A Strong Convergence Theorem under a New Shrinking Projection Method for Finite Families of Nonlinear Mappings in a Hilbert Space

Abstract

In this paper, using a new shrinking projection method, we deal with the strong convergence for finding a common point of the sets of zero points of a maximal monotone mapping, common fixed points of a finite family of demimetric mappings and common zero points of a finite family of inverse strongly monotone mappings in a Hilbert space. Using this result, we get well-known and new strong convergence theorems in a Hilbert space.

1. Introduction

Let H be a real Hilbert space and let C be a nonempty, closed and convex subset of H. Let $T:C\to H$ be a mapping. Then we denote by $F\left(T\right)$ the set of fixed points of T. For a real number t with $0\le t<1$, a mapping $U:C\to H$ is said to be a t-strict pseudo-contraction [1] if

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

for all $x,y\in C$. In particular, if $t=0$, then U is nonexpansive, i.e.,

$$\parallel Ux-Uy\parallel \le \parallel x-y\parallel ,\forall x,y\in C.$$

If U is a t-strict pseudo-contraction and $F\left(U\right)\ne \varnothing$, then we get that, for $x\in C$ and $p\in F\left(U\right)$,

$${\parallel Ux-p\parallel }^2\le {\parallel x-p\parallel }^2+t{\parallel x-Ux\parallel }^2.$$

From this inequality, we get that

$${\parallel Ux-x\parallel }^2{+\parallel x-p\parallel }^2+2\langle Ux-x,x-p\rangle \le {\parallel x-p\parallel }^2+t{\parallel x-Ux\parallel }^2.$$

Then we get that

$$2\langle x-Ux,x-p\rangle \ge {(1-t)\parallel x-Ux\parallel }^2.$$

(1)

A mapping $U:C\to H$ is said to be generalized hybrid [2] if there exist real numbers $\alpha ,\beta$ such that

$${\alpha \parallel Ux-Uy\parallel }^2{+(1-\alpha )\parallel x-Uy\parallel }^2\le {\beta \parallel Ux-y\parallel }^2+(1-\beta ){\parallel x-y\parallel }^2$$

for all $x,y\in C$. Such a mapping U is said to be ($\alpha$, $\beta$)-generalized hybrid. The class of generalized hybrid mappings covers several well-known mappings. A (1,0)-generalized hybrid mapping is nonexpansive. For $\alpha =2$ and $\beta =1$, it is nonspreading [3,4], i.e.,

$${2\parallel Ux-Uy\parallel }^2\le {\parallel Ux-y\parallel }^2+{\parallel Uy-x\parallel }^2,\forall x,y\in C.$$

For $\alpha =\frac{3}{2}$ and $\beta =\frac{1}{2}$, it is also hybrid [5], i.e.,

$${3\parallel Ux-Uy\parallel }^2\le {\parallel x-y\parallel }^2{+\parallel Ux-y\parallel }^2+{\parallel Uy-x\parallel }^2,\forall x,y\in C.$$

In general, nonspreading mappings and hybrid mappings are not continuous; see [6]. If U is a generalized hybrid and $F\left(U\right)\ne \varnothing$, then we get that, for $x\in C$ and $p\in F\left(U\right)$,

$${\alpha \parallel p-Ux\parallel }^2{+(1-\alpha )\parallel p-Ux\parallel }^2\le {\beta \parallel p-x\parallel }^2+(1-\beta ){\parallel p-x\parallel }^2$$

and hence ${\parallel Ux-p\parallel }^2\le {\parallel x-p\parallel }^2.$ From this, we have that

$$2\langle x-p,x-Ux\rangle \ge {\parallel x-Ux\parallel }^2.$$

(2)

We also know that such a mapping exists in a Banach space. Let E be a smooth Banach space and let G be a maximal monotone mapping with $G^{-1}0\ne \varnothing$. Then, for the metric resolvent $J_{\lambda }$ of G for a positive number $\lambda >0$, we obtain from [7,8] that, for $x\in E$ and $p\in G^{-1}0=F\left(J_{\lambda }\right)$,

$$\langle J_{\lambda }x-p,J(x-J_{\lambda }x)\rangle \ge 0.$$

Then we get

$$\langle J_{\lambda }x-x+x-p,J(x-J_{\lambda }x)\rangle \ge 0$$

and hence

$$\langle x-p,J(x-J_{\lambda }x)\rangle \ge {\parallel x-J_{\lambda }x\parallel }^2,$$

(3)

where J is the duality mapping on E. Motivated by (1), (2) and (3), Takahashi [9] introduced a nonlinear mapping in a Banach space as follows: Let C be a nonempty, closed, and convex subset of a smooth Banach E and let $\eta$ be a real number with $\eta \in (-\infty ,1)$. A mapping $U:C\to E$ with $F\left(U\right)\ne \varnothing$ is said to be $\eta$-demimetric if, for $x\in C$ and $p\in F\left(U\right)$,

$$2\langle x-p,J(x-Ux)\rangle \ge {(1-\eta )\parallel x-Ux\parallel }^2.$$

According to this definition, we have that a t-strict pseudo-contraction U with $F\left(U\right)\ne \varnothing$ is t-demimetric, an ($\alpha$, $\beta$)-generalized hybrid mapping U with $F\left(U\right)\ne \varnothing$ is 0-demimetric and the metric resolvent $J_{\lambda }$ with $G^{-1}0\ne \varnothing$ is $(-1)$-demimetric. On the other hand, we know the shrinking projection method which was defined by Takahashi, Takeuchi, and Kubota [10] for finding fixed points of nonexpansive mappings in a Hilbert space. They proved the following strong convergence theorem [10].

Theorem 1

([10]).Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let$U:C\to C$be a nonexpansive mapping. Assume that$F\left(U\right)\ne \varnothing$. For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{y}_n=(1-{\lambda }_n)x_n+{\lambda }_{n}U{x}_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,n=1,2,\dots .,\hfill \end{array}\right.$$

where a real number a and$\left\{{\lambda }_n\right\}\subset (0,\infty )$satisfy the following inequalities:

$$0<a\le {\lambda }_n\le 1,n=1,2,\dots .$$

Then the sequence$\left\{x_n\right\}$converges strongly to$u\in F\left(U\right)$, where$u=P_{F\left(U\right)}{x}_1$and$P_{F\left(U\right)}$is the metric projection of H onto$F\left(U\right)$.

In this paper, using a new shrinking projection method, we prove a strong convergence theorem for finding a common point of the sets of zero points of a maximal monotone mapping, common fixed points for a finite family of demimetric mappings and common zero points of a finite family of inverse strongly monotone mappings in a Hilbert space. Using this result, we obtain well-known and new strong convergence theorems in a Hilbert space. In particular, using the shrinking projection method, we prove a strong convergence theorem for a finite family of generalized hybrid mappings with the variational inequalty problem in a Hilbert space.

