A new relaxed extragradient-like algorithm for approaching common solutions of generalized mixed equilibrium problems, a more general system of variational inequalities and a fixed point problem

Abstract

In this paper, we introduce a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.

1 Introduction

In this paper, we assume that H is a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and the induced norm $\parallel \cdot \parallel$ and C is a nonempty closed convex subset of H. $P_C$ denotes the metric projection of H onto C and $\mathcal{F}(T)$ denotes the fixed points set of a mapping T. The sequence $\{x_n\}$ converges weakly to x which is denoted by $x_n\rightharpoonup x$.

Let $\phi :C\to \mathbb{R}$ be a real-valued function and let $A:C\to H$ be a nonlinear mapping. Suppose that $F:C\times C\to \mathbb{R}$ is a bifunction.

The generalized mixed equilibrium problem is to find $x\in C$ (see, e.g., [1–6]) such that

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

(1.1)

The set of solutions of (1.1) is denoted by $GMEP(F,\phi ,A)$.

If $\phi \equiv 0$, then problem (1.1) reduces to the generalized equilibrium problem, which is to find $x\in C$ (see, e.g., [7–9]) such that

$$F(x,y)+〈Ax,y-x〉\ge 0,\mathrm{\forall }y\in C.$$

(1.2)

The set of solutions of (1.2) is denoted by $GEP(F,A)$.

If $A\equiv 0$, then problem (1.1) reduces to the mixed equilibrium problem, which is to find $x\in C$ (see, e.g., [10–13]) such that

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

(1.3)

The set of solutions of (1.3) is denoted by $MEP(F,\phi )$.

If $F\equiv 0$, then problem (1.1) reduces to the mixed variational inequality of Browder type, which is to find $x\in C$ (see, e.g., [3, 14]) such that

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

(1.4)

The set of solutions of (1.4) is denoted by $MVI(C,\phi ,A)$.

If $\phi \equiv 0$, $A\equiv 0$, then problem (1.1) reduces to the equilibrium problem, which is to find $x\in C$ (see, e.g., [15–17]) such that

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

(1.5)

The set of solutions of (1.5) is denoted by $EP(F)$.

If $F\equiv 0$, $\phi \equiv 0$, then problem (1.1) reduces to the variational inequality, which is to find $x\in C$ (see, e.g., [18–29]) such that

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

(1.6)

The set of solutions of (1.6) is denoted by $VI(C,A)$.

If $F\equiv 0$, $A\equiv 0$, then problem (1.1) reduces to the minimized problem, which is to find $x\in C$ (see, e.g., [18–28]) such that

$$\phi (y)-\phi (x)\ge 0,\mathrm{\forall }y\in C.$$

(1.7)

The set of solutions of (1.7) is denoted by $Argmin(\phi )$.

Let $A,B:C\to H$ be two mappings. Ceng et al. [2] considered the following problem of finding $(x^{\ast },y^{\ast })\in C\times C$ such that

$$\{\begin{array}{c}〈\lambda A{y}^{\ast }+x^{\ast }-y^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈\mu B{x}^{\ast }+y^{\ast }-x^{\ast },x-y^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

(1.8)

which is called a general system of variational inequalities where $\lambda >0$ and $\mu >0$ are two constants. In particular, if $A=B$, $x^{\ast }=y^{\ast }$, then problem (1.8) reduces to the classical variational inequality problem (1.6).

In order to find the common element of the solutions of problem (1.8) and the set of fixed points of one nonexpansive mapping S, Ceng et al. [2] studied the following algorithm: fix $u\in C$, $x_0\in C$, and

$$\{\begin{array}{c}{z}_n=T_{{\lambda }_n}^{(\mathrm{\Theta },\phi )}(x_n-{\lambda }_{n}F{x}_n),\hfill \\ y_n=T_{{\mu }_1}^{G_1}[T_{{\mu }_2}^{G_2}(z_n-{\mu }_2{B}_2{z}_n)-{\mu }_1{B}_1{T}_{{\mu }_2}^{G_2}(z_n-{\mu }_2{B}_2{z}_n)],\hfill \\ x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{y}_n+{\delta }_{n}S{y}_n,n\ge 0.\hfill \end{array}$$

(1.9)

Under appropriate conditions, they obtained one strong convergence theorem.

Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\{A_i\}}_{i=1}^N:C\to H$ be a family of mappings. Cai and Bu [1] considered the following problem of finding $(x_1^{\ast },x_2^{\ast },\dots ,x_N^{\ast })\in C\times C\times \cdots \times C$ such that

$$\{\begin{array}{c}〈{\lambda }_N{A}_N{x}_N^{\ast }+x_1^{\ast }-x_N^{\ast },x-x_1^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\lambda }_{N-1}{A}_{N-1}{x}_{N-1}^{\ast }+x_N^{\ast }-x_{N-1}^{\ast },x-x_N^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ ⋮\hfill \\ 〈{\lambda }_2{A}_2{x}_2^{\ast }+x_3^{\ast }-x_2^{\ast },x-x_3^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈{\lambda }_1{A}_1{x}_1^{\ast }+x_2^{\ast }-x_1^{\ast },x-x_2^{\ast }〉\ge 0,\mathrm{\forall }x\in C.\hfill \end{array}$$

(1.10)

And (1.10) can be rewritten as

$$\{\begin{array}{c}〈x_1^{\ast }-(I-{\lambda }_N{A}_N)x_N^{\ast },x-x_1^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈x_N^{\ast }-(I-{\lambda }_{N-1}{A}_{N-1})x_{N-1}^{\ast },x-x_N^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ ⋮\hfill \\ 〈x_3^{\ast }-(I-{\lambda }_2{A}_2)x_2^{\ast },x-x_3^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \\ 〈x_2^{\ast }-(I-{\lambda }_1{A}_1)x_1^{\ast },x-x_2^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\hfill \end{array}$$

(1.11)

which is called a more general system of variational inequalities in Hilbert spaces, where ${\lambda }_i>0$ for all $i\in \{1,2,\dots ,N\}$. The set of solutions to (1.10) is denoted by Ω. In particular, if $N=2$, $A_1=B$, $A_2=A$, ${\lambda }_1=\mu$, ${\lambda }_2=\lambda$, $x_1^{\ast }=x^{\ast }$, $x_2^{\ast }=y^{\ast }$, then problem (1.10) reduces to problem (1.8).

In order to find a common element of the solutions of problem (1.10) and the common fixed points of a family of strictly pseudocontractive mappings, Cai and Bu [1] studied the following algorithm: pick any $x_0\in H$, set $C_1=C$, $x_1=P_{C_1}{x}_0$, and

$$\{\begin{array}{c}{u}_n=T_{r_{M,n}}^{(F_M,{\phi }_M)}(I-r_{M,n}{B}_M)T_{r_{M-1,n}}^{(F_{M-1},{\phi }_{M-1})}\hfill \\ \phantom{u_n=}\times (I-r_{M-1,n}{B}_{M-1})\cdots T_{r_{1,n}}^{(F_1,{\phi }_1)}(I-r_{1,n}{B}_1)x_n,\hfill \\ y_n=P_C(I-{\lambda }_N{A}_N)P_C(I-{\lambda }_{N-1}{A}_{N-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1)u_n,\hfill \\ z_n={\alpha }_n{y}_n+(1-{\alpha }_n)S_n{y}_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel z_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_0,\mathrm{\forall }n\ge 1.\hfill \end{array}$$

(1.12)

Under suitable conditions, they also obtained one strong convergence theorem.

In this paper, motivated and inspired by the above facts, we study a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.

