Iterative algorithms for quasi-variational inclusions and fixed point problems of pseudocontractions

Abstract

In this paper, quasi-variational inclusions and fixed point problems of pseudocontractions are considered. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

MSC:49J40, 47J20, 47H09, 65J15.

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $A:C\to H$ be a single-valued nonlinear mapping and $B:H\to 2^H$ be a multi-valued mapping. The ‘so called’ quasi-variational inclusion problem is to find an $u\in 2^H$ such that

$$0\in Au+Bu.$$

(1.1)

The set of solutions of (1.1) is denoted by ${(A+B)}^{-1}(0)$. A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see for instance [1–4]. For related work, see [5–10]. The problem (1.1) includes many problems as special cases.

• (1) If $B=\partial \varphi :H\to 2^H$, where $\varphi :H\to R\cup +\mathrm{\infty }$ is a proper convex lower semi-continuous function and ∂ϕ is the subdifferential of ϕ, then the variational inclusion problem (1.1) is equivalent to finding $u\in H$ such that

  $$〈Au,y-u〉+\varphi (y)-\varphi (u)\ge 0,\mathrm{\forall }y\in H,$$

  which is called the mixed quasi-variational inequality (see [11]).

• (2) If $B=\partial {\delta }_C$, where C is a nonempty closed convex subset of H and ${\delta }_C:H\to [0,\mathrm{\infty }]$ is the indicator function of C, i.e.,

  $${\delta }_C=\{\begin{array}{cc}0,\hfill & x\in C,\hfill \\ +\mathrm{\infty },\hfill & x\notin C,\hfill \end{array}$$

then the variational inclusion problem (1.1) is equivalent to finding $u\in C$ such that

$$〈Au,v-u〉\ge 0,\mathrm{\forall }v\in C.$$

This problem is called the Hartman-Stampacchia variational inequality (see [12]).

Let $T:C\to C$ be a nonlinear mapping. The iterative scheme of Mann’s type for approximating fixed points of T is the following: $x_0\in C$ and

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T{x}_n,$$

for all $n\ge 1$, where $\{{\alpha }_n\}$ is a sequence in $[0,1]$; see [13]. For two nonlinear mappings S and T, Takahashi and Tamura [14] considered the following iteration procedure: $x_0\in C$ and

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)S({\beta }_n{x}_n+(1-{\beta }_n)T{x}_n),$$

for all $n\ge 1$, where $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $[0,1]$. Algorithms for finding the fixed points of nonlinear mappings or for finding the zero points of maximal monotone operators have been studied by many authors. The reader can refer to [15–19]. Especially, Takahashi et al. [20] recently gave the following convergence result.

Theorem 1.1 Let C be a closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let $J_{\lambda }^B={(I+\lambda B)}^{-1}$ be the resolvent of B for $\lambda >0$ and let T be a nonexpansive mapping of C into itself, such that $F(T)\cap {(A+B)}^{-1}0\ne \mathrm{\varnothing }$. Let $x_1=x\in C$ and let $\{x_n\}\subset C$ be a sequence generated by

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T({\alpha }_{n}x+(1-{\alpha }_n)J_{{\lambda }_n}^B(x_n-{\lambda }_{n}A{x}_n)),$$

for all $n\ge 0$, where $\{{\lambda }_n\}\subset (0,2\alpha )$, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset (0,1)$ satisfy

$$\begin{array}{c}0<a\le {\lambda }_n\le b<2\alpha ,0<c\le {\beta }_n\le d<1,\hfill \\ \underset{n\to \mathrm{\infty }}{lim}({\lambda }_{n+1}-{\lambda }_n)=0,\underset{n\to \mathrm{\infty }}{lim}{\alpha }_n=0\mathit{\text{and}}\sum _n{\alpha }_n=\mathrm{\infty }.\hfill \end{array}$$

Then $\{x_n\}$ converges strongly to a point of $F(T)\cap {(A+B)}^{-1}0$.

Recently, Zhang et al. [21] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [22] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping.

Motivated and inspired by the works in this field, the purpose of this paper is to consider the quasi-variational inclusions and fixed point problems of pseudocontractions. An iterative algorithm is presented. A strong convergence theorem is demonstrated.

2 Notations and lemmas

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$, respectively. Let C be a nonempty closed convex subset of H. It is well known that in a real Hilbert space H, the following equality holds:

$${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$$

(2.1)

for all $x,y\in H$ and $t\in [0,1]$.

Recall that a mapping $T:C\to C$ is called

(D₁) L-Lipschitzian ⟹ there exists $L>0$ such that $\parallel Tx-Ty\parallel \le L\parallel x-y\parallel$ for all $x,y\in C$; in the case of $L=1$, T is said to be nonexpansive;

(D₂) Firmly nonexpansive ⟹ ${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel (I-T)x-(I-T)y\parallel }^2⟺{\parallel Tx-Ty\parallel }^2\le 〈Tx-Ty,x-y〉$ for all $x,y\in C$;

(D₃) Pseudocontractive ⟹ $〈Tx-Ty,x-y〉\le {\parallel x-y\parallel }^2⟺{\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+{\parallel (I-T)x-(I-T)y\parallel }^2$ for all $x,y\in C$;

(D₄) Strongly monotone ⟹ there exists a positive constant $\tilde{\gamma }$ such that $〈Tx-Ty,x-y〉\ge \tilde{\gamma }\parallel x-y\parallel$ for all $x,y\in C$;

(D₅) Inverse strongly monotone $⟹〈Tx-Ty,x-y〉\ge \alpha {\parallel Tx-Ty\parallel }^2$ for some $\alpha >0$ and for all $x,y\in C$.

