Nemytzki-Edelstein-Meir-Keeler Type Results in b-Metric Spaces

Abstract

We consider some Nemytzki-Edelstein-Meir-Keeler type results in the context of b-metric spaces. In some cases, we assume that the b-metric is continuous. Our results generalize several known ones in existing literature. We also present some examples to illustrate the usability of our results.

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Definitions, Notations, and Preliminaries

Let $\left(X,d\right)$ be a metric space and $T:X\to X$ be a self-mapping. The following Meir-Keeler conditions are well known: For each $\epsilon >\mathrm{0}$, there exists $\delta =\delta \left(\epsilon \right)>\mathrm{0}$ such that$$\begin{array}{}\text{(1)}& \epsilon \le d\left(x,y\right)<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(Tx,Ty\right)<\epsilon ,\end{array}$$or$$\begin{array}{}\text{(2)}& \epsilon <d\left(x,y\right)<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(Tx,Ty\right)\le \epsilon ,\end{array}$$or $T$ is contractive and$$\begin{array}{}\text{(3)}& \epsilon \le d\left(x,y\right)<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(Tx,Ty\right)\le \epsilon .\end{array}$$It is clear that (1) implies (2) and (2) implies (3), while the converse is not true. One says that the mapping $T$ defined on the metric space $\left(X,d\right)$ is contractive if $d\left(Tx,Ty\right)<d\left(x,y\right)$ holds, whenever $x\ne y$. For more details, see [1] (pages 30-33 and 56-58) and [2]. In 1969, Meir-Keeler [2] proved the following.

Theorem 1 ([2], Theorem).

Let $\left(X,d\right)$ be a complete metric space and let $T$ be a self-mapping on $X$ satisfying (1). Then $T\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{\hspace{0.17em}\hspace{0.17em}}$ has a unique fixed point, say $u\in X$, and, for each $x\in X$, ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^{n}x=u$.

For other fixed point results via generalized Meir-Keeler contractions, see [3–5]. Inspired from Meir-Keeler theorem, Ćirić proved the next slightly more general result.

Theorem 2 ([1], Theorem 2.5).

Let $\left(X,d\right)$ be a complete metric space and let $T$ be a self-mapping on $X$ satisfying (2). Then $T\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{\hspace{0.17em}\hspace{0.17em}}$ has a unique fixed point, say $u\in X$, and, for each $x\in X$, ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^{n}x=u.$

The following example shows that Ćirić result is a proper generalization of Meir-Keeler theorem.

Example 3.

Let $X=[\mathrm{0,1}]\cup \left\{\mathrm{2,3}+\mathrm{1}/\mathrm{3},\cdots ,\mathrm{3}n-\mathrm{1,3}n+\mathrm{1}/\mathrm{3}n,\cdots \right\}$ be a subset of reals with the Euclidean metric and let $T$ be a self-mapping on $X$ defined by $$\begin{array}{}\text{(4)}& Tx=\mathrm{0},\hspace{1em}\text{if\hspace{0.17em}\hspace{0.17em}}\mathrm{0}\le x\le \mathrm{1},\hspace{0.17em}\hspace{0.17em}x=\mathrm{2},\cdots ,\mathrm{3}n-\mathrm{1},\cdots \text{(5)}& Tx=\mathrm{1},\hspace{1em}\text{if\hspace{0.17em}\hspace{0.17em}}x=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}},\mathrm{6}+\frac{\mathrm{1}}{\mathrm{6}},\cdots ,\mathrm{3}n+\frac{\mathrm{1}}{\mathrm{3}n},\cdots \end{array}$$Then one can verify that $T$ satisfies (2), while it does not satisfy the Meir-Keeler condition (1). For all details, see [1].

Remark 4.

Theorems 1 and 2 are true if the self-mapping $T:X\to X$ satisfies condition (3). For more details, see [1], pages 30-33.

On the other hand, Bakhtin [6] and Czerwik [7] introduced the concept of b-metric spaces (a generalization of metric spaces) and proved the Banach contraction principle. The definition of a b-metric space is the following.

Definition 5 (Bakhtin [6] and Czerwik [7]).

Let $X$ be a nonempty set and let $s\ge \mathrm{1}$ be a given real number. A function $d:X\times X\to [\mathrm{0},\infty )$ is said to be a b-metric if and only if, for all $x,y,z\in X$, the following conditions are satisfied:

(b1)

$d\left(x,y\right)=\mathrm{0}$ if and only if $x=y.$

(b2)

$d\left(x,y\right)=d\left(y,x\right).$

(b3)

$d\left(x,z\right)\le s\left[d\left(x,y\right)+d\left(y,z\right)\right].$

The triplet $\left(X,d,s\ge \mathrm{1}\right)$ is called a b-metric space with coefficient $s.$

In the last period, many authors obtained several fixed point results for single-valued or set-valued mappings in the context of b-metric spaces. For more details, see [5, 8–35]. It should be noted that the class of b-metric spaces is effectively larger than that of standard metric spaces, since a b-metric is a metric when $s=\mathrm{1}.$ The following example shows that, in general, a b-metric does not necessarily need to be a metric.

Example 6.

Let $\left(X,\rho \right)$ be a metric space and $d\left(x,y\right)={\left(\rho \left(x,y\right)\right)}^p$ where $p>\mathrm{1}$ is a real number. Then $d$ is a b-metric with $s={\mathrm{2}}^{p-\mathrm{1}}$, but $d$ is not a metric on $X.$

The concepts of b-convergence, b-completeness, b-Cauchy, and b-closed set in b-metric spaces have been initiated in [6, 7].

The following two lemmas are very significant in the class of b-metric spaces.

