# The constrained multiple-sets split feasibility problem and its projection algorithms

## Abstract

The projection algorithms for solving the constrained multiple-sets split feasibility problem are presented. The strong convergence results of the algorithms are given under some mild conditions. Especially, the minimum norm solution of the constrained multiple-sets split feasibility problem can be found.

## 1 Introduction

Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let ${C}_{1},{C}_{2},\dots ,{C}_{N}$ be N nonempty closed convex subsets of ${H}_{1}$ and let ${Q}_{1},{Q}_{2},\dots ,{Q}_{M}$ be M nonempty closed convex subsets of ${H}_{2}$. Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator. The multiple-sets split feasibility problem is formulated as follows:

(1.1)

A special case If $N=M=1$, then the multiple-sets split feasibility problem is reduced to the split feasibility problem which is formulated as finding a point x with the property

$x\in C\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Ax\in Q.$

The split feasibility problem 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 [2]. It has been found that the multiple-sets split feasibility problem and the split feasibility problem can be used to model the intensity-modulated radiation therapy [36]. Various algorithms have been invented to solve the multiple-sets split feasibility problem and the split feasibility problem, see, e.g., [724] and references therein.

The popular algorithm that solves the multiple-sets split feasibility problem and the split feasibility problem is Byrne’s CQ algorithm [11] which is found to be a gradient-projection method in convex minimization. Motivated by this idea, in this paper, we present the composite projection algorithms for solving the constrained multiple-sets split feasibility problem. The strong convergence results of the algorithms are given under some mild conditions. Especially, the minimum norm solution of the constrained multiple-sets split feasibility problem can be found.

## 2 Preliminaries

### 2.1 Concepts

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, respectively, and let Ω be a nonempty closed convex subset of H. Recall that the (nearest point or metric) projection from H onto Ω, denoted by ${P}_{\mathrm{\Omega }}$, is defined in such a way that, for each $x\in H$, ${P}_{\mathrm{\Omega }}\left(x\right)$ is the unique point in Ω with the property

$\parallel x-{P}_{\mathrm{\Omega }}\left(x\right)\parallel =min\left\{\parallel x-y\parallel :y\in \mathrm{\Omega }\right\}.$

It is known that ${P}_{\mathrm{\Omega }}$ satisfies

$〈x-y,{P}_{\mathrm{\Omega }}\left(x\right)-{P}_{\mathrm{\Omega }}\left(y\right)〉\ge {\parallel {P}_{\mathrm{\Omega }}\left(x\right)-{P}_{\mathrm{\Omega }}\left(y\right)\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in H.$

Moreover, ${P}_{\mathrm{\Omega }}$ is characterized by the following properties:

$〈x-{P}_{\mathrm{\Omega }}\left(x\right),y-{P}_{\mathrm{\Omega }}\left(x\right)〉\le 0$

for all $x\in H$ and $y\in \mathrm{\Omega }$.

We also recall that a mapping $f:\mathrm{\Omega }\to H$ is said to be ρ-contractive if $\parallel Tx-Ty\parallel \le \rho \parallel x-y\parallel$ for some constant $\rho \in \left[0,1\right)$ and for all $x,y\in \mathrm{\Omega }$. A mapping $T:\mathrm{\Omega }\to \mathrm{\Omega }$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in \mathrm{\Omega }$. A mapping T is called averaged if $T=\left(1-\delta \right)I+\delta U$, where $\delta \in \left(0,1\right)$ and $U:\mathrm{\Omega }\to \mathrm{\Omega }$ is nonexpansive. In this case, we also say that T is δ-averaged. A bounded linear operator B is said to be strongly positive on H if there exists a constant $\alpha >0$ such that

$〈Bx,x〉\ge \alpha {\parallel x\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in H.$

Let A be an operator with domain $D\left(A\right)$ and range $R\left(A\right)$ in H.

1. (i)

A is monotone if for all $x,y\in D\left(A\right)$,

$〈Ax-Ay,x-y〉\ge 0.$
2. (ii)

Given a number $\nu >0$, A is said to be ν-inverse strongly monotone (ν-ism) (or co-coercive) if

$〈Ax-Ay,x-y〉\ge \nu {\parallel Ax-Ay\parallel }^{2},\phantom{\rule{1em}{0ex}}x,y\in H.$

It is easily seen that a projection ${P}_{\mathrm{\Omega }}$ is a 1-ism and hence ${P}_{\mathrm{\Omega }}$ is $\frac{1}{2}$-averaged.

