Modified Mann-Type Subgradient Extragradient Rules for Variational Inequalities and Common Fixed Points Implicating Countably Many Nonexpansive Operators ^†

Abstract

In a real Hilbert space, let the CFPP, VIP, and HFPP denote the common fixed-point problem of countable nonexpansive operators and asymptotically nonexpansive operator, variational inequality problem, and hierarchical fixed point problem, respectively. With the help of the Mann iteration method, a subgradient extragradient approach with a linear-search process, and a hybrid deepest-descent technique, we construct two modified Mann-type subgradient extragradient rules with a linear-search process for finding a common solution of the CFPP and VIP. Under suitable assumptions, we demonstrate the strong convergence of the suggested rules to a common solution of the CFPP and VIP, which is only a solution of a certain HFPP.

1. Introduction

Throughout this paper, we assume that $P_C$ is the metric projection of H onto C, with $\langle ·,·\rangle$ and $\parallel ·\parallel$ denoting the inner product and induced norm of real Hilbert space H and C being a convex and closed set satisfying $\varnothing \ne C\subset H$. Given nonlinear mapping $S:C\to H$, let the $\mathrm{Fix}\left(S\right)$ and $\mathbf{R}$ indicate the fixed-point set of S and the real-number set, respectively. In the fixed point theory, we recall an important class of mappings. A self-mapping S on C is known as being asymptotically nonexpansive iff $\exists {\left\{{\theta }_i\right\}}_{i=1}^{\infty }\subset [0,+\infty )$ s.t. ${lim}_{i\to \infty }{\theta }_i=0$ and

$$\parallel S^{i}u-S^{i}v\parallel \le \parallel u-v\parallel +{\theta }_i\parallel u-v\parallel \forall i\ge 1,u,v\in C.$$

(1)

In particular, whenever ${\theta }_i=0\forall i\ge 1$, S is said to be nonexpansive. In the past several decades, the fixed point theory has played a key role in solving real-world problems such as the time-fractional biological population model [1], fractional multi-dimensional system of boundary value problems on the methylpropane graph [2], traumatic avoidance learning model [3], and so forth.

Given a self-mapping A on H, we consider the classical variational inequality problem (VIP) of finding $u\in C$ s.t. $\langle Au,v-u\rangle \ge 0\forall v\in C$. Its solution set is written as VI($C,A$). To the best of our awareness, one of the most effective techniques for treating the VIP is the extragradient one put forward by Korpelevich [4] in 1976, i.e., for any starting point $u_0\in C$, $\left\{u_i\right\}$ is fabricated below

$$\left\{\begin{array}{cc}& v_i=P_C(u_i-\ell A{u}_i),\hfill \\ \hfill & u_{i+1}=P_C(u_i-\ell A{v}_i)\forall i\ge 0,\hfill \end{array}\right.$$

(2)

where $\ell \in (0,\frac{1}{L})$ and L is Lipschitz constant of A. Whenever $\mathrm{VI}(C,A)\ne \varnothing$, the sequence $\left\{u_i\right\}$ converges weakly to a point in $\mathrm{VI}(C,A)$. At present, the vast literature on Korpelevich’s extragradient technique shows that many authors have paid great attention to it and enhanced this technique in different manners; for details, refer to [5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28] and references therein, to name but a few.

Very recently, Xie et al. [9] suggested the amended inertial extragradient approach with a line-search process for solving the pseudomonotone VIP in H. Let $f:H\to H$ be a contraction with constant $\delta \in [0,1)$ and assume that $\Omega :=\mathrm{VI}(C,A)\ne \varnothing$. Given the sequences $\left\{{\alpha }_i\right\},\left\{{\beta }_i\right\}\subset (0,1]$ such that ${lim}_{i\to \infty }{\beta }_i=0$ and ${\sum }_{i=1}^{\infty }{\beta }_i=\infty$. Their approach is formulated by Algorithm 1 below:

Algorithm 1 Modified inertial extragradient approach (see [9])

Initial Step: Let $\varsigma \in (0,1),\ell \in (0,1),\mu \in (0,1)\gamma \in (0,\infty )$, given any starting points $x_1,x_0$ in H. Iterations: Given the iterates $x_{i-1},x_i(i\ge 1)$, compute $x_{i+1}$ below: Step 1. Set $w_i=x_i+{\alpha }_i(x_i-x_{i-1})$. Step 2. Calculate $v_i=P_C(w_i-{\tau }_{i}A{w}_i)$ and $z_i=P_C(w_i-\varsigma {\tau }_{i}A{v}_i)$, where ${\tau }_i:=\gamma {\ell }^{m_i}$ and $m_i$ is the smallest nonnegative integer m such that $$\gamma {\ell }^m\langle A{w}_i-A{v}_i,z_i-v_i\rangle \le \mu \parallel w_i-v_i\parallel \parallel z_i-v_i\parallel .$$ If $w_i=v_i$ or $A{v}_i=0$, then stop and $v_i$ is an element of $\Omega$. Otherwise, go to Step 3. Step 3. Calculate $x_{i+1}={\beta }_{i}f\left(z_i\right)+(1-{\beta }_i)z_i$. If $w_i=z_i=x_{i+1}$, then $w_i\in \Omega$. Again, set $i:=i+1$ and go to Step 1.

  Under appropriate assumptions, they showed the strong convergence of $\left\{x_i\right\}$ to the solution $p=P_{\Omega }\circ f\left(p\right)$ provided ${lim}_{i\to \infty }\frac{{\alpha }_i}{{\beta }_i}\parallel x_i-x_{i-1}\parallel =0$. In the extragradient technique, two projections onto C have to be calculated per one iteration. In 2018, Thong and Hieu [22] first proposed the inertial subgradient extragradient method, and then proved the weak convergence of this method to an element of $\mathrm{VI}(C,A)$ under mild assumptions. In 2019, Thong and Hieu [17] proposed the inertial-type subgradient extragradient method with a linear-search process for settling the VIP with monotone and Lipschitzian operator A and the fixed-point problem (FPP) of a quasi-nonexpansive operator S with the demiclosedness in H. Assume that $\Omega :=\mathrm{Fix}\left(S\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Given the sequences $\left\{{\alpha }_i\right\}\subset [0,1]$ and $\left\{{\beta }_i\right\}\subset (0,1)$. Their method is formulated by Algorithm 2 below:

Algorithm 2 Inertial-type subgradient extragradient method (see [17])

Initial Step: Let $\nu \in (0,1),l\in (0,1),\gamma \in (0,\infty )$, given any starting points $x_1,x_0$ in H. Iterations: Compute $x_{i+1}$ below: Step 1. Put $w_i=x_i+{\alpha }_i(x_i-x_{i-1})$ and calculate $v_i=P_C(w_i-{\varsigma }_{i}A{w}_i)$, where ${\varsigma }_i$ is picked to be the largest $\varsigma \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\varsigma \parallel A{w}_i-A{v}_i\parallel \le \nu \parallel w_i-v_i\parallel .$$ Step 2. Calculate $z_i=P_{C_i}(w_i-{\varsigma }_{i}A{v}_i)$ with $C_i:=\{v\in H:\langle w_i-{\varsigma }_{i}A{w}_i-v_i,v-v_i\rangle \le 0\}$. Step 3. Calculate $x_{i+1}=(1-{\beta }_i)w_i+{\beta }_{i}S{z}_i$. If $w_i=z_i=x_{i+1}$, then $w_i\in \Omega$. Again, set $i:=i+1$ and go to Step 1.

Under suitable assumptions, it was proven in [17] that $\left\{x_i\right\}$ converges weakly to a point in $\Omega$. Subsequently, Ceng and Shang [25] proposed the hybrid inertial subgradient extragradient rule with a linear-search process for settling the VIP with Lipschitzian pseudomonotonicity operator A and the common fixed-point problem (CFPP) of finite nonexpansive operators ${\left\{S_k\right\}}_{k=1}^N$ and asymptotically nonexpansive operator S on H. Assume that $\Omega :=\mathrm{VI}(C,A)\cap \left({\bigcap }_{k=0}^N\mathrm{Fix}\left(S_k\right)\right)\ne \varnothing$ with $S_0:=S$. Given a $\delta$-contractive map $g:H\to H$ with $\delta \in [0,1)$, and an operator $F:H\to H$ of both $\eta$-strong monotonicity and $\kappa$-Lipschitz continuity, fulfilling $\delta <\zeta :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$ with $0<\mu <\frac{2\eta }{{\kappa }^2}$. Let $\left\{{\beta }_i\right\},\left\{{\alpha }_i\right\}\subset (0,1)$ and $\left\{{ϵ}_i\right\}\subset [0,1]$ s.t. ${\alpha }_i+{\beta }_i<1\forall i\ge 1$. In addition, one writes $S_i:=S_{i\mathrm{mod}N}$ for each $i\ge 1$, where the mod function takes values in $\{1,\dots ,N\}$, that is, whenever $i=qN+j$ for some integers $q\ge 0$ and $0\le j<N$, one has that $S_i=S_N$ in the case of $j=0$ and $S_i=S_j$ in the case of $0<j<N$. Their rule is formulated by Algorithm 3 below:

Algorithm 3 Hybrid inertial subgradient extragradient rule (see [25])

Initial Step: Let $\nu \in (0,1),l\in (0,1),\gamma \in (0,\infty )$, given any starting points $x_1,x_0$ in H. Iterations: Compute $x_{i+1}$ below: Step 1. Put $w_i=S_i{x}_i+{ϵ}_i(S_i{x}_i-S_i{x}_{i-1})$ and calculate $y_i=P_C(w_i-{\varsigma }_{i}A{w}_i)$, with ${\varsigma }_i$ being picked to be the largest $\varsigma \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\varsigma \parallel A{w}_i-A{y}_i\parallel \le \nu \parallel w_i-y_i\parallel .$$ Step 2. Calculate $q_i=P_{C_i}(w_i-{\varsigma }_{i}A{y}_i)$ with $C_i:=\{y\in H:\langle w_i-{\varsigma }_{i}A{w}_i-y_i,y_i-y\rangle \ge 0\}$. Step 3. Calculate $x_{i+1}={\alpha }_{i}g\left(x_i\right)+{\beta }_i{x}_i+((1-{\beta }_i)I-{\alpha }_i\mu F)S^i{q}_i$. Again, set $i:=i+1$ and go to Step 1.

Under appropriate assumptions, it was proven in [25] that, if $S^i{q}_i-S^{i+1}{q}_i\to 0$, then $\left\{x_i\right\}$ converges strongly to $q^{\ast }\in \Omega$ if and only if $x_i-x_{i+1}\to 0$ and $x_i-y_i\to 0$ as $i\to \infty$, with $q^{\ast }\in \Omega$ being only a solution to the hierarchical fixed point problem (HFPP): $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$.

In the rest of this paper, we always assume that the CFPP and HFPP denote the common fixed-point problem of countable nonexpansivity operators ${\left\{S_i\right\}}_{i=1}^{\infty }$ and asymptotical nonexpansivity operator $S_0:=S$ and hierarchical fixed-point problem, respectively. With the help of the Mann iteration method, a subgradient extragradient approach with a linear-search process, and hybrid deepest-descent technique, we construct two amended Mann-type subgradient extragradient rules with a linear-search process for finding a common solution of the CFPP of ${\left\{S_i\right\}}_{i=0}^{\infty }$ and the VIP for pseudomonotone operator A. Via suitable conditions, we show the strong convergence of the proposed rules to a point in $\Omega :=\mathrm{VI}(C,A)\cap \left({\bigcap }_{k=0}^{\infty }\mathrm{Fix}\left(S_k\right)\right)$, which is only a solution of a certain HFPP. In the end, using the main results, we deal with the CFPP and VIP in an illustrated example.

The architecture of this paper is arranged as follows: In Section 2, we recollect certain concepts and basic tools for subsequent applications. In Section 3, we prove the strong convergence of the proposed rules. Finally, in Section 4, the main theorems are exploited to settle the CFPP and VIP in a demonstrated instance. Our rules are more general and more subtle than the above algorithms because they implicate settling the VIP for pseudomonotone operator and the CFPP for countable nonexpansive operators and an asymptotically nonexpansive operator. Our theorems ameliorate and develop the associated theorems pronounced in Xie et al. [9], Ceng and Shang [25], and Thong and Hieu [17].

2. Preliminaries

Given a sequence $\left\{{\upsilon }_i\right\}\subset H$, let ${\upsilon }_i\rightharpoonup \upsilon$ (resp., ${\upsilon }_i\to \upsilon$) represent the weak (resp., strong) convergence of $\left\{{\upsilon }_i\right\}$ to $\upsilon$. A mapping $S:C\to H$ is referred to as being

(a)

of L-Lipschitz continuity (or of L-Lipschitzian property) iff $\exists L>0$ s.t. $L\parallel p-q\parallel \ge \parallel Sp-Sq\parallel \forall p,q\in C$;

(b)

of monotonicity iff $0\le \langle Sp-Sq,p-q\rangle \forall p,q\in C$;

(c)

of pseudomonotonicity iff $\langle Sp,q-p\rangle \ge 0\Rightarrow \langle Sq,q-p\rangle \ge 0\forall p,q\in C$;

(d)

of $\eta$-strong monotonicity iff $\exists \eta >0$ s.t. ${\eta \parallel p-q\parallel }^2\le \langle Sp-Sq,p-q\rangle \forall p,q\in C$;

(e)

of sequential weak continuity iff $\forall \left\{q_i\right\}\subset C$, the relation holds: $q_i\rightharpoonup q\Rightarrow S{q}_i\rightharpoonup Sq$.

Obviously, each monotonicity mapping is of pseudomonotonicity. However, the inverse is false. It is known that $\forall q\in H$, $\exists |$ (nearest point) $P_{C}q\in C$ s.t. $\parallel q-p\parallel \ge \parallel q-P_{C}q\parallel \forall p\in C$. $P_C$ is refereed to as a nearest point (or metric) projection from H onto C. The statements below are valid (see [29]):

(a)

$\parallel P_{C}q-P_C{p\parallel }^2\le \langle q-p,P_{C}q-P_{C}p\rangle \forall q,p\in H$;

(b)

$p=P_{C}q\iff \langle q-p,t-p\rangle \le 0\forall q\in H,t\in C$;

(c)

$\parallel q-P_C{q\parallel }^2+\parallel P_C{q-t\parallel }^2\ge {\parallel q-t\parallel }^2\forall q\in H,t\in C$;

(d)

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

(e)

${\parallel sq+(1-s)p\parallel }^2={s\parallel q\parallel }^2+{(1-s)\parallel p\parallel }^2-s(1-s){\parallel q-p\parallel }^2\forall q,p\in H,s\in [0,1]$.

The following concept and two propositions can be found in [30].

 Definition 1. 

