Iterative algorithms for monotone inclusion problems, fixed point problems and minimization problems

Abstract

We introduce new implicit and explicit iterative schemes for finding a common element of the set of fixed points of k-strictly pseudocontractive mapping and the set of zeros of the sum of two monotone operators in a Hilbert space. Then we establish strong convergence of the sequences generated by the proposed schemes to a common point of two sets, which is a solution of a certain variational inequality. Further, we find the unique solution of the quadratic minimization problem, where the constraint is the common set of two sets mentioned above. As applications, we consider iterative schemes for the Hartmann-Stampacchia variational inequality problem and the equilibrium problem coupled with fixed point problem.

MSC:47H05, 47H09, 47H10, 47J05, 47J07, 47J25, 47J20, 49M05.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H, and let $T:C\to C$ be a self-mapping on C. We denote by $F(T)$ the set of fixed points of T, that is, $F(T):=\{x\in C:Tx=x\}$.

Let $A:C\to H$ be a single-valued nonlinear mapping, and let $B:H\to 2^H$ be a multivalued mapping. Then we consider the monotone inclusion problem (MIP) of finding $x\in H$ such that

$$0\in Ax+Bx.$$

(1.1)

The set of solutions of the MIP (1.1) is denoted by ${(A+B)}^{-1}0$. That is, ${(A+B)}^{-1}0$ is the set of zeros of $A+B$. The MIP (1.1) provides a convenient framework for studying a number of problems arising in structural analysis, mechanics, economics and others; see, for instance [1, 2]. Also, various types of inclusion problems have been extended and generalized, and there are many algorithms for solving variational inclusions. For more details, see [3–5] and the references therein.

The class of pseudocontractive mappings is one of the most important classes of mappings among nonlinear mappings. We recall that a mapping $T:C\to H$ is said to be k-strictly pseudocontractive if there exists a constant $k\in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+k{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

Note that the class of k-strictly pseudocontractive mappings includes the class of nonexpansive mappings as a subclass. That is, T is nonexpansive (i.e., $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$) if and only if T is 0-strictly pseudocontractive. The mapping T is also said to be pseudocontractive if $k=1$, and T is said to be strongly pseudocontractive if there exists a constant $\lambda \in (0,1)$ such that $T-\lambda I$ is pseudocontractive. Clearly, the class of k-strictly pseudocontractive mappings falls into the one between classes of nonexpansive mappings and pseudocontractive mappings. Also, we remark that the class of strongly pseudocontractive mappings is independent of the class of k-strictly pseudocontractive mappings (see [6]). Recently, many authors have been devoting the studies on the problems of finding fixed points for pseudocontractive mappings (see, for example, [7–10] and the references therein).

Recently, in order to study the MIP (1.1) coupled with the fixed point problem, many authors have introduced some iterative schemes for finding a common element of the set of solutions of the MIP (1.1) and the set of fixed points of a countable family of nonexpansive mappings (see [4, 5, 11] and the references therein).

Inspired and motivated by the above-mentioned recent works, in this paper, we introduce new implicit and explicit iterative schemes for finding a common element of the set of the solutions of the MIP (1.1) with a set-valued maximal monotone operator B and an inverse-strongly monotone mapping A and the set of fixed points of a k-strictly pseudocontractive mapping T. Then we establish results of the strong convergence of the sequences generated by the proposed schemes to a common point of two sets, which is a solution of a certain variational inequality. As a direct consequence, we find the unique solution of the quadratic minimization problem:

$${\parallel \tilde{x}\parallel }^2=min\{{\parallel x\parallel }^2:x\in F(T)\cap {(A+B)}^{-1}0\}.$$

Moreover, as applications, we consider iterative algorithms for the Hartmann-Stampacchia variational inequality problem and the equilibrium problem coupled with fixed point problem of nonexpansive mappings.

2 Preliminaries and lemmas

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. In the following, we write $x_n\rightharpoonup x$ to indicate that the sequence $\{x_n\}$ converges weakly to x. $x_n\to x$ implies that $\{x_n\}$ converges strongly to x.

Recall that a mapping $f:C\to C$ is said to be contractive if there exists $l\in [0,1)$ such that

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

A mapping A of C into H is called inverse-strongly monotone if there exists a positive real number α such that

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

for all $x,y\in C$. For such a case, A is called α-inverse-strongly monotone. If A is an α-inverse-strongly monotone mapping of C into H, then it is obvious that A is $\frac{1}{\alpha }$-Lipschitzian and continuous. Let B be a mapping of H into $2^H$. The effective domain of B is denoted by $dom(B)$, that is, $dom(B)=\{x\in H:Bx\ne \mathrm{\varnothing }\}$. A multi-valued mapping B is said to be a monotone operator on H if $〈x-y,u-v〉\ge 0$ for all $x,y\in dom(B)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and $r>0$, we may define a single-valued operator $J_r={(I+rB)}^{-1}:H\to dom(B)$, which is called the resolvent of B for r. Let B be a maximal monotone operator on H, and let $B^{-1}0=\{x\in H:0\in Bx\}$. It is well known that $B^{-1}0=F(J_r)$ for all $r>0$ and the resolvent $J_r$ is firmly nonexpansive, i.e.,

$${\parallel J_{r}x-J_{r}y\parallel }^2\le 〈x-y,J_{r}x-J_{r}y〉,\mathrm{\forall }x,y\in H,$$

(2.1)

and that the resolvent identity

$$J_{\lambda }x=J_{\mu }(\frac{\mu }{\lambda }x+(1-\frac{\mu }{\lambda })J_{\lambda }x)$$

(2.2)

holds for all $\lambda ,\mu >0$ and $x\in H$. It is worth mentioning that the resolvent operator $J_{\lambda }$ is nonexpansive and 1-inverse-strongly monotone, and that a solution of the MIP (1.1) is a fixed point of the operator $J_{\lambda }(I-\lambda A)$ for all $\lambda >0$ (see [11]).

In a real Hilbert space H, we have

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

(2.3)

for all $x,y\in H$ and $\lambda \in \mathbb{R}$. For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel =inf\{\parallel x-y\parallel :y\in C\}.$$

$P_C$ is called the metric projection of H onto C. It is well known that $P_C$ is nonexpansive, and $P_C$ is characterized by the property

$$u=P_{C}x⟺〈x-u,u-y〉\ge 0,\mathrm{\forall }x\in H,y\in C.$$

(2.4)

It is also well known that H satisfies the Opial condition, that is, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$, the inequality

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

holds for every $y\in H$ with $y\ne x$. For these facts, see [12].

We need the following lemmas for the proof of our main results.

Lemma 2.1 In a real Hilbert space H, the following inequality holds:

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉,\mathrm{\forall }x,y\in H.$$

Lemma 2.2 [12]

For all $x,y,z\in H$ and $\alpha ,\beta ,\gamma \in [0,1]$ with $\alpha +\beta +\gamma =1$, the following equality holds:

$${\parallel \alpha x+\beta y+\gamma z\parallel }^2=\alpha {\parallel x\parallel }^2+\beta {\parallel y\parallel }^2+\gamma {\parallel z\parallel }^2-\alpha \beta {\parallel x-y\parallel }^2-\beta \gamma {\parallel y-z\parallel }^2-\gamma \alpha {\parallel z-x\parallel }^2.$$

Lemma 2.3 [13]

Let H be a Hilbert space, let C be a closed convex subset of H. If T is a k-strictly pseudocontractive mapping on C, then the fixed point set $F(T)$ is closed convex, so that the projection $P_{F(T)}$ is well defined, and $F(P_{C}T)=F(T)$.

Lemma 2.4 [13]

Let H be a real Hilbert space, let C be a closed convex subset of H, and let $T:C\to H$ be a k-strictly pseudocontractive mapping. Define a mapping $T:C\to H$ by $Sx=\lambda x+(1-\lambda )Tx$ for all $x\in C$. Then, as $\lambda \in [k,1)$, S is a nonexpansive mapping such that $F(S)=F(T)$.

