In this section, we introduce the concepts of \((\alpha, \beta )\)-generalized convex contraction and \((\alpha, \beta)\)-generalized convex contraction of order 2 and prove the approximate fixed point theorems for such mappings.

### Definition 3.1

Let \((X, d)\) be a metric space. The mapping \(T:X \rightarrow X\) is called an \((\alpha, \beta )\)
*-generalized convex contraction* if there exist mappings \(\alpha, \beta: X \rightarrow[0,\infty)\) and \(a,b \in[0,\infty)\) with \(a+b<1\), satisfying the following condition:

$$ \mbox{for all } x, y \in X,\quad \alpha(x)\beta(y)\geq1 \quad \Longrightarrow\quad d \bigl(T^{2}x,T^{2}y \bigr)\leq ad(Tx,Ty)+bd(x,y). $$

(3.1)

Now, we establish a new approximate fixed point theorem for \((\alpha, \beta)\)-generalized convex contraction mappings in complete metric spaces.

### Theorem 3.2

*Let*
\((X,d)\)
*be a metric space and*
\(T : X \rightarrow X\)
*be an*
\((\alpha , \beta)\)-*generalized convex contraction mapping*. *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0}) \geq1\)
*and*
\(\beta(x_{0}) \geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Proof

First of all, let \(x_{0}\in X\) such that \(\alpha(x_{0}) \geq1\) and \(\beta (x_{0}) \geq1\). Define the sequence \(\{x_{n}\}\) in *X* by \(x_{n+1}=T^{n+1}x_{0}\) for all \(n \in\mathbb{Z}^{+}\cup\{0\}\). If \(x_{n'}=x_{n'+1}\) for some \(n' \in\mathbb{Z}^{+}\cup\{0\}\), then \(x_{n'}\) is a fixed point of *T*. So, we assume that \(x_{n}\neq x_{n+1}\) for all \(n \in\mathbb{Z}^{+} \cup \{0\}\). By the definition of a cyclic \((\alpha,\beta)\)-admissible mapping, we have

$$\begin{aligned}& \alpha(x_{0}) \geq1 \quad\Longrightarrow\quad\beta(x_{1}) \geq1 \quad\Longrightarrow\quad \alpha(x_{2}) \geq 1 \quad\Longrightarrow\quad \cdots, \\& \beta(x_{0}) \geq1 \quad\Longrightarrow\quad\alpha(x_{1}) \geq1 \quad\Longrightarrow\quad \beta(x_{2}) \geq 1 \quad\Longrightarrow\quad \cdots. \end{aligned}$$

Therefore, \(\alpha(x_{n}) \geq1\) and \(\beta(x_{n}) \geq1\) for all \(n \in\mathbb{Z}^{+} \cup\{0\}\). Let us denote

$$v:=d \bigl(Tx_{0},T^{2}x_{0} \bigr)+d(x_{0},Tx_{0}) $$

and

Since \(\alpha(x_{0})\beta(x_{1}) \geq1\), we get

$$\begin{aligned} d \bigl(T^{2}x_{0},T^{3}x_{0} \bigr) \leq& ad \bigl(Tx_{0},T^{2}x_{0} \bigr)+bd(x_{0},Tx_{0}) \\ =& \gamma v. \end{aligned}$$

Also, since \(\alpha(x_{1})\beta(x_{2}) \geq1\), we get

$$\begin{aligned} d \bigl(T^{3}x_{0},T^{4}x_{0} \bigr) \leq& ad \bigl(T^{2}x_{0},T^{3}x_{0} \bigr)+bd \bigl(Tx_{0},T^{2}x_{0} \bigr) \\ \leq& ad \bigl(Tx_{0},T^{2}x_{0} \bigr)+bd(x_{0},Tx_{0})+bd \bigl(Tx_{0},T^{2}x_{0} \bigr) \\ \leq& ad(x_{0},Tx_{0})+ad \bigl(Tx_{0},T^{2}x_{0} \bigr)+bd(x_{0},Tx_{0})+bd \bigl(Tx_{0},T^{2}x_{0} \bigr) \\ =& \gamma v. \end{aligned}$$

