Iterative Algorithms for Pseudomonotone Variational Inequalities and Fixed Point Problems of Pseudocontractive Operators

Abstract

In this paper, we are interested in the pseudomonotone variational inequalities and fixed point problem of pseudocontractive operators in Hilbert spaces. An iterative algorithm has been constructed for finding a common solution of the pseudomonotone variational inequalities and fixed point of pseudocontractive operators. Strong convergence analysis of the proposed procedure is given. Several related corollaries are included.

1. Introduction

Let H be a real Hilbert space endowed with inner product and induced norm denoted by $\langle ·,·\rangle$ and $\parallel ·\parallel$, respectively. Let $\varnothing \ne C\subset H$ be a closed and convex set.

In this article, our study is related to a classical variational inequality (VI) of seeking an element $\tilde{u}\in C$ verifying

$$\langle f\left(\tilde{u}\right),u-\tilde{u}\rangle \ge 0,\forall u\in C,$$

(1)

where $f:H\to H$ is a given operator, under the following assumptions:

(i)

$VI(C,f)$, the solution set of (1), is nonempty;

(ii)

f is pseudomonotone on H, i.e.,

$$\begin{array}{c}\hfill \langle f\left(\tilde{u}\right),u-\tilde{u}\rangle \ge 0\Rightarrow \langle f\left(u\right),u-\tilde{u}\rangle \ge 0,\forall u,\tilde{u}\in H;\end{array}$$

(2)

(iii)

f is $\kappa$-Lipschitz continuous on H (for some $\kappa >0$), i.e.,

$$\begin{array}{c}\hfill \parallel f\left(x\right)-f\left(y\right)\parallel \le \kappa \parallel x-y\parallel ,\forall x,y\in H.\end{array}$$

Numerical iterative methods have been presented, developed and adopted widely as algorithmic solutions to the concept of variational inequalities. This notion, that mainly involves some important operators, plays a key role in applied mathematics, such as obstacle problems, optimization problems, complementarity problems as a unified framework for the study of a large number of significant real-word problems arising in physics, engineering, economics and so on. For more information, the reader can refer to [1,2,3,4,5,6,7,8,9,10,11,12].

For solving VI (1) in which the involved operator f may be monotone, several iterative algorithms have been introduced and studied, see, e.g., [13,14,15,16,17,18]. Among them, the more popular iterative technique is the projected gradient rule ([19,20,21,22,23]): for the fixed previous iteration $x_{n-1}$, calculate the current iteration $x_n$ via the following manner

$$\begin{array}{c}\hfill x_n=P_C[x_{n-1}-\tau f\left(x_{n-1}\right)],n\ge 1,\end{array}$$

(3)

where $P_C$ means the projection operator from H onto C and the positive constant $\tau$ is the step-size.

The projected gradient rule (3) is an effective technique for solving VI (1). However, the involved operator f should be strongly monotone or inverse strongly monotone. In order to overcome this flaw, in [21], Korpelevich put forward an extragradient technique: for the fixed previous iteration $x_{n-1}$, calculate the current iteration $x_n$ via the following manner

$$\left\{\begin{array}{c}{y}_{n-1}=P_C[x_{n-1}-\tau f\left(x_{n-1}\right)],\hfill \\ x_n=P_C[x_{n-1}-\tau f\left(y_{n-1}\right)],n\ge 1,\hfill \end{array}\right.$$

(4)

where the step-size $\tau \in (0,1/\kappa )$.

Korpelevich’s algorithm (4) provides an important idea for solving monotone variational inequality. Please refer to the references [24,25,26,27] for several important extended version of Korpelevich’s algorithm.

The another motivation of this paper is to study the following fixed point equation:

$$\begin{array}{c}\hfill \mathrm{find}x\in C\mathrm{such} \mathrm{that}x=Tx,\end{array}$$

(5)

where $T:C\to C$ is a pseudocontractive operator.

Now, it is well-known that fixed point algorithm of successive approximation is one of the most important techniques in numerical mathematics ([28,29,30,31,32,33,34,35,36,37,38,39,40]). Focusing on the research with pseudocontractive operators originated in their relations with the important class of monotone operators. Algorithmic approximation theories and experiments of pseudocontractive operators have been studied extensively in the literature, see, for example, [41,42,43,44,45,46,47].

Motivated and inspired by the work in this field, the purpose of this paper is to investigate the problem of pseudomonotone variational inequality (1) and fixed point of pseudocontractive operators. We construct an iterative algorithm for seeking a common solution of the pseudomonotone variational inequalities and fixed point of pseudocontractive operators. Strong convergence analysis of the proposed procedure is given. Several related corollaries are included.

2. Preliminaries

Let H be a real Hilbert space. Let $C\subset H$ be a nonempty, closed and convex set. Recall that an operator $f:C\to C$ is said to be monotone if

$$\langle f\left(x\right)-f\left(y\right),x-y\rangle \ge 0,\forall x,y\in C.$$

An operator $T:C\to C$ is said to be pseudocontractive if

$$\parallel Tu-T{u}^{‡}{\parallel }^2 \le \parallel u - u^{‡}{\parallel }^2+ {\parallel (I-T)u-(I-T)u^{‡}\parallel }^2$$

for all $u,u^{‡}\in C$.

Recall that an operator $f:C\to C$ is called weakly sequentially continuous, if for any given sequence $\left\{x_n\right\}\subset C$ satisfying $x_n\rightharpoonup \tilde{x}$, we conclude that $f\left(x_n\right)\rightharpoonup f\left(\tilde{x}\right)$.

