Abstract
We present a fixed-point theorem for a single-valued map in a complete metric space using implicit relation, which is a generalization of several previously stated results including that of Suziki (2008).
1. Introduction
There are a lot of generalizations of Banach fixed-point principle in the literature. See [1–5]. One of the most interesting generalizations is that given by Suzuki [6]. This interesting fixed-point result is as follows.
Theorem 1.1. Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into by Assume that there exists , such that for all , then there exists a unique fixed-point of . Moreover, for all .
Like other generalizations mentioned above in this paper, the Banach contraction principle does not characterize the metric completeness of . However, Theorem 1.1 does characterize the metric completeness as follows.
Theorem 1.2. Define a nonincreasing function as in Theorem 1.1, then for a metric space the following are equivalent: (i) is complete,(ii)Every mapping on satisfying (1.2) has a fixed point.
In addition to the above results, Kikkawa and Suzuki [7] provide a Kannan type version of the theorems mentioned before. In [8], it is provided a Chatterjea type version. Popescu [9] gives a Ciric type version. Recently, Kikkawa and Suzuki also provide multivalued versions which can be found in [10, 11]. Some fixed-point theorems related to Theorems 1.1 and 1.2 have also been proven in [12, 13].
The aim of this paper is to generalize the above results using the implicit relation technique in such a way that for , where is a function as given in Section 2.
2. Implicit Relation
Implicit relations on metric spaces have been used in many papers. See [1, 14–16].
Let denote the nonnegative real numbers, and let be the set of all continuous functions satisfying the following conditions:: is nonincreasing in variables ,: there exists , such that or or implies ,:, for all .
Example 2.1. , where . It is clear that .
Example 2.2. , where .
Let , then we have . Similarly, let , then we have . Again, let , then . Since , is satisfied with . Also , for all . Therefore, .
Example 2.3. , where .
Let , then we have . Similarly, let , then we have . Again, let , then . Thus, is satisfied with . Also , for all . Therefore, .
Example 2.4. , where .
Let , then we have . Similarly, let , then we have . Again, let , then . Since , is satisfied with . Also , for all . Therefore, .
Example 2.5. , where .
Let , then we have . Similarly, let , then we have . Again, let , then . Thus, is satisfied with . Also , for all . Therefore, .
3. Main Result
Theorem 3.1. Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into as in Theorem 1.1. Assume that there exists , such that implies for all , then has a unique fixed-point and holds for every .
Proof. Since , holds for every , by hypotheses, we have
and so from ,
By , we have
for all . Now fix and define a sequence in by . Then from (3.4), we have
This shows that , that is, is Cauchy sequence. Since is complete, converges to some point . Now, we show that
For , there exists , such that for all . Then, we have
Hence, by hypotheses, we have
and so
Letting , we have
and so
By , we have
and this shows that (3.6) is true.
Now, we assume that for all , then from (3.6), we have
for all .
Case 1. Let . In this case, . Now, we show by induction that
for . From (3.4), (3.14) holds for . Assume that (3.14) holds for some with . Since
we have
and so
Therefore, by hypotheses, we have
and so
then
and by , we have
Therefore, (3.14) holds.
Now, from (3.6), we have
This shows that , which contradicts (3.14).
Case 2. Let . In this case, . Again we want to show that (3.14) is true for . From (3.4), (3.14) holds for . Assume that (3.14) holds for some with . Since
we have
and so
Therefore, as in the previous case, we can prove that (3.14) is true for . Again from (3.6), we have
This shows that , which contradicts (3.14).Case 3. Let . In this case, . Note that for , either
or
holds. Indeed, if
then we have
which is a contradiction. Therefore, either
or
holds for every . If
holds, then by hypotheses we have
and so
Letting , we have
which contradicts . If
holds, then by hypotheses we have
and so
Letting , we have
which contradicts .
Therefore, in all the cases, there exists , such that . Since is Cauchy sequence, we obtain . That is, is a fixed point of . The uniqueness of fixed point follows easily from (3.6).
Remark 3.2. If we combine Theorem 3.1 with Examples 2.1, 2.2, 2.3, and 2.4, we have Theorem 2 of [6], Theorem 2.2 of [7], Theorem 3.1 of [7], and Theorem 4 of [8], respectively.
Using Example 2.5, we obtain the following result.
Corollary 3.3. Let be a complete metric space, and let be a mapping on . Define a nonincreasing function from into as in Theorem 1.1. Assume that implies for all , where , then there exists a unique fixed point of .
Remark 3.4. We obtain some new results, if we combine Theorem 3.1 with some examples of .