# $$F(\psi,\varphi)$$-Contractions for α-admissible mappings on M-metric spaces

## Abstract

In this paper, we introduce certain α-admissible mappings which are $$F(\psi,\varphi)$$-contractions on M-metric spaces, and we establish some fixed point results. Our results generalize and extend some well-known results on this topic in the literature.

## Introduction and preliminaries

Geraghty in [10] introduced an interesting class of auxiliary functions to refine the Banach contraction mapping principle. Let $$\mathcal{F}$$ be the function $$\beta:[0,\infty) \to[0,1)$$ which satisfies the condition

$$\lim_{n\to\infty}\beta(t_{n})=1 \quad \mbox{implies}\quad \lim_{n\to\infty}t_{n}=0.$$

By using $$\mathcal{F}$$, Geraghty [10] proved the following theorem.

### Theorem 1.1

([10])

Let $$(X, d)$$ be a complete metric space and $$T : X \to X$$ be an operator. Suppose that there exists $$\beta\in \mathcal{F}$$ satisfying the condition

$$\beta(t_{n})\rightarrow1\quad \textit{implies}\quad t_{n} \rightarrow0.$$

If T satisfies the following inequality

$$d(Tx, Ty) \leq\beta\bigl(d(x, y)\bigr) d(x, y) \quad \textit{for any } x,y \in X,$$
(1)

then T has a unique fixed point.

We now present definitions, lemmas, remarks, and examples that we will use.

### Definition 1.2

([4])

Let $$f:X \to X$$ and $$\alpha:X\times X\to[0, +\infty)$$. We say that f is an α-admissible mapping if $$\alpha (x,y)\geq1$$ implies $$\alpha(fx,fy)\geq1$$ for all $$x,y \in X$$.

### Definition 1.3

([12])

Let Ψ denote all functions $$\psi:[0,\infty)\rightarrow[0,\infty)$$ satisfying:

1. (i)

ψ is strictly increasing and continuous,

2. (ii)

$$\psi ( t ) =0$$ if and only if $$t=0$$.

### Definition 1.4

([5])

An ultra altering distance function is a continuous, nondecreasing mapping $$\varphi:[0,\infty)\rightarrow{}[0,\infty)$$ such that $$\varphi(t)>0$$ for $$t>0$$.

### Remark 1.5

We let Φ denote the class of the ultra altering distance functions.

### Definition 1.6

([5])

A mapping $$F:[0,\infty)^{2}\rightarrow\Bbb{R}$$ is called a C-class function if it is continuous and satisfies the following axioms:

1. 1.

$$F(s,t)\leq s$$;

2. 2.

$$F(s,t)=s$$ implies that either $$s=0$$ or $$t=0$$ for all $$s,t\in [0,\infty)$$.

We denote the C-class functions by $$\mathcal{C}$$.

### Example 1.7

([5])

The following functions are elements of $$\mathcal{C}$$:

1. 1.

$$F(s,t)=s-t$$.

2. 2.

$$F(s,t)=ms$$, $$0 < m < 1$$.

3. 3.

$$F(s,t)=\frac{s}{(1+t)^{r}}$$; $$r\in(0,\infty)$$.

4. 4.

$$F(s,t)=s\beta(s)$$, $$\beta:[0,\infty)\rightarrow(0,1)$$ and is continuous.

5. 5.

$$F(s,t)=s-(\frac{2+t}{1+t})t$$.

6. 6.

$$F(s,t)=\sqrt[n]{\ln(1+s^{n})}$$.

### Definition 1.8

([18], [22, Definition 1.1])

A partial metric on a nonempty set X is a function $$p: X \times X \to\mathbb {R}^{+}$$ such that, for all $$x, y, z \in X$$:

1. (p1)

$$p(x,x)=p(y,y)=p(x,y)\iff x=y$$,

2. (p2)

$$p(x,x)\leq p(x,y)$$,

3. (p3)

$$p(x, y)=p(y,x)$$,

4. (p4)

$$p(x, y)\leq p(x,z)+p(z,y)-p(z,z)$$.

A partial metric space is a pair $$(X, p)$$ such that X is a nonempty set and p is a partial metric on X.

For more details and examples see [1416].

### Definition 1.9

([7])

Let X be a nonempty set. A function $$\mu:X \times X\to\mathbb {R}^{+}$$ is called an m-metric if the following conditions are satisfied:

1. (m1)

$$\mu(x,x)=\mu(y,y)=\mu(x,y)\iff x=y$$,

2. (m2)

$$m_{xy}\leq\mu(x,y)$$,

3. (m3)

$$\mu(x, y)=\mu(y,x)$$,

4. (m4)

$$(\mu(x, y)-m_{xy} )\leq (\mu (x,z)-m_{xz} )+ (\mu(z,y)-m_{zy} )$$,

where

$$m_{xy}:=\min\bigl\{ \mu(x,x), \mu(y,y)\bigr\} .$$

Then the pair $$(X,\mu)$$ is called an M-metric space. The following notation is useful in the sequel:

$$M_{xy}:=\max\bigl\{ \mu(x,x), \mu(y,y)\bigr\} .$$

### Remark 1.10

([7])

For every $$x,y\in X$$,

1. 1.

$$0\leq M_{xy}+m_{xy} =\mu(x,x)+\mu(y,y)$$;

2. 2.

$$0\leq M_{xy}-m_{xy} = |\mu(x,x)-\mu(y,y)|$$;

3. 3.

$$M_{xy}-m_{xy} \leq(M_{xz}-m_{xz}) +(M_{zy}-m_{zy})$$.

## Topology on M-metric space

It is clear that each M-metric m on X generates a $$T_{0}$$ topology $$\tau_{m}$$ on X. The set

$$\bigl\{ B_{\mu}(x, \varepsilon): x \in X, \varepsilon> 0\bigr\} ,$$

where

$$B_{\mu}(x, \varepsilon) =\bigl\{ y \in X : \mu(x, y) < m_{x,y} + \varepsilon\bigr\} ,$$

for all $$x \in X$$ and $$\varepsilon> 0$$, forms the base of $$\tau_{m}$$.

