Convergence theorems for generalized nonexpansive mappings in uniformly convex Banach spaces

Abstract

In this paper, we prove strong and weak convergence theorems for a mapping defined on a bounded, closed and convex subset of a uniformly convex Banach space, satisfying the RCSC condition. This condition was introduced by Karapınar (Dynamical Systems and Methods, 2012). We first establish the demiclosed principle for the mapping satisfying the RCSC condition. Then, using this principle, we establish the weak and strong convergence theorems. Results in the paper extend and improve a number of important results in this literature such as Khan and Suzuki (Nonlinear Anal. 80:211-215, 2013) and Reich (J. Math. Anal. Appl. 67:274-276, 1979).

1 Introduction

Let C be a nonempty closed convex subset of a Banach space X. A mapping $T \colon C\to C$ is said to be nonexpansive if $\Vert Tx-Ty\Vert \leq \Vert x-y\Vert$ for all $x, y \in C$. It is called quasi-nonexpansive [1] if $F(T)\neq\emptyset$ and $\Vert Tx-p\Vert \leq \Vert x-p\Vert$ for all $x \in C$ and for all $p \in F(T)$, where $F(T)$ is the set of fixed points of T, i.e., $F(T) = \{x \in C : Tx = x\}$. Every nonexpansive mapping with $F(T)\neq \emptyset$ is a quasi-nonexpansive mapping.

In 2008, Suzuki [2] introduced a mapping satisfying condition (C). More accurately, a mapping $T \colon C\to C$ is said to satisfy condition (C) if

$$\frac{1}{2}\Vert x-Tx\Vert \leq \Vert x-y\Vert \quad\Longrightarrow\quad \Vert Tx-Ty\Vert \leq \Vert x-y\Vert $$

for all $x,y \in C$. Every nonexpansive mapping satisfies condition (C); also if a mapping satisfies condition (C) and has a fixed point, then it is a quasi-nonexpansive mapping [2].

Fixed point theorems for a mapping satisfying condition (C) were studied by Dhompongsa et al. [3] and Phuengrattana [4]. Khan and Suzuki [5] proved a weak convergence theorem for a mapping satisfying condition (C) in uniformly convex Banach spaces whose dual has the Kadec-Klee property.

In 2013, Karapınar [6] suggested a new modification of mappings satisfying condition (C) to a mapping satisfying (RCSC)-condition.

Definition 1.1

Let T be a mapping on a subset C of a Banach space X. Then T is said to satisfy Reich-Chatterjea-Suzuki-(C) condition ((RCSC)-condition) if

$$\frac{1}{2}\Vert x-Tx\Vert \leq \Vert x-y\Vert \quad\Longrightarrow\quad \Vert Tx-Ty\Vert \leq\frac{1}{3}\bigl(\Vert x-y\Vert +\Vert Tx-y \Vert +\Vert x-Ty\Vert \bigr) $$

for all $x,y \in C$.

Motivated by the above mentioned works, in this paper, we prove some weak and strong convergence theorems for generalized nonexpansive ((RCSC)-condition) mappings in a uniformly convex Banach space, which has the Kadec-Klee property. Our results generalize the results of Khan and Suzuki [5], Reich [7] to the case of a mapping satisfying (RCSC)-condition. For other works in this direction, please see Mogbademu [8], Saluja [9], Thakur [10] and Zheng [11].

2 Preliminaries

Throughout this paper, we denote by $\mathbb{N}$ the set of positive integers and by $\mathbb{R}$ the set of real numbers.

We now recall some definitions and results useful for our main results.

A Banach space X is called uniformly convex [12] if for each $\varepsilon\in(0,2]$ there is $\delta> 0$ such that for $x, y \in X$,

$$\left . \textstyle\begin{array}{r@{}} \Vert x\Vert \leq1 \\ \Vert y\Vert \leq1 \\ \Vert x-y\Vert >\varepsilon \end{array}\displaystyle \right \} \quad\Rightarrow\quad\biggl\Vert \frac{x+y}{2} \biggr\Vert \leq\delta. $$

Lemma 2.1

([12])

Let X be a uniformly convex Banach space. Let $\{x_{n}\}$ and $\{y_{n}\}$ be sequences in X satisfying $\lim_{n\to\infty} \Vert x_{n}\Vert = 1$, $\lim_{n\to\infty} \Vert y_{n}\Vert = 1$ and $\lim_{n\to \infty} \Vert x_{n} + y_{n}\Vert = 2$. Then $\lim_{n\to\infty} \Vert x_{n} - y_{n}\Vert = 0$.

