Strong convergence of viscosity approximation methods for the fixed-point of pseudo-contractive and monotone mappings

Abstract

In this paper, we introduce a viscosity iterative process, which converges strongly to a common element of the set of fixed points of a pseudo-contractive mapping and the set of solutions of a monotone mapping. We also prove that the common element is the unique solution of certain variational inequality. The strong convergence theorems are obtained under some mild conditions. The results presented in this paper extend and unify most of the results that have been proposed for this class of nonlinear mappings.

MSC:47H09, 47H10, 47L25.

1 Introduction

Let C be a closed convex subset of a real Hilbert space H. A mapping $A:C\to H$ is called monotone if and only if

$$〈x-y,Ax-Ay〉\ge 0,\mathrm{\forall }x,y\in C.$$

(1.1)

A mapping $A:C\to H$ is called α-inverse strongly monotone if there exists a positive real number $\alpha >0$ such that

$$〈x-y,Ax-Ay〉\ge \alpha {\parallel Ax-Ay\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.2)

Obviously, the class of monotone mappings includes the class of the α-inverse strongly monotone mappings.

A mapping $T:C\to H$ is called pseudo-contractive if $\mathrm{\forall }x,y\in C$, we have

$$〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^2.$$

(1.3)

A mapping $T:C\to H$ is called κ-strict pseudo-contractive, if there exists a constant $0\le \kappa \le 1$ such that

$$〈x-y,Tx-Ty〉\le {\parallel x-y\parallel }^2-\kappa {\parallel (I-T)x-(I-T)y\parallel }^2,\mathrm{\forall }x,y\in C.$$

(1.4)

A mapping $T:C\to C$ is called non-expansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

(1.5)

Clearly, the class of pseudo-contractive mappings includes the class of strict pseudo-contractive mappings and non-expansive mappings. We denote by $F(T)$ the set of fixed points of T, that is, $F(T)=\{x\in C:Tx=x\}$.

A mapping $f:C\to C$ is called contractive with a contraction coefficient if there exists a constant $\rho \in (0,1)$ such that

$$\parallel f(x)-f(y)\parallel \le \rho \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

(1.6)

For finding an element of the set of fixed points of the non-expansive mappings, Halpern [1] was the first to study the convergence of the scheme in 1967

$$x_{n+1}={\alpha }_{n+1}u+(1-{\alpha }_{n+1})T(x_n).$$

(1.7)

In 2000, Moudafi [2] introduced the viscosity approximation methods and proved the strong convergence of the following iterative algorithm under some suitable conditions

$$x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T(x_n).$$

(1.8)

Viscosity approximation methods are very important, because they are applied to convex optimization, linear programming, monotone inclusions and elliptic differential equations. In a Hilbert space, many authors have studied the fixed points problems of the fixed points for the non-expansive mappings and monotone mappings by the viscosity approximation methods, and obtained a series of good results, see [3–18].

Suppose that A is a monotone mapping from C into H. The classical variational inequality problem is formulated as finding a point $u\in C$ such that $〈v-u,Au〉\ge 0$, $\mathrm{\forall }v\in C$. The set of solutions of variational inequality problems is denoted by $\mathit{VI}(C,A)$.

Takahashi [19, 20] introduced the following scheme and studied the weak and strong convergence theorem of the elements of $F(T)\cap \mathit{VI}(C,A)$, respectively, under different conditions

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_{n+1})T{P}_C(x_n-{\lambda }_n{x}_n),$$

(1.9)

where T is a non-expansive mapping, A is an α-inverse strong monotone operator.

Recently, Zegeye and Shahzad [21] introduced the algorithms and obtained the strong convergence theorems in a Hilbert space, respectively,

$$x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T_n{x}_n$$

(1.10)

and

$$x_{n+1}={\alpha }_{n}f(x_n)+(1-{\alpha }_n)T_{r_n}{F}_{r_n}{x}_n,$$

(1.11)

where $T_n$ are asymptotically non-expansive mappings, and $T_{r_n}$, $F_{r_n}$ are non-expansive mappings.

