Hybrid method for a class of accretive variational inequalities involving nonexpansive mappings

Abstract

In this paper we use contractions to regularize a class of accretive variational inequalities and prove the strongly convergence in Banach spaces. We extend the result of Lu et al. (Nonlinear Anal. 71:1032-1041, 2009) to the framework of Banach spaces.

MSC:47H05, 47H09, 65J15.

1 Introduction

Let H be a Hilbert space, C be a nonempty closed convex subset of H, and $F:C\to H$ a nonlinear mapping. The set of fixed points of F is denoted by $Fix(F)$, i.e., $Fix(F)=\{x\in C:Fx=x\}$. A monotone variational inequality problem is to find a point $x^{\ast }$ with the property

$$x^{\ast }\in C,\text{such that }〈F{x}^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,$$

(1)

where F is a monotone operator.

Recently, Lu et al. [1] were concerned with a special class of variational inequalities in which the mapping F is the complement of a nonexpansive mapping and the constraint set is the set of fixed points of another nonexpansive mapping. Namely, they considered the following type of monotone variational inequality (VI) problem:

$$\text{Find }{x}^{\ast }\in Fix(T),\text{such that }〈(I-V)x^{\ast },x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in Fix(T),$$

(2)

where $T,V:C\to C$ are nonexpansive mappings and $Fix(T)\ne \mathrm{\varnothing }$.

Hybrid methods for solving VI (1) were studied by Yamada [2], where F is Lipschitzian and strongly monotone. However, his methods do not apply to the variational inequality (2) since the mapping $I-V$ fails, in general, to be strongly monotone, though it is Lipschitzian. Therefore, other hybrid methods have to be sought. Recently, Moudafi and Mainge [3] studied VI (2) by regularizing the mapping $tS+(1-t)T$ and defined $\{x_{s,t}\}$ as the unique fixed point of the equation

$$x_{s,t}=sf(x_{s,t})+(1-s)[tS(x_{s,t})+(1-t)T(x_{s,t})],s,t\in (0,1).$$

(3)

Since Moudafi and Mainge’s regularization depends on t, the convergence of the scheme (3) is more complicated. Very recently, Lu et al. [1] studied VI (2) by regularizing the mapping S and defined $\{x_{s,t}\}$ as the unique fixed point of the equation

$$x_{s,t}=s[tf(x_{s,t})+(1-t)S(x_{s,t})]+(1-s)T(x_{s,t}),s,t\in (0,1).$$

(4)

Note that Lu et al.’s regularization (4) no longer depends on t.

Motivated and inspired by the result of Lu et al. [1], we put forward a question: Can this implicit hybrid method [1] in Hilbert spaces be extended to the framework of Banach spaces? In this paper, we give a positive answer.

Throughout this paper, we always assume that E is a real Banach space. Let C be a nonempty closed convex subset of E. Let $F:C\to E$ be a nonlinear mapping.

In this paper, we consider the following type of accretive variational inequality problem:

$$x^{\ast }\in Fix(T),\text{such that }〈(I-S)x^{\ast },j(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in Fix(T),$$

(5)

where $S,T:C\to C$ are two nonexpansive mappings with the set of fixed point $Fix(T)\ne \mathrm{\varnothing }$. Let Ω denote the set of solutions of VI (5) and assume that Ω is nonempty.

2 Preliminaries

Let E be a real Banach space and J be the normalized duality mapping from E into $2^{E^{\ast }}$ given by

$$J(x)=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉=\parallel x\parallel \parallel x^{\ast }\parallel ,\parallel x\parallel =\parallel x^{\ast }\parallel \}$$

for all $x\in E$, where $E^{\ast }$ denotes the dual space of E and $〈\cdot ,\cdot 〉$ the generalized duality pairing between E and $E^{\ast }$.

Let C be a nonempty closed convex subset of a real Banach space E. Recall the following concepts of mappings.