Lemma 2.2

([5])

Let X be a uniformly convex Banach space and let $\{u_{n}\}$, $\{v_{n}\}$ and $\{w_{n}\}$ be sequences in X. Let d and t be real numbers with $d\in(0, \infty)$ and $t \in(0, 1)$. Assume that $\lim_{n\to\infty} \Vert u_{n} - v_{n}\Vert = d$, $\limsup_{n\to \infty} \Vert u_{n}- w_{n}\Vert \leq(1 - t)d$ and $\limsup_{n\to\infty} \Vert v_{n} - w_{n}\Vert \leq td$. Then $\lim_{n\to\infty} \Vert tu_{n} + (1 - t) v_{n} - w_{n}\Vert = 0$.

Proposition 2.1

Let C be a nonempty subset of a Banach space X and $T : C\to C$ be a mapping satisfying (RCSC)-condition. Then T has the following properties:

• (i) If T has a fixed point, then it is a quasi-nonexpansive mapping [6], Proposition 6.

• (ii) If C is closed, then $F(T)$ is closed; further if X is strictly convex and C is convex, then $F(T)$ is also convex [6], Proposition 10.

A Banach space X is said to have the Kadec-Klee property if, for every sequence $\{x_{n}\}$ in X which converges weakly to a point $x \in X$ with $\Vert x_{n}\Vert$ converging to $\Vert x\Vert$, $\{x_{n}\}$ converges strongly to x. Every uniformly convex Banach space has the Kadec-Klee property [13].

Lemma 2.3

([14, 15])

Let X be a reflexive Banach space whose dual has the Kadec-Klee property. Let $\{x_{n}\}$ be a bounded sequence in X and let $y, z\in X$ be weak subsequential limits of $\{x_{n}\}$. Assume that for every $t \in[0, 1]$, $\lim_{n\to\infty} \Vert tx_{n} + (1 - t) y- z\Vert$ exists. Then $y = z$.

Proposition 2.2

Let C be a nonempty subset of a Banach space X and $T \colon C\to C$ be a mapping satisfying (RCSC)-condition. Then

• (1) $\Vert x-Ty\Vert \leq9\Vert Tx-x\Vert + \Vert x-y\Vert$,

• (2) $\Vert y-Ty\Vert \leq9\Vert Tx-x\Vert + 2\Vert x-y\Vert$

hold for all $x, y \in C$.

Proof

(1) follows from [6], Corollary 16.

For (2), it follows from (1) that

$$\begin{aligned} \Vert y - Ty\Vert &\leq \Vert y - x\Vert + \Vert x -Ty\Vert \\ &\leq9\Vert x - Tx\Vert + 2 \Vert x - y\Vert . \end{aligned}$$

Thus we have (2). □

3 Main results

In this section, we prove weak and strong convergence theorems. First, we establish some auxiliary results.

The following lemma is an extension of Lemma 8 of [5] to the case of mappings satisfying (RCSC)-condition.

Lemma 3.1

Let C be a nonempty bounded convex subset of a uniformly convex Banach space X, and let $T \colon C \to C$ be a mapping satisfying (RCSC)-condition. Suppose that for any $\varepsilon> 0$, there exists $\xi(\varepsilon) > 0$ such that $\Vert Tu- u\Vert < \xi (\varepsilon)$, $\Vert Tv - v\Vert <\xi(\varepsilon)$ for some $u,v \in C$. Then, for any $t\in[0, 1]$,

$$\bigl\Vert T\bigl(tu + (1 - t)v\bigr) - \bigl(tu + (1 - t)v\bigr)\bigr\Vert < \varepsilon. $$

Proof

Assume to the contrary that there exist sequences $\{u_{n}\}, \{v_{n}\} \in C$, $\{t_{n}\} \in[0, 1]$ and $\varepsilon> 0$ such that

$$\Vert Tu_{n} - u_{n}\Vert < \frac{1}{n},\qquad \Vert Tv_{n} - v_{n}\Vert < \frac {1}{n} , $$

and

$$\bigl\Vert T\bigl(t_{n}u_{n} + (1 - t_{n})v_{n} \bigr) - \bigl(t_{n}u_{n} + (1 - t_{n})v_{n} \bigr)\bigr\Vert \geq \varepsilon. $$

