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ORIGINAL RESEARCH article

Front. Appl. Math. Stat., 14 June 2024
Sec. Mathematical Physics
This article is part of the Research Topic Approximation Methods and Analytical Modeling Using Partial Differential Equations View all 10 articles

On monoids of metric preserving functions

\r\nViktoriia Bilet
Viktoriia Bilet1*Oleksiy Dovgoshey,Oleksiy Dovgoshey1,2
  • 1Department of Theory of Functions, Institute of Applied Mathematics and Mechanics of NASU, Slovyansk, Ukraine
  • 2Department of Mathematics and Statistics, University of Turku, Turku, Finland

Let X be a class of metric spaces and let PX be the set of all f : [0, ∞) → [0, ∞) preserving X, i.e., (Y, f ∘ ρ) ∈ X whenever (Y, ρ) ∈ X. For arbitrary subset A of the set of all metric preserving functions, we show that the equality PX = A has a solution if A is a monoid with respect to the operation of function composition. In particular, for the set SI of all amenable subadditive increasing functions, there is a class X of metric spaces such that PX = SI holds.

2020 Mathematics Subject Classification: Primary 26A30, Secondary 54E35, 20M20

1 Introduction

The following is a particular case of the concept introduced by Jachymski and Turoboś [1].

Definition 1. Let A be a class of metric spaces. Let us denote by PA the set of all functions f : [0, ∞) → [0, ∞) such that the implication

((X,d)A)((X,fd)A)

is valid for every metric space (X, d).

For mappings F : XY and Φ : YZ, we use the symbol F ∘ Φ to denote the mapping

XFYΦZ.

We also use the following notation:

F, set of functions f : [0, ∞) → [0, ∞);

F0, set of functions fF with f(0) = 0;

Am, set of functions fF0 with f−1(0) = {0};

SI, set of subadditive increasing fAm;

M, class of metric spaces;

U, class of ultrametric spaces;

Dis, class of discrete metric spaces;

M2, class of two-points metric spaces;

M1, class of one-point metric spaces.

The main purpose of this article is to provide a solution to the following problems.

Problem 2. Let APM. Find conditions under which the equation

PX=A    (1)

has a solution XM.

Problem 3. Let APU. Find conditions under which Equation (1) has a solution XU.

In addition, we find all solutions to Equation (1) for A equal to F, F0, or Am and answer the following question.

Question 4. Is there a subclass X of the class M such that

PX=SI?

This question was posed as a challenge in [2] in a different but equivalent form and it was the original motivation for our research.

The article is organized as follows. The next section contains some necessary definitions and facts from the theories of metric spaces and metric preserving functions.

Section 3 provides some definitions from the semigroup theory and describes solutions to Equation (1), for the cases when A is F, F0, or Am. In addition, we show that PX is always a submonoid of (F, ∘). See Theorems 21, 23, 24, and Proposition 27, respectively.

Section 4 provides solutions to Problems 2 and 3, which are given, respectively, in Theorems 30 and 33. Theorem 32 gives a positive answer to Question 4.

2 Preliminaries on metrics and metric preserving functions

Let X be non-empty set. A function d : X × X → [0, ∞) is said to be a metric on the set X if for all x, y, zX we have

(i) d(x, y) ⩾ 0 with equality if and only if x = y, the positivity property;

(ii) d(x, y) = d(y, x), the symmetry property;

(iii) d(x, y) ⩽ d(x, z) + d(z, y), the triangle inequality.

A metric space (X, d) is ultrametric if the strong triangle inequality

d(x,y)max{d(x,z),d(z,y)}

holds for all x, y, zX.

Example 5. Let us denote 0+ by the set (0, ∞). Then the mapping d+:0+×0+[0,),

d+(p,q):={0   if   p=q,max{p,q}   otherwise.

is the ultrametric on 0+ introduced by Delhommé et al. [3].

Definition 6. Let (X, d) be a metric space. The metric d is discrete if there is k ∈ (0, ∞) such that

d(x,y)=k

for all distinct x, yX.

In what follows we will say that a metric space (X, d) is discrete if d is a discrete metric on X. We will denote the class of all discrete metric space by Dis. In addition, for given non-empty set X1, we will denote by DisX1 the subclass of Dis consisting of all metric spaces (X1, d) with discrete d.

Remark 7. All discrete topological spaces can be endowed with a metric which is discrete, but not every metric space with discrete topology is discrete in the sense of Definition 6.

Example 8. Let Mk, for k = 1, 2, be the class of all metric spaces (X, d) satisfying the equality

card(X)=k.

