Journal of Function Spaces
Volume 2017 (2017), Article ID 4168486, 8 pages
https://doi.org/10.1155/2017/4168486

Research Article

Redefinition of -Distance in Metric Spaces

Tomonari Suzuki

Department of Basic Sciences, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan

Correspondence should be addressed to Tomonari Suzuki; suzuki-t@mns.kyutech.ac.jp

Received 10 February 2017; Accepted 21 March 2017; Published 20 April 2017

Academic Editor: Ismat Beg

Copyright © 2017 Tomonari Suzuki. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

The concept of -distance was introduced in 2001; on the other hand, that of -function was introduced by Lin and Du. Strongly inspired by -function, we introduce a new concept, which is a very slight generalization of -distance and is more natural than -distance. So we could say that we redefine -distance in some sense.

1. Introduction

Throughout this paper, we denote by , , and the sets of all positive integers, all rational numbers, and all real numbers, respectively.

In 2001, the concept of -distance was introduced in order to generalize results in [1–8] and others.

Definition 1 (see [9]). Let be a metric space. Then a function from into is called a -distance on if there exists a function from into and the following is satisfied:

() for any .() and for any and , and is concave and continuous in its second variable.() and imply for any .() and imply .() and imply .

We note that the metric is one of -distances on . See [9–15] and references therein for many examples and theorems concerning -distance. For instance, using -distance, Suzuki [14] gave a simple proof of Zhong’s theorem [7].

In 2006, Lin and Du [16] introduced the following very interesting concept of -function, which is similar to that of -distance. However, both are independent.

Definition 2 (Lin and Du [16]). Let be a metric space. Then a function from into is called a -function if the following conditions hold:

()For any , .()If and in with and for some then .()For any sequence in with , and if there exists a sequence in such that , then .()For , and imply .

In this paper, strongly inspired by -function, we introduce a new concept, which is a slight generalization of -distance and which is more natural than -distance.

Before the author wrote the paper [9], he had considered more than twenty definitions. Definition 1 was the most natural definition between them. Therefore, he defined -distance as in Definition 1, though Definition 1 was not the weakest. However, strongly inspired by Definition 2, he finds a more natural definition of “-distance”; see Definition 7. Now he thinks that we have to change the definition of “-distance.”

2. New Definition

In this section, we introduce a new definition of the notion of -distance. Before introducing it, we state something on -distance, which are strongly connected with the new definition.

Definition 3 (see [9]). Let be a -distance on a metric space . Then a sequence in is called -Cauchy if there exist a function from into satisfying (2)–(5) and a sequence in such that .

Lemma 4 (see [9]). Let be a -distance on a metric space . If is a -Cauchy sequence, then is a Cauchy sequence in the usual sense.

Lemma 5 (see [9]). Let be a -distance on a metric space . If a sequence in satisfies for some , then is a -Cauchy sequence. Moreover, if a sequence in also satisfies , then . In particular for , and imply .

Lemma 6 (see [9]). Let be a -distance on a metric space . If a sequence in satisfies , then is a -Cauchy sequence. Moreover, if a sequence in satisfies , then is also a -Cauchy sequence and .

Now we give the new definition.

Definition 7. Let be a metric space and let be a function from into . Then is called a -distance on if the following holds:

() for any .()If and , then . Moreover, if converges to some , then for any .()If , then holds. Moreover, if converges to some , then for any .

The concept of -distance is slightly weaker than that of -distance. However, the author considers that the notions of both are the same. Indeed, we can prove -distance versions of all the existence theorems in [9, 11–14] using the same proof.

Proposition 8. Let be a -distance on a metric space . Then is a -distance.

Proof. () and () are the same. () follows from Lemma 6 and (). () follows from Lemmas 4 and 5 and (3).

Remark 9. A -distance need not be a -distance. However, the only example we have is quite complicated and we have chosen not to include it.

3. Condition (CL)

We have introduced the concept of -Cauchy for -distance; see Definition 3. Instead of this, for -distance, we introduce Condition (CL), which is named after Cauchy and lower semicontinuity.

Definition 10. Let be a -distance on a metric space . Let be a net in . Then is said to satisfy Condition (CL) if the following holds:

(CL1) is a Cauchy net in the usual sense.(CL2)Either of the following holds:

(i) does not converge.(ii)If converges to , then holds for any .

The following lemma is important in this section.