### Definition 2.1

([7])

Let $$(X,\mu)$$ be an M-metric space. Then:

1. 1.

A sequence $$\{x_{n}\}$$ in an M-metric space $$(X, m)$$ converges to a point $$x \in X$$ if

$$\lim_{n\to\infty} \bigl(\mu(x_{n}, x)-m_{x_{n},x}\bigr)=0.$$
(2)
2. 2.

A sequence $$\{x_{n}\}$$ in an M-metric space $$(X, m)$$ is called an m-Cauchy sequence if

$$\lim_{n,m\to \infty}\bigl(\mu(x_{n}, x_{m})-m_{x_{n},x_{m}}\bigr) \quad \mbox{and}\quad \lim _{n,m\to\infty}(M_{x_{n}, x_{m}}-m_{x_{n},x_{m}})$$
(3)

exist (and are finite).

3. 3.

An M-metric space $$(X, m)$$ is said to be complete if every m-Cauchy sequence $$\{x_{n}\}$$ in X converges, with respect to $$\tau _{m}$$, to a point $$x\in X$$ such that

$$\Bigl(\lim_{n\to\infty}\bigl(\mu(x_{n}, x)-m_{x_{n},x}\bigr)=0 \mbox{ and } \lim_{n\to\infty}(M_{x_{n}, x}-m_{x_{n},x})=0 \Bigr).$$

### Lemma 2.2

([7])

Assume that $$x_{n}\to x$$ and $$y_{n}\to y$$ as $$n\to\infty$$ in an M-metric space $$(X, m)$$. Then

$$\lim_{n\to \infty}\bigl(\mu(x_{n}, y_{n})-m_{x_{n},y_{n}} \bigr) = \mu(x, y)-m_{xy}.$$

### Lemma 2.3

([7])

Assume that $$x_{n}\to x$$ as $$n\to\infty$$ in an M-metric space $$(X, m)$$. Then

$$\lim_{n\to\infty}\bigl(\mu(x_{n}, y)-m_{x_{n},y} \bigr) = \mu(x, y)-m_{x,y}$$

for all $$y\in X$$.

### Lemma 2.4

([7])

Assume that $$x_{n}\to x$$ and $$x_{n}\to y$$ as $$n\to\infty$$ in an M-metric space $$(X, m)$$. Then $$\mu(x,y)=m_{xy}$$. Further if $$\mu(x,x)=\mu(y,y)$$, then $$x=y$$.

## Methods

Many authors studied the class of $$\alpha-\psi$$ contractive type mappings and obtained fixed point results for this new class of mappings in metric spaces. Their results contain several well-known fixed point theorems including the Banach contraction principle.

The goal of this article is to introduce the class of $$F(\psi,\varphi )$$-contractions and investigate the existence and uniqueness of fixed points for α-admissible mappings on M-metric spaces.

## Discussion and main results

We start this section with the following main theorem.

### Theorem 4.1

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\bigl( \psi\bigl(\mu(Tx,Ty)\bigr)+l\bigr)^{\alpha(x,Tx)\alpha(y,Ty)}\leq F\bigl(\psi\bigl(\mu (x,y)\bigr),\varphi\bigl(\mu(x,y)\bigr)\bigr)+l$$
(4)

for all $$x,y \in X$$ and $$l\geq1$$, where $$\psi\in\Psi$$, $$\varphi\in \Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Proof

Let $$x_{0} \in X$$ be such that $$\alpha(x_{0},Tx_{0})\geq1$$. Define a sequence $$\{x_{n}\}$$ in X by $$x_{n}=T^{n} x_{0}=Tx_{n-1}$$ for all $$n\in\Bbb{N}$$. Since T is an α-admissible mapping and $$\alpha(x_{0},Tx_{0})\geq 1$$, we deduce that $$\alpha(x_{1},x_{2})=\alpha(Tx_{0},T^{2}x_{0})\geq1$$. Continuing this process, we get $$\alpha(x_{n},Tx_{n})\geq1$$ for all $$n\in\Bbb{N}\cup\{0\}$$. From inequality (4) we have

\begin{aligned} \psi\bigl(\mu(Tx_{n-1},Tx_{n})\bigr)+l \leq& \bigl(\psi \bigl(\mu (Tx_{n-1},Tx_{n})+l\bigr)\bigr)^{\alpha(x_{n-1},Tx_{n-1})\alpha(x_{n},Tx_{n})} \\ \leq& F\bigl(\psi\bigl(\mu(x_{n-1},x_{n})\bigr),\varphi \bigl(\mu(x_{n-1},x_{n})\bigr)\bigr)+l. \end{aligned}

Then we have

$$\psi\bigl(\mu(x_{n},x_{n+1})\bigr)\leq F \bigl(\psi\bigl(\mu(x_{n-1},x_{n})\bigr),\varphi \bigl( \mu(x_{n-1},x_{n})\bigr)\bigr)\leq\psi\bigl( \mu(x_{n-1},x_{n})\bigr).$$
(5)

We want to prove that $$\mu(x_{n},x_{n+1})\to0$$, as $$n\to\infty$$. If $$\mu(x_{n_{0}},x_{n_{0}+1})=0$$, for some $$n_{0}\in\Bbb{N}$$, then by (5)

$$0 \leq \mu(x_{n_{0}+1},x_{n_{0}+2})\leq F\bigl(\psi\bigl(\mu (x_{n_{0}},x_{n_{0}+1})\bigr),\varphi\bigl(\mu(x_{n_{0}},x_{n_{0}+1}) \bigr)\bigr)\leq\psi \bigl(\mu(x_{n_{0}},x_{n_{0}+1})\bigr),$$

hence from the properties of functions F, ψ, and φ we have $$\mu(x_{n_{0}+1},x_{n_{0}+2})=0$$ which means

$$\mu(x_{n},x_{n+1})=0\quad \mbox{for all } n \geq n_{0},\quad \hbox{and thus}\quad \mu(x_{n},x_{n+1}) \to0\quad \mbox{as } n\to \infty.$$

