R E S E A R C H
Open Access
On modulus of continuity of
differentiation operator on weighted Sobolev
classes
Vladislav F Babenko and Oleg V Kovalenko
**Correspondence:
[email protected] Department of Mechanics and Mathematics, O. Gonchar Dnipropetrovsk National University, pr. Gagarina, 72, Dnipropetrovsk, 49010, Ukraine
Abstract
In this paper we investigate the modulus of continuity ofkth order differential operator,k∈N, on the classes of functions defined on half-line that have positive non-increasing continuous majorants of functions and their higher derivatives.
Keywords: inequalities for derivatives; modulus of continuity; Kolmogorov type inequalities; perfect splines
1 Introduction
The following theorem was proved in [] () by Hardy and Littlewood (see also [], Theorem ..).
Theorem A Suppose that a function x(t)is defined for t> and its second derivative x(t)
exists for t> .Let f(t)and g(t)be positive functions(both decreasing or both increasing).
Then(as t→+∞)the following statements hold:
. Ifx(t) =O(f(t)),x(t) =O(g(t)),then
x(t) =Of(t)g(t).
. Ifx(t) =o(f(t)),x(t) =O(g(t)),then
x(t) =of(t)g(t).
. Ifx(t) =O(f(t)),x(t) =o(g(t)),then
x(t) =of(t)g(t).
This theorem had a great impact on formation of the whole field of problems connected with inequalities between derivatives. To confirm this, it is sufficient to note that funda-mental possibility of inequalities for upper bounds of derivatives of functions defined on the whole real line or half-line can be easily derived from this theorem (see [], p.).
In Mordell [] (see also [], Theorem ..) proved the following refinement of Theorem A for non-increasing functions.
Theorem B Let f(t)and g(t)be positive non-increasing on the half-lineR+functions.If a function x(t)is defined on the half-lineR+and for all t> there exists x(t)such that
x(t)≤f(t), x(t)≤g(t),
then for all t>
x(t)≤f(t)g(t).
LetIbe a finite interval, the whole real lineR, or a positive half-lineR+. Denote byCm(I)
(m∈Z+) the set of allm-times continuously differentiable (continuous in the casem= )
functionsx:I→R; byL∞(I) we denote the space of all measurable functionsx:I→R with finite norms
x∞:=ess supx(t):t∈I.
Suppose thatXisC(I) orL∞(I),f∈C(I) is a positive non-increasing function. Forx∈X, set
xX,f:=
xf((··))
X .
For positive functionsf,g∈C(I) and naturalr, set
Lrf,g(I) :=x∈C(I) :xC(I),f<∞,x(r–)∈ACloc,x(r)L∞(I),g<∞
,
Wfr,g(I) :=x∈Lrf,g(I) :x(r)L∞(I),g≤.
In the case whenf ≡ andg≡, we writeLr
∞,∞(I) instead ofLrf,g(I).
Using above notations, the result of Theorem B can be rewritten in the following way:
xC(R
+),√fg≤x
C(R+),fx
L∞(R+),g. ()
In the case whenf(t)≡ andg(t)≡, from () one can obtain Landau’s inequality [] established in :
xC(R
+)≤x
C(R+)x
L∞(R+). ()
Similar to (), sharp inequalities for functions defined on the whole real line are also, in fact, contained in [] (see [], Section .).
Later inequalities of type () for functions defined onRandR+were generalized in many
derivatives; in [] and [] inequalities for derivatives on classes of functions defined on a finite interval are considered; in [] discrete analogues of inequalities are considered; in [] and [] one can find results connected to inequalities for fractional derivatives and further references.
We discuss some of the results for functions defined on the half-line in a more detailed way.
LetTr(t) :=cosrarccost,t∈[–, ], be Chebyshev polynomials of the first kind. Matorin in proved the following theorem (see []).
Theorem C Let k,r∈N,k<r.For arbitrary function x∈Lr
∞,∞(R+),the following inequal-ity holds:
x(k)
∞≤
Tr(k)()
[Tr(r)()]
k r
x– k r
∞ x(r)
k r
∞. ()
In the cases r= and r= ,the inequality above is sharp.
Forr> , inequality () is not sharp. Sharp inequality that estimatesx(k)C(R+) using
xC(R+) andx(r)L∞(R+) for functionsx∈Lr∞,∞(R+) was received by Schoenberg and
Cavaretta (see [, ]) in (see also [], Section .). The function
ω(δ) =ωDk,δ:= sup
x∈Wr
f,g(I),xC(I),f≤δ
x(k)C(I), δ≥, ()
is called modulus of continuity ofkth order differentiation operator on the classWr f,g(I) (k= , , . . . ,r– ), where, as before,Idenotes a finite interval, the whole real lineR, or a positive half-lineR+.
