On the Correlation between Banach Contraction Principle and Caristi’s Fixed Point Theorem in b-Metric Spaces

Abstract

We solve a question posed by E. Karapinar, F. Khojasteh and Z.D. Mitrović in their paper “A Proposal for Revisiting Banach and Caristi Type Theorems in b-Metric Spaces”. We also characterize the completeness of b-metric spaces with the help of a variant of the contractivity condition introduced by the authors in the aforementioned article.

1. Introduction

In order to investigate correlations between the Banach contraction principle and results of Caristi type in the realm of b-metric spaces, Karapinar, Khojasteh and Mitrović proved in [1] (Theorem 1) the following interesting result by using a new type of contractions.

Theorem 1

([1]). Let $\mathcal{T}$ be a self mapping of a complete b-metric space $(\mathcal{X},\mathfrak{b},s)$ such that there is a function $F:\mathcal{X}\to \mathbb{R}$ (the set of real numbers) satisfying the following two conditions:

(c1) F is bounded from below, i.e., there is an $a\in \mathbb{R}$ such that $inf F\left(\mathcal{X}\right)>a$;

(c2) for every $u,v\in \mathcal{X}$:

$\mathfrak{b}(u,\mathcal{T}u)>0\Rightarrow \mathfrak{b}(\mathcal{T}u,\mathcal{T}v)\le \left(F\right(u)-F(\mathcal{T}u\left)\right)\mathfrak{b}(u,v).$

Then $\mathcal{T}$ has a fixed point.

They also gave an example of a complete metric space where we can apply Theorem 1 above but not the Banach contraction principle, and raised the following question [1] (Remark 1): “It is natural to ask if the Banach contraction principle is a consequence of Theorem 1 (over metric spaces)”.

In this note we solve that question in the negative. With the help of a variant of Theorem 1 we also obtain a characterization of complete b-metric spaces which should be compared with the classical result given by Hu in [2], that a necessary and sufficient condition for a metric space to be complete is that every Banach contraction on each of its closed subsets has a fixed point.

Let us recall that many authors have contributed to the development of a consistent theory of fixed point for b-metric spaces (the bibliographies of [1], and [3,4,5] contain a high account of references to this respect). In particular, the Banach contraction principle [6] admits, mutatis mutandis, a full extension to b-metric spaces [7] (Theorem 2.1) (see also [3,8,9]), and regarding the extension of Caristi’s fixed point theorem [10] to b-metric spaces, significant contributions are given, among others, in [11] (Theorem 2.4), as well as in [3] (Corollary 12.1), [7] (Example 2.8) and [12] (Theorem 3.1).

2. Background

In this section we remind some definitions and properties which will be of help to the reader.

The set of non-negative real numbers and the set of natural numbers will be represented by ${\mathbb{R}}^{+}$ and $\mathbb{N}$, respectively.

The notion of a b-metric space has been considered by several authors under different names (see e.g., [13] and [3] (Chapter 12) for details). In our context we adapt that notion as given by Czerwik in [14].

A b-metric space is a triple $(\mathcal{X},\mathfrak{b},s),$ where $\mathcal{X}$ is a set, s is a real number with $s\ge 1,$ and $\mathfrak{b}:\mathcal{X}\times \mathcal{X}\to {\mathbb{R}}^{+}$ is a function satisfying, for every $u,v,w\in \mathcal{X},$ the following conditions:

(b1) $\mathfrak{b}(u,v)=0$ if and only if $u=v;$

(b2) $\mathfrak{b}(u,v)=\mathfrak{b}(v,u);$

(b3) $\mathfrak{b}(u,v)\le s\left(\mathfrak{b}\right(u,w)+\mathfrak{b}(w,v\left)\right).$

If $(\mathcal{X},\mathfrak{b},s)$ is a b-metric space the function $\mathfrak{b}$ is said to be a b-metric on $\mathcal{X}.$ Of course, every metric space is a b-metric space where $s=1$.

