Strong Convergence of a New Iterative Algorithm for Split Monotone Variational Inclusion Problems

Abstract

The main aim of this work is to introduce an implicit general iterative method for approximating a solution of a split variational inclusion problem with a hierarchical optimization problem constraint for a countable family of mappings, which are nonexpansive, in the setting of infinite dimensional Hilbert spaces. Convergence theorem of the sequences generated in our proposed implicit algorithm is obtained under some weak assumptions.

1. Introduction

Let $H_1$ be a real Hilbert space with norm $\parallel ·\parallel$ and inner product $\langle ·,·\rangle$. Suppose that C is a nonempty convex and set in $H_1$, and let $P_C$ be the metric (nearest point) projection from space $H_1$ onto set C. We use $T:C\to H_1$ to denote a mapping on C. Denote by $\mathrm{Fix}(T)$ the set of fixed points of T, i.e., $\mathrm{Fix}(T)=\{x\in C:x=Tx\}$. We use the notations ⇀ and → to indicate the weak convergence and the strong convergence, respectively.

Assume that $B:C\to H_1$ is a nonlinear mapping. The classical monotone variational inequality (VI) is to find $x^{*}\in C$ such that

$$0\le \langle B{x}^{*},x-x^{*}\rangle ,\forall x\in C.$$

(1)

We denote by VI $(C,B)$ the solution set of VI (1). VI (1), which acts as a very powerful and effective research tool, has been applied to study lots of theory problems arising in nonlinear equations, computational mechanics, optimization contact problems in control problems, elasticity, operations research, modern management science, bi-function equilibrium problems in transportation and economics, obstacle, unilateral, moving, etc.; see [1,2,3,4,5,6,7,8,9,10,11,12] and the references therein.

An operator D is said to be a strongly positive operator on $H_1$, if there is a constant $\overline{\xi }>0$ such that

$$\overline{\xi }{\parallel x\parallel }^2\le \langle Dx,x\rangle ,\forall x\in H_1.$$

Solution methods for Lipchitz mappings, in particular, nonexpansive mappings, have widely been applied to investigate minimization problems of various convex functions. A mapping $M:H_1\to 2^{H_1}$ is said to be set-valued monotone if for all $x,y\in H_1,x^{\prime }\in Mx$ and $y^{\prime }\in My$ imply $\langle x-y,x-x^{\prime }\rangle \ge 0$. Recall that $M:H_1\to 2^{H_1}$ is a maximal operator if the graph Gph $(M)$ is not properly contained in the graph of any other monotone operator. As we all know that M is maximal if and only if for $(x,f)\in H_1\times H_1,\langle x-y,f-g\rangle \ge 0$ for every $(y,y^{\prime })\in \mathrm{Gph}(M)$, we have $x^{\prime }\in Mx$.

We now assume that set-valued operator $M:H_1\to 2^{H_1}$ is maximal. We can define a single-valued mapping $J_{\lambda }^M:H_1\to H_1$ by

$$J_{\lambda }^M(x):={(\lambda M+I)}^{-1}(x),\forall x\in H_1,$$

is called the resolvent operator associated with mapping M. It deserves mentioning that it is single-valued and Liptchitz.

Let $H_2$ be another Hilbert space with usual norm ($\parallel ·\parallel$) and inner product ($\langle ·,·\rangle$). Let A be a bounded linear operator from $H_1$ to $A_2$. We consider in this paper the following split variational inclusion problem (SVIP): find $x^{*}\in H_1$ such that

$$0\in B_1(x^{*}),$$

(2)

and

$$A{x}^{*}=y^{*}\in H_2\mathrm{solves}0\in B_2(y^{*}),$$

(3)

where $B_1:H_1\to 2^{H_1}$ and $B_2:H_2\to 2^{H_2}$ are set-valued and maximal monotone. SOLVIP($B_1$) stands for the solution set of (2) and SOLVIP($B_2$) stands for the solution set of (3), respectively. The solution set of SVIP (2)–(3) will be used and denoted by $\mathsf{\Gamma }$. From [13], we know that SVIP (2)–(3) is equivalent to approximating $x^{*}\in H_1$ with $x^{*}=J_{\lambda }^{B_1}(x^{*})$ such that

$$A{x}^{*}=y^{*}\in H_2\mathrm{and}{y}^{*}=J_{\lambda }^{B_2}(y^{*}),$$

holds for any given $\lambda >0$. It is remarkable that if $\mathsf{\Gamma }\ne \varnothing$, then

$$\langle x-J_{\lambda }^{B_1}x,J_{\lambda }^{B_1}x-y\rangle \ge 0,\forall x\in H_1,y\in \mathrm{SOLVIP}(B_1),$$

and

$$\langle v-J_{\lambda }^{B_2}v,J_{\lambda }^{B_2}v-w\rangle \ge 0,\forall v\in H_2,w\in \mathrm{SOLVIP}(B_2).$$

Let ${\{S_i\}}_{i=1}^{\infty }$ be a countable family mappings on $H_1$. We assume that ${\{{\zeta }_i\}}_{i=1}^{\infty }$ is a real sequence in $[0,1]$. For any $n\ge 1$, we give a $W_n$ mapping by:

$$\left\{\begin{array}{c}{U}_{n,n+1}=I,\hfill \\ U_{n,n}=(1-{\zeta }_n)I+{\zeta }_n{S}_n{U}_{n,n+1},\hfill \\ U_{n,n-1}=(1-{\zeta }_{n-1})I+{\zeta }_{n-1}{S}_{n-1}{U}_{n,n},\hfill \\ \cdots \hfill \\ U_{n,k}=(1-{\zeta }_k)I+{\zeta }_k{S}_k{U}_{n,k+1},\hfill \\ U_{n,k-1}=(1-{\zeta }_{k-1})I+{\zeta }_{k-1}{S}_{k-1}{U}_{n,k},\hfill \\ \cdots \hfill \\ U_{n,2}=(1-{\zeta }_2)I+{\zeta }_2{S}_2{U}_{n,3},\hfill \\ W_n=U_{n,1}=(1-{\zeta }_1)I+{\zeta }_1{S}_1{U}_{n,2}.\hfill \end{array}\right.$$

