# Some extensions for Geragthy type contractive mappings

## Abstract

In this paper, we establish some fixed point theorems on some extensions of Geragthy contractive type mappings in the context of b-metric-like spaces.

## Introduction and preliminaries

One of the interesting extensions of the notion of a metric space is the dislocated space, introduced by Hitzler . This notion was rediscovered by Amini-Harandi  and given the name of a metric-like 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):

$$\sigma(x,y)\leq\sigma(x,z)+\sigma(z,y)$$;

and the pair $$(X,\sigma)$$ is called a dislocated (metric-like) space.

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 non-negative reals. First, we recall some basic concepts and notations.

The concept of a b-metric was introduced by Czerwik  as a generalization of the metric (see also Bakhtin  and Bourbaki ) to extend the celebrated Banach contraction mapping principle. Following this initial paper of Czerwik , 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 e.g.  and references therein).

### Definition 1.2



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

In what follows, we recall the notion of b-metric-like space which is an interesting generalization of both b-metric space and metric-like space.

### Definition 1.3



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 b-metric-like if for all $$x, y, z \in X$$ the following conditions are satisfied:

($$bML_{1}$$):

if $$d(x, y) =0$$ then $$x = y$$;

($$bML_{2}$$):

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

($$bML_{3}$$):

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

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

### 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

$$d(f,g) = \max \bigl( \bigl|f(t)-g(t) \bigr| \bigr)^{p}+a,$$

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 b-metric-like space with $$s=2^{p-1}$$. For more examples, see e.g. .

### Remark 1.5

Let $$(X, d)$$ be a b-metric-like 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:

1. (S1)

$$d^{s}(x,x)=0$$ for all $$x \in X$$.

### Definition 1.6



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

1. (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. (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. (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



Let $$(X,d)$$ be a b-metric-like 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. (1)

x is unique.

2. (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



Let $$(X,d)$$ be a b-metric-like space with constant s and $$\{x_{n}\}$$ a sequence in X such that

$$d(x_{n+1},x_{n+2})\leq k d(x_{n},x_{n+1}) ,\quad n=0,1,\ldots,$$

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



Let $$(X,d)$$ be a 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

\begin{aligned} \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). \end{aligned}
(1)

In particular, if $$d(x,x)=0$$, then

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

Notice that, in general, a b-metric-like mapping does not need to be continuous.

The notion of α-admissible and triangular α-admissible mappings were introduced by Samet et al.  and Karapınar et al. , 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 α-admissible mapping if

$$x,y \in X, \quad \alpha(x,y)\geq1 \quad \Rightarrow \quad \alpha (Tx,Ty)\geq1.$$

Moreover, a self-mapping T is called triangular α-admissible if T is α-admissible and

$$x,y,z \in X, \quad \alpha(x,z)\geq1 \mbox{ and } \alpha(z,y)\geq 1 \quad \Rightarrow \quad \alpha(x,y)\geq1.$$

For more details on α-admissible and triangular α-admissible mappings, see e.g. .

Very recently, Popescu  refined the notion of triangular α-orbital admissible as follows.

### Definition 1.11



Let $$T:X\rightarrow X$$ be a mapping and $$\alpha:X\times X\rightarrow [0,\infty)$$ be a function. We say that T is α-orbital admissible if

\begin{aligned} \alpha(x,Tx)\geq1 \quad\Rightarrow\quad\alpha \bigl(Tx,T^{2}x \bigr) \geq1. \end{aligned}

Furthermore, T is called triangular α-orbital admissible if T is α-orbital admissible and

\begin{aligned} \alpha(x,y)\geq1 \mbox{ and } \alpha(y,Ty)\geq1 \quad\Rightarrow\quad \alpha(x,Ty) \geq1. \end{aligned}

As mentioned in  each α-admissible (respectively, triangular α-admissible) mapping is an α-orbital admissible (respectively, triangular α-orbital 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 self-mapping $$T : X \to X$$ such that $$Tx=x$$, for $$x=a,d$$ and

$$Tx=y \quad\mbox{for }(x,y) \in \bigl\{ (b,c),(c,b),(e,f),(f,e),(g,h),(h,g) \bigr\} .$$

Moreover, we define $$\alpha: X \times X \rightarrow \mathbb{R}_{0}^{+}$$, such that

$$\alpha(x, y) = \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1 & \mbox{if } (x, y) \in\{(a,b), (a,c), (b,b), (c,c), (b,c), (c,b), (b,d), (c,d), (d,e) \},\\ 0& \mbox{otherwise}. \end{array}\displaystyle \right .$$

Note that T is α-orbital 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 α-admissible.

### Lemma 1.13



Let $$T:X\to X$$ be a triangular α-orbital 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 α-orbital 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 α-regular in the setting of a b-metric-like space.

### Definition 1.15

(cf. )

Let $$(X,d)$$ be a b-metric-like space, X is said to be α-regular, 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 b-metric-like spaces. The presented results improve, extend, and unify a number of existing results in the literature.

