Cyclic Contractions: An Improvement Result

Abstract

In this chapter, we give an improvement fixed point result for cyclic contractions by weakening the closure assumption that is usually supposed in the literature. As applications, we discuss the existence of solutions to certain systems of functional equations. The main reference of this chapter is the paper [4].

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In this chapter, we give an improvement fixed point result for cyclic contractions by weakening the closure assumption that is usually supposed in the literature. As applications, we discuss the existence of solutions to certain systems of functional equations. The main reference of this chapter is the paper [4].

4.1 Introduction

In [6], Kirk et al. proved the following result.

Theorem 4.1

Let (X, d) be a complete metric space and $(A_i)_{i=1}^p$ be a finite number of nonempty closed subsets of X. Let $T: \bigcup _{i=1}^p A_i\rightarrow \bigcup _{i=1}^p A_i$ be a given mapping. Suppose that the following conditions are satisfied:

• (i) $TA_i\subseteq A_{i+1}$, for $i=1,2,\ldots ,p$, with $A_{p+1}=A_1$.

• (ii) The mapping T satisfies a cyclic contraction; i.e., there exists some constant $k\in (0,1)$ such that

  $$ d(Tx,Ty)\le k d(x,y),\quad (x,y)\in A_i\times A_{i+1},\, i=1,2,\ldots ,p. $$

Then

• (I) $\bigcap _{i=1}^p A_i$ is nonempty.

• (II) T has a unique fixed point in $\bigcap _{i=1}^p A_i$.

Observe that Banach contraction principle follows immediately from Theorem 4.1 by taking $A_i=X$, for every $i=1,2,\ldots ,p$. Observe also that (II) is an immediate consequence of (I) and Banach contraction principle. More precisely, if $\bigcap _{i=1}^p A_i$ is nonempty, from (i), we have $T\left( \bigcap _{i=1}^p A_i\right) \subseteq \bigcap _{i=1}^p A_i$. Moreover, from (ii), we have

$$ d(Tx,Ty)\le k d(x,y),\quad (x,y)\in \bigcap _{i=1}^p A_i\times \bigcap _{i=1}^p A_i. $$

Since $A_i$ is closed for every $i=1,2,\ldots ,p$, and (X, d) is complete, then $\left( \bigcap _{i=1}^p A_i,d\right)$ is a complete metric space. Therefore, applying Banach contraction principle to the mapping $T: \bigcap _{i=1}^p A_i\rightarrow \bigcap _{i=1}^p A_i$, (II) follows.

In this chapter, we address the following question: Is it possible to obtain (I) and (II) of Theorem 4.1 without supposing that $A_i$ is closed for every $i=1,2,\ldots ,p$? Observe that in this case, if $\bigcap _{i=1}^p A_i$ is nonempty, we cannot obtain (II) via Banach contraction principle applied to the mapping $T:\bigcap _{i=1}^p A_i \rightarrow \bigcap _{i=1}^p A_i$, since $\bigcap _{i=1}^p A_i$ is not necessarily complete. We obtain an affirmative answer to the addressed question by supposing only that $A_1$ is closed. Moreover, we consider mappings satisfying a $\varphi$-contraction, which is a contraction involving a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$. An example is provided to illustrate the obtained result. As applications, we give existence results to certain systems of functional equations.

Recall that a functions $\varphi : [0,\infty )\rightarrow [0,\infty )$ is said to be a (c)-comparison function if it satisfies the following conditions:  

($\varphi _1$):

$\varphi$ is a nondecreasing function.

($\varphi _2$):

There exists $k_0=1,2,\ldots$ and $\lambda \in (0,1)$ such that

$$ \varphi ^{k+1}(t)\le \lambda \varphi ^k(t)+v_k, \quad k=k_0,k_0+1,\ldots , $$

for all $t>0$, where $\sum _{k=0}^\infty v_k$ is a convergent series of nonnegative terms. Here, $\varphi ^n$ denotes the nth iterate of $\varphi$.

 

We have the following properties (see [8]).

Lemma 4.1

Let $\varphi : [0,\infty )\rightarrow [0,\infty )$ be a (c)-comparison function. Then

• (i) $\varphi (t)<t$, $t>0$.

• (ii) $\varphi$ is continuous at 0.

• (iii) $\varphi (0)=0$.

