Hybrid Algorithms for Variational Inequalities Involving a Strict Pseudocontraction

Abstract

In a real Hilbert space, we investigate the Tseng’s extragradient algorithms with hybrid adaptive step-sizes for treating a Lipschitzian pseudomonotone variational inequality problem and a strict pseudocontraction fixed-point problem, which are symmetry. By imposing some appropriate weak assumptions on parameters, we obtain a norm solution of the problems, which solves a certain hierarchical variational inequality.

1. Introduction

In a real Hilbert space H, one employs $\langle ·,·\rangle$ and $\parallel ·\parallel$ to stand for its inner product and norm. Let $P_C$ be the projection operator from the space H onto a nonempty convex and closed set C, where $C\subset H$. Let us denote by $\mathrm{Fix}\left(S\right)$ the set of all fixed points of an operator $S:C\to H$. The notations →, $\mathbf{R}$, and ⇀ will be used to stand for the strong convergence, the set of real numbers, and the weak convergence, respectively. A self-operator $S:C\to C$ is named $\varsigma$-strictly pseudocontractive if $\exists \varsigma \in [0,1)$ such that

$${\parallel Su-Sv\parallel }^2\le {\varsigma \parallel (I-S)u-(I-S)v\parallel }^2+{\parallel u-v\parallel }^2\forall u,v\in C.$$

In particular, whenever $\varsigma =0$, S is called nonexpansive. This means that the class of nonexpansive mappings is a proper subclass of the one of strict pseudocontractions. Recall that an operator $S:C\to H$ is called

(i)

Lipschitz with module L if $\exists L>0$ such that

$$\parallel Su-Sv\parallel \le L\parallel u-v\parallel \forall u,v\in C;$$

(ii)

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

(iii)

pseudomonotone if

$$\langle v-u,Su\rangle \ge 0\Rightarrow \langle v-u,Sv\rangle \ge 0\forall u,v\in C;$$

(iv)

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

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

(v)

sequentially weakly continuous if $u_n\rightharpoonup u\Rightarrow S{u}_n\rightharpoonup Su\forall \left\{u_n\right\}\subset C$.

It is not hard to see that the pseudomonotone operators may not be monotone. In addition, recall that the operator $S:C\to C$ is $\varsigma$-strictly pseudocontractive with constant $\varsigma \in [0,1)$ iff the following inequality holds: $2\langle Su-Sv,u-v\rangle \le {2\parallel u-v\parallel }^2-(1-\varsigma ){\parallel (I-S)u-(I-S)v\parallel }^2\forall u,v\in C$. It is obvious that if S is a $\varsigma$-strict pseudocontraction, then S satisfies Lipschitz condition $\parallel Su-Sv\parallel \le \frac{1+\varsigma }{1-\varsigma }\parallel u-v\parallel \forall u,v\in C$. For each point $u\in H$, we know that there exists a unique nearest point in C, denoted by $P_{C}u$, such that $\parallel u-P_{C}u\parallel \le \parallel u-v\parallel \forall v\in C$. The operator $P_C$ is called the metric projection of H onto C.

Consider an operator $A:H\to H$. The classical monotone variational inequality problem (VIP) consists of finding $u^{*}\in C$ s.t. $\langle v-u^{*},A{u}^{*}\rangle \ge 0\forall v\in C$. The solution set of such a VIP is denoted by VI($C,A$). Korpelevich [1] first designed an extragradient method with two projections

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

with $\ell \in (0,\frac{1}{L})$, which has been one of the most popular methods for dealing with the VIP up until now. If $\mathrm{VI}(C,A)\ne \varnothing$, it was shown in [1] that $\left\{x_n\right\}$ weakly converges to a vector in $\mathrm{VI}(C,A)$. The gradient (reduced) type iterative schemes are under the spotlight of investigators of applied mathematicians and engineers in the communities of nonlinear and optimization. Based on this approach, a number of authors have conducted various investigations on efficient iterative algorithms; for examples, see [2,3,4,5,6,7,8,9,10,11].

Let both the operators A and B be inverse-strongly monotone from C to H and the self-mapping $S:C\to C$ be $\varsigma$-strictly pseudocontractive. In 2010, via the extragradient approach, Yao et al. [12] designed an efficient, fast algorithm for obtaining a feasibility point in a common solution set:

$$\left\{\begin{array}{c}{w}_n=P_C(u_n-\mu B{u}_n),\hfill \\ v_n=(1-{\beta }_n)P_C(w_n-\lambda A{w}_n)+{\beta }_{n}f\left(u_n\right),\hfill \\ u_{n+1}={\gamma }_n{P}_C(w_n-\lambda A{w}_n)+{\delta }_{n}S{v}_n+{\sigma }_n{u}_n,\forall n\ge 0,\hfill \end{array}\right.$$

where $f:C\to C$ is a $\delta$-contractive map with $\delta \in [0,\frac{1}{2})$, and $\left\{{\beta }_n\right\},\left\{{\sigma }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\}$ are four sequences in $[0,1]$ s.t. ${\sigma }_n+{\gamma }_n+{\delta }_n=1$, $({\gamma }_n+{\delta }_n)\varsigma \le {\gamma }_n<(1-2\delta ){\delta }_n$, ${\sum }_{n=0}^{\infty }{\beta }_n=\infty$, ${lim\; inf}_{n\to \infty }{\delta }_n>0$, ${lim\; inf}_{n\to \infty }{\sigma }_n>0$, and ${lim}_{n\to \infty }(\frac{{\gamma }_{n+1}}{1-{\sigma }_{n+1}}-\frac{{\gamma }_n}{1-{\sigma }_n})={lim}_{n\to \infty }{\beta }_n=0$. They claimed the strong convergence of the sequence in H.

