Strong convergence theorem of two-step iterative algorithm for split feasibility problems

Abstract

The main purpose of this paper is to introduce a two-step iterative algorithm for split feasibility problems such that the strong convergence is guaranteed. Our result extends and improves the corresponding results of He et al. and some others.

MSC:90C25, 90C30, 47J25.

1 Introduction

The split feasibility problem (SFP) was proposed by Censor and Elfving in [1]. It can be formulated as the problem of finding a point x satisfying the property

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

(1.1)

where A is a given $M\times N$ real matrix, 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 continue to receive great attention (see, for instance, [2–5] and also [6–10]).

We assume that SFP (1.1) 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

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

(1.2)

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

To solve (1.2), Byrne [11] proposed his CQ algorithm which generates a sequence $\{x_n\}$ by

$$x_{n+1}=P_C(x_n-{\tau }_n{A}^{\ast }(I-P_Q)A{x}_n),$$

(1.3)

where ${\tau }_n\in (0,\frac{2}{{\parallel A\parallel }^2})$.

The CQ algorithm (1.3) can be obtained from optimization. In fact, if we introduce the convex objective function

$$f(x)=\frac{1}{2}{\parallel (I-P_Q)Ax\parallel }^2,$$

(1.4)

and analyze the minimization problem

$$\underset{x\in C}{min}f(x),$$

(1.5)

then the CQ algorithm (1.3) comes immediately as a special case of the gradient projection algorithm (GPA). Since the convex objective function $f(x)$ is differentiable and has a Lipschitz gradient, which is given by

$$\mathrm{\nabla }f(x)=A^{\ast }(I-P_Q)Ax,$$

(1.6)

the GPA for solving the minimization problem (1.4) generates a sequence $\{x_n\}$ recursively as

$$x_{n+1}=P_C(x_n-{\tau }_n\mathrm{\nabla }f(x_n)),$$

(1.7)

where ${\tau }_n$ is chosen in the interval $(0,\frac{2}{L})$, and L is the Lipschitz constant of ∇f.

Observe that in algorithms (1.3) and (1.7) mentioned above, in order to implement the CQ algorithm, one has to compute the operator norm $\parallel A\parallel$, which is in general not an easy work in practice. To overcome this difficulty, some authors proposed different adaptive choices of selecting the ${\tau }_n$ (see [11–13]). For instance, López et al. introduced a new way of selecting the stepsize [12] as follows:

$${\tau }_n:=\frac{{\rho }_{n}f(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2},0<{\rho }_n<4.$$

(1.8)

The computation of a projection onto a general closed convex subset is generally difficult. To overcome this difficulty, Fukushima [14] suggested the so-called relaxed projection method to calculate the projection onto a level set of a convex function by computing a sequence of projections onto half-spaces containing the original level set. In the setting of finite-dimensional Hilbert spaces, this idea was followed by Yang [9], who introduced the relaxed CQ algorithms for solving SFP (1.1) where the closed convex subsets C and Q are level sets of convex functions.

Recently, for the purpose of generality, SFP (1.1) has been studied in a more general setting. For instance, see [8, 12]. However, their algorithm has only weak convergence in the setting of infinite-dimensional Hilbert spaces. Very recently, He and Zhao [15] introduced a new relaxed CQ algorithm (1.9) such that the strong convergence is guaranteed in infinite-dimensional Hilbert spaces:

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

(1.9)

Motivated and inspired by the research going on in this section, the purpose of this article is to study a two-step iterative algorithm for split feasibility problems such that the strong convergence is guaranteed in infinite-dimensional Hilbert spaces. Our result extends and improves the corresponding results of He and Zhao [15] and some others.

