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# Iterative common solutions for monotone inclusion problems, fixed point problems and equilibrium problems

Fixed Point Theory and Applications20122012:181

https://doi.org/10.1186/1687-1812-2012-181

• Accepted: 30 September 2012
• Published:

## Abstract

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $\alpha >0$, and let A be an α-inverse strongly-monotone mapping of C into H. Let T be a generalized hybrid mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let $0, and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:

$0<\mu <\frac{2\overline{\gamma }}{{L}^{2}},\phantom{\rule{2em}{0ex}}0<\gamma <\frac{\overline{\gamma }-\frac{{L}^{2}\mu }{2}}{k}.$

Suppose that $F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0\ne \mathrm{\varnothing }$, where $F\left(T\right)$ and ${\left(A+B\right)}^{-1}0$, ${W}^{-1}0$ are the set of fixed points of T and the sets of zero points of $A+B$ and W, respectively. In this paper, we prove a strong convergence theorem for finding a point ${z}_{0}$ of $F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$, where ${z}_{0}$ is a unique fixed point of ${P}_{F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right)$. This point ${z}_{0}\in F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$ is also a unique solution of the variational inequality

$〈\left(V-\gamma g\right){z}_{0},q-{z}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0.$

Using this result, we obtain new and well-known strong convergence theorems in a Hilbert space. In particular, we solve a problem posed by Kurokawa and Takahashi (Nonlinear Anal. 73:1562-1568, 2010).

MSC:47H05, 47H10, 58E35.

## Keywords

• maximal monotone operator
• resolvent
• inverse-strongly monotone mapping
• generalized hybrid mapping
• fixed point
• iteration procedure
• equilibrium problem

## 1 Introduction

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let and be the sets of positive integers and real numbers, respectively. A mapping $T:C\to H$ is called generalized hybrid  if there exist $\alpha ,\beta \in \mathbb{R}$ such that
$\alpha {\parallel Tx-Ty\parallel }^{2}+\left(1-\alpha \right){\parallel x-Ty\parallel }^{2}\le \beta {\parallel Tx-y\parallel }^{2}+\left(1-\beta \right){\parallel x-y\parallel }^{2}$
for all $x,y\in C$. We call such a mapping an (α, β)-generalized hybrid mapping. Kocourek, Takahashi and Yao  proved a fixed point theorem for such mappings in a Hilbert space. Furthermore, they proved a nonlinear mean convergence theorem of Baillon’s type  in a Hilbert space. Notice that the mapping above covers several well-known mappings. For example, an (α, β)-generalized hybrid mapping T is nonexpansive for $\alpha =1$ and $\beta =0$, i.e.,
$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
It is also nonspreading [3, 4] for $\alpha =2$ and $\beta =1$, i.e.,
$2{\parallel Tx-Ty\parallel }^{2}\le {\parallel Tx-y\parallel }^{2}+{\parallel Ty-x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
Furthermore, it is hybrid  for $\alpha =\frac{3}{2}$ and $\beta =\frac{1}{2}$, i.e.,
$3{\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}+{\parallel Tx-y\parallel }^{2}+{\parallel Ty-x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in C.$
We can also show that if $x=Tx$, then for any $y\in C$,
$\alpha {\parallel x-Ty\parallel }^{2}+\left(1-\alpha \right){\parallel x-Ty\parallel }^{2}\le \beta {\parallel x-y\parallel }^{2}+\left(1-\beta \right){\parallel x-y\parallel }^{2},$

and hence $\parallel x-Ty\parallel \le \parallel x-y\parallel$. This means that an (α, β)-generalized hybrid mapping with a fixed point is quasi-nonexpansive. The following strong convergence theorem of Halpern’s type  was proved by Wittmann ; see also .

Theorem 1 Let C be a nonempty closed convex subset of H, and let T be a nonexpansive mapping of C into itself with $F\left(T\right)\ne \mathrm{\varnothing }$. For any ${x}_{1}=x\in C$, define a sequence $\left\{{x}_{n}\right\}$ in C by
${x}_{n+1}={\alpha }_{n}x+\left(1-{\alpha }_{n}\right)T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},$
where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfies
$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}|{\alpha }_{n}-{\alpha }_{n+1}|<\mathrm{\infty }.$

Then $\left\{{x}_{n}\right\}$ converges strongly to a fixed point of T.

Kurokawa and Takahashi  also proved the following strong convergence theorem for nonspreading mappings in a Hilbert space; see also Hojo and Takahashi  for generalized hybrid mappings.

Theorem 2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let T be a nonspreading mapping of C into itself. Let $u\in C$ and define two sequences $\left\{{x}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ in C as follows: ${x}_{1}=x\in C$ and
$\left\{\begin{array}{c}{x}_{n+1}={\alpha }_{n}u+\left(1-{\alpha }_{n}\right){z}_{n},\hfill \\ {z}_{n}=\frac{1}{n}{\sum }_{k=0}^{n-1}{T}^{k}{x}_{n}\hfill \end{array}$

for all $n=1,2,\dots$, where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$. If $F\left(T\right)$ is nonempty, then $\left\{{x}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ converge strongly to Pu, where P is the metric projection of H onto $F\left(T\right)$.

Remark We do not know whether Theorem 1 for nonspreading mappings holds or not; see  and .

In this paper, we provide a strong convergence theorem for finding a point ${z}_{0}$ of $F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$ such that it is a unique fixed point of
${P}_{F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right)$
and a unique solution of the variational inequality
$〈\left(V-\gamma g\right){z}_{0},q-{z}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0,$

where T, A, B, W, g and V denote a generalized hybrid mapping of C into H, an α-inverse strongly-monotone mapping of C into H with $\alpha >0$, maximal monotone operators on H such that the domains of B and W are included in C, a k-contraction of H into itself with $0 and a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$, respectively. Using this result, we obtain new and well-known strong convergence theorems in a Hilbert space. In particular, we solve a problem posed by Kurokawa and Takahashi .