• (i) A mapping $f:C\to C$ is a ρ-contraction if $\rho \in [0,1)$ and the following property is satisfied:

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

• (ii) A mapping $T:C\to C$ is nonexpansive provided

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

• (iii) A mapping $F:C\to E$ is

  • (a) accretive if for any $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

    $$〈Fx-Fy,j(x-y)〉\ge 0;$$

  • (b) strictly accretive if F is accretive and the equality in (a) holds if and only if $x=y$;

  • (c) β-strongly accretive if for any $x,y\in C$ there exists $j(x-y)\in J(x-y)$ such that

    $$〈Fx-Fy,j(x-y)〉\ge \beta {\parallel x-y\parallel }^2$$

for some real constant $\beta >0$.

Let $\phi :[0,\mathrm{\infty }):=R^{+}\to R^{+}$ be a continuous strictly increasing function such that $\phi (0)=0$ and $\phi (t)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$. This function φ is called a gauge function. The duality mapping $J_{\phi }:E\to E^{\ast }$ associated with a gauge function φ is defined by

$$J_{\phi }(x)=\{x^{\ast }\in E^{\ast }:〈x,x^{\ast }〉=\parallel x\parallel \phi (\parallel x\parallel ),\parallel x^{\ast }\parallel =\phi (\parallel x\parallel )\},\mathrm{\forall }x\in E.$$

In the case that $\phi (t)=t$, $J_{\phi }=J$, where J is the normalized duality mapping. Clearly, the relation $J_{\phi }(x)=\frac{\phi (\parallel x\parallel )}{\parallel x\parallel }J(x)$, $\mathrm{\forall }x\ne 0$ holds (see [4]).

Following Browder [4], we say that a Banach space E has a weakly continuous duality mapping if there exists a gauge φ for which the duality mapping $J_{\phi }(x)$ is single valued and weak-to-weak^∗ sequentially continuous (i.e., if $\{x_n\}$ is a sequence in E weakly convergent to a point x, then the sequence $J_{\phi }(x_n)$ converges weakly^∗ to $J_{\phi }(x)$). It is well known that $l^p$ has a weakly continuous duality mapping with a gauge function $\phi (t)=t^{p-1}$ for all $1<p<\mathrm{\infty }$. Set

$$\mathrm{\Phi }(t)={\int }_0^t\phi (\tau )d\tau ,t>0,$$

then

$$J_{\phi }(x)=\partial \mathrm{\Phi }(\parallel x\parallel ),\mathrm{\forall }x\in E,$$

where ∂ denotes the sub-differential in the sense of convex analysis.

Remark 2.1 If $J_{\phi }$ is weak-to-weak^∗ sequentially continuous, then J is strong-to-weak^∗ sequentially continuous.

Indeed, if $x_n\to x$ strongly, then $x_n\to x$ weakly, $J_{\phi }(x_n)$ converges weakly^∗ to $J_{\phi }(x)$ and $\phi (\parallel x_n\parallel )\to \phi (\parallel x\parallel )$ strongly. Since $J_{\phi }(x)\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}=J(x)$, $\mathrm{\forall }x\ne 0$, for any $y\in E$, we have

$$\begin{array}{c}|〈y,J(x_n)〉-〈y,J(x)〉|\hfill \\ =|〈y,J_{\phi }(x_n)\frac{\parallel x_n\parallel }{\phi (\parallel x_n\parallel )}〉-〈y,J_{\phi }(x)\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}〉|\hfill \\ \le |〈y,J_{\phi }(x_n)\frac{\parallel x_n\parallel }{\phi (\parallel x_n\parallel )}〉-〈y,J_{\phi }(x_n)\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}〉|+|〈y,J_{\phi }(x_n)\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}〉-〈y,J_{\phi }(x)\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}〉|\hfill \\ \le \parallel y\parallel \phi (\parallel x_n\parallel )\frac{|\parallel x_n\parallel \phi (\parallel x\parallel )-\parallel x\parallel \phi (\parallel x_n\parallel )|}{\phi (\parallel x_n\parallel )\phi (\parallel x\parallel )}+\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}|〈y,J_{\phi }(x_n)〉-〈y,J_{\phi }(x)〉|\hfill \\ \le \parallel y\parallel \frac{\parallel x_n\parallel |\phi (\parallel x\parallel )-\phi (\parallel x_n\parallel )|+|\parallel x_n\parallel -\parallel x\parallel |\phi (\parallel x_n\parallel )}{\phi (\parallel x\parallel )}\hfill \\ +\frac{\parallel x\parallel }{\phi (\parallel x\parallel )}|〈y,J_{\phi }(x_n)〉-〈y,J_{\phi }(x)〉|.\hfill \end{array}$$