Recall that the metric projection $P_C:H\to C$ is an orthographic projection from H onto C, which possesses the following characteristic: for given $x\in H$,

$$\begin{array}{c}\hfill \langle x-P_C\left[x\right],y-P_C\left[x\right]\rangle \le 0,\forall y\in C.\end{array}$$

(6)

The following symbols will be used in the sequel.

• $u_n\rightharpoonup z^{†}$ denotes the weak convergence of $u_n$ to $z^{†}$.
• $u_n\to z^{†}$ stands for the strong convergence of $u_n$ to $z^{†}$.
• $Fix\left(T\right)$ means the set of fixed points of T.
• ${\omega }_w\left(u_n\right)=\{u^{†}:\exists \left\{u_{n_i}\right\}\subset \left\{u_n\right\}\mathrm{such}\mathrm{that}{u}_{n_i}\rightharpoonup u^{†}(i\to \infty )\}$.

Lemma 1

([1]). Let H be a real Hilbert space. Then, we have

$$\parallel \delta u+(1-\delta )u^{†}{\parallel }^2={\delta \parallel u\parallel }^2+(1-\delta )\parallel u^{†}{\parallel }^2-\delta (1-\delta ){\parallel u-u^{†}\parallel }^2,$$

$\forall u,u^{†}\in H$ and $\forall t\in [0,1]$.

Lemma 2

([45]). Let C a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to C$ be an L-Lipschitz pseudocontractive operator. Let $0<\eta <\frac{1}{\sqrt{1+L^2}+1}$. Then,

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel u^{†}-T((1-\eta )\tilde{u}+\eta T\tilde{u}){\parallel }^2 \le \parallel \tilde{u}-u^{†}{\parallel }^2+(1-\eta ){\parallel \tilde{u}-T((1-\eta )\tilde{u}+\eta T\tilde{u})\parallel }^2,\end{array}\end{array}$$

for all $\tilde{u}\in C$ and $u^{†}\in Fix\left(T\right)$.

Lemma 3

([18]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let $f:H\to H$ be a continuous and pseudomonotone operator. Then $x^{†}\in VI(C,f)$ iff $x^{†}$ solves the following dual variational inequality

$$\begin{array}{c}\hfill \langle f\left(u^{†}\right),u^{†}-x^{†}\rangle \ge 0,\forall u^{†}\in H.\end{array}$$

Lemma 4

([47]). Let H be a real Hilbert space, C a nonempty closed convex subset of H. Let $T:C\to C$ be a continuous pseudocontractive operator. Then

(i) 

$Fix\left(T\right)$ is a closed convex subset of C;

(ii) 

T is demi-closed, i.e., $u_n\rightharpoonup \tilde{u}$ and $T\left(u_n\right)\to u^{†}$ imply that $T\left(\tilde{u}\right)=u^{†}$.

Lemma 5

([15]). Let $\left\{{\mu }_n\right\}\subset (0,\infty )$, $\left\{{\gamma }_n\right\}\subset (0,1)$ and $\left\{{\delta }_n\right\}$ be three real number sequences. If ${\mu }_{n+1}\le (1-{\gamma }_n){\mu }_n+{\delta }_n$ for all $n\ge 0$ with ${\sum }_{n=1}^{\infty }{\gamma }_n=\infty$ and ${lim\; sup}_{n\to \infty }{\delta }_n/{\gamma }_n\le 0$ or ${\sum }_{n=1}^{\infty }\left|{\delta }_n\right|<\infty$, then ${lim}_{n\to \infty }{\mu }_n=0$.

3. Main Results

Let $\varnothing \ne C$ be a convex and closed subset of a real Hilbert space H. Let the operator f be pseudomonotone on H, weakly sequentially continuous and Lipschitz continuous on C with Lipschitz constant $\kappa >0$. Let $T:C\to C$ be an L-Lipschitz pseudocontractive operator with $L\ge 1$.

Next, we first present the following iterative algorithm for solving pseudomonotone variational inequality and fixed point problem of pseudocontractive operator T. In what follows, assume that $\Lambda :=VI(C,f)\cap Fix\left(T\right)\ne \varnothing$.

Remark 1.

By virtue of (6), we know that $u^{†}\in VI(C,f)\iff u^{†}=P_C[u^{†}-\tau f\left(u^{†}\right)]$ for all $\tau >0$. Thus, if at some iterative step $x_n=P_C[x_n-f\left(x_n\right)]$, then $x_n$ is a solution of variational inequality (1) and hence ${\omega }_w\left(x_n\right)\subset VI(C,f)$.

Remark 2.

For given $x_n$, we can find $m\left(x_n\right)$ such that (14) holds. In fact, we can choose $m\left(x_n\right)$ such that ${\gamma }^{m\left(x_n\right)}\le \frac{\theta }{\mu \kappa }$ due to the Lipschitz continuity of f. So, (14) is well-defined. At the same time, there exists a positive $\varsigma >0$ such that ${\displaystyle {\gamma }^{m\left(x_n\right)}\ge \varsigma >0}$ for all $x_n$. As a matter of fact, if $m\left(x_n\right)=0$, then ${\gamma }^{m\left(x_n\right)}=\varsigma =1$. If $m\left(x_n\right)>0$, then we have ${\displaystyle \frac{\kappa \mu {\gamma }^{m\left(x_n\right)}}{\gamma }>\theta }$, which implies that ${\displaystyle 0<\frac{\gamma \theta }{\mu \kappa }<{\gamma }^{m\left(x_n\right)}<1}$ for all n.