Note that in the caser= Theorem B gives an estimate for the modulus of continuity
ω(D,δ)≤δ,δ> .
The result by Schoenberg and Cavaretta gives a sharp Kolmogorov type inequality for arbitrary orders of derivativesk<r. The exact constantC(k,r) in this inequality is given implicitly in terms of a limit of a perfect splines sequence. In the case whenf =g≡, the sharp Kolmogorov type inequality is equivalent to the equalityω(Dk,δ) =C(k,r)δ–kr,δ> .
So we can think that in the case of constantf andgthis result gives the value ofω(Dk,δ) for allδ≥ (although rather implicitly). Theorem C givesω(Dk,δ),δ≥ forr≤ (with explicit constant).
Information about the connection between modulus of continuity of differentiation op-erator and Kolmogorov type inequalities and further references can be found in [], Sec-tion and Chapter .
The aim of this article is to study the functionω(Dk,δ) for arbitraryk,r∈N,k<rand non-increasing continuous positive functionsf andg.
The article is organized in the following way. In Section some auxiliary statements and in Section main statements are given. Sections and are devoted to proofs.
2 Auxiliary results
(a) the derivativeG(r)exists for allt∈(t
i,ti+),i= , , . . . ,n, wheret:=aandtn+:=b; (b) there exists∈ {, –}such that Gg(r()t()t) ≡·(–)ifort∈(t
i,ti+),i= , , . . . ,n.
Denote bynr,g[,a] the set of all perfectg-splinesGdefined on [,a] of orderrwith not more thannknots.
Belowf andgwill denote continuous positive non-increasing on [,∞) functions. The next theorem proves existence and some properties of the perfectg-splineGr,n,f,a∈
r
n,g[,a] that least deviates from zero in · C[,a],f norm.
Theorem Let numbers a> ,r∈N,n∈Z+be given.Then there exists a perfect g-spline Gr,n,f,a∈rn,g[,a]that has n+r+ oscillation points,i.e.,such that there exist n+r+
points≤t<t<· · ·<tn+r+=a such that
Gr,n,f,a(ti) = (–)i+r+Gr,n,f,aC[,a],f·f(ti), i= , , . . . ,n+r+ . ()
For a> ,setϕr,n,f(a) :=Gr,n,f,aC[,a],f.Thenϕr,n,f(a)is a continuous and non-decreasing
function of a∈(,∞).
Remark We do not prove the uniqueness of the splineGr,n,f,a∈rn,g[,a] satisfying (). However, from arguments similar to the ones used in the proof of Theorem , it follows that if two splinesG,G∈nr,g[,a] satisfy (), thenG
(k) () =G
(k)
() for allk= , . . . ,r– .
The role of perfectg-splines becomes clearer due to the following theorem.
Theorem Let r∈N,n∈Z+andδ> be such thatϕr,n,f(a) =δfor some a> .Then,for
k= , , . . . ,r– ,
ωDk,δ≤Gr(k,n),f,aC[,a].
3 Main results
If f(t)≡ andg(t)≡, thenϕr,n,f(∞) :=lima→+∞ϕr,n,f(a) =∞for allr∈N, n∈Z+. In
the case whenf,gare arbitrary positive non-increasing continuous functions, this is not always true.
Setgk(t) := tgk–(s)ds,k= , , . . . ,r, whereg:=g. The following theorem holds.
Theorem Let numbers n∈Z+and r∈Nbe given.ϕr,n,f(∞) <∞if and only if the
fol-lowing conditions hold:
A:=
∞
g(t)dt<∞,
Ak:=
∞
k–
s=
(–)k–s–A
s (k–s– )!t
k–s–+ (–)kg k(t)
dt<∞, k= , . . . ,r–
()
and
sup
t∈[,∞)
|r–
s=(–) r–s–A
s
(r–s–)! tr–s–+ (–)rgr(t)|
Remark From Theorem it follows that for allr∈N,n∈Z+,ϕr,n,f(∞) <∞if and only if
ϕr,,f(∞) <∞.
In the case when conditions () hold, set
Pk(t) := k–
s=
(–)s+A
s (k–s– )!t
k–s–+g
k(t), k= , , . . . ,r. ()
The functionsPkhave the following properties.Pkiskth primitive of the functiong, which does not change sign on [,∞) (positive for evenkand negative for oddk),k= , , . . . ,r;
Pk=Pk–,k= , . . . ,r;Pk() = (–)kAk–,k= , . . . ,r.