It is well-known (see e.g., [13,15,16]) that, as in the metric case, each b-metric $\mathfrak{b}$ on a set $\mathcal{X}$ induces a metrizable topology ${\mathfrak{T}}_{\mathfrak{b}}$ for which a subset $\mathcal{A}$ of $\mathcal{X}$ is declared open provided that for each $u\in \mathcal{A}$ there is an $r>0$ such that $\mathcal{B}(u,r)\subseteq \mathcal{A}$, where $\mathcal{B}(u,r)$= $\{v\in \mathcal{X}:\mathfrak{b}(u,v)<r\}.$

An important consequence is that a sequence ${\left(u_n\right)}_{n\in \mathbb{N}}$ in a b-metric space $(\mathcal{X},\mathfrak{b},s)$ is ${\mathfrak{T}}_{\mathfrak{b}}$-convergent to an $w\in \mathcal{X}$ if and only if ${lim}_n\mathfrak{b}(w,u_n)=0.$

In the sequel all topological properties corresponding to a b-metric space $(\mathcal{X},\mathfrak{b},s)$ will refer to the topology ${\mathfrak{T}}_{\mathfrak{b}}$.

It is appropriate to point out that, unlike the metric case, the set $\mathcal{B}(u,r)$ is not necessarily ${\mathfrak{T}}_{\mathfrak{b}}$-open (see [16] (Example on pages 4310–4311), [17] (Example 3.9)).

Moreover, it is well known that, contrarily to the classical metric case, there exist b-metrics that are not continuous functions (see e.g., [17] (Examples 3.9 and 3.10)).

Finally, we recall that the notions of Cauchy sequence and of complete b-metric space are defined exactly as the corresponding ones that for metric spaces.

3. Results and Examples

We begin this section giving an example that solves the question raised in [1] (Remark 1).

Example 1.

Let $(\mathcal{X},\mathfrak{b},1)$ be the metric space where $\mathcal{X}:={\mathbb{R}}^{+}$ and $\mathfrak{b}$ is the metric on $\mathcal{X}$ given by $\mathfrak{b}(u,u)=0$ for all $u\in \mathcal{X}$, and $\mathfrak{b}(u,v)=max\{u,v\}$ whenever $u\ne v.$

It is clear that $(\mathcal{X},\mathfrak{b},1)$ is complete because the only non-eventually constant Cauchy sequences are those that converge to $0.$

Let $\mathcal{T}$ be the self mapping of $\mathcal{X}$ given by $\mathcal{T}u=u/2$ for all $u\in \mathcal{X}.$

Since $\mathfrak{b}(\mathcal{T}u,\mathcal{T}v)=\mathfrak{b}(u,v)/2$ for all $u,v\in \mathcal{X},$ all conditions of the Banach contraction principle are satisfied.

Next we show that, however, the condition (c2) of Theorem 1 is not fulfilled.

Indeed, let $F:\mathcal{X}\to \mathbb{R}$ be any bounded from below function.

Take $u_0\in \mathcal{X}\setminus \left\{0\right\}.$ Then $\mathfrak{b}({\mathcal{T}}^n{u}_0,{\mathcal{T}}^{n+1}{u}_0)=2^{-n}{u}_0>0$ for all $n\in \mathbb{N}\cup \left\{0\right\}.$ Suppose that the condition (c2) holds. Thus, we have

$$\begin{array}{ccc}\hfill 2^{-(n+1)}{u}_0& =& \mathfrak{b}({\mathcal{T}}^{n+1}{u}_0,{\mathcal{T}}^{n+2}{u}_0)\hfill \\ & \le & (F\left({\mathcal{T}}^n{u}_0\right)-F\left({\mathcal{T}}^{n+1}{u}_0\right))\mathfrak{b}({\mathcal{T}}^n{u}_0,{\mathcal{T}}^{n+1}{u}_0)\hfill \\ & =& (F\left({\mathcal{T}}^n{u}_0\right)-F\left({\mathcal{T}}^{n+1}{u}_0\right))2^{-n}{u}_0,\hfill \end{array}$$

and, hence,

$$F\left({\mathcal{T}}^n{u}_0\right)\ge 2^{-1}+F\left({\mathcal{T}}^{n+1}{u}_0\right),$$

for all $n\in \mathbb{N}\cup \left\{0\right\}.$ Therefore,

$$\begin{array}{ccc}\hfill F\left(u_0\right)& \ge & \frac{1}{2}+F\left(\mathcal{T}{u}_0\right)\ge \frac{1}{2}+\frac{1}{2}+F\left({\mathcal{T}}^2{u}_0\right)\ge \dots \ge \frac{n+1}{2}+F\left({\mathcal{T}}^{n+1}{u}_0\right)\hfill \\ & \ge & \frac{n+1}{2}+infF\left(\mathcal{X}\right),\hfill \end{array}$$

for all $n\in \mathbb{N}\cup \left\{0\right\},$ a contradiction.

