Inertial-Like Subgradient Extragradient Methods for Variational Inequalities and Fixed Points of Asymptotically Nonexpansive and Strictly Pseudocontractive Mappings

Abstract

Let VIP indicate the variational inequality problem with Lipschitzian and pseudomonotone operator and let CFPP denote the common fixed-point problem of an asymptotically nonexpansive mapping and a strictly pseudocontractive mapping in a real Hilbert space. Our object in this article is to establish strong convergence results for solving the VIP and CFPP by utilizing an inertial-like gradient-like extragradient method with line-search process. Via suitable assumptions, it is shown that the sequences generated by such a method converge strongly to a common solution of the VIP and CFPP, which also solves a hierarchical variational inequality (HVI).

1. Introduction

Throughout this paper we assume that C is a nonempty, convex and closed subset of a real Hilbert space $(H,\parallel ·\parallel )$, whose inner product is denoted by $\langle ·,·\rangle$. Moreover, let $P_C$ denote the metric projection of H onto C.

Suppose $A:H\to H$ is a mapping. In this paper, we shall consider the following variational inequality (VI) of finding $x^{*}\in C$ such that

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

(1)

The set of solutions to Equation (1) is denoted by VI($C,A$). In 1976, Korpelevich [1] first introduced an extragradient method, which is one of the most popular approximation ones for solving Equation (1) till now. That is, for any initial $u_0\in C$, the sequence $\left\{u_n\right\}$ is generated 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)

where $\tau$ is a constant in $(0,\frac{1}{L})$ for $L>0$ the Lipschitz constant of mapping A. In the case where $\mathrm{VI}(C,A)\ne \varnothing$, the sequence $\left\{u_n\right\}$ constructed by Equation (2) is weakly convergent to a point in $\mathrm{VI}(C,A)$. Recently, light has been shed on approximation methods for solving problem Equation (1) by many researchers; see, e.g., [2,3,4,5,6,7,8,9,10,11] and references therein, to name but a few.

Let $T:C\to C$ be a mapping. We denote by $\mathrm{Fix}(T)$ the set of fixed points of T, i.e., $\mathrm{Fix}(T)=\{x\in C:x=Tx\}$. T is said to be asymptotically nonexpansive if $\exists \left\{{\theta }_n\right\}\subset [0,+\infty )$ such that ${lim}_{n\to \infty }{\theta }_n=0$ and $\parallel T^{n}u-T^{n}v\parallel \le \parallel u-v\parallel +{\theta }_n\parallel u-v\parallel$, $\forall n\ge 1,$ $u,v\in C$. If ${\theta }_n\equiv 0$, then T is nonexpansive. Also, T is said to be strictly pseudocontractive if $\exists \zeta \in [0,1)$ s.t. ${\parallel Tu-Tv\parallel }^2\le {\parallel u-v\parallel }^2+\zeta {\parallel (I-T)u-(I-T)v\parallel }^2,\forall u,v\in C$. If $\zeta =0$, then T reduces to a nonexpansive mapping. One knows that the class of strict pseudocontractions strictly includes the class of nonexpansive mappings. Both strict pseudocontractions and nonexpansive mappings have been studied extensively by a large number of authors via iteration approximation methods; see, e.g., [12,13,14,15,16,17,18] and references therein.

Let the mappings $A,B:C\to H$ be both inverse-strongly monotone and let the mapping $T:C\to C$ be asymptotically nonexpansive one with a sequence $\left\{{\theta }_n\right\}$. Let $f:C\to C$ be a $\delta$-contraction with $\delta \in [0,1)$. By using a modified extragradient method, Cai et al. [19] designed a viscosity implicit rule for finding a point in the common solution set $\mathsf{\Omega }$ of the VIs for A and B and the FPP of T, i.e., for arbitrarily given $x_1\in C$, $\left\{x_n\right\}$ is the sequence constructed by

$$\left\{\begin{array}{c}{u}_n=s_n{x}_n+(1-s_n)y_n,\hfill \\ y_n=P_C(I-\lambda A)P_C(u_n-\mu B{u}_n),\hfill \\ x_{n+1}=P_C[(T^n{y}_n-{\alpha }_n\rho F{T}^n{y}_n)+{\alpha }_{n}f(x_n)],\hfill \end{array}\right.$$

where $\left\{{\alpha }_n\right\},\left\{s_n\right\}\subset (0,1]$. Under appropriate conditions imposed on $\left\{{\alpha }_n\right\},\left\{s_n\right\}$, they proved that $\left\{x_n\right\}$ is convergent strongly to an element $x^{*}\in \mathsf{\Omega }$ provided ${\sum }_{n=1}^{\infty }\parallel T^{n+1}{y}_n-T^n{y}_n\parallel <\infty$.

In the context of extragradient techniques, one has to compute metric projections two times for each computational step. Without doubt, if C is a general convex and closed set, the computation of the projection onto C might be quite consuming-time. In 2011, inspired by Korpelevich’s extragradient method, Censor et al. [20] first designed the subgradient extragradient method, where a projection onto a half-space is used in place of the second projection onto C. In 2014, Kraikaew and Saejung [21] proposed the Halpern subgradient extragradient method for solving Equation (1), and proved strong convergence of the proposed method to a solution of Equation (1).

In 2018, via the inertial technique, Thong and Hieu [22] studied the inertial subgradient extragradient method, and proved weak convergence of their method to a solution of Equation (1). Very recently, they [23] constructed two inertial subgradient extragradient algorithms with linear-search process for finding a common solution of problem Equation (1) with operator A and the FPP of operator T with demiclosedness property in a real Hilbert space, where A is Lipschitzian and monotone, and T is quasi-nonexpansive. The constructed inertial subgradient extragradient algorithms (Algorithms 1 and 2) are as below:

Algorithm 1: Inertial subgradient extragradient algorithm (I) (see [23], Algorithm 1]).

Initialization: Given $u_0,u_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.  Iterative Steps: Compute $u_{n+1}$ in what follows:  Step 1. Put $v_n={\alpha }_n(u_n-u_{n-1})+u_n$ and calculate $y_n=P_C(v_n-{\tau }_{n}A{v}_n)$, where ${\tau }_n$ is chosen to be the largest $\tau \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $\tau \parallel A{v}_n-A{y}_n\parallel \le \mu \parallel v_n-y_n\parallel$.  Step 2. Calculate $z_n=P_{T_n}(v_n-{\tau }_{n}A{y}_n)$ with $T_n:=\{x\in H:\langle x-y_n,v_n-{\tau }_{n}A{v}_n-y_n\rangle \le 0\}$.  Step 3. Calculate $u_{n+1}={\beta }_{n}T{z}_n+(1-{\beta }_n)v_n$. If $v_n=z_n=u_{n+1}$ then $v_n\in \mathrm{Fix}(T)\cap \mathrm{VI}(C,A)$.   Set  $n:=n+1$ and go to Step 1.

Algorithm 2: Inertial subgradient extragradient algorithm (II) (see [23], Algorithm 2]).

Initialization: Given $u_0,u_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.   Iterative Steps: Calculate $u_{n+1}$ as follows:   Step 1. Put $v_n={\alpha }_n(u_n-u_{n-1})+u_n$ and calculate $y_n=P_C(v_n-{\tau }_{n}A{v}_n)$, where ${\tau }_n$ is chosen to be the largest $\tau \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ satisfying $\tau \parallel A{v}_n-A{y}_n\parallel \le \mu \parallel v_n-y_n\parallel$.   Step 2. Calculate $z_n=P_{T_n}(v_n-{\tau }_{n}A{y}_n)$ with $T_n:=\{x\in H:\langle x-y_n,v_n-{\tau }_{n}A{v}_n-y_n\rangle \le 0\}$.   Step 3. Calculate $u_{n+1}={\beta }_{n}T{z}_n+(1-{\beta }_n)u_n$. If $v_n=z_n=u_n=u_{n+1}$ then   $u_n\in \mathrm{Fix}(T)\cap \mathrm{VI}(C,A)$. Set $n:=n+1$ and go to Step 1.