We will need to use the following notation:

• $Fix\left(T\right)$ stands for the set of fixed points of T;

• ${x}_{n}⇀x$ stands for the weak convergence of $\left\{{x}_{n}\right\}$ to x;

• ${x}_{n}\to x$ stands for the strong convergence of $\left\{{x}_{n}\right\}$ to x.

### 2.2 Mathematical model

Now, we consider the mathematical model of the multiple-sets split feasibility problem. Let $x\in {C}_{1}$. Assume that $Ax\in {Q}_{1}$. Then we get $\left(I-{P}_{{Q}_{1}}\right)Ax=0$, which implies $\gamma {A}^{\ast }\left(I-{P}_{{Q}_{1}}\right)Ax=0$, hence x satisfies the fixed point equation $x=\left(I-\gamma {A}^{\ast }\left(I-{P}_{{Q}_{1}}\right)A\right)x$. At the same time, note that $x\in {C}_{1}$. Thus,

$x={P}_{{C}_{1}}\left(I-\gamma {A}^{\ast }\left(I-{P}_{{Q}_{1}}\right)A\right)x.$

Now, we know x solves the split feasibility problem if and only if x solves the above fixed point equation. This result reminds us that the multiple-sets split feasibility problem is equivalent to a common fixed point problem of finitely many nonexpansive mappings. On the other hand, x solves the multiple-sets split feasibility problem implies that x satisfies two properties:

1. (i)

the distance from x to each ${C}_{i}$ is zero and

2. (ii)

the distance from Ax to each ${Q}_{j}$ is also zero.

First, we consider the following proximity function:

$g\left(x\right)=\frac{1}{2}\sum _{i=1}^{N}{\alpha }_{i}{\parallel x-{P}_{{C}_{i}}x\parallel }^{2}+\frac{1}{2}\sum _{j=1}^{M}{\beta }_{j}{\parallel Ax-{P}_{{Q}_{j}}Ax\parallel }^{2},$

where $\left\{{\alpha }_{i}\right\}$ and $\left\{{\beta }_{j}\right\}$ are positive real numbers, and ${P}_{{C}_{i}}$ and ${P}_{{Q}_{j}}$ are the metric projections onto ${C}_{i}$ and ${Q}_{j}$, respectively. It is clear that the proximity function g is convex and differentiable with the gradient

$\mathrm{\nabla }g\left(x\right)=\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)x+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)Ax.$

We can check that the gradient $\mathrm{\nabla }g\left(x\right)$ is L-Lipschitz continuous with constant

$L=\sum _{i=1}^{N}{\alpha }_{i}+\sum _{j=1}^{M}{\beta }_{j}{\parallel A\parallel }^{2}.$

Note that ${x}^{\ast }$ is a solution of the multiple-sets split feasibility problem (1.1) if and only if $g\left({x}^{\ast }\right)=0$. Since $g\left(x\right)\ge 0$ for all $x\in {H}_{1}$, a solution of the multiple-sets split feasibility problem (1.1) is a minimizer of g over any closed convex subset, with minimum value of zero. This motivates us to consider the following minimization problem:

$\underset{x\in \mathrm{\Omega }}{min}g\left(x\right),$
(2.1)

where Ω is a closed convex subset of ${H}_{1}$ whose intersection with the solution set of the multiple-sets split feasibility problem is nonempty, and get a solution of the so-called constrained multiple-sets split feasibility problem

(2.2)

### 2.3 The well-known lemmas

The following lemmas will be helpful for our main results in the next section.

Lemma 2.1 [25]

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

Lemma 2.2 [26]

Let K be a nonempty closed convex subset of a real Hilbert space H. Let $T:K\to K$ be a nonexpansive mapping with $Fix\left(T\right)\ne \mathrm{\varnothing }$. Then T is demiclosed on K, i.e., if ${x}_{n}⇀x\in K$ weakly and ${x}_{n}-T{x}_{n}\to 0$, then $x=Tx$.

Lemma 2.3 [27]

Assume that $\left\{{a}_{n}\right\}$ is a sequence of nonnegative real numbers such that ${a}_{n+1}\le \left(1-{\gamma }_{n}\right){a}_{n}+{\delta }_{n}$, where $\left\{{\gamma }_{n}\right\}$ is a sequence in $\left(0,1\right)$ and $\left\{{\delta }_{n}\right\}$ is a sequence such that

1. (1)

${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$;

2. (2)

${lim sup}_{n\to \mathrm{\infty }}{\delta }_{n}/{\gamma }_{n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_{n}|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_{n}=0$.