## Main result for b-metric-like 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

$$\lim_{n \to\infty}\beta(t_{n})=\frac{1}{s} \quad \mbox{implies}\quad \lim_{n \to\infty}t_{n}=0,$$
(2)

for some $$s\geq1$$.

### Definition 2.1

Let $$(X,d)$$ be a b-metric-like space with constant $$s\geq1$$, and $$T:X\rightarrow X$$ be a map. We say that T is a generalized almost α-ψ-ϕ-Geraghty 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr)\psi \bigl(M(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(3)

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

\begin{aligned}& M(x,y)=\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty), \frac{d(x,Ty)+d(y,Tx)}{4s} \biggr\} \quad \mbox{and} \end{aligned}
(4)
\begin{aligned}& N(x,y)=\min \bigl\{ d^{s}(x,Ty),d^{s}(y,Tx),d(x,Tx),d(y,Ty) \bigr\} . \end{aligned}
(5)

### 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:

$$\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr) < \frac{1}{s} \quad\mbox{for any } x,y\in X.$$

### Theorem 2.3

Let $$(X,d)$$ be a complete b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized almost α-ψ-ϕ-Geraghty contractive type mapping. We suppose also that

1. (i)

2. (ii)

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

3. (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 α-orbital 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 α-ψ-ϕ-Geraghty contractive type mapping we have

\begin{aligned} \psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)&\leq\alpha(x_{n},x_{n+1}) \psi \bigl(s^{2}d(Tx_{n},Tx_{n+1}) \bigr) \\ &\leq\beta \bigl(\psi \bigl(M(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(M(x_{n},x_{n+1}) \bigr)+L\phi \bigl(N(x_{n},x_{n+1}) \bigr). \end{aligned}
(6)

Thus, we have

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(M(x_{n},x_{n+1}) \bigr)+L \phi \bigl(N(x_{n},x_{n+1}) \bigr),$$
(7)

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

$$M(x_{n},x_{n+1})=\max \biggl\{ d(x_{n},x_{n+1}),d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}), \frac{d(x_{n},x_{n+2})+d(x_{n+1},x_{n+1})}{4s} \biggr\} .$$

Note that

\begin{aligned} \frac{d(x_{n},x_{n+2})+d(x_{n+1},x_{n+1})}{4s} &\leq \frac{s[d(x_{n},x_{n+1})+3 d(x_{n+1},x_{n+2})]}{4s} \\ &= \frac{[d(x_{n},x_{n+1})+3 d(x_{n+1},x_{n+2})]}{4} \\ & \leq\max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}) \bigr\} . \end{aligned}

Consequently, we have

$$M(x_{n},x_{n+1})=\max \bigl\{ d(x_{n},x_{n+1}),d(x_{n+1},x_{n+2}) \bigr\} .$$

If $$M(x_{n},x_{n+1})=d(x_{n+1},x_{n+2})$$, then from (7) we have

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(d(x_{n+1},x_{n+2}) \bigr)\leq \psi \bigl(d(x_{n+1},x_{n+2}) \bigr).$$

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

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(d(x_{n},x_{n+1}) \bigr)\leq\psi \bigl(d(x_{n},x_{n+1}) \bigr).$$

Hence, we get

$$s^{2}d(x_{n+1},x_{n+2})\leq d(x_{n},x_{n+1})\quad\mbox{equivalently}\quad d(x_{n+1},x_{n+2}) \leq \frac{1}{s^{2}}d(x_{n},x_{n+1}) .$$
(8)

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

$$\lim_{n,m\to\infty}d(x_{n},x_{m})=0.$$
(9)

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

$$\lim_{n \to\infty} d(x_{n},x_{n+1})=r,$$
(10)

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

$$\psi \bigl(d(x_{n+1},x_{n+2}) \bigr)< \beta \bigl( \psi \bigl(M(x_{n},x_{n+1}) \bigr) \bigr)\psi \bigl(M(x_{n},x_{n+1}) \bigr)+L\phi \bigl(N(x_{n},x_{n+1}) \bigr),$$
(11)

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

$$\frac{\psi(d(x_{n+1},x_{n+2}))}{\psi(d(x_{n},x_{n+1}))}\leq \beta \bigl(\psi \bigl(d(x_{n},x_{n+1}) \bigr) \bigr) < 1.$$
(12)

By taking the limit as $$n \to\infty$$ in (12) and regarding the continuity of ψ, we get

$$\lim_{n\to\infty} \beta\bigl(\psi\bigl(d(x_{n},x_{n+1})\bigr)\bigr)=1.$$

Hence, we have

$$\lim_{n\to\infty} \psi \bigl(d(x_{n},x_{n+1}) \bigr)=0 \quad\mbox{and so}\quad\lim_{n\to \infty} d(x_{n},x_{n+1})=0 .$$

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

$$d(x_{m_{k}},x_{n_{k}})\geq\varepsilon,$$
(13)

where $$n_{k}$$, $$m_{k}$$ are chosen as the smallest integers satisfying (13), that is,

$$d(x_{m_{k}}, x_{n_{k}-1})< \varepsilon.$$
(14)

