On Strengthened Inertial-Type Subgradient Extragradient Rule with Adaptive Step Sizes for Variational Inequalities and Fixed Points of Asymptotically Nonexpansive Mappings

Abstract

In a real Hilbert space, let the VIP denote a pseudomonotone variational inequality problem with Lipschitz continuity operator, and let the CFPP indicate a common fixed-point problem of finitely many nonexpansive mappings and an asymptotically nonexpansive mapping. On the basis of the Mann iteration method, the viscosity approximation method and the hybrid steepest-descent method, we propose and analyze two strengthened inertial-type subgradient extragradient rules with adaptive step sizes for solving the VIP and CFPP. With the help of suitable restrictions, we show the strong convergence of the suggested rules to a common solution of the VIP and CFPP, which is the unique solution of a hierarchical variational inequality (HVI).

1. Introduction

Let ($H,\langle ·,·\rangle$) be a real Hilbert space with the norm $\parallel ·\parallel$. Given a convex and closed set $C\ne \varnothing$ in H, we denote by $P_C$ the metric projection from H onto C. Suppose that $T:C\to H$ is a nonlinear operator on C. We use the notations $\mathrm{Fix}\left(T\right)$, $\mathbf{R},\to$ and ⇀ to indicate the fixed point set of T, the set of all real numbers, the strong convergence and the weak convergence, respectively. An operator $T:C\to C$ is referred to as being asymptotically nonexpansive if $\exists \left\{{\theta }_k\right\}\subset [0,+\infty )$ s.t. ${lim}_{k\to \infty }{\theta }_k=0$ and

$$\parallel T^{k}u-T^{k}v\parallel \le (1+{\theta }_k)\parallel u-v\parallel \forall k\ge 1,u,v\in C.$$

(1)

In particular, whenever ${\theta }_n=0\forall n\ge 1$, T is referred to as being nonexpansive.

Give an operator $A:H\to H$. The classical variational inequality problem (VIP) is the one of finding $u^{*}\in C$ s.t. $\langle A{u}^{*},u-u^{*}\rangle \ge 0\forall u\in C$. The solution set of the VIP is denoted by VI($C,A$). Currently, one of the most popular methods for solving the VIP is the extragradient method proposed by Korpelevich [1] in 1976, i.e., for any starting point $u_0\in C$, $\left\{u_k\right\}$ the sequence below is constructed

$$\left\{\begin{array}{cc}& v_k=P_C(u_k-\ell A{u}_k),\hfill \\ \hfill & u_{k+1}=P_C(u_k-\ell A{v}_k)\forall k\ge 0,\hfill \end{array}\right.$$

(2)

with $\ell \in (0,\frac{1}{L})$. In case $\mathrm{VI}(C,A)\ne \varnothing$, the sequence $\left\{u_k\right\}$ constructed by (2) converges weakly to an element in $\mathrm{VI}(C,A)$. The literature on the VIP is numerous and Korpelevich’s extragradient method has received great attention given by many authors, who ameliorated it in various aspects; see, e.g., [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26] and references therein, to name but a few.

It is clear that, in the extragradient method, one has to calculate two projections onto C per one iteration. Without doubt, the projection onto a closed convex set C is closely related to a minimum distance problem. In case C there is a general convex and closed set; this might require an outrageous amount of calculation time. In 2011, Censor et al. [6] improved Korpelevich’s extragradient method and first invented the subgradient extragradient method, in which one uses a projection onto a half-space in place of the second projection onto C:

$$\left\{\begin{array}{cc}& v_k=P_C(u_k-\ell A{u}_k),\hfill \\ \hfill & C_k=\{u\in H:\langle u_k-\ell A{u}_k-v_k,u-v_k\rangle \le 0\},\hfill \\ \hfill & u_{k+1}=P_{C_k}(u_k-\ell A{v}_k)\forall k\ge 0,\hfill \end{array}\right.$$

(3)

with $\ell \in (0,\frac{1}{L})$. In 2018, via the inertial approach, Thong and Hieu [19] first suggested the inertial subgradient extragradient method, i.e., for any starting points $x_0,x_1\in H$, $\left\{u_k\right\}$ is the sequence constructed below

$$\left\{\begin{array}{cc}& w_k=u_k+{\alpha }_k(u_k-u_{k-1}),\hfill \\ \hfill & v_k=P_C(w_k-\ell A{w}_k),\hfill \\ \hfill & C_k=\{u\in H:\langle w_k-\ell A{w}_k-v_k,u-v_k\rangle \le 0\},\hfill \\ \hfill & u_{k+1}=P_{C_k}(w_k-\ell A{v}_k)\forall k\ge 1,\hfill \end{array}\right.$$

(4)

with $\ell \in (0,\frac{1}{L})$. Under mild assumptions, they proved that $\left\{u_k\right\}$ converges weakly to an element of $\mathrm{VI}(C,A)$. Very recently, Ceng et al. [27] introduced a modified inertial subgradient extragradient method (i.e., Algorithm 1 specified below) for solving the VIP with pseudomonotone and Lipschitz continuous mapping A and the common fixed-point problem (CFPP) of finitely many nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$ in a real Hilbert space H. Give a contraction $f:H\to H$ with constant $\delta \in [0,1)$, and an $\eta$-strongly monotone and $\kappa$-Lipschitzian mapping $F:H\to H$ with $\delta <\ell :=1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}$ for $\rho \in (0,2\eta /{\kappa }^2)$. Suppose that $\left\{{\beta }_k\right\},\left\{{\gamma }_k\right\},\left\{{ϵ}_k\right\}$ are sequences in $(0,1)$ s.t. ${\beta }_k+{\gamma }_k<1\forall k\ge 1,{\beta }_k\to 0$ and ${ϵ}_k/{\beta }_k\to 0$ as $k\to \infty$. In addition, we write $T_k:=T_{k\mathrm{mod}N}$ for integer $k\ge 1$ with the mod function taking values in the set $\{1,2,\dots ,N\}$, that is, if $k=jN+q$ for some integers $j\ge 0$ and $0\le q<N$, then $T_k=T_N$ if $q=0$ and $T_k=T_q$ if $0<q<N$.

Algorithm 1: A modified inertial subgradient extragradient method (see [27])

Initialization: Give ${\tau }_1>0,\alpha >0,\mu \in (0,1)$. Let $u_0,u_1\in H$ be arbitrary.

Iterative Steps: Calculate $u_{k+1}$ as follows:

Step 1. Given the iterates $u_{k-1}$ and $u_k(k\ge 1)$, choose ${\alpha }_k$ s.t. $0\le {\alpha }_k\le {\tilde{\alpha }}_k$, where $${\tilde{\alpha }}_k=\left\{\begin{array}{cc}& min\{\alpha ,\frac{{ϵ}_k}{\parallel u_k-u_{k-1}\parallel }\},\mathrm{if}{u}_k\ne u_{k-1},\hfill \\ & \alpha ,\mathrm{otherwise}.\hfill \end{array}\right.$$

Step 2. Calculate $w_k=T_k{u}_k+{\alpha }_k(T_k{u}_k-T_k{u}_{k-1})$ and $v_k=P_C(w_k-{\tau }_{k}A{w}_k)$.

Step 3. Construct the half-space $C_k:=\{u\in H:\langle w_k-{\tau }_{k}A{w}_k-v_k,u-v_k\rangle \le 0\}$, and calculate $t_k=P_{C_k}(w_k-{\tau }_{k}A{v}_k)$.

Step 4. Compute $u_{k+1}={\beta }_{k}f\left(u_k\right)+{\gamma }_k{u}_k+((1-{\gamma }_k)I-{\beta }_k\rho F)t_k$, and update $${\tau }_{k+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_k-v_k{\parallel }^2+{\parallel t_k-v_k\parallel }^2}{2\langle A{w}_k-A{v}_k,t_k-v_k\rangle },{\tau }_k\},\mathrm{if}\langle A{w}_k-A{v}_k,t_k-v_k\rangle >0,\hfill \\ & {\tau }_k,\mathrm{otherwise}.\hfill \end{array}\right.$$

Put $k:=k+1$ and return to Step 1.

Under suitable assumptions, it was proven in [27] that $\left\{u_k\right\}$ converges strongly to an element of $\Omega ={\bigcap }_{l=1}^N\mathrm{Fix}\left(T_l\right)\cap \mathrm{VI}(C,A)$. Subsequently, Thong et al. [16] suggested the new inertial subgradient extragradient method (i.e., Algorithm 2 specified below) with adaptive step sizes for solving the VIP with the pseudomonotone and Lipschitz continuous mapping A. Suppose that $\left\{{\lambda }_k\right\}\subset (\lambda ,1)\subset (0,1)$ and $\left\{{ϵ}_k\right\},\left\{{\beta }_k\right\}\subset (0,1)$ are s.t. ${\beta }_k\to 0$ and ${ϵ}_k/{\beta }_k\to 0$ as $k\to \infty$.

Algorithm 2: New inertial subgradient extragradient method (see [16])

Give ${\tau }_1>0,\alpha >0,\mu \in (0,1)$. Let $u_0,u_1\in H$ be arbitrary.

Choose ${\alpha }_k$ s.t.

$0\le {\alpha }_k\le {\tilde{\alpha }}_k:=\left\{\begin{array}{cc}& min\{\alpha ,\frac{{ϵ}_k}{\parallel u_k-u_{k-1}\parallel }\},\mathrm{if}{u}_k\ne u_{k-1},\hfill \\ & \alpha ,\mathrm{otherwise},\hfill \end{array}\right.$

$w_k=(1-{\beta }_k)[u_k+{\alpha }_k(u_k-u_{k-1})]$,

$v_k=P_C(w_k-{\tau }_{k}A{w}_k)$,

$C_k:=\{u\in H:\langle w_k-{\tau }_{k}A{w}_k-v_k,u-v_k\rangle \le 0\}$,

$u_{k+1}=(1-{\lambda }_k)w_k+{\lambda }_k{P}_{C_k}(w_k-{\tau }_{k}A{v}_k)$,

update

${\tau }_{k+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_k-v_k{\parallel }^2+{\parallel t_k-v_k\parallel }^2}{2\langle A{w}_k-A{v}_k,t_k-v_k\rangle },{\tau }_k\},\mathrm{if}\langle A{w}_k-A{v}_k,t_k-v_k\rangle >0,\hfill \\ & {\tau }_k,\mathrm{otherwise},\hfill \end{array}\right.$

where $t_k:=P_{C_k}(w_k-{\tau }_{k}A{v}_k)$.

