Split feasibility and fixed-point problems for asymptotically quasi-nonexpansive mappings

Abstract

The purpose of this paper is to introduce and analyze a weakly convergent theorem by using the regularized method and the relaxed extragradient method for finding a common element of the solution set Γ of the split feasibility problem and $Fix(T)$ of fixed points of asymptotically quasi-nonexpansive mappings T in the setting of infinite-dimensional Hilbert spaces. Consequently, we prove that the sequence generated by the proposed algorithm converges weakly to an element of $Fix(T)\cap \mathrm{\Gamma }$ under mild assumptions.

MSC:47H09, 47J25, 65K10.

1 Introduction

In 1994, Censor and Elfving [1] first introduced the split feasibility problem (SFP) in finite-dimensional Hilbert spaces for modeling inverse problems which arise from phase retrievals and in medical image reconstruction [2]. It was found that the SFP can also be used to model intensity-modulated radiation therapy (IMRT) (see [3–6]). Very recently, Xu [7] considered the SFP in the framework of infinite-dimensional Hilbert spaces. In this setting, the SFP is formulated as the problem of finding a point $x^{\ast }$ with the property

$$x^{\ast }\in C\text{and}A{x}^{\ast }\in Q,$$

(1.1)

where C and Q are the nonempty closed convex subsets of the infinite-dimensional real Hilbert spaces $H_1$ and $H_2$, respectively. Let $A\in B(H_1,H_2)$, where $B(H_1,H_2)$ denotes the family of all bounded linear operators from $H_1$ to $H_2$.

We use Γ to denote the solution set of the SFP, i.e.,

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

Assume that the SFP is consistent (i.e., (1.1) has a solution) so that Γ is closed, convex and nonempty. A special case of the SFP is the following convex constrained linear inverse problem:

$$\text{find }x\in C\text{such that }Ax=b,$$

(1.2)

which has extensively been investigated by using the Landweber iterative method [8]:

$$\begin{array}{c}\text{let }{x}_0\text{ be arbitrary for }n=0,1,\dots ,\text{ let}\hfill \\ x_{n+1}=x_n+\gamma A^T(b-A{x}_n).\hfill \end{array}$$

Comparatively, the SFP has received much less attention so far due to the complexity resulting from the set Q. Therefore, whether various versions of the projected Landweber iterative method [8] can be extended to solve the SFP remains an interesting open topic.

The original algorithm given in [1] involves the computation of the inverse $A^{-1}$ (assuming the existence of the inverse of A):

$$x_{k+1}=A^{-1}{P}_Q(P_{A(C)}(A{x}_k)),k\ge 0,$$

where $C,Q\subset {\mathbb{R}}^n$ are closed convex sets, A is a full rank $n\times n$ matrix and $A(C)=\{y\in {\mathbb{R}}^n|y=Ax,x\in C\}$, and thus has not become popular.

A more popular algorithm that solves the SFP seems to be the CQ algorithm of Byrne [2, 9] which is found to be a gradient-projection method (GPM) in convex minimization. It is also a special case of the proximal forward-backward splitting method [10]. The CQ algorithm only involves the computations of the projections $P_C$ and $P_Q$ onto the sets C and Q, respectively, and is therefore implementable in the case where $P_C$ and $P_Q$ have closed-form expressions (for example, C and Q are closed balls or half-spaces). It remains, however, a challenge on the CQ algorithm in the case where the projection $P_C$ and/or $P_Q$ fail to have closed-form expressions though theoretically we can prove the (weak) convergence of the algorithm.

Recently, Xu [7] gave a continuation of the study on the CQ algorithm and its convergence. He applied Mann’s algorithm to the SFP and proposed an averaged CQ algorithm, which was proved to be weakly convergent to a solution of the SFP. He derived a weak convergence result, which shows that for suitable choices of iterative parameters (including the regularization), the sequence of iterative solutions can converge weakly to an exact solution of the SFP. He also established the strong convergence result, which shows that the minimum-norm solution can be obtained. Later, Deepho and Kumam [11] extended the results of Xu [7] by introducing and studying the modified Halpern iterative scheme for solving the split feasibility problem (SFP) in the setting of infinite-dimensional Hilbert spaces.

Throughout this paper, we always assume that the SFP is consistent, that is, the solution set Γ of the SFP is nonempty. Let $f:H_1\to \mathbb{R}$ be a continuous differentiable function. The minimization problem

$$\underset{x\in C}{min}f(x):=\frac{1}{2}{\parallel Ax-P_{Q}Ax\parallel }^2$$

(1.3)

is ill-posed. Therefore (see [7]), consider the following Tikhonov regularized problem:

$$\underset{x\in C}{min}{f}_{\alpha }(x):=\frac{1}{2}{\parallel Ax-P_{Q}Ax\parallel }^2+\frac{1}{2}\alpha {\parallel x\parallel }^2,$$

(1.4)

where $\alpha >0$ is the regularization parameter.

We observe that the gradient

$$\mathrm{\nabla }{f}_{\alpha }(x)=\mathrm{\nabla }f(x)+\alpha I=A^{\ast }(I-P_Q)A+\alpha I$$

(1.5)

is $(\alpha +{\parallel A\parallel }^2)$-Lipschitz continuous and α-strongly monotone.

