1 Introduction and preliminaries

The generalization of nonexpansive mappings and the study of related fixed point theorems with different practical applications in nonlinear functional analysis have found great importance during the recent decades [1–6, 8–10, 12, 14, 15, 19, 22–24, 26, 27, 32]. Several prominent authors [11, 13, 16, 20, 21, 25, 28, 29, 31] have contributed immensely in this field, and different new classes of mappings with interesting properties have been developed in this context.

In 2008, Suzuki defined a class of generalized nonexpansive mappings on a nonempty subset C of a Banach space X. Such type of mappings was called the class of mappings satisfying the condition (C) (see [17, 18, 30]). For a nonempty bounded and convex subset C, every self-mapping T on C satisfying (C) condition has an almost fixed point sequence [15]. In 2011, Falset et al. [15] introduced two new classes of generalized nonexpansive mappings which are wider than those satisfying (C) condition, but preserving their fixed point properties.

In this paper, we introduce a new class of mappings which is larger than the class satisfying the condition (C). We study the existence of fixed points for this type of mappings with some examples.

First we present some basic concepts.

Definition 1.1

(see [30])

Let C be a nonempty subset of a Banach space X. A mapping \(T:C\longrightarrow X\) is said to be nonexpansive if \(\Vert Tx-Ty \Vert \leq \Vert x-y \Vert \) for all \(x,y\in C\).

Definition 1.2

(see [30])

For a nonempty subset C of a Banach space X, a mapping \(T:C \longrightarrow X\) is called quasi nonexpansive if \(\Vert Tx-z \Vert \leq \Vert x-z \Vert \) for all \(x\in C\) and \(z\in F(T)\) (where \(F(T)\) denotes the set of all fixed points of T).

Definition 1.3

(see [30])

For a nonempty subset C of a Banach space X, a mapping \(T:C \longrightarrow X\) is said to satisfy the condition (C) on C if \(\frac{1}{2} \Vert x-Tx \Vert \leq \Vert x-y \Vert \) implies \(\Vert Tx-Ty \Vert \leq \Vert x-y \Vert \) for all \(x,y \in C \).

Clearly, every nonexpansive mapping satisfies the condition (C) on C. But there are also some noncontinuous mappings satisfying the condition (C) (see [15]).

Definition 1.4

(see [15])

For a nonempty subset C of a Banach space X and \(\lambda\in(0,1)\), a mapping \(T:C\longrightarrow X\) is said to satisfy \((C_{\lambda})\)-condition on C if \(\lambda \Vert x-Tx \Vert \leq \Vert x-y \Vert \) implies \(\Vert Tx-Ty \Vert \leq \Vert x-y \Vert \) for all \(x,y\in C\).

Definition 1.5

(see [15])

If C is a closed convex and bounded subset of X, and a self-mapping T on C is nonexpansive, then there exists a sequence \(x_{n}\) in C such that \(\Vert x_{n}-Tx_{n} \Vert \longrightarrow0\). Such a sequence is called almost fixed point sequence for T.

Definition 1.6

(see [15])

Let C be a nonempty subset of a Banach space X and \(\lbrace x_{n}\rbrace\) be a bounded sequence in X. For each \(x\in X\),

  1. (i)

    asymptotic radius of \(\lbrace x_{n}\rbrace\) at x is defined by \(r(x, \lbrace x_{n}\rbrace) = \lim\sup_{n\longrightarrow \infty} \Vert x_{n} - x \Vert \).

  2. (ii)

    asymptotic radius of \(\lbrace x_{n}\rbrace\) relative to C is defined by \(r(C, \lbrace x_{n}\rbrace) = \inf{r(x,\lbrace x_{n}\rbrace): x\in C}\).

  3. (iii)

    asymptotic center of \(\lbrace x_{n}\rbrace\) relative to C is defined by \(A(C, \lbrace x_{n}\rbrace) = \lbrace x \in C : r(x, \lbrace x_{n}\rbrace) = r(C, \lbrace x_{n}\rbrace) \rbrace\). We note that \(A(C, \lbrace x_{n}\rbrace)\) is nonempty. Again, if X is uniformly convex, then \(A(C, \lbrace x_{n}\rbrace)\) has exactly one point.

Definition 1.7

(see [30])

A Banach space X is said to satisfy the Opial property if, for every sequence \(\lbrace x_{n}\rbrace\) in X with \(x_{n}\longrightarrow z\)(weakly), we have

$$\lim_{n\longrightarrow\infty} \inf \Vert x_{n}-z \Vert < \lim _{n\longrightarrow \infty} \inf \Vert x_{n}-y \Vert $$

whenever \(y\neq z\).

For example, \(l^{p}\) spaces (\(1< p<\infty\)) satisfy this condition.

2 Methods

We apply both analytic as well as fixed point theoretical method to prove our results. Different existing methods in the literature (refer to [15, 20, 30]) as well as some new approaches are also taken.

3 Results and discussion

We construct the following class of mappings.

Definition 3.1

Let C be a nonempty subset of a Banach space X. Let \(\gamma\in[0,1]\) and \(\mu\in[0,\frac{1}{2}]\) such that \(2\mu\leq\gamma\). A mapping \(T:C\longrightarrow X\) is said to satisfy the condition \(B_{\gamma,\mu}\) on C if, for all \(x,y\) in C,

$$\gamma \Vert x-Tx \Vert \leq \Vert x-y \Vert +\mu \Vert y-Ty \Vert $$

implies \(\Vert Tx-Ty \Vert \leq(1-\gamma) \Vert x-y \Vert +\mu( \Vert x-Ty \Vert + \Vert y-Tx \Vert )\).

Clearly, this class includes the class of nonexpansive mappings (for \(\gamma=\mu=0\)).

Also, if a mapping satisfies the condition (C), then it will satisfy the condition \(B_{\gamma,\mu}\) for \(\gamma=\mu=0\).

$$\begin{aligned} &\text{As } \frac{1}{2} \Vert x-Tx \Vert \leq \Vert x-y \Vert \Rightarrow \Vert Tx-Ty \Vert \leq \Vert x-y \Vert \text{ for (C) condition,} \\ &\text{so, clearly, } \Vert Tx-Ty \Vert \leq(1-\gamma) \Vert x-y \Vert + \mu \bigl( \Vert x-Ty \Vert + \Vert y-Tx \Vert \bigr) \text{ for } \gamma= \mu=0. \end{aligned}$$

But the converse is not true.

Example 3.2

Let \(T:[0,2]\longrightarrow\mathbb{R}\) be defined by

$$T(x)= \textstyle\begin{cases} 0& \text{if } x\neq2,\\ 1 & \text{if } x=2. \end{cases} $$

Now, for \(x=1.2, y=2\), \(\frac{1}{2} \Vert Tx-x \Vert =0.6<0.8 = \Vert x-y \Vert \). \(\text{But} \Vert Tx-Ty \Vert =1 \nless0.8= \Vert x-y \Vert \).

So, the condition (C) is not satisfied.

For \(x\neq2,y\neq2\), obviously T satisfies \(B_{\gamma,\mu}\) condition for \(\gamma=1\) and \(\mu=\frac{1}{2}\).

Again, let \(x\neq2,y=2\).

