A Korpelevich-like algorithm for variational inequalities

Abstract

A Korpelevich-like algorithm has been introduced for solving a generalized variational inequality. It is shown that the presented algorithm converges strongly to a special solution of the generalized variational inequality.

MSC:47H05, 47J25.

1 Introduction

Now it is well-known that the variational inequality of finding $x^{\ast }\in C$ such that

$$〈A{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,$$

(1.1)

where C is a nonempty closed convex subset of a real Hilbert space H and $A:C\to H$ is a given mapping, is a fundamental problem in variational analysis and, in particular, in optimization theory. For related works, please see [1–20] and the references contained therein. Especially, Yao, Marino and Muglia [21] presented the following modified Korpelevich method for solving (1.1):

$$\begin{array}{r}{y}_n=P_C[x_n-\lambda A{x}_n-{\alpha }_n{x}_n],\\ x_{n+1}=P_C[x_n-\lambda A{y}_n+\mu (y_n-x_n)],n\ge 0.\end{array}$$

(1.2)

Recently, Aoyama, Iiduka and Takahashi [22] extended the variational inequality (1.1) to Banach spaces as follows:

$$\text{Find }{x}^{\ast }\in C\text{ such that }〈A{x}^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C,$$

(1.3)

where C is a nonempty closed convex subset of a real Banach space E. We use $S(C,A)$ to denote the solution set of (1.3). The generalized variational inequality (1.3) is connected with the fixed point problem for nonlinear mappings. For solving the above generalized variational inequality (1.3), Aoyama, Iiduka and Takahashi [22] introduced the iterative algorithm

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)Q_C[x_n-{\lambda }_{n}A{x}_n],n\ge 0,$$

(1.4)

where $Q_C$ is a sunny nonexpansive retraction from E onto C and $\{{\alpha }_n\}\subset (0,1)$, $\{{\lambda }_n\}\subset (0,\mathrm{\infty })$ are two real number sequences. Motivated by (1.4), Yao and Maruster [23] presented a modification of (1.4) as follows:

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)Q_C[(1-{\alpha }_n)(x_n-\lambda A{x}_n)],n\ge 0.$$

(1.5)

Motivated and inspired by the above algorithms (1.2), (1.4) and (1.5), in this paper, we suggest an extragradient-type method via the sunny nonexpansive retraction for solving the variational inequalities (1.3) in Banach spaces. It is shown that the presented algorithm converges strongly to a special solution of the variational inequality (1.3).

2 Preliminaries

Let C be a nonempty closed convex subset of a real Banach space E. Recall that a mapping A of C into E is said to be accretive if there exists $j(x-y)\in J(x-y)$ such that

$$〈Ax-Ay,j(x-y)〉\ge 0$$

for all $x,y\in C$. A mapping A of C into E is said to be α-strongly accretive if for $\alpha >0$,

$$〈Ax-Ay,j(x-y)〉\ge \alpha {\parallel x-y\parallel }^2$$

for all $x,y\in C$. A mapping A of C into E is said to be α-inverse-strongly accretive if for $\alpha >0$,

$$〈Ax-Ay,j(x-y)〉\ge \alpha {\parallel Ax-Ay\parallel }^2$$

for all $x,y\in C$.

Let $U=\{x\in E:\parallel x\parallel =1\}$. A Banach space E is said to uniformly convex if for each $ϵ\in (0,2]$, there exists $\delta >0$ such that for any $x,y\in U$,

$$\parallel x-y\parallel \ge ϵ\text{implies}\parallel \frac{x+y}{2}\parallel \le 1-\delta .$$

It is known that a uniformly convex Banach space is reflexive and strictly convex. A Banach space E is said to be smooth if the limit

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

(2.1)

exists for all $x,y\in U$. It is also said to be uniformly smooth if the limit (2.1) is attained uniformly for $x,y\in U$. The norm of E is said to be Frechet differentiable if for each $x\in U$, the limit (2.1) is attained uniformly for $y\in U$. And we define a function $\rho :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ called the modulus of smoothness of E as follows:

$$\rho (\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:x,y\in X,\parallel x\parallel =1,\parallel y\parallel =\tau \}.$$

It is known that E is uniformly smooth if and only if ${lim}_{\tau \to 0}\rho (\tau )/\tau =0$. Let q be a fixed real number with $1<q\le 2$. Then a Banach space E is said to be q-uniformly smooth if there exists a constant $c>0$ such that $\rho (\tau )\le c{\tau }^q$ for all $\tau >0$.