Ifϕr,,f(∞) =∞, then, in virtue of Theorems and , for allr∈N,n∈Z+andδ> , there
exists a numberδr,n=δr,n(δ) > such thatGr,n,f,δr,nC[,δr,n],f =δ(if such numberδr,nis not
unique, we can take the minimal value). In this case, for allδ> , the functionω(Dk,δ) is characterized by the following theorem.
Theorem Let r∈Nandϕr,,f(∞) =∞.Then,for allδ> and k= , , . . . ,r– ,
ωDk,δ= lim
n→∞G
(k)
r,n,f,δr,n().
Information about the functionω(Dk,δ) in the case whenϕr,,f(∞) <∞is given by the following theorem.
Theorem Let r∈N,n∈Z+andϕr,,f(∞) <∞.Then,for all k= , , . . . ,r– ,
ωDk,ϕr,n,f(∞)
= lim
a→∞G
(k)
r,n,f,a().
In the case whenϕr,,f(∞) <∞, information about asymptotic behavior of the function
ϕr,n,f(∞) asn→ ∞and fixedris given by the following theorem.
Theorem Let r∈Nandϕr,,f(∞) <∞.limn→∞ϕr,n,f(∞) > if and only iflimt→∞ f(t)
|Pr(t)|<
∞,where the function Pr(t)is defined in().
4 Proofs of the auxiliary results 4.1 Proof of Theorem 1
Proof of existence and uniqueness of the perfectg-splineGr,n,f,auses ideas that were used to prove Theorem .. in monograph [].
In the spaceRn+
consider the sphereSnwith radiusa,i.e.,
Sn=
ξ= (ξ,ξ, . . . ,ξn+) :
n+
i=
|ξi|=a
.
For eachξ∈Sn, consider the partition of the segment [,a] by points
ξ:= , |ξ|, |ξ|+|ξ|, . . . ,
n
i=
|ξi|, n+
i=
SetIk:= (
k–
i=|ξi|,
k
i=|ξi|),k= , , . . . ,n+ . For each of the partitions, consider the func-tion
Ga
ξ(t) :=
(r– )!
a
(t–u)r–
+ gξa(u)du,
wherega
ξ(t) =g(t)sgnξkon each segmentIk,k= , , . . . ,n+ . Then we have (Ga
ξ)(r)=gξaand henceGaξis ag-spline with knots at the points of partition.
LetQξn,+ar–(t) =ni=+r–ai(ξ)tibe the polynomial on whichinfQn+r–Gaξ–Qn+r–C[,a],fover
all polynomials of degree less than or equal ton+r– is attained. Consider the mapping
φ:Sn→Rn,φ(ξ) := (a
r(ξ), . . . ,an+r–(ξ)). From the definition ofφand properties of
poly-nomials of the best approximation, it follows thatφis continuous and odd. Hence from Borsuk’s theorem it follows that there existsξ∈Snsuch thatφ(ξ) = . This means that
the polynomialQξ,a
n+r–has order less than or equal tor– . Therefore, for the function Gr,n,f,a:=Gξ–Q
ξ,a
n+r–, we haveG (r)
r,n,f,a=gξ. Due to the generalization of Chebyshev’s
the-orem about oscillation (see, for example, [, ], Chapter , Section )Gr,n,f,ahasn+r+ oscillation points ≤t<t<· · ·<tn+r+≤aand hence at leastn+rsign changes. Thus,
in view of Rolle’s theorem,G(rr,n),f,a has at leastnsign changes (and hence exactlynsign changes due to construction). This means thatGr,n,f,ais a perfectg-spline with exactlyn nodes, in particular,Gr,n,f,a∈nr,g[,a].
Let us prove thattn+r+=a. Assume the converse. Sincef is non-increasing, we get that Gr,n,f,ahasn+rsign changes and henceGr(r,n),f,a hasn+ sign changes. However, this is impossible. Multiplying, if needed, the functionGr,n,f,aby –, we get a perfectg-spline for which equalities () hold.
The fact that for fixedr∈Nandn∈Z+the functionϕr,n,fis non-decreasing follows from its definition. The continuity of the functionϕr,n,f follows from the continuity of functions
f andg. The theorem is proved.
4.2 Proof of Theorem 2 We need the following lemma.
Lemma Let r∈N,n∈Z+,a> and x∈Lr∞,∞[,a]be given.Assume that a function x has at least n+r sign changes,x(r)has not more than n sign changes and x(r) is non-zero almost everywhere.Then,for all s= , , . . . ,r– ,
sgnx(s)() = –sgnx(s+)(). ()
Remark Notationsgnx(r)() =± means that there existsε> such thatsgnx(r)(t) =±
almost everywhere in the interval (,ε).