By (13), (14), and the triangle inequality, we easily derive that

\begin{aligned} \varepsilon&\leq d(x_{m_{k}},x_{n_{k}}) \leq d(x_{m_{k}},x_{n_{k}-1})+d(x_{n_{k}-1},x_{n_{k}}) < \varepsilon+d(x_{n_{k}-1},x_{n_{k}}). \end{aligned}
(15)

Using (15) and the squeeze theorem we get

$$\lim_{k\to\infty} d(x_{m_{k}},x_{n_{k}})= \varepsilon.$$
(16)

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 α-ψ-ϕ-Geraghty contractive type mapping, we have

\begin{aligned} \psi \bigl( d(x_{m_{k}+1},x_{n_{k}+1}) \bigr)& \leq \alpha(x_{m_{k}},x_{n_{k}})\psi \bigl( d(Tx_{m_{k}},Tx_{n_{k}}) \bigr) \\ & \leq\beta \bigl(\psi \bigl(M(x_{m_{k}},x_{n_{k}}) \bigr) \bigr) \psi \bigl(M(x_{m_{k}},x_{n_{k}}) \bigr)+L\phi \bigl(N(x_{m_{k}},x_{n_{k}}) \bigr), \end{aligned}
(17)

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

\begin{aligned} M(x_{m_{k}},x_{n_{k}})={}&\max \biggl\{ d(x_{m_{k}},x_{n_{k}}),d(x_{m_{k}},x_{m_{k}+1}),d(x_{n_{k}},x_{n_{k}+1}), \\ &{}\frac {d(x_{m_{k}},x_{n_{k}+1})+d(x_{n_{k}},x_{m_{k}+1})}{4} \biggr\} , \end{aligned}
(18)

and

$$N(x_{m_{k}},x_{n_{k}})=\min \bigl\{ d^{s}(x_{m_{k}},x_{n_{k}+1}),d^{s}(x_{n_{k}},x_{m_{k}+1}), d(x_{m_{k}},x_{m_{k}+1}),d(x_{n_{k}},x_{n_{k}+1}) \bigr\} .$$

By taking the limit as $$k\rightarrow\infty$$ in (17) and taking (18), (10) into account, we get

$$\psi(\varepsilon) \leq\lim_{k\to\infty} \beta \bigl( \psi \bigl(M(x_{m_{k}},x_{n_{k}}) \bigr) \bigr) \psi(\varepsilon).$$
(19)

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

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

Since T is continuous,

$$Tu=T \Bigl(\lim_{n\to\infty}x_{n} \Bigr)=\lim _{n\to\infty}Tx_{n}=\lim_{n\to\infty}x_{n+1}=u,$$

and u is a fixed point for T. □

In what follows, we replace the condition of continuity of the operator by the condition of α-regularity of the space.

### Theorem 2.4

Let $$(X,d)$$ be a complete b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized almost α-ψ-ϕ-Geraghty contractive type mapping. We suppose also that:

1. (i)

2. (ii)

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

3. (iii)

X is α-regular 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

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

Since X is α-regular, $$\alpha(x_{n}, x_{n+1})\geq1$$ 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 α-ψ-ϕ-Geraghty contractive type mapping we have

\begin{aligned} \psi \bigl(d(x_{n_{k}+1},Tu) \bigr)&\leq\alpha(x_{n_{k}},u) \psi \bigl(s^{2}d(Tx_{n_{k}},Tu) \bigr) \\ &\leq\beta \bigl(\psi \bigl(M(x_{n_{k}},u) \bigr) \bigr)\psi \bigl(M(x_{n_{k}},u) \bigr)+L\phi \bigl(N(x_{n_{k}},u) \bigr). \end{aligned}

Thus we have

$$\psi \bigl(d(x_{n_{k+1}},Tu) \bigr)< \frac{1}{s} \psi \bigl(M(x_{n_{k}},u) \bigr)+L \phi \bigl(N(x_{n_{k}},u) \bigr),$$
(20)

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,

\begin{aligned} M(x_{n_{k}},u)={}&\max \biggl\{ d(x_{n_{k}},u),d(x_{n_{k}},x_{n_{k}+1}),d(u,Tu), \frac {d(x_{n_{k}},Tu)+d (u,x_{n_{k}+1})}{4s} \biggr\} \\ \leq{}&\max \biggl\{ d(x_{n_{k}},u),d(x_{n_{k}},x_{n_{k}+1}),d(u,Tu),\\ &{}\frac {d(x_{n_{k}},u)+d(u,Tu)+d(u,x_{n_{k}})+d (x_{n_{k}},x_{n_{k}+1})}{4} \biggr\} . \end{aligned}

Hence

$$\lim_{k\to\infty} M(x_{n_{k}},u)\leq\max \biggl\{ 0,d(u,Tu), \frac {d(u,Tu)}{4} \biggr\} =d(u,Tu),$$

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 b-metric-like d, taking the limit as k goes to ∞ on both sides of (20) we have

$$\psi \bigl(d(u,Tu) \bigr)\leq\frac{1}{s}\psi \bigl(d(u,Tu) \bigr).$$

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

$$\psi \bigl(d(x_{n_{k}+1},Tu) \bigr)\leq\beta \bigl(\psi \bigl(M(x_{n_{k}},u) \bigr) \bigr)\psi \bigl(M(x_{n_{k}},u) \bigr)$$

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 α-ψ-ϕ contractive mapping, we will consider the following hypothesis.