2 Preliminaries and lemmas

Throughout the rest of this paper, we assume that H, $H_1$ and $H_2$ all are Hilbert spaces, 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 following notations: → to denote the strong convergence, ⇀ to denote the weak convergence and

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

to denote 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 demiclosed at origin if for any sequence $\{x_n\}\subset H$ with $x_n\rightharpoonup x^{\ast }$ and ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel (I-T)x_n\parallel =0$, 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 demiclosed at origin.

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

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

A function $f:H\to \mathbb{R}$ is said to be weakly lower semi-continuous (w-lsc) at x if $x_n\rightharpoonup x$ implies

$$f(x)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}f(x_n).$$

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 ℝ, 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: $\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel \le {\parallel A\parallel }^2\parallel x-y\parallel$, $x,y\in H_1$.

Proof (i) From the definition of f, we know that f is convex. For any given $x\in H_1$ and for any $v\in H_1$, 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

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

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 $h=1$ in (2.1), we have

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

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 it and then letting $h\to 0+$, 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

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

Now we prove conclusion (ii). Indeed, it follows from (i) that

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

Lemma 2.1 is proved. □

Lemma 2.2 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, we have, for all $x,y\in H$,

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

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. □

Lemma 2.3 [16]

Assume that $\{a_n\}$ is 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 ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.4 [17]

Let X be a real Banach space and $J:X\to 2^{X^{\ast }}$ be the normalized duality mapping, then, for any $x,y\in X$, the following inequality holds:

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

Especially, if X is a real Hilbert space, then we have

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

3 Main results

In this section, we shall prove our main theorem.

Theorem 3.1 Let $H_1$, $H_2$ be two real Hilbert spaces, $A:H_1\to H_2$ be a bounded linear operator. Let $S:H_1\to H_1$ be a firmly nonexpansive mapping, $T_1,T_2:H_2\to H_2$ be two firmly nonexpansive mappings such that $\parallel (I-T_i)x\parallel$ ($i=1,2$) is a convex function from $H_2$ to ℝ with $C:=F(S)\ne \mathrm{\varnothing }$ and $Q:=F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Let $u\in H_1$ and $\{x_n\}$ be a sequence generated by

$$\{\begin{array}{l}{x}_0\in H_1\mathit{\text{ chosen arbitrarily}},\\ x_{n+1}=S({\alpha }_{n}u+(1-{\alpha }_n)(y_n-{\xi }_n\mathrm{\nabla }g(y_n))),\\ y_n={\beta }_{n}u+(1-{\beta }_n)(x_n-{\tau }_n\mathrm{\nabla }f(x_n)),\end{array}$$

(3.1)

where

$$f(x_n)=\frac{1}{2}{\parallel (I-T_1)A{x}_n\parallel }^2,\mathrm{\nabla }f(x_n)=A^{\ast }(I-T_1)A{x}_n,{\tau }_n=\frac{{\rho }_{n}f(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}$$

and

$$g(y_n)=\frac{1}{2}{\parallel (I-T_2)A{y}_n\parallel }^2,\mathrm{\nabla }g(y_n)=A^{\ast }(I-T_2)A{y}_n,{\xi }_n=\frac{{\rho }_{n}g(y_n)}{{\parallel \mathrm{\nabla }g(y_n)\parallel }^2}.$$

If the solution set Γ of SPF (1.1) is not empty, and the sequences $\{{\rho }_n\}\subset (0,4)$ and $\{{\alpha }_n\},\{{\beta }_n\}\subset (0,1)$ satisfy the following conditions:

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

• (ii) the sequence $\{\frac{{\alpha }_n}{{\beta }_n}\}$ is bounded and ${\mathrm{\Sigma }}_{n=1}^{\mathrm{\infty }}{\beta }_n=\mathrm{\infty }$,

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

Proof Since Γ (≠∅) is the solution set of SPF (1.1), 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 2.4 that

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel {\beta }_{n}u+(1-{\beta }_n)(x_n-{\tau }_n\mathrm{\nabla }f(x_n))-p\parallel }^2\\ \le (1-{\beta }_n){\parallel x_n-{\tau }_n\mathrm{\nabla }f(x_n)-p\parallel }^2+2{\beta }_n〈u-p,y_n-p〉.\end{array}$$

