On Mann Viscosity Subgradient Extragradient Algorithms for Fixed Point Problems of Finitely Many Strict Pseudocontractions and Variational Inequalities

Abstract

In a real Hilbert space, we denote CFPP and VIP as common fixed point problem of finitely many strict pseudocontractions and a variational inequality problem for Lipschitzian, pseudomonotone operator, respectively. This paper is devoted to explore how to find a common solution of the CFPP and VIP. To this end, we propose Mann viscosity algorithms with line-search process by virtue of subgradient extragradient techniques. The designed algorithms fully assimilate Mann approximation approach, viscosity iteration algorithm and inertial subgradient extragradient technique with line-search process. Under suitable assumptions, it is proven that the sequences generated by the designed algorithms converge strongly to a common solution of the CFPP and VIP, which is the unique solution to a hierarchical variational inequality (HVI).

1. Introduction and Preliminaries

Throughout this article, we suppose that the real vector space H is a Hilbert one and the nonempty subset C of H is a convex and closed one. An operator $S:C\to H$ is called:

(i) L-Lipschitzian if there exists $L>0$ such that $\parallel Su-Sv\parallel \le L\parallel u-v\parallel \forall u,v\in C$;

(ii) sequentially weakly continuous if for any $\left\{w_n\right\}\subset C$, the following implication holds: $w_n\rightharpoonup w\Rightarrow S{w}_n\rightharpoonup Sw$;

(iii) pseudomonotone if $⟨Su,u-v⟩\le 0\Rightarrow ⟨Sv,u-v⟩\le 0\forall u,v\in C$;

(iv) monotone if $⟨Su-Sv,v-u⟩\le 0\forall u,v\in C$;

(v) $\gamma$-strongly monotone if $\exists \gamma >0$ s.t. $⟨Su-Sw,u-w⟩\ge {\gamma \parallel u-w\parallel }^2\forall u,w\in C$.

It is not difficult to observe that monotonicity ensures the pseudomonotonicity. A self-mapping $S:C\to C$ is called a $\eta$-strict pseudocontraction if the relation holds: $⟨Su-Sv,u-v⟩\le {\parallel u-v\parallel }^2-\frac{1-\eta }{2}{\parallel (I-S)u-(I-S)v\parallel }^2\forall u,v\in C$ for some $\eta \in [0,1)$. By [1] we know that, in the case where S is $\eta$-strictly pseudocontractive, S is Lipschitzian, i.e., $\parallel Su-Sv\parallel \le \frac{1+\eta }{1-\eta }\parallel u-v\parallel \forall u,v\in C$. It is clear that the class of strict pseudocontractions includes the class of nonexpansive operators, i.e., $\parallel Su-Sv\parallel \le \parallel u-v\parallel \forall u,v\in C$. Both classes of nonlinear operators received much attention and many numerical algorithms were designed for calculating their fixed points in Hilbert or Banach spaces; see e.g., [2,3,4,5,6,7,8,9,10,11].

Let A be a self-mapping on H. The classical variational inequality problem (VIP) is to find $z\in C$ such that $⟨Az,y-z⟩\ge 0\forall y\in C$. The solution set of such a VIP is indicated by VI($C,A$). To the best of our knowledge, one of the most effective methods for solving the VIP is the gradient-projection method. Recently, many authors numerically investigated the VIP in finite dimensional spaces, Hilbert spaces or Banach spaces; see e.g., [12,13,14,15,16,17,18,19,20].

In 2014, Kraikaew and Saejung [21] suggested a Halpern-type gradient-like algorithm to deal with the VIP

$$\left\{\begin{array}{c}{v}_k=P_C(u_k-\ell A{u}_k),\hfill \\ C_k=\{v\in H:⟨u_k-\ell A{u}_k-v_k,v_k-v⟩\ge 0\},\hfill \\ w_k=P_{C_k}(u_n-\ell A{v}_k),\hfill \\ u_{k+1}={\varrho }_k{u}_0+(1-{\varrho }_k)w_k\forall k\ge 0,\hfill \end{array}\right.$$

where $\ell \in (0,\frac{1}{L}),\left\{{\varrho }_k\right\}\subset (0,1),{lim}_{k\to \infty }{\varrho }_k=0,{\sum }_{k=1}^{\infty }{\varrho }_k=+\infty$, and established strong convergence theorems for approximation solutions in Hilbert spaces. Later, Thong and Hieu [22] designed an inertial algorithm, i.e., for arbitrarily given $u_0,u_1\in H$, the sequence $\left\{u_k\right\}$ is constructed by

$$\left\{\begin{array}{c}{z}_k=u_k+{\varrho }_k(u_k-u_{k-1}),\hfill \\ v_k=P_C(z_k-\ell A{z}_k),\hfill \\ C_k=\{v\in H:⟨z_k-\ell A{z}_k-v_k,v_k-v⟩\ge 0\},\hfill \\ u_{k+1}=P_{C_k}(z_n-\ell A{v}_k)\forall k\ge 1,\hfill \end{array}\right.$$

with $\ell \in (0,\frac{1}{L})$. Under mild assumptions, they proved that $\left\{u_k\right\}$ converge weakly to a point of $\mathrm{VI}(C,A)$. Very recently, Thong and Hieu [23] suggested two inertial algorithms with linear-search process, to solve the VIP for Lipschitzian, monotone operator A and the FPP for a quasi-nonexpansive operator S satisfying a demiclosedness property in H. Under appropriate assumptions, they proved that the sequences constructed by the suggested algorithms converge weakly to a point of $\mathrm{Fix}\left(S\right)\cap \mathrm{VI}(C,A)$. Further research on common solutions problems, we refer the readers to [24,25,26,27,28,29,30,31,32,33,34,35,36,37,38].

In this paper, we first introduce Mann viscosity algorithms via subgradient extragradient techniques, and then establish some strong convergence theorems in Hilbert spaces. It is remarkable that our algorithms involve line-search process.

The following lemmas are useful for the convergence analysis of our algorithms in the sequel.

Lemma 1.

[39] Let the operator A be pseudomonotone and continuous on C. Given a point $w\in C$. Then the relation holds: $⟨Aw,w-y⟩\le 0\forall y\in C\iff ⟨Ay,w-y⟩\le 0\forall y\in C$.

Lemma 2.