We need the following lemmas for the proof of our main results.

Lemma 2.1 [24]

Let q be a given real number with $1<q\le 2$ and let E be a q-uniformly smooth Banach space. Then

$${\parallel x+y\parallel }^q\le {\parallel x\parallel }^q+q〈y,J_q(x)〉+2{\parallel Ky\parallel }^q$$

for all $x,y\in E$, where K is the q-uniformly smoothness constant of E and $J_q$ is the generalized duality mapping from E into $2^{E^{\ast }}$ defined by

$$J_q(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^q,\parallel f\parallel ={\parallel x\parallel }^{q-1}\},\mathrm{\forall }x\in E.$$

Let D be a subset of C and let Q be a mapping of C into D. Then Q is said to be sunny if

$$Q(Qx+t(x-Qx))=Qx,$$

whenever $Qx+t(x-Qx)\in C$ for $x\in C$ and $t\ge 0$. A mapping Q of C into itself is called a retraction if $Q^2=Q$. If a mapping Q of C into itself is a retraction, then $Qz=z$ for every $z\in R(Q)$, where $R(Q)$ is the range of Q. A subset D of C is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D. We know the following lemma concerning sunny nonexpansive retraction.

Lemma 2.2 [25]

Let C be a closed convex subset of a smooth Banach space E, let D be a nonempty subset of C and Q be a retraction from C onto D. Then Q is sunny and nonexpansive if and only if

$$〈u-Qu,j(y-Qu)〉\le 0$$

for all $u\in C$ and $y\in D$.

Lemma 2.3 [22]

Let C be a nonempty closed convex subset of a smooth Banach space X. Let $Q_C$ be a sunny nonexpansive retraction from X onto C and let A be an accretive operator of C into X. Then for all $\lambda >0$,

$$S(C,A)=F(Q_C(I-\lambda A)),$$

where $S(C,A)=\{x^{\ast }\in C:〈A{x}^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C\}$.

Lemma 2.4 [26]

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space X. Let the mapping $A:C\to X$ be α-inverse-strongly accretive. Then we have

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+2\lambda (K^2\lambda -\alpha ){\parallel Ax-Ay\parallel }^2.$$

In particular, if $0\le \lambda \le \frac{\alpha }{K^2}$, then $I-\lambda A$ is nonexpansive.

Proof Indeed, for all $x,y\in C$, from Lemma 2.1, we have

$$\begin{array}{rcl}{\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2& =& {\parallel (x-y)-\lambda (Ax-Ay)\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2\lambda 〈Ax-Ay,j(x-y)〉+2K^2{\lambda }^2{\parallel Ax-Ay\parallel }^2\\ \le & {\parallel x-y\parallel }^2-2\lambda \alpha {\parallel Ax-Ay\parallel }^2+2K^2{\lambda }^2{\parallel Ax-Ay\parallel }^2\\ =& {\parallel x-y\parallel }^2+2\lambda (K^2\lambda -\alpha ){\parallel Ax-Ay\parallel }^2.\end{array}$$

It is clear that if $0\le \lambda \le \frac{\alpha }{K^2}$, then $I-\lambda A$ is nonexpansive. □

Lemma 2.5 [27]

Let C be a nonempty bounded closed convex subset of a uniformly convex Banach space E and let T be a nonexpansive mapping of C into itself. If $\{x_n\}$ is a sequence of C such that $x_n\to x$ weakly and $x_n-T{x}_n\to 0$ strongly, then x is a fixed point of T.

Lemma 2.6 [28]

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

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

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

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

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

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

3 Main results

In this section, we present our Korpelevich-like algorithm and consequently we will show its strong convergence.