Then all metric spaces belonging to M1M2 are discrete.

Proposition 9. The following statements are equivalent for each metric space (X, d) ∈ M.

(i) (X, d) is discrete.

(ii) Every three-points subspace of (X, d) is discrete.

Proof. The implication (i) ⇒ (ii) is evidently valid.

Suppose that (ii) holds but (X, d) ∉ Dis. Then there are some different points i, j, k, lX such that

d(i,j)d(k,l).    (2)

We write X1 : = {i, j, k} and X2 : = {j, k, l}. Then the spaces (X1, d|X1×X1) and (X2, d|X2×X2) are discrete subspaces of (X, d) by statement (ii). Consequently we have

d(i,j)=d(j,k)    (3)

and

d(j,k)=d(k,l),    (4)

by definition of the class Dis. Now (3) and (4) give us

d(i,j)=d(k,l),

which contradicts (2).

Remark 10. The standard definition of discrete metric can be formulated as follows: “The metric on X is discrete if the distance from each point of X to every other point of X is one.” (see, for example, Searcóid [4]).

Let F be the set of all functions f : [0, ∞) → [0, ∞).

Definition 11. A function fF is metric preserving (ultrametric preserving) if fPM (fPU).

Remark 12. The concept of metric preserving functions can be traced back to Wilson [5]. Similar problems were considered by Blumenthal [6]. The theory of metric preserving functions was developed by Borsík, Doboš, Piotrowski, Vallin, and other mathematicians [719]. See also lectures by Doboš [20] and the introductory paper by Corazza [21]. The study of ultrametric preserving functions began by Pongsriiam and Termwuttipong [22] and was continued in [23, 24].

We will say that fF is amenable if

f-1(0)={0}

holds and the set of all amenable functions from F will be denoted by Am. Let us denote the set of all functions fF satisfying the equality f(0) = 0 by F0. It follows directly from the definition that AmF0F.

Moreover, a function fF is increasing if the implication

(xy)(f(x)f(y))

is valid for all x, y ∈ [0, ∞).

The following theorem was proved in [22].

Theorem 13. A function fF is ultrametric preserving if and only if f is increasing and amenable.

Remark 14. Theorem 13 was generalized in [25] to the special case of the so-called ultrametric distances. These distances were introduced by Priess-Crampe and Ribenboim [26] and were studies by different researchers [2730].

Recall that a function fF is said to be subadditive if

f(x+y)f(x)+f(y)

holds for all x, y ∈ [0, ∞). Let us denote the set of all subadditive increasing functions fAm by SI.

In the next proposition, we restate the equivalence between statements (i) and (ii) of Corollary 36 [2].

Proposition 15. The equality

SI=PUPM

holds.

Remark 16. The metric preserving functions can be considered as a special case of metric products (= metric preserving functions of several variables). See, for example, [3137]. An important special class of ultrametric preserving functions of two variables was first considered in 2009 [38].

3 Preliminaries on semigroups. Solutions to FX = A for A = F, F0, and Am

Let us recall some basic concepts of semigroup theory, see, for example, “Fundamentals of Semigroup Theory” by Howie [39].

A semigroup is a pair (S, *) consisting of a non-empty set S and an associative operation * : S × SS, which is called the multiplication on S. A semigroup S = (S, *) is a monoid if there is eS such that

e*s=s*e=s

for every sS.

Definition 17. Let (S, *) be a semigroup and ∅ ≠ TS. Then T is a subsemigroup of S if a, bTa * bT. If (S, *) is a monoid with the identity e, then T is a submonoid of S if T is a subsemigroup of S and eT.

Example 18. The semigroups (F, ∘), (Am, ∘), (PM, ∘), and (PU, ∘) are monoids, and the identical mapping id:[0, ∞) → [0, ∞), id(x) = x for every x ∈ [0, ∞) is the identity of these monoids.

The following simple lemmas are well-known.

Lemma 19. Let T be a submonoid of a monoid (S, *) and let VT. Then V is a submonoid of (S, *) if and only if V is a submonoid of T.

Lemma 20. Let T1 and T2 be submonoids of a monoid (S, *). Then the intersection T1T2 also is a submonoid of (S, *).

The next theorem describes all solutions to the equation PX = F.

Theorem 21. The following statements are equivalent for every XM.

(i) X is the empty subclass of M.

(ii) The equality

PX=F    (5)

holds.

