Split feasibility problems for total quasi-asymptotically nonexpansive mappings

Abstract

The purpose of this paper is to propose an algorithm for solving the split feasibility problems for total quasi-asymptotically nonexpansive mappings in infinite-dimensional Hilbert spaces. The results presented in the paper not only improve and extend some recent results of Moudafi [Nonlinear Anal. 74:4083-4087, 2011; Inverse Problem 26:055007, 2010], but also improve and extend some recent results of Xu [Inverse Problems 26:105018, 2010; 22:2021-2034, 2006], Censor and Segal [J. Convex Anal. 16:587-600, 2009], Censor et al. [Inverse Problems 21:2071-2084, 2005], Masad and Reich [J. Nonlinear Convex Anal. 8:367-371, 2007], Censor et al. [J. Math. Anal. Appl. 327:1244-1256, 2007], Yang [Inverse Problem 20:1261-1266, 2004] and others.

MSC:47J05, 47H09, 49J25.

1 Introduction

Throughout this paper, we always assume that $H_1$, $H_2$ are real Hilbert spaces, ‘→’, ‘⇀’ denote strong and weak convergence, respectively, and $F(T)$ is a fixed point set of a mapping T.

The split feasibility problem (SFP) in finite-dimensional spaces was first introduced by Censor and Elfving [1] for modeling inverse problems which arise from phase retrievals and in medical image reconstruction [2]. Recently, it has been found that the SFP can also be used in various disciplines such as image restoration, computer tomograph and radiation therapy treatment planning [3–5]. The split feasibility problem in an infinite-dimensional real Hilbert space can be found in [2, 4, 6–10].

The purpose of this paper is to introduce and study the following split feasibility problem for total quasi-asymptotically nonexpansive mappings in the framework of infinite-dimensional real Hilbert spaces:

$$\text{find }{x}^{\ast }\in C\text{ such that }A{x}^{\ast }\in Q,$$

(1.1)

where $A:H_1\to H_2$ is a bounded linear operator, $S:H_1\to H_1$ and $T:H_2\to H_2$ are mappings; $C:=F(S)$ and $Q:=F(T)$. In the sequel, we use Γ to denote the set of solutions of (SFP)-(1.1), i.e.,

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

(1.2)

2 Preliminaries

We first recall some definitions, notations and conclusions which will be needed in proving our main results.

Let E be a Banach space. A mapping $T:E\to E$ is said to be demi-closed at origin if for any sequence $\{x_n\}\subset E$ 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.

• (1) A mapping $G:H\to H$ is said to be a $(\{{\nu }_n\},\{{\mu }_n\},\zeta )$-total quasi-asymptotically nonexpansive mapping if $F(G)\ne \mathrm{\varnothing }$; and there exist nonnegative real sequences $\{{\nu }_n\}$, $\{{\mu }_n\}$ with ${\nu }_n\to 0$ and ${\mu }_n\to 0$ and a strictly increasing continuous function $\zeta :{\mathcal{R}}^{+}\to {\mathcal{R}}^{+}$ with $\zeta (0)=0$ such that for each $n\ge 1$,

  $${\parallel p-G^{n}x\parallel }^2\le {\parallel p-x\parallel }^2+{\nu }_n\zeta (\parallel p-x\parallel )+{\mu }_n,\mathrm{\forall }p\in F(G),x\in H.$$

  (2.1)

Now, we give an example of total quasi-asymptotically nonexpansive mapping.

Let C be a unit ball in a real Hilbert space $l^2$, and let $T:C\to C$ be a mapping defined by

$$T:(x_1,x_2,\dots ,)\to (0,x_1^2,a_2{x}_2,a_3{x}_3,\dots ),(x_1,x_2,\dots ,)\in l^2,$$

where $\{a_i\}$ is a sequence in (0, 1) such that ${\prod }_{i=2}^{\mathrm{\infty }}{a}_i=\frac{1}{2}$.

It is proved in Goebal and Kirk [17] that

• (i) $\parallel Tx-Ty\parallel \le 2\parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$;

• (ii) $\parallel T^{n}x-T^{n}y\parallel \le 2{\prod }_{j=2}^n{a}_j\parallel x-y\parallel$, $\mathrm{\forall }x,y\in C$, $\mathrm{\forall }n\ge 2$.