Lemma 11. Let be a metric space and let be a net in . Assume that is not Cauchy. Let be a sequence in and let be a mapping on . Then there exist a positive number and a sequence in such that

for any .

Proof. Since is not Cauchy, there exist a positive number and mappings and on such that , , and for any . Put and choose with and . Then either

holds. So we can put or such that . Continuing this argument, we can obtain the desired result.

Remark 12. is not necessarily a subnet of .

We do not use the concept of net in Definition 7. However, we can prove something on net as follows.

Lemma 13. Let be a -distance on a metric space . Let be a net in satisfying for some . Then the following holds:

(i) satisfies Condition (CL).(ii)If a net in also satisfies , then holds.

Proof. We choose a sequence in such that for . Arguing by contradiction, we assume is not Cauchy. Let be an arbitrary mapping from into itself. Then by Lemma 11, there exist a positive number and a sequence in satisfying

for any . Since

from (), we have

which is a contradiction. Hence, is Cauchy. Next, we assume converges to . Fix and let with . Then we can choose a sequence in such that

for any . Since converges to , we have from ()

Since is arbitrary, we obtain . Therefore, we have shown (i). Let us prove (ii). Define a directed set and an order ≤ in

Define a net by

Then since , is Cauchy by (i). This implies .

As a corollary of Lemma 13, we obtain the following sequential version. Compare Lemma 14 with Lemma 5.

Lemma 14. Let be a -distance on a metric space . If a sequence in satisfies for some , then satisfies Condition (CL). Moreover, if a sequence in also satisfies , then holds. In particular, for , and imply .

Lemma 15. Let be a directed set and let be a mapping from into itself such that for any . Let be a net in a set . Then is a subnet of .

Proof. Obvious.

Lemma 16. Let be a -distance on a metric space . Let be a directed set such that for any there exists with . Let be a net in satisfying . Then the following holds:

(i) satisfies Condition (CL).(ii)If a net in satisfies , then satisfies Condition (CL) and holds.

Proof. We choose a sequence in such that

for any . We consider the following two cases:

(a)There exists such that for any .(b)For any , there exists with .

In the first case, from the assumption, we can take with . Then since and for any , we have

So

Similarly, . By Lemma 14, we obtain for . Therefore, . Since converges to , thus, is Cauchy. Also, holds. Therefore, we have shown that satisfies Condition (CL) and . As in the above proof, we can prove that satisfies Condition (CL). In the second case, we can define a mapping from into itself satisfying

for any . In fact, for , there exists with . We can choose satisfying and . If there exists such that and , then and hence , which is a contradiction. Similarly implies a contradiction. Therefore, we have defined . We will show that satisfies Condition (CL). Arguing by contradiction, we assume is not Cauchy. Then by Lemma 11, there exist a positive number and a sequence in satisfying

From the definition of , we note for with . Since

from (), we have

So

which is a contradiction. Hence, is Cauchy. Next, we assume converges to . Fix and let with . Then we can choose a sequence in such that

Since , we have from () that

Since is arbitrary, we obtain . Therefore, we have shown that satisfies Condition (CL). Let us prove . Arguing by contradiction, we assume . Fix with . Then we can choose a sequence in such that

Since , we obtain from () that

which is a contradiction. Therefore, we have shown . We have proved (ii). We shall show that satisfies Condition (CL). Fix and let with . Then we can define a mapping from into itself such that

By Lemma 15, we note that is a subnet of . We have

So from (ii), satisfies Condition (CL) and holds. So, is Cauchy. We assume that converges to . Since also converges to , we have

Since is arbitrary, we obtain . Therefore, we have shown that satisfies Condition (CL). We have proved (i).

As a corollary of Lemma 16, we obtain the following sequential version. Compare Lemma 17 with Lemma 6.

Lemma 17. Let be a -distance on a metric space . If a sequence in satisfies , then satisfies Condition (CL). Moreover, if a sequence in satisfies , then satisfies Condition (CL) and holds.

Lemma 18. Let be a -distance on a metric space and let and be sequences in satisfying . Assume that, for any subsequence of the sequence in , there exist subsequences of and of satisfying . Then satisfies Condition (CL) and holds.

Proof. We note that is Cauchy from Lemma 17. Let be an arbitrary subsequence of . Then from the assumption, there exist subsequences of and of satisfying . We note . From (), we have and hence

Since is arbitrary, we obtain . is Cauchy and so is . Next, we assume that converges to , fix , and choose a subsequence of satisfying . Then from the assumption, there exist subsequences of and of satisfying . By (), we obtain

Therefore, satisfies Condition (CL).

