A strong convergence theorem for equilibrium problems and split feasibility problems in Hilbert spaces

Abstract

The main purpose of this paper is to introduce an iterative algorithm for equilibrium problems and split feasibility problems in Hilbert spaces. Under suitable conditions we prove that the sequence converges strongly to a common element of the set of solutions of equilibrium problems and the set of solutions of split feasibility problems. Our result extends and improves the corresponding results of some others.

MSC:90C25, 90C30, 47J25, 47H09.

1 Introduction

Let H be a real Hilbert space with inner product $〈\cdot ,\cdot 〉$ and induced norm $\parallel \cdot \parallel$, let $F:H\times H\to \mathbb{R}$ be a bifunction. Then we consider the following equilibrium problem (EP): find $z\in H$ such that

$$F(z,y)\ge 0,\mathrm{\forall }y\in H.$$

(1.1)

The set of the EP is denoted by Ω, i.e.,

$$\mathrm{\Omega }=\{z\in H:F(z,y)\ge 0,\mathrm{\forall }y\in H\}.$$

The problem (1.1) is very general in the sense that it includes, as special cases, optimization problems, variational inequality problems, the Nash equilibrium problems and others, see, for instance, [1–3]. Some methods have been proposed to solve the EP, see, e.g., [4–6] and [7, 8].

The split feasibility problem (SFP) was proposed by Censer and Elfving in [9]. It can be formulated as the problem of finding a point x satisfying the property:

$$x\in C,Ax\in Q,$$

(1.2)

where A is a given $M\times N$ real matrix, and C and Q are nonempty, closed and convex subsets in ${\mathbb{R}}^N$ and ${\mathbb{R}}^M$, respectively.

Due to its extraordinary utility and broad applicability in many areas of applied mathematics (most notably, fully discretized models of problems in image reconstruction from projections, in image processing, and in intensity-modulated radiation therapy), algorithms for solving convex feasibility problems have been received great attention (see, for instance [10–13] and also [14–18]).

We assume the SFP (1.2) is consistent, and let Γ be the solution set, i.e.,

$$\mathrm{\Gamma }=\{x\in C:Ax\in Q\}.$$

It is not hard to see that Γ is closed convex and $x\in \mathrm{\Gamma }$ if and only if it solves the fixed-point equation

$$x=P_C(I-\gamma A^{\ast }(I-P_Q)A)x,$$

(1.3)

where $P_C$ and $P_Q$ are the orthogonal projection onto C and Q, respectively, $\gamma >0$ is any positive constant and $A^{\ast }$ denotes the adjoint of A.

Recently, for the purpose of generality, the SFP (1.2) has been studied in a more general setting. For instance, see [16, 19]. However, the algorithms in these references have only weak convergence in the setting of infinite-dimensional Hilbert spaces. Very recently, He and Zhao [20] introduce a new relaxed CQ algorithm (1.4) such that the strong convergence is guaranteed in infinite-dimensional Hilbert spaces:

$$x_{n+1}=P_{C_n}({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\tau }_n\mathrm{\nabla }{f}_n(x_n))).$$

(1.4)

Motivated and inspired by the research going on in the sections of equilibrium problems and split feasibility problems, the purpose of this article is to introduce an iterative algorithm for equilibrium problems and split feasibility problems in Hilbert spaces. Under suitable conditions we prove the sequence converges strongly to a common element of the set of solutions of equilibrium problems and the set of solutions of split feasibility problems. Our result extends and improves the corresponding results of He et al. [20] and some others.

2 Preliminaries and lemmas

Throughout this paper, we assume that H, $H_1$ or $H_2$ is a real Hilbert space, A is a bounded linear operator from $H_1$ to $H_2$, and I is the identity operator on H, $H_1$ or $H_2$. If $f:H\to \mathbb{R}$ is a differentiable function, then we denote by ∇f the gradient of the function f. We will also use the notations: → to denote strong convergence, ⇀ to denote weak convergence and to denote by

$$w_{\omega }(x_n)=\{x|\mathrm{\exists }\{x_{n_k}\}\subset \{x_n\}\text{ such that }{x}_{n_k}\rightharpoonup x\}$$

the weak ω-limit set of $\{x_n\}$.

Recall that a mapping $T:H\to H$ is said to be nonexpansive if

$$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,x,y\in H.$$

$T:H\to H$ is said to be firmly nonexpansive if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel (I-T)x-(I-T)y\parallel }^2,x,y\in H.$$

A mapping $T:H\to H$ is said to be demi-closed at origin, if for any sequence $\{x_n\}\subset H$ with $x_n\rightharpoonup x^{\ast }$ and ${lim}_{n\to \mathrm{\infty }}\parallel (I-T)x_n\parallel =0$, then $x^{\ast }=T{x}^{\ast }$.

It is easy to prove that if $T:H\to H$ is a firmly nonexpansive mapping, then T is demi-closed at the origin.