Denote by $k_1^{\frac{1}{2}}=2$, $k_n^{\frac{1}{2}}=2{\prod }_{j=2}^n{a}_j$, $n\ge 2$, then

$$\underset{n\to \mathrm{\infty }}{lim}{k}_n=\underset{n\to \mathrm{\infty }}{lim}{\left(2\prod _{j=2}^n{a}_j\right)}^2=1.$$

Letting ${\nu }_n=(k_n-1)$, $\mathrm{\forall }n\ge 1$, $\zeta (t)=t$, $\mathrm{\forall }t\ge 0$ and $\{{\mu }_n\}$ be a nonnegative real sequence with ${\mu }_n\to 0$, from (i) and (ii), $\mathrm{\forall }x,y\in C$, $n\ge 1$, we have

$${\parallel T^{n}x-T^{n}y\parallel }^2\le {\parallel x-y\parallel }^2+{\nu }_n\zeta \left({\parallel x-y\parallel }^2\right)+{\mu }_n.$$

(2.2)

Again, since $0\in C$ and $0\in F(T)$, this implies that $F(T)\ne \mathrm{\varnothing }$. From (2.2), we have

$${\parallel p-T^{n}y\parallel }^2\le {\parallel p-y\parallel }^2+{\nu }_n\zeta \left({\parallel p-y\parallel }^2\right)+{\mu }_n\mathrm{\forall }p\in F(T),y\in C.$$

(2.3)

This shows that the mapping T defined as above is a total quasi-asymptotically nonexpansive mapping.

• (2) A mapping $G:H\to H$ is said to be $(\{k_n\})$-quasi-asymptotically nonexpansive if $F(G)\ne \mathrm{\varnothing }$; and there exists a sequence $\{k_n\}\subset [1,\mathrm{\infty })$ with $k_n\to 1$ such that for all $n\ge 1$,

  $${\parallel p-G^{n}x\parallel }^2\le k_n{\parallel p-x\parallel }^2,\mathrm{\forall }p\in F(G),x\in H.$$

  (2.4)

• (3) A mapping $G:H\to H$ is said to be quasi-nonexpansive if $F(G)\ne \mathrm{\varnothing }$ such that

  $$\parallel p-Gx\parallel \le \parallel p-x\parallel ,\mathrm{\forall }p\in F(G),x\in H.$$

  (2.5)

Remark 2.3 It is easy to see that every quasi-nonexpansive mapping is a $(\{1\})$-quasi-asymptotically nonexpansive mapping and every $\{k_n\}$-quasi-asymptotically nonexpansive mapping is a $(\{{\nu }_n\},\{{\mu }_n\},\zeta )$-total quasi-asymptotically nonexpansive mapping with $\{{\nu }_n=k_n-1\}$, $\{{\mu }_n=0\}$ and $\zeta (t)=t^2$, $t\ge 0$.

Definition 2.4 (1) A mapping $G:H\to H$ is said to be uniformly L-Lipschitzian if there exists a constant $L>0$ such that

$$\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,\mathrm{\forall }x,y\in H\text{ and }n\ge 1.$$

• (2) A mapping $G:H\to H$ is said to be semi-compact if for any bounded sequence $\{x_n\}\subset H$ with ${lim}_{n\to \mathrm{\infty }}\parallel x_n-G{x}_n\parallel =0$, there exists a subsequence $\{x_{n_i}\}\subset \{x_n\}$ such that $x_{n_i}$ converges strongly to some point $x^{\ast }\in H$.

Proposition 2.5 Let $G:H\to H$ be a $(\{{\nu }_n\},\{{\mu }_n\},\zeta )$-total quasi-asymptotically nonexpansive mapping. Then for each $q\in F(G)$ and for each $x\in H$, the following inequalities are equivalent: for each $n\ge 1$

(2.6)

(2.7)

Proof (I) (2.1) ⇔ (2.6) In fact, since

$$\begin{array}{rcl}{\parallel G^{n}x-q\parallel }^2& =& {\parallel G^{n}x-x+x-q\parallel }^2\\ =& {\parallel G^{n}x-x\parallel }^2+{\parallel x-q\parallel }^2+2〈G^{n}x-x,x-q〉,\mathrm{\forall }x\in H,q\in F(G),\end{array}$$

from (2.1) we have that

$$\begin{array}{c}{\parallel G^{n}x-x\parallel }^2+{\parallel x-q\parallel }^2+2〈G^{n}x-x,x-q〉\hfill \\ \le {\parallel x-q\parallel }^2+{\nu }_n\zeta (\parallel q-x\parallel )+{\mu }_n.\hfill \end{array}$$

Simplifying it, inequality (2.6) is obtained.