Setting $x_{n} = t_{n}u_{n} + (1 - t_{n})v_{n}$ and $w_{n} = Tx_{n}$, from Proposition 2.2(ii), we get

$$\begin{aligned} 0 &< \varepsilon\leq\liminf_{n\to\infty} \Vert Tx_{n}-x_{n} \Vert \\ &\leq\liminf_{n\to\infty}\bigl(9\Vert Tu_{n}-u_{n} \Vert +2\Vert u_{n}-x_{n}\Vert \bigr) \\ &=2\liminf_{n\to\infty} \Vert u_{n}-x_{n}\Vert . \end{aligned}$$

Similarly, we can show that

$$0< \liminf_{n\to\infty} \Vert v_{n}-x_{n}\Vert , $$

and hence

$$0< \liminf_{n\to\infty} \Vert u_{n}-v_{n}\Vert . $$

Since C is bounded and

$$0< \liminf_{n\to\infty} \Vert v_{n}-x_{n}\Vert =\liminf_{n\to\infty }t_{n}\Vert u_{n}-v_{n} \Vert \leq\liminf_{n\to\infty}t_{n}\times\sup _{n\in\mathbb {N}}\Vert u_{n}-v_{n}\Vert , $$

we get $0 < \liminf_{n\to\infty} t_{n}$.

Similarly, we can show that $\limsup_{n\to\infty} t_{n} <1$.

So, without loss of generality, we may assume that $\Vert u_{n}- v_{n}\Vert$ converges to $d\in(0, \infty)$ and $t_{n}$ converges to $t\in(0, 1)$ as $n\to\infty$.

Since $\lim_{n\to\infty} \Vert Tu_{n}-u_{n}\Vert = 0$ and $0 < \liminf_{n\to\infty} \Vert u_{n}-x_{n}\Vert$, we obtain

$$\frac{1}{2}\Vert Tu_{n}-u_{n}\Vert \leq \Vert u_{n}-x_{n}\Vert $$

for sufficiently large $n\in\mathbb{N}$.

Since T satisfies (RCSC)-condition, for sufficiently large $n\in \mathbb{N}$, we have

$$\Vert Tu_{n}-Tx_{n}\Vert \leq\frac{1}{3}\bigl( \Vert u_{n}-x_{n}\Vert +\Vert Tu_{n}-x_{n} \Vert +\Vert u_{n}-Tx_{n}\Vert \bigr). $$

By similar arguments, we have

$$\Vert Tv_{n}-Tx_{n}\Vert \leq\frac{1}{3}\bigl( \Vert v_{n}-x_{n}\Vert +\Vert Tv_{n}-x_{n} \Vert +\Vert v_{n}-Tx_{n}\Vert \bigr) $$

for sufficiently large $n \in\mathbb{N}$.

Now, using the triangular inequality and Proposition 2.2(i), we have

$$\begin{aligned} &\limsup_{n\to\infty} \Vert u_{n}-w_{n}\Vert \\ &\quad\leq\limsup_{n\to \infty }\bigl(\Vert u_{n}-Tu_{n}\Vert +\Vert Tu_{n}-Tx_{n}\Vert \bigr) \\ &\quad\leq\limsup_{n\to\infty} \biggl(\Vert u_{n}-Tu_{n} \Vert +\frac {1}{3} \bigl(\Vert u_{n}-x_{n}\Vert + \Vert Tu_{n}-x_{n}\Vert +\Vert u_{n}-Tx_{n} \Vert \bigr) \biggr) \\ &\quad\leq\limsup_{n\to\infty} \biggl(\Vert u_{n}-Tu_{n} \Vert +\frac {1}{3} \bigl(\Vert u_{n}-x_{n}\Vert +10\Vert u_{n}-Tu_{n}\Vert +2\Vert u_{n}-x_{n} \Vert \bigr) \biggr) \\ &\quad=(1-t)d, \end{aligned}$$

and

$$\begin{aligned} &\limsup_{n\to\infty} \Vert v_{n}-w_{n}\Vert \\ &\quad\leq\limsup_{n\to \infty}\bigl(\Vert v_{n}-Tv_{n}\Vert +\Vert Tv_{n}-Tx_{n}\Vert \bigr) \\ &\quad\leq\limsup_{n\to\infty} \biggl(\Vert v_{n}-Tv_{n} \Vert +\frac {1}{3} \bigl(\Vert v_{n}-x_{n}\Vert + \Vert Tv_{n}-x_{n}\Vert +\Vert v_{n}-Tx_{n} \Vert \bigr) \biggr) \\ &\quad\leq\limsup_{n\to\infty} \biggl(\Vert v_{n}-Tv_{n} \Vert +\frac {1}{3} \bigl(\Vert v_{n}-x_{n}\Vert +10\Vert v_{n}-Tv_{n}\Vert +2\Vert v_{n}-x_{n} \Vert \bigr) \biggr) \\ &\quad=td. \end{aligned}$$

