Multiple-set split feasibility problems for κ-asymptotically strictly pseudo-nonspreading mappings in Hilbert spaces

Abstract

Some weak and strong convergence theorems for solving multiple-set split feasibility problems for κ-asymptotically strictly pseudo-nonspreading mappings in infinite-dimensional Hilbert spaces are proved. The results presented in the paper extend and improve the corresponding results of Xu (Inverse Probl. 22(6):2021-2034, 2006), Osilike and Isiogugu (Nonlinear Anal. 74:1814-1822, 2011), Chang et al. (Abstr. Appl. Anal. 2012:491760, 2012), and others.

MSC:47H05, 47H09, 49M05.

1 Introduction

Throughout this article, we always assume that $H_1$, $H_2$ are real Hilbert spaces; ‘→’ and ‘⇀’ denote strong and weak convergence, respectively.

The split feasibility problem ($SFP$) in finite dimensional spaces was first introduced by Censor and Elfving [1] for modeling inverse problems. The ($SFP$) can be used in various disciplines such as medical image reconstruction [2], image restoration, computer tomography, and radiation therapy treatment planning [3–5]. The multiple-set split feasibility problem ($MSSFP$) was studied in [4–7].

Let $A:H_1\to H_2$ be a bounded linear operator, $S_i:H_1\to H_1$ and $T_i:H_2\to H_2$, $i=1,2,\dots ,N$, be two finite families of mappings, $C:={\bigcap }_{i=1}^{N}F(S_i)$ and $Q:={\bigcap }_{i=1}^{N}F(T_i)$, where $F(S_i)$ and $F(T_i)$ are the sets of fixed points of $S_i$ and $T_i$, respectively.

The so-called multiple set split feasibility problem is

$$\text{to find }{x}^{\ast }\in C\text{ such that }A{x}^{\ast }\in Q.$$

(1.1)

In the sequel, we use Γ to denote the set of solutions of the problem ($MSSFP$) (1.1), that is,

$$\mathrm{\Gamma }=\{x\in C:Ax\in Q\}.$$

(1.2)

Let H be a real Hilbert space and K be a nonempty closed convex subset of H. Following Kohsaka and Takahashi [8–11], a mapping $T:K\to K$ is said to be nonspreading if

$$2{\parallel Tx-Ty\parallel }^2\le {\parallel Tx-y\parallel }^2+{\parallel Ty-x\parallel }^2\text{for all }x,y\in K.$$

It is to see that the above inequality is equivalent to

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+2〈x-Tx,y-Ty〉\text{for all }x,y\in K.$$

In 1967, Browder and Petryshyn [12] introduced the concept of κ-strictly pseudo-nonspreading mapping.

Definition 1.1 [12]

Let H be a real Hilbert space. A mapping $T:D(T)\subset H\to H$ is said to be κ-strictly pseudo-nonspreading if there exists $\kappa \in [0,1)$ such that

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+\kappa {\parallel x-Tx-(y-Ty)\parallel }^2+2〈x-Tx,y-Ty〉,\mathrm{\forall }x,y\in D(T).$$

Clearly, every nonspreading mapping is κ-strictly pseudo-nonspreading.

The class of asymptotically strict pseudo-contractions was introduced by Qihou [13] in 1996. Kim and Xu [14], Inchan and Nammanee [15], Zhou [16] Cho [17], and Ge [18] proved that the class of asymptotically strict pseudo-contractions is demiclosed at the origin and also obtained some weak convergence theorems for the class of mappings. In 2012, Osilike and Isiogugu [19] introduced a class of nonspreading type mappings which is more general than the class studied in [11] in Hilbert spaces and proved some weak and strong convergence theorems in real Hilbert spaces. Recently, Chang et al. [7] studied the multiple-set split feasibility problem for an asymptotically strict pseudo-contraction in the framework of infinite-dimensional Hilbert spaces.

Definition 1.2 [7]

Let H be a real Hilbert space, we say that the mapping $T:D(T)\subset H\to H$ is a κ-asymptotically strict pseudo-contraction if there exists a constant $\kappa \in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ ($n\to \mathrm{\infty }$) such that

$${\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+\kappa {\parallel x-T^{n}x-(y-T^{n}y)\parallel }^2$$

holds for all $x,y\in D(T)$.

In this article we introduce the following class of κ-asymptotically strictly pseudo-nonspreading mappings which is more general than that of κ-strictly pseudo-nonspreading mappings and κ-asymptotically strict pseudo-contractions.

Definition 1.3 Let H be a real Hilbert space. A mapping $T:D(T)\subset H\to H$ is said to be κ-asymptotically strictly pseudo-nonspreading if there exists a constant $\kappa \in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ ($n\to \mathrm{\infty }$) such that

$$\begin{array}{r}{\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+\kappa {\parallel x-T^{n}x-(y-T^{n}y)\parallel }^2+2〈x-T^{n}x,y-T^{n}y〉,\\ \mathrm{\forall }x,y\in D(T).\end{array}$$

(1.3)

Example 1.4 Now, we give an example of κ-asymptotically strict pseudo-contractive mapping.