From conditions of the lemma it follows that the functionx(s)has exactlyn+r–ssign
changes,s= , , . . . ,r. Hence the functionx(s)changes sign on each of its monotonicity
intervals,s= , , . . . ,r– . This implies thatx(s)()= ,s= , , . . . ,r– , and that equalities
() hold. The lemma is proved.
Let us return to the proof of the theorem. Assume the converse, let a function x∈
Wr
f,g(R+) be such thatxC[,∞),f ≤δandx(k)C[,∞)>Gr(k,n),f,aC[,a). We can assume that
(if this is not true, then there exists a point t> such that |x(k)(t)|>G(rk,n),f,aC[,a)
and instead ofx(t) we can consider the functiony(t) :=x(t+t)); theny∈Wfr,g(R+) and
yC[,∞),f ≤δ; moreover, we can assume thatxC(,∞),f <δandx(r)L∞(,∞),g< ; other-wise we can consider the function ( –ε)xinstead of the functionxwithε> so small that inequality () remains true. Further, multiplying, if needed, functionsxandGr,n,f,aby –, we can suppose that
x(k)() >Gr(k,n),f,a() > . ()
Set (t) :=x(t) –Gr,n,f,a(t). Note that in view of construction forg-splines Gr,n,f,a(t), functions(t) andGr,n,f,a(t) have not less thann+rsign changes (Gr,n,f,a(t) has exactlyn+r sign changes); functions(r)(t) andG(rr,n),f,a(t) can have sign changes only in the knots of theg-splineGr,n,f,a(t), and hence not more thannsign changes. From assumptions above it follows that the function(r)(t) is non-zero almost everywhere. From Lemma and () we get
(–)kGr,n,f,a() > . ()
Due to (), (–)kG
r,n,f,a(t) > , wheret is the first oscillation point ofGr,n,f,a. This means that (–)k(t
) < . Since all sign changes of the functionare located inside the
interval (t,a), we get (–)k() < , and hence, in virtue of Lemma , we get(k)() < .
But this contradicts (). The theorem is proved.
5 Proofs of the main results 5.1 Proof of Theorem 3
We prove first that the statement of the theorem is true in the casen= . In the casen= , we writeϕr,f instead ofϕr,,f andGr,f,Minstead ofGr,,f,M. To prove the theorem, we need the following lemma.
Lemma Let conditions()hold.Suppose M> and hm(t)is the mth primitive of the
function g(t)on the interval[,M]that has m zeroes(≤m≤r).Denote byαmthe first
zero of the function hm(t).Then the following inequalities hold:
hm(t)<Pm(t), t∈[,αm]
and
Pm(t) –Pm() +hm()≤hm(t), t∈[,γm],
whereγmis the unique zero of the function Pm(t) –Pm() +hm().
We will proceed using induction onmin order to prove the first inequality.
Let m= . P(t) = –A+g(t). Since conditions () hold and P() = –A, we have P(∞) = . Since the functionh(t) =g(t) +Chas one zero,C∈(–A, ]. This means that
|h(t)|<|P(t)|for allt∈[,α].
To be definite, assume that the numberk+ is even. ThenPk+() > , and in view of
Lemma ,
hk+() > . ()
Assume the converse, let a pointt∈[,αk+) such thathk+(t)≥Pk+(t) exist. Since
the functionhk+(t) hask+ zeroes, the functionhk(k++)(t) =g(t) does not have zeroes, we
have that the functionhk+(t) haskzeroes, and hence, in virtue of induction assumption andPk+(t) =Pk(t), we get that|hk+(t)|<|Pk(t)| ∀t∈[,β], whereβis the first zero of the functionhk+(t). Hence, due to () for allt∈[,β], the following inequality holds:
≥hk+(t) >Pk(t). ()
In view of Rolle’s theorem,β>αk. This means that inequality () holds for allt∈[,αk]. Sincehk+(αk+) = <Pk+(αk+), we havePk+(t) –Pk+(αk+) <hk+(t) –hk+(αk+); on the
other hand, inequality () holds, and hence
Pk+(t) –Pk+(αk+) = – αk+
t
Pk(t)dt> –
αk+
t
hk+(t)dt
=hk+(t) –hk+(αk+).
Contradiction. The first inequality is proved.
