The Split Common Fixed Point Problem for a Family of Multivalued Quasinonexpansive Mappings and Totally Asymptotically Strictly Pseudocontractive Mappings in Banach Spaces

Abstract

In this paper, we introduce an iterative algorithm for solving the split common fixed point problem for a family of multi-valued quasinonexpansive mappings and totally asymptotically strictly pseudocontractive mappings, as well as for a family of totally quasi-$\varphi$-asymptotically nonexpansive mappings and k-quasi-strictly pseudocontractive mappings in the setting of Banach spaces. Our results improve and extend the results of Tang et al., Takahashi, Moudafi, Censor et al., and Byrne et al.

1. Introduction

Let $H_1$ and $H_2$ be two real Hilbert spaces and $A:H_1\to H_2$ be a bounded linear operator. For nonlinear operators $T:H_1⟶H_1$ and $U:H_2⟶H_2$, the split fixed point problem (SFPP) is to find a point:

$$x\in Fix\left(T\right)\mathrm{such}\mathrm{that}Ax\in Fix\left(U\right)$$

(1)

It is often desirable to consider the above problem for finitely many operators. Given n nonlinear operators $T_i:H_1⟶H_1$ and m nonlinear operators $U_j:H_2⟶H_2$, the split common fixed point problem (SCFPP) is to find a point:

$$x\in {\cap }_{i=1}^{n}Fix\left(T_i\right)\mathrm{such}\mathrm{that}Ax\in {\cap }_{j=1}^{m}Fix\left(U_j\right)$$

In particular, if $T_i=P_{C_i}$ and $U_j=P_{Q_j}$, then the SCFPP reduces to the multiple sets split feasibility problem (MSSFP); that is, to find $x\in {\cap }_{i=1}^n{C}_i$, such that $Ax\in {\cap }_{j=1}^m{Q}_j,$ where ${\left\{C_i\right\}}_{i=1}^n$ and ${\left\{Q_j\right\}}_{j=1}^m$ are nonempty closed convex subsets in $H_1$ and $H_2$, respectively.

In the Hilbert space setting, the split feasibility problem and the split common fixed point problem have been studied by several authors; see, for instance, [1,2,3]. In [4], Censor and Segal introduced the iterative scheme:

$$x_{n+1}=U(I-{\rho }_n{A}^{*}(I-T)A)x_n$$

which solves the problem (1) for directed operators. This algorithm was then extended to the case of quasinonexpansive mappings [5], as well as to the case of demicontractive mappings [6]. Recently, Takahashi in [7,8] extended the split feasibility problem in Hilbert spaces to the Banach space setting. Then, Alsulami et al. [1] established some strong convergence theorems for finding a solution of the split feasibility problem in Banach spaces. Using the shrinking projection method of [8], Takahashi proved the strong convergence theorem for finding a solution of the split feasibility problem in Banach spaces. In this direction, Byrne et al. [2] studied the split common null point problem for multi-valued mappings in Hilbert spaces. Consider finitely many multi-valued mappings $F_i:H_1\to 2^{H_1},1\le i\le n$, and $B_j:H_2\to 2^{H_2},1\le j\le m$, and let $A_j:H_1\to H_2$ be bounded linear operators. The split common null point problem is to find a point:

$$z\in H_1\mathrm{such}\mathrm{that}z\in \left({\cap }_{i=1}^n{F}_i^{-1}0\right)\cap ({\cap }_{j=1}^m{A}_j{B}_j^{-1}0)$$

Very recently, using the hybrid method and the shrinking projection method in mathematical programming, Takahashi et al. [9] proved two strong convergence theorems for finding a solution of the split common null point problem in Banach spaces. In [10], Tang et al. proved a theorem regarding the split common fixed point problem for a k-quasi-strictly pseudocontractive mapping and an asymptotical nonexpansive mapping. In this paper, motivated by [11], we use the hybrid method to study the split common fixed point problem for an infinite family of multi-valued quasinonexpansive mappings and an infinite family of L-Lipschitzian continuous and $(k,\left\{{\mu }_n\right\},\left\{{\xi }_n\right\})$-totally asymptotically strictly pseudocontractive mappings. Compared to the Theorem of Tang et al. [10], we remove an extra condition and present a strong convergence theorem, which is more desirable than the weak convergence. The point is that the authors of [10] considered a semi-compact mapping, that is a mapping T on a set X having the property that if $\left\{x_n\right\}$ is a bounded sequence in X such that $\parallel T{x}_n-x_n\parallel$ tends to zero, then $\left\{x_n\right\}$ has a convergent subsequence. We will not assume that our mappings are semi-compact, and at the same time, we propose a different algorithm; instead, we impose some restrictions on the control sequences to get the strong convergence. We also present an algorithm for solving the split common fixed point problem for totally quasi-$\varphi$-asymptotically nonexpansive mappings and for k-quasi-strictly pseudocontractive mappings. Under some mild conditions, we establish the strong convergence of these algorithms in Banach spaces. As applications, we consider the algorithms for a split variational inequality problem and a split common null point problem. Our results improve and generalize the result of Tang et al. [10], Takahashi [12], Moudafi [5], Censor et al. [13] and Byrne et al. [2].

2. Preliminaries

Let E be a real Banach space and C be a nonempty closed convex subset of E. A mapping $T:C\to C$ is said to be $\left\{k_n\right\}$-asymptotically nonexpansive if there exists a sequence $\left\{k_n\right\}\subset [1,\infty )$ with $k_n\to 1$, such that:

$$\parallel T^{n}x-T^{n}y\parallel \le k_n\parallel x-y\parallel ,\forall x,y\in C,n\ge 1$$

The mapping $T:C\to C$ is said to be k-quasi-strictly pseudocontractive if $F\left(T\right)\ne \varnothing$ and there exists a constant $k\in [0,1]$, such that:

$${\parallel Tx-p\parallel }^2\le {\parallel x-p\parallel }^2+k{\parallel x-Tx\parallel }^2\forall x\in C,p\in F\left(T\right)$$

The mapping $T:C\to C$ is said to be $(k,\left\{{\mu }_n\right\},\left\{{\xi }_n\right\})$-totally asymptotically strictly pseudocontractive if there exist a constant $k\in [0,1]$ and null sequences $\left\{{\mu }_n\right\}$ and $\left\{{\xi }_n\right\}$ in $[0,\infty )$ and a continuous strictly increasing function $\zeta :[0,\infty )\to [0,\infty )$ with $\zeta \left(0\right)=0$, such that for all $x,y\in H$ and $n\ge 1$:

$$\parallel T^{n}x-T^n{y\parallel }^2\le {\parallel x-y\parallel }^2{+k\parallel (x-y)-(Tx-Ty)\parallel }^2+{\mu }_n\zeta (\parallel x-y\parallel )+{\xi }_n$$

A Banach space E is said to be uniformly smooth if $\frac{{\rho }_E\left(t\right)}{t}\to 0$ as $t\to 0$, where ${\rho }_E\left(t\right)$ is the modulus of smoothness of E. Let $q>1$; then, E is called q-uniformly smooth if there exists a constant $c>0$, such that ${\rho }_E\left(t\right)\le c{t}^q$ for all $t>0$. Throughout, J will stand for the duality mapping of E. We recall that a Banach space E is smooth if and only if the duality mapping J is single valued.

Lemma 1.

[14] If E is a two-uniformly smooth Banach space, then for each $t>0$ and each $x,y\in E$:

$${\parallel x+ty\parallel }^2\le {\parallel x\parallel }^2+2\langle y,Jx\rangle +2{\parallel ty\parallel }^2$$

For a smooth Banach space E, Alber [15] defined:

$$\varphi (x,y)={\parallel x\parallel }^2-2\langle x,Jy\rangle +{\parallel y\parallel }^2,x,y\in E$$

It follows that $(\parallel x\parallel -{\parallel y\parallel )}^2\le \varphi (x,y)\le (\parallel x\parallel +{\parallel y\parallel )}^2$ for each $x,y\in E$. Moreover, if we denote by ${\mathrm{\Pi }}_{C}x$ the generalized projection from E onto a closed convex subset C in E, then we have:

Lemma 2.

[15] Let E be a smooth, strictly convex and reflexive Banach space and C be a nonempty closed convex subset of E. Then:

(a) 

$\varphi (x,{\mathrm{\Pi }}_{C}y)+\varphi ({\mathrm{\Pi }}_{C}y,y)\le \varphi (x,y)$, for all $x\in C$ and $y\in E$;

(b) 

For $x,y\in E,\varphi (x,y)=0$ if and only if $x=y;$

(c) 

For $x,y,z\in E,\varphi (x,y)\le \varphi (x,z)+\varphi (z,y)+2\langle x-z,Jz-Jy\rangle ;$

(d) 

For $x,y,z\in E,\lambda \in [0,1],\varphi (x,J^{-1}(\lambda Jy+(1-\lambda )Jz))\le \lambda \varphi (x,y)+(1-\lambda )\varphi (x,z).$

Lemma 3.

[16] If E is a uniformly-smooth Banach space and $r>0$, then there exists a continuous, strictly-increasing convex function $g:[0,2r]\to [0,\infty )$, such that $g\left(0\right)=0$ and:

$$\varphi (x,J^{-1}(\lambda Jy+(1-\lambda )Jz))\le \lambda \varphi (x,y)+(1-\lambda )\varphi (x,z)-\lambda (1-\lambda )g(\parallel Jy-Jz\parallel )$$

for all $\lambda \in [0,1],x\in E$ and $y,z\in B_r=\{u\in E:\parallel u\parallel \le r\}.$

We denote by $N\left(C\right)$, $CB\left(C\right)$ and $P\left(C\right)$ the collection of all nonempty subsets, nonempty closed bounded subsets and nonempty proximal bounded subsets of C, respectively. Let $T:E\to N\left(E\right)$ be a multivalued mapping. An element $x\in E$ is said to be a fixed point of T if $x\in Tx$. The set of fixed points of T is denoted by $F\left(T\right)$.

Definition 1.

Let C be a closed convex subset of a smooth Banach space E and $T:C\to N\left(C\right)$ be a multivalued mapping. We set:

$$\mathrm{\Phi }(Tx,Tp)=max\{\underset{q\in Tp}{sup}\underset{y\in Tx}{inf}\varphi (y,q),\underset{y\in Tx}{sup}\underset{q\in Tp}{inf}\varphi (y,q)\}$$

We call T a quasinonexpansive multivalued mapping if $F\left(T\right)\ne \varnothing$ and:

$$\mathrm{\Phi }(Tx,Tp)\le \varphi (x,p),\forall p\in F\left(T\right),\forall x\in C$$

Definition 2.