2 Preliminaries

For solving the equilibrium problem, let us assume that the bifunction F satisfies the following conditions:

• (A1) $F(x,x)=0$ for all $x\in C$;

• (A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for any $x,y\in C$;

• (A3) F is weakly upper semicontinuous, i.e., for each $x,y,z\in C$,

  $$\underset{t\to 0}{lim\hspace{0.17em}sup}F(x+t(z-x),y)\le F(x,y);$$

• (A4) $F(x,\cdot )$ is convex and lower semicontinuous for each $x\in C$;

• (B1) For each $x\in H$ and $r>0$, there exists a bounded subset $D_x\subseteq C$ and $y_x\in C$ such that for any $z\in C\mathrm{\setminus }{D}_x$,

  $$F(z,y_x)+\phi (y_x)-\phi (z)+\frac{1}{r}〈y_x-z,z-x〉<0;$$

• (B2) C is a bounded set.

Let H be a real Hilbert space. It is well known that

$${\parallel x+y\parallel }^2={\parallel x\parallel }^2+{\parallel y\parallel }^2+2〈x,y〉$$

(2.1)

and

$${\parallel x\parallel }^2-{\parallel y\parallel }^2\le \parallel x-y\parallel (\parallel x\parallel +\parallel y\parallel )$$

(2.2)

for all $x,y\in H$.

Definition 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H.

• (1) A mapping $T:C\to C$ is said to be nonexpansive if

  $$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C;$$

• (2) A mapping $T:C\to H$ is said to be L-Lipschitzian if there exists $L>0$ such that

  $$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in C;$$

• (3) A mapping $T:C\to C$ is said to be k-strictly pseudocontractive if there exists a constant $k\in [0,1)$ such that

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

  (2.3)

  It is obvious that $k=0$, then the mapping T is nonexpansive;

• (4) A mapping $T:C\to H$ is said to be monotone if

  $$〈Tx-Ty,x-y〉\ge 0,\mathrm{\forall }x,y\in C;$$

• (5) A mapping $T:C\to H$ is said to be α-inverse-strongly monotone if there exists a positive real number α such that

  $$〈Tx-Ty,x-y〉\ge \alpha {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in C.$$

It is obvious that any α-inverse-strongly monotone mapping T is monotone and $\frac{1}{\alpha }$-Lipschitz continuous.

Definition 2.2 $P_C:H\to C$ is called a metric projection if for every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

In order to prove our main results in the next section, we recall some lemmas.

Lemma 2.1 [30]

Let C be a nonempty closed convex subset of H and let $T:C\to C$ be a k-strictly pseudocontractive mapping, then the following results hold:

• (1) equation (2.3) is equivalent to

  $$〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^2-\frac{1-k}{2}{\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C;$$

  (2.4)

• (2) T is Lipschitz continuous with a constant $\frac{1+k}{1-k}$, i.e.,

  $$\parallel Tx-Ty\parallel \le \frac{1+k}{1-k}\parallel x-y\parallel ,\mathrm{\forall }x,y\in C;$$

  (2.5)

• (3) (Demi-closed principle) $I-T$ is demi-closed on C, that is,

  $$\mathit{\text{if}}{x}_n\rightharpoonup x^{\ast }\in C\mathit{\text{and}}(I-T)x_n\to 0,\mathit{\text{then}}{x}^{\ast }=T{x}^{\ast }.$$

Lemma 2.2 [1]

Let C be a nonempty closed convex subset of H and let $T:C\to H$ be an α-inverse-strongly monotone mapping, then for all $x,y\in C$ and $\lambda >0$, we have

$$\begin{array}{rcl}{\parallel (I-\lambda T)x-(I-\lambda T)y\parallel }^2& =& {\parallel (x-y)-\lambda (Tx-Ty)\parallel }^2\\ =& {\parallel x-y\parallel }^2-2\lambda 〈Tx-Ty,x-y〉+{\lambda }^2{\parallel Tx-Ty\parallel }^2\\ \le & {\parallel x-y\parallel }^2+\lambda (\lambda -2\alpha ){\parallel Tx-Ty\parallel }^2.\end{array}$$

So, if $0<\lambda \le 2\alpha$, then $I-\lambda T$ is a nonexpansive mapping from C to H.

Lemma 2.3 Let C be a nonempty closed convex subset of H and let $P_C:H\to C$ be a metric projection, then

• (1) ${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈x-y,P_{C}x-P_{C}y〉$, $\mathrm{\forall }x,y\in H$;

• (2) moreover, $P_C$ is a nonexpansive mapping, i.e., $\parallel P_{C}x-P_{C}y\parallel \le \parallel x-y\parallel$, $\mathrm{\forall }x,y\in H$;

• (3) $〈x-P_{C}x,y-P_{C}x〉\le 0$, $\mathrm{\forall }x\in H$, $y\in C$;

• (4) ${\parallel x-y\parallel }^2\ge {\parallel x-P_{C}x\parallel }^2+{\parallel y-P_{C}x\parallel }^2$, $\mathrm{\forall }x\in H$, $y\in C$.

Lemma 2.4 [4]

Let C be a nonempty closed convex subset of H. Assume that $F:C\times C\to \mathbb{R}$ satisfies (A1)-(A4) and let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For $r>0$ and $x\in H$, define a mapping $T_r^{(F,\phi )}:H\to C$ as follows:

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

for all $x\in H$. Then the following hold:

• (1) for each $x\in H$, $T_r^{(F,\phi )}(x)\ne \mathrm{\varnothing }$ and $T_r^{(F,\phi )}$ is single-valued;

• (2) $T_r^{(F,\phi )}$ is firmly nonexpansive, that is, for any $x,y\in H$,

  $${\parallel T_r^{(F,\phi )}(x)-T_r^{(F,\phi )}(y)\parallel }^2\le 〈T_r^{(F,\phi )}(x)-T_r^{(F,\phi )}(y),x-y〉;$$

• (3) $\mathcal{F}(T_r^{(F,\phi )})=MEP(F,\phi )$;

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

Lemma 2.5 [3]

Let C be a nonempty closed convex subset of H. Assume that $F:C\times C\to \mathbb{R}$ satisfies (A1)-(A4), $B:C\to H$ is a continuous monotone mapping and let $\phi :C\to \mathbb{R}$ be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For $r>0$ and $x\in H$, define a mapping $K_r^{(F,\phi )}:H\to C$ as follows:

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

for all $x\in H$. Then the following hold:

• (1) for each $x\in H$, $K_r^{(F,\phi )}(x)\ne \mathrm{\varnothing }$ and $K_r^{(F,\phi )}$ is single-valued;

• (2) $K_r^{(F,\phi )}$ is firmly nonexpansive, that is, for any $x,y\in H$,

  $${\parallel K_r^{(F,\phi )}(x)-K_r^{(F,\phi )}(y)\parallel }^2\le 〈K_r^{(F,\phi )}(x)-K_r^{(F,\phi )}(y),x-y〉;$$

• (3) $\mathcal{F}(K_r^{(F,\phi )})=GMEP(F,\phi ,B)$;

• (4) $GMEP(F,\phi ,B)$ is closed and convex.

