R E S E A R C H
Open Access
Ostrowski type inequalities for sets and
functions of bounded variation
Oleg V Kovalenko
**Correspondence: [email protected] Department of Mechanics and Mathematics, O. Honchar Dnipro National University, pr. Gagarina, 72, Dnipro, 49010, Ukraine
Abstract
In this paper we obtain sharp Ostrowski type inequalities for multidimensional sets of bounded variation and multivariate functions of bounded variation.
MSC: Primary 26D15; secondary 41A17; 41A44; 41A63
Keywords: Ostrowski type inequalities; multivariate functions of bounded variation; sets of bounded variation
1 Introduction
In Ostrowski [] proved the following inequality.
Theorem Let f: [, ]→Rbe a differentiable on(, )function with bounded on(, )
derivative.Then,for all x∈[, ],
f(t)dt–f(x)≤
+
x–
sup
t∈(,)
f(t).
The inequality is sharp.
Inequalities that estimate deviation of a function from its mean value using different characteristics of the function are usually called Ostrowski type inequalities. Such inequal-ities have many applications, in particular in the area of numerical methods, and are heav-ily studied. See [] and the references therein for results connected with Ostrowski type inequalities for univariate functions of bounded variation and their applications.
The goal of this article is to obtain sharp Ostrowski type inequalities for multivariate functions and multidimensional sets of bounded variations. There are several ways to ex-tend the notion of bounded variation to multivariate functions, see [] for a review of dif-ferent approaches for functions of two variables; [] for the point of view that is generally accepted in literature now.
We introduce a new definition of bounded variation that is based on the Kronrod-Vitushkin approach []. The introduced variation of a multivariate function has (unlike any of the Kronrod-Vitushkin variations) the following two properties: the variation does not change if the argument of the function is multiplied by a non-zero constant; and the
variation of a multivariate radial function is twice bigger than the variation of the gener-ating one-dimensional function, see Properties and below for rigorous statements of the properties.
The paper is organized as follows. In Section we list the notations used throughout the paper. In Sections and we introduce definitions, justify the correctness of the def-initions, and list some properties of the sets and function variations. Section is devoted to Ostrowski type inequalities.
2 Notations
For a setF⊂Rd, denote by∂F,intFandFits boundary, interior and closure, respectively.
For arbitraryt∈R, set Rd
t :={(x,t)∈Rd: x∈Rd–}. Forx,y∈Rd, by xywe denote the
segment with ends in the pointsxandy, i.e.,xy={( –t)x+ty:t∈[, ]}. Forc∈Rand F⊂Rd, setcF:={cx:x∈F}.
For two setsF,F⊂Rd, setρ(F,F) :=infx∈F,y∈F|x–y|, where|w| denotes the Eu-clidean distance between the pointw∈Rdand zero elementθ ofRd.
Forε≥, two setsF,F⊂Rdare calledε-disjoint ifρ(F,F) >ε. Obviously, two
com-pact setsF,F⊂Rdare -disjoint if and only if they are disjoint. Forε> , a setFis called
ε-connected if there does not exist a partitionF=F∪Fintoε-disjoint non-empty sets
F,F. Some properties ofε-components can be found in Appendix I of [].
For an arbitrary function f: E⊂Rd→Randc∈R, denote by{f ≥c} the set {x∈
E:f(x)≥c}. Similarly, we define the sets{f ≤c}and{f =c}.
Denote bySd–thed– dimensional unit sphere{x∈Rd:|x|= }; forε> andx∈Rd,
Bd(x,ε) :={y∈Rd:|x–y| ≤ε};Bd(ε) :=Bd(θ,ε) andBd:=Bd().
ByPd–we denote thed– dimensional real projective space, i.e., the set of all lines
inRdthat containθ. The distance between two linesr
,r∈Pd–is by definition equal
to the angle betweenr andr. The measure of a set A⊂Pd–is by definition equal to
the spherical measure of the setl∈Al∩Sd–; so that the measure ofPd–is equal to the
measure ofSd–.
For eachr∈Pd–, byd–(r) we denote the hyperplane that containsθand is orthogonal
to the liner;d–(r) is considered as ad– -dimensional space withd– -dimensional
Lebesgue measure and Euclidean metric. For eachβ∈d–(r), byl(r,β) we denote the
line that containsβand is parallel tor.
