In the section, we denote by Ψ the class of functions satisfying the following conditions:
() ψ is an increasing and continuous function in each coordinate;
() for , , and .
Next, we denote by Θ the class of functions satisfying the following conditions:
() φ is continuous and non-decreasing;
() for , and .
And we denote by Φ the class of functions satisfying the following conditions:
() ϕ is continuous;
() for , and .
We now state a new notion of cyclic -contractions in partial metric spaces as follows.
Definition 5 Let be a partial metric space, , be nonempty subsets of X and . An operator is called a cyclic -contraction if
-
(1)
is a cyclic representation of Y with respect to f;
-
(2)
for any , , ,
(2.1)
where , , , and .
Theorem 3 Let be a complete partial metric space, , be nonempty closed subsets of X and . Let be a cyclic -contraction. Then f has a unique fixed point .
Proof Given and let for . If there exists such that , then we finished the proof. Suppose that for any . Notice that for any , there exists such that and .
Step 1. We will prove that
Using (2.1), we have
where
If , then
which implies that , and hence . This contradicts our initial assumption.
From the above argument, we have that for each ,
(2.2)
and
And since the sequence is decreasing, it must converge to some . Taking limit as in (2.2) and by the continuity of φ and ϕ, we get
and so we conclude that and . Thus, we have
(2.3)
By (p2), we also have
(2.4)
Since for all , using (2.3) and (2.4), we obtain that
(2.5)
Step 2. We show that is a Cauchy sequence in the metric space . We claim that the following result holds.
Claim For every , there exists such that if with , then .
Suppose the above statement is false. Then there exists such that for any , there are with with satisfying
Now, we let . Then corresponding to use, we can choose in such a way it is the smallest integer with satisfying and . Therefore, and
Letting , we obtain that
(2.6)
On the other hand, we can conclude that
Letting , we obtain that
(2.7)
Since and using (2.4), (2.6) and (2.7), we have that
(2.8)
and
(2.9)
Since and lie in different adjacently labeled sets and for certain , by using the fact that f is a cyclic -contraction, we have
where
Thus, letting , we can conclude that
which implies , that is, . So, we get a contradiction. Therefore, our claim is proved.
In the sequel, we will show that is a Cauchy sequence in the metric space . Let be given. By our claim, there exists such that if with , then
Since , there exists such that
for any .
Let and . Then there exists such that . Therefore, for , and so we have
Thus, is a Cauchy sequence in the metric space .
Step 3. We show that f has a fixed point ν in .
Since Y is closed, the subspace is complete. Then from Lemma 1, we have that is complete. Thus, there exists such that
And it follows from Lemma 1 that we have
(2.10)
On the other hand, since the sequence is a Cauchy sequence in the metric space , we also have
Since , we can deduce that
(2.11)
Since is a cyclic representation of X with respect to f, the sequence has infinite terms in each for . Now, for all , we may take a subsequence of with and also all converge to ν. Using (2.10) and (2.11), we have
By (2.1),
where
Letting , we have
which implies , that is, . So, .
Step 4. Finally, to prove the uniqueness of the fixed point, suppose that μ, ν are fixed points of f. Then using the inequality (2.1), we obtain that
where
So, we also deduce that
which implies that , and hence , that is, . So, we complete the proof. □
The following provides an example for Theorem 3.
Example 1 Let and , , . We define the partial metric p on X by
and define the function by
Now, we let and be
Then f is a cyclic -contraction and 0 is the unique fixed point.
Proof We claim that f is a cyclic -contraction.
-
(1)
Note that , and . Thus, is a cyclic representation of X with respect to f;
-
(2)
For and (or, and ), without loss of generality, we may assume that , then we have
and
Since
we have
On the other hand, for and , without loss of generality, we may assume that , then it is easy to get the above inequality.
Note that Example 1 satisfies all of the hypotheses of Theorem 3, and we get that 0 is the unique fixed point. □