Lemma 2.6 Let C be a nonempty closed convex subset of H. Let ${\{F_k\}}_{k=1}^M$ be a family of bifunctions from $C\times C$ into ℝ satisfying (A1)-(A4), let ${\{{\phi }_k\}}_{k=1}^M$ be a family of lower semicontinuous functions from C into ℝ, and let ${\{B_k\}}_{k=1}^M$ be a family of ${\beta }_k$-inverse-strongly monotone mappings from C into H. For $F_k$ and ${\phi }_k$, $k=1,2,\dots ,M$, assume that either (B1) or (B2) holds. Let $T:C\to H$ be a mapping defined by

$$T(x)=T_{r_M}^{(F_M,{\phi }_M)}(I-r_M{B}_M)T_{r_{M-1}}^{(F_{M-1},{\phi }_{M-1})}(I-r_{M-1}{B}_{M-1})\cdots T_{r_1}^{(F_1,{\phi }_1)}(I-r_1{B}_1)x,\mathrm{\forall }x\in C.$$

Putting ${\mathrm{\Theta }}^0=I$, where I is an identity mapping,

$${\mathrm{\Theta }}^k=T_{r_k}^{(F_k,{\phi }_k)}(I-r_k{B}_k)T_{r_{k-1}}^{(F_{k-1},{\phi }_{k-1})}(I-r_{k-1}{B}_{k-1})\cdots T_{r_1}^{(F_1,{\phi }_1)}(I-r_1{B}_1),k=1,2,\dots ,M.$$

If $x\in {\bigcap }_{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)$ and $0<r_k\le 2{\beta }_k$, $k=1,2,\dots ,M$, then

• (1) ${\mathrm{\Theta }}^{k}x=x$, $k=1,2,\dots ,M$;

• (2) T is nonexpansive.

Proof (1) Since ${\{B_k\}}_{k=1}^M$ is a family of ${\beta }_k$-inverse-strongly monotone mappings from C into H, so they are continuous monotone mappings. Observe that

$$\begin{array}{r}{T}_{r_k}^{(F_k,{\phi }_k)}(I-r_k{B}_k)x\\ =\{z\in C:F_k(z,y)+{\phi }_k(y)-{\phi }_k(z)+\frac{1}{r_k}〈y-z,z-(I-r_k{B}_k)x〉\ge 0,\mathrm{\forall }y\in C\}\\ =\{z\in C:F_k(z,y)+{\phi }_k(y)-{\phi }_k(z)+〈B_{k}x,y-z〉+\frac{1}{r_k}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}\\ =K_{r_k}^{(F_k,{\phi }_k)}(x).\end{array}$$

By Lemma 2.5, we know that if $x\in {\bigcap }_{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)$ then x is the fixed point of the mapping $K_{r_k}^{(F_k,{\phi }_k)}$, $k=1,2,\dots ,M$, so we have

$$x=K_{r_k}^{(F_k,{\phi }_k)}(x)=T_{r_k}^{(F_k,{\phi }_k)}(I-r_k{B}_k)x,$$

(2.6)

which implies that x is a fixed point of the mapping $T_{r_k}^{(F_k,{\phi }_k)}(I-r_k{B}_k)$. Therefore we get

$${\mathrm{\Theta }}^{k}x=x,k=1,2,\dots ,M.$$

(2) Since $T_r^{(F,\phi )}$ is firmly nonexpansive, then it is obvious that $T_r^{(F,\phi )}$ is nonexpansive. And from Lemma 2.2, we have

$$\begin{array}{rcl}\parallel T(x)-T(y)\parallel & =& \parallel {\mathrm{\Theta }}^{M}x-{\mathrm{\Theta }}^{M}y\parallel \\ =& \parallel T_{r_M}^{(F_M,{\phi }_M)}(I-r_M{B}_M){\mathrm{\Theta }}^{M-1}x-T_{r_M}^{(F_M,{\phi }_M)}(I-r_M{B}_M){\mathrm{\Theta }}^{M-1}y\parallel \\ \le & \parallel (I-r_M{B}_M){\mathrm{\Theta }}^{M-1}x-(I-r_M{B}_M){\mathrm{\Theta }}^{M-1}y\parallel \\ \le & \parallel {\mathrm{\Theta }}^{M-1}x-{\mathrm{\Theta }}^{M-1}y\parallel \le \cdots \le \parallel {\mathrm{\Theta }}^{0}x-{\mathrm{\Theta }}^{0}y\parallel \\ =& \parallel x-y\parallel ,\end{array}$$

which implies T is nonexpansive. □

Lemma 2.7 [1]

Let C be a nonempty closed convex subset of H. Let $A_i$ be ${\alpha }_i$-inverse-strongly monotone from C into H, respectively, where $i\in \{1,2,\dots ,N\}$. Let $G:C\to C$ be a mapping defined by

$$G(x)=P_C(I-{\lambda }_N{A}_N)P_C(I-{\lambda }_{N-1}{A}_{N-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1)x,\mathrm{\forall }x\in C.$$

If $0<{\lambda }_i\le 2{\alpha }_i$, $i=1,2,\dots ,N$, then G is nonexpansive.

Proof Put ${\mathrm{\Omega }}^i=P_C(I-{\lambda }_i{A}_i)P_C(I-{\lambda }_{i-1}{A}_{i-1})\cdots P_C(I-{\lambda }_1{A}_1)$, $i=1,2,\dots ,N$, and ${\mathrm{\Omega }}^0=I$, where I is an identity mapping. Since $P_C$ is nonexpansive and from Lemma 2.2, we have

$$\begin{array}{rcl}\parallel G(x)-G(y)\parallel & =& \parallel {\mathrm{\Omega }}^{N}x-{\mathrm{\Omega }}^{N}y\parallel \\ =& \parallel P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}x-P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}y\parallel \\ \le & \parallel (I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}x-(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}y\parallel \\ \le & \parallel {\mathrm{\Omega }}^{N-1}x-{\mathrm{\Omega }}^{N-1}y\parallel \le \cdots \le \parallel {\mathrm{\Omega }}^{0}x-{\mathrm{\Omega }}^{0}y\parallel \\ \le & \parallel x-y\parallel ,\end{array}$$

which implies G is nonexpansive. □

Lemma 2.8 Let C be a nonempty closed convex subset of H. Let $A_i:C\to H$ be a nonlinear mapping, where $i=1,2,\dots ,N$. For given $x_i^{\ast }\in C$, $i=1,2,\dots ,N$, $(x_1^{\ast },x_2^{\ast },\dots ,x_N^{\ast })$ is a solution of problem (1.10) if and only if

$$x_1^{\ast }=P_C(I-{\lambda }_N{A}_N)x_N^{\ast },x_i^{\ast }=P_C(I-{\lambda }_{i-1}{A}_{i-1})x_{i-1}^{\ast },i=2,3,\dots ,N,$$

(2.7)

that is,

$$x_1^{\ast }=P_C(I-{\lambda }_N{A}_N)P_C(I-{\lambda }_{N-1}{A}_{N-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1)x_1^{\ast }.$$

Proof (⟸) From Lemma 2.3(3), it is obvious that (2.7) is the solution of problem (1.10).

(⟹) Since

$$\begin{array}{r}〈{\lambda }_N{A}_N{x}_N^{\ast }+x_1^{\ast }-x_N^{\ast },x-x_1^{\ast }〉\ge 0,\mathrm{\forall }x\in C\\ \Rightarrow 〈x_1^{\ast }-(I-{\lambda }_N{A}_N)x_N^{\ast },x-x_1^{\ast }〉\ge 0,\mathrm{\forall }x\in C\\ \Rightarrow 〈(I-{\lambda }_N{A}_N)x_N^{\ast }-x_1^{\ast },(I-{\lambda }_N{A}_N)x_N^{\ast }-x_1^{\ast }-(I-{\lambda }_N{A}_N)x_N^{\ast }+x〉\le 0,\mathrm{\forall }x\in C\\ \Rightarrow {\parallel (I-{\lambda }_N{A}_N)x_N^{\ast }-x_1^{\ast }\parallel }^2\le 〈(I-{\lambda }_N{A}_N)x_N^{\ast }-x_1^{\ast },(I-{\lambda }_N{A}_N)x_N^{\ast }-x〉,\mathrm{\forall }x\in C\\ \Rightarrow \parallel (I-{\lambda }_N{A}_N)x_N^{\ast }-x_1^{\ast }\parallel \le \parallel (I-{\lambda }_N{A}_N)x_N^{\ast }-x\parallel ,\mathrm{\forall }x\in C\\ \Rightarrow x_1^{\ast }=P_C(I-{\lambda }_N{A}_N)x_N^{\ast }.\end{array}$$

Similarly, we get

$$x_i^{\ast }=P_C(I-{\lambda }_{i-1}{A}_{i-1})x_{i-1}^{\ast },i=2,3,\dots ,N.$$

Therefore we have

$$x_1^{\ast }=P_C(I-{\lambda }_N{A}_N)P_C(I-{\lambda }_{N-1}{A}_{N-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1)x_1^{\ast },$$

which completes the proof. □

From Lemma 2.8, we know that $x_1^{\ast }=G(x_1^{\ast })$, that is, $x_1^{\ast }$ is a fixed point of the mapping G, where G is defined by Lemma 2.7. Moreover, if we find the fixed point $x_1^{\ast }$, it is easy to solve the other points by (2.7).