Lemma 2.5 [14]

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

$${\parallel (I-rA)x-(I-rA)y\parallel }^2\le {\parallel x-y\parallel }^2+r(r-2\alpha ){\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

In particular, if $0\le r\le 2\alpha$, then $I-rA$ is nonexpansive.

Lemma 2.6 [15]

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

Remark 2.1 Lemma 2.6 implies that ${(A+B)}^{-1}0$ is closed and convex if $B:H\to 2^H$ is a maximal monotone operator and $A:H\to H$ is an inverse-strongly monotone mapping.

The following lemma is a variant of a Minty lemma (see [16]).

Lemma 2.7 Let C be a nonempty closed convex subset of a real Hilbert space H. Assume that the mapping $G:C\to H$ is monotone and weakly continuous along segments, that is, $G(x+ty)\to G(x)$ weakly as $t\to 0$. Then the variational inequality

$$\tilde{x}\in C,〈G\tilde{x},p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in C,$$

is equivalent to the dual variational inequality

$$\tilde{x}\in C,〈Gp,p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in C.$$

Lemma 2.8 [17]

Let $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a real Banach space E, and let $\{{\gamma }_n\}$ be a sequence in $[0,1]$, which satisfies the following condition:

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

Suppose that $x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)z_n$ for all $n\ge 1$, and

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

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

Lemma 2.9 [18]

Let $\{s_n\}$ be a sequence of non-negative real numbers satisfying

$$s_{n+1}\le (1-{\xi }_n)s_n+{\xi }_n{\delta }_n,\mathrm{\forall }n\ge 1,$$

where $\{\xi \}$ and $\{{\delta }_n\}$ satisfy the following conditions:

• (i) $\{{\xi }_n\}\subset [0,1]$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$;

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

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

3 Iterative schemes

Throughout the rest of this paper, we always assume as follows: Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $A:C\to H$ be an α-inverse strongly monotone mapping, and let B be a maximal monotone operator on H such that the domain of B is included in C. Let $J_{\lambda }={(I+\lambda B)}^{-1}$ be the resolvent of B for $\lambda >0$. Let $f:C\to C$ be a contractive mapping with constant $l\in (0,1)$, and let $T:C\to C$ be a k-strictly pseudocontractive mapping. Define a mapping $S:C\to C$ by $Sx=\lambda x+(1-\lambda )Tx$, $\mathrm{\forall }x\in C$, where $\lambda \in [k,1)$. Then, by Lemma 2.4, S is nonexpansive.

In this section, we introduce the following iterative scheme that generates a net $\{x_t\}$ in an implicit way:

$$x_t=tf(x_t)+(1-t)S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t),t\in (0,1),$$

(3.1)

where $0<a\le {\lambda }_t\le b<2\alpha$. We prove strong convergence of $\{x_t\}$, as $t\to 0$, to a point $\tilde{x}$ in $F(T)\cap {(A+B)}^{-1}0$, which is a solution of the following variational inequality:

$$〈(I-f)\tilde{x},p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

(3.2)

Equivalently, $\tilde{x}=P_{F(T)\cap {(A+B)}^{-1}0}(2I-f)\tilde{x}$.

If we take $f\equiv 0$ in (3.1), then we have

$$x_t=(1-t)S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t),t\in (0,1).$$

(3.3)

We also propose the following iterative scheme which generates a sequence $\{x_n\}$ in an explicit way:

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0,$$

(3.4)

where $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$, $\{{\lambda }_n\}\subset (0,2\alpha )$ and $x_0\in C$ is an arbitrary initial guess, and establish the strong convergence of this sequence to a fixed point $\tilde{x}$ of T, which is also a solution of the variational inequality (3.2). If we take $f\equiv 0$ in (3.4), then we have

$$x_{n+1}={\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0.$$

(3.5)

3.1 Strong convergence of the implicit algorithm

For $t\in (0,1)$, consider the following mapping $Q_t$ on C defined by

$$Q_{t}x=tf(x)+(1-t)S{J}_{{\lambda }_t}(x-{\lambda }_{t}Ax),\mathrm{\forall }x\in C.$$

By Lemma 2.5, we have

$$\begin{array}{r}\parallel Q_{t}x-Q_{t}y\parallel \\ =\parallel tf(x)+(1-t)S{J}_{{\lambda }_t}(x-{\lambda }_{t}Ax)-(tf(y)+(1-t)S{J}_{{\lambda }_t}(y-{\lambda }_{t}Ay))\parallel \\ \le t\parallel f(x)-f(y)\parallel +(1-t)\parallel S{J}_{{\lambda }_t}(x-{\lambda }_{t}Ax)-S{J}_{{\lambda }_t}(y-{\lambda }_{t}Ay)\parallel \\ \le tl\parallel x-y\parallel +(1-t)\parallel (I-{\lambda }_{t}A)x-(I-{\lambda }_{t}A)y\parallel \\ \le tl\parallel x-y\parallel +(1-t)\parallel x-y\parallel \\ =[1-(1-l)t]\parallel x-y\parallel .\end{array}$$

Since $0<1-(1-l)t<1$, $Q_t$ is a contractive mapping. Therefore, by the Banach contraction principle, $Q_t$ has a unique fixed point $x_t\in C$, which uniquely solves the fixed point equation

$$x_t=tf(x_t)+(1-t)S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t),t\in (0,1).$$

Now, we prove strong convergence of the sequence $\{x_t\}$, and show the existence of $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$, which solves the variational inequality (3.2).

Theorem 3.1 Suppose that $F(T)\cap {(A+B)}^{-1}0$. Then the net $\{x_t\}$ defined by the implicit method (3.1) converges strongly, as $t\to 0$, to a point $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$, which is the unique solution of the variational inequality (3.2).

Proof First, we can show easily the uniqueness of a solution of the variational inequality (3.2). In fact, if $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$ and $\hat{x}\in F(T)\cap {(A+B)}^{-1}0$ both are solutions to (3.2). Then we have

$$〈(I-f)\tilde{x},\hat{x}-\tilde{x}〉\ge 0,$$

(3.6)

$$〈(I-f)\hat{x},\tilde{x}-\hat{x}〉\ge 0.$$

(3.7)

Adding up (3.6) and (3.7) yields

$$〈(I-f)\tilde{x}-(I-f)\hat{x},\tilde{x}-\hat{x}〉\le 0.$$

This implies that $(1-l){\parallel \tilde{x}-\hat{x}\parallel }^2\le 0$. So $\tilde{x}=\hat{x}$, and the uniqueness is proved. Below, we use $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$ to denote the unique solution of the variational inequality (3.2).

Now, we prove that $\{x_t\}$ is bounded. Set $y_t=J_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)$ for all $t\in (0,1)$. Take $p\in F(T)\cap {(A+B)}^{-1}0$. It is clear that $p=J_{{\lambda }_t}(p-{\lambda }_{t}Ap)=S{J}_{{\lambda }_t}(p-{\lambda }_{t}Ap)$ and $p=Sp$ (by Lemma 2.4). Since $J_{{\lambda }_t}$ is nonexpansive and A is α-inverse-strongly monotone, we have from Lemma 2.5 that

$$\begin{array}{rl}{\parallel y_t-p\parallel }^2& ={\parallel J_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-J_{{\lambda }_t}(p-{\lambda }_{t}Ap)\parallel }^2\\ \le {\parallel x_t-{\lambda }_{t}A{x}_t-(p-{\lambda }_{t}Ap)\parallel }^2\\ \le {\parallel x_t-p\parallel }^2+{\lambda }_t({\lambda }_t-2\alpha ){\parallel A{x}_t-Ap\parallel }^2\\ \le {\parallel x_t-p\parallel }^2.\end{array}$$

(3.8)

So, we have that

$$\parallel y_t-p\parallel \le \parallel x_t-p\parallel .$$

(3.9)

Moreover, from (3.1), it follows that

$$\begin{array}{rl}\parallel x_t-p\parallel & =\parallel tf(x_t)+(I-t)S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-p\parallel \\ \le \parallel t(f(x_t)-f(p))\parallel +t\parallel f(p)-p\parallel +(1-t)\parallel J_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-p\parallel \\ \le tl\parallel x_t-p\parallel +t\parallel f(p)-p\parallel +(1-t)\parallel y_t-p\parallel \\ \le tl\parallel x_t-p\parallel +t\parallel f(p)-p\parallel +(1-t)\parallel x_t-p\parallel \\ \le [1-t(1-l)]\parallel x_t-p\parallel +t\parallel f(p)-p\parallel ,\end{array}$$

(3.10)

that is,

$$\parallel x_t-p\parallel \le \frac{\parallel f(p)-p\parallel }{1-l}.$$

Hence, $\{x_t\}$ is bounded, and so are $\{y_t\}$, $\{f(x_t)\}$, $\{A{x}_t\}$ and $\{S{y}_t\}$.

