Mann-Type Inertial Subgradient Extragradient Rules for Variational Inequalities and Common Fixed Points of Nonexpansive and Quasi-Nonexpansive Mappings

Abstract

Suppose that in a real Hilbert space H, the variational inequality problem with Lipschitzian and pseudomonotone mapping A and the common fixed-point problem of a finite family of nonexpansive mappings and a quasi-nonexpansive mapping with a demiclosedness property are represented by the notations VIP and CFPP, respectively. In this article, we suggest two Mann-type inertial subgradient extragradient iterations for finding a common solution of the VIP and CFPP. Our iterative schemes require only calculating one projection onto the feasible set for every iteration, and the strong convergence theorems are established without the assumption of sequentially weak continuity for A. Finally, in order to support the applicability and implementability of our algorithms, we make use of our main results to solve the VIP and CFPP in two illustrating examples.

1. Introduction

In a real Hilbert space ($H,\parallel ·\parallel$), equipped with the inner product $\langle ·,·\rangle$, we assume that C is a nonempty closed convex subset and $P_C$ is the metric projection of H onto C. If $S:C\to H$ is a mapping on C, then we denote by $\mathrm{Fix}\left(S\right)$ the fixed-point set of S. Moreover, we denote by $\mathbf{R}$ the set of all real numbers. Given a mapping $A:H\to H$. Consider the classical variational inequality problem (VIP) of finding $x^{*}\in C$ such that $\langle A{x}^{*},x-x^{*}\rangle \ge 0$ for all $x\in C$. We denote by VI($C,A$) the solution set of the VIP.

To the best of our knowledge, one of the most efficient methods to deal with the VIP is the extragradient method invented by Korpelevich [1] in 1976, that is, for any given $u_0\in C$, $\left\{u_m\right\}$ is the sequence constructed by

$$\left\{\begin{array}{cc}& v_m=P_C(u_m-\ell A{u}_m),\hfill \\ & u_{m+1}=P_C(u_m-\ell A{v}_m)\forall m\ge 0,\hfill \end{array}\right.$$

(1)

with constant $\ell \in (0,\frac{1}{L})$. If $\mathrm{VI}(C,A)\ne \varnothing$, one knows that this method has only weak convergence, and only requires that A is monotone and L-Lipschitzian. The literature on the VIP is vast, and Korpelevich’s extragradient method has received great attention from many authors, who improved it via various approaches so that some new iterative methods happen to solve the VIP and related optimization problems; see, e.g., [2,3,4,5,6,7,8,9,10,11,12] and the references therein, to name but a few.

It is worth pointing out that the extragradient method needs to calculate two projections onto the feasible set C per iteration. Without question, once one is hard to calculate the projection onto C, the minimum distance problem has to be solved twice per iteration. This perhaps affects the applicability and implementability of the method. To improve Algorithm 1, one has to reduce the number of projections per iteration. In 2011, Censor et al. [13] first suggested the subgradient extragradient method, in which the second projection onto C is replaced by a projection onto a half-space:

$$\left\{\begin{array}{cc}& v_m=P_C(u_m-\ell A{u}_m),\hfill \\ & C_m=\{w\in H:\langle u_m-\ell A{u}_m-v_m,w-v_m\rangle \le 0\},\hfill \\ & u_{m+1}=P_{C_m}(u_m-\ell A{v}_m)\forall m\ge 0,\hfill \end{array}\right.$$

(2)

where A is a L-Lipschitzian monotone mapping and $\ell \in (0,\frac{1}{L})$.

Since then, various modified extragradient-like iterative methods have been investigated by many researchers; see, e.g., [14,15,16,17,18,19]. In 2014, combining the subgradient extragradient method and Halpern’s iteration method, Kraikaew and Saejung [20] proposed the Halpern subgradient extragradient method for solving the VIP, that is, for any given $u_0\in H$, $\left\{u_m\right\}$ is the sequence constructed by

$$\left\{\begin{array}{cc}& v_m=P_C(u_m-\ell A{u}_m),\hfill \\ & C_m=\{v\in H:\langle u_m-\ell A{u}_m-v_m,v-v_m\rangle \le 0\},\hfill \\ & w_m=P_{C_m}(u_m-\ell A{v}_m),\hfill \\ & u_{m+1}={\alpha }_m{u}_0+(1-{\alpha }_m)w_m\forall m\ge 0,\hfill \end{array}\right.$$

(3)

where $\ell \in (0,\frac{1}{L}),\left\{{\alpha }_m\right\}\subset (0,1),{lim}_{m\to \infty }{\alpha }_m=0$ and ${\sum }_{m=1}^{\infty }{\alpha }_m=+\infty$. They proved the strong convergence of $\left\{u_m\right\}$ to $P_{\mathrm{VI}(C,A)}{u}_0$.

In 2018, Thong and Hieu [21] first suggested the inertial subgradient extragradient method, that is, for any given $u_0,u_1\in H$, the sequence $\left\{u_m\right\}$ is generated by

$$\left\{\begin{array}{cc}& w_m=u_m+{\alpha }_m(u_m-u_{m-1}),\hfill \\ & v_m=P_C(w_m-\ell A{w}_m),\hfill \\ & C_m=\{v\in H:\langle w_m-\ell A{w}_m-v_m,v-v_m\rangle \le 0\},\hfill \\ & u_{m+1}=P_{C_m}(w_m-\ell A{v}_m)\forall m\ge 1,\hfill \end{array}\right.$$

(4)

with constant $\ell \in (0,\frac{1}{L})$. Under suitable conditions, they proved the weak convergence of $\left\{u_m\right\}$ to an element of $\mathrm{VI}(C,A)$. Later, Thong and Hieu [22] designed two inertial subgradient extragradient algorithms with linesearch process for solving a VIP with monotone and Lipschitz continuous mapping A and a FPP of quasi-nonexpansive mapping T with a demiclosedness property in H. Under appropriate conditions, they established the weak convergence results for the suggested algorithms.

Suppose that the notations VIP and CFPP represent a variational inequality problem with Lipschitzian and pseudomonotone mapping $A:H\to H$ and a common fixed-point problem of finitely many nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$ and a quasi-nonexpansive mapping T with a demiclosedness property, respectively. Inspired by the research works above, we design two Mann-type inertial subgradient extragradient iterations for finding a common solution of the VIP and CFPP. Our algorithms require only computing one projection onto the feasible set C per iteration, and the strong convergence theorems are established without the assumption of sequentially weak continuity for A on C. Finally, in order to support the applicability and implementability of our algorithms, we make use of our main results to solve the VIP and CFPP in two illustrating examples.

This paper is organized as follows: In Section 2, we recall some definitions and preliminaries for the sequel use. Section 3 deals with the convergence analysis of the proposed algorithms. Finally, in Section 4, in order to support the applicability and implementability of our algorithms, we make use of our main results to find a common solution of the VIP and CFPP in two illustrating examples.

2. Preliminaries

Throughout this paper, we assume that C is a nonempty closed convex subset of a real Hilbert space H. If $\left\{u_m\right\}$ is a sequence in H, then we denote by $u_m\to u$ (respectively, $u_m\rightharpoonup u$) the strong (respectively, weak) convergence of $\left\{u_m\right\}$ to u. A mapping $F:C\to H$ is said to be nonexpansive if $\parallel Fu-Fv\parallel \le \parallel u-v\parallel \forall u,v\in C$. Recall also that $F:C\to H$ is called

(i)

L-Lipschitz continuous (or L-Lipschitzian) if $\exists L>0$ such that $\parallel Fu-Fv\parallel \le L\parallel u-v\parallel \forall u,v\in C$;

(ii)

monotone if $\langle Fu-Fv,u-v\rangle \ge 0\forall u,v\in C$;

(iii)

pseudomonotone if $\langle Fu,v-u\rangle \ge 0\Rightarrow \langle Fv,v-u\rangle \ge 0\forall u,v\in C$;

(iv)

$\alpha$-strongly monotone if $\exists \alpha >0$ such that $\langle Fu-Fv,u-v\rangle \ge {\alpha \parallel u-v\parallel }^2\forall u,v\in C$;

(v)

quasi-nonexpansive if $\mathrm{Fix}\left(F\right)\ne \varnothing$, and $\parallel Fu-p\parallel \le \parallel u-p\parallel \forall u\in C,p\in \mathrm{Fix}\left(F\right)$;

(vi)

sequentially weakly continuous on C if for $\left\{u_m\right\}\subset C$, the relation holds: $u_m\rightharpoonup u\Rightarrow F{u}_m\rightharpoonup Fu$.

It is clear that every monotone operator is pseudomonotone, but the converse is not true. Next, we provide an example of a quasi-nonexpansive mapping which is not nonexpansive.

Example 1.

Let $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. Let $T:H\to H$ be defined as $Tu:=\frac{u}{2}sinu\forall u\in H$. It is clear that $\mathrm{Fix}\left(T\right)=\left\{0\right\}$ and T is quasi-nonexpansive. However, we claim that T is not nonexpansive. Indeed, putting $u=2\pi$ and $v=\frac{3\pi }{2}$, we have $\parallel Tu-Tv\parallel =\parallel \frac{2\pi }{2}sin2\pi -\frac{3\pi }{4}sin\frac{3\pi }{2}\parallel =\frac{3\pi }{4}>\parallel 2\pi -\frac{3\pi }{2}\parallel =\frac{\pi }{2}$.