Remark 1.

Since we are working in the more general context of b-metric spaces, it would be interesting to give an example of a Banach contraction on a non-metric complete b-metric space that does not satisfy condition (c2) of Theorem 1. For it, we proceed to modify Example 1 in the following fashion: Fix $p\in \mathbb{N}\setminus \left\{1\right\}.$ Let $\mathcal{X}$:$={\mathbb{R}}^{+}$ and ${\mathfrak{b}}_p:\mathcal{X}\times \mathcal{X}\to {\mathbb{R}}^{+}$ defined by ${\mathfrak{b}}_p(u,u)=0$ for all $u\in \mathcal{X},$ and ${\mathfrak{b}}_p(u,v)={(max\{u,v\})}^p$ whenever $u\ne v.$ Then $(\mathcal{X},{\mathfrak{b}}_p,2^{p-1})$ is a non-metric complete b-metric space (see e.g., [18] (Example 2.2) or [3] (Example 12.2)). Let $\mathcal{T}$ be the self mapping of $\mathcal{X}$ given in Example 1. Then, it fulfills the conditions of the Banach contraction principle for b-metric spaces ([7] (Theorem 2.1)) with constant of contraction $2^{-p}.$ Analogously to Example 1 we can check that it does not satisfy condition (c2) for any bounded from below function $F:\mathcal{X}\to \mathbb{R}$ because, otherwise, for any $u_0\in \mathcal{X}\setminus \left\{0\right\}$ we should $F\left(u_0\right)\ge 2^{-p}(n+1)+infF\left(\mathcal{X}\right)$ for all $n\in \mathbb{N}\cup \{0\},$ a contradiction.

In the sequel, a self mapping $\mathcal{T}$ of a b-metric space $(\mathcal{X},\mathfrak{b},s)$ such that there is a function $F:\mathcal{X}\to \mathbb{R}$ for which conditions (c1) and (c2) are satisfied will said to be a correlation contraction (on $(\mathcal{X},\mathfrak{b},s)).$

We wonder if Theorem 1 allows us to obtain a characterization of complete b-metric spaces in the style of Hu’s characterization of metric completeness mentioned in Section 1. In this direction, the next is an example of a non-complete metric space such that every correlation contraction on any of its (non necessarily closed) subsets has a fixed point.

Example 2.

Let $\mathfrak{b}$ be the metric on $\mathbb{N}$ defined by $\mathfrak{b}(n,n)=0$ for all $n\in \mathbb{N},$ and $\mathfrak{b}(n,m)=max\{1/n,1/m\}$ whenever $n\ne m.$

Then $(\mathbb{N},\mathfrak{b},1)$ is not complete because ${\left(n\right)}_{n\in \mathbb{N}}$ is a non-convergent Cauchy sequence.

Now let $\mathcal{T}$ be a correlation contraction on a (non-empty) subset $\mathcal{A}$ of $(\mathbb{N},\mathfrak{b},1).$ Then, there is a function $F:\mathcal{A}\to \mathbb{R}$ for which conditions (c1) and (c2) are satisfied.

Suppose that $\mathcal{T}$ has no fixed points. Then $\left|\mathcal{A}\right|\ge 2,$ and $\mathfrak{b}(n,\mathcal{T}n)>0$ for all $n\in \mathcal{A}.$

Choose an $m_0\in \mathcal{A}.$ Since $\mathcal{T}$ has no fixed points we get ${\mathcal{T}}^n{m}_0\ne {\mathcal{T}}^{n+1}{m}_0$, for all $n\in \mathbb{N}\cup \left\{0\right\}.$

Hence, by condition (c2),

$$\mathfrak{b}({\mathcal{T}}^{n+1}{m}_0,\mathcal{T}{m}_0)\le (F\left({\mathcal{T}}^n{m}_0\right)-F\left({\mathcal{T}}^{n+1}{m}_0\right))\mathfrak{b}({\mathcal{T}}^n{m}_0,m_0)$$

for all $n\in \mathbb{N}\cup \left\{0\right\}$.