[40] Suppose that $\left\{s_k\right\}$ is a sequence in $[0,+\infty )$ such that $s_{k+1}\le t_k{b}_k+(1-t_k)s_k\forall k\ge 1$, where $\left\{t_k\right\}$ and $\left\{b_k\right\}$ lie in real line $\mathbf{R}:=(-\infty ,\infty )$, such that:

(a) $\left\{t_k\right\}\subset [0,1]$ and ${\sum }_{k=1}^{\infty }{t}_k=\infty$;

(b) ${lim\; sup}_{k\to \infty }{b}_k\le 0$ or ${\sum }_{k=1}^{\infty }\left|t_k{b}_k\right|<\infty$. Then $s_k\to 0$ as $k\to \infty$.

From Ceng et al. [2] it is not difficult to find that the following lemmas hold.

Lemma 3.

Let Γ be an η-strictly pseudocontractive self-mapping on C. Then $I-\Gamma$ is demiclosed at zero.

Lemma 4.

For $l=1,\dots ,N$, let ${\Gamma }_l$ be an ${\eta }_l$-strictly pseudocontractive self-mapping on C. Then for $l=1,\dots ,N$, the mapping ${\Gamma }_l$ is an η-strict pseudocontraction with $\eta =max\{{\eta }_l:1\le l\le N\}$, such that

$$\parallel {\Gamma }_{l}u-{\Gamma }_{l}v\parallel \le \frac{1+\eta }{1-\eta }\parallel u-v\parallel \forall u,v\in C.$$

Lemma 5.

Let Γ be an η-strictly pseudocontractive self-mapping on C. Given two reals $\gamma ,\beta \in [0,+\infty )$. If $(\gamma +\beta )\eta \le \gamma$, then $\parallel \gamma (u-v)+\beta (\Gamma u-\Gamma v)\parallel \le (\gamma +\beta )\parallel u-v\parallel \forall u,v\in C$.

2. Main Results

Our first algorithm is specified below.

Algorithm 1

Initial Step: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.

Iteration Steps: Compute $x_{n+1}$ below:

Step 1. Put $v_n=x_n-{\sigma }_n(x_{n-1}-x_n)$ and calculate $u_n=P_C(v_n-{\ell }_{n}A{v}_n)$, where ${\ell }_n$ is picked to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\ell \parallel A{v}_n-A{u}_n\parallel \le \mu \parallel v_n-u_n\parallel .$$ (1)

Step 2. Calculate $z_n=(1-{\alpha }_n)P_{C_n}(v_n-{\ell }_{n}A{u}_n)+{\alpha }_{n}f\left(x_n\right)$ with $C_n:=\{v\in H:⟨v_n-{\ell }_{n}A{v}_n-u_n,u_n-v⟩\ge 0\}$.

Step 3. Calculate $$x_{n+1}={\gamma }_n{P}_{C_n}(v_n-{\ell }_{n}A{u}_n)+{\delta }_n{T}_n{z}_n+{\beta }_n{x}_n.$$ (2)

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

In this section, we always suppose that the following hypotheses hold:

$T_k$ is a ${\zeta }_k$-strictly pseudocontractive self-mapping on H for $k=1,\dots ,N$ s.t. $\zeta \in [0,1)$ with $\zeta =max\{{\zeta }_k:1\le k\le N\}$.

A is L-Lipschitzian, pseudomonotone self-mapping on H, and sequentially weakly continuous on C, such that $\Omega :={\cap }_{k=1}^N\mathrm{Fix}\left(T_k\right)\cap \mathrm{VI}(C,A)\ne \varnothing$.

$f:H\to C$ is a $\delta$-contraction with $\delta \in [0,\frac{1}{2})$.

$\left\{{\sigma }_n\right\}\subset [0,1]$ and $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\}\subset (0,1)$ are such that:

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

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

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

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

Following Xu and Kim [40], we denote $T_n:=T_{n\mathrm{mod}N},\forall n\ge 1$, where the mod function takes values in $\{1,2,\dots ,N\}$, i.e., whenever $n=jN+q$ for some $j\ge 0$ and $0\le q<N$, we obtain that $T_n=T_N$ in the case of $q=0$ and $T_n=T_q$ in the case of $0<q<N$.

Lemma 6.

The Armijo-like search rule (1) is well defined, and $min\{\gamma ,\frac{\mu l}{L}\}\le {\ell }_n\le \gamma$.

Proof. 

Obviously, (1) holds for all $\gamma l^m\le \frac{\mu }{L}$. So, ${\ell }_n$ is well defined and ${\ell }_n\le \gamma$. In the case of ${\ell }_n=\gamma$, the inequality is true. In the case of ${\ell }_n<\gamma$, (1) ensures $\parallel A{v}_n-A{P}_C(v_n-\frac{{\ell }_n}{l}A{v}_n)\parallel >\frac{\mu }{\frac{{\ell }_n}{l}}\parallel v_n-P_C(v_n-\frac{{\ell }_n}{l}A{v}_n)\parallel$. The L-Lipschitzian property of A yields ${\ell }_n>\frac{\mu l}{L}$.  □

Lemma 7.

Let $\left\{v_n\right\},\left\{u_n\right\}$ and $\left\{z_n\right\}$ be the sequences constructed by Algorithm 1. Then

$$\begin{array}{cc}\hfill \parallel z_n{-\omega \parallel }^2\le & (1-{\alpha }_n)\parallel v_n{-\omega \parallel }^2+{\alpha }_n\delta \parallel x_n{-\omega \parallel }^2-(1-{\alpha }_n)(1-\mu )[\parallel v_n-u_n{\parallel }^2\hfill \\ & +\parallel h_n-u_n{\parallel }^2]+2{\alpha }_n⟨f\omega -\omega ,z_n-\omega ⟩\forall \omega \in \Omega ,\hfill \end{array}$$

(3)

where $h_n:=P_{C_n}(v_n-{\ell }_{n}A{u}_n)\forall n\ge 1$.

Proof. 

