Hierarchical problems with applications to mathematical programming with multiple sets split feasibility constraints

Abstract

In this paper, we establish a strong convergence theorem for hierarchical problems, an equivalent relation between a multiple sets split feasibility problem and a fixed point problem. As applications of our results, we study the solution of mathematical programming with fixed point and multiple sets split feasibility constraints, mathematical programming with fixed point and multiple sets split equilibrium constraints, mathematical programming with fixed point and split feasibility constraints, mathematical programming with fixed point and split equilibrium constraints, minimum solution of fixed point and multiple sets split feasibility problems, minimum norm solution of fixed point and multiple sets split equilibrium problems, quadratic function programming with fixed point and multiple set split feasibility constraints, mathematical programming with fixed point and multiple set split feasibility inclusions constraints, mathematical programming with fixed point and split minimax constraints.

1 Introduction

The split feasibility problem (SFP) in finite dimensional Hilbert spaces was first introduced by Censor and Elfving [1] for modeling inverse problems which arise from phase retrievals and in medical image reconstruction. Since then, the split feasibility problem (SFP) has received much attention due to its applications in signal processing, image reconstruction, with particular progress in intensity-modulated radiation therapy, approximation theory, control theory, biomedical engineering, communications, and geophysics. For examples, one can refer to [1–5] and related literature. Since then, many researchers have studied (SFP) in finite dimensional or infinite dimensional Hilbert spaces. For example, one can see [2, 6–19].

A special case of problem (SFP) is the convexly constrained linear inverse problem in the finite dimensional Hilbert space [20]:

$$\text{(CLIP)}\text{Find }\overline{x}\in C\text{ such that }A\overline{x}=b,\text{ where }b\in H_2,$$

which has extensively been investigated by using the Landweber iterative method [21]

$$x_{n+1}:=x_n+\gamma A^T(b-A{x}_n),n\in \mathbb{N}.$$

In 2002, Byrne [2] first introduced the so-called CQ algorithm which generates a sequence $\{x_n\}$ by the following recursive procedure:

$$x_{n+1}=P_C(x_n-{\rho }_n{A}^{\ast }(I-P_Q)A{x}_n),$$

(1)

where the stepsize ${\rho }_n$ is chosen in the interval $(0,2/{\parallel A\parallel }^2)$, and $P_C$ and $P_Q$ are the metric projections onto $C\subseteq {\mathbb{R}}^n$ and $Q\subseteq {\mathbb{R}}^m$, respectively. Compared with Censor and Elfving’s algorithm [1] where the matrix inverse A is involved, the CQ algorithm (1) seems more easily executed since it only deals with metric projections with no need to compute matrix inverses.

In 2010, Xu [12] modified Byrne’s CQ algorithm and proved the weak convergence theorem in infinite Hilbert spaces for their modified algorithm.

Let C be a nonempty closed convex subset of a real Hilbert space H with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. A mapping $T:C\to H$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$; T is said to be a quasi-nonexpansive mapping if $Fix(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in C$ and $y\in Fix(T)$, we denote by $Fix(T)=\{x\in C:Tx=x\}$ the set of fixed points of T. $A:C\to H$ is called strongly positive if

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

Let f be a contraction on H and $\{{\alpha }_n\}$ be a sequence in $[0,1]$. In 2004, Xu [22] proved that under some condition on $\{{\alpha }_n\}$, the sequence $\{x_n\}$ generated by

$$x_{n+1}={\alpha }_{n}f{x}_n+(1-{\alpha }_n)T{x}_n$$

strongly converges to $x^{\ast }$ in $Fix(T)$, which is the unique solution of the variational inequality

$$〈(I-f)x^{\ast },x-x^{\ast }〉\ge 0$$

for all $x\in Fix(T)$.

Xu [23] also studied the following minimization problem over the set of fixed points of a nonexpansive operator T on a real Hilbert space H:

$$\underset{x\in Fix(T)}{min}\frac{1}{2}〈Bx,x〉-〈a,x〉,$$

where a is a given point in H and B is a strongly positive bounded linear operator on H. In [23], Xu proved that the sequence $\{x_n\}$ defined by the following iterative method

$$x_{n+1}=(I-{\alpha }_{n}B)T{x}_n+{\alpha }_{n}a$$

converges strongly to the unique solution of the minimization problem of a quadratic function. In [24], Marino et al. considered the following iterative method:

$$x_{n+1}={\alpha }_n\gamma f{x}_n+(I-{\alpha }_{n}A)T{x}_n.$$

(2)

They proved that the sequence generated by (2) converges strongly to the fixed point $x^{\ast }$ of T which solves the following:

$$〈(A-\gamma f)x^{\ast },x-x^{\ast }〉\ge 0$$

for all $x\in Fix(T)$. For some more related works, see [25–27] and the references therein.

In this paper, we establish a strong convergence theorem for hierarchical problems, an equivalent relation between a multiple sets split feasibility problem and a fixed point problem. As applications of our results, we study the solution of mathematical programming with fixed point and multiple sets split feasibility constraints, mathematical programming with fixed point and multiple sets split equilibrium constraints, mathematical programming with fixed point and split feasibility constraints, mathematical programming with fixed point and split equilibrium constraints, minimum solution of fixed point and multiple sets split feasibility problems, minimum norm solution of fixed point and multiple sets split equilibrium problems, quadratic function programming with fixed point and multiple set split feasibility constraints, mathematical programming with fixed point and multiple set split feasibility inclusions constraints, mathematical programming with fixed point and split minimax constraints.

2 Preliminaries

Throughout this paper, let ℕ be the set of positive integers and let ℝ be the set of real numbers, H be a (real) Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively, and let C be a nonempty closed convex subset of H. We denote the strong convergence and the weak convergence of $\{x_n\}$ to $x\in H$ by $x_n\to x$ and $x_n\rightharpoonup x$, respectively. For each $x,y\in H$ and $\lambda \in [0,1]$, we have

$${\parallel \lambda x+(1-\lambda )y\parallel }^2=\lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda ){\parallel x-y\parallel }^2.$$

Hence, we also have

$$2〈x-y,u-v〉={\parallel x-v\parallel }^2+{\parallel y-u\parallel }^2-{\parallel x-u\parallel }^2-{\parallel y-v\parallel }^2$$

(3)

for all $x,y,u,v\in H$.

For $\alpha >0$, a mapping $A:H\to H$ is called α-inverse-strongly monotone (α-ism) if

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

If $0<\lambda \le 2\alpha$, $A:H\to H$ is an α-inverse-strongly monotone mapping, then $I-\lambda A:H\to H$ is nonexpansive. A mapping $T:C\to H$ is said to be a firmly nonexpansive mapping if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel (I-T)x-(I-T)y\parallel }^2$$

for every $x,y\in C$. Let $T:C\to H$ be a mapping. Then $p\in C$ is called an asymptotic fixed point of T [28] if there exists $\{x_n\}\subseteq C$ such that $x_n\rightharpoonup p$, and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. We denote by $F(\hat{T})$ the set of asymptotic fixed points of T. A mapping $T:C\to H$ is said to be demiclosed if it satisfies $F(T)=F(\hat{T})$. A nonlinear operator $V:H\to H$ is called strongly monotone if there exists $\overline{\gamma }>0$ such that $〈x-y,Vx-Vy〉\ge \overline{\gamma }{\parallel x-y\parallel }^2$ for all $x,y\in H$. Such V is also called $\overline{\gamma }$-strongly monotone. A nonlinear operator $V:H\to H$ is called Lipschitzian continuous if there exists $L>0$ such that $\parallel Vx-Vy\parallel \le L\parallel x-y\parallel$ for all $x,y\in H$. Such V is also called L-Lipschitzian continuous.

Let B be a mapping of H into $2^H$. The effective domain of B is denoted by $D(B)$, that is, $D(B)=\{x\in H:Bx\ne \mathrm{\varnothing }\}$. A multi-valued mapping B is said to be a monotone operator on H if $〈x-y,u-v〉\ge 0$ for all $x,y\in D(B)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and $r>0$, we may define a single-valued operator $J_r={(I+rB)}^{-1}:H\to D(B)$, which is called the resolvent of B for r, and let $B^{-1}0=\{x\in H:0\in Bx\}$.

The following lemmas are needed in this paper.

Lemma 2.1 [29]

Let $H_1$ and $H_2$ be two real Hilbert spaces, $A:H_1\to H_2$ be a bounded linear operator, and $A^{\ast }$ be the adjoint of A. Let C be a nonempty closed convex subset of $H_2$, and let $G:H_2\to H_2$ be a firmly nonexpansive mapping. Then $A^{\ast }(I-G)A$ is a $\frac{1}{{\parallel A\parallel }^2}$-ism, that is,

$$\frac{1}{{\parallel A\parallel }^2}{\parallel A^{\ast }(I-G)Ax-A^{\ast }(I-G)Ay\parallel }^2\le 〈x-y,A^{\ast }(I-G)Ax-A^{\ast }(I-G)Ay〉$$

for all $x,y\in H_1$.

Lemma 2.2 [30]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $G:H\to H$ be a firmly nonexpansive mapping. Suppose that $Fix(G)\ne \mathrm{\varnothing }$. Then $〈x-Gx,Gx-w〉\ge 0$ for each $x\in H$ and each $w\in Fix(G)$.

A mapping $T:H\to H$ is said to be averaged if $T=(1-\alpha )I+\alpha S$, where $\alpha \in (0,1)$ and $S:H\to H$ is a nonexpansive mapping. In this case, we also say that T is α-averaged. A firmly nonexpansive mapping is $\frac{1}{2}$-averaged.

Lemma 2.3 [31]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $T:C\to C$ be a mapping. Then the following are satisfied:

• (i) T is nonexpansive if and only if the complement $(I-T)$ is $1/2$-ism.

• (ii) If S is υ-ism, then for $\gamma >0$, γS is $\upsilon /\gamma$-ism.

• (iii) S is averaged if and only if the complement $I-S$ is υ-ism for some $\upsilon >1/2$.

• (iv) If S and T are both averaged, then the product (composite) ST is averaged.

• (v) If the mappings ${\{T_i\}}_{i=1}^n$ are averaged and have a common fixed point, then ${\bigcap }_{i=1}^{n}Fix(T_i)=Fix(T_1\cdots T_n)$.

Lin and Takahashi [39] gave the following results in a Hilbert spaces.

Lemma 2.4 [32]

Let $P_C$ be the metric projection of H onto C, and let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Let $t\ge 0$ satisfy $2\overline{\gamma }>t{L}^2$ and $1>2t\overline{\gamma }$. Then we know that

$$z=P_C(I-tV)z\iff 〈Vz,y-z〉\ge 0\iff z=P_C(I-V)z.$$

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

By Lemma 2.4, we have the following lemma.

Lemma 2.5 Let $V:H\to H$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Let $\theta \in H$ and $V_1:H\to H$ such that $V_{1}x=Vx-\theta$. Then $V_1$ is a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous mapping. Furthermore, there exists a unique fixed point $z_0$ in C satisfying $z_0=P_C(z_0-V{z}_0+\theta )$. This point $z_0\in C$ is also a unique solution of the hierarchical variational inequality

$$〈V{z}_0-\theta ,q-z_0〉\ge 0,\mathrm{\forall }q\in C.$$

Lemma 2.6 [33]

Let B be a maximal monotone mapping on H. Let $J_r$ be the resolvent of B defined by $J_r={(I+rB)}^{-1}$ for each $r>0$. Then the following hold:

• (i) For each $r>0$, $J_r$ is single-valued and firmly nonexpansive;

• (ii) For each $r>0$, $\mathcal{D}(J_r)=H$ and $Fix(J_r)=\{x\in \mathcal{D}(B):0\in Bx\}$;

Lemma 2.7 [33]

Let B be a maximal monotone mapping on H. Let $J_r$ be the resolvent of B defined by $J_r={(I+rB)}^{-1}$ for each $r>0$. Then the following holds:

$$\frac{s-t}{s}〈J_{s}x-J_{t}x,J_{s}x-x〉\ge {\parallel J_{s}x-J_{t}x\parallel }^2$$

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

$$\parallel J_{s}x-J_{t}x\parallel \le \frac{|s-t|}{s}\parallel J_{s}x-x\parallel$$

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

Let $\alpha ,\beta \in \mathbb{R}$, T be a generalized hybrid mapping [34] if $\alpha {\parallel Tx-Ty\parallel }^2+(1-\alpha ){\parallel Ty-x\parallel }^2\le \beta {\parallel Tx-y\parallel }^2+(1-\beta ){\parallel x-y\parallel }^2$ for all $x,y\in C$.

Lemma 2.8 [35]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to H$ be a generalized hybrid mapping, then $F(T)=F(\hat{T})$.

