Demiclosed principle and convergence theorems for asymptotically strictly pseudononspreading mappings and mixed equilibrium problems

Abstract

In this paper, the demiclosed principle for a k-asymptotically strictly pseudononspreading mapping is shown. Meanwhile, an iterative scheme is introduced to approximate a common element of the set of common fixed points of k-asymptotically strictly pseudononspreading mappings and the set of solutions of mixed equilibrium problems in Hilbert spaces, and some weak and strong convergence theorems are proved. The results presented in this paper improve and extend some recent corresponding results.

MSC:47H09, 47J25.

1 Introduction

Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H and $F:C\times C\to R$ be a bifunction, where R is the set of real numbers. The equilibrium problem (for short, EP) is to find $x^{\ast }\in C$ such that

$$F(x^{\ast },y)\ge 0,\mathrm{\forall }y\in C.$$

(1.1)

The set of solutions of EP is denoted by $EP(F)$. Given a mapping $T:C\to C$, let $F(x,y)=〈Tx,y-x〉$ for all $x,y\in C$. Then $x^{\ast }\in EP(F)$ if and only if $x^{\ast }\in C$ is a solution of the variational inequality $〈Tx,y-x〉\ge 0$ for all $y\in C$, i.e., $x^{\ast }$ is a solution of the variational inequality.

Let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a function. The mixed equilibrium problem (for short, $MEP$) is to find $x^{\ast }\in C$ such that

$$F(x^{\ast },y)+\phi (y)-\phi \left(x^{\ast }\right)\ge 0,\mathrm{\forall }y\in C.$$

(1.2)

The set of solutions of $MEP$ is denoted by $MEP(F,\phi )$.

If $\phi =0$, then mixed equilibrium problem (1.2) reduces to (1.1).

If $F=0$, then mixed equilibrium problem (1.2) reduces to the following convex minimization problem:

$$\text{Find }{x}^{\ast }\in C\text{ such that }\phi (y)\ge \phi \left(x^{\ast }\right),\mathrm{\forall }y\in C.$$

(1.3)

The set of solutions of (1.3) is denoted by $CMP(\phi )$.

The mixed equilibrium problem (MEP) includes several important problems arising in physics, engineering, science optimization, economics, transportation, network and structural analysis, Nash equilibrium problems in noncooperative games and others. It has been shown that variational inequalities and mathematical programming problems can be viewed as a special realization of abstract equilibrium problems (e.g., [1, 2]). Many authors have proposed some useful methods to solve the EP, $MEP$; see, for instance, [1–8] and the references therein.

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. Following Kohsaka and Takahashi [9–11], a mapping $T:C\to C$ is said to be nonspreading if

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

It is easy to see that the above inequality is equivalent to

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

In 1967, Browder and Petryshyn [12] introduced the concept of k-strictly pseudononspreading 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 k-strictly pseudononspreading if there exists $k\in [0,1)$ such that

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

Clearly, every nonspreading mapping is k-strictly pseudononspreading.

In 2012, Osilike [13] introduced a class of nonspreading type mappings, which is more general than the mappings studied in [14] in Hilbert spaces, and proved some weak and strong convergence theorems in real Hilbert spaces. Recently, Chang [15] studied the multiple-set split feasibility problem for asymptotically strict pseudocontraction in the framework of infinite-dimensional Hilbert spaces.

Definition 1.2 [15]

Let H be a real Hilbert space. A mapping $T:D(T)\subset H\to H$ is said to be a k-asymptotically strict pseudocontraction if there exist a constant $k\in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ ($n\to \mathrm{\infty }$) such that

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

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

Definition 1.3 Let C be a nonempty subset of a real Hilbert space H. A mapping $T:C\to C$ is said to be k-asymptotically strictly pseudononspreading if there exist a constant $k\in [0,1)$ and a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ ($n\to \mathrm{\infty }$) such that

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

(1.4)

It is easy to see that the class of k-asymptotically strictly pseudononspreading mappings is more general than the classes of k-strictly pseudononspreading mappings and k-asymptotically strict pseudocontractions.

Example 1.4 Let $X=l^2$ with the norm $\parallel \cdot \parallel$ 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 $C=\{x=(x_1,x_2,\dots ,x_n,\dots )|x_i\in R^1,i=1,2,\dots \}$ 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 the mapping $T:C\to C$ by

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

(1.5)

Next we prove that T is a k-asymptotically strictly pseudononspreading mapping.

In fact, for any $x,y\in C$.

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 inequality (1.4) holds for any $k\in [0,1)$.

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

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

Therefore inequality (1.4) holds for any $k\in [0,1)$.

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

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

So, inequality (1.4) holds for any $k\in [0,1)$.

Case 4. 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

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

Thus inequality (1.4) still holds for any $k\in [0,1)$. Therefore the mapping defined by (1.5) is a k-asymptotically strictly pseudononspreading mapping.

A mapping $T:C\to C$ is said to be uniformly L-Lipschitzian if there exists a constant $L>0$ such that for all $(x,y)\in H\times H$,

$$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel .$$

(1.6)

A Banach space E is said to satisfy Opial’s condition if, for any sequence $\{x_n\}$ in E, $x_n\rightharpoonup x$ implies that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n-x\parallel <{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel x_n-y\parallel$ for all $y\in E$ with $y\ne x$. It is well known that every Hilbert space satisfies Opial’s condition.

A mapping T with domain $D(T)$ and range $R(T)$ in E is said to be demiclosed at p if whenever $\{x_n\}$ is a sequence in $D(T)$ such that $\{x_n\}$ converges weakly to $x^{\ast }\in D(T)$ and $\{T{x}_n\}$ converges strongly to p, then $T{x}^{\ast }=p$.

