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A note on the paper ‘Fixed point theorems for cyclic weak contractions in compact metric spaces’
- Zoran Kadelburg^{1},
- Stojan Radenović^{2, 3}Email author and
- Jelena Vujaković^{4}
https://doi.org/10.1186/s13663-016-0537-0
© Kadelburg et al. 2016
- Received: 27 January 2016
- Accepted: 30 March 2016
- Published: 6 April 2016
Abstract
We show that the result on cyclic weak contractions of Harjani et al. (J. Nonlinear Sci. Appl. 6:279-284, 2013) holds without the assumption of compactness of the underlying space, and also without the assumption of continuity of the given mapping.
Keywords
- compact metric space
- cyclic contraction
- weak contraction
- well posed problem
MSC
- 47H10
- 54H25
1 Introduction
In 2003, Kirk et al. introduced in [2] an interesting concept of cyclic contractions in metric spaces, and obtained the corresponding generalizations of Banach’s, as well as some other fixed point results. In subsequent articles, several authors obtained various fixed point theorems, adapting some other known results to their cyclic variants (see, e.g., [1, 3–9]).
The first results for cyclic contractions in compact metric spaces were obtained already in [2]. Recently, Harjani et al. [1] obtained a cyclic fixed point result in compact spaces for weak contractions (in the sense of [10, 11]), thus modifying a theorem of Karapınar [4] and Karapınar and Sadarangani [5].
In this note, we are going to improve the result from [1] showing that it still holds without the assumption of compactness of the underlying space, and also without the assumption of continuity of the given mapping. An application to well posedness of the corresponding problem is also obtained.
2 Preliminaries
The main result in [4, 5] is the following.
Theorem 1
([4], Theorem 6, [5], Theorem 2)
We will use the following auxiliary result.
Lemma 1
([9], Lemma 2.1)
Remark 1
Using the previous lemma, similar to [9], one can prove that there is an equivalence between Theorem 1 and the corresponding fixed point result for non-cyclic weak contraction. In the case when \(Y=X=\bigcup_{i=1}^{p}A_{i}\), no assumption on the closedness of sets \(A_{i}\) is necessary, while in the case when \(Y\ne X\), it is enough to assume that one of these sets is closed (similarly as in [3], Theorem 2.1). However, in this case the cyclic and non-cyclic versions of this assertion are not equivalent anymore.
Theorem 2
([1], Theorem 2.1)
Let \((X,d)\) be a compact metric space and \(T:X\rightarrow X\) a continuous mapping. Suppose that \(X=\bigcup_{i=1}^{p}A_{i}\) is a cyclic representation of X with respect to T. If, for some \(\phi\in\mathcal{F}\), relation (2.2) holds for any \(x\in A_{i}\) and \(y\in A_{i+1}\), \(i=1,2,\ldots,p\) where \(A_{p+1}=A_{1}\), then T has a unique fixed point \(z\in \bigcap_{i=1}^{p}A_{i}\).
Remark 2
Note also that [1], Remark 2.2, is not correct. Namely, Theorem 2 is not a generalization of Theorem 1, since it uses the assumption \(Y=X\), which is not present in Theorem 1.
The following result is a version of [4], Theorem 7, when \((X,d)\) is a compact metric space.
Theorem 3
([1], Theorem 2.3)
Under the assumptions of Theorem 2, the fixed point problem for T is well posed, that is, if there exists a sequence \(\{y_{n}\}\) in X with \(d(y_{n},Ty_{n}) \rightarrow0\) as \(n\rightarrow \infty\), then \(y_{n}\rightarrow z\) as \(n\rightarrow\infty\), where z is the unique fixed point of T.
3 Results
Our main result is the following improvement of Theorem 2, that is, the main result from [1]. Note that compactness of the space is not assumed, nor is the continuity of T. Also, the subsets \(A_{i}\) of X need not be closed. Moreover, the proof is much shorter than the proof of the relevant theorem in [1].
Theorem 4
Let \((X,d)\) be a complete metric space and \(T:X\rightarrow X\) be a mapping. Suppose that \(X=\bigcup_{i=1}^{p}A_{i}\) is a cyclic representation of X with respect to T. If, for some \(\phi\in \mathcal{F}\), relation (2.2) holds for any \(x\in A_{i}\) and \(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where \(A_{p+1}=A_{1}\), then T has a unique fixed point \(z\in X\) and \(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence \(x_{n}=T^{n}x_{0}\), \(x_{0}\in X\) converges to z.
Proof
The uniqueness of the fixed point follows from the contractive condition (2.2), since, again by (2.1), each fixed point of T belongs to \(\bigcap_{i=1}^{p}A_{i}\). □
Taking \(\phi(t)=(1-k)t\), \(t\in[0,+\infty)\), \(k\in[0,1)\) in Theorem 4, we obtain the following version of a Banach-type cyclic contraction result.
