## 1 Introduction and preliminaries

One of the generalizations of the metric spaces are the so-called quasi-metric spaces in which the commutativity condition does not hold in general. Recently, Jleli and Samet [1] obtained a relation between G-metric spaces introduced by Mustafa and Sims [2] and quasi-metric spaces. This work increased the interest to quasi-metric spaces (see [3, 4] for details).

In this paper, we investigate the existence of fixed points of Meir-Keeler type contractions defined on quasi-metric spaces and apply our results to G-metric spaces.

First, we recall the definition of a quasi-metric and quasi-metric space and some topological concepts on these spaces.

### Definition 1

Let X be a nonempty set and $$d:X\times X\rightarrow[0,\infty)$$ be a function which satisfies:

1. (d1)

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

2. (d2)

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

Then d called a quasi-metric and the pair $$(X,d)$$ is called a quasi-metric space.

### Remark 2

Any metric space is a quasi-metric space, but the converse is not true in general.

### Definition 3

Let $$(X,d)$$ be a quasi-metric space.

1. (1)

A sequence $$\{x_{n}\}$$ in X is said to be convergent to x if $$\lim_{n\rightarrow\infty}d(x_{n},x)= \lim_{n\rightarrow\infty}d(x,x_{n})=0$$.

2. (2)

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

3. (3)

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

4. (4)

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

### Remark 4

From Definition 3 it is obvious that a sequence $$\{x_{n}\}$$ in a quasi-metric space is Cauchy if and only if it is both left-Cauchy and right-Cauchy.

### Definition 5

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

1. (1)

$$(X,d)$$ is said to be left-complete if every left-Cauchy sequence in X is convergent.

2. (2)

$$(X,d)$$ is said to be right-complete if every right-Cauchy sequence in X is convergent.

3. (3)

$$(X,d)$$ is said to be complete if every Cauchy sequence in X is convergent.

In the sequel, we shall denote by ℕ the set of nonnegative integers, that is, $$\mathbb{N}=\{0,1,2,\ldots\}$$. We next define the concept of α-admissible mappings which have been recently introduced by Samet [5] and used by many authors to generalize contraction mappings of various types; see [68] for details.

### Definition 6

A mapping $$T:X\to X$$ is called α-admissible if for all $$x,y\in X$$ we have

$$\alpha(x,y)\geq1 \quad\Rightarrow\quad\alpha(Tx,Ty)\geq1,$$
(1.1)

where $$\alpha:X\times X\to[0,\infty)$$ is a given function.

In the existence and uniqueness proofs of fixed points of α-admissible maps, an additional property is required. This property is given below.

### Definition 7

A mapping $$T:X\to X$$ is called triangular α-admissible if it is α-admissible and satisfies

$$\left . \begin{array}{r@{}} \alpha(x,y)\geq1, \\ \alpha(y,z)\geq1 \end{array} \right \} \quad\Rightarrow\quad\alpha(x,z)\geq1,$$
(1.2)

where $$x,y,z\in X$$ and $$\alpha:X\times X\to[0,\infty)$$ is a given function.

The following auxiliary result is going to be used in the proof of existence theorems.

### Lemma 8

[7]

Let $$T:X\to X$$ be a triangular α-admissible mapping. Assume that there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$. If $$x_{n}=T^{n}x_{0}$$, then $$\alpha(x_{m},x_{n})\geq1$$ for all $$m,n\in\mathbb{N}$$.

### Proof

Let $$x_{0}\in X$$ satisfies $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha (Tx_{0},x_{0})\geq1$$. Define the sequence $$\{x_{n}\}$$ in X as $$x_{n+1}=Tx_{n}$$ for $$n\in\mathbb{N}$$. Since T is α-admissible, we have

\begin{aligned}& \alpha(x_{0},Tx_{0})=\alpha(x_{0},x_{1}) \geq1 \quad\Rightarrow\quad\alpha(x_{1},x_{2})\geq 1\quad\Rightarrow\quad\cdots \quad\Rightarrow\quad\alpha(x_{n},x_{n+1})\geq1, \\& \alpha(Tx_{0},x_{0})=\alpha(x_{1},x_{0}) \geq1 \quad\Rightarrow\quad\alpha(x_{2},x_{1})\geq 1\quad\Rightarrow\quad\cdots \quad\Rightarrow\quad\alpha(x_{n+1},x_{n})\geq1, \end{aligned}

for all $$n=0,1,\ldots$$ . On the other hand, since T is triangular α-admissible, we get

$$\left . \begin{array}{r@{}} \alpha(x_{m},x_{m+1})\geq1,\\ \alpha(x_{m+1},x_{m+2})\geq1 \end{array} \right \} \quad\Longrightarrow\quad \alpha(x_{m},x_{m+2})\geq1,$$

and

$$\left . \begin{array}{r@{}} \alpha(x_{m},x_{m-1})\geq1,\\ \alpha(x_{m-1},x_{m-2})\geq1 \end{array} \right \} \quad\Longrightarrow\quad \alpha(x_{m},x_{m-2})\geq1.$$

Similarly,

$$\left . \begin{array}{r@{}} \alpha(x_{m},x_{m+2})\geq1,\\ \alpha(x_{m+2},x_{m+3})\geq1 \end{array} \right \} \quad\Longrightarrow\quad \alpha(x_{m},x_{m+3})\geq1,$$

and

$$\left . \begin{array}{r@{}} \alpha(x_{m},x_{m-2})\geq1,\\ \alpha(x_{m-2},x_{m-3})\geq1 \end{array} \right \} \quad\Longrightarrow\quad \alpha(x_{m},x_{m-3})\geq1.$$

Continuing in this way, we obtain, for all $$m,n \in\mathbb{N}$$,

$$\alpha(x_{m},x_{n})\geq1.$$

□

In this paper we study α-admissible Meir-Keeler (or shortly α-Meir-Keeler) contractions which can be regarded as generalizations of the Meir-Keeler contractions defined in [9]. In fact, we insert α-admissibility into the definition of the original Meir-Keeler contraction.

