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# On hybrid split problem and its nonlinear algorithms

Fixed Point Theory and Applications20132013:47

https://doi.org/10.1186/1687-1812-2013-47

• Accepted: 18 February 2013
• Published:

## Abstract

In this paper, we study a hybrid split problem (HSP for short) for equilibrium problems and fixed point problems of nonlinear operators. Some strong and weak convergence theorems are established.

MSC:47J25, 47H09, 65K10.

## Keywords

• fixed point problem
• equilibrium problem
• hybrid split problem
• iterative algorithm
• strong (weak) convergence theorem

## 1 Introduction

Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$, respectively. Let C be a nonempty subset of H and $T:C\to H$ be a mapping. Denote by $\mathcal{F}\left(T\right)$ the set of fixed points of T. The symbols and are used to denote the sets of positive integers and real numbers, respectively.

Let H be a Hilbert space and C be a closed convex subset of H. Let $f:C×C\to \mathbb{R}$ be a bi-function. The classical equilibrium problem (EP for short) is defined as follows.
(EP)
The symbol $EP\left(f\right)$ is used to denote the set of all solutions of the problem (EP), that is,
$EP\left(f\right)=\left\{u\in K:f\left(u,v\right)\ge 0,\mathrm{\forall }v\in K\right\}.$

It is known that the problem (EP) contains optimization problems, complementary problems, variational inequalities problems, saddle point problems, fixed point problems, bilevel problems, semi-infinite problems and others as special cases and have many applications in physics and economics problems; for detail, one can refer to [13] and references therein.

In last ten years or so, the problem (EP) has been generalized and improved to find a common element of the set of fixed points of a nonlinear operator and the set of solutions of the problem (EP). More precisely, many authors have studied the following problem (FTEP) (see, for instance, [49]):
(FTEP)

where C is a closed convex subset of a Hilbert space H, $f:C×C\to \mathbb{R}$ is a bi-function and $T:C\to C$ is a nonlinear operator.

In this paper, motivated by the problems (EP) and (FTEP), we study the following mathematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short).

Let ${E}_{1}$ and ${E}_{2}$ be two real Banach spaces. Let C be a closed convex subset of ${E}_{1}$ and K be a closed convex subset of ${E}_{2}$. Let $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be two bi-functions, $A:{E}_{1}\to {E}_{2}$ be a bounded linear operator. Let $T:C\to C$ and $S:K\to K$ be two nonlinear operators with $\mathcal{F}\left(T\right)\ne \mathrm{\varnothing }$ and $\mathcal{F}\left(S\right)\ne \mathrm{\varnothing }$. The mathematical model about a hybrid split problem for equilibrium problems and fixed point problems of nonlinear operators (HSP for short) is defined as follows:
(HSP)

In fact, (HSP) contains several important problems as special cases. We give some examples to explain about it.

Example A If T is an identity operator on C, then (HSP) will reduce to the following problem (P1):

(P1) Find $p\in C$ such that $f\left(p,y\right)\ge 0$, $\mathrm{\forall }y\in C$, and $u:=Ap$ satisfying $Su=u\in K$, $g\left(u,v\right)\ge 0$, $\mathrm{\forall }v\in K$.

Example B If S is an identity operator on K, then (HSP) will reduce to the following problem (P2):

(P2) Find $p\in C$ such that $Tp=p$, $f\left(p,y\right)\ge 0$, $\mathrm{\forall }y\in C$, and $u:=Ap\in K$ satisfying $g\left(u,v\right)\ge 0$, $\mathrm{\forall }v\in K$.

Example C If T, S are all identity operators, then (HSP) will reduce to the following split equilibrium problem (P3) which has been considered in [10]:

(P3) Find $p\in C$ such that $f\left(p,y\right)\ge 0$, $\mathrm{\forall }y\in C$, and $u:=Ap\in K$ satisfying $g\left(u,v\right)\ge 0$, $\mathrm{\forall }v\in K$.

Example D If S is an identity operator and $f\left(x,y\right)\equiv 0$ for all $\left(x,y\right)\in C×C$, then (HSP) will reduce to the following problem (P4) which has been studied in [11]:

(P4) Find $p\in C$ such that $Tp=p$ and $u:=Ap\in K$ satisfying $g\left(u,v\right)\ge 0$, $\mathrm{\forall }v\in K$.

In this paper, we introduce some new iterative algorithms for (HSP) and some strong and weak convergence theorems for (HSP) will be established.

## 2 Preliminaries

In what follows, the symbols and → will symbolize weak convergence and strong convergence as usual, respectively. A Banach space $\left(X,\parallel \cdot \parallel \right)$ is said to satisfy Opial’s condition if for each sequence $\left\{{x}_{n}\right\}$ in X which converges weakly to a point $x\in X$, we have
$\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-x\parallel <\underset{n\to \mathrm{\infty }}{lim inf}\parallel {x}_{n}-y\parallel ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in X,y\ne x.$

It is well known that any Hilbert space satisfies Opial’s condition. Let K be a nonempty subset of real Hilbert spaces H. Recall that a mapping $T:K\to K$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in K$.

Let ${H}_{1}$ and ${H}_{2}$ be two Hilbert spaces. Let $A:{H}_{1}\to {H}_{2}$ and $B:{H}_{2}\to {H}_{1}$ be two bounded linear operators. B is called the adjoint operator (or adjoint) of A if for all $z\in {H}_{1}$, $w\in {H}_{2}$, B satisfies $〈Az,w〉=〈z,Bw〉$. It is known that the adjoint operator of a bounded linear operator on a Hilbert space always exists and is bounded linear and unique. Moreover, it is not hard to show that if B is an adjoint operator of A, then $\parallel A\parallel =\parallel B\parallel$.

Example 2.1 ([10])

Let ${H}_{2}=\mathbb{R}$ with the standard norm $|\cdot |$ and ${H}_{1}={\mathbb{R}}^{2}$ with the norm $\parallel \alpha \parallel ={\left({a}_{1}^{2}+{a}_{2}^{2}\right)}^{\frac{1}{2}}$ for some $\alpha =\left({a}_{1},{a}_{2}\right)\in {\mathbb{R}}^{2}$. $〈x,y〉=xy$ denotes the inner product of ${H}_{2}$ for some $x,y\in {H}_{2}$ and $〈\alpha ,\beta 〉={\sum }_{i=1}^{2}{a}_{i}{b}_{i}$ denotes the inner product of ${H}_{1}$ for some $\alpha =\left({a}_{1},{a}_{2}\right),\beta =\left({b}_{1},{b}_{2}\right)\in {H}_{1}$. Let $A\alpha ={a}_{2}-{a}_{1}$ for $\alpha =\left({a}_{1},{a}_{2}\right)\in {H}_{1}$ and $Bx=\left(-x,x\right)$ for $x\in {H}_{2}$, then B is an adjoint operator of A.

Example 2.2 ([10])

Let ${H}_{1}={\mathbb{R}}^{2}$ with the norm $\parallel \alpha \parallel ={\left({a}_{1}^{2}+{a}_{2}^{2}\right)}^{\frac{1}{2}}$ for some $\alpha =\left({a}_{1},{a}_{2}\right)\in {\mathbb{R}}^{2}$ and ${H}_{2}={\mathbb{R}}^{3}$ with the norm $\parallel \gamma \parallel ={\left({c}_{1}^{2}+{c}_{2}^{2}+{c}_{3}^{2}\right)}^{\frac{1}{2}}$ for some $\gamma =\left({c}_{1},{c}_{2},{c}_{3}\right)\in {\mathbb{R}}^{3}$. Let $〈\alpha ,\beta 〉={\sum }_{i=1}^{2}{a}_{i}{b}_{i}$ and $〈\gamma ,\eta 〉={\sum }_{i=1}^{3}{c}_{i}{d}_{i}$ denote the inner product of ${H}_{1}$ and ${H}_{2}$, respectively, where $\alpha =\left({a}_{1},{a}_{2}\right),\beta =\left({b}_{1},{b}_{2}\right)\in {H}_{1}$, $\gamma =\left({c}_{1},{c}_{2},{c}_{3}\right),\eta =\left({d}_{1},{d}_{2},{d}_{3}\right)\in {H}_{2}$. Let $A\alpha =\left({a}_{2},{a}_{1},{a}_{1}-{a}_{2}\right)$ for $\alpha =\left({a}_{1},{a}_{2}\right)\in {H}_{1}$ and $B\gamma =\left({c}_{2}+{c}_{3},{c}_{1}-{c}_{3}\right)$ for $\gamma =\left({c}_{1},{c}_{2},{c}_{3}\right)\in {H}_{2}$. Obviously, B is an adjoint operator of A.