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

Recently, Zhao and Chang [16] proposed the following algorithm for solving k-strictly pseudononspreading mappings and equilibrium problem in Hilbert spaces.

$$\{\begin{array}{l}F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ x_{n+1}={\alpha }_{0,n}{u}_n+{\sum }_{i=1}^{\mathrm{\infty }}{\alpha }_{i,n}{S}_{i,\beta }{u}_n,\end{array}$$

(1.7)

where $S_{i,\beta }:=\beta I+(1-\beta )S_i$, ${\alpha }_{i,n}\subset (0,1)$. Under some suitable conditions, they proved that the sequences $\{x_n\}$, $\{y_n\}$ weakly and strongly converge to a solution of the problem $x^{\ast }\in {\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i)\cap EP(F)$.

For finding a split feasibility problem for k-strictly pseudononspreading mappings in a Hilbert space, in [17], Quan and Chang presented the following iterative method:

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

(1.8)

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 $\varrho =max\{{\varrho }_1,{\varrho }_2,\dots ,{\varrho }_N\}$. Under some suitable conditions, they proved that $\{x_n\}$ weakly and strongly converges to a split fixed point $x^{\ast }\in \mathrm{\Gamma }$.

Inspired and motivated by the recent works of Zhao and Chang [16], Quan and Chang [17], etc., in this paper, we propose an iterative scheme to approximate a common element of the set of solutions of k-asymptotically strictly pseudononspreading mappings and mixed equilibrium problem in infinite-dimensional Hilbert spaces. Some weak and strong convergence theorems are proved. At the same time, the demiclosed principle of a k-asymptotically strictly pseudononspreading mapping is shown. The results presented in this paper improve and extend some recent corresponding results.

2 Preliminaries

Throughout this paper, we denote the strong convergence and weak convergence of a sequence $\{x_n\}$ to a point $x\in X$ by $x_n\to x$, $x_n\rightharpoonup x$, respectively.

Let H be a Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the norm $\parallel \cdot \parallel$, let C be a nonempty closed convex subset of H. For every point $x\in H$, there exists a unique nearest point of C, denoted by $P_{C}x$, such that $\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$ for all $y\in C$. Such a $P_C$ is called the metric projection from H onto C. It is well known that $P_C$ is a firmly nonexpansive mapping from H to C, i.e.,

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉,\mathrm{\forall }x,y\in H.$$

Further, for any $x\in H$ and $z\in C$, $z=P_{C}x$ if and only if

$$〈x-z,z-y〉\ge 0,\mathrm{\forall }y\in C.$$

(2.1)

For solving mixed equilibrium problems, we assume that the bifunction $F:C\times C\to R$ satisfies the following conditions:

(A1) $F(x,x)=0$, $\mathrm{\forall }x\in C$;

(A2) $F(x,y)+F(y,x)\le 0$, $\mathrm{\forall }x,y\in C$;

(A3) For all $x,y,z\in C$, ${lim}_{t\downarrow 0}F(tz+(1-t)x,y)\le F(x,y)$;

(A4) For each $x\in C$, the function $y\mapsto F(x,y)$ is convex and lower semi-continuous.

Lemma 2.1 [18]

Let C be a nonempty closed convex subset of a Hilbert space H. Let F be a bifunction from $C\times C$ to R satisfying (A1)-(A4), and let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a proper lower semi-continuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. For $r>0$ and $x\in C$, define a mapping $T_r:H\to C$ as follows:

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

(2.2)

Then

• (1) For each $x\in H$, $T_r(x)\ne \mathrm{\varnothing }$;

• (2) $T_r$ is single-valued;

• (3) $T_r$ is firmly nonexpansive, that is, $\mathrm{\forall }x,y\in H$,

  $${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉;$$

• (4) $F(T_r)=MEP(F,\phi )$;

• (5) $MEP(F,\phi )$ is closed and convex.

Lemma 2.2 [13]

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

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

The demiclosed principle and the closeness and convexity of the set of fixed points of a nonlinear mapping play very important roles in investigating many nonlinear problems. We now show the demiclosed principle of k-asymptotically strictly pseudononspreading mapping and the closeness and convexity of the set of fixed points of such a mapping, respectively.

Lemma 2.3 Let C be a nonempty closed convex subset of a real Hilbert space H and let $T:C\to C$ be a continuous k-asymptotically strictly pseudononspreading mapping. If $F(T)\ne \mathrm{\varnothing }$, then it is closed and convex.

Proof Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}\subset F(T)$ be a sequence which converges to $x\in C$, we show that $x\in F(T)$.

$$\parallel T^{n}x-x\parallel =\parallel T^{n}x-x_n+x_n-x\parallel \le \parallel T^{n}x-T^n{x}_n\parallel +\parallel x_n-x\parallel .$$

(2.3)

Since T is k-asymptotically strictly pseudononspreading, we have

$$\begin{array}{rl}{\parallel T^{n}x-T^n{x}_n\parallel }^2\le & k_n{\parallel x-x_n\parallel }^2+k{\parallel (T^n{x}_n-x_n)-(T^{n}x-x)\parallel }^2\\ +2〈x-T^{n}x,x_n-T^n{x}_n〉\\ =& k_n{\parallel x_n-x\parallel }^2+k{\parallel T^{n}x-x\parallel }^2\\ \le & {(\sqrt{k_n}\parallel x_n-x\parallel +\sqrt{k}\parallel T^{n}x-x\parallel )}^2.\end{array}$$

(2.4)

Using (2.4) in (2.3), we obtain

$$\parallel T^{n}x-x\parallel \le \sqrt{k_n}\parallel x_n-x\parallel +\sqrt{k}\parallel x-T^{n}x\parallel +\parallel x_n-x\parallel ,$$

so

$$\begin{array}{c}(1-\sqrt{k})\parallel T^{n}x-x\parallel \le (\sqrt{k_n}+1)\parallel x_n-x\parallel ,\hfill \\ \parallel T^{n}x-x\parallel \le \frac{\sqrt{k_n+1}}{1-\sqrt{k}}\parallel x_n-x\parallel .\hfill \end{array}$$

Since $k_n\to 1$ and $\parallel x_n-x\parallel \to 0$ as $n\to \mathrm{\infty }$, we get that ${lim}_{n\to \mathrm{\infty }}\parallel T^{n}x-x\parallel =0$. Since T is continuous, which implies that $x={lim}_{n\to \mathrm{\infty }}{T}^{n}x={lim}_{n\to \mathrm{\infty }}T(T^{n-1}x)=T({lim}_{n\to \mathrm{\infty }}{T}^{n-1}x)=Tx$. Hence, $x\in F(T)$.