Now let

$$\mu(x_{n},x_{n+1})>0 \quad \mbox{for all } n\in\mathbb{N}.$$

Inequality (5) implies that $$\mu(x_{n},x_{n+1})\leq \mu(x_{n-1},x_{n})$$. It follows that the sequence $$\{\mu(x_{n},x_{n+1})\}$$ is decreasing. Thus, there exists $$m\in \mathbb{R_{+}}$$ such that

$$\lim_{n\to\infty}\mu(x_{n},x_{n+1})=m.$$

We want to prove that $$m=0$$. Let $$m > 0$$. From (5) we have

\begin{aligned} \limsup_{n\to\infty}\psi\bigl(\mu(x_{n},x_{n+1}) \bigr) \leq& \limsup_{n\to\infty} F\bigl(\psi\bigl( \mu(x_{n-1},x_{n})\bigr),\varphi\bigl(\mu (x_{n-1},x_{n}) \bigr)\bigr) \\ \leq& \limsup_{n\to\infty} \psi\bigl(\mu(x_{n-1},x_{n}) \bigr). \end{aligned}

Hence we get

$$\psi(m)\leq F\bigl( \psi(m),\varphi(m)\bigr)\leq\psi(m),$$

so

$$F\bigl( \psi(m),\varphi(m)\bigr)=\psi(m).$$

Using the properties of functions F, ψ, and φ, we obtain that $$\psi(m)=0$$ or $$\varphi(m)=0$$, so then $$m=0$$, which is a contradiction. Therefore

$$\mu(x_{n},x_{n+1})\to0 \quad \mbox{as } n\to \infty.$$
(6)

Now we prove that $$\{x_{n}\}$$ is an M-Cauchy sequence in $$(X,\mu)$$. We have

\begin{aligned}& \lim_{n\to\infty}\mu(x_{n},x_{n+1})=0, \\& 0\leq m_{x_{n},x_{n+1}}\leq\mu(x_{n},x_{n+1}) \quad \Rightarrow\quad \lim_{n\to \infty}m_{x_{n},x_{n+1}}=0, \end{aligned}

and

$$m_{x_{n},x_{n+1}}=\min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{n+1},x_{n+1})\bigr\} \quad \Rightarrow\quad \lim _{n\to\infty}\mu(x_{n},x_{n})=0.$$

On the other hand,

$$m_{x_{n},x_{m}}= \min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{m},x_{m})\bigr\} \quad \Rightarrow\quad \lim _{n,m\to\infty}m_{x_{n},x_{m}}=0,$$

so

$$\lim_{n,m\to\infty}(M_{x_{n},x_{m}}-m_{x_{n},x_{m}})=0.$$

We show

$$\lim_{n,m\to\infty}\bigl(\mu(x_{n},x_{m})-m_{x_{n},x_{m}} \bigr)=0.$$

Let

$$M^{*}(x,y):=\mu(x,y)-m_{x,y},\quad \forall x, y \in X.$$

If $$\lim_{n,m\to\infty}M^{*}(x_{n},x_{m})\neq0$$, there exist $$\varepsilon>0$$ and $$\{l_{k}\}\subset\mathbb{N}$$ such that

$$M^{*}(x_{l_{k}},x_{n_{k}})\geq\varepsilon.$$

Suppose that k is the smallest integer which satisfies the above equation such that

$$M^{*}(x_{l_{k}-1},x_{n_{k}})< \varepsilon.$$

Now by (m4) we have

$$\varepsilon\leq M^{*}(x_{l_{k}},x_{n_{k}})\leq M^{*}(x_{l_{k}},x_{l_{k}-1})+M^{*}(x_{l_{k}-1},x_{n_{k}})< M^{*}(x_{l_{k}},x_{l_{k}-1})+\varepsilon.$$

Thus

$$\lim_{k\to\infty}M^{*}(x_{l_{k}},x_{n_{k}})= \varepsilon,$$

which means

$$\lim_{k\to\infty}\bigl(\mu (x_{l_{k}},x_{n_{k}})-m_{x_{l_{k}},x_{n_{k}}} \bigr)=\varepsilon.$$

On the other hand,

$$\lim_{k\to\infty}m_{x_{l_{k}},x_{n_{k}}}=0,$$

so we have

$$\lim_{k\to\infty}\mu(x_{l_{k}},x_{n_{k}})= \varepsilon.$$
(7)

Again by (m4) we have

$$M^{*}(x_{l_{k}},x_{n_{k}})\leq M^{*}(x_{l_{k}},x_{l_{k}+1})+M^{*}(x_{l_{k}+1},x_{n_{k}+1})+ M^{*}(x_{n_{k}+1},x_{n_{k}})$$

and

$$M^{*}(x_{l_{k}+1},x_{n_{k}+1})\leq M^{*}(x_{l_{k}},x_{l_{k}+1})+M^{*}(x_{l_{k}},x_{n_{k}})+ M^{*}(x_{n_{k}+1},x_{n_{k}}),$$

and taking the limit as $$k\to+\infty$$, together with (6) and (7), we have

$$\lim_{k\to\infty}\mu(x_{l_{k}+1},x_{n_{k}+1})= \varepsilon.$$
(8)

Now by (4), (7), and (8) we have

\begin{aligned} \psi\bigl(\mu(x_{m_{k}+1},x_{n_{k}+1})\bigr)+l \leq& \bigl(\psi \bigl(\mu (x_{m_{k}+1},x_{n_{k}+1})\bigr)+l\bigr)^{\alpha(x_{m_{k}},Tx_{m_{k}})\alpha(x_{n_{k}},Tx_{n_{k}})} \\ = & \bigl(\psi\bigl(\mu(Tx_{m_{k}},Tx_{n_{k}})+l\bigr) \bigr)^{\alpha(x_{m_{k}},Tx_{m_{k}}) \alpha(x_{n_{k}},Tx_{n_{k}})} \\ \leq& F\bigl(\psi\bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr),\varphi \bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr)\bigr)+l \\ \leq& \psi\bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr) +l. \end{aligned}

