Approximating fixed points for a reversible semigroup of Lipschitzian mappings in a smooth Banach space

Abstract

In this paper, we approximate a fixed point of the semigroup $\phi =\{T_s:s\in S\}$ of Lipschitzian mappings from a nonempty compact convex subset C of a smooth Banach space E into C with a uniform Lipschitzian condition and with respect to a finite family of sequences ${\{{\mu }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ of left strong regular invariant means defined on an appropriate invariant subspace of $l^{\mathrm{\infty }}(S)$. Our result extends the main results announced by several others.

MSC:47H09, 47H10, 47J25.

1 Introduction

Let E be a real Banach space with the topological dual $E^{\ast }$, and let C be a nonempty closed and convex subset of E. Recall that a mapping T of C into itself is said to be

• (1) Lipschitzian with Lipschitz constant $l>0$ if

  $$\parallel Tx-Ty\parallel \le l\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

• (2) nonexpansive if

  $$\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

• (3) asymptotically nonexpansive if there exists a sequence $\{l_n\}$ of positive numbers such that ${lim}_{n\to \mathrm{\infty }}{l}_n=1$ and

  $$\parallel T^{n}x-T^{n}y\parallel \le l_n\parallel x-y\parallel ,\mathrm{\forall }x,y\in C.$$

A semigroup S is called left reversible if any two closed right ideals of S have non-void intersection, i.e., $aS\cap bS\ne \mathrm{\varnothing }$ for $a,b\in S$. In this case, $(S,⪯)$ is a directed set when the binary relation ⪯ on S is defined by $a⪯b$ if and only if $aS\supset bS$ for $a,b\in S$.

Notation Throughout the rest of this paper, S will always denote a left reversible semigroup with an identity e.

In [1], Lau et al. studied iterative schemes for approximating a fixed point of the semigroup $\phi =\{T(s):s\in S\}$ of nonexpansive mappings on a nonempty compact convex subset C of a smooth (and strictly convex) Banach space and introduced the following iteration process. Let $x_1=x\in C$ and

$$x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)T_{{\mu }_n}{x}_n,n\ge 1,$$

(1)

where ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence of left strong regular invariant means defined on an appropriate invariant subspace of $l^{\mathrm{\infty }}(S)$.

$\phi =\{T(s):s\in S\}$ is called a representation of S as a Lipschitzian mapping on C with Lipschitz constant $\{l(s):s\in S\}$ if $T(s)$ is a Lipschitzian with Lipschitz constant $l(s)$ for each $s\in S$, $T(st)=T(s)T(t)$ for each $t,s\in S$ and $T(e)=I$. φ is called an asymptotically nonexpansive semigroup on C if φ is a representation of S as a Lipschitzian mapping on C with Lipschitz constant $\{l(s);s\in s\}$ and ${lim}_{s}l(s)\le 1$.

In 2008, Saeidi proved the following theorem.

Theorem 1.1 [2]

Let S be a left reversible semigroup and $\phi =\{T_s:s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty compact convex subset C of a smooth Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}K(s)\le 1$, and let f be an α-contraction on C for some $0<\alpha <1$. Let X be a left invariant φ-stable subspace of $L^{\mathrm{\infty }}(\phi )$ containing 1, let ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of left strong regular invariant means defined on X such that ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$, and let ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence defined by

$$c_n=\underset{x,y\in C}{sup}(\parallel T_{{\mu }_n}x-T_{{\mu }_n}y\parallel -\parallel x-y\parallel ),n\ge 1.$$

Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ such that

(C₁) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 1$,

(C₂) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$,

(C₃) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

(C₄) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$,

(C₅) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{c_n}{{\alpha }_n}\le 0$.

If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence generated by $x_1\in C$ and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{T}_{{\mu }_n}{x}_n,n\ge 1,$$

then the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to some $z\in Fix(\phi )$, the set of common fixed points of φ, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(\phi ).$$

Equivalently, one has $z=Pfz$, where P is the unique sunny nonexpansive retraction of C onto $F(\phi )$.

In 2007, Zhang et al. [3] introduced the following composite iteration scheme:

$$\{\begin{array}{c}{y}_n={\beta }_n{x}_n+(1-{\beta }_n)T{t}_n{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n,\hfill \end{array}$$

(2)

where $\{T(t):t\ge 0\}$ is a nonexpansive semigroup from C to C, u is an arbitrary (but fixed) element in C, $\{{\alpha }_n\}\subset (0,1)$ and $\{{\beta }_n\}\subset [0,1]$, $\{t_n\}\subset {\mathbb{R}}^{+}$, and proved some strong convergence theorems of an explicit composite iteration scheme for nonexpansive mappings in the framework of a reflexive Banach space with a uniformly Gâteaux differentiable norm, uniformly smooth Banach space and uniformly convex Banach space with a weakly continuous normalized duality mapping.

Motivated and inspired by Zhang et al. [3] and Saeidi [2], Katchang and Kumam proved the following theorem.

Theorem 1.2 [4]

Let S be a left reversible semigroup, and let $\phi =\{T_s:s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty compact convex subset C of a smooth Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}K(s)\le 1$, and let f be an α-contraction on C for some $0<\alpha <1$. Let X be a left invariant φ-stable subspace of $L^{\mathrm{\infty }}(\phi )$ containing 1, let ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of left strong regular invariant means defined on X such that ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$, and let ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence defined by

$$c_n=\underset{x,y\in C}{sup}(\parallel T_{{\mu }_n}x-T_{{\mu }_n}y\parallel -\parallel x-y\parallel ),n\ge 1.$$

Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ such that

(C₁) ${\alpha }_n+{\beta }_n+{\gamma }_n=1$, $n\ge 1$,

(C₂) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$,

(C₃) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$,

(C₄) $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$,

(C₅) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\frac{c_n}{{\alpha }_n}\le 0$,

(C₆) ${lim}_{n\to \mathrm{\infty }}{\delta }_n=0$.

If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a sequence generated by $x_1\in C$ and

$$\{\begin{array}{c}{y}_n={\delta }_n{x}_n+(1-{\delta }_n)T_{{\mu }_n}{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{y}_n,n\ge 1,\hfill \end{array}$$

(3)

then the sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to some $z\in Fix(\phi )$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in F(\phi ).$$

Equivalently, one has $z=Pfz$, where P is the unique sunny nonexpansive retraction of C onto $F(\phi )$.