It is assumed that product topology is induced on the Cartesian product of a finite num-ber of topological spaces and product measure is induced on the Cartesian product of a finite number of measure spaces. By μk,k∈N, we denote thek-dimensional Lebesgue measure inRk; byμwe denote the spherical measure on the unit sphereSd–and the measure on the projective spacePd–.
3 Variation of a set
3.1 Definition
Definition Denote byN(F) the number of connected components of the setF⊂Rd;
for an empty set, and +∞if the set of connected components is infinite.
Definition For a compact setF⊂Rdand a liner∈Pd–, set
v(F,r) := ess sup β∈d–(r)
NF∩l(r,β).
Definition For a compact setF⊂Rdand a numberp∈[,∞], set
vp(F) :=
⎧ ⎨ ⎩
(μPd–
Pd–vp(F,r)dr)
p, p∈[,∞),
ess supr∈Pd–v(F,r), p=∞.
Remark Ifd= , then for allp∈[,∞] we setvp(F) =N(F).
Definition Let a compact setF⊂Rd,ε≥ andp∈[,∞] be given. Set
Vpε(F) :=sup
n
k=
vp(Fk),
where the supremum is taken over all partitionsF=nk=Fkof the setFinto a finite
num-ber of compact pairwiseε-disjoint subsetsFk.
Remark We writeVp(F) instead ofVp(F). In this case the supremum in the definition
is taken over all partitionsF=nk=Fkof the setFinto a finite number of compact disjoint
subsetsFk.
3.2 Correctness of the definitions
The proofs of measurability of the functions that stay under integral signs (Lemmas -) use ideas from [] (Chapters -) and [] (Chapter ).
Throughout this subsection, we identify each point (r,x) of the space Pd–×Rd–
(r∈Pd–,x∈Rd–) with the linel(r,β), where β∈d–(r) is a point with coordinates
xwith respect to some orthonormal basis ofd–(r) (the basis ofd–(r) is assumed to continuously change asr∈Pd–changes).
We need the following lemma.
Lemma For every Borel set B⊂Rd, the setψ(B)of all(r,x)∈Pd–×Rd– such that
B∩l(r,β)=∅is measurable(Pd–×Rd–).The setϕ(B) :={t∈R: B∩Rdt =∅}is measur-able.
Denote byψone of continuous maps from [, ] ontoSd–; such a map exists due to
the Hahn-Mazurkiewicz theorem; see, for example, Theorem . in []. Define a func-tionψ: [, ]×Rd→Pd–×Rd–using the following rule. For each (t,y)∈[, ]×Rd,
letψ(t,y) be the linel(r,β), wherer∈Pd–is the line that containsψ(t), andβ is such
that the linel(r,β) contains the pointy. It is easy to see that the functionψ is
contin-uous. Thenψ(B) =ψ(B), where˜ B˜ := [, ]×B. This means that ψ(B) is a continuous
image of a Borel setB˜ ⊂Rd+ and hence is measurable (see, for example, Theorem
in []).
Definition For a compact setF⊂Rd andε> , denote by N
ε(F) the number ofε -components of the setF; for an empty set.
Lemma Let a compact set F⊂Rdbe given.Then,for arbitraryε> ,N
ε(F)is finite and limε→Nε(F) =N(F).
With eachε-componentWof the setF, associate a ball with center in an arbitrary point w∈W and radius ε
. Balls that correspond to differentε-components ofFare pairwise
disjoint; hence there can be only a finite number of such balls due to the boundedness ofF.
IfFhas a finite number of connected componentsF, . . . ,Fn, thenε:=mini=jρ(Fi,Fj) >
due to the compactness of components of a compact set. ThenNε(F) =n=N(F) for all
ε<ε.
IfN(F) =∞, then for arbitraryn∈Nwe can choose compact disjoint setsF, . . . ,Fnsuch
thatF=nk=Fk. ThenNε(F)≥nfor allε<mini=jρ(Fi,Fj).
For a compact setF⊂Rdandε> , define functions
vF: Pd–×Rd–→R, vF(r,x) =N
F∩l(r,β), () and
vεF: Pd–×Rd–→R, vεF(r,x) =Nε
F∩l(r,β), NFε:R→R, NFε(t) =Nε
F∩Rdt. The following lemma holds.
Lemma Let a compact set F⊂Rd andε> be given.The function vε
F is measurable
(Pd–×Rd–).The function Nε
F is measurable.
Consider the (countable) setof all closed balls inRdwith rational centers and radii. Let˜ be the set of all finite unions of the balls from. For eachn∈N, definento be the
family of all sets of the formms=Bis, wherem≥n, and{Bis}ms=is a collection of pairwise
ε-disjoint sets from˜. Then the setnis countable for eachn∈N.