Under mild assumptions, they proved that the sequences generated by the proposed algorithms are weakly convergent to a point in $\mathrm{Fix}(T)\cap \mathrm{VI}(C,A)$. Recently, gradient-like methods have been studied extensively by many authors; see, e.g., [24,25,26,27,28,29,30,31,32,33,34,35,36,37,38].

Inspired by the research work of [23], we introduce two inertial-like subgradient algorithms with line-search process for solving Equation (1) with a Lipschitzian and pseudomonotone operator and the common fixed point problem (CFPP) of an asymptotically nonexpansive operator and a strictly pseudocontractive operator in H. The proposed algorithms comprehensively adopt inertial subgradient extragradient method with line-search process, viscosity approximation method, Mann iteration method and asymptotically nonexpansive mapping. Via suitable assumptions, it is shown that the sequences generated by the suggested algorithms converge strongly to a common solution of the VIP and CFPP, which also solves a hierarchical variational inequality (HVI).

2. Preliminaries

Let $x\in H$ and $\left\{x_n\right\}\subset H$. We use the notation $x_n\to x$ (resp., $x_n\rightharpoonup x$) to indicate the strong (resp., weak) convergence of $\left\{x_n\right\}$ to x. Recall that a mapping $T:C\to H$ is said to be:

(i)

L-Lipschitzian (or L-Lipschitz continuous) if $\parallel Tx-Ty\parallel \le L\parallel x-y\parallel ,\forall x,y\in C$ for some $L>0$;

(ii)

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

(iii)

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

(iv)

$\beta$-strongly monotone if $\langle Tu-Tv,u-v\rangle \ge {\beta \parallel u-v\parallel }^2,\forall u,v\in C$ for some $\beta >0$;

(v)

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

For metric projections, it is well known that the following assertions hold:

(i)

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

(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)

${\parallel \tau x+(1-\tau )y\parallel }^2={\tau \parallel x\parallel }^2+{(1-\tau )\parallel y\parallel }^2-\tau (1-\tau ){\parallel x-y\parallel }^2,\forall x,y\in H,\tau \in [0,1]$.

Lemma 1.

[39] Assume that $A:C\to H$ is a continuous pseudomonotone mapping. Then $u^{*}\in C$ is a solution to the VI $\langle A{u}^{*},v-u^{*}\rangle \ge 0,\forall v\in C$, iff $\langle Av,v-u^{*}\rangle \ge 0,\forall v\in C$.

Lemma 2.

[40] Let the real sequence $\left\{t_n\right\}\subset [0,\infty )$ satisfy the conditions: $t_{n+1}\le (1-s_n)t_n+s_n{b}_n,\forall n\ge 1$, where $\left\{s_n\right\}$ and $\left\{b_n\right\}$ are sequences in $(-\infty ,\infty )$ such that (i) $\left\{s_n\right\}\subset [0,1]$ and ${\sum }_{n=1}^{\infty }{s}_n=\infty$, and (ii) ${lim\; sup}_{n\to \infty }{b}_n\le 0$ or ${\sum }_{n=1}^{\infty }\left|s_n{b}_n\right|<\infty$. Then ${lim}_{n\to \infty }{t}_n=0$.

Lemma 3.

[33] Let $T:C\to C$ be a ζ-strict pseudocontraction. If the sequence $\left\{u_n\right\}\subset C$ satisfies $u_n\rightharpoonup u\in C$ and $(I-T)u_n\to 0$, then $u\in \mathrm{Fix}(T)$, where I is the identity operator of H.

Lemma 4.

[33] Let $T:C\to C$ be a ζ-strictly pseudocontractive mapping. Let the real numbers $\gamma ,\delta \ge 0$ satisfy $(\gamma +\delta )\zeta \le \gamma$. Then $\parallel \gamma (x-y)+\delta (Tx-Ty)\parallel \le (\gamma +\delta )\parallel x-y\parallel ,\forall x,y\in C$.

Lemma 5.

[41] Let the Banach space X admit a weakly continuous duality mapping, the subset $C\subset X$ be nonempty, convex and closed, and the asymptotically nonexpansive mapping $T:C\to C$ have a fixed point, i.e., $\mathrm{Fix}(T)\ne \varnothing$. Then $I-T$ is demiclosed at zero, i.e., if the sequence $\left\{u_n\right\}\subset C$ satisfies $u_n\rightharpoonup u\in C$ and $(I-T)u_n\to 0$, then $(I-T)u=0$, where I is the identity mapping of X.

3. Main Results

Unless otherwise stated, we suppose the following.

• $T:H\to H$ is an asymptotically nonexpansive operator with $\left\{{\theta }_n\right\}$ and $S:H\to H$ is a $\zeta$-strictly pseudocontractive mapping.
• $A:H\to H$ is sequentially weakly continuous on C, L-Lipschitzian pseudomonotone on H, and $A(C)$ is bounded.
• $f:H\to C$ is a $\delta$-contraction with $\delta \in [0,\frac{1}{2})$.
• $\mathsf{\Omega }=\mathrm{Fix}(T)\cap \mathrm{Fix}(S)\cap \mathrm{VI}(C,A)\ne \varnothing$.
• $\left\{{\sigma }_n\right\}\subset [0,1]$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\}\subset (0,1)$ such that

  (i)

  ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}<\infty$ and ${\beta }_n+{\gamma }_n+{\delta }_n=1,\forall n\ge 1$;

  (ii)

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

  (iii)

  $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n<(1-2\delta ){\delta }_n,\forall n\ge 1$ and ${lim\; inf}_{n\to \infty }((1-2\delta ){\delta }_n-{\gamma }_n)>0$;

  (iv)

  ${lim\; sup}_{n\to \infty }{\beta }_n<1$, ${lim\; inf}_{n\to \infty }{\beta }_n>0$ and ${lim\; inf}_{n\to \infty }{\delta }_n>0$.

We first introduce an inertial-like subgradient extragradient algorithm (Algorithm 3) with line-search process as follows:

Algorithm 3: Inertial-like subgradient extragradient algorithm (I).

Initialization: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.   Iterative Steps: Compute $x_{n+1}$ in what follows:   Step 1. Put $w_n={\sigma }_n(x_n-x_{n-1})+T^n{x}_n$ and calculate $y_n=P_C(I-{\tau }_{n}A)w_n$, where ${\tau }_n$ is chosen to be the largest $\tau \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ such that $$\tau \parallel A{w}_n-A{y}_n\parallel \le \mu \parallel w_n-y_n\parallel .$$   Step 2. Calculate $z_n=(1-{\alpha }_n)P_{C_n}(w_n-{\tau }_{n}A{y}_n)+{\alpha }_{n}f(x_n)$ with $C_n:=\{x\in H:\langle w_n-{\tau }_{n}A{w}_n-y_n,x-y_n\rangle \le 0\}$.   Step 3. Calculate $$x_{n+1}={\gamma }_n{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)+{\delta }_{n}S{z}_n+{\beta }_n{T}^n{x}_n.$$   Again set $n:=n+1$ and return to Step 1.

Lemma 6.

In Step 1 of Algorithm 3, the Armijo-like search rule

$$\tau \parallel A{w}_n-A{y}_n\parallel \le \mu \parallel w_n-y_n\parallel$$

(3)

is well defined, and the inequality holds: $min\{\gamma ,\frac{\mu l}{L}\}\le {\tau }_n\le \gamma$.

Proof. 

Since A is L-Lipschitzian, we know that Equation (3) holds for all $\gamma l^m\le \frac{\mu }{L}$ and so ${\tau }_n$ is well defined. It is clear that ${\tau }_n\le \gamma$. Next we discuss two cases. In the case where ${\tau }_n=\gamma$, the inequality is valid. In the case where ${\tau }_n<\gamma$, from Equation (3) we derive $\parallel A{w}_n-A{P}_C(w_n-\frac{{\tau }_n}{l}A{w}_n)\parallel >\frac{\mu }{\frac{{\tau }_n}{l}}\parallel w_n-P_C(w_n-\frac{{\tau }_n}{l}A{w}_n)\parallel$. Also, since A is L-Lipschitzian, we get ${\tau }_n>\frac{\mu l}{L}$. Therefore the inequality is true. □

Lemma 7.

