On an open question of Takahashi for nonspreading mappings in Banach spaces

Abstract

In this paper, we first introduce a new class of mappings called asymptotically nonspreading mappings and establish weak and strong convergence theorems of the iterative sequences generated by these mappings in a real Banach space. We modify Halpern’s iterations for finding a fixed point of an asymptotically nonspreading mapping and provide an affirmative answer to an open problem posed by Kurokawa and Takahashi in their final remark of (Kurokawa and Takahashi in Nonlinear Anal. 73:1562-1568, 2010) for nonspreading mappings. Furthermore, we investigate the approximation of common fixed points of asymptotically nonspreading mappings and nonexpansive mappings and derive a strong convergence theorem by a new hybrid method for these mappings. Our results improve and generalize many known results in the current literature.

MSC:47H10, 37C25.

Dedication

Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday

1 Introduction

Throughout this paper, we denote the set of real numbers and the set of positive integers by ℝ and ℕ, respectively. Let E be a Banach space with the norm $\parallel \cdot \parallel$ and the dual space $E^{\ast }$. The modulus δ of convexity of E is denoted by

$$\delta (ϵ)=inf\{1-\frac{\parallel x+y\parallel }{2}:\parallel x\parallel \le 1,\parallel y\parallel \le 1,\parallel x-y\parallel \ge ϵ\}$$

for every ϵ with $0\le ϵ\le 2$. A Banach space E is said to be uniformly convex if $\delta (ϵ)>0$ for every $ϵ>0$. Let $S_E=\{x\in E:\parallel x\parallel =1\}$. The norm of E is said to be Gâteaux differentiable if for each $x,y\in S_E$, the limit

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

(1.1)

exists. In this case, E is called smooth. If the limit (1.1) is attained uniformly in $x,y\in S_E$, then E is called uniformly smooth. The Banach space E is said to be strictly convex if $\parallel \frac{x+y}{2}\parallel <1$ whenever $x,y\in S_E$ and $x\ne y$. It is well known that E is uniformly convex if and only if $E^{\ast }$ is uniformly smooth. It is also known that if E is reflexive, then E is strictly convex if and only if $E^{\ast }$ is smooth; for more details, see [1]. When ${\{x_n\}}_{n\in \mathbb{N}}$ is a sequence in the Banach space E, we denote the strong convergence of ${\{x_n\}}_{n\in \mathbb{N}}$ to $x\in E$ by $x_n\to x$ and the weak convergence by $x_n\rightharpoonup x$. For any sequence ${\{x_n^{\ast }\}}_{n\in \mathbb{N}}$ in $E^{\ast }$, we denote the strong convergence of ${\{x_n^{\ast }\}}_{n\in \mathbb{N}}$ to $x^{\ast }\in E^{\ast }$ by $x_n^{\ast }\to x^{\ast }$, the weak convergence by $x_n^{\ast }\rightharpoonup x^{\ast }$ and the weak-star convergence by $x_n^{\ast }{\rightharpoonup }^{\ast }{x}^{\ast }$. The normalized duality mapping $J:E\to 2^{E^{\ast }}$ is defined by

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

Now, we define a mapping $\rho :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, the modulus of smoothness of E, as follows:

$$\rho (t)=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )-1:x,y\in E,\parallel x\parallel =1,\parallel y\parallel =t\}.$$

It is well known that E is uniformly smooth if and only if ${lim}_{t\to 0}\frac{\rho (t)}{t}=0$. Let $q\in \mathbb{R}$ be such that $1<q\le 2$. Then a Banach space E is said to be q-uniformly smooth if there exists a constant $c_q>0$ such that $\rho (t)\le c_q{t}^q$ for all $t>0$. If a Banach space E admits a sequentially continuous duality mapping J from weak topology to weak-star topology, then J is single-valued and also E is smooth; see [2] for more details. In this case, the normalized duality mapping J is said to be weakly sequentially continuous, i.e., if ${\{x_n\}}_{n\in \mathbb{N}}\subset E$ is a sequence with $x_n\rightharpoonup x\in E$, then $J(x_n){\rightharpoonup }^{\ast }J(x)$ [2]. A Banach space E is said to satisfy the Opial property [3] if for any weakly convergent sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in E with weak limit x,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel$$

for all $y\in E$ with $y\ne x$. It is well known that all Hilbert spaces, all finite dimensional Banach spaces and the Banach spaces $l^p$ ($1\le p<\mathrm{\infty }$) satisfy the Opial property; see, for example, [2, 3]. It is also known that if E admits a weakly sequentially continuous duality mapping, then E is smooth and enjoys the Opial property; see [2] for more details.

Let C be a nonempty subset of a real Banach space E, and let $T:C\to E$ be a mapping. We denote by $F(T)$ the set of fixed points of T, i.e., $F(T)=\{x\in C:Tx=x\}$. A mapping $T:C\to E$ is said to be nonexpansive if $\parallel Tx-Ty\parallel \le \parallel x-y\parallel$ for all $x,y\in C$. A mapping $T:C\to E$ is said to be quasi-nonexpansive if $F(T)\ne \mathrm{\varnothing }$ and $\parallel Tx-y\parallel \le \parallel x-y\parallel$ for all $x\in C$ and $y\in F(T)$. Let C be a nonempty, closed and convex subset of a Hilbert space H and $x\in H$. Then there exists a unique nearest point $z\in C$ such that $\parallel x-z\parallel ={inf}_{y\in C}\parallel x-y\parallel$. We denote such a correspondence by $z=P_{C}x$. The mapping $P_C$ is called metric projection of H onto C.

The concept of nonexpansivity plays an important role in the study of Mann-type iteration for finding fixed points of a mapping $T:C\to C$, where C is a closed and convex subset of a Banach space E. Recall that the Mann-type iteration [4] is given by the following formula

$$x_{n+1}=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,x_1\in C.$$

(1.2)

Here, ${\{{\beta }_n\}}_{n\in \mathbb{N}}$ is a sequence of real numbers in $[0,1]$ satisfying some appropriate conditions. A more general iteration scheme is the Halpern iteration, given by

$$\{\begin{array}{c}u\in E,x_1\in C\text{chosen arbitrarily},\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_{n}T{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n,\hfill \end{array}$$

(1.3)

where the sequences ${\{{\beta }_n\}}_{n\in \mathbb{N}}$ and ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ satisfy some appropriate conditions. In particular, when all ${\alpha }_n=0$, the Halpern iteration (1.3) becomes the standard Mann iteration (1.2). The construction of fixed points of nonexpansive mappings via Halpern’s algorithm [5] has been extensively investigated recently in the current literature (see, for example, [6] and the references therein). Numerous results have been proved on Mann and Halpern’s iterations for nonexpansive mappings in Hilbert and Banach spaces (see, e.g., [6–14]).

