# 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 [514]. 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. (1)

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

2. (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. (1)

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

2. (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 Xus 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. (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. (2)

$$\lim_{n\to\infty}\|x_{n}-p\|$$ exists;

3. (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 Opials 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. □

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## Acknowledgements

The work was supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission (NRU59 Grant No. 59000399). Also, the work was supported by the National Natural Science Foundation of P.R. China (Grant No. 11571095), Program for Innovative Research Team (in Science and Technology) in University of Henan Province (14IRTSTHN023). Yeol Je Cho was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT and future Planning (2014R1A2A2A01002100). Moreover, this work was carried out while Yeol Je Cho was visiting Theoretical and Computational Science Center (TaCS), Science Laboratory Building, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok, Thailand, during 15 January-3 March 2016. He thanks Professor Poom Kumam and the university for their hospitality and support.

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All authors contributed equally and significantly in writing this article. All authors read and approved final manuscript.

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Song, Y., Kumam, P. & Cho, Y.J. Fixed point theorems and iterative approximations for monotone nonexpansive mappings in ordered Banach spaces. Fixed Point Theory Appl 2016, 73 (2016). https://doi.org/10.1186/s13663-016-0563-y