A general iterative algorithm for monotone operators with λ-hybrid mappings in Hilbert spaces

Abstract

Let C be a nonempty closed convex subset of a Hilbert space ℋ, let B, G be two set-valued maximal monotone operators on C into ℋ, and let $g:\mathcal{H}\to \mathcal{H}$ be a k-contraction with $0<k<1$. $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping, $V:\mathcal{H}\to \mathcal{H}$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$, $T:C\to C$ is a λ-hybrid mapping. In this paper, a general iterative scheme for approximating a point of $F(T)\cap {(A+B)}^{-1}0\cap G^{-1}0\ne \mathrm{\varnothing }$ is introduced, where $F(T)$ is the set of fixed points of T, and a strong convergence theorem of the sequence generated by the iterative scheme is proved under suitable conditions. As applications of our strong convergence theorem, the related equilibrium and variational problems are also studied.

MSC:47H05, 47H10, 58E35.

1 Introduction

Throughout this paper, ℋ denotes a real Hilbert space, C a nonempty closed convex subset of ℋ, ℕ the set of all natural numbers and ℝ the set of all real numbers. For a self-mapping T on ℋ, $F(T)$ denotes the set of all fixed points of T.

A set-valued map $B:\mathcal{H}\to 2^{\mathcal{H}}$ with domain $\mathcal{D}(B):=\{x\in \mathcal{H}:Bx\ne \mathrm{\varnothing }\}$ is called monotone if

$$〈x-y,u-v〉\ge 0$$

for all $x,y\in \mathcal{D}(B)$ and for any $u\in Bx$, $v\in By$; B is said to be maximal monotone if its graph $\{(x,u):x\in \mathcal{H},u\in B(x)\}$ is not properly contained in the graph of any other monotone operator. For a positive real number r, the resolvent $J_r^B$ of a monotone operator B for r is a single-valued mapping $J_r^B:\mathcal{H}\to \mathcal{D}(B)$ defined by $z=J_r^B(x)$ if and only if $x\in z+rBz$, that is, $J_r^B(x)={(I+rB)}^{-1}(x)$ for any $x\in \mathcal{H}$, where I is the identity mapping on ℋ. The Yosida approximation $A_r$ of B for $r>0$ is defined as $A_r=\frac{1}{r}(I-J_r^B)$. It is known [1] that $A_{r}x\in B(J_r^{B}x)$ for all $x\in \mathcal{H}$.

The fixed point theory for nonexpansive mappings can be applied to the problem of finding a zero point v of a maximal monotone operator B on ℋ, that is, finding a point $v\in \mathcal{H}$ satisfying $0\in B(v)$. In the sequel, we shall denote the set of all zero points of B by $B^{-1}0$.

A self-mapping V on ℋ is called $\overline{\gamma }$-strongly monotone if there is a positive real number $\overline{\gamma }$ such that

$$〈x-y,Vx-Vy〉\ge \overline{\gamma }{\parallel x-y\parallel }^2,\mathrm{\forall }x,y\in \mathcal{H};$$

V is called L-Lipschitzian if there is a positive real number L such that

$$\parallel Vx-Vy\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in \mathcal{H}.$$

A mapping $A:C\to \mathcal{H}$ is said to be α-inverse strongly monotone if there is a positive real number α such that

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

As easily seen, an α-inverse strongly monotone mapping is $\frac{1}{\alpha }$-Lipschitzian on C.

For $\lambda \in \mathbb{R}$, a mapping $T:C\to \mathcal{H}$ is said to be λ-hybrid if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+\lambda 〈x-Tx,y-Ty〉,\mathrm{\forall }x,y\in C.$$

When $\lambda =2$, T is called nonspreading. It is known that $F(T)$ is closed and convex provided T is a λ-hybrid self-mapping on C, cf. [2].

Recently, Lin and Takahashi [3] introduced an algorithm for finding a point $p_0\in {(A+B)}^{-1}0\cap G^{-1}0$, where A is an α-inverse strongly monotone mapping of C into ℋ, and B, G are two set-valued maximal monotone operators with $\mathcal{D}(B)\subset C$ and $\mathcal{D}(G)\subset C$. More precisely, let g be a k-contraction and V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping. Choose $\mu ,\gamma \in \mathbb{R}$ so that $0<\frac{2\overline{\gamma }}{L^2}$ and $0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}$. Then the algorithm starts with any $x_1\in \mathcal{H}$ and generates a sequence $\{x_n\}$ iteratively by

$$x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)J_{{\sigma }_n}^B(I-{\sigma }_{n}A)J_{r_n}^G{x}_n,n\in \mathbb{N},$$

(1)

where $\{{\alpha }_n\}\subseteq (0,1)$, $\{{\sigma }_n\}\subseteq (0,\mathrm{\infty })$, and $\{r_n\}\subseteq (0,\mathrm{\infty })$ satisfy

$$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0,\sum _{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty },\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_n-{\alpha }_{n+1}|<\mathrm{\infty },0<a\le {\sigma }_n\le 2\alpha ,\hfill \\ \sum _{n=1}^{\mathrm{\infty }}|{\sigma }_n-{\sigma }_{n+1}|<\mathrm{\infty },\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{r}_n>0\text{and}\sum _{n=1}^{\mathrm{\infty }}|r_n-r_{n+1}|<\mathrm{\infty }.\hfill \end{array}$$

They proved that $P_{{(A+B)}^{-1}0\cap G^{-1}0}(I-V+\gamma g)$ has a unique fixed point $p_0$ in Ω, and this $p_0$ is also a unique solution $p_0\in {(A+B)}^{-1}0\cap G^{-1}0$ to the hierarchical variational inequality

$$〈(V-\gamma g)p,q-p〉,\mathrm{\forall }q\in \mathrm{\Omega }.$$

As Lin and Takahashi said in [3], their idea for this algorithm comes from the works of Tian [4].

On the other hand, Manaka and Takahashi [5] used the algorithm

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T(J_{{\sigma }_n}^B(I-{\sigma }_{n}A)x_n)$$

(2)

to find a point $p_0\in F(T)\cap {(A+B)}^{-1}0$ for a nonspreading mapping T, an α-inverse strongly monotone mapping A and a maximal monotone operator B under the conditions

$$0<c\le {\alpha }_n\le d<1\text{and}0<a\le {\sigma }_n\le b<2\alpha ,$$

where $a,b,c,d\in \mathbb{R}$ are fixed. They proved that the sequence $\{x_n\}$ constructed above converges weakly to a point $p\in F(T)\cap {(A+B)}^{-1}0$.

