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On αψMeirKeeler contractive mappings
Fixed Point Theory and Applications volume 2013, Article number: 94 (2013)
Abstract
In this paper, we introduce the notion of αψMeirKeeler contractive mappings via a triangular αadmissible mapping. We discuss the existence and uniqueness of a fixed point of such a mapping in the setting of complete metric spaces. We state a number of examples to illustrate our results.
MSC:46N40, 47H10, 54H25, 46T99.
1 Introduction and preliminaries
Fixedpoint theory is one of the most intriguing research fields in nonlinear analysis. The number of authors have published papers and have increased continuously in the last decades. The main reason for this involvement can be observed easily: Application potential. Fixed point theory has an application in many disciplines such as chemistry, physics, biology, computer science and many branches of mathematics. Banach contraction mapping principle or Banach fixedpoint theorem is the most celebrated and pioneer result in this direction: In a complete metric space, each contraction mapping has a unique fixed point. Following Banach [1], many authors give various generalizations of this principle in various space (see e.g. [2–20]). One of the interesting results was given by Samet et al. [21] by defining αψcontractive mappings via admissible mappings, see also [22].
In this paper, we introduce an αψMeirKeeler contractive mapping in the setting of complete metric spaces via a triangular αadmissible mapping. We prove the existence and uniqueness of a fixed point of such a mapping. We also consider a number of examples to illustrate our results.
Definition 1 Let f:X\to X and \alpha :X\times X\to (\mathrm{\infty},+\mathrm{\infty}). We say that f is a triangular αadmissible mapping if

(T1)
\alpha (x,y)\ge 1 implies \alpha (fx,fy)\ge 1, x,y\in X,

(T2)
\{\begin{array}{c}\alpha (x,z)\ge 1,\hfill \\ \alpha (z,y)\ge 1,\hfill \end{array} implies \alpha (x,y)\ge 1, x,y,z\in X.
Example 2 Let X=\mathbb{R}, fx=\sqrt[3]{x} and \alpha (x,y)={e}^{xy} then f is a triangular αadmissible mapping. Indeed, if \alpha (x,y)={e}^{xy}\ge 1 then x\ge y which implies fx\ge fy. That is, \alpha (fx,fy)={e}^{fxfy}\ge 1. Also, if \{\begin{array}{c}\alpha (x,z)\ge 1,\hfill \\ \alpha (z,y)\ge 1\hfill \end{array} then \{\begin{array}{c}xz\ge 0,\hfill \\ zy\ge 0.\hfill \end{array} That is, xy\ge 0 and so \alpha (x,y)={e}^{xy}\ge 1.
Example 3 Let X=\mathbb{R}, fx={e}^{{x}^{7}} and \alpha (x,y)=\sqrt[5]{xy}+1. Hence, f is a triangular αadmissible mapping. Again, if \alpha (x,y)=\sqrt[5]{xy}+1\ge 1 then x\ge y which implies fx\ge fy. That is, \alpha (fx,fy)\ge 1.
Moreover, if \{\begin{array}{c}\alpha (x,z)\ge 1;\hfill \\ \alpha (z,y)\ge 1,\hfill \end{array} then xy\ge 0, and hence, \alpha (x,y)\ge 1.
Example 4 Let X=[0,+\mathrm{\infty}), fx={x}^{4}+ln({x}^{2}+1) and
Then f is a triangular αadmissible mapping. In fact, if
then x\ge y. Hence, fx\ge fy. That is, \alpha (fx,fy)\ge 1. Also,
Thus, \alpha (x,z)+\alpha (z,y)\le 2\alpha (x,y). Now, if \{\begin{array}{c}\alpha (x,z)\ge 1;\hfill \\ \alpha (z,y)\ge 1,\hfill \end{array} then \alpha (x,y)\ge 1.
Example 5 Let X=\mathbb{R}, fx={x}^{3}+\sqrt[7]{x} and \alpha (x,y)={x}^{5}{y}^{5}+1. Then f is a triangular αadmissible mapping.
Example 6 Let X=[0,+\mathrm{\infty}), fx={x}^{2}+{e}^{x} and
Hence, f is a triangular αadmissible mapping.
Lemma 7 Let f be a triangular αadmissible mapping. Assume that there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0}. Then
Proof Since there exist {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1 then from (T1), we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (f{x}_{0},{f}^{2}{x}_{0})\ge 1. By continuing this process, we get
Suppose that m<n. Since \{\begin{array}{c}\alpha ({x}_{m},{x}_{m+1})\ge 1,\hfill \\ \alpha ({x}_{m+1},{x}_{m+2})\ge 1,\hfill \end{array} then from (T2) we have \alpha ({x}_{m},{x}_{m+2})\ge 1.
Again, since \{\begin{array}{c}\alpha ({x}_{m},{x}_{m+2})\ge 1,\hfill \\ \alpha ({x}_{m+2},{x}_{m+3})\ge 1,\hfill \end{array} then we deduce \alpha ({x}_{m},{x}_{m+3})\ge 1.
By continuing this process, we get \alpha ({x}_{m},{x}_{n})\ge 1. □
Denote with Ψ the family of nondecreasing functions \psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) continuous in t=0 such that

