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

Fixed Point Theory and Applications20182018:22

https://doi.org/10.1186/s13663-018-0647-y

  • Received: 1 June 2018
  • Accepted: 31 August 2018
  • Published:

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.

Keywords

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

MSC

  • 47H10
  • 54H25

1 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.

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

     

2 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 \(x=y\).

3 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.

4 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)
$$u=v. $$
 □

5 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.

6 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].

Declarations

Acknowledgements

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

Availability of data and materials

Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

Funding

We have no funding for this article.

Authors’ contributions

All authors have read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Bilehsavar Branch, Islamic Azad University, Bilehsavar, Iran
(2)
Department of Mathematics, Zanjan Branch, Islamic Azad University, Zanjan, Iran
(3)
Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran
(4)
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway, Ireland

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