Proposition 1.

If $x_n\ne P_C[x_n-f\left(x_n\right)]$, then $x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\ne 0$.

Proof. 

Let $x^{\ast }=P_{\Lambda }\left(u\right)$. Owing to $x_n\in C$ and $y_n\in C$, we have

$$\begin{array}{c}\hfill \langle f\left(x^{\ast }\right),x_n-x^{\ast }\rangle \ge 0,\end{array}$$

(7)

and

$$\begin{array}{c}\hfill \langle f\left(x^{\ast }\right),y_n-x^{\ast }\rangle \ge 0.\end{array}$$

(8)

Applying the pseudomonotonicity (2) of f to (7) and (8), we obtain

$$\begin{array}{c}\hfill \langle f\left(x_n\right),x_n-x^{\ast }\rangle \ge 0,\end{array}$$

(9)

and

$$\begin{array}{c}\hfill \langle f\left(y_n\right),y_n-x^{\ast }\rangle \ge 0.\end{array}$$

(10)

Since $y_n=P_C[x_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right)]$, using the characteristic (6) of projection $P_C$, we have

$$\begin{array}{c}\hfill \langle x_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right)-y_n,y_n-x^{\ast }\rangle \ge 0.\end{array}$$

(11)

Hence,

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \langle x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right),x_n-x^{\ast }\rangle & =\langle x_n-y_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right),x_n-x^{\ast }\rangle +\mu {\gamma }^{m\left(x_n\right)}\langle f\left(x_n\right),x_n-x^{\ast }\rangle \hfill \\ & +\mu {\gamma }^{m\left(x_n\right)}\langle f\left(y_n\right),x_n-y_n\rangle +\mu {\gamma }^{m\left(x_n\right)}\langle f\left(y_n\right),y_n-x^{\ast }\rangle \hfill \\ \hfill \left(\mathit{by}\right(9\left)\mathit{and}\right(10\left)\right)& \ge \langle x_n-y_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right),x_n-x^{\ast }\rangle +\mu {\gamma }^{m\left(x_n\right)}\langle f\left(y_n\right),x_n-y_n\rangle \hfill \\ & =\langle x_n-y_n-\mu {\gamma }^{m\left(x_n\right)}(f\left(x_n\right)-f\left(y_n\right)),x_n-y_n\rangle \hfill \\ & +\langle x_n-y_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right),y_n-x^{\ast }\rangle \hfill \\ \hfill \left(\mathit{by}\right(11\left)\right)& \ge \langle x_n-y_n-\mu {\gamma }^{m\left(x_n\right)}(f\left(x_n\right)-f\left(y_n\right)),x_n-y_n\rangle \hfill \\ & =\parallel x_n-y_n{\parallel }^2-\mu {\gamma }^{m\left(x_n\right)}\langle f\left(x_n\right)-f\left(y_n\right),x_n-y_n\rangle \hfill \\ & \ge \parallel x_n-y_n{\parallel }^2-\mu {\gamma }^{m\left(x_n\right)}\parallel f\left(x_n\right)-f\left(y_n\right)\parallel \parallel x_n-y_n\parallel \hfill \\ & \ge (1-\theta )\parallel x_n-y_n{\parallel }^2\left(\mathit{by}\left(14\right)\right).\hfill \end{array}\end{array}$$

(12)

Since $P_C[x_n-f\left(x_n\right)]\ne x_n$, it follows that $P_C[x_n-\gamma f\left(x_n\right)]\ne x_n$ for all $\gamma >0$. Thus, $y_n\ne x_n$. According to (12), we deduce $x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\ne 0$. □

Remark 3.

In case 1, we have $f\left(x_n\right)\ne 0$ (by Remark 1) and $x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\ne 0$ (by Remark 2) for all $n\ge 0$. According to Proposition 1, the sequence $\left\{u_n\right\}$ is well-defined and hence the sequence $\left\{x_n\right\}$ is well-defined.

Now, in this position, we give the convergence analysis of the iterative sequence $\left\{x_n\right\}$ generated by Algorithm 1.

Algorithm 1: Iterative procedures for VI and FP.

Let $u\in C$ be a fixed point. Let $\left\{{\alpha }_n\right\}$, $\left\{{\sigma }_n\right\}$ and $\left\{{\delta }_n\right\}$ be three real number sequences in $(0,1)$.

Let $\gamma \in (0,1)$, $\mu \in (0,1)$, $\theta \in (0,1)$ and $\tau \in (0,2)$ be four constants.

Step 1. Let $x_0\in C$ be an initial value. Set $n=0$.

Step 2. Assume that the sequence $\left\{x_n\right\}$ has been constructed and then calculate $P_C[x_n-f\left(x_n\right)]$.