2. Preliminaries

Throughout this paper, let H be a real Hilbert space with inner product $〈·,·〉$ and norm $∥·∥$ and let $\mathbb{N}$ and $\mathbb{R}$ be the sets of positive integers and real numbers, respectively. When $\left\{x_n\right\}$ is a sequence in H, we denote by $x_n\to x$ the strong convergence of $\left\{x_n\right\}$ to $x\in H$ and by $x_n\rightharpoonup x$ the weak convergence. We have from [11,12] that, for $x,y\in H$ and $\alpha \in \mathbb{R}$,

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

(4)

Furthermore, we have that, for $x,y,u,v\in H$,

$$2\langle x-y,u-v\rangle ={\parallel x-v\parallel }^2{+\parallel y-u\parallel }^2{-\parallel x-u\parallel }^2-{\parallel y-v\parallel }^2.$$

(5)

Let C be a nonempty, closed and convex subset of H. A mapping $U:C\to H$ with $F\left(U\right)\ne \varnothing$ is said to be quasi-nonexpansive if $\parallel Ux-p\parallel \le \parallel x-p\parallel$ for all $x\in C$ and $p\in F\left(U\right)$. If $U:C\to H$ is quasi-nonexpansive, then $F\left(U\right)$ is closed and convex; see [12,13]. For a nonempty, closed, and convex subset D of H, the nearest point projection of H onto D is denoted by $P_D$, that is,

$$∥x-P_{D}x∥\le ∥x-y∥,\forall x\in H,y\in D.$$

(6)

A mapping $P_D$ is said to be the metric projection of H onto D. The inequality (6) is equivalent to

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

(7)

We obtain from (7) that $P_D$ is firmly nonexpansive, that is,

$${∥P_{D}x-P_{D}y∥}^2\le \langle P_{D}x-P_{D}y,x-y\rangle ,\forall x,y\in H.$$

In fact, from (7) we have that, for $x.y\in H$,

$$〈x-P_{D}y+P_{D}y-P_{D}x,P_{D}y-P_{D}x〉\le 0$$

and hence

$$\begin{array}{cc}\hfill {∥P_{D}x-P_{D}y∥}^2& \le \langle P_{D}x-P_{D}y,x-P_{D}y\rangle \hfill \\ \hfill & =\langle P_{D}x-P_{D}y,x-y+y-P_{D}y\rangle \hfill \\ \hfill & =\langle P_{D}x-P_{D}y,x-y\rangle +\langle P_{D}x-P_{D}y,y-P_{D}y\rangle \hfill \\ \hfill & \le \langle P_{D}x-P_{D}y,x-y\rangle .\hfill \end{array}$$

Furthermore, using (7) and (5), we have that

$$\parallel P_D{x-y\parallel }^2+\parallel P_D{x-x\parallel }^2\le {\parallel x-y\parallel }^2,\forall x\in H,y\in D.$$

(8)

Let C be a nonempty, closed, and convex subset of H. A mapping $A:C\to H$ is said to be $\alpha$-inverse strongly monotone if there exists $\alpha >0$ such that

$$\langle x-y,Ax-Ay\rangle \ge {\alpha \parallel Ax-Ay\parallel }^2,\forall x,y\in C.$$

If A is an $\alpha$-inverse-strongly monotone mapping and $0<\mu \le 2\alpha$, then we obtain from [12] that $I-\mu A:C\to H$ is nonexpansive, i.e.,

$$\parallel (I-\mu A)x-(I-\mu A)y\parallel \le \parallel x-y\parallel ,\forall x,y\in C.$$

(9)

For more results of inverse strongly monotone mappings, see also [12,14,15]. The variational inequalty problem for a nonlinear mapping $A:C\to H$ is to find an element $w\in C$ such that

$$\langle Aw,x-w\rangle \ge 0,\forall x\in C.$$

(10)

The set of solutions of (10) is denoted by $VI(C,A)$. We also have that, for $\mu >0$, $w=P_C(I-\mu A)w$ if and only if $w\in VI(C,A)$. In fact, let $\mu >0$. Then, for $w\in C$,

$$\begin{array}{cc}\hfill w=P_C(I-\mu A)w& ⟺\langle (I-\mu A)w-w,w-y\rangle \ge 0,\forall y\in C\hfill \\ \hfill & ⟺\langle -\mu Aw,w-y\rangle \ge 0,\forall y\in C\hfill \\ \hfill & ⟺\langle Aw,w-y\rangle \le 0,\forall y\in C\hfill \\ \hfill & ⟺\langle Aw,y-w\rangle \ge 0,\forall y\in C\hfill \\ \hfill & ⟺w\in VI(C,A).\hfill \end{array}$$

(11)

Let G be a multi-valued mapping from H into H. The effective domain of G is denoted by $dom\left(G\right)$, i.e., $dom\left(G\right)=\{x\in H:Gx\ne \varnothing \}$. A multi-valued mapping $G\subset H\times H$ is called a monotone mapping on H if $〈x-y,u-v〉\ge 0$ for all $x,y\in dom\left(G\right)$, $u\in Gx$, and $v\in Gy$. A monotone mapping G on H is said to be maximal if its graph is not properly contained in the graph of any other monotone mapping on H. For a maximal monotone mapping G on H, we may define a single-valued mapping $J_r={(I+rG)}^{-1}:H\to dom\left(G\right)$, which is said to be the resolvent of G for $r>0$. We denote by $A_r=\frac{1}{r}(I-J_r)$ the Yosida approximation of G for $r>0$. We get from [8] that

$$A_{r}x\in G{J}_{r}x,\forall x\in H,r>0.$$

(12)

For a maximal monotone mapping G on H, let $G^{-1}0=\{x\in H:0\in Gx\}.$ It is known that $G^{-1}0=F\left(J_r\right)$ for all $r>0$ and the resolvent $J_r$ is firmly nonexpansive:

$$\parallel J_{r}x-J_r{y\parallel }^2\le \langle J_{r}x-J_{r}y,x-y\rangle ,\forall x,y\in H.$$

(13)

Takahashi, Takahashi, and Toyoda [16] proved the following result.

Lemma 1

([16]).Let G be a maximal monotone mapping on a Hilbert space H. For$r>0$and$x\in H$, define the resolvent$J_{r}x$. Then the following inequality holds:

$$\frac{s-t}{s}\langle J_{s}x-J_{t}x,J_{s}x-x\rangle \ge {\parallel J_{s}x-J_{t}x\parallel }^2$$

for all$s,t>0$and$x\in H$.

From Lemma 1, we get that, for $s,t>0$ and $x\in H$,

$$\begin{array}{c}\hfill {\parallel J_{s}x-J_{t}x\parallel }^2\le \frac{|s-t|}{s}\parallel J_{s}x-x\parallel \parallel J_{s}x-J_{t}x\parallel \end{array}$$

and hence

$$\parallel J_{s}x-J_{t}x\parallel \le \frac{|s-t|}{s}\parallel J_{s}x-J_{t}x\parallel .$$

(14)

Using the ideas of [17,18], Alsulami and Takahashi [19] proved the following lemma.