Let C be a unit ball in a real Hilbert $l^2$, and let $T:C\to C$ be a mapping defined by

$$T:(x_1,x_2,\dots )\to (0,x_1^2,a_2{x}_2,a_3{x}_3,\dots ),$$

(1.4)

where $\{a_i\}$ is a sequence in $(0,1)$ such that ${\prod }_{i=2}^{\mathrm{\infty }}{\alpha }_i=\frac{1}{2}$.

It is proved in Goebel and Kirk [20] that

• (i) $\parallel Tx-Ty\parallel \le 2\parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$;

• (ii) $\parallel T^{n}x-T^{n}y\parallel \le 2{\prod }_{i=2}^n{a}_j\parallel x-y\parallel$, $\mathrm{\forall }n\ge 2$ and $x,y\in C$.

Define $k_1^{\frac{1}{2}}=2$, $k_n^{\frac{1}{2}}=2{\prod }_{i=2}^n{a}_j$, $n\ge 2$, then

$$\underset{n\to \mathrm{\infty }}{lim}{k}_n=\underset{n\to \mathrm{\infty }}{lim}{\left(2\prod _{i=2}^n{a}_j\right)}^2=1.$$

Letting $\kappa =0$, then $\mathrm{\forall }x,y\in C$, $n\ge 1$, we have

$$\underline{{\parallel T^{n}x-T^{n}y\parallel }^2\le k_n{\parallel x-y\parallel }^2+\kappa {\parallel x-y-(T^{n}x-T^{n}y)\parallel }^2}.$$

This implies that T is a κ-asymptotically strict pseudo-contractive mapping.

Example 1.5 Now, we give an example of κ-asymptotically strictly pseudo-nonspreading mapping.

Let $X=l^2$ with the norm $\parallel \cdot \parallel$ be defined by

$$\parallel x\parallel =\sqrt{\sum _{i=1}^{\mathrm{\infty }}{x}_i^2},\mathrm{\forall }x=(x_1,x_2,\dots ,x_n,\dots )\in X,$$

and let $C=\{x=(x_1,x_2,\dots ,x_n,\dots )|x_i\in R^1,i=1,2,\dots \}$ be an orthogonal subspace of X (i.e., $\mathrm{\forall }x,y\in C$, we have $〈x,y〉=0$). It is obvious that C is a nonempty closed convex subset of X. For each $x=(x_1,x_2,\dots ,x_n,\dots )\in C$, we define a mapping $T:C\to C$ by

$$Tx=\{\begin{array}{cc}(x_1,x_2,\dots ,x_n,\dots )\hfill & \text{if }{\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0;\hfill \\ (-x_1,-x_2,\dots ,-x_n,\dots )\hfill & \text{if }{\prod }_{i=1}^{\mathrm{\infty }}{x}_i\ge 0.\hfill \end{array}$$

(1.5)

Next we prove that T is a κ-asymptotically strictly pseudo-nonspreading mapping.

In fact, for any $x,y\in C$, we have the following cases.

Case 1. If ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i<0$, then we have $T^{n}x=x$, $T^{n}y=y$, and so then inequality (1.3) holds.

Case 2. If ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i<0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i\ge 0$, then we have that $T^{n}x=x$, $T^{n}y={(-1)}^{n}y$. This implies that

$$\{\begin{array}{c}{\parallel T^{n}x-T^{n}y\parallel }^2={\parallel x-{(-1)}^{n}y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2;\hfill \\ k_n{\parallel x-y\parallel }^2=k_n({\parallel x\parallel }^2+{\parallel y\parallel }^2);\hfill \\ {\parallel x-T^{n}x-(y-T^{n}y)\parallel }^2={[1-{(-1)}^n]}^2{\parallel y\parallel }^2;\hfill \\ 2〈x-T^{n}x,y-T^{n}y〉=0.\hfill \end{array}$$

Therefore inequality (1.3) holds.

Case 3. If ${\prod }_{i=1}^{\mathrm{\infty }}{x}_i\ge 0$ and ${\prod }_{i=1}^{\mathrm{\infty }}{y}_i\ge 0$, then we have $T^{n}x={(-1)}^{n}x$, $T^{n}y={(-1)}^{n}y$. Hence we have

$$\{\begin{array}{c}{\parallel T^{n}x-T^{n}y\parallel }^2={\parallel {(-1)}^{n}x-{(-1)}^{n}y\parallel }^2={\parallel x-y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2;\hfill \\ k_n{\parallel x-y\parallel }^2=k_n({\parallel x\parallel }^2+{\parallel y\parallel }^2);\hfill \\ {\parallel x-T^{n}x-(y-T^{n}y)\parallel }^2={[1-{(-1)}^n]}^2{\parallel x-y\parallel }^2={[1-{(-1)}^n]}^2({\parallel x\parallel }^2+{\parallel y\parallel }^2);\hfill \\ 2〈x-T^{n}x,y-T^{n}y〉=0.\hfill \end{array}$$

Thus inequality (1.3) still holds. Therefore the mapping defined by (1.5) is a κ-asymptotically strictly pseudo-nonspreading mapping.

The purpose of this article is under suitable conditions to prove some weak and strong convergence theorems for solving multiple-set split feasibility problem (1.1) for a κ-asymptotically strictly pseudo-nonspreading mapping in infinite-dimensional Hilbert spaces. The results presented in the paper extend and improve the corresponding results of Xu [6], Osilike and Isiogugu [19], Chang et al. [7], and many others.

