Research  Open  Published:
Some extensions for Geragthy type contractive mappings
Journal of Inequalities and Applicationsvolume 2015, Article number: 303 (2015)
Abstract
In this paper, we establish some fixed point theorems on some extensions of Geragthy contractive type mappings in the context of bmetriclike spaces.
Introduction and preliminaries
One of the interesting extensions of the notion of a metric space is the dislocated space, introduced by Hitzler [1]. This notion was rediscovered by AminiHarandi [2] and given the name of a metriclike space.
Definition 1.1
On a nonempty set X we define a function $\sigma: X\times X\rightarrow[0,\infty)$ such that for all $x,y,z\in X$:
 (σ1):

if $\sigma(x,y)=0$ then $x=y$;
 (σ2):

$\sigma(x,y)=\sigma(y,x)$;
 (σ3):

$\sigma(x,y)\leq\sigma(x,z)+\sigma(z,y)$;
Throughout this paper, we suppose that $\mathbb{N}_{0} = \mathbb{N} \cup\{0\}$ where $\mathbb{N}$ denotes the set of all positive integers. Further, the symbols $\mathbb{R^{+}}$ and $\mathbb{R}^{+}_{0}$ denotes the set of positive reals and the set of nonnegative reals. First, we recall some basic concepts and notations.
The concept of a bmetric was introduced by Czerwik [3] as a generalization of the metric (see also Bakhtin [4] and Bourbaki [5]) to extend the celebrated Banach contraction mapping principle. Following this initial paper of Czerwik [3], a number of researchers in nonlinear analysis investigated the topology of the paper and proved several fixed point theorems in the context of complete bmetric spaces (see e.g. [6–10] and references therein).
Definition 1.2
[3]
Let X be a nonempty set and $s\geq1$ be a given real number. A mapping $d \colon X \times X\to[0, \infty)$ is said to be a bmetric if for all $x, y, z \in X$ the following conditions are satisfied:
 ($bM_{1}$):

$d(x, y) =0$ if and only if $x = y$;
 ($bM_{2}$):

$d(x, y) = d(y,x)$;
 ($bM_{3}$):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.
In what follows, we recall the notion of bmetriclike space which is an interesting generalization of both bmetric space and metriclike space.
Definition 1.3
[11]
Let X be a nonempty set and $s\geq1$ be a given real number. A mapping $d \colon X \times X\to[0, \infty)$ is said to be bmetriclike if for all $x, y, z \in X$ the following conditions are satisfied:
 ($bML_{1}$):

if $d(x, y) =0$ then $x = y$;
 ($bML_{2}$):

$d(x, y) = d(y,x)$;
 ($bML_{3}$):

$d(x, z)\leq s[d(x, y) + d(y, z)]$.
Example 1.4
Let $X=C([0,T])$ be the set of all real continuous functions on the closed interval $[0,T]$. Let $d: X \times X \to\mathbb{R}_{0}^{+}$ be defined
for all $f,g \in X$, $a \in \mathbb{R}_{0}^{+}$, and $p >1$. It is easy to see that $(X,d)$ is a complete bmetriclike space with $s=2^{p1}$. For more examples, see e.g. [11].
Remark 1.5
Let $(X, d)$ be a bmetriclike space with constant $s\geq1$. Then it is clear that $d^{s}(x,y)=2d(x,y)d(x,x)d(y,y)$ satisfies the following:

(S1)
$d^{s}(x,x)=0$ for all $x \in X$.
Definition 1.6
[11]
Let $(X,d)$ be a bmetriclike space. Then:

(1)
a sequence $\{x_{n}\} $ in X is called convergent to $x\in X$ if and only if $\lim_{n\to\infty} d(x_{n},x)=d(x,x)$;

(2)
a sequence $\{x_{n}\} $ in X is called Cauchy sequence if and only if $\lim_{n,m\to\infty} d(x_{n},x_{m})$ exists and finite;

(3)
$(X,d)$ is complete if and only if every Cauchy sequence $\{x_{n}\} $ in X converges to $x\in X $ so that
$$\lim_{n\rightarrow\infty} d(x_{n},x)=d(x,x)=\lim _{m,n\rightarrow \infty} d(x_{n},x_{m}). $$
Proposition 1.7
[11]
Let $(X,d) $ be a bmetriclike space with constant s and let $\{ x_{n}\} $ be a sequence in X such that $\lim_{n\to\infty} d(x_{n},x)=0 $. Then:

(1)
x is unique.