Assume that $\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\}$ are the sequences constructed by Algorithm 3. Then

$$\begin{array}{cc}\parallel z_n{-p\parallel }^2\hfill & \le [1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n-(1-{\alpha }_n)(1-\mu )\times \hfill \\ & \hspace{1em}\times [\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \forall p\in \mathsf{\Omega },\hfill \end{array}$$

(4)

where $u_n:=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$ and ${\mathsf{\Lambda }}_n:={\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]+{\theta }_n(2+{\theta }_n){\parallel x_n-p\parallel }^2$ for all $n\ge 1$.

Proof. 

We observe that

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

So, it follows that $\parallel w_n{-p\parallel }^2-\parallel u_n-w_n{\parallel }^2-2\langle u_n-p,{\tau }_{n}A{y}_n\rangle \ge {\parallel u_n-p\parallel }^2$. Since A is pseudomonotone, we deduce from Equation (3) that $\langle A{y}_n,y_n-p\rangle \ge 0$ and

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

(5)

Since $u_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$ with $C_n:=\{x\in H:0\ge \langle {\tau }_{n}A{w}_n-w_n+y_n,y_n-x\rangle \}$, we have $\langle u_n-y_n,w_n-{\tau }_{n}A{w}_n-y_n\rangle \le 0$, which together with Equation (3), implies that

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

Also, from $w_n={\sigma }_n(x_n-x_{n-1})+T^n{x}_n$ we get

$$\begin{array}{c}\parallel w_n{-p\parallel }^2={\parallel {\sigma }_n(x_n-x_{n-1})+T^n{x}_n-p\parallel }^2\hfill \\ \le [(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}{\parallel ]}^2\hfill \\ ={(1+{\theta }_n)}^2\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]\hfill \\ =\parallel x_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]\hfill \\ =\parallel x_n{-p\parallel }^2+{\mathsf{\Lambda }}_n,\hfill \end{array}$$

where ${\mathsf{\Lambda }}_n:={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]$. Therefore, substituting the last two inequalities for Equation (5), we infer that

$$\begin{array}{cc}\parallel u_n{-p\parallel }^2\hfill & \le \parallel w_n{-p\parallel }^2-(1-\mu )\parallel w_n-y_n{\parallel }^2-(1-\mu ){\parallel u_n-y_n\parallel }^2\hfill \\ & \le {\mathsf{\Lambda }}_n-(1-\mu )\parallel w_n-y_n{\parallel }^2-(1-\mu )\parallel u_n-y_n{\parallel }^2+{\parallel x_n-p\parallel }^2,\forall p\in \mathsf{\Omega }.\hfill \end{array}$$

(6)

In addition, from Algorithm 3 we have

$$z_n-p=(1-{\alpha }_n)(u_n-p)+{\alpha }_n(f-I)p+{\alpha }_n(f(x_n)-f(p)).$$

Since the function $h(t)=t^2,\forall t\in \mathbf{R}$ is convex, from Equation (6) we have

$$\begin{array}{c}\parallel z_n{-p\parallel }^2\hfill \\ \le [{\alpha }_n\delta \parallel x_n-p\parallel +(1-{\alpha }_n)\parallel u_n-p{\parallel ]}^2+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \hfill \\ \le {\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n)[\parallel x_n{-p\parallel }^2+{\mathsf{\Lambda }}_n-(1-\mu )\parallel w_n-y_n{\parallel }^2-(1-\mu )\parallel u_n-y_n{\parallel }^2]\hfill \\ \hspace{1em}+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \hfill \\ =[1-{\alpha }_n(1-\delta )]\parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]\hfill \\ \hspace{1em}+2{\alpha }_n\langle (f-I)p,z_n-p\rangle .\hfill \end{array}$$

This completes the proof. □

Lemma 8.

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

Proof. 

In terms of Algorithm 3, we deduce $w_n-x_n=T^n{x}_n-x_n+{\sigma }_n(x_n-x_{n-1}),\forall n\ge 1$, and hence $\parallel T^n{x}_n-x_n\parallel \le \parallel w_n-x_n\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel \le \parallel w_n-x_n\parallel +\parallel x_n-x_{n-1}\parallel$. Using the conditions $x_n-x_{n+1}\to 0$ and $w_n-x_n\to 0$, we get

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

(7)

Combining the assumptions $w_n-x_n\to 0$ and $w_n-z_n\to 0$ yields

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

Then, from Equation (4) it follows that

$$\begin{array}{c}(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]\hfill \\ \le [1-{\alpha }_n(1-\delta )]\parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n-{\parallel z_n-p\parallel }^2+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \hfill \\ \le \parallel x_n{-p\parallel }^2-\parallel z_n{-p\parallel }^2+{\mathsf{\Lambda }}_n+2{\alpha }_n\parallel (f-I)p\parallel \parallel z_n-p\parallel \hfill \\ \le \parallel x_n-z_n\parallel (\parallel x_n-p\parallel +\parallel z_n-p\parallel )+{\mathsf{\Lambda }}_n+2{\alpha }_n\parallel (f-I)p\parallel \parallel z_n-p\parallel ,\hfill \end{array}$$

where ${\mathsf{\Lambda }}_n:={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]$. Since ${\alpha }_n\to 0,{\mathsf{\Lambda }}_n\to 0$ and $x_n-z_n\to 0$, from the boundedness of $\left\{x_n\right\},\left\{z_n\right\}$ we get

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

Thus as $n\to \infty$,

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

Furthermore, using Algorithm 3 we have $x_{n+1}-z_n={\gamma }_n(u_n-z_n)+{\delta }_n(S{z}_n-z_n)+{\beta }_n(T^n{x}_n-z_n)$, which hence implies

$$\begin{array}{c}{\delta }_n\parallel S{z}_n-z_n\parallel =\parallel x_{n+1}-z_n-{\beta }_n(T^n{x}_n-z_n)-{\gamma }_n(u_n-z_n)\parallel \hfill \\ =\parallel x_{n+1}-x_n+{\delta }_n(x_n-z_n)-{\gamma }_n(u_n-x_n)-{\beta }_n(T^n{x}_n-x_n)\parallel \hfill \\ \le \parallel x_{n+1}-x_n\parallel +\parallel x_n-z_n\parallel +\parallel u_n-x_n\parallel +\parallel T^n{x}_n-x_n\parallel .\hfill \end{array}$$

Note that $x_n-x_{n+1}\to 0,z_n-x_n\to 0,x_n-u_n\to 0,x_n-T^n{x}_n\to 0$ and ${lim\; inf}_{n\to \infty }{\delta }_n>0$. So we obtain

$$\underset{n\to \infty }{lim}\parallel z_n-S{z}_n\parallel =0.$$

(8)

Noticing $y_n=P_C(I-{\tau }_{n}A)w_n$, we have $\langle x-y_n,w_n-{\tau }_{n}A{w}_n-y_n\rangle \le 0,\forall x\in C$, and hence

$$\langle w_n-y_n,x-y_n\rangle +{\tau }_n\langle A{w}_n,y_n-w_n\rangle \le {\tau }_n\langle A{w}_n,x-w_n\rangle ,\forall x\in C.$$

(9)

Since A is Lipschitzian, we infer from the boundedness of $\left\{w_{n_k}\right\}$ that $\left\{A{w}_{n_k}\right\}$ is bounded. From $w_n-y_n\to 0$, we get the boundedness of $\left\{y_{n_k}\right\}$. Taking into account ${\tau }_n\ge min\{\gamma ,\frac{\mu l}{L}\}$, from Equation (9) we have ${lim\; inf}_{k\to \infty }\langle A{w}_{n_k},x-w_{n_k}\rangle \ge 0,\forall x\in C$. Moreover, note that $\langle A{y}_n,x-y_n\rangle =\langle A{y}_n-A{w}_n,x-w_n\rangle +\langle A{w}_n,x-w_n\rangle +\langle A{y}_n,w_n-y_n\rangle$. Since A is L-Lipschitzian, from $w_n-y_n\to 0$ we get $A{w}_n-A{y}_n\to 0$. According to Equation (9) we have ${lim\; inf}_{k\to \infty }\langle A{y}_{n_k},x-y_{n_k}\rangle \ge 0,\forall x\in C$.