Let K be a closed convex subset of a real Hilbert space H. For each point $x\in H$, there exists a unique nearest point in K, denoted by ${P}_{K}x$, such that $\parallel x-{P}_{K}x\parallel \le \parallel x-y\parallel$ $\mathrm{\forall }y\in K$. The mapping ${P}_{K}$ is called the metric projection from H onto K. It is well known that ${P}_{K}$ has the following characteristics:
1. (i)

$〈x-y,{P}_{K}x-{P}_{K}y〉\ge {\parallel {P}_{K}x-{P}_{K}y\parallel }^{2}$ for every $x,y\in H$;

2. (ii)

for $x\in H$ and $z\in K$, $z={P}_{K}\left(x\right)⇔〈x-z,z-y〉\ge 0$, $\mathrm{\forall }y\in K$;

3. (iii)
for $x\in H$ and $y\in K$,
${\parallel y-{P}_{K}\left(x\right)\parallel }^{2}+{\parallel x-{P}_{K}\left(x\right)\parallel }^{2}\le {\parallel x-y\parallel }^{2}.$
(2.1)

Lemma 2.1 (see [1])

Let K be a nonempty closed convex subset of H and F be a bi-function of $K×K$ into satisfying the following conditions:
1. (A1)

$F\left(x,x\right)=0$ for all $x\in K$;

2. (A2)

F is monotone, that is, $F\left(x,y\right)+F\left(y,x\right)\le 0$ for all $x,y\in K$;

3. (A3)

for each $x,y,z\in K$, ${lim sup}_{t↓0}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right)$;

4. (A4)

for each $x\in K$, $y↦F\left(x,y\right)$ is convex and lower semi-continuous.

Let $r>0$ and $x\in H$. Then there exists $z\in K$ such that $F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0$ for all $y\in K$.

Lemma 2.2 (see [12])

Let K be a nonempty closed convex subset of H and let F be a bi-function of $K×K$ into satisfying (A1)-(A4). For $r>0$, define a mapping ${T}_{r}^{F}:H\to K$ as follows:
${T}_{r}^{F}\left(x\right)=\left\{z\in K:F\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in K\right\}$
(2.2)
for all $x\in H$. Then the following hold:
1. (i)

${T}_{r}^{F}$ is single-valued and $\mathcal{F}\left({T}_{r}^{F}\right)=EP\left(F\right)$ for $\mathrm{\forall }r>0$ and $EP\left(F\right)$ is closed and convex;

2. (ii)

${T}_{r}^{F}$ is firmly non-expansive, that is, for any $x,y\in H$, ${\parallel {T}_{r}^{F}x-{T}_{r}^{F}y\parallel }^{2}\le 〈{T}_{r}^{F}x-{T}_{r}^{F}y,x-y〉$.

Lemma 2.3 (see, e.g., [6])

Let H be a real Hilbert space. Then the following hold:
1. (a)

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

2. (b)

${\parallel x-y\parallel }^{2}={\parallel x\parallel }^{2}+{\parallel y\parallel }^{2}-2〈x,y〉$ for all $x,y\in H$;

3. (c)

${\parallel \alpha x+\left(1-\alpha \right)y\parallel }^{2}=\alpha {\parallel x\parallel }^{2}+\left(1-\alpha \right){\parallel y\parallel }^{2}-\alpha \left(1-\alpha \right){\parallel x-y\parallel }^{2}$ for all $x,y\in H$ and $\alpha \in \left[0,1\right]$.

Lemma 2.4 Let ${F}_{r}^{F}$ be the same as in Lemma  2.2. If $\mathcal{F}\left({T}_{r}^{F}\right)=EP\left(F\right)\ne \mathrm{\varnothing }$, then for any $x\in H$ and ${x}^{\ast }\in \mathcal{F}\left({T}_{r}^{F}\right)$, ${\parallel {T}_{r}^{F}x-x\parallel }^{2}\le {\parallel x-{x}^{\ast }\parallel }^{2}-{\parallel {T}_{r}^{F}x-{x}^{\ast }\parallel }^{2}$.

Proof

By (ii) of Lemma 2.2 and (b) of Lemma 2.3,
${\parallel {T}_{r}^{F}x-{x}^{\ast }\parallel }^{2}\le 〈{T}_{r}^{F}x-{x}^{\ast },x-{x}^{\ast }〉=\frac{1}{2}\left({\parallel {T}_{r}^{F}x-{x}^{\ast }\parallel }^{2}+{\parallel x-{x}^{\ast }\parallel }^{2}-{\parallel {T}_{r}^{F}x-x\parallel }^{2}\right),$

which shows that ${\parallel {T}_{r}^{F}x-x\parallel }^{2}\le {\parallel x-{x}^{\ast }\parallel }^{2}-{\parallel {T}_{r}^{F}x-{x}^{\ast }\parallel }^{2}$. □

Lemma 2.5 ([10, 11])

Let the mapping ${T}_{r}^{F}$ be defined as in Lemma 2.2. Then, for $r,s>0$ and $x,y\in H$,
$\parallel {T}_{r}^{F}\left(x\right)-{T}_{s}^{F}\left(y\right)\parallel \le \parallel x-y\parallel +\frac{|s-r|}{s}\parallel {T}_{s}^{F}\left(y\right)-y\parallel .$

In particular, $\parallel {T}_{r}^{F}\left(x\right)-{T}_{r}^{F}\left(y\right)\parallel \le \parallel x-y\parallel$ for any $r>0$ and $x,y\in H$, that is, ${T}_{r}^{F}$ is nonexpansive for any $r>0$.

Remark 2.1 In fact, Lemma 2.5 is motivated by a proof of [[5], Theorem 3.2]. In order to the sake of convenience for proving, we restated the fact and gave its proof in Lemma 2.5 [10, 11].

Lemma 2.6 ([13])

Let $\left\{{a}_{n}\right\}$ be a nonnegative real sequence satisfying the following condition:
${a}_{n+1}\le \left(1-{\lambda }_{n}\right){a}_{n}+{\lambda }_{n}{b}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge {n}_{0},$
where ${n}_{0}$ is some nonnegative integer, $\left\{{\lambda }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{b}_{n}\right\}$ is a sequence in  R such that
1. (i)

${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_{n}=\mathrm{\infty }$;

2. (ii)

${lim sup}_{n\to \mathrm{\infty }}{b}_{n}\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_{n}{b}_{n}$ is convergent. Then ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

Lemma 2.7 ([14])

Let $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ be bounded sequences in a Banach space E and let $\left\{{\beta }_{n}\right\}$ be a sequence in $\left[0,1\right]$ with $0. Suppose ${x}_{n+1}={\beta }_{n}{y}_{n}+\left(1-{\beta }_{n}\right){x}_{n}$ for all integers $n\ge 0$ and ${lim sup}_{n\to \mathrm{\infty }}\left(\parallel {y}_{n+1}-{y}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0$, then ${lim}_{n\to \mathrm{\infty }}\parallel {y}_{n}-{x}_{n}\parallel =0$.

## 3 Weak convergence iterative algorithms for (HSP)

In this section, we will introduce some weak convergence iterative algorithms for the hybrid split problem.

Theorem 3.1 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $T:C\to C$ and $S:K\to K$ be non-expansive mappings and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {y}_{n}=\left(1-\alpha \right){u}_{n}+\alpha T{u}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n},\hfill \\ {x}_{n+1}={P}_{C}\left({y}_{n}+\xi B\left(S{w}_{n}-A{y}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$
(3.1)

where $\alpha \in \left(0,1\right)$, $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in \mathcal{F}\left(T\right)\cap EP\left(f\right):Ap\in \mathcal{F}\left(S\right)\cap EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}⇀q\in \mathrm{\Omega }$ and ${w}_{n}⇀Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$.

