Fixed points of generalized $\alpha$-$\psi$-contractions

Abstract

In this paper, we introduce generalized $\alpha$-$\psi$-contractive mappings and multifunctions and give some results about fixed points of the mappings and multifunctions.

1 Introduction

During the last few decades, there have appeared a lot of papers on fixed points of multifunctions with different methods (see for example [1–9]). One of the most interesting methods is due to Suzuki for fixed points of mappings and multifunctions (see [10] and [11]). Recently, Samet, Vetro and Vetro have introduced the notion of $\alpha$-$\psi$-contractive type mappings [12]. Denote by $\Psi$ the family of nondecreasing functions $\psi :[0,\infty )\rightarrow [0,\infty )$ such that $\sum _{\textit{n}=1}^\infty \psi ^n(t)<\infty$ for all $t>0$, where $\psi ^n$ is the $n$th iterate of $\psi$. It is known that $\psi (t)<t$ for all $t>0$ and $\psi \in \Psi$ [12]. Also, there are a lot of sublinear mappings in $\Psi$ [13]. Let $(X,d)$ be a metric space and $T$ a selfmap on $X$. Then $T$ is called a $\alpha$-$\psi$-contraction mapping whenever there exist $\psi \in \Psi$ and $\alpha :X\times X\rightarrow [0,\infty )$ such that $\alpha (x,y)d(Tx,Ty)\le \psi (d(x,y))$ for all $x,y\in X$ [12]. Also, we say that $T$ is $\alpha$-admissible whenever $\alpha (x,y)\ge 1$ implies $\alpha (Tx,Ty)\ge 1$ [12]. Also, we say that $X$ has the property (B) respect to $\alpha$ if $\{x_n\}$ is a sequence in $X$ such that $\alpha (x_{n},x_{n+1})\ge 1$ for all $n\ge 1$ and $x_{n}\rightarrow x$, then $\alpha (x_{n},x)\ge 1$ for all $n\ge 1$ [12].

Let $(X,d)$ be a complete metric space and $T$ a $\alpha$-admissible $\alpha$-$\psi$-contractive mapping on $X$. Suppose that there exists $x_{0}\in X$ such that $\alpha (x_{0},Tx_{0})\ge 1$. If $T$ is continuous or $X$ has the property (B) respect to $\alpha$, then $T$ has a fixed point ([12]; Theorems 2.1 and 2.2). Finally, we say that $X$ has the property (H) whenever for each $x,y\in X$ there exists $z\in X$ such that $\alpha (x,z)\ge 1$ and $\alpha (y,z)\ge 1$. If $X$ has the property (H) in the Theorems 2.1 and 2.2, then $T$ has a unique fixed point ([12]; Theorem 2.3). It is considerable that the results of Samet, Vetro and Vetro generalize similar ordered results in the literature (see the results of the third section in [12]). Now, by using the main idea of [14], we introduce a new notion. We say that $T$ is a generalized $\alpha$-$\psi$-contractive mapping whenever $\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))$ for all $x,y\in X$, where $M(x,y)=\max \{d(x,y),d(x,Tx),d(y,Ty),\frac{1}{2}\max \{d(x,Ty),d(y,Tx)\}\}$. Throughout the paper, we suppose that $\psi \in \Psi$ is sublinear and $\alpha :X\times X\rightarrow [0,\infty )$ is a mapping.

2 Main results

Now, we are ready to state and prove our main results.

Theorem 2.1

Let $(X,d)$ be a complete metric space and $T$ a continuous generalized $\alpha$-$\psi$-contractive and $\alpha$-admissible selfmap on $X$. If there exists $x_0\in X$ such that $\alpha (x_0,Tx_0)\ge 1$, then $T$ has a fixed point.