Let E be a smooth, strictly convex and reflexive Banach space, and let J be the normalized duality mapping of E. Let C be a nonempty, closed and convex subset of E. The generalized projection ${\mathrm{\Pi }}_C$ from E onto C is denoted by

$${\mathrm{\Pi }}_C(x)=\underset{y\in C}{argmin}\varphi (y,x)$$

for all $x\in E$, where $\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2$ for all $x,y\in E$.

Following Kohsaka and Takahashi [15, 16] (see also [16–21]), a mapping $T:C\to C$ is said to be nonspreading if

$$\varphi (Tx,Ty)+\varphi (Ty,Tx)\le \varphi (Tx,y)+\varphi (Ty,x)$$

for all $x,y\in C$, where $\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2$, $\mathrm{\forall }x,y\in E$. Observe that if E is a real Hilbert space, then J is the identity mapping and $\varphi (x,y)={\parallel x\parallel }^2-2〈x,y〉+{\parallel y\parallel }^2={\parallel x-y\parallel }^2$.

Recently, Kurakawa and Takahashi [17] proved the following fixed point theorem for nonspreading mappings in a Hilbert space.

Theorem 1.1 [17]

Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T:C\to C$ be a nonspreading mapping with $F(T)\ne \mathrm{\varnothing }$. Suppose that ${\{x_n\}}_{n\in \mathbb{N}}$ is a sequence generated by $x_1=x\in C$, $u\in C$ and

$$x_{n+1}={\gamma }_{n}u+(1-{\gamma }_n)\frac{1}{n}\sum _{k=0}^{n-1}{T}^k{x}_n,\mathrm{\forall }n\in \mathbb{N},$$

where $0\le {\gamma }_n\le 1$, ${lim}_{n\to \mathrm{\infty }}{\gamma }_n=0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$. Then ${\{x_n\}}_{n\in \mathbb{N}}$ converges strongly to $P_{F(T)}u$, where $P_{F(T)}$ is the metric projection of H onto $F(T)$.

Kurokawa and Takahashi studied strong convergence theorems for nonspreading mappings and posed the following open problem in their final remark of [17].

Question 1.1 Is there any strong convergence theorem of Halpern type for nonspreading mappings in a Hilbert space H?

By using the iterative schemes proposed by Moudafi [8], Iemoto and Takahashi [18] studied the approximation of common fixed points of nonexpansive mappings and nonspreading mappings in a Hilbert space and proved the following strong convergence theorem.

Theorem 1.2 Let C be a nonempty, closed and convex subset of a Hilbert space H. Let $S:C\to C$ be a nonspreading mapping, and let $T:C\to C$ be a nonexpansive mapping such that $F:=F(S)\cap F(T)\ne \mathrm{\varnothing }$. Define a sequence ${\{x_n\}}_{n\in \mathbb{N}}$ as follows:

$$\{\begin{array}{c}{x}_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n({\beta }_{n}S{x}_n+(1-{\beta }_n)T{x}_n)\hfill \end{array}$$

for all $n\in \mathbb{N}$, where ${\{{\alpha }_n\}}_{n\in \mathbb{N}},{\{{\beta }_n\}}_{n\in \mathbb{N}}\subset [0,1]$. Then the following hold:

• (i) If ${lim}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}(1-{\beta }_n)<\mathrm{\infty }$, then ${\{x_n\}}_{n\in \mathbb{N}}$ converges weakly to $v\in F(S)$;

• (ii) If ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)=\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\beta }_n<\mathrm{\infty }$, then ${\{x_n\}}_{n\in \mathbb{N}}$ converges weakly to $v\in F(T)$;

• (iii) If ${lim}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$ and ${lim}_{n\to \mathrm{\infty }}{\beta }_n(1-{\beta }_n)>0$, then ${\{x_n\}}_{n\in \mathbb{N}}$ converges weakly to $v\in F(S)\cap F(T)$.

Now, we are in a position to introduce the following new class of nonspreading-type mappings in a Banach space.

Definition 1.1 Let E be a real Banach space. A mapping $T:D(T)\subset E\to E$ is said to be asymptotically nonspreading (for short ANS) if

$${\parallel T^{n}x-T^{n}y\parallel }^2\le {\parallel x-y\parallel }^2+2〈x-T^{n}x,J(y-T^{n}y)〉$$

for all $x,y\in D(T)$ and $n\in \mathbb{N}$. The mapping T is called nonspreading if

$${\parallel Tx-Ty\parallel }^2\le {\parallel x-y\parallel }^2+2〈x-Tx,J(y-Ty)〉$$

for all $x,y\in D(T)$, where $D(T)$ is the domain of T and J is the normalized duality mapping of E.

Example 1.1 Let $T:[0,2]\to [0,2]$ be defined by

$$Tx=\{\begin{array}{cc}0\hfill & \text{if }x\in [0,2),\hfill \\ 1\hfill & \text{if }x=2.\hfill \end{array}$$

Then T is an asymptotically nonspreading mapping with $F(T)=\{0\}$. Indeed, for any $x\in [0,2)$ and $y=2$, we have $Tx=0$, $Ty=1$, $T^{n}x=T^{n}y=0$ for all $n\ge 2$. We define the function $f:\mathbb{R}\to \mathbb{R}$ by

$$f(x)={(x-2)}^2+2x,\mathrm{\forall }x\in \mathbb{R}.$$

Then we have

$$f^{\prime }(x)=2(x-2)+2=0⟹x=1,$$

where $f^{\prime }(x)$ is the derivative of f at x. This implies that

$$3=f(1)\le f(x),\mathrm{\forall }x\in \mathbb{R}.$$

Observe now that

$${|Tx-Ty|}^2={|1-0|}^2\le {|x-2|}^2+2x={(x-2)}^2+2x.$$

On the other hand, for any $n\ge 2$, we have

$${|T^{n}x-T^{n}y|}^2=0\le {|x-2|}^2+4x={(x-2)}^2+4x.$$

The other cases can be verified similarly. It is worth mentioning that T is neither nonexpansive nor continuous.

In this paper, we first introduce a new class of asymptotically nonspreading mappings and establish weak and strong convergence theorems of the iterative sequences generated by these mappings in a real Banach space. We modify Mann and Halpern’s iterations for finding a fixed point of an asymptotically nonspreading mapping and provide an affirmative answer to Question 1.1. Furthermore, we study the approximation of common fixed points of asymptotically nonspreading mappings and nonexpansive mappings and derive a strong convergence theorem by a new hybrid method for these mappings. Our results improve and generalize many known results in the current literature; see, for example, [17].