Lemma 7 ([21], Lemma 3.1).

Let $\left\{y_n\right\}$ be a sequence in a b-metric space $\left(X,d,s\ge \mathrm{1}\right)$ such that$$\begin{array}{}\text{(6)}& d\left(y_n,y_{n+\mathrm{1}}\right)\le \lambda d\left(y_{n-\mathrm{1}},y_n\right)\end{array}$$for some $\lambda \in [\mathrm{0},\mathrm{1}/s)$, and each $n=\mathrm{1,2},\cdots$. Then $\left\{y_n\right\}$ is a b-Cauchy sequence in a b-metric space $\left(X,d\right).$

Lemma 8 ([30], Lemma 2.2).

Let $\left\{y_n\right\}$ be a sequence in a b-metric space $\left(X,d,s\ge \mathrm{1}\right)$ such that$$\begin{array}{}\text{(7)}& d\left(y_n,y_{n+\mathrm{1}}\right)\le \lambda d\left(y_{n-\mathrm{1}},y_n\right)\end{array}$$for some $\lambda \in [\mathrm{0,1})$, and each $n=\mathrm{1,2},\cdots$. Then $\left\{y_n\right\}$ is a b-Cauchy sequence in a b-metric space $\left(X,d\right).$

Since in general a b-metric is not continuous, we need the following two lemmas.

Lemma 9 ([36], Lemma 2.1).

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a b-metric space with $s\ge \mathrm{1}$. Suppose that $\left\{x_n\right\}$ and $\left\{y_n\right\}$ are b-convergent to $x$ and $y$, respectively. Then $$\begin{array}{}\text{(8)}& \frac{\mathrm{1}}{s^{\mathrm{2}}}d\left(x,y\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{n\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_n,y_n\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{n\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_n,y_n\right)\le s^{\mathrm{2}}d\left(x,y\right).\end{array}$$In particular, if $x=y$, then we have ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }d\left(x_n,y_n\right)=\mathrm{0}.$ Moreover, for each $z\in X$, we have $$\begin{array}{}\text{(9)}& \frac{\mathrm{1}}{s}d\left(x,z\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{n\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_n,z\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{n\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_n,z\right)\le s^{\mathrm{2}}d\left(x,z\right).\end{array}$$

Lemma 10 (see [37]).

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a b-metric space and $\left\{x_n\right\}$ be a sequence in $X$ such that $$\begin{array}{}\text{(10)}& \underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(x_n,x_{n+\mathrm{1}}\right)=\mathrm{0}.\end{array}$$If $\left\{x_n\right\}$ is not b-Cauchy, then there exist $\epsilon >\mathrm{0}$ and two sequences $\left\{m\left(k\right)\right\}$ and $\left\{n\left(k\right)\right\}$ of positive integers such that, for the following four sequences $$\begin{array}{c}\text{(11)}& d\left(x_{m\left(k\right)},x_{n\left(k\right)}\right),d\left(x_{m\left(k\right)},x_{n\left(k\right)+\mathrm{1}}\right),\\ d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)}\right),\\ d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)+\mathrm{1}}\right),\end{array}$$we have $$\begin{array}{c}\text{(12)}& \epsilon \le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_{m\left(k\right)},x_{n\left(k\right)}\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_{m\left(k\right)},x_{n\left(k\right)}\right)\le \epsilon s,\frac{\epsilon }{s}\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_{m\left(k\right)},x_{n\left(k\right)+\mathrm{1}}\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_{m\left(k\right)},x_{n\left(k\right)+\mathrm{1}}\right)\le \epsilon s^{\mathrm{2}},\\ \frac{\epsilon }{s}\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)}\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)}\right)\le \epsilon s^{\mathrm{2}},\\ \frac{\epsilon }{s^{\mathrm{2}}}\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{i}\mathrm{n}\mathrm{f}}\hspace{0.17em}d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)+\mathrm{1}}\right)\le \mathrm{l}\mathrm{i}\mathrm{m}\underset{k\to \infty }{\mathrm{s}\mathrm{u}\mathrm{p}}\hspace{0.17em}d\left(x_{m\left(k\right)+\mathrm{1}},x_{n\left(k\right)+\mathrm{1}}\right)\le \epsilon s^{\mathrm{3}}.\end{array}$$

Essential to the proofs of fixed point theorems for the most contractive conditions in the context of b-metric spaces are the above two lemmas (see, for example, [3, 5, 9, 13, 17, 23, 25, 28]). However, it is not hard to show that the proofs of the most fixed point theorems in the context of b-metric spaces become simpler and shorter if they are based on Lemma 8.

2. Main Result

To our knowledge, it is not known whether Meir-Keeler and Ćirić theorems hold in the context of b-metric spaces. Also, it is not known that if there are examples such that condition (1) or (2) or (3) holds in the context of b-metric spaces, but $T$ has no fixed point.

Our first result generalizes Lemma 1 of [2]. For some results also see recent paper [38].

Lemma 11.

Let $\left(X,d,s>\mathrm{1}\right)$ be a b-complete b-metric space and $T:X\to X$ such that condition (1) holds. If $\left\{T^{n}x\right\}$ is a b-Cauchy sequence for each $x\in X$, then $T$ has a unique fixed point, say $u\in X$ and $T^{n}x\to u.$

Proof.

Since $\left(X,d,s>\mathrm{1}\right)$ is b-complete, each $\{T^{n}x\}$ has a limit point, say $\eta \left(x\right).$ Since condition (1) implies the continuity of $T$, we have$$\begin{array}{}\text{(13)}& T\left(\eta \left(x\right)\right)=T\left(\underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}{T}^n\left(x\right)\right)=\underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}{T}^{n+\mathrm{1}}\left(x\right)=\eta \left(x\right).\end{array}$$Thus, $\eta \left(x\right)$ is a fixed point, and therefore all $\eta \left(x\right)$ are equal.