3.1 Conditions assumptions

(A1) E is a uniformly convex and 2-uniformly smooth Banach space with a weakly sequentially continuous duality mapping;

(A2) C is a nonempty closed convex subset of E;

(A3) $A:C\to E$ is an α-strongly accretive and L-Lipschitz continuous mapping with $S(C,A)\ne \mathrm{\varnothing }$;

(A4) $Q_C$ is a sunny nonexpansive retraction from E onto C.

3.2 Parameters restrictions

(P1) λ, μ and γ are three positive constants satisfying:

• (i) $\gamma \in (0,1)$, $\lambda \in [a,b]$ for some a, b with $0<a<b<\frac{\alpha }{K^2{L}^2}$;

• (ii) $\frac{\lambda }{\mu }<\frac{\alpha }{K^2{L}^2}$ where K is the smooth constant of E.

(P2) $\{{\alpha }_n\}$ is a sequence in $(0,1)$ such that ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

Algorithm 3.1 For given $x_0\in C$, define a sequence $\{x_n\}$ iteratively by

$$\{\begin{array}{c}{y}_n=Q_C[(1-{\alpha }_n)x_n-\lambda A{x}_n],\hfill \\ x_{n+1}=(1-\gamma )x_n+\gamma Q_C[x_n-\lambda A{y}_n+\mu (y_n-x_n)],n\ge 0.\hfill \end{array}$$

(3.1)

Theorem 3.2 The sequence $\{x_n\}$ generated by (3.1) converges strongly to $Q^{\prime }(0)$, where $Q^{\prime }$ is a sunny nonexpansive retraction of E onto $S(C,A)$.

Proof Let $p\in S(C,A)$. First, from Lemma 2.2, we have $p=Q_C[p-\delta Ap]$ for all $\delta >0$. In particular, $p=Q_C[p-\lambda Ap]=Q_C[{\alpha }_{n}p+(1-{\alpha }_n)(p-\frac{\lambda }{1-{\alpha }_n}Ap)]$ for all $n\ge 0$.

Since $A:C\to E$ is α-strongly accretive and L-Lipschitzian, it must be $\frac{\alpha }{L^2}$-inverse-strongly accretive mapping. Thus, by Lemma 2.4, we have

$${\parallel (I-\lambda A)x-(I-\lambda A)y\parallel }^2\le {\parallel x-y\parallel }^2+2\lambda (K^2\lambda -\frac{\alpha }{L^2}){\parallel Ax-Ay\parallel }^2.$$

Since ${\alpha }_n\to 0$ and $\lambda \in [a,b]\subset (0,\frac{\alpha }{K^2{L}^2})$, we get ${\alpha }_n<1-\frac{K^2{L}^2\lambda }{\alpha }$ for enough large n. Without loss of generality, we may assume that for all $n\in \mathbb{N}$, ${\alpha }_n<1-\frac{K^2{L}^2\lambda }{\alpha }$, i.e., $\frac{\lambda }{1-{\alpha }_n}\in (0,\frac{\alpha }{K^2{L}^2})$. Hence, $I-\frac{\lambda }{1-{\alpha }_n}A$ is nonexpansive.

From (3.1), we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & =& \parallel Q_C[(1-{\alpha }_n)x_n-\lambda A{x}_n]-Q_C[{\alpha }_{n}p+(1-{\alpha }_n)(p-\frac{\lambda }{1-{\alpha }_n}Ap)]\parallel \\ \le & \parallel {\alpha }_n(-p)+(1-{\alpha }_n)[(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)-(p-\frac{\lambda }{1-{\alpha }_n}Ap)]\parallel \\ \le & {\alpha }_n\parallel p\parallel +(1-{\alpha }_n)\parallel (I-\frac{\lambda }{1-{\alpha }_n}A)x_n-(I-\frac{\lambda }{1-{\alpha }_n}A)p\parallel \\ \le & {\alpha }_n\parallel p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel .\end{array}$$

(3.2)

