Mildly Inertial Subgradient Extragradient Method for Variational Inequalities Involving an Asymptotically Nonexpansive and Finitely Many Nonexpansive Mappings

Abstract

In a real Hilbert space, let the notation VIP indicate a variational inequality problem for a Lipschitzian, pseudomonotone operator, and let CFPP denote a common fixed-point problem of an asymptotically nonexpansive mapping and finitely many nonexpansive mappings. This paper introduces mildly inertial algorithms with linesearch process for finding a common solution of the VIP and the CFPP by using a subgradient approach. These fully absorb hybrid steepest-descent ideas, viscosity iteration ideas, and composite Mann-type iterative ideas. With suitable conditions on real parameters, it is shown that the sequences generated our algorithms converge to a common solution in norm, which is a unique solution of a hierarchical variational inequality (HVI).

1. Introduction

Let C be a convex and closed nonempty set in a real Hilbert space $(H,\parallel ·\parallel )$ with inner product $\langle ·,·\rangle$. Let $\mathrm{Fix}\left(S\right)$ indicate the fixed-point set of a non-self operator $S:C\to H$, i.e., $\mathrm{Fix}\left(S\right)=\{u\in C:u=Su\}$. One says that a self operator $T:C\to C$ is asymptotically nonexpansive if and only if $\parallel T^{n}u-T^{n}v\parallel \le (1+{\theta }_n)\parallel u-v\parallel \forall n\ge 1,u,v\in C$, where ${lim}_{n\to \infty }{\theta }_n=0$ is a real sequence. In the case of ${\theta }_n=0\forall n\ge 1$, one says that T is nonexpansive. Both the class of nonexpansive operators and asymptotically nonexpansive operators via various iterative techniques have been studied recently; see, e.g., the works by the authors of [1,2,3,4,5,6,7,8,9,10,11,12,13]. Let $A:H\to H$ be a self operator. Consider the classical variational inequality problem (VIP) of consisting of $u^{*}\in C$ such that

$$\langle A{u}^{*},v-u^{*}\rangle \ge 0\forall v\in C.$$

(1)

The set of solutions of problem (1) is indicated by VI($C,A$). Recently, many authors studied the VIP via mean-valued and projection-based methods; see, e.g., the works by the authors of [14,15,16,17,18,19,20,21]. In 1976, Korpelevich [22] first designed and investigated an extragradient method for a solution of problem (1), that is, for arbitrarily given $u_0\in C$, $\left\{u_n\right\}$ is the sequence constructed by

$$\left\{\begin{array}{c}{v}_n=P_C(u_n-\tau A{u}_n),\hfill \\ u_{n+1}=P_C(u_n-\tau A{v}_n)\forall n\ge 0,\hfill \end{array}\right.$$

(2)

with $\tau \in (0,\frac{1}{L})$. If problem (1) has a solution, then he showed the weak convergence of $\left\{u_n\right\}$ constructed by (2) to a solution of problem (1). Since then, Korpelevich’s extragradient method and its variants have been paid great attention to by many scholars, who improved it in various techniques and approaches; see, e.g., the works by the authors of [23,24,25,26,27,28,29,30,31,32,33,34].

Let ${\left\{T_i\right\}}_{i=1}^N$ be N nonexpansive mappings on H, such that $\Omega ={\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)\ne \varnothing$. Let F be a $\kappa$-Lipschitzian, $\eta$-strongly monotone self-mapping on H, and f be a contractive map with constant $\delta \in (0,1)$. In 2015, Bnouhachem et al. [2] introduced an iterative algorithm for solving a hierarchical fixed point problem (HFPP) for a finite pool ${\left\{T_i\right\}}_{i=1}^N$, i.e., for arbitrarily given $x_0\in H$, the sequence $\left\{x_n\right\}$ is constructed by

$$\left\{\begin{array}{c}{y}_n=(1-{\beta }_n)T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n+{\beta }_n{x}_n,\hfill \\ x_{n+1}={\gamma }_n{x}_n+((1-{\gamma }_n)I-{\alpha }_n\mu F)y_n+{\alpha }_n\rho f\left(y_n\right),\forall n\ge 0,\hfill \end{array}\right.$$

(3)

where $T_{i,n}=(1-{\delta }_{i,n})I+{\delta }_{i,n}{T}_i$ and ${\delta }_{i,n}\in (0,1)$ for integer $i\in \{1,2,\dots ,N\}$. Let the parameters satisfy $0<\mu {\kappa }^2<2\eta$ and $0\le \rho \tau <\nu$, with $\nu =\mu (\eta -\frac{\mu {\kappa }^2}{2})$. Also, suppose that the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ satisfy the following requirements.

(i)

${\sum }_{n=0}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

(ii)

$\left\{{\beta }_n\right\}\subset [\sigma ,1)$ and ${lim}_{n\to \infty }{\beta }_n=\beta <1$;

(iii)

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

(iv)

${lim}_{n\to \infty }|{\delta }_{i,n-1}-{\delta }_{i,n}|=0$ for $i=1,2,\dots ,N$.

They proved the strong convergence of $\left\{x_n\right\}$ to a point $x^{*}\in \Omega$, which is only a solution to the HFPP: $\langle (\mu F-\rho f)x^{*},y-x^{*}\rangle \ge 0\forall y\in \Omega$.

On the other hand, let the mappings $A_1,A_2:C\to H$ be both inverse-strongly monotone and the mapping $T:C\to C$ be asymptotically nonexpansive one with $\left\{{\theta }_n\right\}$. In 2018, by the modified extragradient method, Cai et al. [35] designed a viscosity implicit method for computing a point in the common solution set $\Omega$ of the VIPs for $A_1$ and $A_2$ and the FPP of T, i.e., for arbitrarily given $x_1\in C$, the sequence $\left\{x_n\right\}$ is constructed by

$$\left\{\begin{array}{c}{v}_n=t_n{x}_n+(1-t_n)u_n,\hfill \\ z_n=P_C(v_n-\mu A_2{v}_n),\hfill \\ u_n=P_C(z_n-\lambda A_1{z}_n),\hfill \\ x_{n+1}=P_C[(I-{\alpha }_n\rho F)T^n{u}_n+{\alpha }_{n}f\left(x_n\right)],\hfill \end{array}\right.$$

(4)

where $f:C\to C$ be a $\delta$-contraction with $0\le \delta <1$, and $\left\{{\alpha }_n\right\},\left\{t_n\right\}$ are the sequences in $(0,1]$ satisfying

(i)

${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$, ${lim}_{n\to \infty }{\alpha }_n=0$ and ${\sum }_{n=1}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$;

(ii)

${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

(iii)

$0<ϵ\le t_n\le 1$ and ${\sum }_{n=1}^{\infty }|t_{n+1}-t_n|<\infty$;

(iv)

${\sum }_{n=1}^{\infty }\parallel T^{n+1}{u}_n-T^n{u}_n\parallel <\infty$.

They proved that $\left\{x_n\right\}$ converges strongly to a point $x^{*}\in \Omega$, which is a unique solution to the VIP: $\langle (f-\rho F)x^{*},y-x^{*}\rangle \le 0\forall y\in \Omega$.

Under the setting of extragradient approaches, we must calculate metric projections twice for every iteration. Without doubt, if C is a general convex and closed subset, the computation of the projection onto C might be prohibitively consuming-time. In 2011, motivated by Korpelevich’s extragradient method, Censor et al. [5] first purposed the subgradient extragradient method, where a projection onto a half-space is used in place of the second projection onto C:

$$\left\{\begin{array}{c}{v}_n=P_C(u_n-\ell A{u}_n),\hfill \\ C_n=\{u\in H:\langle u_n-\ell A{u}_n-v_n,u-v_n\rangle \le 0\},\hfill \\ u_{n+1}=P_{C_n}(u_n-\ell A{v}_n)\forall n\ge 0,\hfill \end{array}\right.$$

(5)

with $\ell \in (0,\frac{1}{L})$. In 2014, Kraikaew and Saejung [36] introduced the Halpern subgradient extragradient method for solving VIP (1) and proved that the sequence generated by the proposed method converges strongly to a solution of VIP (1).

In 2018, by virtue of the inertial technique, Thong and Hieu [37] first introduced the inertial subgradient extragradient method and proved weak convergence of the proposed method to a solution of VIP (1). Very recently, Thong and Hieu [37] introduced two inertial subgradient extragradient algorithms with the linesearch process to solve the VIP (1) for Lipschitzian, monotone operator A, and the FPP of quasi-nonexpansive operator S satisfying the demiclosedness in H.

Under mild assumptions, they proved that the sequences defined by the above algorithms converge to a point in $\mathrm{Fix}\left(S\right)\cap \mathrm{VI}(C,A)$ with the aid of dual spaces. Being motivated by the research work [2,37,38] and using the subgradient extragradient technique, this paper designs two mildly inertial algorithms with linesearch process to solve the VIP (1) for Lipschitzian, pseudomonotone operator, and the CFPP of an asymptotically nonexpansive mapping and finitely many nonexpansive mappings in H. Our algorithms fully absorb inertial subgradient extragradient approaches with linesearch process, hybrid steepest-descent algorithms, viscosity iteration techniques, and composite Mann-type iterative methods. Under suitable conditions, it is shown that the sequences constructed by our algorithms converge to a common solution of the VIP and CFPP in norm, which is only a solution of a hierarchical variational inequality (HVI). Finally, we apply our main theorems to deal with the VIP and CFPP in an illustrating example.

The outline of the article is arranged as follows. In Section 2, some concepts and preliminary conclusions are recalled for later use. In Section 3, the convergence criteria of the suggested algorithms are established. In Section 4, our main theorems are used to deal with the VIP and CFPP in an illustrating example. As our algorithms concern solving VIP (1) with Lipschitzian, pseudomonotone operator, and the CFPP of an asymptotically nonexpansive mapping and finitely many nonexpansive mappings, they are more advantageous and more subtle than Algorithms 1 and 2 in [37]. Our theorems strengthen and generalize the corresponding results announced in Bnouhachem et al. [2], Cai et al. [35], Kraikaew and Saejung [36], and Thong and Hieu [37,38].