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

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

Hence from Equation (7) and the assumption $T^n{x}_n-T^{n+1}{x}_n\to 0$ we get

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

(10)

We now choose a sequence $\left\{{\epsilon }_k\right\}\subset (0,1)$ such that ${\epsilon }_k\downarrow 0$ as $k\to \infty$. For each $k\ge 1$, we denote by $m_k$ the smallest natural number satisfying

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

From the decreasing property of $\left\{{\epsilon }_k\right\}$, it is easy to see that $\left\{m_k\right\}$ is increasing. Considering that $\left\{y_{m_k}\right\}\subset C$ implies $A{y}_{m_k}\ne 0,\forall k\ge 1$, we put

$${\mu }_{m_k}=\frac{A{y}_{m_k}}{\parallel A{y}_{m_k}{\parallel }^2}.$$

So we have $\langle A{y}_{m_k},{\mu }_{m_k}\rangle =1,\forall k\ge 1$. Thus, from Equation (9), we have $\langle x+{\epsilon }_k{\mu }_{m_k}-y_{m_k},A{y}_{m_k}\rangle \ge 0,$ $\forall k\ge 1.$ Also, since A is pseudomonotone, we get

$$\langle A(x+{\epsilon }_k{\mu }_{m_k}),x+{\epsilon }_k{\mu }_{m_k}-y_{m_k}\rangle \ge 0,\forall k\ge 1.$$

Consequently,

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

(11)

We show ${lim}_{k\to \infty }{\epsilon }_k{\mu }_{m_k}=0$. In fact, since $w_{n_k}\rightharpoonup z$ and $w_n-y_n\to 0$, we get $y_{n_k}\rightharpoonup z$. So, $\left\{y_n\right\}\subset C$ guarantees $z\in C$. Also, since A is sequentially weakly continuous on C, we deduce that $A{y}_{n_k}\rightharpoonup Az$. So, we get $Az\ne 0$. It follows that $0<\parallel Az\parallel \le {lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel$. Since $\left\{y_{m_k}\right\}\subset \left\{y_{n_k}\right\}$ and ${\epsilon }_k\downarrow 0$ as $k\to \infty$, we obtain that

$$0\le \underset{k\to \infty }{lim\; sup}\parallel {\epsilon }_k{\mu }_{m_k}\parallel =\underset{k\to \infty }{lim\; sup}\frac{{\epsilon }_k}{\parallel A{y}_{m_k}\parallel }\le \frac{{lim\; sup}_{k\to \infty }{\epsilon }_k}{{lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel }=0.$$

Thus ${\epsilon }_k{\mu }_{m_k}\to 0$.

The last step is to show $z\in \mathsf{\Omega }$. Indeed, we have $x_{n_k}\rightharpoonup z$. From Equation (10) we also have $x_{n_k}-T{x}_{n_k}\to 0$. Note that Lemma 5 yields the demiclosedness of $I-T$ at zero. Thus $z\in \mathrm{Fix}(T)$. Moreover, since $w_n-z_n\to 0$ and $w_{n_k}\rightharpoonup z$, we have $z_{n_k}\rightharpoonup z$. From Equation (8) we get $z_{n_k}-S{z}_{n_k}\to 0$. By Lemma 5 we know that $I-S$ is demiclosed at zero, and hence we have $(I-S)z=0$, i.e., $z\in \mathrm{Fix}(S)$. In addition, taking $k\to \infty$, we infer that the right hand side of Equation (11) converges to zero by the Lipschitzian property of A, the boundedness of $\left\{y_{m_k}\right\},\left\{{\mu }_{m_k}\right\}$, and the limit ${lim}_{k\to \infty }{\epsilon }_k{\mu }_{m_k}=0$. Therefore, $\langle Ax,x-z\rangle ={lim\; inf}_{k\to \infty }\langle Ax,x-y_{m_k}\rangle \ge 0,\forall x\in C$. From Lemma 3 we get $z\in \mathrm{VI}(C,A)$, and hence $z\in \mathsf{\Omega }$. This completes the proof. □

Theorem 1.

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

$$x_n\to x^{*}\in \mathsf{\Omega }\iff \left\{\begin{array}{c}{x}_n-x_{n+1}\to 0,\hfill \\ x_n-T^n{x}_n\to 0,\hfill \\ \underset{n\ge 1}{sup}\parallel (T^n-f)x_n\parallel <\infty ,\hfill \end{array}\right.$$

where $x^{*}\in \mathsf{\Omega }$ is only a solution of the HVI: $\langle (f-I)x^{*},p-x^{*}\rangle \le 0,\forall p\in \mathsf{\Omega }$.

Proof. 

Without loss of generality, we may assume that $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$. We can claim that $P_{\mathsf{\Omega }}\circ f$ is a contractive map. Banach’s Contraction Principle ensures that it has a unique fixed point, i.e., $P_{\mathsf{\Omega }}f(x^{*})=x^{*}$. So, there exists a unique solution $x^{*}\in \mathsf{\Omega }$ to the HVI

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

(12)

It is clear that the necessity of the theorem is valid. In fact, if $x_n\to x^{*}\in \mathsf{\Omega }$, then as $n\to \infty$, we obtain that $\parallel x_n-x_{n+1}\parallel \to 0$, $\parallel x_n-T^n{x}_n\parallel \le \parallel x_n-x^{*}\parallel +\parallel x^{*}-T^n{x}_n\parallel \le (2+{\theta }_n)\parallel x_n-x^{*}\parallel \to 0$, and

$$\begin{array}{cc}{\displaystyle \underset{n\ge 1}{sup}}\parallel T^n{x}_n-f(x_n)\parallel \hfill & \le {\displaystyle \underset{n\ge 1}{sup}}(\parallel T^n{x}_n-x^{*}\parallel +\parallel x^{*}-f(x^{*})\parallel +\parallel f(x^{*})-f(x_n)\parallel )\hfill \\ & \le {\displaystyle \underset{n\ge 1}{sup}}[(1+{\theta }_n)\parallel x_n-x^{*}\parallel +\parallel x^{*}-f(x^{*})\parallel +\delta \parallel x^{*}-x_n\parallel ]\hfill \\ & \le {\displaystyle \underset{n\ge 1}{sup}}[(2+{\theta }_n)\parallel x_n-x^{*}\parallel +\parallel x^{*}-f(x^{*})\parallel ]<\infty .\hfill \end{array}$$

We now assume that ${lim}_{n\to \infty }(\parallel x_n-x_{n+1}\parallel +\parallel x_n-T^n{x}_n\parallel )=0$ and ${sup}_{n\ge 1}\parallel (T^n-f)x_n\parallel <\infty$, and prove the sufficiency by the following steps.