Since $1/\mathcal{T}{m}_0\le \mathfrak{b}({\mathcal{T}}^{n+1}{m}_0,\mathcal{T}{m}_0),$ and $\mathfrak{b}({\mathcal{T}}^n{m}_0,m_0)\le 1,$ we deduce that

$$\frac{1}{\mathcal{T}{m}_0}+F\left({\mathcal{T}}^{n+1}{m}_0\right)\le F\left({\mathcal{T}}^n{m}_0\right),$$

for all $n\in \mathbb{N}\cup \left\{0\right\}.$

Therefore,

$$\begin{array}{ccc}\hfill F\left(m_0\right)& \ge & \frac{1}{\mathcal{T}{m}_0}+F\left(\mathcal{T}{m}_0\right)\ge \frac{2}{\mathcal{T}{m}_0}+F\left({\mathcal{T}}^2{m}_0\right)\hfill \\ & \ge & \dots \ge \frac{n}{\mathcal{T}{m}_0}+F\left({\mathcal{T}}^n{m}_0\right)\ge \frac{n}{\mathcal{T}{m}_0}+infF\left(\mathcal{A}\right),\hfill \end{array}$$

for all $n\in \mathbb{N},$ which yields a contradiction.

Motivated by the preceding example, in Definition 1 below we present a modification of the notion of a correlation contraction from which a characterization of b-metric completeness will be obtained via a fixed point result.

To this end, we first recall that a partial order on a set $\mathcal{X}$ is a reflexive, antisymmetric, and transitive binary relation on $\mathcal{X}.$ If ⪯ is a partial order on $\mathcal{X}$, for each $u\in \mathcal{X}$ we denote by $u\uparrow$ the set $\{v\in \mathcal{X}:u⪯v\}.$

On the other hand, given a b-metric space $(\mathcal{X},\mathfrak{b},s)$ we shall denote by $acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right)$ the set of all accumulation points of the metrizable topological space $(\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}}).$ Hence $acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right)$ consists of all points $w\in \mathcal{X}$ for which there is a sequence of distinct points in $\mathcal{X}$ that ${\mathfrak{T}}_{\mathfrak{b}}$-converges to $w.$

Definition 1.

Let $(\mathcal{X},\mathfrak{b},s)$ be a b-metric space. We say that a self mapping $\mathcal{T}$ of $\mathcal{X}$ is a ⪯-correlation contraction (on $(\mathcal{X},\mathfrak{b},s))$ if there is a partial order ⪯ on $\mathcal{X}$ such that the following conditions hold:

(c3) $\mathcal{T}$ is non-decreasing, i.e., $u⪯v\Rightarrow \mathcal{T}u⪯\mathcal{T}v$ for all $u,v\in \mathcal{X};$

(c4) there is $u_0\in \mathcal{X}$ such that $u_0⪯\mathcal{T}{u}_0;$

(c5) there is a bounded from below function $F:\mathcal{X}\to \mathbb{R}$ such that for every $u\in \mathcal{X}$ and $v\in (u\uparrow )\cup acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right),$

$\mathfrak{b}(u,\mathcal{T}u)>0\Rightarrow \mathfrak{b}(\mathcal{T}u,\mathcal{T}v)\le \left(F\right(u)-F(\mathcal{T}u\left)\right)\mathfrak{b}(u,v).$

Remark 2.

The existence of ⪯-correlation contractions on a given b-metric space $(\mathcal{X},\mathfrak{b},s)$ is always guaranteed. Indeed, let i be the identity mapping on $\mathcal{X},$ and ${⪯}_D$ the discrete partial order on $\mathcal{X},$ i.e., $u{⪯}_{D}v\iff u=v.$ It is obvious that conditions (c3)–(c5) are fulfilled for any bounded from below function $F:\mathcal{X}\to \mathbb{R}$ (in particular, (c5) directly follows from the fact that $\mathfrak{b}(u,iu)=0$ for all $u\in \mathcal{X}).$

Furthermore, it follows from Theorem 1 that every correlation contraction $\mathcal{T}$ on a complete b-metric space $(\mathcal{X},\mathfrak{b},s)$ is a ${⪯}_D$-correlation contraction on it: It suffices to observe that conditions (c3) and (c5) are trivially satisfied and, for (c4), notice that every fixed point w of $\mathcal{T}$ obviously verifies $w{⪯}_D\mathcal{T}w.$

We now establish the following variant of Theorem 1.

Theorem 2.

Let $(\mathcal{X},\mathfrak{b},s)$ be a complete b-metric space. Then, every ⪯-correlation contraction on it has a fixed point.

Proof. 