Proof. (i) ⇒ (ii). Let X be the empty subclass of M. Definition 1 implies the inclusion FPX. Let us consider an arbitrary fF. To prove equality (5), it is suffice to show that fPX. Since X is empty, the membership relation (X, d) ∈ X is false for every metric space (X, d). Consequently, the implication

((X,d)X)((X,fd)X)

is valid for every (X, d) ∈ M. It implies fPX by Definition 1. Equality (5) follows.

(ii) ⇒ (i). Let (ii) hold. We must show that X is empty. Suppose contrary that there is a metric space (X, d) ∈ X. Since, by definition, we have X ≠ ∅, there is a point x0X. Consequently, d(x0, x0) = 0 holds. Let c ∈ (0, ∞) and let f : [0, ∞) → [0, ∞) be a constant function,

f(t)=c

for every t ∈ [0, ∞). In particular, we have

f(0)=c>0.    (6)

Equality (5) implies that fd is a metric on X. Thus, we have

0=f(d(x0,x0))=f(0),

which contradicts (6). Statement (i) follows.

Remark 22. Theorem 21 becomes invalid if we allow the empty metric space to be considered. The equality

PX=F

holds if the non-empty class X contains only the empty metric space.

Let us describe now all possible solutions to PX = F0.

Theorem 23. The equality

PX=F0    (7)

holds if and only if X is a non-empty subclass of M1.

Proof. Let XM1 be non-empty. Equality (7) holds if

PXF0    (8)

and

PXF0.    (9)

Here, we prove the validity of (8). Let fF0 be arbitrary. Since every (X, d) ∈ X is a one-point metric space, we have fd = d for all (X, d) ∈ X by the positivity property of metric spaces, Inclusion (8) follows.

Here, we prove (9). The inclusion PXF follows from Definition 1. Thus, if (9) does not hold, then there is f0F such that f0PX,

f0(0)=k   and   k>0.    (10)

Since X is non-empty, there is (X0, d0) ∈ X. Let x0 be a (unique) point of X0. Since f0 belongs to PX, the function f0d0 is a metric on X0. Now, using (10), we obtain

f0(d0(x0,x0))=f0(0)=k>0,

which contradicts the positivity property of metric spaces. Inclusion (9) follows.

Let (7) hold. We must show that X is a non-empty subclass of M1. If X is empty, then

PX=F    (11)

holds by Theorem 21. Equality (11) contradicts equality (7). Hence, X is non-empty. To complete the proof, we must show that

XM1.    (12)

Let us consider the constant function f0 : [0, ∞) → [0, ∞) such that

f0(t)=0,    (13)

for every t ∈ [0, ∞). Then f0 belongs to F0. Hence, for every (X, d) ∈ X, the mapping d0 : = f0d is a metric on X. Now (13) implies d0(x, y) = 0 for all x, yX and (X, d) ∈ X. Hence, card(X) = 1 holds, because the metric space (X, d0) is one-point by the positivity property. Inclusion (12) follows. The proof is completed.

The next theorem gives us all solutions to the equation PX = Am.

Theorem 24. The following statements are equivalent for every XM.

(i) The inclusion

XDis    (14)

holds, and there is (Y, ρ) ∈ X with

card(Y)2,    (15)

and we have

DisX1X    (16)

for every (X1, d1) ∈ X.

(ii) The equality

PX=Am    (17)

holds.

Proof. (i) ⇒ (ii). Let (i) hold. Equality (17) holds if

PXDis    (18)

and

PXDis.    (19)

Here, we prove (18). Inclusion (18) holds if we have

(X1,fd1)X    (20)

for all fAm and (X1, d1) ∈ X. Relation (20) follows from Theorem 23 if (X1, d1) ∈ M1. To see it we only note that AmF0. Let us consider the case when

card(X1)2.

Since (X1, d1) is discrete by (14), Definition 6 implies that there is k1 ∈ (0, ∞) satisfying

d1(x,y)=k1

for all distinct x, yX1. Let fAm be arbitrary. Then f(k1) is strictly positive and

f(d1(x,y))=f(k1)

holds for all distinct x, yX1. Thus, fd1 is a discrete metric on X1, i.e., we have

(X1,fd1)DisX1.    (21)

Now, Equation (20) follows from Equations (16, 21).

Here, we prove (19). To prove, we must show that every fPX is amenable.

Suppose contrary that f belongs to PX but the equality

f(t1)=0    (22)

holds with some t1 ∈ (0, ∞). By statement (i) we can find (Y, ρ) ∈ X such that (15) and

ρ(x,y)=t1

hold for all distinct x, yY. Now fPX and (Y, ρ) ∈ X imply that f ∘ ρ is a metric on Y. Consequently, for all distinct x, yY, we have

f(ρ(x,y))=f(t1)>0,

which contradicts (22). The validity of (19) follows.

