Abstract
We prove a fixed point theorem for a Pata-type map defined on a complete (normal) cone metric space. Our results generalize the recent work of M. Chakraborty and S. K. Samanta. An example demonstrating this fact is also presented.
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Mathematics Subject Classification 47H10 · 54H25 (Primary) · 37C25 (Secondary)
Introduction
The classical Banach fixed point theorem states that if is a complete metric space and is a contraction map, i.e., satisfies
for all and some , then has a unique fixed point. i.e., there exists a unique such that .
In [4], Pata considered a map on the complete metric space that satisfied the condition: for all ,
for every , fixed constants , and , a fixed element , and an increasing function which vanishes at and is continuous at . He proved that the map satisfying (2) has a unique fixed point. Moreover, he also demonstrated that if is a contraction map, then satisfies condition (2), thereby obtaining a generalization of the Banach fixed point theorem.
Another fixed point theorem that is widely popular, is due to Kannan which states that if is a complete metric space and the map is a Kannan contraction, i.e., satisfies
for all and some , then has a unique fixed point.
There are several generalizations of the Kannan fixed point theorem, but the one of particular interest to us is due to Chakraborty and Samanta in [1]. The authors, in [1], consider a map defined on the complete metric space that satisfies the condition: for all ,
for every , fixed constants , and , a fixed element , and a function which vanishes at and is continuous at . The authors prove that the map has a unique fixed point. Moreover, they also demonstrate that if is a Kannan contraction map, then satisfies condition (4).
This article contains a generalization of the main result in [1]. The setting considered is that of complete cone metric spaces, where the underlying cone is normal (see [3]). It is shown that for a map defined on the complete (normal) cone metric space , which satisfies a Pata-type condition, that is an improved version of (4), there exists a unique fixed point. An example to illustrate the main result is also provided.
Preliminaries and the main result
Let E be a real Banach space. A non-empty closed subset P of E is said to be a if
-
(a)
for all .
-
(b)
, where is the zero vector.
The cone is said to be if the interior of , which we will denote by , is non-empty.
Examples of solid cones
-
(1)
Let and .
-
(2)
Let and .
-
(3)
Let and .
The norms on in the examples above are the usual norms.
A cone in a real Banach space E, induces the following partial order on E. For ,
In the case of a solid cone , we will use the notation to denote .
A cone is said to be if for all such that , there exists a constant such that . The examples (1), (2) and (3) above are normal cones with .
Let be a nonempty set, be a real Banach space and be a solid normal cone with normal constant . A map is said to be a cone metric if for all ,
-
(a)
, i.e., .
-
(b)
if and only if .
-
(c)
.
The pair is called acone metric space. It is indeed the case that every metric space is a cone metric space.
Examples of cone metric spaces
-
(1)
([3]) Let and be as in example (2) above, be the map , where is a constant. The pair is a cone metric space.
-
(2)
([2]) Let be a metric space, and be as in example (3) above, be the map It can be verified that is a cone metric space.
Let be a cone metric space. A sequence of points in is said to be Cauchy if for any given , there exists such that , for all , or equivalently, there exists which is independent of such that for all .
The sequence is said to beconvergent, if there exists such that for any given , there exists such that for all .
The cone metric space is said to becomplete if every Cauchy sequence in converges.
The proof of the following lemma can be found in [3].
Lemma 1
Let be a cone metric space and be a sequence in .
-
(1)
is Cauchy if and only if as . i.e., Given and , there exists which is independent of such that , for all .
-
(2)
is convergent to if and only if as .
-
(3)
If converges to , then .
The following is our main result which generalizes Theorem 2.2 in [1].
Theorem 1
Let be a complete cone metric space with normal constant , , , and be fixed constants and be such that . If for every , the map satisfies
for every , where and denotes , then has a unique fixed point.
The proof of the above Theorem is given in Sect. 3. We point out that there is no loss of generality in choosing any such in (5), simply because a change in can essentially be absorbed by assigning a different value to , thanks to the triangle inequality of the cone metric and the sub-additivity of the norm.
Proof of The main result
This section contains a proof of our main result. Although the proof follows a similar pattern as that of Theorem 2.2 in [1], the arguments provided here are different and sometimes simpler.