Proof

Take $x_0\in X$ such that $\alpha (x_0,Tx_0)\ge 1$. Define the sequence $\{x_n\}_{n\ge 0}$ in $X$ by $x_{n+1}=Tx_n$ for all $n\ge 0$. If $x_n=x_{n+1}$ for some $n\ge 0$, then $x^*=x_n$ is a fixed point for $T$. Assume that $x_n\ne x_{n+1}$ for all $n\ge 0$. Since $T$ is $\alpha$-admissible, we get $\alpha (x_n,x_{n+1})\ge 1$ for all $n\ge 1$. But, we have

$$\begin{aligned} d(x_1,x_2)=d(Tx_0,Tx_1)\le \alpha (x_0,x_1)d(Tx_0,Tx_1)\le \psi (M(x_0,x_1)), \end{aligned}$$

where

$$\begin{aligned} M(x_0,x_1)&= \max \{d(x_0,x_1),d(x_0,Tx_0),d(x_1,Tx_1),\frac{1}{2}\max \{d(x_0,Tx_1),d(x_1,Tx_0)\}\} \\&= \max \{d(x_0,x_1),d(x_1,x_2),\frac{1}{2}d(x_0,x_2)\}. \end{aligned}$$

Note that, $M(x_0,x_1)\ne d(x_1,x_2)$ because if $M(x_0,x_1)=d(x_1,x_2)$, then we have $d(x_1,x_2)\!\le \!\psi (d(x_1,x_2))\!<\!d(x_1,x_2)$ which is a contradiction. Thus,

$$\begin{aligned} M(x_0,x_1)=\max \left\{ d(x_0,x_1),\frac{1}{2}d(x_0,x_2)\right\} . \end{aligned}$$

If $M(x_0,x_1)=d(x_0,x_1)$, then $d(x_1,x_2)\le \psi (d(x_0,x_1))$ and if $M(x_0,x_1)=\frac{1}{2}d(x_0,x_2)$, then

$$\begin{aligned} d(x_1,x_2)\le \psi \left( \frac{d(x_0,x_2)}{2}\right) \le \frac{\psi (d(x_0,x_1))+\psi (d(x_1,x_2))}{2} <\frac{1}{2}\psi (d(x_0,x_1))+\frac{1}{2}d(x_1,x_2) \end{aligned}$$

because $\psi$ is sublinear. Hence, $d(x_1,x_2)\le \psi (d(x_0,x_1))$. Now by using induction, we obtain $d(x_n,x_{n+1})\le {\psi }^n(d(x_0,x_1))$ for all $n$. Fix $\varepsilon >0$ and choose $n(\varepsilon )\ge 1$ such that $\sum \nolimits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon$. Let $m>n>n(\varepsilon )$. By using the triangular inequality, we obtain $d(x_n,x_m)\le \sum \nolimits _{k=n}^{m-1}d(x_k,x_{k+1})\le \sum \nolimits _{k=n}^{m-1}{\psi }^k(d(x_0,x_1)) \le \sum \nolimits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon$. Thus, $\{x_n\}_{n\ge 0}$ is a Cauchy sequence. Therefore, there exists $x^*\in X$ such that $x_n\rightarrow x^*$. Thus, $x_{n+1}=Tx_n\rightarrow Tx^*$ and so $x^*$ is a fixed point of $T$. $\square$

Example 2.1

Let $X=[0,+\infty )$ and $d(x,y)=|x-y|$ for all $x,y\in X$. Define the selfmap $T$ on $X$ by $Tx=x+8$ whenever $0\le x\le 1$ and $Tx=9$ whenever $x>1$, and

$$\begin{aligned} \alpha (x,y)= \left\{ \begin{array}{ll} 2 &{}\quad x,y\in [0,1] \,or\, x,y\in [8n,8n+1], for\, some\, n\ge 1,\\ 1 &{}\quad x\in [0,1]\, and\, y\in [8,9],\\ 0 &{}\quad otherwise. \end{array} \right. \end{aligned}$$

If we define $\psi (t)=\frac{t}{2}$ for all $t\ge 0$, then it is easy to check that $T$ is a generalized $\alpha$-$\psi$-contractive mapping. In fact, for each $x,y\in [0,1]$ with $x\le y$ we have

$$\begin{aligned} 2\times |x-y|=\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))=\frac{1}{2}\times 8=4. \end{aligned}$$

For $x\in [0,1]$ and $y\in \left( 1,\frac{81}{19}\right)$ we have

$$\begin{aligned} 0\times |x-9|=\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))=\frac{1}{2}\times |y-9|. \end{aligned}$$

For $x\in [0,1]$ and $y\ge \frac{81}{19}$ we have

$$\begin{aligned} 0\times |x-9|=\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))=\frac{1}{2}|x-y|. \end{aligned}$$

Let $x,y\in [8(n-1),8n]$ for some $n\ge 1$ and $x\le y$. Then

$$\begin{aligned} 2\times 0=\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))=\frac{1}{2}\times |x-9|. \end{aligned}$$