A multivalued mapping T is called demi-closed if ${lim}_{n\to \infty }dist(x_n,T{x}_n)=0$ and $x_n\rightharpoonup w$ imply that $w\in Tw$.

Let C be a nonempty closed convex subset of E and $T:=\left\{T\right(s):0\le s<\infty \}$ be a nonexpansive semigroup on C. We use $Fix\left(T\right)$ to denote the common fixed point set of the semigroup T. It is well known that $Fix\left(T\right)$ is closed and convex. A nonexpansive semigroup T on C is said to be uniformly asymptotically regular (u.a.r.) if for all $h\ge 0$ and any bounded subset D of C:

$$\underset{n\to \infty }{lim}\underset{x\in D}{sup}\parallel T\left(h\right)\left(T\left(t\right)x\right)-T\left(t\right)x\parallel =0$$

For each $h\ge 0$, define ${\sigma }_t\left(x\right)=\frac{1}{t}{\int }_0^{t}T\left(s\right)xds$. Then, ${lim}_{t\to \infty }{sup}_{x\in D}\parallel T\left(h\right)\left({\sigma }_t\left(x\right)\right)-{\sigma }_t\left(x\right)\parallel =0$ provided that D is a closed bounded convex subset of C. It is known that the set $\{{\sigma }_t\left(x\right):t>0\}$ is a u.a.r. nonexpansive semigroup; see [17].

A mapping $T:E\to E$ is said to be α-averaged if $T=(1-\alpha )I+\alpha S$ for some $\alpha \in (0,1)$; here, I is the identity operator, and $S:E\to E$ is a nonexpansive mapping (see [18]). It is known that in a Hilbert space setting, every firmly-nonexpansive mapping (in particular, a projection) is a $\frac{1}{2}$-averaged mapping (see Proposition 11.2 in the book [19]).

Lemma 4.

[20] (i) The composition of finitely many averaged mappings is averaged. In particular, if $T_i$ is ${\alpha }_i$-averaged, where ${\alpha }_i\in (0,1)$ for $i=1,2,$ then the composition $T_1{T}_2$ is α-averaged, where $\alpha ={\alpha }_1+{\alpha }_2-{\alpha }_1{\alpha }_2$. (ii) If the mappings ${\left\{T_i\right\}}_{i=1}^N$ are averaged and have a common fixed point, then ${\cap }_{i=1}^{N}F\left(T_i\right)=F(T_1\cdots T_N)$. (iii) In case E is a uniformly-convex Banach space, every α-averaged mapping is nonexpansive.

Lemma 5.

[21] Let E be a uniformly-convex and smooth Banach space, and let $\left\{x_n\right\}$ and $\left\{y_n\right\}$ be two sequences in E. If $\varphi (x_n,y_n)\to 0$ and either $\left\{x_n\right\}$ or $\left\{y_n\right\}$ is bounded, then $x_n-y_n\to 0$.

Lemma 6.

[15] Let C be a nonempty closed convex subset of a smooth Banach space E and $x\in E$, then $x_0={\mathrm{\Pi }}_{C}x$ if and only if for all $y\in C$, $\langle x_0-y,Jx-J{x}_0\rangle \ge 0.$

Lemma 7.

[22] Let E be a uniformly-convex Banach space, and let $B_r\left(0\right)=\{x\in E:\parallel x\parallel \le r\}$, for $r>0$, then there exists a continuous, strictly-increasing and convex function $g:[0,\infty )\to [0,\infty )$ with $g\left(0\right)=0$, such that, for any given sequence ${\left\{x_n\right\}}_{n=1}^{\infty }\subset B_r\left(0\right)$ and for any given sequence ${\left\{{\alpha }_n\right\}}_{n=1}^{\infty }$ of positive numbers with ${\sum }_{n=1}^{\infty }{a}_n=1$ and for any positive integers $i,j$ with $i<j:$

$$\parallel \sum _{n=1}^{\infty }{\alpha }_n{x}_n{\parallel }^2\le \sum _{n=1}^{\infty }{\alpha }_n\parallel x_n{\parallel }^2-{\alpha }_i{\alpha }_{j}g(\parallel x_i-x_j\parallel ).$$

Lemma 8.

[23] Let $\left\{{\alpha }_n\right\}$ be a sequence in $[0,1]$, ${\delta }_n$ and $\left\{{\gamma }_n\right\}$ be sequences in $\mathbb{R}$, such that (i) ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$, (ii) ${lim\; sup}_{n\to \infty }{\delta }_n\le 0$ and (iii) ${\gamma }_n\ge 0$ and ${\sum }_{n=1}^{\infty }{\gamma }_n<\infty$. If $\left\{a_n\right\}$ is a sequence of nonnegative real numbers, such that $a_{n+1}\le (1-{\alpha }_n)a_n+{\alpha }_n{\delta }_n+{\gamma }_n,$ for each $n\ge 0$, then ${lim}_{n\to \infty }{a}_n=0$.

Lemma 9.

[24] Let $\left\{s_n\right\}$ be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence $\left\{s_{n_i}\right\}$ of $\left\{s_n\right\}$, such that $s_{n_i}\le s_{n_{i+1}}$ for all $i\ge 0$. For every $n\in \mathbb{N}$, define an integer sequence $\left\{\tau \right(n\left)\right\}$ as $\tau \left(n\right)=max\{k\le n:s_k<s_{k+1}\}.$ Then, $\tau \left(n\right)\to \infty$ and $max\{s_{\tau \left(n\right)},s_n\}\le s_{\tau \left(n\right)+1}$.

Lemma 10.

[25] Let $\left\{{\lambda }_n\right\}$ and $\left\{{\gamma }_n\right\}$ be nonnegative and $\left\{{\alpha }_n\right\}$ be positive real numbers, such that ${\lambda }_{n+1}\le {\lambda }_n-{\alpha }_n{\lambda }_n+{\gamma }_n,n\ge 0.$ Let for all $n>1$, $\frac{{\lambda }_n}{{\alpha }_n}\le c_1\mathrm{and}{\alpha }_n\le \alpha .$ Then, ${\lambda }_n\le max\{{\lambda }_1,K^{*}\}$, where $K^{*}=(1+\alpha )c_1$.

Definition 3.

(1) A mapping $T:C\to C$ is said to be a k-quasi-strictly pseudocontractive mapping if there exists $k\in [0,1)$, such that ${\parallel Tx-p\parallel }^2\le {\parallel x-p\parallel }^2+k{\parallel x-Tx\parallel }^2,\forall x\in C,p\in F\left(T\right).$ (2) A mapping $T:C\to C$ is called quasinonexpansive if $F\left(T\right)\ne \varnothing$; and $\varphi (p,Tx)\le \varphi (p,x)\forall x\in C,p\in F\left(T\right).$ (3) A countable family of mappings $\left\{T_i\right\}:C\to C$ is said to be totally uniformly quasi-$\varphi$-asymptotically nonexpansive, if $\Im ={\bigcap }_{i=1}^{\infty }F\left(T_i\right)\ne \varnothing$ and there exist nonnegative real sequences $\left\{{\mu }_n\right\}$,$\left\{{\nu }_n\right\}$ with ${\mu }_n\to 0,{\nu }_n\to 0(asn\to \infty )$ and a strictly-increasing continuous function $\zeta :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ with $\zeta \left(0\right)=0$, such that $\varphi (p,T_i^{n}x)\le \varphi (p,x)+{\nu }_n\zeta \left(\varphi (p,x)\right)+{\mu }_n,n\ge 1,i\ge 1,x\in C,p\in \Im .$ (4) A mapping $T:C\to C$ is said to be uniformly L-Lipschitzian continuous, if there exists a constant $L>0$, such that $\parallel T^{n}x-T^{n}y\parallel \le L\parallel x-y\parallel ,\forall x,y\in C,n\ge 1.$

Lemma 11.

[11] Let E be a real uniformly-smooth and uniformly-convex Banach space and C be a nonempty closed convex subset of E. Let $T:C\to C$ be a closed and totally quasi-$\varphi$-asymptotically nonexpansive mapping with nonnegative real sequences $\left\{{\mu }_n\right\},\left\{{\nu }_n\right\}$ and a strictly-increasing continuous function $\zeta :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$, such that ${\mu }_n\to 0,{\nu }_n\to 0and\zeta \left(0\right)=0$. If ${\mu }_1=0$, then the fixed point set of T is closed and convex.

Lemma 12.

[26] Let C be a nonempty closed convex subset of a real Banach space E, and let $T:C\to C$ be a k-quasi-strictly pseudocontractive mapping. If $F\left(T\right)\ne \varnothing$, then $F\left(T\right)$ is closed and convex.

3. Main Results

This section is devoted to the main results of this paper.

Theorem 1.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<\frac{1}{\sqrt{2}}$, and let $E_2$ be a real smooth Banach space. Let $A:E_1\to E_2$ be a bounded linear operator and $A^{*}$ be its adjoint. Suppose $T:E_2\to E_2$ is a uniformly L-Lipschitzian continuous and $(k,\left\{{\mu }_n\right\},\left\{{\xi }_n\right\})$-totally asymptotically strictly pseudocontractive mapping satisfying the following conditions:

(1) 

${\sum }_{n=1}^{\infty }{\mu }_n<\infty ,{\sum }_{n=1}^{\infty }{\xi }_n<\infty$,

(2) 

$\left\{r_n\right\}$ is a real sequence in $(0,1)$, such that ${\mu }_n=o\left(r_n\right),{\xi }_n=o\left(r_n\right),lim{r}_n=0,{\sum }_{n=1}^{\infty }{r}_n=\infty$,

(3) 

there exist constants $M_0>0,M_1>0$, such that $\zeta \left(\lambda \right)\le M_0{\lambda }^2,\forall \lambda >M_1$.

Let ${\left\{S_n\right\}}_{n=1}^{\infty }:E_1\to CB\left(E_1\right)$ be a family of multivalued quasinonexpansive mappings, such that for each $i\ge 1,S_i$ is demi-closed at zero, and for each $p\in Fix\left(S_i\right),S_i\left(p\right)=\left\{p\right\}$. Suppose:

$$\mathrm{\Omega }=\left\{x\in \bigcap _{i=1}^{\infty }F\left(S_i\right):Ax\in F\left(T\right)\right\}\ne \varnothing$$

and $\left\{x_n\right\}$ is the sequence generated by $x_1\in E_1:$

$$\left\{\begin{array}{c}{u}_n=(1-r_n)x_n\hfill \\ y_n=J_1^{-1}({\alpha }_n{J}_1{u}_n+(1-{\alpha }_n)\gamma A^{*}{J}_2(T^n-I)A{u}_n)\hfill \\ x_{n+1}=J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{w}_{n,i})w_{n,i}\in S_i{y}_n\hfill \end{array}\right.$$

(2)

where $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$; the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_{n,i}\right\}\subset (0,1)$ satisfy the following conditions:

(a) 

${\sum }_{i=0}^{\infty }{\beta }_{n,i}=1,{lim\; inf}_n{\beta }_{n,0}{\beta }_{n,i}>0,$

(b) 

${lim}_{n\to \infty }{\alpha }_n=1,{\sum }_{n=1}^{\infty }(1-{\alpha }_n)<\infty ,(1-{\alpha }_n)=o\left(r_n\right).$

Then, $\left\{x_n\right\}$ converges strongly to an element of Ω.

