Abstract
In this paper, we introduce the notion of a quasi-contraction restricted with a linear bounded mapping in cone metric spaces, and prove a unique fixed point theorem for this quasi-contraction without the normality of the cone. It is worth mentioning that the main result in this paper could not be derived from Ćirić’s result by the scalarization method, and hence indeed improves many recent results in cone metric spaces.
MSC:06A07, 47H10.
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1 Introduction
Let be a complete metric space. Recall that a mapping is called a quasi-contraction, if there exists such that
Ćirić [1] introduced and studied Ćirić’s quasi-contractions as one of the most general classes of contractive-type mappings. He proved the well-known theorem that every Ćirić’s quasi-contraction T has a unique fixed point. Recently, Ilić and Rakočević [2] generalized this notion to cone metric spaces and extended Ćirić’s result to the setting of cone metric spaces, which was then improved by [3, 4]. Afterward, Arandelović and Kečkić [5] considered nonlinear quasi-contractions in cone metric spaces, and proved a fixed point result by the nonlinear scalarization method of Du [6]. It should be pointed out that Theorem 2.1 of [2], Theorem 2.2 of [3], and Theorem 2.1 of [4] could be derived from Ćirić’s result by the nonlinear scalarization method of Du [6]; see (ii) of Remark 2.
In this paper, we introduce the notion of a quasi-contractions restricted with a linear bounded mapping in cone metric spaces. Without using the normality of the cone, we prove the unique existence of fixed point for this quasi-contraction at the expense of
It is worth mentioning that the main result in this paper could not be derived from Ćirić’s result by the scalarization method, and hence it indeed improves the corresponding results of [1–4].
2 Preliminaries
Let E be a topological vector space. A cone of E is a nonempty closed subset P of E such that for each and each , and , where θ is the zero element of E. A cone P of E determines a partial order ⪯ on E by for each . In this case E is called an ordered topological vector space.
A cone P of a topological vector space E is solid if , where intP is the interior of P. For each with , we write . Let P be a solid cone of a topological vector space E. A sequence of E weakly converges [7] to (denote ) if for each , there exists a positive integer such that for all .
A subset D of a topological vector space E is order-convex if for each with , where . An ordered topological vector space E is order-convex if it has a base of neighborhoods of θ consisting of order-convex subsets. In this case the cone P is said to be normal. In the case of a normed vector space, this condition means that the unit ball is order-convex, which is equivalent to the condition that there is some positive number N such that and implies that , and the minimal N is called a normal constant of P. Another equivalent condition is that
Then it is not hard to conclude that P is a non-normal cone of a normed vector space if and only if there exist sequences such that
which implies that the Sandwich theorem does not hold. However, in the sense of weak convergence, the Sandwich theorem still holds even if P is non-normal, and we have the following lemma.
Lemma 1 Let P be a solid cone of a topological vector space E and . If
and there exists some such that and , then .
Proof By and , for each , there exists some positive integer such that for all ,
Thus we have for all , i.e., . The proof is complete. □
The following lemma needed in the further arguments, directly follows from Lemma 1 and Remark 1 of [7].
Lemma 2 Let P be a solid cone of a normed vector space . Then for each sequence , implies . Moreover, if P is normal, then implies .
Let X be a nonempty set and P be a cone of a topological vector space E. A cone metric on X is a mapping such that for each ,
(d1) ;
(d2) ;
(d3) .
The pair is called a cone metric space over P. A cone metric d on X over a solid cone P generates a topology on X which has a base of the family of open d-balls , where for each and each .
Let be a cone metric space over a solid cone P of a topological vector space E. A sequence of X converges [7, 8] to (denote by ) if . A sequence of X is Cauchy [7, 8], if . The cone metric space is complete [7, 8], if each Cauchy sequence of X converges to a point .
3 Main results
Let P be a solid cone of a normed vector space . A mapping is called a quasi-contraction restricted with a linear bounded mapping, if there exists and a linear bounded mapping with such that
Moreover, if A is a contractive mapping (i.e., is a one-to-one mapping such that , is one-to-one and ), then T is reduced to the one considered by Arandelović and Kečkić [5]. In particular when with , T is an Ilić-Rakočević’s quasi-contraction [2], and also an Arandelović-Kečkić’s quasi-contraction.
Remark 1 (i) If is a mapping such that and , then it satisfies (H), and so every contractive mapping A satisfies (H). In fact, let with . By the definition of weak convergence, for each , there exists a positive integer such that , i.e., .
(ii) If P is normal then every linear bounded mapping satisfies (H). In fact, let with . By Lemma 2 and the normality of P, . And so for each linear bounded mapping . Moreover, by Lemma 2, .
The following example shows that there does exist some linear bounded mapping satisfying (H) as P is non-normal.
Example 1 Let with the norm and which is a non-normal solid cone [9]. Let for each and each .
For each , there exists some such that . For each with , there exists a positive integer such that for all , , and hence for each . Thus we have for all , i.e., . By the definition of A, we get , which together with implies that . Thus by Lemma 2, we have . This shows that A satisfies (H).
In the following, without using the normality of P, we show the unique existence of fixed point of quasi-contractions restricted with linear bounded mappings at the expense that (H) is satisfied.
Theorem 1 Let be a complete cone metric space over a solid cone P of a normed vector space and a quasi-contraction restricted with a linear bounded mapping. If the spectral radius and (H) is satisfied, then T has a unique fixed point such that for each , , where for all n.