## 2 Preliminaries

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm , respectively. When $\left\{{x}_{n}\right\}$ is a sequence in H, we denote the strong convergence of $\left\{{x}_{n}\right\}$ to $x\in H$ by ${x}_{n}\to x$ and the weak convergence by ${x}_{n}⇀x$. We have from  that for any $x,y\in H$ and $\lambda \in \mathbb{R}$,
${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,x+y〉$
(2.1)
and
${\parallel \lambda x+\left(1-\lambda \right)y\parallel }^{2}=\lambda {\parallel x\parallel }^{2}+\left(1-\lambda \right){\parallel y\parallel }^{2}-\lambda \left(1-\lambda \right){\parallel x-y\parallel }^{2}.$
(2.2)
Furthermore, we have that for $x,y,u,v\in H$,
$2〈x-y,u-v〉={\parallel x-v\parallel }^{2}+{\parallel y-u\parallel }^{2}-{\parallel x-u\parallel }^{2}-{\parallel y-v\parallel }^{2}.$
(2.3)

if ${x}_{n}⇀u$ and $u\ne v$; see . Let C be a nonempty closed convex subset of a Hilbert space H, and let $T:C\to H$ be a mapping. We denote by $F\left(T\right)$ the set of fixed points for T. A mapping $T:C\to H$ is called quasi-nonexpansive if $F\left(T\right)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in C$ and $y\in F\left(T\right)$. If $T:C\to H$ is quasi-nonexpansive, then $F\left(T\right)$ is closed and convex; see . For a nonempty closed convex subset C of H, the nearest point projection of H onto C is denoted by ${P}_{C}$, that is, for all $x\in H$ and $y\in C$. Such ${P}_{C}$ is called the metric projection of H onto C. We know that the metric projection ${P}_{C}$ is firmly nonexpansive; for all $x,y\in H$. Furthermore, $〈x-{P}_{C}x,y-{P}_{C}x〉\le 0$ holds for all $x\in H$ and $y\in C$; see . The following result is in .

Lemma 3 Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let $T:C\to H$ be a generalized hybrid mapping. Suppose that there exists $\left\{{x}_{n}\right\}\subset C$ such that ${x}_{n}⇀z$ and ${x}_{n}-T{x}_{n}\to 0$. Then $z\in F\left(T\right)$.

Let B be a mapping of H into ${2}^{H}$. The effective domain of B is denoted by $D\left(B\right)$, that is, $D\left(B\right)=\left\{x\in H:Bx\ne \mathrm{\varnothing }\right\}$. A multi-valued mapping B is said to be a monotone operator on H if $〈x-y,u-v〉\ge 0$ for all $x,y\in D\left(B\right)$, $u\in Bx$, 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 ${J}_{r}={\left(I+rB\right)}^{-1}:H\to D\left(B\right)$, which is called the resolvent of B for r. We denote by ${A}_{r}=\frac{1}{r}\left(I-{J}_{r}\right)$ the Yosida approximation of B for $r>0$. We know from  that
${A}_{r}x\in B{J}_{r}x,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H,\phantom{\rule{0.25em}{0ex}}r>0.$
(2.5)
Let B be a maximal monotone operator on H, and let ${B}^{-1}0=\left\{x\in H:0\in Bx\right\}$. It is known that ${B}^{-1}0=F\left({J}_{r}\right)$ for all $r>0$ and the resolvent ${J}_{r}$ is firmly nonexpansive, i.e.,
${\parallel {J}_{r}x-{J}_{r}y\parallel }^{2}\le 〈x-y,{J}_{r}x-{J}_{r}y〉,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H.$
(2.6)

We also know the following lemma from .

Lemma 4 Let H be a real Hilbert space, and let B be a maximal monotone operator on H. For $r>0$ and $x\in H$, define the resolvent ${J}_{r}x$. Then the following holds:
$\frac{s-t}{s}〈{J}_{s}x-{J}_{t}x,{J}_{s}x-x〉\ge {\parallel {J}_{s}x-{J}_{t}x\parallel }^{2}$

for all $s,t>0$ and $x\in H$.

for all $\lambda ,\mu >0$ and $x\in H$; see also [14, 17]. To prove our main result, we need the following lemmas.

Let $\left\{{s}_{n}\right\}$ be a sequence of nonnegative real numbers, let $\left\{{\alpha }_{n}\right\}$ be a sequence of $\left[0,1\right]$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, let $\left\{{\beta }_{n}\right\}$ be a sequence of nonnegative real numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}<\mathrm{\infty }$, and let $\left\{{\gamma }_{n}\right\}$ be a sequence of real numbers with ${lim sup}_{n\to \mathrm{\infty }}{\gamma }_{n}\le 0$. Suppose that
${s}_{n+1}\le \left(1-{\alpha }_{n}\right){s}_{n}+{\alpha }_{n}{\gamma }_{n}+{\beta }_{n}$

for all $n=1,2,\dots$. Then ${lim}_{n\to \mathrm{\infty }}{s}_{n}=0$.

Lemma 6 ()

Let $\left\{{\mathrm{\Gamma }}_{n}\right\}$ be a sequence of real numbers that does not decrease at infinity in the sense that there exists a subsequence $\left\{{\mathrm{\Gamma }}_{{n}_{i}}\right\}$ of $\left\{{\mathrm{\Gamma }}_{n}\right\}$ which satisfies ${\mathrm{\Gamma }}_{{n}_{i}}<{\mathrm{\Gamma }}_{{n}_{i}+1}$ for all $i\in \mathbb{N}$. Define the sequence ${\left\{\tau \left(n\right)\right\}}_{n\ge {n}_{0}}$ of integers as follows:
$\tau \left(n\right)=max\left\{k\le n:{\mathrm{\Gamma }}_{k}<{\mathrm{\Gamma }}_{k+1}\right\},$
where ${n}_{0}\in \mathbb{N}$ such that $\left\{k\le {n}_{0}:{\mathrm{\Gamma }}_{k}<{\mathrm{\Gamma }}_{k+1}\right\}\ne \mathrm{\varnothing }$. Then the following hold:
1. (i)

$\tau \left({n}_{0}\right)\le \tau \left({n}_{0}+1\right)\le \cdots$ and $\tau \left(n\right)\to \mathrm{\infty }$;

2. (ii)

${\mathrm{\Gamma }}_{\tau \left(n\right)}\le {\mathrm{\Gamma }}_{\tau \left(n\right)+1}$ and ${\mathrm{\Gamma }}_{n}\le {\mathrm{\Gamma }}_{\tau \left(n\right)+1}$, $\mathrm{\forall }n\in \mathbb{N}$.

