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Original Research ARTICLE

Front. Appl. Math. Stat., 13 September 2016 |

On Singular Interval-Valued Iteration Groups

  • Institute of Mathematics, Pedagogical University of Cracow, Kraków, Poland

Let I = (a, b) and L be a nowhere dense perfect set containing the ends of the interval I and let φ : I → ℝ be a non-increasing continuous surjection constant on the components of I \ L and the closures of these components be the maximal intervals of constancy of φ. The family {Ft, t ∈ ℝ} of the interval-valued functions Ft(x): = φ−1[t + φ(x)], xI forms a set-valued iteration group. We determine a maximal dense subgroup T ⊊ ℝ such that the set-valued subgroup {Ft, tT} has some regular properties. In particular, the mappings TtFt(x) for tT possess selections ft(x) ∈ Ft(x), which are disjoint group of continuous functions.

1. Introduction

A family of functions {ft : II, t ∈ ℝ} such that ftfs = ft+s, t, s ∈ ℝ is said to be an iteration group, however a family of set-valued functions {Ft : I → 2I, t ∈ ℝ} such that FtFs = Ft+s, t, s ∈ ℝ is said to be a set-valued iteration group (abbreviated to s-v iteration group). The notion of an iteration semigroup of set-valued functions was introduced and studied by Smajdor [1] and then studied in some classes of set-valued functions (see e.g., [2], [3], [4], [5]). The fundamental problem in the theory of multivalued iteration semigroups is the problem of existence and regularity properties of continuous selections. In this note we considered particular set-valued iteration groups whose values are the intervals or singletons. The presented results complete and generalize some of the topics from Zdun [6]. The considered s-v iteration groups have the very irregular properties. For every such s-v iteration group {Ft : I → 2I, t ∈ ℝ} we find a special maximal additive subgroup T ⊂ ℝ such that group {Ft : I → 2I, tT} has several “regular” properties.

2. Materials and Methods

Let I = (a, b) and φ : I → ℝ be a surjection. Define the set-valued functions

Ft(x):=φ1[φ(x)+t],t,xI.    (1)

The surjection φ is said to be the generating function of the family {Ft}.


The family {Ft : I → 2I} is a set-valued iteration group, i.e.,

Ft  Fs=Ft+s,t,s,


Ft  Fs(x)=yFs(x)Ft(y)xI.

Moreover, xFt(x) for t ≠ 0.

Proof. Fix an xI. Let zFtFs(x). Then there exists a yFs(x) such that zFt(y). This means that φ(y) = φ(x) + s and φ(z) = φ(y) + t, which gives that φ(z) = φ(x) + t + s. Hence zFt+s(x). Similarly we prove the converse inclusion.                   □

If φ is a homeomorphism then Equation (1) defines the general form of continuous iteration groups such that F1(x) ≠ x for xI.

If φ is non-injective then s-v iteration group generated by φ has very irregular properties and we will call this group singular. The purpose of this paper is the study of these “singularities.”

Obviously the set-valued functions Ft, t ∈ ℝ pairwise commute. This property is not transferible on the continuous selections of these set-valued mappings.

Let us assume that there exist Fu, Fv with uv which possess homeomorphic commuting selections f and g, that is f(x) ∈ Fu(x) and g(x) ∈ Fv(x) for x ∈ (a, b) and fg = gf. Then the generating function φ satisfies the equations φ(f(x)) = φ(x) + u and φ(g(x)) = φ(x) + v. Note that then f, g are iteratively incommensurable, i.e.,


where fn denotes the n-th iterate of function f and f0 = id. Define

Lf,g:={fn  gm(x),n,m}d.

The set Lf, g does not depend on x and either this set is the interval cl I or Lf, g is a nowhere dense perfect set in I (see Zdun [7]). If the generating function φ is continuous at least at one point of Lf, g then it is continuous and it is monotonic (see [8]).

