Some fixed point results on quasi-b-metric-like spaces

Abstract

In this paper, we investigate the existence and uniqueness of a fixed point of certain operators in the setting of complete quasi-b-metric-like spaces via admissible mappings. Our results improve, extend, and unify several well-known existence results.

1 Introduction and preliminaries

Throughout this paper, we denote $\mathbb{R}^{+}_{0} = [0, +\infty)$ and $\mathbb{N}_{0} = \mathbb{N} \cup\{0\}$, where $\mathbb{N}$ is the set of all positive integers. First, we recall some basic concepts and notation.

The concept of b-metric was introduced by Czerwik [1] as a generalization of metric (see also Bakhtin [2, 3]) to extend the celebrated Banach contraction mapping principle. Following the initial paper of Czerwik [1], a number of researchers in nonlinear analysis investigated the topology of the paper and proved several fixed point theorems in the context of complete b-metric spaces (see [4–8] and references therein).

Definition 1.1

[1]

Let X be a nonempty set, and $s\geq1$ be a given real number. A mapping $d : X \times X\to[0, +\infty)$ is said to be a b-metric if for all $x, y, z \in X$, the following conditions are satisfied:

(b₁):

$d(x, y) =0$ if and only if $x = y$;

(b₂):

$d(x, y) = d(y,x)$;

(b₃):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.

In this case, the pair $(X, d)$ is called a b-metric space (with constant s).

Definition 1.2

[9]

Let X be a nonempty set, and $s\geq1$ be a given real number. A mapping $d : X \times X\to[0, +\infty)$ is said to be a quasi-b-metric if for all $x, y, z \in X$, the following conditions are satisfied:

(bm₁):

$d(x, y) =0$ if and only if $x = y$;

(bm₂):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.

In this case, the pair $(X,d)$ is called a quasi-b-metric space (with constant s).

Definition 1.3

[10]

Let X be a nonempty set, and $s\geq1$ be a given real number. A mapping $d : X \times X\to[0, +\infty)$ is said to be a quasi-b-metric-like if for all $x, y, z \in X$, the following conditions are satisfied:

(bM₁):

$d(x, y) =0$ implies $x = y$;

(bM₂):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.

In this case, the pair $(X, d)$ is called a quasi-b-metric-like space (with constant s).

Example 1.4

Let $X=\{0,\frac{1}{2},\frac{1}{3}\} \cup [1, \infty)$, and let $d : X \times X\to[0, +\infty)$ be defined as

$$d(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 6& \text{if } x=y=0, \\ 3 &\text{if } x=y=\frac{1}{3}, \\ 2 &\text{if } x=0, y=\frac{1}{2}, \\ \frac{1}{2} &\text{if } x=0, y=\frac{1}{3}, \\ \frac{3}{2} &\text{if } x=\frac{1}{3}, y=0, \\ |x-y| &\text{otherwise}. \end{array}\displaystyle \right . $$

It is clear that $(X,d)$ is a quasi-b-metric-like space with constant $s=9$.

Definition 1.5

(see e.g. [10])

Let $(X, d)$ be a quasi-b-metric-like space. Then:

(i)_a :

a sequence $\{x_{n}\}$ in X is called a left-Cauchy sequence if and only if for every $\varepsilon>0$, there exists a positive integer $N=N(\varepsilon)$ such that $d(x_{n}, x_{m})<\varepsilon$ for all $n>m>N$;

(ii)_b :

a sequence $\{x_{n}\}$ in X is called a right-Cauchy sequence if and only if for every $\varepsilon>0$, there exists a positive integer $N=N(\varepsilon)$ such that $d(x_{n}, x_{m})<\varepsilon$ for all $m>n>N$;

(iii)_a :

a quasi-partial metric space is said to be left-complete if every left-Cauchy sequence $\{x_{n}\}$ in X converges with respect to d to a point $u \in X$ such that

$$\lim_{n \to\infty}d( x_{n},u) = d(u,u) =\lim _{n,m\to\infty}d(x_{m}, x_{n})=0,\quad \text{where } m\geq n; $$

(iii)_b :

a quasi-partial metric space is said to be right-complete if every left-Cauchy sequence $\{x_{n}\}$ in X converges with respect to d to a point $u \in X$ such that

$$\lim_{n \to\infty}d(u, x_{n}) = d(u,u) =\lim _{n,m\to\infty}d(x_{n}, x_{m})=0,\quad \text{where } m\geq n. $$

Let $(X, d)$ and $(Y, \alpha)$ be quasi-b-metric-like spaces, and let $f : X \to Y$ be a continuous mapping. Then

$$\lim_{n \to\infty} x_{n}=u \quad \Rightarrow\quad \lim _{n \to\infty} fx_{n}=fu. $$

In 2012, Samet et al. [11] introduced the concept of α-admissible mappings, and in 2013, Karapınar et al. [12] improved this notion as triangular α-admissible mappings.

Definition 1.6

[11, 12]

Let $\alpha: X \times X \rightarrow[0, +\infty)$ be a function. A self-mapping f is called an α-admissible mapping if

$$\alpha(x,y)\geq1 \quad \Rightarrow \quad \alpha(fx,fy)\geq1 $$

for all $x,y \in X$. If, further, f satisfies the condition

$$\alpha(x,z)\geq1 \quad \text{and}\quad \alpha(z,y)\geq1 \quad \Rightarrow \quad \alpha(x,y)\geq1 $$

for all $x,y,z \in X$, then it is called triangular α-admissible mapping.

Very recently, Popescu [13] improved these notions as follows.

