Explicit averaging cyclic algorithm for common fixed points of a finite family of asymptotically strictly pseudocontractive mappings in q-uniformly smooth Banach spaces

Abstract

Let E be a real q-uniformly smooth Banach space which is also uniformly convex and K be a nonempty, closed and convex subset of E. We obtain a weak convergence theorem of the explicit averaging cyclic algorithm for a finite family of asymptotically strictly pseudocontractive mappings of K under suitable control conditions, and elicit a necessary and sufficient condition that guarantees strong convergence of an explicit averaging cyclic process to a common fixed point of a finite family of asymptotically strictly pseudocontractive mappings in q-uniformly smooth Banach spaces. The results of this paper are interesting extensions of those known results.

MSC:47H09, 47H10.

1 Introduction

Let E and $E^{\ast }$ be a real Banach space and the dual space of E, respectively. Let $J_q$ ($q>1$) denote the generalized duality mapping from E into $2^{E^{\ast }}$ given by $J_q(x)=\{f\in E^{\ast }:〈x,f〉={\parallel x\parallel }^q\text{ and }\parallel f\parallel ={\parallel x\parallel }^{q-1}\}$ for all $x\in E$, where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing between E and $E^{\ast }$. In particular, $J_2$ is called the normalized duality mapping and it is usually denoted by J. If E is smooth or $E^{\ast }$ is strictly convex, then $J_q$ is single-valued. In the sequel, we will denote the single-valued generalized duality mapping by $j_q$.

Let K be a nonempty subset of E. A mapping $T:K\to K$ is called asymptotically κ-strictly pseudocontractive with sequence ${\{{\kappa }_n\}}_{n=1}^{\mathrm{\infty }}\subseteq [1,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{\kappa }_n=1$ (see, e.g., [1–3]) if for all $x,y\in K$, there exist a constant $\kappa \in [0,1)$ and $j_q(x-y)\in J_q(x-y)$ such that

$$〈T^{n}x-T^{n}y,j_q(x-y)〉⩽{\kappa }_n{\parallel x-y\parallel }^q-\kappa {\parallel x-y-(T^{n}x-T^{n}y)\parallel }^q,\mathrm{\forall }n⩾1.$$

(1)

If I denotes the identity operator, then (1) can be written in the form

(2)

The class of asymptotically κ-strictly pseudocontractive mappings was first introduced in Hilbert spaces by Qihou [3]. In Hilbert spaces, $j_q$ is the identity, and it is shown by Osilike et al. [2] that (1) (and hence (2)) is equivalent to the inequality

$${\parallel T^{n}x-T^{n}y\parallel }^2⩽{\lambda }_n{\parallel x-y\parallel }^2+\lambda {\parallel x-y-(T^{n}x-T^{n}y)\parallel }^2,$$

where ${lim}_{n\to \mathrm{\infty }}{\lambda }_n={lim}_{n\to \mathrm{\infty }}[1+2({\kappa }_n-1)]=1$, $\lambda =(1-2\kappa )\in [0,1)$.

A mapping T with domain $D(T)$ and range $R(T)$ in E is called strictly pseudocontractive of Browder-Petryshyn type [4] if for all $x,y\in D(T)$, there exist $\kappa \in [0,1)$ and $j_q(x-y)\in J_q(x-y)$ such that

$$〈Tx-Ty,j_q(x-y)〉⩽{\parallel x-y\parallel }^q-\kappa {\parallel x-y-(Tx-Ty)\parallel }^q.$$

(3)

If I denotes the identity operator, then (3) can be written in the form

$$〈(I-T)x-(I-T)y,j_q(x-y)〉⩾\kappa {\parallel (I-T)x-(I-T)y\parallel }^q.$$

(4)

In Hilbert spaces, (3) (and hence (4)) is equivalent to the inequality

$${\parallel Tx-Ty\parallel }^2⩽{\parallel x-y\parallel }^2+k{\parallel x-y-(Tx-Ty)\parallel }^2,k=(1-2\kappa )<1.$$

It is shown in [5] that the class of asymptotically κ-strictly pseudocontractive mappings and the class of κ-strictly pseudocontractive mappings are independent.

A mapping T is said to be uniformly L-Lipschitzian if there exists a constant $L>0$ such that, for all $x,y\in K$,

$$\parallel T^{n}x-T^{n}y\parallel ⩽L\parallel x-y\parallel ,n⩾1.$$

Let ${\{T_j\}}_{j=0}^{N-1}$ be N asymptotically strictly pseudocontractive self-mappings of K, and denote the common fixed points set of ${\{T_j\}}_{j=0}^{N-1}$ by $F:={\bigcap }_{j=0}^{N-1}F(T_j)$, where $F(T_j):=\{x\in K:T_{j}x=x\}$. We consider the following explicit averaging cyclic algorithm.

