Shrinking projection methods for solving split equilibrium problems and fixed point problems for asymptotically nonexpansive mappings in Hilbert spaces

Abstract

In this paper, we propose a new iterative sequence for solving common problems which consist of split equilibrium problems and fixed point problems for asymptotically nonexpansive mappings in the framework of Hilbert spaces and prove some strong convergence theorems of the generated sequence $\{x_{n}\}$ by the shrinking projection method. Our results improve and extend the previous results given in the literature.

1 Introduction

Throughout this paper, let $\mathbb{R}$ and $\mathbb{N}$ denote the set of all real numbers and the set of all positive integers, respectively. Let H be a real Hilbert space and C be a nonempty closed convex subset of H.

A mapping $T:C\times C\to\mathbb{R}$ is said to be asymptotically nonexpansive if there exists a sequence $\{k_{n}\}\subset[1,\infty)$ with $\lim_{n\to\infty}k_{n}=1$ such that

$$\begin{aligned} \bigl\Vert T^{n}x-T^{n}y\bigr\Vert \leq k_{n} \Vert x-y\Vert \end{aligned}$$

for all $x, y\in C$. It is easy to see that, if $k_{n}\equiv1$, then T is said to be nonexpansive. We denote the set of fixed point of T by $F(T)$, that is, $F(T)=\{x\in C:Tx=x\}$. There are many iterative methods for solving a fixed point problem corresponding to an asymptotically nonexpansive mapping (see also [1–3]).

Recall that a Hilbert space H satisfies Opial’s condition [4], that is, for any subsequence $\{x_{n}\}\subset H$ with $x_{n}\rightharpoonup x$, the following inequality

$$\begin{aligned} \liminf_{n\to\infty} \Vert x_{n}-x\Vert < \liminf _{n\to\infty} \Vert x_{n}-y\Vert \end{aligned}$$

holds for all $y\in H$ with $y\neq x$. Furthermore, a Hilbert space H has a Kadec-Klee property, i.e., $x_{n}\rightharpoonup x$ and $\Vert x_{n}\Vert \to \Vert x\Vert$ imply $x_{n}\to x$. In fact, from

$$\begin{aligned} \Vert x_{n}-x\Vert ^{2}=\Vert x_{n}\Vert ^{2}-2\langle x_{n},x\rangle+\Vert x\Vert ^{2}, \end{aligned}$$

we can conclude that a Hilbert space has a Kadec-Klee property.

In 1994, Blum and Oettli [5] introduced the equilibrium problem which is to find $x\in C$ such that

$$\begin{aligned} F(x,y)\geq0 \quad\mbox{for all } y\in C. \end{aligned}$$

(1.1)

They denoted the solution set of problem (1.1) as $EP(F)$. Since the well-known problems were variational problems, complementary problems, fixed point problems, saddle point problems and other problems proposed from the equilibrium problem, it has become the most attractive topic for many mathematicians [6–8]. They have widely spread its applications to other applied disciplines including physics, chemistry, economics and engineering (see, for example, [9–12]).

In 1997, Combettes and Hirstoaga [13] proposed an iterative method for solving problem (1.1) by the assumption that $EP(F)\neq \emptyset$. Moreover, there are many new iteratively generated sequences for solving this problem together with fixed point problems (see [14–17]).

Later, the so-called split equilibrium problem was introduced (shortly, $SEP$). Let $H_{1}$, $H_{2}$ be two real Hilbert spaces. Let C, Q be closed convex subsets of $H_{1}$ and $H_{2}$, respectively, and let $A:H_{1}\to H_{2}$ be a bounded linear operator. Further, let $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ be two bifunctions. The $SEP$ is to find the element $x^{*}\in C$ such that

$$\begin{aligned} F_{1}\bigl(x^{*},y\bigr)\geq0\quad \mbox{for all } y\in C \end{aligned}$$

(1.2)

and such that

$$\begin{aligned} Ax^{*}\in Q \mbox{ solves } F_{2}\bigl(Ax^{*},v\bigr)\geq0\quad \mbox{for all } v\in Q. \end{aligned}$$

(1.3)

The solution sets of problems (1.2) and (1.3) are symbolized by $EP(F_{1})$ and $EP(F_{2})$, respectively. Therefore, we denote $\Omega=\{v\in C: v\in EP(F_{1}) \mbox{ such that } Av\in EP(F_{2})\}$ as the solution set of $SEP$.

Clearly, the $SEP$ contains two equilibrium problems, that is, we find out the solution of one equilibrium problem, i.e., its image under a given bounded linear operator, must be the solution of another equilibrium problem. In order to find a common solution of equilibrium problems, it has been mostly considered in the same spaces. However, we normally found that, in the real-life problems, it may be considered in different spaces. That is how the $SEP$ works very well for this case (see, for example, [18]). Moreover, the split variational inequality problem (shortly, $SVIP$) is its special case, which is to find $x^{*}\in C$ such that

$$\begin{aligned} \bigl\langle f\bigl(x^{*}\bigr),x-x^{*}\bigr\rangle \geq0 \quad\mbox{for all } x\in C, \end{aligned}$$

(1.4)

and corresponding to

$$\begin{aligned} y^{*}=Ax^{*}\in Q \mbox{ solves } \bigl\langle g\bigl(y^{*}\bigr),y-y^{*} \bigr\rangle \geq0 \quad\mbox{for all } y\in Q, \end{aligned}$$

(1.5)

where $f:H_{1}\to H_{1}$ and $g:H_{1}\to H_{2}$ are nonlinear mappings and $A:H_{1}\to H_{2}$ is a bounded linear operator (see [19]).

In 2012, He [18] proposed the new algorithm for solving a split equilibrium problem and investigated the convergence behavior in several ways including both weak and strong convergence. Moreover, they gave some examples and mentioned that there exist many SEPs, and the new methods for solving it further need to be explored in the future. Later, in 2013, Kazmi and Rizvi [20] considered the iterative method to compute the common approximate solution of a split equilibrium problem, a variational inequality problem and a fixed point problem for a nonexpansive mapping in the framework of real Hilbert spaces. They generated the sequence iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=J_{r_{n}}^{F_{1}}(I+\gamma A^{*}(J_{r_{n}}^{F_{2}}-I)A)x_{n},\\ y_{n}=P_{C}(u_{n}-\lambda_{n}Du_{n}),\\ x_{n+1}=\alpha_{n}v+\beta_{n}x_{n}+\gamma_{n}Sy_{n} \end{array}\displaystyle \right . $$

(1.6)

for each $n\geq0$, where $A:H_{1}\to H_{2}$ is a bounded linear operator, $D:C\to H_{1}$ is a τ-inverse strongly monotone mapping, $F_{1}:C\times C\to\mathbb{R}$, $F_{2}:Q\times Q\to\mathbb{R}$ are two bifunctions. They found that, under the sufficient conditions of $r_{n}$, $\lambda_{n}$, γ, $\beta_{n}$ and $\gamma_{n}$, the generated sequence $\{ x_{n}\}$ converges strongly to a common solution of all mentioned problems.