Our concern is now the following: Is it possible to construct a new sequence that converges strongly to a common element of the intersection of the set of fixed points of a pseudo-contractive mapping and the solution set of a variational inequality problem for a monotone mapping?

2 Preliminaries

Let C be a nonempty closed and convex subset of a real Hilbert space H, a mapping $P_C:H\to C$ is called the metric projection, if $\mathrm{\forall }x\in H$, there exists a unique point in C, denote by $P_{C}x$ such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C.$$

It is well known that $P_C$ is a non-expansive mapping, and $P_{C}x$ have the property as follows:

$$〈x-P_{C}x,y-P_{C}x〉\le 0,\mathrm{\forall }x\in H,y\in C,$$

(2.1)

$${\parallel x-y\parallel }^2\ge {\parallel x-P_{C}x\parallel }^2+{\parallel y-P_{C}x\parallel }^2,\mathrm{\forall }x\in H,y\in C.$$

(2.2)

Lemma 2.1 [6]

Let $\{a_n\}$ be a sequence of nonnegative real numbers satisfying the following relation:

$$a_{n+1}\le (1-{\theta }_n)a_n+{\sigma }_n,n\ge 0,$$

where $\{{\theta }_n\}$ is a sequence in (0,1) and $\{{\sigma }_n\}$ is a real sequence such that

• (i) ${\sum }_{n=0}^{\mathrm{\infty }}{\theta }_n=\mathrm{\infty }$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{{\sigma }_n}{{\theta }_n}\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}{\sigma }_n<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.2 [21]

Let C be a closed convex subset of a Hilbert space H. Let $A:C\to H$ be a continuous monotone mapping, let $T:C\to C$ be a continuous pseudo-contractive mapping, define mappings $T_r$ and $F_r$ as follows: $x\in H$, $r\in (0,\mathrm{\infty })$

$$\begin{array}{c}{T}_r(x)=\{z\in C:〈y-z,Tz〉-\frac{1}{r}〈y-z,(1+r)z-x〉\le 0,\mathrm{\forall }y\in C\},\hfill \\ F_r(x)=\{z\in C:〈y-z,Az〉+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}.\hfill \end{array}$$

Then the following hold:

• (i) $T_r$ and $F_r$ are single-valued;

• (ii) $T_r$ and $F_r$ are firmly non-expansive mappings, i.e., ${\parallel T_{r}x-T_{r}y\parallel }^2\le 〈T_{r}x-T_{r}y,x-y〉$, ${\parallel F_{r}x-F_{r}y\parallel }^2\le 〈F_{r}x-F_{r}y,x-y〉$;

• (iii) $F(T_r)=F(T)$, $F(F_r)=\mathit{VI}(C,A)$;

• (iv) $F(T)$ and $\mathit{VI}(C,A)$ are closed convex.

Lemma 2.3 [22]

Let $\{x_n\}$ and $\{z_n\}$ be bounded sequence in a Banach space, and let $\{{\beta }_n\}$ be a sequence in $[0,1]$, which satisfies the following condition:

$$0<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}{\beta }_n<\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\beta }_n<1.$$

Suppose that

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n,n\ge 0$$

and

$$\underset{n\to \mathrm{\infty }}{lim}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Then ${lim}_{n\to \mathrm{\infty }}\parallel z_n-x_n\parallel =0$.

Let C be a closed convex subset of a Hilbert space H. Let $A:C\to H$ be a continuous monotone mapping, let $T:C\to C$ be a continuous pseudo-contractive mapping. Then we define the mappings as follows: for $x\in H$, ${\tau }_n\in (0,\mathrm{\infty })$

$$T_{{\tau }_n}(x)=\{z\in C:〈y-z,Tz〉-\frac{1}{{\tau }_n}〈y-z,(1+{\tau }_n)z-x〉\le 0,\mathrm{\forall }y\in C\},$$

(2.3)

$$F_{{\tau }_n}(x)=\{z\in C:〈y-z,Az〉+\frac{1}{{\tau }_n}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in C\}.$$

(2.4)