## 3 Strong convergence theorems

Let H be a real Hilbert space. A mapping $g:H\to H$ is a contraction if there exists $k\in \left(0,1\right)$ such that $\parallel g\left(x\right)-g\left(y\right)\parallel \le k\parallel x-y\parallel$ for all $x,y\in H$. We call such a mapping g a k-contraction. A nonlinear operator $V:H\to H$ is called strongly monotone if there exists $\overline{\gamma }>0$ such that $〈x-y,Vx-Vy〉\ge \overline{\gamma }{\parallel x-y\parallel }^{2}$ for all $x,y\in H$. Such V is also called $\overline{\gamma }$-strongly monotone. A nonlinear operator $V:H\to H$ is called Lipschitzian continuous if there exists $L>0$ such that $\parallel Vx-Vy\parallel \le L\parallel x-y\parallel$ for all $x,y\in H$. Such V is also called L-Lipschitzian continuous. We know the following three lemmas in a Hilbert space; see Lin and Takahashi .

Lemma 7 ()

Let H be a Hilbert space, and let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator on H with $\overline{\gamma }>0$ and $L>0$. Let $t>0$ satisfy $2\overline{\gamma }>t{L}^{2}$ and $1>2t\overline{\gamma }$. Then $0<1-t\left(2\overline{\gamma }-t{L}^{2}\right)<1$ and $I-tV:H\to H$ is a contraction, where I is the identity operator on H.

Lemma 8 ()

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let ${P}_{C}$ be the metric projection of H onto C, and let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator on H with $\overline{\gamma }>0$ and $L>0$. Let $t>0$ satisfy $2\overline{\gamma }>t{L}^{2}$ and $1>2t\overline{\gamma }$, and let $z\in C$. Then the following are equivalent:
1. (1)

$z={P}_{C}\left(I-tV\right)z$;

2. (2)

$〈Vz,y-z〉\ge 0$, $\mathrm{\forall }y\in C$;

3. (3)

$z={P}_{C}\left(I-V\right)z$.

Such $z\in C$ always exists and is unique.

Lemma 9 ()

Let H be a Hilbert space, and let $g:H\to H$ be a k-contraction with $0. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator on H with $\overline{\gamma }>0$ and $L>0$. Let a real number γ satisfy $0<\gamma <\frac{\overline{\gamma }}{k}$. Then $V-\gamma g:H\to H$ is a $\left(\overline{\gamma }-\gamma k\right)$-strongly monotone and $\left(L+\gamma k\right)$-Lipschitzian continuous mapping. Furthermore, let C be a nonempty closed convex subset of H. Then ${P}_{C}\left(I-V+\gamma g\right)$ has a unique fixed point ${z}_{0}$ in C. This point ${z}_{0}\in C$ is also a unique solution of the variational inequality
$〈\left(V-\gamma g\right){z}_{0},q-{z}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in C.$

Now, we prove the following strong convergence theorem of Halpern’s type  for finding a common solution of a monotone inclusion problem for the sum of two monotone mappings, of a fixed point problem for generalized hybrid mappings and of an equilibrium problem for bifunctions in a Hilbert space.

Theorem 10 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $\alpha >0$, and let A be an α-inverse strongly-monotone mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let ${J}_{\lambda }={\left(I+\lambda B\right)}^{-1}$ and ${T}_{r}={\left(I+rW\right)}^{-1}$ be resolvents of B and W for $\lambda >0$ and $r>0$, respectively. Let S be a generalized hybrid mapping of C into H. Let $0, and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:
$0<\mu <\frac{2\overline{\gamma }}{{L}^{2}},\phantom{\rule{2em}{0ex}}0<\gamma <\frac{\overline{\gamma }-\frac{{L}^{2}\mu }{2}}{k}.$
Suppose $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0\ne \mathrm{\varnothing }$. Let ${x}_{1}=x\in H$, and let $\left\{{x}_{n}\right\}\subset H$ be a sequence generated by
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}\gamma g\left({x}_{n}\right)+\left(I-{\alpha }_{n}V\right)S{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){T}_{{r}_{n}}{x}_{n}\right\}$
for all $n\in \mathbb{N}$, where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ satisfy
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty },\phantom{\rule{2em}{0ex}}0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1,\hfill \\ \underset{n\to \mathrm{\infty }}{lim inf}{r}_{n}>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0

Then $\left\{{x}_{n}\right\}$ converges strongly to ${z}_{0}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$, where ${z}_{0}$ is a unique fixed point in $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$ of ${P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right)$.

