An iterative method for variational inequality problems

Abstract

In this paper, we present some properties of generalized proximity operators andpropose an iterative method of approximating solutions for a class ofgeneralized variational inequalities and show its convergence in uniformlyconvex and smooth Banach spaces.

MSC: 47J20, 46B20, 46N10, 47N10, 49J40.

1 Introduction

Let f be a lower semi-continuous proper convex function from a Hilbert spaceH to $(-\mathrm{\infty },+\mathrm{\infty }]$. The Moreau envelope of the function f isdefined as

$$e^f(x)=\underset{y\in H}{inf}\{f(y)+\frac{1}{2}{\parallel x-y\parallel }^2\}.$$

(1.1)

It is well known that $e^f(x)$ is a continuous convex function, and for every$x\in H$, the infimum in (1.1) is achieved at a unique point${prox}_f(x)$. The operator ${prox}_f$ from H to H, i.e.,

$${prox}_{f}x=\underset{y\in H}{arg\; \hspace{0.17em}min}\{f(y)+\frac{1}{2}{\parallel x-y\parallel }^2\}$$

(1.2)

thus defined, is called the proximity operator of f. When$f={\iota }_K$ is the indicator function of a closed convex setK in H, then ${prox}_f(x)=P_K(x)$ becomes the metric projection operator onK.

In 1994, Alber extended the metric projection operator to uniformly convex anduniformly smooth Banach spaces. Let K be a closed convex subset of auniformly convex and uniformly smooth Banach space X, Alber [1] introduced the generalized projections ${\pi }_K:X^{\ast }\to K$ and ${\mathrm{\Pi }}_K:X\to K$,

$${\pi }_K\left(x^{\ast }\right)=\underset{x\in K}{arg\hspace{0.17em}min}\{{\parallel x^{\ast }\parallel }^2-2〈x^{\ast },x〉+{\parallel x\parallel }^2\}$$

and

$${\mathrm{\Pi }}_K(x)=\underset{y\in K}{arg\hspace{0.17em}min}\{{\parallel Jx\parallel }^2-2〈Jx,y〉+{\parallel y\parallel }^2\},$$

where J is the duality mapping from X to $X^{\ast }$, and studied their properties in detail. In [2], Alber presented some applications of the generalized projections toapproximately solving variational inequalities in Banach spaces. Recently, Li [3] extended the generalized projection operator ${\pi }_K$ from uniformly convex and uniformly smooth Banachspaces to reflexive Banach spaces and studied some properties of the generalizedprojection operator with applications to solving the variational inequality inBanach spaces. By employing the generalized projection operators, Zeng and Yao [4] established some existence results for the variational inequality problemin uniformly convex and uniformly smooth Banach spaces and convergence results forthe variational inequality. In [5], Wu and Huang further introduced and studied a class of generalizedf-projection operators in Banach spaces. As applications, they proposedan iterative method of approximating solutions for the variational inequalityproblem: find $\overline{x}\in K$ such that

$$〈A\overline{x},y-\overline{x}〉+f(y)-f(\overline{x})\ge 0,\mathrm{\forall }y\in K,$$

(1.3)

where K is a nonempty closed convex subset of X,$A:K\to X^{\ast }$ is a mapping and $f:X\to (-\mathrm{\infty },+\mathrm{\infty }]$ is a proper convex, lower semicontinuous andpositively homogeneous function, via

$$x_{n+1}={\pi }_K^f(J{x}_n-{\alpha }_{n}J(x_n-{\pi }_K^f(J{x}_n-\rho A{x}_n))),$$

(1.4)

where

$${\pi }_K^f\left(x^{\ast }\right)=\underset{x\in K}{arg\hspace{0.17em}min}\{2\rho f(x)+{\parallel x^{\ast }\parallel }^2-2〈x^{\ast },x〉+{\parallel x\parallel }^2\},$$

and the parameter sequence $\{{\alpha }_n\}$ satisfies

$$0\le {\alpha }_n\le 1,\sum _{n=0}^{+\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=+\mathrm{\infty },\rho >0,$$

and they proved that $\{x_n\}$ has a subsequence converging to a solution of (1.3)when K is a nonempty compact convex subset of a uniformly convex anduniformly smooth Banach space.