### Definition 9

Let $$(X,d)$$ be a quasi-metric space. Let $$T:X\rightarrow X$$ be a triangular α-admissible mapping. Suppose that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq d(x,y)< \varepsilon+\delta\quad\mbox{implies}\quad \alpha (x,y)d(Tx,Ty)< \varepsilon,$$
(1.3)

for all $$x, y \in X$$. Then T is called α-Meir-Keeler contraction.

### Remark 10

Let T be an α-Meir-Keeler contractive mapping. Then

$$\alpha(x,y)d(Tx,Ty)< d(x,y),$$

for all $$x,y\in X$$ when $$x\neq y$$. Also, if $$x=y$$ then $$d(Tx,Ty)=0$$, i.e.,

$$\alpha(x,y)d(Tx,Ty)\leq d(x,y),$$

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

We also generalize the α-Meir-Keeler contraction by using a more general expression in the contractive condition. Specifically, we define two types of generalized α-Meir-Keeler contraction, say type (I) and type (II) as follows.

### Definition 11

Let $$(X,d)$$ be a quasi-metric space. Let $$T:X\rightarrow X$$ be a triangular α-admissible mapping. Assume that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq M(x,y)< \varepsilon+\delta\quad\mbox{implies}\quad \alpha (x,y)d(Tx,Ty)< \varepsilon,$$
(1.4)

where

$$M(x,y)=\max\bigl\{ d(x,y), d(Tx,x), d(Ty,y)\bigr\} ,$$
(1.5)

for all $$x, y \in X$$. Then T is called a generalized α-Meir-Keeler contraction of type (I).

### Definition 12

Let $$(X,d)$$ be a quasi-metric space. Let $$T:X\rightarrow X$$ be a triangular α-admissible mapping. Assume that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq N(x,y)< \varepsilon+\delta\quad\mbox{implies}\quad \alpha (x,y)d(Tx,Ty)< \varepsilon,$$
(1.6)

where

$$N(x,y)=\max\biggl\{ d(x,y), \frac{1}{2}\bigl[d(Tx,x)+ d(Ty,y) \bigr]\biggr\} ,$$
(1.7)

for all $$x, y \in X$$. Then T is called a generalized α-Meir-Keeler contraction of type (II).

### Remark 13

Let $$T:X\to X$$ be a generalized α-Meir-Keeler contraction of type (I) (respectively, type (II)). Then

$$\alpha(x,y)d(Tx,Ty)< M(x,y) \quad\bigl(\mbox{respectively}, N(x,y)\bigr),$$

for all $$x,y\in X$$ when $$M(x,y)>0$$ (respectively, $$N(x,y)>0$$). Also, if $$M(x,y)=0$$ (respectively, $$N(x,y)=0$$), then $$x=y$$, which implies $$d(x,y)=0$$, i.e.,

$$\alpha(x,y)d(Tx,Ty)\leq M(x,y) \quad\bigl(\mbox{respectively}, N(x,y)\bigr),$$

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

### Remark 14

It is obvious that $$N(x,y)\leq M(x,y)$$ for all $$x,y\in X$$, where $$M(x,y)$$ and $$N(x,y)$$ are defined in (1.5) and (1.7), respectively.

## 2 Main results

Our first result is a fixed point theorem for generalized α-Meir-Keeler contractions of type (I) on quasi-metric spaces.

### Theorem 15

Let $$(X,d)$$ be a complete quasi-metric space and $$T:X\to X$$ be a continuous generalized α-Meir-Keeler contraction of type (I). If $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

### Proof

Let $$x_{0}\in X$$ satisfy $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha (Tx_{0},x_{0})\geq1$$. Define the sequence $$\{x_{n}\}$$ in X as

$$x_{n+1}=Tx_{n} \quad\mbox{for } n\in\mathbb{N}.$$

Notice that if $$x_{n_{0}}=x_{n_{0}+1}$$ for some $$n_{0}>0$$, then $$x_{n_{0}}$$ is a fixed point of T and the proof is done. Assume that $$x_{n}\neq x_{n+1}$$ for all $$n\geq0$$. Since T is α-admissible,

$$\alpha(x_{0},Tx_{0})=\alpha(x_{0},x_{1}) \geq1 \quad\Rightarrow\quad\alpha (Tx_{0},Tx_{1})=\alpha(x_{1},x_{2}) \geq1,$$
(2.1)

and continuing we obtain

$$\alpha(x_{n},x_{n+1})\geq1 \quad\forall n\in \mathbb{N}.$$
(2.2)

Upon substituting $$x=x_{n}$$ and $$y=x_{n+1}$$ in (1.4) we find that for every $$\varepsilon>0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq M(x_{n},x_{n+1})< \varepsilon+ \delta \quad\Longrightarrow\quad\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) <\varepsilon,$$
(2.3)

where

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

In what follows, we examine three cases.