(3.2)

Since $p\in \mathrm{\Gamma }\subset C$, $\mathrm{\nabla }f(p)=0$. Observe that $I-T_1$ is firmly nonexpansive, from Lemma 2.2(ii) we have that

$$\begin{array}{rl}〈\mathrm{\nabla }f(x_n),x_n-p〉& =〈(I-T_1)A{x}_n,A{x}_n-Ap〉\\ \ge {\parallel (I-T_1)A{x}_n\parallel }^2=2f(x_n),\end{array}$$

which implies that

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

Thus, we have

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

(3.3)

and so we have

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& \le (1-{\beta }_n){\parallel x_n-p\parallel }^2+2{\beta }_n〈2(u-p),\frac{(y_n-p)}{2}〉\\ \le (1-{\beta }_n){\parallel x_n-p\parallel }^2+\frac{1}{4}{\beta }_n{\parallel y_n-p\parallel }^2+4{\beta }_n{\parallel u-p\parallel }^2.\end{array}$$

(3.4)

Consequently, we have

$${\parallel y_n-p\parallel }^2\le \frac{1-{\beta }_n}{1-\frac{1}{4}{\beta }_n}{\parallel x_n-p\parallel }^2+\frac{\frac{3}{4}{\beta }_n}{1-\frac{1}{4}{\beta }_n}\frac{16}{3}{\parallel u-p\parallel }^2.$$

It turns out that

$$\parallel y_n-p\parallel \le max\{\parallel x_n-p\parallel ,\frac{16}{3}\parallel u-p\parallel \},$$

and inductively

$$\parallel y_n-p\parallel \le max\{\parallel x_0-p\parallel ,\frac{16}{3}\parallel u-p\parallel \}.$$

This implies that the sequence $\{y_n\}$ is bounded. Since the mapping S is firmly nonexpansive, it follows from Lemma 2.4 that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& ={\parallel S({\alpha }_{n}u+(1-{\alpha }_n)(y_n-{\xi }_n\mathrm{\nabla }g(y_n)))-p\parallel }^2\\ \le {\parallel {\alpha }_n(u-p)+(1-{\alpha }_n)(y_n-{\xi }_n\mathrm{\nabla }g(y_n)-p)\parallel }^2\\ \le (1-{\alpha }_n){\parallel y_n-{\xi }_n\mathrm{\nabla }g(y_n)-p\parallel }^2+2{\alpha }_n〈u-p,x_{n+1}-p〉.\end{array}$$

Since $p\in \mathrm{\Gamma }\subset C$, $\mathrm{\nabla }g(p)=0$. Observe that $I-T_2$ is firmly nonexpansive, it is deduced from Lemma 2.2(ii) that

$$〈\mathrm{\nabla }g(y_n),y_n-p〉=〈(I-T_2)A{y}_n,A{y}_n-Ap〉\ge {\parallel (I-T_2)A{y}_n\parallel }^2=2g(y_n),$$

which implies that

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

Thus, we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & (1-{\alpha }_n){\parallel y_n-{\xi }_n\mathrm{\nabla }g(y_n)-p\parallel }^2+2{\alpha }_n〈u-p,x_{n+1}-p〉\\ \le & (1-{\alpha }_n){\parallel y_n-p\parallel }^2+2{\alpha }_n〈u-p,x_{n+1}-p〉\\ -(1-{\alpha }_n){\rho }_n(4-{\rho }_n)\frac{g^2(y_n)}{{\parallel \mathrm{\nabla }g(y_n)\parallel }^2}\end{array}$$