Remark 12.

If condition (1) holds on $b$-metric spaces $\left(X,d,s>\mathrm{1}\right)$, we do not know whether every sequence $\left\{T^{n}x\right\}$ is b-Cauchy.

However, with a stronger condition than (1), we have a positive response. It will be the subject of Theorem 13.

Now, we announce a Meir-Keeler type result in the context of b-metric spaces.

Theorem 13.

Let $\left(X,d\right)$ be a complete b-metric space and let $T$ be a self-mapping on $X$ satisfying the following condition.

Given $\epsilon >\mathrm{0}$ there exists $\delta >\mathrm{0}$ such that $$\begin{array}{}\text{(14)}& \epsilon \le d\left(x,y\right)<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}{s}^{a}d\left(Tx,Ty\right)<\epsilon ,\end{array}$$where $a>\mathrm{0}$ is given.

Then $T$ has a unique fixed point, say $u\in X$, and, for each $x\in X,{\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^{n}x=u.$

Proof.

It is clear that, for all $x,y\in X$ with $d\left(x,y\right)>\mathrm{0}$, we obtain$$\begin{array}{}\text{(15)}& d\left(Tx,Ty\right)\le \lambda d\left(x,y\right),\end{array}$$where $\lambda =\mathrm{1}/s^a\in [\mathrm{0,1}).$

Let $x_{\mathrm{0}}\in X$ be an arbitrary point. Define the sequence $\left\{x_n\right\}$ by $x_{n+\mathrm{1}}=T{x}_n$ for all $n\ge \mathrm{0}.$ If, for some $n$, $x_n=x_{n+\mathrm{1}}$, then $x_n$ is a fixed point of $T$. From now on, suppose that $x_n\ne x_{n+\mathrm{1}}$ for all $n\ge \mathrm{0}.$ From condition (15), we obtain$$\begin{array}{}\text{(16)}& d\left(x_n,x_{n+\mathrm{1}}\right)\le \lambda d\left(x_{n-\mathrm{1}},x_n\right).\end{array}$$Further, according to ([30], Lemma 2.2.), the sequence $\left\{x_n\right\}$ is b-Cauchy in the b-metric space $\left(X,d\right).$ By b-completeness of $\left(X,d\right)$, there exists $u\in X$ such that$$\begin{array}{}\text{(17)}& \underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}{x}_n=u.\end{array}$$Finally, (15) and (17) imply that $Tu=u$; that is, $u$ is the unique fixed point of $T$ in $X$.

Example 14.

Let $X=\left\{\mathrm{0,1},\mathrm{2}\right\}$ and define $d:X\times X\to [\mathrm{0},+\infty )$ as follows: $d\left(x,x\right)=\mathrm{0}$, $d\left(x,y\right)=d\left(y,x\right)$ for all $x,y\in X$, $d\left(\mathrm{0,1}\right)=\mathrm{1}$, $d\left(\mathrm{0,2}\right)=\mathrm{2.2}$, and $d\left(\mathrm{1,2}\right)=\mathrm{1.1}.$ Then $\left(X,d,\mathrm{22}/\mathrm{21}>\mathrm{1}\right)$ is a b-complete b-metric space, but it is not a metric space. Let $T:X\to X$ be defined by $$\begin{array}{}\text{(18)}& Tx=\left\{\begin{array}{cc}\mathrm{0},\hfill & \text{if\hspace{0.17em}\hspace{0.17em}}x\ne \mathrm{2},\hfill \\ \mathrm{1},\hfill & \text{if\hspace{0.17em}\hspace{0.17em}}x=\mathrm{2}.\hfill \end{array}\right.\end{array}$$We shall check that, for all $x,y\in X$, the contractive condition (15) holds. For this, we distinguish three cases.

(a) $x=\mathrm{0},y=\mathrm{1}\Rightarrow d\left(T\mathrm{0},T\mathrm{1}\right)=d\left(\mathrm{0,0}\right)=\mathrm{0}.$ Obviously, condition (15) holds.

(b) $x=\mathrm{0},y=\mathrm{2}\Rightarrow d\left(T\mathrm{0},T\mathrm{2}\right)=d\left(\mathrm{0,1}\right).$ Since ${\left(\mathrm{22}/\mathrm{21}\right)}^{a}d\left(\mathrm{0,1}\right)\le d\left(\mathrm{0,2}\right)$, i.e., ${\left(\mathrm{22}/\mathrm{21}\right)}^a·\mathrm{1}\le \mathrm{2.2},a>\mathrm{0}$, which is true, hence, again (15) holds.

(c) $x=\mathrm{1},y=\mathrm{2}\Rightarrow d\left(T\mathrm{1},T\mathrm{2}\right)=d\left(\mathrm{0,1}\right)=\mathrm{1}.$ Now, we have ${\left(\mathrm{22}/\mathrm{21}\right)}^a·\mathrm{1}\le \mathrm{1.1}$, i.e., ${\left(\mathrm{22}/\mathrm{21}\right)}^a\le \mathrm{1.1}$, which is also true because $a>\mathrm{0}.$

Therefore, condition (15) holds for each $a>\mathrm{0}.$ However, condition (14) is not true for $a=\mathrm{1}.$ Indeed, for $x=\mathrm{0}$ and $y=\mathrm{2}$, it becomes $$\begin{array}{}\text{(19)}& \epsilon \le d\left(\mathrm{0,2}\right)<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{22}}{\mathrm{21}}<\epsilon ,\end{array}$$or equivalently $$\begin{array}{}\text{(20)}& \epsilon \le \mathrm{2.2}<\epsilon +\delta \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{22}}{\mathrm{21}}<\epsilon .\end{array}$$Take $\epsilon =\mathrm{1}/\mathrm{2}$. Then there exists $\delta =\delta \left(\mathrm{1}/\mathrm{2}\right)>\mathrm{0}$ such that $\mathrm{1}/\mathrm{2}\le \mathrm{2.2}<\mathrm{1}/\mathrm{2}+\delta$ (for example, any $\delta >17/10$). But $\mathrm{22}/\mathrm{21}<\mathrm{1}/\mathrm{2}$ is false.