Step 3. Case 1. If $P_C[x_n-f\left(x_n\right)]\ne x_n$, then calculate the sequence $\left\{y_n\right\}$ by the following manner $$\begin{array}{c}\hfill y_n=P_C[x_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right)],\end{array}$$ (13)  where $m\left(x_n\right)=min\{0,1,2,3,\cdots \}$ and satisfies $$\begin{array}{c}\hfill \mu {\gamma }^{m\left(x_n\right)}\parallel f\left(x_n\right)-f\left(y_n\right)\parallel \le \theta \parallel x_n-y_n\parallel ,\end{array}$$ (14)  and consequently, calculate the sequences $\left\{u_n\right\}$, $\left\{z_n\right\}$ and $\left\{x_{n+1}\right\}$ by the following rule $$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle u_n=P_C\left[x_n-\tau (1-\theta ){\parallel x_n-y_n\parallel }^2\frac{x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\right],}\hfill \\ z_n=(1-{\sigma }_n)u_n+{\sigma }_{n}T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n],\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)z_n.\hfill \end{array}\right.\end{array}$$ (15) Case 2. If $P_C[x_n-f\left(x_n\right)]=x_n$, then calculate the sequence $\left\{x_{n+1}\right\}$ via the following form $$\begin{array}{c}\hfill \left\{\begin{array}{c}{\displaystyle z_n=(1-{\sigma }_n)x_n+{\sigma }_{n}T[(1-{\delta }_n)x_n+{\delta }_{n}T{x}_n],}\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)z_n.\hfill \end{array}\right.\end{array}$$

Step 4. Set $n:=n+1$ and return to Step 2.

Theorem 1.

Suppose that the iterative parameters $\left\{{\alpha }_n\right\}$, $\left\{{\sigma }_n\right\}$ and $\left\{{\delta }_n\right\}$ satisfy the following assumptions:

(C1): 

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

(C2): 

$0<\underline{\sigma }<{\sigma }_n<\overline{\sigma }<{\delta }_n<\overline{\delta }<\frac{1}{\sqrt{1+L^2}+1},\forall n\ge 0$.

Then the sequence $\left\{x_n\right\}$ generated by Algorithm 1 converges strongly to $P_{\Lambda }\left(u\right)$.

Proof. 

Step 1. the sequence $\left\{x_n\right\}$ is bounded. First, we consider Case 1. In this case, from (15) and (12), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel u_n-x^{\ast }{\parallel }^2& \le \parallel x_n-x^{\ast }-\tau (1-\theta ){\parallel x_n-y_n\parallel }^2\frac{x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}{\parallel }^2\hfill \\ & = \parallel x_n-x^{\ast }{\parallel }^2+\frac{{\tau }^2{(1-\theta )}^2{\parallel x_n-y_n\parallel }^4}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\hfill \\ & -\frac{2\tau (1-\theta )\parallel x_n-y_n{\parallel }^2}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\langle x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right),x_n-x^{\ast }\rangle \hfill \\ & \le \parallel x_n-x^{\ast }{\parallel }^2-\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_n-y_n\parallel }^4}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\hfill \\ & \le \parallel x_n-x^{\ast }{\parallel }^2.\hfill \end{array}\end{array}$$

(16)

In the light of (15) and Lemmas 1 and 2, we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel z_n-x^{\ast }{\parallel }^2& = \parallel (1-{\sigma }_n)(u_n-x^{\ast })+{\sigma }_n(T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]-x^{\ast }){\parallel }^2\hfill \\ & =(1-{\sigma }_n)\parallel u_n-x^{\ast }{\parallel }^2-{\sigma }_n(1-{\sigma }_n){\parallel T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]-u_n\parallel }^2\hfill \\ & +{\sigma }_n{\parallel T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]-x^{\ast }\parallel }^2\hfill \\ & \le (1-{\sigma }_n)\parallel u_n-x^{\ast }{\parallel }^2-{\sigma }_n(1-{\sigma }_n){\parallel T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]-u_n\parallel }^2\hfill \\ & +{\sigma }_n(\parallel u_n-x^{\ast }{\parallel }^2+(1-{\delta }_n)\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]{\parallel }^2)\hfill \\ & = \parallel u_n-x^{\ast }{\parallel }^2-{\sigma }_n({\delta }_n-{\sigma }_n){\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]\parallel }^2\hfill \\ & \le \parallel u_n-x^{\ast }{\parallel }^2.\hfill \end{array}\end{array}$$

(17)

By (15), (16) and (17), we get

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }\parallel & = \parallel {\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)(z_n-x^{\ast })\parallel \hfill \\ & \le (1-{\alpha }_n)\parallel z_n-x^{\ast }\parallel +{\alpha }_n\parallel u-x^{\ast }\parallel \hfill \\ & \le (1-{\alpha }_n)\parallel x_n-x^{\ast }\parallel +{\alpha }_n\parallel u-x^{\ast }\parallel .\hfill \end{array}\end{array}$$

By induction, we can deduce that $\parallel x_{n+1}-x^{\ast }\parallel \le max\{\parallel u-x^{\ast }\parallel ,\parallel x_0-x^{\ast }\parallel \}$. Hence, the sequence $\left\{x_n\right\}$ is bounded. It is easy to check that the sequence $\left\{x_n\right\}$ is also bounded in Case 2.