Case 1. Assume that $$M(x_{n},x_{n+1})=d(x_{n},x_{n+1})$$. Then (2.3) becomes

$$\varepsilon\leq d(x_{n},x_{n+1})< \varepsilon+\delta \quad\Longrightarrow\quad\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) <\varepsilon.$$

Therefore, we deduce that

$$d(x_{n+1},x_{n+2})\leq\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) < \varepsilon \leq d(x_{n},x_{n+1}),$$

for all n. That is, $$\{d(x_{n},x_{n+1})\}$$ is a decreasing positive sequence in $$\mathbb{R_{+}}$$ and it converges to some $$r\geq0$$. To show that $$r=0$$, we assume the contrary, that is, $$r>0$$. Then we must have

$$0< r\leq d (x_{n},x_{n+1}) \quad\mbox{for all } n\in\mathbb{N}.$$
(2.4)

Since the condition (2.3) holds for every $$\varepsilon>0$$, we may choose $$\varepsilon=r$$. For this ε, there exists $$\delta (\varepsilon)>0$$ satisfying (2.3). In other words,

$$r\leq d_{n}=d (x_{n},x_{n+1})< r +\delta \quad\Longrightarrow\quad\alpha (x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) < r.$$

However, this implies

\begin{aligned} &r\leq M(x_{n},x_{n+1})=d(x_{n},x_{n+1})< r+ \delta \\ &\quad\Longrightarrow\quad d(x_{n+1},x_{n+2})\leq\alpha (x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) <r, \end{aligned}
(2.5)

which contradicts (2.4). Thus, $$r=0$$, that is,

$$\lim_{n\to\infty} d(x_{n},x_{n+1})=0.$$
(2.6)

Case 2. Assume that $$M(x_{n},x_{n+1})=d(x_{n+1},x_{n})$$. In this case (2.3) becomes

$$\varepsilon\leq d(x_{n+1},x_{n})< \varepsilon+\delta \quad\Longrightarrow\quad\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) <\varepsilon,$$

from which it follows that

$$d(x_{n+1},x_{n+2})\leq\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) < \varepsilon \leq d(x_{n+1},x_{n}).$$

Therefore, we obtain

$$d(x_{n+1},x_{n+2})< d(x_{n+1},x_{n}),$$
(2.7)

for all $$n\in\mathbb{N}$$. Note that by Remark 13, since

$$M(x_{n},x_{n-1})=\max\bigl\{ d(x_{n},x_{n-1}),d(x_{n+1},x_{n}), d(x_{n},x_{n-1})\bigr\} >0,$$

we get

$$d(x_{n+1},x_{n})=d(Tx_{n},Tx_{n-1}) \leq\alpha (x_{n},x_{n-1})d(Tx_{n},Tx_{n-1})< M(x_{n},x_{n-1}),$$
(2.8)

where

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

Then (2.8) becomes

$$d(x_{n+1},x_{n})< \max \bigl\{ d(x_{n},x_{n-1}),d(x_{n+1},x_{n}) \bigr\} ,$$
(2.9)

for all $$n\in\mathbb{N}$$.

Clearly, the case $$\max \{d(x_{n},x_{n-1}),d(x_{n+1},x_{n}) \} =d(x_{n+1},x_{n})$$ is not possible. Indeed, in this case we would get

$$0< d(x_{n+1},x_{n})<d(x_{n+1},x_{n}).$$

Therefore, we should have $$\max \{ d(x_{n},x_{n-1}),d(x_{n+1},x_{n}) \}=d(x_{n},x_{n-1})$$, which implies

$$0< d(x_{n+1},x_{n})< d(x_{n},x_{n-1}),$$
(2.10)

for all $$n\in\mathbb{N}$$. That is, the sequence $$\{ d(x_{n+1},x_{n})\}$$ is decreasing and positive sequence and hence it converges to $$L\geq0$$. In fact, the limit L of this sequence is 0, which can be shown by mimicking the proof of (2.6) done above. In other words, we get

$$\lim_{n\to\infty} d(x_{n+1},x_{n})=0.$$
(2.11)

Finally, taking the limit as $$n\to\infty$$ in (2.7) and using (2.11) we obtain

$$\lim_{n\to\infty} d(x_{n+1},x_{n+2})= 0.$$
(2.12)

Case 3. Assume that $$M(x_{n},x_{n+1})=d(x_{n+2},x_{n+1})$$. In this case (2.3) becomes

$$\varepsilon\leq d(x_{n+2},x_{n+1})< \varepsilon+\delta \quad\Longrightarrow\quad\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) <\varepsilon,$$

or

$$d(x_{n+1},x_{n+2})\leq\alpha(x_{n},x_{n+1})d(Tx_{n},Tx_{n+1}) < \varepsilon \leq d(x_{n+2},x_{n+1}).$$

Therefore, we deduce

$$d(x_{n+1},x_{n+2})< d(x_{n+2},x_{n+1}),$$
(2.13)

for all $$n\in\mathbb{N}$$. By Remark 13, we have

$$d(x_{n+2},x_{n+1})=d(Tx_{n+1},Tx_{n}) \leq\alpha (x_{n+1},x_{n})d(Tx_{n+1},Tx_{n})< M(x_{n+1},x_{n}),$$
(2.14)

where

\begin{aligned} M(x_{n+1},x_{n}) =&\max \bigl\{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}), d(x_{n+1},x_{n}) \bigr\} \\ =&\max \bigl\{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}) \bigr\} \end{aligned}

is clearly positive. Then (2.14) becomes

$$d(x_{n+2},x_{n+1})< \max \bigl\{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}) \bigr\} ,$$
(2.15)

for all $$n\in\mathbb{N}$$.

