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

Fixed Point Theory and Applications20132013:283

https://doi.org/10.1186/1687-1812-2013-283

• Accepted: 18 September 2013
• Published:

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.

Keywords

• hierarchical problems
• multiple sets split feasibility problem
• fixed point problem
• mathematical programming
• minimum norm solution

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 [15] and related literature. Since then, many researchers have studied (SFP) in finite dimensional or infinite dimensional Hilbert spaces. For example, one can see [2, 619].

A special case of problem (SFP) is the convexly constrained linear inverse problem in the finite dimensional Hilbert space [20]:
which has extensively been investigated by using the Landweber iterative method [21]
${x}_{n+1}:={x}_{n}+\gamma {A}^{T}\left(b-A{x}_{n}\right),\phantom{\rule{1em}{0ex}}n\in \mathbb{N}.$
In 2002, Byrne [2] first introduced the so-called CQ algorithm which generates a sequence $\left\{{x}_{n}\right\}$ by the following recursive procedure:
${x}_{n+1}={P}_{C}\left({x}_{n}-{\rho }_{n}{A}^{\ast }\left(I-{P}_{Q}\right)A{x}_{n}\right),$
(1)

where the stepsize ${\rho }_{n}$ is chosen in the interval $\left(0,2/{\parallel A\parallel }^{2}\right)$, 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 $Fix\left(T\right)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in C$ and $y\in Fix\left(T\right)$, we denote by $Fix\left(T\right)=\left\{x\in C:Tx=x\right\}$ the set of fixed points of T. $A:C\to H$ is called strongly positive if
$〈x,Ax〉\ge \alpha {\parallel x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
Let f be a contraction on H and $\left\{{\alpha }_{n}\right\}$ be a sequence in $\left[0,1\right]$. In 2004, Xu [22] proved that under some condition on $\left\{{\alpha }_{n}\right\}$, the sequence $\left\{{x}_{n}\right\}$ generated by
${x}_{n+1}={\alpha }_{n}f{x}_{n}+\left(1-{\alpha }_{n}\right)T{x}_{n}$
strongly converges to ${x}^{\ast }$ in $Fix\left(T\right)$, which is the unique solution of the variational inequality
$〈\left(I-f\right){x}^{\ast },x-{x}^{\ast }〉\ge 0$

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

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:
$\underset{x\in Fix\left(T\right)}{min}\frac{1}{2}〈Bx,x〉-〈a,x〉,$
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 $\left\{{x}_{n}\right\}$ defined by the following iterative method
${x}_{n+1}=\left(I-{\alpha }_{n}B\right)T{x}_{n}+{\alpha }_{n}a$
converges strongly to the unique solution of the minimization problem of a quadratic function. In [24], Marino et al. considered the following iterative method:
${x}_{n+1}={\alpha }_{n}\gamma f{x}_{n}+\left(I-{\alpha }_{n}A\right)T{x}_{n}.$
(2)
They proved that the sequence generated by (2) converges strongly to the fixed point ${x}^{\ast }$ of T which solves the following:
$〈\left(A-\gamma f\right){x}^{\ast },x-{x}^{\ast }〉\ge 0$

for all $x\in Fix\left(T\right)$. For some more related works, see [2527] 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 be the set of positive integers and let 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 $\left\{{x}_{n}\right\}$ to $x\in H$ by ${x}_{n}\to x$ and ${x}_{n}⇀x$, respectively. For each $x,y\in H$ and $\lambda \in \left[0,1\right]$, we have
${\parallel \lambda x+\left(1-\lambda \right)y\parallel }^{2}=\lambda {\parallel x\parallel }^{2}+\left(1-\lambda \right){\parallel y\parallel }^{2}-\lambda \left(1-\lambda \right){\parallel x-y\parallel }^{2}.$
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
$〈x-y,Ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H.$
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
${\parallel Tx-Ty\parallel }^{2}\le {\parallel x-y\parallel }^{2}-{\parallel \left(I-T\right)x-\left(I-T\right)y\parallel }^{2}$

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 $\left\{{x}_{n}\right\}\subseteq C$ such that ${x}_{n}⇀p$, and ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-T{x}_{n}\parallel =0$. We denote by $F\left(\stackrel{ˆ}{T}\right)$ the set of asymptotic fixed points of T. A mapping $T:C\to H$ is said to be demiclosed if it satisfies $F\left(T\right)=F\left(\stackrel{ˆ}{T}\right)$. 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 $D\left(B\right)$, that is, $D\left(B\right)=\left\{x\in H:Bx\ne \mathrm{\varnothing }\right\}$. A multi-valued mapping B is said to be a monotone operator on H if $〈x-y,u-v〉\ge 0$ for all $x,y\in D\left(B\right)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and $r>0$, we may define a single-valued operator ${J}_{r}={\left(I+rB\right)}^{-1}:H\to D\left(B\right)$, which is called the resolvent of B for r, and let ${B}^{-1}0=\left\{x\in H:0\in Bx\right\}$.

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

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 $Fix\left(G\right)\ne \mathrm{\varnothing }$. Then $〈x-Gx,Gx-w〉\ge 0$ for each $x\in H$ and each $w\in Fix\left(G\right)$.

A mapping $T:H\to H$ is said to be averaged if $T=\left(1-\alpha \right)I+\alpha S$, where $\alpha \in \left(0,1\right)$ 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:
1. (i)

T is nonexpansive if and only if the complement $\left(I-T\right)$ is $1/2$-ism.

2. (ii)

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

3. (iii)

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

4. (iv)

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

5. (v)

If the mappings ${\left\{{T}_{i}\right\}}_{i=1}^{n}$ are averaged and have a common fixed point, then ${\bigcap }_{i=1}^{n}Fix\left({T}_{i}\right)=Fix\left({T}_{1}\cdots {T}_{n}\right)$.

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
$z={P}_{C}\left(I-tV\right)z\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}〈Vz,y-z〉\ge 0\phantom{\rule{1em}{0ex}}⇔\phantom{\rule{1em}{0ex}}z={P}_{C}\left(I-V\right)z.$

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 ${z}_{0}={P}_{C}\left({z}_{0}-V{z}_{0}+\theta \right)$. This point ${z}_{0}\in C$ is also a unique solution of the hierarchical variational inequality
$〈V{z}_{0}-\theta ,q-{z}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in C.$

Lemma 2.6 [33]

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

For each $r>0$, ${J}_{r}$ is single-valued and firmly nonexpansive;

2. (ii)

For each $r>0$, $\mathcal{D}\left({J}_{r}\right)=H$ and $Fix\left({J}_{r}\right)=\left\{x\in \mathcal{D}\left(B\right):0\in Bx\right\}$;

Lemma 2.7 [33]

Let B be a maximal monotone mapping on H. Let ${J}_{r}$ be the resolvent of B defined by ${J}_{r}={\left(I+rB\right)}^{-1}$ 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 $\alpha {\parallel Tx-Ty\parallel }^{2}+\left(1-\alpha \right){\parallel Ty-x\parallel }^{2}\le \beta {\parallel Tx-y\parallel }^{2}+\left(1-\beta \right){\parallel x-y\parallel }^{2}$ 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 $F\left(T\right)=F\left(\stackrel{ˆ}{T}\right)$.