Definition 1

([23]). Assume that $T:H\to H$ is a nonlinear operator with $\mathrm{Fix}\left(T\right)\ne \varnothing$. Then $I-T$ is said to be demiclosed at zero if for any $\left\{u_n\right\}$ in H, the implication holds: $u_n\rightharpoonup u$ and $(I-T)u_n\to 0\Rightarrow u\in \mathrm{Fix}\left(T\right)$.

Very recently, Thong and Hieu gave an example to illustrate that there exists a quasi-nonexpansive mapping T, but $I-T$ is not demiclosed at zero; see ([22], Example 2). For each $u\in H$, we know that there exists a unique nearest point in C, denoted by $P_{C}u$, such that $\parallel u-P_{C}u\parallel \le \parallel u-v\parallel \forall v\in C$. $P_C$ is called a metric projection of H onto C.

Lemma 1

([23]). The following hold:

(i) 

$\langle u-v,P_{C}u-P_{C}v\rangle \ge {\parallel P_{C}u-P_{C}v\parallel }^2\forall u,v\in H$;

(ii) 

$\langle u-P_{C}u,v-P_{C}u\rangle \le 0\forall u\in H,v\in C$;

(iii) 

${\parallel u-v\parallel }^2\ge \parallel u-P_C{u\parallel }^2+{\parallel v-P_{C}u\parallel }^2\forall u\in H,v\in C$;

(iv) 

${\parallel u-v\parallel }^2={\parallel u\parallel }^2-{\parallel v\parallel }^2-2\langle u-v,v\rangle \forall u,v\in H$;

(v) 

${\parallel \lambda u+(1-\lambda )v\parallel }^2={\lambda \parallel u\parallel }^2+{(1-\lambda )\parallel v\parallel }^2-\lambda (1-\lambda ){\parallel u-v\parallel }^2\forall u,v\in H,\lambda \in [0,1]$.

Lemma 2

([24]). For all $u\in H$ and $\alpha \ge \beta >0$, the inequalities hold: $\frac{\parallel u-P_C(u-\alpha Au)\parallel }{\alpha }\le \frac{\parallel u-P_C(u-\beta Au)\parallel }{\beta }$ and $\parallel u-P_C(u-\beta Au)\parallel \le \parallel u-P_C(u-\alpha Au)\parallel$.

Lemma 3

([13]). Suppose that $A:C\to H$ is pseudomonotone and continuous. Then $u^{*}\in C$ is a solution to the VIP $\langle A{u}^{*},u-u^{*}\rangle \ge 0\forall u\in C$, if and only if $\langle Au,u-u^{*}\rangle \ge 0\forall u\in C$.

Lemma 4

([25]). Suppose that $\left\{a_m\right\}$ is a sequence of nonnegative numbers satisfying the conditions: $a_{m+1}\le (1-{\lambda }_m)a_m+{\lambda }_m{\gamma }_m\forall m\ge 1$, where $\left\{{\lambda }_m\right\}$ and $\left\{{\gamma }_m\right\}$ lie in $\mathbf{R}=(-\infty ,\infty )$ such that (i) $\left\{{\lambda }_m\right\}\subset [0,1]$ and ${\sum }_{m=1}^{\infty }{\lambda }_m=\infty$, and (ii) ${lim\; sup}_{m\to \infty }{\gamma }_m\le 0$ or ${\sum }_{m=1}^{\infty }\left|{\lambda }_m{\gamma }_m\right|<\infty$. Then ${lim}_{m\to \infty }{a}_m=0$.

Lemma 5

([23]). Suppose that $T:C\to C$ is a nonexpansive mapping with $\mathrm{Fix}\left(T\right)\ne \varnothing$. Then $I-T$ is demiclosed at zero, that is, if $\left\{u_m\right\}$ is a sequence in C such that $u_m\rightharpoonup u\in C$ and $(I-T)u_m\to 0$, then $(I-T)u=0$, where I is the identity mapping of H.

Lemma 6

([25]). Suppose that $\lambda \in (0,1]$, $T:C\to H$ is a nonexpansive mapping, and the mapping $T^{\lambda }:C\to H$ is defined as $T^{\lambda }u:=Tu-\lambda \mu F\left(Tu\right)\forall u\in C$, where $F:H\to H$ is κ-Lipschitzian and η-strongly monotone. Then $T^{\lambda }$ is a contraction provided $0<\mu <\frac{2\eta }{{\kappa }^2}$, that is, $\parallel T^{\lambda }u-T^{\lambda }v\parallel \le (1-\lambda \ell )\parallel u-v\parallel \forall u,v\in C$, where $\ell :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$.

Lemma 7

([26]). Suppose that $\left\{{\Gamma }_m\right\}$ is a sequence of real numbers that does not decrease at infinity in the sense that there exists a subsequence $\left\{{\Gamma }_{m_k}\right\}$ of $\left\{{\Gamma }_m\right\}$ which satisfies ${\Gamma }_{m_k}<{\Gamma }_{m_k+1}$ for each integer $k\ge 1$. Define the sequence ${\left\{\tau \left(m\right)\right\}}_{m\ge m_0}$ of integers as follows:

$$\tau \left(m\right)=max\{k\le m:{\Gamma }_k<{\Gamma }_{k+1}\},$$

where integer $m_0\ge 1$ such that $\{k\le m_0:{\Gamma }_k<{\Gamma }_{k+1}\}\ne \varnothing$. Then, the following conclusions hold:

(i) 

$\tau \left(m_0\right)\le \tau (m_0+1)\le \cdots$ and $\tau \left(m\right)\to \infty$;

(ii) 

${\Gamma }_{\tau \left(m\right)}\le {\Gamma }_{\tau \left(m\right)+1}$ and ${\Gamma }_m\le {\Gamma }_{\tau \left(m\right)+1}\forall m\ge m_0$.

3. Iterative Algorithms and Convergence Criteria

In this section, let the feasible set C be a nonempty closed convex subset of a real Hilbert space H, and assume always that the following hold:

$T_i:H\to H$ is nonexpansive for $i=1,...,N$ and $T:H\to H$ is a quasi-nonexpansive mapping such that $I-T$ is demiclosed at zero;

$A:H\to H$ is L-Lipschitz continuous, pseudomonotone on H, and satisfies the condition that for $\left\{x_n\right\}\subset C$, $x_n\rightharpoonup z\Rightarrow \parallel Az\parallel \le {lim\; inf}_{n\to \infty }\parallel A{x}_n\parallel$;

$\mathsf{\Omega }={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)\ne \varnothing$ with $T_0:=T$;

$f:H\to H$ is a contraction with constant $\delta \in [0,1)$, and $F:H\to H$ is $\eta$-strongly monotone and $\kappa$-Lipschitzian such that $\delta <\tau :=1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}$ for $\rho \in (0,\frac{2\eta }{{\kappa }^2})$; $\left\{{\zeta }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$, and $\left\{{\tau }_n\right\}\subset (0,\infty )$ are such that

(i)

${\beta }_n+{\gamma }_n<1$ and ${\sum }_{n=1}^{\infty }{\beta }_n=\infty$;

(ii)

${lim}_{n\to \infty }{\beta }_n=0$ and ${\tau }_n=\circ \left({\beta }_n\right)$, i.e., ${lim}_{n\to \infty }{\tau }_n/{\beta }_n=0$;

(iii)

$0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$ and $0<{lim\; inf}_{n\to \infty }{\zeta }_n\le {lim\; sup}_{n\to \infty }{\zeta }_n<1$.

In addition, we write $T_n:=T_{n\mathrm{mod}N}$ for integer $n\ge 1$ with the mod function taking values in the set $\{1,2,...,N\}$, i.e., if $n=jN+q$ for some integers $j\ge 0$ and $0\le q<N$, then $T_n=T_N$ if $q=0$ and $T_n=T_q$ if $0<q<N$.

Algorithm 1. Initialization: Let ${\lambda }_1>0,\alpha >0,\mu \in (0,1)$ and $x_0,x_1\in H$ be arbitrary.

Iterative Steps: Calculate $x_{n+1}$ as follows:

Step 1. Given the iterates $x_{n-1}$ and $x_n(n\ge 1)$, choose ${\alpha }_n$ such that $0\le {\alpha }_n\le {\overline{\alpha }}_n$, where

$${\overline{\alpha }}_n=\left\{\begin{array}{cc}& min\{\alpha ,\frac{{\tau }_n}{\parallel x_n-x_{n-1}\parallel }\}\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ & \alpha \mathrm{otherwise}.\hfill \end{array}\right.$$

(5)

Step 2. Compute $w_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and $y_n=P_C(w_n-{\lambda }_{n}A{w}_n)$.

Step 3. Construct the half-space $C_n:=\{z\in H:\langle w_n-{\lambda }_{n}A{w}_n-y_n,z-y_n\rangle \le 0\}$, and compute $z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n)$.

Step 4. Calculate $v_n={\zeta }_n{x}_n+(1-{\zeta }_n)T_n{w}_n$ and $x_{n+1}={\beta }_{n}f\left(x_n\right)+{\gamma }_{n}T{z}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n$, and update

$${\lambda }_{n+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_n-y_n{\parallel }^2+{\parallel z_n-y_n\parallel }^2}{2\langle A{w}_n-A{y}_n,z_n-y_n\rangle },{\lambda }_n\}\mathrm{if}\langle A{w}_n-A{y}_n,z_n-y_n\rangle >0,\hfill \\ & {\lambda }_n\mathrm{otherwise}.\hfill \end{array}\right.$$

(6)

Let $n:=n+1$ and return to Step 1.