Step 2. ${\omega }_w\left(x_n\right)\subset \Lambda$. We firstly discuss Case 1. On account of (15), we achieve

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2& = \parallel {\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)(z_n-x^{\ast }){\parallel }^2\hfill \\ & \le (1-{\alpha }_n){\parallel z_n-x^{\ast }\parallel }^2+2{\alpha }_n\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle .\hfill \end{array}\end{array}$$

(18)

By virtue of (16), (17) and (18), we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2& \le (1-{\alpha }_n)\parallel x_n-x^{\ast }{\parallel }^2-(1-{\alpha }_n){\sigma }_n({\delta }_n-{\sigma }_n){\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]\parallel }^2\hfill \\ & -(1-{\alpha }_n)\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_n-y_n\parallel }^4}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}+2{\alpha }_n\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle \hfill \\ & =(1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n[-(1-{\alpha }_n){\sigma }_n({\delta }_n-{\sigma }_n)\frac{\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]{\parallel }^2}{{\alpha }_n}\hfill \\ & -(1-{\alpha }_n)\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_n-y_n\parallel }^4}{{\alpha }_n{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\parallel }^2}+2\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle ].\hfill \end{array}\end{array}$$

(19)

Write $s_n={\parallel x_n-x^{\ast }\parallel }^2$ and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill t_n& =-(1-{\alpha }_n)\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_n-y_n\parallel }^4}{{\alpha }_n{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\parallel }^2}+2\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle \hfill \\ & -(1-{\alpha }_n){\sigma }_n({\delta }_n-{\sigma }_n)\frac{\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]{\parallel }^2}{{\alpha }_n},\hfill \end{array}\end{array}$$

(20)

for all $n\ge 0$.

We can adapt (19) as

$$\begin{array}{c}\hfill s_{n+1}\le (1-{\alpha }_n)s_n+{\alpha }_n{t}_n\end{array}$$

(21)

for all $n\ge 0$.

Now, we show that ${lim\; sup}_{n\to \infty }{t}_n$ is finite. First, thanks to (20), we deduce that $t_n\le 2\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle \le 2\parallel u-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel$. This together with the boundedness of $\left\{x_n\right\}$ implies that ${lim\; sup}_{n\to \infty }{t}_n$ has a upper bound.

Next, we show that ${lim\; sup}_{n\to \infty }{t}_n$ has a lower bound. As a matter of fact, we can prove that ${lim\; sup}_{n\to \infty }{t}_n\ge -1$. Assume the contrary that ${lim\; sup}_{n\to \infty }{t}_n<-1$. If so, there exists N such that $t_n<-1$ when $n\ge N$. Hence, for all $n\ge N$, from (21), we deduce

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill s_{n+1}& \le (1-{\alpha }_n)s_n+{\alpha }_n{t}_n\hfill \\ & \le (1-{\alpha }_n)s_n-{\alpha }_n\hfill \\ & =s_n-{\alpha }_n(1+s_n)\hfill \\ & \le s_n-{\alpha }_n.\hfill \end{array}\end{array}$$

It follows that $s_{n+1}\le s_N-{\sum }_{k=N}^n{\alpha }_k$, which implies that ${lim\; sup}_{n\to \infty }{s}_n\le s_N-{lim\; sup}_{n\to \infty }{\sum }_{k=N}^n{\alpha }_k=-\infty$. It is a contradiction. So, $-1\le {lim\; sup}_{n\to \infty }{t}_n\le +\infty$. Thus, we can select a subsequence $\left\{x_{n_i}\right\}\subset \left\{x_n\right\}$ (because of the boundedness of $\left\{x_n\right\}$) verifying $x_{n_i}\rightharpoonup x^{†}\in C$ and

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \underset{n\to \infty }{lim\; sup}{t}_n& =\underset{i\to \infty }{lim}{t}_{n_i}=\underset{i\to \infty }{lim}[-\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_{n_i}-y_{n_i}\parallel }^4}{{\alpha }_{n_i}{\parallel x_{n_i}-y_{n_i}+\mu {\gamma }^{m\left(x_{n_i}\right)}f\left(y_{n_i}\right)\parallel }^2}+2\langle u-x^{\ast },x_{n_i+1}-x^{\ast }\rangle \hfill \\ & -{\sigma }_{n_i}({\delta }_{n_i}-{\sigma }_{n_i})\frac{\parallel u_{n_i}-T[(1-{\delta }_{n_i})u_{n_i}+{\delta }_{n_i}T{u}_{n_i}]{\parallel }^2}{{\alpha }_{n_i}}].\hfill \end{array}\end{array}$$

(22)

Based on the boundedness of $\left\{x_{n_i+1}\right\}$, without loss of generality, assume that ${lim}_{i\to \infty }2\langle u-x^{\ast },x_{n_i+1}-x^{\ast }\rangle$ exists. Hence, according to (22), we deduce that the following limit

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \underset{i\to \infty }{lim}\left[\frac{(2-\tau )\tau {(1-\theta )}^2{\parallel x_{n_i}-y_{n_i}\parallel }^4}{{\alpha }_{n_i}{\parallel x_{n_i}-y_{n_i}+\mu {\gamma }^{m\left(x_{n_i}\right)}f\left(y_{n_i}\right)\parallel }^2}+{\sigma }_{n_i}({\delta }_{n_i}-{\sigma }_{n_i})\frac{\parallel u_{n_i}-T[(1-{\delta }_{n_i})u_{n_i}+{\delta }_{n_i}T{u}_{n_i}]{\parallel }^2}{{\alpha }_{n_i}}\right]\end{array}\end{array}$$

(23)

exists.

