# Parallel algorithms for variational inclusions and fixed points with applications

## Abstract

In this paper, we introduce two parallel algorithms for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive mapping in Hilbert spaces and prove some strong convergence theorems of the proposed algorithms. As special cases, we can approach the minimum-norm common element of the zero of the sum of two monotone operators and the fixed point of a nonexpansive mapping without using the metric projection. Further, we give some applications of our main results.

MSC:49J40, 47J20, 47H09, 65J15.

## 1 Introduction

Let H be a real Hilbert space. Let $A:H\to H$ be a single-valued nonlinear mapping and $B:H\to {2}^{H}$ be a set-valued mapping.

Now, we are concerned with the following variational inclusion:

Find a zero $x\in H$ of the sum of two monotone operators A and B such that

$0\in Ax+Bx,$
(1.1)

where 0 is the zero vector in H.

The set of solutions of the problem (1.1) is denoted by ${\left(A+B\right)}^{-1}0$. If $H={R}^{m}$, then the problem (1.1) becomes the generalized equation introduced by Robinson [1]. If $A=0$, then the problem (1.1) becomes the inclusion problem introduced by Rockafellar [2]. It is well known that the problem (1.1) is among the most interesting and intensively studied classes of mathematical problems and has wide applications in the fields of optimization and control, economics and transportation equilibrium, engineering science, and many others. For the past years, many existence results and iterative algorithms for various variational inequality and variational inclusion problems have been extended and generalized in various directions using novel and innovative techniques. A useful and important generalization is called the general variational inclusion involving the sum of two nonlinear operators. Moudafi and Noor [3] studied the sensitivity analysis of variational inclusions by using the technique of resolvent equations. Recently much attention has been given to developing iterative algorithms for solving the variational inclusions. Dong et al. [4] analyzed the solution’s sensitivity for variational inequalities and variational inclusions by using a resolvent operator technique. By using the concept and technique of resolvent operators, Agarwal et al. [5] and Jeong [6] introduced and studied a new system of parametric generalized nonlinear mixed quasi-variational inclusions in a Hilbert space. Lan [7] introduced and studied a stable iteration procedure for a class of generalized mixed quasi-variational inclusion systems in Hilbert spaces. Recently, Zhang et al. [8] introduced a new iterative scheme for finding a common element of the set of solutions to the problem (1.1) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [9] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related work, see [923] and the references therein.

Recently, Takahashi et al. [24] introduced the following iterative algorithm for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive mapping:

${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)S\left({\alpha }_{n}x+\left(1-{\alpha }_{n}\right){J}_{{\lambda }_{n}}^{B}\left({x}_{n}-{\lambda }_{n}A{x}_{n}\right)\right)$
(1.2)

for all $n\ge 0$. Under some assumptions, they proved that the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point of $F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

Remark 1.1 We note that the algorithm (1.2) cannot be used to find the minimum-norm element due to the facts that $x\in C$ and S is a self-mapping of C. However, there exist a large number of problems for which one needs to find the minimum-norm solution (see, for example, [2529]). A useful path to circumvent this problem is to use projection. Bauschke and Browein [30] and Censor and Zenios [31] provide reviews of the field. The main difficulty is in the computation. Hence it is an interesting problem to find the minimum-norm element without using the projection.

Motivated and inspired by the works in this field, we first suggest the following two algorithms without using projection:

${x}_{t}=\left(1-\kappa \right)S{x}_{t}+\kappa {J}_{\lambda }^{B}\left(t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right)$

for all $t\in \left(0,1\right)$ and

${x}_{n+1}=\left(1-\kappa \right)S{x}_{n}+\kappa {J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)$

for all $n\ge 0$. Notice that these two algorithms are indeed well defined (see the next section). We show that the suggested algorithms converge strongly to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$ which solves the following variational inequality:

$〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},\stackrel{˜}{x}-z〉\ge 0$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

As special cases, we can approach the minimum-norm element in $F\left(S\right)\cap {\left(A+B\right)}^{-1}0$ without using the metric projection and give some applications.

## 2 Preliminaries

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H.

1. (1)

A mapping $S:C\to C$ is said to be nonexpansive if

$\parallel Sx-Sy\parallel \le \parallel x-y\parallel$

for all $x,y\in C$. We denote by $F\left(S\right)$ the set of fixed points of S.

2. (2)

A mapping $A:C\to H$ is said to be α-inverse strongly monotone if there exists $\alpha >0$ such that

$〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^{2}$

for all $x,y\in C$.

It is well known that, if A is α-inverse strongly monotone, then $\parallel Ax-Ay\parallel \le \frac{1}{\alpha }\parallel x-y\parallel$ for all $x,y\in C$.

Let B be a mapping from H into ${2}^{H}$. The effective domain of B is denoted by $dom\left(B\right)$, that is, $dom\left(B\right)=\left\{x\in H:Bx\ne \mathrm{\varnothing }\right\}$.

1. (3)

A multi-valued mapping B is said to be a monotone operator on H if

$〈x-y,u-v〉\ge 0$

for all $x,y\in dom\left(B\right)$, $u\in Bx$, and $v\in By$.

1. (4)

A monotone operator B on H is said to be maximal if its graph is not strictly contained in the graph of any other monotone operator on H.

Let B be a maximal monotone operator on H and ${B}^{-1}0=\left\{x\in H:0\in Bx\right\}$. For a maximal monotone operator B on H and $\lambda >0$, we may define a single-valued operator ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}:H\to dom\left(B\right)$, which is called the resolvent of B for λ. It is well known that the resolvent ${J}_{\lambda }^{B}$ is firmly nonexpansive, i.e.,

${\parallel {J}_{\lambda }^{B}x-{J}_{\lambda }^{B}y\parallel }^{2}\le 〈{J}_{\lambda }^{B}x-{J}_{\lambda }^{B}y,x-y〉$

for all $x,y\in C$ and ${B}^{-1}0=F\left({J}_{\lambda }^{B}\right)$ for all $\lambda >0$. The following resolvent identity is well known: for any $\lambda >0$ and $\mu >0$, the following identity holds:

${J}_{\lambda }^{B}x={J}_{\mu }^{B}\left(\frac{\mu }{\lambda }x+\left(1-\frac{\mu }{\lambda }\right){J}_{\lambda }^{B}x\right)$
(2.1)

for all $x\in H$.

We use the notation that ${x}_{n}⇀x$ stands for the weak convergence of $\left({x}_{n}\right)$ to x and ${x}_{n}\to x$ stands for the strong convergence of $\left({x}_{n}\right)$ to x, respectively.

We need the following lemmas for the next section.

Lemma 2.1 ([32])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be an α-inverse strongly monotone mapping and $\lambda >0$ be a constant. Then we have

${\parallel \left(I-\lambda A\right)x-\left(I-\lambda A\right)y\parallel }^{2}\le {\parallel x-y\parallel }^{2}+\lambda \left(\lambda -2\alpha \right){\parallel Ax-Ay\parallel }^{2}$

for all $x,y\in C$. In particular, if $0\le \lambda \le 2\alpha$, then $I-\lambda A$ is nonexpansive.

Lemma 2.2 ([33])

Let C be a closed convex subset of a Hilbert space H. Let $S:C\to C$ be a nonexpansive mapping. Then $F\left(S\right)$ is a closed convex subset of C and the mapping $I-S$ is demiclosed at 0, i.e. whenever $\left\{{x}_{n}\right\}\subset C$ is such that ${x}_{n}⇀x$ and $\left(I-S\right){x}_{n}\to 0$, then $\left(I-S\right)x=0$.

Lemma 2.3 ([1])

Let C be a nonempty closed convex subset of a real Hilbert space H. Assume that the mapping $F:C\to H$ is monotone and weakly continuous along segments, that is, $F\left(x+ty\right)\to F\left(x\right)$ weakly as $t\to 0$. Then the variational inequality

${x}^{\ast }\in C,\phantom{\rule{1em}{0ex}}〈F{x}^{\ast },x-{x}^{\ast }〉\ge 0$

for all $x\in C$ is equivalent to the dual variational inequality

${x}^{\ast }\in C,\phantom{\rule{1em}{0ex}}〈Fx,x-{x}^{\ast }〉\ge 0$

for all $x\in C$.