Proof. 

Since ζ is continuous, ζ attains its maximum in $[0,M_1]$, and by assumption, $\zeta \left(\lambda \right)\le M_0{\lambda }^2,$$\forall \lambda >M_1$. In either case, we have $\zeta \left(\lambda \right)\le M+M_0{\lambda }^2,\forall \lambda \in [0,\infty )$. Let $p\in \mathrm{\Omega }$, then:

$$\begin{array}{c}\hfill \varphi (p,u_n)\le (1-r_n)\varphi (p,x_n)+r_n{\parallel p\parallel }^2\end{array}$$

(3)

From (2) and Lemma 2(d,c), we have:

$$\begin{array}{cc}\hfill \varphi (p,y_n)& \le {\alpha }_n\varphi (p,u_n)+(1-{\alpha }_n)\varphi (p,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & \le {\alpha }_n\varphi (p,u_n)+(1-{\alpha }_n)[\varphi (p,u_n)+\varphi (u_n,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & +2\langle p-u_n,J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle ]\hfill \\ & =\varphi (p,u_n)+(1-{\alpha }_n)[\parallel u_n{\parallel }^2+{\gamma }^2{\parallel A\parallel }^2{\parallel (T^n-I)A{u}_n\parallel }^2-2\langle u_n,\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle \hfill \\ & +2\langle p-u_n,J_1{u}_n\rangle +2\langle p-u_n,\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle ]\hfill \\ & \le \varphi (p,u_n)+(1-{\alpha }_n){[\parallel p\parallel }^2+{\gamma }^2{\parallel A\parallel }^2{\parallel (T^n-I)A{u}_n\parallel }^2-2\langle u_n,\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle \hfill \\ & +2\langle p-u_n,\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle ]\hfill \end{array}$$

(4)

From Lemma 1, we have:

$$\begin{array}{cc}\hfill -2\langle u_n,\gamma A^{*}{J}_2(T_n-I)A{u}_n\rangle & \le \parallel \gamma A^{*}{J}_2(T^n-I)A{u}_n{\parallel }^2+2\parallel t{u}_n{\parallel }^2-{\parallel u_n+\gamma A^{*}{J}_2(T_n-I)A{u}_n\parallel }^2\hfill \\ & \le {\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+{\parallel u_n\parallel }^2\hfill \\ & ={\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+4\parallel \frac{1}{2}{u}_n-\frac{1}{2}p+\frac{1}{2}p{\parallel }^2\hfill \\ & \le {\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+4(\frac{1}{2}\parallel u_n{-p\parallel }^2+\frac{1}{2}{\parallel p\parallel }^2)\hfill \\ & ={\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+2\parallel u_n{-p\parallel }^2+{2\parallel p\parallel }^2)\hfill \end{array}$$

(5)

Since $Ap\in F\left(T\right)$ and T is a totally quasi-asymptotically strictly pseudocontractive mapping, we obtain:

$$\begin{array}{cc}\hfill \langle u_n-p,\gamma A^{*}{J}_2(T^n-I)A{u}_n& =\gamma \langle A(u_n-p),J_2(T^n-I)A{u}_n\rangle \hfill \\ & =\gamma \langle A(u_n-p)+(T^n-I)A{u}_n-(T^n-I)A{u}_n,J_2(T^n-I)A{u}_n\rangle \hfill \\ & =\gamma (\langle T^{n}A\left(u_n\right)-Ap,J_2(T^n-I)A{u}_n\rangle -\parallel (T^n-I)A{u}_n{\parallel }^2)\hfill \\ & \le \gamma (\frac{1}{2}[\parallel (T^n-I)A{u}_n{\parallel }^2+2{\parallel t(T^{n}A{u}_n-Ap)\parallel }^2\hfill \\ & -\parallel Ap-A{u}_n{\parallel }^2]-\parallel (T^n-I)A{u}_n{\parallel }^2)\hfill \\ & \le \gamma (\frac{1}{2}[\parallel (T^n-I)A{u}_n{\parallel }^2+{\parallel (T^{n}A{u}_n-Ap)\parallel }^2\hfill \\ & -\parallel Ap-A{u}_n{\parallel }^2]-\parallel (T^n-I)A{u}_n{\parallel }^2)\hfill \\ & \le \gamma (\frac{1}{2}[\parallel A{u}_n{-Ap\parallel }^2+k\parallel (T^n-I)A{u}_n{\parallel }^2+{\mu }_n\zeta (\parallel A{u}_n-Ap\parallel )+{\xi }_n\left]\right)\hfill \\ & -\frac{1}{2}(\parallel (T^n-I)A{u}_n{\parallel }^2+\parallel Ap-A{u}_n{\parallel }^2)\hfill \\ & =\gamma (\frac{k-1}{2}\parallel (T^n-I)A{u}_n{\parallel }^2+\frac{{\mu }_n}{2}[M+M_0\parallel A{u}_n{-Ap\parallel }^2]+\frac{{\xi }_n}{2})\hfill \end{array}$$

(6)

Substituting (5) and (6) into (4), we have:

$$\begin{array}{cc}\hfill \varphi (p,y_n)& \le {\alpha }_n\varphi (p,u_n)+(1-{\alpha }_n)\varphi (p,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & \le \varphi (p,u_n)+(1-{\alpha }_n){[3\parallel p\parallel }^2+2{\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+2{\parallel u_n-p\parallel }^2\hfill \\ & +\gamma (k-1)\parallel (T^n-I)A{u}_n{\parallel }^2+\gamma {\mu }_n[M+M_0{\parallel A\parallel }^2\parallel u_n-p{\parallel }^2]+\gamma {\xi }_n\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\parallel u_n{-p\parallel }^2+\gamma {\xi }_n\hfill \end{array}$$

(7)

From Lemma 1 and the fact that $0<t<\frac{1}{\sqrt{2}}$, we have:

$$\begin{array}{cc}\hfill \varphi (p,y_n)& \le {\alpha }_n\varphi (p,u_n)+(1-{\alpha }_n)\varphi (p,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\parallel u_n{-p\parallel }^2+\gamma {\xi }_n\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2\left)\right[\parallel u_n{\parallel }^2-\langle p,J{u}_n\rangle +{2\parallel tp\parallel }^2]+\gamma {\xi }_n\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\varphi (p,u_n)+\gamma {\xi }_n\hfill \end{array}$$

(8)

Putting (3) and (8) into (2), we obtain:

$$\begin{array}{cc}\hfill \varphi (p,x_{n+1})& =\varphi (p,J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{w}_{n,i}))\hfill \\ & \le {\beta }_{n,0}\varphi (p,y_n)+\sum _{i=1}^{\infty }{\beta }_{n,i}\varphi (p,w_{n,i})\hfill \\ & ={\beta }_{n,0}\varphi (p,y_n)+\sum _{i=1}^{\infty }{\beta }_{n,i}\underset{t\in S_i\left(p\right)}{inf}\varphi (p,w_{n,i})\hfill \\ & \le {\beta }_{n,0}\varphi (p,y_n)+\sum _{i=1}^{\infty }{\beta }_{n,i}\mathrm{\Phi }(p,w_{n,i})=\varphi (p,y_n)\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\varphi (p,u_n)+\gamma {\xi }_n\hfill \\ & \le (1-r_n)\varphi (p,x_n)+r_n{\parallel p\parallel }^2+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2\left)\right((1-r_n)\varphi (p,x_n)+r_n{\parallel p\parallel }^2)+\gamma {\xi }_n\hfill \\ & \le \varphi (p,x_n)-(r_n-\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)(1-r_n)\varphi (p,x_n)\hfill \\ & +(3(1-{\alpha }_n)+r_n+{\mu }_n\gamma M_0{\parallel A\parallel }^2{r}_n{)\parallel p\parallel }^2+\gamma {\mu }_{n}M+\gamma {\xi }_n\hfill \\ & \le \varphi (p,x_n)-(r_n-(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2\left)\right)(1-r_n)\varphi (p,x_n)+{\sigma }_n\hfill \end{array}$$

(9)

where ${\sigma }_n=(3(1-{\alpha }_n)+r_n+{\mu }_n\gamma M_0{\parallel A\parallel }^2{r}_n{)\parallel p\parallel }^2+{\mu }_n\gamma M+\gamma {\xi }_n$. Since ${\mu }_n=o\left(r_n\right),(1-{\alpha }_n)=o\left(r_n\right)$ and ${\xi }_n=o\left(r_n\right)$, we may assume without loss of generality that there exist constants $k_0\in (0,1)$ and $M_2>0$, such that for all $n\ge 1$:

$$\frac{{\mu }_n}{r_n}\le \frac{r_n(1-k_0+2)-2}{r_n(1-r_n)\gamma M_0{\parallel A\parallel }^2}and\frac{{\sigma }_n}{r_n}\le M_2$$

Thus, we obtain:

$$\begin{array}{c}\hfill \varphi (p,x_{n+1})\le \varphi (p,x_n)-r_n{k}_0\varphi (p,x_n)+{\sigma }_n\end{array}$$

(10)

According to Lemma 10, $\varphi (p,x_{n+1})\le max\{\varphi (p,x_1),(1+k_0)M_2\}$. Therefore, $\left\{\varphi (p,x_n)\right\}$ and $\left\{x_n\right\}$ are bounded. Furthermore, the sequences $\left\{y_n\right\}$ and $\left\{u_n\right\}$ are bounded, as well. We now consider two cases.

Case 1.

