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Fixed points of generalized MeirKeeler contraction mappings in bmetriclike spaces
Fixed Point Theory and Applications volume 2016, Article number: 34 (2016)
Abstract
In this paper, we introduce the notion of generalized MeirKeeler contraction mappings in the setup of bmetriclike spaces. Then we establish some fixed point results for this class of contractions. We also provide some examples to verify the effectiveness and applicability of our main results.
Introduction and preliminaries
In 1969, Meir and Keeler [1] proved the following very attractive fixed point theorem, which is a generalization of the Banach contraction principle [2].
Definition 1.1
[1]
Let \((X,d)\) be a metric space. Then a mapping T on X is said to be a MeirKeeler contraction (MKC, for short) if for any \(\varepsilon>0\), there exists \(\delta>0\) such that
for all \(x,y\in X\).
Theorem 1.2
[1]
Let \((X,d)\) be a complete metric space. If \(T:X\rightarrow X \) is a MeirKeeler contraction, then T has a unique fixed point.
Alsulami et al. [3] defined two types of generalized αadmissible [4] MeirKeeler contractions and proved some fixed point theorems for these kinds of mappings. MeirKeeler contraction has many generalizations in the area studied by some scholars (cf. e.g. [5–8]).
On the other hand, AminiHarandi [9] presented a new extension of the concept of the partial metric space [10], called a metriclike space. The concept of bmetriclike space which generalizes the notions of partial metric space, metriclike space and bmetric space [11] was introduced by Alghamdi et al. in [12]. They established some fixed point theorems in partial metric spaces, bmetric spaces and bmetriclike spaces. It is well known that all these spaces are generalization of the usual metric spaces. There are several types of generalized metric spaces [13, 14], introduced by modifying and improving metric axioms. These generalized metric spaces often appear to be metrizable and the contraction conditions may be preserved under special transforms. Hence the fixed point theory in such spaces may be a consequence of the fixed point theory in certain metric spaces. However, it is not true that all generalized fixed point results become evident in this way. More precisely, these results are based on some contractive conditions, and some of these conditions do not remain authentic when one considers the problem in the associated metric space [15].
In the present work, using the concepts of MeirKeeler contractions and bmetriclike spaces, we define a new concept of generalized MeirKeeler contraction on a bmetriclike space. Then we investigate some fixed point results for this class of contractions. We give an example that shows that our results in bmetriclike spaces may not be deduced from certain ones in bmetric spaces. We also provide some examples to support the usability of our results.
It is convenient and, more importantly helpful to recall some basic definitions and facts which will be used further on. Throughout this paper, we denote by \(\mathbb{R}^{+}\) the set of nonnegative real numbers.
Definition 1.3
[11]
Let X be a nonempty set, and let the function \(d : X \times X\rightarrow\mathbb{R}^{+}\) satisfies:
 (b1):

\(d(x,y) = 0\) if and only if \(x = y\),
 (b2):

\(d(x,y) = d(y,x)\) for all \(x,y\in X\),
 (b3):

there exists a real number \(s \geq1\) such that \(d(x,z) \leq s[d(x,y) + d(y,z)]\) for all \(x,y,z\in X\).
Definition 1.4
[16]
A partial bmetric on a nonempty set X is a mapping \(p_{b}:X\times X\rightarrow\mathbb{R}^{+}\) such that for some real number \(s\ge1\) and all \(x,y,z\in X\):
 (p_{ b }1):

\(x=y\) if and only if \(p_{b}(x,x)=p_{b}(x,y)=p_{b}(y,y)\),
 (p_{ b }2):

\(p_{b}(x,x)\leq p_{b}(x,y)\),
 (p_{ b }3):

\(p_{b}(x,y)=p_{b}(y,x)\),
 (p_{ b }4):

\(p_{b}(x,y)\leq s[p_{b}(x,z)+p_{b}(z,y)]p_{b}(z,z)\).
A pair \((X,p_{b})\) is called a partial bmetric space, if X is a nonempty set and \(p_{b}\) is a partial bmetric on X. The number s is called the coefficient of \((X,p_{b})\).
It is clear that if in Definitions 1.3 and 1.4 \(s=1\), then they are the usual metric and partial metric space, respectively.
Definition 1.5
[9]
A metriclike on a nonempty set X is a mapping \(\sigma:X\times X\rightarrow\mathbb{R}^{+}\) such that for all \(x,y,z\in X\):
 (σ1):

\(\sigma(x,y)=0\) implies \(x=y\),
 (σ2):

\(\sigma(x,y)=\sigma(y,x)\),
 (σ3):

