# Functions of Boundedκφ-Variation in the Sense of Riesz-Korenblum

Authors: Mariela Castillo; Sergio Rivas; María Sanoja; Iván Zea
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We present the space of functions of boundedκφ-variation in the sense of Riesz-Korenblum, denoted by κBVφ[a,b], which is a combination of the notions of bounded φ-variation in the sense of Riesz and bounded κ-variation in the sense of Korenblum. Moreover, we prove that the space generated by this class of functions is a Banach space with a given norm and we prove that the uniformly bounded composition operator satisfies Matkowski's weak condition.


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## Body

## 1. Introduction

The concept of functions of bounded variation has been well known since Jordan [1] gave the complete characterization of functions of a bounded variation as the difference of two increasing functions in 1881. This class of functions immediately proved to be important in connection with the rectification of curves and with Dirichlet’s theorem on the convergence of Fourier series. Functions of a bounded variation exhibit many interesting properties that make them a suitable class of functions in a variety of contexts with wide applications in pure and applied mathematics (see [2–4]).Riesz [5] in 1910 generalized the notion of Jordan and introduced the concept of bounded p-variation (1<p<∞) and showed that, for 1<p<∞, this class coincides with the class of functions absolutely continuous with the derivative in the space Lp. On the other hand, this notion of bounded p-variation was generalized by Medvedev [6] in 1953 who introduced the concept of bounded φ-variation in the sense of Riesz and also showed a Riesz’s lemma for this class of functions.Korenblum [7] in 1975 introduced the notion of bounded κ-variation. This concept differs from others due to the fact that it introduces a distortion function κ that measures intervals in the domain of the function and not in the range. In 1985, Cyphert and Kelingos [8] showed that a function u is of bounded κ-variation if it can be written as the difference of two κ-decreasing functions. In 1986, S. K. Kim and J. Kim [9] and Park [10], in 2010, introduced the notion of functions of κϕ-bounded variation on compact interval [a,b]⊂ℝ which is a combination of concepts of bounded κ-variation and bounded ϕ-variation in the sense of Schramm [11], and in 2011 Aziz et al. [12] showed that the space of bounded κ-variation satisfies Matkowski’s weak condition.Recently in [13] Castillo et al.  introduce the notion of bounded κ-variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded p-variation in the sense of Riesz and bounded κ-variation in the sense of Korenblum.The purpose of this paper is twofold. First, to introduce the concept of boundedκφ-variation in the sense of Riesz-Korenblum, which is a combination of the notions of bounded φ-variation in the sense of Riesz and bounded κ-variation in the sense of Korenblum. We prove some properties of this class of functions and its relation with the functions of bounded κ-variation and bounded φ-variation in the sense of Riesz. Second we prove that the space generated by this class of functions is a Banach space with a given norm and that the uniformly bounded composition operator satisfies Matkowski’s weak condition in this space. The Matkowski property has been studied by several authors (see [14–16]), and for Matkowski’s weak property, see also [3, 17–21]. In [22–24] Matkowski, Merentes, and others authors have been studying a weaker condition on the composition operator such as uniformly bounded and uniformly continuous composition operators.


## 2. Preliminaries

In this section we present some definitions and preliminary results related to the notion of functions of boundedκφ-variation in the sense of Riesz-Korenblum.Definition 1.
Letu:[a,b]→ℝ be a function. For each partition π:a=t0<t1<⋯<tn=b of the interval [a,b], we define
(1)V(u;[a,b])≔supπ∑i=1n‍|u(ti)-u(ti-1)|,
where the supremum is taken over all partitions π of the interval [a,b]. If  V(u;[a,b])<∞, we say that u has bounded variation. We denote by BV[a,b] the collection of all functions of bounded variation on [a,b].Now, we will give some well-known properties of the space of functionsBV[a,b]. (1)
If the functionu is monotone, then V(u;[a,b])=|u(b)-u(a)|.(2)
Ifu∈BV[a,b], then u is bounded on [a,b].(3)
A functionu has bounded variation in an interval [a,b] if and only if it can be decomposed as a difference of increasing functions.(4)
Every function of bounded variation has left- and right-hand limits at each point of its domain.(5)
Ifu∈Lip[a,b], then u∈BV[a,b] that is, Lip[a,b]⊂BV[a,b].(6)
BV[a,b] is a Banach space endowed with the norm
(2)∥u∥BV=|u(a)|+V(u;[a,b]),u∈BV[a,b].In 1937, Young (see [25]) introduced the definition of φ-function as follows.Definition 2.
A functionφ:[0,∞)→[0,∞) is said to be a φ-function if it satisfies the following properties.(a)
φ is continuous on [0,∞).(b)
φ(t)=0 if and only if t=0.(c)
φ is strictly increasing.(d)
limt→∞φ(t)=∞.Definition 3 (conditions∞1 and Δ2).
Letφ be a convex φ-function, then(a)
φ satisfies the condition ∞1 if limt→∞(φ(t)/t)=∞,(b)
φ satisfies the condition Δ2(∞) if there is C>0, x0>0 such that
(3)φ(2t)≤Cφ(t),t≥x0.The notion of bounded variation due to Jordan (Definition1) was generalized by Medvedev (see [6]) as follows.Definition 4.