Recently, many authors studied fixed point results for a nonlinear semigroup mapping, for example, [5–8].

In this paper, motivated and inspired by Qianglian et al. [9], Lau et al. [1], Zhang et al. [3], Saeidi [2], Katchang and Kumam [4], Sunthrayuth and Kumam [10, 11] and Wattanawitoon and Kumam [12], we introduce the composite explicit viscosity iterative schemes as follows:

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_{1,n})+{\beta }_n{x}_n+{\gamma }_n{T}_{{\mu }_{1,n}}{y}_{1,n},\\ y_{i,n}={\delta }_{i,n}{y}_{i+1,n}+(I-{\delta }_{i,n})T_{{\mu }_{i,n}}{y}_{i+1,n},i=1,2,\dots ,m,\\ y_{m+1,n}=x_n\end{array}$$

(4)

for an asymptotically nonexpansive semigroup $\phi =\{T_s:s\in S\}$ on a compact convex subset C of a smooth Banach space E with respect to a finite family of left regular sequences ${\{{\mu }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ of invariant means defined on an appropriate invariant subspace of $l^{\mathrm{\infty }}(S)$. We prove, under certain appropriate assumptions on the sequences ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\delta }_n\}}_{i=1,n=1}^{m,\mathrm{\infty }}$, that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ defined by (4) converge strongly to $z\in Fix(\phi )$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(\phi ).$$

Our result improves and extends many previous results (e.g., [1, 2, 4, 13–15] and many others).

2 Preliminaries

Let $E^{\ast }$ be the topological dual of a real Banach space E. The value of $j\in E^{\ast }$ at $x\in E$ will be denoted by $〈x,j〉$ or $j(x)$. With each $x\in E$, we associate the set

$$J(x)=\{j\in E^{\ast }:〈x,j〉={\parallel x\parallel }^2={\parallel j\parallel }^2\}.$$

Using the Hahn-Banach theorem, it is immediately clear that $J(x)\ne \varphi$ for each $x\in E$. The multi-valued mapping J from E into $E^{\ast }$ is said to be the (normalized) duality mapping. A Banach space E is said to be smooth if the duality mapping J is single-valued. As is well known, the duality mapping is norm to weak-star continuous when E is smooth, see [16].

Let $B(S)$ be the Banach space of all bounded real-valued functions defined on S with supremum norm. For each $s\in S$, we define the left and right translation operators $l(s)f$ and $r(s)f$ on $B(S)$ by

$$(l(s)f)(t)=f(st),\text{and}(r(s)f)(t)=f(ts),$$

for each $s\in S$ and $f\in B(S)$, respectively. Let X be a subspace of $B(S)$ containing 1 and let $X^{\ast }$ be its topological dual. An element μ of $X^{\ast }$ is said to be a mean on X if $\parallel \mu \parallel =\mu (1)=1$. Let $\mu \in X^{\ast }$. Then we define $r{(s)}^{\ast }\mu ,l{(s)}^{\ast }\mu \in X^{\ast }$ by $(r{(s)}^{\ast }\mu )f=\mu (r(s)f)$, $(l{(s)}^{\ast }\mu )f=\mu (l(s)f)$ for each $f\in X$ and $s\in S$. It is easy to see that if μ is a mean on X, then $r{(s)}^{\ast }\mu$ and $l{(s)}^{\ast }\mu$ are also. We often write ${\mu }_t(f(t))$ instead of $\mu (f)$ for $\mu \in X^{\ast }$ and $f\in X$. Let X be left invariant (resp. right invariant), i.e., $l(s)(X)\subset X$ (resp. $r(s)(X)\subset X$) for each $s\in S$. A mean μ on X is said to be left invariant (resp. right invariant) if $l{(s)}^{\ast }\mu =\mu$ (resp. $r{(s)}^{\ast }\mu =\mu$) for each $s\in S$ and $f\in X$. X is said to be left (resp. right) amenable if X has a left (resp. right) invariant mean. The semigroup S is amenable (i.e., S is both left and right amenable) when S is a commutative semigroup or a solvable group. However, the free group (or semigroup) on two generators is not left amenable. If a semigroup S is left amenable, then S is left reversible, but the converse is not true see [17], [[18], p.335]. A net $\{{\mu }_{\alpha }\}$ of means on X is said to be strongly left regular if

$$\underset{\alpha }{lim}\parallel l{(s)}^{\ast }{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0,$$

for each $s\in S$, where $l{(s)}^{\ast }$ is the adjoint operator of $l_s$.

Let $\phi =\{T(s):s\in S\}$ be a representation of S as a Lipschitzian mapping on C with Lipschitz constant $\{l(s):s\in S\}$. By $Fix(\phi )$ we denote the set of common fixed points of φ, i.e.,

$$Fix(\phi )=\bigcap _{s\in S}\{x\in C:T(s)x=x\}.$$

We denote by $C_a$ the set of almost periodic elements in C, i.e., all $x\in C$ such that $\{T(s)x:s\in S\}$ is relatively compact in the norm topology of E. Let X be a subspace of $B(S)$ such that the functions (i) $s\to 〈T(s)x,x^{\ast }〉$ and (ii) $s\to \parallel T(s)x-y\parallel$ on S are in X for all $x,y\in C$ and $x^{\ast }\in E^{\ast }$. We will call a subspace X of $B(S)$ satisfying (i) and (ii) φ-stable. We know that if X is a subspace of $B(S)$ containing 1 and the function $s\to 〈T(s)x,x^{\ast }〉$ on S is in X for all $x\in C$ and $x^{\ast }\in E^{\ast }$, then there exists a unique point $x_0\in E$ such that $\mu 〈T(\cdot )x,x^{\ast }〉=〈x_0,x^{\ast }〉$ for a mean μ on X, $x\in C$ and $x^{\ast }\in E$. We denote such a point $x_0\in E$ by $T_{\mu }x$. See [19] for more details.