In the extragradient approach, one has to compute two projection operators onto C. It is clear that the projection operator onto the convex set C is closely related to a minimum distance problem. In the case where C is a general convex set, the computation of two projections might be prohibitively time-consuming. Via Korpelevich’s extragradient approach, Censor et al. [13] suggested a subgradient algorithm, in which the second projection operator onto the subset C is changed onto a half-space. Recently, numerous methods of reduced-gradient-type are focused and extensively investigated in both infinite and infinite dimensional spaces; see, for example [14,15,16,17,18,19,20,21,22,23]. Based on inertial effects, Thong and Hieu [24] proposed an inertial subgradient method, and also proved the weak convergence of their algorithms. In addition, the authors [25] investigated subgradient-based fast algorithms with inertial effects.

Inspired by the above research works in [12,24,25,26,27], we are concerned with hybrid-adaptive step-sizes Tseng’s extragradient algorithms, that are more advantageous and more subtle than the above iterative algorithms because they involve solving the VIP with Lipschitzian, pseudomonotone operators, and the common fixed-point problem of a finite family of strict pseudocontractions in Hilbert spaces. By imposing some appropriate weak assumptions on parameters, one obtains a norm solution of the problems, which solves a certain hierarchical variational inequality. The outline of this article is organized below. In Section 2, a toolbox containing definitions and preliminary results is provided. In Section 3, we propose and investigate the iterative algorithms and their convergence criteria. In Section 4, theorems of norm solutions are employed as illustrating examples to support the convergence criteria.

2. Preliminaries

Lemma 1

([28]). Let $S:C\to C$ be a ς-strict pseudocontraction. If $\left\{u_n\right\}$ is a sequence in C such that $(Id-S)u_n\to 0$, where $Id$ is the identity operator on H, and $u_n\rightharpoonup u\in C$, then $u=Su$. Further, S is $\frac{1+\zeta }{1-\zeta }$ Lipschitz continuous.

Lemma 2

([29]). Let $S:C\to C$ be a ς-strict pseudocontraction, and let γ and β be real numbers in $[0,+\infty )$. Then, $\parallel \gamma (u-v)+\beta (Sv-Su)\parallel \le (\beta +\gamma )\parallel v-u\parallel \forall v,u\in C$ provided that $\gamma \ge \varsigma (\beta +\gamma )$.

Lemma 3

([30]). Let f be a pseudomonotone mapping from C into H which is continuous on finite-dimensional subspaces. Then, $x\in C$ is a solution of $\langle u-x,f\left(x\right)\rangle \ge 0$, $\forall u\in C$ iff $\langle u-x,f\left(u\right)\rangle \ge 0$, $\forall u\in C.$

Lemma 4

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

3. Results

From now on, one can always assume that our feasibility set $\mathrm{\Omega }=\mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$ is consistent.

Put $n:=n+1$ and return to Step 1 (Algorithm 1), where $\left\{{ϵ}_n\right\}\subset (0,1]$ and $\left\{{\beta }_n\right\},\left\{{\sigma }_n\right\},\left\{{\gamma }_n\right\},\left\{{\delta }_n\right\}\subset (0,1)$ are such that ${\sigma }_n+{\gamma }_n+{\delta }_n=1$; ${lim}_{n\to \infty }\frac{{ϵ}_n}{{\beta }_n}={lim}_{n\to \infty }{\beta }_n=0$; ${lim\; inf}_{n\to \infty }{\delta }_n>0$; ${lim\; inf}_{n\to \infty }{\sigma }_n>0$; ${lim\; inf}_{n\to \infty }((1-2\delta ){\delta }_n-{\gamma }_n)>0$; ${lim\; sup}_{n\to \infty }{\sigma }_n<1$; ${\sum }_{n=1}^{\infty }{\beta }_n=\infty$; $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n<(1-2\delta ){\delta }_n$; the pseudomonotone self-operator A is Lipschitz continuous with module L and sequentially weakly continuous on H; T is a $\zeta$-strictly pseudocontractive self-operator on H; and $f:$ is a $\delta$-contraction operator, where $\delta \in [0,\frac{1}{2})$, from H to C.

Algorithm 1: Initial Step: Fix two initials $x_0$ and $x_1$ in H and set $\alpha >0,{\tau }_1>0,\mu \in (0,1)$.

Iteration Steps: calculate iterative sequence $x_{n+1}$ as follows: Step 1. Given the iterates $x_{n-1}$ and $x_n(n\ge 1)$, choose ${\alpha }_n$ s.t. $0\le {\alpha }_n\le {\overline{\alpha }}_n$, where $${\overline{\alpha }}_n=\left\{\begin{array}{c}min\{\alpha ,\frac{{ϵ}_n}{\parallel x_n-x_{n-1}\parallel }\}\mathrm{if}{x}_n\ne x_{n-1},\hfill \\ \alpha \mathrm{otherwise}.\hfill \end{array}\right.$$ (1) Step 2. Let $w_n=x_n+{\alpha }_n(x_n-x_{n-1})$ and calculate $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$. Step 3. Calculate $x_{n+1}={\sigma }_n{x}_n+{\gamma }_n(y_n-{\tau }_n(A{y}_n-A{w}_n))+{\delta }_{n}T{z}_n$, where $z_n={\beta }_{n}f\left(x_n\right)+(1-{\beta }_n)(y_n-{\tau }_n(A{y}_n-A{w}_n))$. Update $${\tau }_{n+1}=\left\{\begin{array}{c}min\{\frac{\mu \parallel w_n-y_n\parallel }{\parallel A{w}_n-A{y}_n\parallel },{\tau }_n\}\mathrm{if}A{w}_n-A{y}_n\ne 0,\hfill \\ {\tau }_n\mathrm{otherwise}.\hfill \end{array}\right.$$ (2)

Remark 1.

We show${lim}_{n\to \infty }({\alpha }_n\parallel x_n-x_{n-1}\parallel )/{\beta }_n=0$. It follows from (1) that${ϵ}_n\ge {\alpha }_n\parallel x_n-x_{n-1}\parallel \forall n\ge 1$. Since${lim}_{n\to \infty }\frac{{ϵ}_n}{{\beta }_n}=0$, one sees that$0={lim}_{n\to \infty }\frac{{ϵ}_n}{{\beta }_n}\ge {lim\; sup}_{n\to \infty }\frac{{\alpha }_n\parallel x_n-x_{n-1}\parallel }{{\beta }_n}$.

