Iterative algorithm for a family of split equilibrium problems and fixed point problems in Hilbert spaces with applications

Abstract

In this paper, we propose an iterative algorithm and, by using the proposed algorithm, prove some strong convergence theorems for finding a common element of the set of solutions of a finite family of split equilibrium problems and the set of common fixed points of a countable family of nonexpansive mappings in Hilbert spaces. An example is given to illustrate the main result of this paper. As an application, we construct an algorithm to solve an optimization problem.

1 Introduction

Throughout this paper, let $\mathbb{R}$ denote the set of all real numbers, $\mathbb{N}$ denote the set of all positive integer numbers, H be a real Hilbert space and C be a nonempty closed convex subset of H. A mapping $S: C\to C$ is said to nonexpansive if

$$\Vert Sx-Sy\Vert \leq \Vert x-y\Vert $$

for all $x,y\in C$. The set of fixed points of S is denoted by $\operatorname {Fix}(S)$. It is known that the set $\operatorname {Fix}(S)$ is closed and convex.

Let $F: C\times C\to\mathbb{R}$ be a bifunction. The equilibrium problem for F is to find $z\in C$ such that

$$ F(z,y)\geq0 $$

(1.1)

for all $y\in C$. The set of all solutions of the problem (1.1) is denoted by $\operatorname {EP}(F)$, i.e.,

$$\operatorname {EP}(F)=\bigl\{ z\in C: F(z,y)\geq0, \forall y\in C\bigr\} . $$

From the problem (1.1), we can consider some related problems, that is, variational inequality problems, complementarity problems, fixed point problems, game theory and other problems. Also, many problems in physics, optimization, and economics can be reduced to finding a solution of the problem (1.1) (see [1–4]).

In 1997, Combettes and Hirstoaga [5] introduced an iterative scheme of finding a solution of the problem (1.1) under the assumption that $\operatorname {EP}(F)$ is nonempty. Later on, many iterative algorithms were considered to find a common element of the set of $\operatorname {Fix}(S)\cap \operatorname {EP}(F)$ (see [6–11]).

Recently, some new problems called split variational inequality problems were considered by some authors. Especially, Censor et al. [12] initially studied this class of split variational inequality problems.

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces. Given the operators $f: H_{1}\to H_{1}$ and $g: H_{2}\to H_{2}$, bounded linear operator $A:H_{1}\to H_{2}$, and nonempty closed convex subsets $C\subset H_{1}$ and $Q\subset H_{2}$, the split variational inequality problem is formulated as follows:

Find a point $x^{*}\in C$ such that

$$\bigl\langle f\bigl(x^{*}\bigr),x-x^{*} \bigr\rangle \geq0 $$

for all $x\in C$ and such that

$$y^{*}=Ax^{*}\in Q \quad \mbox{solves}\quad \bigl\langle g\bigl(y^{*}\bigr),y-y^{*}\bigr\rangle \geq0 $$

for all $y\in Q$.

After investigating the algorithm of Censor et al., Moudafi [13] introduced a new iterative scheme to solve the following split monotone variational inclusion:

Find $x^{*}\in H_{1}$ such that

$$0\in f\bigl(x^{*}\bigr)+B_{1}\bigl(x^{*}\bigr) $$

and such that

$$y^{*}=Ax^{*}\in H_{2} \quad \mbox{solves}\quad 0\in g\bigl(y^{*}\bigr)+B_{2} \bigl(y^{*}\bigr), $$

where $B_{1}: H_{i}\to2^{H_{i}}$ is a set-valued mappings for $i=1,2$.

In 2013, Kazmi and Rizvi [14] considered a new class of split equilibrium problems. Let $F_{1}: C\times C\to\mathbb{R}$ and $F_{2}: Q\times Q\to\mathbb{R}$ be two bifunctions and $A: H_{1}\to H_{2}$ be a bounded linear operator. The split equilibrium problem is as follows:

Find $x^{*}\in C$ such that

$$ F_{1}\bigl(x^{*},x\bigr)\geq0 $$

(1.2)

for all $x\in C$ and such that

$$ y^{*}=Ax^{*}\in Q\quad \mbox{solves}\quad F_{2}\bigl(y^{*},y\bigr)\geq0 $$

(1.3)

for all $y\in Q$. The set of all solutions of the problems (1.2) and (1.3) is denoted by Ω, i.e.,

$$\Omega=\bigl\{ z\in C: z\in \operatorname {EP}(F_{1}) \mbox{ such that } Az\in \operatorname {EP}(F_{2})\bigr\} . $$

For more details as regards the split equilibrium problems, refer to [15, 16], in which the author gave an iterative algorithm to find a common element of the sets of solutions of the split equilibrium problem and hierarchical fixed point problem.

In this paper, inspired by the results in [14] and [16], we propose an iterative algorithm to find a common element of the set of solutions for a family of split equilibrium problems and the set of common fixed points of a countable family of nonexpansive mappings. In particular, we use some new methods to prove the main result of this paper. As an application, we propose an iterative algorithm to solve a split variational inequality problem.

2 Preliminaries

Let H be a Hilbert space and C be a nonempty closed subset of H. For each point $x\in H$, there exists a unique nearest point of C, denoted by $P_{C}x$, such that

$$\Vert x-P_{C}x\Vert \leq \Vert x-y\Vert $$

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 onto C, i.e.,

$$\Vert P_{C}x-P_{C}y\Vert ^{2}\leq\langle P_{C}x-P_{C}y,x-y\rangle $$

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

$$\langle x-z,z-y \rangle\geq0 $$

for all $y\in C$.

A mapping $B:C\to H$ is called α-inverse strongly monotone if there exists $\alpha>0$ such that

$$\langle x-y,Bx-By\rangle\geq\alpha \Vert Bx-By\Vert ^{2} $$

for all $x,y\in H$. For each $\lambda\in(0, 2\alpha]$, $I-\lambda B$ is a nonexpansive mapping of C into H (see [17]).

Consider the following variational inequality for an inverse strongly monotone mapping B:

Find $u\in C$ such that

$$\langle v-u,Bu \rangle\geq0 $$

for all $v\in C$. The set of solutions of the variational inequality is denoted $\operatorname {VI}(C,B)$. It is well known that

$$u\in \operatorname {VI}(C,B)\quad \Longleftrightarrow \quad u=P_{C}(u-\lambda Bu) $$

for any $\lambda>0$. By this property, we can use a simple method to show that $u\in \operatorname {VI}(C,B)$. In fact, let $\{x_{n}\}$ be a sequence in C with $x_{n}\rightharpoonup u$. If $x_{n}-P_{C}(I-\lambda B)x_{n}\to0$, then, by the demiclosedness principle, it follows that $u=P_{C}(I-\lambda B)$, i.e., $u\in \operatorname {VI}(C,B)$. In Section 3, we use this method to show the conclusions of our main results in this paper.

Let $S: C\to C$ be a mapping. It is well known that S is nonexpansive if and only if the complement $I-S$ is $\frac {1}{2}$-inverse strongly monotone (see [18]). Assume that $\operatorname {Fix}(S)\neq\emptyset$. Then we have

$$ \Vert Sx-x\Vert ^{2}\leq2 \langle x-Sx,x-\hat{x}\rangle $$

(2.1)

for all $x\in C$ and $\hat{x}\in \operatorname {Fix}(S)$, which is obtained directly from

$$\begin{aligned} \Vert x-\hat{x}\Vert ^{2}&\geq \Vert Sx-S\hat{x}\Vert ^{2}=\Vert Sx-\hat{x}\Vert ^{2}=\bigl\Vert Sx-x+(x-\hat {x})\bigr\Vert ^{2} \\ &=\Vert Sx-x\Vert ^{2}+\Vert x-\hat{x}\Vert ^{2}+2 \langle Sx-x, x-\hat{x}\rangle. \end{aligned} $$

Let F be a bifunction of $C\times C$ into $\mathbb{R}$ satisfying the following conditions:

• (A1) $F(x, x)=0$ for all $x\in C$;

• (A2) F is monotone, i.e., $F(x,y)+F(y,x)\leq0$ for all $x,y\in C$;

• (A3) for each $x,y,z\in C$, $\lim_{t\downarrow0}F(tz+(1-t)x,y)\leq F(x,y)$;

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

Lemma 2.1

[19]

Let C be a nonempty closed convex subset of a Hilbert space H and $F: C\times C\to\mathbb{R}$ be a bifunction which satisfies the conditions (A1)-(A4). For any $x\in H$ and $r >0$, define a mapping $T_{r}:H\to C$ by

$$ T_{r}^{F}(x)=\biggl\{ z\in C:F(z,y)+\frac{1}{r}\langle y-z,z-x\rangle\geq0, \forall y\in C\biggr\} . $$

(2.2)

Then $T_{r}^{F}$ is well defined and the following hold:

• (1) $T_{r}^{F}$ is single-valued;

• (2) $T_{r}^{F}$ is firmly nonexpansive, i.e., for any $x,y\in H$,

  $$\bigl\Vert T_{r}^{F}x-T_{r}^{F}y \bigr\Vert ^{2}\leq\bigl\langle T_{r}^{F}x-T_{r}^{F}y,x-y \bigr\rangle ; $$

• (3) $\operatorname {Fix}(T_{r}^{F})=\operatorname {EP}(F)$;

• (4) $\operatorname {EP}(F)$ is closed and convex.

Lemma 2.2

[20]

Let $F: C\times C\to\mathbb{R}$ be a bifunction satisfying the conditions (A1)-A(4). Let $T^{F}_{r}$ and $T^{F}_{s}$ be defined as in Lemma  2.1 with $r,s>0$. Then, for any $x,y\in H$, one has

$$\bigl\Vert T^{F}_{r}x-T^{F}_{s}y \bigr\Vert \leq \vert x-y\vert + \biggl\vert 1-\frac{s}{r} \biggr\vert \bigl\Vert T^{F}_{r}x-x\bigr\Vert . $$

Remark 2.1

In [20], some other conditions are required besides the conditions (A1)-(A4). In fact, the conditions (A1)-(A4) are enough for Lemma 2.2. For the proof, refer to [9, 20].

Lemma 2.3

[9]

Let $F: C\times C\to\mathbb{R}$ be a functions satisfying the conditions (A1)-(A4) and $T^{F}_{s}$, $T^{F}_{t}$ be defined as in Lemma  2.1 with $s,t>0$. Then the following holds:

$$\bigl\Vert T_{s}^{F}x-T_{t}^{F}x \bigr\Vert ^{2}\leq\frac{s-t}{s}\langle T_{s}x-T_{t}x, T_{s}x-x\rangle $$

for all $x\in H$.