Define the Picard iterates

$$x_{n+1}^{\alpha }=P_C(I-\gamma (A^{\ast }(I-P_Q)A+\alpha I))x_n^{\alpha }.$$

(1.6)

Xu [7] showed that if SFP (1.1) is consistent, then as $n\to \mathrm{\infty }$, $x_n^{\alpha }\to x_{\alpha }$ and consequently the strong ${lim}_{\alpha \to 0}{x}_{\alpha }$ exists and is the minimum-norm solution of the SFP. Note that (1.6) is double-step iteration. Xu [7] further suggested the following single step regularized method:

$$x_{n+1}=P_C(I-\gamma \mathrm{\nabla }{f}_{{\alpha }_n})x_n=P_C((1-{\alpha }_n{\gamma }_n)x_n-{\gamma }_n{A}^{\ast }(I-P_Q)A{x}_n).$$

(1.7)

He proved that the sequence $\{x_n\}$ generated by (1.7) converges in norm to the minimum-norm solution of the SFP provided the parameters $\{{\alpha }_n\}$ and $\{{\gamma }_n\}$ satisfy the following conditions:

• (i) ${\alpha }_n\to 0$ and $0<{\gamma }_n<\frac{{\alpha }_n}{{\parallel A\parallel }^2+{\alpha }_n}$;

• (ii) ${\sum }_n{\alpha }_n{\gamma }_n=\mathrm{\infty }$;

• (iii) $\frac{|{\gamma }_{n+1}-{\gamma }_n|+{\gamma }_n|{\alpha }_{n+1}-{\alpha }_n|}{{({\alpha }_{n+1}{\gamma }_{n+1})}^2}\to 0$.

Motivated by the idea of the relaxed extragradient method and Xu’s regularization, Ceng, Ansari and Yao [12] presented the following relaxed extragradient method with regularization for finding a common element of the solution set of the split feasibility problem and the set $Fix(S)$ of fixed points of a nonexpansive mapping S:

$$\{\begin{array}{c}{x}_0=x\in C\text{chosen arbitrarily},\hfill \\ y_n=(1-{\beta }_n)+{\beta }_n{P}_C(x_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)),\hfill \\ x_{n+1}={\gamma }_n{x}_n+(1-{\gamma }_n)S{P}_C(y_n-\lambda \mathrm{\nabla }{f}_{{\alpha }_n}(y_n)),\mathrm{\forall }n\ge 0.\hfill \end{array}$$

(1.8)

They only obtained the weak convergence of iterative algorithm (1.8).

The purpose of this paper to study and analyze an relaxed extragradient method with regularization for finding a common element of the solution set Γ of the SFP and the set solutions of fixed points for asymptotically quasi-nonexpansive mappings and a Lipschitz continuous mapping in a real Hilbert space. We prove that the sequence generated by the proposed method converges weakly to an element $\hat{x}$ in $Fix(T)\cap \mathrm{\Gamma }$.

2 Preliminaries

We first recall some definitions, notations, and conclusions which will be needed in proving our main results. Let H be a real Hilbert space with the inner product $〈\cdot ,\cdot 〉$ and $\parallel \cdot \parallel$ and let C be a closed and convex subset of H. Let E be a Banach space. A mapping $T:E\to E$ is said to be demi-closed at origin if for any sequences $\{x_n\}\subset E$ with $x_n\rightharpoonup x^{\ast }$ and $\parallel (I-T)x_n\parallel \to 0$, $x^{\ast }=T{x}^{\ast }$. A Banach space E is said to have the Opial property if for any sequence $\{x_n\}$ with $x_n\rightharpoonup x^{\ast }$,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-x^{\ast }\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel x_n-y\parallel ,\mathrm{\forall }y\in E\text{ with }y\ne x^{\ast }.$$

Remark 2.1 It is well known that each Hilbert space possesses the Opial property.

Definition 2.2 Let H be a real Hilbert space, let C be a nonempty and closed convex subset. We denote by $Fix(T)$ the set of fixed points of T, that is, $Fix(T)=\{x\in C:x=Tx\}$. Then T is said to be

• (i) nonexpansive if$\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$;

• (ii) quasi-nonexpansive if$\parallel Tx-p\parallel \le \parallel x-p\parallel$ for all $x\in C$ and $p\in F(T)$;

• (iii) asymptotically nonexpansive if there exist a sequence $k_n\ge 1$ and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

  $$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel$$

for all $x,y\in C$ and $n\ge 1$;

• (iv) asymptotically quasi-nonexpansive if there exist a sequence $k_n\ge 1$ and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$ such that

  $$\parallel T^{n}x-p\parallel \le k_n\parallel x-p\parallel$$

for all $x\in C$, $p\in F(T)$ and $n\ge 1$;

• (v) uniformlyL-Lipschitzian if there exists a constant $L>0$ such that

  $$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel$$

for all $x,y\in C$ and $n\ge 1$.

Remark 2.3 By the above definitions, it is clear that:

• (i) a nonexpansive mapping is an asymptotically quasi-nonexpansive mapping;

• (ii) a quasi-nonexpansive mapping is an asymptotically-quasi nonexpansive mapping;

• (iii) an asymptotically nonexpansive mapping is an asymptotically quasi-nonexpansive mapping.

Proposition 2.4 (see [9])

We have the following assertions.

• (i) Tis nonexpansive if and only if the complement$I-T$is$\frac{1}{2}$-ism.

• (ii) IfTisν-ism and$\gamma >0$, thenγTis$\frac{\nu }{\gamma }$-ism.

• (iii) Tis averaged if and only if the complement$I-T$isν-ism for some$\nu >\frac{1}{2}$.

Indeed, for$\alpha \in (0,1)$, Tisα-averaged if and only if$I-T$is$\frac{1}{2\alpha }$-ism.

Proposition 2.5 (see [9, 13])

We have the following assertions.

• (i) If$T=(1-\alpha )S+\alpha V$for some$\alpha \in (0,1)$, Sis averaged andVis nonexpansive, thenTis averaged.

• (ii) Tis firmly nonexpansive if and only if the complement$I-T$is firmly nonexpansive.

• (iii) If$T=(1-\alpha )S+\alpha V$for some$\alpha \in (0,1)$, Sis firmly nonexpansive andVis nonexpansive, thenTis averaged.

