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Some fixed point theorems for $(\alpha,\psi)$rational type contractive mappings
Fixed Point Theory and Applicationsvolume 2015, Article number: 97 (2015)
Abstract
In this paper, we introduce the concept of $(\alpha,\psi)$rational type contractive mappings and provide sufficient conditions for the existence and uniqueness of a fixed point for such class of generalized nonlinear contractive mappings in the setting of generalized metric spaces. We also deduce several interesting corollaries.
Introduction and preliminaries
Fixed point theory has gained very large impetus due to its wide range of applications in various fields such as engineering, economics, computer science, and many others. It is well known that the contractive conditions are very indispensable in the study of fixed point theory, and Banach’s fixed point theorem [1] for contraction mappings is one of the pivotal result in analysis. This theorem has been extended and generalized by various authors (see, e.g., [2–28]) in various abstract spaces, one of which is generalized metric space.
As pointed out in [3], the topology of a generalized metric space has some disadvantages:

(T1)
A generalized metric does not need to be continuous.

(T2)
A convergent sequence in generalized metric spaces does not need to be Cauchy.

(T3)
A generalized metric space does not need to be Hausdorff, and hence the uniqueness of the limits cannot be guaranteed.
In this paper, we introduce the concept of $(\alpha,\psi)$rational type contractive mappings and provide sufficient conditions for the existence and uniqueness of fixed points for such class of generalized nonlinear contractive mappings in the framework of generalized metric spaces by caring the problems (T1)(T3) mentioned above. We also deduce several interesting corollaries. The proved results generalize and extend various wellknown results in the literature. The techniques used in this paper have been studied and improved by various authors (see [3–9, 16] and references cited therein).
To start with, we give some notations and introduce some definitions which will be used in the sequel.
Definition 1.1
[2]
Let X be a nonempty set and $d : X\times X \to[0,\infty)$ satisfy the following conditions, for all $x, y \in X$ and all distinct $u, v \in X$ each of which is different from x and y:

(GMS1)
$d(x, y) = 0$ if and only if $x = y$,

(GMS2)
$d(x, y) = d(y, x)$,

(GMS3)
$d(x, y) \le d(x,u) + d(u, v) + d(v, y)$.
Then the map d is called a generalized metric and abbreviated as GM. Here, the pair $(X, d)$ is called a generalized metric space and abbreviated as GMS.
In the above definition, if d satisfies only (GMS1) and (GMS2), then it is called a semimetric (see, e.g., [18]).
A sequence $\{x_{n}\}$ in a GMS $(X,d)$ is GMS convergent to a limit x if and only if $d(x_{n}, x)\to0$ as $n\to\infty$.
A sequence $\{x_{n}\}$ in a GMS $(X, d)$ is GMS Cauchy if and only if for every $\epsilon> 0$ there exists a positive integer $N(\epsilon)$ such that $d(x_{n}, x_{m}) < \epsilon$, for all $n > m > N(\epsilon)$.
A GMS $(X,d)$ is called complete if every GMS Cauchy sequence in X is GMS convergent.
A mapping $T : (X, d)\to(X, d)$ is continuous if for any sequence $\{ x_{n}\}$ in X such that $d(x_{n}, x)\to0$ as $n\to\infty$, we have $d(Tx_{n}, Tx)\to0$ as $n\to\infty$.
The following assumption was suggested by Wilson [18] to replace the triangle inequality with the weakened condition.