Motivated and inspired by the above works, we continue to study some properties ofgeneralized proximity operators and propose an iterative method of approximatingsolutions for the following generalized variational inequality problem: find$\overline{x}\in domf$ such that

$$〈A\overline{x},y-\overline{x}〉+f(y)-f(\overline{x})\ge 0,\mathrm{\forall }y\in domf,$$

(1.5)

where $f:X\to (-\mathrm{\infty },+\mathrm{\infty }]$ is a proper convex and lower semicontinuous function,$A:X\to X^{\ast }$ is a norm-to-weak continuous operator. Our iterativemethod is different from that given in [5]. We also prove a convergence result for this iterative method in smoothand uniformly convex Banach spaces. Let K be a nonempty closed convex setof X. If we replace f by $f+I_K$ in (1.5), where $I_K$ is the indicator function of K, then (1.5)reduces to (1.3).

2 Preliminaries

Let X be a reflexive, smooth and strictly convex Banach space with the dualspace $X^{\ast }$. We denote by $x_n\to x$ and $x_n\rightharpoonup x$ the strong and the weak convergence to x ofa sequence $\{x_n\}$ in a Banach space X, respectively. Let${\mathrm{\Gamma }}_0(X)$ denote the class of all lower semi-continuous properconvex functions from X to $(-\mathrm{\infty },+\mathrm{\infty }]$. Let $B(x,\delta )$ denote the closed ball of $x\in X$ and radius $\delta >0$. Let $S(X)=\{x\in X:\parallel x\parallel =1\}$ be the unit sphere.

A Banach space X is said to be strictly convex if $\frac{1}{2}\parallel x+y\parallel <1$ for all $x,y\in S(X)$ and $x\ne y$. The Banach space X is said to be smoothprovided

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

exists for each $x,y\in S(X)$. We recall that uniform convexity of X meansthat for any given $ϵ>0$, there exists $\delta >0$ such that for all $x,y\in X$ with $\parallel x\parallel \le 1$, $\parallel y\parallel \le 1$, and $\parallel x-y\parallel =ϵ$, the inequality

$$\parallel x+y\parallel \le 2(1-\delta )$$

holds.

A subset C of X is called boundedly compact if for any$\delta >0$ the intersection $C\cap B(0,\delta )$ is empty or compact.

The duality mapping $J:X\rightrightarrows X^{\ast }$ is defined by

$$J(x)=\{x^{\ast }\in X^{\ast }|〈x^{\ast },x〉={\parallel x^{\ast }\parallel }^2={\parallel x\parallel }^2\},\mathrm{\forall }x\in X.$$

The following basic results concerning the duality mapping are well known [2, 6, 7]:

• (1) X is reflexive if and only if J is surjective;

• (2) X is strictly convex if and only if J is injective;

• (3) X is smooth if and only if J is single-valued;

• (4) if X is smooth, then J is norm-to-weak star continuous;

• (5) J is monotone, i.e., $〈Jx-Jy,x-y〉\ge 0$, $\mathrm{\forall }x,y\in X$;

• (6) if X is strictly convex and smooth, then $〈Jx-Jy,x-y〉=0\Rightarrow x=y$, $\mathrm{\forall }x,y\in X$;

• (7) if a Banach space X is reflexive strictly convex and smooth, then the duality mapping $J^{\ast }$ from $X^{\ast }$ into X is the inverse of J, that is, $J^{-1}=J^{\ast }$.