A function $f:H\to \mathbb{R}$ is called convex if

$$f(\lambda x+(1-\lambda )y)\le \lambda f(x)+(1-\lambda )f(y),\mathrm{\forall }\lambda \in (0,1),\mathrm{\forall }x,y\in H.$$

Lemma 2.1 Let $T:H_2\to H_2$ be a firmly nonexpansive mapping such that $\parallel (I-T)x\parallel$ is a convex function from $H_2$ to $\overline{\mathbb{R}}=[-\mathrm{\infty },+\mathrm{\infty }]$. Let $A:H_1\to H_2$ be a bounded linear operator and

$$f(x):=\frac{1}{2}{\parallel (I-T)Ax\parallel }^2,\mathrm{\forall }x\in H_1.$$

Then

• (i) $\mathrm{\nabla }f(x)=A^{\ast }(I-T)Ax$, $x\in H_1$.

• (ii) ∇f is ${\parallel A\parallel }^2$-Lipschitz, i.e., $\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel \le {\parallel A\parallel }^2\parallel x-y\parallel$, $x,y\in H_1$.

Proof (i) From the definition of f, we know that f is convex. First we prove that the limit

$$〈\mathrm{\nabla }f(x),v〉=\underset{h\to 0+}{lim}\frac{f(x+hv)-f(x)}{h}$$

exists in $\overline{\mathcal{R}}:=\{-\mathrm{\infty }\}\cup \mathbb{R}\cup \{+\mathrm{\infty }\}$ and satisfies

$$〈\mathrm{\nabla }f(x),v〉\le f(x+v)-f(x),\mathrm{\forall }v\in H_1.$$

If fact, if $0<h_1\le h_2$, then

$$f(x+h_{1}v)-f(x)=f(\frac{h_1}{h_2}(x+h_{2}v)+(1-\frac{h_1}{h_2})x)-f(x).$$

Since f is convex and $\frac{h_1}{h_2}\le 1$, it follows that

$$f(x+h_{1}v)-f(x)\le \frac{h_1}{h_2}f(x+h_{2}v)+(1-\frac{h_1}{h_2})f(x)-f(x),$$

and hence that

$$\frac{f(x+h_{1}v)-f(x)}{h_1}\le \frac{f(x+h_{2}v)-f(x)}{h_2}.$$

This shows that this difference quotient is increasing, therefore it has a limit in $\overline{\mathcal{R}}$ as $h\to 0+$:

$$〈\mathrm{\nabla }f(x),v〉=\underset{h>0}{inf}\frac{f(x+hv)-f(x)}{h}=\underset{h\to 0+}{lim}\frac{f(x+hv)-f(x)}{h}.$$

(2.1)

This implies that f is differential. Taking $h=1$, (2.1) implies that

$$〈\mathrm{\nabla }f(x),v〉\le f(x+v)-f(x).$$

Next we prove that

$$\mathrm{\nabla }f(x)=A^{\ast }(I-T)Ax,x\in H_1.$$

In fact, since

$$\underset{h\to 0+}{lim}\frac{f(x+hv)-f(x)}{h}=\underset{h\to 0+}{lim}\frac{{\parallel Ax+hAv-TA(x+hv)\parallel }^2-{\parallel (I-T)Ax\parallel }^2}{2h}$$

(2.2)

and

$$\begin{array}{r}{\parallel Ax+hAv-TA(x+hv)\parallel }^2-{\parallel (I-T)Ax\parallel }^2\\ ={\parallel Ax\parallel }^2+h^2{\parallel Av\parallel }^2+2h〈A^{\ast }Ax,v〉+{\parallel TA(x+hv)\parallel }^2-{\parallel Ax\parallel }^2-{\parallel TAx\parallel }^2\\ -2〈Ax,TA(x+hv)-TAx〉-2h〈A^{\ast }TA(x+hv),v〉.\end{array}$$

(2.3)

Substituting (2.3) into (2.2), simplifying and then letting $h\to 0+$ and taking the limit we have

$$\begin{array}{rl}\underset{h\to 0+}{lim}\frac{f(x+hv)-f(x)}{h}& =\underset{h\to 0+}{lim}\frac{2h\{〈A^{\ast }Ax,v〉-〈A^{\ast }TA(x+hv),v〉\}}{2h}\\ =〈A^{\ast }(I-T)Ax,v〉,\mathrm{\forall }v\in H_1.\end{array}$$

It follows from (2.1) that

$$\mathrm{\nabla }f(x)=A^{\ast }(I-T)Ax,x\in H_1.$$

(ii) From (i) we have

$$\begin{array}{rl}\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel & =\parallel A^{\ast }(I-T)Ax-A^{\ast }(I-T)Ay\parallel \\ =\parallel A^{\ast }[(I-T)Ax-(I-T)Ay]\parallel \\ \le \parallel A\parallel \parallel Ax-Ay\parallel \le {\parallel A\parallel }^2\parallel x-y\parallel ,x,y\in H_1.\end{array}$$

 □

Lemma 2.2 (See, for example, [21])

Let $T:H\to H$ be an operator. The following statements are equivalent.