Then \(\Vert Tx-Ty \Vert =1\) and

$$\begin{aligned} &(1-\gamma) \Vert x-y \Vert +\mu( \Vert x-Ty \Vert + \Vert y-Tx \Vert \\ &\quad =\frac{1}{2} \Vert x-1 \Vert +1 \quad\biggl(\text{for } \gamma=1,\mu= \frac{1}{2}\biggr) \\ &\quad >1= \Vert Tx-Ty \Vert . \end{aligned}$$

For \(x=2,y\neq2\),

$$\begin{aligned} &\Vert Tx-Ty \Vert =1 < (1-\gamma) \Vert x-y \Vert +\mu\bigl( \Vert x-Ty \Vert + \Vert y-Tx \Vert \bigr) \\ &\quad \biggl(\text{for } \gamma=1, \mu=\frac{1}{2}\biggr) \text{ as above.} \end{aligned}$$

For \(x=2,y=2\), \(B_{\gamma,\mu}\) condition (for \(\gamma=1,\mu=\frac {1}{2}\)) is obviously satisfied by T.

The following lemma shows that T satisfying \(B_{\gamma,\mu}\) condition is also quasi-nonexpansive.

Lemma 3.3

For a nonempty subset C of a Banach space X, let \(T:C\longrightarrow X\) be a mapping satisfying \(B_{\gamma,\mu }\) condition. If z is a fixed point of T on C, then for all \(x\in C\),

$$\Vert z-Tx \Vert \leq \Vert z-x \Vert . $$

Proof

We have

$$\begin{aligned} \gamma \Vert z-Tz \Vert =0 \leq \Vert z-x \Vert +\mu \Vert x-Tx \Vert . \end{aligned}$$

By \(B_{\gamma,\mu}\) condition,

$$\begin{aligned} &\Vert Tz-Tx \Vert \leq(1-\gamma) \Vert z-x \Vert +\mu\bigl( \Vert x-Tz \Vert + \Vert z-Tx \Vert \bigr) \\ &\phantom{\Vert Tz-Tx \Vert }=(1-\gamma) \Vert z-x \Vert +\mu\bigl( \Vert x-z \Vert + \Vert z-Tx \Vert \bigr) \\ &\quad\Rightarrow \quad \Vert z-Tx \Vert \leq\biggl(\frac{1-\gamma+\mu}{1-\mu}\biggr) \Vert z-x \Vert \leq \Vert z-x \Vert \quad(\text{as } 2\mu\leq\gamma) \end{aligned}$$

showing that T is quasi-nonexpansive. □

However, the converse of Lemma 3.3 does not hold in general.

Example 3.4

Let T be a mapping on \([0,4]\) defined by

$$T(x)= \textstyle\begin{cases} 0& \text{if } x\neq4,\\ 3& \text{if } x=4. \end{cases} $$

Clearly T has a fixed point at \(x=0\), and also \(\Vert T(x) \Vert \leq \Vert x \Vert \ \forall x\in[0,4]\).

Hence, T is quasi-nonexpansive.

We show that T does not satisfy the condition \(B_{\gamma,\mu}\).

For \(x=4, y=3\),

$$\gamma \Vert x-Tx \Vert =\gamma\leq1+3\mu= \Vert 4-3 \Vert +\mu \bigl\Vert 3-T(3)\bigr\Vert . $$

But

$$\begin{aligned} \Vert Tx-Ty \Vert = \bigl\Vert T(4)-T(3) \bigr\Vert =3 \end{aligned}$$

and

$$\begin{aligned} & (1-\gamma) \Vert x-y \Vert + \mu\bigl( \Vert x-Ty \Vert + \Vert y-Tx \Vert \bigr) \\ &\quad =1-\gamma+4\mu \\ &\quad \leq1-\gamma+ 2\gamma \quad(2 \mu\leq\gamma) \\ &\quad < 3 \quad\bigl(\text{as } \gamma\in[0,1]\bigr) \\ &\quad = \Vert Tx-Ty \Vert . \end{aligned}$$

So, \(B_{\gamma,\mu}\) condition is not satisfied.

The following are some basic properties of mappings which satisfy the condition \(B_{\gamma,\mu}\) on C.

Proposition 3.5

Let C be a nonempty subset of a Banach space X. Let \(T:C\longrightarrow C\) satisfy the condition \(B_{\gamma,\mu}\) on C. Then, for all \(x,y \in C\) and for \(c\in[0,1]\),

  1. (i)

    \(\Vert Tx-T^{2}x \Vert \leq \Vert x-Tx \Vert \),

  2. (ii)

    at least one of the following ((a) and (b)) holds:

    1. (a)

      \(\frac{c}{2} \Vert x-Tx \Vert \leq \Vert x-y \Vert \)

    2. (b)

      \(\frac{c}{2} \Vert Tx-T^{2}x \Vert \leq \Vert Tx-y \Vert \).

    The condition (a) implies \(\Vert Tx-Ty \Vert \leq(1-\frac{c}{2}) \Vert x-y \Vert +\mu ( \Vert x-Ty \Vert + \Vert y-Tx \Vert )\) and the condition (b) implies \(\Vert T^{2}x-Ty \Vert \leq (1-\frac{c}{2}) \Vert Tx-y \Vert +\mu( \Vert Tx-Ty \Vert + \Vert y-T^{2}x \Vert )\).

  3. (iii)

    \(\Vert x-Ty \Vert \leq(3-c) \Vert x-Tx \Vert +(1-\frac{c}{2}) \Vert x-y \Vert +\mu (2 \Vert x-Tx \Vert + \Vert x-Ty \Vert +\Vert y-Tx \Vert + 2 \Vert Tx-T^{2}x \Vert )\).

Proof

(i) We have, for all \(x\in C\),

$$\gamma \Vert x-Tx \Vert \leq \Vert x-Tx \Vert +\mu \bigl\Vert Tx-T^{2}x \bigr\Vert . $$

So, by the condition \(B_{\gamma,\mu}\) (replacing y by Tx),

$$\begin{aligned} &\bigl\Vert Tx-T^{2}x \bigr\Vert \leq(1-\gamma) \Vert x-Tx \Vert +\mu \bigl\Vert x-T^{2}x \bigr\Vert \\ &\phantom{\bigl\Vert Tx-T^{2}x \bigr\Vert }\leq(1-\gamma) \Vert x-Tx \Vert +\mu \Vert x-Tx \Vert +\mu \bigl\Vert Tx-T^{2}x \bigr\Vert \\ &\quad\Rightarrow \quad\bigl\Vert Tx-T^{2}x \bigr\Vert \leq\frac{1-\gamma+\mu}{1-\mu} \Vert x-Tx \Vert \leq \Vert x-Tx \Vert . \end{aligned}$$

(ii) We assume on the contrary that \(\frac{c}{2} \Vert x-Tx \Vert > \Vert x-y \Vert \) and \(\frac{c}{2} \Vert Tx-T^{2}x \Vert > \Vert Tx-y \Vert \) for some \(x,y\in C\).

Now,

$$\begin{aligned} \Vert x-Tx \Vert &\leq \Vert x-y \Vert + \Vert y-Tx \Vert \\ &< \frac{c}{2} \Vert x-Tx \Vert +\frac{c}{2} \bigl\Vert Tx-T^{2}x \bigr\Vert \\ &\leq\frac{c}{2} \Vert x-Tx \Vert +\frac{c}{2} \Vert x-Tx \Vert \quad\bigl(\text{by }(i)\bigr) \\ &\leq \Vert x-Tx \Vert \quad(\text{since }c\leq1 ), \end{aligned}$$

i.e.,

$$\begin{aligned} \Vert x-Tx \Vert < \Vert x-Tx \Vert , \end{aligned}$$

which is impossible.