Now, we give an example supporting Theorem 13.

Example 15.

Let $X=\left[\mathrm{0,1}\right],d\left(x,y\right)={\left(x-y\right)}^{\mathrm{2}}.$ Then $\left(X,d,\mathrm{2}\right)$ is a b-complete b-metric space. Let $T:X\to X$ be defined as $Tx=\left(\mathrm{1}/\mathrm{4}\right)x^{\mathrm{2}}$. Taking $\delta =\epsilon$, we get, for $x$ and $y$ satisfying $\epsilon \le d\left(x,y\right)<\epsilon +\delta =\mathrm{2}\epsilon$, $$\begin{array}{}\text{(21)}& s·d\left(Tx,Ty\right)=\mathrm{2}·d\left(Tx,Ty\right)=\mathrm{2}·\frac{\mathrm{1}}{\mathrm{16}}{\left(x-y\right)}^{\mathrm{2}}{\left(x+y\right)}^{\mathrm{2}}\le \frac{\mathrm{1}}{\mathrm{2}}{\left(x-y\right)}^{\mathrm{2}}<\epsilon .\end{array}$$Hence, all the conditions of Theorem 13 are satisfied. The mapping $T$ has a unique fixed point, which is $u=\mathrm{0}.$

Let ${\mathcal{G}}_s$ be the class of all mappings $g:[\mathrm{0},\infty )\to [\mathrm{0},\mathrm{1}/s)$ which satisfy the condition: $g\left(t_n\right)\to \mathrm{1}/s$, whenever $t_n\to \mathrm{0}.$ Note that ${\mathcal{G}}_s\ne \varnothing .$ As an example, consider the mapping $g:[\mathrm{0},\infty )\to [\mathrm{0},\mathrm{1}/s)$ given by $g\left(t\right)=\left(\mathrm{1}/s\right)e^{-t}$ for $t>\mathrm{0}$ and $g\left(\mathrm{0}\right)\in [\mathrm{0},\mathrm{1}/s).$

The following is Geraghty type result in the context of b-metric spaces (see, for instance, [17], where authors use Lemma 1.4.).

Theorem 16.

Let $\left(X,d,s>\mathrm{1}\right)$ be a complete b-metric space. Suppose that the mapping $T:X\to X$ satisfies the condition $$\begin{array}{}\text{(22)}& d\left(Tx,Ty\right)\le g\left(d\left(x,y\right)\right)d\left(x,y\right)\end{array}$$for all $x,y\in X$ and some $g\in {\mathcal{G}}_s.$ Then $T$ has a unique fixed point $u\in X$ and for each $x\in X$ the Picard sequence $\left\{T^{n}x\right\}$ converges to $u$ in $\left(X,d,s>\mathrm{1}\right).$

Proof.

Since $g:[\mathrm{0},\infty )\to [\mathrm{0},\mathrm{1}/s)$, we get$$\begin{array}{}\text{(23)}& d\left(Tx,Ty\right)\le \frac{\mathrm{1}}{s}d\left(x,y\right)=\lambda d\left(x,y\right).\end{array}$$In view of $\lambda \in \left(\mathrm{0,1}\right)$, the result follows according to Lemma 8. and condition (15).

It is well known that, in compact metric spaces, fixed point results can be obtained under the strict contractive condition ($d\left(Tx,Ty\right)<d\left(x,y\right)$ whenever $x\ne y$). In the case of b-metric spaces with a continuous b-metric, the following results of Nemytzki and Edelstein can be obtained in the same way as in the metric case (see [1], pages 56-58).

Theorem 17.

Let $\left(X,d,s>\mathrm{1}\right)$ be a compact b-metric space with continuous b-metric $d$ and let $T:X\to X$ be a self-mapping. Suppose that the following condition holds: $$\begin{array}{}\text{(24)}& d\left(Tx,Ty\right)<d\left(x,y\right)\hspace{1em}\text{for\hspace{0.17em}\hspace{0.17em}}x\ne y.\end{array}$$Then $T$ has a unique fixed point, say $u\in X$, and, for each $x\in X,{\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^{n}x=u.$

Proof.

Define a function $f:X\to [\mathrm{0},+\infty )$ by$$\begin{array}{}\text{(25)}& f\left(x\right)=d\left(x,Tx\right).\end{array}$$Since $T$ is continuous, $f$ is also continuous. So, as $\left(X,d,s>\mathrm{1}\right)$ is a compact b-metric space, there exists a point $u\in X$ such that$$\begin{array}{}\text{(26)}& f\left(u\right)=d\left(u,Tu\right)=\underset{x\in X}{\mathrm{m}\mathrm{i}\mathrm{n}}\left\{d\left(x,Tx\right)\right\}.\end{array}$$If we assume that $u\ne Tu$, then as $T$ is contractive ($d\left(Tx,Ty\right)<d\left(x,y\right)$ whether $x\ne y$), one writes$$\begin{array}{}\text{(27)}& f\left(Tu\right)=d\left(Tu,TTu\right)<d\left(u,Tu\right)=f\left(u\right),\end{array}$$which is a contradiction. Therefore, $u$ is a fixed point of $T.$ The uniqueness is obvious.