Algorithm 1: of Thong and Hieu [37]

1Initial Step: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$. 2 Iteration Steps: Compute $x_{n+1}$ in what follows,  Step 1. Put $u_n=x_n-{\alpha }_n(x_{n-1}-x_n)$ and calculate $y_n=P_C(u_n-{\ell }_{n}A{u}_n)$, where ${\ell }_n$ is chosen to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $\mu \parallel u_n-y_n\parallel \ge \ell \parallel A{u}_n-A{y}_n\parallel$.  Step 2. Calculate $z_n=P_{C_n}(u_n-{\ell }_{n}A{y}_n)$ with $C_n:=\{u\in H:\langle u_n-{\ell }_{n}A{u}_n-y_n,u-y_n\rangle \le 0\}$.  Step 3. Calculate $x_{n+1}=(1-{\beta }_n)u_n+{\beta }_{n}S{z}_n$. If $u_n=z_n=x_{n+1}$ then $u_n\in \mathrm{Fix}\left(S\right)\cap \mathrm{VI}(C,A)$. Put $n:=n+1$ and return to Step 1.

Algorithm 2: of Thong and Hieu [37]

1Initial Step: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$. 2 Iteration Steps: Compute $x_{n+1}$ in what follows,  Step 1. Put $u_n=x_n-{\alpha }_n(x_{n-1}-x_n)$ and calculate $y_n=P_C(u_n-{\ell }_{n}A{u}_n)$, where ${\ell }_n$ is chosen to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $\mu \parallel u_n-y_n\parallel \ge \ell \parallel A{u}_n-A{y}_n\parallel$.  Step 2. Calculate $z_n=P_{C_n}(u_n-{\ell }_{n}A{y}_n)$ with $C_n:=\{u\in H:\langle u_n-{\ell }_{n}A{u}_n-y_n,u-y_n\rangle \le 0\}$.  Step 3. Calculate $x_{n+1}=(1-{\beta }_n)x_n+{\beta }_{n}S{z}_n$. If $u_n=z_n=x_n=x_{n+1}$ then $x_n\in \mathrm{Fix}\left(S\right)\cap \mathrm{VI}(C,A)$. Put $n:=n+1$ and return to Step 1.

2. Preliminaries

Given a sequence $\left\{u_n\right\}$ in H. We use the notations $u_n\to u$ and $u_n\rightharpoonup u$ to indicate the strong convergence of $\left\{u_n\right\}$ to u and weak convergence of $\left\{u_n\right\}$ to u, respectively. An operator $T:C\to H$ is said to be

(i)

L-Lipschitz continuous (or L-Lipschitzian) iff $\exists L>0$ s.t.

$$\parallel Tu-Tv\parallel \le L\parallel u-v\parallel \forall u,v\in C;$$

(ii)

monotone iff

$$\langle Tu-Tv,u-v\rangle \ge 0\forall u,v\in C;$$

(iii)

pseudomonotone iff

$$\langle Tu,v-u\rangle \ge 0\Rightarrow \langle Tv,v-u\rangle \ge 0\forall u,v\in C;$$

(iv)

$\beta$-strongly monotone if $\exists \beta >0$ s.t.

$$\langle Tu-Tv,u-v\rangle \ge {\beta \parallel u-v\parallel }^2\forall u,v\in C;$$

(v)

sequentially weakly continuous if $\forall \left\{u_n\right\}\subset C$, the relation holds: $u_n\rightharpoonup u\Rightarrow T{u}_n\rightharpoonup Tu$.

It is clear that every monotone mapping is pseudomonotone but the converse is not valid; e.g., take $Tx:=\frac{a}{a+x},x,a\in (0,+\infty )$.

For every $u\in H$, we know that there is only a nearest point in C, indicated by $P_{C}u$, s.t. $\parallel u-P_{C}u\parallel \le \parallel u-v\parallel \forall v\in C$. The operator $P_C$ is said to be the metric projection from H to C.

Proposition 1.

The following hold in real Hilbert spaces:

(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) 

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

(iv) 

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

(v) 

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

An operator $S:H\to H$ is called an averaged one if $\exists \alpha \in (0,1)$ s.t. $S=(1-\alpha )I+\alpha T$, where I is the identity operator of H and $T:H\to H$ is a nonexpansive operator. In this case, S is also called $\alpha$-averaged. It is clear that the averaged operator S is also nonexpansive and $\mathrm{Fix}\left(S\right)=\mathrm{Fix}\left(T\right)$.

Lemma 1.

[2] If the mappings ${\left\{T_i\right\}}_{i=1}^N$ defined on H are averaged and have a common fixed point, then ${\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)=\mathrm{Fix}(T_1{T}_2\cdots T_N)$.

The next result immediately follows from the subdifferential inequality of the function ${\parallel ·\parallel }^2/2$.

Lemma 2.

The following inequality holds,

$${\parallel u+v\parallel }^2-{\parallel u\parallel }^2\le 2\langle v,u+v\rangle \forall u,v\in H.$$

Lemma 3.

[39] Assume that the mapping A is pseudomonotone and continuous on C. Given a point $u\in C$. Then the relation holds: $\langle Au,v-u\rangle \ge 0\forall v\in C\iff \langle Av,v-u\rangle \ge 0\forall v\in C$.

Lemma 4.

[40] Let $\left\{t_n\right\}$ be a sequence in $[0,+\infty )$ satisfying the condition $t_{n+1}\le s_n{b}_n+(1-s_n)t_n\forall n\ge 1$, where $\left\{s_n\right\}$ and $\left\{b_n\right\}$ lie in $\mathbf{R}:=(-\infty ,\infty )$ s.t. (a) $\left\{s_n\right\}\subset [0,1]$ and ${\sum }_{n=1}^{\infty }{s}_n=\infty$, and (b) ${lim\; sup}_{n\to \infty }{b}_n\le 0$ or ${\sum }_{n=1}^{\infty }\left|s_n{b}_n\right|<\infty$. Then $t_n\to 0$ as $n\to \infty$.

Definition 1.

An operator $S:C\to H$ is called ζ-strictly pseudocontractive iff $\exists \zeta \in [0,1)$ s.t. ${\parallel Su-Sv\parallel }^2{-\zeta \parallel (I-S)u-(I-S)v\parallel }^2\le {\parallel u-v\parallel }^2\forall u,v\in C$.

Lemma 5.

[41] Assume that $S:C\to H$ is ζ-strictly pseudocontractive. Define $T:C\to H$ by $Tu=\mu Su+(1-\mu )u\forall u\in C$. If $\mu \in [\zeta ,1),T$ is nonexpansive such that $\mathrm{Fix}\left(T\right)=\mathrm{Fix}\left(S\right)$.

Lemma 6.

[42] Let $\ell \in (0,1]$, $S:C\to H$ be nonexpansive, and $S^{\ell }:C\to H$ be defined as $S^{\ell }u:=Su-\ell \mu F\left(Su\right)\forall u\in C$, where F is κ-Lipschitzian and η-strongly monotone self-mapping on H. Then, $S^{\ell }$ is a contractive map provided $0<\mu <\frac{2\eta }{{\kappa }^2}$, i.e., $\parallel S^{\ell }u-S^{\ell }v\parallel \le (1-\ell \tau )\parallel u-v\parallel \forall u,v\in C$, where $\tau =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$.

Lemma 7.

[43] Assume that the Banach space X admits a weakly continuous duality mapping; the subset $C\subset X$ is nonempty, convex, and closed; and the asymptotically nonexpansive mapping $S:C\to C$ has a fixed point. Then, $I-S$ is demiclosed at zero, i.e., if the sequence $\left\{u_n\right\}\subset C$ satisfies $u_n\rightharpoonup u\in C$ and $u_n-S{u}_n\to 0$, then $u\in \mathrm{Fix}\left(S\right)$.

3. Main Results

In this section, we always suppose the following conditions.

• T is an asymptotically nonexpansive operator on H with $\left\{{\theta }_n\right\}$ and ${\left\{T_i\right\}}_{i=1}^N$ are N nonexpansive operators on H.
• A is L-Lipschitzian, pseudomonotone on H, and sequentially weakly continuous on C, s.t. $\Omega ={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)\ne \varnothing$ with $T_0:=T$.
• f is a contractive map on H with coefficient $\delta \in [0,1)$, and F is $\kappa$-Lipschitzian, $\eta$-strongly monotone on H.
• $\nu \delta <\tau :=1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}$ for $\nu \ge 0$ and $\rho \in (0,\frac{2\eta }{{\kappa }^2})$.
• $T_{i,n}:=(1-{\delta }_{i,n})I+{\delta }_{i,n}{T}_i\mathrm{and}{\delta }_{i,n}\in (0,1)\mathrm{for}i=1,2,\dots ,N.$
• $\left\{{\sigma }_n\right\}\subset [0,1]$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ such that

  (i)

  ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}<\infty$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$;

  (ii)

  ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$ and ${lim}_{n\to \infty }{\alpha }_n=0$;

  (iii)

  $\left\{{\beta }_n\right\}\subset [\sigma ,1)$ and ${lim}_{n\to \infty }{\beta }_n=\beta <1$;

  (iv)

  ${lim\; sup}_{n\to \infty }{\gamma }_n<1$, ${lim\; inf}_{n\to \infty }{\gamma }_n>0$ and ${\alpha }_n+{\gamma }_n\le 1\forall n\ge 1$. For example, take

  $${\alpha }_n=\frac{1}{n+1},{\sigma }_n=\frac{1}{{(n+1)}^2}={\theta }_n,{\beta }_n=\frac{n}{2(n+1)},{\gamma }_n=\frac{n}{4(n+1)}.$$

Remark 1.