4. Examples

In this section, in order to show that the concept of -distance is more natural than that of -distance, we give some examples. Compare them with Propositions and in [9] and Proposition in [11]. We note that we have not yet proved -distance versions of Propositions 19–21 below.

The following is connected with the result in Zhong [7]. See also Turinici [17].

Proposition 19. Let be a -distance on a metric space . Let be a nonincreasing function from into with ; and let be a function from into with for . Then a function from into defined by

for is also a -distance on .

Proof. Using (), we have

for any and hence holds. Define a function from into by

for . From the assumption of , is well defined. If , then we have from the definition of . So,

Hence, we note that

Using (31), we next show . We assume and . Without loss of generality, we may assume and for any . Then for with , we have

and hence

This implies that is bounded. From the definition of , is also bounded and so is . We put

Then we have by (31)

By (), we obtain . Moreover, if converges to some , then for any . This yields for any . Therefore, holds. We can prove as in the proof of .

We call that a nonempty subset of is called -bounded if . The following is connected with the result in Ume [18].

Proposition 20. Let be a -distance on a metric space . Let be a set-valued mapping on such that is a nonempty -bounded subset of for any . Assume that , , and imply . Then a function from into defined by

for any is also a -distance.

Proof. It is obvious that is -bounded for any . For nonempty -bounded subsets and of , we put

Then . Indeed, for , we have

Using , we can write

From (), we have

for nonempty -bounded subsets , , and of . So, for , we obtain

This is . We next show . We assume and . We choose a subsequence of the sequence such that for any . Since , we can choose a sequence in such that

for . We note

Let be an arbitrary subsequence of and choose a subsequence of such that for any . Then we have

and hence from Lemma 18. We also have

and hence from Lemma 18. Thus, holds. If converges to some , then and also converge to . So from the assumption, holds. Also, holds for any from Lemma 18. We obtain

We have shown . Let us prove . We assume . Fix . Then since

satisfies Condition (CL) by Lemma 14. We assume that converges to . We choose a sequence in such that

for . Since

we have by Lemma 14. Hence, converges to . From the assumption, holds. As in the proof of (), we can prove for any .

We finally prove the following. We note that our proof is very easy while the proof of Proposition in [11] is difficult.

Proposition 21. Let be a metric space. Let be a set of -distances on . If a function defined by

is a real-valued function, then is a -distance on .

Proof. We have

for any . We next show . We assume and . Then since for any , and hold. Thus, from , holds. If converges to some , then holds for any and by using again. Using it, we have

for any . We have shown . We can similarly prove .

5. Kirk and Saliga’s Fixed Point Theorem

Kirk and Saliga generalized Caristi’s fixed point theorem [2, 3]; see Theorem in [19]. The author thinks that the proof in [19] is splendid. In this section, we generalize Kirk and Saliga’s fixed point theorem.

Let be an ordinal number. We denote by and the successor and the predecessor of , respectively. We recall that is isolated if exists. is limit if and does not exist. For a set , we denote by cardinal number of .

Theorem 22. Let be a complete metric space with a -distance . Let be a mapping on . Let be a function from into bounded from below such that whenever a net satisfies the following:

(a) converges to .(b)For any , there exists satisfying .(c).(d) for with .

Assume for any , and there exists such that

for any . Fix . Let be the first uncountable ordinal number and put . Define a net by

Then is well defined and there exists satisfying and .

Proof. Define a function form into by

where is the identity mapping on . It is obvious that

holds for any . We shall show by transfinite induction that

(a) is well defined,(b) for ,

for . It is obvious that is true. Fix with , and assume that is true for . In the case where is isolated, it is obvious that is well defined. We have

For , we have

Hence, is true. In the other case, where is limit, since is a nonincreasing and is bounded from below, converges and hence it is Cauchy. We have

By Lemma 16, satisfies Condition (CL). Since is complete, converges. Hence, is well defined. For , we have

Hence, is true. Therefore, by transfinite induction, is true for any . Arguing by contradiction, we assume the following:

(a) for any .

Then there exists satisfying . So we have

which is a contradiction. Therefore, there exists satisfying . We have

Hence, and hold. By Lemma 14, holds. Therefore, holds.

Conflicts of Interest

The author declares that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

The author is supported in part by Grants-in-Aid for Scientific Research from the Japanese Ministry of Education, Culture, Sports, Science and Technology.

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