Fixed point theorems and iterative approximations for monotone nonexpansive mappings in ordered Banach spaces

Abstract

In this paper, we prove some existence theorems of fixed points of a monotone nonexpansive mapping T in a Banach space E with the partial order ‘≤’, where a such mapping may be discontinuous. In particular, in finite dimensional spaces, such a mapping T has a fixed point in E if and only if the sequence $\{T^{n}0\}$ is bounded in E. In order to find a fixed point of such a mapping T, we prove the weak convergence of the Mann iteration scheme under the condition $\sum_{n=1}^{\infty}\beta_{n}(1-\beta_{n})=\infty$, which entails $\beta _{n}=\frac{1}{n+1}$ as a special case.

1 Introduction

Let T be a mapping with domain $D(T)$ and range $R(T)$ in a Banach space E. Then T is called nonexpansive if

$$\|Tx-Ty\|\leq\|x-y\| $$

for all $x,y\in D(T)$. The fixed point set of T is denoted by $F(T):= \{x\in K; T x = x\}$.

In 2010, Aoyama et al. [1] introduced a class of λ-hybrid mappings, that is, a mapping T is called a λ-hybrid mapping in Hilbert space H if

$$\|Tx-Ty\|^{2}\leq\|x-y\|^{2}+2(1-\lambda)\langle x-Tx,y-Ty \rangle $$

for all $x,y\in D(T)$. They showed a fixed point theorem and an ergodic theorem for such a mapping. Clearly, a nonexpansive mapping is a 1-hybrid mapping. In 2011, Aoyama and Kohsaka [2] also introduced the concept of α-nonexpansive mapping, that is, a mapping T is called α-nonexpansive if $\alpha<1$ and

$$\|Tx-Ty\|^{2} \leq\alpha\|Tx-y\|+\alpha\|Ty-x\|+(1-2\alpha)\|x-y\| $$

for all $x,y\in D(T)$. Obviously, a nonexpansive mapping is 0-nonexpansive and a λ-hybrid mapping is $\frac{1-\lambda }{2-\lambda}$-nonexpansive if $\lambda<2$ in a Hilbert space H (for more details, see [2]).

The following classical result for nonexpansive mappings was showed to still hold for α-nonexpansive mappings in a uniformly convex Banach space E.

Theorem 1.1

([2])

Let C be a nonempty and closed convex subset of uniformly convex Banach space E and $T:C\to C$ be an α-nonexpansive mapping. Then $F(T)\ne\emptyset$ if and only if $\{T^{n}x\}$ is bounded for some $x \in C$.

Very recently, Bachar and Khamsi [3] introduced the concept of a monotone nonexpansive mapping in a Banach space E endowed with the partial order ‘≤’ and investigated common approximate fixed points of monotone nonexpansive semigroups. A mapping $T:D(T)\to R(T)$ is called monotone nonexpansive if T is monotone ($Tx\leq Ty$ whenever $x\leq y$) and

$$\|Tx-Ty\|\leq\|x-y\| $$

for all $x,y\in D(T)$ with $x\leq y$. Clearly, a monotone nonexpansive mapping may be discontinuous.

In this paper, we show the following existence theorem of fixed points for a monotone nonexpansive mapping T.

Theorem 1.2

Let K be a nonempty and closed convex subset of a uniformly convex Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that there exists $x\in K$ such that $x\leq Tx$ (or $Tx\leq x$) and the sequence $\{T^{n}x\}$ is bounded. Then $F(T)\ne\emptyset$ and $x\leq y^{*}$ (or $y^{*}\leq x$) for some $y^{*}\in F(T)$.