Proof Let $z\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$. We have that $z=Sz$, $z={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right)z$ and $z={T}_{{r}_{n}}z$. Putting ${w}_{n}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){T}_{{r}_{n}}{x}_{n}$ and ${u}_{n}={T}_{{r}_{n}}{x}_{n}$, we obtain that
Put $\tau =\overline{\gamma }-\frac{{L}^{2}\mu }{2}$. Using ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, we have that for any $x,y\in H$,
Since $1-{\alpha }_{n}\tau >0$, we obtain that for any $x,y\in H$,
$\parallel \left(I-{\alpha }_{n}V\right)x-\left(I-{\alpha }_{n}V\right)y\parallel \le \left(1-{\alpha }_{n}\tau \right)\parallel x-y\parallel .$
(3.3)
Putting ${y}_{n}={\alpha }_{n}\gamma g\left({x}_{n}\right)+\left(I-{\alpha }_{n}V\right)S{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){T}_{{r}_{n}}{x}_{n}$, from $z={\alpha }_{n}Vz+z-{\alpha }_{n}Vz$, (3.1) and (3.3) we have that
Putting $K=max\left\{\parallel {x}_{1}-z\parallel ,\frac{\parallel \gamma g\left(z\right)-Vz\parallel }{\tau -\gamma k}\right\}$, we have that $\parallel {x}_{n}-z\parallel \le K$ for all $n\in \mathbb{N}$. Then $\left\{{x}_{n}\right\}$ is bounded. Furthermore, $\left\{{u}_{n}\right\}$, $\left\{{w}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are bounded. Using Lemma 9, we can take a unique ${z}_{0}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$ such that
${z}_{0}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right){z}_{0}.$
From the definition of $\left\{{x}_{n}\right\}$, we have that
${x}_{n+1}-{x}_{n}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}\gamma g\left({x}_{n}\right)+\left(I-{\alpha }_{n}V\right)S{w}_{n}\right\}-{x}_{n}$
and hence
$\begin{array}{rcl}{x}_{n+1}-{x}_{n}-\left(1-{\beta }_{n}\right){\alpha }_{n}\gamma g\left({x}_{n}\right)& =& {\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left(I-{\alpha }_{n}V\right)S{w}_{n}-{x}_{n}\\ =& \left(1-{\beta }_{n}\right)\left\{\left(I-{\alpha }_{n}V\right)S{w}_{n}-{x}_{n}\right\}\\ =& \left(1-{\beta }_{n}\right)\left(S{w}_{n}-{x}_{n}-{\alpha }_{n}VS{w}_{n}\right).\end{array}$
From (2.3) and (3.1), we have that
$\begin{array}{rcl}2〈{x}_{n}-S{w}_{n},{x}_{n}-{z}_{0}〉& =& {\parallel {x}_{n}-{z}_{0}\parallel }^{2}+{\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-{\parallel S{w}_{n}-{z}_{0}\parallel }^{2}\\ \ge & {\parallel {x}_{n}-{z}_{0}\parallel }^{2}+{\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-{\parallel {x}_{n}-{z}_{0}\parallel }^{2}\\ =& {\parallel S{w}_{n}-{x}_{n}\parallel }^{2}.\end{array}$
(3.5)
From (3.4) and (3.5), we also have that
$\begin{array}{rcl}-2〈{x}_{n}-{x}_{n+1},{x}_{n}-{z}_{0}〉& =& 2\left(1-{\beta }_{n}\right){\alpha }_{n}〈\gamma g\left({x}_{n}\right),{x}_{n}-{z}_{0}〉\\ -2\left(1-{\beta }_{n}\right)〈{x}_{n}-S{w}_{n},{x}_{n}-{z}_{0}〉-2\left(1-{\beta }_{n}\right){\alpha }_{n}〈VS{w}_{n},{x}_{n}-{z}_{0}〉\\ \le & 2\left(1-{\beta }_{n}\right){\alpha }_{n}〈\gamma g\left({x}_{n}\right),{x}_{n}-{z}_{0}〉\\ -\left(1-{\beta }_{n}\right){\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-2\left(1-{\beta }_{n}\right){\alpha }_{n}〈VS{w}_{n},{x}_{n}-{z}_{0}〉.\end{array}$
(3.6)
Furthermore, using (2.3) and (3.6), we have that
$\begin{array}{c}{\parallel {x}_{n+1}-{z}_{0}\parallel }^{2}-{\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}-{\parallel {x}_{n}-{z}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\beta }_{n}\right){\alpha }_{n}〈\gamma g\left({x}_{n}\right),{x}_{n}-{z}_{0}〉\hfill \\ \phantom{\rule{2em}{0ex}}-\left(1-{\beta }_{n}\right){\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-2\left(1-{\beta }_{n}\right){\alpha }_{n}〈VS{w}_{n},{x}_{n}-{z}_{0}〉.\hfill \end{array}$
Setting ${\mathrm{\Gamma }}_{n}={\parallel {x}_{n}-{z}_{0}\parallel }^{2}$, we have that
Noting that
$\begin{array}{rcl}\parallel {x}_{n+1}-{x}_{n}\parallel & =& \parallel \left(1-{\beta }_{n}\right){\alpha }_{n}\left(\gamma g\left({x}_{n}\right)-VS{w}_{n}\right)+\left(1-{\beta }_{n}\right)\left(S{w}_{n}-{x}_{n}\right)\parallel \\ \le & \left(1-{\beta }_{n}\right)\left(\parallel S{w}_{n}-{x}_{n}\parallel +{\alpha }_{n}\parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel \right),\end{array}$
(3.8)
we have that
$\begin{array}{rcl}{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}& \le & {\left(1-{\beta }_{n}\right)}^{2}{\left(\parallel S{w}_{n}-{x}_{n}\parallel +{\alpha }_{n}\parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel \right)}^{2}\\ =& {\left(1-{\beta }_{n}\right)}^{2}{\parallel S{w}_{n}-{x}_{n}\parallel }^{2}+{\left(1-{\beta }_{n}\right)}^{2}2{\alpha }_{n}\parallel S{w}_{n}-{x}_{n}\parallel \parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel \\ +{\left(1-{\beta }_{n}\right)}^{2}{\alpha }_{n}^{2}{\parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel }^{2}.\end{array}$
(3.9)
Thus, we have from (3.7) and (3.9) that
$\begin{array}{rcl}{\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_{n}& \le & {\parallel {x}_{n}-{x}_{n+1}\parallel }^{2}+2\left(1-{\beta }_{n}\right){\alpha }_{n}〈\gamma g\left({x}_{n}\right),{x}_{n}-{z}_{0}〉\\ -\left(1-{\beta }_{n}\right){\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-2\left(1-{\beta }_{n}\right){\alpha }_{n}〈VS{w}_{n},{x}_{n}-{z}_{0}〉\\ \le & {\left(1-{\beta }_{n}\right)}^{2}{\parallel S{w}_{n}-{x}_{n}\parallel }^{2}+{\left(1-{\beta }_{n}\right)}^{2}2{\alpha }_{n}\parallel S{w}_{n}-{x}_{n}\parallel \parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel \\ +{\left(1-{\beta }_{n}\right)}^{2}{\alpha }_{n}^{2}{\parallel \gamma g\left({x}_{n}\right)-VS{w}_{n}\parallel }^{2}+2\left(1-{\beta }_{n}\right){\alpha }_{n}〈\gamma g\left({x}_{n}\right),{x}_{n}-{z}_{0}〉\\ -\left(1-{\beta }_{n}\right){\parallel S{w}_{n}-{x}_{n}\parallel }^{2}-2\left(1-{\beta }_{n}\right){\alpha }_{n}〈VS{w}_{n},{x}_{n}-{z}_{0}〉\end{array}$

We divide the proof into two cases.