Now, we show that $F(T)$ is convex.

For $x,y\in F(T)$ and $t\in (0,1)$, put $z=tx+(1-t)y$. We show that $z=Tz$. In fact, we have

$$\begin{array}{rl}{\parallel z-T^{n}z\parallel }^2=& {\parallel z\parallel }^2-2〈z,T^{n}z〉+{\parallel T^{n}z\parallel }^2\\ =& {\parallel z\parallel }^2-2〈tx+(1-t)y,T^{n}z〉+{\parallel T^{n}z\parallel }^2\\ =& {\parallel z\parallel }^2-2t〈x,T^{n}z〉-2(1-t)〈y,T^{n}z〉+{\parallel T^{n}z\parallel }^2\\ =& {\parallel z\parallel }^2+t{\parallel x\parallel }^2-2t〈x,T^{n}z〉+t{\parallel T^{n}z\parallel }^2+(1-t){\parallel y\parallel }^2\\ -2(1-t)〈y,T^{n}z〉+(1-t){\parallel T^{n}z\parallel }^2-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2\\ =& {\parallel z\parallel }^2+t{\parallel x-T^{n}z\parallel }^2+(1-t){\parallel y-T^{n}z\parallel }^2-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2\\ \le & {\parallel z\parallel }^2+t[k_n{\parallel x-z\parallel }^2+k{\parallel x-T^{n}x-(z-T^{n}z)\parallel }^2+2〈x-T^{n}x,z-T^{n}z〉]\\ +(1-t)[k_n{\parallel y-z\parallel }^2+k{\parallel y-T^{n}y-(z-T^{n}z)\parallel }^2\\ +2〈y-T^{n}y,z-T^{n}z〉]-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2\\ \le & {\parallel z\parallel }^2+t{k}_n{\parallel x-z\parallel }^2+tk{\parallel z-T^{n}z\parallel }^2+(1-t)k_n{\parallel y-z\parallel }^2\\ +(1-t)k{\parallel z-T^{n}z\parallel }^2-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2\\ =& {\parallel z\parallel }^2+t{k}_n{\parallel x-z\parallel }^2+(1-t)k_n{\parallel y-z\parallel }^2\\ +k{\parallel z-T^{n}z\parallel }^2-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2.\end{array}$$

So,

$$\begin{array}{rl}(1-k){\parallel z-T^{n}z\parallel }^2\le & {\parallel z\parallel }^2+t{k}_n〈x-z,x-z〉\\ +(1-t)k_n〈y-z,y-z〉-t{\parallel x\parallel }^2-(1-t){\parallel y\parallel }^2\\ \le & (k_n-1)[t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2+{\parallel z\parallel }^2].\end{array}$$

Since $k_n\to 1$ as $n\to \mathrm{\infty }$, we obtain that ${lim}_{n\to \mathrm{\infty }}{\parallel z-T^{n}z\parallel }^2=0$, which implies that ${lim}_{n\to \mathrm{\infty }}{T}^{n}z=z$, $z={lim}_{n\to \mathrm{\infty }}{T}^{n}z=T{lim}_{n\to \mathrm{\infty }}(T^{n-1}z)=Tz$. Hence, $z\in F(T)$, which means that $F(T)$ is convex. □

Lemma 2.4 Let C be a nonempty closed convex subset of a real Hilbert space H, and let $T:C\to C$ be a k-asymptotically strictly pseudononspreading and uniformly L-Lipschitzian mapping. Then, for any sequence $\{x_n\}$ in C converging weakly to a point p and $\{\parallel x_n-T{x}_n\parallel \}$ converging strongly to 0, we have $p=Tp$.

Proof Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$, by induction we can prove that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T^m{x}_n\parallel =0\text{for each }m\ge 1.$$

In fact, it is obvious that the conclusion is true for $m=1$. Suppose that the conclusion holds for $m>1$, now we prove that the conclusion is also true for $m+1$.

Indeed, since T is uniformly L-Lipschitzian, we have

$$\parallel x_n-T^{m+1}{x}_n\parallel \le \parallel x_n-T^m{x}_n\parallel +\parallel T^m{x}_n-T^{m+1}{x}_n\parallel \le \parallel x_n-T^m{x}_n\parallel +L\parallel x_n-T{x}_n\parallel .$$

So, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^{m+1}{x}_n\parallel =0$.

For each $x\in H$, define $f:H\to [0,\mathrm{\infty })$ by

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

(2.5)

Then from Lemma 2.2 we have

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

(2.6)

Thus, for any $x\in H$, $f(x)=f(p)+{\parallel p-x\parallel }^2$ and

$$f\left(T^{m}p\right)=f(p)+{\parallel p-T^{m}p\parallel }^2,$$

(2.7)

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

(2.8)

It follows from (2.7) and (2.8) that

$$(1-k)\parallel p-T^{m}p\parallel \le (k_m-1)\parallel f(p)\parallel .$$

(2.9)

That is,

$$\underset{m\to \mathrm{\infty }}{lim}\parallel p-T^{m}p\parallel =0.$$

(2.10)

Hence we have

$$\begin{array}{rl}\parallel Tp-p\parallel & \le \parallel Tp-T^{m}p\parallel +\parallel T^{m}p-p\parallel \\ \le L\parallel p-T^{m-1}p\parallel +\parallel T^{m}p-p\parallel .\end{array}$$

This is $p=Tp$, as desired. The proof is completed. □

Lemma 2.5 [19]