Recently, in 2014, Bnouhachem [21] introduced a new iterative method for solving split equilibrium problem and hierarchical fixed point problems by defining the sequence $\{x_{n}\}$ as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}(I+\gamma A^{*}(T_{r_{n}}^{F_{2}}-I)A)x_{n},\\ y_{n}=\beta_{n}Sx_{n}+(1-\beta_{n})u_{n},\\ x_{n+1}=P_{C}[\alpha_{n}\rho U(x_{n})+(I-\alpha_{n}\mu F)(T(y_{n}))] \end{array}\displaystyle \right . $$

(1.7)

for each $n\geq0$, where S, T are nonexpansive mappings, $F:C\to C$ is a k-Lipschitz mapping and η-strongly monotone, $U:C\to C$ is a τ-Lipschitz mapping. Also, they proved some strong convergence theorems for the proposed iteration under some appropriate conditions.

In this paper, motivated and inspired by the results [18, 20, 21] and the recent works in this field, we introduce the shrinking projection method for solving split equilibrium problems and fixed point problems for asymptotically nonexpansive mappings in the framework of Hilbert spaces and prove some strong convergence theorems for the proposed new iterative method. In fact, our results improve and extend the results given by some authors.

2 Preliminaries

In this section, we recall some concepts including the assumption which will be needed for the proof of our main result.

Let H be a Hilbert space and C be a nonempty closed convex subset of H. For each $x\in H$, there exists a unique nearest point of C, denoted by $P_{C}x$, such that

$$\begin{aligned} \Vert x-P_{c}x\Vert \leq \Vert x-y\Vert \end{aligned}$$

for all $y\in C$. $P_{C}$ is called the metric projection from H onto C. It is well known that $P_{C}$ is a firmly nonexpansive mapping from H onto C, that is,

$$\begin{aligned} \Vert P_{C}x-P_{C}y\Vert ^{2}\leq\langle P_{C}x-P_{C}y,x-y\rangle \end{aligned}$$

for all $x,y\in H$. Furthermore, for any $x\in H$ and $z\in C$, $z=P_{C}x$ if and only if

$$\begin{aligned} \langle x-z,z-y\rangle\geq0 \end{aligned}$$

for all $y\in C$. A mapping $A:C\to H$ is called α-inverse strongly monotone if there exists $\alpha>0$ such that

$$\begin{aligned} \langle x-y, Ax-Ay\rangle\geq\alpha \Vert Ax-Ay\Vert ^{2} \end{aligned}$$

for all $x,y\in H$. Moreover, we can investigate that, for each $\lambda \in(0,2\alpha]$, $I-\lambda A$ is a nonexpansive mapping of C into H (see [22]).

Lemma 2.1

In a Hilbert space H, the following identity holds:

$$\begin{aligned} \bigl\Vert \lambda x+(1-\lambda)y\bigr\Vert ^{2}=\lambda \Vert x \Vert ^{2}+(1-\lambda)\Vert y\Vert ^{2}-\lambda (1- \lambda)\Vert x-y\Vert ^{2} \end{aligned}$$

for all $x,y\in H$ and $\lambda\in[0,1]$.

Lemma 2.2

[23]

Let T be an asymptotically nonexpansive mapping defined on a bounded closed convex subset C of a Hilbert space H. Assume that $\{x_{n}\}$ is a sequence in C with the following properties:

• (1) $x_{n}\rightharpoonup z$;

• (2) $Tx_{n}-x_{n}\rightarrow0$.

Then $z\in F(T)$.

Assumption 2.3

[24]

Let $F:C\times C\to\mathbb{R}$ be a bifunction satisfying the following conditions:

• (A1) $F(x,x)=0$ for all $x\in C$;

• (A2) F is monotone, i.e., $F(x,y)+F(y,x)\leq0$ for all $x,y\in C$;

• (A3) for each $x,y,z\in C$, $\lim_{t\downarrow 0}F(tz+(1-t)x,y)\leq F(x,y)$;

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

Lemma 2.4

[24]

Let C be a nonempty closed convex subset of a Hilbert space H and $F:C\times C\to\mathbb{R}$ be a bifunction which satisfies conditions (A1)-(A4). For any $x\in H$ and $r>0$, define a mapping $T_{r}^{F}:H\to C$ by

$$\begin{aligned} T_{r}^{F}(x)=\biggl\{ z\in C: F(z,y)+\frac{1}{r} \langle y-z,z-x\rangle\geq0, \forall y\in C\biggr\} . \end{aligned}$$

Then $T_{r}^{F}$ is well defined and the following hold:

• (1) $T_{r}^{F}$ is single-valued;

• (2) $T_{r}^{F}$ is firmly nonexpansive, i.e., for any $x,y\in H$,

  $$\begin{aligned} \bigl\Vert T_{r}^{F}x-T_{r}^{F}y \bigr\Vert ^{2}\leq\bigl\langle T_{r}^{F}x-T_{r}^{F}y,x-y \bigr\rangle ; \end{aligned}$$

• (3) $F(T_{r}^{F})=EP(F)$;

• (4) $EP(F)$ is closed and convex.

3 Main results

In this section, we prove some strong convergence theorems of an iterative algorithm for solving a split equilibrium together with a fixed point problem revolving an asymptotically nonexpansive mapping in the framework of Hilbert spaces.