Proof Let $p\in \mathrm{\Omega }$, the following several inequalities can be proved easily:
$\parallel {y}_{n}-p\parallel \le \parallel {u}_{n}-p\parallel \le \parallel {x}_{n}-p\parallel ,\phantom{\rule{2em}{0ex}}\parallel {w}_{n}-Ap\parallel \le \parallel A{y}_{n}-Ap\parallel .$
(3.2)
By Lemma 2.4, ${\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-A{y}_{n}\parallel }^{2}\le {\parallel A{y}_{n}-Ap\parallel }^{2}-{\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-Ap\parallel }^{2}$, hence
$\begin{array}{rl}{\parallel S{w}_{n}-Ap\parallel }^{2}& ={\parallel S{T}_{{r}_{n}}^{g}A{y}_{n}-Ap\parallel }^{2}\le {\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-Ap\parallel }^{2}\\ \le {\parallel A{y}_{n}-Ap\parallel }^{2}-{\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-A{y}_{n}\parallel }^{2}.\end{array}$
(3.3)
By (b) of Lemma 2.3 and (3.3), for each $n\in \mathbb{N}$, we have
(3.4)
On the other hand, ${\parallel B\left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\le {\parallel B\parallel }^{2}{\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}$, so from (3.1)-(3.4), we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-p\parallel }^{2}& =& {\parallel {P}_{C}\left({y}_{n}+\xi B\left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\right)-p\parallel }^{2}\le {\parallel {y}_{n}+\xi B\left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}-p\parallel }^{2}\\ =& {\parallel {y}_{n}-p\parallel }^{2}+{\parallel \xi B\left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}+2\xi 〈{y}_{n}-p,B\left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}〉\\ \le & {\parallel {y}_{n}-p\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ -\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ =& {\parallel {y}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{f}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{f}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}.\end{array}$
(3.5)
Since $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$, $\xi \left(1-\xi {\parallel B\parallel }^{2}\right)>0$, by (3.2) and (3.5), we have
$\parallel {x}_{n+1}-p\parallel \le \parallel {y}_{n}-p\parallel \le \parallel {u}_{n}-p\parallel \le \parallel {x}_{n}-p\parallel$
(3.6)
and
$\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}+\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {x}_{n+1}-p\parallel }^{2}.$
(3.7)
The inequality (3.6) implies ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ exists. Further, from (3.6) and (3.7), we get
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-p\parallel ,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {w}_{n}-A{y}_{n}\parallel =0.\end{array}$
(3.8)
The inequality (3.8) also implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-{w}_{n}\parallel =0.$
(3.9)
Using Lemma 2.4 and (3.8), we have
$\begin{array}{rl}{\parallel {u}_{n}-{x}_{n}\parallel }^{2}& ={\parallel {T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}\parallel }^{2}\le {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {T}_{{r}_{n}}^{f}{x}_{n}-p\parallel }^{2}\\ ={\parallel {x}_{n}-p\parallel }^{2}-{\parallel {u}_{n}-p\parallel }^{2}\to 0.\end{array}$
(3.10)
Notice that
$\begin{array}{rcl}{\parallel {y}_{n}-p\parallel }^{2}& =& \left(1-\alpha \right){\parallel {u}_{n}-p\parallel }^{2}+\alpha {\parallel T{u}_{n}-p\parallel }^{2}-\alpha \left(1-\alpha \right){\parallel T{u}_{n}-{u}_{n}\parallel }^{2}\\ \le & {\parallel {u}_{n}-p\parallel }^{2}-\alpha \left(1-\alpha \right){\parallel T{u}_{n}-{u}_{n}\parallel }^{2},\end{array}$
hence,
$\underset{n\to \mathrm{\infty }}{lim}\parallel T{u}_{n}-{u}_{n}\parallel =0.$
(3.11)
From (3.10) and (3.11), we also have
(3.12)

The existence of ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-p\parallel$ implies that $\left\{{x}_{n}\right\}$ is bounded, hence $\left\{{x}_{n}\right\}$ has a weak convergence subsequence $\left\{{x}_{{n}_{j}}\right\}$. Assume that ${x}_{{n}_{j}}⇀q$ for some $q\in C$, then ${y}_{{n}_{j}}⇀q$, $A{y}_{{n}_{j}}⇀Aq\in K$ and ${w}_{{n}_{j}}={T}_{{r}_{{n}_{j}}}^{g}A{y}_{{n}_{j}}⇀Aq$ by (3.12) and (3.8).

We say $q\in \mathrm{\Omega }$, in other words, $q\in \mathcal{F}\left(T\right)\cap EP\left(f\right)$ and $Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. By (3.10), we also obtain ${u}_{{n}_{j}}⇀q$. If $Tq\ne q$, then, by Opial’s condition and (3.11), we get
$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim inf}\parallel {u}_{{n}_{j}}-q\parallel & <& \underset{j\to \mathrm{\infty }}{lim inf}\parallel {u}_{{n}_{j}}-Tq\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim inf}\parallel {u}_{{n}_{j}}-T{u}_{{n}_{j}}+T{u}_{{n}_{j}}-Tq\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim inf}\parallel {u}_{{n}_{j}}-q\parallel ,\end{array}$
which is a contradiction. Hence $Tq=q$ or $q\in \mathcal{F}\left(T\right)$. On the other hand, from Lemma 2.2, we know $EP\left(f\right)=\mathcal{F}\left({T}_{r}^{f}\right)$ for any $r>0$. Hence, if ${T}_{r}^{f}q\ne q$ for $r>0$, then by Opial’s condition and (3.10) and Lemma 2.5, we have
$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-q\parallel & <& \underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-{T}_{r}^{f}q\parallel \\ =& \underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-{T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}+{T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}-{T}_{r}^{f}q\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim inf}\left\{\parallel {x}_{{n}_{j}}-{T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}\parallel +\parallel {T}_{r}^{f}q-{T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}\parallel \right\}\\ \le & \underset{j\to \mathrm{\infty }}{lim inf}\left\{\parallel {x}_{{n}_{j}}-{T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}\parallel +\parallel {x}_{{n}_{j}}-q\parallel +\frac{|r-{r}_{{n}_{j}}|}{{r}_{{n}_{j}}}\parallel {T}_{{r}_{{n}_{j}}}^{f}{x}_{{n}_{j}}-{x}_{{n}_{j}}\parallel \right\}\\ =& \underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-q\parallel ,\end{array}$

which is also a contradiction. So, for each $r>0$, ${T}_{r}^{f}q=q$, namely $q\in EP\left(f\right)$. Thus, we have proved $q\in \mathcal{F}\left(T\right)\cap EP\left(f\right)$. Similarly, we can also prove $Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. Hence, $q\in \mathrm{\Omega }$.

Finally, we prove $\left\{{x}_{n}\right\}$ converges weakly to $q\in \mathrm{\Omega }$. Otherwise, if there exists another subsequence of $\left\{{x}_{n}\right\}$, which is denoted by $\left\{{x}_{{n}_{l}}\right\}$, such that ${x}_{{n}_{l}}⇀\overline{x}\in \mathrm{\Omega }$ with $\overline{x}\ne q$, then by Opial’s condition,
$\underset{l\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{l}}-\overline{x}\parallel <\underset{l\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{l}}-q\parallel =\underset{j\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{j}}-q\parallel <\underset{l\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{l}}-\overline{x}\parallel .$

This is a contradiction. Hence $\left\{{x}_{n}\right\}$ converges weakly to an element $q\in \mathrm{\Omega }$. Together with $\parallel {u}_{n}-{x}_{n}\parallel \to 0$ (see (3.10)), we also get ${u}_{n}⇀q$.

Finally, we prove $\left\{{w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n}\right\}$ converges weakly to $Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. From (3.12), we have ${y}_{n}⇀q$, so $A{y}_{n}⇀Aq$. Thus, from (3.8) we have ${w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n}⇀Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. The proof is completed. □

If $T=I$ or $S=I$, where I denotes an identity operator, then the following corollaries follow from Theorem 3.1.