We have more


If f and g are commuting iteratively incommensurable homeomorphisms, then there exist infinitely many s-v iteration groups {Ft, t ∈ ℝ} of type (1) such that f(x) ∈ F1(x) and g(x) ∈ Fs(x) for an s ∉ ℚ, but the only one of them has a monotonic generating function φ. Then the generating function φ is continuous and φ[Lf,g] = ℝ.

The proof is a simple consequence of Theorem 2 and Corollary 1 in Zdun [8].

The family {Ft, t ∈ ℝ} is a single-valued iteration group if and only if Lf, g = [a, b]. Then φ is strictly monotonic (see Zdun [8]).

In this paper we consider the case where Lf, g ≠ [a, b], that is {Ft : t ∈ ℝ} is a proper set-valued iteration group.

In the next section we will consider the more general case.

3. Results

Assume the following general hypothesis:

(H)φ:Iis a non-decreasing and non-injective surjection.

Then the function φ is continuous and the values of Ft are closed intervals or singletons. Denote by {Iα, α ∈ A} a family of the intervals of constancy of φ. These intervals are closed. Put



L:=IαAIntIα.    (2)

Note that φ|L* is strictly increasing, φ[Iα] are singletons and if Iα < Iβ then φ[Iα] < φ[Iβ].

It is easy to verify that the s-v iteration group {Ft : Icc[I], t ∈ ℝ} generated by φ has the following properties.


(i) For every xI Ft(x) either is a closed proper interval Iα or a singleton belonging to L*;

(ii) for every xI the s-v function tFt(x) is strictly decreasing, i.e., if s < t then for every uFs(x) and vFt(x), u < v;

(iii) for every xI tFt(x)=I;

(iv) every s-v function Ft is constant on the intervals Iα;

(v) if st then Ft(x) ∩ Fs(x) = Ø for xI, that is the group {Ft, t ∈ ℝ} is disjoint.

The conditions (i), (ii), (iii) characterize the interval-valued iteration groups. We have the following.


If an s-v iteration group {Ft, t ∈ ℝ} satisfies conditions (i), (ii), and (iii), where {Iα, α ∈ A} is a given family of closed disjoint proper intervals, then there exists a function φ satisfying (H) such that Ft are given by the formula (1).

Proof. Define


Let x0I and put h(t):=Ft(x0). Note that h is a bijection from ℝ onto X. Define φ by the following way: if xIα for an α ∈ A then φ(x):=h-1(Iα), if xL* then φ(x): = h({x}). It is easy to see that φ is a non-decreasing surjection of I onto ℝ constant on the intervals Iα and


Since Ft  Fs(x0)=Ft+s(x0) we have




Let xI. Then, by (iii), there exists an s ∈ ℝ such that xh(s). Hence φ(x) ∈ φ[h(s)] = s, thus φ(x) = s. This gives that φ[Ft(x)] ⊂ φ[Ft(h(s)] = φ(x) + t, so

φ[Ft(x)]=φ(x)+t.    (3)

Since Ft(x) ⊂ φ−1[φ[Ft(x)]] we have Ft(x) ⊂ φ−1[φ(x) + t]. Note that φ−1[φ(x) + t] is a singleton or equals to one of the intervals Iα. If Ft(x) is a singleton then, by (i), Ft(x)Iα for any α ∈ A. Thus φ−1[φ(x) + t] is not any of the intervals Iα, so it is a singleton. If Ft(x) is an interval Iα, then φ−1[φ(x + t)] must be also the same interval. This gives equality Ft(x) = φ−1[φ(x + t)].                   □


Let a family of set-valued function Ft be given by (1), where φ satisfies (H). Define


for t ∈ ℝ, xI. Then

(i) the families {f-t,t} and {f+t,t} are iteration groups;

(ii) f-t and f+t for t ∈ ℝ are non-decreasing discontinuous functions constant on the intervals of constancy of φ;

(iii) the mappings tf±t(x) are strictly decreasing;

(iv) f-t[I]L, f+t[I]L, t ∈ ℝ;

(v) Ft(x)=[f-t(x),f+t(x)], t ∈ ℝ.