Remark 1.

It is easy to see that, from (5) we get ${lim}_{n\to \infty }\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel =0$. Indeed, we have ${\alpha }_n\parallel x_n-x_{n-1}\parallel \le {\tau }_n\forall n\ge 1$, which together with ${lim}_{n\to \infty }\frac{{\tau }_n}{{\beta }_n}=0$ implies that $\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \le \frac{{\tau }_n}{{\beta }_n}\to 0$ as $n\to \infty$.

Lemma 8.

Let $\left\{{\lambda }_n\right\}$ be generated by (6). Then $\left\{{\lambda }_n\right\}$ is a nonincreasing sequence with ${\lambda }_n\ge \lambda :=min\{{\lambda }_1,\frac{\mu }{L}\}\forall n\ge 1$, and ${lim}_{n\to \infty }{\lambda }_n\ge \lambda :=min\{{\lambda }_1,\frac{\mu }{L}\}$.

Proof. 

First, from (6) it is clear that ${\lambda }_n\ge {\lambda }_{n+1}\forall n\ge 1$. Furthermore, observe that

$$\left.\begin{array}{cc}& \frac{1}{2}(\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\ge \parallel w_n-y_n\parallel \parallel z_n-y_n\parallel \hfill \\ & \langle A{w}_n-A{y}_n,z_n-y_n\rangle \le L\parallel w_n-y_n\parallel \parallel z_n-y_n\parallel \hfill \end{array}\right\}\Rightarrow {\lambda }_{n+1}\ge min\{{\lambda }_n,\frac{\mu }{L}\}.$$

□

Remark 2.

In terms of Lemmas 2 and 8, we claim that if $w_n=y_n$ or $A{y}_n=0$, then $y_n$ is an element of $\mathrm{VI}(C,A)$. Indeed, if $w_n=y_n$ or $A{y}_n=0$, then $0=\parallel y_n-P_C(y_n-{\lambda }_{n}A{y}_n)\parallel \ge \parallel y_n-P_C(y_n-\lambda A{y}_n)\parallel$. Thus, the assertion is valid.

The following lemmas are quite helpful for the convergence analysis of our algorithms.

Lemma 9.

Let $\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\}$ be the sequences generated by Algorithm 1. Then

$$\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel w_n-y_n{\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}){\parallel z_n-y_n\parallel }^2\forall p\in \mathsf{\Omega }.$$

(7)

Proof. 

First, by the definition of $\left\{{\lambda }_n\right\}$ we claim that

$$2\langle A{w}_n-A{y}_n,z_n-y_n\rangle \le \frac{\mu }{{\lambda }_{n+1}}\parallel w_n-y_n{\parallel }^2+\frac{\mu }{{\lambda }_{n+1}}{\parallel z_n-y_n\parallel }^2\forall n\ge 1.$$

(8)

Indeed, if $\langle A{w}_n-A{y}_n,z_n-y_n\rangle \le 0$, then inequality (8) holds. Otherwise, from (6) we get (8). Furthermore, observe that for each $p\in \mathsf{\Omega }\subset C\subset C_n$,

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2\hfill & =\parallel P_{C_n}(w_n-{\lambda }_{n}A{y}_n)-P_{C_n}{p\parallel }^2\le \langle z_n-p,w_n-{\lambda }_{n}A{y}_n-p\rangle \hfill \\ & =\frac{1}{2}\parallel z_n{-p\parallel }^2+\frac{1}{2}\parallel w_n{-p\parallel }^2-\frac{1}{2}{\parallel z_n-w_n\parallel }^2-\langle z_n-p,{\lambda }_{n}A{y}_n\rangle ,\hfill \end{array}$$

which hence yields

$$\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-{\parallel z_n-w_n\parallel }^2-2\langle z_n-p,{\lambda }_{n}A{y}_n\rangle .$$

(9)

From $p\in \mathrm{VI}(C,A)$, we get $\langle Ap,x-p\rangle \ge 0\forall x\in C$. By the pseudomonotonicity of A on C we have $\langle Ax,x-p\rangle \ge 0\forall x\in C$. Putting $x:=y_n\in C$ we get $\langle A{y}_n,p-y_n\rangle \le 0$. Thus,

$$\langle A{y}_n,p-z_n\rangle =\langle A{y}_n,p-y_n\rangle +\langle A{y}_n,y_n-z_n\rangle \le \langle A{y}_n,y_n-z_n\rangle .$$

(10)

Substituting (10) for (9), we obtain

$$\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-\parallel z_n-y_n{\parallel }^2-{\parallel y_n-w_n\parallel }^2+2\langle w_n-{\lambda }_{n}A{y}_n-y_n,z_n-y_n\rangle .$$

(11)

Since $z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n)$, we get $z_n\in C_n:=\{z\in H:\langle w_n-{\lambda }_{n}A{w}_n-y_n,z-y_n\rangle \le 0\}$, and hence

$$\begin{array}{cc}2\langle w_n-{\lambda }_{n}A{y}_n-y_n,z_n-y_n\rangle \hfill & =2\langle w_n-{\lambda }_{n}A{w}_n-y_n,z_n-y_n\rangle +2{\lambda }_n\langle A{w}_n-A{y}_n,z_n-y_n\rangle \hfill \\ & \le 2{\lambda }_n\langle A{w}_n-A{y}_n,z_n-y_n\rangle ,\hfill \end{array}$$

which together with (8), implies that

$$2\langle w_n-{\lambda }_{n}A{y}_n-y_n,z_n-y_n\rangle \le \mu \frac{{\lambda }_n}{{\lambda }_{n+1}}\parallel w_n-y_n{\parallel }^2+\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}{\parallel z_n-y_n\parallel }^2.$$

Therefore, substituting the last inequality for (11), we infer that inequality (7) holds. □

Lemma 10.

Suppose that $\left\{w_n\right\},\left\{x_n\right\},\left\{y_n\right\}$, and $\left\{z_n\right\}$ are bounded sequences generated by Algorithm 1. If $x_n-x_{n+1}\to 0,w_n-y_n\to 0,w_n-z_n\to 0,z_n-T_n{z}_n\to 0$, and $\exists \left\{w_{n_k}\right\}\subset \left\{w_n\right\}$ s.t. $w_{n_k}\rightharpoonup z\in H$, then $z\in \mathsf{\Omega }$.

Proof. 

Utilizing the similar arguments to those in the proof of Lemma 3.3 of [12], we can derive the desired result. □

Lemma 11.

Assume that $\left\{w_n\right\},\left\{x_n\right\},\left\{y_n\right\},\left\{z_n\right\}$ are the sequences generated by Algorithm 1. Then they all are bounded.

Proof. 

Since $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$ and $0<{lim\; inf}_{n\to \infty }{\zeta }_n\le {lim\; sup}_{n\to \infty }$${\zeta }_n<1$, we may assume, without loss of generality, that

$$\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)\mathrm{and}\left\{{\zeta }_n\right\}\subset [c,d]\subset (0,1).$$

(12)

Choose a fixed $p\in \mathsf{\Omega }$ arbitrarily. Then we obtain $Tp=p$ and $T_{n}p=p$ for all $n\ge 1$, and (7) holds. Noticing ${lim}_{n\to \infty }(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})=1-\mu >0$, we might assume that $1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}>0$ for all $n\ge 1$. So it follows from (7) that for all $n\ge 1$,

$$\parallel z_n-p\parallel \le \parallel w_n-p\parallel .$$

(13)

Furthermore, note that

$$\parallel w_n-p\parallel \le \parallel x_n-p\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel =\parallel x_n-p\parallel +{\beta }_n·\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel .$$

(14)

In terms of Remark 1, one has $\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \to 0$ as $n\to \infty$. Hence we deduce that $\exists M_1>0$ s.t.

$$M_1\ge \frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \forall n\ge 1.$$

(15)

Using (13)–(15), we obtain that for all $n\ge 1$,

$$\parallel z_n-p\parallel \le \parallel w_n-p\parallel \le \parallel x_n-p\parallel +{\beta }_n{M}_1.$$

(16)