• (i) T is firmly nonexpansive.

• (ii) ${\parallel Tx-Ty\parallel }^2\le 〈x-y,Tx-Ty〉$, $\mathrm{\forall }x,y\in H$.

• (iii) $I-T$ is firmly nonexpansive.

Proof (i) ⇒ (ii): Since T is firmly nonexpansive, for all $x,y\in H$ we have

$$\begin{array}{rl}{\parallel Tx-Ty\parallel }^2& \le {\parallel x-y\parallel }^2-{\parallel (I-T)x-(I-T)y\parallel }^2\\ ={\parallel x-y\parallel }^2-{\parallel x-y\parallel }^2-{\parallel Tx-Ty\parallel }^2+2〈x-y,Tx-Ty〉\\ =2〈x-y,Tx-Ty〉-{\parallel Tx-Ty\parallel }^2,\end{array}$$

hence

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

(ii) ⇒ (iii): From (ii), we know that for all $x,y\in H$

$$\begin{array}{rl}{\parallel (I-T)x-(I-T)y\parallel }^2& ={\parallel (x-y)-(Tx-Ty)\parallel }^2\\ ={\parallel x-y\parallel }^2-2〈x-y,Tx-Ty〉+{\parallel Tx-Ty\parallel }^2\\ \le {\parallel x-y\parallel }^2-〈x-y,Tx-Ty〉\\ =〈x-y,(I-T)x-(I-T)y〉.\end{array}$$

This implies that $I-T$ is firmly nonexpansive.

(iii) ⇒ (i): From (iii) we immediately know that T is firmly nonexpansive.

Let C be a nonempty closed convex subset of H. Recall that for every point $x\in H$, there exists a unique nearest point of C, denoted by $P_{C}x$, such that $\parallel x-P_{C}x\parallel \le \parallel x-y\parallel$ for all $y\in C$. Such a $P_C$ is called the metric projection from H onto C. We know that $P_C$ is a firmly nonexpansive mapping from H onto C, i.e.,

$${\parallel P_{C}x-P_{C}y\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉,\mathrm{\forall }x,y\in H.$$

Further, for any $x\in H$ and $z\in C$, $z=P_{C}x$ if and only if

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

(2.4)

Throughout this paper, let us assume that a bifunction $F:H\times H\to \mathbb{R}$ satisfies the following conditions:

(A1) $F(x,x)=0$, $\mathrm{\forall }x\in H$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$, $\mathrm{\forall }x,y\in H$;

(A3) ${lim}_{t\downarrow 0}F(tz+(1-t)x,y)\le F(x,y)$, $\mathrm{\forall }x,y,z\in H$;

(A4) for each $x\in H$, $y\mapsto F(x,y)$ is convex and lower semicontinuous.

 □

Lemma 2.3 ([1, 4])

Let H be a Hilbert space and let $F:H\times H\to \mathbb{R}$ satisfy (A1), (A2), (A3), and (A4). Then, for any $r>0$ and $x\in H$, there exists $z\in H$ such that

$$F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H.$$

Furthermore, if

$$T_{r}x=\{z\in H:F(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\mathrm{\forall }y\in H\},$$

then the following hold:

• (1) $T_r$ is single-valued;

• (2) $T_r$ is firmly nonexpansive;

• (3) $F(T_r)=\mathrm{\Omega }$;

• (4) Ω is closed and convex.

The following results play an important role in this paper.

Lemma 2.4 ([22])

Let X be a real Hilbert space, then we have

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,x+y〉,\mathrm{\forall }x,y\in X.$$

Lemma 2.5 ([23])

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences in a Banach space X. Let $\{{\beta }_n\}$ be a sequence in $[0,1]$ satisfying $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$. Suppose that

$$x_{n+1}=(1-{\beta }_n)y_n+{\beta }_n{x}_n$$

for all integer $n\ge 0$ and

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

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

Lemma 2.6 ([24])

Let $\{a_n\}$ be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\gamma }_n)a_n+{\gamma }_n{\sigma }_n,n=0,1,2,\dots ,$$

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

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

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

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

3 Main results

We are now in a position to prove the following theorem.

Theorem 3.1 Let $H_1$, $H_2$ be two real Hilbert spaces, $F:H_1\times H_1\to \mathbb{R}$ be a bifunction satisfying (A1), (A2), (A3), and (A4). Let $A:H_1\to H_2$ be a bounded linear operator, $S:H_1\to H_1$ be a firmly nonexpansive mapping, and let $T:H_2\to H_2$ be a firmly nonexpansive mapping such that $\parallel (I-T)x\parallel$ is a convex function from $H_2$ to ℝ. Assume that $C:=F(S)\cap \mathrm{\Omega }\ne \mathrm{\varnothing }$ and $Q:=F(T)\ne \mathrm{\varnothing }$. Let $u\in H_1$ and $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{x}_0\in H_1\mathit{\text{ chosen arbitrarily}},\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)y_n,\hfill \\ F(y_n,x)+\frac{1}{{\lambda }_n}〈x-y_n,y_n-z_n〉\ge 0,\mathrm{\forall }x\in H_1,\hfill \\ z_n=S({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))),\hfill \end{array}$$