• (iv) The composite of finite many averaged mappings is averaged. That is, if each of the mappings${\{T_i\}}_{i=1}^n$is averaged, then so is the composite$T_1\circ T_2\circ \cdots \circ T_N$. In particular, if$T_1$is${\alpha }_1$-averaged and$T_2$is${\alpha }_2$-averaged, where${\alpha }_1,{\alpha }_2\in (0,1)$, then the composite$T_1\circ T_2$isα-averaged, where$\alpha ={\alpha }_1+{\alpha }_2-{\alpha }_1{\alpha }_2$.

• (v) If the mappings${\{T_i\}}_{i=1}^n$are averaged and have a common fixed point, then

  $$\bigcap _{i=1}^{n}Fix(T_i)=Fix(T_1\cdots T_N).$$

Lemma 2.6 (see [14], demiclosedness principle)

LetCbe a nonempty closed and convex subset of a real Hilbert spaceHand let$S:C\to C$be a nonexpansive mapping with$Fix(S)\ne \mathrm{\varnothing }$. If the sequence$\{x_n\}\subseteq C$converges weakly toxand the sequence$\{(I-S)x_n\}$converges strongly toy, then$(I-S)x=y$; in particular, if$y=0$, then$x\in Fix(S)$.

Lemma 2.7 (see [15])

Let the sequences $\{a_n\}$ and $\{u_n\}$ of real numbers satisfy

$$a_{n+1}\le (1+u_n)a_n,\mathrm{\forall }n\ge 1,$$

where$a_n\ge 0$, $u_n\ge 0$and${\sum }_{n=1}^{\mathrm{\infty }}{u}_n<\mathrm{\infty }$. Then

• (1) ${lim}_{n\to \mathrm{\infty }}{a}_n$exists;

• (2) if${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n=0$, then${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

The following lemma gives some characterizations and useful properties of the metric projection $P_C$ in a Hilbert space.

For every point $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}x$, such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\mathrm{\forall }y\in C,$$

(2.1)

where $P_C$ is called the metric projection ofH onto C. We know that $P_C$ is a nonexpansive mapping of H onto C.

Proposition 2.8For given$x\in H$and$z\in C$:

• (i) $z=P_{C}x$if and only if$〈x-z,y-z〉\le 0$for all$y\in C$.

• (ii) $z=P_{C}x$if and only if${\parallel x-z\parallel }^2\le {\parallel x-y\parallel }^2-{\parallel y-z\parallel }^2$for all$y\in C$.

• (iii) For all$y\in H$, $〈P_{C}x-P_{C}y,x-y〉\ge {\parallel P_{C}x-P_{C}y\parallel }^2$.

Lemma 2.9 (see [16])

LetHbe a real Hilbert space. Then the following equations hold:

• (i) ${\parallel x-y\parallel }^2={\parallel x\parallel }^2-{\parallel y\parallel }^2-2〈x-y,y〉$for all$x,y\in H$;

• (ii) ${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2$for all$t\in [0,1]$and$x,y\in H$.

Let K be a nonempty closed convex subset of a real Hilbert space H and let $F:K\to H$ be a monotone mapping. The variational inequality problem (VIP) is to find $x\in K$ such that

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

The solution set of the VIP is denoted by $VIP(K,F)$. It is well known that

$$x\in VI(K,F)\iff x=P_K(x-\lambda Fx),\mathrm{\forall }\lambda >0.$$

A set-valued mapping $T:H\to 2^H$ is called monotone if for all $x,y\in H$, $f\in Tx$ and $g\in Ty$ imply

$$〈x-y,f-g〉\ge 0.$$

A monotone mapping $T:H\to 2^H$ is called maximal if its graph $G(T)$ is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if for $(x,f)\in H\times H$, $〈x-y,f-g〉\ge 0$ for every $(y,g)\in G(T)$ implies $f\in Tx$. Let $F:K\to H$ be a monotone and k-Lipschitz continuous mapping and let $N_{K}v$ be the normal cone to K at $v\in K$, that is,

$$N_{K}v=\{w\in H:〈v-u,w〉\ge 0,\mathrm{\forall }u\in K\}.$$

Define

$$Tv=\{\begin{array}{cc}Fv+N_{K}v\hfill & \text{if }v\in K,\hfill \\ \mathrm{\varnothing }\hfill & \text{if }v\notin K.\hfill \end{array}$$

Then T is maximal monotone and $0\in Tv$ if and only if $v\in VI(K,F)$; see [15] for more details.

We can use fixed point algorithms to solve the SFP on the basis of the following observation.

Let $\lambda >0$ and assume that $x^{\ast }\in \mathrm{\Gamma }$. Then $A{x}^{\ast }\in Q$, which implies that $(I-P_Q)A{x}^{\ast }=0$, and thus $\lambda A^{\ast }(I-P_Q)A{x}^{\ast }=0$. Hence, we have the fixed point equation $(I-\lambda A^{\ast }(I-P_Q)A)x^{\ast }=x^{\ast }$. Requiring that $x^{\ast }\in C$, we consider the fixed point equation

$$P_C(I-\lambda \mathrm{\nabla }f)x^{\ast }=P_C(I-\lambda A^{\ast }(I-P_Q)A)x^{\ast }=x^{\ast }.$$

(2.2)

It is proved in [[7], Proposition 3.2] that the solutions of fixed point equation (2.2) are exactly the solutions of the SFP; namely, for given $x^{\ast }\in H_1$, $x^{\ast }$ solves the SFP if and only if $x^{\ast }$ solves fixed point equation (2.2).

Proposition 2.10 (see [12])

Given$x^{\ast }\in H_1$, the following statements are equivalent.

• (i) $x^{\ast }$solves the SFP;

• (ii) $x^{\ast }$solves fixed point equation (2.2);

• (iii) $x^{\ast }$solves the variational inequality problem (VIP) of finding$x^{\ast }\in C$such that

  $$〈\mathrm{\nabla }f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,$$

  (2.3)

where$\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$and$A^{\ast }$is the adjoint ofA.

Proof (i) ⇔ (ii). See the proof in [[7], Proposition 3.2].