Remark 2.1 If T is a generalized hybrid mapping with $Fix(T)\ne \mathrm{\varnothing }$. By the definition of T and Lemma 2.8, we have that T is a quasi-nonexpansive mapping with $F(T)=F(\hat{T})$.

Lemma 2.9 [36]

Let $\{a_n\}$ be a sequence of real numbers such that there exists a subsequence $\{n_i\}$ of $\{n\}$ such that $a_{n_i}<a_{n_i+1}$ for all $i\in \mathbb{N}$. Then there exists a nondecreasing sequence $\{m_k\}\subseteq \mathbb{N}$ such that $m_k\to \mathrm{\infty }$ and the following properties are satisfied for all (sufficiently large) numbers $k\in \mathbb{N}$:

$$a_{m_k}\le a_{m_k+1}\mathit{\text{and}}{a}_k\le a_{m_k+1}.$$

In fact, $m_k=max\{j\le k:a_j<a_{j+1}\}$.

Lemma 2.10 [37]

Let ${\{a_n\}}_{n\in \mathbb{N}}$ be a sequence of nonnegative real numbers, $\{{\alpha }_n\}$ be a sequence of real numbers in $[0,1]$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$, $\{u_n\}$ be a sequence of nonnegative real numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{u}_n<\mathrm{\infty }$, $\{t_n\}$ be a sequence of real numbers with $lim\hspace{0.17em}sup{t}_n\le 0$. Suppose that $a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{t}_n+u_n$ for each $n\in \mathbb{N}$. Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

We know that the equilibrium problem is to find $z\in C$ such that

$$\text{(EP)}g(z,y)\ge 0\text{for each }y\in C,$$

where $g:C\times C\to \mathbb{R}$ is a bifunction. This problem includes fixed point problems, optimization problems, variational inequality problems, Nash equilibrium problems, minimax inequalities, and saddle point problems as special cases. (For examples, one can see [38] and related literatures.)

The solution set of equilibrium problem (EP) is denoted by $EP(g)$. For solving the equilibrium problem, let us assume that the bifunction $g:C\times C\to \mathbb{R}$ satisfies the following conditions:

• (A1) $g(x,x)=0$ for each $x\in C$;

• (A2) g is monotone, i.e., $g(x,y)+g(y,x)\le 0$ for any $x,y\in C$;

• (A3) for each $x,y,z\in C$, ${lim}_{t\downarrow 0}g(tz+(1-t)x,y)\le g(x,y)$;

• (A4) for each $x\in C$, the scalar function $y\to g(x,y)$ is convex and lower semicontinuous.

We have the following result from Blum and Oettli [38].

Theorem 2.1 [38]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $g:C\times C\to \mathbb{R}$ be a bifunction which satisfies conditions (A1)-(A4). Then, for each $r>0$ and each $x\in H$, there exists $z\in C$ such that

$$g(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0$$

for all $y\in C$.

In 2005, Combettes and Hirstoaga [39] established the following important properties of a resolvent operator.

Theorem 2.2 [39]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let $g:C\times C\to \mathbb{R}$ be a function satisfying conditions (A1)-(A4). For $r>0$, define $T_r^g:H\to C$ by

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

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

• (i) $T_r^g$ is single-valued;

• (ii) $T_r^g$ is firmly nonexpansive, that is, ${\parallel T_r^{g}x-T_r^{g}y\parallel }^2\le 〈x-y,T_r^{g}x-T_r^{g}y〉$ for all $x,y\in H$;

• (iii) $\{x\in H:T_r^{g}x=x\}=\{x\in C:g(x,y)\ge 0,\mathrm{\forall }y\in C\}$;

• (iv) $\{x\in C:g(x,y)\ge 0,\mathrm{\forall }y\in C\}$ is a closed and convex subset of C.

We call such $T_r^g$ the resolvent of g for $r>0$.

3 Convergence theorems of hierarchical problems

Let H be a real Hilbert space, and let I be an identity mapping on H, C be a nonempty closed convex subset of H. For each $i=1,2$, let ${\kappa }_i>0$ and let $B_i$ be a ${\kappa }_i$-inverse-strongly monotone mapping of C into H. Let $G_i$ be a maximal monotone mapping on H such that the domain of $G_i$ is included in C for each $i=1,2$. Let $J_{\lambda }={(I+\lambda G_1)}^{-1}$ and $T_r={(I+r{G}_2)}^{-1}$ for each $\lambda >0$ and $r>0$. Let $\{{\theta }_n\}\subset H$ be a sequence. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$. Throughout this paper, we use these notations and assumptions unless specified otherwise.

The following strong convergence theorem for hierarchical problems is one of our main results of this paper.

Theorem 3.1 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$ such that $F(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0\ne \mathrm{\varnothing }$. Take $\mu \in \mathbb{R}$ as follows:

$$0<\mu <\frac{2\overline{\gamma }}{L^2}.$$

Let $\{x_n\}\subset H$ be defined by

$$(3.1)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=J_{{\lambda }_n}(I-{\lambda }_n{B}_1)T_{r_n}(I-r_n{B}_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<2{\kappa }_1$, and $0<a\le r_n\le b<2{\kappa }_2$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=\theta$ for some $\theta \in H$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0}(\overline{x}-V\overline{x}+\theta )$. This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:

$$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0.$$

Proof Take any $\overline{x}\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0$ and let $\overline{x}$ be fixed. Then $\overline{x}=J_{{\lambda }_n}(I-{\lambda }_n{B}_1)\overline{x}$ and $\overline{x}=T_{r_n}(I-r_n{B}_2)\overline{x}$. Let $u_n=T_{r_n}(I-r_n{B}_2)x_n$. For each $n\in \mathbb{N}$, we have

$$\begin{array}{c}{\parallel u_n-\overline{x}\parallel }^2\hfill \\ ={\parallel T_{r_n}(I-r_n{B}_2)x_n-T_{r_n}(I-r_n{B}_2)\overline{x}\parallel }^2\hfill \\ \le {\parallel (x_n-\overline{x})-r_n(B_2{x}_n-B_2\overline{x})\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2-2r_n〈x_n-\overline{x},B_2{x}_n-B_2\overline{x}〉+r_n^2{\parallel B_2{x}_n-B_2\overline{x}\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2-2r_n{\kappa }_2{\parallel B_2{x}_n-B_2\overline{x}\parallel }^2+r_n^2{\parallel B_2{x}_n-B_2\overline{x}\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2-r_n(2{\kappa }_2-r_n){\parallel B_2{x}_n-B_2\overline{x}\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2,\hfill \end{array}$$

(4)

and

$$\begin{array}{c}{\parallel y_n-\overline{x}\parallel }^2\hfill \\ ={\parallel J_{{\lambda }_n}(I-{\lambda }_n{B}_1)u_n-J_{{\lambda }_n}(I-{\lambda }_n{B}_1)\overline{x}\parallel }^2\hfill \\ \le {\parallel (u_n-\overline{x})-{\lambda }_n(B_1{u}_n-B_1\overline{x})\parallel }^2\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2-2{\lambda }_n〈u_n-\overline{x},B_1{u}_n-B_1\overline{x}〉+{\lambda }_n^2{\parallel B_1{u}_n-B_1\overline{x}\parallel }^2\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2-2{\lambda }_n{\kappa }_1{\parallel B_1{u}_n-B_1\overline{x}\parallel }^2+{\lambda }_n^2{\parallel B_1{u}_n-B_1\overline{x}\parallel }^2\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2-{\lambda }_n(2{\kappa }_1-{\lambda }_n){\parallel B_1{u}_n-B_1\overline{x}\parallel }^2\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2.\hfill \end{array}$$

(5)

Since T is a quasi-nonexpansive mapping, we obtain that

$$\parallel s_n-\overline{x}\parallel =\parallel T{y}_n-\overline{x}\parallel \le \parallel y_n-\overline{x}\parallel \le \parallel u_n-\overline{x}\parallel \le \parallel x_n-\overline{x}\parallel .$$

(6)

Let $z_n={\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n$, we have that

$$\begin{array}{rcl}\parallel z_n-\overline{x}\parallel & =& \parallel {\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n-\overline{x}\parallel \\ \le & {\beta }_n\parallel {\theta }_n-V\overline{x}\parallel +\parallel (I-{\beta }_{n}V)(s_n-\overline{x})\parallel \\ \le & {\beta }_n\parallel {\theta }_n-V\overline{x}\parallel +\parallel (I-{\beta }_{n}V)s_n-(I-{\beta }_{n}V)\overline{x}\parallel .\end{array}$$

(7)

Put $\tau =\overline{\gamma }-\frac{L^2\mu }{2}$, we have that

$$\begin{array}{c}{\parallel (I-{\beta }_{n}V)s_n-(I-{\beta }_{n}V)\overline{x}\parallel }^2\hfill \\ ={\parallel s_n-\overline{x}\parallel }^2-2{\beta }_n〈s_n-\overline{x},V{s}_n-V\overline{x}〉+{\beta }_n^2{\parallel V{s}_n-V\overline{x}\parallel }^2\hfill \\ \le {\parallel s_n-\overline{x}\parallel }^2-2{\beta }_n\overline{\gamma }{\parallel s_n-\overline{x}\parallel }^2+{\beta }_n^2{L}^2{\parallel s_n-\overline{x}\parallel }^2\hfill \\ \le (1-2{\beta }_n\overline{\gamma }+{\beta }_n^2{L}^2){\parallel s_n-\overline{x}\parallel }^2\hfill \\ \le (1-2{\beta }_n\tau -{\beta }_n(L^2\mu -{\beta }_n{L}^2)){\parallel s_n-\overline{x}\parallel }^2\hfill \\ \le (1-2{\beta }_n\tau +{\beta }_n^2{\tau }^2){\parallel s_n-\overline{x}\parallel }^2\hfill \\ \le {(1-{\beta }_n\tau )}^2{\parallel x_n-\overline{x}\parallel }^2.\hfill \end{array}$$

Since $1-{\beta }_n\tau >0$, we obtain that

$$\parallel (I-{\beta }_{n}V)s_n-(I-{\beta }_{n}V)\overline{x}\parallel \le (1-{\beta }_n\tau )\parallel x_n-\overline{x}\parallel .$$

(8)

We have from (7) and (8) that

$$\parallel z_n-\overline{x}\parallel \le {\beta }_n\parallel {\theta }_n-V\overline{x}\parallel +(1-{\beta }_n\tau )\parallel x_n-\overline{x}\parallel .$$

(9)

Thus, we obtain from the definition of $x_n$ and (9) that

$$\begin{array}{rcl}\parallel x_{n+1}-\overline{x}\parallel & =& \parallel {\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n)-\overline{x}\parallel \\ \le & {\alpha }_n\parallel x_n-\overline{x}\parallel +(1-{\alpha }_n)\parallel ({\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n)-\overline{x}\parallel \\ \le & {\alpha }_n\parallel x_n-\overline{x}\parallel +(1-{\alpha }_n)[{\beta }_n\parallel {\theta }_n-V\overline{x}\parallel +(1-{\beta }_n\tau )\parallel x_n-\overline{x}\parallel ]\\ \le & [1-(1-{\alpha }_n){\beta }_n\tau ]\parallel x_n-\overline{x}\parallel +{\beta }_n(1-{\alpha }_n)\tau \frac{\parallel {\theta }_n-V\overline{x}\parallel }{\tau }\\ \le & max\{\parallel x_n-\overline{x}\parallel ,\frac{\parallel {\theta }_n-V\overline{x}\parallel }{\tau }\}\\ \le & max\{\parallel x_n-\overline{x}\parallel ,M\},\end{array}$$

where $M=max\{\frac{\parallel {\theta }_n-V\overline{x}\parallel }{\tau },n\in \mathbb{N}\}$. By induction, we deduce

$$\parallel x_n-\overline{x}\parallel \le max\{\parallel x_1-\overline{x}\parallel ,M\}.$$

This implies that the sequence $\{x_n\}$ is bounded. Furthermore, $\{u_n\}$, $\{z_n\}$, $\{y_n\}$ and $\{s_n\}$ are bounded.