Under appropriate assumptions, it was proven in [16] that $\left\{u_k\right\}$ converges strongly to an element of $\mathrm{VI}(C,A)$. In a real Hilbert space H, let the VIP denote a pseudomonotone variational inequality problem with Lipschitz continuity operator A, and let the CFPP indicate a common fixed-point problem of finitely many nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$ and an asymptotically nonexpansive mapping $T_0:=T$. On the basis of Mann iteration method, viscosity approximation method and hybrid steepest-descent method, we propose and analyze two strengthened inertial-type subgradient extragradient rules (i.e., Algorithms 3 and 4 formulated in Section 3) with adaptive step sizes for solving the VIP and CFPP. With the help of suitable restrictions, we show strong convergence of the suggested rules to a common solution of the VIP and CFPP, which is the unique solution of a hierarchical variational inequality (HVI) defined on the common solution set $\Omega :={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$. In the end, our main results are applied to solve the VIP and CFPP in an illustrated example.

This article is arranged below: In Section 2, we present some concepts and basic tools for further use. Section 3 treats the convergence analysis of the suggested rules. In the end, Section 4 applies our main results to solve the VIP and CFPP in an illustrated example. Our results improve and extend the corresponding results announced by some others, e.g., Kraikaew and Saejung [20], Ceng et al. [27], Thong et al. [16] and Ceng and Shang [22].

2. Preliminaries

Suppose that C is a nonempty closed convex subset of a real Hilbert space H. Given $u\in H$ and $\left\{u_k\right\}\subset H$, we use the $u_k\to u$ (resp., $u_k\rightharpoonup u$) to indicate the strong (resp., weak) convergence of $\left\{u_k\right\}$ to u. An operator $\Gamma :C\to H$ is referred to as being

(i)

L-Lipschitz continuous (or L-Lipschitzian) in case $\exists L>0$ s.t. $\parallel \Gamma u-\Gamma v\parallel \le L\parallel u-v\parallel \forall u,v\in C$;

(ii)

monotone in case $\langle \Gamma u-\Gamma v,u-v\rangle \ge 0\forall u,v\in C$;

(iii)

pseudomonotone in case $\langle \Gamma u,v-u\rangle \ge 0\Rightarrow \langle \Gamma v,v-u\rangle \ge 0\forall u,v\in C$;

(iv)

$\alpha$-strongly monotone in case $\exists \alpha >0$ s.t. $\langle \Gamma u-\Gamma v,u-v\rangle \ge {\alpha \parallel u-v\parallel }^2\forall u,v\in C$;

(v)

sequentially weakly continuous in case $\forall \left\{u_k\right\}\subset C$, the relation holds: $u_k\rightharpoonup u\Rightarrow \Gamma u_k\rightharpoonup \Gamma u$.

Obviously, every monotone mapping is pseudomonotone but the converse is not true. For every $u\in H$, one knows that there is only a nearest point in C, denoted by $P_{C}u$, s.t. $\parallel u-P_{C}u\parallel \le \parallel u-v\parallel \forall v\in C$. $P_C$ is referred to as a metric projection of H onto C.

Lemma 1

(see [28]). The following hold:

(i) 

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

(ii) 

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

(iii) 

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

(iv) 

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

(v) 

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

Lemma 2

(see [10]). For any $u\in H$ and $\lambda \ge \eta >0$ the inequalities hold: $\frac{\parallel u-P_C(u-\lambda Au)\parallel }{\lambda }\le \frac{\parallel u-P_C(u-\eta Au)\parallel }{\eta }$ and $\parallel u-P_C(u-\eta Au)\parallel \le \parallel u-P_C(u-\lambda Au)\parallel$.

Lemma 3

(see [6], Lemma 2.1). Suppose that $A:C\to H$ is pseudomonotone and continuous. Then $u^{*}\in C$ is a solution to the VIP $\langle A{u}^{*},v-u^{*}\rangle \ge 0\forall v\in C$, if and only if $\langle Av,v-u^{*}\rangle \ge 0\forall v\in C$.

Lemma 4

(see [29]). Suppose that $\left\{a_k\right\}$ is a sequence of nonnegative reals s.t. $a_{k+1}\le (1-{\lambda }_k)a_k+{\lambda }_k{\gamma }_k\forall k\ge 1$, where $\left\{{\lambda }_k\right\}$ and $\left\{{\gamma }_k\right\}$ are sequences of reals s.t. (a) $\left\{{\lambda }_k\right\}\subset [0,1]$ and ${\sum }_{k=1}^{\infty }{\lambda }_k=\infty$, and (b) ${lim\; sup}_{k\to \infty }{\gamma }_k\le 0$ or ${\sum }_{k=1}^{\infty }\left|{\lambda }_k{\gamma }_k\right|<\infty$. Then ${lim}_{k\to \infty }{a}_k=0$.

Lemma 5

(see [30]). Assume that X is a Banach space which admits a weakly continuous duality mapping, C is a nonempty, convex and closed set in X, and $T:C\to C$ is an asymptotically nonexpansive mapping with $\mathrm{Fix}\left(T\right)\ne \varnothing$. Then $I-T$ is demiclosed at zero, i.e., for any $\left\{u_k\right\}\subset C$ with $u_k\rightharpoonup u\in C$, the relation holds: $(I-T)u_k\to 0\Rightarrow u\in \mathrm{Fix}\left(T\right)$, where I is the identity mapping of X.

Lemma 6

(see [29], Lemma 3.1). Give $\lambda \in (0,1]$. Suppose that the mapping $T:C\to H$ is nonexpansive, and the operator $T^{\lambda }:C\to H$ is formulated by $T^{\lambda }u:=Tu-\lambda \mu F\left(Tu\right)\forall u\in C$, where $F:H\to H$ is κ-Lipschitzian and η-strongly monotone. Then $T^{\lambda }$ is a contraction provided $0<\mu <\frac{2\eta }{{\kappa }^2}$, i.e., $\parallel T^{\lambda }u-T^{\lambda }v\parallel \le (1-\lambda \ell )\parallel u-v\parallel \forall u,v\in C$, with $\ell =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}\in (0,1]$.

The following lemma will play an important role in the convergence analysis of the suggested rules in this paper.

Lemma 7

(see [31]). Suppose that $\left\{{\Gamma }_k\right\}$ is a sequence in $\mathbf{R}$, which does not decrease at infinity in the sense that $\exists \left\{{\Gamma }_{k_l}\right\}\subset \left\{{\Gamma }_k\right\}$ s.t. ${\Gamma }_{k_l}<{\Gamma }_{k_l+1}\forall l\ge 1$. Let the sequence ${\{\wp \left(k\right)\}}_{k\ge k_0}$ of integers be defined as

$$\wp \left(k\right)=max\{l\le k:{\Gamma }_l<{\Gamma }_{l+1}\},$$

where integer $k_0\ge 1$ s.t. $\{l\le k_0:{\Gamma }_l<{\Gamma }_{l+1}\}\ne \varnothing$. Then the statements hold below:

(i) 

$\wp \left(k_0\right)\le \wp (k_0+1)\le \cdots$ and $\wp \left(k\right)\to \infty$;

(ii) 

${\Gamma }_{\wp \left(k\right)}\le {\Gamma }_{\wp \left(k\right)+1}$ and ${\Gamma }_k\le {\Gamma }_{\wp \left(k\right)+1}\forall k\ge k_0$.

3. Main Results

Suppose that the feasible set C is nonempty, convex and closed in a real Hilbert space H. In what follows, we assume always that the conditions hold below.

$T:H\to H$ is asymptotically nonexpansive and $T_l:H\to H$ is nonexpansive for $l=1,\dots ,N$ s.t. ${\left\{T_n\right\}}_{n=1}^{\infty }$ is formulated as in Algorithm 1 (Such hypothesis will be imposed on Algorithms 3 and 4 specified below).

$A:H\to H$ is L-Lipschitz continuous, pseudomonotone on H, s.t. $\parallel Au\parallel \le {lim\; inf}_{n\to \infty }\parallel A{u}_k\parallel$$\forall \left\{u_k\right\}\subset C$ with $u_k\rightharpoonup u$, and $\Omega ={\bigcap }_{l=0}^N\mathrm{Fix}\left(T_l\right)\cap \mathrm{VI}(C,A)\ne \varnothing$ with $T_0:=T$.

$f:H\to H$ is a contraction with coefficient $\delta \in [0,1)$, and $F:H\to H$ is $\eta$-strongly monotone and $\kappa$-Lipschitzian s.t. $\delta <\ell :=1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}$ for $\rho \in (0,\frac{2\eta }{{\kappa }^2})$.

$\left\{{\lambda }_n\right\},\left\{{ϵ}_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\}\subset (0,1)$ with ${\beta }_n+{\gamma }_n<1\forall n\ge 1$, s.t.

(i)

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

(ii)

${lim}_{n\to \infty }\frac{{ϵ}_n}{{\beta }_n}=0$ and ${lim}_{n\to \infty }\frac{{\theta }_n}{{\beta }_n}=0$;

(iii)

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

(iv)

$\left\{{\lambda }_n\right\}\subset [\underline{\lambda },\overline{\lambda }]\subset (0,1)$.

Algorithm 3: Strengthened inertial-type subgradient extragradient rules

Initialization: Give ${\tau }_1>0,\alpha >0,\mu \in (0,1)$. Let $x_0,x_1\in H$ be arbitrary and choose ${\alpha }_n$ s.t. $$0\le {\alpha }_n\le {\tilde{\alpha }}_n:=\left\{\begin{array}{cc}\hfill & min\{\alpha ,\frac{{ϵ}_n}{\parallel x_n-x_{n-1}\parallel }\},\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ \hfill & \alpha ,\mathrm{otherwise}.\hfill \end{array}\right.$$ (5)

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

Step 1. Set $t_n=x_n+{\alpha }_n(x_n-x_{n-1})$, and compute

$w_n={\beta }_{n}f\left(x_n\right)+{\gamma }_n{x}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T_n{t}_n$

and $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$.

Step 2. Compute $x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_n{T}^n{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)$ with

$C_n:=\{x\in H:\langle w_n-{\tau }_{n}A{w}_n-y_n,x-y_n\rangle \le 0\}$.

Step 3. Update $${\tau }_{n+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_n-y_n{\parallel }^2+{\parallel z_n-y_n\parallel }^2}{2\langle A{w}_n-A{y}_n,z_n-y_n\rangle },{\tau }_n\},\mathrm{if}\langle A{w}_n-A{y}_n,z_n-y_n\rangle >0,\hfill \\ \hfill & {\tau }_n,\mathrm{otherwise},\hfill \end{array}\right.$$ (6)

with $z_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$. Again set $n:=n+1$ and go to Step 1.