• (ii) ⇔ (iii). Observe that

  $$\begin{array}{rcl}{P}_C(I-\lambda A^{\ast }(I-P_Q)A)x^{\ast }=x^{\ast }& \iff & 〈(I-\lambda A^{\ast }(I-P_Q)A)x^{\ast }-x^{\ast },x-x^{\ast }〉\le 0,\mathrm{\forall }x\in C\\ \iff & -\lambda 〈A^{\ast }(I-P_Q)A{x}^{\ast },x-x^{\ast }〉\le 0,\mathrm{\forall }x\in C\\ \iff & 〈\mathrm{\nabla }f\left(x^{\ast }\right),x-x^{\ast }〉\ge 0,\mathrm{\forall }x\in C,\end{array}$$

where $\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$. □

Remark 2.11 It is clear from Proposition 2.10 that

$$\mathrm{\Gamma }:=Fix(P_C(I-\lambda \mathrm{\nabla }f))=VI(C,\mathrm{\nabla }f)$$

for any $\lambda >0$, where $Fix(P_C(I-\lambda \mathrm{\nabla }f))$ and $VI(C,\mathrm{\nabla }f)$ denote the set of fixed points of $P_C(I-\lambda \mathrm{\nabla }f)$ and the solution set of VIP.

3 Main result

Theorem 3.1LetCbe a nonempty, closed, and convex subset of a real Hilbert spaceHand let$T:C\to C$be a uniformlyL-Lipschitzian and asymptotically quasi-nonexpansive mappings with$Fix(T)\cap \mathrm{\Gamma }\ne \mathrm{\varnothing }$and$\{k_n\}\subset [1,\mathrm{\infty })$for all$n\in \mathbb{N}$such that${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$. Let$\{x_n\}$and$\{y_n\}$be the sequences inCgenerated by the following algorithm:

$$\{\begin{array}{c}{x}_0=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n,\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(3.1)

where$\mathrm{\nabla }{f}_{{\alpha }_n}=\mathrm{\nabla }f+{\alpha }_{n}I=A^{\ast }(I-P_Q)A+{\alpha }_{n}I$, and three sequences$\{{\alpha }_n\}$, $\{{\lambda }_n\}$, and$\{{\beta }_n\}$satisfy the conditions:

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$,

• (ii) $\{{\lambda }_n\}\subset [a,b]$for some$a,b\in (0,\frac{1}{{\parallel A\parallel }^2})$and${\sum }_{n=1}^{\mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|<\mathrm{\infty }$,

• (iii) $\{{\beta }_n\}\subset [c,d]$for some$c,d\in (0,1)$.

Then the sequences$\{x_n\}$and$\{y_n\}$converge weakly to an element$\hat{x}\in Fix(T)\cap \mathrm{\Gamma }$.

Proof We first show that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ζ-averaged for each ${\lambda }_n\in (0,\frac{2}{\alpha +{\parallel A\parallel }^2})$, where

$$\zeta =\frac{2+\lambda (\alpha +{\parallel A\parallel }^2)}{4}.$$

Indeed, it is easy to see that $\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$ is $\frac{1}{{\parallel A\parallel }^2}$-ism, that is,

$$〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\ge \frac{1}{{\parallel A\parallel }^2}{\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2.$$

Observe that

$$\begin{array}{c}(\alpha +{\parallel A\parallel }^2)〈\mathrm{\nabla }{f}_{\alpha }(x)-\mathrm{\nabla }{f}_{\alpha }(y),x-y〉\hfill \\ =(\alpha +{\parallel A\parallel }^2)[\alpha {\parallel x-y\parallel }^2+〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉]\hfill \\ ={\alpha }^2{\parallel x-y\parallel }^2+\alpha 〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\hfill \\ +\alpha {\parallel A\parallel }^2{\parallel x-y\parallel }^2+{\parallel A\parallel }^2〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉\hfill \\ \ge {\alpha }^2{\parallel x-y\parallel }^2+2\alpha 〈\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y),x-y〉+{\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2\hfill \\ ={\parallel \alpha (x-y)+\mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2\hfill \\ ={\parallel \mathrm{\nabla }f(x)-\mathrm{\nabla }f(y)\parallel }^2.\hfill \end{array}$$

Hence, it follows that $\mathrm{\nabla }{f}_{\alpha }=\alpha I+A^{\ast }(I-P_Q)A$ is $\frac{1}{\alpha +{\parallel A\parallel }^2}$-ism. Thus, $\lambda \mathrm{\nabla }{f}_{\alpha }$ is $\frac{1}{\lambda (\alpha +{\parallel A\parallel }^2)}$-ism. By Proposition 2.4(iii) the composite $(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is $\frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}$-averaged. Therefore, noting that $P_C$ is $\frac{1}{2}$-averaged and utilizing Proposition 2.5(iv), we know that for each $\lambda \in (0,\frac{2}{\alpha +{\parallel A\parallel }^2})$, $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is ζ-averaged with

$$\zeta =\frac{1}{2}+\frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}-\frac{1}{2}\cdot \frac{\lambda (\alpha +{\parallel A\parallel }^2)}{2}=\frac{2+\lambda (\alpha +{\parallel A\parallel }^2)}{4}\in (0,1).$$

This shows that $P_C(I-\lambda \mathrm{\nabla }{f}_{\alpha })$ is nonexpansive. Furthermore, for $\{{\lambda }_n\}\in [a,b]$ with $a,b\in (0,\frac{1}{{\parallel A\parallel }^2})$, utilizing the fact that ${lim}_{n\to \mathrm{\infty }}\frac{1}{{\alpha }_n+{\parallel A\parallel }^2}=\frac{1}{{\parallel A\parallel }^2}$, we may assume that

$$0<a\le {\lambda }_n\le b<\frac{1}{{\parallel A\parallel }^2},\mathrm{\forall }n\ge 0.$$

Consequently, it follows that for each integer $n\ge 0$, $P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})$ is ${\zeta }_n$-averaged with

$${\zeta }_n=\frac{1}{2}+\frac{{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{2}-\frac{1}{2}\cdot \frac{{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{2}=\frac{2+{\lambda }_n({\alpha }_n+{\parallel A\parallel }^2)}{4}\in (0,1).$$

This immediately implies that $P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})$ is nonexpansive for all $n\ge 0$.