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

Lemma 2.9 [36]

Let $\left\{{a}_{n}\right\}$ be a sequence of real numbers such that there exists a subsequence $\left\{{n}_{i}\right\}$ of $\left\{n\right\}$ such that ${a}_{{n}_{i}}<{a}_{{n}_{i}+1}$ for all $i\in \mathbb{N}$. Then there exists a nondecreasing sequence $\left\{{m}_{k}\right\}\subseteq \mathbb{N}$ such that ${m}_{k}\to \mathrm{\infty }$ and the following properties are satisfied for all (sufficiently large) numbers $k\in \mathbb{N}$:
${a}_{{m}_{k}}\le {a}_{{m}_{k}+1}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{a}_{k}\le {a}_{{m}_{k}+1}.$

In fact, ${m}_{k}=max\left\{j\le k:{a}_{j}<{a}_{j+1}\right\}$.

Lemma 2.10 [37]

Let ${\left\{{a}_{n}\right\}}_{n\in \mathbb{N}}$ be a sequence of nonnegative real numbers, $\left\{{\alpha }_{n}\right\}$ be a sequence of real numbers in $\left[0,1\right]$ with ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_{n}=\mathrm{\infty }$, $\left\{{u}_{n}\right\}$ be a sequence of nonnegative real numbers with ${\sum }_{n=1}^{\mathrm{\infty }}{u}_{n}<\mathrm{\infty }$, $\left\{{t}_{n}\right\}$ be a sequence of real numbers with $lim sup{t}_{n}\le 0$. Suppose that ${a}_{n+1}\le \left(1-{\alpha }_{n}\right){a}_{n}+{\alpha }_{n}{t}_{n}+{u}_{n}$ 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

where $g:C×C\to \mathbb{R}$ 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 $EP\left(g\right)$. For solving the equilibrium problem, let us assume that the bifunction $g:C×C\to \mathbb{R}$ satisfies the following conditions:
1. (A1)

$g\left(x,x\right)=0$ for each $x\in C$;

2. (A2)

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

3. (A3)

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

4. (A4)

for each $x\in C$, the scalar function $y\to g\left(x,y\right)$ 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 $g:C×C\to \mathbb{R}$ 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
$g\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0$

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 $g:C×C\to \mathbb{R}$ be a function satisfying conditions (A1)-(A4). For $r>0$, define ${T}_{r}^{g}:H\to C$ by
${T}_{r}^{g}x=\left\{z\in C:g\left(z,y\right)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\right\}$
for all $x\in H$. Then the following hold:
1. (i)

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

2. (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 $x,y\in H$;

3. (iii)

$\left\{x\in H:{T}_{r}^{g}x=x\right\}=\left\{x\in C:g\left(x,y\right)\ge 0,\mathrm{\forall }y\in C\right\}$;

4. (iv)

$\left\{x\in C:g\left(x,y\right)\ge 0,\mathrm{\forall }y\in C\right\}$ 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 ${J}_{\lambda }={\left(I+\lambda {G}_{1}\right)}^{-1}$ and ${T}_{r}={\left(I+r{G}_{2}\right)}^{-1}$ for each $\lambda >0$ and $r>0$. Let $\left\{{\theta }_{n}\right\}\subset H$ 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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$ such that $F\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0\ne \mathrm{\varnothing }$. Take $\mu \in \mathbb{R}$ as follows:
$0<\mu <\frac{2\overline{\gamma }}{{L}^{2}}.$
Let $\left\{{x}_{n}\right\}\subset H$ be defined by
$\left(3.1\right)\phantom{\rule{1em}{0ex}}\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{0.25em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right){T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right){x}_{n},\hfill \\ {s}_{n}=T{y}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}\right)\hfill \end{array}$
for each $n\in \mathbb{N}$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$, $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$. Assume that:
1. (i)

$0<{lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\alpha }_{n}<1$;

2. (ii)

${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$, and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}=\mathrm{\infty }$;

3. (iii)

$0, and $0;

4. (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 $\overline{x}={P}_{Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0}\left(\overline{x}-V\overline{x}+\theta \right)$. This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:
$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0.$
Proof Take any $\overline{x}\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0$ and let $\overline{x}$ be fixed. Then $\overline{x}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right)\overline{x}$ and $\overline{x}={T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right)\overline{x}$. Let ${u}_{n}={T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right){x}_{n}$. For each $n\in \mathbb{N}$, we have