Proof. (i) Fix an xI. Note that f-t(x), f+t(x)Ft(x) since the sets Ft(x) are closed. Hence, by Equation (1),

φ(f±t(x))=φ(x)+t,    (4)

so φ(f±t(f±s(x)))=φ(x)+t+s=φ(f±t+s(x)). This implies that


for an α ∈ A or both belong to L*, since Iα for α ∈ A are the intervals of constancy of φ. Obviously, in the second case, both values are equal. However, at the first case, f+t(f+s(x))sup Iα=f+t+s(x) and f-t+s(x)=inf Iαf-t(f-s(x)). On the other hand, putting f+s(x)=:y we have that f+t(y)Iα and f+t(y)Ft(y). Hence Ft(y)=Iα and f+t(y)=sup Iαf+t+s(x). This gives that


Similarly we prove that


(iv) Proving (i) we have shown that f±t(x) either belong to L* or equals to one of the ends of the interval Iα which belong to L. Both cases give that f±t(x)L.

The remaining assertions are the simple consequences of formula (Equation 1).               □

Let φ be non-decreasing and non-injective surjection. Define the following family of functions


The index cf is uniquely determined. This allows us to define


As a particular case of Proposition 2.2 in Farzadfard and Zdun [9] we get the following


If f ∈ Realm(φ) then the following conditions are equivalent:

(i) φ[L*] = φ[L*] + ind f;

(ii) φ[I \ L*] = φ[I \ L*] + ind f;

(iii) f[L*] = L*;

(iv) f maps each Iα into another one; moreover for every Iβ there exists Iα such that f[Iα] ⊂ Iβ.

Let φ satisfy (H) and define

T:={t:φ[IL]+t=φ[IL]}.    (5)

If T ≠ {0}, then T is a countable Abelian subgroup of group (ℝ, +).

In fact, since φ is constant in the intervals Iα, we have φ[I \ L*]={φ[Iα],αA}. It is easy to see that this set is unbounded above and below thus it is infinite and, consequently, countable since the intervals {Iα, α ∈ A} are pairwise disjoint.


A subgroup T given by Equation (5) is said to be a supporting group of the s-v iteration group {Ft : t ∈ ℝ}.


Let T ≠ {0} be a supporting group of s-v iteration group {Ft : t ∈ ℝ} generated by a function φ satisfying (H). Then

(i) if tT then for every xL* Ft(x) is a single point and Ft(x) ∈ L*;

(ii) if tT then for every α ∈ A there exists β ∈ A such that Ft(x)=Iβ for xIα;

(iii) if tT then for every β ∈ A there exists α ∈ A such that Ft(x)=Iβ for xIα;

(iv) if Ft[L*] = L* then tT.

Proof. (i) By Equation (2) f-t,f+tRealm(φ), indf±t=t for t ∈ ℝ and φ(f-t(x))=φ(f+t(x)). By Lemma 1 f±t(x)L* for xL*. Since φ|Iα is injective f-t(x)=f+t(x) for xL*. Thus, by Proposition 3 (v), Ft(x) is a singleton belonging to L*.

(ii) Let xIα. By Lemma 1 f±t(x)Iβ for a β ∈ A. Thus Ft(x)Iβ. If Ft(x) is a singleton then, by Proposition 1 (i), Ft(x) belongs to L*, so f±t(x)L*, but this is a contradiction. Thus Ft(x) is a proper interval, so Ft(x)=Iβ.

(iii) Fix a β ∈ A. Since φ[Iβ] is a singleton and φ is a surjection from I onto ℝ there exists an xI such that φ[Iβ] = t+φ(x), that is Ft(x)=Iβ. Suppose xL*. Then, by Lemma 1, f±t(x)L*, but this is a contradiction since f±t(x)Ft(x)=Iβ, so there exists an α ∈ A such that xIα.