For Step 2 in Algorithm 3, the composite mapping $T_{N,n}{T}_{N-1,n}\cdots T_{1,n}$ with $T_{i,n}:=(1-{\delta }_{i,n})I+{\delta }_{i,n}{T}_i$ and ${\delta }_{i,n}\in (0,1)$ for $i=1,2,\dots ,N$, has the following property,

$${\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)={\cap }_{i=1}^N\mathrm{Fix}\left(T_{i,n}\right)=\mathrm{Fix}(T_{N,n}{T}_{N-1,n}\cdots T_{1,n})\forall n\ge 1,$$

due to Lemmas 1 and 5.

Algorithm 3: MISEA I

1Initial Step: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$. 2 Iteration Steps: Compute $x_{n+1}$ in what follows.  Step 1. Put $u_n=x_n-{\sigma }_n(x_{n-1}-x_n)$ and calculate $y_n=P_C(u_n-{\ell }_{n}A{u}_n)$, where ${\ell }_n$ is chosen to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $$\ell \parallel A{u}_n-A{y}_n\parallel \le \mu \parallel u_n-y_n\parallel .$$ (6)  Step 2. Calculate $z_n={\beta }_n{x}_n+(1-{\beta }_n)T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{P}_{C_n}(u_n-{\ell }_{n}A{y}_n)$ with $C_n:=\{u\in H:\langle u_n-{\ell }_{n}A{u}_n-y_n,u-y_n\rangle \le 0\}$.  Step 3. Calculate $$x_{n+1}={\alpha }_n\nu f\left(x_n\right)+{\gamma }_n{x}_n+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n.$$ (7)  Update $n:=n+1$ and return to Step 1.

Lemma 8.

The Armijo-like search rule (6) is defined well, and the following holds: $min\{\gamma ,\frac{\mu l}{L}\}\le {\ell }_n\le \gamma$.

Proof. 

As A is L-Lipschitzian, we get $\frac{\mu }{L}\parallel A{u}_n-A{P}_C(u_n-\gamma l^{m}A{u}_n)\parallel \le \mu \parallel u_n-P_C(u_n-\gamma l^{m}A{u}_n)\parallel$. Therefore, (6) is valid for $\gamma l^m\le \frac{\mu }{L}$. This means that ${\ell }_n$ is defined well. It is clear that ${\ell }_n\le \gamma$. In the case of ${\ell }_n=\gamma$, the inequality holds. In the case of ${\ell }_n<\gamma$, from (6) it follows that $\parallel A{u}_n-A{P}_C(u_n-\frac{{\ell }_n}{l}A{u}_n)\parallel >\frac{\mu }{\frac{{\ell }_n}{l}}\parallel u_n-P_C(u_n-\frac{{\ell }_n}{l}A{u}_n)\parallel$. Thus, from the L-Lipschitzian property of A, we get ${\ell }_n>\frac{\mu l}{L}$. Consequently, the inequality holds. □

Lemma 9.

Let $\left\{u_n\right\},\left\{y_n\right\},\left\{z_n\right\}$ be the sequences generated by Algorithm 3. Then

$$\begin{array}{cc}\parallel z_n{-\omega \parallel }^2\hfill & \le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n){\parallel u_n-\omega \parallel }^2\hfill \\ \hfill & -(1-{\beta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\forall \omega \in \Omega ,n\ge 1,\hfill \end{array}$$

(8)

where $v_n:=P_{C_n}(u_n-{\ell }_{n}A{y}_n)$.

Proof. 

First, take an arbitrary $p\in \Omega \subset C\subset C_n$. We note that

$$\begin{array}{cc}2\parallel v_n{-p\parallel }^2\hfill & =2\parallel P_{C_n}(u_n-{\ell }_{n}A{y}_n)-P_{C_n}{p\parallel }^2\le 2\langle v_n-p,u_n-{\ell }_{n}A{y}_n-p\rangle \hfill \\ \hfill & =\parallel v_n{-p\parallel }^2+\parallel u_n{-p\parallel }^2-{\parallel v_n-u_n\parallel }^2-2\langle v_n-p,{\ell }_{n}A{y}_n\rangle .\hfill \end{array}$$

So, it follows that $\parallel v_n{-p\parallel }^2\le \parallel u_n{-p\parallel }^2-{\parallel v_n-u_n\parallel }^2-2\langle v_n-p,{\ell }_{n}A{y}_n\rangle$, which together with (6) and the pseudomonotonicity of A, deduces that $\langle A{y}_n,y_n-p\rangle \ge 0$ and

$$\begin{array}{cc}\parallel v_n{-p\parallel }^2\hfill & \le \parallel u_n{-p\parallel }^2-{\parallel v_n-u_n\parallel }^2+2{\ell }_n(\langle A{y}_n,p-y_n\rangle +\langle A{y}_n,y_n-v_n\rangle )\hfill \\ \hfill & \le \parallel u_n{-p\parallel }^2-{\parallel v_n-u_n\parallel }^2+2{\ell }_n\langle A{y}_n,y_n-v_n\rangle \hfill \\ \hfill & =\parallel u_n{-p\parallel }^2-\parallel v_n-y_n{\parallel }^2-{\parallel y_n-u_n\parallel }^2+2\langle u_n-{\ell }_{n}A{y}_n-y_n,v_n-y_n\rangle .\hfill \end{array}$$

(9)

As $v_n=P_{C_n}(u_n-{\ell }_{n}A{y}_n)$ with $C_n:=\{u\in H:\langle u_n-{\ell }_{n}A{u}_n-y_n,u-y_n\rangle \le 0\}$, we have $\langle u_n-{\ell }_{n}A{u}_n-y_n,v_n-y_n\rangle \le 0$, which together with (6), implies that

$$\begin{array}{cc}2\langle u_n-{\ell }_{n}A{y}_n-y_n,v_n-y_n\rangle \hfill & =2\langle u_n-{\ell }_{n}A{u}_n-y_n,v_n-y_n\rangle +2{\ell }_n\langle A{u}_n-A{y}_n,v_n-y_n\rangle \hfill \\ \hfill & \le 2\mu \parallel u_n-y_n\parallel \parallel v_n-y_n\parallel \le \mu (\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2).\hfill \end{array}$$

Therefore, substituting the last inequality for (9), we obtain

$$\parallel v_n{-p\parallel }^2\le \parallel u_n{-p\parallel }^2-(1-\mu )\parallel u_n-y_n{\parallel }^2-(1-\mu ){\parallel v_n-y_n\parallel }^2\forall p\in \Omega ,$$

(10)

which together with Algorithm 3 and $\mathrm{Fix}(T_{N,n}{T}_{N-1,n}\cdots T_{1,n})={\cap }_{i=1}^N\mathrm{Fix}\left(T_{i,n}\right)={\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)$, due to Lemmas 1 and 5 implies that for all $\omega \in \Omega$,

$$\begin{array}{c}\parallel z_n{-\omega \parallel }^2\le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n){\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-\omega \parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n){\parallel v_n-\omega \parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n)[\parallel u_n{-\omega \parallel }^2-(1-\mu )\parallel u_n-y_n{\parallel }^2-(1-\mu )\parallel v_n-y_n{\parallel }^2]\hfill \\ ={\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n)\parallel u_n{-\omega \parallel }^2-(1-{\beta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2].\hfill \end{array}$$

This completes the proof. □

Lemma 10.

Let $\left\{u_n\right\},\left\{x_n\right\},\left\{y_n\right\},$ and $\left\{z_n\right\}$ be bounded vector sequences generated by Algorithm 3. If $T^n{x}_n-T^{n+1}{x}_n\to 0,x_n-x_{n+1}\to 0,u_n-y_n\to 0,u_n-z_n\to 0$ and $\exists \left\{w_{n_k}\right\}\subset \left\{u_n\right\}$ such that $w_{n_k}\rightharpoonup z\in H$, then $z\in \Omega$.

Proof. 

From Algorithm 3, we get $u_n-x_n={\sigma }_n(x_n-x_{n-1})\forall n\ge 1$, and therefore $\parallel u_n-x_n\parallel ={\sigma }_n\parallel x_n-x_{n-1}\parallel \le \parallel x_n-x_{n-1}\parallel$. Utilizing the assumption $x_n-x_{n+1}\to 0$, we have $u_n-x_n\to 0$. So, it follows from the assumption $u_n-y_n\to 0$, that $\parallel x_n-y_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0(n\to \infty )$. Therefore, according to the assumption $u_n-z_n\to 0$, we get $\parallel x_n-z_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-z_n\parallel \to 0(n\to \infty )$. Furthermore, in terms of Lemma 9 we deduce that for each $\omega \in \Omega$,

$$\begin{array}{c}(1-{\beta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\hfill \\ \le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n)\parallel u_n{-\omega \parallel }^2-{\parallel z_n-\omega \parallel }^2\hfill \\ \le {\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n)(\parallel x_n-\omega \parallel +\parallel x_n-x_{n-1}{\parallel )}^2-{\parallel z_n-\omega \parallel }^2\hfill \\ ={\beta }_n\parallel x_n{-\omega \parallel }^2+(1-{\beta }_n)[\parallel x_n{-\omega \parallel }^2+\parallel x_n-x_{n-1}\parallel (2\parallel x_n-\omega \parallel +\parallel x_n-x_{n-1}\parallel \left)\right]-\parallel z_n{-\omega \parallel }^2\hfill \\ =\parallel x_n{-\omega \parallel }^2-\parallel z_n{-\omega \parallel }^2+(1-{\beta }_n)\parallel x_n-x_{n-1}\parallel (2\parallel x_n-\omega \parallel +\parallel x_n-x_{n-1}\parallel )\hfill \\ \le (\parallel x_n-\omega \parallel +\parallel z_n-\omega \parallel )\parallel x_n-z_n\parallel +\parallel x_n-x_{n-1}\parallel (2\parallel x_n-\omega \parallel +\parallel x_n-x_{n-1}\parallel ).\hfill \end{array}$$