Conversely, from (2.6) the inequality (2.1) can be obtained immediately.

• (II) (2.6) ⇔ (2.7) In fact, since

  $$\begin{array}{rcl}〈x-G^{n}x,x-q〉& =& 〈x-G^{n}x,x-G^{n}x+G^{n}x-q〉\\ =& {\parallel x-G^{n}x\parallel }^2+〈x-G^{n}x,G^{n}x-q〉\end{array}$$

it follows from (2.6) that

$$2({\parallel x-G^{n}x\parallel }^2+〈x-G^{n}x,G^{n}x-q〉)\ge {\parallel x-G^{n}x\parallel }^2-{\nu }_n\zeta (\parallel q-x\parallel )-{\mu }_n.$$

Simplifying it, the inequality (2.7) is obtained.

Conversely, from (2.7) the inequality (2.6) can be obtained immediately.

This completes the proof of Proposition 2.5. □

Lemma 2.6 [11]

Let $\{a_n\}$, $\{b_n\}$ and $\{{\delta }_n\}$ be sequences of nonnegative real numbers satisfying

$$a_{n+1}\le (1+{\delta }_n)a_n+b_n,\mathrm{\forall }n\ge 1.$$

If ${\sum }_{i=1}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$ and ${\sum }_{i=1}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, then the limit ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists.

3 Split feasibility problem

For solving the split feasibility problem (1.1), let us assume that the following conditions are satisfied:

• 1. $H_1$ and $H_2$ are two real Hilbert spaces, $A:H_1\to H_2$ is a bounded linear operator;

• 2. $S:H_1\to H_1$ and $T:H_2\to H_2$ are two uniformly L-Lipschitzian and ($\{{\nu }_n\},\{{\mu }_n\},\zeta$)-total quasi-asymptotically nonexpansive mappings satisfying the following conditions:

• (i) T and S both are demi-closed at origin;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}({\mu }_n+{\nu }_n)<\mathrm{\infty }$;

• (iii) there exist positive constants M and $M^{\ast }$ such that $\zeta (t)\le \zeta (M)+M^{\ast }{t}^2$, $\mathrm{\forall }t\ge 0$.

We are now in a position to give the following result.

Theorem 3.1 Let $H_1$, $H_2$, A, S, T, L, $\{{\mu }_n\}$, $\{{\nu }_n\}$, ζ be the same as above. Let $\{x_n\}$ be the sequence generated by:

$$\{\begin{array}{c}{x}_1\in H_1\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}^n(u_n),\hfill \\ u_n=x_n+\gamma A^{\ast }(T^n-I)A{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.1)

where $\{{\alpha }_n\}$ is a sequence in $[0,1]$, and $\gamma >0$ is a constant satisfying the following conditions:

• (iv) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$; and $\gamma \in (0,\frac{1}{{\parallel A\parallel }^2})$,

• (I) If $\mathrm{\Gamma }\ne \mathrm{\varnothing }$ (where Γ is the set of solutions to ((SFP)-(1.1)), then $\{x_n\}$ converges weakly to a point $x^{\ast }\in \mathrm{\Gamma }$.

• (II) In addition, if S is also semi-compact, then $\{x_n\}$ and $\{u_n\}$ both converge strongly to $x^{\ast }\in \mathrm{\Gamma }$.