For a given $x_0\in K$, and a real sequence ${\{{\alpha }_n\}}_{n=0}^{\mathrm{\infty }}\subseteq (0,1)$, the sequence ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is generated as follows:

$$\begin{array}{c}{x}_1={\alpha }_0{x}_0+(1-{\alpha }_0)T_0{x}_0,\hfill \\ x_2={\alpha }_1{x}_1+(1-{\alpha }_1)T_1{x}_1,\hfill \\ ⋮\hfill \\ x_N={\alpha }_{N-1}{x}_{N-1}+(1-{\alpha }_{N-1})T_{N-1}{x}_{N-1},\hfill \\ x_{N+1}={\alpha }_N{x}_N+(1-{\alpha }_N)T_0^2{x}_N,\hfill \\ x_{N+2}={\alpha }_{N+1}{x}_{N+1}+(1-{\alpha }_{N+1})T_1^2{x}_{N+1},\hfill \\ ⋮\hfill \\ x_{2N}={\alpha }_{2N-1}{x}_{2N-1}+(1-{\alpha }_{2N-1})T_{N-1}^2{x}_{2N-1},\hfill \\ x_{2N+1}={\alpha }_{2N}{x}_{2N}+(1-{\alpha }_{2N})T_0^3{x}_{2N},\hfill \\ x_{2N+2}={\alpha }_{2N+1}{x}_{2N+1}+(1-{\alpha }_{2N+1})T_1^3{x}_{2N+1},\hfill \\ ⋮\hfill \end{array}$$

The algorithm can be expressed in a compact form as

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T_{i(n)}^{k(n)}{x}_n,n⩾0,$$

(5)

where $n=(k-1)N+i$ with $i=i(n)\in I=\{0,1,2,\dots ,N-1\}$, $k=k(n)⩾1$ a positive integer and ${lim}_{n\to \mathrm{\infty }}k(n)=\mathrm{\infty }$. The cyclic algorithm was first studied by Acedo and Xu [6] for the iterative approximation of common fixed points of a finite family of strictly pseudocontractive mappings in Hilbert spaces, and it is better than implicit iteration methods.

In [7] Xiaolong Qin et al. proved the following theorem in a Hilbert space.

Theorem QCKS Let K be a closed and convex subset of a Hilbert space H and $N⩾1$ be an integer. Let, for each $1⩽i⩽N$, $T_i:K\to K$ be an asymptotically ${\kappa }_i$-strictly pseudo-contractive mapping for some $0⩽{\kappa }_i<1$ and a sequence $\{k_{n,i}\}$ such that ${\sum }_{n=0}^{\mathrm{\infty }}(k_{n,i}-1)<\mathrm{\infty }$. Let $\kappa =max\{{\kappa }_i:1⩽i⩽N\}$ and ${\kappa }_n=max\{{\kappa }_{n,i}:1⩽i⩽N\}$. Assume that $F\ne \mathrm{\varnothing }$. For any $x_0\in K$, let $\{x_n\}$ be the sequence generated by the cyclic algorithm (5). Assume that the control sequence $\{{\alpha }_n\}$ is chosen such that $\kappa +ϵ⩽{\alpha }_n⩽1-ϵ$ for all $n⩾0$ and a small enough constant $ϵ\in (0,1)$. Then $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_i\}}_{i=1}^N$.

Osilike and Shehu [8] extended the result of Theorem QCKS from a Hilbert space to 2-uniformly smooth Banach spaces which are also uniformly convex. They proved the following theorem.

Theorem OS Let E be a real 2-uniformly smooth Banach space which is also uniformly convex, and K be a nonempty, closed and convex subset of E. Let ${\{T_j\}}_{j=0}^{N-1}$ be N asymptotically ${\lambda }_j$-strictly pseudocontractive self-mappings of K for some $0⩽{\lambda }_j<1$ with a sequence ${\{{\kappa }_n^{(j)}\}}_{n=0}^{\mathrm{\infty }}\subset [1,\mathrm{\infty })$ such that ${\sum }_{n=0}^{\mathrm{\infty }}({\kappa }_n^{(j)}-1)<\mathrm{\infty }$, $\mathrm{\forall }j\in J=\{0,1,2,\dots ,N-1\}$, and $F\ne \mathrm{\varnothing }$. Let $\{{\alpha }_n\}$ satisfy the conditions

$$\begin{array}{ll}({\mathrm{i}}^{\ast })& 0⩽{\alpha }_n<1,n⩾0,\\ (\mathrm{i}{\mathrm{i}}^{\ast })& 0<a⩽1-{\alpha }_n⩽b<\frac{2\lambda }{C_2},\end{array}$$

where $\lambda ={min}_{j\in J}\{{\lambda }_j\}$ and $C_2$ is the constant appearing in the inequality (7) with $q=2$. Let $\{x_n\}$ be the sequence generated by the cyclic algorithm (5). Then $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$.

We would like to point out that the condition (${\mathrm{ii}}^{\ast }$) in Theorem OS excludes the natural choice $1-\frac{1}{n}$ for ${\alpha }_n$. This is overcome by this paper. Moreover, we improve and extend the result of Theorem OS from 2-uniformly smooth Banach spaces to q-uniformly smooth Banach spaces which are also uniformly convex. We prove that if $\{{\alpha }_n\}$ satisfies the conditions

$$\begin{array}{ll}(\mathrm{i})& \mu ⩽{\alpha }_n<1,n⩾0,\\ (\mathrm{i}\mathrm{i})& \sum _{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]=\mathrm{\infty },\end{array}$$

(6)

where $\mu =max\{0,1-{(\frac{q\lambda }{C_q})}^{\frac{1}{q-1}}\}$, $\lambda ={min}_{j\in J}\{{\lambda }_j\}$, then the iterative sequence (5) converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$.