Theorem 18.

Let $\left(X,d,s>\mathrm{1}\right)$ be a b-metric space with a continuous b-metric $d$ and let $T:X\to X$ be a self-map. Suppose that the following condition holds: $$\begin{array}{}\text{(28)}& d\left(Tx,Ty\right)<d\left(x,y\right)\hspace{1em}\text{for\hspace{0.17em}\hspace{0.17em}}x\ne y.\end{array}$$If there exists a point $x_{\mathrm{0}}\in X$ such that the sequence $\left\{T^n{x}_{\mathrm{0}}\right\}$ contains a convergent subsequence $\left\{T^{n_i}{x}_{\mathrm{0}}\right\}$ to $u$, then $u$ is the unique fixed point of $T.$

Proof.

Consider the real sequence $\left\{d\left(T^n{x}_{\mathrm{0}},T^{n+\mathrm{1}}{x}_{\mathrm{0}}\right)\right\}.$ If $T^{k+\mathrm{1}}{x}_{\mathrm{0}}=T^k{x}_{\mathrm{0}}$ for some $k\in \mathbb{N}$, then $\left\{T^n{x}_{\mathrm{0}}\right\}$ for $n\ge k$ is a stationary sequence and so $T^k{x}_{\mathrm{0}}=u.$ Thus, $T^{k+\mathrm{1}}{x}_{\mathrm{0}}=T^k{x}_{\mathrm{0}}$ implies $Tu=u.$ Assume now that $T^{n+\mathrm{1}}{x}_{\mathrm{0}}\ne T^n{x}_{\mathrm{0}}$ for all $n\in \mathbb{N}.$ Then as $T$ is contractive, $\left\{d\left(T^n{x}_{\mathrm{0}},T^{n+\mathrm{1}}{x}_{\mathrm{0}}\right)\right\}$ is a strictly decreasing sequence of positive reals. Therefore, it converges. Since $T^{n_i}{x}_{\mathrm{0}}\to u$ as $i\to \infty$ and $T$ is continuous, we have$$\begin{array}{c}\text{(29)}& \underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}{T}^{n_i+\mathrm{1}}{x}_{\mathrm{0}}=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}T{T}^{n_i}{x}_{\mathrm{0}}=Tu;\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}{T}^{n_i+\mathrm{2}}{x}_{\mathrm{0}}=T^{\mathrm{2}}u.\end{array}$$Thus $$\begin{array}{c}\text{(30)}& \underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+\mathrm{1}}{x}_{\mathrm{0}}\right)=d\left(u,Tu\right),\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i+\mathrm{1}}{x}_{\mathrm{0}},T^{n_i+\mathrm{2}}{x}_{\mathrm{0}}\right)=d\left(Tu,T^{\mathrm{2}}u\right).\end{array}$$Since $\{d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+\mathrm{1}}{x}_{\mathrm{0}}\right)\}$ and $\{d\left(T^{n_i+\mathrm{1}}{x}_{\mathrm{0}},T^{n_i+\mathrm{2}}{x}_{\mathrm{0}}\right)\}$ are subsequences of the convergent sequence $\left\{d\left(T^n{x}_{\mathrm{0}},T^{n+\mathrm{1}}{x}_{\mathrm{0}}\right)\right\}$, they have the same limit. Therefore,$$\begin{array}{}\text{(31)}& d\left(u,Tu\right)=d\left(Tu,T^{\mathrm{2}}u\right).\end{array}$$Hence $Tu=u.$ If not, as $T$ is contractive, we have$$\begin{array}{}\text{(32)}& d\left(Tu,T^{\mathrm{2}}u\right)<d\left(u,Tu\right),\end{array}$$which is a contradiction.

Remark 19.

The two previous theorems are known in literature as Nemytzki and Edelstein theorems, respectively. It is clear that Edelstein theorem extends the result of Nemytzki.

In the sequel, we consider $\epsilon -$contractive mappings in the context of b-metric spaces. Namely, we first introduce the following.

Definition 20.

A mapping $T$ of a b-metric space $\left(X,d,s\ge \mathrm{1}\right)$ into itself is said to be $\epsilon -$contractive, if and only if there exists $\epsilon >\mathrm{0}$ such that$$\begin{array}{}\text{(33)}& \mathrm{0}<d\left(x,y\right)<\epsilon \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(Tx,Ty\right)<d\left(x,y\right).\end{array}$$The following results extend ones from standard metric spaces to b-metric spaces, with a continuous b-metric $d.$

Theorem 21.

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a b-metric space with continuous b-metric $d$ and $T$ an $\epsilon -$contractive self-mapping on $X.$ If, for some $x_{\mathrm{0}}\in X$, the sequence $\left\{T^n{x}_{\mathrm{0}}\right\}$ has a convergent subsequence $\{T^{n_i}{x}_{\mathrm{0}}\}$ to $u\in X$, then $u$ is a periodic point; that is, there exists a positive integer $k$ such that $T^{k}u=u.$

Proof.