Noticing ${\beta }_n+{\gamma }_n<1\forall n\ge 1$, we have $\frac{{\beta }_n}{1-{\gamma }_n}<1$ for all $n\ge 1$. So, using Lemma 6 and (16) we deduce that

$$\begin{array}{cc}\parallel v_n-p\parallel \hfill & \le {\zeta }_n\parallel x_n-p\parallel +(1-{\zeta }_n)\parallel T_n{w}_n-p\parallel \hfill \\ & \le {\zeta }_n\parallel x_n-p\parallel +(1-{\zeta }_n)\parallel w_n-p\parallel \hfill \\ & \le {\zeta }_n(\parallel x_n-p\parallel +{\beta }_n{M}_1)+(1-{\zeta }_n)(\parallel x_n-p\parallel +{\beta }_n{M}_1)\hfill \\ & =\parallel x_n-p\parallel +{\beta }_n{M}_1,\hfill \end{array}$$

and hence

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & =\parallel {\beta }_{n}f\left(x_n\right)+{\gamma }_{n}T{z}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n-p\parallel \hfill \\ & \le {\beta }_n\parallel f\left(x_n\right)-p\parallel +{\gamma }_n\parallel T{z}_n-p\parallel \hfill \\ & +(1-{\beta }_n-{\gamma }_n)\parallel (\frac{1-{\gamma }_n}{1-{\beta }_n-{\gamma }_n}I-\frac{{\beta }_n}{1-{\beta }_n-{\gamma }_n}\rho F)v_n-p\parallel \hfill \\ & \le {\beta }_n(\parallel f\left(x_n\right)-f\left(p\right)\parallel +\parallel f\left(p\right)-p\parallel )+{\gamma }_n\parallel z_n-p\parallel \hfill \\ & +(1-{\beta }_n-{\gamma }_n)\parallel (\frac{1-{\gamma }_n}{1-{\beta }_n-{\gamma }_n}I-\frac{{\beta }_n}{1-{\beta }_n-{\gamma }_n}\rho F)v_n-p\parallel \hfill \\ & \le {\beta }_n(\delta \parallel x_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\gamma }_n\parallel z_n-p\parallel \hfill \\ & +(1-{\gamma }_n)\parallel (I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n-(1-\frac{{\beta }_n}{1-{\gamma }_n})p\parallel \hfill \\ & ={\beta }_n(\delta \parallel x_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\gamma }_n\parallel z_n-p\parallel \hfill \\ & +(1-{\gamma }_n)\parallel (I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)p+\frac{{\beta }_n}{1-{\gamma }_n}(I-\rho F)p\parallel \hfill \\ & \le {\beta }_n(\delta \parallel x_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\gamma }_n\parallel z_n-p\parallel \hfill \\ & +(1-{\gamma }_n)[(1-\frac{{\beta }_n}{1-{\gamma }_n}\tau )\parallel v_n-p\parallel +\frac{{\beta }_n}{1-{\gamma }_n}\parallel (I-\rho F)p\parallel ]\hfill \\ & ={\beta }_n(\delta \parallel x_n-p\parallel +\parallel f\left(p\right)-p\parallel )+{\gamma }_n\parallel z_n-p\parallel \hfill \\ & +(1-{\gamma }_n-{\beta }_n\tau )\parallel v_n-p\parallel +{\beta }_n\parallel (I-\rho F)p\parallel \hfill \\ & \le {\beta }_n\delta (\parallel x_n-p\parallel +{\beta }_n{M}_1)+{\beta }_n\parallel f\left(p\right)-p\parallel +{\gamma }_n(\parallel x_n-p\parallel +{\beta }_n{M}_1)\hfill \\ & +(1-{\gamma }_n-{\beta }_n\tau )(\parallel x_n-p\parallel +{\beta }_n{M}_1)+{\beta }_n\parallel (I-\rho F)p\parallel \hfill \\ & \le [1-{\beta }_n(\tau -\delta )]\parallel x_n-p\parallel +{\beta }_n(M_1+\parallel f\left(p\right)-p\parallel +\parallel (I-\rho F)p\parallel )\hfill \\ & =[1-{\beta }_n(\tau -\delta )]\parallel x_n-p\parallel +{\beta }_n(\tau -\delta )·\frac{M_1+\parallel f\left(p\right)-p\parallel +\parallel (I-\rho F)p\parallel }{\tau -\delta }\hfill \\ & \le max\{\parallel x_n-p\parallel ,\frac{M_1+\parallel f\left(p\right)-p\parallel +\parallel (I-\rho F)p\parallel }{\tau -\delta }\}.\hfill \end{array}$$

By induction, we obtain $\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{M_1+\parallel f\left(p\right)-p\parallel +\parallel (I-\rho F)p\parallel }{\tau -\delta }\}\forall n\ge 1$. Thus, $\left\{x_n\right\}$ is bounded, and so are the sequences $\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{T{z}_n\right\},\left\{F{v}_n\right\},\left\{T_n{w}_n\right\}$. □

Theorem 1.

Let the sequence $\left\{x_n\right\}$ be constructed by Algorithm 1. Then $\left\{x_n\right\}$ converges strongly to the unique solution $x^{*}\in \mathsf{\Omega }$ of the following VIP:

$$\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }.$$

Proof. 

First, it is not difficult to show that $P_{\mathsf{\Omega }}(f+I-\rho F)$ is a contraction. In fact, by Lemma 6 and the Banach contraction mapping principle, we obtain that $P_{\mathsf{\Omega }}(f+I-\rho F)$ has a unique fixed point. Say $x^{*}\in H$, i.e., $x^{*}=P_{\mathsf{\Omega }}(f+I-\rho F)x^{*}$. Thus, the following VIP has only a solution $x^{*}\in \mathsf{\Omega }$:

$$\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }.$$

(17)

□

We now claim that

$${\gamma }_n(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]\le \parallel x_n-x^{*}{\parallel }^2-{\parallel x_{n+1}-x^{*}\parallel }^2+{\beta }_n{M}_4,$$

for some $M_4>0$. In fact, observe that

$$\begin{array}{cc}& x_{n+1}-x^{*}={\beta }_n(f\left(x_n\right)-x^{*})+{\gamma }_n(T{z}_n-x^{*})+(1-{\beta }_n-{\gamma }_n)\left\{\frac{1-{\gamma }_n}{1-{\beta }_n-{\gamma }_n}\right[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n\hfill \\ & -(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)x^{*}]+\frac{{\beta }_n}{1-{\beta }_n-{\gamma }_n}(I-\rho F)x^{*}\}\hfill \\ & ={\beta }_n(f\left(x_n\right)-f\left(x^{*}\right))+{\gamma }_n(T{z}_n-x^{*})+(1-{\gamma }_n)[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)x^{*}]\hfill \\ & +{\beta }_n(f-\rho F)x^{*}.\hfill \end{array}$$

Using Lemma 6 and the convexity of the function $h\left(t\right)=t^2\forall t\in \mathbf{R}$, we have

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & \le \parallel {\beta }_n(f\left(x_n\right)-f\left(x^{*}\right))+{\gamma }_n(T{z}_n-x^{*})+(1-{\gamma }_n)[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)x^{*}]{\parallel }^2\hfill \\ & +2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n\parallel z_n-x^{*}{\parallel }^2+(1-{\beta }_n\tau -{\gamma }_n){\parallel v_n-x^{*}\parallel }^2+{\beta }_n{M}_2\hfill \end{array}$$

(18)

where $M_2\ge {sup}_{n\ge 1}2\parallel f-\rho F)x^{*}\parallel \parallel x_n-x^{*}\parallel$ for some $M_2>0$. From (7) and (17), we have

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n[\parallel w_n-x^{*}{\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel w_n-y_n{\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel z_n-y_n{\parallel }^2]\hfill \\ & +(1-{\beta }_n\tau -{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel w_n-x^{*}{\parallel }^2]+{\beta }_n{M}_2.\hfill \end{array}$$

(19)

Again from (16), we obtain

$$\parallel w_n-x^{*}{\parallel }^2\le (\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2\le {\parallel x_n-x^{*}\parallel }^2+{\beta }_n{M}_3,$$

(20)

where $M_3\ge {sup}_{n\ge 1}(2M_1\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1^2)$ for some $M_3>0$. Using (19) and (20), we get

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & \le [1-{\beta }_n(\tau -\delta )](\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{M}_3)-{\gamma }_n(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]+{\beta }_n{M}_2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2-{\gamma }_n(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]+{\beta }_n{M}_4,\hfill \end{array}$$

where $M_4:=M_2+M_3$. Consequently,

$${\gamma }_n(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]\le \parallel x_n-x^{*}{\parallel }^2-{\parallel x_{n+1}-x^{*}\parallel }^2+{\beta }_n{M}_4.$$

(21)

Next we claim that

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le [1-{\beta }_n(\tau -\delta )]{\parallel x_n-x^{*}\parallel }^2\hfill \\ & +{\beta }_n(\tau -\delta )[\frac{2}{\tau -\delta }\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle +\frac{3M}{\tau -\delta }·\frac{{\alpha }_n}{{\beta }_n}·\parallel x_n-x_{n-1}\parallel ]\hfill \end{array}$$

for some $M>0$. In fact, it is easy to see that

$$\parallel w_n-x^{*}{\parallel }^2\le \parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel [2\parallel x_n-x^{*}\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel ].$$

(22)

Using (16), (18), and (22), we get

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n\parallel w_n-x^{*}{\parallel }^2+(1-{\beta }_n\tau -{\gamma }_n)[{\zeta }_n{\parallel x_n-x^{*}\parallel }^2\hfill \\ & +(1-{\zeta }_n)\parallel w_n-x^{*}{\parallel }^2]+2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n[\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-x^{*}\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel \left)\right]\hfill \\ & +(1-{\beta }_n\tau -{\gamma }_n)\{{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)[\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-x^{*}\parallel \hfill \\ & +{\alpha }_n\parallel x_n-x_{n-1}\parallel \left)\right]\}+2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le [1-{\beta }_n(\tau -\delta )]\parallel x_n-x^{*}{\parallel }^2+{\beta }_n(\tau -\delta )·[\frac{2\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle }{\tau -\delta }+\frac{3M}{\tau -\delta }·\frac{{\alpha }_n}{{\beta }_n}·\parallel x_n-x_{n-1}\parallel ],\hfill \end{array}$$

(23)

where $M\ge {sup}_{n\ge 1}\{\parallel x_n-x^{*}\parallel ,{\alpha }_n\parallel x_n-x_{n-1}\parallel \}$ for some $M>0$.