Lemma 2.4

[21]

Let $\{a_{n}\}$ be a sequence in $[0,1]$ such that $\sum_{n=1}^{\infty}a_{n}=1$. Then we have the following:

$$\Biggl\Vert \sum_{n=1}^{\infty}a_{n}x_{n} \Biggr\Vert ^{2}\leq\sum _{n=1}^{\infty}a_{n}\Vert x_{n} \Vert ^{2} $$

for any bounded sequence $\{x_{n}\}$ in a Hilbert space H.

Lemma 2.5

(Demiclosedness principle)

Let T be a nonexpansive mapping on a closed convex subset C of a real Hilbert space H. Then $I-T$ is demiclosed at any point $y\in H$, that is, if $x_{n}\rightharpoonup x$ and $x_{n}-Tx_{n}\to y\in H$, then $x-Tx=y$.

Lemma 2.6

[22]

Assume that $\{a_{n}\}$ is a sequence of nonnegative numbers such that

$$a_{n+1}\leq(1-\gamma_{n})a_{n}+\delta_{n} $$

for each $n\geq0$, where $\{\gamma_{n}\}$ is a sequence in $(0,1)$ and $\{\delta_{n}\}$ is a sequence in $\mathbb{R}$ such that

• (1) $\sum_{n=1}^{\infty}\gamma_{n}=\infty$;

• (2) $\limsup_{n\to\infty}\delta_{n}/\gamma_{n}\leq0$ or $\sum_{=1}^{\infty} \vert \delta_{n}\vert <\infty$.

Then $\lim_{n\to\infty}a_{n}=0$.

Lemma 2.7

[23, 24]

Let U and V be nonexpansive mappings. For $\sigma\in(0,1)$, define $S=\sigma U+(1-\sigma)V$. Suppose that $\operatorname {Fix}(U)\cap \operatorname {Fix}(V)\neq \emptyset$. Then $\operatorname {Fix}(U)\cap \operatorname {Fix}(V)=\operatorname {Fix}(S)$.

From [24] we can see that Lemma 2.7 holds whenever U and V are self or non-self mappings.

Lemma 2.8

[24]

Let C be a nonempty closed convex subset of a Hilbert space H and $T: C\to H$ be a nonexpansive mapping with $\operatorname {Fix}(T)\neq\emptyset$. Let $P_{C}$ be the metric projection from H onto C. Then $\operatorname {Fix}(P_{C}T)=\operatorname {Fix}(T)=\operatorname {Fix}(TP_{C})$.

Remark 2.2

Let $S_{1},S_{2}: C\to H$ be two nonexpansive mappings with $\operatorname {Fix}(S_{1})\cap \operatorname {Fix}(S_{2})\neq\emptyset$. Let $\sigma\in(0,1)$ and define the mapping $S: C\to H$ by $S=\sigma S_{1}+(1-\sigma)S_{2}$. By Lemmas 2.7 and 2.8, it is easy to see that $\operatorname {Fix}(P_{C}S)=\operatorname {Fix}(P_{C}S_{1})\cap \operatorname {Fix}(P_{C}S_{2})$.

From Remark 2.2, we get the following result.

Lemma 2.9

Let $\{B_{i}\}_{i=1}^{N}$ be a finite family of inverse strongly monotone mappings from C to H with the constants $\{\beta_{i}\}_{i=1}^{N}$ and assume that $\bigcap_{i=1}^{N}\operatorname {VI}(C,B_{i})\neq\emptyset$. Let $B=\sum_{i=1}^{N} \alpha_{i}B_{i}$ with $\{\alpha_{i}\}_{i=1}^{N}\subset(0,1)$ and $\sum_{i=1}^{N}\alpha_{i}=1$. Then $B: C\to H$ is a β-inverse strongly monotone mapping with $\beta=\min\{\beta_{1},\ldots, \beta_{N}\}$ and $\operatorname {VI}(C,B)=\bigcap_{i=1}^{N} \operatorname {VI}(C,B_{i})$.

Proof

It is easy to show that B is a β-inverse strongly monotone mapping. In fact, for all $x,y\in C$, by Lemma 2.4, we have

$$\begin{aligned} \beta \Vert Bx-By\Vert ^{2}&=\beta\Biggl\Vert \sum _{i=1}^{N}\alpha_{i} (B_{i}x- B_{i}y)\Biggr\Vert ^{2} \\ &\leq\beta\sum_{i=1}^{N} \alpha_{i}\Vert B_{i}x-B_{i}y\Vert ^{2} \\ &\leq \sum_{i=1}^{N}\alpha_{i} \beta_{i}\Vert B_{i}x-B_{i}y\Vert ^{2} \\ &\leq\sum_{i=1}^{N}\alpha_{i} \langle x-y, B_{i}x-B_{i}y\rangle \\ &=\langle x-y, Bx-By\rangle, \end{aligned} $$

which implies that B is a β-inverse strongly monotone mapping.

Next, we prove that $\operatorname {VI}(C,B)=\bigcap_{i=1}^{N} \operatorname {VI}(C,B_{i})$. Obviously, we have

$$\bigcap_{i=1}^{N} \operatorname {VI}(C,B_{i})\subset \operatorname {VI}(C,B). $$

Now, for any $w\in \operatorname {VI}(C,B)$, we show that $w\in\bigcap_{i=1}^{N} \operatorname {VI}(C,B_{i})$. Take a constant $\lambda\in(0,2\beta]$. Then $I-\lambda B$ is nonexpansive. Note that $I-\lambda B=\sum_{i=1}^{N}\alpha_{i}(I-\lambda B_{i})$ and each $I-\lambda B_{i}$ is nonexpansive. From Remark 2.2, it follows that

$$\operatorname {Fix}\bigl(P_{C}(I-\lambda B)\bigr)=\bigcap_{i=1}^{N} \operatorname {Fix}\bigl(P_{C}(I-\lambda B_{i})\bigr). $$

Thus we have

$$w\in \operatorname {VI}(C,B)\quad \Longleftrightarrow\quad w=P_{C}(I-\lambda B)w=P_{C}(I-\lambda B_{i})w \quad \Longleftrightarrow\quad w\in \operatorname {VI}(C,B_{i}) $$

for each $i=1,\ldots, N$. Therefore, $w\in\bigcap_{i=1}^{N} \operatorname {VI}(C,B_{i})$. This completes the proof. □

3 Main result

Now, we give the main results of this paper.

Theorem 3.1

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and $C\subset H_{1}$, $Q\subset H_{2}$ be nonempty closed convex subsets. Let $A_{i}: H_{1}\to H_{2}$ be a bounded linear operator for each $i=1,\ldots ,N_{1}$ with $N_{1}\in\mathbb{N}$ and $B_{i}: C\to H_{1}$ be a $\beta_{i}$-inverse strongly monotone operator for each $i=1,\ldots,N_{2}$ with $N_{2}\in\mathbb{N}$. Assume that $F:C\times C\to\mathbb{R}$ satisfies (A1)-(A4), $F_{i}: Q\times Q\to\mathbb{R}$ ($i=1,\ldots, N_{1}$) satisfies (A1)-(A4). Let $\{S_{n}\}$ be a countable family of nonexpansive mappings from C into C. Assume that $\Theta=\Gamma \cap\Omega\cap \operatorname {VI}\neq\emptyset$, where $\Gamma=\bigcap_{n=1}^{\infty} \operatorname {Fix}(S_{n})$, $\Omega=\{z\in C: z\in \operatorname {EP}(F)\ \textit{and}\ A_{i}z\in \operatorname {EP}(F_{i}), i=1,\ldots ,N_{1} \}$ and $\operatorname {VI}=\bigcap_{i=1}^{N_{2}}\operatorname {VI}(C,B_{i})$. Let $\{\gamma_{1},\ldots, \gamma_{N_{2}}\}\subset (0,1)$ with $\sum_{i=1}^{N_{2}}\gamma_{i}=1$. Take $v,x_{1}\in C$ arbitrarily and define an iterative scheme in the following manner:

$$ \textstyle\begin{cases} u_{i,n}=T_{r_{n}}^{F}(I-\gamma A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i}) x_{n}, \quad i=1,\ldots , N_{1},\\ y_{n}=P_{C} (I-\lambda_{n} (\sum_{i=1}^{N_{2}}\gamma_{i} B_{i} ) ) (\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} ),\\ x_{n+1}=\alpha_{n} v+ \sum_{i=1}^{n}( \alpha_{i-1} - \alpha_{i} ) S_{i}y_{n}, \end{cases} $$

(3.1)

for each $i=1,\ldots,N_{1}$ and $n\in\mathbb{N}$, where $\{r_{n}\}\subset (r,\infty)$ with $r>0$, $\{\lambda_{n}\}\subset(0,2\beta)$ with $\beta =\min\{\beta_{1},\ldots,\beta_{N_{2}}\}$ and $\gamma\subset(0,1/L^{2}]$, $L=\max\{L_{1},\ldots,L_{N_{1}}\}$ and $L_{i}$ is the spectral radius of the operator $A_{i}^{*}A_{i}$ and $A_{i}^{*}$ is the adjoint of $A_{i}$ for each $i\in\{ 1,\ldots, N_{1}\}$, and $\{\alpha_{n}\}\subset(0,1)$ is a strictly decreasing sequence. Let $\alpha_{0}=1$ and assume that the control sequences $\{\alpha_{n}\}$, $\{\lambda_{n}\}$, $\{r_{n}\}$ satisfy the following conditions:

• (1) $\lim_{n\to\infty}\alpha_{n}=0$ and $\sum_{n=1}^{\infty}\alpha _{n}=\infty$;

• (2) $\sum_{n=1}^{\infty} \vert r_{n+1}-r_{n}\vert <\infty$ and $\sum_{n=1}^{\infty} \vert \lambda_{n+1}-\lambda_{n}\vert <\infty$;

• (3) $\lim_{n\to\infty} \lambda_{n}=\lambda>0$.

Then the sequence $\{x_{n}\}$ defined by (3.1) converges strongly to a point $z=P_{\Theta}v$.