Since $\left\{T^n{x}_{\mathrm{0}}\right\}$ has a cluster point, there exist positive integers $p$ and $q$ with $p<q$ such that $d\left(T^p{x}_{\mathrm{0}},T^q{x}_{\mathrm{0}}\right)<\epsilon$; that is, $d\left(T^p{x}_{\mathrm{0}},T^k{T}^p{x}_{\mathrm{0}}\right)<\epsilon$, where $k=q-p.$ Then the sequence ${\left\{d\left(T^n{x}_{\mathrm{0}},T^{n+k}{x}_{\mathrm{0}}\right)\right\}}_{n=p}^{+\infty }$ is nonincreasing due to the fact that $T$ is $\epsilon -$contractive. Thus, this sequence converges and so$$\begin{array}{}\text{(34)}& \underset{n\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^n{x}_{\mathrm{0}},T^{n+k}{x}_{\mathrm{0}}\right)=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+k}{x}_{\mathrm{0}}\right)=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i+\mathrm{1}}{x}_{\mathrm{0}},T^{n_i+k+\mathrm{1}}{x}_{\mathrm{0}}\right).\end{array}$$Since $T$ and the b-metric $d$ are both continuous, we have$$\begin{array}{}\text{(35)}& d\left(u,T^{k}u\right)=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+k}{x}_{\mathrm{0}}\right)\le d\left(T^p{x}_{\mathrm{0}},T^k{T}^p{x}_{\mathrm{0}}\right)<\epsilon ,\end{array}$$and$$\begin{array}{}\text{(36)}& d\left(Tu,T{T}^{k}u\right)=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i+\mathrm{1}}{x}_{\mathrm{0}},T^{n_i+k+\mathrm{1}}{x}_{\mathrm{0}}\right).\end{array}$$Thus,$$\begin{array}{}\text{(37)}& d\left(Tu,T{T}^{k}u\right)=d\left(u,T^{k}u\right)<\epsilon .\end{array}$$Hence, $T^{k}u=u$. Otherwise, as $T$ is $\epsilon -$contractive, we would have $d\left(Tu,T{T}^{k}u\right)<d\left(u,T^{k}u\right)$.

Now, consider a class of mappings $T$ of a b-metric space $\left(X,d,s\ge \mathrm{1}\right)$ into itself which satisfy the following condition.

For every $x,y\in X$ there exists a positive integer $k\left(x,y\right)$ such that$$\begin{array}{}\text{(38)}& \mathrm{0}<d\left(x,y\right)\text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(T^{n}x,T^{n}y\right)<d\left(x,y\right)\hspace{1em}\forall n\ge k\left(x,y\right).\end{array}$$A mapping $T$ satisfying (38) is called an eventually contractive mapping.

It is obvious that any contractive mapping is eventually contractive (it satisfies (38) with $k\left(x,y\right)=\mathrm{1}$), but the implication is not reverse.

Contractive and $\epsilon -$contractive mappings are continuous. However, eventually contractive mappings need not be continuous nor orbitally continuous. Recall that a mapping $T$ is said to be orbitally continuous if, for each $x\in X$, $T^{n_i}x\to u\in X$ implies $T{T}^{n_i}x\to Tu.$

Now, we announce the next result.

Theorem 22.

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a b-metric space with continuous b-metric $d$ and $T:X\to X$ be an eventually contractive and orbitally continuous mapping. If, for some $x_{\mathrm{0}}\in X$, the sequence of iterates $\left\{T^n{x}_{\mathrm{0}}\right\}$ has a subsequence $\{T^{n_i}{x}_{\mathrm{0}}\}$ converging to $u\in X$, then $u$ is the unique fixed point of $T$ and ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^n{x}_{\mathrm{0}}=u.$

Proof.

If $d\left(T^{n_{\mathrm{0}}}{x}_{\mathrm{0}},T^{n_{\mathrm{0}}+\mathrm{1}}{x}_{\mathrm{0}}\right)=\mathrm{0}$ for some $n_{\mathrm{0}}\in \mathbb{N}$, then $T^{n}x=T^{n_{\mathrm{0}}}{x}_{\mathrm{0}}$ for all $n\ge n_{\mathrm{0}}.$ Thus, $T^{n_{\mathrm{0}}}{x}_{\mathrm{0}}=u$ and so $T^{n_{\mathrm{0}}+\mathrm{1}}{x}_{\mathrm{0}}=T^{n_{\mathrm{0}}}{x}_{\mathrm{0}}$ implies that $Tu=u.$