(3.1)

where

$$\begin{array}{c}f(x_n)=\frac{1}{2}{\parallel (I-T)A{x}_n\parallel }^2,\mathrm{\nabla }f(x_n)=A^{\ast }(I-T)A{x}_n\ne 0\mathrm{\forall }n\ge 1,\hfill \\ {\xi }_n=\frac{{\rho }_{n}f(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}.\hfill \end{array}$$

If the solution set Γ of SPF (1.2) is not empty, and the sequences $\{{\rho }_n\}\subset (0,4)$, $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ satisfy the following conditions:

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

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

• (iii) ${\lambda }_n\in (a,b)\subset (0,+\mathrm{\infty })$ and ${lim}_{n\to \mathrm{\infty }}({\lambda }_{n+1}-{\lambda }_n)=0$,

then the sequence $\{x_n\}$ converges strongly to $P_{\mathrm{\Gamma }}u$.

Proof Since the solution set Ω of EP and the solution set of SPF (1.2) are both closed and convex, Γ (≠∅) is closed and convex. Thus, the metric projection $P_{\mathrm{\Gamma }}$ is well defined.

Letting $p=P_{\mathrm{\Gamma }}u$, it follows from Lemma 2.3 that $y_n=T_{{\lambda }_n}{z}_n$ and

$$\parallel y_n-p\parallel =\parallel T_{{\lambda }_n}{z}_n-T_{{\lambda }_n}p\parallel \le \parallel z_n-p\parallel .$$

(3.2)

Observing that S is firmly nonexpansive, we have

$$\begin{array}{rl}\parallel z_n-p\parallel & =\parallel S({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)))-p\parallel \\ \le \parallel {\alpha }_n(u-p)+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p)\parallel \\ \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p\parallel .\end{array}$$

(3.3)

Since $p\in \mathrm{\Gamma }\subset C$, $\mathrm{\nabla }f(p)=0$. Observe that $I-T$ is firmly nonexpansive, from Lemma 2.2(ii) we have

$$\begin{array}{rl}〈\mathrm{\nabla }f(x_n),x_n-p〉& =〈(I-T)A{x}_n,A{x}_n-Ap〉\\ \ge {\parallel (I-T)A{x}_n\parallel }^2=2f(x_n).\end{array}$$

(3.4)

This implies that

$$\begin{array}{rl}{\parallel x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p\parallel }^2& ={\parallel x_n-p\parallel }^2+{\parallel {\xi }_n\mathrm{\nabla }f(x_n)\parallel }^2-2{\xi }_n〈\mathrm{\nabla }f(x_n),x_n-p〉\\ \le {\parallel x_n-p\parallel }^2+{\xi }_n^2{\parallel \mathrm{\nabla }f(x_n)\parallel }^2-4{\xi }_{n}f(x_n)\\ ={\parallel x_n-p\parallel }^2-{\rho }_n(4-{\rho }_n)\frac{f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}\\ \le {\parallel x_n-p\parallel }^2.\end{array}$$

(3.5)

Substituting (3.5) into (3.3), we get

$$\parallel z_n-p\parallel \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel .$$

(3.6)

Thus, from (3.2) and (3.6) we have

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\beta }_n{x}_n+(1-{\beta }_n)y_n-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel y_n-p\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel z_n-p\parallel \\ \le (1-{\alpha }_n(1-{\beta }_n))\parallel x_n-p\parallel +{\alpha }_n(1-{\beta }_n)\parallel u-p\parallel .\end{array}$$

It turns out that

$$\parallel x_{n+1}-p\parallel \le max\{\parallel x_n-p\parallel ,\parallel u-p\parallel \}.$$

By induction, we have

$$\parallel x_n-p\parallel \le max\{\parallel x_0-p\parallel ,\parallel u-p\parallel \}.$$

This implies that the sequence $\{x_n\}$ is bounded. From (3.2) and (3.6) we know that $\{y_n\}$ and $\{z_n\}$ both are bounded.

From Lemma 2.4 and (3.5), we have

$$\begin{array}{rl}{\parallel z_n-p\parallel }^2=& {\parallel S({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)))-p\parallel }^2\\ \le & {\parallel {\alpha }_n(u-p)+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p)\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p\parallel }^2+2{\alpha }_n〈u-p,z_n-p〉\\ \le & (1-{\alpha }_n){\parallel x_n-p\parallel }^2+2{\alpha }_n〈u-p,z_n-p〉\\ -(1-{\alpha }_n){\rho }_n(4-{\rho }_n)\frac{f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}.\end{array}$$

(3.7)