For each $n\ge 0$, we set

$$\begin{array}{cc}& {\Gamma }_n={\parallel x_n-x^{*}\parallel }^2,\hfill \\ & {\epsilon }_n={\beta }_n(\tau -\delta ),\hfill \\ & {\vartheta }_n={\alpha }_n\parallel x_n-x_{n-1}\parallel 3M+2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle .\hfill \end{array}$$

Then (23) can be rewritten as the following formula:

$${\Gamma }_{n+1}\le (1-{\epsilon }_n){\Gamma }_n+{\vartheta }_n\forall n\ge 0.$$

(24)

We next show the convergence of $\left\{{\Gamma }_n\right\}$ to zero by the following two cases:

Case 1.Suppose that there exists an integer $n_0\ge 1$ such that $\left\{{\Gamma }_n\right\}$ is non-increasing. Then

$${\Gamma }_n-{\Gamma }_{n+1}\to 0.$$

From (21), we get

$${\gamma }_n(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]\le {\Gamma }_n-{\Gamma }_{n+1}+{\beta }_n{M}_4.$$

Since ${\beta }_n\to 0,{\Gamma }_n-{\Gamma }_{n+1}\to 0,1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}}\to 1-\mu$ and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$, we have

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

(25)

Using Lemma 1 (v), we deduce from (16) that

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & =\parallel {\beta }_{n}f\left(x_n\right)+{\gamma }_{n}T{z}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n-x^{*}{\parallel }^2\hfill \\ & =\parallel {\beta }_n(f\left(x_n\right)-\rho F{v}_n)+{\gamma }_n(T{z}_n-x^{*})+(1-{\gamma }_n)(v_n-x^{*}){\parallel }^2\hfill \\ & \le \parallel {\gamma }_n(T{z}_n-x^{*})+(1-{\gamma }_n)(v_n-x^{*}){\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & ={\gamma }_n\parallel T{z}_n-x^{*}{\parallel }^2+(1-{\gamma }_n)\parallel v_n-x^{*}{\parallel }^2-{\gamma }_n(1-{\gamma }_n){\parallel T{z}_n-v_n\parallel }^2\hfill \\ & +2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & ={\gamma }_n\parallel T{z}_n-x^{*}{\parallel }^2+(1-{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel T_n{w}_n-x^{*}{\parallel }^2-{\zeta }_n(1-{\zeta }_n)\parallel x_n-T_n{w}_n{\parallel }^2]\hfill \\ & -{\gamma }_n(1-{\gamma }_n){\parallel T{z}_n-v_n\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & \le {\gamma }_n\parallel z_n-x^{*}{\parallel }^2+(1-{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel w_n-x^{*}{\parallel }^2-{\zeta }_n(1-{\zeta }_n)\parallel x_n-T_n{w}_n{\parallel }^2]\hfill \\ & -{\gamma }_n(1-{\gamma }_n){\parallel T{z}_n-v_n\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & \le {\gamma }_n(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2+(1-{\gamma }_n)(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-(1-{\gamma }_n){\zeta }_n(1-{\zeta }_n){\parallel x_n-T_n{w}_n\parallel }^2\hfill \\ & -{\gamma }_n(1-{\gamma }_n)\parallel T{z}_n-v_n{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & =(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-(1-{\gamma }_n){\zeta }_n(1-{\zeta }_n){\parallel x_n-T_n{w}_n\parallel }^2\hfill \\ & -{\gamma }_n(1-{\gamma }_n)\parallel T{z}_n-v_n{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel ,\hfill \end{array}$$

which immediately yields

$$\begin{array}{cc}& (1-{\gamma }_n){\zeta }_n(1-{\zeta }_n)\parallel x_n-T_n{w}_n{\parallel }^2+{\gamma }_n(1-{\gamma }_n){\parallel T{z}_n-v_n\parallel }^2\hfill \\ & \le (\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-\parallel x_{n+1}-x^{*}{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & ={\Gamma }_n-{\Gamma }_{n+1}+{\beta }_n{M}_1(2\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1)+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel .\hfill \end{array}$$

Since ${\beta }_n\to 0,{\Gamma }_n-{\Gamma }_{n+1}\to 0$, $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ and $\left\{{\zeta }_n\right\}\subset [c,d]\subset (0,1)$, we have

$$\underset{n\to \infty }{lim}\parallel x_n-T_n{w}_n\parallel =\underset{n\to \infty }{lim}\parallel T{z}_n-v_n\parallel =0.$$

(26)

Using Lemma 1 (v) again, we have

$$\begin{array}{cc}\parallel T{z}_n-v_n{\parallel }^2\hfill & =\parallel {\zeta }_n(T{z}_n-x_n)+(1-{\zeta }_n)(T{z}_n-T_n{w}_n){\parallel }^2\hfill \\ & ={\zeta }_n\parallel T{z}_n-x_n{\parallel }^2+(1-{\zeta }_n)\parallel T{z}_n-T_n{w}_n{\parallel }^2-{\zeta }_n(1-{\zeta }_n){\parallel T_n{w}_n-x_n\parallel }^2.\hfill \end{array}$$

So it follows from (26) and $\left\{{\zeta }_n\right\}\subset [c,d]\subset (0,1)$ that

$$\underset{n\to \infty }{lim}\parallel T{z}_n-x_n\parallel =\underset{n\to \infty }{lim}\parallel T{z}_n-T_n{w}_n\parallel =0.$$

(27)

Therefore, from (25)–(27), we conclude that

$$\parallel w_n-z_n\parallel \le \parallel w_n-y_n\parallel +\parallel y_n-z_n\parallel \to 0(n\to \infty ),$$

(28)

$$\begin{array}{cc}\parallel z_n-T_n{z}_n\parallel \hfill & \le \parallel z_n-w_n\parallel +\parallel w_n-x_n\parallel +\parallel x_n-T_n{w}_n\parallel +\parallel T_n{w}_n-T_n{z}_n\parallel \hfill \\ & \le 2\parallel z_n-w_n\parallel +\parallel w_n-x_n\parallel +\parallel x_n-T_n{w}_n\parallel \to 0(n\to \infty ),\hfill \end{array}$$

(29)

and

$$\begin{array}{cc}& \parallel x_{n+1}-x_n\parallel =\parallel {\beta }_{n}f\left(x_n\right)+{\gamma }_{n}T{z}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n-x_n\parallel \hfill \\ & =\parallel {\beta }_n(f\left(x_n\right)-\rho F{v}_n)+{\gamma }_n(T{z}_n-x_n)+(1-{\gamma }_n)(v_n-x_n)\parallel \hfill \\ & \le {\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel +{\gamma }_n\parallel T{z}_n-x_n\parallel +(1-{\gamma }_n)\parallel v_n-x_n\parallel \hfill \\ & \le {\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel \rho F{v}_n\parallel )+{\gamma }_n\parallel T{z}_n-x_n\parallel +(1-{\gamma }_n)(\parallel v_n-T{z}_n\parallel +\parallel T{z}_n-x_n\parallel )\hfill \\ & \le {\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel \rho F{v}_n\parallel )+\parallel T{z}_n-x_n\parallel +\parallel v_n-T{z}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(30)

Next, by the boundedness of $\left\{x_n\right\}$, we know that $\exists \left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ s.t.

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_n-x^{*}\rangle =\underset{k\to \infty }{lim}\langle (f-\rho F)x^{*},x_{n_k}-x^{*}\rangle .$$

(31)

Further we might assume that $x_{n_k}\rightharpoonup \widehat{x}$. So, from (31) we have

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_n-x^{*}\rangle =\langle (f-\rho F)x^{*},\widehat{x}-x^{*}\rangle .$$

(32)

Noticing $w_n-x_n\to 0$ and $x_{n_k}\rightharpoonup \widehat{x}$, we obtain $w_{n_k}\rightharpoonup \widehat{x}$. Since $x_n-x_{n+1}\to 0,w_n-y_n\to 0,w_n-z_n\to 0,z_n-T_n{z}_n\to 0$ (due to (25) and (28)–(30)) and $w_{n_k}\rightharpoonup \widehat{x}$, by Lemma 10 we get $\widehat{x}\in \mathsf{\Omega }$. So it follows from (17) and (32) that

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_n-x^{*}\rangle =\langle (f-\rho F)x^{*},\widehat{x}-x^{*}\rangle \le 0,$$

(33)

which hence yields

$$\begin{array}{cc}& {\displaystyle \underset{n\to \infty }{lim\; sup}}\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}[\parallel (f-\rho F)x^{*}\parallel \parallel x_{n+1}-x_n\parallel +\langle (f-\rho F)x^{*},x_n-x^{*}\rangle ]\le 0.\hfill \end{array}$$

(34)

Since $\left\{{\beta }_n(\tau -\delta )\right\}\subset [0,1],{\sum }_{n=1}^{\infty }{\beta }_n(\tau -\delta )=\infty$, and

$$\underset{n\to \infty }{lim\; sup}[\frac{2\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle }{\tau -\delta }+\frac{3M}{\tau -\delta }·\frac{{\alpha }_n}{{\beta }_n}·\parallel x_n-x_{n-1}\parallel ]\le 0,$$

by Lemma 4 we conclude from (23) that ${lim}_{n\to 0}\parallel x_n-x^{*}\parallel =0$.