In order to finding a fixed point of a nonexpansive mapping T, Mann [4] introduced the following iteration scheme which is referred to as the Mann iteration: for any $x_{1}\in D(T)$,

$$ x_{n+1}=\beta_{n}x_{n}+(1- \beta_{n})Tx_{n} $$

(1.1)

for each $n\geq1$, where $\beta_{n}\in[0,1]$ is a sequence with some conditions. Subsequently, many mathematical workers have been investigated the convergence of the Mann iteration and its modified version for nonexpansive mappings and pseudo-contractions. For example, see [5–14]. However, there are not many convergence theorems of such an iteration in an ordered Banach space $(E,\leq)$. Recently, Dehaish and Khamsi [15] obtained the weak convergence of the Mann iteration for a monotone nonexpansive mapping provided $\alpha _{n}\in[a,b]\subset(0,1)$. But their results do not entail $\beta _{n}=\frac{1}{n+1}$.

Motivated by the above results, we consider the weak convergence of the Mann iteration scheme for a monotone nonexpansive mapping T under the condition

$$\sum_{n=1}^{\infty}\beta_{n}(1- \beta_{n})=\infty, $$

which contain $\beta_{n}=\frac{1}{n+1}$ as a special case.

2 Preliminaries and basic results

Let P be a closed convex cone of a real Banach space E. A partial order ‘≤’ with respect to P in E is defined as follows:

$$x\leq y \quad (x< y )\quad \mbox{if and only if} \quad y-x\in P\quad ( y-x\in P \mbox{ and }x\ne y) $$

for all $x,y\in E$.

Throughout this paper, let E be a Banach space with the norm ‘$\| \cdot\|$’ and the partial order ‘≤’. Let $F(T)=\{x\in H: Tx=x\}$ denote the set of all fixed points of a mapping T. An order interval $[x,y]$ for all $x,y\in E$ is given by

$$ [x,y]=\{z\in E: x\leq z\leq y\}. $$

(2.1)

Obviously, the order interval $[x,y]$ is closed and convex. In fact, let $z_{1},z_{2}\in[x,y]$. Then $z_{1}-x\in P$, $z_{2}-x\in P$, $y-z_{1}\in P$, and $y-z_{2}\in P$; and so, for any $t\in(0,1)$,

$$\begin{aligned}& tz_{1}+(1-t)z_{2}-x=t(z_{1}-x)+(1-t) (z_{2}-x)\in P, \\& y-\bigl(tz_{1}+(1-t)z_{2}\bigr)=t(y-z_{1})+(1-t) (y-z_{2})\in P. \end{aligned}$$

Thus $tz_{1}+(1-t)z_{2}\in[x,y]$, that is, $[x,y]$ is convex. Let $\{z_{n}\} \subset[x,y]$ with $\lim_{n\to\infty}z_{n}=z$. Then, for each $n\geq 1$, $z_{n}-x\in P$ and $y-z_{n}\in P$, and hence we have

$$\lim_{n\to\infty}z_{n}-x= z-x\in P,\qquad \lim _{n\to\infty}y-z_{n}= y-z\in P, $$

that is, $x\leq z\leq y$ and so $z\in[x,y]$, that is, $[x,y]$ is closed. Then the convexity of the order interval $[x,y]$ implies that

$$ x\leq tx+(1-t)y\leq y $$

(2.2)

for all $x,y\in E$ with $x\leq y$.

Definition 2.1

Let K be a nonempty closed and convex subset of a Banach space E. A mapping $T : K \to E$ is said to be:

• (1) monotone [3] if $Tx\leq Ty$ for all $x, y \in K$ with $x\leq y$;

• (2) monotone nonexpansive [3] if T is monotone and

  $$\|Tx-Ty\|\leq\|x-y\| $$

  for all $x, y \in K$ with $x\leq y$.

A Banach space E is said to be:

• (1) strictly convex if $\|\frac{x+y}{2}\|<1$ for all $x,y\in E$ with $\|x\|=\|y\|=1$ and $x\neq y$;

• (2) uniformly convex if, for all $\varepsilon\in (0,2]$, there exists $\delta>0$ such that $\frac{\|x+y\|}{2}<1-\delta$ for all $x,y\in E$ with $\|x\|=\|y\|=1$ and $\|x-y\|\geq\varepsilon$.