Let ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }\subset [0,1]$ and suppose that ${\left\{S_i\right\}}_{i=1}^{\infty }$ is a sequence of nonexpansive operators from C into itself. For any $k\ge 1$, the self-mapping $W_k$ on C is constructed as follows:

$$\left\{\begin{array}{cc}& U_{k,k+1}=I,\hfill \\ \hfill & U_{k,k}={\xi }_k{S}_k{U}_{k,k+1}+(1-{\xi }_k)I,\hfill \\ \hfill & U_{k,k-1}={\xi }_{k-1}{S}_{k-1}{U}_{k,k}+(1-{\xi }_{k-1})I,\hfill \\ \hfill & \cdots \hfill \\ \hfill & U_{k,i}={\xi }_i{S}_i{U}_{k,i+1}+(1-{\xi }_i)I,\hfill \\ \hfill & \cdots \hfill \\ \hfill & U_{k,2}={\xi }_2{S}_2{U}_{k,3}+(1-{\xi }_2)I,\hfill \\ \hfill & W_k=U_{k,1}={\xi }_1{S}_1{U}_{k,2}+(1-{\xi }_1)I.\hfill \end{array}\right.$$

(3)

Then, $W_k$ is refereed to as a W-operator fabricated by $S_k,\dots ,S_2,S_1$ and ${\xi }_k,\dots ,{\xi }_2,{\xi }_1$.

 Proposition 1. 

Let ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }\subset (0,1]$ and suppose that ${\left\{S_i\right\}}_{i=1}^{\infty }$ is a sequence of nonexpansive operators from C into itself, such that ${\bigcap }_{i=1}^{\infty }\mathrm{Fix}\left(S_i\right)\ne \varnothing$. Then,

 (a) 

$W_k$is of nonexpansivity and${\bigcap }_{i=1}^k\mathrm{Fix}\left(S_i\right)=\mathrm{Fix}\left(W_k\right)\forall k\ge 1$;

 (b) 

$\forall q\in C,i\ge 1$, ${lim}_{k\to \infty }{U}_{k,i}q$ exists;

 (c) 

the operator W, formulated as $Wq:={lim}_{k\to \infty }{W}_{k}q={lim}_{k\to \infty }{U}_{k,1}q\forall q\in C$, is a nonexpansive operator s.t. ${\bigcap }_{i=1}^{\infty }\mathrm{Fix}\left(S_i\right)=\mathrm{Fix}\left(W\right)$, and it is refereed to as the W-operator fabricated by $S_1,S_2,\dots$ and ${\xi }_1,{\xi }_2,\dots$.

 Proposition 2. 

Let ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }\subset (0,\varsigma ]$ for certain $\varsigma \in (0,1)$ and suppose that ${\left\{S_i\right\}}_{i=1}^{\infty }$ is a sequence of nonexpansive operators from C into itself, such that ${\bigcap }_{i=1}^{\infty }\mathrm{Fix}\left(S_i\right)\ne \varnothing$. Then, ${lim}_{k\to \infty }{sup}_{p\in D}\parallel W_{k}p-Wp\parallel =0$ for each bounded set $D\subset C$.

Throughout this paper, we always assume that ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }\subset (0,\varsigma ]$ for some $\varsigma \in (0,1)$. Later on, we will make use of the following lemmas to demonstrate our main results.

 Lemma 1 

([28]). Let $H_1$ and $H_2$ be two real Hilbert spaces. Suppose that $F:H_1\to H_2$ is of uniform continuity on each boundedness subset of $H_1$ and D is of boundedness in $H_1$. Then, $F\left(D\right)$ is of boundedness.

It is clear that the relation below holds for the inner product in H:

$$2\langle p,q+p\rangle +{\parallel q\parallel }^2\ge {\parallel q+p\parallel }^2\forall q,p\in H.$$

(4)

 Lemma 2 

([31]). Each Hilbert space fulfills Opial’s condition, that is, $\forall \left\{q_n\right\}\subset H$ with $q_n\rightharpoonup q$, the relation ${lim\; inf}_{n\to \infty }\parallel q_n-p\parallel >{lim\; inf}_{n\to \infty }\parallel q_n-q\parallel \forall p\in H,p\ne q$ is true.

 Lemma 3 

([9]). Suppose that $F:C\to H$ of both pseudomonotonicity and continuity, given $u\in C$. Then, the relation holds: $\langle Fu,\upsilon -u\rangle \ge 0\forall \upsilon \in C\iff \langle F\upsilon ,\upsilon -u\rangle \ge 0\forall \upsilon \in C$.

 Lemma 4 

([32]). Suppose that the sequence $\left\{a_{\iota }\right\}\subset [0,\infty )$ is such that ${\lambda }_{\iota }{\gamma }_{\iota }+(1-{\lambda }_{\iota })a_{\iota }\ge a_{\iota +1}\forall \iota \ge 1$, where the real sequences $\left\{{\lambda }_{\iota }\right\}$ and $\left\{{\gamma }_{\iota }\right\}$ satisfy the conditions: (a) $\left\{{\lambda }_{\iota }\right\}\subset [0,1]$ and ${\sum }_{\iota =1}^{\infty }{\lambda }_{\iota }=\infty$, and (b) ${lim\; sup}_{\iota \to \infty }{\gamma }_{\iota }\le 0$ or ${\sum }_{\iota =1}^{\infty }\left|{\lambda }_{\iota }{\gamma }_{\iota }\right|<\infty$. Then, ${lim}_{\iota \to \infty }{a}_{\iota }=0$.

 Lemma 5 

([33]). Let E be a Banach space and admit a duality mapping of weak continuity. Suppose that C is convex and closed such that $\varnothing \ne C\subset E$, and that S is asymptotically nonexpansive self-mapping on C such that $\mathrm{Fix}\left(S\right)\ne \varnothing$. Then, $I-S$ is demiclosed at zero, i.e., for $\left\{q_{\iota }\right\}\subset C$ satisfying both $q_{\iota }\rightharpoonup q\in C$ and $(I-S)q_{\iota }\to 0$, one has $(I-S)q=0$, with I being the identity operator of E.

The following lemmas are very crucial to the convergence analysis of our designed rules.

 Lemma 6 

([34]). Suppose that $\left\{{\Phi }_m\right\}$ is a sequence in $\mathbf{R}$, which does not decrease at infinity, that is, $\exists \left\{{\Phi }_{m_{\iota }}\right\}\subset \left\{{\Phi }_m\right\}$ s.t. ${\Phi }_{m_{\iota }}<{\Phi }_{m_{\iota }+1}\forall \iota \ge 1$. The sequence ${\left\{\phi \left(m\right)\right\}}_{m\ge m_0}$ of integers is formulated below:

$$\phi \left(m\right)=max\{\iota \le m:{\Phi }_{\iota }<{\Phi }_{\iota +1}\},$$

where $m_0\ge 1$ s.t. $\{\iota \le m_0:{\Phi }_{\iota }<{\Phi }_{\iota +1}\}\ne \varnothing$. Then, the statements hold below:

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

(b) ${\Phi }_{\phi \left(m\right)}\le {\Phi }_{\phi \left(m\right)+1}$ and ${\Phi }_m\le {\Phi }_{\phi \left(m\right)+1}\forall m\ge m_0$.

 Lemma 7 

([32]). Given a number λ in $(0,1]$, suppose that S is a nonexpansive self-mapping on C, and $S^{\lambda }:C\to H$ is the operator formulated as $S^{\lambda }p:=Sp-\lambda \rho F\left(Sp\right)\forall p\in C$, with $F:C\to H$ being of both κ-Lipschitz continuity and η-strong monotonicity. Then, $S^{\lambda }$ is a contractive map for $\rho \in (0,\frac{2\eta }{{\kappa }^2})$, that is, $\parallel S^{\lambda }q-S^{\lambda }p\parallel \le (1-\lambda \zeta )\parallel q-p\parallel \forall q,p\in C$, with $\zeta =1-\sqrt{1-\rho (2\eta -\rho {\kappa }^2)}\in (0,1]$.

3. Criteria of Strong Convergence

In what follows, let us suppose that the conditions are valid below.

${\left\{S_i\right\}}_{i=1}^{\infty }$ is a sequence of nonexpansive operators on H and S is asymptotically nonexpansive operator on H with $\left\{{\theta }_i\right\}$.

$W_n$ is the W-operator constructed by $S_n,S_{n-1},\dots ,S_1$ and ${\xi }_n,{\xi }_{n-1},\dots ,{\xi }_1$, with ${\left\{{\xi }_i\right\}}_{i=1}^{\infty }\subset (0,\varsigma ]$ for certain $\varsigma \in (0,1)$.

A is of both pseudomonotonicity and L-Lipschitz continuity on H, s.t. $\parallel Au\parallel \le {lim\; inf}_{n\to \infty }$$\parallel A{\upsilon }_n\parallel$ for each $\left\{{\upsilon }_n\right\}\subset C$ with ${\upsilon }_n\rightharpoonup u$.

g is a $\delta$-contractive map on H with $\delta \in [0,1)$, and F is of $\eta$-strong monotonicity and $\kappa$-Lipschitz continuity on H s.t. $\delta <\zeta :=1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}$ with $0<\mu <\frac{2\eta }{{\kappa }^2})$.

$\Omega =\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)\ne \varnothing$ where $S_0:=S$.

$\left\{{ϵ}_n\right\},\left\{{\sigma }_n\right\}\subset [0,1]$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}\subset (0,1)$ with ${\alpha }_n+{\beta }_n<1\forall n\ge 1$, s.t.

(i)

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

(ii)

${sup}_{n\ge 1}({ϵ}_n/{\alpha }_n)<\infty$ and ${lim}_{n\to \infty }({\theta }_n/{\alpha }_n)=0$;

(iii)

$1>{lim\; sup}_{n\to \infty }{\sigma }_n\ge {lim\; inf}_{n\to \infty }{\sigma }_n>0$;

(iv)

$1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$.

 Lemma 8. 

The linear-search process (6) in the following Algorithm 4 is well formulated, and the relation holds: $\gamma \ge {\varsigma }_n\ge min\{\frac{\nu l}{L},\gamma \}$.

 Proof. 

Note that $\parallel A{u}_n-A{P}_C(u_n-\gamma l^{m}A{u}_n)\parallel \le L\parallel u_n-P_C(u_n-\gamma l^{m}A{u}_n)\parallel$. Then, (6) is valid for each $\gamma l^m\le \frac{\nu }{L}$ and ${\varsigma }_n$ is well defined. Clearly, ${\varsigma }_n\le \gamma$. When ${\varsigma }_n=\gamma$, the conclusion is true. When ${\varsigma }_n<\gamma$, from (6), one obtains $\parallel A{u}_n-A{P}_C(u_n-\frac{{\varsigma }_n}{l}A{u}_n)\parallel >\frac{\nu }{({\varsigma }_n/l)}\parallel u_n-P_C(u_n-\frac{{\varsigma }_n}{l}A{u}_n)\parallel$, which immediately yields ${\varsigma }_n>\frac{\nu l}{L}$. Thus, the conclusion is true.    □

 Lemma 9. 

Suppose that the sequences $\left\{w_n\right\},\left\{u_n\right\},\left\{y_n\right\},\left\{q_n\right\}$ are constructed in Algorithm 4. Then,

$$\begin{array}{cc}\parallel q_n{-p\parallel }^2\hfill & \le \parallel w_n{-p\parallel }^2-(1-\nu )\parallel y_n-q_n{\parallel }^2-(1-\nu ){\parallel y_n-u_n\parallel }^2\hfill \\ \hfill & -{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2\forall p\in \Omega .\hfill \end{array}$$

(5)

Algorithm 4 The 1st modified Mann-type subgradient extragradient rule

Initial Steps: Let $l\in (0,1),\nu \in (0,1),\gamma \in (0,\infty )$, given any starting points $x_1,x_0$ in H. Iterations: Calculate $x_{n+1}(n\ge 1)$ below: Step 1. Set $w_n=x_n+{ϵ}_n(x_n-x_{n-1})$ and $u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n$, and calculate $y_n=P_C(u_n-{\varsigma }_{n}A{u}_n)$, with ${\varsigma }_n$ being picked to be the largest $\varsigma \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\varsigma \parallel A{u}_n-A{y}_n\parallel \le \nu \parallel u_n-y_n\parallel .$$ (6) Step 2. Calculate $q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n)$ with $C_n:=\{y\in H:\langle u_n-{\varsigma }_{n}A{u}_n-y_n,y_n-y\rangle \ge 0\}$. Step 3. Calculate $$x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n.$$ (7) Put $n:=n+1$ and return to Step 1.

 Proof. 

It is clear that $C_n\supset C\supset \Omega$. Observe that $\forall p\in \Omega$,

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

Thus, one has

$$\parallel q_n{-p\parallel }^2\le \parallel u_n{-p\parallel }^2-{\parallel q_n-u_n\parallel }^2-2{\varsigma }_n\langle q_n-p,A{y}_n\rangle .$$

(8)

Thanks to $q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n)$ where $C_n:=\{y\in H:\langle u_n-{\varsigma }_{n}A{u}_n-y_n,y_n-y\rangle \ge 0\}$, one obtains $\langle u_n-{\varsigma }_{n}A{u}_n-y_n,y_n-q_n\rangle \ge 0$. Using the pseudomonotonicity of A, from (8) and (6), we deduce that

$$\begin{array}{cc}\parallel q_n{-p\parallel }^2\hfill & \le \parallel u_n{-p\parallel }^2-{\parallel q_n-u_n\parallel }^2-2{\varsigma }_n\langle A{y}_n,y_n-p+q_n-y_n\rangle \hfill \\ \hfill & \le \parallel u_n{-p\parallel }^2-{\parallel q_n-u_n\parallel }^2-2{\varsigma }_n\langle A{y}_n,q_n-y_n\rangle \hfill \\ \hfill & =\parallel u_n{-p\parallel }^2-\parallel q_n-y_n{\parallel }^2-{\parallel y_n-u_n\parallel }^2+2\langle u_n-{\varsigma }_{n}A{y}_n-y_n,q_n-y_n\rangle \hfill \\ \hfill & =\parallel u_n{-p\parallel }^2-\parallel q_n-y_n{\parallel }^2-{\parallel y_n-u_n\parallel }^2+2\langle u_n-{\varsigma }_{n}A{u}_n-y_n,q_n-y_n\rangle \hfill \\ \hfill & +2{\varsigma }_n\langle A{u}_n-A{y}_n,q_n-y_n\rangle \hfill \\ \hfill & \le \parallel u_n{-p\parallel }^2-\parallel q_n-y_n{\parallel }^2-\parallel y_n-u_n{\parallel }^2+2\nu \parallel u_n-y_n\parallel \parallel q_n-y_n\parallel \hfill \\ \hfill & \le \parallel u_n{-p\parallel }^2-\parallel q_n-y_n{\parallel }^2-\parallel y_n-u_n{\parallel }^2+\nu (\parallel u_n-y_n{\parallel }^2+\parallel q_n-y_n{\parallel }^2)\hfill \\ \hfill & =\parallel u_n{-p\parallel }^2-(1-\nu )\parallel y_n-q_n{\parallel }^2-(1-\nu ){\parallel y_n-u_n\parallel }^2.\hfill \end{array}$$

(9)

Owing to $u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n$, one has

$$\begin{array}{cc}\parallel u_n{-p\parallel }^2\hfill & =(1-{\sigma }_n)\parallel w_n{-p\parallel }^2+{\sigma }_n\parallel W_n{w}_n{-p\parallel }^2-{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2\hfill \\ & \le (1-{\sigma }_n)\parallel w_n{-p\parallel }^2+{\sigma }_n\parallel w_n{-p\parallel }^2-{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2\hfill \\ & =\parallel w_n{-p\parallel }^2-{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2.\hfill \end{array}$$

Consequently, this, together with (9), ensures that inequality (5) is true.    □

 Lemma 10. 