Proof

We first show that, for each $i=1,\ldots, N_{1}$ and $n\in \mathbb{N}$, $A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i}$ is a $\frac {1}{2L_{i}^{2}}$-inverse strongly monotone mapping. In fact, since $T_{r_{n}}^{F_{i}}$ is (firmly) nonexpansive and $I-T_{r_{n}}^{F_{i}}$ is $\frac {1}{2}$-inverse strongly monotone, we have

$$\begin{aligned} &\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}x-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}y\bigr\Vert ^{2} \\ &\quad =\bigl\langle A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}}\bigr) (A_{i}x-A_{i}y), A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr) (A_{i}x-A_{i}y)\bigr\rangle \\ &\quad =\bigl\langle (I-T_{r_{n}}) (A_{i}x-A_{i}y),A_{i}A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr) (A_{i}x-A_{i}y) \bigr\rangle \\ &\quad \leq L_{i}^{2}\bigl\langle \bigl(I-T_{r_{n}}^{F_{i}} \bigr) (A_{i}x-A_{i}y), \bigl(I-T_{r_{n}}^{F_{i}} \bigr) (A_{i}x-A_{i}y)\bigr\rangle \\ &\quad = L_{i}^{2}\bigl\Vert \bigl(I-T_{r_{n}}^{F_{i}} \bigr) (A_{i}x-A_{i}y)\bigr\Vert ^{2} \\ &\quad \leq2L_{i}^{2}\bigl\langle A_{i}x-A_{i}y, \bigl(I-T_{r_{n}}^{F_{i}}\bigr) (A_{i}x-A_{i}y) \bigr\rangle \\ &\quad =2L_{i}^{2} \bigl\langle x-y, A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr) A_{i}x-A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr) A_{i} y\bigr\rangle \end{aligned}$$

for all $x,y\in H_{1}$, which implies that $A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i}$ is a $\frac {1}{2L_{i}^{2}}$-inverse strongly monotone mapping. Note that $\gamma\in (0,\frac{1}{L_{i}^{2}}]$. Thus $I-\gamma A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i}$ is nonexpansive for each $i=1,\ldots, N_{1}$ and $n\in\mathbb{N}$.

Now, we complete the proof by the next steps.

Step 1. $\{x_{n}\}$ is bounded.

Let $p\in\Theta$. Then $p=T_{r_{n}}^{F_{i}}p$ and $(I-\gamma A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i})p=p$. Thus we have

$$\begin{aligned} \Vert u_{i,n}-p\Vert &=\bigl\Vert T_{r_{n}}^{F } \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F } \bigr)A_{i}\bigr)x_{n}-T_{r_{n}}^{F_{i}}\bigl(I- \gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}\bigr)p\bigr\Vert \\ &\leq\bigl\Vert \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}\bigr)x_{n}-\bigl(I-\gamma A^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr)p\bigr\Vert \\ &\leq \Vert x_{n}-p\Vert . \end{aligned}$$

(3.2)

Let $B=\sum_{i=1}^{N_{2}}\gamma_{i} B_{i}$. Then B is a β-inverse strongly monotone mapping. Since $\{\lambda_{n}\}\subset(0,2\beta)$, $I-\lambda_{n} B$ is nonexpansive. Thus from (3.2), we have

$$\begin{aligned} \Vert y_{n}-p\Vert &= \Biggl\Vert P_{C}(I- \lambda_{n} B) \frac{1}{N_{1}} \sum_{i=1}^{N_{1}}u_{i,n}-P_{C}(I- \lambda_{n} B)p \Biggr\Vert \\ &\leq \Biggl\Vert (I-\lambda_{n} B)\frac{1}{N_{1}} \sum _{i=1}^{N_{1}}u_{i,n} -(I-\lambda_{n} B)p \Biggr\Vert \\ &\leq \Biggl\Vert \frac{1}{N_{1}} \sum_{i=1}^{N_{1}}u_{i,n}-p \Biggr\Vert \\ &\leq\frac{1}{N_{1}} \sum_{i=1}^{N_{1}} \Vert u_{i,n}-p\Vert \\ &\leq \Vert x_{n}-p\Vert . \end{aligned}$$

(3.3)

Thus from (3.3), it follows that

$$\begin{aligned} \Vert x_{n+1}-p\Vert &=\Biggl\Vert \alpha_{n}(v-p)+ \sum_{i=1}^{n}(\alpha_{i-1}-\alpha _{i}) (S_{i}y_{n}-S_{i}p) \Biggr\Vert \\ &\leq\alpha_{n}\Vert v-p\Vert + \sum_{i=1}^{n}( \alpha_{i-1}-\alpha_{i})\Vert y_{n}- p \Vert \\ &\leq\alpha_{n}\Vert v-p\Vert + \sum_{i=1}^{n}( \alpha_{i-1}-\alpha_{i})\Vert x_{n}- p \Vert \\ &=\alpha_{n}\Vert v-p\Vert + (1-\alpha_{n})\Vert x_{n}- p \Vert \\ &\leq \max\bigl\{ \Vert v-p\Vert ,\Vert x_{n}-p\Vert \bigr\} \end{aligned} $$

for all $n\in\mathbb{N}$, which implies that $\{x_{n}\}$ is bounded and so are $\{u_{i,n}\}$ ($i=1,\ldots,N_{1}$) and $\{y_{n}\}$.

Step 2. $\lim_{n\to\infty} \Vert x_{n+1}-x_{n}\Vert =0$ and $\lim_{n\to\infty }\Vert u_{i,n+1}-u_{i,n}\Vert =0$ for each $i=1,\ldots, N_{1}$.

Since the mappings $I-\gamma A^{*}(I-T_{r_{n}}^{F_{i}})A$ are nonexpansive, by Lemmas 2.2 and 2.3, we have

$$\begin{aligned} & \Vert u_{i,n+1}-u_{i,n}\Vert \\ &\quad=\bigl\Vert T_{r_{n+1}}^{F}\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n+1}}^{F_{i}}\bigr)A_{i}\bigr)x_{n+1}-T_{r_{n}}^{F} \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}\bigr)x_{n}\bigr\Vert \\ &\quad\leq\bigl\Vert \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n+1}}^{F_{i}} \bigr)A_{i}\bigr)x_{n+1}-\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr)x_{n} \bigr\Vert \\ &\qquad{}+\frac{\vert r_{n+1}-r_{n}\vert }{r_{n+1}}\bigl\Vert T_{r_{n+1}}^{F}\bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n+1}}^{F_{i}}\bigr)A_{i} \bigr)x_{n+1}-\bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n+1}}^{F_{i}} \bigr)A_{i}\bigr)x_{n+1}\bigr\Vert \\ &\quad\leq \Vert x_{n+1}-x_{n}\Vert +\bigl\Vert \bigl(I- \gamma A_{i}^{*}\bigl(I-T_{r_{n+1}}^{F_{i}} \bigr)A_{i}\bigr)x_{n}-\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr)x_{n} \bigr\Vert \\ &\qquad{}+\frac{\vert r_{n+1}-r_{n}\vert }{r_{n+1}}\delta_{n+1} \\ &\quad=\Vert x_{n+1}-x_{n}\Vert +\bigl\Vert \gamma A_{i}^{*}\bigl( T_{r_{n+1}}^{F_{i}} A_{i} x_{n}- T_{r_{n}}^{F_{i}} A_{i} x_{n} \bigr)\bigr\Vert +\frac{\vert r_{n+1}-r_{n}\vert }{r_{n+1}}\delta_{n+1} \\ &\quad\leq \Vert x_{n+1}-x_{n}\Vert + \gamma\bigl\Vert A_{i}^{*}\bigr\Vert \biggl[\frac {\vert r_{n+1}-r_{n}\vert }{r_{n+1}} \bigl\vert \bigl\langle T_{r_{n+1}}^{F_{i}} A_{i} x_{n}- T_{r_{n}}^{F_{i}} A_{i} x_{n}, T_{r_{n+1}}^{F_{i}} A_{i} x_{n}- A_{i} x_{n}\bigr\rangle \bigr\vert \biggr]^{\frac{1}{2}} \\ &\qquad{} +\frac{\vert r_{n+1}-r_{n}\vert }{r }\delta_{n+1} \\ &\quad\leq \Vert x_{n+1}-x_{n}\Vert + \gamma\bigl\Vert A_{i}^{*}\bigr\Vert \biggl[\frac{\vert r_{n+1}-r_{n}\vert }{r }\sigma_{n+1} \biggr]^{\frac{1}{2}} +\frac{\vert r_{n+1}-r_{n}\vert }{r }\delta _{n+1} \\ &\quad\leq \Vert x_{n+1}-x_{n}\Vert +\eta_{i,n+1}, \end{aligned}$$

(3.4)

where

$$\begin{aligned}& \sigma_{n+1}=\sup_{n\in\mathbb{N}} \bigl\vert \bigl\langle T_{r_{n+1}}^{F_{i}} A_{i} x_{n}- T_{r_{n}}^{F_{i}} A_{i} x_{n}, T_{r_{n+1}}^{F_{i}} A_{i} x_{n}- A_{i} x_{n}\bigr\rangle \bigr|, \\& \delta_{n+1}=\sup_{n\in\mathbb{N}}\bigl\Vert T_{r_{n+1}}^{F}\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n+1}}^{F_{i}}\bigr)A_{i}\bigr)x_{n+1}- \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n+1}}^{F_{i}} \bigr)A_{i}\bigr)x_{n+1}\bigr\Vert , \end{aligned}$$

and

$$\eta_{i,n+1}=\gamma\bigl\Vert A_{i}^{*}\bigr\Vert \biggl[ \frac{\vert r_{n+1}-r_{n}\vert }{r }\sigma _{n+1} \biggr]^{\frac{1}{2}}+\frac{\vert r_{n+1}-r_{n}\vert }{r } \delta_{n+1}. $$

Note that

$$ \begin{aligned}[b] & \Biggl\Vert (I-\lambda_{n+1}B)\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}u_{i,n+1} -(I- \lambda_{n}B)\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} \Biggr\Vert \\ &\quad= \Biggl\Vert (I-\lambda_{n+1}B)\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}u_{i,n+1} -(I-\lambda_{n+1}B) \frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} +(\lambda _{n}-\lambda_{n+1})Bw_{n} \Biggr\Vert \\ &\quad\leq \Biggl\Vert (I-\lambda_{n+1}B)\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}u_{i,n+1} -(I-\lambda_{n+1}B) \frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} \Biggr\Vert \\ &\qquad{} +\vert \lambda_{n}-\lambda_{n+1}\vert \Vert Bw_{n}\Vert \\ &\quad\leq\frac{1}{N_{1}}\sum_{i=1}^{N_{1}} \Vert u_{i,n+1}-u_{i,n}\Vert +\vert \lambda _{n}- \lambda_{n+1}\vert \Vert Bw_{n}\Vert , \end{aligned} $$