The following inequality was showed by Xu [16] in a uniformly convex Banach space E, which is known as Xu’s inequality.

Lemma 2.2

(Xu [16], Theorem 2)

For any real numbers $q>1$ and $r>0$, a Banach space E is uniformly convex if and only if there exists a continuous strictly increasing convex function $g:[0,+\infty)\to[0,+\infty)$ with $g(0)=0$ such that

$$ \bigl\Vert t x+(1-t)y\bigr\Vert ^{q}\leq t \Vert x \Vert ^{q}+(1-t)\Vert y\Vert ^{q}-\omega(q,t)g\bigl( \Vert x-y\Vert \bigr) $$

(2.3)

for all $x,y\in B_{r}(0)=\{x\in E; \|x\|\leq r\}$ and $t\in[0,1]$, where $\omega(q,t)=t^{q}(1-t)+t(1-t)^{q}$. In particular, take $q=2$ and $t=\frac{1}{2}$,

$$ \biggl\Vert \frac{x+y}{2}\biggr\Vert ^{2}\leq \frac{1}{2}\Vert x\Vert ^{2}+\frac{1}{2}\Vert y\Vert ^{2}-\frac{1}{4}g\bigl(\Vert x-y\Vert \bigr). $$

(2.4)

The following conclusion is well known.

Lemma 2.3

(Takahashi [17], Theorem 1.3.11)

Let K be a nonempty closed convex subset of a reflexive Banach space E. Assume that $\varphi:K\to R$ is a proper convex lower semi-continuous and coercive function. Then the function φ attains its minimum on K, that is, there exists $x\in K$ such that

$$\varphi(x)=\inf_{y\in K} \varphi(y). $$

3 Main results

3.1 Existence of fixed points

In this section, we prove some existence theorems of fixed points of a monotone nonexpansive mapping in a uniformly convex Banach space $(E,\leq)$.

Theorem 3.1

Let K be a nonempty and closed convex subset of a uniformly convex Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that there exists $x\in K$ such that $x\leq Tx$, the sequence $\{T^{n}x\}_{n=1}^{\infty}$ is bounded. Then $F(T)\ne\emptyset$ and $y'\geq x$ for some $y'\in F(T)$.

Proof

Let $x_{1}=x$ and $x_{n+1}=Tx_{n}=T^{n}x$. Then $x_{1}=x\leq Tx=x_{2}$, and so,

$$x_{2}=Tx_{1}=Tx\leq Tx_{2}=T^{2}x=x_{3}. $$

By analogy, we must have

$$x=x_{1}\leq x_{2}\leq x_{3}\leq\cdots\leq x_{n}\leq x_{n+1}\leq\cdots. $$

Let $K_{n}=\{z\in K:x_{n}\leq z\}$ for all $n\geq1$. Clearly, for each $n\geq1$, $K_{n}$ is closed convex and $y\in K_{n}$ and so $K_{n}$ is nonempty too. Let $K^{*}=\bigcap_{n=1}^{\infty}K_{n}$. Then $K^{*}$ is a nonempty closed convex subset of K. Since $\{x_{n}\}$ is bounded, we can define a function $\varphi:K^{*}\to[0,+\infty)$ as follows:

$$\varphi(z)=\limsup_{n\to\infty}\|x_{n}-z\|^{2} $$

for all $z\in K^{*}$. From Lemma 2.3, it follows that there exists $y^{*}\in K_{1}$ such that

$$ \varphi\bigl(y^{*}\bigr)=\inf_{z\in K^{*}} \varphi(z). $$

(3.1)

Now, we show $y^{*}=Ty^{*}$. In fact, by the definition of $K^{*}$, we obtain

$$x_{1}\leq x_{2}\leq x_{3}\leq\cdots\leq x_{n}\leq x_{n+1}\leq\cdots\leq y^{*}. $$