So, at least one of (a) and (b) holds.

(iii) \(\Vert x-Ty \Vert \leq \Vert x-Tx \Vert + \Vert Tx-Ty \Vert \)

If (ii)(a) holds,

$$\begin{aligned} \Vert x-Ty \Vert \leq{}& \Vert x-Tx \Vert +\biggl(1-\frac{c}{2} \biggr) \Vert x-y \Vert +\mu\bigl( \Vert x-Ty \Vert + \Vert y-Tx \Vert \bigr) \\ \leq{}&(3-c) \Vert x-Tx \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert x-y \Vert \\ &{}+\mu\bigl(2 \Vert x-Tx \Vert +\Vert x-Ty \Vert + \Vert y-Tx \Vert +2 \bigl\Vert Tx-T^{2}x \bigr\Vert \bigr). \end{aligned}$$

If (ii)(b) holds,

$$\begin{aligned} \Vert x-Ty \Vert \leq{}& \Vert x-Tx \Vert + \bigl\Vert Tx-T^{2}x \bigr\Vert + \bigl\Vert T^{2}x-Ty \bigr\Vert \\ \leq{}& \Vert x-Tx \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert Tx-x \Vert +\mu\bigl( \Vert Tx-Tx \Vert + \bigl\Vert x-T^{2}x \bigr\Vert \bigr) \\ &{}+\biggl(1-\frac{c}{2}\biggr) \Vert Tx-y \Vert +\mu \bigl( \Vert Tx-Ty \Vert + \bigl\Vert y-T^{2}x \bigr\Vert \bigr) \\ ={}&(3-c) \Vert x-Tx \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert x-y \Vert \\ &{}+\mu\bigl( \bigl\Vert x-T^{2}x \bigr\Vert + \Vert Tx-Ty \Vert + \bigl\Vert y-T^{2}x \bigr\Vert \bigr) \\ \leq{}&(3-c) \Vert x-Tx \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert x-y \Vert +\mu\bigl( \Vert x-Tx \Vert + \bigl\Vert Tx-T^{2}x \bigr\Vert \\ &{}+ \Vert x-Tx \Vert + \Vert x-Ty \Vert + \Vert y-Tx \Vert + \bigl\Vert Tx-T^{2}x \bigr\Vert \bigr) \\ ={}&(3-c) \Vert x-Tx \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert x-y \Vert \\ &{}+\mu\bigl(2 \Vert x-Tx \Vert + \Vert x-Ty \Vert + \Vert y-Tx \Vert +2 \bigl\Vert Tx-T^{2}x \bigr\Vert \bigr). \end{aligned}$$

 □

Proposition 3.6

Let C be a nonempty convex and bounded subset of a Banach space X and T be a self-mapping on C. We assume that T satisfies the condition \(B_{\gamma,\mu}\) on C. For \(x_{0}\in C\), let a sequence \(\lbrace x_{n} \rbrace\) in C be defined by

$$\begin{aligned} x_{n+1}=\lambda Tx_{n}+(1-\lambda)x_{n}, \end{aligned}$$
(3.1)

where \(\lambda\in[\gamma,1)-\lbrace0 \rbrace, n\in\mathbb {N} \cup\lbrace0\rbrace\). Then \(\Vert Tx_{n}-x_{n} \Vert \longrightarrow0\) as \(n\longrightarrow\infty\).

Proof

Since \(\lambda\geq\gamma\), we have

$$\begin{aligned} \gamma \Vert x_{n}-Tx_{n} \Vert \leq{}&\lambda \Vert x_{n}-Tx_{n} \Vert \\ ={}& \Vert x_{n}-x_{n+1} \Vert \quad \bigl(\text{by (3.1)}\bigr) \end{aligned}$$

i.e.,

$$\begin{aligned} \gamma \Vert x_{n}-Tx_{n} \Vert \leq \Vert x_{n}-x_{n+1} \Vert +\mu \Vert x_{n+1}-Tx_{n+1} \Vert . \end{aligned}$$

So, by the condition \(B_{\gamma,\mu}\) (for \(y=x_{n+1}\)),

$$\begin{aligned} &\Vert Tx_{n}-Tx_{n+1} \Vert \leq (1-\gamma) \Vert x_{n}-x_{n+1} \Vert +\mu \bigl( \Vert x_{n}-Tx_{n+1} \Vert + \Vert x_{n+1}-Tx_{n} \Vert \bigr) \\ &\quad \Rightarrow\quad \biggl\Vert Tx_{n}-\frac{1}{\lambda} \bigl(x_{n+2}-(1-\lambda )x_{n+1}\bigr) \biggr\Vert \\ &\phantom{\quad \Rightarrow\quad}\quad\leq \Vert x_{n}-x_{n+1} \Vert +\mu\biggl( \biggl\Vert x_{n}-\frac{1}{\lambda }\bigl(x_{n+2} -(1-\lambda)x_{n+1}\bigr) \biggr\Vert + \Vert x_{n+1}-Tx_{n} \Vert \biggr), \end{aligned}$$

from which we get

$$\begin{aligned} &(1-\mu)\lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert \leq0 \\ &\quad\Rightarrow\quad \lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0 \quad(\text{since } \mu\neq1). \end{aligned}$$

 □

Corollary 3.7

Let C be a closed convex and bounded subset of a Banach space X. Let T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\). Then the sequence \(\lbrace x_{n} \rbrace\) as defined above is an almost fixed point sequence.

For a nonempty compact convex subset C of X, we have the following fixed point result.

Theorem 3.8

Let C be a compact and convex subset of a Banach space X. Let T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\). For \(x_{0}\in C\), let \(\lbrace x_{n}\rbrace\) be a sequence in C as defined in Proposition 3.6, where γ is sufficiently small. Then \(\lbrace x_{n}\rbrace\) converges strongly to a fixed point of T.

Proof

Since C is compact, there exists a subsequence \(\lbrace x_{n_{j}} \rbrace\) of \(\lbrace x_{n} \rbrace\) and \(z\in C\) such that \(\lbrace x_{n_{j}} \rbrace\) converges to z (see [31]).

Now, by Proposition 3.5(ii), for \(\gamma=\frac{c}{2}, c\in[0,1]\)

$$\begin{aligned} &\gamma \Vert x_{n_{j}}-Tx_{n_{j}} \Vert \leq \Vert x_{n_{j}}-z \Vert \\ &\quad \Rightarrow\quad\gamma \Vert x_{n_{j}}-Tx_{n_{j}} \Vert \leq \Vert x_{n_{j}}-z \Vert +\mu \Vert z-Tz \Vert . \end{aligned}$$

So, by the condition \(B_{\gamma,\mu}\),

$$\begin{aligned} \Vert Tx_{n_{j}}-Tz \Vert \leq(1-\gamma) \Vert x_{n_{j}}-z \Vert +\mu \bigl( \Vert x_{n_{j}}-Tz \Vert + \Vert z-Tx_{n_{j}} \Vert \bigr). \end{aligned}$$
(3.2)

