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Fixed points of mappings with a contractive iterate at a point in partial metric spaces
Fixed Point Theory and Applications volume 2013, Article number: 335 (2013)
In 1994, Matthews introduced and studied the concept of partial metric space and obtained a Banach-type fixed point theorem on complete partial metric spaces. In this paper we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results generalize and unify some results of Sehgal, Guseman and Ćirić for mappings with a generalized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.
In 1922, Banach proved the following famous fixed point theorem . Let be a complete metric space. Let T be a contractive mapping on X, that is, one for which exists satisfying
for all . Then there exists a unique fixed point of T. This theorem, called the Banach contraction principle, is a forceful tool in nonlinear analysis. This principle has many applications and has been extended by a great number of authors. For the convenience of the reader, let us recall the following results [2–4].
In 1969, Sehgal  proved the following interesting generalization of the contraction mapping principle.
Theorem 1.1 ()
Let be a complete metric space, and be a continuous mapping. If for each there exists a positive integer such that
for all , then T has a unique fixed point . Moreover, for any , .
In 1970, Guseman  (see also ) generalized the result of Sehgal to mappings which are both necessarily continuous and which have a contractive iterate at each point in a (possibly proper) subset of the space.
In 1983, Ćirić , among other things, proved the following interesting generalization of Sehgal’s result.
Theorem 1.2 ()
Let be a complete metric space, and . If for each there exists a positive integer such that
holds for all , then T has a unique fixed point . Moreover, for every , .
Partial metric spaces were introduced in  by Matthews as part of the study of denotational semantics of dataflow networks and served as a device to solve some difficulties in domain theory of computer science (see also [7, 8]), in particular the ones which arose in the modeling of a parallel computation program given in . The concept has since proved extremely useful in domain theory (see, e.g., [10–13]) and in constructing models in the theory of computation (see, e.g., [14–17]).
On the other hand, fixed point theory of mappings defined on partial metric spaces since the first results obtained in  has flourished in the meantime, a fact evidenced by quite a number of papers dedicated to this subject (see, e.g., [18–47]). The task seems to have been set forth of determining what known fixed point results from the usual metric setting remain valid - after adequate modifications that should as much as possible reflect the nature of the concept of partial metric - when formulated in the partial metric setting.
The potentially nonzero self-distance, built into Matthew’s definition of partial metrics, was taken into account in  in an essential way by a rather mild variation of the classical Banach contractive condition, and in  further considerations in this direction were carried out which were in turn generalized by Chi et al. in . This paper represents a continuation of the previous work by the authors. Now we study fixed point results of new mappings with a contractive iterate at a point in partial metric spaces. Our results generalize and unify some results of Sehgal, Guseman and Ćirić for mappings with a generalized contractive iterate at a point to partial metric spaces. We give some generalized versions of the fixed point theorem of Matthews. The theory is illustrated by some examples.
Throughout this paper the letters ℝ and ℕ will denote the set of real numbers and positive integers, respectively.
Let us recall  that a nonnegative mapping , where X is a nonempty set, is said to be a partial metric on X if for any the following four conditions hold true:
if , then ,
The pair is then called a partial metric space. A sequence of elements of X is called p-Cauchy if the limit exists and is finite. The partial metric space is called complete if for each p-Cauchy sequence there is some such that
Observe that condition (P4) is a strengthening of the triangle inequality and that implies as in the case of an ordinary metric.
It can be shown that if is a partial metric space, then by , for , a metric is defined on the set X such that converges to with respect to if and only if (2.1) holds. Also is a complete partial metric space if and only if is a complete metric space. For proofs of these facts, see [7, 44].
A paradigm for partial metric spaces is the pair where and for . Below we give two more examples of partial metrics both of which are taken from .
Example 2.1 If , then defines a partial metric p on X.
Example 2.2 Let , where is the set of nonnegative integers.
By denote the set if for some , and the set if . Then a partial metric is defined on X by
In  Matthews proved the following extension of the Banach contraction principle to the setting of partial metric spaces.
Theorem 2.1 Let be a complete partial metric space, and be a given mapping. Suppose that for each the following condition holds
Then there is a unique such that . Also and for each the sequence converges with respect to the metric to z.
A variant of the result above concerning the so-called dualistic partial metric spaces was later given in . Altun et al.  further generalized the result of Matthews as well as extended to partial metric spaces several other well-known results about fixed points of mappings on metric spaces.