If $x\in [0,1]$ and $y\in [8,9]$, then

$$\begin{aligned} 1\times |x-1|=\alpha (x,y)d(Tx,Ty)\le \psi (M(x,y))=\frac{1}{2}\max \{|x-y|,8\}. \end{aligned}$$

Also, for $x_0=0$ we have $\alpha (0,T0)=\alpha (0,8)=1$. Obviously $T$ is continuous and so it remains to show that $T$ is $\alpha$-admissible. If $x,y\in [0,1]$ or $x,y\in [8n,8n+1]$ for some $n\ge 1$, then $\alpha (x,y)=\alpha (Tx,Ty)=2$. If $x\in [0,1]$ and $y\in [8,9]$, then $\alpha (x,y)=1$ and $\alpha (Tx,Ty)=2$. Hence, $T$ is $\alpha$-admissible. Now, note that $T$ has the fixed point $x_0=9$. Finally, note that $\alpha (x,y)d(Tx,Ty)\nleq \psi (d(x,y))$ for all $x,y\in [0,1]$. Thus, the last result is a generalization of Theorem 2.1 in [12].

Now, we state multifunction version of our results. For $A, B \in CB(X)$, let

$$\begin{aligned} H(A, B) = \max \left\{ \mathop {\sup }\limits _{x \in A} \,d(x, B), \,\mathop {\sup }\limits _{y \in B} \,d(y, A)\right\} , \end{aligned}$$

where $d(x, B) = \inf _{y \in B} d(x, y)$. It is well known that $H$ is a metric on $CB(X)$. Such a map $H$ is called Hausdorff metric induced by $d$. Similar to mapping case, we say that the multifunction $T:X\rightarrow CB(X)$ on a metric space $X$ is a generalized $\alpha$-$\psi$-contraction whenever $\alpha (x,y)H(Tx,Ty)\le \psi (M(x,y))$ for all $x,y\in X$, where $M(x,y)=\max \{d(x,y),d(x,$ $Tx), d(y,Ty),\frac{1}{2}\max \{d(x,Ty),d(y,Tx)\}\}$.

Theorem 2.2

Let $(X,d)$ be a complete metric space, $\alpha :X\times X\rightarrow [0,+\infty )$ a mapping, $\psi \in \Psi$ and $T:X\rightarrow CB(X)$ a generalized $\alpha$-$\psi$-contractive multifunction such that $\alpha (x,y)\ge 1$ implies $\alpha (u,v)\ge 1$ for all $u\in Tx$ and $v\in Ty$. Suppose that there exists $x_0\in X$ and $x_1\in Tx_0$ such that $\alpha (x_0,x_1)\ge 1$. If $X$ has the property (B) respect to $\alpha$, then, $T$ has a fixed point.

Proof

Take $x_0\in X$ and $x_1\in Tx_0$ such that $\alpha (x_0,x_1)\ge 1$. If $x_0=x_1$, then $x_0$ is a fixed point for $T$. Let $x_0\ne x_1$. Then,

$$\begin{aligned} H(Tx_0,Tx_1)\le \alpha (x_0,x_1)H(Tx_0,Tx_1)\le \psi (M(x_0,x_1)), \end{aligned}$$

where $M(x_0,x_1)=\max \{d(x_0,x_1),d(x_1,Tx_1),\frac{1}{2}d(x_0,Tx_1)\}$ because $x_1\in Tx_0$. Let $x_1\notin Tx_1$. Then $M(x_0,x_1)\ne d(x_1,Tx_1)$. Thus, $M(x_0,x_1)=\max \{d(x_0,x_1),\frac{1}{2}d(x_0,Tx_1)\}$. If $M(x_0,x_1)=d(x_0,x_1)$, then $d(x_1,Tx_1)\le \psi (d(x_0,x_1))$. If $M(x_0,x_1)=\frac{1}{2}d(x_0,Tx_1)$, then $d(x_1,Tx_1)\le \psi (\frac{1}{2}d(x_0,Tx_1))\le \frac{1}{2}\psi (d(x_0,x_1))+\frac{1}{2}\psi (d(x_1,Tx_1))$ and so

$$\begin{aligned} d(x_1,Tx_1)<\frac{1}{2}\psi (d(x_0,x_1))+\frac{1}{2}d(x_1,Tx_1)\Rightarrow d(x_1,Tx_1)\le \psi (d(x_0,x_1)) \end{aligned}$$