It then follows from Lemma 2.2 that

$$0 < \varepsilon\leq\lim_{n\to\infty} \Vert x_{n}-w_{n} \Vert = 0, $$

which is a contradiction, and this completes the proof. □

We now establish the demiclosed principle for the mapping satisfying (RCSC)-condition.

Proposition 3.1

Let T be a mapping on a bounded and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition. Then $I-T$ is demiclosed at zero. That is, if $\{x_{n}\} \in C$ converges weakly to $x_{0}\in C$ and $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert = 0$, then $Tx_{0} = x_{0}$.

Proof

Let $\xi\colon(0, \infty) \to(0, \infty)$ be a function satisfying the conclusion of Lemma 3.1. Let $\{x_{n}\}$ be a sequence converging weakly to $x_{0}\in C$ and $\lim_{n\to \infty} \Vert Tx_{n}-x_{n}\Vert = 0$. For arbitrarily chosen $\varepsilon> 0$, define a strictly decreasing sequence $\{\varepsilon_{n}\}$ in $(0, \infty)$ by

$$\varepsilon_{1} = \varepsilon \quad\mbox{and}\quad \varepsilon_{n+1} = \frac{\min\{\varepsilon_{n},\xi(\varepsilon_{n})\} }{2}. $$

It is obvious that $\varepsilon_{n+1} < \xi(\varepsilon_{n})$. Choose a subsequence $\{x_{f(n)}\}$ of $\{x_{n}\}$ such that $\Vert x_{f(n)}-Tx_{f(n)}\Vert < \xi(\varepsilon_{n})$. Since $x_{0}$ belongs to the closed convex hull of $\{x_{f(n)} : n \in\mathbb{N}\}$, it is a weak limit of $\{x_{f(n)}\}$. Hence, there exist $y\in C$ and $v\in\mathbb{N}$ such that $\Vert y-x_{0}\Vert < \varepsilon$ and y belongs to the convex hull of $\{x_{f(n)} : n = 1, 2,\ldots, v\}$. Using Lemma 3.1, we have $\Vert Ty- y\Vert < \varepsilon$. Using Proposition 2.2(ii), we obtain

$$\Vert Tx_{0}- x_{0}\Vert \leq9 \Vert Ty- y\Vert + 2 \Vert y- x_{0}\Vert < 11\varepsilon. $$

Since $\varepsilon> 0$ is arbitrary, we obtain $Tx_{0} = x_{0}$. □

Lemma 3.2

Let T be a mapping on a bounded and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition. For arbitrary $x_{1} \in C$ and a real number $\alpha\in[1/2, 1)$, construct a sequence $\{x_{n}\}$ in C by

$$ x_{n+1} = \alpha T x_{n} + (1 - \alpha) x_{n}. $$

(3.1)

If $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0$, then $\lim_{n\to \infty }\Vert tx_{n} + (1-t) p-q\Vert$ exists, where $p, q \in F(T)$ and $t \in[0, 1]$.

Proof

Since T satisfies (RCSC)-condition, by Proposition 2.1, it is quasi-nonexpansive. Let $S = \alpha T + (1-\alpha)I$, then S is a self-mapping on C, and $F(S) = F(T)$ also S is quasi-nonexpansive, and

$$x_{n+1} = \alpha Tx_{n} + (1-\alpha)x_{n} = Sx_{n} = S^{n}x_{1}. $$

Thus, for any $q \in F(S)$, we have

$$\begin{aligned} \Vert x_{n+1}-q\Vert &= \Vert Sx_{n}-q\Vert \\ &\leq \Vert x_{n}-q\Vert , \end{aligned}$$

hence the sequence $\{\Vert x_{n}-q\Vert \}$ is nonincreasing and bounded below. Therefore, it converges.

