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A fixed point theorem for Meir-Keeler type contraction via Gupta-Saxena expression
Fixed Point Theory and Applications volume 2015, Article number: 115 (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.
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
Let \((X,d)\) be a complete metric space and let f be a mapping from X into itself satisfying the following condition:
Then f has a unique fixed point \(u \in X\). Moreover, for all \(x \in X\), the sequence \(\{f^{n}(x)\}\) converges to u.
As pointed out in [11], it is easy to observe that the conclusion of Banach theorem holds for the contraction in Theorem 1.1 which is called a strict contraction, that is, it satisfies
In 1984, Gupta and Saxena proved the following fixed point result.
Theorem 1.2
Let \((X,d)\) be a complete metric space and let f be a continuous mapping from X into itself satisfying
for all \(x,y \in X\), \(x\neq y\), where \(\alpha_{1}\), \(\alpha _{2}\), \(\alpha_{3}\) are constants with \(\alpha_{1},\alpha_{2},\alpha _{3}> 0\) and \(\alpha_{1}+\alpha_{2}+\alpha_{3}< 1\). Then f has a unique fixed point \(u\in X\). Moreover, for all \(x \in X\), the sequence \(\{f^{n}(x)\}\) converges to u.
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.
For any \(\epsilon> 0\), there exists \(\delta(\epsilon)> 0\) such that
for all \(x,y \in X\) with \(x\neq y\). Then f has a unique fixed point \(u \in X\). Moreover, \(\lim_{n \to\infty}f^{n}(x_{0})= u\) for any \(x_{0} \in X\).
Proof
It is easy to observe that condition (1) implies that
Let \(x_{0} \in X\) and consider the sequence \(\{x_{n}\}= \{ f^{n}(x_{0})\}_{n\geq0}\). We will prove that \(\{x_{n}\}\) is a Cauchy sequence in X. If there exists \(l_{0}\in\mathbb{N}\) such that \(x_{l_{0}}=x_{l_{0+1}}\), then clearly \(x_{l_{0}}\) is a fixed point of f. Now assume that \(x_{k}\neq x_{k+1}\) for all \(k\in\mathbb{N}\). Define
Following (2), we obtain that
This results in
that is, the sequence \(\{s_{n}\}\) is decreasing. Then \(s_{n}\) converges to some \(s \geq 0 \); and, moreover, \(s_{n} \geq s\), \(\forall n \geq0\). We also have \(2s_{n} + s_{n-1} \rightarrow3s\) as \(n \rightarrow+\infty\). From (1), if \(s>0\), there exists \(\delta(s)>0\) such that
implies
which contradicts \(s_{n}\geq s\). Thus, we deduce that
Now, let
By the convergence of the sequence \(\{d(x_{n},x_{n+1})\}\) to 0, there exists \(k_{0}\in\mathbb{N}\) such that
Now, we define the set Ω by
We will prove that
Clearly, for \(\gamma\in\Omega\), there exists \(p \geq k_{0}\) such that \(\gamma= x_{p}\) and \(d(x_{p},x_{k_{0}})< \frac{3\epsilon}{7} + \frac{\delta '(\epsilon)}{3} \). If \(p= k_{0}\), we have \(f(\gamma)= x_{k_{0}+1} \in \Omega\) by (3). Then we will assume that \(p > k_{0}\). We distinguish two cases as follows.
(1) First case: Assume that
First, we will show that
From (5), we have
Moreover, by using (3) and (5), we get
Then we obtain
From (7) and (8), we deduce that (6) is satisfied. In this case, the inequality
implies by (1) that
Now, using the triangular inequality together with (3) and (9), we obtain that
This implies that \(f(\gamma)= f(x_{p})= x_{p+1}\in\Omega\).
(2) Second case: Suppose that
From (2), we infer that
On the other hand, from (3) we have
We consider the following two situations.
-
(i)
If \(d(x_{k_{0}},x_{k_{0}+1})\leq d(x_{k_{0}},x_{p})\), then (11) gives
$$d\bigl(f(x_{p}),x_{k_{0}}\bigr)< \frac{1}{3}d(x_{p},x_{p+1})+ \frac{1}{3}d(x_{p},x_{p+1}) + \frac {1}{3}d(x_{p},x_{k_{0}})+ \frac{4}{3}d(x_{k_{0}+1},x_{k_{0}}). $$From (3) and (10), we deduce that
$$\begin{aligned} d\bigl(f(x_{p}),x_{k_{0}}\bigr) < & \frac{2}{3}\biggl( \frac{\delta'(\epsilon)}{6}\biggr)+ \frac {1}{3}\biggl( \frac{3\epsilon}{7}\biggr)+ \frac {4}{3}\biggl( \frac{\delta'(\epsilon)}{6}\biggr) \\ =& \frac{\delta'(\epsilon)}{3}+ \frac {\epsilon}{7} \\ < & \frac{\delta'(\epsilon)}{3}+ \frac {3\epsilon}{7}. \end{aligned}$$ -
(ii)
If \(d(x_{k_{0}},x_{k_{0}+1})> d(x_{k_{0}},x_{p})\), then
$$\begin{aligned} d\bigl(f(x_{p}),x_{k_{0}}\bigr) \leq&d(x_{p+1},x_{p})+ d(x_{p},x_{k_{0}}) \\ < &d(x_{p+1},x_{p})+ d(x_{k_{0}},x_{k_{0}+1}) \\ < & \frac{\delta'(\epsilon)}{6} + \frac {\delta'(\epsilon)}{6} \\ =& \frac{\delta'(\epsilon)}{3} \\ < & \frac{\delta'(\epsilon)}{3}+ \frac {3\epsilon}{7}. \end{aligned}$$
In both situations (i) and (ii), we have \(f(\gamma)= f(x_{p})= x_{p+1} \in\Omega\). Thus, (4) holds and
Now, \(\forall m, n \in\mathbb{N}\) satisfying \(m > n> k_{0}\), by (12), we have
Therefore, \(\{x_{n}\}\) is a Cauchy sequence in X.
