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:

$$\text{Find an }x\in \bigcap _{i=1}^N{C}_i\text{ such that }Ax\in \bigcap _{j=1}^M{Q}_j.$$

(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

$$x\in C\text{and}Ax\in Q.$$

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 [3–6]. Various algorithms have been invented to solve the multiple-sets split feasibility problem and the split feasibility problem, see, e.g., [7–24] 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 $x\in H$, $P_{\mathrm{\Omega }}(x)$ is the unique point in Ω with the property

$$\parallel x-P_{\mathrm{\Omega }}(x)\parallel =min\{\parallel x-y\parallel :y\in \mathrm{\Omega }\}.$$

It is known that $P_{\mathrm{\Omega }}$ satisfies

$$〈x-y,P_{\mathrm{\Omega }}(x)-P_{\mathrm{\Omega }}(y)〉\ge {\parallel P_{\mathrm{\Omega }}(x)-P_{\mathrm{\Omega }}(y)\parallel }^2,\mathrm{\forall }x,y\in H.$$

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

$$〈x-P_{\mathrm{\Omega }}(x),y-P_{\mathrm{\Omega }}(x)〉\le 0$$

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 $\rho \in [0,1)$ 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 $T=(1-\delta )I+\delta U$, where $\delta \in (0,1)$ 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

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

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

• (i) A is monotone if for all $x,y\in D(A)$,

  $$〈Ax-Ay,x-y〉\ge 0.$$

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

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

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:

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

• $x_n\rightharpoonup x$ stands for the weak convergence of $\{x_n\}$ to x;

• $x_n\to x$ stands for the strong convergence of $\{x_n\}$ 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 $(I-P_{Q_1})Ax=0$, which implies $\gamma A^{\ast }(I-P_{Q_1})Ax=0$, hence x satisfies the fixed point equation $x=(I-\gamma A^{\ast }(I-P_{Q_1})A)x$. At the same time, note that $x\in C_1$. Thus,

$$x=P_{C_1}(I-\gamma A^{\ast }(I-P_{Q_1})A)x.$$

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:

• (i) the distance from x to each $C_i$ is zero and

• (ii) the distance from Ax to each $Q_j$ is also zero.

First, we consider the following proximity function:

$$g(x)=\frac{1}{2}\sum _{i=1}^N{\alpha }_i{\parallel x-P_{C_i}x\parallel }^2+\frac{1}{2}\sum _{j=1}^M{\beta }_j{\parallel Ax-P_{Q_j}Ax\parallel }^2,$$

where $\{{\alpha }_i\}$ and $\{{\beta }_j\}$ 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

$$\mathrm{\nabla }g(x)=\sum _{i=1}^N{\alpha }_i(I-P_{C_i})x+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})Ax.$$

We can check that the gradient $\mathrm{\nabla }g(x)$ 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 $g(x^{\ast })=0$. Since $g(x)\ge 0$ 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:

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

(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

$$x^{\ast }\in \mathrm{\Omega }\text{ such that }{x}^{\ast }\text{ solves (1.1)}.$$

(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 $\{x_n\}$ and $\{z_n\}$ be bounded sequences in a Banach space X and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}=(1-{\beta }_n)z_n+{\beta }_n{x}_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. 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 $Fix(T)\ne \mathrm{\varnothing }$. Then T is demiclosed on K, i.e., if $x_n\rightharpoonup x\in K$ weakly and $x_n-T{x}_n\to 0$, then $x=Tx$.

Lemma 2.3 [27]

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

• (1) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n/{\gamma }_n\le 0$ 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 $\{x_n\}$ iteratively by

$$\begin{array}{rl}{x}_{n+1}=& P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))\\ \times P_{\mathrm{\Omega }}({\xi }_n\sigma f+(I-{\xi }_{n}B))x_n,\end{array}$$

(3.1)

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

Fact 3.2 The mapping $I-\gamma ({\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A)$ is $\frac{\gamma L}{2}$-averaged.

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

Lemma 3.3 (Baillon-Haddad) [28]

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 $S:H\to H$.

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

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

• (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 $S,T,V:H\to H$.

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

• (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.

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

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

Proof of Fact 3.2 Since gradient $\mathrm{\nabla }g(x)={\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})x+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})Ax$ 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 $\gamma ({\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A)$ is $\frac{1}{\gamma L}$-ism. Again, from Lemma 3.4(iii), we deduce that $I-\gamma ({\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A)$ is $\frac{\gamma L}{2}$-averaged. □

Now, we prove the convergence of the sequence $\{x_n\}$.

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

• (i) ${lim}_{n\to \mathrm{\infty }}{\xi }_n=0$ and

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$.