Lemma 2.4 ([34])

Let $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space X and $\left\{{\beta }_{n}\right\}$ be a sequence in $\left[0,1\right]$ with

$0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1.$

Suppose that ${x}_{n+1}=\left(1-{\beta }_{n}\right){y}_{n}+{\beta }_{n}{x}_{n}$ for all $n\ge 0$ and

$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

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

Lemma 2.5 ([35])

Assume that $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that

${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n}{\gamma }_{n}$

for all $n\ge 1$, where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that

1. (a)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (b)

${lim sup}_{n\to \mathrm{\infty }}{\delta }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}{\gamma }_{n}|<\mathrm{\infty }$.

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

## 3 Main results

In this section, we prove our main results.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H. Let $f:C\to H$ be a ρ-contraction and γ be a constant such that $0<\gamma <\frac{1}{\rho }$. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ and S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. Let λ and κ be two constants satisfying $a\le \lambda \le b$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$ and $\kappa \in \left(0,1\right)$. For any $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$, let $\left\{{x}_{t}\right\}\subset C$ be a net generated by

${x}_{t}=\left(1-\kappa \right)S{x}_{t}+\kappa {J}_{\lambda }^{B}\left(t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right).$
(3.1)

Then the net $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0+$ to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$, which solves the following variational inequality:

$〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},\stackrel{˜}{x}-z〉\ge 0$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

Proof First, we show that the net $\left\{{x}_{t}\right\}$ is well defined. For any $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$, we define a mapping $W:=\left(1-\kappa \right)S+\kappa {J}_{\lambda }^{B}\left(t\gamma f+\left(1-t\right)I-\lambda A\right)$. Note that ${J}_{\lambda }^{B}$, S, and $I-\frac{\lambda }{1-t}A$ (see Lemma 2.1) are nonexpansive. For any $x,y\in C$, we have

$\begin{array}{r}\parallel Wx-Wy\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel \left(1-\kappa \right)\left(Sx-Sy\right)+\kappa \left({J}_{\lambda }^{B}\left(t\gamma f\left(x\right)+\left(1-t\right)\left(I-\frac{\lambda }{1-t}A\right)x\right)\\ \phantom{\rule{2em}{0ex}}-{J}_{\lambda }^{B}\left(t\gamma f\left(y\right)+\left(1-t\right)\left(I-\frac{\lambda }{1-t}A\right)y\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right)\parallel Sx-Sy\parallel \\ \phantom{\rule{2em}{0ex}}+\kappa \parallel t\gamma \left(f\left(x\right)-f\left(y\right)\right)+\left(1-t\right)\left[\left(I-\frac{\lambda }{1-t}A\right)x-\left(I-\frac{\lambda }{1-t}A\right)y\right]\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right)\parallel x-y\parallel +\kappa t\gamma \parallel f\left(x\right)-f\left(y\right)\parallel \\ \phantom{\rule{2em}{0ex}}+\left(1-t\right)\kappa \parallel \left(I-\frac{\lambda }{1-t}A\right)x-\left(I-\frac{\lambda }{1-t}A\right)y\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right)\parallel x-y\parallel +t\kappa \gamma \rho \parallel x-y\parallel +\left(1-t\right)\kappa \parallel x-y\parallel \\ \phantom{\rule{1em}{0ex}}=\left[1-\left(1-\gamma \rho \right)\kappa t\right]\parallel x-y\parallel ,\end{array}$

which implies the mapping W is a contraction on C. We use ${x}_{t}$ to denote the unique fixed point of W in C. Therefore, $\left\{{x}_{t}\right\}$ is well defined. Set ${y}_{t}={J}_{\lambda }^{B}{u}_{t}$ and ${u}_{t}=\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}$ for all $t>0$. Taking $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$, it is obvious that $z=Sz={J}_{\lambda }^{B}\left(z-\lambda Az\right)$ for all $\lambda >0$ and so

$z=Sz={J}_{\lambda }^{B}\left(z-\lambda Az\right)={J}_{\lambda }^{B}\left(tz+\left(1-t\right)\left(I-\frac{\lambda }{1-t}A\right)z\right)$

for all $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$. From (3.1), it follows that

$\begin{array}{rl}\parallel {x}_{t}-z\parallel & =\parallel \left(1-\kappa \right)\left(S{x}_{t}-z\right)+\kappa \left({y}_{t}-z\right)\parallel \\ \le \left(1-\kappa \right)\parallel S{x}_{t}-z\parallel +\kappa \parallel {y}_{t}-z\parallel \\ \le \left(1-\kappa \right)\parallel {x}_{t}-z\parallel +\kappa \parallel {y}_{t}-z\parallel .\end{array}$

Hence we get $\parallel {x}_{t}-z\parallel \le \parallel {y}_{t}-z\parallel$. Since ${J}_{\lambda }^{B}$ is nonexpansive, we have

$\begin{array}{r}\parallel {y}_{t}-z\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {J}_{\lambda }^{B}\left(t\gamma f\left({x}_{t}\right)+\left(1-t\right)\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)\right)-{J}_{\lambda }^{B}\left(tz+\left(1-t\right)\left(z-\frac{\lambda }{1-t}Az\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left(t\gamma f\left({x}_{t}\right)+\left(1-t\right)\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)\right)-\left(tz+\left(1-t\right)\left(z-\frac{\lambda }{1-t}Az\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel \left(1-t\right)\left(\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\right)+t\left(\gamma f\left({x}_{t}\right)-z\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right)\parallel \left(I-\frac{\lambda }{1-t}A\right){x}_{t}-\left(I-\frac{\lambda }{1-t}A\right)z\parallel +t\gamma \parallel f\left({x}_{t}\right)-f\left(z\right)\parallel +t\parallel \gamma f\left(z\right)-z\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right)\parallel {x}_{t}-z\parallel +t\gamma \rho \parallel {x}_{t}-z\parallel +t\parallel \gamma f\left(z\right)-z\parallel .\end{array}$
(3.2)

Thus it follows that

$\parallel {x}_{t}-z\parallel \le \frac{1}{1-\gamma \rho }\parallel \gamma f\left(z\right)-z\parallel .$

Therefore, $\left\{{x}_{t}\right\}$ is bounded. We deduce immediately that $\left\{f\left({x}_{t}\right)\right\}$, $\left\{A{x}_{t}\right\}$, $\left\{S{x}_{t}\right\}$, $\left\{{u}_{t}\right\}$, and $\left\{{y}_{t}\right\}$ are also bounded. By using the convexity of $\parallel \cdot \parallel$ and the α-inverse strong monotonicity of A, from (3.2), we derive

$\begin{array}{r}{\parallel {x}_{t}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {y}_{t}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel \left(1-t\right)\left(\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\right)+t\left(\gamma f\left({x}_{t}\right)-z\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right){\parallel \left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-t\right){\parallel \left({x}_{t}-z\right)-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-t\right)\left({\parallel {x}_{t}-z\parallel }^{2}-\frac{2\lambda }{1-t}〈A{x}_{t}-Az,{x}_{t}-z〉+\frac{{\lambda }^{2}}{{\left(1-t\right)}^{2}}{\parallel A{x}_{t}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right)\left({\parallel {x}_{t}-z\parallel }^{2}-\frac{2\alpha \lambda }{1-t}{\parallel A{x}_{t}-Az\parallel }^{2}+\frac{{\lambda }^{2}}{{\left(1-t\right)}^{2}}{\parallel A{x}_{t}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-t\right)\left({\parallel {x}_{t}-z\parallel }^{2}+\frac{\lambda }{{\left(1-t\right)}^{2}}\left(\lambda -2\left(1-t\right)\alpha \right){\parallel A{x}_{t}-Az\parallel }^{2}\right)+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right){\parallel {x}_{t}-z\parallel }^{2}+\frac{\lambda }{\left(1-t\right)}\left(\lambda -2\left(1-t\right)\alpha \right){\parallel A{x}_{t}-Az\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\end{array}$
(3.3)

and so

$\frac{\lambda }{\left(1-t\right)}\left(2\left(1-t\right)\alpha -\lambda \right){\parallel A{x}_{t}-Az\parallel }^{2}\le t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}-t{\parallel {x}_{t}-z\parallel }^{2}\to 0.$

By the assumption, we have $2\left(1-t\right)\alpha -\lambda >0$ for all $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$ and so we obtain

$\underset{t\to 0+}{lim}\parallel A{x}_{t}-Az\parallel =0.$
(3.4)

Next, we show $\parallel {x}_{t}-S{x}_{t}\parallel \to 0$. By using the firm nonexpansivity of ${J}_{\lambda }^{B}$, we have