Let the number sequences $\{a_n\}$ and $\{{\alpha }_n\}$ satisfy

$$a_{n+1}\le (1+{\alpha }_n)a_n,\mathrm{\forall }n\ge 1,$$

where $a_n\ge 0$, ${\alpha }_n\ge 0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$. Then

• (1) ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists;

• (2) if ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n=0$, then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

3 Main results

Theorem 3.1 Let C be a nonempty and closed convex subset of a real Hilbert space H, let F be a bifunction from $C\times C$ to R satisfying (A1)-(A4), and let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a proper lower semi-continuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. Let $T_i:C\to C$ be a uniformly $L_i$-Lipschitzian and ${\tau }_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S_i:C\to C$ be a uniformly ${\tilde{L}}_i$-Lipschitzian and $l_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$, $i=1,2,\dots ,N$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ F(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}_{n(modN)}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}^n{y}_n,\end{array}$$

(3.1)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, $k=max\{{\tau }_1,{\tau }_2,\dots ,{\tau }_N\}\in (0,1)$, and the sequence $\{r_n\}\subset (0,\mathrm{\infty })$ satisfies that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$. If $\mathrm{\Gamma }:={\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\cap MEP(F,\phi )\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$.

Proof The proof is divided into four steps.

Step 1. Firstly, we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for any $p\in \mathrm{\Gamma }$.

Taking $p\in \mathrm{\Gamma }$ and putting $\rho =max\{l_1,l_2,\dots ,l_N\}\in (0,1)$, it follows from Lemma 2.1 that $u_n=T_{r_n}{x}_n$, $p=T_{r_n}p$, we have

$$\parallel u_n-p\parallel \le \parallel T_{r_n}{x}_n-T_{r_n}p\parallel \le \parallel x_n-p\parallel ,$$

(3.2)

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

(3.3)

Since

$$\begin{array}{rl}{\parallel S_{n(modN)}^n{y}_n-p\parallel }^2& ={\parallel S_{n(modN)}^n{y}_n-S_{n(modN)}^{n}p\parallel }^2\\ \le {\rho }_n{\parallel y_n-p\parallel }^2+\rho {\parallel (I-S_{n(modN)}^n)y_n\parallel }^2\end{array}$$

(3.4)

and

$$\begin{array}{rl}{\parallel S_{n(modN)}^n{y}_n-p\parallel }^2=& {\parallel S_{n(modN)}^n{y}_n-y_n+y_n-p\parallel }^2\\ =& {\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2+{\parallel y_n-p\parallel }^2\\ +2〈S_{n(modN)}^n{y}_n-y_n,y_n-p〉,\end{array}$$

(3.5)

from (3.4) and (3.5) we have

$$2〈S_{n(modN)}^n{y}_n-y_n,y_n-p〉\le ({\rho }_n-1){\parallel y_n-p\parallel }^2+(\rho -1){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2.$$

(3.6)

Substituting (3.6) into (3.3) and simplifying, we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & {\parallel y_n-p\parallel }^2+{\alpha }_n({\rho }_n-1){\parallel y_n-p\parallel }^2\\ +{\alpha }_n(\rho -1){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2+{\alpha }_n^2{\parallel y_n-S_{n(modN)}^n{y}_n\parallel }^2\\ =& [1+{\alpha }_n({\rho }_n-1)]{\parallel y_n-p\parallel }^2-{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2.\end{array}$$

(3.7)

On the other hand,

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

(3.8)

Since $T_i$ is a ${\tau }_i$-asymptotically strictly pseudononspreading mapping, we have

$${\parallel T_{n(modN)}^n{u}_n-p\parallel }^2\le k_n{\parallel u_n-p\parallel }^2+k{\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2.$$

(3.9)

Again since

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

(3.10)

so we have

$$\begin{array}{r}2〈T_{n(modN)}^n{u}_n-u_n\cdot u_n-p〉\\ \le (k_n-1){\parallel u_n-p\parallel }^2+(k-1){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2.\end{array}$$

(3.11)

From (3.8) and (3.11), we have

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2\le & {\parallel u_n-p\parallel }^2+{\beta }_n(k_n-1){\parallel u_n-p\parallel }^2+{\beta }_n(k-1){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2\\ +{\beta }_n^2{\parallel T_{n(modN)}^n-I)u_n\parallel }^2\\ \le & [1+{\beta }_n(k_n-1)]{\parallel u_n-p\parallel }^2-{\beta }_n(1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2.\end{array}$$

(3.12)

By using (3.7) and (3.12), we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & [1+{\alpha }_n({\rho }_n-1)]\{[1+{\beta }_n(k_n-1)]{\parallel u_n-p\parallel }^2\\ -{\beta }_n(1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2\}\\ -{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2\\ =& [1+{\alpha }_n({\rho }_n-1)][1+{\beta }_n(k_n-1)]{\parallel u_n-p\parallel }^2\\ -{\beta }_n[1+{\alpha }_n({\rho }_n-1)](1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2\\ -{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2\\ \le & \{1+{\beta }_n(k_n-1)+{\alpha }_n({\rho }_n-1)[1+{\beta }_n(k_n-1)]\}{\parallel x_n-p\parallel }^2\\ -{\beta }_n[1+{\alpha }_n({\rho }_n-1)](1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2\\ -{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2.\end{array}$$

(3.13)

Let $M_n:={\beta }_n(k_n-1)+{\alpha }_n({\rho }_n-1)[1+{\beta }_n(k_n-1)]$. Since $\sum ({\rho }_n-1)<\mathrm{\infty }$ and $\sum (k_n-1)<\mathrm{\infty }$, so $\sum M_n<\mathrm{\infty }$, we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & (1+M_n){\parallel x_n-p\parallel }^2-{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2\\ -{\beta }_n[1+{\alpha }_n({\rho }_n-1)](1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2\\ \le & (1+M_n){\parallel x_n-p\parallel }^2.\end{array}$$

(3.14)

Using Lemma 2.5, we show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. Further, it follows from (3.2) and (3.12) that $\{y_n\}$ and $\{u_n\}$ are bounded.