(3.5)

where $w_{n}=\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n}$. Let $M_{1}= \sup_{n\in\mathbb{N}}\Vert Bw_{n}\Vert$. By (3.1), (3.4), and (3.5), we have

$$ \begin{aligned}[b] \Vert y_{n+1}-y_{n}\Vert &= \Biggl\Vert P_{C}(I-\lambda_{n+1}B)\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}u_{i,n+1}-P_{C}(I- \lambda_{n}B)\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} \Biggr\Vert \\ &\leq \Biggl\Vert (I-\lambda_{n+1}B)\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}u_{i,n+1}-(I-\lambda_{n}B) \frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n} \Biggr\Vert \\ &\leq\frac{1}{N_{1}}\sum_{i=1}^{N_{1}} \Vert u_{i,n+1}-u_{i,n}\Vert +\vert \lambda _{n}- \lambda_{n+1}\vert \Vert Bw_{n}\Vert \\ &\leq \Vert x_{n+1}-x_{n}\Vert +\frac{1}{N_{1}}\sum _{i=1}^{N_{1}}\eta _{i,n+1}+\vert \lambda_{n}-\lambda_{n+1}\vert M_{1}. \end{aligned} $$

(3.6)

Since $\{\alpha_{n}\}$ is strictly decreasing, by using (3.6), we have

$$\begin{aligned} &\Vert x_{n+1}-x_{n}\Vert \\ &\quad=\Biggl\Vert (\alpha_{n}-\alpha_{n-1})v+\sum _{i=1}^{n-1}(\alpha_{i-1}-\alpha _{i}) (S_{i}y_{n}-S_{i}y_{n-1}) +(\alpha_{n-1}-\alpha_{n})S_{n}y_{n}\Biggr\Vert \\ &\quad\leq( \alpha_{n-1}-\alpha_{n} )\Vert v\Vert +\sum _{i=1}^{n-1}(\alpha _{i-1}- \alpha_{i})\Vert S_{i}y_{n}-S_{i}y_{n-1} \Vert +(\alpha_{n-1}-\alpha_{n})\Vert S_{n}y_{n} \Vert \\ &\quad\leq( \alpha_{n-1}-\alpha_{n} )\Vert v\Vert +\sum _{i=1}^{n-1}(\alpha _{i-1}- \alpha_{i})\Vert y_{n}- y_{n-1}\Vert +( \alpha_{n-1}-\alpha_{n})\Vert S_{n}y_{n} \Vert \\ &\quad= (\alpha_{n-1}-\alpha_{n})\Vert v\Vert + (1- \alpha_{n-1})\Vert y_{n}- y_{n-1}\Vert +( \alpha_{n-1}-\alpha_{n})\Vert S_{n}y_{n} \Vert \\ &\quad\leq(1-\alpha_{n-1})\Vert x_{n}-x_{n-1}\Vert + \frac{1}{N_{1}}\sum_{i=1}^{N_{1}}\eta _{i,n}+\vert \lambda_{n-1}-\lambda_{n}\vert M_{1}+(\alpha_{n-1}-\alpha_{n})M_{2}, \end{aligned} $$

where $M_{2}=\sup\{\Vert S_{n}y_{n}\Vert +\Vert v\Vert :n\in\mathbb{N}\}$. By (i) and (ii) and Lemma 2.6, we conclude that

$$ \lim_{n\to\infty} \Vert x_{n+1}-x_{n}\Vert =0. $$

(3.7)

Further, by (3.4) and (3.6), we have

$$ \lim_{n\to\infty} \Vert y_{n+1}-y_{n}\Vert =0,\qquad \lim_{n\to\infty} \Vert u_{i,n+1}-u_{i,n}\Vert =0, \quad i \in\{1,\ldots,N_{1}\}. $$

(3.8)

Step 3. $\lim_{n\to\infty} \Vert S_{i}x_{n}-x_{n}\Vert \to0$ for each $i\in \mathbb{N}$.

First, we show that $\lim_{n\to\infty} \Vert u_{i,n}-x_{n}\Vert =0$ for each $i\in \{1,\ldots,N_{1}\}$. Since each $A_{i}^{*}(I-T_{r_{n}}^{F_{i}})A_{i}$ is $\frac{1}{ 2L_{i}^{2}}$-inverse strongly monotone, by (3.1), we have

$$\begin{aligned} \Vert u_{i,n}-p\Vert ^{2} =&\bigl\Vert T_{r_{n}}^{F}\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr)x_{n}-T_{r_{n}}^{F} \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}\bigr)p\bigr\Vert ^{2} \\ \leq&\bigl\Vert \bigl(I-\gamma A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i}\bigr)x_{n}-\bigl(I-\gamma A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr)p\bigr\Vert ^{2} \\ =&\bigl\Vert (x_{n}-p)-\gamma\bigl(A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n}-A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} p\bigr)\bigr\Vert ^{2} \\ =&\Vert x_{n}-p\Vert ^{2}-2\gamma\bigl\langle x_{n}-p, A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n}-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A _{i}p\bigr\rangle \\ &{}+\gamma^{2}\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n}-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} p\bigr\Vert ^{2} \\ \leq&\Vert x_{n}-p\Vert ^{2}- \frac{ \gamma}{L_{i}^{2}}\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n}-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} p\bigr\Vert ^{2} \\ &{}+\gamma^{2}\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n}-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} p\bigr\Vert ^{2} \\ =&\Vert x_{n}-p\Vert ^{2}+\gamma\biggl(\gamma- \frac{ 1 }{L_{i}^{2}}\biggr)\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n}-A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} p\bigr\Vert ^{2} \\ =&\Vert x_{n}-p\Vert ^{2}+\gamma\biggl(\gamma- \frac{1 }{L_{i}^{2}}\biggr)\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n} \bigr\Vert ^{2}. \end{aligned}$$

(3.9)

From Lemma 2.4 and (3.9), it follows that

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} =&\Biggl\Vert \alpha_{n}(v-p)+\sum_{i=1}^{n}( \alpha_{i-1}-\alpha _{i}) (S_{i}y_{n}-p) \Biggr\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+\sum _{i=1}^{n}(\alpha_{i-1}-\alpha_{i}) \Vert S_{i}y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+\sum _{i=1}^{n}(\alpha_{i-1}-\alpha_{i}) \Vert y_{n}-p\Vert ^{2} \\ =&\alpha_{n}\Vert v-p\Vert ^{2}+ (1- \alpha_{n})\Vert y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+(1- \alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\Vert u_{i,n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}\\ &{}+ (1- \alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\biggl[\Vert x_{n}-p\Vert ^{2} +\gamma\biggl(\gamma-\frac{1 }{L_{i}^{2}}\biggr)\bigl\Vert A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n} \bigr\Vert ^{2}\biggr] \\ =&\alpha_{n}\Vert v-p\Vert ^{2}+(1-\alpha_{n}) \Vert x_{n}-p\Vert ^{2} \\ &{}+ (1-\alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\gamma\biggl(\gamma-\frac{1 }{L_{i}^{2}}\biggr)\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n} \bigr\Vert ^{2} \\ \leq& \alpha_{n}\Vert v-p\Vert ^{2}+ \Vert x_{n}-p\Vert ^{2} \\ &{}+ (1-\alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\gamma\biggl(\gamma-\frac{1 }{L_{i}^{2}}\biggr)\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n} \bigr\Vert ^{2}. \end{aligned}$$

Since $\gamma<\frac{1}{L^{2}}=\max\{\frac{1}{L_{1}^{2}},\ldots, \frac {1}{L_{N_{1}}^{2}}\}$, we have

$$\begin{aligned} &(1-\alpha_{n})\frac{1}{N_{1}}\gamma \biggl( \frac{1 }{L_{i}^{2}}-\gamma \biggr)\bigl\Vert A_{i}^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n} \bigr\Vert ^{2} \\ &\quad \leq(1-\alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\gamma \biggl( \frac{1 }{L_{i}^{2}}-\gamma \biggr)\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i} x_{n} \bigr\Vert ^{2} \\ &\quad \leq\alpha_{n}\Vert v-p\Vert ^{2}+\Vert x_{n}-p\Vert ^{2}-\Vert x_{n+1}-p\Vert ^{2} \\ &\quad \leq\alpha_{n}\Vert v-p\Vert ^{2}+\Vert x_{n}-x_{n+1}\Vert \bigl(\Vert x_{n}-p\Vert + \Vert x_{n+1}-p\Vert \bigr). \end{aligned} $$

Since $\alpha_{n}\to0$, by (3.7), we have

$$ \lim_{n\to\infty}\bigl\Vert A_{i}^{*}\bigl(I-T_{r_{n}}^{F_{i}} \bigr)A _{i}x_{n} \bigr\Vert =0 $$

(3.10)

for each $i\in\{1,\ldots, N_{1}\}$, which implies that

$$ \lim_{n\to\infty}\bigl\Vert \bigl(I-T_{r_{n}}^{F_{i}} \bigr)A_{i} x_{n} \bigr\Vert =0 $$