Since ${lim}_{i\to \infty }{\alpha }_{n_i}=0$ and ${lim\; inf}_{i\to \infty }{\sigma }_{n_i}({\delta }_{n_i}-{\sigma }_{n_i})>0$, it follows from (23) that

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim}\frac{\parallel x_{n_i}-y_{n_i}{\parallel }^4}{\parallel x_{n_i}-y_{n_i}+\mu {\gamma }^{m\left(x_{n_i}\right)}f\left(y_{n_i}\right){\parallel }^2}=0\end{array}$$

(24)

and

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim}\parallel u_{n_i}-T[(1-{\delta }_{n_i})u_{n_i}+{\delta }_{n_i}T{u}_{n_i}]\parallel =0.\end{array}$$

(25)

Note that $\parallel x_{n_i}-y_{n_i}+\mu {\gamma }^{m\left(x_{n_i}\right)}f\left(y_{n_i}\right)\parallel$ is bounded. In virtue of this fact and (24), we derive

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim}\parallel x_{n_i}-y_{n_i}\parallel =0\end{array}$$

(26)

Combining (14) and (26), we obtain

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim}\parallel f\left(x_{n_i}\right)-f\left(y_{n_i}\right)\parallel =0.\end{array}$$

(27)

As a result of (15), we have the following estimate

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel u_n-x_n\parallel & = \parallel P_C\left[x_n-\tau (1-\theta ){\parallel x_n-y_n\parallel }^2\frac{x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\right]-P_C\left[x_n\right]\parallel \hfill \\ & \le \frac{\tau (1-\theta )\parallel x_n-y_n{\parallel }^2}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)\parallel }.\hfill \end{array}\end{array}$$

This together with (24) implies that

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim}\parallel u_{n_i}-x_{n_i}\parallel =0.\end{array}$$

(28)

Applying the characterization (6) of projection $P_C$, we have

$$\begin{array}{c}\hfill \langle x_{n_i}-\mu {\gamma }^{m\left(x_{n_i}\right)}f\left(x_{n_i}\right)-y_{n_i},y_{n_i}-x^{†}\rangle \ge 0,\forall x^{†}\in C.\end{array}$$

It yields

$$\begin{array}{c}\hfill \langle f\left(x_{n_i}\right),x^{†}-x_{n_i}\rangle \ge \langle f\left(x_{n_i}\right),y_{n_i}-x_{n_i}\rangle +\frac{1}{\mu {\gamma }^{m\left(x_{n_i}\right)}}\langle y_{n_i}-x^{†},y_{n_i}-x_{n_i}\rangle ,\forall x^{†}\in C.\end{array}$$

(29)

Noting that $\left\{f\left(x_{n_i}\right)\right\}$ and $\left\{y_{n_i}\right\}$ are bounded, $\frac{\gamma \theta }{\kappa }<\mu {\gamma }^{m\left(x_{n_i}\right)}\le \mu$ due to Remark 2, in view of (26) and (29), we obtain

$$\begin{array}{c}\hfill \underset{i\to \infty }{lim\; inf}\langle f\left(x_{n_i}\right),x^{†}-x_{n_i}\rangle \ge 0,\forall x^{†}\in C.\end{array}$$

(30)

Thanks to (30), we can choose a positive real numbers sequence $\left\{{ϵ}_j\right\}$ satisfying ${lim}_{j\to \infty }{ϵ}_j=0$. For each ${ϵ}_j$, there exists the smallest positive integer $k_i$ such that

$$\begin{array}{c}\hfill \langle f(x_{n_{i_j}}),x^{†}-x_{n_{i_j}}\rangle +{ϵ}_j\ge 0,\forall j\ge k_i.\end{array}$$

(31)

Moreover, for each $j>0$, $f(x_{n_{i_j}})\ne 0$ (by Remark 3), letting $w(x_{n_{i_j}})=\frac{f(x_{n_{i_j}})}{\parallel f(x_{n_{i_j}}){\parallel }^2}$, then $\langle f(x_{n_{i_j}}),w(x_{n_{i_j}})\rangle =1$. By virtue of (31), we have

$$\begin{array}{c}\hfill \langle f(x_{n_{i_j}}),x^{†}+{ϵ}_{j}w(x_{n_{i_j}})-x_{n_{i_j}}\rangle \ge 0,\end{array}$$

which implies, together with the pseudomonotonicity of f on H, that

$$\begin{array}{c}\hfill \langle f(x^{†}+{ϵ}_{j}w(x_{n_{i_j}})),x^{†}+{ϵ}_{j}w(x_{n_{i_j}})-x_{n_{i_j}}\rangle \ge 0.\end{array}$$

It follows that

$$\begin{array}{c}\hfill \langle f(x^{†}),x^{†}-x_{n_{i_j}}\rangle \ge \langle f(x^{†})-f(x^{†}+{ϵ}_{j}w(x_{n_{i_j}})),x^{†}+{ϵ}_{j}w(x_{n_{i_j}})-x_{n_{i_j}}\rangle +\langle f(x^{†}),-{ϵ}_{j}w(x_{n_{i_j}})\rangle .\end{array}$$

(32)

Since the sequence $\{x_{n_{i_j}}\}$ is bounded, without loss of generality, we assume that $x_{n_{i_j}}\rightharpoonup v\in C$ as $j\to \infty$. Furthermore, $f(x_{n_{i_j}})\rightharpoonup f\left(v\right)$ due to the weakly sequentially continuity of f. Assume that $f\left(v\right)\ne 0$ (otherwise, $v\in VI(C,f)$ and ${\omega }_w\left(x_n\right)\subset VI(C,f)$). Thus, we have

$$\begin{array}{c}\hfill \underset{j\to \infty }{lim\; inf}\parallel f(x_{n_{i_j}})\parallel \ge \parallel f\left(v\right)\parallel ,\end{array}$$

and consequently,

$$\begin{array}{c}\hfill \underset{j\to \infty }{lim}\parallel {ϵ}_{j}w(x_{n_{i_j}})\parallel =\underset{j\to \infty }{lim}\frac{{ϵ}_j}{\parallel f(x_{n_{i_j}})\parallel }=0.\end{array}$$