• (iv) $\displaystyle \sum _{n=0}^\infty \varphi ^n(t)<\infty$, $t>0$.

In [8], the authors extended Theorem 4.1 to the class of cyclic $\varphi$-contractions, where $\varphi : [0,\infty )\rightarrow [0,\infty )$ is a (c)-comparison function. Moreover, they provided error estimates for approximating the fixed point. The obtained result in [8] is the following.

Theorem 4.2

Let (X, d) be a complete metric space and $(A_i)_{i=1}^p$ be a finite number of nonempty closed subsets of X. Let $T: \bigcup _{i=1}^p A_i\rightarrow \bigcup _{i=1}^p A_i$ be a given mapping. Suppose that the following conditions are satisfied:

• (i) $TA_i\subseteq A_{i+1}$, for $i=1,2,\ldots ,p$, with $A_{p+1}=A_1$.

• (ii) The mapping T satisfies a cyclic $\varphi$-contraction; i.e., there exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ d(Tx,Ty)\le \varphi (d(x,y)),\quad (x,y)\in A_i\times A_{i+1},\, i=1,2,\ldots ,p. $$

Then

• (I) T has a unique fixed point $x^*\in \bigcap _{i=1}^p A_i$. For any $x_0\in \bigcup _{i=1}^p A_i$, the Picard sequence $\{T^nx_0\}$ converges to $x^*$.

• (II) The following estimates hold:

  $$\begin{aligned} d(T^nx_0,x^*)\le & {} s\left( \varphi ^n(d(x_0,Tx_0))\right) ,\,\, n=1,2,\ldots ,\\ d(T^nx_0,x^*)\le & {} s\left( \varphi (d(T^nx_0,T^{n+1}x_0))\right) ,\,\, n=1,2,\ldots . \end{aligned}$$

• (III) For any $x\in \bigcup _{i=1}^p A_i$,

  $$ d(x,x^*)\le s(d(x,Tx)), $$

  where $s(t)=\displaystyle \sum _{k=0}^\infty \varphi ^k(t)$, $t\ge 0$.

In this chapter, we shall prove that the results of Theorem 4.2 hold true by assuming only that $A_1$ is closed. We present also an example where our result can be used; however, Theorem 4.2 cannot be applied.

Remark 4.1

Theorem 4.2 is a cyclical-type generalization of the following ordinary fixed point theorem.

Theorem 4.3

Let (X, d) be a complete metric space and $T: X\rightarrow X$ be a given mapping. Suppose that there exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

$$ d(Tx,Ty)\le \varphi (d(x,y)),\quad (x,y)\in X\times X. $$

Then

• (I) T has a unique fixed point $x^*\in X$. For any $x_0\in X$, the Picard sequence $\{T^nx_0\}$ converges to $x^*$.

• (II) The following estimates hold:

  $$\begin{aligned} d(T^nx_0,x^*)\le & {} s\left( \varphi ^n(d(x_0,Tx_0))\right) ,\,\, n=1,2,\ldots ,\\ d(T^nx_0,x^*)\le & {} s\left( d(T^nx_0,T^{n+1}x_0)\right) ,\,\, n=1,2,\ldots . \end{aligned}$$

• (III) For any $x\in X$,

  $$ d(x,x^*)\le s(d(x,Tx)). $$

Note that in [11], the author claimed that Theorems 4.2 and 4.3 are equivalent. In fact, he claimed that by applying Theorem 4.3 to the mapping $T: \bigcap _{i=1}^p A_i \rightarrow \bigcap _{i=1}^p A_i$, we retrieve the results in Theorem 4.2. Obviously, such claim is not true. At first, in Theorem 4.2(I), for any $x_0\in \bigcup _{i=1}^p A_i$, the Picard sequence $\{T^nx_0\}$ converges to the fixed point of T. However, by applying Theorem 4.3 with $X=\bigcap _{i=1}^p A_i$, the convergence holds only for $x_0\in \bigcap _{i=1}^p A_i$. The same remark holds for the estimates given by (II) and (III) in Theorem 4.2.

For other results related to cyclic and generalized cyclic contractions, we refer the reader to [1,2,3, 5, 7, 9, 10, 12] and the references therein.