## 3 Main results

Let ${H}_{1}$ and ${H}_{2}$ be two real Hilbert spaces. Let ${C}_{1},{C}_{2},\dots ,{C}_{N}$ be N nonempty closed convex subsets of ${H}_{1}$ and let ${Q}_{1},{Q}_{2},\dots ,{Q}_{M}$ be M nonempty closed convex subsets of ${H}_{2}$. Let $A:{H}_{1}\to {H}_{2}$ be a bounded linear operator. Assume that the multiple-sets split feasibility problem is consistent, i.e., it is solvable. Now, we are devoted to solving the constrained multiple-set split feasibility problem (2.2).

For solving (2.2), we introduce the following iterative algorithm.

Algorithm 3.1 Let $f:{H}_{1}\to {H}_{1}$ be a ρ-contraction. Let $B:{H}_{1}\to {H}_{1}$ be a self-adjoint, strongly positive bounded linear operator with coefficient $\alpha >0$. Let σ and γ be two constants such that $0<\gamma <\frac{2}{L}$ and $0<\sigma \rho <\alpha$. For arbitrary initial point ${x}_{0}\in {H}_{1}$, we define a sequence $\left\{{x}_{n}\right\}$ iteratively by

$\begin{array}{rl}{x}_{n+1}=& {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right)\\ ×{P}_{\mathrm{\Omega }}\left({\xi }_{n}\sigma f+\left(I-{\xi }_{n}B\right)\right){x}_{n},\end{array}$
(3.1)

for all $n\ge 0$, where $\left\{{\xi }_{n}\right\}$ is a real sequence in $\left(0,1\right)$.

Fact 3.2 The mapping $I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)$ is $\frac{\gamma L}{2}$-averaged.

In order to check Fact 3.2, we need the following lemmas.

If $h:H\to R$ has an L-Lipschitz continuous gradient h, then h is $\frac{1}{L}$-ism.

Lemma 3.4 Given $T:H\to H$ and let $V=I-T$ be the complement of T. Given also $S:H\to H$.

1. (i)

T is nonexpansive if and only if V is $\frac{1}{2}$-inverse strongly monotone (in short, $\frac{1}{2}$-ism).

2. (ii)

If S is ν-ism, then for $\gamma >0$, γS is $\frac{\nu }{\gamma }$-ism.

3. (iii)

S is averaged if and only if the complement $I-S$ is ν-ism for some $\nu >\frac{1}{2}$.

Lemma 3.5 Given operators $S,T,V:H\to H$.

1. (i)

If $S=\left(1-\alpha \right)T+\alpha V$ for some $\alpha \in \left(0,1\right)$ and if T is averaged and V is nonexpansive, then S is averaged.

2. (ii)

S is firmly nonexpansive if and only if the complement $I-S$ is firmly nonexpansive. If S is firmly nonexpansive, then S is averaged.

3. (iii)

If $S=\left(1-\alpha \right)T+\alpha V$ for some $\alpha \in \left(0,1\right)$, T is firmly nonexpansive and V is nonexpansive, then S is averaged.

4. (iv)

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

Proof of Fact 3.2 Since gradient $\mathrm{\nabla }g\left(x\right)={\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)x+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)Ax$ has an L-Lipschitz constant $L={\sum }_{i=1}^{N}{\alpha }_{i}+{\sum }_{j=1}^{M}{\beta }_{j}{\parallel A\parallel }^{2}$, by Lemma 3.4, g is $\frac{1}{L}$-ism and $\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)$ is $\frac{1}{\gamma L}$-ism. Again, from Lemma 3.4(iii), we deduce that $I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)$ is $\frac{\gamma L}{2}$-averaged. □

Now, we prove the convergence of the sequence $\left\{{x}_{n}\right\}$.