It is worth mentioning that, putting ${\gamma }_n=0\forall n\ge 1$, $f=0$ and $T_l=T=\rho F=I$ for $l=1,\dots ,N$, we transform Algorithm 3 into Algorithm 2. It is clear that, from (5) one has $\frac{{\alpha }_n}{{\beta }_n}\parallel x_{n-1}-x_n\parallel \to 0$ as $n\to \infty$. In fact, from ${\alpha }_n\parallel x_{n-1}-x_n\parallel \le {ϵ}_n\forall n\ge 1$, it follows that $\frac{{\alpha }_n}{{\beta }_n}\parallel x_{n-1}-x_n\parallel \le \frac{{ϵ}_n}{{\beta }_n}\to 0$ (due to condition (ii)).

Lemma 8.

Suppose $\left\{{\tau }_n\right\}$ is constructed by (6). Then $\left\{{\tau }_n\right\}$ is nonincreasing s.t. ${\tau }_n\ge \tau :=min\{{\tau }_1,\frac{\mu }{L}\}\forall n\ge 1$, and ${lim}_{n\to \infty }{\tau }_n\ge \tau :=min\{{\tau }_1,\frac{\mu }{L}\}$.

Proof. 

From (6), we observe that ${\tau }_n\ge {\tau }_{n+1}\forall n\ge 1$. Moreover, it is clear that

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

□

It is worth pointing out that, if $w_n=y_n$ or $A{y}_n=0$, then $y_n\in \mathrm{VI}(C,A)$ (due to Lemmas 2 and 8). In fact, when $w_n=y_n$ or $A{y}_n=0$, one has $0=\parallel y_n-P_C(y_n-{\tau }_{n}A{y}_n)\parallel \ge \parallel y_n-P_C(y_n-\tau A{y}_n)\parallel$. So, one gets $y_n\in \mathrm{VI}(C,A)$.

Lemma 9.

Under the assumptions in Algorithm 3, there exists an integer $m_0\ge 1$ s.t. $\forall n\ge m_0$, $1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}>0$ and

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

(7)

Proof. 

First, let us show that

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

(8)

In fact, in case $\langle A{w}_n-A{y}_n,z_n-y_n\rangle \le 0$, then (8) holds. Otherwise, from (6) one gets (8). Note that for every $p\in \Omega \subset C\subset C_n$,

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

which hence arrives at

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

(9)

Owing to $p\in \mathrm{VI}(C,A)$, one gets $\langle Ap,u-p\rangle \ge 0\forall u\in C$. By the pseudomonotonicity of A on C one has $\langle Au,u-p\rangle \ge 0\forall u\in C$. Setting $u:=y_n\in C$ one gets $\langle A{y}_n,p-y_n\rangle \le 0$. Thus,

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

(10)

Combining (9) and (10), one obtains

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

(11)

Noticing $z_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$ and $C_n:=\{x\in H:\langle w_n-{\tau }_{n}A{w}_n-y_n,x-y_n\rangle \le 0\}$, one has

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

which along with (8), yields

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

This along with (11), leads to

$$\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2).$$

(12)

In addition, from Algorithm 3 one gets

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le (1-{\lambda }_n)\parallel w_n{-p\parallel }^2+{\lambda }_n{(1+{\theta }_n)}^2{\parallel z_n-p\parallel }^2\hfill \\ & \le (1-{\lambda }_n)\parallel w_n{-p\parallel }^2+{\lambda }_n\parallel z_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2,\hfill \end{array}$$

which along with (12), yields

$$\parallel x_{n+1}{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}){\lambda }_n(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2.$$

Since $1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}\to 1-\mu >0$ as $n\to \infty$, obviously there exists an integer $m_0\ge 1$ such that for each $n\ge m_0$, $1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}>0$ and (7) holds. □

Lemma 10.

Suppose that $\left\{x_n\right\},\left\{z_n\right\}$ are bounded sequences constructed in Algorithm 3 such that $x_n-x_{n+1}\to 0,x_n-z_n\to 0,x_n-T_n{x}_n\to 0$ and $x_n-T^n{x}_n\to 0$. If $T^n{x}_n-T^{n+1}{x}_n\to 0$ and $\exists \left\{x_{n_k}\right\}\subset \left\{x_n\right\}$ s.t. $x_{n_k}\rightharpoonup z\in H$, then $z\in \Omega$.

Proof. 

From Algorithm 3 and the hypotheses $x_n-x_{n+1}\to 0$ and $x_n-T_n{x}_n\to 0$, we obtain that

$$\parallel t_n-x_n\parallel ={\alpha }_n\parallel x_n-x_{n-1}\parallel \to 0(n\to \infty ),$$

and hence

$$\begin{array}{cc}\parallel t_n-T_n{t}_n\parallel \hfill & \le \parallel t_n-x_n\parallel +\parallel x_n-T_n{x}_n\parallel +\parallel T_n{x}_n-T_n{t}_n\parallel \hfill \\ & \le 2\parallel t_n-x_n\parallel +\parallel x_n-T_n{x}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

It is clear that $\left\{t_n\right\}$ is bounded. Note that

$$\begin{array}{cc}\parallel w_n-x_n\parallel \hfill & =\parallel (1-{\gamma }_n)(T_n{t}_n-x_n)+{\beta }_n(f\left(x_n\right)-\rho F{T}_n{t}_n)\parallel \hfill \\ & =\parallel (1-{\gamma }_n)[T_n{t}_n-t_n+t_n-x_n]+{\beta }_n(f\left(x_n\right)-\rho F{T}_n{t}_n)\parallel \hfill \\ & \le (1-{\gamma }_n)\parallel T_n{t}_n-t_n+t_n-x_n\parallel +{\beta }_n\parallel f\left(x_n\right)-\rho F{T}_n{t}_n\parallel \hfill \\ & \le \parallel T_n{t}_n-t_n\parallel +\parallel t_n-x_n\parallel +{\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel \rho F{T}_n{t}_n\parallel ).\hfill \end{array}$$

Since ${\beta }_n\to 0$, $\parallel t_n-x_n\parallel \to 0$ and $\parallel t_n-T_n{t}_n\parallel \to 0$, by the boundedness of $\left\{x_n\right\}$ and $\left\{T_n{t}_n\right\}$ we deduce that

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

(13)

It is clear that $\left\{w_n\right\}$ is bounded. From (12) it follows that

$$\begin{array}{cc}(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill & \le \parallel w_n{-p\parallel }^2-{\parallel z_n-p\parallel }^2\hfill \\ & \le \parallel w_n-z_n\parallel (\parallel w_n-p\parallel +\parallel z_n-p\parallel )\hfill \\ & \le (\parallel w_n-x_n\parallel +\parallel x_n-z_n\parallel \left)\right(\parallel w_n-p\parallel +\parallel z_n-p\parallel ).\hfill \end{array}$$

Since $1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}\to 1-\mu >0$ and $\parallel x_n-z_n\parallel \to 0$ (due to the hypothesis), from (13) one gets

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

It is clear that $\left\{y_n\right\}$ is bounded. Moreover, thanks to the hypotheses $x_n-z_n\to 0,x_n-T^n{x}_n\to 0$ and $T^n{x}_n-T^{n+1}{x}_n\to 0$, one has that

$$\begin{array}{cc}\parallel z_n-T^n{z}_n\parallel \hfill & \le \parallel z_n-x_n\parallel +\parallel x_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T^n{z}_n\parallel \hfill \\ & \le (2+{\theta }_n)\parallel z_n-x_n\parallel +\parallel x_n-T^n{x}_n\parallel \to 0(n\to \infty ),\hfill \end{array}$$

and hence

$$\begin{array}{cc}& \parallel z_n-T{z}_n\parallel \le \parallel z_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T^{n+1}{x}_n\parallel +\parallel T^{n+1}{x}_n-T{z}_n\parallel \hfill \\ \hfill & \le (2+{\theta }_1)\parallel z_n-T^n{x}_n\parallel +\parallel T^n{x}_n-T^{n+1}{x}_n\parallel \hfill \\ \hfill & \le (2+{\theta }_1)(\parallel z_n-T^n{z}_n\parallel +\parallel T^n{z}_n-T^n{x}_n\parallel )+\parallel T^n{x}_n-T^{n+1}{x}_n\parallel \hfill \\ \hfill & \le (2+{\theta }_1)\{\parallel z_n-T^n{z}_n\parallel +(1+{\theta }_n)\parallel z_n-x_n\parallel \}+\parallel T^n{x}_n-T^{n+1}{x}_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(14)

We claim that ${lim}_{n\to \infty }\parallel x_n-T_l{x}_n\parallel =0$ for $l=1,\dots ,N$. In fact, observe that for $l=1,\dots ,N$,

$$\begin{array}{cc}\parallel x_n-T_{n+l}{x}_n\parallel \hfill & \le \parallel x_n-T_{n+l}{x}_{n+l}\parallel +\parallel T_{n+l}{x}_{n+l}-T_{n+l}{x}_n\parallel \hfill \\ & \le \parallel x_n-T_{n+l}{x}_{n+l}\parallel +\parallel x_{n+l}-x_n\parallel \hfill \\ & \le \parallel T_{n+l}{x}_{n+l}-x_{n+l}\parallel +2\parallel x_{n+l}-x_n\parallel .\hfill \end{array}$$

So, using the hypotheses $\parallel x_n-T_n{x}_n\parallel \to 0$ and $\parallel x_n-x_{n+1}\parallel \to 0$, one gets ${lim}_{n\to \infty }\parallel x_n-T_{n+l}{x}_n\parallel =0$ for $l=1,\dots ,N$. This hence ensures that

$$\underset{n\to \infty }{lim}\parallel x_n-T_l{x}_n\parallel =0\forall l\in \{1,2,\dots ,N\}.$$

(15)

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

$$\frac{1}{{\tau }_{n_k}}\langle w_{n_k}-y_{n_k},u-y_{n_k}\rangle +\langle A{w}_{n_k},y_{n_k}-w_{n_k}\rangle \le \langle A{w}_{n_k},u-w_{n_k}\rangle \forall u\in C.$$

(16)