\psi (t)=0 if and only if t=0,

\psi (t+s)\le \psi (t)+\psi (s).
2 Main results
Definition 8 Let (X,d) be a metric space and \psi \in \mathrm{\Psi}. Suppose that f:X\to X is a triangular αadmissible mapping satisfying the following condition: for each \epsilon >0 there exists \delta >0 such that
for all x,y\in X. Then f is called an αψMeirKeeler contractive mapping.
Remark 9 Let f be an αψMeirKeeler contractive mapping. Then
for all x,y\in X when x\ne y. Also, if x=y then d(fx,fy)=0. i.e.,
for all x,y\in X.
Theorem 10 Let (X,d) be a complete metric space. Suppose that f is a continuous αψMeirKeeler contractive mapping and that there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1, then f has a fixed point.
Proof Let {x}_{0}\in X and define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}\cup \{0\}, then obviously f has a fixed point. Hence, we suppose that
for all n\in \mathbb{N}\cup \{0\}. We have d({x}_{n},{x}_{n+1})>0 for all n\in \mathbb{N}\cup \{0\}. Now define {s}_{n}=\psi (d({x}_{n},{x}_{n+1})). By Remark 9, we deduce that for all n\in \mathbb{N}\cup \{0\},
Then by applying Lemma 7
Hence, the sequence \{{s}_{n}\} is decreasing in {\mathbb{R}}_{+} and so it is convergent to s\in {\mathbb{R}}_{+}. We will show that s=0. Suppose, to the contrary, that s>0. Hence, we have
Let \epsilon =s>0. Then by hypothesis, there exists a \delta (\epsilon )>0 such that (2.1) holds. On the other hand, by the definition of ε, there exists {n}_{0}\in \mathbb{N} such that
Now by (2.1), we have
which is a contradiction. Hence, s=0, that is, {lim}_{n\to +\mathrm{\infty}}{s}_{n}=0. Now, by the continuity of ψ in t=0, we have {lim}_{n\to +\mathrm{\infty}}d({x}_{n},{x}_{n+1})=0. For given \epsilon >0, by the hypothesis, there exists a \delta =\delta (\epsilon )>0 such that (2.1) holds. Without loss of generality, we assume \delta <\epsilon. Since s=0, then there exists N\in \mathbb{N} such that
We will prove that for any fixed k\ge {N}_{0},
holds. Note that (2.5), by (2.4), holds for l=1. Suppose the condition (2.1) is satisfied for some m\in \mathbb{N}. For l=m+1, by (2.4), we get
If \psi (d({x}_{k1},{x}_{k+m}))\ge \epsilon, then by (2.1) we get
and hence (2.5) holds.
If \psi (d({x}_{k1},{x}_{k+m}))<\epsilon, by Remark 9, we get
Consequently, (2.5) holds for l=m+1. Hence, \psi (d({x}_{k},{x}_{k+l}))\le \epsilon for all k\ge {N}_{0} and l\ge 1, which means
Hence \{{x}_{n}\} is a Cauchy sequence. Since (X,d) is complete, there exists z\in X such that {x}_{n}\to z as n\to \mathrm{\infty}. Now, since, f is continuous then
that is, f has a fixed point. □
Theorem 11 Let (X,d) be a complete metric space and let f be a αψMeirKeeler contractive mapping. If the following conditions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1,