Theorem 3.6 Suppose that $S\ne \mathrm{\varnothing }$. Assume that the sequence $\left\{{\xi }_{n}\right\}$ satisfies the control conditions:

1. (i)

${lim}_{n\to \mathrm{\infty }}{\xi }_{n}=0$ and

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by (3.1) converges to a solution ${x}^{\ast }$ of (2.2), where ${x}^{\ast }$ also solves the following VI:

${x}^{\ast }\in S\phantom{\rule{0.1em}{0ex}}\mathit{\text{such that}}\phantom{\rule{0.1em}{0ex}}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },\stackrel{˜}{x}-{x}^{\ast }〉\le 0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}\stackrel{˜}{x}\in S,$
(3.2)

where S is the set of solutions of (2.2).

Proof Let ${x}^{\ast }\in S$. Since B is strongly positive bounded linear operator with coefficient $\alpha >0$, we have $\parallel I-{\xi }_{n}B\parallel \le 1-\alpha {\xi }_{n}$ (without loss of generality, we may assume ${\xi }_{n}\le \frac{1}{\alpha }$). Thus, by (3.1), we have

$\begin{array}{r}\parallel {x}_{n+1}-{x}^{\ast }\parallel \\ \phantom{\rule{1em}{0ex}}=\parallel {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right)\\ \phantom{\rule{2em}{0ex}}×{P}_{\mathrm{\Omega }}\left({\xi }_{n}\sigma f+\left(I-{\xi }_{n}B\right)\right){x}_{n}-{x}^{\ast }\parallel \\ \phantom{\rule{1em}{0ex}}\le \parallel {\xi }_{n}\sigma f\left({x}_{n}\right)+\left(I-{\xi }_{n}B\right){x}_{n}-{x}^{\ast }\parallel \\ \phantom{\rule{1em}{0ex}}\le {\xi }_{n}\sigma \parallel f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\parallel +\parallel I-{\xi }_{n}B\parallel \parallel {x}_{n}-{x}^{\ast }\parallel +{\xi }_{n}\parallel \sigma f\left({x}^{\ast }\right)-B{x}^{\ast }\parallel \\ \phantom{\rule{1em}{0ex}}\le {\xi }_{n}\sigma \rho \parallel {x}_{n}-{x}^{\ast }\parallel +\left(1-{\xi }_{n}\alpha \right)\parallel {x}_{n}-{x}^{\ast }\parallel +{\xi }_{n}\parallel \sigma f\left({x}^{\ast }\right)-B{x}^{\ast }\parallel \\ \phantom{\rule{1em}{0ex}}=\left[1-\left(\alpha -\sigma \rho \right){\xi }_{n}\right]\parallel {x}_{n}-{x}^{\ast }\parallel +\left(\alpha -\sigma \rho \right){\xi }_{n}\parallel f\left({x}^{\ast }\right)-B{x}^{\ast }\parallel /\left(\alpha -\sigma \rho \right).\end{array}$

An induction yields

$\begin{array}{rcl}\parallel {x}_{n+1}-{x}^{\ast }\parallel & \le & max\left\{\parallel {x}_{n}-{x}^{\ast }\parallel ,\frac{\parallel f\left({x}^{\ast }\right)-B{x}^{\ast }\parallel }{\alpha -\sigma \rho }\right\}\\ \le & max\left\{\parallel {x}_{0}-{x}^{\ast }\parallel ,\frac{\parallel f\left({x}^{\ast }\right)-B{x}^{\ast }\parallel }{\alpha -\sigma \rho }\right\}.\end{array}$

Hence, $\left\{{x}_{n}\right\}$ is bounded.

It is well-known that the metric projection ${P}_{\mathrm{\Omega }}$ is firmly nonexpansive, hence averaged. By Fact 3.2, $I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)$ is $\frac{\gamma L}{2}$-averaged. From Lemma 3.5, the composite of three averaged mappings is averaged. So, ${P}_{\mathrm{\Omega }}\left(I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){P}_{\mathrm{\Omega }}$ is an averaged mapping. Thus, there must exist a positive constant $\delta \in \left(0,1\right)$ such that

${P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){P}_{\mathrm{\Omega }}=\left(1-\delta \right)I+\delta U,$

where U is a nonexpansive mapping. Set ${y}_{n}={\xi }_{n}\sigma f\left({x}_{n}\right)+\left(I-{\xi }_{n}B\right){x}_{n}$ for all $n\ge 0$. Then we have