Theorem 3.1

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and C, Q be nonempty closed convex subsets of Hilbert spaces $H_{1}$ and $H_{2}$, respectively. Let $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ be two bifunctions satisfying conditions (A1)-(A4) and $F_{2}$ be upper semi-continuous in the first argument. Let $T:C\to C$ be an asymptotically nonexpansive mapping and $A:H_{1}\to H_{2}$ be a bounded linear operator. Suppose that $F(T)\cap\Omega\neq\emptyset$, where $\Omega=\{v\in C: v\in EP(F_{1})\textit{ such that }Av\in EP(F_{2})\}$, and let $x_{0}\in C$. For $C_{1}=C$ and $x_{1}=P_{C_{1}}x_{0}$, define a sequence $\{x_{n}\}$ iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)x_{n},\\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})T^{n}u_{n},\\ C_{n+1}=\{z\in C_{n}:\Vert y_{n}-z\Vert ^{2}\leq \Vert x_{n}-z\Vert ^{2}+\theta_{n}\},\\ x_{n+1}=P_{C_{n+1}}x_{0} \end{array}\displaystyle \right . $$

(3.1)

for each $n\geq1$, where $\theta_{n}=(1-\alpha_{n})(k_{n}^{2}-1)\sup\{\Vert x_{n}-z\Vert ^{2}:z\in\Omega\}$, $0\leq\alpha_{n}\leq a<1$ for all $n\in\mathbb{N}$, $0< b\leq r_{n}<\infty$, $\gamma\in(0,1/L)$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A. Then the sequence $\{x_{n}\}$ generated by (3.1) strongly converges to a point $z_{0}\in F(T)\cap\Omega$.

Proof

First of all, we investigate that, for each $n\in\mathbb{N}$, $A^{*}(I-T_{r_{n}}^{F_{2}})A$ is a $\frac{1}{2L}$-inverse strongly monotone mapping. Since $T_{r_{n}}^{F_{2}}$ is firmly nonexpansive and $(I-T_{r_{n}}^{F_{2}})$ is $\frac{1}{2}$-inverse strongly monotone, it follows that

$$\begin{aligned} &\bigl\Vert A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax-A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ay\bigr\Vert ^{2} \\ &\quad=\bigl\langle A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay),A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay)\bigr\rangle \\ &\quad=\bigl\langle \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay),AA^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay)\bigr\rangle \\ &\quad\leq L\bigl\langle \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay), \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay)\bigr\rangle \\ &\quad=L\bigl\Vert \bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay)\bigr\Vert ^{2} \\ &\quad\leq 2L\bigl\langle x-y, A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr) (Ax-Ay)\bigr\rangle \end{aligned}$$

for all $x,y\in H$, from which it can be concluded that $A^{*}(I-T_{r_{n}}^{F_{2}})A$ is a $\frac{1}{2L}$-inverse strongly monotone mapping. Moreover, we claim that since $\gamma\in(0,\frac{1}{L})$, $I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A$ is nonexpansive.

Next, we show that $F(T)\cap\Omega\subset C_{n+1}$ for all $n\in\mathbb {N}$. Let $p\in F(T)\cap\Omega$, i.e., $T_{r_{n}}^{F_{1}}p=p$ and $(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)p=p$. By mathematical induction, we have $p\in C=C_{1}$ and hence $F(T)\cap\Omega\subset C_{1}$. Let $F(T)\cap \Omega\subset C_{k}$ for some $k\in\mathbb{N}$. It follows that

$$\begin{aligned} \Vert u_{k}-p\Vert =&\bigl\| T_{r_{k}}^{F_{1}} \bigl(I-\gamma A^{*}\bigl(I-T_{r_{k}}^{F_{2}}\bigr)A \bigr)x_{k}-T_{r_{k}}^{F_{1}}\bigl(I-\gamma A^{*} \bigl(I-T_{r_{k}}^{F_{2}}\bigr)A\bigr)p\bigr\| \\ \leq&\bigl\| \bigl(I-\gamma A^{*}\bigl(I-T_{r_{k}}^{F_{2}}\bigr)A \bigr)x_{k}-\bigl(I-\gamma A^{*}\bigl(I-T_{r_{k}}^{F_{2}} \bigr)A\bigr)p\bigr\| \\ \leq&\Vert x_{k}-p\Vert \end{aligned}$$

(3.2)

and

$$\begin{aligned} &\Vert y_{k}-p\Vert ^{2} \\ &\quad=\bigl\Vert \alpha_{k}x_{k}+(1- \alpha_{k})T^{n}u_{k}-p\bigr\Vert ^{2} \\ &\quad\leq\alpha_{k}\Vert x_{k}-p\Vert ^{2}+(1-\alpha_{k})\bigl\Vert T^{n}u_{k}-T^{n}-p \bigr\Vert ^{2}-\alpha _{k}(1-\alpha_{k})\bigl\Vert x_{k}-p-\bigl(T^{n}u_{k}-T^{n}p \bigr)\bigr\Vert ^{2} \\ &\quad\leq\alpha_{k}\Vert x_{k}-p\Vert ^{2}+(1-\alpha_{k})\Vert u_{k}-p\Vert ^{2}-\alpha _{k}(1-\alpha_{k})\bigl\Vert x_{k}-T^{n}u_{k}\bigr\Vert ^{2} \\ &\quad\leq\alpha_{k}\Vert x_{k}-p\Vert ^{2}+(1-\alpha_{k})k_{k}^{2}\Vert x_{k}-p\Vert ^{2} \\ &\quad=\Vert x_{k}-p\Vert ^{2}+(1- \alpha_{k}) \bigl(k_{k}^{2}-1\bigr)\Vert x_{k}-p\Vert ^{2} \\ &\quad\leq \Vert x_{k}-p\Vert ^{2}+(1- \alpha_{k}) \bigl(k_{k}^{2}-1\bigr)M_{k}^{2} \\ &\quad=\Vert x_{k}-p\Vert ^{2}+\theta_{k}, \end{aligned}$$

(3.3)

where $M_{k}=\sup\{\Vert x_{k}-z\Vert :z\in\Omega\}$ and $\theta_{k}=(1-\alpha _{k})(k_{k}^{2}-1)M_{k}^{2}$. It can be concluded that $p\in C_{k+1}$ and $F(T)\cap\Omega\subset C_{k+1}$ and, further, $F(T)\cap\Omega\subset C_{n}$ for all $n\in\mathbb{N}$.