Therefore we get

\begin{aligned} \psi\bigl(\mu(x_{m_{k}+1},x_{n_{k}+1})\bigr) \leq & F\bigl(\psi \bigl(\mu (x_{m_{k}},x_{n_{k}})\bigr),\varphi\bigl( \mu(x_{m_{k}},x_{n_{k}})\bigr)\bigr) \\ \leq& \psi\bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr). \end{aligned}

Letting $$k\to\infty$$ in the above inequality, we get

$$\psi(\varepsilon) \leq F\bigl(\psi(\varepsilon),\varphi(\varepsilon)\bigr) \leq \psi(\varepsilon),$$

so

$$F\bigl(\psi(\varepsilon),\varphi(\varepsilon)\bigr)=\psi(\varepsilon).$$

Using the properties of F, ψ, and φ, we obtain $$\psi(\varepsilon)=0$$ or $$\varphi(\varepsilon)=0$$, and then $$\varepsilon=0$$, which is a contradiction. Therefore $$\{x_{n}\}$$ is an M-Cauchy sequence. Now, by the completeness of X, $$x_{n}\to x$$ for some $$x\in X$$ in the $$\tau_{m}$$ topology, i.e.,

$$\lim_{n\to\infty}\bigl(\mu(x_{n},x)-m_{x_{n},x} \bigr)=0$$

and

$$\lim_{n\to\infty}(M_{(x_{n},x)}-m_{x_{n},x})=0.$$

However, $$\lim_{n\to\infty} m_{x_{n},x} =0$$, hence $$\lim_{n\to\infty}\mu (x_{n},x)=0$$, and by Remark 1.10

$$\mu(x,x)=0.$$

Now suppose (a) holds. Then T is continuous and we have

$$\lim_{n\to\infty}\bigl(\mu(Tx_{n},Tx)-m_{Tx_{n},Tx} \bigr)=0,$$

i.e.,

$$\lim_{n\to\infty}\bigl(\mu(x_{n+1},Tx)-m_{x_{n+1},Tx} \bigr)=0,$$

and similar to the above, we have $$\lim_{n\to\infty} m_{x_{n+1},Tx} =0$$. Hence $$\lim_{n\to \infty}\mu(x_{n+1},Tx)=0$$ and by Remark 1.10, $$\mu(Tx,Tx)=0$$. On the other hand, $$x_{n}\to x$$ as $$n\to\infty$$ so by Lemma 2.3, we get

$$\bigl(\mu(x_{n},Tx)- m_{x_{n},Tx}\bigr)\to\bigl( \mu(x,Tx)-m_{x,Tx}\bigr) = \mu(x,Tx) \quad \mbox{as } n\to\infty,$$

but we have

$$\bigl(\mu(x_{n},Tx)- m_{x_{n},Tx}\bigr)\to0 \quad \mbox{as } n\to \infty.$$

Thus

$$\mu(x,Tx)=0,$$

therefore $$\mu(x,Tx)=\mu(Tx,Tx)=\mu(x,x)=0$$ and by (m1) we get

$$Tx=x.$$

Next suppose (b) holds. Then $$\alpha(x,Tx)\geq1$$. Now by (4) we have

\begin{aligned} \psi\bigl(\mu(Tx_{n},Tx)\bigr)+l \leq& \bigl(\psi\bigl( \mu(Tx_{n},Tx)\bigr)+l\bigr)^{\alpha(x_{n},Tx_{n})\alpha(x,Tx)} \\ \leq& F\bigl( \psi\bigl(\mu(x_{n},x)\bigr),\varphi\bigl( \mu(x_{n},x)\bigr)\bigr)+l, \end{aligned}

that is, $$\psi(\mu(Tx_{n},Tx))\leq F( \psi(\mu(x_{n},x)),\varphi(\mu (x_{n},x))) \leq\psi(\mu(x_{n},x))$$, and so we get

$$\mu(Tx_{n},Tx)\to0\quad \mbox{as } n\to\infty.$$

On the other hand,

$$0\leq m_{Tx_{n},Tx}\leq \mu(Tx_{n},Tx)\to0\quad \mbox{as } n\to \infty.$$

Thus $$Tx_{n} \to Tx$$ in the $$\tau_{m}$$ topology.

The proof of $$Tx=x$$ follows as in (a). □

### Theorem 4.2

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\bigl(\alpha(x,Tx)\alpha(y,Ty)+1\bigr)^{\psi(\mu(Tx,Ty))} \leq2^{F (\psi(\mu (x,y)),\varphi(\mu(x,y)))}$$
(9)

for all $$x,y \in X$$, where $$\psi\in\Psi$$, $$\varphi\in\Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Proof

Let $$x_{0} \in X$$ be such that $$\alpha(x_{0},Tx_{0})\geq1$$. Define a sequence $$\{x_{n}\}$$ in X by $$x_{n}=T^{n} x_{0}=Tx_{n-1}$$ for all $$n\in\mathbb{N}$$. Since T is an α-admissible mapping and $$\alpha(x_{0},Tx_{0})\geq 1$$, we deduce that $$\alpha(x_{1},x_{2})=\alpha(Tx_{0},T^{2}x_{0})\geq1$$. Continuing this process, we get $$\alpha(x_{n},Tx_{n})\geq1$$ for all $$n\in\mathbb{N}\cup\{ 0\}$$. From inequality (9) we have

\begin{aligned} 2^{\psi( \mu(Tx_{n-1},Tx_{n}))} \leq&\bigl(\alpha(x_{n-1},Tx_{n-1})\alpha (x_{n},Tx_{n})+1\bigr)^{\psi( \mu(Tx_{n-1},Tx_{n}))} \\ \leq& 2^{F( \psi(\mu(x_{n-1},x_{n})),\varphi (\mu(x_{n-1},x_{n})))}. \end{aligned}