Very recently, Liu et al. [6] modified the iterative scheme (2) to approximate a point $p\in F(T)\cap {(A+B)}^{-1}0$ for a nonspreading mapping T, an α-inverse strongly monotone mapping and a maximal monotone operator B. For any $u\in \mathcal{H}$, they put $x_1$ to be any point of ℋ and define recursively for all $n\in \mathbb{N}$,

$$\{\begin{array}{l}{z}_n=J_{{\alpha }_n}^B(I-{\sigma }_{n}A)x_n,\\ y_n=\frac{1}{n}{\sum }_{i=0}^{n-1}{T}^i{z}_n,\\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n,\end{array}$$

(3)

where $\{{\alpha }_n\}$ is a suitable sequence in $[0,1]$.

Motivated by the above works, in this paper we introduce a general iterative scheme for approximating a point of $F(T)\cap {(A+B)}^{-1}0\cap G^{-1}0\ne \mathrm{\varnothing }$, where T is a λ-hybrid self-mapping on C, $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping and B and G are two maximal monotone operators. A strong convergence theorem of the sequence generated by our iterative scheme is proved under suitable conditions. Our result improves and generalizes the main theorem of Lin and Takahashi [3]. As applications of our strong convergence theorem, the related equilibrium and variational problems are also studied.

2 Preliminaries

In order to facilitate our investigation, in what follows we recall some basic facts. A mapping $T:C\to \mathcal{H}$ is said to be

• (i) nonexpansive if

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

• (ii) firmly nonexpansive if

  $${\parallel Tx-Ty\parallel }^2\le 〈x-y,Tx-Ty〉,\mathrm{\forall }x,y\in C.$$

The metric projection $P_C$ from ℋ onto C is the mapping that assigns each $x\in \mathcal{H}$ the unique point $P_{C}x$ in C with the property

$$\parallel x-P_{C}x\parallel =\underset{y\in C}{min}\parallel y-x\parallel .$$

It is known that $P_C$ is nonexpansive and characterized by the inequality: for any $x\in \mathcal{H}$,

$$〈x-P_{C}x,y-P_{C}x〉\le 0,\mathrm{\forall }y\in C,$$

(4)

cf. [7].

For any $x,y\in \mathcal{H}$, one has

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉.$$

(5)

Lemma 2.1 (Demiclosedness principle) [8]

Let T be a nonexpansive self-mapping on a nonempty closed convex subset C of ℋ, and suppose that $\{x_n\}$ is a sequence in C such that $\{x_n\}$ converges weakly to some $z\in C$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. Then $Tz=z$.

For $s>0$, the resolvent $J_s^B$ of the maximal monotone operator B on ℋ has the following properties, cf. [9].

Lemma 2.2 Let B be a maximal monotone operator on ℋ. Then, for any $s>0$,

• (a) $J_s^B$ is single-valued and firmly nonexpansive;

• (b) $\mathcal{D}(J_s^B)=\mathcal{H}$ and $F(J_s^B)=B^{-1}0$.

The following lemma can be derived easily from the resolvent identity of a monotone operator B:

$$J_s^{B}x=J_t^B(\frac{t}{s}x+(1-\frac{t}{s})J_s^{B}x)\text{ for any }s,t>0\text{ and any }x\in \mathcal{H}.$$

Lemma 2.3 Let B be a monotone operator on ℋ. Then, for any $s,t\in \mathbb{R}$ with $s,t>0$ and for any $x\in \mathcal{H}$,

$$\parallel J_s^{B}x-J_t^{B}x\parallel \le \frac{|s-t|}{s}\parallel x-J_s^{B}x\parallel .$$

When B is maximal monotone, a different proof may be found in Takahashi et al. [9].

Lemma 2.4 [5]

Let $A:C\to \mathcal{H}$ be an α-inverse strongly monotone mapping, and let B be a maximal monotone operator on ℋ with $\mathcal{D}(B)\subseteq C$. Then, for any $\sigma >0$, one has ${(A+B)}^{-1}0=F(J_{\sigma }^B(I-\sigma A))$.

Lemma 2.5 [3]

Let $A:C\to \mathcal{H}$ be an α-inverse strongly monotone mapping. Then, for any $\sigma \in (0,2\alpha ]$, $(I-\sigma A)$ is nonexpansive.

Lemma 2.6 [3]

Let $g:\mathcal{H}\to \mathcal{H}$ be a k-contraction with $0<k<1$, let $V:\mathcal{H}\to \mathcal{H}$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$, and let γ be a real number satisfying $0<\gamma <\frac{\overline{\gamma }}{k}$. Then $V-\gamma g$ is a $(\overline{\gamma }-\gamma k)$-strongly monotone and $(L+\gamma k)$-Lipschitzian mapping. Furthermore, for any nonempty closed convex subset Ω of ℋ, $P_{\mathrm{\Omega }}(I-V+\gamma g)$ has a unique fixed point $p_0\in \mathrm{\Omega }$, which is also a unique solution of the variational inequality

$$〈(V-\gamma g)z,q-z〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

Lemma 2.7 [10]

Let $\{s_n\}$ be a sequence of nonnegative real numbers satisfying

$$s_{n+1}\le (1-{\alpha }_n)s_n+{\alpha }_n{\mu }_n+{\nu }_n,n\in \mathbb{N},$$

where $\{{\alpha }_n\}$, $\{{\mu }_n\}$ and $\{{\nu }_n\}$ verify the following conditions:

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

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

• (iii) $\{{\nu }_n\}\subseteq [0,\mathrm{\infty })$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\nu }_n<\mathrm{\infty }$.

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

3 Strong convergence theorems

We begin the proof of the main result of this paper. As the proof is rather lengthy, we divide the proof into many assertions.

Theorem 3.1 Suppose that

(3.1.1) $G:C\to 2^{\mathcal{H}}$ and $B:C\to 2^{\mathcal{H}}$ are two maximal monotone operators with $\mathcal{D}(G)\subseteq C$ and $\mathcal{D}(B)\subseteq C$;

(3.1.2) $g:\mathcal{H}\to \mathcal{H}$ is a k-contraction, $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping, and $V:\mathcal{H}\to \mathcal{H}$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$;

(3.1.3) $T:C\to C$ is a λ-hybrid mapping;

(3.1.4) $\mathrm{\Omega }:=F(T)\cap {(A+B)}^{-1}0\cap G^{-1}0\ne \mathrm{\varnothing }$;

(3.1.5) μ and γ are two real numbers satisfying $0<\mu <\frac{2\overline{\gamma }}{L^2}$ and $0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}$.