Lemma 2

([19]).Let C be a nonempty, closed and convex subset of a Hilbert space H. Let$G\subset H\times H$be a maximal monotone mapping and let$J_{\lambda }={(I+\lambda G)}^{-1}$be the resolvent of G for$\lambda >0$. Let$\kappa >0$and let$U:C\to H$be a κ-inverse strongly monotone mapping. Suppose that$G^{-1}0\cap U^{-1}0\ne \varnothing$. Let$\lambda ,r>0$and$z\in C$. Then the following are equivalent:

(i) 

$z=J_{\lambda }(I-rU)z$;

(ii) 

$0\in Uz+Gz$;

(iii) 

$z\in G^{-1}0\cap U^{-1}0$.

When a Banach space E is a Hilbert space, the definition of a demimetric mapping is as follows: Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let $\eta \in (-\infty ,1)$. A mapping $U:C\to H$ with $F\left(U\right)\ne \varnothing$ is said to be $\eta$-demimetric [9] if, for $x\in C$ and $q\in F\left(U\right)$,

$$\langle x-q,x-Ux\rangle \ge \frac{1-\eta }{2}{\parallel x-Ux\parallel }^2.$$

The following lemma which was essentially proved in [9] is important and crucial in the proof of the main result. For the sake of completeness, we give the proof.

Lemma 3

([9]).Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let η be a real number with$\eta \in (-\infty ,1)$and let U be an η-demimetric mapping of C into H. Then$F\left(U\right)$is closed and convex.

Proof. 

Let us show that $F\left(U\right)$ is closed. For a sequence $\left\{q_n\right\}$ such that $q_n\to q$ and $q_n\in F\left(U\right)$, we have from the definition of U that

$$2\langle q-q_n,q-Uq\rangle \ge (1-\eta ){\parallel q-Uq\parallel }^2.$$

From $q_n\to q$, we have $0\ge {(1-\eta )\parallel q-Uq\parallel }^2$. From $1-\eta >0$, we have $\parallel q-Uq\parallel =0$ and hence $q=Uq$. This implies that $F\left(U\right)$ is closed.

Let us prove that $F\left(U\right)$ is convex. Let $p,q\in F\left(U\right)$ and set $z=\alpha p+(1-\alpha )q$, where $\alpha \in [0,1]$. Then we have that

$$2\langle z-p,z-Uz\rangle \ge {(1-\eta )\parallel z-Uz\parallel }^2\mathrm{and}2\langle z-q,z-Uz\rangle \ge (1-\eta ){\parallel z-Uz\parallel }^2.$$

From $\alpha \ge 0$ and $1-\alpha \ge 0$, we also have that

$$2\langle \alpha z-\alpha p,z-Uz\rangle \ge {\alpha (1-\eta )\parallel z-Uz\parallel }^2$$

and $2\langle (1-\alpha )z-(1-\alpha )q,z-Uz\rangle \ge {(1-\alpha )(1-\eta )\parallel z-Uz\parallel }^2.$ > From these inequalities, we get that

$$0=2\langle z-z,z-Uz\rangle \ge {(1-\eta )\parallel z-Uz\parallel }^2.$$

From $1-\eta >0$ we get that $\parallel z-Uz\parallel =0$ and hence $z=Uz$. This means that $F\left(U\right)$ is convex. ☐

Takahashi, Wen, and Yao [20] proved the following lemma which is also used in the proof of the main result.

Lemma 4

([20]).Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let$\eta \in (-\infty ,1)$and let a mapping$T:C\to H$with$F\left(T\right)\ne \varnothing$be η-demimetric. Let μ be a real number with$0<\mu \le 1-\eta$and define$U=(1-\mu )I+\mu T$. Then U is a quasi-nonexpansive mapping of C into H.

3. Main Result

In this section, using a new shrinking projection method, we obtain a strong convergence theorem for finding a common point of the sets of zero points of a maximal monotone mapping, common fixed points for a finite family of demimetric mappings and common zero points of a finite family of inverse strongly monotone mappings in a Hilbert space. Let C be a nonempty, closed and convex subset of a Hilbert space H. Then a mapping $T:C\to H$ is said to be demiclosed if, for a sequence $\left\{x_n\right\}$ in C such that $x_n\rightharpoonup w$ and $x_n-T{x}_n\to 0$, $w=Tw$ holds; see [21].

Theorem 2.

Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let$\{k_1,\dots ,k_M\}\subset (-\infty ,1)$and$\{{\mu }_1,\dots ,{\mu }_N\}\subset (0,\infty )$. Let${\left\{T_j\right\}}_{j=1}^M$be a finite family of$k_j$-demimetric and demiclosed mappings of C into itself and let${\left\{B_i\right\}}_{i=1}^N$be a finite family of${\mu }_i$-inverse strongly monotone mappings of C into H. Let A and G be maximal monotone mappings on H and let$J_r={(I+rA)}^{-1}$and$Q_{\lambda }={(I+\lambda G)}^{-1}$be the resolvents of A and G for$r>0$ and $\lambda >0$, respectively. Assume that

$$\mathsf{\Omega }=A^{-1}0\cap \left({\cap }_{j=1}^{M}F\left(T_j\right)\right)\cap \left({\cap }_{i=1}^N{(B_i+G)}^{-1}0\right)\ne \varnothing .$$

For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{y}_n=\sum _{j=1}^M{\xi }_j((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n,\hfill \\ z_n=\sum _{i=1}^N{\sigma }_i{Q}_{{\eta }_n}(I-{\eta }_n{B}_i)y_n,\hfill \\ u_n=J_{r_n}{z}_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel ,\parallel z_n-z\parallel \le \parallel y_n-z\parallel \hfill \\ and\langle z_n-z,z_n-u_n\rangle \ge {\parallel z_n-u_n\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$\left\{{\lambda }_n\right\},\left\{{\eta }_n\right\},\left\{r_n\right\}\subset (0,\infty )$,$\{{\xi }_1,\dots ,{\xi }_M\},\{{\sigma }_1,\dots ,{\sigma }_N\}\subset (0,1)$and$a,b,c\in \mathbb{R}$satisfy the following:

(1) 

$0<a\le {\lambda }_n\le min\{1-k_1,\dots ,1-k_M\},\forall n\in \mathbb{N}$;

(2) 

$0<b\le {\eta }_n\le 2min\{{\mu }_1,\dots ,{\mu }_N\},\forall n\in \mathbb{N}$;

(3) 

$0<c\le r_n,\forall n\in \mathbb{N}$;

(4) 

${\sum }_{j=1}^M{\xi }_j=1$and${\sum }_{i=1}^N{\sigma }_i=1$.