(W)
For each pair of (distinct) points u, v, there is a number $r_{u,v} > 0$ such that for every $z \in X$, $r_{u,v} < d(u, z) + d(z, v)$.
Proposition 1.1
[20]
In a semimetric space, the assumption (W) is equivalent to the assertion that the limits are unique.
Proposition 1.2
[20]
Suppose that $\{x_{n}\}$ is a Cauchy sequence in a GMS $(X,d)$ with $\lim_{n\to\infty} d(x_{n},u) = 0$, where $u\in X$. Then $\lim_{n\to \infty} d(x_{n}, z) = d(u, z)$, for all $z\in X$. In particular, the sequence $\{x_{n}\}$ does not converge to z if $z \neq u$.
Definition 1.2
Let X be a nonempty set, $T :X\to X$ and $\alpha : X\times X\to[0,\infty)$ be two mappings. We say that T is an αadmissible mapping if $\alpha(x, y)\ge1$ implies $\alpha (Tx, Ty)\ge1$, for all $x, y \in X$.
Definition 1.3
Let $(X,d)$ be a GMS and $\alpha:X\times X\to[0,\infty)$. X is called αregular GMS if, for a sequence $\{x_{n}\}$ in X such that $x_{n}\to x$ and $\alpha(x_{n}, x_{n+1})\ge 1$, there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $\alpha(x_{n_{k}}, x)\ge 1$ for all $k \in\mathbb{N}$.
Throughout the paper, $F(T)$ denotes the set of fixed points of the mapping T.
Main results
The contraction mappings considered in this paper are constructed via auxiliary functions defined below. Let Ψ be a family of functions $\psi:[0,\infty) \to[0,\infty)$ satisfying the following properties:

(i)
ψ is upper semicontinuous, strictly increasing;

(ii)
$\{\psi^{n}(t)\}_{n\in\mathbb{N}}$ converges to 0 as $n\to\infty $, for all $t > 0$;

(iii)
$\psi(t) < t$, for every $t >0$.
Definition 2.1
Let $(X,d)$ be a GMS and $\alpha: X\times X\to[0,\infty)$. A self mapping $T:X\to X$ is said to be $(\alpha,\psi)$rational typeI contractive mapping if there exists a function $\psi\in\Psi$, such that for all $x,y \in X$ the following condition holds:
where
Next, we state and prove an existence and uniqueness theorem for fixed point of $(\alpha,\psi)$rational typeI contractive mappings.
Theorem 2.1
Let $(X, d)$ be a complete GMS, $T: X\to X$ be a self mapping and $\alpha:X\times X \to[0,\infty)$ a given function. Suppose that the following conditions are satisfied:

(i)
T is an αadmissible mapping;

(ii)
T is an $(\alpha,\psi)$rational typeI contractive mapping;

(iii)
there exists $x_{0}\in X$ such that $\alpha( x_{0}, T x_{0}) \ge1$, $\alpha( x_{0}, T^{2} x_{0}) \ge1$;