By continuing this process, we get

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr) \leq \textstyle\begin{cases} \gamma^{(m-1)/2} v &\mbox{if } m = 1,3,5,\ldots,\\ \gamma^{m/2} v &\mbox{if } m = 2,4,6,\ldots. \end{cases}\displaystyle \end{aligned}$$

Thus it follows that \(d(T^{m}x_{0},T^{m+1}x_{0})\rightarrow0\) as \(m \rightarrow\infty\). So, *T* is an asymptotically regular at a point \(x_{0}\in X\). By using Lemma 2.7, we see that *T* has the approximate fixed point property.

Next, we will show that *T* has a fixed point provided that \((X,d)\) is a complete metric space and *T* is continuous. Now, we will prove that \(\{x_{n}\}\) is a Cauchy sequence in *X*. Without loss of generality, we may assume that \(m,n \in\mathbb{Z}^{+}\) such that \(n>m>1\). We distinguish two cases as follows.

*Case* 1: Let *m* be an even number. That is, \(m=2l\), where \(l\in\mathbb{Z}^{+}\). Therefore,

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq& d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr)+d \bigl(T^{m+1}x_{0},T^{m+2}x_{0} \bigr)+ \cdots +d \bigl(T^{n-1}x_{0},T^{n}x_{0} \bigr) \\ \leq& \gamma^{l}v+\gamma^{l}v+\gamma^{l+1}v+ \gamma^{l+1}v+\gamma ^{l+2}v+\cdots \\ \leq& 2\gamma^{l}v+2\gamma^{l+1}v+2\gamma^{l+2}v+ \cdots \\ =& \frac{2\gamma^{l}v}{1-\gamma}. \end{aligned}$$

*Case* 2: Let *m* be an odd number. That is, \(m=2l+1\), where \(l\in\mathbb{Z}^{+}\). Therefore,

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq& d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr)+d \bigl(T^{m+1}x_{0},T^{m+2}x_{0} \bigr)+ \cdots +d \bigl(T^{n-1}x_{0},T^{n}x_{0} \bigr) \\ \leq& \gamma^{l}v+\gamma^{l+1}v+\gamma^{l+1}v+ \gamma^{l+2}v+\gamma ^{l+2}v+\cdots \\ \leq& 2\gamma^{l}v+2\gamma^{l+1}v+2\gamma^{l+2}v+ \cdots \\ =& \frac{2\gamma^{l}v}{1-\gamma}. \end{aligned}$$

It follows that

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq \textstyle\begin{cases} \frac{2\gamma^{m/2}v}{1-\gamma} &\mbox{if } m = 2,4,6,\ldots,\\ \frac{2\gamma^{(m-1)/2}v}{1-\gamma} &\mbox{if } m = 3,5,7,\ldots. \end{cases}\displaystyle \end{aligned}$$

Therefore, \(d(T^{m}x_{0},T^{n}x_{0}) \rightarrow0\) as \(m,n\rightarrow\infty\), that is, \(\{x_{n}\}\) is a Cauchy sequence in *X*. By using the completeness of *X*, there exists \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) as \(n\rightarrow\infty\). Since *T* is continuous, we obtain

$$x^{*}=\lim_{n\rightarrow\infty} x_{n+1}=\lim_{n\rightarrow\infty}Tx_{n} =Tx^{*} $$

and thus *T* has a fixed point. This completes the proof. □

Next we give an example to illustrate the usability of Theorem 3.2.