2 Preliminaries

In this section, we collect some lemmas which will be used in the proofs for the main results in the next sections.

Let C and D be nonempty subsets of a real Banach space E with $D\subset C$. A mapping $Q_D:C\to D$ is said to be sunny if

$$Q_D(Q_{D}x+t(x-Q_{D}x))=Q_{D}x$$

for each $x\in E$ and $t\ge 0$. A mapping $Q_D:C\to D$ is said to be a retraction if $Q_{D}x=x$ for each $x\in C$.

Lemma 2.1 [22]

Let C and D be nonempty subsets of a real Banach space E with $D\subset C$, and let $Q_D:C\to D$ be a retraction from C into D. Then $Q_D$ is sunny and nonexpansive if and only if

$$〈z-Q_D(z),J(y-Q_D(z))〉\le 0$$

for all $z\in C$ and $y\in D$, where J is the normalized duality mapping of E.

Lemma 2.2 [22]

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

$${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2〈y,J(x+y)〉$$

for all $x,y\in E$.

Proposition 2.1 [19]

Let C be a nonempty, closed and convex subset of a real Hilbert space H, and let $T:C\to C$ be a nonspreading mapping. If $F(T)\ne \mathrm{\varnothing }$, then it is closed and convex.

Let C be a nonempty, closed and convex subset of a Banach space E, and let ${\{x_n\}}_{n\in \mathbb{N}}$ be a bounded sequence in E. For any $x\in E$, we set

$$r(x,{\{x_n\}}_{n\in \mathbb{N}})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x-x_n\parallel .$$

The asymptotic radius of ${\{x_n\}}_{n\in \mathbb{N}}$ relative to C is defined by

$$r(C,{\{x_n\}}_{n\in \mathbb{N}})=inf\{r(x,{\{x_n\}}_{n\in \mathbb{N}}):x\in C\}.$$

The asymptotic center of ${\{x_n\}}_{n\in \mathbb{N}}$ relative to C is the set

$$A(C,{\{x_n\}}_{n\in \mathbb{N}})=\{x\in C:r(x,{\{x_n\}}_{n\in \mathbb{N}})=r(C,{\{x_n\}}_{n\in \mathbb{N}})\}.$$

It is well known that, in a uniformly convex Banach space E, $A(C,{\{x_n\}}_{n\in \mathbb{N}})$ consists of exactly one point; see [3, 22].

Lemma 2.3 [23]

Let ${\{s_n\}}_{n\in \mathbb{N}}$ be a sequence of nonnegative real numbers satisfying the inequality

$$s_{n+1}\le (1-{\gamma }_n)s_n+{\gamma }_n{\delta }_n,\mathrm{\forall }n\ge 1,$$

where ${\{{\gamma }_n\}}_{n\in \mathbb{N}}$ and ${\{{\delta }_n\}}_{n\in \mathbb{N}}$ satisfy the conditions:

• (i) ${\{{\gamma }_n\}}_{n\in \mathbb{N}}\subset [0,1]$ and ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n=\mathrm{\infty }$ or, equivalently, ${\prod }_{n=1}^{\mathrm{\infty }}(1-{\gamma }_n)=0$;

• (ii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\delta }_n\le 0$, or

(ii′) ${\sum }_{n=1}^{\mathrm{\infty }}{\gamma }_n{\delta }_n<\mathrm{\infty }$.

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

Lemma 2.4 [24]

Let ${\{a_n\}}_{n\in \mathbb{N}}$ be a sequence of real numbers such that there exists a subsequence ${\{n_i\}}_{i\in \mathbb{N}}$ of ${\{n\}}_{n\in \mathbb{N}}$ such that $a_{n_i}<a_{n_i+1}$ for all $i\in \mathbb{N}$. Then there exists a subsequence ${\{m_k\}}_{k\in \mathbb{N}}\subset \mathbb{N}$ such that $m_k\to \mathrm{\infty }$ and the following properties are satisfied by all (sufficiently large) numbers $k\in \mathbb{N}$:

$$a_{m_k}\le a_{m_k+1}\mathit{\text{and}}{a}_k\le a_{m_k+1}.$$

In fact, $m_k=max\{j\le k:a_j<a_{j+1}\}$.

Lemma 2.5 [25, 26]

Let E be a uniformly convex Banach space and $B_r:=\{x\in E:\parallel x\parallel \le r\}$, $r>0$. Then there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $g(0)=0$ such that

$${\parallel \lambda x+\beta y+\gamma z\parallel }^2\le \lambda {\parallel x\parallel }^2+\beta {\parallel y\parallel }^2+\gamma {\parallel z\parallel }^2-\lambda \beta g(\parallel x-y\parallel )$$

for all $x,y,z\in B_r$ and all $\lambda ,\beta ,\gamma \in [0,1]$ with $\lambda +\beta +\gamma =1$.

3 Fixed point theorems

In the following, we present the existence theorems of fixed points of asymptotically nonspreading mappings in a Banach space.

Theorem 3.1 Let C be a nonempty, closed and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an asymptotically nonspreading mapping. Then the following assertions are equivalent.

• (1) The fixed point set $F(T)\ne \mathrm{\varnothing }$.

• (2) There exists a bounded sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in C such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

Proof The implication (1) ⟹ (2) is obvious. For the converse implication, suppose that there exists a bounded sequence ${\{x_n\}}_{n\in \mathbb{N}}$ in C such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. Consequently, there is a bounded subsequence ${\{T{x}_{n_k}\}}_{k\in \mathbb{N}}$ of ${\{T{x}_n\}}_{n\in \mathbb{N}}$ such that ${lim}_{k\to \mathrm{\infty }}\parallel T{x}_{n_k}-x_{n_k}\parallel =0$. Suppose $A(C,{\{x_{n_k}\}}_{k\in \mathbb{N}})=\{z\}$. Let $M_1=sup\{\parallel x_{n_k}\parallel ,\parallel T{x}_{n_k}\parallel ,\parallel z\parallel ,\parallel Tz\parallel :k\in \mathbb{N}\}<\mathrm{\infty }$. Since T is an asymptotically nonspreading mapping, we obtain

$$\begin{array}{rl}{\parallel x_{n_k}-Tz\parallel }^2\le & {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+{\parallel T{x}_{n_k}-Tz\parallel }^2+2\parallel x_{n_k}-T{x}_{n_k}\parallel \parallel T{x}_{n_k}-Tz\parallel \\ \le & {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+{\parallel T{x}_{n_k}-Tz\parallel }^2+2M_1\parallel x_{n_k}-T{x}_{n_k}\parallel \\ \le & {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+{\parallel x_{n_k}-z\parallel }^2\\ +2〈x_{n_k}-T{x}_{n_k},J(z-Tz)〉+2M_1\parallel x_{n_k}-T{x}_{n_k}\parallel \\ \le & {\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+{\parallel x_{n_k}-z\parallel }^2+6M_1\parallel x_{n_k}-T{x}_{n_k}\parallel .\end{array}$$

