to X$ is L-Lipschitzian mapping. Assume $0<\rho <{\left(\frac{q\eta }{{\kappa }_q{k}^q}\right)}^{\frac{1}{q-1}},0<{\mu }_i<{\left(\frac{q{\alpha }_i}{{\kappa }_q}\right)}^{\frac{1}{q-1}},i=1,2$, and $0\le \gamma L<\tau$, where $\tau =\rho (\eta -\frac{{\kappa }_q{\rho }^{q-1}{k}^q}{q})$. Recently, Song and Ceng [25] proposed and considered a very general iterative scheme by the modified relaxed extragradient method, i.e., for arbitrary initial $x_0\in C$, we generate $\left\{x_n\right\}$ by

$$\left\{\begin{array}{c}{y}_n=(1-{\beta }_n)x_n+{\beta }_n{\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(I-{\mu }_2{A}_2)x_n,\hfill \\ x_{n+1}={\Pi }_C[((1-{\gamma }_n)I-{\alpha }_n\rho F)S_n{y}_n+\gamma {\alpha }_{n}f\left(x_n\right)+{\gamma }_n{x}_n],\forall n\ge 0,\hfill \end{array}\right.$$

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ satisfying the conditions: (i) ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$ and ${\alpha }_n\to 0$; (ii) ${lim\; sup}_{n\to \infty }{\gamma }_n<1$, ${lim\; inf}_{n\to \infty }{\gamma }_n>0$, ${\sum }_{n=0}^{\infty }|{\gamma }_{n+1}-{\gamma }_n|<\infty$; and (iii) ${\sum }_{n=0}^{\infty }|{\beta }_{n+1}-{\beta }_n|<\infty$, ${lim\; inf}_{n\to \infty }{\beta }_n>0$. The authors claimed convergence of $\left\{x_n\right\}$ to $x^{*}\in \Omega$, which deals with the variational inequality: $⟨(\rho F-\gamma f)x^{*},J(x^{*}-p)⟩\le 0,\forall p\in \Omega$.

On the other hand, Let $A:X\to X$ be an $\alpha$-inverse-strongly accretive operator, $B:X\to 2^X$ be an m-accretive operator, $f:X\to X$ be a contraction with constant $\delta \in (0,1)$. Assume that the inclusion of finding $x^{*}\in X$ such that $0\in (A+B)x^{*}$, has a solution, i.e., $\Omega ={(A+B)}^{-1}0\ne \varnothing$. In 2017, Chang et al. [26] introduced and studied a generalized viscosity implicit rule, i.e., for arbitrary initial $x_0\in X$, we generate $\left\{x_n\right\}$ by

$$x_{n+1}=(1-{\alpha }_n)J_{\lambda }^B(I-\lambda A)(t_n{x}_n+(1-t_n)x_{n+1})+{\alpha }_{n}f\left(x_n\right),\forall n\ge 0,$$

where $J_{\lambda }^B={(I+\lambda B)}^{-1}$, $\left\{t_n\right\},\left\{{\alpha }_n\right\}\subset [0,1]$ and $\lambda \in (0,\infty )$ satisfying the conditions: (i) ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$; (ii) ${\sum }_{n=0}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$; (iii) $0<\epsilon \le t_n\le t_{n+1}<1,\forall n\ge 0$; and (iv) $0<\lambda \le {\left(\frac{\alpha q}{{\kappa }_q}\right)}^{\frac{1}{q-1}}$. These authors studied and proved convergence of $\left\{x_n\right\}$ to $x^{*}\in \Omega$, solving the inequality: $⟨(I-f)x^{*},J(x^{*}-p)⟩\le 0,\forall p\in \Omega$. For recent results, we refer the reader to [27,28,29,30,31,32,33,34]. The purpose of this work is to approximate a common solution of GSVI (1), a variational inclusion and a common fixed point problem of a countable family of nonexpansive mappings in spaces with uniformly convex and q-uniformly smooth structures. This paper introduces and considers a generalized Mann viscosity implicit rule, based on the Korpelevich’s extragradient method, the implicit approximation method and the Mann’s iteration method. We investigate norm convergence of the sequences generated by the generalized Mann viscosity implicit rule to a common solution of the GSVI, VI and CFPP, which solves a hierarchical variational inequality. Our results improve and extend the results reported recently, e.g., Ceng et al. [16], Song and Ceng [25] and Chang et al. [26].

Next, for simplicity, we employ $x_n\rightharpoonup x$ (resp., $x_n\to x$) to present the weak (resp., strong) convergence of the sequence $\left\{x_n\right\}$ to x. It is known that $J\left(tx\right)=tJ\left(x\right)$ and $J(-x)=-J\left(x\right)$ for all $t>0$ and $x\in X$. Then the convex modulus of X is defined by ${\delta }_X\left(ϵ\right)=inf\{1-\parallel \frac{1}{2}(x+y)\parallel :x,y\in U,\parallel x-y\parallel \ge ϵ\},\forall ϵ\in [0,2]$. X is said to be uniformly convex if ${\delta }_X\left(0\right)=0$, and ${\delta }_X\left(ϵ\right)>0$ for each $ϵ\in (0,2]$. Let q be a fixed real number with $q>1$. Then a Banach space X is said to be q-uniformly convex if ${\delta }_X\left(t\right)\ge c{t}^q,\forall t\in (0,2]$, where $c>0$. Each Hilbert space H is 2-uniformly convex, while $L^p$ and ${\ell }_p$ spaces are $max\{2,p\}$-uniformly convex for each $p>1$.

Proposition 1.

[35] Let X be space with smooth and uniformly convex structures, and $r>0$. Then $g\left(0\right)=0$ and $g(\parallel x-y\parallel )\le {\parallel x\parallel }^2-2⟨x,J\left(y\right)⟩+{\parallel y\parallel }^2$ for all $x,y\in B_r=\{y\in X:\parallel y\parallel \le r\}$, where $g:[0,2r]\to \mathbf{R}$ is a continuous, strictly increasing, and convex function.

Let ${\rho }_X:[0,\infty )\to [0,\infty )$ be the smooth modulus of X defined by

$${\rho }_X\left(t\right)=sup\{\frac{\parallel x+y\parallel +\parallel x-y\parallel -2}{2}:\parallel x\parallel =1,\parallel y\parallel \le t\}.$$

A Banach space X is said to be q-uniformly smooth if ${\rho }_X\left(t\right)\le c{t}^q,\forall t>0$, where $c>0$. It is known that each Hilbert, $L^p$ and ${\ell }_p$ spaces are uniformly smooth where $p>1$. More precisely, each Hilbert space H is 2-uniformly smooth, while $L^p$ and ${\ell }_p$ spaces are $min\{2,p\}$-uniformly smooth for each $p>1$. Let $q>1$. $J_q:X\to 2^{X^{*}}$, the duality mapping, is defined by

$$J_q\left(x\right):=\{\varphi \in X^{*}:⟨x,\varphi ⟩={\parallel x\parallel }^q\mathrm{and}\parallel \varphi \parallel ={\parallel x\parallel }^{q-1}\},\forall x\in X.$$

It is quite easy to see that $J_q\left(x\right)=J\left(x\right){\parallel x\parallel }^{q-2}$, and if $X=H$, then $J_2=J=I$ the identity mapping of H.

Proposition 2.

[35] Let $q\in (1,2]$ a given real number and let X be uniformly smooth with order q. Then ${\parallel x+y\parallel }^q-q⟨y,J_q\left(x\right)⟩\le {\parallel x\parallel }^q+{\kappa }_q{\parallel y\parallel }^q,\forall x,y\in X$, where ${\kappa }_q$ is the real smooth constant. In particular, if X is uniformly smooth with order 2, then ${\parallel x+y\parallel }^2-2⟨y,J\left(x\right)⟩\le {\parallel x\parallel }^2+{\kappa }_2{\parallel y\parallel }^2,\forall x,y\in X$.

Using the structures of subdifferentials, we obtain the following tool.

Lemma 1.

Let $q>1$ and X be a real normed space with the generalized duality mapping $J_q$. Then, for any given $x,y\in X$, ${\parallel x+y\parallel }^q-q⟨y,j_q(x+y)⟩\le {\parallel x\parallel }^q,\forall j_q(x+y)\in J_q(x+y)$.

Let D be a set in set C and let $\Pi$ map C into D. We say that $\Pi$ is sunny if $\Pi \left(x\right)=\Pi \left[t\right(x-\Pi \left(x\right))+\Pi (x\left)\right]$, whenever $\Pi \left(x\right)+t(x-\Pi (x\left)\right)\in C$ for $x\in C$ and $t\ge 0$. We say $\Pi$ is a retraction if $\Pi ={\Pi }^2$. We say that a subset D of C is a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D.

Proposition 3.

[36] Let X be smooth, D be a non-empty set in C and Π be a retraction onto D. (i) Π is nonexpansive sunny; (ii) $⟨x-y,J(\Pi \left(x\right)-\Pi \left(y\right))⟩\ge {\parallel \Pi \left(x\right)-\Pi \left(y\right)\parallel }^2$, $\forall x,y\in C$; (iii) $⟨x-\Pi \left(x\right),J(y-\Pi (x\left)\right)⟩\le 0$, $\forall x\in C,y\in D$. Then the above relations are equivalent to each other.

Let $A:C\to 2^X$ be a set-valued operator with $Ax\ne \varnothing ,\forall x\in C$. Let $q>1$. An operator A is accretive if for each $x,y\in C$, there exists $j_q(x-y)\in J_q(x-y)$ such that $⟨j_q(x-y),u-v⟩\ge 0,$ $\forall u\in Ax,v\in Ay$. An accretive operator A is inverse-strongly accretive of order q, i.e., $\alpha$-inverse-strongly accretive, if for each $x,y\in C$, there exist $\alpha >0$ such that $⟨u-v,j_q(x-y)⟩\ge \alpha {\parallel Ax-Ay\parallel }^q,$ $\forall u\in Ax,v\in Ay$, where $j_q(x-y)\in J_q(x-y)$. In a Hilbert space H, $A:C\to H$ is called $\alpha$-inverse-strongly monotone.

Operator A is said to be m-accretive if and only if $(I+\lambda A)C=X$ for all $\lambda >0$ and A is accretive. One defines the mapping $J_{\lambda }^A:(I+\lambda A)C\to C$ by $J_{\lambda }^A={(I+\lambda A)}^{-1}$ with real constant $\lambda >0$. Such $J_{\lambda }^A$ is called the resolvent mapping of A for each $\lambda >0$.