Next, we show that $C_{n}$ is closed and convex for all $n\in\mathbb{N}$. First, it is obvious that $C_{1}=C$ is closed and convex. By induction, we suppose that $C_{k}$ is closed and convex for some $k\in\mathbb{N}$. Let $z_{m}\in C_{k+1}\subset C_{k}$ with $z_{m}\to z$. Since $C_{k}$ is closed, it follows that $x\in C_{k}$ and $\Vert y_{k}-z_{m}\Vert ^{2} \leq \Vert z_{m}-x_{k}\Vert ^{2}+\theta _{k}$. Then we have

$$\begin{aligned} \Vert y_{k}-z\Vert ^{2} =&\Vert y_{k}-z_{m}+z_{m}-z \Vert ^{2} \\ =&\Vert y_{k}-z_{m}\Vert ^{2}+\Vert z_{m}-z\Vert ^{2}+2\langle y_{k}-z_{m},z_{m}-z \rangle \\ \leq&\Vert z_{m}-x_{k}\Vert ^{2}+ \theta_{k}+\Vert z_{m}-z\Vert ^{2}+2\Vert y_{k}-z_{m}\Vert \Vert z_{m}-z\Vert . \end{aligned}$$

Letting $m\to\infty$, we have

$$\begin{aligned} \Vert y_{k}-z\Vert ^{2}\leq \Vert z-x_{k} \Vert ^{2}+\theta_{k}, \end{aligned}$$

which means that $z\in C_{k+1}$. Let $x,y\in C_{k+1}\subset C_{k}$ and $z=\alpha x+(1-\alpha)y$ for any $\alpha\in[0,1]$. Since $C_{k}$ is convex, $z\in C_{k}$, $\Vert y_{k}-x\Vert ^{2}\leq \Vert x-x_{k}\Vert ^{2}+\theta_{k}$ and $\Vert y_{k}-y\Vert ^{2}\leq \Vert x-x_{k}\Vert ^{2}+\theta_{k}$ and so

$$\begin{aligned} \Vert y_{k}-z\Vert ^{2} =&\bigl\Vert y_{k}-\bigl(\alpha x+(1-\alpha)y\bigr)\bigr\Vert ^{2} \\ =&\bigl\Vert \alpha(y_{k}-x)+(1-\alpha) (y_{k}-y)\bigr\Vert ^{2} \\ =&\alpha \Vert y_{k}-x\Vert ^{2}+(1-\alpha)\Vert y_{k}-y\Vert ^{2}-\alpha(1-\alpha)\bigl\Vert y_{k}-x-(y_{k}-y)\bigr\Vert ^{2} \\ \leq&\alpha\bigl(\Vert x-x_{k}\Vert ^{2}+ \theta_{k}\bigr)+(1-\alpha) \bigl(\Vert y-x_{k}\Vert ^{2}+\theta _{k}\bigr)-\alpha(1-\alpha)\Vert y-x\Vert ^{2} \\ \leq&\alpha \Vert x-x_{k}\Vert ^{2}+ \theta_{k}+(1-\alpha)\Vert y-x_{k}\Vert ^{2}- \alpha(1-\alpha )\bigl\Vert (x_{k}-x)-(x_{k}-y)\bigr\Vert ^{2} \\ =&\alpha \Vert x-x_{k}\Vert ^{2}+(1-\alpha)\Vert y-x_{k}\Vert ^{2}-\alpha(1-\alpha)\bigl\Vert (x_{k}-x)-(x_{k}-y)\bigr\Vert ^{2}+ \theta_{k} \\ =&\bigl\Vert \alpha(x_{k}-x)+(1-\alpha) (x_{k}-y)\bigr\Vert ^{2}+\theta_{k} \\ =&\Vert x_{k}-z\Vert ^{2}+\theta_{k}. \end{aligned}$$

Therefore, $z\in C_{k+1}$ and hence $C_{k+1}$ is closed and convex. It is immediately concluded that $C_{n}$ is closed and convex for all $n\in \mathbb{N}$, which implies that $\{x_{n}\}$ is well defined.

Next, from $x_{n}=P_{C_{n}}x_{0}$, we have

$$\begin{aligned} \langle x_{0}-x_{n},x_{n}-y\rangle\geq0 \end{aligned}$$

for all $y\in C_{n}$. Since $p\in F(T)\cap\Omega$, we have

$$\begin{aligned} \langle x_{0}-x_{n},x_{n}-p\rangle\geq0 \end{aligned}$$

for all $p\in F(T)\cap\Omega$, that is, we have

$$\begin{aligned} 0 \leq&\langle x_{0}-x_{n},x_{n}-p\rangle \leq-\Vert x_{n}-x_{0}\Vert ^{2}+\Vert x_{0}-x_{n}\Vert \Vert x_{0}-p\Vert . \end{aligned}$$

This implies that

$$\begin{aligned} \Vert x_{n}-x_{0}\Vert \leq \Vert x_{0}-p\Vert \end{aligned}$$

(3.4)

for all $n\in\mathbb{N}$. From $x_{n}=P_{C_{n}}x_{0}$ and $x_{n+1}=P_{C_{n+1}}x_{0}\in C_{n+1}\subset C_{n}$, we also have

$$\begin{aligned} \langle x_{0}-x_{n},x_{n}-x_{n+1} \rangle\geq0 \end{aligned}$$

(3.5)

for all $n\in\mathbb{N}$, and so we have

$$\begin{aligned} 0 \leq&\langle x_{0}-x_{n},x_{n}-x_{n+1} \rangle \\ =&\langle x_{0}-x_{n},x_{n}-x_{0}+x_{0}-x_{n+1} \rangle \\ \leq&-\Vert x_{n}-x_{0}\Vert ^{2}+\Vert x_{0}-x_{n}\Vert \Vert x_{0}-x_{n+1} \Vert . \end{aligned}$$

Hence we have

$$\begin{aligned} \Vert x_{n}-x_{0}\Vert ^{2}\leq \Vert x_{0}-x_{n}\Vert \Vert x_{0}-x_{n+1} \Vert , \end{aligned}$$

(3.6)

that is, $\Vert x_{n}-x_{0}\Vert \leq \Vert x_{0}-x_{n+1}\Vert$ for all $n\in\mathbb{N}$. From (3.4), it follows that $\{x_{n}\}$ is bounded and $\lim_{n\to \infty} \Vert x_{n}-x_{0}\Vert$ exists.