As ${lim}_{n\to \infty }(1-{\beta }_n)=(1-\beta )>0,\mu \in (0,1),x_n-x_{n+1}\to 0$ and $x_n-z_n\to 0$, from the boundedness of $\left\{x_n\right\},\left\{z_n\right\}$ we get

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

Thus we obtain that $\parallel x_n-v_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel +\parallel y_n-v_n\parallel \to 0(n\to \infty )$. □

Now, according to (7) in Algorithm 3, we have

$$\begin{array}{cc}{x}_{n+1}-x_n\hfill & =(1-{\gamma }_n)(T^n{z}_n-x_n)-{\alpha }_n\rho F{T}^n{z}_n+{\alpha }_n\nu f\left(x_n\right)\hfill \\ \hfill & =(1-{\gamma }_n)(T^n{z}_n-T^n{x}_n)+(1-{\gamma }_n)(T^n{x}_n-x_n)-{\alpha }_n\rho F{T}^n{z}_n+{\alpha }_n\nu f\left(x_n\right).\hfill \end{array}$$

So it follows that

$$\begin{array}{c}(1-{\gamma }_n)\parallel T^n{x}_n-x_n\parallel =\parallel x_{n+1}-x_n-{\alpha }_n\nu f\left(x_n\right)-(1-{\gamma }_n)(T^n{z}_n-T^n{x}_n)+{\alpha }_n\rho F{T}^n{z}_n\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\alpha }_n(\parallel \nu f\left(x_n\right)\parallel +\parallel \rho F{T}^n{z}_n\parallel )+(1-{\gamma }_n)\parallel T^n{z}_n-T^n{x}_n\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +{\alpha }_n(\parallel \nu f\left(x_n\right)\parallel +\parallel \rho F{T}^n{z}_n\parallel )+(1+{\theta }_n)\parallel z_n-x_n\parallel .\hfill \end{array}$$

Since ${lim\; inf}_{n\to \infty }(1-{\gamma }_n)>0,{\alpha }_n\to 0,{\theta }_n\to 0,x_n-x_{n+1}\to 0$ and $x_n-z_n\to 0$, from the boundedness of $\left\{x_n\right\},\left\{z_n\right\}$ and the Lipschitz continuity of $f,F,T$, we infer that

$$\underset{n\to \infty }{lim}\parallel x_n-T^n{x}_n\parallel =0.$$

(11)

Also, let the mapping $W:H\to H$ be defined as $Wx:=\beta x+(1-\beta )T_{N,n}{T}_{N-1,n}\cdots T_{1,n}x$, where $\beta \in [\sigma ,1)$. By Lemma 5 we know that W is nonexpansive self-mapping on H and $\mathrm{Fix}\left(W\right)={\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)$. We observe that

$$\begin{array}{c}\parallel W{x}_n-x_n\parallel \le \parallel W{x}_n-z_n\parallel +\parallel z_n-x_n\parallel \hfill \\ =\parallel (\beta -{\beta }_n)(x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n)+(1-{\beta }_n)(T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n-T_N^n{T}_{N-1}^n\cdots T_1^n{v}_n)\parallel \hfill \\ +\parallel z_n-x_n\parallel \hfill \\ \le |\beta -{\beta }_n|\parallel x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\parallel +(1-{\beta }_n)\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n-T_N^n{T}_{N-1}^n\cdots T_1^n{v}_n\parallel \hfill \\ +\parallel z_n-x_n\parallel \hfill \\ \le |\beta -{\beta }_n|\parallel x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\parallel +\parallel x_n-v_n\parallel +\parallel z_n-x_n\parallel .\hfill \end{array}$$

Since $\left\{x_n\right\}$ is bounded and the composite $T_{N,n}{T}_{N-1,n}\cdots T_{1,n}$ is nonexpansive, from ${lim}_{n\to \infty }{\beta }_n=\beta ,x_n-v_n\to 0$ and $x_n-z_n\to 0$ we deduce that

$$\underset{n\to \infty }{lim}\parallel x_n-W{x}_n\parallel =0.$$

(12)

Noticing $y_n=P_C(u_n-{\ell }_{n}A{u}_n)$, we get $\langle u_n-{\ell }_{n}A{u}_n-y_n,x-y_n\rangle \le 0\forall x\in C$, and hence

$$\frac{1}{{\ell }_n}\langle u_n-y_n,x-y_n\rangle +\langle A{u}_n,y_n-u_n\rangle \le \langle A{u}_n,x-u_n\rangle \forall x\in C.$$

(13)

Then, by the boundedness of $\left\{u_{n_k}\right\}$ and Lipschitzian property of A, we know that $\left\{A{u}_{n_k}\right\}$ is bounded. Also, from $u_n-y_n\to 0$, we have that $\left\{y_{n_k}\right\}$ is bounded as well. Observe that ${\ell }_n\ge min\{\gamma ,\frac{\mu l}{L}\}$. So, from (13), it follows that ${lim\; inf}_{k\to \infty }\langle A{u}_{n_k},x-u_{n_k}\rangle \ge 0\forall x\in C$. Moreover, note that $\langle A{y}_n,x-y_n\rangle =\langle A{y}_n-A{u}_n,x-u_n\rangle +\langle A{u}_n,x-u_n\rangle +\langle A{y}_n,u_n-y_n\rangle$. Since $u_n-y_n\to 0$, from L-Lipschitzian property of A we get $A{u}_n-A{y}_n\to 0$, which together with (13) arrives at ${lim\; inf}_{k\to \infty }\langle A{y}_{n_k},x-y_{n_k}\rangle \ge 0\forall x\in C$.

We below claim that $x_n-T{x}_n\to 0$. Indeed, observe that

$$\begin{array}{cc}\parallel T{x}_n-x_n\parallel \hfill & \le \parallel T{x}_n-T^{n+1}{x}_n\parallel +\parallel T^{n+1}{x}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-x_n\parallel \hfill \\ \hfill & \le (1+{\theta }_1)\parallel x_n-T^n{x}_n\parallel +\parallel T^{n+1}{x}_n-T^n{x}_n\parallel +\parallel T^n{x}_n-x_n\parallel \hfill \\ \hfill & =(2+{\theta }_1)\parallel x_n-T^n{x}_n\parallel +\parallel T^{n+1}{x}_n-T^n{x}_n\parallel .\hfill \end{array}$$

Therefore, from (11) and the assumption $T^n{x}_n-T^{n+1}{x}_n\to 0$, we get

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

(14)

We now select a sequence $\left\{{ϵ}_k\right\}\subset (0,1)$ s.t. ${ϵ}_k\downarrow 0$ as $k\to \infty$. For every $k\ge 1$, we indicate by $m_k$ the smallest natural number s.t.

$$\langle A{y}_{n_j},x-y_{n_j}\rangle +{ϵ}_k\ge 0\forall j\ge m_k.$$

(15)

As $\left\{{ϵ}_k\right\}$ is decreasing, $\left\{m_k\right\}$ obviously is increasing. Considering that $\left\{y_{m_k}\right\}\subset C$ ensures $A{y}_{m_k}\ne 0\forall k\ge 1$, we put ${\nu }_{m_k}=\frac{A{y}_{m_k}}{\parallel A{y}_{m_k}{\parallel }^2}$, we have $\langle A{y}_{m_k},{\nu }_{m_k}\rangle =1\forall k\ge 1$. Therefore, from (15), we have $\langle A{y}_{m_k},x+{ϵ}_k{\nu }_{m_k}-y_{m_k}\rangle \ge 0\forall k\ge 1$. Also, from the pseudomonotonicity of A we get $\langle A(x+{ϵ}_k{\nu }_{m_k}),x+{ϵ}_k{\nu }_{m_k}-y_{m_k}\rangle \ge 0\forall k\ge 1$. This means that

$$\langle Ax,x-y_{m_k}\rangle \ge \langle Ax-A(x+{ϵ}_k{\nu }_{m_k}),x+{ϵ}_k{\nu }_{m_k}-y_{m_k}\rangle -{ϵ}_k\langle Ax,{\nu }_{m_k}\rangle \forall k\ge 1.$$

(16)

We show that ${lim}_{k\to \infty }{ϵ}_k{\nu }_{m_k}=0$. In fact, from $u_{n_k}\rightharpoonup z$ and $u_n-y_n\to 0$, we get $y_{n_k}\rightharpoonup z$. Hence, $\left\{y_n\right\}\subset C$ ensures $z\in C$. Also, since A is sequentially weakly continuous, we infer that $A{y}_{n_k}\rightharpoonup Az$. So, we get $Az\ne 0$ (otherwise, z is a solution). Utilizing the sequentially weak lower semicontinuity of the norm $\parallel ·\parallel$, we have $0<\parallel Az\parallel \le {lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel$. Since $\left\{y_{m_k}\right\}\subset \left\{y_{n_k}\right\}$ and ${ϵ}_k\downarrow 0$ as $k\to \infty$, we deduce that $0\le {lim\; sup}_{k\to \infty }\parallel {ϵ}_k{\nu }_{m_k}\parallel ={lim\; sup}_{k\to \infty }\frac{{ϵ}_k}{\parallel A{y}_{m_k}\parallel }\le \frac{{lim\; sup}_{k\to \infty }{ϵ}_k}{{lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel }=0$. Thus we have ${ϵ}_k{\mu }_{m_k}\to 0$.