Corollary 3.1 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $S:K\to K$ be a non-expansive mapping and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{u}_{n},\hfill \\ {x}_{n+1}={P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$

where $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in EP\left(f\right):Ap\in \mathcal{F}\left(S\right)\cap EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}⇀q\in \mathrm{\Omega }$ and ${w}_{n}⇀Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$.

Corollary 3.2 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $T:C\to C$ be a non-expansive mapping and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {y}_{n}=\left(1-\alpha \right){u}_{n}+\alpha T{u}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n},\hfill \\ {x}_{n+1}={P}_{C}\left({y}_{n}+\xi B\left({w}_{n}-A{y}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$

where $\alpha \in \left(0,1\right)$, $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in \mathcal{F}\left(T\right)\cap EP\left(f\right):Ap\in EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}⇀q\in \mathrm{\Omega }$ and ${w}_{n}⇀Aq\in EP\left(g\right)$.

Corollary 3.3 Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{u}_{n},\hfill \\ {x}_{n+1}={P}_{C}\left({u}_{n}+\xi B\left({w}_{n}-A{u}_{n}\right)\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$

where $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in EP\left(f\right):Ap\in EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}⇀q\in \mathrm{\Omega }$ and ${w}_{n}⇀Aq\in EP\left(g\right)$.

## 4 Strong convergence iterative algorithms for (HSP)

In this section, we introduce two strong convergence algorithms for (HSP); see Theorem 4.1 and Theorem 4.2.

Theorem 4.1 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $T:C\to C$ and $S:K\to K$ be non-expansive mappings and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in {C}_{1}:=C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {y}_{n}=\left(1-\alpha \right){u}_{n}+\alpha T{u}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n},\hfill \\ {z}_{n}={P}_{C}\left({y}_{n}+\xi B\left(S{w}_{n}-A{y}_{n}\right)\right),\hfill \\ {C}_{n+1}=\left\{v\in {C}_{n}:\parallel {z}_{n}-v\parallel \le \parallel {y}_{n}-v\parallel \le \parallel {x}_{n}-v\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$
(4.1)

where $\alpha \in \left(0,1\right)$, $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in \mathcal{F}\left(T\right)\cap EP\left(f\right):Ap\in \mathcal{F}\left(S\right)\cap EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$.

Proof We claim that ${C}_{n}$ is a nonempty closed convex set for $n\in \mathbb{N}$. In fact, let $p\in \mathrm{\Omega }$, it follows from (3.4) that
$2\xi 〈{y}_{n}-p,B\left(S{w}_{n}-A{y}_{n}\right)〉\le -\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{x}_{n}\parallel }^{2}-\xi {\parallel S{w}_{n}-A{y}_{n}\parallel }^{2}.$
(4.2)
By (3.2), (4.1) and (4.2), we obtain
$\begin{array}{rcl}{\parallel {z}_{n}-p\parallel }^{2}& \le & {\parallel {y}_{n}+\xi B\left(S{w}_{n}-A{y}_{n}\right)-p\parallel }^{2}\\ =& {\parallel {y}_{n}-p\parallel }^{2}+{\parallel \xi B\left(S{w}_{n}-A{y}_{n}\right)\parallel }^{2}+2\xi 〈{y}_{n}-p,B\left(S{w}_{n}-A{y}_{n}\right)〉\\ \le & {\parallel {y}_{n}-p\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n}-A{y}_{n}\parallel }^{2}-\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel S{w}_{n}-A{y}_{n}\parallel }^{2}\\ =& {\parallel {y}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ \le & {\parallel {u}_{n}-p\parallel }^{2}-\left(1-\alpha \right)\alpha {\parallel {u}_{n}-T{u}_{n}\parallel }^{2}\\ -\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}-\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}\\ -\xi {\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel }^{2}-\left(1-\alpha \right)\alpha {\parallel {u}_{n}-T{u}_{n}\parallel }^{2}.\end{array}$
(4.3)
Notice $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$, $\xi \left(1-\xi {\parallel B\parallel }^{2}\right)>0$. It follows from (4.3) that

hence $p\in {C}_{n}$, which yields that $\mathrm{\Omega }\subset {C}_{n}$ and ${C}_{n}\ne \mathrm{\varnothing }$ for $n\in \mathbb{N}$.

It is not hard to verify that ${C}_{n}$ is closed for $n\in \mathbb{N}$, so it suffices to verify ${C}_{n}$ is convex for $n\in \mathbb{N}$. Indeed, let ${w}_{1},{w}_{2}\in {C}_{n+1}$ and $\gamma \in \left[0,1\right]$, we have

namely $\parallel {z}_{n}-\left(\gamma {w}_{1}+\left(1-\gamma \right){w}_{2}\right)\parallel \le \parallel {y}_{n}-\left(\gamma {w}_{1}+\left(1-\gamma \right){w}_{2}\right)\parallel$. Similarly, $\parallel {y}_{n}-\left(\gamma {w}_{1}+\left(1-\gamma \right){w}_{2}\right)\parallel \le \parallel {x}_{n}-\left(\gamma {w}_{1}+\left(1-\gamma \right){w}_{2}\right)\parallel$, which implies $\gamma {w}_{1}+\left(1-\gamma \right){w}_{2}\in {C}_{n+1}$ and ${C}_{n+1}$ is a convex set, $n\in \mathbb{N}$.

Notice that ${C}_{n+1}\subset {C}_{n}$ and ${x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right)\subset {C}_{n}$, then $\parallel {x}_{n+1}-{x}_{1}\parallel \le \parallel {x}_{n}-{x}_{1}\parallel$ for $n>1$. It follows that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{1}\parallel$ exists. Hence $\left\{{x}_{n}\right\}$ is bounded, which yields that $\left\{{z}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are bounded. For some $k,n\in \mathbb{N}$ with $k>n>1$, from ${x}_{k}={P}_{{C}_{k}}\left({x}_{1}\right)\subset {C}_{n}$ and (2.1), we have
$\begin{array}{rcl}{\parallel {x}_{n}-{x}_{k}\parallel }^{2}+{\parallel {x}_{1}-{x}_{k}\parallel }^{2}& =& {\parallel {x}_{n}-{P}_{{C}_{k}}\left({x}_{1}\right)\parallel }^{2}+{\parallel {x}_{1}-{P}_{{C}_{k}}\left({x}_{1}\right)\parallel }^{2}\\ \le & {\parallel {x}_{n}-{x}_{1}\parallel }^{2}.\end{array}$
(4.4)

By ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{1}\parallel$ exists and (4.4), we have ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{x}_{k}\parallel =0$, so $\left\{{x}_{n}\right\}$ is a Cauchy sequence.

Let ${x}_{n}\to q$, then $q\in \mathrm{\Omega }$. Firstly, by ${x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right)\in {C}_{n+1}\subset {C}_{n}$, from (4.1) we have
$\begin{array}{r}\parallel {z}_{n}-{x}_{n}\parallel \le \parallel {z}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel \le 2\parallel {x}_{n+1}-{x}_{n}\parallel \to 0,\\ \parallel {y}_{n}-{x}_{n}\parallel \le \parallel {y}_{n}-{x}_{n+1}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel \le 2\parallel {x}_{n+1}-{x}_{n}\parallel \to 0.\end{array}$
(4.5)
Setting $\rho =\xi \left(1-\xi {\parallel B\parallel }^{2}\right)$, by (4.3) again, we have
(4.6)
So,
$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel T{u}_{n}-{u}_{n}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {w}_{n}-A{y}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel \left({T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel =0,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-A{y}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel \left(S{T}_{{r}_{n}}^{g}-I\right)A{y}_{n}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-{w}_{n}\parallel =0.\end{array}$
(4.7)
Notice that ${lim}_{n\to \mathrm{\infty }}\parallel T{u}_{n}-{u}_{n}\parallel =0$ and $\parallel {y}_{n}-{u}_{n}\parallel =\alpha \parallel T{u}_{n}-{u}_{n}\parallel$, so
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{u}_{n}\parallel =0.$
(4.8)
Further, from (4.5) and (4.8),
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{u}_{n}\parallel =0.$
(4.9)
Since ${x}_{n}\to q$, we have ${u}_{n}\to q$ by (4.9). Thus
$\parallel Tq-q\parallel \le \parallel Tq-T{u}_{n}\parallel +\parallel T{u}_{n}-{u}_{n}\parallel +\parallel {u}_{n}-q\parallel \to 0,$
namely $Tq=q$ and $q\in \mathcal{F}\left(T\right)$. On the other hand, for $r>0$, by Lemma 2.5, we have
$\begin{array}{rcl}\parallel {T}_{r}^{f}q-q\parallel & \le & \parallel {T}_{r}^{f}q-{T}_{{r}_{n}}^{f}{x}_{n}+{T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}+{x}_{n}-q\parallel \\ \le & \parallel {x}_{n}-q\parallel +\frac{|{r}_{n}-r|}{{r}_{n}}\parallel {T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}\parallel +\parallel {T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}\parallel +\parallel {x}_{n}-q\parallel \to 0,\end{array}$

which yields $q\in \mathcal{F}\left({T}_{r}^{f}\right)=EP\left(f\right)$. We have verified $q\in \mathcal{F}\left(T\right)\cap EP\left(f\right)$.