Lemma 2.

[37] The following statements hold:

(i) 

the resolvent identity: $J_{\lambda }^{A}x=J_{\mu }(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{\lambda }^{A}x),\forall \lambda ,\mu >0,x\in X$;

(ii) 

if $J_{\lambda }^A$ is a resolvent of A for $\lambda >0$, then $J_{\lambda }^A$ is a single-valued nonexpansive mapping with $\mathrm{Fix}\left(J_{\lambda }^A\right)=A^{-1}0$, where $A^{-1}0=\{x\in C:0\in Ax\}$;

(iii) 

in a Hilbert space H, an operator A is maximal monotone iff it is m-accretive.

Let $A:C\to X$ be an $\alpha$-inverse-strongly accretive mapping and $B:C\to 2^X$ be an m-accretive operator. In the sequel, one will use the notation $T_{\lambda }:=J_{\lambda }^B(I-\lambda A)={(I+\lambda B)}^{-1}(I-\lambda A),\forall \lambda >0$. The following statements (see [38]) hold:

(i)

$\mathrm{Fix}\left(T_{\lambda }\right)={(A+B)}^{-1}0,\forall \lambda >0$;

(ii)

$\parallel x-T_{\lambda }x\parallel \le 2\parallel x-T_{s}x\parallel$ for $0<\lambda \le s$ and $x\in X$.

Proposition 4.

[38] Let X be a Banach space with the uniformly convex and q-uniformly smooth structures with $1<q\le 2$. Assume that $A:C\to X$ is a α-inverse-strongly accretive single-valued mapping and $B:C\to 2^X$ is an m-accretive operator. Then

$$\parallel T_{\lambda }x-T_{\lambda }{y\parallel }^q\le {\parallel x-y\parallel }^q-\lambda (\alpha q-{\lambda }^{q-1}{\kappa }_q){\parallel Ax-Ay\parallel }^q-\varphi (\parallel (I-J_{\lambda }^B)(I-\lambda A)x-(I-J_{\lambda }^B)(I-\lambda A)y\parallel ),$$

for all $x,y\in {\tilde{B}}_r:=\{x\in C:\parallel x\parallel \le r\}$, where $\varphi :{\mathbf{R}}^{+}\to {\mathbf{R}}^{+}$ with $\varphi \left(0\right)=0$ is a convex, strictly increasing and continuous function, λ and r two positive real constants, ${\kappa }_q$ is the real smooth constant of X, and $T_{\lambda }$ and $J_{\lambda }^B$ are resolvent operators defined as above. In particular, if $0<\lambda \le {\left(\frac{\alpha q}{{\kappa }_q}\right)}^{\frac{1}{q-1}}$, then $T_{\lambda }$ is nonexpansive.

Lemma 3.

[39] Let X be uniformly smooth, T be single-valued nonexpansivitity on C with $\mathrm{Fix}\left(T\right)\ne \varnothing$, and $f:C\to C$ be a any contraction. For each $t\in (0,1)$, one employs $z_t\in C$ to present the unique fixed point of the new contraction $C\ni z\mapsto tf\left(z\right)+(1-t)Tz$ on C, i.e., $z_t=(1-t)T{z}_t+tf\left(z_t\right)$. Then $\left\{z_t\right\}$ converges to $x^{*}\in \mathrm{Fix}\left(T\right)$ in norm, which deals with the variational inequality: $⟨(I-f)x^{*},J(x^{*}-p)⟩\le 0,\forall p\in \mathrm{Fix}\left(T\right)$.

Lemma 4.

[25] Let X be a uniformly smooth with order q. Suppose that ${\Pi }_C$ is a sunny nonexpansive retraction from X onto C. Let the mapping $A_i:C\to X$ be ${\alpha }_i$-inverse-strongly accretive of order q for $i=1,2$. Let the mapping $G:C\to C$ be defined as $Gx:={\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(I-{\mu }_2{A}_2),\forall x\in C$. If $0<{\mu }_i\le {\left(\frac{q{\alpha }_i}{{\kappa }_q}\right)}^{\frac{1}{q-1}}$ for $i=1,2$, then $G:C\to C$ is nonexpansive. For given $(x^{*},y^{*})\in C\times C$, $(x^{*},y^{*})$ is a solution of GSVI (1) if and only if $x^{*}={\Pi }_C(y^{*}-{\mu }_1{A}_1{y}^{*})$ where $y^{*}={\Pi }_C(x^{*}-{\mu }_2{A}_2{x}^{*})$, i.e., $x^{*}=G{x}^{*}$.

Lemma 5.

[40] Let ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a mapping sequence on C. Suppose that ${\sum }_{n=1}^{\infty }sup\{\parallel S_{n}x-S_{n-1}x\parallel :x\in C\}<\infty$. Then $\left\{S_{n}x\right\}$ converges to some point of C in norm for each $x\in C$. Besides, we present S, a self-mapping, on C by $Sx={lim}_{n\to \infty }{S}_{n}x,\forall x\in C$. Then ${lim}_{n\to \infty }sup\{\parallel S_{n}x-Sx\parallel :x\in C\}=0$.

Lemma 6.

[41] Let X be Banach space. Let $\left\{{\alpha }_n\right\}$ be a real sequence in $(0,1)$ with ${lim\; sup}_{n\to \infty }{\alpha }_n<1$ and ${lim\; inf}_{n\to \infty }{\alpha }_n>0$. Let $x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)y_n,\forall n\ge 0$ and ${lim\; sup}_{n\to \infty }(\parallel y_n-y_{n+1}\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$, where $\left\{x_n\right\}$ and $\left\{y_n\right\}$ be bounded sequences in X. Then ${lim}_{n\to \infty }\parallel y_n-x_n\parallel =0$.

Lemma 7.

[42] Let X be strictly convex, and ${\left\{T_n\right\}}_{n=0}^{\infty }$ be a sequence of nonexpansive mappings on C. Suppose that ${\cap }_{n=0}^{\infty }\mathrm{Fix}\left(T_n\right)\ne \varnothing$. Let $\left\{{\lambda }_n\right\}$ be a sequence of positive numbers with ${\sum }_{n=0}^{\infty }{\lambda }_n=1$. Then a mapping S on C defined by $Sx={\sum }_{n=0}^{\infty }{\lambda }_n{T}_{n}x$ for $x\in C$ is defined well, nonexpansive and $\mathrm{Fix}\left(S\right)={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(T_n\right)$ holds.

Lemma 8.

[43] Let $\left\{a_n\right\}$ be a non-negative number sequence of with $a_{n+1}\le a_n(1-{\lambda }_n)+{\lambda }_n{\gamma }_n,\forall n\ge 1$, where $\left\{{\gamma }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are sequences such that (a) ${\sum }_{n=1}^{\infty }\left|{\lambda }_n{\gamma }_n\right|<\infty$ (or ${lim\; sup}_{n\to \infty }{\gamma }_n\le 0$ ) and (b) $\left\{{\lambda }_n\right\}\subset [0,1]$ and ${\sum }_{n=1}^{\infty }{\lambda }_n=\infty$. Then $a_n$ goes to zero as n goes to the infinity.

Lemma 9.

[7,35] Let X be uniformly convex, and the ball $B_r=\{x\in X:\parallel x\parallel \le r\},r>0$. Then

$${\parallel \alpha x+\beta y+\gamma y\parallel }^2+\alpha \beta g(\parallel x-y\parallel )\le {\alpha \parallel x\parallel }^2+{\beta \parallel y\parallel }^2+\gamma {\parallel z\parallel }^2$$

for all $x,y,z\in B_r$ and $\alpha ,\beta ,\gamma \in [0,1]$ with $\alpha +\beta +\gamma =1$, where $g:[0,\infty )\to [0,\infty )$ is a convex, continuous and strictly increasing function.

3. Iterative Algorithms and Convergence Criteria

Space X presents a real Banach space and its topological dual is $X^{*}$, and C is a non-empty convex and closed set in space X. We are now ready to state and prove the main results in this paper.

Theorem 1.

Let X be uniformly convex and uniformly smooth with the constant $1<q\le 2$. Let ${\Pi }_C$ be a nonexpansive sunny retraction from X onto C. Assume that the mappings $A,A_i:C\to X$ are inverse-strongly accretive of order q and ${\alpha }_i$-inverse-strongly accretive of order q, respectively for $i=1,2$. Let $B:C\to 2^X$ be an m-accretive operator, and let ${\left\{S_n\right\}}_{n=0}^{\infty }$ be a countable family of nonexpansive single-valued self-mappings on C such that $\Omega ={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap {(A+B)}^{-1}0\ne \varnothing$ where $\mathrm{GSVI}(C,A_1,A_2)$ is the fixed point set of $G:={\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(I-{\mu }_2{A}_2)$ with $0<{\mu }_i<{\left(\frac{q{\alpha }_i}{{\kappa }_q}\right)}^{\frac{1}{q-1}}$ for $i=1,2$. Let $f:C\to C$ be a contraction with constant $\delta \in (0,1)$. For arbitrary initial $x_0\in C$, $\left\{x_n\right\}$ is a sequence generated by

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+{\gamma }_n{J}_{{\lambda }_n}^B(I-{\lambda }_{n}A)(t_n{x}_n+(1-t_n)y_n)+{\beta }_n{x}_n,\hfill \\ v_n={\Pi }_C(I-{\mu }_1{A}_1){\Pi }_C(y_n-{\mu }_2{A}_2{y}_n),\hfill \\ x_{n+1}=(1-{\delta }_n)S_n{v}_n+{\delta }_n{x}_n,n\ge 0,\hfill \end{array}\right.$$

where $\left\{{\lambda }_n\right\}\subset (0,{\left(\frac{q\alpha }{{\kappa }_q}\right)}^{\frac{1}{q-1}})$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\},\left\{t_n\right\}\subset (0,1)$ satisfy

(i) 

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, ${lim}_{n\to \infty }{\alpha }_n=0$ and ${\alpha }_n+{\beta }_n+{\gamma }_n=1$;

(ii) 

${lim}_{n\to \infty }|{\beta }_n-{\beta }_{n-1}|={lim}_{n\to \infty }|{\gamma }_n-{\gamma }_{n-1}|=0$;

(iii) 

${lim}_{n\to \infty }|t_n-t_{n-1}|=0$ and ${lim\; inf}_{n\to \infty }{\gamma }_n(1-t_n)>0$;

(iv) 

${lim\; inf}_{n\to \infty }{\beta }_n{\gamma }_n>0$, ${lim\; sup}_{n\to \infty }{\delta }_n<1$ and ${lim\; inf}_{n\to \infty }{\delta }_n>0$;

(v) 

$0<\overline{\lambda }\le {\lambda }_n,\forall n\ge 0$ and ${lim}_{n\to \infty }{\lambda }_n=\lambda <{\left(\frac{q\alpha }{{\kappa }_q}\right)}^{\frac{1}{q-1}}$.