Start with any $x_1\in \mathcal{H}$ and define a sequence $\{x_n\}$ iteratively by

$$\{\begin{array}{l}{z}_n=J_{{\sigma }_n}^B(I-{\sigma }_{n}A)J_{r_n}^G{x}_n,n\in \mathbb{N},\\ y_n=\frac{1}{n}{\sum }_{i=0}^{n-1}{T}^i{z}_n,n\in \mathbb{N},\\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\end{array}$$

(6)

where the sequences $\{{\alpha }_n\}$, $\{{\sigma }_n\}$ and $\{r_n\}$ verify the following conditions:

(3.1.6) $\{{\alpha }_n\}$ is a sequence in $[0,1]$ with ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(3.1.7) $\{{\sigma }_n\}$ and $\{r_n\}$ are sequences in $(0,\mathrm{\infty })$ so that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and there are $a,b\in \mathbb{R}$ with $0<a\le {\sigma }_n\le b<2\alpha$ for all $n\in \mathbb{N}$.

Then the sequence $\{x_n\}$ constructed by algorithm (6) converges strongly to a point $p_0\in \mathrm{\Omega }$, where $p_0$ is the unique fixed point of $P_{\mathrm{\Omega }}(I-V+\gamma g)$, and this point $p_0$ is also a unique solution of the hierarchical inequality

$$〈(V-\gamma g)p,q-p〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

(7)

Proof In what follows, p is a point in Ω, $\tau =\overline{\gamma }-\frac{L^2\mu }{2}$, and $u_n=J_{r_n}^G{x}_n$ for all $n\in \mathbb{N}$.

• Assertion (A): There is$N\in \mathbb{N}$such that

  $$\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \le (1-{\alpha }_n\tau )\parallel y_n-p\parallel .$$

Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, there is $N\in \mathbb{N}$ such that $1-{\alpha }_n\tau >0$ and $\mu -{\alpha }_n>0$ for all $n\ge N$. Then, as V is $\overline{\gamma }$-strongly monotone and L-Lipschitzian, we have that for all $n\ge N$,

$$\begin{array}{c}{\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2\hfill \\ ={\parallel (y_n-p)-{\alpha }_n(V{y}_n-Vp)\parallel }^2\hfill \\ ={\parallel y_n-p\parallel }^2-2{\alpha }_n〈y_n-p,V{y}_n-Vp〉+{\alpha }_n^2{\parallel V{y}_n-Vp\parallel }^2\hfill \\ \le {\parallel y_n-p\parallel }^2-2{\alpha }_n\overline{\gamma }{\parallel y_n-p\parallel }^2+{\alpha }_n^2{L}^2{\parallel y_n-p\parallel }^2\hfill \\ =(1-2{\alpha }_n\tau -{\alpha }_n{L}^2\mu +{\alpha }_n^2{L}^2){\parallel y_n-p\parallel }^2\hfill \\ \le (1-2{\alpha }_n\tau +{\alpha }_n^2{\tau }^2){\parallel y_n-p\parallel }^2-{\alpha }_n{L}^2(\mu -{\alpha }_n){\parallel y_n-p\parallel }^2\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2,\hfill \end{array}$$

(8)

and so the assertion holds.

• Assertion (B): The sequences$\{x_n\}$, $\{y_n\}$, $\{z_n\}$, $\{u_n\}$, $\{A{u}_n\}$, $\{V{y}_n\}$, $\{g(x_n)\}$and$\{T^n{z}_n\}$are bounded.

We firstly show that

$$\parallel y_n-p\parallel \le \parallel z_n-p\parallel \le \parallel x_n-p\parallel ,\mathrm{\forall }n\in \mathbb{N}.$$

(9)

On account of $p=J_{{\sigma }_n}^B(I-{\sigma }_{n}A)p$ by Lemma 2.4, $p=J_{r_n}^{G}p$ by assumption and the facts that a resolvent is nonexpansive and A is α-inverse strongly monotone, we have

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2& ={\parallel J_{{\sigma }_n}^B(I-{\sigma }_{n}A)J_{r_n}^G{x}_n-J_{{\sigma }_n}^B(I-{\sigma }_{n}A)J_{r_n}^{G}p\parallel }^2\\ \le {\parallel (I-{\sigma }_{n}A)J_{r_n}^G{x}_n-(I-{\sigma }_{n}A)J_{r_n}^{G}p\parallel }^2\\ ={\parallel J_{r_n}^G{x}_n-J_{r_n}^{G}p-{\sigma }_n(A{J}_{r_n}^G{x}_n-A{J}_{r_n}^{G}p)\parallel }^2\\ ={\parallel J_{r_n}^G{x}_n-J_{r_n}^{G}p\parallel }^2-2{\sigma }_n〈u_n-p,A{u}_n-Ap〉+{\sigma }_n^2{\parallel A{u}_n-Ap\parallel }^2\end{array}$$

(10)

$$\begin{array}{r}\le {\parallel x_n-p\parallel }^2-2{\sigma }_n\alpha {\parallel A{u}_n-Ap\parallel }^2+{\sigma }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ ={\parallel x_n-p\parallel }^2-{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2\\ \le {\parallel x_n-p\parallel }^2,\end{array}$$

(11)

where the last inequality follows from the hypothesis that $2\alpha \ge {\sigma }_n$. Therefore,

$$\parallel z_n-p\parallel \le \parallel x_n-p\parallel .$$

Since T is λ-hybrid, we have, for any $n\in \mathbb{N}$,

$${\parallel T{z}_n-p\parallel }^2={\parallel T{z}_n-Tp\parallel }^2\le {\parallel z_n-p\parallel }^2+\lambda 〈z_n-T{z}_n,p-Tp〉={\parallel z_n-p\parallel }^2,$$

and so, by induction, it comes easily that

$$\parallel T^i{z}_n-p\parallel \le \parallel z_n-p\parallel \text{for all }i,n\in \mathbb{N}.$$

(12)

Consequently,

$$\begin{array}{rl}\parallel y_n-p\parallel & =\parallel \frac{1}{n}\sum _{i=0}^{n-1}{T}^i{z}_n-p\parallel \\ \le \frac{1}{n}\sum _{i=0}^{n-1}\parallel T^i{z}_n-p\parallel \\ \le \frac{1}{n}\sum _{i=0}^{n-1}\parallel z_n-p\parallel \\ =\parallel z_n-p\parallel \le \parallel x_n-p\parallel .\end{array}$$