Therefore, from Lemma 2.6 and (3.2), (3.7) we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2=& {\parallel {\beta }_n{x}_n+(1-{\beta }_n)y_n-p\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel z_n-p\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)(1-{\alpha }_n){\parallel x_n-p\parallel }^2+2{\alpha }_n(1-{\beta }_n)〈u-p,z_n-p〉\\ -(1-{\alpha }_n)(1-{\beta }_n){\rho }_n(4-{\rho }_n)\frac{f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}\\ =& {\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\beta }_n){\parallel x_n-p\parallel }^2+2{\alpha }_n(1-{\beta }_n)〈u-p,z_n-p〉\\ -(1-{\alpha }_n)(1-{\beta }_n){\rho }_n(4-{\rho }_n)\frac{f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}.\end{array}$$

(3.8)

On the other hand, without loss of generality, we may assume that there is a constant $\sigma >0$ such that

$$(1-{\alpha }_n)(1-{\beta }_n){\rho }_n(4-{\rho }_n)>\sigma ,\mathrm{\forall }n\ge 1.$$

Setting $s_n={\parallel x_n-p\parallel }^2$, we get the following inequality:

$$s_{n+1}-s_n+{\alpha }_n(1-{\beta }_n)s_n+\frac{\sigma f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}\le 2{\alpha }_n(1-{\beta }_n)〈u-p,z_n-p〉.$$

(3.9)

Now, we prove $s_n\to 0$ by employing the technique studied by Maingé [25]. For the purpose we consider two cases.

Case 1: $\{s_n\}$ is eventually decreasing, i.e., there exists a sufficient large positive integer $k\ge 1$ such that $s_n>s_{n+1}$ holds for all $n\ge k$. In this case, $\{s_n\}$ must be convergent, and from (3.9) it follows that

$$\frac{\sigma f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}\le (s_n-s_{n+1})+{\alpha }_n(1-{\beta }_n)M,$$

(3.10)

where M is a constant such that $M\ge 2\parallel z_n-p\parallel \parallel u-p\parallel$ for all $n\in \mathbb{N}$. Using the condition (i) and (3.10), we have

$$\frac{f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}\to 0(n\to \mathrm{\infty }).$$

(3.11)

Moreover, it follows from Lemma 2.1(ii) that for all $n\in \mathbb{N}$

$$\parallel \mathrm{\nabla }f(x_n)\parallel =\parallel \mathrm{\nabla }f(x_n)-\mathrm{\nabla }f(p)\parallel \le {\parallel A\parallel }^2\parallel x_n-p\parallel .$$

This implies that $\{\parallel \mathrm{\nabla }f(x_n)\parallel \}$ is bounded. From (3.11) it yields $f(x_n)\to 0$, namely

$$\parallel (I-T)A{x}_n\parallel \to 0.$$

(3.12)

Furthermore, we have

$$\underset{n\to \mathrm{\infty }}{lim}{\xi }_n=0.$$

(3.13)

For any $x^{\ast }\in w_{\omega }(x_n)$, and if $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$ such that $x_{n_k}\rightharpoonup x^{\ast }\in H_1$, then

$$A{x}_{n_k}\rightharpoonup A{x}^{\ast }.$$

(3.14)

On the other hand, from (3.12), we have

$$\parallel (I-T)A{x}_{n_k}\parallel \to 0.$$

(3.15)

Since T is demi-closed at origin, from (3.14) and (3.15) we have $A{x}^{\ast }\in F(T)$, i.e., $A{x}^{\ast }\in Q$.

In order to prove $x^{\ast }\in C=F(S)\cap \mathrm{\Omega }$, we need to prove ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-z_n\parallel =0$. In fact, from (3.1) we have

$$F(y_n,x)+\frac{1}{{\lambda }_n}〈x-y_n,y_n-z_n〉\ge 0,\mathrm{\forall }x\in H_1.$$

Taking $x=y_{n+1}$, we get

$$F(y_n,y_{n+1})+\frac{1}{{\lambda }_n}〈y_{n+1}-y_n,y_n-z_n〉\ge 0.$$

Similarly, we also have

$$F(y_{n+1},y_n)+\frac{1}{{\lambda }_{n+1}}〈y_n-y_{n+1},y_{n+1}-z_{n+1}〉\ge 0.$$

Adding up the above two inequalities, we get

$$F(y_n,y_{n+1})+F(y_{n+1},y_n)+〈y_{n+1}-y_n,\frac{y_n-z_n}{{\lambda }_n}-\frac{y_{n+1}-z_{n+1}}{{\lambda }_{n+1}}〉\ge 0.$$

From (A2), we have

$$〈y_{n+1}-y_n,\frac{y_n-z_n}{{\lambda }_n}-\frac{y_{n+1}-z_{n+1}}{{\lambda }_{n+1}}〉\ge 0.$$

Multiplying the above inequality by ${\lambda }_n$ and simplifying, we have

$$〈y_{n+1}-y_n,y_n-y_{n+1}+y_{n+1}-z_n-\frac{{\lambda }_n}{{\lambda }_{n+1}}(y_{n+1}-z_{n+1})〉\ge 0.$$