This implies that

$$\begin{array}{rl}\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-Tz\parallel }^2\le & \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-T{x}_{n_k}\parallel }^2+\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_{n_k}-z\parallel }^2\\ +6M_1\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-T{x}_{n_k}\parallel .\end{array}$$

Thus we have

$$r(Tz,{\{x_{n_k}\}}_{k\in \mathbb{N}})=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-Tz\parallel \le \underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_{n_k}-z\parallel =r(z,{\{x_{n_k}\}}_{k\in \mathbb{N}}).$$

This means that $Tz\in A(C,{\{x_{n_k}\}}_{k\in \mathbb{N}})$. By the uniform convexity of E, we conclude that $Tz=z$, which completes the proof. □

The following result is an immediate consequence of Theorem 3.1.

Proposition 3.1 (Demiclosedness principle)

Let C be a nonempty, closed and convex subset of a real uniformly convex Banach space E. Suppose that $T:C\to E$ is an asymptotically nonspreading mapping with $F(T)\ne \mathrm{\varnothing }$. If ${\{x_n\}}_{n\in \mathbb{N}}$ is a sequence in C that converges weakly to x and if ${\{(I-T)x_n\}}_{n\in \mathbb{N}}$ converges strongly to 0, then $x\in F(T)$.

Theorem 3.2 Let C be a nonempty, closed and convex subset of a uniformly convex Banach space E. Let $T:C\to C$ be an asymptotically nonspreading mapping which is uniformly asymptotically regular, i.e., ${lim}_{n\to \mathrm{\infty }}\parallel T^{n}x-T^{n+1}x\parallel =0$ for all $x\in C$. Then the following assertions are equivalent.

• (1) The fixed point set $F(T)\ne \mathrm{\varnothing }$.

• (2) There exists $x\in C$ such that the sequence ${\{T^{n}x\}}_{n\in \mathbb{N}}$ is bounded.

Proof The implication (1) ⟹ (2) is obvious. For the converse implication, suppose that there exists $x\in C$ such that the sequence ${\{T^{n}x\}}_{n\in \mathbb{N}}$ is bounded. Setting $u_n=T^{n}x$ for all $n\in \mathbb{N}$, the uniformly asymptotical regularity of T assures that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T{u}_n-u_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel T^{n+1}x-T^{n}x\parallel =0.$$

Since ${\{u_n\}}_{n\in \mathbb{N}}$ is bounded, in view of Theorem 3.1, we conclude that $F(T)\ne \mathrm{\varnothing }$, which completes the proof. □

Theorem 3.3 Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $T:C\to C$ be a nonspreading mapping. Then the following assertions are equivalent.

• (1) The fixed point set $F(T)\ne \mathrm{\varnothing }$.

• (2) There exists $x\in C$ such that the sequence ${\{T^{n}x\}}_{n\in \mathbb{N}}$ is bounded.

Proof It is obvious that (1) implies (2). Now, suppose that there exists $x\in C$ such that the sequence ${\{T^{n}x\}}_{n\in \mathbb{N}}$ is bounded. Put $x_{n+1}=T^{n}x=T{x}_n$ and $z_n=\frac{1}{n}{\sum }_{k=1}^n{T}^{k-1}x=\frac{1}{n}{\sum }_{k=1}^n{x}_k$ for all $n\in \mathbb{N}$. Continuing the same process as in the proof of Theorem 3.1 in [17], we conclude that $z_n\rightharpoonup z\in F(T)$ as $n\to \mathrm{\infty }$, which completes the proof. □

4 Weak and strong convergence theorems

In this section, we prove weak and strong convergence theorems for asymptotically nonspreading mappings in a Banach space.

Lemma 4.1 Let C be a nonempty, closed and convex subset of a real Banach space E. Let $T:C\to C$ be an asymptotically nonspreading mapping. Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence in C such that $\parallel x_n-x_{n+1}\parallel \to 0$ and $\parallel x_n-T^n{x}_n\parallel \to 0$ as $n\to \mathrm{\infty }$. Then ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^m{x}_n\parallel =0$ for all $m\in \mathbb{N}$.

Proof We divide the proof into several steps.

Step 1. We claim that the following statements hold:

• (a) ${lim}_{n\to \mathrm{\infty }}\parallel T^{n+1}{x}_n-T^{n+1}{x}_{n+1}\parallel =0$;

• (b) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^{n+1}{x}_n\parallel =0$;

• (c) ${lim}_{n\to \mathrm{\infty }}\parallel T^n{x}_n-T^{n+1}{x}_n\parallel =0$.

Since T is an asymptotically nonspreading mapping, we obtain

$$\begin{array}{rl}{\parallel T^{n+1}{x}_n-T^{n+1}{x}_{n+1}\parallel }^2& \le {\parallel x_n-x_{n+1}\parallel }^2+2〈x_n-T^{n+1}{x}_n,J(x_{n+1}-T^{n+1}{x}_{n+1})〉\\ \le {\parallel x_n-x_{n+1}\parallel }^2+2\parallel x_n-T^{n+1}{x}_n\parallel \parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel .\end{array}$$

Due to the boundedness of ${\{x_n\}}_{n\in \mathbb{N}}$, we deduce that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^{n+1}{x}_n-T^{n+1}{x}_{n+1}\parallel =0.$$

Observe now that

$$\parallel x_n-T^{n+1}{x}_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-T^{n+1}{x}_{n+1}\parallel +\parallel T^{n+1}{x}_{n+1}-T^{n+1}{x}_n\parallel .$$

Thus we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T^{n+1}{x}_n\parallel =0.$$

This implies that

$$\parallel T^n{x}_n-T^{n+1}{x}_n\parallel \le \parallel T^n{x}_n-x_n\parallel +\parallel x_n-T^{n+1}{x}_n\parallel \to 0$$

as $n\to \mathrm{\infty }$.

Step 2. We prove the following assertions:

• (d) ${lim}_{n\to \mathrm{\infty }}\parallel T^{n+1}{x}_n-T{x}_n\parallel =0$;

• (e) ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$.