Let B be a mapping of H into $2^H$. The effective domain of B is denoted by $dom(B)$, that is, $dom(B)=\{x\in H:Bx\ne \mathrm{\varnothing }\}$. A multi-valued mapping B is said to be a monotone operator on H iff

$$〈x-y,u-v〉\ge 0$$

for all $x,y\in dom(B)$, $u\in Bx$, and $v\in By$. A monotone operator B on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone operator on H. Let B be a maximal monotone operator on H and let $B^{-1}0=\{x\in H:0\in Bx\}$.

For a maximal monotone operator B on H and $\lambda >0$, we may define a single-valued operator $J_{\lambda }^B={(I+\lambda B)}^{-1}:H\to dom(B)$, which is called the resolvent of B for λ. It is known that the resolvent $J_{\lambda }^B$ is firmly nonexpansive, i.e.,

$${\parallel J_{\lambda }^{B}x-J_{\lambda }^{B}y\parallel }^2\le 〈J_{\lambda }^{B}x-J_{\lambda }^{B}y,x-y〉$$

for all $x,y\in C$ and $B^{-1}0=Fix(J_{\lambda }^B)$ for all $\lambda >0$.

Usually, the convergence of fixed point algorithms requires some additional smoothness properties of the mapping T such as demi-closedness.

Recall that a mapping T is said to be demiclosed if, for any sequence $\{x_n\}$ which weakly converges to $\tilde{x}$, and if the sequence $\{T(x_n)\}$ strongly converges to z, then $T(\tilde{x})=z$. For the pseudocontractions, the following demiclosed principle is well known.

Lemma 2.1 ([23])

Let H be a real Hilbert space, C a closed convex subset of H. Let $U:C\to C$ be a continuous pseudo-contractive mapping. Then

• (i) $Fix(U)$ is a closed convex subset of C,

• (ii) $(I-U)$ is demiclosed at zero.

Lemma 2.2 ([24])

Let $\{r_n\}$ be a sequence of real numbers. Assume $\{r_n\}$ does not decrease at infinity, that is, there exists at least a subsequence $\{r_{n_k}\}$ of $\{r_n\}$ such that $r_{n_k}\le r_{n_k+1}$ for all $k\ge 0$. For every $n\ge N$, define an integer sequence $\{\tau (n)\}$ as

$$\tau (n)=max\{i\le n:r_{n_i}<r_{n_i+1}\}.$$

Then $\tau (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, and for all $n\ge N$

$$max\{r_{\tau (n)},r_n\}\le r_{\tau (n)+1}.$$

Lemma 2.3 ([25])

Assume $\{a_n\}$ is a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\delta }_n{\gamma }_n,$$

where $\{{\gamma }_n\}$ is a sequence in $(0,1)$ and $\{{\delta }_n\}$ is a sequence such that

• (1) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$;

• (2) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$ or ${\sum }_{n=1}^{\mathrm{\infty }}|{\delta }_n{\gamma }_n|<\mathrm{\infty }$.

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

In the sequel we shall use the following notations:

• 1. ${\omega }_w(u_n)=\{x:\mathrm{\exists }{u}_{n_j}\to x\text{ weakly}\}$ denote the weak ω-limit set of $\{u_n\}$;

• 2. $u_n\rightharpoonup x$ stands for the weak convergence of $\{u_n\}$ to x;

• 3. $u_n\to x$ stands for the strong convergence of $\{u_n\}$ to x;

• 4. $Fix(T)$ stands for the set of fixed points of T.

3 Main results

In this section, we consider a strong convergence theorem for quasi-variational inclusions and fixed point problems of pseudocontractive mappings in a Hilbert space.

Algorithm 3.1 Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let $J_{\lambda }^B={(I+\lambda B)}^{-1}$ be the resolvent of B for λ. Let $F:C\to H$ be an $L_1$-Lipschitzian and ς strongly monotone mapping and $f:C\to C$ be a ρ-contraction such that $\rho <max\{1,\varsigma /2\}$. Let $T:C\to C$ be an $L_2(>1)$-Lipschitzian pseudocontraction. For $x_0\in C$, define a sequence $\{x_n\}$ as follows:

$$\{\begin{array}{c}{z}_n=J_{\lambda }^B(I-\lambda A)x_n,\hfill \\ y_n=\nu z_n+(1-\nu )T((1-\zeta )z_n+\zeta T{z}_n),\hfill \\ x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}f(x_n)+(I-{\beta }_{n}F)y_n),\hfill \end{array}$$

(3.1)

for all $n\in \mathbb{N}$, where λ, ν and ζ are three constants, $\{{\alpha }_n\}$ and $\{{\beta }_n\}$ are two sequences in $[0,1]$.