Lemma 2.9 [31]

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X and let $\{{\beta }_n\}$ be a sequence in $[0,1]$ with $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)y_n$ for all integers $n\ge 0$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.10 [30]

Let $T:C\to C$ be a nonexpansive mapping with $\mathcal{F}(T)\ne \mathrm{\varnothing }$. If $x_n\rightharpoonup x^{\ast }$ and $(I-T)x_n\to y^{\ast }$, then $(I-T)x^{\ast }=y^{\ast }$.

Lemma 2.11 [30]

Assume that $\{{\alpha }_n\}$ is a sequence of nonnegative real numbers such that

$${\alpha }_{n+1}\le (1-{\gamma }_n){\alpha }_n+{\delta }_n,\mathrm{\forall }n\ge 1,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

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

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\delta }_n}{{\gamma }_n}\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n|<\mathrm{\infty }$.

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

3 Main results

In this section, we state and verify our main results. We have the following theorem.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let ${\{F_k\}}_{k=1}^M$ be a family of bifunctions from $C\times C$ into ℝ satisfying (A1)-(A4), let ${\{{\phi }_k\}}_{k=1}^M:C\to \mathbb{R}$ be a family of lower semicontinuous and convex functions and let ${\{B_k\}}_{k=1}^M$ be a family of ${\beta }_k$-inverse-strongly monotone mappings from C into H. Let $A_i$ be ${\alpha }_i$-inverse-strongly monotone from C into H, respectively, where $i\in \{1,2,\dots ,N\}$. Let S be a δ-strict pseudocontractive mapping from C into itself such that $\mathcal{F}=[{\bigcap }_{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)]\cap \mathcal{F}(G)\cap \mathcal{F}(S)\ne \mathrm{\varnothing }$, where G is defined by Lemma 2.7. For $F_k$ and ${\phi }_k$, $k=1,2,\dots ,M$, assume that either (B1) or (B2) holds. Pick any $x_0\in C$, let $\{x_n\}\subset C$ be a sequence generated by

$$\{\begin{array}{c}{z}_n=T_{r_{M,n}}^{(F_M,{\phi }_M)}(I-r_{M,n}{B}_M)T_{r_{M-1,n}}^{(F_{M-1},{\phi }_{M-1})}\hfill \\ \phantom{z_n=}\times (I-r_{M-1,n}{B}_{M-1})\cdots T_{r_{1,n}}^{(F_1,{\phi }_1)}(I-r_{1,n}{B}_1)x_n,\hfill \\ y_n=P_C(I-{\lambda }_N{A}_N)P_C(I-{\lambda }_{N-1}{A}_{N-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1)z_n,\hfill \\ x_{n+1}=a_n{x}_0+b_n{x}_n+c_n{y}_n+d_{n}S{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(3.1)

where ${\lambda }_i\in (0,2{\alpha }_i)$, $i=1,2,\dots ,N$, $\delta \in (0,1)$. $\{a_n\},\{b_n\},\{c_n\},\{d_n\}\subset [0,1]$ satisfy the following conditions:

• (i) $a_n+b_n+c_n+d_n=1$ and $(c_n+d_n)\delta \le c_n$ for all $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{a}_n=0$ and ${\sum }_{n=0}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{b}_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{b}_n<1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{d}_n>0$;

• (iv) ${lim}_{n\to \mathrm{\infty }}(\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n})=0$;

• (v) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{r}_{k,n}<2{\beta }_k$, $k=1,2,\dots ,M$.

Then $\{x_n\}\subset C$ converges strongly to $P_{\mathcal{F}}{x}_0$.

Proof Putting

$$\begin{array}{rcl}{\mathrm{\Theta }}_n^k& =& T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k)T_{r_{k-1,n}}^{(F_{k-1},{\phi }_{k-1})}(I-r_{k-1,n}{B}_{k-1})\cdots T_{r_{1,n}}^{(F_1,{\phi }_1)}(I-r_{1,n}{B}_1),\\ \mathrm{\forall }k\in \{1,\dots ,M\},n\in \mathbb{N},\end{array}$$

and

$${\mathrm{\Omega }}^i=P_C(I-{\lambda }_i{A}_i)P_C(I-{\lambda }_{i-1}{A}_{i-1})\cdots P_C(I-{\lambda }_2{A}_2)P_C(I-{\lambda }_1{A}_1),\mathrm{\forall }i\in \{1,2,\dots ,N\},$$

${\mathrm{\Theta }}_n^0={\mathrm{\Omega }}^0=I$, where I is the identity mapping on H. Then we have that $z_n={\mathrm{\Theta }}_n^M{x}_n$ and $y_n={\mathrm{\Omega }}^N{z}_n$. From Lemma 2.6 and Lemma 2.7, it can be seen easily that ${\mathrm{\Theta }}_n^k$ and ${\mathrm{\Omega }}^i$ are nonexpansive, where $k\in \{1,2,\dots ,M\}$, $i\in \{1,2,\dots ,N\}$. We divide the proof into six steps.

Step 1. Firstly, we show that $\{x_n\}$ is bounded.

Indeed, take $p\in \mathcal{F}$ arbitrarily. Since $p={\mathrm{\Theta }}_n^{k}p=Sp$, $\mathrm{\forall }k\in \{1,2,\dots ,M\}$, $\mathrm{\forall }n\in \mathbb{N}$. By Lemma 2.6, we have

$$\parallel z_n-p\parallel =\parallel {\mathrm{\Theta }}_n^M{x}_n-{\mathrm{\Theta }}_n^{M}p\parallel \le \parallel x_n-p\parallel .$$

(3.2)

It follows from Lemma 2.7 and (3.2) that

$$\parallel y_n-p\parallel =\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel \le \parallel z_n-p\parallel \le \parallel x_n-p\parallel .$$

(3.3)

Furthermore, from (3.1), we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel a_n{x}_0+b_n{x}_n+c_n{y}_n+d_{n}S{y}_n-p\parallel \\ =& \parallel a_n(x_0-p)+b_n(x_n-p)+c_n(y_n-p)+d_n(S{y}_n-p)\parallel \\ \le & a_n\parallel x_0-p\parallel +b_n\parallel x_n-p\parallel +\parallel c_n(y_n-p)+d_n(S{y}_n-p)\parallel .\end{array}$$

(3.4)