Letφ be a φ-function and u:[a,b]→ℝ be a function. For each partition π:a=t0<t1<⋯<tn=b of the interval [a,b], we define
(4)VφR(u)=VφR(u;[a,b])≔supπ∑i=1n‍φ(|u(ti)-u(ti-1)||ti-ti-1|)|ti-ti-1|,
where the supremum is taken over all partitions π of the interval [a,b]. If  VφR(u;[a,b])<∞, we say that u has bounded φ-variation in the sense of Riesz. We denote by VφR[a,b] the class of all functions of bounded φ-variation in the sense of Riesz on [a,b].Now we give some properties of the class of functionsVφR[a,b]. (1)
Ifu∈Lip[a,b], then u∈VφR[a,b]; that is,  Lip[a,b]⊂VφR[a,b].(2)
Ifφ is convex then VφR[a,b]⊂BV[a,b] and if limt→∞(φ(t)/t)=γ<∞, then VφR[a,b] = BV[a,b].(3)
Ifφ satisfies the condition ∞1 and VφR(u;[a,b])<∞, then u is bounded on [a,b].(4)
VφR[a,b] is a symmetric convex set.(5)
Ifφ is convex, then κVφR[a,b] is a vector space if and only if φ satisfies the condition Δ2(∞).(6)
Ifφ is convex, then the set RVφ[a,b]={u:[a,b]→ℝ:∃λ>0,VφR(λu)<∞} is a Banach space endowed with the norm
(5)||u||φ=|u(a)|+inf{ε>0:Vφ(uε)≤1}.(7)
Letφ be a convex φ-function such that it satisfies the condition ∞1. A function u has bounded φ-variation in an interval [a,b] if and only if u is absolutely continuous in [a,b] and ∫ab‍φ(|u′(t)|)dt<∞. Also,
(6)VφR(u;[a,b])=∫ab‍φ(|u′(t)|)dt.Other generalization of the notion of bounded variation was introduced by Korenblum. Korenblum employed a functionκ:[0,1]→[0,1] called κ-function. This function κ can be viewed as a rescaling of lengths of subintervals of [a,b] such that the length of [a,b] is 1 if κ(1)=1.Definition 5.
A functionκ:[0,1]→[0,1] is said to be a κ-function if it satisfies the following properties:(a)
κ is continuous with κ(0)=0 and κ(1)=1,(b)
κ is concave (down), increasing, and(c)
limt→0+(κ(t)/t)=∞.The set of allκ-functions will be denoted by 𝒦. Note that, every κ-function κ is subadditive; that is,
(7)κ(t1+t2)≤κ(t1)+κ(t2),t1,t2∈[0,1].Then, for all partitionπ:a=t0<t1<⋯<tn=b of [a,b], we have
(8)1=κ(1)=κ(∑i=1n‍ti-ti-1b-a)≤∑i=1n‍κ(ti-ti-1b-a).Korenblum (see [7]) introduces the definition of bounded κ-variation as follows.Definition 6.
A real functionu on [a,b] is said to be of bounded κ-variation, if
(9)κV(u)=κV(u;[a,b])κV(u)≔supπ∑i=1n|u(ti)-u(ti-1)|∑i=1nκ((ti-ti-1)/(b-a))<∞,
where the supremum is taken over all partitions π of the interval [a,b]. We denote by κBV[a,b] the collection of all functions of bounded κ-variation on [a,b].Next, some properties of the spaceκBV[a,b] are exposed (see [8]). (1)
If the functionu is monotone, then κV(u;[a,b])=|u(b)-u(a)|.(2)
Ifu∈κBV[a,b], then u is bounded on [a,b].(3)
Ifu∈BV[a,b], then u∈κBV[a,b]; that is, BV[a,b]⊂κBV[a,b].(4)
A functionu has bounded κ-variation in an interval [a,b] if and only if it can be decomposed as a difference of κ-decreasing functions.(5)
Every function of boundedκ-variation has left- and right-hand limits at each point of its domain.(6)
κBV[a,b] is a Banach space endowed with the norm
(10)||u||κ=|u(a)|+κV(u;[a,b]),u∈κBV[a,b].


## 3. Main Results

In this section we present the principal results of this paper. Next, we introduce the definition of function of boundedκφ-variation in the sense of Riesz-Korenblum for the function u:[a,b]→ℝ.Definition 7.
Letφ be a φ-function, κ∈𝒦, and u:[a,b]→ℝ be a function. For each partition π:a=t0<t1<⋯<tn=b of the interval [a,b], we define
(11)κVφR(u)=κVφR(u;[a,b])≔supπ∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a)),
where the supremum is taken over all partitions π of the interval [a,b]. If κVφR(u;[a,b])<∞, we say that u has bounded κφ-variation in the sense of Riesz-Korenblum. We will denote by κVφR[a,b] the class of all functions of bounded κφ-variation in the sense of Riesz-Korenblum on [a,b].Remark 8.
Note that the classκVφR[a,b] is not empty since for an affine function u:[a,b]→ℝ is defined by u(t):=dt+e,t∈[a,b], where d,e are fixed real numbers. For a given partition π:a=t0<t1<⋯<tn=b of [a,b] we have
(12)∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=φ(|d|)(b-a)∑i=1nκ((ti-ti-1)/(b-a)).
Taking the supremum over all partitionsπ of the interval [a,b], the greater value of the right side of the above expression is obtain for the partition π:a=t0<t1=b and in this case we get
(13)φ(|d|)(b-a)κ((t1-t0)/(b-a))=φ(|d|)(b-a)κ(1)=φ(|d|)(b-a).
Therefore,(14)κVφR(u;[a,b])=φ(|d|)(b-a).In the following proposition, we prove two important properties of the spaceκRVφ[a,b].Proposition 9.
Letφ be a convex φ-function, then (a)
κVφR[a,b]⊂κBV[a,b].(b)
Iflimt→∞(φ(t)/t)=γ<∞, then κVφR[a,b]=κBV[a,b].Proof.
(a) Letu∈κVφR[a,b],π:a=t0<t1<⋯<tn=b, be a partition of the interval [a,b] and
(15)σ≔{i=1,…,n:|u(ti)-u(ti-1)||ti-ti-1|≤1}.
Sinceφ is a convex φ-function and φ(0)=0, we have
(16)φ(1)=φ(tt)≤1tφ(t),t≥1,
so t≤φ(t)/φ(1).
Hence, fori∉σ we get
(17)|u(ti)-u(ti-1)||ti-ti-1|≤φ(|u(ti)-u(ti-1)|/|ti-ti-1|)φ(1)
multiplied by |ti-ti-1| and applying the sum on both sides of the above inequality, we have
(18)∑i∉σn‍|u(ti)-u(ti-1)||ti-ti-1||ti-ti-1|≤1φ(1)∑i∉σn‍φ(|u(ti)-u(ti-1)||ti-ti-1|)|ti-ti-1|,
then
(19)∑i=1n|u(ti)-u(ti-1)|∑i=1nκ((ti-ti-1)/(b-a))=∑i=1n(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=∑i∈σ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))+∑i∉σ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤∑i∈σ|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))+1φ(1).∑i∉σφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤(b-a)∑i=1nκ((ti-ti-1)/(b-a))+1φ(1)κVφR(u)≤(b-a)+1φ(1)κVφR(u).