3 Main results

Theorem 3.1 Let C be a nonempty closed convex subset of a Hilbert space H. Let $T:C\to C$ be a continuous pseudo-contractive mapping, let $A:C\to H$ be a continuous monotone mapping such that $F=F(T)\cap \mathit{VI}(C,A)\ne \mathrm{\varnothing }$, let $f:C\to C$ be a contraction with a contraction coefficient $\rho \in (0,1)$. The mappings $T_{{\tau }_n}$ and $F_{{\tau }_n}$ are defined as (2.3) and (2.4), respectively. Let $\{x_n\}$ be a sequence generated by $x_0\in C$

$$\{\begin{array}{c}{y}_n={\lambda }_n{x}_n+(1-{\lambda }_n)F_{{\tau }_n}{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{T}_{{\tau }_n}{y}_n,\hfill \end{array}$$

(3.1)

where ${\lambda }_n\in [0,1]$, let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ be sequences of nonnegative real numbers in $[0,1]$ and

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<1$;

• (iv) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\tau }_n>0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\tau }_{n+1}-{\tau }_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}=P_{F}f(\overline{x})$, and also $\overline{x}$ is the unique solution of the variational inequality

$$〈f(\overline{x})-\overline{x},y-\overline{x}〉\le 0,\mathrm{\forall }y\in F.$$

(3.2)

Proof First, we prove that $\{x_n\}$ is bounded. Take $p\in F$, then we have from Lemmas 2.2 that

$$\parallel y_n-p\parallel \le {\lambda }_n\parallel x_n-p\parallel +(1-{\lambda }_n)\parallel F_{{\tau }_n}{x}_n-F_{{\tau }_n}p\parallel \le \parallel x_n-p\parallel .$$

(3.3)

For $n\ge 0$, because $T_{{\tau }_n}$ and $F_{{\tau }_n}$ are non-expansive, and f is contractive, we have

$$\begin{array}{r}\parallel x_{n+1}-p\parallel \\ =\parallel {\alpha }_n(f(x_n)-p)+{\beta }_n(x_n-p)+{\gamma }_n(T_{{\tau }_n}{y}_n-p)\parallel \\ \le {\alpha }_n\parallel f(x_n)-f(p)\parallel +{\alpha }_n\parallel f(p)-p\parallel +{\beta }_n\parallel x_n-p\parallel +{\gamma }_n\parallel T_{{\tau }_n}{y}_n-p\parallel \\ \le \rho {\alpha }_n\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ \le [1-(1-\rho ){\alpha }_n]\parallel x_n-p\parallel +{\alpha }_n\parallel f(p)-p\parallel \\ \le max\left\{\parallel x_0-p,\frac{f(p)-p}{1-\rho }\parallel \right\}.\end{array}$$

Therefore, $\{x_n\}$ is bounded. Consequently, we get that $\{F_{{\tau }_n}{x}_n\}$, $\{T_{{\tau }_n}{y}_n\}$ and $\{y_n\}$, $\{f(x_n)\}$ are bounded.

Next, we show that $\parallel x_{n+1}-x_n\parallel \to 0$.

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel \le & {\lambda }_{n+1}\parallel x_{n+1}-x_n\parallel +(1-{\lambda }_{n+1})\parallel F_{{\tau }_{n+1}}{x}_{n+1}-F_{{\tau }_n}{x}_n\parallel \\ +|{\lambda }_{n+1}-{\lambda }_n|\parallel x_n-F_{{\tau }_n}{x}_n\parallel .\end{array}$$

(3.4)

Let $v_n=F_{{\tau }_n}{x}_n$, $v_{n+1}=F_{{\tau }_{n+1}}{x}_{n+1}$, by the definition of the mapping $F_{{\tau }_n}$, we have that

$$〈y-v_n,A{v}_n〉+\frac{1}{{\tau }_n}〈y-v_n,v_n-x_n〉\ge 0,\mathrm{\forall }y\in C,$$