\(\sigma(x,y)\leq\sigma(x,z)+\sigma(z,y)\).
The pair \((X,\sigma)\) is called a metriclike space.
Example 1.6
[17]
Let \(X=[0,1]\). Then the mapping \(\sigma_{1}:X\times X\rightarrow\mathbb{R}^{+}\) defined by \(\sigma_{1}(x,y)=x+yxy\) is a metriclike on X.
Example 1.7
[17]
Let \(X=\mathbb{R}\), then the mappings \(\sigma_{i}:X\times X\rightarrow\mathbb{R}^{+}\) (\(i\in\{2,3,4\}\)) defined by
are metricslike on X, where \(a\geq0\) and \(b\in\mathbb{R}\).
Definition 1.8
[12]
Let X be a nonempty set and \(s\geq1\) be a given real number. A function \(\sigma_{b}:X\times X\rightarrow\mathbb{R}^{+}\) is a bmetriclike if, for all \(x,y,z\in X\), the following conditions are satisfied:
 (\(\sigma_{b}1\)):

\(\sigma_{b}(x,y)=0\) implies \(x=y\),
 (\(\sigma_{b}2\)):

\(\sigma_{b}(x,y)=\sigma_{b}(y,x)\),
 (\(\sigma_{b}3\)):

\(\sigma_{b}(x,y)\leq s[\sigma_{b}(x,z)+\sigma_{b}(z,y)]\).
A bmetriclike space is a pair \((X,\sigma_{b})\) such that X is a nonempty set and \(\sigma_{b}\) is a bmetriclike on X. The number s is called the coefficient of \((X,\sigma_{b})\).
Some examples of bmetriclike spaces can be constructed with the help of the following proposition.
Proposition 1.9
[18]
Let \((X,\sigma)\) be a metriclike space and \(\sigma_{b}(x,y)=[\sigma(x,y)]^{p}\), where \(p>1\). Then \(\sigma_{b}\) is a bmetriclike with coefficient \(s=2^{p1}\).
Every partial bmetric space is a bmetriclike space with the same coefficient s and every bmetric space is also a bmetriclike space with the same coefficient s. However, the converses of these facts need not hold. For instance, assume that \(p>1\), then \(\sigma_{1}^{p}\) and \(\sigma_{4}^{p}\) are bmetricslike, but \(\sigma_{1}^{p}\) is not bmetric and \(\sigma_{4}^{p}\) is not partial bmetric.
Each bmetriclike \(\sigma_{b}\) on a nonempty set X generates a topology \(\tau_{\sigma_{b}}\) on X whose base is the family of open \(\sigma_{b}\)balls \(\{B_{\sigma_{b}}(x,\varepsilon):x\in X,\varepsilon >0\}\) where \(B_{\sigma_{b}}(x,\varepsilon)=\{y\in X:\sigma_{b}(x,y)\sigma_{b}(x,x)<\varepsilon\}\) for all \(x\in X\) and \(\varepsilon>0\).
Now, we recall the concepts of Cauchy sequence and convergent sequence in the framework of bmetriclike spaces.
Definition 1.10
[12]
Let \((X,\sigma_{b})\) be a bmetriclike space with coefficient s, \(\{x_{n}\}\) be any sequence in X and \(x\in X\). Then:

(i)
The sequence \(\{x_{n}\}\) is said to be convergent to x with respect to \(\tau_{\sigma_{b}}\) if \(\lim_{n\rightarrow \infty}\sigma_{b}(x_{n},x)=\sigma_{b}(x,x)\).

(ii)
The sequence \(\{x_{n}\}\) is said to be a Cauchy sequence in \((X,\sigma_{b})\), if \(\lim_{n,m\rightarrow \infty}\sigma_{b}(x_{n},x_{m})\) exists and is finite.