1. (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

\begin{aligned} \psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr)&\leq\psi \bigl(s^{2}d \bigl(Tx^{*},Ty^{*} \bigr) \bigr) \\ &\leq\alpha \bigl(x^{*},y^{*} \bigr)\psi \bigl(s^{2} d \bigl(Tx^{*},Ty^{*} \bigr) \bigr) \\ &\leq\beta \bigl(\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr) \bigr)\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr) \\ &< \frac{1}{s}\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr), \end{aligned}

### Definition 2.6

Let $$(X,d)$$ be a b-metric-like space with constant $$s\geq1$$, $$T:X\rightarrow X$$ be a map, we say that T is a generalized rational α-ψ-ϕ-Geraghty 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl( \psi \bigl(K(x,y) \bigr) \bigr)\psi \bigl(K(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(21)

for all $$x,y \in X$$, where $$N(x,y)$$ is defined as in (5) and

$$K(x,y)=\max \biggl\{ d(x,y), \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}, \frac {d(x,Tx)d(y,Ty)}{1+d(Tx,Ty)} \biggr\} .$$
(22)

### Definition 2.7

Let $$(X,d)$$ be a b-metric-like space with constant $$s\geq1$$, $$T:X\rightarrow X$$ be a map, we say that T is a generalized rational α-ψ-ϕ-Geraghty 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl( \psi \bigl(Q(x,y) \bigr) \bigr)\psi \bigl(Q(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(23)

for all $$x,y \in X$$, where $$N(x,y)$$ is defined as in (5) and

$$Q(x,y)=\max \biggl\{ d(x,y), \frac {d(x,Tx)d(y,Ty)+d^{s}(x,Ty)d^{s}(y,Tx)}{1+s(d(x,Tx)+d(y,Ty))} \biggr\} .$$
(24)

### Theorem 2.8

Let $$(X,d)$$ be a complete b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized rational α-ψ-ϕ-Geraghty contractive mapping of type (I) such that:

1. (i)

2. (ii)

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

3. (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 α-ψ-ϕ-Geraghty contractive of type (I) we have

\begin{aligned} \psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)&\leq\alpha(x_{n},x_{n+1}) \psi \bigl(s^{2}d(Tx_{n},Tx_{n+1}) \bigr) \\ &\leq\beta \bigl(\psi \bigl(K(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(K(x_{n},x_{n+1}) \bigr)+L\phi \bigl(N(x_{n},x_{n+1}) \bigr). \end{aligned}
(25)

Since $$N(x_{n},x_{n+1})=0$$, the above inequality implies that

$$\psi \bigl(s^{2} d(x_{n+1},x_{n+2}) \bigr)\leq\beta \bigl(\psi \bigl(K(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(K(x_{n},x_{n+1}) \bigr),$$
(26)

where

$$K(x_{n},x_{n+1})=\max \biggl\{ d(x_{n},x_{n+1}), \frac {d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n},x_{n+1})}, \frac {d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n+1},x_{n+2})} \biggr\} .$$

On the other hand, we have

$$\frac{d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n+1},x_{n+2})}\leq\frac {d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{d(x_{n+1},x_{n+2})} =d(x_{n},x_{n+1})$$

and

$$\frac{d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n},x_{n+1})}\leq\frac {d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{d(x_{n},x_{n+1})} =d(x_{n+1},x_{n+2}).$$

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

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(d(x_{n+1},x_{n+2}) \bigr).$$

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

$$\psi \bigl( d(x_{n+1},x_{n+2}) \bigr) \leq\psi \bigl(s^{2} d(x_{n+1},x_{n+2}) \bigr)\leq\beta \bigl( \psi \bigl(K(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(d(x_{n},x_{n+1}) \bigr).$$
(27)

By Remark 2.2, we get

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \psi \bigl(d(x_{n},x_{n+1}) \bigr) \mbox{ and hence}, d(x_{n+1},x_{n+2})< \frac{1}{s^{2}}d(x_{n},x_{n+1}).$$
(28)

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

$$\lim_{n,m\to\infty}d(x_{n},x_{m})=0.$$
(29)

Case (ii): $$s=1$$. Since $$\{d(x_{n},x_{n+1})\}$$ is a decreasing sequence, there exists $$r\geq0$$ such that

$$\lim_{n \to\infty} d(x_{n},x_{n+1})=r,$$
(30)

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

$$\psi( r) \leq\lim_{n \to\infty} \beta \bigl(\psi \bigl(K(x_{n},x_{n+1}) \bigr) \bigr) \psi(r).$$

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

$$d(x_{m_{k}},x_{n_{k}})\geq\varepsilon,$$
(31)

where $$n_{k}$$, $$m_{k}$$ are chosen as the smallest integers satisfying (31), that is,

$$d(x_{m_{k}}, x_{n_{k}-1})< \varepsilon.$$
(32)