Suppose that $\left\{w_n\right\},\left\{u_n\right\},\left\{x_n\right\},\left\{q_n\right\}$ are boundedness sequences constructed in Algorithm 4. Assume that $S^n{x}_n-S^{n+1}{x}_n\to 0,u_n-x_n\to 0,w_n-q_n\to 0$ and $x_n-x_{n+1}\to 0$. Then, ${\omega }_w\left(\left\{x_n\right\}\right)\subset \Omega$, where ${\omega }_w\left(\left\{x_n\right\}\right)=\{q\in H:x_{n_l}\rightharpoonup q\mathrm{for}\mathrm{some}\left\{x_{n_l}\right\}\subset \left\{x_n\right\}\}$.

 Proof. 

Take a fixed $q\in {\omega }_w\left(\left\{x_n\right\}\right)$ arbitrarily. Then, $\exists \left\{x_{n_l}\right\}\subset \left\{x_n\right\}$ s.t. $x_{n_l}\rightharpoonup q\in H$. Thanks to $u_n-x_n\to 0$, we know that $\exists \left\{u_{n_l}\right\}\subset \left\{u_n\right\}$ s.t. $u_{n_l}\rightharpoonup q\in H$. In what follows, we claim $q\in \Omega$. In fact, by Lemma 9, we obtain that, for each $p\in \Omega$,

$$\begin{array}{cc}& (1-\nu )\parallel y_n-q_n{\parallel }^2+(1-\nu )\parallel y_n-u_n{\parallel }^2+{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2\hfill \\ & \le \parallel w_n{-p\parallel }^2-\parallel q_n{-p\parallel }^2\le \parallel w_n-q_n\parallel (\parallel w_n-p\parallel +\parallel q_n-p\parallel ).\hfill \end{array}$$

Since $w_n-q_n\to 0,\nu \in (0,1)$ and $1>{lim\; sup}_{n\to \infty }{\sigma }_n\ge {lim\; inf}_{n\to \infty }{\sigma }_n>0$, from boundedness of $\left\{w_n\right\},\left\{q_n\right\}$, we deduce that

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

This immediately yields

$$\parallel u_n-q_n\parallel \le \parallel u_n-y_n\parallel +\parallel y_n-q_n\parallel \to 0(n\to \infty ).$$

Clearly, one has $\parallel w_n-x_n\parallel ={ϵ}_n\parallel x_n-x_{n-1}\parallel \to 0$ (due to ${sup}_{n\ge 1}({ϵ}_n/{\alpha }_n)<\infty$). Hence, we have

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

(10)

Noticing $x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n$, we obtain $x_{n+1}-S^n{q}_n={\alpha }_{n}g\left(x_n\right)+{\beta }_n(x_n-S^n{q}_n)-{\alpha }_n\mu F{S}^n{q}_n$, which immediately yields

$$\begin{array}{cc}\parallel x_n-S^n{q}_n\parallel \hfill & \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-S^n{q}_n\parallel \hfill \\ & \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel g\left(x_n\right)\parallel +{\beta }_n\parallel x_n-S^n{q}_n\parallel +{\alpha }_n\parallel \mu F{S}^n{q}_n\parallel .\hfill \end{array}$$

Thus, it follows that

$$(1-{\beta }_n)\parallel x_n-S^n{q}_n\parallel \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel ).$$

Since $x_{n+1}-x_n\to 0,{\alpha }_n\to 0,{lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$ and $\left\{x_n\right\},\left\{q_n\right\}$ are of boundedness, one obtains

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

We claim $\parallel x_n-S{x}_n\parallel \to 0(n\to \infty )$. Indeed, using the asymptotical nonexpansivity of S, one deduces that

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

Since $u_n-x_n\to 0$, $u_n-q_n\to 0$ and $x_n-S^n{q}_n\to 0$, we obtain

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

(11)

In addition, let us show that ${lim}_{n\to \infty }\parallel x_n-W{x}_n\parallel =0$. In fact, note that

$$\parallel W{x}_n-x_n\parallel \le \parallel W{x}_n-W_n{x}_n\parallel +\parallel W_n{x}_n-x_n\parallel \le \underset{u\in D}{sup}\parallel Wu-W_{n}u\parallel +\parallel W_n{x}_n-x_n\parallel ,$$

where $D=\{x_n:n\ge 1\}$. Using Proposition 2, from (10), we obtain

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

(12)

   □

In what follows, we claim $q\in \mathrm{VI}(C,A)$. Indeed, noticing $u_n-y_n\to 0$ and $u_{n_l}\rightharpoonup q$, we have $y_{n_l}\rightharpoonup q$. In addition, noticing $\left\{y_n\right\}\subset C$ and $y_{n_l}\rightharpoonup q$, by the convexity and closedness of C, one obtains $q\in C$. Next, we discuss two situations. When $Aq=0$, it is readily known that $q\in \mathrm{VI}(C,A)$ (due to $\langle Aq,y-q\rangle \ge 0\forall y\in C$).

Let $Aq\ne 0$. Since $y_{n_l}\rightharpoonup q$ as $l\to \infty$, using the hypothesis on A, one obtains ${lim\; inf}_{l\to \infty }\parallel A{y}_{n_l}\parallel$$\ge \parallel Aq\parallel >0$. Hence, one might assume $\parallel A{y}_{n_l}\parallel \ne 0\forall l\ge 1$. Moreover, using $y_n=P_C(u_n-{\varsigma }_{n}A{u}_n)$, one has $\langle u_n-{\varsigma }_{n}A{u}_n-y_n,y-y_n\rangle \le 0\forall y\in C$, and hence

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

(13)

Since A is uniform continuous, $\left\{A{u}_n\right\}$ is of boundedness (by Lemma 1). Noticing the boundedness of $\left\{y_n\right\}$, by Lemma 8 and (13), one obtains ${lim\; inf}_{l\to \infty }\langle A{u}_{n_l},$$y-u_{n_l}\rangle \ge 0\forall y\in C$. In addition, it is readily known that $\langle A{y}_n,y-y_n\rangle =\langle A{y}_n-A{u}_n,y-u_n\rangle +\langle A{u}_n,y-u_n\rangle +\langle A{y}_n,u_n-y_n\rangle$. Note that $u_n-y_n\to 0$ and A is uniform continuous. Thus, one obtains $A{y}_n-A{u}_n\to 0$. This hence arrives at ${lim\; inf}_{l\to \infty }\langle A{y}_{n_l},y-y_{n_l}\rangle \ge 0\forall y\in C$.

In order to demonstrate $q\in \mathrm{VI}(C,A)$, one chooses $\left\{{\kappa }_l\right\}\subset (0,1)$ s.t. ${\kappa }_l\downarrow 0(l\to \infty )$. For each l, one denotes by $m_l$ the smallest natural number satisfying

$$\langle A{y}_{n_i},y-y_{n_i}\rangle +{\kappa }_l\ge 0\forall i\ge m_l.$$

(14)

Note that $\left\{{\kappa }_l\right\}$ is of decreasement. Thus, it is readily known that $\left\{m_l\right\}$ is an increasing. Using $A{y}_{m_l}\ne 0\forall l\ge 1$ (owing to $\left\{A{y}_{m_l}\right\}\subset \left\{A{y}_{n_l}\right\}$), we set ${\upsilon }_{m_l}=\frac{A{y}_{m_l}}{\parallel A{y}_{m_l}{\parallel }^2}$, and obtain $\langle A{y}_{m_l},{\upsilon }_{m_l}\rangle =1\forall l\ge 1$. Thus, from (14), one obtains $\langle A{y}_{m_l},y+{\kappa }_l{\upsilon }_{m_l}-y_{m_l}\rangle \ge 0\forall l\ge 1$. In addition, by the pseudomonotonicity of A, one has $\langle A(y+{\kappa }_l{\upsilon }_{m_l}),y+{\kappa }_l{\upsilon }_{m_l}-y_{m_l}\rangle \ge 0\forall l\ge 1$. This immediately arrives at

$$\langle Ay,y-y_{m_l}\rangle \ge \langle Ay-A(y+{\kappa }_l{\upsilon }_{m_l}),y+{\kappa }_l{\upsilon }_{m_l}-y_{m_l}\rangle -{\kappa }_l\langle Ay,{\upsilon }_{m_l}\rangle \forall l\ge 1.$$

(15)

We show that ${lim}_{l\to \infty }{\kappa }_l{\upsilon }_{m_l}=0$. In fact, from $u_{n_l}\rightharpoonup q\in C$ and $u_n-y_n\to 0$, we obtain $y_{n_l}\rightharpoonup q$. Note that $\left\{y_{m_l}\right\}\subset \left\{y_{n_l}\right\}$ and ${\kappa }_l\downarrow 0(l\to \infty )$. Thus, one deduces that $0\le {lim\; sup}_{l\to \infty }\parallel {\kappa }_l{\upsilon }_{m_l}\parallel ={lim\; sup}_{l\to \infty }\frac{{\kappa }_l}{\parallel A{y}_{m_l}\parallel }\le \frac{{lim\; sup}_{l\to \infty }{\kappa }_l}{{lim\; inf}_{l\to \infty }\parallel A{y}_{n_l}\parallel }=0$. Therefore, one obtains ${\kappa }_l{\upsilon }_{m_l}\to 0(l\to \infty )$. Note that A is uniformly continuous, the sequences $\left\{y_{m_l}\right\},\left\{{\upsilon }_{m_l}\right\}$ are of boundedness, and ${lim}_{l\to \infty }{\kappa }_l{\upsilon }_{m_l}=0$. Consequently, letting $l\to \infty$, one concludes that $\langle Ay,y-q\rangle ={lim\; inf}_{l\to \infty }\langle Ay,y-y_{m_l}\rangle \ge 0\forall y\in C$. By Lemma 3, one has $q\in \mathrm{VI}(C,A)$.

Next, we show that $q\in \Omega$. In fact, since (11) guarantees $x_{n_l}-S{x}_{n_l}\to 0$, by Lemma 5, we obtain the demiclosedness of $I-S$ at zero. Thus, from $x_{n_l}\rightharpoonup q$, one obtains $(I-S)q=0$, that is, $q\in \mathrm{Fix}\left(S\right)$. In addition, we claim $q\in \mathrm{Fix}\left(W\right)={\bigcap }_{i=1}^{\infty }\mathrm{Fix}\left(S_i\right)$. Conversely, we suppose that $q\notin \mathrm{Fix}\left(W\right)$, that is, $Wq\ne q$. Using Lemma 2 and Proposition 1 (c), we obtain

$$\underset{l\to \infty }{lim\; inf}\parallel x_{n_l}-q\parallel <\underset{l\to \infty }{lim\; inf}\parallel x_{n_l}-Wq\parallel \le \underset{l\to \infty }{lim\; inf}(\parallel x_{n_l}-W{x}_{n_l}\parallel +\parallel W{x}_{n_l}-Wq\parallel ),$$

which together with (12) yields ${lim\; inf}_{l\to \infty }\parallel x_{n_l}-q\parallel <{lim\; inf}_{l\to \infty }\parallel x_{n_l}-q\parallel$, which leads to a contradiction. Thus, one has $q\in {\bigcap }_{i=1}^{\infty }\mathrm{Fix}\left(S_i\right)$. Consequently, $q\in {\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\cap \mathrm{VI}(C,A)$, that is, $q\in \Omega$.

 Theorem 1. 

Suppose that $\left\{x_n\right\}$ is the sequence constructed in Algorithm 4. Then,

$$x_n\to q^{\ast }\in \Omega \iff \left\{\begin{array}{cc}& S^n{x}_n-S^{n+1}{x}_n\to 0,\hfill \\ & \underset{n\ge 1}{sup}\parallel x_{n-1}-x_n\parallel <\infty ,\hfill \end{array}\right.$$

with $q^{\ast }\in \Omega$ being only a solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$.

 Proof. 

Because $1>{lim\; sup}_{n\to \infty }{\sigma }_n\ge {lim\; inf}_{n\to \infty }{\sigma }_n>0$ and ${lim}_{n\to \infty }{\theta }_n/{\alpha }_n=0$, we might suppose that $\left\{{\sigma }_n\right\}\subset [\overline{a},\overline{b}]\subset (0,1)$ and ${\beta }_n(\zeta -\delta )/2\ge {\theta }_n$ for all n. Let us show that $P_{\Omega }(I-\mu F+g):H\to H$ is the contractive map on H. Indeed, using Lemma 7, one has

$$\parallel P_{\Omega }(I-\mu F+g)u-P_{\Omega }(I-\mu F+g)\upsilon \parallel \le [1-(\zeta -\delta )]\parallel u-\upsilon \parallel \forall u,\upsilon \in H.$$

This ensures that $P_{\Omega }(I-\mu F+g)$ is a contractive map. Thus, it is readily known that there exists $q^{\ast }\in H$, which is only a fixed point of $P_{\Omega }(I-\mu F+g)$, that is, $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. That is, there exists $q^{\ast }\in \Omega$, which is only a solution to the following VIP:

$$\langle (\mu F-g)q^{\ast },q-q^{\ast }\rangle \ge 0\forall q\in \Omega .$$

(16)

We first show the necessity of the theorem. In fact, when $x_n\to q^{\ast }\in \Omega$, we know that $q^{\ast }=S{q}^{\ast }$ and

$$\begin{array}{cc}\parallel S^n{x}_n-S^{n+1}{x}_n\parallel \hfill & \le \parallel S^n{x}_n-q^{\ast }\parallel +\parallel q^{\ast }-S^{n+1}{x}_n\parallel \hfill \\ & \le (1+{\theta }_n)\parallel x_n-q^{\ast }\parallel +(1+{\theta }_{n+1})\parallel q^{\ast }-x_n\parallel \hfill \\ & =(2+{\theta }_n+{\theta }_{n+1})\parallel x_n-q^{\ast }\parallel \to 0(n\to \infty ).\hfill \end{array}$$

Since $\parallel x_{n+1}-q^{\ast }\parallel +\parallel q^{\ast }-x_n\parallel \ge \parallel x_{n+1}-x_n\parallel$, one has

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =0.$$

This immediately yields ${sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$.

In what follows, we claim the sufficiency of the theorem. To the goal, under the assumption $S^n{x}_n-S^{n+1}{x}_n\to 0$ with ${sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, we divide the remainder of the proof into several claims.    □

 Claim 1. 