Next, we show that $\Vert x_{n}-x_{n+1}\Vert \to0$. From (3.5), we have

$$\begin{aligned} &\Vert x_{n}-x_{n+1}\Vert ^{2} \\ &\quad=\bigl\Vert (x_{n}-x_{0})+(x_{0}-x_{n+1}) \bigr\Vert ^{2} \\ &\quad=\Vert x_{n}-x_{0}\Vert ^{2}+2\langle x_{n}-x_{0},x_{0}-x_{n+1}\rangle+\Vert x_{0}-x_{n+1}\Vert ^{2} \\ &\quad=\Vert x_{n}-x_{0}\Vert ^{2}+2\langle x_{n}-x_{0}, x_{0}-x_{n}\rangle+2 \langle x_{n}-x_{0}, x_{n}-x_{n+1}\rangle+ \Vert x_{0}-x_{n+1}\Vert ^{2} \\ &\quad\leq \Vert x_{n}-x_{0}\Vert ^{2}-2\Vert x_{n}-x_{0}\Vert ^{2}+\Vert x_{0}-x_{n+1}\Vert ^{2} \\ &\quad=\Vert x_{0}-x_{n+1}\Vert ^{2}-\Vert x_{n}-x_{0}\Vert ^{2}. \end{aligned}$$

Since the limit of $\{\Vert x_{n}-x_{0}\Vert \}$ exists, we have

$$\begin{aligned} \lim_{n\to\infty} \Vert x_{n}-x_{n+1} \Vert =0. \end{aligned}$$

(3.7)

Thus, by (3.7) and (3.14), we have

$$\begin{aligned} \Vert y_{n}-x_{n}\Vert \leq \Vert y_{n}-x_{n+1}\Vert +\Vert x_{n+1}-x_{n} \Vert \to0 \end{aligned}$$

(3.8)

as $n\to\infty$. Furthermore, since $T_{r_{n}}^{F_{1}}$ is firmly nonexpansive, we have

$$\begin{aligned} \Vert u_{n}-p\Vert ^{2} =& \bigl\| T_{r_{n}}^{F_{1}}\bigl(x_{n}-\gamma A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n}\bigr)-T_{r_{n}}^{F_{1}} \bigl(p-\gamma A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ap\bigr) \bigr\| ^{2}\\ \leq&\bigl\| \bigl(I-\gamma A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)A \bigr)x_{n}-\bigl(I-\gamma A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)A\bigr)p\bigr\| ^{2}\\ &{}-\bigl\| \bigl(I-T_{r_{n}}^{F_{1}}\bigr) \bigl(I-\gamma A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n}\bigr)- \bigl(I-T_{r_{n}}^{F_{1}}\bigr) \bigl(I-\gamma A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ap\bigr)\bigr\| ^{2}\\ =& \bigl\| x_{n}-p-\gamma \bigl(A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}-A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ap\bigr) \bigr\| ^{2}-\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n}\bigr\| ^{2}\\ =& \|x_{n}-p\|^{2}-2\gamma\bigl\langle x_{n}-p, A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n}-A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ap \bigr\rangle \\ &{}+\gamma^{2}\bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}-A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ap\bigr\| ^{2}-\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n}\bigr\| ^{2}\\ \leq&\|x_{n}-p\|^{2}+\gamma\biggl(\gamma- \frac{1}{L}\biggr)\bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}\bigr\| ^{2}-\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n} \bigr\| ^{2}, \end{aligned}$$

where $z_{n}=(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)x_{n}$. Moreover,

$$\begin{aligned} \|y_{n}-p\|^{2} =&\bigl\| \alpha_{n}x_{n}+(1-\alpha_{n})T^{n}u_{n}-p \bigr\| ^{2} \\ \leq&\alpha_{n}\|x_{n}-p\|^{2}+(1- \alpha_{n})k_{n}^{2} \biggl[\|x_{n}-p \|^{2} \\ &{}+\gamma\biggl(\gamma-\frac{1}{L}\biggr)\bigl\| A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n}\bigr\| ^{2}-\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n}\bigr\| ^{2} \biggr] \\ =&\alpha_{n}\|x_{n}-p\|^{2}+(1- \alpha_{n})k_{n}^{2}\|x_{n}-p \|^{2}-(1-\alpha _{n})k_{n}^{2} \bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n}\bigr\| ^{2} \\ &{}+(1-\alpha_{n})k_{n}^{2}\gamma\biggl( \gamma-\frac{1}{L}\biggr)\bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}\bigr\| ^{2}, \end{aligned}$$

which leads to

$$\begin{aligned} &(1-\alpha_{n})k_{n}^{2} \biggl[\gamma\biggl(\frac{1}{L}-\gamma\biggr)\bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}\bigr\| ^{2}+\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n} \bigr\| ^{2} \biggr] \\ &\quad\leq\bigl(\alpha_{n}+(1-\alpha_{n})k_{n}^{2} \bigr)\|x_{n}-p\|^{2}-\|y_{n}-p\|^{2}. \end{aligned}$$

(3.9)

Letting $\rho_{n}=k_{n}-1$. Then it is clear that $\rho_{n}\to0$ as $n\to \infty$ and, by (3.9), we exactly have

$$\begin{aligned} (1-\alpha_{n})k_{n}^{2} \biggl[\gamma \biggl(\frac{1}{L} -&\gamma\biggr)\bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n})\bigr\| ^{2}+\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n} \bigr\| ^{2} \biggr] \\ \leq&\alpha_{n}\|x_{n}-p\|^{2}+(1- \alpha_{n}) (\rho_{n}+1)^{2}\|x_{n}-p\| ^{2}-\|y_{n}-p\|^{2} \\ \leq&\|x_{n}-p\|^{2}-\|y_{n}-p \|^{2}+(1-\alpha_{n}) \bigl(\rho_{n}^{2}+2 \rho_{n}\bigr)\| x_{n}-p\|^{2} \\ \leq&\bigl(\|x_{n}-p\|+\|y_{n}-p\|\bigr) \|x_{n}-y_{n}\|+(1-\alpha_{n}) \bigl(\rho _{n}^{2}+2\rho_{n}\bigr)\|x_{n}-p \|^{2}. \end{aligned}$$

By (3.8) and $\rho_{n}\to0$ as $n\to\infty$, we have

$$\begin{aligned} \bigl\| A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n} \bigr\| \to0, \qquad\bigl\| z_{n}-T_{r_{n}}^{F_{1}}z_{n}\bigr\| \to0 \end{aligned}$$

(3.10)

as $n\to\infty$. Furthermore, since A is linear bounded and so is $A^{*}$, we can conclude that

$$\begin{aligned} \lim_{n\to\infty}\bigl\| \bigl(I-T_{r_{n}}^{F_{2}} \bigr)Ax_{n}\bigr\| =0. \end{aligned}$$

(3.11)