Next, we show that $\{\parallel x_n-p\parallel \}$ is bounded. Indeed, as $p=(I-{\alpha }_{n}V)p+{\alpha }_{n}Vp$, we have

$$\begin{array}{c}\parallel x_{n+1}-p\parallel \hfill \\ =\parallel {\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n-{\alpha }_{n}Vp-(I-{\alpha }_{n}V)p\parallel \hfill \\ \le \parallel {\alpha }_n(\gamma g(x_n)-Vp)\parallel +\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \hfill \\ \le {\alpha }_n\parallel \gamma g(x_n)-\gamma g(p)\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel \hfill \\ \le {\alpha }_n\gamma k\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel ,\hfill \end{array}$$

which together with Assertion (A) implies that for all $n\ge N$,

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & \le {\alpha }_n\gamma k\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel +(1-{\alpha }_n\tau )\parallel y_n-p\parallel \\ \le (1-{\alpha }_n(\tau -\gamma k))\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma g(p)-Vp\parallel \\ \le \parallel x_n-p\parallel +\parallel \gamma g(p)-Vp\parallel ,\end{array}$$

from which we inductively deduce that

$$\parallel x_n-p\parallel \le \parallel x_1-p\parallel +\parallel \gamma g(p)-Vp\parallel .$$

Thus, $\{x_n\}$ is bounded, and so are $\{y_n\}$, $\{z_n\}$ by (9). And then the boundedness of $\{T^n{z}_n\}$ follows from (12). The fact that $\{V{y}_n\}$ and $\{g(x_n)\}$ are bounded is due to the fact that V and g are L-Lipschitzian and k-contraction, respectively.

Finally, from $\parallel u_n-p\parallel =\parallel J_{{\sigma }_n}^G{x}_n-J_{{\sigma }_n}^{G}p\parallel \le \parallel x_n-p\parallel$, we deduce that $\{u_n\}$ is bounded, and $\{A{u}_n\}$ is bounded comes from A is $\frac{1}{\alpha }$-Lipschitzian.

• Assertion (C): ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

We at first show that ${lim}_{n\to \mathrm{\infty }}\parallel y_{n+1}-y_n\parallel =0$. By Assertion (B), we can choose a positive real number M so that

$$max\{\underset{n\in \mathbb{N}}{sup}\parallel T^n{z}_n\parallel ,\underset{n\in \mathbb{N}}{sup}\parallel y_n\parallel \}\le M.$$

Then, for all $n\in \mathbb{N}$,

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel & =\parallel \frac{1}{n+1}\sum _{i=0}^n{T}^i{z}_n-\frac{1}{n}\sum _{i=0}^{n-1}{T}^i{z}_n\parallel \\ \le \frac{1}{n+1}\parallel T^n{z}_n\parallel +(\frac{1}{n}-\frac{1}{n+1})\parallel \sum _{i=0}^{n-1}{T}^i{z}_n\parallel \\ =\frac{1}{n+1}\parallel T^n{z}_n\parallel +\frac{1}{n+1}\parallel y_n\parallel \le \frac{2M}{n+1}.\end{array}$$

Hence ${lim}_{n\to \mathrm{\infty }}\parallel y_{n+1}-y_n\parallel =0$.

Now, for all $n\ge N$, we have from Assertion (A) that

$$\begin{array}{c}\parallel x_{n+2}-x_{n+1}\parallel \hfill \\ =\parallel {\alpha }_{n+1}\gamma g(x_{n+1})+(I-{\alpha }_{n+1}V)y_{n+1}-{\alpha }_n\gamma g(x_n)-(I-{\alpha }_{n}V)y_n\parallel \hfill \\ \le \parallel {\alpha }_{n+1}\gamma g(x_{n+1})-{\alpha }_{n+1}\gamma g(x_n)\parallel +\parallel {\alpha }_{n+1}\gamma g(x_n)-{\alpha }_n\gamma g(x_n)\parallel \hfill \\ +\parallel (I-{\alpha }_{n+1}V)y_{n+1}-(I-{\alpha }_{n+1}V)y_n\parallel +\parallel (I-{\alpha }_{n+1}V)y_n-(I-{\alpha }_{n}V)y_n\parallel \hfill \\ \le {\alpha }_{n+1}\gamma k\parallel x_{n+1}-x_n\parallel +\gamma |{\alpha }_{n+1}-{\alpha }_n|\parallel g(x_n)\parallel +(1-{\alpha }_{n+1}\tau )\parallel y_{n+1}-y_n\parallel \hfill \\ +|{\alpha }_{n+1}-{\alpha }_n|\parallel V{y}_n\parallel .\hfill \end{array}$$

Consequently, using condition (3.1.6), Assertion (B) and ${lim}_{n\to \mathrm{\infty }}\parallel y_{n+1}-y_n\parallel =0$, we get ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

• Assertion (D): ${lim}_{n\to \mathrm{\infty }}\parallel A{u}_n-Ap\parallel =0$.

Using (5), it follows from (8), (9) and (11) that

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ ={\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p+{\alpha }_n\gamma g(x_n)-Vp\parallel }^2\hfill \\ \le {\parallel (I-{\alpha }_{n}V)y_n-(I-{\alpha }_{n}V)p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (5)}\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\parallel y_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (8)}\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\parallel z_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (9)}\hfill \\ \le {(1-{\alpha }_n\tau )}^2({\parallel x_n-p\parallel }^2-{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2)\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (11)}\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\hfill \\ \le {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\hfill \end{array}$$

(13)

Hence, on account of $0<a\le {\sigma }_n\le b<2\alpha$ for all $n\in \mathbb{N}$, we have

$$\begin{array}{c}{(1-{\alpha }_n\tau )}^{2}a(2\alpha -b){\parallel A{u}_n-Ap\parallel }^2\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\hfill \\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉,\hfill \end{array}$$

which together with Assertions (B) and (C) implies that ${lim}_{n\to \mathrm{\infty }}\parallel A{u}_n-Ap\parallel =0$.

• Assertion (E): ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$.

From

$$\begin{array}{rl}\parallel x_n-y_n\parallel & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-y_n\parallel \\ =\parallel x_n-x_{n+1}\parallel +\parallel {\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n-y_n\parallel \\ =\parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel \gamma g(x_n)-V{y}_n\parallel ,\end{array}$$

condition (3.1.6) and Assertions (B) and (C), we conclude that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$.