One claims the boundedness of $\left\{x_n\right\}$. In fact, picking a $q\in \Omega$ arbitrarily, one has that $Sq=q$, $W_{n}q=q$, and (5) leads to

$$\begin{array}{cc}\parallel q_n{-q\parallel }^2\hfill & \le \parallel w_n{-q\parallel }^2-(1-\nu )\parallel y_n-q_n{\parallel }^2-(1-\nu ){\parallel y_n-u_n\parallel }^2\hfill \\ \hfill & -{\sigma }_n(1-{\sigma }_n){\parallel w_n-W_n{w}_n\parallel }^2\forall q\in \Omega ,\hfill \end{array}$$

(17)

which hence yields

$$\parallel q_n-q\parallel \le \parallel w_n-q\parallel .$$

(18)

By the formulation of $w_n$, one obtains

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

(19)

Noticing ${sup}_{n\ge 1}({ϵ}_n/{\alpha }_n)<\infty$ and ${sup}_{n\ge 1}\parallel x_n-x_{n-1}\parallel <\infty$, one obtains ${sup}_{n\ge 1}({ϵ}_n/{\alpha }_n)\parallel x_n-x_{n-1}\parallel <\infty$, which guarantees that $\exists M_1>0$ s.t.

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

(20)

From (18)–(20), one obtains

$$\parallel q_n-q\parallel \le \parallel w_n-q\parallel \le \parallel x_n-q\parallel +{\alpha }_n{M}_1.$$

(21)

In addition, observe that

$$\parallel u_n-q\parallel \le (1-{\sigma }_n)\parallel w_n-q\parallel +{\sigma }_n\parallel W_n{w}_n-q\parallel \le \parallel w_n-q\parallel ,$$

(22)

which, together with (9) and (21), yields

$$\parallel q_n-q\parallel \le \parallel u_n-q\parallel \le \parallel w_n-q\parallel \le \parallel x_n-q\parallel +{\alpha }_n{M}_1.$$

(23)

Thus, using (23) and ${\alpha }_n+{\beta }_n<1\forall n\ge 1$, from Lemma 7, we obtain

$$\begin{array}{cc}& \parallel x_{n+1}-q\parallel =\parallel {\alpha }_{n}g\left(x_n\right)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n-q\parallel \hfill \\ & =\parallel {\alpha }_n(g\left(x_n\right)-q)+{\beta }_n(x_n-q)+(1-{\alpha }_n-{\beta }_n)\{\frac{1-{\beta }_n}{1-{\alpha }_n-{\beta }_n}\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]+\frac{{\alpha }_n}{1-{\alpha }_n-{\beta }_n}(I-\mu F)q\}\parallel \hfill \\ & =\parallel {\alpha }_n(g\left(x_n\right)-g\left(p\right))+{\beta }_n(x_n-q)+(1-{\beta }_n)\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]+{\alpha }_n(f-\mu F)q\parallel \hfill \\ & \le {\alpha }_n\parallel g\left(x_n\right)-g\left(q\right)\parallel +{\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n)\hfill \\ & \times \parallel (I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q\parallel +{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le {\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n)\hfill \\ & \times (1-\frac{{\alpha }_n}{1-{\beta }_n}\zeta )(1+{\theta }_n)\parallel q_n-q\parallel +{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le {\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n(\parallel x_n-q\parallel +{\alpha }_n{M}_1)+(1-{\beta }_n-{\alpha }_n\zeta )\hfill \\ & \times (\parallel x_n-q\parallel +{\alpha }_n{M}_1)+{\theta }_n\parallel q_n-q\parallel +{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le [{\alpha }_n\delta +{\beta }_n+(1-{\beta }_n-{\alpha }_n\zeta )]\parallel x_n-q\parallel +{\alpha }_n{M}_1\hfill \\ & +\frac{{\alpha }_n(\zeta -\delta )(\parallel x_n-q\parallel +{\alpha }_n{M}_1)}{2}+{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le [1-\frac{{\alpha }_n(\zeta -\delta )}{2}]\parallel x_n-q\parallel +{\alpha }_n(2M_1+\parallel (g-\mu F)q\parallel )\hfill \\ & =[1-\frac{{\alpha }_n(\zeta -\delta )}{2}]\parallel x_n-q\parallel +\frac{{\alpha }_n(\zeta -\delta )}{2}·\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\hfill \\ & \le max\{\parallel x_n-q\parallel ,\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\},\hfill \end{array}$$

which immediately arrives at

$$\parallel x_n-q\parallel \le max\{\parallel x_1-q\parallel ,\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\}\forall n\ge 1.$$

Therefore, one obtains the boundedness of $\left\{x_n\right\}$. This ensures that $\left\{w_n\right\},\left\{u_n\right\},\left\{y_n\right\},\left\{q_n\right\},$$\left\{g\left(x_n\right)\right\}$, $\left\{A{y}_n\right\}$ and $\left\{W_n{w}_n\right\},\left\{S^n{q}_n\right\}$ are bounded.

 Claim 2. 

One claims that

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+({\alpha }_n+{\theta }_n)M_4,\hfill \end{array}$$

for certain $M_4>0$. In fact, one has

$$\begin{array}{cc}{x}_{n+1}-q\hfill & ={\alpha }_n(g\left(x_n\right)-q)+{\beta }_n(x_n-q)+(1-{\alpha }_n-{\beta }_n)\{\frac{1-{\beta }_n}{1-{\alpha }_n-{\beta }_n}\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]+\frac{{\alpha }_n}{1-{\alpha }_n-{\beta }_n}(I-\mu F)q\}\hfill \\ & ={\alpha }_n(g\left(x_n\right)-g\left(q\right))+{\beta }_n(x_n-q)+(1-{\beta }_n)\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]+{\alpha }_n(g-\mu F)q.\hfill \end{array}$$

Using the convex property of $\varphi \left(s\right)=s^2\forall s\in \mathbf{R}$, one obtains

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le \parallel {\alpha }_n(g\left(x_n\right)-g\left(q\right))+{\beta }_n(x_n-q)+(1-{\beta }_n)\hfill \\ \hfill & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]{\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le [{\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n)\hfill \\ \hfill & \times (1-\frac{{\alpha }_n}{1-{\beta }_n}\zeta )(1+{\theta }_n)\parallel q_n-q{\parallel ]}^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & =[{\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n\parallel x_n-q\parallel +(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\parallel q_n-q{\parallel ]}^2\hfill \\ \hfill & +2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel x_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n){\parallel q_n-q\parallel }^2+{\alpha }_n{M}_2\hfill \end{array}$$

(24)

(because of ${\alpha }_n\delta +{\beta }_n+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\le 1+{\alpha }_n(\delta -\zeta )+{\theta }_n\le 1-\frac{{\alpha }_n(\zeta -\delta )}{2}$), with ${sup}_{n\ge 1}2\parallel (g-\mu F)q\parallel \parallel x_n-q\parallel \le M_2$ for certain $M_2>0$. Combining (17) and (24), one obtains

$$\begin{array}{cc}& \parallel x_{n+1}{-q\parallel }^2\le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel x_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[\parallel w_n{-q\parallel }^2\hfill \\ \hfill & -(1-\nu )\parallel y_n-q_n{\parallel }^2-(1-\nu )\parallel y_n-u_n{\parallel }^2-{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2]+{\alpha }_n{M}_2.\hfill \end{array}$$

(25)

In addition, from (23), we have

$$\begin{array}{cc}\parallel w_n{-q\parallel }^2\hfill & \le (\parallel x_n-q{\parallel +{\alpha }_n{M}_1)}^2\hfill \\ \hfill & =\parallel x_n{-q\parallel }^2+{\alpha }_n(2M_1\parallel x_n-q\parallel +{\alpha }_n{M}_1^2)\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2+{\alpha }_n{M}_3,\hfill \end{array}$$

(26)

where ${sup}_{n\ge 1}(2M_1\parallel x_n-q\parallel +{\alpha }_n{M}_1^2)\le M_3$ for certain $M_3>0$. From (25) and (26), one obtains

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n[\parallel x_n{-q\parallel }^2+{\alpha }_n{M}_3]+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[\parallel x_n{-q\parallel }^2\hfill \\ & +{\alpha }_n{M}_3]-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[(1-\nu )\parallel y_n-q_n{\parallel }^2+(1-\nu ){\parallel y_n-u_n\parallel }^2\hfill \\ & +{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2]+{\alpha }_n{M}_2\hfill \\ & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[(1-\nu )(\parallel y_n-q_n{\parallel }^2\hfill \\ & +\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2]+({\alpha }_n+{\theta }_n)M_4\hfill \\ & \le \parallel x_n{-q\parallel }^2-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)\hfill \\ & +{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2]+({\alpha }_n+{\theta }_n)M_4,\hfill \end{array}$$

where ${sup}_{n\ge 1}(\parallel x_n-q{\parallel }^2+M_3+M_2)\le M_4$ for certain $M_4>0$. Consequently,

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2\}\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+({\alpha }_n+{\theta }_n)M_4.\hfill \end{array}$$

(27)

 Claim 3. 

One claims that

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q{\parallel }^2+{\alpha }_n(\zeta -\delta )\{\frac{2}{\zeta -\delta }\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right\}\hfill \end{array}$$

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

$$\begin{array}{cc}\parallel w_n{-q\parallel }^2\hfill & \le (\parallel x_n-q\parallel +{ϵ}_n\parallel x_n-x_{n-1}{\parallel )}^2\hfill \\ \hfill & =\parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel +{ϵ}_n\parallel x_n-x_{n-1}\parallel ).\hfill \end{array}$$

(28)

Using (23), (24) and (28), one obtains

$$\begin{array}{cc}& \parallel x_{n+1}{-q\parallel }^2\hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel x_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n){\parallel q_n-q\parallel }^2\hfill \\ \hfill & +2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel x_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta ){\parallel w_n-q\parallel }^2\hfill \\ \hfill & +{\theta }_n{\parallel q_n-q\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+(1-{\alpha }_n\zeta )[\parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel \hfill \\ \hfill & +{ϵ}_n\parallel x_n-x_{n-1}\parallel \left)\right]+{\theta }_n{\parallel q_n-q\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel \hfill \\ \hfill & +{ϵ}_n\parallel x_n-x_{n-1}\parallel )+{\theta }_n{\parallel q_n-q\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+({ϵ}_n\parallel x_n-x_{n-1}\parallel 3+{\theta }_n)M+2{ϵ}_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & =[1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+{\alpha }_n(\zeta -\delta )[\frac{2\langle (g-\mu F)q,x_{n+1}-q\rangle }{\zeta -\delta }+\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right]\hfill \end{array}$$

(29)

with ${sup}_{n\ge 1}\{\parallel x_n-q\parallel ,{ϵ}_n\parallel x_n-x_{n-1}\parallel ,\parallel q_n-q{\parallel }^2\}\le M$ for certain $M>0$.

 Claim 4. 

One claims that $x_n\to q^{\ast }\in \Omega$, which is only a solution to the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. In fact, using (29) with $q=q^{\ast }$, one obtains

$$\begin{array}{cc}\parallel x_{n+1}-q^{\ast }{\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q^{\ast }{\parallel }^2+{\alpha }_n(\zeta -\delta )[\frac{2\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle }{\zeta -\delta }\hfill \\ \hfill & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right].\hfill \end{array}$$

(30)

Putting ${\Phi }_n={\parallel x_n-q^{\ast }\parallel }^2$, one demonstrates ${\Phi }_n\to 0(n\to \infty )$ in both aspects below.

 Aspect 1. 

Suppose that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Phi }_n\right\}$ is non-increasing. It is clear that the limit ${lim}_{n\to \infty }{\Phi }_n=d<\infty$ and ${lim}_{n\to \infty }({\Phi }_n-{\Phi }_{n+1})=0$. Setting $q=q^{\ast }$, by (27) and $\left\{{\sigma }_n\right\}\subset [\overline{a},\overline{b}]\subset (0,1)$ one obtains

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+\overline{a}(1-\overline{b})\parallel w_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le \parallel x_n-q^{\ast }{\parallel }^2-{\parallel x_{n+1}-q^{\ast }\parallel }^2+({\alpha }_n+{\theta }_n)M_4\le {\Phi }_n-{\Phi }_{n+1}+({\alpha }_n+{\theta }_n)M_4.\hfill \end{array}$$

Noticing ${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$, ${\alpha }_n\to 0,{\theta }_n\to 0$ and ${\Phi }_n-{\Phi }_{n+1}\to 0$, one has

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

(31)

Thus, it follows that

$$\parallel u_n-q_n\parallel \le \parallel u_n-y_n\parallel +\parallel y_n-q_n\parallel \to 0(n\to \infty ).$$

(32)

Noticing $w_n-x_n={ϵ}_n(x_n-x_{n-1})$ and $u_n-w_n={\sigma }_n(W_n{w}_n-w_n)$, we obtain

$$\begin{array}{cc}\parallel u_n-x_n\parallel \hfill & \le \parallel u_n-w_n\parallel +\parallel w_n-x_n\parallel \hfill \\ & ={\sigma }_n\parallel W_n{w}_n-w_n\parallel +{ϵ}_n\parallel x_n-x_{n-1}\parallel \hfill \\ & \le \parallel W_n{w}_n-w_n\parallel +{\alpha }_n·\frac{{ϵ}_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \hfill \\ & \le \parallel W_n{w}_n-w_n\parallel +{\alpha }_n·{\displaystyle \underset{n\ge 1}{sup}}\frac{{ϵ}_n}{{\alpha }_n}·{\displaystyle \underset{n\ge 1}{sup}}\parallel x_n-x_{n-1}\parallel .\hfill \end{array}$$

Since ${sup}_{n\ge 1}\frac{{ϵ}_n}{{\alpha }_n}<\infty$, ${sup}_{n\ge 1}\parallel x_n-x_{n-1}\parallel <\infty$ and ${\alpha }_n\to 0$, using (31), one has

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

(33)

Moreover, noticing $x_{n+1}-q^{\ast }={\beta }_n(x_n-q^{\ast })+(1-{\beta }_n)(S^n{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n)$, we obtain from (23) that

$$\begin{array}{cc}& \parallel x_{n+1}-q^{\ast }{\parallel }^2={\parallel {\beta }_n(x_n-q^{\ast })+(1-{\beta }_n)(S^n{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n)\parallel }^2\hfill \\ & \le \parallel {\beta }_n(x_n-q^{\ast })+(1-{\beta }_n)(S^n{q}_n-q^{\ast }){\parallel }^2+2\langle {\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n),x_{n+1}-q^{\ast }\rangle \hfill \\ & \le {\beta }_n\parallel x_n-q^{\ast }{\parallel }^2+(1-{\beta }_n)\parallel S^n{q}_n-q^{\ast }{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-S^n{q}_n\parallel }^2\hfill \\ & +2\parallel {\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n)\parallel \parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & \le {\beta }_n\parallel x_n-q^{\ast }{\parallel }^2+(1-{\beta }_n){(1+{\theta }_n)}^2\parallel q_n-q^{\ast }{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-S^n{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & \le {\beta }_n{(1+{\theta }_n)}^2(\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2+(1-{\beta }_n){(1+{\theta }_n)}^2(\parallel x_n-q^{\ast }{\parallel +{\alpha }_n{M}_1)}^2\hfill \\ & -{\beta }_n(1-{\beta }_n)\parallel x_n-S^n{q}_n{\parallel }^2+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & ={(1+{\theta }_n)}^2(\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2-{\beta }_n(1-{\beta }_n){\parallel x_n-S^n{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & ={(1+{\theta }_n)}^2\parallel x_n-q^{\ast }{\parallel }^2+{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1]\hfill \\ & -{\beta }_n(1-{\beta }_n)\parallel x_n-S^n{q}_n{\parallel }^2+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel .\hfill \end{array}$$