Since $(c_n+d_n)\delta \le c_n$, (2.3) and (2.4), we have

$$\begin{array}{r}{\parallel c_n(y_n-p)+d_n(S{y}_n-p)\parallel }^2\\ =c_n^2{\parallel y_n-p\parallel }^2+d_n^2{\parallel S{y}_n-p\parallel }^2+2c_n{d}_n〈S{y}_n-p,y_n-p〉\\ \le c_n^2{\parallel y_n-p\parallel }^2+d_n^2[{\parallel y_n-p\parallel }^2+\delta {\parallel y_n-S{y}_n\parallel }^2]\\ +2c_n{d}_n[{\parallel y_n-p\parallel }^2-\frac{1-\delta }{2}{\parallel y_n-S{y}_n\parallel }^2]\\ ={(c_n+d_n)}^2{\parallel y_n-p\parallel }^2+[d_n^2\delta -(1-\delta )c_n{d}_n]{\parallel y_n-S{y}_n\parallel }^2\\ ={(c_n+d_n)}^2{\parallel y_n-p\parallel }^2+d_n[(c_n+d_n)\delta -c_n]{\parallel y_n-S{y}_n\parallel }^2\\ \le {(c_n+d_n)}^2{\parallel y_n-p\parallel }^2,\end{array}$$

which implies that

$$\parallel c_n(y_n-p)+d_n(S{y}_n-p)\parallel \le (c_n+d_n)\parallel y_n-p\parallel .$$

(3.5)

From (3.2)-(3.5) it follows that

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & a_n\parallel x_0-p\parallel +b_n\parallel x_n-p\parallel +\parallel c_n(y_n-p)+d_n(S{y}_n-p)\parallel \\ \le & a_n\parallel x_0-p\parallel +b_n\parallel x_n-p\parallel +(c_n+d_n)\parallel y_n-p\parallel \\ \le & a_n\parallel x_0-p\parallel +b_n\parallel x_n-p\parallel +(c_n+d_n)\parallel x_n-p\parallel \\ =& a_n\parallel x_0-p\parallel +(1-a_n)\parallel x_n-p\parallel .\end{array}$$

So, we have

$$\parallel x_{n+1}-p\parallel \le max\{\parallel x_0-p\parallel ,\parallel x_n-p\parallel \},\mathrm{\forall }n\ge 0.$$

By induction, we obtain that

$$\parallel x_n-p\parallel \le \parallel x_0-p\parallel ,\mathrm{\forall }n\ge 0.$$

Hence, $\{x_n\}$ is bounded. Consequently, we deduce immediately that $\{z_n\}$, $\{y_n\}$, $\{S{y}_n\}$ are bounded.

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

Indeed, define $x_{n+1}=b_n{x}_n+(1-b_n)w_n$ for all $n\ge 0$. It follows that

$$\begin{array}{rcl}{w}_{n+1}-w_n& =& \frac{x_{n+2}-b_{n+1}{x}_{n+1}}{1-b_{n+1}}-\frac{x_{n+1}-b_n{x}_n}{1-b_n}\\ =& \frac{a_{n+1}{x}_0+c_{n+1}{y}_{n+1}+d_{n+1}S{y}_{n+1}}{1-b_{n+1}}-\frac{a_n{x}_0+c_n{y}_n+d_{n}S{y}_n}{1-b_n}\\ =& \frac{a_{n+1}{x}_0}{1-b_{n+1}}-\frac{a_n{x}_0}{1-b_n}+\frac{c_{n+1}(y_{n+1}-y_n)+d_{n+1}(S{y}_{n+1}-S{y}_n)}{1-b_{n+1}}\\ +(\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n})y_n+(\frac{d_{n+1}}{1-b_{n+1}}-\frac{d_n}{1-b_n})S{y}_n.\end{array}$$

(3.6)

Observe that

$$\begin{array}{r}{\parallel c_{n+1}(y_{n+1}-y_n)+d_{n+1}(S{y}_{n+1}-S{y}_n)\parallel }^2\\ =c_{n+1}^2{\parallel y_{n+1}-y_n\parallel }^2+d_{n+1}^2{\parallel S{y}_{n+1}-S{y}_n\parallel }^2+2c_{n+1}{d}_{n+1}〈S{y}_{n+1}-S{y}_n,y_{n+1}-y_n〉\\ \le c_{n+1}^2{\parallel y_{n+1}-y_n\parallel }^2+d_{n+1}^2[{\parallel y_{n+1}-y_n\parallel }^2+\delta {\parallel (y_{n+1}-S{y}_{n+1})-(y_n-S{y}_n)\parallel }^2]\\ +2c_{n+1}{d}_{n+1}[{\parallel y_{n+1}-y_n\parallel }^2-\frac{1-\delta }{2}{\parallel (y_{n+1}-S{y}_{n+1})-(y_n-S{y}_n)\parallel }^2]\\ ={(c_{n+1}+d_{n+1})}^2{\parallel y_{n+1}-y_n\parallel }^2+[d_{n+1}^2\delta -(1-\delta )c_{n+1}{d}_{n+1}]{\parallel (y_{n+1}-S{y}_{n+1})-(y_n-S{y}_n)\parallel }^2\\ ={(c_{n+1}+d_{n+1})}^2{\parallel y_{n+1}-y_n\parallel }^2+d_{n+1}[(c_{n+1}+d_{n+1})\delta -c_{n+1}]\\ \times {\parallel (y_{n+1}-S{y}_{n+1})-(y_n-S{y}_n)\parallel }^2\\ \le {(c_{n+1}+d_{n+1})}^2{\parallel y_{n+1}-y_n\parallel }^2,\end{array}$$

which implies that

$$\parallel c_{n+1}(y_{n+1}-y_n)+d_{n+1}(S{y}_{n+1}-S{y}_n)\parallel \le (c_{n+1}+d_{n+1})\parallel y_{n+1}-y_n\parallel .$$

(3.7)

Since

$$\parallel z_{n+1}-z_n\parallel =\parallel {\mathrm{\Theta }}_n^M{x}_{n+1}-{\mathrm{\Theta }}_n^M{x}_n\parallel \le \parallel x_{n+1}-x_n\parallel ,$$

(3.8)

then we have

$$\parallel y_{n+1}-y_n\parallel =\parallel {\mathrm{\Omega }}^N{z}_{n+1}-{\mathrm{\Omega }}^N{z}_n\parallel \le \parallel z_{n+1}-z_n\parallel \le \parallel x_{n+1}-x_n\parallel .$$

(3.9)

Hence it follows from (3.6), (3.7), (3.9) and $\frac{c_{n+1}+d_{n+1}}{1-b_{n+1}}=\frac{1-a_{n+1}-b_{n+1}}{1-b_{n+1}}<1$ that

$$\begin{array}{rl}\parallel w_{n+1}-w_n\parallel \le & (\frac{a_{n+1}}{1-b_{n+1}}+\frac{a_n}{1-b_n})\parallel x_0\parallel +\frac{\parallel c_{n+1}(y_{n+1}-y_n)+d_{n+1}(S{y}_{n+1}-S{y}_n)\parallel }{1-b_{n+1}}\\ +|\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n}|\parallel y_n\parallel +|\frac{d_{n+1}}{1-b_{n+1}}-\frac{d_n}{1-b_n}|\parallel S{y}_n\parallel \\ \le & (\frac{a_{n+1}}{1-b_{n+1}}+\frac{a_n}{1-b_n})\parallel x_0\parallel +\frac{c_{n+1}+d_{n+1}}{1-b_{n+1}}\parallel x_{n+1}-x_n\parallel \\ +|\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n}|\parallel y_n\parallel +|\frac{a_{n+1}+c_{n+1}}{1-b_{n+1}}-\frac{a_n+c_n}{1-b_n}|\parallel S{y}_n\parallel \\ \le & \frac{a_{n+1}}{1-b_{n+1}}(\parallel x_0\parallel +\parallel S{y}_n\parallel )+\frac{a_n}{1-b_n}(\parallel x_0\parallel +\parallel S{y}_n\parallel )+\parallel x_{n+1}-x_n\parallel \\ +|\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n}|(\parallel y_n\parallel +\parallel S{y}_n\parallel ).\end{array}$$

(3.10)

Consequently, it follows from (3.10), conditions (ii), (iv) and $\{y_n\}$, $\{S{y}_n\}$ are bounded that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel w_{n+1}-w_n\parallel -\parallel x_{n+1}-x_n\parallel )\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\frac{a_{n+1}}{1-b_{n+1}}(\parallel x_0\parallel +\parallel S{y}_n\parallel )+\frac{a_n}{1-b_n}(\parallel x_0\parallel +\parallel S{y}_n\parallel )\\ +|\frac{c_{n+1}}{1-b_{n+1}}-\frac{c_n}{1-b_n}|(\parallel y_n\parallel +\parallel S{y}_n\parallel )\}=0.\end{array}$$

Hence, by Lemma 2.9, we get ${lim}_{n\to \mathrm{\infty }}\parallel w_n-x_n\parallel =0$. Thus, from condition (iii), we have

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

(3.11)

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel =0$, $k=1,2,\dots ,M$ and ${lim}_{n\to \mathrm{\infty }}\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel =0$, $i=1,2,\dots ,N$.