Finally, we claim $z\in \Omega$. In fact, from $u_n-x_n\to 0$ and $u_{n_k}\rightharpoonup z$, we have $x_{n_k}\rightharpoonup z$. By (14) we get $x_{n_k}-T{x}_{n_k}\to 0$. Because Lemma 7 ensures the demiclosedness of $I-T$ at zero, we have $z\in \mathrm{Fix}\left(T\right)$. Moreover, using $u_n-x_n\to 0$ and $u_{n_k}\rightharpoonup z$, we have $x_{n_k}\rightharpoonup z$. Using (12) we get $x_{n_k}-W{x}_{n_k}\to 0$. Using Lemma 7 we deduce that $I-W$ has the demiclosedness at zero. So, we have $(I-W)z=0$, i.e., $z\in \mathrm{Fix}\left(W\right)={\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)$. In addition, taking $k\to \infty$, we conclude that the right hand side of (16) tends to zero according to the Lipschitzian property of A, the boundedness of $\left\{y_{m_k}\right\},\left\{{\nu }_{m_k}\right\}$ and the limit ${lim}_{k\to \infty }{ϵ}_k{\nu }_{m_k}=0$. Consequently, we get $\langle Ay,y-z\rangle ={lim\; inf}_{k\to \infty }\langle Ay,y-y_{m_k}\rangle \ge 0\forall y\in C$. By Lemma 3 we have $z\in \mathrm{VI}(C,A)$. So, $z\in {\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)=\Omega$.

Remark 2.

It is clear that the boundedness assumption of the generated sequences in Lemma 10 can be disposed with when T is the identity.

Theorem 1.

Assume that the sequence $\left\{x_n\right\}$ constructed by Algorithm 3 satisfies $T^n{x}_n-T^{n+1}{x}_n\to 0$. Then

$$x_n\to x^{*}\in \Omega \iff \left\{\begin{array}{c}{x}_n-x_{n+1}\to 0,\hfill \\ x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\to 0\hfill \end{array}\right.$$

where $x^{*}\in \Omega$ is only a solution to the HVI: $\langle (\nu f-\rho F)x^{*},\omega -x^{*}\rangle \le 0\forall \omega \in \Omega$.

Proof. 

We first note that ${lim\; sup}_{n\to \infty }{\gamma }_n<1$ and ${lim\; inf}_{n\to \infty }{\gamma }_n>0$. Then, we may suppose that $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$. We show that $P_{\Omega }(\nu f+I-\rho F)$ is a contractive map. In fact, using Lemma 6 we get

$$\begin{array}{c}\parallel P_{\Omega }(\nu f+I-\rho F)u-P_{\Omega }(\nu f+I-\rho F)v\parallel \le \nu \parallel f\left(u\right)-f\left(v\right)\parallel +\parallel (I-\rho F)u-(I-\rho F)v\parallel \hfill \\ \le \nu \delta \parallel u-v\parallel +(1-\tau )\parallel u-v\parallel =[1-(\tau -\nu \delta \left)\right]\parallel u-v\parallel \forall u,v\in H.\hfill \end{array}$$

This means that $P_{\Omega }(\nu f+I-\rho F)$ has only a fixed point $x^{*}\in H$, i.e., $x^{*}=P_{\Omega }(\nu f+I-\rho F)x^{*}$. Accordingly, there is only a solution $x^{*}\in \Omega ={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ to the VIP

$$\langle (\nu f-\rho F)x^{*},\omega -x^{*}\rangle \le 0\forall \omega \in \Omega .$$

(17)

It is now easy to see that the necessity of the theorem is valid. Indeed, if $x_n\to x^{*}\in \Omega ={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$, then $T_1{x}^{*}=x^{*},...,T_N{x}^{*}=x^{*}$, which together with ${\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)={\cap }_{i=1}^N\mathrm{Fix}\left(T_{i,n}\right)=\mathrm{Fix}(T_N^n{T}_{N-1}^n\cdots T_1^n)$ (due to Lemmas 1 and 5), imply that $\parallel x_n-x_{n+1}\parallel \le \parallel x_n-x^{*}\parallel +\parallel x_{n+1}-x^{*}\parallel \to 0(n\to \infty )$, and

$$\begin{array}{cc}\parallel x_n-T_N^n{T}_{N-1}^n\cdots T_1^n{x}_n\parallel \hfill & \le \parallel x_n-x^{*}\parallel +\parallel T_N^n{T}_{N-1}^n\cdots T_1^n{x}_n-x^{*}\parallel \hfill \\ \hfill & \le \parallel x_n-x^{*}\parallel +\parallel x_n-x^{*}\parallel =2\parallel x_n-x^{*}\parallel \to 0(n\to \infty ).\hfill \end{array}$$

We below claim the sufficiency of the theorem. For this purpose, we suppose ${lim}_{n\to \infty }(\parallel x_n-x_{n+1}\parallel +\parallel x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\parallel )=0$ and prove the sufficiency by the following steps. □

Step 1. We claim the boundedness of $\left\{x_n\right\}$. In fact, noticing ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$, we know that ${\theta }_n\le \frac{{\alpha }_n(\tau -\nu \delta )}{2}\forall n\ge n_0$ for some $n_0\ge 1$. Therefore, we have that for all $n\ge n_0$,

$${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-{\alpha }_n(\tau -\nu \delta )+{\theta }_n\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}.$$

Let p be an arbitrary point in $\Omega ={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$. Then $Tp=p,T_{i}p=p,i=1,\dots ,N$, and (10) is true, that is,

$$\parallel v_n{-p\parallel }^2+(1-\mu )\parallel u_n-y_n{\parallel }^2+(1-\mu )\parallel v_n-y_n{\parallel }^2\le {\parallel u_n-p\parallel }^2.$$

(18)

Thus, we obtain

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

(19)

From the definition of $u_n$, we have

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

(20)

From ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}<\infty$ and ${sup}_{n\ge 1}\parallel x_n-x_{n-1}\parallel <\infty$, we infer that ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel <\infty$, which immediately yields that $\exists M_1>0$ s.t.

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

(21)

Using (19)–(21), we obtain

$$\parallel v_n-p\parallel \le \parallel u_n-p\parallel \le \parallel x_n-p\parallel +{\alpha }_n{M}_1\forall n\ge 1.$$

(22)

Accordingly, by Algorithm 3, Lemma 6 and (22) we conclude that for all $n\ge n_0$,

$$\begin{array}{c}\parallel z_n-p\parallel =\parallel (1-{\beta }_n)(T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-p)+{\beta }_n(x_n-p)\parallel \hfill \\ \le (1-{\beta }_n)\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-p\parallel +{\beta }_n\parallel x_n-p\parallel \hfill \\ \le (1-{\beta }_n)(\parallel x_n-p\parallel +{\alpha }_n{M}_1)+{\beta }_n\parallel x_n-p\parallel \le \parallel x_n-p\parallel +{\alpha }_n{M}_1,\hfill \end{array}$$

(23)

and therefore

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & =\parallel {\gamma }_n(x_n-p)+{\alpha }_n(\nu f\left(x_n\right)-\rho Fp)+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ \hfill & -((1-{\gamma }_n)I-{\alpha }_n\rho F)p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel +{\gamma }_n\parallel x_n-p\parallel \hfill \\ \hfill & +\parallel ((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n-((1-{\gamma }_n)I-{\alpha }_n\rho F)p\parallel \hfill \\ \hfill & ={\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel +{\gamma }_n\parallel x_n-p\parallel \hfill \\ \hfill & +(1-{\gamma }_n)\parallel (I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)T^n{z}_n-(I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel +{\gamma }_n\parallel x_n-p\parallel \hfill \\ \hfill & +(1-{\gamma }_n)(1-\frac{{\alpha }_n}{1-{\gamma }_n}\tau )(1+{\theta }_n)\parallel z_n-p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel \hfill \\ \hfill & +{\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)(\parallel x_n-p\parallel +{\alpha }_n{M}_1)\hfill \\ \hfill & =[{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]\parallel x_n-p\parallel \hfill \\ \hfill & +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n){\alpha }_n{M}_1+{\alpha }_n\parallel (\nu f-\rho F)p\parallel \hfill \\ \hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n-p\parallel +\frac{{\alpha }_n(\tau -\nu \delta )}{2}·\frac{2(M_1+\parallel (\nu f-\rho F)p\parallel )}{\tau -\nu \delta }\hfill \\ \hfill & \le max\{\frac{2(M_1+\parallel (\nu f-\rho F)p\parallel )}{\tau -\nu \delta },\parallel x_n-p\parallel \}.\hfill \end{array}$$

By induction, we conclude that $\parallel x_n-p\parallel \le max\{\frac{2(M_1+\parallel (\rho F-\nu f)p\parallel )}{\tau -\nu \delta },\parallel x_{n_0}-p\parallel \}\forall n\ge n_0$. Therefore, we get the boundedness of vector sequence $\left\{x_n\right\}$.