(iv)
either T is continuous, or X is αregular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$. Further, if for all $x, y\in F(T)$, we have $\alpha(x, y)\ge1$ then T has a unique fixed point in X.
Proof
Let $x_{0}\in X$ satisfies $\alpha( x_{0}, T x_{0}) \ge1$ and $\alpha( x_{0}, T^{2} x_{0}) \ge1$. We construct the sequence $\{ x_{n}\}$ in X as $x_{n}= T^{n} x_{0} = T x_{ n1}$, for $n\in\mathbb{N}$. It is obvious that if $x_{n_{0}}=x_{n_{0}+1}$, for some $n_{0}\in\mathbb {N}$, then $x_{n_{0}}$ is a fixed point of T. Consequently, we suppose that $x_{n}\ne x_{n+1}$ for all $n \in\mathbb{N}$.
Since T is αadmissible, $\alpha(x_{0},Tx_{0})=\alpha(x_{0},x_{1})\ge 1 \Longrightarrow$ $\alpha(Tx_{0},Tx_{1})=\alpha(x_{1},x_{2})\ge1 \Longrightarrow$ and thus, $\alpha(Tx_{1},Tx_{2})=\alpha(x_{2},x_{3})\ge1 \ldots$ , and hence by induction, we get $\alpha(x_{n},x_{n+1})\ge1$ for all $n\ge0$.
By similar arguments, since $\alpha( x_{0}, T^{2} x_{0}) \ge1$, we have $\alpha(x_{0},x_{2})=\alpha(x_{0},T^{2}x_{0})\ge1$, $\alpha(Tx_{0},Tx_{2})=\alpha(x_{1},x_{3})\ge1$. By induction, we get $\alpha (x_{n},x_{n+2})\ge1$ for all $n\ge0$. Consider (2.1) with $x=x_{n}$ and $y=x_{n+1}$. Clearly, we have
where
since $\frac{d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n},x_{n+1})}\leq d(x_{n+1},x_{n+2})$ and $\frac{d(x_{n},x_{n+1})d(x_{n+1},x_{n+2})}{1+d(x_{n+1},x_{n+2})}\leq d(x_{n},x_{n+1})$. If for some n, we have $M(x_{n},x_{n+1})=d(x_{n+1},x_{n+2})$, then
which is impossible. Hence, $M(x_{n},x_{n+1})=d(x_{n},x_{n+1})$, for all $n\in\mathbb{N}$,
From the property (iii) of ψ, we conclude that
for every $n \in\mathbb{N}$. Combining (2.4) and (2.5), we deduce $d(x_{n+1},x_{n+2}) \le\psi^{n}(d ( x_{0} , x_{1} ))$, for all $n \in\mathbb{N}$. Using the property (ii) of ψ, it is clear that
Consider now (2.1) with $x=x_{n1}$ and $y=x_{n+1}$. We have
where
From (2.5) we have $d(x_{n+1},x_{n+2})< d(x_{n1},x_{n})$. Define $a_{n} = d(x_{n}, x_{n+2})$ and $b_{n} = d(x_{n}, x_{n+1})$. Then
If $M(x_{n1},x_{n+1})= b_{n1}$, or $\frac {b_{n1}b_{n+1}}{1+a_{n1}}$ or $\frac{b_{n1}b_{n+1}}{1+a_{n}}$ then taking lim sup as $n\to\infty$ in (2.7) and using (2.6) and upper semicontinuity of ψ we get
and hence,
If $M(x_{n1},x_{n+1})= a_{n1}$, then (2.7) implies
due to the property (iii) of ψ. In other words, the sequence $\{ a_{n}\}$ is positive monotone decreasing, and hence, it converges to some $t\ge0$. Assume that $t > 0$. Now, by (2.7), we get
which is a contradiction. Therefore,
Now, we shall prove that $x_{n}\neq x_{m}$, for all $n \neq m$. Assume on the contrary that $x_{n}= x_{m}$, for some $m,n\in\mathbb{N}$ with $n \neq m$. Since $d(x_{p}, x_{p+1}) > 0$, for each $p \in\mathbb{N}$, without loss of generality, we may assume that $m > n + 1$. Substitute again $x=x_{n}=x_{m}$ and $y=x_{n+1}=x_{m+1}$ in (2.1), which yields
where
If $M(x_{m1},x_{m})=d(x_{m1},x_{m})$, then (2.10) implies
If on the other hand $M(x_{m1},x_{m})=d(x_{m},x_{m+1})$, then from (2.10) we have
Using the property (iii) of ψ, the two inequalities (2.12) and (2.13) imply
which is impossible.
Now, we shall prove that $\{ x_{n}\}$ is a Cauchy sequence, that is, $\lim_{n\to\infty}d(x_{n},x_{n+k})=0$, for all $k\in\mathbb{N}$. We have already proved the cases for $k=1$ and $k=2$ in (2.6) and (2.9), respectively. Take arbitrary $k\ge3$. We discuss two cases.
Case 1. Suppose that $k=2m+1$, where $m\ge1$. Using the quadrilateral inequality (GMS3), we have
Case 2. Suppose that $k=2m$, where $m\ge2$. Again, by applying the quadrilateral inequality, we have
since $\lim_{n\to\infty}=0$ because of (2.9). In both of the above cases, we have $\lim_{n\to\infty }d(x_{n},x_{n+k})=0$, for all $k\ge3$. Hence we conclude that $\{x_{n}\}$ is a Cauchy sequence in $(X,d)$. Since $(X,d)$ is complete, there exists $x^{*} \in X$ such that
We will show next that the limit $x^{*}$ of the sequence $\{x_{n}\}$ is a fixed point of T. First, we suppose that T is continuous. Then from (2.16) we have
Due to Proposition 1.2, we conclude that
that is, $x^{*}$ is a fixed point of T.
Now, we suppose that X is αregular. Then there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $\alpha(x_{n_{k}1}, x^{*}) \ge1$ for all $k \in\mathbb {N}$. Now, from inequality (2.1) with $x = x_{n_{k}}$ and $y=x^{*}$, we obtain
where
Letting $k\to\infty$ in (2.18), we obtain $M(x_{n_{k}},x^{*})=d(x^{*},Tx^{*})$. Therefore, upon taking the limit as $k\to\infty$, in inequality (2.17), we have $d(x^{*},Tx^{*})\le\psi(d(x^{*},Tx^{*}))< d(x^{*},Tx^{*})$, which implies $x^{*}=Tx^{*}$, that is, $x^{*}$ is a fixed point of T.
Finally, suppose that $x^{*}$ and $y^{*}$ are two fixed points of T such that $x^{*}\ne y^{*}$. Then by the hypothesis, $\alpha(x^{*}, y^{*})\ge1$. Hence, from (2.1) with $x=x^{*}$ and $y=y^{*}$ we have
where
Hence, we get $d(x^{*},y^{*})\le\psi(d(x^{*},y^{*}))< d(x^{*},y^{*})$, which is possible only if $d(x^{*},y^{*})=0$, that is, $x^{*}=y^{*}$. Hence T has a unique fixed point. □
Definition 2.2
Let $(X,d)$ be a generalized metric space and $\alpha: X\times X\to \mathbb{R}^{+}$. A mapping $T: X\to X$ is said to be $(\alpha,\psi )$rational typeII contractive mapping if there exists a $\psi\in\Psi $, such that, for all $x,y \in X$, the following condition holds:
where
For this class of mappings we state a similar existence and uniqueness theorem.
Theorem 2.2
Let $(X, d)$ be a complete generalized metric space, $T: X\to X$ be a self mapping, and $\alpha:X\times X \to\mathbb{R}$. Suppose that the following conditions are satisfied:

(i)
T is an αadmissible mapping;

(ii)
T is an $(\alpha,\psi)$rational typeII contractive mapping;

(iii)
there exists $x_{0}\in X$ such that $\alpha( x_{0}, T x_{0}) \ge1$ and $\alpha( x_{0}, T^{2} x_{0}) \ge1$;

(iv)
either T is continuous, or X is αregular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$. Further, if for all $x, y\in F(T)$, we have $\alpha(x, y)\ge1$, then T has a unique fixed point in X.
Proof
The proof can be done by following the lines of the proof of Theorem 2.1. □
The following example illustrating Theorem 2.1 is inspired by [4].
Example 2.1
Let X be a finite set defined as $X=\{1,2,3,4\}$. Define $d:X\times X\to[0,\infty)$ as:
The function d is not a metric on X. Indeed, note that
that is, the triangle inequality is not satisfied. However, d is a generalized metric on X and, moreover, $(X,d)$ is a complete generalized metric space. Define $T: X \to X$ as
$\alpha(x,y)$ as $\alpha(x,y)=1$ and $\psi(t)=\frac{t}{2}$. Then, for $x=1,2,3$ and $y=1,2,3$, we have
On the other hand, for $x=1,2,3$ and $y=4$ we obtain
and
and hence
For $x=4$, $y=4$, the contraction condition is obvious. Clearly, T satisfies the conditions of Theorem 2.1 and has a unique fixed point $x=2$.
Some consequences
In this section we give some consequences of the main results presented above. Specifically, we apply our results to generalized metric spaces endowed with a partial order.
Definition 3.1
Let $(X,\preceq)$ be a partially ordered set. A mapping $T:X\to X$ is said to be nondecreasing with respect to ⪯ if for every $x,y\in X$ $x\preceq y$ implies $Tx\preceq Ty$.
Definition 3.2
Let $(X,d,\preceq)$ be a partially ordered GMS. X is called regular GMS if, whenever $\{x_{n}\}$ is a sequence in X such that $x_{n}\to x$ and $x_{n}\preceq x_{n+1}$, then there exists a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $x_{n_{k}}\preceq x$ for all $k \in\mathbb{N}$.
Theorem 3.1
Let $(X, d, \preceq)$ be a partially ordered complete generalized metric space and $T:X\to X$ be a nondecreasing self mapping. Suppose that the following conditions are satisfied:

(i)
There exists a function $\psi\in\Psi$ for which
$$ d(Tx,Ty)\le\psi\bigl(M(x,y)\bigr), $$(3.1)where
$$M(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac {d(x,Tx)d(y,Ty)}{1+d(x,y)},\frac{d(x,Tx)d(y,Ty)}{1+d(Tx,Ty)} \biggr\} $$for all $x,y\in X$ with $x\preceq y$.