(3.5)

$$〈y-v_{n+1},A{v}_{n+1}〉+\frac{1}{{\tau }_{n+1}}〈y-v_{n+1},v_{n+1}-x_{n+1}〉\ge 0,\mathrm{\forall }y\in C.$$

(3.6)

Putting $y:=v_{n+1}$ in (3.5), and letting $y:=v_n$ in (3.6), we have that

$$〈v_{n+1}-v_n,A{v}_n〉+\frac{1}{{\tau }_n}〈v_{n+1}-v_n,v_n-x_n〉\ge 0,$$

(3.7)

$$〈v_n-v_{n+1},A{v}_{n+1}〉+\frac{1}{{\tau }_{n+1}}〈v_n-v_{n+1},v_{n+1}-x_{n+1}〉\ge 0.$$

(3.8)

Adding (3.7) and (3.8), we have that

$$〈v_{n+1}-v_n,A{v}_n-A{v}_{n+1}〉+〈v_{n+1}-v_n,\frac{v_n-x_n}{{\tau }_n}-\frac{v_{n+1}-x_{n+1}}{{\tau }_{n+1}}〉\ge 0.$$

Since A is a monotone mapping, which implies that

$$〈v_{n+1}-v_n,\frac{v_n-x_n}{{\tau }_n}-\frac{v_{n+1}-x_{n+1}}{{\tau }_{n+1}}〉\ge 0.$$

Therefore, we have that

$$〈v_{n+1}-v_n,v_n-x_n-\frac{{\tau }_n(v_{n+1}-x_{n+1})}{{\tau }_{n+1}}+v_{n+1}-v_{n+1}〉\ge 0,$$

i.e.,

$$\begin{array}{rl}{\parallel v_{n+1}-v_n\parallel }^2& \le 〈v_{n+1}-v_n,x_{n+1}-x_n+(1-\frac{{\tau }_n}{{\tau }_{n+1}})(v_{n+1}-x_{n+1})〉\\ \le \parallel v_{n+1}-v_n\parallel \{\parallel x_{n+1}-x_n\parallel +|1-\frac{{\tau }_n}{{\tau }_{n+1}}|\parallel v_{n+1}-x_{n+1}\parallel \}.\end{array}$$

(3.9)

Without loss of generality, let b be a real number such that ${\tau }_n>b>0$, $\mathrm{\forall }n\in N$, then we have that

$$\begin{array}{rl}\parallel v_{n+1}-v_n\parallel & \le \parallel x_{n+1}-x_n\parallel +|1-\frac{{\tau }_n}{{\tau }_{n+1}}|\parallel v_{n+1}-x_{n+1}\parallel \\ \le \parallel x_{n+1}-x_n\parallel +\frac{1}{b}|{\tau }_{n+1}-{\tau }_n|K,\end{array}$$

(3.10)

where $K=sup\parallel v_{n+1}-x_{n+1}\parallel$. Then we have from (3.10) and (3.4) that

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & \le & \parallel x_{n+1}-x_n\parallel +\frac{(1-{\lambda }_{n+1})|{\tau }_{n+1}-{\tau }_n|}{b}K\\ +|{\lambda }_{n+1}-{\lambda }_n|\parallel x_n-F_{{\tau }_n}{x}_n\parallel .\end{array}$$

(3.11)

On the other hand, let $u_n=T_{{\tau }_n}{y}_n$, $u_{n+1}=T_{{\tau }_{n+1}}{y}_{n+1}$, we have that

$$〈y-u_n,T{u}_n〉-\frac{1}{{\tau }_n}〈y-u_n,(1+{\tau }_n)u_n-y_n〉\le 0,\mathrm{\forall }y\in C,$$

(3.12)

$$〈y-u_{n+1},T{u}_{n+1}〉-\frac{1}{{\tau }_{n+1}}〈y-u_{n+1},(1+{\tau }_{n+1})u_{n+1}-y_{n+1}〉\le 0,\mathrm{\forall }y\in C.$$

(3.13)