From (3.8) and (3.10), we have

$$\begin{array}{rl}{(1-tl)}^2{\parallel x_t-p\parallel }^2\le & {[(1-t)\parallel y_t-p\parallel +t\parallel f(p)-p\parallel ]}^2\\ =& {(1-t)}^2{\parallel y_t-p\parallel }^2+t^2{\parallel f(p)-p\parallel }^2\\ +2(1-t)t\parallel f(p)-p\parallel \parallel y_t-p\parallel \\ \le & {\parallel y_t-p\parallel }^2+t{M}_1\\ \le & {\parallel x_t-p\parallel }^2+{\lambda }_t({\lambda }_t-2\alpha ){\parallel A{x}_t-Ap\parallel }^2+t{M}_1,\end{array}$$

(3.11)

where $M_1>0$ is an appropriate constant. This means that

$$\begin{array}{rl}a(2\alpha -b){\parallel A{x}_t-Ap\parallel }^2& \le {\lambda }_t(2\alpha -{\lambda }_t){\parallel A{x}_t-Ap\parallel }^2\\ \le t(2l-t{l}^2){\parallel x_t-p\parallel }^2+t{M}_1\to 0\text{as }t\to 0.\end{array}$$

Since $a(2\alpha -b)>0$, we deduce that

$$\underset{t\to 0}{lim}\parallel A{x}_t-Ap\parallel =0.$$

(3.12)

From (2.1) and (2.3), we also obtain

$$\begin{array}{r}{\parallel y_t-p\parallel }^2\\ ={\parallel J_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-J_{{\lambda }_t}(p-{\lambda }_{t}Ap)\parallel }^2\\ \le 〈(x_t-{\lambda }_{t}A{x}_t)-(p-{\lambda }_{t}Ap),y_t-p〉\\ =\frac{1}{2}({\parallel (x_t-{\lambda }_{t}A{x}_t)-(p-{\lambda }_{t}Ap)\parallel }^2+{\parallel y_t-p\parallel }^2\\ -{\parallel (x_t-p)-{\lambda }_t(A{x}_t-Ap)-(y_t-p)\parallel }^2)\\ \le \frac{1}{2}({\parallel x_t-p\parallel }^2+{\parallel y_t-p\parallel }^2-{\parallel (x_t-y_t)-{\lambda }_t(A{x}_t-Ap)\parallel }^2)\\ =\frac{1}{2}({\parallel x_t-p\parallel }^2+{\parallel y_t-p\parallel }^2-{\parallel x_t-y_t\parallel }^2+2{\lambda }_t〈x_t-y_t,A{x}_t-Ap〉-{\lambda }_t^2{\parallel A{x}_t-Ap\parallel }^2).\end{array}$$

So, we get

$$\begin{array}{rl}{\parallel y_t-p\parallel }^2\le & {\parallel x_t-p\parallel }^2-{\parallel x_t-y_t\parallel }^2\\ +2{\lambda }_t〈x_t-y_t,A{x}_t-Ap〉-{\lambda }_t^2{\parallel A{x}_t-Ap\parallel }^2.\end{array}$$

(3.13)

Since ${\parallel \cdot \parallel }^2$ is a convex function, by (3.13), we have

$$\begin{array}{rl}{\parallel x_t-p\parallel }^2=& {\parallel t(f(x_t)-p)+(1-t)(S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-p)\parallel }^2\\ \le & t{(\parallel f(x_t)-f(p)\parallel +\parallel f(p)-p\parallel )}^2+(1-t){\parallel S{y}_t-Sp\parallel }^2\\ \le & t{(l\parallel x_t-p\parallel +\parallel f(p)-p\parallel )}^2+(1-t){\parallel y_t-p\parallel }^2\\ \le & t{(l\parallel x_t-p\parallel +\parallel f(p)-p\parallel )}^2\\ +(1-t)({\parallel x_t-p\parallel }^2-{\parallel x_t-y_t\parallel }^2+2{\lambda }_t〈x_t-y_t,A{x}_t-Ap〉).\end{array}$$

(3.14)

We deduce from (3.14) that

$$(1-t){\parallel x_t-y_t\parallel }^2\le (t+\parallel A{x}_t-Ap\parallel )M_2,$$

(3.15)

where $M_2>0$ is an appropriate constant. Since $t\to 0$ and $\parallel A{x}_t-Ap\parallel \to 0$, we have

$$\underset{t\to \mathrm{\infty }}{lim}\parallel x_t-y_t\parallel =0.$$

(3.16)

Observing that

$$\begin{array}{rl}\parallel S{y}_t-x_t\parallel & =\parallel S{y}_t-(tf(x_t)+(1-t)S{y}_t)\parallel \\ =t\parallel S{y}_t-f(x_t)\parallel \to 0\text{as }t\to 0,\end{array}$$

by (3.16), we obtain

$$\begin{array}{rl}\parallel S{x}_t-x_t\parallel & \le \parallel S{x}_t-S{y}_t\parallel +\parallel S{y}_t-x_t\parallel \\ \le \parallel x_t-y_t\parallel +\parallel S{y}_t-x_t\parallel \to 0\text{as }t\to 0.\end{array}$$

(3.17)

Let $\{t_n\}\subset (0,1)$ be a sequence such that $t_n\to 0$ as $n\to \mathrm{\infty }$. Put $x_n:=x_{t_n}$, $y_n:=y_{t_n}$ and ${\lambda }_n:={\lambda }_{t_n}$. Since $\{x_n\}$ is bounded, there exists a subsequence $\{x_{n_i}\}$ of $\{x_n\}$, which converges weakly to $\tilde{x}$.

Next, we show that $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$. Since C is closed and convex, C is weakly closed. So, we have $\tilde{x}\in C$. Let us show $\tilde{x}\in F(T)$. Assume that $\tilde{x}\notin F(T)$ ($=F(S)$). Since $x_{n_i}\rightharpoonup \tilde{x}$ and $\tilde{x}\ne S\tilde{x}$, it follows from the Opial condition and (3.17) that

$$\begin{array}{rl}\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-\tilde{x}\parallel & <\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-S\tilde{x}\parallel \\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(\parallel x_{n_i}-S{x}_{n_i}\parallel +\parallel S{x}_{n_i}-S\tilde{x}\parallel )\\ \le \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_i}-\tilde{x}\parallel ,\end{array}$$

which is a contradiction. So we get $\tilde{x}\in F(T)$.