$\begin{array}{c}{\parallel {u}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right){x}_{n}-{T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right)\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel \left({x}_{n}-\overline{x}\right)-{r}_{n}\left({B}_{2}{x}_{n}-{B}_{2}\overline{x}\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-2{r}_{n}〈{x}_{n}-\overline{x},{B}_{2}{x}_{n}-{B}_{2}\overline{x}〉+{r}_{n}^{2}{\parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-2{r}_{n}{\kappa }_{2}{\parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel }^{2}+{r}_{n}^{2}{\parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{r}_{n}\left(2{\kappa }_{2}-{r}_{n}\right){\parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2},\hfill \end{array}$
(4)
and
$\begin{array}{c}{\parallel {y}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right){u}_{n}-{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right)\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel \left({u}_{n}-\overline{x}\right)-{\lambda }_{n}\left({B}_{1}{u}_{n}-{B}_{1}\overline{x}\right)\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}-2{\lambda }_{n}〈{u}_{n}-\overline{x},{B}_{1}{u}_{n}-{B}_{1}\overline{x}〉+{\lambda }_{n}^{2}{\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}-2{\lambda }_{n}{\kappa }_{1}{\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}+{\lambda }_{n}^{2}{\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}-{\lambda }_{n}\left(2{\kappa }_{1}-{\lambda }_{n}\right){\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}.\hfill \end{array}$
(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 ${z}_{n}={\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}$, we have that
$\begin{array}{rcl}\parallel {z}_{n}-\overline{x}\parallel & =& \parallel {\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}-\overline{x}\parallel \\ \le & {\beta }_{n}\parallel {\theta }_{n}-V\overline{x}\parallel +\parallel \left(I-{\beta }_{n}V\right)\left({s}_{n}-\overline{x}\right)\parallel \\ \le & {\beta }_{n}\parallel {\theta }_{n}-V\overline{x}\parallel +\parallel \left(I-{\beta }_{n}V\right){s}_{n}-\left(I-{\beta }_{n}V\right)\overline{x}\parallel .\end{array}$
(7)
Put $\tau =\overline{\gamma }-\frac{{L}^{2}\mu }{2}$, we have that
$\begin{array}{c}{\parallel \left(I-{\beta }_{n}V\right){s}_{n}-\left(I-{\beta }_{n}V\right)\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\parallel {s}_{n}-\overline{x}\parallel }^{2}-2{\beta }_{n}〈{s}_{n}-\overline{x},V{s}_{n}-V\overline{x}〉+{\beta }_{n}^{2}{\parallel V{s}_{n}-V\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {s}_{n}-\overline{x}\parallel }^{2}-2{\beta }_{n}\overline{\gamma }{\parallel {s}_{n}-\overline{x}\parallel }^{2}+{\beta }_{n}^{2}{L}^{2}{\parallel {s}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-2{\beta }_{n}\overline{\gamma }+{\beta }_{n}^{2}{L}^{2}\right){\parallel {s}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-2{\beta }_{n}\tau -{\beta }_{n}\left({L}^{2}\mu -{\beta }_{n}{L}^{2}\right)\right){\parallel {s}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-2{\beta }_{n}\tau +{\beta }_{n}^{2}{\tau }^{2}\right){\parallel {s}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\left(1-{\beta }_{n}\tau \right)}^{2}{\parallel {x}_{n}-\overline{x}\parallel }^{2}.\hfill \end{array}$
Since $1-{\beta }_{n}\tau >0$, we obtain that
$\parallel \left(I-{\beta }_{n}V\right){s}_{n}-\left(I-{\beta }_{n}V\right)\overline{x}\parallel \le \left(1-{\beta }_{n}\tau \right)\parallel {x}_{n}-\overline{x}\parallel .$
(8)
We have from (7) and (8) that
$\parallel {z}_{n}-\overline{x}\parallel \le {\beta }_{n}\parallel {\theta }_{n}-V\overline{x}\parallel +\left(1-{\beta }_{n}\tau \right)\parallel {x}_{n}-\overline{x}\parallel .$
(9)
Thus, we obtain from the definition of ${x}_{n}$ and (9) that
$\begin{array}{rcl}\parallel {x}_{n+1}-\overline{x}\parallel & =& \parallel {\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}\right)-\overline{x}\parallel \\ \le & {\alpha }_{n}\parallel {x}_{n}-\overline{x}\parallel +\left(1-{\alpha }_{n}\right)\parallel \left({\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}\right)-\overline{x}\parallel \\ \le & {\alpha }_{n}\parallel {x}_{n}-\overline{x}\parallel +\left(1-{\alpha }_{n}\right)\left[{\beta }_{n}\parallel {\theta }_{n}-V\overline{x}\parallel +\left(1-{\beta }_{n}\tau \right)\parallel {x}_{n}-\overline{x}\parallel \right]\\ \le & \left[1-\left(1-{\alpha }_{n}\right){\beta }_{n}\tau \right]\parallel {x}_{n}-\overline{x}\parallel +{\beta }_{n}\left(1-{\alpha }_{n}\right)\tau \frac{\parallel {\theta }_{n}-V\overline{x}\parallel }{\tau }\\ \le & max\left\{\parallel {x}_{n}-\overline{x}\parallel ,\frac{\parallel {\theta }_{n}-V\overline{x}\parallel }{\tau }\right\}\\ \le & max\left\{\parallel {x}_{n}-\overline{x}\parallel ,M\right\},\end{array}$
where $M=max\left\{\frac{\parallel {\theta }_{n}-V\overline{x}\parallel }{\tau },n\in \mathbb{N}\right\}$. By induction, we deduce
$\parallel {x}_{n}-\overline{x}\parallel \le max\left\{\parallel {x}_{1}-\overline{x}\parallel ,M\right\}.$