Taking a different approach to the way in which contractive condition (2.2) can be generalized for partial metrics, we [29, 30] have obtained other extensions of Theorem 2.1. To state one of them, we will use the following notation. Given a partial metric space set and . Notice that may be empty and that if p is a metric, then, clearly, and .
Theorem 2.2 (Theorem 3.1 of )
Let be a complete partial metric space, and be a given mapping. Suppose that for each the following condition holds
Then the set is nonempty. There is a unique such that . For each , the sequence converges with respect to the metric to u.
Remark 2.1 Although Theorem 2.2 does not imply uniqueness of the fixed point, it is easy to see that, under the assumptions made, if u and v are both fixed points satisfying , then .
Remark 2.2 Completeness of a partial metric does not necessarily entail that is nonempty. A (class of) counterexample(s) is easily constructed as follows.
Let be a partial metric space, and be an arbitrary mapping. If are such that , define and . Then is a partial metric space, as is easily verified.
Now if , then, given a sequence , we have and whenever . Thus there are no nonstationary p-Cauchy sequences. Hence is complete. But whenever .
If condition (2.3) is replaced by the somewhat stronger condition below, then the uniqueness of the fixed point is guaranteed.
Theorem 2.3 (Theorem 3.2 of )
Let be a complete partial metric space, and be a given mapping. Suppose that for each the following condition holds
Then there is a unique such that . Furthermore, and for each the sequence converges with respect to the metric to z.
3 Auxiliary results
We now introduce the two types of contractive conditions that we shall be considering in this paper. Let us remark that if , then we write for the identity mapping , i.e., , .
Definition 3.1 Let be a partial metric space, and .
We say that T is a -operator on X if for each there is some such that for each there holds(3.1)
for some .
We say that T is a C-operator on X if for each there is some such that for each there holds(3.2)
for some .
For a -operator T on a partial metric space and define the supporting sequence at the point x as the sequence , where and . Clearly, this is a strictly increasing sequence. Also set .
Lemma 3.1 Let T be a -operator on a partial metric space , , be the supporting sequence at x and and be given integers. Then we must have
Proof Case 1. Suppose . By (3.1) we know that if
Now if (3.4) were to hold, then (3.5) and (P2) would imply that (3.6) is true as well. So (3.7) also holds and thus
meaning (by (P3)) that . But this contradicts the assumption , so (3.4) must be true.
Case 2. Suppose . Since , the assertion here follows from the previous case.
Case 3. Suppose now . Assume that
since otherwise there is nothing to prove. Then by (3.1) we must have . But then, by the previous case, there is some such that
and we are done.
For , the argument carries on by induction. Suppose that (3.3) holds for some . If for all , then we must have . By the induction hypothesis, there is some such that
But since , the last inequality clashes with our assumption. □
To shorten the foregoing considerations, we introduce some auxiliary notions as follows. Fix . For integers and , use Lemma 3.1 repeatedly to fix integers , and such that, putting , there holds
for all , where
We shall refer to and as the -descent and the -signature at x, respectively. Set . We shall say that x is of type 1 if there are sequences of positive integers and , the first one strictly increasing, such that for all we have and ; here and henceforth, for a finite set P, we denote by the number of its elements. We shall say that x is of type 2 if x is not of type 1, i.e., if there are such that for all and all there holds .
To make the proof of our main result more transparent, we have extracted from it several parts and presented them first in form of the next five lemmas.
Lemma 3.2 Let T be a -operator on a partial metric space , , and let be the supporting sequence at x. Then:
If is the -descent at x, then
where we set .
If is such that , then for some there must hold
Proof Regarding Lemma 3.1, we get for all j such that , recursively. Now (a) follows directly.
To prove (b) simply observe that implies that the set is a subset of with elements so that there must be some with . Hence
where we used (a) and the fact that . □
Lemma 3.3 If T is a -operator on a partial metric space and , then there is some such that for all we must have .
Proof If , then this is obvious. Thus suppose and set . If , then it is certainly true that for all . Now suppose that the same is valid for some .
If for some , then by the induction hypothesis there holds . Otherwise, we must have
Now using (3.1)
for some . Hence we either have
or, using (3.8) and the fact that , it must be that
By induction the desired conclusion follows. □
Lemma 3.4 Let T be a -operator on a partial metric space and . If x is of type 1, then .
Proof Fix . If is the -descent, then by (a) of Lemma 3.2 we have
Since , this implies .
Now given choose such that and such that for all it holds . Let be arbitrary.
Suppose first that . Then
Suppose now that . For , we have , so by (b) of Lemma 3.2 there must be some such that .