2 Preliminaries

In the sequel, we first recall some definitions, notations, and conclusions which will be needed in proving our main results.

Let E be a real Banach space. A mapping T with domain $D(T)$ and range $R(T)$ in E is said to be demiclosed at origin if whenever $\{x_n\}$ is a sequence in $D(T)$ converging weakly to a point $x^{\ast }\in D(T)$ and $\parallel (I-T)x_n\parallel$ converging strongly to 0, then $T{x}^{\ast }=x^{\ast }$.

A Banach space E is said to have the $Opial$ property if, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x^{\ast }$, we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x^{\ast }\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel$$

for all $y\in E$ with $y\ne x^{\ast }$.

It is well known that each Hilbert space possesses the Opial property.

A mapping $T:K\to K$ is said to be semicompact if for any bounded sequence $\{x_n\}\subset K$ with ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$, there exists a subsequence $\{x_{n_i}\}\subset \{x_n\}$ such that $\{x_{n_i}\}$ converges strongly to some point $x^{\ast }\in K$.

A mapping $T:K\to K$ is said to be uniformly L-Lipschitzian if there exists a constant $L>0$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in K.$$

Let K be a nonempty closed convex subset of a real Hilbert space H. The metric projection $P_K:H\to K$ is a mapping such that for each $x\in H$, $P_{K}x$ is the unique point in K such that $\parallel x-P_{K}x\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }y\in K$. It is known that for each $x\in H$,

$$〈x-P_{K}x,y-P_{K}x〉\le 0,\mathrm{\forall }y\in K.$$

Lemma 2.1 Let H be a real Hilbert space, then the following results hold:

• (i) For all $x,y\in H$ and for all $t\in [0,1]$,

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

• (ii) ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉$.

• (iii) If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence in H which converges weakly to $z\in H$, then

  $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-y\parallel }^2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-z\parallel }^2+{\parallel z-y\parallel }^2,\mathrm{\forall }y\in H.$$

Lemma 2.2 Let K be a nonempty closed convex subset of a real Hilbert space H, and let $T:K\to K$ be a continuous κ-asymptotically strictly pseudo-nonspreading mapping. If $F(T)\ne \mathrm{\varnothing }$, then it is a closed and convex subset.

Proof Let $\{x_n\}\subset F(T)$ be a sequence such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x^{\ast }\in K$. Now we prove that $x^{\ast }\in F(T)$. In fact, since T is κ-asymptotically strictly pseudo-nonspreading, for each $m\ge 1$, we have

$$\begin{array}{rl}{\parallel T^m{x}^{\ast }-x_n\parallel }^2=& {\parallel T^m{x}^{\ast }-T^m{x}_n\parallel }^2\\ \le & k_m{\parallel x_n-x^{\ast }\parallel }^2+2〈x^{\ast }-T^m{x}^{\ast },x_n-T^m{x}_n〉\\ +\kappa {\parallel x^{\ast }-T^m{x}^{\ast }-(x_n-T^m{x}_n)\parallel }^2\\ =& k_m{\parallel x_n-x^{\ast }\parallel }^2+\kappa {\parallel x^{\ast }-T^m{x}^{\ast }\parallel }^2.\end{array}$$

Taking the limit as $n\to \mathrm{\infty }$ in the above inequality, we have

$${\parallel T^m{x}^{\ast }-x^{\ast }\parallel }^2\le \kappa {\parallel x^{\ast }-T^m{x}^{\ast }\parallel }^2.$$

Since $\kappa \in (0,1)$, we have $\parallel T^m{x}^{\ast }-x^{\ast }\parallel =0$ for each $m\ge 1$. Hence $T{x}^{\ast }=x^{\ast }$. This shows that $F(T)$ is closed.

Now we prove that $F(T)$ is convex. In fact, let $p_1,p_2\in F(T)$, and $z=\lambda p_1+(1-\lambda )p_2$, we prove that $z\in F(T)$. Since $p_1-z=(1-\lambda )(p_1-p_2)$ and $p_2-z=\lambda (p_2-p_1)$, by using Lemma 2.1(i), we have

$$\begin{array}{rcl}{\parallel z-T^{m}z\parallel }^2& =& {\parallel \lambda (p_1-T^{m}z)+(1-\lambda )(p_2-T^{m}z)\parallel }^2\\ =& \lambda {\parallel p_1-T^{m}z\parallel }^2+(1-\lambda ){\parallel p_2-T^{m}z\parallel }^2-\lambda (1-\lambda ){\parallel p_1-p_2\parallel }^2\\ \le & \lambda (k_m{\parallel p_1-z\parallel }^2+\kappa {\parallel p_1-T^m{p}_1-(z-T^{m}z)\parallel }^2+2〈p_1-T^m{p}_1,z-T^{m}z〉)\\ +(1-\lambda )(k_m{\parallel p_2-z\parallel }^2+\kappa {\parallel p_2-T^m{p}_2-(z-T^{m}z)\parallel }^2\\ +2〈p_2-T^m{p}_2,z-T^{m}z〉)-\lambda (1-\lambda ){\parallel p_1-p_2\parallel }^2\\ =& \lambda (k_m{\parallel p_1-z\parallel }^2+\kappa {\parallel z-T^{m}z\parallel }^2)+(1-\lambda )(k_m{\parallel p_2-z\parallel }^2+\kappa {\parallel z-T^{m}z\parallel }^2)\\ -\lambda (1-\lambda ){\parallel p_1-p_2\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}$ on both sides of the above inequality, we have

