The paper deals with a class of contracting mappings that includes the Boyd-Wong and the Matkowski classes. Some of the new fixed point theorems obtained here unify and extend well-known results also in the case of metric spaces.
In the simplest case, the condition with a contracting mapping has the form , where is a metric space, f is a mapping on X, and is such that , . Boyd and Wong  assumed that φ is upper semicontinuous from the right (i.e., , ); Matkowski in  assumed that φ is nondecreasing and such that for each . We assume that all sequences such that , , converge to zero. It appears that the classes of Boyd-Wong’s and Matkowski’s mappings are included in this new class (the problem of for Boyd-Wong mappings is meaningless for contractions). The main results are Theorems 21, 22, 23, and theorems extending the well-known classical results: Theorem 30 (Matkowski’s theorem) and Theorem 31 (covering the theorems of Romaguera and Boyd-Wong).
Let us recall the notions of a partial metric space due to Matthews [, Definition 3.1] and of a dualistic partial metric due to Oltra and Valero  and O’Neill .
Definition 1 A dualistic partial metric is a mapping such that
If p is nonnegative, then it is a partial metric.
If p is a dualistic partial metric on X, then defined by
A dualistic partial metric space is 0-complete if for every sequence in X such that , there exists in .
In fact Definition 3 is too abstract. The condition in means that (see (6), (2)). If, in addition, , then in (see [, Lemma 2.2] or [, Proposition 1.4]).
Corollary 4A dualistic partial metric spaceis 0-complete iff every sequencesuch thatconverges into a point . Ifpis a metric, then 0-completeness is identical with completeness.
There exist 0-complete partial metric spaces which are not complete . Some criterions of 0-completeness can be found in [, Section 4].
Proposition 5Letbe a partial metric space. Thenconverges intoiffand iff we havein .
Proof A sequence converges to x in iff (see [, Lemma 2.2] or [, Proposition 1.4]). Assume . Then, for nonnegative p, condition (2) yields and , i.e., and in . We also have
and consequently, , i.e., and converges in to x. □
Let be the family of all subsets of Y. We say that is a (multivalued) mapping if for all .
Now, let us investigate the concept of a 0-closed graph.
Definition 6 Let be a dualistic partial metric space. A mapping has a 0-closed graph if for all sequences , in X the following condition is satisfied:
Proposition 5 and [, Proposition 1.4] yield the following.
Corollary 7For all sequences , with , condition (8) can be replaced by
Ifpis nonnegative, then (8) is equivalent to (9).
Clearly, if has a closed graph in , then (9) is satisfied as Kerp with topology induced by p is a closed metric subspace of [, Lemma 2.2].
For a partial metric space , a nonempty set , and , let us adopt
The family generates a topology , and we get a topological space . Assume that p is nonnegative and
holds. Then, for all sequences , such that and , , there exist , with . We have
and consequently, in (Proposition 5). If in , then holds.
For a partial metric space and nonempty , let us adopt
If p is a metric, then P is the Hausdorff metric of the metric p, whenever A and C are nonempty closed and bounded subsets of X. In general, P is not a partial metric (see [, Proposition 2.2(i), Proposition 2.3(h3)]).
Clearly, for and if (10) holds, then is continuous on Kerp.
Corollary 8Letbe a partial metric space. Assume that for each , the mappingsatisfies (10) andin . ThenFhas a 0-closed graph (and conditions (8), (9) are equivalent).
Proposition 9Letbe a dualistic partial metric space. Ifhas a 0-closed graph, thenis closed in .
Proof If , , then for , and each such that , condition (8) yields , i.e., is closed in Kerp (which with topology induced by p is a closed metric subspace of [, Lemma 2.2]). □
Now, let us investigate a ‘contraction’ condition.
Let Φ be a class of mappings such that , ; iff and .
Proposition 10Letbe a partial metric space. Assume thatsatisfy
for . Then ; ifHhas a 0-closed graph, thenholds.
Proof For and nonnegative p, conditions (2), (12) yield
i.e., and . What is more, from our inequality it follows that there exist such that . Now, (8) for , yields . □
Propositions 9, 10 yield the following.
Theorem 11Letbe a 0-complete partial metric space, and letℱbe a family of mappingswith 0-closed graphs. Assume that ; some and at least all differentsatisfy (12). Then all members ofℱhave the same set of fixed points; this set is closed inand contained in Kerp.