Then we have

$$\psi\bigl(\mu(x_{n},x_{n+1})\bigr)\leq F\bigl( \psi\bigl(\mu(x_{n-1},x_{n})\bigr),\varphi\bigl(\mu (x_{n-1},x_{n})\bigr)\bigr)\leq\psi\bigl(\mu(x_{n-1},x_{n}) \bigr).$$
(10)

Now similar to the proof in Theorem 4.1, we get

$$\mu(x_{n},x_{n+1})\to0 \quad \mbox{as } n\to \infty.$$
(11)

Now we prove that $$\{x_{n}\}$$ is an M-Cauchy sequence in $$(X,\mu)$$. We have

\begin{aligned}& \lim_{n\to\infty}\mu(x_{n},x_{n+1})=0, \\& 0\leq m_{x_{n},x_{n+1}}\leq\mu(x_{n},x_{n+1})\quad \Rightarrow \quad \lim_{n\to \infty}m_{x_{n},x_{n+1}}=0, \end{aligned}

and

$$m_{x_{n},x_{n+1}}=\min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{n+1},x_{n+1})\bigr\} \quad \Rightarrow\quad \lim _{n\to\infty}\mu(x_{n},x_{n})=0.$$

On the other hand,

$$m_{x_{n},x_{m}}= \min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{m},x_{m})\bigr\} \quad \Rightarrow\quad \lim _{n,m\to\infty}m_{x_{n},x_{m}}=0,$$

so

$$\lim_{n,m\to\infty}(M_{x_{n},x_{m}}-m_{x_{n},x_{m}})=0.$$

We show

$$\lim_{n,m\to\infty}\bigl(\mu(x_{n},x_{m})-m_{x_{n},x_{m}} \bigr)=0.$$

Let

$$M^{*}(x,y):=\mu(x,y)-m_{x,y},\quad \forall x, y \in X.$$

If $$\lim_{n,m\to\infty}M^{*}(x_{n},x_{m})\neq0$$, there exist $$\varepsilon>0$$ and $$\{l_{k}\}\subset\mathbb{N}$$ such that

$$M^{*}(x_{l_{k}},x_{n_{k}})\geq\varepsilon.$$

Suppose that k is the smallest integer which satisfies the above equation such that

$$M^{*}(x_{l_{k}-1},x_{n_{k}})< \varepsilon.$$

Again as in the proof in Theorem 4.1, we obtain that

$$\lim_{k\to\infty}\mu(x_{m_{k}},x_{n_{k}})= \varepsilon$$
(12)

and

$$\lim_{k\to\infty}\mu(x_{l_{k}+1},x_{n_{k}+1})= \varepsilon.$$
(13)

Now by (9), (12), and (13) we have

\begin{aligned} 2^{\psi(\mu(x_{m_{k}+1},x_{n_{k}+1}))} \leq& \bigl(\alpha (x_{m_{k}},Tx_{m_{k}}) \alpha(x_{n_{k}},Tx_{n_{k}})+1\bigr)^{\psi(\mu (x_{m_{k}+1},x_{n_{k}+1}))} \\ \leq& 2^{F(\psi(\mu(x_{m_{k}},x_{n_{k}})),\varphi (\mu(x_{m_{k}},x_{n_{k}})))}. \end{aligned}

Therefore we get

$$\psi\bigl(\mu(x_{m_{k}+1},x_{n_{k}+1})\bigr) \leq F\bigl(\psi\bigl(\mu (x_{m_{k}},x_{n_{k}})\bigr),\varphi\bigl(\mu(x_{m_{k}},x_{n_{k}}) \bigr)\bigr) \leq \psi\bigl(\mu (x_{m_{k}},x_{n_{k}})\bigr).$$

Letting $$k\to\infty$$ in the above inequality, we get

$$\psi(\varepsilon) \leq F\bigl(\psi(\varepsilon),\varphi(\varepsilon)\bigr) \leq \psi(\varepsilon),$$

so

$$F\bigl( \psi(\varepsilon),\varphi(\varepsilon)\bigr)=\psi(\varepsilon).$$

Using the properties of functions F, ψ, and φ, we obtain that $$\psi(\varepsilon)=0$$, or $$\varphi(\varepsilon)=0$$, and then $$\varepsilon=0$$, which is a contradiction. Therefore $$\{x_{n}\}$$ is an M-Cauchy sequence.

Now, by the completeness of X, $$x_{n}\to x$$ for some $$x\in X$$ in the $$\tau_{m}$$ topology, i.e.,

$$\lim_{n\to\infty}\bigl(\mu(x_{n},x)-m_{x_{n},x} \bigr)=0$$

and

$$\lim_{n\to\infty}(M_{(x_{n},x)}-m_{x_{n},x})=0.$$

However, $$\lim_{n\to\infty} m_{x_{n},x} =0$$, hence $$\lim_{n\to\infty}\mu (x_{n},x)=0$$ and by Remark 1.10

$$\mu(x,x)=0.$$

Now suppose (a) holds. Then, as in the proof in Theorem 4.1, we have $$Tx=x$$. Next suppose (b) holds. Then $$\alpha(x,Tx)\geq1$$. From (9) we have

\begin{aligned} 2^{\psi(\mu(Tx_{n},Tx))} \leq& \bigl(\alpha(x_{n},Tx_{n})\alpha (x,Tx)+1\bigr)^{\psi(\mu(Tx_{n},Tx))} \\ \leq& 2^{ F(\psi(\mu(x_{n},x)),\varphi(\mu(x_{n},x)))}, \end{aligned}

that is, $$\psi(\mu(Tx_{n},Tx))\leq F( \psi(\mu(x_{n},x)),\varphi(\mu (x_{n},x))) \leq\psi(\mu(x_{n},x))$$, and so we get

$$\mu(Tx_{n},Tx)\to0 \quad \mbox{as } n\to\infty.$$

On the other hand,

$$0\leq m_{Tx_{n},Tx}\leq \mu(Tx_{n},Tx)\to0 \quad \mbox{as } n\to \infty.$$

Thus $$Tx_{n} \to Tx$$ in the $$\tau_{m}$$ topology.