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel z-T^{m}z\parallel }^2\le \underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\kappa {\parallel z-T^{m}z\parallel }^2.$$

Since $\kappa <1$, we have

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel T^{m}z-z\parallel }^2=0,$$

and so ${lim}_{m\to \mathrm{\infty }}{T}^{m}z=z$, i.e., $Tz=z$. This completes the proof. □

Lemma 2.3 Let K be a nonempty closed convex subset of a real Hilbert space H, and let $T:K\to K$ be a continuous κ-asymptotically strictly pseudo-nonspreading mapping. Then $(I-T)$ is demiclosed at 0, that is, if $x_n\rightharpoonup x^{\ast }$ and ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel (I-T^m)x_n\parallel =0$, then $\parallel (I-T)x^{\ast }\parallel =0$.

Proof Since $\{x_n\}$ is weak convergence, $\{x_n\}$ is bounded. For each $x\in H$, define $f:H\to [0,\mathrm{\infty })$ by

$$f(x):=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x\parallel }^2,x\in H.$$

From Lemma 2.1(iii), we have

$$f(x)=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-x^{\ast }\parallel }^2+{\parallel x^{\ast }-x\parallel }^2,x\in H.$$

Thus we have

$$f(x)=f\left(x^{\ast }\right)+{\parallel x-x^{\ast }\parallel }^2,x\in H.$$

In particular, for each $m\ge 1$,

$$f\left(T^m{x}^{\ast }\right)=f\left(x^{\ast }\right)+{\parallel T^m{x}^{\ast }-x^{\ast }\parallel }^2.$$

(2.1)

On the other hand, we have

$$\begin{array}{rl}f\left(T^m{x}^{\ast }\right)& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-T^m{x}^{\ast }\parallel }^2\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-T^m{x}_n+T^m{x}_n-T^m{x}^{\ast }\parallel }^2\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}({\parallel x_n-T^m{x}_n\parallel }^2+2〈x_n-T^m{x}_n,T^m{x}_n-T^m{x}^{\ast }〉+{\parallel T^m{x}_n-T^m{x}^{\ast }\parallel }^2).\end{array}$$

Since ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel (I-T^m)x_n\parallel =0$ and T is a κ-asymptotically strictly pseudo-nonspreading mapping, taking ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}$ on both sides of the above equality, we get

$$\begin{array}{rl}\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}f\left(T^m{x}^{\ast }\right)\le & \underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel T^m{x}_n-T^m{x}^{\ast }\parallel }^2\\ \le & \underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(k_m{\parallel x_n-x^{\ast }\parallel }^2+\kappa {\parallel x_n-T^m{x}_n-(x^{\ast }-T^m{x}^{\ast })\parallel }^2\\ +2〈x_n-T^m{x}_n,x^{\ast }-T^m{x}^{\ast }〉).\end{array}$$

By virtue of ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel (I-T^m)x_n\parallel =0$ and $k_m\to 1$ ($m\to \mathrm{\infty }$), we have

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}f\left(T^m{x}^{\ast }\right)\le f\left(x^{\ast }\right)+\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\kappa {\parallel x^{\ast }-T^m{x}^{\ast }\parallel }^2.$$

(2.2)

On the other hand, it follows from (2.1) that

$$\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}f\left(T^m{x}^{\ast }\right)=f\left(x^{\ast }\right)+\underset{m\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel T^m{x}^{\ast }-x^{\ast }\parallel }^2,\mathrm{\forall }x\in H.$$

(2.3)

Since $\kappa <1$, it follows from (2.2) and (2.3) that ${lim\hspace{0.17em}sup}_{m\to \mathrm{\infty }}{\parallel T^m{x}^{\ast }-x^{\ast }\parallel }^2=0$. So ${lim}_{m\to \mathrm{\infty }}{T}^m{x}^{\ast }=x^{\ast }$ and $T{x}^{\ast }=x^{\ast }$. This completes the proof. □

3 Main results

Theorem 3.1 Let $H_1$, $H_2$, A, $\{S_i\}$, $\{T_i\}$, C, Q be the same as in multiple set split feasibility problem (1.1). For each $i=1,2,\dots ,N$, let $T_i$ be a uniformly $\widetilde{L_i}$-Lipschitzian and ${\kappa }_i$-asymptotically strictly pseudo-nonspreading mapping, $S_i$ be a uniformly $L_i$-Lipschitzian and ${\varrho }_i$-asymptotically strictly pseudo-nonspreading mapping. Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{x}_1\in H_1\mathit{\text{ chosen arbitrarily}},\hfill \\ u_n=x_n+\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n,\hfill \\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}_{n(modN)}^n{u}_n,\hfill \end{array}$$

(3.1)

where γ is a constant and $\gamma \in (0,\frac{1-\kappa }{\lambda })$, λ is the spectral of the operator $A^{\ast }A$, $\kappa =max\{{\kappa }_1,{\kappa }_2,\dots ,{\kappa }_N\}$ and $\{{\alpha }_n\}$ is a sequence in $(0,1-\varrho ]$ with $\varrho =max\{{\varrho }_1,{\varrho }_2,\dots ,{\varrho }_N\}$. If $\mathrm{\Gamma }\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$.