(ii)
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 as n\to +\mathrm{\infty}, then \alpha ({x}_{n},x)\ge 1 for all n.
Then f has a fixed point.
Proof Following the proof of the Theorem 10, we say that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and that there exist z\in X such that {x}_{n}\to z as n\to +\mathrm{\infty}. Hence, from (ii) \alpha ({x}_{n},z)\ge 1. By Remark 9, we have
By taking limit as n\to +\mathrm{\infty}, in the above inequality, we get \psi (d(fz,z))\le 0, that is, d(fz,z)=0. Hence, fz=z. □
Next, we give some examples to validate our main result.
Example 12 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
and \psi (t)=\frac{1}{4}t. Clearly, (X,d) is a complete metric space. We show that f is a triangular αadmissible mapping. Let x,y\in X, if \alpha (x,y)\ge 1 then x,y\in [0,1]. On the other hand, for all x,y\in [0,1] we have fx\le 1 and fy\le 1. It follows that \alpha (fx,fy)\ge 1. Also, if \alpha (x,z)\ge 1 and \alpha (z,y)\ge 1 then x,y,z\in [0,1] and hence, \alpha (x,y)\ge 1. Thus, the assertion holds by the same arguments. Notice that \alpha (0,f0)\ge 1.
Now, if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, then \{{x}_{n}\}\subset [0,1] and hence x\in [0,1]. This implies that \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}. Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =\epsilon the condition (2.1) holds. Otherwise, \alpha (x,y)=1. Hence, for given \epsilon >0 we have \alpha (x,y)\psi (d(fx,fy))\le 0<\epsilon. Hence, conditions of Theorem 11 holds and f has a fixed point. But, if x,y\in (1,\mathrm{\infty}) and
where \epsilon >0 and \delta >0. Then
That is, the MeirKeeler theorem cannot applied for this example.
Example 13 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
and \psi (t)=\frac{1}{2}t,
Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =\epsilon the condition (2.1) holds. Otherwise, \alpha (x,y)=1. Hence, for given \epsilon >0 we have \alpha (x,y)\psi (d(fx,fy))\le 0<\epsilon. Hence, conditions of Theorem 11 holds and f has a fixed point.
Denote with {\mathrm{\Psi}}_{st} the family of strictly nondecreasing functions {\psi}_{st}:[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) continuous in t=0 such that

{\psi}_{st}(t)=0 if and only if t=0,

{\psi}_{st}(t+s)\le {\psi}_{st}(t)+{\psi}_{st}(s).
Definition 14 Let (X,d) be a metric space and {\psi}_{st}\in {\mathrm{\Psi}}_{st}. Suppose that f:X\to X is a triangular αadmissible mapping satisfying the following condition: for each \epsilon >0 there exists \delta >0 such that
for all x,y\in X where
Then f is called generalized an α{\psi}_{st}MeirKeeler contractive mapping.
Remark 15 Let f be a generalized α{\psi}_{st}MeirKeeler contractive mapping. Then
for all x,y\in X when M(x,y)>0. Also, if M(x,y)=0 then x=y which implies \psi (d(fx,fy))=0, i.e.,
for all x,y\in X.
Proposition 16 Let (X,d) be a metric space and f:X\to X a generalized α{\psi}_{st}MeirKeeler contractive mapping, if there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1 . Then {lim}_{n\to \mathrm{\infty}}d({f}^{n+1}{x}_{0},{f}^{n}{x}_{0})=0.
Proof Define a sequence \{{x}_{n}\} by {x}_{n}={f}^{n}{x}_{0} for all n\in \mathbb{N}. If {x}_{{n}_{0}}={x}_{{n}_{0}+1} for some {n}_{0}\in \mathbb{N}\cup \{0\}, then obviously f has a fixed point. Hence, we suppose that
for all n\in \mathbb{N}\cup \{0\}. Then we have M({x}_{n+1},{x}_{n})>0 for every n\ge 0. Then by Lemma 7 and Remark 15, we have
Now, since {\psi}_{st} is strictly nondecreasing then, we get
Hence, the case where
is not possible. Therefore, we deduce that
for all n. That is, {\{d({x}_{n+1},{x}_{n})\}}_{n=0}^{\mathrm{\infty}} is a decreasing sequence in {\mathbb{R}}_{+} and it converges to \epsilon \in {\mathbb{R}}_{+}, that is,
Notice that \epsilon =inf\{p({x}_{n},{x}_{n+1}):n\in \mathbb{N}\}. Let us prove that \epsilon =0. Suppose, to the contrary, that \epsilon >0. Then \psi (\epsilon )>0. Regarding (2.11) together with the assumption that f is a generalized α{\psi}_{st}MeirKeeler contractive mapping, for {\psi}_{st}(\epsilon ), there exists \delta >0 and a natural number m such that
implies that
Now, since {\psi}_{st} is strictly nondecreasing then we get
which is a contradiction since \epsilon =inf\{p({x}_{n},{x}_{n+1}):n\in \mathbb{N}\}. Then \epsilon =0 and so {lim}_{n\to \mathrm{\infty}}d({x}_{n+1},{x}_{n})=0. □
Theorem 17 Let (X,d) be a complete metric space and f:X\to X a orbitally continuous generalized α{\psi}_{st}MeirKeeler contractive mapping, if there exist {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Then, f has a fixed point.
Proof Define {x}_{n+1}={f}^{n+1}{x}_{0} for all n\ge 0. We want to prove that {lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0. If this is not, then there exist \epsilon >0 and a subsequence \{{x}_{n(i)}\} of \{{x}_{n}\} such that
For this \epsilon >0, there exists \delta >0 such that \epsilon \le {\psi}_{st}(M(x,y))<\epsilon +\delta implies that \alpha (x,y){\psi}_{st}(d(fx,fy))<\epsilon. Put r=min\{\epsilon ,\delta \} and {s}_{n}=d({x}_{n},{x}_{n+1}) for all n\ge 1. From Proposition 16, there exists {n}_{0} such that
for all n\ge {n}_{0}. Let n(i)>{n}_{0}. We get n(i)\le n(i+1)1. If d({x}_{n(i)},{x}_{n(i+1)1})\le \epsilon +\frac{r}{2}, then
which contradicts the assumption (2.12). Therefore, there are values of k such that n(i)\le k\le n(i+1) and d({x}_{n(i)},{x}_{k})>\epsilon +\frac{r}{2}. Now if d({x}_{n(i)},{x}_{n(i)+1})\ge \epsilon +\frac{r}{2}, then
which is a contradiction with (2.13). Hence, there are values of k with n(i)\le k\le n(i+1) such that d({x}_{n(i)},{x}_{k})<\epsilon +\frac{r}{2}. Choose smallest integer k with k\ge n(i) such that d({x}_{n(i)},{x}_{k})\ge \epsilon +\frac{r}{2}. Thus, d({x}_{n(i)},{x}_{k1})<\epsilon +\frac{r}{2} and so
Now, we can choose a natural number k satisfying n(i)\le k\le n(i+1) such that
Therefore, we obtain
and
Thus, we have
Now, the inequalities (2.15)(2.18) imply that M({x}_{n(i)},{x}_{k})<\epsilon +r\le \epsilon +\delta and so {\psi}_{st}(M({x}_{n(i)},{x}_{k}))<{\psi}_{st}(\epsilon +\delta )\le {\psi}_{st}(\epsilon )+{\psi}_{st}(\delta ) the fact that f is a generalized α{\psi}_{st}MeirKeeler contractive mapping yields that,
Then d({x}_{n(i)+1},{x}_{k+1})<\epsilon. We deduce,
Hence, from this with (2.14), (2.16) and (2.17), we obtain
which is a contradiction. We obtained that {lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0 and so \{{x}_{n}={f}^{n}{x}_{0}\} is a Cauchy sequence. Since, X is complete, then there exists z\in X such that {f}^{n}{x}_{0}\to z as n\to \mathrm{\infty}. Now, since f is orbitally continuous, then z=fz. □
Example 18 Let X=[0,\mathrm{\infty}) and d(x,y)=xy be a metric on X. Define f:X\to X by
and {\psi}_{st}(t)=\frac{1}{2}t,
Clearly, f is a triangular αadmissible mapping and orbitally continuous. Let x,y\in [0,1]. Without loss of generality, take x\le y. Then
Clearly, by taking \delta =3\epsilon the condition (2.8) holds. Otherwise, \alpha (x,y)=0. Hence, for given \epsilon >0, we have 0=\alpha (x,y){\psi}_{st}(d(fx,fy))<\epsilon. Hence, condition of Theorem 17 is held and f has a fixed point.
Theorem 19 Assume that all the hypotheses of Theorem 10 (11 and 17) hold. Adding the following conditions:

(iii)
for all x\ne y\in X there exists v\in X such that \alpha (x,v)\ge 1 and \alpha (v,y)\ge 1,
we obtain the uniqueness of the fixed point of f.
Proof Suppose that z and {z}^{\ast} are two fixed points of f such that z\ne {z}^{\ast}. Then \alpha (z,v)\ge 1 and \alpha (v,{z}^{\ast})\ge 1. Hence, from (T2), we have \alpha (z,{z}^{\ast})\ge 1. Now, by Remark 9, we get
which is a contradiction and so z={z}^{\ast}. Similarly, for Theorem 17, we can observe that f has a unique fixed point. □
We can obtain the following corollaries intermediately.
Corollary 20 Let (X,d) be a complete metric space and f:X\to X is selfmapping. Suppose that for each \epsilon >0, there exists \delta >0 such that
where \psi \in \mathrm{\Psi} and L\ge 1. Then f has a unique fixed points.
Corollary 21 Let (X,d) be a complete metric space and f:X\to X a orbitally continuous selfmapping. Suppose that for each \epsilon >0 there exists \delta >0 such that
where {\psi}_{st}\in {\mathrm{\Psi}}_{st}, L\ge 1 and
Then f has unique fixed points.
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Acknowledgements
The authors would like to thank the referees for their valuable comments and suggestions. Also, the second author would like to thank the Commission on Higher Education, the Thailand Research Fund and the King Mongkut’s University of Technology Thonburi (Grant No. MRG5580213) for financial support during the preparation of this manuscript. The third author is thankful for support of Astara Branch, Islamic Azad University, during this research.
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Karapınar, E., Kumam, P. & Salimi, P. On αψMeirKeeler contractive mappings. Fixed Point Theory Appl 2013, 94 (2013). https://doi.org/10.1186/16871812201394
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DOI: https://doi.org/10.1186/16871812201394
Keywords
 MeirKeeler contractive mappings
 triangular αadmissible mappings
 fixed points