Case 1: Suppose that ${\mathrm{\Gamma }}_{n+1}\le {\mathrm{\Gamma }}_{n}$ for all $n\in \mathbb{N}$. In this case, ${lim}_{n\to \mathrm{\infty }}{\mathrm{\Gamma }}_{n}$ exists and then ${lim}_{n\to \mathrm{\infty }}\left({\mathrm{\Gamma }}_{n+1}-{\mathrm{\Gamma }}_{n}\right)=0$. Using $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$ and ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, we have from (3.10) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-{x}_{n}\parallel =0.$
(3.11)
Using (3.8), we also have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0.$
(3.12)
Since ${x}_{n+1}-{x}_{n}=\left(1-{\beta }_{n}\right)\left({y}_{n}-{x}_{n}\right)$, we have from (3.12) that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0.$
(3.13)
We also have from (2.6) that
$\begin{array}{rcl}2{\parallel {u}_{n}-{z}_{0}\parallel }^{2}& =& 2{\parallel {T}_{{r}_{n}}{x}_{n}-{T}_{{r}_{n}}{z}_{0}\parallel }^{2}\\ \le & 2〈{x}_{n}-{z}_{0},{u}_{n}-{z}_{0}〉\\ =& {\parallel {x}_{n}-{z}_{0}\parallel }^{2}+{\parallel {u}_{n}-{z}_{0}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\end{array}$
and hence
${\parallel {u}_{n}-{z}_{0}\parallel }^{2}\le {\parallel {x}_{n}-{z}_{0}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}.$
(3.14)
From (3.1) we have that
${\parallel S{w}_{n}-{z}_{0}\parallel }^{2}\le {\parallel {u}_{n}-{z}_{0}\parallel }^{2}\le {\parallel {x}_{n}-{z}_{0}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}$
and hence
${\parallel {u}_{n}-{x}_{n}\parallel }^{2}\le {\parallel {x}_{n}-{z}_{0}\parallel }^{2}-{\parallel S{w}_{n}-{z}_{0}\parallel }^{2}\le M{\parallel S{w}_{n}-{x}_{n}\parallel }^{2},$
where $M=sup\left\{\parallel {x}_{n}-{z}_{0}\parallel +\parallel S{w}_{n}-{z}_{0}\parallel :n\in \mathbb{N}\right\}$. Thus, from (3.11) we have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{x}_{n}\parallel =0.$
(3.15)
We show ${lim}_{n\to \mathrm{\infty }}\parallel S{w}_{n}-{w}_{n}\parallel =0$. Since ${\parallel \cdot \parallel }^{2}$ is a convex function, we have that
From ${z}_{0}={\alpha }_{n}V{z}_{0}+{z}_{0}-{\alpha }_{n}V{z}_{0}$ and (2.1), we also have that
Then we have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel A{u}_{n}-A{z}_{0}\parallel =0.$
(3.20)
Since ${J}_{{\lambda }_{n}}$ is firmly nonexpansive, we have that
Since $\parallel S{w}_{n}-{w}_{n}\parallel \le \parallel S{w}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-{w}_{n}\parallel$, we have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-{w}_{n}\parallel =0.$
(3.24)
Take ${\lambda }_{0}\in \mathbb{R}$ with $0 arbitrarily. Put ${s}_{n}=\left(I-{\lambda }_{n}A\right){u}_{n}$. Using ${u}_{n}={T}_{{r}_{n}}{x}_{n}$ and ${w}_{n}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){u}_{n}$, we have from Lemma 4 that
$\begin{array}{rcl}\parallel {J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{n}-{w}_{n}\parallel & =& \parallel {J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{n}-{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){u}_{n}\parallel \\ =& \parallel {J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{n}-{J}_{{\lambda }_{0}}\left(I-{\lambda }_{n}A\right){u}_{n}\\ +{J}_{{\lambda }_{0}}\left(I-{\lambda }_{n}A\right){u}_{n}-{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){u}_{n}\parallel \\ \le & \parallel \left(I-{\lambda }_{0}A\right){u}_{n}-\left(I-{\lambda }_{n}A\right){u}_{n}\parallel +\parallel {J}_{{\lambda }_{0}}{s}_{n}-{J}_{{\lambda }_{n}}{s}_{n}\parallel \\ \le & |{\lambda }_{0}-{\lambda }_{n}|\parallel A{u}_{n}\parallel +\frac{|{\lambda }_{0}-{\lambda }_{n}|}{{\lambda }_{0}}\parallel {J}_{{\lambda }_{0}}{s}_{n}-{s}_{n}\parallel .\end{array}$
(3.25)
We also have from (3.25) that
$\parallel {u}_{n}-{J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{n}\parallel \le \parallel {u}_{n}-{w}_{n}\parallel +\parallel {w}_{n}-{J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{n}\parallel .$
(3.26)

We will use (3.25) and (3.26) later.