Proof The proof is divided into five steps.

(I) We first prove the limit ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for any $p\in \mathrm{\Gamma }$.

Since $p\in \mathrm{\Gamma }$, we have $p\in C:={\bigcap }_{i=1}^{N}F(S_i)$ and $Ap\in Q:={\bigcap }_{i=1}^{N}F(T_i)$. It follows from (3.1) that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2=& {\parallel u_n-p+{\alpha }_n(S_{n(modN)}^n{u}_n-u_n)\parallel }^2\\ =& {\parallel u_n-p\parallel }^2+2{\alpha }_n〈u_n-p,S_{n(modN)}^n{u}_n-u_n〉\\ +{\alpha }_n^2{\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2.\end{array}$$

(3.2)

Because $S_i$ is a ${\varrho }_i$-asymptotically strictly pseudo-nonspreading mapping, for any $v\in H_1$, we have

$$\begin{array}{r}{\parallel S_{n(modN)}^n{u}_n-S_{n(modN)}^{n}v\parallel }^2\\ \le {\parallel u_n-v\parallel }^2+\varrho {\parallel u_n-S_{n(modN)}^n{u}_n-(v-S_{n(modN)}^{n}v)\parallel }^2\\ +2〈u_n-S_{n(modN)}^n{u}_n,v-S_{n(modN)}^{n}v〉.\end{array}$$

Taking $v=p$, we have

$${\parallel S_{n(modN)}^n{u}_n-p\parallel }^2\le {\parallel u_n-p\parallel }^2+\varrho {\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2.$$

Therefore we have

$$\begin{array}{rl}{\parallel S_{n(modN)}^n{u}_n-p\parallel }^2& ={\parallel S_{n(modN)}^n{u}_n-u_n+u_n-p\parallel }^2\\ ={\parallel S_{n(modN)}^n{u}_n-u_n\parallel }^2+2〈S_{n(modN)}^n{u}_n-u_n,u_n-p〉+{\parallel u_n-p\parallel }^2\\ \le {\parallel u_n-p\parallel }^2+\varrho {\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2.\end{array}$$

Simplifying the above inequality, we have that

$$2{\alpha }_n〈S_{n(modN)}^n{u}_n-u_n,u_n-p〉\le {\alpha }_n(\varrho -1){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2.$$

(3.3)

It follows from (3.2) and (3.3) that

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ \le {\parallel u_n-p\parallel }^2+{\alpha }_n(\varrho -1){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2+{\alpha }_n^2{\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2\\ ={\parallel u_n-p\parallel }^2-{\alpha }_n(1-\varrho -{\alpha }_n){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2.\end{array}$$

(3.4)

On the other hand,

$$\begin{array}{rl}{\parallel u_n-p\parallel }^2=& {\parallel x_n-p+\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+2\gamma 〈x_n-p,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ +{\gamma }^2{\parallel A^{\ast }(T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+2\gamma 〈x_n-p,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ +{\gamma }^2〈A^{\ast }(T_{n(modN)}^n-I)A{x}_n,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ =& {\parallel x_n-p\parallel }^2+2\gamma 〈x_n-p,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ +{\gamma }^2〈A{A}^{\ast }(T_{n(modN)}^n-I)A{x}_n,(T_{n(modN)}^n-I)A{x}_n〉\\ \le & {\parallel x_n-p\parallel }^2+2\gamma 〈x_n-p,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ +{\gamma }^2{\parallel A\parallel }^2{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2.\end{array}$$

(3.5)

Since $T_i$ is a ${\kappa }_i$-asymptotically strictly pseudo-nonspreading mapping and noting $Ap\in {\bigcap }_{i-1}^{N}F(T_i)$, we have

$$\begin{array}{rl}{\parallel T_{n(modN)}^{n}A{x}_n-Ap\parallel }^2& ={\parallel T_{n(modN)}^{n}A{x}_n-T_{n(modN)}^{n}Ap\parallel }^2\\ \le {\parallel A{x}_n-Ap\parallel }^2+\kappa {\parallel T_{n(modN)}^{n}A{x}_n-A{x}_n\parallel }^2.\end{array}$$

(3.6)

Again since

$$\begin{array}{rl}{\parallel T_{n(modN)}^{n}A{x}_n-Ap\parallel }^2=& {\parallel T_{n(modN)}^{n}A{x}_n-A{x}_n+A{x}_n-Ap\parallel }^2\\ =& {\parallel T_{n(modN)}^{n}A{x}_n-A{x}_n\parallel }^2+{\parallel A{x}_n-Ap\parallel }^2\\ +2〈T_{n(modN)}^{n}A{x}_n-A{x}_n,A{x}_n-Ap〉,\end{array}$$

(3.7)

hence from (3.6) and (3.7) we have that

$$2〈T_{n(modN)}^{n}A{x}_n-A{x}_n,A{x}_n-Ap〉\le (\kappa -1){\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2.$$

(3.8)