Case 2.Suppose that $\exists \left\{{\Gamma }_{n_k}\right\}\subset \left\{{\Gamma }_n\right\}$ s.t. ${\Gamma }_{n_k}<{\Gamma }_{n_k+1}\forall k\in \mathbf{N}$, where $\mathbf{N}$ is the set of all positive integers. Define the mapping $\tau :\mathbf{N}\to \mathbf{N}$ by

$$\tau \left(n\right):=max\{k\le n:{\Gamma }_k<{\Gamma }_{k+1}\}.$$

Using Lemma 7, we have

$${\Gamma }_{\tau \left(n\right)}\le {\Gamma }_{\tau \left(n\right)+1}\mathrm{and}{\Gamma }_n\le {\Gamma }_{\tau \left(n\right)+1}.$$

Putting ${\Gamma }_n={\parallel x_n-x^{*}\parallel }^2\forall n\in \mathbf{N}$ and using the same inference as in Case 1, we can obtain

$$\underset{n\to \infty }{lim}\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}\parallel =0$$

(35)

and

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_{\tau \left(n\right)+1}-x^{*}\rangle \le 0.$$

(36)

Because of ${\Gamma }_{\tau \left(n\right)}\le {\Gamma }_{\tau \left(n\right)+1}$ and ${\beta }_{\tau \left(n\right)}>0$, we conclude from (23) that

$$\begin{array}{cc}\parallel x_{\tau \left(n\right)}-x^{*}{\parallel }^2\hfill & \le \frac{2}{\tau -\delta }\langle (f-\rho F)x^{*},x_{\tau \left(n\right)+1}-x^{*}\rangle +\frac{3M}{\tau -\delta }·\frac{{\alpha }_{\tau \left(n\right)}}{{\beta }_{\tau \left(n\right)}}·\parallel x_{\tau \left(n\right)}-x_{\tau \left(n\right)-1}\parallel ,\hfill \end{array}$$

and hence

$$\underset{n\to \infty }{lim\; sup}{\parallel x_{\tau \left(n\right)}-x^{*}\parallel }^2\le 0.$$

Thus, we have

$$\underset{n\to \infty }{lim}{\parallel x_{\tau \left(n\right)}-x^{*}\parallel }^2=0.$$

Using (35), we obtain

$$\begin{array}{cc}& \parallel x_{\tau \left(n\right)+1}-x^{*}{\parallel }^2-{\parallel x_{\tau \left(n\right)}-x^{*}\parallel }^2\hfill \\ & =2\langle x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)},x_{\tau \left(n\right)}-x^{*}\rangle +{\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}\parallel }^2\hfill \\ & \le 2\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel +\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}{\parallel }^2\to 0(n\to \infty ).\hfill \end{array}$$

Taking into account ${\Gamma }_n\le {\Gamma }_{\tau \left(n\right)+1}$, we have

$$\begin{array}{cc}& \parallel x_n-x^{*}{\parallel }^2\le {\parallel x_{\tau \left(n\right)+1}-x^{*}\parallel }^2\hfill \\ & \le \parallel x_{\tau \left(n\right)}-x^{*}{\parallel }^2+2\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}\parallel \parallel x_{\tau \left(n\right)}-x^{*}\parallel +\parallel x_{\tau \left(n\right)+1}-x_{\tau \left(n\right)}{\parallel }^2.\hfill \end{array}$$

It is easy to see from (35) that $x_n\to x^{*}$ as $n\to \infty$. This completes the proof.

Next, we introduce another Mann-type inertial subgradient extragradient algorithm.

Algorithm 2. Initialization: Let ${\lambda }_1>0,\alpha >0,\mu \in (0,1)$ and $x_0,x_1\in H$ be arbitrary.

Iterative Steps: Calculate $x_{n+1}$ as follows:

Step 1. Given the iterates $x_{n-1}$ and $x_n(n\ge 1)$, choose ${\alpha }_n$ such that $0\le {\alpha }_n\le {\overline{\alpha }}_n$, where

$${\overline{\alpha }}_n=\left\{\begin{array}{cc}& min\{\alpha ,\frac{{\tau }_n}{\parallel x_n-x_{n-1}\parallel }\}\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ & \alpha \mathrm{otherwise}.\hfill \end{array}\right.$$

(37)

Step 2. Compute $w_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and $y_n=P_C(w_n-{\lambda }_{n}A{w}_n)$.

Step 3. Construct the half-space $C_n:=\{z\in H:\langle w_n-{\lambda }_{n}A{w}_n-y_n,z-y_n\rangle \le 0\}$, and compute $z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n)$.

Step 4. Calculate $v_n={\zeta }_n{x}_n+(1-{\zeta }_n)T{z}_n$ and $x_{n+1}={\beta }_{n}f\left(x_n\right)+{\gamma }_n{T}_n{w}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n$, and update

$${\lambda }_{n+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_n-y_n{\parallel }^2+{\parallel z_n-y_n\parallel }^2}{2\langle A{w}_n-A{y}_n,z_n-y_n\rangle },{\lambda }_n\}\mathrm{if}\langle A{w}_n-A{y}_n,z_n-y_n\rangle >0,\hfill \\ & {\lambda }_n\mathrm{otherwise}.\hfill \end{array}\right.$$

(38)

Let $n:=n+1$ and return to Step 1.

It is worth pointing out that Lemmas 8–11 are still valid for Algorithm 2.

Theorem 2.

Let the sequence $\left\{x_n\right\}$ be constructed by Algorithm 2. Then $\left\{x_n\right\}$ converges strongly to the unique solution $x^{*}\in \mathsf{\Omega }$ of the following VIP:

$$\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \mathsf{\Omega }.$$

Proof. 

Utilizing the same arguments as in the proof of Theorem 1, we deduce that there exists a unique solution $x^{*}\in \mathsf{\Omega }={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ to the VIP (17). □

We now claim that

$$(1-{\beta }_n\tau -{\gamma }_n)(1-{\zeta }_n)(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]\le \parallel x_n-x^{*}{\parallel }^2-{\parallel x_{n+1}-x^{*}\parallel }^2+{\beta }_n{M}_4,$$

(39)

for some $M_4>0$. In fact, observe that

$$\begin{array}{cc}{x}_{n+1}-x^{*}\hfill & ={\beta }_n(f\left(x_n\right)-f\left(x^{*}\right))+{\gamma }_n(T_n{w}_n-x^{*})\hfill \\ & +(1-{\gamma }_n)[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)v_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)x^{*}]+{\beta }_n(f-\rho F)x^{*},\hfill \end{array}$$

where $v_n:={\zeta }_n{x}_n+(1-{\zeta }_n)T{z}_n$. Using the similar arguments to those of (19) and (20), we have

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n\parallel w_n-x^{*}{\parallel }^2+(1-{\beta }_n\tau -{\gamma }_n)\{{\zeta }_n{\parallel x_n-x^{*}\parallel }^2\hfill \\ & +(1-{\zeta }_n)[\parallel w_n-x^{*}{\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel w_n-y_n{\parallel }^2-(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel z_n-y_n{\parallel }^2\left]\right\}+{\beta }_n{M}_2.\hfill \end{array}$$

and

$$\parallel w_n-x^{*}{\parallel }^2\le (\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2\le {\parallel x_n-x^{*}\parallel }^2+{\beta }_n{M}_3,$$

where $M_2\ge {sup}_{n\ge 1}2\parallel (f-\rho F)x^{*}\parallel \parallel x_n-x^{*}\parallel$ for some $M_2>0$ and $M_3\ge {sup}_{n\ge 1}(2M_1\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1^2)$ for some $M_3>0$. Combining the last inequalities, we obtain

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n(\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{M}_3)+(1-{\beta }_n\tau -{\gamma }_n)(\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{M}_3)\hfill \\ & -(1-{\beta }_n\tau -{\gamma }_n)(1-{\zeta }_n)[(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel w_n-y_n{\parallel }^2+(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})\parallel z_n-y_n{\parallel }^2]+{\beta }_n{M}_2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2-(1-{\beta }_n\tau -{\gamma }_n)(1-{\zeta }_n)(1-\mu \frac{{\lambda }_n}{{\lambda }_{n+1}})[\parallel w_n-y_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2]+{\beta }_n{M}_4,\hfill \end{array}$$

where $M_4:=M_2+M_3$. This ensures that (39) holds.

Next we claim that

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le [1-{\beta }_n(\tau -\delta )]{\parallel x_n-x^{*}\parallel }^2\hfill \\ & +{\beta }_n(\tau -\delta )[\frac{2}{\tau -\delta }\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle +\frac{3M}{\tau -\delta }·\frac{{\alpha }_n}{{\beta }_n}·\parallel x_n-x_{n-1}\parallel ]\hfill \end{array}$$

(40)

for some $M>0$. In fact, using the similar arguments to those of (22) and (23), we have

$$\parallel w_n-x^{*}{\parallel }^2\le \parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel [2\parallel x_n-x^{*}\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel ],$$

and

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n\parallel w_n-x^{*}{\parallel }^2+(1-{\beta }_n\tau -{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel z_n-x^{*}{\parallel }^2]\hfill \\ & +2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+(1-{\beta }_n\tau )[\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-x^{*}\parallel \hfill \\ & +{\alpha }_n\parallel x_n-x_{n-1}\parallel \left)\right]+2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le [1-{\beta }_n(\tau -\delta )]\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-x^{*}\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel )\hfill \\ & +2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ & \le [1-{\beta }_n(\tau -\delta )]\parallel x_n-x^{*}{\parallel }^2+{\beta }_n(\tau -\delta )·[\frac{2\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle }{\tau -\delta }+\frac{3M}{\tau -\delta }·\frac{{\alpha }_n}{{\beta }_n}·\parallel x_n-x_{n-1}\parallel ],\hfill \end{array}$$

(41)

where $M\ge {sup}_{n\ge 1}\{\parallel x_n-x^{*}\parallel ,{\alpha }_n\parallel x_n-x_{n-1}\parallel \}$ for some $M>0$.