Lemma 2.1 [20]

Let S be a semigroup and C be a nonempty closed convex subset of a reflexive Banach space E. Let $\phi =\{T_t:t\in S\}$ be a nonexpansive semigroup on H such that $\{T_{t}x:t\in S\}$ is bounded for some $x\in C$, let X be a subspace of $B(S)$ such that $1\in X$ and the mapping $t\to 〈T_{t}x,y^{\ast }〉$ is an element of X for each $x\in C$ and $y^{\ast }\in E^{\ast }$, and μ is a mean on X. If we write $T_{\mu }x$ instead of $\int T_{t}xd\mu (t)$, then the following hold:

• (i) $T_{\mu }$ is a nonexpansive mapping from C into C.

• (ii) $T_{\mu }x=x$ for each $x\in Fix(\phi )$.

• (iii) $T_{\mu }x\in \overline{\mathit{co}}\{T_{t}x:t\in S\}$ for each $x\in C$.

Lemma 2.2 [21]

Let $\phi =\{T(s):s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty weakly compact convex subset C of a Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}l(s)\le 1$ on the Lipschitz constant of mappings. Let X be a left invariant and φ-stable subspace of $B(S)$, and let ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ be an asymptotically left invariant sequence of means on X. If $z\in C_a$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel T_{{\mu }_n}z-z\parallel =0$, then z is a common fixed point of φ.

Lemma 2.3 [2]

Let $\phi =\{T(s):s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty weakly compact convex subset C of a Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}l(s)\le 1$ on the Lipschitz constant of mappings. Let X be a left invariant subspace of $B(S)$ containing 1 such that the mapping $s\to 〈T(s)x,x^{\ast }〉$ on S is in X for all $x\in C$ and $x^{\ast }\in E^{\ast }$, and ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ is an asymptotically left invariant sequence of means on X. Then

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\left(\underset{x,y\in C}{sup}(\parallel T_{{\mu }_n}x-T_{{\mu }_n}y\parallel -\parallel x-y\parallel )\right)\le 0.$$

Let D be a subset of B, where B is a subset of a Banach space E, and let P be a retraction of B onto D, that is, $Px=x$ for each $x\in D$. Then P is said to be sunny [22] if for each $x\in B$ and $t\ge 0$ with $Px+t(x-Px)\in B$, $P(Px+t(x-Px))=Px$. A subset D of B is said to be a sunny nonexpansive retract of B if there exists a sunny nonexpansive retraction P of B into D.

Lemma 2.4 [1]

Let $\phi =\{T(s):s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty compact convex subset C of a smooth Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}l(s)\le 1$ on the Lipschitz constant of mappings. Let X be a left invariant and φ-stable subspace of $B(S)$ containing 1 and μ be a left invariant mean on X. Then $Fix(\phi )$ is a sunny nonexpansive retract of C, and the sunny nonexpansive retraction of C onto $Fix(\phi )$ is unique.

Lemma 2.5 [16]

Let C be a nonempty convex subset of a smooth Banach space E, let D be a nonempty subset of C, and let $P:C\to D$ be a retraction. Then the following are equivalent:

• (a) P is sunny nonexpansive.

• (b) $〈x-Px,J(y-Px)〉\le 0$ for all $x\in C$ and $y\in D$.

• (c) $〈x-y,J(Px-Py)〉\ge {\parallel Px-Py\parallel }^2$ for all $x,y\in C$.

Lemma 2.6 [3]

Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ be bounded sequences in a Banach space X, and let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in $[0,1]$ such that $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$. Suppose $x_{n+1}={\alpha }_n{x}_n+(1-{\alpha }_n)y_n$ for all integers $n\ge 0$ and

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Then ${lim}_{n\to \mathrm{\infty }}\parallel y_n-x_n\parallel =0$.

Lemma 2.7 [19]

Let E be a real smooth Banach space and J be the duality mapping. Then

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

Lemma 2.8 [23]

Let ${\{a_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-b_n)a_n+b_n{c}_n,n\ge 0,$$

where ${\{b_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{c_n\}}_{n=1}^{\mathrm{\infty }}$ are sequences of real numbers satisfying the following conditions:

• (i) ${\{b_n\}}_{n=1}^{\mathrm{\infty }}\subset (0,1)$, ${\sum }_{n=0}^{\mathrm{\infty }}{b}_n=\mathrm{\infty }$,

• (ii) either ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{c}_n\le 0$ or ${\sum }_{n=0}^{\mathrm{\infty }}|b_n{c}_n|<\mathrm{\infty }$.

Then ${lim}_{n\to \mathrm{\infty }}{a}_n=0$.

Lemma 2.9 [16]

Let $(X,d)$ be a metric space. A subset C of X is compact if and only if every sequence in C contains a convergent subsequence with limit in C.

3 The main result

In this section, we establish a strong convergence theorem for finding a common fixed point of an asymptotically nonexpansive semigroup in a smooth Banach space.

Theorem 3.1 Let $\phi =\{T(s):s\in S\}$ be a representation of S as a Lipschitzian mapping from a nonempty compact convex subset C of a smooth Banach space E into C, with uniform Lipschitzian constant ${lim}_{s}l(s)\le 1$ on the Lipschitz constant of mappings, such that $Fix(\phi )\ne \mathrm{\varnothing }$, and let f be a contraction of C into itself with constant $\alpha \in (0,1)$. Let X be a left invariant and φ-stable subspace of $B(S)$ containing 1 and the function $t\to 〈T_{t}x,y〉$ is an element of X for each $x\in C$ and $y\in H$, and let ${\{{\mu }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ be a finite family of left regular sequences of invariant means on X such that for $i=1,2,\dots ,m$, ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{i,n+1}-{\mu }_{i,n}\parallel =0$. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying the condition

(${\mathrm{C}}_5^{\prime }$) ${lim}_{n\to \mathrm{\infty }}{\delta }_{i,n}=1$, $i=1,2,\dots ,m$.

If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ are sequences generated by $x_1\in C$ and

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_{1,n})+{\beta }_n{x}_n+{\gamma }_n{T}_{{\mu }_{1,n}}{y}_{1,n},\\ y_{i,n}={\delta }_{i,n}{y}_{i+1,n}+(I-{\delta }_{i,n})T_{{\mu }_{i,n}}{y}_{i+1,n},i=1,2,\dots ,m,\\ y_{m+1,n}=x_n,\end{array}$$

(5)

then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converge strongly to $z\in Fix(\phi )$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(\phi ).$$

(6)

Equivalently, $z=Pf(z)$, where P denotes the unique sunny nonexpansive retraction of C onto $Fix(\phi )$.