The previous result becomes a little more interesting if a mapping has a fixed point.
Let us assume that the following condition is satisfied for a partial metric space , P defined by (11), and :
Then, for any , we have
i.e., condition (13) yields condition (12). The subsequent two propositions enable us to strengthen condition (12).
Proposition 12Letbe a partial metric space, and letbe compact in . Then, for any , there existssuch that .
Proof Let be a sequence in C such that
There exists a subsequence of and such that in , i.e., [, Lemma 2.2]
Corollary 13Letbe a partial metric space, and letbe mappings withHcompact valued in . Then condition (12) is equivalent to
Condition (14) extends the idea of α-step mappings [, Definition 17].
The next example shows that even a ‘good’ mapping φ in condition (14) does not guarantee the existence of a fixed point.
Example 14 Let us consider a continuous mapping defined by for , and for . An easy computation proves that φ is increasing on and, therefore, φ is nondecreasing on . Let us consider and , . For , we have
Now, for , we obtain
i.e., (14) is satisfied for , , . Still it is clear that has no fixed point.
It is a good idea suggested by  to gather together the properties of φ. Let us recall that Φ is a class of mappings such that , ; and iff and .
Proposition 15Assume that . Then every sequencesuch that , (in particular , ) is nonincreasing; if, in addition, φis nondecreasing, then , holds.
Proof From , , it follows that . Similarly, for nondecreasing φ, we get . □
Let us present some subclasses of .
Let consist of mappings for which every sequence such that , converges to zero.
Proposition 16We have . For , the sequenceis nonincreasing and it converges to 0, . If a mappingsatisfies
Proof Assume that . Suppose that for . Then , , is a good counterexample (the sequence does not converge to 0). Therefore holds. Suppose . Then, for , , , we obtain a divergent sequence such that , . Consequently, every satisfies . The sequence is nonincreasing, , and it converges to 0 as . Assume . Then any sequence such that (), is nonincreasing and therefore it converges, say, to . Suppose . Then (15) yields
a contradiction. □
Let consist of mappings φ upper semicontinuous from the right, call them Boyd-Wong mappings. Proposition 16 yields .
In turn, let consist of nondecreasing mappings such that , (Matkowski mappings). It is well known [, Lemma] that . Moreover, , , yields (Proposition 15) and hence . Consequently, holds. Let us note that φ from Example 14 belongs to .
The following is a kind of the reverse condition.
Proposition 17Letbe a sequence convergent to zero and such that , , for a . Then there exists a mappingsuch that , and , , .
Proof Let us adopt and , . Clearly, ψ is nondecreasing and continuous from the right as . Therefore, ψ is also upper semicontinuous. The sequence is nonincreasing as , . Let us adopt . For , we have
and the case of is trivial as then . Consequently, we get (if ), (if ), and so on. Finally, we obtain , (the possible case of some ). □
The following modification of φ is useful.
Proposition 18Assume thatis a mapping. Thenψdefined by
belongs to Φ iff ; iff ; φsatisfies (15) iffψsatisfies (15). In addition, if φis nondecreasing, thenψis increasing.
Proof It is clear that holds iff . If φ is nondecreasing, then for we obtain
i.e., ψ is increasing. The remaining part of the proof is also trivial. □
The next two propositions can be helpful in proving fixed point theorems.
Proposition 19Assume thatis a partial metric space, and let . If mappingssatisfy (12) andHhas a 0-closed graph, then for , whereψis defined by (16), condition (14) holds.
Proof For , there exist and such that . If , then (2) yields , i.e., x is a fixed point of G, and by Proposition 10, . □
Proposition 20Letbe a partial metric space, letℱbe a nonempty family of at most two mappingswith 0-closed graphs, and let . Assume that all members ofℱeither satisfy (12) andφhas property (15) or satisfy (14). Then there exists a sequencesuch that , , , and .
Proof If satisfies (15) and (12) holds, then for ψ as in (16) and , condition (14) is satisfied (Proposition 19), and (Propositions 18, 16). Thus, it is sufficient to consider the case of and condition (14). Let be arbitrary, , and let be such that . If is defined, then is such that , and, similarly, satisfies . Thus, for , we have , , and converges to zero as . □
As was shown in Example 14, conditions (12), (14) are too weak to guarantee the existence of a fixed point, even for . The next two theorems, with stronger assumptions, are general results.