Let $y:=u_{n+1}$ in (3.12), and let $y:=u_n$ in (3.13), we have that

$$〈u_{n+1}-u_n,T{u}_n〉-\frac{1}{{\tau }_n}〈u_{n+1}-u_n,(1+{\tau }_n)u_n-y_n〉\le 0,$$

(3.14)

$$〈u_n-u_{n+1},T{u}_{n+1}〉-\frac{1}{{\tau }_{n+1}}〈u_n-u_{n+1},(1+{\tau }_{n+1})u_{n+1}-y_{n+1}〉\le 0.$$

(3.15)

Adding (3.14) and (3.15), and because T is pseudo-contractive, we have that

$$〈u_{n+1}-u_n,\frac{u_n-y_n}{{\tau }_n}-\frac{u_{n+1}-y_{n+1}}{{\tau }_{n+1}}〉\ge 0.$$

Therefore, we have

$$〈u_{n+1}-u_n,u_n-y_n-\frac{{\tau }_n(u_{n+1}-y_{n+1})}{{\tau }_{n+1}}+u_{n+1}-u_{n+1}〉\ge 0.$$

Hence we have that

$$\parallel u_{n+1}-u_n\parallel \le \parallel y_{n+1}-y_n\parallel +\frac{1}{b}|{\tau }_{n+1}-{\tau }_n|M,$$

(3.16)

where $M=sup\{\parallel u_n-y_n\parallel :n\in N\}$.

Let $x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)z_n$, hence we have that

$$\begin{array}{rl}{z}_{n+1}-z_n=& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(f(x_{n+1})-f(x_n))+(\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n})f(x_n)\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}(u_{n+1}-u_n)+(\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n})u_n.\end{array}$$

(3.17)

Hence we have from (3.17), (3.16), (3.11) and condition (iii) that

$$\begin{array}{r}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le \frac{(\rho -1){\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\{\parallel f(x_n)\parallel +\parallel u_n\parallel \}\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\frac{|{\tau }_{n+1}-{\tau }_n|}{b}((1-{\lambda }_{n+1})K+M)+\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}|{\lambda }_{n+1}-{\lambda }_n|\parallel x_n-F_{{\tau }_n}{x}_n\parallel .\end{array}$$

(3.18)

Notice conditions (ii) and (iv), we have that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )=0.$$

(3.19)

Hence we have from Lemma 2.3 that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel z_n-x_n\parallel =0.$$

(3.20)

Therefore, we have that

$$\parallel x_{n+1}-x_n\parallel =|1-{\beta }_n|\parallel z_n-x_n\parallel \to 0.$$

(3.21)

Hence we have from (3.10) and (3.16) that

$$\parallel y_{n+1}-y_n\parallel \to 0,\parallel u_{n+1}-u_n\parallel \to 0,\parallel v_{n+1}-v_n\parallel \to 0.$$

(3.22)

Since $x_n={\alpha }_{n-1}f(x_{n-1})+{\beta }_{n-1}{x}_{n-1}+{\gamma }_{n-1}{v}_{n-1}$, so we have that

$$\begin{array}{rl}\parallel x_n-v_n\parallel & \le \parallel x_n-v_{n-1}\parallel +\parallel v_{n-1}-v_n\parallel \\ \le [1-{\alpha }_{n-1}(1-\rho )]\parallel x_{n-1}-v_{n-1}\parallel +\parallel v_{n-1}-v_n\parallel .\end{array}$$

According to Lemma 2.1, we have that

$$\parallel x_n-v_n\parallel \to 0.$$

In the same way, we have that

$$\parallel x_n-y_n\parallel \to 0,\parallel x_n-u_n\parallel \to 0.$$

Consequently, we have that

$$\parallel y_n-u_n\parallel \to 0.$$

Now, we show that ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈f(\overline{x})-\overline{x},x_{n+1}-\overline{x}〉\le 0$.

Since sequence $\{x_n\}$ is bounded, then there exists a sub-sequence $\{x_{nk}\}$ of $\{x_n\}$ and $w\in C$ such that $x_{nk}\rightharpoonup w$. Next, we show that $w\in F=F(T)\cap \mathit{VI}(C,A)$.