Let us show that ${lim sup}_{n\to \mathrm{\infty }}〈\left(V-\gamma g\right){z}_{0},{x}_{n}-{z}_{0}〉\ge 0$. Put
$A=\underset{n\to \mathrm{\infty }}{lim sup}〈\left(V-\gamma g\right){z}_{0},{x}_{n}-{z}_{0}〉.$
Without loss of generality, we may assume that there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that $A={lim}_{i\to \mathrm{\infty }}〈\left(V-\gamma g\right){z}_{0},{x}_{{n}_{i}}-{z}_{0}〉$ and $\left\{{x}_{{n}_{i}}\right\}$ converges weakly to some point $w\in H$. From $\parallel {x}_{n}-{w}_{n}\parallel \to 0$ and $\parallel {x}_{n}-{u}_{n}\parallel \to 0$, we also have that $\left\{{w}_{{n}_{i}}\right\}$ and $\left\{{u}_{{n}_{i}}\right\}$ converge weakly to $w\in C$. On the other hand, from $\left\{{\lambda }_{{n}_{i}}\right\}\subset \left[a,b\right]$ there exists a subsequence $\left\{{\lambda }_{{n}_{{i}_{j}}}\right\}$ of $\left\{{\lambda }_{{n}_{i}}\right\}$ such that ${\lambda }_{{n}_{{i}_{j}}}\to {\lambda }_{0}$ for some ${\lambda }_{0}\in \left[a,b\right]$. Without loss of generality, we assume that ${w}_{{n}_{i}}\to w$, ${u}_{{n}_{i}}\to w$ and ${\lambda }_{{n}_{i}}\to {\lambda }_{0}$. From (3.24) we know ${lim}_{n\to \mathrm{\infty }}\parallel S{w}_{n}-{w}_{n}\parallel =0$. Thus, we have from Lemma 3 that $w=Sw$. Since W is a monotone operator and $\frac{{x}_{{n}_{i}}-{u}_{{n}_{i}}}{{r}_{{n}_{i}}}\in W{u}_{{n}_{i}}$, we have that for any $\left(u,v\right)\in W$,
$〈u-{u}_{{n}_{i}},v-\frac{{x}_{{n}_{i}}-{u}_{{n}_{i}}}{{r}_{{n}_{i}}}〉\ge 0.$
Since ${lim inf}_{n\to \mathrm{\infty }}{r}_{n}>0$, ${u}_{{n}_{i}}⇀w$ and ${x}_{{n}_{i}}-{u}_{{n}_{i}}\to 0$, we have
$〈u-w,v〉\ge 0.$
Since W is a maximal monotone operator, we have $0\in Ww$ and hence $w\in {W}^{-1}0$. Since ${\lambda }_{{n}_{i}}\to {\lambda }_{0}$, we have from (3.25) that
$\parallel {J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{{n}_{i}}-{w}_{{n}_{i}}\parallel \to 0.$
Furthermore, we have from (3.26) that
$\parallel {u}_{{n}_{i}}-{J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right){u}_{{n}_{i}}\parallel \to 0.$
Since ${J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right)$ is nonexpansive, we have that $w={J}_{{\lambda }_{0}}\left(I-{\lambda }_{0}A\right)w$. This means that $0\in Aw+Bw$. Thus, we have
$w\in F\left(T\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0.$
Then we have
$A=\underset{i\to \mathrm{\infty }}{lim}〈\left(V-\gamma g\right){z}_{0},{x}_{{n}_{i}}-{z}_{0}〉=〈\left(V-\gamma g\right){z}_{0},w-{z}_{0}〉\ge 0.$
(3.27)
Since ${y}_{n}-{z}_{0}={\alpha }_{n}\left(\gamma g\left({x}_{n}\right)-V{z}_{0}\right)+\left(I-{\alpha }_{n}V\right)S{w}_{n}-\left(I-{\alpha }_{n}V\right){z}_{0}$, we have
By (3.27) and Lemma 5, we obtain that ${x}_{n}\to {z}_{0}$, where
${z}_{0}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right){z}_{0}.$
Case 2: Suppose that there exists a subsequence $\left\{{\mathrm{\Gamma }}_{{n}_{i}}\right\}\subset \left\{{\mathrm{\Gamma }}_{n}\right\}$ such that ${\mathrm{\Gamma }}_{{n}_{i}}<{\mathrm{\Gamma }}_{{n}_{i}+1}$ for all $i\in \mathbb{N}$. In this case, we define $\tau :\mathbb{N}\to \mathbb{N}$ by
$\tau \left(n\right)=max\left\{k\le n:{\mathrm{\Gamma }}_{k}<{\mathrm{\Gamma }}_{k+1}\right\}.$
Then we have from Lemma 6 that ${\mathrm{\Gamma }}_{\tau \left(n\right)}<{\mathrm{\Gamma }}_{\tau \left(n\right)+1}$. Thus, we have from (3.10) that for all $n\in \mathbb{N}$,
Using ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$ and $0<{lim inf}_{n\to \mathrm{\infty }}{\beta }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\beta }_{n}<1$, we have from (3.28) and Lemma 6 that
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{\tau }\left(n\right)-{x}_{\tau }\left(n\right)\parallel =0.$
(3.29)
As in the proof of Case 1, we also have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{\tau \left(n\right)+1}-{x}_{\tau \left(n\right)}\parallel =0$
(3.30)
and
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel =0.$
(3.31)
Furthermore, we have that ${lim}_{n\to \mathrm{\infty }}\parallel {u}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)}\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel A{u}_{\tau \left(n\right)}-A{z}_{0}\parallel =0$, and . From these we have that ${lim}_{n\to \mathrm{\infty }}\parallel S{w}_{\tau \left(n\right)}-{w}_{\tau \left(n\right)}\parallel =0$. As in the proof of Case 1, we can show that
$\underset{n\to \mathrm{\infty }}{lim sup}〈\left(V-\gamma g\right){z}_{0},{x}_{\tau \left(n\right)}-{z}_{0}〉\ge 0.$
From ${\mathrm{\Gamma }}_{\tau \left(n\right)}<{\mathrm{\Gamma }}_{\tau \left(n\right)+1}$, we have that
Since $\left(1-{\beta }_{\tau \left(n\right)}\right){\alpha }_{\tau \left(n\right)}>0$, we have that
and hence $\parallel {x}_{\tau \left(n\right)}-{z}_{0}\parallel \to 0$ as $n\to \mathrm{\infty }$. Since ${x}_{\tau \left(n\right)}-{x}_{\tau \left(n\right)+1}\to 0$, we have $\parallel {x}_{\tau \left(n\right)+1}-{z}_{0}\parallel \to 0$ as $n\to \mathrm{\infty }$. Using Lemma 6 again, we obtain that
$\parallel {x}_{n}-{z}_{0}\parallel \le \parallel {x}_{\tau \left(n\right)+1}-{z}_{0}\parallel \to 0$

