#### Abstract

Without the continuity and nondecreasing property of the comparison function, we in this paper prove some fixed point theorems of generalized contractions of rational type in ordered partial metric spaces, which generalize and improve the corresponding results of Luong and Thuan. An example is given to support the usability of our results.

#### 1. Introduction and Preliminaries

Throughout this paper, and will denote the set of nonnegative real numbers and the set of all positive integer numbers.

It is well known that the Banach contraction principle is one of the pivotal results of analysis. Generalizations of this principle have been obtained in several directions. In [1], Jaggi introduced a contraction of rational type in a metric space and proved the unique existence of fixed point as the contraction is continuous and the metric space is complete, which was then extended to the case of ordered metric spaces by Harjani et al. [2]. In [3], Luong and Thuan considered weak contractions of rational type in ordered metric spaces and proved the following fixed point theorem.

Theorem 1 (see [3]). Let be a complete ordered metric space and a nondecreasing mapping such that there exists a function with such that for all with and , where . Assume that is lower semicontinuous, and has the following property:(A1) if is a nondecreasing sequence in such that , then .If there exists such that , then has a fixed point.

In [4], Matthews introduced the partial metric space and extended the Banach contraction principle to the case of partial metric spaces, which was then improved by Oltra and Valero [5]. In [68], the authors studied the unique existence of fixed point of generalized contractions in partial metric spaces. Recall that a mapping is called a -generalized contraction (-GC) if there exists a comparison function such that where is a partial metric space, , and . Under several different assumptions made on , they obtained the following fixed point results.

Theorem 2 (see Theorem 1 of [6], Theorem 1 of [7], and Theorems 3 and 4 of [8]). Let be a complete partial metric space. Assume the following.

(i) is a -GC and one of the following conditions is satisfied:(H1) is continuous and nondecreasing, and for all ;(H2) is nondecreasing, and the series is convergent for all ( denotes the th iterate of );(H3) is upper semicontinuous from the right, and for all .

Or (ii) is a -GC and the following condition is satisfied:(H4) is nondecreasing, and for all .

Then has a unique fixed point.

For other references concerned with various fixed point results and common fixed point results for contractions in the setting of metric-like, partial metric, and ordered partial metric spaces, we refer the readers to [924].

In this paper, we establish some fixed point theorems for generalized contractions of rational type in ordered partial metric spaces, which generalize and improve Theorems 1 and 2. An example is given to support the usability of our results. Even in the setting of metric spaces, the results presented in this paper are still new since the comparison function is not necessarily assumed to be upper semicontinuous from the right or nondecreasing.

Following [4, 5], a partial metric on a set is a function such that, for each ,(p1) if and only if ;(p2); (p3);(p4).Observe that, if , then . A partial metric space is a pair such that is a set and is a partial metric on . Each partial metric on induces a topology on which has as a base of the family of open balls , where for all and .

Let be a partial metric space and a sequence of . The sequence converges, with respect to , to a point (denoted by ) if . Define a function by . Then is a metric on . The sequence converges, with respect to , to a point (denoted by ) if and only if The sequence is called a Cauchy sequence if exists and is finite; is called complete if every Cauchy sequence converges, with respect to , to a point such that . In particular, is called a 0-Cauchy sequence if ;   is called 0-complete if every 0-Cauchy sequence converges, with respect to , to a point such that . Every complete partial metric space is 0-complete, but the converse may not be true; see [17].

Remark 3 (see [4, 5]). A partial metric space is complete if and only if is complete.

#### 2. Fixed Point Theorems

Let be an ordered partial metric space and . For all with , set A mapping is said to be a -generalized contraction of rational type (-GCRT), if there exists a comparison function such that for all with and .