Let $\mathcal{T}$ be a ⪯-correlation contraction on $(\mathcal{X},\mathfrak{b},s)$. Then, there is a partial order ⪯ on $\mathcal{X}$ and a bounded from below function $F:\mathcal{X}\to \mathbb{R}$ for which conditions (c3)–(c5) are fulfilled.

Let $u_0\in \mathcal{X}$ be such that $u_0⪯\mathcal{T}{u}_0.$ Therefore, by (c3), ${\mathcal{T}}^n{u}_0⪯{\mathcal{T}}^{n+1}{u}_0$ for all $n\in \mathbb{N}\cup \left\{0\right\}.$

If ${\mathcal{T}}^n{u}_0={\mathcal{T}}^{n+1}{u}_0$ for some $n,$${\mathcal{T}}^n{u}_0$ is a fixed point of $\mathcal{T}.$

So, we assume that ${\mathcal{T}}^n{u}_0\ne {\mathcal{T}}^{n+1}{u}_0$ for all $n\in \mathbb{N}\cup \left\{0\right\}.$ Thus $\mathfrak{b}({\mathcal{T}}^n{u}_0,{\mathcal{T}}^{n+1}{u}_0)>0$ for all $n\in \mathbb{N}\cup \left\{0\right\},$ and we can apply condition (c5), which implies that

$$\mathfrak{b}({\mathcal{T}}^{n+1}{u}_0,{\mathcal{T}}^{n+2}{u}_0)\le (F\left({\mathcal{T}}^n{u}_0\right)-F\left({\mathcal{T}}^{n+1}{u}_0\right))\mathfrak{b}({\mathcal{T}}^n{u}_0,{\mathcal{T}}^{n+1}{u}_0),$$

for all $n\in \mathbb{N}\cup \left\{0\right\}.$

Thus $F\left({\mathcal{T}}^{n+1}{u}_0\right)<F\left({\mathcal{T}}^n{u}_0\right)$ for all $n\in \mathbb{N}\cup \left\{0\right\},$ so $(F{\left({\mathcal{T}}^n{u}_0\right)}_{n\in \mathbb{N}}$ is a strictly decreasing sequence in $\mathbb{R}$. Hence it converges to the real number ${inf}_n{\mathcal{T}}^n\left(u_0\right)$ (recall that F is bounded from below), and consequently it is a Cauchy sequence in $\mathbb{R}.$

Now, by repeating the argument given by the authors in their proof of Theorem 1 ([1], lines 12–22 of page 2 and line 1 of page 3), we deduce that ${\left({\mathcal{T}}^n{u}_0\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(\mathcal{X},\mathfrak{b},s).$ Therefore, there exists $w\in \mathcal{X}$ such that ${\left({\mathcal{T}}^n{u}_0\right)}_{n\in \mathbb{N}}$ ${\mathfrak{T}}_{\mathfrak{b}}$-converges to $w.$ Thus $w\in acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right),$ and again we can apply (c5) to deduce that

$$\mathfrak{b}({\mathcal{T}}^{n+1}{u}_0,\mathcal{T}w)\le (F\left({\mathcal{T}}^n{u}_0\right)-F\left({\mathcal{T}}^{n+1}{u}_0\right))\mathfrak{b}({\mathcal{T}}^n{u}_0,w),$$

for all $n\in \mathbb{N}\cup \left\{0\right\}.$

Since $(F{\left({\mathcal{T}}^n{u}_0\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}.$ and ${lim}_n\mathfrak{b}({\mathcal{T}}^n{u}_0,w)=0,$ we conclude that ${lim}_n\mathfrak{b}({\mathcal{T}}^{n+1}{u}_0,\mathcal{T}w)=0,$ so $w=\mathcal{T}w.$ This completes the proof. □

We turn our attention to the relationship between Theorems 1 and 2. In connection with this, and as we have point out above, in Example 1 of [1], the authors presented an instance of a complete metric space where we can apply Theorem 1 but not the Banach contraction principle. By the second part of Remark 2, we also can apply Theorem 2 to every correlation contraction in [1] (Example 1). The following is an example where we can apply Theorem 2 but not Theorem 1.

Example 3.