Letting $n\to \mathrm{\infty }$, we have

$$\underset{n\to \mathrm{\infty }}{lim}〈y,J(x_n)〉=〈y,J(x)〉,$$

i.e., J is strong-to-weak^∗ sequentially continuous.

Lemma 2.1 ([[5], Lemma 2.1])

Assume that a Banach space E has a weakly continuous duality mapping $J_{\phi }$ with a gauge φ. For all $x,y\in E$, the following inequality holds:

$$\mathrm{\Phi }(\parallel x+y\parallel )\le \mathrm{\Phi }(\parallel x\parallel )+〈y,J_{\phi }(x+y)〉.$$

In particular, for all $x,y\in E$,

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x+y)〉.$$

Lemma 2.2 (see [6])

Let C be a nonempty closed convex subset of a real Banach space E. Assume that $F:C\to E$ is accretive and weakly continuous along segments; that is $F(x+ty)\rightharpoonup F(x)$ as $t\to 0$. Then the variational inequality

$$x^{\ast }\in C,〈F{x}^{\ast },j(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C$$

is equivalent to the dual variational inequality

$$x^{\ast }\in C,〈Fx,j(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in C.$$

3 Main results

In this section, we introduce an implicit algorithm and prove this algorithm converges strongly to $x^{\ast }$ which solves VI (5). Let C be a nonempty closed convex subset of a real Banach space E. Let $f:C\to C$ be a contraction and $S,T:C\to C$ be two nonexpansive mappings. For $s,t\in (0,1)$, we define the following mapping:

$$x\mapsto W_{s,t}x:=s[tf(x)+(1-t)Sx]+(1-s)Tx.$$

It is obvious that $W_{s,t}:C\to C$ is a contraction. So the contraction $W_{s,t}$ has a unique fixed point which is denoted $x_{s,t}$. Namely,

$$x_{s,t}=s[tf(x_{s,t})+(1-t)S(x_{s,t})]+(1-s)T(x_{s,t}),s,t\in (0,1).$$

(6)

Theorem 3.1 Let C be a nonempty closed convex subset of a reflexive Banach space E which has a weakly continuous duality map $J_{\phi }(x)$ with the gauge φ. Let $f:C\to C$ be a contraction with constant $\rho >0$ and $S,T:C\to C$ be two nonexpansive mappings with $Fix(T)\ne \mathrm{\varnothing }$. Suppose that the solution set Ω of VI (5) is nonempty. Let, for each $(s,t)\in {(0,1)}^2$, $\{x_{s,t}\}$ be defined implicitly by (6). Then, for each fixed $t\in (0,1)$, the net $\{x_{s,t}\}$ converges in norm, as $s\to 0$, to a point $x_t\in Fix(T)$. Moreover, as $t\to 0$, the net $\{x_t\}$ converges in norm to the unique solution $x^{\ast }$ of the following variational inequality:

$$x^{\ast }\in \mathrm{\Omega },〈(I-f)x^{\ast },J(x-x^{\ast })〉\ge 0,\mathrm{\forall }x\in \mathrm{\Omega }.$$

(7)

Hence, for each null sequence $\{t_n\}$ in $(0,1)$, there exists another null sequence $\{s_n\}$ in $(0,1)$, such that the sequence $x_{s_n,t_n}\to x^{\ast }$ in norm as $n\to \mathrm{\infty }$.

Proof Step 1. For each fixed $t\in (0,1)$, the net $\{x_{s,t}\}$ is bounded.