By virtue of (3.8) we have

$$\begin{array}{r}〈T_{n(modN)}^{n}A{x}_n-A{x}_n,T_{n(modN)}^{n}A{x}_n-Ap〉\\ =〈T_{n(modN)}^{n}A{x}_n-A{x}_n,T_{n(modN)}^{n}A{x}_n-Ap+A{x}_n-A{x}_n〉\\ ={\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2+〈T_{n(modN)}^{n}A{x}_n-A{x}_n,A{x}_n-Ap〉\\ \le {\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2+\frac{\kappa -1}{2}{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ =\frac{\kappa +1}{2}{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2.\end{array}$$

(3.9)

It follows from (3.9) that

$$\begin{array}{r}2\gamma 〈x_n-p,A^{\ast }(T_{n(modN)}^n-I)A{x}_n〉\\ =2\gamma 〈A(x_n-p),(T_{n(modN)}^n-I)A{x}_n〉\\ =2\gamma 〈A(x_n-p)+(T_{n(modN)}^n-I)A{x}_n-(T_{n(modN)}^n-I)A{x}_n,(T_{n(modN)}^n-I)A{x}_n〉\\ =2\gamma 〈T_{n(modN)}^{n}A{x}_n-Ap,(T_{n(modN)}^n-I)A{x}_n〉-2\gamma {\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ \le [\gamma (1+\kappa )-2\gamma ]{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ =\gamma (\kappa -1){\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2.\end{array}$$

(3.10)

Substituting (3.10) into (3.5) and then substituting the resulting inequality into (3.4), we have

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ \le {\parallel x_n-p\parallel }^2+{\gamma }^2{\parallel A\parallel }^2{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2+[\gamma (\kappa -1)]{\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ -{\alpha }_n(1-\kappa -{\alpha }_n){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-\gamma (1-\kappa -\gamma {\parallel A\parallel }^2){\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2\\ -{\alpha }_n(1-\kappa -{\alpha }_n){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2\\ \le {\parallel x_n-p\parallel }^2.\end{array}$$

(3.11)

This shows that the limit ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists.

(II) Now we prove that the limit ${lim}_{n\to \mathrm{\infty }}\parallel u_n-p\parallel$ exists.

By (3.11) we have

$$\begin{array}{rl}\gamma (1-\kappa -\gamma {\parallel A\parallel }^2)& {\parallel (T_{n(modN)}^n-I)A{x}_n\parallel }^2+{\alpha }_n(1-\kappa -{\alpha }_n){\parallel u_n-S_{n(modN)}^n{u}_n\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2.\end{array}$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (T_{n(modN)}^n-I)A{x}_n\parallel =0,$$

(3.12)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-S_{n(modN)}^n{u}_n\parallel =0.$$

(3.13)

It follows from (3.5), (3.12), and (3.13) that the limit ${lim}_{n\to \mathrm{\infty }}\parallel u_n-p\parallel$ exists and

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

(III) Now, we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel u_{n+1}-u_n\parallel =0$.

In fact, it follows from (3.1) that

$$\begin{array}{r}\parallel x_{n+1}-x_n\parallel \\ =\parallel (1-{\alpha }_n)u_n+{\alpha }_n{S}_{n(modN)}^n{u}_n-x_n\parallel \\ =\parallel (1-{\alpha }_n)(x_n+\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n)+{\alpha }_n{S}_{n(modN)}^n{u}_n-x_n\parallel \\ =\parallel (1-{\alpha }_n)\left(\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\right)+{\alpha }_n(S_{n(modN)}^n{u}_n-x_n)\parallel \\ =\parallel (1-{\alpha }_n)\left(\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\right)+{\alpha }_n(S_{n(modN)}^n{u}_n-u_n)+{\alpha }_n(u_n-x_n)\parallel \\ =\parallel (1-{\alpha }_n)\left(\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\right)+{\alpha }_n(S_{n(modN)}^n{u}_n-u_n)\\ +{\alpha }_n\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\parallel \\ =\parallel \gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n+{\alpha }_n(S_{n(modN)}^n{u}_n-u_n)\parallel .\end{array}$$

(3.14)

This together with (3.12) and (3.13) shows that

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

(3.15)

Similarly, it follows from (3.1), (3.12), and (3.15) that

$$\begin{array}{r}\parallel u_{n+1}-u_n\parallel \\ =\parallel x_{n+1}+\gamma A^{\ast }(T_{n+1(modN)}^{n+1}-I)A{x}_{n+1}-[x_n+\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n]\parallel \\ \le \parallel x_{n+1}-x_n\parallel +\parallel \gamma A^{\ast }(T_{n+1(modN)}^{n+1}-I)A{x}_{n+1}\parallel +\parallel \gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\parallel \\ \to 0(\text{as }n\to \mathrm{\infty }).\end{array}$$

(3.16)

(IV) We prove that, for each $j=1,2,\dots ,N$,

$$\parallel u_{iN+j}-S_j{u}_{iN+j}\parallel \to 0,\parallel A{x}_{iN+j}-T_{j}A{x}_{iN+j}\parallel \to 0(i\to \mathrm{\infty }).$$

(3.17)

In fact, it follows from (3.13) that

$$\parallel u_{iN+j}-S_j^{iN+j}{u}_{iN+j}\parallel \to 0(i\to \mathrm{\infty }).$$

(3.18)