From Lemma it follows thatsgnhm() =sgnPm(). From the proved part of the lemma it now follows that the functionPm(t) –Pm() +hm() has exactly one zero. The second inequality in the statement of the lemma can be proved using arguments similar to the ones used in the proof of the first inequality. The lemma is proved.
Let us return to the proof of the theorem in the case whenn= . Let conditions () and () hold. Set Kr :=supt∈[,∞)|
Pr(t)|
f(t) . Assume the converse, let ϕr,f(∞) =∞. This means that there existsM> such thatϕr,f(M) >Kr. Due to Lemma the inequality|Gr,f,M(t)|<|Pr(t)|holds on the interval [,αrM]. Moreover,|Pr(t)| ≤Krf(t) <
ϕr,f(M)f(t). Thus, on the interval [,αMr ], the inequality|Gr,f,M(t)|<ϕr,f(M)f(t) holds. However, in this caseGr,f,M(t) has not more thanroscillating points. Contradiction. Suf-ficiency is proved.
Let now ϕr,f(∞) <∞. This means that for all a≥,t∈[,a] we have |Gr,f,a(t)| ≤
ϕr,f(∞)f(t)≤ϕr,f(∞)f(). Passing to the limit asa→ ∞, we get existence of bounded on [,∞) primitive Qr of orderr of the function g(t). Really, for eacha> , we have
Gr,f,a(t) =gr(t) +
r–
k=ck(a)tk, whereckare real functions,k= , . . . ,r– , andgris some (fixed) primitive of orderrof the functiong. Fora,t≥, setR(a;t) :=rk–=ck(a)tk. From Markov’s inequality for polynomials we get that fork= , . . . ,r– there exists a constant
Nkindependent ofasuch that
ck(a)=R(k)(a; )≤max t∈[,]
R(k)(a;t)≤Nkmax t∈[,]
R(a;t)
≤Nk
max
t∈[,]
gr(t)+ϕr,f(∞)f()
.
limn→∞ck(an),k= , . . . ,r– . For the functionQr(t) :=gr(t) +
r–
k=cktk, fixedt≥ and all naturalnsuch thatan>t, we have
Qr(t)=
gr(t) +
r–
k=
ck(an)tk+ r–
k= cktk–
r–
k= ck(an)tk
≤gr(t) + r–
k= ck(an)tk
+ r– k=
ck–ck(an)
tk
=Gr,f,an(t)+ r– k=
ck–ck(an)
tk
.
Sincencan be arbitrarily large and all the functionsGr,n,f,a,a> are bounded on [,a], by an absolute constantϕr,f(∞)f(), we get thatQris a bounded on [,∞) primitive of order
rof the functiong.
Since functionsf(t) andg(t) are bounded, we get that all functionsQr(k)(t),k= , . . . ,r– , are also bounded on [,∞). Note that the only bounded on [,∞) primitives of the functiong(t) of orderk∈Nare functionsPk(t) +Ck, whereCk∈R, and only in the case when corresponding conditions () hold. This means that conditions () hold. Necessity of conditions () are proved.
Note that from arguments above it follows that the following lemma holds.
Lemma Let r∈N,r≥andϕr,f(∞) <∞.Then|Gr(r,f–,kM–)()| →AkandαkM+→ ∞when
M→ ∞,whereαMk+is the first zero of the function G(rr,f–,Mk–),k= , , . . . ,r– .
We will prove that condition () also holds. Iff(∞) > , then condition () holds always when conditions () hold. Below we will assume that
f(∞) = . ()
From Lemma we get
α
M r–
Pr–(t)dt
≥Gr,f,M() –Gr,f,M
αrM–
≥
γM
Pr–(t) –Pr–() +Gr,f,M()
dt,
whereγMis zero of the functionPr–(t) –Pr–() +Gr,f,M(). In view of Lemma (withk=r– ,M→ ∞) we get
αrM–
Pr–(t)dt →
∞
Pr–(t)dt
=Ar–; –Pr–() +Gr,f,M()→,
hence
γM→ ∞ and
γM
Pr–(t) –Pr–() +Gr,f,M()
From Lemma and equality () we getGr,f,M(αMr–)→ and hence
Gr,f,M()→Ar–, M→ ∞. ()
We will show thatϕr,f(∞)≥supt∈[,∞)Pfr((tt)). Assume the converse. Suppose that a point tsuch that|Pr(t)|>ϕr,f(∞)f(t) exists. We can chooseε> in such a way that
Pr(t) –εsgn
Pr()>ϕr,f(∞)f(t). ()
In virtue of () and Lemma , we can chooseM> big enough, so that
Pr(t) –εsgn
Pr()<Gr,f,M(t)<Pr(t) ∀t∈[,γ],
whereγ is zero of the functionPr(t) –εsgn[Pr()]. Since inequality () holds, we have
γ >t, and hence|Gr,f,M(t)|>ϕr,f(∞)f(t). Contradiction. Thus condition () is proved.