Since the sequence $\{\Vert p-q\Vert \}$ obviously converges, we see that $\lim_{n\to\infty} \Vert tx_{n}+(1-t)p-q\Vert$ exists for $t=1$ and $t=0$. Thus it remains to consider $t\in(0, 1)$.

Let $\lim_{n\to\infty} \Vert x_{n}-p\Vert = d$. If $d = 0$, there is nothing to prove. Take $d > 0$. We have

$$\begin{aligned} \liminf_{m,n\to\infty}\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert &\geq\liminf_{m,n\to\infty} \bigl(\Vert x_{n}-p\Vert -\bigl\Vert p-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \bigr) \\ &\geq\liminf_{m,n\to\infty} \bigl(\Vert x_{n}-p\Vert - \bigl\Vert p-\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \bigr) \\ =& (1-t)d>0 \end{aligned}$$

for all $l\in\mathbb{N}\cup\{0\}$, where $S^{0}$ is the identity mapping on C. Then there exists $\nu\in\mathbb{N}$ such that

$$\frac{1}{2}\Vert x_{n}-Tx_{n}\Vert \leq\bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert $$

for all $l\geq0$ and $m, n\geq\nu$. Since T satisfies (RCSC)-condition and Proposition 2.2(i), we obtain

$$\begin{aligned} &\bigl\Vert Tx_{n}-T\circ S^{l}\bigl(tx_{m}+(1-t)p \bigr)\bigr\Vert \\ &\quad \leq\frac{1}{3}\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert +\frac {1}{3}\bigl\Vert Tx_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\qquad{} +\frac{1}{3}\bigl\Vert x_{n}-T\circ S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert , \end{aligned}$$

and hence

$$\begin{aligned} &\bigl\Vert x_{n+1}-S^{l+1}\bigl(tx_{m}+(1-t)p \bigr)\bigr\Vert \\ & \quad= \bigl\Vert Sx_{n}-S\circ S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\quad \leq\bigl\| \alpha Tx_{n} + (1-\alpha)x_{n}- \alpha T\circ S^{l}\bigl(tx_{m} + (1-t) p\bigr) \\ &\qquad{} -(1-\alpha)S^{l}\bigl(tx_{m} + (1-t)p\bigr) \bigr\| \\ & \quad=\bigl\| \alpha\bigl(Tx_{n}-T\circ S^{l} \bigl(tx_{m} + (1-t)p\bigr)\bigr) \\ & \qquad{}+(1-\alpha) \bigl(x_{n}-S^{l} \bigl(tx_{m} + (1-t)p\bigr)\bigr)\bigr\| \\ & \quad\leq\alpha\bigl\Vert Tx_{n}-T\circ S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\qquad{} +(1-\alpha)\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ & \quad\leq\alpha \biggl\{ \frac{1}{3} \bigl(\bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert +\bigl\Vert Tx_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\qquad{} +\bigl\Vert x_{n}-T\circ S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \bigr) \biggr\} \\ &\qquad{} +(1-\alpha)\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\quad \leq\alpha \biggl\{ \frac{1}{3} \bigl(\bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert +\bigl\Vert Tx_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\qquad{} +9\Vert Tx_{n}-x_{n}\Vert +\bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \bigr) \biggr\} \\ &\qquad{} +(1-\alpha)\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\quad \leq\alpha \biggl\{ \bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr) \bigr\Vert +\frac {10}{3}\Vert Tx_{n}-x_{n}\Vert \biggr\} \\ &\qquad{} +(1-\alpha)\bigl\Vert x_{n}-S^{l} \bigl(tx_{m}+(1-t)p\bigr)\bigr\Vert \\ &\quad =\bigl\Vert x_{n}-S^{l}\bigl(tx_{m}+(1-t)p\bigr) \bigr\Vert +\frac{10}{3}\Vert Tx_{n}-x_{n}\Vert \end{aligned}$$

for all $l\geq0$ and $m, n \geq\nu$.