The case $$\max \{d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}) \} =d(x_{n+2},x_{n+1})$$ is impossible, since it yields

$$0< d(x_{n+2},x_{n+1})<d(x_{n+2},x_{n+1}).$$

The other case, that is, $$\max \{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}) \}=d(x_{n+1},x_{n})$$ implies

$$0< d(x_{n+2},x_{n+1})< d(x_{n+1},x_{n}),$$
(2.16)

for all $$n\in\mathbb{N}$$. As in Case 2, the sequence $$\{ d(x_{n+1},x_{n})\}$$ is decreasing and positive sequence and hence it converges to $$L=0$$. Finally, taking the limit as $$n\to\infty$$ in (2.13) we end up with

$$\lim_{n\to\infty} d(x_{n+1},x_{n+2})= 0.$$
(2.17)

As a result, we see that in all three cases, the sequence $$\{d_{n}\}$$ defined by $$d_{n}:=d(x_{n},x_{n+1})$$ converges to 0 as $$n\to\infty$$. Using similar arguments, it can be shown that the sequence $$\{f_{n}\}$$ where $$f_{n}:=d(x_{n+1},x_{n})$$ also converges to 0. We first note that

$$\alpha(Tx_{0},x_{0})=\alpha(x_{1},x_{0}) \geq1 \quad\Rightarrow\quad\alpha (Tx_{1},Tx_{0})=\alpha(x_{2},x_{1}) \geq1,$$
(2.18)

and continuing in this way, we obtain

$$\alpha(x_{n+1},x_{n})\geq1 \quad\forall n\in \mathbb{N}.$$
(2.19)

Substituting $$x=x_{n+1}$$ and $$y=x_{n}$$ in (1.4) we find that for every $$\varepsilon>0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq M(x_{n+1},x_{n})< \varepsilon+ \delta \quad\Longrightarrow\quad\alpha(x_{n+1},x_{n})d(Tx_{n+1},Tx_{n}) <\varepsilon,$$
(2.20)

where

\begin{aligned} M(x_{n+1},x_{n}) =&\max \bigl\{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}), d(x_{n+1},x_{n}) \bigr\} \\ =&\max \bigl\{ d(x_{n+1},x_{n}),d(x_{n+2},x_{n+1}) \bigr\} . \end{aligned}

We need to examine two cases.

Case 1. Assume that $$M(x_{n+1},x_{n})=d(x_{n+1},x_{n})$$. Then (2.20) becomes

$$\varepsilon\leq d(x_{n+1},x_{n})< \varepsilon+\delta \quad\Longrightarrow\quad\alpha(x_{n+1},x_{n})d(Tx_{n+1},Tx_{n}) <\varepsilon.$$

Then we have

$$d(x_{n+2},x_{n+1})\leq\alpha(x_{n+1},x_{n})d(Tx_{n+1},Tx_{n}) < \varepsilon\leq d(x_{n+1},x_{n}),$$

for all n. That is, $$\{d(x_{n+1},x_{n})\}$$ is a decreasing positive sequence in $$\mathbb{R_{+}}$$ and it converges to $$L\geq0$$. As above, it can be shown that $$L=0$$.

Case 2. Assume that $$M(x_{n+1},x_{n})=d(x_{n+2},x_{n+1})$$. In this case (2.20) becomes

$$\varepsilon\leq d(x_{n+2},x_{n+1})< \varepsilon+\delta \quad\Longrightarrow\quad\alpha(x_{n+1},x_{n})d(Tx_{n+1},Tx_{n}) <\varepsilon,$$

or

$$d(x_{n+2},x_{n+1})\leq\alpha(x_{n+1},x_{n})d(Tx_{n+1},Tx_{n}) < \varepsilon\leq d(x_{n+2},x_{n+1}),$$

which results in

$$0< d(x_{n+2},x_{n+1})< d(x_{n+2},x_{n+1}),$$
(2.21)

for all $$n\in\mathbb{N}$$ and is not possible.

Thus, we have only one possibility, $$M(x_{n+1},x_{n})=d(x_{n+1},x_{n})$$, which leads to the fact that the sequence $$\{f_{n}\}=\{d(x_{n+1},n_{n})\}$$ converges to 0.

We next show that the sequence $$\{x_{n}\}$$ is both right and left Cauchy. First, we show that $$\{x_{n}\}$$ is a right-Cauchy sequence in $$(X,d)$$. We will prove that for every $$\varepsilon>0$$ there exists $$N\in\mathbb{N}$$ such that

$$d(x_{l},x_{l+k})< \varepsilon,$$
(2.22)

for all $$l\geq N$$ and $$k\in\mathbb{N}$$. Since the sequences $$\{d_{n}\}$$ and $$\{f_{n}\}$$ both converge to 0 as $$n\to\infty$$, for every $$\delta>0$$ there exist $$N_{1},N_{2} \in\mathbb {N}$$ such that

$$d(x_{n},x_{n+1})< \delta\quad\mbox{for all } n \geq N_{1}\in\mathbb{N} \quad\mbox{and}\quad d(x_{n+1},x_{n})< \delta \quad\mbox{for all } n\geq N_{2}\in\mathbb{N}.$$
(2.23)