Again,

$$\begin{aligned} & \Vert x_{n_{j}}-Tz \Vert \\ &\quad\leq \Vert x_{n_{j}}-Tx_{n_{j}} \Vert + \Vert Tx_{n_{j}}-Tz \Vert \\ &\quad\leq \Vert x_{n_{j}}-Tx_{n_{j}} \Vert +(1-\gamma) \Vert x_{n_{j}}-z \Vert +\mu \bigl( \Vert x_{n_{j}}-Tz \Vert + \Vert z-Tx_{n_{j}} \Vert \bigr) \quad \bigl(\text{by (3.2)}\bigr) \\ &\quad\leq \Vert x_{n_{j}}-Tx_{n_{j}} \Vert +(1-\gamma) \Vert x_{n_{j}}-z \Vert +\mu \bigl( \Vert x_{n_{j}}-Tz \Vert + \Vert z-x_{n_{j}} \Vert + \Vert x_{n_{j}}-Tx_{n_{j}} \Vert \bigr). \end{aligned}$$

So, taking \(n_{j}\longrightarrow\infty\) and using Proposition 3.6, we get

$$\begin{aligned} &(1-\mu) \Vert z-Tz \Vert \leq0 \\ &\quad \Rightarrow\quad Tz=z, \quad(\text{since } \mu\neq1) \end{aligned}$$

showing that z is a fixed point for T. Now,

$$\begin{aligned} \Vert x_{n+1}-z \Vert &\leq\lambda \Vert Tx_{n}-z \Vert +(1-\lambda) \Vert x_{n}-z \Vert \\ &\leq\lambda \Vert x_{n}-z \Vert +(1-\lambda) \Vert x_{n}-z \Vert \quad (\text{by Lemma 3.3}) \\ &= \Vert x_{n}-z \Vert \quad\text{for all } n\in\mathbb{N}\cup\lbrace0 \rbrace. \end{aligned}$$

Thus, \(\lbrace \Vert x_{n}-z \Vert \rbrace\) is a monotonically decreasing sequence of nonnegative real numbers and will converge to some real, say u. Now,

$$\begin{aligned} &\Vert x_{n}-Tz \Vert \leq \Vert x_{n}-Tx_{n} \Vert +(1-\gamma) \Vert x_{n}-z \Vert +\mu \bigl( \Vert x_{n}-Tz \Vert + \Vert z-x_{n} \Vert + \Vert x_{n}-Tx_{n} \Vert \bigr) \\ &\quad \Rightarrow\quad \Vert x_{n}-z \Vert \leq \Vert x_{n}-Tx_{n} \Vert +(1-\gamma) \Vert x_{n}-z \Vert +\mu \bigl(2 \Vert x_{n}-z \Vert + \Vert x_{n}-Tx_{n} \Vert \bigr). \end{aligned}$$

Taking limit as \(n\longrightarrow\infty\), we get \(u\leq(1-\gamma)u+\mu(2u) \Rightarrow(\gamma-2\mu)u\leq0\), which is possible only for \(u=0\), since \(2\mu\leq\gamma\). Hence, \(\lbrace x_{n} \rbrace\) converges strongly to z. □

Example 3.9

On the subset \(C=[0,4]\) of the Banach space \(\mathbb{R}\), define T by

$$T(x)= \textstyle\begin{cases} 0& \text{if } x\neq4,\\ 2& \text{if } x=4. \end{cases} $$

Then T satisfies the condition \(B_{\gamma,\mu}\). Let \(\lbrace x_{n}\rbrace\) be a sequence in C defined as in Proposition 3.6.

Suppose, \(x_{0}= 3 \). Then we have \(x_{n}=3(1-\lambda)^{n}\), converging to 0 as \(n\longrightarrow\infty\). Clearly, 0 is the fixed point of T.

Theorem 3.10

Let C be a weakly compact and convex subset of a uniformly convex Banach space X. Let T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\). Then T has a fixed point.

Proof

Consider the sequence \(\lbrace x_{n}\rbrace\) in C as defined in Proposition 3.6. Then \(\lim\sup_{n} \Vert Tx_{n}-x_{n} \Vert =0\). As in [15], let g be a continuous convex function from C into \([0,\infty)\) defined by

$$g(x)= \lim_{n \longrightarrow\infty}\sup \Vert x_{n}-x \Vert $$

for all \(x\in C\).

Again, since C is weakly compact and g is weakly lower semi-continuous, there is \(z\in C\) such that

$$g(z)=\min\bigl\lbrace g(x):x\in C\bigr\rbrace . $$

Now, by Proposition 3.5(iii) (for \(\gamma=\frac{c}{2}\)),

$$\begin{aligned} \Vert x_{n}-Tz \Vert \leq{}& (3-2\gamma)\Vert x_{n}-Tx_{n} \Vert +(1-\gamma) \Vert x_{n}-z \Vert +\mu \bigl(2 \Vert x_{n}-Tx_{n} \Vert \\ &{}+ \Vert x_{n}-Tz \Vert + \Vert x_{n}-z \Vert + \Vert x_{n}-Tx_{n} \Vert +2 \Vert x_{n}-Tx_{n} \Vert \bigr). \end{aligned}$$

So,

$$\begin{aligned} & (1-\mu) \lim_{n\longrightarrow\infty}\sup \Vert x_{n}-Tz \Vert \leq(1-\gamma+\mu)\lim_{n\longrightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\quad \Rightarrow\quad \lim_{n\longrightarrow\infty}\sup \Vert x_{n}-Tz \Vert \leq\frac {(1-\gamma+\mu)}{(1-\mu)} \lim_{n\longrightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\quad\Rightarrow\quad g(Tz) \leq g(z). \end{aligned}$$

Since \(g(z)\) is the minimum, \(g(Tz)=g(z)\).

Now, if \(Tz \neq z\), then as g is strictly quasi-convex, we have \(g(z) \leq g(\lambda Tz+(1-\lambda)z)< \max\lbrace g(z), g(Tz)\rbrace =g(z)\), which is a contradiction.

Hence, \(Tz=z\). □

Next, we consider the Banach space X with the Opial property.

Theorem 3.11

Let C be a nonempty subset of a Banach space X having the Opial property. Let T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\). If \(\lbrace x_{n}\rbrace\) is a sequence in X such that

  1. (i)

    \(\lbrace x_{n} \rbrace\) converges weakly to z,

  2. (ii)

    \(\lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0\),

then \(Tz=z\).

Proof

By Proposition 3.5(ii), (for \(\gamma=\frac {c}{2},c\in[0,1]\))

$$\gamma \Vert x_{n}-Tx_{n} \Vert \leq \Vert x_{n}-z \Vert \leq \Vert x_{n}-z \Vert +\mu \Vert z-Tz \Vert . $$

So, by the condition \(B_{\gamma,\mu}\),

$$\begin{aligned} \Vert Tx_{n}-Tz \Vert &\leq(1-\gamma) \Vert x_{n}-z \Vert +\mu \bigl( \Vert x_{n}-Tz \Vert + \Vert z-Tx_{n} \Vert \bigr). \end{aligned}$$
(3.3)

Now,

$$\begin{aligned} \Vert x_{n}-Tz \Vert \leq{}& \Vert x_{n}-Tx_{n} \Vert + \Vert Tx_{n}-Tz \Vert \\ \leq{}& \Vert x_{n}-Tx_{n} \Vert +(1-\gamma) \Vert x_{n}-z \Vert \\ &{}+\mu \bigl( \Vert x_{n}-Tz \Vert + \Vert z-x_{n} \Vert + \Vert x_{n}-Tx_{n} \Vert \bigr) \quad\bigl(\text{by (3.3)} \bigr). \end{aligned}$$

So, taking limit as \(n\longrightarrow\infty\) and using (ii), we get

$$\begin{aligned} \Vert x_{n}-Tz \Vert &\leq\frac{1-\gamma +\mu}{1-\mu} \Vert x_{n}-z \Vert \\ &\leq \Vert x_{n}-z \Vert . \end{aligned}$$

So,

$$\begin{aligned} \lim_{n\longrightarrow\infty}\inf \Vert x_{n}-Tz \Vert \leq\lim _{n\longrightarrow\infty}\inf \Vert x_{n}-z \Vert . \end{aligned}$$
(3.4)

Let \(Tz\neq z\). Since \(x_{n}\longrightarrow z\) (weakly), by the Opial property, we have

$$\lim_{n\longrightarrow\infty}\inf \Vert x_{n}-z \Vert < \lim _{n\longrightarrow \infty}\inf \Vert x_{n}-Tz \Vert , $$

which is a contradiction to (3.4).