4.2 Main Result

We deal with the following problem: Find $x\in X$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} Tx=x,\\ x\in \bigcap _{i=1}^p A_i, \end{array} \right. \end{aligned}$$

(4.1)

where (X, d) is a complete metric space, $A_1$ is a nonempty closed subset of X, $A_i$, $i=2,3,\ldots ,p$ are arbitrary nonempty subsets of X (nonnecessarily closed), and $T: X\rightarrow X$ is a mapping satisfying a cyclic $\varphi$-contraction.

We have the following result.

Theorem 4.4

Let (X, d) be a complete metric space. Let $(A_i)_{i=1}^p$ be a finite number of nonempty subsets of X. Let $T: X\rightarrow X$ be a given mapping. Suppose that the following conditions are satisfied:

• (i) $A_1$ is closed.

• (ii) $TA_i\subseteq A_{i+1}$ for all $i=1,2,\ldots ,p$ with $A_{p+1}=A_1$.

• (iii) There exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ d(Tx,Ty)\le \varphi (d(x,y)),\quad (x,y)\in A_i\times A_{i+1},\, i=1,2,\ldots ,p. $$

Then

• (I) For any $x_0\in \bigcup _{i=1}^p A_i$, the Picard sequence $\{T^nx_0\}$ converges to $x^*\in X$, the unique solution to (4.1).

• (II) The following estimates hold:

  $$\begin{aligned} d(T^nx_0,x^*)\le & {} s\left( \varphi ^n(d(x_0,Tx_0))\right) ,\,\, n\in \mathbb {N},\\ d(T^nx_0,x^*)\le & {} s\left( d(T^nx_0,T^{n+1}x_0)\right) ,\,\, n\in \mathbb {N}. \end{aligned}$$

• (III) For any $x\in \bigcup _{i=1}^p A_i$, we have

  $$ d(x,x^*)\le s(d(x,Tx)). $$

Proof

Let $x_0\in \bigcup _{i=1}^p A_i$ be an arbitrary point. Without restriction of the generality, we may assume that $x_0\in A_1$. Let $\{x_n\}\subset X$ be the Picard sequence defined by

$$ x_{n}=T^nx_0,\quad n\in \mathbb {N}. $$

We argue exactly as in the proof of Theorem 4.2 in [8] to obtain that

$$ d(x_n,x_{n+m}) \le \sum _{i=n}^{\infty }\varphi ^i(d(x_0,x_1)), $$

for any $(n,m)\in \mathbb {N}\times \mathbb {N}\backslash \{0\}$. From Lemma 4.1, the series $\sum _{i=0}^{\infty }\varphi ^i(d(x_0,x_1))$ is convergent, which implies that

$$ \sum _{i=n}^{\infty }\varphi ^i(d(x_0,x_1))\rightarrow 0 \text{ as } n\rightarrow \infty . $$

As consequence, $\{x_n\}$ is a Cauchy sequence in (X, d). Since (X, d) is a complete metric space, there exists some $x^*\in X$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }d(x_n,x^*)=0. \end{aligned}$$

(4.2)

On the other hand, from (ii), we obtain

$$\begin{aligned} x_{sp+r-1}\in A_r,\quad r\in \{1,2,\ldots ,p\},\,\, s\in \mathbb {N}. \end{aligned}$$

(4.3)

Using (4.3), we obtain

$$ \{x_{np}\}\subset A_1. $$

Since $A_1$ is closed, it follows from (4.2) that

$$\begin{aligned} x^*\in A_1. \end{aligned}$$

(4.4)

Again, from (4.3), we know that

$$\begin{aligned} \left\{ x_{np+1}\right\} \subset A_2. \end{aligned}$$

(4.5)

Now, (4.4) and (4.5) yield

$$ (x^*,x_{np+1})\in A_1\times A_2,\quad n\in \mathbb {N}. $$

Then by (iii), we obtain

$$ d(Tx^*,x_{np+2})=d(Tx^*,Tx_{np+1})\le \varphi (d(x^*,x_{np+1})),\quad n\in \mathbb {N}. $$

Note that from Lemma 4.1, we know that

$$ \lim _{t\rightarrow 0^+}\varphi (t)=0. $$

Using this fact, and passing to the limit as $n\rightarrow \infty$ in the above inequality, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }d(x_{np+2},Tx^*)=0. \end{aligned}$$

(4.6)

Now, it follows immediately from (4.2), (4.6) and the uniqueness of the limit that

$$\begin{aligned} x^*=Tx^*. \end{aligned}$$

(4.7)

