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Fixed point results for generalized (\alpha ,\psi )MeirKeeler contractive mappings and applications
Journal of Inequalities and Applications volume 2014, Article number: 68 (2014)
Abstract
In this paper, we introduce a new type of a generalized(\alpha ,\psi )MeirKeeler contractive mapping and establish some interesting theorems on the existence of fixed points for such mappings via admissible mappings. Applying our results, we derive fixed point theorems in ordinary metric spaces and metric spaces endowed with an arbitrary binary relation.
MSC:47H10, 54H25.
1 Introduction
Fixed points and fixed point theorems have countless applications and have become a major theoretical tool in many fields such as differential equations, mathematical economics, game theory, dynamics, optimal control, functional analysis, and operator theory. In particular, the wellknown Banach contraction principle is one of the forceful tools in nonlinear analysis, which states that every contraction selfmapping T on complete metric spaces (X,d) (i.e., d(Tx,Ty)\le kd(x,y) for all x,y\in X, where k\in [0,1)) has a unique fixed point. Due to its simplicity and importance, this classical principle has been generalized by several authors in different directions.
In 1969 Meir and Keeler [1] established a fixed point theorem in a metric space (X,d) for mappings satisfying the condition that for each \epsilon >0 there exists \delta (\epsilon )>0 such that
for all x,y\in X. This condition is called the MeirKeeler contractive type condition. Afterward, many authors extended and improved this condition and established fixed point results for new generalized conditions, see Maiti and Pal [2], Park and Rhoades [3], Mongkolkeha and Kumam [4] and others.
On the other hand, Samet et al. [5] introduced the notions of α, ψ contractive and αadmissible mapping in metric spaces. They also proved a fixed point theorem for α, ψ contractive mappings in complete metric spaces using the concept of αadmissible mapping.
Theorem 1 ([5])
Let (X,d) be a complete metric space and T:X\to X be (\alpha ,\psi )contractive mapping. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii)
T is continuous.
Then there exists u\in X such that Tu=u.
Theorem 2 ([5])
Let (X,d) be a complete metric space and T:X\to X be (\alpha ,\psi )contractive mapping. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x\in X as n\to \mathrm{\infty} then \alpha ({x}_{n},x)\ge 1 for all n.
Then, there exists u\in X such that Tu=u.
These results can be used as an efficient tool to study various fixed point results such as fixed point results in partially ordered spaces, fixed point results for cyclic mappings, multidimensional fixed point results (coupled fixed point results, tripled fixed point results, quadrupled fixed point etc.). Moreover, such type of fixed point results are helpful to solve several problems and equations like the boundary value problem, differential equations, nonlinear integral equations etc. In the recent literature, a wideranging discussion of fixed point theorems for admissible mappings had the interest of several mathematicians, for example, see [6–12].