Choose δ such as $$\delta<\varepsilon$$. We will prove (2.22) by using induction on k. For $$k=1$$, (2.22) becomes

$$d(x_{l},x_{l+1})< \varepsilon,$$
(2.24)

and clearly holds for all $$l\geq N=\max\{N_{1},N_{2}\}$$ due to (2.23) and the choice of δ. Assume that the inequality (2.22) holds for some $$k=m$$, that is,

$$d(x_{l},x_{l+m})< \varepsilon, \quad\mbox{for all } l\geq N.$$
(2.25)

For $$k=m+1$$ we have to show that $$d(x_{l},x_{l+m+1})< \varepsilon$$ for all $$l\geq N$$. From the triangle inequality, we have

$$d(x_{l-1},x_{l+m})< d(x_{l-1},x_{l})+d(x_{l},x_{l+m})< \delta +\varepsilon,$$
(2.26)

for all $$l\geq N$$. If $$d (x_{l-1},x_{l+m})\geq\varepsilon$$, then for

$$M(x_{l-1},x_{l+m})= \max \bigl\{ d (x_{l-1},x_{l+m}), d (x_{l},x_{l-1}), d (x_{l+m+1},x_{l+m}) \bigr\} ,$$

we have

\begin{aligned} \varepsilon \leq& d (x_{l-1},x_{l+m}) \leq M(x_{l-1},x_{l+m}) \\ =& \max \bigl\{ d (x_{l-1},x_{l+m}), d (x_{l},x_{l-1}), d (x_{l+m+1},x_{l+m}) \bigr\} < \{\varepsilon+\delta,\delta, \delta \}\leq\varepsilon +\delta, \end{aligned}

and because of Lemma 8, the contractive condition (1.4) with $$x=x_{l-1}$$ and $$y=x_{l+m}$$ yields

\begin{aligned} &\varepsilon\leq M(x_{l-1},x_{l+m}) < \delta+\varepsilon\\ &\quad\Longrightarrow\quad d (x_{l},x_{l+m+1})\leq \alpha (x_{l-1},x_{l+m})d(x_{l},x_{l+m+1}) = \alpha(x_{l-1},x_{l+m})d(Tx_{l-1},Tx_{l+m})< \varepsilon, \end{aligned}

and hence (2.22) holds for $$k=m+1$$.

If $$d (x_{l-1},x_{l+m})< \varepsilon$$, then

\begin{aligned} M(x_{l-1},x_{l+m}) =& \max \bigl\{ d (x_{l-1},x_{l+m}), d (x_{l},x_{l-1}), d (x_{l+m+1},x_{l+m}) \bigr\} \\ < & \{\varepsilon,\delta,\delta \}\leq\varepsilon. \end{aligned}

Regarding Remark 13, we deduce

\begin{aligned} d (x_{l},x_{l+m+1}) \leq& \alpha(x_{l-1},x_{l+m})d(x_{l},x_{l+m+1}) \\ \leq& M(x_{l-1},x_{l+m})< \varepsilon, \end{aligned}

that is, inequality (2.22) holds for $$k=m+1$$. Hence $$d(x_{l},x_{l+k})<\varepsilon$$ for all $$l\geq N$$ and $$k\geq1$$, which means

$$d(x_{n},x_{m})< \varepsilon, \quad\mbox{for all } m\geq n \geq N.$$
(2.27)

Consequently, $$\{x_{n}\}$$ is a right-Cauchy sequence in $$(X, d)$$. Due to the similarity, the proof that $$\{x_{n}\}$$ is a left-Cauchy sequence in $$(X, d)$$ is omitted. By Remark 4, we deduce that $$\{x_{n}\}$$ is a Cauchy sequence in complete quasi-metric space $$(X, d)$$. Therefore, there exists $$z\in X$$ such that

$$\lim_{n\rightarrow\infty}d(x_{n},z)=\lim _{n\rightarrow\infty}d(z, x_{n})=0.$$
(2.28)

Employing the property (d1) and the continuity of T we get

$$\lim_{n\rightarrow\infty}d(x_{n},Tz)=\lim _{n\rightarrow\infty}d(Tx_{n-1},Tz)=0$$
(2.29)

and

$$\lim_{n\rightarrow\infty}d(Tz, x_{n})=\lim _{n\rightarrow\infty}d(Tz, Tx_{n-1})=0.$$
(2.30)

Combining (2.29) and (2.30), we deduce

$$\lim_{n\rightarrow\infty}d(x_{n},Tz)=\lim _{n\rightarrow\infty}d(Tz, x_{n})=0.$$
(2.31)

From (2.28) and (2.31), due to the uniqueness of the limit, we conclude that $$z=Tz$$, that is, z is a fixed point of T. □

Below we state an existence theorem for fixed point of generalized α-Meir-Keeler contraction of type (II). Taking Remark 14 into account, we observe that the proof of this theorem is similar to the proof of Theorem 15.

### Theorem 16

Let $$(X,d)$$ be a complete quasi-metric space and $$T:X\to X$$ be a continuous generalized α-Meir-Keeler contraction of type (II). If $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

One advantage of α-admissibility is that the continuity of the contraction is not required whenever the following condition is satisfied.

1. (A)

If $$\{x_{n}\}$$ is a sequence in X which converges to x and satisfies $$\alpha(x_{n+1},x_{n})\geq1$$ and $$\alpha(x_{n},x_{n+1})\geq 1$$ for all n then there exists a subsequence $$\{x_{n(k)}\}$$ of $$\{ x_{n}\}$$ such that $$\alpha(x,x_{n(k)})\geq1$$ and $$\alpha (x_{n(k)},x)\geq1$$ for all k.