$\begin{array}{rl}{\parallel {y}_{t}-z\parallel }^{2}=& {\parallel {J}_{\lambda }^{B}\left(t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right)-z\parallel }^{2}\\ =& {\parallel {J}_{\lambda }^{B}\left(t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right)-{J}_{\lambda }^{B}\left(z-\lambda Az\right)\parallel }^{2}\\ \le & 〈t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}-\left(z-\lambda Az\right),{y}_{t}-z〉\\ =& \frac{1}{2}\left({\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}-\left(z-\lambda Az\right)\parallel }^{2}+{\parallel {y}_{t}-z\parallel }^{2}\\ -{\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda \left(A{x}_{t}-Az\right)-{y}_{t}\parallel }^{2}\right).\end{array}$

Thus it follows that

$\begin{array}{rl}{\parallel {y}_{t}-z\parallel }^{2}\le & {\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}-\left(z-\lambda Az\right)\parallel }^{2}\\ -{\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda \left(A{x}_{t}-Az\right)-{y}_{t}\parallel }^{2}.\end{array}$

By the nonexpansivity of $I-\frac{\lambda }{1-t}A$, we have

$\begin{array}{r}{\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda A{x}_{t}-\left(z-\lambda Az\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left(1-t\right)\left(\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\right)+t\left(\gamma f\left({x}_{t}\right)-z\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right){\parallel \left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-t\right){\parallel {x}_{t}-z\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\end{array}$

and thus

$\begin{array}{rl}{\parallel {x}_{t}-z\parallel }^{2}\le & {\parallel {y}_{t}-z\parallel }^{2}\\ \le & \left(1-t\right){\parallel {x}_{t}-z\parallel }^{2}+t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ -{\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda \left(A{x}_{t}-Az\right)-{y}_{t}\parallel }^{2}.\end{array}$

Hence it follows that

${\parallel t\gamma f\left({x}_{t}\right)+\left(1-t\right){x}_{t}-\lambda \left(A{x}_{t}-Az\right)-{y}_{t}\parallel }^{2}\le t{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\to 0.$

Since $\parallel A{x}_{t}-Az\parallel \to 0$, we deduce ${lim}_{t\to 0+}\parallel {x}_{t}-{y}_{t}\parallel =0$, which implies that

$\underset{t\to 0+}{lim}\parallel {x}_{t}-S{x}_{t}\parallel =0.$
(3.5)

From (3.2), we have

$\begin{array}{r}{\parallel {y}_{t}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel \left(1-t\right)\left(\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\right)+t\left(\gamma f\left({x}_{t}\right)-z\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\left(1-t\right)}^{2}{\parallel \left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2t\left(1-t\right)〈\gamma f\left({x}_{t}\right)-z,\left({x}_{t}-\frac{\lambda }{1-t}A{x}_{t}\right)-\left(z-\frac{\lambda }{1-t}Az\right)〉+{t}^{2}{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\left(1-t\right)}^{2}{\parallel {x}_{t}-z\parallel }^{2}+2t\left(1-t\right)〈\gamma f\left({x}_{t}\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉\\ \phantom{\rule{2em}{0ex}}+{t}^{2}{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\left(1-t\right)}^{2}{\parallel {x}_{t}-z\parallel }^{2}+2t\left(1-t\right)\gamma 〈f\left({x}_{t}\right)-f\left(z\right),{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉\\ \phantom{\rule{2em}{0ex}}+2t\left(1-t\right)〈\gamma f\left(z\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉+{t}^{2}{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}.\end{array}$

Note that $\parallel {x}_{t}-z\parallel \le \parallel {y}_{t}-z\parallel$. Then we obtain

$\begin{array}{rl}{\parallel {x}_{t}-z\parallel }^{2}\le & {\left(1-t\right)}^{2}{\parallel {x}_{t}-z\parallel }^{2}+2t\left(1-t\right)\gamma \parallel f\left({x}_{t}\right)-f\left(z\right)\parallel \left(\parallel {x}_{t}-z\parallel +\parallel \frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)\parallel \right)\\ +2t\left(1-t\right)〈\gamma f\left(z\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉+{t}^{2}{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \le & {\left(1-t\right)}^{2}{\parallel {x}_{t}-z\parallel }^{2}+2t\left(1-t\right)\gamma \rho {\parallel {x}_{t}-z\parallel }^{2}+2t\lambda \gamma \rho \parallel {x}_{t}-z\parallel \parallel A{x}_{t}-Az\parallel \\ +2t\left(1-t\right)〈\gamma f\left(z\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉+{t}^{2}{\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}\\ \le & \left[1-2\left(1-\gamma \rho \right)t\right]{\parallel {x}_{t}-z\parallel }^{2}+2t\left[\left(1-t\right)〈\gamma f\left(z\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉\\ +\frac{t}{2}\left({\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}+{\parallel {x}_{t}-z\parallel }^{2}\right)+\lambda \gamma \rho \parallel {x}_{t}-z\parallel \parallel A{x}_{t}-Az\parallel \right].\end{array}$

Thus it follows that

$\begin{array}{rl}{\parallel {x}_{t}-z\parallel }^{2}\le & \frac{1}{1-\gamma \rho }\left(〈\gamma f\left(z\right)-z,{x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z〉\\ +\frac{t}{2}\left({\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}+{\parallel {x}_{t}-z\parallel }^{2}\right)+t\parallel \gamma f\left(z\right)-z\parallel \parallel {x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z\parallel \\ +\lambda \gamma \rho \parallel {x}_{t}-z\parallel \parallel A{x}_{t}-Az\parallel \right)\\ \le & \frac{1}{1-\gamma \rho }〈\gamma f\left(z\right)-z,{x}_{t}-z〉+\left(t+\parallel A{x}_{t}-Az\parallel \right)M,\end{array}$
(3.6)

where M is some constant such that

$\begin{array}{r}sup\frac{1}{1-\gamma \rho }\left\{\frac{1}{2}\left({\parallel \gamma f\left({x}_{t}\right)-z\parallel }^{2}+{\parallel {x}_{t}-z\parallel }^{2}\right)+\parallel \gamma f\left(z\right)-z\parallel \parallel {x}_{t}-\frac{\lambda }{1-t}\left(A{x}_{t}-Az\right)-z\parallel ,\\ \phantom{\rule{1em}{0ex}}\lambda \gamma \rho \parallel {x}_{t}-z\parallel :t\in \left(0,1-\frac{\lambda }{2\alpha }\right)\right\}\le M.\end{array}$

Next, we show that $\left\{{x}_{t}\right\}$ is relatively norm-compact as $t\to 0+$. Assume that $\left\{{t}_{n}\right\}\subset \left(0,1-\frac{\lambda }{2\alpha }\right)$ is such that ${t}_{n}\to 0+$ as $n\to \mathrm{\infty }$. Put ${x}_{n}:={x}_{{t}_{n}}$. From (3.6), we have

${\parallel {x}_{n}-z\parallel }^{2}\le \frac{1}{1-\gamma \rho }〈\gamma f\left(z\right)-z,{x}_{n}-z〉+\left({t}_{n}+\parallel A{x}_{n}-Az\parallel \right)M.$
(3.7)

Since $\left\{{x}_{n}\right\}$ is bounded, without loss of generality, we may assume that ${x}_{{n}_{j}}⇀\stackrel{˜}{x}\in C$. Hence ${y}_{{n}_{j}}⇀\stackrel{˜}{x}$ because of $\parallel {x}_{n}-{y}_{n}\parallel \to 0$. From (3.5), we have

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-S{x}_{n}\parallel =0.$
(3.8)

We can use Lemma 2.2 to (3.8) to deduce $\stackrel{˜}{x}\in F\left(S\right)$. Further, we show that $\stackrel{˜}{x}$ is also in ${\left(A+B\right)}^{-1}0$. Let $v\in Bu$. Note that ${y}_{n}={J}_{\lambda }^{B}\left({t}_{n}\gamma f\left({x}_{n}\right)+\left(1-{t}_{n}\right){x}_{n}-\lambda A{x}_{n}\right)$ for all $n\ge 1$. Then we have

$\begin{array}{r}{t}_{n}\gamma f\left({x}_{n}\right)+\left(1-{t}_{n}\right){x}_{n}-\lambda A{x}_{n}\in \left(I+\lambda B\right){y}_{n}\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\frac{{t}_{n}\gamma f\left({x}_{n}\right)}{\lambda }+\frac{1-{t}_{n}}{\lambda }{x}_{n}-A{x}_{n}-\frac{{y}_{n}}{\lambda }\in B{y}_{n}.\end{array}$