Now, we demonstrate the convergence analysis of the algorithm (3.1).

Theorem 3.2 Suppose $\mathrm{\Gamma }:=Fix(T)\cap {(A+B)}^{-1}(0)\ne \mathrm{\varnothing }$. Assume the following conditions are satisfied:

(C1) ${\alpha }_n\in [a,b]\subset (0,1)$;

(C2) ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$;

(C3) $\lambda \in (0,2\alpha )$ and $0<1-\nu \le \zeta <\frac{1}{\sqrt{1+L_2^2}+1}$.

Then the sequence $\{x_n\}$ defined by (3.1) converges strongly to $u=P_{\mathrm{\Gamma }}(I-F+f)u$.

Proof Let $x^{\ast }\in Fix(T)\cap {(A+B)}^{-1}(0)$. Then, we get $x^{\ast }=J_{\lambda }^B(I-\lambda A)x^{\ast }=T{x}^{\ast }$. From (3.1), we have

$$\begin{array}{rl}{\parallel z_n-x^{\ast }\parallel }^2& ={\parallel J_{\lambda }^B(I-\lambda A)x_n-J_{\lambda }^B(I-\lambda A)x^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }-\lambda (A{x}_n-A{x}^{\ast })\parallel }^2\\ ={\parallel x_n-x^{\ast }\parallel }^2-2\lambda 〈A{x}_n-A{x}^{\ast },x_n-x^{\ast }〉+{\lambda }^2{\parallel A{x}_n-A{x}^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2-2\lambda \alpha {\parallel A{x}_n-A{x}^{\ast }\parallel }^2+{\lambda }^2{\parallel A{x}_n-A{x}^{\ast }\parallel }^2\\ ={\parallel x_n-x^{\ast }\parallel }^2-\lambda (2\alpha -\lambda ){\parallel A{x}_n-A{x}^{\ast }\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.2)

It follows that

$$\parallel z_n-x^{\ast }\parallel \le \parallel x_n-x^{\ast }\parallel .$$

(3.3)

Since $x^{\ast }\in Fix(T)$, we have from (D₃) that

$${\parallel Tx-x^{\ast }\parallel }^2\le {\parallel x-x^{\ast }\parallel }^2+{\parallel Tx-x\parallel }^2,$$

(3.4)

for all $x\in C$.

Thus,

$$\begin{array}{rl}{\parallel T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\le & {\parallel (1-\zeta )(z_n-x^{\ast })+\zeta (T{z}_n-x^{\ast })\parallel }^2\\ +{\parallel ((1-\zeta )I+\zeta T)z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2.\end{array}$$

(3.5)

By (3.4), (3.5), and (2.1), we obtain

$$\begin{array}{r}{\parallel T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\\ \le {\parallel (1-\zeta )(z_n-x^{\ast })+\zeta (T{z}_n-x^{\ast })\parallel }^2\\ +{\parallel ((1-\zeta )I+\zeta T)z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2\\ ={\parallel (1-\zeta )(z_n-T((1-\zeta )I+\zeta T)z_n)+\zeta (T{z}_n-T((1-\zeta )I+\zeta T)z_n)\parallel }^2\\ +{\parallel (1-\zeta )(z_n-x^{\ast })+\zeta (T{z}_n-x^{\ast })\parallel }^2\\ =(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2+\zeta {\parallel T{z}_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2\\ -\zeta (1-\zeta ){\parallel z_n-T{z}_n\parallel }^2+(1-\zeta ){\parallel z_n-x^{\ast }\parallel }^2+\zeta {\parallel T{z}_n-x^{\ast }\parallel }^2-\zeta (1-\zeta ){\parallel z_n-T{z}_n\parallel }^2\\ \le (1-\zeta ){\parallel z_n-x^{\ast }\parallel }^2+\zeta ({\parallel z_n-x^{\ast }\parallel }^2+{\parallel z_n-T{z}_n\parallel }^2)\\ -2\zeta (1-\zeta ){\parallel z_n-T{z}_n\parallel }^2+(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2\\ +\zeta {\parallel T{z}_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2.\end{array}$$

Noting that T is $L_2$-Lipschitzian and $z_n-((1-\zeta )I+\zeta T)z_n=\zeta (z_n-T{z}_n)$, we have

$$\begin{array}{r}{\parallel T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\\ \le (1-\zeta ){\parallel z_n-x^{\ast }\parallel }^2+\zeta ({\parallel z_n-x^{\ast }\parallel }^2+{\parallel z_n-T{z}_n\parallel }^2)\\ -2\zeta (1-\zeta ){\parallel z_n-T{z}_n\parallel }^2+(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2+{\zeta }^3{L}_2^2{\parallel z_n-T{z}_n\parallel }^2\\ ={\parallel z_n-x^{\ast }\parallel }^2+(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2\\ -\zeta (1-2\zeta -{\zeta }^2{L}_2^2){\parallel z_n-T{z}_n\parallel }^2.\end{array}$$

(3.6)