We have thus shown that must hold for all . Therefore if , then
The previous analysis proves . □
Lemma 3.5 Let T be a -operator on a partial metric space and . If x is of type 2, then the sequence is p-Cauchy.
Proof Let be the supporting sequence at x.
We first show . Suppose that this is not true and pick a real θ with . Let and be positive integers, where , such that
The fact that implies that is defined and since , we have . For , we have so, by (b) of Lemma 3.2, there is some such that
By the preceding part and since , with as in Lemma 3.3, we have .
Let us prove
Given , take such that for all . Let and be arbitrary. For , we have , so for some it must be
and we are done.
From (3.9) it now immediately follows that
Indeed, given , consider as in (3.9) and let be arbitrary. Then
we now only need to show that
Let be arbitrary and let be as in (3.10). We claim that holds for all . This would prove (3.12) since .
Suppose to the contrary that there is some with . Put . implies . If z is of type 1, then by Lemma 3.4 we have , so is p-Cauchy and we are finished. Suppose now that z is of type 2, so that by what we have proved thus far we know that and also that (3.10) holds with z taken instead of x. It cannot be because this would mean that , so using Lemma 3.1, it would follow for some , i.e., , giving , a contradiction. So . The argument actually shows that holds for every , where is the supporting sequence at the point z. So . Now use the fact that and (3.10) (with z taken instead of x and instead of ε of course) to find such that
As and , there is some with . Now, using , we obtain
which is not possible. □
Lemma 3.6 Let and let be any mapping satisfying (P3). Suppose that is such that holds for some positive integer k, and that there exists such that
Proof From , , we have
hence . But (3.14) now gives
so . □
4 Main results
Having made the necessary preparations, we are now able to prove fixed point results for -operators on complete partial metric spaces. But first we prove a proposition giving some insight into the structure of this type of mappings.
Proposition 4.1 If T is a -operator on a complete partial metric space , then
for each , the sequence -converges to some ;
for all , there holds .
Proof The existence of such points is assured by Lemmas 3.4 and 3.5 and completeness if , and is self-evident if .
To prove (2), let be arbitrary and suppose that . If , then (by (P2) and (P3)) and we are done. Thus assume that and let be arbitrary such that . There is some such that for all there holds
For we have
Fix any and set . By (3.1) there is some such that
Now is just , which is false by our choice of ε. This leaves us with the only other possibility: , i.e., .
From the preceding analysis it follows that , which by (P2) actually means that . □
Theorem 4.1 If T is a -operator on a complete partial metric space , then there is a fixed point of T such that , where are as in Proposition 4.1.
Proof For put (this is consistent with the notation of Lemma 3.5). Set . For pick such that for all it holds
(You can first pick such that , then choose such that for all there holds and finally put .)
Notice that if and are nonnegative integers for , then we have
First we prove that .
For let be such that holds for all .
Fix and let be the supporting sequence at . Let be any integer such that . We have
First suppose that for all , and for all . Then, by repeated use of (3.1), we obtain
and continuing in this manner finally
On the other hand, if for some , or for some , then by (4.1) we must have .
From the above considerations and from , it is now clear that .
So by completeness there is some such that
Let be the supporting sequence at u.
Let us show by induction on k that if is such that , for all , then
Suppose first that .
so the desired conclusion immediately follows. Now suppose that the assertion is true for some , take any such that , , and proceed as follows.
for all , so that taking the limit above as m approaches infinity and using (4.3) (which is justified since ) it follows that .
Now for each , since , there must be some such that
Using , we proceed to obtain
Now we have and also and for all . Hence, in view of the induction hypothesis, we finally arrive at .
Using (4.3) it is straightforward to see that for all there holds : indeed this follows by letting tend to infinity in
Thus . But by definition of I we must actually have .
We now claim that there are positive integers such that
Assume this is not the case. Then can hold for at most one . As we have for all , our assumption implies in particular that . Thus we can take some such that . The assumption also allows us to find some such that for all k with we have , and such that for all it holds (remember that ).
Take any k with . Then . There is some positive such that . Let i be the smallest integer with such that , and let be the greatest integer such that . So . By (3.1) there is some such that
Clearly, we have and . The minimality of i and m and the fact that can now easily be used to deduce that . Therefore and this cannot be true by the choice of ε.
So we have proved that there are positive integers such that . Since , we must have (by (P2) and (P3)), i.e., for . As and , by Lemma 3.6 it follows . Of course . □
Remark 4.1 To ensure uniqueness of the fixed point, we can strengthen condition (3.1) as follows. Given a partial metric space , call a -operator if for each there is some such that for each there holds
for some . Evidently, each -operator is a -operator as well so that if is complete, the conclusion of Theorem 4.1 holds. But now, in addition, if and , then
so that either or , meaning that in any case we must have .
