On hybrid split problem and its nonlinear algorithms

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.

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 $\mathcal{F}(T)$ the set of fixed points of T. The symbols ℕ and ℝ 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 $f:C\times C\to \mathbb{R}$ be a bi-function. The classical equilibrium problem (EP for short) is defined as follows.

$$\text{Find }p\in C\text{ such that }f(p,y)\ge 0,\mathrm{\forall }y\in C.$$

(EP)

The symbol $EP(f)$ is used to denote the set of all solutions of the problem (EP), that is,

$$EP(f)=\{u\in K:f(u,v)\ge 0,\mathrm{\forall }v\in K\}.$$

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 [1–3] 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, [4–9]):

$$\text{Find }p\in C\text{ such that }Tp=p\text{ and }f(p,y)\ge 0,\mathrm{\forall }y\in C,$$

(FTEP)

where C is a closed convex subset of a Hilbert space H, $f:C\times C\to \mathbb{R}$ 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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $\mathcal{F}(T)\ne \mathrm{\varnothing }$ and $\mathcal{F}(S)\ne \mathrm{\varnothing }$. 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:

$$\begin{array}{r}\text{Find }p\in C\text{ such that }Tp=p,f(p,y)\ge 0,\mathrm{\forall }y\in C,\text{and}\\ u:=Ap\text{ satisfying }Su=u\in K,g(u,v)\ge 0,\mathrm{\forall }v\in K.\end{array}$$

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

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

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

Example D If S is an identity operator and $f(x,y)\equiv 0$ for all $(x,y)\in C\times C$, 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 $g(u,v)\ge 0$, $\mathrm{\forall }v\in K$.

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 $(X,\parallel \cdot \parallel )$ is said to satisfy Opial’s condition if for each sequence $\{x_n\}$ in X which converges weakly to a point $x\in X$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel ,\mathrm{\forall }y\in X,y\ne x.$$

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 $\parallel \alpha \parallel ={(a_1^2+a_2^2)}^{\frac{1}{2}}$ for some $\alpha =(a_1,a_2)\in {\mathbb{R}}^2$. $〈x,y〉=xy$ 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 $\alpha =(a_1,a_2),\beta =(b_1,b_2)\in H_1$. Let $A\alpha =a_2-a_1$ for $\alpha =(a_1,a_2)\in H_1$ and $Bx=(-x,x)$ 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 $\parallel \alpha \parallel ={(a_1^2+a_2^2)}^{\frac{1}{2}}$ for some $\alpha =(a_1,a_2)\in {\mathbb{R}}^2$ and $H_2={\mathbb{R}}^3$ with the norm $\parallel \gamma \parallel ={(c_1^2+c_2^2+c_3^2)}^{\frac{1}{2}}$ for some $\gamma =(c_1,c_2,c_3)\in {\mathbb{R}}^3$. 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 $\alpha =(a_1,a_2),\beta =(b_1,b_2)\in H_1$, $\gamma =(c_1,c_2,c_3),\eta =(d_1,d_2,d_3)\in H_2$. Let $A\alpha =(a_2,a_1,a_1-a_2)$ for $\alpha =(a_1,a_2)\in H_1$ and $B\gamma =(c_2+c_3,c_1-c_3)$ for $\gamma =(c_1,c_2,c_3)\in H_2$. 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:

• (i) $〈x-y,P_{K}x-P_{K}y〉\ge {\parallel P_{K}x-P_{K}y\parallel }^2$ for every $x,y\in H$;

• (ii) for $x\in H$ and $z\in K$, $z=P_K(x)\iff 〈x-z,z-y〉\ge 0$, $\mathrm{\forall }y\in K$;

• (iii) for $x\in H$ and $y\in K$,

  $${\parallel y-P_K(x)\parallel }^2+{\parallel x-P_K(x)\parallel }^2\le {\parallel x-y\parallel }^2.$$

  (2.1)

Lemma 2.1 (see [1])

Let K be a nonempty closed convex subset of H and F be a bi-function of $K\times K$ into ℝ satisfying the following conditions:

• (A1) $F(x,x)=0$ for all $x\in K$;

• (A2) F is monotone, that is, $F(x,y)+F(y,x)\le 0$ for all $x,y\in K$;

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

• (A4) for each $x\in K$, $y\mapsto F(x,y)$ is convex and lower semi-continuous.

Let $r>0$ and $x\in H$. Then there exists $z\in K$ such that $F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0$ 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 $K\times K$ into ℝ satisfying (A1)-(A4). For $r>0$, define a mapping $T_r^F:H\to K$ as follows:

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

(2.2)

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

• (i) $T_r^F$ is single-valued and $\mathcal{F}(T_r^F)=EP(F)$ for $\mathrm{\forall }r>0$ and $EP(F)$ is closed and convex;

• (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:

• (a) ${\parallel x+y\parallel }^2\le {\parallel y\parallel }^2+2〈x,x+y〉$;