Replacing the continuity of the contraction in Theorem 16 by the condition (A) on the space $$(X,d)$$ we deduce another existence theorem.

### Theorem 17

Let $$(X,d)$$ be a complete quasi-metric space and $$T:X\to X$$ be a generalized α-Meir-Keeler contraction of type (II) and let $$(X,d)$$ satisfies the condition (A). If $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

### Proof

Following the lines of the proof of Theorem 15, we know that the sequence $$\{x_{n}\}$$ defined by $$x_{n+1} = Tx_{n}$$, where $$x_{0}\in X$$ satisfies $$\alpha(x_{0},Tx_{0})\geq 1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$, converges to some $$z\in X$$. From (2.2) and condition (A), there exists a subsequence $$\{ x_{n(k)}\}$$ of $$\{x_{n}\}$$ such that $$\alpha(z, x_{n(k)})\geq1$$ and $$\alpha(x_{n(k)},z)\geq1$$ for all k. Regarding Remark 13 we have for all $$k\in\mathbb{N}$$

\begin{aligned} &d(Tz,x_{n(k)+1})=d(Tz,Tx_{n(k)})\leq \alpha (z,x_{n(k)})d(Tz,Tx_{n(k)})\leq N(z,x_{n(k)}), \\ &d(x_{n(k)+1},Tz)=d(Tx_{n(k)},Tz)\leq\alpha (x_{n(k)},z)d(Tx_{n(k)},Tz) \leq N(x_{n(k)},z), \end{aligned}
(2.32)

where

\begin{aligned} &N(z,x_{n(k)})=\max\biggl\{ d(z,x_{n(k)}),\frac {1}{2}\bigl[d(Tz,z)+d(Tx_{n(k)},x_{n(k)}) \bigr]\biggr\} , \\ &N(x_{n(k)},z)=\max\biggl\{ d(x_{n(k)},z), \frac {1}{2} \bigl[d(Tx_{n(k)},x_{n(k)})+d(Tz,z)\bigr]\biggr\} . \end{aligned}
(2.33)

Letting $$k \rightarrow\infty$$ in (2.33) we obtain

\begin{aligned}& \lim_{k\to\infty}N(z,x_{n(k)})=\max\biggl\{ 0, \frac {d(Tz,z)}{2}\biggr\} =\frac{d(Tz,z)}{2}, \\& \lim_{k\to\infty}N(x_{n(k)},z)=\max\biggl\{ 0, \frac {d(Tz,z)}{2}\biggr\} =\frac{d(Tz,z)}{2}. \end{aligned}

Thus, upon taking the limit in (2.32) as $$k\to\infty$$, we conclude

\begin{aligned} &0\leq d(Tz,z)\leq \frac{d(Tz,z)}{2} , \\ &0\leq d(z,Tz)\leq \frac{d(Tz,z)}{2}. \end{aligned}
(2.34)

The first inequality implies $$d(Tz,z)=0$$, and hence $$Tz=z$$, which completes the proof. □

We next consider a particular case of the main theorems in which the mapping is an α-Meir-Keeler contractive mapping, that is, it satisfies Definition 9.

### Corollary 18

Let $$(X,d)$$ be a complete quasi-metric space and $$T:X\to X$$ be a continuous, α-Meir-Keeler contraction, that is, for every $$\varepsilon>0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq d(x,y)< \varepsilon+\delta\quad\Longrightarrow\quad\alpha (x,y)d(Tx,Ty) <\varepsilon,$$
(2.35)

holds for all $$x,y\in X$$. If $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha (Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

### Proof

It is obvious that if (2.35) holds, then using the fact that

$$d(x,y)\leq M(x,y)=\max\bigl\{ d(x,y),d(Tx,x),d(Ty,y)\bigr\} ,$$

we conclude

$$\varepsilon\leq M(x,y)< \varepsilon+\delta\quad\Longrightarrow\quad \varepsilon \leq d(x,y)<\varepsilon+\delta\quad\Longrightarrow\quad\alpha(x,y)d(Tx,Ty) < \varepsilon,$$
(2.36)

for all $$x,y\in X$$. In other words, T satisfies the conditions in the statement of Theorem 15 and hence has a fixed point. □

Finally, we replace the continuity of the contraction in Corollary 18 by the condition (A) on the space $$(X,d)$$, which results in the following existence theorem the proof of which is identical to the proof of Theorem 17.

### Corollary 19

Let $$(X,d)$$ be a complete quasi-metric space satisfying the condition (A) and let $$T:X\to X$$ be an α-Meir-Keeler contractive mapping. If there exists $$x_{0} \in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$ and $$\alpha(Tx_{0},x_{0})\geq1$$, then T has a fixed point.

We end this section with an example of an α-Meir-Keeler contraction defined on a quasi-metric space.