For any $z\in Fix(T)$, we have

$$\begin{array}{c}\parallel s[t(f(x_{s,t})-f(z))+(1-t)(S(x_{s,t})-S(z))]+(1-s)(T(x_{s,t})-z)\parallel \hfill \\ \le s\parallel t(f(x_{s,t})-f(z))+(1-t)(S(x_{s,t})-S(z))\parallel +(1-s)\parallel T(x_{s,t})-Tz\parallel \hfill \\ \le st\parallel f(x_{s,t})-f(z)\parallel +(1-t)s\parallel S(x_{s,t})-S(z)\parallel +(1-s)\parallel x_{s,t}-z\parallel \hfill \\ \le st\rho \parallel x_{s,t}-z\parallel +(1-t)s\parallel x_{s,t}-z\parallel +(1-s)\parallel x_{s,t}-z\parallel \hfill \\ =(1-st(1-\rho ))\parallel x_{s,t}-z\parallel .\hfill \end{array}$$

Combining the above inequality and Lemma 2.1, we obtain

$$\begin{array}{rcl}\mathrm{\Phi }(\parallel x_{s,t}-z\parallel )& =& \mathrm{\Phi }(\parallel s[t(f(x_{s,t})-f(z))+(1-t)(S(x_{s,t})-S(z))]\\ +(1-s)(T(x_{s,t})-z)+st(f(z)-z)+s(1-t)(S(z)-z)\parallel )\\ \le & \mathrm{\Phi }\left(\parallel s[t(f(x_{s,t})-f(z))+(1-t)(S(x_{s,t})-S(z))]+(1-s)(T(x_{s,t})-z)\parallel \right)\\ +st〈f(z)-z,J_{\phi }(x_{s,t}-z)〉+(1-t)s〈S(z)-z,J_{\phi }(x_{s,t}-z)〉\\ \le & \mathrm{\Phi }((1-st(1-\rho ))\parallel x_{s,t}-z\parallel )+st〈f(z)-z,J_{\phi }(x_{s,t}-z)〉\\ +(1-t)s〈S(z)-z,J_{\phi }(x_{s,t}-z)〉\\ \le & (1-st(1-\rho ))\mathrm{\Phi }(\parallel x_{s,t}-z\parallel )+st〈f(z)-z,J_{\phi }(x_{s,t}-z)〉\\ +(1-t)s〈S(z)-z,J_{\phi }(x_{s,t}-z)〉,\end{array}$$

which implies that

$$\mathrm{\Phi }(\parallel x_{s,t}-z\parallel )\le \frac{t}{t(1-\rho )}〈f(z)-z,J_{\phi }(x_{s,t}-z)〉+\frac{1-t}{t(1-\rho )}〈S(z)-z,J_{\phi }(x_{s,t}-z)〉.$$

(8)

Taking $\phi (t)=t$, then $J_{\phi }=J$ and $\mathrm{\Phi }(t)=\frac{t^2}{2}$, from (8) we have

$$\begin{array}{rcl}{\parallel x_{s,t}-z\parallel }^2& \le & \frac{2t}{t(1-\rho )}〈f(z)-z,J(x_{s,t}-z)〉+\frac{2(1-t)}{t(1-\rho )}〈S(z)-z,J(x_{s,t}-z)〉\\ \le & \frac{2t}{t(1-\rho )}\parallel f(z)-z\parallel \parallel x_{s,t}-z\parallel +\frac{2(1-t)}{t(1-\rho )}\parallel S(z)-z\parallel \parallel x_{s,t}-z\parallel ,\end{array}$$

(9)

which implies that

$$\parallel x_{s,t}-z\parallel \le \frac{2}{t(1-\rho )}max\{\parallel f(z)-z\parallel ,\parallel S(z)-z\parallel \}.$$

So for each fixed $t\in (0,1)$, $\{x_{s,t}\}$ is bounded, furthermore $\{f(x_{s,t})\}$, $\{S(x_{s,t})\}$ and $\{T(x_{s,t})\}$ are all bounded.

Step 2. $x_{s,t}\to x_t\in Fix(T)$ as $s\to 0$.

