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Fixed point theorems for MeirKeeler type mappings in Mmetric spaces with applications
Fixed Point Theory and Applicationsvolume 2015, Article number: 210 (2015)
Abstract
In this paper, we establish some fixed point theorems for a MeirKeeler type contraction in Mmetric spaces via GuptaSaxena type contraction. Also, we extend and improve very recent results in fixed point theory.
Introduction and preliminaries
Ekeland formulated a variational principle that is the foundation of modern variational calculus, having applications in many branches of mathematics, including optimization and fixed point theory [1] and applications in nonlinear analysis, since it entails the existence of approximate solutions of minimization problems for a lower semicontinuous function that is bounded from below on complete metric spaces. Also, Ekeland’s variational principle is a fruitful tool in simplifying and unifying the proofs of already known theorems and has many generalizations; see Borwein and Zhu [2].
Matthews in 1994 [3] introduced a partial metric space and proved the contraction principle of Banach in this new framework. Afterward, by several mathematicians many fixed point theorems were founded in partial metric spaces. Recently Haghi et al. [4] published a paper which stated that we should ‘be careful on partial metric fixed point results’ along with very some results therein. They showed that fixed point generalizations to partial metric spaces can be obtained from the corresponding results in metric spaces.
In 2014, Asadi et al. [5] introduced the Mmetric space, which extends the pmetric space and certain fixed point theorems obtained therein.
In this paper, we establish some of the fixed point theorem for a MeirKeeler type contraction in Mmetric spaces via a GuptaSaxena type contraction. Also, we extend and improve very recent results in fixed point theory.
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.
Notation
The following notations are useful in the sequel:

(i)
$m_{xy}:=\min\{m(x,x), m(y,y)\}=m(x,x)\vee m(y,y)$,

(ii)
$M_{xy}:=\max\{m(x,x), m(y,y)\}=m(x,x)\wedge m(y,y)$.
Now we want to extend Definition 1.1 as follows.
Definition 1.2
Let X be a nonempty set. A function $m :X \times X\to\mathbb{R}^{+}$ is called a mmetric if the following conditions are satisfied:

(m1)
$m(x,x)=m(y,y)=m(x,y)\iff x=y$,

(m2)
$m_{xy}\leq m(x,y)$,

(m3)
$m(x, y)=m(y,x)$,

(m4)
$(m(x, y)m_{xy} )\leq (m(x,z)m_{xz} )+ (m(z,y)m_{zy} )$.
Then the pair $(X,m)$ is called an Mmetric space.
According to the above, our definition of the condition (p1) in the definition [3] changes to (m1) and (p2) for $p(x,x)$ is expressed by just $p(y,y)=0$; we may have $p(y,y)\neq0$, so we improved that condition by replacing it by $\min\{p(x,x), p(y,y)\}\leq p(x,y)$, and also we improved the condition (p4) to the form (m4). In the sequel we present an example that holds for the mmetric, but not for the pmetric.
Remark 1.1
For every $x,y\in X$:

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

(ii)
$0\leq M_{xy}m_{xy} = m(x,x)m(y,y)$,

(iii)
$M_{xy}m_{xy} \leq(M_{xz}m_{xz}) +(M_{zy}m_{zy})$.
The next examples state that $m^{s}$ and $m^{w}$ are ordinary metrics.
Example 1.1
Let m be a mmetric. Put:

(i)
$m^{w}(x,y)=m(x,y)2m_{xy}+M_{xy}$,

(ii)
$m^{s}(x,y)=m(x,y)m_{xy}$ when $x\neq y$ and $m^{s}(x,y)=0$ if $x=y$.
Then $m^{w}$ and $m^{s}$ are ordinary metrics.
Proof
If $m^{w}(x,y)=0$, then
But from (1) and $m_{xy}\leq m(x,y)$ we get $m_{xy}=M_{xy}=m(x,x)=m(y,y)$, so by (1) we obtain $m(x,y)=m(x,x)=m(y,y)$, and therefore $x=y$. For the triangle inequality it is enough that we consider Remark 1.1 and (m4). □
In the following example, we present an example of a mmetric which is not a pmetric.
Remark 1.2
For every $x,y\in X$:

(i)
$m(x,y)M_{xy}\leq m^{w}(x,y)\leq m(x,y)+M_{xy}$,

(ii)
$(m(x,y)M_{xy})\leq m^{s}(x,y)\leq m(x,y)$.
Example 1.2
Let $X=\{1,2,3\}$. Define
so m is an mmetric but m is not a pmetric. Since $m(2,2) \nleq m(1,2)$, m is not a pmetric. If $D(x,y)=m(x,y)m_{x,y}$ then $m(1,2)=m_{1,2}=8$ but it means $D(1,2)=0$, while $1\neq2$ means D is not a metric.
Example 1.3
([5])
Let $(X,d)$ be a metric space, $\phi:[0,\infty)\to[\phi(0),\infty)$ be a one to one and nondecreasing or strictly increasing mapping with $\phi (0)$, defined such that
Then $m(x,y)=\phi(d(x,y))$ is an mmetric.
Example 1.4
Let $(X,d)$ be a metric space. Then $m(x,y)=ad(x,y)+b$ where $a,b>0$ is an mmetric, because we can put $\phi(t)=at+b$.
Remark 1.3
According to the Example 1.4, by the Banach contraction
we have
which does not imply that we have the ordinary Banach contraction
for all selfmaps T on X. So this states that if the mmetric m and the ordinary metric d even have the same topology, but the Banach contraction of an mmetric, this does not imply the Banach contraction of the ordinary metric d.
Lemma 1.1
([5])
Every pmetric is an mmetric.
Topology for Mmetric space
It is clear that each mmetric p on X generates a $T_{0}$ topology $\tau_{m}$ on X. The set
where
for all $x \in X$ and $\varepsilon> 0$, forms the base of $\tau_{m}$.
Definition 2.1
Let $(X,m)$ be an Mmetric space. Then:

(1)
A sequence $\{x_{n}\}$ in an Mmetric space $(X, m)$ converges to a point $x \in X$ if and only if
$$ \lim_{n\to\infty} \bigl(m(x_{n}, x)m_{x_{n},x}\bigr)=0. $$(2) 
(2)
A sequence $\{x_{n}\}$ in an Mmetric space $(X, m)$ is called an mCauchy sequence if
$$ \lim_{n,m\to \infty}\bigl(m(x_{n}, x_{m})m_{x_{n},x_{m}}\bigr)\quad \mbox{and}\quad \lim _{n,m\to\infty}(M_{x_{n}, x_{m}}m_{x_{n},x_{m}}) $$(3)in this space exist (and are finite).

(3)
An Mmetric space $(X, m)$ is said to be complete if every mCauchy sequence $\{x_{n}\}$ in X converges, with respect to $\tau _{m}$, to a point $x\in X$ such that
$$\Bigl(\lim_{n\to\infty}\bigl(m(x_{n}, x)m_{x_{n},x} \bigr)=0 \mbox{ and } \lim_{n\to\infty}(M_{x_{n}, x}m_{x_{n},x})=0 \Bigr). $$
Lemma 2.1
Let $(X, m)$ be an Mmetric space. Then:

(1)
$\{x_{n}\}$ is a mCauchy sequence in $(X, m)$ if and only if it is a Cauchy sequence in the metric space $(X, m^{w})$.

(2)
An Mmetric space $(X, m)$ is complete if and only if the metric space $(X, m^{w})$ is complete. Furthermore,
$$\lim_{n\to\infty}m^{w}(x_{n}, x) = 0\ \iff\ \Bigl(\lim_{n\to\infty}\bigl(m(x_{n}, x)m_{x_{n},x}\bigr)=0 \textit{ and } \lim_{n\to\infty}(M_{x_{n}, x}m_{x_{n},x})=0 \Bigr). $$
Likewise the above definition holds also for $m^{s}$.
Lemma 2.2
Assume that $x_{n}\to x$ and $y_{n}\to y$ as $n\to\infty$ in an Mmetric space $(X, m)$. Then
Proof
We have
□
From Lemma 2.2 we can deduce the following lemma.
Lemma 2.3
Assume that $x_{n}\to x$ as $n\to\infty$ in an Mmetric space $(X, m)$. Then
for all $y\in X$.
Lemma 2.4
Assume that $x_{n}\to x$ and $x_{n}\to y$ as $n\to\infty$ in an Mmetric space $(X, m)$. Then $m(x,y)=m_{xy}$. Further if $m(x,x)=m(y,y)$, then $x=y$.
Proof
By Lemma 2.2 we have
□
Lemma 2.5
Let $\{x_{n}\}$ be a sequence in an Mmetric space $(X,m)$, such that
Then