Assume now that $d\left(T^{n_{\mathrm{0}}}{x}_{\mathrm{0}},T^{n_{\mathrm{0}}+\mathrm{1}}{x}_{\mathrm{0}}\right)>\mathrm{0}$ for all $n\in \mathbb{N}.$ Consider $u={\mathrm{l}\mathrm{i}\mathrm{m}}_{i\to \infty }{T}^{n_i}{x}_{\mathrm{0}}.$ Since $T$ is orbitally continuous, for any fixed positive integer $p$, we have$$\begin{array}{}\text{(39)}& \underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i+p}{x}_{\mathrm{0}},T^{n_i+p+\mathrm{1}}{x}_{\mathrm{0}}\right)=d\left(T^{p}u,T^{p+\mathrm{1}}u\right).\end{array}$$Assume that $Tu\ne u.$ As $T$ is eventually contractive, there exists $r\in \mathbb{N}$ such that$$\begin{array}{}\text{(40)}& d\left(T^{r}u,T^r{T}^u\right)<d\left(u,Tu\right).\end{array}$$Hence ${\epsilon }_{\mathrm{0}}=\left(\mathrm{1}/\mathrm{2}\right)\left[d\left(u,Tu\right)-d\left(T^{r}u,T^r{T}^u\right)\right]>\mathrm{0}.$ Since$$\begin{array}{}\text{(41)}& \underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i+r}{x}_{\mathrm{0}},T^{n_i+r+\mathrm{1}}{x}_{\mathrm{0}}\right)=d\left(T^{r}u,TTu\right),\end{array}$$for arbitrary $\epsilon >\mathrm{0}$, there exists a sufficiently large $n_i=q$ such that$$\begin{array}{}\text{(42)}& d\left(T^{q+r}{x}_{\mathrm{0}},T^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)<d\left(T^{r}u.T^{\mathrm{r}}Tu\right)+\epsilon .\end{array}$$For $\epsilon ={\epsilon }_{\mathrm{0}}$, we have$$\begin{array}{}\text{(43)}& d\left(T^{q+r}{x}_{\mathrm{0}},T^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)<\frac{\mathrm{1}}{\mathrm{2}}\left[d\left(u,Tu\right)+d\left(T^{r}u,T^{r}Tu\right)\right].\end{array}$$Since $d\left(T^{n_{\mathrm{0}}}{x}_{\mathrm{0}},T^{n_{\mathrm{0}}+\mathrm{1}}{x}_{\mathrm{0}}\right)>\mathrm{0}$ for all $n\in \mathbb{N}$ and $T$ is eventually contractive, there exists some positive integer $k=k\left(T^{q+r}{x}_{\mathrm{0}},T^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)$ such that$$\begin{array}{}\text{(44)}& d\left(T^n{T}^{q+r}{x}_{\mathrm{0}},T^n{T}^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)<d\left(T^{q+r}{x}_{\mathrm{0}},T^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)\hspace{1em}\forall n\ge k;\end{array}$$that is, $d\left(T^n{x}_{\mathrm{0}},T^{n+\mathrm{1}}{x}_{\mathrm{0}}\right)<d\left(T^{q+r}{x}_{\mathrm{0}},T^{q+r+\mathrm{1}}{x}_{\mathrm{0}}\right)$ for all $n\ge k+q+r.$ So, we obtain that$$\begin{array}{}\text{(45)}& d\left(T^n{x}_{\mathrm{0}},T^{n+\mathrm{1}}{x}_{\mathrm{0}}\right)<\frac{\mathrm{1}}{\mathrm{2}}\left[d\left(u,Tu\right)+d\left(T^{r}u,T^{r}Tu\right)\right]\hspace{1em}\forall n\ge k+q+r.\end{array}$$Hence$$\begin{array}{}\text{(46)}& d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+\mathrm{1}}{x}_{\mathrm{0}}\right)<\frac{\mathrm{1}}{\mathrm{2}}\left[d\left(u,Tu\right)+d\left(T^{r}u,T^{r}Tu\right)\right]\hspace{1em}\forall n_i\ge k+q+r.\end{array}$$Thus we get$$\begin{array}{}\text{(47)}& d\left(u,Tu\right)=\underset{i\to \infty }{\mathrm{l}\mathrm{i}\mathrm{m}}\hspace{0.17em}d\left(T^{n_i}{x}_{\mathrm{0}},T^{n_i+\mathrm{1}}{x}_{\mathrm{0}}\right)\le \frac{\mathrm{1}}{\mathrm{2}}\left[d\left(u,Tu\right)+d\left(T^{r}u,T^{r}Tu\right)\right].\end{array}$$Hence$$\begin{array}{}\text{(48)}& d\left(u,Tu\right)\le d\left(T^{r}u,T^{r}Tu\right),\end{array}$$which contradicts the choice of $r.$ Therefore, $d\left(u,Tu\right)=\mathrm{0}$; that is, $Tu=u.$

Now, we show that ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\infty }{T}^n{x}_{\mathrm{0}}=u.$ Let $\epsilon >\mathrm{0}$ be arbitrary. Since $T^{n_i}{x}_{\mathrm{0}}\to u$ as $i\to \infty$, there exists a sufficiently large $n_i=l$ such that $d\left(u,T^l{x}_{\mathrm{0}}\right)<\epsilon$. If $T^{n_i}{x}_{\mathrm{0}}\ne u$, then $T^{l+\mathrm{1}}{x}_{\mathrm{0}}=Tu=u$ and hence $T^n{x}_{\mathrm{0}}=u$ for all $n\ge l.$ Assume that $d\left(u,T^l{x}_{\mathrm{0}}\right)>\mathrm{0}.$ As $T$ is eventually $\epsilon -$contractive and $d\left(u,T^l{x}_{\mathrm{0}}\right)<\epsilon$, there is $k=k\left(u,T^l{x}_{\mathrm{0}}\right)\in \mathbb{N}$ such that$$\begin{array}{}\text{(49)}& d\left(T^{n}u,T^n{T}^l{x}_{\mathrm{0}}\right)<d\left(u,T^l{x}_{\mathrm{0}}\right)\hspace{1em}\forall n\ge k.\end{array}$$Since $Tu=u$, we have$$\begin{array}{}\text{(50)}& d\left(u,T^n{x}_{\mathrm{0}}\right)=d\left(T^{n-l}u,T^{n-l}{T}^l{x}_{\mathrm{0}}\right)<d\left(u,T^{l}u\right)<\epsilon \hspace{1em}\forall n\ge k+l.\end{array}$$Therefore, ${\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^n{x}_{\mathrm{0}}=u.$

Corollary 23.