Furthermore, we elicit a necessary and sufficient condition that guarantees strong convergence of the iterative sequence (5) to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$ in q-uniformly smooth Banach spaces.

We will use the notation:

• 1. ⇀ for weak convergence.

• 2. ${\omega }_{\mathcal{W}}(x_n)=\{x:\mathrm{\exists }{x}_{n_j}\rightharpoonup x\}$ denotes the weak ω-limit set of $\{x_n\}$.

2 Preliminaries

Let E be a real Banach space. The modulus of smoothness of E is the function ${\rho }_E:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ defined by

$${\rho }_E(\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:\parallel x\parallel ⩽1,\parallel y\parallel ⩽\tau \}.$$

E is uniformly smooth if and only if ${lim}_{\tau \to 0}[{\rho }_E(\tau )/\tau ]=0$.

Let $q>1$. E is said to be q-uniformly smooth (or to have a modulus of smoothness of power type $q>1$) if there exists a constant $c>0$ such that ${\rho }_E(\tau )⩽c{\tau }^q$. Hilbert spaces, $L_p$ (or $l_p$) spaces ($1<p<\mathrm{\infty }$) and the Sobolev spaces $W_m^p$ ($1<p<\mathrm{\infty }$) are q-uniformly smooth. Hilbert spaces are 2-uniformly smooth while

$$L_p(\text{or }{l}_p)\text{or}{W}_m^p\text{is}\{\begin{array}{cc}p\text{-uniformly smooth}\hfill & \text{if }1<p⩽2,\hfill \\ 2\text{-uniformly smooth}\hfill & \text{if }p⩾2.\hfill \end{array}$$

Theorem HKX ([[9], p.1130])

Let $q>1$ and let E be a real q-uniformly smooth Banach space. Then there exists a constant $C_q>0$ such that, for all $x,y\in E$,

$${\parallel x+y\parallel }^q⩽{\parallel x\parallel }^q+q〈y,j_q(x)〉+C_q{\parallel y\parallel }^q.$$

(7)

E is said to have a Fréchet differentiable norm if, for all $x\in U=\{x\in E:\parallel x\parallel =1\}$,

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists and is attained uniformly in $y\in U$. In this case, there exists an increasing function $b:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with ${lim}_{t\to 0^{+}}[b(t)/t]=0$ such that, for all $x,h\in E$,

$$\frac{1}{2}{\parallel x\parallel }^2+〈h,j(x)〉⩽\frac{1}{2}{\parallel x+h\parallel }^2⩽\frac{1}{2}{\parallel x\parallel }^2+〈h,j(x)〉+b(\parallel h\parallel ).$$

(8)

It is well known (see, for example, [[10], p.107]) that a q-uniformly smooth Banach space has a Fréchet differentiable norm.

Lemma 2.1 ([[5], p.1338])

Let E be a real q-uniformly smooth Banach space which is also uniformly convex. Let K be a nonempty, closed and convex subset of E and $T:K\to K$ be an asymptotically κ-strictly pseudocontractive mapping with a nonempty fixed point set. Then $(I-T)$ is demiclosed at zero, that is, if whenever $\{x_n\}\subset D(T)$ such that $\{x_n\}$ converges weakly to $x\in D(T)$ and $\{(I-T)x_n\}$ converges strongly to 0, then $Tx=x$.

Lemma 2.2 ([[2], p.80])

Let ${\{a_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{b_n\}}_{n=0}^{\mathrm{\infty }}$, ${\{{\delta }_n\}}_{n=0}^{\mathrm{\infty }}$ be sequences of nonnegative real numbers satisfying the following inequality:

$$a_{n+1}⩽(1+{\delta }_n)a_n+b_n,\mathrm{\forall }n⩾0.$$

If ${\sum }_{n=0}^{\mathrm{\infty }}{\delta }_n<\mathrm{\infty }$ and ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n<\mathrm{\infty }$, then ${lim}_{n\to \mathrm{\infty }}{a}_n$ exists. If, in addition, ${\{a_n\}}_{n=0}^{\mathrm{\infty }}$ has a subsequence which converges strongly to zero, then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.3 ([[2], p.78])

Let E be a real Banach space, K be a nonempty subset of E and $T:K\to K$ be an asymptotically κ-strictly pseudocontractive mapping. Then T is uniformly L-Lipschitzian.

Lemma 2.4 Let E be a real q-uniformly smooth Banach space which is also uniformly convex, and let K be a nonempty, closed and convex subset of E. Let, for each $0⩽j⩽N-1$, $T_j:K\to K$ be an asymptotically ${\lambda }_j$-strictly pseudocontractive mapping with $F\ne \mathrm{\varnothing }$. Let ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ be the sequence satisfying the following conditions:

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists for every $p\in F$;

• (b) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_j{x}_n\parallel =0$, for each $0⩽j⩽N-1$;

• (c) ${lim}_{n\to \mathrm{\infty }}\parallel t{x}_n+(1-t)p_1-p_2\parallel$ exists for all $t\in [0,1]$ and for all $p_1,p_2\in F$.