We divide the remainder of the proof into several steps.

Step 1. We prove that $\{x_n\}$ is bounded. Indeed, we take a fixed $p\in Fix(T)\cap \mathrm{\Gamma }$ arbitrarily. Then we get $P_C(I-{\lambda }_n\mathrm{\nabla }f)p=p$ for ${\lambda }_n\in (0,\frac{2}{{\parallel A\parallel }^2})$. Since $P_C$ and $(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})$ are nonexpansive mappings, then we have

$$\begin{array}{rcl}\parallel y_n-p\parallel & =& \parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n-P_C(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ \le & \parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p\parallel \\ +\parallel P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p-P_C(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ \le & \parallel x_n-p\parallel +\parallel (I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})p-(I-{\lambda }_n\mathrm{\nabla }f)p\parallel \\ =& \parallel x_n-p\parallel +\parallel {\lambda }_n\mathrm{\nabla }fp-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}p\parallel \\ =& \parallel x_n-p\parallel +{\lambda }_n\parallel \mathrm{\nabla }fp-\mathrm{\nabla }{f}_{{\alpha }_n}p\parallel \\ =& \parallel x_n-p\parallel +{\lambda }_n\parallel \mathrm{\nabla }fp-\mathrm{\nabla }fp-{\alpha }_{n}p\parallel \\ \le & \parallel x_n-p\parallel +{\alpha }_n{\lambda }_n\parallel p\parallel .\end{array}$$

(3.2)

Observe that

$$\begin{array}{rcl}\parallel x_{n+1}-p\parallel & =& \parallel {\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n-p\parallel \\ \le & {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)\parallel T^n{y}_n-p\parallel \\ \le & {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)k_n\parallel y_n-p\parallel \\ \le & {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)k_n(\parallel x_n-p\parallel +{\lambda }_n{\alpha }_n\parallel p\parallel )\\ =& {\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n)k_n\parallel x_n-p\parallel +(1-{\beta }_n)k_n{\alpha }_n{\lambda }_n\parallel p\parallel \\ =& (1+(k_n-1)(1-{\beta }_n))\parallel x_n-p\parallel +(1-{\beta }_n)k_n{\alpha }_n{\lambda }_n\parallel p\parallel .\end{array}$$

(3.3)

Since ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$, according to Lemma 2.7 and (i), (ii) and (3.3), we obtain that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel \text{ exists for each }p\in Fix(T)\cap \mathrm{\Gamma }.$$

(3.4)

This implies that $\{x_n\}$ is bounded and $\{y_n\}$ is also bounded.

It follows that

$$\parallel T^n{x}_n-p\parallel \le k_n\parallel x_n-p\parallel .$$

Hence $\{T^n{x}_n-p\}$ is bounded.

Step 2. We prove that

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

In fact, it follows from (3.2) that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {(\parallel x_n-p\parallel +{\alpha }_n{\lambda }_n\parallel p\parallel )}^2\\ \le & {\parallel x_n-p\parallel }^2+2{\alpha }_n{\lambda }_n\parallel p\parallel \parallel x_n-p\parallel +{\alpha }_n^2{\lambda }_n^2{\parallel p\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+{\alpha }_n(2{\lambda }_n\parallel p\parallel \parallel x_n-p\parallel +{\alpha }_n{\lambda }_n^2{\parallel p\parallel }^2)\\ =& {\parallel x_n-p\parallel }^2+{\alpha }_{n}M,\end{array}$$

where $M={sup}_{n\ge 0}\{2{\lambda }_n\parallel p\parallel \parallel x_n-p\parallel +{\alpha }_n{\lambda }_n^2{\parallel p\parallel }^2\}<\mathrm{\infty }$.

It follows that

$$\begin{array}{rcl}{\parallel T^n{y}_n-p\parallel }^2& \le & {(k_n\parallel y_n-p\parallel )}^2\\ =& k_n^2{\parallel y_n-p\parallel }^2\\ =& k_n^2{\parallel x_n-p\parallel }^2+{\alpha }_n{k}_n^{2}M.\end{array}$$

Also, observe that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n-p\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel T^n{y}_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)(k_n^2{\parallel x_n-p\parallel }^2+{\alpha }_n{k}_n^{2}M)-{\beta }_n(1-{\beta }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ =& {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)k_n^2{\parallel x_n-p\parallel }^2+(1-{\beta }_n)k_n^2{\alpha }_{n}M\\ -{\beta }_n(1-{\beta }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ =& (k_n^2-{\beta }_n(k_n^2-1)){\parallel x_n-p\parallel }^2+(1-{\beta }_n)k_n^2{\alpha }_{n}M-{\beta }_n(1-{\beta }_n){\parallel T^n{y}_n-x_n\parallel }^2.\end{array}$$

Hence, we have

$$\begin{array}{r}{\beta }_n(1-{\beta }_n){\parallel T^n{y}_n-x_n\parallel }^2\\ \le (k_n^2-{\beta }_n(k_n^2-1)){\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2+(1-{\beta }_n)k_n^2{\alpha }_{n}M.\end{array}$$

(3.5)

By the conditions (i), (iii) and ${lim}_{n\to \mathrm{\infty }}{k}_n=1$, we can conclude that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^n{y}_n-x_n\parallel =0.$$

(3.6)

Consider that since $y_n=P_C(x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n)$ and by Proposition 2.8(ii), we have