Let $h \colon\mathbb{N} \to[0, \infty)$ be a function defined by

$$h(n)= \bigl\Vert tx_{n}+(1-t)p-q\bigr\Vert . $$

Take two subsequences $\{f(n)\}$ and $\{g(n)\}$ of $\{n\}$ such that $\nu< f(1)$, $f(n) < g(n)$ for each $n\in\mathbb{N}$ and

$$\lim_{n\to\infty}h\bigl(f(n)\bigr) = \liminf_{n\to\infty}h(n),\qquad \lim_{n\to\infty}h\bigl(g(n)\bigr) = \limsup_{n\to\infty}h(n). $$

Set $u_{n} = x_{g(n)}$, $v_{n} = p$ and $w_{n} = S^{g(n)-f(n)}(tx_{f(n)} + (1 - t) p)$. Then we have

$$\begin{aligned}& \lim_{n\to\infty} \Vert u_{n}- v_{n}\Vert = d, \end{aligned}$$

(3.2)

$$\begin{aligned}& \begin{aligned}[b] \limsup_{n\to\infty} \Vert u_{n}-w_{n}\Vert ={}& \limsup_{n\to\infty }\bigl\Vert x_{g(n)}-S^{g(n)-f(n)} \bigl(tx_{f(n)}+(1-t)p\bigr)\bigr\Vert \\ \leq{}&\limsup_{n\to\infty}\bigl\Vert x_{f(n)}- \bigl(tx_{f(n)}+(1-t)p\bigr)\bigr\Vert \\ &{} + \frac{10}{3}\limsup_{n\to\infty} \Vert x_{n}-Tx_{n}\Vert \\ ={}&(1-t)\limsup_{n\to\infty} \Vert x_{f(n)}-p\Vert \\ ={}&(1-t)d, \end{aligned} \end{aligned}$$

(3.3)

and

$$ \limsup_{n\to\infty} \Vert v_{n}-w_{n} \Vert \leq td. $$

(3.4)

By (3.2), (3.3), (3.4) and Lemma 2.2, we have

$$\lim_{n\to\infty} \bigl\Vert tu_{n} + (1 - t) v_{n} - w_{n}\bigr\Vert = 0. $$

Substituting the value of $u_{n}$, $v_{n}$ and $w_{n}$, we have

$$\lim_{n\to\infty}\bigl\Vert tx_{g(n)} + (1 - t)p - S^{g(n)-f(n)}\bigl(tx_{f(n)} + (1-t)p\bigr)\bigr\Vert = 0. $$

Using the quasi-nonexpansiveness of S, we get

$$\begin{aligned} \limsup_{n\to\infty}h(n)={}&\lim_{n\to\infty}h\bigl(g(n) \bigr) \\ \leq{}&\limsup_{n\to\infty} \bigl(\bigl\Vert tx_{g(n)}+(1-t)p-S^{g(n)-f(n)} \bigl(tx_{f(n)}+(1-t)p\bigr)\bigr\Vert \\ &{} +\bigl\Vert S^{g(n)-f(n)}\bigl(tx_{f(n)}+(1-t)p \bigr)-q\bigr\Vert \bigr) \\ ={}& \limsup_{n\to\infty}\bigl\Vert S^{g(n)-f(n)} \bigl(tx_{f(n)}+(1-t)p\bigr)-q\bigr\Vert \\ \leq{}&\limsup_{n\to\infty}\bigl\Vert \bigl(tx_{f(n)}+(1-t)p \bigr)-q\bigr\Vert \\ ={}&\lim_{n\to\infty}h\bigl(f(n)\bigr) \\ ={}&\liminf_{n\to\infty}h(n). \end{aligned}$$

Thus $\lim_{n\to\infty} h(n) = \lim_{n\to\infty} \Vert tx_{n} + (1 - t) p - q\Vert$ exists. □

Now, we prove a weak convergence theorem.

Theorem 3.1

Let X be a uniformly convex Banach space whose dual has the Kadec-Klee property. Let T be a mapping on a bounded, closed and convex subset C of X. Assume that T satisfies (RCSC)-condition and define a sequence $\{x_{n}\}$ in C by (3.1). If $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0$, then $\{x_{n}\}$ converges weakly to a fixed point of T.

Proof

Let W be the set of all weak subsequential limits of $\{x_{n}\}$. Since $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert$ is equal to 0, by Proposition 3.1 we have $W \subset F(T)$. Using Lemma 2.3 and Lemma 3.2, W is singleton. But X is a uniformly convex Banach space, hence reflexive. So every sequence $\{x_{n}\}$ has a subsequence converging weakly to the unique element of W. Since W is singleton, therefore $\{x_{n}\}$ itself converges weakly to the unique element of W. □

Remark 1

Theorem 3.1 is a generalization of Theorem 11 of [5].

Since the dual of a reflexive Banach space with Fréchet differentiable norm has the Kadec-Klee property [16], as a direct consequence of Theorem 3.1, we get the following result.