Lemma 5.

Let$\left\{{\tau }_n\right\}$be generated by (2). Then,$\left\{{\tau }_n\right\}$is a nonincreasing sequence with${\tau }_n\ge \tilde{\tau }:=min\{{\tau }_1,\frac{\mu }{L}\}\forall n\ge 1$and${lim}_{n\to \infty }{\tau }_n\ge \tilde{\tau }:=min\{{\tau }_1,\frac{\mu }{L}\}$.

Proof. 

By borrowing (2), one concludes that ${\tau }_n\ge {\tau }_{n+1}\forall n\ge 1$. One also has

$$\parallel A{w}_n-A{y}_n\parallel \le L\parallel w_n-y_n\parallel \Rightarrow {\tau }_{n+1}\ge min\{{\tau }_n,\frac{\mu }{L}\}.$$

Note that ${\tau }_1\ge \tilde{\tau }:=min\{{\tau }_1,\frac{\mu }{L}\}$. So, ${\tau }_n\ge \tilde{\tau }:=min\{{\tau }_1,\frac{\mu }{L}\}\forall n\ge 1$. □

Lemma 6.

Let$\left\{y_n\right\},\left\{w_n\right\}$, and$\left\{z_n\right\}$be three iterative vector sequences defined by Algorithm 1. We have

$$\begin{array}{ll}\parallel p-z_n{\parallel }^2& \le {\beta }_n\delta \parallel p-x_n{\parallel }^2+(1-{\beta }_n)\parallel p-w_n{\parallel }^2-(1-{\beta }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2}){\parallel w_n-y_n\parallel }^2\\ & +2{\beta }_n\langle (f-I)p,z_n-p\rangle ,\forall p\in \mathrm{\Omega },\end{array}$$

(3)

where$u_n:=y_n-{\tau }_n(A{y}_n-A{w}_n)$.

Proof. 

Fixing $p\in \mathrm{\Omega }=\mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$ arbitrarily, one asserts $\langle Ap,y_n-p\rangle \ge 0$ and $Tp=p$. This yields

$$\begin{array}{c}\parallel p-u_n{\parallel }^2=\parallel p-y_n{\parallel }^2+{\tau }_n^2{\parallel A{y}_n-A{w}_n\parallel }^2-2{\tau }_n\langle y_n-p,A{y}_n-A{w}_n\rangle \hfill \\ =\parallel p-w_n{\parallel }^2+\parallel w_n-y_n{\parallel }^2+2\langle y_n-w_n,w_n-p\rangle +{\tau }_n^2{\parallel A{y}_n-A{w}_n\parallel }^2\hfill \\ -2{\tau }_n\langle y_n-p,A{y}_n-A{w}_n\rangle \hfill \\ =\parallel p-w_n{\parallel }^2-\parallel w_n-y_n{\parallel }^2+2\langle y_n-w_n,y_n-p\rangle +{\tau }_n^2{\parallel A{y}_n-A{w}_n\parallel }^2\hfill \\ -2{\tau }_n\langle y_n-p,A{y}_n-A{w}_n\rangle .\hfill \end{array}$$

Thanks to $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$, we have

$$\langle y_n-w_n,y_n-p\rangle \le -{\tau }_n\langle A{w}_n,y_n-p\rangle .$$

This ensures that

$$\begin{array}{ll}\parallel p-u_n{\parallel }^2& \le \parallel p-w_n{\parallel }^2-\parallel w_n-y_n{\parallel }^2-2{\tau }_n\langle A{w}_n,y_n-p\rangle +{\tau }_n^2{\parallel A{y}_n-A{w}_n\parallel }^2\\ & -2{\tau }_n\langle y_n-p,A{y}_n-A{w}_n\rangle \\ & =\parallel p-w_n{\parallel }^2-\parallel w_n-y_n{\parallel }^2+{\tau }_n^2{\parallel A{y}_n-A{w}_n\parallel }^2-2{\tau }_n\langle A{y}_n,y_n-p\rangle .\end{array}$$

By using the fact that $\langle Ap,y_n-p\rangle \ge 0$, one obtains that $\langle A{y}_n,y_n-p\rangle \ge 0$. Hence,

$$\parallel p-u_n{\parallel }^2\le \parallel p-w_n{\parallel }^2+{\tau }_n^2\parallel A{y}_n-A{w}_n{\parallel }^2-{\parallel w_n-y_n\parallel }^2.$$

(4)

Moreover, from (2), it follows that

$${\tau }_{n+1}\parallel A{w}_n-A{y}_n\parallel \le \mu \parallel w_n-y_n\parallel \forall n\ge 1.$$

(5)

Combining (4) and (5), we obtain

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

(6)

On the other hand,

$$\begin{array}{ll}{z}_n-p& =(1-{\beta }_n)(y_n-{\tau }_n(A{y}_n-A{w}_n))-p+{\beta }_{n}f\left(x_n\right)\\ & =(1-{\beta }_n)(u_n-p)+{\beta }_n(f-I)p+{\beta }_n(f\left(x_n\right)-f\left(p\right)).\end{array}$$

Using the convexity of the norm function, we get

$$\begin{array}{ll}\parallel z_n{-p\parallel }^2& \le 2{\beta }_n\langle fp-p,z_n-p\rangle +{\parallel (1-{\beta }_n)(u_n-p)-{\beta }_n(f\left(p\right)-f\left(x_n\right))\parallel }^2\\ & \le 2{\beta }_n\langle (fp-p),z_n-p\rangle +[(1-{\beta }_n)\parallel p-u_n\parallel +{\beta }_n\delta \parallel p-x_n{\parallel ]}^2\\ & \le 2{\beta }_n\langle (fp-p),z_n-p\rangle +(1-{\beta }_n)\parallel p-u_n{\parallel }^2+{\beta }_n\delta {\parallel p-x_n\parallel }^2\\ & \le (1-{\beta }_n)[\parallel p-w_n{\parallel }^2-(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})\parallel w_n-y_n{\parallel }^2]+{\beta }_n\delta {\parallel p-x_n\parallel }^2\\ & +2{\beta }_n\langle (fp-p),z_n-p\rangle \\ & =(1-{\beta }_n)\parallel p-w_n{\parallel }^2-(1-{\beta }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})\parallel w_n-y_n{\parallel }^2+{\beta }_n\delta {\parallel p-x_n\parallel }^2\\ & +2{\beta }_n\langle (fp-p),z_n-p\rangle .\end{array}$$

This completes the proof. □

Lemma 7.