(2)
$\frac{1}{s} d(x,y) \leq\lim_{n\to\infty} d(x_{n},y) \leq s d(x,y) $ for all $y\in X$.
Lemma 1.8
[11]
Let $(X,d)$ be a bmetriclike space with constant s and $\{x_{n}\} $ a sequence in X such that
where $0\leq k $ and $s k < 1$. Then $\{x_{n}\}$ is a Cauchy sequence in X and $\lim_{n,m\to\infty} d(x_{n},x_{m})=0 $.
Lemma 1.9
[12]
Let $(X,d)$ be a bmetriclike space with constant s and assume that $\{x_{n}\}$ and $\{y_{n}\}$ are sequences in X converging to x and y, respectively. Then
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
In particular, if $d(x,x)=0$, then
Notice that, in general, a bmetriclike mapping does not need to be continuous.
The notion of αadmissible and triangular αadmissible mappings were introduced by Samet et al. [13] and Karapınar et al. [14], respectively.
Definition 1.10
Let $T:X\rightarrow X$ be a mapping and $\alpha: X \times X \rightarrow [0, \infty)$ be a function. We say that T is an αadmissible mapping if
Moreover, a selfmapping T is called triangular αadmissible if T is αadmissible and
For more details on αadmissible and triangular αadmissible mappings, see e.g. [13–17].
Very recently, Popescu [18] refined the notion of triangular αorbital admissible as follows.
Definition 1.11
[18]
Let $T:X\rightarrow X$ be a mapping and $\alpha:X\times X\rightarrow [0,\infty)$ be a function. We say that T is αorbital admissible if
Furthermore, T is called triangular αorbital admissible if T is αorbital admissible and
As mentioned in [18] each αadmissible (respectively, triangular αadmissible) mapping is an αorbital admissible (respectively, triangular αorbital admissible) mapping. In the following example we shall show that the converse is not true.
Example 1.12
Let $X = \{a,b,c,d,e,f,g,h\}$. We define a selfmapping $T : X \to X $ such that $Tx=x$, for $x=a,d$ and
Moreover, we define $\alpha: X \times X \rightarrow \mathbb{R}_{0}^{+}$, such that
Note that T is αorbital admissible, since $\alpha(b,Tb) = \alpha(b,c) = 1 $ and $\alpha(c,Tc) = \alpha(c,b) =1 $. On the other hand, we have $\alpha(d,e) = 1 $, but $\alpha(Td,Te)=\alpha(d,f)=0$. Hence, T is not αadmissible.
Lemma 1.13
[18]
Let $T:X\to X$ be a triangular αorbital admissible mapping. Assume that there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq1$. Define a sequence $\{x_{n}\}$ by $x_{n+1}=Tx_{n}$ for each $n\in\mathbb {N}_{0}$. Then we have $\alpha(x_{n},x_{m})\geq1$ for all $m,n \in\mathbb{N}$ with $n< m$.
Lemma 1.14
Let $T:X\to X$ be a triangular αorbital admissible mapping. Assume that there exists $x_{0}\in X$ such that $\alpha(Tx_{0},x_{0})\geq1$. Define a sequence $\{x_{n}\}$ by $x_{n+1}=Tx_{n}$ for each $n\in\mathbb {N}_{0}$. Then we have $\alpha(x_{m},x_{n})\geq1$ for all $m,n \in\mathbb{N}$ with $n< m$.
We characterize the notion of αregular in the setting of a bmetriclike space.
Definition 1.15
(cf. [18])
Let $(X,d)$ be a bmetriclike space, X is said to be αregular, if for every sequence $\{x_{n}\}$ in X such that $\alpha(x_{n},x_{n+1})\geq1$ (respectively, $\alpha(x_{n+1},x_{n})\geq1$) 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 $\alpha(x_{n_{k}},x)\geq1$ (respectively, $\alpha(x,x_{n_{{k}}})\geq1$) for all k.
In this paper, we shall prove the existence and uniqueness of a fixed point for certain operators in the setting of bmetriclike spaces. The presented results improve, extend, and unify a number of existing results in the literature.
Main result for bmetriclike spaces
In this section, we shall state and prove our main results. First, we recall the following classes of auxiliary functions. Let Ψ be the set of all increasing and continuous functions $\psi: [0,\infty) \to [0,\infty)$ with $\psi^{1}(\{0\})=\{0\}$. Let $\mathcal{F}_{s}$ be the family of all functions $\beta: [0,\infty) \to[0,\frac{1}{s})$ which satisfy the condition
for some $s\geq1$.
Definition 2.1
Let $(X,d)$ be a bmetriclike space with constant $s\geq1$, and $T:X\rightarrow X$ be a map. We say that T is a generalized almost αψϕGeraghty contractive type mapping if there exist a function $\alpha:X\times X\to[0,\infty)$, $\psi,\phi \in\Psi$, $\beta\in\mathcal{F}_{s}$, and some $L\geq0$ such that
for all $x,y \in X$, where
Remark 2.2
Since the functions belonging to $\mathcal{F}_{s}$ are strictly smaller than $\frac{1}{s}$, for some $s \geq1$, the expression $\beta(\psi(M(x,y))) $ in (3) can be estimated from above as follows:
Theorem 2.3
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized almost αψϕGeraghty contractive type mapping. We suppose also that