Consider the following envelope function:

$$e_V^f\left(x^{\ast }\right)=\underset{x\in X}{inf}\{f(x)+\frac{1}{2}V(x^{\ast },x)\},$$

(2.1)

where $V(x^{\ast },x)={\parallel x^{\ast }\parallel }^2-2〈x^{\ast },x〉+{\parallel x\parallel }^2$. Since the function $(x,x^{\ast })\to f(x)+\frac{1}{2}V(x^{\ast },x)$ is lower semicontinuous convex, one sees that$e_V^f(x^{\ast })$ is lower semicontinuous and convex byProposition 4.4 in [8].

For every $x^{\ast }\in X^{\ast }$, the infimum in (2.1) is achieved at a unique point${\pi }_f(x^{\ast })$, i.e.,

$${\pi }_f\left(x^{\ast }\right):=\underset{x\in K}{arg\; \hspace{0.17em}min}\{f(x)+\frac{1}{2}V(x^{\ast },x)\}.$$

The operator ${\pi }_f$ is called the generalized proximity operator. It canbe characterized by the inclusion

$$x^{\ast }-J{\pi }_f\left(x^{\ast }\right)\in \partial f\left({\pi }_f\left(x^{\ast }\right)\right),$$

(2.2)

equivalently,

$${\pi }_f={(J+\partial f)}^{-1}.$$

(2.3)

From (2.3), we easily know that ${\pi }_f$ is maximal monotone by Theorem 2.6.2 in [6]. Observe that when $\rho =1$,

$${\pi }_K^f\left(x^{\ast }\right)={\pi }_{f+I_K}\left(x^{\ast }\right).$$

If, in addition, $domf=K$, then ${\pi }_K^f(x^{\ast })={\pi }_f(x^{\ast })$.

Lemma 2.1 ([9])

LetXbe a smooth, strictly convex and reflexive Banach space,let$\{x_n\}$be a sequence inX, and$x\in X$. If$〈x_n-x,J{x}_n-Jx〉\to 0$, then$x_n\rightharpoonup x$, $J{x}_n\rightharpoonup Jx$and$\parallel x_n\parallel \to \parallel x\parallel$.

Lemma 2.2 ([10])

Let$r>0$be a fixed real number. Then a Banach spaceXis uniformly convex if and only if there is a continuous, strictlyincreasing and convex function$g:R^{+}\to R^{+}$with$g(0)=0$such that

$${\parallel \lambda x+(1-\lambda )y\parallel }^2\le \lambda {\parallel x\parallel }^2+(1-\lambda ){\parallel y\parallel }^2-\lambda (1-\lambda )g(\parallel x-y\parallel ),\mathrm{\forall }x,y\in B_r,0\le \lambda \le 1,$$

where$B_r=\{x\in X:\parallel x\parallel \le r\}$.

3 Main results

Proposition 3.1Let$f\in {\mathrm{\Gamma }}_0(X)$. Then the following hold:

• (i) $〈{\pi }_f(x^{\ast })-{\pi }_f(y^{\ast }),(x^{\ast }-J{\pi }_f(x^{\ast }))-(y^{\ast }-J{\pi }_f(y^{\ast }))〉\ge 0$, $\mathrm{\forall }{x}^{\ast },y^{\ast }\in X^{\ast }$;

• (ii) ${\pi }_f$is bounded on each nonempty bounded subset of$C\subset X^{\ast }$;

• (iii) if$\{x_n^{\ast }\}$is a sequence in$X^{\ast }$such that$x_n^{\ast }\to x^{\ast }$, then${\pi }_f(x_n^{\ast })\rightharpoonup {\pi }_f(x^{\ast })$, $J{\pi }_f(x_n^{\ast })\rightharpoonup J{\pi }_f(x^{\ast })$and$\parallel {\pi }_f(x_n^{\ast })\parallel \to \parallel {\pi }_f(x^{\ast })\parallel$;

• (iv) if domfis a nonempty boundedly compact convex subset, then${\pi }_f$is weak-to-norm continuous, that is, if$x_n^{\ast }\rightharpoonup x^{\ast }$, then${\pi }_f(x_n^{\ast })\to {\pi }_f(x^{\ast })$.