The proof of conclusion (I) (1) First, we prove that for each $p\in \mathrm{\Gamma }$, the following limits exist:

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-p\parallel .$$

(3.2)

In fact, since $p\in \mathrm{\Gamma }$, we have $p\in C:=F(S)$ and $Ap\in Q:=F(T)$. It follows from (3.1) and (2.4) that

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& =& {\parallel u_n-p-{\alpha }_n(u_n-S^n{u}_n)\parallel }^2\\ =& {\parallel u_n-p\parallel }^2-2{\alpha }_n〈u_n-p,u_n-S^n{u}_n〉+{\alpha }_n^2{\parallel u_n-S^n{u}_n\parallel }^2\\ \le & {\parallel u_n-p\parallel }^2-{\alpha }_n\{{\parallel u_n-S^n{u}_n\parallel }^2-{\nu }_n\zeta (\parallel u_n-p\parallel )-{\mu }_n\}\\ +{\alpha }_n^2{\parallel u_n-S^n{u}_n\parallel }^2\left(\text{by (2.6)}\right)\\ =& {\parallel u_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2+{\alpha }_n({\nu }_n\zeta (\parallel u_n-p\parallel )+{\mu }_n).\end{array}$$

(3.3)

On the other hand, by condition (iii), we have

$$\zeta (\parallel u_n-p\parallel )\le \zeta (M)+M^{\ast }{\parallel u_n-p\parallel }^2.$$

(3.4)

Substituting (3.4) into (3.3) and simplifying, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & (1+{\alpha }_n{\nu }_n{M}^{\ast }){\parallel u_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2\\ +{\alpha }_n({\nu }_n\zeta (M)+{\mu }_n)\\ \le & (1+{\nu }_n{M}^{\ast }){\parallel u_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2+{\nu }_n\zeta (M)+{\mu }_n.\end{array}$$

(3.5)

On the other hand,

$$\begin{array}{rcl}{\parallel u_n-p\parallel }^2& =& {\parallel x_n-p+\gamma A^{\ast }(T^n-I)A{x}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+{\gamma }^2{\parallel A^{\ast }(T^n-I)A{x}_n\parallel }^2+2\gamma 〈x_n-p,A^{\ast }(T^n-I)A{x}_n〉,\end{array}$$

(3.6)

and

$$\begin{array}{rcl}{\gamma }^2{\parallel A^{\ast }(T^n-I)A{x}_n\parallel }^2& =& {\gamma }^2〈A^{\ast }(T^n-I)A{x}_n,A^{\ast }(T^n-I)A{x}_n〉\\ =& {\gamma }^2〈A{A}^{\ast }(T^n-I)A{x}_n,(T^n-I)A{x}_n〉\\ \le & {\gamma }^2{\parallel A\parallel }^2{\parallel T^{n}A{x}_n-A{x}_n\parallel }^2,\end{array}$$

(3.7)

and

(3.8)

In (2.5), taking $x=A{x}_n$, $G^n=T^n$, $q=Ap$, and noting $Ap\in F(T)$, from (2.7) and condition (iii), we have

(3.9)

Substituting (3.9) into (3.8) and simplifying it, we have

(3.10)

Substituting (3.7) and (3.10) into (3.6) after simplifying, we have

$$\begin{array}{rcl}{\parallel u_n-p\parallel }^2& \le & (1+\gamma {\nu }_n{M}^{\ast }{\parallel A\parallel }^2){\parallel x_n-p\parallel }^2+\gamma ({\nu }_n\zeta (M)+{\mu }_n)\\ -\gamma (1-\gamma {\parallel A\parallel }^2){\parallel (T^n-I)A{x}_n\parallel }^2.\end{array}$$

(3.11)

Substituting (3.11) into (3.5) and simplifying it, we have

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^2& \le & (1+{\nu }_n{M}^{\ast })\{(1+\gamma {\nu }_n{M}^{\ast }{\parallel A\parallel }^2){\parallel x_n-p\parallel }^2\\ +\gamma ({\nu }_n\zeta (M)+{\mu }_n)-\gamma (1-\gamma {\parallel A\parallel }^2){\parallel (T^n-I)A{x}_n\parallel }^2\}\\ -{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2+{\nu }_n\zeta (M)+{\mu }_n\\ \le & (1+{\xi }_n){\parallel x_n-p\parallel }^2+{\eta }_n-\gamma (1-\gamma {\parallel A\parallel }^2){\parallel (T^n-I)A{x}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2,\end{array}$$