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

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

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

Proof

By (ii) of Lemma 2.2 and (b) of Lemma 2.3,

$${\parallel T_r^{F}x-x^{\ast }\parallel }^2\le 〈T_r^{F}x-x^{\ast },x-x^{\ast }〉=\frac{1}{2}({\parallel T_r^{F}x-x^{\ast }\parallel }^2+{\parallel x-x^{\ast }\parallel }^2-{\parallel T_r^{F}x-x\parallel }^2),$$

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 $x,y\in H$,

$$\parallel T_r^F(x)-T_s^F(y)\parallel \le \parallel x-y\parallel +\frac{|s-r|}{s}\parallel T_s^F(y)-y\parallel .$$

In particular, $\parallel T_r^F(x)-T_r^F(y)\parallel \le \parallel x-y\parallel$ 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 $\{a_n\}$ be a nonnegative real sequence satisfying the following condition:

$$a_{n+1}\le (1-{\lambda }_n)a_n+{\lambda }_n{b}_n,\mathrm{\forall }n\ge n_0,$$

where $n_0$ is some nonnegative integer, $\{{\lambda }_n\}$ is a sequence in $(0,1)$ and $\{b_n\}$ is a sequence in  R such that

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

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n\le 0$ 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 $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space E and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<lim\hspace{0.17em}inf{\beta }_n\le lim\hspace{0.17em}sup{\beta }_n<1$. Suppose $x_{n+1}={\beta }_n{y}_n+(1-{\beta }_n)x_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$, 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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ y_n=(1-\alpha )u_n+\alpha T{u}_n,\hfill \\ w_n=T_{r_n}^{g}A{y}_n,\hfill \\ x_{n+1}=P_C(y_n+\xi B(S{w}_n-A{y}_n)),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

(3.1)

where $\alpha \in (0,1)$, $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in \mathcal{F}(T)\cap EP(f):Ap\in \mathcal{F}(S)\cap EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\rightharpoonup q\in \mathrm{\Omega }$ and $w_n\rightharpoonup Aq\in \mathcal{F}(S)\cap EP(g)$.

Proof Let $p\in \mathrm{\Omega }$, the following several inequalities can be proved easily:

$$\parallel y_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\parallel w_n-Ap\parallel \le \parallel A{y}_n-Ap\parallel .$$

(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, ${\parallel B(S{T}_{r_n}^g-I)A{y}_n\parallel }^2\le {\parallel B\parallel }^2{\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2$, so from (3.1)-(3.4), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel P_C(y_n+\xi B(S{T}_{r_n}^g-I)A{y}_n)-p\parallel }^2\le {\parallel y_n+\xi B(S{T}_{r_n}^g-I)A{y}_n-p\parallel }^2\\ =& {\parallel y_n-p\parallel }^2+{\parallel \xi B(S{T}_{r_n}^g-I)A{y}_n\parallel }^2+2\xi 〈y_n-p,B(S{T}_{r_n}^g-I)A{y}_n〉\\ \le & {\parallel y_n-p\parallel }^2+{\xi }^2{\parallel B\parallel }^2{\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2-\xi {\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2\\ -\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^f-I)A{y}_n\parallel }^2-\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^f-I)A{y}_n\parallel }^2-\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2.\end{array}$$

(3.5)

Since $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$, $\xi (1-\xi {\parallel B\parallel }^2)>0$, 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

$$\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2+\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2.$$

(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

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-p\parallel ,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel (T_{r_n}^g-I)A{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-A{y}_n\parallel =0.\end{array}$$

(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

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& (1-\alpha ){\parallel u_n-p\parallel }^2+\alpha {\parallel T{u}_n-p\parallel }^2-\alpha (1-\alpha ){\parallel T{u}_n-u_n\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2-\alpha (1-\alpha ){\parallel T{u}_n-u_n\parallel }^2,\end{array}$$

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

$$\begin{array}{rl}\parallel y_n-x_n\parallel & \le \parallel y_n-u_n\parallel +\parallel u_n-x_n\parallel \\ =\alpha \parallel T{u}_n-u_n\parallel +\parallel u_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.12)

The existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ implies that $\{x_n\}$ is bounded, hence $\{x_n\}$ has a weak convergence subsequence $\{x_{n_j}\}$. Assume that $x_{n_j}\rightharpoonup q$ for some $q\in C$, then $y_{n_j}\rightharpoonup q$, $A{y}_{n_j}\rightharpoonup Aq\in K$ and $w_{n_j}=T_{r_{n_j}}^{g}A{y}_{n_j}\rightharpoonup Aq$ by (3.12) and (3.8).