Step 2. We claim that $\exists M_4>0$ s.t. $\forall n\ge n_0$,

$$(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_4.$$

In fact, using Lemma 6, Lemma 9, and the convexity of ${\parallel ·\parallel }^2$, from ${\alpha }_n+{\gamma }_n\le 1$, we obtain that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}-p{\parallel }^2=\parallel {\alpha }_n\nu (f\left(x_n\right)-f\left(p\right))+{\gamma }_n(x_n-p)+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ -((1-{\gamma }_n)I-{\alpha }_n\rho F)p+{\alpha }_n(\nu f-\rho F){p\parallel }^2\hfill \\ \le \parallel {\alpha }_n\nu (f\left(x_n\right)-f\left(p\right))+{\gamma }_n(x_n-p)+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ -((1-{\gamma }_n)I-{\alpha }_n\rho F){p\parallel }^2+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ =\parallel {\alpha }_n\nu (f\left(x_n\right)-f\left(p\right))+{\gamma }_n(x_n-p)+(1-{\gamma }_n)[(I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)T^n{z}_n\hfill \\ -(I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)p{]\parallel }^2+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le [{\alpha }_n\nu \delta \parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n)(1-\frac{{\alpha }_n}{1-{\gamma }_n}\tau )(1+{\theta }_n)\parallel z_n-p{\parallel ]}^2\hfill \\ +2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ +2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\beta }_n{\parallel x_n-p\parallel }^2\hfill \\ +(1-{\beta }_n)\parallel u_n{-p\parallel }^2-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2,\hfill \end{array}$$

(24)

where ${sup}_{n\ge 1}2\parallel (\nu f-\rho F)p\parallel \parallel x_{n+1}-p\parallel \le M_2$ for some $M_2>0$. Also, from (22), we get

$$\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\alpha }_n(2M_1\parallel x_n-p\parallel +{\alpha }_n{M}_1^2)\le \parallel x_n{-p\parallel }^2+{\alpha }_n{M}_3,$$

(25)

where ${sup}_{n\ge 1}\{2M_1\parallel x_n-p\parallel +{\alpha }_n{M}_1^2\}\le M_3$ for some $M_3>0$. Note that ${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}$ for all $n\ge n_0$. Substituting (25) for (24), we deduce that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\beta }_n{\parallel x_n-p\parallel }^2\hfill \\ +(1-{\beta }_n)(\parallel x_n{-p\parallel }^2+{\alpha }_n{M}_3)-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2\hfill \\ \le [{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]{\parallel x_n-p\parallel }^2\hfill \\ +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\alpha }_n{M}_3-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2\hfill \\ \le (1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}){\parallel x_n-p\parallel }^2-(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )\times \hfill \\ \times [\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]+{\alpha }_n{M}_2+{\alpha }_n{M}_3\hfill \\ \le \parallel x_n{-p\parallel }^2-(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]+{\alpha }_n{M}_4,\hfill \end{array}$$

where $M_4:=M_2+M_3$. This immediately implies that for all $n\ge n_0$,

$$(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_4.$$

(26)

Step 3. We claim that $\exists M>0$ s.t. $\forall n\ge n_0$,

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n-p{\parallel }^2+\frac{{\alpha }_n(\tau -\nu \delta )}{2}[\frac{4}{\tau -\nu \delta }\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \hfill & +\frac{{\sigma }_n}{{\alpha }_n}·\frac{2M}{\tau -\nu \delta }\parallel x_n-x_{n-1}\parallel ].\hfill \end{array}$$

In fact, we get

$$\parallel u_n{-p\parallel }^2\le (\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}{\parallel )}^2\le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M,$$

(27)

with ${sup}_{n\ge 1}\{2\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel \}\le M$ for some $M>0$. Note that ${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}$ for all $n\ge n_0$. Thus, combining (24) and (27), we have that for all $n\ge n_0$,

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[\parallel x_n{-p\parallel }^2\hfill \\ \hfill & +{\sigma }_n\parallel x_n-x_{n-1}\parallel M]+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \hfill & =[{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]{\parallel x_n-p\parallel }^2\hfill \\ \hfill & +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n){\sigma }_n\parallel x_n-x_{n-1}\parallel M+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]{\parallel x_n-p\parallel }^2+\frac{{\alpha }_n(\tau -\nu \delta )}{2}[\frac{4\langle (\nu f-\rho F)p,x_{n+1}-p\rangle }{\tau -\nu \delta }+\frac{{\sigma }_n}{{\alpha }_n}·\frac{\parallel x_n-x_{n-1}\parallel 2M}{\tau -\nu \delta }].\hfill \end{array}$$

(28)

Step 4. We claim that $x_n\to x^{*}\in \Omega$, which is only a solution to the VIP (17). In fact, setting $p=x^{*}$, we obtain from (28) that

$$\begin{array}{cc}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n-x^{*}{\parallel }^2+\frac{{\alpha }_n(\tau -\nu \delta )}{2}[\frac{4}{\tau -\nu \delta }\langle (\nu f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \hfill \\ \hfill & +\frac{{\sigma }_n}{{\alpha }_n}·\frac{2M}{\tau -\nu \delta }\parallel x_n-x_{n-1}\parallel ].\hfill \end{array}$$

(29)

According to Lemma 4, it is sufficient to prove that ${lim\; sup}_{n\to \infty }\langle (\nu f-\rho F)x^{*},x_{n+1}-x^{*}\rangle \le 0$. As $x_n-x_{n+1}\to 0,{\alpha }_n\to 0,{\beta }_n\to \beta <1$ and ${\theta }_n\to 0$, from (26) and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$, we have

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}(1-b-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\hfill \\ \le {\displaystyle \underset{n\to \infty }{lim\; sup}}(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\hfill \\ \le {\displaystyle \underset{n\to \infty }{lim\; sup}}(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel =0.\hfill \end{array}$$

This immediately implies that

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

(30)

In addition, it is clear that $\parallel u_n-x_n\parallel ={\sigma }_n\parallel x_n-x_{n-1}\parallel \le \parallel x_n-x_{n-1}\parallel \to 0(n\to \infty )$, and hence $\parallel x_n-y_n\parallel \le \parallel x_n-u_n\parallel +\parallel u_n-y_n\parallel \to 0(n\to \infty )$. So it follows from (30) that $\parallel x_n-v_n\parallel \le \parallel x_n-y_n\parallel +\parallel y_n-v_n\parallel \to 0(n\to \infty )$. Thus, from Algorithm 3 and the assumption $x_n-T_N^n{T}_{N-1}^n\cdots T_1^n{x}_n\to 0$, we obtain

$$\begin{array}{c}\parallel z_n-x_n\parallel =(1-{\beta }_n)\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-x_n\parallel \le \parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-x_n\parallel \hfill \\ \le \parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{v}_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\parallel +\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n-x_n\parallel \hfill \\ \le \parallel v_n-x_n\parallel +\parallel T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n-x_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(31)

As $x_n-y_n\to 0,x_n-z_n\to 0$ and $u_n-x_n\to 0$, we deduce that as $n\to \infty$,

$$\parallel u_n-y_n\parallel \le \parallel u_n-x_n\parallel +\parallel x_n-y_n\parallel \to 0\mathrm{and}\parallel u_n-z_n\parallel \le \parallel u_n-x_n\parallel +\parallel x_n-z_n\parallel \to 0.$$

(32)

On the other hand, from the boundedness of $\left\{x_n\right\}$, it follows that $\exists \left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ s.t.

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

(33)

Utilizing the reflexivity of H and the boundedness of $\left\{x_n\right\}$, one may suppose that $x_{n_k}\rightharpoonup \tilde{x}$. Therefore, one gets from (33),

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

(34)

It is easy to see from $u_n-x_n\to 0$ and $x_{n_k}\rightharpoonup \tilde{x}$ that $w_{n_k}\rightharpoonup \tilde{x}$. Since $T^n{x}_n-T^{n+1}{x}_n\to 0,x_n-x_{n+1}\to 0,u_n-y_n\to 0,u_n-z_n\to 0$ and $w_{n_k}\rightharpoonup \tilde{x}$, from Lemma 10 we get $\tilde{x}\in \Omega$. Therefore, from (17) and (34), we infer that

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

which together with $x_n-x_{n+1}\to 0$, implies that

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

(35)

Observe that $\left\{\frac{{\alpha }_n(\tau -\nu \delta )}{2}\right\}\subset [0,1],{\sum }_{n=1}^{\infty }\frac{{\alpha }_n(\tau -\nu \delta )}{2}=\infty$, and

$$\underset{n\to \infty }{lim\; sup}[\frac{4}{\tau -\nu \delta }\langle (\nu f-\rho F)x^{*},x_{n+1}-x^{*}\rangle +\frac{{\sigma }_n}{{\alpha }_n}·\frac{2M}{\tau -\nu \delta }\parallel x_n-x_{n-1}\parallel ]\le 0.$$

(36)

Consequently, by Lemma 4 we obtain from (29) that $\parallel x_n-x^{*}\parallel \to 0$ as $n\to \infty$.

Next, we introduce another mildly inertial subgradient extragradient algorithm with line-search process.

It is remarkable that Lemmas 8 and 9 remain true for Algorithm 4.

Algorithm 4: MISEA II

1Initial Step: Given $x_0,x_1\in H$ arbitrary. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$. 2 Iteration Steps: Compute $x_{n+1}$ in what follows:  Step 1. Put $u_n=x_n-{\sigma }_n(x_{n-1}-x_n)$ and calculate $y_n=P_C(u_n-{\ell }_{n}A{u}_n)$, where ${\ell }_n$ is chosen to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $$\ell \parallel A{u}_n-A{y}_n\parallel \le \mu \parallel u_n-y_n\parallel .$$ (37)  Step 2. Calculate $z_n={\beta }_n{x}_n+(1-{\beta }_n)T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{P}_{C_n}(u_n-{\ell }_{n}A{y}_n)$ with $C_n:=\{u\in H:\langle u_n-{\ell }_{n}A{u}_n-y_n,u-y_n\rangle \le 0\}$.  Step 3. Calculate $$x_{n+1}={\gamma }_n{u}_n+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n+{\alpha }_n\nu f\left(x_n\right).$$ (38)  Update $n:=n+1$ and return to Step 1.

Theorem 2.

Assume that the sequence $\left\{x_n\right\}$ constructed by Algorithm 4 satisfies $T^n{x}_n-T^{n+1}{x}_n\to 0$. Then,

$$x_n\to x^{*}\in \Omega \iff \left\{\begin{array}{c}{x}_n-x_{n+1}\to 0,\hfill \\ x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\to 0\hfill \end{array}\right.$$

where $x^{*}\in \Omega$ is only a solution to the HVI: $\langle (\nu f-\rho F)x^{*},\omega -x^{*}\rangle \le 0\forall \omega \in \Omega$.

Proof. 