Proof (i) Take $x^{\ast },y^{\ast }\in X^{\ast }$. Then (2.2) yields

$$〈x^{\ast }-J{\pi }_f\left(x^{\ast }\right),{\pi }_f\left(y^{\ast }\right)-{\pi }_f\left(x^{\ast }\right)〉\le f\left({\pi }_f\left(y^{\ast }\right)\right)-f\left({\pi }_f\left(x^{\ast }\right)\right)$$

and

$$〈y^{\ast }-J{\pi }_f\left(y^{\ast }\right),{\pi }_f\left(x^{\ast }\right)-{\pi }_f\left(y^{\ast }\right)〉\le f\left({\pi }_f\left(x^{\ast }\right)\right)-f\left({\pi }_f\left(y^{\ast }\right)\right).$$

Adding these two inequalities, we obtain

$$〈{\pi }_f\left(x^{\ast }\right)-{\pi }_f\left(y^{\ast }\right),(x^{\ast }-J{\pi }_f\left(x^{\ast }\right))-(y^{\ast }-J{\pi }_f\left(y^{\ast }\right))〉\ge 0.$$

• (ii) Suppose that ${\pi }_f$ is not bounded on some nonempty bounded subset of C. Then there exists a bounded sequence $\{x_n^{\ast }\}\subset C$ such that $\parallel {\pi }_f(x_n^{\ast })\parallel \to \mathrm{\infty }$. Fix $x^{\ast }\in X^{\ast }$. From (i), we obtain the following:

  $$\begin{array}{rcl}\parallel {\pi }_f\left(x_n^{\ast }\right)-{\pi }_f\left(x^{\ast }\right)\parallel \parallel x_n^{\ast }-x^{\ast }\parallel & \ge & 〈{\pi }_f\left(x_n^{\ast }\right)-{\pi }_f\left(x^{\ast }\right),x_n^{\ast }-x^{\ast }〉\\ \ge & 〈{\pi }_f\left(x_n^{\ast }\right)-{\pi }_f\left(x^{\ast }\right),J{\pi }_f\left(x_n^{\ast }\right)-J{\pi }_f\left(x^{\ast }\right)〉\\ =& \frac{1}{2}(V(J{\pi }_f\left(x_n^{\ast }\right),{\pi }_f\left(x^{\ast }\right))+V(J{\pi }_f\left(x^{\ast }\right),{\pi }_f\left(x_n^{\ast }\right)))\\ \ge & {(\parallel {\pi }_f\left(x_n^{\ast }\right)\parallel -\parallel {\pi }_f\left(x^{\ast }\right)\parallel )}^2.\end{array}$$

So, we have $\parallel x_n^{\ast }\parallel \to \mathrm{\infty }$. This is a contradiction.

• (iii) Let $\{x_n^{\ast }\}$ be a sequence in $X^{\ast }$ such that $x_n^{\ast }\to x^{\ast }$. It follows from (ii) that $\{{\pi }_f(x_n^{\ast })\}$ is bounded. From (i), we have

  $$0\le 〈{\pi }_f\left(x_n^{\ast }\right)-{\pi }_f\left(x^{\ast }\right),J{\pi }_f\left(x_n^{\ast }\right)-J{\pi }_f\left(x^{\ast }\right)〉\le \parallel {\pi }_f\left(x_n^{\ast }\right)-{\pi }_f\left(x^{\ast }\right)\parallel \parallel x_n^{\ast }-x^{\ast }\parallel \to 0.$$

Thus, Lemma 2.1 implies that ${\pi }_f(x_n^{\ast })\rightharpoonup {\pi }_f(x^{\ast })$, $J{\pi }_f(x_n^{\ast })\rightharpoonup J{\pi }_f(x^{\ast })$ and $\parallel {\pi }_f(x_n^{\ast })\parallel \to \parallel {\pi }_f(x^{\ast })\parallel$.