Then(20)∑i=1n|u(ti)-u(ti-1)|∑i=1nκ((ti-ti-1)/(b-a))≤(b-a)+1φ(1)κVφR(u).
Considering the supremum over all partitionsπ of the interval [a,b] in the above expression, we get
(21)κV(u)≤b-a+1φ(1)κVφR(u),
therefore κVφR[a,b]⊂κBV[a,b].
Now, we will show part (b). If(22)0<limt→∞φ(t)t=γ<∞,
then there exist x0>0 and c>0, such that
(23)φ(t)≤ct,t≥x0.
Let us consider the partitionπ:a=t0<t1<⋯<tn=b of the interval [a,b],u∈κBV[a,b], and
(24)αx0≔{i=1,…,n:|u(ti)-u(ti-1)||ti-ti-1|>x0},
then
(25)∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=∑i∈αx0φ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))+∑i∉αx0φ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤∑i∈αx0c|u(ti)-u(ti-1)|∑i=1nκ((ti-ti-1)/(b-a))+∑i∉αx0φ(x0)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a)),
thus,
(26)∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤c∑i∈αx0|u(ti)-u(ti-1)|∑i=1nκ((ti-ti-1)/(b-a))+φ(x0)∑i∉αx0|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤c·κV(u)+φ(x0)(b-a)∑i=1nκ((ti-ti-1)/(b-a))≤c·κV(u)+φ(x0)(b-a).
Then by considering the supremum over all partitionsπ of the interval [a,b] of the left side, we get
(27)κVφR(u)≤c·κV(u)+φ(x0)(b-a),
that is,
(28)κBV[a,b]⊂κVφR[a,b].
Therefore, from part (a) and (28) we have κVφR[a,b]=κBV[a,b].Proposition 10.
LetLip[a,b] be the Banach space of all Lipschitz functions u:[a,b]→ℝ. Then Lip[a,b]⊂κVφR[a,b].Proof.
Letπ:a=t0<t1<⋯<tn=b be a partition of the interval [a,b] and u∈Lip[a,b], then there exists c∈ℝ such that for any x,y∈[a,b] we have |u(x)-u(y)|≤c|x-y|. Hence,
(29)∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤∑i=1nφ(c|ti-ti-1|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=∑i=1nφ(c)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=φ(c)(b-a)∑i=1nκ((ti-ti-1)/(b-a))≤φ(c)(b-a),
considering the supremum over all partitions π of the interval [a,b], we get
(30)κVφR(u)≤φ(c)(b-a)<∞.
Therefore,Lip[a,b]⊂κVφR[a,b].The class of functions of a boundedκφ-variation has many interesting properties as the following proposition showes.Proposition 11.
Letφ be a φ-function, κ∈𝒦 and u:[a,b]→ℝ be a function, then(a)
The functionκVφR(·):κVφR[a,b]→ℝ is an even function, that is, κVφR(u)=κVφR(-u).(b)
κVφR(u)=0 if and only if u is constant.(c)
IfκVφR(u)<∞ then u is bounded on [a,b].(d)
φ is convex if and only if the function κVφR(·):κVφR[a,b]→[0,∞) defined by
(31)κVφR(u):=κVφR(u;[a,b]),u∈κVφR[a,b]is convex.Proof.
(a) From Definition4, we have
(32)κVφR(u)=supπ∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=supπ∑i=1nφ(|-u(ti)-(-u(ti-1))|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))=κVφR(-u).
(b) Ifu is constant, then κVφR(u)=0. Now if κVφR(u)=0, we get that
(33)0=κVφR(u)≥φ(|u(t)-u(a)|/|t-a|)|t-a|κ((t-a)/(b-a))+κ((b-t)/(b-a)),
for some t∈(a,b]. From Definitions 2 and 5, we have
(34)0≥φ(|u(t)-u(a)|/|t-a|)|t-a|κ((t-a)/(b-a))+κ((b-t)/(b-a))≥0,
hence
(35)φ(|u(t)-u(a)||t-a|)|t-a|=0,t∈(a,b].
Since φ(0)=0 if and only if t=0, we get
(36)|u(t)-u(a)||t-a|=0,
therefore, u(t)=u(a)t∈[a,b]. So, u is constant.
(c) Suppose thatu∈κVφR[a,b] is unbounded on [a,b]. Then, there exists a sequence {tn}n≥1, tn∈[a,b], for all n≥1, such that limn→∞|u(tn)|=∞. Let {tnk}nk≥1 be a subsequence of {tn}n≥1 such that {tnk}nk≥1 converges to point x∈[a,b]. Then, {u(tnk)}nk≥1 is a subsequence of {u(tn)}n≥1. So, limk→∞|u(tnk)|=∞.
Case  1. Suppose that x≠a. Since
(37)φ(|u(tnk)-u(a)|/|tnk-a|)|tnk-a|κ((tnk-a)/(b-a))+κ((b-tnk)/(b-a))≤κVφR(u)<∞,
for all nk≥1 and since φ is continuous, we have
(38)φ(limk→∞(|u(tnk)-u(a)|/|x-a|))|x-a|κ((x-a)/(b-a))+κ((b-x)/(b-a))=limk→∞φ(|u(tnk)-u(a)|/|tnk-a|)|tnk-a|κ((tnk-a)/(b-a))+κ((b-tnk)/(b-a))≤κVφR(u).
On the other hand, |u(tnk)-u(a)| tends to infinity as k→∞. Then, since φ(t)→∞ as t→∞, we get
(39)limk→∞φ(|u(tnk)-u(a)|/|x-a|)|x-a|κ((x-a)/(b-a))+κ((b-x)/(b-a))=∞,
then κVφR(u)=∞, which is a contradiction.