Since T is an asymptotically nonspreading mapping, we get

$$\begin{array}{rl}{\parallel T^{n+1}{x}_n-T{x}_n\parallel }^2& ={\parallel T\left(T^n{x}_n\right)-T{x}_n\parallel }^2\\ \le {\parallel T^n{x}_n-x_n\parallel }^2+2〈T^n{x}_n-T^{n+1}{x}_n,J(x_n-T{x}_n)〉\\ \le {\parallel T^n{x}_n-x_n\parallel }^2+2\parallel T^n{x}_n-T^{n+1}{x}_n\parallel \parallel x_n-T{x}_n\parallel .\end{array}$$

Due to the boundedness of ${\{x_n\}}_{n\in \mathbb{N}}$ and in view of Step 1(c), we deduce that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^{n+1}{x}_n-T{x}_n\parallel =0.$$

Observe now that

$$\parallel x_n-T{x}_n\parallel \le \parallel x_n-T^{n+1}{x}_n\parallel +\parallel T^{n+1}{x}_n-T{x}_n\parallel \to 0$$

as $n\to \mathrm{\infty }$.

Step 3. We show that ${lim}_{n\to \mathrm{\infty }}\parallel T^{m-1}{x}_n-T^m{x}_n\parallel =0$ for all $m\in \mathbb{N}$.

To this end, we apply the principle of mathematical induction. In view of Step 2(e), for $m=1$, we deduce that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. Now, suppose that for $m\in \mathbb{N}$,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^{m-1}{x}_n-T^m{x}_n\parallel =0.$$

We prove that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T^m{x}_n-T^{m+1}{x}_n\parallel =0.$$

Since T is an asymptotically nonspreading mapping, we have

$$\begin{array}{rl}{\parallel T^m{x}_n-T^{m+1}{x}_n\parallel }^2& ={\parallel T\left(T^{m-1}{x}_n\right)-T\left(T^m{x}_n\right)\parallel }^2\\ \le {\parallel T^{m-1}{x}_n-T^m{x}_n\parallel }^2+2〈T^{m-1}{x}_n-T^m{x}_n,J(T^m{x}_n-T^{m+1}{x}_n)〉\\ \le {\parallel T^{m-1}{x}_n-T^m{x}_n\parallel }^2+2\parallel T^{m-1}{x}_n-T^m{x}_n\parallel \parallel T^m{x}_n-T^{m+1}{x}_n\parallel .\end{array}$$

Thus we have ${lim}_{n\to \mathrm{\infty }}\parallel T^{m-1}{x}_n-T^m{x}_n\parallel =0$ for all $m\in \mathbb{N}$.

By the triangle inequality, we see that for any $m\in \mathbb{N}$,

$$\parallel x_n-T^m{x}_n\parallel \le \parallel x_n-T{x}_n\parallel +\parallel T{x}_n-T^2{x}_n\parallel +\cdots +\parallel T^{m-1}{x}_n-T^m{x}_n\parallel .$$

In view of Steps 2 and 3, we conclude that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^m{x}_n\parallel =0$ for all $m\in \mathbb{N}$. This completes the proof. □

Theorem 4.1 Let C be a nonempty, closed and convex subset of a uniformly convex Banach space E with Opial property, and let $T:C\to C$ be an asymptotically nonspreading mapping such that $F(T)\ne \mathrm{\varnothing }$. Assume that ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ is a sequence in $(0,1)$ such that $0<\delta \le {\alpha }_n\le 1-\delta <1$. Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence in C generated by the modified Mann iteration process

$$x_{n+1}=(1-{\alpha }_n)x_n+{\alpha }_n{T}^n{x}_n,\mathrm{\forall }n\in \mathbb{N}.$$

(4.1)

Then the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ generated by algorithm (4.1) converges weakly to an element of $F(T)$.

Proof Take any $p\in F(T)$ arbitrarily chosen. In view of Lemma 2.5, there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $g(0)=0$ such that

$$\begin{array}{rl}{\parallel x_{n+1}-p\parallel }^2& ={\parallel (1-{\alpha }_n)(x_n-p)+{\alpha }_n(T^n{x}_n-p)\parallel }^2\\ \le (1-{\alpha }_n){\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel T^n{x}_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ \le (1-{\alpha }_n){\parallel x_n-p\parallel }^2+{\alpha }_n{\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ ={\parallel x_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ \le {\parallel x_n-p\parallel }^2-{\delta }^{2}g\left(\parallel x_n-T^n{x}_n\parallel \right).\end{array}$$

(4.2)

Since $\delta >0$, we have from (4.2) that

$$\parallel x_{n+1}-p\parallel \le \parallel x_n-p\parallel ,\mathrm{\forall }n\in \mathbb{N}.$$

This implies that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-p\parallel$ exists and hence ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded. Setting

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-p\parallel =d,$$

it follows from (4.2) that

$${\delta }^{2}g\left(\parallel x_n-T^n{x}_n\parallel \right)\le {\parallel x_n-p\parallel }^2-{\parallel x_{n+1}-p\parallel }^2,$$

which yields that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T^n{x}_n\parallel =0$. In view of (4.1), we see that

$$\parallel x_{n+1}-x_n\parallel ={\alpha }_n\parallel x_n-T^n{x}_n\parallel \le (1-\delta )\parallel x_n-T^n{x}_n\parallel ,\mathrm{\forall }n\in \mathbb{N}.$$

Thus we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_{n+1}\parallel =0$. Employing Proposition 3.1 and Lemma 4.1, we conclude that there exists $x\in F(T)$ such that $x_n\rightharpoonup x$ as $n\to \mathrm{\infty }$, which completes the proof. □

Theorem 4.2 Let E be a real uniformly convex Banach space which admits the weakly sequentially continuous duality mapping J, and let C be a nonempty, closed and convex subset of E. Let $T:C\to C$ be an asymptotically nonspreading mapping such that $F:=F(T)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ and ${\{{\beta }_n\}}_{n\in \mathbb{N}}$ be two sequences in $[0,1]$ satisfying the following control conditions:

• (a) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

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

Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence generated by

$$\{\begin{array}{c}u\in C,x_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n.\hfill \end{array}$$

(4.3)

Then the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ defined in (4.3) converges strongly to $Q_{F}u$, where $Q_F$ is the sunny nonexpansive retraction from E onto F.

Proof We divide the proof into several steps.

Since T is a quasi-nonexpansive mapping, so we have F is closed and convex. Set

$$z=Q_{F}u.$$

Step 1. We prove that the sequences ${\{x_n\}}_{n\in \mathbb{N}}$, ${\{y_n\}}_{n\in \mathbb{N}}$ and ${\{T^n{x}_n\}}_{n\in \mathbb{N}}$ are bounded.

We first show that ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded.