Then we have $x_{n+1}=Tx_{n}\leq Ty^{*}$ by the monotonicity of T and hence, for each $n\geq1$, $x_{n}\leq Ty^{*}$. So we have $Ty^{*}\in K^{*}$. From the convexity of $K^{*}$, it follows that $\frac{y^{*}+Ty^{*}}{2}\in K^{*}$ and so, by (3.1), we have

$$ \varphi\bigl(y^{*}\bigr)\leq\varphi \biggl(\frac{y^{*}+Ty^{*}}{2} \biggr),\qquad \varphi \bigl(y^{*}\bigr)\leq\varphi\bigl(Ty^{*}\bigr). $$

(3.2)

On the other hand, we have

$$\begin{aligned} \varphi\bigl(Ty^{*}\bigr)&=\limsup_{n\to\infty}\bigl\Vert x_{n+1}-Ty^{*}\bigr\Vert ^{2} \\ &=\limsup_{n\to\infty}\bigl\Vert Tx_{n}-Ty^{*}\bigr\Vert ^{2} \\ &\leq\limsup_{n\to\infty}\bigl\Vert x_{n}-y^{*}\bigr\Vert ^{2} \\ &=\varphi\bigl(y^{*}\bigr). \end{aligned}$$

(3.3)

Combining (3.2) and (3.3), we have

$$ \varphi\bigl(Ty^{*}\bigr)=\varphi\bigl(y^{*}\bigr). $$

(3.4)

It follows from Lemma 2.2 ($q=2$ and $t=\frac{1}{2}$) and (3.4) that

$$\begin{aligned} \varphi \biggl(\frac{y^{*}+Ty^{*}}{2} \biggr)&= \limsup_{n\to\infty}\biggl\Vert x_{n}-\frac{y^{*}+Ty^{*}}{2}\biggr\Vert ^{2} \\ &=\limsup_{n\to\infty}\biggl\Vert \frac{x_{n}-y^{*}}{2}+ \frac{x_{n}-Ty^{*}}{2} \biggr\Vert ^{2} \\ &\leq\limsup_{n\to\infty} \biggl(\frac{1}{2}\bigl\Vert x_{n}-y^{*}\bigr\Vert ^{2}+\frac{1}{2}\bigl\Vert x_{n}-Ty^{*}\bigr\Vert ^{2}-\frac{1}{4}g\bigl(\bigl\Vert y^{*}-Ty^{*}\bigr\Vert \bigr) \biggr) \\ &\leq\frac{1}{2}\varphi\bigl(y^{*}\bigr)+\frac{1}{2}\varphi\bigl(Ty^{*} \bigr)-\frac{1}{4}g\bigl(\bigl\Vert y^{*}-Ty^{*}\bigr\Vert \bigr) \\ &= \varphi\bigl(y^{*}\bigr)-\frac{1}{4}g\bigl(\bigl\Vert y^{*}-Ty^{*}\bigr\Vert \bigr). \end{aligned}$$

Noticing (3.2), we have

$$g\bigl(\bigl\Vert y^{*}-Ty^{*}\bigr\Vert \bigr)\leq\varphi\bigl(y^{*}\bigr)- \varphi \biggl(\frac{y^{*}+Ty^{*}}{2} \biggr)\leq0 $$

and so $g(\|y^{*}-Ty^{*}\|)=0$. Thus we have $y^{*}=Ty^{*}$ by the property of g. This yields the desired conclusion. This completes the proof. □

Theorem 3.2

Let K be a nonempty and closed convex subset of a uniformly convex Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that there exists $x\in K$ such that $Tx\leq x$, the sequence $\{T^{n}x\}_{n=1}^{\infty}$ is bounded and all $n\geq1$. Then $F(T)\ne\emptyset$ and $y'\leq x$ for some $y'\in F(T)$.