For each $n\ge 0$, we set

$$\begin{array}{cc}& {\Gamma }_n={\parallel x_n-x^{*}\parallel }^2,\hfill \\ & {\epsilon }_n={\beta }_n(\tau -\delta ),\hfill \\ & {\vartheta }_n={\alpha }_n\parallel x_n-x_{n-1}\parallel 3M+2{\beta }_n\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle .\hfill \end{array}$$

Then (41) can be rewritten as the following formula:

$${\Gamma }_{n+1}\le (1-{\epsilon }_n){\Gamma }_n+{\vartheta }_n\forall n\ge 0.$$

(42)

We next show the convergence of $\left\{{\Gamma }_n\right\}$ to zero by the following two cases:

Case 3.Suppose that there exists an integer $n_0\ge 1$ such that $\left\{{\Gamma }_n\right\}$ is non-increasing. Then

$${\Gamma }_n-{\Gamma }_{n+1}\to 0.$$

Using the similar arguments to those of (25), we have

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

(43)

Using Lemma 1 (v), we get

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ & =\parallel {\beta }_n(f\left(x_n\right)-\rho F{v}_n)+{\gamma }_n(T_n{w}_n-x^{*})+(1-{\gamma }_n)(v_n-x^{*}){\parallel }^2\hfill \\ & \le \parallel {\gamma }_n(T_n{w}_n-x^{*})+(1-{\gamma }_n)(v_n-x^{*}){\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & ={\gamma }_n\parallel T_n{w}_n-x^{*}{\parallel }^2+(1-{\gamma }_n)\parallel v_n-x^{*}{\parallel }^2-{\gamma }_n(1-{\gamma }_n){\parallel T_n{w}_n-v_n\parallel }^2\hfill \\ & +2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & ={\gamma }_n\parallel T_n{w}_n-x^{*}{\parallel }^2+(1-{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel T{z}_n-x^{*}{\parallel }^2-{\zeta }_n(1-{\zeta }_n)\parallel x_n-T{z}_n{\parallel }^2]\hfill \\ & -{\gamma }_n(1-{\gamma }_n){\parallel T_n{w}_n-v_n\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & \le {\gamma }_n\parallel w_n-x^{*}{\parallel }^2+(1-{\gamma }_n)[{\zeta }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\zeta }_n)\parallel z_n-x^{*}{\parallel }^2-{\zeta }_n(1-{\zeta }_n)\parallel x_n-T{z}_n{\parallel }^2]\hfill \\ & -{\gamma }_n(1-{\gamma }_n){\parallel T_n{w}_n-v_n\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{v}_n,x_{n+1}-x^{*}\rangle \hfill \\ & \le {\gamma }_n(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2+(1-{\gamma }_n)(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-(1-{\gamma }_n){\zeta }_n(1-{\zeta }_n){\parallel x_n-T{z}_n\parallel }^2\hfill \\ & -{\gamma }_n(1-{\gamma }_n)\parallel T_n{w}_n-v_n{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & =(\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-(1-{\gamma }_n){\zeta }_n(1-{\zeta }_n){\parallel x_n-T{z}_n\parallel }^2\hfill \\ & -{\gamma }_n(1-{\gamma }_n)\parallel T_n{w}_n-v_n{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel ,\hfill \end{array}$$

which immediately yields

$$\begin{array}{cc}& (1-{\gamma }_n){\zeta }_n(1-{\zeta }_n)\parallel x_n-T{z}_n{\parallel }^2+{\gamma }_n(1-{\gamma }_n){\parallel T_n{w}_n-v_n\parallel }^2\hfill \\ & \le (\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1{)}^2-\parallel x_{n+1}-x^{*}{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel \hfill \\ & ={\Gamma }_n-{\Gamma }_{n+1}+{\beta }_n{M}_1(2\parallel x_n-x^{*}\parallel +{\beta }_n{M}_1)+2{\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel \parallel x_{n+1}-x^{*}\parallel .\hfill \end{array}$$

Since ${\beta }_n\to 0,{\Gamma }_n-{\Gamma }_{n+1}\to 0,\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$ and $\left\{{\zeta }_n\right\}\subset [c,d]\subset (0,1)$, we have

$$\underset{n\to \infty }{lim}\parallel x_n-T{z}_n\parallel =\underset{n\to \infty }{lim}\parallel T_n{w}_n-v_n\parallel =0.$$

(44)

Note that

$$\begin{array}{cc}\parallel T_n{w}_n-v_n{\parallel }^2\hfill & =\parallel {\zeta }_n(T_n{w}_n-x_n)+(1-{\zeta }_n)(T_n{w}_n-T{z}_n){\parallel }^2\hfill \\ & ={\zeta }_n\parallel T_n{w}_n-x_n{\parallel }^2+(1-{\zeta }_n)\parallel T_n{w}_n-T{z}_n{\parallel }^2-{\zeta }_n(1-{\zeta }_n){\parallel T{z}_n-x_n\parallel }^2.\hfill \end{array}$$

Hence, from (44) we have

$$\underset{n\to \infty }{lim}\parallel T_n{w}_n-x_n\parallel =\underset{n\to \infty }{lim}\parallel T_n{w}_n-T{z}_n\parallel =0.$$

(45)

So, from (43)–(45) we infer that

$$\parallel w_n-z_n\parallel \le \parallel w_n-y_n\parallel +\parallel y_n-z_n\parallel \to 0(n\to \infty ),$$

(46)

$$\begin{array}{cc}\parallel z_n-T_n{z}_n\parallel \hfill & \le \parallel z_n-w_n\parallel +\parallel w_n-x_n\parallel +\parallel x_n-T_n{w}_n\parallel +\parallel T_n{w}_n-T_n{z}_n\parallel \hfill \\ & \le 2\parallel z_n-w_n\parallel +\parallel w_n-x_n\parallel +\parallel x_n-T_n{w}_n\parallel \to 0(n\to \infty ),\hfill \end{array}$$

(47)

and

$$\begin{array}{cc}& \parallel x_{n+1}-x_n\parallel =\parallel {\beta }_{n}f\left(x_n\right)+{\gamma }_n{T}_n{w}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)v_n-x_n\parallel \hfill \\ & =\parallel {\beta }_n(f\left(x_n\right)-\rho F{v}_n)+{\gamma }_n(T_n{w}_n-x_n)+(1-{\gamma }_n)(v_n-x_n)\parallel \hfill \\ & \le {\beta }_n\parallel f\left(x_n\right)-\rho F{v}_n\parallel +{\gamma }_n\parallel T_n{w}_n-x_n\parallel +(1-{\gamma }_n)\parallel v_n-x_n\parallel \hfill \\ & \le {\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel \rho F{v}_n\parallel )+{\gamma }_n\parallel T_n{w}_n-x_n\parallel +(1-{\gamma }_n)(\parallel v_n-T_n{w}_n\parallel +\parallel T_n{w}_n-x_n\parallel )\hfill \\ & \le {\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel \rho F{v}_n\parallel )+\parallel T_n{w}_n-x_n\parallel +\parallel v_n-T_n{w}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(48)

In addition, using the similar arguments to those of (33) and (34), we have

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_n-x^{*}\rangle \le 0,$$

and hence

$$\underset{n\to \infty }{lim\; sup}\langle (f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \le 0.$$

Consequently, applying Lemma 4 to (41), we have ${lim}_{n\to 0}\parallel x_n-x^{*}\parallel =0$.

Case 4.Suppose that $\exists \left\{{\Gamma }_{n_k}\right\}\subset \left\{{\Gamma }_n\right\}$ s.t. ${\Gamma }_{n_k}<{\Gamma }_{n_k+1}\forall k\in \mathbf{N}$, where $\mathbf{N}$ is the set of all positive integers. Define the mapping $\tau :\mathbf{N}\to \mathbf{N}$ by $\tau \left(n\right):=max\{k\le n:{\Gamma }_k<{\Gamma }_{k+1}\}.$ In the remainder of the proof, using the same arguments as in Case 2 of the proof of Theorem 1, we obtain the desired assertion. This completes the proof.

It is markable that our results improve and extend the corresponding results of Kraikaew and Saejung [20] and Ceng et al. [11], in the following aspects.

(i) Our problem of finding an element of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ includes as a special case the problem of finding an element of $\mathrm{VI}(C,A)$ in [20], where $T_1,...,T_N$ are nonexpansive and $T_0=T$ is quasi-nonexpansive. It is worth mentioning that Halpern’s subgradient extragradient method for solving the VIP in [20] is extended to develop our Mann-type inertial subgradient extragradient rule for solving the VIP and CFPP, in which A is L-Lipschitz continuous, pseudomonotone on H, but it is not required to be sequentially weakly continuous on C.