Since B is monotone, we have, for all $\left(u,v\right)\in B$,

$\begin{array}{r}〈\frac{{t}_{n}\gamma f\left({x}_{n}\right)}{\lambda }+\frac{1-{t}_{n}}{\lambda }{x}_{n}-A{x}_{n}-\frac{{y}_{n}}{\lambda }-v,{y}_{n}-u〉\ge 0\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}〈{t}_{n}\gamma f\left({x}_{n}\right)+\left(1-{t}_{n}\right){x}_{n}-\lambda A{x}_{n}-{y}_{n}-\lambda v,{y}_{n}-u〉\ge 0\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}〈A{x}_{n}+v,{y}_{n}-u〉\le \frac{1}{\lambda }〈{x}_{n}-{y}_{n},{y}_{n}-u〉-\frac{{t}_{n}}{\lambda }〈{x}_{n}-\gamma f\left({x}_{n}\right),{y}_{n}-u〉\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}〈A\stackrel{˜}{x}+v,{y}_{n}-u〉\le \frac{1}{\lambda }〈{x}_{n}-{y}_{n},{y}_{n}-u〉-\frac{{t}_{n}}{\lambda }〈{x}_{n}-\gamma f\left({x}_{n}\right),{y}_{n}-u〉\\ \phantom{⟹〈A\stackrel{˜}{x}+v,{y}_{n}-u〉}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}+〈A\stackrel{˜}{x}-A{x}_{n},{y}_{n}-u〉\\ \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}〈A\stackrel{˜}{x}+v,{y}_{n}-u〉\le \frac{1}{\lambda }\parallel {x}_{n}-{y}_{n}\parallel \parallel {y}_{n}-u\parallel +\frac{{t}_{n}}{\lambda }\parallel {x}_{n}-\gamma f\left({x}_{n}\right)\parallel \parallel {y}_{n}-u\parallel \\ \phantom{⟹〈A\stackrel{˜}{x}+v,{y}_{n}-u〉}\phantom{\rule{1em}{0ex}}\phantom{\rule{2em}{0ex}}+\parallel A\stackrel{˜}{x}-A{x}_{n}\parallel \parallel {y}_{n}-u\parallel .\end{array}$

Thus it follows that

$\begin{array}{rl}〈A\stackrel{˜}{x}+v,\stackrel{˜}{x}-u〉\le & \frac{1}{\lambda }\parallel {x}_{{n}_{j}}-{y}_{{n}_{j}}\parallel \parallel {y}_{{n}_{j}}-u\parallel +\frac{{t}_{{n}_{j}}}{\lambda }\parallel {x}_{{n}_{j}}-\gamma f\left({x}_{{n}_{j}}\right)\parallel \parallel {y}_{{n}_{j}}-u\parallel \\ +\parallel A\stackrel{˜}{x}-A{x}_{{n}_{j}}\parallel \parallel {y}_{{n}_{j}}-u\parallel +〈A\stackrel{˜}{x}+v,\stackrel{˜}{x}-{y}_{{n}_{j}}〉.\end{array}$
(3.9)

Since $〈{x}_{{n}_{j}}-\stackrel{˜}{x},A{x}_{{n}_{j}}-A\stackrel{˜}{x}〉\ge \alpha {\parallel A{x}_{{n}_{j}}-A\stackrel{˜}{x}\parallel }^{2}$, $A{x}_{{n}_{j}}\to Az$, and ${x}_{{n}_{j}}⇀\stackrel{˜}{x}$, it follows that $A{x}_{{n}_{j}}\to A\stackrel{˜}{x}$. We also observe that ${t}_{n}\to 0$ and $\parallel {y}_{n}-{x}_{n}\parallel \to 0$. Then, from (3.9), we can derive $〈A\stackrel{˜}{x}+v,\stackrel{˜}{x}-u〉\le 0$, that is, $〈-A\stackrel{˜}{x}-v,\stackrel{˜}{x}-u〉\ge 0$. Since B is maximal monotone, we have $-A\stackrel{˜}{x}\in B\stackrel{˜}{x}$. This shows that $0\in \left(A+B\right)\stackrel{˜}{x}$. Hence we have $\stackrel{˜}{x}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. Therefore, substituting $\stackrel{˜}{x}$ for z in (3.7), we get

${\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\le \frac{1}{1-\gamma \rho }〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉+\left({t}_{n}+\parallel A{x}_{n}-A\stackrel{˜}{x}\parallel \right)M.$

Consequently, the weak convergence of $\left\{{x}_{n}\right\}$ to $\stackrel{˜}{x}$ actually implies that ${x}_{n}\to \stackrel{˜}{x}$. This proved the relative norm-compactness of the net $\left\{{x}_{t}\right\}$ as $t\to 0+$.

Now, we return to (3.7) and, taking the limit as $n\to \mathrm{\infty }$, we have

${\parallel \stackrel{˜}{x}-z\parallel }^{2}\le \frac{1}{1-\gamma \rho }〈\gamma f\left(z\right)-z,\stackrel{˜}{x}-z〉$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. In particular, $\stackrel{˜}{x}$ solves the following variational inequality:

$\stackrel{˜}{x}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0,\phantom{\rule{1em}{0ex}}〈\gamma f\left(z\right)-z,\stackrel{˜}{x}-z〉\ge 0$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$ or the equivalent dual variational inequality (see Lemma 2.3):

$\stackrel{˜}{x}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0,\phantom{\rule{1em}{0ex}}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},\stackrel{˜}{x}-z〉\ge 0$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. Hence $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$. Clearly, this is sufficient to conclude that the entire net $\left\{{x}_{t}\right\}$ converges to $\stackrel{˜}{x}$. This completes the proof. □

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H. Let $f:C\to H$ be a ρ-contraction and γ be a constant such that $0<\gamma <\frac{1}{\rho }$. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ and S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. For any ${x}_{0}\in C$, let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by

${x}_{n+1}=\left(1-\kappa \right)S{x}_{n}+\kappa {J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)$
(3.10)

for all $n\ge 0$, where $\kappa \in \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,2\alpha \right)$ and $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfy the following conditions:

1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_{n+1}}{{\alpha }_{n}}=1$ and ${\sum }_{n}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$a\left(1-{\alpha }_{n}\right)\le {\lambda }_{n}\le b\left(1-{\alpha }_{n}\right)$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}-{\lambda }_{n}}{{\alpha }_{n+1}}=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$, which solves the following variational inequality:

$〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},\stackrel{˜}{x}-z〉\ge 0$

for all $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

Proof Set ${y}_{n}={J}_{{\lambda }_{n}}^{B}{u}_{n}$, ${u}_{n}={\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}$ for all $n\ge 0$. Pick up $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. It is obvious that

$z=Sz={J}_{{\lambda }_{n}}^{B}\left(z-{\lambda }_{n}Az\right)={J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}z+\left(1-{\alpha }_{n}\right)\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right)$

for all $n\ge 0$. Since ${J}_{{\lambda }_{n}}^{B}$, S, and $I-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A$ are nonexpansive for all $\lambda >0$ and $n\ge 1$, we have

$\begin{array}{r}\parallel {y}_{n}-z\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)-z\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)\right)\\ \phantom{\rule{2em}{0ex}}-{J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}z+\left(1-{\alpha }_{n}\right)\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel \left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)\right)\\ \phantom{\rule{2em}{0ex}}-\left({\alpha }_{n}z+\left(1-{\alpha }_{n}\right)\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right){\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\parallel \left(1-{\alpha }_{n}\right)\left(\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right)+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-z\right)\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right)\parallel {x}_{n}-z\parallel +{\alpha }_{n}\parallel \gamma f\left({x}_{n}\right)-\gamma f\left(z\right)\parallel +{\alpha }_{n}\parallel \gamma f\left(z\right)-z\parallel \\ \phantom{\rule{1em}{0ex}}\le \left[1-\left(1-\gamma \rho \right){\alpha }_{n}\right]\parallel {x}_{n}-z\parallel +{\alpha }_{n}\parallel \gamma f\left(z\right)-z\parallel .\end{array}$
(3.11)