(ii)
There exists $x_{0}\in X$ such that $x_{0}\preceq Tx_{0}$ and $x_{0}\preceq T^{2}x_{0}$.

(iii)
Either T is continuous, or X is regular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$.
Proof
Define a mapping $\alpha: X\times X\to[0,\infty)$ as follows.
Then the existence conditions of Theorem 2.1 hold and hence T has a fixed point which is the limit of the sequence $\{T^{n}x_{0}\}$. □
Theorem 3.2
Let $(X, d, \preceq)$ be a partially ordered complete generalized metric space and $T:X\to X$ be a nondecreasing self mapping. Suppose that the following conditions are satisfied:

(i)
There exist a function $\psi\in\Psi$ for which
$$ d(Tx,Ty)\le\psi\bigl(M(x,y)\bigr), $$(3.2)where
$$\begin{aligned} M(x,y)={}&\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty), \\ &{} \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)+d(x,Ty)+d(y,Tx)}, \frac{d(x,Ty)d(x,y)}{1+d(x,Tx)+d(y,Tx)+d(y,Ty)}\biggr\} \end{aligned}$$for all $x,y\in X$ with $x\preceq y$;

(ii)
there exists $x_{0}\in X$ such that $x_{0}\preceq Tx_{0}$ and $x_{0}\preceq T^{2}x_{0}$;

(iii)
either T is continuous, or X is regular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$.
Proof
Employing again a mapping $\alpha: X\times X\to[0,\infty)$ defined as
we observe that the existence conditions of Theorem 2.2 hold and hence, T has a fixed point which is the limit of the sequence $\{ T^{n}x_{0}\}$. □
Several particular cases can also be deduced from the above results.
Corollary 3.1
Let $(X, d)$ be a complete generalized metric space, T be a self mapping, $T: X\to X$, and $\alpha:X\times X \to\mathbb{R}$. Suppose that the following conditions are satisfied:

(i)
T is an αadmissible mapping;

(ii)
T satisfies
$$ d(Tx,Ty)\le kM(x,y), $$(3.3)where
$$\begin{aligned} M(x,y)={}&\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty), \\ & {} \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)+d(x,Ty)+d(y,Tx)}, \frac{d(x,Ty)d(x,y)}{1+d(x,Tx)+d(y,Tx)+d(y,Ty)}\biggr\} \end{aligned}$$for some $k\in[0,1)$;

(iii)
there exists $x_{0}\in X$ such that $\alpha( x_{0}, T x_{0}) \ge1$, $\alpha(x_{0}, T^{2}x_{0}) \ge1$;

(iv)
either T is continuous, or X is αregular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$. Further, if, for all $x, y\in F(T)$, we have $\alpha(x, y)\ge1$, then T has a unique fixed point in X.
Proof
Define $\psi(t)=kt$. Clearly, $\psi\in\Psi$. By Theorem 2.2, T has a unique fixed point. □
Corollary 3.2
Let $(X, d,\preceq)$ be a partially ordered complete generalized metric space and $T: X\to X$ be a nondecreasing mapping. Suppose that the following conditions are satisfied:

(i)
$$ d(Tx,Ty)\le kM(x,y), $$(3.4)
where
$$\begin{aligned} M(x,y)={}&\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty), \\ & {} \frac{d(x,Tx)d(y,Ty)}{1+d(x,y)+d(x,Ty)+d(y,Tx)}, \frac{d(x,Ty)d(x,y)}{1+d(x,Tx)+d(y,Tx)+d(y,Ty)}\biggr\} \end{aligned}$$for all $x,y\in X$ with $x\preceq y$ and some $k\in[0,1)$;

(ii)
there exists $x_{0}\in X$ such that $x_{0}\preceq Tx_{0}$ and $x_{0}\preceq T^{2}x_{0}$;

(iii)
either T is continuous, or X is regular.
Then T has a fixed point $x^{*}\in X$ and $\{T^{n}x_{0}\}$ converges to $x^{*}$.
Proof
Define $\alpha: X\times X\to[0,\infty)$ as
Corollary 3.1 implies that T has a fixed point. □
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MSC
 47H10
 54H25
Keywords
 fixed point
 αadmissible
 contraction mappings
 metric space