Step 1. We claim the boundedness of $\left\{x_n\right\}$. In fact, take a fixed $p\in \mathsf{\Omega }$ arbitrarily. From Equation (6) we get

$$\parallel w_n{-p\parallel }^2-(1-\mu )\parallel w_n-y_n{\parallel }^2-(1-\mu )\parallel u_n-y_n{\parallel }^2\ge {\parallel u_n-p\parallel }^2,$$

(13)

which hence yields

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

(14)

By the definition of $w_n$, we have

$$\begin{array}{cc}\parallel w_n-p\parallel \hfill & \le (1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel \hfill \\ & =(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel .\hfill \end{array}$$

(15)

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

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

(16)

From Equations (14)–(16), we obtain

$$\parallel u_n-p\parallel \le \parallel w_n-p\parallel \le (1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1,\forall n\ge 1.$$

(17)

Note that $A(C)$ is bounded, $y_n=P_C(I-{\tau }_n)A{w}_n,f(H)\subset C\subset C_n$ and $u_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$. Hence, we know that $\left\{A{y}_n\right\}$ is a bounded sequence. So, from ${sup}_{n\ge 1}\parallel (T^n-f)x_n\parallel <\infty$, it follows that

$$\begin{array}{cc}\parallel u_n-f(x_n)\parallel \hfill & =\parallel P_{C_n}(w_n-{\tau }_{n}A{y}_n)-P_{C_n}f(x_n)\parallel \le \parallel w_n-{\tau }_{n}A{y}_n-f(x_n)\parallel \hfill \\ & \le \parallel w_n-T^n{x}_n\parallel +\parallel T^n{x}_n-f(x_n)\parallel +{\tau }_n\parallel A{y}_n\parallel \hfill \\ & \le \parallel x_n-x_{n-1}\parallel +\parallel (T^n-f)x_n\parallel +\gamma \parallel A{y}_n\parallel \le M_0,\hfill \end{array}$$

where ${sup}_{n\ge 1}(\parallel x_n-x_{n-1}\parallel +\parallel (T^n-f)x_n\parallel +\gamma \parallel A{y}_n\parallel )\le M_0$ for some $M_0>0$. Taking into account ${lim}_{n\to \infty }\frac{{\theta }_n(2+{\theta }_n)}{{\alpha }_n(1-{\beta }_n)}=0$, we know that $\exists n_0\ge 1$ such that

$${\theta }_n(2+{\theta }_n)\le \frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}(\le \frac{{\alpha }_n(1-\delta )}{2}),\forall n\ge n_0.$$

So, from Algorithm 3 and Equation (17) it follows that for all $n\ge n_0$,

$$\begin{array}{cc}\parallel z_n-p\parallel \hfill & \le {\alpha }_n\delta \parallel x_n-p\parallel +(1-{\alpha }_n)\parallel u_n-p\parallel +{\alpha }_n\parallel (f-I)p\parallel \hfill \\ & \le [1-{\alpha }_n(1-\delta )+{\theta }_n]\parallel x_n-p\parallel +{\alpha }_n(M_1+\parallel (f-I)p\parallel )\hfill \\ & \le [1-\frac{{\alpha }_n(1-\delta )}{2}]\parallel x_n-p\parallel +{\alpha }_n(M_1+\parallel (f-I)p\parallel ),\hfill \end{array}$$

which together with Lemma 4 and $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, implies that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}-p\parallel =\parallel {\beta }_n(T^n{x}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel \hfill \\ \le {\beta }_n(1+{\theta }_n)\parallel x_n-p\parallel +(1-{\beta }_n)\parallel z_n-p\parallel +{\gamma }_n{\alpha }_n\parallel u_n-f(x_n)\parallel \hfill \\ \le {\beta }_n(1+{\theta }_n)\parallel x_n-p\parallel +(1-{\beta }_n)[(1-\frac{{\alpha }_n(1-\delta )}{2})\parallel x_n-p\parallel +{\alpha }_n(M_0+M_1+\parallel (f-I)p\parallel )]\hfill \\ \le [1-\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}+{\beta }_n\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}]\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)(M_0+M_1+\parallel (f-I)p\parallel )\hfill \\ =[1-\frac{{\alpha }_n{(1-{\beta }_n)}^2(1-\delta )}{2}]\parallel x_n-p\parallel +\frac{{\alpha }_n{(1-{\beta }_n)}^2(1-\delta )}{2}·\frac{2(M_0+M_1+\parallel (f-I)p\parallel )}{(1-\delta )(1-{\beta }_n)}.\hfill \end{array}$$

By induction, we obtain $\parallel x_n-p\parallel \le max\{\parallel x_{n_0}-p\parallel ,\frac{2(M_0+M_1+\parallel (f-I)p\parallel )}{(1-\delta )(1-b)}\},\forall n\ge n_0$. Therefore, we derive the boundedness of $\left\{x_n\right\}$ and hence the one of sequences $\left\{u_n\right\},\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{f(x_n)\right\},\left\{S{z}_n\right\},\left\{T^n{x}_n\right\}$.

Step 2. We claim that $\exists M_4>0$ s.t.

$$(1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\alpha }_n{M}_4,\forall n\ge n_0.$$

In fact, using Lemmas 4 and 7 and the convexity of ${\parallel ·\parallel }^2$, we get

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2={\parallel {\beta }_n(T^n{x}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel }^2\hfill \\ \le {\beta }_n\parallel T^n{x}_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)]\parallel }^2\hfill \\ \hspace{1em}+2(1-{\beta }_n){\alpha }_n\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\beta }_n\parallel T^n{x}_n{-p\parallel }^2+(1-{\beta }_n)\{[1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n\hfill \\ \hspace{1em}-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \}\hfill \\ \hspace{1em}+2(1-{\beta }_n){\alpha }_n\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\beta }_n\parallel T^n{x}_n{-p\parallel }^2+(1-{\beta }_n)\{[1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n\hfill \\ \hspace{1em}-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_2\},\hfill \end{array}$$

(18)

where

$${\mathsf{\Lambda }}_n:={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ],$$

and

$$\underset{n\ge 1}{sup}2(\parallel (f-I)p\parallel \parallel z_n-p\parallel +\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel )\le M_2$$

for some $M_2>0$. Also, from Equation (16) we have

$$\begin{array}{cc}{\mathsf{\Lambda }}_n\hfill & ={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]\hfill \\ & \le {\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\alpha }_n{M}_1\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1]\hfill \\ & ={\alpha }_n\{\frac{{\theta }_n}{{\alpha }_n}(2+{\theta }_n)\parallel x_n{-p\parallel }^2+M_1\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1\left]\right\}\le {\alpha }_n{M}_3,\hfill \end{array}$$

(19)

where

$$\underset{n\ge 1}{sup}\{\frac{{\theta }_n}{{\alpha }_n}(2+{\theta }_n)\parallel x_n{-p\parallel }^2+M_1\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1\left]\right\}\le M_3$$

for some $M_3>0$. Note that

$${\theta }_n(2+{\theta }_n)\le \frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2},\forall n\ge n_0.$$

Substituting Equation (19) for Equation (18), we obtain that for all $n\ge n_0$,

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le {\beta }_n{(1+{\theta }_n)}^2\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{[1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\alpha }_n{M}_3\hfill \\ & \hspace{1em}-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_2\}\hfill \\ & \le [1-\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}]{\parallel x_n-p\parallel }^2+{\alpha }_n{M}_3\hfill \\ & \hspace{1em}-(1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_2\hfill \\ & \le \parallel x_n{-p\parallel }^2-(1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_4,\hfill \end{array}$$

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

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

(20)

Step 3. We claim that $\exists M>0$ s.t.

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ \hspace{1em}+\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f(p)-p,x_n-p\rangle \hfill \\ \hspace{1em}+\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}(\frac{{\theta }_n}{{\alpha }_n}·\frac{2M^2}{1-b}+\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M)\}.\hfill \end{array}$$

In fact, we get

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\le [(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}{\parallel ]}^2\hfill \\ =\parallel x_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]\hfill \\ \le \parallel x_n{-p\parallel }^2+{\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,\hfill \end{array}$$

(21)

where $M\ge {sup}_{n\ge 1}\{(1+{\theta }_n)\parallel x_n-p\parallel ,{\sigma }_n\parallel x_n-x_{n-1}\parallel \}$ for some $M>0$. From Algorithm 3 and the convexity of ${\parallel ·\parallel }^2$, we have

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2={\parallel {\beta }_n(T^n{x}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel }^2\hfill \\ \le \parallel {\beta }_n(T^n{x}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p){\parallel }^2+2{\gamma }_n{\alpha }_n\langle u_n-f(x_n),x_{n+1}-p\rangle \hfill \\ \le {\beta }_n\parallel T^n{x}_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)]\parallel }^2\hfill \\ \hspace{1em}+2{\gamma }_n{\alpha }_n\langle u_n-p,x_{n+1}-p\rangle +2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle ,\hfill \end{array}$$

which together with Lemma 4, leads to

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\le {\beta }_n{(1+{\theta }_n)}^2\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\parallel z_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel u_n-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \hspace{1em}+2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle \hfill \\ \le {\beta }_n{(1+{\theta }_n)}^2\parallel x_n{-p\parallel }^2+(1-{\beta }_n)[(1-{\alpha }_n)\parallel u_n-p{\parallel }^2+2{\alpha }_n\langle f(x_n)-p,z_n-p\rangle ]\hfill \\ \hspace{1em}+{\gamma }_n{\alpha }_n(\parallel u_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)+2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle .\hfill \end{array}$$