This together with (32) and f being Lipschitz continuous, we deduce

$$\begin{array}{c}\hfill \langle f\left(x^{†}\right),x^{†}-v\rangle \ge 0.\end{array}$$

(33)

It follows from Lemma 3 that $v\in VI(C,f)$ and hence ${\omega }_w\left(x_n\right)\subset VI(C,f)$.

Since T is L-Lipschitzian, we have

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel u_n-T{u}_n\parallel & \le \parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]\parallel +\parallel T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]-T{u}_n\parallel \hfill \\ & \le \parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]\parallel +L{\delta }_n\parallel u_n-T{u}_n\parallel ,\hfill \end{array}\end{array}$$

which yields

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \parallel u_n-T{u}_n \parallel \le \frac{1}{1-{\delta }_{n}L}\parallel u_n-T[(1-{\delta }_n)u_n+{\delta }_{n}T{u}_n]\parallel .\end{array}\end{array}$$

(34)

On the basis of (25), (28) and (34), we derive

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \underset{j\to \infty }{lim}\parallel x_{n_{i_j}}-T{x}_{n_{i_j}}\parallel =0.\end{array}\end{array}$$

(35)

Consequently, applying Lemma 4 to (35) to deduce that $v\in Fix\left(T\right)$. Thus, $v\in VI(C,f)\cap Fix\left(T\right)=\Lambda$.

In case 2, we have $x_n\in VI(C,f)$ and the following estimate (by the similar argument as (19))

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2& \le (1-{\alpha }_n)\parallel x_n-x^{\ast }{\parallel }^2-(1-{\alpha }_n){\sigma }_n({\delta }_n-{\sigma }_n){\parallel x_n-T[(1-{\delta }_n)x_n+{\delta }_{n}T{x}_n]\parallel }^2\hfill \\ & +2{\alpha }_n\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle \hfill \\ & =(1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+{\alpha }_n[-(1-{\alpha }_n){\sigma }_n({\delta }_n-{\sigma }_n)\frac{\parallel x_n-T[(1-{\delta }_n)x_n+{\delta }_{n}T{x}_n]{\parallel }^2}{{\alpha }_n}\hfill \\ & +2\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle ].\hfill \end{array}\end{array}$$

Consequently, there exists a subsequence $\left\{x_{n_j}\right\}\subset \left\{x_n\right\}$ such that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \underset{j\to \infty }{lim}{\sigma }_{n_j}({\delta }_{n_j}-{\sigma }_{n_j})\frac{\parallel x_{n_j}-T[(1-{\delta }_{n_i})x_{n_j}+{\delta }_{n_j}T{x}_{n_j}]{\parallel }^2}{{\alpha }_{n_j}}=0.\end{array}\end{array}$$

It follows that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \underset{j\to \infty }{lim}\parallel x_{n_j}-T{x}_{n_j}]\parallel =0.\end{array}\end{array}$$

Thus, we also deduce that ${\omega }_w\left(x_n\right)\subset \Lambda$.

Step 3. $x_n\to P_{\Lambda }\left(u\right)$.

In Case 1 or Case 2, we have

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim\; sup}\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle =\langle u-x^{\ast },v-x^{\ast }\rangle \le 0.\end{array}$$

(36)

From (16), (17) and (18), we obtain

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \parallel x_{n+1}-x^{\ast }{\parallel }^2& = \parallel {\alpha }_n(u-x^{\ast })+(1-{\alpha }_n)(z_n-x^{\ast }){\parallel }^2\hfill \\ & \le (1-{\alpha }_n){\parallel x_n-x^{\ast }\parallel }^2+2{\alpha }_n\langle u-x^{\ast },x_{n+1}-x^{\ast }\rangle .\hfill \end{array}\end{array}$$

(37)

Finally, applying Lemma 5 with (36) to (37), we conclude that $x_n\to x^{\ast }$. This completes the proof. □

Remark 4.

We assume that f is κ-Lipschitz continuous. However, the information of κ is not necessary priority to be known. That is, we need not to estimate the value of κ.

Remark 5.

It is obvious that monotonicity implies pseudomonotonicity. Hence, our theorem holds when the involved operator f is monotone.

Assume that the above Algorithm 2 does not terminate in a finite iterations.

Algorithm 2: Iterative procedures for VI.

Step 1. Fixed four constants $\gamma \in (0,1)$, $\mu \in (0,1)$, $\theta \in (0,1)$ and $\tau \in (0,2)$. Let $x_0\in C$ be an initial value. Set $n=0$.

Step 2. Assume that the sequence $\left\{x_n\right\}$ has been constructed and then calculate $P_C[x_n-f\left(x_n\right)]$. If $P_C[x_n-f\left(x_n\right)]=x_n$, then stop. Otherwise, continuously proceed the following steps.