We say $q\in \mathrm{\Omega }$, in other words, $q\in \mathcal{F}(T)\cap EP(f)$ and $Aq\in \mathcal{F}(S)\cap EP(g)$. By (3.10), we also obtain $u_{n_j}\rightharpoonup q$. If $Tq\ne q$, then, by Opial’s condition and (3.11), we get

$$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-q\parallel & <& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-Tq\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-T{u}_{n_j}+T{u}_{n_j}-Tq\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-q\parallel ,\end{array}$$

which is a contradiction. Hence $Tq=q$ or $q\in \mathcal{F}(T)$. On the other hand, from Lemma 2.2, we know $EP(f)=\mathcal{F}(T_r^f)$ 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

$$\begin{array}{rcl}\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_j}-q\parallel & <& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_j}-T_r^{f}q\parallel \\ =& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_j}-T_{r_{n_j}}^f{x}_{n_j}+T_{r_{n_j}}^f{x}_{n_j}-T_r^{f}q\parallel \\ \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel x_{n_j}-T_{r_{n_j}}^f{x}_{n_j}\parallel +\parallel T_r^{f}q-T_{r_{n_j}}^f{x}_{n_j}\parallel \}\\ \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel x_{n_j}-T_{r_{n_j}}^f{x}_{n_j}\parallel +\parallel x_{n_j}-q\parallel +\frac{|r-r_{n_j}|}{r_{n_j}}\parallel T_{r_{n_j}}^f{x}_{n_j}-x_{n_j}\parallel \}\\ =& \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_j}-q\parallel ,\end{array}$$

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

Finally, we prove $\{x_n\}$ converges weakly to $q\in \mathrm{\Omega }$. Otherwise, if there exists another subsequence of $\{x_n\}$, which is denoted by $\{x_{n_l}\}$, such that $x_{n_l}\rightharpoonup \overline{x}\in \mathrm{\Omega }$ with $\overline{x}\ne q$, then by Opial’s condition,

$$\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_l}-\overline{x}\parallel <\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_l}-q\parallel =\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_j}-q\parallel <\underset{l\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_l}-\overline{x}\parallel .$$

This is a contradiction. Hence $\{x_n\}$ 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 $u_n\rightharpoonup q$.

Finally, we prove $\{w_n=T_{r_n}^{g}A{y}_n\}$ converges weakly to $Aq\in \mathcal{F}(S)\cap EP(g)$. From (3.12), we have $y_n\rightharpoonup q$, so $A{y}_n\rightharpoonup Aq$. Thus, from (3.8) we have $w_n=T_{r_n}^{g}A{y}_n\rightharpoonup Aq\in \mathcal{F}(S)\cap EP(g)$. 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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ w_n=T_{r_n}^{g}A{u}_n,\hfill \\ x_{n+1}=P_C(u_n+\xi B(S{w}_n-A{u}_n)),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in EP(f):Ap\in \mathcal{F}(S)\cap EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\rightharpoonup q\in \mathrm{\Omega }$ and $w_n\rightharpoonup Aq\in \mathcal{F}(S)\cap EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ y_n=(1-\alpha )u_n+\alpha T{u}_n,\hfill \\ w_n=T_{r_n}^{g}A{y}_n,\hfill \\ x_{n+1}=P_C(y_n+\xi B(w_n-A{y}_n)),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\alpha \in (0,1)$, $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in \mathcal{F}(T)\cap EP(f):Ap\in EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\rightharpoonup q\in \mathrm{\Omega }$ and $w_n\rightharpoonup Aq\in EP(g)$.

Corollary 3.3 Let $C\subset H_1$ and $K\subset H_2$ be two nonempty closed convex sets. Let $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ w_n=T_{r_n}^{g}A{u}_n,\hfill \\ x_{n+1}=P_C(u_n+\xi B(w_n-A{u}_n)),\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in EP(f):Ap\in EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\rightharpoonup q\in \mathrm{\Omega }$ and $w_n\rightharpoonup Aq\in EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C_1:=C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ y_n=(1-\alpha )u_n+\alpha T{u}_n,\hfill \\ w_n=T_{r_n}^{g}A{y}_n,\hfill \\ z_n=P_C(y_n+\xi B(S{w}_n-A{y}_n)),\hfill \\ C_{n+1}=\{v\in C_n:\parallel z_n-v\parallel \le \parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),n\in \mathbb{N},\hfill \end{array}$$

(4.1)

where $\alpha \in (0,1)$, $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in \mathcal{F}(T)\cap EP(f):Ap\in \mathcal{F}(S)\cap EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\to q\in \mathrm{\Omega }$ and $w_n\to Aq\in \mathcal{F}(S)\cap EP(g)$.

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

$$2\xi 〈y_n-p,B(S{w}_n-A{y}_n)〉\le -\xi {\parallel (T_{r_n}^g-I)A{x}_n\parallel }^2-\xi {\parallel S{w}_n-A{y}_n\parallel }^2.$$

(4.2)

By (3.2), (4.1) and (4.2), we obtain

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& \le & {\parallel y_n+\xi B(S{w}_n-A{y}_n)-p\parallel }^2\\ =& {\parallel y_n-p\parallel }^2+{\parallel \xi B(S{w}_n-A{y}_n)\parallel }^2+2\xi 〈y_n-p,B(S{w}_n-A{y}_n)〉\\ \le & {\parallel y_n-p\parallel }^2+{\xi }^2{\parallel B\parallel }^2{\parallel S{w}_n-A{y}_n\parallel }^2-\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2-\xi {\parallel S{w}_n-A{y}_n\parallel }^2\\ =& {\parallel y_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2-\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2-(1-\alpha )\alpha {\parallel u_n-T{u}_n\parallel }^2\\ -\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2-\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel }^2\\ -\xi {\parallel (T_{r_n}^g-I)A{y}_n\parallel }^2-(1-\alpha )\alpha {\parallel u_n-T{u}_n\parallel }^2.\end{array}$$