We shall show that $\tilde{x}\in {(A+B)}^{-1}0$. Since $\parallel x_t-y_t\parallel \to 0$ as $t\to 0$, it follows that $\{y_{n_i}\}$ converges weakly to $\tilde{x}$. We choose a subsequence $\{{\lambda }_{n_i}\}$ of $\{{\lambda }_n\}$ such that ${\lambda }_{n_i}\to \lambda$. Let $v\in Bu$. Noting that

$$y_t=J_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)={(I+{\lambda }_{t}B)}^{-1}(x_t-{\lambda }_{t}A{x}_t),$$

we have that

$$x_t-{\lambda }_{t}A{x}_t\in y_t+{\lambda }_{t}B{y}_t,$$

and so,

$$\frac{x_t-y_t}{{\lambda }_t}-A{x}_t\in B{y}_t.$$

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

$$〈\frac{x_t-y_t}{{\lambda }_t}-A{x}_t-v,y_t-u〉\ge 0.$$

(3.18)

Since $〈x_t-\tilde{x},A{x}_t-A\tilde{x}〉\ge \alpha {\parallel A{x}_t-A\tilde{x}\parallel }^2$ and $x_{n_i}\rightharpoonup \tilde{x}$, we have $A{x}_{n_i}\to A\tilde{x}$. Then by (3.16) and (3.18), we obtain

$$〈-A\tilde{x}-v,\tilde{x}-u〉\ge 0.$$

Since B is maximal monotone, this implies that $-A\tilde{x}\in B\tilde{x}$, that is, $0\in (A+B)\tilde{x}$. Hence, we have $\tilde{x}\in {(A+B)}^{-1}0$. Thus, we conclude that $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$.

On the one hand, we note that for $p\in F(T)\cap {(A+B)}^{-1}0$,

$$x_t-p=t(f(x_t)-f(p))+t(f(p)-p)+(1-t)(S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-p).$$

Then it follows that

$$\begin{array}{rl}{\parallel x_t-p\parallel }^2=& 〈x_t-p,x_t-p〉\\ =& 〈t(f(x_t)-f(p)),x_t-p〉+t〈f(p)-p,x_t-p〉\\ +(1-t)〈S{J}_{{\lambda }_t}(x_t-{\lambda }_{t}A{x}_t)-p,x_t-p〉\\ \le & tl{\parallel x_t-p\parallel }^2+t〈f(p)-p,x_t-p〉+(1-t){\parallel x_t-p\parallel }^2\\ =& (1-(1-l)t){\parallel x_t-p\parallel }^2+t〈f(p)-p,x_t-p〉.\end{array}$$

Hence, we have

$${\parallel x_t-p\parallel }^2\le \frac{1}{1-l}〈f(p)-p,x_t-p〉.$$

(3.19)

In particular,

$${\parallel x_{n_i}-p\parallel }^2\le \frac{1}{1-l}〈f(p)-p,x_{n_i}-p〉.$$

(3.20)

Since $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$, by (3.20), we obtain

$$\parallel x_{n_i}-\tilde{x}\parallel \frac{1}{1-l}〈f(\tilde{x})-\tilde{x},x_{n_i}-\tilde{x}〉.$$

(3.21)

Since $x_{n_i}\rightharpoonup \tilde{x}$, it follows from (3.21) that $x_{n_i}\to \tilde{x}$ as $i\to \mathrm{\infty }$.

Now, we return to (3.20) and take the limit as $i\to \mathrm{\infty }$ to get

$${\parallel \tilde{x}-p\parallel }^2\le \frac{1}{1-l}〈(I-f)p,p-\tilde{x}〉.$$

(3.22)

In particular, $\tilde{x}$ solves the following variational inequality

$$\tilde{x}\in F(T)\cap {(A+B)}^{-1}0,〈(I-f)p,p-\tilde{x}〉\ge 0,p\in F(T)\cap {(A+B)}^{-1}0,$$

or the equivalent dual variational inequality (see Lemma 2.7)

$$\tilde{x}\in F(T)\cap {(A+B)}^{-1}0,〈(I-f)\tilde{x},p-\tilde{x}〉\ge 0,p\in F(T)\cap {(A+B)}^{-1}0.$$

(3.23)

Finally, we show that the net $\{x_t\}$ converges strongly, as $t\to 0$, to $\tilde{x}$. To this end, let $\{s_k\}\subset (0,1)$ be another sequence such that $s_k\to 0$ as $k\to \mathrm{\infty }$. Put $x_k:=x_{s_k}$, $y_k:=y_{s_k}$ and ${\lambda }_k:={\lambda }_{s_k}$. Let $\{x_{k_j}\}$ be a subsequence of $\{x_k\}$, and assume that $x_{k_j}\rightharpoonup \hat{x}$. By the same proof as the one above, we have $\hat{x}\in F(T)\cap {(A+B)}^{-1}0$. Moreover, it follows from (3.23) that

$$〈(I-f)\tilde{x},\tilde{x}-\hat{x}〉\le 0.$$

(3.24)

Interchanging $\tilde{x}$ and $\hat{x}$, we obtain

$$〈(I-f)\hat{x},\hat{x}-\tilde{x}〉\le 0.$$

(3.25)

Adding up (3.24) and (3.25) yields

$$〈(I-f)\tilde{x}-(I-f)\hat{x},\tilde{x}-\hat{x}〉\le 0.$$

Hence,

$${\parallel \tilde{x}-\hat{x}\parallel }^2\le 〈f(\tilde{x})-f(\hat{x}),\tilde{x}-\hat{x}〉\le l{\parallel \tilde{x}-\hat{x}\parallel }^2,$$

that is, $(1-l){\parallel \tilde{x}-\hat{x}\parallel }^2\le 0$. Since $l\in (0,1)$, we have $\tilde{x}=\hat{x}$. Therefore, we conclude that $x_t\to \tilde{x}$ as $t\to 0$.

Note that $P_{F(T)\cap {(A+B)}^{-1}0}$ is well defined by Lemma 2.3 and Remark 2.1. The variational inequality (3.2) can be rewritten as

$$〈(2I-f)\tilde{x}-\tilde{x},p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

By (2.4), this is equivalent to the fixed point equation

$$\tilde{x}=P_{F(T)\cap {(A+B)}^{-1}0}(2I-f)\tilde{x}.$$

This completes the proof. □

From Theorem 3.1, we can deduce the following result.

Corollary 3.1 Suppose that $F(T)\cap {(A+B)}^{-1}0\ne \mathrm{\varnothing }$. Then the net $\{x_t\}$ defined by the implicit method (3.3) converges strongly, as $t\to 0$, to $\tilde{x}$, which solves the following minimum norm problem: find $\tilde{x}\in F(T)\cap {(A+B)}^{-1}0$ such that

$$\parallel \tilde{x}\parallel =\underset{x\in F(T)\cap {(A+B)}^{-1}0}{min}\parallel x\parallel .$$

(3.26)

Proof From (3.22) with $f\equiv 0$ and $l=0$, we have

$${\parallel \tilde{x}-p\parallel }^2\le 〈p,p-\tilde{x}〉,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

Equivalently,

$$〈\tilde{x},p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

This obviously implies that

$${\parallel \tilde{x}\parallel }^2\le 〈p,\tilde{x}〉\le \parallel p\parallel \parallel \tilde{x}\parallel ,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

It turns out that $\parallel \tilde{x}\parallel \le \parallel p\parallel$ for all $p\in F(T)\cap {(A+B)}^{-1}0$. Therefore, $\tilde{x}$ is the minimum-norm point of $F(T)\cap {(A+B)}^{-1}0$. □

3.2 Strong convergence of the explicit algorithm

Now, using Theorem 3.1, we establish the strong convergence of an explicit iterative scheme for finding a solution of the variational inequality (3.2), where the constraint set is the common set of the fixed point set $F(T)$ of the k-strictly pseudocontractive mapping T and the solution set ${(A+B)}^{-1}0$ of the MIP (1.1).

Theorem 3.2 Suppose that $F(T)\cap {(A+B)}^{-1}0\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset (0,2\alpha )$ satisfy the following conditions:

• (C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<c\le {\beta }_n\le d<1$;

• (C4) $0<a\le {\lambda }_n\le b<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_n-{\lambda }_{n+1})=0$.