(3.12)

where

$$\begin{array}{c}{\xi }_n={\nu }_n(M^{\ast }+\gamma M^{\ast }{\parallel A\parallel }^2+\gamma {\nu }_n{M}^{\ast }{\parallel A\parallel }^2),\hfill \\ {\eta }_n=[(1+{\nu }_n{M}^{\ast })\gamma +1]({\nu }_n\zeta (M)+{\mu }_n).\hfill \end{array}$$

By condition (iii), we have

$$\sum _{n=1}^{\mathrm{\infty }}{\xi }_n<\mathrm{\infty },\text{and}\sum _{n=1}^{\mathrm{\infty }}{\eta }_n<\mathrm{\infty }.$$

By condition (iv), $(1-\gamma {\parallel A\parallel }^2)>0$. Hence, from (3.12), we have

$${\parallel x_{n+1}-p\parallel }^2\le (1+{\xi }_n){\parallel x_n-p\parallel }^2+{\eta }_n,\mathrm{\forall }n\ge 1.$$

By Lemma 2.6, the following limit exists:

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel .$$

(3.13)

Now, we rewrite (3.12) as follows:

$$\begin{array}{c}\gamma (1-\gamma {\parallel A\parallel }^2){\parallel (T^n-I)A{x}_n\parallel }^2+{\alpha }_n(1-{\alpha }_n){\parallel u_n-S^n{u}_n\parallel }^2\hfill \\ \le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2\hfill \\ +{\xi }_n{\parallel x_n-p\parallel }^2+{\eta }_n\to 0(\text{as }n\to \mathrm{\infty }).\hfill \end{array}$$

This together with the condition (iv) implies that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-S^n{u}_n\parallel =0;$$

(3.14)

and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel (T^n-I)A{x}_n\parallel =0.$$

(3.15)

It follows from (3.6), (3.14) and (3.15) that the limit ${lim}_{n\to \mathrm{\infty }}\parallel u_n-p\parallel$ exists and

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-p\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel .$$

The conclusion (3.2) is proved.

• (2) Next, we prove that

  $$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =0\text{and}\underset{n\to \mathrm{\infty }}{lim}\parallel u_{n+1}-u_n\parallel =0.$$

  (3.16)

In fact, it follows from (3.1) that

$$\begin{array}{rcl}\parallel x_{n+1}-x_n\parallel & =& \parallel (1-{\alpha }_n)u_n+{\alpha }_n{S}^n(u_n)-x_n\parallel \\ =& \parallel (1-{\alpha }_n)(x_n+\gamma A^{\ast }(T^n-I)A{x}_n)+{\alpha }_n{S}^n(u_n)-x_n\parallel \\ =& \parallel (1-{\alpha }_n)\gamma A^{\ast }(T^n-I)A{x}_n+{\alpha }_n(S^n(u_n)-x_n)\parallel \\ =& \parallel (1-{\alpha }_n)\gamma A^{\ast }(T^n-I)A{x}_n+{\alpha }_n(S^n(u_n)-u_n)+{\alpha }_n(u_n-x_n)\parallel \\ =& \parallel (1-{\alpha }_n)\gamma A^{\ast }(T^n-I)A{x}_n+{\alpha }_n(S^n(u_n)-u_n)+{\alpha }_n\gamma A^{\ast }(T^n-I)A{x}_n\parallel \\ =& \parallel \gamma A^{\ast }(T^n-I)A{x}_n+{\alpha }_n(S^n(u_n)-u_n)\parallel .\end{array}$$

In view of (3.14) and (3.15), we have that

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

(3.17)

Similarly, it follows from (3.1), (3.15) and (3.17) that

$$\begin{array}{rcl}\parallel u_{n+1}-u_n\parallel & =& \parallel x_{n+1}+\gamma A^{\ast }(T^{n+1}-I)A{x}_{n+1}-(x_n+\gamma A^{\ast }(T^n-I)A{x}_n)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\gamma \parallel A^{\ast }(T^{n+1}-I)A{x}_{n+1}\parallel \\ +\gamma \parallel A^{\ast }(T^n-I)A{x}_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).\end{array}$$

(3.18)

The conclusion (3.16) is proved.