### Example 3.3

Let \(X=[1,\infty)\) and \(d:X \times X \rightarrow\mathbb{R}\) be defined by \(d(x,y)=|x-y|\) for all \(x,y\in X\). Define \(T:X \rightarrow X\) and \(\alpha, \beta: X \rightarrow[0,\infty )\) by

$$\begin{aligned}& Tx= \textstyle\begin{cases} \frac{x+6}{4}, & \mbox{if } x \in[1,2],\\ \sqrt{2x}, & \mbox{if } x \in(2,8),\\ x^{2}-8x+4, & \mbox{otherwise}, \end{cases}\displaystyle \qquad \alpha(x)= \textstyle\begin{cases} \frac{x+2}{2} ,& \mbox{if } x \in[1,2],\\ 0 ,& \mbox{otherwise}, \end{cases}\displaystyle \end{aligned}$$

and

$$\begin{aligned} \beta(x)= \textstyle\begin{cases} \frac{x+4}{3} ,& \mbox{if } x \in[1,2],\\ 0 ,& \mbox{otherwise}. \end{cases}\displaystyle \end{aligned}$$

It is easy to see that \((X,d)\) is a complete metric space and *T* is continuous.

Now, we show that Theorem 3.2 can guarantee the existence of fixed point of *T*. First of all, we will show that *T* is an \((\alpha, \beta )\)-generalized convex contraction mapping with \(a=\frac{2}{3}\) and \(b=\frac{1}{9}\).

For \(x,y \in X\) with \(\alpha(x)\beta(y) \geq1\), we have \(x,y \in [1,2]\) and thus

$$\begin{aligned} d \bigl(T^{2}x,T^{2}y \bigr) =& \biggl| T \biggl( \frac{x+6}{4} \biggr) - T \biggl(\frac {y+6}{4} \biggr) \biggr| \\ =& \biggl| \frac{x+30}{16}- \frac{y+30}{16} \biggr| \\ =& \frac{1}{16} |x-y| \\ \leq& \frac{5}{18} |x-y| \\ =& \frac{2}{3} \biggl| \frac{x+3}{4}- \frac{y+3}{4} \biggr| + \frac {1}{9} |x-y| \\ =& ad(Tx,Ty)+bd(x,y). \end{aligned}$$

This shows that *T* is an \((\alpha, \beta)\)-generalized convex contraction mapping with \(a=\frac{2}{3}\) and \(b=\frac{1}{9}\). Clearly, *T* is a cyclic \((\alpha,\beta)\)-admissible mapping. It is easy to see that there is \(x_{0} = 1 \in X\) such that

$$\begin{aligned} \alpha(x_{0}) = \alpha(1) =1.5 \geq1 \quad\mbox{and} \quad \beta(x_{0}) = \beta(1) =5/3 \geq1. \end{aligned}$$

(3.2)

By using Theorem 3.2, we see that *T* has a fixed point in *X*.

### Corollary 3.4

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$ \alpha(x)\beta(y)d \bigl(T^{2}x,T^{2}y \bigr) \leq ad(Tx,Ty)+bd(x,y) $$

(3.3)

*for all*
\(x,y \in X\), *where*
\(a,b \in[0,1)\)
*with*
\(a+b<1\). *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0}) \geq1\)
*and*
\(\beta (x_{0})\geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Proof

We will show that *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. Suppose that \(x,y \in X\) with \(\alpha(x)\beta(y) \geq1\) and then

$$\begin{aligned} d \bigl(T^{2}x,T^{2}y \bigr) \leq& \alpha(x)\beta(y)d \bigl(T^{2}x,T^{2}y \bigr) \\ \leq& ad(Tx,Ty)+bd(x,y). \end{aligned}$$

This implies that *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. By Theorem 3.2, we get the desired result. □

### Corollary 3.5

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$\begin{aligned} \bigl[d \bigl(T^{2}x,T^{2}y \bigr)+\tau \bigr]^{\alpha(x)\beta(y)} \leq ad(Tx,Ty)+bd(x,y)+\tau \end{aligned}$$

(3.4)

*for all*
\(x,y \in X\), *where*
\(a,b \in[0,1)\)
*with*
\(a+b<1\)
*and*
\(\tau\geq1\). *Assume that*
*T*
*is an*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0})\geq1\)
*and*
\(\beta(Tx_{0})\geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Proof