Case  2. Suppose that x=a. Then, since
(40)φ(|u(b)-u(tnk)|/|b-tnk|)|b-tnk|κ((tnk-a)/(b-a))+κ((b-tnk)/(b-a))≤κVφR(u)<∞,
for all nk≥1, and since φ is continuous, we have
(41)φ(limk→∞|u(b)-u(tnk)|/|b-x|)|b-x|κ((x-a)/(b-a))+κ((b-x)/(b-a))=limk→∞φ(|u(b)-u(tnk)|/|b-tnk|)|b-tnk|κ((tnk-a)/(b-a))+κ((b-tnk)/(b-a))≤κVφR(u).
Since φ(t)→∞ as t→∞ and limk→∞|u(b)-u(tnk)|=∞, we get
(42)limk→∞φ(|u(b)-u(tnk)|/|b-x|)|b-x|κ((x-a)/(b-a))+κ((b-x)/(b-a))=∞,
so κVφR=∞ which is a contradiction. In both cases a contradiction is reached. Hence u is bounded.
(d) Letu,v:[a,b]→ℝ and α,β∈[0,1] such that α+β=1. Suppose that φ is convex, then
(43)ακVφR(u)+βκVφR(v)=αsupπ∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))+βsupπ∑i=1nφ(|v(ti)-v(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≥supπ(∑i=1n[αφ(|u(ti)-u(ti-1)||ti-ti-1|)|ti-ti-1|+βφ(|v(ti)-v(ti-1)||ti-ti-1|)|ti-ti-1|])×(∑i=1nκ(ti-ti-1b-a))-1,
where the supremum is taken over all partitions π of the interval [a,b].
Sinceφ is convex, we have
(44)ακVφR(u)+βκVφR(v)≥supπ(∑i=1nφ(|α(u(ti)-u(ti-1))+β(v(ti)-v(ti-1))||ti-ti-1|)×|ti-ti-1||α(u(ti)-u(ti-1))+β(v(ti)-v(ti-1))||ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1=supπ(∑i=1nφ(|(αu+βv)(ti)-(αu+βv)(ti-1)||ti-ti-1|)×|ti-ti-1||(αu+βv)(ti)-(αu+βv)(ti-1)||ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1=κVφR(αu+βv).
So κVφR(αu+βv)≤ακVφR(u)+βκVφR(v) for all α,β∈[0,1] such that α+β=1, and therefore κVφR(·) is convex.
In order to prove the order direction, let us consider the following functions.u:[a,b]→ℝ defined by u(t):=xt and v:[a,b]→ℝ defined by v(t):=yt,t∈[a,b]. Therefore, we have
(45)κVφR(u)=φ(x)(b-a),κVφR(v)=φ(y)(b-a),κVφR(αu+βv)=φ(αx+βy)(b-a).
Since κVφR(·) is convex, we get
(46)φ(αx+βy)(b-a)=κVφR(αu+βv)φ(αx+βy)(b-a)≤ακVφR(u)+βκVφR(v)φ(αx+βy)(b-a)=αφ(x)(b-a)+βφ(y)(b-a),
which implies that
(47)φ(αx+βy)≤αφ(x)+βφ(y).
Therefore, φ is convex, and the proof of the theorem is completed.Remark 12.
The part (c) of Proposition11 is a consequence of the part (a) of Proposition 9 if the φ-function is convex.Proposition 13.
Letφ be a φ-function, κ∈𝒦, and u,v:[a,b]→ℝ be functions, then
(48)κVφR(αu+βv)≤κVφR(u)+κVφR(v)α,β∈[0,1],α+β=1.Proof.
Letα,β∈[0,1] such that α+β=1 and x,y∈[0,∞). Since φ is nondecreasing and nonnegative and αx+βy is one of the segment joining point x with y, then we have
(49)φ(αx+βy)≤max{φ(x),φ(y)}≤φ(x)+φ(y).
From the inequality above, we deduce that
(50)κVφR(u)+κVφR(v)=supπ∑i=1nφ(|u(ti)-u(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))+supπ∑i=1nφ(|v(ti)-v(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≥supπ(∑i=1nφ(|α(u(ti)-u(ti-1))+β(v(ti)-v(ti-1))||ti-ti-1|)×|ti-ti-1||α(u(ti)-u(ti-1))+β(v(ti)-v(ti-1))||ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1,
thus
(51)κVφR(u)+κVφR(v)=supπ(∑i=1nφ(|(αu+βv)(ti)-(αu+βv)(ti-1)||ti-ti-1|)×|ti-ti-1||(αu+βv)(ti)-(αu+βv)(ti-1)||ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1=κVφR(αu+βv).
Therefore, κVφR(αu+βv)≤κVφR(u)+κVφR(v) for all α,β∈[0,1] such that α+β=1.Remark 14.
From Propositions11 and 13, we have κVφR[a,b] to be convex and symmetric.The following lemma allows us to give a characterization of the spaceκRVφ[a,b].Lemma 15.
LetX be a vector space and A⊂X a nonempty convex and symmetric set. Then one has the following.(a)
0∈A.(b)
The vector space associated byA is iqual to:
(52)〈A〉={x∈X:∃λ>0,λx∈A}=⋃λ>0‍λA.Proof.
(a) For allx∈A, we have that -x∈A0=(1/2)x+(1/2)(-x)∈A.
(b) By definition,A⊂⋃λ>0‍λA⊂〈A〉. In order to show the other inclusion, we have to prove that ⋃λ>0‍λA is a vector space. Indeed, if α∈ℝ, we get
(53)α(λx)∈⋃λ>0‍λA,α>0,α(λx)=0∈⋃λ>0‍λA,α=0,α(λx)=(-αλ)(-x)∈⋃λ>0‍λA,α<0.
Also, ifx,y∈A and α>0, β>0, then
(54)αx+βy=(α+β)(αα+βx+βα+βy)∈(α+β)A.
Therefore,⋃λ>0‍λA is a vector space.
Let us now prove that⋃λ>0‍λA={x∈X:∃λ>0,λx∈A}. Indeed, if z=λx,x∈A,λ>0, then 1/z=x∈A. Conversely, if λx∈A, for some λ>0, then x=(1/λ)(λx)∈(1/λ)A.As a consequence of Lemma15 and since κVφR is a convex and symmetric set, we have the following corollary.Corollary 16.
