We start this section with the following definition, which can be seen in [9, 16, 17, 30].
Definition 2.1 Let be a partial metric space. A mapping is said to be continuous at if for every , there exists such that .
Definition 2.2 [25]
The function is called an altering distance function, if the following properties are satisfied:
-
(1)
φ is continuous and nondecreasing;
-
(2)
if and only if .
Lemma 2.1 [31]
Let be a partial metric space, be a given mapping. Suppose that T is continuous at . Then, for each sequence in X, in in holds.
Theorem 2.1 Let be a partially ordered set and suppose that there exists a partial metric p on X such that is complete. Let be a continuous nondecreasing mapping. Suppose that for comparable , we have
(2.1)
where
ψ
and
φ
are altering distance functions with
for all , and is a continuous function with if and only if . If there exists such that , then f has a fixed point.
Proof If , then is a fixed point of f. Suppose that , we can choose such that . Since f is a nondecreasing function, we have
Continuing this process, we can construct a sequence in X such that with
It is clear that if for some , then f has a fixed point. Taking for all , now let us prove the following inequality:
(2.3)
Suppose this is not true, then for some , that is,
(2.4)
From (2.1) and (2.4), we obtain that
this together with (2.2) shows that
Using the property of ϕ, we have
(2.5)
Since
applying (2.5), we get
(2.6)
From the property of ψ, we have , which contradicts with for all ; hence (2.3) holds. Therefore, is a nonincreasing sequence, and thus there exists such that
Using (2.1), we obtain
it means that
Letting in the above inequality, we get
the continuity of ϕ guarantees that
and the property of ϕ gives that
(2.8)
Since
on taking inferior limit in the above inequalities and using (2.8), we obtain that and so , therefore,
moreover, we have
Now, we claim that is a Cauchy sequence in the metric space (and so also in the space by Lemma 1.1). For this, it is sufficient to show that is a Cauchy sequence in . Suppose that this is not the case, then using Lemma 1.1 we have that is not a Cauchy sequence in . By Lemma 1.3, we obtain that there exist and two sequences and of positive integers such that and sequences in (1.2) tend to ε when . For two comparable elements and , we can obtain, from (2.1), that
Taking in (2.9), we get
which implies that , hence , a contradiction. Thus, is a Cauchy sequence in and so is a Cauchy sequence both in and in . Since is complete then the sequence converges to some , that is
(2.10)
Moreover, since is a Cauchy sequence in , we have . By and , we have . Then (2.10) yields that
(2.11)
Applying the triangular inequality, we have
taking in the above inequalities, then the continuity of f and Lemma 2.1 give that
hence
By combining (2.1) and (2.12), we have
which yields that , and thus , that is . Therefore, z is a fixed point of f. □
Theorem 2.2 Suppose that X, f, ψ, φ, and ϕ are the same as in Theorem 2.1 except the continuity of f. Suppose that for a nondecreasing sequence in X with , we have for all . If there exists such that , then f has a fixed point.
Proof As in the proof of Theorem 2.1, we have a Cauchy sequence in X. Since is complete, there exists such that , that is,
due to the hypothesis, we get . Similar to the proof of Theorem 2.1, we have that
From (2.1), we obtain that
Letting in the above inequalities, and by Lemma 1.2, we have
which implies, from (2.2), that , hence , and thus . Therefore, f has a fixed point. □
Theorem 2.3 Let be a complete partial metric space, f and g be self-mappings on X. Suppose that for all
(2.13)
where
ψ
and
φ
are altering distance functions with
for all , and is a continuous function with if and only if .
Then f and g have a unique common fixed point.
Proof Let be an arbitrary point in X. One can choose such that . Also, one can choose such that . Continuing this process, one can construct a sequence in X such that
(2.15)
Now, we discuss the following two cases.
Case 1. If for some , then f and g have at least one common fixed point. In fact, if for some , that is , which implies that . If (), then . Using (2.13), we have
(2.16)
With the help of (2.14) and (2.16), we conclude that , hence, using the property of ϕ, we get , that is . By similar arguments, we obtain , and so on. Thus, becomes a constant from , that is,
(2.17)
Equations (2.15) and (2.17) yield that
(2.18)
which implies that is the common fixed point of f and g. Similarly, one can show that if (), then f and g have at least one common fixed point. Therefore, we have proved that if for some , then f and g have at least one common fixed point.