Definition 1.7

[13]

Let $\alpha:X\times X\rightarrow[0,\infty)$ be a function. If $f: X\to X$ satisfies the condition

$$ (\mathrm{T}1)'\quad \alpha(x,fx)\geq1\quad \Rightarrow\quad \alpha \bigl(fx,f^{2}x\bigr) \geq1 $$

for all $x\in X$, then it is called a right-α-orbital admissible mapping. If f satisfies the condition

$$ (\mathrm{T}1)'' \quad \alpha(fx,x)\geq1 \quad \Rightarrow\quad \alpha\bigl(f^{2}x,fx\bigr) \geq1 $$

for all $x\in X$, then it is called a left-α-orbital admissible mapping. Furthermore, if f is both right-α-orbital admissible and left-α-orbital admissible, then f is called an α-orbital admissible mapping.

Triangular α-admissible mappings defined by Popescu [13] impose the following definitions.

Definition 1.8

[13]

Let $f:X\rightarrow X$ be a self-mapping, and $\alpha:X\times X\rightarrow[0,\infty)$ be a function. Then f is said to be triangular right-α-orbital admissible if f is right-α-orbital admissible and

$$ (\mathrm{T}2)'\quad \alpha(x,y)\geq1\quad \text{and}\quad \alpha(y,fy)\geq 1 \quad \Rightarrow\quad \alpha(x,fy) \geq1 $$

and is said to be triangular left-α-orbital admissible if f is α-orbital admissible and

$$ (\mathrm{T}2)'' \quad \alpha(fx,x)\geq1 \quad \text{and} \quad \alpha(x,y)\geq 1\quad \Rightarrow\quad \alpha(fx,y) \geq1. $$

If T satisfies both (T2)′ and (T2)″, then it is called triangular α-orbital admissible.

It is easy to conclude that each α-admissible mapping is an α-orbital admissible mapping and each triangular α-admissible mapping is a triangular α-orbital admissible mapping. However, the converses of the statements are false. In the following example, we see that a mapping that is triangular α-orbital admissible need not be triangular α-admissible.

Example 1.9

Let $X = \{x_{i}: i=1,\ldots, n\}$ for some $n\geq4$, and $d : X \times X \to\mathbb{R}_{0}^{+}$ with $d(x, y) = |x-y|$. We define a self-mapping $f : X \to X$ such that $fx_{i}=x_{i}$ for $i=1,2$, $fx_{i}=x_{j}$ for $i,j \in\{ 3,4\}$, $i\neq j$, $fx_{i}=x_{i+1}$ for $i \in\{5,\ldots, n-1\}$, and $fx_{n}=fx_{5}$. Moreover, let $\alpha: X \times X \to\mathbb{R}_{0}^{+}$ be such that

$$\alpha(x, y) = \left \{ \textstyle\begin{array}{l@{\quad}l} 1 & \text{if } (x, y) \in \{(x_{1},x_{3}), (x_{1},x_{4}), (x_{3},x_{3}), (x_{4},x_{4}), \\ &\hphantom{\text{if } (x, y) \in{}}{}(x_{3},x_{4}), (x_{4},x_{3}), (x_{3},x_{2}), (x_{4},x_{2}) \}, \\ 0 & \text{otherwise}. \end{array}\displaystyle \right . $$

Note that f is α-orbital admissible since $\alpha(x_{3},fx_{3}) = \alpha(x_{3},x_{4}) = 1$ and $\alpha(x_{4},fx_{4}) = \alpha(x_{4},x_{3}) =1$. On the other hand, we have $\alpha(x_{1},x_{3}) = \alpha(x_{3},x_{2}) = 1$, but $\alpha(x_{1},x_{2})=0$. Hence, T is not triangular α-admissible.

Definition 1.10

[13]

Let $(X,d)$ be a quasi-b-metric-like space. Then X is said to be α-regular if for every sequence $\{x_{n}\}$ in X such that $\alpha(x_{n},x_{n+1})\geq1$ for all n and $x_{n}\rightarrow x\in X$ as $n\rightarrow\infty$, there exists a subsequence $\{x_{n(k)}\}$ of $\{x_{n}\}$ such that $\alpha(x_{n(k)},x)\geq1$ for all k.

2 Main result

The notion of $(b)$-comparison was introduced by Berinde [14] in order to extend the notion of $(c)$-comparison.

Definition 2.1

[14]

Let $s\geq1$ be a real number. A mapping $\psi:\mathbb{R}_{0}^{+}\to\mathbb{R}_{0}^{+}$ is called a $(b)$-comparison function if the following conditions are fulfilled:

• (1) ψ is monotone increasing;

• (2) there exist $k_{0}\in\mathbb{N}$, $a\in(0,1)$, and a convergent series of nonnegative terms $\sum_{k=1}^{\infty}v_{k}$ such that $s^{k+1}\psi ^{k+1}(t)\leq a s^{k}\psi^{k}(t)+v_{k}$ for all $k\geq k_{0}$ and $t\in[0,\infty)$.

The class of $(b)$-comparison functions will be denoted by $\Psi_{b}$. Notice that the notion of a $(b)$-comparison function reduces to the concept of a $(c)$-comparison function if $s=1$.

The following lemma will be used in the proof of our main result.

Lemma 2.2

[15, 16]

Let $s\geq1$ be a real number. If $\psi :\mathbb{R}_{0}^{+}\to\mathbb{R}_{0}^{+}$ is a $(b)$-comparison function, then:

• (1) the series $\sum_{k=0}^{\infty}s^{k}\psi^{k}(t)$ converges for any $t\in\mathbb{R}_{0}^{+}$;

• (2) the function $p_{s}:[0,\infty)\to[0,\infty)$ defined by

  $$p_{s}(t)= \sum_{k=0}^{\infty}s^{k}\psi^{k}(t)\quad \textit{for all } t\in[0,\infty) $$

  is increasing and continuous at 0.

Remark 2.3

It is easy to see that if $\psi(t) \in\Psi _{b}$, then $\psi(t)< t$ for all $t>0$. In fact, if there is a $t^{*}>0$ such that $\psi(t^{*})\geq t^{*}$, then we have $\psi^{2}(t^{*})\geq\psi(t^{*})\geq t^{*}$ (since ψ is increasing). Continuing in the same manner, we get $\psi^{n}(t^{*})\geq t^{*}>0$, $n\in \mathbb{N}$. This contradicts Lemma 2.2.