Let $p\in F$ be fixed. In view of Lemma 2.5, there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $g(0)=0$ such that

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2& ={\parallel (1-{\beta }_n)x_n+{\beta }_n{T}^n{x}_n-p\parallel }^2\\ \le (1-{\beta }_n){\parallel x_n-p\parallel }^2+{\beta }_n{\parallel T^n{x}_n-p\parallel }^2-{\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ \le (1-{\beta }_n){\parallel x_n-p\parallel }^2+{\beta }_n{\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ ={\parallel x_n-p\parallel }^2-{\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\\ \le {\parallel x_n-p\parallel }^2.\end{array}$$

(4.4)

This implies that

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-p\parallel \\ \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel y_n-p\parallel \\ \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ \le max\{\parallel u-p\parallel ,\parallel x_n-p\parallel \}.\end{array}$$

By induction, we obtain

$$\parallel x_{n+1}-p\parallel \le max\{\parallel u-p\parallel ,\parallel x_1-p\parallel \}$$

for all $n\in \mathbb{N}$. This implies that the sequence ${\{\parallel x_n-p\parallel \}}_{n\in \mathbb{N}}$ is bounded and hence the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded. This, together with (4.3), implies that the sequences ${\{y_n\}}_{n\in \mathbb{N}}$ and ${\{T^n{x}_n\}}_{n\in \mathbb{N}}$ are bounded too.

Step 2. We prove that for any $n\in \mathbb{N}$,

$${\parallel x_{n+1}-z\parallel }^2\le (1-{\alpha }_n){\parallel x_n-z\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉.$$

(4.5)

Let us show (4.5). For each $n\in \mathbb{N}$, in view of (4.4), we obtain

$${\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2-{\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right).$$

This implies that

$$\begin{array}{rl}{\parallel x_{n+1}-z\parallel }^2& ={\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-z\parallel }^2\\ \le {\alpha }_n{\parallel u-z\parallel }^2+(1-{\alpha }_n){\parallel y_n-z\parallel }^2\\ \le {\alpha }_n{\parallel u-z\parallel }^2+(1-{\alpha }_n)[{\parallel x_n-z\parallel }^2-{\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)].\end{array}$$

(4.6)

Let $M_2:=sup\{|{\parallel u-z\parallel }^2-{\parallel x_n-z\parallel }^2|+{\beta }_n(1-{\beta }_n)g(\parallel x_n-T^n{x}_n\parallel ):n\in \mathbb{N}\}$. It follows from (4.6) that

$${\beta }_n(1-{\beta }_n)g\left(\parallel x_n-T^n{x}_n\parallel \right)\le {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+{\alpha }_n{M}_2.$$

(4.7)

In view of Lemma 2.2 and (4.4), we obtain

$$\begin{array}{rl}{\parallel x_{n+1}-z\parallel }^2& ={\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-z\parallel }^2\\ ={\parallel {\alpha }_n(u-z)+(1-{\alpha }_n)(y_n-z)\parallel }^2\\ \le {\parallel (1-{\alpha }_n)(y_n-z)\parallel }^2+2〈{\alpha }_n(u-z),J(x_{n+1}-z)〉\\ ={(1-{\alpha }_n)}^2{\parallel y_n-z\parallel }^2+2〈{\alpha }_n(u-z),J(x_{n+1}-z)〉\\ \le (1-{\alpha }_n){\parallel y_n-z\parallel }^2+2〈{\alpha }_n(u-z),J(x_{n+1}-z)〉\\ =(1-{\alpha }_n){\parallel x_n-z\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉.\end{array}$$

Step 3. We prove that $x_n\to z$ as $n\to \mathrm{\infty }$.

We discuss the following two possible cases.

Case 1. If ${\{\parallel x_n-z\parallel \}}_{n\in \mathbb{N}}$ is eventually decreasing, then there exists $n_0\in \mathbb{N}$ such that the sequence ${\{\parallel x_n-z\parallel \}}_{n=n_0}^{\mathrm{\infty }}$ is decreasing. Thus, the sequence ${\{\parallel x_n-z\parallel \}}_{n\in \mathbb{N}}$ is convergent and hence ${\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2\to 0$ as $n\to \mathrm{\infty }$. This, together with condition (c) and (4.7), implies that

$$\underset{n\to \mathrm{\infty }}{lim}g\left(\parallel x_n-T^n{x}_n\parallel \right)=0.$$

From the properties of g, it follows that

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

(4.8)

On the other hand, we have

$$x_n-y_n={\beta }_n(x_n-T^n{x}_n)\text{and}{x}_{n+1}-y_n={\alpha }_n(u-y_n).$$

This implies that

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

(4.9)

By the triangle inequality, we conclude that

$$\parallel x_{n+1}-x_n\parallel \le \parallel x_{n+1}-y_n\parallel +\parallel y_n-x_n\parallel .$$

It follows from (4.9) that

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

(4.10)

Exploiting Lemma 4.1, (4.8) and (4.10), we obtain

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

(4.11)

Since ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded, there exists a subsequence ${\{x_{n_i}\}}_{i\in \mathbb{N}}$ of ${\{x_n\}}_{n\in \mathbb{N}}$ such that $x_{n_i+1}\rightharpoonup y\in C$ as $i\to \mathrm{\infty }$. In view of Proposition 3.1 and (4.11), we conclude that $y\in F(T)$. This, together with Lemma 2.1, implies that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-z,J(x_{n+1}-z)〉& =\underset{i\to \mathrm{\infty }}{lim}〈u-z,J(x_{n_i+1}-z)〉\\ =〈u-z,J(y-z)〉\\ \le 0.\end{array}$$

(4.12)

Thus we have the desired result by Lemma 2.3.

Case 2. If ${\{\parallel x_n-z\parallel \}}_{n\in \mathbb{N}}$ is not eventually decreasing, then there exists a subsequence ${\{n_i\}}_{i\in \mathbb{N}}$ of ${\{n\}}_{n\in \mathbb{N}}$ such that

$$\parallel x_{n_i}-z\parallel <\parallel x_{n_i+1}-z\parallel$$

for all $i\in \mathbb{N}$. In view of Lemma 2.4, there exists a nondecreasing sequence ${\{m_k\}}_{k\in \mathbb{N}}\subset \mathbb{N}$ such that $m_k\to \mathrm{\infty }$

$$\parallel x_{m_k}-z\parallel <\parallel x_{m_k+1}-z\parallel \text{and}\parallel x_k-z\parallel \le \parallel x_{m_k+1}-z\parallel$$

for all $k\in \mathbb{N}$. This, together with (4.7), implies that

$${\beta }_{m_k}(1-{\beta }_{m_k})g\left(\parallel x_{m_k}-T^{m_k}{x}_{m_k}\parallel \right)\le {\parallel x_{m_k}-z\parallel }^2-{\parallel x_{m_k+1}-z\parallel }^2+{\alpha }_{m_k}{M}_2\le {\alpha }_{m_k}{M}_2$$

for all $k\in \mathbb{N}$. Then, by conditions (a) and (c) and the properties of g, we get