(3.5)

and

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& \le (1-{\alpha }_n){\parallel y_n-p\parallel }^2+2{\alpha }_n〈u-p,x_{n+1}-p〉\\ \le (1-{\alpha }_n){\parallel y_n-p\parallel }^2+\frac{1}{4}{\alpha }_n{\parallel x_{n+1}-p\parallel }^2+4{\alpha }_n{\parallel u-p\parallel }^2.\end{array}$$

Consequently, we have

$${\parallel x_{n+1}-p\parallel }^2\le \frac{1-{\alpha }_n}{1-\frac{1}{4}{\alpha }_n}{\parallel y_n-p\parallel }^2+\frac{\frac{3}{4}{\alpha }_n}{1-\frac{1}{4}{\alpha }_n}\frac{16}{3}{\parallel u-p\parallel }^2.$$

It turns out that

$$\parallel x_{n+1}-p\parallel \le max\{\parallel y_n-p\parallel ,\frac{16}{3}\parallel u-p\parallel \}.$$

Since $\{y_n\}$ is bounded, so is $\{x_n\}$. Since ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n=0$, without loss of generality, we may assume that there is $\sigma >0$ such that

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

Substituting (3.3) into (3.5), we have

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

(3.6)

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

$$\begin{array}{r}{s}_{n+1}-s_n+{\beta }_n{s}_n+\frac{\sigma f^2(x_n)}{{\parallel \mathrm{\nabla }f(x_n)\parallel }^2}+\frac{\sigma g^2(y_n)}{{\parallel \mathrm{\nabla }g(y_n)\parallel }^2}\\ \le 2{\alpha }_n〈u-p,x_{n+1}-p〉+2{\beta }_n〈u-p,y_n-p〉.\end{array}$$

(3.7)

Now, we prove $s_n\to 0$. For the purpose, we consider two cases.

Case 1: $\{s_n\}$ is eventually decreasing, i.e., there exists a sufficiently 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 (3.7) it follows that

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

(3.8)

where M is a constant such that $M\ge 2(\parallel x_{n+1}-p\parallel +\parallel y_n-p\parallel )\parallel u-p\parallel$ for all $n\in \mathbb{N}$. Using condition (i) and (3.8), we have that

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

To verify that $f(x_n)\to 0$ and $g(y_n)\to 0$, it suffices to show that $\{\parallel \mathrm{\nabla }f(x_n)\parallel \}$ and $\{\parallel \mathrm{\nabla }g(y_n)\parallel \}$ are bounded. In fact, it follows from Lemma 2.1(ii) that for all $n\in \mathbb{N}$,

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

and

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

These imply that $\{\parallel \mathrm{\nabla }f(x_n)\parallel \}$ and $\{\parallel \mathrm{\nabla }g(y_n)\parallel \}$ are bounded. It yields $f(x_n)\to 0$ and $g(y_n)\to 0$, namely

$$\parallel (I-T_1)A{x}_n\parallel \to 0,\parallel (I-T_2)A{y}_n\parallel \to 0.$$

(3.9)

From (3.1) we have

$$\begin{array}{rl}\parallel x_n-y_n\parallel & =\parallel {\beta }_{n}u+(1-{\beta }_n)(x_n-{\tau }_n\mathrm{\nabla }f(x_n))-x_n\parallel \\ =\parallel {\beta }_n(u-x_n+{\tau }_n\mathrm{\nabla }f(x_n))\parallel \\ ={\beta }_n\parallel u-x_n+{\tau }_n\mathrm{\nabla }f(x_n)\parallel .\end{array}$$

Noting that $\{x_n\}$, $\{\parallel \mathrm{\nabla }f(x_n)\parallel \}$ are bounded and ${\beta }_n\to 0$, ${\tau }_n\to 0$, we get

$$\parallel x_n-y_n\parallel \to 0(n\to \mathrm{\infty }).$$

(3.10)

For any $x^{\ast }\in w_{\omega }(x_n)$, and $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$ such that $x_{n_k}\rightharpoonup x^{\ast }\in H_1$, then it follows from (3.10) that $y_{n_k}\rightharpoonup x^{\ast }$. Thus we have

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

(3.11)

On the other hand, from (3.9) we have

$$\parallel (I-T_1)A{x}_{n_k}\parallel \to 0,\parallel (I-T_2)A{y}_{n_k}\parallel \to 0.$$

(3.12)

Since $T_1$, $T_2$ are demiclosed at origin, from (3.11) and (3.12) we have that $A{x}^{\ast }\in F(T_1)\cap F(T_2)$, i.e., $A{x}^{\ast }\in Q$.