(4)

If each $S_i$ is nonexpansive, then $W_n$ is Lipschitz. Indeed, it is also nonexpansive and called a W-mapping defined by $S_n,S_{n-1},\dots ,S_1$ and ${\zeta }_n,{\zeta }_{n-1},\dots ,{\zeta }_1$. From [14], we know that $W_n$ is a nonexpansive mapping with the relation $\mathrm{Fix}(W_n)={\bigcap }_{i=1}^n\mathrm{Fix}(S_i)$, for each $n\ge 1$; for each $x\in H_1$ and for each positive integer k, the ${lim}_{n\to \infty }{U}_{n,k}x$ exists; W is defined by

$$Wx:=\underset{n\to \infty }{lim}{W}_{n}x=\underset{n\to \infty }{lim}{U}_{n,1}x,\forall x\in H_1$$

has the nonexpansivity and it satisfies $\mathrm{Fix}(W)={\bigcap }_{i=1}^{\infty }\mathrm{Fix}(S_i)$ (We call a W-mapping generated by $S_1,S_2,\dots$ and ${\zeta }_1,{\zeta }_2,\dots$). Recently, common fixed-point problems, which finds applications in signal process and medical image restoration, have been studied based on mean-valued or projection methods; see [14,15,16,17,18,19] and references cited therein.

In this present work, we investigate an implicit general iterative method for computing a solution of the SVIP with a hierarchical optimization problem constraint for a countable family of mappings, which will be assumed to have the nonexpansivity, in the framework of real Hilbert spaces. Norm convergence theorems of the sequences generated by our implicit general algorithm are established under some suitable assumptions. Our results extend, unify, develop and improve the corresponding ones in the recent literature.

2. Preliminaries

Now we list some basic notations and facts. $H_1$ will be assumed to be a real Hilbert space and C will be assume to be a closed nonempty convex subset in $H_1$. A mapping $F:C\to H_1$ is called a $\kappa$-Lipschitzian mapping if there is a number $\kappa >0$ with $\kappa \parallel x-y\parallel \ge \parallel F(x)-F(y)\parallel$, $\forall x,y\in C$. In particular, if $\kappa =1$, then F is said to be a nonexpansive operator. If $\kappa <1$, then F is said to be a contractive operator. A mapping $F:C\to H_1$ is said to be $\eta$-strongly monotone if there exists a number $\eta >0$ such that ${\eta \parallel x-y\parallel }^2\le \langle x-y,Fx-Fy\rangle ,$ $\forall x,y\in C$. In all Hilbert spaces, we known

$${\lambda \parallel x\parallel }^2+{(1-\lambda )\parallel y\parallel }^2{-\lambda (1-\lambda )\parallel x-y\parallel }^2={\parallel \lambda x+(1-\lambda )y\parallel }^2,$$

for all $x,y\in H_1$ and $\lambda \in [0,1]$.

Fixing $x\in H_1$, we see that there exists a unique nearest point in closed convex set C. We denote it by $P_{C}x$. $\parallel x-y\parallel \ge \parallel x-P_{C}x\parallel ,$ $\forall y\in C$. The mapping $P_C$ is called the metric or nearest point projection of $H_1$ onto C. We know that $P_C$ is an nonexpansive operator from space $H_1$ onto set C. In addition, we know that

$$\langle x-y,P_{C}x-P_{C}y\rangle \ge {\parallel P_{C}y-P_{C}x\parallel }^2,\forall x,y\in H_1.$$

(5)

$P_{C}x$ also enjoys

$$\langle x-P_{C}x,y-P_{C}x\rangle \le 0,$$

(6)

for all $x\in H_1$ and $y\in C$. It is not too hard to see that (6) is equivalent to the following relation

$${\parallel x-y\parallel }^2-\parallel x-P_C{x\parallel }^2\ge {\parallel y-P_{C}x\parallel }^2,\forall x\in H_1,y\in C.$$

(7)

It is also not hard to find that every nonexpansive operator $S:H_1\to H_1$ satisfies the following relation

$$\langle (I-S)x-(I-S)y,Sy-Sx\rangle \le \frac{1}{2}{\parallel (I-S)x-(I-S)y\parallel }^2,\forall (x,y)\in H_1\times H_1.$$

(8)

In particular, one has

$$\langle (I-S)x,y-Sx\rangle \le \frac{1}{2}{\parallel (I-S)x\parallel }^2,\forall (x,y)\in H_1\times \mathrm{Fix}(S).$$

(9)

Let $T:H_1\to H_1$ be a self mapping. It is said to be an averaged operator if it is a combination of the identity operator I and a nonexpansivity operator, that is, $T\equiv (1-\alpha )I+\alpha S,$ where $\alpha \in (0,1)$ and $S:H_1\to H_1$ is an nonexpansive operator. We mention that the class of averaged mappings are of course nonexpansive. Also, mappings, which are firmly nonexpansive are averaged. Indeed, projections on convex nonempty closed sets and resolvent operators of set-valued monotone operators. Some important properties and relations of averaged mappings are gathered in the following lemma; see e.g., [20,21,22,23,24,25] and references cited therein.

Lemma 1.

For any given $\lambda >0$, let the mapping $G:H_1\to H_1$ be defined as $G:=J_{\lambda }^{B_1}(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$ where $\gamma \in (0,\frac{1}{L})$, L is the spectral radius of the operator $A^{*}A$ and $A^{*}$ is the adjoint of A. Then G is a nonexpansive mapping. If $\mathsf{\Gamma }\ne \varnothing$, then $\mathsf{\Gamma }=\mathrm{Fix}(G)$.