As $J_{r_n}^G$ is firmly nonexpansive, one has

$$\begin{array}{rl}2{\parallel u_n-p\parallel }^2& =2{\parallel J_{r_n}^G{x}_n-J_{r_n}^{G}p\parallel }^2\\ \le 2〈x_n-p,u_n-p〉\\ ={\parallel x_n-p\parallel }^2+{\parallel u_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2,\end{array}$$

and so

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2.$$

(14)

In addition, since A is α-inverse strongly monotone, we see from (10) that

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2& \le {\parallel J_{r_n}^G{x}_n-J_{r_n}^{G}p\parallel }^2-2{\sigma }_n〈u_n-p,A{u}_n-Ap〉+{\sigma }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ ={\parallel u_n-p\parallel }^2-2{\sigma }_n〈u_n-p,A{u}_n-Ap〉+{\sigma }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ \le {\parallel u_n-p\parallel }^2-2{\sigma }_n\alpha {\parallel A{u}_n-Ap\parallel }^2+{\sigma }_n^2{\parallel A{u}_n-Ap\parallel }^2\\ ={\parallel u_n-p\parallel }^2-{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2.\end{array}$$

(15)

Then (13), (14) and (15) give us that

$$\begin{array}{c}{\parallel x_{n+1}-p\parallel }^2\hfill \\ \le {(1-{\alpha }_n\tau )}^2{\parallel z_n-p\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (13)}\hfill \\ \le {(1-{\alpha }_n\tau )}^2({\parallel u_n-p\parallel }^2-{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2)\text{by (15)}\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\hfill \\ \le {(1-{\alpha }_n\tau )}^2({\parallel x_n-p\parallel }^2-{\parallel u_n-x_n\parallel }^2)-{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2\hfill \\ +2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\text{by (14)}\hfill \\ \le {\parallel x_n-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2-{(1-{\alpha }_n\tau )}^2{\parallel u_n-x_n\parallel }^2\hfill \\ -{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\hfill \end{array}$$

Therefore

$$\begin{array}{c}{(1-{\alpha }_n\tau )}^2{\parallel u_n-x_n\parallel }^2\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\hfill \\ -{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉\hfill \\ \le \parallel x_n-x_{n+1}\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )+{\alpha }_n^2{\tau }^2{\parallel x_n-p\parallel }^2\hfill \\ -{(1-{\alpha }_n\tau )}^2{\sigma }_n(2\alpha -{\sigma }_n){\parallel A{u}_n-Ap\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-Vp,x_{n+1}-p〉.\hfill \end{array}$$

By Assertions (C), (D) and condition (3.1.6), we obtain

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

• Assertion (F): Ω is a nonempty closed convex subset of ℋ.

Since Ω is nonempty by assumption, it suffices to show that Ω is closed and convex. From Lemma 2.4, we have that for any $\sigma >0$,

$${(A+B)}^{-1}0=F(J_{\sigma }^B(I-\sigma A)).$$

In case $0<\sigma \le 2\alpha$, we have that $I-\sigma A$ is nonexpansive by Lemma 2.5. Then $J_{\sigma }^B(I-\sigma A)$ is nonexpansive, and so ${(A+B)}^{-1}0=F(J_{\sigma }^B(I-\sigma A))$ is closed and convex. In the like manner, for any $r>0$, $G^{-1}0=F(J_r^G)$ is closed and convex. Besides, it is shown in [2] that $F(T)$ is closed and convex. Hence comes the conclusion for Ω.

• Assertion (G): $P_{\mathrm{\Omega }}(I-V+\gamma g)$has a unique fixed point$p_0$in Ω, and this$p_0$is also a unique solution$p_0\in \mathrm{\Omega }$to the hierarchical variational inequality (7)

  $$〈(V-\gamma g)p,q-p〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

Taking into account that Ω is a nonempty closed convex subset, the conclusion follows from Lemma 2.6.

• Assertion (H): Let$p_0$be the unique solution to the hierarchical variational inequality (7). Then

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(V-\gamma g)p_0,x_n-p_0〉\le 0.$$

Choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ so that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(V-\gamma g)p_0,x_n-p_0〉=\underset{i\to \mathrm{\infty }}{lim}〈(V-\gamma g)p_0,x_{n_i}-p_0〉$$

and $\{x_{n_i}\}$ converges weakly to $w\in \mathcal{H}$. In view of Assertion (E), we see that both of the sequences $\{u_{n_i}\}$ and $\{y_{n_i}\}$ converge weakly to w. We at first show that $w\in F(T)$. Since T is λ-hybrid, one has ${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+\lambda 〈x-Tx,y-Ty〉$ for all $x,y\in C$. Express λ as $\lambda =2(1-\delta )$. Then, for all $x,y\in C$, we have

$$\begin{array}{c}{\parallel Tx-Ty\parallel }^2\hfill \\ \le {\parallel x-Ty+Ty-y\parallel }^2+2(1-\delta )〈x-Tx,y-Ty〉\hfill \\ ={\parallel x-Ty\parallel }^2+2〈x-Ty,Ty-y〉+{\parallel Ty-y\parallel }^2+2(1-\delta )〈x-Tx,y-Ty〉\hfill \\ ={\parallel x-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈x-(1-\delta )x+(1-\delta )Tx-Ty,Ty-y〉\hfill \\ ={\parallel x-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈\delta x+(1-\delta )Tx-Ty,Ty-y〉,\hfill \end{array}$$

that is,

$$0\le {\parallel x-Ty\parallel }^2-{\parallel Tx-Ty\parallel }^2+{\parallel Ty-y\parallel }^2+2〈\delta x+(1-\delta )Tx-Ty,Ty-y〉.$$

In particular, for all $i\in \mathbb{N}$, all $z_n$ and all $y\in C$,

$$\begin{array}{rcl}0& \le & {\parallel T^i{z}_n-Ty\parallel }^2-{\parallel T^{i+1}{z}_n-Ty\parallel }^2+{\parallel Ty-y\parallel }^2\\ +2〈\delta T^i{z}_n+(1-\delta )T^{i+1}{z}_n-Ty,Ty-y〉.\end{array}$$

Summing these inequalities from $i=0$ to $n-1$ and dividing by n, we obtain

$$\begin{array}{rcl}0& \le & \frac{1}{n}({\parallel z_n-Ty\parallel }^2-{\parallel T^n{z}_n-Ty\parallel }^2)+{\parallel Ty-y\parallel }^2\\ +2〈\delta y_n+(1-\delta )(\frac{n+1}{n}{y}_{n+1}-\frac{z_n}{n})-Ty,Ty-y〉.\end{array}$$

(16)

Replacing n with $n_i$ in (16) and letting $i\to \mathrm{\infty }$, we get via Assertion (B) that

$$0\le {\parallel Ty-y\parallel }^2+2〈\delta w+(1-\delta )w-Ty,Ty-y〉.$$

(17)

Putting $y=w$ in (17), we arrive at $0\le -{\parallel Tw-w\parallel }^2$. This shows that $w\in F(T)$.