Then the sequence $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$.

Proof Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists, then $\{x_n\}$ is bounded. By (b) and Lemma 2.1, we have ${\omega }_{\mathcal{W}}(x_n)\subset F$. Assume that $p_1,p_2\in {\omega }_{\mathcal{W}}(x_n)$ and that $\{x_{n_i}\}$ and $\{x_{m_j}\}$ are subsequences of $\{x_n\}$ such that $x_{n_i}\rightharpoonup p_1$ and $x_{m_j}\rightharpoonup p_2$, respectively. Since E is a real q-uniformly smooth Banach space, which is also uniformly convex, then E has a Fréchet differentiable norm. Set $x=p_1-p_2$, $h=t(x_n-p_1)$ in (8), we obtain

$$\begin{array}{rcl}\frac{1}{2}{\parallel p_1-p_2\parallel }^2+t〈x_n-p_1,j(p_1-p_2)〉& ⩽& \frac{1}{2}{\parallel t{x}_n+(1-t)p_1-p_2\parallel }^2\\ ⩽& \frac{1}{2}{\parallel p_1-p_2\parallel }^2+b(t\parallel x_n-p_1\parallel )\\ +t〈x_n-p_1,j(p_1-p_2)〉,\end{array}$$

where b is an increasing function. Since $\parallel x_n-p_1\parallel ⩽M$, $\mathrm{\forall }n⩾0$, for some $M>0$, then

$$\begin{array}{rcl}\frac{1}{2}{\parallel p_1-p_2\parallel }^2+t〈x_n-p_1,j(p_1-p_2)〉& ⩽& \frac{1}{2}{\parallel t{x}_n+(1-t)p_1-p_2\parallel }^2\\ ⩽& \frac{1}{2}{\parallel p_1-p_2\parallel }^2+b(tM)+t〈x_n-p_1,j(p_1-p_2)〉.\end{array}$$

Therefore,

$$\begin{array}{rcl}\frac{1}{2}{\parallel p_1-p_2\parallel }^2+t\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈x_n-p_1,j(p_1-p_2)〉& ⩽& \frac{1}{2}\underset{n\to \mathrm{\infty }}{lim}{\parallel t{x}_n+(1-t)p_1-p_2\parallel }^2\\ ⩽& \frac{1}{2}{\parallel p_1-p_2\parallel }^2+b(tM)\\ +t\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}〈x_n-p_1,j(p_1-p_2)〉.\end{array}$$

Hence, ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}〈x_n-p_1,j(p_1-p_2)〉⩽{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}〈x_n-p_1,j(p_1-p_2)〉+b(tM)/t$. Since ${lim}_{t\to 0^{+}}[b(tM)/t]=0$, then ${lim}_{n\to \mathrm{\infty }}〈x_n-p_1,j(p_1-p_2)〉$ exists. Since ${lim}_{n\to \mathrm{\infty }}〈x_n-p_1,j(p_1-p_2)〉=〈p-p_1,j(p_1-p_2)〉$, for all $p\in {\omega }_{\mathcal{W}}(x_n)$. Set $p=p_2$. We have $〈p_2-p_1,j(p_1-p_2)〉=0$, that is, $p_2=p_1$. Hence, ${\omega }_{\mathcal{W}}(x_n)$ is a singleton, so that $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$. □

3 Main results

Theorem 3.1 Let E be a real q-uniformly smooth Banach space which is also uniformly convex and K be a nonempty, closed and convex subset of E. Let $N⩾1$ be an integer and $J=\{0,1,2,\dots ,N-1\}$. Let, for each $j\in J$, $T_j:K\to K$ be an asymptotically ${\lambda }_j$-strictly pseudocontractive mapping for some $0⩽{\lambda }_j<1$ with sequences ${\{{\kappa }_{n,j}\}}_{n=0}^{\mathrm{\infty }}\subset [1,\mathrm{\infty })$ such that ${\sum }_{n=0}^{\mathrm{\infty }}({\kappa }_n-1)<\mathrm{\infty }$, where ${\kappa }_n={max}_{j\in J}\{{\kappa }_{n,j}\}$, and $F:={\bigcap }_{j=0}^{N-1}F(T_j)\ne \mathrm{\varnothing }$. Let $\lambda ={min}_{j\in J}\{{\lambda }_j\}$. Let $\{{\alpha }_n\}$ satisfy the conditions (6) and $\{x_n\}$ be the sequence generated by the cyclic algorithm (5). Then $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$.