This hence arrives at

$$\begin{array}{cc}& {\beta }_n(1-{\beta }_n)\parallel x_n-S^n{q}_n{\parallel }^2\le {(1+{\theta }_n)}^2\parallel x_n-q^{\ast }{\parallel }^2-{\parallel x_{n+1}-q^{\ast }\parallel }^2\hfill \\ & +{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1]+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & \le {(1+{\theta }_n)}^2{\Phi }_n-{\Phi }_{n+1}+{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2{\Phi }_n^{\frac{1}{2}}+{\alpha }_n{M}_1]+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel ){\Phi }_{n+1}^{\frac{1}{2}}.\hfill \end{array}$$

Since $1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$, ${\theta }_n\to 0,{\alpha }_n\to 0,{\Phi }_n-{\Phi }_{n+1}\to 0$ and ${lim}_{n\to \infty }{\Phi }_n=d<+\infty$, from the boundedness of $\left\{g\left(x_n\right)\right\},\left\{S^n{q}_n\right\}$, we infer that

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

Thus, it follows that

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+(1-{\beta }_n)(S^n{q}_n-x_n)-{\alpha }_n\mu F{S}^n{q}_n\parallel \hfill \\ \hfill & \le (1-{\beta }_n)\parallel S^n{q}_n-x_n\parallel +{\alpha }_n\parallel g\left(x_n\right)-\mu F{S}^n{q}_n\parallel \hfill \\ \hfill & \le \parallel S^n{q}_n-x_n\parallel +{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\to 0(n\to \infty ).\hfill \end{array}$$

(34)

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

$$\underset{n\to \infty }{lim\; sup}\langle (g-\mu F)q^{\ast },x_n-q^{\ast }\rangle =\underset{\iota \to \infty }{lim}\langle (g-\mu F)q^{\ast },x_{n_{\iota }}-q^{\ast }\rangle .$$

(35)

Noticing the reflexivity of H and boundedness of $\left\{x_n\right\}$, one might suppose that $x_{n_{\iota }}\rightharpoonup \tilde{q}$. Hence, using (35), we obtain

$$\begin{array}{cc}{\displaystyle \underset{n\to \infty }{lim\; sup}}\langle (g-\mu F)q^{\ast },x_n-q^{\ast }\rangle \hfill & ={\displaystyle \underset{\iota \to \infty }{lim}}\langle (g-\mu F)q^{\ast },x_{n_{\iota }}-q^{\ast }\rangle \hfill \\ \hfill & =\langle (g-\mu F)q^{\ast },\tilde{q}-q^{\ast }\rangle .\hfill \end{array}$$

(36)

Note that $\parallel w_n-x_n\parallel ={ϵ}_n\parallel x_n-x_{n-1}\parallel \to 0$ (due to ${sup}_{n\ge 1}({ϵ}_n/{\alpha }_n)<\infty$). Thus, we obtain

$$\parallel q_n-w_n\parallel \le \parallel q_n-u_n\parallel +\parallel u_n-x_n\parallel +\parallel x_n-w_n\parallel \to 0(n\to \infty ).$$

Noticing $x_{n+1}-x_n\to 0,u_n-x_n\to 0,w_n-q_n\to 0$ and $S^n{x}_n-S^{n+1}{x}_n\to 0$, from Lemma 10, one obtains $\tilde{q}\in \in {\omega }_w\left(\left\{x_n\right\}\right)\subset \Omega$. Thus, using (36) and (16), one has

$$\underset{n\to \infty }{lim\; sup}\langle (g-\mu F)q^{\ast },x_n-q^{\ast }\rangle =\langle (g-\mu F)q^{\ast },\tilde{q}-q^{\ast }\rangle \le 0,$$

(37)

which, together with (34), yields

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

(38)

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

$$\underset{n\to \infty }{lim\; sup}[\frac{2\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle }{\zeta -\delta }+\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right]\le 0,$$

by the application of Lemma 4 to (30), one has ${lim}_{n\to \infty }{\parallel x_n-q^{\ast }\parallel }^2=0$.

 Aspect 2. 

Suppose that $\exists \left\{{\Phi }_{n_{\iota }}\right\}\subset \left\{{\Phi }_n\right\}$ s.t. ${\Phi }_{n_{\iota }}<{\Phi }_{n_{\iota }+1}\forall \iota \in \mathcal{N}$, with $\mathcal{N}$ being the set of all natural numbers. The self-mapping $\phi$ on $\mathcal{N}$ is formulated as

$$\phi \left(n\right):=max\{\iota \le n:{\Phi }_{\iota }<{\Phi }_{\iota +1}\}.$$

Using Lemma 6, one obtains

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

Putting $q=q^{\ast }$, from (27), we have

$$\begin{array}{cc}& (1-{\beta }_{\phi \left(n\right)}-{\alpha }_{\phi \left(n\right)}\zeta )(1+{\theta }_{\phi \left(n\right)})\{(1-\nu )(\parallel y_{\phi \left(n\right)}-q_{\phi \left(n\right)}{\parallel }^2\hfill \\ \hfill & +\parallel y_{\phi \left(n\right)}-u_{\phi \left(n\right)}{\parallel }^2)+\overline{a}(1-\overline{b})\parallel w_{\phi \left(n\right)}-W_{\phi \left(n\right)}{w}_{\phi \left(n\right)}{\parallel }^2\}\hfill \\ \hfill & \le (1-{\beta }_{\phi \left(n\right)}-{\alpha }_{\phi \left(n\right)}\zeta )(1+{\theta }_{\phi \left(n\right)})\{(1-\nu )(\parallel y_{\phi \left(n\right)}-q_{\phi \left(n\right)}{\parallel }^2\hfill \\ \hfill & +\parallel y_{\phi \left(n\right)}-u_{\phi \left(n\right)}{\parallel }^2)+{\sigma }_{\phi \left(n\right)}(1-{\sigma }_{\phi \left(n\right)})\parallel w_{\phi \left(n\right)}-W_{\phi \left(n\right)}{w}_{\phi \left(n\right)}{\parallel }^2\}\hfill \\ \hfill & \le \parallel x_{\phi \left(n\right)}-q^{\ast }{\parallel }^2-{\parallel x_{\phi \left(n\right)+1}-q^{\ast }\parallel }^2+({\alpha }_{\phi \left(n\right)}+{\theta }_{\phi \left(n\right)})M_4\hfill \\ \hfill & ={\Phi }_{\phi \left(n\right)}-{\Phi }_{\phi \left(n\right)+1}+({\alpha }_{\phi \left(n\right)}+{\theta }_{\phi \left(n\right)})M_4,\hfill \end{array}$$

(39)

which immediately yields

$$\underset{n\to \infty }{lim}\parallel w_{\phi \left(n\right)}-W_{\phi \left(n\right)}{w}_{\phi \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel y_{\phi \left(n\right)}-u_{\phi \left(n\right)}\parallel =\underset{n\to \infty }{lim}\parallel y_{\phi \left(n\right)}-q_{\phi \left(n\right)}\parallel =0.$$

Using the similar arguments to those of Aspect 1, one obtains

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

and

$$\underset{n\to \infty }{lim\; sup}\langle (g-\mu F)q^{\ast },x_{\phi \left(n\right)+1}-q^{\ast }\rangle \le 0.$$

(40)

On the other hand, by (30), one has

$$\begin{array}{cc}{\alpha }_{\phi \left(n\right)}(\zeta -\delta ){\Phi }_{\phi \left(n\right)}\hfill & \le {\Phi }_{\phi \left(n\right)}-{\Phi }_{\phi \left(n\right)+1}+{\alpha }_{\phi \left(n\right)}(\zeta -\delta )[\frac{2\langle (g-\mu F)q^{\ast },x_{\phi \left(n\right)+1}-q^{\ast }\rangle }{\zeta -\delta }\hfill \\ & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}3\parallel x_{\phi \left(n\right)}-x_{\phi \left(n\right)-1}\parallel +\frac{{\theta }_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}\left)\right]\hfill \\ & \le {\alpha }_{\phi \left(n\right)}(\zeta -\delta )[\frac{2\langle (g-\mu F)q^{\ast },x_{\phi \left(n\right)+1}-q^{\ast }\rangle }{\zeta -\delta }\hfill \\ & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}3\parallel x_{\phi \left(n\right)}-x_{\phi \left(n\right)-1}\parallel +\frac{{\theta }_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}\left)\right],\hfill \end{array}$$

$$\underset{n\to \infty }{lim\; sup}{\Phi }_{\phi \left(n\right)}\le \underset{n\to \infty }{lim\; sup}[\frac{2\langle (g-\mu F)q^{\ast },x_{\phi \left(n\right)+1}-q^{\ast }\rangle }{\zeta -\delta }+\frac{M}{\zeta -\delta }(\frac{{ϵ}_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}3\parallel x_{\phi \left(n\right)}-x_{\phi \left(n\right)-1}\parallel +\frac{{\theta }_{\phi \left(n\right)}}{{\alpha }_{\phi \left(n\right)}}\left)\right]\le 0.$$

Thus, ${lim}_{n\to \infty }{\parallel x_{\phi \left(n\right)}-q^{\ast }\parallel }^2=0$. In addition, note that

$$\begin{array}{cc}& \parallel x_{\phi \left(n\right)+1}-q^{\ast }{\parallel }^2-{\parallel x_{\phi \left(n\right)}-q^{\ast }\parallel }^2\hfill \\ \hfill & =2\langle x_{\phi \left(n\right)+1}-x_{\phi \left(n\right)},x_{\phi \left(n\right)}-q^{\ast }\rangle +{\parallel x_{\phi \left(n\right)+1}-x_{\phi \left(n\right)}\parallel }^2\hfill \\ \hfill & \le 2\parallel x_{\phi \left(n\right)+1}-x_{\phi \left(n\right)}\parallel \parallel x_{\phi \left(n\right)}-q^{\ast }\parallel +\parallel x_{\phi \left(n\right)+1}-x_{\phi \left(n\right)}{\parallel }^2.\hfill \end{array}$$

(41)

Owing to ${\Phi }_n\le {\Phi }_{\phi \left(n\right)+1}$, one obtains

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

This means that $x_n\to q^{\ast }$ as $n\to \infty$.

In particular, when S is a nonexpansive operator, it is also asymptotically nonexpansive. In this case, the power $S^n$ in Algorithm 4 can be simplified into S. In this way, we can obtain the following Theorem 2.

 Theorem 2. 

Suppose that S is of nonexpansivity on H and $\left\{x_n\right\}$ is constructed in the modification of Algorithm 4, i.e., for any starting points $x_1,x_0$ in H,

$$\left\{\begin{array}{cc}& w_n=x_n+{ϵ}_n(x_n-x_{n-1}),\hfill \\ \hfill & u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S{q}_n\forall n\ge 1,\hfill \end{array}\right.$$

(42)

with $C_n$ and ${\varsigma }_n$ being picked as in Algorithm 4. Then, $x_n\to q^{\ast }\in \Omega \iff {sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, with $q^{\ast }\in \Omega$ being only a solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$.

 Proof. 

We first pick a $q\in \Omega$ arbitrarily. Obviously, the necessity holds. Next, it is sufficient to demonstrate the sufficiency. To this goal, under the condition ${sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, one divides the remainder of the proof into several claims.    □

 Claim 1. 

One claims the boundedness of $\left\{x_n\right\}$. In fact, using the similar inferences to those of Claim 1 in the proof of the above theorem, one obtains the claim.

 Claim 2. 

One claims that

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel w_n-W_n{w}_n{\parallel }^2\}\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+{\alpha }_n{M}_4,\hfill \end{array}$$

(43)

for some $M_4>0$. In fact, using the similar inferences to those of Step 2 in the proof of the above theorem, one obtains the claim.

 Claim 3. 

One claims that

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q{\parallel }^2+{\alpha }_n(\zeta -\delta )\{\frac{2}{\zeta -\delta }\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ & +\frac{3M}{\zeta -\delta }·\frac{{ϵ}_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \}\hfill \end{array}$$

for some $M>0$. In fact, using the similar inferences to those of Claim 3 in the proof of the above theorem, one obtains the claim.

 Claim 4. 

One claims that $x_n\to q^{\ast }\in \Omega$, which is only a solution to the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. In fact, setting $q=q^{\ast }$, by Claim 3, one obtains

$$\begin{array}{cc}\parallel x_{n+1}-q^{\ast }{\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]{\parallel x_n-q^{\ast }\parallel }^2+{\alpha }_n(\zeta -\delta )\hfill \\ \hfill & \times \{\frac{2}{\zeta -\delta }\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle +\frac{3M}{\zeta -\delta }·\frac{{ϵ}_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \}.\hfill \end{array}$$

(44)

Setting ${\Phi }_n={\parallel x_n-q^{\ast }\parallel }^2$, one demonstrates ${\Phi }_n\to 0(n\to \infty )$ in both aspects below.

 Aspect 1. 