Corollary 3.1

Let X be a uniformly convex Banach space whose norm is Fréchet differentiable. Let T be a mapping on a bounded, closed and convex subset C of X. Assume that T satisfies (RCSC)-condition and define a sequence $\{x_{n}\}$ in C by (3.1). If $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0$, then $\{x_{n}\}$ converges weakly to a fixed point of T.

Recall that a mapping $T \colon C \to C$ is said to satisfy condition (I) [17] if there exists a nondecreasing function $f :[0,\infty) \to[0,\infty)$ with $f(0) = 0$, $f(r) > 0$ for all $r \in(0, \infty)$ such that $d(x, Tx) \geq f(d(x, F(T)))$ for all $x \in C$, where $d(x, F(T)) = \inf_{p\in F(T)} d(x, p)$.

We now establish a strong convergence theorem.

Theorem 3.2

Let T be a mapping on a bounded, closed and convex subset C of a uniformly convex Banach space X. Assume that T satisfies (RCSC)-condition and define a sequence $\{x_{n}\}$ in C by (3.1). If $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0$ and T satisfies condition (I), then $\{x_{n}\}$ converges strongly to a fixed point of T.

Proof

By Lemma 3.2, we know that $\lim_{n\to\infty} \Vert x_{n}-p\Vert$ exists for all $p\in F(T)$, and hence $\lim_{n\to\infty} d(x_{n}, F(T))$ exists. Assume that $\lim_{n\to\infty} \Vert x_{n}-p\Vert = r$ for some $r \geq0$.

If $r = 0$, then $\{x_{n}\}$ converges strongly to p and the result follows.

Suppose $r > 0$. From the hypothesis and condition (I), we have $\lim_{n\to\infty} \Vert Tx_{n}-x_{n}\Vert =0$ and $f(d(x_{n}, F(T))) \leq \Vert Tx_{n}-x_{n}\Vert$. This gives $\lim_{n \to\infty}f(d(x_{n}, F(T))) = 0$. Since f is a nondecreasing function, we have $\lim_{n\to\infty} d(x_{n}, F(T)) =0$. Thus, there exist a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ and a sequence $\{y_{k}\} \subset F(T)$ such that

$$\Vert x_{n_{k}} - y_{k}\Vert < \frac{1}{2^{k}} \quad\mbox{for all } k \in\mathbb{N} . $$

Again, we see that

$$\begin{aligned} \Vert x_{n+1}-y_{k}\Vert &= \bigl\Vert \alpha Tx_{n} + (1-\alpha )x_{n}-y_{k}\bigr\Vert \\ &\leq\alpha \Vert Tx_{n}-y_{k}\Vert +(1-\alpha)\Vert x_{n}-y_{k}\Vert \\ &\leq \Vert x_{n}-y_{k}\Vert \\ &< \frac{1}{2^{k}}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert y_{k+1}-y_{k}\Vert & \leq \Vert y_{k+1}-x_{k+1}\Vert + \Vert x_{k+1} - y_{k}\Vert \\ & \leq\frac{1}{2^{k+1}} + \frac{1}{2^{k}} \\ & < \frac{1}{2^{k-1}} \rightarrow0 \quad\mbox{as } n \rightarrow \infty. \end{aligned}$$

This shows that $\{y_{k}\}$ is a Cauchy sequence in a complete space, and hence it converges to a point $p \in X$. Since $F(T)$ is closed, therefore $p \in F(T)$ and then $\{x_{n_{k}}\}$ converges strongly to p. Since $\lim_{n\to\infty} \Vert x_{n}- p\Vert$ exists, $x_{n} \to p \in F(T)$. This completes the proof. □

We now give an example of mapping T which satisfies (RCSC)-condition but fails to satisfy condition (C).

Example 1

Let $X=\mathbb{R}$ with usual metric and $C=[0,1]\subset X$. Define a mapping $T\colon C\to C$ by the rule

$$ Tx= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0, & x \in [0 , \frac{4}{5} ),\\ \frac{x}{2}, & x\in [\frac{4}{5} , 1 ] . \end{array}\displaystyle \right . $$

Set $x=\frac{9}{10}$ and $y=\frac{3}{5}$, we see that

$$\frac{1}{2}\vert x-Tx\vert =\frac{9}{40}< \frac{3}{10}= \vert x-y\vert , $$

and

$$\vert Tx-Ty\vert = \frac{9}{20} > \frac{3}{10} =\vert x-y \vert , $$

i.e.,

$$\frac{1}{2}\vert x-Tx\vert \leq \vert x-y\vert \quad\nRightarrow \quad \vert Tx-Ty\vert \leq \vert x-y\vert , $$

hence, T fails to satisfy condition (C).