(3.11)

for each $i\in\{1,\ldots, N_{1}\}$. Since $T_{r_{n}}^{F }$ is firmly nonexpansive and $I-\gamma A_{i} ^{*}(I-T_{r_{n}}^{F_{i}})A_{i}$ is nonexpansive, by (3.1), we have

$$\begin{aligned} \Vert u_{i,n}-p \Vert ^{2} =& \bigl\Vert T_{r_{n}}^{F } \bigl(x_{n}+\gamma A_{i} ^{*}\bigl(T_{r_{n}}^{F_{i}}-I\bigr)A_{i} x_{n} \bigr)-T_{r_{n}}^{F }(p) \bigr\Vert ^{2} \\ \leq& \bigl\langle u_{i,n}-p,x_{n}+\gamma A _{i}^{*}\bigl(T_{r_{n}}^{i}-I\bigr)A_{i} x_{n}-p \bigr\rangle \\ =&\frac{1}{2} \bigl\{ \Vert u_{i,n}-p \Vert ^{2}+ \bigl\Vert x_{n}+\gamma A _{i} ^{*} \bigl(T_{r_{n}}^{F_{i}}-I\bigr)A _{i}x_{n}-p \bigr\Vert ^{2} \\ &{}- \bigl\Vert u_{i,n}-p-\bigl[x_{n}+\gamma A _{i}^{*}\bigl(T_{r_{n}}^{F_{i}}-I\bigr)A_{i} x_{n}-p\bigr] \bigr\Vert ^{2} \bigr\} \\ =&\frac{1}{2} \bigl\{ \Vert u_{i,n}-p \Vert ^{2}+ \bigl\Vert \bigl(I-\gamma A_{i} ^{*} \bigl(I-T_{r_{n}}^{F_{i}}\bigr)A_{i}\bigr) x_{n}-\bigl(I-\gamma A _{i} ^{*}\bigl(I-T_{r_{n}}^{F_{2}} \bigr)A_{i}\bigr)p \bigr\Vert ^{2} \\ &{}- \bigl\Vert u_{i,n}-x_{n}-\gamma A _{i}^{*} \bigl(T_{r_{n}}^{F_{i}}-I\bigr)A _{i}x_{n} \bigr\Vert ^{2} \bigr\} \\ \leq&\frac{1}{2} \bigl\{ \Vert u_{i,n}-p \Vert ^{2}+ \Vert x_{n}-p \Vert ^{2} - \bigl\Vert u_{i,n}-x_{n}-\gamma A _{i}^{*}\bigl(T_{r_{n}}^{F_{i}}-I \bigr)A_{i} x_{n} \bigr\Vert ^{2} \bigr\} \\ =&\frac{1}{2} \bigl\{ \Vert u_{i,n}-p \Vert ^{2}+ \Vert x_{n}-p \Vert ^{2}- \bigl[ \Vert u_{i,n}-x_{n} \Vert ^{2}+\gamma^{2} \bigl\Vert A_{i} ^{*}\bigl(T_{r_{n}}^{F_{i}}-I \bigr)A_{i} x_{n} \bigr\Vert ^{2} \\ &{} -2\gamma\bigl\langle u_{i,n}-x_{n},A_{i} ^{*} \bigl(T_{r_{n}}^{F_{i}}-I\bigr)A_{i} x_{n}\bigr\rangle \bigr] \bigr\} , \end{aligned}$$

which implies that

$$ \Vert u_{i,n}-p \Vert ^{2}\leq \Vert x_{n}-p \Vert ^{2}- \Vert u_{i,n}-x_{n} \Vert ^{2}+2\gamma \Vert u_{i,n}-x_{n} \Vert \bigl\Vert A_{i}^{*}\bigl(T_{r_{n}}^{F_{i}}-I \bigr)A_{i} x_{n}\bigr\Vert . $$

(3.12)

Now, from (3.1) and (3.12), it follows that

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} \leq& \alpha_{n}\Vert v-p\Vert ^{2}+ (1-\alpha_{n}) \Vert y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+(1- \alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}}\Vert u_{i,n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+ (1- \alpha_{n})\sum_{i=1}^{N_{1}} \frac{1}{N_{1}} \bigl(\Vert x_{n}-p \Vert ^{2}- \Vert u_{i,n}-x_{n} \Vert ^{2} \\ &{}+2\gamma \Vert u_{i,n}-x_{n} \Vert \bigl\Vert A_{i}^{*}\bigl(T_{r_{n}}^{F_{i}}-I\bigr)A_{i} x_{n}\bigr\Vert \bigr) \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+ \Vert x_{n}-p \Vert ^{2}- (1-\alpha_{n})\sum _{i=1}^{N_{1}}\frac {1}{N_{1}}\Vert u_{i,n}-x_{n} \Vert ^{2} \\ &{}+2\gamma\sum_{i=1}^{N_{1}}\frac{1}{N_{1}} \Vert u_{i,n}-x_{n} \Vert \bigl\Vert A_{i}^{*} \bigl(T_{r_{n}}^{F_{i}}-I\bigr)A_{i} x_{n}\bigr\Vert , \end{aligned}$$

and so

$$\begin{aligned} (1-\alpha_{n})\frac{1}{N_{1}} \Vert u_{i,n}-x_{n} \Vert ^{2} \leq&(1-\alpha_{n})\sum _{i=1}^{N_{1}}\frac{1}{N_{1}} \Vert u_{i,n}-x_{n}\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+ \Vert x_{n}-x_{n+1}\Vert \bigl(\Vert x_{n}-p\Vert + \Vert x_{n+1}-p\Vert \bigr) \\ &{}+ 2\gamma\sum_{i=1}^{N_{1}}\frac{1}{N_{1}} \bigl(\Vert u_{i,n}\Vert +\Vert x_{n} \Vert \bigr)\bigl\Vert A_{i}^{*}\bigl(T_{r_{n}}^{F_{i}}-I \bigr)A_{i} x_{n}\bigr\Vert ). \end{aligned}$$

Since $\alpha_{n}\to0$, both $\{u_{i,n}\}$ and $\{x_{n}\}$ are bounded, by (3.7) and (3.10), we have

$$ \lim_{n\to\infty} \Vert u_{i,n}-x_{n}\Vert =0 $$

(3.13)

for each $i\in\{1,\ldots,N_{1}\}$.

Next, we show that $\lim_{n\to\infty} \Vert y_{n}-u_{n}\Vert =0$, where $u_{n}=\frac {1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n}$. Note that $p={P_{C} (I-\lambda_{n}B)p}$. By (3.1), we have

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} \leq& \alpha_{n}\Vert v-p\Vert ^{2}+ (1-\alpha_{n}) \Vert y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}+(1- \alpha_{n}) \bigl\Vert u_{n}-p-\lambda_{n}(Bu_{n}-Bp) \bigr\Vert ^{2} \\ =&\alpha_{n}\Vert v-p\Vert ^{2}\\ &{}+(1-\alpha_{n}) \bigl(\Vert u_{n}-p\Vert ^{2} -2\lambda_{n}\langle u_{n}-p, Bu_{n}-Bp \rangle+ \lambda_{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2}\bigr) \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}\\ &{}+(1- \alpha_{n}) \bigl(\Vert u_{n}-p\Vert ^{2} -2\lambda_{n}\beta \Vert Bu_{n}-Bp \Vert ^{2}+ \lambda_{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2}\bigr) \\ \leq&\alpha_{n}\Vert v-p\Vert ^{2}\\ &{}+(1- \alpha_{n}) \bigl(\Vert x_{n}-p\Vert ^{2} -2\lambda_{n}\beta \Vert Bu_{n}-Bp \Vert ^{2}+ \lambda_{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2}\bigr) \\ =&\alpha_{n}\Vert v-p\Vert ^{2}+(1-\alpha_{n}) \Vert x_{n}-p\Vert ^{2} \\ &{} +(1-\alpha_{n}) \lambda_{n}( \lambda_{n}-2\beta) \Vert Bu_{n}-Bp \Vert ^{2} \end{aligned}$$

and so

$$\begin{aligned} &(1-\alpha_{n}) \lambda_{n}( 2\beta- \lambda_{n})\Vert Bu_{n}-Bp \Vert ^{2} \\ &\quad \leq\alpha_{n}\Vert v-p\Vert ^{2}+ \Vert x_{n}-x_{n+1}\Vert \bigl(\Vert x_{n}-p\Vert + \Vert x_{n+1}-p\Vert \bigr). \end{aligned} $$

Since $\alpha_{n}\to0$ and $0<\lim_{n\to\infty} \lambda_{n}=\lambda<2\beta$, by (3.7), we have

$$ \lim_{n\to\infty} \Vert Bu_{n}-Bp \Vert =0. $$

(3.14)

Since $P_{C}$ is firmly nonexpansive and $(I-\lambda_{n} B)$ is nonexpansive, by (3.1), we have

$$\begin{aligned} \Vert y_{n}-p\Vert ^{2} ={}&\bigl\Vert P_{C}(u_{n}-\lambda_{n}Bu_{n})-P_{C}(p- \lambda_{n}Bp)\bigr\Vert ^{2} \\ \leq{}&\bigl\langle y_{n}-p,u_{n}-\lambda_{n}Bu_{n}-(p- \lambda_{n}Bp)\bigr\rangle \\ ={}& \frac{1}{2} \bigl(\Vert y_{n}-p\Vert ^{2}+ \bigl\Vert (I-\lambda_{n} B)u_{n}-(I-\lambda_{n} B)p\bigr\Vert ^{2} -\bigl\Vert y_{n}-u_{n}+ \lambda_{n}(Bu_{n}-Bp)\bigr\Vert ^{2} \bigr) \\ \leq{}& \frac{1}{2} \bigl(\Vert y_{n}-p\Vert ^{2}+ \Vert u_{n}- p\Vert ^{2} -\bigl\Vert y_{n}-u_{n}+\lambda _{n}(Bu_{n}-Bp)\bigr\Vert ^{2} \bigr) \\ ={}&\frac{1}{2} \bigl(\Vert y_{n}-p\Vert ^{2}+ \Vert u_{n}- p\Vert ^{2} -\Vert y_{n}-u_{n} \Vert ^{2}-\lambda_{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2} \\ & {}-2\lambda_{n}\langle y_{n}-u_{n},Bu_{n}-Bp \rangle \bigr) \\ \leq{}&\frac{1}{2} \bigl(\Vert y_{n}-p\Vert ^{2}+ \Vert u_{n}- p\Vert ^{2} -\Vert y_{n}-u_{n} \Vert ^{2}-\lambda _{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2} \\ & {}+2\lambda_{n}\Vert y_{n}-u_{n}\Vert \Vert Bu_{n}-Bp\Vert \bigr) \end{aligned}$$

and so

$$ \begin{aligned}[b] \Vert y_{n}-p\Vert ^{2}\leq{}&\Vert u_{n}- p\Vert ^{2} -\Vert y_{n}-u_{n} \Vert ^{2}-\lambda_{n}^{2}\Vert Bu_{n}-Bp\Vert ^{2} \\ &{}+2\lambda_{n}\Vert y_{n}-u_{n}\Vert \Vert Bu_{n}-Bp\Vert \\ \leq{}&\Vert x_{n}- p\Vert ^{2} -\Vert y_{n}-u_{n}\Vert ^{2} +2\lambda_{n} \Vert y_{n}-u_{n}\Vert \Vert Bu_{n}-Bp\Vert . \end{aligned} $$

(3.15)

From (3.1) and (3.15), we have

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2}\leq{}& \alpha_{n}\Vert v-p\Vert ^{2}+ (1-\alpha_{n}) \Vert y_{n}-p\Vert ^{2} \\ \leq{}&\alpha_{n}\Vert v-p\Vert ^{2}\\ &{}+ (1- \alpha_{n}) \bigl( \Vert x_{n}- p\Vert ^{2} - \Vert y_{n}-u_{n}\Vert ^{2} +2\lambda_{n}\Vert y_{n}-u_{n}\Vert \Vert Bu_{n}-Bp\Vert \bigr) \\ \leq{}&\alpha_{n}\Vert v-p\Vert ^{2}+\Vert x_{n}-p\Vert ^{2}- (1-\alpha_{n})\Vert y_{n}-u_{n}\Vert ^{2} \\ &{}+2 (1-\alpha_{n})\lambda_{n}\Vert y_{n}-u_{n} \Vert \Vert Bu_{n}-Bp\Vert . \end{aligned} $$