On the other hand, from (3.14) we have

$$\begin{array}{r}{\beta }_n[1+{\alpha }_n({\rho }_n-1)](1-k-{\beta }_n){\parallel u_n-T_{n(modN)}^n{u}_n\parallel }^2+{\alpha }_n(1-\rho -{\alpha }_n){\parallel S_{n(modN)}^n{y}_n-y_n\parallel }^2\\ \le (1+M_n){\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2.\end{array}$$

(3.15)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists and by the fact that $M_n\to 0$, taking limit on both sides of inequality (3.15), we get

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

(3.16)

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

(3.17)

Step 2. Now, we prove that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$, ${lim}_{n\to \mathrm{\infty }}\parallel y_{n+1}-y_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u_n\parallel =0$.

It follows from Lemma 2.1 that $u_n=T_{r_n}{x}_n$, $p=T_{r_n}p$, so

$$\begin{array}{rl}{\parallel u_n-p\parallel }^2& ={\parallel T_{r_n}{x}_n-T_{r_n}p\parallel }^2\\ \le 〈x_n-p,u_n-p〉\\ =\frac{1}{2}({\parallel x_n-p\parallel }^2+{\parallel u_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2).\end{array}$$

(3.18)

This shows that

$${\parallel u_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2.$$

(3.19)

By (3.13) and (3.19), we obtain

$${\parallel x_{n+1}-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_n-u_n\parallel }^2+M_n{\parallel x_n-p\parallel }^2.$$

(3.20)

So,

$${\parallel x_n-u_n\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+M_n{\parallel x_n-p\parallel }^2.$$

(3.21)

Thus, we obtain

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

(3.22)

In fact, it follows from (3.1) that

$$\begin{array}{rl}\parallel x_{n+1}-x_n\parallel =& \parallel (1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}^n{y}_n-x_n\parallel \\ =& \parallel (1-{\alpha }_n)(u_n+{\beta }_n(T_{n(modN)}^n-I)u_n)+{\alpha }_n{S}_{n(modN)}^n{y}_n-x_n\parallel \\ =& \parallel (1-{\alpha }_n){\beta }_n(T_{n(modN)}^n-I)u_n+{\alpha }_n(S_{n(modN)}^n{y}_n-u_n)+(u_n-x_n)\parallel \\ =& \parallel (1-{\alpha }_n){\beta }_n(T_{n(modN)}^n-I)u_n+{\alpha }_n(S_{n(modN)}^n{y}_n-y_n)\\ +{\alpha }_n(y_n-u_n)+(u_n-x_n)\parallel \\ =& \parallel (1-{\alpha }_n){\beta }_n(T_{n(modN)}^n-I)u_n+{\alpha }_n(S_{n(modN)}^n{y}_n-y_n)\\ +{\alpha }_n{\beta }_n(T_{n(modN)}^n-I)u_n+(u_n-x_n)\parallel \\ =& \parallel {\beta }_n(T_{n(modN)}^n-I)u_n+{\alpha }_n(S_{n(modN)}^n{y}_n-y_n)\parallel +\parallel u_n-x_n\parallel \\ \le & \parallel {\beta }_n(T_{n(modN)}^n-I)u_n\parallel +{\alpha }_n\parallel S_{n(modN)}^n{y}_n-y_n\parallel +\parallel u_n-x_n\parallel .\end{array}$$

(3.23)

From (3.16), (3.17) and (3.22) we have

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

(3.24)

Similarly, it follows from (3.1) that

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel =& \parallel u_{n+1}+{\beta }_{n+1}(T_{n+1(modN)}^{n+1}-I)u_{n+1}-u_n+{\beta }_n(T_{n(modN)}^n-I)u_n\parallel \\ \le & \parallel u_{n+1}-u_n\parallel +{\beta }_{n+1}\parallel (T_{n+1(modN)}^{n+1}-I)u_{n+1}\parallel \\ +{\beta }_n\parallel (T_{n(modN)}^n-I)u_n\parallel ,\end{array}$$

(3.25)

where

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & =& \parallel T_{r_{n+1}}{x}_{n+1}-T_{r_n}{x}_n\parallel \\ \le & \parallel T_{r_{n+1}}{x}_{n+1}-T_{r_{n+1}}{x}_n\parallel +\parallel T_{r_{n+1}}{x}_n-T_{r_n}{x}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\parallel T_{r_{n+1}}{x}_n-T_{r_n}{x}_n\parallel .\end{array}$$

(3.26)

On the other hand, it follows from Lemma 2.1 that $u_n=T_{r_n}{x}_n$ and $u_{n+1}=T_{r_{n+1}}{x}_{n+1}$. We have

$$F(u_{n+1},y)+\phi (y)-\phi (u_{n+1})+\frac{1}{r_{n+1}}〈y-u_{n+1},u_{n+1}-x_{n+1}〉\ge 0,\mathrm{\forall }y\in C,$$

and

$$F(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

Particularly, we have

$$F(u_{n+1},u_n)+\phi (u_n)-\phi (u_{n+1})+\frac{1}{r_{n+1}}〈u_n-u_{n+1},u_{n+1}-x_{n+1}〉\ge 0,\mathrm{\forall }y\in C,$$

(3.27)

and

$$F(u_n,u_{n+1})+\phi (u_{n+1})-\phi (u_n)+\frac{1}{r_n}〈u_{n+1}-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C.$$

(3.28)

Summing up (3.27) and (3.28) and using (A2), we obtain

$$\frac{1}{r_{n+1}}〈u_n-u_{n+1},u_{n+1}-x_{n+1}〉+\frac{1}{r_n}〈u_{n+1}-u_n,u_n-x_n〉\ge 0.$$