Next, we prove $Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. Since ${x}_{n}\to q$ and ${x}_{n}-{y}_{n}\to 0$ by (4.8) and (4.9) and ${w}_{n}-A{y}_{n}\to 0$ by (4.7), we have ${y}_{n}\to q$ and $A{y}_{n}\to Aq$ and ${w}_{n}\to Aq$. So,
$\parallel SAq-Aq\parallel \le \parallel SAq-S{w}_{n}\parallel +\parallel S{w}_{n}-{w}_{n}\parallel +\parallel {w}_{n}-Aq\parallel \to 0,$
namely $SAq=Aq$ and $Aq\in \mathcal{F}\left(S\right)$. On the other hand, for $r>0$, by Lemma 2.5 again, we have
$\begin{array}{rcl}\parallel {T}_{r}^{g}Aq-Aq\parallel & \le & \parallel {T}_{r}^{g}Aq-{T}_{{r}_{n}}^{g}A{y}_{n}+{T}_{{r}_{n}}^{g}A{y}_{n}-A{y}_{n}+A{y}_{n}-Aq\parallel \\ \le & \parallel A{y}_{n}-Aq\parallel +\frac{|{r}_{n}-r|}{{r}_{n}}\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-A{y}_{n}\parallel \\ +\parallel {T}_{{r}_{n}}^{g}A{y}_{n}-A{y}_{n}\parallel +\parallel A{y}_{n}-Aq\parallel \to 0,\end{array}$

which implies that $Aq\in \mathcal{F}\left({T}_{r}^{g}\right)=EP\left(g\right)$. We have verified $Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$.

So, we have obtained $q\in \mathrm{\Omega }$ and ${x}_{n},{u}_{n}\to q$ and ${w}_{n}\to Aq$, the proof is completed. □

If $T=I$ or $S=I$, where I denotes an identity operator, then the following corollaries follow from Theorem 4.1.

Corollary 4.1 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4) and $S:K\to K$ be a non-expansive mapping. Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in {C}_{1}:=C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{u}_{n},\hfill \\ {z}_{n}={P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)\right),\hfill \\ {C}_{n+1}=\left\{v\in {C}_{n}:\parallel {z}_{n}-v\parallel \le \parallel {u}_{n}-v\parallel \le \parallel {x}_{n}-v\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$

where $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in EP\left(f\right):Ap\in \mathcal{F}\left(S\right)\cap EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$.

Corollary 4.2 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $T:C\to C$ be a non-expansive mapping and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in {C}_{1}:=C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {y}_{n}=\left(1-\alpha \right){u}_{n}+\alpha T{u}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{y}_{n},\hfill \\ {z}_{n}={P}_{C}\left({y}_{n}+\xi B\left({w}_{n}-A{y}_{n}\right)\right),\hfill \\ {C}_{n+1}=\left\{v\in {C}_{n}:\parallel {z}_{n}-v\parallel \le \parallel {y}_{n}-v\parallel \le \parallel {x}_{n}-v\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$

where, $\alpha \in \left(0,1\right)$, $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in \mathcal{F}\left(T\right)\cap EP\left(f\right):Ap\in EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in EP\left(g\right)$.

Corollary 4.3 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in {C}_{1}:=C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\phantom{\rule{2em}{0ex}}{w}_{n}={T}_{{r}_{n}}^{g}A{u}_{n},\hfill \\ {z}_{n}={P}_{C}\left({y}_{n}+\xi B\left({w}_{n}-A{u}_{n}\right)\right),\hfill \\ {C}_{n+1}=\left\{v\in {C}_{n}:\parallel {z}_{n}-v\parallel \le \parallel {u}_{n}-v\parallel \le \parallel {x}_{n}-v\parallel \right\},\hfill \\ {x}_{n+1}={P}_{{C}_{n+1}}\left({x}_{1}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$

where $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$ with ${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C. Suppose that $\mathrm{\Omega }=\left\{p\in EP\left(f\right):Ap\in EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in EP\left(g\right)$.

It is well known that the viscosity iterative method is always applied to study the iterative solution for the fixed point problem of nonlinear operators, for example, [5, 6, 8, 15, 16]. Similarly, the viscosity iterative method can also be used to study the hybrid split problem (HSP). So, at the end of this paper, we introduce a viscosity iterative algorithm which can converge strongly to a solution of (HSP).

Theorem 4.2 Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let $C\subset {H}_{1}$ and $K\subset {H}_{2}$ be two nonempty closed convex sets. Let $h:C\to C$ be a α-contraction mapping, $T:C\to C$ and $S:K\to K$ be non-expansive mappings and $f:C×C\to \mathbb{R}$ and $g:K×K\to \mathbb{R}$ be bi-functions satisfying the conditions (A1)-(A4). Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator with its adjoint B. Let ${x}_{1}\in C$, $\left\{{x}_{n}\right\}$ and $\left\{{u}_{n}\right\}$ be sequences generated by
$\left\{\begin{array}{c}{u}_{n}={T}_{{r}_{n}}^{f}{x}_{n},\hfill \\ {w}_{n}={T}_{{r}_{n}}^{g}A{u}_{n},\hfill \\ {y}_{n}={P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)\right),\hfill \\ {z}_{n}=\left(1-r\right){x}_{n}+rT{y}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}h\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right){z}_{n},\phantom{\rule{1em}{0ex}}n\in \mathbb{N},\hfill \end{array}$
(4.10)
where $r\in \left(0,1\right)$, $\xi \in \left(0,\frac{1}{{\parallel B\parallel }^{2}}\right)$ and $\left\{{r}_{n}\right\}\subset \left(0,+\mathrm{\infty }\right)$, ${P}_{C}$ is a projection operator from ${H}_{1}$ into C, and the coefficients $\left\{{\alpha }_{n}\right\}$ and $\left\{{r}_{n}\right\}$ satisfy the following conditions:
1. (1)

$\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, ${lim}_{n\to \mathrm{\infty }}{\alpha }_{n}=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$;

2. (2)

${lim inf}_{n\to +\mathrm{\infty }}{r}_{n}>0$, ${lim}_{n\to \mathrm{\infty }}|{r}_{n+1}-{r}_{n}|=0$.

Suppose that $\mathrm{\Omega }=\left\{p\in \mathcal{F}\left(T\right)\cap EP\left(f\right):Ap\in \mathcal{F}\left(S\right)\cap EP\left(g\right)\right\}\ne \mathrm{\varnothing }$, then ${x}_{n},{u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$, where $q={P}_{\mathrm{\Omega }}h\left(q\right)$.