Lemma 4. Let be an ordered partial metric space and a nondecreasing -GCRT. Assume that (H5) and for all .For each such that , let for all . If for all , then

Proof. Note that is nondecreasing and ; then is nondecreasing. Since for all , then for all . By (4) and (p4), for all , we have Thus by the nondecreasing property of , (7), and (8), for all . Now, we claim that, for all , Suppose, on the contrary, that there exists some such that . Then by (8), (9), and for all , we have This is a contradiction and hence (10) is true. Consequently, is a decreasing sequence of positive real numbers. This yields that there exists such that In the following, we will show that . Suppose, on the contrary, that . It follows from (8), (9), and (10) that, for all , Letting in (13), by (12), (H5), and , we get a contradiction . Hence , and consequently by (12) Now, we show that conclusion (6) is true. If otherwise, there exist some and subsequences and in , with , such that, for all , From (15) we may assume that, without loss of generality, for all , Then by (p4), for all , we have Letting in the above inequality, by (14), we get Also by (p4), for all , we have Letting in the above inequality, by (14) and (18), we get Similarly, we can obtain By the nondecreasing property of , (5), and (15), we get, for all , where , , . By (14) and (18), we have which together with (14), (18), and (21) implies that there exists such that, for all , where . It follows from (15), (18), and (21) that By (15), (22), and (24), we have, for all , Letting in (26), by (20), (25), (H5), and , we get a contradiction . Hence conclusion (6) is true. The proof is complete.

Remark 5. It is easy to see that Lemma 4 is still valid for -GCRT and -GC.

Theorem 6. Let be a 0-complete ordered partial metric space and a nondecreasing -GCRT. Assume that(H6) and ,  for all , and has the following property:(A2) if is a nondecreasing sequence in such that , then .If there exists such that , then has a fixed point.

Proof. Let for all . If there exists some such that , then is a fixed point and hence the proof is complete. Therefore, we may assume that for all . By Lemma 4, is 0-Cauchy. So by the 0-completeness of , there exists such that , and consider That is, . By (A2), , and so, for all , Now, we claim that, for all , If otherwise, there exists some such that ; then for all since is nondecreasing. This contradicts with the assumption for all , and hence (29) is true. Since is nondecreasing, then , for all , and hence by (A2) Let for all . Then is a nondecreasing sequence since is nondecreasing. We may assume that, for all , If otherwise, there exists some such that ; then is a fixed point and hence the proof is complete. By Lemma 4 and the 0-completeness of , there exists such that , and consider That is, . By (A2), we have, for all , It follows from , , and the continuity of the metric that That is, which together with (27) and (32) implies that By (p4), we have, , Letting in the above two inequalities, by (27), (32), and (36), we get In what follows, we will show . Suppose, on the contrary, that ; then . Clearly, and for all by (29), (30), and (31). Consequently, for all . Thus by (5), we have, for all , where , , . By (27), (32), and (36), we have which together with (27), (32), (36), and (38) implies that there exists such that, for all , where . It follows from (36) and (38) that By (39), (40), and (42), we have, for all , Letting in (44), by (36), (43), (H6), and , we get a contradiction . Hence , and consequently . From (30) and (33), it follows that . This together with yields that . The proof is complete.

In particular when is a -GCRT, condition (H6) could be weakened, and we have the following result.

Theorem 7. Let be a 0-complete ordered partial metric space and a nondecreasing -GCRT. Assume that (H5) is satisfied and has property (A2). If there exists such that , then has a fixed point.

Proof. Following the proof of Theorem 6, we find that (27)–(36) and (39) still hold by Remark 5. Instead of (40), by (5), (6), (39), and , we have, for all , where , . By (28), (32), (36), and , there exists such that, for all , which together with (45) implies that, for all , That is, is decreasing. Moreover by (36), we have Instead of (44), by (39), (45), and (46), for all , Letting in (49), by (48), (H5), and , we get a contradiction . Hence , and consequently . From (30) and (33), it follows that . This together with yields that . The proof is complete.

Corollary 8. Let be a 0-complete ordered partial metric space and a nondecreasing mapping such that there exists a function with such that for all with and . Assume that, for all , and has property (A2). If there exists such that , then has a fixed point.