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\parallel x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}(x_n)-p\parallel }^2-{\parallel x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}(x_n)-y_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),p-y_n〉\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n(〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n)-\mathrm{\nabla }{f}_{{\alpha }_n}(p),p-x_n〉\\ +〈\mathrm{\nabla }{f}_{{\alpha }_n}(p),p-x_n〉+〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),x_n-y_n〉)\\ \le & {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n(〈\mathrm{\nabla }{f}_{{\alpha }_n}(p),p-x_n〉+〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),x_n-y_n〉)\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n[〈({\alpha }_{n}I+\mathrm{\nabla }f)p,p-x_n〉+〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),x_n-y_n〉]\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n[{\alpha }_n〈p,p-x_n〉+〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),x_n-y_n〉]\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n〈p,p-x_n〉+2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),x_n-y_n〉\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n〈p,p-x_n〉-2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),y_n-x_n〉\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n〈p,p-x_n〉-2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),y_n-p+p-x_n〉\\ =& {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n〈p,p-x_n〉-2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),y_n-p〉\\ -2{\lambda }_n〈\mathrm{\nabla }{f}_{{\alpha }_n}(x_n),p-x_n〉\\ \le & {\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel p-x_n\parallel -2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel y_n-p\parallel \\ -2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel p-x_n\parallel .\end{array}$$

(3.7)

Consequently, utilizing Lemma 2.9(ii) and (3.7), we conclude that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel {\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n-p\parallel }^2\\ =& {\parallel {\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n-({\beta }_n+(1-{\beta }_n))p\parallel }^2\\ =& {\parallel {\beta }_n{x}_n+(1-{\beta }_n)T^n{y}_n-{\beta }_{n}p-(1-{\beta }_n)p\parallel }^2\\ =& {\parallel {\beta }_n(x_n-p)+(1-{\beta }_n)(T^n{y}_n-p)\parallel }^2\\ =& {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n){\parallel T^n{y}_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ \le & {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)k_n^2{\parallel y_n-p\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ =& {\beta }_n{\parallel x_n-p\parallel }^2+(1-{\beta }_n)k_n^2[{\parallel x_n-p\parallel }^2-{\parallel x_n-y_n\parallel }^2+2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel p-x_n\parallel \\ -2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel y_n-p\parallel -2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel p-x_n\parallel ]\\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ =& ({\beta }_n+(1-{\beta }_n)k_n^2){\parallel x_n-p\parallel }^2-(1-{\beta }_n)k_n^2{\parallel x_n-y_n\parallel }^2\\ +2(1-{\beta }_n)k_n^2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel p-x_n\parallel \\ -2(1-{\beta }_n)k_n^2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel y_n-p\parallel \\ -2(1-{\beta }_n)k_n^2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel p-x_n\parallel \\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ =& (k_n^2-{\beta }_n(k_n^2-1)){\parallel x_n-p\parallel }^2-(1-{\beta }_n)k_n^2{\parallel x_n-y_n\parallel }^2\\ +2(1-{\beta }_n)k_n^2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel p-x_n\parallel \\ -2(1-{\beta }_n)k_n^2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel y_n-p\parallel \\ -2(1-{\beta }_n)k_n^2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel \parallel p-x_n\parallel \\ -{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2.\end{array}$$

It follows that we get

$$\begin{array}{r}(1-{\beta }_n)k_n^2{\parallel x_n-y_n\parallel }^2-2(1-{\beta }_n)k_n^2{\lambda }_n{\alpha }_n\parallel p\parallel \parallel p-x_n\parallel \\ +2(1-{\beta }_n)k_n^2{\lambda }_n\parallel \mathrm{\nabla }{f}_{{\alpha }_n}(x_n)\parallel (\parallel y_n-p\parallel +\parallel p-x_n\parallel )+{\beta }_n(1-{\beta }_n){\parallel x_n-T^n{y}_n\parallel }^2\\ \le (k_n^2-{\beta }_n(k_n^2-1)){\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2.\end{array}$$

(3.8)

So, taking $n\to \mathrm{\infty }$, since ${lim}_{n\to 0}{k}_n=1$, (i)-(iii), (3.6) and (3.8), we can conclude that

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

(3.9)

Consider

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel {\beta }_n{x}_n-x_n+(1-{\beta }_n)T^n{y}_n\parallel \\ =& \parallel -(1-{\beta }_n)x_n+(1-{\beta }_n)T^n{y}_n\parallel \\ \le & (1-{\beta }_n)\parallel T^n{y}_n-x_n\parallel .\end{array}$$

(3.10)

From (3.6) we obtain

$$\parallel x_{n+1}-x_n\parallel \le (1-{\beta }_n)\parallel T^n{y}_n-x_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

(3.11)

Observe that

$$\begin{array}{rcl}\parallel T^n{y}_n-y_n\parallel & =& \parallel T^n{y}_n-x_n+x_n-y_n\parallel \\ \le & \parallel T^n{y}_n-x_n\parallel +\parallel x_n-y_n\parallel .\end{array}$$

So, from (3.6) and (3.9), we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^n{y}_n-y_n\parallel =0.$$

(3.12)