Proof Pick a $p\in F$. We firstly show that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists. To see this, using (2) and (7), we obtain

$$\begin{array}{rcl}{\parallel x_{n+1}-p\parallel }^q& =& {\parallel x_n-p-(1-{\alpha }_n)[x_n-p-(T_{i(n)}^{k(n)}{x}_n-p)]\parallel }^q\\ ⩽& {\parallel x_n-p\parallel }^q+C_q{(1-{\alpha }_n)}^q{\parallel x_n-p-(T_{i(n)}^{k(n)}{x}_n-p)\parallel }^q\\ -q(1-{\alpha }_n)〈x_n-p-(T_{i(n)}^{k(n)}{x}_n-p),j_q(x_n-p)〉\\ ⩽& {\parallel x_n-p\parallel }^q+C_q{(1-{\alpha }_n)}^q{\parallel x_n-p-(T_{i(n)}^{k(n)}{x}_n-p)\parallel }^q\\ -q(1-{\alpha }_n)\{{\lambda }_{i(n)}{\parallel x_n-p-(T_{i(n)}^{k(n)}{x}_n-p)\parallel }^q\\ -({\kappa }_{k(n),i(n)}-1){\parallel x_n-p\parallel }^q\}\\ =& [1+q(1-{\alpha }_n)({\kappa }_{k(n),i(n)}-1)]{\parallel x_n-p\parallel }^q\\ -(1-{\alpha }_n)[q{\lambda }_{i(n)}-C_q{(1-{\alpha }_n)}^{q-1}]{\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel }^q\\ ⩽& [1+q(1-{\alpha }_n)({\kappa }_{k(n)}-1)]{\parallel x_n-p\parallel }^q\\ -(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]{\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel }^q,\end{array}$$

(9)

where ${\kappa }_{k(n)}={max}_{i\in J}\{{\kappa }_{k(n),i(n)}\}$. Since $\mu ⩽{\alpha }_n<1$ for all n, where $\mu =max\{0,1-{(\frac{q\lambda }{C_q})}^{\frac{1}{q-1}}\}$, we get $(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]⩾0$. Therefore, (9) implies

$${\parallel x_{n+1}-p\parallel }^q⩽[1+q(1-{\alpha }_n)({\kappa }_{k(n)}-1)]{\parallel x_n-p\parallel }^q.$$

(10)

Let ${\delta }_n=1+q(1-{\alpha }_n)({\kappa }_{k(n)}-1)$. Since ${\sum }_{n=0}^{\mathrm{\infty }}({\kappa }_n-1)<\mathrm{\infty }$, we have

$$\sum _{n=0}^{\mathrm{\infty }}({\delta }_n-1)=q\sum _{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)({\kappa }_{k(n)}-1)⩽qN\sum _{n=1}^{\mathrm{\infty }}({\kappa }_n-1)<\mathrm{\infty },$$

then (10) implies ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists by Lemma 2.2 (and hence the sequence $\{\parallel x_n-p\parallel \}$ is bounded, that is, there exists a constant $M>0$ such that $\parallel x_n-p\parallel <M$).

Then we prove ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_j{x}_n\parallel =0$, $\mathrm{\forall }j\in J$. In fact, it follows from (9) that

$$\begin{array}{rcl}(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]{\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel }^q& ⩽& {\parallel x_n-p\parallel }^q-{\parallel x_{n+1}-p\parallel }^q\\ +q(1-{\alpha }_n)({\kappa }_{k(n)}-1){\parallel x_n-p\parallel }^q.\end{array}$$

Then

$$\sum _{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]{\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel }^q<{\parallel x_0-p\parallel }^q+M^q\sum _{n=0}^{\mathrm{\infty }}({\delta }_{k(n)}-1)<\mathrm{\infty }.$$

(11)

Since ${\sum }_{n=0}^{\mathrm{\infty }}(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]=\mathrm{\infty }$, then (11) implies that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel =0$. Thus ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel =0$.

For all $n>N$, we have $k(n)-1=k(n-N)$ and $i(n)=i(n-N)$. By Lemma 2.3, we know that $T_j$ is uniformly $L_j$-Lipschitzian, then there exists a constant $L={max}_{j\in J}\{L_j\}$, such that

$$\parallel T_j^{n}x-T_j^{n}y\parallel ⩽L\parallel x-y\parallel ,\mathrm{\forall }n⩾0,\mathrm{\forall }x,y\in K\text{ and }\mathrm{\forall }j\in J.$$

Thus

$$\begin{array}{rcl}\parallel x_n-T_{i(n)}{x}_n\parallel & ⩽& \parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel +\parallel T_{i(n)}^{k(n)}{x}_n-T_{i(n)}{x}_n\parallel \\ ⩽& \parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel +L\parallel T_{i(n)}^{k(n)-1}{x}_n-x_n\parallel \\ ⩽& \parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel +L\parallel T_{i(n)}^{k(n)-1}{x}_n-T_{i(n-N)}^{k(n)-1}{x}_{n-N}\parallel \\ +L\parallel T_{i(n-N)}^{k(n)-1}{x}_{n-N}-x_{n-N-1}\parallel +L\parallel x_{n-N-1}-x_n\parallel \\ ⩽& \parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel +L^2\parallel x_n-x_{n-N}\parallel \\ +L\parallel T_{i(n-N)}^{k(n-N)}{x}_{n-N}-x_{n-N-1}\parallel +L\parallel x_{n-N-1}-x_n\parallel .\end{array}$$