From Equations (17) and (21) we know that

$$\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M.$$

Hence, we have

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n-p{\parallel }^2+{\beta }_n{\theta }_{n}2M^2+(1-{\beta }_n)(1-{\alpha }_n)({\theta }_{n}2M^2\hfill \\ & \hspace{1em}+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)+2{\alpha }_n{\delta }_n\langle f(x_n)-p,z_n-p\rangle +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2\hfill \\ & \hspace{1em}+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n){\alpha }_n({\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)\hfill \\ & \hspace{1em}+2{\gamma }_n{\alpha }_n\langle f(x_n)-p,z_n-x_{n+1}\rangle \hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & \hspace{1em}+2{\alpha }_n{\delta }_n\delta \parallel x_n{-p\parallel }^2+2{\alpha }_n{\delta }_n\langle f(p)-p,x_n-p\rangle +2{\alpha }_n{\delta }_n\parallel f(x_n)-p\parallel \parallel z_n-x_n\parallel \hfill \\ & \hspace{1em}+{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n)(\frac{{\theta }_{n}2M^2}{1-{\beta }_n}+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M),\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ \hspace{1em}+\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f(p)-p,x_n-p\rangle \hfill \\ \hspace{1em}+\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}(\frac{{\theta }_n}{{\alpha }_n}·\frac{2M^2}{1-b}+\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M)\}.\hfill \end{array}$$

(22)

Step 4. We claim the strong convergence of $\left\{x_n\right\}$ to a unique solution $x^{*}\in \mathsf{\Omega }$ to the HVI Equation (12). In fact, setting $p=x^{*}$, from Equation (22) we know that

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n-x^{*}{\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-x^{*}\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ \hspace{1em}+\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-x^{*}\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f(x^{*})-x^{*},x_n-x^{*}\rangle \hfill \\ \hspace{1em}+\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}(\frac{{\theta }_n}{{\alpha }_n}·\frac{2M^2}{1-b}+\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M)\}.\hfill \end{array}$$

According to Lemma 4, it is sufficient to prove that ${lim\; sup}_{n\to \infty }\langle (f-I)x^{*},x_n-x^{*}\rangle \le 0$. Since $x_n-x_{n+1}\to 0,{\alpha }_n\to 0$ and $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$, from Equation (20) we get

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}(1-{\alpha }_n)(1-b)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]\hfill \\ \le {\displaystyle \underset{n\to \infty }{lim\; sup}}[\parallel x_n{-p\parallel }^2-\parallel x_{n+1}-p{\parallel }^2+{\alpha }_n{M}_4]\hfill \\ \le {\displaystyle \underset{n\to \infty }{lim\; sup}}(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel =0,\hfill \end{array}$$

which hence leads to

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

(23)

Obviously, the assumptions $\parallel x_n-x_{n+1}\parallel \to 0$ and $\parallel x_n-T^n{x}_n\parallel \to 0$ guarantee that $\parallel w_n-x_n\parallel \le \parallel T^n{x}_n-x_n\parallel +\parallel x_n-x_{n-1}\parallel \to 0(n\to \infty )$. Thus,

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

Since $z_n=(1-{\alpha }_n)u_n+{\alpha }_{n}f(x_n)$ with $u_n:=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$, from Equation (23) and the boundedness of $\left\{x_n\right\},\left\{u_n\right\}$, we get

$$\begin{array}{c}\parallel z_n-y_n\parallel \le {\alpha }_n(\parallel f(x_n)\parallel +\parallel u_n\parallel )+\parallel u_n-y_n\parallel \to 0,(n\to \infty ),\hfill \end{array}$$

(24)

and hence

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

Obviously, combining Equations (23) and (24) guarantees that

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

Since $\left\{x_n\right\}$ is bounded, we know that $\exists \left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ s.t.

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

(25)

Next, we may suppose that $x_{n_k}\rightharpoonup \tilde{x}$. Hence from Equation (25) we get

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

(26)

From $w_n-x_n\to 0$ and $x_{n_k}\rightharpoonup \tilde{x}$ it follows that $w_{n_k}\rightharpoonup \tilde{x}$.

Since $T^n{x}_n-T^{n+1}{x}_n\to 0,x_n-x_{n+1}\to 0,w_n-x_n\to 0,w_n-z_n\to 0$ and $w_{n_k}\rightharpoonup \tilde{x}$, from Lemma 8 we conclude that $\tilde{x}\in \mathsf{\Omega }$. Therefore, from Equations (12) and (26) we infer that

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

Note that

$$\sum _{n=0}^{\infty }\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n=\infty .$$

It is clear that

$$\begin{array}{c}{\displaystyle \underset{n\to \infty }{lim\; sup}}\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-x^{*}\parallel \parallel z_n-x_{n+1}\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-x^{*}\parallel \parallel z_n-x_n\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f(x^{*})-x^{*},x_n-x^{*}\rangle +\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}(\frac{{\theta }_n}{{\alpha }_n}·\frac{2M^2}{1-b}+\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M)\}\le 0.\hfill \end{array}$$

Consequently, all conditions of Lemma 4 are satisfied, and hence we immediately deduce that $x_n\to x^{*}$. This completes the proof. □

Next, we introduce another inertial-like subgradient extragradient algorithm (Algorithm 4) with line-search process as the following.

It is remarkable that Lemmas 6–8 are still valid for Algorithm 4.

Algorithm 4: Inertial-like subgradient extragradient algorithm (II).

Initialization: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.   Iterative Steps: Compute $x_{n+1}$ in what follows:   Step 1. Put $w_n={\sigma }_n(x_n-x_{n-1})+T^n{x}_n$ and calculate $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$, where ${\tau }_n$ is chosen to be the largest $\tau \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ such that $$\tau \parallel A{w}_n-A{y}_n\parallel \le \mu \parallel w_n-y_n\parallel .$$   Step 2. Calculate $z_n=(1-{\alpha }_n)P_{C_n}(w_n-{\tau }_{n}A{y}_n)+{\alpha }_{n}f(x_n)$ with $C_n:=\{x\in H:\langle w_n-{\tau }_{n}A{w}_n-y_n,x-y_n\rangle \le 0\}$.   Step 3. Calculate $$x_{n+1}={\gamma }_n{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)+{\delta }_{n}S{z}_n+{\beta }_n{T}^n{w}_n.$$   Again set $n:=n+1$ and return to Step 1.

Theorem 2.

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

$$x_n\to x^{*}\in \mathsf{\Omega }\iff \left\{\begin{array}{c}{x}_n-x_{n+1}\to 0,\hfill \\ x_n-T^n{x}_n\to 0,\hfill \\ {sup}_{n\ge 1}\parallel (T^n-f)x_n\parallel <\infty ,\hfill \end{array}\right.$$

where $x^{*}\in \mathsf{\Omega }$ is only a solution of the HVI: $\langle (I-f)x^{*},p-x^{*}\rangle \ge 0,\forall p\in \mathsf{\Omega }$.

Proof. 

Using the same reasoning as in the proof of Theorem 1, we know that there is only a solution $x^{*}\in \mathsf{\Omega }$ of Equation (12), and that the necessity of the theorem is true.

We claim the sufficiency of the theorem below. For the purpose, we suppose that ${lim}_{n\to \infty }(\parallel x_n-x_{n+1}\parallel +\parallel x_n-T^n{x}_n\parallel )=0$ and ${sup}_{n\ge 1}\parallel (T^n-f)x_n\parallel <\infty$. Then we prove the sufficiency by the following steps.