• (3) Next, we prove that

  $$\parallel u_n-S{u}_n\parallel \to 0\text{and}\parallel A{x}_n-TA{x}_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

  (3.19)

In fact, from (3.14), we have

$${\zeta }_n:=\parallel u_n-S^n{u}_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

(3.20)

Since S is uniformly L-Lipschitzian continuous, it follows from (3.16) and (3.20) that

$$\begin{array}{rcl}\parallel u_n-S{u}_n\parallel & \le & \parallel u_n-S^n{u}_n\parallel +\parallel S^n{u}_n-S{u}_n\parallel \\ \le & {\zeta }_n+L\parallel S^{n-1}{u}_n-u_n\parallel \\ \le & {\zeta }_n+L\{\parallel S^{n-1}{u}_n-S^{n-1}{u}_{n-1}\parallel \\ +\parallel S^{n-1}{u}_{n-1}-u_n\parallel \}\\ \le & {\zeta }_n+L^2\parallel u_n-u_{n-1}\parallel \\ +L\parallel S^{n-1}{u}_{n-1}-u_{n-1}+u_{n-1}-u_n\parallel \\ \le & {\zeta }_n+L(1+L)\parallel u_n-u_{n-1}\parallel +L{\zeta }_{n-1}\to 0(\text{as }n\to \mathrm{\infty }).\end{array}$$

Similarly, from (3.15), we have

$$\parallel A{x}_n-T^{n}A{x}_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

(3.21)

Since T is uniformly L-Lipschitzian continuous, by the same way as above, from (3.16) and (3.21), we can also prove that

$$\parallel A{x}_n-TA{x}_n\parallel \to 0(\text{as }n\to \mathrm{\infty }).$$

(3.22)

• (4) Finally, we prove that $x_n\rightharpoonup x^{\ast }$ and $u_n\rightharpoonup x^{\ast }$, which is a solution of (SFP)-(1.1).

Since $\{u_n\}$ is bounded, there exists a subsequence $\{u_{n_i}\}\subset \{u_n\}$ such that $u_{n_i}\rightharpoonup x^{\ast }$ (some point in $H_1$). From (3.19), we have

$$\parallel u_{n_i}-S{u}_{n_i}\parallel \to 0(\text{as }{n}_i\to \mathrm{\infty }).$$

(3.23)

By the assumption that S is demi-closed at zero, we get that $x^{\ast }\in F(S)$.

Moreover, from (3.1) and (3.15), we have

$$x_{n_i}=u_{n_i}-\gamma A^{\ast }(T^{n_i}-I)A{x}_{n_i}\rightharpoonup x^{\ast }.$$

Since A is a linear bounded operator, we get $A{x}_{n_i}\rightharpoonup A{x}^{\ast }$. In view of (3.19), we have

$$\parallel A{x}_{n_i}-TA{x}_{n_i}\parallel \to 0(\text{as }{n}_i\to \mathrm{\infty }).$$

Since T is demi-closed at zero, we have $A{x}^{\ast }\in F(T)$. Summing up the above argument, it is clear that $x^{\ast }\in \mathrm{\Gamma }$, i.e., $x^{\ast }$ is a solution to the (SFP)-(1.1).

Now, we prove that $x_n\rightharpoonup x^{\ast }$ and $u_n\rightharpoonup x^{\ast }$.

Suppose, to the contrary, that if there exists another subsequence $\{u_{n_j}\}\subset \{u_n\}$ such that $u_{n_j}\rightharpoonup y^{\ast }\in \mathrm{\Gamma }$ with $y^{\ast }\ne x^{\ast }$, then by virtue of (3.2) and the Opial property of Hilbert space, we have

$$\begin{array}{rcl}\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-x^{\ast }\parallel & <& \underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-y^{\ast }\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-y^{\ast }\parallel \\ =& \underset{n_j\to \mathrm{\infty }}{lim}\parallel u_{n_j}-y^{\ast }\parallel <\underset{n_j\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_j}-x^{\ast }\parallel \\ =& \underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x^{\ast }\parallel =\underset{n_i\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\parallel u_{n_i}-x^{\ast }\parallel .\end{array}$$