By (31), (32), and the triangle inequality, we easily derive that

\begin{aligned} \varepsilon&\leq d(x_{m_{k}},x_{n_{k}}) \leq d(x_{m_{k}},x_{n_{k}-1})+d(x_{n_{k}-1},x_{n_{k}}) \\ & < \varepsilon+d(x_{n_{k}-1},x_{n_{k}}). \end{aligned}
(33)

Using (33) and the squeeze theorem we get

$$\lim_{k\to\infty} d(x_{m_{k}},x_{n_{k}})= \varepsilon.$$
(34)

Regarding that T is generalized rational α-ψ-ϕ-Geraghty contractive mapping of type (I), we have

\begin{aligned} \psi\bigl( d(x_{m_{k}+1},x_{n_{k}+1} )\bigr)& \leq \alpha(x_{m_{k}},x_{n_{k}})\psi \bigl( d(Tx_{m_{k}},Tx_{n_{k}}) \bigr) \\ & \leq\beta \bigl(\psi \bigl(K(x_{m_{k}},x_{n_{k}}) \bigr) \bigr) \psi \bigl(K(x_{m_{k}},x_{n_{k}}) \bigr)+L\phi \bigl(N(x_{m_{k}},x_{n_{k}}) \bigr), \end{aligned}
(35)

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

\begin{aligned} K(x_{m_{k}},x_{n_{k}})={}&\max\biggl\{ d(x_{m_{k}},x_{n_{k}}), \frac {d(x_{m_{k}},x_{m_{k}+1})d(x_{n_{k}},x_{n_{k}+1})}{1+d(x_{m_{k}},x_{n_{k}})}, \\ &{}\frac{d(x_{m_{k}},x_{m_{k}+1})d(x_{n_{k}},x_{n_{k}+1})}{1+d(x_{m_{k}+1},x_{n_{k}+1})}\biggr\} \end{aligned}
(36)

and

$$N(x_{m_{k}},x_{n_{k}})=\min \bigl\{ d^{s}(x_{m_{k}},x_{n_{k}+1}),d^{s}(x_{n_{k}},x_{m_{k}+1}), d(x_{m_{k}},x_{m_{k}+1}),d(x_{n_{k}},x_{n_{k}+1}) \bigr\} .$$

It is clear that

$$\lim_{k\to\infty} K(x_{m_{k}},x_{n_{k}})= \varepsilon \quad\mbox{and}\quad \lim_{k\to\infty} N(x_{m_{k}},x_{n_{k}})=0.$$

By taking the limit as $$k\rightarrow\infty$$ in (35) and taking (36), (30) into account, we get

\begin{aligned} \psi({\varepsilon}) &\leq\lim_{k\to\infty} \beta \bigl( \psi \bigl(K(x_{m_{k}},x_{n_{k}}) \bigr) \bigr) \lim _{k\to\infty} \psi \bigl(K(x_{m_{k}},x_{n_{k}}) \bigr) \\ &\leq\lim_{k\to\infty} \beta \bigl(\psi \bigl(K(x_{m_{k}},x_{n_{k}}) \bigr) \bigr)\psi ( \varepsilon). \end{aligned}
(37)

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

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

Now, if T is continuous, then

$$Tu=T \Bigl(\lim_{n\to\infty}x_{n} \Bigr)=\lim _{n\to\infty}Tx_{n}=\lim_{n\to\infty}x_{n+1}=u,$$

and u is a fixed point for T. □

### Theorem 2.9

Let $$(X,d)$$ be a complete b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized rational α-ψ-ϕ-Geraghty contractive of mapping type (I) such that:

1. (i)

2. (ii)

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

3. (iii)

X is α-regular 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 α-regular, then, as in the proof of Theorem 2.4, there exists a subsequence $$\{x_{n_{k}}\}$$ of $$\{x_{n}\}$$ such that

$$\psi \bigl(d(x_{n_{k}+1},Tu) \bigr)< \frac{1}{s} \psi \bigl(K(x_{n_{k}},u) \bigr)+L \phi \bigl(N(x_{n_{k}},u) \bigr),$$
(38)

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

\begin{aligned} \psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr)&\leq\psi \bigl(s^{2}d \bigl(Tx^{*},Ty^{*} \bigr) \bigr) \\ &\leq\alpha \bigl(x^{*},y^{*} \bigr)\psi \bigl(s^{2} d \bigl(Tx^{*},Ty^{*} \bigr) \bigr) \\ &\leq\beta \bigl(\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr) \bigr)\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr) \\ &< \frac{1}{s}\psi \bigl(d \bigl(x^{*},y^{*} \bigr) \bigr), \end{aligned}

### Theorem 2.11

Let $$(X,d)$$ be a complete b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized rational α-ψ-ϕ-Geraghty contractive mapping of type (II) such that:

1. (i)

2. (ii)