Noticing the boundedness of $\left\{w_{n_k}\right\}$ and Lipschitz continuity of A, one gets the boundedness of $\left\{A{w}_{n_k}\right\}$. Owing to ${\tau }_n\ge \tau :=min\{{\tau }_1,\frac{\mu }{L}\}$ and $w_n-y_n\to 0$, from (16) one gets

$$\underset{k\to \infty }{lim\; inf}\langle A{w}_{n_k},u-w_{n_k}\rangle \ge 0\forall u\in C.$$

Next, let us show $z\in \mathrm{VI}(C,A)$. In fact, since $w_n-x_n\to 0$, $w_n-y_n\to 0$ and $x_{n_k}\rightharpoonup z$, we obtain that $x_n-y_n\to 0$ and $y_{n_k}\rightharpoonup z$. Since C is convex and closed, from $\left\{y_n\right\}\subset C$ one has $z\in C$. In what follows, we consider two cases. If $Az=0$, then it is clear that $z\in \mathrm{VI}(C,A)$ because $\langle Az,u-z\rangle \ge 0\forall u\in C$. Assume that $Az\ne 0$. Then, by the hypothesis on A, instead of the sequentially weak continuity of A, we get $0<\parallel Az\parallel \le {lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel$. Note that combining $w_n-y_n\to 0$ and L-Lipschitz continuity of A attains $A{w}_n-A{y}_n\to 0$. So it follows that

$$0<\parallel Az\parallel \le \underset{k\to \infty }{lim\; inf}\parallel A{y}_{n_k}\parallel =\underset{k\to \infty }{lim\; inf}(\parallel A{y}_{n_k}\parallel -\parallel A{y}_{n_k}-A{w}_{n_k}\parallel )\le \underset{k\to \infty }{lim\; inf}\parallel A{w}_{n_k}\parallel .$$

So, we may assume, without loss of generality, that $\parallel A{w}_{n_k}\parallel \ne 0\forall k\ge 1$.

We now pick a sequence $\left\{{\delta }_k\right\}\subset (0,1)$ s.t. ${\delta }_k\downarrow 0$ as $k\to \infty$. For every $k\ge 1$, let the $m_k$ indicate the smallest positive integer s.t.

$$\langle A{w}_{n_j},u-w_{n_j}\rangle +{\delta }_k\ge 0\forall j\ge m_k.$$

(17)

Noticing that $\left\{{\delta }_k\right\}$ is decreasing, one knows that $\left\{m_k\right\}$ is increasing. Since $\left\{w_{m_k}\right\}\subset \left\{w_{n_k}\right\}$ and $\parallel A{w}_{n_k}\parallel \ne 0\forall k\ge 1$, we put ${\hslash }_{m_k}=\frac{A{w}_{m_k}}{\parallel A{w}_{m_k}{\parallel }^2}$, and get $\langle A{w}_{m_k},{\hslash }_{m_k}\rangle =1\forall k\ge 1$. So, from (17) one has $\langle A{w}_{m_k},u+{\delta }_k{\hslash }_{m_k}-w_{m_k}\rangle \ge 0\forall k\ge 1$. Moreover, using the pseudomonotonicity of A one has $\langle A(u+{\delta }_k{\hslash }_{m_k}),u+{\delta }_k{\hslash }_{m_k}-w_{m_k}\rangle \ge 0\forall k\ge 1$. This immediately arrives at

$$\langle Au,u-w_{m_k}\rangle \ge \langle Au-A(u+{\delta }_k{\hslash }_{m_k}),u+{\delta }_k{\hslash }_{m_k}-w_{m_k}\rangle -{\delta }_k\langle Au,{\hslash }_{m_k}\rangle \forall k\ge 1.$$

(18)

Let us show that ${lim}_{k\to \infty }{\delta }_k{\hslash }_{m_k}=0$. In fact, since $0<\parallel Az\parallel \le {lim\; inf}_{k\to \infty }\parallel A{w}_{n_k}\parallel$, $\left\{w_{m_k}\right\}\subset \left\{w_{n_k}\right\}$ and ${\delta }_k\downarrow 0$, we deduce that $0\le {lim\; sup}_{k\to \infty }\parallel {\delta }_k{\hslash }_{m_k}\parallel ={lim\; sup}_{k\to \infty }\frac{{\delta }_k}{\parallel A{w}_{m_k}\parallel }$$\le \frac{{lim\; sup}_{k\to \infty }{\delta }_k}{{lim\; inf}_{k\to \infty }\parallel A{w}_{n_k}\parallel }=0$. So, one has ${\delta }_k{\hslash }_{m_k}\to 0$ as $k\to \infty$.

In the end, we claim that $z\in \Omega$. In fact, using (15) one has $x_{n_k}-T_l{x}_{n_k}\to 0$ for $l=1,\dots ,N$. Since Lemma 5 ensures the demiclosedness of $I-T_l$ at zero, from $x_{n_k}\rightharpoonup z$ one has $z\in \mathrm{Fix}\left(T_l\right)\forall l\in \{1,\dots ,N\}$. So, $z\in {\bigcap }_{l=1}^N\mathrm{Fix}\left(T_l\right)$. Meanwhile, from $x_n-z_n\to 0$ and $x_{n_k}\rightharpoonup z$, one gets $z_{n_k}\rightharpoonup z$. Using (14) one has $z_{n_k}-T{z}_{n_k}\to 0$. From Lemma 5 one knows the demiclosedness of $I-T$ at zero, and hence gets $z\in \mathrm{Fix}\left(T\right)$. Additionally, letting $k\to \infty$, we infer that the right hand side of (18) tends to zero by the uniform continuity of A, the boundedness of $\left\{w_{m_k}\right\},\left\{{\hslash }_{m_k}\right\}$ and the limit ${lim}_{k\to \infty }{\delta }_k{\hslash }_{m_k}=0$. Consequently, $\langle Au,u-z\rangle ={lim\; inf}_{k\to \infty }\langle Au,u-w_{m_k}\rangle \ge 0\forall u\in C$. Using Lemma 3 one has $z\in \mathrm{VI}(C,A)$. Therefore, $z\in {\bigcap }_{l=0}^N\mathrm{Fix}\left(T_l\right)\cap \mathrm{VI}(C,A)=\Omega$. □

Theorem 1.

Suppose that $\left\{x_n\right\}$ is the sequence constructed in Algorithm 3. Then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \Omega$ provided $T^n{x}_n-T^{n+1}{x}_n\to 0$, where $x^{*}\in \Omega$ is the unique solution to the HVI: $\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \Omega$.

Proof. 

Since ${lim}_{n\to \infty }\frac{{\theta }_n}{{\beta }_n}=0$ and $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$, we might assume that ${\theta }_n\le \frac{{\beta }_n(\ell -\delta )}{2}\forall n\ge 1$ and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$. Let us show that $P_{\Omega }(f-\rho F+I)$ is a contraction. In fact, using Lemma 6 one has

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

which ensures that $P_{\Omega }(f-\rho F+I)$ is a contraction. Using Banach’s Contraction Mapping Principle we know that $P_{\Omega }(f-\rho F+I)$ has a unique fixed point. Say $x^{*}\in H$, i.e., $x^{*}=P_{\Omega }(f-\rho F+I)x^{*}$. Therefore, $\exists |x^{*}\in \Omega ={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ solving the HVI

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

(19)

In what follows we claim that the conclusion of the theorem is valid. To the goal, we divide the remainder of the proof into several steps.

Claim 1. We prove that $\left\{x_n\right\}$ is bounded. In fact, for an arbitrary $p\in \Omega ={\bigcap }_{l=0}^N\mathrm{Fix}\left(T_l\right)\cap \mathrm{VI}(C,A)$, we obtain that $Tp=p,T_{n}p=p\forall n\ge 1$, and (12) holds, i.e.,

$$\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2).$$

Since ${lim}_{n\to \infty }(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})=1-\mu >0$, we may assume, without loss of generality, that $1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}>0\forall n\ge 1$. Hence, one gets

$$\parallel z_n-p\parallel \le \parallel w_n-p\parallel \forall n\ge 1.$$

(20)

From the definition of $t_n$, we get

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

(21)

Since $\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \to 0$ as $n\to \infty$, one knows that $\exists K_1>0$ s.t.

$$\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \le K_1\forall n\ge 1,$$

which together (21), yields

$$\parallel t_n-p\parallel \le \parallel x_n-p\parallel +{\beta }_n{K}_1\forall n\ge 1.$$

(22)

So, from ${\beta }_n+{\gamma }_n<1$, Lemma 6 and (20) it follows that

$$\begin{array}{cc}& \parallel z_n-p\parallel \le {\beta }_n\parallel f\left(x_n\right)-f\left(p\right)\parallel +{\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n)\hfill \\ \hfill & \times \parallel (I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)T_n{t}_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)p+\frac{{\beta }_n}{1-{\gamma }_n}(f-\rho F)p\parallel \hfill \\ \hfill & \le {\beta }_n\delta \parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +(1-{\gamma }_n-{\beta }_n\ell )\parallel t_n-p\parallel +{\beta }_n\parallel (f-\rho F)p\parallel \hfill \\ \hfill & \le {\beta }_n\delta (\parallel x_n-p\parallel +{\beta }_n{K}_1)+{\gamma }_n(\parallel x_n-p\parallel +{\beta }_n{K}_1)\hfill \\ \hfill & +(1-{\gamma }_n-{\beta }_n\ell )(\parallel x_n-p\parallel +{\beta }_n{K}_1)+{\beta }_n\parallel (f-\rho F)p\parallel \hfill \\ \hfill & \le \parallel x_n-p\parallel +{\beta }_n(K_1+\parallel (f-\rho F)p\parallel ).\hfill \end{array}$$

(23)

Noticing $x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_n{T}^n{z}_n$, we infer from (23) that

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & \le (1-{\lambda }_n)\parallel w_n-p\parallel +{\lambda }_n(1+{\theta }_n)\parallel z_n-p\parallel \hfill \\ & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n-p\parallel +{\beta }_n(K_1+\parallel (f-\rho F)p\parallel )\hfill \\ & +\frac{{\beta }_n(\ell -\delta )}{2}·[\parallel x_n-p\parallel +{\beta }_n(K_1+\parallel (f-\rho F)p\parallel \left)\right]\hfill \\ & \le [1-\frac{{\beta }_n(\ell -\delta )}{2}]\parallel x_n-p\parallel +\frac{{\beta }_n(\ell -\delta )}{2}·\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\hfill \\ & \le max\{\parallel x_n-p\parallel ,\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\}.\hfill \end{array}$$

By induction, we obtain $\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\}\forall n\ge 1$. Thus, $\left\{x_n\right\}$ is bounded, and so are the sequences $\left\{t_n\right\},\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{f\left(x_n\right)\right\},\left\{T_n{t}_n\right\},\left\{T^n{z}_n\right\}$.