Definition 2.4

Let $(X,d)$ be a complete quasi-b-metric-like space with a constant $s \geq1$. A self-mapping $f: X \to X$ is called $(\alpha ,\psi)$-contractive mapping if there exist two functions $\psi\in \Psi_{b}$ and $\alpha: X \times X \to[0,\infty)$ satisfying the following condition:

$$ \alpha(x,y)d(fx,fy) \leq\psi\bigl(d(x,y)\bigr) $$

(2.1)

for all $x,y \in X$.

Theorem 2.5

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be an $(\alpha,\psi)$-contractive mapping. Suppose also that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) f is continuous.

Then f has a fixed point u in X, and $d(u,u)=0$.

Proof

By (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0})\geq1$ and $\alpha(fx_{0},x_{0})\geq1$. Define the iterative sequence $\{x_{n}\}$ in X by $x_{n+1}=fx_{n}$ for all $n\in\mathbb{N}_{0}$. Note that if there exists $n_{0} \in\mathbb{N}_{0}$ such that $x_{n_{0}}= x_{n_{0}+1}$, then $x_{n_{0}}$ becomes a fixed point, which completes the proof. Hence, throughout the proof, we suppose that $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}_{0}$. Regarding the fact that f is α-orbital admissible, from (ii) we derive that

$$\alpha(x_{0},x_{1})= \alpha(x_{0}, fx_{0}) \geq1 \quad \Rightarrow \quad \alpha(fx_{0},fx_{1})= \alpha(x_{1},x_{2}) \geq1. $$

Inductively, we get that

$$ \alpha(x_{n},x_{n+1})\geq1\quad \text{for all } n\in\mathbb{N}_{0}. $$

(2.2)

Analogously, again by (ii) and the fact that f is α-orbital admissible we find that

$$\alpha(x_{1},x_{0})= \alpha(fx_{0}, x_{0}) \geq1 \quad \Rightarrow \quad \alpha(fx_{1},fx_{0})= \alpha(x_{2},x_{1}) \geq1. $$

Consequently, we observe that

$$ \alpha(x_{n+1},x_{n})\geq1 \quad \text{for all } n\in\mathbb{N}_{0}. $$

(2.3)

From (2.1), by taking $x=x_{n}$ and $y=x_{n-1}$, we find that

$$\begin{aligned} d(x_{n+1},x_{n})&= d(fx_{n},fx_{n-1}) \\ &\leq\alpha(x_{n},x_{n-1})d(fx_{n},fx_{n-1}) \\ & \leq\psi\bigl(d(x_{n},x_{n-1})\bigr). \end{aligned}$$

In view of Remark 2.3, we get that

$$ d(x_{n+1},x_{n}) \leq\psi\bigl(d(x_{n},x_{n-1}) \bigr) < d(x_{n},x_{n-1}) \quad \text{for all } n\in\mathbb {N}. $$

(2.4)

By analogy, again by (2.1) and by substituting $x=x_{n-1}$ and $y=x_{n}$, we have

$$\begin{aligned} d(x_{n},x_{n+1})&= d(fx_{n-1},fx_{n}) \\ &\leq\alpha(x_{n-1},x_{n })d(fx_{n-1},fx_{n }) \\ & \leq\psi\bigl(d(x_{n-1},x_{n})\bigr). \end{aligned}$$

Consequently,

$$ d(x_{n},x_{n+1})\leq\psi\bigl(d(x_{n-1},x_{n}) \bigr) < d(x_{n-1},x_{n}) \quad \text{for all } n\in\mathbb {N}. $$

(2.5)

From (2.4) and (2.5) we derive that

$$ d(x_{n},x_{n+1})\leq\psi^{n} \bigl(d(x_{0},x_{1})\bigr)\quad \text{and}\quad d(x_{n+1},x_{n})\leq\psi^{n}\bigl(d(x_{1},x_{0}) \bigr) \quad \text{for all } n\in\mathbb {N}. $$

(2.6)

By Lemma 2.2(1) and letting $n \to\infty$ in (2.6), we have $\lim_{n\rightarrow\infty}d(x_{n},x_{n+1})=\lim_{n\rightarrow \infty}d(x_{n+1}, x_{n})=0$.

We further prove that the sequence $\{x_{n}\}$ is right-Cauchy and left-Cauchy. For all $n,p\in\mathbb{N}$, we have

$$\begin{aligned} d(x_{n},x_{n+p})&\leq \sum_{i=1}^{p-1} s^{i}d(x_{n+i-1},x_{n+i})+s^{p-1}d(x_{n+p-1},x_{n+p}) \\ &< \sum_{i=1}^{p} s^{i} \psi^{n+i-1}\bigl(d(x_{0},x_{1})\bigr) \\ &= \frac{1}{s^{n-1} } \sum^{n+p-1}_{k=n} s^{k} \psi^{k}\bigl(d(x_{0},x_{1}) \bigr). \end{aligned}$$

By letting $n,p \to\infty$ we get that

$$ \lim_{n,p\to\infty} d(x_{n},x_{n+p})=0, $$

that is, the sequence $\{x_{n}\}$ is right-Cauchy.