Let$\left\{z_n\right\},\left\{y_n\right\}$, and$\left\{x_n\right\}$be three iterative sequences, which are bounded, defined by Algorithm 1. Suppose that there exists a subsequence$\left\{w_{n_k}\right\}$of the weakly convergent sequence$\left\{w_n\right\}$such that$w_{n_k}\rightharpoonup z\in H$. If$\parallel x_n-x_{n+1}\parallel \to 0,\parallel w_n-y_n\parallel \to 0,\parallel w_n-z_n\parallel \to 0$, then$z\in \mathrm{\Omega }$.

Proof. 

Algorithm 1 shows $\parallel x_n-w_n\parallel ={\alpha }_n\parallel x_{n-1}-x_n\parallel$. Utilizing Remark 1, we have ${lim}_{n\to \infty }\parallel w_n-x_n\parallel =0$. This, together with the assumption $w_n-z_n\to 0$, yields that

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

Since $\left\{x_n\right\}$ is bounded and ${\alpha }_n(x_n-x_{n-1})\to 0$, one asserts that $\left\{w_n\right\}$ is bounded. Note that (4) yields

$$\parallel u_n{-p\parallel }^2\le \parallel w_n{-p\parallel }^2+{\tau }_1^2{L}^2{\parallel y_n-w_n\parallel }^2.$$

Hence, $\left\{u_n\right\}$ is bounded, where $u_n:=y_n-{\tau }_n(A{y}_n-A{w}_n)$. By Algorithm 1, we also get

$$z_n-x_n={\beta }_{n}f\left(x_n\right)+u_n-x_n-{\beta }_n{u}_n.$$

So, it follows from the boundedness of $\left\{x_n\right\}$ and $\left\{u_n\right\}$ that

$$\parallel u_n-x_n\parallel =\parallel z_n-x_n-{\beta }_{n}f\left(x_n\right)+{\beta }_n{u}_n\parallel \le \parallel z_n-x_n\parallel +{\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel u_n\parallel ),$$

which indicates $u_n-x_n$ tends to 0 as n tends to the infinity. Using Algorithm 1 again, we get

$$\begin{array}{ll}{x}_{n+1}-z_n& ={\sigma }_n(x_n-z_n)+{\gamma }_n(u_n-z_n)+{\delta }_n(T{z}_n-z_n)\\ & ={\sigma }_n(x_n-z_n)+{\gamma }_n(u_n-x_n+x_n-z_n)+{\delta }_n(T{z}_n-z_n)\\ & =(1-{\delta }_n)(x_n-z_n)+{\gamma }_n(u_n-x_n)+{\delta }_n(T{z}_n-z_n),\end{array}$$

which immediately leads to

$$\begin{array}{ll}{\delta }_n\parallel z_n-T{z}_n\parallel & =\parallel x_{n+1}-{\gamma }_n(u_n-x_n)-z_n-(1-{\delta }_n)(x_n-z_n)\parallel \\ & =\parallel x_{n+1}-(1-{\delta }_n)(x_n-z_n)-x_n+x_n-z_n-{\gamma }_n(u_n-x_n)\parallel \\ & =\parallel x_{n+1}-x_n+{\delta }_n(x_n-z_n)-{\gamma }_n(u_n-x_n)\parallel \\ & \le \parallel x_{n+1}-x_n\parallel +\parallel x_n-z_n\parallel +\parallel u_n-x_n\parallel .\end{array}$$

Since $x_n-x_{n+1},$$z_n-x_n$, and $u_n-x_n$ tend to 0 as n tends to the infinity and ${lim\; inf}_{n\to \infty }{\delta }_n>0$, we obtain

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

which, together with Lemma 1, ensures that

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

(7)

From the restriction on the operator A, we have

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

(8)

Using the boundedness of $\left\{w_{n_k}\right\}$ and Lipschitzian property of A, we get the boundedness of $\left\{A{w}_{n_k}\right\}$. Note that ${\tau }_n\ge \tilde{\tau }:=min\{{\tau }_1,\frac{\mu }{L}\}$ and the boundedness of $\left\{y_{n_k}\right\}$. Inequality (8) deduces ${lim\; inf}_{k\to \infty }\langle w_{n_k}-x,A{w}_{n_k}\rangle \le 0\forall x\in C$. Borrowing the facts that ${lim}_{n\to \infty }\parallel w_n-y_n\parallel =0$ and A is Lipschitz continuous with moudle L, one concludes that ${lim}_{n\to \infty }\parallel A{y}_n-A{w}_n\parallel =0$, which combines with (8) and sends us to the situation ${lim\; inf}_{k\to \infty }\langle y_{n_k}-x,A{y}_{n_k}\rangle \le 0\forall x\in C$.