Next, we show that $\|u_{n}-x_{n}\|\to0$. We investigate the following:

$$\begin{aligned} \|u_{n}-x_{n}\| =& \bigl\| T_{r_{n}}^{F_{1}}z_{n}-x_{n}\bigr\| \\ \leq&\bigl\| T_{r_{n}}^{F_{1}}z_{n}-z_{n} \bigr\| +\|z_{n}-x_{n}\| \\ =&\bigl\| T_{r_{n}}^{F_{1}}z_{n}-z_{n}\bigr\| +\bigl\| \bigl(I-\gamma A^{*}\bigl(I-T_{r_{n}}^{F_{2}}\bigr)A \bigr)x_{n}-x_{n}\bigr\| \\ =&\bigl\| T_{r_{n}}^{F_{1}}z_{n}-z_{n}\bigr\| +\gamma\bigl\| A^{*} \bigl(I-T_{r_{n}}^{F_{2}}\bigr)Ax_{n}\bigr\| . \end{aligned}$$

(3.12)

Consequently, by (3.12), we can conclude that

$$\begin{aligned} \|u_{n}-x_{n}\|\to0. \end{aligned}$$

(3.13)

Next, we show that $\|T^{n}x_{n}-x_{n}\|\to0$. We first consider

$$\begin{aligned} \|y_{n}-x_{n}\|=\bigl\| \alpha_{n}x_{n}+(1- \alpha_{n})T^{n}u_{n}\bigr\| =(1-\alpha_{n})\bigl\| T^{n}u_{n}-x_{n}\bigr\| , \end{aligned}$$

and since $x_{n+1}\in C_{n+1}\subset C_{n}$, we have

$$\begin{aligned} \|y_{n}-x_{n+1}\|^{2}\leq\|x_{n}-x_{n+1} \|^{2}+\theta_{n}, \end{aligned}$$

which means that

$$\begin{aligned} \|y_{n}-x_{n+1}\|\leq\|x_{n}-x_{n+1} \|+\sqrt{\theta_{n}}. \end{aligned}$$

(3.14)

Hence,

$$\begin{aligned}[b] \bigl\| T^{n}u_{n}-x_{n}\bigr\| &=\frac{1}{1-\alpha_{n}} \|y_{n}-x_{n}\| \\ &\leq\frac{1}{1-a}\bigl(\|y_{n}-x_{n+1}\|+ \|x_{n+1}-x_{n}\|\bigr) \\ &\leq\frac{1}{1-a}\bigl(\|x_{n}-x_{n+1}\|+\sqrt{ \theta_{n}}\bigr)+\frac{1}{1-a}\| x_{n+1}-x_{n}\|, \end{aligned} $$

and so $\|T^{n}u_{n}-x_{n}\|\to0$. Consider

$$\begin{aligned} \bigl\| T^{n}x_{n}-x_{n}\bigr\| \leq&\bigl\| T^{n}x_{n}-T^{n}u_{n} \bigr\| +\bigl\| T^{n}u_{n}-x_{n}\bigr\| \\ \leq&k_{n}\|x_{n}-u_{n}\|+\bigl\| T^{n}u_{n}-x_{n} \bigr\| . \end{aligned}$$

Therefore, we have $\|T^{n}x_{n}-x_{n}\|\to0$ as $n\to\infty$. Putting $k_{\infty}=\sup\{k_{n}:n\geq1\}<\infty$, we deduce that

$$\begin{aligned} \bigl\| Tx_{n}-x_{n}\bigr\| \leq&\bigl\| Tx_{n}-T^{n+1}x_{n} \bigr\| +\bigl\| T^{n+1}x_{n}-T^{n+1}x_{n+1}\bigr\| \\ &{}+\bigl\| T^{n+1}x_{n+1}-x_{n+1}\bigr\| +\|x_{n+1}-x_{n} \| \\ \leq&k_{\infty}\bigl\| x_{n}-T^{n}x_{n} \bigr\| +(1+k_{\infty})\|x_{n}-x_{n+1}\|+\bigl\| T^{n+1}x_{n+1}-x_{n+1} \bigr\| . \end{aligned}$$

Hence we have $\|Tx_{n}-x_{n}\|\to0$ as $n\to\infty$. Without loss of generality, since $\{x_{n}\}$ is bounded, we may assume that $x_{n}\rightharpoonup x^{*}$. It is easy to see that $x^{*}\in C_{n}$ for all $n\geq1$. On the other hand, we have

$$\begin{aligned} \|x_{n}-x_{0}\|\leq\bigl\| x^{*}-x_{0}\bigr\| . \end{aligned}$$

It follows that

$$\begin{aligned} \bigl\| x^{*}-x_{0}\bigr\| \leq\liminf_{n\to\infty}\|x_{n}-x_{0} \|\leq\limsup_{n\to\infty}\| x_{n}-x_{0}\|\leq \bigl\| x^{*}-x_{0}\bigr\| \end{aligned}$$

and so

$$\begin{aligned} \lim_{n\to\infty}\|x_{n}-x_{0}\|= \bigl\| x^{*}-x_{0}\bigr\| . \end{aligned}$$

Hence $\|x_{n}\|\to\|x^{*}\|$. Since every Hilbert space has the Kadec-Klee property, we immediately have $x_{n}\to x^{*}$.

Finally, we prove that $x^{*}\in F(T)\cap\Omega$. Since $x_{n}\to x^{*}$ and $x_{n}-Tx_{n}\to0$ as $n\to\infty$, consider

$$\begin{aligned} \bigl\| x^{*}-Tx^{*}\bigr\| \leq&\bigl\| x^{*}-x_{n}\bigr\| +\|x_{n}-Tx_{n}\|+ \bigl\| Tx_{n}-Tx^{*}\bigr\| \\ \leq&(1+k_{1})\bigl\| x^{*}-x_{n}\bigr\| +\|x_{n}-Tx_{n} \|. \end{aligned}$$

We can see that $\|x^{*}-Tx^{*}\|=0$ and, further, $x^{*}\in F(T)$. Therefore, we have $x^{*}\in F(T)$.