Analogously,

$$ \lim_{n,p\to\infty} d(x_{n+p},x_{n})=0, $$

that is, the sequence $\{x_{n}\}$ is left-Cauchy. As a result, the sequence $\{x_{n}\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists a point $u \in X$ such that

$$ \lim_{n \to\infty}d(u, x_{n})=\lim _{n \to\infty}d( x_{n},u) = d(u,u) =\lim_{n,m\to\infty}d(x_{n}, x_{m})= \lim_{n,m\to\infty}d(x_{m}, x_{n})=0. $$

(2.7)

Since f is continuous, we have

$$u=\lim_{n \to\infty} x_{n+1}= \lim_{n \to\infty} fx_{n}=fu. $$

 □

Example 2.6

Let $(X,d)$ be a quasi b-metric like space defined in Example 1.4, and let the mapping $f:X\mapsto X$ be defined as

$$fx=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{3} &\text{if } x=0, \\ \frac{1}{2} &\text{if } x=\frac{1}{3}, \\ \frac{1}{2} &\text{if } x=\frac{1}{2}, \\ x+1 &\text{if } x\geq1. \end{array}\displaystyle \right . $$

Let $\psi(t)=\frac{t}{2}$, $t\geq0$, and let $\alpha:X\times X\to [0,\infty)$ be defined as

$$\alpha(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\text{if } x,y\in\{0,\frac{1}{3},\frac{1}{2}\}, \\ 0 &\text{otherwise}. \end{array}\displaystyle \right . $$

Then $\psi\in\Psi_{b}$, and f is an $(\alpha,\psi)$-contractive mapping. Since the conditions of Theorem 2.5 are satisfied, it follows that f has a fixed point in X.

It is possible to remove the heavy condition of continuity of the self-mapping f in Theorem 2.5. For this purpose, we need the following result, which is inspired from the results in [17].

Lemma 2.7

Let $(X,d)$ be a quasi-b-metric-like space with constant s and assume that $\{x_{n}\}$ and $\{y_{n}\}$ are sequences in X converging to x and y, respectively. Then

$$\begin{aligned} \frac{1}{s^{2}} d(x,y)-\frac{1}{s} d(x,x)-d(y,y) &\leq\liminf _{n\rightarrow\infty} d(x_{n},y_{n}) \leq \limsup _{n\rightarrow\infty} d(x_{n},y_{n}) \\ &\leq s d(x,x) + s^{2} d(y,y) + s^{2} d(x,y) . \end{aligned}$$

In particular, if $d(x,y)=0$, then $\lim_{n\rightarrow \infty} d(x_{n},y_{n})=0$.

Moreover, for each $z\in X$, we have

$$ \frac{1}{s} d(x,z)- d(x,x) \leq\liminf_{n\rightarrow\infty} d(x_{n},z) \leq\limsup_{n\rightarrow\infty} d(x_{n},z) \leq s d(x,z) + s d(x,x). $$

(2.8)

If $d(x,x)=0$, then

$$\frac{1}{s} d(x,z) \leq\liminf_{n\rightarrow\infty} d(x_{n},z) \leq\limsup_{n\rightarrow\infty} d(x_{n},z) \leq s d(x,z). $$

Theorem 2.8

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be an $(\alpha,\psi)$-contractive mapping. Suppose also that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) X is α-regular.

Then f has a fixed point u in X, and $d(u,u)=0$.

Proof

By verbatim of the proof of Theorem 2.5 we find an iterative sequence $\{x_{n}\}$ that converges to a point $u \in X$ such that (2.7) holds.

Since $d(u,u)=0$, by Lemma 2.7 we have

$$\begin{aligned} \frac{1}{s} d(u,fu) &\leq\liminf_{n\rightarrow\infty} d(x_{n+1},fu) \\ &\leq\limsup_{n\rightarrow\infty} d(x_{n+1},fu) \\ &=\limsup_{n\rightarrow\infty} d(fx_{n},fu) \\ &\leq\limsup_{n\rightarrow\infty}\alpha (x_{n},u)d(fx_{n},fu) \\ & \leq\limsup_{n\rightarrow\infty}\psi\bigl(d(x_{n},u)\bigr). \end{aligned}$$

By letting $n \to\infty$ in these inequalities we derive that $\frac {1}{s} d(u,fu) =0$ and hence $fu=u$. □

It is natural to consider the uniqueness of a fixed point of an $(\alpha,\psi)$-contractive mapping. We notice that we need to add an additional condition to guarantee the uniqueness.

• (U) For all $x,y\in\operatorname{Fix}(f)$, either $\alpha(x,y)\geq1$ or $\alpha (y,x)\geq1$.

Here, $\operatorname{Fix}(f)$ denotes the set of all fixed points of f.

Theorem 2.9

Adding condition (U) to hypotheses of Theorem  2.5 (or Theorem  2.8), we obtain the uniqueness of a fixed point of f.

Proof

Suppose that $x^{*}$ and $y^{*}$ are two distinct fixed points of f, so that $d(x^{*},y^{*})>0$.

If, for example, $\alpha(x^{*},y^{*})\geq1$, then

$$\begin{aligned} d\bigl(x^{*},y^{*}\bigr)&= d\bigl(fx^{*},fy^{*}\bigr) \\ &\leq\alpha\bigl(x^{*},y^{*}\bigr)d\bigl(fx^{*},fy^{*}\bigr) \\ &\leq\psi\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr) \\ &< d\bigl(x^{*},y^{*}\bigr), \end{aligned}$$

which is a contradiction. □

Definition 2.10

Let $(X,d)$ be a complete quasi-b-metric-like space with a constant $s \geq1$. A self-mapping $f: X \to X$ is called a generalized $(\alpha,\psi)$-contractive mapping of type $(A)$ if there exist two functions $\psi\in\Psi_{b}$ and $\alpha: X \times X \to[0,\infty)$ satisfying the following condition:

$$ \alpha(x,y) d(fx,fy) \leq\psi\bigl(M(x,y)\bigr) $$

(2.9)

for all $x,y \in X$, where

$$ M(x,y)=\max \bigl\{ d(x,y),d(x,fx),d(y,fy) \bigr\} . $$

(2.10)

Theorem 2.11

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a generalized $(\alpha,\psi)$-contractive mapping of type $(A)$. Assume that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) f is continuous.

Then f has a fixed point u in X, and $d(u,u)=0$.