We will show that *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. Suppose that \(x,y \in X\) with \(\alpha(x)\beta(y) \geq1\) and hence

$$\begin{aligned} d \bigl(T^{2}x,T^{2}y \bigr)+\tau \leq& \bigl[d \bigl(T^{2}x,T^{2}y \bigr)+\tau \bigr]^{\alpha(x)\beta(y)} \\ \leq& ad(Tx,Ty)+bd(x,y)+\tau. \end{aligned}$$

This implies that

$$d \bigl(T^{2}x,T^{2}y \bigr) \leq ad(Tx,Ty)+bd(x,y), $$

that is, *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. By Theorem 3.2, we get the desired result. □

### Corollary 3.6

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$\begin{aligned} \bigl[\tau-1+\alpha(x)\beta(y) \bigr]^{d(T^{2}x,T^{2}y)} \leq\tau^{ad(Tx,Ty)+bd(x,y)} \end{aligned}$$

(3.5)

*for all*
\(x,y \in X\), *where*
\(a,b \in[0,1)\)
*with*
\(a+b<1\)
*and*
\(\tau>1\). *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0})\beta(Tx_{0})\geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Proof

We will show that *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. Suppose that \(x,y \in X\) with \(\alpha(x)\beta(y) \geq1\) and hence

$$\begin{aligned} \tau^{d(T^{2}x,T^{2}y)} \leq& \bigl[\tau-1+\alpha(x)\beta(y) \bigr]^{d(T^{2}x,T^{2}y)} \\ \leq& \tau^{ad(Tx,Ty)+bd(x,y)}. \end{aligned}$$

This implies that

$$d \bigl(T^{2}x,T^{2}y \bigr) \leq ad(Tx,Ty)+bd(x,y), $$

that is, *T* is an \((\alpha,\beta)\)-generalized convex contraction mapping. By Theorem 3.2, we get the desired result. □

We observe that the Banach contractive condition due to Banach [1] implies the contractive condition (3.1) whenever \(\alpha,\beta:X \rightarrow[0,\infty)\) defined by \(\alpha(x) = \beta(x) = 1\) for all \(x\in X\). From the previous observation, we get the following result.

### Corollary 3.7

*Let*
\((X,d)\)
*be a metric space and*
\(T : X \rightarrow X\)
*be a Banach contraction mapping*, *i*.*e*., *there exists*
\(k \in[0,1)\)
*such that*

$$\begin{aligned} d(Tx,Ty) \leq kd(x,y) \end{aligned}$$

(3.6)

*for all*
\(x,y \in X\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

Next, we introduce the concept of an \((\alpha, \beta)\)-generalized convex contraction of order 2 and also establish a new approximate fixed point theorem for such mappings in complete metric spaces.

### Definition 3.8

Let \((X, d)\) be a metric space. The mapping \(T:X \rightarrow X\) is called an \((\alpha, \beta )\)
*-generalized convex contraction of order 2* if there exist mappings \(\alpha, \beta: X \rightarrow[0,\infty)\) and \(a_{1},a_{2},b_{1},b_{2}\in[0,1)\) with \(a_{1}+a_{2}+b_{1}+b_{2}<1\), satisfying the following condition:

$$\begin{aligned} &\mbox{for all } x, y \in X, \quad\alpha(x)\beta(y)\geq1 \\ &\quad\Longrightarrow \quad d \bigl(T^{2}x,T^{2}y \bigr)\leq a_{1}d(x,Tx)+a_{2}d \bigl(Tx,T^{2}x \bigr)+b_{1}d(y,Ty)+b_{2}d \bigl(Ty,T^{2}y \bigr). \end{aligned}$$

(3.7)

### Theorem 3.9

*Let*
\((X,d)\)
*be a metric space and*
\(T : X \rightarrow X\)
*be an*
\((\alpha , \beta)\)-*generalized convex contraction mapping of order* 2. *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0}) \geq1\)
*and*
\(\beta(x_{0}) \geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Proof