Therefore, we have

$$\begin{aligned} (1-\alpha_{n})\Vert y_{n}-u_{n}\Vert ^{2}\leq {}&\alpha_{n}\Vert v-p\Vert ^{2}+\Vert x_{n}-x_{n+1}\Vert \bigl(\Vert x_{n+1}-p\Vert + \Vert x_{n}-p\Vert \bigr) \\ &{}+2 (1-\alpha_{n})\lambda_{n}\bigl(\Vert y_{n} \Vert +\Vert u_{n}\Vert \bigr)\Vert Bu_{n}-Bp\Vert . \end{aligned} $$

Since $\lim_{n\to\infty}\alpha_{n}=0$ and both $\{y_{n}\}$ and $\{u_{n}\}$ are bounded, by (3.7) and (3.14), we have

$$ \lim_{n\to\infty} \Vert y_{n}-u_{n}\Vert =0. $$

(3.16)

Further, from (3.7), (3.13), (3.16), and

$$\begin{aligned} \Vert x_{n+1}-y_{n}\Vert &\leq \Vert x_{n+1}-x_{n}\Vert +\Vert x_{n}-u_{n} \Vert +\Vert u_{n}-y_{n}\Vert \\ &\leq \Vert x_{n+1}-x_{n}\Vert +\sum _{i=1}^{N_{1}}\frac{1}{N_{1}}\Vert x_{n}-u_{i,n}\Vert +\Vert u_{n}-y_{n} \Vert , \end{aligned} $$

it follows that

$$ \lim_{n\to\infty} \Vert x_{n+1}-y_{n}\Vert =0. $$

(3.17)

Now, from (3.1), it follows that

$$ \sum_{i=1}^{n}(\alpha_{i-1}- \alpha_{i}) (S_{i}y_{n}-y_{n})=x_{n+1}-y_{n}- \alpha _{n}(v-y_{n}). $$

(3.18)

Since $\{\alpha_{n}\}$ is strictly decreasing, for each $i\in\mathbb{N}$, by (2.1) and (3.18), we have

$$\begin{aligned} (\alpha_{i-1}-\alpha_{i})\Vert S_{i}y_{n}-y_{n}\Vert ^{2}&\leq\sum _{i=1}^{n}(\alpha _{i-1}- \alpha_{i})\Vert S_{i}y_{n}-y_{n} \Vert ^{2} \\ &\leq2\sum_{i=1}^{n}(\alpha_{i-1}- \alpha_{i})\langle S_{i}y_{n}-y_{n}, p-y_{n}\rangle \\ &=2\langle x_{n+1}-y_{n},y_{n}-p\rangle-2 \alpha_{n}\langle v-y_{n},p-y_{n} \rangle \\ &\leq2\Vert x_{n+1}-y_{n}\Vert \Vert y_{n}-p \Vert +2\alpha_{n}\Vert v-y_{n}\Vert \Vert y_{n}-p \Vert . \end{aligned}$$

Since $\lim_{n\to\infty}\alpha_{n}=0$ and $\{y_{n}\}$ is bounded, by (3.17), one has

$$ \lim_{n\to\infty} \Vert S_{i}y_{n}-y_{n} \Vert =0 $$

(3.19)

for all $i\in\mathbb{N}$. Further, since

$$\begin{aligned} \Vert S_{i}x_{n}-x_{n}\Vert &\leq \Vert S_{i}x_{n}-S_{i}y_{n}\Vert + \Vert S_{i}y_{n}-y_{n}\Vert +\Vert y_{n}-x_{n}\Vert \\ &\leq2\Vert y_{n}-x_{n}\Vert +\Vert S_{i}y_{n}-y_{n}\Vert \\ &\leq2\Vert y_{n}-x_{n+1}\Vert +2\Vert x_{n+1}-x_{n}\Vert +\Vert S_{i}y_{n}-y_{n} \Vert , \end{aligned} $$

by (3.7), (3.17), and (3.19), we obtain

$$ \lim_{n\to\infty} \Vert S_{i}x_{n}-x_{n} \Vert =0 $$

(3.20)

for all $i\in\mathbb{N}$.

Step 4. $\limsup_{n\to\infty}\langle v-z,x_{n}-z\rangle\leq0$.

Let $z=P_{\Theta}v$. Since $\{x_{n}\}$ is bounded, we can choose a subsequence $\{x_{n_{j}}\}$ of $\{x_{n}\}$ such that

$$\limsup_{n\to\infty}\langle v-z,x_{n}-z\rangle=\lim _{j\to\infty}\langle v-z,x_{n_{j}}-z \rangle. $$

Since $\{x_{n_{j}}\}$ is bounded, there exists a subsequence $\{ x_{n_{j_{i}}}\}$ of $\{x_{n_{j}}\}$ converging weakly to a point $w\in C$. Without loss of generality, we can assume that $x_{n_{j}}\rightharpoonup w$.

Now, we show that $w\in\Theta$. First of all, we prove that $w\in \Gamma=\bigcap_{i=1}^{\infty} \operatorname {Fix}(S_{i})$. In fact, since $x_{n}-S_{i}x_{n}\to0$ for each $i\in\mathbb{N}$ and $x_{n_{j}}\rightharpoonup w$, by Lemma 2.5, we obtain $w\in\bigcap_{i=1}^{\infty} \operatorname {Fix}(S_{i})=\Gamma$.

Next, we show that $w\in\Omega$, i.e., $w\in \operatorname {EP}(F )$ and $A_{i}w\in \operatorname {EP}(F_{i})$ for each $i=1,\ldots, N_{1}$.

Let $w_{i,n}=(I-A_{i}^{*}(I-T_{r_{n}}^{F_{i}}))A_{i}x_{n}$ for each $i=1,\ldots, N_{1}$. By (3.10) and (3.13) we see that $w_{i,n}-x_{n}\to0$ and $T_{r_{n}}^{F}w_{i,n}-w_{i,n}\to0$ as $n\to\infty$. By Lemma 2.2 we see that $\Vert T_{r_{n}}^{F}w_{i,n}-T_{r}^{F}w_{i,n}\Vert \leq \vert 1-\frac{r}{r_{n}}\vert \Vert T_{r_{n}}^{F}w_{i,n}-w_{i,n}\Vert \to0$ as $n\to\infty$. Hence $T_{r}^{F}w_{i,n}-w_{i,n}\to0$ as $n\to\infty$ for each $i=1,\ldots,N_{1}$. Since $T_{r}^{F}$ is non-expansive and $\{w_{i,n}\}$ converges weakly to w, by Lemma 2.5 we get $w=T_{r}^{F} w$, i.e., $w\in \operatorname {EP}(F)$. On the other hand, since $(I-\gamma A^{*}_{i}(I-T_{r_{n}}^{F_{i}})A_{i} )x_{n}-x_{n}\to0$ (by (3.13)) and $I-\gamma A^{*}_{i}(I-T_{r_{n}}^{F_{i}})A_{i}$ is non-expansive, from Lemmas 2.2 and 2.5 it follows that $w=(I-\gamma A^{*}_{i}(I-T_{r}^{F_{i}})A_{i} )w$, i.e., $w=T_{r}^{F} A_{i}w$. Therefore, $w\in\Omega$.

Finally, we prove that $w\in \operatorname {VI}=\bigcap_{i=1}^{N_{2}} \operatorname {VI}(C,B_{i})$ by demiclosedness principle. Obviously, we only need to show that $w=P_{C}(w-\lambda B_{i}w)$, where $\lambda=\lim_{n\to\infty}\lambda_{n}$. By (3.1) and (3.16), one has $\Vert u_{n}-P_{C}(I-\lambda_{n} B)u_{n}\Vert \to0$, where $u_{n}=\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}u_{i,n}$. Then we have

$$\begin{aligned} \bigl\Vert u_{n}-P_{C}(I-\lambda B)u_{n}\bigr\Vert &\leq\bigl\Vert u_{n}-P_{C}(I- \lambda_{n} B)u_{n}\bigr\Vert +\bigl\Vert P_{C}(I-\lambda_{n} B)u_{n}-P_{C}(I- \lambda B)u_{n}\bigr\Vert \\ &\leq\bigl\Vert u_{n}-P_{C}(I-\lambda_{n} B)u_{n}\bigr\Vert +\bigl\Vert (I-\lambda_{n} B)u_{n}- (I-\lambda B)u_{n}\bigr\Vert \\ &\leq\bigl\Vert u_{n}-P_{C}(I-\lambda_{n} B)u_{n}\bigr\Vert + \vert \lambda-\lambda_{n}\vert \Vert Bu_{n}\Vert . \end{aligned} $$

Since $\lambda_{n}\to\lambda>0$, $\{Bu_{n}\}$ is bounded and $\Vert u_{n}-P_{C}(I-\lambda B)u_{n}\Vert \to0$, we have

$$\lim_{n\to\infty}\bigl\Vert u_{n}-P_{C}(I- \lambda B)u_{n}\bigr\Vert =0. $$

On the other hand, since $\{\lambda_{n}\}\subset(0,2\beta)$, one has $\lambda\in(0,2\beta]$. Thus $I-\lambda B$ is nonexpansive and, further, $P_{C}(I-\lambda B)$ is nonexpansive. Noting that $u_{n_{j}}\rightharpoonup w$ as $j\to\infty$, by Lemma 2.5, we obtain $w=P_{C}(I-\lambda B)w$. By Lemma 2.8, we get $w\in \operatorname {VI}=\bigcap_{i=1}^{N_{2}}\operatorname {VI}(C,B_{i})$. Therefore, $w\in\Theta$. By the property on $P_{C}$, we have

$$ \limsup_{n\to\infty}\langle v-z,x_{n}-z\rangle=\lim _{j\to\infty}\langle v-z,x_{n_{j}}-z\rangle=\langle v-z,w-z \rangle\leq0. $$

(3.21)

Step 5. $x_{n}\to z=P_{\Theta}v$ as $n\to\infty$.