Next, from (ii) and (4.7), we obtain

$$\begin{aligned} x^*\in \bigcap _{i=1}^p A_i. \end{aligned}$$

(4.8)

Then from (4.7) and (4.8), we deduce that $x^*\in X$ is a solution to (4.1). In order to prove the uniqueness of solutions to (4.1), suppose that $y^*\in X$ is a solution to (4.1) with $d(x^*,y^*)>0$. Using (iii), we get

$$ d(x^*,y^*)=d(Tx^*,Ty^*)\le \varphi (d(x^*,y^*)). $$

Since $d(x^*,y^*)>0$, using Lemma 4.1, we have

$$ \varphi (d(x^*,y^*))<d(x^*,y^*). $$

Then

$$ d(x^*,y^*)<d(x^*,y^*), $$

which is a contradiction. As consequence, $d(x^*,y^*)=0$, i.e., $x^*=y^*$. This proves that $x^*\in X$ is the unique solution to (4.1). Therefore, (I) is proved. The estimates given by (II) and (III) follow using exactly the same arguments as in the proof of Theorem 4.2 in [8].

Using the same argument as that used in the proof of Theorem 4.4, we obtain the following result.

Theorem 4.5

Let (X, d) be a complete metric space. Let $(A_i)_{i=1}^p$ be a finite number of nonempty subsets of X. Let $T: \bigcup _{i=1}^pA_i\rightarrow \bigcup _{i=1}^pA_i$ be a given mapping. Suppose that the following conditions are satisfied:

• (i) $A_1$ is closed.

• (ii) $TA_i\subseteq A_{i+1}$ for all $i=1,2,\ldots ,p$ with $A_{p+1}=A_1$.

• (iii) There exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ d(Tx,Ty)\le \varphi (d(x,y)),\quad (x,y)\in A_i\times A_{i+1},\, i=1,2,\ldots ,p. $$

Then

• (I) T has a unique fixed point $x^*\in \bigcap _{i=1}^p A_i$. For any $x_0\in \bigcup _{i=1}^p A_i$, the Picard sequence $\{T^nx_0\}$ converges to $x^*$.

• (II) The following estimates hold:

  $$\begin{aligned} d(T^nx_0,x^*)\le & {} s\left( \varphi ^n(d(x_0,Tx_0))\right) ,\,\, n\in \mathbb {N},\\ d(T^nx_0,x^*)\le & {} s\left( \varphi (d(T^nx_0,T^{n+1}x_0))\right) ,\,\, n\in \mathbb {N}. \end{aligned}$$

• (III) For any $x\in \bigcup _{i=1}^p A_i$,

  $$ d(x,x^*)\le s(d(x,Tx)), $$

  where $s(t)=\sum _{k=0}^\infty \varphi ^k(t)$, $t\ge 0$.

The following simple example shows that Theorem 4.5 is more general than Theorem 4.2.

Example 4.1

Let $X=\mathbb {R}$. The set X is equipped with the standard metric

$$\begin{aligned} d(x,y)=|x-y|,\quad (x,y)\in X\times X. \end{aligned}$$

(4.9)

Then (X, d) is a complete metric space. Let us consider the two subsets $A_1=[0,2]$ and $A_2=(1,\infty )$. Define the mapping $T: A_1\cup A_2=[0,\infty )\rightarrow A_1\cup A_2$ by

$$ Tx=2,\quad x\in A_1\cup A_2. $$

Clearly, we have

$$ TA_1=\{2\}\subset A_2\quad \text{ and }\quad TA_2=\{2\}\subset A_1. $$

Moreover, for any (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$, we have

$$ d(Tx,Ty)\le \varphi (d(x,y)),\quad (x,y)\in A_1\times A_2. $$

Therefore, by Theorem 4.5, T has a unique fixed point $x^*\in A_1\cap A_2=(1,2]$. In this case, we have $x^*=2$. Observe that Theorem 4.2 cannot be applied in this case since $A_2=(1,\infty )$ is an open subset of X.

Note that under the assumptions of Theorem 4.4, the mapping $T: X\rightarrow X$ has at least one fixed point in X, which means that a fixed point of T in X is not necessarily unique. But from result (I), the mapping T has a unique fixed point in $\bigcap _{i=1}^p A_i$. The following simple example illustrates this fact.