Hence we have

$\begin{array}{rl}\parallel {x}_{n+1}-z\parallel & \le \left(1-\kappa \right)\parallel S{x}_{n}-z\parallel +\kappa \parallel {y}_{n}-z\parallel \\ \le \left(1-\kappa \right)\parallel {x}_{n}-z\parallel +\kappa \left[1-\left(1-\gamma \rho \right){\alpha }_{n}\right]\parallel {x}_{n}-z\parallel +\kappa {\alpha }_{n}\parallel \gamma f\left(z\right)-z\parallel \\ =\left[1-\left(1-\gamma \rho \right)\kappa {\alpha }_{n}\right]\parallel {x}_{n}-z\parallel +\kappa {\alpha }_{n}\parallel \gamma f\left(z\right)-z\parallel .\end{array}$

By induction, we have

$\parallel {x}_{n+1}-z\parallel \le max\left\{\parallel {x}_{0}-z\parallel ,\frac{1}{1-\gamma \rho }\parallel \gamma f\left(z\right)-z\parallel \right\}.$

Therefore, $\left\{{x}_{n}\right\}$ is bounded. Since A is α-inverse strongly monotone, it is $\frac{1}{\alpha }$-Lipschitz continuous. We deduce immediately that $\left\{f\left({x}_{n}\right)\right\}$, $\left\{S{x}_{n}\right\}$, $\left\{A{x}_{n}\right\}$, $\left\{{u}_{n}\right\}$, and $\left\{{y}_{n}\right\}$ are also bounded. By using the convexity of $\parallel \cdot \parallel$ and the α-inverse strong monotonicity of A, it follows from (3.11) that

$\begin{array}{r}{\parallel \left(1-{\alpha }_{n}\right)\left(\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right)+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-z\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right){\parallel \left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right){\parallel \left({x}_{n}-z\right)-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}\left(A{x}_{n}-Az\right)\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-z\parallel }^{2}-\frac{2{\lambda }_{n}}{1-{\alpha }_{n}}〈A{x}_{n}-Az,{x}_{n}-z〉+\frac{{\lambda }_{n}^{2}}{{\left(1-{\alpha }_{n}\right)}^{2}}{\parallel A{x}_{n}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-z\parallel }^{2}-\frac{2\alpha {\lambda }_{n}}{1-{\alpha }_{n}}{\parallel A{x}_{n}-Az\parallel }^{2}+\frac{{\lambda }_{n}^{2}}{{\left(1-{\alpha }_{n}\right)}^{2}}{\parallel A{x}_{n}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-z\parallel }^{2}+\frac{{\lambda }_{n}}{{\left(1-{\alpha }_{n}\right)}^{2}}\left({\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \right){\parallel A{x}_{n}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}.\end{array}$
(3.12)

By the condition (c), we get ${\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \le 0$ for all $n\ge 0$. Then, from (3.11) and (3.12), we obtain

$\begin{array}{rl}{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\le & \left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-z\parallel }^{2}+\frac{{\lambda }_{n}}{{\left(1-{\alpha }_{n}\right)}^{2}}\left({\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \right){\parallel A{x}_{n}-Az\parallel }^{2}\right)\\ +{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}.\end{array}$
(3.13)

From (3.10), it follows that

$\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel }^{2}& ={\parallel \left(1-\kappa \right)\left(S{x}_{n}-z\right)+\kappa \left({J}_{{\lambda }_{n}}^{B}{u}_{n}-z\right)\parallel }^{2}\\ \le \left(1-\kappa \right){\parallel {x}_{n}-z\parallel }^{2}+\kappa {\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}.\end{array}$
(3.14)

Next, we estimate $\parallel {x}_{n+1}-{x}_{n}\parallel$. In fact, we have

$\begin{array}{rl}\parallel {x}_{n+2}-{x}_{n+1}\parallel & =\parallel \left(1-\kappa \right)\left(S{x}_{n+1}-S{x}_{n}\right)+\kappa \left({y}_{n+1}-{y}_{n}\right)\parallel \\ \le \left(1-\kappa \right)\parallel {x}_{n+1}-{x}_{n}\parallel +\kappa \parallel {y}_{n+1}-{y}_{n}\parallel \end{array}$

and

$\begin{array}{rl}\parallel {y}_{n+1}-{y}_{n}\parallel =& \parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n+1}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ \le & \parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n+1}-{J}_{{\lambda }_{n+1}}^{B}{u}_{n}\parallel +\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ \le & \parallel \left({\alpha }_{n+1}\gamma f\left({x}_{n+1}\right)+\left(1-{\alpha }_{n+1}\right){x}_{n+1}-{\lambda }_{n+1}A{x}_{n+1}\right)\\ -\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)\parallel +\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ =& \parallel {\alpha }_{n+1}\gamma \left(f\left({x}_{n+1}\right)-f\left({x}_{n}\right)\right)+\left({\alpha }_{n+1}-{\alpha }_{n}\right)\gamma f\left({x}_{n}\right)\\ +\left(1-{\alpha }_{n+1}\right)\left[\left(I-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A\right){x}_{n+1}-\left(I-\frac{{\lambda }_{n+1}}{1-{\alpha }_{n+1}}A\right){x}_{n}\right]\\ +\left({\alpha }_{n}-{\alpha }_{n+1}\right){x}_{n}+\left({\lambda }_{n}-{\lambda }_{n+1}\right)A{x}_{n}\parallel +\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ \le & {\alpha }_{n+1}\gamma \rho \parallel {x}_{n+1}-{x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\left(\parallel \gamma f\left({x}_{n}\right)\parallel +\parallel {x}_{n}\parallel \right)\\ +\left(1-{\alpha }_{n+1}\right)\parallel {x}_{n+1}-{x}_{n}\parallel +|{\lambda }_{n}-{\lambda }_{n+1}|\parallel A{x}_{n}\parallel +\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ =& \left[1-\left(1-\gamma \rho \right){\alpha }_{n+1}\right]\parallel {x}_{n+1}-{x}_{n}\parallel +|{\alpha }_{n+1}-{\alpha }_{n}|\left(\parallel \gamma f\left({x}_{n}\right)\parallel +\parallel {x}_{n}\parallel \right)\\ +|{\lambda }_{n}-{\lambda }_{n+1}|\parallel A{x}_{n}\parallel +\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel .\end{array}$

By the resolvent identity (2.1), we have

${J}_{{\lambda }_{n+1}}^{B}{u}_{n}={J}_{{\lambda }_{n}}^{B}\left(\frac{{\lambda }_{n}}{{\lambda }_{n+1}}{u}_{n}+\left(1-\frac{{\lambda }_{n}}{{\lambda }_{n+1}}\right){J}_{{\lambda }_{n+1}}^{B}{u}_{n}\right).$

Thus it follows that

$\begin{array}{rl}\parallel {J}_{{\lambda }_{n+1}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel & =\parallel {J}_{{\lambda }_{n}}^{B}\left(\frac{{\lambda }_{n}}{{\lambda }_{n+1}}{u}_{n}+\left(1-\frac{{\lambda }_{n}}{{\lambda }_{n+1}}\right){J}_{{\lambda }_{n+1}}^{B}{u}_{n}\right)-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \\ \le \parallel \left(\frac{{\lambda }_{n}}{{\lambda }_{n+1}}{u}_{n}+\left(1-\frac{{\lambda }_{n}}{{\lambda }_{n+1}}\right){J}_{{\lambda }_{n+1}}^{B}{u}_{n}\right)-{u}_{n}\parallel \\ \le \frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\lambda }_{n+1}}\parallel {u}_{n}-{J}_{{\lambda }_{n+1}}^{B}{u}_{n}\parallel \end{array}$

and so

$\begin{array}{r}\parallel {x}_{n+2}-{x}_{n+1}\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right)\parallel {x}_{n+1}-{x}_{n}\parallel +\kappa \parallel {y}_{n+1}-{y}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right)\parallel {x}_{n+1}-{x}_{n}\parallel +\kappa \left[1-\left(1-\gamma \rho \right){\alpha }_{n+1}\right]\parallel {x}_{n+1}-{x}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\kappa |{\alpha }_{n+1}-{\alpha }_{n}|\left(\parallel \gamma f\left({x}_{n}\right)\parallel +\parallel {x}_{n}\parallel \right)+\kappa |{\lambda }_{n}-{\lambda }_{n+1}|\parallel A{x}_{n}\parallel \\ \phantom{\rule{2em}{0ex}}+\kappa \frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\lambda }_{n+1}}\parallel {u}_{n}-{J}_{{\lambda }_{n+1}}^{B}{u}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}\le \left[1-\left(1-\gamma \rho \right)\kappa {\alpha }_{n+1}\right]\parallel {x}_{n+1}-{x}_{n}\parallel +\left(1-\gamma \rho \right)\kappa {\alpha }_{n+1}\left[\frac{|{\alpha }_{n+1}-{\alpha }_{n}|}{{\alpha }_{n+1}}\frac{\parallel \gamma f\left({x}_{n}\right)\parallel +\parallel {x}_{n}\parallel }{1-\gamma \rho }\\ \phantom{\rule{2em}{0ex}}+\frac{|{\lambda }_{n}-{\lambda }_{n+1}|}{{\alpha }_{n+1}}\frac{\parallel A{x}_{n}\parallel }{1-\gamma \rho }+\frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\alpha }_{n+1}{\lambda }_{n+1}}\frac{\parallel {u}_{n}-{J}_{{\lambda }_{n+1}}^{B}{u}_{n}\parallel }{1-\gamma \rho }\right].\end{array}$