Thus,

$$〈u_{n+1}-u_n,\frac{u_n-x_n}{r_n}-\frac{u_{n+1}-x_{n+1}}{r_{n+1}}〉\ge 0,$$

which implies that

$$\begin{array}{rcl}0& \le & 〈u_{n+1}-u_n,u_n-x_n-\frac{r_n}{r_{n+1}}(u_{n+1}-x_{n+1})〉\\ =& 〈u_{n+1}-u_n,u_n-u_{n+1}+u_{n+1}-x_n-\frac{r_n}{r_{n+1}}(u_{n+1}-x_{n+1})〉.\end{array}$$

Therefore,

$$\begin{array}{rcl}{\parallel u_{n+1}-u_n\parallel }^2& \le & 〈u_{n+1}-u_n,x_{n+1}-x_n+(1-\frac{r_n}{r_{n+1}})(u_{n+1}-x_{n+1})〉\\ \le & \parallel u_{n+1}-u_n\parallel \cdot [\parallel x_{n+1}-x_n\parallel +|1-\frac{r_n}{r_{n+1}}|\cdot \parallel u_{n+1}-x_{n+1}\parallel ].\end{array}$$

Thus, we have

$$\parallel u_{n+1}-u_n\parallel \le \parallel x_{n+1}-x_n\parallel +|1-\frac{r_n}{r_{n+1}}|\cdot \parallel u_{n+1}-x_{n+1}\parallel .$$

(3.29)

It follows from (3.16), (3.17), (3.24), (3.25) and (3.29) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_{n+1}-u_n\parallel =0$$

(3.30)

and

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

(3.31)

Put $L=\{L_1,L_2,\dots ,L_N,{\tilde{L}}_1,{\tilde{L}}_2,\dots ,{\tilde{L}}_N\}$. Since

$$\begin{array}{rl}\parallel u_n-T_{n(modN)}{u}_n\parallel \le & \parallel u_n-T_{n(modN)}^n{u}_n\parallel +\parallel T_{n(modN)}^n{u}_n-T_{n(modN)}{u}_n\parallel \\ \le & \parallel u_n-T_{n(modN)}^n{u}_n\parallel +L\parallel T_{n(modN)}^{n-1}{u}_n-u_n\parallel \\ \le & \parallel u_n-T_{n(modN)}^n{u}_n\parallel +L[\parallel T_{n-1(modN)}^{n-1}{u}_n-T_{n-1(modN)}^{n-1}{u}_{n-1}\parallel \\ +\parallel T_{n-1(modN)}^{n-1}{u}_{n-1}-u_{n-1}\parallel +\parallel u_{n-1}-u_n\parallel ]\\ \le & \parallel u_n-T_{n(modN)}^n{u}_n\parallel +L^2\parallel u_n-u_{n-1}\parallel \\ +L\parallel T_{n-1(modN)}^{n-1}{u}_{n-1}-u_{n-1}\parallel +L\parallel u_{n-1}-u_n\parallel ,\end{array}$$

(3.32)

from (3.16), (3.22), (3.30) and (3.32) we get

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

(3.33)

Similarly, we have

$$\begin{array}{rl}\parallel y_n-S_{n(modN)}{y}_n\parallel \le & \parallel y_n-S_{n(modN)}^n{y}_n\parallel +\parallel S_{n(modN)}^n{y}_n{y}_n-S_{n(modN)}{y}_n\parallel \\ \le & \parallel y_n-S_{n(modN)}^n{y}_n\parallel +L[\parallel S_{n-1(modN)}^{n-1}{y}_n-S_{n-1(modN)}^{n-1}{y}_{n-1}\parallel \\ +\parallel S_{n-1(modN)}^{n-1}{y}_{n-1}-y_{n-1}\parallel +\parallel y_{n-1}-y_n\parallel ]\\ \le & \parallel y_n-T_{n(modN)}^n{y}_n\parallel +L^2\parallel y_n-y_{n-1}\parallel \\ +L\parallel T_{n-1(modN)}^{n-1}{y}_{n-1}-y_{n-1}\parallel +L\parallel y_{n-1}-y_n\parallel .\end{array}$$

This implies that

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

(3.34)

Since $\parallel x_{n+1}-y_n\parallel ={\alpha }_n\parallel y_n-S_{n(modN)}^n{y}_n\parallel$, so

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

(3.35)

By (3.24) and (3.35), we have

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

(3.36)

It follows from (3.22) and (3.36) that

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

(3.37)

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for any $p\in \mathrm{\Gamma }$ and $\parallel x_n-p\parallel -\parallel x_n-y_n\parallel \le \parallel y_n-p\parallel \le \parallel x_n-p\parallel +\parallel x_n-y_n\parallel$, it follows from (3.36) that ${lim}_{n\to \mathrm{\infty }}\parallel y_n-p\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ holds. Similarly, ${lim}_{n\to \mathrm{\infty }}\parallel u_n-p\parallel ={lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ holds for any $p\in \mathrm{\Gamma }$.

Step 3. We show that $x^{\ast }\in \mathrm{\Gamma }:={\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\cap MEP(F,\phi )$.

Firstly, we show that $x^{\ast }\in {\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)$.

In fact, since $\{y_n\}$ is bounded, there exists a subsequence $\{y_{n_i}\}\subset \{y_n\}$ such that $\{y_{n_i}\}\rightharpoonup x^{\ast }\in C$. 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 $\{y_{n_i(j)}\}\rightharpoonup x^{\ast }$. Again, by (3.34) we know that $\parallel y_{iN+j}-S_j{u}_{iN+j}\parallel \to 0$ as $i\to \mathrm{\infty }$, therefore we have that ${lim}_{n_{i(j)}\to \mathrm{\infty }}\parallel y_{n_i(j)}-S_j{y}_{n_i(j)}\parallel =0$.