$\begin{array}{rcl}{x}_{n+1}& =& \left(\left(1-\delta \right)I+\delta U\right)\left({\xi }_{n}\sigma f\left({x}_{n}\right)+\left(I-{\xi }_{n}B\right){x}_{n}\right)\\ =& \left(1-\delta \right){x}_{n}+{\xi }_{n}\left(1-\delta \right)\left(\sigma f\left({x}_{n}\right)-B{x}_{n}\right)+\delta U{y}_{n}\\ =& \left(1-\delta \right){x}_{n}+\delta \left(\frac{1-\delta }{\delta }{\xi }_{n}\left(\sigma f\left({x}_{n}\right)-B{x}_{n}\right)+U{y}_{n}\right)\\ =& \left(1-\delta \right){x}_{n}+\delta {z}_{n},\end{array}$

where

${z}_{n}=\frac{\left(1-\delta \right){\xi }_{n}}{\delta }\left(\sigma f\left({x}_{n}\right)-B{x}_{n}\right)+U{y}_{n}.$

By virtue of ${\xi }_{n}\to 0$ (as $n\to \mathrm{\infty }$) and the boundedness of the sequences $\left\{f\left({x}_{n}\right)\right\}$ and $\left\{B{x}_{n}\right\}$, we firstly observe that

$\underset{n\to \mathrm{\infty }}{lim}\parallel {y}_{n}-{x}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel =0,$

and

$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-U{y}_{n}\parallel =\underset{n\to \mathrm{\infty }}{lim}\frac{\left(1-\delta \right){\xi }_{n}}{\delta }\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel =0.$

Next, we estimate $\parallel {z}_{n+1}-{z}_{n}\parallel$. Note that

${z}_{n+1}-{z}_{n}=\frac{\left(1-\delta \right){\xi }_{n+1}}{\delta }\left(\sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\right)+U{y}_{n+1}-\frac{\left(1-\delta \right){\xi }_{n}}{\delta }\left(\sigma f\left({x}_{n}\right)-B{x}_{n}\right)-U{y}_{n}.$

It follows that

$\begin{array}{rcl}\parallel {z}_{n+1}-{z}_{n}\parallel & \le & \frac{1-\delta }{\delta }\left({\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \right)+\parallel U{y}_{n+1}-U{y}_{n}\parallel \\ \le & \frac{1-\delta }{\delta }\left({\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \right)+\parallel {y}_{n+1}-{y}_{n}\parallel .\end{array}$

Since ${y}_{n+1}-{y}_{n}={\xi }_{n+1}\sigma f\left({x}_{n+1}\right)+\left(I-{\xi }_{n+1}B\right){x}_{n+1}-{\xi }_{n}\sigma f\left({x}_{n}\right)-\left(I-{\xi }_{n}B\right){x}_{n}$, we get

$\begin{array}{rcl}\parallel {z}_{n+1}-{z}_{n}\parallel & \le & \parallel {\xi }_{n+1}\sigma f\left({x}_{n+1}\right)+\left(I-{\xi }_{n+1}B\right){x}_{n+1}-{\xi }_{n}\sigma f\left({x}_{n}\right)-\left(I-{\xi }_{n}B\right){x}_{n}\parallel \\ +\frac{1-\delta }{\delta }\left({\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \right)\\ \le & \parallel {x}_{n+1}-{x}_{n}\parallel +{\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \\ +\frac{1-\delta }{\delta }\left({\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \right).\end{array}$

It follows that

$\begin{array}{rcl}\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel & \le & {\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \\ +\frac{1-\delta }{\delta }\left({\xi }_{n+1}\parallel \sigma f\left({x}_{n+1}\right)-B{x}_{n+1}\parallel +{\xi }_{n}\parallel \sigma f\left({x}_{n}\right)-B{x}_{n}\parallel \right).\end{array}$

Since ${lim}_{n\to \mathrm{\infty }}{\xi }_{n}=0$ and the sequences $\left\{f\left({x}_{n}\right)\right\}$, $\left\{B{x}_{n}\right\}$ are bounded, we deduce

$\underset{n\to \mathrm{\infty }}{lim sup}\left(\parallel {z}_{n+1}-{z}_{n}\parallel -\parallel {x}_{n+1}-{x}_{n}\parallel \right)\le 0.$

By Lemma 2.1, we get

$\underset{n\to \mathrm{\infty }}{lim}\parallel {z}_{n}-{x}_{n}\parallel =0.$

Therefore,

$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel U{x}_{n}-{x}_{n}\parallel \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}\parallel {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){P}_{\mathrm{\Omega }}\left({x}_{n}\right)-{x}_{n}\parallel =0.\end{array}$