Since $\zeta <\frac{1}{\sqrt{1+L_2^2}+1}$, we have $1-2\zeta -{\zeta }^2{L}_2^2>0$. From (3.6), we can deduce

$${\parallel T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\le {\parallel z_n-x^{\ast }\parallel }^2+(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2.$$

(3.7)

Hence,

$$\begin{array}{rl}{\parallel y_n-x^{\ast }\parallel }^2=& {\parallel \nu z_n+(1-\nu )T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\\ =& {\parallel \nu (z_n-x^{\ast })+(1-\nu )(T((1-\zeta )I+\zeta T)z_n-x^{\ast })\parallel }^2\\ =& \nu {\parallel z_n-x^{\ast }\parallel }^2+(1-\nu ){\parallel T((1-\zeta )I+\zeta T)z_n-x^{\ast }\parallel }^2\\ -\nu (1-\nu ){\parallel T((1-\zeta )I+\zeta T)z_n-z_n\parallel }^2\\ \le & \nu {\parallel z_n-x^{\ast }\parallel }^2+(1-\nu )[{\parallel z_n-x^{\ast }\parallel }^2+(1-\zeta ){\parallel z_n-T((1-\zeta )I+\zeta T)z_n\parallel }^2]\\ -\nu (1-\nu ){\parallel T((1-\zeta )I+\zeta T)z_n-z_n\parallel }^2\\ =& {\parallel z_n-x^{\ast }\parallel }^2+(1-\nu )(1-\zeta -\nu ){\parallel T((1-\zeta )I+\zeta T)z_n-z_n\parallel }^2.\end{array}$$

(3.8)

By (C3) and (3.8), we obtain

$$\parallel y_n-x^{\ast }\parallel \le \parallel z_n-x^{\ast }\parallel .$$

(3.9)

Let $u_n={\beta }_{n}f(x_n)+(I-{\beta }_{n}F)y_n$ for all $n\ge 0$. Then, we have

$$\begin{array}{rl}\parallel u_n-x^{\ast }\parallel =& \parallel {\beta }_{n}f(x_n)+(I-{\beta }_{n}F)y_n-x^{\ast }\parallel \\ \le & {\beta }_n\parallel f(x_n)-F{x}^{\ast }\parallel +\parallel (I-{\beta }_{n}F)y_n-(I-{\beta }_{n}F)x^{\ast }\parallel \\ \le & {\beta }_n\parallel f(x_n)-f\left(x^{\ast }\right)\parallel +{\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel \\ +\parallel (I-{\beta }_{n}F)y_n-(I-{\beta }_{n}F)x^{\ast }\parallel \\ \le & {\beta }_n\rho \parallel x_n-x^{\ast }\parallel +{\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel \\ +\parallel (I-{\beta }_{n}F)y_n-(I-{\beta }_{n}F)x^{\ast }\parallel .\end{array}$$

(3.10)

Since F is $L_1$-Lipschitzian and ς strongly monotone, we have

$$\begin{array}{r}{\parallel (I-{\beta }_{n}F)y_n-(I-{\beta }_{n}F)x^{\ast }\parallel }^2\\ ={\parallel (y_n-x^{\ast })-{\beta }_n(F{y}_n-F{x}^{\ast })\parallel }^2\\ ={\parallel y_n-x^{\ast }\parallel }^2-2{\beta }_n〈F{y}_n-F{x}^{\ast },y_n-x^{\ast }〉+{\beta }_n^2{\parallel F{y}_n-F{x}^{\ast }\parallel }^2\\ \le {\parallel y_n-x^{\ast }\parallel }^2-2{\beta }_n\varsigma {\parallel y_n-x^{\ast }\parallel }^2+{\beta }_n^2{L}_1^2{\parallel y_n-x^{\ast }\parallel }^2\\ =(1-2{\beta }_n\varsigma +{\beta }_n^2{L}_1^2){\parallel y_n-x^{\ast }\parallel }^2.\end{array}$$

(3.11)

Noting that $L_1\ge \varsigma$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, without loss of generality, we assume that ${\beta }_n<\frac{\varsigma }{L_1^2-\varsigma /4}$ for all $n\ge 0$. Thus, $1-2{\beta }_n\varsigma +{\beta }_n^2{L}_1^2\le {(1-{\beta }_n\frac{\varsigma }{2})}^2$. So,

$$\parallel (I-{\beta }_{n}F)y_n-(I-{\beta }_{n}F)x^{\ast }\parallel \le (1-{\beta }_n\frac{\varsigma }{2})\parallel y_n-x^{\ast }\parallel .$$

(3.12)

We have from (3.9), (3.10), and (3.12)

$$\begin{array}{rl}\parallel u_n-x^{\ast }\parallel & \le {\beta }_n\rho \parallel x_n-x^{\ast }\parallel +{\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel +(1-{\beta }_n\frac{\varsigma }{2})\parallel x_n-x^{\ast }\parallel \\ =[1-(\frac{\varsigma }{2}-\rho ){\beta }_n]\parallel x_n-x^{\ast }\parallel +{\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel .\end{array}$$

(3.13)