Proof From Lemma 2.1 and the definition of ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$, for every $z\in Fix(\phi )$, we have

$$\begin{array}{rl}\parallel y_{i,n}-z\parallel & =\parallel {\delta }_{i,n}{y}_{i+1,n}+(1-{\delta }_{i,n})T_{{\mu }_{i,n}}{y}_{i+1,n}-z\parallel \\ \le {\delta }_{i,n}\parallel y_{i+1,n}-z\parallel +(1-{\delta }_{i,n})\parallel T_{{\mu }_{i,n}}{y}_{i+1,n}-T_{{\mu }_{i,n}}z\parallel \\ ={\delta }_{i,n}\parallel y_{i+1,n}-z\parallel +(1-{\delta }_{i,n})\parallel y_{i+1,n}-z\parallel \\ =\parallel y_{i+1,n}-z\parallel .\end{array}$$

Therefore, we have

$$\parallel y_{1,n}-z\parallel \le \parallel y_{2,n}-z\parallel \le \cdots \le \parallel y_{m,n}-z\parallel \le \parallel x_n-z\parallel .$$

(7)

We shall divide the proof into several steps.

Step 1. Let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in C. Then

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_{{\mu }_{i,n+1}}{t}_n-T_{{\mu }_{i,n}}{t}_n\parallel =0,i=1,2,\dots ,m.$$

Proof of Step 1. This assertion is proved in [24, 25].

Step 2. ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+1}-x_n\parallel =0$.

Proof of Step 2. From the definition of ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$, we have

$$\begin{array}{r}\parallel y_{i,n+1}-y_{i,n}\parallel \\ =\parallel {\delta }_{i,n+1}{y}_{i+1,n+1}+(1-{\delta }_{i,n+1})T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}\\ -{\delta }_{i,n}{y}_{i+1,n}-(1-{\delta }_{i,n})T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel \\ =\parallel {\delta }_{i,n+1}{y}_{i+1,n+1}-{\delta }_{i,n+1}{y}_{i+1,n}+{\delta }_{i,n+1}{y}_{i+1,n}\\ +(1-{\delta }_{i,n+1})T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}-{\delta }_{i,n}{y}_{i+1,n}-(1-{\delta }_{i,n})T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel \\ \le {\delta }_{i,n+1}\parallel y_{i+1,n+1}-y_{i+1,n}\parallel +|{\delta }_{i,n+1}-{\delta }_{i,n}|\parallel y_{i+1,n}\parallel \\ +(1-{\delta }_{i,n+1})\parallel T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}\parallel +(1-{\delta }_{i,n})\parallel T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel \\ \le \parallel y_{i+1,n+1}-y_{i+1,n}\parallel +|{\delta }_{i,n+1}-{\delta }_{i+1,n}|\parallel y_{i+1,n}\parallel \\ +(1-{\delta }_{i,n+1})\parallel T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}\parallel +(1-{\delta }_{i,n})\parallel T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel ,\end{array}$$

which implies that

$$\begin{array}{r}\parallel y_{i,n+1}-y_{i,n}\parallel \\ \le \parallel x_{n+1}-x_n\parallel +\sum _{j=i}^m(|{\delta }_{j,n+1}-{\delta }_{j,n}|\parallel y_{j+1,n}\parallel \\ +(1-{\delta }_{i,n+1})\parallel T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}\parallel +(1-{\delta }_{i,n})\parallel T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel ).\end{array}$$

(8)

Define

$$x_{n+1}=(1-{\beta }_n)z_n+{\beta }_n{x}_n,n\ge 1.$$

(9)

Observe that from the definition of $z_n$, we obtain

$$\begin{array}{rl}{z}_{n+1}-z_n=& \frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}f(y_{1,n+1})+{\gamma }_{n+1}{T}_{{\mu }_{1,n+1}}{y}_{1,n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n}f(y_{1,n})+{\gamma }_n{T}_{{\mu }_{1,n}}{y}_{1,n}}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(f(y_{1,n+1})-T_{{\mu }_{1,n+1}}{y}_{1,n+1})-\frac{{\alpha }_n}{1-{\beta }_n}(f(y_{1,n})-T_{{\mu }_{1,n}}{y}_{1,n})\\ +T_{{\mu }_{1,n+1}}{y}_{1,n+1}-T_{{\mu }_{1,n}}{y}_{1,n}.\end{array}$$

It follows that

$$\begin{array}{r}\parallel z_{n+1}-z_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(y_{1,n+1})-T_{{\mu }_{1,n+1}}{y}_{1,n+1})\parallel \\ +\frac{{\alpha }_n}{1-{\beta }_n}\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{y}_{1,n}\parallel \\ +\parallel y_{1,n+1}-y_{1,n}\parallel +\parallel T_{{\mu }_{1,n+1}}{y}_{1,n}-T_{{\mu }_{1,n}}{y}_{1,n}\parallel .\end{array}$$

(10)

Substituting (8) into (10), we obtain

$$\begin{array}{r}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel f(y_{1,n+1})-T_{{\mu }_{1,n+1}}{y}_{1,n+1}\parallel \\ +\frac{{\alpha }_n}{1-{\beta }_n}\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{y}_{1,n}\parallel \\ +\parallel T_{{\mu }_{1,n+1}}{y}_{1,n}-T_{{\mu }_{1,n}}{y}_{1,n}\parallel +\sum _{j=1}^m(|{\delta }_{j,n+1}-{\delta }_{j,n}|\parallel y_{j+1,n}\parallel \\ +(1-{\delta }_{i,n+1})\parallel T_{{\mu }_{i,n+1}}{y}_{i+1,n+1}\parallel +(1-{\delta }_{i,n})\parallel T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel ).\end{array}$$

(11)

It follows from Step 1, conditions (C₂) and (${\mathrm{C}}_5^{\prime }$) that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Applying Lemma 2.6 to (9), we get

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

Consequently,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \mathrm{\infty }}{lim}(1-{\beta }_n)\parallel x_n-z_n\parallel =0.$$

Step 3. We claim that $\omega ({\{x_n\}}_{n=1}^{\mathrm{\infty }})\subset Fix(\phi )$, where

$$\omega \left({\{x_n\}}_{n=1}^{\mathrm{\infty }}\right):=\{x\in C:{\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}\subset {\{x_n\}}_{n=1}^{\mathrm{\infty }},\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-x\parallel =0\}.$$

Proof of Step 3. From Lemma 2.9, we get $\omega ({\{x_n\}}_{n=1}^{\mathrm{\infty }})\ne \mathrm{\varnothing }$.