Suppose that there exists $n_0\in \mathbb{N}$, such that ${\left\{\varphi (p,x_n)\right\}}_{n=n_0}^{\infty }$ is nonincreasing. Then, ${\left\{\varphi (p,x_n)\right\}}_{n=1}^{\infty }$ converges, and $\varphi (p,x_n)-\varphi (p,x_{n+1})\to 0$ as $n\to \infty$. Since $E_1$ is a uniformly smooth Banach space, it follows from Lemma 3 and Equations (8) and (10) that:

$$\begin{array}{cc}\hfill \varphi (p,x_{n+1})& \le \varphi (p,y_n)\hfill \\ & \le {\alpha }_n\varphi (p,u_n)+(1-{\alpha }_n)\varphi (p,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))-{\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel )\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\varphi (p,u_n)+\gamma {\xi }_n-{\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel \hfill \\ & \le \varphi (p,x_n)-(r_n-(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2\left)\right)\varphi (p,u_n)+(3(1-{\alpha }_n)+r_n){\parallel p\parallel }^2\hfill \\ & +\gamma {\xi }_n{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2-{\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel )\hfill \\ & \le \varphi (p,x_n)-r_n{k}_0\varphi (p,x_n)+{\sigma }_n{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & -{\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel )\hfill \end{array}$$

(11)

Hence, from (10), we have:

$${\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel )\le \varphi (p,x_n)-\varphi (p,x_{n+1})-r_n{k}_0\varphi (p,x_n)+{\sigma }_n$$

and:

$${\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\le \varphi (p,x_n)-\varphi (p,x_{n+1})-r_n{k}_0\varphi (p,x_n)+{\sigma }_n$$

Therefore, ${\alpha }_n(1-{\alpha }_n)g(\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel )$ and ${\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2$ tend to zero as $n\to \infty$. Since $lim\; inf{\alpha }_n(1-{\alpha }_n)>0$ and $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$, we obtain:

$$\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel ⟶0n\to \infty$$

(12)

$$\parallel (T^n-I)A{u}_n{\parallel }^2⟶0n\to \infty$$

(13)

Furthermore, we observe that $\parallel J_1{y}_n-J_1{u}_n\parallel =(1-{\alpha }_n)\parallel J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\parallel \to 0$. Since $J_1^{-1}$ is uniformly norm-to-norm continuous on bounded subsets, we conclude that:

$$\underset{n\to \infty }{lim}\parallel y_n-u_n\parallel =0$$

(14)

Using (7) and Lemma 3 in (2), we have:

$$\begin{array}{cc}\hfill \varphi (p,x_{n+1})& =\varphi (p,J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{w}_{n,i}))\hfill \\ & \le {\beta }_{n,0}\varphi (p,y_n)+\sum _{i=1}^{\infty }{\beta }_{n,i}\varphi (p,w_{n,i})-{\beta }_{n,0}{\beta }_{n,i}g(\parallel J_1{y}_n-J_1{w}_{n,i}\parallel )\hfill \\ & \le \varphi (p,y_n)-{\beta }_{n,0}{\beta }_{n,i}g(\parallel J_1{y}_n-J_1{w}_{n,i}\parallel )\hfill \\ & \le \varphi (p,u_n)+3(1-{\alpha }_n){\parallel p\parallel }^2{-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T^n-I)A{u}_n{\parallel }^2\hfill \\ & +\gamma {\mu }_{n}M+(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2)\parallel u_n{-p\parallel }^2+\gamma {\xi }_n-{\beta }_{n,0}{\beta }_{n,i}g(\parallel J_1{y}_n-J_1{w}_{n,i}\parallel )\hfill \end{array}$$

(15)

It now follows from (3) and $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$ that:

$$\begin{array}{cc}\hfill {\beta }_{n,0}{\beta }_{n,i}g(\parallel J_1{y}_n-J_1{w}_{n,i}\parallel )& \le \varphi (p,x_n)-\varphi (p,x_{n+1})-(r_n-(\gamma {\mu }_n{M}_0{\parallel A\parallel }^2+2\left)\right)\varphi (p,u_n)\hfill \\ & +(3(1-{\alpha }_n)+r_n){\parallel p\parallel }^2+\gamma {\xi }_n\hfill \\ & \le \varphi (p,x_n)-\varphi (p,x_{n+1})-r_n{k}_0\varphi (p,x_n)+{\sigma }_n\hfill \end{array}$$

From Condition (a), we have ${lim}_{n\to \infty }g(\parallel J_1{y}_n-J_1{w}_{n,i}\parallel )=0$. Since g is continuous and $g\left(0\right)=0$, we obtain ${lim}_{n\to \infty }\parallel J_1{y}_n-J_1{w}_{n,i}\parallel =0$. Since $J_1^{-1}$ is uniformly norm-to-norm continuous on bounded subsets, we have:

$$\underset{n\to \infty }{lim}\parallel y_n-w_{n,i}\parallel =0\forall i\in \mathbb{N}$$

(16)

which implies that ${lim}_{n\to \infty }dist(y_n,S_i{y}_n)\le {lim}_{n\to \infty }\parallel y_n-w_{n,i}\parallel =0,\forall i\in \mathbb{N}$. From (2), we obtain:

$$\parallel J_1{x}_{n+1}-J_1{y}_n\parallel =(1-{\beta }_{n,0})\parallel J_1{y}_n-J_1{w}_{n,i}\parallel ⟶0n\to \infty$$

Since J is uniformly norm-to-norm continuous on bounded subsets, we have:

$$\parallel x_{n+1}-y_n\parallel ⟶0n\to \infty$$

(17)

From (14), (17) and ${lim}_{n\to \infty }{r}_n=0$, we have:

$$\begin{array}{cc}\hfill \parallel x_{n+1}-x_n\parallel & \le \parallel x_{n+1}-y_n\parallel +\parallel y_n-u_n\parallel +\parallel u_n-x_n\parallel \hfill \\ & =\parallel x_{n+1}-y_n\parallel +\parallel y_n-u_n\parallel +r_n\parallel x_n\parallel ⟶0n\to \infty \hfill \end{array}$$

Consequently:

$$\begin{array}{cc}\hfill \parallel u_{n+1}-u_n\parallel & =\parallel (1-r_{n+1})x_{n+1}-(1-r_n)x_n)\parallel \hfill \\ & \le |r_{n+1}-r_n|\parallel x_{n+1}\parallel +(1-r_n)\parallel x_{n+1}-x_n\parallel ⟶0n\to \infty \hfill \end{array}$$

(18)

Using the fact that T is uniformly L-Lipschitzian, we have:

$$\begin{array}{cc}\hfill \parallel TA{u}_n-A{u}_n\parallel & \le \parallel TA{u}_n-T^{n+1}A{u}_n\parallel +\parallel T^{n+1}A{u}_n-T^{n+1}A{u}_{n+1}\parallel \hfill \\ & +\parallel T^{n+1}A{u}_{n+1}-A{u}_{n+1}\parallel +\parallel A{u}_{n+1}-A{u}_n\parallel \hfill \\ & \le L\parallel A{u}_n-T^{n}A{u}_n\parallel +(1+L)\parallel A{u}_{n+1}-A{u}_n\parallel +\parallel T^{n+1}A{u}_{n+1}-A{u}_{n+1}\parallel \hfill \\ & \le L\parallel A{u}_n-T^{n}A{u}_n\parallel +(1+L)\parallel A\parallel \parallel u_{n+1}-u_n\parallel +\parallel T^{n+1}A{u}_{n+1}-A{u}_{n+1}\parallel \hfill \end{array}$$

From (13) and (18), we obtain:

$$\parallel (T-I)A{u}_n\parallel ⟶0,n\to \infty$$

(19)

Since $\left\{x_n\right\}$ is bounded, there exists a subsequence $\left\{x_{n_j}\right\}$ of $\left\{x_n\right\}$, such that $x_{n_j}\rightharpoonup z$. Using the fact that $x_{n_j}\rightharpoonup z$ and $\parallel y_n-x_n\parallel \to 0$, $n\to \infty$, we have that $y_{n_j}\rightharpoonup z$. Similarly, $u_{n_j}\rightharpoonup z$, since $\parallel u_n-x_n\parallel \to 0,n\to \infty$. Now, we show that $z\in \mathrm{\Omega }$. Since $y_{n_j}\rightharpoonup z$ and ${lim}_{n\to \infty }dist(y_n,S_i\left(y_n\right))=0$ and by the demi-closedness of each $S_i$, we have $z\in {\bigcap }_{i\in \mathbb{N}}F\left(S_i\right)$. On the other hand, since A is a bounded operator, it follows from $u_{n_j}\rightharpoonup z$ that $A{u}_{n_j}\rightharpoonup Az$. Hence, from (13), we have $\parallel TA{u}_{n_j}-A{u}_{n_j}\parallel \to 0$ as $j\to \infty$. Since T is demi-closed at zero, we have that $Az\in F\left(T\right)$. Hence, $z\in \mathrm{\Omega }$. Next, we prove that $\left\{x_n\right\}$ converges strongly to z. From (7), Lemma 1 and $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$, we have:

$$\begin{array}{cc}\hfill \varphi (z,x_{n+1})& \le \varphi (z,y_n)\le {\alpha }_n\varphi (z,u_n)+(1-{\alpha }_n)\varphi (z,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & \le {\alpha }_n\varphi (z,u_n)+(1-{\alpha }_n)[\varphi (z,u_n)+\varphi (u_n,J_1^{-1}\left(\gamma A^{*}{J}_2(T^n-I)A{u}_n\right))\hfill \\ & +2\langle z-u_n,J_1{u}_n-\gamma A^{*}{J}_2(T^n-I)A{u}_n\rangle ]\hfill \\ & \le \varphi (z,u_n)+(1-{\alpha }_n){[\parallel z\parallel }^2+\parallel u_n{-z+z\parallel }^2+2{\gamma }^2{\parallel A\parallel }^2{\parallel (T^n-I)A{u}_n\parallel }^2\hfill \\ & +\gamma (k-1)\parallel (T^n-I)A{u}_n{\parallel }^2+\gamma {\mu }_n[M+M_0{\parallel A\parallel }^2\parallel u_n-z{\parallel }^2]+\gamma {\xi }_n\hfill \\ & \le \varphi (z,u_n)+(1-{\alpha }_n){[\parallel z\parallel }^2+\parallel u_n{-z\parallel }^2+{\parallel z\parallel }^2+2\langle u_n-z,Jz\rangle \hfill \\ & +2{\gamma }^2{\parallel A\parallel }^2\parallel (T^n-I)A{u}_n{\parallel }^2+\gamma (k-1){\parallel (T^n-I)A{u}_n\parallel }^2\hfill \\ & +\gamma {\mu }_n[M+M_0{\parallel A\parallel }^2\parallel u_n-z{\parallel }^2]+\gamma {\xi }_n\hfill \\ & \le \varphi (z,u_n)+(1-{\alpha }_n)(\parallel u_n-z\parallel +2\langle u_n,J_{1}z\rangle )+{\mu }_n{M}^{*}+\gamma {\xi }_n\hfill \\ & \le (1-r_n)\varphi (z,x_n)-2r_n\langle x_n-z,J_{1}z\rangle +(1-{\alpha }_n)(\parallel u_n-z\parallel \hfill \\ & +2\langle x_n,J_{1}z\rangle )+{\mu }_n{M}^{*}+\gamma {\xi }_n\hfill \\ & \le (1-r_n)\varphi (z,x_n)-2r_n\langle x_n-z,J_{1}z\rangle +(1-{\alpha }_n)(\parallel u_n{-z\parallel }^2\hfill \\ & +2\langle x_n,J_{1}z\rangle +{\mu }_n{M}^{*}+\gamma {\xi }_n\hfill \end{array}$$

(20)

where $M^{*}>\gamma {sup}_{n\ge 0}(M+M_0{\parallel A\parallel }^2\parallel u_n-z{\parallel }^2)>0$. It is clear that $-2\langle u_n-z,z\rangle \to 0,n\to \infty$, and ${\sum }_{n=1}^{\infty }{M}^{*}{\mu }_n<\infty ,{\sum }_{n=1}^{\infty }\gamma {\xi }_n<\infty$ and ${\sum }_{n=1}^{\infty }(1-{\alpha }_n)(\parallel u_n{-z\parallel }^2+2\langle x_n,J_{1}z\rangle <\infty$. Now, using Lemma 8 in (20), we have $\varphi (z,x_n)\to 0$. Therefore, $x_n\to z$ as $n\to \infty$.