The proof of $$Tx=x$$ follows as in (a). □

### Theorem 4.3

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\alpha(x,Tx)\alpha(y,Ty)\psi\bigl(\mu(Tx,Ty)\bigr) \leq F\bigl( \psi\bigl(\mu(x,y)\bigr),\varphi\bigl(\mu(x,y)\bigr)\bigr)$$
(14)

for all $$x,y \in X$$, where $$\psi\in\Psi$$, $$\varphi\in\Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Proof

Let $$x_{0} \in X$$ be such that $$\alpha(x_{0},Tx_{0})\geq1$$. Define a sequence $$\{x_{n}\}$$ in X by $$x_{n}=T^{n} x_{0}=Tx_{n-1}$$ for all $$n\in\mathbb{N}$$. Since T is an α-admissible mapping and $$\alpha(x_{0},Tx_{0})\geq 1$$, we deduce that $$\alpha(x_{1},x_{2})=\alpha(Tx_{0},T^{2}x_{0})\geq1$$. Continuing this process, we get $$\alpha(x_{n},Tx_{n})\geq1$$ for all $$n\in\mathbb{N}\cup\{ 0\}$$. From inequality (14) we have

\begin{aligned} \psi\bigl( \mu(Tx_{n-1},Tx_{n})\bigr) \leq& \alpha(x_{n-1},Tx_{n-1})\alpha (x_{n},Tx_{n}){ \psi\bigl( \mu(Tx_{n-1},Tx_{n})\bigr)} \\ \leq& F\bigl( \psi\bigl(\mu(x_{n-1},x_{n})\bigr),\varphi \bigl(\mu(x_{n-1},x_{n})\bigr)\bigr). \end{aligned}

Then we have

$$\psi\bigl(\mu(x_{n},x_{n+1})\bigr)\leq F\bigl( \psi\bigl(\mu(x_{n-1},x_{n})\bigr),\varphi\bigl(\mu (x_{n-1},x_{n})\bigr)\bigr)\leq\psi\bigl(\mu(x_{n-1},x_{n}) \bigr).$$
(15)

Now, similar to the proof in Theorem 4.1, we get

$$\mu(x_{n},x_{n+1})\to0 \quad \mbox{as } n\to \infty.$$
(16)

Now we prove that $$\{x_{n}\}$$ is an M-Cauchy sequence in $$(X,\mu)$$. We have

\begin{aligned}& \lim_{n\to\infty}\mu(x_{n},x_{n+1})=0, \\& 0\leq m_{x_{n},x_{n+1}}\leq\mu(x_{n},x_{n+1}) \quad \Rightarrow \quad \lim_{n\to \infty}m_{x_{n},x_{n+1}}=0, \end{aligned}

and

$$m_{x_{n},x_{n+1}}=\min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{n+1},x_{n+1})\bigr\} \quad \Rightarrow\quad \lim _{n\to\infty}\mu(x_{n},x_{n})=0.$$

On the other hand,

$$m_{x_{n},x_{m}}= \min\bigl\{ \mu(x_{n},x_{n}), \mu(x_{m},x_{m})\bigr\} \quad \Rightarrow\quad \lim _{n,m\to\infty}m_{x_{n},x_{m}}=0,$$

so

$$\lim_{n,m\to\infty}(M_{x_{n},x_{m}}-m_{x_{n},x_{m}})=0.$$

We show

$$\lim_{n,m\to\infty}\bigl(\mu(x_{n},x_{m})-m_{x_{n},x_{m}} \bigr)=0.$$

Let

$$M^{*}(x,y):=\mu(x,y)-m_{x,y},\quad \forall x, y \in X.$$

If $$\lim_{n,m\to\infty}M^{*}(x_{n},x_{m})\neq0$$, there exist $$\varepsilon>0$$ and $$\{l_{k}\}\subset\mathbb{N}$$ such that

$$M^{*}(x_{l_{k}},x_{n_{k}})\geq\varepsilon.$$

Suppose that k is the smallest integer which satisfies the above equation such that

$$M^{*}(x_{l_{k}-1},x_{n_{k}})< \varepsilon.$$

Again as in the proof in Theorem 4.1, we obtain that

$$\lim_{k\to\infty}\mu(x_{m_{k}},x_{n_{k}})= \varepsilon$$
(17)

and

$$\lim_{k\to\infty}\mu(x_{l_{k}+1},x_{n_{k}+1})= \varepsilon.$$
(18)

Now by (14), (17), and (18) we have

\begin{aligned} \psi\bigl(\mu(x_{m_{k}+1},x_{n_{k}+1})\bigr) \leq& \alpha (x_{m_{k}},Tx_{m_{k}})\alpha(x_{n_{k}},Tx_{n_{k}}) \psi\bigl(\mu (x_{m_{k}+1},x_{n_{k}+1})\bigr) \\ \leq& F\bigl(\psi\bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr),\varphi \bigl(\mu(x_{m_{k}},x_{n_{k}})\bigr)\bigr). \end{aligned}

Therefore we get

$$\psi\bigl(\mu(x_{m_{k}+1},x_{n_{k}+1})\bigr) \leq F\bigl(\psi\bigl(\mu (x_{m_{k}},x_{n_{k}})\bigr),\varphi\bigl(\mu(x_{m_{k}},x_{n_{k}}) \bigr)\bigr) \leq \psi\bigl(\mu (x_{m_{k}},x_{n_{k}})\bigr).$$

Letting $$k\to\infty$$ in the above inequality, we get

$$\psi(\varepsilon) \leq F\bigl(\psi(\varepsilon),\varphi(\varepsilon)\bigr) \leq \psi(\varepsilon),$$

so

$$F\bigl( \psi(\varepsilon),\varphi(\varepsilon)\bigr)=\psi(\varepsilon).$$

Using the properties of functions F, ψ, and φ, we obtain that $$\psi(\varepsilon)=0$$, or $$\varphi(\varepsilon)=0$$, then $$\varepsilon=0$$, which is a contradiction. Therefore $$\{x_{n}\}$$ is an M-Cauchy sequence.