because $\psi$ is sublinear. Thus there exists $x_2\in Tx_1$ such that $d(x_1,x_2)\!\le \!\psi (d(x_0,x_1))$. Let $x_1\!\ne \! x_2$. Then, $H(Tx_1,Tx_2)\!\le \!\alpha (x_1,x_2)H(Tx_1,Tx_2)\!\le \!\psi (M(x_1,x_2))$ and $\alpha (x_1,x_2)\ge 1$, where $M(x_1,x_2)=\max \{d(x_1,x_2),d(x_2,Tx_2),\frac{1}{2}d(x_1,Tx_2)\}$ because $x_2\in Tx_1$. Let $x_2\notin Tx_2$. Thus, $M(x_1,x_2)=\max \{d(x_1,x_2),\frac{1}{2}d(x_1,Tx_2)\}$. If $M(x_1,x_2)=d(x_1,x_2)$, then $d(x_2,Tx_2)\le \psi (d(x_1,x_2))$. If $M(x_1,x_2)\!=\!\frac{1}{2}d(x_1,Tx_2)$, then $d(x_2,Tx_2)\!<\!\frac{1}{2}\psi (d(x_1,x_2))\!+\!\frac{1}{2}d(x_2,Tx_2)$ because $\psi$ is sublinear. Thus, we get $d(x_2,Tx_2)\le \psi (d(x_1,x_2))$. This implies that there exists $x_3\!\in \! Tx_2$ such that $d(x_2,x_3)\!\le \!\psi (d(x_1,x_2))$ and so $d(x_2,x_3)\!\le \!{\psi }^2(d(x_0,x_1))$. By continuing this steps, we obtain a sequence $\{x_n\}_{n\!\ge \! 0}$ in $X$ such that $x_{n+1}\in Tx_n$, $\alpha (x_n,x_{n+1})\ge 1$ and $d(x_n,x_{n+1})\le {\psi }^n(d(x_0,x_1))$ for all $n\ge 0$. Fix $\varepsilon >0$ and choose $n(\varepsilon )\ge 1$ such that $\sum \nolimits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon$. Let $m>n>n(\varepsilon )$. By using the triangular inequality, we obtain

$$\begin{aligned} d(x_n,x_m)\le \sum \limits _{k=n}^{m-1}d(x_k,x_{k+1})\le \sum \limits _{k=n}^{m-1}{\psi }^k(d(x_0,x_1)) \le \sum \limits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon . \end{aligned}$$

Thus, $\{x_n\}_{n\ge 0}$ is a Cauchy sequence. Hence, there exists $x^*\in X$ such that $x_n\rightarrow x^*$. Note that, $\alpha (x_n,x^*)\ge 1$ for all $n$. Moreover, we have

$$\begin{aligned} H(Tx_n,Tx^*)\le \alpha (x_n,x^*)H(Tx_n,Tx^*)\le \psi (M(x_n,x^*)) \end{aligned}$$

for all $n$, where

$$\begin{aligned} M(x_n,x^*)=\max \{d(x_n,x^*),d(x_n,Tx_n),d(x^*,Tx^*),\frac{1}{2}\max \{d(x_n,Tx^*),d(x^*,Tx_n)\}\}. \end{aligned}$$

If $M(x_n,x^*)=d(x_n,x^*)$, then $d(x_{n+1},Tx^*)\le H(Tx_n,Tx^*)\le \psi (d(x_n,x^*))$. In the case $M(x_n,x^*)=d(x_n,Tx_n)$, we have

$$\begin{aligned} d(x_{n+1},Tx^*)\le H(Tx_n,Tx^*)\le \psi (d(x_n,Tx_n))\le \psi (d(x_n,x_{n+1})). \end{aligned}$$

If $M(x_n,x^*)=d(x^*,Tx^*)$, then $x^*\in Tx^*$. In fact, if $x^*\notin Tx^*$, then $d(x^*,Tx^*)>0$ and so $d(x_{n+1},Tx^*)\le H(Tx_n,Tx^*)\le \psi (d(x^*,Tx^*))$. Hence,

$$\begin{aligned} d(x^*,Tx^*)=\lim \limits _{n\rightarrow +\infty }d(x_{n+1},Tx^*)\le \psi (d(x^*,Tx^*))<d(x^*,Tx^*) \end{aligned}$$

which is a contradiction. If $M(x_n,x^*)=\frac{1}{2}d(x_n,Tx^*)$, then

$$\begin{aligned} d(x_{n+1},Tx^*)\le H(Tx_n,Tx^*)\le \psi \left( \frac{1}{2}d(x_n,Tx^*)\right) \le \frac{1}{2}d(x_n,Tx^*). \end{aligned}$$