Proof By , the inverse of exists, denote it by . Moreover, by Neumann’s formula,
which together with implies that . It follows from and Gelfand’s formula that there exist and some positive integer such that
We claim that for all ,
In the following we shall show this claim by induction.
If , then , and so the claim is trivial.
Assume that (4) is true for n. To prove that (4) holds for , it suffices to show that
By (1),
where
Consider the case that .
If , then by (2), (4), (6), and ,
i.e., (5) holds.
If , then by (6) and ,
i.e., (5) holds.
If , then by (6) and ,
which implies that
Act on the above inequality with , then by ,
i.e., (5) holds.
If , then by (2), (4), (6), and ,
i.e., (5) holds.
If , then set , and so
Consider the case that .
If , or , or , then by (2), (4), (6), and ,
i.e., (5) holds.
If , or , then set , or , respectively, and so (7) follows.
From the above discussions of both cases, we have the result that either (5) holds, and so the proof of our claim is complete, or there exists such that (7) holds. For the latter situation, continue in a similar way, and we will have the result that either
which together with (7) forces that
i.e., (5) holds, and so the proof of our claim is complete, or there exists such that
If the above procedure ends by the k th step with , that is, there exist integers such that
then by (2) and ,
i.e., (5) holds, and so the proof of our claim is complete.
If the above procedure continues more than n steps, then there exist integers such that
It is clear that implies that there exist two integers with such that , then by (8),
and so
Note that , then by Neumann’s formula and , is invertible (denote its inverse by ), and . Acting (9) with , we get , and hence (5) holds by (9). The proof of our claim is complete.
For all and each , set
From (1), it follows that, for each , there exists some such that . Consequently for all , there exists () such that
Note that , which together with (4) implies that
Thus by (10),
It follows from (3) that (), and hence (), which together with Lemma 2 implies that (). Moreover by (11) and Lemma 1, we get
i.e., is a Cauchy sequence of X. Therefore by the completeness of X, there exists some such that (), i.e.,
By (1),
where .
If , or , or , then by (12), (13), (14), (H), and Lemma 1, we get and hence .
If , then by (14),
and hence by (13), for each ,
which implies that
Acting (16) with , by we get and hence .
If , then by (14), we have
and so
Thus it follows from (13) and (H) that (15) holds for each . Consequently, we obtain (16). Acting (16) with , by we get and hence . This shows that is a fixed point of T.
If x is another fixed point of T, then by (1),
where . If ,or , then , and hence . If , or or , then we must have , and hence . Acting on it with , by we have . This shows is the unique fixed point of T. The proof is complete. □
Following (i) of Remark 1 and the proof of Theorem 1, we have the following result.
Corollary 1 (see [[4], Theorem 2.1])
Let be a complete cone metric space over a solid cone P of a topological vector space E and an Ilić-Rakočević’s quasi-contraction. Then T has a unique fixed point such that for each , , where for all n.
Remark 2 (i) Theorem 2.1 of [2], Theorem 2.2 of [3], and Theorem 2.1 of [4] are special cases of Theorem 1 with for .
(ii) Let , where is defined by
for some . Then is a metric on X by Theorem 2.1 of [6]. If T is a quasi-contraction with , then applying Lemma 1.1 of [6], we have
and hence Theorem 2.1 of [2], Theorem 2.2 of [3], and Theorem 2.1 of [4] directly follow from Ćirić’s result by Theorem 2.3 of [6].
-
(iii)
If T is a quasi-contraction restricted with a linear bounded mapping, one may not be sure that there exists some such that (17) is satisfied, and so Theorem 1 cannot be derived from Ćirić’s result. Therefore Theorem 1 indeed improves the corresponding result of [1–4].
-
(iv)
Compared with Corollary 1 of [5], the mapping A is no longer necessarily assumed to be contractive in Theorem 1.
Remark 3 It is not hard to see from the proof of Theorem 1 that Theorem 1 remains valid in the setting of partial cone metric spaces.
Remark 4 If A is a linear unbounded mapping, then may not be invertible and (3) is not satisfied, and hence the method used in Theorem 1 becomes invalid.
The following example shows the usability of Theorem 1.
Example 2 Let E and P be the same ones as those in Example 1 and . Define a mapping by
Then is a complete cone metric space. Let for each and each . Then from Example 1 we find that A satisfies (H). It is clear that
i.e., (1) is satisfied. Since for each , then . Moreover, from we have
which implies that . By Gelfand’s formula, since . Thus all the assumptions of Theorem 1 are satisfied, and hence T has a unique fixed point. In fact, θ is the unique fixed point of T.
Note that if T is not an Ilić-Rakočević’s quasi-contraction and A is not a contractive linear bounded mapping (let for each , then for each , and so , which implies that ), then the fixed point results of [2–5] are not applicable in this case.
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Acknowledgements
The work was supported by Natural Science Foundation of China (11161022), Natural Science Foundation of Jiangxi Province (20114BAB211006, 20122BAB201015), Educational Department of Jiangxi Province (GJJ12280, GJJ13297) and Program for Excellent Youth Talents of JXUFE (201201). The authors express their gratitude to the referees for their helpful suggestions which improved the proof of Theorem 1.
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Li, Z., Jiang, S. Quasi-contractions restricted with linear bounded mappings in cone metric spaces. Fixed Point Theory Appl 2014, 87 (2014). https://doi.org/10.1186/1687-1812-2014-87
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DOI: https://doi.org/10.1186/1687-1812-2014-87