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
By (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq1$. Define a sequence $\{x_{n}\}\subset X$ by $x_{n+1}=Tx_{n}$ for all $n\in \mathbb{N}_{0}$. As T is triangular αorbital admissible, by Lemma 1.13 we have $\alpha(x_{n},x_{n+1})\geq1$ for all $n\in\mathbb{N}_{0}$. Throughout the proof, we suppose that $x_{n}\neq x_{n+1}$ for all $n\in \mathbb{N}_{0}$. Indeed, if there exists $n_{0}$ such that $x_{n_{0}}=x_{n_{0}+1}$, then $x_{n_{0}}$ becomes the fixed point of T, which completes the proof.
Since T is a generalized almost αψϕGeraghty contractive type mapping we have
Thus, we have
where $N(x_{n},x_{n+1})=\min\{ d^{s}(x_{n},x_{n+2}), d^{s}(x_{n+1},x_{n+1}), d(x_{n},x_{n+1}), d(x_{n+1},x_{n+2}) \}=0$, and
Note that
Consequently, we have
If $M(x_{n},x_{n+1})=d(x_{n+1},x_{n+2})$, then from (7) we have
Since ψ is increasing, we derive that $s^{2}d(x_{n+1},x_{n+2})< d(x_{n+1},x_{n+2})$, which is a contradiction as $s\geq 1$. Thus, $M(x_{n},x_{n+1})=d(x_{n},x_{n+1})$. Again by (7), we find
Hence, we get
Case (i): $s>1$. Since $\frac{1}{s^{2}}>0$ and $s\frac{1}{s^{2}}=\frac {1}{s}<1$, by Lemma 1.8, the sequence $\{x_{n}\}$ is Cauchy and
Case (ii): $s=1$. From (8), we have $d(x_{n+1},x_{n+2})\leq d(x_{n},x_{n+1})$ for all n. Thus, we conclude that
for some $r\geq0$. We shall prove that $r=0$. Suppose, on the contrary, that $r>0$. Note that, for $s=1$, the inequality (6) turns into
where $N(x_{n},x_{n+1})=0$ and $M(x_{n},x_{n+1})=d(x_{n},x_{n+1})$ as evaluated above. Thus, (11) yields
By taking the limit as $n \to\infty$ in (12) and regarding the continuity of ψ, we get
Hence, we have
Consequently $r=0$. In what follows, we shall prove that $\{x_{n}\}$ is a Cauchy sequence. Indeed we will prove that $\lim_{m,n\rightarrow\infty }d(x_{n},x_{m})=0 $. Suppose, on the contrary, that there exist $\varepsilon>0$ and corresponding subsequences $\{n_{k}\} $ and $\{m_{k}\}$ of $\mathbb{N}$ satisfying $n_{k}>m_{k}>k$ for which
where $n_{k}$, $m_{k}$ are chosen as the smallest integers satisfying (13), that is,
By (13), (14), and the triangle inequality, we easily derive that
Using (15) and the squeeze theorem we get
In a similar way, we can prove that $\lim_{k \rightarrow\infty }d(x_{m_{k}},x_{n_{k}+1})=0$, $\lim_{k \rightarrow\infty }d(x_{n_{k}},x_{m_{k}+1})=0$.
Regarding that T is a generalized almost αψϕGeraghty contractive type mapping, we have
for all $x,y \in X$, where
and
By taking the limit as $k\rightarrow\infty$ in (17) and taking (18), (10) into account, we get
Since β is a Geraghty function, we derive that $\psi (M(x_{m_{k}},x_{n_{k}})) \to0$. Consequently, we have $d(x_{m_{k}},x_{n_{k}}) \to0$, which is a contradiction. Hence, we conclude that $\lim_{m,n\rightarrow\infty }d(x_{n},x_{m})=0 $, and the sequence $\{x_{n}\}$ is Cauchy for any $s\geq1$.
By completeness of $(X,d)$, there exists $u\in X$ such that
Since T is continuous,
and u is a fixed point for T. □
In what follows, we replace the condition of continuity of the operator by the condition of αregularity of the space.
Theorem 2.4
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized almost αψϕGeraghty contractive type mapping. We suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
X is αregular and d is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
Following the lines of the proof of Theorem 2.3, we conclude that there exists $u\in X$ such that
Since X is αregular, $\alpha(x_{n}, x_{n+1})\geq1$ for all n. Due to the fact that $\lim_{n\to\infty}x_{n}=u$, there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $\alpha(x_{n_{k}},u)\geq 1$ for all k. To prove that u is a fixed point for T, suppose on the contrary that $d(u,Tu)>0$.
Now, by using the properties of ψ and as T is a generalized almost αψϕGeraghty contractive type mapping we have
Thus we have
where $N(x_{n_{k}},u)=\min\{ d^{s}(x_{n_{k}},Tu),d^{s}(u,x_{n_{k}+1}),d(x_{n_{k}},x_{n_{k}+1}),d(u,Tu) \}$, and note that $\lim_{k\to\infty}N(x_{n_{k}},u)=0$. Moreover,
Hence
and by the definition of $M(x_{n_{k}},u)$ we have $\lim_{k\to\infty } M(x_{n_{k}},u)=d(u,Tu)$.
By the continuity of ψ and the bmetriclike d, taking the limit as k goes to ∞ on both sides of (20) we have
Thus $1=\frac{\psi(d(u,Tu))}{\psi(d(u,Tu))}\leq\frac{1}{s}$, which is a contradiction in the case $s>1$. Hence $d(u,Tu)=0$; therefore $Tu=u$. In the case $s=1$ we take the limit as k goes to ∞ on both sides of
and get $\lim_{k\rightarrow\infty} \beta(\psi(M(x_{n_{k}},u)))=1$ and as $\beta\in\mathcal{F}_{1}$ so we have $\lim_{k\rightarrow \infty} \psi(M(x_{n_{k}},u))=0$. Thus we have $d(u,Tu)=0$; therefore $Tu=u$. □
For the uniqueness of a fixed point of a generalized αψϕ contractive mapping, we will consider the following hypothesis.