Proof. 

Since $J_{\lambda }^{B_1}$ and $J_{\lambda }^{B_2}$ are mappings enjoys the firm nonexpansivity, they, of course, are averaged. For $L\gamma \in (0,1)$, the mapping $(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$ is averaged. So $G:=J_{\lambda }^{B_1}(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$ is a averaged operator and hence a nonexpansive operator.

Next, let us show that if $\mathsf{\Gamma }\ne \varnothing$ then $\mathsf{\Gamma }=\mathrm{Fix}(G)$. Indeed, it is clear that $\mathsf{\Gamma }\subseteq \mathrm{Fix}(G)$. Conversely, we take $p\in \mathrm{Fix}(G)$ and $q\in \mathsf{\Gamma }$ arbitrarily. Then $J_{\lambda }^{B_1}(p+\gamma A^{*}(J_{\lambda }^{B_2}-I)Ap)=p$. Hence,

$$\langle (p+\gamma A^{*}(J_{\lambda }^{B_2}-I)Ap)-p,p-u\rangle \ge 0,\forall u\in \mathrm{SOLVIP}(B_1),$$

which immediately yields

$$\langle J_{\lambda }^{B_2}Ap-Ap,Ap-Au\rangle \ge 0,\forall u\in \mathrm{SOLVIP}(B_1).$$

One has

$$\langle J_{\lambda }^{B_2}Ap-Ap,v-J_{\lambda }^{B_2}Ap\rangle \ge 0,\forall v\in \mathrm{SOLVIP}(B_2).$$

Using the last two inequalities, we obtain

$$\langle J_{\lambda }^{B_2}Ap-Ap,v-J_{\lambda }^{B_2}Ap+Ap-Au\rangle \ge 0,\forall u\in \mathrm{SOLVIP}(B_1),v\in \mathrm{SOLVIP}(B_2),$$

which immediately leads to

$$\langle J_{\lambda }^{B_2}Ap-Ap,v-Au\rangle \ge {\parallel J_{\lambda }^{B_2}Ap-Ap\parallel }^2,\forall u\in \mathrm{SOLVIP}(B_1),v\in \mathrm{SOLVIP}(B_2).$$

(10)

Taking into account $q\in \mathsf{\Gamma }$, one knows that $q\in \mathrm{SOLVIP}(B_1)$ and $Aq\in \mathrm{SOLVIP}(B_2)$. So it follows from (10) that $J_{\lambda }^{B_2}Ap=Ap$, i.e., $Ap\in \mathrm{SOLVIP}(B_2)$. Also, from $p\in \mathrm{Fix}(G)$ we get

$$p=J_{\lambda }^{B_1}(p+\gamma A^{*}(J_{\lambda }^{B_2}-I)Ap)=J_{\lambda }^{B_1}p.$$

Hence, $p\in \mathrm{SOLVIP}(B_1)$. Consequently, $p\in \mathsf{\Gamma }$. This completes the proof. □

Lemma 2.

[26], Let ${\{S_i\}}_{i=1}^{\infty }$ be a countable family on a real Hilbert space $H_1$ with the restriction ${\bigcap }_{i=1}^{\infty }\mathrm{Fix}(S_i)\ne \varnothing$. ${\{{\zeta }_i\}}_{i=1}^{\infty }$ will be assumed to be a sequence in $(0,l]$ for some $l\in (0,1]$. If C is any bounded set in $H_1$ and each $S_i$ is the self nonexpansivity, then ${lim}_{n\to \infty }{sup}_{x\in C}Wx-W_{n}x=0.$

Through the rest of this paper, ${\{{\zeta }_i\}}_{i=1}^{\infty }$ will be assumed to be in $(0,l]$ for some $l\in (0,1)$.

Lemma 3.

[27], Assume that both $\{x_n\}$ and $\{z_n\}$ are bounded real sequences in infinite dimensional space either Banach or Hilbert. We support that $\{{\beta }_n\}$ is a sequence with the restriction that it is bounded away from $[0,1]$, that is, $0<{lim\; inf}_{n\to \infty }{\beta }_n\le {lim\; sup}_{n\to \infty }{\beta }_n<1$. We assume $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$ $\forall n\ge 0$ and ${lim\; sup}_{n\to \infty }(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$ Hence, ${lim}_{n\to \infty }\parallel x_n-z_n\parallel =0$.

Lemma 4.

[28], Let C be a closed nonempty convex set in a real Hilbert space $H_1$, and let $B:C\to H_1$ be a monotone and hemicontinuous mapping. We the following:

(i) $\mathrm{VI}(C,B)=\{x^{*}\in C:\langle By,y-x^{*}\rangle \ge 0,\forall y\in C\}$;

(ii) $\mathrm{VI}(C,B)=\mathrm{Fix}(P_C(I-\lambda B))$ for all $\lambda >0$;

(iii) $\mathrm{VI}(C,B)$ is singleton, if B is strongly monotone and Lipschitz continuous.

Lemma 5.

[29], All Hilbert spaces satisfies the well known Opial condition: the inequality ${lim\; inf}_{n\to \infty }\parallel x_n-y\parallel \ge {lim\; inf}_{n\to \infty }\parallel x_n-x\parallel$ holds for every $y\ne x$ and for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$.

Lemma 6.

[30], Assume that S is a nonexpansive self-mapping on a closed convex nonempty set C in $H_1$. If S is fixed-point free, then $I-S$ is demi-closed at zero, i.e., if $\{x_n\}$ is a sequence in C weakly converging to some x in the set and the sequence $\{(S-I)x_n\}$ converges strongly to zero, then $(S-I)x=0$, where I stands for the identity operator.

Lemma 7.