Corollary 1
Let \((X,d)\) be a complete metric space and \(T:X\rightarrow X\) be a mapping. Suppose that \(X=\bigcup_{i=1}^{p}A_{i}\) is a cyclic representation of X with respect to T. If, for some \(k\in[0,1)\), the inequality \(d(Tx,Ty)\leq kd(x,y)\) holds for any \(x\in A_{i}\) and \(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where \(A_{p+1}=A_{1}\), then T has a unique fixed point \(z\in X\) and \(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence \(x_{n}=T^{n}x_{0}\), \(x_{0}\in X\), converges to z.
Example 1
(1) \(x\in(-\infty,-2]\), (2) \(x\in(-2,0)\), (3) \(x\in[0,2)\), as well as: (a) \(y\in(-2,0)\), (b) \(y\in[0,2)\), (c) \(y\in[2,+\infty)\).
Case 1(a). \(Tx=-\frac{1}{3}x<-\frac{1}{2}x\), \(Ty=-\frac{1}{2}y\), and \(d(Tx,Ty)<\frac{1}{2}(-x-(-y))=\frac{1}{2}(y-x)=\frac{1}{2}d(x,y)\).
Case 1(b). \(Tx=-\frac{1}{3}x<-\frac{1}{2}x\), \(Ty=-\frac{1}{2}y\), \(d(Tx,Ty)<\frac{1}{2}(-x-(-y))=\frac{1}{2}(y-x)=\frac{1}{2}d(x,y)\).
Case 1(c). \(Tx=-\frac{1}{3}x\), \(Ty=-\frac{1}{3}y\), \(d(Tx,Ty)=\frac{1}{3}(-x-(-y))=\frac{1}{3}(y-x)<\frac{1}{2}d(x,y)\).
Case 2(a). \(Tx=-\frac{1}{2}x\), \(Ty=-\frac{1}{2}y\), \(d(Tx,Ty)=\frac{1}{2}|x-y|=\frac{1}{2}d(x,y)\).
Case 2(b). \(Tx=-\frac{1}{2}x\), \(Ty=-\frac{1}{2}y\), \(d(Tx,Ty)=\frac{1}{2}(x-y)=\frac{1}{2}d(x,y)\).
The cases 2(c), 3(a), 3(b), and 3(c) are symmetric to the cases 1(b), 2(b), 2(a), and 1(a), respectively. On the other hand, \(\phi(t)\geq\frac{1}{2}t\) for all \(t\geq0\).
- 1.
X is not compact;
- 2.
T is not continuous;
- 3.
\(A_{i}\), \(i\in\{1,2\}\), are not closed,
The following modification of Theorem 4 can be proved nearly in the same way.
Theorem 5
Let X be a nonempty set, and \(A_{1}\), \(A_{2}\), …, \(A_{p}\) be its nonempty subsets, with at least one of them being closed. Let \(Y=\bigcup_{i=1}^{p}A_{i}\) be a cyclic representation of Y with respect to a mapping \(T:Y\to Y\). If, for some \(\phi\in \mathcal{F}\), relation (2.2) holds for any \(x\in A_{i}\) and \(y\in A_{i+1}\), \(i=1,2,\ldots,p\), where \(A_{p+1}=A_{1}\), then T has a unique fixed point \(z\in Y\) and \(z\in\bigcap_{i=1}^{p}A_{i}\). Moreover, each Picard sequence \(x_{n}=T^{n}x_{0}\), \(x_{0}\in Y\) converges to z.
Proof
Suppose, without loss of generality, that the subset \(A_{1}\) is closed (and hence complete) in X. The only difference in the proof, compared with Theorem 4, is that, after proving that \(\{x_{n}\}\) is a Cauchy sequence (in Y), we conclude that there is \(z\in A_{1} ({}\subseteq Y)\) such that \(x_{n}\to z\), as \(n\to\infty\). See also the proof of [3], Theorem 2.1. □
As another consequence of Theorem 4, we obtain the following improvement of Theorem 3.
Corollary 2
Under the assumptions of Theorem 4, the fixed point problem for T is well posed, that is, if \(\{y_{n}\} \) is a sequence in X satisfying \(d( y_{n},Ty_{n}) \rightarrow0\) as \(n\rightarrow\infty\), then \(y_{n}\rightarrow z\) as \(n\rightarrow \infty\), where z is the unique fixed point of T (whose existence is guaranteed by Theorem 4).
Proof
Since, according to Theorem 4, it follows that the unique fixed point z of T belongs to \(\bigcap_{i=1}^{p}A_{i}\), the rest of the proof is the same as in [1], p. 282. Namely, this proof uses neither compactness of X, nor continuity of T. □
Remark 3
As a kind of conclusion, we state once more that there are two kinds of cyclic fixed point results - those treating mappings \(T:Y\to Y\), where \(Y=\bigcup_{i=1}^{p}A_{i}\subset X\) (possibly with \(Y\ne X\)) and those where \(Y=X\). In results of the first kind, it is enough to assume that one of the sets \(A_{i}\) is closed, while in results of the second kind, no closedness assumption is needed. In both cases, if the considered mapping T is continuous, all the results reduce to the case when all \(A_{i}\)’s are closed.
Declarations
Acknowledgements
The authors are indebted to the referees of this paper who helped us to improve the text. The first author is thankful to the Ministry of Education, Science and Technological Development of Serbia, Grant No. 174002.
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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