Observe that

$$\parallel x_n-x_{n+1}\parallel =(1-{\alpha }_n)\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Consequently,

$$\parallel x_n-x_{n+l}\parallel \to 0\text{as }n\to \mathrm{\infty },\text{ for all integer }l.$$

Observe also that

$$\parallel x_{n-1}-T_{i(n)}^{k(n)}{x}_n\parallel ⩽\parallel x_n-x_{n-1}\parallel +\parallel x_n-T_{i(n)}^{k(n)}{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Hence,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{i(n)}{x}_n\parallel =0.$$

Consequently, for all $j\in J$, we have

$$\parallel x_n-T_{n+j}{x}_n\parallel ⩽\parallel x_n-x_{n+j}\parallel +\parallel x_{n+j}-T_{n+j}{x}_{n+j}\parallel +L\parallel x_n-x_{n+j}\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_j{x}_n\parallel =0,\mathrm{\forall }j\in J.$$

Now we prove that for all $p_1,p_2\in F$, ${lim}_{n\to \mathrm{\infty }}\parallel t{x}_n+(1-t)p_1-p_2\parallel$ exists for all $t\in [0,1]$. Let $a_n(t)=\parallel t{x}_n+(1-t)p_1-p_2\parallel$. It is obvious that ${lim}_{n\to \mathrm{\infty }}{a}_n(0)=\parallel p_1-p_2\parallel$ and ${lim}_{n\to \mathrm{\infty }}{a}_n(1)={lim}_{n\to \mathrm{\infty }}\parallel x_n-p_2\parallel$ exist. So, we only need to consider the case of $t\in (0,1)$. Define $A_n:K\to K$ by

$$A_{n}x={\alpha }_{n}x+(1-{\alpha }_n)T_{i(n)}^{k(n)}x,x\in K.$$

Then for all $x,y\in K$

$$\begin{array}{rcl}{\parallel A_{n}x-A_{n}y\parallel }^q& ⩽& {\parallel x-y\parallel }^q-q(1-{\alpha }_n)〈(I-T_{i(n)}^{k(n)})x-(I-T_{i(n)}^{k(n)})y,j_q(x-y)〉\\ +C_q{(1-{\alpha }_n)}^q{\parallel x-y-(T_{i(n)}^{k(n)}x-T_{i(n)}^{k(n)}y)\parallel }^q\\ ⩽& [1+q(1-{\alpha }_n)({\kappa }_{k(n)}-1)]{\parallel x-y\parallel }^q\\ -(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]{\parallel x-y-(T_{i(n)}^{k(n)}x-T_{i(n)}^{k(n)}y)\parallel }^q.\end{array}$$

By the choice of ${\alpha }_n$, we have $(1-{\alpha }_n)[q\lambda -C_q{(1-{\alpha }_n)}^{q-1}]⩾0$, so it follows that ${\parallel A_{n}x-A_{n}y\parallel }^q⩽[1+q(1-{\alpha }_n)({\kappa }_{k(n)}-1)]{\parallel x-y\parallel }^q={\delta }_n{\parallel x-y\parallel }^q$. For the convenience of the following discussion, set ${\eta }_n={({\delta }_n)}^{\frac{1}{q}}$, then $\parallel A_{n}x-A_{n}y\parallel ⩽{\eta }_n\parallel x-y\parallel$.

Set $S_{n,m}=A_{n+m-1}{A}_{n+m-2}\cdots A_n$, $m⩾1$. We have

$$\parallel S_{n,m}x-S_{n,m}y\parallel ⩽\left(\prod _{j=n}^{n+m-1}{\eta }_j\right)\parallel x-y\parallel \text{for all }x,y\in K,$$

and

$$S_{n,m}{x}_n=x_{n+m},S_{n,m}p=p\text{for all }p\in F.$$

Set $b_{n,m}=\parallel S_{n,m}(t{x}_n+(1-t)p_1)-t{S}_{n,m}{x}_n-(1-t)S_{n,m}{p}_1\parallel$. If $\parallel x_n-p_1\parallel =0$ for some $n_0$, then $x_n=p_1$ for any $n⩾n_0$ so that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p_1\parallel =0$, in fact, $\{x_n\}$ converges strongly to $p_1\in F$. Thus we may assume $\parallel x_n-p_1\parallel >0$ for any $n⩾0$. Let δ denote the modulus of convexity of E. It is well known (see, for example, [[11], p.108]) that

$$\begin{array}{rcl}\parallel tx+(1-t)y\parallel & ⩽& 1-2min\{t,(1-t)\}\delta (\parallel x-y\parallel )\\ ⩽& 1-2t(1-t)\delta (\parallel x-y\parallel )\end{array}$$

(12)

for all $t\in [0,1]$ and for all $x,y\in E$ such that $\parallel x\parallel ⩽1$, $\parallel y\parallel ⩽1$. Set

$$w_{n,m}=\frac{S_{n,m}{p}_1-S_{n,m}(t{x}_n+(1-t)p_1)}{t({\prod }_{j=n}^{n+m-1}{\eta }_j)\parallel x_n-p_1\parallel },z_{n,m}=\frac{S_{n,m}(t{x}_n+(1-t)p_1)-S_{n,m}{x}_n}{(1-t)({\prod }_{j=n}^{n+m-1}{\eta }_j)\parallel x_n-p_1\parallel }.$$