(4.3)

Notice $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$, $\xi (1-\xi {\parallel B\parallel }^2)>0$. It follows from (4.3) that

$$\parallel z_n-p\parallel \le \parallel y_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel \text{for all }n\in \mathbb{N},$$

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 $\gamma \in [0,1]$, we have

namely $\parallel z_n-(\gamma w_1+(1-\gamma )w_2)\parallel \le \parallel y_n-(\gamma w_1+(1-\gamma )w_2)\parallel$. Similarly, $\parallel y_n-(\gamma w_1+(1-\gamma )w_2)\parallel \le \parallel x_n-(\gamma w_1+(1-\gamma )w_2)\parallel$, which implies $\gamma w_1+(1-\gamma )w_2\in C_{n+1}$ and $C_{n+1}$ is a convex set, $n\in \mathbb{N}$.

Notice that $C_{n+1}\subset C_n$ and $x_{n+1}=P_{C_{n+1}}(x_1)\subset C_n$, 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 $\{x_n\}$ is bounded, which yields that $\{z_n\}$ and $\{y_n\}$ are bounded. For some $k,n\in \mathbb{N}$ with $k>n>1$, from $x_k=P_{C_k}(x_1)\subset C_n$ and (2.1), we have

$$\begin{array}{rcl}{\parallel x_n-x_k\parallel }^2+{\parallel x_1-x_k\parallel }^2& =& {\parallel x_n-P_{C_k}(x_1)\parallel }^2+{\parallel x_1-P_{C_k}(x_1)\parallel }^2\\ \le & {\parallel x_n-x_1\parallel }^2.\end{array}$$

(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 $\{x_n\}$ is a Cauchy sequence.

Let $x_n\to q$, then $q\in \mathrm{\Omega }$. Firstly, by $x_{n+1}=P_{C_{n+1}}(x_1)\in C_{n+1}\subset C_n$, 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 $\rho =\xi (1-\xi {\parallel B\parallel }^2)$, by (4.3) again, we have

(4.6)

So,

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel T{u}_n-u_n\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-A{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel (T_{r_n}^g-I)A{y}_n\parallel =0,\\ \underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_n-A{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel (S{T}_{r_n}^g-I)A{y}_n\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_n-w_n\parallel =0.\end{array}$$

(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 $q\in \mathcal{F}(T)$. 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 $q\in \mathcal{F}(T_r^f)=EP(f)$. We have verified $q\in \mathcal{F}(T)\cap EP(f)$.

Next, we prove $Aq\in \mathcal{F}(S)\cap EP(g)$. 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 $Aq\in \mathcal{F}(S)$. 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 $Aq\in \mathcal{F}(T_r^g)=EP(g)$. We have verified $Aq\in \mathcal{F}(S)\cap EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C_1:=C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ w_n=T_{r_n}^{g}A{u}_n,\hfill \\ z_n=P_C(u_n+\xi B(S{w}_n-A{u}_n)),\hfill \\ C_{n+1}=\{v\in C_n:\parallel z_n-v\parallel \le \parallel u_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in EP(f):Ap\in \mathcal{F}(S)\cap EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\to q\in \mathrm{\Omega }$ and $w_n\to Aq\in \mathcal{F}(S)\cap EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C_1:=C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ y_n=(1-\alpha )u_n+\alpha T{u}_n,\hfill \\ w_n=T_{r_n}^{g}A{y}_n,\hfill \\ z_n=P_C(y_n+\xi B(w_n-A{y}_n)),\hfill \\ C_{n+1}=\{v\in C_n:\parallel z_n-v\parallel \le \parallel y_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),n\in \mathbb{N},\hfill \end{array}$$

where, $\alpha \in (0,1)$, $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in \mathcal{F}(T)\cap EP(f):Ap\in EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\to q\in \mathrm{\Omega }$ and $w_n\to Aq\in EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C_1:=C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,w_n=T_{r_n}^{g}A{u}_n,\hfill \\ z_n=P_C(y_n+\xi B(w_n-A{u}_n)),\hfill \\ C_{n+1}=\{v\in C_n:\parallel z_n-v\parallel \le \parallel u_n-v\parallel \le \parallel x_n-v\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}(x_1),n\in \mathbb{N},\hfill \end{array}$$

where $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, $P_C$ is a projection operator from $H_1$ into C. Suppose that $\mathrm{\Omega }=\{p\in EP(f):Ap\in EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\to q\in \mathrm{\Omega }$ and $w_n\to Aq\in EP(g)$.