Let the sequence $\{x_n\}$ be generated iteratively by (3.4):

where $x_0\in C$ is an arbitrary initial guess. Then the sequence $\{x_n\}$ converges strongly to a point $\tilde{x}$ in $F(T)\cap {(A+B)}^{-1}0$, which is the unique solution of the variational inequality (3.2).

Proof First, from condition (C1), without loss of generality, we assume that $\frac{2(1-l){\alpha }_n}{1-{\alpha }_{n}l}<1$, and we note that $F(T)=F(S)$. From now, we put $y_n=J_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)$.

We divide the proof several steps as follows.

Step 1. We show that $\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-l}\}$ for all $n\ge 0$ and all $p\in F(T)\cap {(A+B)}^{-1}0$ ($=F(S)\cap {(A+B)}^{-1}0$). Indeed, let $p\in F(T)\cap {(A+B)}^{-1}0$. From $p=J_{{\lambda }_n}(p-{\lambda }_{n}Ap)=S{J}_{{\lambda }_n}(p-{\lambda }_{n}Ap)$, $Sp=p$ and Lemma 2.5, we get

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel J_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-J_{{\lambda }_n}(p-rAp)\parallel }^2\\ \le {\parallel (x_n-{\lambda }_{n}A{x}_n)-(p-{\lambda }_{n}Ap)\parallel }^2\\ ={\parallel (x_n-p)-{\lambda }_n(A{x}_n-Ap)\parallel }^2\\ ={\parallel x_n-p\parallel }^2-2{\lambda }_n〈x_n-p,A{x}_n-Ap〉+{\lambda }_n^2{\parallel A{x}_n-Ap\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-2{\lambda }_n\alpha {\parallel A{x}_n-Ap\parallel }^2+{\lambda }_n^2{\parallel A{x}_n-Ap\parallel }^2\\ ={\parallel x_n-p\parallel }^2+{\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-Ap\parallel }^2\\ \le {\parallel x_n-p\parallel }^2.\end{array}$$

(3.27)

Using (3.27), we have

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel =& \parallel {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-p\parallel \\ =& \parallel {\alpha }_n(f(x_n)-p)+{\beta }_n(x_n-p)+(1-{\alpha }_n-{\beta }_n)(S{y}_n-Sp)\parallel \\ \le & {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel x_n-p\parallel \\ +(1-{\alpha }_n-{\beta }_n)\parallel y_n-p\parallel \\ \le & {\alpha }_{n}l\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\alpha }_n-{\beta }_n)\parallel x_n-p\parallel \\ =& (1-(1-l){\alpha }_n)\parallel x_n-p\parallel +(1-l){\alpha }_n\frac{\parallel f(p)-p\parallel }{1-l}.\end{array}$$

Using an induction, we have

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-l}\}.$$

Hence, $\{x_n\}$ is bounded, and so are $\{y_n\}$, $\{A{x}_n\}$, $\{f(x_n)\}$ and $\{S{y}_n\}$.

Step 2. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$. Put $u_n=x_n-{\lambda }_{n}A{x}_n$, and define

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n,\mathrm{\forall }n\ge 0.$$

(3.28)

Then we have

$$\begin{array}{r}{z}_{n+1}-z_n\\ =\frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}\\ =\frac{{\alpha }_{n+1}f(x_{n+1})+{\beta }_{n+1}{x}_{n+1}+(1-{\alpha }_{n+1}-{\beta }_{n+1})S{J}_{{\lambda }_{n+1}}{u}_{n+1}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}{u}_n-{\beta }_n{x}_n}{1-{\beta }_n}\\ =\frac{{\alpha }_{n+1}f(x_{n+1})+(1-{\alpha }_{n+1}-{\beta }_{n+1})S{J}_{{\lambda }_{n+1}}{u}_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_{n}f(x_n)+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}{u}_n}{1-{\beta }_n}\\ =\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}f(x_{n+1})-\frac{{\alpha }_n}{1-{\beta }_n}f(x_n)+S{J}_{{\lambda }_{n+1}}{u}_{n+1}-S{J}_{{\lambda }_n}{u}_n\\ -\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}S{J}_{{\lambda }_{n+1}}{u}_{n+1}+\frac{{\alpha }_n}{1-{\beta }_n}S{J}_{{\lambda }_n}{u}_n\\ =\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(f(x_{n+1})-S{J}_{{\lambda }_{n+1}}{u}_{n+1})+\frac{{\alpha }_n}{1-{\beta }_n}(S{J}_{{\lambda }_n}{u}_n-f(x_n))\\ +S{J}_{{\lambda }_{n+1}}{u}_{n+1}-S{J}_{{\lambda }_n}{u}_n.\end{array}$$

(3.29)

Since $I-{\lambda }_{n+1}A$ is nonexpansive for ${\lambda }_{n+1}\in (0,2\alpha )$ (by Lemma 2.5), we have

$$\parallel (I-{\lambda }_{n+1}A)x_{n+1}-(I-{\lambda }_{n+1}A)x_n\parallel \le \parallel x_{n+1}-x_n\parallel .$$

(3.30)

By the resolvent identity (2.2) and (3.30), we get

$$\begin{array}{r}\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-J_{{\lambda }_n}{u}_n\parallel \\ =\parallel J_{{\lambda }_n}(\frac{{\lambda }_n}{{\lambda }_{n+1}}{u}_{n+1}+(1-\frac{{\lambda }_n}{{\lambda }_{n+1}})J_{{\lambda }_{n+1}}{u}_{n+1})-J_{{\lambda }_n}{u}_n\parallel \\ \le \parallel \frac{{\lambda }_n}{{\lambda }_{n+1}}{u}_{n+1}+(1-\frac{{\lambda }_n}{{\lambda }_{n+1}})J_{{\lambda }_{n+1}}{u}_{n+1}-u_n\parallel \\ \le \frac{{\lambda }_n}{{\lambda }_{n+1}}\parallel u_{n+1}-u_n\parallel +|1-\frac{{\lambda }_n}{{\lambda }_{n+1}}|\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-u_n\parallel \\ \le \parallel u_{n+1}-u_n\parallel +|1-\frac{{\lambda }_n}{{\lambda }_{n+1}}|(\parallel u_{n+1}-u_n\parallel +\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-u_n\parallel )\\ \le \parallel (x_{n+1}-{\lambda }_{n+1}A{x}_{n+1})-(x_n-{\lambda }_{n}A{x}_n)\parallel \\ +\left|\frac{{\lambda }_{n+1}-{\lambda }_n}{a}\right|(\parallel u_{n+1}-u_n\parallel +\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-u_n\parallel )\\ \le \parallel (I-{\lambda }_{n+1}A)x_{n+1}-(I-{\lambda }_{n+1}A)x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|\parallel A{x}_n\parallel \\ +|{\lambda }_{n+1}-{\lambda }_n|\frac{1}{a}(\parallel u_{n+1}-u_n\parallel +\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-u_n\parallel )\\ \le \parallel x_{n+1}-x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|M_1,\end{array}$$

(3.31)

where $M_1>0$ is an appropriate constant. Hence, from (3.29) and (3.31), we obtain

$$\begin{array}{r}\parallel z_{n+1}-z_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\parallel f(x_{n+1})\parallel +\parallel S{J}_{{\lambda }_{n+1}}{u}_{n+1}\parallel )+\frac{{\alpha }_n}{1-{\beta }_n}(\parallel S{J}_{{\lambda }_n}{u}_n\parallel +\parallel f(x_n)\parallel )\\ +\parallel S{J}_{{\lambda }_{n+1}}{u}_{n+1}-S{J}_{{\lambda }_n}{u}_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\parallel f(x_{n+1})\parallel +\parallel S{J}_{{\lambda }_{n+1}}{u}_{n+1}\parallel )+\frac{{\alpha }_n}{1-{\beta }_n}(\parallel S{J}_{{\lambda }_n}{u}_n\parallel +\parallel f(x_n)\parallel )\\ +\parallel J_{{\lambda }_{n+1}}{u}_{n+1}-J_{{\lambda }_n}{u}_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\parallel f(x_{n+1})\parallel +\parallel S{J}_{{\lambda }_{n+1}}{u}_{n+1}\parallel )+\frac{{\alpha }_n}{1-{\beta }_n}(\parallel S{J}_{{\lambda }_n}{u}_n\parallel +\parallel f(x_n)\parallel )\\ +\parallel x_{n+1}-x_n\parallel +|{\lambda }_{n+1}-{\lambda }_n|M_1.\end{array}$$