Case 2.

Assume that there exists a subsequence $\{x_{n_j}\}$ of $\left\{x_n\right\}$, such that $\varphi (z,x_{n_j})<\varphi (z,x_{n_j+1}),$$\forall j\in \mathbb{N}$. By Lemma 9, there exists a nondecreasing sequence $\left\{\tau \right(n\left)\right\}$ of $\mathbb{N}$, such that for all $n\ge n_0$ (for some $n_0$ large enough) $\tau \left(n\right)\to \infty$ as $n\to \infty$ and such that the following inequalities hold:

$$\varphi (z,x_n)<\varphi (z,x_{\tau \left(n\right)+1}),\varphi (z,x_{\tau \left(n\right)})<\varphi (z,x_{\tau \left(n\right)+1})$$

By a similar argument as in Case 1, we obtain:

$$\begin{array}{cc}\hfill \varphi (z,x_{\tau \left(n\right)+1})& \le (1-r_{\tau \left(n\right)})\varphi (z,x_{\tau \left(n\right)})-2r_{\tau \left(n\right)}\langle x_{\tau \left(n\right)}-z,J_{1}z\rangle \hfill \\ & +(1-{\alpha }_{\tau \left(n\right)})(\parallel u_{\tau \left(n\right)}-z{\parallel }^2+2\langle x_{\tau \left(n\right)},J_{1}z\rangle )+\gamma {\mu }_{\tau \left(n\right)}{M}^{*}+\gamma {\xi }_{\tau \left(n\right)}\hfill \end{array}$$

(21)

and $lim\langle x_{\tau \left(n\right)}-z,J_{1}z\rangle =0$. Since $\varphi (z,x_{\tau \left(n\right)})\le \varphi (z,x_{\tau \left(n\right)+1})$, we have:

$$\begin{array}{cc}\hfill r_{\tau \left(n\right)}\varphi (z,x_{\tau \left(n\right)})& \le \varphi (z,x_{\tau \left(n\right)})-\varphi (z,x_{\tau \left(n\right)+1})-2r_{\tau \left(n\right)}\langle x_{\tau \left(n\right)}-z,J_{1}z\rangle \hfill \\ & +(1-{\alpha }_{\tau \left(n\right)})(\parallel u_{\tau \left(n\right)}-z{\parallel }^2+2\langle x_{\tau \left(n\right)},J_{1}z\rangle )+\gamma {\mu }_{\tau \left(n\right)}{M}^{*}+\gamma {\xi }_{\tau \left(n\right)}\hfill \end{array}$$

By our assumption that $r_{\tau \left(n\right)}>0$, we obtain:

$$\varphi (z,x_{\tau \left(n\right)})\le -2r_{\tau \left(n\right)}\langle x_{\tau \left(n\right)}-z,J_{1}z\rangle +(1-{\alpha }_{\tau \left(n\right)})(\parallel u_{\tau \left(n\right)}-z{\parallel }^2+2\langle x_{\tau \left(n\right)},J_{1}z\rangle )+\gamma {\mu }_{\tau \left(n\right)}{M}^{*}+\gamma {\xi }_{\tau \left(n\right)}$$

which implies that ${lim}_{n\to \infty }\varphi (\overline{x},x_{\tau \left(n\right)})=0$. It now follows from (21) that ${lim}_{n\to \infty }\varphi (\overline{x},x_{\tau \left(n\right)+1})=0$. Now, since $\varphi (\overline{x},x_n)<\varphi (\overline{x},x_{\tau \left(n\right)+1})$, we obtain that $\varphi (\overline{x},x_n)\to 0$. Finally, we conclude from Lemma 5 that $\left\{x_n\right\}$ converges strongly to $\overline{x}$.

☐

Theorem 2.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<\frac{1}{\sqrt{2}}$, and let $E_2$ be a real smooth Banach space. Let $A:E_1\to E_2$ be a bounded linear operator and $A^{*}$ be its adjoint. Let $T_i:E_2\to E_2(i\in \mathbb{N})$ be an infinite family of k-quasi-strict pseudocontractive mappings and ${\left\{S_i\right\}}_{i=1}^{\infty }:E_1\to E_1$ be an infinite family of uniformly $L_i$-Lipschitzian continuous and totally quasi-$\varphi$-asymptotically nonexpansive mappings. Let $\left\{x_n\right\}$ be the sequence generated by $x_1\in E_1$:

$$\left\{\begin{array}{c}{u}_n=J_1^{-1}({\alpha }_{n,0}{J}_1{x}_n+\sum _{i=1}^{\infty }{\alpha }_{n,i}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right))\hfill \\ y_{n,m}=J_1^{-1}({\beta }_n{J}_1{x}_1+(1-{\beta }_n)J_1{S}_m^n{x}_n)\hfill \\ C_{n+1}=\{z\in C_n:su{p}_{m\ge 1}\varphi (z,y_{n,m})\le {\beta }_n\varphi (z,x_1)+(1-{\beta }_n)(\varphi (z,x_n)+\parallel x_n{\parallel }^2+{\parallel z\parallel }^2)+{\xi }_n\}\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\hfill \end{array}\right.$$

(22)

where ${\xi }_n={\nu }_n{sup}_{z\in \mathrm{\Omega }}\zeta \left(\varphi (z,u_n)\right)+{\mu }_n,\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$, and ${\mathrm{\Pi }}_{C_{n+1}}$ is the generalized projection of E onto $C_{n+1}$; and the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_{n,i}\right\}\subset (0,1)$ and satisfy the following conditions:

(a) 

$\left\{{\beta }_n\right\}\subset [0,1]and{lim}_{n\to \infty }{\beta }_n=0$

(b) 

$\left\{{\alpha }_{n,i}\right\}\subseteq [0,1],{\sum }_{i=0}^{\infty }{\alpha }_{n,i}=1and{lim}_{n\to \infty }{\alpha }_{n,0}=1$

If $\mathrm{\Omega }=\{x\in {\cap }_{m=1}^{\infty }F\left(S_m\right):Ax\in {\cap }_{i=1}^{\infty }F\left(T_i\right)\}$ is nonempty and bounded and ${\mu }_1=0$, then $\left\{x_n\right\}$ converges strongly to: ${\mathrm{\Pi }}_{\mathrm{\Omega }}u$.

Proof. 

(I) Both Ω and $C_n$, $n\ge 1$, are closed and convex.

We know from Lemma 11 and Lemma 12 that $F\left(T_i\right)$ and $F\left(S_i\right)$, $i\ge 1$, are closed and convex. This implies that Ω is closed and convex. Again, by the assumption, $C_1=E_1$ is closed and convex. Now, suppose that $C_n$ is closed and convex for some $n\ge 1$. In view of the definition of $\varphi$, we have:

$$\begin{array}{cc}\hfill C_{n+1}& =\{z\in C_n:\underset{m\ge 1}{sup}\varphi (z,y_{n,m})\le {\beta }_n\varphi (z,x_1)+(1-{\beta }_n)(\varphi (z,x_n)+2\langle z,J_1{x}_n\rangle )+{\xi }_n\}\hfill \\ & ={\cap }_{m\ge 1}\{z\in E_1:\varphi (z,y_{n,m})\le {\beta }_n\varphi (z,x_1)+(1-{\beta }_n)(\varphi (z,x_n)+2\langle z,J_1{x}_n\rangle )+{\xi }_n\}\cap C_n\hfill \\ \hfill =& {\cap }_{m\ge 1}\{z\in E_1:2{\beta }_n\langle z,J_1{x}_1\rangle +2(1-{\beta }_n)\langle z,J_1{x}_n\rangle -2\langle z,y_{n,m}\rangle \le {\beta }_n\parallel x_1{\parallel }^2+2(1-{\beta }_n){\parallel x_n\parallel }^2\hfill \\ & -\parallel y_{n,m}{\parallel }^2+{\parallel z\parallel }^2\}\cap C_n\hfill \end{array}$$

from which, it follows that $C_{n+1}$ is closed and convex.

(II) $\mathrm{\Omega }\subset C_n,n\ge 1$.