[31], Assume that $\{a_n\}$ be a real iterative sequence with the conditions $a_{n+1}-(1-{\lambda }_n)a_n\le {\lambda }_n{\gamma }_n,$ $\forall n\ge 0,$ where $\{{\lambda }_n\}$ and $\{{\gamma }_n\}$ are real sequences with the restrictionis $\{{\lambda }_n\}\subset [0,1]$ and ${\sum }_{n=0}^{\infty }{\lambda }_n=\infty$, ${lim\; sup}_{n\to \infty }{\gamma }_n\le 0$. Then ${lim}_{n\to \infty }{a}_n=0$.

3. Main Results

Theorem 1.

Let $A:H_1\to H_2$, where $H_1$ and $H_2$ are two different Hilbert spaces, be a linearly bounded operator. Suppose that $B_1:H_1\to 2^{H_1}$ and $B_2:H_2\to 2^{H_2}$ are maximal monotone mappings. Let $f:H_1\to H_1$ be a contraction mapping with contractive coefficient $\alpha \in (0,1)$ and let the linearly bounded operator $D:H_1\to H_1$ be strongly positive with coefficient $\overline{\xi }>0$ and $0<\xi <\frac{\overline{\xi }}{\alpha }$. Let the mapping $G:H_1\to H_1$ be defined as $G:=J_{\lambda }^{B_1}(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$, where $\lambda >0$, $\gamma \in (0,\frac{1}{L}),$ L be the spectral radius of $A^{*}A$ and $A^{*}$ is the adjoint operator of A. Assume that $\mathsf{\Omega }:=({\bigcap }_{i=1}^{\infty }\mathrm{Fix}(S_i))\cap \mathsf{\Gamma }\ne \varnothing$. For an arbitrary $x_1\in H_1$, we define $\{x_n\}$ and $\{y_n\}$ by

$$\left\{\begin{array}{c}{y}_n=G((1-{\gamma }_n)W_n{y}_n+{\gamma }_n{x}_n),\hfill \\ x_{n+1}={\alpha }_n\xi f(x_n)+{\beta }_n{x}_n+[(1-{\beta }_n)I-{\alpha }_{n}D]y_n,\forall n\ge 1,\hfill \end{array}\right.$$

(11)

where $\{W_n\}$ is defined in (4), and $\{{\alpha }_n\},\{{\beta }_n\}$ and $\{{\gamma }_n\}$ are real number sequences in $(0,1)$. Suppose the parameter sequences satisfy the following three restrictions:

(C1) ${\{{\beta }_n\}}_{n=1}^{\infty }\subset [a,b]$ for some $a,b\in (0,1)$;

(C2) ${\alpha }_n\to 0$ as $n\to \infty$ and ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$;

(C3) $1>{lim\; sup}_{n\to \infty }{\gamma }_n\ge {lim\; inf}_{n\to \infty }{\gamma }_n>0$ and ${lim}_{n\to \infty }|{\gamma }_{n+1}-{\gamma }_n|=0$.

Then $\{x_n\}$ converges to a point $z\in \mathsf{\Omega }$ in norm and z is a solution to

$$\langle (D-\xi f)z,z-p\rangle \le 0,\forall p\in \mathsf{\Omega },$$

that is, $P_{\mathsf{\Omega }}(z-Dz+\xi f(z))=z$.

Proof. 

First of all, taking into account that ${\alpha }_n\to 0$ as $n\to \infty$ and $1>{lim\; sup}_{n\to \infty }{\gamma }_n\ge {lim\; inf}_{n\to \infty }{\gamma }_n>0$, we can suppose $\{{\alpha }_n(\overline{\xi }-\xi \alpha )\}\subset (0,1)$ and $\{{\gamma }_n\}\subset [c,d]\subset (0,1)$ for some $c,d\in (0,1)$. Please note that the mapping $G:H_1\to H_1$ is defined as $G:=J_{\lambda }^{B_1}(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$, where $\lambda >0$, $L\gamma \in (0,1),L$ is the radius of the operator $A^{*}A$. By virtue of Lemma 1, we get that G is is nonexpansivity. It is easy to see that there exists an element $y_n\in H_1$, which is unique, such that

$$y_n=G({\gamma }_n{x}_n+(1-{\gamma }_n)W_n{y}_n).$$

(12)

Define a mapping $F_n$ by

$$F_{n}x=G({\gamma }_n{x}_n+(1-{\gamma }_n)W_{n}x),\forall x\in H_1.$$

Since each $W_n:H_1\to H_1$ is a nonexpansive mapping, we deduce that, all $x,y\in H_1$,

$$\begin{array}{cc}\parallel F_{n}x-F_{n}y\parallel \hfill & =\parallel G({\gamma }_n{x}_n+(1-{\gamma }_n)W_{n}x)-G({\gamma }_n{x}_n+(1-{\gamma }_n)W_{n}y)\parallel \hfill \\ & \le (1-{\gamma }_n)\parallel W_{n}y-W_{n}x\parallel \le (1-{\gamma }_n)\parallel x-y\parallel .\hfill \end{array}$$

Also, from $\{{\gamma }_n\}\subset [c,d]\subset (0,1)$, we get $1>1-{\gamma }_n>0$ for all $n\ge 1$. Thus, $F_n$ is a contraction operator. This shows that there exists an element $y_n\in C$, satisfying (12). Indeed, $y_n$ is also unique. So, it can be readily seen that the general implicit iterative scheme (11) can be rewritten as

$$\left\{\begin{array}{c}{u}_n={\gamma }_n{x}_n+(1-{\gamma }_n)W_n{y}_n,\hfill \\ y_n=J_{\lambda }^{B_1}(\gamma A^{*}(J_{\lambda }^{B_2}-I)A{u}_n+u_n),\hfill \\ x_{n+1}={\alpha }_n\xi f(x_n)+[(1-{\beta }_n)I-{\alpha }_{n}D]y_n+{\beta }_n{x}_n,\forall n\ge 1.\hfill \end{array}\right.$$

(13)

Next, we divide the rest of our proofs into some steps to prove this theorem.