By (3.1) and (3.2), we have

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & \le & (1-\gamma )\parallel x_n-p\parallel +\gamma \parallel Q_C[x_n-\lambda A{y}_n+\mu (y_n-x_n)]-p\parallel \\ =& (1-\gamma )\parallel x_n-p\parallel +\gamma \parallel Q_C[(1-\mu )x_n+\mu (y_n-\frac{\lambda }{\mu }A{y}_n)]\\ -Q_C[(1-\mu )p+\mu (p-\frac{\lambda }{\mu }Ap)]\parallel \\ \le & (1-\gamma )\parallel x_n-p\parallel \\ +\gamma \parallel (1-\mu )(x_n-p)+\mu [(y_n-\frac{\lambda }{\mu }A{y}_n)-(p-\frac{\lambda }{\mu }Ap)]\parallel \\ \le & (1-\gamma )\parallel x_n-p\parallel +(1-\mu )\gamma \parallel x_n-p\parallel \\ +\mu \gamma \parallel (y_n-\frac{\lambda }{\mu }A{y}_n)-(p-\frac{\lambda }{\mu }Ap)\parallel \\ \le & (1-\mu \gamma )\parallel x_n-p\parallel +\mu \gamma \parallel y_n-p\parallel \\ \le & (1-\mu \gamma )\parallel x_n-p\parallel +\mu \gamma {\alpha }_n\parallel p\parallel +\mu \gamma (1-{\alpha }_n)\parallel x_n-p\parallel \\ =& (1-\mu \gamma {\alpha }_n)\parallel x_n-p\parallel +\mu \gamma {\alpha }_n\parallel p\parallel \\ \le & max\{\parallel x_n-p\parallel ,\parallel p\parallel \}\\ ⋮\\ \le & max\{\parallel x_0-p\parallel ,\parallel p\parallel \}.\end{array}$$

(3.3)

Hence, $\{x_n\}$ is bounded.

Set $z_n=Q_C[x_n-\lambda A{y}_n+\mu (y_n-x_n)]$. From (3.1), we have $x_{n+1}=(1-\gamma )x_n+\gamma z_n$ for all $n\ge 0$. Then we have

$$\begin{array}{rcl}\parallel y_n-y_{n-1}\parallel & =& \parallel Q_C[(1-{\alpha }_n)x_n-\lambda A{x}_n]-Q_C[(1-{\alpha }_{n-1})x_{n-1}-\lambda A{x}_{n-1}]\parallel \\ \le & \parallel (1-{\alpha }_n)(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)-(1-{\alpha }_{n-1})(x_{n-1}-\frac{\lambda }{1-{\alpha }_{n-1}}A{x}_{n-1})\parallel \\ \le & (1-{\alpha }_n)\parallel (x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)-(x_{n-1}-\frac{\lambda }{1-{\alpha }_n}A{x}_{n-1})\parallel \\ +|{\alpha }_n-{\alpha }_{n-1}|\parallel x_{n-1}\parallel \\ \le & (1-{\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel x_{n-1}\parallel ,\end{array}$$

and thus

$$\begin{array}{rcl}\parallel z_n-z_{n-1}\parallel & =& \parallel Q_C[x_n-\lambda A{y}_n+\mu (y_n-x_n)]-Q_C[x_{n-1}-\lambda A{y}_{n-1}+\mu (y_{n-1}-x_{n-1})]\parallel \\ \le & (1-\mu )\parallel x_n-x_{n-1}\parallel +\mu \parallel (y_n-\frac{\lambda }{\mu }A{y}_n)-(y_{n-1}-\frac{\lambda }{\mu }A{y}_{n-1})\parallel \\ \le & (1-\mu )\parallel x_n-x_{n-1}\parallel +\mu \parallel y_n-y_{n-1}\parallel \\ \le & (1-\mu {\alpha }_n)\parallel x_n-x_{n-1}\parallel +|{\alpha }_n-{\alpha }_{n-1}|\parallel x_{n-1}\parallel .\end{array}$$

It follows that

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

This together with Lemma 2.6 implies that

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

From (3.2), we have

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\parallel {\alpha }_n(-p)+(1-{\alpha }_n)[(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)-(p-\frac{\lambda }{1-{\alpha }_n}Ap)]\parallel }^2\\ \le & {\alpha }_n{\parallel p\parallel }^2+(1-{\alpha }_n){\parallel (x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)-(p-\frac{\lambda }{1-{\alpha }_n}Ap)\parallel }^2\\ \le & {\alpha }_n{\parallel p\parallel }^2+(1-{\alpha }_n){\parallel x_n-p\parallel }^2+2\lambda (\frac{K^2\lambda }{1-{\alpha }_n}-\frac{\alpha }{L^2}){\parallel A{x}_n-Ap\parallel }^2.\end{array}$$