It is clear that $\mathrm{\Omega }\subset E_1$. Suppose that $\mathrm{\Omega }\subset C_n$ for some $n\ge 1$. Let $u\in \mathrm{\Omega }\subset C_n$, then we have:

$$\begin{array}{cc}\hfill \varphi (u,u_n)& =\varphi (u,J_1^{-1}({\alpha }_{n,0}{J}_1{x}_n+\sum _{i=1}^{\infty }{\alpha }_{n,i}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right)))\hfill \\ & \le {\alpha }_{n,0}\varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}\varphi (u,J_1^{-1}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right))\hfill \\ & \le \varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}\left[\varphi \right(x_n,J_1^{-1}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right)\hfill \\ & +2\langle u-x_n,J_1{x}_n-\gamma A^{*}{J}_2(T_i-I)A{x}_n\rangle \hfill \\ & \le \varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}[\parallel x_n{\parallel }^2+2\langle u-x_n,J_1{x}_n\rangle +{\gamma }^2{\parallel A\parallel }^2{\parallel (T_i-I)A{x}_n\parallel }^2\hfill \\ & -2\langle x_n,J_1^{-1}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right)\rangle +2\langle u-x_n,\gamma A^{*}{J}_2(T_i-I)A{x}_n\rangle \hfill \end{array}$$

(23)

From Lemma 1, we have:

$$\begin{array}{cc}\hfill -2\langle x_n,\gamma A^{*}{J}_2(T_i-I)A{x}_n\rangle & \le \parallel \gamma A^{*}{J}_2(T_i-I)A{x}_n{\parallel }^2+2\parallel t{x}_n{\parallel }^2-{\parallel x_n+\gamma A^{*}{J}_2(T_i-I)A{x}_n\parallel }^2\hfill \\ & \le {\gamma }^2{\parallel A\parallel }^2\parallel (T_i-I)A{x}_n{\parallel }^2+{\parallel x_n\parallel }^2\hfill \end{array}$$

(24)

Since $Au\in {\cap }_{i=1}^{\infty }F\left(T_i\right)$ and $T_i$ is a k-quasi-strictly pseudocontractive mapping:

$$\begin{array}{cc}\hfill \langle x_n-u,\gamma A^{*}{J}_2(T_i-I)A{x}_n\rangle & =\gamma \langle A(x_n-u),J_2(T_i-I)A{x}_n\rangle \hfill \\ & =\gamma \langle A(x_n-u)+(T_i-I)A{x}_n-(T_i-I)A{x}_n,J_2(T_i-I)A{x}_n\rangle \hfill \\ & =\gamma (\langle T_{i}A\left(x_n\right)-Au,J_2(T_i-I)A{x}_n\rangle -\parallel (T_i-I)A{x}_n{\parallel }^2)\hfill \\ & \le \gamma (\frac{1}{2}(\parallel T_{i}A{x}_n{-Au\parallel }^2+\parallel (T_i-I)A{x}_n{\parallel }^2\left)\right)-\gamma \parallel (T_i-I)A{x}_n{\parallel }^2\hfill \\ & =\frac{\gamma }{2}(\parallel T_{i}A{x}_n{-Au\parallel }^2-\parallel (T_i-I)A{x}_n{\parallel }^2)\hfill \\ & \le \frac{\gamma }{2}(\parallel A{x}_n{-Au\parallel }^2+(k-1)\parallel (T_i-I)A{x}_n{\parallel }^2)\hfill \\ & \le \frac{1}{2}\parallel x_n{-u\parallel }^2+\frac{\gamma }{2}(k-1){\parallel (T_i-I)A{x}_n\parallel }^2\hfill \end{array}$$

(25)

Substituting (24) and (25) into (23), we obtain:

$$\begin{array}{cc}\hfill \varphi (u,u_n)& \le {\alpha }_{n,0}\varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}\varphi (u,J_1^{-1}\left(\gamma A^{*}{J}_2(T_i-I)A{x}_n\right))\hfill \\ & \le \varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}[2\langle u,J_1{x}_n\rangle {-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T_i-I)A{x}_n{\parallel }^2+\parallel x_n-u{\parallel }^2]\hfill \\ & \le \varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_n{\parallel }^2+{\parallel u\parallel }^2{)-\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T_i-I)A{x}_n{\parallel }^2\hfill \end{array}$$

(26)

It now follows from Lemma 2(d) and Equation (22):

$$\begin{array}{cc}\hfill \varphi (u,y_{n,m})& \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)\varphi (u,S_n^m{u}_n)\hfill \\ & \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)[\varphi (u,u_n)+{\nu }_n\zeta \left(\varphi (u,u_n)\right)+{\mu }_n]\hfill \\ & \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)[\varphi (u,u_n)+{\nu }_n\underset{u\in \mathrm{\Omega }}{sup}\zeta \left(\varphi (u,u_n)\right)+{\mu }_n]\hfill \\ & ={\beta }_n\varphi (u,x_1)+(1-{\beta }_n)(\varphi (u,u_n)+{\xi }_n)\forall m\ge 1\hfill \\ & \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)(\varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_n{\parallel }^2+{\parallel u\parallel }^2)\hfill \\ & +{\xi }_n{)-\gamma (1-2\gamma \parallel A\parallel }^2)\parallel (T_i-I)A{x}_n{\parallel }^2\forall m\ge 1\hfill \\ & \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)(\varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_n{\parallel }^2+{\parallel u\parallel }^2)+{\xi }_n)\forall m\ge 1\hfill \end{array}$$

(27)

Therefore, we have:

$$\begin{array}{cc}\hfill \underset{m\ge 1}{sup}\varphi (u,y_{n,m})& \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)(\varphi (u,x_n)+\sum _{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_n{\parallel }^2+{\parallel u\parallel }^2)+{\xi }_n)\hfill \\ & \le {\beta }_n\varphi (u,x_1)+(1-{\beta }_n)(\varphi (u,x_n)+\parallel x_n{\parallel }^2+{\parallel u\parallel }^2+{\xi }_n)\hfill \end{array}$$

(28)

This argument shows that $u\in C_{n+1}$, and so, $F\subset C_{n+1}$.

(III) $\left\{x_n\right\}$ converges strongly to some point $p^{*}\in E_1$.

Since $x_n={\mathrm{\Pi }}_{C_n}{x}_1$, from Lemma 6, we have $\langle x_n-y,J_1{x}_1-J_1{x}_n\rangle \ge 0,\forall y\in C_n$. Again, since $\mathrm{\Omega }\subset C_n$, we obtain $\langle x_n-u,J_1{x}_1-J_1{x}_n\rangle \ge 0,\forall u\in \mathrm{\Omega }$. It now follows from Lemma 2(a) that for each $u\in \mathrm{\Omega }$ and each $n\ge 1$:

$$\varphi (x_n,x_1)=\varphi ({\mathrm{\Pi }}_{C_n}{x}_1,x_1)\le \varphi (u,x_1)-\varphi (u,x_n)\le \varphi (u,x_1)$$

(29)

Therefore, $\left\{\varphi (x_n,x_1)\right\}$ is bounded, and so is $\left\{x_n\right\}$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_1$ and $x_{n+1}={\mathrm{\Pi }}_{C_n+1}{x}_1\in C_{n+1}\subset C_n$, we have $\varphi (x_n,x_1)\le \varphi (x_{n+1},x_1),n\ge 1$. This implies that $\left\{\varphi (x_n,x_1)\right\}$ is nondecreasing. Hence, ${lim}_{n\to \infty }\varphi (x_n,x_1)$ exists. Since E is reflexive, there exists a subsequence $x_{n_i}\subset x_n$, such that $x_{n_i}\rightharpoonup p^{*}$ (some point in $E_1$). Since $C_n$ is closed and convex and $C_{n+1}\subset C_n$, it follows that $C_n$ is weakly closed and $p^{*}\in C_n$ for each $n\ge 1$. Now, in view of $x_{n_i}={\mathrm{\Pi }}_{C_{n_i}}{x}_1$, we have $\varphi (x_{n_i},x_1)\le \varphi (p^{*},x_1),\forall n_i\ge 1.$ Since the norm $\parallel .\parallel$ is weakly lower semicontinuous, we have:

$$\underset{n_i\to \infty }{lim\; inf}\varphi (x_{n_i},x_1)=\underset{n_i\to \infty }{lim\; inf}\{\parallel x_{n_i}{\parallel }^2+\parallel x_1{\parallel }^2-2\langle x_{n_i},J_1{x}_1\rangle \}\ge \parallel p^{*}{\parallel }^2+{\parallel x_1\parallel }^2-2\langle p^{*},x_1\rangle =\varphi (p^{*},x_1)$$

and so:

$$\varphi (p^{*},x_1)\le \underset{n_i\to \infty }{lim\; inf}\varphi (x_{n_i},x_1)\le \underset{n_i\to \infty }{lim\; sup}\varphi (x_{n_i},x_1)\le \varphi (p^{*},x_1)$$

This implies that ${lim}_{n_i}\varphi (x_{n_i},x_1)=\varphi (x_1,p^{*}),$ and so, $\parallel x_{n_i}\parallel \to \parallel p^{*}\parallel$. Since $x_{n_i}\rightharpoonup p^{*}$ and $E_1$ is uniformly convex, we obtain ${lim}_{n_i\to \infty }{x}_{n_i}=p^{*}.$ Now, the convergence of $\left\{\varphi (x_n,x_1)\right\}$, together with ${lim}_{n_i\to \infty }\varphi (x_{n_i},x_1)=\varphi (p^{*},x_1)$, implies that ${lim}_{n\to \infty }\varphi (x_n,x_1)=\varphi (p^{*},x_1).$ If there exists some subsequence $\{x_{n_j}\}\subset \left\{x_n\right\}$, such that $x_{n_j}\to q$, then from Lemma 2(a), we have:

$$\begin{array}{cc}\hfill \varphi (p^{*},q)& =\underset{n_i,n_j\to \infty }{lim}\varphi (x_{n_i},x_{n_j})=\underset{n_i,n_j\to \infty }{lim}\varphi (x_{n_i},{\mathrm{\Pi }}_{C_j}{x}_1)\le \underset{n_i,n_j\to \infty }{lim}(\varphi (x_{n_i},x_1)-\varphi ({\mathrm{\Pi }}_{C_j}{x}_1,x_1))\hfill \\ & \le \underset{n_i,n_j\to \infty }{lim}(\varphi (x_{n_i},x_1)-\varphi (x_{n_j},x_1))=\varphi (p^{*},q)-\varphi (p^{*},q)=0\hfill \end{array}$$

i.e., $p^{*}=q$, and so:

$$\underset{n\to \infty }{lim}{x}_n=p^{*}$$

(30)

By the way, it follows from from (26) that $\varphi (u,u_n)$ is bounded, so:

$$\underset{n\to \infty }{lim}{\xi }_n=\underset{n\to \infty }{lim}\{{\nu }_n\underset{p\in \mathrm{\Omega }}{sup}\zeta \left(\varphi (p,u_n)\right)+{\mu }_n\}=0$$

(31)

(IV) $p^{*}\in \mathrm{\Omega }$. Since $x_{n+1}\in C_{n+1}$, from (28), (30) and (31):

$$\underset{m\ge 1}{sup}\varphi (x_{n+1},y_{n,m})\le {\beta }_n\varphi (x_{n+1},x_1)+(1-{\beta }_n)[\varphi (x_{n+1},x_n)+{\sum }_{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_n{\parallel }^2+\parallel x_{n+1}{\parallel }^2)+{\xi }_n]\to 0$$

(32)

Since $x_{n+1}\in C_{n+1}$, from (27) and (32) we have:

$$\begin{array}{cc}\hfill {\gamma (1-k-2\gamma \parallel A\parallel }^2)\parallel (T_i-I)A{x}_n{\parallel }^2& \le {\beta }_n\varphi (x_{n+1},x_1)+(1-{\beta }_n)(\varphi (x_{n+1},x_n)\hfill \\ & +\sum _{i=1}^{\infty }{\alpha }_{n,i}(\parallel x_{n+1}{\parallel }^2+\parallel x_n{\parallel }^2)+{\xi }_n)-\varphi (x_{n+1},y_{n,m})\to 0n\to \infty \hfill \end{array}$$

(33)