This implies that the sequence $\left\{{x}_{n}\right\}$ is bounded. Furthermore, $\left\{{u}_{n}\right\}$, $\left\{{z}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{s}_{n}\right\}$ are bounded.

By the definition of $\left\{{x}_{n}\right\}$, we have that
$\begin{array}{rcl}{x}_{n+1}-{x}_{n}& =& {\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}\right)-{x}_{n}\\ =& \left(1-{\alpha }_{n}\right)\left[\left({\beta }_{n}{\theta }_{n}+\left(I-{\beta }_{n}V\right){s}_{n}\right)-{x}_{n}\right]\\ =& \left(1-{\alpha }_{n}\right)\left[{\beta }_{n}{\theta }_{n}-{\beta }_{n}V{s}_{n}+{s}_{n}-{x}_{n}\right].\end{array}$
(10)
By (10), we have that
$\begin{array}{c}〈{x}_{n+1}-{x}_{n},{x}_{n}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈\left(1-{\alpha }_{n}\right)\left[{\beta }_{n}{\theta }_{n}-{\beta }_{n}V{s}_{n}+{s}_{n}-{x}_{n}\right],{x}_{n}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}=\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉+\left(1-{\alpha }_{n}\right)〈{s}_{n}-{x}_{n},{x}_{n}-\overline{x}〉.\hfill \end{array}$
(11)
By (3) and (11), we have that
$\begin{array}{c}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}-{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=2\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\left(1-{\alpha }_{n}\right)\left[{\parallel {s}_{n}-\overline{x}\parallel }^{2}-{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {s}_{n}-{x}_{n}\parallel }^{2}\right].\hfill \end{array}$
(12)
By (5) and (12), we have that
$\begin{array}{c}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}-{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(1-{\alpha }_{n}\right){\parallel {s}_{n}-{x}_{n}\parallel }^{2}.\hfill \end{array}$
(13)
By (10), we obtain that
$\begin{array}{c}{\parallel {x}_{n+1}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\left(1-{\alpha }_{n}\right)}^{2}{\left[{\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel +\parallel {s}_{n}-{x}_{n}\parallel \right]}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel {\theta }_{n}-V{s}_{n}\parallel }^{2}+{\parallel {s}_{n}-{x}_{n}\parallel }^{2}+2{\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel \parallel {s}_{n}-{x}_{n}\parallel \right].\hfill \end{array}$
(14)
By (13) and (14), we have that
$\begin{array}{c}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}-{\parallel {x}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉-\left(1-{\alpha }_{n}\right){\parallel {s}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel {\theta }_{n}-V{s}_{n}\parallel }^{2}+{\parallel {s}_{n}-{x}_{n}\parallel }^{2}+2{\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel \parallel {s}_{n}-{x}_{n}\parallel \right]\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉-\left(1-{\alpha }_{n}\right){\alpha }_{n}{\parallel {s}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel {\theta }_{n}-V{s}_{n}\parallel }^{2}+2{\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel \parallel {s}_{n}-{x}_{n}\parallel \right].\hfill \end{array}$
Hence, we obtain that
$\begin{array}{c}{\parallel {x}_{n+1}-\overline{x}\parallel }^{2}-{\parallel {x}_{n}-\overline{x}\parallel }^{2}+\left(1-{\alpha }_{n}\right){\alpha }_{n}{\parallel {s}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{n}\right){\beta }_{n}〈{\theta }_{n},{x}_{n}-\overline{x}〉-2\left(1-{\alpha }_{n}\right){\beta }_{n}〈V{s}_{n},{x}_{n}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\left(1-{\alpha }_{n}\right)}^{2}\left[{\beta }_{n}^{2}{\parallel {\theta }_{n}-V{s}_{n}\parallel }^{2}+2{\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel \parallel {s}_{n}-{x}_{n}\parallel \right].\hfill \end{array}$
(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
$\parallel {z}_{n}-{s}_{n}\parallel \le \parallel {\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}-{s}_{n}\parallel \le {\beta }_{n}\parallel {\theta }_{n}-V{s}_{n}\parallel .$
(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
${\parallel {s}_{n}-\overline{x}\parallel }^{2}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{r}_{n}\left(2{\kappa }_{2}-{r}_{n}\right){\parallel {B}_{2}{u}_{n}-{B}_{2}\overline{x}\parallel }^{2}.$
Therefore,
$\begin{array}{rcl}{r}_{n}\left(2{\kappa }_{2}-{r}_{n}\right){\parallel {B}_{2}{u}_{n}-{B}_{2}\overline{x}\parallel }^{2}& \le & {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {s}_{n}-\overline{x}\parallel }^{2}\\ \le & \parallel {x}_{n}-{s}_{n}\parallel \left(\parallel {s}_{n}-\overline{x}\parallel +\parallel {x}_{n}-\overline{x}\parallel \right).\end{array}$
(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
$\begin{array}{c}2{\parallel {u}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=2{\parallel {T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right){x}_{n}-{T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right)\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2〈{u}_{n}-\overline{x},\left(I-{r}_{n}{B}_{2}\right){x}_{n}-\left(I-{r}_{n}{B}_{2}\right)\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le 2〈{u}_{n}-\overline{x},{x}_{n}-\overline{x}〉-2{r}_{n}〈{u}_{n}-\overline{x},{B}_{2}{x}_{n}-{B}_{2}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-2{r}_{n}〈{x}_{n}-\overline{x},{B}_{2}{x}_{n}-{B}_{2}\overline{x}〉-2{r}_{n}〈{u}_{n}-{x}_{n},{B}_{2}{x}_{n}-{B}_{2}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-2{\lambda }_{n}{\kappa }_{2}{\parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel }^{2}+2{r}_{n}〈{x}_{n}-{u}_{n},{B}_{2}{x}_{n}-{B}_{2}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {u}_{n}-\overline{x}\parallel }^{2}+{\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{u}_{n}\parallel \parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel .\hfill \end{array}$
(23)
By (6) and (23), we have that
$\begin{array}{rcl}{\parallel {s}_{n}-\overline{x}\parallel }^{2}& \le & {\parallel {u}_{n}-\overline{x}\parallel }^{2}\\ \le & {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {u}_{n}-{x}_{n}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{u}_{n}\parallel \parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel .\end{array}$
Therefore,
$\begin{array}{c}{\parallel {u}_{n}-{x}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {s}_{n}-\overline{x}\parallel }^{2}+2{r}_{n}\parallel {x}_{n}-{u}_{n}\parallel \parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{s}_{n}\parallel \left(\parallel {x}_{n}-\overline{x}\parallel +\parallel {s}_{n}-\overline{x}\parallel \right)+2{r}_{n}\parallel {x}_{n}-{u}_{n}\parallel \parallel {B}_{2}{x}_{n}-{B}_{2}\overline{x}\parallel .\hfill \end{array}$
(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
${\parallel {s}_{n}-\overline{x}\parallel }^{2}\le {\parallel {y}_{n}-\overline{x}\parallel }^{2}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\lambda }_{n}\left(2{\kappa }_{1}-{\lambda }_{n}\right){\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}.$
Therefore,
$\begin{array}{rcl}{\lambda }_{n}\left(2{\kappa }_{1}-{\lambda }_{n}\right){\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}& \le & {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {s}_{n}-\overline{x}\parallel }^{2}\\ \le & \parallel {x}_{n}-{s}_{n}\parallel \left(\parallel {s}_{n}-\overline{x}\parallel +\parallel {x}_{n}-\overline{x}\parallel \right).\end{array}$