By the definition of $\{x_n\}$, we have that

$$\begin{array}{rcl}{x}_{n+1}-x_n& =& {\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n)-x_n\\ =& (1-{\alpha }_n)[({\beta }_n{\theta }_n+(I-{\beta }_{n}V)s_n)-x_n]\\ =& (1-{\alpha }_n)[{\beta }_n{\theta }_n-{\beta }_{n}V{s}_n+s_n-x_n].\end{array}$$

(10)

By (10), we have that

$$\begin{array}{c}〈x_{n+1}-x_n,x_n-\overline{x}〉\hfill \\ =〈(1-{\alpha }_n)[{\beta }_n{\theta }_n-{\beta }_{n}V{s}_n+s_n-x_n],x_n-\overline{x}〉\hfill \\ =(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉+(1-{\alpha }_n)〈s_n-x_n,x_n-\overline{x}〉.\hfill \end{array}$$

(11)

By (3) and (11), we have that

$$\begin{array}{c}{\parallel x_{n+1}-\overline{x}\parallel }^2-{\parallel x_n-\overline{x}\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ =2(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-2(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉\hfill \\ +(1-{\alpha }_n)[{\parallel s_n-\overline{x}\parallel }^2-{\parallel x_n-\overline{x}\parallel }^2-{\parallel s_n-x_n\parallel }^2].\hfill \end{array}$$

(12)

By (5) and (12), we have that

$$\begin{array}{c}{\parallel x_{n+1}-\overline{x}\parallel }^2-{\parallel x_n-\overline{x}\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ \le 2(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-2(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉\hfill \\ -(1-{\alpha }_n){\parallel s_n-x_n\parallel }^2.\hfill \end{array}$$

(13)

By (10), we obtain that

$$\begin{array}{c}{\parallel x_{n+1}-x_n\parallel }^2\hfill \\ \le {(1-{\alpha }_n)}^2{[{\beta }_n\parallel {\theta }_n-V{s}_n\parallel +\parallel s_n-x_n\parallel ]}^2\hfill \\ ={(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel {\theta }_n-V{s}_n\parallel }^2+{\parallel s_n-x_n\parallel }^2+2{\beta }_n\parallel {\theta }_n-V{s}_n\parallel \parallel s_n-x_n\parallel ].\hfill \end{array}$$

(14)

By (13) and (14), we have that

$$\begin{array}{c}{\parallel x_{n+1}-\overline{x}\parallel }^2-{\parallel x_n-\overline{x}\parallel }^2\hfill \\ \le 2(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-2(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉-(1-{\alpha }_n){\parallel s_n-x_n\parallel }^2\hfill \\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel {\theta }_n-V{s}_n\parallel }^2+{\parallel s_n-x_n\parallel }^2+2{\beta }_n\parallel {\theta }_n-V{s}_n\parallel \parallel s_n-x_n\parallel ]\hfill \\ \le 2(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-2(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉-(1-{\alpha }_n){\alpha }_n{\parallel s_n-x_n\parallel }^2\hfill \\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel {\theta }_n-V{s}_n\parallel }^2+2{\beta }_n\parallel {\theta }_n-V{s}_n\parallel \parallel s_n-x_n\parallel ].\hfill \end{array}$$

Hence, we obtain that

$$\begin{array}{c}{\parallel x_{n+1}-\overline{x}\parallel }^2-{\parallel x_n-\overline{x}\parallel }^2+(1-{\alpha }_n){\alpha }_n{\parallel s_n-x_n\parallel }^2\hfill \\ \le 2(1-{\alpha }_n){\beta }_n〈{\theta }_n,x_n-\overline{x}〉-2(1-{\alpha }_n){\beta }_n〈V{s}_n,x_n-\overline{x}〉\hfill \\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel {\theta }_n-V{s}_n\parallel }^2+2{\beta }_n\parallel {\theta }_n-V{s}_n\parallel \parallel s_n-x_n\parallel ].\hfill \end{array}$$

(15)

We will divide the proof into two cases as follows.

Case 1: There exists a natural number N such that $\parallel x_{n+1}-\overline{x}\parallel \le \parallel x_n-\overline{x}\parallel$ for each $n\ge N$. So, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\overline{x}\parallel$ exists. Hence, it follows from (15), (i), and (ii) that

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

(16)

By (14), (16), (i), and (ii), we have that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(17)

We also have that

$$\parallel z_n-s_n\parallel \le \parallel {\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n-s_n\parallel \le {\beta }_n\parallel {\theta }_n-V{s}_n\parallel .$$

(18)

By (18), (iv), and (ii), we have that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-s_n\parallel =0.$$

(19)

By (16) and (19), we have that

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

(20)

By (10) and (6), we have that

$${\parallel s_n-\overline{x}\parallel }^2\le {\parallel u_n-\overline{x}\parallel }^2\le {\parallel x_n-\overline{x}\parallel }^2-r_n(2{\kappa }_2-r_n){\parallel B_2{u}_n-B_2\overline{x}\parallel }^2.$$

Therefore,

$$\begin{array}{rcl}{r}_n(2{\kappa }_2-r_n){\parallel B_2{u}_n-B_2\overline{x}\parallel }^2& \le & {\parallel x_n-\overline{x}\parallel }^2-{\parallel s_n-\overline{x}\parallel }^2\\ \le & \parallel x_n-s_n\parallel (\parallel s_n-\overline{x}\parallel +\parallel x_n-\overline{x}\parallel ).\end{array}$$

(21)

Thus, by (16), (21), and (iii), we have that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_2{u}_n-B_2\overline{x}\parallel =0.$$

(22)

Since $T_{r_n}$ is firmly nonexpansive, we have from (3) that

$$\begin{array}{c}2{\parallel u_n-\overline{x}\parallel }^2\hfill \\ =2{\parallel T_{r_n}(I-r_n{B}_2)x_n-T_{r_n}(I-r_n{B}_2)\overline{x}\parallel }^2\hfill \\ \le 2〈u_n-\overline{x},(I-r_n{B}_2)x_n-(I-r_n{B}_2)\overline{x}〉\hfill \\ \le 2〈u_n-\overline{x},x_n-\overline{x}〉-2r_n〈u_n-\overline{x},B_2{x}_n-B_2\overline{x}〉\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2+{\parallel x_n-\overline{x}\parallel }^2-{\parallel u_n-x_n\parallel }^2\hfill \\ -2r_n〈x_n-\overline{x},B_2{x}_n-B_2\overline{x}〉-2r_n〈u_n-x_n,B_2{x}_n-B_2\overline{x}〉\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2+{\parallel x_n-\overline{x}\parallel }^2-{\parallel u_n-x_n\parallel }^2\hfill \\ -2{\lambda }_n{\kappa }_2{\parallel B_2{x}_n-B_2\overline{x}\parallel }^2+2r_n〈x_n-u_n,B_2{x}_n-B_2\overline{x}〉\hfill \\ \le {\parallel u_n-\overline{x}\parallel }^2+{\parallel x_n-\overline{x}\parallel }^2-{\parallel u_n-x_n\parallel }^2+2r_n\parallel x_n-u_n\parallel \parallel B_2{x}_n-B_2\overline{x}\parallel .\hfill \end{array}$$

(23)

By (6) and (23), we have that

$$\begin{array}{rcl}{\parallel s_n-\overline{x}\parallel }^2& \le & {\parallel u_n-\overline{x}\parallel }^2\\ \le & {\parallel x_n-\overline{x}\parallel }^2-{\parallel u_n-x_n\parallel }^2+2r_n\parallel x_n-u_n\parallel \parallel B_2{x}_n-B_2\overline{x}\parallel .\end{array}$$

Therefore,

$$\begin{array}{c}{\parallel u_n-x_n\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2-{\parallel s_n-\overline{x}\parallel }^2+2r_n\parallel x_n-u_n\parallel \parallel B_2{x}_n-B_2\overline{x}\parallel \hfill \\ \le \parallel x_n-s_n\parallel (\parallel x_n-\overline{x}\parallel +\parallel s_n-\overline{x}\parallel )+2r_n\parallel x_n-u_n\parallel \parallel B_2{x}_n-B_2\overline{x}\parallel .\hfill \end{array}$$

(24)

Thus, by (16), (22), and (24), we have that

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

(25)

By (5) and (6), we have that

$${\parallel s_n-\overline{x}\parallel }^2\le {\parallel y_n-\overline{x}\parallel }^2\le {\parallel x_n-\overline{x}\parallel }^2-{\lambda }_n(2{\kappa }_1-{\lambda }_n){\parallel B_1{u}_n-B_1\overline{x}\parallel }^2.$$

Therefore,

$$\begin{array}{rcl}{\lambda }_n(2{\kappa }_1-{\lambda }_n){\parallel B_1{u}_n-B_1\overline{x}\parallel }^2& \le & {\parallel x_n-\overline{x}\parallel }^2-{\parallel s_n-\overline{x}\parallel }^2\\ \le & \parallel x_n-s_n\parallel (\parallel s_n-\overline{x}\parallel +\parallel x_n-\overline{x}\parallel ).\end{array}$$

(26)

Thus, by (16), (26), and (iii), we have that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_1{u}_n-B_1\overline{x}\parallel =0.$$

(27)

Since $J_{{\lambda }_n}$ is firmly nonexpansive, we have from (3) that

$$\begin{array}{c}2{\parallel y_n-\overline{x}\parallel }^2\hfill \\ =2{\parallel J_{{\lambda }_n}(I-{\lambda }_n{B}_1)u_n-J_{{\lambda }_n}(I-{\lambda }_n{B}_1)\overline{x}\parallel }^2\hfill \\ \le 2〈y_n-\overline{x},(I-{\lambda }_n{B}_1)u_n-(I-{\lambda }_n{B}_1)\overline{x}〉\hfill \\ \le 2〈y_n-\overline{x},u_n-\overline{x}〉-2{\lambda }_n〈y_n-\overline{x},B_1{u}_n-B_1\overline{x}〉\hfill \\ \le {\parallel y_n-\overline{x}\parallel }^2+{\parallel u_n-\overline{x}\parallel }^2-{\parallel y_n-u_n\parallel }^2\hfill \\ -2{\lambda }_n〈u_n-\overline{x},B_1{u}_n-B_1\overline{x}〉-2{\lambda }_n〈y_n-u_n,B_1{u}_n-B_1\overline{x}〉\hfill \\ \le {\parallel y_n-\overline{x}\parallel }^2+{\parallel u_n-\overline{x}\parallel }^2-{\parallel y_n-u_n\parallel }^2\hfill \\ -2{\lambda }_n{\kappa }_1{\parallel B_1{u}_n-B_1\overline{x}\parallel }^2+2{\lambda }_n〈u_n-y_n,B_1{u}_n-B_1\overline{x}〉\hfill \\ \le {\parallel y_n-\overline{x}\parallel }^2+{\parallel u_n-\overline{x}\parallel }^2-{\parallel y_n-u_n\parallel }^2+2{\lambda }_n\parallel u_n-y_n\parallel \parallel B_1{u}_n-B_1\overline{x}\parallel .\hfill \end{array}$$

(28)

By (6) and (28), we have that

$$\begin{array}{rcl}{\parallel s_n-\overline{x}\parallel }^2& \le & {\parallel y_n-\overline{x}\parallel }^2\\ \le & {\parallel u_n-\overline{x}\parallel }^2-{\parallel y_n-u_n\parallel }^2+2{\lambda }_n\parallel u_n-y_n\parallel \parallel B_1{u}_n-B_1\overline{x}\parallel \\ \le & {\parallel x_n-\overline{x}\parallel }^2-{\parallel y_n-u_n\parallel }^2+2{\lambda }_n\parallel u_n-y_n\parallel \parallel B_1{u}_n-B_1\overline{x}\parallel .\end{array}$$

Therefore,

$$\begin{array}{c}{\parallel y_n-u_n\parallel }^2\hfill \\ \le {\parallel x_n-\overline{x}\parallel }^2-{\parallel s_n-\overline{x}\parallel }^2+2{\lambda }_n\parallel u_n-y_n\parallel \parallel B_1{u}_n-B_1\overline{x}\parallel \hfill \\ \le \parallel x_n-s_n\parallel (\parallel x_n-\overline{x}\parallel +\parallel s_n-\overline{x}\parallel )+2{\lambda }_n\parallel u_n-y_n\parallel \parallel B_1{u}_n-B_1\overline{x}\parallel .\hfill \end{array}$$

(29)

Thus, by (16), (27), and (29), we have that

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

(30)

Since $Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0$ is a nonempty closed convex subset of H, by Lemma 2.5, we can take ${\overline{x}}_0\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0$ such that

$${\overline{x}}_0=P_{Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0}({\overline{x}}_0-V{\overline{x}}_0+\theta ).$$

This point ${\overline{x}}_0$ is also a unique solution of the hierarchical variational inequality

$$〈V{\overline{x}}_0-\theta ,q-{\overline{x}}_0〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0.$$

(31)

We want to show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_n-{\overline{x}}_0〉\ge 0.$$

Without loss of generality, there exists a subsequence $\{z_{n_k}\}$ of $\{z_n\}$ such that $z_{n_k}\rightharpoonup w$ for some $w\in H$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_n-{\overline{x}}_0〉=\underset{k\to \mathrm{\infty }}{lim}〈(V{\overline{x}}_0-\theta ,z_{n_k}-{\overline{x}}_0〉.$$