Claim 2. We prove that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+({\beta }_n+{\theta }_n)K_4,$$

for some $K_4>0$. In fact, one has

$$\begin{array}{cc}{w}_n-p\hfill & ={\beta }_n(f\left(x_n\right)-f\left(p\right))+{\gamma }_n(x_n-p)+(1-{\gamma }_n)[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)T_n{t}_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)p]\hfill \\ & +{\beta }_n(f-\rho F)p.\hfill \end{array}$$

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

$$\begin{array}{cc}& \parallel w_n{-p\parallel }^2\hfill \\ \hfill & \le \parallel {\beta }_n(f\left(x_n\right)-f\left(p\right))+{\gamma }_n(x_n-p)+(1-{\gamma }_n)[(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)T_n{t}_n-(I-\frac{{\beta }_n}{1-{\gamma }_n}\rho F)p]{\parallel }^2\hfill \\ \hfill & +2{\beta }_n\langle (f-\rho F)p,w_n-p\rangle \hfill \\ \hfill & \le [{\beta }_n\delta \parallel x_n-p\parallel +{\gamma }_n\parallel x_n-p\parallel +(1-{\beta }_n\ell -{\gamma }_n)\parallel t_n-p{\parallel ]}^2+2{\beta }_n\langle (f-\rho F)p,w_n-p\rangle \hfill \\ \hfill & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel t_n-p\parallel }^2+2{\beta }_n\langle (f-\rho F)p,w_n-p\rangle \hfill \\ \hfill & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel t_n-p\parallel }^2+{\beta }_n{K}_2\hfill \end{array}$$

(24)

(due to ${\beta }_n\delta +{\gamma }_n+(1-{\beta }_n\ell -{\gamma }_n)=1-{\beta }_n(\ell -\delta )\le 1$), where ${sup}_{n\ge 1}2\parallel (f-\rho F)p\parallel \parallel w_n-p\parallel \le K_2$ for some $K_2>0$. Noticing $x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_n{T}^n{z}_n$, from (12) we have

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le (1-{\lambda }_n)\parallel w_n{-p\parallel }^2+{\lambda }_n\parallel z_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ \hfill & \le \parallel w_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2.\hfill \end{array}$$

(25)

Moreover, from (22) we have

$$\begin{array}{cc}\parallel t_n{-p\parallel }^2\hfill & \le \parallel x_n{-p\parallel }^2+{\beta }_n(2K_1\parallel x_n-p\parallel +{\beta }_n{K}_1^2)\hfill \\ \hfill & \le \parallel x_n{-p\parallel }^2+{\beta }_n{K}_3,\hfill \end{array}$$

(26)

where ${sup}_{n\ge 1}(2K_1\parallel x_n-p\parallel +{\beta }_n{K}_1^2)\le K_3$ for some $K_3>0$. Combining (24)–(26), we obtain

$$\begin{array}{cc}& \parallel x_{n+1}{-p\parallel }^2\hfill \\ & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel t_n-p\parallel }^2+{\beta }_n{K}_2\hfill \\ & -{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & +{\beta }_n{K}_3+{\beta }_n{K}_2+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ & \le \parallel x_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+({\beta }_n+{\theta }_n)K_4,\hfill \end{array}$$

where ${sup}_{n\ge 1}[K_2+K_3+(2+{\theta }_n)\parallel z_n-p{\parallel }^2]\le K_4$. This immediately implies that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+({\theta }_n+{\beta }_n)K_4.$$

(27)

Claim 3. We prove that

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-{\beta }_n(\ell -\delta )]{\parallel x_n-p\parallel }^2\hfill \\ & +{\beta }_n(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_n}{3{\beta }_n}+\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel )+\frac{2\langle (f-\rho F)p,w_n-p\rangle }{\ell -\delta }]\hfill \end{array}$$

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

$$\begin{array}{cc}\parallel t_n{-p\parallel }^2\hfill & \le \parallel x_n{-p\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel [2\parallel x_n-p\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel ].\hfill \end{array}$$

(28)

Combining (20), (24) and (28), one has

$$\begin{array}{cc}& \parallel x_{n+1}{-p\parallel }^2\hfill \\ & \le (1-{\lambda }_n)\parallel w_n{-p\parallel }^2+{\lambda }_n{(1+{\theta }_n)}^2\parallel z_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel w_n-p\parallel }^2\hfill \\ & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel t_n-p\parallel }^2+2{\beta }_n\langle (f-\rho F)p,w_n-p\rangle \hfill \\ & +{\theta }_n(2+{\theta }_n){\parallel w_n-p\parallel }^2\hfill \\ & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n{-p\parallel }^2+{\beta }_n(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_n}{3{\beta }_n}+\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel )+\frac{2\langle (f-\rho F)p,w_n-p\rangle }{\ell -\delta }],\hfill \end{array}$$

where ${sup}_{n\ge 1}\{\parallel x_n-p\parallel ,{\alpha }_n\parallel x_n-x_{n-1}\parallel ,(2+{\theta }_n)\parallel w_n-p{\parallel }^2\}\le K$ for some $K>0$.

Claim 4. We prove that $\left\{x_n\right\}$ converges strongly to the unique solution $x^{*}\in \Omega$ of the HVI (19). In fact, setting $p=x^{*}$, one obtains from the last inequality that

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

(29)

Putting ${\Gamma }_n={\parallel x_n-x^{*}\parallel }^2$, we demonstrate ${\Gamma }_n\to 0(n\to \infty )$ by considering the following cases.

Case 1. Assume that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Gamma }_n\right\}$ is nonincreasing. Then the limit ${lim}_{n\to \infty }{\Gamma }_n=\varsigma <+\infty$ and ${lim}_{n\to \infty }({\Gamma }_n-{\Gamma }_{n+1})=0$.

Using (27), ${\Gamma }_n-{\Gamma }_{n+1}\to 0,{\beta }_n\to 0,{\theta }_n\to 0,1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}\to 1-\mu$ and $\left\{{\lambda }_n\right\}\subset [\underline{\lambda },\overline{\lambda }]\subset (0,1)$, one has

$$\begin{array}{cc}& {\displaystyle \underset{n\to \infty }{lim\; sup}}\underline{\lambda }(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}[\parallel x_n-x^{*}{\parallel }^2-\parallel x_{n+1}-x^{*}{\parallel }^2+({\beta }_n+{\theta }_n)K_4]\hfill \\ & ={\displaystyle \underset{n\to \infty }{lim\; sup}}({\Gamma }_n-{\Gamma }_{n+1}+({\beta }_n+{\theta }_n)K_4)=0.\hfill \end{array}$$

This immediately ensures that

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

Hence, one gets

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

(30)

Noticing $w_n-x^{*}={\gamma }_n(x_n-x^{*})+(1-{\gamma }_n)(T_n{t}_n-x^{*})+{\beta }_n(f\left(x_n\right)-\rho F{T}_n{t}_n)$, we deduce from (25) and (26) that

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2\le \parallel w_n-x^{*}{\parallel }^2+{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & \le \parallel {\gamma }_n(x_n-x^{*})+(1-{\gamma }_n)(T_n{t}_n-x^{*}){\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-\rho F{T}_n{t}_n,w_n-x^{*}\rangle \hfill \\ & +{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & \le {\gamma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\gamma }_n)\parallel T_n{t}_n-x^{*}{\parallel }^2-{\gamma }_n(1-{\gamma }_n){\parallel x_n-T_n{t}_n\parallel }^2\hfill \\ & +2{\beta }_n\parallel f\left(x_n\right)-\rho F{T}_n{t}_n\parallel \parallel w_n-x^{*}\parallel +{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & \le {\gamma }_n(\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3)+(1-{\gamma }_n)(\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3)-{\gamma }_n(1-{\gamma }_n){\parallel x_n-T_n{t}_n\parallel }^2\hfill \\ & +2{\beta }_n\parallel f\left(x_n\right)-\rho F{T}_n{t}_n\parallel \parallel w_n-x^{*}\parallel +{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & =\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3-{\gamma }_n(1-{\gamma }_n)\parallel x_n-T_n{t}_n{\parallel }^2+2{\beta }_n\parallel f\left(x_n\right)-\rho F{T}_n{t}_n\parallel \parallel w_n-x^{*}\parallel \hfill \\ & +{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2,\hfill \end{array}$$

which together with $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$, yields

$$\begin{array}{cc}& a(1-b)\parallel x_n-T_n{t}_n{\parallel }^2\le {\gamma }_n(1-{\gamma }_n){\parallel x_n-T_n{t}_n\parallel }^2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2-\parallel x_{n+1}-x^{*}{\parallel }^2+{\beta }_n{K}_3+2{\beta }_n\parallel f\left(x_n\right)-\rho F{T}_n{t}_n\parallel \parallel w_n-x^{*}\parallel \hfill \\ & +{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & \le {\Gamma }_n-{\Gamma }_{n+1}+{\beta }_n[K_3+2(\parallel f\left(x_n\right)\parallel +\parallel \rho F{T}_n{t}_n\parallel )\parallel w_n-x^{*}\parallel ]+{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2.\hfill \end{array}$$

Since ${\Gamma }_n-{\Gamma }_{n+1}\to 0,{\beta }_n\to 0$ and ${\theta }_n\to 0$, by the boundedness of $\left\{w_n\right\},\left\{x_n\right\},\left\{z_n\right\},\left\{T_n{t}_n\right\}$ one has

$$\underset{n\to \infty }{lim}\parallel x_n-T_n{t}_n\parallel =0,$$

and hence

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

This together with (30), leads to

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

(31)

and hence

$$\begin{array}{cc}& \parallel x_n-T_n{x}_n\parallel \le \parallel x_n-T_n{t}_n\parallel +\parallel T_n{t}_n-T_n{x}_n\parallel \hfill \\ \hfill & \le \parallel x_n-T_n{t}_n\parallel +\parallel t_n-x_n\parallel =\parallel x_n-T_n{t}_n\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(32)