Then$\left\{x_n\right\}$converges strongly to a point$z_0\in \mathsf{\Omega }$, where$z_0=P_{\mathsf{\Omega }}{x}_1$.

Proof. 

Since a mapping $B_i$ is ${\mu }_i$-inverse strongly monotone for all $i\in \{1,\dots ,N\}$ and $0<b\le {\eta }_n\le 2{\mu }_i$, we have that $Q_{{\eta }_n}(I-{\eta }_n{B}_i)$ is nonexpansive and

$$F\left(Q_{{\eta }_n}(I-{\eta }_n{B}_i)\right)={(B_i+G)}^{-1}0$$

is closed and convex. Furthermore, we have from Lemma 3 that $F\left(T_j\right)$ is closed and convex. We also know that $A^{-1}0$ is closed and convex. Then,

$$\mathsf{\Omega }=A^{-1}0\cap \left({\cap }_{j=1}^{M}F\left(T_j\right)\right)\cap \left({\cap }_{i=1}^N{(B_i+G)}^{-1}0\right)$$

is nonempty, closed, and convex. Therefore, $P_{\mathsf{\Omega }}$ is well defined.

We have that

$$\begin{array}{cc}\hfill \parallel y_n-z\parallel \le \parallel x_n-z\parallel ⟺& {\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2\hfill \\ \hfill ⟺& {\parallel y_n\parallel }^2-{\parallel x_n\parallel }^2-2\langle y_n-x_n,z\rangle \le 0.\hfill \end{array}$$

Similarly, we have that

$$\begin{array}{cc}\hfill \parallel z_n-z\parallel \le \parallel y_n-z\parallel ⟺& {\parallel z_n\parallel }^2-{\parallel y_n\parallel }^2-2\langle z_n-y_n,z\rangle \le 0.\hfill \end{array}$$

Thus $\{z\in C:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \mathrm{and}\parallel z_n-z\parallel \le \parallel y_n-z\parallel \}$ is closed and convex. We also have that $\{z\in C:\langle z_n-z,z_n-u_n\rangle \ge \parallel z_n-u_n{\parallel }^2\}$ is closed and convex. Then $C_n$ is closed and convex for all $n\in \mathbb{N}$. Let us show that $\mathsf{\Omega }\subset C_n$ for all $n\in \mathbb{N}$. We have that $\mathsf{\Omega }\subset C_1=C.$ Assume that $\mathsf{\Omega }\subset C_k$ for some $k\in \mathbb{N}$. From Lemma 4 we have that, for $z\in \mathsf{\Omega }$,

$$\begin{array}{cc}\hfill \parallel y_k-z\parallel & =\parallel \sum _{j=1}^M{\xi }_j((1-{\lambda }_k)I+{\lambda }_k{T}_j)x_k-z\parallel \hfill \\ \hfill & \le \sum _{j=1}^M{\xi }_j\parallel ((1-{\lambda }_k)I+{\lambda }_k{T}_j)x_k-z\parallel \hfill \\ \hfill & \le \sum _{j=1}^M{\xi }_j\parallel x_k-z\parallel =\parallel x_k-z\parallel .\hfill \end{array}$$

(15)

Furthermore, since $Q_{{\eta }_k}(I-{\eta }_k{B}_i)$ is nonexpansive and hence quasi-nonexpansive, we have that, for $z\in \mathsf{\Omega }$,

$$\begin{array}{cc}\hfill \parallel z_k-z\parallel & =\parallel \sum _{i=1}^N{\sigma }_i{Q}_{{\eta }_k}(I-{\eta }_k{B}_i)y_k-z\parallel \hfill \\ \hfill & \le \sum _{i=1}^N{\sigma }_i\parallel Q_{{\eta }_k}(I-{\eta }_k{B}_i)y_k-z\parallel \hfill \\ \hfill & \le \sum _{i=1}^N{\sigma }_i\parallel y_k-z\parallel =\parallel y_k-z\parallel .\hfill \end{array}$$

(16)

Since $J_{r_k}$ is the resolvent of A and $u_k=J_{r_k}{z}_k$, we also have that

$$\langle z_k-J_{r_k}{z}_k,J_{r_k}{z}_k-z\rangle \ge 0,\forall z\in \mathsf{\Omega }.$$

From $\langle z_k-J_{r_k}{z}_k,J_{r_k}{z}_k-z_k+z_k-z\rangle \ge 0$, we have that

$$\langle z_k-J_{r_k}{z}_k,z_k-z\rangle \ge {\parallel z_k-J_{r_k}{z}_k\parallel }^2.$$

This implies that

$$\langle z_k-u_k,z_k-z\rangle \ge {\parallel z_k-u_k\parallel }^2.$$

From these, we have that $\mathsf{\Omega }\subset C_{k+1}$. Therefore, we have by mathematical induction that $\mathsf{\Omega }\subset C_n$ for all $n\in \mathbb{N}$. Thus $x_{n+1}=P_{C_{n+1}}{x}_1$ is well defined.

Since $\mathsf{\Omega }$ is nonempty, closed, and convex, there exists $z_0\in \mathsf{\Omega }$ such that $z_0=P_{\mathsf{\Omega }}{x}_1$. By $x_{n+1}=P_{C_{n+1}}{x}_1$, we get that

$$\parallel x_1-x_{n+1}\parallel \le \parallel x_1-z\parallel$$

for all $z\in C_{n+1}$. From $z_0\in \mathsf{\Omega }\subset C_{n+1}$ we obtain that

$$\parallel x_1-x_{n+1}\parallel \le \parallel x_1-z_0\parallel .$$

(17)

This implies that $\left\{x_n\right\}$ is bounded. Since $x_n=P_{C_n}{x}_1$ and $x_{n+1}\in C_{n+1}\subset C_n$, we get that

$$\parallel x_1-x_n\parallel \le \parallel x_1-x_{n+1}\parallel .$$

Thus $\{\parallel x_1-x_n\parallel \}$ is bounded and nondecreasing. Then the limit of $\{\parallel x_1-x_n\parallel \}$ exists. Put ${lim}_{n\to \infty }\parallel x_n-x_1\parallel =c$. For any $m,n\in \mathbb{N}$ with $m\ge n$, we have $C_m\subset C_n$. >From $x_m=P_{C_m}{x}_1\in C_m\subset C_n$ and (8), we have that

$$\parallel x_m-P_{C_n}{x}_1{\parallel }^2+\parallel P_{C_n}{x}_1-x_1{\parallel }^2\le {\parallel x_1-x_m\parallel }^2.$$

This implies that

$$\parallel x_m-x_n{\parallel }^2\le \parallel x_1-x_m{\parallel }^2-\parallel x_n-x_1{\parallel }^2\le c^2-{\parallel x_n-x_1\parallel }^2.$$

(18)

Since $c^2-{\parallel x_n-x_1\parallel }^2\to 0$ as $n\to \infty$, we have that $\left\{x_n\right\}$ is a Caushy sequence. Since H is complete and C is closed, there exists a point $u\in C$ such that ${lim}_{n\to \infty }{x}_n=u$.