(32)

By (20) and (25), we have that

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

and $u_{n_k}\rightharpoonup w$. On the other hand, since $0<a\le {\lambda }_n\le b<2{\kappa }_1$, there exists a subsequence $\{{\lambda }_{n_{k_j}}\}$ of $\{{\lambda }_{n_k}\}$ such that $\{{\lambda }_{n_{k_j}}\}$ converges to a number $\overline{\lambda }\in [a,b]$. By (30) and Lemma 2.7, we have that

$$\begin{array}{c}\parallel u_{n_{k_j}}-J_{\overline{\lambda }}(I-\overline{\lambda }{B}_1)u_{n_{k_j}}\parallel \hfill \\ \le \parallel u_{n_{k_j}}-J_{{\lambda }_{n_{k_j}}}(I-{\lambda }_{n_{k_j}}{B}_1)u_{n_{k_j}}\parallel +\parallel J_{{\lambda }_{n_{k_j}}}(I-\overline{\lambda }{B}_1)u_{n_{k_j}}-J_{\overline{\lambda }}(I-\overline{\lambda }{B}_1)u_{n_{k_j}}\parallel \hfill \\ +\parallel J_{{\lambda }_{n_{k_j}}}(I-{\lambda }_{n_{k_j}}{B}_1)u_{n_{k_j}}-J_{{\lambda }_{n_{k_j}}}(I-\overline{\lambda }{B}_1)u_{n_{k_j}}\parallel \hfill \\ \le \parallel u_{n_{k_j}}-y_{n_{k_j}}\parallel +|{\lambda }_{n_{k_j}}-\overline{\lambda }|\parallel B_1{u}_{n_{k_j}}\parallel \hfill \\ +\frac{|{\lambda }_{n_{k_j}}-\overline{\lambda }|}{\overline{\lambda }}\parallel J_{\overline{\lambda }}(I-\overline{\lambda }{B}_1)u_{n_{k_j}}-(I-\overline{\lambda }{B}_1)u_{n_{k_j}}\parallel \to 0.\hfill \end{array}$$

(33)

By (33), $u_{n_{k_j}}\rightharpoonup w$, Lemma 2.6 and 2.7, $w\in Fix(J_{\overline{\lambda }}(I-\overline{\lambda }{B}_1))={(B_1+G_1)}^{-1}0$. Without loss of generality and $0<a\le r_n\le b<2{\kappa }_2$, there exists a subsequence $\{r_{n_{k_j}}\}$ of $\{r_{n_k}\}$ such that $\{r_{n_{k_j}}\}$ converges to a number $\overline{r}\in [a,b]$. By (25) and Lemma 2.7, we have that

$$\begin{array}{c}\parallel x_{n_{k_j}}-T_{\overline{r}}(I-\overline{r}{B}_2)x_{n_{k_j}}\parallel \hfill \\ \le \parallel x_{n_{k_j}}-T_{r_{n_{k_j}}}(I-r_{n_{k_j}}{B}_2)x_{n_{k_j}}\parallel +\parallel T_{r_{n_{k_j}}}(I-r_{n_{k_j}}{B}_2)x_{n_{k_j}}-T_{r_{n_{k_j}}}(I-\overline{r}{B}_2)x_{n_{k_j}}\parallel \hfill \\ +\parallel T_{r_{n_{k_j}}}(I-\overline{r}{B}_2)u_{n_{k_j}}-T_{\overline{r}}(I-\overline{r}{B}_2)u_{n_{k_j}}\parallel \hfill \\ \le \parallel x_{n_{k_j}}-u_{n_{k_j}}\parallel +|r_{n_{k_j}}-\overline{r}|\parallel B_2{u}_{n_{k_j}}\parallel \hfill \\ +\frac{|r_{n_{k_j}}-\overline{r}|}{\overline{r}}\parallel T_{\overline{r}}(I-\overline{r}{B}_2)u_{n_{k_j}}-(I-\overline{r}{B}_2)u_{n_{k_j}}\parallel \to 0.\hfill \end{array}$$

(34)

By (34), $x_{n_{k_j}}\rightharpoonup w$, Lemma 2.8, we have that $w\in Fix(T_{\overline{r}}(I-\overline{r}{B}_2))={(B_2+G_2)}^{-1}0$. From (16), (25), and (30), we have that

$$\parallel T{y}_n-y_n\parallel =\parallel s_n-y_n\parallel \le \parallel s_n-x_n\parallel +\parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0.$$

Since $Fix(T)=Fix(\hat{T})$, we have from $\parallel T{y}_{n_j}-y_{n_j}\parallel \to 0$ and $y_{n_{k_j}}\rightharpoonup w$ that $w\in F(T)$. Hence, $w\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0$. So, we have from (31) and (32) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_n-{\overline{x}}_0〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_0-\theta ,z_{n_k}-{\overline{x}}_0〉=〈V{\overline{x}}_0-\theta ,w-{\overline{x}}_0〉\ge 0.$$

(35)

Let $z_n={\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n$. Then it follows from (7) that

$$\begin{array}{rcl}{\parallel z_n-{\overline{x}}_0\parallel }^2& =& {\parallel {\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n-{\overline{x}}_0\parallel }^2\\ =& {\parallel {\beta }_n({\theta }_n-V{\overline{x}}_0)+(1-{\beta }_{n}V)(s_n-{\overline{x}}_0)\parallel }^2\\ \le & {\parallel (1-{\beta }_{n}V)(s_n-{\overline{x}}_0)\parallel }^2+2{\beta }_n〈{\theta }_n-V{\overline{x}}_0,z_n-{\overline{x}}_0〉\\ \le & {(1-{\beta }_n\tau )}^2{\parallel x_n-{\overline{x}}_0\parallel }^2+2{\beta }_n〈{\theta }_n-V{\overline{x}}_0,z_n-{\overline{x}}_0〉.\end{array}$$

(36)

Thus, we obtain from the definition of $x_n$ and (36) that

$$\begin{array}{c}{\parallel x_{n+1}-{\overline{x}}_0\parallel }^2\hfill \\ ={\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)-{\overline{x}}_0\parallel }^2\hfill \\ \le {\alpha }_n{\parallel x_n-{\overline{x}}_0\parallel }^2+(1-{\alpha }_n){\parallel ({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)-{\overline{x}}_0\parallel }^2\hfill \\ \le {\alpha }_n{\parallel x_n-{\overline{x}}_0\parallel }^2+(1-{\alpha }_n)({(1-{\beta }_n\tau )}^2{\parallel x_n-{\overline{x}}_0\parallel }^2+2{\beta }_n〈{\theta }_n-V{\overline{x}}_0,z_n-{\overline{x}}_0〉)\hfill \\ \le [{\alpha }_n+(1-{\alpha }_n){(1-{\beta }_n\tau )}^2]{\parallel x_n-{\overline{x}}_0\parallel }^2+2{\beta }_n(1-{\alpha }_n)〈{\theta }_n-V{\overline{x}}_0,z_n-{\overline{x}}_0〉\hfill \\ \le [1-(1-{\alpha }_n)(2{\beta }_n\tau -{({\beta }_n\tau )}^2)]{\parallel x_n-{\overline{x}}_0\parallel }^2+2{\beta }_n(1-{\alpha }_n)〈{\theta }_n-V{\overline{x}}_0,z_n-{\overline{x}}_0〉\hfill \\ \le [1-2(1-{\alpha }_n){\beta }_n\tau ]{\parallel x_n-{\overline{x}}_0\parallel }^2+(1-{\alpha }_n){({\beta }_n\tau )}^2{\parallel x_n-{\overline{x}}_0\parallel }^2\hfill \\ +2{\beta }_n(1-{\alpha }_n)〈{\theta }_n-\theta ,z_n-{\overline{x}}_0〉+2{\beta }_n(1-{\alpha }_n)〈\theta -V{\overline{x}}_0,z_n-{\overline{x}}_0〉\hfill \\ \le [1-2(1-{\alpha }_n){\beta }_n\tau ]{\parallel x_n-{\overline{x}}_0\parallel }^2+2(1-{\alpha }_n){\beta }_n\tau (\frac{{\beta }_n\tau {\parallel x_n-{\overline{x}}_0\parallel }^2}{2}\hfill \\ +\frac{〈{\theta }_n-\theta ,z_n-{\overline{x}}_0〉}{\tau }+\frac{〈\theta -V{\overline{x}}_0,z_n-{\overline{x}}_0〉}{\tau }).\hfill \end{array}$$

(37)

By (35), (37), assumptions, and Lemma 2.10, we know that ${lim}_{n\to \mathrm{\infty }}{x}_n={\overline{x}}_0$, where

$${\overline{x}}_0=P_{Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0}({\overline{x}}_0-V{\overline{x}}_0+\theta ).$$

Case 2: Suppose that there exists $\{n_i\}$ of $\{n\}$ such that $\parallel x_{n_i}-\overline{x}\parallel \le \parallel x_{n_i+1}-\overline{x}\parallel$ for all $i\in \mathbb{N}$. By Lemma 2.9, there exists a nondecreasing sequence $\{m_j\}$ in ℕ such that $m_j\to \mathrm{\infty }$ and

$$\parallel x_{m_j}-\overline{x}\parallel \le \parallel x_{m_j+1}-\overline{x}\parallel \text{and}\parallel x_j-\overline{x}\parallel \le \parallel x_{m_j+1}-\overline{x}\parallel .$$

(38)

Hence, it follows from (15) and (38) that

$$\begin{array}{c}(1-{\alpha }_{m_j}){\alpha }_{m_j}{\parallel s_{m_j}-x_{m_j}\parallel }^2\hfill \\ \le 2(1-{\alpha }_{m_j}){\beta }_{m_j}〈{\theta }_{m_j},x_{m_j}-\overline{x}〉-2(1-{\alpha }_{m_j}){\beta }_{m_j}〈V{s}_{m_j},x_{m_j}-\overline{x}〉\hfill \\ +{(1-{\alpha }_{m_j})}^2[{\beta }_{m_j}^2{\parallel {\theta }_{m_j}-V{s}_{m_j}\parallel }^2+2{\beta }_{m_j}\parallel {\theta }_{m_j}-V{s}_{m_j}\parallel \parallel s_{m_j}-x_{m_j}\parallel ]\hfill \end{array}$$

(39)

for each $j\in \mathbb{N}$. Hence, it follows from (39), (i), and (ii) that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel s_{m_j}-x_{m_j}\parallel =0.$$

(40)

We want to show that

$$\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_{m_j}-{\overline{x}}_0〉\ge 0.$$

Without loss of generality, there exists a subsequence $\{z_{m_{j_k}}\}$ of $\{z_{m_j}\}$ such that $z_{m_{j_k}}\rightharpoonup w$ for some $w\in H$ and

$$\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_{m_j}-{\overline{x}}_0〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_0-\theta ,z_{m_{j_k}}-{\overline{x}}_0〉.$$

(41)

With the similar argument as in the proof of Case 1, we have $w\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0$. So, we have from (41) and (32) that

$$\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈V{\overline{x}}_0-\theta ,z_{m_j}-{\overline{x}}_0〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_0-\theta ,z_{m_{j_k}}-{\overline{x}}_0〉=〈V{\overline{x}}_0-\theta ,w-{\overline{x}}_0〉\ge 0.$$

(42)

With the similar argument as in the proof of Case 1, we have

$$\begin{array}{c}{\parallel x_{m_j+1}-{\overline{x}}_0\parallel }^2\hfill \\ \le [1-2(1-{\alpha }_{m_j}){\beta }_{m_j}\tau ]{\parallel x_{m_j}-{\overline{x}}_0\parallel }^2+(1-{\alpha }_{m_j}){({\beta }_{m_j}\tau )}^2{\parallel x_{m_j}-{\overline{x}}_0\parallel }^2\hfill \\ +2{\beta }_{m_j}(1-{\alpha }_{m_j})〈{\theta }_n-\theta ,z_{m_j}-{\overline{x}}_0〉+2{\beta }_{m_j}(1-{\alpha }_{m_j})〈\theta -V{\overline{x}}_0,z_{m_j}-{\overline{x}}_0〉.\hfill \end{array}$$

(43)

From $\parallel x_{m_j}-\overline{x}\parallel \le \parallel x_{m_j+1}-\overline{x}\parallel$, we have that

$$\begin{array}{c}2(1-{\alpha }_{m_j}){\beta }_{m_j}\tau {\parallel x_{m_j}-{\overline{x}}_0\parallel }^2\hfill \\ \le (1-{\alpha }_{m_j}){({\beta }_{m_j}\tau )}^2{\parallel x_{m_j}-{\overline{x}}_0\parallel }^2+2{\beta }_{m_j}(1-{\alpha }_{m_j})〈{\theta }_n-\theta ,z_{m_j}-{\overline{x}}_0〉\hfill \\ +2{\beta }_{m_j}(1-{\alpha }_{m_j})〈\theta -V{\overline{x}}_0,z_{m_j}-{\overline{x}}_0〉.\hfill \end{array}$$