Proof of Theorem 1
First we prove uniqueness of the fixed point. Suppose that are such that , and . Letting in inequality (5) yields, , a contradiction to the fact that . The uniqueness follows.
To show the existence of a fixed point, consider the sequence . Without loss of generality, we assume that , for all Since for each , and ,
it follows that there exists no , for which . For otherwise, it would mean that there exists some such that for every ,
Letting yields, , i.e., , a contradiction. Thus, for each , letting in inequality (6) yields,
Iterating we obtain
for all
Next, we show that the sequence is bounded above by . This is certainly the case when . Assume that . The claim follows from induction, if we show that . Using (8), it follows that
Thus, for all ,
Consider the monotonically decreasing sequence [see (7)]. Since it is bounded below by , it is convergent. Let . Clearly . We will in fact, prove that . For each , it follows from (6) and (9) that
where . Rearranging the above inequality and using the fact that the sequence is monotonically decreasing yields,
for each . Fixing such an and letting in (10), we obtain
Multiplying both sides by and simplifying yields, for each ,
Letting yields, . Thus, in fact,
Next, we show that the sequence is Cauchy. In view of Lemma 1 (i), it suffices to show that, given and , there exists which is independent of , such that , for all .
By (11), choose such that
for all .
Consider for all . Letting in (5) yields,
If or , then it follows from (12) that
If , then the normality of the underlying cone and an application of the triangle inequality yield,
From a rearrangement of terms, it follows from (12) that
Thus, it follows from (13) and (14) that given and , by choosing , which is independent of , one has that
for all . i.e., the sequence is Cauchy in the cone metric space . By completeness, there exists a unique such that as .
We complete the proof by showing that this is a fixed point of . Let be arbitrary, . Choose such that
Choose such that
Consider . From (5), it follows that
Rearranging the terms in the above inequality yields,
It follows from (15) and (16) that
Since is arbitrary, the proof is complete.
Applications and examples
This section contains applications of Theorem 1, presented as corollaries. Recall the definition of and the notation from Theorem 1.
Corollary 1
Let be a complete cone metric space with normal constant and . If the map satisfies
for all , then has a unique fixed point.
Proof
It suffices to show that satisfies condition (5). The result then follows from Theorem 1. Fix . Observe that
It follows from (17) and a Bernoulli inequality argument similar to Sect. 3 in [1], that for all , ,
where . Comparing (18) with (5), one sees that satisfies (5) with , , .
Corollary 2
(The Kannan Fixed Point Theorem) Let be a complete metric space and . If the map satisfies
for all , then has a unique fixed point.
Proof
Observe that and satisfies the hypothesis of Corollary 1. The result follows.
We end with an example which demonstrates that Theorem 1 indeed generalizes the main result in [1]. Observe that it suffices to produce a complete metric space and a map which satisfies
-
(A)
for some and ,
-
(B)
for some .
Example
Let with being the usual metric. It is clear that is a complete cone metric space with . Define the map by if and . Observe that for , and
Similarly, for , and
Since, by choice, , it follows that . Moreover, for all . Putting all this together implies that satisfies condition (A) above with . Since , it follows that also satisfies condition (B) above with and . From the proof of Corollary 1, it follows that satisfies (5). However, does not satisfy (4), as it satisfies condition (B) above. This can be seen by setting in (4). It is immediate that has a unique fixed point.
References
Chakraborty, M., Samanta, S.K.: A fixed point theorem for kannan-type maps in metric spaces, pre-print (2012), arXiv:1211.7331v2 [math.GN], Nov 2012
Haghi, R.H., Rezapour, Sh.: Fixed points of multifunctions on regular cone metric spaces. Expositiones Mathematicae 28(1), 71–77 (2010)
Huang, L.-G., Zhang, X.: Cone metric spaces and fixed point theorems of contractive mappings. J. Math. Anal. Appl. 332, 1468–1476 (2007)
Pata, V.: A fixed point theorem in metric spaces. J. Fixed Point Theory Appl. 10, 299–305 (2011)
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Balasubramanian, S. A Pata-type fixed point theorem. Math Sci 8, 65–69 (2014). https://doi.org/10.1007/s40096-014-0127-4
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DOI: https://doi.org/10.1007/s40096-014-0127-4