Since $v_n=F_{{\tau }_n}{x}_n$, we have that

$$〈y-v_{nk},A{v}_{nk}〉+\frac{1}{{\tau }_{nk}}〈y-v_{nk},v_{nk}-x_{nk}〉\ge 0,\mathrm{\forall }y\in C.$$

Let $v_t=tv+(1-t)w$, $t\in (0,1)$, $v\in C$, then we have that

$$〈v_t-v_{nk},A{v}_t〉\ge 〈v_t-v_{nk},A{v}_t-A{v}_{nk}〉-〈v_t-v_{nk},\frac{v_{nk}-x_{nk}}{{\tau }_{nk}}〉.$$

Since $x_{nk}-v_{nk}\to 0$, and also A is monotone, we have that

$$0\le \underset{k\to \mathrm{\infty }}{lim}〈v_t-v_{nk},A{v}_t〉=〈v_t-w,A{v}_t〉.$$

Consequently,

$$〈v_t-w,A{v}_t〉\ge 0.$$

If $t\to 0$, by the continuity of A, we have $〈v-w,Aw〉\ge 0$, $\mathrm{\forall }v\in C$. Thus, $w\in \mathit{VI}(C,A)$.

In addition, since $u_n=F_{{\tau }_n}{y}_n$, we have that

$$〈y-u_{nk},T{u}_{nk}〉-\frac{1}{{\tau }_{nk}}〈y-u_{nk},(1+{\tau }_{nk})u_{nk}-y_{nk}〉\le 0,\mathrm{\forall }y\in C.$$

Let $v_t=tv+(1-t)w$, $t\in (0,1)$, $v\in C$, then we have that

$$\begin{array}{rl}〈u_{nk}-v_t,T{v}_t〉& \ge 〈v_t-u_{nk},T{u}_{nk}-T{v}_t〉-〈v_t-u_{nk},\frac{1+{\tau }_{nk}}{{\tau }_{nk}}{u}_{nk}-y_{nk}〉\\ \ge -〈v_t-u_{nk},v_t〉-\frac{1}{{\tau }_{nk}}〈v_t-u_{nk},u_{nk}-y_{nk}〉.\end{array}$$

Since $y_{nk}-u_{nk}\to 0$, if $t\to 0$, by the continuity of T, we have that

$$-〈v-w,Tw〉\ge -〈v-w,w〉,\mathrm{\forall }v\in C.$$

Let $v=Tw$, we have $w=Tw$, thus, $w\in F(T)$. Consequently, we conclude that $w\in F=F(T)\cap \mathit{VI}(C,A)$.

Because $\overline{x}=P_{F}f(\overline{x})$, we have from (2.1) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈f(\overline{x})-\overline{x},x_n-\overline{x}〉=〈f(\overline{x})-P_{F}f(\overline{x}),w-P_{F}f(\overline{x})〉\le 0.$$

(3.23)