On the other hand, from (20), (24) and (26) it follows that

$$\begin{array}{cc}& \parallel x_{n+1}-x^{*}{\parallel }^2=(1-{\lambda }_n)\parallel w_n-x^{*}{\parallel }^2+{\lambda }_n\parallel T^n{z}_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le (1-{\lambda }_n)\parallel w_n-x^{*}{\parallel }^2+{\lambda }_n{(1+{\theta }_n)}^2\parallel z_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le \parallel w_n-x^{*}{\parallel }^2+{\theta }_n(2+{\theta }_n)\parallel z_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le {\beta }_n\delta \parallel x_n-x^{*}{\parallel }^2+{\gamma }_n\parallel x_n-x^{*}{\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel t_n-x^{*}\parallel }^2+{\beta }_n{K}_2\hfill \\ & +{\theta }_n(2+{\theta }_n)\parallel z_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le {\beta }_n\delta (\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3)+{\gamma }_n(\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3)+(1-{\beta }_n\ell -{\gamma }_n)(\parallel x_n-x^{*}{\parallel }^2\hfill \\ & +{\beta }_n{K}_3)+{\beta }_n{K}_2+{\theta }_n(2+{\theta }_n)\parallel z_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & =[1-{\beta }_n(\ell -\delta )](\parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3)+{\beta }_n{K}_2+{\theta }_n(2+{\theta }_n){\parallel z_n-x^{*}\parallel }^2\hfill \\ & -{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2+{\beta }_n{K}_3+{\beta }_n{K}_2+{\theta }_n(2+{\theta }_n)\parallel z_n-x^{*}{\parallel }^2-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2+({\beta }_n+{\theta }_n)K_4-{\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2,\hfill \end{array}$$

which together with $\left\{{\lambda }_n\right\}\subset [\underline{\lambda },\overline{\lambda }]\subset (0,1)$, arrives at

$$\begin{array}{cc}\underline{\lambda }(1-\overline{\lambda }){\parallel w_n-T^n{z}_n\parallel }^2\hfill & \le {\lambda }_n(1-{\lambda }_n){\parallel w_n-T^n{z}_n\parallel }^2\hfill \\ & \le \parallel x_n-x^{*}{\parallel }^2-{\parallel x_{n+1}-x^{*}\parallel }^2+({\beta }_n+{\theta }_n)K_4\hfill \\ & ={\Gamma }_n-{\Gamma }_{n+1}+({\beta }_n+{\theta }_n)K_4.\hfill \end{array}$$

Since ${\Gamma }_n-{\Gamma }_{n+1}\to 0,{\beta }_n\to 0$ and ${\theta }_n\to 0$, one obtains

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

Therefore, we have

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & =\parallel (1-{\lambda }_n)(w_n-x_n)+{\lambda }_n(T^n{z}_n-x_n)\parallel \hfill \\ \hfill & =\parallel (1-{\lambda }_n)(w_n-x_n)+{\lambda }_n(T^n{z}_n-w_n+w_n-x_n)\parallel \hfill \\ \hfill & =\parallel w_n-x_n+{\lambda }_n(T^n{z}_n-w_n)\parallel \hfill \\ \hfill & \le \parallel w_n-x_n\parallel +\parallel T^n{z}_n-w_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(33)

and

$$\begin{array}{cc}\parallel x_n-T^n{x}_n\parallel \hfill & \le \parallel x_n-w_n\parallel +\parallel w_n-T^n{z}_n\parallel +\parallel T^n{z}_n-T^n{x}_n\parallel \hfill \\ \hfill & \le \parallel x_n-w_n\parallel +\parallel w_n-T^n{z}_n\parallel +(1+{\theta }_n)\parallel z_n-x_n\parallel \to 0(n\to \infty ).\hfill \end{array}$$

(34)

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

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

Since H is reflexive and $\left\{x_n\right\}$ is bounded, we may assume, without loss of generality, that $x_{n_k}\rightharpoonup \tilde{x}$. Hence we get

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

(35)

Note that $x_n-x_{n+1}\to 0,x_n-z_n\to 0,x_n-T_n{x}_n\to 0$ and $x_n-T^n{x}_n\to 0$ (due to (31)–(34)). Since $T^n{x}_n-T^{n+1}{x}_n\to 0$ and $x_{n_k}\rightharpoonup \tilde{x}$ with $\left\{x_{n_k}\right\}\subset \left\{x_n\right\}$, by Lemma 10 we infer that $\tilde{x}\in \Omega$. Thus, using (19) and (35) one has

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

(36)

which immediately leads to

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

(37)

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

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

Consequently, applying Lemma 4 to (29), we have ${lim}_{n\to 0}{\parallel x_n-x^{*}\parallel }^2=0$.

Case 2. Assume that $\exists \left\{{\Gamma }_{n_l}\right\}\subset \left\{{\Gamma }_n\right\}$ s.t. ${\Gamma }_{n_l}<{\Gamma }_{n_l+1}\forall l\in \mathcal{N}$, where $\mathcal{N}$ is the set of all positive integers. Define the mapping $\wp :\mathcal{N}\to \mathcal{N}$ by

$$\wp \left(n\right):=max\{l\le n:{\Gamma }_l<{\Gamma }_{l+1}\}.$$

In terms of Lemma 7, we obtain

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

Putting $p=x^{*}$, from (27) we have

$$\begin{array}{cc}& {\displaystyle \underset{n\to \infty }{lim\; sup}}\underline{\lambda }(1-\mu \frac{{\tau }_{\wp \left(n\right)}}{{\tau }_{\wp \left(n\right)+1}})(\parallel y_{\wp \left(n\right)}-w_{\wp \left(n\right)}{\parallel }^2+\parallel z_{\wp \left(n\right)}-y_{\wp \left(n\right)}{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}{\lambda }_{\wp \left(n\right)}(1-\mu \frac{{\tau }_{\wp \left(n\right)}}{{\tau }_{\wp \left(n\right)+1}})(\parallel y_{\wp \left(n\right)}-w_{\wp \left(n\right)}{\parallel }^2+\parallel z_{\wp \left(n\right)}-y_{\wp \left(n\right)}{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}({\Gamma }_{\wp \left(n\right)}-{\Gamma }_{\wp \left(n\right)+1}+({\beta }_{\wp \left(n\right)}+{\theta }_{\wp \left(n\right)})K_4)=0.\hfill \end{array}$$

This Hence ensures that

$$\underset{n\to \infty }{lim}\parallel y_{\wp \left(n\right)}-w_{\wp \left(n\right)}\parallel =0\mathrm{and}\underset{n\to \infty }{lim}\parallel z_{\wp \left(n\right)}-y_{\wp \left(n\right)}\parallel =0.$$

Using the same inferences as in the proof of Case 1, we deduce that ${lim}_{n\to \infty }\parallel x_{\wp \left(n\right)}-z_{\wp \left(n\right)}\parallel =0$,

$$\underset{n\to \infty }{lim}\parallel x_{\wp \left(n\right)}-T_{\wp \left(n\right)}{x}_{\wp \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel x_{\wp \left(n\right)+1}-x_{\wp \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel x_{\wp \left(n\right)}-T^{\wp \left(n\right)}{x}_{\wp \left(n\right)}\parallel =0,$$

and

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

On the other hand, from (29) we obtain

$$\begin{array}{cc}& {\beta }_{\wp \left(n\right)}(\ell -\delta ){\Gamma }_{\wp \left(n\right)}\hfill \\ & \le {\Gamma }_{\wp \left(n\right)}-{\Gamma }_{\wp \left(n\right)+1}+{\beta }_{\wp \left(n\right)}(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_{\wp \left(n\right)}}{3{\beta }_{\wp \left(n\right)}}+\frac{{\alpha }_{\wp \left(n\right)}}{{\beta }_{\wp \left(n\right)}}\parallel x_{\wp \left(n\right)}-x_{\wp \left(n\right)-1}\parallel )+\frac{2\langle (f-\rho F)x^{*},w_{\wp \left(n\right)}-x^{*}\rangle }{\ell -\delta }]\hfill \\ & \le {\beta }_{\wp \left(n\right)}(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_{\wp \left(n\right)}}{3{\beta }_{\wp \left(n\right)}}+\frac{{\alpha }_{\wp \left(n\right)}}{{\beta }_{\wp \left(n\right)}}\parallel x_{\wp \left(n\right)}-x_{\wp \left(n\right)-1}\parallel )+\frac{2\langle (f-\rho F)x^{*},w_{\wp \left(n\right)}-x^{*}\rangle }{\ell -\delta }],\hfill \end{array}$$

which hence yields

$$\underset{n\to \infty }{lim\; sup}{\Gamma }_{\wp \left(n\right)}\le \underset{n\to \infty }{lim\; sup}[\frac{3K}{\ell -\delta }(\frac{{\theta }_{\wp \left(n\right)}}{3{\beta }_{\wp \left(n\right)}}+\frac{{\alpha }_{\wp \left(n\right)}}{{\beta }_{\wp \left(n\right)}}\parallel x_{\wp \left(n\right)}-x_{\wp \left(n\right)-1}\parallel )+\frac{2\langle (f-\rho F)x^{*},w_{\wp \left(n\right)}-x^{*}\rangle }{\ell -\delta }]\le 0.$$

Thus, ${lim}_{n\to \infty }{\parallel x_{\wp \left(n\right)}-x^{*}\parallel }^2=0$. Moreover, note that

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

Owing to ${\Gamma }_n\le {\Gamma }_{\wp \left(n\right)+1}$, we get

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

That is, $x_n\to x^{*}$ as $n\to \infty$. This completes the proof. □

Theorem 2.

Suppose that $T:H\to H$ is nonexpansive and $\left\{x_n\right\}$ is the sequence constructed by the modified version of Algorithm Section 3, i.e., for any starting points $x_0,x_1\in H$,

$$\left\{\begin{array}{cc}& t_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & w_n={\beta }_{n}f\left(x_n\right)+{\gamma }_n{x}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T_n{t}_n,\hfill \\ & y_n=P_C(w_n-{\tau }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n),\hfill \\ & x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_{n}T{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

where for each $n\ge 1$, ${\alpha }_n$, ${\tau }_n$ and $C_n$ are chosen as in Algorithm 3. Then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \Omega$, which is the unique solution to the HVI: $\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \Omega$.

Proof. 

We divide the proof of the theorem into several steps.

Claim 1. We prove that $\left\{x_n\right\}$ is bounded. In fact, using the same arguments as in Claim 1 of the proof of Theorem 1, we obtain the desired assertion.

Claim 2. We prove that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\beta }_n{K}_4$$

for some $K_4>0$. In fact, using the same argument as in Claim 2 of the proof of Theorem 1, we obtain the desired assertion.

Claim 3. We prove that

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

for some $K>0$. In fact, using the same argument as in Claim 3 of the proof of Theorem 1, we obtain the desired assertion.