(44)

Since $(1-{\alpha }_{m_j}){\beta }_{m_j}>0$, we have that

$$2\tau {\parallel x_{m_j}-{\overline{x}}_0\parallel }^2\le {\beta }_{m_j}\tau {\parallel x_{m_j}-{\overline{x}}_0\parallel }^2+2〈{\theta }_n-\theta ,z_{m_j}-{\overline{x}}_0〉+2〈\theta -V{\overline{x}}_0,z_{m_j}-{\overline{x}}_0〉.$$

(45)

By (42), (45), and assumptions, we know that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{m_j}-{\overline{x}}_0\parallel =0.$$

By (14), (40), and assumptions, we know that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{m_j+1}-x_{m_j}\parallel =0.$$

Thus, we have that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{m_j+1}-{\overline{x}}_0\parallel =0.$$

(46)

By (38) and (46), we have that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_j-{\overline{x}}_0\parallel \le \underset{j\to \mathrm{\infty }}{lim}\parallel x_{m_j+1}-{\overline{x}}_0\parallel =0.$$

Therefore, the proof is completed. □

Let C and Q be nonempty closed convex subsets of real Hilbert spaces $H_1$ and $H_2$, respectively. Let I denote the identy mapping on $H_1$ and on $H_2$. Let $G_i$ be a maximal monotone mapping on $H_1$ such that the domain of $G_i$ is included in C for each $i=1,2$. Let $J_{\lambda }={(I+\lambda G_1)}^{-1}$ and $T_r={(I+r{G}_2)}^{-1}$ for each $\lambda >0$ and $r>0$. Let $A_i:H_1\to H_2$ be a bounded linear operator, and let $A_i^{\ast }$ be the adjoint of $A_i$ for each $i=1,2$. Now, we recall the following multiple sets split feasibility problem:

(MSFP_FF) Find $\overline{x}\in H_1$ such that $\overline{x}\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $A_1\overline{x}\in Fix(F_1)$, and $A_2\overline{x}\in Fix(F_2)$ for each $n\in \mathbb{N}$.

In order to study the convergence theorems for the solution set of multiple sets split feasibility problem (MSFP_FF), we must give an essential result in this paper.

Theorem 3.2 Given any $\overline{x}\in H_1$.

• (i) If $\overline{x}$ is a solution of (MSFP_FF), then $J_{{\lambda }_n}(I-{\rho }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-{\sigma }_n{A}_2^{\ast }(I-F_2)A_2)\overline{x}=\overline{x}$ for each $n\in \mathbb{N}$.

• (ii) Suppose that $J_{{\lambda }_n}(I-{\rho }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-{\sigma }_n{A}_2^{\ast }(I-F_2)A_2)\overline{x}=\overline{x}$ with $0<{\rho }_n<\frac{2}{{\parallel A_1\parallel }^2+2}$, $0<{\sigma }_n<\frac{2}{{\parallel A_2\parallel }^2+2}$ for each $n\in \mathbb{N}$ and the solution set of (MSFP_FF) is nonempty. Then $\overline{x}$ is a solution of (MSFP_FF).

Proof (i) Suppose that $\overline{x}\in H_1$ is a solution of (MSFP_FF). Then $\overline{x}\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $A_1\overline{x}\in Fix(F_1)$, and $A_2\overline{x}\in Fix(F_2)$ for each $n\in \mathbb{N}$. It is easy to see that

$$J_{{\lambda }_n}(I-{\rho }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-{\sigma }_n{A}_2^{\ast }(I-F_2)A_2)\overline{x}=\overline{x}$$

for each $n\in \mathbb{N}$.

(ii) Since the solution set of (MSFP_FF) is nonempty, there exists $\overline{w}\in H_1$ such that $\overline{w}\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $A_1\overline{w}\in Fix(F_1)$, and $A_2\overline{w}\in Fix(F_2)$. So,

$$\overline{w}\in Fix(J_{{\lambda }_n})\cap Fix(I-{\rho }_n{A}_1^{\ast }(I-F_1)A_1)\cap Fix(T_{r_n})\cap Fix(I-{\sigma }_n{A}_2^{\ast }(I-F_2)A_2)\ne \mathrm{\varnothing }.$$

(47)

By Lemma 2.1, we have that

$$A_1^{\ast }(I-F_1)A_1\text{ is }\frac{1}{{\parallel A_1\parallel }^2}\text{-ism}.$$

(48)

For each $n\in \mathbb{N}$, by (48), $0<{\rho }_n<\frac{2}{{\parallel A_1\parallel }^2+2}$, and Lemma 2.3(ii), (iii), we know that

$$I-{\rho }_n{A}_1^{\ast }(I-F_1)A_1\text{ is averaged.}$$

(49)

By Lemma 2.1 again, we have that

$$A_2^{\ast }(I-F_2)A_2\text{ is }\frac{1}{{\parallel A_2\parallel }^2}\text{-ism}.$$

(50)

For each $n\in \mathbb{N}$, by (50), $0<{\sigma }_n<\frac{2}{{\parallel A_2\parallel }^2+2}$, and Lemma 2.3(ii), (iii), we know that

$$I-{\sigma }_n{A}_2^{\ast }(I-F_2)A_2\text{ is averaged.}$$

(51)

On the other hand, for each $n\in \mathbb{N}$, since $J_{{\lambda }_n}$, and $T_{r_n}$ are firmly nonexpansive mappings, it is easy to see that

$$J_{{\lambda }_n}\text{ and }{T}_{r_n}\text{ are }\frac{1}{2}\text{ averaged}.$$

(52)

Hence, by (49), (51), (52), and Lemma 2.3(v), we have that for each $n\in \mathbb{N}$,

$$\begin{array}{c}\overline{x}\in Fix\left(J_{{\lambda }_n}(I-\rho A_1^{\ast }(I-F_1)A_1)T_{r_n}(I-\sigma A_2^{\ast }(I-F_2)A_2)\right)\hfill \\ =Fix(J_{{\lambda }_n})\cap Fix(I-\rho A_1^{\ast }(I-F_1)A_1)\cap Fix(T_{r_n})\cap Fix(I-\sigma A_2^{\ast }(I-F_2)A_2).\hfill \end{array}$$

This implies that for each $n\in \mathbb{N}$,

$$\overline{x}=J_{{\lambda }_n}(I-\rho A_1^{\ast }(I-F_1)A_1)\overline{x}\text{and}\overline{x}=T_{r_n}(I-\sigma A_2^{\ast }(I-F_2)A_2)\overline{x}.$$

By Lemma 2.2, for each $n\in \mathbb{N}$,

$$\begin{array}{c}〈(\overline{x}-\rho A_1^{\ast }(I-F_1)A_1\overline{x})-\overline{x},\overline{x}-w〉\ge 0\text{for each }w\in Fix(J_{{\lambda }_n}),\hfill \\ 〈(\overline{x}-\rho A_2^{\ast }(I-F_2)A_2\overline{x})-\overline{x},\overline{x}-w〉\ge 0\text{for each }w\in Fix(T_{r_n}).\hfill \end{array}$$

That is, for each $n\in \mathbb{N}$,

$$\begin{array}{r}〈A_1^{\ast }(I-F_1)A_1\overline{x},\overline{x}-w〉\le 0\text{for each }w\in Fix(J_{{\lambda }_n}),\\ 〈A_2^{\ast }(I-F_2)A_2\overline{x},\overline{x}-w〉\le 0\text{for each }w\in Fix(T_{r_n}).\end{array}$$

(53)

For each $n\in \mathbb{N}$, by (53) and the fact that $A_i^{\ast }$ is the adjoint of $A_i$ for each $i=1,2$,

$$\begin{array}{r}〈A_1\overline{x}-F_1{A}_1\overline{x},A_1\overline{x}-A_{1}w〉\le 0\text{for each }w\in Fix(J_{{\lambda }_n}),\\ 〈A_2\overline{x}-F_2{A}_2\overline{x},A_2\overline{x}-A_{2}w〉\le 0\text{for each }w\in Fix(T_{r_n}).\end{array}$$

(54)

On the other hand, by Lemma 2.2 again,

$$\begin{array}{r}〈A_1\overline{x}-F_1{A}_1\overline{x},v_1-F_{1}A\overline{x}〉\le 0\text{for each }{v}_1\in Fix(F_1),\\ 〈A_2\overline{x}-F_2{A}_2\overline{x},v_2-F_2{A}_2\overline{x}〉\le 0\text{for each }{v}_2\in Fix(F_2).\end{array}$$

(55)

For each $n\in \mathbb{N}$, by (54) and (55),

$$\begin{array}{r}〈A_1\overline{x}-F_1{A}_1\overline{x},v_1-F_1{A}_1\overline{x}+A_1\overline{x}-A_{1}w〉\le 0,\\ 〈A_2\overline{x}-F_2{A}_2\overline{x},v_2-F_2{A}_2\overline{x}+A_2\overline{x}-Aw〉\le 0\end{array}$$

(56)

for each $w\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $v_2\in Fix(F_2)$, and $v_1\in Fix(F_1)$.

That is, for each $n\in \mathbb{N}$,

$$\begin{array}{r}{\parallel A_1\overline{x}-F_1{A}_1\overline{x}\parallel }^2\le 〈A_1\overline{x}-F_1{A}_1\overline{x},A_{1}w-v_1〉,\\ {\parallel A_2\overline{x}-F_2{A}_2\overline{x}\parallel }^2\le 〈A_2\overline{x}-F_2{A}_2\overline{x},A_{2}w-v_2〉\end{array}$$

(57)

for each $w\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $v_1\in Fix(F_1)$, and $v_2\in Fix(F_2)$.

Since $\overline{w}$ is a solution of multiple sets split feasibility problem (MSFP_FF), we know that $\overline{w}\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n})$, $A_1\overline{w}\in Fix(F_1)$, and $A_2\overline{w}\in Fix(F_2)$ for each $n\in \mathbb{N}$. So, it follows from (57) that $A_1\overline{x}=Fix(F_1)$ and $A_2\overline{x}=Fix(F_2)$. Furthermore, $\overline{x}\in Fix(J_{{\lambda }_n})$ and $\overline{x}\in Fix(T_{r_n})$ for each $n\in \mathbb{N}$. Therefore, $\overline{x}$ is a solution of (MSFP_FF). □

Applying Theorem 3.1 and Theorem 3.2, we can find the solution of the following hierarchical problem.

Theorem 3.3 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. Let C and Q be two nonempty closed convex subsets of real Hilbert spaces $H_1$ and $H_2$, respectively. For each $i=1,2$, let $F_i$ be a firmly nonexpansive mapping of $H_2$ into $H_2$, let $A_i:H_1\to H_2$ be a bounded linear operator, and let $A_i^{\ast }$ be the adjoint of $A_i$. Suppose that the solution set of (MSFP_FF) is Ω and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(3.3)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$, and $0<a\le r_n\le b<\frac{2}{{\parallel A_2\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=\theta$ for some $\theta \in H$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x}+\theta )$. This point $\overline{x}$ is also a unique solution of the following hierarchical problem: Find $\overline{x}\in Fix(T)\cap \mathrm{\Omega }$ such that

$$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap \mathrm{\Omega }.$$

Proof Since $F_i$ is firmly nonexpansive, it follows from Lemma 2.1 that we have that $A_i^{\ast }(I-F_i)A_i:C_1\to H_1$ is $\frac{1}{{\parallel A_i\parallel }^2}$-ism for each $i=1,2$. For each $i=1,2$, put $B_i=A_i^{\ast }(I-F_i)A_i$ in Theorem 3.3. Then algorithm (3.1) in Theorem 3.1 follows immediately from algorithm (3.3) in Theorem 3.3.