Proof

Let $x_{1}=x$, $x_{n+1}=Tx_{n}=T^{n}x$, and let $K_{n}=\{z\in K:z\leq x_{n}\}$ for all $n\geq1$. Using the same proof technique of Theorem 3.1, it is easy to obtain

$$x_{n+1}\leq x_{n} $$

for all $n\geq1$ and $K^{*}=\bigcap_{n=1}^{\infty}K_{n}$ is a nonempty closed convex subset of K. The remainder of the proof is the same as ones of Theorem 3.1 and so we omit it. □

Theorem 3.3

Let E be a uniformly convex Banach space with the partial order ‘≤’ with respect to closed convex cone P and $T : P\to P$ be a monotone nonexpansive mapping. Assume that the sequence $\{T^{n}0\}_{n=1}^{\infty}$ is bounded. Then $F(T)\ne\emptyset$.

Proof

It follows from the definition of the partial order ‘≤’ that $0\leq T0$. Then the conclusions directly follow from Theorem 3.1. □

Denote $\mathbb{R}^{m}=\{(r_{1}, r_{2},\ldots, r_{m}): r_{i}\in \mathbb{R}, i=1,2,\ldots,m\}$ and $\mathbb{R}^{m}_{+}=\{(r_{1}, r_{2},\ldots, r_{m}): r_{i}\geq0, i=1,2,\ldots,m\}$, where $\mathbb{R}$ is the set of all real numbers.

Theorem 3.4

Let $T : \mathbb{R}^{m}_{+}\to\mathbb {R}^{m}_{+}$ be a monotone nonexpansive mapping. Assume that the sequence $\{T^{n}0\}_{n=1}^{\infty}$ is bounded. Then $F(T)\ne\emptyset$.

Proof

Let $T^{n}0=(r^{(n)}_{1},r^{(n)}_{2},\ldots,r^{(n)}_{m})\in\mathbb {R}^{m}_{+}$. It follows from the boundedness of the sequence $\{T^{n}0\}$ that there exist a positive real number r such that $r^{(n)}_{i}\leq r$ for all n and $i=1,2,\ldots,m$. Take $y=(r,r,\ldots,r)$. So the conclusions directly follow from Theorem 3.3. □

Theorem 3.5

Let K be a nonempty and closed convex subset of a Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that $F(T)\ne\emptyset$ and there exist $x\in K$ and $p\in F(T)$ such that $p\leq x$ (or $x\leq p$). Then the sequence $\{T^{n}x\}$ is bounded.

Proof

Let $x_{1}=x$ and $x_{n+1}=Tx_{n}=T^{n}x$. Then it follows from the conditions $p=Tp$ and $p\leq x$ (or $x\leq p$) that $p=Tp\leq Tx_{n}=x_{n+1}$ (or $x_{n+1}=Tx_{n}\leq Tp=p$) for all $n\geq1$ and so

$$\begin{aligned}& \|x_{2}-p\|=\|Tx_{1}-Tp\| \leq\|x_{1}-p\|=\|x-p \|, \\& \|x_{3}-p\|=\|Tx_{2}-Tp\|\leq\|x_{2}-p\|\leq\|x-p \|, \\& \ldots, \\& \|x_{n}-p\|=\|Tx_{n-1}-Tp\| \leq\|x_{n-1}-p\|\leq\|x-p \|, \\& \|x_{n+1}-p\|=\|Tx_{n}-Tp\| \leq\|x_{n}-p\|\leq\|x-p \|, \\& \ldots \end{aligned}$$

and so $\|x_{n}-p\|\leq\|x-p\|$ for all $n\geq1$ and hence the sequence $\{T^{n}x\}$ is bounded. This completes the proof. □

Theorem 3.6

Let E be a Banach space with the partial order ‘≤’ with respect to closed convex cone P and $T : P\to P$ be a monotone nonexpansive mapping. Assume that $F(T)\ne \emptyset$. Then the sequence $\{T^{n}0\}$ is bounded. Furthermore, the sequence $\{T^{n}x\}$ is bounded for all $x\in P$.