(A)
$\lim_{n\to\infty}m(x_{n},x_{n1})=0$,

(B)
$\lim_{n\to\infty}m(x_{n},x_{n})=0$,

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

(D)
$\{x_{n}\}$ is an mCauchy sequence.
Proof
From (4) we have,
thus
which implies (A).
To prove (B), from (m2) and (A) we have
That is, (B) holds.
Clearly, (C) holds, since $\lim_{n\to\infty}m(x_{n},x_{n})=0$. □
Theorem 2.1
The topology $\tau_{m}$ is not Hausdorff.
Theorem 2.2
Let $(X,m)$ be a complete Mmetric space and $T: X\to X$ be mapping satisfying the following condition:
Then T has a unique fixed point.
Theorem 2.3
Let $(X,m)$ be a complete Mmetric space and $T: X\to X$ be mapping satisfying the following condition:
Then T has a unique fixed point.
Main result and fixed point theorems
The following definition is new version of the definition in [7] for an Mmetric space.
Definition 3.1
A MeirKeeler mapping is a mapping $T:M\to M$ on an Mmetric space $(X,M)$ such that
Theorem 3.1
Let $(X,m)$ be a complete Mmetric space and let T be a mapping from X into itself satisfying the following condition:
Then T has a unique fixed point $u \in X$. Moreover, for all $x\in X$, the sequence $\{T_{n} (x)\}$ converges to u.
Proof
We first observe that (7) trivially implies that T is a strict contraction, i.e.,
Let $x_{0}\in X$ and $x_{n}:=Tx_{n1}$, so we have
So the sequence $\{m(x_{n},x_{n1})\}$ is bounded below and decreasing; thus $m(x_{n},x_{n1})\to m$ for some $m\in\Bbb{R}^{+}$. Let $m>0$, therefore $m(x_{n},x_{n1})\geq m$. On the other hand for $m>0$ there exists $\delta(m)>0$ such that
which implies that it is contradiction; so $m=0$, i.e.,
and
since, $\lim_{n\to\infty}m(x_{n},x_{n})=0$. Now we want to show that $\lim_{m,n\to\infty}m(x_{m},x_{n})=0$. Let it be untrue. So for some $\varepsilon>0$ we have $\limsup_{m,n\to\infty }m(x_{m},x_{n})>2\varepsilon$. Also, by hypothesis, there exists a $\delta > 0$, such that
which remains true with δ replaced by $\delta' =\min\{\delta ,\varepsilon\}$. Now by (10)
and for $m,n>N$, $m(x_{m},x_{n})>2\varepsilon$. This implies, since
that there exists i with $m< i< n$ with
However, for all m and i,
which contradicts (12). So by (11) and $\lim_{m,n\to\infty }m(x_{m},x_{n})=0$ we see that the sequence $\{x_{n}\}$ is a Cauchy sequence and by completeness of X, $x_{n}\to x^{*}$ in m for some $x^{*}\in X$, i.e.,
But $m_{x_{n},x^{*}}\to0$ because $m(x_{n}, x_{n})\to0$ so $m(x_{n}, x^{*})\to 0$. Thus, by hypothesis, $m(Tx_{n}, Tx^{*})\leq m(x_{n},x^{*}) \to0$. Hence by (m2) $m_{Tx_{n},Tx^{*}}\leq m(Tx_{n},Tx^{*})\to0$, so by (2) $Tx_{n}\to Tx^{*}$.
Equation (10) implies that $m(x_{n},Tx_{n})\to0$. Since $m_{x_{n},Tx_{n}}\to0$, by Lemma 2.2, we get $m(x^{*},Tx^{*})=m_{x^{*},Tx^{*}}$.
On the other hand, by Lemma 2.2 and
we have
thus $m(x^{*},x^{*})=m_{x^{*},Tx^{*}}=m(Tx^{*},Tx^{*})$ and since
now by (m1) $x^{*}=Tx^{*}$. Uniqueness by the contraction (8) is clear. □
Put
Theorem 3.2
Let $(X,m)$ be a complete Mmetric space and let T be a continuous mapping from X into itself satisfying the following condition:
for some $0< k<\frac{1}{3}$. Then T has a unique fixed point $u \in X$. Moreover, for all $x\in X$, the sequence $\{T_{n} (x)\}$ converges to u.
Proof
We first observe that (14) trivially implies that T is a strict contraction, i.e.,
Let $x_{0}\in X$ and $x_{n}:=Tx_{n1}$ so we have
therefore
where $r=\frac{k}{12k}<1$. Now by Lemma 2.5, $\{x_{n}\}$ is a Cauchy sequence, and by completeness of X, $Tx_{n1}=x_{n}\to x^{*}$ in m for some $x^{*}\in X$. Since T is a continuous mapping, so $x_{n}=Tx_{n1}\to Tx^{*}$, in m now by Lemma 2.4 we find
by Lemma 2.2 and
So $x^{*}=Tx^{*}$. Uniqueness by the contraction (15) is clear. □
Corollary 3.1
(Gupta and Saxena [8])
Let $(X,d)$ be a complete metric space and T be a continuous mapping from X into itself. Assume that T satisfies
where $k \in(0, \frac{1}{3})$ is a constant. Then T has a unique fixed point $u \in X$. Moreover, for all $x\in X$, the sequence $\{T_{n} (x)\}$ converges to u.
Applications
In this section, after an idea of Samet et al. [9], we shall state an integral version of the GuptaSaxena result.
Theorem 4.1
Let $(X,m)$ be an Mmetric space and let T be a selfmapping defined on X. Assume that there exists a function $\varphi:[0,\infty)\to [0,\infty)$ satisfying the following:

(1)
$\varphi(0)=0$ and $t>0\Rightarrow\varphi(t)>0$;

(2)
φ is nondecreasing and right continuous;

(3)
for every $\varepsilon>0$, there exists $\delta>0$ such that
$$ \varepsilon\leq\varphi\bigl(kC(x,y)\bigr)< \varepsilon+\delta \Rightarrow\varphi \bigl(m(Tx,Ty)\bigr)< \varepsilon, $$(17)for some $0< k<\frac{1}{3}$ and for all $x,y\in X$ with $x\neq y$.
Then (14) is satisfied.
Proof
Fix $\varepsilon>0$, so $\varphi(\varepsilon) > 0$. Hence by (17) there exists $\delta_{1}>0$ such that
According to the right continuity of φ
Now for $x,y\in X$ with $x\neq y$ and fixed
since φ is a nondecreasing mapping, we have
So we get
which implies that $m(Tx,Ty)<\varepsilon$. □
Corollary 4.1
Let $(X,m)$ be an Mmetric space and let T be a selfmapping defined on X. Let $h :[0,\infty)\to[0,\infty)$ be a locally integrable function such that

(1)
$t>0\Rightarrow\int_{0}^{t} h(s)\,ds>0$;

(2)
for every $\varepsilon>0$, there exists $\delta>0$ such that
$$\frac{1}{k}\varepsilon\leq\int_{0}^{C(x,y)}h(s)\,ds< \frac {1}{k}\varepsilon+\delta\Rightarrow\int_{0}^{ \frac {1}{k}m(Tx,Ty)}h(s)\,ds< \frac{1}{k}\varepsilon, $$(20)for some $0< k<\frac{1}{3}$ and for all $x,y\in X$ with $x\neq y$.
Then (14) is satisfied.
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Acknowledgements
This research was supported by the Zanjan Branch, Islamic Azad University, Zanjan, Iran. The author would like to acknowledge this support. The author expresses deep gratitude to the referee for his/her valuable comments and suggestions.
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MSC
 47H10
 54H25
Keywords
 fixed point
 partial metric space
 Mmetric space