Step 1. We prove that $\{x_n\},\{y_n\},\{u_n\},\{f(x_n)\}$ and $\{W_n{y}_n\}$ are bounded sequence in $H_1$. By arbitrarily taking an element $p\in \mathsf{\Omega }={\bigcap }_{n=1}^{\infty }\mathrm{Fix}(S_n)\cap \mathsf{\Gamma }$, we get $p=J_{\lambda }^{B_1}p,Ap=J_{\lambda }^{B_2}(Ap)$ and $W_{n}p=p$ $\forall n\ge 1$. Since each $W_n:H_1\to H_1$ is a nonexpansive operator, it follows that

$$\begin{array}{cc}\parallel u_n-p\parallel \hfill & \le {\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n)\parallel p-W_n{y}_n\parallel \hfill \\ & \le {\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n)\parallel p-y_n\parallel .\hfill \end{array}$$

(14)

Note that

$$\begin{array}{c}\parallel y_n{-p\parallel }^2\hfill \\ \le \parallel u_n+\gamma A^{*}(J_{\lambda }^{B_2}-I)A{u}_n{-p\parallel }^2\hfill \\ \le \parallel u_n{-p\parallel }^2+{\gamma }^2\langle (J_{\lambda }^{B_2}-I)A{u}_n,A{A}^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle +2\gamma \langle u_n-p,A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle .\hfill \end{array}$$

(15)

Please note that

$$\begin{array}{cc}L{\gamma }^2{\parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2& =L{\gamma }^2\langle (J_{\lambda }^{B_2}-I)A{u}_n,(J_{\lambda }^{B_2}-I)A{u}_n\rangle \hfill \\ & \ge {\gamma }^2\langle (J_{\lambda }^{B_2}-I)A{u}_n,A{A}^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle .\hfill \end{array}$$

(16)

By considering item $2\gamma \langle u_n-p,A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle$ and by using (9), we have

$$\begin{array}{c}2\gamma \langle u_n-p,A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle =2\gamma \langle A(u_n-p),(J_{\lambda }^{B_2}-I)A{u}_n\rangle \hfill \\ =2\gamma \{\langle Ap-J_{\lambda }^{B_2}A{u}_n,A{u}_n-J_{\lambda }^{B_2}A{u}_n\rangle -\parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2\}\hfill \\ \le 2\gamma \{\frac{1}{2}\parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2-\parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2\}\hfill \\ =-\gamma \parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2.\hfill \end{array}$$

(17)

Using inequalities (15), (16) and (17), we obtain

$$\begin{array}{cc}\parallel y_n{-p\parallel }^2\hfill & \le \parallel u_n{-p\parallel }^2+L{\gamma }^2\parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2-\gamma {\parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2\hfill \\ & =\parallel u_n{-p\parallel }^2+\gamma (L\gamma -1){\parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2.\hfill \end{array}$$

(18)

From $(L\gamma )\in (0,1)$, we get

$$\parallel y_n-p\parallel \le \parallel u_n-p\parallel ,\forall n\ge 1.$$

(19)

Substituting (19) for (14), we have

$$\parallel u_n-p\parallel \le (1-{\gamma }_n)\parallel u_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel ,$$

which combining (19) yields that

$$\parallel y_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel ,\forall n\ge 1.$$

(20)

Thanks to the two restrictions (C1) and (C2), we can suppose that ${\alpha }_n\le (1-{\beta }_n){\parallel D\parallel }^{-1}$, $\forall n\ge 1$. Since D is linearly strongly positive bounded, we can easily get that

$$1-{\beta }_n-{\alpha }_n\overline{\xi }\ge \parallel (1-{\beta }_n)I-{\alpha }_{n}D\parallel .$$

(21)

In view of (13), (20) and (21), one has that

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & \le {\alpha }_n\xi \parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel \xi f(p)-Dp\parallel +{\beta }_n\parallel x_n-p\parallel \hfill \\ & +\parallel [(1-{\beta }_n)I-{\alpha }_{n}D](y_n-p)\parallel \hfill \\ & \le {\alpha }_n\xi \alpha \parallel x_n-p\parallel +{\alpha }_n\parallel \xi f(p)-Dp\parallel +{\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n-{\alpha }_n\overline{\xi })\parallel x_n-p\parallel \hfill \\ & \le max\{\parallel p-x_1\parallel ,\frac{\parallel \xi f(p)-Dp\parallel }{\overline{\xi }-\xi \alpha }\},\forall n\ge 1.\hfill \end{array}$$

It immediately yields that $\{x_n\}$ is a bounded sequence in $H_1$. Indeed, $\{y_n\},\{u_n\},\{f(x_n)\},\{W_n{y}_n\}$ and $\{D{y}_n)\}$ (due to (20) and the Lipschitz continuity of $W_n,D$ and f) are bounded sequences. From this, we fix a bounded subset $C\subset H_1$ with the restriction

$$u_n,x_n,y_n\in C,\forall n\ge 1.$$

(22)

Step 2. We aim that $\parallel x_{n+1}-x_n\parallel \to 0$ and $\parallel y_{n+1}-y_n\parallel \to 0$ as $n\to \infty$. Indeed, we set

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)v_n,\forall n\ge 1.$$

(23)

This shows that

$$\begin{array}{cc}{v}_n\hfill & =\frac{1}{1-{\beta }_n}\{{\alpha }_n\xi f(x_n)+{\beta }_n{x}_n+y_n-{\beta }_n{y}_n-{\alpha }_{n}D{y}_n\}-\frac{{\beta }_n}{1-{\beta }_n}{x}_n\hfill \\ & =\frac{{\alpha }_n}{1-{\beta }_n}(\xi f(x_n)-D{y}_n)+y_n.\hfill \end{array}$$

(24)

Hence,

$$\begin{array}{c}\parallel v_{n+1}-v_n\parallel \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel \xi f(x_{n+1})-D{y}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \xi f(x_n)-D{y}_n\parallel +\parallel y_{n+1}-y_n\parallel .\hfill \end{array}$$