It follows from Lemma 2.6 that

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2& ={\parallel {\mathrm{\Theta }}_n^M{x}_n-{\mathrm{\Theta }}_n^{M}p\parallel }^2\le {\parallel {\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k}p\parallel }^2\\ ={\parallel T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}p\parallel }^2\\ \le {\parallel (I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}p\parallel }^2\\ \le {\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\beta }_k){\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2\\ \le {\parallel x_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\beta }_k){\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2.\end{array}$$

(3.12)

By Lemma 2.3 and Lemma 2.2, we have

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p\parallel }^2\\ \le & {\parallel (I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p\parallel }^2\\ \le & {\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^{N-1}p\parallel }^2+{\lambda }_N({\lambda }_N-2{\alpha }_N){\parallel A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p\parallel }^2.\end{array}$$

By induction, we get

$${\parallel y_n-p\parallel }^2\le {\parallel z_n-p\parallel }^2+\sum _{i=1}^N{\lambda }_i({\lambda }_i-2{\alpha }_i){\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel }^2.$$

(3.13)

From condition (i) and (3.5), we get

$$\begin{array}{r}{\parallel x_{n+1}-p\parallel }^2\\ =〈a_n(x_0-p)+b_n(x_n-p)+c_n(y_n-p)+d_n(S{y}_n-p),x_{n+1}-p〉\\ =a_n〈x_0-p,x_{n+1}-p〉+b_n〈x_n-p,x_{n+1}-p〉+〈c_n(y_n-p)+d_n(S{y}_n-p),x_{n+1}-p〉\\ \le a_n〈x_0-p,x_{n+1}-p〉+b_n\parallel x_n-p\parallel \parallel x_{n+1}-p\parallel \\ +\parallel c_n(y_n-p)+d_n(S{y}_n-p)\parallel \parallel x_{n+1}-p\parallel \\ \le a_n〈x_0-p,x_{n+1}-p〉+b_n\parallel x_n-p\parallel \parallel x_{n+1}-p\parallel +(c_n+d_n)\parallel y_n-p\parallel \parallel x_{n+1}-p\parallel \\ \le a_n〈x_0-p,x_{n+1}-p〉+\frac{b_n}{2}({\parallel x_n-p\parallel }^2+{\parallel x_{n+1}-p\parallel }^2)\\ +\frac{c_n+d_n}{2}({\parallel y_n-p\parallel }^2+{\parallel x_{n+1}-p\parallel }^2),\end{array}$$

that is,

$${\parallel x_{n+1}-p\parallel }^2\le \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2+\frac{c_n+d_n}{1+a_n}{\parallel y_n-p\parallel }^2.$$

(3.14)

So, in terms of (3.12) and (3.13), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}\{{\parallel z_n-p\parallel }^2+\sum _{i=1}^N{\lambda }_i({\lambda }_i-2{\alpha }_i){\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel }^2\}\\ \le & \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}\{{\parallel x_n-p\parallel }^2+r_{k,n}(r_{k,n}-2{\beta }_k){\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2\\ +\sum _{i=1}^N{\lambda }_i({\lambda }_i-2{\alpha }_i){\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel }^2\}\\ \le & \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{1-a_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}\{r_{k,n}(r_{k,n}-2{\beta }_k){\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2\\ +\sum _{i=1}^N{\lambda }_i({\lambda }_i-2{\alpha }_i){\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel }^2\}.\end{array}$$

Therefore,

$$\begin{array}{r}{r}_{k,n}(2{\beta }_k-r_{k,n}){\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2+\sum _{i=1}^N{\lambda }_i(2{\alpha }_i-{\lambda }_i){\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ \le \frac{2a_n}{c_n+d_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{1-a_n}{c_n+d_n}({\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2)\\ \le \frac{2a_n}{c_n+d_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{1-a_n}{c_n+d_n}\parallel x_{n+1}-x_n\parallel (\parallel x_{n+1}-p\parallel +\parallel x_n-p\parallel ).\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{a}_n=0$, $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_{k,n}\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{r}_{k,n}<2{\beta }_k$, $k=1,2,\dots ,M$, ${\lambda }_i\in (0,2{\alpha }_i)$, $i=1,2,\dots ,N$, $\delta \in (0,1)$, ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}(c_n+d_n)>0$ and $\{x_n\}$ is bounded, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel =0,k=1,2,\dots ,M,$$

(3.15)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel =0,i=1,2,\dots ,N.$$

(3.16)

Step 4. We prove that ${lim}_{n\to \mathrm{\infty }}\parallel S{y}_n-y_n\parallel =0$.

Indeed, utilizing firmly nonexpansive of $T_{r_k}^{(F_k,{\phi }_k)}$ and Lemma 2.2, we have

$$\begin{array}{rcl}{\parallel {\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k}p\parallel }^2& =& {\parallel T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k)p\parallel }^2\\ \le & 〈(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-(I-r_{k,n}{B}_k)p,{\mathrm{\Theta }}_n^k{x}_n-p〉\\ =& \frac{1}{2}({\parallel (I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-(I-r_{k,n}{B}_k)p\parallel }^2+{\parallel {\mathrm{\Theta }}_n^k{x}_n-p\parallel }^2\\ -{\parallel (I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n-(I-r_{k,n}{B}_k)p-({\mathrm{\Theta }}_n^k{x}_n-p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2+{\parallel {\mathrm{\Theta }}_n^k{x}_n-p\parallel }^2\\ -{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n-r_{k,n}(B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p)\parallel }^2),\end{array}$$

which implies

$$\begin{array}{rcl}{\parallel {\mathrm{\Theta }}_n^k{x}_n-p\parallel }^2& \le & {\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n-r_{k,n}(B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p)\parallel }^2\\ =& {\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2-r_{k,n}^2{\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel }^2\\ +2r_{k,n}〈{\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n,B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p〉\\ \le & {\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2\\ +2r_{k,n}〈{\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n,B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p〉.\end{array}$$

(3.17)

From (3.14), (3.3), Lemma 2.6 and (3.17), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2+\frac{c_n+d_n}{1+a_n}{\parallel z_n-p\parallel }^2\\ \le & \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2+\frac{c_n+d_n}{1+a_n}\parallel {\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k}p\parallel \\ \le & \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}[{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2\\ +2r_{k,n}〈{\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n,B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p〉]\\ \le & \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}[{\parallel x_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2\\ +2r_{k,n}〈{\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n,B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p〉]\\ \le & \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}[{\parallel x_n-p\parallel }^2-{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2\\ +2r_{k,n}\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel \parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel ].\end{array}$$