Note that the functionsvεF andNFεtake only non-negative integer values. The sets{vεF≥ }=Pd–×Rd–and{NFε≥}=Rare measurable. Suppose nownis a natural number. Below we prove that
vεF≥n= B=ms=Bis∈n
m
s=
ψ(Bis∩F)
m s=
ψ(∂Bis∩F)
()
and
NFε≥n= B=ms=Bis∈n
m
s=
ϕ(Bis∩F)
m s=
ϕ(∂Bis∩F)
,
If (r,x) belongs to the right-hand side of (), then there exists a setB=ms=Bis∈n such thatl(r,β)∩Bis∩Fis a non-empty set strictly insideBis,s= , . . . ,m. Since the sets that constituteBareε-disjoint, by the definition ofn, we obtain thatvεF(r,x)≥m≥n
and hence (r,x)∈ {vε
F≥n}.
Let (r,x)∈ {vεF≥n}andF, . . . ,Fm,m≥nbe allε-components of the setl(r,β)∩F. Since
mini=jρ(Fi,Fj) >ε, there existsδ> such thatmini=jρ(Fi,Fj) >ε+ δ. Consider a finite
coverCof the compact setl(r,β)∩Fby open balls with rational centers and radii such that each ball has diameter less thanδand contains some point froml(r,β)∩F. Denote by Bk the union of closures of all balls from the coverC that intersectFk,k= , . . . ,m.
Thenmk=Bk belongs tonby construction, and hence (r,x) belongs to the right-hand
side of ().
SinceFis a closed set andnis a countable set, Lemma implies that the sets{vεF≥n}
and{Nε
F≥n}are measurable; hence the functionsvεF andNFεare measurable.
Lemma The function vF defined by()is measurable(Pd–×Rd–)for each compact
F⊂Rd.
For each fixed (r,x)∈Pd–×Rd–, one haslim
ε→vεF(r,x) =vF(r,x) due to Lemma .
Hence the measurability ofvF is a consequence of the measurability ofvεF,ε> , stated
in Lemma .
Tonelli’s theorem and Lemma imply that the functionvp(F) is well defined for every
≤p≤ ∞and compactF⊂Rd; hence the functionsVε
p(F) are also well defined for all
ε≥.
3.3 Some properties of the sets variation
The following property is a direct consequence of the definitions.
Property Letp∈[,∞],ε≥ andF⊂Rdbe a compact set. Thenv
p(F)≤Vpε(F). IfF
isε-connected, thenvp(F) =Vpε(F).
Property Letp∈[,∞] andF⊂Rdbe a compact set. Then
Vp(F) =lim
ε→V
ε
p(F). ()
From the definition it follows thatVε
p (F)≤Vpε(F) wheneverε>ε. This implies that
limε→Vpε(F) exists and does not exceedVp(F).
Assume thatVp(F) <∞. Then, for arbitraryδ> , there exists a partitionF=
n(δ) k=Fk
of the setFinto pairwise disjoint compact setsFksuch thatVp(F)≤
n(δ)
k=vp(Fk) +δ. Set
ε:=mini=jρ(Fi,Fj). Thenε> and for allε<εVpε(F)≥Vp(F) –δ. This implies ().
The case whenVε
p(F) =∞can be considered in a similar way.
Property Letn∈Nand pairwise disjoint compact setsF, . . . ,Fn⊂Rdbe given. Then,
for allp∈[,∞],vp(
n k=Fk)≤
n
k=vp(Fk).
It is sufficient to prove the property in the casen= . SetF:=F∪F.
SinceFandFare compact disjoint sets, we haveρ(F,F) > , and hence, for arbitrary
r∈Pd– andβ ∈d–(r), one hasN(F∩l(r,β)) =N(F
implies thatv(F,r)≤v(F,r) +v(F,r) for allr∈Pd–, and hencevp(F)≤vp(F) +vp(F) for
allp∈[,∞].
Property Assumen∈N,ε≥ and pairwiseε-disjoint compact setsF, . . . ,Fn⊂Rdare
given. Then, for allp∈[,∞],
Vpε
n
k=
Fk
=
n
k=
Vpε(Fk).
It is sufficient to prove the property in the casen= . SetF:=F∪F.
Let{Tk}mk=,m∈N, be a partition of the setFinto compact pairwiseε-disjoint subsets.