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 $f:C\times C\to \mathbb{R}$ and $g:K\times K\to \mathbb{R}$ 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 $x_1\in C$, $\{x_n\}$ and $\{u_n\}$ be sequences generated by

$$\{\begin{array}{c}{u}_n=T_{r_n}^f{x}_n,\hfill \\ w_n=T_{r_n}^{g}A{u}_n,\hfill \\ y_n=P_C(u_n+\xi B(S{w}_n-A{u}_n)),\hfill \\ z_n=(1-r)x_n+rT{y}_n,\hfill \\ x_{n+1}={\alpha }_{n}h(x_n)+(1-{\alpha }_n)z_n,n\in \mathbb{N},\hfill \end{array}$$

(4.10)

where $r\in (0,1)$, $\xi \in (0,\frac{1}{{\parallel B\parallel }^2})$ and $\{r_n\}\subset (0,+\mathrm{\infty })$, $P_C$ is a projection operator from $H_1$ into C, and the coefficients $\{{\alpha }_n\}$ and $\{r_n\}$ satisfy the following conditions:

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

• (2) ${lim\hspace{0.17em}inf}_{n\to +\mathrm{\infty }}{r}_n>0$, ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$.

Suppose that $\mathrm{\Omega }=\{p\in \mathcal{F}(T)\cap EP(f):Ap\in \mathcal{F}(S)\cap EP(g)\}\ne \mathrm{\varnothing }$, then $x_n,u_n\to q\in \mathrm{\Omega }$ and $w_n\to Aq\in \mathcal{F}(S)\cap EP(g)$, where $q=P_{\mathrm{\Omega }}h(q)$.

Proof Let $p\in \mathrm{\Omega }$. The following inequalities are easily verified:

$$\parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\parallel w_n-Ap\parallel \le \parallel A{u}_n-Ap\parallel .$$

(4.11)

By Lemma 2.4,

(4.12)

From (4.10) and (4.12), we have

(4.13)

and

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel P_C(u_n+\xi B(S{w}_n-A{u}_n)-P_{C}p\parallel }^2\\ \le & {\parallel u_n-p+\xi B(S{w}_n-A{u}_n)\parallel }^2\\ =& {\parallel u_n-p\parallel }^2+{\parallel \xi B(S{w}_n-A{u}_n)\parallel }^2+2\xi 〈u_n-p,B(S{w}_n-A{u}_n)〉\\ \le & {\parallel u_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_n-A{u}_n\parallel }^2-\xi {\parallel T_{r_n}^{g}A{u}_n-A{u}_n\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_n-A{u}_n\parallel }^2-\xi {\parallel T_{r_n}^{g}A{u}_n-A{u}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_n-A{u}_n\parallel }^2-\xi {\parallel w_n-A{u}_n\parallel }^2.\end{array}$$

(4.14)

So, from (4.10)-(4.11) and (4.14), we have

$$\parallel y_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\parallel z_n-p\parallel \le \parallel x_n-p\parallel .$$

(4.15)

We say $\{x_n\}$ is bounded. In fact, from (4.10) and (4.15), we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(z_n-p)\parallel \le (1-{\alpha }_n)\parallel z_n-p\parallel +{\alpha }_n\parallel f(x_n)-p\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-p\parallel +{\alpha }_n\alpha \parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ =& (1-{\alpha }_n(1-\alpha ))\parallel x_n-p\parallel +{\alpha }_n(1-\alpha )\frac{\parallel f(p)-p\parallel }{1-\alpha },\end{array}$$

which implies that

$$\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{\parallel f(p)-p\parallel }{1-\alpha }\},\mathrm{\forall }n\in \mathbb{N},$$

(4.16)

so $\{x_n\}$ is bounded. Further, $\{u_n\}$, $\{w_n\}$ and $\{y_n\}$ are also bounded by (4.11).

By Lemma 2.5, from (4.10) we have

$$\begin{array}{r}{\parallel u_{n+1}-u_n\parallel }^2={\parallel T_{r_{n+1}}^f{x}_{n+1}-T_{r_n}^f{x}_n\parallel }^2\\ \phantom{{\parallel u_{n+1}-u_n\parallel }^2}\le {(\parallel x_{n+1}-x_n\parallel +\frac{|r_n-r_{n+1}|}{r_n}\parallel T_{r_n}^f{x}_n-x_n\parallel )}^2\\ \phantom{{\parallel u_{n+1}-u_n\parallel }^2}\le {\parallel x_{n+1}-x_n\parallel }^2+\frac{|r_n-r_{n+1}|}{r_n}{M}_1,\\ {\parallel w_{n+1}-w_n\parallel }^2={\parallel T_{r_{n+1}}^{g}A{u}_{n+1}-T_{r_n}^{g}A{u}_n\parallel }^2\\ \phantom{{\parallel w_{n+1}-w_n\parallel }^2}\le {(\parallel A{u}_{n+1}-A{u}_n\parallel +\frac{|r_n-r_{n+1}|}{r_n}\parallel T_{r_n}^{g}A{u}_n-A{u}_n\parallel )}^2\\ \phantom{{\parallel w_{n+1}-w_n\parallel }^2}\le {\parallel A{u}_{n+1}-A{u}_n\parallel }^2+\frac{|r_n-r_{n+1}|}{r_n}{M}_1\end{array}$$