Recall that a sequence in a partial metric space is called 0-Cauchy with respect to p (see, e.g., ) if . We say that is 0-complete if every 0-Cauchy sequence in X -converges to some (for which we then necessarily must have ). Note that every 0-Cauchy sequence in is Cauchy in , and that every complete partial metric space is 0-complete.
Remark 4.2 Recently a very interesting paper by Haghi, Rezapour and Shahzad  showed up in which the authors associated to each partial metric space a metric space by setting and if , and proved that is 0-complete if and only if is complete. They then proceeded to demonstrate how using the associated metric d some of the fixed point results in partial metric spaces can easily be deduced from the corresponding known results in metric spaces.
Let us point out that these considerations cannot apply to -operators since the terms and on the right-hand side of (3.1) are not multiplied by α. Thus our Theorem 4.1 cannot follow from the result of Ćirić it generalizes.
If we completely neglect the role of self-distances in (3.1), we can easily verify that the statement of Theorem 1.2 remains valid upon substituting the words ‘partial metric’ for ‘metric’ and ‘0-complete’ for ‘complete’. We will prove this using the approach of Haghi, Rezapour and Shahzad  that will allow us to deduce Theorem 4.2 directly from Ćirić’s result (Theorem 1.2).
Theorem 4.2 If T is a C-operator on a 0-complete partial metric space , then there is a unique fixed point z of T. Furthermore, we have and for each the sequence -converges to z.
Proof Let d be defined as in Remark 4.2. So is a complete metric space (see Proposition 2.1 of ). Observe that we have for all . For let be as in (3.2), and for set
and , . We thus have that
for all . We check that for all it holds that , so that Theorem 1.2 can immediately be applied.
Case 1. There is some such that and . Here we have
Case 2. For all we have that . So it must be , in particular, and hence , i.e., . But by our assumption it now follows that . Similarly, from , we obtain and consequently . Now . □
Remark 4.3 It should be pointed out, however, that even though the results of Haghi et al. can deduce the same fixed point as the corresponding partial metric fixed point result, using the partial metric version computers evaluate faster since many nonsense terms are omitted. This is very important from the aspect of computer science due to its cost and explains the vast body of partial metric fixed point results found in literature.
Given a -operator and a point x, one may ask what the minimal value of is for which inequality (3.1) holds true. In the following example, for an arbitrary positive integer m, we construct a -operator on a complete partial metric space such that for some it must be .
Example 4.1 Denote by the set of all sequences and for by the set of all n-tuples of positive integers. Put . For set
and define (thus if , then and ). Here ‘’ stands for the domain of the function x. Then is a partial metric space (see ) and a complete one as can easily be verified.
Fix and define as follows. For let .
If , then set . If , then define by , and the following two conditions:
if is finite and has at most elements, then if , and else;
if is either infinite or finite with more than elements, then if , and else.
Let us show that T is a -operator with for all . Before we proceed, observe that if k is a nonnegative integer such that and for , then for we must have for all .
Case 1. There is a nonnegative integer i with such that . Denote by k the least such nonnegative integer.
If , then .
If , then since we must actually have and thus . Hence .
Case 2. for some and . Here .
Case 3. for some and . Here .
Condition (4.4) fails because the fixed point is not unique. So T is not a -operator, hence not a C-operator either.
Now suppose that is an arbitrary positive integer and take such that for all , and , for all .
We have , for , and . So we see that for this particular choice of x and y, substituting for in (3.1) makes the inequality false.
Let us use this very example to illustrate Proposition 4.1. Let be defined by for all . For let be defined by for .
If , we clearly have . Similarly, if , then . So because and . Also
Remark 4.4 It should be noted that if in Theorem 4.1 we require to be equal to 1 for all , then Theorem 2.2 is obtained as a corollary. On the other hand, as already pointed out, if in Theorem 4.2 p is a complete (ordinary) metric on X, then the result of Ćirić (Theorem 1.2) is recovered.
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This work was supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.
The authors declare that they have no competing interests.
All authors read and approved the final manuscript.
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Ilić, D., Pavlović, V. & Rakočević, V. Fixed points of mappings with a contractive iterate at a point in partial metric spaces. Fixed Point Theory Appl 2013, 335 (2013). https://doi.org/10.1186/1687-1812-2013-335
- fixed point
- partial metric space
- contractive iterate at a point