First of all, let \(x_{0}\in X\) such that \(\alpha(x_{0}) \geq1\) and \(\beta (x_{0}) \geq1\). Define the sequence \(\{x_{n}\}\) in *X* by \(x_{n+1}=T^{n+1}x_{0}\) for all \(n \in\mathbb{Z}^{+} \cup\{0\}\). If \(x_{n'}=x_{n'+1}\) for some \(n' \in\mathbb{Z}^{+}\cup\{0\}\), then \(x_{n'}\) is a fixed point of *T*. So, we may assume that \(x_{n}\neq x_{n+1}\) for all \(n \in\mathbb{Z}^{+} \cup\{0\}\). It follows from *T* being a cyclic \((\alpha,\beta)\)-admissible mapping that

$$\begin{aligned}& \alpha(x_{0}) \geq1\quad \Longrightarrow\quad\beta(x_{1}) \geq1 \quad\Longrightarrow\quad \alpha(x_{2}) \geq 1 \quad\Longrightarrow\quad \cdots, \\& \beta(x_{0}) \geq1\quad\Longrightarrow\quad\alpha(x_{1}) \geq1 \quad\Longrightarrow\quad \beta(x_{2}) \geq 1 \quad\Longrightarrow\quad \cdots. \end{aligned}$$

Therefore, \(\alpha(x_{n}) \geq1\) and \(\beta(x_{n}) \geq1\) for all \(n \in\mathbb{Z}^{+} \cup\{0\}\). Let us denote

$$\begin{aligned}& w:=d \bigl(Tx_{0},T^{2}x_{0} \bigr)+d(x_{0},Tx_{0}), \\& \delta:=1-b_{2}, \end{aligned}$$

and

$$\epsilon:=a_{1}+a_{2}+b_{1}. $$

Since \(\alpha(x_{0})\beta(x_{1}) \geq1\), we get

$$\begin{aligned} d \bigl(T^{2}x_{0},T^{3}x_{0} \bigr) \leq& a_{1}d(x_{0},Tx_{0})+a_{2}d \bigl(Tx_{0},T^{2}x_{0} \bigr)+b_{1}d \bigl(Tx_{0},T^{2}x_{0} \bigr)+b_{2}d \bigl(T^{3}x_{0},T^{2}x_{0} \bigr) \\ \leq& a_{1}w+(a_{2}+b_{1})w+b_{2}d \bigl(T^{3}x_{0},T^{2}x_{0} \bigr). \end{aligned}$$

This implies that \(d(T^{2}x_{0},T^{3}x_{0})\leq\frac{\epsilon}{\delta}w\). Also, since \(\alpha(x_{1})\beta(x_{2}) \geq1\), we get

$$\begin{aligned} d \bigl(T^{3}x_{0},T^{4}x_{0} \bigr) \leq& a_{1}d \bigl(Tx_{0},T^{2}x_{0} \bigr)+a_{2}d \bigl(T^{2}x_{0},T^{3}x_{0} \bigr)\\ &{}+b_{1}d \bigl(T^{2}x_{0},T^{3}x_{0} \bigr)+b_{2}d \bigl(T^{3}x_{0},T^{4}x_{0} \bigr) \\ \leq& a_{1}w+(a_{2}+b_{1})\frac{\epsilon}{\delta}w+b_{2}d \bigl(T^{3}x_{0},T^{4}x_{0} \bigr) \\ \leq& a_{1}w+(a_{2}+b_{1})w+b_{2}d \bigl(T^{3}x_{0},T^{2}x_{0} \bigr). \end{aligned}$$

It means that \(d(T^{3}x_{0},T^{4}x_{0})\leq\frac{\epsilon}{\delta}w\). By continuing this process, we get

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr) \leq \textstyle\begin{cases} (\frac{\epsilon}{\delta} )^{(m-1)/2} v &\mbox{if } m = 1,3,5,\ldots,\\ (\frac{\epsilon}{\delta} )^{m/2} v &\mbox{if } m = 2,4,6,\ldots. \end{cases}\displaystyle \end{aligned}$$

This follows that \(d(T^{m}x_{0},T^{m+1}x_{0}) \rightarrow0\) as \(m \rightarrow\infty\) and thus *T* is an asymptotically regular at a point \(x_{0}\in X\). By using Lemma 2.7, we see that *T* has the approximate fixed point property.