as $n\to \mathrm{\infty }$. This completes the proof. □

## 4 Applications

In this section, using Theorem 10, we can obtain well-known and new strong convergence theorems in a Hilbert space. Let H be a Hilbert space, and let f be a proper lower semicontinuous convex function of H into $\left(-\mathrm{\infty },\mathrm{\infty }\right]$. Then the subdifferential ∂f of f is defined as follows:
$\partial f\left(x\right)=\left\{z\in H:f\left(x\right)+〈z,y-x〉\le f\left(y\right),\mathrm{\forall }y\in H\right\}$
for all $x\in H$. From Rockafellar , we know that ∂f is a maximal monotone operator. Let C be a nonempty closed convex subset of H, and let ${i}_{C}$ be the indicator function of C, i.e.,
${i}_{C}\left(x\right)=\left\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ \mathrm{\infty },\hfill & x\notin C.\hfill \end{array}$
Then, ${i}_{C}$ is a proper lower semicontinuous convex function on H. So, we can define the resolvent ${J}_{\lambda }$ of $\partial {i}_{C}$ for $\lambda >0$, i.e.,
${J}_{\lambda }x={\left(I+\lambda \partial {i}_{C}\right)}^{-1}x$

for all $x\in H$. We know that ${J}_{\lambda }x={P}_{C}x$ for all $x\in H$ and $\lambda >0$; see .

Theorem 11 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a generalized hybrid mapping of C into C. Suppose $F\left(S\right)\ne \mathrm{\varnothing }$. Let $u,{x}_{1}\in C$, and let $\left\{{x}_{n}\right\}\subset C$ be a sequence generated by
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}u+\left(1-{\alpha }_{n}\right)S{x}_{n}\right\}$
for all $n\in \mathbb{N}$, where $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$ and $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ satisfy
$\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$
and
$0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1.$

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${z}_{0}\in F\left(S\right)$, where ${z}_{0}={P}_{F\left(S\right)}u$.

Proof Put $A=0$, $B=W=\partial {i}_{C}$ and ${\lambda }_{n}={r}_{n}=1$ for all $n\in \mathbb{N}$ in Theorem 10. Then we have ${J}_{{\lambda }_{n}}={T}_{{r}_{n}}={P}_{C}$ for all $n\in \mathbb{N}$. Furthermore, put $g\left(x\right)=u$ and $V\left(x\right)=x$ for all $x\in H$. Then we can take $\overline{\gamma }=L=1$. Thus, we can take $\mu =1$. On the other hand, since $\parallel g\left(x\right)-g\left(y\right)\parallel =0\le \frac{1}{3}\parallel x-y\parallel$ for all $x,y\in H$, we can take $k=\frac{1}{3}$. So, we can take $\gamma =1$. Then for $u,{x}_{1}\in C$, we get that
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}u+\left(I-{\alpha }_{n}\right)S{x}_{n}\right\}$
for all $n\in \mathbb{N}$. So, we have $\left\{{x}_{n}\right\}\subset C$. We also have
${z}_{0}={P}_{F\left(S\right)\cap C}\left(I-V+\gamma g\right){z}_{0}={P}_{F\left(S\right)}\left({z}_{0}-{z}_{0}+1\cdot u\right)={P}_{F\left(S\right)}u.$

Thus, we obtain the desired result by Theorem 10. □

Theorem 11 solves the problem posed by Kurokawa and Takahashi . The following result is a strong convergence theorem of Halpern’s type  for finding a common solution of a monotone inclusion problem for the sum of two monotone mappings, of a fixed point problem for nonexpansive mappings and of an equilibrium problem for bifunctions in a Hilbert space.

Theorem 12 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let $\alpha >0$, and let A be an α-inverse strongly-monotone mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let ${J}_{\lambda }={\left(I+\lambda B\right)}^{-1}$ and ${T}_{r}={\left(I+rW\right)}^{-1}$ be resolvents of B and W for $\lambda >0$ and $r>0$, respectively. Let S be a nonexpansive mapping of C into H. Let $0, and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:
$0<\mu <\frac{2\overline{\gamma }}{{L}^{2}},\phantom{\rule{2em}{0ex}}0<\gamma <\frac{\overline{\gamma }-\frac{{L}^{2}\mu }{2}}{k}.$
Suppose $F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0\ne \mathrm{\varnothing }$. Let ${x}_{1}=x\in H$, and let $\left\{{x}_{n}\right\}\subset H$ be a sequence generated by
${x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}\gamma g\left({x}_{n}\right)+\left(I-{\alpha }_{n}V\right)S{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}A\right){T}_{{r}_{n}}{x}_{n}\right\}$
for all $n\in \mathbb{N}$, where $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ satisfy
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty },\phantom{\rule{2em}{0ex}}0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1,\hfill \\ \underset{n\to \mathrm{\infty }}{lim inf}{r}_{n}>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}0

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${z}_{0}\in F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0$, where ${z}_{0}={P}_{F\left(S\right)\cap {\left(A+B\right)}^{-1}0\cap {W}^{-1}0}\left(I-V+\gamma g\right){z}_{0}$.