Proof

It follows from the definition of T that $0\leq p$ for all $p\in F(T)$. Then the conclusion that $\{T^{n}0\}$ is bounded directly follows from Theorem 3.5. For each $x\in P$, it is obvious that $0\leq x$ and hence, by the monotonicity of T, we have

$$T0\leq Tx, T^{2}0\leq T^{2}x, \ldots, T^{n}0\leq T^{n}x, \ldots. $$

It follows from the definition of a monotone nonexpansive mapping that

$$\begin{aligned}& \Vert Tx-T0\Vert \leq \Vert x-0\Vert =\Vert x\Vert , \\& \bigl\Vert T^{2}x-T^{2}0\bigr\Vert \leq \Vert Tx-T0 \Vert \leq \Vert x\Vert , \\& \ldots, \\& \bigl\Vert T^{n}x-T^{n}0\bigr\Vert \leq\bigl\Vert T^{n-1}x-T^{n-1}0\bigr\Vert \leq \Vert x\Vert , \\& \bigl\Vert T^{n+1}x-T^{n+1}0\bigr\Vert \leq\bigl\Vert T^{n}x-T^{n}0\bigr\Vert \leq \Vert x\Vert , \\& \ldots \end{aligned}$$

and so the sequence $\{T^{n}x\}$ is bounded. The desired conclusion follows. This completes the proof. □

Theorem 3.7

Let $T : \mathbb{R}^{m}_{+}\to\mathbb {R}^{m}_{+}$ be a monotone nonexpansive mapping. Then $F(T)\ne\emptyset$ if and only if the sequence $\{T^{n}0\}$ is bounded.

Proof

The conclusions directly follow from Theorems 3.4 and 3.6. □

3.2 The convergence of the Mann iteration

In this section, for a monotone nonexpansive mapping T, we consider the Mann iteration sequence defined by

$$ x_{n+1}=\beta_{n}x_{n}+(1- \beta_{n})Tx_{n} $$

(3.5)

for each $n\geq1$, where $\{\beta_{n}\}$ in $(0,1)$ satisfies the following condition:

$$\sum_{n=1}^{\infty}\beta_{n}(1- \beta_{n})=\infty. $$

Clearly, the above condition contains $\beta_{n}=\frac{1}{n+1}$ as a special case.

The following lemma is showed by Dehaish and Khamsi [15], where the conclusion (3) is obtained from the proof of Lemma 3.1 in [15].

Lemma 3.8

(Dehaish and Khamsi [15], Lemmas 3.1 and 3.2)

Let K be a nonempty and closed convex subset of a Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that the sequence $\{x_{n}\}$ is defined by (3.5) and $x_{1}\leq Tx_{1}$ (or $Tx_{1}\leq x_{1}$). If $F(T)\ne\emptyset$ and $p\leq x_{1}$ (or $x_{1}\leq p$) for some $p\in F(T)$, then

• (1) $\{x_{n}\}$ is bounded and $x_{n}\leq x_{n+1}\leq Tx_{n}$ (or $Tx_{n}\leq x_{n+1}\leq x_{n}$);

• (2) $\lim_{n\to\infty}\|x_{n}-p\|$ exists;

• (3) $x_{n}\leq x$ (or $x\leq x_{n}$) for all $n\geq1$ provided $\{x_{n}\}$ weakly converges to a point $x\in K$.