Since $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$, we have:

$$\parallel (T_i-I)A{x}_n\parallel \to 0n\to \infty$$

(34)

Since $x_n\to p^{*}$, it follows from (32) and Lemma 5 that for each $m\ge 1$:

$$\underset{n\to \infty }{lim}{y}_{n,m}=p^{*}$$

(35)

Since $\left\{x_n\right\}$ is a bounded sequence and ${\left\{S_m\right\}}_{m=1}^{\infty }$ is uniformly totally quasi-asymptotically nonexpansive, ${\left\{S_m^n{x}_n\right\}}_{m,n=1}^{\infty }$ is uniformly bounded. In view of ${\beta }_n\to 0$ and (22), we conclude that for each $m\ge 1$:

$$\parallel J_1{y}_{n,m}-J_1{S}_m^n{x}_n\parallel =\underset{n\to \infty }{lim}{\beta }_n\parallel J_1{x}_1-J_1{S}_m^n{x}_n\parallel =0$$

(36)

Since for each $m\ge 1,J_1{y}_{n,m}\to J_1{p}^{*}$, it follows that for each $m\ge 1$, ${lim}_{n\to \infty }{J}_1{S}_m^n{x}_n=J_1{p}^{*}$. Since $J_1$ is continuous on each bounded subset of $E_1$, for each $m\ge 1$:

$$\underset{n\to \infty }{lim}{S}_m^n{x}_n=p^{*}$$

(37)

On the other hand, by the assumption that for each $m\ge 1$, $S_m$ is uniformly $L_m$-Lipschitzian continuous, we have:

$$\begin{array}{cc}\hfill \parallel S_m^{n+1}{x}_n-S_m^n{x}_n\parallel & \le \parallel S_m^{n+1}{x}_n-S_m^{n+1}{x}_{n+1}\parallel +\parallel S_m^{n+1}{x}_{n+1}-x_{n+1}\parallel +\parallel x_{n+1}-x_n\parallel +\parallel x_n-S_m^n{x}_n\parallel \hfill \\ & \le (L_m+1)\parallel x_{n+1}-x_n\parallel +\parallel S_m^{n+1}{x}_{n+1}-x_{n+1}\parallel +\parallel x_n-S_m^n{x}_n\parallel \hfill \end{array}$$

(38)

From (37) and $x_n\to p^{*}$, we have that ${lim}_{n\to \infty }\parallel S_m^{n+1}{x}_n-S_m^n{x}_n\parallel =0$ and ${lim}_{n\to \infty }{S}_m^{n+1}{x}_n=p^{*}$, i.e., ${lim}_{n\to \infty }{S}_m{S}_m^n{x}_n=p^{*}$. In view of the closedness of $S_m$, it follows that $S_m{p}^{*}=p^{*}$, i.e., for each $m\ge 1,p^{*}\in F\left(S_m\right)$. By the arbitrariness of $m\ge 1$, we have $p^{*}\in {\cap }_{m=1}^{\infty }F\left(S_m\right)$. On the other hand, since A is bounded, it follows from $x_{n_i}\rightharpoonup p^{*}$ that $A{x}_{n_i}\rightharpoonup A{p}^{*}$. Hence, from (34), we have that:

$$\parallel T_{i}A{x}_{n_i}-A{x}_{n_i}\parallel ⟶0,i\to \infty$$

Since $T_i$ is demi-closed at zero, we have that $Az\in F\left(T_i\right)$. Hence, $z\in \mathrm{\Omega }$.

(V) Finally, $p^{*}\in {\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_1$, and so, $x_n\to {\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_1$.

Let $w={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_1$. Since $w\in \mathrm{\Omega }\subset C_n$ and $x_n={\mathrm{\Pi }}_{C_n}{x}_1$, we have $\varphi (x_n,x_1)\le \varphi (w,x_1),n\ge 1$. This implies that $\varphi (p^{*},x_1)={lim}_{n\to \infty }\varphi (x_n,x_{1)}\le \varphi (w,x_1).$ Since $w={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_1$, it follows that $p^{*}=w$, and hence, $x_n\to p^{*}={\mathrm{\Pi }}_{\mathrm{\Omega }}{x}_1$. ☐

Corollary 1.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<\frac{1}{\sqrt{2}}$, and let $E_2$ be a real smooth Banach space. Let $A:E_1\to E_2$ be a bounded linear operator and $A^{*}$ be its adjoint. Let $T:E_2\to E_2$ be a k-quasi-strict pseudocontractive mapping and T be demi-closed at zero. Let ${\left\{S_n\right\}}_{n=1}^{\infty }:E_1\to CB\left(E_1\right)$ be a family of multivalued quasinonexpansive mappings, such that for each $i\ge 1,S_i$ is demi-closed at zero. Assume that for each $p\in Fix\left(S_i\right),S_i\left(p\right)=\left\{p\right\}$. Let $\left\{x_n\right\}$ be the sequence generated by $x_1\in E_1$:

$$\left\{\begin{array}{c}{u}_n=(1-r_n)x_n\hfill \\ y_n=J_1^{-1}({\alpha }_n{J}_1{u}_n+(1-{\alpha }_n)\gamma A^{*}{J}_2(T-I)A{u}_n)\hfill \\ x_{n+1}=J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{w}_{n,i})w_{n,i}\in S_i{y}_n\hfill \end{array}\right.$$

where $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$; the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_{n,i}\right\}\subset (0,1)$ satisfy the following conditions:

(a) 

${\sum }_{i=0}^{\infty }{\beta }_{n,i}=1and{lim\; inf}_n{\beta }_{n,0}{\beta }_{n,i}>0$,

(b) 

${lim}_{n\to \infty }{\alpha }_n=1,{\sum }_{n=1}^{\infty }(1-{\alpha }_n)<\infty and(1-{\alpha }_n)=o\left(r_n\right).$

Then, $\left\{x_n\right\}$ converges strongly to an element of Ω.

Proof. 

Since every k-quasi-strictly pseudocontractive mapping is clearly $(k,0,0)$-totally asymptotically strictly pseudocontractive, the result follows. ☐

Corollary 2.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<\frac{1}{\sqrt{2}}$, and let $E_2$ be a real smooth Banach space. Let $A:E_1\to E_2$ be a bounded linear operator and $A^{*}$ be its adjoint. Let $T:E_2\to E_2$ be a uniformly L-Lipschitzian continuous and $(k,\left\{{\mu }_n\right\},\left\{{\xi }_n\right\})$-totally asymptotically strictly pseudocontractive mapping satisfying the following conditions:

(a) 

${\sum }_{n=1}^{\infty }{\mu }_n<\infty ,{\sum }_{n=1}^{\infty }{\xi }_n<\infty$,

(b) 

$\left\{r_n\right\}$ is a real sequence in $(0,1)$, such that ${\mu }_n=o\left(r_n\right),{\xi }_n=o\left(r_n\right),lim{r}_n=0,{\sum }_{n=1}^{\infty }{r}_n=\infty$,

(c) 

there exist constants $M_0>0,M_1>0$, such that $\zeta \left(\lambda \right)\le M_0{\lambda }^2,\forall \lambda >M_1$.

Let $\mathfrak{F}=\left\{S\right(t):0\le t<\infty \}$ be a one-parameter nonexpansive semigroup on $E_1$. Suppose further that $\mathrm{\Omega }=\{x\in {\cap }_{t\ge 0}F\left(S\left(t\right)\right):Ax\in F\left(T\right)\}\ne \varnothing$, and $\left\{x_n\right\}$ is the sequence generated by $x_1\in E_1$:

$$\left\{\begin{array}{c}{u}_n=(1-r_n)x_n\hfill \\ y_n=J_1^{-1}({\alpha }_n{J}_1{u}_n+(1-{\alpha }_n)\gamma A^{*}{J}_2(T^n-I)A{u}_n)\hfill \\ x_{n+1}=J_1^{-1}({\beta }_n{J}_1{y}_n+(1-{\beta }_n)\left(\frac{1}{t_n}{\int }_0^{t_n}S\left(u\right)du{J}_1{y}_n\right)\hfill \end{array}\right.$$

where $\gamma \in (0,\frac{1-k}{{2\parallel A\parallel }^2})$; the sequence $\left\{{\alpha }_n\right\}\subset (0,1),0<ϵ\le {\beta }_n\le b<1$, and ${lim}_{n\to \infty }{\alpha }_n=1,$ ${\sum }_{n=1}^{\infty }(1-{\alpha }_n)<\infty and(1-{\alpha }_n)=o\left(r_n\right)$. Then, $\left\{x_n\right\}$ converges strongly to to an element of Ω.

Proof. 

Since $\{{\sigma }_t\left(x\right)=\frac{1}{t}{\int }_0^{t}S\left(u\right)xdu:t\ge 0\}$ is a u.a.r. nonexpansive semigroup, the result follows from Corollary 1. ☐

In the following, we shall provide an example to illustrate the main result of this paper.

Example 1.

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

$$T(x_1,x_2,...)=(0,x_1,a_2{x}_2,a_3{x}_3,...)$$

where $\left\{a_i\right\}$ is a sequence in $(0,1)$, such that ${\prod }_{i=2}^{\infty }{a}_i=\frac{1}{2}$. It was shown in [27] that T is a $(0,k_n-1,{\xi }_n)-$totally asymptotically strictly pseudocontractive mapping and $F\left(T\right)=\left\{0\right\}$, where $k_n=2{\prod }_{i=2}^n{a}_i$. Let B be the unit interval in $\mathbb{R}$, and let $S_i:B\to B$ be a mapping defined by:

$$S_i\left(x\right)=\{\begin{array}{c}{\displaystyle \frac{1}{2^i}xx\in [0,\frac{1}{2}]}\hfill \\ {\displaystyle 0x\in (\frac{1}{2},1]}\hfill \end{array}$$

Then, ${\cap }_{i=1}^{\infty }Fix\left(S_i\right)=\left\{0\right\}$ and:

$$\begin{array}{c}\hfill |S_{i}x-0|=|\frac{1}{2^i}x-0|=\frac{1}{2^i}\left|x\right|\le \left|x\right|\end{array}$$

Therefore, each $S_i$ is a quasinonexpansive mapping. Let $A:B\to C$ be the linear operator defined by:

$$A\left(x\right)=(0,x,a_{2}x,a_3{a}_{2}x,a_4{a}_3{a}_{2}x,...),x\in B\subset \mathbb{R}.$$

Then, A is bounded and $\parallel A\parallel =1+a_2^2+{\left(a_3{a}_2\right)}^2+{\left(a_4{a}_3{a}_2\right)}^2+\cdots$. It now follows that:

$$A^{*}:C\to B,A^{*}(x_1,x_2,\cdots )=x_2+a_2{x}_3+a_3{a}_2{x}_4+a_4{a}_3{a}_2{x}_5+\cdots .$$