Hence we have

$$\begin{array}{rl}{\parallel y_{n+1}-y_n\parallel }^2& \le 〈y_{n+1}-y_n,y_{n+1}-z_n-\frac{{\lambda }_n}{{\lambda }_{n+1}}(y_{n+1}-z_{n+1})〉\\ =〈y_{n+1}-y_n,z_{n+1}-z_n+(1-\frac{{\lambda }_n}{{\lambda }_{n+1}})(y_{n+1}-z_{n+1})〉\\ \le \parallel y_{n+1}-y_n\parallel (\parallel z_{n+1}-z_n\parallel +|1-\frac{{\lambda }_n}{{\lambda }_{n+1}}|\cdot \parallel y_{n+1}-z_{n+1}\parallel )\end{array}$$

and hence

$$\begin{array}{rl}\parallel y_{n+1}-y_n\parallel & \le \parallel z_{n+1}-z_n\parallel +|1-\frac{{\lambda }_n}{{\lambda }_{n+1}}|\parallel y_{n+1}-z_{n+1}\parallel \\ \le \parallel z_{n+1}-z_n\parallel +\frac{1}{a}|{\lambda }_{n+1}-{\lambda }_n|\cdot \parallel y_{n+1}-z_{n+1}\parallel .\end{array}$$

By (3.1) we have

$$\begin{array}{rl}\parallel z_{n+1}-z_n\parallel =& \parallel S({\alpha }_{n+1}u+(1-{\alpha }_{n+1})(x_{n+1}-{\xi }_{n+1}\mathrm{\nabla }f(x_{n+1})))\\ -S({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)))\parallel \\ \le & \parallel ({\alpha }_{n+1}-{\alpha }_n)u\\ +(1-{\alpha }_{n+1})\{(x_{n+1}-{\xi }_{n+1}\mathrm{\nabla }f(x_{n+1}))-(x_n-{\xi }_n\mathrm{\nabla }f(x_n))\}\\ -({\alpha }_{n+1}-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))\parallel \\ \le & (1-{\alpha }_{n+1})\parallel x_{n+1}-x_n\parallel +N_n\le \parallel x_{n+1}-x_n\parallel +N_n,\end{array}$$

(3.16)

where

$$\begin{array}{rl}{N}_n=& |{\alpha }_{n+1}-{\alpha }_n|\cdot \parallel u\parallel +(1-{\alpha }_{n+1})({\xi }_{n+1}\parallel \mathrm{\nabla }f(x_{n+1})\parallel +{\xi }_n\parallel \mathrm{\nabla }f(x_n)\parallel )\\ +|{\alpha }_{n+1}-{\alpha }_n|\cdot \parallel x_n-{\xi }_n\mathrm{\nabla }f(x_n)\parallel \to 0(n\to \mathrm{\infty }).\end{array}$$

(3.17)

This implies that

$$\parallel y_{n+1}-y_n\parallel \le \parallel x_{n+1}-x_n\parallel +\frac{1}{a}|{\lambda }_{n+1}-{\lambda }_n|\cdot \parallel y_{n+1}-z_{n+1}\parallel +N_n.$$

It follows that

$$\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel \le \frac{1}{a}|{\lambda }_{n+1}-{\lambda }_n|\cdot \parallel y_{n+1}-z_{n+1}\parallel +N_n.$$

In view of condition (iii) and (3.17) we get

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

By Lemma 2.5, we obtain

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

(3.18)

Consequently

$$\begin{array}{rl}\parallel x_{n+1}-x_n\parallel & =\parallel {\beta }_n{x}_n+(1-{\beta }_n)y_n-x_n\parallel \\ =(1-{\beta }_n)\parallel y_n-x_n\parallel \to 0(n\to \mathrm{\infty }).\end{array}$$

(3.19)

Since S is firmly nonexpansive, it follows from (3.1) that

$$\begin{array}{rl}2{\parallel z_n-p\parallel }^2=& 2{\parallel S({\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)))-Sp\parallel }^2\\ \le & 2〈{\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))-p,z_n-p〉\\ =& {\parallel {\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))-p\parallel }^2+{\parallel z_n-p\parallel }^2\\ -{\parallel {\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))-p-z_n+p\parallel }^2\\ =& {\parallel {\alpha }_n(u-p)+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)-p)\parallel }^2+{\parallel z_n-p\parallel }^2\\ -{\parallel {\alpha }_n(u-z_n)+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)-z_n)\parallel }^2\\ \le & (1-{\alpha }_n){\parallel x_n-p\parallel }^2+{\parallel z_n-p\parallel }^2-{\parallel x_n-z_n\parallel }^2+M_n,\end{array}$$

where

$$\begin{array}{rl}{M}_n:=& {\alpha }_n{\parallel u-p\parallel }^2+(1-{\alpha }_n){\parallel {\xi }_n\mathrm{\nabla }f(x_n)\parallel }^2-2(1-{\alpha }_n){\xi }_n〈x_n-p,\mathrm{\nabla }f(x_n)〉\\ -{\alpha }_n{\parallel u-z_n\parallel }^2-(1-{\alpha }_n)\{{\parallel {\xi }_n\mathrm{\nabla }f(x_n)\parallel }^2-2〈x_n-z_n,{\xi }_n\mathrm{\nabla }f(x_n)〉\}\\ +{\alpha }_n{\parallel x_n-z_n\parallel }^2+{\alpha }_n(1-{\alpha }_n){\parallel x_n-u-{\xi }_n\mathrm{\nabla }f(x_n)\parallel }^2\\ \to & 0(\text{as }n\to \mathrm{\infty }).\end{array}$$