Proof

As in the proof of Theorem 2.5, we construct an iterative sequence $x_{n+1}=fx_{n}$, $n\in\mathbb{N}_{0}$, where the existence of $x_{0}\in X$ is guaranteed by (ii). By the same reason as in the proof of Theorem 2.5, we may assume that $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}_{0}$, and we can conclude that

$$ \alpha(x_{n},x_{n+1})\geq1 \quad \text{and} \quad \alpha (x_{n+1},x_{n})\geq1\quad \text{for all } n\in \mathbb{N}_{0}. $$

(2.11)

From (2.9) we have

$$\begin{aligned} \begin{aligned} d(x_{n},x_{n+1})&= d(fx_{n-1},fx_{n}) \\ &\leq\alpha(x_{n-1},x_{n})d(fx_{n-1},fx_{n}) \\ &\leq\psi\bigl(M(x_{n-1},x_{n})\bigr) \end{aligned} \end{aligned}$$

for all $n \in\mathbb{N}$, where

$$\begin{aligned} M(x_{n-1},x_{n})&=\max \bigl\{ d(x_{n-1},x_{n }),d(x_{n},fx_{n}),d(x_{n-1},fx_{n-1}) \bigr\} \\ &=\max \bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} . \end{aligned}$$

If $M(x_{n-1},x_{n})=d(x_{n},x_{n+1})$, then since we assumed that $x_{n}\neq x_{n+1}$,

$$d(x_{n},x_{n+1})\leq\psi\bigl(d(x_{n},x_{n+1}) \bigr)< d(x_{n},x_{n+1}), $$

which is a contradiction. It allows us to conclude that $M(x_{n-1},x_{n})=d(x_{n-1},x_{n})$, $n\in\mathbb{N}$.

Thus,

$$d(x_{n},x_{n+1})\leq\psi\bigl(d(x_{n-1},x_{n}) \bigr)< d(x_{n-1},x_{n})\quad \text{for all } n\in\mathbb{N} $$

and

$$ d(x_{n},x_{n+1})\leq\psi^{n} \bigl(d(x_{0},x_{1})\bigr)\quad \text{for all } n\in \mathbb{N}. $$

(2.12)

Analogously, letting $x=x_{n}$ and $y=x_{n-1}$ in (2.9), we get

$$\begin{aligned} d(x_{n+1},x_{n}) =& d(fx_{n},fx_{n-1}) \\ \leq& \alpha(x_{n},x_{n-1})d(fx_{n},fx_{n-1}) \\ \leq& \psi\bigl(M(x_{n},x_{n-1})\bigr) \end{aligned}$$

(2.13)

for all $n \in\mathbb{N}$, where

$$\begin{aligned} M(x_{n},x_{n-1}) =&\max \bigl\{ d(x_{n},x_{n-1}),d(x_{n},fx_{n}),d(x_{n-1},fx_{n-1}) \bigr\} \\ =&\max \bigl\{ d(x_{n},x_{n-1}),d(x_{n},x_{n+1}),d(x_{n-1},x_{n}) \bigr\} . \end{aligned}$$

For the estimation of $d(x_{n+1},x_{n})$, we will consider three different cases.

Case 1. If $M(x_{n},x_{n-1})=d(x_{n-1},x_{n})$, then, by (2.13),

$$ d(x_{n+1},x_{n})\leq\psi\bigl(d(x_{n-1},x_{n}) \bigr). $$

(2.14)

Case 2. If $M(x_{n},x_{n-1})=d(x_{n},x_{n+1})$, then

$$ d(x_{n+1},x_{n})\leq\psi\bigl(d(x_{n},x_{n+1}) \bigr). $$

By Remark 2.3 we find that

$$d(x_{n+1},x_{n})\leq\psi\bigl(d(x_{n},x_{n+1}) \bigr)< \psi^{n+1} \bigl(d(x_{0},x_{1}) \bigr). $$

Case 3. Otherwise, $M(x_{n},x_{n-1})=d(x_{n},x_{n-1})$ and

$$ d(x_{n+1},x_{n})\leq\psi\bigl(d(x_{n},x_{n-1}) \bigr). $$

(2.15)

Observing (2.14) and (2.15), it follows that, for any $n\in \mathbb{N}$,

$$ d(x_{n+1},x_{n})\leq\max \bigl\{ \psi^{n} \bigl(d(x_{0},x_{1})\bigr), \psi ^{n} \bigl(d(x_{1},x_{0})\bigr) \bigr\} . $$

(2.16)

Obviously, in all considered cases, we deduce that

$$\lim_{n\rightarrow\infty}d(x_{n+1},x_{n})=\lim _{n\rightarrow\infty }d(x_{n},x_{n+1}) =0. $$

Since ψ is an increasing function, let

$$v=\max \bigl\{ d(x_{0},x_{1}), d(x_{1},x_{0}) \bigr\} . $$

Consequently, we have that $d(x_{n+1},x_{n})\leq\psi^{n}(v)$ and $d(x_{n},x_{n+1})\leq\psi^{n}(v)$. By applying (bM₂) for any $n,p\in\mathbb{N}$ it follows that

$$\begin{aligned} d(x_{n},x_{n+p}) \leq&\sum_{i=1}^{p-1} s^{i}d(x_{n+i-1},x_{n+i})+s^{p-1}d(x_{n+p-1},x_{n+p}) \\ \leq&\sum_{i=1}^{p} s^{i}d(x_{n+i-1},x_{n+i}) \\ \leq&\sum_{i=1}^{p} s^{i} \psi^{n+i-1}(v) \\ =&\frac{1}{s^{n-1}}\sum_{i=1}^{p} s^{n+i-1}\psi^{n+i-1}(v). \end{aligned}$$

Therefore, $\lim_{n,p\to\infty} d(x_{n},x_{n+p})=0$ and, likewise, $\lim_{n,p\to\infty} d(x_{n+p},x_{n})=0$. Since, X is complete, there exists $u\in X$ such that $\lim_{n\rightarrow \infty}x_{n}=u$ and

$$ \lim_{n\rightarrow\infty}d(u,x_{n})=\lim _{n\rightarrow\infty }d(x_{n},u)=d(u,u)=0. $$

(2.17)

Furthermore, f is a continuous mapping, and hence $u=\lim_{n\rightarrow\infty} x_{n}=\lim_{n\rightarrow\infty}fx_{n-1}=fu$. □

Theorem 2.12

Adding condition (U) to hypotheses of Theorem  2.11, we obtain the uniqueness of a fixed point of T.