By the definition of the sequence $\left\{{x}_{n}\right\}$, we know that ${x}_{n}\in \mathrm{\Omega }$. Hence, ${P}_{\mathrm{\Omega }}\left({x}_{n}\right)={x}_{n}$. So,

$\underset{n\to \mathrm{\infty }}{lim}\parallel {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){x}_{n}-{x}_{n}\parallel =0.$

Next we prove

$\underset{n\to \mathrm{\infty }}{lim sup}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\le 0.$

In order to get this inequality, we need to prove the following:

$\underset{n\to \mathrm{\infty }}{lim sup}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{x}_{n}-{x}^{\ast }〉\le 0,$

where ${x}^{\ast }$ is the unique solution of VI(3.2). For this purpose, we choose a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that

$\underset{n\to \mathrm{\infty }}{lim sup}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{x}_{n}-{x}^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{x}_{{n}_{i}}-{x}^{\ast }〉.$

Since $\left\{{x}_{{n}_{i}}\right\}$ is bounded, there exists a subsequence of $\left\{{x}_{{n}_{i}}\right\}$ which converges weakly to a point $\stackrel{˜}{x}$. Without loss of generality, we may assume that $\left\{{x}_{{n}_{i}}\right\}$ converges weakly to $\stackrel{˜}{x}$. Since ${P}_{\mathrm{\Omega }}\left(I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right)$ is nonexpansive, by Lemma 2.2, we have ${x}_{{n}_{i}}⇀\stackrel{˜}{x}\in Fix\left({P}_{\mathrm{\Omega }}\left(I-\gamma \left({\sum }_{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+{\sum }_{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right)\right)$. Therefore,

$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim sup}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{x}_{n}-{x}^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{x}_{{n}_{i}}-{x}^{\ast }〉\\ =& 〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },\stackrel{˜}{x}-{x}^{\ast }〉\le 0.\end{array}$

Since $\parallel {x}_{n}-{P}_{\mathrm{\Omega }}\left({y}_{n}\right)\parallel =\parallel {P}_{\mathrm{\Omega }}\left({x}_{n}\right)-{P}_{\mathrm{\Omega }}\left({y}_{n}\right)\parallel \le \parallel {x}_{n}-{y}_{n}\parallel \to 0$, we obtain

$\underset{n\to \mathrm{\infty }}{lim sup}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\le 0.$

Note that

${\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}=〈{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{y}_{n},{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉+〈{y}_{n}-{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉.$

From the property of the metric ${P}_{\mathrm{\Omega }}$, we have $〈{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{y}_{n},{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\le 0$. Hence,

$\begin{array}{rcl}{\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}& \le & 〈{y}_{n}-{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ =& 〈{\xi }_{n}\sigma \left(f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\right)+\left(I-{\xi }_{n}B\right)\left({x}_{n}-{x}^{\ast }\right),{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ +{\xi }_{n}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ \le & \left({\xi }_{n}\sigma \parallel f\left({x}_{n}\right)-f\left({x}^{\ast }\right)\parallel +\parallel I-{\xi }_{n}B\parallel \parallel {x}_{n}-{x}^{\ast }\parallel \right)\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel \\ +{\xi }_{n}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ \le & \left(1-{\xi }_{n}\left(\alpha -\sigma \rho \right)\right)\parallel {x}_{n}-{x}^{\ast }\parallel \parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel \\ +{\xi }_{n}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ \le & \frac{1-{\xi }_{n}\left(\alpha -\sigma \rho \right)}{2}{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+\frac{1}{2}{\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ +{\xi }_{n}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉.\end{array}$

It follows that

${\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}\le \left[1-\left(\alpha -\sigma \rho \right){\xi }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+2{\xi }_{n}〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉.$