One now focuses on $z\in \mathrm{Fix}\left(T\right)$. Thanks to the weak convergence $w_{n_k}\rightharpoonup z$, as $k\to \infty$, one reaches $x_{n_k}\rightharpoonup z$. Without loss of generality, we may assume $l=n_k\mathrm{mod}N$ for all k. Since by the assumption $x_n-x_{n+1}\to 0$ we have $x_{n_k+j}\rightharpoonup z$ for all $j\ge 1$, we deduce from (7) that

$$\parallel x_{n_k+j}-T_{l+j}{x}_{n_k+j}\parallel =\parallel T_{n_k+j}{x}_{n_k+j}-x_{n_k+j}\parallel \to 0$$

as $k\to \infty$. An application of Lemma 1 is to yield $z\in \mathrm{Fix}\left(T_{l+j}\right)$ for all j. This amounts to

$$z\in \mathrm{Fix}\left(T\right).$$

(9)

Let $\left\{{\epsilon }_k\right\}$ be a decreasing real sequence in $(0,1)$ converging to 0 and let

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

(10)

for all $j\ge m_k,$ where $m_k$ is the smallest integer satisfying the above inequality. Note that sequence $\left\{m_k\right\}$ is increasing and $A{y}_{m_k}\ne 0$. It follows that $\langle A{y}_{m_k},h_{m_k}\rangle =1,$ where $h_{m_k}:=\frac{A{y}_{m_k}}{\parallel A{y}_{m_k}{\parallel }^2}$. This sends us to $\langle y_{m_k}-x-{\epsilon }_k{h}_{m_k},A{y}_{m_k}\rangle \le 0$, which guarantees

$$\langle y_{m_k}-x-{\epsilon }_k{h}_{m_k},A({\epsilon }_k{h}_{m_k}+x)\rangle \le 0.$$

This sends us to

$$\langle Ax,y_{m_k}-x\rangle \le \langle y_{m_k}-x-{\epsilon }_k{h}_{m_k},Ax-A({\epsilon }_k{h}_{m_k}+x)\rangle -{\epsilon }_k\langle Ax,h_{m_k}\rangle .$$

(11)

On the other hand, one has ${lim}_{n\to \infty }\parallel w_n-y_n\parallel =0$ and $w_{n_k}\rightharpoonup z$ as $k\to \infty$. This infers $w_{n_k}\rightharpoonup z$, which lies in $C,$ as k goes to the infinity. So, $A{y}_{n_k}\rightharpoonup Az$ as k goes to the infinity. This shows that z is not a solution. In the sense of norms, one obtains ${lim\; inf}_{k\to \infty }\parallel A{y}_{n_k}\parallel \ge \parallel Az\parallel$. This further concludes that

$$\begin{array}{ll}0& ={\displaystyle \underset{k\to \infty }{lim\; sup}}{\epsilon }_k/{\displaystyle \underset{k\to \infty }{lim\; inf}}\parallel A{y}_{n_k}\parallel \\ & \ge {\displaystyle \underset{k\to \infty }{lim\; sup}}{\epsilon }_k/\parallel A{y}_{m_k}\parallel \\ & \ge {\displaystyle \underset{k\to \infty }{lim\; sup}}\parallel h_{m_k}{\epsilon }_k\parallel \\ & \ge 0,\end{array}$$

which reaches that $h_{m_k}{\epsilon }_k\to 0$ as $k\to \infty$.

Finally, one focuses on the desired point z. (11), the boundedness of sequences $\left\{h_{m_k}\right\}$ and $\left\{y_{m_k}\right\}$, and the fact that $h_{m_k}{\epsilon }_k\to 0$ as $k\to \infty$, yield that $\langle z-x,Ax\rangle ={lim\; inf}_{k\to \infty }\langle y_{m_k}-x,Ax\rangle \le 0$ for all x in C. Lemma 3 asserts that the desired point z is a solution to the VIP, e.g., $z\in \mathrm{VI}(C,A)$. Therefore, we have from (9) that $z\in \mathrm{\Omega }:=\mathrm{VI}(C,A)\cap \mathrm{Fix}\left(T\right)$. The proof is complete. □

Theorem 1.

Let$\left\{x_n\right\}$be a vector sequence constructed by Algorithm 1 and let$A\left(H\right)$be bounded. Suppose that$x^{*}$is in Ω, which uniquely solves $\langle x^{*}-f{x}^{*},x^{*}-x\rangle \le 0$, $\forall x\in \mathrm{\Omega }$. Then,

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

Proof. 

Noticing condition (iv) on $\left\{{\sigma }_n\right\}$, one may assume that $\left\{{\sigma }_n\right\}\subset [a,b]$, which is a subset of $(0,1)$. Using the Banach Fixed Point Theory, one deduces that a unique point $x^{*}$ in H s.t. $x^{*}=P_{\mathrm{\Omega }}f\left(x^{*}\right)$. Hence, there is a solution $x^{*}\in \mathrm{\Omega }=\mathrm{Fix}\left(T\right)\cap \mathrm{VI}(C,A)$ to the HVI problem

$$\langle x^{*}-x,x^{*}-f{x}^{*}\rangle \le 0$$

(12)

for any point x in $\mathrm{\Omega }$. If ${lim}_{n\to \infty }\parallel x_n-x^{*}\parallel =0$, then

$$\underset{n\ge 1}{sup}(\parallel x^{*}-x_n\parallel +\parallel f\left(x^{*}\right)-f\left(x_n\right)\parallel +\parallel f\left(x^{*}\right)-x^{*}\parallel )\ge \underset{n\ge 1}{sup}\parallel x_n-f\left(x_n\right)\parallel$$

and

$$\parallel x_{n+1}-x^{*}\parallel +\parallel x^{*}-x_n\parallel \ge \parallel x_{n+1}-x_n\parallel \ge 0.$$

So, $x_{n+1}-x_n\to 0$ as $n\to \infty .$ In order to prove the sufficiency of the theorem, one supposes $\parallel x_n-x_{n+1}\parallel \to 0$ and ${sup}_{n\ge 1}\parallel (I-f)x_n\parallel <\infty$. Then, we divide the proof of the sufficiency into several steps. □

Step 1. One proves the boundedness of $\left\{x_n\right\}$. In fact, taking an arbitrary $p\in \mathrm{\Omega }$, one has $p=Tp$ and (6), that is,

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

(13)

Since ${lim}_{n\to \infty }(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})=1-{\mu }^2>0$, there exists an integer $n_0\ge 1$ with

$$1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2}>0\forall n\ge n_0.$$

Using (13), we have

$$\parallel p-u_n\parallel \le \parallel p-w_n\parallel ,\forall n\ge n_0.$$

(14)

So,

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

(15)

From Remark 1, we have $\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel \to 0(n\to \infty )$. This ensures that $\exists M_1>0$ s.t.