Theorem 3.9

Let K be a nonempty and closed convex subset of a uniformly convex Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that the sequence $\{x_{n}\}$ is defined by (3.5) and $x_{1}\leq Tx_{1}$ (or $Tx_{1}\leq x_{1}$). If $F(T)\ne\emptyset$ and $p\leq x_{1}$ (or $x_{1}\leq p$) for some $p\in F(T)$, then

$$\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. $$

Proof

It follows from Lemma 3.8 that

$$p\leq x_{1}\leq x_{n} \quad (\mbox{or }\,\,x_{n}\leq x_{1}\leq p) $$

for all $n\geq1$. Then it follows from the nonexpansiveness of T, $p=Tp$, and an application of Lemma 2.2 ($q=2$ and $t=\beta _{n}$) that

$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2}&=\bigl\Vert \beta_{n}(x_{n}-p)+(1-\beta_{n}) (Tx_{n}-Tp)\bigr\Vert ^{2} \\ &\leq\beta_{n}\Vert x_{n}-p\Vert ^{2}+(1- \beta_{n})\Vert Tx_{n}-Tp\Vert ^{2}- \beta_{n}(1-\beta _{n})g\bigl(\Vert x_{n}-Tx_{n} \Vert \bigr) \\ &\leq \Vert x_{n}-p\Vert ^{2}-\beta_{n}(1- \beta_{n})g\bigl(\Vert x_{n}-Tx_{n}\Vert \bigr), \end{aligned}$$

and so

$$\beta_{n}(1-\beta_{n})g\bigl(\Vert x_{n}-Tx_{n} \Vert \bigr)\leq \Vert x_{n}-p\Vert ^{2}-\Vert x_{n+1}-p\Vert ^{2}. $$

Therefore, we have

$$ \sum_{n=1}^{\infty}\beta_{n}(1-\beta_{n})g\bigl(\Vert x_{n}-Tx_{n} \Vert \bigr)\leq \Vert x_{1}-p\Vert ^{2}< +\infty. $$

(3.6)

Now, we claim that there exists a subsequence $\{x_{n_{k}}\}$ such that

$$ \lim_{k\to\infty}g\bigl(\Vert x_{n_{k}}-Tx_{n_{k}} \Vert \bigr)=0. $$

(3.7)

Suppose that the conclusion is not true. Then, for all subsequences $\{ x_{n_{k}}\}$ such that $\lim_{k\to\infty}g(\|x_{n_{k}}-Tx_{n_{k}}\|)>0$, we have

$$\liminf_{n\to\infty}g\bigl(\Vert x_{n}-Tx_{n} \Vert \bigr)>0. $$

Thus there exists a positive number a and a positive integer N such that $g(\|x_{n}-Tx_{n}\|)>a>0$ for all $n>N$. Consequently, we have

$$\beta_{n}(1-\beta_{n})g\bigl(\Vert x_{n}-Tx_{n} \Vert \bigr)\geq a\beta_{n}(1-\beta_{n}) $$

and hence, by the condition $\sum_{n=1}^{\infty}\beta_{n}(1-\beta _{n})=+\infty$, we obtain

$$\sum_{n=1}^{\infty}\beta_{n}(1- \beta_{n})g\bigl(\Vert x_{n}-Tx_{n}\Vert \bigr)=+\infty. $$

This contradicts (3.6). So (3.7) holds and hence, by the property of g, we have

$$\lim_{k\to\infty}\|x_{n_{k}}-Tx_{n_{k}}\|=0. $$

On the other hand, we have

$$\begin{aligned} \Vert x_{n+1}-Tx_{n+1}\Vert &= \bigl\Vert \beta_{n}(x_{n}-Tx_{n})+(Tx_{n}-Tx_{n+1}) \bigr\Vert \\ &\leq\beta_{n}\Vert x_{n}-Tx_{n}\Vert +\Vert x_{n+1}-x_{n}\Vert \\ &= \beta_{n}\Vert x_{n}-x_{n}-Tx_{n} \Vert +(1-\beta_{n})\Vert x_{n}-Tx_{n}\Vert \\ &= \Vert x_{n}-Tx_{n}\Vert . \end{aligned}$$

Therefore, the sequence $\{\|x_{n}-Tx_{n}\|\}$ is monotonically non-increasing and hence it follows that $\lim_{n\to\infty}\| x_{n}-Tx_{n}\|$ exists. This yields the desired conclusion. This completes the proof. □

Recall that a Banach space E is said to satisfy Opial’s condition [12] if a sequence $\{x_{n}\}$ with $\{x_{n}\}$ weakly converges to a point $x\in E$ implies

$$\limsup_{n\rightarrow\infty}\|x_{n}-x\|< \limsup _{n\rightarrow \infty}\|x_{n}-y\| $$

for all $y\in E$ with $y\neq x$.