Theorem 21Letbe a 0-complete partial metric space, and letbe a mapping with 0-closed graph. Assume that for athere exists a sequenceinXsuch that
and . Theninis a fixed point ofF, and .
Proof Clearly, converges to zero as . Therefore, is a Cauchy sequence (see (2)) and it converges in , say, to (X is 0-complete). There exist such that . Now, from
and condition (8) it follows that . □
Theorem 21, with , (for ), is an extension of the Nadler theorem on multivalued contractions [, Theorem 5].
Now, Theorem 21 and Theorem 11 yield the following.
Theorem 22Letbe a 0-complete partial metric space, and letℱbe a family of mappingswith 0-closed graphs. Assume that ; some and at least all differentsatisfy (12). If for anthere exists a sequencesuch thatand (17) holds, then all members ofℱhave the same nonempty set of fixed points; this set is closed inand contained in Kerp.
A simple consequence of Theorem 21 is the following one.
Theorem 23Letbe a 0-complete partial metric space, and letbe a mapping with a 0-closed graph (e.g., is continuous). Assume that for aandthe following condition is satisfied:
Theninis a fixed point off, and .
Proof For (and ), we have (see (2))
and condition (17) holds. □
The next proposition shows that Theorem 23 is related to the well-known theorem of Matkowski [, Theorem 1.2, p.8].
Proposition 24Letbe a partial metric space, and letbe a bounded mapping satisfying
for a nondecreasing mapping . Then condition (18) holds.
Proof Let us adopt and , . For each , we have
as φ is nondecreasing, and we get (18). □
Let be a partial metric space, and let be a mapping. Let us recall the conditions used by Romaguera in :
For nondecreasing φ, Ćirić’s condition (21) is more general than (20), and (20) is more general than condition (19). All these conditions are used to prove that f has a fixed point.
Let us note that for conditions (19), (20), (21) can be arbitrary as on the right-hand side of any of these inequalities (for ) means that x is a fixed point of f.
Many fixed point theorems use more sophisticated conditions than (19), (20), (21) (see, e.g., , ) or the spaces under consideration have a richer structure (see, e.g., , ). We are interested in extending the most classical results.
The subsequent two lemmas are proved for condition (21), and the reasonings for (20) and (19) as well can be easily deduced. The next lemma (for condition (21)) has much in common with Romaguera’s Lemma 1 and Lemma 2 .
Lemma 25Letbe a partial metric space, and letbe a mapping satisfying condition (19), (20), or (21) for a . Then, for any , the condition
is satisfied, and .
Proof For notational simplicity, let us adopt , . For , condition (21) has the following form:
holds, i.e., (Lemma 1 ). This last equality and condition (21) for yield
i.e., (). This contradiction proves that must hold, and then condition (21) yields (see Romaguera’s Lemma 2 ). Now, it is clear that for arbitrary and the sequence , where , , converges to zero as and holds. □
The next lemma is also helpful in proving fixed point theorems.
Lemma 26Letbe a 0-complete partial metric space, and letbe a mapping satisfying condition (19), (20), or (21) for a . If for , holds, thenconverges into a unique fixed point off, and this point belongs to Kerp.
Proof From it follows that converges to a point in (Corollary 4). We get
Suppose . Then from
and it follows that
for large n. Now we get
a contradiction. Clearly, means that (see (2)), and (see (7)). If x, y are fixed points of f, then (see (2), (21)) means that . □
The next result extends Romaguera’s Theorem 4 , and consequently, an earlier celebrated result due to Matkowski [, Theorem 1.2, p.8].
Theorem 27Letbe a 0-complete partial metric space, and letbe a mapping satisfying condition (19) or (20) for asuch that
holds (e.g., ifφis nondecreasing). Thenfhas a unique fixed point; if , thenandin , .