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_{m_k}-T^{m_k}{x}_{m_k}\parallel =0.$$

By the same argument, as in Case 1, we arrive at

$$\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-z,J(x_{m_k}-z)〉=\underset{k\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-z,J(x_{m_k+1}-z)〉\le 0.$$

Next, it follows from (4.5) that

$${\parallel x_{m_k+1}-z\parallel }^2\le (1-{\alpha }_{m_k}){\parallel x_{m_k}-z\parallel }^2+{\alpha }_{m_k}〈u-z,J(x_{m_k+1}-z)〉.$$

(4.13)

Since $\parallel x_{m_k}-z\parallel \le \parallel x_{m_k+1}-z\parallel$, we conclude that

$$\begin{array}{rl}{\alpha }_{m_k}{\parallel x_{m_k}-z\parallel }^2& \le {\parallel x_{m_k}-z\parallel }^2-{\parallel x_{m_k+1}-z\parallel }^2+{\alpha }_{m_k}〈u-z,J(x_{m_k+1}-z)〉\\ \le 2{\alpha }_{m_k}〈u-z,J(x_{m_k+1}-z)〉.\end{array}$$

(4.14)

In particular, since ${\alpha }_{m_k}>0$, we obtain

$${\parallel x_{m_k}-z\parallel }^2\le 〈u-z,J(x_{m_k+1}-z)〉$$

and hence

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_{m_k}-z\parallel =0.$$

This, together with (4.13), implies that

$$\underset{k\to \mathrm{\infty }}{lim}\parallel x_{m_k+1}-z\parallel =0.$$

On the other hand, we have $\parallel x_k-z\parallel \le \parallel x_{m_k+1}-z\parallel$ for all $k\in \mathbb{N}$, which implies that $x_k\to z$ as $k\to \mathrm{\infty }$. Thus, we have $x_n\to z$ as $n\to \mathrm{\infty }$, which completes the proof. □

Let C be a nonempty, closed and convex subset of a Hilbert space H, and let $T:C\to C$ be a nonspreading mapping such that $F(T)\ne \mathrm{\varnothing }$. For any real number $\beta \in (0,1)$, we define a mapping $T_{\beta }:C\to C$ by

$$T_{\beta }x=(1-\beta )Ix+\beta Tx(x\in C),$$

(4.15)

where I is the identity mapping on H. It is easy to verify that $T_{\beta }$ is a nonspreading mapping and $F(T_{\beta })=F(T)$. Therefore, in view of Proposition 2.1, $F(T_{\beta })$ is closed and convex. The following strong convergence result provides an affirmative answer to open Question 1.1 in the case where the mapping T is nonspreading. It is worth mentioning that our method of proof is different from that in [19] and can be applied in uniformly convex Banach spaces. In fact, an answer will be given for more general spaces than Hilbert spaces.

Corollary 4.1 Let E be a real uniformly convex Banach space which admits the weakly sequentially continuous duality mapping J, and let C be a nonempty, closed and convex subset of E. Let $T:C\to C$ be a nonspreading mapping such that $F:=F(T)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ be a sequence in $[0,1]$ satisfying the following control conditions:

• (a) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

For any real number $\beta \in (0,1)$, let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence generated by

$$\{\begin{array}{c}u\in C,x_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n=(1-\beta )x_n+\beta T{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n.\hfill \end{array}$$

Then the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ converges strongly to $Q_{F}u$, where $Q_F$ is the sunny nonexpansive retraction from E onto F.

Corollary 4.2 Let C be a nonempty, closed and convex subset of a Hilbert space H, and let $T:C\to C$ be a nonspreading mapping such that $F:=F(T)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ be a sequence in $[0,1]$ satisfying the following control conditions:

• (a) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$.

For any real number $\beta \in (0,1)$, let $T_{\beta }:C\to C$ be defined by (4.15). Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence generated by

$$\{\begin{array}{c}u\in C,x_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)T_{\beta }{x}_n.\hfill \end{array}$$

Then the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ converges strongly to $P_{F}u$, where $P_F$ is the metric projection from H onto F.

Theorem 4.3 Let E be a uniformly convex Banach space which admits the weakly sequentially continuous duality mapping J, and let C be a nonempty, closed and convex subset of E. Let $T_1:C\to C$ be an asymptotically nonspreading mapping, and let $T_2:C\to C$ be a nonexpansive mapping such that $F:=F(T_1)\cap F(T_2)\ne \mathrm{\varnothing }$. Let ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$, ${\{{\beta }_{n,1}\}}_{n\in \mathbb{N}}$, ${\{{\beta }_{n,2}\}}_{n\in \mathbb{N}}$, ${\{{\beta }_{n,3}\}}_{n\in \mathbb{N}}$ be sequences in $[0,1]$ satisfying the following control conditions:

• (a) ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$;

• (b) ${\sum }_{n=1}^{\mathrm{\infty }}{\alpha }_n=\mathrm{\infty }$;

• (c) ${\beta }_{n,1}+{\beta }_{n,2}+{\beta }_{n,3}=1$, $\mathrm{\forall }n\in \mathbb{N}$;

• (d) ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_{n,j}{\beta }_{n,3}>0$, $j=1,2$.

Let ${\{x_n\}}_{n\in \mathbb{N}}$ be a sequence generated by

$$\{\begin{array}{c}u\in C,x_1\in C\mathit{\text{chosen arbitrarily}},\hfill \\ y_n={\beta }_{n,1}{T}_1{x}_n+{\beta }_{n,2}{T}_2{x}_n+{\beta }_{n,3}{x}_n,\hfill \\ x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)y_n.\hfill \end{array}$$

(4.16)

Then the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ defined in (4.16) converges strongly to $Q_{F}u$, where $Q_F$ is a sunny nonexpansive retraction from E onto F.

Proof We divide the proof into several steps.

Since T is a quasi-nonexpansive mapping, so we have F is closed and convex. Set

$$z=Q_{F}u.$$

Step 1. We prove that the sequences ${\{x_n\}}_{n\in \mathbb{N}}$, ${\{y_n\}}_{n\in \mathbb{N}}$, ${\{T_1{x}_n\}}_{n\in \mathbb{N}}$ and ${\{T_2{x}_n\}}_{n\in \mathbb{N}}$ are bounded.

We first show that ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded.