Letφ be a φ-function and κ∈𝒦. Then, the vector space generated by the class κVφR[a,b] is equal to the following
(55)κRVφ[a,b]={u:[a,b]→ℝ:∃λ>0,κVφR(λu)<∞}.Theorem 17.
Letφ be a φ-function and κ∈𝒦, then
(56)RVφ[a,b]⊂κRVφ[a,b]⊂κBV[a,b].Proof.
First we prove thatRVφ[a,b]⊂κRVφ[a,b]. Let u∈RVφ[a,b], then there exists λ>0 such that VφR(λu)<∞ and π:a=t0<t1<⋯<tn=b is a partition of the interval [a,b], then from inequality (8) we have
(57)∑i=1n‍φ(|λu(ti)-λu(ti-1)||ti-ti-1|)|ti-ti-1|≤VφR(λu)=VφR(λu)κ(∑i=1n‍ti-ti-1b-a)≤VφR(λu)∑i=1n‍κ(ti-ti-1b-a).
Thus,(58)∑i=1nφ(|λu(ti)-λu(ti-1)|/|ti-ti-1|)|ti-ti-1|∑i=1nκ((ti-ti-1)/(b-a))≤VφR(λu).
Then, considering the supremum of the left side, we get(59)κVφR(λu)≤VφR(λu),
therefore, u∈κRVφ[a,b] and RVφ[a,b]⊂κRVφ[a,b]. On the other hand, by part (a) of Proposition 9 we get that κRVφ[a,b]⊂κBV[a,b].Theorem 18.
Letφ be a convex φ-function, then
(60)Λ={u∈κRVφ[a,b]:κVφR(u)≤1}
is a symmetric convex absorbent subset of κRVφ[a,b].Proof.
First we show the convexity. Letα,β∈[0,1] such that α+β=1 and u,v∈Λ. Then, by Proposition 11 we get
(61)κVφR(αu+βv)≤ακVφR(u)+βκVφR(v)≤α·1+β·1=1,
thus αu+βv∈Λ.
Now letu∈Λ and 0<|λ|≤1. If 0<λ≤1 by Proposition 11, we have
(62)κVφR(λu)≤λκVφR(u)≤λ·1≤1.
For the case-1≤λ<0, by the symmetric and convexity of the functional κVφR(·) given in Proposition 11 we get that
(63)κVφR(λu)=κVφR(-λ(-u))≤-λκVφR(-u)=|λ|κVφR(u).
Hence, we have shown thatΛ is balance. Now we will show that Λ is absorbent. Let u∈κRVφ[a,b] then there exist α>0 such that κVφR(αu)<∞. If κVφR(αu)≤1 then αu∈Λ. On the other hand, if κVφR(αu)>1 we have
(64)κVφR(αuκVφR(αu))≤1κVφR(αu)κVφR(αu)=1,
in this case, (α/κVφR(αu))u∈Λ. Hence, Λ is absorbent.Remark 19.
As a consequence of Theorem18, the Minkowski functional associated to the set Λ defines a seminorm on κRVφ[a,b] and is defined by
(65)μΛ(u)=inf{λ>0:κVφR(uλ)≤1}.Theorem 20.
Letφ be a convex φ-function. Then, (κRVφ[a,b],||·||κφ), where the functional ||·||κφ:κRVφ[a,b]→ℝ, defined by
(66)||u||κφ=|u(a)|+μΛ(u),u∈κRVφ[a,b],
is a normed space.Proof.
Letu,v∈κRVφ[a,b], α∈ℝ. Then, we have the following.(a)
||u||κφ≥0 since |u(a)|≥0 and μΛ(u)≥0.(b)
(67)||αu||κφ=|αu(a)|+|α|μΛ(u)||αu||κφ=|α||u(a)|+|α|μΛ(u)||αu||κφ=|α|(|u(a)|+μΛ(u))||αu||κφ=|α|||u||κφ.Therefore,||αu||κφ=|α|||u||κφ.(c)
(68)||u+v||κφ=|(u+v)(a)|+μΛ(u+v)||u+v||κφ=|(u)(a)+v(a)|+μΛ(u+v)||u+v||κφ≤|u(a)|+|v(a)|+μΛ(u)+μΛ(v)||u+v||κφ=||u||κφ+||v||κφ.Thus,||u+v||κφ≤||u||κφ+||v||κφ.(d)
Let us now prove that||u||κφ=0 if and only if u=0. Indeed, suppose that ||u||κφ=0, that is,
(69)|u(a)|+μΛ(u)=0,then|u(a)|=0 and 0=μΛ(u)=inf{λ>0:κVφR(u/λ)≤1}. This implies that for each positive integer n, there exists λn such that 1/λn>λn>0 and κVφR(u/λn)≤1. As the function κVφR(·) is convex, we have κVφR(u)≤λn. Taking the limit as n→∞, we get κVφR(u)=0. Moreover, by part (b) of Proposition 11 we have that u is constant; that is, u(t)=u(a),t∈[a,b], therefore u=0.Now, suppose thatu=0. Then |u(a)|=0 and μΛ(u)=0. Hence ||u||κφ=0. Therefore (κRVφ[a,b],||·||κφ) is a normed space.Lemma 21.
Letu∈κRVφR[a,b]. If λ>0, then μΛ(u)≤λ if and only if κVφR(u/λ;[a,b])≤1.Proof.
Case  1. Let u∈κRVφR[a,b] and μΛ(u)<λ. Then, by the infimum property, there is 0<k<λ such that κVφR(u/k)≤1. Hence, by the convexity of κVφR(·),
(70)κVφR(uλ)=κVφR(ukkλ)≤kλκVφR(uk)≤kλ≤1.
Case  2. Let u∈κRVφR[a,b] and μΛ(u)=λ. Then, by the infimum property, there exists a sequence {λn}n∈ℕ such that
(71)λn>λ,κVφR(uk)≤1,n∈ℕ,limn→∞λn=λ.