From (6) and the boundedness of the sequences $\{f(x_{s,t})\}$, $\{S(x_{s,t})\}$ and $\{T(x_{s,t})\}$, for each fixed $t\in (0,1)$ we have

$$\parallel x_{s,t}-T{x}_{s,t}\parallel =s\parallel tf(x_{s,t})+(1-t)S(x_{s,t})-T{x}_{s,t}\parallel \to 0(s\to 0).$$

(10)

Assume that $\{s_n\}\subset (0,1)$ is such that $s_n\to 0$ ($n\to \mathrm{\infty }$). From (8), for any $z\in Fix(T)$, we have

$$\mathrm{\Phi }(\parallel x_{s_n,t}-z\parallel )\le \frac{t}{t(1-\rho )}〈f(z)-z,J_{\phi }(x_{s_n,t}-z)〉+\frac{1-t}{t(1-\rho )}〈S(z)-z,J_{\phi }(x_{s_n,t}-z)〉.$$

(11)

Since $\{x_{s_n,t}\}$ is bounded, without loss of generality, we may assume that $\{x_{s_n,t}\}$ converges weakly to a point $x_t$ as $n\to \mathrm{\infty }$. This together with (10) implies that $x_t\in Fix(T)$. Taking $z=x_t$ in (11), we have

$$\mathrm{\Phi }(\parallel x_{s_n,t}-x_t\parallel )\le \frac{1}{t(1-\rho )}〈tf(x_t)+(1-t)S(x_t)-x_t,J_{\phi }(x_{s_n,t}-x_t)〉.$$

(12)

Since $J_{\phi }$ is weakly continuous, it follows from (12) that $\mathrm{\Phi }(\parallel x_{s_n,t}-x_t\parallel )\to 0$ as $n\to \mathrm{\infty }$, which implies that $x_{s_n,t}\to x_t$ strongly. This has proved the relative norm compactness of the net $\{x_{s,t}\}$ as $s\to 0$.

Taking $s=s_n$ in (9), we have

$${\parallel x_{s_n,t}-z\parallel }^2\le \frac{2t}{t(1-\rho )}〈f(z)-z,J(x_{s_n,t}-z)〉+\frac{2(1-t)}{t(1-\rho )}〈S(z)-z,J(x_{s_n,t}-z)〉.$$

Since $J_{\phi }$ is weakly continuous, then by Remark 2.1, J is strong-to-weak^∗ sequentially continuous. Let $s_n\to 0$ in the above inequality, we have

$${\parallel x_t-z\parallel }^2\le \frac{2t}{t(1-\rho )}〈f(z)-z,J(x_t-z)〉+\frac{2(1-t)}{t(1-\rho )}〈S(z)-z,J(x_t-z)〉.$$

Hence we obtain

$$x_t\in Fix(T),〈tf(z)+(1-t)S(z)-z,J(x_t-z)〉\ge 0,\mathrm{\forall }z\in Fix(T).$$

This together with Lemma 2.2, we have

$$x_t\in Fix(T),〈tf(x_t)+(1-t)S(x_t)-x_t,J(x_t-z)〉\ge 0,\mathrm{\forall }z\in Fix(T).$$

(13)

Next, we prove that the entire net $\{x_{s,t}\}$ converges strongly to $x_t$ as $s\to 0$. We assume that $x_{s_n^{\prime },t}\to x_t^{\prime }$ where $s_n^{\prime }\to 0$. Similar to the above proof, we have $x_t^{\prime }\in Fix(T)$ and

$$x_t^{\prime }\in Fix(T),〈tf\left(x_t^{\prime }\right)+(1-t)S\left(x_t^{\prime }\right)-x_t^{\prime },J(x_t^{\prime }-z)〉\ge 0,\mathrm{\forall }z\in Fix(T).$$

(14)

Taking $z=x^{\prime }(t)$ and $z=x_t$ in (13) and (14), respectively, we have

$$\begin{array}{c}t〈f(x_t)-x_t,J(x_t-x_t^{\prime })〉+(1-t)〈S(x_t)-x_t,J(x_t-x_t^{\prime })〉\ge 0,\hfill \\ t〈f\left(x_t^{\prime }\right)-x_t^{\prime },J(x_t^{\prime }-x_t)〉+(1-t)〈S\left(x_t^{\prime }\right)-x_t^{\prime },J(x_t^{\prime }-x_t)〉\ge 0.\hfill \end{array}$$