Since $0<a\le {\sigma }_{n_i}\le b<2\alpha$, $\{{\sigma }_{n_i}\}$ has a convergent subsequence. For simplicity, we assume that $\{{\sigma }_{n_i}\}$ converges to a number $\sigma \in [a,b]$. Note that for all $n\in \mathbb{N}$,

$$\begin{array}{c}\parallel J_{\sigma }^B(I-\sigma A)u_n-z_n\parallel \hfill \\ \le \parallel J_{\sigma }^B(I-\sigma A)u_n-J_{\sigma }^B(I-{\sigma }_{n}A)u_n\parallel +\parallel J_{\sigma }^B(I-{\sigma }_{n}A)u_n-z_n\parallel \hfill \\ \le \parallel (I-\sigma A)u_n-(I-{\sigma }_{n}A)u_n\parallel +\parallel J_{\sigma }^B(I-{\sigma }_{n}A)u_n-J_{{\sigma }_n}^B(I-{\sigma }_{n}A)u_n\parallel \hfill \\ \le |{\sigma }_n-\sigma |\parallel A{u}_n\parallel +\frac{|{\sigma }_n-\sigma |}{\sigma }\parallel J_{\sigma }^B{v}_n-v_n\parallel ,\hfill \end{array}$$

(18)

where $v_n=(I-{\sigma }_{n}A)u_n$. Furthermore, we have for all $n\in \mathbb{N}$,

$$\begin{array}{rl}\parallel J_{\sigma }^B{v}_n-v_n\parallel & \le \parallel J_{\sigma }^B{v}_n-p\parallel +\parallel v_n-p\parallel \\ =\parallel J_{\sigma }^B{v}_n-J_{\sigma }^{B}p\parallel +\parallel v_n-p\parallel \\ \le 2\parallel v_n-p\parallel \end{array}$$

and

$$\begin{array}{rl}\parallel v_n-p\parallel & =\parallel (I-{\sigma }_{n}A)u_n-(I-{\sigma }_{n}A)p-{\sigma }_{n}Ap\parallel \\ \le \parallel u_n-p\parallel +b\parallel Ap\parallel \text{by Lemma 2.5}.\end{array}$$

Hence, $\{J_{\sigma }^B{v}_n-v_n\}$ is bounded. Now, replace n with $n_i$ in (18) and note that both $\{A{u}_{n_i}\}$ and $\{J_{\sigma }^B{v}_{n_i}-v_{n_i}\}$ are bounded. Letting $i\to \mathrm{\infty }$, we get that

$$\parallel J_{\sigma }^B(I-\sigma A)u_{n_i}-u_{n_i}\parallel \to 0.$$

Applying Lemma 2.1 to the nonexpansive mapping $J_{\sigma }^B(I-\sigma A)$, we conclude that $w=J_{\sigma }^B(I-\sigma A)w$, that is, $w\in {(A+B)}^{-1}0$.

We now show that $w\in G^{-1}0$. Since G is a maximal monotone operator, the Yosida approximation $A_{r}x=\frac{x-J_r^{G}x}{r}$ of G for $r>0$ is in $F(J_r^G(x))$. So, for any $(z,v)\in G$, one has

$$〈z-u_{n_i},v-\frac{x_{n_i}-u_{n_i}}{r_{n_i}}〉\ge 0.$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, $\{u_{n_i}\}$ converges weakly to w and $x_{n_i}-u_{n_i}\to 0$, we have

$$〈z-w,v〉\ge 0,$$

and then the maximality of T shows that $0\in Gw$, that is, $0\in G^{-1}0$.

In summary, we have shown that $w\in \mathrm{\Omega }$, and so, by Assertion (G), we conclude that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(V-\gamma g)p_0,x_n-p_0〉& =\underset{i\to \mathrm{\infty }}{lim}〈(V-\gamma g)p_0,x_{n_i}-p_0〉\\ =〈(V-\gamma g)p_0,w-p_0〉\ge 0.\end{array}$$

• Assertion (I): The sequence$\{x_n\}$converges strongly to$p_0$.

Replacing p with $p_0$ in (13), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p_0\parallel }^2& \le & {(1-{\alpha }_n\tau )}^2{\parallel z_n-p_0\parallel }^2+2{\alpha }_n〈\gamma g(x_n)-V{p}_0,x_{n+1}-p_0〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel z_n-p_0\parallel }^2+2{\alpha }_n\gamma k\parallel x_n-p_0\parallel \parallel x_{n+1}-p_0\parallel \\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ \le & {(1-{\alpha }_n\tau )}^2{\parallel x_n-p_0\parallel }^2+{\alpha }_n\gamma k({\parallel x_n-p_0\parallel }^2+{\parallel x_{n+1}-p_0\parallel }^2)\\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\text{by (9)}\\ =& ({(1-{\alpha }_n\tau )}^2+{\alpha }_n\gamma k){\parallel x_n-p_0\parallel }^2+{\alpha }_n\gamma k{\parallel x_{n+1}-p_0\parallel }^2\\ +2{\alpha }_n〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉.\end{array}$$

Hence

$$\begin{array}{rcl}{\parallel x_{n+1}-p_0\parallel }^2& \le & \frac{1-2{\alpha }_n\tau +{({\alpha }_n\tau )}^2+{\alpha }_n\gamma k}{1-{\alpha }_n\gamma k}{\parallel x_n-p_0\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ =& (1-\frac{2(\tau -\gamma k){\alpha }_n}{1-{\alpha }_n\gamma k}){\parallel x_n-p_0\parallel }^2+\frac{{({\alpha }_n\tau )}^2}{1-{\alpha }_n\gamma k}{\parallel x_n-p_0\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉\\ =& (1-{\beta }_n){\parallel x_n-p_0\parallel }^2\\ +{\beta }_n(\frac{{\alpha }_n{\tau }^2{\parallel x_n-p_0\parallel }^2}{2(\tau -\gamma k)}+\frac{1}{\tau -\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉),\end{array}$$

where ${\beta }_n=\frac{2(\tau -\gamma k){\alpha }_n}{1-{\alpha }_n\gamma k}$. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$, and since

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\frac{{\alpha }_n{\tau }^2{\parallel x_n-p_0\parallel }^2}{2(\tau -\gamma k)}+\frac{1}{\tau -\gamma k}〈\gamma g(p_0)-V{p}_0,x_{n+1}-p_0〉)\le 0,$$

by Assertion (H), it follows from Lemma 2.7 that $\{x_n\}$ converges strongly to $p_0$. This completes the proof. □

When T is the identity mapping, the theorem reduces to the following corollary.