First, taking an arbitrary $p\in \Omega \subset C\subset C_n$, we observe that

$$\begin{array}{cc}\hfill 2\parallel h_n{-p\parallel }^2& \le 2⟨h_n-p,v_n-{\ell }_{n}A{u}_n-p⟩\hfill \\ & =\parallel h_n{-p\parallel }^2+\parallel v_n{-p\parallel }^2-{\parallel h_n-v_n\parallel }^2-2⟨{\ell }_{n}A{u}_n,h_n-p⟩.\hfill \end{array}$$

So, it follows that $\parallel v_n{-p\parallel }^2-2⟨h_n-p,{\ell }_{n}A{u}_n⟩-\parallel h_n-v_n{\parallel }^2\ge {\parallel h_n-p\parallel }^2$, which together with (1), we deduce that $0\ge ⟨p-u_n,A{u}_n⟩$ and

$$\begin{array}{cc}\hfill \parallel h_n{-p\parallel }^2& \le \parallel v_n{-p\parallel }^2-{\parallel h_n-v_n\parallel }^2+2{\ell }_n(⟨A{u}_n,p-u_n⟩+⟨A{u}_n,u_n-h_n⟩)\hfill \\ & \le \parallel v_n{-p\parallel }^2-\parallel u_n-h_n{\parallel }^2-{\parallel v_n-u_n\parallel }^2+2⟨u_n-v_n+{\ell }_{n}A{u}_n,u_n-h_n⟩.\hfill \end{array}$$

(4)

Since $h_n=P_{C_n}(v_n-{\ell }_{n}A{u}_n)$ with $C_n:=\{v\in H:⟨u_n-v_n+{\ell }_{n}A{v}_n,u_n-v⟩\le 0\}$, we have $⟨u_n-v_n+{\ell }_{n}A{v}_n,u_n-h_n⟩\le 0$, which together with (1), implies that

$$\begin{array}{cc}\hfill 2⟨u_n-v_n+{\ell }_{n}A{u}_n,u_n-h_n⟩& =2⟨u_n-v_n+{\ell }_{n}A{v}_n,u_n-h_n⟩+2{\ell }_n⟨A{v}_n-A{u}_n,h_n-u_n⟩\hfill \\ & \le 2\mu \parallel u_n-v_n\parallel \parallel u_n-h_n\parallel \le \mu (\parallel v_n-u_n{\parallel }^2+\parallel h_n-u_n{\parallel }^2).\hfill \end{array}$$

Therefore, substituting the last inequality for (4), we infer that

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

(5)

In addition, we have

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

Using the convexity of the function $h\left(t\right)=t^2\forall t\in \mathbf{R}$, from (5) we get

$$\begin{array}{cc}\hfill \parallel z_n{-p\parallel }^2& \le [{\alpha }_n\delta \parallel x_n-p\parallel +(1-{\alpha }_n)\parallel h_n-p{\parallel ]}^2+2{\alpha }_n⟨(f-I)p,z_n-p⟩\hfill \\ & \le {\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\parallel h_n-p\parallel }^2+2{\alpha }_n⟨(f-I)p,z_n-p⟩\hfill \\ & \le {\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n)\parallel v_n{-p\parallel }^2-(1-{\alpha }_n)(1-\mu )[\parallel v_n-u_n{\parallel }^2\hfill \\ & +\parallel h_n-u_n{\parallel }^2]+2{\alpha }_n⟨(f-I)p,z_n-p⟩.\hfill \end{array}$$

 □

Lemma 8.

Let $\left\{x_n\right\},\left\{u_n\right\},$ and $\left\{z_n\right\}$ be bounded sequences constructed by Algorithm 1. If $x_n-x_{n+1}\to 0,v_n-u_n\to 0,v_n-z_n\to 0$ and $\exists \left\{v_{n_i}\right\}\subset \left\{v_n\right\}$ s.t. $v_{n_i}\rightharpoonup z\in H$, then $z\in \Omega$.

Proof. 

According to Algorithm 1, we get ${\sigma }_n(x_n-x_{n-1})=v_n-x_n\forall n\ge 1$, and hence $\parallel x_n-x_{n-1}\parallel \ge \parallel v_n-x_n\parallel$. Using the assumption $x_n-x_{n+1}\to 0$, we have

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

(6)

So,

$$\parallel z_n-x_n\parallel \le \parallel v_n-z_n\parallel +\parallel v_n-x_n\parallel \to 0.$$

Since $\left\{x_n\right\}$ is bounded, from $v_n=x_n-{\sigma }_n(x_{n-1}-x_n)$ we know that $\left\{v_n\right\}$ is a bounded vector sequence. According to (5), we obtain that $h_n:=P_{C_n}(v_n-{\ell }_{n}A{u}_n)$ is a bounded vector sequence. Also, by Algorithm 1 we get ${\alpha }_{n}f\left(x_n\right)+h_n-x_n-{\alpha }_n{h}_n=z_n-x_n$. So, the boundedness of $\left\{x_n\right\},\left\{h_n\right\}$ guarantees that as $n\to \infty$,

$$\parallel h_n-x_n\parallel =\parallel z_n-x_n-{\alpha }_{n}f\left(x_n\right)+{\alpha }_n{h}_n\parallel \le \parallel z_n-x_n\parallel +{\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel h_n\parallel )\to 0.$$

It follows that

$$x_{n+1}-z_n={\gamma }_n(h_n-x_n)+{\delta }_n(T_n{z}_n-z_n)+(1-{\delta }_n)(x_n-z_n),$$

which immediately yields

$$\begin{array}{cc}\hfill {\delta }_n\parallel T_n{z}_n-z_n\parallel & =\parallel x_{n+1}-x_n+x_n-z_n-(1-{\delta }_n)(x_n-z_n)-{\gamma }_n(h_n-x_n)\parallel \hfill \\ & =\parallel x_{n+1}-x_n+{\delta }_n(x_n-z_n)-{\gamma }_n(h_n-x_n)\parallel \hfill \\ & \le \parallel x_{n+1}-x_n\parallel +\parallel x_n-z_n\parallel +\parallel h_n-x_n\parallel .\hfill \end{array}$$

Since $x_n-x_{n+1}\to 0,z_n-x_n\to 0,h_n-x_n\to 0$ and ${lim\; inf}_{n\to \infty }{\delta }_n>0$, we obtain $\parallel z_n-T_n{z}_n\parallel \to 0$ as $n\to \infty$. This further implies that

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

(7)

We have $⟨v_n-{\ell }_{n}A{v}_n-u_n,v-u_n⟩\le 0\forall v\in C$, and

$$⟨v_n-u_n,v-u_n⟩+{\ell }_n⟨A{v}_n,u_n-v_n⟩\le {\ell }_n⟨A{v}_n,v-v_n⟩\forall v\in C.$$

(8)