### Example 20

Let $$X=[0,\infty)$$. Define

$$d(x,y)=\left \{ \begin{array}{@{}l@{\quad}l} x-y &\mbox{if } x\geq y, \\ \frac{y-x}{2} & \mbox{if } x< y. \end{array} \right .$$

The function $$d(x,y)$$ is a quasi-metric but not a metric on X. Indeed, note that $$d(1,3)=1\neq d(3,1)=2$$. The space $$(X,d)$$ is a complete quasi-metric space. Define the mappings $$T: X\to X$$ and $$\alpha: X\times X\to[0,\infty)$$ as follows:

$$Tx=\left \{ \begin{array}{@{}l@{\quad}l} \frac{x^{2}}{8} &\mbox{if } x\in[0,1],\\ 3x & \mbox{if } x\in(1,\infty), \end{array} \right .\qquad \alpha(x,y)=\left \{ \begin{array}{@{}l@{\quad}l} 1&\mbox{if } x,y\in[0,1], \\ -1 & \mbox{otherwise}. \end{array} \right .$$

It is easy to see that T is triangular α-admissible. Note that if $$\alpha(x,y)\geq1$$, then $$x,y\in [0,1]$$ and hence both Tx and Ty are also in $$[0,1]$$. Thus, $$\alpha(Tx,Ty)=\alpha (\frac{x^{2}}{8},\frac{y^{2}}{8})=1\geq1$$. Also, if $$\alpha(x,z)\geq1$$ and $$\alpha(z,y)\geq1$$, then $$x,y,z\in[0,1]$$ and thus, $$\alpha(x,y)=1\geq1$$. The map T is not continuous, however, the condition (A) holds on X. More precisely, if the sequence $$\{x_{n}\}\subset X$$ satisfies $$\alpha(x_{n},x_{n+1})\geq1$$ and $$\alpha(x_{n+1},x_{n})\geq1$$, and if $$\lim_{n\to\infty}x_{n}=x$$, then $$\{x_{n}\}\subset[0,1]$$, and hence $$x\in[0,1]$$. Then $$\alpha(x_{n},x)\geq1$$.

Note that for $$x,y\in[0,1]$$, we have $$x+y\leq2$$.

If $$x\geq y$$ then for $$\varepsilon>0$$ we choose $$\delta=3\varepsilon$$ so that $$\varepsilon\leq d(x,y)=x-y <\varepsilon+\delta$$ implies $$\alpha(x,y)d(Tx,Ty)= \frac {x^{2}-y^{2}}{8}= \frac{(x-y)(x+y)}{8}< \frac {2(\varepsilon+\delta)}{8}=\varepsilon$$.

If $$x< y$$ then for $$\varepsilon>0$$ we choose again $$\delta=3\varepsilon$$ so that $$\varepsilon\leq d(x,y)=\frac{y-x}{2} <\varepsilon+\delta$$ implies $$\alpha(x,y)d(Tx,Ty)= \frac {x^{2}/8-y^{2}/8}{2}= \frac{(x-y)(x+y)}{16}< \frac {2(\varepsilon+\delta)}{16}= \frac{\varepsilon}{2}$$.

In other words, for every $$\varepsilon>0$$, there exists δ which is actually $$\delta=3\varepsilon$$. Therefore, the map T is an α-Meir-Keeler contraction. Finally, note that $$\alpha(0,T0)\geq1$$ and $$\alpha(T0,0)\geq1$$. All conditions of Corollary 19 are satisfied and T has a fixed point $$x=0$$.

## 3 Consequences: G-metric spaces

In this section, we present some results which show that several fixed point theorems on G-metric spaces are in fact direct consequences of the existence theorems given in the previous section.

First, we briefly recollect some basic notions of G-metric and G-metric space [2].

### Definition 21

Let X be a nonempty set, $$G:X\times X\times X\rightarrow{}[ 0,\infty )$$ be a function satisfying the following conditions:

1. (G1)

$$G(x,y,z)=0$$ if $$x=y=z$$,

2. (G2)

$$0< G(x,x,y)$$ for all $$x,y\in X$$ with $$x\neq y$$,

3. (G3)

$$G(x,x,y)\leq G(x,y,z)$$ for all $$x,y,z\in X$$ with $$z\neq y$$,

4. (G4)

$$G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots$$ (symmetry in all variables),

5. (G5)

$$G(x,y,z)\leq G(x,a,a)+G(a,y,z)$$ for all $$x,y,z,a\in X$$ (rectangle inequality).

Then the function G is called a G-metric on X and the pair $$( X,G )$$ is called a G-metric space.

It is obvious that for every G-metric on the set X, the expression

$$d_{G}(x,y)=G(x,x,y)+G(x,y,y)$$

is a standard metric on X.

### Definition 22

(see [2])

Let $$(X,G)$$ be a G-metric space and let $$\{x_{n}\}$$ be a sequence in X.

1. (1)

A point $$x\in X$$ is said to be the limit of the sequence $$\{x_{n}\}$$ if

$$\lim_{n,m\rightarrow\infty}G(x,x_{n},x_{m})=0$$

and the sequence $$\{x_{n}\}$$ is said to be G-convergent to x.

2. (2)

A sequence $$\{x_{n}\}$$ is called a G-Cauchy sequence if for every $$\varepsilon>0$$, there is a positive integer N such that $$G(x_{n},x_{m},x_{l})<\varepsilon$$ for all $$n,m,l\geq$$ N; that is, if $$G(x_{n},x_{m},x_{l})\rightarrow0$$ as $$n,m,l\rightarrow\infty$$.

3. (3)

$$(X,G)$$ is said to be G-complete (or a complete G-metric space) if every G-Cauchy sequence in $$(X,G)$$ is G-convergent in X.