Finally, we show that ${x}_{n}\to {x}^{\ast }$. From (3.1), we have

$\begin{array}{rcl}{\parallel {x}_{n+1}-{x}^{\ast }\parallel }^{2}& =& {\parallel {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \le & {\parallel {P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }\parallel }^{2}\\ \le & \left[1-\left(\alpha -\sigma \rho \right){\xi }_{n}\right]{\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}\\ +\left(\alpha -\sigma \rho \right){\xi }_{n}\frac{2}{\alpha -\sigma \rho }〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\\ =& \left(1-{\gamma }_{n}\right){\parallel {x}_{n}-{x}^{\ast }\parallel }^{2}+{\delta }_{n},\end{array}$

where ${\gamma }_{n}=\left(\alpha -\sigma \rho \right){\xi }_{n}$ and ${\delta }_{n}=\left(\alpha -\sigma \rho \right){\xi }_{n}\frac{2}{\alpha -\sigma \rho }〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉$. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_{n}=\mathrm{\infty }$ and ${lim sup}_{n\to \mathrm{\infty }}\frac{{\delta }_{n}}{{\gamma }_{n}}={lim sup}_{n\to \mathrm{\infty }}\frac{2}{\alpha -\sigma \rho }〈\sigma f\left({x}^{\ast }\right)-B{x}^{\ast },{P}_{\mathrm{\Omega }}\left({y}_{n}\right)-{x}^{\ast }〉\le 0$, all conditions of Lemma 2.3 are satisfied. Therefore, we immediately deduce that ${x}_{n}\to {x}^{\ast }$. This completes the proof. □

From (3.1) and Theorem 3.6, we can deduce easily the following results.

Algorithm 3.7 For an arbitrary initial point ${x}_{0}\in {H}_{1}$, we define a sequence $\left\{{x}_{n}\right\}$ iteratively by

$\begin{array}{rl}{x}_{n+1}=& {P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right)\\ ×{P}_{\mathrm{\Omega }}\left({\xi }_{n}\sigma f\left({x}_{n}\right)+\left(1-{\xi }_{n}\right){x}_{n}\right),\end{array}$
(3.3)

for all $n\ge 0$, where $\left\{{\xi }_{n}\right\}$ is a real sequence in $\left(0,1\right)$.

Corollary 3.8 Suppose that $S\ne \mathrm{\varnothing }$. Assume that the sequence $\left\{{\xi }_{n}\right\}$ satisfies the conditions

1. (i)

${lim}_{n\to \mathrm{\infty }}{\xi }_{n}=0$ and

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by (3.3) converges to a point ${x}^{\ast }$, which solves the following variational inequality:

${x}^{\ast }\in S\phantom{\rule{0.1em}{0ex}}\mathit{\text{such that}}\phantom{\rule{0.1em}{0ex}}〈\sigma f\left({x}^{\ast }\right)-{x}^{\ast },\stackrel{˜}{x}-{x}^{\ast }〉\le 0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}\stackrel{˜}{x}\in S.$

Algorithm 3.9 For an arbitrary initial point ${x}_{0}$, we define a sequence $\left\{{x}_{n}\right\}$ iteratively by

${x}_{n+1}={P}_{\mathrm{\Omega }}\left(I-\gamma \left(\sum _{i=1}^{N}{\alpha }_{i}\left(I-{P}_{{C}_{i}}\right)+\sum _{j=1}^{M}{\beta }_{j}{A}^{\ast }\left(I-{P}_{{Q}_{j}}\right)A\right)\right){P}_{\mathrm{\Omega }}\left(\left(1-{\xi }_{n}\right){x}_{n}\right),$
(3.4)

for all $n\ge 0$, where $\left\{{\xi }_{n}\right\}$ is a real sequence in $\left(0,1\right)$.

Corollary 3.10 Suppose that $S\ne \mathrm{\varnothing }$. Assume that the sequence $\left\{{\xi }_{n}\right\}$ satisfies the conditions

1. (i)

${lim}_{n\to \mathrm{\infty }}{\xi }_{n}=0$ and

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_{n}=\mathrm{\infty }$.

Then the sequence $\left\{{x}_{n}\right\}$ generated by (3.4) converges to a point ${x}^{\ast }\in S$ which is the minimum norm element in S.

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## Acknowledgements

Jinwei Shi was supported in part by the scientific research fund of the Educational Commission of Hebei Province of China (No. 936101101) and the National Natural Science Foundation of China (No. 51077053). Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.

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Correspondence to Jinwei Shi.

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The authors declare that they have no competing interests.

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Zheng, Y., Shi, J. & Liou, YC. The constrained multiple-sets split feasibility problem and its projection algorithms. J Inequal Appl 2013, 272 (2013). https://doi.org/10.1186/1029-242X-2013-272

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• DOI: https://doi.org/10.1186/1029-242X-2013-272

### Keywords

• Nonexpansive Mapping
• Real Hilbert Space
• Common Fixed Point
• Nonempty Closed Convex Subset
• Projection Algorithm