In this paper, we introduce new type of contractive mapping based on MeirKeeler type contractive condition. For such mappings, we study and establish fixed point theorems via admissible mappings. Moreover, we present some applications of our new results.
2 Preliminaries
In the sequel, ℕ denote the set of positive integers. Let Ψ stands for the family of nondecreasing functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) such that {\sum}_{n=1}^{\mathrm{\infty}}{\psi}^{n}(t)<\mathrm{\infty} for each t>0, where {\psi}^{n} is the n th iterate of ψ.
Remark 3 For every function \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) the following holds:
if ψ is nondecreasing, then for each t>0,
Therefore, if \psi \in \mathrm{\Psi}, then for each t>0, \psi (t)<t and \psi (0)=0.
Example 4 Let {\psi}_{1},{\psi}_{2}:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) be defined in the following way:
It is clear that {\psi}_{1},{\psi}_{2}\in \mathrm{\Psi}. Notice that {\psi}_{1}, {\psi}_{2} are examples of continuous and discontinuous functions in Ψ.
Definition 5 ([5])
Let (X,d) be a metric space and T:X\to X be a given mapping. We say that T is an (\alpha ,\psi )contractive mapping if there exist two functions \alpha :X\times X\to [0,\mathrm{\infty}) and \psi \in \mathrm{\Psi} such that
for all x,y\in X.
Remark 6 If T:X\to X satisfies the Banach contraction principle in a metric space (X,d), then T is an (\alpha ,\psi )contractive mapping, where \alpha (x,y)=1 for all x,y\in X and \psi (t)=kt for all t>0, where k\in [0,1).
Definition 7 ([5])
Let T:X\to X and \alpha :X\times X\to [0,\mathrm{\infty}). We say that T is αadmissible when if x,y\in X such that \alpha (x,y)\ge 1 then we have \alpha (Tx,Ty)\ge 1.
Example 8 Let X=(0,\mathrm{\infty}). Define T:X\to X and \alpha :X\times X\to [0,\mathrm{\infty}) by Tx=ln(x+1) for all x\in X and
Then, T is αadmissible.
Example 9 Let X=[1,\mathrm{\infty}). Define T:X\to X and \alpha :X\times X\to [0,\mathrm{\infty}) by Tx={x}^{2} for all x\in X and
Then T is αadmissible.
Remark 10 ([5])
Every nondecreasing selfmapping T is αadmissible.
3 Main results
In this section, introducing the class of (\alpha ,\psi )MeirKeeler contractive mappings and the class of generalized(\alpha ,\psi )MeirKeeler contractive mappings, we study the existence of fixed points for mappings via admissible mappings.
Definition 11 Let (X,d) be a metric space and T:X\to X. The mapping T is called an (\alpha ,\psi )MeirKeeler contractive mapping if there exist two functions \psi \in \mathrm{\Psi} and \alpha :X\times X\to [0,\mathrm{\infty}) satisfying the following condition:
for each \epsilon >0 there exists \delta (\epsilon )>0 such that
Remark 12 It is easily shown that if T:X\to X is an (\alpha ,\psi )MeirKeeler type contraction, then
for all x,y\in X when x\ne y. Also, if x=y then d(Tx,Ty)=0 and thus \alpha (x,y)d(Tx,Ty)\le \psi (d(x,y)). Therefore, if T:X\to X is a (\alpha ,\psi )MeirKeeler type contraction, then
for all x,y\in X.
Definition 13 Let (X,d) be a metric space and T:X\to X. The mapping T is called a generalized(\alpha ,\psi )MeirKeeler contractive mapping if there exist two functions \psi \in \mathrm{\Psi} and \alpha :X\times X\to [0,\mathrm{\infty}) satisfying the following condition:
for each \epsilon >0 there exists \delta (\epsilon )>0 such that
where
Remark 14 If T:X\to X is a generalized(\alpha ,\psi )MeirKeeler type contraction, then