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

$$x^{\ast }\in S\mathit{\text{such that}}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },\tilde{x}-x^{\ast }〉\le 0\mathit{\text{for all}}\tilde{x}\in S,$$

(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

$$\begin{array}{r}\parallel x_{n+1}-x^{\ast }\parallel \\ =\parallel P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))\\ \times P_{\mathrm{\Omega }}({\xi }_n\sigma f+(I-{\xi }_{n}B))x_n-x^{\ast }\parallel \\ \le \parallel {\xi }_n\sigma f(x_n)+(I-{\xi }_{n}B)x_n-x^{\ast }\parallel \\ \le {\xi }_n\sigma \parallel f(x_n)-f\left(x^{\ast }\right)\parallel +\parallel I-{\xi }_{n}B\parallel \parallel x_n-x^{\ast }\parallel +{\xi }_n\parallel \sigma f\left(x^{\ast }\right)-B{x}^{\ast }\parallel \\ \le {\xi }_n\sigma \rho \parallel x_n-x^{\ast }\parallel +(1-{\xi }_n\alpha )\parallel x_n-x^{\ast }\parallel +{\xi }_n\parallel \sigma f\left(x^{\ast }\right)-B{x}^{\ast }\parallel \\ =[1-(\alpha -\sigma \rho ){\xi }_n]\parallel x_n-x^{\ast }\parallel +(\alpha -\sigma \rho ){\xi }_n\parallel f\left(x^{\ast }\right)-B{x}^{\ast }\parallel /(\alpha -\sigma \rho ).\end{array}$$

An induction yields

$$\begin{array}{rcl}\parallel x_{n+1}-x^{\ast }\parallel & \le & max\{\parallel x_n-x^{\ast }\parallel ,\frac{\parallel f(x^{\ast })-B{x}^{\ast }\parallel }{\alpha -\sigma \rho }\}\\ \le & max\{\parallel x_0-x^{\ast }\parallel ,\frac{\parallel f(x^{\ast })-B{x}^{\ast }\parallel }{\alpha -\sigma \rho }\}.\end{array}$$

Hence, $\{x_n\}$ is bounded.

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

$$P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))P_{\mathrm{\Omega }}=(1-\delta )I+\delta U,$$

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

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

where

$$z_n=\frac{(1-\delta ){\xi }_n}{\delta }(\sigma f(x_n)-B{x}_n)+U{y}_n.$$

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

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel =0,$$

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel z_n-U{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\frac{(1-\delta ){\xi }_n}{\delta }\parallel \sigma f(x_n)-B{x}_n\parallel =0.$$

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

$$z_{n+1}-z_n=\frac{(1-\delta ){\xi }_{n+1}}{\delta }(\sigma f(x_{n+1})-B{x}_{n+1})+U{y}_{n+1}-\frac{(1-\delta ){\xi }_n}{\delta }(\sigma f(x_n)-B{x}_n)-U{y}_n.$$

It follows that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \frac{1-\delta }{\delta }({\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel )+\parallel U{y}_{n+1}-U{y}_n\parallel \\ \le & \frac{1-\delta }{\delta }({\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel )+\parallel y_{n+1}-y_n\parallel .\end{array}$$

Since $y_{n+1}-y_n={\xi }_{n+1}\sigma f(x_{n+1})+(I-{\xi }_{n+1}B)x_{n+1}-{\xi }_n\sigma f(x_n)-(I-{\xi }_{n}B)x_n$, we get

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel & \le & \parallel {\xi }_{n+1}\sigma f(x_{n+1})+(I-{\xi }_{n+1}B)x_{n+1}-{\xi }_n\sigma f(x_n)-(I-{\xi }_{n}B)x_n\parallel \\ +\frac{1-\delta }{\delta }({\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel )\\ \le & \parallel x_{n+1}-x_n\parallel +{\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel \\ +\frac{1-\delta }{\delta }({\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel ).\end{array}$$

It follows that

$$\begin{array}{rcl}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel & \le & {\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel \\ +\frac{1-\delta }{\delta }({\xi }_{n+1}\parallel \sigma f(x_{n+1})-B{x}_{n+1}\parallel +{\xi }_n\parallel \sigma f(x_n)-B{x}_n\parallel ).\end{array}$$

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

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

By Lemma 2.1, we get

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

Therefore,

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim}\parallel U{x}_n-x_n\parallel \\ =\underset{n\to \mathrm{\infty }}{lim}\parallel P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))P_{\mathrm{\Omega }}(x_n)-x_n\parallel =0.\end{array}$$

By the definition of the sequence $\{x_n\}$, we know that $x_n\in \mathrm{\Omega }$. Hence, $P_{\mathrm{\Omega }}(x_n)=x_n$. So,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))x_n-x_n\parallel =0.$$

Next we prove

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉\le 0.$$

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

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },x_n-x^{\ast }〉\le 0,$$

where $x^{\ast }$ is the unique solution of VI(3.2). For this purpose, we choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },x_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },x_{n_i}-x^{\ast }〉.$$