Assume that ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel (S_n-S_{n-1})x\parallel <\infty$ for any bounded set D, which is subset of C and let S be a self-mapping $Sx={lim}_{n\to \infty }{S}_{n}x,\forall x\in C$ and suppose that $\mathrm{Fix}\left(S\right)={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)$. Then $x_n\to x^{*}\in \Omega$, which solves $⟨(I-f)x^{*},J(x^{*}-p)⟩\le 0,\forall p\in \Omega$ uniquely.

Proof. 

Set $u_n={\Pi }_C(y_n-{\mu }_2{A}_2{y}_n).$ It is not hard to find that our scheme can be re-written by

$$\left\{\begin{array}{cc}\hfill & y_n={\alpha }_{n}f\left(y_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n(t_n{x}_n+(1-t_n)y_n),\hfill \\ \hfill & x_{n+1}=(1-{\delta }_n)S_{n}G{y}_n+{\delta }_n{x}_n,n\ge 0,\hfill \end{array}\right.$$

(2)

where $T_n:=J_{{\lambda }_n}^B(I-{\lambda }_{n}A),\forall n\ge 0$. By condition (v) and Proposition 4, one observes that $T_n:C\to C$ is a nonexpansive mapping for each $n\ge 0$. Since ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, we know that

$${\alpha }_n\delta +{\gamma }_n(1-t_n)+{\beta }_n+{\gamma }_n{t}_n={\alpha }_n\delta +{\gamma }_n+{\beta }_n=1-{\alpha }_n(1-\delta ),\forall n\ge 0.$$

One first claims that the sequence $\left\{x_n\right\}$ generated by (2) is well defined. Indeed, for each fixed $x_n\in C$, one defines a mapping $F_n:C\to C$ by $F_n\left(x\right)={\alpha }_{n}f\left(x\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n(t_n{x}_n+(1-t_n)x),\forall x\in C$. Then, one gets, for any $x,y\in C$,

$$\begin{array}{c}\parallel F_n\left(x\right)-F_n\left(y\right)\parallel \le {\alpha }_n\parallel f\left(x\right)-f\left(y\right)\parallel +{\gamma }_n\parallel T_n(t_n{x}_n+(1-t_n)x)-T_n(t_n{x}_n+(1-t_n)y)\parallel \hfill \\ \le {\alpha }_n\delta \parallel x-y\parallel +{\gamma }_n(1-t_n)\parallel x-y\parallel =({\alpha }_n\delta +{\gamma }_n(1-t_n))\parallel x-y\parallel \le (1-{\alpha }_n(1-\delta ))\parallel x-y\parallel .\hfill \end{array}$$

This implies that $F_n$ is a strictly contraction operator. Hence the Banach fixed-point theorem ensures that there is a unique fixed point $y_n\in C$ satisfying

$$y_n={\alpha }_{n}f\left(y_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n(t_n{x}_n+(1-t_n)y_n).$$

Next, one claims that $\left\{x_n\right\}$ is bounded. Indeed, arbitrarily take a fixed $p\in \Omega ={\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap {(A+B)}^{-1}0$. One knows that $S_{n}p=p,Gp=p$ and $T_{n}p=p$. Moreover, by using the nonexpansivity of $T_n$, we have

$$\begin{array}{cc}\parallel y_n-p\parallel \hfill & \le {\alpha }_n(\parallel f\left(y_n\right)-f\left(p\right)\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel T_n(t_n{x}_n+(1-t_n)y_n)-p\parallel \hfill \\ & \le {\alpha }_n(\delta \parallel y_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\beta }_n\parallel x_n-p\parallel +{\gamma }_n[t_n\parallel x_n-p\parallel +(1-t_n)\parallel y_n-p\parallel ]\hfill \\ & =({\alpha }_n\delta +{\gamma }_n(1-t_n))\parallel y_n-p\parallel +({\beta }_n+{\gamma }_n{t}_n)\parallel x_n-p\parallel +{\alpha }_n\parallel p-f\left(p\right)\parallel ,\hfill \end{array}$$

which hence implies that

$$\begin{array}{cc}\parallel y_n-p\parallel \hfill & \le \frac{{\gamma }_n{t}_n+{\beta }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \hfill \\ \hfill & =\frac{1-{\alpha }_n(1-\delta )-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \hfill \\ \hfill & =(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel .\hfill \end{array}$$

(3)

Thus, from (2) and (3), we have

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \le {\delta }_n\parallel x_n-p\parallel +(1-{\delta }_n)\parallel S_{n}G{y}_n-p\parallel \le (1-{\delta }_n)\parallel y_n-p\parallel +{\delta }_n\parallel x_n-p\parallel \hfill \\ \le {\delta }_n\parallel x_n-p\parallel +(1-{\delta }_n)\{(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_n-p\parallel +\frac{{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \}\hfill \\ =[1-\frac{(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{\alpha }_n]\parallel x_n-p\parallel +\frac{(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{\alpha }_n\frac{\parallel f\left(p\right)-p\parallel }{1-\delta }.\hfill \end{array}$$

By induction, we get that $\left\{x_n\right\}$ is bounded. Please note that G is non-expansive thanks to Lemma 4. Using (3) and the nonexpansivity of $I-{\mu }_1{A}_1,I-{\mu }_2{A}_2,S_n,T_n$ and G, it is guaranteed that $\left\{u_n\right\},\left\{v_n\right\},\left\{y_n\right\},\left\{G{y}_n\right\},\left\{S_n{v}_n\right\}$ and $\left\{T_n{z}_n\right\}$ are bounded too, where $u_n:={\Pi }_C(I-{\mu }_2{A}_2)y_n,v_n:={\Pi }_C(I-{\mu }_1{A}_1)u_n$ and $z_n:=t_n{x}_n+(1-t_n)y_n$ for all $n\ge 0$. Thanks to (2), we have

$$\left\{\begin{array}{c}{z}_n=t_n(x_n-y_n)+y_n,\hfill \\ z_{n-1}=t_{n-1}(x_{n-1}-y_{n-1})+y_{n-1},\forall n\ge 1,\hfill \end{array}\right.$$

and

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n{z}_n,\hfill \\ y_{n-1}={\alpha }_{n-1}f\left(y_{n-1}\right)+{\beta }_{n-1}{x}_{n-1}+{\gamma }_{n-1}{T}_{n-1}{z}_{n-1},\forall n\ge 1,\hfill \end{array}\right.$$

Simple calculations show that

$$z_n-z_{n-1}=(t_n-t_{n-1})(x_{n-1}-y_{n-1})+(1-t_n)(y_n-y_{n-1})+t_n(x_n-x_{n-1}),$$

and

$$\begin{array}{cc}{y}_n-y_{n-1}\hfill & =({\alpha }_n-{\alpha }_{n-1})f\left(y_{n-1}\right)+{\alpha }_n(f\left(y_n\right)-f\left(y_{n-1}\right))+{\beta }_n(x_n-x_{n-1})\hfill \\ \hfill & +({\beta }_n-{\beta }_{n-1})x_{n-1}+{\gamma }_n(T_n{z}_n-T_{n-1}{z}_{n-1})+({\gamma }_n-{\gamma }_{n-1})T_{n-1}{z}_{n-1}.\hfill \end{array}$$

(4)

It follows from the resolvent identity that

$$\begin{array}{cc}\hfill & \parallel T_n{z}_n-T_{n-1}{z}_{n-1}\parallel \le \parallel T_n{z}_n-T_n{z}_{n-1}\parallel +\parallel T_n{z}_{n-1}-T_{n-1}{z}_{n-1}\parallel \hfill \\ \hfill & \le \parallel z_n-z_{n-1}\parallel +\parallel J_{{\lambda }_n}^B(I-{\lambda }_{n}A)z_{n-1}-J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n-1}A)z_{n-1}\parallel \hfill \\ \hfill & \le \parallel z_n-z_{n-1}\parallel +\parallel J_{{\lambda }_n}^B(I-{\lambda }_{n}A)z_{n-1}-J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n}A)z_{n-1}\parallel \hfill \\ \hfill & +\parallel J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n}A)z_{n-1}-J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n-1}A)z_{n-1}\parallel \hfill \\ \hfill & =\parallel z_n-z_{n-1}\parallel +\parallel J_{{\lambda }_{n-1}}^B(\frac{{\lambda }_{n-1}}{{\lambda }_n}I+(1-\frac{{\lambda }_{n-1}}{{\lambda }_n})J_{{\lambda }_n}^B)(I-{\lambda }_{n}A)z_{n-1}-J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n}A)z_{n-1}\parallel \hfill \\ \hfill & +\parallel J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n}A)z_{n-1}-J_{{\lambda }_{n-1}}^B(I-{\lambda }_{n-1}A)z_{n-1}\parallel \hfill \\ \hfill & \le t_n\parallel x_n-x_{n-1}\parallel +|t_n-t_{n-1}|\parallel x_{n-1}-y_{n-1}\parallel +(1-t_n)\parallel y_n-y_{n-1}\parallel \hfill \\ \hfill & +|1-\frac{{\lambda }_{n-1}}{{\lambda }_n}|\parallel J_{{\lambda }_n}^B(I-{\lambda }_{n}A)z_{n-1}-(I-{\lambda }_{n}A)z_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|\parallel A{z}_{n-1}\parallel \hfill \\ \hfill & \le t_n\parallel x_n-x_{n-1}\parallel +|t_n-t_{n-1}|\parallel x_{n-1}-y_{n-1}\parallel +(1-t_n)\parallel y_n-y_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|M_1,\hfill \end{array}$$