If $M(x_n,x^*)=\frac{1}{2}d(x^*,Tx_n)$, then we have

$$\begin{aligned} d(x_{n+1},Tx^*)\le H(Tx_n,Tx^*)\le \psi \left( \frac{1}{2}d(x^*,Tx_n)\right) \le \frac{1}{2}\psi (d(x^*,x_{n+1})). \end{aligned}$$

Since $\psi$ is continuous at $t=0$, we get $d(x^*,Tx^*)=0$ and so $x^*\in Tx^*$. $\square$

Example 2.2

Let $X=[-2,-1]\cup \{0\}\cup [1,\frac{5}{2}]$ and $d(x,y)=|x-y|$ for all $x,y\in X$. Define the multivalued $T:X\rightarrow CB(X)$ by

$$\begin{aligned} Tx=\left\{ \begin{array}{ll} [-\frac{x}{4}+2,\frac{5}{2}] &{}\quad x\in [-2,-\frac{3}{2})\\ \{0\} &{}\quad x\in \{-1,0,1\}\cup (2,\frac{5}{2}]\cup [-\frac{3}{2},-1)\\ \left[ -{\frac{3}{2}}, - {\frac{x}{4}}-1\right] &{}\quad x \in (1,2], \end{array} \right. \end{aligned}$$

$\psi (t)=\frac{4t}{5}$ for all $t\ge 0$ and $\alpha :X\times X\rightarrow [0,+\infty )$ by

$$\begin{aligned} \alpha (x,y)=\left\{ \begin{array}{ll} 2 &{}\quad x,y\in \Big [-2,-\frac{3}{2}\Big ) \text{ or } x,y \in (1,2] \\ 1 &{}\quad x\in \Big [-2,-\frac{3}{2}\Big )\cup (1,2]~and~y\in \{-1,0,1\}\cup \Big (2,\frac{5}{2}\Big ]\cup \Big [-\frac{3}{2},-1\Big )\\ \frac{3}{2} &{}\quad x,y\in \{-1,0,1\}\cup \Big (2,\frac{5}{2}\Big ]\cup \Big [-\frac{3}{2},-1\Big )\\ \frac{1}{2} &{}\quad x\in \Big [-2,-\frac{3}{2}\Big )~and~y\in (1,2]\\ \end{array} \right. \end{aligned}$$

with $\alpha (x,y)=\alpha (y,x)$ for all $x,y\in X$. One can check that $(X,d)$ is a complete metric space, $X$ has the property (B) respect to $\alpha$ and $T$ is a closed and bounded valued generalized $\alpha$-$\psi$-contractive multifunction on $X$. Note that, for $x_0=-2$ we have $Tx_0=\{\frac{5}{2}\}$ and $\alpha \left( -2,\frac{5}{2}\right) =1$. Therefore, $T$ satisfies the conditions Theorem 2.2.

Now by mixing our idea with the Suzuki’s idea, we give the following result. We say that the multifunction $T:X\rightarrow CB(X)$ on a metric spaces $X$ is a Suzuki-generalized $\alpha$-$\psi$-contraction if $\theta (r)d(x,Tx)\le d(x,y)$ implies $\alpha (x,y)H(Tx,Ty)\le \psi (M(x,y))$ for all $x,y\in X$, where $M(x,y)=\max \{d(x,y),\frac{1}{2}\max \{d(x,Tx),d(y,Ty)\},\frac{d(x,Ty)+d(y,Tx)}{2}\}$ and

$$\begin{aligned} \theta (r)=\left\{ \begin{array}{ll} \frac{1}{2} &{}\quad 0\le r\le \frac{1}{2}(\sqrt{5}-1),\\ \frac{1-r}{2r^2} &{}\quad \frac{1}{2}(\sqrt{5}-1)\le r\le \frac{1}{\sqrt{2}},\\ \frac{1}{1+2r} &{}\quad \frac{1}{\sqrt{2}}\le r<1. \end{array} \right. \end{aligned}$$

Finally, we say that $X$ has the property (C) respect to $\alpha$ whenever for each sequence $\{x_n\}$ in $X$ and $x\in X$ such that $\alpha (x_n,x)\ge 1$ for all $n$ and $x_n\rightarrow x^*\in X$ we have $\alpha (x^*,x)\ge 1$.