Claim 4. We prove that $\left\{x_n\right\}$ converges strongly to the unique solution $x^{*}\in \Omega$ of the HVI (19), with $T_0=T$ being a nonexpansive mapping. In fact, setting $p=x^{*}$, one deduces from Claim 3 that

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

(38)

Putting ${\Gamma }_n={\parallel x_n-x^{*}\parallel }^2$, we demonstrate ${\Gamma }_n\to 0(n\to \infty )$ by considering the following cases.

Case 1. Assume that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Gamma }_n\right\}$ is nonincreasing. Then the limit ${lim}_{n\to \infty }{\Gamma }_n=\varsigma <+\infty$ and ${lim}_{n\to \infty }({\Gamma }_n-{\Gamma }_{n+1})=0$.

Using Claim 2, ${\Gamma }_n-{\Gamma }_{n+1}\to 0,{\beta }_n\to 0,1-\mu \frac{{\tau }_n}{{\tau }_{n+1}}\to 1-\mu$ and $\left\{{\lambda }_n\right\}\subset [\underline{\lambda },\overline{\lambda }]\subset (0,1)$, one has

$$\begin{array}{cc}& {\displaystyle \underset{n\to \infty }{lim\; sup}}\underline{\lambda }(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & \le {\displaystyle \underset{n\to \infty }{lim\; sup}}({\Gamma }_n-{\Gamma }_{n+1}+{\beta }_n{K}_4)=0,\hfill \end{array}$$

which immediately yields

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

Hence, one gets

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

(39)

Using the same inferences as in Case 1 of the proof of Theorem 1, we deduce that 0,

$$\underset{n\to \infty }{lim}\parallel x_n-z_n\parallel =\underset{n\to \infty }{lim}\parallel x_n-T_n{x}_n\parallel =0,$$

(40)

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \infty }{lim}\parallel x_n-T^n{x}_n\parallel =0,$$

(41)

and

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

(42)

As a result, applying Lemma 4 to (38), one conclude that ${lim}_{n\to \infty }{\parallel x_n-x^{*}\parallel }^2=0$.

Case 2. Assume that $\exists \left\{{\Gamma }_{n_l}\right\}\subset \left\{{\Gamma }_n\right\}$ s.t. ${\Gamma }_{n_l}<{\Gamma }_{n_l+1}\forall l\in \mathcal{N}$, with $\mathcal{N}$ being the set of all natural numbers. Let the mapping $\wp :\mathcal{N}\to \mathcal{N}$ be defined as

$$\wp \left(n\right):=max\{l\le n:{\Gamma }_l<{\Gamma }_{l+1}\}.$$

From Lemma 7, we have

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

In the rest of the proof, using the same inferences as in Case 2 of the proof of Theorem 1, we derive the desired result. This completes the proof. □

Next, we formulate another strengthened inertial-type subgradient extragradient rule below.

It is worth pointing out that Lemmas 8–10 are still valid for Algorithm 4.

Theorem 3.

Suppose that $\left\{x_n\right\}$ is the sequence constructed in Algorithm 4. Then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \Omega$ provided $T^n{x}_n-T^{n+1}{x}_n\to 0$, where $x^{*}\in \Omega$ is the unique solution to the HVI: $\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \Omega$.

Proof. 

Since ${lim}_{n\to \infty }\frac{{\theta }_n}{{\beta }_n}=0$ and $0<{lim\; inf}_{n\to \infty }{\gamma }_n\le {lim\; sup}_{n\to \infty }{\gamma }_n<1$, we might assume that ${\theta }_n\le \frac{{\beta }_n(\ell -\delta )}{2}\forall n\ge 1$ and $\left\{{\gamma }_n\right\}\subset [a,b]\subset (0,1)$. Using the same arguments as in the proof of Theorem 1, we deduce that there exists the unique solution $x^{*}\in \Omega ={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ of the VIP (20).

Next we show the conclusion of the theorem. To the goal, we divide the remainder of the proof into several steps.

Algorithm 4: Strengthened inertial-type subgradient extragradient rules

Initialization: Give ${\tau }_1>0,\alpha >0,\mu \in (0,1)$. Let $x_0,x_1\in H$ be arbitrary and choose ${\alpha }_n$ s.t.

$$0\le {\alpha }_n\le {\tilde{\alpha }}_n:=\left\{\begin{array}{cc}\hfill & min\{\alpha ,\frac{{ϵ}_n}{\parallel x_n-x_{n-1}\parallel }\},\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ \hfill & \alpha ,\mathrm{otherwise}.\hfill \end{array}\right.$$ (43)

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

Step 1. Set $t_n=x_n+{\alpha }_n(x_n-x_{n-1})$, and calculate

$w_n={\beta }_{n}f\left(x_n\right)+{\gamma }_n{t}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T_n{x}_n$

and $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$.

Step 2. Calculate $x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_n{T}^n{P}_{C_n}(w_n-{\tau }_{n}A{y}_n)$ with

$C_n:=\{x\in H:\langle w_n-{\tau }_{n}A{w}_n-y_n,x-y_n\rangle \le 0\}$.

Step 3. Update $${\tau }_{n+1}=\left\{\begin{array}{cc}& min\{\mu \frac{\parallel w_n-y_n{\parallel }^2+{\parallel z_n-y_n\parallel }^2}{2\langle A{w}_n-A{y}_n,z_n-y_n\rangle },{\tau }_n\},\mathrm{if}\langle A{w}_n-A{y}_n,z_n-y_n\rangle >0,\hfill \\ \hfill & {\tau }_n,\mathrm{otherwise},\hfill \end{array}\right.$$ (44)

with $z_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n)$. Again set $n:=n+1$ and go to Step 1.

Claim 1. We prove that $\left\{x_n\right\}$ is bounded. In fact, using the same arguments as in Step 1 of the proof of Theorem 1, one obtains that inequalities (20)–(22) hold. So, from ${\beta }_n+{\gamma }_n<1$, Lemma 6 and (22) it follows that

$$\begin{array}{cc}\parallel z_n-p\parallel \hfill & \le {\beta }_n\delta \parallel x_n-p\parallel +{\gamma }_n\parallel t_n-p\parallel +(1-{\gamma }_n-{\beta }_n\ell )\parallel x_n-p\parallel +{\beta }_n\parallel (f-\rho F)p\parallel \hfill \\ & \le {\beta }_n\delta (\parallel x_n-p\parallel +{\beta }_n{K}_1)+{\gamma }_n(\parallel x_n-p\parallel +{\beta }_n{K}_1)+(1-{\gamma }_n-{\beta }_n\ell )\hfill \\ & \times (\parallel x_n-p\parallel +{\beta }_n{K}_1)+{\beta }_n\parallel (f-\rho F)p\parallel \hfill \\ & \le \parallel x_n-p\parallel +{\beta }_n(K_1+\parallel (f-\rho F)p\parallel ).\hfill \end{array}$$

Noticing $x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_n{T}^n{z}_n$, from the last inequality one gets

$$\begin{array}{cc}\parallel x_{n+1}-p\parallel \hfill & \le \parallel w_n-p\parallel +{\theta }_n\parallel w_n-p\parallel \hfill \\ & \le [1-\frac{{\beta }_n(\ell -\delta )}{2}]\parallel x_n-p\parallel +\frac{{\beta }_n(\ell -\delta )}{2}·\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\hfill \\ & \le max\{\parallel x_n-p\parallel ,\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\}.\hfill \end{array}$$

By induction, we obtain $\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{3(K_1+\parallel (f-\rho F)p\parallel )}{\ell -\delta }\}\forall n\ge 1$. Thus, $\left\{x_n\right\}$ is bounded, and so are the sequences $\left\{t_n\right\},\left\{w_n\right\},\left\{y_n\right\},\left\{z_n\right\},\left\{f\left(x_n\right)\right\},\left\{T_n{x}_n\right\},\left\{T^n{z}_n\right\}$.

Claim 2. We prove that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+({\theta }_n+{\beta }_n)K_4$$

for some $K_4>0$. Using the similar arguments to that of (24), one gets

$$\begin{array}{c}\parallel w_n{-p\parallel }^2\le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel t_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel x_n-p\parallel }^2+{\beta }_n{K}_2,\hfill \end{array}$$

(45)

where ${sup}_{n\ge 1}2\parallel (f-\rho F)p\parallel \parallel w_n-p\parallel \le K_2$ for some $K_2>0$. Utilizing the same arguments as those of (25) and (26), we have

$$\parallel x_{n+1}{-p\parallel }^2\le \parallel w_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2,$$

(46)

and

$$\parallel t_n{-p\parallel }^2\le {\parallel x_n-p\parallel }^2+{\beta }_n{K}_3,$$

(47)

where ${sup}_{n\ge 1}(2K_1\parallel x_n-p\parallel +{\beta }_n{K}_1^2)\le K_3$ for some $K_3>0$. From (45)–(47), we get

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n(\parallel x_n{-p\parallel }^2+{\beta }_n{K}_3)+(1-{\beta }_n\ell -{\gamma }_n){\parallel x_n-p\parallel }^2+{\beta }_n{K}_2\hfill \\ & -{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\hfill \\ & +{\beta }_n{K}_3+{\beta }_n{K}_2+{\theta }_n(2+{\theta }_n){\parallel z_n-p\parallel }^2\hfill \\ & \le \parallel x_n{-p\parallel }^2-{\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)+({\theta }_n+{\beta }_n)K_4,\hfill \end{array}$$

where ${sup}_{n\ge 1}[K_2+K_3+(2+{\theta }_n)\parallel z_n-p{\parallel }^2]\le K_4$. This immediately implies that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+({\theta }_n+{\beta }_n)K_4.$$

(48)

Claim 3. We prove that

$$\begin{array}{cc}\parallel x_{n+1}{-p\parallel }^2\hfill & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n{-p\parallel }^2+{\beta }_n(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_n}{3{\beta }_n}+\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel )\hfill \\ & +\frac{2\langle (f-\rho F)p,w_n-p\rangle }{\ell -\delta }]\hfill \end{array}$$

for some $K>0$. In fact, one has

$$\parallel t_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel [2\parallel x_n-p\parallel +{\alpha }_n\parallel x_n-x_{n-1}\parallel ].$$

(49)

Combining (20), (45) and (49), we have

$$\begin{array}{cc}\hfill & \parallel x_{n+1}{-p\parallel }^2\hfill \\ \hfill & \le {\beta }_n\delta \parallel x_n{-p\parallel }^2+{\gamma }_n\parallel t_n{-p\parallel }^2+(1-{\beta }_n\ell -{\gamma }_n){\parallel x_n-p\parallel }^2+2{\beta }_n\langle (f-\rho F)p,w_n-p\rangle \hfill \\ \hfill & +{\theta }_n(2+{\theta }_n){\parallel w_n-p\parallel }^2\hfill \\ \hfill & \le [1-{\beta }_n(\ell -\delta )]\parallel x_n{-p\parallel }^2+{\beta }_n(\ell -\delta )[\frac{3K}{\ell -\delta }(\frac{{\theta }_n}{3{\beta }_n}+\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel )+\frac{2\langle (f-\rho F)p,w_n-p\rangle }{\ell -\delta }],\hfill \end{array}$$

(50)

where ${sup}_{n\ge 1}\{\parallel x_n-p\parallel ,{\alpha }_n\parallel x_n-x_{n-1}\parallel ,(2+{\theta }_n)\parallel w_n-p{\parallel }^2\}\le K$ for some $K>0$.