Now, by the completeness of X, $$x_{n}\to x$$ for some $$x\in X$$ in the $$\tau_{m}$$ topology, i.e.,

$$\lim_{n\to\infty}\bigl(\mu(x_{n},x)-m_{x_{n},x} \bigr)=0$$

and

$$\lim_{n\to\infty}(M_{(x_{n},x)}-m_{x_{n},x})=0.$$

However, $$\lim_{n\to\infty} m_{x_{n},x} =0$$, hence $$\lim_{n\to\infty}\mu (x_{n},x)=0$$ and by Remark 1.10

$$\mu(x,x)=0.$$

Now suppose (a) holds. Then, as in the proof in Theorem 4.1, we have $$Tx=x$$. Next suppose (b) holds. Then $$\alpha(x,Tx)\geq1$$. From (14) we have

\begin{aligned} \psi\bigl(\mu(Tx_{n},Tx)\bigr) \leq& \alpha(x_{n},Tx_{n}) \alpha(x,Tx)\psi\bigl(\mu (Tx_{n},Tx)\bigr) \\ \leq& F\bigl(\psi\bigl(\mu(x_{n},x)\bigr),\varphi\bigl( \mu(x_{n},x)\bigr)\bigr), \end{aligned}

that is, $$\psi(\mu(Tx_{n},Tx))\leq F( \psi(\mu(x_{n},x)),\varphi(\mu (x_{n},x))) \leq\psi(\mu(x_{n},x))$$, and so we get

$$\mu(Tx_{n},Tx)\to0 \quad \mbox{as } n\to\infty.$$

On the other hand,

$$0\leq m_{Tx_{n},Tx}\leq \mu(Tx_{n},Tx)\to0\quad \mbox{as } n\to \infty.$$

Thus $$Tx_{n} \to Tx$$ in the $$\tau_{m}$$ topology.

The proof of $$Tx=x$$ follows as in (a). □

### Theorem 4.4

Assume that all of the hypotheses of Theorems 4.1 or 4.2 or 4.3 hold. In addition, suppose the following condition is satisfied:

1. (c)

if $$Tx=x$$ then $$\alpha(x,Tx)\geq1$$.

Then the fixed point of T is unique.

### Proof

Suppose that $$u,v\in X$$ are two fixed points of T such that $$u\neq v$$. Then $$\alpha(u,Tu)\geq1$$ and $$\alpha(v,Tv)\geq1$$.

For Theorem 4.1, we have

\begin{aligned}& \begin{aligned}[b] \psi\bigl(d(Tu,Tv)\bigr)+l &\leq \bigl(\psi\bigl(d(Tu,Tv)\bigr)+l \bigr)^{\alpha(u,Tu)\alpha (v,Tv)} \\ &\leq F\bigl(\psi\bigl(d(u,v)\bigr),\varphi \bigl(d(u,v)\bigr)\bigr)+l, \end{aligned} \end{aligned}
(19)
\begin{aligned}& \begin{aligned}[b] \psi\bigl(d(Tu,Tu)\bigr)+l &\leq\bigl( \psi\bigl(d(Tu,Tu)\bigr)+l \bigr)^{\alpha(u,Tu)\alpha (u,Tu)} \\ &\leq F\bigl(\psi\bigl(d(u,u)\bigr),\varphi\bigl( d(u,u)\bigr)\bigr)+l. \end{aligned} \end{aligned}
(20)

For Theorem 4.2, we have

\begin{aligned}& \begin{aligned}[b] 2^{\psi(\mu(Tu,Tv))}&\leq \bigl(\alpha(u,Tu)\alpha(v,Tv)+1 \bigr)^{\psi(\mu (Tu,Tv))} \\ &\leq 2^{F (\psi(\mu(u,v)),\varphi(\mu(u,v)))}, \end{aligned} \end{aligned}
(21)
\begin{aligned}& \begin{aligned}[b] 2^{\psi(\mu(Tu,Tu))} &\leq \bigl(\alpha(u,Tu)\alpha(u,Tu)+1 \bigr)^{\psi(\mu (Tu,Tu))} \\ &\leq 2^{F (\psi(\mu(u,u)),\varphi(\mu(u,u)))}. \end{aligned} \end{aligned}
(22)

For Theorem 4.3, we have

\begin{aligned}& \begin{aligned}[b] \psi\bigl(\mu(Tu,Tv)\bigr) &\leq \bigl(\alpha(u,Tu) \alpha(v,Tv)+1\bigr)\psi\bigl(\mu (Tu,Tv)\bigr) \\ &\leq F \bigl(\psi\bigl(\mu(u,v)\bigr),\varphi\bigl(\mu(u,v)\bigr)\bigr), \end{aligned} \end{aligned}
(23)
\begin{aligned}& \begin{aligned}[b] \psi\bigl(\mu(Tu,Tu)\bigr) &\leq \bigl(\alpha(u,Tu)\alpha(u,Tu)+1\bigr) \psi\bigl(\mu(Tu,Tu)\bigr) \\ &\leq F \bigl(\psi\bigl(\mu(u,u)\bigr),\varphi\bigl(\mu(u,u)\bigr)\bigr). \end{aligned} \end{aligned}
(24)

Therefore equations (19), (20), (21), (22), (23), and (24) imply that

\begin{aligned}& F \bigl(\psi\bigl(\mu(u,v)\bigr),\varphi\bigl(\mu(u,v)\bigr)\bigr)=\psi\bigl( \mu(Tu,Tv)\bigr)=\psi\bigl(\mu(u,v)\bigr), \\& F \bigl(\psi\bigl(\mu(u,u)\bigr),\varphi\bigl(\mu(u,u)\bigr)\bigr)=\psi\bigl( \mu(Tu,Tu)\bigr)=\psi\bigl(\mu(u,u)\bigr), \\& F \bigl(\psi\bigl(\mu(v,v)\bigr),\varphi\bigl(\mu(v,v)\bigr)\bigr)=\psi\bigl( \mu(Tv,Tv)\bigr)=\psi\bigl(\mu(v,v)\bigr), \end{aligned}

and so from the properties of functions F, ψ, and φ, we have

$$\mu(u,v)=\mu(u,u)=\mu(v,v)=0.$$

Therefore by (m1)

$$u=v.$$

□

## Consequences

From Theorems 4.1, 4.2, and 4.3 we obtain the following corollaries as an extension of several known results in the literature.