Let $x\in \omega ({\{x_n\}}_{n=1}^{\mathrm{\infty }})$. Then there exists a subsequence ${\{x_{n_j}\}}_{j=1}^{\mathrm{\infty }}$ of ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ such that

$$\underset{j\to \mathrm{\infty }}{lim}\parallel x_{n_j}-x\parallel =0.$$

(12)

Observe that

$$\begin{array}{r}\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T_{{\mu }_{1,n}}{x}_n\parallel \\ =\parallel x_n-x_{n+1}\parallel +\parallel {\alpha }_n(f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n)+{\beta }_n(x_n-T_{{\mu }_{1,n}}{x}_n)\\ +{\gamma }_n(T_{{\mu }_{1,n}}{y}_{1,n}-T_{{\mu }_{1,n}}{x}_n)\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n\parallel +{\beta }_n\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ +{\gamma }_n\parallel T_{{\mu }_{1,n}}{y}_{1,n}-T_{{\mu }_{1,n}}{x}_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n\parallel +{\beta }_n\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ +{\gamma }_n\parallel y_{1,n}-x_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n\parallel +{\beta }_n\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ +{\gamma }_n\sum _{i=1}^m\parallel y_{i,n}-y_{i+1,n}\parallel \\ =\parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n\parallel +{\beta }_n\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ +{\gamma }_n\sum _{i=1}^m(1-{\delta }_{i,n})\parallel y_{i+1,n}-T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel .\end{array}$$

Therefore, we have

$$\begin{array}{r}(1-{\beta }_n)\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel \\ \le \parallel x_n-x_{n+1}\parallel +{\alpha }_n\parallel f(y_{1,n})-T_{{\mu }_{1,n}}{x}_n\parallel \\ +{\gamma }_n\sum _{i=1}^m(1-{\delta }_{i,n})\parallel y_{i+1,n}-T_{{\mu }_{i,n}}{y}_{i+1,n}\parallel .\end{array}$$

(13)

From condition (C₄), it follows that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}(1-{\beta }_n)>0.$$

(14)

By conditions (C₂) and (${\mathrm{C}}_5^{\prime }$), Step 2, (13) and (14), we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_{{\mu }_{1,n}}{x}_n\parallel =0.$$

(15)

Indeed, observe that

$$\begin{array}{rl}\underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x-T_{{\mu }_{1,n_j}}x\parallel \le & \underset{j\to \mathrm{\infty }}{lim\hspace{0.17em}sup}(\parallel x-x_{n_j}\parallel +\parallel x_{n_j}-T_{{\mu }_{1,n_j}}{x}_{n_j}\parallel \\ +\parallel T_{{\mu }_{1,n_j}}{x}_{n_j}-T_{{\mu }_{1,n_j}}x\parallel ).\end{array}$$

Thus, due to (15), Lemma 2.2 and Lemma 2.3, we get $x\in Fix(\phi )$.

Step 4. ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $z=Pf(z)$.

Proof of Step 4. From Lemma 2.4 there exists a unique sunny nonexpansive retraction P of C onto $Fix(\phi )$. Since f is a contraction of C into itself, therefore Pf is a contraction. Then the Banach contraction guarantees that Pf has a unique fixed point z. By Lemma 2.5, z is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(\phi ).$$

(16)

Let us show that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-I)z,J(x_n-z)〉\le 0.$$

Indeed, we can choose a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-I)z,J(x_n-z)〉=\underset{k\to \mathrm{\infty }}{lim}〈(f-I)z,J(x_{n_k}-z)〉.$$

(17)

Since C is compact, we may assume, with no loss of generality, that $\{x_{n_k}\}$ converges strongly to some $y\in C$. By Step 3, $y\in Fix(\phi )$. Because the duality mapping J is norm to weak-star continuous from (16) and (17), we have

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈(f-I)z,J(x_n-z)〉\le 0.$$

(18)

Using Lemma 2.1, Lemma 2.7 and relation (7), we have

$$\begin{array}{r}{\parallel x_{n+1}-z\parallel }^2\\ ={\parallel {\alpha }_n(f(y_{1,n})-z)+{\beta }_n(x_n-z)+{\gamma }_n(T_{{\mu }_{1,n}}{y}_{1,n}-z)\parallel }^2\\ \le {\parallel {\beta }_n(x_n-z)+{\gamma }_n(T_{{\mu }_{1,n}}{y}_{1,n}-z)\parallel }^2+2{\alpha }_n〈f(y_{1,n})-z,J(x_{n+1}-z)〉\\ \le {[{\beta }_n\parallel x_n-z\parallel +{\gamma }_n\parallel T_{{\mu }_{1,n}}{y}_{1,n}-z\parallel ]}^2+2{\alpha }_n〈f(y_{1,n})-f(z),J(x_{n+1}-z)〉\\ +2{\alpha }_n〈f(z)-z,J(x_{n+1}-z)〉\\ \le {[{\beta }_n\parallel x_n-z\parallel +{\gamma }_n\parallel y_{1,n}-z\parallel ]}^2+2{\alpha }_n\alpha \parallel y_{1,n}-z\parallel \parallel J(x_{n+1}-z)\parallel \\ +2{\alpha }_n〈f(z)-z,J(x_{n+1}-z)〉\\ \le {[{\beta }_n\parallel x_n-z\parallel +{\gamma }_n\parallel x_n-z\parallel ]}^2+2{\alpha }_n\alpha \parallel x_n-z\parallel \parallel x_{n+1}-z\parallel \\ +2{\alpha }_n〈f(z)-z,J(x_{n+1}-z)〉\\ \le {(1-{\alpha }_n)}^2{\parallel x_n-z\parallel }^2+{\alpha }_n\alpha ({\parallel x_n-z\parallel }^2+{\parallel x_{n+1}-z\parallel }^2)\\ +2{\alpha }_n〈f(z)-z,J(x_{n+1}-z)〉,\end{array}$$