We compute that

$$\begin{array}{rcl}\parallel y_{n+1}-y_n\parallel & =& \parallel P_C(x_{n+1}-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}}{x}_{n+1})-P_C(x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n)\parallel \\ =& \parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})x_{n+1}-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n\parallel \\ \le & \parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})x_{n+1}-P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})x_n\parallel \\ +\parallel P_C(I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})x_n-P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\parallel (I-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}})x_n-(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel x_n-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}}{x}_n-(x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n)\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel {\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n-{\lambda }_{n+1}\mathrm{\nabla }{f}_{{\alpha }_{n+1}}{x}_n\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel {\lambda }_n(\mathrm{\nabla }f+{\alpha }_n)x_n-{\lambda }_{n+1}(\mathrm{\nabla }f+{\alpha }_{n+1})x_n\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel {\lambda }_n\mathrm{\nabla }f{x}_n+{\lambda }_n{\alpha }_n{x}_n-({\lambda }_{n+1}\mathrm{\nabla }f{x}_n+{\lambda }_{n+1}{\alpha }_{n+1}{x}_n)\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel ({\lambda }_n-{\lambda }_{n+1})\mathrm{\nabla }f{x}_n+{\lambda }_n{\alpha }_n{x}_n-{\lambda }_{n+1}{\alpha }_{n+1}{x}_n\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel ({\lambda }_n-{\lambda }_{n+1})\mathrm{\nabla }f{x}_n+{\lambda }_n{\alpha }_n{x}_n-{\lambda }_n{\alpha }_{n+1}{x}_n\\ +{\lambda }_n{\alpha }_{n+1}{x}_n-{\lambda }_{n+1}{\alpha }_{n+1}{x}_n\parallel \\ =& \parallel x_{n+1}-x_n\parallel +\parallel ({\lambda }_n-{\lambda }_{n+1})\mathrm{\nabla }f{x}_n+{\lambda }_n({\alpha }_n-{\alpha }_{n+1})x_n\\ +({\lambda }_n-{\lambda }_{n+1}){\alpha }_{n+1}{x}_n\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +|{\lambda }_n-{\lambda }_{n+1}|\parallel \mathrm{\nabla }f{x}_n\parallel +{\lambda }_n|{\alpha }_n-{\alpha }_{n+1}|\parallel x_n\parallel \\ +{\alpha }_{n+1}|{\lambda }_n-{\lambda }_{n+1}|\parallel x_n\parallel .\end{array}$$

From the conditions (i), (ii) and (3.11), we obtain that

$$\parallel y_{n+1}-y_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

(3.13)

Since T is uniformly L-Lipschitzian continuous, then

$$\begin{array}{rcl}\parallel y_n-T{y}_n\parallel & \le & \parallel y_n-y_{n+1}\parallel +\parallel y_{n+1}-T^{n+1}{y}_{n+1}\parallel +\parallel T^{n+1}{y}_{n+1}-T^{n+1}{y}_n\parallel +\parallel T^{n+1}{y}_n-T{y}_n\parallel \\ \le & \parallel y_n-y_{n+1}\parallel +\parallel y_{n+1}-T^{n+1}{y}_{n+1}\parallel +L\parallel y_n-y_{n+1}\parallel +L\parallel T^n{y}_n-y_n\parallel .\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel y_{n+1}-y_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel y_n-T^n{y}_n\parallel =0$, it follows that

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

(3.14)

Step 3. We show that $\hat{x}\in Fix(T)\cap \mathrm{\Gamma }$.

Since $\mathrm{\nabla }f=A^{\ast }(I-P_Q)A$ is Lipschitz continuous, from (3.9), we have

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

Since $\{x_n\}$ is bounded, there is a subsequence $\{x_{n_i}\}$ of $\{x_n\}$ that converges weakly to some $\hat{x}$.

First, we show that $\hat{x}\in \mathrm{\Gamma }$. Since $\parallel x_n-y_n\parallel \to 0$, it is known that $y_{n_i}\rightharpoonup \hat{x}$.

Put

$$S{w}_1=\{\begin{array}{cc}\mathrm{\nabla }f{w}_1+N_C{w}_1\hfill & \text{if }{w}_1\in C,\hfill \\ \mathrm{\varnothing }\hfill & \text{if }{w}_1\notin C,\hfill \end{array}$$

where $N_C{w}_1=\{z\in H_1:〈w_1-u,z〉\ge 0,\mathrm{\forall }u\in C\}$. Then S is maximal monotone and $0\in S{w}_1$ if and only if $w_1\in VI(C,\mathrm{\nabla }f)$; (see [17]) for more details. Let $(w_1,z)\in G(S)$, we have

$$z\in S{w}_1=\mathrm{\nabla }f{w}_1+N_C{w}_1,$$

and hence

$$z-\mathrm{\nabla }f{w}_1\in N_C{w}_1.$$

So, we have

$$〈w_1-u,z-\mathrm{\nabla }f{w}_1〉\ge 0,\mathrm{\forall }u\in C.$$

On the other hand, from

$$y_n=P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n\text{and}{w}_1\in C,$$

we have

$$〈x_n-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n-y_n,y_n-w_1〉\ge 0,$$

and

$$〈w_1-y_n,\frac{y_n-x_n}{{\lambda }_n}+\mathrm{\nabla }{f}_{{\alpha }_n}{x}_n〉\ge 0.$$

Therefore, from $z-\mathrm{\nabla }f{w}_1\in N_C{w}_1$ and $y_{n_i}\in C$, it follows that

$$\begin{array}{rcl}〈w_1-y_{n_i},z〉& \ge & 〈w_1-y_{n_i},\mathrm{\nabla }f{w}_1〉\\ \ge & 〈w_1-y_{n_i},\mathrm{\nabla }f{w}_1〉-〈w_1-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}+\mathrm{\nabla }{f}_{{\alpha }_{n_i}}{x}_{n_i}〉\\ =& 〈w_1-y_{n_i},\mathrm{\nabla }f{w}_1〉-〈w_1-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}+\mathrm{\nabla }f{x}_{n_i}〉\\ -{\alpha }_{n_i}〈w_1-y_{n_i},x_{n_i}〉\\ =& 〈w_1-y_{n_i},\mathrm{\nabla }f{w}_1-\mathrm{\nabla }f{y}_{n_i}〉+〈w_1-y_{n_i},\mathrm{\nabla }f{y}_{n_i}-\mathrm{\nabla }f{x}_{n_i}〉\\ -〈w_1-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉-{\alpha }_{n_i}〈w_1-y_{n_i},x_{n_i}〉\\ \ge & 〈w_1-y_{n_i},\mathrm{\nabla }f{y}_{n_i}-\mathrm{\nabla }f{x}_{n_i}〉-〈w_1-y_{n_i},\frac{y_{n_i}-x_{n_i}}{{\lambda }_{n_i}}〉\\ -{\alpha }_{n_i}〈w_1-y_{n_i},x_{n_i}〉.\end{array}$$

Hence, we obtain

$$〈w_1-\hat{x},z〉\ge 0\text{as }i\to \mathrm{\infty }.$$

Since S is maximal monotone, we have $\hat{x}\in S^{-1}0$, and hence $\hat{x}\in VI(C,\mathrm{\nabla }f)$. Thus, it is clear that $\hat{x}\in \mathrm{\Gamma }$.