Then $\parallel w_{n,m}\parallel ⩽1$ and $\parallel z_{n,m}\parallel ⩽1$ so that it follows from (12) that

$$2t(1-t)\delta (\parallel w_{n,m}-z_{n,m}\parallel )⩽1-\parallel t{w}_{n,m}+(1-t)z_{n,m}\parallel .$$

(13)

Observe that

$$\parallel w_{n,m}-z_{n,m}\parallel =\frac{b_{n,m}}{t(1-t)({\prod }_{j=n}^{n+m-1}{\eta }_j)\parallel x_n-p_1\parallel }$$

and

$$\parallel t{w}_{n,m}+(1-t)z_{n,m}\parallel =\frac{\parallel S_{n,m}{x}_n-S_{n,m}{p}_1\parallel }{({\prod }_{j=n}^{n+m-1}{\eta }_j)\parallel x_n-p_1\parallel },$$

it follows from (13) that

(14)

Since E is uniformly convex, then $\delta (s)/s$ is nondecreasing, and since $({\prod }_{j=n}^{n+m-1}{\eta }_j)\parallel x_n-p_1\parallel ⩽({\prod }_{j=n}^{n+m-1}{\eta }_j){\eta }_{n-1}\parallel x_{n-1}-p_1\parallel ⩽\cdots ⩽({\prod }_{j=n}^{n+m-1}{\eta }_j)({\prod }_{j=0}^{n-1}{\eta }_j)\parallel x_0-p_1\parallel =({\prod }_{j=0}^{n+m-1}{\eta }_j)\parallel x_0-p_1\parallel$, hence it follows from (14) that

$$\begin{array}{c}\frac{({\prod }_{j=0}^{n+m-1}{\eta }_j)\parallel x_0-p_1\parallel }{2}\delta \left(\frac{4}{({\prod }_{j=0}^{n+m-1}{\eta }_j)\parallel x_0-p_1\parallel }{b}_{n,m}\right)\hfill \\ ⩽\left(\prod _{j=n}^{n+m-1}{\eta }_j\right)\parallel x_n-p_1\parallel -\parallel x_{n+m}-p_1\parallel (\text{since }t(1-t)⩽\frac{1}{4}\text{ for all }t\in [0,1]).\hfill \end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}{\prod }_{j=0}^{n+m-1}{\eta }_j$ exits and ${lim}_{n\to \mathrm{\infty }}{\prod }_{j=0}^{n+m-1}{\eta }_j\ne 0$. Also since ${lim}_{n\to \mathrm{\infty }}{\prod }_{j=n}^{n+m-1}{\eta }_j=1$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p_1\parallel$ exists, then the continuity of δ and $\delta (0)=0$ yield ${lim}_{n\to \mathrm{\infty }}{b}_{n,m}=0$ uniformly for all $m⩾1$. Observe that

$$\begin{array}{rcl}{a}_{n+m}(t)& ⩽& \parallel t{x}_{n+m}+(1-t)p_1-p_2+(S_{n,m}(t{x}_n+(1-t)p_1)\\ -t{S}_{n,m}{x}_n-(1-t)S_{n,m}{p}_1)\parallel \\ +\parallel S_{n,m}(t{x}_n+(1-t)p_1)-t{S}_{n,m}{x}_n-(1-t)S_{n,m}{p}_1\parallel \\ =& \parallel S_{n,m}(t{x}_n+(1-t)p_1)-S_{n,m}{p}_2\parallel +b_{n,m}\\ ⩽& \left(\prod _{j=n}^{n+m-1}{\eta }_j\right)\parallel t{x}_n+(1-t)p_1-p_2\parallel +b_{n,m}=\left(\prod _{j=n}^{n+m-1}{\eta }_j\right)a_n(t)+b_{n,m}.\end{array}$$

Hence ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{a}_n(t)⩽{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{a}_n(t)$, this ensures that ${lim}_{n\to \mathrm{\infty }}{a}_n(t)$ exists for all $t\in (0,1)$.

Now apply Lemma 2.4 to conclude that $\{x_n\}$ converges weakly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$. □

Theorem 3.2 Let E be a real q-uniformly smooth Banach space, and let K be a nonempty, closed and convex subset of E. Let $N⩾1$ be an integer and $J=\{0,1,2,\dots ,N-1\}$. Let, for each $j\in J$, $T_j:K\to K$ be an asymptotically ${\lambda }_j$-strictly pseudocontractive mapping for some $0⩽{\lambda }_j<1$ with sequences ${\{{\kappa }_{n,j}\}}_{n=0}^{\mathrm{\infty }}\subset [1,\mathrm{\infty })$ such that ${\sum }_{n=0}^{\mathrm{\infty }}({\kappa }_n-1)<\mathrm{\infty }$, where ${\kappa }_n={max}_{j\in J}\{{\kappa }_{n,j}\}$, and $F:={\bigcap }_{j=0}^{N-1}F(T_j)\ne \mathrm{\varnothing }$. Let $\lambda ={min}_{j\in J}\{{\lambda }_j\}$. Let $\{{\alpha }_n\}$ satisfy the conditions (6) and $\{x_n\}$ be the sequence generated by the cyclic algorithm (5). Then $\{x_n\}$ converges strongly to a common fixed point of the family ${\{T_j\}}_{j=0}^{N-1}$ if and only if

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}d(x_n,F)=0,$$

where $d(x_n,F)={inf}_{p\in F}\parallel x_n-p\parallel$.