Using the similar inference to that in the proof of Theorem 1, we obtain that there is only a solution $x^{*}\in \Omega ={\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ to the HVI (17), and that the necessity of the theorem is true.

We claim the sufficiency of the theorem below. For this purpose, we suppose ${lim}_{n\to \infty }(\parallel x_n-x_{n+1}\parallel +\parallel x_n-T_{N,n}{T}_{N-1,n}\cdots T_{1,n}{x}_n\parallel )=0$ and prove the sufficiency by the following steps.

Step 1. We claim the boundedness of $\left\{x_n\right\}$. In fact, using the similar reasoning to that in Step 1 for the proof of Theorem 1, we know that inequalities (18)–(23) hold. Taking into account ${lim}_{n\to \infty }\frac{{\theta }_n}{{\alpha }_n}=0$, we know that ${\theta }_n\le \frac{{\alpha }_n(\tau -\nu \delta )}{2}\forall n\ge n_0$ for some $n_0\ge 1$. Hence we deduce that for all $n\ge n_0$,

$${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-{\alpha }_n(\tau -\nu \delta )+{\theta }_n\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}.$$

Also, from Algorithm 4, Lemma 6, and (22) and (23) we obtain

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & =\parallel {\gamma }_n(u_n-p)+{\alpha }_n(\nu f\left(x_n\right)-\rho Fp)+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ \hfill & -((1-{\gamma }_n)I-{\alpha }_n\rho F)p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel +{\gamma }_n\parallel u_n-p\parallel \hfill \\ \hfill & +(1-{\gamma }_n)\parallel (I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)T^n{z}_n-(I-\frac{{\alpha }_n}{1-{\gamma }_n}\rho F)p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel +{\gamma }_n\parallel u_n-p\parallel \hfill \\ \hfill & +(1-{\gamma }_n)(1-\frac{{\alpha }_n}{1-{\gamma }_n}\tau )(1+{\theta }_n)\parallel z_n-p\parallel \hfill \\ \hfill & \le {\alpha }_n\nu \delta \parallel x_n-p\parallel +{\alpha }_n\parallel (\nu f-\rho F)p\parallel \hfill \\ \hfill & +{\gamma }_n(\parallel x_n-p\parallel +{\alpha }_n{M}_1)+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)(\parallel x_n-p\parallel +{\alpha }_n{M}_1)\hfill \\ \hfill & =[{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]\parallel x_n-p\parallel \hfill \\ \hfill & +[{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]{\alpha }_n{M}_1+{\alpha }_n\parallel (\nu f-\rho F)p\parallel \hfill \\ \hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n-p\parallel +\frac{{\alpha }_n(\tau -\nu \delta )}{2}·\frac{2(M_1+\parallel (\nu f-\rho F)p\parallel )}{\tau -\nu \delta }\hfill \\ \hfill & \le max\{\frac{2(M_1+\parallel (\nu f-\rho F)p\parallel )}{\tau -\nu \delta },\parallel x_n-p\parallel \}.\hfill \end{array}$$

By induction, we conclude that $\parallel x_n-p\parallel \le max\{\frac{2(M_1+\parallel (\rho F-\nu f)p\parallel )}{\tau -\nu \delta },\parallel x_{n_0}-p\parallel \}\forall n\ge n_0$. Therefore, we obtain the boundedness of vector sequence $\left\{x_n\right\}$.

Step 2. One claims $\exists M_4>0$ s.t. $\forall n\ge n_0$,

$$(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_4.$$

In fact, using Lemma 6, Lemma 9, and the convexity of ${\parallel ·\parallel }^2$, from ${\alpha }_n+{\gamma }_n\le 1$, we obtain that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}-p{\parallel }^2=\parallel {\gamma }_n(u_n-p)+{\alpha }_n\nu (f\left(x_n\right)-f\left(p\right))+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ -((1-{\gamma }_n)I-{\alpha }_n\rho F)p+{\alpha }_n(\nu f-\rho F){p\parallel }^2\hfill \\ \le \parallel {\alpha }_n\nu (f\left(x_n\right)-f\left(p\right))+{\gamma }_n(u_n-p)+((1-{\gamma }_n)I-{\alpha }_n\rho F)T^n{z}_n\hfill \\ -((1-{\gamma }_n)I-{\alpha }_n\rho F){p\parallel }^2+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le [{\alpha }_n\nu \delta \parallel x_n-p\parallel +{\gamma }_n\parallel u_n-p\parallel +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\parallel z_n-p{\parallel ]}^2\hfill \\ +2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel u_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ +2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le {\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel u_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\beta }_n{\parallel x_n-p\parallel }^2\hfill \\ +(1-{\beta }_n)\parallel u_n{-p\parallel }^2-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2,\hfill \end{array}$$

(39)

where ${sup}_{n\ge 1}2\parallel (\nu f-\rho F)p\parallel \parallel x_{n+1}-p\parallel \le M_2$ for some $M_2>0$. Also, from (22) we have

$$\parallel u_n{-p\parallel }^2\le {\parallel x_n-p\parallel }^2+{\alpha }_n{M}_3,$$

(40)

where ${sup}_{n\ge 1}\{2M_1\parallel x_n-p\parallel +{\alpha }_n{M}_1^2\}\le M_3$ for some $M_3>0$. Note that ${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}$ for all $n\ge n_0$. Substituting (40) for (39), we deduce that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\gamma }_n(\parallel x_n{-p\parallel }^2+{\alpha }_n{M}_3)+{\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[\parallel x_n{-p\parallel }^2\hfill \\ +{\alpha }_n{M}_3-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2\hfill \\ =[{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]{\parallel x_n-p\parallel }^2+{\gamma }_n{\alpha }_n{M}_3\hfill \\ +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\alpha }_n{M}_3-(1-{\beta }_n)(1-\mu )(\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2\left)\right]+{\alpha }_n{M}_2\hfill \\ \le (1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}){\parallel x_n-p\parallel }^2-(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )\times \hfill \\ \times [\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]+{\alpha }_n{M}_2+{\alpha }_n{M}_3\hfill \\ \le \parallel x_n{-p\parallel }^2-(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]+{\alpha }_n{M}_4,\hfill \end{array}$$

where $M_4:=M_2+M_3$. This immediately implies that for all $n\ge n_0$,

$$(1-{\gamma }_n-{\alpha }_n\tau )(1-{\beta }_n)(1+{\theta }_n)(1-\mu )[\parallel u_n-y_n{\parallel }^2+\parallel v_n-y_n{\parallel }^2]\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_4.$$

(41)

Step 3. One claims that $\exists M>0$ s.t. $\forall n\ge n_0$,

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n-p{\parallel }^2+\frac{{\alpha }_n(\tau -\nu \delta )}{2}[\frac{4}{\tau -\nu \delta }\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \hfill & +\frac{{\sigma }_n}{{\alpha }_n}·\frac{2M}{\tau -\nu \delta }\parallel x_n-x_{n-1}\parallel ].\hfill \end{array}$$

In fact, we get

$$\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M,$$

(42)

where ${sup}_{n\ge 1}\{2\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel \}\le M$ for some $M>0$. Observe that ${\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)\le 1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}$ for all $n\ge n_0$. Thus, combining (39) and (42), we have that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le {\gamma }_n(\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M)+{\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[{\beta }_n{\parallel x_n-p\parallel }^2\hfill \\ +(1-{\beta }_n)(\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M\left)\right]+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le {\gamma }_n(\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M)+{\alpha }_n\nu \delta \parallel x_n{-p\parallel }^2+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)[\parallel x_n{-p\parallel }^2\hfill \\ +{\sigma }_n\parallel x_n-x_{n-1}\parallel M]+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ =[{\alpha }_n\nu \delta +{\gamma }_n+(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n)]\parallel x_n{-p\parallel }^2+{\gamma }_n{\sigma }_n\parallel x_n-x_{n-1}\parallel M\hfill \\ +(1-{\gamma }_n-{\alpha }_n\tau )(1+{\theta }_n){\sigma }_n\parallel x_n-x_{n-1}\parallel M+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ \le [1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel M+2{\alpha }_n\langle (\nu f-\rho F)p,x_{n+1}-p\rangle \hfill \\ =[1-\frac{{\alpha }_n(\tau -\nu \delta )}{2}]{\parallel x_n-p\parallel }^2+\frac{{\alpha }_n(\tau -\nu \delta )}{2}[\frac{4\langle (\nu f-\rho F)p,x_{n+1}-p\rangle }{\tau -\nu \delta }+\frac{{\sigma }_n}{{\alpha }_n}·\frac{\parallel x_n-x_{n-1}\parallel 2M}{\tau -\nu \delta }].\hfill \end{array}$$

(43)

Step 4. One claims that $x_n\to x^{*}\in \Omega$, which is only a solution to the VIP (17). In fact, using the similar inference to that in Step 4 for the proof of Theorem 1, one derives the desired conclusion. □

Example 1.

We can get an example of T satisfying the condition assumed in Theorems 1 and 2. As a matter of fact, we put $H=\mathbf{R}$, whose inner product and induced norm are defined by $\langle a,b\rangle =ab$ and $\parallel ·\parallel =|·|$ indicate, respectively. Let $T:H\to H$ be defined as $Tx:=sin\left(\frac{7}{8}x\right)\forall x\in H$. Then T is a contraction with constant $\frac{7}{8}$, and hence a nonexpansive mapping. Thus, T is an asymptotically nonexpansive mapping. As

$$\parallel T^{n}x-T^{n}y\parallel \le \frac{7}{8}\parallel T^{n-1}x-T^{n-1}y\parallel \le \cdots \le {\left(\frac{7}{8}\right)}^n\parallel x-y\parallel \forall x,y\in H,$$

we know that for any sequence $\left\{x_n\right\}\subset H$,

$$\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \le {\left(\frac{7}{8}\right)}^{n-1}\parallel T^2{x}_n-T{x}_n\parallel ={\left(\frac{7}{8}\right)}^{n-1}\parallel sin\left(\frac{7}{8}T{x}_n\right)-sin\left(\frac{7}{8}{x}_n\right)\parallel \le 2{\left(\frac{7}{8}\right)}^{n-1}\to 0$$

as $n\to \infty$. That is, $T^n{x}_n-T^{n+1}{x}_n\to 0(n\to \infty )$.