(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
$\begin{array}{c}2{\parallel {y}_{n}-\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=2{\parallel {J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right){u}_{n}-{J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right)\overline{x}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2〈{y}_{n}-\overline{x},\left(I-{\lambda }_{n}{B}_{1}\right){u}_{n}-\left(I-{\lambda }_{n}{B}_{1}\right)\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le 2〈{y}_{n}-\overline{x},{u}_{n}-\overline{x}〉-2{\lambda }_{n}〈{y}_{n}-\overline{x},{B}_{1}{u}_{n}-{B}_{1}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {y}_{n}-\overline{x}\parallel }^{2}+{\parallel {u}_{n}-\overline{x}\parallel }^{2}-{\parallel {y}_{n}-{u}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-2{\lambda }_{n}〈{u}_{n}-\overline{x},{B}_{1}{u}_{n}-{B}_{1}\overline{x}〉-2{\lambda }_{n}〈{y}_{n}-{u}_{n},{B}_{1}{u}_{n}-{B}_{1}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {y}_{n}-\overline{x}\parallel }^{2}+{\parallel {u}_{n}-\overline{x}\parallel }^{2}-{\parallel {y}_{n}-{u}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-2{\lambda }_{n}{\kappa }_{1}{\parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel }^{2}+2{\lambda }_{n}〈{u}_{n}-{y}_{n},{B}_{1}{u}_{n}-{B}_{1}\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {y}_{n}-\overline{x}\parallel }^{2}+{\parallel {u}_{n}-\overline{x}\parallel }^{2}-{\parallel {y}_{n}-{u}_{n}\parallel }^{2}+2{\lambda }_{n}\parallel {u}_{n}-{y}_{n}\parallel \parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel .\hfill \end{array}$
(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,
$\begin{array}{c}{\parallel {y}_{n}-{u}_{n}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\parallel {x}_{n}-\overline{x}\parallel }^{2}-{\parallel {s}_{n}-\overline{x}\parallel }^{2}+2{\lambda }_{n}\parallel {u}_{n}-{y}_{n}\parallel \parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{n}-{s}_{n}\parallel \left(\parallel {x}_{n}-\overline{x}\parallel +\parallel {s}_{n}-\overline{x}\parallel \right)+2{\lambda }_{n}\parallel {u}_{n}-{y}_{n}\parallel \parallel {B}_{1}{u}_{n}-{B}_{1}\overline{x}\parallel .\hfill \end{array}$
(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 $Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0$ is a nonempty closed convex subset of H, by Lemma 2.5, we can take ${\overline{x}}_{0}\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0$ such that
${\overline{x}}_{0}={P}_{Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0}\left({\overline{x}}_{0}-V{\overline{x}}_{0}+\theta \right).$
This point ${\overline{x}}_{0}$ is also a unique solution of the hierarchical variational inequality
$〈V{\overline{x}}_{0}-\theta ,q-{\overline{x}}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0.$
(31)
We want to show that
$\underset{n\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{n}-{\overline{x}}_{0}〉\ge 0.$
Without loss of generality, there exists a subsequence $\left\{{z}_{{n}_{k}}\right\}$ of $\left\{{z}_{n}\right\}$ such that ${z}_{{n}_{k}}⇀w$ for some $w\in H$ and
$\underset{n\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{n}-{\overline{x}}_{0}〉=\underset{k\to \mathrm{\infty }}{lim}〈\left(V{\overline{x}}_{0}-\theta ,{z}_{{n}_{k}}-{\overline{x}}_{0}〉.$
(32)
By (20) and (25), we have that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {u}_{n}-{z}_{n}\parallel =0$
and ${u}_{{n}_{k}}⇀w$. On the other hand, since $0, there exists a subsequence $\left\{{\lambda }_{{n}_{{k}_{j}}}\right\}$ of $\left\{{\lambda }_{{n}_{k}}\right\}$ such that $\left\{{\lambda }_{{n}_{{k}_{j}}}\right\}$ converges to a number $\overline{\lambda }\in \left[a,b\right]$. By (30) and Lemma 2.7, we have that
$\begin{array}{c}\parallel {u}_{{n}_{{k}_{j}}}-{J}_{\overline{\lambda }}\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {u}_{{n}_{{k}_{j}}}-{J}_{{\lambda }_{{n}_{{k}_{j}}}}\left(I-{\lambda }_{{n}_{{k}_{j}}}{B}_{1}\right){u}_{{n}_{{k}_{j}}}\parallel +\parallel {J}_{{\lambda }_{{n}_{{k}_{j}}}}\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}-{J}_{\overline{\lambda }}\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\parallel {J}_{{\lambda }_{{n}_{{k}_{j}}}}\left(I-{\lambda }_{{n}_{{k}_{j}}}{B}_{1}\right){u}_{{n}_{{k}_{j}}}-{J}_{{\lambda }_{{n}_{{k}_{j}}}}\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {u}_{{n}_{{k}_{j}}}-{y}_{{n}_{{k}_{j}}}\parallel +|{\lambda }_{{n}_{{k}_{j}}}-\overline{\lambda }|\parallel {B}_{1}{u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{|{\lambda }_{{n}_{{k}_{j}}}-\overline{\lambda }|}{\overline{\lambda }}\parallel {J}_{\overline{\lambda }}\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}-\left(I-\overline{\lambda }{B}_{1}\right){u}_{{n}_{{k}_{j}}}\parallel \to 0.\hfill \end{array}$
(33)
By (33), ${u}_{{n}_{{k}_{j}}}⇀w$, Lemma 2.6 and 2.7, $w\in Fix\left({J}_{\overline{\lambda }}\left(I-\overline{\lambda }{B}_{1}\right)\right)={\left({B}_{1}+{G}_{1}\right)}^{-1}0$. Without loss of generality and $0, there exists a subsequence $\left\{{r}_{{n}_{{k}_{j}}}\right\}$ of $\left\{{r}_{{n}_{k}}\right\}$ such that $\left\{{r}_{{n}_{{k}_{j}}}\right\}$ converges to a number $\overline{r}\in \left[a,b\right]$. By (25) and Lemma 2.7, we have that
$\begin{array}{c}\parallel {x}_{{n}_{{k}_{j}}}-{T}_{\overline{r}}\left(I-\overline{r}{B}_{2}\right){x}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{{n}_{{k}_{j}}}-{T}_{{r}_{{n}_{{k}_{j}}}}\left(I-{r}_{{n}_{{k}_{j}}}{B}_{2}\right){x}_{{n}_{{k}_{j}}}\parallel +\parallel {T}_{{r}_{{n}_{{k}_{j}}}}\left(I-{r}_{{n}_{{k}_{j}}}{B}_{2}\right){x}_{{n}_{{k}_{j}}}-{T}_{{r}_{{n}_{{k}_{j}}}}\left(I-\overline{r}{B}_{2}\right){x}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\parallel {T}_{{r}_{{n}_{{k}_{j}}}}\left(I-\overline{r}{B}_{2}\right){u}_{{n}_{{k}_{j}}}-{T}_{\overline{r}}\left(I-\overline{r}{B}_{2}\right){u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel {x}_{{n}_{{k}_{j}}}-{u}_{{n}_{{k}_{j}}}\parallel +|{r}_{{n}_{{k}_{j}}}-\overline{r}|\parallel {B}_{2}{u}_{{n}_{{k}_{j}}}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{|{r}_{{n}_{{k}_{j}}}-\overline{r}|}{\overline{r}}\parallel {T}_{\overline{r}}\left(I-\overline{r}{B}_{2}\right){u}_{{n}_{{k}_{j}}}-\left(I-\overline{r}{B}_{2}\right){u}_{{n}_{{k}_{j}}}\parallel \to 0.\hfill \end{array}$
(34)
By (34), ${x}_{{n}_{{k}_{j}}}⇀w$, Lemma 2.8, we have that $w\in Fix\left({T}_{\overline{r}}\left(I-\overline{r}{B}_{2}\right)\right)={\left({B}_{2}+{G}_{2}\right)}^{-1}0$. 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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$, we have from $\parallel T{y}_{{n}_{j}}-{y}_{{n}_{j}}\parallel \to 0$ and ${y}_{{n}_{{k}_{j}}}⇀w$ that $w\in F\left(T\right)$. Hence, $w\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0$. So, we have from (31) and (32) that