Since $S_j$ is uniformly $L_j$-Lipschitzian continuous, it follows from (3.16) and (3.18) that

$$\begin{array}{r}\parallel u_{iN+j}-S_j{u}_{iN+j}\parallel \\ \le \parallel u_{iN+j}-S_j^{iN+j}{u}_{iN+j}\parallel +\parallel S_j^{iN+j}{u}_{iN+j}-S_j{u}_{iN+j}\parallel \\ \le \parallel u_{iN+j}-S_j^{iN+j}{u}_{iN+j}\parallel +L_j\parallel S_j^{iN+j-1}{u}_{iN+j}-u_{iN+j}\parallel \\ \le \parallel u_{iN+j}-S_j^{iN+j}{u}_{iN+j}\parallel +L_j[\parallel S_j^{iN+j-1}{u}_{iN+j}-S_j^{iN+j-1}{u}_{iN+j-1}\parallel \\ +\parallel S_j^{iN+j-1}{u}_{iN+j-1}-u_{iN+j}\parallel ]\\ \le \parallel u_{iN+j}-S_j^{iN+j}{u}_{iN+j}\parallel +L_j^2\parallel u_{iN+j}-u_{iN+j-1}\parallel \\ +L_j[\parallel S_j^{iN+j-1}{u}_{iN+j-1}-u_{iN+j-1}\parallel +\parallel u_{iN+j-1}-u_{iN+j}\parallel ]\\ \to 0(\text{as }n\to \mathrm{\infty }).\end{array}$$

Similarly, we can prove that for each $i=1,2,\dots ,N$,

$$\parallel A{x}_{iN+j}-T_j^{iN+j}A{x}_{iN+j}\parallel \to 0(i\to \mathrm{\infty }).$$

(3.19)

Since $T_j$ is uniformly $\widetilde{L_j}$-Lipschitzian continuous, in the same way as above, we can also prove that

$$\parallel A{x}_{iN+j}-T_{j}A{x}_{iN+j}\parallel \to 0(\text{as }i\to \mathrm{\infty }).$$

(V) Finally, we prove that $x_n\rightharpoonup x^{\ast }$, $u_n\rightharpoonup x^{\ast }$, and it is a solution of problem ($MSSFP$) (1.1).

In fact, since $\{u_n\}$ is bounded, there exists a subsequence $\{u_{n_i}\}\subset \{u_n\}$ such that $u_{n_i}\rightharpoonup x^{\ast }\in H_1$. Hence, for any positive integer $j=1,2,\dots ,N$, there exists a subsequence $n_i(j)\subset n_i$ with $n_i(j)modN=j$ such that $u_{n_i(j)}\rightharpoonup x^{\ast }$. Again from (3.17) we have that

$$\parallel u_{n_i(j)}-S_j{u}_{n_i(j)}\parallel \to 0,n_{i(j)}\to \mathrm{\infty }.$$

(3.20)

Since $S_j$ is demiclosed at zero, it follows that $x^{\ast }\in F(S_j)$. By the arbitrariness of $j=1,2,\dots ,N$, we have

$$x^{\ast }\in C:=\bigcap _{i=1}^{N}F(S_i).$$

Moreover, from (3.1) and (3.13) we have $x_{n_i}=u_{n_i}-\gamma A^{\ast }(T_{n_i(modN)}^{n_i}-I)A{x}_{n_i}\rightharpoonup x^{\ast }$. Since A is a linear bounded operator, it follows that $A{x}_{n_i}\rightharpoonup A{x}^{\ast }$. For any positive integer $k=1,2,\dots ,N$, there exists a subsequence $x_{n_i(k)}\subset x_{n_i}$ with $n_i(k)(modN)=k$ such that $A{x}_{n_i(k)}\rightharpoonup A{x}^{\ast }$ and $\parallel A{x}_{n_i(k)}-T_{k}A{x}_{n_i(k)}\parallel \to 0$. Since $T_k$ is demiclosed at zero, we have $A{x}^{\ast }\in F(T_k)$. By the arbitrariness of k, it follows that $A{x}^{\ast }\in Q:={\bigcap }_{k=1}^{N}F(T_k)$. This together with $x^{\ast }\in C$ shows that $x^{\ast }\in \mathrm{\Gamma }$, that is, $x^{\ast }$ is a solution to the problem ($MSSFP$) (1.1).

Next we prove that $x_n\rightharpoonup x^{\ast }$ and $u_n\rightharpoonup x^{\ast }$.

In fact, assume that there exists another subsequence $u_{n_l}\subset u_n$ such that $u_{n_l}\rightharpoonup y^{\ast }\in \mathrm{\Gamma }$ with $y^{\ast }\ne x^{\ast }$. Consequently, by virtue of the existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ and the Opial property of a Hilbert space, we have

$$\begin{array}{rl}\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-x^{\ast }\parallel & <\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-y^{\ast }\parallel \\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_n-y^{\ast }\parallel \underset{n_j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-y^{\ast }\parallel \\ <\underset{n_j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-x^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_n-x^{\ast }\parallel \\ =\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-x^{\ast }\parallel .\end{array}$$

This is a contradiction. Therefore, $u_n\rightharpoonup x^{\ast }$. By (3.1) and (3.13), we have

$$x_n=u_n-\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n\rightharpoonup x^{\ast }.$$