Then, by Property ,
m
k=
vp(Tk) = m
k=
vp
(Tk∩F)∪(Tk∩F)
≤
m
k=
vp(Tk∩F) + m
k=
vp(Tk∩F)
≤Vpε(F) +Vpε(F)
since{Tk∩F}mk= and{Tk∩F}mk= are partitions of the setsF andF into pairwiseε
-disjoint compact subsets. This implies that
Vpε(F)≤Vpε(F) +Vpε(F).
On the other hand, for arbitrary partitions of the setsFandFinto compactε-disjoint
sets{Tk}sk=and{Tk}mk=respectively,s,m∈N,
Tksk=∪Tkmk=
is a partition of the setFinto compactε-disjoint sets, and hence
Vε
p(F)≥Vpε(F) +Vpε(F).
Property If ε≥ andF⊂Rd is a compact set that has exactly n∈Nε-connected
componentsF, . . . ,Fn, then, for allp∈[,∞],Vpε(F) =
n
k=vp(Fk).
Note that eachε-connected component of a compact set is compact. Hence, by Proper-ties and ,Vε
p(F) =
n
k=Vpε(Fk) =nk=vp(Fk).
Property IfF⊂Rdis a compact set, α= andαF:={αx:x∈F}, then for arbitrary
p∈[,∞]vp(F) =vp(αF) andVp(F) =Vp(αF).
The property follows from the observation that for arbitraryr∈Pd–andβ∈d–(r) one
4 Variation of a function
4.1 Definition
Definition Let a setE⊂Rdand a functionf:E→Rbe given. Fort∈R, the set
L(f;t) :=x∈E:f(x) =t
is called a level set of the functionf.
The variation of a continuous function is given by the following definition.
Definition LetE⊂Rd,f:E→Rbe a continuous on a compact subsetF⊂Efunction,
andp∈[,∞]. Set
vp(f;F) :=
∞
–∞vp
F∩L(f;t)dt.
IfFis locally connected, then set
Vp(f;F) :=
∞
–∞Vp
F∩L(f;t)dt.
IfF=E, then we writevp(f) andVp(f) instead ofvp(f;E) andVp(f;E), respectively.
4.2 Correctness of the definitions
We need to prove that the functions under the integral signs are measurable.
Lemma Let E⊂Rd,f: E→Rbe a continuous on a compact subset F⊂E function,and
p∈[,∞].Then the function vp(F∩L(f;·))is measurable.
Without loss of generality, we may assume thatRd⊃F=E, and we need to prove that
the functionvp(L(f;·)) is measurable. Consider the graph
(f) :=(x,t)∈Rd+:x∈E,f(x) =t ()
of the functionf and two functionsv(f): Pd×Rd→Randvf:R×Pd–×Rd–→R
defined by the formulavf(t,r,x) =vL(f;t)(r,x) (see () for the definition of the functionvF).
Since the setEis compact and the functionf is continuous onE, the set(f)⊂Rd+is
compact. This, by Lemma , implies that the functionv(f)is measurable (Pd×Rd).
Recall thatRd+={(x, )∈Rd+:x∈Rd}. The function
φ:R×Pd–×Rd–→Pd×Rd∩(R,y)∈Pd×Rd: R⊂Rd+
that maps a point (t,r,x) to the point (R,y) withR={(z, ),z∈r}andy= (x,t) is continuous and has continuous inverse. Moreover, for arbitrary (t,r,x)∈R×Pd–×Rd–, we obtain
vf(t,r,x) =v(f)(φ(t,r,x)). Hence, for arbitraryc∈R, the functionφmaps the set
(t,r,x)∈R×Pd–×Rd–: vf(t,r,x)≥c
and the set
(R,y)∈Pd×Rd:v(f)(R,y)≥c
∩(R,y)∈Pd×Rd:R⊂Rd+.
The latter is an intersection of a measurable (due to measurability ofv(f)) set and a closed
set; hence the former is also a measurable set. This means that the functionvf is
mea-surable (R×Pd–×Rd–), and hence the statement of the lemma is true due to Tonelli’s
theorem.
The following result is well known (see, for example, [], Lemma , or [], Lemma in Chapter ).
Lemma For an arbitrary function f: E→R,denote by Textrthe set of t∈Rsuch that
L(f;t)contains an extremum point.Then Textris at most countable.
Lemma Let E⊂Rd,f:E→Rbe a continuous on a compact locally connected subset
F⊂E function,and p∈[,∞].Then the function Vp(F∩L(f;·))is measurable.