By (3.1), we have

$$\begin{aligned} \Vert x_{n+1}-z\Vert ^{2}&=\Biggl\Vert \alpha_{n} v+\sum_{i=1}^{n}( \alpha_{i-1}-\alpha _{i})S_{i}y_{n}-z \Biggr\Vert ^{2} \\ &=\alpha_{n}\langle v-z,x_{n+1}-z\rangle+ \sum _{i=1}^{n}(\alpha_{i-1}-\alpha _{i})\langle S_{i}y_{n}-z,x_{n+1}-z \rangle \\ &\leq\alpha_{n}\langle v-z,x_{n+1}-z\rangle+ \frac{\sum_{i=1}^{n}(\alpha _{i-1}-\alpha_{i})}{2}\bigl(\Vert S_{i}y_{n}-z\Vert ^{2}+\Vert x_{n+1}-z\Vert ^{2}\bigr) \\ &\leq \alpha_{n}\langle v-z,x_{n+1}-z\rangle+ \frac{\sum_{i=1}^{n}(\alpha _{i-1}-\alpha_{i})}{2}\bigl(\Vert x_{n}-z\Vert ^{2}+\Vert x_{n+1}-z\Vert ^{2}\bigr) \\ &= \alpha_{n}\langle v-z,x_{n+1}-z\rangle+\frac{1-\alpha_{n} }{2} \bigl( \Vert x_{n}-z\Vert ^{2}+\Vert x_{n+1}-z \Vert ^{2}\bigr) \\ &\leq\alpha_{n}\langle v-z,x_{n+1}-z\rangle+ \frac{1-\alpha_{n} }{2} \Vert x_{n}-z\Vert ^{2}+ \frac{1 }{2}\Vert x_{n+1}-z\Vert ^{2}, \end{aligned} $$

which implies that

$$\Vert x_{n+1}-z\Vert ^{2}\leq(1-\alpha_{n}) \Vert x_{n}-z\Vert ^{2}+2\alpha_{n}\langle v-z,x_{n+1}-z\rangle. $$

By Lemma 2.6 and (3.21), we can conclude that $\lim_{n\to\infty} \Vert x_{n}-z\Vert =0$. This completes the proof. □

The following results follow directly from Theorem 3.1.

Corollary 3.2

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and $C\subset H_{1}$, $Q\subset H_{2}$ be nonempty closed convex subsets. Let $A: H_{1}\to H_{2}$ be a bounded linear operator and $B: C\to H_{1}$ be a β-inverse strongly monotone operator. Assume that $F :C\times C\to\mathbb{R}$, $F_{1}: Q\times Q\to\mathbb{R}$ are bifunctions satisfying the conditions (A1)-(A4). Let $\{S_{n}\}$ be countable family of nonexpansive mappings from C into C. Assume that $\Theta =\Gamma\cap\Omega\cap \operatorname {VI}(C,B)\neq\emptyset$, where $\Gamma=\bigcap_{n=1}^{\infty} \operatorname {Fix}(S_{n})$ and $\Omega=\{z\in C: z\in \operatorname {EP}(F)\ \textit{and}\ Az\in \operatorname {EP}(F_{1}) \}$. Take $v\in C$ arbitrarily and define an iterative scheme in the following manner:

$$ \textstyle\begin{cases} u_{n}=T_{r_{n}}^{F}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{1}})A )x_{n},\\ y_{n}=P_{C}(u_{n}-\lambda_{n} Bu_{n}),\\ x_{n+1}=\alpha_{n} v+ \sum_{i=1}^{n}(\alpha_{i-1}-\alpha_{i}) S_{i} y_{n}, \end{cases} $$

(3.22)

for all $n\in\mathbb{N}$, where $\{r_{n}\}\subset(r,\infty)$ with $r>0$, $\{\lambda_{n}\}\subset(0,2\beta)$, and $\gamma\subset(0,1/L^{2}]$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A, $\alpha_{0}=1$, and $\{\alpha_{n}\}\subset(0,1)$ is a strictly decreasing sequence. Assume that the control sequences $\{\alpha_{n}\}$, $\{\lambda_{n}\}$, and $\{r_{n}\}$ satisfy the following conditions:

• (1) $\lim_{n\to\infty}\alpha_{n}=0$ and $\sum_{n=1}^{\infty}\alpha _{n}=\infty$;

• (2) $\sum_{n=1}^{\infty} \vert r_{n+1}-r_{n}\vert <\infty$ and $\sum_{n=1}^{\infty} \vert \lambda_{n+1}-\lambda_{n}\vert <\infty$;

• (3) $\lim_{n\to\infty} \lambda_{n}=\lambda\in(0,2\beta)$.

Then the sequence $\{x_{n}\}$ defined by (3.22) converges strongly to a point $z=P_{\Theta}v$.

Corollary 3.3

Let $H_{1}$, $H_{2}$ be two real Hilbert spaces and $C\subset H_{1}$, $Q\subset H_{2}$ be nonempty closed convex subsets. Let $A: H_{1}\to H_{2}$ be a bounded linear operator and $B: C\to H_{1}$ be a β-inverse strongly monotone operator. Assume that $F :C\times C\to\mathbb{R}$, $F_{1}: Q\times Q\to\mathbb{R}$ are the bifunctions satisfying the conditions (A1)-(A4). Let $S: C\to C$ be a nonexpansive mapping. Assume that $\Theta=\operatorname {Fix}(S) \cap\Omega\cap \operatorname {VI}(C,B)\neq\emptyset$, where $\Omega=\{z\in C: z\in \operatorname {EP}(F)\ \textit{and}\ Az\in \operatorname {EP}(F_{1}) \}$. Take $v\in C$ arbitrarily and define an iterative scheme in the following manner:

$$ \textstyle\begin{cases} u_{n}=T_{r_{n}}^{F}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{1}})A )x_{n},\\ y_{n}=P_{C}(u_{n}-\lambda_{n} Bu_{n}),\\ x_{n+1}=\alpha_{n} v+ (1-\alpha_{n}) S y_{n} \end{cases} $$

(3.23)

for all $n\in\mathbb{N}$, where $\{r_{n}\}\subset(r,\infty)$ with $r>0$, $\{\lambda_{n}\}\subset(0,2\beta)$, and $\gamma\subset(0,1/L^{2}]$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A, $\{\alpha_{n}\}\subset(0,1)$ is a sequence. Assume that the control sequences $\{\alpha_{n}\}$, $\{\lambda_{n}\}$, and $\{r_{n}\}$ satisfy the following conditions:

• (1) $\lim_{n\to\infty}\alpha_{n}=0$ and $\sum_{n=1}^{\infty}\alpha _{n}=\infty$;

• (2) $\sum_{n=1}^{\infty} \vert r_{n+1}-r_{n}\vert <\infty$ and $\sum_{n=1}^{\infty} \vert \lambda_{n+1}-\lambda_{n}\vert <\infty$;

• (3) $\lim_{n\to\infty} \lambda_{n}=\lambda\in(0,2\beta)$.

Then the sequence $\{x_{n}\}$ defined by (3.23) converges strongly to a point $z=P_{\Theta}v$.

Remark 3.4

Theorem 3.1 and Corollary 3.3 extend the corresponding one of Kazmi and Rizvi [14] from a nonexpansive mapping to a finite of family of nonexpansive mappings and from a split equilibrium problem to a finite of family of split equilibrium problems. It is a little simple to prove that $w\in \operatorname {VI}$ by the demiclosedness principle in Theorem 3.1.

We give an example to illustrate Theorem 3.1 as follows.

Example 3.5

Let $H_{1}=\mathbb{R}$ and $H_{2}=\mathbb{R}^{2}$, $C=[0,1]$, and $Q=[0,1]\times[0,1]$. Let $A_{1}: H_{1}\to H_{2}$ and $A_{2}: H_{1}\to H_{2}$ defined by $A_{1}x=(x,x)^{T}$ and $A_{2}x=(\frac{x}{2},\frac {x}{2})^{T}$ for each $x\in H_{1}$. Then $A_{1}^{*}y=y_{1}+y_{2}$ and $A_{2}^{*}y=\frac{y_{1}+y_{2}}{2}$ for each $y=(y_{1},y_{2})^{T}\in H_{2}$. Then $L_{1}=2$ and $L_{2}=\frac{1}{2}$, where $L_{1}$ and $L_{2}$ are the spectral radius of $A_{1}^{*}A_{1}$ and $A_{2}^{*}A_{2}$, respectively.

Let $B_{1}= 2(x-1)$ and $B_{2}= -4$ for all $x\in C$. Then it is easy to see that $B_{1}$ and $B_{2}$ are $\frac{1}{2}$ and 1-inverse strongly monotone operators from C into $H_{1}$. Find that $\operatorname {VI}=\operatorname {VI}(C,B_{1})\cap \operatorname {VI}(C,B_{2})=\{1\}$. For each $n\in\mathbb{N}$, let $S_{n}:C\to C$ defined by $S_{n}(x)=x+\frac{1}{3n}$ for each $x\in[0,\frac{1}{2}]$ and $S_{n}(x)=x$ for each $x\in(\frac{1}{2},1]$. Then $\{S_{n}\}$ is a countable family of nonexpansive mappings from C into C and it is easy to see that $\Gamma=\bigcap_{n=1}^{\infty} \operatorname {Fix}(S_{n})=(\frac{1}{2},1]$. For each $x,y\in C$, define the bifunction $F: C\times C\to\mathbb{R}$ by $F(x,y)=x-y$ for all $x,y\in C$. For each $u=(u_{1},u_{2})^{T}$ and $v=(v_{1},v_{2})^{T}\in Q$, define $F_{1}: Q\times Q\to\mathbb{R}$ and $F_{2}:Q\times Q\to\mathbb{R}$ by

$$F_{1}(u,v)=u_{1}+u_{2}-v_{1}-v_{2} $$

and

$$F_{2}(u,v)= \textstyle\begin{cases} 0, & \mbox{if } u=v,\\ 2 , & \mbox{if } u=(1,1) \mbox{ or } (\frac{1}{2},\frac{1}{2} ) \mbox{ and } v\neq(1,1) \mbox{ or } (\frac{1}{2},\frac{1}{2} ),\\ -2 ,& \mbox{if } v=(1,1) \mbox{ or } (\frac{1}{2},\frac{1}{2} ) \mbox{ and } u\neq(1,1) \mbox{ or } (\frac{1}{2},\frac{1}{2} ),\\ u_{1}^{2}+u_{2}^{2}-v_{1}-v_{2}, & \mbox{otherwise}. \end{cases} $$

It is easy to check that the bifunctions F, $F_{1}$, and $F_{2}$ satisfy the conditions (A1)-(A4) and $F_{1}$. Moreover, $\Omega=\{1\}$, where $\Omega=\{z\in C: z\in \operatorname {EP}(F), A_{1}z\in \operatorname {EP}(F_{1}) \mbox{ and } A_{2}z\in \operatorname {EP}(F_{2}) \}$. Therefore, $\Theta=\Gamma\cap \mathrm{VI}\cap \Omega=\{1\}$.

