$F(\psi,\varphi)$-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.

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:

• (i) ψ is strictly increasing and continuous,

• (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. $F(s,t)\leq s$;

• 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. $F(s,t)=s-t$.

• 2. $F(s,t)=ms$, $0 < m < 1$.

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

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

• 5. $F(s,t)=s-(\frac{2+t}{1+t})t$.

• 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 $x, y, z \in X$:

• (p1) $p(x,x)=p(y,y)=p(x,y)\iff x=y$,

• (p2) $p(x,x)\leq p(x,y)$,

• (p3) $p(x, y)=p(y,x)$,

• (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 [14–16].

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:

• (m1) $\mu(x,x)=\mu(y,y)=\mu(x,y)\iff x=y$,

• (m2) $m_{xy}\leq\mu(x,y)$,

• (m3) $\mu(x, y)=\mu(y,x)$,

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

• 2. $0\leq M_{xy}-m_{xy} = |\mu(x,x)-\mu(y,y)|$;

• 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. 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. 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. 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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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:

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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

• (a) T is continuous,

  or

• (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 [1–4, 6, 8–12, 19–21, 23].