We now put, for $n\in \mathbb{N}$, ${\alpha }_n=\frac{1}{3}$, $r_n=\frac{1}{n}$, ${\beta }_{n,0}=\frac{1}{2}$, ${\beta }_{n,0}=\frac{1}{3^i}$ and $\lambda =\frac{1}{4}(1+a_2^2+\cdots +{(a_n\cdots a_2)}^2)$. Furthermore, we have:

$$\mathrm{\Omega }=\{x\in F\left(T\right):Ax\in {\cap }_{i=1}^{\infty }F\left(S_i\right)\}=\left\{0\right\}$$

Now, all of the assumptions in Theorem 1 are satisfied. Let us consider the following numerical algorithm:

$$T^n(x_1,x_2,...)=(0,0,...,0,a_n...a_2{x}_1^2,a_{n+1}...a_2{x}_2,...)$$

$$T^n\left(A{u}_n\right)-A{u}_n=(0,-u_n,-a_2{u}_n,-a_3{a}_2{u}_n,...,-a_n...a_2{u}_n,0,0,...$$

$$A^{*}(T^n\left(A{u}_n\right)-A{u}_n)=-u_n(1+a_2^2+{\left(a_3{a}_2\right)}^2+...+{(a_n...a_2)}^2)$$

$$y_n=\frac{1}{6}{u}_n=\frac{1}{6}(1-\frac{1}{n})x_n,x_{n+1}=\frac{1}{2}{y}_n+\sum _{i=1}^{\infty }\frac{1}{3^i}(\frac{1}{2^i}{y}_n)=\frac{1}{10}{y}_n$$

$$x_{n+1}=\frac{1}{60}(1-\frac{1}{n})x_n$$

By Theorem 1, the sequence $\left\{x_n\right\}$ converges to the unique element of Ω.

4. Application

Let E be a uniformly-smooth Banach space, $E^{*}$ be the dual of E, J be the duality mapping on E and $F:E\to 2^{E^{*}}$ be a multi-valued operator. Recall that F is called monotone if $\langle u-v,x-y\rangle \ge 0$, for any $(x,u),(y,v)\in G\left(F\right)$, where $G\left(F\right)=\left\{\right(x,u):x\in D(F),u\in F(x\left)\right\}$. A monotone operator F is said to be maximally monotone if its graph $G\left(F\right)$ is not properly contained in the graph of any other monotone operator. For a maximally-monotone operator $F:E\to 2^{E^{*}}$ and $r>0$, we can define a single-valued operator:

$$J_r^F={(J+rF)}^{-1}J:E\to E$$

It is known that for any $r>0,J_r^F$ is firmly nonexpansive, and its domain is all of E, also $0\in F\left(x\right)$ if and only if $x\in Fix\left(J_r^F\right)$.

Theorem 3.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<1/\sqrt{2}$, and let $E_2$ be a real smooth Banach space and $T:E_1\to E_2$ be a bounded linear operator. Let $A:E_2\to 2^{E_2^{*}}$ and $B_i:E_1\to 2^{E_1^{*}}$, for $i=1,2,...$, be maximal monotone mappings, such that $A^{-1}0\ne \varnothing$ and ${\cap }_{i=1}^{\infty }{B}_i^{-1}0\ne \varnothing$. Suppose:

$$\mathrm{\Omega }=\{x\in E_1:0\in {\cap }_{i=1}^{\infty }{B}_i\left(x\right)\mathrm{such}\mathrm{that}0\in A\left(Tx\right)\}\ne \varnothing$$

Let $\left\{x_n\right\}$ be a sequence generated by $x_0\in E_1$ and:

$$\left\{\begin{array}{c}{u}_n=(1-r_n)x_n\hfill \\ y_n=J_1^{-1}({\alpha }_n{J}_1{u}_n+(1-{\alpha }_n)\gamma T^{*}{J}_2(J_r^{A}T{u}_n-T{u}_n)\hfill \\ x_{n+1}=J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{J}_{\mu }^{B_i}{y}_n\hfill \end{array}\right.$$

where $r,\mu >0,\gamma \in (0,\frac{1-k}{{2\parallel T\parallel }^2})$, and the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_{n,i}\right\}\subset (0,1)$ satisfy the following conditions:

(1) 

${\sum }_{i=0}^{\infty }{\beta }_{n,i}=1and{lim\; inf}_n{\beta }_{n,0}{\beta }_{n,i}>0$,

(2) 

${lim}_{n\to \infty }{\alpha }_n=1,{\sum }_{n=1}^{\infty }(1-{\alpha }_n)<\infty and(1-{\alpha }_n)=o\left(r_n\right).$

Then, $\left\{x_n\right\}$ converges strongly to an element of Ω.

Proof. 

Since $J_r^A$ and $J_{\mu }^{B_i}$ are nonexpansive, the result follows from Corollary 1. ☐

Remark 1.

Set $S_i=J_r^{B_i}$ in Corollary 1, where $B_i$ is a maximal monotone mapping, then Corollary 1 improves Theorem 4.2 of Takahashi et al. [12].

Moudafi [28] introduced the split monotone variational inclusion (SMVIP) in Hilbert spaces. We present the SMVIP in a Banach space. Let $E_1$ and $E_2$ be two real Banach spaces and $J_1$ and $J_2$ be the duality mapping of $E_1$ and $E_2$, respectively. Given the operators $f:E_1\to E_1,g:E_2\to E_2$, a bounded linear operator $A:E_1\to E_2$ and two multi-valued mappings $B_1:E_1\to 2^{E_1^{*}}$ and $B_2:E_2\to 2^{E_2^{*}}$, the SMVI is formulated as follows:

$$\mathrm{find}\mathrm{a}\mathrm{point}x\in C\mathrm{such}\mathrm{that}0\in J_1\left(f\left(x\right)\right)+B_1\left(x\right)$$

$$\mathrm{and}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{point}:$$

$$y=A\left(x\right)\in E_2\mathrm{solves}0\in J_2\left(g\left(y\right)\right)+B_2\left(y\right)$$

Note that if C and Q are nonempty closed convex subsets of $E_1$ and $E_2$, (resp.) and $B_1=N_C$ and $B_2=N_Q$, where $N_C$ and $N_Q$ are normal cones to C and Q (resp.), then the split monotone variational inclusion problem reduces to the split variational inequality problem (SVIP), which is formulated as follows: find a point:

$$x\in C\mathrm{such}\mathrm{that}\langle J_1\left(f\left(x\right)\right),w-x\rangle \ge 0\mathrm{for}\mathrm{all}w\in C$$

$$\mathrm{and}\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{point}:$$

$$y=Ax\in Q\mathrm{solves}\langle J_2\left(g\left(y\right)\right),z-y\rangle \ge 0\mathrm{for}\mathrm{all}z\in Q$$

SVIP is quite general and enables the split minimization between two spaces in such a way that the image of a solution of one minimization problem, under a given bounded linear operator, is a solution of another minimization problem.

Let $h:C\to E$ be an operator, and let $C\subset E$. The operator h is called inverse strongly monotone with constant $\beta >0$, or in brief $(\beta -ism)$, on E if:

$$\langle h\left(x\right)-h\left(y\right),Jx-Jy\rangle \ge {\beta \parallel h\left(x\right)-h\left(y\right)\parallel }^2,\forall x,y\in C$$

Remark 2.

If $h:E\to E$ is an $\alpha -ism$ operator on E and $B:E\to 2^{E^{*}}$ is a maximal monotone mapping, then $J_{\lambda }^B(I-\lambda h)$ is averaged for each $\lambda \in (0,2\alpha )$.

Theorem 4.

Let $E_1$ be a real uniformly-convex and two-uniformly-smooth Banach space with the best smoothness constant t satisfying $0<t<1/\sqrt{2}$, and let $E_2$ be a real smooth Banach space and $T:E_1\to E_2$ be a bounded linear operator. Let $A:E_2\to 2^{E_2^{*}}$ and, for $i=1,2,...$, $B_i:E_1\to 2^{E_1^{*}}$ be maximal monotone mappings, such that $A^{-1}0\ne \varnothing$ and ${\cap }_{i=1}^{\infty }{B}_i^{-1}0\ne \varnothing$; and that $h:E_2\to E_2$ is an $\alpha -ism$ operator and $g_i:E_1\to E_1$ is a ${\gamma }_i-ism$ operator. Assume that $\rho =\alpha in{f}_{i\in \mathbb{N}}{\gamma }_i>0$ and $\tau \in (0,2\rho )$. Suppose SMVI:

$$\left\{\begin{array}{c}x\in {\cap }_{i=1}^{\infty }{B}_i^{-1}00\in J_1\left(g_i\left(x\right)\right)+B_i\left(x\right)\forall i\in \mathbb{N}\hfill \\ Tx\in A^{-1}00\in J_2\left(h\left(Tx\right)\right)+A\left(Tx\right)\hfill \end{array}\right.$$

has a nonempty solution set Ω. Let $\left\{x_n\right\}$ be a sequence generated by $x_0\in E_1$ and:

$$\left\{\begin{array}{c}{u}_n=(1-r_n)x_n\hfill \\ y_n=J_1^{-1}({\alpha }_n{J}_1{u}_n+(1-{\alpha }_n)\gamma T^{*}{J}_2\left((J_r^A(I-\tau h)-I)T{u}_n\right))\hfill \\ x_{n+1}=J_1^{-1}({\beta }_{n,0}{J}_1{y}_n+\sum _{i=1}^{\infty }{\beta }_{n,i}{J}_1{J}_{\mu }^{B_i}(I-\tau g_i)y_n)\hfill \end{array}\right.$$

where $\gamma \in (0,\frac{1-k}{{2\parallel T\parallel }^2})$; the sequences $\left\{{\alpha }_n\right\},\left\{{\beta }_{n,i}\right\}\subset (0,1)$ satisfy the following conditions:

(1) 

${\sum }_{i=0}^{\infty }{\beta }_{n,i}=1and{lim\; inf}_n{\beta }_{n,0}{\beta }_{n,i}>0$,

(2) 

${lim}_{n\to \infty }{\alpha }_n=1,{\sum }_{n=1}^{\infty }(1-{\alpha }_n)<\infty and(1-{\alpha }_n)=o\left(r_n\right).$

Then, $\left\{x_n\right\}$ converges strongly to an element of Ω.

Proof. 

The results follow from Remark 2, Lemma 4(iii) and Corollary 1. ☐

We mention in passing that the above theorem improves and extends Theorems 6.3 and 6.5 of [13] to Banach spaces. Indeed, we removed an extra condition and obtained a strong convergence theorem, which is more desirable than the weak convergence already obtained by the authors.