Note that ${\ell }_n\ge min\{\gamma ,\frac{\mu l}{L}\}$. So, ${lim\; inf}_{i\to \infty }⟨A{v}_{n_i},v-v_{n_i}⟩\ge 0\forall v\in C$. This yields ${lim\; inf}_{i\to \infty }⟨A{u}_{n_i},v-u_{n_i}⟩\ge 0\forall v\in C$. Since $v_n-x_n\to 0$ and $v_{n_i}\rightharpoonup z$, we get $x_{n_i}\rightharpoonup z$. We may assume $k=n_i\mathrm{mod}N$ for all i. By the assumption $x_n-x_{n+k}\to 0$, we have $x_{n_i+j}\rightharpoonup z$ for all $j\ge 1$. Hence, $\parallel x_{n_i+j}-T_{k+j}{x}_{n_i+j}\parallel =\parallel x_{n_i+j}-T_{n_i+j}{x}_{n_i+j}\parallel \to 0$. Then the demiclosedness principle implies that $z\in \mathrm{Fix}\left(T_{k+j}\right)$ for all j. This ensures that

$$z\in \bigcap _{k=1}^N\mathrm{Fix}\left(T_k\right).$$

(9)

We now take a sequence $\left\{{\varsigma }_i\right\}\subset (0,1)$ satisfying ${\varsigma }_i\downarrow 0$ as $i\to \infty$. For all $i\ge 1$, we denote by $m_i$ the smallest natural number satisfying

$$⟨A{u}_{n_j},v-u_{n_j}⟩+{\varsigma }_i\ge 0\forall j\ge m_i.$$

(10)

Since $\left\{{\varsigma }_i\right\}$ is decreasing, it is clear that $\left\{m_i\right\}$ is increasing. Noticing that $\left\{u_{m_i}\right\}\subset C$ ensures $A{u}_{m_i}\ne 0\forall i\ge 1$, we set $e_{m_i}=\frac{A{u}_{m_i}}{\parallel A{u}_{m_i}{\parallel }^2}$, we get $⟨A{u}_{m_i},e_{m_i}⟩=1\forall i\ge 1$. So, from (10) we get $⟨A{u}_{m_i},v+{\varsigma }_i{e}_{m_i}-u_{m_i}⟩\ge 0\forall i\ge 1$. Also, the pseudomonotonicity of A implies $⟨A(v+{\varsigma }_i{e}_{m_i}),v+{\varsigma }_i{e}_{m_i}-u_{m_i}⟩\ge 0\forall i\ge 1$. This immediately leads to

$$⟨Av-A(v+{\varsigma }_i{h}_{m_i}),v+{\varsigma }_i{e}_{m_i}-u_{m_i}⟩-{\varsigma }_i⟨Av,h_{m_i}⟩\le ⟨Av,v-u_{m_i}⟩\forall i\ge 1.$$

(11)

We claim ${lim}_{i\to \infty }{\varsigma }_i{e}_{m_i}=0$. Indeed, from $v_{n_i}\rightharpoonup z$ and $v_n-u_n\to 0$, we obtain $u_{n_i}\rightharpoonup z$. So, $\left\{u_n\right\}\subset C$ ensures $z\in C$. Also, the sequentially weak continuity of A guarantees that $A{u}_{n_i}\rightharpoonup Az$. Thus, we have $Az\ne 0$ (otherwise, z is a solution). Moreover, the sequentially weak lower semicontinuity of $\parallel ·\parallel$ ensures $0<\parallel Az\parallel \le {lim\; inf}_{i\to \infty }\parallel A{u}_{n_i}\parallel$. Since $\left\{u_{m_i}\right\}\subset \left\{u_{n_i}\right\}$ and ${\varsigma }_i\downarrow 0$ as $i\to \infty$, we deduce that $0\le {lim\; sup}_{i\to \infty }\parallel {\varsigma }_i{e}_{m_i}\parallel ={lim\; sup}_{i\to \infty }\frac{{\varsigma }_i}{\parallel A{u}_{m_i}\parallel }\le \frac{{lim\; sup}_{i\to \infty }{\varsigma }_i}{{lim\; inf}_{i\to \infty }\parallel A{u}_{n_i}\parallel }=0$. Hence we get ${\varsigma }_i{e}_{m_i}\to 0$.

Finally we claim $z\in \Omega$. In fact, letting $i\to \infty$, we conclude that the right hand side of (11) tends to zero by the Lipschitzian property of A, the boundedness of $\left\{u_{m_i}\right\},\left\{h_{m_i}\right\}$ and the limit ${lim}_{i\to \infty }{\varsigma }_i{e}_{m_i}=0$. Thus, we get $⟨Av,v-z⟩={lim\; inf}_{i\to \infty }⟨Av,v-u_{m_i}⟩\ge 0\forall v\in C$. So, $z\in \mathrm{VI}(C,A)$. Therefore, from (9) we have $z\in {\cap }_{k=1}^N\mathrm{Fix}\left(T_k\right)\cap \mathrm{VI}(C,A)=\Omega$.  □

Theorem 1.

Assume $A\left(C\right)$ is bounded. Let $\left\{x_n\right\}$ be constructed by Algorithm 1. Then

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

where $x^{*}\in \Omega$ is the unique solution to the hierarchical variational inequality (HVI):  $⟨(I-f)x^{*},x^{*}-\omega ⟩\le 0,\forall \omega \in \Omega$.

Proof. 

Taking into account condition (iv) on $\left\{{\gamma }_n\right\}$, we may suppose that $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$. Applying Banach’s Contraction Principle, we obtain existence and uniqueness of a fixed point $x^{*}\in H$ for the mapping $P_{\Omega }\circ f$, which means that $x^{*}=P_{\Omega }f\left(x^{*}\right)$. Hence, the HVI

$$⟨(I-f)x^{*},x^{*}-\omega ⟩\le 0,\forall \omega \in \Omega$$

(12)

has a unique solution $x^{*}\in \Omega :={\cap }_{k=1}^N\mathrm{Fix}\left(T_k\right)\cap \mathrm{VI}(C,A)$

It is now obvious that the necessity of the theorem is true. In fact, if $x_n\to x^{*}\in \Omega$, then we get ${sup}_{n\ge 1}\parallel x_n-f\left(x_n\right)\parallel \le {sup}_{n\ge 1}(\parallel x_n-x^{*}\parallel +\parallel x^{*}-f\left(x^{*}\right)\parallel +\parallel f\left(x^{*}\right)-f\left(x_n\right)\parallel )<\infty$ and

$$\parallel x_n-x_{n+1}\parallel \le \parallel x_n-x^{*}\parallel +\parallel x_{n+1}-x^{*}\parallel \to 0(n\to \infty ).$$

For the sufficient condition, let us suppose $x_n-x_{n+1}\to 0$ and ${sup}_{n\ge 1}\parallel (I-f)x_n\parallel <\infty$. The sufficiency of our conclusion is proved in the following steps.  □

Step 1. We show the boundedness of $\left\{x_n\right\}$. In fact, let p be an arbitrary point in $\Omega$. Then $T_{n}p=p\forall n\ge 1$, and

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

(13)

which hence leads to

$$\parallel v_n-p\parallel \ge \parallel h_n-p\parallel \forall n\ge 1.$$

(14)

By the definition of $v_n$, we have

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

(15)

Noticing ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}<\infty$ and ${sup}_{n\ge 1}\parallel x_n-x_{n-1}\parallel <\infty$, we obtain that ${sup}_{n\ge 1}\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel <\infty$. This ensures that $\exists M_1>0$ s.t.