(25)

By using Lemma 1, we know that $G:=J_{\lambda }^{B_1}(I+\gamma A^{*}(J_{\lambda }^{B_2}-I)A)$ is Lipchitz. Indeed, it is nonexpansive. Hence, we obtain from (13) that

$$\parallel y_{n+1}-y_n\parallel =\parallel G{u}_{n+1}-G{u}_n\parallel \le \parallel u_{n+1}-u_n\parallel .$$

(26)

However, we have that

$$\begin{array}{c}{\displaystyle \underset{x\in C}{sup}}[\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel ]+\parallel u_{n+1}-u_n\parallel \hfill \\ \ge \parallel W_{n+1}{y}_{n+1}-W_n{y}_{n+1}\parallel +\parallel W_n{y}_{n+1}-W_n{y}_n\parallel \hfill \\ \ge \parallel W_{n+1}{y}_{n+1}-W_n{y}_n\parallel ,\hfill \end{array}$$

(27)

where C stands for the bounded subset in $H_1$ defined by (22). Simple calculations show that

$$\begin{array}{cc}\parallel u_{n+1}-u_n\parallel \hfill & \le (1-{\gamma }_{n+1})\parallel W_{n+1}{y}_{n+1}-W_n{y}_n\parallel +{\gamma }_{n+1}\parallel x_{n+1}-x_n\parallel \hfill \\ & +|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-W_n{y}_n\parallel \hfill \\ & \le {\gamma }_{n+1}\parallel x_{n+1}-x_n\parallel +(1-{\gamma }_{n+1})\{{\displaystyle \underset{x\in C}{sup}}[\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel ]\hfill \\ & +\parallel u_{n+1}-u_n\parallel \}+|{\gamma }_{n+1}-{\gamma }_n|\parallel x_n-W_n{y}_n\parallel .\hfill \end{array}$$

(28)

So it yields from $\{{\gamma }_n\}\subset [c,d]$ that

$$\parallel u_{n+1}-u_n\parallel \le \parallel x_{n+1}-x_n\parallel +\frac{1}{c}{\displaystyle \underset{x\in C}{sup}}[\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel ]+|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-W_n{y}_n\parallel }{c}.$$

(29)

Thus, from (25), (26) and (29) we deduce that

$$\begin{array}{l}\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel \xi f(x_{n+1})-D{y}_{n+1}\parallel +\frac{{\alpha }_n}{1-{\beta }_n}\parallel \xi f(x_n)-D{y}_n\parallel \\ +\frac{1}{c}{\displaystyle \underset{x\in C}{sup}}[\parallel W_{n+1}x-Wx\parallel +\parallel Wx-W_{n}x\parallel ]+|{\gamma }_{n+1}-{\gamma }_n|\frac{\parallel x_n-W_n{y}_n\parallel }{c}\\ \ge \parallel v_{n+1}-v_n\parallel -\parallel x_{n+1}-x_n\parallel .\end{array}$$

Thanks to the three assumptions (C1), (C2), (C3), and Lemma 2,

$$\underset{n\to \infty }{lim\; sup}(\parallel v_{n+1}-v_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

From Lemma 3, we thus obtain that

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

(30)

This in turn implies that

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

(31)

This together with (26) and (29), implies that

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

(32)

Step 3. We aim to prove $\parallel x_n-u_n\parallel \to 0,\parallel x_n-y_n\parallel \to 0,\parallel y_n-W_n{y}_n\parallel \to 0$ and $\parallel x_n-G{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, we set $f_n=\xi f(x_n)-D{y}_n$ for all $n\ge 1$. For any $p\in \mathsf{\Omega }$, we observe that

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le \parallel {\beta }_n(x_n-p)+(1-{\beta }_n)(y_n-p){\parallel }^2+2\langle {\alpha }_n{f}_n,x_{n+1}-p\rangle \hfill \\ & \le (1-{\beta }_n)\parallel y_n{-p\parallel }^2+{\beta }_n\parallel x_n{-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-y_n\parallel }^2\hfill \\ & +2{\alpha }_n\parallel f_n\parallel \parallel x_{n+1}-p\parallel \hfill \\ & \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n){\parallel y_n-p\parallel }^2+2{\alpha }_n{M}^2,\hfill \end{array}$$

(33)

where $M=max\{{sup}_{n\ge 1}\parallel f_n\parallel ,{sup}_{n\ge 1}\parallel x_n-p\parallel \}$. Substituting (18) for (33), we obtain from (20) that $\parallel p-x_{n+1}{\parallel }^2\le \parallel p-x_n{\parallel }^2-\gamma (1-{\beta }_n)(1-L\gamma ){\parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2+2{\alpha }_n{M}^2.$ Therefore,

$$\underset{n\to \infty }{lim}\parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel =0.$$

(34)

From the assumption that $J_{\lambda }^{B_1}$ is a firmly nonexpansive mapping, we have

$$\begin{array}{cc}2\parallel y_n{-p\parallel }^2\hfill & \le 2\langle y_n-p,u_n+\gamma A^{*}(J_{\lambda }^{B_2}-I)A{u}_n-p\rangle \hfill \\ & =\parallel y_n{-p\parallel }^2+{\parallel u_n-p\parallel }^2+2\gamma \langle u_n-p,A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\rangle \hfill \\ & +{\gamma }^2\parallel A^{*}(J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2-{\parallel y_n-u_n-\gamma A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2\hfill \\ & \le \parallel y_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-\gamma \parallel (J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2+{\gamma }^2{\parallel A^{*}(J_{\lambda }^{B_2}-I)A{u}_n\parallel }^2\hfill \\ & -\parallel y_n-u_n-\gamma A^{*}(J_{\lambda }^{B_2}-I)A{u}_n{\parallel }^2\hfill \\ & \le \parallel y_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-\parallel y_n-u_n{\parallel }^2+2\gamma \parallel A(y_n-u_n)\parallel \parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel .\hfill \end{array}$$