(5)

where

$$\underset{n\ge 1}{sup}\{\frac{\parallel J_{{\lambda }_n}^B(I-{\lambda }_{n}A)z_{n-1}-(I-{\lambda }_{n}A)z_{n-1}\parallel }{\overline{\lambda }}+\parallel A{z}_{n-1}\parallel \}\le M_1$$

for some $M_1>0$. This together with (4), implies that

$$\begin{array}{cc}\parallel y_n-y_{n-1}\parallel \hfill & \le |{\alpha }_n-{\alpha }_{n-1}|\parallel f\left(y_{n-1}\right)\parallel +{\beta }_n\parallel x_{n-1}-x_n\parallel +{\alpha }_n\parallel f\left(y_n\right)-f\left(y_{n-1}\right)\parallel \hfill \\ & +{\gamma }_n\parallel T_n{z}_n-T_{n-1}{z}_{n-1}\parallel +|{\gamma }_n-{\gamma }_{n-1}|\parallel T_{n-1}{z}_{n-1}\parallel +|{\beta }_{n-1}-{\beta }_n|\parallel x_{n-1}\parallel \hfill \\ & \le {\alpha }_n\delta \parallel y_n-y_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel f\left(y_{n-1}\right)\parallel +{\beta }_n\parallel x_n-x_{n-1}\parallel \hfill \\ & +|{\beta }_n-{\beta }_{n-1}|\parallel x_{n-1}\parallel +{\gamma }_n\{t_n\parallel x_n-x_{n-1}\parallel +|t_n-t_{n-1}|\parallel x_{n-1}-y_{n-1}\parallel \hfill \\ & +(1-t_n)\parallel y_n-y_{n-1}\parallel +|{\lambda }_n-{\lambda }_{n-1}|M_1\}+|{\gamma }_n-{\gamma }_{n-1}|\parallel T_{n-1}{z}_{n-1}\parallel \hfill \\ & \le ({\alpha }_n\delta +{\gamma }_n(1-t_n))\parallel y_n-y_{n-1}\parallel +({\beta }_n+{\gamma }_n{t}_n)\parallel x_n-x_{n-1}\parallel +\left(\right|{\alpha }_n-{\alpha }_{n-1}|\hfill \\ & +|{\gamma }_{n-1}-{\gamma }_n|+|{\beta }_{n-1}-{\beta }_n|+|t_n-t_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2,\hfill \end{array}$$

where

$$\underset{n\ge 0}{sup}\{\parallel f\left(y_n\right)\parallel +\parallel x_n\parallel +\parallel y_n\parallel +M_1+\parallel T_n{z}_n\parallel \}\le M_2$$

for some $M_2>0$. So it follows that

$$\begin{array}{cc}\parallel y_n-y_{n-1}\parallel \hfill & \le \frac{{\beta }_n+{\gamma }_n{t}_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel x_{n-1}-x_n\parallel +\frac{1}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\left(\right|{\alpha }_n-{\alpha }_{n-1}|\hfill \\ & +|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_{n-1}-{\gamma }_n|+|t_n-t_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2\hfill \\ & =(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_{n-1}-x_n\parallel +\frac{1}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\left(\right|{\alpha }_n-{\alpha }_{n-1}|\hfill \\ & +|{\beta }_n-{\beta }_{n-1}|+|{\gamma }_n-{\gamma }_{n-1}|+|t_n-t_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2\hfill \\ & \le \parallel x_n-x_{n-1}\parallel +\frac{1}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\left(\right|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\hfill \\ & +|{\gamma }_{n-1}-{\gamma }_n|+|t_n-t_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2.\hfill \end{array}$$

Hence we get

$$\begin{array}{c}\parallel S_{n}G{y}_n-S_{n-1}G{y}_{n-1}\parallel \le \parallel S_{n}G{y}_n-S_{n}G{y}_{n-1}\parallel +\parallel S_{n-1}G{y}_{n-1}-S_{n}G{y}_{n-1}\parallel \hfill \\ \le \parallel y_n-y_{n-1}\parallel +\parallel S_{n}G{y}_{n-1}-S_{n-1}G{y}_{n-1}\parallel \hfill \\ \le \parallel x_n-x_{n-1}\parallel +\frac{1}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\left(\right|{\beta }_{n-1}-{\beta }_n|+|{\alpha }_n-{\alpha }_{n-1}|\hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|+|t_{n-1}-t_n|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2+\parallel S_{n}G{y}_{n-1}-S_{n-1}G{y}_{n-1}\parallel .\hfill \end{array}$$

Consequently,

$$\begin{array}{c}\parallel S_{n}G{y}_n-S_{n-1}G{y}_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel \le \frac{1}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\left(\right|{\alpha }_n-{\alpha }_{n-1}|+|{\beta }_n-{\beta }_{n-1}|\hfill \\ +|{\gamma }_n-{\gamma }_{n-1}|+|t_n-t_{n-1}|+|{\lambda }_n-{\lambda }_{n-1}\left|\right)M_2+\parallel S_{n}G{y}_{n-1}-S_{n-1}G{y}_{n-1}\parallel .\hfill \end{array}$$

Since ${\sum }_{n=1}^{\infty }{sup}_{x\in D}\parallel (S_n-S_{n-1})x\parallel <\infty$ for bounded subset $D=\{G{y}_n:n\ge 0\}$ of C (due to the assumption), we know that ${lim}_{n\to \infty }\parallel (S_{n}G-S_{n-1}G)y_{n-1}\parallel =0$. Please note that ${lim}_{n\to \infty }{\alpha }_n=0,{lim}_{n\to \infty }{\lambda }_n=\lambda$ and ${lim\; inf}_{n\to \infty }{\gamma }_n(1-t_n)>0$. Thus, from $|{\beta }_n-{\beta }_{n-1}|\to 0,|{\gamma }_n-{\gamma }_{n-1}|\to 0$ and $|t_n-t_{n-1}|\to 0$ as $n\to \infty$ (due to conditions (ii), (iii)), we get

$$\underset{n\to \infty }{lim\; sup}(\parallel S_{n}G{y}_n-S_{n-1}G{y}_{n-1}\parallel -\parallel x_n-x_{n-1}\parallel )\le 0.$$

So it follows from condition (iv) and Lemma 6 that ${lim}_{n\to \infty }\parallel S_{n}G{y}_n-x_n\parallel =0$. Hence we obtain

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \infty }{lim}(1-{\delta }_n)\parallel S_{n}G{y}_n-x_n\parallel =0.$$

(6)

Let $\overline{p}:={\Pi }_C(I-{\mu }_2{A}_2)p$. Please note that $u_n={\Pi }_C(I-{\mu }_2{A}_2)y_n$ and $v_n={\Pi }_C(I-{\mu }_1{A}_1)u_n$. Then $v_n=G{y}_n$. From Proposition 4 (see also Lemma 2.13 in [25]), we have

$$\begin{array}{cc}\parallel u_n-\overline{p}{\parallel }^q\hfill & =\parallel {\Pi }_C(I-{\mu }_2{A}_2)y_n-{\Pi }_C(I-{\mu }_2{A}_2){p\parallel }^q\hfill \\ \hfill & \le \parallel (I-{\mu }_2{A}_2)y_n-(I-{\mu }_2{A}_2){p\parallel }^q\hfill \\ \hfill & \le \parallel y_n{-p\parallel }^q-{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1}){\parallel A_2{y}_n-A_{2}p\parallel }^q,\hfill \end{array}$$

(7)

and

$$\begin{array}{cc}\parallel v_n{-p\parallel }^q\hfill & =\parallel {\Pi }_C(I-{\mu }_1{A}_1)u_n-{\Pi }_C(I-{\mu }_1{A}_1)\overline{p}{\parallel }^q\hfill \\ \hfill & \le \parallel (I-{\mu }_1{A}_1)u_n-(I-{\mu }_1{A}_1)\overline{p}{\parallel }^q\hfill \\ \hfill & \le \parallel u_n-\overline{p}{\parallel }^q-{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1}){\parallel A_1{u}_n-A_1\overline{p}\parallel }^q.\hfill \end{array}$$

(8)

Substituting (7) to (8), we obtain

$$\begin{array}{cc}\parallel v_n{-p\parallel }^q\hfill & \le \parallel y_n{-p\parallel }^q-{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1}){\parallel A_2{y}_n-A_{2}p\parallel }^q\hfill \\ \hfill & -{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1}){\parallel A_1{u}_n-A_1\overline{p}\parallel }^q.\hfill \end{array}$$

(9)

According to Proposition 4, we obtain from (2) that $\parallel z_n{-p\parallel }^q\le t_n\parallel x_n{-p\parallel }^q+(1-t_n){\parallel y_n-p\parallel }^q$, and hence

$$\begin{array}{c}\parallel y_n{-p\parallel }^q={\parallel {\beta }_n(p-x_n)+{\alpha }_n(f\left(p\right)-f\left(y_n\right))+{\gamma }_n(p-T_n{z}_n)+{\alpha }_n(p-f\left(p\right))\parallel }^q\hfill \\ \le \parallel {\alpha }_n(f\left(y_n\right)-f\left(p\right))+{\beta }_n(x_n-p)+{\gamma }_n(T_n{z}_n-p){\parallel }^q+q{\alpha }_n⟨f\left(p\right)-p,J_q(y_n-p)⟩\hfill \\ \le {\alpha }_n\parallel f\left(y_n\right)-f\left(p\right){\parallel }^q+{\beta }_n\parallel x_n{-p\parallel }^q+{\gamma }_n{\parallel T_n{z}_n-p\parallel }^q+q{\alpha }_n⟨f\left(p\right)-p,J_q(y_n-p)⟩\hfill \\ \le {\alpha }_n\delta \parallel y_n{-p\parallel }^q+{\beta }_n\parallel x_n{-p\parallel }^q+{\gamma }_n[t_n\parallel x_n{-p\parallel }^q+(1-t_n)\parallel y_n-p{\parallel }^q]\hfill \\ +q{\alpha }_n\parallel f\left(p\right)-p\parallel \parallel y_n{-p\parallel }^{q-1},\hfill \end{array}$$

which immediately yields

$$\parallel y_n{-p\parallel }^q\le (1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_n{-p\parallel }^q+\frac{q{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n{-p\parallel }^{q-1}.$$

This, together with the convexity of ${\parallel ·\parallel }^q$ and (9), leads to