Claim 4. We prove that $\left\{x_n\right\}$ converges strongly to a unique solution $x^{*}\in \Omega$ of the HVI (19). In fact, using the same arguments as in Claim 4 of the proof of Theorem 1, we obtain the desired assertion. □

Theorem 4.

Suppose that $T:H\to H$ is nonexpansive and $\left\{x_n\right\}$ is the sequence constructed by the modified version of Algorithm 4, i.e., for any starting points $x_0,x_1\in H$,

$$\left\{\begin{array}{cc}& t_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ & w_n={\beta }_{n}f\left(x_n\right)+{\gamma }_n{t}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T_n{x}_n,\hfill \\ & y_n=P_C(w_n-{\tau }_{n}A{w}_n),\hfill \\ & z_n=P_{C_n}(w_n-{\tau }_{n}A{y}_n),\hfill \\ & x_{n+1}=(1-{\lambda }_n)w_n+{\lambda }_{n}T{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

where for each $n\ge 1$, ${\alpha }_n$, ${\tau }_n$ and $C_n$ are chosen as in Algorithm 4. Then $\left\{x_n\right\}$ converges strongly to $x^{*}\in \Omega$, which is the unique solution to the HVI: $\langle (\rho F-f)x^{*},p-x^{*}\rangle \ge 0\forall p\in \Omega$.

Proof. 

We divide the proof of the theorem into several steps.

Claim 1. We prove that $\left\{x_n\right\}$ is bounded. In fact, using the same arguments as in Claim 1 of the proof of Theorem 2, we obtain the desired assertion.

Claim 2. We prove that

$${\lambda }_n(1-\mu \frac{{\tau }_n}{{\tau }_{n+1}})(\parallel y_n-w_n{\parallel }^2+\parallel z_n-y_n{\parallel }^2)\le \parallel x_n{-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+{\theta }_n{k}_4$$

for some $K_4>0$. In fact, using the same arguments as in Claim 2 of the proof of Theorem 2, we obtain the desired assertion.

Claim 3. We prove that

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

for some $K>0$. In fact, using the same arguments as in Claim 3 of the proof of Theorem 2, we obtain the desired assertion.

Claim 4. We prove that $\left\{x_n\right\}$ converges strongly to a unique solution $x^{*}\in \Omega$ of the HVI (19), with $T_0=T$ a nonexpansive mapping. In fact, using the same arguments as in Claim 4 of the proof of Theorem 2, we obtain the desired assertion. This completes the proof. □

Remark 1.

Compared with the corresponding results in Kraikaew and Saejung [20], Ceng et al. [27], Thong et al. [16] and Ceng and Shang [22], our results improve and extend them in the following aspects.

(i) 

The problem of finding an element of $\mathrm{VI}(C,A)$ in [20] is extended to develop our problem of finding an element of ${\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ where $T_i$ is nonexpansive for $i=1,\dots ,N$ and $T_0=T$ is asymptotically nonexpansive. The Halpern subgradient extragradient method for solving the VIP in [20] is extended to develop our strengthened inertial-type subgradient extragradient rule with adaptive step sizes for solving the VIP and CFPP on the basis of Mann iteration method, viscosity approximation method and hybrid steepest-descent method.

(ii) 

The problem of finding an element of $\mathrm{VI}(C,A)$ in [16] is extended to develop our problem of finding an element of ${\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ where $T_i$ is nonexpansive for $i=1,\dots ,N$ and $T_0=T$ is asymptotically nonexpansive. The inertial subgradient extragradient method for solving the VIP in [16] is extended to develop our strengthened inertial-type subgradient extragradient rule for solving the VIP and the CFPP of finitely many nonexpansive mappings and an asymptotically nonexpansive mapping on the basis of Mann iteration method, viscosity approximation method and hybrid steepest-descent method.

(iii) 

The problem of finding an element of ${\bigcap }_{i=1}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ (where $T_i$ is nonexpansive for $i=1,\dots ,N$) in [27] is extended to develop our problem of finding an element of ${\cap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ where $T_i$ is nonexpansive for $i=1,\dots ,N$ and $T_0=T$ is asymptotically nonexpansive. The modified inertial subgradient extragradient method in [27] is extended to develop our strengthened inertial-type subgradient extragradient rule involving asymptotically nonexpansive mapping. It is worth pointing out that the modified inertial subgradient extragradient method in [27] only concerns nonexpansive mappngs.

(iv) 

While the problem of finding an element of ${\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$ in [22] is still studied in this paper, the hybrid inertial subgradient extragradient method with line-search process in [22] is extended to develop our strengthened inertial-type subgradient extragradient rule with adaptive step sizes, e.g., the original inertial technique $w_n=T_n{x}_n+{\alpha }_n(T_n{x}_n-T_n{x}_{n-1})$ is extended to develop our strengthened inertial-type iteration approach $t_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and $w_n={\beta }_{n}f\left(x_n\right)+{\gamma }_n{x}_n+((1-{\gamma }_n)I-{\beta }_n\rho F)T_n{t}_n$. Our strong convergence theorems are more beneficial and more subtle than the corresponding theorems [22] because the conclusion $x_n\to x^{*}\in \Omega \iff \parallel x_n-x_{n+1}\parallel +\parallel x_n-y_n\parallel \to 0(n\to \infty )$ in the corresponding theorems [22] is updated by our conclusion $x_n\to x^{*}\in \Omega$.

4. Applicability and Implementability of Rules

In this section, our main results are exploited to solve the VIP and CFPP in an illustrated example. Put ${\tau }_1=\alpha =\mu =\frac{1}{2},{\lambda }_n=\frac{1}{3},{\beta }_n=\frac{1}{3(n+1)},{\gamma }_n=\frac{n}{3(n+1)}$ and ${ϵ}_n=\frac{1}{3{(n+1)}^2}$.

We next provide an example of Lipschitz continuous and pseudomonotone mapping A, asymptotically nonexpansive mapping T and nonexpansive mapping $T_1$ with $\Omega =\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-2,3]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. The starting points $x_0,x_1$ are randomly chosen in H. Take $\rho =2$ and $f\left(x\right)=F\left(x\right)=\frac{1}{2}x\forall x\in H$. Then $\delta =\kappa =\eta =\frac{1}{2}$, $\rho =2\in (0,\frac{2\eta }{{\kappa }^2})=(0,4)$ and

$$\delta =\frac{1}{2}<\ell =1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}=1.$$

Let $A:H\to H$ and $T,T_1:H\to H$ be defined as $Ax:=\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|}$, $Tx:=\frac{2}{3}sinx$ and $T_{1}x:=sinx$ for all $x\in H$. Let us show that A is pseudomonotone and Lipschitz continuous. In fact, for all $x,y\in H$ we have

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

This ensures that A is of Lipschitz continuity on H. Moreover, we claim that A is pseudomonotone. Actually, it is easy to see that for all $x,y\in H$,

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

Moreover, it is easy to verify that T is asymptotically nonexpansive with ${\theta }_n={\left(\frac{2}{3}\right)}^n\forall n\ge 1$, such that $\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \to 0$ as $n\to \infty$. In fact, we note that

$$\parallel T^{n}x-T^{n}y\parallel \le \frac{2}{3}\parallel T^{n-1}x-T^{n-1}y\parallel \le \cdots \le {\left(\frac{2}{3}\right)}^n\parallel x-y\parallel \le (1+{\theta }_n)\parallel x-y\parallel ,$$

and

$$\parallel T^{n+1}{x}_n-T^n{x}_n\parallel \le {\left(\frac{2}{3}\right)}^{n-1}\parallel T^2{x}_n-T{x}_n\parallel ={\left(\frac{2}{3}\right)}^{n-1}\parallel \frac{2}{3}sin\left(T{x}_n\right)-\frac{2}{3}sin{x}_n\parallel \le 2{\left(\frac{2}{3}\right)}^n\to 0.$$

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

$$\underset{n\to \infty }{lim}\frac{{\theta }_n}{{\beta }_n}=\underset{n\to \infty }{lim}\frac{{(2/3)}^n}{1/3(n+1)}=0.$$

In addition, it is clear that $T_1$ is nonexpansive and $\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$. Therefore, $\Omega =\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 3 can be rewritten below:

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

(51)

(with $(1-{\gamma }_n)I-{\beta }_n\rho F=\frac{2}{3}I$), where for each $n\ge 1$, ${\alpha }_n$, ${\tau }_n$ and $C_n$ are chosen as in Algorithm 3. Then, by Theorem 1, we know that $\left\{x_n\right\}$ converges to $0\in \Omega =\mathrm{Fix}\left(T_1\right)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$.

In particular, since $Tx:=\frac{2}{3}sinx$ is also nonexpansive, we consider the modified version of Algorithm 3, that is,

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

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

5. Conclusions

In this article, we have put forward two strengthened inertial-type subgradient extragradient rules with adaptive step sizes for solving the VIP and CFPP in Hilbert spaces, where the VIP denotes a pseudomonotone variational inequality problem with Lipschitz continuity operator A, and the CFPP indicates a common fixed-point problem of finitely many nonexpansive mappings ${\left\{T_i\right\}}_{i=1}^N$ and an asymptotically nonexpansive mapping $T_0:=T$. Under the lack of the sequential weak continuity and Lipschitz constant of the cost operator A, we have demonstrated the strong convergence of the suggested rules to a common solution of the VIP and CFPP, which is the unique solution of a hierarchical variational inequality (HVI) defined on the common solution set $\Omega :={\bigcap }_{i=0}^N\mathrm{Fix}\left(T_i\right)\cap \mathrm{VI}(C,A)$. In addition, an illustrated example is provided to demonstrate the applicability and implementability of our proposed rules.