Step 1. We claim the boundedness of $\left\{x_n\right\}$. In fact, using the same reasoning as in Step 1 of the proof of Theorem 1, we obtain that inequalities Equations (13)–(17) hold. Noticing ${lim}_{n\to \infty }\frac{{\theta }_n(2+{\theta }_n)}{{\alpha }_n(1-{\beta }_n)}=0$, we infer that $\exists n_0\ge 1$ s.t.

$${\theta }_n(2+{\theta }_n)\le \frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}(\le \frac{{\alpha }_n(1-\delta )}{2}),\forall n\ge n_0.$$

So, from Algorithm 4 and Equation (17) it follows that for all $n\ge n_0$,

$$\begin{array}{cc}\parallel z_n-p\parallel \hfill & \le {\alpha }_n\delta \parallel x_n-p\parallel +(1-{\alpha }_n)[(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1]+{\alpha }_n\parallel (f-I)p\parallel \hfill \\ & \le [1-\frac{{\alpha }_n(1-\delta )}{2}]\parallel x_n-p\parallel +{\alpha }_n(M_1+\parallel (f-I)p\parallel ),\hfill \end{array}$$

which together with Lemma 4 and $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, implies that for all $n\ge n_0$,

$$\begin{array}{c}\parallel x_{n+1}-p\parallel =\parallel {\beta }_n(T^n{w}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel \hfill \\ \le {\beta }_n(1+{\theta }_n)\parallel w_n-p\parallel +(1-{\beta }_n)\parallel z_n-p\parallel +{\gamma }_n{\alpha }_n\parallel u_n-f(x_n)\parallel \hfill \\ \le [1-\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}+{\beta }_n{\theta }_n(2+{\theta }_n)]\parallel x_n-p\parallel +{\beta }_n(1+{\theta }_n){\alpha }_n{M}_1\hfill \\ \hspace{1em}+{\alpha }_n(1-{\beta }_n)(M_0+M_1+\parallel (f-I)p\parallel )\hfill \\ \le [1-\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}+{\beta }_n\frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2}]\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)(M_0+M_1\frac{1+{\theta }_n}{1-{\beta }_n}+\parallel (f-I)p\parallel )\hfill \\ =[1-\frac{{\alpha }_n{(1-{\beta }_n)}^2(1-\delta )}{2}]\parallel x_n-p\parallel +\frac{{\alpha }_n{(1-{\beta }_n)}^2(1-\delta )}{2}·\frac{2(M_0+M_1\frac{1+{\theta }_n}{1-{\beta }_n}+\parallel (f-I)p\parallel )}{(1-\delta )(1-{\beta }_n)}.\hfill \end{array}$$

Hence,

$$\parallel x_n-p\parallel \le max\{\parallel x_{n_0}-p\parallel ,\frac{2(M_0+M_1\frac{2}{1-b}+\parallel (f-I)p\parallel )}{(1-\delta )(1-b)}\},\forall n\ge n_0.$$

Thus, sequence $\left\{x_n\right\}$ is bounded.

Step 2. We claim that for all $n\ge n_0$,

$$\parallel x_n{-p\parallel }^2-\parallel x_{n+1}{-p\parallel }^2+{\alpha }_n{M}_4\ge (1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2],$$

with constant $M_4>0$. Indeed, utilizing Lemmas 4 and 7 and the convexity of ${\parallel ·\parallel }^2$, one reaches

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2={\parallel {\beta }_n(T^n{w}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel }^2\hfill \\ \le {\beta }_n\parallel T^n{w}_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)]\parallel }^2\hfill \\ \hspace{1em}+2(1-{\beta }_n){\alpha }_n\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\beta }_n{(1+{\theta }_n)}^2\parallel w_n{-p\parallel }^2+(1-{\beta }_n)\{[1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n\hfill \\ \hspace{1em}-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+2{\alpha }_n\langle (f-I)p,z_n-p\rangle \}\hfill \\ \hspace{1em}+2(1-{\beta }_n){\alpha }_n\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel \hfill \\ \le {\beta }_n{(1+{\theta }_n)}^2(\parallel x_n{-p\parallel }^2+{\mathsf{\Lambda }}_n)+(1-{\beta }_n)\{[1-{\alpha }_n(1-\delta )]{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\mathsf{\Lambda }}_n\hfill \\ \hspace{1em}-(1-{\alpha }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_2\},\hfill \end{array}$$

(27)

where ${\mathsf{\Lambda }}_n:={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]$, and ${sup}_{n\ge 1}2(\parallel (f-I)p\parallel \parallel z_n-p\parallel +\parallel u_n-f(x_n)\parallel \parallel x_{n+1}-p\parallel )\le M_2$ for some $M_2>0$. Also, from Equation (16) we have

$$\begin{array}{cc}{\mathsf{\Lambda }}_n\hfill & ={\theta }_n(2+{\theta }_n)\parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel ]\hfill \\ & \le {\alpha }_n\{\frac{{\theta }_n}{{\alpha }_n}(2+{\theta }_n)\parallel x_n{-p\parallel }^2+M_1\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1\left]\right\}\le {\alpha }_n{M}_3,\hfill \end{array}$$

(28)

where ${sup}_{n\ge 1}\{\frac{{\theta }_n}{{\alpha }_n}(2+{\theta }_n)\parallel x_n{-p\parallel }^2+M_1\parallel [2(1+{\theta }_n)\parallel x_n-p\parallel +{\alpha }_n{M}_1\left]\right\}\le M_3$ for some $M_3>0$. Note that ${\theta }_n(2+{\theta }_n)\le \frac{{\alpha }_n(1-{\beta }_n)(1-\delta )}{2},\forall n\ge n_0$. Substituting Equation (28) for Equation (27), we obtain that for all $n\ge n_0$,

$$\begin{array}{ll}\parallel x_{n+1}{-p\parallel }^2& \le [1-{\alpha }_n(1-{\beta }_n)(1-\delta )+{\beta }_n{\theta }_n(2+{\theta }_n)]{\parallel x_n-p\parallel }^2+{\beta }_n{(1+{\theta }_n)}^2{\alpha }_n{M}_3\\ & \hspace{1em}+(1-{\beta }_n)(1-{\alpha }_n){\alpha }_n{M}_3-(1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2\\ & \hspace{1em}+\parallel u_n-y_n{\parallel }^2]+(1-{\beta }_n){\alpha }_n{M}_2\\ & \le \parallel x_n{-p\parallel }^2-(1-{\alpha }_n)(1-{\beta }_n)(1-\mu )[\parallel w_n-y_n{\parallel }^2+\parallel u_n-y_n{\parallel }^2]+{\alpha }_n{M}_4,\end{array}$$

where $M_4:=M_2+4M_3$. This immediately implies that for all $n\ge n_0$,

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

Step 3. We claim that $\exists M>0$ s.t.

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ \hspace{1em}+\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f(x_n)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f(p)-p,x_n-p\rangle \hfill \\ \hspace{1em}+\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}(\frac{{\theta }_n}{{\alpha }_n}·\frac{2M^2(1+b{(1+{\theta }_n)}^2)}{1-b}+\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \frac{3M(1+b{\theta }_n(2+{\theta }_n))}{1-b})\}.\hfill \end{array}$$

(29)

In fact, we get

$$\parallel w_n{-p\parallel }^2\le [(1+{\theta }_n)\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}{\parallel ]}^2\le \parallel x_n{-p\parallel }^2+{\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,$$

(30)

where $\exists M>0$ s.t. ${sup}_{n\ge 1}\{(1+{\theta }_n)\parallel x_n-p\parallel ,{\sigma }_n\parallel x_n-x_{n-1}\parallel \}\le M$. From Algorithm 4 and the convexity of ${\parallel ·\parallel }^2$, we have