and consequently,

$$\begin{array}{r}{\parallel x_{n+1}-z\parallel }^2\\ \le \frac{{(1-{\alpha }_n)}^2+{\alpha }_n\alpha }{1-{\alpha }_n\alpha }{\parallel x_n-z\parallel }^2+\frac{2{\alpha }_n}{1-{\alpha }_n\alpha }〈f(z)-z,J(x_{n+1}-z)〉\\ \le (1-\frac{2{\alpha }_n(1-{\alpha }_n)}{1-{\alpha }_n\alpha }){\parallel x_n-z\parallel }^2\\ +\frac{2{\alpha }_n(1-{\alpha }_n)}{1-{\alpha }_n\alpha }(\frac{1}{1-\alpha }〈f(z)-z,J(x_{n+1}-z)〉+\frac{{\alpha }_n}{2(1-\alpha )}{\parallel x_n-z\parallel }^2).\end{array}$$

Then we have

$${\parallel x_{n+1}-z\parallel }^2\le (1-b_n){\parallel x_n-z\parallel }^2+b_n{c}_n,$$

(19)

where $b_n=\frac{2{\alpha }_n(1-{\alpha }_n)}{1-{\alpha }_n\alpha }$ and

$$c_n=\frac{1}{1-\alpha }〈f(z)-z,J(x_{n+1}-z)〉+\frac{{\alpha }_n}{2(1-\alpha )}{\parallel x_n-z\parallel }^2.$$

It follows from conditions (C₂), (C₃) and (18) that

$$\sum _{n=1}^{\mathrm{\infty }}{b}_n=\mathrm{\infty },\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{c}_n\le 0.$$

Therefore, applying Lemma 2.8 to (19), we have that ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ converges strongly to $z=Pf(z)$ and since for $i=1,2,\dots ,m$, $\parallel y_{i,n}-z\parallel \le \parallel x_n-z\parallel$, therefore ${\{y_n\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converges strongly to $z=Pf(z)$. This completes the proof. □

4 Applications

Let ${\{g_{n,j}\}}_{n,j=0}^{\mathrm{\infty }}$ be a family of real numbers. Then $\{g_{n,j}\}$ is said to be the strongly regular summation method [26, 27] if $\{g_{n,j}\}$ satisfies the following conditions:

(S₁) $g_{n,j}\ge 0$,

(S₂) ${\sum }_{j=0}^{\mathrm{\infty }}{g}_{n,j}=1$ for every n,

(S₃) ${lim}_{n\to \mathrm{\infty }}{g}_{n,j}=0$ for every j,

(S₄) ${lim}_{n\to \mathrm{\infty }}{\sum }_{j=0}^{\mathrm{\infty }}|g_{n,j+1}-g_{n,j}|=0$.

Corollary 4.1 Let C be a compact convex subset of a smooth Banach space E, and let f be a contraction of C into itself with constant $\alpha \in (0,1)$. Let T be an asymptotically nonexpansive mapping of C into itself with Lipschitz constants $\{k(j)\}$, and for $i=1,2,\dots ,m$, let ${\{g_{(i,n),j}\}}_{n,j=0}^{\mathrm{\infty }}$ be a finite family of strongly regular summation methods such that

$$\underset{n\to \mathrm{\infty }}{lim}\sum _{j=0}^{\mathrm{\infty }}|g_{(i,n),j+1}-g_{(i,n),j}|=0\mathit{\text{and}}\sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{g}_{(i,n),j}(k(j)-1)<\mathrm{\infty }.$$

Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ are sequences generated by $x_1\in C$ and

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_{1,n})+{\beta }_n{x}_n+{\gamma }_n\sum _{j=0}^{\mathrm{\infty }}{g}_{(1,n),j}T(j)y_{1,n},\\ y_{i,n}={\delta }_{i,n}{y}_{i+1,n}+(I-{\delta }_{i,n})\sum _{j=0}^{\mathrm{\infty }}{g}_{(i,n),j}T(j)y_{i+1,n},i=1,2,\dots ,m,\\ y_{m+1,n}=x_n,\end{array}$$

then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(T).$$

Equivalently, $z=Pf(z)$, where P denotes the unique sunny nonexpansive retraction of C onto $Fix(T)$.

Proof Denote by ${\mathbb{Z}}_{+}=(\mathbb{Z},+)$ the semigroup of nonnegative integers. It is obvious that $\phi =\{T^j:j\in \mathbb{Z}\}$ is an asymptotically nonexpansive semigroup on C. For every $n\in {\mathbb{Z}}_{+}$ and $f\in l^{\mathrm{\infty }}({\mathbb{Z}}_{+})$, define

$${\mu }_{i,n}f=\sum _{j=0}^{\mathrm{\infty }}{g}_{(i,n),j}f(j).$$

Hence ${\{{\mu }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ is a strongly regular sequence of means on $l^{\mathrm{\infty }}({\mathbb{Z}}_{+})$ and ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{i,n+1}-{\mu }_{i,n}\parallel =0$ [28]. Further, for each $y\in C$, we have

$$T_{{\mu }_{i,n}}=\sum _{j=0}^{\mathrm{\infty }}{g}_{(i,n),j}T(j)y.$$

By Theorem 3.1, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$. This completes the proof. □

Example 4.2 Let C be a compact convex subset of a smooth Banach space E such that $0\in C$, and let f be a contraction of C into itself with constant $\alpha \in (0,1)$. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). Let ${\{t_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ with $t_1>t_2>\cdots$ , ${lim}_{n\to \mathrm{\infty }}{t}_n=0$, ${lim}_{n\to \mathrm{\infty }}\frac{t_{n+1}}{t_n}=1$ and ${\sum }_{n=0}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$. Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences generated by $x_1\in C$ and

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n{t}_n\sum _{j=0}^{\mathrm{\infty }}{(1-t_n)}^j\sqrt[n]{{(1+\frac{1}{n^2})}^j}{y}_n,\\ y_n={\delta }_n{x}_n+(I-{\delta }_n)t_n\sum _{j=0}^{\mathrm{\infty }}{(1-t_n)}^j\sqrt[n]{{(1+\frac{1}{n^2})}^j}{x}_n.\end{array}$$

Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to $z\in C$.