(H)
For all $x,y\in\operatorname{Fix}(T)$, either $\alpha(x,y)\geq1$ or $\alpha (y,x)\geq1$.
Here, $\operatorname{Fix}(T)$ denotes the set of fixed points of T.
Theorem 2.5
Adding condition (H) to the hypotheses of Theorem 2.3 (or Theorem 2.4), we obtain the uniqueness of the fixed point of T.
Proof
Suppose that $x^{*}$ and $y^{*}$ are two fixed points of T. Then it is obvious that $M(x^{*},y^{*})= d(x^{*},y^{*})$ and $N(x^{*},y^{*})=0$. So, we have
which is a contradiction. □
Definition 2.6
Let $(X,d)$ be a bmetriclike space with constant $s\geq1$, $T:X\rightarrow X$ be a map, we say that T is a generalized rational αψϕGeraghty contractive mapping of type (I) if there exist a function $\alpha:X\times X\to[0,\infty)$, $\psi,\phi\in \Psi$, $\beta\in\mathcal{F}_{s}$, and some $L\geq0$ such that
for all $x,y \in X$, where $N(x,y)$ is defined as in (5) and
Definition 2.7
Let $(X,d)$ be a bmetriclike space with constant $s\geq1$, $T:X\rightarrow X$ be a map, we say that T is a generalized rational αψϕGeraghty contractive of type (II) mapping if there exist a function $\alpha:X\times X\to[0,\infty)$, $\psi,\phi\in \Psi$, $\beta\in\mathcal{F}_{s}$ and some $L\geq0$ such that
for all $x,y \in X$, where $N(x,y)$ is defined as in (5) and
Theorem 2.8
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized rational αψϕGeraghty contractive mapping of type (I) such that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
We shall use the same techniques as in the proof of Theorem 2.3. First of all, we shall construct a sequence $\{x_{n}\}\subset X$ where $x_{n+1}=Tx_{n}$ for which $\alpha(x_{n},x_{n+1})\geq1$ and $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}_{0}$.
Since T is generalized rational αψϕGeraghty contractive of type (I) we have
Since $N(x_{n},x_{n+1})=0$, the above inequality implies that
where
On the other hand, we have
and
Consequently, we get $K(x_{n},x_{n+1})\leq\max\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2})\}$.
If $\max\{d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2})\}=d(x_{n+1},x_{n+2})$, then from (26) together with Remark 2.2, we have
This is a contradiction since ψ is increasing. Thus, we have $\max\{d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2})\} =d(x_{n},x_{n+1})$ and by the definition of $K(x_{n},x_{n+1})$ we shall have $K(x_{n},x_{n+1})=d(x_{n},x_{n+1})$. Consequently, the inequality (26) turns into
By Remark 2.2, we get
Case (i): $s>1$. Since $\frac{1}{s^{2}}>0$ and $s\frac{1}{s^{2}}=\frac {1}{s}<1$, by Lemma 1.8, the sequence $\{x_{n}\}$ is Cauchy and
Case (ii): $s=1$. Since $\{d(x_{n},x_{n+1})\}$ is a decreasing sequence, there exists $r\geq0$ such that
for some $r\geq0$. We shall prove that $r=0$. Suppose, on the contrary, that $r>0$. By letting $n \to\infty$ in (27) we find
It yields $1=\lim_{n \to\infty} \beta(\psi(K(x_{n},x_{n+1})))$. Since $\beta\in\mathcal{F}_{1}$, we get $\psi(K(x_{n},x_{n+1})) \to0$, which implies that $d(x_{n},x_{n+1}) \to0$, that is, $r=0$.
In what follows, we shall prove that $\{x_{n}\}$ is a Cauchy sequence. Indeed we will prove that $\lim_{m,n\rightarrow\infty }d(x_{n},x_{m})=0 $. Suppose, on the contrary, that there exist $\varepsilon>0$ and corresponding subsequences $\{n_{k}\} $ and $\{m_{k}\}$ of $\mathbb{N}$ satisfying $n_{k}>m_{k}>k$ for which
where $n_{k}$, $m_{k}$ are chosen as the smallest integers satisfying (31), that is,
By (31), (32), and the triangle inequality, we easily derive that
Using (33) and the squeeze theorem we get
Regarding that T is generalized rational αψϕGeraghty contractive mapping of type (I), we have
for all $x,y \in X$, where
and
It is clear that
By taking the limit as $k\rightarrow\infty$ in (35) and taking (36), (30) into account, we get
Since β is a Geraghty function, we derive that $\psi (K(x_{m_{k}},x_{n_{k}})) \to0$. Consequently, we have $d(x_{m_{k}},x_{n_{k}}) \to0$, which is a contradiction. Hence $\lim_{m,n\rightarrow\infty }d(x_{n},x_{m})=0 $, and the sequence $\{x_{n}\}$ is Cauchy for any $s \geq1$.
By completeness of $(X,d)$, there exists $u\in X$ such that
Now, if T is continuous, then
and u is a fixed point for T. □
Theorem 2.9
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized rational αψϕGeraghty contractive of mapping type (I) such that:

(i)
T is triangular αorbital admissible;

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

(iii)
X is αregular and d is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
Verbatim of the proof of Theorem 2.8, we conclude that the iterative sequence $\{x_{n}\}$ is Cauchy and converges to $u \in X$. Since X is αregular, then, as in the proof of Theorem 2.4, there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that
where $K(x_{n_{k}},u)=\max\{d(x_{n_{k}},u),\frac {d(x_{n_{k}},x_{n_{k}+1})d(u,Tu)}{1+d(x_{n_{k}},u)}, \frac {d(x_{n_{k}},x_{n_{k}+1})d(u,Tu)}{1+d(x_{n_{k}+1},Tu)}\}$.
Hence $\lim_{k\to\infty} K(x_{n_{k}},u)=0$ and as in the proof of Theorem 2.4 $\lim_{k\to\infty}N(x_{n_{k}},u)=0$.
Thus taking the limit as $k \rightarrow\infty$ on both sides of (38) and keeping in mind that ψ and d are continuous we have $\psi(d(u,Tu))\leq0$. Hence $d(u,Tu)=0$; therefore $Tu=u$. □
Theorem 2.10
Adding condition (H) to the hypotheses of Theorem 2.8 (or Theorem 2.9), we obtain uniqueness of the fixed point of T.
Proof
As in the proof of Theorem 2.5, we suppose that $x^{*}$ and $y^{*}$ are two fixed points of T. Then, clearly, we have $K(x^{*},y^{*})= d(x^{*},y^{*})$ and $N(x^{*},y^{*})=0$. So, we have
which is a contradiction. □
Theorem 2.11
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized rational αψϕGeraghty contractive mapping of type (II) such that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
Verbatim of the lines in the proof of Theorem 2.3, we construct a sequence $\{x_{n}\}\subset X$ where $x_{n+1}=Tx_{n}$ for which $\alpha (x_{n},x_{n+1})\geq1$ and $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}_{0}$. Moreover, by using the fact that T is a generalized rational αψϕGeraghty contractive mapping of type (II) and the property of ψ we have
By using the same arguments as in the proof of Theorem 2.3, we derive that
where
Since $d(x_{n},x_{n+2})\leq s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]$, we have
Hence, we get $Q(x_{n},x_{n+1})=d(x_{n},x_{n+1})$.
By using (40) we get $\psi(s^{2}d(x_{n+1},x_{n+2}))\leq\psi (d(x_{n},x_{n+1}))$. Since ψ is increasing we have $d(x_{n+1},x_{n+2})<\frac {1}{s^{2}}d(x_{n},x_{n+1})$. If $s>1$ then, as in the proof of Theorem 2.3, by using Lemma 1.8, we conclude that $\{x_{n}\}$ is a Cauchy sequence and $\lim_{n,m\to\infty }d(x_{n},x_{m})=0$. If $s=1$, by verbatim of the proof of Theorem 2.3, we deduce that $\{x_{n}\}$ is a Cauchy sequence.
Since $(X,d)$ is complete, there exists $u\in X$ such that $0=\lim_{n,m\to\infty}d(x_{n},x_{m})= \lim_{n\to\infty}d(x_{n},u)=d(u,u)$. Now, since T is continuous, $Tu=T(\lim_{n\to\infty}x_{n})=\lim_{n\to\infty}Tx_{n}=\lim_{n\to\infty}x_{n+1}=u$ and u is a fixed point for T. □
Theorem 2.12
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a generalized rational αψϕGeraghty contractive mapping of type (II) such that:

(i)
T is triangular αorbital admissible;