(3.4)

From (3.1), (3.3) and (3.4), we obtain

Therefore, we have

$$\begin{array}{rcl}0& \le & -2\gamma \lambda \mu (\frac{K^2\lambda }{1-{\alpha }_n}-\frac{\alpha }{L^2}){\parallel A{x}_n-Ap\parallel }^2-2\gamma \lambda \mu (\frac{K^2\lambda }{\mu }-\frac{\alpha }{L^2}){\parallel A{y}_n-Ap\parallel }^2\\ \le & {\alpha }_n\gamma \mu {\parallel p\parallel }^2+{\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\\ =& {\alpha }_n\gamma \mu {\parallel p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )(\parallel x_n-p\parallel -\parallel x_{n+1}-p\parallel )\\ \le & {\alpha }_n\gamma \mu {\parallel p\parallel }^2+(\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel .\end{array}$$

Since ${\alpha }_n\to 0$ and $\parallel x_n-x_{n+1}\parallel \to 0$ , we obtain

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

It follows that

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

Since A is α-strongly accretive, we deduce

$$\parallel A{y}_n-A{x}_n\parallel \ge \alpha \parallel y_n-x_n\parallel ,$$

which implies that

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

that is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel Q_C[(1-{\alpha }_n)x_n-\lambda A{x}_n]-x_n\parallel =0.$$

It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel Q_C[x_n-\lambda A{x}_n]-x_n\parallel =0.$$

(3.5)

Next, we show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(x_n-Q^{\prime }(0))〉\ge 0.$$

(3.6)

To show (3.6), since $\{x_n\}$ is bounded, we can choose a sequence $\{x_{n_i}\}$ of $\{x_n\}$ converging weakly to z such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(x_n-Q^{\prime }(0))〉=\underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(x_{n_i}-Q^{\prime }(0))〉.$$

(3.7)

We first prove $z\in S(C,A)$. It follows that

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

(3.8)

By Lemma 2.5 and (3.8), we have $z\in F(Q_C(I-\lambda A))$, it follows from Lemma 2.3 that $z\in S(C,A)$.

Now, from (3.7) and Lemma 2.2, we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(x_n-Q^{\prime }(0))〉& =& \underset{i\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(x_{n_i}-Q^{\prime }(0))〉\\ =& 〈Q^{\prime }(0),j(z-Q^{\prime }(0))〉\\ \ge & 0.\end{array}$$

Noticing that $\parallel x_n-y_n\parallel \to 0$, we deduce that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈Q^{\prime }(0),j(y_n-Q^{\prime }(0))〉\ge 0.$$

Since $y_n=Q_C[(1-{\alpha }_n)(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)]$ and $Q^{\prime }(0)=Q_C[{\alpha }_n{Q}^{\prime }(0)+(1-{\alpha }_n)(Q^{\prime }(0)-\frac{\lambda }{1-{\alpha }_n}A{Q}^{\prime }(0))]$ for all $n\ge 0$, we can deduce from Lemma 2.2 that

$$〈Q_C[(1-{\alpha }_n)(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)]-[(1-{\alpha }_n)(x_n-\frac{\lambda }{1-{\alpha }_n}A{x}_n)],j(y_n-Q^{\prime }(0))〉\le 0$$

and

Therefore, we have

which implies that

$${\parallel y_n-Q^{\prime }(0)\parallel }^2\le (1-{\alpha }_n){\parallel x_n-Q^{\prime }(0)\parallel }^2+2{\alpha }_n〈-Q^{\prime }(0),j(y_n-Q^{\prime }(0))〉.$$

(3.9)

Finally, we will prove that the sequence $x_n\to Q^{\prime }(0)$. As a matter of fact, from (3.1) and (3.9), we have

Applying Lemma 2.6 to the last inequality, we conclude that $x_n$ converges strongly to $Q^{\prime }(0)$. This completes the proof. □