Adding up the above two inequalities yields

$$t〈(I-f)x_t-(I-f)x_t^{\prime },J(x_t-x_t^{\prime })〉+(1-t)〈(I-S)x_t-(I-S)x_t^{\prime },J(x_t-x_t^{\prime })〉\le 0.$$

Since

$$\begin{array}{c}〈(I-f)x_t-(I-f)x_t^{\prime },J(x_t-x_t^{\prime })〉\ge (1-\rho ){\parallel x_t-x_t^{\prime }\parallel }^2,\hfill \\ 〈(I-S)x_t-(I-S)x_t^{\prime },J(x_t-x_t^{\prime })〉\ge 0,\hfill \end{array}$$

we obtain

$$(1-\rho ){\parallel x_t-x_t^{\prime }\parallel }^2\le 0,$$

i.e., $x_t=x_t^{\prime }$. So the entire net $\{x_{s,t}\}$ converges in norm to $x_t\in Fix(T)$ as $s\to 0$.

Step 3. The net $\{x_t\}$ is bounded.

For any $y\in \mathrm{\Omega }$, taking $z=y$ in (13), we have

$$〈tf(x_t)+(1-t)S(x_t)-x_t,J(x_t-y)〉\ge 0,$$

which together with the fact of $y\in \mathrm{\Omega }$ implies that

$$\begin{array}{c}t〈(I-f)x_t-(I-f)y,J(x_t-y)〉+(1-t)〈(I-S)x_t-(I-S)y,J(x_t-y)〉\hfill \\ \le t〈f(y)-y,J(x_t-y)〉+(1-t)〈Sy-y,J(x_t-y)〉\hfill \\ \le t〈f(y)-y,J(x_t-y)〉.\hfill \end{array}$$

(15)

Since $I-f$ is strongly accretive and $I-S$ is accretive, we obtain

$$〈(I-f)x_t-(I-f)y,J(x_t-y)〉\ge (1-\rho ){\parallel x_t-y\parallel }^2,$$

(16)

$$〈(I-S)x_t-(I-S)y,J(x_t-y)〉\ge 0.$$

(17)

It follows from (15)-(17) that

$$\begin{array}{rcl}{\parallel x_t-y\parallel }^2& \le & \frac{1}{1-\rho }〈f(y)-y,J(x_t-y)〉\\ \le & \frac{1}{1-\rho }\parallel f(y)-y\parallel \parallel x_t-y\parallel .\end{array}$$

(18)

Hence we have

$$\parallel x_t-y\parallel \le \frac{1}{1-\rho }\parallel f(y)-y\parallel ,\mathrm{\forall }t\in (0,1).$$

Step 4. The net $x_t\to x^{\ast }\in \mathrm{\Omega }$ which solves VI (7).

First, the uniqueness of the solution of VI (7) is obvious. We denote the unique solution by $x^{\ast }$.

Next we prove that ${\omega }_w(x_t)\subset \mathrm{\Omega }$, i.e., if $\{t_n\}$ is a null sequence in $(0,1)$ such that $x_{t_n}\to x^{\prime }$ weakly as $n\to \mathrm{\infty }$, then $x^{\prime }\in \mathrm{\Omega }$. Indeed, since $\{x_t\}\subset Fix(T)$, then $x^{\prime }\in Fix(T)$. Since $I-S$ is accretive, for any $z\in Fix(T)$ we have

$$〈(I-S)z,J(z-x_t)〉\ge 〈(I-S)x_t,J(z-x_t)〉.$$

(19)

It follows from (13) that

$$〈(I-S)x_t,J(z-x_t)〉\ge \frac{t}{1-t}〈(I-f)x_t,J(x_t-z)〉.$$

(20)

By virtue of (19) and (20), we have

$$〈(I-S)z,J(z-x_t)〉\ge \frac{t}{1-t}〈(I-f)x_t,J(x_t-z)〉,$$

furthermore, we get

$$\begin{array}{rcl}〈(I-S)z,J_{\phi }(z-x_t)〉& \ge & \frac{t}{(1-t)}〈(I-f)x_t,J_{\phi }(x_t-z)〉,\\ \ge & -\frac{t}{(1-t)}(\parallel x_t\parallel +\parallel f(x_t)\parallel )\phi (\parallel x_t-z\parallel ).\end{array}$$