Corollary 3.2 Suppose that

(3.2.1) $G:C\to 2^{\mathcal{H}}$ and $B:C\to 2^{\mathcal{H}}$ are two maximal monotone operators with $\mathcal{D}(G)\subseteq C$ and $\mathcal{D}(B)\subseteq C$;

(3.2.2) $g:\mathcal{H}\to \mathcal{H}$ is a k-contraction, $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping, and $V:\mathcal{H}\to \mathcal{H}$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$;

(3.2.3) $\mathrm{\Omega }:={(A+B)}^{-1}0\cap G^{-1}0\ne \mathrm{\varnothing }$;

(3.2.4) μ and γ are two real numbers satisfying $0<\mu <\frac{2\overline{\gamma }}{L^2}$ and $0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}$.

Start with any $x_1\in \mathcal{H}$ and define a sequence $\{x_n\}$ iteratively by

$$\{\begin{array}{l}{y}_n=J_{{\sigma }_n}^B(I-{\sigma }_{n}A)J_{r_n}^G{x}_n,n\in \mathbb{N},\\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\end{array}$$

where the sequences $\{{\alpha }_n\}$, $\{{\sigma }_n\}$ and $\{r_n\}$ verify the following conditions:

(3.2.5) $\{{\alpha }_n\}$ is a sequence in $[0,1]$ with ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(3.2.6) $\{{\sigma }_n\}$ and $\{r_n\}$ are sequences in $(0,\mathrm{\infty })$ so that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and there are $a,b\in \mathbb{R}$ with $0<a\le {\sigma }_n\le b<2\alpha$ for all $n\in \mathbb{N}$.

Then $P_{\mathrm{\Omega }}(I-V+\gamma g)$ has a unique fixed point $p_0$ in Ω, and this $p_0$ is also a unique solution $p_0\in \mathrm{\Omega }$ to the hierarchical variational inequality

$$〈(V-\gamma g)p,q-p〉,\mathrm{\forall }q\in \mathrm{\Omega }.$$

Here we would like to remark that Corollary 3.2 is related to Theorem 8 in Lin and Takahashi [3], although our conditions (3.2.5) and (3.2.6) are different from the corresponding ones in [3].

4 Applications

In this section, we shall apply Theorem 3.1 to study the related equilibrium problem. Let $f:C\times C\to \mathbb{R}$ and $A:C\to \mathcal{H}$. Then a generalized equilibrium problem is the problem of finding $\hat{x}\in C$ such that

$$f(\hat{x},y)+〈A\hat{x},y-\hat{x}〉\ge 0,\mathrm{\forall }y\in C.$$

(19)

The solution set for Eq. (19) is denoted by $EP(f,A)$, that is,

$$EP(f,A)=\{\hat{x}\in C:f(\hat{x},y)+〈A\hat{x},y-\hat{x}〉\ge 0,\mathrm{\forall }y\in C\}.$$

In case $A=0$, problem (19) reduces to the equilibrium problem of finding $\hat{x}\in C$ such that

$$f(\hat{x},y)\ge 0,\mathrm{\forall }y\in C,$$

whose solution set is denoted by $EP(f)$. When $f=0$, the generalized equilibrium problem becomes the variational problem of finding $\hat{x}\in C$ such that

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

whose solution set is denoted by $VI(C,A)$. For solving an equilibrium problem, we assume that the function f satisfies the following conditions:

• (A1) $f(x,x)=0$, $\mathrm{\forall }x\in C$;

• (A2) f is monotone, that is, $f(x,y)+f(y,x)\le 0$, $\mathrm{\forall }x\in C$;

• (A3) for all $x,y,z\in C$, ${lim\hspace{0.17em}sup}_{t\downarrow 0}f((1-t)x+tz,y)\le f(x,y)$;

• (A4) for all $x\in C$, $f(x,\cdot )$ is convex and lower semicontinuous.

The following Lemma 4.1 appears implicitly in Blum and Oettli [11] and is proved in detail by Aoyama et al. [12], while Lemma 4.2 is Lemma 2.12 of Combettes and Hirstoaga [13].

Lemma 4.1 [11, 12]

Let $f:C\times C\to \mathbb{R}$ be a function satisfying conditions (A1)-(A4), and let $r>0$ and $x\in \mathcal{H}$. Then there exists a unique $z\in C$ such that

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

Lemma 4.2 [13]

Let $f:C\times C\to \mathbb{R}$ be a function satisfying conditions (A1)-(A4). For $r>0$, define $J_r^f:\mathcal{H}\to C$ by

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

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

• (a) $J_r^f$ is single-valued;

• (b) $J_r^f$ is firmly nonexpansive;

• (c) $F(J_r^f)=EP(f)$;

• (d) $EP(f)$ is closed and convex.

We call $J_r^f$ the resolvent of f for $r>0$. Using Lemmas 4.1 and 4.2, Takahashi et al. [9] established the lemma below.

Lemma 4.3 [9]

Let $f:C\times C\to \mathbb{R}$ be a function satisfying conditions (A1)-(A4) and define a set-valued mapping of ℋ into itself by

$$G_f(x)=\{\begin{array}{ll}\{z\in \mathcal{H}:f(x,y)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\},& \mathrm{\forall }x\in C,\\ \mathrm{\varnothing },& \mathrm{\forall }x\notin C.\end{array}$$

Then the following hold:

• (a) $G_f$ is a maximal monotone operator with $\mathcal{D}(G_f)\subseteq C$;

• (b) $EP(f)=G_f^{-1}0$;

• (c) $J_r^{G_f}x=J_r^{f}x$ for all $x\in \mathcal{H}$.