Proof We define

$$\begin{array}{r}T:C\to C,\\ Tx=\sqrt[n]{(1+\frac{1}{n^2})}x.\end{array}$$

Obviously, T is an asymptotically nonexpansive mapping with Lipschitz constants $l_n=(1+\frac{1}{n^2})$. Define $g_{n,j}=t_n{(1-t_n)}^j$. Then it follows that $\{q_{n,j}\}$ is a strongly regular summation method [28]. We also have

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{g}_{n,j}(l_n-1)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{t}_n{(1-t_n)}^j(1+\frac{1}{n^2}-1)\\ =\sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{t}_n{(1-t_n)}^j\frac{1}{n^2}\\ \le \sum _{n=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{t}_n\frac{1}{n^2}\\ =\sum _{n=0}^{\mathrm{\infty }}{t}_n\sum _{j=0}^{\mathrm{\infty }}\frac{1}{n^2}<\mathrm{\infty }.\end{array}$$

Therefore, by taking $y_{1,n}=y_n$ and ${\delta }_{1,n}={\delta }_n$ in Corollary 4.1, we complete the proof. □

Let $Q={\{q_{n,m}\}}_{n,m=0}^{\mathrm{\infty }}$ be a matrix satisfying the following conditions:

• (a) ${sup}_{n\ge 0}{\sum }_{m=0}^{\mathrm{\infty }}|q_{n,m}|<\mathrm{\infty }$,

• (b) ${\sum }_{m=0}^{\mathrm{\infty }}{q}_{n,m}=1$ for every $n\in \mathbb{N}$,

• (c) ${lim}_{n\to \mathrm{\infty }}{\sum }_{m=0}^{\mathrm{\infty }}|q_{n,m+1}-q_{n,m}|=0$.

Such a matrix Q is called strongly regular in the sense of Lorentz [29]. If Q is a strongly regular matrix, then for each $m\in \mathbb{N}$, we have ${lim}_{n\to \mathrm{\infty }}{q}_{n,m}=0$, see [30]. Strongly regular matrices were used in the context of nonlinear ergodic theory in [31] and [32].

Corollary 4.3 Let C be a compact convex subset of a smooth Banach space E. Let T be an asymptotically nonexpansive mapping of C into itself, and let $Q={\{q_{n,m}\}}_{n,m=0}^{\mathrm{\infty }}$ be a strongly regular matrix. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ are sequences generated by $x_1\in C$ and

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_{1,n})+{\beta }_n{x}_n+{\gamma }_n\sum _{m=0}^{\mathrm{\infty }}{q}_{(1,n),m}{T}^m{y}_{1,n},\\ y_{i,n}={\delta }_{i,n}{y}_{i+1,n}+(I-{\delta }_{i,n})\sum _{m=0}^{\mathrm{\infty }}{q}_{(i,n),m}{T}^m{y}_{i+1,n},i=1,2,\dots ,m,\\ y_{m+1,n}=x_n,\end{array}$$

then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(T).$$

Equivalently, $z=Pf(z)$, where P denotes the unique sunny nonexpansive retraction of C onto $Fix(T)$.

Proof Let $x\in C$. For each $n\in \mathbb{N}$, define

$${\mu }_{i,n}f=\sum _{m=0}^{\mathrm{\infty }}{q}_{(i,n),m}f(m)$$

for each $f\in L^{\mathrm{\infty }}(\mathbb{N})$. Hence ${\{{\mu }_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ is a strongly regular sequence of means on $l^{\mathrm{\infty }}({\mathbb{Z}}_{+})$ and ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{i,n+1}-{\mu }_{i,n}\parallel =0$ [33]. Further, for each $y\in C$, we have

$$T_{{\mu }_{i,n}}=\sum _{m=0}^{\mathrm{\infty }}{q}_{(i,n),m}{T}^{m}y.$$

By Theorem 3.1, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_{i,n}\}}_{i=1,n=1}^{m,\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$. This completes the proof. □

Example 4.4 Let C be a compact convex subset of a smooth Banach space E such that $0\in C$, and let f be a contraction of C into itself with constant $\alpha \in (0,1)$. Let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). Let ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences generated by $x_1\in C$ and

$$\begin{array}{r}{x}_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{x}_n+{\gamma }_n\sum _{m=0}^n\frac{2m}{n(n-1)}\sqrt[n]{{(1-\frac{1}{n^2})}^m}{y}_n,\\ y_n={\delta }_n{x}_n+(I-{\delta }_n)\sum _{m=0}^n\frac{2m}{n(n-1)}\sqrt[n]{{(1-\frac{1}{n^2})}^m}{x}_n.\end{array}$$

Then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to $z\in C$.

Proof We define

$$\begin{array}{r}T:C\to C,\\ Tx=\sqrt[n]{(1-\frac{1}{n^2})}x.\end{array}$$

Obviously, T is an asymptotically nonexpansive mapping with Lipschitz constants $l_n=(1-\frac{1}{n^2})$. Define

$$q_{n,m}=\{\begin{array}{cc}\frac{2m}{n(n-1)},\hfill & 0\le m\le n,\hfill \\ 0,\hfill & m>n.\hfill \end{array}$$

Then it follows that $\{q_{n,m}\}$ is a strongly regular matrix. Further, we have

$$\sum _{m=0}^{\mathrm{\infty }}{q}_{n,m}=\sum _{m=0}^n\frac{2m}{n(n-1)}=\frac{2}{n(n-1)}\frac{n(n-1)}{2}=1.$$

Therefore

$$\underset{n\ge 0}{sup}\sum _{m=0}^{\mathrm{\infty }}{q}_{n,m}<\mathrm{\infty }.$$