Proof. Let . Clearly, for all by . For all , it follows from (51) that Then the conclusion follows immediately from Theorem 7. The proof is complete.

Remark 9. If is lower semicontinuous from the right and , then, for all , Therefore Theorem 1 follows immediately from Corollary 8.

Now we illustrate Theorems 6 and 7 by the following example.

Example 10. Let with the partial metric for all . Clearly, and hence, by Remark 3, is a complete partial metric space since is a complete metric space. Define a partial order on by where and is the usual order of reals. Let It is easy to see that is nondecreasing, has property (A2), and for all . Direct calculations give that That is, (H6) is satisfied. In particular, (H5) is satisfied. For each with and , we must have and hence . For each with and , we have two cases:(i)if and, then ;(ii)if and , then .
Thus by (56), . This shows that is a -GCRT(), and consequently the existence of fixed point follows immediately from Theorem 6 or Theorem 7. In fact, is a fixed point of .

In what follows, we will extend Theorem 2 to the case of ordered partial metric space under weaker conditions.

Theorem 11. Let be a 0-complete ordered partial metric space and a nondecreasing mapping such that, for each with , (2) is satisfied for . Assume that either (H4) or (H5) is satisfied and has the following property:(A3) if is a nondecreasing sequence in such that , then for all .
If there exists such that , then has a fixed point.

Proof.
Case 1 ((H5) is satisfied). Following the proof of Theorem 6, we get (27) and (28) by Remark 5 and (A3). By (p4), for all , Letting in (58), by (27), we get Now, we show that is a fixed point of . If otherwise, then . By (p4), (2), and (28), we have, for all , where , . It follows from (27) and (59) that which together with (27) implies that there exist and such that, for all , Thus, for all , and hence, by (60), Letting in (64), by (27), we get a contradiction since for all . Hence is a fixed point of .
Case 2 ((H4) is satisfied). By the same method used in Lemma 4 and [8, Theorem 4], we can show that (27) and (59) still hold. Then by analogy to Case 1, is a fixed point of . The proof is complete.

Remark 12. If is a generalized contraction of rational type, then the method used in Theorem 11 fails to work since there exists such that for all unless , which could not be done. Therefore the existence of fixed point of generalized contractions of rational type could not be obtained under the weaker condition (A3).

By the methods used in Theorem 11 and [8, Theorem 4], we have the following fixed point result.

Theorem 13. Let be a 0-complete partial metric space and a -GC. If either (H4) or (H5) is satisfied, then has a unique fixed point.

Remark 14. (i) In the case that (H5) is satisfied, the continuity and nondecreasing property of necessarily assumed in Theorem 2 is removed in Theorem 13. Note that (H3) implies (H5); then Theorem 13 improves Theorem1 of [6], Theorem 3 of [8], and Theorem 2.3 of [23].
(ii) In the case that (H4) is satisfied, Theorem 13 is generalization of Theorem 4 of [8] to the case of -GC and hence improves Theorem 1 of [7] since (H2) implies (H4).

Example 15. Let with the partial metric for all . By analogy to Example 10, we find from Remark 3 that is a complete partial metric space. Let and is given by (56). It is easy to check that is a -GC and (H5) is satisfied. Therefore the unique existence of fixed point follows immediately from Theorem 13. In fact, is the unique fixed point of .
For each , we have . This implies that is not upper semicontinuous from the right at . Meanwhile, it is clear that is decreasing on . Therefore, we cannot invoke Theorem 2 to show the unique existence of fixed point.

Remark 16. It is worth mentioning that, even in the setting of metric spaces, the main results in this paper are still new since the continuity and nondecreasing property of the comparison function necessarily assumed in [24] and other relating references is removed.

#### Acknowledgments

This work was supported by the Natural Science Foundation of China (11161022), the Natural Science Foundation of Jiangxi Province (20114BAB211006, 20122BAB201015), the Educational Department of Jiangxi Province (GJJ12280, GJJ13297), and the Program for Excellent Youth Talents of JXUFE(201201).