This is a contradiction. Therefore, $u_n\rightharpoonup x^{\ast }$. By using (3.1) and (3.15), we have

$$x_n=u_n-\gamma A^{\ast }(T_n^n-I)A{x}_n\rightharpoonup x^{\ast }.$$

 □

The proof of conclusion (II) By the assumption that S is semi-compact, it follows from (3.23) that there exists a subsequence of $\{u_{n_i}\}$ (without loss of generality, we still denote it by $\{u_{n_i}\}$) such that $u_{n_i}\to u^{\ast }\in H$ (some point in H). Since $u_{n_i}\rightharpoonup x^{\ast }$. This implies that $x^{\ast }=u^{\ast }$, and so $u_{n_i}\to x^{\ast }\in \mathrm{\Gamma }$. By virtue of (3.2), we know that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x^{\ast }\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x^{\ast }\parallel =0$, i.e., $\{u_n\}$ and $\{x_n\}$ both converge strongly to $x^{\ast }\in \mathrm{\Gamma }$.

This completes the proof of Theorem 3.1. □

Theorem 3.2 Let $H_1$, $H_2$ and A be the same as in Theorem  3.1. Let $S:H_1\to H_1$ and $T:H_2\to H_2$ be two $(\{k_n\})$-quasi-asymptotically nonexpansive mappings with $\{k_n\}\subset [1,\mathrm{\infty })$, $k_n\to 1$ satisfying the following conditions:

• (i) T and S both are demi-closed at origin;

• (ii) ${\sum }_{n=1}^{\mathrm{\infty }}(k_n-1)<\mathrm{\infty }$.

Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{x}_1\in H_1\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}^n(u_n),\hfill \\ u_n=x_n+\gamma A^{\ast }(T^n-I)A{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.24)

where $\{{\alpha }_n\}$ is a sequence in $[0,1]$ and $\gamma >0$ is a constant satisfying the condition (iv) in Theorem  3.1. Then the conclusions in Theorem  3.1 still hold.

Proof By assumptions, $S:H_1\to H_1$ and $T:H_2\to H_2$ both are $(\{k_n\})$-quasi-asymptotically nonexpansive mappings with $\{k_n\}\subset [1,\mathrm{\infty })$, $k_n\to 1$; by Remark 2.3, S and T both are uniformly L-Lipschitzian (where $L={sup}_{n\ge 1}{k}_n$) and ($\{{\nu }_n\},\{{\mu }_n\},\zeta$)-total quasi-asymptotically nonexpansive mapping with $\{{\nu }_n=k_n-1\}$, $\{{\mu }_n=0\}$ and $\zeta (t)=t^2$, $t\ge 0$. Therefore, all conditions in Theorem 3.1 are satisfied. The conclusions of Theorem 3.2 can be obtained from Theorem 3.1 immediately. □

Theorem 3.3 Let $H_1$, $H_2$ and A be the same as in Theorem  3.1. Let $S:H_1\to H_1$ and $T:H_2\to H_2$ be two quasi-nonexpansive mappings and demi-closed at origin. Let $\{x_n\}$ be the sequence generated by

$$\{\begin{array}{c}{x}_1\in H_1\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}=(1-{\alpha }_n)u_n+{\alpha }_n{S}^n(u_n),\hfill \\ u_n=x_n+\gamma A^{\ast }(T^n-I)A{x}_n,\mathrm{\forall }n\ge 1,\hfill \end{array}$$

(3.25)

where $\{{\alpha }_n\}$ is a sequence in $[0,1]$ and $\gamma >0$ is a constant satisfying the condition (iv) in Theorem  3.1. Then the conclusions in Theorem  3.1 still hold.

Proof By the assumptions, $S:H_1\to H_1$ and $T:H_2\to H_2$ are quasi-nonexpansive mappings. By Remark 2.3, S and T both are uniformly L-Lipschitzian (where $L=1$) and ($\{1\}$)- quasi-asymptotically nonexpansive mappings. Therefore, all conditions in Theorem 3.2 are satisfied. The conclusions of Theorem 3.3 can be obtained from Theorem 3.2 immediately. □

Remark 3.4 Theorems 3.1, 3.2 and 3.3 not only improve and extend the corresponding results of Moudafi [12, 13], but also improve and extend the corresponding results of Censor et al. [4, 5], Yang [7], Xu [14], Censor and Segal [15], Masad and Reich [16] and others.