Next, we show that *T* has a fixed point provided that \((X,d)\) is a complete metric space and *T* is continuous. First, we will claim that \(\{x_{n}\}\) is a Cauchy sequence in *X*. Let \(m,n \in\mathbb{Z}^{+}\) such that \(n>m>1\). Now, we distinguish the following cases.

*Case* 1: *m* is even number such that \(m=2l\), where \(l\in \mathbb{Z}^{+}\). Therefore,

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq& d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr)+d \bigl(T^{m+1}x_{0},T^{m+2}x_{0} \bigr)+ \cdots +d \bigl(T^{n-1}x_{0},T^{n}x_{0} \bigr) \\ \leq& \biggl(\frac{\epsilon}{\delta} \biggr)^{l}w+ \biggl( \frac{\epsilon}{\delta } \biggr)^{l}w+ \biggl(\frac{\epsilon}{\delta} \biggr)^{l+1}w+ \biggl(\frac{\epsilon}{\delta } \biggr)^{l+1}w+ \biggl( \frac{\epsilon}{\delta} \biggr)^{l+2}w+\cdots \\ \leq& 2 \biggl(\frac{\epsilon}{\delta} \biggr)^{l}w+2 \biggl( \frac{\epsilon}{\delta } \biggr)^{l+1}w+2 \biggl(\frac{\epsilon}{\delta} \biggr)^{l+2}w+\cdots \\ =& \frac{2(\frac{\epsilon}{\delta})^{l}w}{1-\frac{\epsilon}{\delta}}. \end{aligned}$$

*Case* 2: *m* is odd number such that \(m=2l+1\), where \(l\in \mathbb{Z}^{+}\). Therefore,

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq& d \bigl(T^{m}x_{0},T^{m+1}x_{0} \bigr)+d \bigl(T^{m+1}x_{0},T^{m+2}x_{0} \bigr)+ \cdots +d \bigl(T^{n-1}x_{0},T^{n}x_{0} \bigr) \\ \leq& \biggl(\frac{\epsilon}{\delta} \biggr)^{l}w+ \biggl( \frac{\epsilon}{\delta } \biggr)^{l+1}w+ \biggl(\frac{\epsilon}{\delta} \biggr)^{l+1}w+ \biggl(\frac{\epsilon}{\delta } \biggr)^{l+2}w+ \biggl( \frac{\epsilon}{\delta} \biggr)^{l+2}w+\cdots \\ \leq& 2 \biggl(\frac{\epsilon}{\delta} \biggr)^{l}w+2 \biggl( \frac{\epsilon}{\delta } \biggr)^{l+1}w+2 \biggl(\frac{\epsilon}{\delta} \biggr)^{l+2}w+\cdots \\ =& \frac{2(\frac{\epsilon}{\delta})^{l}w}{1-\frac{\epsilon}{\delta}}. \end{aligned}$$

Therefore, we can conclude that

$$\begin{aligned} d \bigl(T^{m}x_{0},T^{n}x_{0} \bigr) \leq \textstyle\begin{cases} \frac{2(\frac{\epsilon}{\delta})^{m/2}w}{1-\frac{\epsilon}{\delta}} &\mbox{if } m = 2,4,6,\ldots, \\ \frac{2(\frac{\epsilon}{\delta})^{(m-1)/2}w}{1-\frac{\epsilon}{\delta }} &\mbox{if } m = 3,5,7,\ldots. \end{cases}\displaystyle \end{aligned}$$

This implies that \(\{x_{n}\}\) is a Cauchy sequence in *X*. By the completeness of *X*, there exists \(x^{*}\in X\) such that \(x_{n}\rightarrow x^{*}\) as \(n\rightarrow\infty\). As follows from *T* being continuous, we get \(x_{n+1}=Tx_{n}\rightarrow Tx^{*}\) as \(n \rightarrow\infty\). By the uniqueness of the limit \(\{x_{n}\}\), we obtain \(Tx^{*}=x^{*}\) and thus *T* has a fixed point. This completes the proof. □