On the other hand,

$$\begin{array}{rl}\sum _{m=0}^{\mathrm{\infty }}|q_{n,m+1}-q_{n,m}|& =\sum _{m=0}^n|q_{n,m+1}-q_{n,m}|\\ =\sum _{m=0}^{n-1}|q_{n,m+1}-q_{n,m}|+|q_{n,n+1}-q_{n,n}|\\ =\sum _{m=0}^{n-1}|\frac{2(m+1)}{n(n-1)}-\frac{2m}{n(n-1)}|+|0-\frac{2n}{n(n-1)}|\\ =\sum _{m=0}^{n-1}|\frac{2(m+1)}{n(n-1)}-\frac{2m}{n(n-1)}|+\frac{2n}{n(n-1)}\\ =\sum _{m=0}^{n-1}\frac{2}{n(n-1)}+\frac{2n}{n(n-1)}\\ =\frac{4n}{n(n-1)}\to 0\end{array}$$

as $n\to \mathrm{\infty }$. By taking $y_{1,n}=y_n$ and ${\delta }_{1,n}={\delta }_n$ in Corollary 4.3, we complete the proof. □

Corollary 4.5 Let C be a compact convex subset of a smooth Banach space E such that $0\in C$, and let T be an asymptotically nonexpansive mapping of C into itself with Lipschitz constants $\{k(j)\}$ satisfying ${\sum }_{j=0}^{\mathrm{\infty }}(k(j)-1)<\mathrm{\infty }$. Let f be a contraction of C into itself with constant $\alpha \in (0,1)$, let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ are sequences generated by $x_1\in C$ and

$$\begin{array}{r}{y}_n={\delta }_n{x}_n+\frac{2(1-{\delta }_n)}{{(n+1)}^2}\sum _{j=0}^{{(n+1)}^2}{T}^j{x}_n,\\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{x}_n+\frac{2{\gamma }_n}{{(n+1)}^2}\sum _{j=0}^{{(n+1)}^2}{T}^j{y}_n,\end{array}$$

then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$, which is the unique solution of the variational inequality

$$〈(f-I)z,J(y-z)〉\le 0,\mathrm{\forall }y\in Fix(T).$$

(20)

Equivalently, $z=Pf(z)$, where P denotes the unique sunny nonexpansive retraction of C onto $Fix(T)$.

Proof Denote by ${\mathbb{Z}}_{+}=(\mathbb{Z},+)$ the semigroup of nonnegative integers. It is obvious that $\phi =\{T^j:j\in \mathbb{Z}\}$ is an asymptotically nonexpansive semigroup on C. For every $n\in {\mathbb{Z}}_{+}$ and $f\in l^{\mathrm{\infty }}({\mathbb{Z}}_{+})$, define

$${\mu }_{n}f=\frac{1}{{(n+1)}^2}\sum _{j=0}^{\mathrm{\infty }}f(j).$$

Hence ${\{{\mu }_n\}}_{n=1}^{\mathrm{\infty }}$ is a strongly regular sequence of means on $l^{\mathrm{\infty }}({\mathbb{Z}}_{+})$ and ${lim}_{n\to \mathrm{\infty }}\parallel {\mu }_{n+1}-{\mu }_n\parallel =0$ [28]. Further, for each $y\in C$, we have

$$T_{{\mu }_n}=\frac{1}{{(n+1)}^2}\sum _{j=0}^{\mathrm{\infty }}{T}^{j}y.$$

By Theorem 3.1, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to $z\in Fix(T)$. This completes the proof. □

Example 4.6 Let C be a compact convex subset of a smooth Banach space E such that $0\in C$. Let f be a contraction of C into itself with constant $\alpha \in (0,1)$, let ${\{{\alpha }_n\}}_{n=1}^{\mathrm{\infty }}$, ${\{{\beta }_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{{\gamma }_n\}}_{n=1}^{\mathrm{\infty }}$ be sequences in $(0,1)$ satisfying conditions (C₁)-(C₄), and let ${\{{\delta }_n\}}_{n=1}^{\mathrm{\infty }}$ be a sequence in $(0,1)$ satisfying condition (${\mathrm{C}}_5^{\prime }$). If ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ are sequences generated by $x_1\in C$ and

$$\begin{array}{r}{y}_n={\delta }_n{x}_n+\frac{2(1-{\delta }_n)}{{(n+1)}^2}\sum _{j=0}^{{(n+1)}^2}\sqrt[n]{{(1+\frac{1}{n{ln}^{2}n})}^j}{x}_n,\\ x_{n+1}={\alpha }_{n}f(y_n)+{\beta }_n{x}_n+\frac{2{\gamma }_n}{{(n+1)}^2}\sum _{j=0}^{{(n+1)}^2}\sqrt[n]{{(1+\frac{1}{n{ln}^{2}n})}^j}{y}_n,\end{array}$$

then ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ and ${\{y_n\}}_{n=1}^{\mathrm{\infty }}$ converge strongly to $z\in C$.

Proof We define

$$\begin{array}{r}T:C\to C,\\ T(x)=\sqrt[n]{(1+\frac{1}{n{ln}^{2}n})}x.\end{array}$$

Obviously, T is an asymptotically nonexpansive mapping with Lipschitz constants $l_n(T)=(1+\frac{1}{n{ln}^{2}n})$. Moreover,

$$\sum _{n=0}^{\mathrm{\infty }}(l_n-1)=\sum _{n=0}^{\mathrm{\infty }}(1+\frac{1}{n{ln}^{2}n}-1)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n{ln}^{2}n}<\mathrm{\infty }.$$

Therefore, applying Corollary 4.5, the result follows. □

Remark 4.7 For deducing some more applications, we refer to [13, 19, 24, 25, 28, 30].

Remark 4.8 Theorem 3.1 improves and extends [[4], Theorem 3.1] and [[2], Theorem 3.1] in the following aspects.

• (1) Theorem 3.1 extends [[4], Theorem 3.1] and [[2], Theorem 3.1] from one sequence of means to a finite family of sequences of means.

• (2) In Theorem 3.1, by taking $T_{{\mu }_{i,n}}=I$ for $i=1,2,\dots ,m-1$, $T_{{\mu }_{m,n}}=T_{{\mu }_n}$ and $y_{m,n}=y_n$, one can see that [[4], Theorem 3.1] is a special case of Theorem 3.1.

• (3) In Theorem 3.1, by taking $T_{{\mu }_{i,n}}=I$ for $i=1,2,\dots ,m$, one can see that [[2], Theorem 3.1] is a special case of Theorem 3.1.

• (4) Theorem 3.1 gives all consequences of [[4], Theorem 3.1] and [[2], Theorem 3.1] without assumption $C_5$ used in [[4], Theorem 3.1] and [[2], Theorem 3.1].