Let $p\in F$ be fixed. In view of Lemma 2.5, there exists a continuous, strictly increasing and convex function $g:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ with $g(0)=0$ such that

$$\begin{array}{rl}{\parallel y_n-p\parallel }^2=& {\parallel {\beta }_{n,1}{T}_1{x}_n+{\beta }_{n,2}{T}_2{x}_n+{\beta }_{n,3}{x}_n-p\parallel }^2\\ \le & {\beta }_{n,1}{\parallel T_1{x}_n-p\parallel }^2+{\beta }_{n,2}{\parallel T_2{x}_n-p\parallel }^2+{\beta }_{n,3}{\parallel x_n-p\parallel }^2\\ -{\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel )\\ \le & {\beta }_{n,1}{\parallel x_n-p\parallel }^2+{\beta }_{n,2}{\parallel x_n-p\parallel }^2+{\beta }_{n,3}{\parallel x_n-p\parallel }^2\\ -{\beta }_{n,j}{\beta }_{n,3}{\parallel x_n-T_j{x}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2-{\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel )\\ \le & {\parallel x_n-p\parallel }^2,j=1,2.\end{array}$$

(4.17)

This implies that

$$\begin{array}{rl}\parallel x_{n+1}-p\parallel & =\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-p\parallel \\ \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel y_n-p\parallel \\ \le {\alpha }_n\parallel u-p\parallel +(1-{\alpha }_n)\parallel x_n-p\parallel \\ \le max\{\parallel u-p\parallel ,\parallel x_n-p\parallel \}.\end{array}$$

By induction, we obtain

$$\parallel x_{n+1}-p\parallel \le max\{\parallel u-p\parallel ,\parallel x_1-p\parallel \}$$

for all $n\in \mathbb{N}$. This implies that the sequence ${\{\parallel x_n-p\parallel \}}_{n\in \mathbb{N}}$ is bounded and hence the sequence ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded. This, together with (4.16), implies that the sequences ${\{y_n\}}_{n\in \mathbb{N}}$, ${\{T_1{x}_n\}}_{n\in \mathbb{N}}$ and ${\{T_2{x}_n\}}_{n\in \mathbb{N}}$ are bounded too.

Step 2. We prove that for any $n\in \mathbb{N}$,

$${\parallel x_{n+1}-z\parallel }^2\le (1-{\alpha }_n){\parallel x_n-z\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉.$$

(4.18)

Let us show (4.18). For each $n\in \mathbb{N}$ and $j=1,2$, in view of (4.17), we obtain

$${\parallel y_n-z\parallel }^2\le {\parallel x_n-z\parallel }^2-{\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel ).$$

This implies that

$$\begin{array}{rl}{\parallel x_{n+1}-z\parallel }^2& ={\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-z\parallel }^2\\ \le {\alpha }_n{\parallel u-z\parallel }^2+(1-{\alpha }_n){\parallel y_n-z\parallel }^2\\ \le {\alpha }_n{\parallel u-z\parallel }^2+(1-{\alpha }_n)[{\parallel x_n-z\parallel }^2-{\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel )].\end{array}$$

(4.19)

Let $M_3:=sup\{|{\parallel u-z\parallel }^2-{\parallel x_n-z\parallel }^2|+{\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel ):n\in \mathbb{N},j=1,2\}$. It follows from (4.19) that

$${\beta }_{n,j}{\beta }_{n,3}g(\parallel x_n-T_j{x}_n\parallel )\le {\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2+{\alpha }_n{M}_3,j=1,2.$$

(4.20)

In view of Lemma 2.2 and (4.17), we obtain

$$\begin{array}{rl}{\parallel x_{n+1}-z\parallel }^2& ={\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-z\parallel }^2\\ \le {\parallel {\alpha }_{n}u+(1-{\alpha }_n)y_n-z-{\alpha }_n(u-z)\parallel }^2+2〈{\alpha }_n(u-z),J(x_{n+1}-z)〉\\ ={\parallel (1-{\alpha }_n)(y_n-z)\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉\\ =(1-{\alpha }_n){\parallel y_n-z\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉\\ \le (1-{\alpha }_n){\parallel x_n-z\parallel }^2+2{\alpha }_n〈u-z,J(x_{n+1}-z)〉.\end{array}$$

Step 3. We prove that $x_n\to z$ as $n\to \mathrm{\infty }$.

We discuss the following two possible cases.

Case 1. If ${\{\parallel x_n-z\parallel \}}_{n\in \mathbb{N}}$ is eventually decreasing, then there exists $n_0\in \mathbb{N}$ such that the sequence ${\{\parallel x_n-z\parallel \}}_{n=n_0}^{\mathrm{\infty }}$ is decreasing. Thus, the sequence ${\{\parallel x_n-z\parallel \}}_{n\in \mathbb{N}}$ is convergent and hence ${\parallel x_n-z\parallel }^2-{\parallel x_{n+1}-z\parallel }^2\to 0$ as $n\to \mathrm{\infty }$. This, together with condition (d) and (4.20), implies that

$$\underset{n\to \mathrm{\infty }}{lim}g(\parallel x_n-T_j{x}_n\parallel )=0,j=1,2.$$

From the properties of g, it follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-T_j{x}_n\parallel =0,j=1,2.$$

(4.21)

On the other hand, we have

$$y_n-x_n={\beta }_{n,1}(T_1{x}_n-x_n)+{\beta }_{n,2}(T_2{x}_n-x_n)\text{and}{x}_{n+1}-y_n={\alpha }_n(u-y_n).$$

This implies that

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

(4.22)

By the triangle inequality, we conclude that

$$\parallel x_{n+1}-x_n\parallel \le \parallel x_{n+1}-y_n\parallel +\parallel y_n-x_n\parallel .$$

It follows from (4.22) that

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

(4.23)

Since ${\{x_n\}}_{n\in \mathbb{N}}$ is bounded, there exists a subsequence ${\{x_{n_i}\}}_{i\in \mathbb{N}}$ of ${\{x_n\}}_{n\in \mathbb{N}}$ such that $x_{n_i+1}\rightharpoonup y\in C$ as $i\to \mathrm{\infty }$. In view of Proposition 3.1 and (4.21), we conclude that $y\in F$. This, together with Lemma 2.1, implies that

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}〈u-z,J(x_{n+1}-z)〉& =\underset{i\to \mathrm{\infty }}{lim}〈u-z,J(x_{n_i+1}-z)〉\\ =〈u-z,J(y-z)〉\\ \le 0.\end{array}$$

(4.24)

Thus we have the desired result by Lemma 2.3.

Case 2. By the same method, as in the proof of Theorem 4.2, we can prove that $x_n\to z$ as $n\to \mathrm{\infty }$. This completes the proof. □

Remark 4.1 (1) Note that [[18], Theorem 4.1] is a weak convergence result and that our Theorem 4.3 is a strong convergence result. However, it is worth pointing out that the method of proving Theorem 4.3 is very different from the method of proving Theorem 4.1 of [18].

(2) In most cases, strong convergence is more desirable than weak convergence.