for all x,y\in X when x\ne y. Also, if x=y then d(Tx,Ty)=0 and thus \alpha (x,y)d(Tx,Ty)\le \psi (M(x,y)). Therefore, if T:X\to X is a generalized(\alpha ,\psi )MeirKeeler type contraction, then
for all x,y\in X.
Theorem 15 Let (X,d) be a complete metric space and T:X\to X be a generalized(\alpha ,\psi )MeirKeeler contractive mapping. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii)
T is continuous.
Then there exists u\in X such that Tu=u.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},T{x}_{0})\ge 1. Note that such a point {x}_{0} exists due to condition (ii). We define the sequence \{{x}_{n}\} in X by {x}_{n+1}=T{x}_{n} for all n\ge 0. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}, then clearly {x}_{{n}_{0}} is a fixed point of T. Hence, throughout the proof, we suppose that {x}_{n}\ne {x}_{n+1} for all n\in \mathbb{N}. Since T is αadmissible, we have
By induction, we obtain
From (3.7) and Remark (14), it follows that for all n\in \mathbb{N}, we have
If max\{d({x}_{n},{x}_{n1}),d({x}_{n+1},{x}_{n})\}=d({x}_{n+1},{x}_{n}), from (3.8) and Remark 3, we have
which is a contradiction. So we have max\{d({x}_{n},{x}_{n1}),d({x}_{n+1},{x}_{n})\}=d({x}_{n},{x}_{n1}). From (3.8), we get
for all n\in \mathbb{N}. Inductively, for each n\in \mathbb{N}, we obtain
Now we show that \{{x}_{n}\} is a Cauchy sequence. Take \epsilon >0 and N(\epsilon )\in \mathbb{N} in such a way that {\sum}_{n\ge N(\epsilon )}{\psi}^{n}(d({x}_{1},{x}_{0}))\le \epsilon. Let n,m\in \mathbb{N} with m>n>N(\epsilon ). Due to the triangle inequality, we have
Hence, we conclude that \{{x}_{n}\} is a Cauchy sequence in the complete metric space (X,d). Thus, there exists u\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=u. Since T is continuous,
that is, u is a fixed point of T. This completes the proof. □
Corollary 16 Let (X,d) be a complete metric space and T:X\to X be an (\alpha ,\psi )MeirKeeler contractive mapping. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii)
T is continuous.
Then there exists u\in X such that Tu=u.
We obtain the following fixed point result without any continuity assumption for the mapping T.
Theorem 17 Let (X,d) be a complete metric space and T:X\to X be a generalized(\alpha ,\psi )MeirKeeler contractive mapping such that ψ is continuous. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii′)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x\in X as n\to \mathrm{\infty} then \alpha ({x}_{n},x)\ge 1 for all n.
Then there exists u\in X such that Tu=u.
Proof Following the proof of Theorem 15, we obtain the sequence \{{x}_{n}\} in X defined by {x}_{n+1}=T{x}_{n} for all n\ge 0 and which converges for some u\in X. From (3.7) and condition (iii), we have \alpha ({x}_{n},u)\ge 1 for all n\in \mathbb{N}. Next, we suppose that d(u,Tu)\ne 0. Applying Remark 12, for each n\in \mathbb{N}, we have
Letting n\to \mathrm{\infty} in the above equality and keeping the continuity of ψ in mind, we get
which is a contradiction. Thus, we have d(u,Tu)=0, that is, u=Tu. Therefore, u is a fixed point of T. This completes the proof. □
In the next corollary, we can omit the continuity hypothesis at every point of ψ.
Corollary 18 Let (X,d) be a complete metric space and T:X\to X be an (\alpha ,\psi )MeirKeeler contractive mapping. Suppose that