Suppose that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Phi }_n\right\}$ is non-increasing. Then, the limit ${lim}_{n\to \infty }{\Phi }_n=d<\infty$ and ${lim}_{n\to \infty }({\Phi }_n-{\Phi }_{n+1})=0$. Using the similar inferences to those of Aspect 1 of Claim 4 in the proof of the above theorem, one obtains

$$\underset{n\to \infty }{lim}\parallel w_n-W_n{w}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-y_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-q_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-x_n\parallel =0.$$

(45)

From (4) and (23), one has

$$\begin{array}{cc}& \parallel x_{n+1}-q^{\ast }{\parallel }^2={\parallel {\beta }_n(x_n-q^{\ast })+(1-{\beta }_n)(S{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu FS{q}_n)\parallel }^2\hfill \\ & \le \parallel {\beta }_n(x_n-q^{\ast })+(1-{\beta }_n)(S{q}_n-q^{\ast }){\parallel }^2+2{\alpha }_n\langle g\left(x_n\right)-\mu FS{q}_n,x_{n+1}-q^{\ast }\rangle \hfill \\ & \le {\beta }_n\parallel x_n-q^{\ast }{\parallel }^2+(1-{\beta }_n)\parallel S{q}_n-q^{\ast }{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-S{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n\langle g\left(x_n\right)-\mu FS{q}_n,x_{n+1}-q^{\ast }\rangle \hfill \\ & \le (\parallel x_n-q^{\ast }\parallel {\alpha }_n{M}_1{)}^2-{\beta }_n(1-{\beta }_n){\parallel x_n-S{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \parallel x_{n+1}-q^{\ast }\parallel .\hfill \end{array}$$

This hence arrives at

$${\beta }_n(1-{\beta }_n)\parallel x_n-S{q}_n{\parallel }^2\le (\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2-\parallel x_{n+1}-q^{\ast }{\parallel }^2+2{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \parallel x_{n+1}-q^{\ast }\parallel .$$

Since $1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$, ${\alpha }_n\to 0$ and ${lim}_{n\to \infty }{\Phi }_n=d<+\infty$, from the boundedness of $\left\{g\left(x_n\right)\right\},\left\{S{q}_n\right\}$, we infer that

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

Therefore,

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+(1-{\beta }_n)(S{q}_n-x_n)-{\alpha }_n\mu FS{q}_n\parallel \hfill \\ \hfill & \le (1-{\beta }_n)\parallel S{q}_n-x_n\parallel +{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \hfill \\ \hfill & \le \parallel S{q}_n-x_n\parallel +{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu FS{q}_n\parallel )\to 0(n\to \infty ).\hfill \end{array}$$

(46)

Again utilizing the similar inferences to those of Aspect 1 of Claim 4 in the proof of the above theorem, one obtains ${lim}_{n\to \infty }{\parallel x_n-q^{\ast }\parallel }^2=0$.

 Aspect 2. 

Suppose that $\exists \left\{{\Phi }_{n_{\iota }}\right\}\subset \left\{{\Phi }_n\right\}$ s.t. ${\Phi }_{n_{\iota }}<{\Phi }_{n_{\iota }+1}\forall \iota \in \mathcal{N}$, with $\mathcal{N}$ being the set of all natural numbers. The self-mapping $\phi$ on $\mathcal{N}$ is formulated as

$$\phi \left(n\right):=max\{\iota \le n:{\Phi }_{\iota }<{\Phi }_{\iota +1}\}.$$

From Lemma 6, one obtains

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

Finally, by the similar inferences to those of Aspect 2 of Claim 4 in the proof of the above theorem, one can obtain the claim.

On the other hand, we put forward another modification of a Mann-type subgradient extragradient rule.

It is worth mentioning that (9) and Lemmas 8–10 remain true for Algorithm 5:

Algorithm 5 The 2nd modified Mann-type subgradient extragradient rule

Initial Step: Let $l\in (0,1),\nu \in (0,1),\gamma \in (0,\infty )$, given any starting points $x_1,x_0$ in H. Iterations: Compute $x_{n+1}(n\ge 1)$ below: Step 1. Set $w_n=x_n+{ϵ}_n(x_n-x_{n-1})$ and $u_n=(1-{\sigma }_n)x_n+{\sigma }_n{W}_n{w}_n$, and calculate $y_n=P_C(u_n-{\varsigma }_{n}A{u}_n)$, with ${\varsigma }_n$ being picked to be the largest $\varsigma \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\varsigma \parallel A{u}_n-A{y}_n\parallel \le \nu \parallel u_n-y_n\parallel .$$ (47) Step 2. Calculate $q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n)$, where $C_n:=\{y\in H:\langle u_n-{\varsigma }_{n}A{u}_n-y_n,y_n-y\rangle \ge 0\}$. Step 3. Calculate $$x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{u}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n.$$ (48) Put $n:=n+1$ and return to Step 1.

 Theorem 3. 

Suppose that $\left\{x_n\right\}$ is the sequence constructed in Algorithm 5. Then,

$$x_n\to q^{\ast }\in \Omega \iff \left\{\begin{array}{cc}& S^n{x}_n-S^{n+1}{x}_n\to 0,\hfill \\ & \underset{n\ge 1}{sup}\parallel x_{n-1}-x_n\parallel <\infty ,\hfill \end{array}\right.$$

with $q^{\ast }\in \Omega$ being only a solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$.

 Proof. 

By the similar inferences to those in the proof of the first theorem, one obtains that $\exists q^{\ast }\in \Omega$, which is only a solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. Obviously, the necessity holds.

In what follows, one claims the sufficiency. To the goal, under the assumption $S^n{x}_n-S^{n+1}{x}_n\to 0$ with ${sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, one divides the claim of the sufficiency into several claims. □

 Claim 1. 

One claims the boundedness of $\left\{x_n\right\}$. In fact, using the similar inferences to those of Claim 1 in the proof of the first theorem, one has that (19) and (20) hold. It is easy to see from (9) that

$$\parallel q_n-q\parallel \le \parallel u_n-q\parallel \le (1-{\sigma }_n)\parallel x_n-q\parallel +{\sigma }_n\parallel w_n-q\parallel \le \parallel x_n-q\parallel +{\alpha }_n{M}_1\forall n\ge 1.$$

(49)

Hence, using ${\alpha }_n+{\beta }_n<1$, Lemma 7, and (49), we obtain

$$\begin{array}{cc}& \parallel x_{n+1}-q\parallel =\parallel {\alpha }_n(g\left(x_n\right)-g\left(q\right))+{\beta }_n(u_n-q)+(1-{\beta }_n)\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q+\frac{{\alpha }_n}{1-{\beta }_n}(g-\mu F)q]\parallel \hfill \\ & \le {\alpha }_n\parallel g\left(x_n\right)-g\left(q\right)\parallel +{\beta }_n\parallel u_n-q\parallel +(1-{\beta }_n)\hfill \\ & \times \parallel (I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q+\frac{{\alpha }_n}{1-{\beta }_n}(g-\mu F)q\parallel \hfill \\ & \le {\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n\parallel u_n-q\parallel +(1-{\beta }_n)\hfill \\ & \times (1-\frac{{\alpha }_n}{1-{\beta }_n}\zeta )(1+{\theta }_n)\parallel q_n-q\parallel +{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le {\alpha }_n\delta \parallel x_n-q\parallel +{\beta }_n(\parallel x_n-q\parallel +{\alpha }_n{M}_1)+(1-{\beta }_n-{\alpha }_n\zeta )\hfill \\ & \times (\parallel x_n-q\parallel +{\alpha }_n{M}_1)+{\theta }_n\parallel q_n-q\parallel +{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le [{\alpha }_n\delta +{\beta }_n+(1-{\beta }_n-{\alpha }_n\zeta )]\parallel x_n-q\parallel +{\alpha }_n{M}_1\hfill \\ & +\frac{{\alpha }_n(\zeta -\delta )(\parallel x_n-q\parallel +{\alpha }_n{M}_1)}{2}+{\alpha }_n\parallel (g-\mu F)q\parallel \hfill \\ & \le [1-\frac{{\alpha }_n(\zeta -\delta )}{2}]\parallel x_n-q\parallel +{\alpha }_n(2M_1+\parallel (g-\mu F)q\parallel )\hfill \\ & =[1-\frac{{\alpha }_n(\zeta -\delta )}{2}]\parallel x_n-q\parallel +\frac{{\alpha }_n(\zeta -\delta )}{2}·\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\hfill \\ & \le max\{\parallel x_n-q\parallel ,\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\},\hfill \end{array}$$

which immediately yields

$$\parallel x_n-q\parallel \le max\{\parallel x_1-q\parallel ,\frac{2(2M_1+\parallel (g-\mu F)q\parallel )}{\zeta -\delta }\}\forall n\ge 1.$$

Therefore, we show the boundedness of $\left\{x_n\right\}$. This ensures that the sequences $\left\{w_n\right\},\left\{u_n\right\},\left\{y_n\right\},\left\{q_n\right\},\left\{g\left(x_n\right)\right\},\left\{A{y}_n\right\},\left\{W_n{w}_n\right\},\left\{S^n{q}_n\right\}$ are bounded.

 Claim 2. 

One claims that

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+({\alpha }_n+{\theta }_n)M_4,\hfill \end{array}$$

for some $M_4>0$—in fact, since

$$\begin{array}{cc}{x}_{n+1}-q\hfill & ={\alpha }_n(g\left(x_n\right)-g\left(q\right)){\beta }_n(u_n-q)+(1-{\beta }_n)\hfill \\ & \times [(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)S^n{q}_n-(I-\frac{{\alpha }_n}{1-{\beta }_n}\mu F)q]+{\alpha }_n(g-\mu F)q.\hfill \end{array}$$

Using the same inferences as those of (24), one has

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n){\parallel q_n-q\parallel }^2\hfill \\ \hfill & +2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n){\parallel q_n-q\parallel }^2+{\alpha }_n{M}_2,\hfill \end{array}$$

(50)

with $M_2\ge {sup}_{n\ge 1}2\parallel (g-\mu F)q\parallel \parallel x_n-q\parallel$ for certain $M_2>0$. In addition, using (19) and (20), one obtains

$$\parallel w_n{-q\parallel }^2\le (\parallel x_n-q\parallel +{\alpha }_n{M}_1{)}^2\le {\parallel x_n-q\parallel }^2+{\alpha }_n{M}_3,$$

with $M_3\ge {sup}_{n\ge 1}(2M_1\parallel x_n-q\parallel +{\alpha }_n{M}_1^2)$ for certain $M_3>0$. Moreover, using (49), we deduce that

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

which, together with (9) and (50), leads to

$$\begin{array}{cc}& \parallel x_{n+1}{-q\parallel }^2\le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)[\parallel u_n{-q\parallel }^2\hfill \\ \hfill & -(1-\nu )\parallel y_n-q_n{\parallel }^2-(1-\nu )\parallel y_n-u_n{\parallel }^2]+{\alpha }_n{M}_2\hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-{\sigma }_n){\parallel x_n-q\parallel }^2\hfill \\ \hfill & +{\sigma }_n\parallel W_n{w}_n{-q\parallel }^2-{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2-(1-\nu )(\parallel y_n-q_n{\parallel }^2\hfill \\ \hfill & +\parallel y_n-u_n{\parallel }^2\left)\right\}+{\alpha }_n{M}_2\hfill \\ \hfill & \le {\alpha }_n\delta (\parallel x_n{-q\parallel }^2+{\alpha }_n{M}_3)+{\beta }_n(\parallel x_n-q{\parallel }^2+{\alpha }_n{M}_3)+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\hfill \\ \hfill & \times \{(1-{\sigma }_n)(\parallel x_n{-q\parallel }^2+{\alpha }_n{M}_3)+{\sigma }_n(\parallel x_n{-q\parallel }^2+{\alpha }_n{M}_3)-{\sigma }_n(1-{\sigma }_n){\parallel x_n-W_n{w}_n\parallel }^2\hfill \\ \hfill & -(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2\left)\right\}+{\alpha }_n{M}_2\hfill \\ \hfill & =[{\alpha }_n\delta +{\beta }_n+(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)](\parallel x_n-q{\parallel }^2+{\alpha }_n{M}_3)-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\hfill \\ \hfill & \times \left\{(1-\nu )\right(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}+{\alpha }_n{M}_2\hfill \\ \hfill & \le [{\alpha }_n\delta +{\beta }_n+(1-{\beta }_n-{\alpha }_n\zeta )+{\theta }_n](\parallel x_n-q{\parallel }^2+{\alpha }_n{M}_3)-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\hfill \\ \hfill & \times \left\{(1-\nu )\right(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}+{\alpha }_n{M}_2\hfill \\ \hfill & =[1-{\alpha }_n(\zeta -\delta )+{\theta }_n](\parallel x_n-q{\parallel }^2+{\alpha }_n{M}_3)-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\hfill \\ \hfill & \times \left\{(1-\nu )\right(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}+{\alpha }_n{M}_2\hfill \\ \hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+{\theta }_n{\parallel x_n-q\parallel }^2+{\alpha }_n{M}_3-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\hfill \\ \hfill & \times \left\{(1-\nu )\right(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}+{\alpha }_n{M}_2\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2-(1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)\hfill \\ \hfill & +{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}+({\alpha }_n+{\theta }_n)M_4,\hfill \end{array}$$

(51)

where $M_4\ge {sup}_{n\ge 1}(\parallel x_n-q{\parallel }^2+M_3+M_2)$ for certain $M_4>0$. Thus, we obtain

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+({\alpha }_n+{\theta }_n)M_4.\hfill \end{array}$$

(52)

 Claim 3. 

One claims that

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q{\parallel }^2+{\alpha }_n(\zeta -\delta )\{\frac{2}{\zeta -\delta }\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right\}\hfill \end{array}$$

for certain $M>0$. In fact, one has

$$\parallel w_n{-q\parallel }^2\le \parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel +{ϵ}_n\parallel x_n-x_{n-1}\parallel ),$$

and hence

$$\begin{array}{cc}& \parallel u_n{-q\parallel }^2\le (1-{\sigma }_n)\parallel x_n{-q\parallel }^2+{\sigma }_n{\parallel W_n{w}_n-q\parallel }^2\hfill \\ \hfill & \le (1-{\sigma }_n)\parallel x_n{-q\parallel }^2+{\sigma }_n\{\parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel +{ϵ}_n\parallel x_n-x_{n-1}\parallel \left)\right\}\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel +{ϵ}_n\parallel x_n-x_{n-1}\parallel ).\hfill \end{array}$$

(53)

From (49), (50) and (53), one obtains

$$\begin{array}{cc}& \parallel x_{n+1}{-q\parallel }^2\hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+{\beta }_n\parallel u_n{-q\parallel }^2+(1-{\beta }_n-{\alpha }_n\zeta ){\parallel q_n-q\parallel }^2\hfill \\ \hfill & +{\theta }_n{\parallel q_n-q\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le {\alpha }_n\delta \parallel x_n{-q\parallel }^2+(1-{\alpha }_n\zeta )[\parallel x_n{-q\parallel }^2+{ϵ}_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-q\parallel \hfill \\ \hfill & +{ϵ}_n\parallel x_n-x_{n-1}\parallel \left)\right]+{\theta }_n{\parallel q_n-q\parallel }^2+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+({ϵ}_n\parallel x_n-x_{n-1}\parallel 3+{\theta }_n)M+2{\alpha }_n\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ \hfill & =[1-{\alpha }_n(\zeta -\delta )]\parallel x_n{-q\parallel }^2+{\alpha }_n(\zeta -\delta )[\frac{2\langle (g-\mu F)q,x_{n+1}-q\rangle }{\zeta -\delta }+\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right],\hfill \end{array}$$

(54)

where $M\ge {sup}_{n\ge 1}\{\parallel x_n-q\parallel ,{ϵ}_n\parallel x_n-x_{n-1}\parallel ,\parallel q_n-q{\parallel }^2\}$ for certain $M>0$.

 Claim 4. 

One claims that $x_n\to q^{\ast }\in \Omega$, which is only a solution to the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. In fact, using (54) with $q=q^{\ast }$, one obtains

$$\begin{array}{cc}\parallel x_{n+1}-q^{\ast }{\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q^{\ast }{\parallel }^2+{\alpha }_n(\zeta -\delta )[\frac{2\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle }{\zeta -\delta }\hfill \\ \hfill & +\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right].\hfill \end{array}$$

(55)

Setting ${\Phi }_n={\parallel x_n-q^{\ast }\parallel }^2$, one demonstrates ${\Phi }_n\to 0(n\to \infty )$ in both aspects below.