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

(16)

Combining (14)–(16), we get

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

(17)

Note that $A\left(C\right)$ is bounded, $u_n=P_C(v_n-{\ell }_{n}A{v}_n)$, $f\left(H\right)\subset C\subset C_n$ and $h_n=P_{C_n}(v_n-{\ell }_{n}A{u}_n)$. Hence we know that $\left\{A{u}_n\right\}$ is bounded. So, from ${sup}_{n\ge 1}\parallel (I-f)x_n\parallel <\infty$, it follows that

$$\begin{array}{cc}\hfill \parallel h_n-f\left(x_n\right)\parallel & \le \parallel v_n-{\ell }_{n}A{u}_n-f\left(x_n\right)\parallel \hfill \\ & \le \parallel x_n-x_{n-1}\parallel +\parallel x_n-f\left(x_n\right)\parallel +\gamma \parallel A{u}_n\parallel \le M_0,\hfill \end{array}$$

where $\exists M_0>0$ s.t. $M_0\ge {sup}_{n\ge 1}(\parallel x_n-x_{n-1}\parallel +\parallel x_n-f\left(x_n\right)\parallel +\gamma \parallel A{u}_n\parallel )$ (due to the assumption $x_n-x_{n+1}\to 0$). Consequently,

$$\begin{array}{cc}\hfill \parallel z_n-p\parallel & \le {\alpha }_n\delta \parallel x_n-p\parallel +(1-{\alpha }_n)\parallel h_n-p\parallel +{\alpha }_n\parallel (f-I)p\parallel \hfill \\ & \le (1-{\alpha }_n(1-\delta ))\parallel x_n-p\parallel +{\alpha }_n(M_1+\parallel (f-I)p\parallel ),\hfill \end{array}$$

which together with $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, yields

$$\begin{array}{cc}\hfill \parallel x_{n+1}-p\parallel & \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p)]\parallel +{\gamma }_n\parallel h_n-z_n\parallel \hfill \\ & \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)[(1-{\alpha }_n(1-\delta ))\parallel x_n-p\parallel +{\alpha }_n(M_0+M_1+\parallel (f-I)p\parallel \left)\right]\hfill \\ & =[1-{\alpha }_n(1-{\beta }_n)(1-\delta )]\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)(1-\delta )\frac{M_0+M_1+\parallel (f-I)p\parallel }{1-\delta }.\hfill \end{array}$$

This shows that $\parallel x_n-p\parallel \le max\{\parallel x_1-p\parallel ,\frac{M_0+M_1+\parallel (I-f)p\parallel }{1-\delta }\}\forall n\ge 1$. Thus, $\left\{x_n\right\}$ is bounded, and so are the sequences $\left\{h_n\right\},\left\{v_n\right\},\left\{u_n\right\},\left\{z_n\right\},\left\{T_n{z}_n\right\}$.

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

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

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

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le \parallel {\beta }_n(x_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p){\parallel }^2+2{\gamma }_n{\alpha }_n⟨h_n-f\left(x_n\right),x_{n+1}-p⟩\hfill \\ & \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\parallel z_n{-p\parallel }^2+2(1-{\beta }_n){\alpha }_n\parallel h_n-f\left(x_n\right)\parallel \parallel x_{n+1}-p\parallel \hfill \\ & \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\parallel v_n-p\parallel }^2\hfill \\ & -(1-{\alpha }_n)(1-\mu )[\parallel v_n-u_n{\parallel }^2+\parallel h_n-u_n{\parallel }^2]+{\alpha }_n{M}_2\},\hfill \end{array}$$

(18)

where $\exists M_2>0$ s.t. $M_2\ge {sup}_{n\ge 1}2(\parallel (f-I)p\parallel \parallel z_n-p\parallel +\parallel u_n-f\left(x_n\right)\parallel \parallel x_{n+1}-p\parallel )$. Also,

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

(19)

where $\exists M_3>0$ s.t. $M_3\ge {sup}_{n\ge 1}(2M_1\parallel x_n-p\parallel +{\beta }_n{M}_1^2)$. Substituting (19) for (18), we have

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

(20)

where $M_4:=M_2+M_3$. This immediately implies that

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

(21)

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

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}⟨f\left(p\right)-p,x_n-p⟩\hfill \\ +\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M\}.\hfill \end{array}$$

In fact, we get

$$\begin{array}{cc}\hfill \parallel v_n{-p\parallel }^2& \le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel )\hfill \\ & \le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,\hfill \end{array}$$

(22)

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

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le \parallel {\beta }_n(x_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p){\parallel }^2+2{\gamma }_n{\alpha }_n⟨h_n-f\left(x_n\right),x_{n+1}-p⟩\hfill \\ & \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p)]\parallel }^2\hfill \\ & +2{\gamma }_n{\alpha }_n⟨h_n-p,x_{n+1}-p⟩+2{\gamma }_n{\alpha }_n⟨p-f\left(x_n\right),x_{n+1}-p⟩,\hfill \end{array}$$

which leads to

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le {\beta }_n\parallel x_n{-p\parallel }^2+(1-{\beta }_n)[(1-{\alpha }_n)\parallel h_n-p{\parallel }^2+2{\alpha }_n⟨f\left(x_n\right)-p,z_n-p⟩]\hfill \\ & +{\gamma }_n{\alpha }_n(\parallel h_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)+2{\gamma }_n{\alpha }_n⟨p-f\left(x_n\right),x_{n+1}-p⟩.\hfill \end{array}$$

Using (17) and (22) we obtain that $\parallel h_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M$. Hence,