From (3.1) and (3.13), we have

$$\begin{array}{rl}\parallel x_{n+1}-x^{\ast }\parallel =& \parallel {\alpha }_n(x_n-x^{\ast })+(1-{\alpha }_n)(u_n-x^{\ast })\parallel \\ \le & (1-{\alpha }_n)([1-(\frac{\varsigma }{2}-\rho ){\beta }_n]\parallel x_n-x^{\ast }\parallel +{\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel )\\ +{\alpha }_n\parallel x_n-x^{\ast }\parallel \\ =& [1-(\frac{\varsigma }{2}-\rho )(1-{\alpha }_n){\beta }_n]\parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n){\beta }_n\parallel f\left(x^{\ast }\right)-F{x}^{\ast }\parallel .\end{array}$$

(3.14)

By the definition of $x_n$, we have

$$\begin{array}{rl}{x}_{n+1}-x_n& ={\alpha }_n{x}_n+(1-{\alpha }_n)({\beta }_{n}f(x_n)+(I-{\beta }_{n}F)y_n)-x_n\\ =(1-{\alpha }_n)[{\beta }_{n}f(x_n)-{\beta }_{n}F{y}_n+y_n-x_n].\end{array}$$

(3.15)

Hence,

$$\begin{array}{rl}〈x_{n+1}-x_n,x_n-x^{\ast }〉=& 〈(1-{\alpha }_n)[{\beta }_{n}f(x_n)-{\beta }_{n}F{y}_n+y_n-x_n],x_n-x^{\ast }〉\\ =& (1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ +(1-{\alpha }_n)〈y_n-x_n,x_n-x^{\ast }〉.\end{array}$$

(3.16)

Since $2〈x_{n+1}-x_n,x_n-x^{\ast }〉={\parallel x_{n+1}-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2$ and $2〈y_n-x_n,x_n-x^{\ast }〉={\parallel y_n-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2-{\parallel y_n-x_n\parallel }^2$, it follows from (3.16), (3.3), and (3.9) that

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x_n\parallel }^2\\ =2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ +(1-{\alpha }_n)[{\parallel y_n-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2-{\parallel y_n-x_n\parallel }^2]\\ \le 2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ -(1-{\alpha }_n){\parallel y_n-x_n\parallel }^2.\end{array}$$

(3.17)

By (3.15), we obtain

$$\begin{array}{rl}{\parallel x_{n+1}-x_n\parallel }^2\le & {(1-{\alpha }_n)}^2{[{\beta }_n\parallel f(x_n)-F{y}_n\parallel +\parallel y_n-x_n\parallel ]}^2\\ =& {(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel f(x_n)-F{y}_n\parallel }^2+{\parallel y_n-x_n\parallel }^2\\ +2{\beta }_n\parallel f(x_n)-F{y}_n\parallel \parallel y_n-x_n\parallel ].\end{array}$$

(3.18)

Combining (3.17) and (3.18) to deduce

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2\\ \le 2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ -(1-{\alpha }_n){\parallel y_n-x_n\parallel }^2+{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel f(x_n)-F{y}_n\parallel }^2\\ +{\parallel y_n-x_n\parallel }^2+2{\beta }_n\parallel f(x_n)-F{y}_n\parallel \parallel y_n-x_n\parallel ]\\ \le 2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉-(1-{\alpha }_n){\alpha }_n{\parallel y_n-x_n\parallel }^2\\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel f(x_n)-F{y}_n\parallel }^2+2{\beta }_n\parallel f(x_n)-F{y}_n\parallel \parallel y_n-x_n\parallel ].\end{array}$$

Hence, we obtain

$$\begin{array}{r}{\parallel x_{n+1}-x^{\ast }\parallel }^2-{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\alpha }_n{\parallel y_n-x_n\parallel }^2\\ \le 2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉-2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel f(x_n)-F{y}_n\parallel }^2+2{\beta }_n\parallel f(x_n)-F{y}_n\parallel \parallel y_n-x_n\parallel ].\end{array}$$

It follows that, hence, we obtain

$$\begin{array}{r}(1-{\alpha }_n){\alpha }_n{\parallel y_n-x_n\parallel }^2\\ \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel x_{n+1}-x^{\ast }\parallel }^2+2(1-{\alpha }_n){\beta }_n〈f(x_n),x_n-x^{\ast }〉\\ -2(1-{\alpha }_n){\beta }_n〈F{y}_n,x_n-x^{\ast }〉\\ +{(1-{\alpha }_n)}^2[{\beta }_n^2{\parallel f(x_n)-F{y}_n\parallel }^2+2{\beta }_n\parallel f(x_n)-F{y}_n\parallel \parallel y_n-x_n\parallel ].\end{array}$$

(3.19)

Next we divide our proof into two possible cases.

