Integral Type F-Contractions in Partial Metric Spaces

Abstract

Partial metric spaces were introduced as a generalization of usual metric spaces where the self-distance for any point need not be equal to zero. In this work, we defined generalized integral type F-contractions and proved common fixed point theorems for four mappings satisfying this type (Branciari type) of contractions in partial metric spaces.

1. Introduction and Preliminaries

Let be a nonempty set and satisfy :  , :  , :  , :  

for all and . Then the pair is called a partial metric space (in short PMS) and is called a partial metric on ([1]).

Let be a PMS. Then, the functions given byare (usual) metrics on . It is clear that and are equivalent ([1]).

Definition 1 (see [1]). (i)A sequence in a PMS converges to if and only if (ii)A sequence in a PMS is called a Cauchy sequence if and only if exists (and finite).(iii)A PMS is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that (iv)A mapping is said to be continuous at if for every , there exists such that

Lemma 2 (see [1]). (i)A sequence is Cauchy in a PMS if and only if is Cauchy in a metric space .(ii)A PMS is complete if and only if the metric space is complete. Moreover,  where is a limit of in .

Remark 3 (see [2]). Let be a PMS. Therefore,(i)if , then ;(ii)if , then .

Lemma 4 (see [3]). Assume as in a PMS such that . Then for every .

In literature, there are many generalizations of Banach contraction principle in metric and generalized metric spaces. One of them is integral type contraction which was defined by Brianciari ([4]). On the other hand, Wardowski [5] introduced contraction in metric spaces as a generalization Banach contraction principle. For more details, you can see [5–9]. In this work, we will introduce generalized integral type contraction in partial metric spaces and prove common fixed point theorems.

Definition 5 (see [5]). Let a mapping satisfy the following:(F1) is strictly increasing, i.e., for all such that , ;(F2)for each sequence , ;(F3)there exists such that .

Definition 6 (see [5]). A mapping is said to be contraction if there exists such that

Theorem 7 (see [5]). Let be a complete metric space and let be an contraction. Then T has a unique fixed point in .

Example 8 (see [5]). Let be given by . satisfies (F1), (F2), and (F3). Each mapping is an -contraction such that, for all in and ,It is clear that for such that the inequality also holds; i.e., is a Banach contraction.

Definition 9 (see [10]). The mappings are said to be weakly compatible if and commute at each coincidence point; i.e., for some .

2. Main Results

Theorem 10. Let be a complete partial metric space and , are mappings satisfying and . Suppose there exist and such that for all satisfying whereand is a Lebesgue integrable mapping which is summable, nonnegative and for each If(i) , , or is closed,(ii) is continuous,(iii) and are weakly compatible,then the pairs , and have a unique common fixed point.

Proof. Let be arbitrary. Define a sequence for byStep I. Prove that as .
By (6),whereIf , then it follows from (10) which is a contradiction (as ). Thus From (10),Continuing this way, we haveUsing (14) and (15),AndThen, it follows . By and (F2), we haveStep II. Now, we prove that is Cauchy sequence. By and (F3), there exits such thatBy (16) and (17),and Using the above inequalities and (19),Therefore, there exists such that for all , or Let with ; using triangular inequality, we have As , the series converges, soThus is a Cauchy sequence in . Therefore, is a Cauchy sequence in . Since is complete partial metric space, then is complete metric space. Then, there exists a such that . Moreover Since , then , converge to .
Step III. We will prove that , and have a coincidence point.
Suppose is closed, there exists such that We shall show that Then from (6),wherePassing to limit as , This is a contradiction with . Thus we have Therefore . Since and are weakly compatible .
Now we show that .wherePassing to the limit as and using continuity of , we have which is a contradiction. Therefore ; that is, is a fixed point of and .
Now we show that is a fixed point of and Since , there exists a point such that . Suppose that , then whereThuswhich is a contradiction. Thus . By weak compatibility of and , . Finally we show that . From (6), whereThus,and we have .
So is a common fixed point of , , and .
Step IV. We show uniqueness of common fixed point. Let be another common fixed point of and and .
From (6), we have whereHence,which is a contradiction. So

Corollary 11. Let be a complete partial metric space and , are two mappings. Suppose there exist and such that for all satisfying whereand is a Lebesgue integrable mapping which is summable, nonnegative and for each If(i) or is closed,(ii) is continuous,then the pairs and have a unique common fixed point.

Corollary 12. Let be a complete partial metric space and are two mappings. Suppose there exist and such that for all satisfying And is a Lebesgue integrable mapping which is summable, nonnegative and for each If(i) or is closed,(ii) is continuous,then the pairs and have a unique common fixed point.

Theorem 13. Let be a complete partial metric space and be mappings. Suppose there exist and such that for all satisfying where And is a Lebesgue integrable mapping which is summable, nonnegative and for each If(i) or is continuous, or(ii) is continuous,then the pairs and have a unique common fixed point.

Proof. Let be arbitrary. Define a sequence for byStep I. Prove that as .
By (47),whereThen the proof is similar proof of Theorem 10.
We will prove that and have common fixed point. Since is complete partial metric space, then is complete metric space. Then, there exists such that . Moreover We consider two cases.
Case 1. Suppose is continuous. Then, . Thus is a fixed point of .
Now we prove is a fixed point of . On the contrary, we assume . From (47), where Letting , This is a contradiction with . Thus we have .
Similarly, we have the same results when is continuous.
Case 2. Now, we suppose that is continuous. We can assume there exists such that (i.e., ) for all . Then from (47) we havewhereThen there exists such that, for all , we haveThus, we have for all Since is continuous, taking the limit as , we get which is a contradiction. Therefore and is a fixed point of
Now we show that is a fixed point of whereThus Hence, .
So is a common fixed point of and .
Now we prove uniqueness of common fixed point. We assume that is another common fixed point of and and .
From (47), we have where Hence, which is a contradiction with . So .

Corollary 14. Let be a complete partial metric space and two mappings. Suppose there exist and such that for all satisfying and is a Lebesgue integrable mapping which is summable, nonnegative and for each If(i) or is continuous, or(ii) is continuous, then the pairs and have a unique common fixed point.

Example 15. Let , and for all . Then is complete partial metric space. Let and Consider in Example 8. Then all conditions of Theorem 10 and the contractive condition (6) are satisfied for some and for .
If ,If ,Therefore 0 is a fixed point of , and .

Data Availability

The author did not use any data set.

Conflicts of Interest

The author declares that there are no conflicts of interest.