The theorem is proved in the case whenn= . Letnbe an arbitrary natural number now.
We will prove that for allr,n∈N,ϕr,n,f(∞) <∞if and only ifϕr,f(∞) <∞.
It is clear that ϕr,f(M) ≥ ϕr,n,f(M) for all M> , and hence ϕr,f(∞) < ∞ implies
ϕr,n,f(∞) <∞.
Assume thatϕr,n,f(∞) <∞. Denote bytnM,k thekth knot of theg-splineGr,n,f,M(t),k= , , . . . ,n. SettnM,:= ,tnM,n+:=M. Let ≤k≤n+ be the smallest number of the knots of theg-splineGr,n,f,M(t) for which the set{tnM,k:M> }is unbounded. We can choose an increasing sequence{Ml}∞l=,Ml→ ∞asl→ ∞such thattnM,sl→tn,s<∞,s≤k– and
tMl
n,k→ ∞asl→ ∞.
Denote byGKr,f,M(t) the primitive of orderrof the functiongthat least deviates from zero in norm · C[K,K+M],f. Set
ϕrK,f(M) :=GKr,f,MC[K,K+M],f.
Then, for alll,Gr,n,f,MlC[tMl
n,k–,tMln,k],f ≥
ϕrtn,f,k–(tMl
n,k–t Ml
n,k–). Passing to the limit asl→ ∞,
we getϕtn,k–
r,f (∞)≤ϕr,n,f(∞) <∞. Note that from the case whenn= proved above, it follows that for allK> ,ϕrK,f(∞) <∞if and only ifϕr,f(∞) <∞. The theorem is proved.
Remark In the case whenf ≡, for all naturalr,ϕr,f(∞) <∞impliesϕr–,f(∞) <∞. At the same time not for all functionsf,ϕr,f(∞) <∞impliesϕr–,f(∞) <∞. In particular this implies that for non-constant functionsf,
sup
x∈Wfr,g(R+),xC(R+),f≤δ
x(k)C(R +),f
may be infinite for allδ> .
Really, in the case whenf ≡, condition () holds always when conditions () hold. If
g(t) =– + (t+√)e–(t+
√
and
f(t) =e–(t+
√
) ,
then
P(t) = –
(t+√)e–(t+
√
) ,
P(t) =e– (t+√)
,
and condition () holds whenr= and does not hold whenr= .
5.2 Proof of Theorem 4
To prove Theorem , it is sufficient to prove that for allδ> there exists ag-splineGr,δ= Gr,δ(·,f,g) of orderrdefined on [,∞) with infinite number of knotsyk(k= , , . . .), =
y<y<· · ·<yk<· · ·,yk→ ∞(k→ ∞), with the following properties:
. Gr,δC[,∞),f=δand eitherGr(r,δ)≡gorG
(r)
r,δ≡–gon the intervals(yk,yk+) (k= , , , . . .).
. For allc> , the sequences{G(rk,n),f,δr,n}∞n=(k= , , . . . ,r– ) (whose elements are defined on[,c]for big enoughn) converge toGr(k,δ)uniformly on[,c].
Really, from Theorem it will follow thatω(Dk,δ) =|G(k)
r,δ()|, and from condition it
will follow that
lim
n→∞G
(k)
r,n,f,δr,n()=G (k)
r,δ().
{δr,n}∞n=is a non-decreasing sequence. Moreover, this sequence is unbounded because
otherwise we would get a perfectg-splineGwith arbitrarily close oscillating points; this is impossible because the functionsGandG(r)(and henceG) are bounded.
Denote bytn,k(k= , . . . ,n,n= , , . . .) the knots of theg-splineGr,n,f,δr,n. We can choose
a sequencens(ns→ ∞ass→ ∞) such that every sequence{tns,k}∞ns≥k(k= , , . . .) has a (finite or infinite) limit.
Let ≤y<y<· · ·be all distinct finite limits of these sequences, ordered in an
ascend-ing way. The number of the nodesykis infinite since from the statement of the theorem, we haveϕr,n,f(∞) =∞for alln∈N.