Hence, we obtain

$$\parallel y_n{-p\parallel }^2\le -\parallel y_n-u_n{\parallel }^2+\parallel u_n{-p\parallel }^2+2\gamma \parallel A(y_n-u_n)\parallel \parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel .$$

(35)

Substituting (35) for (33), one concludes from (20) that

$$\begin{array}{cc}(1-{\beta }_n){\parallel y_n-u_n\parallel }^2\hfill & \le \parallel x_n-p\parallel \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-p\parallel \parallel x_n-x_{n+1}\parallel \hfill \\ & +2\gamma (1-{\beta }_n)\parallel A(y_n-u_n)\parallel \parallel (J_{\lambda }^{B_2}-I)A{u}_n\parallel +2{\alpha }_n{M}^2.\hfill \end{array}$$

(C1), (C2), (31), and (34) send us to

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

(36)

Also, according to (11) and (19) we have

$$\begin{array}{cc}\parallel p-u_n{\parallel }^2\hfill & \le {\gamma }_n\langle u_n-p,x_n-p\rangle +(1-{\gamma }_n)\parallel p-W_n{y}_n\parallel \parallel p-u_n\parallel \hfill \\ & \le {\gamma }_n\langle u_n-p,x_n-p\rangle +(1-{\gamma }_n){\parallel p-u_n\parallel }^2,\hfill \end{array}$$

which immediately leads to

$$\parallel u_n{-p\parallel }^2\le \frac{1}{2}[\parallel u_n{-p\parallel }^2-\parallel x_n-u_n{\parallel }^2+\parallel x_n-p{\parallel }^2].$$

It follows from (19) and (33) that

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\beta }_n\parallel p-x_n{\parallel }^2+(1-{\beta }_n)\parallel p-y_n{\parallel }^2-{\beta }_n(1-{\beta }_n)\parallel x_n-y_n{\parallel }^2+2{\alpha }_n\parallel f_n\parallel \parallel p-x_{n+1}\parallel \hfill \\ \le {\beta }_n\parallel p-x_n{\parallel }^2+(1-{\beta }_n)\parallel p-y_n{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-y_n\parallel }^2+2{\alpha }_n{M}^2\hfill \\ \le {\beta }_n\parallel p-x_n{\parallel }^2+(1-{\beta }_n)[\parallel p-x_n{\parallel }^2-\parallel x_n-u_n{\parallel }^2]-{\beta }_n(1-{\beta }_n){\parallel x_n-y_n\parallel }^2+2{\alpha }_n{M}^2\hfill \\ =\parallel x_n{-p\parallel }^2-(1-{\beta }_n)\parallel x_n-u_n{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-y_n\parallel }^2+2{\alpha }_n{M}^2.\hfill \end{array}$$

This implies that

$$\begin{array}{c}(1-{\beta }_n)\parallel x_n-u_n{\parallel }^2+{\beta }_n(1-{\beta }_n){\parallel x_n-y_n\parallel }^2\hfill \\ \le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+2{\alpha }_n{M}^2\hfill \\ \le \parallel x_n-p\parallel \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-p\parallel \parallel x_n-x_{n+1}\parallel +2{\alpha }_n{M}^2.\hfill \end{array}$$

(C1), (C2), and (3.21) send us to

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

(37)

Noticing that $\parallel u_n-x_n\parallel =(1-{\gamma }_n)\parallel W_n{y}_n-x_n\parallel \ge (1-d)\parallel W_n{y}_n-x_n\parallel$,

$$\parallel y_n-W_n{y}_n\parallel \le \parallel y_n-x_n\parallel +\parallel x_n-W_n{y}_n\parallel ,$$

and

$$\parallel x_n-G{x}_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel +\parallel G{u}_n-G{x}_n\parallel \le 2\parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel ,$$

we deduce from (36) and (37) that

$$\underset{n\to \infty }{lim}\parallel x_n-W_n{y}_n\parallel =0,\underset{n\to \infty }{lim}\parallel y_n-W_n{y}_n\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel x_n-G{x}_n\parallel =0.$$

(38)

Step 4. We aims to ${lim\; sup}_{n\to \infty }\langle (\xi f-D)z,x_n-z\rangle \le 0$, where z denotes the fixed-point of mapping $P_{\mathsf{\Omega }}(I-D+\xi f)$. Indeed, we first show that $\mathrm{VI}(\mathsf{\Omega },D-\xi f)$ consists of one point. As a matter of fact, we note that linear bounded operator D is strongly positive with its coefficient $\overline{\xi }>0$ and $0<\xi \alpha <\overline{\xi }$. Then for any $x,y\in H_1$, we have

$$\langle (D-\xi f)x-(D-\xi f)y,x-y\rangle \ge \overline{\xi }{\parallel x-y\parallel }^2{-\xi \alpha \parallel x-y\parallel }^2=(\overline{\xi }-\xi \alpha ){\parallel x-y\parallel }^2.$$

Hence we knows that monotone operator $D-\xi f$ is strongly and the coefficient satisfies $\overline{\xi }-\xi \alpha >0$. It is also clear that $D-\xi f$ is Lipschitzian. Therefore, by Lemma 4 (iii) we deduce that $\mathrm{VI}(\mathsf{\Omega },D-\xi f)$ is a single-point set. Say $z\in H_1$, that is, $\mathrm{VI}(\mathsf{\Omega },D-\xi f)=\{z\}$. Also, by Lemma 4 (ii) we have $z=P_{\mathsf{\Omega }}(z-Dz+\xi f(z))$. Since $\{x_n\}$ is a bounded sequence in $H_1$, without loss of generality, we may choose a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ such that

$$\underset{n\to \infty }{lim\; sup}\langle (\xi f-D)z,x_n-z\rangle =\underset{i\to \infty }{lim}\langle (\xi f-D)z,x_{n_i}-z\rangle .$$

(39)

We have proved that sequence $\{x_{n_i}\}$ is bounded, it is not too hard to see its a subsequence $\{x_{n_{i_j}}\}$ of $\{x_{n_i}\}$ converges weakly to w. Let suppose that $x_{n_i}\rightharpoonup w$. From (37), we obtain that $y_{n_i}\rightharpoonup w$.