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

3. (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 α-ψ-ϕ-Geraghty contractive mapping of type (II) and the property of ψ we have

\begin{aligned} \psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)&\leq\alpha(x_{n},x_{n+1}) \psi \bigl(s^{2}d(Tx_{n},Tx_{n+1}) \bigr) \\ &\leq\beta \bigl(\psi \bigl(Q(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(Q(x_{n},x_{n+1}) \bigr)+L\phi \bigl(N(x_{n},x_{n+1}) \bigr). \end{aligned}
(39)

By using the same arguments as in the proof of Theorem 2.3, we derive that

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(Q(x_{n},x_{n+1}) \bigr),$$
(40)

where

$$Q(x_{n},x_{n+1})=\max \biggl\{ d(x_{n},x_{n+1}), \frac {d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})+d^{s}(x_{n}, x_{n+2})d^{s}(x_{n+1},x_{n+1})}{1+s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]} \biggr\} .$$

Since $$d(x_{n},x_{n+2})\leq s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]$$, we have

\begin{aligned} d(x_{n},x_{n+1})&\geq\frac {d(x_{n},x_{n+1})d(x_{n},x_{n+2})}{1+d(x_{n},x_{n+2})} \\ &\geq\frac {d(x_{n},x_{n+1})d(x_{n},x_{n+2})}{1+s[d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})]}. \end{aligned}

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 b-metric-like space with constant $$s\geq1$$ and $$T:X\rightarrow X$$ be a generalized rational α-ψ-ϕ-Geraghty contractive mapping of type (II) such that:

1. (i)

2. (ii)

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

3. (iii)

X is α-regular 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 α-regular and by following the lines of the proof of Theorem 2.4 there exists a subsequence $$\{x_{n_{k}}\}$$ of $$\{x_{n}\}$$ such that

$$\psi \bigl(d(x_{n_{k}+1},Tu) \bigr)< \frac{1}{s}\psi \bigl(Q(x_{n_{k}},u) \bigr)+L\phi \bigl(N(x_{n_{k}},u) \bigr),$$
(41)

where

$$Q(x_{n_{k}},u)=\max \biggl\{ d(x_{n_{k}},u),\frac {d(x_{n_{k}},x_{n_{k}+1})d(u,Tu)+d^{s}(x_{n_{k}},Tu) d^{s}(u,x_{n_{k}+1})}{1+s[d(x_{n_{k}},x_{n_{k}+1})+d(u,Tu)]} \biggr\} .$$

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 b-metric-like 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2}d(Tx,Ty) \bigr)\leq\beta \bigl( \psi \bigl(d(x,y) \bigr) \bigr)\psi \bigl(d(x,y) \bigr), \end{aligned}
(42)

for all $$x,y \in X$$. Suppose also that:

1. (i)

2. (ii)

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

3. (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 α-orbital 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

\begin{aligned} \psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)&\leq \alpha(x_{n},x_{n+1})\psi \bigl(s^{2}d(Tx_{n},Tx_{n+1}) \bigr) \\ &\leq\beta \bigl(\psi \bigl(d(x_{n},x_{n+1}) \bigr) \bigr) \psi \bigl(d(x_{n},x_{n+1}) \bigr). \end{aligned}

Thus, we have

$$\psi \bigl(s^{2}d(x_{n+1},x_{n+2}) \bigr)< \frac{1}{s}\psi \bigl(d(x_{n},x_{n+1}) \bigr)< \psi \bigl(d(x_{n},x_{n+1}) \bigr).$$
(43)

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

$$\psi \bigl(d(x_{n+1},x_{n+2}) \bigr) \leq\beta \bigl(\psi \bigl(d(x_{n},x_{n+1}) \bigr) \bigr)\psi \bigl(d(x_{n},x_{n+1}) \bigr)$$

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

$$\lim_{n\rightarrow\infty}d(x_{n},x_{n+1})=0.$$
(44)

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

$$d(x_{m_{k}},x_{n_{k}})\geq\varepsilon,$$
(45)

where $$n_{k}$$, $$m_{k}$$ are chosen as the smallest integers satisfying (45), that is,

$$d(x_{m_{k}}, x_{n_{k}-1})< \varepsilon.$$
(46)

By (45), (46), and the triangle inequality, we easily derive that

\begin{aligned} \varepsilon\leq d(x_{m_{k}},x_{n_{k}}) & \leq d(x_{m_{k}},x_{n_{k}-1})+d(x_{n_{k}-1},x_{n_{k}}) < \varepsilon+d(x_{n_{k}-1},x_{n_{k}}). \end{aligned}
(47)

Using (47) and the squeeze theorem we get

$$\lim_{k\to\infty} d(x_{m_{k}},x_{n_{k}})= \varepsilon.$$
(48)

As T satisfies (42), we have

\begin{aligned} \psi\bigl(d(x_{m_{k}+1},x_{n_{k}+1})\bigr)&\leq \alpha(x_{m_{k}},x_{n_{k}}) \psi \bigl(d(Tx_{m_{k}},Tx_{n_{k}}) \bigr) \\ &\leq\beta \bigl(\psi \bigl(d(x_{m_{k}},x_{n_{k}}) \bigr) \bigr)\psi \bigl(d(x_{m_{k}},x_{n_{k}}) \bigr). \end{aligned}
(49)