Since $\{x_{n_i}\}$ is bounded, there exists a subsequence of $\{x_{n_i}\}$ which converges weakly to a point $\tilde{x}$. Without loss of generality, we may assume that $\{x_{n_i}\}$ converges weakly to $\tilde{x}$. Since $P_{\mathrm{\Omega }}(I-\gamma ({\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))$ is nonexpansive, by Lemma 2.2, we have $x_{n_i}\rightharpoonup \tilde{x}\in Fix(P_{\mathrm{\Omega }}(I-\gamma ({\sum }_{i=1}^N{\alpha }_i(I-P_{C_i})+{\sum }_{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A)))$. Therefore,

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },x_n-x^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },x_{n_i}-x^{\ast }〉\\ =& 〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },\tilde{x}-x^{\ast }〉\le 0.\end{array}$$

Since $\parallel x_n-P_{\mathrm{\Omega }}(y_n)\parallel =\parallel P_{\mathrm{\Omega }}(x_n)-P_{\mathrm{\Omega }}(y_n)\parallel \le \parallel x_n-y_n\parallel \to 0$, we obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉\le 0.$$

Note that

$${\parallel P_{\mathrm{\Omega }}(y_n)-x^{\ast }\parallel }^2=〈P_{\mathrm{\Omega }}(y_n)-y_n,P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉+〈y_n-x^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉.$$

From the property of the metric $P_{\mathrm{\Omega }}$, we have $〈P_{\mathrm{\Omega }}(y_n)-y_n,P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉\le 0$. Hence,

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

It follows that

$${\parallel P_{\mathrm{\Omega }}(y_n)-x^{\ast }\parallel }^2\le [1-(\alpha -\sigma \rho ){\xi }_n]{\parallel x_n-x^{\ast }\parallel }^2+2{\xi }_n〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉.$$

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

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& {\parallel P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))P_{\mathrm{\Omega }}(y_n)-x^{\ast }\parallel }^2\\ \le & {\parallel P_{\mathrm{\Omega }}(y_n)-x^{\ast }\parallel }^2\\ \le & [1-(\alpha -\sigma \rho ){\xi }_n]{\parallel x_n-x^{\ast }\parallel }^2\\ +(\alpha -\sigma \rho ){\xi }_n\frac{2}{\alpha -\sigma \rho }〈\sigma f\left(x^{\ast }\right)-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉\\ =& (1-{\gamma }_n){\parallel x_n-x^{\ast }\parallel }^2+{\delta }_n,\end{array}$$

where ${\gamma }_n=(\alpha -\sigma \rho ){\xi }_n$ and ${\delta }_n=(\alpha -\sigma \rho ){\xi }_n\frac{2}{\alpha -\sigma \rho }〈\sigma f(x^{\ast })-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉$. Since ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}={lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{2}{\alpha -\sigma \rho }〈\sigma f(x^{\ast })-B{x}^{\ast },P_{\mathrm{\Omega }}(y_n)-x^{\ast }〉\le 0$, 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 $\{x_n\}$ iteratively by

$$\begin{array}{rl}{x}_{n+1}=& P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))\\ \times P_{\mathrm{\Omega }}({\xi }_n\sigma f(x_n)+(1-{\xi }_n)x_n),\end{array}$$

(3.3)

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

Corollary 3.8 Suppose that $S\ne \mathrm{\varnothing }$. Assume that the sequence $\{{\xi }_n\}$ satisfies the conditions

• (i) ${lim}_{n\to \mathrm{\infty }}{\xi }_n=0$ and

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$.

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

$$x^{\ast }\in S\mathit{\text{such that}}〈\sigma f\left(x^{\ast }\right)-x^{\ast },\tilde{x}-x^{\ast }〉\le 0\mathit{\text{for all}}\tilde{x}\in S.$$

Algorithm 3.9 For an arbitrary initial point $x_0$, we define a sequence $\{x_n\}$ iteratively by

$$x_{n+1}=P_{\mathrm{\Omega }}(I-\gamma (\sum _{i=1}^N{\alpha }_i(I-P_{C_i})+\sum _{j=1}^M{\beta }_j{A}^{\ast }(I-P_{Q_j})A))P_{\mathrm{\Omega }}((1-{\xi }_n)x_n),$$

(3.4)

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

Corollary 3.10 Suppose that $S\ne \mathrm{\varnothing }$. Assume that the sequence $\{{\xi }_n\}$ satisfies the conditions

• (i) ${lim}_{n\to \mathrm{\infty }}{\xi }_n=0$ and

• (ii) ${\sum }_{n=0}^{\mathrm{\infty }}{\xi }_n=\mathrm{\infty }$.

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