(iii)
\((X,\sigma_{b})\) is said to be a complete bmetriclike space if for every Cauchy sequence \(\{x_{n}\}\) in X there exists \(x\in X \) such that
$$ \lim_{n,m\rightarrow\infty }\sigma_{b}(x_{n},x_{m})= \lim_{n\rightarrow\infty }\sigma_{b}(x_{n},x)= \sigma_{b}(x,x). $$
Note that in a bmetriclike space the limit of a convergent sequence may not be unique (since already partial metric spaces share this property).
Definition 1.11
[19]
Suppose that \((X,\sigma_{b})\) is a bmetriclike space. A mapping \(T:X \rightarrow X\) is said to be continuous at a point \(x\in X\), if for every \(\varepsilon>0\) there exists a \(\delta>0\) such that \(T(B_{\sigma_{b}}(x,\delta))\subseteq B_{\sigma_{b}}(Tx,\varepsilon)\). The mapping T is continuous on X if it is continuous at each point x in X.
Note that if \(T:X \rightarrow X\) is a continuous mapping and \(\{x_{n}\}\) is a sequence in X with \(\lim_{n\rightarrow \infty}\sigma_{b}(x_{n},x)=\sigma_{b}(x,x)\), then \(\lim_{n\rightarrow \infty}\sigma_{b}(Tx_{n},Tx)=\sigma_{b}(Tx,Tx)\).
Samet et al. in [4] introduced the concept of αadmissible mappings and established some new fixed point theorems for these mappings. Thereafter, many researchers improved and generalized fixed point results by using this notion for single valued and multivalued mappings (cf. e.g. [3, 7, 20] for details).
Definition 1.12
[3]
Let X be a nonempty set, \(T:X\rightarrow X\) be a mapping and \(\alpha:X\times X\rightarrow[0,\infty)\) be a function. Then f is said to be αadmissible if for all \(x,y\in X\) we have
Definition 1.13
[3]
A mapping \(T:X\rightarrow X\) is called triangular αadmissible if it is αadmissible and, moreover, it satisfies the following implication:
where \(x,y,z\in X\).
The following lemma is useful in proving our main results, stated and proved according to [7], Lemma 7.
Lemma 1.14
Let \((X,\sigma_{b})\) be a bmetriclike space and \(T:X\rightarrow X\) be a triangular αadmissible mapping. Assume that there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\) and \(\alpha(Tx_{0},x_{0})\geq 1\). If \(x_{n}=T^{n}x_{0}\), then \(\alpha(x_{m},x_{n})\geq1\) for all \(m,n\in\mathbb{N}\).
Main results
In this section, first we describe the concept of generalized MeirKeeler contraction on a bmetriclike space which can be regarded as an extension of the MeirKeeler contractions defined in [1]. Then we demonstrate some fixed point results for this class of contractions.
Definition 2.1
Suppose that \((X,\sigma_{b})\) is a bmetriclike space with coefficient s. A triangular αadmissible mapping \(T:X\rightarrow X\) is said to be generalized MeirKeeler contraction if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
for all \(x,y\in X\) where \(\beta:[0,\infty)\rightarrow(0,\frac{1}{s})\) is a given function.
Remark 2.2
Let T be a generalized MeirKeeler contractive mapping. Then it is intuitively clear that
for all \(x,y\in X\) when \(x\neq y\).
We are now in a position to define two types of generalized MeirKeeler contractions on bmetriclike spaces, say type (I) and type (II).
Definition 2.3
Let \((X,\sigma_{b})\) be a bmetriclike space with coefficient s. A triangular αadmissible mapping \(T:X\rightarrow X\) is said to be generalized MeirKeeler contraction of type (I) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
where
for all \(x,y\in X\).
Definition 2.4
Let \((X,\sigma_{b})\) be a bmetriclike space with coefficient s. A triangular αadmissible mapping \(T:X\rightarrow X\) is said to be generalized MeirKeeler contraction of type (II) if for every \(\varepsilon>0\) there exists \(\delta>0\) such that
where
for all \(x,y\in X\).
In the following, we illustrate two important properties concerned with these new generalized MeirKeeler contractions, which we will require in our subsequent arguments.
Remark 2.5
Suppose that \(T:X\rightarrow X\) is a generalized MeirKeeler contraction of type (I) (respectively, type (II)). Then
for all \(x,y\in X\) when \(M(x,y)>0\) (respectively, \(N(x,y)>0\)).
Remark 2.6
It is readily verified that \(N(x,y)\leq M(x,y)\) for all \(x,y\in X\), where \(M(x,y)\) and \(N(x,y)\) are defined in (3) and (5), respectively.
Next, we establish a fixed point theorem for generalized MeirKeeler type contractions via a rational expression. The presented theorem is a generalization of the result of Samet et al. [8].
Theorem 2.7
Let \((X,\sigma_{b})\) be a complete bmetriclike space and \(T:X \rightarrow X\) be a triangular αadmissible mapping. Suppose that the following conditions hold:

(a)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\), \(\alpha(Tx_{0},x_{0})\geq1\),

(b)
if \(\{x_{n}\}\) is a sequence in X such that \(x_{n}\rightarrow z\) as \(n\rightarrow\infty\) and \(\alpha(x_{n},x_{m})\geq1\) for all \(n,m\in\mathbb{N}\), then \(\alpha(x_{n},z)\geq1\) for all \(n\in\mathbb{N}\),