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

(16)

Combining (14), (15), and (16), we have

$$\parallel u_n-p\parallel \le \parallel w_n-p\parallel \le \parallel x_n-p\parallel +{\beta }_n{M}_1\forall n\ge n_0.$$

(17)

Note that $A\left(H\right)$ is bounded, $y_n=P_C(w_n-{\tau }_{n}A{w}_n)$, $f\left(H\right)\subset C$, and

$$u_n=y_n-{\tau }_n(A{y}_n-A{w}_n).$$

Hence, we know that $\left\{A{y}_n\right\}$ and $\left\{A{w}_n\right\}$ are both bounded. From ${sup}_{n\ge 1}\parallel (I-f)x_n\parallel <\infty$ and ${\alpha }_n\parallel x_n-x_{n-1}\parallel \to 0$, we conclude that

$$\begin{array}{ll}\parallel u_n-f\left(x_n\right)\parallel & =\parallel y_n-{\tau }_n(A{y}_n-A{w}_n)-P_{C}f\left(x_n\right)\parallel \\ & \le \parallel P_C(w_n-{\tau }_{n}A{w}_n)-P_{C}f\left(x_n\right)\parallel +{\tau }_n\parallel A{y}_n-A{w}_n\parallel \\ & \le {\alpha }_n\parallel x_n-x_{n-1}\parallel +\parallel x_n-f\left(x_n\right)\parallel +{\tau }_1(\parallel A{w}_n\parallel +\parallel A{y}_n-A{w}_n\parallel )\le M_0,\end{array}$$

where

$$\underset{n\ge 1}{sup}\{{\alpha }_n\parallel x_n-x_{n-1}\parallel +\parallel x_n-f\left(x_n\right)\parallel +{\tau }_1(\parallel A{w}_n\parallel +\parallel A{y}_n-A{w}_n\parallel )\}\le M_0$$

for some $M_0>0$. By using (17), one concludes

$$\begin{array}{ll}\parallel p-z_n\parallel & \le (1-{\beta }_n)\parallel p-u_n\parallel +{\beta }_n\parallel fp-p\parallel +{\beta }_n\delta \parallel p-x_n\parallel \\ & \le (1-{\beta }_n)(\parallel p-x_n\parallel +{\beta }_n{M}_1)+{\beta }_n\parallel fp-p\parallel +{\beta }_n\delta \parallel p-x_n\parallel \\ & \le (1-{\beta }_n(1-\delta ))\parallel p-x_n\parallel +{\beta }_n(\parallel fp-p\parallel +M_1),\end{array}$$

which, together with Lemma 2 and $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$, yields

$$\begin{array}{ll}\parallel p-x_{n+1}\parallel & \le (1-{\sigma }_n)\parallel \frac{1}{1-{\sigma }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T{z}_n-p)]\parallel +{\gamma }_n\parallel u_n-z_n\parallel +{\sigma }_n\parallel p-x_n\parallel \\ & \le (1-{\sigma }_n)\parallel p-z_n\parallel +{\gamma }_n{\beta }_n\parallel u_n-f\left(x_n\right)\parallel +{\sigma }_n\parallel p-x_n\parallel \\ & \le (1-{\sigma }_n)[(1-{\beta }_n(1-\delta ))\parallel p-x_n\parallel +{\beta }_n(M_0+M_1+\parallel (f-I)p\parallel \left)\right]+{\sigma }_n\parallel p-x_n\parallel \\ & =[1-{\beta }_n(1-{\sigma }_n)(1-\delta )]\parallel p-x_n\parallel +{\beta }_n(1-{\sigma }_n)(1-\delta )\frac{M_1+M_0+\parallel (I-f)p\parallel }{1-\delta }\\ & \le max\{\frac{M_0+M_1+\parallel (f-I)p\parallel }{1-\delta },\parallel p-x_n\parallel \}.\end{array}$$

By induction, we obtain

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

This indicates that all the vector sequence $\left\{x_n\right\},\left\{u_n\right\},\left\{w_n\right\},\left\{y_n\right\},$ and $\left\{z_n\right\}$ are bounded sequences.

Step 2. We claim

$$(1-{\beta }_n)(1-{\sigma }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})\parallel y_n-w_n{\parallel }^2\le \parallel p-x_n{\parallel }^2+{\alpha }_n{M}_4-{\parallel p-x_{n+1}\parallel }^2,\forall n\ge n_0,$$

for some $M_4>0$. Indeed, using Lemma 2, Lemma 6, and the convexity of ${\parallel ·\parallel }^2$, we have from $({\gamma }_n+{\delta }_n)\zeta \le {\gamma }_n$ that $\forall n\ge n_0$,

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

(18)

where ${sup}_{n\ge 1}2(\parallel (f-I)p\parallel \parallel z_n-p\parallel +\parallel u_n-f\left(x_n\right)\parallel \parallel x_{n+1}-p\parallel )\le M_2$ for some $M_2>0$. In addition, from (17) we get

$$\begin{array}{ll}\parallel p-w_n{\parallel }^2& ={\beta }_n(2M_1\parallel p-x_n\parallel +{\beta }_n{M}_1^2)+\parallel p-x_n{\parallel }^2\\ & \le {\beta }_n{M}_3+{\parallel p-x_n\parallel }^2,\end{array}$$

(19)

where ${sup}_{n\ge 1}({\beta }_n{M}_1^2+2M_1\parallel p-x_n\parallel )\le M_3$ for some $M_3>0$. Substituting (19) for (18), we obtain that for all $n\ge n_0$,

$$\begin{array}{ll}\parallel p-x_{n+1}{\parallel }^2& \le {\sigma }_n\parallel p-x_n{\parallel }^2+(1-{\sigma }_n)\{\delta {\beta }_n\parallel p-x_n{\parallel }^2+(1-{\beta }_n)[\parallel p-x_n{\parallel }^2+{\beta }_n{M}_3]\\ & -(1-{\beta }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})\parallel y_n-w_n{\parallel }^2+{\beta }_n{M}_2\}\\ & =[1-(1-{\sigma }_n)(1-\delta ){\beta }_n]{\parallel p-x_n\parallel }^2+(1-{\sigma }_n){\beta }_n(1-{\beta }_n)M_3\\ & -(1-{\beta }_n)(1-{\sigma }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2}){\parallel y_n-w_n\parallel }^2+(1-{\sigma }_n){\beta }_n{M}_2\\ & \le \parallel p-x_n{\parallel }^2-(1-{\beta }_n)(1-{\sigma }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2}){\parallel y_n-w_n\parallel }^2+{\beta }_n{M}_4,\end{array}$$

