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- Open Access
A fixed point theorem for Meir-Keeler type contraction via Gupta-Saxena expression
- Najeh Redjel^{1, 2},
- Abdelkader Dehici^{1, 2} and
- İnci M Erhan^{3}Email author
https://doi.org/10.1186/s13663-015-0363-9
© Redjel et al. 2015
- Received: 9 March 2015
- Accepted: 23 June 2015
- Published: 15 July 2015
Abstract
In this paper, following the idea of Samet et al. (J. Nonlinear. Sci. Appl. 6:162-169, 2013), we establish a new fixed point theorem for a Meir-Keeler type contraction via Gupta-Saxena rational expression which enables us to extend and generalize their main result (Gupta and Saxena in Math. Stud. 52:156-158, 1984). As an application we derive some fixed points of mappings of integral type.
Keywords
- complete metric space
- fixed point
- Meir-Keeler mapping
- Gupta-Saxena rational expression
MSC
- 47H10
- 54H25
1 Introduction
It is well known that the contraction mapping principle of Banach [1] was the starting point of great discoveries and advances in mathematics, in particular in nonlinear analysis. This principle was the subject of several extensions by means of various generalized contractions (see, for example, [2–10]). Among the most relevant results in this direction one can give that of Meir and Keeler [11] who proved the following fixed point result.
Theorem 1.1
Theorem 1.2
For more details on this theorem, we refer, e.g., to [12, 13].
In this paper, we establish a new fixed point theorem of Meir-Keeler type involving Gupta-Saxena expression which extends Theorem 1.2 in the case where \(\alpha_{1},\alpha _{2},\alpha_{3} \in\, ]0, \frac{1}{3}[\). We also apply our theoretical results to contractions of integral type.
2 Main results
Our main result is the following theorem.
Theorem 2.1
Let \((X,d) \) be a complete metric space and let \(f: X \rightarrow X\) be a continuous mapping. Assume that the following condition holds.
Proof
- (i)If \(d(x_{k_{0}},x_{k_{0}+1})\leq d(x_{k_{0}},x_{p})\), then (11) givesFrom (3) and (10), we deduce that$$d\bigl(f(x_{p}),x_{k_{0}}\bigr)< \frac{1}{3}d(x_{p},x_{p+1})+ \frac{1}{3}d(x_{p},x_{p+1}) + \frac {1}{3}d(x_{p},x_{k_{0}})+ \frac{4}{3}d(x_{k_{0}+1},x_{k_{0}}). $$$$\begin{aligned} d\bigl(f(x_{p}),x_{k_{0}}\bigr) < & \frac{2}{3}\biggl( \frac{\delta'(\epsilon)}{6}\biggr)+ \frac {1}{3}\biggl( \frac{3\epsilon}{7}\biggr)+ \frac {4}{3}\biggl( \frac{\delta'(\epsilon)}{6}\biggr) \\ =& \frac{\delta'(\epsilon)}{3}+ \frac {\epsilon}{7} \\ < & \frac{\delta'(\epsilon)}{3}+ \frac {3\epsilon}{7}. \end{aligned}$$
- (ii)If \(d(x_{k_{0}},x_{k_{0}+1})> d(x_{k_{0}},x_{p})\), then$$\begin{aligned} d\bigl(f(x_{p}),x_{k_{0}}\bigr) \leq&d(x_{p+1},x_{p})+ d(x_{p},x_{k_{0}}) \\ < &d(x_{p+1},x_{p})+ d(x_{k_{0}},x_{k_{0}+1}) \\ < & \frac{\delta'(\epsilon)}{6} + \frac {\delta'(\epsilon)}{6} \\ =& \frac{\delta'(\epsilon)}{3} \\ < & \frac{\delta'(\epsilon)}{3}+ \frac {3\epsilon}{7}. \end{aligned}$$
Since \((X,d)\) is a complete metric space, then there exists \(u\in X\) such that \(x_{n} \rightarrow u\) as \(n \rightarrow +\infty\). The fact that \(x_{n+1}= f(x_{n})\) and the continuity of f imply that \(u= f(u)\), that is, u is a fixed point of f.
Now, we show that the result of Gupta and Saxena [12], where \(\alpha_{1}, \alpha_{2}, \alpha_{3} \in\, ]0, \frac{1}{3}[\), is a particular case of Theorem 2.1.
Corollary 2.2
(Gupta and Saxena [12])
Proof
Notice that the contraction mapping of Gupta and Saxena is a not a strict contraction, but k-contraction. Therefore, Theorem 2.1 is an extension of Gupta-Saxena result.
3 Applications
In this section, following the idea of Samet et al. [14], we will give an integral version of Gupta-Saxena result.
We start with the following theorem.
Theorem 3.1
- (i)
\(\rho(0)=0\) and \(\rho(t)>0\) for every \(t>0\);
- (ii)
ρ is nondecreasing and right continuous;
- (iii)for every \(\epsilon> 0\), there exists \(\delta (\epsilon)>0\) such thatfor all \(x,y \in X\) with \(x\neq y\).$$\begin{aligned}& 3\epsilon\leq\rho \biggl( \frac {(1+d(x,f(x)))d(y,f(y))}{1+d(x,y)}+ \frac {d(x,f(x))d(y,f(y))}{d(x,y)}+d(x,y) \biggr)< 3 \epsilon+ \delta (\epsilon) \\& \quad \Longrightarrow\quad \rho \bigl(3d\bigl(f(x),f(y)\bigr) \bigr) < 3\epsilon \end{aligned}$$
Proof
- (i)
h is continuous and nondecreasing;
- (ii)
\(h(0)=0\) and \(h(t)> 0\) for all \(t > 0\).
Corollary 3.2
Proof
This follows immediately from Theorem 3.1 since every continuous function \(h: [0,+\infty[\, \rightarrow[0,+\infty[\) is right continuous. □
As a consequence of this corollary, we have another result.
Corollary 3.3
Now, we are able to obtain an integral version of Gupta-Saxena result.
Corollary 3.4
Let \((X,d)\) be a complete metric space and let f be a continuous mapping from X into itself. Let φ be a locally integrable function from \([0, +\infty[\) into itself such that \(\int_{0}^{t} \varphi(s)\, ds > 0\) for all \(t > 0\). Assume that f satisfies the following condition.
Proof
Let \(\epsilon>0\). It is easy to observe that (15) is satisfied for \(\delta(\epsilon)= 3\epsilon( \frac {1}{\mu}-1)\). Then (1) holds and this completes the proof. □
Remark 3.5
Declarations
Acknowledgements
This work was elaborated within the framework of the scientific stay of the first two authors at Atılım University (Turkey). They thank all the staff of the Department of Mathematics, Atılım University and, in particular, Professor Karapınar for offering a pleasant environment of work and for his helpful discussions.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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