Without loss of generalization, we may assume thatF=E. Taking into account Prop-erty , it is sufficient to prove that each of the functionsVν
p(L(f;·)) is measurable,ν> .
Let ν > be fixed. The function Nf(t) := Nν(L(f;t)) =Nν(f)(t) is measurable due to Lemma ((f) is defined by ()). Due to Lemma , for eachk= , , . . . , the setTk:=
{Nf=k} \Textris measurable; moreover, obviously these sets are pairwise disjoint.
Letc∈Rbe given. Then
VpνL(f;·)≤c\Textr=
k∈N
t∈Tk:Vpν
L(f;t)≤c
and it is sufficient to prove that for eachk∈Nthe set{t∈Tk:Vpν(L(f;t))≤c}is
measur-able.
Let somek∈Nandt∗∈Tkbe fixed. The setL(f;t∗) contains exactlyk ν-connected
componentsF, . . . ,Fk. Each of the components is a compact set, hence there existsε>
such that the setsUs(ε) ={x:ρ(x,Fs) <ε},s= , . . . ,k, are pairwiseν-disjoint. SetU(ε) :=
k s=Us(ε).
There existsδ> such thatL(f;t)⊂U(ε) for allt∈(t∗–δ,t∗+δ). Really, assume the contrary, suppose that there exists a sequencean,an→ asn→ ∞, such that each of the
level setsL(f;t∗+an) contains a pointxn∈/U(ε). Switching to a subsequence, if needed,
we may assume that the sequencexnconverges to somex∈Rd. SinceU(ε) is an open set,
x∈/U(ε). However, this is impossible sincef is continuous, and hencef(x) =t∗.
For eachs= , . . . ,k, consider an arbitrary pointxs∈Fs. Since the level setL(f,t∗) does
not contain extremums,Fis locally connected andf is continuous; for small enoughεs>
, the setf(Bd(x
s,εs)) contains a neighborhood (t∗–δs,t∗+δs) oft∗ (δs> ). Hence, for
arbitraryt∈(t∗–δs,t∗+δs), the level setL(f;t) contains at least oneν-component inside
Us(ε). This means that there existsδ=δ(t∗) > such that each of the level setsL(f,t),
t∈(t∗–δ,t∗+δ) contains at least oneν-component insideUs(ε),s= , . . . ,k. Hence, for all
s= , . . . ,k. This implies that for allt∈Tk∩(t∗–δ,t∗+δ)
VpνL(f;t)=
k
s=
vp
L(f;t)∩Us(ε)
()
due to Property .
For eacht∗∈Tk, setW(t∗) := (t∗–δ(t∗),t∗+δ(t∗)); the setsW(t∗),t∗∈Tk, constitute
an open cover of the setTk. SinceRis a Lindelöf space, we can find a countable subcover
W,W, . . . ofTk. SetW˜:=W∩Tk,W˜m:= (Wm\
m–
s= Ws)∩Tk,m= , , . . . . We obtain
a countable partition of the setTkinto pairwise disjoint measurable subsetsW˜m,m∈N,
such that on each of the setsW˜m we have representation () of the functionVpν(L(f;·))
(the setsUs(ε),s= , . . . ,k, might be different for differentm∈N). Due to Lemma , this
implies the measurability of the set
t∈Tk:Vpν
L(f;t)≤c=
m∈N
˜ Wm∩
k
s=
vp
L(f;t)∩Us(ε)
≤c
and hence the lemma is proved.
4.3 Some properties of the function variation
Below we list some properties of the function variation. Everywherep∈[,∞] and a com-pact setF⊂Rdare fixed,f is continuous onFfunction. For properties ofV
p(f), the setF
is further assumed to be locally connected.
Property vp(f) andVp(f) are non-negative. Iff is constant, thenvp(f) =Vp(f) = .
The fact that variations are non-negative follows from the definition. Iff is constant, then it has exactly one non-empty level set.
Property Ifα,β∈R, thenvp(α·f+β) =|α|vp(f) andVp(α·f +β) =|α|Vp(f).
Note thatL(α·f+β;t) =L(f;α–·(t–β)) for allα= andβ,t∈R. Making substitution s=α–·(t–β) in the integrals from Definition , we obtain the required equalities. In the caseα= , the property follows from Property .
Property For arbitraryα= ,vp(f;F) =vp(f(α·);α–F) and
Vp(f;F) =Vp
f(α·);α–F.
The property follows from Property .