Case 2. If for some , then f and g have at least one common fixed point. Indeed, if (), then . Hence, , due to (2.13), we have
(2.19)
Applying (2.14) and (2.19), we obtain . Using the property of ϕ, we have
(2.20)
From (2.20) and using , we get that
(2.21)
which implies that , and thus . Hence we obtain that f and g have at least one common fixed point from case 1. Similarly, it is easy to show that if for some (), then f and g have at least one common fixed point, this completes the proof of case 2.
Taking and for all . Now we prove that for every , we have
(2.22)
Suppose this is not true, then for some , that is,
Using (2.13) and (2.15), we obtain that
(2.23)
Equations (2.14) and (2.23) give that . Using the property of ϕ, we get , which contradicts with for , hence (2.22) holds.
Similarly, one can show that for every , the following inequality holds.
(2.24)
Equations (2.22) and (2.24) imply that the sequence is nonincreasing, and consequently there exists some such that
(2.25)
By (2.25) and the following inequalities,
we get that is bounded, and hence it has some subsequence converging to some , that is,
(2.26)
Taking (P2) into account, we have
which combining with (2.25) shows that is bounded, and hence there exists subsequence of such that converges to some , that is,
(2.27)
By (2.13), we have
(2.28)
Letting in (2.28), and using (2.25)-(2.27), we obtain that
(2.29)
which means that , hence and .
Since
taking the limit as , we have , which implies that , that is,
(2.30)
Now, we claim that is a Cauchy sequence in the metric space (and so also in the space by Lemma 1.1). For this, it is sufficient to show that is a Cauchy sequence in . Suppose that this is not the case, then using Lemma 1.1, we have that is not a Cauchy sequence in . By Lemma 1.3, we obtain that there exist and two sequences and of positive integers such that and sequences in (1.2) tend to ε when .
From (2.13), we get that
Letting in the above inequalities and using the continuity of ψ, φ and ϕ, we get that
therefore, we get that . Hence, which is a contradiction. Thus, is a Cauchy sequence in , and is also a Cauchy sequence in . Since is complete, then the sequence converges to some , that is,
Moreover, the sequence and converge to , that is,
and
Using the fact that is a Cauchy sequence in , we have , which together with yields that . Hence, we have
By substituting , in (2.13), we get that
letting and applying Lemma 1.2, we conclude that
which yields that ; hence, , and thus . Similarly, one can easily show that , therefore, z is the common fixed point of f and g.
Now we prove the uniqueness of common fixed point. Let us suppose that u is also the common fixed point of f and g. Since
which means that ; hence, , and so . Thus, the uniqueness of the common fixed point is proved. □
By taking in Theorems 2.1-2.3, respectively, we have the following results.
Corollary 2.1 Let be a partially ordered set and suppose that there exists a partial metric p on X such that is complete. Let be a continuous nondecreasing mapping. Suppose that for comparable , we have
where ψ is an altering distance function and is a continuous function with if and only if . If there exists such that , then f has a fixed point.
Corollary 2.2 Suppose that X, f, ψ, and ϕ are the same as in Corollary 2.1 except the continuity of f. Suppose that for a nondecreasing sequence in X with , we have for all . If there exists such that , then f has a fixed point.
Corollary 2.3 Let be a complete partial metric space, f and g be self-mappings on X. Suppose that there exist functions ψ and ϕ such that for all
where ψ is an altering distance function and is a continuous function with if and only if .
Then f and g have a unique common fixed point.
Remark 2.1 If we replace the partial metric p by (usual) metric d in Corollaries 2.1-2.3, then we get Theorems 2.1-2.3 of [27].
Now, we introduce an example to support the usability of our results.
Example 2.1 Let be endowed with the usual partial metric defined by . It is easy to show that the partial metric space is complete. Also, define the mappings by and , respectively. Let us take such that and , respectively, and take such that . If , then
and
So, we have
If , then
and
So, we have
From the above arguments, we conclude that (2.13) holds; hence, all the required hypotheses of Theorem 2.3 are satisfied. Thus, we deduce the existence and uniqueness of a common fixed point of f and g. Here, 0 is the unique common fixed point.