Since the solution set of (MSFP_FF) is nonempty, by (47), we have for each $n\in \mathbb{N}$

$$\overline{w}\in Fix\left(J_{{\lambda }_n}\left((I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)\right)\right)\cap Fix\left(T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)\right)\ne \mathrm{\varnothing }.$$

(58)

This implies that for each $n\in \mathbb{N}$,

$$\overline{w}\in Fix(J_{{\lambda }_n}(I-{\lambda }_n{B}_1))\cap Fix(T_{r_n}(I-r_n{B}_2))\ne \mathrm{\varnothing }.$$

(59)

So,

$$\overline{w}\in {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0\ne \mathrm{\varnothing }.$$

(60)

It follows from Theorem 3.1 that ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where

$$\overline{x}=P_{Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0}(\overline{x}-V\overline{x}+\theta ).$$

This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:

$$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap {(B_1+G_1)}^{-1}0\cap {(B_2+G_2)}^{-1}0,$$

that is, for each $n\in \mathbb{N}$,

$$\overline{x}=J_{{\lambda }_n}(I-{\lambda }_n{B}_1)\overline{x}=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)\overline{x}$$

(61)

and

$$\overline{x}=T_{r_n}(I-r_n{B}_2)\overline{x}=T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)\overline{x}.$$

(62)

This implies that for each $n\in \mathbb{N}$,

$$\overline{x}=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)\overline{x}.$$

(63)

By assumptions, (63), and Theorem 3.2(ii), we know that $\overline{x}$ is a solution of (MSFP_FF). Furthermore, $\overline{x}\in Fix(T)$. Therefore, $\overline{x}\in Fix(T)\cap \mathrm{\Omega }$. By the same argument as (61), (62), and (63), we also have

$$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap \mathrm{\Omega }.$$

Therefore, the proof is completed. □

Remark 3.1 In Theorem 3.3, we establish a strong convergence theorem for hierarchical problem (MSFP_FF) without calculating the inverse of the operator we consider.

4 Applications to mathematical programming with multiple sets split feasibility constraints

By Theorem 3.3, we obtain mathematical programming with fixed point and multiple sets split feasibility constraints.

Theorem 4.1 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. In Theorem  3.3, let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Let

$$\mathrm{\Omega }=\{q\in H_1:q\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n}),A_{1}q\in Fix(F_1),A_{2}q\in Fix(F_2),n\in \mathbb{N}\}.$$

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and multiple sets split feasibility constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}h(q)$.

Proof Put $\theta =0$ in Theorem 3.3. Then, by Theorem 3.3, there exists $\overline{x}\in Fix(T)\cap \mathrm{\Omega }$ such that

$$〈V\overline{x},q-\overline{x}〉\ge 0,\mathrm{\forall }q\in F(T)\cap \mathrm{\Omega }.$$

(64)

Since $h:C\to \mathbb{R}$ is a convex Gâteaux differential function with Gâteaux derivative V, we obtain that

$$\begin{array}{rcl}〈V\overline{x},y-\overline{x}〉& =& \underset{t\to 0}{lim}\frac{h(\overline{x}+t(y-\overline{x}))-h(\overline{x})}{t}\\ =& \underset{t\to 0}{lim}\frac{h((1-t)\overline{x}+ty)-h(\overline{x})}{t}\\ \le & \underset{t\to 0}{lim}\frac{(1-t)h(\overline{x})+th(y)-h(\overline{x})}{t}\\ =& h(y)-h(\overline{x})\end{array}$$

(65)

for all $y\in C$. By (64) and (65), it is easy to see that $h(\overline{x})\le h(q)$ for all $q\in Fix(T)\cap \mathrm{\Omega }$. □

We can apply Theorem 4.1 to study the mathematical programming of a quadratic function with fixed point and multiple sets split feasibility constraints.

Theorem 4.2 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. In Theorem  3.3, let $B:C\to C$ be a strongly positive self-adjoint bounded linear operator and $a\in H$. Let

$$\mathrm{\Omega }=\{q\in H_1:q\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n}),A_{1}q\in Fix(F_1),A_{2}q\in Fix(F_2),n\in \mathbb{N}\}.$$

Let $\{x_n\}\subset H$ be defined by

$$(4.2)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(s_n-{\beta }_n(B(s_n)-a)))\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$, and $0<a\le r_n\le b<\frac{2}{{\parallel A_2\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$. This point $\overline{x}$ is also a unique solution of the mathematical programming of a quadratic function with fixed point and multiple sets split feasibility constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}\frac{1}{2}〈Bq,q〉-〈a,q〉$.

Proof Let $h:C\to \mathbb{R}$ be defined by

$$h(x)=\frac{1}{2}〈Bx,x〉-〈a,x〉.$$

It is easy to see that h is a convex function. Since B is a strongly positive self-adjoint operator, there exists $\eta >0$ such that $〈Bx,x〉\ge \eta {\parallel x\parallel }^2$. This implies that

$$〈Bx-By,x-y〉=〈B(x-y),x-y〉\ge \eta {\parallel x-y\parallel }^2.$$

(66)

Therefore,

$$〈Bx,x〉+〈By,y〉\ge 〈Bx,y〉+〈By,x〉.$$

(67)

From this we can show that h is a convex function. Indeed, for any $x,y\in C$ and any $\lambda \in [0,1]$. It follows from (67) that

$$\begin{array}{c}h(\lambda x+(1-\lambda )y)\hfill \\ =\frac{1}{2}〈B(\lambda x+(1-\lambda )y),\lambda x+(1-\lambda )y〉-〈a,\lambda x+(1-\lambda )y〉\hfill \\ =\frac{1}{2}〈\lambda Bx+(1-\lambda )By,\lambda x+(1-\lambda )y〉-〈a,\lambda x+(1-\lambda )y〉\hfill \\ =\frac{1}{2}{\lambda }^2〈Bx,x〉+\frac{1}{2}\lambda (1-\lambda )〈Bx,y〉+\frac{1}{2}\lambda (1-\lambda )〈By,x〉+\frac{1}{2}{(1-\lambda )}^2〈By,y〉\hfill \\ -\lambda 〈a,x〉-(1-\lambda )〈a,y〉\hfill \\ \le \frac{1}{2}{\lambda }^2〈Bx,x〉+\frac{1}{2}\lambda (1-\lambda )(〈Bx,x〉+〈By,y〉)+\frac{1}{2}{(1-\lambda )}^2〈By,y〉\hfill \\ -\lambda 〈a,x〉-(1-\lambda )〈a,y〉\hfill \\ \le \frac{1}{2}\lambda 〈Bx,x〉+\frac{1}{2}(1-\lambda )〈By,y〉-\lambda 〈a,x〉-(1-\lambda )〈a,y〉\hfill \\ \le \lambda h(x)+(1-\lambda )h(y).\hfill \end{array}$$

Let $V(x)=B(x)-a$ for all $x\in C$. It is easy to see that V is the Gâteaux derivative of h. Indeed, for any $u\in H$, $x\in C$ and any $t\in [0,1]$. Since B is a self-adjoint bounded linear operator, we see that for each $u\in H$,

$$\begin{array}{c}{h}^{\prime }(x)(u)\hfill \\ =\underset{t\to 0}{lim}\frac{h(x+tu)-h(x)}{t}\hfill \\ =\underset{t\to 0}{lim}\frac{\frac{1}{2}〈B(x+tu),x+tu〉-〈a,x+tu〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ =\underset{t\to 0}{lim}\frac{\frac{1}{2}〈tBu+Bx,tu+x〉-〈a,tu+x〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ =\underset{t\to 0}{lim}\frac{\frac{1}{2}[t^2〈Bu,u〉+t〈Bu,x〉+t〈Bx,u〉+〈Bx,x〉]-t〈a,u〉-〈a,x〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ =\underset{t\to 0}{lim}\frac{1}{2}[t〈Bu,u〉+〈Bu,x〉+〈Bx,u〉]-〈a,u〉\hfill \\ =〈Bx,u〉-〈a,u〉=〈Bx-a,u〉=〈Vx,u〉.\hfill \end{array}$$

Therefore, V is the Gâteaux derivative of h. Since B is a strongly positive bounded linear operator in H, we have that

$$\parallel Vx-Vy\parallel =\parallel Bx-a-By+a\parallel =\parallel Bx-By\parallel \le \parallel B\parallel \parallel x-y\parallel .$$

This implies that V is Lipschitz, and we have that

$$〈Vx-Vy,x-y〉=〈Bx-a-By+a,x-y〉=〈B(x-y),x-y〉\ge \eta {\parallel x-y\parallel }^2.$$

This implies that V is strongly monotone. Therefore, Theorem 4.2 follows from Theorem 4.1. □

Theorem 4.3 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. In Theorem  3.3, let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Let ${\mathrm{\Phi }}_i$ be a maximal monotone mapping on $H_2$ such that the domain of ${\mathrm{\Phi }}_i$ is included in Q for each $i=1,2$, where Q is a closed convex subset of $H_2$. Let

$$\mathrm{\Omega }=\{q\in H_1:q\in G_1^{-1}0\cap G_2^{-1}0,A_{1}q\in {\mathrm{\Phi }}_1^{-1}0,A_{2}q\in {\mathrm{\Phi }}_2^{-1}0\}.$$

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and multiple sets split feasibility problem constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}h(q)$.

Proof Let $J_{\lambda }={(I+\lambda G_1)}^{-1}$, $T_r={(I+r{G}_2)}^{-1}$, $F_1={(I+\lambda {\mathrm{\Phi }}_1)}^{-1}$, and $F_2={(I+r{\mathrm{\Phi }}_2)}^{-1}$ for each $\lambda >0$ and $r>0$ in Theorem 4.2. Then $Fix(J_{\lambda })=G_1^{-1}0$, $Fix(T_r)=G_2^{-1}0$, $Fix(F_1)={\mathrm{\Phi }}_1^{-1}0$, and $Fix(F_2)={\mathrm{\Phi }}_2^{-1}0$. Therefore, Theorem 4.3 follows from Theorem 4.2. □

Takahashi et al. [40] showed the following result.

Lemma 4.1 [40]

Let C be a nonempty closed convex subset of a Hilbert space H, and let $g:C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (A1)-(A4). Define $A_g$ as follows:

$$\text{(L4.1)}{A}_{g}x=\{\begin{array}{cc}\{z\in H:g(x,y)\ge 〈y-x,z〉,\mathrm{\forall }y\in C\},\hfill & \mathrm{\forall }x\in C,\hfill \\ \mathrm{\varnothing },\hfill & \mathrm{\forall }x\notin C.\hfill \end{array}$$

Then $EP(g)=A_g^{-1}0$ and $A_g$ is a maximal monotone operator with the domain of $A_g\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r^g$ of g coincides with the resolvent of $A_g$, i.e., $T_r^{g}x={(I+r{A}_g)}^{-1}x$.

Now, we consider the following multiple sets split equilibrium problem:

(MSEP) Find $\overline{x}\in H_1$ such that $\overline{x}\in EP(g_1)\cap EP(g_2)$, $A_1\overline{x}\in EP(f_1)$ and $A_2\overline{x}\in EP(f_2)$.

Applying Theorems 2.2 and 4.1, Lemma 4.1, we can find the minimum norm solution of (MSEP) and mathematical programming with fixed point and multiple sets split equilibrium constraints.

Theorem 4.4 Let $T:C\to H$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. Let C and Q be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $g_1:C\times C\to \mathbb{R}$, $g_2:C\times C\to \mathbb{R}$, $f_1:Q\times Q\to \mathbb{R}$, and $f_2:Q\times Q\to \mathbb{R}$ be bifunctions satisfying conditions (A1)-(A4), and let $T_{{\lambda }_n}^{g_1}$, $T_{r_n}^{g_2}$, $T_{{\lambda }_n}^{f_1}$, $T_{r_n}^{f_2}$ be the resolvent of $g_1$, $g_2$, $f_1$, $f_2$, respectively, for ${\lambda }_n>0$, $r_n>0$. For $i=1,2$, let $A_i:H_1\to H_2$ be a bounded linear operator, and let $A_i^{\ast }$ be the adjoint of $A_i$. Let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Suppose that Ω is the solution set of (MSEP) and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(4.2)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=T_{{\lambda }_n}^{g_1}(I-{\lambda }_n{A}_1^{\ast }(I-T_{{\lambda }_n}^{f_1})A_1)T_{r_n}^{g_2}(I-r_n{A}_2^{\ast }(I-T_{r_n}^{f_2})A_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$, and $0<a\le r_n\le b<\frac{2}{{\parallel A_2\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and multiple sets split equilibrium constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}h(q)$.

Proof Define $A_g$ as (L4.1). By Lemma 4.1, we know that $EP(g)=A_g^{-1}0$ and $A_g$ is a maximal monotone operator with the domain of $A_g\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r^g$ of g coincides with the resolvent of $A_g$, i.e., $T_r^{g}x={(I+r{A}_g)}^{-1}x$. By Theorem 2.2, $T_{{\lambda }_n}^{f_1}$, $T_{r_n}^{f_2}$ are firmly nonexpansive mappings.