Letting $t=t_n\to 0$ ($n\to \mathrm{\infty }$) in the above inequality, since $\{x_t\}$ is bounded and φ is a continuous strictly increasing function, we have

$$〈(I-S)z,J_{\phi }(z-x^{\prime })〉\ge 0,\mathrm{\forall }z\in Fix(T).$$

This implies that

$$〈(I-S)z,J(z-x^{\prime })〉\ge 0,\mathrm{\forall }z\in Fix(T),$$

hence from the above inequality and Lemma 2.2, we have

$$〈(I-S)x^{\prime },J(z-x^{\prime })〉\ge 0,\mathrm{\forall }z\in Fix(T),$$

i.e., $x^{\prime }\in \mathrm{\Omega }$.

Next we show that $x^{\prime }$ is the solution of VI (7). Taking $y=x^{\prime }$ and $t=t_n$ in (18), we obtain

$$\begin{array}{rcl}{\parallel x_{t_n}-x^{\prime }\parallel }^2& \le & \frac{1}{1-\rho }〈f\left(x^{\prime }\right)-x^{\prime },J(x_{t_n}-x^{\prime })〉\\ =& \frac{1}{1-\rho }〈f\left(x^{\prime }\right)-x^{\prime },J_{\phi }(x_{t_n}-x^{\prime })〉\frac{\parallel x_{t_n}-x^{\prime }\parallel }{\phi (\parallel x_{t_n}-x^{\prime }\parallel )},\end{array}$$

which implies that

$$\parallel x_{t_n}-x^{\prime }\parallel \phi \left(\parallel x_{t_n}-x^{\prime }\parallel \right)\le \frac{1}{1-\rho }〈f\left(x^{\prime }\right)-x^{\prime },J_{\phi }(x_{t_n}-x^{\prime })〉.$$

(21)

Since $x_{t_n}\to x^{\prime }$ weakly and $J_{\phi }$ is weakly continuous, let $t_n\to 0$ in (21), we get

$$\parallel x_{t_n}-x^{\prime }\parallel \phi \left(\parallel x_{t_n}-x^{\prime }\parallel \right)\to 0(n\to \mathrm{\infty }),$$

which together with the property of φ implies that $x_{t_n}\to x^{\prime }$ in norm. It follows from (15) and (17) that

$$〈(I-f)x_t,J(x_t-y)〉\le 0.$$

(22)

Since J is strong-to-weak^∗ sequentially continuous and f is a contraction, we have

$$\begin{array}{c}|〈(I-f)x_{t_n},J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x^{\prime }-y)〉|\hfill \\ \le |〈(I-f)x_{t_n},J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x_{t_n}-y)〉|\hfill \\ +|〈(I-f)x^{\prime },J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x^{\prime }-y)〉|\hfill \\ \le |〈(I-f)x_{t_n},J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x_{t_n}-y)〉|\hfill \\ +|〈(I-f)x^{\prime },J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x^{\prime }-y)〉|\hfill \\ \le (1+\rho )\parallel x_{t_n}-x^{\prime }\parallel \parallel x_{t_n}-y\parallel \hfill \\ +|〈(I-f)x^{\prime },J(x_{t_n}-y)〉-〈(I-f)x^{\prime },J(x^{\prime }-y)〉|\to 0(n\to \mathrm{\infty }).\hfill \end{array}$$

(23)

Letting $t=t_n\to 0$ ($n\to \mathrm{\infty }$) in (22) and combining (23) we have

$$〈(I-f)x^{\prime },J(x^{\prime }-y)〉\le 0,\mathrm{\forall }y\in \mathrm{\Omega }.$$

So $x^{\prime }$ is the solution of VI (7). By uniqueness, we have $x^{\prime }=x^{\ast }$. Therefore, $x_t\to x^{\ast }$ in norm as $t\to 0$. The proof is complete. □