$\underset{n\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{n}-{\overline{x}}_{0}〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_{0}-\theta ,{z}_{{n}_{k}}-{\overline{x}}_{0}〉=〈V{\overline{x}}_{0}-\theta ,w-{\overline{x}}_{0}〉\ge 0.$
(35)
Let ${z}_{n}={\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}$. Then it follows from (7) that
$\begin{array}{rcl}{\parallel {z}_{n}-{\overline{x}}_{0}\parallel }^{2}& =& {\parallel {\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}-{\overline{x}}_{0}\parallel }^{2}\\ =& {\parallel {\beta }_{n}\left({\theta }_{n}-V{\overline{x}}_{0}\right)+\left(1-{\beta }_{n}V\right)\left({s}_{n}-{\overline{x}}_{0}\right)\parallel }^{2}\\ \le & {\parallel \left(1-{\beta }_{n}V\right)\left({s}_{n}-{\overline{x}}_{0}\right)\parallel }^{2}+2{\beta }_{n}〈{\theta }_{n}-V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉\\ \le & {\left(1-{\beta }_{n}\tau \right)}^{2}{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+2{\beta }_{n}〈{\theta }_{n}-V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉.\end{array}$
(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 \\ \phantom{\rule{1em}{0ex}}={\parallel {\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}\right)-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+\left(1-{\alpha }_{n}\right){\parallel \left({\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}\right)-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le {\alpha }_{n}{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+\left(1-{\alpha }_{n}\right)\left({\left(1-{\beta }_{n}\tau \right)}^{2}{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+2{\beta }_{n}〈{\theta }_{n}-V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[{\alpha }_{n}+\left(1-{\alpha }_{n}\right){\left(1-{\beta }_{n}\tau \right)}^{2}\right]{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+2{\beta }_{n}\left(1-{\alpha }_{n}\right)〈{\theta }_{n}-V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[1-\left(1-{\alpha }_{n}\right)\left(2{\beta }_{n}\tau -{\left({\beta }_{n}\tau \right)}^{2}\right)\right]{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+2{\beta }_{n}\left(1-{\alpha }_{n}\right)〈{\theta }_{n}-V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[1-2\left(1-{\alpha }_{n}\right){\beta }_{n}\tau \right]{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+\left(1-{\alpha }_{n}\right){\left({\beta }_{n}\tau \right)}^{2}{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2{\beta }_{n}\left(1-{\alpha }_{n}\right)〈{\theta }_{n}-\theta ,{z}_{n}-{\overline{x}}_{0}〉+2{\beta }_{n}\left(1-{\alpha }_{n}\right)〈\theta -V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[1-2\left(1-{\alpha }_{n}\right){\beta }_{n}\tau \right]{\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}+2\left(1-{\alpha }_{n}\right){\beta }_{n}\tau \left(\frac{{\beta }_{n}\tau {\parallel {x}_{n}-{\overline{x}}_{0}\parallel }^{2}}{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{〈{\theta }_{n}-\theta ,{z}_{n}-{\overline{x}}_{0}〉}{\tau }+\frac{〈\theta -V{\overline{x}}_{0},{z}_{n}-{\overline{x}}_{0}〉}{\tau }\right).\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
${\overline{x}}_{0}={P}_{Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0}\left({\overline{x}}_{0}-V{\overline{x}}_{0}+\theta \right).$
Case 2: Suppose that there exists $\left\{{n}_{i}\right\}$ of $\left\{n\right\}$ 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 $\left\{{m}_{j}\right\}$ in such that ${m}_{j}\to \mathrm{\infty }$ and
$\parallel {x}_{{m}_{j}}-\overline{x}\parallel \le \parallel {x}_{{m}_{j}+1}-\overline{x}\parallel \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\parallel {x}_{j}-\overline{x}\parallel \le \parallel {x}_{{m}_{j}+1}-\overline{x}\parallel .$
(38)
Hence, it follows from (15) and (38) that
$\begin{array}{c}\left(1-{\alpha }_{{m}_{j}}\right){\alpha }_{{m}_{j}}{\parallel {s}_{{m}_{j}}-{x}_{{m}_{j}}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\left(1-{\alpha }_{{m}_{j}}\right){\beta }_{{m}_{j}}〈{\theta }_{{m}_{j}},{x}_{{m}_{j}}-\overline{x}〉-2\left(1-{\alpha }_{{m}_{j}}\right){\beta }_{{m}_{j}}〈V{s}_{{m}_{j}},{x}_{{m}_{j}}-\overline{x}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+{\left(1-{\alpha }_{{m}_{j}}\right)}^{2}\left[{\beta }_{{m}_{j}}^{2}{\parallel {\theta }_{{m}_{j}}-V{s}_{{m}_{j}}\parallel }^{2}+2{\beta }_{{m}_{j}}\parallel {\theta }_{{m}_{j}}-V{s}_{{m}_{j}}\parallel \parallel {s}_{{m}_{j}}-{x}_{{m}_{j}}\parallel \right]\hfill \end{array}$
(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
$\underset{j\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{{m}_{j}}-{\overline{x}}_{0}〉\ge 0.$
Without loss of generality, there exists a subsequence $\left\{{z}_{{m}_{{j}_{k}}}\right\}$ of $\left\{{z}_{{m}_{j}}\right\}$ such that ${z}_{{m}_{{j}_{k}}}⇀w$ for some $w\in H$ and
$\underset{j\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{{m}_{j}}-{\overline{x}}_{0}〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_{0}-\theta ,{z}_{{m}_{{j}_{k}}}-{\overline{x}}_{0}〉.$
(41)
With the similar argument as in the proof of Case 1, we have $w\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0$. So, we have from (41) and (32) that
$\underset{j\to \mathrm{\infty }}{lim sup}〈V{\overline{x}}_{0}-\theta ,{z}_{{m}_{j}}-{\overline{x}}_{0}〉=\underset{k\to \mathrm{\infty }}{lim}〈V{\overline{x}}_{0}-\theta ,{z}_{{m}_{{j}_{k}}}-{\overline{x}}_{0}〉=〈V{\overline{x}}_{0}-\theta ,w-{\overline{x}}_{0}〉\ge 0.$
(42)
With the similar argument as in the proof of Case 1, we have
$\begin{array}{c}{\parallel {x}_{{m}_{j}+1}-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[1-2\left(1-{\alpha }_{{m}_{j}}\right){\beta }_{{m}_{j}}\tau \right]{\parallel {x}_{{m}_{j}}-{\overline{x}}_{0}\parallel }^{2}+\left(1-{\alpha }_{{m}_{j}}\right){\left({\beta }_{{m}_{j}}\tau \right)}^{2}{\parallel {x}_{{m}_{j}}-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2{\beta }_{{m}_{j}}\left(1-{\alpha }_{{m}_{j}}\right)〈{\theta }_{n}-\theta ,{z}_{{m}_{j}}-{\overline{x}}_{0}〉+2{\beta }_{{m}_{j}}\left(1-{\alpha }_{{m}_{j}}\right)〈\theta -V{\overline{x}}_{0},{z}_{{m}_{j}}-{\overline{x}}_{0}〉.\hfill \end{array}$
(43)
From $\parallel {x}_{{m}_{j}}-\overline{x}\parallel \le \parallel {x}_{{m}_{j}+1}-\overline{x}\parallel$, we have that
$\begin{array}{c}2\left(1-{\alpha }_{{m}_{j}}\right){\beta }_{{m}_{j}}\tau {\parallel {x}_{{m}_{j}}-{\overline{x}}_{0}\parallel }^{2}\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-{\alpha }_{{m}_{j}}\right){\left({\beta }_{{m}_{j}}\tau \right)}^{2}{\parallel {x}_{{m}_{j}}-{\overline{x}}_{0}\parallel }^{2}+2{\beta }_{{m}_{j}}\left(1-{\alpha }_{{m}_{j}}\right)〈{\theta }_{n}-\theta ,{z}_{{m}_{j}}-{\overline{x}}_{0}〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2{\beta }_{{m}_{j}}\left(1-{\alpha }_{{m}_{j}}\right)〈\theta -V{\overline{x}}_{0},{z}_{{m}_{j}}-{\overline{x}}_{0}〉.\hfill \end{array}$
(44)
Since $\left(1-{\alpha }_{{m}_{j}}\right){\beta }_{{m}_{j}}>0$, 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 ${J}_{\lambda }={\left(I+\lambda {G}_{1}\right)}^{-1}$ and ${T}_{r}={\left(I+r{G}_{2}\right)}^{-1}$ 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:

(MSFPFF) Find $\overline{x}\in {H}_{1}$ such that $\overline{x}\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${A}_{1}\overline{x}\in Fix\left({F}_{1}\right)$, and ${A}_{2}\overline{x}\in Fix\left({F}_{2}\right)$ for each $n\in \mathbb{N}$.

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

Theorem 3.2 Given any $\overline{x}\in {H}_{1}$.
1. (i)

If $\overline{x}$ is a solution of (MSFPFF), then ${J}_{{\lambda }_{n}}\left(I-{\rho }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{\sigma }_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}=\overline{x}$ for each $n\in \mathbb{N}$.

2. (ii)

Suppose that ${J}_{{\lambda }_{n}}\left(I-{\rho }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{\sigma }_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}=\overline{x}$ 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 (MSFPFF) is nonempty. Then $\overline{x}$ is a solution of (MSFPFF).

Proof (i) Suppose that $\overline{x}\in {H}_{1}$ is a solution of (MSFPFF). Then $\overline{x}\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${A}_{1}\overline{x}\in Fix\left({F}_{1}\right)$, and ${A}_{2}\overline{x}\in Fix\left({F}_{2}\right)$ for each $n\in \mathbb{N}$. It is easy to see that
${J}_{{\lambda }_{n}}\left(I-{\rho }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{\sigma }_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}=\overline{x}$

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

(ii) Since the solution set of (MSFPFF) is nonempty, there exists $\overline{w}\in {H}_{1}$ such that $\overline{w}\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${A}_{1}\overline{w}\in Fix\left({F}_{1}\right)$, and ${A}_{2}\overline{w}\in Fix\left({F}_{2}\right)$. So,
$\overline{w}\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left(I-{\rho }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right)\cap Fix\left({T}_{{r}_{n}}\right)\cap Fix\left(I-{\sigma }_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\ne \mathrm{\varnothing }.$
(47)
By Lemma 2.1, we have that
(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
(49)
By Lemma 2.1 again, we have that
(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
(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
(52)
Hence, by (49), (51), (52), and Lemma 2.3(v), we have that for each $n\in \mathbb{N}$,
$\begin{array}{c}\overline{x}\in Fix\left({J}_{{\lambda }_{n}}\left(I-\rho {A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-\sigma {A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left(I-\rho {A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right)\cap Fix\left({T}_{{r}_{n}}\right)\cap Fix\left(I-\sigma {A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right).\hfill \end{array}$
This implies that for each $n\in \mathbb{N}$,
$\overline{x}={J}_{{\lambda }_{n}}\left(I-\rho {A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right)\overline{x}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\overline{x}={T}_{{r}_{n}}\left(I-\sigma {A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}.$
By Lemma 2.2, for each $n\in \mathbb{N}$,
That is, for each $n\in \mathbb{N}$,
(53)
For each $n\in \mathbb{N}$, by (53) and the fact that ${A}_{i}^{\ast }$ is the adjoint of ${A}_{i}$ for each $i=1,2$,
(54)
On the other hand, by Lemma 2.2 again,
(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 $w\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${v}_{2}\in Fix\left({F}_{2}\right)$, and ${v}_{1}\in Fix\left({F}_{1}\right)$.

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 $w\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${v}_{1}\in Fix\left({F}_{1}\right)$, and ${v}_{2}\in Fix\left({F}_{2}\right)$.

Since $\overline{w}$ is a solution of multiple sets split feasibility problem (MSFPFF), we know that $\overline{w}\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right)$, ${A}_{1}\overline{w}\in Fix\left({F}_{1}\right)$, and ${A}_{2}\overline{w}\in Fix\left({F}_{2}\right)$ for each $n\in \mathbb{N}$. So, it follows from (57) that ${A}_{1}\overline{x}=Fix\left({F}_{1}\right)$ and ${A}_{2}\overline{x}=Fix\left({F}_{2}\right)$. Furthermore, $\overline{x}\in Fix\left({J}_{{\lambda }_{n}}\right)$ and $\overline{x}\in Fix\left({T}_{{r}_{n}}\right)$ for each $n\in \mathbb{N}$. Therefore, $\overline{x}$ is a solution of (MSFPFF). □

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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$. 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 (MSFPFF) is Ω and $Fix\left(T\right)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$. Let $\left\{{x}_{n}\right\}\subset H$ be defined by
$\left(3.3\right)\phantom{\rule{1em}{0ex}}\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{0.25em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{r}_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right){x}_{n},\hfill \\ {s}_{n}=T{y}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left(1-{\beta }_{n}V\right){s}_{n}\right)\hfill \end{array}$
for each $n\in \mathbb{N}$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$, $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$. Assume that:
1. (i)

$0<{lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\alpha }_{n}<1$;

2. (ii)

${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$, and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}=\mathrm{\infty }$;

3. (iii)

$0, and $0;

4. (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 $\overline{x}={P}_{Fix\left(T\right)\cap \mathrm{\Omega }}\left(\overline{x}-V\overline{x}+\theta \right)$. This point $\overline{x}$ is also a unique solution of the following hierarchical problem: Find $\overline{x}\in Fix\left(T\right)\cap \mathrm{\Omega }$ such that
$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in Fix\left(T\right)\cap \mathrm{\Omega }.$

Proof Since ${F}_{i}$ is firmly nonexpansive, it follows from Lemma 2.1 that we have that ${A}_{i}^{\ast }\left(I-{F}_{i}\right){A}_{i}:{C}_{1}\to {H}_{1}$ is $\frac{1}{{\parallel {A}_{i}\parallel }^{2}}$-ism for each $i=1,2$. For each $i=1,2$, put ${B}_{i}={A}_{i}^{\ast }\left(I-{F}_{i}\right){A}_{i}$ 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 (MSFPFF) is nonempty, by (47), we have for each $n\in \mathbb{N}$
$\overline{w}\in Fix\left({J}_{{\lambda }_{n}}\left(\left(I-{\lambda }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right)\right)\right)\cap Fix\left({T}_{{r}_{n}}\left(I-{r}_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\right)\ne \mathrm{\varnothing }.$
(58)
This implies that for each $n\in \mathbb{N}$,
$\overline{w}\in Fix\left({J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right)\right)\cap Fix\left({T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right)\right)\ne \mathrm{\varnothing }.$
(59)
So,
$\overline{w}\in {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0\ne \mathrm{\varnothing }.$
(60)
It follows from Theorem 3.1 that ${lim}_{n\to \mathrm{\infty }}{x}_{n}=\overline{x}$, where
$\overline{x}={P}_{Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0}\left(\overline{x}-V\overline{x}+\theta \right).$
This point $\overline{x}$ is also a unique solution of the following hierarchical variational inequality:
$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in Fix\left(T\right)\cap {\left({B}_{1}+{G}_{1}\right)}^{-1}0\cap {\left({B}_{2}+{G}_{2}\right)}^{-1}0,$
that is, for each $n\in \mathbb{N}$,
$\overline{x}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{B}_{1}\right)\overline{x}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right)\overline{x}$
(61)
and
$\overline{x}={T}_{{r}_{n}}\left(I-{r}_{n}{B}_{2}\right)\overline{x}={T}_{{r}_{n}}\left(I-{r}_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}.$
(62)
This implies that for each $n\in \mathbb{N}$,
$\overline{x}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{r}_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right)\overline{x}.$
(63)
By assumptions, (63), and Theorem 3.2(ii), we know that $\overline{x}$ is a solution of (MSFPFF). Furthermore, $\overline{x}\in Fix\left(T\right)$. Therefore, $\overline{x}\in Fix\left(T\right)\cap \mathrm{\Omega }$. By the same argument as (61), (62), and (63), we also have
$〈V\overline{x}-\theta ,q-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in Fix\left(T\right)\cap \mathrm{\Omega }.$

Therefore, the proof is completed. □

Remark 3.1 In Theorem 3.3, we establish a strong convergence theorem for hierarchical problem (MSFPFF) 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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$. In Theorem  3.3, let $h:C\to \mathbb{R}$ be a convex Gâteaux differential function with Gâteaux derivative V. Let
$\mathrm{\Omega }=\left\{q\in {H}_{1}:q\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right),{A}_{1}q\in Fix\left({F}_{1}\right),{A}_{2}q\in Fix\left({F}_{2}\right),n\in \mathbb{N}\right\}.$