Using (18), we have ${lim}_{n\to \infty }\parallel x_{n+1}-x_n\parallel =0$. By $x_{n+1}\in C_{n+1}$, we get that

$$\begin{array}{cc}\hfill \parallel y_n-x_n\parallel & \le \parallel y_n-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel \hfill \\ \hfill & \le 2\parallel x_n-x_{n+1}\parallel .\hfill \end{array}$$

(19)

This implies that

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

(20)

Furthermore, we have from $x_{n+1}\in C_{n+1}$ that $\parallel z_n-x_{n+1}\parallel \le \parallel y_n-x_{n+1}\parallel$. We get from $\parallel y_n-x_{n+1}\parallel \to 0$ that $\parallel z_n-x_{n+1}\parallel \to 0$. From

$$\parallel y_n-z_n\parallel \le \parallel y_n-x_{n+1}\parallel +\parallel x_{n+1}-z_n\parallel$$

we have that

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

(21)

Let us show $\parallel z_n-u_n\parallel \to 0$. We have from $x_{n+1}\in C_{n+1}$ that

$$\langle z_n-x_{n+1},z_n-u_n\rangle \ge {\parallel z_n-u_n\parallel }^2.$$

Since $\parallel z_n-x_{n+1}\parallel \parallel z_n-u_n\parallel \ge \langle z_n-x_{n+1},z_n-u_n\rangle \ge {\parallel z_n-u_n\parallel }^2$, we have that $\parallel z_n-x_{n+1}\parallel \ge \parallel z_n-u_n\parallel$. Then we get from $\parallel z_n-x_{n+1}\parallel \to 0$ that

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

(22)

Since $T_j$ is $k_j$-demimetric for all $j\in \{1,\dots ,M\}$, we get that, for $z\in {\cap }_{j=1}^{M}F\left(T_j\right)$,

$$\begin{array}{cc}\hfill \langle x_n-z,& x_n-y_n\rangle =\langle x_n-z,x_n-\sum _{j=1}^M{\xi }_j((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n\rangle \hfill \\ \hfill & =\sum _{j=1}^M{\xi }_j\langle x_n-z,x_n-((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n\rangle \hfill \\ \hfill & =\sum _{j=1}^M{\xi }_j{\lambda }_n\langle x_n-z,x_n-T_j{x}_n\rangle \hfill \\ \hfill & \ge \sum _{j=1}^M{\xi }_j{\lambda }_n\frac{1-k_j}{2}{\parallel x_n-T_j{x}_n\parallel }^2\hfill \\ \hfill & \ge \sum _{j=1}^M{\xi }_{j}a\frac{1-k_j}{2}{\parallel x_n-T_j{x}_n\parallel }^2.\hfill \end{array}$$

We have from ${lim}_{n\to \infty }\parallel y_n-x_n\parallel =0$ that

$$\underset{n\to \infty }{lim}\parallel x_n-T_j{x}_n\parallel =0,\forall j\in \{1,\dots ,M\}.$$

Since $T_j$ are demiclosed for all $j\in \{1,\dots ,M\}$ and ${lim}_{n\to \infty }{x}_n=u$, we have that $u\in {\cap }_{j=1}^{M}F\left(T_j\right)$. Let us show that $u\in {\cap }_{i=1}^N{(B_i+G)}^{-1}0$. Since $Q_{{\eta }_n}(I-{\eta }_n{B}_i)$ is nonexpansive for all $i\in \{1,\dots ,N\}$, we get that, for $z\in {\cap }_{i=1}^N{(B_i+G)}^{-1}0$,

$$\begin{array}{cc}\hfill \langle y_n-z,& y_n-z_n\rangle =\langle y_n-z,y_n-\sum _{i=1}^N{\sigma }_i{Q}_{{\eta }_n}(I-{\eta }_n{B}_i)y_n\rangle \hfill \\ \hfill & =\sum _{i=1}^N{\sigma }_i\langle y_n-z,y_n-Q_{{\eta }_n}(I-{\eta }_n{B}_i)y_n\rangle \hfill \\ \hfill & \ge \sum _{i=1}^N{\sigma }_i\frac{1}{2}{\parallel y_n-Q_{{\eta }_n}(I-{\eta }_n{B}_i)y_n\parallel }^2.\hfill \end{array}$$

We have from ${lim}_{n\to \infty }\parallel y_n-z_n\parallel =0$ that

$$\underset{n\to \infty }{lim}\parallel y_n-Q_{{\eta }_n}(I-{\eta }_n{B}_i)y_n\parallel =0,\forall i\in \{1,\dots ,N\}.$$

Since $\left\{{\eta }_n\right\}$ is bounded, we get that there exists a subsequence $\left\{{\eta }_{n_l}\right\}$ of $\left\{{\eta }_n\right\}$ such that ${lim}_{l\to \infty }{\eta }_{n_l}=\eta$ and $0<b\le \eta \le 2min\{{\mu }_1,\dots ,{\mu }_N\}$. For such $\eta$, we get that, for $i\in \{1,\dots ,N\}$ and a subsequence $\left\{y_{n_l}\right\}$ of $\left\{y_n\right\}$ corresponding to the sequence $\left\{{\eta }_{n_l}\right\}$,

$$\begin{array}{cc}\hfill \parallel y_{n_l}-Q_{\eta }(I-\eta B_i)y_{n_l}\parallel & \le \parallel y_{n_l}-Q_{{\eta }_{n_l}}(I-{\eta }_{n_l}{B}_i)y_{n_l}\parallel \hfill \\ \hfill & +\parallel Q_{{\eta }_{n_l}}(I-{\eta }_{n_l}{B}_i)y_{n_l}-Q_{{\eta }_{n_l}}(I-\eta B_i)y_{n_l}\parallel \hfill \\ \hfill & +\parallel Q_{{\eta }_{n_l}}(I-\eta B_i)y_{n_l}-Q_{\eta }(I-\eta B_i)y_{n_l}\parallel \hfill \\ \hfill & \le \parallel y_{n_l}-Q_{{\eta }_{n_l}}(I-{\eta }_{n_l}{B}_i)y_{n_l}\parallel \hfill \\ \hfill & +\parallel (I-{\eta }_{n_l}{B}_i)y_{n_l}-(I-\eta B_i)y_{n_l}\parallel \hfill \\ \hfill & +\parallel Q_{{\eta }_{n_l}}(I-\eta B_i)y_{n_l}-Q_{\eta }(I-\eta B_i)y_{n_l}\parallel \hfill \\ \hfill & \le \parallel y_{n_l}-Q_{{\eta }_{n_l}}(I-{\eta }_{n_l}{B}_i)y_{n_l}\parallel +|{\eta }_{n_l}-\eta |\parallel B_i{y}_{n_l}\parallel \hfill \\ \hfill & +\frac{|{\eta }_{n_l}-\eta |}{\eta }\parallel Q_{\eta }(I-\eta B_i)y_{n_l}-(I-\eta B_i)y_{n_l}\parallel .\hfill \end{array}$$