(4.17)

and

$$\begin{array}{rcl}{\parallel y_{n+1}-y_n\parallel }^2& \le & {\parallel u_{n+1}+\xi B(S{w}_{n+1}-A{u}_{n+1})-u_n-\xi B(S{w}_n-A{u}_n)\parallel }^2\\ =& {\parallel u_{n+1}-u_n+\xi B(S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n))\parallel }^2\\ =& {\parallel u_{n+1}-u_n\parallel }^2+{\parallel \xi B(S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n))\parallel }^2\\ +2\xi 〈u_{n+1}-u_n,B(S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n))〉\\ \le & {\parallel u_{n+1}-u_n\parallel }^2+{\xi }^2{\parallel B\parallel }^2{\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +2\xi 〈A(u_{n+1}-u_n),S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)〉\\ =& {\parallel u_{n+1}-u_n\parallel }^2+{\xi }^2{\parallel B\parallel }^2{\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +2\xi 〈A(u_{n+1}-u_n)+S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n),S{w}_{n+1}\\ -A{u}_{n+1}-(S{w}_n-A{u}_n)〉\\ -2\xi 〈S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n),S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)〉\\ =& {\parallel u_{n+1}-u_n\parallel }^2+{\xi }^2{\parallel B\parallel }^2{\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +2\xi 〈S{w}_{n+1}-S{w}_n,S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)〉\\ -2\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\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}-(S{w}_n-A{u}_n)\parallel }^2\\ +2\xi \frac{1}{2}\{{\parallel S{w}_{n+1}-S{w}_n\parallel }^2+{\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ -{\parallel A{u}_{n+1}-A{u}_n\parallel }^2\}\\ -2\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\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}-(S{w}_n-A{u}_n)\parallel }^2\\ +\xi \{{\parallel S{w}_{n+1}-S{w}_n\parallel }^2-{\parallel A{u}_{n+1}-A{u}_n\parallel }^2\}\\ -\xi {\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ \le & {\parallel u_{n+1}-u_n\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +\xi \{{\parallel w_{n+1}-w_n\parallel }^2-{\parallel A{u}_{n+1}-A{u}_n\parallel }^2\}\\ \le & {\parallel u_{n+1}-u_n\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +\xi \{{\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\}\\ =& {\parallel u_{n+1}-u_n\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +\xi \frac{|r_n-r_{n+1}|}{r_n}{M}_1\\ \le & {\parallel x_{n+1}-x_n\parallel }^2-\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_{n+1}-A{u}_{n+1}-(S{w}_n-A{u}_n)\parallel }^2\\ +\frac{|r_n-r_{n+1}|}{r_n}(\xi M_1+M_1),\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 ${\beta }_n=1-(1-{\alpha }_n)(1-r)$ and $v_n=\frac{x_{n+1}-x_n+{\beta }_n{x}_n}{{\beta }_n}$, namely $v_n=\frac{{\alpha }_{n}f(x_n)+(1-{\alpha }_n)rT{y}_n}{{\beta }_n}$. Let $M_2$ be a constant satisfying ${sup}_{n\in \mathbb{N}}\{\parallel \frac{f(x_{n+1})}{{\beta }_{n+1}}\parallel ,\parallel \frac{f(x_n)}{{\beta }_n}\parallel ,\parallel T{y}_n\parallel \}\le M_2$ for all $n\in \mathbb{N}$. Then

$$\begin{array}{rcl}\parallel v_{n+1}-v_n\parallel & =& \parallel \frac{{\alpha }_{n+1}f(x_{n+1})+(1-{\alpha }_{n+1})rT{y}_{n+1}}{{\beta }_{n+1}}-\frac{{\alpha }_{n}f(x_n)+(1-{\alpha }_n)rT{y}_n}{{\beta }_n}\parallel \\ \le & {\alpha }_{n+1}\parallel \frac{f(x_{n+1})}{{\beta }_{n+1}}\parallel +{\alpha }_n\parallel \frac{f(x_n)}{{\beta }_n}\parallel +r\parallel \frac{(1-{\alpha }_{n+1})T{y}_{n+1}}{{\beta }_{n+1}}-\frac{(1-{\alpha }_n)T{y}_n}{{\beta }_n}\parallel \\ \le & ({\alpha }_{n+1}+{\alpha }_n)M_2+r\parallel \frac{(1-{\alpha }_{n+1})(T{y}_{n+1}-T{y}_n)}{{\beta }_{n+1}}\\ +\frac{(1-{\alpha }_{n+1})T{y}_n}{{\beta }_{n+1}}-\frac{(1-{\alpha }_n)T{y}_n}{{\beta }_n}\parallel \\ \le & ({\alpha }_{n+1}+{\alpha }_n)M_2+r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}+|\frac{(1-{\alpha }_{n+1})}{{\beta }_{n+1}}-\frac{(1-{\alpha }_n)}{{\beta }_n}|M_2\\ =& ({\alpha }_{n+1}+{\alpha }_n)M_2+r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}\\ +\left|\frac{(1-r)({\alpha }_n-{\alpha }_{n+1})+{\beta }_{n+1}{\alpha }_n-{\beta }_n{\alpha }_{n+1}}{{\beta }_n{\beta }_{n+1}}\right|M_2\\ \le & ({\alpha }_{n+1}+{\alpha }_n)M_2+r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}+2\frac{{\alpha }_n+{\alpha }_{n+1}}{{\beta }_n{\beta }_{n+1}}{M}_2\\ :=& {\rho }_n+r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}.\end{array}$$