(20)

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

$$(1-{\beta }_n)(1-{\sigma }_n)(1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2})\parallel y_n-x_n{\parallel }^2\le \parallel p-x_n{\parallel }^2+{\beta }_n{M}_4-{\parallel p-x_{n+1}\parallel }^2.$$

(21)

Step 3. One proves

$$\begin{array}{c}\parallel p-x_{n+1}{\parallel }^2\hfill \\ \le [1-\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\beta }_n{\gamma }_n}{\beta }_n]\parallel x_n{-p\parallel }^2+\frac{[(1-2\delta ){\delta }_n-{\gamma }_n]{\beta }_n}{1-{\beta }_n{\gamma }_n}·\{\frac{2{\gamma }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_{n+1}\parallel \hfill \\ +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\parallel f\left(x_n\right)-p\parallel \parallel z_n-x_n\parallel +\frac{2{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}\langle f\left(p\right)-p,x_n-p\rangle \hfill \\ +\frac{{\gamma }_n+{\delta }_n}{(1-2\delta ){\delta }_n-{\gamma }_n}·\frac{{\alpha }_n}{{\beta }_n}\parallel x_n-x_{n-1}\parallel 3M\},\forall n\ge n_0,\hfill \end{array}$$

where M is some appropriate constant.

$$\begin{array}{ll}3M{\alpha }_n\parallel x_n-x_{n-1}\parallel +\parallel p-x_n{\parallel }^2& \ge \parallel p-x_n{\parallel }^2+{\alpha }_n({\alpha }_n\parallel x_n-x_{n-1}\parallel +2\parallel p-x_n\parallel )\parallel x_n-x_{n-1}\parallel \\ & =(\parallel p-x_n\parallel +{\alpha }_n\parallel x_n-x_{n-1}{\parallel )}^2\\ & \ge \parallel p-w_n{\parallel }^2,\end{array}$$

(22)

where $M\ge {sup}_{n\ge 1}\{{\alpha }_n\parallel x_n-x_{n-1}\parallel ,\parallel p-x_n\parallel \}$. From the convexity of ${\parallel ·\parallel }^2$, one arrives at

$$\begin{array}{ll}\parallel x_{n+1}{-p\parallel }^2& =\parallel {\sigma }_n(x_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T{z}_n-p)+{\gamma }_n(u_n-z_n){\parallel }^2\\ & \le \parallel {\sigma }_n(x_n-p)+{\gamma }_n(z_n-p)+{\delta }_n(T{z}_n-p){\parallel }^2+2{\gamma }_n{\beta }_n\langle u_n-f\left(x_n\right),x_{n+1}-p\rangle \\ & \le {\sigma }_n\parallel x_n{-p\parallel }^2+(1-{\sigma }_n){\parallel \frac{1}{1-{\sigma }_n}[{\gamma }_n(z_n-p)+{\delta }_n(T{z}_n-p)]\parallel }^2\\ & +2{\gamma }_n{\beta }_n\langle u_n-p,x_{n+1}-p\rangle +2{\gamma }_n{\beta }_n\langle p-f\left(x_n\right),x_{n+1}-p\rangle ,\end{array}$$

which yields that

$$\begin{array}{ll}\parallel x_{n+1}{-p\parallel }^2& \le (1-{\sigma }_n)\parallel p-z_n{\parallel }^2+{\sigma }_n\parallel p-x_n{\parallel }^2+2{\gamma }_n{\beta }_n\parallel p-u_n\parallel \parallel p-x_{n+1}\parallel \\ & +2{\gamma }_n{\beta }_n\langle p-f\left(x_n\right),x_{n+1}-p\rangle \\ & \le {\sigma }_n\parallel p-x_n{\parallel }^2+(1-{\sigma }_n)[(1-{\beta }_n)\parallel p-u_n{\parallel }^2+2{\beta }_n\langle f\left(x_n\right)-p,z_n-p\rangle ]\\ & +{\gamma }_n{\beta }_n(\parallel p-u_n{\parallel }^2+\parallel p-x_{n+1}{\parallel }^2)+2{\gamma }_n{\beta }_n\langle p-f\left(x_n\right),x_{n+1}-p\rangle .\end{array}$$

From (17) and (22) we know that $\parallel u_n{-p\parallel }^2\le \parallel x_n{-p\parallel }^2+{\alpha }_n\parallel x_n-x_{n-1}\parallel 3M\forall n\ge n_0$. Hence, we have, $\forall n\ge n_0$, that

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

which immediately yields

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

(23)

Step 4. We claim strong convergence of vector sequence $\left\{x_n\right\}$ to the unique solution of HVI (12), $x^{*}\in \mathrm{\Omega }$. One lets $p=x^{*}$, and use (23) to obtain

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

(24)

From (21), $x_n-x_{n+1}\to 0,{\beta }_n\to 0,1-{\mu }^2\frac{{\tau }_n^2}{{\tau }_{n+1}^2}\to 1-{\mu }^2$, and $\left\{{\sigma }_n\right\}\subset [a,b]\subset (0,1)$, we obtain

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

This immediately implies that $w_n-y_n\to 0$. From the Lipschitzian property of A, we have $\parallel u_n-y_n\parallel ={\tau }_n\parallel A{y}_n-A{w}_n\parallel \le {\tau }_{1}L\parallel y_n-w_n\parallel$. Consequently,

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

(25)