Let $(\mathcal{X},\mathfrak{b},1)$ be the complete metric space where $\mathcal{X}:=\mathbb{N}\cup \left\{0\right\}$ and $\mathfrak{b}$ is the metric on $\mathcal{X}$ given by $\mathfrak{b}(u,u)=0$ for all $u\in \mathcal{X},$$\mathfrak{b}(0,n)=\mathfrak{b}(n,0)=1/n$ for all $n\in \mathbb{N},$ and $\mathfrak{b}(n,m)=max\{1/n,1/m\}$ whenever $n,m\in \mathbb{N}$ with $n\ne m.$

Clearly $acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right)=\left\{0\right\}.$

Let $\mathcal{T}$ be the self mapping of $\mathcal{X}$ defined by $\mathcal{T}0=0$ and $\mathcal{T}n=n(n+1)$ for all $n\in \mathbb{N},$ and let ≤ be the usual order on $\mathcal{X}.$ Then $\mathcal{T}u\le \mathcal{T}v$ whenever $u\le v,$ and $u\le \mathcal{T}u$ for all $u,v\in \mathcal{X},$ so conditions (c3) and (c4) hold.

Now take $F:\mathcal{X}\to \mathbb{R}$ defined by $F\left(0\right)=0$ and $F\left(n\right)=1/n$ for all $n\in \mathbb{N}$. We have $infF\left(\mathcal{X}\right)=0.$ Furthermore, for each $u,v\in \mathcal{X}$, with $u\ne v,$ such that $\mathfrak{b}(u,\mathcal{T}u)>0$ and $v\in (u\uparrow )\cup acc\left((\mathcal{X},{\mathfrak{T}}_{\mathfrak{b}})\right),$ we get

$$\begin{array}{ccc}\hfill \mathfrak{b}(\mathcal{T}u,\mathcal{T}v)& =& \frac{1}{\mathcal{T}u}=\frac{1}{u(u+1)}=\left(\frac{1}{u}-\frac{1}{u(u+1)}\right)\frac{1}{u}\hfill \\ & =& \left(F\right(u)-F(\mathcal{T}u\left)\right)\mathfrak{b}(u,v).\hfill \end{array}$$

We have shown that $\mathcal{T}$ is a ≤-correlation contraction on $(\mathcal{X},\mathfrak{b},1).$

However $\mathcal{T}$ is not a correlation contraction on $(\mathcal{X},\mathfrak{b},1)$: Otherwise, its restriction to $\mathbb{N}$ would also be a correlation contraction on $(\mathbb{N},\mathfrak{b},1)$ and, by Example 2 it would have, at least, a fixed point belonging to $\mathbb{N}.$

We finish the paper with our promised characterization of b-metric completeness and with two observations related to it.

The following lemma, that provides a full b-metric generalization of the corresponding result for metric spaces, will be useful in the proof of the ‘only if’ part of our characterization.

Lemma 1.

If $\mathcal{C}$ is a closed subset of a complete b-metric space $(\mathcal{X},\mathfrak{b},s)$, then $(\mathcal{C},\mathfrak{b}{\mid }_{\mathcal{C}},s)$ is also a complete b-metric space.

Proof. 

Let ${\left(x_n\right)}_{n\in \mathbb{N}}$ be a Cauchy sequence in $(\mathcal{C},\mathfrak{b}{\mid }_{\mathcal{C}},s)$. Then ${\left(x_n\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(\mathcal{X},\mathfrak{b},s).$ Therefore there exists $x\in \mathcal{X}$ such that ${\left(x_n\right)}_{n\in \mathbb{N}}$${\mathfrak{T}}_{\mathfrak{b}}$-converges to $x.$ Since $\mathcal{C}$ is closed we get that $x\in \mathcal{C}.$ Hence $(\mathcal{C},\mathfrak{b}{\mid }_{\mathcal{C}},s)$ is complete. □

Theorem 3.

A b-metric space is complete if and only if every ⪯-correlation contraction on any of its closed subsets has a fixed point.

Proof. 

Let $\mathcal{C}$ be a closed subset of a complete b-metric space $(\mathcal{X},\mathfrak{b},s)$ and let $\mathcal{T}$ be a ⪯-correlation contraction on $\mathcal{C}$ endowed with the restriction of $\mathfrak{b}.$ By Lemma 1, $(\mathcal{C},\mathfrak{b}{\mid }_{\mathcal{C}},s)$ is complete. We deduce from Theorem 2 that $\mathcal{T}$ has a fixed point (in $\mathcal{C}$).