Proof Let $p\in \mathrm{\Omega }$. The following inequalities are easily verified:
$\parallel {u}_{n}-p\parallel \le \parallel {x}_{n}-p\parallel ,\phantom{\rule{2em}{0ex}}\parallel {w}_{n}-Ap\parallel \le \parallel A{u}_{n}-Ap\parallel .$
(4.11)
By Lemma 2.4,
(4.12)
From (4.10) and (4.12), we have
(4.13)
and
$\begin{array}{rcl}{\parallel {y}_{n}-p\parallel }^{2}& =& {\parallel {P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)-{P}_{C}p\parallel }^{2}\\ \le & {\parallel {u}_{n}-p+\xi B\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ =& {\parallel {u}_{n}-p\parallel }^{2}+{\parallel \xi B\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}+2\xi 〈{u}_{n}-p,B\left(S{w}_{n}-A{u}_{n}\right)〉\\ \le & {\parallel {u}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n}-A{u}_{n}\parallel }^{2}-\xi {\parallel {T}_{{r}_{n}}^{g}A{u}_{n}-A{u}_{n}\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n}-A{u}_{n}\parallel }^{2}-\xi {\parallel {T}_{{r}_{n}}^{g}A{u}_{n}-A{u}_{n}\parallel }^{2}\\ =& {\parallel {x}_{n}-p\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n}-A{u}_{n}\parallel }^{2}-\xi {\parallel {w}_{n}-A{u}_{n}\parallel }^{2}.\end{array}$
(4.14)
So, from (4.10)-(4.11) and (4.14), we have
$\parallel {y}_{n}-p\parallel \le \parallel {u}_{n}-p\parallel \le \parallel {x}_{n}-p\parallel ,\phantom{\rule{2em}{0ex}}\parallel {z}_{n}-p\parallel \le \parallel {x}_{n}-p\parallel .$
(4.15)
We say $\left\{{x}_{n}\right\}$ is bounded. In fact, from (4.10) and (4.15), we have
$\begin{array}{rcl}\parallel {x}_{n+1}-p\parallel & =& \parallel {\alpha }_{n}\left(f\left({x}_{n}\right)-p\right)+\left(1-{\alpha }_{n}\right)\left({z}_{n}-p\right)\parallel \le \left(1-{\alpha }_{n}\right)\parallel {z}_{n}-p\parallel +{\alpha }_{n}\parallel f\left({x}_{n}\right)-p\parallel \\ \le & \left(1-{\alpha }_{n}\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\alpha \parallel {x}_{n}-p\parallel +{\alpha }_{n}\parallel f\left(p\right)-p\parallel \\ =& \left(1-{\alpha }_{n}\left(1-\alpha \right)\right)\parallel {x}_{n}-p\parallel +{\alpha }_{n}\left(1-\alpha \right)\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha },\end{array}$
which implies that
$\parallel {x}_{n}-p\parallel \le max\left\{\parallel {x}_{1}-p\parallel ,\frac{\parallel f\left(p\right)-p\parallel }{1-\alpha }\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N},$
(4.16)

so $\left\{{x}_{n}\right\}$ is bounded. Further, $\left\{{u}_{n}\right\}$, $\left\{{w}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ are also bounded by (4.11).