So, \(Tz=z\). □

Example 3.12

We consider the set \(C(\subset l^{p}, 1< p<\infty)\) where

$$C=\biggl\lbrace \lbrace x_{n}\rbrace\in l^{p}: \vert x_{1} \vert \leq\frac{1}{2}, x_{j}=0\ \forall j \neq1\biggr\rbrace . $$

Let \(\lbrace a_{n}\rbrace\) be a sequence in C such that

$$a_{1}=\biggl\lbrace \frac{1}{2},0,0,0\ldots\biggr\rbrace ,\qquad a_{2}=\biggl\lbrace \frac{2}{3},0,0,\ldots\biggr\rbrace \quad \cdots\quad a_{n+1}=\biggl\lbrace \frac{n}{n+1},0,0\ldots\biggr\rbrace . $$

Then \(\lbrace a_{n}\rbrace\) converges to \(z=\lbrace1,0,0\ldots\rbrace \).

Let \(T:C\longrightarrow C\) be such that \(T(\lbrace a_{n}\rbrace )=T(\lbrace x_{1},0,0\ldots\rbrace)=\lbrace x_{1}^{2},0,0\ldots\rbrace \).

Now, let \(X_{1}=\lbrace x_{1},0,0\ldots\rbrace\), \(Y_{1}=\lbrace y_{1},0,0\ldots\rbrace\in C\). Then

$$\begin{aligned} & \Vert TX_{1}-TY_{1} \Vert _{p} \\ &\quad = \bigl\Vert \bigl\lbrace x_{1}^{2}-y_{1}^{2},0,0, \ldots\bigr\rbrace \bigr\Vert _{p} \\ &\quad = \bigl\vert x_{1}^{2}-y_{1}^{2} \bigr\vert \leq\bigl( \vert x_{1} \vert + \vert y_{1} \vert \bigr) \vert x_{1}-y_{1} \vert \\ &\quad \leq \vert x_{1}-y_{1} \vert = \Vert X_{1}-Y_{1} \Vert _{p}. \end{aligned}$$

So, T is nonexpansive and hence satisfies the condition \(B_{\gamma,\mu}\).

Again,

$$\begin{aligned} & \lim_{n\longrightarrow\infty} \Vert TX_{n}-X_{n} \Vert _{p} \\ &\quad = \lim_{n\longrightarrow\infty} \biggl\Vert \biggl\lbrace \biggl( \frac{n}{n+1}\biggr)^{2}-\biggl(\frac {n}{n+1}\biggr),0,0\ldots \ldots\biggr\rbrace \biggr\Vert _{p} \\ &\quad = \lim_{n\longrightarrow\infty} \biggl\vert \biggl(\frac{n}{n+1} \biggr)^{2}-\biggl(\frac{n}{n+1}\biggr) \biggr\vert \\ &\quad =0. \end{aligned}$$

Thus, all the conditions of Theorem 3.11 are satisfied. Hence, \(Tz=z\). (Clearly, \(z=\lbrace1,0,0\ldots\rbrace\) is a fixed point of T.)

In the following result, we take C as a weakly compact and convex subset.

Theorem 3.13

Let C be a weakly compact convex subset of a Banach space X with the Opial property, T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\), and the sequence \(\lbrace x_{n}\rbrace\) in C be as defined in Proposition 3.6. Then \(\lbrace x_{n} \rbrace\) converges weakly to a fixed point of T.

Proof

By Proposition 3.6, \(\Vert Tx_{n}-x_{n} \Vert \longrightarrow0\) as \(n\longrightarrow\infty\).

Since C is weakly compact, there exists a subsequence \(\lbrace x_{n_{j}}\rbrace\) of \(\lbrace x_{n} \rbrace\) and \(z\in C\) such that \(\lbrace x_{n_{j}}\rbrace\) converges weakly to z. Now, by Theorem 3.11, z is a fixed point of T.

We assume that \(\lbrace x_{n} \rbrace\) does not converge weakly to z. Then there is a subsequence \(\lbrace x_{n_{k}}\rbrace\) of \(\lbrace x_{n}\rbrace\) and \(u\in C\) such that \(\lbrace x_{n_{k}}\rbrace\) converges weakly to u and \(u\neq z\). Again, \(Tu=u\) (by Theorem 3.11).

Now,

$$\begin{aligned} &\lim_{n\longrightarrow\infty}\inf \Vert x_{n}-z \Vert \\ &\quad =\lim _{n_{j}\longrightarrow\infty}\inf \Vert x_{n_{j}}-z \Vert \\ &\quad< \lim_{n_{j}\longrightarrow\infty}\inf \Vert x_{n_{j}}-u \Vert \quad (\text{by Opial property}) \\ &\quad=\lim_{n_{k}\longrightarrow\infty}\inf \Vert x_{n_{k}}-u \Vert \\ &\quad< \lim_{n_{k}\longrightarrow\infty}\inf \Vert x_{n_{k}}-z \Vert \\ &\quad=\lim_{n\longrightarrow\infty}\inf \Vert x_{n}-z \Vert , \end{aligned}$$

which is a contradiction.

So, \(\lbrace x_{n}\rbrace\) converges weakly to z. □

Example 3.14

We consider the same set C and the self-mapping T as in Example 3.12. For \(X_{0}=\lbrace\frac {1}{2},0,0\ldots\rbrace\in C\), let the sequence \(\lbrace X_{n} \rbrace\) in C be defined by

$$X_{n+1}=\lambda TX_{n}+(1-\lambda)X_{n}, \quad n\in \mathbb{N}\cup\lbrace 0\rbrace, \lambda\in[\gamma,1)-\lbrace0\rbrace, $$

which converges weakly to a fixed point of T by Theorem 3.13.

Now

$$\begin{aligned} &X_{1}=\biggl\lbrace \frac{\lambda}{2^{2}}+(1-\lambda) \frac {1}{2},0,0,\ldots\biggr\rbrace , \\ &X_{2}=\biggl\lbrace \lambda\biggl(\frac{\lambda}{2^{2}}+(1-\lambda) \frac {1}{2}\biggr)^{2}+(1-\lambda) \biggl(\frac{\lambda}{2^{2}}+(1- \lambda)\frac {1}{2}\biggr),0,0,\ldots\biggr\rbrace , \quad\text{etc}. \end{aligned}$$

Clearly, \(\lbrace X_{n} \rbrace\) converges to \(\lbrace 0,0,0,\ldots\rbrace\) which is a fixed point of T in C.