Put $G_1=A_{g_1}$, $G_2=A_{g_2}$, $F_1=T_{{\lambda }_n}^{f_1}$ and $F_2=T_{r_n}^{f_2}$ in Theorem 3.3. Then $J_{{\lambda }_n}x={(I+{\lambda }_n{A}_{g_1})}^{-1}x=T_{{\lambda }_n}^{g_1}x$, $T_{r_n}x={(I+r_n{A}_{g_2})}^{-1}x=T_{r_n}^{g_2}x$. By Theorem 2.2, we have that $Fix(J_{{\lambda }_n})=Fix(T_{{\lambda }_n}^{g_1})=EP(g_1)$, $Fix(T_{r_n})=Fix(T_{r_n}^{g_2})=EP(g_2)$, $Fix(F_1)=Fix(T_{{\lambda }_n}^{f_1})=EP(f_1)$ and $Fix(F_2)=Fix(T_{r_n}^{f_2})=EP(f_2)$. Therefore, the solution set of (MSEP) coincides with the solution set of (MSFP_FF). Therefore, by Theorem 4.1, we get the result. □

The following unique minimum norm common solution of a fixed point problem and multiple sets split equilibrium problem is a special case of Theorem 4.4.

Corollary 4.1 Let C and Q be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $g_1:C\times C\to \mathbb{R}$, $g_2:C\times C\to \mathbb{R}$, $f_1:Q\times Q\to \mathbb{R}$, and $f_2:Q\times Q\to \mathbb{R}$ be bifunctions satisfying conditions (A1)-(A4), and let $T_{{\lambda }_n}^{g_1}$, $T_{r_n}^{g_2}$, $T_{{\lambda }_n}^{f_1}$, $T_{r_n}^{f_2}$ be the resolvent of $g_1$, $g_2$, $f_1$, $f_2$, respectively, for ${\lambda }_n>0$, $r_n>0$. For $i=1,2$, let $A_i:H_1\to H_2$ be a bounded linear operator, and let $A_i^{\ast }$ be the adjoint of $A_i$. Suppose that Ω is the solution set of (MSEP) and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(4.2)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=T_{{\lambda }_n}^{g_1}(I-{\lambda }_n{A}_1^{\ast }(I-T_{{\lambda }_n}^{f_1})A_1)T_{r_n}^{g_2}(I-r_n{A}_2^{\ast }(I-T_{r_n}^{f_2})A_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_n)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$, and $0<a\le r_n\le b<\frac{2}{{\parallel A_2\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}0$. This point $\overline{x}$ is also a unique minimum norm solution of the fixed point and multiple sets split equilibrium constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}\parallel q\parallel$.

Proof Let $h(x)=\frac{1}{2}{\parallel x\parallel }^2$, and let V be the Gâteaux derivative of h. It is easy to see $V(x)=x$ for each $x\in H$. Then Corollary 4.1 follows immediately from Theorem 4.4. □

Now, we consider the following split equilibrium problem:

(SEP) Find $\overline{x}\in C$ such that $\overline{x}\in EP(g_1)$ and $A_1\overline{x}\in EP(f_1)$.

Applying Theorems 2.2 and 4.1, Lemma 4.1, we can find the unique minimum norm common solution of fixed point and split equilibrium constraints and the solution of mathematical programming with fixed point and split equilibrium constraints.

Theorem 4.5 Let C and Q be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $g_1:C\times C\to \mathbb{R}$ and $f_1:Q\times Q\to \mathbb{R}$ be bifunctions satisfying conditions (A1)-(A4), and let $T_{{\lambda }_n}^{g_1}$, $T_{{\lambda }_n}^{f_1}$ be the resolvent of $g_1$, $f_1$, respectively, for ${\lambda }_n>0$, $r_n>0$. Let $A_1:H_1\to H_2$ be a bounded linear operator, and let $A_1^{\ast }$ be the adjoint of $A_1$. Let $T:C\to C$ be a quasi-nonexpansive mapping with $F(T)=F(\hat{T})$. Let $V:C\to C$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$ and $\{{\theta }_n\}\subseteq C$. Let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Suppose that Ω is the solution set of (SEP) and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(4.3)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=T_{{\lambda }_n}^{g_1}(I-{\lambda }_n{A}_1^{\ast }(I-T_{{\lambda }_n}^{f_1})A_1)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and split equilibrium constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}h(q)$.

Proof Define $A_g$ as (L4.1). By Lemma 4.1, we know that $EP(g)=A_g^{-1}0$ and $A_g$ is a maximal monotone operator with the domain of $A_g$ included in C. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r^g$ of g coincides with the resolvent of $A_g$, i.e., $T_r^{g}x={(I+r{A}_g)}^{-1}x$. By Theorem 2.2, $T_{{\lambda }_n}^{f_1}$ is a firmly nonexpansive mapping; we also know that the identity mapping I is a firmly nonexpansive mapping.

Put $G_1=A_{g_1}$, $G_2=A_{g_2}$, $F_1=T_{{\lambda }_n}^{f_1}$, and $F_2=I$ in Theorem 3.3. Then $J_{{\lambda }_n}x={(I+{\lambda }_n{A}_{g_1})}^{-1}x=T_{{\lambda }_n}^{g_1}x$, $T_{r_n}x={(I+r_n{A}_{g_2})}^{-1}x=T_{r_n}^{g_2}x$. Let $g_2(x,y)=0$, $\mathrm{\forall }x,y\in C$. Then $T_{r_n}x={(I+r_n{A}_{g_2})}^{-1}x=T_{r_n}^{g_2}x=P_{C}x$. By Theorem 2.2, we have that $Fix(J_{{\lambda }_n})=Fix(T_{{\lambda }_n}^{g_1})=EP(g_1)$, $Fix(T_{r_n})=Fix(P_C)=C$, $Fix(F_1)=Fix(T_{{\lambda }_n}^{f_1})=EP(f_1)$, and $Fix(F_2)=Fix(I)=H_2$. So, we have that the solution set of (SEP) coincides with the solution set of (MSFP_FF). On the other hand, we have that $T_{r_n}^{g_2}(I-r_n{A}_1^{\ast }(I-F_2)A_1)x_n=P_C{x}_n=x_n$. Then algorithm (3.3) in Theorem 3.3 follows immediately from algorithm (4.3) in Theorem 4.5. Therefore, by Theorems 3.3 and 4.1, we get the result. □

Put $h(x)=\frac{1}{2}{\parallel x\parallel }^2$ for each $x\in H$ in Theorem 4.5, we obtain a unique minimum norm common solution of a fixed point problem and a split equilibrium constraints.

Corollary 4.2 Let C and Q be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $g_1:C\times C\to \mathbb{R}$ and $f_1:Q\times Q\to \mathbb{R}$ be bifunctions satisfying conditions (A1)-(A4), and let $T_{{\lambda }_n}^{g_1}$, $T_{{\lambda }_n}^{f_1}$ be the resolvent of $g_1$, $f_1$, respectively, for ${\lambda }_n>0$, $r_n>0$. Let $A_1:H_1\to H_2$ be a bounded linear operator, and let $A_1^{\ast }$ be the adjoint of $A_1$. Let $T:C\to C$ be a quasi-nonexpansive mapping with $F(T)=F(\hat{T})$. Let $V:C\to C$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$ and $\{{\theta }_n\}\subseteq C$. Suppose that the solution set of (SEP) is Ω and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(4.3)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=T_{{\lambda }_n}^{g_1}(I-{\lambda }_n{A}_1^{\ast }(I-T_{{\lambda }_n}^{f_1})A_1)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(0)$. This point $\overline{x}$ is also a unique minimum norm common solution of fixed point and split equilibrium constraints: Find ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}\parallel q\parallel$.

Now, we recall the following split feasibility problem:

(SFP) Find $\overline{x}\in H_1$ such that $\overline{x}\in C$ and $A\overline{x}\in Q$.

Applying Theorem 4.5, we can find a unique minimum norm common solution with fixed point and split feasibility constraints, and the solution of mathematical programming with fixed point and split feasibility constraints.

Theorem 4.6 Let C and Q be nonempty closed convex subsets of Hilbert spaces $H_1$ and $H_2$, respectively. Let $A_1:H_1\to H_2$ be a bounded linear operator, and let $A_1^{\ast }$ be the adjoint of $A_i$. Let $T:C\to C$ be a quasi-nonexpansive mapping with $F(T)=F(\hat{T})$. Let $V:C\to C$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$ and $\{{\theta }_n\}\subseteq C$. Let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Suppose that the solution set of (SFP) is Ω and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}\subset H$ be defined by

$$(4.4)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=P_C(I-{\lambda }_n{A}_1^{\ast }(I-P_Q)A_1)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and split feasibility constraints: ${min}_{q\in Fix(T)\cap \mathrm{\Omega }}h(q)$.

Proof Put $g_1(x,y)=0$, $\mathrm{\forall }x,y\in C$ and $f_1(x,y)=0$, $\mathrm{\forall }x,y\in Q$ in Theorem 4.5. Then $T_{{\lambda }_n}^{g_1}=P_C$, $T_{r_n}^{f_1}=P_Q$. Therefore, algorithm (4.3) in Theorem 4.5 follows immediately from algorithm (4.4) in Theorem 4.6.

By Theorem 2.2, we have that $EP(g_1)=Fix(T_{{\lambda }_n}^{g_1})=Fix(P_C)=C$ and $EP(f_1)=Fix(T_{r_n}^{f_1})=Fix(P_Q)=Q$. So, the solution set of (SEP) coincides with the solution set of (SFP). Therefore, by Theorem 4.5, we get the result. □

For each $i=1,2$, let $X_i$, $Y_i$ be two Hilbert spaces, a function $F:X_i\times Y_i\to \mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ is said to be convex-concave iff it is convex in the variable x and concave in the variable y. To such a function, Rockafellar associated the operator $T_F$, defined by $T_F={\partial }_{1}F\times {\partial }_2(-F)$, where ${\partial }_1$ (resp. ${\partial }_2$) stands for the subdifferential of F with respect to the first (resp. the second) variable. $T_F$ is a maximal monotone operator if and only if F is closed and proper in the Rockafellar sense (see [41]). Moreover, it is well known that $({\overline{x}}_1,{\overline{y}}_1)\in H_1=X_1\times Y_1$ is a saddlepoint of F, namely

$$F({\overline{x}}_1,y)\le F({\overline{x}}_1,{\overline{y}}_1)\le F(x,{\overline{y}}_1)\text{for all }x\in X_i,y\in Y_i,$$

if and only if the following monotone variational inclusion holds true, that is, $(0,0)\in T_F({\overline{x}}_1,{\overline{y}}_1)$. If $({\overline{x}}_1,{\overline{y}}_1)\in H_1=X_1\times Y_1$ is a saddlepoint of F, then

$$\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}F(x,y)=F({\overline{x}}_1,{\overline{y}}_1)=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}F(x,y).$$

For each $i=1,2$, let ${\phi }_i:X_1\times Y_1\to \mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ and ${\psi }_i:X_2\times Y_2\to \mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ be convex-concave functions. Now, we consider the following multiple sets split minimax problem:

(MSMMP) Find $({\overline{x}}_1,{\overline{y}}_1)\in H_1=X_1\times Y_1$ such that for each $i=1,2$,

$$\begin{array}{c}\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}{\phi }_i(x,y)={\phi }_i({\overline{x}}_1,{\overline{y}}_1)=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}{\phi }_i(x,y)\text{and}\hfill \\ \underset{u\in X_2}{inf}\underset{v\in Y_2}{sup}{\psi }_i(u,v)={\psi }_i(A_i({\overline{x}}_1,{\overline{y}}_1))=\underset{v\in Y_2}{sup}\underset{u\in X_2}{inf}{\psi }_i(u,v).\hfill \end{array}$$

By Theorem 3.3, we can find the unique minimum norm common solution of fixed point and multiple sets split minimax problems (MSMMP) and the solution of mathematical programming with fixed point and multiple sets split minimax (MSMMP) constraints.

Theorem 4.7 Let $T:C_1\times C_2\to X_1\times X_2$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. For each $i=1,2$, let $A_1:X_1\times Y_1\to X_2\times Y_2$, $A_2:X_1\times Y_1\to X_2\times Y_2$ be a bounded linear operator, $A_i^{\ast }$ be the adjoint of $A_i$. For each $i=1,2$, let ${\phi }_i:X_1\times Y_1\to \mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ and ${\psi }_i:X_2\times Y_2\to \mathbb{R}\cup \{-\mathrm{\infty },\mathrm{\infty }\}$ be convex-concave functions which are proper and closed in the Rockafellar sense. Let $C_1$, $C_2$ be closed convex subsets of Hilbert spaces $X_1$, $Y_1$, and let $C=C_1\times C_2$ be a closed convex subset of $H_1=X_1\times Y_1$. Let $J_{{\lambda }_n}={(I+{\lambda }_n{T}_{{\phi }_1})}^{-1}$, $T_{r_n}={(I+r_n{T}_{{\phi }_2})}^{-1}$, $F_1={(I+{\lambda }_n{T}_{{\psi }_1})}^{-1}$, $F_2={(I+{\lambda }_n{T}_{{\psi }_2})}^{-1}$. Let $h:C_1\times C_2\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Let V be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$ and suppose that the solution of multiple sets split minimax problem (MSMMP) is Ω and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be defined by

$$(4.7)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)T_{r_n}(I-r_n{A}_2^{\ast }(I-F_2)A_2)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, and $\{r_n\}\subset (0,\mathrm{\infty })$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$, and $0<a\le r_n\le b<\frac{2}{{\parallel A_2\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and multiple split minimax constraints: ${min}_{(q_1,q_2)\in Fix(T)\cap \mathrm{\Omega }}h(q_1,q_2)$.