• (iv) From (2.2), we know that

  $$〈x_n^{\ast }-J{\pi }_f\left(x_n^{\ast }\right),y-{\pi }_f\left(x_n^{\ast }\right)〉\le f(y)-f\left({\pi }_f\left(x_n^{\ast }\right)\right),\mathrm{\forall }y\in domf.$$

  (3.1)

Since $x_n^{\ast }\rightharpoonup x^{\ast }$, $\{x_n^{\ast }\}$ is bounded. It follows from (ii) that$\{{\pi }_f(x_n^{\ast })\}$ is bounded. Since domf is boundedly compact,there exists a subsequence $\{x_{n_i}^{\ast }\}$ of $\{x_n^{\ast }\}$ such that

$${\pi }_f\left(x_{n_i}^{\ast }\right)\to \overline{x}\in domf\text{as }i\to +\mathrm{\infty }.$$

Since J is norm-to-weak star continuous and f is lowersemicontinuous, we obtain that

$$〈x^{\ast }-J\overline{x},y-\overline{x}〉\le f(y)-f(\overline{x}),\mathrm{\forall }y\in domf.$$

(3.2)

Then, by (2.2), we have $\overline{x}={\pi }_f(x^{\ast })$. Similar to the above arguments, we know that${\pi }_f(x^{\ast })$ is the unique limit point of $\{{\pi }_f(x_n^{\ast })\}$. Hence, ${\pi }_f(x_n^{\ast })\to {\pi }_f(x^{\ast })$. □

With the help of the operator ${\pi }_f$, we can show that the envelope function$e_V^f$ is Gâteaux differentiable.

Proposition 3.2Let$f\in {\mathrm{\Gamma }}_0(X)$. Then$e_V^f$is Gâteaux differentiable and$\mathrm{\nabla }{e}_V^f(x^{\ast })=J^{\ast }{x}^{\ast }-{\pi }_f(x^{\ast })$.

Proof For any $h\in X^{\ast }$, by definitions of $e_V^f$ and ${\pi }_f$, we have

$$\begin{array}{c}\frac{e_V^f(x^{\ast }+th)-e_V^f(x^{\ast })}{t}\hfill \\ =\frac{f({\pi }_f(x^{\ast }+th))+\frac{1}{2}V({\pi }_f(x^{\ast }+th),x^{\ast }+th)-f({\pi }_f(x^{\ast }))-\frac{1}{2}V({\pi }_f(x^{\ast }),x^{\ast })}{t}\hfill \\ \le \frac{f({\pi }_f(x^{\ast }))+\frac{1}{2}V({\pi }_f(x^{\ast }),x^{\ast }+th)-f({\pi }_f(x^{\ast }))-\frac{1}{2}V({\pi }_f(x^{\ast }),x^{\ast })}{t}\hfill \\ =\frac{\frac{1}{2}{\parallel x^{\ast }+th\parallel }^2-\frac{1}{2}{\parallel x^{\ast }\parallel }^2-〈{\pi }_f(x^{\ast }),th〉}{t}.\hfill \end{array}$$

Since $\mathrm{\nabla }({\parallel z\parallel }^2)=2J^{\ast }(z)$, for any $z\in X^{\ast }$, we get that

$$\begin{array}{rcl}\underset{t\to 0}{lim\hspace{0.17em}sup}\frac{e_V^f(x^{\ast }+th)-e_V^f(x^{\ast })}{t}& \le & \underset{t\to 0}{lim}\frac{\frac{1}{2}{\parallel x^{\ast }+th\parallel }^2-\frac{1}{2}{\parallel x^{\ast }\parallel }^2}{t}-〈{\pi }_f\left(x^{\ast }\right),h〉\\ =& 〈J^{\ast }{x}^{\ast }-{\pi }_f\left(x^{\ast }\right),h〉.\end{array}$$