Next, we show that $x^{*}\in\Omega$. By (3.1), $u_{n}=T_{r_{n}}^{F_{1}}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}}))$, that is,

$$\begin{aligned} F_{1}(u_{n},y)+\frac{1}{r_{n}}\langle y-u_{n},u_{n}-x_{n} \rangle-\frac {1}{r_{n}}\bigl\langle y-u_{n},\gamma A^{*} \bigl(T_{r_{n}}^{F_{2}}-I\bigr)Ax_{n}\bigr\rangle \geq0 \end{aligned}$$

for all $y\in C$. From (A2), it follows that

$$\begin{aligned} -\frac{1}{r_{n}}\bigl\langle y-u_{n},\gamma A^{*}\bigl(T_{r_{n}}^{F_{2}}-I \bigr)Ax_{n}\bigr\rangle +\frac {1}{r_{n}}\langle y-u_{n},u_{n}-x_{n} \rangle\geq F_{1}(y,u_{n}) \end{aligned}$$

for all $y\in C$. Since $\|A^{*}(T_{r_{n}}^{F_{2}}-I)Ax_{n}\|\to0$, $\| u_{n}-x_{n}\|\to0$ and $\|x_{n}-x^{*}\|\to0$ as $n\to\infty$, we have

$$\begin{aligned} F_{1}\bigl(y,x^{*}\bigr)\leq0 \end{aligned}$$

for all $y\in C$. Let $y_{t}=ty+(1-t)x^{*}$ for any $0< t\leq1$ and $y\in C$. It means that $y_{t}\in C$ and hence

$$\begin{aligned} 0=F_{1}(y_{t},y_{t})\leq tF_{1}(y_{t},y)+(1-t)F_{1} \bigl(y_{t},x^{*}\bigr)\leq tF_{1}(y_{t},y), \end{aligned}$$

and then $F_{1}(y_{t},y)\geq0$. Letting $t\to0$, we immediately have $F_{1}(x^{*},y)\geq0$, i.e., $x^{*}\in EP(F_{1})$.

Next, we show that $Ax^{*}\in EP(F_{2})$. Since A is a bounded linear operator and (3.11), we have

$$\begin{aligned} \bigl\| T_{r_{n}}^{F_{2}}Ax_{n}-Ax^{*}\bigr\| \leq\bigl\| T_{r_{n}}^{Ax_{n}} \bigr\| -\bigl\| Ax_{n}-Ax^{*}\bigr\| \to0 \end{aligned}$$

as $n\to\infty$, which yields that $T_{r_{n}}^{F_{2}}Ax_{n}\to Ax^{*}$. By the definition of $T_{r_{n}}^{F_{2}}$, we have

$$\begin{aligned} F_{2}\bigl(T_{r_{n}}^{F_{2}}Ax_{n},y \bigr)+\frac{1}{r_{n}}\bigl\langle y-T_{r_{n}}^{F_{2}}Ax_{n},T_{r_{n}}^{F_{2}}Ax_{n}-Ax_{n} \bigr\rangle \geq0 \end{aligned}$$

(3.15)

for all $y\in C$. Since $F_{2}$ is upper semi-continuous in the first argument, taking lim sup in (3.15), it follows that

$$\begin{aligned} F_{2}\bigl(Ax^{*},y\bigr)\geq0 \end{aligned}$$

for all $x,y\in C$, from which it can be concluded that $Ax^{*}\in EP(F_{2})$. Consequently, $x^{*}\in\Omega$. This completes the proof. □

In Theorem 3.1, if the mapping T is a nonexpansive mapping, then we immediately have the following.

Corollary 3.2

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and C, Q be nonempty closed convex subsets of Hilbert spaces $H_{1}$ and $H_{2}$, respectively. Let $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ be two bifunctions satisfying conditions (A1)-(A4) and $F_{2}$ be upper semi-continuous in the first argument. Let $T:C\to C$ be a nonexpansive mapping and $A:H_{1}\to H_{2}$ be a bounded linear operator. Suppose that $F(T)\cap\Omega\neq\emptyset$, where $\Omega=\{v\in C: v\in EP(F_{1})\textit{ such that }Av\in EP(F_{2})\}$, and let $x_{0}\in C$. For $C_{1}=C$ and $x_{1}=P_{C_{1}}x_{0}$, define a sequence $\{x_{n}\}$ iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)x_{n},\\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})Tu_{n},\\ C_{n+1}=\{z\in C_{n}:\|y_{n}-z\|^{2}\leq\|x_{n}-z\|^{2}+\theta_{n}\},\\ x_{n+1}=P_{C_{n+1}}x_{0} \end{array}\displaystyle \right . $$

(3.16)

for each $n\in\mathbb{N}$, where $M_{n}=\sup\{\|x_{n}-z\|:x\in\Omega\}$ and $\theta_{n}=(1-\alpha_{n})(k_{n}^{2}-1)M_{n}^{2}$, $0\leq\alpha_{n}\leq a<1$ for all $n\in\mathbb{N}$, $0< b\leq r_{n}<\infty$, $\gamma\in(0,1/L)$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A. Then the sequence $\{x_{n}\}$ generated by (3.16) strongly converges to a point $z_{0}\in F(T)\cap\Omega$.

If $H_{1}=H_{2}$, $C=Q$ and $A=I$ in Theorem 3.1, then we have the following.

Corollary 3.3

Let H be a real Hilbert space and C be a nonempty closed convex subset of a Hilbert space H. Let $F_{1},F_{2}:C\times C\to\mathbb{R}$ be bifunctions satisfying conditions (A1)-(A4) and $F_{2}$ be upper semi-continuous in the first argument. Let $T:C\to C$ be an asymptotically nonexpansive mapping. Suppose that $F(T)\cap\Omega\neq \emptyset$, where $\Omega=\{v\in C: v\in EP(F_{1})\cap EP(F_{2})\}$, and let $x_{0}\in C$. For $C_{1}=C$ and $x_{1}=P_{C_{1}}x_{0}$, define a sequence $\{ x_{n}\}$ iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}T_{r_{n}}^{F_{2}}x_{n},\\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})T^{n}u_{n},\\ C_{n+1}=\{z\in C_{n}:\|y_{n}-z\|^{2}\leq\|x_{n}-z\|^{2}+\theta_{n}\},\\ x_{n+1}=P_{C_{n+1}}x_{0} \end{array}\displaystyle \right . $$

(3.17)

for each $n\in\mathbb{N}$, where $M_{n}=\sup\{\|x_{n}-z\|:z\in\Omega\}$ and $\theta_{n}=(1-\alpha_{n})(k_{n}^{2}-1)(M_{n})^{2}$, $0\leq\alpha_{n}\leq a<1$ and $0< b\leq r_{n}<\infty$ for all $n\in\mathbb{N}$. Then the sequence $\{x_{n}\}$ generated by (3.17) strongly converges to a point $z_{0}\in F(T)\cap\Omega$.