### Corollary 3.10

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$ \alpha(x)\beta(y)d \bigl(T^{2}x,T^{2}y \bigr) \leq a_{1}d(x,Tx)+a_{2}d \bigl(Tx,T^{2}x \bigr)+b_{1}d(y,Ty)+b_{2}d \bigl(Ty,T^{2}y \bigr) $$

(3.8)

*for all*
\(x,y \in X\), *where*
\(a_{1},a_{2},b_{1},b_{2}\in[0,1)\)
*with*
\(a_{1}+a_{2}+b_{1}+b_{2}<1\). *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0})\geq1 \)
*and*
\(\beta(x_{0}) \geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Corollary 3.11

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$\begin{aligned} &\bigl[d \bigl(T^{2}x,T^{2}y \bigr)+\tau \bigr]^{\alpha(x)\beta(y)} \\ &\quad \leq a_{1}d(x,Tx)+a_{2}d \bigl(Tx,T^{2}x \bigr)+b_{1}d(y,Ty)+b_{2}d \bigl(Ty,T^{2}y \bigr)+ \tau \end{aligned}$$

(3.9)

*for all*
\(x,y \in X\), *where*
\(a_{1},a_{2},b_{1},b_{2}\in[0,1)\)
*with*
\(a_{1}+a_{2}+b_{1}+b_{2}<1\)
*and*
\(\tau\geq1\). *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0}\in X\)
*such that*
\(\alpha(x_{0}) \geq1\)
*and*
\(\beta (x_{0}) \geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

### Corollary 3.12

*Let*
\((X,d)\)
*be a metric space*, \(\alpha,\beta: X \rightarrow[0,\infty )\)
*be two mappings and*
\(T:X\rightarrow X\)
*be a mapping such that*

$$\begin{aligned} \bigl[\tau-1+\alpha(x)\beta(y) \bigr]^{d(T^{2}x,T^{2}y)} \leq\tau ^{a_{1}d(x,Tx)+a_{2}d(Tx,T^{2}x)+b_{1}d(y,Ty)+b_{2}d(Ty,T^{2}y)} \end{aligned}$$

(3.10)

*for all*
\(x,y \in X\), *where*
\(a_{1},a_{2},b_{1},b_{2}\in[0,1)\)
*with*
\(a_{1}+a_{2}+b_{1}+b_{2}<1\)
*and*
\(\tau>1\). *Assume that*
*T*
*is a cyclic*
\((\alpha,\beta)\)-*admissible mapping and there exists*
\(x_{0} \in X\)
*such that*
\(\alpha(x_{0})\geq1\)
*and*
\(\beta(x_{0}) \geq1\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

The following result is a special case of Theorem 3.9 because the Kannan contractive condition due to Kannan [6] implies the contractive condition (3.7) if \(\alpha,\beta:X \rightarrow[0,\infty)\) defined by \(\alpha(x) = \beta (x) = 1\) for all \(x\in X\).

### Corollary 3.13

*Let*
\((X,d)\)
*be a metric space and*
\(T : X \rightarrow X\)
*be a Kannan contraction mapping*, *i*.*e*., *there exists*
\(k \in[0,1/2)\)
*such that*

$$\begin{aligned} d(Tx,Ty) \leq k \bigl[d(x,Tx)+d(y,Ty) \bigr] \end{aligned}$$

(3.11)

*for all*
\(x,y \in X\). *Then*
*T*
*has the approximate fixed point property*. *Moreover*, *if*
*T*
*is continuous and*
\((X,d)\)
*is a complete metric space*, *then*
*T*
*has a fixed point*.

Corollaries 3.7 and 3.13, are interesting for defining the concepts of other classes of nonlinear mappings which are generalizations of several well-known mappings due to Chatterjea [10], Ćirić [11], Geraghty [12], Meir and Keeler [13], Mizoguchi and Takahashi [14], Suzuki [15], *etc.*