On the other hand,

$$\begin{array}{c}\frac{e_V^f(x^{\ast }+th)-e_V^f(x^{\ast })}{t}\hfill \\ =\frac{f({\pi }_f(x^{\ast }+th))+\frac{1}{2}V({\pi }_f(x^{\ast }+th),x^{\ast }+th)-f({\pi }_f(x^{\ast }))-\frac{1}{2}V({\pi }_f(x^{\ast }),x^{\ast })}{t}\hfill \\ \ge \frac{f({\pi }_f(x^{\ast }+th))+\frac{1}{2}V({\pi }_f(x^{\ast }+th),x^{\ast }+th)-f({\pi }_f(x^{\ast }+th))-\frac{1}{2}V({\pi }_f(x^{\ast }+th),x^{\ast })}{t}\hfill \\ =\frac{\frac{1}{2}{\parallel x^{\ast }+th\parallel }^2-\frac{1}{2}{\parallel x^{\ast }\parallel }^2}{t}-〈{\pi }_f(x^{\ast }+th),h〉.\hfill \end{array}$$

By Proposition 3.1(iii), we have ${\pi }_f(x^{\ast }+th)\rightharpoonup {\pi }_f(x^{\ast })$ as $t\to 0$. Hence, we get that

$$\begin{array}{rcl}\underset{t\to 0}{lim\hspace{0.17em}inf}\frac{e_V^f(x^{\ast }+th)-e_V^f(x^{\ast })}{t}& \ge & \underset{t\to 0}{lim}\frac{\frac{1}{2}{\parallel x^{\ast }+th\parallel }^2-\frac{1}{2}{\parallel x^{\ast }\parallel }^2}{t}-〈{\pi }_f(x^{\ast }+th),h〉\\ =& 〈J^{\ast }{x}^{\ast }-{\pi }_f\left(x^{\ast }\right),h〉.\end{array}$$

This implies that

$$\underset{t\to 0}{lim}\frac{e_V^f(x^{\ast }+th)-e_V^f(x^{\ast })}{t}=〈J^{\ast }{x}^{\ast }-{\pi }_f\left(x^{\ast }\right),h〉.$$

Hence $e_V^f$ is Gâteaux differentiable and$\mathrm{\nabla }{e}_V^f(x^{\ast })=J^{\ast }{x}^{\ast }-{\pi }_f(x^{\ast })$. □

In the following, we propose a modification of the iterative method given in [5] and prove that the iterative sequence has a subsequence converging to asolution of (1.5) when X is a smooth and uniformly convex Banach space andf is not necessarily positively homogeneous.

By (2.2), we can easily prove the following result.

Proposition 3.3Let$f\in {\mathrm{\Gamma }}_0(X)$. Then the point$\overline{x}\in domf$is a solution of the variational inequality

$$〈Ax,y-x〉+f(y)-f(x)\ge 0,\mathrm{\forall }y\in domf$$

if and only if$\overline{x}\in domf$is a solution of the following inclusion:

$$x={\pi }_f(Jx-Ax).$$

The following lemma will be used in proving the convergence of the iterative methodfor variational inequality problem (1.5).

Lemma 3.1Let$f\in {\mathrm{\Gamma }}_0(X)$. If$f(x)\ge 0$for all$x\in domf$and$f(0)=0$, then

$$\parallel {\pi }_f\left(x^{\ast }\right)\parallel \le \parallel x^{\ast }\parallel .$$

(3.3)

Proof From (2.2), we know that

$$〈x^{\ast }-J{\pi }_f\left(x^{\ast }\right),y-{\pi }_f\left(x^{\ast }\right)〉\le f(y)-f\left({\pi }_f\left(x^{\ast }\right)\right),\mathrm{\forall }y\in domf.$$

Noticing that $f(x)\ge 0$ for all $x\in domf$ and $f(0)=0$, it follows that

$$〈x^{\ast }-J{\pi }_f\left(x^{\ast }\right),-{\pi }_f\left(x^{\ast }\right)〉\le -f\left({\pi }_f\left(x^{\ast }\right)\right)\le 0.$$