Similar result as in Theorem 3.13 can be obtained by taking C as a closed, convex, and bounded subset of a Banach space X with the Opial property.

Next we discuss the convergence of some iteration schemes to the fixed point of T.

For a Banach space X, the Mann iteration scheme (see [29]) is defined by: \(x_{0} \in X\), \(x_{n+1}=(1-\alpha_{n})x_{n}+\alpha_{n}Tx_{n}, n\in\mathbb{N}\cup \lbrace0\rbrace, 0\leq\alpha_{n}\leq1\), for each \(n, \sum\alpha _{n}=\infty\).

Using Theorem 3.8, we can see the convergence of the above iteration scheme to the fixed point of T.

Here, we discuss the convergence of the following type of extended Mann iteration scheme:

$$\begin{aligned} \begin{aligned} &x_{0} \in X, \\ &y_{n}=\alpha_{n} Tx_{n}+(1-\alpha_{n})x_{n}, \\ &x_{n+1}= \beta_{n}Ty_{n}+(1-\beta_{n})y_{n}, \\ &n\in\mathbb{N}\cup\lbrace0\rbrace, 0\leq\alpha_{n}, \beta_{n}\leq 1, \text{ for each } n, \text{ and } \sum \alpha_{n}=\infty, \sum\beta_{n} =\infty. \end{aligned} \end{aligned}$$
(3.5)

Lemma 3.15

Let C be a nonempty closed and convex subset of a Banach space X. Let T be a self-mapping on C satisfying the condition \(B_{\gamma,\mu}\). For \(x_{0}\in C\), let \(\lbrace x_{n}\rbrace\) be the sequence in C defined by the above iteration scheme (3.5). Then \(\lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert \) exists for all \(z\in F(T)\).

Proof

Let \(F(T)\neq\phi\), and let \(z\in F(T)\).

By Lemma 3.3, \(\Vert z-Tx_{n} \Vert \leq \Vert z-x_{n} \Vert \ \forall n\in\mathbb {N} \cup\lbrace0\rbrace\).

Now,

$$\begin{aligned} \Vert x_{n+1}-z \Vert ={}& \bigl\Vert \beta_{n}Ty_{n}+(1- \beta _{n})y_{n}-z \bigr\Vert \\ \leq{}&\beta_{n} \Vert Ty_{n}-z \Vert +(1- \beta_{n}) \Vert y_{n}-z \Vert \\ \leq{}&\beta_{n} \Vert y_{n}-z \Vert +(1- \beta_{n}) \Vert y_{n}-z \Vert \\ ={}& \Vert y_{n}-z \Vert \\ ={}& \bigl\Vert \alpha_{n} Tx_{n}+(1-\alpha_{n})x_{n}-z \bigr\Vert \\ \leq{}& \alpha_{n} \Vert x_{n}-z \Vert +(1- \alpha_{n}) \Vert x_{n}-z \Vert \\ ={}& \Vert x_{n}-z \Vert , \end{aligned}$$

i.e., \(\lbrace \Vert x_{n}-z \Vert \rbrace\) is a nonincreasing and bounded sequence.

Thus \(\lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert \) exists for all \(z\in F(T)\). □

Lemma 3.16

([2])

Let X be a uniformly convex Banach space. Let \(\lbrace\lambda_{n}\rbrace\) be a sequence of real numbers such that \(0< a \leq\lambda_{n}\leq b<1 \ \forall n\in \mathbb{N}\), and let \(\lbrace x_{n}\rbrace\) and \(\lbrace y_{n}\rbrace\) be sequences in X such that \(\lim_{n\longrightarrow\infty}\sup \Vert x_{n} \Vert \leq r, \lim_{n\longrightarrow\infty}\sup \Vert y_{n} \Vert \leq r\), and \(\lim_{n\longrightarrow\infty} \Vert \lambda_{n}x_{n}+(1-\lambda_{n})y_{n} \Vert =r\) for some \(r\geq0\). Then \(\lim_{n\longrightarrow\infty} \Vert x_{n}-y_{n} \Vert =0\).

Theorem 3.17

Let C be a nonempty closed convex subset of a uniformly convex Banach space X. Let T be a self-mapping on C satisfying \(B_{\gamma,\mu}\) condition. Let \(\lbrace x_{n}\rbrace\) be a sequence in C defined by the iteration scheme (3.5) where \(\alpha_{n},\beta_{n}\in(0,1)\). Then \(F(T)\neq\phi\) if and only if \(\lbrace x_{n}\rbrace\) is bounded and \(\lim_{n\longrightarrow \infty} \Vert Tx_{n}-x_{n} \Vert =0\).

Proof

Let \(F(T)\neq\phi\) and \(z\in F(T)\).

By Lemma 3.15, \(\lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert \) exists and \(\lbrace x_{n}\rbrace\) is bounded.

$$\begin{aligned} \lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert =p \quad (\text{say}). \end{aligned}$$
(3.6)

Now,

$$\begin{aligned} \Vert Ty_{n}-z \Vert \leq{}& \Vert y_{n}-z \Vert \quad (\text{by Lemma 3.3}) \\ \leq{}& \Vert x_{n}-z \Vert . \end{aligned}$$

So,

$$\begin{aligned} \lim_{n\longrightarrow\infty}\sup \Vert Ty_{n}-z \Vert \leq\lim_{n\longrightarrow\infty}\sup \Vert y_{n}-z \Vert \leq\lim _{n\longrightarrow \infty}\sup \Vert x_{n}-z \Vert =p. \end{aligned}$$
(3.7)

Also,

$$\begin{aligned} &\lim_{n\longrightarrow\infty} \bigl\Vert \beta _{n}(Ty_{n}-z)+(1-\beta_{n}) (y_{n}-z) \bigr\Vert \\ &\quad=\lim_{n\longrightarrow\infty} \Vert x_{n+1}-z \Vert =p. \end{aligned}$$

So, by Lemma 3.16, \(\lim_{n\longrightarrow\infty} \Vert Ty_{n}-y_{n} \Vert =0\).

Again, by Lemma 3.3 we have

$$\begin{aligned} \lim_{n\longrightarrow\infty}\sup \Vert Tx_{n}-z \Vert \leq\lim _{n\longrightarrow \infty}\sup \Vert x_{n}-z \Vert =p. \end{aligned}$$
(3.8)

Now,

$$\begin{aligned} &\Vert x_{n+1}-z \Vert \leq \beta_{n} \Vert Ty_{n}-z \Vert +(1-\beta_{n}) \Vert y_{n}-z \Vert \\ &\phantom{\Vert x_{n+1}-z \Vert }\leq\beta_{n} \Vert y_{n}-z \Vert +(1- \beta_{n}) \Vert y_{n}-z \Vert \\ &\phantom{\Vert x_{n+1}-z \Vert }= \Vert y_{n}-z \Vert \ \forall n\in\mathbb{N}\cup\lbrace0\rbrace \\ &\quad\Rightarrow\quad p\leq\lim_{n\longrightarrow\infty}\inf \Vert y_{n}-z \Vert \\ &\phantom{\quad\Rightarrow\quad p. }\leq \lim_{n\longrightarrow\infty}\sup \Vert y_{n}-z \Vert \leq p \quad\bigl(\text{using (3.7)}\bigr) \\ &\quad \Rightarrow\quad \lim_{n\longrightarrow\infty} \Vert y_{n}-z \Vert =p. \end{aligned}$$

So,

$$\begin{aligned} \lim_{n\longrightarrow\infty} \bigl\Vert \alpha _{n}(Tx_{n}-z)+(1- \alpha_{n}) (x_{n}-z) \bigr\Vert =\lim_{n\longrightarrow\infty} \Vert y_{n}-z \Vert =p. \end{aligned}$$
(3.9)

Using (3.6), (3.8), and (3.9) with Lemma 3.16, we get

$$\begin{aligned} &\lim_{n\longrightarrow\infty} \bigl\Vert (Tx_{n}-z)-(x_{n}-z) \bigr\Vert =0 \\ &\quad \Rightarrow\quad \lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0. \end{aligned}$$

Conversely, let \(\lbrace x_{n}\rbrace\) be bounded and \(\lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0\).