Sinceu/λn pointwise converges to u/λ on [a,b] as n→∞, by the lower semicontinuity of κVφR(·), we obtain that
(72)κVφR(uλ)≤limn→∞κVφR(uλn)≤1.
As, by the definition of the infimum, the converse is obvious, this completes the proof.In the next theorem, we prove that(κRVφ[a,b],||·||κφ) is a Banach space.Theorem 22.
Letφ be a convex φ-function such that φ satisfies the condition ∞1, then the space (κRVφ[a,b],||·||κφ) is a Banach space.Proof.
Let{un}n∈ℕ be a Cauchy sequence in (κRVφ[a,b],||·||κφ), then given that 0<ε<1, there is N∈ℕ, such that for n,m≥N we have
(73)||un-um||κφ<ε,n,m≥N,
that is,
(74)|(un-um)(a)|+μΛ(un-um)<ε,n,m≥N.
Then,(75)|un(a)-um(a)|<ε,μΛ(un-um)<εn,m≥N,
by Lemma 21 and the last inequality, we have
(76)κVφR(un-umε;[a,b])<1.
Then, for a partitionπ:a≤x<y≤b we get that
(77)φ(|(un-um)(y)-(un-um)(x)|ε|y-x|)|y-x|×(κ(x-ab-a)+κ(y-xb-a)+κ(b-yb-a))-1≤κVφR(un-umε;[a,b])<1.
Hence, we have
(78)φ(|(un-um)(y)-(un-um)(x)|ε|y-x|)|y-x|<κ(x-ab-a)+κ(y-xb-a)+κ(b-yb-a)≤3·κ(1)=3,
therefore
(79)|(un-um)(y)-(un-um)(x)|<φ-1(3|y-x|)|y-x|ε.
Sinceφ satisfies the condition ∞1 and φ-1 is continuous, we obtain that
(80)φ-1(3|y-x|)|y-x|,x,y∈[a,b],x≠y,
is bounded. Let M>0 be a upper bound of (80), then
(81)|(un-um)(y)-(un-um)(x)|<Mε.
As a consequence, the sequence{un}n∈ℕ is a uniformly Cauchy sequence, on the interval [a,b], and by the completeness of ℝlimn→∞un(t) exist for all t∈[a,b].
We defined on[a,b] the function u(t):=limn→∞un(t). We claim that u∈κRVp[a,b]. In fact, let π:a=t1<t2<⋯<tn=b be a partition of the interval [a,b]. Then, for n≥N and ε>0 one has
(82)κVφR(un-uε)=supπ(∑i=1nφ(|(un-u)(ti)-(un-u)(ti-1)|ε|ti-ti-1|)×|ti-ti-1||(un-u)(ti)-(un-u)(ti-1)|ε|ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1=supπ(∑i=1nφ(|(un-limm→∞um)(ti)-(un-limm→∞um)(ti-1)|(ε|ti-ti-1|)-1)×|ti-ti-1||(un-u)(ti)-(un-u)(ti-1)|ε|ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1=supπlimm→∞(∑i=1nφ(|(un-um)(ti)-(un-um)(ti-1)|ε|ti-ti-1|)×|ti-ti-1||(un-u)(ti)-(un-u)(ti-1)|ε|ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1.
Since forn,m≥N,
(83)1≥κVφR(un-umε)1≥(∑i=1n‍φ(|(un-um)(ti)-(un-um)(ti-1)|ε|ti-ti-1|)|ti-ti-1|)1×(∑i=1n‍κ(ti-ti-1b-a))-1,
then
(84)limm→∞(∑i=1nφ(|(un-um)(ti)-(un-um)(ti-1)|ε|ti-ti-1|)×|ti-ti-1||(un-um)(ti)-(un-um)(ti-1)|ε|ti-ti-1|)(∑i=1nκ(ti-ti-1b-a))-1≤1,n≥N.
Thus,(85)κVφR(un-uε)≤1.
Therefore,(un-u)/ε∈κRVφ[a,b] for all n≥N. As un∈κRVφ[a,b] for all n≥N, and κRVφ[a,b] is a vector space, then u∈κRVφ[a,b].
Finally, let us prove that{un}n≥1 converge in norm to u. Let ε>0 be arbitrary, then
(86)||un-u||κφ=||un-limm→∞um||κφ||un-u||κφ=limm→∞||un-um||κφ<ε,∀n≥N.
Therefore, the sequence{un}n≥1 converges to u in the norm ||·||κφ and thus κRVφ[a,b] is a Banach space.


### 3.1. Uniformly Bounded Composition Operator

In this section, we present the other main result of this paper; namely, we show that any uniformly bounded composition operator that maps the space theκRVφ[a,b] into itself necessarily satisfies the so-called Matkowski’s weak condition.Definition 23.
For a given functionh:[a,b]×ℝ→ℝ, the composition operator H:ℝ[a,b]→ℝ[a,b] generated by h is defined by
(87)Hu(t)=h(t,u(t)),u∈ℝ[a,b],t∈[a,b].
Here ℝ[a,b] denotes the family of all functions u:[a,b]→ℝ.Remark 24.
SinceκRVφ[a,b]⊂κBV[a,b], then every function of bounded κφ-variation in the sense of Riesz-Korenblum has left- and right-hand limits at each point of its domain (see [8]).Now, we will give the definition of left regularization of a function.Definition 25.
Letu∈κRVφ[a,b], one defined its left regularization u-:(a,b]→ℝ of mapping u by the following:
(88)u-(t)≔{lims→t-u(s),t∈(a,b],u(a),t=a.We will denote byκRVφ-[a,b] the subset in κRVφ[a,b] which consists of those functions that are left continuous on (a,b].Lemma 26.
Ifu∈κRVφ[a,b], then u-∈κRVφ-[a,b].Thus, if a functionu has bounded κφ-variation in the sense of Riesz-Korenblum, then its left regularization is a left continuous function.Definition 27 (see [24, Definition  1]).
Let𝕏 and 𝕐 be two metric (or normed) spaces. One says that a mapping H:𝕏→𝕐 is uniformly bounded if, for any t>0, there exists a nonnegative real number γ(t) such that for any nonempty set B⊂𝕏 we have
(89)diamB≤t⟹diamH(B)≤γ(t).Theorem 28.