Hence,

$${\parallel {\pi }_f\left(x^{\ast }\right)\parallel }^2\le 〈x^{\ast },{\pi }_f\left(x^{\ast }\right)〉,$$

and hence

$$\parallel {\pi }_f\left(x^{\ast }\right)\parallel \le \parallel x^{\ast }\parallel .$$

 □

Proposition 3.4LetXbe a smooth and uniformly convex Banach space. Let$A:X\to X^{\ast }$be a norm-to-weak continuous operator. Supposethat$f\in {\mathrm{\Gamma }}_0(X)$and domfis nonempty boundedly compact convex. Suppose that

• (i) $f(x)\ge 0$for all$x\in domf$and$f(0)=0$;

• (ii) for any$x\in domf$,

  $$\parallel Jx-Ax\parallel \le \parallel x\parallel .$$

Let$x_0\in domf$and the sequence$\{x_n\}$be generated by the following iteration scheme:

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{\pi }_f(J{x}_n-A{x}_n),$$

where$\{{\alpha }_n\}$satisfies the conditions:

• (a) $0\le {\alpha }_n\le 1$for all$n=0,1,2,\dots$ ;

• (b) ${\sum }_{n=0}^{+\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=+\mathrm{\infty }$.

Then generalized variational inequality (1.5) has a solution$\overline{x}\in domf$, and there exists a subsequence$\{x_{n_i}\}$of$\{x_n\}$such that$x_{n_i}\to \overline{x}$as$i\to \mathrm{\infty }$.

Proof By (3.3), we have

$$\parallel {\pi }_f(J{x}_n-A{x}_n)\parallel \le \parallel J{x}_n-A{x}_n\parallel .$$

(3.4)

By (3.4) and condition (ii), we obtain

$$\parallel x_{n+1}\parallel \le (1-{\alpha }_n)\parallel x_n\parallel +{\alpha }_n\parallel {\pi }_f(J{x}_n-A{x}_n)\parallel \le \parallel x_n\parallel .$$

Then $\{x_n\}$ and $\{{\pi }_f(J{x}_n-A{x}_n)\}$ are bounded. Hence, by Lemma 2.2, there exists acontinuous, strictly increasing and convex function $g:R^{+}\to R^{+}$ with $g(0)=0$ such that

$$\begin{array}{rcl}{\parallel x_{n+1}\parallel }^2& =& {\parallel (1-{\alpha }_n)x_n+{\alpha }_n{\pi }_f(J{x}_n-A{x}_n)\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n\parallel }^2+{\alpha }_n{\parallel {\pi }_f(J{x}_n-A{x}_n)\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right).\end{array}$$

(3.5)

It follows from (3.5), (3.4) and condition (ii) that

$${\parallel x_{n+1}\parallel }^2\le (1-{\alpha }_n){\parallel x_n\parallel }^2+{\alpha }_n{\parallel x_n\parallel }^2-{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right).$$

That is,

$${\parallel x_{n+1}\parallel }^2\le {\parallel x_n\parallel }^2-{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right).$$

(3.6)

Taking the sum for $n=0,1,2,\dots ,m$ in (3.6), we get

$$\begin{array}{c}\sum _{n=0}^m{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right)\hfill \\ \le {\parallel x_0\parallel }^2-{\parallel x_{m+1}\parallel }^2\hfill \\ \le {\parallel x_0\parallel }^2.\hfill \end{array}$$

Hence,

$$\sum _{n=0}^{+\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right)<+\mathrm{\infty }.$$

(3.7)

Due to the condition ${\sum }_{n=0}^{+\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=+\mathrm{\infty }$, we may assume, without loss of generality, that

$$g\left(\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \right)\to 0\text{as }n\to +\mathrm{\infty }.$$