Case 1. There exists an integer number m such that $\parallel x_{n+1}-x^{\ast }\parallel \le \parallel x_n-x^{\ast }$ for all $n\ge m$. In this case, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel$ exists. Since ${\alpha }_n\in [a,b]\subset (0,1)$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, by (3.19), we derive

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

(3.20)

This together with (3.18) implies that

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

(3.21)

Note that

$$\begin{array}{rl}\parallel u_n-y_n\parallel & =\parallel {\beta }_{n}f(x_n)+(I-{\beta }_{n}F)y_n-y_n\parallel \\ \le {\beta }_n\parallel f(x_n)-F{y}_n\parallel .\end{array}$$

So,

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

(3.22)

By (3.20) and (3.22), we obtain

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

(3.23)

From (3.2) and (3.9), we have

$${\parallel y_n-x^{\ast }\parallel }^2\le {\parallel z_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-\lambda (2\alpha -\lambda ){\parallel A{x}_n-A{x}^{\ast }\parallel }^2.$$

Hence,

$$\begin{array}{rl}\lambda (2\alpha -\lambda ){\parallel A{x}_n-A{x}^{\ast }\parallel }^2& \le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel y_n-x^{\ast }\parallel }^2\\ \le \parallel x_n-y_n\parallel (\parallel x_n-x^{\ast }\parallel +\parallel y_n-x^{\ast }\parallel ).\end{array}$$

Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_n-A{x}^{\ast }\parallel =0.$$

(3.24)

Since $J_{\lambda }^B$ is firmly nonexpansive and A is monotone, we have

$$\begin{array}{rcl}{\parallel z_n-x^{\ast }\parallel }^2& =& {\parallel J_{\lambda }^B(I-\lambda A)x_n-J_{\lambda }^B(I-\lambda A)x^{\ast }\parallel }^2\\ \le & 〈(I-\lambda A)x_n-(I-\lambda A)x^{\ast },z_n-x^{\ast }〉\\ =& 〈z_n-x^{\ast },x_n-x^{\ast }〉-\lambda 〈z_n-x^{\ast },A{x}_n-A{x}^{\ast }〉\\ =& \frac{1}{2}({\parallel z_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x_n\parallel }^2)\\ -\lambda 〈x_n-x^{\ast },A{x}_n-A{x}^{\ast }〉-\lambda 〈z_n-x_n,A{x}_n-A{x}^{\ast }〉\\ \le & \frac{1}{2}({\parallel z_n-x^{\ast }\parallel }^2+{\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x_n\parallel }^2)\\ +\lambda \parallel z_n-x_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel .\end{array}$$

It follows that

$${\parallel z_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x_n\parallel }^2+2\lambda \parallel z_n-x_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel .$$

(3.25)

By (3.25) and (3.9), we deduce

$${\parallel y_n-x^{\ast }\parallel }^2\le {\parallel z_n-x^{\ast }\parallel }^2\le {\parallel x_n-x^{\ast }\parallel }^2-{\parallel z_n-x_n\parallel }^2+2\lambda \parallel z_n-x_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel .$$

Therefore,

$$\begin{array}{rcl}{\parallel z_n-x_n\parallel }^2& \le & {\parallel x_n-x^{\ast }\parallel }^2-{\parallel y_n-x^{\ast }\parallel }^2+2\lambda \parallel z_n-x_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel \\ \le & \parallel x_n-y_n\parallel (\parallel x_n-x^{\ast }\parallel +\parallel y_n-x^{\ast }\parallel )+2\lambda \parallel z_n-x_n\parallel \parallel A{x}_n-A{x}^{\ast }\parallel .\end{array}$$

(3.26)

Equations (3.20), (3.24), and (3.26) imply that

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

(3.27)

Notice that $F-f$ is $(\varsigma -\rho )$ strongly monotone. Thus, the variational inequality of finding $y\in \mathrm{\Gamma }$ such that $〈(F-f)y,x-y〉\ge 0$ for all $x\in \mathrm{\Gamma }$ has a unique solution, denoted by $x^{\ast }$, that is, $x^{\ast }=P_{\mathrm{\Gamma }}(I-V+F)(x^{\ast })$. Next, we prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_n-x^{\ast }〉\le 0.$$

Since $u_n$ is bounded, without loss of generality, we assume that there exists a subsequence $\{z_{n_i}\}$ of $\{u_n\}$ such that $u_{n_i}\rightharpoonup \tilde{x}$ for some $\tilde{x}\in H$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_n-x^{\ast }〉=\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_{n_i}-x^{\ast }〉.$$

Thus, we have that $x_{n_i}\rightharpoonup \tilde{x}$ and

$$\underset{i\to \mathrm{\infty }}{lim}\parallel J_{\lambda }^B(I-\lambda A)x_{n_i}-x_{n_i}\parallel =0.$$

Therefore, $\tilde{x}\in Fix(J_{\lambda }^B(I-\lambda A))={(A+B)}^{-1}(0)$.

Next we show that $\tilde{x}\in Fix(T)$. First, we show that $Fix(T)=Fix(T((1-\zeta )I+\zeta T))$. As a matter of fact, $Fix(T)\subset Fix(T((1-\zeta )I+\zeta T))$ is obvious. Next, we show that $Fix(T((1-\zeta )I+\zeta T))\subset Fix(T)$.