Step 3. Calculate $$\begin{array}{c}\hfill y_n=P_C[x_n-\mu {\gamma }^{m\left(x_n\right)}f\left(x_n\right)],\end{array}$$  where $m\left(x_n\right)=min\{0,1,2,3,\cdots \}$ and satisfies $$\begin{array}{c}\hfill \mu {\gamma }^{m\left(x_n\right)}\parallel f\left(x_n\right)-f\left(y_n\right)\parallel \le \theta \parallel x_n-y_n\parallel .\end{array}$$

Step 4. Let $u\in C$ be a fixed point. Let $\left\{{\alpha }_n\right\}$ be a real number sequence in $(0,1)$. Compute the sequence $\left\{x_{n+1}\right\}$ via the following form $$\begin{array}{c}\hfill {\displaystyle x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)P_C\left[x_n-\tau (1-\theta ){\parallel x_n-y_n\parallel }^2\frac{x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right)}{\parallel x_n-y_n+\mu {\gamma }^{m\left(x_n\right)}f\left(y_n\right){\parallel }^2}\right].}\end{array}$$

Step 5. Set $n:=n+1$ and return to Step 2.

Corollary 1.

Suppose that $VI(C,f)\ne \varnothing$. Assume that the iterative parameter $\left\{{\alpha }_n\right\}$ satisfies condition (C1) in Theorem 1. Then the sequence $\left\{x_n\right\}$ generated by Algorithm 2 converges strongly to $P_{VI(C,f)}\left(u\right)$.

Corollary 2.

Suppose that $Fix\left(T\right)\ne \varnothing$. Assume that the iterative parameters $\left\{{\alpha }_n\right\}$, $\left\{{\sigma }_n\right\}$ and $\left\{{\delta }_n\right\}$ satisfy the conditions (C1) and (C2) in Theorem 1. Then the sequence $\left\{x_n\right\}$ generated by Algorithm 3 converges strongly to $P_{Fix\left(T\right)}\left(u\right)$.

Algorithm 3: Iterative procedures for FP.

Step 1. Let $x_0\in C$ be an initial value. Set $n=0$.

Step 2. Assume that the sequence $\left\{x_n\right\}$ has been constructed. Let $u\in C$ be a fixed point. Let $\left\{{\alpha }_n\right\}$, $\left\{{\sigma }_n\right\}$ and $\left\{{\delta }_n\right\}$ be three real number sequences in $(0,1)$. Compute the sequences $\left\{z_n\right\}$ and $\left\{x_{n+1}\right\}$ via the following iterations $$\begin{array}{c}\hfill \left\{\begin{array}{c}{z}_n=(1-{\sigma }_n)x_n+{\sigma }_{n}T[(1-{\delta }_n)x_n+{\delta }_{n}T{x}_n],\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)z_n.\hfill \end{array}\right.\end{array}$$

4. Applications

Let $\varnothing \ne C$ be a convex and closed subset of a real Hilbert space H. Recall that an operator $T:C\to C$ is said to be $\alpha$-strictly pseudocontractive if there exists a constant $\alpha \in (0,1)$ satisfying

$$\parallel Tz-T{z}^{†}{\parallel }^2 \le \parallel z-z^{†}{\parallel }^2+ \alpha {\parallel (I-T)z-(I-T)z^{†}\parallel }^2$$

for all $z,z^{†}\in C$.

Remark 6.

It is easy to check that the class of pseudocontractive operators strictly includes the class of strictly pseudocontractive operators.

Proposition 2

([48]). Let $\varnothing \ne C$ be a convex and closed subset of a real Hilbert space H. Let $T:C\to C$ is said to be an α-strictly pseudocontractive operator. Then,

(i) 

T is $\frac{1+\alpha }{1-\alpha }$-Lipschitz;

(ii) 

$I-T$ is demi-closed at 0.

Now, by using Remark 6 and Proposition 2, we can apply Theorem 1 for solving pseudomonotone variational inequalities and fixed point problem of strictly pseudocontractive operators.

Theorem 2.

Let $\varnothing \ne C$ be a convex and closed subset of a real Hilbert space H. Let the operator f be pseudomonotone on H, weakly sequentially continuous and Lipschitz continuous on C with Lipschitz constant $\kappa >0$. Let $T:C\to C$ be an α-strictly pseudocontractive operator. Suppose that the iterative parameters $\left\{{\alpha }_n\right\}$, $\left\{{\sigma }_n\right\}$ and $\left\{{\delta }_n\right\}$ satisfy the following assumptions:

(C1): 

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

(C2): 

$0<\underline{\sigma }<{\sigma }_n<\overline{\sigma }<{\delta }_n<\overline{\delta }<\frac{1}{\sqrt{1+L^2}+1}(\forall n\ge 0)$ where $L=\frac{1+\alpha }{1-\alpha }$.

Then the sequence $\left\{x_n\right\}$ generated by Algorithm 1 converges strongly to $P_{\Lambda }\left(u\right)$.

Remark 7.

In [49], Anh and Phuong introduced an iteration algorithm for solving pseudomonotone variational inequalities and fixed point problem of strictly pseudocontractive operators. Theorem 2 extends the main result of ([49] Theorem 3.3) from weak convergence to strong convergence.

Remark 8.

In [50], Strodiot, Nguyen and Vuong presented a shrinking projection algorithm for solving variational inequalities and fixed point problem of strictly pseudocontractive operators. Note that the the computation of projection $P_{C_{n+1}}$ (([50] Algorithm 1-VI) is expensive. Our Algorithm 1 is more applicable.