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^q={\parallel {\delta }_n(x_n-p)+(1-{\delta }_n)(S_{n}G{y}_n-p)\parallel }^q\hfill \\ \le {\delta }_n\parallel x_n{-p\parallel }^q+(1-{\delta }_n){\parallel S_n{v}_n-p\parallel }^q\hfill \\ \le {\delta }_n\parallel x_n{-p\parallel }^q+(1-{\delta }_n)\{\parallel y_n{-p\parallel }^q-{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1}){\parallel A_2{y}_n-A_{2}p\parallel }^q\hfill \\ -{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1})\parallel A_1{u}_n-A_1\overline{p}{\parallel }^q\}\hfill \\ \le {\delta }_n\parallel x_n{-p\parallel }^q+(1-{\delta }_n)\{(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}){\parallel x_n-p\parallel }^q+\frac{q\parallel f\left(p\right)-p\parallel \parallel y_n{-p\parallel }^{q-1}}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{\alpha }_n\hfill \\ -{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1})\parallel A_2{y}_n-A_2{p\parallel }^q-{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1})\parallel A_1{u}_n-A_1\overline{p}{\parallel }^q\}\hfill \\ =(1-\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}){\parallel x_n-p\parallel }^q+\frac{q(1-{\delta }_n)\parallel f\left(p\right)-p\parallel \parallel y_n{-p\parallel }^{q-1}}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{\alpha }_n\hfill \\ -(1-{\delta }_n)[{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1})\parallel A_2{y}_n-A_2{p\parallel }^q+{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1})\parallel A_1{u}_n-A_1\overline{p}{\parallel }^q]\hfill \\ \le \parallel x_n{-p\parallel }^q-(1-{\delta }_n)[{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1}){\parallel A_2{y}_n-A_{2}p\parallel }^q\hfill \\ +{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1})\parallel A_1{u}_n-A_1\overline{p}{\parallel }^q]+{\alpha }_n{M}_3,\hfill \end{array}$$

(10)

where

$$\underset{n\ge 0}{sup}\{\frac{q(1-{\delta }_n)}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p{\parallel }^{q-1}\}\le M_3$$

for some $M_3>0$. So it follows from (10) and Proposition 2 that

$$\begin{array}{c}(1-{\delta }_n)[{\mu }_2(q{\alpha }_2-{\kappa }_q{\mu }_2^{q-1})\parallel A_2{y}_n-A_2{p\parallel }^q+{\mu }_1(q{\alpha }_1-{\kappa }_q{\mu }_1^{q-1})\parallel A_1{u}_n-A_1\overline{p}{\parallel }^q]\hfill \\ \le \parallel x_n{-p\parallel }^q-{\parallel x_{n+1}-p\parallel }^q+{\alpha }_n{M}_3\hfill \\ \le q\parallel x_n-x_{n+1}\parallel \parallel x_{n+1}{-p\parallel }^{q-1}+{\kappa }_q{\parallel x_n-x_{n+1}\parallel }^q+{\alpha }_n{M}_3.\hfill \end{array}$$

Since $0<{\mu }_i<{\left(\frac{q{\alpha }_i}{{\kappa }_q}\right)}^{\frac{1}{q-1}}$ for $i=1,2$, from conditions (i), (iv) and (6) we get

$$\underset{n\to \infty }{lim}\parallel A_2{y}_n-A_{2}p\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel A_1{u}_n-A_1\overline{p}\parallel =0.$$

(11)

Using Proposition 1, we have

$$\begin{array}{c}\parallel u_n-\overline{p}{\parallel }^2={\parallel {\Pi }_C(I-{\mu }_2{A}_2)y_n-{\Pi }_C(I-{\mu }_2{A}_2)p\parallel }^2\hfill \\ \le ⟨(I-{\mu }_2{A}_2)y_n-(I-{\mu }_2{A}_2)p,J(u_n-\overline{p})⟩\hfill \\ =⟨y_n-p,J(u_n-\overline{p})⟩+{\mu }_2⟨A_{2}p-A_2{y}_n,J(u_n-\overline{p})⟩\hfill \\ \le \frac{1}{2}[\parallel y_n{-p\parallel }^2+\parallel u_n-\overline{p}{\parallel }^2-g_1(\parallel y_n-u_n-(p-\overline{p})\parallel \left)\right]+{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel ,\hfill \end{array}$$

where $g_1$ is given by Proposition 1. This yields

$$\parallel u_n-\overline{p}{\parallel }^2\le \parallel y_n{-p\parallel }^2-g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )+2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel .$$

(12)

In the same way, we derive

$$\begin{array}{cc}\parallel v_n{-p\parallel }^2\hfill & =\parallel {\Pi }_C(I-{\mu }_1{A}_1)u_n-{\Pi }_C(I-{\mu }_1{A}_1)\overline{p}{\parallel }^2\hfill \\ & \le ⟨(I-{\mu }_1{A}_1)u_n-(I-{\mu }_1{A}_1)\overline{p},J(v_n-p)⟩\hfill \\ & =⟨u_n-\overline{p},J(v_n-p)⟩+{\mu }_1⟨A_1\overline{p}-A_1{u}_n,J(v_n-p)⟩\hfill \\ & \le \frac{1}{2}[\parallel u_n-\overline{p}{\parallel }^2+\parallel v_n{-p\parallel }^2-g_2(\parallel u_n-v_n+(p-\overline{p})\parallel \left)\right]+{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel ,\hfill \end{array}$$

where $g_2$ is given by Proposition 1. This yields

$$\parallel v_n{-p\parallel }^2\le \parallel u_n-\overline{p}{\parallel }^2-g_2(\parallel u_n-v_n+(p-\overline{p})\parallel )+2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel .$$

(13)

Substituting (12) for (13), we get

$$\begin{array}{cc}\parallel v_n{-p\parallel }^2\hfill & \le \parallel y_n{-p\parallel }^2-g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )-g_2(\parallel u_n-v_n+(p-\overline{p})\parallel )\hfill \\ & +2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel .\hfill \end{array}$$

(14)

Please note that ${\parallel ·\parallel }^2$ is convex. Using Proposition 1, Lemmas 1 and 9, one concludes

$$\begin{array}{cc}\parallel y_n{-p\parallel }^2\hfill & \le \parallel {\beta }_n(p-x_n)+{\alpha }_n(f\left(p\right)-f\left(y_n\right))+{\gamma }_n(p-T_n{z}_n){\parallel }^2+2{\alpha }_n⟨f\left(p\right)-p,J(y_n-p)⟩\hfill \\ & \le {\alpha }_n\parallel f\left(p\right)-f\left(y_n\right){\parallel }^2+{\beta }_n\parallel x_n{-p\parallel }^2+{\gamma }_n\parallel T_n{z}_n{-p\parallel }^2-{\beta }_n{\gamma }_n{g}_3(\parallel x_n-T_n{z}_n\parallel )\hfill \\ & +2{\alpha }_n⟨f\left(p\right)-p,J(y_n-p)⟩\hfill \\ & \le {\alpha }_n\delta \parallel p-y_n{\parallel }^2+{\beta }_n\parallel x_n{-p\parallel }^2+{\gamma }_n(t_n\parallel x_n{-p\parallel }^2+(1-t_n)\parallel y_n-p{\parallel }^2)\hfill \\ & +2{\alpha }_n\parallel f\left(p\right)-p\parallel \parallel p-y_n\parallel -{\beta }_n{\gamma }_n{g}_3(\parallel x_n-T_n{z}_n\parallel ),\hfill \end{array}$$

which immediately sends

$$\begin{array}{cc}\parallel y_n{-p\parallel }^2\hfill & \le (1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_n{-p\parallel }^2+\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel \hfill \\ & -\frac{{\beta }_n{\gamma }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{g}_3(\parallel x_n-T_n{z}_n\parallel ).\hfill \end{array}$$

This together with (2) and (14) leads to

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2={\parallel {\delta }_n(x_n-p)+(1-{\delta }_n)(S_{n}G{y}_n-p)\parallel }^2\hfill \\ \le {\delta }_n\parallel p-x_n{\parallel }^2+(1-{\delta }_n){\parallel S_n{v}_n-p\parallel }^2\hfill \\ \le {\delta }_n\parallel p-x_n{\parallel }^2+(1-{\delta }_n)\{\parallel y_n{-p\parallel }^2-g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )-g_2(\parallel u_n-v_n+(p-\overline{p})\parallel )\hfill \\ +2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel \}\hfill \\ \le {\delta }_n\parallel x_n{-p\parallel }^2+(1-{\delta }_n)\{(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}){\parallel x_n-p\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel -\frac{{\beta }_n{\gamma }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}{g}_3(\parallel x_n-T_n{z}_n\parallel )\hfill \\ -g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )-g_2(\parallel u_n-v_n+(p-\overline{p})\parallel )\hfill \\ +2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel \}\hfill \\ \le (1-\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))})\parallel x_n{-p\parallel }^2+\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel \hfill \\ -(1-{\delta }_n)[\frac{{\beta }_n{\gamma }_n{g}_3(\parallel x_n-T_n{z}_n\parallel )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}+g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )+g_2(\parallel u_n-v_n+(p-\overline{p})\parallel \left)\right]\hfill \\ +2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel \hfill \\ \le \parallel x_n{-p\parallel }^2-(1-{\delta }_n)[\frac{{\beta }_n{\gamma }_n{g}_3(\parallel x_n-T_n{z}_n\parallel )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}+g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )\hfill \\ +g_2(\parallel u_n-v_n+(p-\overline{p})\parallel \left)\right]+2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel \hfill \\ +\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel ,\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}(1-{\delta }_n)[\frac{{\beta }_n{\gamma }_n{g}_3(\parallel x_n-T_n{z}_n\parallel )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}+g_1(\parallel y_n-u_n-(p-\overline{p})\parallel )+g_2(\parallel u_n-v_n+(p-\overline{p})\parallel \left)\right]\hfill \\ \le \parallel x_n{-p\parallel }^2-\parallel x_{n+1}{-p\parallel }^2+2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel \hfill \\ +\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel \hfill \\ \le (\parallel p-x_n\parallel +\parallel p-x_{n+1}\parallel )\parallel x_n-x_{n+1}\parallel +2{\mu }_2\parallel A_{2}p-A_2{y}_n\parallel \parallel u_n-\overline{p}\parallel \hfill \\ +2{\mu }_1\parallel A_1\overline{p}-A_1{u}_n\parallel \parallel v_n-p\parallel +\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\parallel f\left(p\right)-p\parallel \parallel y_n-p\parallel .\hfill \end{array}$$

Using (6) and (11), from ${lim\; inf}_{n\to \infty }{\beta }_n{\gamma }_n>0,$ and ${lim\; inf}_{n\to \infty }(1-{\delta }_n)>0$, we have

$$\underset{n\to \infty }{lim}{g}_1(\parallel y_n-u_n-(p-\overline{p})\parallel )=\underset{n\to \infty }{lim}{g}_2(\parallel u_n-v_n+(p-\overline{p})\parallel )=\underset{n\to \infty }{lim}{g}_3(\parallel x_n-T_n{z}_n\parallel )=0.$$