Remark 3.

Compared with the corresponding results in Bnouhachem et al. [2], Cai et al. [35], Kraikaew and Saejung [36], and Thong and Hieu [37,38], our results improve and extend them in what follows.

(i) 

The problem of obtaining a point of $\mathrm{VI}(C,A)$ in the work by the authors of [36] is extendable to the development of our problem of obtaining a point of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$, where $T_0:=T$ is asymptotically nonexpansive and ${\left\{T_i\right\}}_{i=1}^N$ is a pool of nonexpansive maps. The Halpern subgradient method for solving the VIP in the work by the authors of [36] is extendable to the development of our mildly inertial subgradient algorithms with linesearch process for solving the VIP and CFPP.

(ii) 

The problem of obtaining a point of $\mathrm{VI}(C,A)$ in the work by the authors of [37] is extendable to the development of our problem of finding a point of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$, where $T_0:=T$ is asymptotically nonexpansive and ${\left\{T_i\right\}}_{i=1}^N$ is a pool of nonexpansive maps. The inertial subgradient method with weak convergence for solving the VIP in the work by the authors of [37] is extendable to the development of our mildly inertial subgradient algorithms with linesearch process (which are convergent in norm) for solving the VIP and CFPP.

(iii) 

The problem of obtaining a point of $\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(T\right)$ (where A is monotone and T is quasi-nonexpansive) in the work by the authors of [38] is extendable to the development of our problem of obtaining a point of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$, where $T_0:=T$ is asymptotically nonexpansive and ${\left\{T_i\right\}}_{i=1}^N$ is a pool of nonexpansive maps. The inertial subgradient extragradient method with linesearch (which is weakly convergent) for solving the VIP and FPP in the work by the authors of [38] is extendable to the development of our mildly inertial subgradient algorithms with linesearch process (which are convergent in norm) for solving the VIP and CFPP. It is worth mentioning that the inertial subgradient method with linesearch process in the work by the authors of [38] combines the inertial subgradient approaches [37] with the Mann method.

(iv) 

The problem of obtaining a point in the common fixed-point set ${\cap }_{i=1}^N\mathrm{Fix}\left(T_i\right)$ of N nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$ in the work by the authors of [2], is extendable to the development of our problem of obtaining a point of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$, where $T_0:=T$ is asymptotically nonexpansive and ${\left\{T_i\right\}}_{i=1}^N$ is a pool of nonexpansive maps. The iterative algorithm for hierarchical FPPs for finitely many nonexpansive mappings in the work by the authors of [2] (i.e., iterative scheme (3) in this paper), is extendable to the development of our mildly inertial subgradient algorithms with linesearch process for solving the VIP and CFPP. Meantime, the restrictions ${lim\; sup}_{n\to \infty }{\gamma }_n<1$, ${lim\; inf}_{n\to \infty }{\gamma }_n>0$ and ${lim}_{n\to \infty }|{\delta }_{n-1}^i-{\delta }_{i,n}|=0$ for $i=1,\dots ,N$ imposed on (3), are dropped, where $0<{lim\; inf}_{n\to \infty }{\gamma }_n<{lim\; sup}_{n\to \infty }{\gamma }_n<1$ is weakened to the condition $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$.

(v) 

The problem of obtaining a point in the common solution set Ω of the VIPs for two inverse-strongly monotone mappings and the FPP of an asymptotically nonexpansive mapping in the work by the authors of [35], is extendable to the development of our problem of obtaining a point of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ where $T_0:=T$ is asymptotically nonexpansive and ${\left\{T_i\right\}}_{i=1}^N$ is a pool of nonexpansive maps. The viscosity implicit rule involving a modified extragradient method in the work by the authors of [35] (i.e., iterative scheme (4) in this paper), is extendable to the development of our mildly inertial subgradient algorithms with linesearch process for solving the VIP and CFPP. Moreover, the conditions ${\sum }_{n=1}^{\infty }|{\alpha }_{n+1}-{\alpha }_n|<\infty$ and ${\sum }_{n=1}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$ imposed on (4), are deleted where ${\sum }_{n=1}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$ is weakened to the assumption $\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \to 0(n\to \infty )$.

4. Applications

In this section, our main theorems are used to deal with the VIP and CFPP in an illustrating example. The initial point $x_0=x_1$ is randomly chosen in $\mathbf{R}$. Take $\nu f\left(x\right)=F\left(x\right)=\frac{1}{2}x,\gamma =l=\mu =\frac{1}{2},{\sigma }_n={\alpha }_n=\frac{1}{n+1},{\beta }_n=\frac{1}{3},{\gamma }_n=\frac{1}{2},\nu =\frac{3}{4},f=\frac{2}{3}I$ and $\rho =2$. Then, we know that ${\alpha }_n+{\gamma }_n\le 1\forall n\ge 1$, $\nu \delta =\kappa =\eta =\frac{1}{2}$, and

$$\tau =1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}=1-\sqrt{1-2(2·\frac{1}{2}-2{\left(\frac{1}{2}\right)}^2)}=1\in (0,1].$$

We first provide an example of a Lipschitzian, pseudomonotone operator A, asymptotically nonexpansive operator T, and nonexpansive operator $T_1$ with $\Omega =\mathrm{Fix}\left(T\right)\cap \mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-1,3]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. Let $A,T,T_1,T_1^n:H\to H$ be defined as $Ax:=\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|}$, $Tx:=\frac{4}{5}sinx$, $Tx:=sinx$ and $T_1^{n}x:=\frac{3}{8}x+\frac{5}{8}sinx\forall x\in H,n\ge 1$. Then it is clear that $T_1$ is a nonexpansive mapping on H. Moreover, from Lemma 5 we know that $\mathrm{Fix}\left(T_1^n\right)=\mathrm{Fix}\left(T_1\right)=\left\{0\right\}\forall n\ge 1$. Now, we first show that A is Lipschitzian, pseudomonotone operator with $L=2$. In fact, for all $x,y\in H$ we get

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

This means that A is Lipschitzian with $L=2$. We below claim that A is pseudomonotone. For any given $x,y\in H$, it is clear that the relation holds:

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

Furthermore, it is easy to see that T is asymptotically nonexpansive with ${\theta }_n={\left(\frac{4}{5}\right)}^n\forall n\ge 1$, such that $\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, we observe that

$$\parallel T^{n}x-T^{n}y\parallel \le \frac{4}{5}\parallel T^{n-1}x-T^{n-1}y\parallel \le \cdots \le {\left(\frac{4}{5}\right)}^n\parallel x-y\parallel \le (1+{\theta }_n)\parallel x-y\parallel ,$$

and

$$\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \le {\left(\frac{4}{5}\right)}^{n-1}\parallel T^2{x}_n-T{x}_n\parallel ={\left(\frac{4}{5}\right)}^{n-1}\parallel \frac{4}{5}sin\left(T{x}_n\right)-\frac{4}{5}sin{x}_n\parallel \le 2{\left(\frac{4}{5}\right)}^n\to 0(n\to \infty ).$$

It is clear that $\mathrm{Fix}\left(T\right)=\left\{0\right\}$ and

$$\underset{n\to \infty }{lim}\frac{{\theta }_n}{{\alpha }_n}=\underset{n\to \infty }{lim}\frac{{(4/5)}^n}{1/(n+1)}=0.$$

Therefore, $\Omega =\mathrm{Fix}\left(T\right)\cap \mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 3 can be rewritten as follows,

$$\left\{\begin{array}{c}{u}_n=x_n+\frac{1}{n+1}(x_n-x_{n-1}),\hfill \\ y_n=P_C(u_n-{\ell }_{n}A{u}_n),\hfill \\ z_n=\frac{1}{3}{x}_n+\frac{2}{3}{T}_1^n{P}_{C_n}(u_n-{\ell }_{n}A{y}_n),\hfill \\ x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{2}{x}_n+(\frac{n}{n+1}-\frac{1}{2})T^n{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

(44)

where for every $n\ge 1$, $C_n$ and ${\ell }_n$ are picked up as in Algorithm 3. Then, by Theorem 1, we know that $\left\{x_n\right\}$ converges to $0\in \Omega =\mathrm{Fix}\left(T\right)\cap \mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)$ if and only if $|x_n-x_{n+1}|+|x_n-T_1^n{x}_n|\to 0$ as $n\to \infty$.

On the other hand, Algorithm 4 can be rewritten as follows,

$$\left\{\begin{array}{c}{u}_n=x_n+\frac{1}{n+1}(x_n-x_{n-1}),\hfill \\ y_n=P_C(u_n-{\ell }_{n}A{u}_n),\hfill \\ z_n=\frac{1}{3}{x}_n+\frac{2}{3}{T}_1^n{P}_{C_n}(u_n-{\ell }_{n}A{y}_n),\hfill \\ x_{n+1}=\frac{1}{n+1}·\frac{1}{2}{x}_n+\frac{1}{2}{u}_n+(\frac{n}{n+1}-\frac{1}{2})T^n{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

(45)

where for every $n\ge 1$, $C_n$ and ${\ell }_n$ are picked up as in Algorithm 4. Then, by Theorem 2, we know that $\left\{x_n\right\}$ converges to $0\in \Omega =\mathrm{Fix}\left(T\right)\cap \mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)$ if and only if $|x_n-x_{n+1}|+|x_n-T_1^n{x}_n|\to 0$ as $n\to \infty$.