Letφ be a φ-function, κ∈𝒦, and h:[a,b]×ℝ→ℝ be a function continuous with respect to the second variable. Suppose that the composition operator H generated by h maps κRVφ[a,b] into itself and satisfies the following inequality:
(90)||H(u)-H(v)||κφ≤γ(||u-v||κφ),u,v∈κRVφ[a,b]
for some function γ:[0,∞)→[0,∞). Then, there exist functions α,β∈κRVφ-[a,b] such that
(91)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ,
where h-(·,x):(a,b]→ℝ is the left regularization of h(·,x) for all x∈ℝ.Proof.
By hypothesis, forx∈ℝ fixed the constant function u(t)=x,t∈[a,b] belongs to κRVφ[a,b]. Since H maps κRVφ[a,b] into itself, we have that the function (Hu)(t)=h(t,u(t))=h(t,x)∈κRVφ[a,b]. By Lemma 26, the left regularization h-(·,x)∈κRVφ- for every x∈ℝ.
From inequality (90) and the definition of the norm ||·||κφ we obtain for u1,u2∈κRVφ[a,b] that
(92)μΛ(H(u1)-H(u2))≤||H(u1)-H(u2)||κφμΛ(H(u1)-H(u2))≤γ(||u1-u2||κφ).
From the inequality (92) and Lemma 21, if γ(||u1-u2||κφ)>0, then
(93)κVφR(H(u1)-H(u2)γ(||u1-u2||κφ))≤1.
On the other hand, ifa≤r<s≤b, then from the definitions of the operator H, the functional κVφR, and inequality (93), we have
(94)φ(|H(u1)(s)-H(u2)(s)-H(u1)(r)+H(u2)(r)|γ(||u1-u2||κφ)|s-r|)×|s-r|(κ(r-ab-a)+κ(s-rb-a)+κ(b-sb-a))-1≤κVφR(H(u1)-H(u2)γ(||u1-u2||κφ))≤1,
whence
(95)φ(|h(s,u1(s))-h(s,u2(s))-h(r,u1(r))+h(r,u2(r))|γ(||u1-u2||κφ)|s-r|)×|s-r|(κ(r-ab-a)+κ(s-rb-a)+κ(b-sb-a))-1≤1.
Letx1,x2∈ℝ,x1<x2 and consider the functions u1,u2:[a,b]→ℝ defined by
(96)uk(t):=x1-x22(s-r)(t-r)+xk+x22,t∈[a,b],k=1,2
then uk∈κRVφ[a,b],k=1,2. From (96), we have
(97)u1-u2=x1-x22,
therefore,
(98)||u1-u2||κφ=|x1-x22|.
Moreover,(99)u1(s)=x1,u1(r)=x1+x22=u2(s),u2(r)=x2.
From (95) we get
(100)φ(|h(s,x1)-h(s,x1+x22)-h(r,x1+x22)+h(r,x2)|×(γ(|(x1-x2)/2|)|s-r|)-1x1+x22)≤3|s-r|,
then
(101)|h(s,x1)-h(s,x1+x22)-h(r,x1+x22)+h(r,x2)|≤φ-1(3|s-r|)γ(|x1-x22|)|s-r|.
Sinceφ satisfies the condition ∞1, we have
(102)limt→0φ-1(1t)t=0,
then from the definition of h- and letting r tend to s in (101), we obtain
(103)|h-(s,x1)-2h-(s,x1+x22)+h-(s,x2)|≤0,
therefore
(104)h-(s,x1+x22)=h-(s,x1)+h-(s,x2)2,s∈(a,b],x1,x2∈ℝ
which proves that for every fixed s∈(a,b] the function h-(s,·) satisfies the Jensen functional equation in ℝ (see [26, page 315]). The continuity of h with respect to the second variable implies that for every t∈[a,b] there exist α(t),β(t)∈ℝ such that
(105)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.
Sinceh-(·,x)∈κRVφ-[a,b] for all x∈ℝ in particular for x=0,
(106)β(t):=h-(t,0)t∈[a,b],
therefore β∈κRVφ-[a,b].
On the other hand, consideringx=1 we obtain that
(107)α(t)=h-(t,x)-β(t),t∈[a,b],
as κRVφ[a,b] is a vector space which implies that α∈κRVφ-[a,b].Theorem 29.
Letφ be a function and κ∈𝒦. Suppose that a function h:[a,b]×ℝ→ℝ is continuous with respect to the second variable. If the composition operator H generated by h maps the space κRVφ[a,b] into itself and is uniformly bounded, then there exist the functions α,β∈κRVφ-[a,b], such that
(108)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.Proof.
Take anyt≥0 and u,v∈κRVφ[a,b] such that
(109)||u-v||κφ≤t.
Sincediam{u,v}≤t, by the uniform boundedness of H, we have
(110)diamH({u,v})≤γ(t),
that is,
(111)||H(u)-H(v)||κφ=diamH({u,v})≤γ(||u-v||κφ),
therefore, by the Theorem 28 we have
(112)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.Remark 30.
Observe that similar results hold for the right regularizationh+ of h defined by
(113)h+(t,x):=lims→t+h(s,x)t∈[a,b).


## 3.1. Uniformly Bounded Composition Operator

In this section, we present the other main result of this paper; namely, we show that any uniformly bounded composition operator that maps the space theκRVφ[a,b] into itself necessarily satisfies the so-called Matkowski’s weak condition.Definition 23.
For a given functionh:[a,b]×ℝ→ℝ, the composition operator H:ℝ[a,b]→ℝ[a,b] generated by h is defined by
(87)Hu(t)=h(t,u(t)),u∈ℝ[a,b],t∈[a,b].
Here ℝ[a,b] denotes the family of all functions u:[a,b]→ℝ.Remark 24.
SinceκRVφ[a,b]⊂κBV[a,b], then every function of bounded κφ-variation in the sense of Riesz-Korenblum has left- and right-hand limits at each point of its domain (see [8]).Now, we will give the definition of left regularization of a function.Definition 25.