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+(1-{\beta }_n)(1-{\alpha }_n){\sigma }_n\parallel x_n-x_{n-1}\parallel 3M\hfill \\ & +2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,z_n-p⟩+{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)\hfill \\ & +(1-{\beta }_n){\alpha }_n{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M+2{\gamma }_n{\alpha }_n⟨f\left(x_n\right)-p,z_n-x_{n+1}⟩\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & +2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,x_n-p⟩+2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,z_n-x_n⟩\hfill \\ & +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n){\sigma }_n\parallel x_n-x_{n-1}\parallel 3M\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & +2{\alpha }_n{\delta }_n\delta \parallel x_n{-p\parallel }^2+2{\alpha }_n{\delta }_n⟨f\left(p\right)-p,x_n-p⟩+2{\alpha }_n{\delta }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel \hfill \\ & +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+(1-{\beta }_n){\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}⟨f\left(p\right)-p,x_n-p⟩\hfill \\ +\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M\}.\hfill \end{array}$$

(23)

Step 4. We show that $x_n\to x^{*}\in \Omega$, where $x^{*}$ is the unique solution of (12). Indeed, putting $p=x^{*}$, we infer from (23) that

$$\begin{array}{c}\parallel x_{n+1}-x^{*}{\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n-x^{*}{\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-x^{*}\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-x^{*}\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}⟨f\left(x^{*}\right)-x^{*},x_n-x^{*}⟩\hfill \\ +\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M\}.\hfill \end{array}$$

(24)

It is sufficient to show that ${lim\; sup}_{n\to \infty }⟨(f-I)x^{*},x_n-x^{*}⟩\le 0$. From (21), $x_n-x_{n+1}\to 0,{\alpha }_n\to 0$ and $\left\{{\beta }_n\right\}\subset [a,b]\subset (0,1)$, we get

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

This ensures that

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

(25)

Consequently,

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

Since $z_n={\alpha }_{n}f\left(x_n\right)+(1-{\alpha }_n)h_n$ with $h_n:=P_{C_n}(v_n-{\ell }_{n}A{u}_n)$, we get

$$\begin{array}{cc}\hfill \parallel z_n-u_n\parallel & =\parallel {\alpha }_{n}f\left(x_n\right)-{\alpha }_n{h}_n+h_n-u_n\parallel \hfill \\ & \le {\alpha }_n(\parallel f\left(x_n\right)\parallel +\parallel h_n\parallel )+\parallel h_n-u_n\parallel \to 0(n\to \infty ),\hfill \end{array}$$

(26)

and hence

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

(27)

Obviously, combining (25) and (26), guarantees that

$$\parallel v_n-z_n\parallel \le \parallel v_n-u_n\parallel +\parallel u_n-z_n\parallel \to 0(n\to \infty ).$$

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

$$\underset{n\to \infty }{lim\; sup}⟨(f-I)x^{*},x_n-x^{*}⟩=\underset{i\to \infty }{lim}⟨(f-I)x^{*},x_{n_i}-x^{*}⟩.$$

(28)

Since $\left\{x_n\right\}$ is bounded, we may suppose that $x_{n_i}\rightharpoonup \tilde{x}$. Hence from (28) we get

$$\underset{n\to \infty }{lim\; sup}⟨(f-I)x^{*},x_n-x^{*}⟩=\underset{i\to \infty }{lim}⟨(f-I)x^{*},x_{n_i}-x^{*}⟩=⟨(f-I)x^{*},\tilde{x}-x^{*}⟩.$$

(29)

It is easy to see from $v_n-x_n\to 0$ and $x_{n_i}\rightharpoonup \tilde{x}$ that $v_{n_i}\rightharpoonup \tilde{x}$. Since $x_n-x_{n+1}\to 0,v_n-u_n\to 0,v_n-z_n\to 0$ and $v_{n_i}\rightharpoonup \tilde{x}$, by Lemma 8 we infer that $\tilde{x}\in \Omega$. Therefore, from (12) and (29) we conclude that

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

(30)

Note that ${lim\; inf}_{n\to \infty }\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}>0$. It follows that ${\sum }_{n=0}^{\infty }\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n=\infty$. It is clear that

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

(31)

Therefore, by Lemma 2 we immediately deduce that $x_n\to x^{*}$.

Next, we introduce another Mann viscosity algorithm with line-search process by the subgradient extragradient technique.

Algorithm 2

Initial Step: Given $x_0,x_1\in H$ arbitrarily. Let $\gamma >0,l\in (0,1),\mu \in (0,1)$.

Iteration Steps: Compute $x_{n+1}$ below:

Step 1. Put $v_n=x_n-{\sigma }_n(x_{n-1}-x_n)$ and calculate $u_n=P_C(v_n-{\ell }_{n}A{v}_n)$, where ${\ell }_n$ is picked to be the largest $\ell \in \{\gamma ,\gamma l,\gamma l^2,\dots \}$ s.t. $$\ell \parallel A{v}_n-A{u}_n\parallel \le \mu \parallel v_n-u_n\parallel .$$ (32)

Step 2. Calculate $z_n=(1-{\alpha }_n)P_{C_n}(v_n-{\ell }_{n}A{u}_n)+{\alpha }_{n}f\left(x_n\right)$ with $C_n:=\{v\in H:⟨v_n-{\ell }_{n}A{v}_n-u_n,u_n-v⟩\ge 0\}$.

Step 3. Calculate $$x_{n+1}={\gamma }_n{P}_{C_n}(v_n-{\ell }_{n}A{u}_n)+{\delta }_n{T}_n{z}_n+{\beta }_n{v}_n.$$ (33)

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

It is remarkable that Lemmas 6, 7 and 8 remain true for Algorithm 2.

Theorem 2.

Assume $A\left(C\right)$ is bounded. Let $\left\{x_n\right\}$ be constructed by Algorithm 2. Then

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

where $x^{*}\in \Omega$ is the unique solution of the HVI: $⟨(I-f)x^{*},x^{*}-\omega ⟩\le 0,\forall \omega \in \Omega$.

Proof. 

For the necessity of our proof, we can observe that, by a similar approach to that in the proof of Theorem 1, we obtain that there is a unique solution $x^{*}\in \Omega$ of (12).