It follows that

$$\begin{array}{r}\frac{c_n+d_n}{1+a_n}{\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel }^2\\ \le \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{1-a_n}{1+a_n}{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ +\frac{2r_{k,n}(c_n+d_n)}{1+a_n}\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel \parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel \\ \le \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ +\frac{2r_{k,n}(c_n+d_n)}{1+a_n}\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel \parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel \\ \le 2a_n\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\parallel x_{n+1}-x_n\parallel (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\\ +\frac{2r_{k,n}(c_n+d_n)}{1+a_n}\parallel {\mathrm{\Theta }}_n^{k-1}{x}_n-{\mathrm{\Theta }}_n^k{x}_n\parallel \parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel .\end{array}$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\frac{c_n+d_n}{1+a_n}>0$, $a_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel B_k{\mathrm{\Theta }}_n^{k-1}{x}_n-B_{k}p\parallel \to 0$, we conclude that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel {\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k-1}{x}_n\parallel =0,k=1,2,\dots ,M.$$

(3.18)

Therefore we get

$$\begin{array}{rl}\parallel x_n-z_n\parallel & =\parallel {\mathrm{\Theta }}_n^0{x}_n-{\mathrm{\Theta }}_n^M{x}_n\parallel \\ \le \parallel {\mathrm{\Theta }}_n^0{x}_n-{\mathrm{\Theta }}_n^1{x}_n\parallel +\parallel {\mathrm{\Theta }}_n^1{x}_n-{\mathrm{\Theta }}_n^2{x}_n\parallel +\cdots +\parallel {\mathrm{\Theta }}_n^{M-1}{x}_n-{\mathrm{\Theta }}_n^M{x}_n\parallel \to 0\\ \phantom{\le }\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.19)

From Lemma 2.3(1), we obtain

$$\begin{array}{rcl}{\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel }^2& =& {\parallel P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-P_C(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p\parallel }^2\\ \le & 〈(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p,{\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p〉\\ =& \frac{1}{2}({\parallel (I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p\parallel }^2+{\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel }^2\\ -{\parallel (I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}{z}_n-(I-{\lambda }_N{A}_N){\mathrm{\Omega }}^{N-1}p-({\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p)\parallel }^2)\\ \le & \frac{1}{2}({\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^{N-1}p\parallel }^2+{\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel }^2\\ -\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p\\ -{\lambda }_N(A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p){\parallel }^2),\end{array}$$

which implies that

$$\begin{array}{r}{\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel }^2\\ \le {\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^{N-1}p\parallel }^2\\ -{\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p-{\lambda }_N(A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p)\parallel }^2\\ ={\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^{N-1}p\parallel }^2\\ -{\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p\parallel }^2-{\lambda }_N^2{\parallel A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p)\parallel }^2\\ +2{\lambda }_N〈{\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p,A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p〉\\ \le {\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^{N-1}p\parallel }^2-{\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p\parallel }^2\\ +2{\lambda }_N\parallel {\mathrm{\Omega }}^{N-1}{z}_n-{\mathrm{\Omega }}^N{z}_n+{\mathrm{\Omega }}^{N}p-{\mathrm{\Omega }}^{N-1}p\parallel \parallel A_N{\mathrm{\Omega }}^{N-1}{z}_n-A_N{\mathrm{\Omega }}^{N-1}p\parallel .\end{array}$$

(3.20)

By induction, we have

$$\begin{array}{rcl}{\parallel {\mathrm{\Omega }}^N{z}_n-{\mathrm{\Omega }}^{N}p\parallel }^2& \le & {\parallel z_n-p\parallel }^2-\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ +2\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel \\ \le & {\parallel x_n-p\parallel }^2-\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{N-1}p\parallel }^2\\ +2\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel ,\end{array}$$

that is,

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2-\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ +2\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel .\end{array}$$

(3.21)

From (3.14) and (3.21),

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & \frac{2a_n}{1+a_n}〈x_0-p,x_{n+1}-p〉+\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}[{\parallel x_n-p\parallel }^2-\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ +2\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel ]\\ \le & \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{b_n}{1+a_n}{\parallel x_n-p\parallel }^2\\ +\frac{c_n+d_n}{1+a_n}[{\parallel x_n-p\parallel }^2-\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ +2\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel ].\end{array}$$

It follows that

$$\begin{array}{r}\frac{c_n+d_n}{1+a_n}\sum _{i=1}^N{\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel }^2\\ \le \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\frac{1-a_n}{1+a_n}{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ +\frac{2(c_n+d_n)}{1+a_n}\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel \\ \le \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ +\frac{2(c_n+d_n)}{1+a_n}\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel \\ \le \frac{2a_n}{1+a_n}\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\parallel x_{n+1}-x_n\parallel (\parallel x_{n+1}-p\parallel +\parallel x_n-p\parallel )\\ +\frac{2(c_n+d_n)}{1+a_n}\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel \\ \le 2a_n\parallel x_0-p\parallel \parallel x_{n+1}-p\parallel +\parallel x_{n+1}-x_n\parallel (\parallel x_{n+1}-p\parallel +\parallel x_n-p\parallel )\\ +\frac{2(c_n+d_n)}{1+a_n}\sum _{i=1}^N{\lambda }_i\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel .\end{array}$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\frac{c_n+d_n}{1+a_n}>0$, $a_n\to 0$, $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel A_i{\mathrm{\Omega }}^{i-1}{z}_n-A_i{\mathrm{\Omega }}^{i-1}p\parallel \to 0$, we conclude that

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^N\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel =0.$$

(3.22)

Therefore, we get

$$\begin{array}{rcl}\parallel z_n-y_n\parallel & =& \parallel {\mathrm{\Omega }}^0{z}_n-{\mathrm{\Omega }}^N{z}_n\parallel \\ \le & \sum _{i=1}^N\parallel {\mathrm{\Omega }}^{i-1}{z}_n-{\mathrm{\Omega }}^i{z}_n+{\mathrm{\Omega }}^{i}p-{\mathrm{\Omega }}^{i-1}p\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.23)

Thus from (3.19) and (3.23), we have

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

(3.24)

Observe that

$$\begin{array}{rcl}{d}_n\parallel S{y}_n-x_n\parallel & =& \parallel d_{n}S{y}_n-d_n{x}_n\parallel \\ =& \parallel x_{n+1}-a_n{x}_0-b_n{x}_n-c_n{y}_n-d_n{x}_n\parallel \\ =& \parallel x_{n+1}-x_n+(1-b_n-d_n)x_n-c_n{y}_n-a_n{x}_0\parallel \\ =& \parallel x_{n+1}-x_n+a_n(x_n-x_0)+c_n(x_n-y_n)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +a_n\parallel x_n-x_0\parallel +c_n\parallel x_n-y_n\parallel .\end{array}$$

Since ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{d}_n>0$, $\parallel x_{n+1}-x_n\parallel \to 0$, $a_n\to 0$ and $\parallel x_n-y_n\parallel \to 0$, we have

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

(3.25)

From (3.24) and (3.25), we conclude that

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

(3.26)

Step 5. In this step, we prove that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x_0-\overline{x},x_n-\overline{x}〉\le 0$, where $\overline{x}=P_{\mathcal{F}}{x}_0$.