(ii) ⇒ (i). Let X satisfy equality (17). Since AmF holds, the class X is non-empty by Theorem 21. Moreover, using Theorem 23, we see that X contains a metric space (X, d) with card(X) ⩾ 2, because AmF0.

If the inequality

card(Y)2

holds for every (Y, ρ) ∈ X, then all metric spaces belonging to X are discrete (see Example 8). Using the definitions of Dis and Am, it is easy to prove that for each (X1, d1) ∈ Dis and every (X1, d) ∈ DisX1 there exists fAm such that d = fd1. Hence to complete the proof, it is suffice to show that every (X, d) ∈ X is discrete when

card(X)3.    (23)

Let us consider arbitrary (X, d) ∈ X satisfying (23). Suppose that (X, d) ∉ Dis. Then by Proposition 9 there are distinct a, b, c, ∈ X such that

d(b,c){d(a,b),d(c,a)}.    (24)

Let c1 and c2 be points of (0, ∞) such that

c2>2c1.    (25)

Now we can define fAm as

f(t):={0   if   t=0,c2   if   t=d(b,c),c1   otherwise.    (26)

Equality (17) implies that fd is a metric on X. Consequently, we have

f(d(b,c))f(d(b,a))+f(d(b,c))    (27)

by triangle inequality. Now using Equations (24, 26) we can rewrite Equation (27) as

c2c1+c1,

which contradicts Equation (25). It implies (X, d) ∈ Dis. The proof is completed.

Corollary 25. The equalities

PDis=PM2=Am

hold.

Remark 26. The equality

PM2=Am

is known, see, for example, Remark 1.2 in the article [13]. This article also contains “constructive” characterizations of the smallest bilateral ideal and the largest subgroup of the monoid PM.

Proposition 27. Let X be a subclass of M. Then PX is a submonoid of (F, ∘).

Proof. It follows directly from Definition 1 that

PXF

holds and that the identity mapping id:[0, ∞) → [0, ∞) belongs to PX. Hence, by Lemma 19, it is suffice to prove

fgPX    (28)

for all f, gPX.

Let us consider arbitrary fPX and gPX. Then, using Definition 1, we see that (X, gd) belongs to X for every (X, d) ∈ X. Consequently,

(X,f(gd))X    (29)

holds. Since the composition of functions is always associative, the equality

(fg)d=f(gd)    (30)

holds for every (X, d) ∈ X. Now Equation (28) follows from Equations (29, 30).

The above proposition implies the following corollary.

Corollary 28. If the equation

PX=A

has a solution, then A is a submonoid of F.

The following example shows that the converse of Corollary 28 is, generally speaking, false.

Example 29. Let us define A1F as

A1={f1,id},

where f1F is defined such that

f1(t):={1   if   t=0,0   if   t=1,t   otherwise    (31)

and id is the identical mapping of [0, ∞). The equalities f1f1 = id, f1 ∘ id = f1 = id ∘ f1 show that A1 is a submonoid of (F, ∘). Suppose that there is X1M satisfying the equality

PX1=A1.    (32)

Then using Theorem 21, we see that X1 is non-empty because A1F holds. Let (X1, d1) be an arbitrary metric space from A1. Since X1 is non-empty, we can find x1X1. Then (32) implies that f1d1 is metric on X1. Now using (31), we obtain

f1(d1(x1,x1))=f1(0)=1,

which contradicts the positivity property of metrics.

4 Submonoids of monoids PM and PU

The following theorem provides a solution to Problem 2.

Theorem 30. Let A be a non-empty subset of the set PM of all metric preserving functions. Then the following statements are equivalent.

(i) The equation

PX=A    (33)

has a solution XM.

(ii) A is a submonoid of (F, ∘).

(iii) A is a submonoid of (PM, ∘).

Proof. (i) ⇒ (ii). Suppose that there is XM such that (33) holds.

Then A is a submonoid of (F, ∘) by Proposition 27.

(ii) ⇒ (iii). Let A be a submonoid of (F, ∘). By Proposition 27, the monoid (PM, ∘) also is a submonoid of (F, ∘). Then using the inclusion APM, we obtain that A is a submonoid of (PM, ∘) by Lemma 19.

(iii) ⇒ (i). Let A be a submonoid of (PM, ∘). We must prove that (33) has a solution XM.