4 Applications

4.1 Applications to split variational inequality problems

Firstly, we point out the so-called variational inequality problem (shortly, $VIP$), which is to find a point $x^{*}\in C$ which satisfies the following inequality:

$$\bigl\langle Ax^{*},z-x^{*}\bigr\rangle \geq0 $$

for all $z\in C$. Its solution set is symbolized by $VI(A,C)$.

In 2012, Censor et al. [19] proposed the split variational inequality problem (shortly, $SVIP$) which is formulated as follows:

$$\mbox{Find a point }x^{*}\in C\mbox{ such that }\bigl\langle f\bigl(x^{*}\bigr), x-x^{*}\bigr\rangle \geq0\quad\mbox{for all }x\in C $$

and such that

$$y^{*}=Ax^{*}\in Q\mbox{ solves }\bigl\langle g\bigl(y^{*}\bigr),y-y^{*}\bigr\rangle \geq0 \quad\mbox{for all }y\in Q, $$

where $A:C\to C$ is a bounded linear operator. The solution set of split variational inequality problem is denoted by the $SVIP$.

Setting $F_{1}(x,y)=\langle f(x),y-x\rangle$ and $F_{2}(x,y)=\langle g(x),y-x\rangle$, it is clear that $F_{1}$, $F_{2}$ satisfy conditions (A1)-(A4), where f and g are $\eta_{1}$- and $\eta_{2}$-inverse strongly monotone mappings, respectively. Then, by Theorem 3.1, we get the following.

Theorem 4.1

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and C, Q be nonempty closed convex subsets of Hilbert spaces $H_{1}$ and $H_{2}$, respectively. Let f and g be $\eta_{1}$- and $\eta_{2}$-inverse strongly monotone mappings, respectively. Let $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ be two bifunctions satisfying conditions (A1)-(A4), which are defined by f and g, and $F_{2}$ be upper semi-continuous in the first argument. Let $T:C\to C$ be an asymptotically nonexpansive mapping and $A:H_{1}\to H_{2}$ be a bounded linear operator. Suppose that $F(T)\cap\Omega\neq\emptyset$, where $\Omega=\{v\in C: v\in EP(F_{1})\textit{ such that }Av\in EP(F_{2})\}$, and let $x_{0}\in C$. For $C_{1}=C$ and $x_{1}=P_{C_{1}}x_{0}$, define a sequence $\{x_{n}\}$ iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)x_{n},\\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})T^{n}u_{n},\\ C_{n+1}=\{z\in C_{n}:\|y_{n}-z\|^{2}\leq\|x_{n}-z\|^{2}+\theta_{n}\},\\ x_{n+1}=P_{C_{n+1}}x_{0} \end{array}\displaystyle \right . $$

(4.1)

for each $n\in\mathbb{N}$, where $M_{n}=\sup\{\|x_{n}-z\|:z\in\Omega\}$ and $\theta_{n}=(1-\alpha_{n})(k_{n}^{2}-1)M_{n}^{2}$, $0\leq\alpha_{n}\leq a<1$ for all $n\in\mathbb{N}$, $0< b\leq r_{n}<\infty$, $\gamma\in(0,1/L)$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A. Then the sequence $\{x_{n}\}$ generated by (4.1) strongly converges to a point $z_{0}\in F(T)\cap\Omega$.

Proof

The desired result can be proved directly through Theorem 3.1. □

4.2 Applications to split optimization problems

In this section, we mention applications to the split optimization problem, which is to find $x^{*}\in C$ such that

$$\begin{aligned} f\bigl(x^{*}\bigr)\geq f(x) \mbox{ for all } x\in C \mbox{ satisfying } Ax^{*}=y^{*} \in Q \mbox{ solves } g\bigl(y^{*}\bigr)\geq g(y) \end{aligned}$$

(4.2)

for all $y\in Q$. We symbolize Γ for the solution set of the split optimization problem.

Let $f:C\to\mathbb{R}$ and $g:Q\to\mathbb{R}$ be two functions satisfying the following assumption:

• (1) for each $x,y\in C$, $f(tx+(1-t)y)\leq f(y)$, and for each $u,v\in Q$, $g(tu+(1-t)v)\leq g(v)$;

• (2) $f(x)$ is concave and upper semi-continuous for all $x\in C$ and $g(u)$ is concave and upper semi-continuous for all $u\in Q$.

Let $F_{1}(x,y)=f(x)-f(y)$ for all $x,y\in C$ and $F_{2}(u,v)=g(u)-g(v)$ for all $u,v\in Q$. If f and g satisfy conditions (1) and (2), then it is clear that $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to \mathbb{R}$ are two bifunctions satisfying conditions (A1)-(A4). Therefore, by Theorem 3.1, we have the following.

Theorem 4.2

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and C, Q be nonempty closed convex subsets of Hilbert spaces $H_{1}$ and $H_{2}$, respectively. Let $f:C\to\mathbb{R}$ and $g:Q\to\mathbb{R}$ be two functions satisfying conditions (1) and (2). Let $F_{1}:C\times C\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ be two bifunctions satisfying conditions (A1)-(A4) and $F_{2}$ be upper semi-continuous in the first argument. Let $A:H_{1}\to H_{2}$ be a bounded linear operator. Suppose that $F(T)\cap\Gamma\neq\emptyset$ and let $x_{0}\in C$. For $C_{1}=C$ and $x_{1}=P_{C_{1}}x_{0}$, define a sequence $\{x_{n}\}$ iteratively as follows:

$$ \left \{ \textstyle\begin{array}{@{}l} u_{n}=T_{r_{n}}^{F_{1}}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{2}})A)x_{n},\\ y_{n}=\alpha_{n}x_{n}+(1-\alpha_{n})u_{n},\\ C_{n+1}=\{z\in C_{n}:\|y_{n}-z\|\leq\|x_{n}-z\|\},\\ x_{n+1}=P_{C_{n+1}}x_{0} \end{array}\displaystyle \right . $$

(4.3)

for each $n\in\mathbb{N}$, where $0\leq\alpha_{n}\leq a<1$, $0< b\leq r_{n}<\infty$, and $\gamma\in(0,1/L)$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A. Then the sequence $\{x_{n}\}$ generated by (4.3) strongly converges to a point $z_{0}\in F(T)\cap\Gamma$.

Proof

The desired result can be proved directly through Theorem 3.1. □