We show the sufficiency below. To this aim, we suppose $x_n-x_{n+1}\to 0$ and ${sup}_{n\ge 1}\parallel (I-f)x_n\parallel <\infty$, and prove the sufficiency by the following steps.  □

Step 1. We show the boundedness of $\left\{x_n\right\}$. In fact, by the similar inference to that in Step 1 for the proof of Theorem 1, we obtain that (13)–(17) hold. So, using Algorithm 2 and (17) we obtain

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

which together with $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, yields

$$\begin{array}{cc}\hfill \parallel x_{n+1}-p\parallel & \le {\beta }_n\parallel v_n-p\parallel +(1-{\beta }_n)\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p)]\parallel +{\gamma }_n\parallel h_n-z_n\parallel \hfill \\ & \le {\beta }_n(\parallel x_n-p\parallel +{\alpha }_n{M}_1)+(1-{\beta }_n)[(1-{\alpha }_n(1-\delta ))\parallel x_n-p\parallel \hfill \\ & +{\alpha }_n(M_0+M_1+\parallel (f-I)p\parallel \left)\right]\hfill \\ & =[1-{\alpha }_n(1-{\beta }_n)(1-\delta )]\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)(1-\delta )\frac{M_0+\frac{1}{1-{\beta }_n}{M}_1+\parallel (f-I)p\parallel }{1-\delta }.\hfill \end{array}$$

Therefore, we get the boundedness of $\left\{x_n\right\}$ and hence the one of sequences $\left\{h_n\right\},\left\{v_n\right\},\left\{u_n\right\},\left\{z_n\right\},\left\{T_n{z}_n\right\}$.

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

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

In fact, by Lemma 7 and the convexity of ${\parallel ·\parallel }^2$, we get

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le \parallel {\beta }_n(v_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p){\parallel }^2+2{\gamma }_n{\alpha }_n⟨h_n-f\left(x_n\right),x_{n+1}-p⟩\hfill \\ & \le {\beta }_n\parallel v_n{-p\parallel }^2+(1-{\beta }_n)\parallel z_n{-p\parallel }^2+2(1-{\beta }_n){\alpha }_n\parallel h_n-f\left(x_n\right)\parallel \parallel x_{n+1}-p\parallel \hfill \\ & \le {\beta }_n\parallel v_n{-p\parallel }^2+(1-{\beta }_n)\{{\alpha }_n\delta \parallel x_n{-p\parallel }^2+(1-{\alpha }_n){\parallel v_n-p\parallel }^2\hfill \\ & -(1-{\alpha }_n)(1-\mu )[\parallel v_n-u_n{\parallel }^2+\parallel h_n-u_n{\parallel }^2]+{\alpha }_n{M}_2\},\hfill \end{array}$$

(34)

where $\exists M_2>0$ s.t. $M_2\ge {sup}_{n\ge 1}2(\parallel (f-I)p\parallel \parallel z_n-p\parallel +\parallel u_n-f\left(x_n\right)\parallel \parallel x_{n+1}-p\parallel )$. Also,

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

(35)

where $\exists M_3>0$ s.t. $M_3\ge {sup}_{n\ge 1}(2M_1\parallel x_n-p\parallel +{\beta }_n{M}_1^2)$. Substituting (35) for (34), we have

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

(36)

where $M_4:=M_2+M_3$. This ensures that

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

(37)

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

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}⟨f\left(p\right)-p,x_n-p⟩\hfill \\ +\frac{1}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M\}.\hfill \end{array}$$

In fact, we get

$$\begin{array}{cc}\hfill \parallel v_n{-p\parallel }^2& \le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel (2\parallel x_n-p\parallel +{\sigma }_n\parallel x_n-x_{n-1}\parallel )\hfill \\ & \le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,\hfill \end{array}$$

(38)

where $\exists M>0$ s.t. $M\ge {sup}_{n\ge 1}\{\parallel x_n-p\parallel ,{\sigma }_n\parallel x_n-x_{n-1}\parallel \}$. Using Algorithm 1 and the convexity of ${\parallel ·\parallel }^2$, we get

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le \parallel {\beta }_n(v_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p){\parallel }^2+2{\gamma }_n{\alpha }_n⟨h_n-f\left(x_n\right),x_{n+1}-p⟩\hfill \\ & \le {\beta }_n\parallel v_n{-p\parallel }^2+(1-{\beta }_n){\parallel \frac{1}{1-{\beta }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T_n{z}_n-p)]\parallel }^2\hfill \\ & +2{\gamma }_n{\alpha }_n⟨h_n-p,x_{n+1}-p⟩+2{\gamma }_n{\alpha }_n⟨p-f\left(x_n\right),x_{n+1}-p⟩,\hfill \end{array}$$

which leads to

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& {\beta }_n\parallel v_n{-p\parallel }^2+(1-{\beta }_n)[(1-{\alpha }_n)\parallel h_n-p{\parallel }^2+2{\alpha }_n⟨f\left(x_n\right)-p,z_n-p⟩]\hfill \\ & +{\gamma }_n{\alpha }_n(\parallel h_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)+2{\gamma }_n{\alpha }_n⟨p-f\left(x_n\right),x_{n+1}-p⟩.\hfill \end{array}$$

Using (17) and (38) we deduce that $\parallel h_n{-p\parallel }^2\le \parallel v_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M$. Hence,

$$\begin{array}{cc}\hfill \parallel x_{n+1}{-p\parallel }^2& \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+[1-{\alpha }_n(1-{\beta }_n)]{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M\hfill \\ & +2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,z_n-p⟩+{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}-p{\parallel }^2)\hfill \\ & +(1-{\beta }_n){\alpha }_n{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M+2{\gamma }_n{\alpha }_n⟨f\left(x_n\right)-p,z_n-x_{n+1}⟩\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & +2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,x_n-p⟩+2{\alpha }_n{\delta }_n⟨f\left(x_n\right)-p,z_n-x_n⟩\hfill \\ & +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M\hfill \\ & \le [1-{\alpha }_n(1-{\beta }_n)]\parallel x_n{-p\parallel }^2+2{\gamma }_n{\alpha }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ & +2{\alpha }_n{\delta }_n\delta \parallel x_n{-p\parallel }^2+2{\alpha }_n{\delta }_n⟨f\left(p\right)-p,x_n-p⟩+2{\alpha }_n{\delta }_n\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel \hfill \\ & +{\gamma }_n{\alpha }_n(\parallel x_n{-p\parallel }^2+\parallel x_{n+1}{-p\parallel }^2)+{\sigma }_n\parallel x_n-x_{n-1}\parallel 3M,\hfill \end{array}$$

which immediately yields

$$\begin{array}{c}\parallel x_{n+1}{-p\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\alpha }_n{\gamma }_n}{\alpha }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\alpha }_n}{1-{\alpha }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}⟨f\left(p\right)-p,x_n-p⟩\hfill \\ +\frac{1}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\sigma }_n}{{\alpha }_n}\parallel x_n-x_{n-1}\parallel 3M\}.\hfill \end{array}$$

(39)

Step 4. In order to show that $x_n\to x^{*}\in \Omega$, which is the unique solution of (12), we can follow a similar method to that in Step 4 for the proof of Theorem 1.

