# Locally Defined Operators in the Space of Functions of Bounded Riesz-Variation

Authors: W. Aziz; J. A. Guerrero; K. Maldonado; N. Merentes
Journal: Journal of Mathematics
 (2015)

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## Abstract

We study the locally defined operator on the spaces of bounded Rieszp-variation functions and we prove that those operators are the Nemytskii operator.


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## Body

## 1. Introduction

We have an closed intervalI of the real line and let X=X(I),Y=Y(I) be function spaces f:I→R. An operator K:X→Y is called alocally defined operator, or (X,Y)-local operator, briefly, alocal operator [1], if for every open interval J⊂R and for all functions f,g∈X, the implication(1)fJ∩I=gJ∩I⟹KfJ∩I=KgJ∩Iis true, where f|J∩I denotes the restriction of f to J∩I.There is a vast literature on the problem treated here, mainly compiled of definitions of locally defined operators involving a measure space (cf., e.g., [2–5]). Also we proved that, in general, K is a composition (or Nemytskii) operator of the form K(f)(x)=h(x,f(x)) for a two-variable function h. Assuming additionally that K is continuous in measure, the generating function h can be replaced by a function satisfying the Caratheodory conditions (cf. [6]). The present paper concerns topological aspects of locally defined operators (cf. [1, 7–10]). For more knowledge on theory of the composition operators, see Appell and Zabrejko [11]. In [7] it was done is the case when X=Cn(I) and Y=C(I) or Y=C1(I). Subsequently, this result has been extended by several authors: [8, 9, 12] (for spaces of Whitney differentiable functions), [10, 13] (for space of Hölder functions), [14] (for continuous and monotone functions), and [1] (for functions of bounded φ-variation in the sense Wiener). In the present paper we are interested in such operators in the context of bounded Riesz-variation functions. In particular, we show that if the operator K maps the space RVp(I,R) into itself and is locally defined, then K is a Nemytskii composition operator.


## 2. Notation and Preliminaries

In this section we present some necessary notations and definitions and recall some knowledge concerning the bounded Riesz-variation.In the sequel,N, N0, and R denote, respectively, the set of positive integers, nonnegative integers, and the set of real numbers.LetI=[a,b]⊂R; π={ti}i=0m be partition of I, defined by π:a=t0<t1<⋯<tm=b. As usually, RI denote the family of all functions f:I→R.Given1≤p<∞,f∈RI and a partition π of I, we define(2)vpRf=vpRf,I≔supπ∑i=1mfti-fti-1pti-ti-1p-1,where the supremum is taken over all partitions π of I. vpR(f) is the classical p-variation of f in the sense of Riesz [15] in I. A function f is said to be of bounded p-variation if vpR(f,I)<∞. By RVp(I) we denote the Banach space of all functions f∈RI of bounded p-variation equipped with the norm(3)fp≔fa+vpRf1/p,f∈RVpI.Lemma 1.
LetI=[a,b]⊂R be an interval and let (x0,y0)∈I×R, x0<sup(I) be fixed. Then for every sequence (xk,yk)∈I×R satisfying the condition(4)limk→∞xk,yk=x0,y0,xk+1<xk;yk+1<yk,k∈Nthere exists a function ψ∈RVp(I) such that, for all k∈N0,(5)ψxk=yk.Proof.
Take an arbitrary sequence(xk,yk)∈I×R satisfying (4) and define a sequence of functions ψk:I→R, k∈N, by(6)ψkx≔y0,forx∈a,x0;yk-y0xk-x0x-x0+y0,forx∈x0,xk;yi-yi-1xi-xi-1x-xi+yi,forx∈xi,xi-1,i∈2,…,k;y1,forx∈x1,b.Let us observe that(7)ψkx0=y0,ψkxk=ψk+lxk=yk,k,l∈N,and for every x∈I∖{xk:k∈N0} there exist k0∈N such that(8)ψkx=ψk0x,k≥k0,k∈N.Put(9)ψx=limk→∞ψkx,x∈I.
From (7) and (8), the function ψ is well defined. Moreover, ψ is nondecreasing and(10)ψxk=yk,∀k∈N0,and by (9), for each ϵ>0, we obtain(11)ψkx-ψx<ϵ,∀x∈I,so ψk-ψ∞≤ϵ. Thus the sequence (ψk)k∈N tends uniformly to ψ.
Now asψk∈RVp(I) for all k∈N and ψk tends uniformly to ψ, then(12)ψxi-ψxi-1pxi-xi-1p-1=limk→∞ψkxi-ψkxi-1pxi-xi-1p-1≤limk→∞vpRψk,I<∞;thus vpR(ψ,I)<∞ and therefore ψ∈RVp(I).Similarly, we can get the following.Remark 2.
If(x0,y0)∈I×R, where x0>inf⁡(I) and (xk,yk)∈I×R is a sequence satisfying the condition (13)limk→∞xk,yk=x0,y0,xk<xk+1;yk≤yk+1,k∈N,then there exists a function ψ∈RVp(I) such that, for all k∈N0,(14)ψxk=yk.