Since $S_j$ is demiclosed at zero, it follows from Lemma 2.4 that $x^{\ast }\in F(S_j)$. By the arbitrariness of $j=1,2,\dots ,N$, we have

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

On the other hand, since ${lim}_{n\to \mathrm{\infty }}\parallel y_n-u_n\parallel =0$, we know that $u_{n_i}\rightharpoonup x^{\ast }$, too. Similarly, it follows from (3.33) and Lemma 2.4 that $x^{\ast }\in F(T_j)$. By the arbitrariness of $j=1,2,\dots ,N$, we have

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

Now, we show that $x^{\ast }\in MEP(F,\phi )$.

By Lemma 2.1, since $u_n=T_{r_n}{x}_n$, we have

$$F(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in K.$$

(3.38)

From (A2), we obtain

$$\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge -F(u_n,y)\ge F(y,u_n),$$

(3.39)

and hence

$$\phi (y)-\phi (u_{n_i})+\frac{1}{r_{n_i}}〈y-u_{n_i},u_{n_i}-x_{n_i}〉\ge F(y,u_{n_i}).$$

(3.40)

By ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$, we have ${lim}_{i\to \mathrm{\infty }}\frac{\parallel u_{n_i}-x_{n_i}\parallel }{r_{n_i}}=0$. Since $u_{n_i}\rightharpoonup x^{\ast }$, it follows from (A4) and the weak lower semicontinuity of φ that

$$F(y,x^{\ast })-\phi (y)+\phi \left(x^{\ast }\right)\le 0.$$

(3.41)

Put $z_t=ty+(1-t)x^{\ast }$ for all $t\in (0,1]$ and $y\in C$. Consequently, we get $z_t\in C$. Hence

$$F(z_t,p)-\phi (z_t)+\phi \left(x^{\ast }\right)\le 0.$$

(3.42)

From (A1) and (A4), and the convexity of φ, we have

$$\begin{array}{rcl}0& =& F(z_t,z_t)-\phi (z_t)+\phi (z_t)\\ \le & tF(z_t,y)+(1-t)\phi (z_t,x^{\ast })+t\phi (y)+(1-t)\phi \left(x^{\ast }\right)-\phi (z_t)\\ \le & t[F(z_t,y)+\phi (y)-\phi (z_t)].\end{array}$$

Therefore

$$F(z_t,y)+\phi (y)-\phi (z_t)\ge 0.$$

Letting $t\to 0$, and from the weak lower semicontinuity of φ, we have

$$F(x^{\ast },y)+\phi (y)-\phi \left(x^{\ast }\right)\ge 0.$$

This implies that $x^{\ast }\in MEP(F,\phi )$. Hence $x^{\ast }\in \mathrm{\Gamma }$.

Step 4. Finally, we prove that $x_n\rightharpoonup x^{\ast }$ and $u_n\rightharpoonup x^{\ast }$, $x^{\ast }\in \mathrm{\Gamma }$.

Due to $u_{n_i}\rightharpoonup x^{\ast }$, we know that $x_{n_i}\rightharpoonup x^{\ast }$ from (3.37). Suppose that there exists another subsequence $\{x_{n_j}\}$ of $\{x_n\}$ such that $\{x_{n_j}\}\rightharpoonup y^{\ast }\in \mathrm{\Gamma }$ with $y^{\ast }\ne x^{\ast }$. Using the same proof method as in Step 3, we know that $y^{\ast }\in \mathrm{\Gamma }$. Consequently, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y^{\ast }\parallel$ exists. By using Opial’s property of a Hilbert space, we have

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

This is a contradiction. Therefore $x_n\rightharpoonup x^{\ast }$. By (3.1) and (3.22), we have $u_n\rightharpoonup x^{\ast }$. Therefore, the conclusion follows.

This completes the proof of Theorem 3.1. □

Taking $\phi =0$, $N=1$ in Theorem 3.1, we have the following result.

Corollary 3.2 Let C be a nonempty and closed convex subset of a real Hilbert space H, let F be a bifunction from $C\times C$ to R satisfying (A1)-(A4), and let $T:C\to C$ be a uniformly L-Lipschitzian and k-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S:C\to C$ be a uniformly $\tilde{L}$-Lipschitzian and ρ-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}^n{y}_n,\end{array}$$

(3.43)

where $k\in (0,1)$, $\rho \in (0,1)$, $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$ and the sequence $\{r_n\}\subset (0,\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$. If $\mathrm{\Gamma }:=F(S)\cap F(T)\cap EP(F)\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$.

Corollary 3.3 Let C be a nonempty and closed convex subset of a real Hilbert space H, let F be a bifunction from $C\times C$ to R satisfying (A1)-(A4), and let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a proper lower semi-continuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. Let $T_i:C\to C$ be a uniformly $L_i$-Lipschitzian and ${\tau }_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S_i:C\to C$ be a uniformly ${\tilde{L}}_i$-Lipschitzian and $l_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$, $i=1,2,\dots ,N$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ F(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}_{n(modN)}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}{y}_n,\end{array}$$

(3.44)

where $k=max\{{\tau }_1,{\tau }_2,\dots ,{\tau }_N\}\in (0,1)$, $\rho =max\{l_1,l_2,\dots ,l_N\}\in (0,1)$, $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$ and the sequence $\{r_n\}\subset (0,\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$. If $\mathrm{\Gamma }:={\bigcap }_{i=1}^{\mathrm{\infty }}F(S_i){\bigcap }_{i=1}^{\mathrm{\infty }}F(T_i)\cap MEP(F,\phi )\ne \mathrm{\varnothing }$, and there exists a positive integer j such that $S_j$ is semi-compact, 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 semi-compact. It follows from (3.34) that

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

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

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

That is, $\{x_n\}$, $\{u_n\}$ and $\{y_n\}$ converge strongly to the point $x^{\ast }\in \mathrm{\Gamma }$. This completes the proof. □

4 Applications

4.1 Application to a convex minimization problem

It is well known that mixed equilibrium problem (1.2) reduces to the convex minimization problem as $F=0$. Therefore, Theorem 3.1 can be used to solve convex minimization problem (1.3), and the following result can be directly deduced from Theorem 3.1.