Next, we turn to prove $x^{\ast }\in C$. For convenience, we set $v_n:={\alpha }_{n}u+(1-{\alpha }_n)(y_n-{\xi }_n\mathrm{\nabla }g(y_n))$. Since S is firmly nonexpansive, it follows from Lemma 2.4 that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& ={\parallel S{v}_n-Sp\parallel }^2={\parallel S{v}_n-S{y}_n+S{y}_n-Sp\parallel }^2\\ \le {\parallel S{y}_n-Sp\parallel }^2+2〈S{v}_n-S{y}_n,x_{n+1}-p〉\\ \le {\parallel S{y}_n-p\parallel }^2+2\parallel S{v}_n-S{y}_n\parallel \parallel x_{n+1}-p\parallel \\ \le {\parallel y_n-p\parallel }^2-{\parallel (I-S)y_n\parallel }^2+2\parallel v_n-y_n\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

(3.13)

In view of the definition of $v_n$, we have

$$\parallel v_n-y_n\parallel \le {\alpha }_n\parallel y_n-u\parallel +{\xi }_n\parallel \mathrm{\nabla }g(y_n)\parallel ={\alpha }_n\parallel y_n-u\parallel +\frac{{\rho }_{n}g(y_n)}{\parallel \mathrm{\nabla }g(y_n)\parallel }.$$

(3.14)

From (3.4), (3.13) and (3.14), we have

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2\le & (1-{\beta }_n){\parallel x_n-p\parallel }^2+\frac{1}{4}{\beta }_n{\parallel y_n-p\parallel }^2+4{\beta }_n{\parallel u-p\parallel }^2\\ -{\parallel (I-S)y_n\parallel }^2+({\alpha }_n+\frac{g(y_n)}{\parallel \mathrm{\nabla }g(y_n)\parallel })N,\end{array}$$

(3.15)

where $N\ge 2\parallel x_{n+1}-p\parallel (\parallel y_n-u\parallel +2{\rho }_n)$ is a suitable constant. Clearly, from (3.15) we have

$$\begin{array}{rl}{\parallel (I-S)y_n\parallel }^2\le & s_n-s_{n+1}+\frac{1}{4}{\beta }_n{\parallel y_n-p\parallel }^2+4{\beta }_n{\parallel u-p\parallel }^2\\ +({\alpha }_n+\frac{g(y_n)}{\parallel \mathrm{\nabla }g(y_n)\parallel })N.\end{array}$$

(3.16)

Thus, we assert that $\parallel (I-S)y_n\parallel \to 0$. In view of $y_{n_k}\rightharpoonup x^{\ast }$ and S is demiclosed at origin, we get $x^{\ast }\in F(S)$, i.e., $x^{\ast }\in C$. Consequently, $w_w(x_n)\subset \mathrm{\Gamma }$. Furthermore, we have

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

and

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

From (3.7) we have

$$\begin{array}{rl}{s}_{n+1}& \le (1-{\beta }_n)s_n+2{\alpha }_n〈u-p,x_{n+1}-p〉+2{\beta }_n〈u-p,y_n-p〉\\ =(1-{\beta }_n)s_n+{\beta }_n(\frac{2{\alpha }_n}{{\beta }_n}〈u-p,x_{n+1}-p〉+2〈u-p,y_n-p〉).\end{array}$$

(3.17)

From condition (ii) and Lemma 2.3, we obtain $s_n\to 0$.