Property Lett∈Rbe such thatF\L(f;t) has exactlyn≥ connected components F,F, . . . ,Fn. Assume that for allk= , . . . ,n,Fk\Fk⊂L(f;t). Thenvp(f;F)≤
n
k=vp(f;Fk)
andVp(f;F) =
n
k=Vp(f;Fk).
Consider arbitrarys∈R,s=t. Fork= , . . . ,n, setWk:=Fk∩L(f;s). Then the setWkis
closed,k= , . . . ,n, andWk⊂Fk due to conditions of the property and the fact that
disjoint sets. From Properties and it follows thatVp(L(f;s)) =
n
k=Vp(Fk∩L(f;s)) and
vp(L(f;s))≤
n
k=vp(Fk∩L(f;s)). The statement of the property now follows from
Defini-tion .
Property Ifd= andf: [, ]→Ris a continuous function, then for allp∈[,∞]
vp
f; [, ]=Vp
f; [, ]=
f.
Remark fis the classical variation of a univariate functionfon [, ]. We allowf to be +∞in the case whenf is not a function of bounded variation.
The Banach indicatrix theorem [] states thatf is equal to the integral overt∈R of number of points inL(f;t). In the definition ofvp(f; [, ]), the number of components
ofL(f;t) is integrated overt∈R. Each component of a compact set inRis a point or a segment. The family of level setsL(f;t) that contain a segment as a connected component is at most countable because each of such level sets contains extremum (see Lemma ). Hencevp(f; [, ]) =
f. It is easy to see thatvp(f; [, ]) =Vp(f; [, ]).
Property Letϕ: [, ]→Rbe a continuous function and d∈N. Letfϕ: Bd→R, fϕ(x) =ϕ(|x|). Then, for allp∈[,∞],
vp
fϕ;Bd
=Vp
fϕ;Bd
= ·
ϕ.
In the cased= , the property follows from Property , so we can assume thatd≥. Let arbitraryt=ϕ() be fixed. For arbitraryr∈Pd–andβ∈d–(r) the numberN(L(fϕ;t)∩ l(r,β)) can be obtained by the following procedure: consider the liner=l(r,θ) and mark points of the setL(fϕ;t)∩l(r,θ); cut the interval (–|β|,|β|) from the line and stick the points –|β|and|β|together; the number of components of marked points on the obtained ‘cut’ line is equal toN(L(fϕ;t)∩l(r,β)). This shows that for arbitraryβ
NL(fϕ;t)∩l(r,β)
≤NL(fϕ;t)∩l(r,θ)
. ()
From the choice oftit follows that
θ∈/L(fϕ;t), ()
and hence there exists ε> such that B(ε)∩L(fϕ;t) =∅. This implies that the set L(fϕ;t)∩l(r,θ) does not contain pointsxwith|x|<εand hence for allβsuch that|β|<ε () becomes equality. This implies thatv(L(fϕ;t),r) =N(L(fϕ;t)∩l(r,θ)). From () it follows thatN(L(fϕ;t)∩l(r,θ)) = ·N(L(ϕ;t)). Equalityvp(fϕ;B) = ·
ϕfollows from Property
now. Equalityvp(fϕ;B) =Vp(fϕ;B) follows from the geometry of the level sets offϕ.
5 Ostrowski type inequalities
5.1 Auxiliary results
Lemma Let d∈N,d≥,ε> ,x∈Rd,r∈Pd–and a measurable set F⊂Bd(x,ε)be
and r such that
μd–β∈d–(r) :F∩l(r,β)=∅>A·μd–Bd–(ε) wheneverμdF>α·μdBd(ε).
The fact thatαdoes not depend onεfollows from the observation that
μdF
μdBd(ε)=
μd(
εF)
μdBd
and
μd–{β∈d–(r) :F∩l(r,β)=∅}
μd–Bd–(ε) =
μd–{β∈d–(r) :
εF∩l(r,β)=∅}
μd–Bd– .
The fact thatαis independent ofxandris obvious. The existence ofαfollows from the equality
μdF=
d–(r)∩Bd(y,ε)
μl(r,β)∩Fμd–(dβ), ()
where y∈d–(r) is such that the linel(r,y) containsx and equality μd–(d–(r)∩ Bd(y,ε)) =μd–Bd–.
Lemma Let p∈[,∞),A> and B∈[,A]be given.Then
A
B+ p(A–B)≥
– B A
p
.