By Lemma 2.5, from (4.10) we have
$\begin{array}{r}{\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}={\parallel {T}_{{r}_{n+1}}^{f}{x}_{n+1}-{T}_{{r}_{n}}^{f}{x}_{n}\parallel }^{2}\\ \phantom{{\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}}\le {\left(\parallel {x}_{n+1}-{x}_{n}\parallel +\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}\parallel {T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}\parallel \right)}^{2}\\ \phantom{{\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}}\le {\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}+\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}{M}_{1},\\ {\parallel {w}_{n+1}-{w}_{n}\parallel }^{2}={\parallel {T}_{{r}_{n+1}}^{g}A{u}_{n+1}-{T}_{{r}_{n}}^{g}A{u}_{n}\parallel }^{2}\\ \phantom{{\parallel {w}_{n+1}-{w}_{n}\parallel }^{2}}\le {\left(\parallel A{u}_{n+1}-A{u}_{n}\parallel +\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}\parallel {T}_{{r}_{n}}^{g}A{u}_{n}-A{u}_{n}\parallel \right)}^{2}\\ \phantom{{\parallel {w}_{n+1}-{w}_{n}\parallel }^{2}}\le {\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}+\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}{M}_{1}\end{array}$
(4.17)
and
$\begin{array}{rcl}{\parallel {y}_{n+1}-{y}_{n}\parallel }^{2}& \le & {\parallel {u}_{n+1}+\xi B\left(S{w}_{n+1}-A{u}_{n+1}\right)-{u}_{n}-\xi B\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ =& {\parallel {u}_{n+1}-{u}_{n}+\xi B\left(S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\right)\parallel }^{2}\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\parallel \xi B\left(S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\right)\parallel }^{2}\\ +2\xi 〈{u}_{n+1}-{u}_{n},B\left(S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\right)〉\\ \le & {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +2\xi 〈A\left({u}_{n+1}-{u}_{n}\right),S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)〉\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +2\xi 〈A\left({u}_{n+1}-{u}_{n}\right)+S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right),S{w}_{n+1}\\ -A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)〉\\ -2\xi 〈S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right),S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)〉\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +2\xi 〈S{w}_{n+1}-S{w}_{n},S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)〉\\ -2\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +2\xi \frac{1}{2}\left\{{\parallel S{w}_{n+1}-S{w}_{n}\parallel }^{2}+{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ -{\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}\right\}\\ -2\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}+{\xi }^{2}{\parallel B\parallel }^{2}{\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +\xi \left\{{\parallel S{w}_{n+1}-S{w}_{n}\parallel }^{2}-{\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}\right\}\\ -\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ \le & {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +\xi \left\{{\parallel {w}_{n+1}-{w}_{n}\parallel }^{2}-{\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}\right\}\\ \le & {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +\xi \left\{{\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}+\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}{M}_{1}-{\parallel A{u}_{n+1}-A{u}_{n}\parallel }^{2}\right\}\\ =& {\parallel {u}_{n+1}-{u}_{n}\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +\xi \frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}{M}_{1}\\ \le & {\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}-\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n+1}-A{u}_{n+1}-\left(S{w}_{n}-A{u}_{n}\right)\parallel }^{2}\\ +\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}\left(\xi {M}_{1}+{M}_{1}\right),\end{array}$
(4.18)
where ${M}_{1}$ is a constant satisfying
Proving $\parallel {x}_{n+1}-{x}_{n}\parallel \to 0$ as $n\to \mathrm{\infty }$. Setting ${\beta }_{n}=1-\left(1-{\alpha }_{n}\right)\left(1-r\right)$ and ${v}_{n}=\frac{{x}_{n+1}-{x}_{n}+{\beta }_{n}{x}_{n}}{{\beta }_{n}}$, namely ${v}_{n}=\frac{{\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)rT{y}_{n}}{{\beta }_{n}}$. Let ${M}_{2}$ be a constant satisfying ${sup}_{n\in \mathbb{N}}\left\{\parallel \frac{f\left({x}_{n+1}\right)}{{\beta }_{n+1}}\parallel ,\parallel \frac{f\left({x}_{n}\right)}{{\beta }_{n}}\parallel ,\parallel T{y}_{n}\parallel \right\}\le {M}_{2}$ for all $n\in \mathbb{N}$. Then
$\begin{array}{rcl}\parallel {v}_{n+1}-{v}_{n}\parallel & =& \parallel \frac{{\alpha }_{n+1}f\left({x}_{n+1}\right)+\left(1-{\alpha }_{n+1}\right)rT{y}_{n+1}}{{\beta }_{n+1}}-\frac{{\alpha }_{n}f\left({x}_{n}\right)+\left(1-{\alpha }_{n}\right)rT{y}_{n}}{{\beta }_{n}}\parallel \\ \le & {\alpha }_{n+1}\parallel \frac{f\left({x}_{n+1}\right)}{{\beta }_{n+1}}\parallel +{\alpha }_{n}\parallel \frac{f\left({x}_{n}\right)}{{\beta }_{n}}\parallel +r\parallel \frac{\left(1-{\alpha }_{n+1}\right)T{y}_{n+1}}{{\beta }_{n+1}}-\frac{\left(1-{\alpha }_{n}\right)T{y}_{n}}{{\beta }_{n}}\parallel \\ \le & \left({\alpha }_{n+1}+{\alpha }_{n}\right){M}_{2}+r\parallel \frac{\left(1-{\alpha }_{n+1}\right)\left(T{y}_{n+1}-T{y}_{n}\right)}{{\beta }_{n+1}}\\ +\frac{\left(1-{\alpha }_{n+1}\right)T{y}_{n}}{{\beta }_{n+1}}-\frac{\left(1-{\alpha }_{n}\right)T{y}_{n}}{{\beta }_{n}}\parallel \\ \le & \left({\alpha }_{n+1}+{\alpha }_{n}\right){M}_{2}+r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}+|\frac{\left(1-{\alpha }_{n+1}\right)}{{\beta }_{n+1}}-\frac{\left(1-{\alpha }_{n}\right)}{{\beta }_{n}}|{M}_{2}\\ =& \left({\alpha }_{n+1}+{\alpha }_{n}\right){M}_{2}+r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}\\ +|\frac{\left(1-r\right)\left({\alpha }_{n}-{\alpha }_{n+1}\right)+{\beta }_{n+1}{\alpha }_{n}-{\beta }_{n}{\alpha }_{n+1}}{{\beta }_{n}{\beta }_{n+1}}|{M}_{2}\\ \le & \left({\alpha }_{n+1}+{\alpha }_{n}\right){M}_{2}+r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}+2\frac{{\alpha }_{n}+{\alpha }_{n+1}}{{\beta }_{n}{\beta }_{n+1}}{M}_{2}\\ :=& {\rho }_{n}+r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}.\end{array}$
(4.19)
From (4.18) and (4.19), we have
$\begin{array}{rcl}{\parallel {v}_{n+1}-{v}_{n}\parallel }^{2}& \le & {\left({\rho }_{n}+r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}\right)}^{2}\\ =& {\rho }_{n}^{2}+2{\rho }_{n}r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}+{r}^{2}\frac{{\left(1-{\alpha }_{n+1}\right)}^{2}{\parallel {y}_{n+1}-{y}_{n}\parallel }^{2}}{{\beta }_{n+1}^{2}},\\ \le & {\rho }_{n}^{2}+2{\rho }_{n}r\frac{\left(1-{\alpha }_{n+1}\right)\parallel {y}_{n+1}-{y}_{n}\parallel }{{\beta }_{n+1}}+{r}^{2}\frac{{\left(1-{\alpha }_{n+1}\right)}^{2}}{{\beta }_{n+1}^{2}}{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\\ +{r}^{2}\frac{{\left(1-{\alpha }_{n+1}\right)}^{2}}{{\beta }_{n+1}^{2}}\frac{|{r}_{n}-{r}_{n+1}|}{{r}_{n}}\left(1+\xi \right){M}_{1}.\end{array}$
(4.20)
By the conditions (1) and (2) and (4.20), we obtain
$\underset{n\to \mathrm{\infty }}{lim sup}\left\{{\parallel {v}_{n+1}-{v}_{n}\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\right\}\le 0.$
(4.21)
Notice ${\parallel {v}_{n+1}-{v}_{n}\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}=\left(\parallel {v}_{n+1}-{v}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\left(\parallel {v}_{n+1}-{v}_{n}\parallel +\parallel {x}_{n+1}-{x}_{n}\parallel \right)$, hence from (4.21) we have
$\underset{n\to \mathrm{\infty }}{lim sup}\left\{\parallel {v}_{n+1}-{v}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right\}\le 0.$
(4.22)
By Lemma 2.7 and (4.22), we have ${lim}_{n\to \mathrm{\infty }}\parallel {v}_{n}-{x}_{n}\parallel =0$, which implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0$
(4.23)
by the definition of ${v}_{n}$. Since $\parallel {x}_{n+1}-{z}_{n}\parallel \to 0$, together with (4.23), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{z}_{n}\parallel =0.$
(4.24)
Using (4.10), (4.12) and (4.15),
$\begin{array}{rcl}{\parallel {x}_{n+1}-p\parallel }^{2}& =& {\parallel {\alpha }_{n}\left(f\left({x}_{n}\right)-p\right)+\left(1-{\alpha }_{n}\right)\left({z}_{n}-p\right)\parallel }^{2}\\ \le & \left(1-{\alpha }_{n}\right){\parallel {z}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ \le & \left(1-r\right){\parallel {x}_{n}-p\parallel }^{2}+r{\parallel {u}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}-r{\parallel {u}_{n}-{x}_{n}\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2},\end{array}$
(4.25)
which yields
$\begin{array}{rcl}r{\parallel {u}_{n}-{x}_{n}\parallel }^{2}& \le & {\parallel {x}_{n}-p\parallel }^{2}-{\parallel {x}_{n+1}-p\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ =& \left(\parallel {x}_{n}-p\parallel +\parallel {x}_{n+1}-p\parallel \right)\left(\parallel {x}_{n}-p\parallel -\parallel {x}_{n+1}-p\parallel \right)+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ \le & \left(\parallel {x}_{n}-p\parallel +\parallel {x}_{n+1}-p\parallel \right)\parallel {x}_{n}-{x}_{n+1}\parallel +{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}.\end{array}$
(4.26)
From (4.26) we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{{r}_{n}}^{f}{x}_{n}-{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{x}_{n}\parallel =0.$
(4.27)
Again, applying (4.25), (4.15) and (4.14), we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-p\parallel }^{2}& \le & \left(1-{\alpha }_{n}\right){\parallel {z}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ \le & \left(1-r\right){\parallel {x}_{n}-p\parallel }^{2}+r{\parallel {y}_{n}-p\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2}\\ \le & {\parallel {x}_{n}-p\parallel }^{2}-r\xi \left(1-\xi {\parallel B\parallel }^{2}\right){\parallel S{w}_{n}-A{u}_{n}\parallel }^{2}\\ -r\xi {\parallel {w}_{n}-A{u}_{n}\parallel }^{2}+{\alpha }_{n}{\parallel f\left({x}_{n}\right)-p\parallel }^{2},\end{array}$
(4.28)
which implies that
(4.29)
From (4.29) we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}_{{r}_{n}}^{g}A{u}_{n}-A{u}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel {w}_{n}-A{u}_{n}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-A{u}_{n}\parallel =0$
(4.30)
and
$\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_{n}-{w}_{n}\parallel =0.$
(4.31)
Notice ${y}_{n}={P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)\right)$ and ${u}_{n}\in C$ for all $n\in \mathbb{N}$, so
$\begin{array}{rcl}\parallel {y}_{n}-{u}_{n}\parallel & =& \parallel {P}_{C}\left({u}_{n}+\xi B\left(S{w}_{n}-A{u}_{n}\right)\right)-{P}_{C}{u}_{n}\parallel \le \parallel \xi B\left(S{w}_{n}-A{u}_{n}\right)\parallel \\ \le & \xi \parallel B\parallel \parallel S{w}_{n}-A{u}_{n}\parallel ,\end{array}$
so
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{u}_{n}\parallel =0.$
(4.32)
Further, from (4.27), (4.32) and (4.24), we have
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{z}_{n}\parallel =0$
(4.33)
and
$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-T{y}_{n}\parallel =0\phantom{\rule{1em}{0ex}}\text{by (4.10), (4.24) and (4.33)}.$
(4.34)
Let $q={P}_{\mathrm{\Omega }}f\left(q\right)$. Choose a subsequence $\left\{{x}_{{n}_{k}}\right\}$ such that
$\underset{n\to \mathrm{\infty }}{lim sup}〈f\left(q\right)-q,{x}_{n}-q〉=\underset{k\to \mathrm{\infty }}{lim}〈f\left(q\right)-q,{x}_{{n}_{k}}-q〉.$
(4.35)
Since $\left\{{x}_{n}\right\}$ is bounded, $\left\{〈f\left(q\right)-q,{x}_{n}-q〉\right\}$ is bounded. Hence ${lim sup}_{n\to \mathrm{\infty }}〈f\left(q\right)-q,{x}_{n}-q〉$ is a constant, namely ${lim}_{n\to \mathrm{\infty }}〈f\left(q\right)-q,{x}_{{n}_{k}}-q〉$ exists, which implies (4.35) is well defined. Because $\left\{{x}_{n}\right\}$ is bounded, $\left\{{x}_{{n}_{k}}\right\}$ has a weak convergence subsequence which is still denoted by $\left\{{x}_{{n}_{k}}\right\}$. Suppose ${x}_{{n}_{k}}⇀{x}^{\ast }$, we say ${x}^{\ast }\in \mathrm{\Omega }$. When ${x}_{{n}_{k}}⇀{x}^{\ast }$, from (4.30), (4.32) and (4.33), we have
${u}_{{n}_{k}}⇀{x}^{\ast },\phantom{\rule{2em}{0ex}}{y}_{{n}_{k}}⇀{x}^{\ast },\phantom{\rule{2em}{0ex}}{z}_{{n}_{k}}⇀{x}^{\ast },\phantom{\rule{2em}{0ex}}A{u}_{{n}_{k}}⇀A{x}^{\ast },\phantom{\rule{2em}{0ex}}{w}_{{n}_{k}}⇀A{x}^{\ast }.$
(4.36)
If $T{x}^{\ast }\ne {x}^{\ast }$, then by (4.34) and (4.36) and Opial’s condition, we have
$\begin{array}{rcl}\underset{k\to \mathrm{\infty }}{lim inf}\parallel {y}_{{n}_{k}}-{x}^{\ast }\parallel & <& \underset{k\to \mathrm{\infty }}{lim inf}\parallel {y}_{{n}_{k}}-T{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim inf}\left\{\parallel {y}_{{n}_{k}}-T{y}_{{n}_{k}}\parallel +\parallel T{y}_{{n}_{k}}-T{x}^{\ast }\parallel \right\}\\ \le & \underset{k\to \mathrm{\infty }}{lim inf}\left\{\parallel {y}_{{n}_{k}}-T{y}_{{n}_{k}}\parallel +\parallel {y}_{{n}_{k}}-{x}^{\ast }\parallel \right\}=\underset{k\to \mathrm{\infty }}{lim inf}\parallel {y}_{{n}_{k}}-{x}^{\ast }\parallel ,\end{array}$
(4.37)
which is a contradiction, so $T{x}^{\ast }={x}^{\ast }$ and ${x}^{\ast }\in \mathcal{F}\left(T\right)$. Since for each $r>0$, $EP\left(f\right)=\mathcal{F}\left({T}_{r}^{f}\right)$ by Lemma 2.2, we have ${x}^{\ast }\in \mathcal{F}\left({T}_{r}^{f}\right)$. Otherwise, if there exists $r>0$ such that ${T}_{r}^{f}{x}^{\ast }\ne {x}^{\ast }$, then by (4.27) and Lemma 2.5 and Opial’s condition, we have
$\begin{array}{rcl}\underset{k\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{k}}-{x}^{\ast }\parallel & <& \underset{k\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{k}}-{T}_{r}^{f}{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim inf}\left\{\parallel {x}_{{n}_{k}}-{T}_{{n}_{k}}^{f}{x}_{{n}_{k}}\parallel +\parallel {T}_{{n}_{k}}^{f}{x}_{{n}_{k}}-{T}_{r}^{f}{x}^{\ast }\parallel \right\}\\ =& \underset{k\to \mathrm{\infty }}{lim inf}\parallel {T}_{{n}_{k}}^{f}{x}_{{n}_{k}}-{T}_{r}^{f}{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim inf}\left\{\parallel {x}_{{n}_{k}}-{x}^{\ast }\parallel +\frac{|{r}_{{n}_{k}}-r|}{{r}_{{n}_{k}}}\parallel {T}_{{n}_{k}}^{f}{x}_{{n}_{k}}-{x}_{{n}_{k}}\parallel \right\}\\ =& \underset{k\to \mathrm{\infty }}{lim inf}\parallel {x}_{{n}_{k}}-{x}^{\ast }\parallel ,\end{array}$
(4.38)
which is also a contradiction, so ${T}_{r}^{f}{x}^{\ast }={x}^{\ast }$ and ${x}^{\ast }\in \mathcal{F}\left({T}_{r}^{f}\right)=EP\left(f\right)$. Up to now, we have proved ${x}^{\ast }\in \mathcal{F}\left(T\right)\cap EP\left(f\right)$. Similarly, we can also prove $A{x}^{\ast }\in \mathcal{F}\left(S\right)\cap EP\left(g\right)$. Hence ${x}^{\ast }\in \mathrm{\Omega }$, because of this, we can also obtain
(4.39)
Finally, we prove the conclusion of this theorem is right. For $q={P}_{\mathrm{\Omega }}f\left(q\right)$, from (4.10) we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-q\parallel }^{2}& =& {\parallel {\alpha }_{n}\left(h\left({x}_{n}\right)-q\right)+\left(1-{\alpha }_{n}\right)\left({z}_{n}-q\right)\parallel }^{2}\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {z}_{n}-q\parallel }^{2}+2{\alpha }_{n}〈h\left({x}_{n}\right)-q,{x}_{n+1}-q〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+2{\alpha }_{n}〈h\left({x}_{n}\right)-h\left(q\right)+h\left(q\right)-q,{x}_{n+1}-q〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+2{\alpha }_{n}\alpha \parallel {x}_{n}-q\parallel \parallel {x}_{n+1}-q\parallel +2{\alpha }_{n}〈h\left(q\right)-q,{x}_{n+1}-q〉\\ \le & {\left(1-{\alpha }_{n}\right)}^{2}{\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈h\left(q\right)-q,{x}_{n+1}-q〉\\ =& \left(1-2{\alpha }_{n}\right){\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}^{2}{\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n}-q\parallel }^{2}+{\alpha }_{n}\alpha {\parallel {x}_{n+1}-q\parallel }^{2}\\ +2{\alpha }_{n}〈h\left(q\right)-q,{x}_{n+1}-q〉.\end{array}$
(4.40)
From (4.40) we have
$\begin{array}{rcl}{\parallel {x}_{n+1}-q\parallel }^{2}& \le & \left(1-{\alpha }_{n}\frac{2-2\alpha }{1-{\alpha }_{n}\alpha }\right){\parallel {x}_{n}-q\parallel }^{2}+\frac{{\alpha }_{n}^{2}}{1-{\alpha }_{n}\alpha }{\parallel {x}_{n}-q\parallel }^{2}\\ +2\frac{{\alpha }_{n}}{1-{\alpha }_{n}\alpha }〈h\left(q\right)-q,{x}_{n+1}-q〉,\end{array}$
(4.41)