(4.19)

From (4.18) and (4.19), we have

$$\begin{array}{rcl}{\parallel v_{n+1}-v_n\parallel }^2& \le & {({\rho }_n+r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}})}^2\\ =& {\rho }_n^2+2{\rho }_{n}r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}+r^2\frac{{(1-{\alpha }_{n+1})}^2{\parallel y_{n+1}-y_n\parallel }^2}{{\beta }_{n+1}^2},\\ \le & {\rho }_n^2+2{\rho }_{n}r\frac{(1-{\alpha }_{n+1})\parallel y_{n+1}-y_n\parallel }{{\beta }_{n+1}}+r^2\frac{{(1-{\alpha }_{n+1})}^2}{{\beta }_{n+1}^2}{\parallel x_{n+1}-x_n\parallel }^2\\ +r^2\frac{{(1-{\alpha }_{n+1})}^2}{{\beta }_{n+1}^2}\frac{|r_n-r_{n+1}|}{r_n}(1+\xi )M_1.\end{array}$$

(4.20)

By the conditions (1) and (2) and (4.20), we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{{\parallel v_{n+1}-v_n\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2\}\le 0.$$

(4.21)

Notice ${\parallel v_{n+1}-v_n\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2=(\parallel v_{n+1}-v_n\parallel -\parallel x_{n+1}-x_n\parallel )(\parallel v_{n+1}-v_n\parallel +\parallel x_{n+1}-x_n\parallel )$, hence from (4.21) we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\parallel v_{n+1}-v_n\parallel -\parallel x_{n+1}-x_n\parallel \}\le 0.$$

(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),

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\alpha }_n(f(x_n)-p)+(1-{\alpha }_n)(z_n-p)\parallel }^2\\ \le & (1-{\alpha }_n){\parallel z_n-p\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ \le & (1-r){\parallel x_n-p\parallel }^2+r{\parallel u_n-p\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-r{\parallel u_n-x_n\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2,\end{array}$$

(4.25)

which yields

$$\begin{array}{rcl}r{\parallel u_n-x_n\parallel }^2& \le & {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ =& (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )(\parallel x_n-p\parallel -\parallel x_{n+1}-p\parallel )+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ \le & (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +{\alpha }_n{\parallel f(x_n)-p\parallel }^2.\end{array}$$

(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

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & (1-{\alpha }_n){\parallel z_n-p\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ \le & (1-r){\parallel x_n-p\parallel }^2+r{\parallel y_n-p\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2\\ \le & {\parallel x_n-p\parallel }^2-r\xi (1-\xi {\parallel B\parallel }^2){\parallel S{w}_n-A{u}_n\parallel }^2\\ -r\xi {\parallel w_n-A{u}_n\parallel }^2+{\alpha }_n{\parallel f(x_n)-p\parallel }^2,\end{array}$$

(4.28)

which implies that

(4.29)

From (4.29) we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{r_n}^{g}A{u}_n-A{u}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-A{u}_n\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel S{w}_n-A{u}_n\parallel =0$$

(4.30)

and

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

(4.31)

Notice $y_n=P_C(u_n+\xi B(S{w}_n-A{u}_n))$ and $u_n\in C$ for all $n\in \mathbb{N}$, so

$$\begin{array}{rcl}\parallel y_n-u_n\parallel & =& \parallel P_C(u_n+\xi B(S{w}_n-A{u}_n))-P_C{u}_n\parallel \le \parallel \xi B(S{w}_n-A{u}_n)\parallel \\ \le & \xi \parallel B\parallel \parallel S{w}_n-A{u}_n\parallel ,\end{array}$$

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

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

(4.33)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-T{y}_n\parallel =0\text{by (4.10), (4.24) and (4.33)}.$$

(4.34)

Let $q=P_{\mathrm{\Omega }}f(q)$. Choose a subsequence $\{x_{n_k}\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,x_n-q〉=\underset{k\to \mathrm{\infty }}{lim}〈f(q)-q,x_{n_k}-q〉.$$

(4.35)

Since $\{x_n\}$ is bounded, $\{〈f(q)-q,x_n-q〉\}$ is bounded. Hence ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(q)-q,x_n-q〉$ is a constant, namely ${lim}_{n\to \mathrm{\infty }}〈f(q)-q,x_{n_k}-q〉$ exists, which implies (4.35) is well defined. Because $\{x_n\}$ is bounded, $\{x_{n_k}\}$ has a weak convergence subsequence which is still denoted by $\{x_{n_k}\}$. Suppose $x_{n_k}\rightharpoonup x^{\ast }$, we say $x^{\ast }\in \mathrm{\Omega }$. When $x_{n_k}\rightharpoonup x^{\ast }$, from (4.30), (4.32) and (4.33), we have

$$u_{n_k}\rightharpoonup x^{\ast },y_{n_k}\rightharpoonup x^{\ast },z_{n_k}\rightharpoonup x^{\ast },A{u}_{n_k}\rightharpoonup A{x}^{\ast },w_{n_k}\rightharpoonup A{x}^{\ast }.$$