Proof

Suppose that $fx^{*}=x^{*}$ and $fy^{*}=y^{*}$. Then

$$\begin{aligned} d\bigl(x^{*},y^{*}\bigr)&= d\bigl(fx^{*},fy^{*}\bigr) \\ &\leq\alpha\bigl(x^{*},y^{*}\bigr)\psi\bigl( d\bigl(fx^{*},fy^{*}\bigr)\bigr) \\ &\leq\psi\bigl(M\bigl(x^{*},y^{*}\bigr)\bigr) \\ &=\psi\bigl(d\bigl(x^{*},y^{*}\bigr)\bigr), \end{aligned}$$

so that $d(x^{*},y^{*})=0 \Rightarrow x^{*}=y^{*}$. □

In the following example, we show the existence of a function satisfying conditions of Theorem 2.11 but not satisfying conditions of Theorem 2.5.

Example 2.13

Let $(X,d)$ be a quasi-b-metric-like space described in Example 1.4, and $f:X\mapsto X$ the mapping

$$fx=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{2} &\text{if } x=0, \\ 0, &\text{if } x=\frac{1}{3}, \\ \frac{1}{2} &\text{if } x=\frac{1}{2}, \\ x+1 &\text{if } x\geq1. \end{array}\displaystyle \right . $$

Let $\psi(t)=\frac{t}{2}$, $t\geq0$, and let $\alpha:X\times X\to [0,\infty)$ be defined as

$$\alpha(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\text{if } (x,y)\in \{(0,\frac{1}{3}),(0,\frac {1}{2}),(\frac{1}{2},0),(\frac{1}{2},\frac{1}{3}),(\frac {1}{2},\frac{1}{2}) \}, \\ 0 &\text{otherwise}. \end{array}\displaystyle \right . $$

Then (2.1) does not hold, for example, for $x=0$ and $y=\frac {1}{3}$, but (2.9) holds, f has a unique fixed point $u=\frac {1}{2}$, and $d(u,u)=0$.

Definition 2.14

Let $(X,d)$ be a complete quasi-b-metric-like space with a constant $s \geq1$. A self-mapping $f: X \to X$ is called a generalized $(\alpha,\psi)$-contractive mapping of type $(B)$ if there exist two functions $\psi\in\Psi_{b}$ and $\alpha: X \times X \to[0,\infty)$ satisfying the following condition:

$$ \alpha(x,y) d(fx,fy) \leq\psi\bigl(N(x,y)\bigr) $$

(2.18)

for all $x,y \in X$, where

$$ N(x,y)=\max \biggl\{ d(x,y), \frac{d(x,fx)+d(y,fy)}{2} \biggr\} . $$

(2.19)

The following theorem can be deduced from the inequality $N(x,y) \leq M(x,y)$ for all x, y, together with the monotonicity of ψ.

Theorem 2.15

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a generalized $(\alpha,\psi)$-contractive mapping of type $(B)$. Assume that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) f is continuous.

Then f has a fixed point u in X, and $d(u,u)=0$.

Definition 2.16

Let $(X,d)$ be a complete quasi-b-metric-like space with a constant $s \geq1$. A self-mapping $f: X \to X$ is called a generalized $(\alpha,\psi)$-contractive mapping of type $(C)$ if there exist two functions $\psi\in\Psi_{b}$ and $\alpha: X \times X \to[0,\infty)$ satisfying the following condition:

$$ s\alpha(x,y) d(fx,fy) \leq\psi\bigl(M(x,y)\bigr) $$

(2.20)

for all $x,y \in X$, where

$$ M(x,y)=\max \bigl\{ d(x,y),d(x,fx),d(y,fy) \bigr\} . $$

(2.21)

The following theorem is easily observed from Theorem 2.11 since inequality (2.9) can be easily derived from inequality (2.20).

Theorem 2.17

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a generalized $(\alpha,\psi)$-contractive mapping of type $(C)$. Assume that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) f is continuous.

Then f has a fixed point u in X, and $d(u,u)=0$.

In the next theorems, we establish a fixed point result for a generalized $(\alpha,\psi)$-contractive mapping of type $(C)$ without any continuity assumption on the mapping f.

Theorem 2.18

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a generalized $(\alpha,\psi)$-contractive mapping of type $(C)$. Suppose that

• (i) f is α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) X is α-regular.

Then f has a fixed point u in X, and $d(u,u)=0$.

Proof

As in the proof of Theorem 2.5, we consider an iterative sequence $\{x_{n}\}$, and we obtain the existence of $u\in X$ such that (2.17) holds. By Lemma 2.7 we get

$$\begin{aligned} d(u,fu) \leq& s \liminf_{n\to\infty} d(x_{n+1},fu) \\ \leq& s\limsup_{n\to\infty} d(x_{n+1},fu) \\ \leq& s\limsup_{n\to\infty} \alpha(x_{n},u) d(fx_{n},fu) \\ \leq& \limsup_{n\to\infty} \psi\bigl(M(x_{n},u)\bigr), \end{aligned}$$

where

$$M(x_{n},u)=\max \bigl\{ d(x_{n-1},u), d(x_{n-1},x_{n}),d(u,fu) \bigr\} . $$

According to (2.17) and the fact that $\lim_{n\rightarrow\infty }d(x_{n-1},x_{n})=0$, it remains to discuss only the case $M(x_{n},u)=d(u,fu)$ because otherwise it follows $d(u,fu)=0 \Rightarrow u=fu$.

Notice that, under this assumption, $d(u,fu)\leq\psi(d(u,fu))$ also implies $d(u,fu)=0$ since $\psi(t)< t$ for any $t>0$. Hence, u is a fixed point of the mapping f. □

Theorem 2.19

Adding condition (U) to hypotheses of Theorem  2.17 (or Theorem  2.18), we obtain the uniqueness of a fixed point of T.