Let \(z\in A(C,\lbrace x_{n}\rbrace)\). By Proposition 3.5(iii), for, \(\gamma=\frac{c}{2}, c\in[0,1]\),

$$\begin{aligned} &\Vert x_{n}-Tz \Vert \leq (3-c) \Vert x_{n}-Tx_{n} \Vert +\biggl(1-\frac{c}{2}\biggr) \Vert x_{n}-z \Vert +\mu \bigl(2 \Vert x_{n}-Tx_{n} \Vert \\ &\phantom{\Vert x_{n}-Tz \Vert \leq}{}+ \Vert x_{n}-Tz \Vert + \Vert z-Tx_{n} \Vert +2 \bigl\Vert Tx_{n}-T^{2}x_{n} \bigr\Vert \bigr) \\ &\phantom{\Vert x_{n}-Tz \Vert }\leq (3-c) \Vert x_{n}-Tx_{n} \Vert +\biggl(1- \frac{c}{2}\biggr) \Vert x_{n}-z \Vert +\mu\bigl(2 \Vert x_{n}-Tx_{n} \Vert \\ &\phantom{\Vert x_{n}-Tz \Vert \leq}{}+ \Vert x_{n}-Tz \Vert + \Vert x_{n}-z \Vert + \Vert x_{n}-Tx_{n} \Vert +2 \Vert x_{n}-Tx_{n} \Vert \bigr) \\ &\phantom{\Vert x_{n}-Tz \Vert \leq} \bigl(\text{by Proposition 3.5(i)}\bigr) \\ &\quad \Rightarrow \quad(1-\mu) \lim_{n\rightarrow\infty}\sup \Vert x_{n}-Tz \Vert \leq \biggl(1-\frac{c}{2}+\mu\biggr)\lim_{n\rightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\quad \Rightarrow\quad \lim_{n\rightarrow\infty}\sup \Vert x_{n}-Tz \Vert \leq \frac {1-\frac{c}{2}+\mu}{1-\mu} \lim_{n\rightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\phantom{\quad \Rightarrow\quad\lim_{n\rightarrow\infty}\sup \Vert x_{n}-Tz \Vert .}\leq \lim_{n\rightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\phantom{\quad \Rightarrow\quad}\quad\biggl( \text{as } \frac{1-\frac{c}{2}+\mu}{1-\mu} \leq1, \text{ for } 2\mu\leq \gamma=\frac{c}{2} \biggr) \\ &\quad \Rightarrow\quad r\bigl(Tz,\lbrace x_{n}\rbrace\bigr) \leq r\bigl(z, \lbrace x_{n} \rbrace\bigr). \end{aligned}$$

So, \(Tz\in A(C,\lbrace x_{n}\rbrace)\).

Since X is uniformly convex, so \(Tz=z\), i.e., \(z\in F(T)\).

So, \(F(T)\neq\phi\). □

Example 3.18

Let \(T:[0,5]\longrightarrow[0,5]\) be defined by

$$T(x)= \textstyle\begin{cases} 0& \text{if } x\neq5,\\ 2& \text{if } x=5. \end{cases} $$

Then T satisfies the condition \(B_{\gamma,\mu}\).

For \(x_{0}=2\), we construct the sequence \(\lbrace x_{n} \rbrace\) by iteration scheme (3.5), where \(\alpha_{n}=\alpha\in(0,1)\) and \(\beta_{n} = \beta\in(0,1), n\in\mathbb{N}\cup\lbrace0\rbrace\).

Then we get \(x_{n}=2(1-\alpha)^{n}(1-\beta)^{n}, n\geq1\). Clearly \(\lbrace x_{n}\rbrace\) is bounded.

$$\begin{aligned} \text{Now,} \lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert &=\lim_{n\longrightarrow\infty} \bigl\Vert T\bigl(2(1- \alpha)^{n}(1-\beta)^{n}\bigr)-2(1-\alpha )^{n}(1- \beta)^{n} \bigr\Vert \\ &=\lim_{n\longrightarrow\infty} \bigl\vert 2(1-\alpha) (1-\beta) \bigr\vert \\ &=0. \end{aligned}$$

Thus, by Theorem 3.17, \(F(T)\neq\phi\). Clearly, \(0\in F(T)\) in this case.

Next we consider the following type of extended Picard–Mann hybrid iteration scheme:

$$\begin{aligned} \begin{aligned} &x_{0}\in X, \\ &y_{n}=(1-b_{n})x_{n}+b_{n}Tx_{n}, \\ &z_{n}=(1-a_{n})x_{n}+a_{n}Ty_{n}, \\ &x_{n+1}=Tz_{n}, \\ &0\leq a_{n},b_{n}\leq1 \text{ for each } n, \quad \sum _{n} a_{n}=\infty,\qquad \sum_{n} b_{n}=\infty. \end{aligned} \end{aligned}$$
(3.10)

Lemma 3.19

Let T be a self-mapping on a nonempty closed and convex subset C of a Banach space X. Let T satisfy the condition \(B_{\gamma,\mu}\) on C. For \(x_{0}\in C\), we define a sequence \(\lbrace x_{n} \rbrace\) in C by the iteration scheme (3.10), where \(0< a_{n},b_{n}<1\). Then \(\lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert \) exists for all \(z\in F(T)\).

Proof

Similar as in Lemma 3.15. □

Theorem 3.20

Let T be a self-mapping on a nonempty closed and convex subset C of a uniformly convex Banach space X. Let T satisfy the condition \(B_{\gamma,\mu}\) on C. Let \(\lbrace x_{n} \rbrace\) be a sequence in C defined by the iteration scheme (3.10), where \(0< a_{n},b_{n}<1\) and \(\lim_{n\longrightarrow\infty } a_{n}=k\ (\neq0)\). Then \(F(T)\neq\phi\) if and only if \(\lbrace x_{n}\rbrace\) is bounded and \(\lim_{n\rightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0\).