It is sufficient to prove that the functionϕ(x) = p+ ( – p)x– ( –x)pis non-negative
on [, ]. Sinceϕ() =ϕ() = , the functionϕhas at least one zero on (, ). The function
ϕ(x) =p( –x)p–+ – pis decreasing on [, ], hence has at most one zero on (, ). This
implies thatϕ() > and hence the functionϕis increasing on [,x∗] and is decreasing on [x∗, ], wherex∗is zero ofϕon (, ); henceϕis non-negative on [, ].
5.2 Main results
The following theorem is the main tool to prove Ostrowski type inequalities for functions and sets of bounded variation below.
Theorem Let d∈Nand two sets F,W⊂Bdbe given.Assume that the following
proper-ties hold:
. Fis measurable andθ∈/F;
. Wis closed andθ∈/W;and
. Ifx∈Fandy∈Bd\F,thenxy∩W=∅.
Then,for all p∈[,∞],
μdF≤μ
dBd
The inequality is sharp in the sense that for arbitraryε> there exist sets F and W that satisfy conditions above and such that
μdF>
μdBd –ε
vp(W).
If ()becomes equality,thenμdF= .
We will prove Theorem in the next subsections. Here we state two consequences of this theorem, which can be considered as Ostrowski type inequalities.
Theorem Let d∈Nand a continuous function f: Bd→Rbe given.Then,for all p∈ [,∞],
μdBd
Bd
f(x)dx–f(θ)≤vp(f) .
The inequality is sharp.It becomes equality only in the case when f is constant.
Due to Property , we can assume thatf(θ) = , and it is sufficient to prove that
Bdf(x)dx≤
μdBd
vp(f). ()
Consider a set
:=(x,t)∈Bd×[,∞) :f(x)≥t.
Then
Bd
f(x)dx≤μd+=
t≥
μd∩Rdt+dt. ()
For eacht> , consider the setsF:=∩Rd+
t andW:=(f)∩Rdt+(see () for the
defi-nition of(f)). BothFandWare closed sets that do not containθsincef(θ) = . Ifx∈F andy∈Bd\F, thenf(x)≥tandf(y) <tand hence the segmentxycontains a pointzwith f(z) =t, i.e.,xy∩W=∅. This means that all the conditions of Theorem are satisfied and hence
μd∩Rdt+=μd(F)≤μ
dBd
vp(W) =
μdBd
vp
L(f;t)
with equality possible only in the case whenμdF= . Taking into account (), we obtain
μd+≤μ
dBd
t≥
vp
L(f;t)dt≤μ
dBd
t∈Rvp
L(f;t)dt=μ
dBd
vp(f)
For allε> , consider the functionϕε: [, ]→R,ϕε(t) = fort≥ε,ϕε() = andϕεis linear on [,ε]. Due to Property , for the radial functionfε(x) :Bd→R,fε(x) =ϕε(|x|), and arbitraryp∈[,∞]vp(fε) = ; moreover,
Bdfε(x)dx→μdBdasε→. This proves the sharpness of the stated inequality.
Theorem Let d∈Nand a closed set F⊂Bdbe given.Ifθ∈/F,then for all p∈[,∞]
μdF≤μ
dBd
vp(F).
The inequality is sharp.If equality holds,thenμdF= .
It is enough to apply Theorem with W =F; all three conditions of Theorem are satisfied.
For arbitraryε> , consider a setFε:={x∈Bd: |x| ≥ε}. For allp∈[,∞],vp(Fε) = ;
μdF
ε→μdBdasε→. This proves that the stated inequality is sharp.
Remark In all three theorems variationvpcan be substituted byVpdue to Property .
The inequalities will remain sharp.
Remark Properties and state thatVpis additive. This gives motivation to callVpa
variation, rather thanvp.
5.3 More auxiliary results
Denote byF˜the set of all pointsx∈Fsuch thatlimδ→+μ
d(F∩Bd(x,δ))
μdBd(δ) = . ThenF˜∩Sd–=∅ and, by the Lebesgue density theorem,
μdF˜=μdF. ()
Lemma Assume that the conditions of Theoremhold.If r∈Pd–is such that v(W,r) = ,then for arbitraryβ∈d–eitherF˜⊃intBd∩l(r,β),orF˜∩l(r,β) =∅.