Theorem 2.3

Let $(X,d)$ be a complete metric space, $\alpha :X\times X\rightarrow [0,+\infty )$ a mapping, $\psi \in \Psi$ and $T:X\rightarrow CB(X)$ a Suzuki-generalized $\alpha$-$\psi$-contractive multifunction such that $\alpha (x,y)\ge 1$ implies $\alpha (u,v)\ge 1$ for all $u\in Tx$ and $v\in Ty$. Suppose that there exists $x_0\in X$ and $x_1\in Tx_0$ such that $\alpha (x_0,x_1)\ge 1$. If $X$ has the property (C) respect to $\alpha$, then $T$ has a fixed point.

Proof

Take $x_0\in X$ and $x_1\in Tx_0$ such that $\alpha (x_0,x_1)\ge 1$. If $x_0=x_1$, then $x_0$ is a fixed point of $T$. Let $x_1\ne x_0$. Since $\theta (r)\le 1$, $\theta (r)d(x_0,Tx_0)\le d(x_0,Tx_0)\le d(x_0,x_1)$ and so $d(x_1,Tx_1)\le H(Tx_0,Tx_1)\le \alpha (x_0,x_1)H(Tx_0,Tx_1)\le \psi (M(x_0,x_1))$, where $M(x_0,x_1)=\max \{d(x_0,x_1),\frac{1}{2}d(x_1,Tx_1),\frac{1}{2}d(x_0,Tx_1)\}$ because $x_1\in Tx_0$. Now, let $x_1\notin Tx_1$. Then $M(x_0,x_1)\ne \frac{1}{2}d(x_1,Tx_1)$. If $M(x_0,x_1)=d(x_0,x_1)$, then $d(x_1,Tx_1)\le H(Tx_0,Tx_1)\le \psi (d(x_0,x_1))$. If $M(x_0,x_1)=\frac{1}{2}d(x_0,Tx_1)$, then

$$\begin{aligned} d(x_1,Tx_1)\le H(Tx_0,Tx_1)\le \psi (\frac{1}{2}d(x_0,Tx_1))\le \frac{1}{2}\psi (d(x_0,x_1))+\frac{1}{2}\psi (d(x_1,Tx_1)) \end{aligned}$$

and so $d(x_1,Tx_1)<\frac{1}{2}\psi (d(x_0,x_1))+\frac{1}{2}d(x_1,Tx_1)$. Hence, $d(x_1,Tx_1)\le \psi (d(x_0,x_1))$. Thus, there exists $x_2\in Tx_1$ such that $d(x_1,x_2)\le \psi (d(x_0,x_1))$. By continuing this process, we obtain a sequence $\{x_n\}_{n\ge 0}$ in $X$ such that $x_{n+1}\in Tx_n$, $\alpha (x_n,x_{n+1})\ge 1$ and $d(x_n,x_{n+1})\!\le \! {\psi }^n(d(x_0,x_1))$ for all $n\!\ge \! 0$. Fix $\varepsilon >0$ and choose $n(\varepsilon )\!\ge \!1$ such that $\sum \nolimits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon$. Let $m>n>n(\varepsilon )$. By using the triangular inequality, we obtain

$$\begin{aligned} d(x_n,x_m)\le \sum \limits _{k=n}^{m-1}d(x_k,x_{k+1})\le \sum \limits _{k=n}^{m-1}{\psi }^k(d(x_0,x_1)) \le \sum \limits _{n\ge n(\varepsilon )}{\psi }^n(d(x_0,x_1))<\varepsilon . \end{aligned}$$