Theorem 4.4 Suppose that

(4.4.1) $f:C\times C\to \mathbb{R}$ is a function satisfying conditions (A1)-(A4) and $B:C\to 2^{\mathcal{H}}$ is a maximal monotone operator with $\mathcal{D}(B)\subseteq C$;

(4.4.2) $g:\mathcal{H}\to \mathcal{H}$ is a k-contraction, $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping, and $V:\mathcal{H}\to \mathcal{H}$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$;

(4.4.3) $T:C\to C$ is a λ-hybrid mapping;

(4.4.4) $\mathrm{\Omega }:=F(T)\cap {(A+B)}^{-1}0\cap EP(f)\ne \mathrm{\varnothing }$;

(4.4.5) μ and γ are two real numbers satisfying $0<\mu <\frac{2\overline{\gamma }}{L^2}$ and $0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}$.

Start with any $x_1\in \mathcal{H}$ and define a sequence $\{x_n\}$ iteratively by

$$\{\begin{array}{l}f(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ z_n=J_{{\sigma }_n}^B(I-{\sigma }_{n}A)u_n,n\in \mathbb{N},\\ y_n=\frac{1}{n}{\sum }_{i=0}^{n-1}{T}^i{z}_n,n\in \mathbb{N},\\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\end{array}$$

(20)

where the sequences $\{{\alpha }_n\}$, $\{{\sigma }_n\}$ and $\{r_n\}$ verify the following conditions:

(4.4.6) $\{{\alpha }_n\}$ is a sequence in $[0,1]$ with ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(4.4.7) $\{{\sigma }_n\}$ and $\{r_n\}$ are sequences in $(0,\mathrm{\infty })$ so that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and there are two $a,b\in \mathbb{R}$ with $0<a\le {\sigma }_n\le b<2\alpha$ for all $n\in \mathbb{N}$.

Then the sequence $\{x_n\}$ constructed by algorithm (20) converges strongly to a point $p_0\in \mathrm{\Omega }$, where $p_0$ is the unique fixed point of $P_{\mathrm{\Omega }}(I-V+\gamma g)$, and this point $p_0$ is also a unique solution of the hierarchical inequality

$$〈(V-\gamma g)p,q-p〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

Proof The set-valued mapping $G_f$ associated with f defined in Lemma 4.3 is a maximal monotone operator with $\mathcal{D}(G_f)\subseteq C$, and it follows from Lemmas 4.2 and 4.3 that $F(J_r^{G_f})=F(J_r^f)=EP(f)=G_f^{-1}0$ for any $r>0$. Putting $G=G_f$ in Theorem 3.1, we see that $u_n=J_{r_n}^{G_f}{x}_n$, and so the conclusion follows from Theorem 3.1. □

Since C is a nonempty closed convex subset of ℋ, the indicator function ${\iota }_C$ defined by

$${\iota }_C(x)=\{\begin{array}{ll}0,& x\in C,\\ \mathrm{\infty },& x\notin C\end{array}$$

is a proper lower semicontinuous convex function, and its subdifferential $\partial {\iota }_C$ defined by

$$\partial {\iota }_C(x)=\{z\in \mathcal{H}:〈y-x,z〉\le {\iota }_C(y)-{\iota }_C(x),\mathrm{\forall }y\in \mathcal{H}\}$$

is a maximal monotone operator, cf. Rockafellar [14]. As shown in Lin and Takahashi [3] the resolvent $J_r^{\partial {\iota }_C}$ of $\partial {\iota }_C$ for $r>0$ is the same as the metric projection $P_C$.

Theorem 4.5 Suppose that

(4.5.1) $f:C\times C\to \mathbb{R}$ is a function satisfying conditions (A1)-(A4);

(4.5.2) $g:\mathcal{H}\to \mathcal{H}$ is a k-contraction, $A:C\to \mathcal{H}$ is an α-inverse strongly monotone mapping, and $V:\mathcal{H}\to \mathcal{H}$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian mapping with $\overline{\gamma }>0$ and $L>0$;

(4.5.3) $T:C\to C$ is a λ-hybrid mapping;

(4.5.4) $\mathrm{\Omega }:=F(T)\cap VI(C,A)\ne \mathrm{\varnothing }$;

(4.5.5) μ and γ are two real numbers satisfying $0<\mu <\frac{2\overline{\gamma }}{L^2}$ and $0<\gamma <\frac{\overline{\gamma }-\frac{L^2\mu }{2}}{k}$.

Start with any $x_1\in \mathcal{H}$ and define a sequence $\{x_n\}$ iteratively by

$$\{\begin{array}{l}{z}_n=P_C(I-{\sigma }_{n}A)P_C{x}_n,n\in \mathbb{N},\\ y_n=\frac{1}{n}{\sum }_{i=0}^{n-1}{T}^i{z}_n,n\in \mathbb{N},\\ x_{n+1}={\alpha }_n\gamma g(x_n)+(I-{\alpha }_{n}V)y_n,\end{array}$$

(21)

where the sequences $\{{\alpha }_n\}$, $\{{\sigma }_n\}$ and $\{r_n\}$ verify the following conditions:

(4.5.6) $\{{\alpha }_n\}$ is a sequence in $[0,1]$ with ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

(4.5.7) $\{{\sigma }_n\}$ is a sequence in $(0,\mathrm{\infty })$ so that there are two $a,b\in \mathbb{R}$ with $0<a\le {\sigma }_n\le b<2\alpha$ for all $n\in \mathbb{N}$.

Then the sequence $\{x_n\}$ constructed by algorithm (21) converges strongly to a point $p_0\in \mathrm{\Omega }$, where $p_0$ is the unique fixed point of $P_{\mathrm{\Omega }}(I-V+\gamma g)$, and this point $p_0$ is also a unique solution of the hierarchical inequality

$$〈(V-\gamma g)p,q-p〉\ge 0,\mathrm{\forall }q\in \mathrm{\Omega }.$$

Proof Put $B=G=\partial {\iota }_C$ in Theorem 3.1. Then, for ${\sigma }_n>0$ and $r_n>0$, we have that $J_{{\sigma }_n}^B=J_{r_n}^{\partial {\iota }_C}=P_C$. Furthermore, as shown in Theorem 12 in Lin and Takahashi [3], we have

$${(\partial {\iota }_C)}^{-1}0=C\text{and}{(A+\partial {\iota }_C)}^{-1}0=VI(C,A).$$

Thus we obtain the desired results from Theorem 3.1. □