On the other hand, we get that, for a fixed $y\in C$ and $i\in \{1,\dots ,N\}$,

$$\begin{array}{cc}\hfill b\parallel B_i{y}_n\parallel & \le {\eta }_n\parallel B_i{y}_n\parallel =\parallel {\eta }_n{B}_i{y}_n\parallel \hfill \\ \hfill & =\parallel y_n-(y-{\eta }_n{B}_{i}y)+y-{\eta }_n{B}_{i}y-(y_n-{\eta }_n{B}_i{y}_n)\parallel \hfill \\ \hfill & \le \parallel y_n-y\parallel +{\eta }_n\parallel B_{i}y\parallel +\parallel (I-{\eta }_n{B}_i)y-(I-{\eta }_n{B}_i)y_n\parallel \hfill \\ \hfill & \le \parallel y_n-y\parallel +2min\{{\mu }_1,\dots ,{\mu }_N\}\parallel B_{i}y\parallel +\parallel y-y_n\parallel .\hfill \end{array}$$

Since $\left\{y_n\right\}$ is bounded, we have that $\left\{B_i{y}_n\right\}$ is bounded for all $i\in \{1,\dots ,N\}$. Thus we get that

$$\underset{l\to \infty }{lim}\parallel x_{n_l}-Q_{\eta }(I-\eta B_i)x_{n_l}\parallel =0,\forall i\in \{1,\dots ,N\}.$$

Since ${lim}_{l\to \infty }{x}_{n_l}=u$ and $Q_{\eta }(I-\eta B_i)$ are demiclosed for all $i\in \{1,\dots ,N\}$, we get $u\in {\cap }_{i=1}^N{(B_i+G)}^{-1}0$. Let us show $u\in A^{-1}0$. We have from (22) that

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

Using $r_n\ge c$, we get

$${\displaystyle \underset{n\to \infty }{lim}\frac{1}{r_n}\parallel z_n-u_n\parallel =0.}$$

Therefore, we have

$$\underset{n\to \infty }{lim}\parallel A_{r_n}{z}_n\parallel =\underset{n\to \infty }{lim}\frac{1}{r_n}\parallel z_n-u_n\parallel =0.$$

For $(p,p^{*})\in A$, from the monotonicity of A, we have $\langle p-u_n,p^{*}-A_{r_n}{z}_n\rangle \ge 0$ for all $n\in \mathbb{N}$. Since $z_n\to u$ and hence $u_n\to u$, we get $\langle p-u,p^{*}\rangle \ge 0$. From the maximallity of A, we have $u\in A^{-1}0$. Therefore, we have $u\in \mathsf{\Omega }$.

Since $z_0=P_{\mathsf{\Omega }}{x}_1$, $u\in \mathsf{\Omega }$ and $x_n\to u$, we have from (17) that

$$\begin{array}{cc}\hfill \parallel x_1-z_0\parallel \le \parallel x_1-u\parallel & =\underset{n\to \infty }{lim}\parallel x_1-x_n\parallel \le \parallel x_1-z_0\parallel .\hfill \end{array}$$

Then $u=z_0.$ Therefore, we have $x_n\to u=z_0$. This completes the proof. ☐

4. Applications

In this section, using Theorem 2, we obtain well-known and new strong convergence theorems in Hilbert spaces. We know the following lemma proved by Marino and Xu [22]; see also [23]. For the sake of completeness, we give the proof.

Lemma 5

([22,23]).Let C be a nonempty, closed and convex subset of a Hilbert space H. Let k be a real number with$0\le k<1$and let$U:C\to H$be a k-strict pseudo-contraction. If$x_n\rightharpoonup u$and$x_n-U{x}_n\to 0$, then$u\in F\left(U\right)$.

Proof. 

Let us show that a nonexpansive mapping $T:C\to H$ is demiclosed. Let $\left\{x_n\right\}$ be a sequence in C such that $x_n\rightharpoonup u$ and $x_n-T{x}_n\to 0$. We have that

$$\begin{array}{cc}\hfill \parallel u-& {Tu\parallel }^2={\parallel u-x_n+x_n-Tu\parallel }^2\hfill \\ \hfill & =\parallel u-x_n{\parallel }^2+{\parallel x_n-Tu\parallel }^2+2\langle u-x_n,x_n-Tu\rangle \hfill \\ \hfill & =\parallel u-x_n{\parallel }^2+{\parallel x_n-T{x}_n+T{x}_n-Tu\parallel }^2+2\langle u-x_n,x_n-u+u-Tu\rangle \hfill \\ \hfill & =\parallel u-x_n{\parallel }^2+\parallel x_n-T{x}_n{\parallel }^2+{\parallel T{x}_n-Tu\parallel }^2+2\langle x_n-T{x}_n,T{x}_n-Tu\rangle \hfill \\ \hfill & -2\parallel u-x_n{\parallel }^2+2\langle u-x_n,u-Tu\rangle \hfill \\ \hfill & \le \parallel u-x_n{\parallel }^2+\parallel x_n-T{x}_n{\parallel }^2+{\parallel x_n-u\parallel }^2+2\langle x_n-T{x}_n,T{x}_n-Tu\rangle \hfill \\ \hfill & -2\parallel u-x_n{\parallel }^2+2\langle u-x_n,u-Tu\rangle \hfill \\ \hfill & =\parallel x_n-T{x}_n{\parallel }^2+2\langle x_n-T{x}_n,T{x}_n-Tu\rangle +2\langle u-x_n,u-Tu\rangle \to 0.\hfill \end{array}$$

Then, $u=Tu$. It is obvious that a mapping $B=I-U:C\to H$ is $\frac{1-k}{2}$-inverse strongly monotone. Put $\alpha =\frac{1-k}{2}$. We have that

$${\alpha \parallel Bx-By\parallel }^2\le \langle x-y,Bx-By\rangle ,\forall x,y\in C.$$

(23)

From $U=I-B$ and (9), we have that

$$I-2\alpha B=I-2\alpha (I-U)=(1-2\alpha )I+2\alpha U$$

is nonexpansive. If $x_n\rightharpoonup u$ and $x_n-U{x}_n\to 0$, then

$$x_n-((1-2\alpha )I+2\alpha U)x_n=2\alpha (I-U)x_n\to 0.$$

Since $(1-2\alpha )I+2\alpha U$ is nonexpansive, we have $u\in F\left(\right(1-2\alpha )I+2\alpha U)=F\left(U\right)$. This implies that U is demiclosed. ☐

Furthermore, we know the following lemma from Kocourek, Takahashi, and Yao [2]; see also [24].