Using the properties of $g_1,g_2$ and $g_3$, we deduce that

$$\underset{n\to \infty }{lim}\parallel y_n-u_n-(p-\overline{p})\parallel =\underset{n\to \infty }{lim}\parallel u_n-v_n+(p-\overline{p})\parallel =\underset{n\to \infty }{lim}\parallel x_n-T_n{z}_n\parallel =0.$$

(15)

From (15) we get

$$\parallel y_n-G{y}_n\parallel =\parallel y_n-v_n\parallel \le \parallel y_n-u_n-(p-\overline{p})\parallel +\parallel u_n-v_n+(p-\overline{p})\parallel \to 0(n\to \infty ).$$

(16)

Meantime, again from (2) we have $y_n-x_n={\alpha }_n(f\left(y_n\right)-x_n)+{\gamma }_n(T_n{z}_n-x_n)$. Hence from (15) we get $\parallel y_n-x_n\parallel \le {\alpha }_n\parallel f\left(y_n\right)-x_n\parallel +\parallel T_n{z}_n-x_n\parallel \to 0(n\to \infty )$. This together with (16), implies that

$$\begin{array}{cc}\parallel x_n-G{x}_n\parallel \hfill & \le \parallel x_n-y_n\parallel +\parallel y_n-G{y}_n\parallel +\parallel G{y}_n-G{x}_n\parallel \hfill \\ \hfill & \le \parallel y_n-G{y}_n\parallel +2\parallel x_n-y_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(17)

Next, one claims that $\parallel x_n-S{x}_n\parallel \to 0$, $\parallel x_n-T_{\lambda }{x}_n\parallel \to 0$ and $\parallel x_n-W{x}_n\parallel \to 0$ as $n\to \infty$, where $Sx={lim}_{n\to \infty }{S}_{n}x,\forall x\in C$, $T_{\lambda }=J_{\lambda }^B(I-\lambda A)$ and $Wx={\theta }_{1}Sx+{\theta }_{2}Gx+{\theta }_3{T}_{\lambda }x,\forall x\in C$ for constants ${\theta }_1,{\theta }_2,{\theta }_3\in (0,1)$ satisfying ${\theta }_1+{\theta }_2+{\theta }_3=1$. Indeed, since $x_{n+1}={\delta }_n{x}_n+(1-{\delta }_n)S_{n}G{y}_n$ leads to $\parallel S_{n}G{y}_n-x_n\parallel =\frac{1}{1-{\delta }_n}\parallel x_{n+1}-x_n\parallel$, we deduce from (17), ${lim\; inf}_{n\to \infty }(1-{\delta }_n)>0$ and $x_n-y_n\to 0$ that

$$\begin{array}{cc}\parallel S_n{x}_n-x_n\parallel \hfill & \le \parallel S_n{x}_n-S_{n}G{x}_n\parallel +\parallel S_{n}G{x}_n-S_{n}G{y}_n\parallel +\parallel S_{n}G{y}_n-x_n\parallel \hfill \\ & \le \parallel x_n-G{x}_n\parallel +\parallel x_n-y_n\parallel +\frac{1}{1-{\delta }_n}\parallel x_{n+1}-x_n\parallel \to 0(n\to \infty ),\hfill \end{array}$$

which implies that

$$\parallel S{x}_n-x_n\parallel \le \parallel S{x}_n-S_n{x}_n\parallel +\parallel S_n{x}_n-x_n\parallel \to 0(n\to \infty ).$$

(18)

Furthermore, using the similar arguments to those of (5), we obtain

$$\begin{array}{cc}\parallel T_n{z}_n-T_{\lambda }{z}_n\parallel \hfill & \le |1-\frac{\lambda }{{\lambda }_n}|\parallel J_{{\lambda }_n}^B(I-{\lambda }_{n}A)z_n-(I-{\lambda }_{n}A)z_n\parallel +|{\lambda }_n-\lambda |\parallel A{z}_n\parallel \hfill \\ & =|1-\frac{\lambda }{{\lambda }_n}|\parallel T_n{z}_n-(I-{\lambda }_{n}A)z_n\parallel +|{\lambda }_n-\lambda |\parallel A{z}_n\parallel .\hfill \end{array}$$

Since ${lim}_{n\to \infty }{\lambda }_n=\lambda$ and the sequences $\left\{z_n\right\},\left\{T_n{z}_n\right\},\left\{A{z}_n\right\}$ are bounded, we get

$$\underset{n\to \infty }{lim}\parallel T_n{z}_n-T_{\lambda }{z}_n\parallel =0.$$

(19)

Taking into account condition (v), i.e., $0<\overline{\lambda }\le {\lambda }_n,\forall n\ge 0$ and ${lim}_{n\to \infty }{\lambda }_n=\lambda <{\left(\frac{q\alpha }{{\kappa }_q}\right)}^{\frac{1}{q-1}}$, we know that $0<\overline{\lambda }\le \lambda <{\left(\frac{q\alpha }{{\kappa }_q}\right)}^{\frac{1}{q-1}}$. So it follows from Proposition 4 that $\mathrm{Fix}\left(T_{\lambda }\right)={(A+B)}^{-1}0$ and $T_{\lambda }:C\to C$ is nonexpansive. Therefore, we deduce from (15), (19) and $x_n-y_n\to 0$ that

$$\begin{array}{cc}\parallel T_{\lambda }{x}_n-x_n\parallel \hfill & \le \parallel T_{\lambda }{x}_n-T_{\lambda }{z}_n\parallel +\parallel T_{\lambda }{z}_n-T_n{z}_n\parallel +\parallel T_n{z}_n-x_n\parallel \hfill \\ & \le \parallel x_n-z_n\parallel +\parallel T_{\lambda }{z}_n-T_n{z}_n\parallel +\parallel T_n{z}_n-x_n\parallel \hfill \\ & \le \parallel x_n-y_n\parallel +\parallel T_{\lambda }{z}_n-T_n{z}_n\parallel +\parallel T_n{z}_n-x_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(20)

We now define the mapping $Wx={\theta }_{1}Sx+{\theta }_{2}Gx+{\theta }_3{T}_{\lambda }x,\forall x\in C$ for constants ${\theta }_1,{\theta }_2,{\theta }_3\in (0,1)$ satisfying ${\theta }_1+{\theta }_2+{\theta }_3=1$. So by using Lemma 7, we know that $\mathrm{Fix}\left(W\right)=\mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(G\right)\cap \mathrm{Fix}\left(T_{\lambda }\right)=\Omega$. One observes that

$$\begin{array}{cc}\parallel x_n-W{x}_n\parallel \hfill & =\parallel {\theta }_1(x_n-S{x}_n)+{\theta }_2(x_n-G{x}_n)+{\theta }_3(x_n-T_{\lambda }{x}_n)\parallel \hfill \\ \hfill & \le {\theta }_1\parallel x_n-S{x}_n\parallel +{\theta }_2\parallel x_n-G{x}_n\parallel +{\theta }_3\parallel x_n-T_{\lambda }{x}_n\parallel .\hfill \end{array}$$

(21)

From (17), (18), (20) and (21), we get

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

The next step is to claim

$$\underset{n\to \infty }{lim\; sup}⟨J(x_n-x^{*}),f\left(x^{*}\right)-x^{*}⟩\le 0,$$

(22)

with $x^{*}=$s-${lim}_{n\to \infty }{x}_t$, where $x_t$ is a fixed point of $x\mapsto tf\left(x\right)+(1-t)Wx$ for each $t\in (0,1)$. Please note that the existence of $x^{*}$ ($x^{*}\in \mathrm{Fix}\left(W\right)$) is from Lemma 3. Indeed, the Banach contraction mapping principle guarantees that for each $t\in (0,1)$, $x_t$ satifies $x_t=tf\left(x_t\right)+(1-t)W{x}_t$. Hence we have $\parallel x_t-x_n\parallel =\parallel (W{x}_t-x_n)(1-t)+(f\left(x_t\right)-x_n)t\parallel$. Using the known subdifferential inequality (see [7]), we conclude that

$$\begin{array}{cc}\parallel x_n-x_t{\parallel }^2\hfill & \le 2t⟨x_n-f\left(x_t\right),J(x_n-x_t)⟩+{(1-t)}^2{\parallel W{x}_t-x_n\parallel }^2\hfill \\ \hfill & \le 2t⟨x_n-f\left(x_t\right),J(x_n-x_t)⟩+{(1-t)}^2(\parallel W{x}_n-x_n\parallel +\parallel W{x}_t-W{x}_n{\parallel )}^2\hfill \\ \hfill & \le 2t⟨x_n-f\left(x_t\right),J(x_n-x_t)⟩+{(1-t)}^2(\parallel x_n-x_t\parallel +\parallel x_n-W{x}_n{\parallel )}^2\hfill \\ \hfill & =(t^2-2t+1)\parallel x_n-x_t{\parallel }^2+2t⟨x_t-f\left(x_t\right),J(x_n-x_t)⟩+f_n\left(t\right)+2t{\parallel x_n-x_t\parallel }^2,\hfill \end{array}$$

(23)

where

$$f_n\left(t\right)=(\parallel x_n-W{x}_n\parallel +2\parallel x_n-x_t\parallel )\parallel x_n-W{x}_n{\parallel (1-t)}^2\to 0(n\to \infty ).$$

(24)

It follows from (23) that

$$⟨x_t-f\left(x_t\right),J(x_t-x_n)⟩\le \frac{t}{2}{\parallel x_t-x_n\parallel }^2+\frac{1}{2t}{f}_n\left(t\right).$$

(25)

Using both (25) and (24), we derive

$$\underset{n\to \infty }{lim\; sup}⟨x_t-f\left(x_t\right),J(x_t-x_n)⟩\le \frac{t}{2}{M}_4,$$

(26)

where $sup\{\parallel x_t-x_n{\parallel }^2:t\in (0,1)\mathrm{and}n\ge 0\}\le M_4$ for some $M_4>0$. Taking $t\to 0$ in (26), we have

$$\underset{t\to 0}{lim\; sup}\underset{n\to \infty }{lim\; sup}⟨f\left(x_t\right)-x_t,J(x_n-x_t)⟩\le 0.$$