Theorem 4.1 Let C be a nonempty and closed convex subset of a real Hilbert space H, let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a proper lower semi-continuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. Let $T_i:C\to C$ be a uniformly $L_i$-Lipschitzian and ${\tau }_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S_i:C\to C$ be a uniformly ${\tilde{L}}_i$-Lipschitzian and $l_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$, $i=1,2,\dots ,N$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ \phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}_{n(modN)}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}^n{y}_n,\end{array}$$

(4.1)

where $k=max\{{\tau }_1,{\tau }_2,\dots ,{\tau }_N\}\in (0,1)$, $\rho =max\{l_1,l_2,\dots ,l_N\}\in (0,1)$, $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, and the sequence $\{r_n\}\subset (0,\mathrm{\infty })$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$. If ${\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\cap CMP(\phi )\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in {\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\cap CMP(\phi )$.

4.2 Application to a convex feasibility problem

The so-called convex feasibility problem for a family of mappings ${\{T_i\}}_{i=1}^{\omega }$ (where ω may be a finite positive integer or +∞) is to find a point of the nonempty intersection ${\bigcap }_{i=1}^{\omega }{C}_i$, where $C_i$ is the fixed point set of mapping $T_i$, $i=1,2,\dots ,\omega$.

In Theorem 3.1 if $F=0$, $\phi =0$, then the condition ‘$u_n\in C$ such that $\mathrm{\forall }y\in C$, $〈y-u_n,u_n-x_n〉\ge 0$’ is equivalent to $u_n=P_C(x_n)$. Therefore, the following result can be directly obtained from Theorem 3.1.

Theorem 4.2 Let C be a nonempty and closed convex subset of a real Hilbert space H, let $T_i:C\to C$ be a uniformly $L_i$-Lipschitzian and ${\tau }_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S_i:C\to C$ be a uniformly ${\tilde{L}}_i$-Lipschitzian and $l_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$, $i=1,2,\dots ,N$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ u_n=P_C(x_n),\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}_{n(modN)}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}^n{y}_n,\end{array}$$

(4.2)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$ and $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, $k=max\{{\tau }_1,{\tau }_2,\dots ,{\tau }_N\}\in (0,1)$. If $\mathrm{\Gamma }:={\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$, which is a solution of the convex feasibility problem for mappings ${\{T_i\}}_{i=1}^N$ and ${\{S_i\}}_{i=1}^N$.

4.3 Application to the mixed variational inequality problem of Browder type

A variational inequality problem (VIP) is formulated as a problem of finding a point $x^{\ast }$ with property $x^{\ast }\in C$, $〈A{x}^{\ast },z-x^{\ast }〉\ge 0$, $\mathrm{\forall }z\in C$. We will denote the solution set of VIP by $VI(A,C)$. We know that given a mapping $T:C\to C$, let $F(x,y)=〈Tx,y-x〉$ for all $x,y\in C$. Then $x^{\ast }\in EP(F)$ if and only if $x^{\ast }\in C$ is a solution of the variational inequality $〈Tx,y-x〉\ge 0$ for all $y\in C$, i.e., $x^{\ast }$ is a solution of the variational inequality.

In [20], the mixed variational inequality of Browder type (VI) is shown to be equivalent to finding a point $u\in C$ such that

$$〈Au,y-u〉+\phi (y)-\phi (u)\ge 0,\mathrm{\forall }y\in C.$$

We will denote the solution set of a mixed variational inequality of Browder type by $VI(A,C,\phi )$.

A mapping $A:C\to H$ is said to be an α-inverse-strongly monotone mapping if there exists a constant $\alpha >0$ such that $〈Ax-Ay,x-y〉\ge \alpha {\parallel Ax-Ay\parallel }^2$ for any $x,y\in C$. Setting $F(x,y)=〈Ax,y-x〉$, it is easy to show that F satisfies conditions (A1)-(A4) as A is an α-inverse-strongly monotone mapping. Then it follows from Theorem 3.1 that the following result holds.

Theorem 4.3 Let C be a nonempty and closed convex subset of a real Hilbert space H, let $A:C\to H$ be an α-inverse-strongly monotone mapping, and let $\phi :C\to R\cup \{+\mathrm{\infty }\}$ be a proper lower semi-continuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. Let $T_i:C\to C$ be a uniformly $L_i$-Lipschitzian and ${\tau }_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{k_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, let $S_i:C\to C$ be a uniformly ${\tilde{L}}_i$-Lipschitzian and $l_i$-asymptotically strictly pseudononspreading mapping with the sequence $\{{\rho }_n\}\subset [1,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}({\rho }_n-1)<\mathrm{\infty }$, $i=1,2,\dots ,N$. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}\mathrm{\forall }{x}_1\in C,\\ 〈A{u}_n,y-u_n〉+\phi (y)-\phi (u_n)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\mathrm{\forall }y\in C,\\ y_n=(1-{\beta }_n)u_n+{\beta }_n{T}_{n(modN)}^n{u}_n,\\ x_{n+1}=(1-{\alpha }_n)y_n+{\alpha }_n{S}_{n(modN)}^n{y}_n,\end{array}$$

(4.3)

where $\{{\alpha }_n\}$ is a sequence in $(0,1)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n>0$, $\{{\beta }_n\}$ is a sequence in $(0,1-k)$ with ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n>0$, $k=max\{{\tau }_1,{\tau }_2,\dots ,{\tau }_N\}\in (0,1)$, and the sequence $\{r_n\}\subset (0,\mathrm{\infty })$ satisfies ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n>0$ and ${lim}_{n\to \mathrm{\infty }}|r_{n+1}-r_n|=0$. If $\mathrm{\Gamma }:={\bigcap }_{i=1}^{N}F(S_i){\bigcap }_{i=1}^{N}F(T_i)\cap VI(A,C,\phi )\ne \mathrm{\varnothing }$, then the sequence $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$.