 Aspect 1. 

Suppose that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Phi }_n\right\}$ is non-increasing. Then, the limit ${lim}_{n\to \infty }{\Phi }_n=d<\infty$ and ${lim}_{n\to \infty }({\Phi }_n-{\Phi }_{n+1})=0$. Using (52) with $q=q^{\ast }$ and $\left\{{\sigma }_n\right\}\subset [\overline{a},\overline{b}]\subset (0,1)$, one obtains

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+\overline{a}(1-\overline{b})\parallel x_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le (1-{\beta }_n-{\alpha }_n\zeta )(1+{\theta }_n)\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}\hfill \\ & \le {\Phi }_n-{\Phi }_{n+1}+({\alpha }_n+{\theta }_n)M_4.\hfill \end{array}$$

Noticing ${lim\; inf}_{n\to \infty }(1-{\beta }_n)>0$, ${\alpha }_n\to 0,{\theta }_n\to 0$ and ${\Phi }_n-{\Phi }_{n+1}\to 0$, one has

$$\underset{n\to \infty }{lim}\parallel x_n-W_n{w}_n\parallel =\underset{n\to \infty }{lim}\parallel y_n-u_n\parallel =\underset{n\to \infty }{lim}\parallel y_n-q_n\parallel =0.$$

(56)

Hence, one obtains

$$\parallel u_n-q_n\parallel \le \parallel u_n-y_n\parallel +\parallel y_n-q_n\parallel \to 0(n\to \infty ).$$

(57)

Since $w_n-x_n={ϵ}_n(x_n-x_{n-1})$ and $u_n-x_n={\sigma }_n(W_n{w}_n-x_n)$, we obtain

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

(58)

Moreover, noticing $x_{n+1}-q^{\ast }={\beta }_n(u_n-q^{\ast })+(1-{\beta }_n)(S^n{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n)$, we obtain from (49) that

$$\begin{array}{cc}& \parallel x_{n+1}-q^{\ast }{\parallel }^2={\parallel {\beta }_n(u_n-q^{\ast })+(1-{\beta }_n)(S^n{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu F{S}^n{q}_n)\parallel }^2\hfill \\ & \le {\beta }_n\parallel u_n-q^{\ast }{\parallel }^2+(1-{\beta }_n){(1+{\theta }_n)}^2\parallel q_n-q^{\ast }{\parallel }^2-{\gamma }_n(1-{\gamma }_n){\parallel u_n-S^n{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n\langle g\left(x_n\right)-\mu F{S}^n{q}_n,x_{n+1}-q^{\ast }\rangle \hfill \\ & \le {(1+{\theta }_n)}^2(\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2-{\beta }_n(1-{\beta }_n){\parallel u_n-S^n{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & ={(1+{\theta }_n)}^2\parallel x_n-q^{\ast }{\parallel }^2+{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1]\hfill \\ & -{\beta }_n(1-{\beta }_n)\parallel u_n-S^n{q}_n{\parallel }^2+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel .\hfill \end{array}$$

This hence arrives at

$$\begin{array}{cc}& {\beta }_n(1-{\beta }_n)\parallel u_n-S^n{q}_n{\parallel }^2\le {(1+{\theta }_n)}^2\parallel x_n-q^{\ast }{\parallel }^2-{\parallel x_{n+1}-q^{\ast }\parallel }^2\hfill \\ & +{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1]+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\parallel x_{n+1}-q^{\ast }\parallel \hfill \\ & \le {(1+{\theta }_n)}^2{\Phi }_n-{\Phi }_{n+1}+{(1+{\theta }_n)}^2{\alpha }_n{M}_1[2{\Phi }_n^{\frac{1}{2}}+{\alpha }_n{M}_1]+2{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel ){\Phi }_{n+1}^{\frac{1}{2}}.\hfill \end{array}$$

Since $1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$, ${\theta }_n\to 0,{\alpha }_n\to 0,{\Phi }_n-{\Phi }_{n+1}\to 0$ and ${lim}_{n\to \infty }{\Phi }_n=d<+\infty$, from the boundedness of $\left\{g\left(x_n\right)\right\},\left\{S^n{q}_n\right\}$, we infer that

$$\underset{n\to \infty }{lim}\parallel u_n-S^n{q}_n\parallel =0.$$

Thus, it follows from Algorithm 5 that

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+{\beta }_n(u_n-x_n)+(1-{\beta }_n)(S^n{q}_n-x_n)-{\alpha }_n\mu F{S}^n{q}_n\parallel \hfill \\ \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+(u_n-x_n)+(1-{\beta }_n)(S^n{q}_n-u_n)-{\alpha }_n\mu F{S}^n{q}_n\parallel \hfill \\ \hfill & \le \parallel u_n-x_n\parallel +\parallel S^n{q}_n-u_n\parallel +{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu F{S}^n{q}_n\parallel )\to 0(n\to \infty ).\hfill \end{array}$$

(59)

Utilizing the same inferences as those of (38), one obtains

$$\underset{n\to \infty }{lim\; sup}\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle \le 0.$$

(60)

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

$$\underset{n\to \infty }{lim\; sup}[\frac{2\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle }{\zeta -\delta }+\frac{M}{\zeta -\delta }(\frac{{ϵ}_n}{{\alpha }_n}3\parallel x_n-x_{n-1}\parallel +\frac{{\theta }_n}{{\alpha }_n}\left)\right]\le 0.$$

Therefore, by the application of Lemma 4 to (55), one has ${lim}_{n\to \infty }{\parallel x_n-q^{\ast }\parallel }^2=0$.

 Aspect 2.  

Suppose that $\exists \left\{{\Phi }_{n_{\iota }}\right\}\subset \left\{{\Phi }_n\right\}$ s.t. ${\Phi }_{n_{\iota }}<{\Phi }_{n_{\iota }+1}\forall \iota \in \mathcal{N}$, with $\mathcal{N}$ being the set of all natural numbers. The self-mapping $\phi$ on $\mathcal{N}$ is formulated as

$$\phi \left(n\right):=max\{\iota \le n:{\Phi }_{\iota }<{\Phi }_{\iota +1}\}.$$

From Lemma 6, one obtains

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

Finally, by the similar inferences to those of Aspect 2 of Claim 4 in the proof of the first theorem, one can derive the claim.

In particular, when S is a nonexpansive operator, it is also asymptotically nonexpansive. In this case, the power $S^n$ in Algorithm 5 can be simplified into S. In this way, we can obtain the following Theorem 3.

 Theorem 4. 

Suppose that S is of nonexpansivity on H and $\left\{x_n\right\}$ is constructed in the modification of Algorithm 5, i.e., for any starting points $x_1,x_0$ in H,

$$\left\{\begin{array}{cc}& w_n=x_n+{ϵ}_n(x_n-x_{n-1}),\hfill \\ \hfill & u_n=(1-{\sigma }_n)x_n+{\sigma }_n{W}_n{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{u}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S{q}_n\forall n\ge 1,\hfill \end{array}\right.$$

(61)

with $C_n$ and ${\varsigma }_n$ being picked as in Algorithm 5. Then, $x_n\to q^{\ast }\in \Omega \iff {sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, with $q^{\ast }\in \Omega$ being only a solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$.

 Proof. 

We first pick a $q\in \Omega$ arbitrarily. Obviously, the necessity holds. Next, it is sufficient to demonstrate the sufficiency. To the goal, under the condition ${sup}_{n\ge 1}\parallel x_{n-1}-x_n\parallel <\infty$, one divides the surplus of the proof into several claims. □

 Claim 1. 

One claims the boundedness of $\left\{x_n\right\}$. In fact, using the similar inferences to those of Claim 1 in the proof of the third theorem, one obtains the claim.

 Claim 2. 

One claims that

$$\begin{array}{cc}& (1-{\beta }_n-{\alpha }_n\zeta )\{(1-\nu )(\parallel y_n-q_n{\parallel }^2+\parallel y_n-u_n{\parallel }^2)+{\sigma }_n(1-{\sigma }_n)\parallel x_n-W_n{w}_n{\parallel }^2\}\hfill \\ \hfill & \le \parallel x_n{-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+{\alpha }_n{M}_4,\hfill \end{array}$$

(62)

for certain $M_4>0$. In fact, using the similar inferences to those of Claim 2 in the proof of the third theorem, one obtains the claim.

 Claim 3. 

One claims that

$$\begin{array}{cc}\parallel x_{n+1}{-q\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]\parallel x_n-q{\parallel }^2+{\alpha }_n(\zeta -\delta )\{\frac{2}{\zeta -\delta }\langle (g-\mu F)q,x_{n+1}-q\rangle \hfill \\ & +\frac{3M}{\zeta -\delta }·\frac{{ϵ}_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \}\hfill \end{array}$$

for certain $M>0$. In fact, using the similar inferences to those of Claim 3 in the proof of the third theorem, one obtains the claim.

 Claim 4. 

One claims that $x_n\to q^{\ast }\in \Omega$, which is only a solution to the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$. In fact, setting $q=q^{\ast }$, by Claim 3, one obtains

$$\begin{array}{cc}\parallel x_{n+1}-q^{\ast }{\parallel }^2\hfill & \le [1-{\alpha }_n(\zeta -\delta )]{\parallel x_n-q^{\ast }\parallel }^2+{\alpha }_n(\zeta -\delta )\hfill \\ \hfill & \times \{\frac{2}{\zeta -\delta }\langle (g-\mu F)q^{\ast },x_{n+1}-q^{\ast }\rangle +\frac{3M}{\zeta -\delta }·\frac{{ϵ}_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel \}.\hfill \end{array}$$

(63)

Putting ${\Phi }_n={\parallel x_n-q^{\ast }\parallel }^2$, one shows ${\Phi }_n\to 0(n\to \infty )$ in both aspects below.

 Aspect 1. 

Suppose that ∃ (integer) $n_0\ge 1$ s.t. $\left\{{\Phi }_n\right\}$ is non-increasing. Then, the limit ${lim}_{n\to \infty }{\Phi }_n=d<+\infty$ and ${lim}_{n\to \infty }({\Phi }_n-{\Phi }_{n+1})=0$. Using the similar inferences to those of Aspect 1 of Claim 4 in the proof of the third theorem, one obtains

$$\underset{n\to \infty }{lim}\parallel x_n-W_n{w}_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-y_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-q_n\parallel =\underset{n\to \infty }{lim}\parallel u_n-x_n\parallel =0.$$

(64)

From (4) and (49), one has

$$\begin{array}{cc}& \parallel x_{n+1}-q^{\ast }{\parallel }^2={\parallel {\beta }_n(u_n-q^{\ast })+(1-{\beta }_n)(S{q}_n-q^{\ast })+{\alpha }_n(g\left(x_n\right)-\mu FS{q}_n)\parallel }^2\hfill \\ & \le \parallel {\beta }_n(u_n-q^{\ast })+(1-{\beta }_n)(S{q}_n-q^{\ast }){\parallel }^2+2{\alpha }_n\langle g\left(x_n\right)-\mu FS{q}_n,x_{n+1}-q^{\ast }\rangle \hfill \\ & \le {\beta }_n\parallel u_n-q^{\ast }{\parallel }^2+(1-{\beta }_n)\parallel S{q}_n-q^{\ast }{\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel u_n-S{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n\langle g\left(x_n\right)-\mu FS{q}_n,x_{n+1}-q^{\ast }\rangle \hfill \\ & \le (\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2-{\beta }_n(1-{\beta }_n){\parallel u_n-S{q}_n\parallel }^2\hfill \\ & +2{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \parallel x_{n+1}-q^{\ast }\parallel ,\hfill \end{array}$$

which immediately yields

$${\beta }_n(1-{\beta }_n)\parallel u_n-S{q}_n{\parallel }^2\le (\parallel x_n-q^{\ast }\parallel +{\alpha }_n{M}_1{)}^2-\parallel x_{n+1}-q^{\ast }{\parallel }^2+2{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \parallel x_{n+1}-q^{\ast }\parallel .$$

Since $1>{lim\; sup}_{n\to \infty }{\beta }_n\ge {lim\; inf}_{n\to \infty }{\beta }_n>0$, ${\alpha }_n\to 0$ and ${lim}_{n\to \infty }{\Phi }_n=d<+\infty$, from the boundedness of $\left\{g\left(x_n\right)\right\},\left\{S{q}_n\right\}$, we infer that

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

Therefore,

$$\begin{array}{cc}\parallel x_{n+1}-x_n\parallel \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+{\beta }_n(u_n-x_n)+(1-{\beta }_n)(S{q}_n-x_n)-{\alpha }_n\mu FS{q}_n\parallel \hfill \\ \hfill & =\parallel {\alpha }_{n}g\left(x_n\right)+(u_n-x_n)+(1-{\beta }_n)(S{q}_n-u_n)-{\alpha }_n\mu FS{q}_n\parallel \hfill \\ \hfill & \le \parallel u_n-x_n\parallel +\parallel S{q}_n-u_n\parallel +{\alpha }_n\parallel g\left(x_n\right)-\mu FS{q}_n\parallel \hfill \\ \hfill & \le \parallel u_n-x_n\parallel +\parallel S{q}_n-u_n\parallel +{\alpha }_n(\parallel g\left(x_n\right)\parallel +\parallel \mu FS{q}_n\parallel )\to 0(n\to \infty ).\hfill \end{array}$$

(65)

Again utilizing the similar inferences to those of Aspect 1 of Claim 4 in the proof of the third theorem, one obtains ${lim}_{n\to \infty }{\parallel x_n-q^{\ast }\parallel }^2=0$.

 Aspect 2. 

Suppose that $\exists \left\{{\Phi }_{n_{\iota }}\right\}\subset \left\{{\Phi }_n\right\}$ s.t. ${\Phi }_{n_{\iota }}<{\Phi }_{n_{\iota }+1}\forall \iota \in \mathcal{N}$, with $\mathcal{N}$ being the set of all natural numbers. The self-mapping $\phi$ on $\mathcal{N}$ is formulated as

$$\phi \left(n\right):=max\{\iota \le n:{\Phi }_{\iota }<{\Phi }_{\iota +1}\}.$$

Using Lemma 6, one obtains

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

Finally, by the similar inferences to those of Aspect 2 of Claim 4 in the proof of the third theorem, one can obtain the claim.

It is remarkable that, in comparison with the associated theorems in Xie et al. [9], Ceng and Shang [25], and Thong and Hieu [17], our theorems ameliorate and develop them in the aspects below.

(i)

The issue for one to find a point in $\mathrm{VI}(C,A)$ (see [9]) is developed into the issue for us to find a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)$ with both each $S_i$ being of nonexpansivity and $S_0=S$ being of asymptotical nonexpansivity. The modified inertial extragradient rule with a linear-search process for settling the VIP in [9] is developed into our modified Mann-type subgradient extragradient rule with a linear-search process for settling the CFPP and VIP, which is on the basis of the Mann iteration method, subgradient extragradient approach with a linear-search process, and the hybrid deepest-descent technique.