(4.36)

If $T{x}^{\ast }\ne x^{\ast }$, then by (4.34) and (4.36) and Opial’s condition, we have

$$\begin{array}{rcl}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_k}-x^{\ast }\parallel & <& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_k}-T{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel y_{n_k}-T{y}_{n_k}\parallel +\parallel T{y}_{n_k}-T{x}^{\ast }\parallel \}\\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel y_{n_k}-T{y}_{n_k}\parallel +\parallel y_{n_k}-x^{\ast }\parallel \}=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel y_{n_k}-x^{\ast }\parallel ,\end{array}$$

(4.37)

which is a contradiction, so $T{x}^{\ast }=x^{\ast }$ and $x^{\ast }\in \mathcal{F}(T)$. Since for each $r>0$, $EP(f)=\mathcal{F}(T_r^f)$ by Lemma 2.2, we have $x^{\ast }\in \mathcal{F}(T_r^f)$. 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

$$\begin{array}{rcl}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_k}-x^{\ast }\parallel & <& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_k}-T_r^f{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel x_{n_k}-T_{n_k}^f{x}_{n_k}\parallel +\parallel T_{n_k}^f{x}_{n_k}-T_r^f{x}^{\ast }\parallel \}\\ =& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel T_{n_k}^f{x}_{n_k}-T_r^f{x}^{\ast }\parallel \\ \le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\{\parallel x_{n_k}-x^{\ast }\parallel +\frac{|r_{n_k}-r|}{r_{n_k}}\parallel T_{n_k}^f{x}_{n_k}-x_{n_k}\parallel \}\\ =& \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_{n_k}-x^{\ast }\parallel ,\end{array}$$

(4.38)

which is also a contradiction, so $T_r^f{x}^{\ast }=x^{\ast }$ and $x^{\ast }\in \mathcal{F}(T_r^f)=EP(f)$. Up to now, we have proved $x^{\ast }\in \mathcal{F}(T)\cap EP(f)$. Similarly, we can also prove $A{x}^{\ast }\in \mathcal{F}(S)\cap EP(g)$. Hence $x^{\ast }\in \mathrm{\Omega }$, because of this, we can also obtain

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(q)-q,x_n-q〉& =& \underset{k\to \mathrm{\infty }}{lim}〈f(q)-q,x_{n_k}-q〉\\ =& 〈f(q)-q,x^{\ast }-q〉\le 0,\text{where }q=P_{C}f(q).\end{array}$$

(4.39)

Finally, we prove the conclusion of this theorem is right. For $q=P_{\mathrm{\Omega }}f(q)$, from (4.10) we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& =& {\parallel {\alpha }_n(h(x_n)-q)+(1-{\alpha }_n)(z_n-q)\parallel }^2\\ \le & {(1-{\alpha }_n)}^2{\parallel z_n-q\parallel }^2+2{\alpha }_n〈h(x_n)-q,x_{n+1}-q〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+2{\alpha }_n〈h(x_n)-h(q)+h(q)-q,x_{n+1}-q〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+2{\alpha }_n\alpha \parallel x_n-q\parallel \parallel x_{n+1}-q\parallel +2{\alpha }_n〈h(q)-q,x_{n+1}-q〉\\ \le & {(1-{\alpha }_n)}^2{\parallel x_n-q\parallel }^2+{\alpha }_n\alpha {\parallel x_n-q\parallel }^2+{\alpha }_n\alpha {\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈h(q)-q,x_{n+1}-q〉\\ =& (1-2{\alpha }_n){\parallel x_n-q\parallel }^2+{\alpha }_n^2{\parallel x_n-q\parallel }^2+{\alpha }_n\alpha {\parallel x_n-q\parallel }^2+{\alpha }_n\alpha {\parallel x_{n+1}-q\parallel }^2\\ +2{\alpha }_n〈h(q)-q,x_{n+1}-q〉.\end{array}$$

(4.40)

From (4.40) we have

$$\begin{array}{rcl}{\parallel x_{n+1}-q\parallel }^2& \le & (1-{\alpha }_n\frac{2-2\alpha }{1-{\alpha }_n\alpha }){\parallel x_n-q\parallel }^2+\frac{{\alpha }_n^2}{1-{\alpha }_n\alpha }{\parallel x_n-q\parallel }^2\\ +2\frac{{\alpha }_n}{1-{\alpha }_n\alpha }〈h(q)-q,x_{n+1}-q〉,\end{array}$$

(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 $w_n\to Aq\in F(S)\cap EP(f)$, respectively. The proof is completed. □

Remark

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

• (2) Obviously, if $H_1=H_2$ in this paper, these weak and strong convergence theorems are also true;

• (3) In this paper, if T is a nonexpansive mapping from $H_1$ into $H_1$ and $f(x,y)$ is a bi-function from $H_1\times H_1$ into ℝ with the conditions (A1)-(A4), S is a nonexpansive mapping from $H_2$ into $H_2$ and $g(u,v)$ is a bi-function from $H_2\times H_2$ into ℝ with the conditions (A1)-(A4), then we may obtain a series of similar algorithms.