By the assumptions, we know that $\frac{|{\alpha }_{n+1}-{\alpha }_{n}|}{{\alpha }_{n+1}}\to 0$ and $\frac{|{\lambda }_{n+1}-{\lambda }_{n}|}{{\alpha }_{n+1}}\to 0$. Then, from Lemma 2.5, we get

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.15)

Thus, from (3.13) and (3.14), it follows that

$\begin{array}{r}{\parallel {x}_{n+1}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right){\parallel {x}_{n}-z\parallel }^{2}+\kappa {\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \kappa \left[\left(1-{\alpha }_{n}\right)\left({\parallel {x}_{n}-z\parallel }^{2}+\frac{{\lambda }_{n}}{{\left(1-{\alpha }_{n}\right)}^{2}}\left({\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \right){\parallel A{x}_{n}-Az\parallel }^{2}\right)\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\right]+\left(1-\kappa \right){\parallel {x}_{n}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left[1-\kappa {\alpha }_{n}\right]{\parallel {x}_{n}-z\parallel }^{2}+\frac{\kappa {\lambda }_{n}}{1-{\alpha }_{n}}\left({\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \right){\parallel A{x}_{n}-Az\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-z\parallel }^{2}+\frac{\kappa {\lambda }_{n}}{1-{\alpha }_{n}}\left({\lambda }_{n}-2\left(1-{\alpha }_{n}\right)\alpha \right){\parallel A{x}_{n}-Az\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\end{array}$

and so

$\begin{array}{r}\frac{\kappa {\lambda }_{n}}{\left(1-{\alpha }_{n}\right)}\left(2\left(1-{\alpha }_{n}\right)\alpha -{\lambda }_{n}\right){\parallel A{x}_{n}-Az\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n+1}-z\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-z\parallel -\parallel {x}_{n+1}-z\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel +\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}.\end{array}$

Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n+1}-{x}_{n}\parallel =0$, and ${lim inf}_{n\to \mathrm{\infty }}\frac{\kappa {\lambda }_{n}}{\left(1-{\alpha }_{n}\right)}\left(2\left(1-{\alpha }_{n}\right)\alpha -{\lambda }_{n}\right)>0$, we have

$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_{n}-Az\parallel =0.$
(3.16)

Next, we show $\parallel {x}_{n}-S{x}_{n}\parallel \to 0$. By using the firm nonexpansivity of ${J}_{{\lambda }_{n}}^{B}$, we have

$\begin{array}{rl}{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}=& {\parallel {J}_{{\lambda }_{n}}^{B}\left({\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)-{J}_{{\lambda }_{n}}^{B}\left(z-{\lambda }_{n}Az\right)\parallel }^{2}\\ \le & 〈{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}-\left(z-{\lambda }_{n}Az\right),{J}_{{\lambda }_{n}}^{B}{u}_{n}-z〉\\ =& \frac{1}{2}\left({\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}-\left(z-{\lambda }_{n}Az\right)\parallel }^{2}+{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\\ -{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}\left(A{x}_{n}-Az\right)-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\right).\end{array}$

From the condition (c) and the α-inverse strongly monotonicity of A, we know that $I-{\lambda }_{n}A/\left(1-{\alpha }_{n}\right)$ is nonexpansive. Hence it follows that

$\begin{array}{r}{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}-\left(z-{\lambda }_{n}Az\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}={\parallel \left(1-{\alpha }_{n}\right)\left(\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\right)+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-z\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right){\parallel \left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(z-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}Az\right)\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\end{array}$

and thus

$\begin{array}{rl}{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\le & \frac{1}{2}\left(\left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}+{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\\ -{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}-{\lambda }_{n}\left(A{x}_{n}-Az\right)\parallel }^{2}\right),\end{array}$

that is,

$\begin{array}{r}{\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-z\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}-{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}-{\lambda }_{n}\left(A{x}_{n}-Az\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}-{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}〈{\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n},A{x}_{n}-Az〉-{\lambda }_{n}^{2}{\parallel A{x}_{n}-Az\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+{\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}-{\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \parallel A{x}_{n}-Az\parallel .\end{array}$

This together with (3.14) implies that

$\begin{array}{rl}{\parallel {x}_{n+1}-z\parallel }^{2}\le & \left(1-\kappa \right){\parallel {x}_{n}-z\parallel }^{2}+\kappa \left(1-{\alpha }_{n}\right){\parallel {x}_{n}-z\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ -\kappa {\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\kappa \parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \parallel A{x}_{n}-Az\parallel \\ =& \left[1-\kappa {\alpha }_{n}\right]{\parallel {x}_{n}-z\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ -\kappa {\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\\ +2{\lambda }_{n}\kappa \parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \parallel A{x}_{n}-Az\parallel \end{array}$

and hence

$\begin{array}{r}\kappa {\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-z\parallel }^{2}-{\parallel {x}_{n+1}-z\parallel }^{2}-\kappa {\alpha }_{n}{\parallel {x}_{n}-z\parallel }^{2}+\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}\kappa \parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \parallel A{x}_{n}-Az\parallel \\ \phantom{\rule{1em}{0ex}}\le \left(\parallel {x}_{n}-z\parallel +\parallel {x}_{n+1}-z\parallel \right)\parallel {x}_{n+1}-{x}_{n}\parallel +\kappa {\alpha }_{n}{\parallel \gamma f\left({x}_{n}\right)-z\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\lambda }_{n}\kappa \parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel \parallel A{x}_{n}-Az\parallel .\end{array}$

Since $\parallel {x}_{n+1}-{x}_{n}\parallel \to 0$, ${\alpha }_{n}\to 0$, and $\parallel A{x}_{n}-Az\parallel \to 0$ (by (3.16)), we deduce

$\underset{n\to \mathrm{\infty }}{lim}\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel =0.$

This implies that

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{J}_{{\lambda }_{n}}^{B}{u}_{n}\parallel =0.$
(3.17)

Combining (3.10), (3.15), and (3.17), we get

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-S{x}_{n}\parallel =0.$
(3.18)

Put $\stackrel{˜}{x}={lim}_{t\to 0+}{x}_{t}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$, where $\left\{{x}_{t}\right\}$ is the net defined by (3.1).

Finally, we show that ${x}_{n}\to \stackrel{˜}{x}$. Taking $z=\stackrel{˜}{x}$ in (3.16), we get $\parallel A{x}_{n}-A\stackrel{˜}{x}\parallel \to 0$. First, we prove ${lim sup}_{n\to \mathrm{\infty }}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉\le 0$. We take a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that

$\underset{n\to \mathrm{\infty }}{lim sup}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉=\underset{i\to \mathrm{\infty }}{lim}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{{n}_{i}}-\stackrel{˜}{x}〉.$

There exists a subsequence $\left\{{x}_{{n}_{{i}_{j}}}\right\}$ of $\left\{{x}_{{n}_{i}}\right\}$ which converges weakly to a point $w\in C$. Hence $\left\{{y}_{{n}_{{i}_{j}}}\right\}$ also converges weakly to w because of $\parallel {x}_{{n}_{{i}_{j}}}-{y}_{{n}_{{i}_{j}}}\parallel \to 0$. By the demi-closedness principle of the nonexpansive mapping (see Lemma 2.2) and (3.18), we deduce $w\in F\left(S\right)$. Furthermore, by a similar argument to that of Theorem 3.1, we can show that w is also in ${\left(A+B\right)}^{-1}0$. Hence we have $w\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. This implies that