For the converse, suppose that $(\mathcal{X},\mathfrak{b},s)$ is a non-complete b-metric space for which every ⪯-correlation contraction on any of its closed subsets has a fixed point. Then, there exists a non-convergent Cauchy sequence ${\left(u_n\right)}_{n\in \mathbb{N}}$ in $(\mathcal{X},\mathfrak{b},s),$ with $u_n\ne u_m$ whenever $n\ne m.$

From standard arguments we can find a sequence ${\left(j\left(n\right)\right)}_{n\in \mathbb{N}}$ in $\mathbb{N}.$ such that the following properties are fulfilled:

(P1) $j\left(1\right)>1$, $j(n+1)>max\{n+1,j(n\left)\right\}$ for all $n\in \mathbb{N};$

and

(P2) for each $n\in \mathbb{N},$$\mathfrak{b}(u_{j\left(n\right)},u_k)<2^{-(n+1)}b(u_n,u_m)$ whenever $k\ge j\left(n\right)$ and $m\ne n.$

Put $\mathcal{C}=\{u_n:n\in \mathbb{N}\}$ and define a self mapping $\mathcal{T}$ on $\mathcal{C}$ by $\mathcal{T}{u}_n=u_{j\left(n\right)}$ for all $n\in \mathbb{N}.$ Of course $\mathcal{T}$ has no fixed points because $j\left(n\right)>n$ and thus $u_n\ne u_{j\left(n\right)}$ for all $n\in \mathbb{N}.$

Finally, we going to check that $\mathcal{T}$ is a ⪯-correlation contraction on the closed subset $\mathcal{C}$ of $(\mathcal{X},\mathfrak{b},s).$

Let ⪯ be the partial order on $\mathcal{C}$ defined by

$u_n⪯u_m\iff n\le m.$

Condition (c3) is clearly verified: Indeed, if $u_n⪯u_m$ we deduce that $n\le m,$ so $j\left(n\right)\le j\left(m\right)$ and consequently $\mathcal{T}{u}_n=u_{j\left(n\right)}⪯u_{j\left(m\right)}=\mathcal{T}{u}_m.$

Moreover $u_n⪯\mathcal{T}{u}_n$ for all $n\in \mathbb{N}$ because, by (P1), $n<j\left(n\right).$ so condition (c4) is also fulfilled.

Let now $F:\mathcal{C}\to \mathbb{R}$ defined by $F\left(u_n\right)=2^{-n}$ for all $n\in \mathbb{N}$. Thus $infF\left(\mathcal{C}\right)=0.$

Pick $u_n\in \mathcal{C}.$ Then $\mathfrak{b}(u_n,\mathcal{T}{u}_n)>0.$ For each $u_m\in \mathcal{C}\setminus \left\{u_n\right\}$ such that $u_m\in u_n\uparrow ,$ we have $n<m,$ and hence, $j\left(n\right)<j\left(m\right)$, by (P1). Then, from (P2) we deduce that

$$\begin{array}{ccc}\hfill \mathfrak{b}(\mathcal{T}{u}_n,\mathcal{T}{u}_m)& =& \mathfrak{b}(u_{j\left(n\right)},u_{j\left(m\right)})<2^{-(n+1)}\mathfrak{b}(u_n,u_m)\hfill \\ & \le & (2^{-n}-2^{-j\left(n\right)})\mathfrak{b}(u_n,u_m)\hfill \\ & =& (F\left(u_n\right)-F\left(\mathcal{T}{u}_n\right))\mathfrak{b}(u_n,u_m).\hfill \end{array}$$

Hence, condition (c5) is also satisfied (note that $acc\left((\mathcal{C},{\mathfrak{T}}_{\mathfrak{b}}{\mid }_{\mathcal{C}})\right)$ is the empty set).

We conclude that $(\mathcal{X},\mathfrak{b},s)$ is complete. □

Remark 3.

Although the function F constructed in the proof of Theorem 3 satisfies $F\left(u_n\right)>0$ for all $n\in \mathbb{N},$ we could have selected it to fulfill $F\left(u_n\right)<0$ for all $n\in \mathbb{N}.$ For instance, by defining $F\left(u_n\right)=2^{-n}-1$ for all $n\in \mathbb{N}.$

Remark 4.

The metric space $(\mathbb{N},\mathfrak{b},1)$ constructed in Example 2 is not complete. Hence, by Theorem 3, it has closed subsets endowed with ⪯-correlation contractions that are free of fixed points. In fact, the restriction to $\mathbb{N}.$ of the ⪯-correlation contraction $\mathcal{T}$ constructed in Example 3 provides an instance of this situation.