This completes the proof of Theorem 3.1. □

Theorem 3.2 Let $H_1$, $H_2$, A, $\{S_i\}$, $\{T_i\}$, C, Q be the same as in Theorem 3.1. For each $i=1,2,\dots ,N$, let $T_i$ be a uniformly $\widetilde{L_i}$-Lipschitzian and ${\kappa }_i$-asymptotically strictly pseudo-nonspreading mapping, $S_i$ be a uniformly $L_i$-Lipschitzian and ${\varrho }_i$-asymptotically strictly pseudo-nonspreading mapping. Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{l}{x}_1\in H_1\mathit{\text{ chosen arbitrarily}},\\ u_n=x_n+\gamma A^{\ast }(T_{n(modN)}^n-I)A{x}_n,\\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}_{n(modN)}^n{u}_n,\end{array}$$

where γ is a constant and $\gamma \in (0,\frac{1-\kappa }{\lambda })$, λ is the spectral of the operator $A^{\ast }A$, $\kappa =max\{{\kappa }_1,{\kappa }_2,\dots ,{\kappa }_N\}$ and $\{{\alpha }_n\}$ is a sequence in $(0,1-\varrho ]$ with $\varrho =max\{{\varrho }_1,{\varrho }_2,\dots ,{\varrho }_N\}$. If $\mathrm{\Gamma }\ne \mathrm{\varnothing }$ and if there exists a positive integer j such that $S_j$ is semicompact, then the sequence $\{x_n\}$ converges strongly to a point $x^{\ast }\in \mathrm{\Gamma }$.

Proof Without loss of generality, we can assume that $S_1$ is semicompact. It follows from (3.17) that

$$\parallel u_{n_i(1)}-S_1{u}_{n_i(1)}\parallel \to 0,n_{i(1)}\to \mathrm{\infty }.$$

Therefore, there exists a subsequence of $\{u_{n_i(1)}\}$, which (for the sake of convenience) we still denote by $\{u_{n_i(1)}\}$, such that $u_{n_i(1)}\to u^{\ast }\in H_1$. Since $u_{n_i(1)}\rightharpoonup x^{\ast }$, $x^{\ast }=u^{\ast }$, and so $u_{n_i(1)}\to x^{\ast }\in \mathrm{\Gamma }$. By virtue of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists, we know that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x^{\ast }\parallel =0,\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-x^{\ast }\parallel =0,$$

that is, $\{u_n\}$ and $\{x_n\}$ both converge strongly to the point $x^{\ast }\in \mathrm{\Gamma }$. This completes the proof of Theorem 3.2. □

4 Applications

In this section we shall utilize the results presented in Section 3 to study the hierarchical variational inequality problem.

Let H be a real Hilbert space, $S_i$, $i=1,2,\dots ,N$, be uniformly $L_i$-Lipschitzian and ${\varrho }_i$-asymptotically strictly pseudo-nonspreading mappings with $\mathcal{F}:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\ne \mathrm{\varnothing }$. Let $T:H\to H$ be a nonspreading mapping. The so-called hierarchical variational inequality problem for a finite family of mappings $\{S_i\}$ with respect to the mapping T is to find an $x^{\ast }\in \mathcal{F}$ such that

$$〈x^{\ast }-T{x}^{\ast },x^{\ast }-x〉\le 0,\mathrm{\forall }x\in \mathcal{F}.$$

(4.1)

It is easy to see that (4.1) is equivalent to the following fixed point problem:

$$\text{find }{x}^{\ast }\in \mathcal{F}\text{ such that }{x}^{\ast }=P_{\mathcal{F}}T{x}^{\ast },$$

(4.2)

where $P_{\mathcal{F}}$ is the metric projection from H onto ℱ. Letting $C=\mathcal{F}$ and $Q=F(P_{\mathcal{F}}T)$ (the fixed point set of $P_{\mathcal{F}}T$) and $A=I$ (the identity mapping on H), problem (4.2) is equivalent to the following multi-set split feasibility problem:

$$\text{find }{x}^{\ast }\in C\text{ such that }{x}^{\ast }\in Q.$$

(4.3)

Hence from Theorem 3.1 we have the following theorem.

Theorem 4.1 Let H, $\{S_i\}$, T, C, Q be the same as above. Let $\{x_n\}$, $\{u_n\}$ be the sequences defined by

$$\{\begin{array}{c}{x}_1\in H_1\mathit{\text{ chosen arbitrarily}},\hfill \\ u_n=x_n+\gamma (T-I)x_n,n\ge 1,\hfill \\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}_{n(modN)}^n{u}_n,\hfill \end{array}$$

(4.4)

where γ is a constant and $\gamma \in (0,1)$, and $\{{\alpha }_n\}$ is a sequence in $(0,1-\varrho ]$ with $\varrho =max\{{\varrho }_1,{\varrho }_2,\dots ,{\varrho }_N\}$. If $\mathrm{\Gamma }\ne \mathrm{\varnothing }$, then $\{x_n\}$ converges weakly to a solution of hierarchical variational inequality problem (4.1).

Proof In fact, by the assumption that T is a nonspreading mapping, T is κ-strictly pseudo-nonspreading with $\kappa =0$. Taking $N=1$ and $A=I$ in Theorem 3.1, by the same method as that given in Theorem 3.1, we can prove that $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$, which is a solution of hierarchical variational inequality problem (4.1) immediately. □