Finally, we apply our main results to solve the VIP and common fixed point problem (CFPP) in the following illustrating example.

The starting point $x_0=x_1$ is randomly picked in the real line. Put $f\left(u\right)=\frac{1}{8}sinu,\gamma =l=\mu =\frac{1}{2},{\sigma }_n={\alpha }_n=\frac{1}{n+1},{\beta }_n=\frac{1}{3},{\gamma }_n=\frac{1}{6}$ and ${\delta }_n=\frac{1}{2}$.

We first provide an example of Lipschitzian, pseudomonotone self-mapping A satisfying the boundedness of $A\left(C\right)$ and strictly pseudocontractive self-mapping $T_1$ with $\Omega =\mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-1,2]$ and H be the real line with the inner product $⟨a,b⟩=ab$ and induced norm $\parallel ·\parallel =|·|$. Then f is a $\delta$-contractive map with $\delta =\frac{1}{8}\in [0,\frac{1}{2})$ and $f\left(H\right)\subset C$ because $\parallel f\left(u\right)-f\left(v\right)\parallel =\frac{1}{8}\parallel sinu-sinv\parallel \le \frac{1}{8}\parallel u-v\parallel$ for all $u,v\in H$.

Let $A:H\to H$ and $T_1:H\to H$ be defined as $Au:=\frac{1}{1+|sinu|}-\frac{1}{1+\left|u\right|}$, and $T_{1}u:=\frac{1}{2}u-\frac{3}{8}sinu$ for all $u\in H$. Now, we first show that A is L-Lipschitzian, pseudomonotone operator with $L=2$, such that $A\left(C\right)$ is bounded. In fact, for all $u,v\in H$ we get

$$\begin{array}{cc}\hfill \parallel Au-Av\parallel & \le |\frac{1}{1+\parallel u\parallel }-\frac{1}{1+\parallel v\parallel }|+|\frac{1}{1+\parallel sinu\parallel }-\frac{1}{1+\parallel sinv\parallel }|\hfill \\ & =|\frac{\parallel v\parallel -\parallel u\parallel }{(1+\parallel u\parallel )(1+\parallel v\parallel )}|+|\frac{\parallel sinv\parallel -\parallel sinu\parallel }{(1+\parallel sinu\parallel )(1+\parallel sinv\parallel )}|\hfill \\ & \le \frac{\parallel u-v\parallel }{(1+\parallel u\parallel )(1+\parallel v\parallel )}+\frac{\parallel sinu-sinv\parallel }{(1+\parallel sinu\parallel )(1+\parallel sinv\parallel )}\hfill \\ & \le 2\parallel u-v\parallel .\hfill \end{array}$$

This implies that A is 2-Lipschitzian. Next, we show that A is pseudomonotone. For any given $u,v\in H$, it is clear that the relation holds:

$$⟨Au,u-v⟩=(\frac{1}{1+|sinu|}-\frac{1}{1+\left|u\right|})(u-v)\le 0\Rightarrow ⟨Av,u-v⟩=(\frac{1}{1+|sinv|}-\frac{1}{1+\left|v\right|})(u-v)\le 0.$$

Furthermore, it is easy to see that $T_1$ is strictly pseudocontractive with constant ${\zeta }_1=\frac{1}{4}$. In fact, we observe that for all $u,v\in H$,

$$\parallel T_{1}u-T_{1}v\parallel \le \frac{1}{2}\parallel u-v\parallel +\frac{3}{8}\parallel sinu-sinv\parallel \le \parallel u-v\parallel +\frac{1}{4}\parallel (I-T_1)u-(I-T_1)v\parallel .$$

It is clear that $({\gamma }_n+{\delta }_n){\zeta }_1=(\frac{1}{6}+\frac{1}{2})·\frac{1}{4}\le \frac{1}{6}={\gamma }_n<(1-2\delta ){\delta }_n=(1-2·\frac{1}{8})\frac{1}{2}=\frac{3}{8}$ for all $n\ge 1$. In addition, it is clear that $\mathrm{Fix}\left(T_1\right)=\left\{0\right\}$ and $A0=0$ because the derivative $d\left(T_{1}u\right)/du=\frac{1}{2}-\frac{3}{8}cosu>0$. Therefore, $\Omega =\left\{0\right\}\ne \varnothing$. In this case, Algorithm 1 can be rewritten below:

$$\left\{\begin{array}{c}{v}_n=x_n-\frac{1}{n+1}(x_{n-1}-x_n),\hfill \\ u_n=P_C(v_n-{\ell }_{n}A{v}_n),\hfill \\ z_n=\frac{1}{n+1}f\left(x_n\right)+\frac{n}{n+1}{P}_{C_n}(v_n-{\ell }_{n}A{u}_n),\hfill \\ x_{n+1}=\frac{1}{3}{x}_n+\frac{1}{6}{P}_{C_n}(v_n-{\ell }_{n}A{u}_n)+\frac{1}{2}{T}_1{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

with $\left\{C_n\right\}$ and $\left\{{\ell }_n\right\}$, selected as in Algorithm 1. Then, by Theorem 1, we know that $x_n\to 0\in \Omega$ iff $x_n-x_{n+1}\to 0(n\to \infty )$ and ${sup}_{n\ge 1}|x_n-\frac{1}{8}sin{x}_n|<\infty$.

On the other hand, Algorithm 2 can be rewritten below:

$$\left\{\begin{array}{c}{v}_n=x_n-\frac{1}{n+1}(x_{n-1}-x_n),\hfill \\ u_n=P_C(v_n-{\ell }_{n}A{v}_n),\hfill \\ z_n=\frac{1}{n+1}f\left(x_n\right)+\frac{n}{n+1}{P}_{C_n}(v_n-{\ell }_{n}A{u}_n),\hfill \\ x_{n+1}=\frac{1}{3}{v}_n+\frac{1}{6}{P}_{C_n}(v_n-{\ell }_{n}A{u}_n)+\frac{1}{2}{T}_1{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

with $\left\{C_n\right\}$ and $\left\{{\ell }_n\right\}$, selected as in Algorithm 2. Then, by Theorem 2 , we know that $x_n\to 0\in \Omega$ iff $x_n-x_{n+1}\to 0(n\to \infty )$ and ${sup}_{n\ge 1}|x_n-\frac{1}{8}sin{x}_n|<\infty$.