Indeed, take a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_0-\overline{x},x_n-\overline{x}〉=\underset{n\to \mathrm{\infty }}{lim}〈x_0-\overline{x},x_{n_i}-\overline{x}〉.$$

Since $\{x_n\}$ is bounded, there exists a subsequence of $\{x_n\}$ which converges weakly to $x^{\ast }$. Without loss of generality, we may assume that $x_{n_i}\rightharpoonup x^{\ast }$. From (3.18) and (3.24), we have ${\mathrm{\Theta }}_{n_i}^k{x}_{n_i}\rightharpoonup x^{\ast }$, $y_{n_i}\rightharpoonup x^{\ast }$, where $k\in \{1,2,\dots ,M\}$. From (3.26) and Lemma 2.1, we have $x^{\ast }=S{x}^{\ast }$ that is $x^{\ast }\in \mathcal{F}(S)$. Utilizing Lemma 2.7, we known that G is nonexpansive. And from (3.23), we obtain

$$\parallel y_{n_i}-G{y}_{n_i}\parallel =\parallel G{z}_{n_i}-G{y}_{n_i}\parallel \le \parallel z_{n_i}-y_{n_i}\parallel \to 0\text{as }i\to \mathrm{\infty }.$$

According to Lemma 2.10, we obtain $(I-G)x^{\ast }=0$, that is, $x^{\ast }\in \mathcal{F}(G)$.

Next we prove that $x^{\ast }\in {\bigcap }_{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)$. Since

$${\mathrm{\Theta }}_n^k{x}_n=T_{r_{k,n}}^{(F_k,{\phi }_k)}(I-r_{k,n}{B}_k){\mathrm{\Theta }}_n^{k-1}{x}_n,n\ge 1,k\in \{1,2,\dots ,M\}.$$

For all $y\in C$, we have

$$\begin{array}{r}{F}_k({\mathrm{\Theta }}_n^k{x}_n,y)+{\phi }_k(y)-{\phi }_k\left({\mathrm{\Theta }}_n^k{x}_n\right)+〈B_k{\mathrm{\Theta }}_n^{k-1}{x}_n,y-{\mathrm{\Theta }}_n^k{x}_n〉\\ +\frac{1}{r_{k,n}}〈y-{\mathrm{\Theta }}_n^k{x}_n,{\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k-1}{x}_n〉\ge 0.\end{array}$$

By (A2), we have

$$\begin{array}{r}{\phi }_k(y)-{\phi }_k\left({\mathrm{\Theta }}_n^k{x}_n\right)+〈B_k{\mathrm{\Theta }}_n^{k-1}{x}_n,y-{\mathrm{\Theta }}_n^k{x}_n〉+\frac{1}{r_{k,n}}〈y-{\mathrm{\Theta }}_n^k{x}_n,{\mathrm{\Theta }}_n^k{x}_n-{\mathrm{\Theta }}_n^{k-1}{x}_n〉\\ \ge F_k(y,{\mathrm{\Theta }}_n^k{x}_n).\end{array}$$

Replacing n by $n_i$ in the above inequality, we have

$$\begin{array}{r}{\phi }_k(y)-{\phi }_k\left({\mathrm{\Theta }}_{n_i}^k{x}_{n_i}\right)+〈B_k{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i},y-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}〉+\frac{1}{r_{k,n_i}}〈y-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}〉\\ \ge F_k(y,{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}).\end{array}$$

Let $z_t=ty+(1-t)x^{\ast }$ for all $t\in (0,1]$ and $y\in C$. This implies that $z_t\in C$. Then we have

$$\begin{array}{r}〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},B_k{z}_t〉\\ \ge 〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},B_k{z}_t〉+{\phi }_k\left({\mathrm{\Theta }}_{n_i}^k{x}_{n_i}\right)-{\phi }_k(z_t)\\ -〈B_k{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i},z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}〉-\frac{1}{r_{k,n_i}}〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}〉+F_k(z_t,{\mathrm{\Theta }}_{n_i}^k)\\ ={\phi }_k\left({\mathrm{\Theta }}_{n_i}^k{x}_{n_i}\right)-{\phi }_k(z_t)+〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},B_k{z}_t-B_k{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}〉\\ +〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},B_k{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-B_k{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}〉\\ -〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},\frac{{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}}{r_{k,n_i}}〉+F_k(z_t,{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}).\end{array}$$

By (3.18) we have $\parallel B_k{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-B_k{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}\parallel \to 0$ as $i\to \mathrm{\infty }$. Furthermore, by the monotonicity of $B_k$, we obtain $〈z_t-{\mathrm{\Theta }}_{n_i}^k{x}_{n_i},B_k{z}_t-B_k{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}〉\ge 0$. Then from (A4), the lower semicontinuity of φ and

$$\frac{{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}-{\mathrm{\Theta }}_{n_i}^{k-1}{x}_{n_i}}{r_{k,n_i}}\to 0,{\mathrm{\Theta }}_{n_i}^k{x}_{n_i}\rightharpoonup x^{\ast },$$

we obtain that

$$〈z_t-x^{\ast },B_k{z}_t〉\ge {\phi }_k\left(x^{\ast }\right)-{\phi }_k(z_t)+F_k(z_t,x^{\ast }).$$

(3.27)

Using (A1), (A4) and (3.27), we have

$$\begin{array}{rcl}0& =& F_k(z_t,z_t)+{\phi }_k(z_t)-{\phi }_k(z_t)\\ \le & t{F}_k(z_t,y)+(1-t)F_k(z_t,x^{\ast })+t{\phi }_k(y)+(1-t){\phi }_k\left(x^{\ast }\right)-t{\phi }_k(z_t)-(1-t){\phi }_k(z_t)\\ =& t[F_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)]+(1-t)[F_k(z_t,x^{\ast })+{\phi }_k\left(x^{\ast }\right)-{\phi }_k(z_t)]\\ \le & t[F_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)]+(1-t)〈z_t-x^{\ast },B_k{z}_t〉\\ =& t[F_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)]+(1-t)t〈y-x^{\ast },B_k{z}_t〉,\end{array}$$

and hence

$$0\le F_k(z_t,y)+{\phi }_k(y)-{\phi }_k(z_t)+(1-t)〈y-x^{\ast },B_k{z}_t〉.$$

Letting $t\to 0$, we have, for each $y\in C$,

$$0\le F_k(x^{\ast },y)+{\phi }_k(y)-{\phi }_k\left(x^{\ast }\right)+〈y-x^{\ast },B_k{x}^{\ast }〉.$$

This implies that $x^{\ast }\in GMEP(F_k,{\phi }_k,B_k)$. Hence $x^{\ast }\in {\bigcap }_{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)$. Therefore,

$$x^{\ast }\in \mathcal{F}=[\bigcap _{k=1}^{M}GMEP(F_k,{\phi }_k,B_k)]\cap \mathcal{F}(G)\cap \mathcal{F}(S).$$

This together with the property of metric projection implies that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_0-\overline{x},x_n-\overline{x}〉=\underset{n\to \mathrm{\infty }}{lim}〈x_0-\overline{x},x_{n_i}-\overline{x}〉=〈x_0-\overline{x},x^{\ast }-\overline{x}〉\le 0.$$

(3.28)

Step 6. Finally, we can easily show that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$.

Indeed, from (3.14) and (3.3), we have

$$\begin{array}{rcl}{\parallel x_{n+1}-\overline{x}\parallel }^2& \le & \frac{2a_n}{1+a_n}〈x_0-\overline{x},x_{n+1}-\overline{x}〉+\frac{b_n}{1+a_n}{\parallel x_n-\overline{x}\parallel }^2+\frac{c_n+d_n}{1+a_n}{\parallel x_n-\overline{x}\parallel }^2\\ =& \frac{2a_n}{1+a_n}〈x_0-\overline{x},x_{n+1}-\overline{x}〉+(1-\frac{2a_n}{1+a_n}){\parallel x_n-\overline{x}\parallel }^2.\end{array}$$

It is clear that ${\sum }_{n=1}^{\mathrm{\infty }}\frac{2a_n}{1+a_n}=\mathrm{\infty }$. Hence, applying (3.28) and Lemma 2.11, we obtain immediately that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$. This completes the proof. □