(ii)

The issue for ones to find a point in $\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)$ with a quasi-nonexpansive operator S in [17] is developed into the issue for us to find a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)$ with both $S_i$ being of nonexpansivity and $S_0=S$ being of asymptotical nonexpansivity. The inertial subgradient extragradient rule with a linear-search process for settling the VIP and FPP in [17] is developed into our modified Mann-type subgradient extragradient rule with a linear-search process for settling the CFPP and VIP, which is on the basis of the Mann iteration method, subgradient extragradient approach with a linear-search process, and the hybrid deepest-descent technique.

(iii)

The issue for one to find a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^N\mathrm{Fix}\left(S_i\right)\right)$ with finite nonexpansive operators ${\left\{S_i\right\}}_{i=1}^N$ (see [25]) is developed into the issue for us to find a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)$ with countable nonexpansive operators ${\left\{S_i\right\}}_{i=1}^{\infty }$. The hybrid inertial subgradient extragradient rule with a linear-search process in [25] is developed into our modified Mann-type subgradient extragradient rule with a linear-search process, e.g., the original inertial step $w_n=S_n{x}_n+{ϵ}_n(S_n{x}_n-S_n{x}_{n-1})$ is developed into the modified Mann iteration step: $w_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and $u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n$. In addition, it was shown in [25] that, under the condition $S^n{q}_n-S^{n+1}{q}_n\to 0$, the relation holds:

$$x_n\to q^{\ast }\in \Omega \iff \parallel x_n-y_n\parallel +\parallel x_n-x_{n+1}\parallel \to 0\mathrm{with}{q}^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }.$$

In this paper, using Lemma 6, we show that, under the condition $S^n{x}_n-S^{n+1}{x}_n\to 0$, the relation holds:

$$x_n\to q^{\ast }\in \Omega \iff \underset{n\ge 1}{sup}\parallel x_{n-1}-x_n\parallel <\infty \mathrm{with}{q}^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }.$$

4. Implementability and Applicability of Rules

In what follows, we provide an illustrated instance to demonstrate the implementability and applicability of proposed rules. Put $\mu =2,\gamma =1,\nu =l=\frac{1}{2},{\sigma }_n=\frac{1}{3},{ϵ}_n={\alpha }_n=\frac{1}{3(n+1)}$ and ${\beta }_n=\frac{n}{3(n+1)}$. First, we construct an example of $\Omega =\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)\ne \varnothing$ with $S_0:=S$, where $A:H\to H$ is of both pseudomonotonicity and Lipschitz continuity, $S:H\to H$ is of asymptotical nonexpansivity and each $S_i:H\to H$ is of nonexpansivity. We put $H=\mathbf{R}$ and use the $\langle r,s\rangle =rs$ and $\parallel ·\parallel =|·|$ to denote its inner product and induced norm, respectively. Moreover, we set $C=[-2,5]$. The starting points $x_1,x_0$ are arbitrarily picked in $[-2,5]$. Let $g\left(x\right)=F\left(x\right)=\frac{1}{2}x\forall x\in H$ with

$$\delta =\frac{1}{2}<\zeta =1-\sqrt{1-\mu (2\eta -\mu {\kappa }^2)}=1-\sqrt{1-2(2·\frac{1}{2}-2{\left(\frac{1}{2}\right)}^2)}=1.$$

Let $A:H\to H$ and $S,S_i:H\to H$ be formulated by $Ax=1/(1+|sinx\left|\right)-1/(1+|x\left|\right)$, $Sx=3sinx/5$ and $S_{i}x=Tx=sinx\forall x\in H,i\ge 1$, respectively. We now claim that A is pseudomonotone and Lipschitz continuous. In fact, one has

$$\parallel Ax-Ay\parallel \le \frac{\parallel y-x\parallel }{(1+\parallel y\parallel )(1+\parallel x\parallel )}+\frac{\parallel siny-sinx\parallel }{(1+\parallel siny\parallel )(1+\parallel sinx\parallel )}\le 2\parallel x-y\parallel \forall x,y\in H$$

This means that A is of Lipschitz continuity. In addition, one shows that A is of pseudomonotonicity. It can be readily seen that

$$\begin{array}{cc}& \langle Ax,y-x\rangle =(1/(1+|sinx|)-1/(1+\left|x\right|\left)\right)(y-x)\ge 0\hfill \\ & \Rightarrow \langle Ay,y-x\rangle =(1/(1+|siny|)-1/(1+\left|y\right|\left)\right(y-x)\ge 0\forall x,y\in H.\hfill \end{array}$$

Meanwhile, it is easily known that S is of asymptotical nonexpansivity with ${\theta }_n={\left(\frac{3}{5}\right)}^n\forall n\ge 1$, such that $\parallel S^{n+1}{x}_n-S^n{x}_n\parallel \to 0$ as $n\to \infty$. Indeed, we observe that

$$\parallel S^{n}x-S^{n}y\parallel \le \frac{3}{5}\parallel S^{n-1}x-S^{n-1}y\parallel \le \cdots \le {\left(\frac{3}{5}\right)}^n\parallel x-y\parallel \le (1+{\theta }_n)\parallel x-y\parallel ,$$

and

$$\parallel S^{n+1}{x}_n-S^n{x}_n\parallel \le {\left(\frac{3}{5}\right)}^{n-1}\parallel S^2{x}_n-S{x}_n\parallel ={\left(\frac{3}{5}\right)}^{n-1}\parallel \frac{3}{5}sin\left(S{x}_n\right)-\frac{3}{5}sin{x}_n\parallel \le 2{\left(\frac{3}{5}\right)}^n\to 0.$$

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

$$\underset{n\to \infty }{lim}\frac{{\theta }_n}{{\alpha }_n}=\underset{n\to \infty }{lim}\frac{{(3/5)}^n}{1/3(n+1)}=0.$$

In addition, it is easy to see that $S_i=T$ is of nonexpansivity and $\mathrm{Fix}\left(S_i\right)=\left\{0\right\}$. Thus, $\Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(T\right)\cap \mathrm{Fix}\left(S\right)=\left\{0\right\}$.

 Example 1. 

Noticing $W_n=T$ and $(1-{\beta }_n)I-{\alpha }_n\mu F=(1-\frac{n}{3(n+1)})I-\frac{1}{3(n+1)}2·\frac{1}{2}I=\frac{2}{3}I$, we rewrite Algorithm 4 as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+\frac{1}{3(n+1)}(x_n-x_{n-1}),\hfill \\ \hfill & u_n=\frac{2}{3}{w}_n+\frac{1}{3}T{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{x}_n+\frac{2}{3}{S}^n{q}_n,\hfill \end{array}\right.$$

(66)

with $C_n$ and ${\varsigma }_n$ being picked as in Algorithm 4 for every n. Hence, using Theorem 1, one has that $x_n\to 0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$ if and only if ${sup}_{n\ge 1}|x_n-x_{n-1}|<\infty$.

 Example 2. 

From the nonexpansivity of $Sx:=\frac{3}{5}sinx$, one obtains the following modification of Algorithm 4:

$$\left\{\begin{array}{cc}& w_n=x_n+\frac{1}{3(n+1)}(x_n-x_{n-1}),\hfill \\ \hfill & u_n=\frac{2}{3}{w}_n+\frac{1}{3}T{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{x}_n+\frac{2}{3}S{q}_n,\hfill \end{array}\right.$$

(67)

with $C_n$ and ${\varsigma }_n$ being picked in the above way. Thus, using Theorem 2, one knows that $x_n\to 0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$ if and only if ${sup}_{n\ge 1}|x_n-x_{n-1}|<\infty$.

 Example 3. 

Noticing $W_n=T$ and $(1-{\beta }_n)I-{\alpha }_n\mu F=(1-\frac{n}{3(n+1)})I-\frac{1}{3(n+1)}2·\frac{1}{2}I=\frac{2}{3}I$, we rewrite Algorithm 5 as follows:

$$\left\{\begin{array}{cc}& w_n=x_n+\frac{1}{3(n+1)}(x_n-x_{n-1}),\hfill \\ \hfill & u_n=\frac{2}{3}{x}_n+\frac{1}{3}T{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{u}_n+\frac{2}{3}{S}^n{q}_n,\hfill \end{array}\right.$$

(68)

with $C_n$ and ${\varsigma }_n$ being picked as in Algorithm 5 for every n. Hence, using Theorem 3, one has that $x_n\to 0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$ if and only if ${sup}_{n\ge 1}|x_n-x_{n-1}|<\infty$.

 Example 4. 

From the nonexpansivity of $Sx:=\frac{3}{5}sinx$, one obtains the following modification of Algorithm 5:

$$\left\{\begin{array}{cc}& w_n=x_n+\frac{1}{3(n+1)}(x_n-x_{n-1}),\hfill \\ \hfill & u_n=\frac{2}{3}{x}_n+\frac{1}{3}T{w}_n,\hfill \\ \hfill & y_n=P_C(u_n-{\varsigma }_{n}A{u}_n),\hfill \\ \hfill & q_n=P_{C_n}(u_n-{\varsigma }_{n}A{y}_n),\hfill \\ \hfill & x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{u}_n+\frac{2}{3}S{q}_n,\hfill \end{array}\right.$$

(69)

with $C_n$ and ${\varsigma }_n$ being picked in the above way. Thus, using Theorem 4, one knows that $x_n\to 0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$ if and only if ${sup}_{n\ge 1}|x_n-x_{n-1}|<\infty$.

It is noteworthy that the above two modified Mann-type subgradient extragradient algorithms with a linear-search process (i.e., Algorithms 4 and 5) are both applied for finding a point in the common solution set $\Omega =\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)$ with countable nonexpansive operators ${\left\{S_i\right\}}_{i=1}^{\infty }$ and asymptotically nonexpansive operator S. Under the same conditions imposed on the parameter sequences, we show the strong convergence of these two different algorithms to an element $q^{\ast }\in \Omega$, which is also a unique solution of the HFPP: $q^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }$; see Theorems 1 and 3 for more details. Note that Algorithm 4 is very similar to Algorithm 5 because these two different algorithms belong to the same class of modified Mann-type subgradient extragradient rules with a linear-search process. It is not difficult to find that Algorithm 4 is extended to develop Algorithm 5, e.g., (i) the original Mann iterative step $u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n$ in Algorithm 4 is developed into the modified Mann iterative step $u_n=(1-{\sigma }_n)x_n+{\sigma }_n{W}_n{w}_n$ in Algorithm 5, and (ii) the original viscosity hybrid deepest-descent step $x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{x}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n$ in Algorithm 4 is developed into the modified viscosity hybrid deepest-descent step $x_{n+1}={\alpha }_{n}g\left(x_n\right)+{\beta }_n{u}_n+((1-{\beta }_n)I-{\alpha }_n\mu F)S^n{q}_n$ in Algorithm 5. In the above Examples 1 and 3, the iterative schemes (66) and (68) are numerical examples of Algorithms 4 and 5, respectively, and both are applied for finding $0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$. Compared with scheme (66), the scheme (68) improves and develops it in the following aspects:

(i)

The original Mann iterative step $u_n=\frac{2}{3}{w}_n+\frac{1}{3}T{w}_n$ in (66) is developed into the modified Mann iterative step $u_n=\frac{2}{3}{x}_n+\frac{1}{3}T{w}_n$ in (68);

(ii)

The original viscosity hybrid deepest-descent step $x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{x}_n+\frac{2}{3}{S}^n{q}_n$ in (66) is developed into the modified viscosity hybrid deepest-descent step $x_{n+1}=\frac{1}{3(n+1)}·\frac{1}{2}{x}_n+\frac{n}{3(n+1)}{u}_n+\frac{2}{3}{S}^n{q}_n$ in (68).

Finally, applying Theorems 1 and 3 to schemes (66) and (68), respectively, we obtain that $x_n\to 0\in \Omega =\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(S\right)\cap \mathrm{Fix}\left(T\right)$ if and only if ${sup}_{n\ge 1}|x_n-x_{n-1}|<\infty$.

5. Conclusions

In real Hilbert spaces, we have designed two modified Mann-type subgradient extragradient rules with a linear-search process for settling the variational inequality problem (VIP) for a Lipschitz continuity and pseudomonotonicity operator A, and the common fixed-point problem (CFPP) for countable nonexpansivity operators ${\left\{S_i\right\}}_{i=1}^{\infty }$ and an asymptotical nonexpansivity operator $S_0:=S$. Under the lack of the sequential weak continuity and Lipschitz constant of the cost operator A, we have demonstrated the strong convergence of the constructed algorithms to a common element of the solution set of the VIP and the common fixed-point set of operators ${\left\{S_i\right\}}_{i=0}^{\infty }$, which is only a solution of a certain hierarchical fixed-point problem (HFPP). In addition, an illustrated example is provided to demonstrate the implementability and applicability of our proposed rules.

It is worth pointing out that there are our contributions to the research area of finding a common solution of the VIP and CFPP in three aspects below:

First, we extend the problem considered in [25], that is, the problem of finding a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^N\mathrm{Fix}\left(S_i\right)\right)$ with finite nonexpansive operators ${\left\{S_i\right\}}_{i=1}^N$ is developed into the problem of finding a point in $\mathrm{VI}(C,A)\cap \left({\bigcap }_{i=0}^{\infty }\mathrm{Fix}\left(S_i\right)\right)$ with countable nonexpansive operators ${\left\{S_i\right\}}_{i=1}^{\infty }$.

Second, we improve the rules proposed in [25], that is, the hybrid inertial subgradient extragradient rule with a linear-search process in [25] is developed into our modified Mann-type subgradient extragradient rule with a linear-search process, e.g., the original inertial step $w_n=S_n{x}_n+{ϵ}_n(S_n{x}_n-S_n{x}_{n-1})$ is developed into the modified Mann iteration step: $w_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and $u_n=(1-{\sigma }_n)w_n+{\sigma }_n{W}_n{w}_n$.

Finally, we weaken the convergence criteria presented in [25]. Indeed, it was shown in [25] that, under the condition $S^n{q}_n-S^{n+1}{q}_n\to 0$, the relation holds:

$$x_n\to q^{\ast }\in \Omega \iff \parallel x_n-y_n\parallel +\parallel x_n-x_{n+1}\parallel \to 0\mathrm{with}{q}^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }.$$

In this article, using Lemma 6 (i.e., Maingé’s lemma [34]), we show that, under the condition $S^n{x}_n-S^{n+1}{x}_n\to 0$, the relation holds:

$$x_n\to q^{\ast }\in \Omega \iff \underset{n\ge 1}{sup}\parallel x_{n-1}-x_n\parallel <\infty \mathrm{with}{q}^{\ast }=P_{\Omega }(I-\mu F+g)q^{\ast }.$$

In addition, it is worth mentioning that part of our future research is aimed at acquiring the strong convergence results for the modifications of our proposed rules with a Nesterov inertial extrapolation step and adaptive stepsizes.