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.
To show the uniqueness, we assume that \(u'\) is another fixed point of f. From (2) it follows that
which is a contradiction. This proves the uniqueness of the fixed point and completes the proof of the theorem. □
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])
Let \((X,d)\) be a complete metric space and f be a continuous mapping from X into itself. Assume that f satisfies
where \(k\in\, ]0, \frac{1}{3}[\) is a constant. Then f has a unique fixed point \(u \in X\). Moreover, \(\forall x \in X\), the sequence \(\{f^{n}(x)\}\) converges to u.
Proof
Let \(\epsilon> 0\). If we take
then, whenever
Notice that since \(k<\frac{1}{3}\), then \(\frac{\epsilon}{k}>3\epsilon\). Thus the condition (1) of Theorem 2.1 is satisfied, which completes the 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
Let \((X,d)\) be a metric space and let f be a self-mapping defined on X. Assume that there exists a function ρ from \([0,+\infty[\) into itself satisfying the following:
-
(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 that
$$\begin{aligned}& 3\epsilon\leq\rho \biggl( \frac {(1+d(x,f(x)))d(y,f(y))}{1+d(x,y)}+ \frac {d(x,f(x))d(y,f(y))}{d(x,y)}+d(x,y) \biggr)< 3 \epsilon+ \delta (\epsilon) \\& \quad \Longrightarrow\quad \rho \bigl(3d\bigl(f(x),f(y)\bigr) \bigr) < 3\epsilon \end{aligned}$$for all \(x,y \in X\) with \(x\neq y\).
Then (1) is satisfied.
Proof
Fix \(\epsilon> 0\). Since \(\rho(3\epsilon) > 0\), by (iii), for \(\rho(3\epsilon)\) there exists \(\theta>0\) such that
From the right continuity of ρ, there exists \(\delta> 0\) such that \(\rho(3\epsilon+ \delta) < \rho(3\epsilon) +\theta\). Fix \(x,y \in X\), \(x\neq y\) such that
Since ρ is nondecreasing, we deduce
Then, by (14), we have
which implies that \(d(f(x),f(y))<\epsilon\). Then (1) is satisfied and this completes the proof. □
Now, we denote by Ξ the set of all mappings \(h: [0,+\infty[ \, \rightarrow[0, +\infty[\) satisfying:
-
(i)
h is continuous and nondecreasing;
-
(ii)
\(h(0)=0\) and \(h(t)> 0\) for all \(t > 0\).
Corollary 3.2
Let \((X,d)\) be a metric space and let f be a mapping from X into itself. Assume that for each \(\epsilon> 0\), there exists \(\delta (\epsilon)\) such that
for all \(x,y \in X\), with \(x\neq y\), where \(h \in\Xi\) is a given function. Then (1) is satisfied.
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
Let \((X,d)\) be a metric space and let f be a 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 for each \(\epsilon>0\) there exists \(\delta(\epsilon)\) such that
Then (1) is satisfied.
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.
For all \(x,y \in X\), \(x\neq y\),
where \(\mu\in\, ]0,1[\). Then f has a unique fixed point \(u \in X\). Moreover, for any \(x\in X\), the sequence \(\{f^{n}(x)\}\) converges to u.
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
Note that the result of Corollary 2.2 can be established from Corollary 3.4 by taking \(\varphi\equiv1\) and \(\mu= 3k\), \(k \in\, ]0, \frac {1}{3}[\). Clearly, for this choice, (16) becomes
which is exactly the contractive condition of Corollary 2.2.
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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.
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Redjel, N., Dehici, A. & Erhan, İ.M. A fixed point theorem for Meir-Keeler type contraction via Gupta-Saxena expression. Fixed Point Theory Appl 2015, 115 (2015). https://doi.org/10.1186/s13663-015-0363-9
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DOI: https://doi.org/10.1186/s13663-015-0363-9