(ii) Our problem of finding an element of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ includes as a special case the problem of finding an element of ${\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ in [11], where in [11], A is required to be L-Lipschitz continuous, pseudomonotone on H, and sequentially weakly continuous on C. The modified inertial subgradient extragradient method for solving the VIP and CFPP in [11] is extended to develop our Mann-type inertial subgradient extragradient rule for solving the VIP and CFPP, where $T_i$ is nonexpansive for $i=1,...,N$ and $T_0=T$ is quasi-nonexpansive.

4. Applicability and Implementability of Algorithms

In this section, in order to support the applicability and implementability of our Algorithms 1 and 2, we make use of our main results to find a common solution of the VIP and CFPP in two illustrating examples.

Example 2.

Let $C=[-1,1]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. Let $x_0,x_1\in H$ be arbitrary. Put $f\left(x\right)=F\left(x\right)=\frac{1}{2}x,{\beta }_n=\frac{1}{n+1},{\tau }_n={\beta }_n^2,\mu =0.2,\alpha ={\lambda }_1=0.1,{\gamma }_n={\zeta }_n=\frac{1}{3},\rho =2$, and

$${\alpha }_n=\left\{\begin{array}{cc}& min\{\frac{{\beta }_n^2}{\parallel x_n-x_{n-1}\parallel },\alpha \}\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ & \alpha \mathrm{otherwise}.\hfill \end{array}\right.$$

Then we know that $\kappa =\eta =\frac{1}{2}$ and $\tau =1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}=1\in (0,1]$. For $N=1$, we now present Lipschitz continuous and pseudomonotone mapping A, quasi-nonexpansive mapping T and nonexpansive mapping $T_1$ such that $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Indeed, let $A,T,T_1:H\to H$ be defined as $Ax:=\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|},T_{1}x:=sinx$ and $Tx:=\frac{x}{2}sinx$ for all $x\in H$. We first show that A is pseudomonotone and L-Lipschitz continuous with $L=2$. Indeed, it is easy to see that for all $x,y\in H$,

$$\begin{array}{cc}\parallel Ax-Ay\parallel \hfill & =|\frac{1}{1+\parallel sinx\parallel }-\frac{1}{1+\parallel x\parallel }-\frac{1}{1+\parallel siny\parallel }+\frac{1}{1+\parallel y\parallel }|\hfill \\ & \le |\frac{\parallel y\parallel -\parallel x\parallel }{(1+\parallel x\parallel )(1+\parallel y\parallel )}|+|\frac{\parallel siny\parallel -\parallel sinx\parallel }{(1+\parallel sinx\parallel )(1+\parallel siny\parallel )}|\hfill \\ & \le \frac{\parallel x-y\parallel }{(1+\parallel x\parallel )(1+\parallel y\parallel )}+\frac{\parallel sinx-siny\parallel }{(1+\parallel sinx\parallel )(1+\parallel siny\parallel )}\hfill \\ & \le 2\parallel x-y\parallel ,\hfill \end{array}$$

and

$$\langle Ax,y-x\rangle =(\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|})(y-x)\ge 0\Rightarrow \langle Ay,y-x\rangle =(\frac{1}{1+|siny|}-\frac{1}{1+\left|y\right|})(y-x)\ge 0.$$

Furthermore, it is clear that $\mathrm{Fix}\left(T\right)=\left\{0\right\}$, T is quasi-nonexpansive but not nonexpansive. Meantime, $I-T$ is demiclosed at 0 due to the continuity of T. In addition, it is clear that $T_1$ is nonexpansive and $\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$. Therefore, $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 1 can be rewritten as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & y_n=P_C(w_n-{\lambda }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n),\hfill \\ & v_n=\frac{1}{3}{x}_n+\frac{2}{3}{T}_1{w}_n,\hfill \\ & x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{3}T{z}_n+(\frac{n}{n+1}-\frac{1}{3})v_n\forall n\ge 1,\hfill \end{array}\right.$$

(49)

where for each $n\ge 1$, $C_n$ and ${\lambda }_n$ are chosen as in Algorithm 1. So, using Theorem 1, we know that $\left\{x_n\right\}$ converges to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$. Meanwhile, Algorithm 2 can be rewritten as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & y_n=P_C(w_n-{\lambda }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n),\hfill \\ & v_n=\frac{1}{3}{x}_n+\frac{2}{3}T{z}_n,\hfill \\ & x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{3}{T}_1{w}_n+(\frac{n}{n+1}-\frac{1}{3})v_n\forall n\ge 1,\hfill \end{array}\right.$$

(50)

where, for each $n\ge 1$, $C_n$ and ${\lambda }_n$ are chosen as in Algorithm 2. So, using Theorem 2, we know that $\left\{x_n\right\}$ converges to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$.

Example 3.

Let $H=L^2\left([0,1]\right)$ with the inner product and induced norm defined by

$$\langle x,y\rangle ={\int }_0^{1}x\left(t\right)y\left(t\right)dt\mathrm{and}\parallel x\parallel =({\int }_0^1{\left|x\left(t\right)\right|}^2{dt)}^{1/2}\forall x,y\in H,$$

respectively. Then $(H,\langle ·,·\rangle )$ is a Hilbert space. Let $C:=\{x\in H:\parallel x\parallel \le 1\}$ be the unit closed ball of H. It is known that

$$P_C\left(x\right)=\left\{\begin{array}{cc}& \frac{x}{\parallel x\parallel }\mathrm{if}\parallel x\parallel >1,\hfill \\ & x\mathrm{if}\parallel x\parallel \le 1.\hfill \end{array}\right.$$

Let $x_0,x_1\in H$ be arbitrary. Put $f\left(x\right)=F\left(x\right)=\frac{1}{2}x,{\beta }_n=\frac{1}{n+1},{\tau }_n={\beta }_n^2,\mu =0.2,\alpha ={\lambda }_1=0.1,{\gamma }_n={\zeta }_n=\frac{1}{3},\rho =2$, and

$${\alpha }_n=\left\{\begin{array}{cc}& min\{\frac{{\beta }_n^2}{\parallel x_n-x_{n-1}\parallel },\alpha \}\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ & \alpha \mathrm{otherwise}.\hfill \end{array}\right.$$

Then we know that $\kappa =\eta =\frac{1}{2}$ and $\tau =1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}=1\in (0,1]$. For $N=1$, we now present Lipschitz continuous and pseudomonotone mapping A, quasi-nonexpansive mapping T and nonexpansive mapping $T_1$ such that $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Indeed, let $A,T,T_1:H\to H$ be defined as $\left(Ax\right)\left(t\right):=max\{0,x\left(t\right)\},\left(T_{1}x\right)\left(t\right):=\frac{1}{2}x\left(t\right)-\frac{1}{2}sinx\left(t\right)$ and $\left(Tx\right)\left(t\right):=\frac{1}{2}x\left(t\right)+\frac{1}{2}sinx\left(t\right)$ for all $x\in H$. It can be easily verified (see, e.g., [8,9]) that A is monotone and L-Lipschitz continuous with $L=1$, and the solution set of the VIP for A is given by

$$\mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing .$$

We next show that T and $T_1$ are nonexpansive and $\mathrm{Fix}\left(T\right)=\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$. Indeed, it is easy to see that for all $x,y\in H$,

$$\begin{array}{cc}\parallel Tx-Ty\parallel \hfill & =({\int }_0^1|\frac{1}{2}(x\left(t\right)-y\left(t\right))+\frac{1}{2}(sinx\left(t\right)-siny\left(t\right)){{|}^{2}dt)}^{1/2}\hfill \\ & \le ({\int }_0^1(\frac{1}{2}|x\left(t\right)-y\left(t\right)|+\frac{1}{2}{|x\left(t\right)-y\left(t\right)|)}^2{dt)}^{1/2}\hfill \\ & =({\int }_0^1{|x\left(t\right)-y\left(t\right)|}^2{dt)}^{1/2}\hfill \\ & =\parallel x-y\parallel .\hfill \end{array}$$

Similarly, we get $\parallel T_{1}x-T_{1}y\parallel \le \parallel x-y\parallel \forall x,y\in H$. Moreover, it is clear that $\mathrm{Fix}\left(T\right)=\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$. Therefore, $\mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 1 can be rewritten as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & y_n=P_C(w_n-{\lambda }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n),\hfill \\ & v_n=\frac{1}{3}{x}_n+\frac{2}{3}{T}_1{w}_n,\hfill \\ & x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{3}T{z}_n+(\frac{n}{n+1}-\frac{1}{3})v_n\forall n\ge 1,\hfill \end{array}\right.$$

(51)

where for each $n\ge 1$, $C_n$ and ${\lambda }_n$ are chosen as in Algorithm 1. So, using Theorem 1, we know that $\left\{x_n\right\}$ converges strongly to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$. Meantime, Algorithm 2 can be rewritten as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & y_n=P_C(w_n-{\lambda }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\lambda }_{n}A{y}_n),\hfill \\ & v_n=\frac{1}{3}{x}_n+\frac{2}{3}T{z}_n,\hfill \\ & x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{3}{T}_1{w}_n+(\frac{n}{n+1}-\frac{1}{3})v_n\forall n\ge 1,\hfill \end{array}\right.$$

(52)

where for each $n\ge 1$, $C_n$ and ${\lambda }_n$ are chosen as in Algorithm 2. So, using Theorem 2, we know that $\left\{x_n\right\}$ converges strongly to $0\in \mathsf{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$.