Take any $x^{\ast }\in Fix(T((1-\zeta )I+\zeta T))$. We have $T((1-\zeta )I+\zeta T)x^{\ast }=x^{\ast }$. Set $S=(1-\zeta )I+\zeta T$. We have $TS{x}^{\ast }=x^{\ast }$. Write $S{x}^{\ast }=y^{\ast }$. Then, $T{y}^{\ast }=x^{\ast }$. Now we show $x^{\ast }=y^{\ast }$. In fact,

$$\begin{array}{rl}\parallel x^{\ast }-y^{\ast }\parallel & =\parallel T{y}^{\ast }-S{x}^{\ast }\parallel =\parallel T{y}^{\ast }-(1-\zeta )x^{\ast }-\zeta T{x}^{\ast }\parallel \\ =\zeta \parallel T{y}^{\ast }-T{x}^{\ast }\parallel \le \zeta L_2\parallel y^{\ast }-x^{\ast }\parallel .\end{array}$$

Since, $\zeta <\frac{1}{\sqrt{1+L_2^2}+1}<\frac{1}{L_2}$, we deduce $y^{\ast }=x^{\ast }\in Fix(S)=Fix(T)$. Thus, $x^{\ast }\in Fix(T)$. Hence, $Fix(T((1-\zeta )I+\zeta T))\subset Fix(T)$. Therefore, $Fix(T((1-\zeta )I+\zeta T))=Fix(T)$.

By (3.1), (3.20), and (3.27), we deduce

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T((1-\zeta )I+\zeta T)x_n-x_n\parallel =0.$$

(3.28)

Next we prove that $T((1-\zeta )I+\zeta T)-I$ is demiclosed at 0. Let the sequence $\{w_n\}\subset H_2$ satisfying $w_n\rightharpoonup x^{†}$ and $w_n-T((1-\zeta )I+\zeta T)w_n\to 0$. Next, we will show that $x^{†}\in Fix(T((1-\zeta )I+\zeta T))=Fix(T)$.

Since T is $L_2$-Lipschizian, we have

$$\begin{array}{rl}\parallel w_n-T{w}_n\parallel & \le \parallel w_n-T((1-\zeta )I+\zeta T)w_n\parallel +\parallel T((1-\zeta )I+\zeta T)w_n-T{w}_n\parallel \\ \le \parallel w_n-T((1-\zeta )I+\zeta T)w_n\parallel +\zeta L\parallel w_n-T{w}_n\parallel .\end{array}$$

It follows that

$$\parallel w_n-T{w}_n\parallel \le \frac{1}{1-\zeta L}\parallel w_n-T((1-\zeta )I+\zeta T)w_n\parallel .$$

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-T{w}_n\parallel =0.$$

Since $T-I$ is demiclosed at 0 by Lemma 2.1, we immediately deduce $x^{†}\in Fix(T)=Fix(T((1-\zeta )I+\zeta T))$. Therefore, $T((1-\zeta )I+\zeta T)-I$ is demiclosed at 0. By (3.28), we deduce $\tilde{x}\in Fix(T)$. Hence, $\tilde{x}\in \mathrm{\Gamma }$. So,

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_n-x^{\ast }〉& =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_{n_i}-x^{\ast }〉\\ =& 〈(f-F)x^{\ast },\tilde{x}-x^{\ast }〉\\ \le & 0.\end{array}$$

(3.29)

Note that

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& =& {\parallel {\beta }_n(f(x_n)-f\left(x^{\ast }\right))+{\beta }_n(f\left(x^{\ast }\right)-F{x}^{\ast })+(I-{\beta }_{n}F)(y_n-x^{\ast })\parallel }^2\\ \le & {\parallel (I-{\beta }_{n}F)(y_n-x^{\ast })\parallel }^2+2{\beta }_n〈f(x_n)-f\left(x^{\ast }\right),u_n-x^{\ast }〉\\ +2{\beta }_n〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉\\ \le & {(1-{\beta }_n\frac{\varsigma }{2})}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\beta }_n\rho \parallel x_n-x^{\ast }\parallel \parallel u_n-x^{\ast }\parallel \\ +2{\beta }_n〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉\\ \le & {(1-{\beta }_n\frac{\varsigma }{2})}^2{\parallel x_n-x^{\ast }\parallel }^2+2{\beta }_n\rho {\parallel x_n-x^{\ast }\parallel }^2+\frac{1}{2}{\parallel u_n-x^{\ast }\parallel }^2\\ +2{\beta }_n〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉.\end{array}$$

It follows that

$$\begin{array}{rcl}{\parallel u_n-x^{\ast }\parallel }^2& \le & [1-2(\frac{\varsigma }{2}-\rho ){\beta }_n]{\parallel x_n-x^{\ast }\parallel }^2+{\beta }_n^2{\frac{\varsigma }{4}}^2{\parallel x_n-x^{\ast }\parallel }^2\\ +4{\beta }_n〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉.\end{array}$$