Proof Since $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. There exists $({\overline{a}}_1,{\overline{b}}_1)\in H_1=X_1\times Y_1$ such that for each $i=1,2$,

$$\begin{array}{r}\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}{\phi }_i(x,y)={\phi }_i({\overline{a}}_1,{\overline{b}}_1)=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}{\phi }_i(x,y)\text{and}\\ \underset{u\in X_2}{inf}\underset{v\in Y_2}{sup}{\psi }_i(u,v)={\psi }_i(A_i({\overline{a}}_1,{\overline{b}}_1)=\underset{v\in Y_2}{sup}\underset{u\in X_2}{inf}{\psi }_i(u,v).\end{array}$$

(68)

That is,

$$\begin{array}{c}(0,0)\in T_{{\phi }_1}({\overline{a}}_1,{\overline{b}}_1)\cap T_{{\phi }_2}({\overline{a}}_1,{\overline{b}}_1),(0,0)\in T_{{\psi }_1}({\overline{u}}_1,{\overline{v}}_1)\text{and}(0,0)\in T_{{\psi }_2}({\overline{u}}_2,{\overline{v}}_2),\hfill \\ \text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1)\text{ and }({\overline{u}}_2,{\overline{v}}_2)=A_2({\overline{a}}_1,{\overline{b}}_1).\hfill \end{array}$$

(69)

That is,

$$\begin{array}{c}({\overline{a}}_1,{\overline{b}}_1)\in T_{{\phi }_1}^{-1}(0,0)\cap T_{{\phi }_2}^{-1}(0,0),({\overline{u}}_1,{\overline{v}}_1)\in T_{{\psi }_1}^{-1}(0,0),\text{and}({\overline{u}}_2,{\overline{v}}_2)\in T_{{\psi }_2}^{-1}(0,0),\hfill \\ \text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1)\text{ and }({\overline{u}}_2,{\overline{v}}_2)=A_2({\overline{a}}_1,{\overline{b}}_1).\hfill \end{array}$$

(70)

That is,

$$\begin{array}{c}({\overline{a}}_1,{\overline{b}}_1)\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n}),({\overline{u}}_1,{\overline{v}}_1)\in Fix(F_1),\text{and}({\overline{u}}_2,{\overline{v}}_2)\in Fix(F_2),\hfill \\ \text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1)\text{ and }({\overline{u}}_2,{\overline{v}}_2)=A_2({\overline{a}}_1,{\overline{b}}_1).\hfill \end{array}$$

(71)

That is,

$$\begin{array}{c}({\overline{a}}_1,{\overline{b}}_1)\in Fix(J_{{\lambda }_n})\cap Fix(T_{r_n}),A_1({\overline{a}}_1,{\overline{b}}_1)\in Fix(F_1),\text{and}{A}_2({\overline{a}}_1,{\overline{b}}_1)\in Fix(F_2),\hfill \\ \text{ where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1)\text{ and }({\overline{u}}_2,{\overline{v}}_2)=A_2({\overline{a}}_1,{\overline{b}}_1).\hfill \end{array}$$

(72)

This implies that $Fix(T)\cap {\mathrm{\Omega }}_1\ne \mathrm{\varnothing }$, where ${\mathrm{\Omega }}_1$ is the solution set of (MSFP_FF).

By Theorem 3.3, we have that ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap {\mathrm{\Omega }}_1}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:

$$〈V\overline{x},q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap {\mathrm{\Omega }}_1.$$

By (68), (69), (70), (71), and (72), and Theorem 4.1, we get the result. □

Now, we recall the following split minimax problem:

(SMMP) Find $(\overline{x},\overline{y})\in H_1=X_1\times Y_1$ such that

$$\begin{array}{c}\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}{\phi }_1(x,y)={\phi }_1(\overline{x},\overline{y})=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}{\phi }_1(x,y)\text{and}\hfill \\ \underset{u\in X_2}{inf}\underset{v\in Y_2}{sup}{\psi }_1(u,v)={\psi }_1(A_1(\overline{x},\overline{y}))=\underset{v\in Y_2}{sup}\underset{u\in X_2}{inf}{\psi }_1(u,v).\hfill \end{array}$$

By Theorem 3.3, we can find the solution of split minimax problem (SMMP) and mathematical programming with fixed point and split minimax problem (SMMP) constraints.

Theorem 4.8 Let $T:C_1\times C_2\to X_1\times X_2$ be a quasi-nonexpansive mapping with $Fix(T)=Fix(\hat{T})$. Let $A_1:X_1\times Y_1\to X_2\times Y_2$ be a bounded linear operator. Let $A_1^{\ast }$ be the adjoint of $A_1$. Let $A_1$, $A_1^{\ast }$, T, V be defined as in Theorem  4.7. Let ${\phi }_1$, ${\psi }_1$ be defined as in Theorem  4.7. Let $C_1$, $C_2$ closed convex subsets of Hilbert spaces $X_1$, $X_2$, let $C=C_1\times C_2$ be a closed convex subset of $H_1=X_1\times Y_1$, and let $J_{{\lambda }_n}={(I+{\lambda }_n{T}_{{\phi }_1})}^{-1}$, $F_1={(I+{\lambda }_n{T}_{{\psi }_1})}^{-1}$. Let $V:C\to C$ be a $\overline{\gamma }$-strongly monotone and L-Lipschitzian continuous operator with $\overline{\gamma }>0$ and $L>0$ and $\{{\theta }_n\}\subseteq C$. Let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Suppose that the solution of split minimax problem (SMMP) is Ω and $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be defined by

$$(4.8)\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=J_{{\lambda }_n}(I-{\lambda }_n{A}_1^{\ast }(I-F_1)A_1)x_n,\hfill \\ s_n=T{y}_n,\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_n{\theta }_n+(1-{\beta }_{n}V)s_n)\hfill \end{array}$$

for each $n\in \mathbb{N}$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$, $\{{\alpha }_n\}\subset (0,1)$, and $\{{\beta }_n\}\subset (0,1)$. Assume that:

• (i) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

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

• (iii) $0<a\le {\lambda }_n\le b<\frac{2}{{\parallel A_1\parallel }^2+2}$;

• (iv) ${lim}_{n\to \mathrm{\infty }}{\theta }_n=0$.

Then ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the mathematical programming with fixed point and split minimax problem constraints: ${min}_{(q_1,q_2)\in Fix(T)\cap \mathrm{\Omega }}h(q_1,q_2)$.

Proof Define $A_g$ as (L4.1). By Lemma 4.1, we know that $EP(g)=A_g^{-1}0$ and $A_g$ is a maximal monotone operator with the domain of $A_g\subset C$. Furthermore, for any $x\in H$ and $r>0$, the resolvent $T_r^g$ of g coincides with the resolvent of $A_g$, i.e., $T_r^{g}x={(I+r{A}_g)}^{-1}x$. By Theorem 2.2, $T_{{\lambda }_n}^{f_1}$ is a firmly nonexpansive mapping, we also know that the identity mapping I is a firmly nonexpansive mapping.

Put $G_2=A_{g_2}$ and $F_2=I$ in Theorem 3.3. Then $T_{r_n}x={(I+r_n{A}_{g_2})}^{-1}x=T_{r_n}^{g_2}x$. Let $g_2(x,y)=0$, $\mathrm{\forall }x,y\in C$. Then $T_{r_n}x={(I+r_n{A}_{g_2})}^{-1}x=T_{r_n}^{g_2}x=P_{C}x$. So, we have that $Fix(T_{r_n})=Fix(P_C)=C$ and $Fix(F_2)=Fix(I)=H_2$. So, we have that $T_{r_n}^{g_2}(I-r_n{A}_1^{\ast }(I-F_2)A_1)x_n=P_C{x}_n=x_n$. Then algorithm (3.3) in Theorem 3.3 follows immediately from algorithm (4.8) in Theorem 4.8.

Since $Fix(T)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$, there exists $({\overline{a}}_1,{\overline{b}}_1)\in H_1=X_1\times Y_1$ such that

$$\begin{array}{r}\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}{\phi }_1(x,y)={\phi }_1({\overline{a}}_1,{\overline{b}}_1)=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}{\phi }_1(x,y)\text{and }\\ \underset{u\in X_2}{inf}\underset{v\in Y_2}{sup}{\psi }_1(u,v)={\psi }_1(A_1({\overline{a}}_1,{\overline{b}}_1))=\underset{v\in Y_2}{sup}\underset{u\in X_2}{inf}{\psi }_1(u,v).\end{array}$$

(73)

That is,

$$(0,0)\in T_{{\phi }_1}({\overline{a}}_1,{\overline{b}}_1)\text{and}(0,0)\in T_{{\psi }_1}({\overline{u}}_1,{\overline{v}}_1),\text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1).$$

(74)

That is,

$$({\overline{a}}_1,{\overline{b}}_1)\in T_{{\phi }_1}^{-1}(0,0)\text{and}({\overline{u}}_1,{\overline{v}}_1)\in T_{{\psi }_1}^{-1}(0,0),\text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1).$$

(75)

That is,

$$({\overline{a}}_1,{\overline{b}}_1)\in Fix(J_{{\lambda }_n})\text{and}({\overline{u}}_1,{\overline{v}}_1)\in Fix(F_1),\text{where }({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{a}}_1,{\overline{b}}_1).$$

(76)

That is,

$$({\overline{a}}_1,{\overline{b}}_1)\in Fix(J_{{\lambda }_n})\text{and}{A}_1({\overline{a}}_1,{\overline{b}}_1)\in Fix(F_1).$$

(77)

This implies that $Fix(T)\cap {\mathrm{\Omega }}_1\ne \mathrm{\varnothing }$, where ${\mathrm{\Omega }}_1$ is the solution set of (SFP_FF).

By Theorem 3.3, we have that ${lim}_{n\to \mathrm{\infty }}{x}_n=\overline{x}$, where $\overline{x}=P_{Fix(T)\cap \mathrm{\Omega }}(\overline{x}-V\overline{x})$. This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:

$$〈V\overline{x},q-\overline{x}〉\ge 0,\mathrm{\forall }q\in Fix(T)\cap \mathrm{\Omega }.$$

By (73), (74), (75), (76), and (77), and Theorem 4.1, we get the result. □

Now, we recall the following multiple split minimax-equilibrium problem:

(MSMMP) Find $({\overline{x}}_1,{\overline{y}}_1)\in H_1=X_1\times Y_1$ such that for each $i=1,2$, $({\overline{a}}_1,{\overline{b}}_1)\in H_1=X_1\times Y_1$ such that

$$\begin{array}{c}\underset{x\in X_1}{inf}\underset{y\in Y_1}{sup}{\phi }_i(x,y)={\phi }_i({\overline{x}}_1,{\overline{y}}_1)=\underset{y\in Y_1}{sup}\underset{x\in X_1}{inf}{\phi }_i(x,y)\text{and}\hfill \\ ({\overline{u}}_1,{\overline{v}}_1)=A_1({\overline{x}}_1,{\overline{y}}_1)\in EP(f_1)\text{and}\hfill \\ ({\overline{u}}_2,{\overline{v}}_2)=A_2({\overline{x}}_1,{\overline{y}}_1)\in EP(f_2).\hfill \end{array}$$

By the same argument as in Theorems 4.4 and 4.7, we can find the solution of multiple split minimax problem (MSMMEP) and mathematical programming with fixed point and multiple split minimax problem (MSMMEP) constraints.

By the same argument as in Theorems 4.5 and 4.8, we can find the solution of the split minimax-equilibrium problem (SMMEP) and mathematical programming with fixed point and multiple split minimax-equilibrium problem (SMMEP) constraints.

5 Concluding remark

Applying Theorems 4.3-4.8 and following the same arguments as in Theorem 4.2, we can study the mathematical programming of a quadratic function with various types of fixed point and multiple split feasibility constraints.