Thus, $\{x_n\}_{n\ge 0}$ is a Cauchy sequence. Therefore, there exists $x^*\in X$ such that $x_n\rightarrow x^*$. By using the assumption, we get $\alpha (x_n,x^*)\ge 1$ for all $n$. Now, we show that $d(x^*,Tx)\le \psi (d(x^*,x))$ for all $x\in X\setminus \{x^*\}$ with $\alpha (x_n,x)\ge 1$ for all $n$. Suppose that $x\in X\setminus \{x^*\}$ with $\alpha (x_n,x)\ge 1$ for all $n$. Since $x_n\rightarrow x^*$, there exists $n_0\in N$ such that $d(x_n,x^*)\le \frac{1}{3}d(x,x^*)$ for all $n\ge n_0$. Then,

$$\begin{aligned} \theta (r)d(x_n,Tx_n)&\le d(x_n,Tx_n)\le d(x_n,x_{n+1})\le d(x_n,x^*)+d(x^*,x_{n+1})\\&= d(x_n,x^*)+d(x^*,x_{n+1})\le \frac{2}{3}d(x,x^*)=d(x,x^*)-\frac{1}{3}d(x,x^*)\\&\le d(x,x^*)-d(x^*,x_n)\le d(x,x_n) \quad (*) \end{aligned}$$

and so $H(Tx_n,Tx)\le \alpha (x_n,x)H(Tx_n,Tx)\le \psi (M(x_n,x))$, where

$$\begin{aligned} M(x_n,x)=\max \Big \{d(x_n,x),\frac{1}{2}\max \{d(x_n,Tx_n),d(x,Tx)\},\frac{d(x,Tx_n)+d(x_n,Tx)}{2}\Big \}. \end{aligned}$$

Thus, $d(x_{n+1},Tx)\le \psi (M(x_n,x))$ for all $n\ge n_0$. Therefore, if $M(x_n,x)=d(x_n,x)$ or $M(x_n,x)=\frac{1}{2}d(x_n,Tx_n)$, then by using (*) we have $d(x_{n+1},Tx)\le \psi (d(x_n,x))$ and so $\lim \nolimits _{n\rightarrow \infty }d(x_{n+1},Tx)\le \lim \nolimits _{n\rightarrow \infty }\psi (d(x_n,x))\Rightarrow d(x^*,Tx)\le \psi (d(x^*,x))$. Now, note that if $M(x_n,x)=\frac{1}{2}d(x,Tx)$, then $d(x_{n+1},Tx)\le \psi (\frac{1}{2}d(x,Tx))\le \frac{1}{2}\psi (d(x,x_n))+\frac{1}{2}d(x_n,Tx)$ and so $\lim \nolimits _{n\rightarrow \infty }d(x_{n+1},Tx)\le \frac{1}{2}\lim \nolimits _{n\rightarrow \infty }\psi (d(x,x_n))+\frac{1}{2} \lim \nolimits _{n\rightarrow \infty }d(x_n,Tx)$. Hence, we obtain $d(x^*,Tx)\le \psi (d(x^*,x))$. If $M(x_n,x)=\frac{d(x,Tx_n)+d(x_n,Tx)}{2}$, then

$$\begin{aligned} d(x_{n+1},Tx)\le \psi \left( \frac{d(x,Tx_n)+d(x_n,Tx)}{2}\right) \le \frac{1}{2}\psi (d(x,x_{n+1}))+\frac{1}{2}d(x_n,Tx) \end{aligned}$$

and so $d(x^*,Tx)\le \psi (d(x^*,x))$. Therefore, we prove the claim. Again by using the assumption, we get $\alpha (x,x^*)\ge 1$ and so $H(Tx,Tx^*)\le \alpha (x,x^*)H(Tx,Tx^*)$. Now, we show that $H(Tx,Tx^*)\le \psi (M(x,x^*))$. If $x\ne x^*$, then we get three cases. First, suppose that $0\le r\le \frac{1}{2}(\sqrt{5}-1)$. Then, $\theta (r)=\frac{1}{2}$ and

$$\begin{aligned} d(x,Tx)\le d(x,x^*)+d(x^*,Tx)\le d(x,x^*)+\psi (d(x^*,x))<2d(x,x^*) \end{aligned}$$