Next, we show the weak convergence of the sequence $\{x_{n}\}$ defined by (3.5). The proof is similar to the ones of Dehaish and Khamsi [15], but, for more details, we give the proof.

Theorem 3.10

Let K be a nonempty and closed convex subset of a uniformly convex Banach space $(E,\leq)$ and $T : K\to K$ be a monotone nonexpansive mapping. Assume that E satisfies Opial’s condition and the sequence $\{x_{n}\}$ is defined by (3.5) with $x_{1}\leq Tx_{1}$ (or $Tx_{1}\leq x_{1}$). If $F(T)\ne\emptyset$ and $p\leq x_{1}$ (or $x_{1}\leq p$) for some $p\in F(T)$, then $\{x_{n}\}$ weakly converges to a fixed point $x^{*}$ of T.

Proof

It follows from Lemma 3.8 and Theorem 3.9 that $\{x_{n}\}$ is bounded and

$$\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. $$

Then there exists a subsequence $\{x_{n_{k}}\}\subset\{x_{n}\}$ such that $\{x_{n_{k}}\}$ weakly converges to a point $x^{*}\in K$. Following Lemma 3.8, we have $x_{1}\leq x_{n_{k}}\leq x^{*}$ (or $x^{*}\leq x_{n_{k}}\leq x_{1}$) for all $k\geq1$. In particular, we have

$$\lim_{k\to\infty}\|x_{n_{k}}-Tx_{n_{k}}\|=0. $$

Now, we claim $x^{*}=Tx^{*}$. In fact, suppose that this is not true. Then, from the nonexpansiveness of T and Opial’s condition, it follows that

$$\begin{aligned} \limsup_{k\rightarrow\infty}\bigl\Vert x_{n_{k}}-x^{*}\bigr\Vert &< \limsup_{k\rightarrow \infty}\bigl\Vert x_{n_{k}}-Tx^{*}\bigr\Vert \\ &\leq\limsup_{k\rightarrow\infty}\bigl(\Vert x_{n_{k}}-Tx_{n_{k}} \Vert +\bigl\Vert Tx_{n_{k}}-Tx^{*}\bigr\Vert \bigr) \\ &\leq\limsup_{k\rightarrow\infty}\bigl\Vert x_{n_{k}}-x^{*}\bigr\Vert , \end{aligned}$$

which is a contradiction. Thus, by Lemma 3.8(2), it follows that the limit $\lim_{n\to\infty}\|x_{n}-x^{*}\|$ exists.

Now, we show that $\{x_{n}\}$ weakly converges to the point $x^{*}$. Suppose that this is not true. Then there exists a subsequence $\{x_{n_{j}}\}$ that converges weakly to a point $z\in K$ and $z\ne x^{*}$. Similarly, it follows that $z=Tz$ and $\lim_{n\to\infty}\|x_{n}-z\|$ exists. It follows from Opial’s condition that

$$\lim_{n\to\infty} \Vert x_{n}-z\Vert < \lim _{n\to\infty}\bigl\Vert x_{n}-x^{*}\bigr\Vert =\limsup _{i\rightarrow\infty}\bigl\Vert x_{n_{i}}-x^{*}\bigr\Vert < \lim _{n\to\infty} \Vert x_{n}-z\Vert . $$

This is a contradiction and hence $x^{*}=z$. This completes the proof. □