Lemma 6

([2,24]).Let C be a nonempty, closed and convex subset of a Hilbert space H and let$U:C\to H$be generalized hybrid. If$x_n\rightharpoonup u$and$x_n-U{x}_n\to 0$, then$u\in F\left(U\right)$.

We prove a strong convergence theorem for a finite family of strict pseudo-contractions in a Hilbert space.

Theorem 3.

Let C be a nonempty, closed and convex subset of a Hilbert space H. Let$\{k_1,\dots ,k_M\}\subset [0,1)$and let${\left\{T_j\right\}}_{j=1}^M$be a finite family of$k_j$-strict pseudo-contractions of C into itself. Assume that${\cap }_{j=1}^{M}F\left(T_j\right)\ne \varnothing$. For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{y}_n={\sum }_{j=1}^M{\xi }_j((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$a\in \mathbb{R}$, $\left\{{\lambda }_n\right\}\subset (0,\infty )$and$\{{\xi }_1,\dots ,{\xi }_M\}\subset (0,1)$satisfy the following:

(1) 

$0<a\le {\lambda }_n\le min\{1-k_1,\dots ,1-k_M\},\forall n\in \mathbb{N}$;

(2) 

${\sum }_{j=1}^M{\xi }_j=1$.

Then$\left\{x_n\right\}$converges strongly to a point$z_0\in {\cap }_{j=1}^{M}F\left(T_j\right)$, where$z_0=P_{{\cap }_{j=1}^{M}F\left(T_j\right)}{x}_1$.

Proof. 

Since $T_j$ is a $k_j$-strict pseudo-contraction of C into itself with $F\left(T_j\right)\ne \varnothing$, from (1), $T_j$ is a $k_j$-demimetric mapping. Furthermore, we have from Lemma 5 that $T_j$ is demiclosed. We also have that if $B_i=0$ for all $i\in \{1,\dots ,N\}$ in Theorem 2, then $B_i$ is a 1-inverse strongly monotone mapping. Putting ${\eta }_n=1$ for all $n\in \mathbb{N}$ in Theorem 2, we have that $z_n=y_n$ for all $n\in \mathbb{N}$. Furthermore, putting $A=G=0$ and ${\eta }_n=r_n=1$ for all $n\in \mathbb{N}$ in Theorem 2, we have that

$$Q_{{\nu }_n}=J_{r_n}=I,\forall {\nu }_n>0,r_n>0.$$

Then we have that $u_n=z_n=y_n$ for all $n\in \mathbb{N}$. Thus, we get the desired result from Theorem 2. ☐

As a direct result of Theorem 3, we have Theorem 1 in Introduction. We can also prove the following strong convergence theorem for a finite family of inverse strongly monotone mappings in a Hilbert space. Let g be a proper, lower semicontinuous and convex function of a Hilbert space H into $(-\infty ,\infty ]$. The subdifferential $\partial g$ of g is defined as follows:

$$\partial g\left(x\right)=\{z\in H:g(x)+\langle z,y-x\rangle \le g(y),\forall y\in H\}$$

for all $x\in H$. We have from Rockafellar [25] that $\partial g$ is a maximal monotone mapping. Let D be a nonempty, closed, and convex subset of a Hilbert space H and let $i_D$ be the indicator function of D, i.e.,

$$i_D\left(x\right)=\left\{\begin{array}{cc}0,\hfill & x\in D,\hfill \\ \infty ,\hfill & x\notin D.\hfill \end{array}\right.$$

Then $i_D$ is a proper, lower semicontinuous and convex function on H and then the subdifferential $\partial i_D$ of $i_D$ is a maximal monotone mapping. Thus we define the resolvent $J_{\lambda }$ of $\partial i_D$ for $\lambda >0$, i.e.,

$$J_{\lambda }x={(I+\lambda \partial i_D)}^{-1}x$$

for all $x\in H$. We get that, for $x\in H$ and $u\in D$,

$$\begin{array}{cc}\hfill u=& J_{\lambda }x⟺x\in u+\lambda \partial i_{D}u⟺x\in u+\lambda N_{D}u\hfill \\ \hfill & ⟺x-u\in \lambda N_{D}u\hfill \\ \hfill & ⟺\frac{1}{\lambda }\langle x-u,v-u\rangle \le 0,\forall v\in D\hfill \\ \hfill & ⟺\langle x-u,v-u\rangle \le 0,\forall v\in D\hfill \\ \hfill & ⟺u=P_{D}x,\hfill \end{array}$$

where $N_{D}u$ is the normal cone to D at u, i.e.,

$$N_{D}u=\{z\in H:\langle z,v-u\rangle \le 0,\forall v\in D\}.$$

Theorem 4.

Let C be a nonempty, closed and convex subset of a Hilbert space H. Let$\{{\mu }_1,\dots ,{\mu }_N\}\subset (0,\infty )$. Let${\left\{B_i\right\}}_{i=1}^N$be a finite family of${\mu }_i$-inverse strongly monotone mappings of C into H. Assume that${\cap }_{i=1}^{N}VI(C,B_i)\ne \varnothing$. Let$x_1\in C$and$C_1=C$. Let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{z}_n=\sum _{i=1}^N{\sigma }_i{P}_C(I-{\eta }_n{B}_i)x_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel z_n-z\parallel \le \parallel x_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$b\in \mathbb{R}$, $\left\{{\eta }_n\right\}\subset (0,\infty )$and$\{{\sigma }_1,\dots ,{\sigma }_N\}\subset (0,1)$satisfy the following:

(1) 

$0<b\le {\eta }_n\le 2min\{{\mu }_1,\dots ,{\mu }_N\},\forall n\in \mathbb{N}$;

(2) 

${\sum }_{i=1}^N{\sigma }_i=1$.

Then$\left\{x_n\right\}$converges strongly to$z_0\in {\cap }_{i=1}^{N}VI(C,B_i)$, where$z_0=P_{{\cap }_{i=1}^{N}VI(C,B_i)}{x}_1$.

Proof. 

Putting $G=\partial i_C$ in Theorem 2, we get that for ${\eta }_n>0$, $J_{{\eta }_n}=P_C.$ Furthermore, we have ${(\partial i_C)}^{-1}0=C$ and ${(B_i+\partial i_C)}^{-1}0=VI(C,B_i)$. In fact, we get that, for $z\in C$,

$$\begin{array}{cc}\hfill z\in (& B_i+\partial i_C{)}^{-1}0⟺0\in B_{i}z+\partial i_{C}z\hfill \\ \hfill & ⟺0\in B_{i}z+N_{C}z⟺-B_{i}z\in N_{C}z\hfill \\ \hfill & ⟺\langle -B_{i}z,v-z\rangle \le 0,\forall v\in C\hfill \\ \hfill & ⟺\langle B_{i}z,v-z\rangle \ge 0,\forall v\in C\hfill \\ \hfill & ⟺z\in VI(C,B_i).\hfill \end{array}$$

The identity mapping I is a $\frac{1}{2}$-demimetric mapping of C into H. Put $T_j=I$ for all $j\in \{1,\dots ,M\}$ and ${\lambda }_n=\frac{1}{2}$ for all $n\in \mathbb{N}$ in Theorem 2. Then we get that $y_n=x_n$ for all $n\in \mathbb{N}$. Furthermore, putting $A=0$, we have $u_n=z_n$. Thus, we get the desired result from Theorem 2. ☐

We prove a strong convergence theorem for a finite family of generalized hybrid mappings and a finite family of inverse strongly monotone mappings in a Hilbert space.