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

Proof Put $\theta =0$ in Theorem 3.3. Then, by Theorem 3.3, there exists $\overline{x}\in Fix\left(T\right)\cap \mathrm{\Omega }$ such that
$〈V\overline{x},q-\overline{x}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }q\in F\left(T\right)\cap \mathrm{\Omega }.$
(64)
Since $h:C\to \mathbb{R}$ is a convex Gâteaux differential function with Gâteaux derivative V, we obtain that
$\begin{array}{rcl}〈V\overline{x},y-\overline{x}〉& =& \underset{t\to 0}{lim}\frac{h\left(\overline{x}+t\left(y-\overline{x}\right)\right)-h\left(\overline{x}\right)}{t}\\ =& \underset{t\to 0}{lim}\frac{h\left(\left(1-t\right)\overline{x}+ty\right)-h\left(\overline{x}\right)}{t}\\ \le & \underset{t\to 0}{lim}\frac{\left(1-t\right)h\left(\overline{x}\right)+th\left(y\right)-h\left(\overline{x}\right)}{t}\\ =& h\left(y\right)-h\left(\overline{x}\right)\end{array}$
(65)

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

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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$. In Theorem  3.3, let $B:C\to C$ be a strongly positive self-adjoint bounded linear operator and $a\in H$. Let
$\mathrm{\Omega }=\left\{q\in {H}_{1}:q\in Fix\left({J}_{{\lambda }_{n}}\right)\cap Fix\left({T}_{{r}_{n}}\right),{A}_{1}q\in Fix\left({F}_{1}\right),{A}_{2}q\in Fix\left({F}_{2}\right),n\in \mathbb{N}\right\}.$
Let $\left\{{x}_{n}\right\}\subset H$ be defined by
$\left(4.2\right)\phantom{\rule{1em}{0ex}}\left\{\begin{array}{c}{x}_{1}\in C\phantom{\rule{0.25em}{0ex}}\mathit{\text{chosen arbitrarily}},\hfill \\ {y}_{n}={J}_{{\lambda }_{n}}\left(I-{\lambda }_{n}{A}_{1}^{\ast }\left(I-{F}_{1}\right){A}_{1}\right){T}_{{r}_{n}}\left(I-{r}_{n}{A}_{2}^{\ast }\left(I-{F}_{2}\right){A}_{2}\right){x}_{n},\hfill \\ {s}_{n}=T{y}_{n},\hfill \\ {x}_{n+1}={\alpha }_{n}{x}_{n}+\left(1-{\alpha }_{n}\right)\left({\beta }_{n}{\theta }_{n}+\left({s}_{n}-{\beta }_{n}\left(B\left({s}_{n}\right)-a\right)\right)\right)\hfill \end{array}$
for each $n\in \mathbb{N}$, $\left\{{\lambda }_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$, $\left\{{\alpha }_{n}\right\}\subset \left(0,1\right)$, $\left\{{\beta }_{n}\right\}\subset \left(0,1\right)$, and $\left\{{r}_{n}\right\}\subset \left(0,\mathrm{\infty }\right)$. Assume that:
1. (i)

$0<{lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\le {lim sup}_{n\to \mathrm{\infty }}{\alpha }_{n}<1$;

2. (ii)

${lim}_{n\to \mathrm{\infty }}{\beta }_{n}=0$, and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_{n}=\mathrm{\infty }$;

3. (iii)

$0, and $0;

4. (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: ${min}_{q\in Fix\left(T\right)\cap \mathrm{\Omega }}\frac{1}{2}〈Bq,q〉-〈a,q〉$.

Proof Let $h:C\to \mathbb{R}$ be defined by
$h\left(x\right)=\frac{1}{2}〈Bx,x〉-〈a,x〉.$
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
$〈Bx-By,x-y〉=〈B\left(x-y\right),x-y〉\ge \eta {\parallel x-y\parallel }^{2}.$
(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 $\lambda \in \left[0,1\right]$. It follows from (67) that
$\begin{array}{c}h\left(\lambda x+\left(1-\lambda \right)y\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}〈B\left(\lambda x+\left(1-\lambda \right)y\right),\lambda x+\left(1-\lambda \right)y〉-〈a,\lambda x+\left(1-\lambda \right)y〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}〈\lambda Bx+\left(1-\lambda \right)By,\lambda x+\left(1-\lambda \right)y〉-〈a,\lambda x+\left(1-\lambda \right)y〉\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}{\lambda }^{2}〈Bx,x〉+\frac{1}{2}\lambda \left(1-\lambda \right)〈Bx,y〉+\frac{1}{2}\lambda \left(1-\lambda \right)〈By,x〉+\frac{1}{2}{\left(1-\lambda \right)}^{2}〈By,y〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\lambda 〈a,x〉-\left(1-\lambda \right)〈a,y〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}{\lambda }^{2}〈Bx,x〉+\frac{1}{2}\lambda \left(1-\lambda \right)\left(〈Bx,x〉+〈By,y〉\right)+\frac{1}{2}{\left(1-\lambda \right)}^{2}〈By,y〉\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\lambda 〈a,x〉-\left(1-\lambda \right)〈a,y〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{2}\lambda 〈Bx,x〉+\frac{1}{2}\left(1-\lambda \right)〈By,y〉-\lambda 〈a,x〉-\left(1-\lambda \right)〈a,y〉\hfill \\ \phantom{\rule{1em}{0ex}}\le \lambda h\left(x\right)+\left(1-\lambda \right)h\left(y\right).\hfill \end{array}$
Let $V\left(x\right)=B\left(x\right)-a$ 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 $t\in \left[0,1\right]$. Since B is a self-adjoint bounded linear operator, we see that for each $u\in H$,
$\begin{array}{c}{h}^{\prime }\left(x\right)\left(u\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{t\to 0}{lim}\frac{h\left(x+tu\right)-h\left(x\right)}{t}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{t\to 0}{lim}\frac{\frac{1}{2}〈B\left(x+tu\right),x+tu〉-〈a,x+tu〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{t\to 0}{lim}\frac{\frac{1}{2}〈tBu+Bx,tu+x〉-〈a,tu+x〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{t\to 0}{lim}\frac{\frac{1}{2}\left[{t}^{2}〈Bu,u〉+t〈Bu,x〉+t〈Bx,u〉+〈Bx,x〉\right]-t〈a,u〉-〈a,x〉-\frac{1}{2}〈Bx,x〉+〈a,x〉}{t}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{t\to 0}{lim}\frac{1}{2}\left[t〈Bu,u〉+〈Bu,x〉+〈Bx,u〉\right]-〈a,u〉\hfill \\ \phantom{\rule{1em}{0ex}}=〈Bx,u〉-〈a,u〉=〈Bx-a,u〉=〈Vx,u〉.\hfill \end{array}$
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
$〈Vx-Vy,x-y〉=〈Bx-a-By+a,x-y〉=〈B\left(x-y\right),x-y〉\ge \eta {\parallel x-y\parallel }^{2}.$

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 $Fix\left(T\right)=Fix\left(\stackrel{ˆ}{T}\right)$. 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
$\mathrm{\Omega }=\left\{q\in {H}_{1}:q\in {G}_{1}^{-1}0\cap {G}_{2}^{-1}0,{A}_{1}q\in {\mathrm{\Phi }}_{1}^{-1}0,{A}_{2}q\in {\mathrm{\Phi }}_{2}^{-1}0\right\}.$

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: ${min}_{q\in Fix\left(T\right)\cap \mathrm{\Omega }}h\left(q\right)$.

Proof Let ${J}_{\lambda }={\left(I+\lambda {G}_{1}\right)}^{-1}$,