$\underset{n\to \mathrm{\infty }}{lim sup}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉=\underset{j\to \mathrm{\infty }}{lim}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{{n}_{{i}_{j}}}-\stackrel{˜}{x}〉=〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},w-\stackrel{˜}{x}〉.$

Note that $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(\gamma f\left(\stackrel{˜}{x}\right)\right)$. Then we have

$〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},w-\stackrel{˜}{x}〉\le 0$

for all $w\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$. Therefore, it follows that

$\underset{n\to \mathrm{\infty }}{lim sup}〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉\le 0.$

From (3.10), we have

$\begin{array}{r}{\parallel {x}_{n+1}-\stackrel{˜}{x}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa {\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-\stackrel{˜}{x}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa {\parallel {J}_{{\lambda }_{n}}^{B}{u}_{n}-{J}_{{\lambda }_{n}}^{B}\left(\stackrel{˜}{x}-{\lambda }_{n}A\stackrel{˜}{x}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa {\parallel {u}_{n}-\left(\stackrel{˜}{x}-{\lambda }_{n}A\stackrel{˜}{x}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa {\parallel {\alpha }_{n}\gamma f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}-\left(\stackrel{˜}{x}-{\lambda }_{n}A\stackrel{˜}{x}\right)\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\kappa {\parallel \left(1-{\alpha }_{n}\right)\left(\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(\stackrel{˜}{x}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A\stackrel{˜}{x}\right)\right)+{\alpha }_{n}\left(\gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+\left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\\ \phantom{\rule{1em}{0ex}}=\left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa \left({\left(1-{\alpha }_{n}\right)}^{2}{\parallel \left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(\stackrel{˜}{x}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A\stackrel{˜}{x}\right)\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\gamma f\left({x}_{n}\right)-\stackrel{˜}{x},\left({x}_{n}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A{x}_{n}\right)-\left(\stackrel{˜}{x}-\frac{{\lambda }_{n}}{1-{\alpha }_{n}}A\stackrel{˜}{x}\right)〉\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}^{2}{\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa \left({\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+2{\alpha }_{n}{\lambda }_{n}〈\gamma f\left({x}_{n}\right)-\stackrel{˜}{x},A{x}_{n}-A\stackrel{˜}{x}〉\\ \phantom{\rule{2em}{0ex}}+2{\alpha }_{n}\left(1-{\alpha }_{n}\right)\gamma 〈f\left({x}_{n}\right)-f\left(\stackrel{˜}{x}\right),{x}_{n}-\stackrel{˜}{x}〉+2{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉\\ \phantom{\rule{2em}{0ex}}+{\alpha }_{n}^{2}{\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}\le \left(1-\kappa \right){\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+\kappa \left({\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+2{\alpha }_{n}{\lambda }_{n}\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel \parallel A{x}_{n}-A\stackrel{˜}{x}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\alpha }_{n}\left(1-{\alpha }_{n}\right)\gamma \rho {\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+2{\alpha }_{n}\left(1-{\alpha }_{n}\right)〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉+{\alpha }_{n}^{2}{\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}\le \left[1-2\kappa \left(1-\gamma \rho \right){\alpha }_{n}\right]{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}+2{\alpha }_{n}\kappa {\lambda }_{n}\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel \parallel A{x}_{n}-A\stackrel{˜}{x}\parallel \\ \phantom{\rule{2em}{0ex}}+2{\alpha }_{n}\kappa \left(1-{\alpha }_{n}\right)〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉+\kappa {\alpha }_{n}^{2}\left({\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}+{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\right)\\ \phantom{\rule{1em}{0ex}}=\left[1-2\kappa \left(1-\gamma \rho \right){\alpha }_{n}\right]{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\\ \phantom{\rule{2em}{0ex}}+2\kappa \left(1-\gamma \rho \right){\alpha }_{n}\left[\frac{{\lambda }_{n}}{1-\gamma \rho }\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel \parallel A{x}_{n}-A\stackrel{˜}{x}\parallel \\ \phantom{\rule{2em}{0ex}}+\frac{1-{\alpha }_{n}}{1-\gamma \rho }〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉+\frac{{\alpha }_{n}}{2\left(1-\gamma \rho \right)}\left({\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}+{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\right)\right].\end{array}$

It is clear that ${\sum }_{n}2\kappa \left(1-\gamma \rho \right){\alpha }_{n}=\mathrm{\infty }$ and

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim sup}\left\{\frac{{\lambda }_{n}}{1-\gamma \rho }\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel \parallel A{x}_{n}-A\stackrel{˜}{x}\parallel +\frac{1-{\alpha }_{n}}{1-\gamma \rho }〈\gamma f\left(\stackrel{˜}{x}\right)-\stackrel{˜}{x},{x}_{n}-\stackrel{˜}{x}〉\\ \phantom{\rule{1em}{0ex}}+\frac{{\alpha }_{n}}{2\left(1-\gamma \rho \right)}\left({\parallel \gamma f\left({x}_{n}\right)-\stackrel{˜}{x}\parallel }^{2}+{\parallel {x}_{n}-\stackrel{˜}{x}\parallel }^{2}\right)\right\}\le 0.\end{array}$

Therefore, we can apply Lemma 2.5 to conclude that ${x}_{n}\to \stackrel{˜}{x}$. This completes the proof. □

Remark 3.3 One quite often seeks a particular solution of a given nonlinear problem, in particular, the minimum-norm element. For instance, given a closed convex subset C of a Hilbert space ${H}_{1}$ and a bounded linear operator $W:{H}_{1}\to {H}_{2}$, where ${H}_{2}$ is another Hilbert space. The C-constrained pseudoinverse of W, ${W}_{C}^{†}$, is then defined as the minimum-norm solution of the constrained minimization problem

${W}_{C}^{†}\left(b\right):=arg\underset{x\in C}{min}\parallel Wx-b\parallel ,$

which is equivalent to the fixed point problem

$u={proj}_{C}\left(u-\mu {W}^{\ast }\left(Wu-b\right)\right),$

where ${W}^{\ast }$ is the adjoint of W and $\mu >0$ is a constant, and $b\in {H}_{2}$ is such that ${P}_{\overline{W\left(C\right)}}\left(b\right)\in W\left(C\right)$. From Theorems 3.1 and 3.2, we get the following corollaries which can find the minimum-norm element in $F\left(S\right)\cap {\left(A+B\right)}^{-1}0$; that is, find $\stackrel{˜}{x}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0$ such that

$\stackrel{˜}{x}=arg\underset{x\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0}{min}\parallel x\parallel .$

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ and S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. Let λ and κ be two constants satisfying $a\le \lambda \le b$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$ and $\kappa \in \left(0,1\right)$. For any $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$, let $\left\{{x}_{t}\right\}\subset C$ be a net generated by

${x}_{t}=\left(1-\kappa \right)S{x}_{t}+\kappa {J}_{\lambda }^{B}\left(\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right).$

Then the net $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0+$ to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(0\right)$ which is the minimum-norm element in $F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

Corollary 3.5 Let C be a closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H and let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ such that ${\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. Let λ be a constant satisfying $a\le \lambda \le b$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$. For any $t\in \left(0,1-\frac{\lambda }{2\alpha }\right)$, let $\left\{{x}_{t}\right\}\subset C$ be a net generated by

${x}_{t}={J}_{\lambda }^{B}\left(\left(1-t\right){x}_{t}-\lambda A{x}_{t}\right).$

Then the net $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0+$ to a point $\stackrel{˜}{x}={P}_{{\left(A+B\right)}^{-1}0}\left(0\right)$, which is the minimum-norm element in ${\left(A+B\right)}^{-1}0$.

Corollary 3.6 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H. Let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ and let S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. For any ${x}_{0}\in C$, let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by

${x}_{n+1}=\left(1-\kappa \right)S{x}_{n}+\kappa {J}_{{\lambda }_{n}}^{B}\left(\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)$

for all $n\ge 0$, where $\kappa \in \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,2\alpha \right)$, and $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfy the following conditions:

1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_{n+1}}{{\alpha }_{n}}=1$, and ${\sum }_{n}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$a\left(1-{\alpha }_{n}\right)\le {\lambda }_{n}\le b\left(1-{\alpha }_{n}\right)$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}-{\lambda }_{n}}{{\alpha }_{n}}=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0}\left(0\right)$, which is the minimum-norm element in $F\left(S\right)\cap {\left(A+B\right)}^{-1}0$.