Example 2.20

Let $(X,d)$ be a quasi b-metric like space defined in Example 1.4, and let the mapping $f:X\mapsto X$ be defined as

$$fx=\left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\text{if } x=0, \\ \frac{1}{2} &\text{if } x=\frac{1}{3}, \\ \frac{1}{2} &\text{if } x=\frac{1}{2}, \\ x+1 &\text{if } x\geq1. \end{array}\displaystyle \right . $$

Let $\psi(t)=\frac{t}{2}$, $t\geq0$, and $\alpha:X\times X\to [0,\infty)$ be defined as

$$\alpha(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\text{if } (x,y)\in \{(\frac{1}{3},\frac {1}{2}),(\frac{1}{2},\frac{1}{2}) \}, \\ \frac{1}{9} &\text{if } (x,y)\in \{0,\frac{1}{3},\frac {1}{2} \}\times \{0,\frac{1}{3},\frac{1}{2} \} \setminus \{(\frac{1}{3},\frac{1}{2}),(\frac{1}{2},\frac {1}{2}) \}, \\ 0 &\text{otherwise}. \end{array}\displaystyle \right . $$

Then $\psi\in\Psi_{b}$, and f is a generalized $(\alpha,\psi )$-contractive mapping of type $(C)$. Since the conditions of Theorem 2.18 are satisfied, it follows that f has a fixed point in X.

Definition 2.21

Let $(X,d)$ be a complete quasi-b-metric-like space with a constant $s \geq1$. A self-mapping $f: X \to X$ is called a generalized $(\alpha,\psi)$-contractive mapping of type $(D)$ if there exist two functions $\psi\in\Psi_{b}$ and $\alpha: X \times X \to[0,\infty)$ satisfying the following condition:

$$ \alpha(x,y) d(fx,fy) \leq\psi\bigl(L(x,y)\bigr) $$

(2.22)

for all $x,y \in X$, where

$$ L(x,y)=\max \biggl\{ d(x,y), \frac{d(x,fx)+d(y,fy)}{2s} \biggr\} . $$

(2.23)

Theorem 2.22

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a generalized $(\alpha,\psi)$-contractive mapping of type $(D)$. Suppose that

• (i) f is triangular α-orbital admissible;

• (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},fx_{0}) \geq1$ and $\alpha(fx_{0},x_{0})\geq1$;

• (iii) X is α-regular.

Then f has a fixed point in X, that is, there exists $u \in X$ such that $fu=u$ and $d(u,u)=0$.

Proof

As in the proof of Theorem 2.11, we consider an iterative sequence $\{x_{n}\}$ and obtain the existence of $u\in X$ such that (2.17) holds. By Lemma 2.7 we get

$$\begin{aligned} \frac{ 1}{s} d(u,fu) \leq& \liminf_{n\to\infty} d(x_{n+1},fu) \\ \leq& \limsup_{n\to\infty} d(x_{n+1},fu) \\ \leq& \limsup_{n\to\infty} \alpha(x_{n},u) d(fx_{n},fu) \\ \leq& \limsup_{n\to\infty} \psi\bigl(N(x_{n},u)\bigr), \end{aligned}$$

where

$$N(x_{n},u)=\max \biggl\{ d(x_{n-1},u), \frac {d(x_{n-1},x_{n})+d(u,fu)}{2s} \biggr\} . $$

If $N(x_{n},u)=d(x_{n-1},u)$, then we conclude the result due to (2.17). Taking $\lim_{n\rightarrow\infty}d(x_{n-1}, x_{n})=0$ into account, we deduce that $\lim_{n\rightarrow\infty} N(x_{n},u)=\frac{ d(u,fu)}{2s}$. Notice that, under this assumption, $\frac{ 1}{s}d(u,fu)\leq\psi (\frac{ d(u,fu)}{2s})$ also implies $d(u,fu)=0$ since $\psi(t)< t$ for any $t>0$. Hence, u is a fixed point of the mapping f. □

Theorem 2.23

Adding condition (U) to hypotheses of Theorem  2.15 (and respectively, Theorem  2.22), we obtain the uniqueness of a fixed point of T.

3 Consequences

In this section, we will list some consequences of our main results.

3.1 For standard quasi-b-metric-like

Corollary 3.1

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a mapping such that

$$ d(fx,fy) \leq\psi\bigl(\max \bigl\{ d(x,y),d(x,fx),d(y,fy) \bigr\} \bigr) $$

(3.1)

for all $x,y \in X$, where $\psi\in\Psi_{b}$. If f is continuous, then f has a fixed point u in X, and $d(u,u)=0$.

Proof

The proof of Corollary 3.1 follows from Theorem 2.12 by taking $\alpha(x,y)=1$ for all $x,y \in X$, so (ii) is satisfied for any $x_{0}\in X$, f is obviously an α-orbital admissible, and (U) holds. Inequality (3.1) allows us to conclude that f is a generalized $(\alpha,\psi)$-contractive mapping of type $(A)$. □

Corollary 3.2

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a continuous mapping such that

$$ d(fx,fy) \leq\psi \biggl(\max \biggl\{ d(x,y),\frac {d(x,fx)+d(y,fy)}{2} \biggr\} \biggr) $$

(3.2)

for all $x,y \in X$, where $\psi\in\Psi_{b}$. Then f has a fixed point u in X, and $d(u,u)=0$.

Proof

The proof of Corollary 3.2 follows from Theorem 2.15 by taking $\alpha(x,y)=1$ for all $x,y \in X$ since then (2.18) follows from (3.2). □

Notice that the continuity condition of f in Corollary 3.1 can be removed by adding an extra term s.