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

$$\lim_{k\rightarrow\infty} d(x_{m_{k}},x_{n_{k}})=0.$$

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 b-metric-like 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2}d(Tx,Ty) \bigr)\leq\beta \bigl( \psi \bigl(d(x,y) \bigr) \bigr)\psi \bigl(d(x,y) \bigr), \end{aligned}
(50)

for all $$x,y \in X$$. Suppose also that:

1. (i)

2. (ii)

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

3. (iii)

X is α-regular 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 b-metric-like 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

\begin{aligned} \alpha(x,y)\psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr)\psi \bigl(M(x,y) \bigr), \end{aligned}
(51)

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

$$M(x,y)=\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty), \frac{d(x,Ty)+d(y,Tx)}{4s} \biggr\} .$$
(52)

Suppose also that:

1. (i)

2. (ii)

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

3. (iii)

T is continuous or (iii)′ X is α-regular 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 b-metric-like 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:

1. (a)

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

2. (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:

1. (i)

2. (ii)

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

3. (iii)

T is continuous or (iii)′ X is α-regular 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 b-metric-like 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:

1. (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:

1. (i)

2. (ii)

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

3. (iii)

T is continuous or (iii)′ X is α-regular 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 b-metric-like 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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr)\psi \bigl(M(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(53)

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

\begin{aligned}& M(x,y)=\max \biggl\{ d(x,y),d(x,Tx),d(y,Ty), \frac{d(x,Ty)+d(y,Tx)}{4s} \biggr\} \quad \textit{and} \end{aligned}
(54)
\begin{aligned}& N(x,y)=\min \bigl\{ d^{s}(x,Ty),d^{s}(y,Tx),d(x,Tx),d(y,Ty) \bigr\} . \end{aligned}
(55)

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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(K(x,y) \bigr) \bigr)\psi \bigl(K(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(56)

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

\begin{aligned}& K(x,y)=\max \biggl\{ d(x,y), \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}, \frac {d(x,Tx) d(y,Ty)}{1+d(Tx,Ty)} \biggr\} \quad \textit{and} \end{aligned}
(57)
\begin{aligned}& N(x,y)=\min \bigl\{ d^{s}(x,Ty),d^{s}(y,Tx),d(x,Tx),d(y,Ty) \bigr\} . \end{aligned}
(58)

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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(Q(x,y) \bigr) \bigr)\psi \bigl(Q(x,y) \bigr)+L\phi \bigl(N(x,y) \bigr), \end{aligned}
(59)

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

\begin{aligned}& Q(x,y)=\max \biggl\{ d(x,y), \frac {d(x,Tx)d(y,Ty)+d^{s}(x,Ty)d^{s}(y,Tx)}{1+s(d(x,Tx)+d(y,Ty))} \biggr\} \quad \textit{and} \end{aligned}
(60)
\begin{aligned}& N(x,y)=\min \bigl\{ d^{s}(x,Ty),d^{s}(y,Tx),d(x,Tx),d(y,Ty) \bigr\} . \end{aligned}
(61)

Then T has a fixed point $$u \in X$$ with $$d(u,u)=0$$.

If take $$L=0$$ in Corollaries 3.4-3.6, we get three more consequences. Regarding the volume of the paper, we skip the details.

### Corollary 3.7

Let $$(X,d)$$ be a complete b-metric-like 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

\begin{aligned} \psi \bigl(s^{2}d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(d(x,y) \bigr) \bigr)\psi \bigl(d(x,y) \bigr), \end{aligned}
(62)

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 b-metric-like spaces endowed with a partial order

In this section, from our main results, we shall derive easily various fixed point results on a b-metric-like 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

$$x,y\in X, \quad x\preceq y \quad\Longrightarrow\quad Tx\preceq Ty.$$

### 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 b-metric-like 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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr)\psi \bigl(M(x,y) \bigr), \end{aligned}
(63)

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:

1. (i)

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

2. (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

\begin{aligned} \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 . \end{aligned}

Clearly, T satisfies (51), that is,

$$\alpha(x,y)\psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(M(x,y) \bigr) \bigr)\psi \bigl(M(x,y) \bigr),$$

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

$$\alpha(x,y)\geq1\quad \Longrightarrow\quad x\succeq y \mbox{ or } x\preceq y \quad\Longrightarrow \quad Tx\succeq Ty \mbox{ or } Tx\preceq Ty \quad \Longrightarrow\quad\alpha(Tx,Ty)\geq1.$$

Hence, the self-mapping T is α-admissible. Similarly, we can prove that T is triangular α-admissible and so triangular α-orbital 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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(K(x,y) \bigr) \bigr)\psi \bigl(K(x,y) \bigr), \end{aligned}
(64)

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:

1. (i)

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

2. (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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2} d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(Q(x,y) \bigr) \bigr)\psi \bigl(Q(x,y) \bigr), \end{aligned}
(65)

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:

1. (i)