Corollary 3.7 Let C be a closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping from C into H and let B be a maximal monotone operator on H such that the domain of B is included in C. Let ${J}_{\lambda }^{B}={\left(I+\lambda B\right)}^{-1}$ be the resolvent of B for any $\lambda >0$ such that ${\left(A+B\right)}^{-1}0\ne \mathrm{\varnothing }$. For any ${x}_{0}\in C$, let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by

${x}_{n+1}=\left(1-\kappa \right){x}_{n}+\kappa {J}_{{\lambda }_{n}}^{B}\left(\left(1-{\alpha }_{n}\right){x}_{n}-{\lambda }_{n}A{x}_{n}\right)$

for all $n\ge 0$, where $\kappa \in \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,2\alpha \right)$, and $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfy the following conditions:

1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_{n+1}}{{\alpha }_{n}}=1$, and ${\sum }_{n}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$a\left(1-{\alpha }_{n}\right)\le {\lambda }_{n}\le b\left(1-{\alpha }_{n}\right)$, where $\left[a,b\right]\subset \left(0,2\alpha \right)$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}-{\lambda }_{n}}{{\alpha }_{n}}=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point $\stackrel{˜}{x}={P}_{{\left(A+B\right)}^{-1}0}\left(0\right)$, which is the minimum-norm element in ${\left(A+B\right)}^{-1}0$.

Remark 3.8 The present paper provides some methods which do not use projection for finding the minimum-norm solution problem.

## 4 Applications

Next, we consider the problem for finding the minimum-norm solution of a mathematical model related to equilibrium problems. Let C be a nonempty closed convex subset of a Hilbert space and $G:C×C\to R$ be a bifunction satisfying the following conditions:

(E1) $G\left(x,x\right)=0$ for all $x\in C$;

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

(E3) for all $x,y,z\in C$, ${lim sup}_{t↓0}G\left(tz+\left(1-t\right)x,y\right)\le G\left(x,y\right)$;

(E4) for all $x\in C$, $G\left(x,\cdot \right)$ is convex and lower semicontinuous.

Then the mathematical model related to the equilibrium problem (with respect to C) is as follows:

Find $\stackrel{˜}{x}\in C$ such that

$G\left(\stackrel{˜}{x},y\right)\ge 0$
(4.1)

for all $y\in C$. The set of such solutions $\stackrel{˜}{x}$ is denoted by $EP\left(G\right)$.

The following lemma appears implicitly in Blum and Oettli [36].

Lemma 4.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4). Then, for any $r>0$ and $x\in H$, there exists $z\in C$ such that

$G\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0$

for all $y\in C$.

The following lemma was given in Combettes and Hirstoaga [37].

Lemma 4.2 Assume that G is a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4). For any $r>0$ and $x\in H$, define a mapping ${T}_{r}:H\to C$ as follows:

${T}_{r}\left(x\right)=\left\{z\in C:G\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$

for all $x\in H$. Then the following hold:

1. (a)

${T}_{r}$ is single-valued;

2. (b)

${T}_{r}$ is a firmly nonexpansive mapping, i.e., for all $x,y\in H$,

${\parallel {T}_{r}x-{T}_{r}y\parallel }^{2}\le 〈{T}_{r}x-{T}_{r}y,x-y〉;$
3. (c)

$F\left({T}_{r}\right)=EP\left(G\right)$;

4. (d)

$EP\left(G\right)$ is closed and convex.

We call such a ${T}_{r}$ the resolvent of G for any $r>0$. Using Lemmas 4.1 and 4.2, we have the following lemma (see [38] for a more general result).

Lemma 4.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4). Let ${A}_{G}$ be a multi-valued mapping from H into itself defined by

${A}_{G}x=\left\{\begin{array}{cc}\left\{z\in H:G\left(x,y\right)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\right\},\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$

Then $EP\left(G\right)={A}_{G}^{-1}\left(0\right)$ and ${A}_{G}$ is a maximal monotone operator with $dom\left({A}_{G}\right)\subset C$. Further, for any $x\in H$ and $r>0$, the resolvent ${T}_{r}$ of G coincides with the resolvent of ${A}_{G}$, i.e.,

${T}_{r}x={\left(I+r{A}_{G}\right)}^{-1}x.$

Form Lemma 4.3 and Theorems 3.1 and 3.2, we have the following results.

Theorem 4.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4) and ${T}_{r}$ be the resolvent of G for any $r>0$. Let S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap EP\left(G\right)\ne \mathrm{\varnothing }$. For any $t\in \left(0,1\right)$, let $\left\{{x}_{t}\right\}\subset C$ be a net generated by

${x}_{t}=\left(1-\kappa \right)S{x}_{t}+\kappa {T}_{r}\left(\left(1-t\right){x}_{t}\right).$

Then the net $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0+$ to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap EP\left(G\right)}\left(0\right)$, which is the minimum-norm element in $F\left(S\right)\cap EP\left(G\right)$.

Proof From Lemma 4.3, we know ${A}_{G}$ is maximal monotone. Thus, in Theorem 3.1, we can set ${J}_{\lambda }^{B}={T}_{r}$. At the same time, in Theorem 3.1, we can choose $f=0$ and $A=0$, and (3.1) reduces to

${x}_{t}=\left(1-\kappa \right)S{x}_{t}+\kappa {T}_{r}\left(\left(1-t\right){x}_{t}\right).$

Consequently, from Theorem 3.1, we get the desired result. This completes the proof. □

Corollary 4.5 Let C be a nonempty closed convex subset of a real Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4) and ${T}_{r}$ be the resolvent of G for any $r>0$. Suppose that $EP\left(G\right)\ne \mathrm{\varnothing }$. For any $t\in \left(0,1\right)$, let $\left\{{x}_{t}\right\}\subset C$ be a net generated by

${x}_{t}={T}_{r}\left(\left(1-t\right){x}_{t}\right).$

Then the net $\left\{{x}_{t}\right\}$ converges strongly as $t\to 0+$ to a point $\stackrel{˜}{x}={P}_{EP\left(G\right)}\left(0\right)$, which is the minimum-norm element in $EP\left(G\right)$.

Theorem 4.6 Let C be a nonempty closed and convex subset of a real Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4) and ${T}_{\lambda }$ be the resolvent of G for any $\lambda >0$. Let S be a nonexpansive mapping from C into itself such that $F\left(S\right)\cap EP\left(G\right)\ne \mathrm{\varnothing }$. For any ${x}_{0}\in C$, let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by

${x}_{n+1}=\left(1-\kappa \right)S{x}_{n}+\kappa {T}_{{\lambda }_{n}}\left(\left(1-{\alpha }_{n}\right){x}_{n}\right)$

for all $n\ge 0$, where $\kappa \in \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$, and $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfy the conditions:

1. (a)

${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${lim}_{n\to \mathrm{\infty }}\frac{{\alpha }_{n+1}}{{\alpha }_{n}}=1$, and ${\sum }_{n}{\alpha }_{n}=\mathrm{\infty }$;

2. (b)

$a\le {\lambda }_{n}\le b$, where $\left[a,b\right]\subset \left(0,\mathrm{\infty }\right)$ and ${lim}_{n\to \mathrm{\infty }}\frac{{\lambda }_{n+1}-{\lambda }_{n}}{{\alpha }_{n}}=0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to a point $\stackrel{˜}{x}={P}_{F\left(S\right)\cap EP\left(G\right)}\left(0\right)$, which is the minimum-norm element in $F\left(S\right)\cap EP\left(G\right)$.

Proof From Lemma 4.3, we know ${A}_{G}$ is maximal monotone. Thus, in Theorem 3.2, we can set ${J}_{{\lambda }_{n}}^{B}={T}_{{\lambda }_{n}}$. At the same time, in Theorem 3.2, we can choose $f=0$ and $A=0$, and (3.10) reduces to

${x}_{n+1}=\left(1-\kappa \right)S{x}_{n}+\kappa {T}_{{\lambda }_{n}}\left(\left(1-{\alpha }_{n}\right){x}_{n}\right)$

for all $n\ge 0$. Consequently, from Theorem 3.2, we get the desired result. This completes the proof. □

Corollary 4.7 Let C be a nonempty closed convex subset of a real Hilbert space H. Let G be a bifunction from $C×C$ into R satisfying the conditions (E1)-(E4) and ${T}_{\lambda }$ be the resolvent of G for any $\lambda >0$. Suppose $EP\left(G\right)\ne \mathrm{\varnothing }$. For any ${x}_{0}\in C$, let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by

${x}_{n+1}$