Proof

Let \(F(T)\neq\phi\) and \(z\in F(T)\). Then, by Lemma 3.19, \(\lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert \) exists and \(\lbrace x_{n}\rbrace\) is bounded.

$$\begin{aligned} \lim_{n\longrightarrow\infty} \Vert x_{n}-z \Vert = p \quad (\text{say}). \end{aligned}$$
(3.11)

As in Theorem 3.17, we have

$$\begin{aligned} \Vert Ty_{n}-z \Vert \leq \Vert y_{n}-z \Vert \leq \Vert x_{n}-z \Vert . \end{aligned}$$

So,

$$\begin{aligned} \lim_{n\longrightarrow\infty}\sup \Vert Ty_{n}-z \Vert \leq\lim _{n\longrightarrow \infty}\sup \Vert y_{n}-z \Vert \leq\lim _{n\longrightarrow\infty}\sup \Vert x_{n}-z \Vert = p. \end{aligned}$$
(3.12)

Again,

$$\begin{aligned} &\Vert x_{n+1}-z \Vert \\ &\quad = \Vert Tz_{n}-z \Vert \leq \Vert z_{n}-z \Vert \\ &\quad = \bigl\Vert (1-a_{n})x_{n}+a_{n}Ty_{n}-z \bigr\Vert \\ &\quad \leq a_{n} \Vert Ty_{n}-z \Vert +(1-a_{n}) \Vert x_{n}-z \Vert \\ &\quad \leq a_{n} \Vert y_{n}-z \Vert +(1-a_{n}) \Vert x_{n}-z \Vert \\ &\quad = a_{n} \bigl\Vert (1-b_{n})x_{n}+b_{n}Tx_{n}-z \bigr\Vert +(1-a_{n}) \Vert x_{n}-z \Vert \\ &\quad \leq a_{n}\bigl(b_{n} \Vert Tx_{n}-z \Vert +(1-b_{n}) \Vert x_{n}-z \Vert \bigr)+(1-a_{n}) \Vert x_{n}-z \Vert \\ &\quad \leq a_{n}b_{n} \Vert x_{n}-z \Vert +a_{n} \Vert x_{n}-z \Vert -a_{n}b_{n} \Vert x_{n}-z \Vert + \Vert x_{n}-z \Vert -a_{n} \Vert x_{n}-z \Vert \\ &\quad = \Vert x_{n}-z \Vert . \end{aligned}$$
(3.13)

Thus,

$$\begin{aligned} \Vert x_{n+1}-z \Vert \leq \Vert x_{n}-z \Vert . \end{aligned}$$
(3.14)

Again, by Lemma 3.3, we have

$$\begin{aligned} &\Vert Tx_{n}-z \Vert \leq \Vert x_{n}-z \Vert \quad \text{for all } n\in\mathbb {N}\cup\lbrace0\rbrace \\ &\quad \Rightarrow\quad \lim_{n\longrightarrow\infty}\sup \Vert Tx_{n}-z \Vert \leq\lim_{n\longrightarrow\infty}\sup \Vert x_{n}-z \Vert =p. \end{aligned}$$
(3.15)

Now, from equation (3.13), we have

$$\begin{aligned} &\Vert x_{n+1}-z \Vert \leq a_{n} \Vert y_{n}-z \Vert +(1-a_{n}) \Vert x_{n}-z \Vert \\ &\quad\Rightarrow\quad \Vert x_{n+1}-z \Vert - (1-a_{n}) \Vert x_{n}-z \Vert \leq a_{n} \Vert y_{n}-z \Vert \\ &\quad \Rightarrow\quad p-(1-k)p\leq k\lim_{n\longrightarrow\infty}\inf \Vert y_{n}-z \Vert \\ &\quad \Rightarrow\quad p\leq\lim_{n\longrightarrow\infty}\inf \Vert y_{n}-z \Vert \lim_{n\longrightarrow\infty}\sup \Vert y_{n}-z \Vert \leq p \quad \bigl(\text{by (3.12)}\bigr) \\ &\quad \Rightarrow\quad \lim_{n\longrightarrow\infty} \Vert y_{n}-z \Vert =p. \end{aligned}$$

Thus,

$$\begin{aligned} \lim_{n\longrightarrow\infty}& \bigl\Vert b_{n}(Tx_{n}-z)+(1-b_{n}) (x_{n}-z) \bigr\Vert =\lim_{n\longrightarrow\infty} \Vert y_{n}-z \Vert =p. \end{aligned}$$
(3.16)

By (3.11), (3.15), (3.16) and Lemma 3.16, we have \(\lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0\).

For the converse part, let \(\lbrace x_{n}\rbrace\) be bounded and \(\lim_{n\longrightarrow\infty} \Vert Tx_{n}-x_{n} \Vert =0\).

Let \(z\in A(C,\lbrace x_{n}\rbrace)\). Then, using Proposition 3.5(iii) and following Theorem 3.17, for \(\gamma=\frac{c}{2}, c\in[0,1]\), we have

$$\begin{aligned} &\lim_{n\rightarrow\infty}\sup \Vert x_{n}-Tz \Vert \leq \frac{1-\frac {c}{2}+\mu}{1-\mu} \lim_{n\rightarrow\infty}\sup \Vert x_{n}-z \Vert \\ &\phantom{\lim_{n\rightarrow\infty}\sup \Vert x_{n}-Tz \Vert .}\leq\lim_{n\rightarrow\infty}\sup \Vert x_{n}-z \Vert \quad\biggl(\text{as } \frac{1-\frac{c}{2}+\mu}{1-\mu} \leq1, \text{ for } 2\mu \leq \gamma=\frac{c}{2} \biggr) \\ &\quad \Rightarrow \quad r\bigl(Tz,\lbrace x_{n}\rbrace\bigr) \leq r\bigl(z, \lbrace x_{n} \rbrace\bigr). \end{aligned}$$

Hence, \(Tz\in A(C,\lbrace x_{n}\rbrace)\).

X being uniformly convex, \(Tz=z\), i.e., \(z\in F(T)\).

Hence, \(F(T)\neq\phi\). □

Example 3.21

We consider the same set C with the self-mapping T as in Example 3.12.

For \(X_{0}= \lbrace\frac{1}{3},0,0,0\ldots\rbrace\in C\), we construct the sequence \(\lbrace X_{n}\rbrace\) in C by the iteration scheme (3.10).

Let \(a_{n}=a\) and \(b_{n}=b \ \forall n\in\mathbb{N}\cup \lbrace0\rbrace\). Then

$$\begin{aligned} &Y_{0}=(1-b)\biggl\lbrace \frac{1}{3},0,0,0\ldots\biggr\rbrace +b\biggl\lbrace \frac {1}{3^{2}},0,0,0\ldots\biggr\rbrace , \\ &Z_{0}=(1-a)\biggl\lbrace \frac{1}{3},0,0,0\ldots\biggr\rbrace +a\biggl\lbrace \biggl((1-b)\frac {1}{3}+b\frac{1}{3^{2}} \biggr)^{2},0,0,0\ldots\biggr\rbrace , \\ &X_{1}= \biggl\lbrace \biggl((1-a)\frac{1}{3}+a \biggl((1-b) \frac{1}{3}+b\frac {1}{3^{2}} \biggr)^{2} \biggr)^{2},0,0, \ldots\biggr\rbrace , \quad\text{and so on.} \end{aligned}$$

Thus, it can be seen that \(\lbrace X_{n} \rbrace\) is bounded and \(\lim_{n\longrightarrow\infty} \Vert TX_{n}-X_{n} \Vert =0\). Hence, by Theorem 3.20, \(F(T)\neq\phi\) (here, clearly, \(\lbrace0,0,\ldots\rbrace\in F(T)\)).

4 Conclusion

Throughout the paper, we have discussed some fixed point results for the class of mappings with \(B_{\gamma,\mu}\) condition. In 2010, Harandi and Emami (see [7]) studied some fixed point theorems for contraction type mappings in partially ordered metric spaces with applications in solving ordinary differential equations. In this context, the study of fixed point theory in partially ordered metric spaces for the class of mappings with \(B_{\gamma,\mu}\) condition with different practical applications is a scope for future study.