Theorem 5.

Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let$\{{\mu }_1,\dots ,{\mu }_N\}\subset (0,\infty )$. Let${\left\{T_j\right\}}_{j=1}^M$be a finite family of generalized hybrid mappings of C into itself and let${\left\{B_i\right\}}_{i=1}^N$be a finite family of${\mu }_i$-inverse strongly monotone mappings of C into H. Suppose that

$${\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}VI(C,B_i)\right)\ne \varnothing .$$

For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{y}_n=\sum _{j=1}^M{\xi }_j((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n,\hfill \\ z_n=\sum _{i=1}^N{\sigma }_i{P}_C(I-{\eta }_n{B}_i)y_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel and\parallel z_n-z\parallel \le \parallel y_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$a,b,c\in \mathbb{R}$,$\left\{{\lambda }_n\right\},\left\{{\eta }_n\right\}\subset (0,\infty )$,$\{{\xi }_1,\dots ,{\xi }_M\},\{{\sigma }_1,\dots ,{\sigma }_N\}\subset (0,1)$and$\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$satisfy the following conditions:

(1) 

$0<a\le {\lambda }_n\le 1,\forall n\in \mathbb{N}$;

(2) 

$0<b\le {\eta }_n\le 2min\{{\mu }_1,\dots ,{\mu }_N\},\forall n\in \mathbb{N}$;

(3) 

${\sum }_{j=1}^M{\xi }_j=1$and${\sum }_{i=1}^N{\sigma }_i=1$.

Then$\left\{x_n\right\}$converges strongly to a point$z_0\in {\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}VI(C,B_i)\right)$, where$z_0=P_{{\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}VI(C,B_i)\right)}{x}_1$.

Proof. 

Since $T_j$ is a generalized hybrid mapping of C into itself such that $F\left(T_j\right)\ne \varnothing$, from (2), $T_j$ is 0-demimetric. Furthermore, from Lemma 6, $T_j$ is demiclosed. Furtheremore, put $G=\partial i_C$ as in the proof of Theorem 4. Then we have that $Q_{{\eta }_n}(I-{\eta }_n{B}_i)=P_C(I-{\eta }_n{B}_i)$ in Theorem 2. We also have that if $A=0$, then $J_{r_n}=I$ and $u_n=z_n$. Therefore, we get the desired result from Theorem 2.

We prove a strong convergence theorem for a finite family of generalized hybrid mappings and a finite family of nonexpansive mappings in a Hilbert space.

Theorem 6.

Let C be a nonempty, closed, and convex subset of a Hilbert space H. Let${\left\{T_j\right\}}_{j=1}^M$be a finite family of generalized hybrid mappings of C into itself and let${\left\{U_i\right\}}_{i=1}^N$be a finite family of nonexpansive mappings of C into H. Suppose that${\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}F\left(U_i\right)\right)\ne \varnothing$. For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{y}_n={\sum }_{j=1}^M{\xi }_j((1-{\lambda }_n)I+{\lambda }_n{T}_j)x_n,\hfill \\ z_n={\sum }_{i=1}^N{\sigma }_i((1-{\eta }_n)I+{\eta }_n{U}_i)y_n,\hfill \\ C_{n+1}=\{z\in C_n:\parallel y_n-z\parallel \le \parallel x_n-z\parallel and\parallel z_n-z\parallel \le \parallel y_n-z\parallel \},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$a,b\in \mathbb{R}$, $\left\{{\lambda }_n\right\},\left\{{\eta }_n\right\}\subset (0,\infty )$and$\{{\xi }_1,\dots ,{\xi }_M\},\{{\sigma }_1,\dots ,{\sigma }_N\}\subset (0,1)$satisfy the following conditions:

(1) 

$0<a\le {\lambda }_n\le 1,\forall n\in \mathbb{N}$;

(2) 

$0<b\le {\eta }_n\le 1,\forall n\in \mathbb{N}$;

(3) 

${\sum }_{j=1}^M{\xi }_j=1$and${\sum }_{i=1}^N{\sigma }_i=1$.

Then$\left\{x_n\right\}$converges strongly to a point$z_0\in {\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}F\left(U_i\right)\right)$, where$z_0=P_{{\cap }_{j=1}^{M}F\left(T_j\right)\cap \left({\cap }_{i=1}^{N}F\left(U_i\right)\right)}{x}_1$.

Proof. 

As in the proof of Theorem 5, $T_j$ is 0-demimetric and demiclosed. Since $U_i$ is nonexpansive, $B_i=I-U_i$ is a $\frac{1}{2}$-inverse strongly monotone mapping. Furthermore, we get that

$$I-{\eta }_n{B}_i=I-{\eta }_n(I-U_i)=(1-{\eta }_n)I+{\eta }_n{U}_i.$$

Putting $A=G=0$, we get the desired result from Theorem 2.

We finally prove a strong convergence theorem for resolvents of a maximal monotone mapping in a Hilbert space.

Theorem 7.

Let H be a Hilbert space. Let A be a maximal monotone mapping on H and let$J_r={(I+rA)}^{-1}$be the resolvents of A for$r>0$. Suppose that$A^{-1}0\ne \varnothing .$For$x_1\in C$and$C_1=C$, let$\left\{x_n\right\}$be a sequence defined by

$$\left\{\begin{array}{c}{u}_n=J_{r_n}{x}_n,\hfill \\ C_{n+1}=\{z\in C_n:\langle x_n-z,x_n-u_n\rangle \ge \parallel x_n-u_n{\parallel }^2\},\hfill \\ x_{n+1}=P_{C_{n+1}}{x}_1,\forall n\in \mathbb{N},\hfill \end{array}\right.$$

where$c\in \mathbb{R}$and$\left\{r_n\right\}\subset (0,\infty )$satisfy the following:

$$0<c\le r_n,\forall n\in \mathbb{N}.$$

Then$\left\{x_n\right\}$converges strongly to a point$z_0\in A^{-1}0$, where$z_0=P_{A^{-1}0}{x}_1$.

Proof. 

Put $T_j=I$ and $B_i=0$ for all $j\in \{1,2,\dots ,M\}$ and $i\in \{1,2,\dots ,N\}$ in Theorem 2. Furthermore, put $G=0$. Then we have that $x_n=y_n=z_n$. Thus we get the desired result from Theorem 2.