Next, we show that $\hat{x}\in Fix(T)$. Indeed, since $y_{n_i}\rightharpoonup \hat{x}$ and $\parallel y_{n_i}-T{y}_{n_i}\parallel \to 0$ by (3.14) and Lemma 2.6, we get $\hat{x}\in Fix(T)$. Therefore, we have $\hat{x}\in Fix(T)\cap \mathrm{\Gamma }$.

Let $\{x_{n_j}\}$ be another subsequence of $\{x_n\}$ such that $\{x_{n_j}\}\rightharpoonup \overline{x}$. Then $\overline{x}\in Fix(T)\cap \mathrm{\Gamma }$. Let us show that $\hat{x}=\overline{x}$. Assume that $\hat{x}\ne \overline{x}$. From the Opial condition [18], we have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-\hat{x}\parallel & =& \underset{n_i\to \mathrm{\infty }}{lim}inf\parallel x_{n_i}-\hat{x}\parallel \\ <& \underset{n_i\to \mathrm{\infty }}{lim}inf\parallel x_{n_i}-\overline{x}\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-\overline{x}\parallel \\ =& \underset{n_j\to \mathrm{\infty }}{lim}inf\parallel x_{n_j}-\overline{x}\parallel \\ <& \underset{n_j\to \mathrm{\infty }}{lim}inf\parallel x_{n_j}-\hat{x}\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel x_n-\hat{x}\parallel .\end{array}$$

This is a contradiction. Thus, we have $\hat{x}=\overline{x}$. This implies

$$x_n\rightharpoonup \hat{x}\in Fix(T)\cap \mathrm{\Gamma }.$$

Further, from $\parallel x_n-y_n\parallel \to 0$, it follows that $y_n\rightharpoonup \hat{x}$. This shows that both sequences $\{x_n\}$ and $\{y_n\}$ converge weakly to $\hat{x}\in Fix(T)\cap \mathrm{\Gamma }$. This completes the proof. □

Utilizing Theorem 3.1, we have the following new results in the setting of real Hilbert spaces.

Take $T^n\equiv I(\mathit{identity}\mathit{mappings})$ in Theorem 3.1. Therefore the conclusion follows.

Corollary 3.2LetCbe a nonempty, closed, and convex subset of a real Hilbert spaceH. Suppose that$\mathrm{\Gamma }\ne \mathrm{\varnothing }$. Let$\{x_n\}$be a sequence inCgenerated by the following algorithm:

$$\{\begin{array}{c}{x}_0=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)P_C(I-{\lambda }_n\mathrm{\nabla }{f}_{{\alpha }_n})x_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(3.15)

where$\mathrm{\nabla }{f}_{{\alpha }_n}=\mathrm{\nabla }f+{\alpha }_{n}I=A^{\ast }(I-P_Q)A+{\alpha }_{n}I$, and the sequences$\{{\alpha }_n\}$, $\{{\lambda }_n\}$, and$\{{\beta }_n\}$satisfy the conditions:

• (i) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n<\mathrm{\infty }$,

• (ii) $\{{\lambda }_n\}\subset [a,b]$for some$a,b\in (0,\frac{1}{{\parallel A\parallel }^2})$and${\sum }_{n=1}^{\mathrm{\infty }}|{\lambda }_{n+1}-{\lambda }_n|<\mathrm{\infty }$,

• (iii) $\{{\beta }_n\}\subset [c,d]$for some$c,d\in (0,1)$.

Then the sequence$\{x_n\}$converges weakly to an element$\hat{x}\in \mathrm{\Gamma }$.

Take $P_C\equiv I(\mathit{identity}\mathit{mappings})$ in Theorem 3.1. Therefore the conclusion follows.

Corollary 3.3LetCbe a nonempty, closed, and convex subset of a real Hilbert spaceHand let$T:C\to C$be a uniformlyL-Lipschitzian and quasi-nonexpansive mapping with$Fix(T)\ne \mathrm{\varnothing }$and$\{k_n\}\subset [1,\mathrm{\infty })$for all$n\in \mathbb{N}$such that${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$. Let$\{x_n\}$be the sequence inCgenerated by the following algorithm:

$$\{\begin{array}{c}{x}_0=x\in C\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)T^n{x}_n,\mathrm{\forall }n\ge 0,\hfill \end{array}$$

(3.16)

and let the sequence$\{{\beta }_n\}$satisfy the condition$\{{\beta }_n\}\subset [c,d]$for some$c,d\in (0,1)$. Then the sequence$\{x_n\}$converges weakly to an element$\hat{x}\in Fix(T)$.

Remark 3.4 Theorem 3.1 improves and extends [[7], Theorem 5.7] in the following aspects:

• (a) The iterative algorithm [[7], Theorem 5.7] is extended for developing our relaxed extragradient algorithm with regularization in Theorem 3.1.

• (b) The technique of proving weak convergence in Theorem 3.1 is different from that in [[7], Theorem 5.7] because of our technique to use asymptotically quasi-nonexpansive mappings and the property of maximal monotone mappings.

• (c) The problem of finding a common element of $Fix(T)\cap \mathrm{\Gamma }$ for asymptotically quasi-nonexpansive mappings which is more general than that for nonexpansive mappings and the problem of finding a solution of the SFP in [[7], Theorem 5.7].