Example 4.2

Let $X=\mathbb {R}$ and d be the metric on X given by (4.9). Let $T: X\rightarrow X$ be the mapping defined by

$$\begin{aligned} Tx=\left\{ \begin{array}{lll} -1 &{} \text{ if } &{} x <0,\\ 2 &{} \text{ if } &{} x\ge 0. \end{array} \right. \end{aligned}$$

Let $A_1=[0,2]$ and $A_2=(1,\infty )$. Observe that

$$ TA_1=\{2\}\subset A_2\quad \text{ and }\quad TA_2=\{2\}\subset A_1. $$

Moreover, for all $(x,y)\in A_1\times A_2$, we have

$$ d(Tx,Ty)=d(2,2)=0\le \varphi (d(x,y)), $$

for any (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$. Then all the required conditions of Theorem 4.4 are satisfied. In this case, we observe that $x^*=2$ is the unique solution to (4.1) with $p=2$. However, the mapping T has two fixed points in $X=\mathbb {R}$, $x^*=2$, and $y^*=-1$.

4.3 Applications

Motivated by the suggestion of Kirk et al. [6] “Of course it would even be nicer to have applications,” we present in this section some possible applications of the main result of this chapter.

Let (X, d) be a complete metric space, $T: X\rightarrow X$ be a given mapping, and $\alpha : X\rightarrow \mathbb {R}$ be a given function. We are concerned with the study of the existence of solutions to the following problem: Find $x\in X$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} Tx&{}=&{} x,\\ \alpha (x)&{}=&{}0. \end{array} \right. \end{aligned}$$

(4.10)

We have the following result.

Theorem 4.6

Suppose that the following conditions are satisfied:

• (i) $\alpha$ is lower semi-continuous.

• (ii) There exists some $x_0\in X$ such that $\alpha (x_0)\le 0$.

• (iii) For every $x\in X$, we have

  $$ \alpha (x)\alpha (Tx)\le 0. $$

• (iv) There exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ \alpha (x)\alpha (y)\le 0 \Longrightarrow d(Tx,Ty)\le \varphi (d(x,y)). $$

Then (4.10) has a unique solution.

Proof

Set

$$ A_1=\{x\in X:\, \alpha (x)\le 0\} $$

and

$$ A_2=\{x\in X:\, \alpha (x)\ge 0\}. $$

From (ii), the set $A_1$ is nonempty (since $x_0\in A_1$). From (iii), we have $TA_1\subseteq A_2$ and $TA_2\subseteq A_1$. Moreover, since $\alpha$ is lower semi-continuous, then $A_1$ is a closed subset of X. Now, from (iv), for every pair of elements $(x,y)\in A_1\times A_2$, we have

$$ d(Tx,Ty)\le \varphi (d(x,y)). $$

Applying Theorem 4.4, we obtain the existence of a unique solution to (4.10).

Remark 4.2

The result of Theorem 4.6 is still valid if we replace condition (i) by

(i’) $\alpha$ is upper semi-continuous. In this case, we set

$$ A_1=\{x\in X:\, \alpha (x)\ge 0\} $$

and

$$ A_2=\{x\in X:\, \alpha (x)\le 0\}. $$

Since $\alpha$ is upper semi-continuous, then $A_1$ is a closed subset of X.

Next, we consider the problem: Find $x\in X$ such that

$$\begin{aligned} \left\{ \begin{array}{lll} Tx&{}=&{} x,\\ \alpha (x)&{}=&{}0,\\ \beta (x)&{}=&{} 0, \end{array} \right. \end{aligned}$$

(4.11)

where $\alpha ,\beta : X\rightarrow \mathbb {R}$ are given functions.

Theorem 4.7

Suppose that the following conditions are satisfied:

• (i) $\alpha$ and $\beta$ are lower semi-continuous.

• (ii) There exists some $x_0\in X$ such that $\alpha (x_0)\le 0$ and $\beta (x_0)\le 0$.