Next, we show that $x_n\to \overline{x}$. From formula (3.3), we have that

$$\begin{array}{rl}{\parallel x_{n+1}-\overline{x}\parallel }^2=& {\parallel {\alpha }_n(f(x_n)-\overline{x})+{\beta }_n(x_n-\overline{x})+{\gamma }_n(u_n-\overline{x})\parallel }^2\\ \le & {\alpha }_n^2{\parallel f(x_n)-\overline{x}\parallel }^2+{\parallel {\beta }_n(x_n-\overline{x})+{\gamma }_n(u_n-\overline{x})\parallel }^2\\ +2{\alpha }_n〈f(x_n)-\overline{x},{\beta }_n(x_n-\overline{x})+{\gamma }_n(u_n-\overline{x})〉\\ \le & {\alpha }_n^2{\parallel f(x_n)-\overline{x}\parallel }^2+{(1-{\alpha }_n)}^2{\parallel x_n-\overline{x}\parallel }^2+2{\alpha }_n{\beta }_n〈f(x_n)-\overline{x},x_n-\overline{x}〉\\ +2{\alpha }_n{\gamma }_n\parallel f(x_n)-\overline{x}\parallel \parallel u_n-\overline{x}\parallel \\ \le & {\alpha }_n^2{\parallel f(x_n)-\overline{x}\parallel }^2+{(1-{\alpha }_n)}^2{\parallel x_n-\overline{x}\parallel }^2+2{\alpha }_n{\beta }_n〈f(\overline{x})-\overline{x},x_n-\overline{x}〉\\ +2{\alpha }_n{\beta }_n〈f(x_n)-f(\overline{x}),x_n-\overline{x}〉+2{\alpha }_n{\gamma }_n\parallel f(x_n)-\overline{x}\parallel \parallel u_n-\overline{x}\parallel \\ =& [{(1-{\alpha }_n)}^2+2\rho {\alpha }_n{\beta }_n]{\parallel x_n-\overline{x}\parallel }^2+{\alpha }_n^2{\parallel f(x_n)-\overline{x}\parallel }^2+2{\alpha }_n{\beta }_n〈f(\overline{x})-\overline{x},x_n-\overline{x}〉\\ +2{\alpha }_n{\gamma }_n\parallel f(x_n)-\overline{x}\parallel \parallel u_n-\overline{x}\parallel \\ =& (1-2{\alpha }_n(1-\rho {\beta }_n)){\parallel x_n-\overline{x}\parallel }^2+{\alpha }_n^2[{\parallel f(x_n)-\overline{x}\parallel }^2+{\parallel x_n-\overline{x}\parallel }^2]\\ +2{\alpha }_n{\gamma }_n\parallel f(x_n)-\overline{x}\parallel \parallel u_n-\overline{x}\parallel +2{\alpha }_n{\beta }_n〈f(\overline{x})-\overline{x},x_n-\overline{x}〉.\end{array}$$

Let ${\theta }_n=2{\alpha }_n(1-\rho {\beta }_n)$, ${\sigma }_n={\alpha }_n^2[{\parallel f(x_n)-\overline{x}\parallel }^2+{\parallel x_n-\overline{x}\parallel }^2]+2{\alpha }_n{\gamma }_n\parallel f(\overline{x})-\overline{x}\parallel \parallel u_n-\overline{x}\parallel +2{\alpha }_n{\beta }_n〈f(\overline{x})-\overline{x},x_n-\overline{x}〉$. According to Lemma 2.1 and formula (3.23), we have that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-\overline{x}\parallel =0$, i.e., the sequence $\{x_n\}$ converges strongly to $\overline{x}\in F$.

According to formula (3.23), we conclude that $\overline{x}$ is the solution of the variational inequality (3.2). Now, we show that $\overline{x}$ is the unique solution of the variational inequality (3.2).

Suppose that $\overline{y}\in F$ is another solution of the variational inequality (3.2). Because $\overline{x}$ is the solution of the variational inequality (3.2), i.e., $〈f(\overline{x})-\overline{x},y-\overline{x}〉\le 0$, $\mathrm{\forall }y\in F$. Because $\overline{y}\in F$, then we have

$$〈f(\overline{x})-\overline{x},\overline{y}-\overline{x}〉\le 0.$$

(3.24)

On the other hand, to the solution $\overline{y}\in F$, since $\overline{x}\in F$, so

$$〈f(\overline{y})-\overline{y},\overline{x}-\overline{y}〉\le 0.$$

(3.25)

Adding (3.24) and (3.25), we have that

$$〈\overline{x}-\overline{y}-(f(\overline{x})-f(\overline{y})),\overline{x}-\overline{y}〉\le 0,$$

i.e.,

$$〈\overline{x}-\overline{y}-(f(\overline{x})-f(\overline{y})),\overline{x}-\overline{y}〉\le 〈f(\overline{x})-f(\overline{y}),\overline{x}-\overline{y}〉.$$

Hence

$${\parallel \overline{x}-\overline{y}\parallel }^2\le 〈f(\overline{x})-f(\overline{y}),\overline{x}-\overline{y}〉\le \rho {\parallel \overline{x}-\overline{y}\parallel }^2.$$