Hence, $\theta (r)d(x,Tx)\le d(x,x^*)$ and so $H(Tx,Tx^*)\le \psi (M(x,x^*))$. Now, suppose that $\frac{1}{2}(\sqrt{5}-1)\le r\le \frac{1}{\sqrt{2}}$. Then, $\theta (r)=\frac{1-r}{2r^2}$. In this case, for each $n\ge 1$ there exists $y_n\in Tx$ such that $d(x^*,y_n)\le d(x^*,Tx)+ \frac{1}{n} d(x^*,x)$. Thus,

$$\begin{aligned} d(x,Tx)\le d(x,y_n)&\le d(x,x^*)+d(x^*,y_n)\le d(x,x^*)+d(x^*,Tx)+\frac{1}{n} d(x^*,x)\\&\le d(x,x^*)+ \psi (d(x,x^*))+\frac{1}{n}d(x^*,x)< \left( 2+\frac{1}{n}\right) d(x^*,x) \end{aligned}$$

for all $n\ge 1$ and so

$$\begin{aligned} \theta (r)d(x,Tx) = \frac{1}{2}d(x,Tx)\le d(x^*,x)\Rightarrow H(Tx,Tx^*)\le \psi (M(x,x^*)). \end{aligned}$$

Finally, suppose that $\frac{1}{\sqrt{2}}\le r<1$. Then, $\theta (r)=\frac{1}{1+2r}$. For each $n\ge 1$, there exists $z_{n} \in Tx$ such that $d(x^{*}, z_{n})\le d(x^{*}, Tx)+ \left( \frac{1}{4}+\frac{1}{n}\right) d(x^{*}, x)$. Hence,

$$\begin{aligned} d(x,Tx)&\le d(x,z_{n})\le d(x,x^*)+d(x^{*},z_{n}) \le d(x, x^*)+ d(x^{*},Tx)\!+\! \left( \frac{1}{4}\!+\!\frac{1}{n}\right) d(x^*,x)\\&\le d(x,x^*) + \psi (d(x,x^*)) \!+\!\left( \frac{1}{4}\!+\!\frac{1}{n}\right) d(x^{*},x) < \left( 2+\left( \frac{1}{4}\!+\!\frac{1}{n}\right) \right) d(x^*,x) \end{aligned}$$

for all $n$. Thus, $d(x,Tx)\le \left( 2+\frac{1}{4}\right) d(x^*,x)=\frac{9}{4} d(x^*,x)$. This implies that

$$\begin{aligned} \theta (r) d(x,Tx)= \frac{1}{1+2r}d(x,Tx) \le \frac{4}{9}d(x,Tx) \le d(x^*,x) \end{aligned}$$

and   so $H(Tx,Tx^{*})\le \psi (M(x,x^{*}))$. Thus, $d(x^*,Tx^*)=\lim \nolimits _{n\rightarrow \infty }d(x_{n+1},Tx^*)\le \lim \nolimits _{n\rightarrow \infty } H(Tx_n,Tx^*)\le \lim \nolimits _{n\rightarrow \infty } \psi (M(x_n,x^*))$. If $M(x_n,x^*)=d(x_n,x^*)$, then $d(x^*,Tx^*)\le \lim \nolimits _{n\rightarrow \infty }\psi (d(x_n,x^*))=0$ and so we get $d(x^*,Tx^*)=0$. If $M(x_n,x^*)=\frac{1}{2}d(x_n,Tx_n)$, then

$$\begin{aligned} d(x^*,Tx^*)\le \lim \limits _{n\rightarrow \infty }\psi \left( \frac{1}{2}d(x_n,Tx_n)\right) \le \lim \limits _{n\rightarrow \infty }\psi (d(x_n,x_{n+1}))=0. \end{aligned}$$

If $M(x_n,x^*)=\frac{1}{2}d(x^*,Tx^*)$, then $d(x^*,Tx^*)\le \lim \nolimits _{n\rightarrow \infty }\psi (\frac{1}{2}d(x^*,Tx^*))<\frac{1}{2}d(x^*,Tx^*)$ which is a contradiction. If $M(x_n,x^*)=\frac{d(x_n,Tx^*)+d(x^*,Tx_n)}{2}$, then

$$\begin{aligned}&d(x^*,Tx^*)\le \lim \limits _{n\rightarrow \infty }\psi \left( \frac{d(x_n,Tx^*)+d(x^*,Tx_n)}{2}\right) \le \frac{1}{2} \lim \limits _{n\rightarrow \infty } \psi (d(x_{n},x^*))\\&\quad +\frac{1}{2} \lim \limits _{n\rightarrow \infty } \psi (d(x^*,Tx^*)) +\frac{1}{2} \lim \limits _{n\rightarrow \infty } \psi (d(x^*,x_{n+1}))\le \frac{1}{2} \psi (d(x^*,Tx^*)). \end{aligned}$$

Therefore, $d(x^*,Tx^*)=0$ and so $x^*\in Tx^*$. $\square$