Thus, we get

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

Since $z_n={\beta }_{n}f\left(x_n\right)+(1-{\beta }_n)u_n$ with $u_n:=y_n-{\tau }_n(A{y}_n-A{w}_n)$, from (25) and the boundedness of $\left\{x_n\right\},\left\{u_n\right\}$, we get

$$\begin{array}{ll}\parallel z_n-y_n\parallel & =\parallel {\beta }_{n}f\left(x_n\right)-{\beta }_n{u}_n+u_n-y_n\parallel \\ & \le {\beta }_n(\parallel f\left(x_n\right)\parallel +\parallel u_n\parallel )+\parallel u_n-y_n\parallel \to 0(n\to \infty ),\end{array}$$

(26)

and hence,

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

(27)

Obviously, combining (25) and (26) guarantees that $\parallel w_n-z_n\parallel \le \parallel y_n-w_n\parallel +\parallel z_n-y_n\parallel ,$ which indicates that ${lim}_{n\to \infty }\parallel w_n-z_n\parallel =0.$ Let vector sequence $\left\{x_{n_k}\right\}$ be a subsequence of original sequence $\left\{x_n\right\}$. From its boundedness, one asserts that

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

(28)

Without loss of generality, one lets $x_{n_k}\rightharpoonup \tilde{x}$. (28) implies

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

(29)

On the other hand, one has ${lim}_{k\to \infty }\parallel w_{n_k}-x_{n_k}=0$. This indicates $w_{n_k}\rightharpoonup \tilde{x}$. Lemma 7 guarantees that $\tilde{x}$ is in $\mathrm{\Omega }$. Therefore, (12) and (29) amount to ${lim\; sup}_{n\to \infty }\langle x_n-x^{*},(f-I)x^{*}\rangle =\langle \tilde{x}-x^{*},(f-I)x^{*}\rangle \le 0.$ Note that ${lim\; inf}_{n\to \infty }\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\beta }_n{\gamma }_n}>0$. It follows that ${\sum }_{n=1}^{\infty }\frac{(1-2\delta ){\delta }_n-{\gamma }_n}{1-{\beta }_n{\gamma }_n}{\beta }_n=\infty$. It is clear that

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

By utilizing Lemma 4, one concludes $x_n\to x^{*}$ easily. The proof is complete.

4. Applications

In this section, our main results are applied to solve the VIP and CFPP in an illustrating example. The initial point $x_0=x_1$ is randomly chosen in $\mathbf{R}$. Take $f\left(x\right)=\frac{sinx}{4},\alpha ={\tau }_1=\mu =\frac{1}{2},{ϵ}_n=\frac{1}{n^2},{\beta }_n=\frac{1}{n+1},{\sigma }_n=\frac{1}{3},{\gamma }_n=\frac{1}{6}$, and ${\delta }_n=\frac{1}{2}$.

We first provide an example of Lipschitzian, pseudomonotone operator A satisfying the boundedness of $A\left(H\right)$ and strictly pseudocontractive operator $T_1$ with $\mathrm{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)\ne \varnothing$. Let $C=[-1,1]$ and $H=\mathbf{R}$ with the inner product $\langle a,b\rangle =ab$ and induced norm $\parallel ·\parallel =|·|$. Then, f is a $\delta$-contractive map with $\delta =\frac{1}{4}\in [0,\frac{1}{2})$ and $f\left(H\right)\subset C$ because $\parallel f\left(x\right)-f\left(y\right)\parallel =\frac{1}{4}\parallel sinx-siny\parallel \le \frac{1}{4}\parallel x-y\parallel$ for all $x,y\in H$.

Let $A:H\to H$ and $T_1:H\to H$ be defined as $Ax:=\frac{1}{1+|sinx|}-\frac{1}{1+\left|x\right|}$ and $T_{1}x:=\frac{5}{8}x-\frac{1}{4}sinx$ for all $x\in H$. Now, we first show that A is Lipschitzian, pseudomonotone operator with $L=2$, such that $A\left(H\right)$ is bounded. Indeed, for all $x,y\in H$, we have

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

This implies that A is Lipschitzian operator with $L=2$. Next, we verify that A is pseudomonotone. For any given $x,y\in H$, it is clear that the relation holds:

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

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

$$\parallel Tx-Ty\parallel \le \frac{5}{8}\parallel x-y\parallel +\frac{1}{4}\parallel sinx-siny\parallel \le \parallel x-y\parallel +\frac{1}{4}\parallel (I-T)x-(I-T)y\parallel .$$

It is clear that $({\gamma }_n+{\delta }_n){\zeta }_1=(\frac{1}{6}+\frac{1}{2})\times \frac{1}{4}\le \frac{1}{6}={\gamma }_n<(1-2\delta ){\delta }_n=(1-2\times \frac{1}{4})\frac{1}{2}=\frac{1}{4}$ 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 $d\left(T_{1}u\right)/du=\frac{5}{8}-\frac{1}{4}cosu>0$. Therefore, $\mathrm{\Omega }=\mathrm{Fix}\left(T_1\right)\cap \mathrm{VI}(C,A)=\left\{0\right\}\ne \varnothing$. In this case, Algorithm 1 can be rewritten below:

$$\left\{\begin{array}{c}{w}_n=x_n+{\alpha }_n(x_n-x_{n-1}),\hfill \\ y_n=P_C(w_n-{\tau }_{n}A{w}_n),\hfill \\ z_n=\frac{1}{n+1}f\left(x_n\right)+\frac{n}{n+1}(y_n-{\tau }_n(A{y}_n-A{w}_n)),\hfill \\ x_{n+1}=\frac{1}{3}{x}_n+\frac{1}{6}(y_n-{\tau }_n(A{y}_n-A{w}_n))+\frac{1}{2}{T}_1{z}_n\forall n\ge 1,\hfill \end{array}\right.$$

where, for each $n\ge 1$, ${\overline{\alpha }}_n(={\alpha }_n)$ and ${\tau }_n$ are chosen as in Algorithm 1. Then, by Theorem 1, we know that $x_n\to 0\in \mathrm{\Omega }$ iff $x_n-x_{n+1}\to 0(n\to \infty )$ and ${sup}_{n\ge 1}\left|(I-f)x_n\right|<\infty$.