Assume that for someβ∈d–(r) there existx∈ ˜F∩l(r,β) andy∈(intBd∩l(r,β))\ ˜F. From the definition ofF˜it follows that there exista> and a sequenceρn→ asn→ ∞
such thatμd(Bd(y,ρ
n)\F)≥a·μdBd(ρn) for alln∈N. From () (withF substituted by
Bd(y,ρ
n)\F) it follows that there existsA> such that
μd–(ρn) >A·μd–Bd–(ρn) ()
for alln∈N, where
(ρn) =
β∈d–(r) :Bd(y,ρn)\F
∩l(r,β)=∅.
Sincex∈ ˜F, there existsδ> such that for allρ<δone hasμd(Bd(x,ρ)∩F)≥α( –A)·
μdBd(ρ) (the numberα( –A) is defined in Lemma ). Lemma implies that
for allρ≤δ, where
(ρ) =
β∈d–(r) :Bd(x,ρ)∩F∩l(r,β)=∅.
Choosenso big thatρn<δ. Then
μd–(ρn) +μd–(ρn) >μd–Bd–(ρn) ()
due to () and (). Moreover, sincex,y∈l(r,β), we receive that
(ρn),(ρn)⊂d–(r)∩Bd(β,ρn) ()
and
μd–d–(r)∩Bd(β,ρn)
=μd–Bd–(ρn)
. ()
Set=(ρn)∩(ρn). Then, due to (), () and (),μd–> . But each linel(r,β),
β∈, contains a point fromW due to Condition of Theorem and the definitions of the sets(ρn) and(ρn); this contradicts assumptionv(W,r) = of the lemma.
Lemma Assume that the conditions of Theorem hold. Let R⊂ Pd– be such that
v(W,r) = for all r∈R.If R contains d lines that are not contained in any d– -dimensional hyperplane,thenμd(F) = .
Due to () it is enough to prove thatF˜ =∅. Letr, . . . ,rdbe the lines from the
state-ment of the lemma, and let ρ, . . . ,ρd be unit vectors parallel to these lines. Set P :=
{d
k=tkρk:tk∈(–, ),k= , . . . ,d}, thenPis an open inRdset.
Consider arbitraryx∈intBd. Chooseε> such thatx+εP⊂Bd. Then, for all points yfrom the segmentθx,Py:=y+εP⊂Bd.y∈θxPyis an open cover of a compact setθx,
hence it contains a finite subcoverP,P, . . . ,Pm,m∈N. From Lemma it follows that for
eachs= , . . . ,meitherPs⊂ ˜F, or
Ps∩ ˜F=∅. ()
Sinceθ∈ ˜/F, we obtain that () holds for eachs= , . . . ,mand hencex∈ ˜/F.
5.4 Proof of Theorem 1
Ifv(W,r)≥ for almost allr∈Pd–, thenv
p(W)≥ and inequality () holds. It is strict
because Condition of Theorem holds. If there is a setR⊂Pd–of positive measure such
thatv(W,r) = for allr∈R, thenμdF= due to Lemma , and inequality () holds.
Assume that there exists R⊂Pd–, μR> , such that v(W,r) = for all r∈R and v(W,r)≥ for almost allr∈Pd–\R. Then
vp(W)≥ –
μR
Really, ifp=∞, thenv∞(W)≥ in the caseμR<μPd–andv∞(W) = in the caseμR=
μPd–=μSd–. In both cases () holds. In the casep∈[,∞)
vp(W)≥
μSd–
μR+ p·μSd––μR p
≥ – μR
μSd–
due to Lemma .
Conditions and of the theorem imply that there existsε> such thatBd(ε)∩W=∅
andBd(ε)∩F=∅. Set:=
r∈R(r∩Bd). Below we prove that
μd(∩F) <μ
d
. ()
In order to prove (), it is enough to show that
μd(∩ ˜F) <μ
d
()
due to (). Consider arbitraryr∈R. Then all points of the intersectionr∩ ˜Fare from one side ofr∩Bd(ε). This fact can be proved using arguments similar to the proof of Lemma .
Denote byχthe characteristic function of the set∩ ˜F. Thenχ(x) = for all|x|<εand
χ(x) +χ(–x)≤ for allx∈. This implies (). Finally, having (), we can write
μdF≤μd(F∩) +μdBd\
<μdBd– μ
d
=μdBd– ·
μdBd
μSd–μR
= μ
dBd
– μR
μSd–
.
The latter together with () proves ().
The same example as in Theorem shows that inequality () is sharp.
Acknowledgements
The author was partially supported by grant 0117U001208. Competing interests
The author declares that they have no competing interests. Author’s contributions
All authors made an equal contribution to the paper, have read and approved the final manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 16 March 2017 Accepted: 16 June 2017
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