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

(iii)
X is αregular and d is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
By following the proof of Theorem 2.11 line by line, we see that $\{x_{n}\}$ converges to $u \in X$. Due to the fact that X is αregular and by following the lines of the proof of Theorem 2.4 there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that
where
Note that $\lim_{k\to\infty}Q(x_{n_{k}},u)=0$ and $\lim_{k\to\infty}N(x_{n_{k}},u)=0$. Thus taking the limit as $n \rightarrow\infty$ on both sides of (41) and keeping in mind that ψ and d are continuous we have $\psi(d(u,Tu))=0$, and so $d(u,Tu)=0$. Thus $u=Tu$. □
Theorem 2.13
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a mapping. Suppose that there exist a function $\alpha:X\times X\to[0,\infty)$, $\psi\in\Psi$, $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$. Suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Proof
By (ii) there exists $x_{0}\in X$ such that $\alpha(x_{0},Tx_{0})\geq1$. Define a sequence $\{x_{n}\}\subset X$ by $x_{n+1}=Tx_{n}$ for all $n\in \mathbb{N}_{0}$. As T is triangular αorbital admissible, by Lemma 1.13 we have $\alpha(x_{n},x_{n+1})\geq1$ for all $n\in\mathbb{N}_{0}$. Notice that if there exists a natural number $n_{0}$ such that $x_{n_{0}}= x_{n_{0}+1}$, then the proof is complete. To avoid this trivial case, from now on, we assume that $x_{n}\neq x_{n+1}$ for all $n\in\mathbb{N}_{0}$.
Since T satisfies (42) we have
Thus, we have
Since ψ is increasing, we have $d(x_{n+1},x_{n+2}) < \frac{1}{s^{2} }d(x_{n},x_{n+1})$.
Case (i): If $s>1$, then, since $\frac{1}{s^{2}}>0$ and $s\frac {1}{s^{2}}=\frac{1}{s}<1$, by Lemma 1.8, $\{x_{n}\}$ is a Cauchy sequence and $\lim_{n,m\to\infty}d(x_{n},x_{m})=0$.
Case (ii): If $s=1$, then as $d(x_{n+1},x_{n+2}) < d(x_{n},x_{n+1})$, there exists $r\geq0$ such that $\lim_{n\rightarrow\infty }d(x_{n},x_{n+1})=r$. If $r>0$ then taking the limit as $n\rightarrow \infty$ on both sides of
we get $\lim_{n\rightarrow\infty}\beta(\psi(d(x_{n},x_{n+1})))=1$ and as β is a Geraghty function, we derive that
That is, $r=0$. In what follows we shall prove that $\{x_{n}\}$ is a Cauchy sequence. Indeed we will prove that $\lim_{m,n\rightarrow\infty }d(x_{n},x_{m})=0 $. Suppose, on the contrary, that there exist $\varepsilon>0$ and corresponding subsequences $\{n_{k}\} $ and $\{m_{k}\}$ of $\mathbb{N}$ satisfying $n_{k}>m_{k}>k$ for which
where $n_{k}$, $m_{k}$ are chosen as the smallest integers satisfying (45), that is,
By (45), (46), and the triangle inequality, we easily derive that
Using (47) and the squeeze theorem we get
As T satisfies (42), we have
Taking the limit as $k\rightarrow\infty$ for (49) we get $\lim_{k\rightarrow\infty}\beta(\psi (d(x_{m_{k}},x_{n_{k}})))=1$. Thus
Therefore, $\{x_{n}\}$ is a Cauchy sequence for $s\geq1$. By completeness of $(X,d)$, there exists $u\in X$ such that $0=\lim_{n,m\to\infty}d(x_{n},x_{m})=\lim_{n\to\infty}d(x_{n},u)=d(u,u)$.
Since T is continuous, $Tu=T(\lim_{n\to\infty}x_{n})=\lim_{n\to\infty}Tx_{n}=\lim_{n\to\infty}x_{n+1}=u$ and u is a fixed point for T. □
Theorem 2.14
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a mapping. Suppose that there exist a function $\alpha:X\times X\to[0,\infty)$, $\psi\in\Psi$, and $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$. Suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
X is αregular and d is continuous.
Then T has a fixed point, $u \in X$ with $d(u,u)=0$.
Theorem 2.15
Adding condition (H) to the hypotheses of Theorem 2.13 (or Theorem 2.14), we obtain the uniqueness of the fixed point of T.
Remark 2.16
Notice that we get several corollaries by replacing the auxiliary functions ψ and β in a proper way. In particular, by taking $\psi(t)=t$ we find the extended version of several existing results.
Expected consequences
In this section, we shall consider some immediate consequences of our main results.
The following result is obtained by letting $L=0$ in Theorem 2.3 or 2.4.
Corollary 3.1
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ and $T:X\rightarrow X$ be a mapping. Suppose that there exist $\alpha:X\times X\to[0,\infty)$, $\psi\in \Psi$, $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$, where
Suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous or (iii)′ X is αregular and d is continuous.
Then T has a fixed point.
Adding condition (H) to the hypothesis of Corollary 3.1, we guarantee the uniqueness of the fixed point.
Again by letting $L=0$ in Theorem 2.5 and Theorem 2.10 we get two more corollaries as Corollary 3.1. We skip the details regarding the volume of the paper.
Corollary 3.2
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq 1$, $T:X\rightarrow X$ be a map and $\alpha: X\times X \rightarrow [0,\infty) $ be a function. Suppose that T satisfies at least one of the following conditions:

(a)
$\alpha(x,y) d(Tx,Ty) \leq\frac{1}{2 s^{3}} M(x,y)$;

(b)
$\alpha(x,y) d(Tx,Ty) \leq\frac{1}{2 s^{3}} K(x,y)$;
where $M(x,y)$, $K(x,y)$ are defined as in (4), (22). Suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous or (iii)′ X is αregular and d is continuous.
Then T has a unique fixed point $u \in X$ with $d(u,u)=0$.
Proof
It is sufficient to take $L=0$, $\psi(t)=t$, and $\beta(t)= \frac{1}{2 s}$ in Theorem 2.5 and Theorem 2.10 (and thus, Theorem 2.3 or Theorem 2.4, Theorem 2.8 or Theorem 2.9, respectively). □
Adding condition (H) to the hypothesis of Corollary 3.2, we guarantee the uniqueness of the fixed point.
Corollary 3.3
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq 1$, $T:X\rightarrow X$ be a map, and $\alpha: X\times X \rightarrow [0,\infty) $ be a function. Suppose that T satisfies at least one of the following conditions:

(c)
$\alpha(x,y) d(Tx,Ty) \leq\frac{1}{2 s^{3}} Q(x,y)$,
where $M(x,y)$, $K(x,y)$, $Q(x,y)$ are defined as in (24). Suppose also that:

(i)
T is triangular αorbital admissible;

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

(iii)
T is continuous or (iii)′ X is αregular and d is continuous.
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
Proof
It is sufficient to take $L=0$, $\psi(t)=t$, and $\beta(t)= \frac{1}{2 s}$ in Theorem 2.11 or Theorem 2.12, respectively. □
For standard bmetriclike spaces
If we set $\alpha(x,y)=1$ for all $x,y \in X$ in Theorem 2.5, then we derive the following results.
Corollary 3.4
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ such that d is continuous and $T:X\rightarrow X$ be a mapping. Suppose that there exist $\psi,\phi\in\Psi$, $\beta\in\mathcal{F}_{s}$, and some $L\geq0$ such that
for all $x,y \in X$, where
Then T has a unique fixed point $u \in X$ with $d(u,u)=0$.
If we set $\alpha(x,y)=1$ for all $x,y \in X$ in Theorem 2.10, then we derive the following results.
Corollary 3.5
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ such that d is continuous and $T:X\rightarrow X$ be a mapping. Suppose that there exist $\psi,\phi\in\Psi$, $\beta\in\mathcal{F}_{s}$, and some $L\geq0$ such that
for all $x,y \in X$, where
Then T has a unique fixed point $u \in X$ with $d(u,u)=0$.
If we set $\alpha(x,y)=1$ for all $x,y \in X$ in Theorem 2.11, then we derive the following results.
Corollary 3.6
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ such that d is continuous and $T:X\rightarrow X$ be a mapping. Suppose that there exist $\psi,\phi\in\Psi$, $\beta\in\mathcal{F}_{s}$, and some $L\geq0$ such that
for all $x,y \in X$, where
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
If take $L=0$ in Corollaries 3.43.6, we get three more consequences. Regarding the volume of the paper, we skip the details.
Corollary 3.7
Let $(X,d)$ be a complete bmetriclike space with constant $s\geq1$ such that d is continuous and $T:X\rightarrow X$ be mapping. Suppose that there exist $\psi\in\Psi$ and $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$. Then T has a unique fixed point $u \in X$ with $d(u,u)=0$.
Proof
It follows from Theorem 2.15 by $\alpha(x,y)=1$ for all $x,y \in X$. □
For bmetriclike spaces endowed with a partial order
In this section, from our main results, we shall derive easily various fixed point results on a bmetriclike space endowed with a partial order. We, first, recall some notions.
Definition 3.8
Let $(X,\preceq)$ be a partially ordered set and $T: X\to X$ be a given mapping. We say that T is nondecreasing with respect to ⪯ if
Definition 3.9
Let $(X,\preceq)$ be a partially ordered set. A sequence $\{x_{n}\} \subset X$ is said to be nondecreasing (respectively, nonincreasing) with respect to ⪯ if $x_{n}\preceq x_{n+1}$ (respectively, $x_{n+1}\preceq x_{n} $ for all n).
Definition 3.10
Let $(X,\preceq)$ be a partially ordered set and d be a bmetriclike on X. We say that $(X,\preceq,d)$ is regular if for every nondecreasing (respectively, nonincreasing) sequence $\{x_{n}\}\subset X$ 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, $x_{n_{k}}\succeq x$) for all k.
We have the following result.
Corollary 3.11
Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset) and d be a bmetriclike on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $T: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exist $\psi\in\Psi$, $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$ with $x\succeq y$ or $y\succeq x$ where $M(x,y)$ is defined as in (4). Suppose also that the following conditions hold:

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

(ii)
T is continuous or (ii)′ $(X,\preceq,d)$ is regular and d is continuous.
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
Proof
Define the mapping $\alpha: X\times X\to[0,\infty)$ by
Clearly, T satisfies (51), that is,
for all $x,y\in X$. From condition (i), we have $\alpha(x_{0},Tx_{0})\geq 1$. Moreover, for all $x,y\in X$, from the monotone property of T, we have
Hence, the selfmapping T is αadmissible. Similarly, we can prove that T is triangular αadmissible and so triangular αorbital admissible. Now, if T is continuous, the existence of a fixed point follows from Corollary 3.1. Suppose now 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 $n\rightarrow\infty$. From the regularity hypothesis and as 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 Corollary 3.1. □
In an analogous way, we derive the following results from Theorem 2.8 and Theorem 2.11, respectively.
Corollary 3.12
Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset) and d be a bmetriclike on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $T: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exist $\psi\in\Psi$, $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$ with $x\succeq y$ or $y\succeq x$ where $M(x,y)$ is defined as in (22). Suppose also that the following conditions hold:

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

(ii)
T is continuous or (ii)′ $(X,\preceq,d)$ is regular and d is continuous.
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
Corollary 3.13
Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset) and d be a bmetriclike on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $T: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exist $\psi\in\Psi$, $\beta\in\mathcal{F}_{s}$ such that
for all $x,y \in X$ with $x\succeq y$ or $y\succeq x$ where $M(x,y)$ is defined as in (24). Suppose also that the following conditions hold:

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

(ii)
T is continuous or (ii)′ $(X,\preceq,d)$ is regular and d is continuous.
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
Corollary 3.14
Let $(X,\preceq)$ be a partially ordered set (which does not contain an infinite totally unordered subset) and d be bmetriclike on X with constant $s\geq1$ such that $(X,d)$ is complete. Let $T: X\to X$ be a nondecreasing mapping with respect to ⪯. Suppose that there exist $\psi \in\Psi$, $\beta\in\mathcal{F}_{s}$ such that
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 $x_{0}\preceq Tx_{0}$;

(ii)
T is continuous or (ii)′ $(X,\preceq,d)$ is regular and d is continuous.
Then T has a fixed point $u \in X$ with $d(u,u)=0$.
Example 3.15
Let $X=[0,\infty)$ define $d:X \times X \rightarrow[0,\infty) $ by $d(x,y)=xy^{2}$. Then $(X,d)$ is complete bmetric (so bmetriclike) space with constant $s=2$. Define $T:X \rightarrow X$ and $\alpha(x,y): X\times X \to [0,\infty) $ as follows:
and
It is clear that T is triangular αorbital admissible and we have $\alpha(0,T0) \geq1 $. Moreover, X is αregular and d is continuous.
Let $\psi(t)=t$, $\beta(t)=\frac{1}{4}$ then clearly $\psi\in\Psi$ and $\beta\in\mathcal{F}_{2}$. Moreover, T satisfies (50) for the following reason: if $x,y \in[0,1]$, then
Otherwise,
Therefore, by Theorem 2.14, T has a fixed point $x=0$.
Example 3.16
Let $X=\{ 0,1,2\}$ define $d:X \times X \rightarrow[0,\infty) $ by $d(x,y)=(\max\{x,y\})^{\frac{3}{2}}$. Then $(X,d)$ is complete bmetriclike space with constant $s=2^{\frac{3}{2}1}=2^{\frac {1}{2}}$ such that d is continuous. Define $T:X \rightarrow X$ by $T=\{(0,0), (1,0), (2,0)\}$.
Let $\psi(t)=t$, $\beta(t)=\frac{1}{2^{\frac{1}{2}}} e^{t}$ or $\beta(t)=\frac{1}{2^{\frac{1}{2}}+t}$, then clearly $\psi\in\Psi$ and $\beta\in\mathcal{F}_{2^{\frac{1}{2}}}$. Note that $K(0,1)=1$, $K(0,2)=K(1,2)=2^{\frac{3}{2}}$, and clearly T satisfies (56) with $L=0$ Therefore, by Corollary 3.5, T has a fixed point $x=0$.
Conclusion
It is clear that we can list several more results by replacing the bmetriclike space, with some other abstract space, such as a bmetric space, a metric space, a metriclike space, a partial metric space, and so on.
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DOI
MSC
 46T99
 46N40
 47H10
 54H25
Keywords
 bmetriclike space
 fixed point
 αadmissible
 contractive mapping