For alli∈Nand for all small enoughε> , there existsN=N(i,ε) such that for every
n>N(i,ε),Gr(r,n),f,δr,n≡gorGr(r,n),f,δr,n≡–gonIi(ε) := (yi–+ε,yi–ε). In other words, for each
i∈Nstarting with somen=N(i,ε), the restriction of theg-splineGr,n,f,δr,nto the interval Ii(ε) is a primitive of orderrofgor –g. Sinceε> is arbitrary, on each interval (yi–,yi) we get existence of point-wise limit limn→∞Gr,n,f,δr,n=:Gr,δ; moreover, on the intervals
(yi,yi+),Gr(r,δ)≡gorG
(r)
r,δ≡–g(i= , , . . .). It is clear thatGr,δC[,∞),f =δ. Using arguments similar to the ones used to prove thatlimn→∞δr,n= +∞, we can prove thatyk→ ∞(k→ ∞).
Let us fix somec> . Starting with somen, allg-splinesGr,n,f,δr,n are defined on [,c].
FromGr,n,f,δr,nC[,δr,n],f =δand the fact thatf is non-increasing (and hence is bounded) it follows that the sequence{Gr,n,f,δr,n}∞n=is uniformly bounded on [,c]; from|G
(r)
bounded) it follows that sequences{G(rk,n),f,δr,n}∞n=,k= , . . . ,r– , are uniformly bounded on [,c] and equicontinuous. The later implies uniform convergence on [,c] of the sequence
Gr,n,f,δr,ntoGr,f. The theorem is proved.
5.3 Proof of Theorem 5
Letn≥. We can choose an increasing sequence{Mk}k∞=,Mk→ ∞ask→ ∞in such a way that all sequencestMk
n,s, ≤s≤n(as above,tMn,skis thesth knot of theg-splineGr,n,f,Mk)
have limits (finite or infinite). Lettn,<· · ·<tn,mbe all distinct finite limits of these se-quences in ascending order. Analogously to the proof of Theorem , we get uniform on each segment [,c],c> convergence of the sequenceGr,n,f,Mk to theg-splinePr,n,f,{Mk}
withmknots (defined on the whole half-line) together with all derivatives up to the order
r– inclusively. For brevity we will writePr,n,f instead ofPr,n,f,{Mk}.
Let the functionx(t) be such that
xC[,∞),f≤ϕr,n,f(∞),
x(r)L∞(,∞),g≤.
()
We will show that for alls= , , . . . ,r– ,
x(s)C[,∞)≤Pr(s,n),fC[,∞). ()
Assume the converse, let for somes,x(s)
C[,∞)>Pr(s,n),fC[,∞). Then there existsε>
such thatx(s)
C[,∞)> ( +ε)P(rs,n),fC[,∞). We can suppose that|x(s)()|> ( +ε)|P(rs,)n,f()|
(if this is not true, then there exists a pointt> such that|x(s)(t)|>( +ε)Pr(s,n),fC[,∞)
and instead of the functionx(t) we can considery(t) :=x(t+t). Moreover, since the
func-tions f andg are non-increasing, conditions () and () are not broken and uniform norms of the functionxand its derivatives do not increase). Moreover, we can assume that
x(s)() > ( +ε)Pr(s,n),f() > ()
(if this is not true, we can multiplyxand (or)Pr,n,fby –). Setk(t) :=x(t)–(+ε)Pr,n,f,Mk(t), t∈[,Mk]. We can choosekso big that
x(s)() > ( +ε)Pr(s,n),f,M
k() ()
and
( +ε)ϕr,n,f(Mk) >ϕr,n,f(∞). ()
From Lemma we get
(–)sPr,n,f,Mk() > . ()
In view of () and () (–)s
k() < , and hence due to Lemma we get(ks)() < . However, this contradicts ().
In virtue of property () proved above, the limitlimMk→∞|G (k)
r,n,f,Mk()|does not depend
5.4 Proof of Theorem 6
We will need the following lemmas.
Lemma Suppose thatϕr,n,f(∞) <∞and Qr,n∈Lr∞,∞(R+)is a g-spline of rth order de-fined on half-line[,∞)with n∈Z+knots =:t<t<· · ·<tn.Then there exists∈ {–, }
such that for s= , , . . . ,r,
Qr(s,)n(t) =Pr(s)(t), t≥tn. ()
We will prove the lemma using induction. In the cases=r, equality () holds. Let it be true fors=k≥. We will prove that it is true fors=k– too. In view of the induction assumptionQ(rk,n)(t) =Pr(k)(t),t≥tn. Moreover,Qr(k,n–)(∞) =Pr(k–)(∞) = . Then, fort≥tn,
–Q(rk,n–)(t) =
∞
t
Q(rk,n)(s)ds=
∞
t
Pr(k)(s)ds= –P(rk–)(t).