So,

$$\begin{array}{rcl}{\parallel x_{n+1}-x^{\ast }\parallel }^2& =& {\parallel {\alpha }_n(x_n-x^{\ast })+(1-{\alpha }_n)(u_n-x^{\ast })\parallel }^2\\ \le & {\alpha }_n{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\parallel u_n-x^{\ast }\parallel }^2\\ \le & [1-2(\frac{\varsigma }{2}-\rho )(1-{\alpha }_n){\beta }_n]{\parallel x_n-x^{\ast }\parallel }^2+(1-{\alpha }_n){\beta }_n^2\frac{{\varsigma }^2}{4}{\parallel x_n-x^{\ast }\parallel }^2\\ +4(1-{\alpha }_n){\beta }_n〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉\\ =& [1-(\varsigma -2\rho )(1-{\alpha }_n){\beta }_n]{\parallel x_n-x^{\ast }\parallel }^2\\ +(\varsigma -2\rho )(1-{\alpha }_n){\beta }_n\{{\beta }_n\frac{{\varsigma }^2}{4(\varsigma -2\rho )}{\parallel x_n-x^{\ast }\parallel }^2\\ +\frac{4}{\varsigma -2\rho }〈f\left(x^{\ast }\right)-F{x}^{\ast },u_n-x^{\ast }〉\}.\end{array}$$

(3.30)

Applying Lemma 2.3 to (3.30) we deduce $x_n\to x^{\ast }$.

Case 2. Assume there exists an integer $n_0$ such that $\parallel x_{n_0}-x^{\ast }\parallel \le \parallel x_{n_0+1}-x^{\ast }\parallel$. In this case, we set ${\omega }_n=\{\parallel x_n-x^{\ast }\parallel \}$. Then, we have ${\omega }_{n_0}\le {\omega }_{n_0+1}$. Define an integer sequence $\{{\tau }_n\}$ for all $n\ge n_0$ as follows:

$$\tau (n)=max\{l\in \mathbb{N}|n_0\le l\le n,{\omega }_l\le {\omega }_{l+1}\}.$$

It is clear that $\tau (n)$ is a non-decreasing sequence satisfying

$$\underset{n\to \mathrm{\infty }}{lim}\tau (n)=\mathrm{\infty }$$

and

$${\omega }_{\tau (n)}\le {\omega }_{\tau (n)+1},$$

for all $n\ge n_0$. From (3.19), we get

$$\begin{array}{r}(1-{\alpha }_{\tau (n)}){\alpha }_{\tau (n)}{\parallel y_{\tau (n)}-x_{\tau (n)}\parallel }^2\\ \le {\parallel x_{\tau (n)}-x^{\ast }\parallel }^2-{\parallel x_{\tau (n)+1}-x^{\ast }\parallel }^2+2(1-{\alpha }_{\tau (n)}){\beta }_{\tau (n)}〈f(x_{\tau (n)}),x_{\tau (n)}-x^{\ast }〉\\ -2(1-{\alpha }_{\tau (n)}){\beta }_{\tau (n)}〈F{y}_{\tau (n)},x_{\tau (n)}-x^{\ast }〉\\ +{(1-{\alpha }_{\tau (n)})}^2[{\beta }_{\tau (n)}^2{\parallel f(x_{\tau (n)})-F{y}_{\tau (n)}\parallel }^2\\ +2{\beta }_{\tau (n)}\parallel f(x_{\tau (n)})-F{y}_{\tau (n)}\parallel \parallel y_{\tau (n)}-x_{\tau (n)}\parallel ].\end{array}$$

(3.31)

It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_{\tau (n)}-x_{\tau (n)}\parallel =0.$$

By a similar argument to that of (3.29) and (3.30), we can prove that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-F)x^{\ast },u_{\tau (n)}-x^{\ast }〉\le 0,$$

(3.32)

and

$$\begin{array}{rcl}{\omega }_{\tau (n)+1}^2& \le & [1-2(\frac{\varsigma }{2}-\rho )(1-{\alpha }_{\tau (n)}){\beta }_{\tau (n)}]{\omega }_{\tau (n)}^2\\ +(1-{\alpha }_{\tau (n)}){\beta }_{\tau (n)}^2\frac{{\varsigma }^2}{4}{\omega }_{\tau (n)}^2\\ +4(1-{\alpha }_{\tau (n)}){\beta }_{\tau (n)}〈f\left(x^{\ast }\right)-F{x}^{\ast },u_{\tau (n)}-x^{\ast }〉.\end{array}$$

(3.33)

Since ${\omega }_{\tau (n)}\le {\omega }_{\tau (n)+1}$, we have from (3.33)

$${\omega }_{\tau (n)}^2\le \frac{16}{4({\varsigma }^2-2\rho )-{\varsigma }^2{\beta }_{\tau (n)}}〈f\left(x^{\ast }\right)-F{x}^{\ast },u_{\tau }(n)-x^{\ast }〉.$$

(3.34)

Combining (3.33) and (3.34), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\omega }_{\tau (n)}\le 0,$$

and hence

$$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau (n)}=0.$$

(3.35)

From (3.33), we also obtain

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\omega }_{\tau (n)+1}\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\omega }_{\tau (n)}.$$

This together with (3.35) imply that

$$\underset{n\to \mathrm{\infty }}{lim}{\omega }_{\tau (n)+1}=0.$$

Applying Lemma 2.2 to get

$$0\le {\omega }_n\le max\{{\omega }_{\tau (n)},{\omega }_{\tau (n)+1}\}.$$

Therefore, ${\omega }_n\to 0$. That is, $x_n\to x^{\ast }$. This completes the proof. □