Proof In view of Lemmas 25, 26 it is sufficient to prove that is a Cauchy sequence for , . Suppose that there are infinitely many such that . Let be the smallest numbers satisfying this inequality. For simplicity let us adopt and , . We have
which for means that
as we have . Now, for , condition (20) yields
and we obtain (from (19) as well)
for large k. Now, , , and condition (22) yield
a contradiction. Therefore, is a Cauchy sequence. □
Let be continuous mappings. Then such that on and on is a member of and φ satisfies conditions (15), (22); clearly, φ does not necessarily belong to or to .
Paesano and Vetro  have proved some theorems on coincidences and common fixed points. Maybe Theorem 27 can be extended to that case.
The next result extends Romaguera’s Theorem 3 , and consequently, an earlier celebrated result due to Boyd-Wong [, Theorem 1]. Let us recall that does not spoil the generality of condition (21). The proof is a modification of the one presented by Romaguera.
Theorem 28Letbe a 0-complete partial metric space, and letbe a mapping satisfying condition (21) for a mappingand satisfying (15) (e.g., withφupper semicontinuous from the right). Thenfhas a unique fixed point; if , thenandin , .
Proof We follow the initial part of the proof of Theorem 27 preceding the sentence with condition (20) (Proposition 16 yields ). For large and , we obtain
and . Now, condition (21) yields
and we obtain (see (15))
a contradiction. Therefore, is a Cauchy sequence. □
The next lemma enables us to extend the preceding two theorems.
Lemma 29Letbe a mapping such thatforhas a unique fixed point, say, x. Thenxis the unique fixed point off. If, in addition, , , then , holds.
Proof If x is a fixed point of , then means that is a fixed point of and the uniqueness yields . If are fixed points of f, then we get , , and as has a unique fixed point. If holds for each , then we also obtain , which means that . □
Theorem 30Letbe a 0-complete partial metric space, and letbe a mapping satisfying condition (19) or (20) withfreplaced byfor a , and ahaving property (22) (e.g., withφnondecreasing). Thenfhas a unique fixed point; if , thenandin , .
Proof Clearly, all the assumptions of Theorem 27 are satisfied for f replaced by . Now, we apply Lemma 29. □
The next theorem is a consequence of Theorem 28 and of Lemma 29.
Theorem 31Letbe a 0-complete partial metric space, and letbe a mapping satisfying the condition
for , and a mappingsatisfying (15) (e.g., withφupper semicontinuous from the right). Thenfhas a unique fixed point; if , thenandin , .
Now, we present the respective versions of conditions (19), (20), (21) for a multivalued mapping , where P is defined by (11).
For nondecreasing φ, condition (26) is more general than (25), and (25) is more general than condition (24).
The subsequent two lemmas are proved for condition (26), and the reasonings for (25) and (24) as well can be easily deduced.
At first, let us present the following extension of Lemma 25.
Lemma 32Letbe a partial metric space, and letbe a mapping. Then
holds, and condition (26) for anyyields , , ( in place offor (25)), and
holds (e.g., Fsatisfies (28) and is compact valued), then there exists , , such that
Proof For , we have (see (2))
Consequently, we get
as , holds. Clearly,
is satisfied and hence we get (27). Suppose for a . Then (see (26))
means that (as ), and . This contradiction proves (28).
Let , be arbitrary. If , are known, then let be such that (see (29) or (28) and Proposition 12) . If , then for we get . □
Now, let us prove an analog of Lemma 26.
Lemma 33Letbe a 0-complete partial metric space, and letbe a mapping with 0-closed graph satisfying condition (24), (25), or (26) for a . Ifand , thenconverges into a fixed point ofF, and this point belongs to Kerp.
Proof From it follows that converges to a point in (Corollary 4). We get
Suppose . Then from
and it follows that
for large n. Now we get
a contradiction. Clearly, means that (see (2)). There exist such that , and now,
together with condition (8) yield . □
It should be noted that a partial metric p defines metric δ in the following way: iff , and for . The topology of is clearly larger than the topology of (see (7)). Moreover, is 0-complete iff is complete , [, Proposition 2.1].
If the proof of a theorem is based on , , then it works for and the theorem is an immediate consequence of the respective result (if known) for metric spaces. Numerous examples can be found in .
Let us add that Corollary 4 and Proposition 5 show that for a 0-complete partial metric space , if we prove that , then the remaining part of the proof concerns the metric space (see (7)). Let us also recall that Kerp with its partial metric topology is a closed metric subspace of [, Lemma 2.2].
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