Proof It follows from (10) that

$${\parallel x_{n+1}-p\parallel }^q⩽{\delta }_n{\parallel x_n-p\parallel }^q.$$

Thus ${[d(x_{n+1}-p)]}^q⩽{\delta }_n{[d(x_n-p)]}^q$, and it follows from Lemma 2.2 that ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ exists.

Now if $\{x_n\}$ converges strongly to a common fixed point p of the family ${\{T_j\}}_{j=0}^{N-1}$, then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel =0$. Since

$$0⩽d(x_n,F)⩽\parallel x_n-p\parallel ,$$

we have ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F)=0$.

Conversely, suppose ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}d(x_n,F)=0$, then the existence of ${lim}_{n\to \mathrm{\infty }}d(x_n,F)$ implies that ${lim}_{n\to \mathrm{\infty }}d(x_n,F)=0$. Thus, for arbitrary $ϵ>0$, there exists a positive integer $n_0$ such that $d(x_n,F)<\frac{ϵ}{2}$ for any $n⩾n_0$.

From (10), we have

$${\parallel x_{n+1}-p\parallel }^q⩽{\parallel x_n-p\parallel }^q+M^q({\delta }_n-1),n⩾0,$$

and for some $M>0$, $\parallel x_n-p\parallel <M$. Now, an induction yields

$$\begin{array}{rcl}{\parallel x_n-p\parallel }^q& ⩽& {\parallel x_{n-1}-p\parallel }^q+M^q({\delta }_{n-1}-1)\\ ⩽& {\parallel x_{n-2}-p\parallel }^q+M^q({\delta }_{n-2}-1)+M^q({\delta }_{n-1}-1)\\ ⩽& \cdots ⩽{\parallel x_l-p\parallel }^q+M^q\sum _{j=l}^{n-1}({\delta }_j-1),n-1⩾l⩾0.\end{array}$$

Since ${\sum }_{n=0}^{\mathrm{\infty }}({\delta }_n-1)<\mathrm{\infty }$, then there exists a positive integer $n_1$ such that ${\sum }_{j=n}^{\mathrm{\infty }}({\delta }_j-1)<{(\frac{ϵ}{2M})}^q$, $\mathrm{\forall }n⩾n_1$. Choose $N=max\{n_0,n_1\}$, then for all $n,m⩾N+1$ and for all $p\in F$, we have

$$\begin{array}{rcl}\parallel x_n-x_m\parallel & ⩽& \parallel x_n-p\parallel +\parallel x_m-p\parallel \\ ⩽& {[{\parallel x_N-p\parallel }^q+M^q\sum _{j=N}^{n-1}({\delta }_j-1)]}^{\frac{1}{q}}+{[{\parallel x_N-p\parallel }^q+M^q\sum _{j=N}^{m-1}({\delta }_j-1)]}^{\frac{1}{q}}\\ ⩽& {[{\parallel x_N-p\parallel }^q+M^q\sum _{j=N}^{\mathrm{\infty }}({\delta }_j-1)]}^{\frac{1}{q}}+{[{\parallel x_N-p\parallel }^q+M^q\sum _{j=N}^{\mathrm{\infty }}({\delta }_j-1)]}^{\frac{1}{q}}\\ =& 2{[{\parallel x_N-p\parallel }^q+M^q\sum _{j=N}^{\mathrm{\infty }}({\delta }_j-1)]}^{\frac{1}{q}}.\end{array}$$

Taking infimum over all $p\in F$, we obtain

$$\begin{array}{rcl}\parallel x_n-x_m\parallel & ⩽& 2{\{{[d(x_N,F)]}^q+M^q\sum _{j=N}^{\mathrm{\infty }}({\delta }_j-1)\}}^{\frac{1}{q}}\\ <& 2{[{\left(\frac{ϵ}{2}\right)}^q+M^q{\left(\frac{ϵ}{2M}\right)}^q]}^{\frac{1}{q}}<2ϵ.\end{array}$$

Thus ${\{x_n\}}_{n=0}^{\mathrm{\infty }}$ is Cauchy. Suppose ${lim}_{n\to }{x}_n=u$. Then for all $j\in J$ we have

$$0⩽\parallel u-T_{j}u\parallel ⩽\parallel u-x_n\parallel +\parallel x_n-T_j{x}_n\parallel +L\parallel x_n-u\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

Thus $u\in F(T_j)$, $\mathrm{\forall }j\in J$, and hence $u\in F$. □