Therefore we have

$${\parallel z_n-p\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel x_n-z_n\parallel }^2+M_n.$$

This together with (3.8) shows that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& \le {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel z_n-p\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-(1-{\beta }_n){\parallel x_n-z_n\parallel }^2+(1-{\beta }_n)M_n.\end{array}$$

Then we obtain

$$\begin{array}{rl}(1-{\beta }_n){\parallel x_n-z_n\parallel }^2& \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+(1-{\beta }_n)M_n\\ =s_n-s_{n+1}+(1-{\beta }_n)M_n.\end{array}$$

Therefore, we get

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

(3.20)

By virtue of (3.18), we have

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

(3.21)

Now, we turn to a proof that $x^{\ast }\in C=F(S)\cap \mathrm{\Omega }$. For this purpose, we denote

$$v_n:={\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n)).$$

In view of condition (i) and (3.13) we have

$$\begin{array}{rl}\parallel v_n-x_n\parallel & =\parallel {\alpha }_{n}u+(1-{\alpha }_n)(x_n-{\xi }_n\mathrm{\nabla }f(x_n))-x_n\parallel \\ =\parallel {\alpha }_n(u-x_n)-(1-{\alpha }_n){\xi }_n\mathrm{\nabla }f(x_n)\parallel \\ \le {\alpha }_n\parallel u-x_n\parallel +(1-{\alpha }_n){\xi }_n\parallel \mathrm{\nabla }f(x_n)\parallel \to 0.\end{array}$$

(3.22)

Since S is firmly nonexpansive (and so it is also nonexpansive), it follows from Lemma 2.4 that

$$\begin{array}{rl}{\parallel z_{n+1}-p\parallel }^2& ={\parallel S{v}_{n+1}-S{x}_n+S{x}_n-Sp\parallel }^2\\ \le {\parallel S{x}_n-Sp\parallel }^2+2〈S{v}_{n+1}-S{x}_{n+1}+S{x}_{n+1}-S{x}_n,z_{n+1}-p〉\\ \le {\parallel x_n-p\parallel }^2-{\parallel (I-S)x_n\parallel }^2+2(\parallel v_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel )\parallel z_{n+1}-p\parallel .\end{array}$$

Thus, we have

$$\begin{array}{r}{\parallel (I-S)x_n\parallel }^2\\ \le {\parallel x_n-p\parallel }^2-{\parallel z_{n+1}-p\parallel }^2+2(\parallel v_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel )\parallel z_{n+1}-p\parallel \\ \le {\parallel x_n-p\parallel }^2-{(\parallel z_{n+1}-x_{n+1}\parallel -\parallel x_{n+1}-p\parallel )}^2\\ +2(\parallel v_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel )\parallel z_{n+1}-p\parallel \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2-{\parallel z_{n+1}-x_{n+1}\parallel }^2+2\parallel x_{n+1}-p\parallel \cdot \parallel z_{n+1}-x_{n+1}\parallel \\ +2(\parallel v_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel )\parallel z_{n+1}-p\parallel \\ =s_n-s_{n+1}-{\parallel z_{n+1}-x_{n+1}\parallel }^2+2\parallel x_{n+1}-p\parallel \cdot \parallel z_{n+1}-x_{n+1}\parallel \\ +2(\parallel v_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel )\parallel z_{n+1}-p\parallel .\end{array}$$

(3.23)

It follows from (3.19), (3.20), and (3.22) that $\parallel (I-S)x_n\parallel \to 0$. In view of $x_{n_k}\rightharpoonup x^{\ast }$ and that S is demi-closed at origin, we get $x^{\ast }\in F(S)$.

On the other hand, from $x_{n_k}\rightharpoonup x^{\ast }$ and (3.18), we obtain $y_{n_k}\rightharpoonup x^{\ast }$. From (3.1), for any $x\in H_1$, we have

$$F(y_n,x)+\frac{1}{{\lambda }_n}〈x-y_n,y_n-z_n〉\ge 0.$$

From (A2), we have

$$\frac{1}{{\lambda }_n}〈x-y_n,y_n-z_n〉\ge F(x,y_n),\mathrm{\forall }x\in H_1.$$

Replacing n by $n_k$, we have

$$〈x-y_{n_k},\frac{y_{n_k}-z_{n_k}}{{\lambda }_{n_k}}〉\ge F(x,y_{n_k}),\mathrm{\forall }x\in H_1.$$