On the other hand, we have

$$\begin{array}{c}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩=⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩-⟨f\left(x^{*}\right)-x^{*},J(x_n-x_t)⟩\hfill \\ +⟨f\left(x^{*}\right)-x^{*},J(x_n-x_t)⟩-⟨f\left(x^{*}\right)-x_t,J(x_n-x_t)⟩+⟨f\left(x^{*}\right)-x_t,J(x_n-x_t)⟩\hfill \\ -⟨f\left(x_t\right)-x_t,J(x_n-x_t)⟩+⟨f\left(x_t\right)-x_t,J(x_n-x_t)⟩\hfill \\ =⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})-J(x_n-x_t)⟩+⟨x_t-x^{*},J(x_n-x_t)⟩\hfill \\ +⟨x_t-f\left(x_t\right),J(x_t-x_n)⟩+⟨f\left(x^{*}\right)-f\left(x_t\right),J(x_n-x_t)⟩.\hfill \end{array}$$

So it follows that

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩\le {\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})-J(x_n-x_t)⟩\hfill \\ +(1+\delta )\parallel x_t-x^{*}\parallel {\displaystyle \underset{n\to \infty }{lim\; sup}}\parallel x_n-x_t\parallel +{\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x_t\right)-x_t,J(x_n-x_t)⟩.\hfill \end{array}$$

Taking into account that $x_t\to x^{*}$ yield

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩={\displaystyle \underset{t\to 0}{lim\; sup}\underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩\hfill \\ \le {\displaystyle \underset{t\to 0}{lim\; sup}\underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})-J(x_n-x_t)⟩.\hfill \end{array}$$

Using the property on nonlinear mapping J yields (22). Please note that $x_n-y_n\to 0$ implies $J(y_n-x^{*})-J(x_n-x^{*})\to 0$. Thus, we conclude from (22) that

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩={\displaystyle \underset{n\to \infty }{lim\; sup}}⟨f\left(x^{*}\right)-x^{*},J(x_n-x^{*})⟩\le 0.\hfill \end{array}$$

(27)

One observes that

$$\begin{array}{c}\parallel y_n-x^{*}{\parallel }^2={\parallel {\alpha }_n(f\left(y_n\right)-f\left(x^{*}\right))+{\beta }_n(x_n-x^{*})+{\gamma }_n(T_n{z}_n-x^{*})+{\alpha }_n(f\left(x^{*}\right)-x^{*})\parallel }^2\hfill \\ \le {\alpha }_n\parallel f\left(y_n\right)-f\left(x^{*}\right){\parallel }^2+{\beta }_n\parallel x_n-x^{*}{\parallel }^2+{\gamma }_n{\parallel z_n-x^{*}\parallel }^2+2{\alpha }_n⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩\hfill \\ \le {\alpha }_n\delta \parallel y_n-x^{*}{\parallel }^2+{\beta }_n\parallel x_n-x^{*}{\parallel }^2+{\gamma }_n(t_n\parallel x_n-x^{*}{\parallel }^2+(1-t_n)\parallel y_n-x^{*}{\parallel }^2)\hfill \\ +2{\alpha }_n⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩,\hfill \end{array}$$

which hence yields

$$\begin{array}{cc}\parallel y_n-x^{*}{\parallel }^2\hfill & \le (1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}){\parallel x_n-x^{*}\parallel }^2+\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩.\hfill \end{array}$$

(28)

By the convexity of ${\parallel ·\parallel }^2$, the nonexpansivity of $S_{n}G$ and (28), we get

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^2={\parallel (x_n-x^{*}){\delta }_n+(S_{n}G{y}_n-x^{*})(1-{\delta }_n)\parallel }^2\hfill \\ \le {\delta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\delta }_n)\{(1-\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}){\parallel x_n-x^{*}\parallel }^2\hfill \\ +\frac{2{\alpha }_n}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩\}\hfill \\ =\left(1-\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\right){\parallel x_n-x^{*}\parallel }^2+\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}·\frac{2⟨f\left(x^{*}\right)-x^{*},J(y_n-x^{*})⟩}{1-\delta }.\hfill \end{array}$$

(29)

Since ${lim\; inf}_{n\to \infty }\frac{(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}>0,\left\{\frac{{\alpha }_n(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\right\}\subset (0,1)$ and ${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$, we know that $\left\{\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}\right\}\subset (0,1)$ and ${\sum }_{n=0}^{\infty }\frac{{\alpha }_n(1-{\delta }_n)(1-\delta )}{1-({\alpha }_n\delta +{\gamma }_n(1-t_n))}=\infty$. Using (27) and Lemma 8, we conclude from (29) that $\parallel x_n-x^{*}\parallel \to 0$ as $n\to \infty$. This proof is now complete.

Remark 1.

From the related associated results in Ceng et al. [16], Song and Ceng [25], our obtained results extend and improve and them in the following ways:

(i) The approximating problem of ${\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)$ in [[16], Theorem 3.1] is moved to devise our approximating problem ${\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap {(A+B)}^{-1}0$ where ${(A+B)}^{-1}0$ is the solution set of the VI: $0\in (A+B)x$. The implicit (two-step) relaxed extragradient method in [[16], Theorem 3.1] is extended to develop our generalized Mann viscosity implicit rule in Theorem 1. That is, two iterative steps $y_n=(1-{\alpha }_n)G{x}_n+{\alpha }_{n}f\left(y_n\right)$ and $x_{n+1}=(1-{\beta }_n)S_n{y}_n+{\beta }_n{x}_n$ in [[16], Theorem 3.1] is refined to develop our two iterative steps $y_n={\alpha }_{n}f\left(y_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n(t_n{x}_n+(1-t_n)y_n)$ and $x_{n+1}={\delta }_n{x}_n+(1-{\delta }_n)S_{n}G{y}_n$, where $T_n=J_{{\lambda }_n}^B(I-{\lambda }_{n}A)$. In addition, uniformly convex and 2-uniformly smooth restructures in [[16], Theorem 3.1] is generalized to the structures of uniformly convex and q-uniformly smooth for $1<q\le 2$.

(ii) The problem of finding an element of ${\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)$ in [[25], Theorem 3.1] is generalized to devise our approximating problem on the element in ${\cap }_{n=0}^{\infty }\mathrm{Fix}\left(S_n\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap {(A+B)}^{-1}0$, where ${(A+B)}^{-1}0$ is the solution set of the VI: $0\in (A+B)x$. The modified relaxed extragradient method in [[25], Theorem 3.1] is extended to develop our generalized Mann viscosity implicit rule in Theorem 1. That is, two iterative steps $y_n=(1-{\beta }_n)x_n+{\beta }_{n}G{x}_n$ and $x_{n+1}={\Pi }_C[{\alpha }_n\gamma f\left(x_n\right)+{\gamma }_n{x}_n+((1-{\gamma }_n)I-{\alpha }_n\rho F)S_n{y}_n]$ in [[25], Theorem 3.1] is extended to develop our two iterative steps $y_n={\alpha }_{n}f\left(y_n\right)+{\beta }_n{x}_n+{\gamma }_n{T}_n(t_n{x}_n+(1-t_n)y_n)$ and $x_{n+1}={\delta }_n{x}_n+(1-{\delta }_n)S_{n}G{y}_n$, where $T_n=J_{{\lambda }_n}^B(I-{\lambda }_{n}A)$.

Next, Theorem 1 is applied to solve the GSVI, VIP and FPP in an illustrating example. Let $C=[-2,2]$ and $H=\mathbf{R}$ with the inner product $⟨a,b⟩=ab$ and induced norm $\parallel ·\parallel =|·|$. The initial point $x_0$ is randomly chosen in C. We define $f\left(x\right)=\frac{1}{2}x,Sx=sinx$ and $A_{1}x=A_{2}x=Ax=\frac{2}{3}x+\frac{1}{4}sinx$ for all $x\in C$. Then f is $\frac{1}{2}$-contraction, S is a nonexpansive self-mapping on C with $\mathrm{Fix}\left(S\right)=\left\{0\right\}$ and A is $\frac{11}{12}$-Lipschitzian and $\frac{5}{12}$-strongly monotone mapping. Indeed, we observe that

$$\parallel Ax-Ay\parallel \le \frac{2}{3}\parallel x-y\parallel +\frac{1}{4}\parallel sinx-siny\parallel \le (\frac{2}{3}+\frac{1}{4})\parallel x-y\parallel =\frac{11}{12}\parallel x-y\parallel ,$$

and

$$⟨Ax-Ay,x-y⟩=\frac{2}{3}⟨x-y,x-y⟩+\frac{1}{4}⟨sinx-siny,x-y⟩\ge (\frac{2}{3}-\frac{1}{4}){\parallel x-y\parallel }^2=\frac{5}{12}{\parallel x-y\parallel }^2.$$

This ensures that $⟨Ax-Ay,x-y⟩\ge \frac{60}{121}{\parallel Ax-Ay\parallel }^2$. So it follows that $A_1=A_2=A$ is $\frac{60}{121}$-inverse-strongly monotone, and hence ${\alpha }_1={\alpha }_2=\alpha =\frac{60}{121}$. Therefore, it is easy to see that $\Omega =\mathrm{Fix}\left(S\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. Let ${\mu }_1={\mu }_2=\alpha =\frac{60}{121}$. Putting ${\alpha }_n=\frac{1}{2(n+2)},{\beta }_n=\frac{1}{2}-\frac{1}{2(n+2)},{\gamma }_n=\frac{1}{2},{\delta }_n=\frac{1}{2},t_n=\frac{1}{2}$ and ${\lambda }_n=\overline{\lambda }=\alpha =\frac{60}{121}$, we know that the conditions (i)–(v) on the parameter sequences $\left\{{\lambda }_n\right\}\subset (0,2\alpha )$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\},\left\{t_n\right\}\subset (0,1)$ all are satisfied. In this case, the iterative scheme in Theorem 1 can be rewritten as follows:

$$\left\{\begin{array}{c}{y}_n=\frac{1}{2(n+2)}·\frac{1}{2}{y}_n+(\frac{1}{2}-\frac{1}{2(n+2)})x_n+\frac{1}{2}{P}_C(I-\frac{60}{121}A)\left(\frac{x_n+y_n}{2}\right),\hfill \\ x_{n+1}=\frac{x_n+SG{y}_n}{2},\hfill \end{array}\right.$$

where $G:=P_C(I-\frac{60}{121}A)P_C(I-\frac{60}{121}A)$. Then, by Theorem 1, we know that $\left\{x_n\right\}$ converges to $0\in \Omega =\mathrm{Fix}\left(S\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap \mathrm{VI}(C,A)$.

4. Applications

In this section, we will apply the main result of this paper for solving some important optimization problems in the setting of Hilbert spaces.