Because $\rho \in (0,1)$, hence we conclude that $\overline{x}=\overline{y}$, the uniqueness of the solution is obtained. □

Theorem 3.2 Let C be a nonempty closed convex subset of a Hilbert space H. Let $T:C\to C$ be a continuous pseudo-contractive mapping, let $A:C\to H$ be a continuous monotone mapping such that $F=F(T)\cap \mathit{VI}(C,A)\ne \mathrm{\varnothing }$, let $f:C\to C$ be a contraction with a contraction coefficient $\rho \in (0,1)$. The mappings $T_{{\tau }_n}$ and $F_{{\tau }_n}$ are defined as (2.3) and (2.4), respectively. Let $\{x_n\}$ be a sequence generated by $x_0\in C$

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{T}_{{\tau }_n}{F}_{{\tau }_n}{x}_n,$$

(3.26)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ are sequences of nonnegative real numbers in $[0,1]$ and

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\tau }_n>0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\tau }_{n+1}-{\tau }_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}=P_{F}f(\overline{x})$, and also $\overline{x}$ is the unique solution of the variational inequality

$$〈f(\overline{x})-\overline{x},y-\overline{x}〉\le 0,\mathrm{\forall }y\in F.$$

(3.27)

Proof Putting ${\lambda }_n=0$ in Theorem 3.1, we can obtain the result. □

If in Theorem 3.1 and Theorem 3.2, let $f:\equiv u\in C$ be a constant mapping, we have the following theorems.

Theorem 3.3 Let C be a nonempty closed convex subset of a Hilbert space H. Let $T:C\to C$ be a continuous pseudo-contractive mapping, let $A:C\to H$ be a continuous accretive mapping such that $F=F(T)\cap \mathit{VI}(C,A)\ne \mathrm{\varnothing }$. The mappings $T_{{\tau }_n}$ and $F_{{\tau }_n}$ are defined as (2.3) and (2.4), respectively. Let $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{c}{x}_0=x\in C,\hfill \\ y_n={\lambda }_n{x}_n+(1-{\lambda }_n)F_{{\tau }_n}{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{T}_{{\tau }_n}{y}_n,\hfill \end{array}$$

(3.28)

where ${\lambda }_n\in [0,1]$, and let $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ be sequences of nonnegative real numbers in $[0,1]$ and

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\lambda }_n<{lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\lambda }_n<1$;

• (iv) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\tau }_n>0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\tau }_{n+1}-{\tau }_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges strongly to $\overline{x}=P_{F}u$, and also $\overline{x}$ is the unique solution of the variational inequality

$$〈u-\overline{x},y-\overline{x}〉\le 0,\mathrm{\forall }y\in F.$$

(3.29)

Theorem 3.4 Let C be a nonempty closed convex subset of a Hilbert space H. Let $T:C\to C$ be a continuous pseudo-contractive mapping, let $A:C\to H$ be a continuous monotone mapping such that $F=F(T)\cap \mathit{VI}(C,A)\ne \mathrm{\varnothing }$. The mappings $T_{{\tau }_n}$ and $F_{{\tau }_n}$ are defined as (2.3) and (2.4), respectively. Let $\{x_n\}$ be a sequence generated by $x_0\in C$

$$x_{n+1}={\alpha }_{n}u+{\beta }_n{x}_n+{\gamma }_n{T}_{{\tau }_n}{F}_{{\tau }_n}{x}_n,$$

(3.30)

where $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{{\gamma }_n\}$ are sequences of nonnegative real numbers in $[0,1]$ and

• (i) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 0$;

• (ii) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$, ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (iii) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\tau }_n>0$, ${\sum }_{n=1}^{\mathrm{\infty }}|{\tau }_{n+1}-{\tau }_n|<\mathrm{\infty }$.

Then the sequence $\{x_n\}$ converges to $\overline{x}=P_{F}u$, and also $\overline{x}$ is the unique solution of the variational inequality

$$〈u-\overline{x},y-\overline{x}〉\le 0,\mathrm{\forall }y\in F.$$

(3.31)