To verify that T satisfies condition (RCSC), consider the following cases.

Case-I: Let $x, y \in [0 , \frac{4}{5} )$, then we have

$$\vert Tx-Ty\vert = 0 \leq\frac{1}{3} \bigl[\vert x-y\vert +\vert Tx-y\vert +\vert x-Ty\vert \bigr], $$

$x, y \in [0 , \frac{4}{5} )$.

Case-II: Let $x, y \in [\frac{4}{5} , 1 ]$, then

$$\vert Tx-Ty\vert =\biggl\vert \frac{x}{2}-\frac{y}{2}\biggr\vert . $$

Since

$$\begin{aligned}& \vert x-y\vert >\biggl\vert \frac{x}{2}-\frac{y}{2}\biggr\vert =\vert Tx-Ty\vert , \\& \vert Tx-y\vert =\biggl\vert \frac{x}{2}-y\biggr\vert >\biggl\vert \frac {x}{2}-\frac {y}{2}\biggr\vert =\vert Tx-Ty\vert \end{aligned}$$

and

$$\vert x-Ty\vert =\biggl\vert x-\frac{y}{2}\biggr\vert >\biggl\vert \frac {x}{2}-\frac {y}{2}\biggr\vert =\vert Tx-Ty\vert , $$

which implies that

$$\vert Tx-Ty\vert < \frac{1}{3} \bigl[ \vert x-y\vert +\vert Tx-y \vert +\vert x-Ty\vert \bigr] $$

for all $x, y \in [\frac{4}{5} , 1 ]$.

Case-III: Let $x \in [0 , \frac{4}{5} )$ and $y \in [\frac{4}{5} , 1 ]$ or $x \in [\frac {4}{5} , 1 ]$ and $y \in [0 , \frac{4}{5} )$. Then

$$\vert Tx-Ty\vert = \frac{y}{2}. $$

Also,

$$ \frac{1}{3} \bigl[ \vert x-y\vert +\vert Tx-y\vert + \vert x-Ty\vert \bigr] = \frac {1}{3} \biggl[y-x + y + \biggl\vert x- \frac{y}{2}\biggr\vert \biggr]. $$

(3.5)

We now have two subcases as follows.

Case-III(A): $x \geq\frac{y}{2}$, then $\vert x-\frac{y}{2}\vert = x-\frac{y}{2}$, and by (3.5) we have

$$\begin{aligned} \frac{1}{3} \bigl[ \vert x-y\vert +\vert Tx-y\vert +\vert x-Ty \vert \bigr] &= \frac {y}{2} =\vert Tx-Ty\vert . \end{aligned}$$

Case-III(B): $x < \frac{y}{2}$, then $\vert x-\frac{y}{2}\vert = \frac{y}{2}-x$, and by (3.5) we have

$$\begin{aligned} \frac{1}{3} \bigl[ \vert x-y\vert +\vert Tx-y\vert +\vert x-Ty \vert \bigr] &= \frac {1}{3} \biggl[\frac{5y}{2}-2x \biggr] > \frac{1}{3} \biggl[\frac{5y}{2}-y \biggr] = \frac{y}{2} = \vert Tx-Ty\vert . \end{aligned}$$

Hence $\vert Tx-Ty\vert \leq\frac{1}{3} [ \vert x-y\vert +\vert Tx-y\vert +\vert x-Ty\vert ]$ for all $x \in [0 , \frac{4}{5} )$ and $y \in [\frac{4}{5} , 1 ]$.

Case-IV: Let $x \in [\frac{4}{5} , 1 ]$ and $y \in [0 , \frac{4}{5} )$. By interchanging the role of x and y in Case-III, we can see that

$$\vert Tx-Ty\vert \leq\frac{1}{3} \bigl[ \vert x-y\vert +\vert Tx-y\vert +\vert x-Ty\vert \bigr] $$

for all $x \in [\frac{4}{5} , 1 ]$ and $y \in [0 , \frac{4}{5} )$.

In view of Case-I to Case-IV, we can say that T satisfies condition (RCSC) for all $x,y\in C$.