Since $\parallel \frac{y_{n_k}-z_{n_k}}{{\lambda }_{n_k}}\parallel \to 0$ and $y_{n_k}\rightharpoonup x^{\ast }$, from (A4) we have

$$F(x,x^{\ast })\le 0,\mathrm{\forall }x\in H_1.$$

(3.24)

Put $w_t=tx+(1-t)x^{\ast }$ for all $t\in (0,1]$ and $x\in H_1$. Then we get $w_t\in H_1$. So, from (3.24) we have

$$F(w_t,x^{\ast })\le 0,\mathrm{\forall }x\in H_1.$$

From (A4), we have

$$\begin{array}{rl}0=F(w_t,w_t)& \le tF(w_t,x)+(1-t)F(w_t,x^{\ast })\\ \le tF(w_t,x),\end{array}$$

and hence $F(w_t,x)\ge 0$. Letting $t\to 0$, we have

$$F(x^{\ast },x)\ge 0,\mathrm{\forall }x\in H_1.$$

This implies $x^{\ast }\in \mathrm{\Omega }$. Consequently, $x^{\ast }\in C$, and hence $w_w(x_n)\subset \mathrm{\Gamma }$. Furthermore, in view of (3.20) we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,z_n-p〉& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,x_n-p〉\\ =\underset{w\in w_w(x_n)}{max}〈u-P_{\mathrm{\Gamma }}u,w-P_{\mathrm{\Gamma }}u〉\le 0.\end{array}$$

On the other hand, from (3.9), we have

$$s_{n+1}\le (1-{\alpha }_n(1-{\beta }_n))s_n+2{\alpha }_n(1-{\beta }_n)〈u-p,z_n-p〉.$$

(3.25)

Applying Lemma 2.6 to (3.25), from the condition (i) we obtain $s_n\to 0$, that is, $x_n\to p$.

Case 2: $\{s_n\}$ is not eventually decreasing, that is, we can find a positive integer $n_0$ such that $s_{n_0}\le s_{n_0+1}$. Now we define

$$U_n:=\{n_0\le k\le n:s_k\le s_{k+1}\},n>n_0.$$

It easy to see that $U_n$ is nonempty and satisfies $U_n\subseteq U_{n+1}$. Let

$$\psi (n):=max{U}_n,n>n_0.$$

It is clear that $\psi (n)\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$ (otherwise, $\{s_n\}$ is eventually decreasing). It is also clear that $s_{\psi (n)}\le s_{\psi (n)+1}$ for all $n>n_0$. Moreover, we prove that

$$s_n\le s_{\psi (n)+1},\mathrm{\forall }n>n_0.$$

(3.26)

In fact, if $\psi (n)=n$, then the inequality (3.26) is trivial; if $\psi (n)<n$, from the definition of $\psi (n)$, there exists some $i\in \mathbb{N}$ such that $\psi (n)+i=n$, we deduce that

$$s_{\psi (n)+1}>s_{\psi (n)+2}>\cdots >s_{\psi (n)+i}=s_n,$$

and the inequality (3.26) holds again. Since $s_{\psi (n)}\le s_{\psi (n)+1}$ for all $n>n_0$, it follows from (3.10) that

$$\frac{\sigma f^2(x_{\psi (n)})}{{\parallel \mathrm{\nabla }f(x_{\psi (n)})\parallel }^2}\le {\alpha }_{\psi (n)}(1-{\beta }_{\psi (n)})M\to 0.$$

Noting that $\{\parallel \mathrm{\nabla }f(x_{\psi (n)})\parallel \}$ is bounded, we get $f(x_{\psi (n)})\to 0$. By the same argument to the proof in case 1, we have $w_w(x_{\psi (n)})\subset \mathrm{\Gamma }$. From (3.19) we have

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

(3.27)

Furthermore, in view of (3.20), we can deduce that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,z_{\psi (n)}-p〉\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,x_{\psi (n)}-p〉\\ =\underset{w\in w_w(x_{\psi (n)})}{max}〈u-P_{\mathrm{\Gamma }}u,w-P_{\mathrm{\Gamma }}u〉\le 0.\end{array}$$

(3.28)

Since $s_{\psi (n)}\le s_{\psi (n)+1}$, it follows from (3.9) that

$$s_{\psi (n)}\le 2〈u-p,z_{\psi (n)}-p〉,n>n_0.$$

(3.29)

Combining (3.28) and (3.29) we have

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

(3.30)

and hence $s_{\psi (n)}\to 0$, which together with (3.27) implies that

$$\begin{array}{rl}\sqrt{s_{\psi (n)+1}}& \le \parallel (x_{\psi (n)}-p)+(x_{\psi (n)+1}-x_{\psi (n)})\parallel \\ \le \sqrt{s_{\psi (n)}}+\parallel x_{\psi (n)+1}-x_{\psi (n)}\parallel \to 0.\end{array}$$

Noting the inequality (3.26), this shows that $s_n\to 0$, that is, $x_n\to p$. This completes the proof of Theorem 3.1. □