by (4.41) and Lemma 2.6, we have ${x}_{n}\to q\in \mathrm{\Omega }$. Again, from (4.27) and (4.30), we have ${u}_{n}\to q\in \mathrm{\Omega }$ and ${w}_{n}\to Aq\in F\left(S\right)\cap EP\left(f\right)$, respectively. The proof is completed. □

Remark
1. (1)

In this paper, the iterative coefficient α or r can be replaced with the sequence $\left\{{\zeta }_{n}\right\}$ if $\left\{{\zeta }_{n}\right\}$ satisfies $\left\{{\zeta }_{n}\right\}\subset \left[\varrho ,\vartheta \right]$, where $\varrho ,\vartheta \in \left(0,1\right)$;

2. (2)

Obviously, if ${H}_{1}={H}_{2}$ in this paper, these weak and strong convergence theorems are also true;

3. (3)

In this paper, if T is a nonexpansive mapping from ${H}_{1}$ into ${H}_{1}$ and $f\left(x,y\right)$ is a bi-function from ${H}_{1}×{H}_{1}$ into with the conditions (A1)-(A4), S is a nonexpansive mapping from ${H}_{2}$ into ${H}_{2}$ and $g\left(u,v\right)$ is a bi-function from ${H}_{2}×{H}_{2}$ into with the conditions (A1)-(A4), then we may obtain a series of similar algorithms.

## Declarations

### Acknowledgements

The first author was supported by the Natural Science Foundation of Yunnan Province (2010ZC152). The second author was supported partially by grant No. NSC 101-2115-M-017-001 of the National Science Council of the Republic of China.

## Authors’ Affiliations

(1)
Department of Mathematics, Honghe University, Yunnan, 661100, China
(2)
Department of Mathematics, National Kaohsiung Normal University, Kaohsiung, 824, Taiwan

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