4.1. Variational Inequality Problems

Let $A:C\to H$ be a single-valued nonself mapping. Recall the monotone variational inequality of getting the desired vector $x^{*}\in C$ with $⟨A{x}^{*},x-x^{*}⟩\ge 0,\forall x\in C$, whose solution set of is $\mathrm{VI}(C,A)$. Let $I_C$ be an indicator operator of C given by

$$I_{C}y=\left\{\begin{array}{c}0\mathrm{if}y\in C,\hfill \\ \infty \mathrm{if}y\notin C.\hfill \end{array}\right.$$

We denote the normal cone of C at u by $N_C\left(u\right)$, i.e., $N_C\left(u\right)$ is a set consists of such points which solve $⟨w,v-u⟩\le 0,\forall v\in C$. It is known that $I_C$ is a convex, lower semi-continuous and proper function and the subdifferential $\partial I_C$ is maximally monotone. For $\lambda >0$, the resolvent mapping of $\partial I_C$ is denoted by $J_{\lambda }^{\partial I_C}$, i.e., $J_{\lambda }^{\partial I_C}={(I+\lambda \partial I_C)}^{-1}$. Please note that

$$\begin{array}{cc}\partial I_C\left(u\right)\hfill & =\{w\in H:I_C\left(v\right)+⟨w,v-u⟩\le I_C\left(u\right),\forall v\in C\}\hfill \\ & =\{w\in H:⟨w,v-u⟩\le 0\forall v\in C\}=N_C\left(u\right),\forall u\in C.\hfill \end{array}$$

So we know that $u=J_{\lambda }^{\partial I_C}\left(x\right)\iff x-u\in \lambda N_C\left(u\right)\iff ⟨x-u,v-u⟩\le 0,\forall v\in C\iff u=P_C\left(x\right)$. Hence we get ${(A+\partial I_C)}^{-1}0=\mathrm{VI}(C,A)$.

Next, putting $B=\partial I_C$ in Theorem 1, we can obtain the following result.

Theorem 2.

Let non-empty set C be a convex close in a Hilbert space X stated as Theorem 1. For $i=1,2$, mappings $A,A_i:C\to H$ are α-inverse-strongly monotone and ${\alpha }_i$-inverse-strongly monotone, respectively. Let S be a nonexpansive singled-valued self-mapping on C. Suppose $\Omega =\mathrm{Fix}\left(S\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap \mathrm{VI}(C,A)\ne \varnothing$, where $\mathrm{GSVI}(C,A_1,A_2)$ is the fixed-point set of $G:=P_C(I-{\mu }_1{A}_1)P_C(I-{\mu }_2{A}_2)$ with $0<{\mu }_i<2{\alpha }_i$ for $i=1,2$. Let $f:C\to C$ be a strictly contraction with constant $\delta \in (0,1)$. For arbitrary initial $x_0\in C$, define $\left\{x_n\right\}$ by

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+{\gamma }_n{P}_C(I-{\lambda }_{n}A)(t_n{x}_n+(1-t_n)y_n)+{\beta }_n{x}_n,\hfill \\ x_{n+1}=(1-{\delta }_n)SG{y}_n+{\delta }_n{x}_n,n\ge 0,\hfill \end{array}\right.$$

where $0<\overline{\lambda }\le {\lambda }_n,\forall n\ge 0$ and ${lim}_{n\to \infty }{\lambda }_n=\lambda <2\alpha$, and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\},\left\{t_n\right\}\subset (0,1)$ satisfy conditions (i)–(iii) as in Theorem 1 in Section 2. Then $x_n\to x^{*}\in \Omega$, which is the unique solution to the variational inequality: $⟨(I-f)x^{*},x^{*}-p⟩\le 0,\forall p\in \Omega$.

4.2. Convex Minimization Problems

Let $g:H\to \mathbf{R}$ and $h:H\to \mathbf{R}$ be two functions, where g is convex smooth and h is proper convex and lower semicontinuous. The associated minima problem is to find $x^{*}\in H$ such that

$$g\left(x^{*}\right)+h\left(x^{*}\right)=\underset{x\in H}{min}\{g\left(x\right)+h\left(x\right)\}.$$

(30)

By Fermat’s rule, we know that the problem (30) is equivalent to the fact that finds $x^{*}\in H$ such that $0\in \nabla g\left(x^{*}\right)+\partial h\left(x^{*}\right)$ with $\nabla g$ being the gradient of function g and $\partial h$ being the subdifferential function of function h. It is also known that if $\nabla g$ is $\frac{1}{\alpha }$-Lipschitz continuous, then it is also $\alpha$-inverse-strongly monotone. Next, putting $A=\nabla g$ and $B=\partial h$ in Theorem 1, we can obtain the following result.

Theorem 3.

Let $g:H\to \mathbf{R}$ be a convex and differentiable function whose gradient $\nabla g$ is $\frac{1}{\alpha }$-Lipschitz continuous and $h:H\to \mathbf{R}$ be a convex and lower semi-continuous function. $A_i:C\to H$ are supposed to be ${\alpha }_i$-inverse-strongly monotone for $i=1,2$. Let S be a nonexpansive single-valued self-mapping on C such that $\Omega =\mathrm{Fix}\left(S\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap {(\nabla g+\partial h)}^{-1}0\ne \varnothing$ where ${(\nabla g+\partial h)}^{-1}0$ is the set of minima attained by $g+h$, and $\mathrm{GSVI}(C,A_1,A_2)$ is the fixed point set of $G:=P_C(I-{\mu }_1{A}_1)P_C(I-{\mu }_2{A}_2)$ with $0<{\mu }_i<2{\alpha }_i$ for $i=1,2$. Let $f:C\to C$ be a strictly contraction with constant $\delta \in (0,1)$. For arbitrary initial $x_0\in C$, define $\left\{x_n\right\}$ by

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+{\gamma }_n{J}_{{\lambda }_n}^{\partial h}(I-{\lambda }_n\nabla g)(t_n{x}_n+(1-t_n)y_n)+{\beta }_n{x}_n,\hfill \\ x_{n+1}=(1-{\delta }_n)SG{y}_n+{\delta }_n{x}_n,n\ge 0,\hfill \end{array}\right.$$

where $0<\overline{\lambda }\le {\lambda }_n,\forall n\ge 0$ and ${lim}_{n\to \infty }{\lambda }_n=\lambda <2\alpha$, and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\},\left\{t_n\right\}\subset (0,1)$ satisfy conditions (i)–(iii) as in Theorem 1 in Section 2. Then $x_n\to x^{*}\in \Omega$, which uniquely solves $⟨(I-f)x^{*},x^{*}-p⟩\le 0,\forall p\in \Omega$.

4.3. Split Feasibility Problems

Let C and Q be non-empty convex closed sets in Hilbert spaces $H_1$ and $H_2$, respectively. Let $T:H_1\to H_2$ be a linearly bounded operator with its adjoint $T^{*}$. Consider the split feasibility problem (SFP) of obtaining a desired point $x^{*}\in C\mathrm{and}T{x}^{*}\in Q$. The SFP can be borrowed to model the radiation therapy. It is clear that the set of solutions of the SFP is $C\cap T^{-1}Q$. To solve the SFP, we can rewrite it as the following convexly constrained minimization problem:

$$\underset{x\in C}{min}g\left(x\right):=\frac{1}{2}{\parallel Tx-P_{Q}Tx\parallel }^2.$$

Please note that the function g is differentiable convex whose Lipschitz gradient is given by $\nabla g=T^{*}(I-P_Q)T$. Furthermore, $\nabla g$ is $\frac{1}{{\parallel T\parallel }^2}$-inverse-strongly monotone, where ${\parallel T\parallel }^2$ is the spectral radius of $T^{*}T$. Thus, $x^{*}$ solves the SFP if and only if $x^{*}\in H_1$ such that

$$\begin{array}{cc}0\in \nabla g\left(x^{*}\right)+\partial I_C\left(x^{*}\right)\hfill & \iff x^{*}-\lambda \nabla g\left(x^{*}\right)\in (I+\lambda \partial I_C)x^{*}\hfill \\ & \iff x^{*}=J_{\lambda }^{\partial I_C}(x^{*}-\lambda \nabla g\left(x^{*}\right))\hfill \\ & \iff x^{*}=P_C(x^{*}-\lambda \nabla g\left(x^{*}\right)).\hfill \end{array}$$

Next, putting $A=\nabla g$ and $B=\partial I_C$ in Theorem 1, we can obtain the following result:

Theorem 4.

Let C and Q be nonempty closed convex subsets of $H_1$ and $H_2$, respectively. Let $T:H_1\to H_2$ be a bounded linear operator with its adjoint $T^{*}$. Let the mapping $A_i:C\to H_1$ be ${\alpha }_i$-inverse-strongly monotone for $i=1,2$. Let S be a nonexpansive self-mapping on C such that $\Omega =\mathrm{Fix}\left(S\right)\cap \mathrm{GSVI}(C,A_1,A_2)\cap (C\cap T^{-1}Q)\ne \varnothing$ where $\mathrm{GSVI}(C,A_1,A_2)$ is the fixed point set of $G:=P_C(I-{\mu }_1{A}_1)P_C(I-{\mu }_2{A}_2)$ with $0<{\mu }_i<2{\alpha }_i$ for $i=1,2$. Let $f:C\to C$ be a δ-contraction with constant $\delta \in (0,1)$. For arbitrarily given $x_0\in C$, let $\left\{x_n\right\}$ be a sequence generated by

$$\left\{\begin{array}{c}{y}_n={\alpha }_{n}f\left(y_n\right)+{\gamma }_n{P}_C(I-{\lambda }_n{T}^{*}(I-P_Q)T)(t_n{x}_n+(1-t_n)y_n)+{\beta }_n{x}_n,\hfill \\ x_{n+1}={\delta }_n{x}_n+(1-{\delta }_n)SG{y}_n,n\ge 0,\hfill \end{array}\right.$$

where $0<\overline{\lambda }\le {\lambda }_n,\forall n\ge 0$ and ${lim}_{n\to \infty }{\lambda }_n=\lambda <\frac{2}{{\parallel T\parallel }^2}$, and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\},\left\{t_n\right\}\subset (0,1)$ satisfy conditions (i)–(iii) as in Theorem 1 in Section 2. Then $x_n\to x^{*}\in \Omega$, which uniquely solves $⟨(I-f)x^{*},x^{*}-p⟩\le 0,\forall p\in \Omega$.