(3.32)

It follows from conditions (C1) and (C4) that

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

Thus, by Lemma 2.8, we have

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

(3.33)

Consequently, we obtain

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

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel A{x}_n-Ap\parallel =0$ for $p\in F(T)\cap {(A+B)}^{-1}0$. From (3.4), (3.27) and Lemma 2.2, we have

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ ={\parallel {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-p\parallel }^2\\ ={\parallel {\alpha }_n(f(x_n)-p)+{\beta }_n(x_n-p)+(1-{\alpha }_n-{\beta }_n)(S{y}_n-p)\parallel }^2\\ ={\alpha }_n{\parallel f(x_n)-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n-{\beta }_n){\parallel S{y}_n-p\parallel }^2\\ -{\alpha }_n{\beta }_n{\parallel f(x_n)-x_n\parallel }^2-{\beta }_n(1-{\alpha }_n-{\beta }_n){\parallel x_n-S{y}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n-{\beta }_n){\parallel S{y}_n-f(x_n)\parallel }^2\\ \le {\alpha }_n{(\parallel f(x_n)-f(p)\parallel +\parallel f(p)-p\parallel )}^2+{\beta }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n-{\beta }_n){\parallel y_n-p\parallel }^2\\ \le {\alpha }_n(l^2{\parallel x_n-p\parallel }^2+2l\parallel x_n-p\parallel \parallel f(p)-p\parallel +{\parallel f(p)-p\parallel }^2)+{\beta }_n{\parallel x_n-p\parallel }^2\\ +(1-{\alpha }_n-{\beta }_n){\parallel x_n-p\parallel }^2+(1-{\alpha }_n-{\beta }_n){\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-Ap\parallel }^2\\ \le (1-(1-l){\alpha }_n){\parallel x_n-p\parallel }^2+(1-{\alpha }_n-{\beta }_n){\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-Ap\parallel }^2\\ +{\alpha }_n(2l\parallel x_n-p\parallel \parallel f(p)-p\parallel +{\parallel f(p)-p\parallel }^2)\\ \le {\parallel x_n-p\parallel }^2+(1-{\alpha }_n-{\beta }_n){\lambda }_n({\lambda }_n-2\alpha ){\parallel A{x}_n-Ap\parallel }^2+{\alpha }_n{M}_2,\end{array}$$

(3.34)

where $M_2>0$ is an appropriate constant. From (3.34) and conditions (C3) and (C4), we deduce that

$$\begin{array}{rl}(1-{\alpha }_n-d)a(2\alpha -b){\parallel A{x}_n-Ap\parallel }^2& \le (1-{\alpha }_n-{\beta }_n){\lambda }_n(2\alpha -{\lambda }_n){\parallel A{x}_n-Ap\parallel }^2\\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+{\alpha }_n{M}_2.\end{array}$$

Since ${\alpha }_n\to 0$ (by condition (C1)) and $\parallel x_{n+1}-x_n\parallel \to 0$ (by Step 2), we conclude that

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

Step 4. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$. First, from (2.1) and (2.3), we get for $p\in F(T)\cap {(A+B)}^{-1}0$,

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2=& {\parallel J_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-p\parallel }^2\\ =& {\parallel J_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-J_{{\lambda }_n}(p-{\lambda }_{n}Ap)\parallel }^2\\ \le & 〈(x_n-{\lambda }_{n}A{x}_n)-(p-{\lambda }_{n}Ap),y_n-p〉\\ =& \frac{1}{2}({\parallel (x_n-{\lambda }_{n}A{x}_n)-(p-{\lambda }_{n}Ap)\parallel }^2+{\parallel y_n-p\parallel }^2\\ -{\parallel (x_n-{\lambda }_{n}A{x}_n)-(p-{\lambda }_{n}Ap)-(y_n-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel x_n-y_n-{\lambda }_n(A{x}_n-Ap)\parallel }^2)\\ =& \frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel y_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2\\ +2{\lambda }_n〈x_n-y_n,A{x}_n-Ap〉-{\lambda }_n^2{\parallel A{x}_n-Ap\parallel }^2).\end{array}$$

So, we have

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2\le & {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n〈x_n-y_n,A{x}_n-Ap〉\\ -{\lambda }_n^2{\parallel A{x}_n-Ap\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n〈x_n-y_n,A{x}_n-Ap〉.\end{array}$$

(3.35)

Using (3.34) and (3.35), we obtain

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ \le {\alpha }_n(l^2{\parallel x_n-p\parallel }^2+2l\parallel x_n-p\parallel \parallel f(p)-p\parallel +{\parallel f(p)-p\parallel }^2)+{\beta }_n{\parallel x_n-p\parallel }^2\\ +(1-{\alpha }_n-{\beta }_n){\parallel y_n-p\parallel }^2\\ \le {\alpha }_{n}l{\parallel x_n-p\parallel }^2+{\alpha }_n{M}_2+{\beta }_n{\parallel x_n-p\parallel }^2\\ +(1-{\alpha }_n-{\beta }_n)({\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n〈x_n-y_n,A{x}_n-Ap〉)\\ =(1-(1-l){\alpha }_n){\parallel x_n-p\parallel }^2-(1-{\alpha }_n-{\beta }_n){\parallel x_n-y_n\parallel }^2\\ +2{\lambda }_n〈x_n-y_n,A{x}_n-Ap〉+{\alpha }_n{M}_2\\ \le {\parallel x_n-p\parallel }^2-(1-{\alpha }_n-{\beta }_n){\parallel x_n-y_n\parallel }^2+2b{M}_3\parallel A{x}_n-Ap\parallel +{\alpha }_n{M}_2,\end{array}$$

(3.36)

where $M_2,M_3>0$ are appropriate constants. This implies that

$$\begin{array}{r}(1-{\alpha }_n-d){\parallel x_n-y_n\parallel }^2\\ \le (1-{\alpha }_n-{\beta }_n){\parallel x_n-y_n\parallel }^2\\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_{n+1}-p\parallel +\parallel x_n-p\parallel )+2b{M}_3\parallel A{x}_n-Ap\parallel +{\alpha }_n{M}_2.\end{array}$$

Thus, from condition (C1), Step 2 and Step 3, we deduce that

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

Step 5. We show that ${lim}_{n\to \mathrm{\infty }}\parallel S{x}_n-x_n\parallel =0$. First, by (3.4), we have

$$\begin{array}{rl}\parallel S{y}_n-x_n\parallel & \le \parallel S{y}_n-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel \\ =\parallel S{y}_n-({\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{y}_n)\parallel +\parallel x_{n+1}-x_n\parallel \\ \le {\alpha }_n\parallel S{y}_n-f(x_n)\parallel +{\beta }_n\parallel x_n-S{y}_n\parallel +\parallel x_{n+1}-x_n\parallel ,\end{array}$$

and so,

$$\parallel S{y}_n-x_n\parallel \le \frac{1}{1-{\beta }_n}({\alpha }_n\parallel S{y}_n-f(x_n)\parallel +\parallel x_{n+1}-x_n\parallel ).$$