Corollary 3.3

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a mapping such that

$$ d(fx,fy) \leq s \psi\bigl(\max \bigl\{ d(x,y),d(x,fx),d(y,fy) \bigr\} \bigr) $$

(3.3)

for all $x,y \in X$, where $\psi\in\Psi_{b}$. Then f has a fixed point u in X such that $d(u,u)=0$.

Proof

The proof of Corollary 3.3 follows from Theorem 2.18 by taking $\alpha(x,y)=1$ for all $x,y \in X$. Then f is an α-orbital admissible mapping, and both inequalities in (ii) hold for any $x_{0}\in X$. Notice that since $\alpha(x,y)=1$, any constructive sequence turns to be regular, and thus X is α-regular. □

Corollary 3.4

Let $(X,d)$ be a complete quasi-b-metric-like space, and let $f: X \to X$ be a mapping such that

$$ d(fx,fy) \leq\psi\bigl(d(x,y)\bigr) $$

(3.4)

for all $x,y \in X$, where $\psi\in\Psi_{b}$. Then f has a fixed point u in X such that $d(u,u)=0$.

Proof

The proof of Corollary 3.4 follows from Theorem 2.8 by taking $\alpha(x,y)=1$ for all $x,y \in X$ and observing that X is α-regular and that (i) and (ii) hold. □

3.2 For standard quasi-b-metric-like spaces with a partial order

In this section, we deduce various fixed point results on a quasi-b-metric-like space endowed with a partial order. We, first, recollect some basic notions and notation.

Definition 3.5

Let $(X,\preceq)$ be a partially ordered set, and $f: X\to X$ be a given mapping. We say that f is nondecreasing with respect to ⪯ if for all $x,y\in X$,

$$x\preceq y\quad \Rightarrow \quad fx\preceq fy. $$

Definition 3.6

Let $(X,\preceq)$ be a partially ordered set. A sequence $\{x_{n}\} \subseteq X$ is said to be nondecreasing (respectively, nonincreasing) with respect to ⪯ if $x_{n}\preceq x_{n+1}$, $n\in\mathbb{N}$ (respectively, $x_{n+1}\preceq x_{n}$, $n\in\mathbb{N}$).

Definition 3.7

Let $(X,\preceq)$ be a partially ordered set, and d be a b-metric-like on X. We say that $(X,\preceq,d)$ is regular if for every nondecreasing (respectively, nonincreasing) sequence $\{x_{n}\} \subseteq X$ such that $x_{n}\to x\in X$ as $n\to\infty$, there exists a subsequence $\{ x_{n_{k}}\}$ of $\{x_{n}\}$ such that $x_{n_{k}}\preceq x$ (respectively, $x\preceq x_{n_{k}}$) for all k.

We have the following result.

Corollary 3.8

Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset), and d be a b-metric-like on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $f: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exists $\psi\in\Psi_{b}$ such that

$$ d(fx,fy) \leq\psi\bigl(M(x,y)\bigr) $$

(3.5)

for all $x,y \in X$ with $x\preceq y$ or $y\preceq x$, where $M(x,y)$ is defined as in (2.10). Suppose also that the following conditions hold:

(i):

there exists $x_{0}\in X$ such that $x_{0}\preceq fx_{0}$ and $fx_{0}\preceq x_{0}$;

(ii):

f is continuous or

(ii)′:

$(X,\preceq,d)$ is regular, and d is continuous.

Then f has a fixed point $u \in X$ with $d(u,u)=0$. Moreover, if for all $x,y\in X$, there exists $z\in X$ such that $x\preceq z$ and $y\preceq z$, then f has a unique fixed point.

Proof

Define the mapping $\alpha: X\times X\to[0,\infty)$ by

$$ \alpha(x,y)=\left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\mbox{if } x\preceq y \mbox{ or } x\succeq y, \\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . $$

Clearly, f satisfies (2.20), that is,

$$\alpha(x,y)s d(fx, fy)\leq\psi\bigl(M(x,y)\bigr) $$

for all $x,y\in X$. From condition (i) we have $x_{0}\preceq fx_{0}$ and $fx_{0}\preceq x_{0}$. Moreover, for all $x,y\in X$, from the monotone property of f we have

$$\begin{aligned}& \alpha(x,y)\geq1\quad \Rightarrow \quad x\succeq y\quad \mbox{or} \\& x \preceq y \quad \Rightarrow \quad fx\succeq fy \quad \mbox{or} \\& fx\preceq fy \quad \Rightarrow \quad \alpha(fx,fy)\geq1. \end{aligned}$$

Hence, the self-mapping f is α-admissible. Similarly, we can prove that f is triangular α-admissible and so triangular α-orbital admissible. Now, if f is continuous, then the existence of a fixed point follows from Theorem 2.17.

Suppose that $(X,\preceq,d)$ is regular. Let $\{x_{n}\}$ be a sequence in X such that $\alpha(x_{n},x_{n+1})\geq1$ for all n and $x_{n} \rightarrow x\in X$ as $n\rightarrow\infty$. By the regularity hypothesis, since X does not contain an infinite totally unordered subset, there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $x_{n_{k}}\preceq x$ or $x\preceq x_{n_{k}}$ for all k.

This implies from the definition of α that $\alpha (x_{n_{k}},x)\geq1$ for all k. In this case, the existence of a fixed point follows again from Theorem 2.18. □

Corollary 3.9

Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset), and d be a b-metric-like on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $f: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exists $\psi\in\Psi_{b}$ such that

$$ d(fx,fy) \leq\psi\bigl(d(x,y)\bigr) $$

(3.6)

for all $x,y \in X$ with $x\succeq y$ or $y\succeq x$. Suppose also that the following conditions hold:

• (i) there exists $x_{0}\in X$ such that $x_{0}\preceq fx_{0}$ and $fx_{0}\preceq x_{0}$;

• (ii) f is continuous.

Then T has a fixed point $u \in X$ with $d(u,u)=0$. Moreover, if for all $x,y\in X$, there exists $z\in X$ such that $x\preceq z$ and $y\preceq z$, we have the uniqueness of a fixed point.