(c)
for each \(\varepsilon>0\), there exists \(\delta>0\) satisfying the following condition:
$$ \begin{aligned} &2s\varepsilon\leq \sigma_{b}(y,Ty)\frac{1+\sigma _{b}(x,Tx)}{1+M(x,y)}+N(x,y)< s(2 \varepsilon+\delta)\quad \textit{implies} \\ &\alpha(x,y) \sigma_{b}(Tx,Ty)< \varepsilon. \end{aligned} $$(6)
Then T has a fixed point in X.
Proof
It is easy to observe that condition (6) implies that \(x\neq y\) or \(y\neq Ty \) and also
Let \(x_{0}\in X\) be such that condition (a) holds and define a sequence \(\{x_{n}\}\) in X such that \(x_{1}=Tx_{0}\), \(x_{n+1}=Tx_{n}\) for all \(n\in\mathbb{N}\). We may suppose that \(x_{n+1}\neq x_{n}\) for all \(n\in\mathbb{N}\cup\{0\}\), otherwise T has obviously a fixed point. Now, since T is αadmissible, then
and repeatedly using (8) we obtain
Replace x by \(x_{n}\) and y by \(x_{n+1}\) in (7) and taking into account equation (9), we deduce
where
We distinguish two following cases:
Case 1. Assume that \(M(x_{n},x_{n+1})=\sigma_{b}(x_{n+2},x_{n+1})\). By virtue of Remark 2.6 and also (7) we observe that
which gives a contradiction.
Case 2. Assume that \(M(x_{n},x_{n+1})=\sigma_{b}(x_{n},x_{n+1})\). Then \(N(x_{n},x_{n+1})=\sigma_{b}(x_{n},x_{n+1})\), too. Applying Remark 2.6, we get
Therefore
for all n. That is, \(\{\sigma_{b}(x_{n},x_{n+1})\}\) is a strictly decreasing positive sequence in \(\mathbb{R}^{+}\) and it converges to some \(r\geq0\). We claim that \(r=0\). To support the claim, let it be untrue. Then we have \(r>0\). We assert that
Since the condition (6) holds for every \(\varepsilon>0\), we may choose \(\varepsilon=\frac{r}{s}\) and let δ be such that satisfying condition (6). We know that \({\lim_{n\rightarrow\infty}}[\sigma_{b}(x_{n+1},x_{n+2})+\sigma _{b}(x_{n},x_{n+1})]=2r\). Hence there exists \(N_{0}\in\mathbb{N}\) such that \(2r<\sigma_{b}(x_{N_{0}+1},x_{N_{0}+2})+\sigma _{b}(x_{N_{0}},x_{N_{0}+1})<2r+\delta\). Consequently,
which leads to a contradiction with the condition (10). Thus, \(r=0\), that is,
We claim that the sequence \(\{x_{n}\}\) is a Cauchy sequence. Let \(\varepsilon>0\). Let \(\delta^{\prime}=\min\{\delta,\varepsilon,1\}\). From (11) there exists \(k\in\mathbb{N}\) such that
Now, we define the set \(\Lambda\subset X\) by
We will show that \(T(\Lambda)\subset\Lambda\). Let \(\lambda\in\Lambda\). There exists \(p\geq k\) such that \(\lambda=x_{p}\) and \(\sigma_{b}(x_{p},x_{k})< s(2\varepsilon+\frac{\delta^{\prime}}{2})\).
If \(p=k\), then \(T(\lambda)=x_{k+1}\in\Lambda\) by (12). We will assume that \(p>k\). First, we suppose that \(2s\varepsilon\leq\sigma_{b}(x_{p},x_{k})\), so
Let us prove that
We know \(\sigma_{b}(x_{p},x_{k})\leq N(x_{p},x_{k})\), then from (13) we get
Regarding (12) and since \(2s\varepsilon\leq\sigma_{b}(x_{p},x_{k})\), then \(M(x_{p},x_{k})=N(x_{p},x_{k})= \sigma_{b}(x_{p},x_{k})\). So
Therefore
It follows from (15) and (16) that (14) holds. Then
and from (9), (6) we conclude that
Now, using (\(\sigma_{b}3\)) together with (18) and (12) we obtain
This implies that \(T\lambda=Tx_{p}=x_{p+1}\in\Lambda\).
Next, we suppose that \(\sigma_{b}(x_{p},x_{k})<2s\varepsilon\). From (7) we derive
On the other hand, applying (12), we have
Then
In turn this proves \(T\lambda=Tx_{p}=x_{p+1}\in \Lambda\). Hence \(T(\Lambda)\subset\Lambda\) so
Now, for all \(m,n\in\mathbb{N}\) such that \(m>n>k\), by (19) we get
It follows that \(\lim_{m,n\rightarrow\infty}\sigma_{b}(x_{m},x_{n})=0\). Hence \(\{ x_{n}\}\) is a Cauchy sequence in X and since X is complete there exists \(z\in X\) such that
Finally, from (7) we observe that
Applying the definition of \(N(z,x_{n})\), the righthand side of the above inequality tends to \(\frac{1}{4}\sigma_{b}(Tz,z)\) when n tends to infinity. It implies that \(\sigma_{b}(Tz,z)=0\) and \(Tz=z\). □
The following example reveals the usefulness of Theorem 2.7.
Example 2.8
Let \(X=\{0,1,2,3\}\). Define \(\sigma_{b}:X\times X\rightarrow\mathbb{R}^{+}\) as follows:
Clearly, \((X,\sigma_{b})\) is a complete bmetriclike space with \(s=2\). Consider \(T:X\rightarrow X\) defined by \(T0=0\), \(T1=1\), \(T2=2\), and \(T3=1\). Also, define \(\alpha:X\times X\rightarrow\mathbb{R}^{+}\) as follows:
It easily can be shown that T is triangular αadmissible. In order to check the condition (6), we choose \(\delta=4\varepsilon\) so that \(4\varepsilon\leq \sigma_{b}(y,Ty)\frac{1+\sigma_{b}(x,Tx)}{1+M(x,y)}+N(x,y) <8\varepsilon\) which implies \(\alpha(x,y)\sigma_{b}(Tx,Ty)<\varepsilon\).
Note that \(\alpha(1,T1)\geq1\), \(\alpha(T1,1)\geq1\). Now, all conditions of Theorem 2.7 are satisfied and so T has a fixed point.
On the other hand, let \(d_{\sigma_{b}}\) be the bmetric associated to bmetriclike \(\sigma_{b}\) defined by \(d_{\sigma_{b}}(x,y)=0\) if \(x=y\) and \(d_{\sigma_{b}}(x,y)=\sigma_{b}(x,y)\), elsewhere. Then condition (6) does not hold in bmetric space \((X,d_{\sigma_{b}})\). Let \(\varepsilon=\frac{1}{4}\), \(x=0\), and \(y=2\). Then \(1=4\varepsilon\leq d_{\sigma_{b}}(2,T2)\frac{1+d_{\sigma_{b}}(0,T0)}{1+M(0,2)}+N(0,2)=1 <4\varepsilon+2\delta=1+2\delta\), for each \(\delta>0\). But \(\alpha(0,2)d_{\sigma_{b}}(T0,T2)=\frac{1}{3} \nless\frac{1}{4}\).
Theorem 2.9
Let \((X,\sigma_{b})\) be a complete bmetriclike space and \(T:X \rightarrow X\) be an αadmissible mapping. Assume that there exists a function \(\theta:\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}\) satisfying the following conditions:

(a)
\(\theta(0)=0\) and \(\theta(t)>0\) for every \(t>0\),

(b)
θ is nondecreasing and right continuous,

(c)
for every \(\varepsilon>0\), there exists \(\delta>0\) such that
$$\begin{aligned}& 2\varepsilon\leq\theta \biggl(\frac{1}{s} \sigma_{b}(y,Ty) \frac{1+\sigma_{b}(x,Tx)}{1+M(x,y)}+\frac {1}{s}N(x,y) \biggr)< 2\varepsilon+\delta \quad \textit{implies} \\& \theta\bigl(2\alpha(x,y) \sigma_{b}(Tx,Ty) \bigr)< 2\varepsilon, \end{aligned}$$for all \(x,y\in X\), then (6) is satisfied.
Proof
Fix \(\varepsilon>0\). Since \(\theta(2\varepsilon)>0\) by (c), there exists \(\delta>0\) such that
In view of the fact that θ is right continuous, then there exists \(\delta^{\prime}>0\) such that \(\theta(2\varepsilon+\delta^{\prime})<\theta(2\varepsilon)+\delta\). Fix \(x,y\in X\) such that
Since θ is nondecreasing we get
In the light of (20), we conclude that
It enforces that \(\alpha(x,y) \sigma_{b}(Tx,Ty)<\varepsilon\), i.e., (6) is satisfied. □
Corollary 2.10
Let \((X,\sigma_{b})\) be a complete bmetriclike space and \(T:X \rightarrow X\) be a mapping. Let φ be a locally integrable function from \(\mathbb{R}^{+}\) into itself such that \(\int_{0}^{t}\varphi(s)\, ds>0\) for all \(t>0\). Assume that conditions (a) and (b) of Theorem 2.7 hold and also T fulfills the following condition for all \(x,y \in X\):
where \(c\in(0,\frac{1}{2s})\) is a constant. Then T has a fixed point.
Proof
As a result of Theorem 2.9 if for each \(\varepsilon>0\), there exists \(\delta>0\) such that
then (6) is satisfied.
Fix \(\varepsilon>0\). Take \(\delta=2\varepsilon(\frac{1}{2c}1)\), then
□
Now, we establish an existence of fixed point of mapping satisfying generalized MeirKeeler contractions of type (I) in the setup of bmetriclike spaces. For this purpose, we need the following definition.
Definition 2.11
Let \((X,\sigma_{b})\) be a bmetriclike space, and let T be a selfmapping on X. T is called orbitally continuous whenever
for each \(x,z\in X\).
It is clear that continuous mappings are orbitally continuous. But the converse may not be true. To show this, let \(([0,1],\sigma_{b})\) be the bmetriclike space, where \(\sigma_{b}(x,y)=[\max\{x,y\}]^{q}\) (\(q\geq1\)). Consider \(T:X\rightarrow X\) defined by
Clearly T is not continuous but it is orbitally continuous.
Theorem 2.12
Let \((X,\sigma_{b})\) be a complete bmetriclike space with coefficient s and \(T:X \rightarrow X\) be a mapping. Suppose that the following conditions hold:

(a)
T is an orbitally continuous generalized MeirKeeler contraction of type (I),

(b)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\), \(\alpha(Tx_{0},x_{0})\geq1\),