If we let $$\varphi(t)=\psi(t)=t$$, we get the following three corollaries.

### Corollary 5.1

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\bigl( \mu(Tx,Ty)+l\bigr))^{\alpha(x,Tx)\alpha(y,Ty)}\leq F\bigl(\mu(x,y)\bigr), \mu(x,y))+l$$
(25)

for all $$x,y \in X$$ and $$l\geq1$$, where $$\psi\in\Psi$$, $$\varphi\in \Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Corollary 5.2

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\bigl(\alpha(x,Tx)\alpha(y,Ty)+1\bigr)^{\mu(Tx,Ty)} \leq2^{F (\mu(x,y)),\mu(x,y))}$$
(26)

for all $$x,y \in X$$, where $$\psi\in\Psi$$, $$\varphi\in\Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Corollary 5.3

Let $$(X,\mu)$$ be a complete M-metric space and $$T: X\to X$$ be an α-admissible mapping. Suppose that the following condition is satisfied:

$$\alpha(x,Tx)\alpha(y,Ty)\mu(Tx,Ty) \leq F\bigl(\mu(x,y),\mu(x,y)\bigr)$$
(27)

for all $$x,y \in X$$, where $$\psi\in\Psi$$, $$\varphi\in\Phi$$, and $$F\in\mathcal{C}$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Lemma 5.4

([7])

Every p-metric and metric is an M-metric.

If we let $$\beta\in\mathcal{F}$$, $$\varphi(t)=\psi(t)=t$$ and $$F(s,t)=\beta(s)s$$, by Lemma 5.4 we get three results of Hussein et al. [13] (they are the immediate consequences of our results).

### Corollary 5.5

([13, Theorem 4])

Let $$(X,d)$$ be a complete metric space and $$T: X\to X$$ be an α-admissible mapping. Assume that there exists a function $$\beta:\mathbb{R}^{+}\to[0,1]$$ such that, for any bounded sequence $$\{t_{n}\}$$ of positive reals, $$\beta(t_{n})\to1$$ implies $$t_{n}\to0$$ and

$$\bigl( d(Tx,Ty)+l\bigr)^{\alpha(x,Tx)\alpha(y,Ty)}\leq\beta\bigl(d(x,y)\bigr)d(x,y)+l$$
(28)

for all $$x,y \in X$$ where $$l\geq1$$. Suppose that either

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Corollary 5.6

([13, Theorem 6])

Let $$(X,d)$$ be a complete metric space and $$T: X\to X$$ be an α-admissible mapping. Assume that there exists a function $$\beta:\mathbb{R}^{+}\to[0,1]$$ such that, for any bounded sequence $$\{t_{n}\}$$ of positive reals, $$\beta(t_{n})\to1$$ implies $$t_{n}\to0$$ and

$$\bigl(\alpha(x,Tx)\alpha(y,Ty)+1\bigr)^{ d(Tx,Ty)}\leq2^{\beta(d(x,y))d(x,y)}$$
(29)

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

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

### Corollary 5.7

([13, Theorem 8])

Let $$(X,d)$$ be a complete metric space and $$T: X\to X$$ be an α-admissible mapping. Assume that there exists a function $$\beta:\mathbb{R}^{+}\to[0,1]$$ such that, for any bounded sequence $$\{t_{n}\}$$ of positive reals, $$\beta(t_{n})\to1$$ implies $$t_{n}\to0$$ and

$$\bigl(\alpha(x,Tx)\alpha(y,Ty)\bigr)d(Tx,Ty)\leq\beta\bigl(d(x,y)\bigr)d(x,y)$$
(30)

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

1. (a)

T is continuous,

or

2. (b)

if $$\{x_{n}\}$$ is a sequence in X such that $$x_{n}\to x$$, $$\alpha(x_{n},x_{n+1})\geq1$$ for all n, then $$\alpha (x,Tx)\geq1$$.

If there exists $$x_{0}\in X$$ such that $$\alpha(x_{0},Tx_{0})\geq1$$, then T has a fixed point.

## Conclusion

Recently, the authors in [17] introduced the class of α-ψ contractive type mappings and obtained a fixed point result for this new class of mappings in the set-up of metric spaces. Their result contains several well-known fixed point theorems including the Banach contraction principle. Matthews (1994) in [18] established fixed point theorems in partial metric spaces. The authors in [7] introduced M-metric spaces which extend p-metric spaces and the authors established some new fixed point theorems.

In this paper, we introduce the class of $$F(\psi,\varphi)$$-contractions and investigate the existence and uniqueness of fixed points for α-admissible mappings on M-metric spaces. We also show that the fixed point results in [13] and Geraghty’s theorem [10] (Theorem 1.1) are immediate consequences of our results. For further results, we refer the reader to [14, 6, 812, 1921, 23].

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### Acknowledgements

The authors would like to thank the anonymous referee for his/her comments that helped us improve this article.

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Monfared, H., Asadi, M., Azhini, M. et al. $$F(\psi,\varphi)$$-Contractions for α-admissible mappings on M-metric spaces. Fixed Point Theory Appl 2018, 22 (2018). https://doi.org/10.1186/s13663-018-0647-y

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### Keywords

• Fixed point
• M-Metric space
• C-Class function
• $$F(\psi,\varphi)$$-Contraction