Next, let us pay our focus to $w\in {\bigcap }_{i=1}^{\infty }\mathrm{Fix}(S_i)=\mathrm{Fix}(W)$. Supposing on the contrary that, $w\notin \mathrm{Fix}(W)$, i.e., $Ww\ne w$, we see from Lemma 5 that

$$\begin{array}{cc}{\displaystyle \underset{i\to \infty }{lim\; inf}}\parallel y_{n_i}-w\parallel \hfill & <{\displaystyle \underset{i\to \infty }{lim\; inf}}\parallel y_{n_i}-Ww\parallel \hfill \\ & \le {\displaystyle \underset{i\to \infty }{lim\; inf}}\{\parallel W{y}_{n_i}-Ww\parallel +\parallel W{y}_{n_i}-y_{n_i}\parallel \}\hfill \\ & \le {\displaystyle \underset{i\to \infty }{lim\; inf}}\{\parallel W{y}_{n_i}-y_{n_i}\parallel +\parallel y_{n_i}-w\parallel \}.\hfill \end{array}$$

(40)

On the other hand, we have

$$\parallel W{y}_n-y_n\parallel \le \parallel W{y}_n-W_n{y}_n\parallel +\parallel W_n{y}_n-y_n\parallel \le \underset{x\in C}{sup}\parallel Wx-W_{n}x\parallel +\parallel W_n{y}_n-y_n\parallel .$$

By using Lemma 3 and (38), we obtain that ${lim}_{i\to \infty }\parallel W{y}_n-y_n\parallel =0$, which together with (40), yields ${lim\; inf}_{i\to \infty }\parallel y_{n_i}-w\parallel >{lim\; inf}_{i\to \infty }\parallel y_{n_i}-w\parallel .$ This reaches a contraction, and hence we have $w\in \mathrm{Fix}(W)={\bigcap }_{i=1}^{\infty }\mathrm{Fix}(S_i)$. Please $G:H_1\to H_1$ is a nonexpansive mapping. Since $x_{n_i}\rightharpoonup w$ and ${lim}_{n\to \infty }\parallel x_n-G{x}_n\parallel =0$ (due to (38)), by Lemma 6, we get that $w\in \mathrm{Fix}(G)$. From Lemma 1, we get that $w\in \mathsf{\Gamma }$. Therefore, $w\in \mathsf{\Omega }$. Since z is a fixed point of mapping $P_{\mathsf{\Omega }}(I-D+\xi f)$ and $w\in \mathsf{\Omega }$, we have

$$\begin{array}{cc}{\displaystyle \underset{n\to \infty }{lim\; sup}}\langle (\xi f-D)z,x_n-z\rangle \hfill & ={\displaystyle \underset{i\to \infty }{lim}}\langle (\xi f-D)z,x_{n_i}-z\rangle \hfill \\ & =\langle (\xi f-D)z,w-z\rangle \hfill \\ & =\langle (z-Dz+\xi f(z))-z,w-z\rangle \le 0.\hfill \end{array}$$

(41)

Step 5. We aim to $x_n\to z$ and $y_n\to z$ as $n\to \infty$. Indeed, by (3.10) and (3.11) we have

$$\begin{array}{cc}\parallel x_{n+1}{-z\parallel }^2\hfill & ={\alpha }_n\langle (\xi f-D)z,x_{n+1}-z\rangle +{\beta }_n\langle x_n-z,x_{n+1}-z\rangle \hfill \\ & +\langle [(1-{\beta }_n)I-{\alpha }_{n}D]y_n-[(1-{\beta }_n)I-{\alpha }_{n}D]z,x_{n+1}-z\rangle \hfill \\ & \le {\alpha }_n\langle (\xi f-D)z,x_{n+1}-z\rangle +{\beta }_n\langle x_n-z,x_{n+1}-z\rangle \hfill \\ & +\parallel [(1-{\beta }_n)I-{\alpha }_{n}D](y_n-z)\parallel \parallel x_{n+1}-z\parallel \hfill \\ & \le {\alpha }_n\langle (\xi f-D)z,x_{n+1}-z\rangle +\frac{1}{2}{\beta }_n(\parallel x_n{-z\parallel }^2+\parallel x_{n+1}-z{\parallel }^2)\hfill \\ & +(1-{\beta }_n-{\alpha }_n\overline{\xi })\parallel y_n-z\parallel \parallel x_{n+1}-z\parallel \hfill \\ & \le {\alpha }_n\langle (\xi f-D)z,x_{n+1}-z\rangle +\frac{1}{2}(1-{\alpha }_n\overline{\xi })(\parallel x_n{-z\parallel }^2+\parallel x_{n+1}-z{\parallel }^2).\hfill \end{array}$$

This immediately implies that

$$\parallel x_{n+1}{-z\parallel }^2\le (1-{\alpha }_n\overline{\xi }){\parallel x_n-z\parallel }^2+2{\alpha }_n\langle (\xi f-D)z,x_{n+1}-z\rangle .$$

By using Lemma 7, we infer that $|x_n-z\parallel \to 0$ as $n\to \infty$. This completes the proof. □

4. Conclusions

In this paper, we studied an implicit general iterative method for approximating a solution of a split variational inclusion problem with a hierarchical optimization problem constraint for a countable family of mappings, which are nonexpansive, in the setting of infinite dimensional Hilbert spaces. Convergence theorem of the sequences generated in our proposed implicit algorithm is obtained without compact assumptions.