## 3. Locally Defined Operators

Now we can introduce the definition of the local defined operators of typeK:RVp(I)→C(I).Definition 3 (see [1]).
An operatorK:RVp(I)→C(I) is said to be locally defined, if, for every two functions f,g∈RVp(I) and for every open interval J⊂R,(15)fJ∩I=gJ∩I⟹KfJ∩I=KgJ∩I.Theorem 4.
Let1<p<∞. If a locally defined operator K maps RVp(I) into C(I), then there exists a unique function h:I×R→R such that, for all f∈RVp(I),(16)Kft=ht,ft,t∈I.Proof.
We begin by showing that, for everyf,g∈RVp(I) and, for every x0∈int⁡(I), the condition(17)fx0=gx0implies that(18)Kfx0=Kgx0.To this end choose arbitrary x0∈int⁡(I) and take an arbitrary pair of functions f,g∈RVp(I) which fulfil (17) (i.e., f(x0)=g(x0)). The function φ:I→R, defined by(19)φx=fx,forx∈a,x0;gx,forx∈x0,b,belongs to RVp(I). Indeed, define the functions f1,g1:I→R by(20)f1x=fx-fx0,forx∈a,x0;0,forx∈x0,b,g1x=0,forx∈a,x0;gx-gx0,forx∈x0,b.
Sincef,g∈RVp(I), vpR(f)<∞ and vpR(g)<∞. Let π={xi}i=0m be a partition of I such that xl-1≤x0<xl for some 1≤l≤m. Then(21)∑i=1mf1xi-f1xi-1pxi-xi-1p-1≤vpRf.Hence vpR(f1)<∞. By a similar reasoning, we have vpR(g1)<∞. Finally f1+g1∈RVp(I), as RVp(I) is a linear space. Thus(22)vpRf1+g1<∞.Since, for all x,x′∈I(23)f1+g1x-f1+g1x′=ψx-ψx′,the condition (22) implies that φ∈RVp(I).
As(24)f-∞,x0∩I=φ-∞,x0∩I,gx0,∞∩I=ψx0,∞∩I,by definition of a local operator, we get(25)Kf-∞,x0∩I=Kφ-∞,x0∩I,Kgx0,∞∩I=Kφx0,∞∩I.Therefore, by the continuity of K(f), K(g) and K(φ) at x0, we obtain(26)Kfx0=Kφx0=Kgx0.
Suppose now thatx0 is the left endpoint of the interval I (i.e., x0=a). There exists a sequence (xk,yk)∈I×R such that x0<xk+1<xk, y0≤yk+1<yk,k∈N, and by the continuity of f and g at x0(27)limk→∞xk,yk=x0,y0.By Lemma 1 there exists a function ψ∈RVp(I) such that ψ(xk)=yk for all k∈N0.
There is no loss of generality in supposing thatf(x0)=g(x0)=y0, ψ(x2k-1)=y2k-1=g(x2k-1) and ψ(x2k)=y2k=f(x2k), k∈N.
According to the first part of the proof, we have(28)Kψx2k-1=Kgx2k-1,Kψx2k=Kfx2k,k∈N.Hence, by continuity of K(ψ), K(g), and K(f) at x0, letting k→∞, we get(29)Kfx0=Kψx0=Kgx0.
Whenx0 is the right endpoint of I, the argument is similar.
To define the functionh:I×R→R and fix arbitrarily an y0∈R, let us define a function Py0:I→R by(30)Py0x≔y0,x∈I.
Of coursePy0, as a constant function, belongs to RVp(I). For x0∈I, y0∈R, put (31)hx0,y0≔KPy0x0.
Since, by (30), for all functions f,(32)fx0=Pfx0x0,according to what has already been proved, we have(33)Kfx0=KPfx0x0=hx0,fx0.
To prove the uniqueness ofh, assume that h¯:I×R→R is such that(34)Kfx=h¯x,fxfor all f∈RVp(I) and x∈I. To show that h=h¯ let us fix arbitrarily x∈I, y∈R and take f∈RVp(I) with f(x)=y. From (33), we have(35)h¯x,y=h¯x,fx=Kfx=hx,fx=hx,y,which proves the uniqueness of h.Definition 5.
LetX⊂R and a function h:X×R→R be fixed. The mapping H:RX→RX, given by(36)Hfx≔hx,fx,f∈RX,x∈X,is said to be composition (Nemytskii or superposition) operator. The function h is referred to as the generator of the operator H.As an immediate consequence of Theorem4 we get the following.Corollary 6.
Let1≤p<∞. If a local operator K maps RVp(I) into C(I), then it is a Nemytskii operator.Note that if a local operatorK maps RVp(I) into itself then, obviously, K maps RVp(I) into C(I). Therefore, by Theorem 4, we get the following.Theorem 7.
Let1≤p<∞. If a local operator K maps RVp(I) into itself, then there exists a unique function h:I×R→R such that, for all f∈RVp(I), (37)Kfx=hx,fx,x∈I.Corollary 8.
Let1≤p<∞. If a local operator K maps RVp(I) into itself, then it is a Nemytskii operator.Under the additional assumption that the locally defined operator is uniformly continuous, we get a complete characterization of its generating functionh. Namely, we have the following.Theorem 9.
Let1≤p<∞. If a local operator K:RVp(I)→RVp(I) is uniformly continuous, then there exists f1,f2∈RVp(I) such that (38)Kfx=f1xfx+f2x,f∈RVp(I),x∈I.Proof.
From Theorem7 there exists a unique function h:I×R→R such that K(f)(x)=h(x,f(x)) for all f∈RVp(I), x∈I. Fix (x0,y0)∈I×R, take an arbitrary sequence xn∈I with xn→x0, and let Py0:I→R be defined by Py0(x)=y0, x∈I. Since h(x0,y0)=K(Py0)(x0),(39)hxn,y0-hx0,y0=hxn,Py0xn-hx0,Py0x0=KPy0xn-KPy0x0;applying the continuity of K(Py0) at x0, we get the continuity of h with respect to the first variable. Thus, by [16, Theorem 1] (with φ(x)=ψ(x)=xp),(40)hx,y=f1xy+f2x,x∈I,y∈R,for some f1,f2:I→R. Since h(·,y0)=K(Py0)(·)∈RVp(I) and f2(x)=h(x,0), f1(x)=h(x,1)-f2(x), the functions f1,f2∈RVp(I).



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*Source: 925091-2015-07-06.xml*