Let (X, d) be a metric space such that

{d(x,y):x,yX}=[0,).    (34)
X:={(X,fd):fA}.    (35)

Thus, a metric space (Y, ρ) belongs to X if and only if Y = X and there is fA such that ρ = fd.

We claim that Equation (33) holds if X is defined by Equality (35). To prove it, we note that Equation (33) holds if

APX    (36)

and

APX.    (37)

Here, we prove Inclusion (36). This inclusion holds if for every fA and each (Y, ρ) ∈ X we have (Y, f ∘ ρ) ∈ X. Let us consider arbitrary (Y, ρ) ∈ X and fA. Then, using Equation (35), we can find gA such that

X=Y   and   ρ=gd.    (38)

Since A is a monoid, the membership relations fA and gA imply gfA. Hence, we have

(X,gfd)X    (39)

by Equality (35). Now (Y, f ∘ ρ) ∈ X follows from Equations (38, 39).

Here, we prove Inclusion (37). Let g1 belongs to PX and let (X, d) be the same as in (35). Then (X, g1d) belongs to X and, using (35), we can find f1A such that

(X,g1d)=(X,f1d).    (40)

Equality (40) implies

g1(d(x,y))=f1(d(x,y)),    (41)

for all x, yX. Consequently, g1(t) = f1(t) holds for every t ∈ [0, ∞) by Equation (34, 41). Thus, we have g1 = f1. That implies g1A. Inclusion (37) follows. The proof is completed.

Remark 31. A reviewer of the article noted that condition (34) can be neatly expressed in terms of center distances which stems from article [40].

Let us turn now to Question 4. Proposition 15 and Lemma 20 provide the following result.

Theorem 32. There is XM such that

PX=SI.    (42)

Proof. By Proposition 27, the monoids (PM, ∘) and (PU, ∘) are submonoids of (F, ∘). The equality

SI=PMPU    (43)

holds by Proposition 15. Using Equality (43) and Lemma 20 with T1 = PM, T2 = PU, and S = F, we see that SI also is a submonoid of F. Consequently, Theorem 30 with A = SI implies that there is XM such that (42) holds.

The next theorem is an ultrametric analog of Theorem 30 and it gives us a solution to Problem 3.

Theorem 33. Let A be a non-empty subset of the set PU of all ultrametric preserving functions. Then the following statements are equivalent.

(i) The equation PX = A has a solution XU.

(ii) A is a submonoid of (F, ∘).

(iii) A is a submonoid of (PU, ∘).

A proof of Theorem 33 can be obtained by a simple modification of the proof of Theorem 30. We only note that the ultrametric space defined in Example 5 satisfies equality (34) with X=0+ and d = d+.

5 Two conjectures

Conjecture 34. The equation

PX=A

has a solution XM for every submonoid A of the monoid Am.

Example 29 shows that we cannot replace Am with F in Conjecture 34, but we hope that the following is valid.

Conjecture 35. For every submonoid A of the monoid F, there exists XM such that PX and A are isomorphic submonoids.

Data availability statement

The original contributions presented in the study are included in the article/supplementary material, further inquiries can be directed to the corresponding author.

Author contributions

VB: Writing – original draft, Writing – review & editing. OD: Methodology, Project administration, Supervision, Validation, Writing – original draft, Writing – review & editing.

Funding

The author(s) declare financial support was received for the research, authorship, and/or publication of this article. VB was partially supported by a grant from the Simons Foundation (Award 1160640, Presidential Discretionary-Ukraine Support Grants, VB). OD was supported by grant 359772 of the Academy of Finland.

Acknowledgments

We express our deep gratitude toward the reviewers for the careful study of our paper. Their elaborate, constructive criticism allowed us to improve this paper and suggested further directions of the research.

Conflict of interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

Publisher's note

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Keywords: metric preserving function, monoid, subadditive function, ultrametric space, ultrametric preserving function

Citation: Bilet V and Dovgoshey O (2024) On monoids of metric preserving functions. Front. Appl. Math. Stat. 10:1420671. doi: 10.3389/fams.2024.1420671

Received: 20 April 2024; Accepted: 20 May 2024;
Published: 14 June 2024.

Edited by:

Kateryna Buryachenko, Humboldt University of Berlin, Germany

Reviewed by:

Ravindra K. Bisht, National Defence Academy, India
Filip Turoboś, Lodz University of Technology, Poland

Copyright © 2024 Bilet and Dovgoshey. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.

*Correspondence: Viktoriia Bilet, viktoriiabilet@gmail.com

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