By conditions (C1) and (C3) and Step 2, we obtain

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

This together with Step 4 yields that

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

Step 6. We show that

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

where $\tilde{x}={lim}_{t\to 0}{x}_t$ with $x_t$ being defined by (3.1). We note that from Theorem 3.1, $\tilde{x}\in Fix(T)\cap {(A+B)}^{-1}0$, and $\tilde{x}$ is the unique solution of the variational inequality (3.2). To show this, we can choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

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

Since $\{x_{n_i}\}$ is bounded, there exists a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$, which converges weakly to w. Without loss of generality, we can assume that $x_{n_i}\rightharpoonup w$. By the same argument as in the proof of Theorem 3.1 together with Step 5, we have $w\in F(T)\cap {(A+B)}^{-1}0$. Since $\tilde{x}=P_{F(T)\cap {(A+B)}^{-1}0}(2I-f)\tilde{x}$ is the unique solution of the variational inequality (3.2), we deduce that

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

Step 7. We show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\tilde{x}\parallel =0$, where $\tilde{x}={lim}_{t\to 0}{x}_t$ with $x_t$ being defined by (3.1), and $\tilde{x}$ is the unique solution of the variational inequality (3.2). Indeed, from (3.4), we note that

$$\begin{array}{rl}{x}_{n+1}-\tilde{x}& ={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{J}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n)-\tilde{x}\\ ={\alpha }_n(f(x_n)-\tilde{x})+{\beta }_n(x_n-\tilde{x})+(1-{\alpha }_n-{\beta }_n)(S{J}_{{\lambda }_n}{y}_n-\tilde{x}).\end{array}$$

Applying Lemma 2.1, we have

$$\begin{array}{rl}{\parallel x_{n+1}-\tilde{x}\parallel }^2\le & {\parallel {\beta }_n(x_n-\tilde{x})+(1-{\alpha }_n-{\beta }_n)(S{J}_{{\lambda }_n}{y}_n-\tilde{x})\parallel }^2\\ +2{\alpha }_n〈f(x_n)-f(\tilde{x}),x_{n+1}-\tilde{x}〉+2{\alpha }_n〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉\\ \le & {({\beta }_n\parallel x_n-\tilde{x}\parallel +(1-{\alpha }_n-{\beta }_n)\parallel y_n-\tilde{x}\parallel )}^2\\ +2{\alpha }_{n}l\parallel x_n-\tilde{x}\parallel \parallel x_{n+1}-\tilde{x}\parallel +2{\alpha }_n〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-\tilde{x}\parallel }^2+{\alpha }_{n}l({\parallel x_n-\tilde{x}\parallel }^2+{\parallel x_{n+1}-\tilde{x}\parallel }^2)\\ +2{\alpha }_n〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉.\end{array}$$

This implies that

$$\begin{array}{rl}{\parallel x_{n+1}-\tilde{x}\parallel }^2\le & \frac{1-(2-l){\alpha }_n+{\alpha }_n^2}{1-{\alpha }_{n}l}{\parallel x_n-\tilde{x}\parallel }^2+\frac{2{\alpha }_n}{1-{\alpha }_{n}l}〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉\\ =& \frac{1-(2-l){\alpha }_n}{1-{\alpha }_{n}l}{\parallel x_n-\tilde{x}\parallel }^2+\frac{{\alpha }_n^2}{1-{\alpha }_{n}l}{\parallel x_n-\tilde{x}\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_{n}l}〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉\\ =& (1-\frac{2(1-l){\alpha }_n}{1-{\alpha }_{n}l}){\parallel x_n-\tilde{x}\parallel }^2+\frac{{\alpha }_n^2}{1-{\alpha }_{n}l}{\parallel x_n-\tilde{x}\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_{n}l}〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉\\ \le & (1-\frac{2(1-l){\alpha }_n}{1-{\alpha }_{n}l}){\parallel x_n-\tilde{x}\parallel }^2\\ +\frac{2(1-l){\alpha }_n}{1-{\alpha }_{n}l}(\frac{{\alpha }_n{M}_4}{2(1-l)}+\frac{1}{1-l}〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉)\\ =& (1-{\xi }_n){\parallel x_n-\tilde{x}\parallel }^2+{\xi }_n{\delta }_n,\end{array}$$

where $M_4>0$ is an appropriate constant, ${\xi }_n=\frac{2(1-l){\alpha }_n}{1-{\alpha }_{n}l}$ and

$${\delta }_n=\frac{{\alpha }_n{M}_4}{2(1-l)}+\frac{1}{1-l}〈f(\tilde{x})-\tilde{x},x_{n+1}-\tilde{x}〉.$$

From conditions (C1) and (C2) and Step 6, it is easy to see that ${\xi }_n\to 0$, ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$. Hence, by Lemma 2.9, we conclude that $x_n\to \tilde{x}$ as $n\to \mathrm{\infty }$. This completes the proof. □

From Theorem 3.2, we deduce immediately the following result.

Corollary 3.2 Suppose that $F(T)\cap {(A+B)}^{-1}0\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset (0,2\alpha )$ satisfy the following conditions:

• (C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<c\le {\beta }_n\le d<1$;

• (C4) $0<a\le {\lambda }_n\le b<2\alpha$.

Let the sequence $\{x_n\}$ be generated iteratively by (3.5):

where $x_0\in C$ is an arbitrary initial guess. Then the sequence $\{x_n\}$ converges strongly to a point $\tilde{x}$ in $F(T)\cap {(A+B)}^{-1}0$, which is the unique solution of the minimum norm problem (3.26).

Proof The variational inequality (3.2) is reduced to the inequality

$$〈\tilde{x},p-\tilde{x}〉\ge 0,\mathrm{\forall }p\in F(T)\cap {(A+B)}^{-1}0.$$

This is equivalent to ${\parallel \tilde{x}\parallel }^2\le 〈p,\tilde{x}〉\le \parallel p\parallel \parallel \tilde{x}\parallel$ for all $p\in F(T)\cap {(A+B)}^{-1}0$. It turns out that $\parallel \tilde{x}\parallel \le \parallel p\parallel$ for all $p\in F(T)\cap {(A+B)}^{-1}0$ and $\tilde{x}$ is the minimum-norm point of $F(T)\cap {(A+B)}^{-1}0$. □

Remark 3.1 It is worth pointing out that our iterative schemes (3.1) and (3.4) are new ones different from those in the literature. The iterative schemes (3.3) and (3.5) are also new ones different from those in the literature (see [5, 11] and the references therein).

4 Applications

Let H be a real Hilbert space, and let g be a proper lower semicontinuous convex function of H into $(-\mathrm{\infty },\mathrm{\infty }]$. Then the subdifferential ∂g of g is defined as follows:

$$\partial g(x)=\{z\in H\mid g(x)+〈z,y-x〉\le g(y),y\in H\}$$

for all $x\in H$. From Rockafellar [19], we know that ∂g is maximal monotone. Let C be a closed convex subset of H, and let $i_C$ be the indicator function of C, i.e.,

$$i_C(x)=\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ \mathrm{\infty },\hfill & x\notin C.\hfill \end{array}$$

(4.1)

Since $i_C$ is a proper lower semicontinuous convex function on H, the subdifferential $\partial i_C$ of $i_C$ is a maximal monotone operator. It is well known that if $B=\partial i_C$, then the MIP (1.1) is equivalent to find $u\in C$ such that

$$〈Au,v-u〉\ge 0,\mathrm{\forall }v\in C.$$

(4.2)

This problem is called Hartman-Stampacchia variational inequality (see [20]). The set of solutions of the variational inequality (4.2) is denoted by $VI(C,A)$.

The following result is proved by Takahashi et al. [11].

Lemma 4.1 [11]

Let C be a nonempty closed convex subset of a real Hilbert space H, let $P_C$ be the metric projection from H onto C, let $\partial i_C$ be the subdifferential of $i_C$, and let $J_{\lambda }$ be the resolvent of $\partial i_C$ for $\lambda >0$, where $i_C$ is defined by (4.1) and $J_{\lambda }={(I+\lambda \partial i_C)}^{-1}$. Then

$$u=J_{\lambda }x⟺u=P_{C}x,\mathrm{\forall }x\in H,y\in C.$$

Applying Theorem 3.2, we can obtain a strong convergence theorem for finding a common element of the set of solutions to the variational inequality (4.2) and the set of fixed points of a nonexpansive mapping.