Letu∈κRVφ[a,b], one defined its left regularization u-:(a,b]→ℝ of mapping u by the following:
(88)u-(t)≔{lims→t-u(s),t∈(a,b],u(a),t=a.We will denote byκRVφ-[a,b] the subset in κRVφ[a,b] which consists of those functions that are left continuous on (a,b].Lemma 26.
Ifu∈κRVφ[a,b], then u-∈κRVφ-[a,b].Thus, if a functionu has bounded κφ-variation in the sense of Riesz-Korenblum, then its left regularization is a left continuous function.Definition 27 (see [24, Definition  1]).
Let𝕏 and 𝕐 be two metric (or normed) spaces. One says that a mapping H:𝕏→𝕐 is uniformly bounded if, for any t>0, there exists a nonnegative real number γ(t) such that for any nonempty set B⊂𝕏 we have
(89)diamB≤t⟹diamH(B)≤γ(t).Theorem 28.
Letφ be a φ-function, κ∈𝒦, and h:[a,b]×ℝ→ℝ be a function continuous with respect to the second variable. Suppose that the composition operator H generated by h maps κRVφ[a,b] into itself and satisfies the following inequality:
(90)||H(u)-H(v)||κφ≤γ(||u-v||κφ),u,v∈κRVφ[a,b]
for some function γ:[0,∞)→[0,∞). Then, there exist functions α,β∈κRVφ-[a,b] such that
(91)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ,
where h-(·,x):(a,b]→ℝ is the left regularization of h(·,x) for all x∈ℝ.Proof.
By hypothesis, forx∈ℝ fixed the constant function u(t)=x,t∈[a,b] belongs to κRVφ[a,b]. Since H maps κRVφ[a,b] into itself, we have that the function (Hu)(t)=h(t,u(t))=h(t,x)∈κRVφ[a,b]. By Lemma 26, the left regularization h-(·,x)∈κRVφ- for every x∈ℝ.
From inequality (90) and the definition of the norm ||·||κφ we obtain for u1,u2∈κRVφ[a,b] that
(92)μΛ(H(u1)-H(u2))≤||H(u1)-H(u2)||κφμΛ(H(u1)-H(u2))≤γ(||u1-u2||κφ).
From the inequality (92) and Lemma 21, if γ(||u1-u2||κφ)>0, then
(93)κVφR(H(u1)-H(u2)γ(||u1-u2||κφ))≤1.
On the other hand, ifa≤r<s≤b, then from the definitions of the operator H, the functional κVφR, and inequality (93), we have
(94)φ(|H(u1)(s)-H(u2)(s)-H(u1)(r)+H(u2)(r)|γ(||u1-u2||κφ)|s-r|)×|s-r|(κ(r-ab-a)+κ(s-rb-a)+κ(b-sb-a))-1≤κVφR(H(u1)-H(u2)γ(||u1-u2||κφ))≤1,
whence
(95)φ(|h(s,u1(s))-h(s,u2(s))-h(r,u1(r))+h(r,u2(r))|γ(||u1-u2||κφ)|s-r|)×|s-r|(κ(r-ab-a)+κ(s-rb-a)+κ(b-sb-a))-1≤1.
Letx1,x2∈ℝ,x1<x2 and consider the functions u1,u2:[a,b]→ℝ defined by
(96)uk(t):=x1-x22(s-r)(t-r)+xk+x22,t∈[a,b],k=1,2
then uk∈κRVφ[a,b],k=1,2. From (96), we have
(97)u1-u2=x1-x22,
therefore,
(98)||u1-u2||κφ=|x1-x22|.
Moreover,(99)u1(s)=x1,u1(r)=x1+x22=u2(s),u2(r)=x2.
From (95) we get
(100)φ(|h(s,x1)-h(s,x1+x22)-h(r,x1+x22)+h(r,x2)|×(γ(|(x1-x2)/2|)|s-r|)-1x1+x22)≤3|s-r|,
then
(101)|h(s,x1)-h(s,x1+x22)-h(r,x1+x22)+h(r,x2)|≤φ-1(3|s-r|)γ(|x1-x22|)|s-r|.
Sinceφ satisfies the condition ∞1, we have
(102)limt→0φ-1(1t)t=0,
then from the definition of h- and letting r tend to s in (101), we obtain
(103)|h-(s,x1)-2h-(s,x1+x22)+h-(s,x2)|≤0,
therefore
(104)h-(s,x1+x22)=h-(s,x1)+h-(s,x2)2,s∈(a,b],x1,x2∈ℝ
which proves that for every fixed s∈(a,b] the function h-(s,·) satisfies the Jensen functional equation in ℝ (see [26, page 315]). The continuity of h with respect to the second variable implies that for every t∈[a,b] there exist α(t),β(t)∈ℝ such that
(105)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.
Sinceh-(·,x)∈κRVφ-[a,b] for all x∈ℝ in particular for x=0,
(106)β(t):=h-(t,0)t∈[a,b],
therefore β∈κRVφ-[a,b].
On the other hand, consideringx=1 we obtain that
(107)α(t)=h-(t,x)-β(t),t∈[a,b],
as κRVφ[a,b] is a vector space which implies that α∈κRVφ-[a,b].Theorem 29.
Letφ be a function and κ∈𝒦. Suppose that a function h:[a,b]×ℝ→ℝ is continuous with respect to the second variable. If the composition operator H generated by h maps the space κRVφ[a,b] into itself and is uniformly bounded, then there exist the functions α,β∈κRVφ-[a,b], such that
(108)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.Proof.
Take anyt≥0 and u,v∈κRVφ[a,b] such that
(109)||u-v||κφ≤t.
Sincediam{u,v}≤t, by the uniform boundedness of H, we have
(110)diamH({u,v})≤γ(t),
that is,
(111)||H(u)-H(v)||κφ=diamH({u,v})≤γ(||u-v||κφ),
therefore, by the Theorem 28 we have
(112)h-(t,x)=α(t)x+β(t),t∈[a,b],x∈ℝ.Remark 30.
Observe that similar results hold for the right regularizationh+ of h defined by
(113)h+(t,x):=lims→t+h(s,x)t∈[a,b).



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*Source: 718507-2013-03-31.xml*