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2={\parallel {\beta }_n(T^n{w}_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)+{\gamma }_n(u_n-z_n)\parallel }^2\hfill \\ \le {\beta }_n\parallel T^n{w}_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(S{z}_n-p)]\parallel }^2\hfill \\ \hspace{1em}+2{\gamma }_n{\alpha }_n\langle u_n-p,x_{n+1}-p\rangle +2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle ,\hfill \end{array}$$

which together with Lemma 4, leads to

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\le {\beta }_n{(1+{\theta }_n)}^2\parallel w_n{-p\parallel }^2+(1-{\beta }_n)\parallel z_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel u_n-p\parallel \parallel x_{n+1}-p\parallel \hfill \\ \hspace{1em}+2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle \hfill \\ \le {\beta }_n{(1+{\theta }_n)}^2(\parallel x_n{-p\parallel }^2+{\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)+(1-{\beta }_n)[(1-{\alpha }_n){\parallel u_n-p\parallel }^2\hfill \\ \hspace{1em}+2{\alpha }_n\langle f(x_n)-p,z_n-p\rangle ]+{\gamma }_n{\alpha }_n(\parallel u_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)\hfill \\ \hspace{1em}+2{\gamma }_n{\alpha }_n\langle p-f(x_n),x_{n+1}-p\rangle .\hfill \end{array}$$

By Step 3 of Algorithm 4, and from Equation (30) we know that $\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M$. Hence, we have

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n-p{\parallel }^2+{\beta }_n{\theta }_{n}2M^2+(1-{\beta }_n)(1-{\alpha }_n)({\theta }_{n}2M^2\hfill \\ & \hspace{1em}+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)+2{\alpha }_n{\delta }_n\langle f(x_n)-p,z_n-p\rangle +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2\hfill \\ & \hspace{1em}+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n){\alpha }_n({\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)\hfill \\ & \hspace{1em}+2{\gamma }_n{\alpha }_n\langle f(x_n)-p,z_n-x_{n+1}\rangle +{\beta }_n{(1+{\theta }_n)}^2({\theta }_{n}2M^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M)\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & \hspace{1em}+2{\alpha }_n{\delta }_n\langle f(x_n)-p,x_n-p\rangle +2{\alpha }_n{\delta }_n\langle f(x_n)-p,z_n-x_n\rangle \hfill \\ & \hspace{1em}+{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n)[{\theta }_n\frac{2M^2(1+{\beta }_n{(1+{\theta }_n)}^2)}{1-{\beta }_n}\hfill \\ & \hspace{1em}+{\sigma }_n\parallel x_n-x_{n-1}\parallel \frac{3M(1+{\beta }_n{\theta }_n(2+{\theta }_n))}{1-{\beta }_n}]\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f(x_n)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & \hspace{1em}+2{\alpha }_n{\delta }_n\delta \parallel x_n{-p\parallel }^2+2{\alpha }_n{\delta }_n\langle f(p)-p,x_n-p\rangle +2{\alpha }_n{\delta }_n\parallel f(x_n)-p\parallel \parallel z_n-x_n\parallel \hfill \\ & \hspace{1em}+{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n)[{\theta }_n\frac{2M^2(1+b{(1+{\theta }_n)}^2)}{1-b}\hfill \\ & \hspace{1em}+{\sigma }_n\parallel x_n-x_{n-1}\parallel \frac{3M(1+b{\theta }_n(2+{\theta }_n))}{1-b}],\hfill \end{array}$$

which immediately yields Equation (29).

Step 4. We claim the strong convergence of $\left\{x_n\right\}$ to a unique solution $x^{*}\in \mathsf{\Omega }$ of HVI Equation (12). In fact, using the same reasoning as in Step 4 of the proof of Theorem 1, we derive the desired conclusion. This completes the proof. □

Next, we shall show how to solve the VIP and CFPP in the following illustrating example.

The initial point $x_0=x_1$ is randomly chosen in $\mathbf{R}=(-\infty ,\infty )$. Take $f(x)=\frac{1}{4}sinx,\gamma =l=\mu =\frac{1}{2},{\sigma }_n={\alpha }_n=\frac{1}{n+1},{\beta }_n=\frac{1}{3},{\gamma }_n=\frac{1}{6}$, and ${\delta }_n=\frac{1}{2}$. Then we know that $\delta =\frac{1}{4}$ and $f(\mathbf{R})\subset [-\frac{1}{4},\frac{1}{4}]$.

We first provide an example of Lipschitz continuous and pseudomonotone mapping A, asymptotically nonexpansive mapping T and strictly pseudocontractive mapping S with $\mathsf{\Omega }=\mathrm{Fix}(T)\cap \mathrm{Fix}(S)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-1.5,1]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. Let $A,T,S:H\to H$ be defined as $Ax:=\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|}$, $Tx:=\frac{4}{5}sinx$ and $Sx:=\frac{1}{3}x+\frac{1}{2}sinx$ for all $x\in H$. Now, we first show that A is pseudomonotone and Lipschitz continuous with $L=2$ such that $A(C)$ is bounded. Indeed, it is clear that $A(C)$ is bounded. Moreover, for all $x,y\in H$ we have

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

This implies that A is Lipschitz continuous with $L=2$. Next, we show 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 ${\theta }_n={(\frac{4}{5})}^n,\forall n\ge 1$, such that $\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, we observe that

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

and

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

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

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

Moreover, it is readily seen that ${sup}_{n\ge 1}|(T^n-f)x_n|={sup}_{n\ge 1}|\frac{4}{5}sin(T^{n-1}{x}_n)-\frac{1}{4}sin{x}_n|\le \frac{21}{20}<\infty$. In addition, it is clear that S is strictly pseudocontractive with constant $\zeta =\frac{1}{4}$. Indeed, we observe that for all $x,y\in H$,

$${\parallel Sx-Sy\parallel }^2\le [\frac{1}{3}\parallel x-y\parallel +\frac{1}{2}{\parallel sinx-siny\parallel ]}^2\le {\parallel x-y\parallel }^2+\frac{1}{4}{\parallel (I-S)x-(I-S)y\parallel }^2.$$

It is clear that $({\gamma }_n+{\delta }_n)\zeta =(\frac{1}{6}+\frac{1}{2})·\frac{1}{4}\le \frac{1}{6}={\gamma }_n<(1-2\delta ){\delta }_n=(1-2·\frac{1}{4})·\frac{1}{2}=\frac{1}{4}$ for all $n\ge 1$. Therefore, $\mathsf{\Omega }=\mathrm{Fix}(T)\cap \mathrm{Fix}(S)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 3 can be rewritten as follows:

$$\left\{\begin{array}{c}{w}_n=T^n{x}_n+\frac{1}{n+1}(x_n-x_{n-1}),\hfill \\ y_n=P_C(w_n-{\tau }_{n}A{w}_n),\hfill \\ z_n=\frac{1}{n+1}f(x_n)+\frac{n}{n+1}{P}_{C_n}(w_n-{\tau }_{n}A{y}_n),\hfill \\ x_{n+1}=\frac{1}{3}{T}^n{x}_n+\frac{1}{6}{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)+\frac{1}{2}S{z}_n,\forall n\ge 1,\hfill \end{array}\right.$$

where $C_n$ and ${\tau }_n$ are picked up as in Algorithm 3. Thus, by Theorem 1, we know that $\left\{x_n\right\}$ converges to $0\in \mathsf{\Omega }$ if and only if $|x_n-x_{n+1}|+|x_n-T^n{x}_n|\to 0,(n\to \infty )$.

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

$$\left\{\begin{array}{c}{w}_n=T^n{x}_n+\frac{1}{n+1}(x_n-x_{n-1}),\hfill \\ y_n=P_C(w_n-{\tau }_{n}A{w}_n),\hfill \\ z_n=\frac{1}{n+1}f(x_n)+\frac{n}{n+1}{P}_{C_n}(w_n-{\tau }_{n}A{y}_n),\hfill \\ x_{n+1}=\frac{1}{3}{T}^n{w}_n+\frac{1}{6}{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)+\frac{1}{2}S{z}_n,\forall n\ge 1,\hfill \end{array}\right.$$

where $C_n$ and ${\tau }_n$ are picked up as in Algorithm 4. Thus, by Theorem 2, we know that $\left\{x_n\right\}$ converges to $0\in \mathsf{\Omega }$ if and only if $|x_n-x_{n+1}|+|x_n-T^n{x}_n|\to 0,(n\to \infty )$.