Theorem 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H, and let S be a nonexpansive mapping of C into itself such that $F(S)\cap VI(C,A)\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset (0,2\alpha )$ satisfy the following conditions:

• (C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<c\le {\beta }_n\le d<1$;

• (C4) $0<a\le {\lambda }_n\le b<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_n-{\lambda }_{n+1})=0$.

Let the sequence $\{x_n\}$ be generated iteratively by

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{P}_C(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0,$$

where $x_0\in C$ is an arbitrary initial guess. Then the sequence $\{x_n\}$ converges strongly to a point $\tilde{x}$ in $F(S)\cap VI(C,A)$.

Proof Put $B=\partial i_C$. It is easy to show that $VI(C,A)={(A+\partial i_C)}^{-1}0$. In fact,

$$\begin{array}{rl}x\in {(A+\partial i_C)}^{-1}0& ⟺0\in Ax+\partial i_{C}x\\ ⟺-Ax\in \partial i_{C}x\\ ⟺〈Ax,u-x〉\ge 0(\mathrm{\forall }u\in C)\\ ⟺x\in VI(C,A).\end{array}$$

From Lemma 4.1, we get $J_{{\lambda }_n}=P_C$ for all ${\lambda }_n$. Hence, the desired result follows from Theorem 3.2. □

As in [11, 21], we consider the problem for finding a common element of the set of solutions of a mathematical model related to equilibrium problems and the set of fixed points of a nonexpansive mapping in a Hilbert space.

Let C be a nonempty closed convex subset of a Hilbert space H, and let us assume that a bifunction $\mathrm{\Theta }:C\times C\to \mathbb{R}$ satisfies the following conditions:

• (A1) $\mathrm{\Theta }(x,x)=0$ for all $x\in C$;

• (A2) Θ is monotone, that is, $\mathrm{\Theta }(x,y)+\mathrm{\Theta }(y,x)\le 0$ for all $x,y\in C$;

• (A3) for each $x,y,z\in C$,

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

• (A4) for each $x\in C$, $y\mapsto \mathrm{\Theta }(x,y)$ is convex and lower semicontinuous.

Then the mathematical model related to the equilibrium problem (with respect to C) is find $\hat{x}\in C$ such that

$$\mathrm{\Theta }(\hat{x},y)\ge 0$$

(4.3)

for all $y\in C$. The set of such solutions $\hat{x}$ is denoted by $EP(\mathrm{\Theta })$. The following lemma was given in [22, 23].

Lemma 4.2 [22, 23]

Let C be a nonempty closed convex subset of H, and let Θ be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Then for any $r>0$ and $x\in H$, there exists $z\in C$ such that

$$\mathrm{\Theta }(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C.$$

Moreover, if we define $T_r:H\to C$ as follows:

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

for all $x\in H$, then the following hold:

• (1) $T_r$ is single-valued;

• (2) $T_r$ is firmly nonexpansive, that is, for any $x,y\in H$,

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

• (3) $F(T_r)=EP(\mathrm{\Theta })$;

• (4) $EP(\mathrm{\Theta })$ is closed and convex.

We call such $T_r$ the resolvent of Θ for $r>0$. The following lemma was given in Takahashi et al. [11].

Lemma 4.3 [11]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let Θ be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $A_{\mathrm{\Theta }}$ be a multivalued mapping of H into itself defined by

$$A_{\mathrm{\Theta }}x=\{\begin{array}{cc}\{z\in H:\mathrm{\Theta }(x,y)\ge 〈y-x,z〉\},\hfill & x\in C,\hfill \\ \mathrm{\varnothing },\hfill & x\notin C.\hfill \end{array}$$

Then $EP(\mathrm{\Theta })=A_{\mathrm{\Theta }}^{-1}0$, and $A_{\mathrm{\Theta }}$ is a maximal monotone operator with $dom(A_{\mathrm{\Theta }})\subset C$. Moreover, for any $x\in H$ and $r>0$, the resolvent $T_r$ of Θ coincides with the resolvent of $A_{\mathrm{\Theta }}$; i.e.,

$$T_{r}x={(I+r{A}_{\mathrm{\Theta }})}^{-1}x.$$

Applying Lemma 4.3 and Theorem 3.2, we can obtain the following results.

Theorem 4.2 Let C be a nonempty closed convex subset of a real Hilbert space H, and let Θ be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $A_{\mathrm{\Theta }}$ be a maximal monotone operator with $dom(A_{\mathrm{\Theta }})\subset C$ defined as in Lemma  4.3, and let $T_{\lambda }$ be the resolvent of Θ for $\lambda >0$. Let A be an α-inverse strongly monotone mapping of C into H, and let S be a nonexpansive mapping from C into itself such that $F(S)\cap {(A+A_{\mathrm{\Theta }})}^{-1}0\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset (0,2\alpha )$ satisfy the following conditions:

• (C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<c\le {\beta }_n\le d<1$;

• (C4) $0<a\le {\lambda }_n\le b<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_n-{\lambda }_{n+1})=0$.

Let the sequence $\{x_n\}$ be generated iteratively by

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+(1-{\alpha }_n-{\beta }_n)S{T}_{{\lambda }_n}(x_n-{\lambda }_{n}A{x}_n),\mathrm{\forall }n\ge 0,$$

where $x_0\in C$ is an arbitrary initial guess. Then the sequence $\{x_n\}$ converges strongly to a point $\tilde{x}$ in $F(S)\cap {(A+A_{\mathrm{\Theta }})}^{-1}0$.

Theorem 4.3 Let C be a nonempty closed convex subset of a real Hilbert space H, and let Θ be a bifunction of $C\times C$ into ℝ satisfying (A1)-(A4). Let $A_{\mathrm{\Theta }}$ be a maximal monotone operator with $dom(A_{\mathrm{\Theta }})\subset C$ defined as in Lemma  4.3, and let $T_{\lambda }$ be the resolvent of Θ for $\lambda >0$, and let S be a nonexpansive mapping from C into itself such that $F(S)\cap EP(\mathrm{\Theta })\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ and $\{{\lambda }_n\}\subset (0,2\alpha )$ satisfy the following conditions:

• (C1) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (C2) ${\sum }_{n=0}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (C3) $0<c\le {\beta }_n\le d<1$;

• (C4) $0<a\le {\lambda }_n\le b<2\alpha$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_n-{\lambda }_{n+1})=0$.

Let the sequence $\{x_n\}$ be generated iteratively by

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

where $x_0\in C$ is an arbitrary initial guess. Then the sequence $\{x_n\}$ converges strongly to a point $\tilde{x}$ in $F(S)\cap EP(\mathrm{\Theta })$.

Proof Put $A=0$ in Theorem 4.2. From Lemma 4.3, we also know that $J_{{\lambda }_n}=T_{{\lambda }_n}$ for all $n\ge 0$. Hence, the desired result follows from Theorem 4.2. □

Remark 4.1 (1) As in Corollary 3.2, if we take $f\equiv 0$ in Theorems 4.1, 4.2 and 4.3, then we can obtain the minimum-norm point of $F(S)\cap VI(C,A)$, $F(S)\cap {(A+A_{\mathrm{\Theta }})}^{-1}0$ and $F(S)\cap EP(\mathrm{\Theta })$, respectively.

(2) For several iterative schemes for zeros of monotone operators, variational inequality problems, generalized equilibrium problems, convex minimization problems, and fixed point problems, we can also refer to [24–29] and the references therein. By combining our methods in this paper and methods in [24–29], we will consider new iterative schemes for the above-mentioned problems coupled with the fixed point problems of nonlinear operators.