Applying the properties of g, we can deduce that

$$\parallel x_n-{\pi }_f(J{x}_n-A{x}_n)\parallel \to 0\text{as }n\to +\mathrm{\infty }.$$

(3.8)

Since domf is boundedly compact, there exists a subsequence$\{x_{n_i}\}$ of $\{x_n\}$ such that

$$x_{n_i}\to \overline{x}\in domf\text{as }i\to +\mathrm{\infty }.$$

Since A is norm-to-weak continuous and J is norm-to-weak starcontinuous, we get that

$$J{x}_{n_i}-A{x}_{n_i}\rightharpoonup J\overline{x}-A\overline{x}\text{as }i\to +\mathrm{\infty }.$$

Since ${\pi }_f$ is weak-to-norm continuous byProposition 3.1(iv),

$${\pi }_f(J{x}_{n_i}-A{x}_{n_i})\to {\pi }_f(J\overline{x}-A\overline{x})\text{as }i\to +\mathrm{\infty }.$$

Hence, (3.8) yields

$$\overline{x}={\pi }_f(J\overline{x}-A\overline{x}).$$

Now it follows from Proposition 3.3 that $\overline{x}$ is a solution of generalized variationalinequality (1.5). □

4 Application

Let $f\in {\mathrm{\Gamma }}_0(X)$ and let $g:X\to \mathbb{R}$ be a convex and Gâteaux differentiable function.Consider the optimization problem

$$\underset{x\in X}{min}f(x)+g(x).$$

(P)

We denote by $Sol(\mathrm{P})$ the solution set of problem (P). Despite itssimplicity, problem (P) has been shown to cover a wide range of apparently unrelatedsignal recovery formulations (see [11, 12]).

Notice that

$$\begin{array}{rcl}\overline{x}\in Sol(\mathrm{P})& \iff & 0\in \partial f(\overline{x})+\mathrm{\nabla }g(\overline{x})\\ \iff & -\mathrm{\nabla }g(\overline{x})\in \partial f(\overline{x})\\ \iff & 〈\mathrm{\nabla }g(\overline{x}),y-\overline{x}〉+f(y)-f(\overline{x})\ge 0,\mathrm{\forall }y\in X.\end{array}$$

Note that if g is convex and Gâteaux differentiable, then∇g is norm-to-weak continuous from X to$X^{\ast }$ by Corollary 3.1 in [13]. Therefore, as an application of Proposition 3.4, we have thefollowing result.

Proposition 4.1LetXbe a smooth and uniformly convex Banach space. Let$g:X\to \mathbb{R}$be convex and Gâteaux differentiable. Suppose that$f\in {\mathrm{\Gamma }}_0(X)$and domfis a nonempty boundedly compact convex subset ofX. Suppose that

• (i) $f(x)\ge 0$for all$x\in domf$and$f(0)=0$;

• (ii) for any$x\in domf$,

  $$\parallel Jx-\mathrm{\nabla }g(x)\parallel \le \parallel x\parallel .$$

Let$x_0\in domf$and the sequence$\{x_n\}$be generated by the following iteration scheme:

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{\pi }_f(J{x}_n-\mathrm{\nabla }g(x_n)),$$

where$\{{\alpha }_n\}$satisfies the conditions:

• (a) $0\le {\alpha }_n\le 1$for all$n=0,1,2,\dots$ ;

• (b) ${\sum }_{n=0}^{+\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=+\mathrm{\infty }$.

Then problem (P) has a solution$\overline{x}$and there exists a subsequence$\{x_{n_i}\}$of$\{x_n\}$such that$x_{n_i}\to \overline{x}$as$i\to \mathrm{\infty }$.

5 Concluding remark

This paper has improved the iterative method of Wu and Huang [5] for solving generalized variational inequality problem (1.5), severalresults regarding the generalized proximity operator and its relations with theenvelope function are presented. In addition, it is shown that under an appropriateassumption some optimization problem can be transformed into (1.5) and then theiterative method can be applied.