### Theorem 23

[1]

Let $$(X,G)$$ be a G-metric space. Let $$d:X\times X \rightarrow [0,\infty)$$ be the function defined by $$d(x,y)=G(x,y,y)$$. Then

1. (1)

$$(X,d)$$ is a quasi-metric space;

2. (2)

$$\{x_{n}\}\subset X$$ is G-convergent to $$x\in X$$ if and only if $$\{x_{n}\}$$ is convergent to x in $$(X,d)$$;

3. (3)

$$\{x_{n}\}\subset X$$ is G-Cauchy if and only if $$\{ x_{n}\}$$ is Cauchy in $$(X,d)$$;

4. (4)

$$(X,G)$$ is G-complete if and only if $$(X,d)$$ is complete.

Admissible mappings in the context of G-metric spaces can be defined as follows [10].

### Definition 24

A mapping $$T:X\to X$$ is called β-admissible if for all $$x,y\in X$$ we have

$$\beta(x,y,y)\geq1 \quad\Rightarrow\quad\beta(Tx,Ty,Ty)\geq1,$$
(3.1)

where $$\beta:X\times X\times X\to[0,\infty)$$ is a given function. If in addition,

$$\left . \begin{array}{r@{}} \beta(x,y,y)\geq1, \\ \beta(y,z,z)\geq1 \end{array} \right \} \quad\Rightarrow\quad\beta(x,z,z)\geq1,$$
(3.2)

for all $$x,y,z\in X$$, then T is called triangular β-admissible.

### Definition 25

Let $$(X,G)$$ be a G-metric space. Let $$T:X\rightarrow X$$ be a triangular β-admissible mapping. Suppose that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq G(x,y,y)< \varepsilon+\delta\quad\mbox{implies}\quad \beta (x,y,y)G(Tx,Ty,Ty)< \varepsilon,$$
(3.3)

for all $$x, y \in X$$. Then T is called β-Meir-Keeler contraction.

For more details on β-admissible maps on G-metric spaces we refer the reader to [10].

### Definition 26

(I) Let $$(X,G)$$ be a G-metric space. Let $$T:X\rightarrow X$$ be a triangular β-admissible mapping. Suppose that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq M(x,y,y)< \varepsilon+\delta\quad\mbox{implies}\quad \beta (x,y,y)G(Tx,Ty,Ty)< \varepsilon,$$
(3.4)

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

$$M(x,y,y)=\max\bigl\{ G(x,y,y),G(Tx,x,x),G(Ty,y,y)\bigr\} .$$
(3.5)

Then T is called a generalized β-Meir-Keeler contraction of type (I).

(II) Let $$(X,G)$$ be a G-metric space. Let $$T:X\rightarrow X$$ be a triangular β-admissible mapping. Suppose that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq N(x,y,y)< \varepsilon+\delta\quad\mbox{implies}\quad \beta (x,y,y)G(Tx,Ty,Ty)< \varepsilon,$$
(3.6)

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

$$M(x,y,y)=\max\biggl\{ G(x,y,y),\frac{G(Tx,x,x)+G(Ty,y,y)}{2}\biggr\} .$$
(3.7)

Then T is called a generalized β-Meir-Keeler contraction of type (II).

### Lemma 27

Let $$T:X \rightarrow X$$ where X is nonempty set. Then T is β-admissible on $$(X, G)$$ if and only if T is α-admissible on $$(X,d)$$.

### Proof

The proof is obvious by taking $$\alpha(x,y)=\beta(x,y,y)$$. □

The next theorem is a consequence of Corollary 18.

### Theorem 28

Let $$(X,G)$$ be a complete G-metric space and $$T:X\to X$$ be a continuous, β-Meir-Keeler contraction. If $$\beta(x_{0},Tx_{0}, Tx_{0})\geq1$$ and $$\beta(Tx_{0},Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

### Proof

Consider the quasi-metric $$d(x,y)=G(x,y,y)$$ for all $$x,y\in X$$. Due to Lemma 27 and (3.3), we find that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

$$\varepsilon\leq d(x,y)< \varepsilon+\delta\quad\mbox{implies}\quad \alpha (x,y)d(Tx,Ty)< \varepsilon,$$
(3.8)

for all $$x, y \in X$$. Then the proof follows from Corollary 18. □

Our last theorem is a consequence of Theorems 15 and 16.

### Theorem 29

Let $$(X,G)$$ be a complete G-metric space and $$T:X\to X$$ be a continuous, generalized β-Meir-Keeler contraction of type (I) or (II). If $$\beta(x_{0},Tx_{0}, Tx_{0})\geq1$$ and $$\alpha(Tx_{0},Tx_{0},x_{0})\geq1$$ for some $$x_{0}\in X$$, then T has a fixed point in X.

### Proof

Since the function $$d(x,y)=G(x,y,y)$$ is a quasi-metric on X, employing Lemma 27 and (3.4) (respectively (3.6)), we see that for every $$\varepsilon> 0$$ there exists $$\delta>0$$ such that

\begin{aligned} &\varepsilon\leq M(x,y)< \varepsilon+\delta \quad\mbox{implies}\quad \alpha (x,y)d(Tx,Ty)< \varepsilon,\quad\mbox{or}\\ &\varepsilon\leq N(x,y)< \varepsilon+\delta\quad\mbox{implies}\quad \alpha(x,y)d(Tx,Ty)< \varepsilon, \end{aligned}
(3.9)

for all $$x, y \in X$$ for the generalized α-Meir-Keeler mappings of types (I) or (II), respectively. Then the proof follows from Theorem 15 (respectively, Theorem 16). □