Proof We know that a nonexpansive mapping T of C into H is a $\left(1,0\right)$-generalized hybrid mapping. So, we obtain the desired result by Theorem 10. □

Let $f:C×C\to \mathbb{R}$ be a bifunction. The equilibrium problem (with respect to C) is to find $\stackrel{ˆ}{x}\in C$ such that
$f\left(\stackrel{ˆ}{x},y\right)\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
(4.1)
The set of such solutions $\stackrel{ˆ}{x}$ is denoted by $EP\left(f\right)$, i.e.,
$EP\left(f\right)=\left\{\stackrel{ˆ}{x}\in C:f\left(\stackrel{ˆ}{x},y\right)\ge 0,\mathrm{\forall }y\in C\right\}.$

For solving the equilibrium problem, let us assume that the bifunction $f:C×C\to \mathbb{R}$ satisfies the following conditions:

(A1) $f\left(x,x\right)=0$ for all $x\in C$;

(A2) f is monotone, i.e., $f\left(x,y\right)+f\left(y,x\right)\le 0$ for all $x,y\in C$;

(A3) for all $x,y,z\in C$,
$\underset{t↓0}{lim sup}f\left(tz+\left(1-t\right)x,y\right)\le f\left(x,y\right);$

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

The following lemmas were given in Combettes and Hirstoaga  and Takahashi, Takahashi and Toyoda ; see also [24, 25].

Lemma 13 ()

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Assume that $f:C×C\to \mathbb{R}$ satisfies (A1)-(A4). For $r>0$ and $x\in H$, define a mapping ${T}_{r}:H\to C$ as follows:
${T}_{r}x=\left\{z\in C:f\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$
for all $x\in H$. Then the following hold:
1. (1)

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

2. (2)
${T}_{r}$ is a firmly nonexpansive mapping, i.e., for all $x,y\in H$,
${\parallel {T}_{r}x-{T}_{r}y\parallel }^{2}\le 〈{T}_{r}x-{T}_{r}y,x-y〉;$

3. (3)

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

4. (4)

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

We call such ${T}_{r}$ the resolvent of f for $r>0$.

Lemma 14 ()

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let $f:C×C\to \mathbb{R}$ satisfy (A1)-(A4). Let ${A}_{f}$ be a set-valued mapping of H into itself defined by
${A}_{f}x=\left\{\begin{array}{cc}\left\{z\in H:f\left(x,y\right)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\right\},\hfill & \mathrm{\forall }x\in C,\hfill \\ \mathrm{\varnothing },\hfill & \mathrm{\forall }x\notin C.\hfill \end{array}$
Then $EP\left(f\right)={A}_{f}^{-1}0$ and ${A}_{f}$ is a maximal monotone operator with $D\left({A}_{f}\right)\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent ${T}_{r}$ of f coincides with the resolvent of ${A}_{f}$, i.e.,
${T}_{r}x={\left(I+r{A}_{f}\right)}^{-1}x.$

Using Lemmas 13, 14 and Theorem 10, we also obtain the following result for generalized hybrid mappings of C into H with equilibrium problem in a Hilbert space; see also .

Theorem 15 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a generalised hybrid mapping of C into H. Let f be a bifunction of $C×C$ into satisfying (A1)-(A4). Let $0, and let g be a k-contraction of H into itself. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator of H into itself with $\overline{\gamma }>0$ and $L>0$. Take $\mu ,\gamma \in \mathbb{R}$ as follows:
$0<\mu <\frac{2\overline{\gamma }}{{L}^{2}},\phantom{\rule{2em}{0ex}}0<\gamma <\frac{\overline{\gamma }-\frac{{L}^{2}\mu }{2}}{k}.$
Suppose that $F\left(S\right)\cap EP\left(f\right)\ne \mathrm{\varnothing }$. Let ${x}_{1}=x\in H$, and let $\left\{{x}_{n}\right\}\subset H$ be a sequence generated by
$\begin{array}{c}f\left({u}_{n},y\right)+\frac{1}{{r}_{n}}〈y-{u}_{n},{u}_{n}-{x}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C,\hfill \\ {x}_{n+1}={\beta }_{n}{x}_{n}+\left(1-{\beta }_{n}\right)\left\{{\alpha }_{n}\gamma g\left({x}_{n}\right)+\left(I-{\alpha }_{n}V\right)S{u}_{n}\right\}\hfill \end{array}$
for all $n\in \mathbb{N}$, where $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$ satisfy
$\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{n}=0,\phantom{\rule{2em}{0ex}}\sum _{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty },\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim inf}{r}_{n}>0,\hfill \\ \text{and}\phantom{\rule{1em}{0ex}}0<\underset{n\to \mathrm{\infty }}{lim inf}{\beta }_{n}\le \underset{n\to \mathrm{\infty }}{lim sup}{\beta }_{n}<1.\hfill \end{array}$

Then the sequence $\left\{{x}_{n}\right\}$ converges strongly to ${z}_{0}\in F\left(S\right)\cap EP\left(f\right)$, where ${z}_{0}={P}_{F\left(S\right)\cap EP\left(f\right)}\left(I-V+\gamma g\right){z}_{0}$.

Proof Put $A=0$ and $B=\partial {i}_{C}$ in Theorem 10. Furthermore, for the bifunction $f:C×C\to \mathbb{R}$, define ${A}_{f}$ as in Lemma 14. Put $W={A}_{f}$ in Theorem 10, and let ${T}_{{r}_{n}}$ be the resolvent of ${A}_{f}$ for ${r}_{n}>0$. Then we obtain that the domain of ${A}_{f}$ is included in C and ${T}_{{r}_{n}}{x}_{n}={u}_{n}$ for all $n\in \mathbb{N}$. Thus, we obtain the desired result by Theorem 10. □

## Declarations

### Acknowledgements

The first author was partially supported by Grant-in-Aid for Scientific Research No. 23540188 from Japan Society for the Promotion of Science. The second and the third authors were partially supported by the grant Taiwan NSC 99-2115-M-110-007-MY3 and the grant Taiwan NSC 99-2115-M-037-002-MY3, respectively.

## Authors’ Affiliations

(1)
Department of Mathematical and Computing Sciences, Tokyo Institute of Technology, Tokyo 152-8552, Japan
(2)
Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung, 80424, Taiwan
(3)
Center for General Education, Kaohsiung Medical University, Kaohsiung, 80702, Taiwan

## References 