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

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

(3.18)

In fact, if $\psi (n)=n$, then inequality (3.18) 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 inequality (3.18) holds again. Since $s_{\psi (n)}\le s_{\psi (n)+1}$ for all $n>n_0$, it follows from (3.8) that

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

Noting that $\{\parallel \mathrm{\nabla }f(x_{\psi (n)})\parallel \}$ and $\{\parallel \mathrm{\nabla }g(y_{\psi (n)})\parallel \}$ are both bounded, we get

$$f(x_{\psi (n)})\to 0,g(y_{\psi (n)})\to 0.$$

By the same argument to the proof in case 1, we have $w_w(x_{\psi (n)})\subset \mathrm{\Gamma }$ and

$$\parallel x_{\psi (n)}-y_{\psi (n)}\parallel \to 0.$$

(3.19)

On the other hand, noting $s_{\psi (n)}\le s_{\psi (n)+1}$ again, we have from (3.14) and (3.16) that

$$\begin{array}{r}\parallel x_{\psi (n)+1}-y_{\psi (n)}\parallel \\ =\parallel S{v}_{\psi (n)}-S{y}_{\psi (n)}+S{y}_{\psi (n)}-y_{\psi (n)}\parallel \\ \le \parallel S{v}_{\psi (n)}-S{y}_{\psi (n)}\parallel +\parallel S{y}_{\psi (n)}-y_{\psi (n)}\parallel \\ \le \parallel v_{\psi (n)}-y_{\psi (n)}\parallel +\parallel (I-S)y_{\psi (n)}\parallel \\ \le {\alpha }_{\psi (n)}\parallel y_{\psi (n)}-u\parallel +\frac{{\rho }_{\psi (n)}g(y_{\psi (n)})}{\parallel \mathrm{\nabla }g(y_{\psi (n)})\parallel }\\ +\sqrt{\frac{1}{4}{\beta }_{\psi (n)}{\parallel y_{\psi (n)}-p\parallel }^2+4{\beta }_{\psi (n)}{\parallel u-p\parallel }^2+({\alpha }_{\psi (n)}+\frac{g(y_{\psi (n)})}{\parallel \mathrm{\nabla }g(y_{\psi (n)})\parallel })N}.\end{array}$$

(3.20)

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

$$\parallel x_{\psi (n)+1}-y_{\psi (n)}\parallel \to 0.$$

(3.21)

From (3.19) and (3.21), we have

$$\parallel x_{\psi (n)}-x_{\psi (n)+1}\parallel \le \parallel x_{\psi (n)}-y_{\psi (n)}\parallel +\parallel y_{\psi (n)}-x_{\psi (n)+1}\parallel \to 0(n\to \mathrm{\infty }).$$

(3.22)

Furthermore, we can deduce that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,x_{\psi (n)+1}-p〉& =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-p,x_{\psi (n)}-p〉\\ =\underset{w\in w_w(x_{\psi (n)})}{max}〈u-P_{\mathrm{\Gamma }}u,w-P_{\mathrm{\Gamma }}u〉\le 0\end{array}$$

(3.23)

and

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

(3.24)

Since $s_{\psi (n)}\le s_{\psi (n)+1}$, it follows from (3.7) that

$$s_{\psi (n)}\le 2〈u-p,y_{\psi (n)}-p〉+\frac{2{\alpha }_{\psi (n)}}{{\beta }_{\psi (n)}}〈u-p,x_{\psi (n)+1}-p〉,n>n_0.$$

(3.25)

Combining (3.23), (3.24) and (3.25), we have

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

(3.26)

and hence $s_{\psi (n)}\to 0$, which together with (3.22) implies that

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

(3.27)

Noting inequality (3.18), this shows that $s_n\to 0$, that is, $x_n\to p$. This completes the proof of Theorem 3.1. □