(iv) Since φ satisfies relation Equation (3) we have φ[L*] = φ[Ft[L*]] = φ[L*] + t, so, by Lemma 1, tT.           □

Directly by Theorem 3 we get the following


Let T ≠ {0} be the supporting group of the s-v group {Ft : t ∈ ℝ} with generating function satisfying (H). Then

(i) T={t:ωA ω¯A Ft[Iω]=Iω¯};

(ii) T={t:xL* Ft(x) is a singleton};

(iii) T = {t ∈ ℝ:Ft[L*] = L*}.


A family of continuous mappings {ft : II, tT} such that ftfs = ft + s for t, sT is said to be a T-iteration group.

Now we consider the problems connected with continuous selections of s-v iteration groups. The iteration groups {f-t,t} and {f+t,t} are the monotonic selections of s-v group {Ft, t ∈ ℝ} that is f±t(x)Ft(x), but they are discontinuous.

Let φ satisfies (H) and Iα = :[aα, bα] for α ∈ A be the intervals of constancy of φ. For tT define the affine mappings qt, α : [aα, bα] → I such that


For every tT define the following mapping

qt(x):={qt,α(x),xIαf+t(x),xL.    (6)


If T ≠ {0} is the supporting group of s-v group {Ft : t ∈ ℝ} generated by a function satisfying condition (H), then {qt : II, tT} is a T-iteration group of continuous functions. Moreover, qt(x) ∈ Ft(x) for tT and xI.

Proof. Note that qt,α[Iα]=Ft[Iα] and Ft[Iα1]<Ft[Iα2] if Iα1 < Iα2. Hence, by Theorem 3, it follows that the mappings qt are strictly increasing surjections and, consequently, they are continuous.

It follows that for every t, sT, qt  qs[Iα]=qt[Fs[Iα]]=Ft[Fs[Iα]]=Ft+s[Iα]=qt+s[Iα]. Since the composition of affine functions is an affine function and there exists a unique increasing affine function mapping Iα onto the interval Ft+s[Iα] we get that qtqs = qt + s on Iα. Now it is easy to see that Proposition 3 implies our assertion.               □


If s-v group {Ft : t ∈ ℝ} generated by a function satisfying condition (H) has a non trivial supporting group T, then there exists infinitely many disjoint T-iteration groups {ft, tT} of continuous functions such that ft(x) ∈ Ft(x) for tT and xI. T is a maximal additive group with this property.

Proof. Let γ : II be a homeomorphism such that γ(x) = x for xL and for every α ∈ A γ[Iα] = Iα. Put

ft:=γ1  qt  γ,tT.

It follows, by Lemma 2, that {ft, tT} is a T-iteration group and ft(x) ∈ Ft(x).

Let Ft have a continuous and strictly increasing selection f. Since for every α ∈ A, f[Iα] is a proper interval, Ft[Iα] is also an interval. Thus, by Corollary 1, tT.                         □

Let us make the following assumptions.

(i) Let L be a Cantor set in I, that is L is a nowhere dense perfect set in I = (a, b) and a, bL.

(ii) Let Iω, ω ∈ ℚ be open pairwise disjoint intervals such that

IL=:ω Iω.

(iii) Let φ:I → ℝ be a Lebesgue function which lives on a set L that is φ is a continuous non-increasing surjection constant on cl Iω and, let cl Iω be the maximal intervals of constancy of φ.

The conditions (i), (ii), and (iii) imply that φ is continuous and



Let T be the supporting group of s-v group {Ft : t ∈ ℝ} generated by a function φ satisfying condition (H). If the group T is acyclic then the set L defined by (2) is a Cantor set and φ is a Lebesgue function which lives on L.

Proof. By Lemma 2 the family of mappings {qt, tT} defined by Equation (6) is a disjoint T-iteration group. Denote by LT the set of limit points of the orbits O(x) = {qt(x) : tT}, i.e., LT=O(x)d. In Zdun [10] (see Th.1) it is proved that the set LT does not depend on x and LT is either a Cantor set in I or LT = [a, b] or LT = {a, b}. Moreover, LT = {a, b} if and only if {qt, tT} is a cyclic group (see [10] Theorem 2).