(i)
T is αadmissible;

(ii)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1;

(iii)
if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x\in X as n\to \mathrm{\infty} then \alpha ({x}_{n},x)\ge 1 for all n.
Then there exists u\in X such that Tu=u.
Proof Since the proof of the existence of a fixed point is straightforward by following the same lines as in the proof of Theorem 17, in order to avoid repetition, the details are omitted. □
Now, we present the following example in support of our main result.
Example 19 Let X=\mathbb{R} with the usual metric d(x,y)=xy. Then {d}_{2}((x,y)(u,v))=d(x,u)+d(y,v) becomes a metric on {X}^{2}. Define {T}_{F}:{X}^{2}\to {X}^{2} as follows: {T}_{F}(x,y)=(F(x,y),F(y,x)) where F:{X}^{2}\to X is defined by
for all (x,y)\in {X}^{2}. Let us take
It is clear that {T}_{F} is an (\alpha ,\psi )MeirKeeler contractive mapping with \psi (t)=\frac{t}{2}. Indeed, for all (x,y),(u,v)\in {X}^{2}, we have
On the other hand, by elementary calculations and taking into account that {T}_{F} is αadmissible, we get
with \frac{1}{4}\delta (\epsilon )<\epsilon.
Remark 20 Note that the main theorem of [11] is not applicable to this example. Hence, our result is stronger than the main result of [11]. Indeed, we take x=u in the statement of the mentioned theorem, we get
On the other hand, by elementary calculations, we get
Thus, we get
which is a contradiction.
4 Applications
4.1 Fixed point results on an ordinary metric space
We have the following fixed point results in ordinary metric space.
Theorem 21 Let (X,d) be a complete metric space and T:X\to X be continuous mapping and there exists \psi \in \mathrm{\Psi} satisfying the following condition:
for each \epsilon >0 there exists \delta (\epsilon )>0 such that
where
Then there exists u\in X such that Tu=u.
Proof Consider the mapping \alpha :X\times X\to [0,\mathrm{\infty}) defined by
From the definition of α, it easy to see that T is αadmissible and also it is a generalized(\alpha ,\psi )MeirKeeler contractive mapping. Moreover, all the hypotheses of Theorem 15 (or Theorem 17) are satisfied and so the existence of the fixed point of T follows from Theorem 15 (or Theorem 17). □
Taking \psi (t)=kt, where k\in (0,1), we get the following result.
Corollary 22 Let (X,d) be a complete metric space and T:X\to X be continuous mapping satisfying the following condition:
for each \epsilon >0 there exists \delta (\epsilon )>0 such that
for all x,y\in X, where k\in (0,1) and
Then there exists u\in X such that Tu=u.
4.2 Fixed point results on a metric space endowed with an arbitrary binary relation
In this section, we present fixed point theorems on a metric space endowed with an arbitrary binary relation. The following notions and definition are needed.
Let (X,d) be a metric space and ℛ be a binary relation over X. Denote
this is the symmetric relation attached to ℛ. Clearly,
Definition 23 We say that T:X\to X is a comparative mapping if T maps comparable elements into comparable elements, that is,
for all x,y\in X.
Definition 24 Let (X,d) be a metric space, be a symmetric relation attached to binary relation ℛ over X and T:X\to X. The mapping T is called a generalizedψMeirKeeler contractive mapping with respect to if there exists a function \psi \in \mathrm{\Psi} satisfying the following condition:
for each \epsilon >0 there exists \delta (\epsilon )>0 such that for x,y\in X for which x\mathcal{S}y,
where
Theorem 25 Let (X,d) be a complete metric space, ℛ be a binary relation over X and T:X\to X be a comparative generalized(\alpha ,\psi )MeirKeeler contractive mapping. Suppose that

(i)
T is comparative mapping;

(ii)
there exists {x}_{0}\in X such that {x}_{0}\mathcal{S}(T{x}_{0});

(iii)
T is continuous.
Then there exists u\in X such that Tu=u.
Proof Consider the mapping \alpha :X\times X\to [0,\mathrm{\infty}) defined by
From condition (ii), we get \alpha ({x}_{0},T{x}_{0})\ge 1. It follows from T is comparative mapping that T is an αadmissible mapping. Since T is a generalizedψMeirKeeler contractive mapping with respect to , we have, for all x,y\in X,
This implies that T is a generalized(\alpha ,\psi )MeirKeeler contractive mapping. Now all the hypotheses of Theorem 15 are satisfied and so the existence of the fixed point of T follows from Theorem 15. □
In order to remove the continuity of T, we need the following condition:
() if \{{x}_{n}\} is the sequence in X such that {x}_{n}\mathcal{R}{x}_{n+1} for all n\in \mathbb{N} and it converges to the point x\in X, then {x}_{n}\mathcal{S}u.
Theorem 26 Let (X,d) be a complete metric space, be a symmetric relation attached to binary relation ℛ over X and T:X\to X be a generalized(\alpha ,\psi )MeirKeeler contractive mapping with respect to such that ψ is continuous. Suppose that

(i)
T is comparative mapping;

(ii)
there exists {x}_{0}\in X such that {x}_{0}\mathcal{S}(T{x}_{0});

(iii)
the condition () holds.
Then there exists u\in X such that Tu=u.
Proof The result follows from Theorem 17 by considering the mapping α given by (4.4) and by observing that condition () implies condition (iii′). □
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Acknowledgements
This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks the DSR for technical and financial support. The authors thank the referees for their valuable comments and suggestions.
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Latif, A., Gordji, M.E., Karapınar, E. et al. Fixed point results for generalized (\alpha ,\psi )MeirKeeler contractive mappings and applications. J Inequal Appl 2014, 68 (2014). https://doi.org/10.1186/1029242X201468
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DOI: https://doi.org/10.1186/1029242X201468
Keywords
 αadmissible mapping
 binary relation
 generalized (\alpha ,\psi )MeirKeeler contractive mapping