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

2. (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 b-metric-like 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

\begin{aligned} \psi \bigl(s^{2}d(Tx,Ty) \bigr)\leq\beta \bigl(\psi \bigl(d(x,y) \bigr) \bigr)\psi \bigl(d(x,y) \bigr), \end{aligned}
(66)

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

1. (i)

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

2. (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 $$d(x,y)=|x-y|^{2}$$. Then $$(X,d)$$ is complete b-metric (so b-metric-like) space with constant $$s=2$$. Define $$T:X \rightarrow X$$ and $$\alpha(x,y): X\times X \to [0,\infty)$$ as follows:

$$Tx=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{x}{4} & \mbox{if } x \in[0,1],\\ \frac{x^{2}}{x+1} & \mbox{if } x \in(1,\infty), \end{array}\displaystyle \right .$$

and

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

It is clear that T is triangular α-orbital admissible and we have $$\alpha(0,T0) \geq1$$. Moreover, X is α-regular 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

$$\alpha(x,y) \psi \bigl(2^{2} d(Tx,Ty) \bigr)= 2^{2} \biggl(\frac {x}{4}-\frac{y}{4} \biggr)^{2}= \frac{(x-y)^{2}}{4}=\beta \bigl(\psi \bigl(d(x,y) \bigr) \bigr) \psi(d(x,y).$$

Otherwise,

$$\alpha(x,y) \psi \bigl(2^{2} d(Tx,Ty) \bigr)=0 \leq\beta \bigl(\psi \bigl(d(x,y) \bigr) \bigr) \psi(d(x,y).$$

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 b-metric-like 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 b-metric-like space, with some other abstract space, such as a b-metric space, a metric space, a metric-like space, a partial metric space, and so on.

## References

1. Hitzler, P: Generalized metrics and topology in logic programming semantics. Ph.D. thesis, School of Mathematics, Applied Mathematics and Statistics, National University Ireland, University College Cork (2001)

2. Amini-Harandi, A: Metric-like spaces, partial metric spaces and fixed points. Fixed Point Theory Appl. 2012, Article ID 204 (2012)

3. Czerwik, S: Contraction mappings in b-metric spaces. Acta Math. Inform. Univ. Ostrav. 1(1), 5-11 (1993)

4. Bakhtin, IA: The contraction mapping principle in quasimetric spaces. In: Functional Analysis, vol. 30, pp. 26-37. Ul’yanovsk. Gos. Ped. Inst., Ul’yanovsk (1989)

5. Bourbaki, N: General Topology. Hermann, Paris (1974)

6. Aydi, H, Bota, M-F, Karapınar, E, Moradi, S: A common fixed point for weak ϕ-contractions on b-metric spaces. Fixed Point Theory 13(2), 337-346 (2012)

7. Bota, M-F, Chifu, C, Karapınar, E: Fixed point theorems for generalized (α-ψ)-Ćirić-type contractive multivalued operators in b-metric spaces. Abstr. Appl. Anal. 2014, Article ID 246806 (2014)

8. Bota, M-F, Karapınar, E, Mlesnite, O: Ulam-Hyers stability results for fixed point problems via alpha-psi-contractive mapping in b-metric space. Abstr. Appl. Anal. 2013, Article ID 825293 (2013)

9. Bota, M-F, Karapınar, E: A note on ‘Some results on multi-valued weakly Jungck mappings in b-metric space’. Cent. Eur. J. Math. 11(9), 1711-1712 (2013)

10. Kutbi, MA, Karapınar, E, Ahmed, J, Azam, A: Some fixed point results for multi-valued mappings in b-metric spaces. J. Inequal. Appl. 2014, Article ID 126 (2014)

11. Alghamdi, MA, Hussain, N, Salimi, P: Fixed point and coupled fixed point theorems on b-metric-like spaces. J. Inequal. Appl. 2013, Article ID 402 (2013)

12. Hussain, N, Roshan, JR, Parvaneh, V, Kadelburg, Z: Fixed points of contractive mappings in b-metric-like spaces. Sci. World J. 2014, Article ID 471827 (2014)

13. Samet, B, Vetro, C, Vetro, P: α-ψ-Contractive type mappings. Nonlinear Anal. 75, 2154-2165 (2012)

14. Karapınar, E, Kuman, P, Salimi, P: On α-ψ-Meir-Keeler contractive mappings. Fixed Point Theory Appl. 2013, Article ID 94 (2013)

15. Karapınar, E, Samet, B: Generalized (α-ψ) contractive type mappings and related fixed point theorems with applications. Abstr. Appl. Anal. 2012, Article ID 793486 (2012)

16. Karapınar, E: α-ψ-Geraghty contraction type mappings and some related fixed point results. Filomat 28(1), 37-48 (2014)

17. Karapınar, E: A discussion on ‘α-ψ-Geraghty contraction type mappings’. Filomat 28(4), 761-766 (2014)

18. Popescu, O: Some new fixed point theorems for α-Geraghty contractive type maps in metric spaces. Fixed Point Theory Appl. 2014, Article ID 190 (2014)

## Author information

Authors

### Corresponding author

Correspondence to Erdal Karapınar.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

## Rights and permissions 