If $\left(X,d,s\ge \mathrm{1}\right)$ is a compact b-metric space with a continuous b-metric $d$ and $T$ is a continuous and eventually contractive self-mapping on $\left(X,d\right)$, then $T$ has a unique fixed point, say $u\in X$. Also, for every $x\in X,{\mathrm{l}\mathrm{i}\mathrm{m}}_{n\to \infty }{T}^{n}x=u.$

A mapping $T$ of a b-metric space $\left(X,d,s\ge \mathrm{1}\right)$ into itself is said to be an eventually $\epsilon -$contractive mapping if there exists $\epsilon >\mathrm{0}$ such that, for every $x,y\in X$, there is a positive integer $k\left(x,y\right)$ such that $$\begin{array}{}\text{(51)}& \mathrm{0}<d\left(x,y\right)<\epsilon \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(T^{n}x,T^{n}y\right)<d\left(x,y\right)\hspace{1em}\forall n\ge k\left(x,y\right).\end{array}$$For eventually $\epsilon -$contractive mappings in the context of b-metric spaces, we announce the next result.

Theorem 24.

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a b-metric space with a continuous b-metric $d$ and $T:X\to X$ be an eventually $\epsilon -$contractive and orbitally continuous self-map on $\left(X,d\right).$ If, for some $x_{\mathrm{0}}\in X$, the sequence $\left\{T^n{x}_{\mathrm{0}}\right\}$ has a convergent subsequence $\{T^{n_i}{x}_{\mathrm{0}}\}$ to $u\in X$, then $u$ is a periodic point of $T.$

Proof.

The proof is very similar to the ones in the previous theorem. Therefore, it is omitted.

Corollary 25.

If $\left(X,d,s\ge \mathrm{1}\right)$ is a compact b-metric space with a continuous b-metric $d$ and $T:X\to X$ is an eventually $\epsilon -$contractive and continuous self-map on $\left(X,d\right)$, then the set of periodic points of $T$ is not empty.

The following two results extend ones from standard metric spaces to b-metric spaces (see [15]).

Theorem 26.

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a compact b-metric space with continuous b-metric $d$ and $T$ a continuous self-map on $X$ such that for every $x,y\in X$ there exists a positive integer $n\left(x,y\right)$ such that $$\begin{array}{}\text{(52)}& \begin{array}{c}\mathrm{0}<d\left(x,y\right)\text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(T^{n\left(x,y\right)}x,T^{n\left(x,y\right)}y\right)<d\left(x,y\right).\end{array}\end{array}$$Then $T$ has a unique fixed point.

Proof.

Define $f:X\to [\mathrm{0},+\infty )$ by $f\left(x\right)=d\left(x,Tx\right).$ Since $d$ and $T$ are continuous, $f$ is also continuous. Therefore, $f$ attains its minimum on $X$ in some point, say $u\in X.$ Assume that $f\left(u\right)=d\left(u,Tu\right)>\mathrm{0}.$ Then, by hypothesis, there exists a positive integer $n\left(u,Tu\right)$ such that$$\begin{array}{}\text{(53)}& d\left(T^{n\left(u,Tu\right)}u,T^{n\left(u,Tu\right)}Tu\right)<d\left(u,Tu\right);\end{array}$$that is, $f\left(T^{n\left(u,Tu\right)}u\right)<f\left(u\right)$, which contradicts the choice of $u.$ Therefore, $Tu=u.$

Theorem 27.

Let $\left(X,d,s\ge \mathrm{1}\right)$ be a compact b-metric space with a continuous b-metric $d$ and $T$ be a continuous self-map on $X$ such that, for every $x,y\in X$, $$\begin{array}{}\text{(54)}& \mathrm{0}<d\left(x,y\right)<\epsilon \text{\hspace{0.17em}\hspace{0.17em}implies\hspace{0.17em}\hspace{0.17em}}d\left(T^{n\left(x,y\right)}x,T^{n\left(x,y\right)}y\right)<d\left(x,y\right);\hspace{1em}n\left(x,y\right)\in \mathbb{N}.\end{array}$$Then the set of periodic points of $T$ is not empty.

Proof.

Let $x_{\mathrm{0}}\in X.$ Since $\left(X,d\right)$ is compact, $\left\{T^n{x}_{\mathrm{0}}\right\}$ has a cluster point. Therefore, there exist positive integers $p$ and $k$ such that$$\begin{array}{}\text{(55)}& d\left(T^p{x}_{\mathrm{0}},T^k{T}^p{x}_{\mathrm{0}}\right)=d\left(T^p{x}_{\mathrm{0}},T^{k+p}{x}_{\mathrm{0}}\right)<\epsilon .\end{array}$$Define $f\left(x_{\mathrm{0}}\right)=d\left(x_{\mathrm{0}},T^k{x}_{\mathrm{0}}\right).$ Since $T$ is continuous, $T^k$ is continuous and so $f$ is continuous. Therefore, there is some $u\in X$ such that $f\left(u\right)=\mathrm{m}\mathrm{i}\mathrm{n}\left\{f\left(x\right):x\in X\right\}.$ Then$$\begin{array}{}\text{(56)}& f\left(u\right)=d\left(u,T^{k}u\right)=\mathrm{m}\mathrm{i}\mathrm{n}\left\{d\left(x,T^{k}x\right):x\in X\right\}\le d\left(T^p{x}_{\mathrm{0}},T^k{T}^p{x}_{\mathrm{0}}\right)<\epsilon .\end{array}$$Assume that $T^{k}u\ne u.$ Then, as $d\left(u,T^{k}u\right)<\epsilon$, by hypothesis there exists $n=n\left(u,T^{k}u\right)\in \mathbb{N}$ such that$$\begin{array}{}\text{(57)}& d\left(T^{n}u,T^n{T}^{k}u\right)<d\left(u,T^{k}u\right).\end{array}$$Thus, $f\left(T^{n}u\right)<f\left(u\right)$, which contradicts the choice of $u.$ Therefore, $T^{k}u=u.$

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding the publication of this paper.

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