Since qt(x) ∈ Ft(x) we have φ(qt(x)) = φ(x) + t for xI. LT ≠ [a, b]. In fact, suppose that LT = [a, b]. Fix an xI and an interval Iα. By the density of the orbit O(x) there exist u, v ∈ ℝ such that uv and qu(x),qv(x)Iα. Hence φ(x) + u = φ(qu(x)) = φ(qv(x)) = φ(x) + v what is a contradiction.

By Proposition 1 (ii) and Lemma 2 the mapping Φ(t): = qt is an isomorphism of T onto the group {qt, tT}. Thus T is cyclic if and only if {qt, tT} is cyclic, so T is cyclic if and only if LT = {a, b}. Hence T is acyclic if and only if LT is a Cantor set.

If T is acyclic then φ lives on LT. Let xLT and tT. Then qt(x)=f+t(x)L. Thus O(x) ⊂ L and, consequently, LTL, so L is also a Cantor set. By the assumption φ lives on L, however by the definition of qt φ lives on LF. Thus we get LF = L.               □


If f, g are commuting, iteratively incommensurable homeomorphisms and Lf, g ≠ cl I, then f and g are embeddable in a non-extensible disjoint T-iteration group {ft, tT}, where T is a dense, countable subgroup of ℝ.

Proof. By Theorem 2 there exists an s-v iteration group {Ft : t ∈ ℝ} with continuous non-decreasing generating function φ such that f(x) ∈ F1(x) and g(x) ∈ Fs(x) for an s ∉ ℚ and φ[Lf, g] = ℝ. Since Lf, g ≠ cl I, φ is a Lebesgue function which lives on Lf, g. Define T by Equation (5). By Theorem 5 f and g are embeddable in a T-iteration group {ft, tT}. Since 1, sT the group T is dense.               □

4. Discussion

In this note we consider the relation between the iteration groups of monotonic functions and the interval-valued iteration groups. These groups are still poorly investigated.

In Section 2 we indicate a desirability of the generalization of classical iteration groups in the real case. It is known that not all commutable iteratively incommensurable homeomorphisms are embeddable in an iteration group. However, Theorem 2 shows that the embeddabilty is always possible for s-v iteration groups.

Propositions 1 and 2 characterize s-v iteration groups of the form Equation (1). It is shown that, in our investigations, the form Equation (1) of s-v iteration groups are quite natural. Proposition 3 shows how s-v iteration groups of the form Equation (1) determine iterations groups of non-decreasing functions which are not injective.

A key concept of the paper is the supporting group T defined by Equation (5). If T is non-trivial additive group then it is countable and the set of all intervals of constancy of the generating function φ is also countable. Theorem 3 and Corollary 1 explain the meaning of the supporting group T. The restricted s-v group {Ft : tT} has a property that s-v functions Ft transform the intervals of constancy of the generating function φ onto itself and the points from its complement, that is the set L*, onto singletons in L*. Moreover, Theorem 4 and Corollary 1 show that each s-v function Ft for tT has continuous selection ft such that family {ft : tT} forms a group. Moreover, any Ft for tT has no continuous selection.

We have also proved that supporting group T is acyclic if and only if the generating function φ is a Lebesgue function which lives in a Cantor set.

The presented results may be helpful in the constructions of different iteration groups of non-decreasing functions.

Author Contributions

MZ conceived the study and prepared the manuscript.

Conflict of Interest Statement

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


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Keywords: iteration group, set-valued functions, simultaneous functional equations, Cantor set, singular Lebesgue function

Citation: Zdun MC (2016) On Singular Interval-Valued Iteration Groups. Front. Appl. Math. Stat. 2:13. doi: 10.3389/fams.2016.00013

Received: 24 May 2016; Accepted: 30 August 2016;
Published: 13 September 2016.

Edited by:

Witold Jarczyk, University of Zielona Góra, Poland

Reviewed by:

Ludwig Reich, University of Graz, Austria
Krzysztof Cieplinski, AGH University of Science and Technology, Poland

Copyright © 2016 Zdun. 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) or licensor 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: Marek C. Zdun,