• (iii) For every $x\in X$, we have

  $$ \alpha (x)\le 0,\, \beta (x)\le 0 \Longrightarrow \alpha (Tx)\ge 0,\, \beta (Tx)\ge 0. $$

• (iv) For every $x\in X$, we have

  $$ \alpha (x)\ge 0,\,\beta (x)\ge 0 \Longrightarrow \alpha (Tx)\le 0,\, \beta (Tx)\le 0. $$

• (v) There exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ \alpha (x)\le 0,\, \beta (x)\le 0, \, \alpha (y)\ge 0,\,\beta (y)\ge 0 \Longrightarrow d(Tx,Ty)\le \varphi (d(x,y)). $$

Then (4.11) has a unique solution.

Proof

We argue as in the proof of Theorem 4.6 by considering the sets

$$ A_1=\{x\in X:\, \alpha (x)\le 0,\,\beta (x)\le 0\} $$

and

$$ A_2=\{x\in X:\, \alpha (x)\ge 0,\,\beta (x)\ge 0\}. $$

Remark 4.3

The result of Theorem 4.7 is still valid if we replace condition (i) by

(i’) $\alpha$ and $\beta$ are upper semi-continuous.

Theorem 4.8

Suppose that the following conditions are satisfied:

• (i) $\alpha$ is lower semi-continuous and $\beta$ is upper semi-continuous.

• (ii) There exists some $x_0\in X$ such that $\alpha (x_0)\le 0$ and $\beta (x_0)\ge 0$.

• (iii) For every $x\in X$, we have

  $$ \alpha (x)\le 0,\, \beta (x)\ge 0 \Longrightarrow \alpha (Tx)\ge 0,\, \beta (Tx)\le 0. $$

• (iv) For every $x\in X$, we have

  $$ \alpha (x)\ge 0,\,\beta (x)\le 0 \Longrightarrow \alpha (Tx)\le 0,\, \beta (Tx)\ge 0. $$

• (v) There exists a (c)-comparison function $\varphi : [0,\infty )\rightarrow [0,\infty )$ such that

  $$ \alpha (x)\le 0,\, \beta (x)\ge 0, \, \alpha (y)\ge 0,\,\beta (y)\le 0 \Longrightarrow d(Tx,Ty)\le \varphi (d(x,y)). $$

Then (4.11) has a unique solution.

Proof

We argue as in the proof of Theorem 4.6 by considering the sets

$$ A_1=\{x\in X:\, \alpha (x)\le 0,\,\beta (x)\ge 0\} $$

and

$$s A_2=\{x\in X:\, \alpha (x)\ge 0,\,\beta (x)\le 0\}. $$

Note that since $\alpha$ is lower semi-continuous and $\beta$ is upper semi-continuous, then $A_1$ is a closed subset of X.

We end this chapter with the following example.

Example 4.3

Let $X=[-1,1]$ be the set endowed with the standard metric

$$ d(x,y)=|x-y|,\quad (x,y)\in X\times X. $$

Let us consider the function $\alpha : X\rightarrow \mathbb {R}$ defined by

$$ \alpha (x)=\left\{ \begin{array}{lll} -3 &{} \text{ if } &{} x=-1,\\ \\ x(x^2+1)&{} \text{ if } &{} x\in (-1,1]. \end{array} \right. $$

Clearly, $\alpha$ is a lower semi-continuous function, since

$$ \alpha (-1)=-3\le \liminf _{x\rightarrow -1}\alpha (x)=-2. $$

Let us consider the mapping $T: X\rightarrow X$ defined by

$$ Tx=-\frac{x}{3},\quad x\in X. $$

For $x_0=-1$, we have $\alpha (x_0)=-3<0$.

For $x=-1$, we have

$$ \alpha (x)\alpha (Tx)=\alpha (-1)\alpha (T(-1))=-\left( \frac{1}{9}+1\right) <0. $$

For $x\in (-1,1]$, we have

$$ \alpha (x)\alpha (Tx)=-\frac{x^2}{3} (x^2+1)\left( \frac{x^2}{9}+1\right) \le 0. $$

Moreover, for all $(x,y)\in X\times X$, we have

$$ d(Tx,Ty)\le \varphi (d(x,y)), $$

where $\varphi (t)=\frac{t}{3}$, $t\ge 0$. Therefore, all the assumptions of Theorem 4.6 are satisfied. Then there is a unique $x^*\in X$ such that

$$ \left\{ \begin{array}{lll} Tx^*&{}=&{}x^*,\\ \varphi (x^*)&{}=&{}0. \end{array} \right. $$

Obviously, in this example, we have $x^*=0$.