(c)
if \(\{x_{n}\}\) is a sequence in X such that \(x_{n}\rightarrow z\) as \(n\rightarrow\infty\) and \(\alpha(x_{n},x_{m})\geq1\) for all \(n,m\in\mathbb{N}\), then \(\alpha(z,z)\geq1\),

(d)
\(s>1\) or β is a continuous function.
Then T has a fixed point in X.
Proof
Let \(x_{0}\in X\) be such that condition (b) holds and define a sequence \(\{x_{n}\}\) in X so that \(x_{1}=Tx_{0}\), \(x_{n+1}=Tx_{n}\) for all \(n\in \mathbb{N}\). Without loss of generality, we may suppose that \(x_{n+1}\neq x_{n}\) for all \(n\in\mathbb{N}\cup\{0\}\). Since T is αadmissible, then
Replace x by \(x_{n}\) and y by \(x_{n+1}\) in (2); we observe that for every \(\varepsilon>0\) there exists \(\delta>0\) such that
where
Next, we distinguish two following cases:
Case 1. Assume that \(M(x_{n},x_{n+1})=\sigma_{b}(x_{n+2},x_{n+1})\). In this case equation (22) becomes
and using (21) we have
Hence, we deduce
for all \(n\in\mathbb{N}\), which gives a contradiction.
Case 2. Assume that \(M(x_{n},x_{n+1})=\sigma_{b}(x_{n},x_{n+1})\). Since \(M(x_{n},x_{n+1})>0\), for all n, due to Remark 2.5, we get
for all n. That is, \(\{\sigma_{b}(x_{n},x_{n+1})\}\) is a strictly decreasing positive sequence in \(\mathbb{R}^{+}\) and it converges to some \(r\geq0\). We declare that \(r=0\). To support the claim, let it be untrue. Then we have \(r>0\). We assert that
First, suppose that \(s>1\). Applying (23), we have
By taking the limit as n tends to infinity we get \(r\leq\frac {1}{s}r< r\), which is a contradiction and so \(r=0\). Next, assume that β is a continuous function. It is an easy verification that \(\{\beta(\sigma _{b}(x_{n},x_{n+1}))\sigma_{b}(x_{n},x_{n+1})\}\) is a strictly decreasing positive sequence in \(\mathbb{R}^{+}\) and it converges to some \(r'\geq0\). We consider the two following cases:
(I) \(r'=0\).
Since \(\lim_{n\rightarrow\infty}\sigma_{b}(x_{n},x_{n+1})\neq0\), so we have
Now, let \(\varepsilon'>0\) be given. Since \(\lim_{k\rightarrow\infty} \beta(\sigma_{b}(x_{n_{k}},x_{n_{k}+1}))\sigma_{b}(x_{n_{k}},x_{n_{k}+1})=0\), therefore using (25) we derive
It enforces that \(\lim_{k\rightarrow\infty}\beta(\sigma _{b}(x_{n_{k}},x_{n_{k}+1}))= 0\). Now, continuity of β implies that \(\beta(r)=0\), which is a contradiction.
(II) \(r'>0\).
If \(r< r'\), then \(\beta(\sigma_{b}(x_{n},x_{n+1}))\sigma _{b}(x_{n},x_{n+1})<\frac{1}{s}\sigma_{b}(x_{n},x_{n+1})\), and by taking the limit as n tends to infinity we get \(r'\leq\frac {r}{s}\leq r\), which is in contradiction with our assumption. Now, we suppose that \(r\geq r'\). Let \(\delta>0\) be such that satisfying (2) whenever \(\varepsilon=r'\). We know that there exists \(N_{0}\in\mathbb{N}\) such that
Thus, \(\sigma_{b}(x_{N_{0}+1},x_{N_{0}+2})\leq\alpha (x_{N_{0}},x_{N_{0}+1})\sigma_{b}(Tx_{N_{0}},Tx_{N_{0}+1})< r'\leq r\), which leads to a contradiction with the condition (24). Thus, \(r=0\) and so
Next, we intend to show that the sequence \(\{x_{n}\}\) is a Cauchy sequence in \((X,\sigma_{b})\). For this purpose, we will prove that for every \(\varepsilon>0\) there exists \(N\in\mathbb{N}\) such that
for all \(l\geq N\) and \(k\in\mathbb{N}\). Since the sequence \(\{\sigma_{b}(x_{n},x_{n+1})\}\) converges to 0 as \(n\rightarrow\infty\), for every \(\delta>0\) there exists \(N\in\mathbb{N}\) such that
Choose δ such that \(\delta<\varepsilon\). We will prove (26) by using induction on k. For \(k=1 \), (26) becomes
and it clearly holds for all \(l\geq N\) (due to the choice of δ). Assume that the inequality (26) holds for some \(k=m\), that is,
For \(k=m+1\) we have to show that \(\sigma_{b}(x_{l},x_{l+m+1})<\varepsilon\) for all \(l\geq N\). Employing the condition (\(\sigma_{b}3\)), we get
for all \(l\geq N\). If \(\beta(\sigma_{b}(x_{l1},x_{l+m}))\sigma _{b}(x_{l1},x_{l+m})\geq \varepsilon\), then we deduce
and according to Lemma 1.14, on using the contractive condition (2) with \(x=x_{l1}\), \(y=x_{l+m}\) we find
which in turn implies that
and hence (26) holds for \(k=m+1\). If \(\beta(\sigma_{b}(x_{l1},x_{l+m}))\sigma _{b}(x_{l1},x_{l+m})<\varepsilon\), then
Regarding Remark 2.5, we get
that is, (26) holds for \(k=m+1\). Note that \(M(x_{l1},x_{l+m})>0\), otherwise \(\sigma_{b}(x_{l},x_{l1})=0\), and hence \(x_{l}=x_{l1}\), which is in contradiction with our assumption. Thus \(\sigma_{b}(x_{l},x_{l+k})<\varepsilon\) for all \(l\geq N\) and \(k\geq1\), it means
Therefore \(\lim_{n,m\rightarrow \infty}\sigma_{b}(x_{n},x_{m})=0 \) and since X is a complete bmetriclike space, there exists \(z\in X\) such that
Next, we will show that z is a fixed point of T. We aim to show that \(\sigma_{b}(Tz,z)=0\). Assume that \(\sigma_{b}(Tz,z)>0\). Thus we have \(M(z,z)\geq\sigma_{b}(Tz,z)>0\) and applying orbitally continuity of T it follows that
So \(\{x_{n+1}\}\) converges to Tz. Using (\(\sigma_{b}3\)), we have
Therefore we deduce
which is a contradiction. Consequently, \(Tz=z\). □
By Remark 2.6 we know \(N(x,y)\leq M(x,y)\), so a slight change in the proof of Theorem 2.12 shows actually the following theorem holds.
Theorem 2.13
Let \((X,\sigma_{b})\) be a complete bmetriclike space, \(T:X \rightarrow X\) be a mapping. Suppose that the following conditions hold:

(a)
T is an orbitally continuous generalized MeirKeeler contraction of type (II),

(b)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\), \(\alpha(Tx_{0},x_{0})\geq1\),

(c)
if \(\{x_{n}\}\) is a sequence in X such that \(x_{n}\rightarrow z\) as \(n\rightarrow\infty\) and \(\alpha(x_{n},x_{m})\geq1\) for all \(n,m\in\mathbb{N}\), then \(\alpha(z,z)\geq1\),

(d)
\(s>1\) or β is a continuous function.
Then T has a fixed point in X.
There is an analogous result for the generalized MeirKeeler contraction. The proof is an easy adaptation of the one given in Theorem 2.12.
Proposition 2.14
Under the hypotheses of Theorem 2.12 consider a particular case, T is a generalized MeirKeeler contraction, then T has a fixed point in X.
It is useful to seek a suitable replacement for the orbitally continuity of the contraction T. The next theorem indicates how this can be achieved. In fact with the aid of αadmissibility of the contraction we will show that orbitally continuity assumption is not required whenever the following condition is satisfied.

(A)
If \(\{x_{n}\}\) is a sequence in X which converges to z with respect to \(\tau_{\sigma_{b}}\) and satisfies \(\alpha(x_{n+1},x_{n})\geq1\) and \(\alpha(x_{n},x_{n+1})\geq 1\) for all n, then there exists a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that \(\alpha(z,x_{n_{k}})\geq1\) and \(\alpha (x_{n_{k}},z)\geq 1\) for all k.
Theorem 2.15
Let \((X,\sigma_{b})\) be a complete bmetriclike space with coefficient s and satisfies the condition (A). Also, let \(T:X \rightarrow X\) be a mapping. Suppose that the following conditions hold:

(a)
\(T:X \rightarrow X\) is a generalized MeirKeeler contraction of type (II),

(b)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\), \(\alpha(Tx_{0},x_{0})\geq1\),

(c)
\(s>1\) or β is a continuous function.
Then T has a fixed point in X.
Proof
Following the proof of Theorem 2.7, we observe that the sequence \(\{x_{n}\}\) defined by \(x_{1}=Tx_{0}\) and \(x_{n+1}=Tx_{n}\) (\(n\in\mathbb{N}\)), converges to some \(z\in X \) with \(\sigma_{b}(z,z)=0\). By condition (A), there exists a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that \(\alpha(z,x_{n_{k}})\geq1\) and \(\alpha(x_{n_{k}},z)\geq1\) for all k. Note that if \(N(z,x_{n_{k}})=0\), then \(Tz=z\) and the proof is done. Regarding Remark 2.5, for all \(k \in \mathbb{N}\) we have
where
Now on taking the limit \(k\rightarrow\infty\) and applying (\(\sigma_{b}3\)) we obtain
Thus, we conclude that
Applying again (\(\sigma_{b}3\)), we have
and passing to the limit \(k\rightarrow\infty\) in the above relation we obtain
This inequality implies \(\sigma_{b}(Tz,z)=0\) and hence \(Tz=z\), which completes the proof. □
Proposition 2.16
Let \((X,\sigma_{b})\) be a complete bmetriclike space with coefficient s and satisfies the condition (A). Let \(T:X \rightarrow X\) be a generalized MeirKeeler contraction. Also, suppose that the following conditions hold:

(a)
there exists \(x_{0}\in X\) such that \(\alpha(x_{0},Tx_{0})\geq1\), \(\alpha(Tx_{0},x_{0})\geq1\),

(b)
\(s>1\) or β is a continuous function.
Then T has a fixed point in X.
The usability of these results is demonstrated by the two following examples.
Example 2.17
Let \((X,\sigma_{b})\) and α be as in Example 2.8. Consider \(T:X\rightarrow X\) defined by \(T0=T2=0\) and \(T1=2\). Also, define \(\beta:[0,+\infty)\rightarrow (0,\frac{1}{s})\) as follows:
In order to check the condition (2), we choose \(\delta=\varepsilon\) so that \(\varepsilon\leq\beta(\sigma_{b}(x,y)) M(x,y)<\varepsilon+\delta=2\varepsilon\), which implies \(\alpha (x,y)\sigma_{b}(Tx,Ty)<\varepsilon\).
Therefore, the map T is a generalized MeirKeeler contraction of type (I). Note that T is continuous with respect to \(\tau_{\sigma_{b}}\) and \(\alpha(0,T0)\geq1\), \(\alpha(T0,0)\geq1\). Now, all conditions of Theorem 2.12 are satisfied and so T has a fixed point.
Example 2.18
Let \(X=\mathbb{R}^{+}\) equipped with the bmetriclike \(\sigma_{b}:X\times X\rightarrow\mathbb{R}^{+}\) defined by
It is easy to see that \((X,\sigma_{b})\) is a complete bmetriclike space, with \(s=2\). Define the selfmapping \(T:X\rightarrow X\) and the functions \(\alpha :X\times X\rightarrow[0,+\infty)\), \(\beta:[0,+\infty)\rightarrow (0,\frac{1}{s})\) as follows:
Then the mapping T is triangular αadmissible. On the other hand, the condition (A) holds on X. More precisely, if the sequence \(\{x_{n}\}\subset X\) satisfies \(\alpha(x_{n},x_{n+1})\geq1\), \(\alpha (x_{n+1},x_{n})\geq1\), and \({\lim_{n\rightarrow\infty}}x_{n}=x\) with respect to \(\tau_{\sigma _{b}}\), for some \(x\in X\), then \(\{x_{n}\}\subset[0,1]\) and, moreover, \(\lim_{n\rightarrow \infty}(x_{n}^{2}+x^{2})^{2}=4x^{4}\), which gives us \(x=0\). Hence \(\alpha(x_{n},x)\geq1\) and \(\alpha(x,x_{n})\geq1\).
Next we prove that T is a generalized MeirKeeler contraction. We show this in the three following steps.
Step 1. If \(x\notin[0,1]\) or \(y\notin[0,1]\).
In this case, \(\alpha(x,y)=0\) and evidently (1) holds.
Step 2. Let \(x,y\in[0,1]\) with \(\sigma_{b}(x,y)\in[0,1]\).
Let \(\varepsilon>0\) be given and choose \(\delta=3\varepsilon\). Now if \(\varepsilon\leq \beta(\sigma_{b}(x,y))\sigma_{b}(x,y)=\frac{1}{4}(x^{2}+y^{2})^{2}<\varepsilon +\delta=4\varepsilon\), then
Step 3. Let \(x,y\in[0,1]\) with \(\sigma_{b}(x,y)\notin[0,1]\).
Take \(\delta=3\varepsilon\). Then the inequality \(\varepsilon\leq \beta(\sigma_{b}(x,y))\sigma_{b}(x,y)=\frac {(x^{2}+y^{2})^{4}}{3(x^{2}+y^{2})^{2}+1}<4\varepsilon\), implies that \(\alpha (x,y)\sigma_{b}(Tx,Ty)=\frac{1}{16}(x^{2}+y^{2})^{2}<\varepsilon\).
Also, notice that \(\alpha(0,T0)\geq1\) and \(\alpha(T0,0)\geq1\). We conclude that all of the assumptions of Proposition 2.16 are satisfied. Moreover, T has fixed points \(x=0\) and \(x=2\).
A remarkable fact concerning Example 2.18 is that the restriction of T to the interval \([0,1]\) is orbitally continuous and so by the definition of α that example fulfills all conditions of Theorem 2.12, too.
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MSC
 47H10
 54H25
Keywords
 αadmissible
 fixed point
 MeirKeeler contraction
 bmetriclike