Let $\alpha_{0}=1$, $\gamma_{1}=\gamma_{2}=\frac{1}{2}$, and $\gamma=\frac {1}{4}$. For each $n\in\mathbb{N}$, let $r_{n}= 2$, $\lambda_{n}=\frac {1}{4}$, $\alpha_{n}=\frac{1}{n}$. Then the sequences $\{\alpha_{n}\}$, $\{ \lambda_{n}\}$, $\{r_{n}\}$ satisfy the conditions (1)-(3) in Theorem 3.1.

For each $x\in C$ and each $n\in\mathbb{N}$, we compute $T_{r_{n}}^{F_{1}}A_{1}x$, i.e., $T_{r_{n}}^{F_{1}}(x,x)$. Find $z=(1,1)$ such that

$$\begin{aligned} F_{1}(z,y)+\frac{1}{r_{n}}\langle y-z,z-A_{1}x \rangle&=2-(y_{1}+y_{2})+ \frac {1}{2} \bigl[(y_{1}-1) (1-x)+(y_{2}-1) (1-x)\bigr] \\ &=2-(y_{1}+y_{2})+ \frac{1}{2}(1-x) ( y_{1} + y_{2}-2) \\ &=\bigl[2-(y_{1}+y_{2})\bigr] \biggl[1- \frac{1}{2}(1-x) \biggr] \\ &\geq0 \end{aligned} $$

for all $y=(y_{1},y_{2})\in Q$. Thus, from Lemma 2.1(1), it follows that $T_{r_{n}}^{F_{1}}A_{1}x=(1,1)$ for each $x\in C$. Similarly, for each $x\in [0,1]$, we can find $z=(1,1)$ such that, for $y=(\frac{1}{2}, \frac{1}{2})$,

$$\begin{aligned} F_{2}(z,y)+\frac{1}{r_{n}}\langle y-z,z-A_{2}x \rangle&=1-\frac {1}{2}(1-x)=\frac{1 }{2}+\frac{x}{4}\geq0; \end{aligned} $$

for $y=(1, 1)$,

$$F_{2}(z,y)+\frac{1}{r_{n}}\langle y-z,z-A_{2}x\rangle=0; $$

for $y\in Q\setminus\{(1,1),(\frac{1}{2},\frac{1}{2})\}$,

$$F_{2}(z,y)+\frac{1}{r_{n}}\langle y-z,z-A_{2}x\rangle=2+ \frac{1}{2} \biggl[(y_{1}-1) \biggl(1-\frac{x}{2} \biggr)+(y_{2}-1) \biggl(1-\frac{x}{2} \biggr) \biggr] \geq0. $$

Thus $z=(1,1)=T_{r_{n}}^{F_{2}}A_{2}x$ for all $x\in C$ by Lemma 2.1(1).

Now, take $v=\frac{1}{2}$ and $x_{1}=\frac{1}{4}$ and define the sequence $\{x_{n}\}$ defined by (3.1). Since each $x_{n}\in C$, from the statement above we get $T_{r_{n}}^{F_{i}}A_{i}x_{n}=(1,1)$ for each $i=1,2$. Furthermore, we can get

$$\begin{aligned} \bigl(I-\gamma A_{1}^{*}\bigl(I-T_{r_{n}}^{F_{1}} \bigr)A_{1}\bigr) x_{n}&=\bigl(x_{n}-\gamma A_{1}^{*}\bigl(A_{1}x_{n}-T_{r_{n}}^{F_{1}}A_{1}x_{n} \bigr)\bigr) \\ &=\bigl(x_{n}-\gamma A_{1}^{*}\bigl((x_{n},x_{n})-(1,1) \bigr)\bigr) \\ &=x_{n}-2\gamma( x_{n}-1) \\ &=\frac{1+x_{n}}{2}. \end{aligned} $$

Note that

$$\begin{aligned} F(1,y)+\frac{1}{r_{n}} \biggl\langle y-z,z-\frac{1+x_{n}}{2} \biggr\rangle &=1-y+\frac{1}{2}(y-1) \biggl(1-\frac{1+x_{n}}{2} \biggr) \\ &=(1-y) \biggl(1-\frac{1}{2} \biggl(1-\frac{1+x_{n}}{2} \biggr) \biggr) \\ &=(1-y) \biggl( \frac{1}{2}+\frac{1+x_{n}}{4} \biggr) \\ &\geq0 \end{aligned} $$

for all $y\in C$. Thus $u_{1,n}=1$ by Lemma 2.1(1) for each $n\in \mathbb{N}$. Similarly, we can conclude that $u_{2,n}=1$ for each $n\in \mathbb{N}$.

Next, we compute the sequence $\{y_{n}\}$. By the definition of $\{y_{n}\}$, we see that

$$\begin{aligned} y_{n}&=P_{C} \biggl[ \biggl(I- \lambda_{n}\frac{B_{1}+B_{2}}{2} \biggr)\frac {u_{1,n}+u_{2,n}}{2} \biggr] \\ &= P_{C} \biggl(1+\frac{2}{4} \biggr) =1 \end{aligned} $$

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

Finally, we compute the sequence $\{x_{n}\}$ by the following iteration:

$$\begin{aligned} x_{n+1}&=\alpha_{n}v+\sum _{i=1}^{n}(\alpha_{i-1}-\alpha_{i})S_{i}y_{n} \\ &=\alpha_{n}v+1-\alpha_{n} \\ &= 1-\frac{1}{2n} \\ &\to1=P_{\Theta}v=P_{\{1\}}\frac{1}{2} \end{aligned} $$

as $n\to\infty$ as shown by Theorem 3.1.

4 Applications

In this section, let $H_{1}$, $H_{2}$ be two real Hilbert spaces and C, Q be two nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $f: C\to \mathbb{R}$, $g: Q\to \mathbb{R}$ be two operators and $A: H_{1}\to H_{2}$ be a bounded linear operator.

We consider the following optimization problem:

$$ \begin{aligned} \mbox{find } x^{*}\in C& \mbox{ such that } f\bigl(x^{*}\bigr)\leq f(x),\quad \forall x\in C, \\ &\mbox{ and } y^{*}=Ax^{*} \mbox{ such that } g\bigl(y^{*}\bigr)\leq g(y), \quad \forall y\in Q. \end{aligned} $$

(4.1)

We denote the set of solutions of (4.1) by Θ and assume that $\Theta\neq\emptyset$. Let $F(x,y)=f(y)-f(x)$ for all $x,y\in C$ and $F_{1}(x,y)=g(y)-g(x)$ for all $x,y\in Q$. Then $F(x,y)$ and $G(x,y)$ satisfy the conditions (A1)-(A4) in Section 2 provided that f is convex and lower semicontinuous on C and g is convex and lower semicontinuous on Q. Let $\Omega=\{z\in C: z\in \operatorname {EP}(F)\mbox{ and }Az\in \operatorname {EP}(F_{1}) \}$. Obviously, $\Theta=\Omega$.

By Corollary 3.3 with $B=I$ and $S=I$, we have the following iterative algorithm, which strongly converges to a point $z=P_{\Theta}v$, which solves the optimization problem (4.1):

$$ \textstyle\begin{cases} u_{n}=T_{r_{n}}^{F}(I-\gamma A^{*}(I-T_{r_{n}}^{F_{1}})A )x_{n},\\ y_{n}=P_{C}(u_{n}-\lambda_{n}u_{n}),\\ x_{n+1}=\alpha_{n} v+ (1-\alpha_{n})y_{n}, \end{cases} $$

(4.2)

where $\{r_{n}\}\subset(r,\infty)$ with $r>0$, $\{\lambda_{n}\}\subset (0,2)$, and $\gamma\subset(0,1/L^{2}]$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A, $\{\alpha_{n}\} \subset(0,1)$ is a sequence. Assume that the control sequences $\{\alpha _{n}\}$, $\{\lambda_{n}\}$, and $\{r_{n}\}$ satisfy the following conditions:

• (1) $\lim_{n\to\infty}\alpha_{n}=0$ and $\sum_{n=1}^{\infty}\alpha _{n}=\infty$;

• (2) $\sum_{n=1}^{\infty} \vert r_{n+1}-r_{n}\vert <\infty$, $\sum_{n=1}^{\infty} \vert \alpha_{n+1}-\alpha_{n}\vert <\infty$, and $\sum_{n=1}^{\infty} \vert \lambda _{n+1}-\lambda_{n}\vert <\infty$;

• (3) $\lim_{n\to\infty} \lambda_{n}=\lambda\in(0,2)$.

For the special case with $H_{1}=H_{2}$ and $C=Q$, we consider the following multi-objective optimization problem:

$$ \textstyle\begin{cases} \min\{f(x), g(x)\},\\ x\in C. \end{cases} $$

(4.3)

We denote the set of solution of (4.3) by Γ and assume that $\Gamma\neq\emptyset$. In (4.2), setting $A=I$ we get the following algorithm, which strongly converges to the solution of multi-objective optimization problem (4.3):

$$\textstyle\begin{cases} u_{n}=T_{r_{n}}^{F}(I-\gamma(I-T_{r_{n}}^{F_{1}}))x_{n},\\ y_{n}=P_{C}(u_{n}-\lambda_{n}u_{n}),\\ x_{n+1}=\alpha_{n} v+ (1-\alpha_{n})y_{n}, \end{cases} $$

where $\gamma\subset(0,1/L^{2}]$, L is the spectral radius of the operator $I^{*}I$ and $I^{*}$ is the adjoint of I, other parameters such as $\{\alpha_{n}\}$, $\{\lambda_{n}\}$, and $\{r_{n}\}$ satisfy the same conditions (1)-(3).

5 Conclusion

In this paper, we construct an iterative algorithm to find a common element of the set of solutions of a finite family of split equilibrium problems and the set of common fixed points of a countable family of nonexpansive mappings in Hilbert spaces. In the proof methods, we use the inverse strong monotonicity of each $A^{*}(I-T_{r_{n}})A$, which is such that the proof is simple and is different from the ones given in [14–16]. Also, in the results of this paper, we do not assume that each $F_{i}$ is upper semi-continuous in the first argument for each $i=1,\ldots,N_{1}$, which is required in the result in [14–16]. As an application, we solve an optimization problem by the result of this paper.