The lemma is proved.
Lemma Letϕr,n,f(∞) <∞andlimt→∞Pfr(t(t))=∞.Then the number of oscillation points of the g-spline Pr,n,f tends to infinity as n→ ∞.
Let somen∈Nbe fixed. Suppose thatM> is such that for allt>M, Pfr((tt)) >
ϕr,n,f(∞).
Let theg-splinePr,n,f havekoscillating points ≤a<a<· · ·<ak. Denote by ≤b< b<· · ·<bn+r+all oscillation points of theg-splineGr,n,f,K, whereKis chosen so big that
sgnGr,n,f,K(bs) =sgnPr,n,f(bs), s= , , . . . ,k,bk+>max{ak,M} andϕr,n,f(K) > ϕr,n,f(∞). Chooseε> so small that ( –ε)ϕr,n,f(K) >ϕr,n,f(∞).
Set(t) :=Pr,n,f(t) – ( –ε)Gr,n,f,K(t). Thensgn(as) =sgnPr,n,f(as),s= , , . . . ,k, since
Pr,n,f(as)=ϕr,n,f(∞)f(as) > ( –ε)ϕr,n,f(K)f(as)≥( –ε)Gr,n,f,K(as).
Hence the function(t) hask– sign changes on the interval [,ak]. Moreover, fors=
k+ , . . . ,n+r+ ,
Pr,n,f(bs)<
ϕr,n,f(∞)
f(bs) < ( –ε)ϕr,n,f(K)f(bs) = ( –ε)Gr,n,f,K(bs).
This means thatsgn(bs) =sgnGr,n,f,K(bs),s=k+ , . . . ,n+r+ . Hence the function(t) hasn+r–ksign changes on the interval [bk+,a], and hence at leastn+r– sign changes
on the whole interval [,K]. This means that the function(r)(t) has at leastn– sign
changes, and hence theg-splinePr,n,f has at leastn– knots. LettK
n,s,s= , . . . ,n, be the knots of theg-splineGr,n,f,K. Note that for alls= , , . . . ,n, the
g-splineGr,n,f,K has at leastsoscillating points on the interval [,tkn,s]. Really, assume the converse, suppose for some ≤s≤nthat theg-splineGr,n,f,Khas less thansoscillation points on the interval [,tK
n,s]. Then theg-splineGr,n,f,Khas more thann+r+–soscillation points on the interval (tK
n,s,K], and hence more thann+r–ssign changes. This means that the functionG(rr,n),f,Khas more thann–ssign changes on the interval (tK
n,s,K]. However, this is impossible.
Lemma For all s= , , . . . ,r and all t≥ (almost everywhere is the case when s=r),the following inequality holds:
Pr(s,n),f(t)≤P(rs)(t). ()
We will prove the statement of the lemma using induction. In the cases=r, inequality () holds with equality sign. Let inequality () hold withs=k≥. We will prove that it is true fors=k– .
Assume the converse. Let
T:=t∈[,∞) :P(rk,n–),f (t)>Pr(k–)(t)=∅.
Denote by <t<· · ·<tlall knots of theg-splinePr,n,f. Then, due to Lemma ,T⊂[,tl) and
P(k–)
r,n,f (tl)=P(rk–)(tl). ()
Leta∈T. Then
Pr(k–)(tl) –P(rk–)(a)=
atlPr(k)(t)dt=
tl
a
Pr(k)(t)dt
≥
tl
a
P(rk,n),f(t)dt≥
tl
a
P(rk,n),f(t)dt
=P(rk,n–),f (tl) –Pr(k,n–),f (a);
this is impossible in virtue of (), and the facts thatsgnP(rk–)(tl) =sgnP(rk–)(a), function |P(rk–)|is non-increasing anda∈T. Contradiction. HenceT=∅and the lemma is proved.
Let us return to the proof of the theorem. Let
lim
t→∞
f(t) |Pr(t)|
=:c<∞. ()
Then, in view of Theorem ,c> . We will show thatϕr,n,f(∞)≥c ∀n. Assume the con-verse, let a numbernsuch that
ϕr,n,f(∞) <
c ()
exist. Denote by <t<t<· · ·<tkall knots ofPr,n,f. Then, due to Lemma ,Pr,n,f(t) =
±Pr(t),t≥tk. In virtue of () we have
Pr,n,f(t)< f(t)
c
for allt≥.f(∞) = since () holds. ThenPr,n,f(∞) = , and hence
t≥tk. But then|Pr(t)|<f(tc),i.e.,
f(t) |Pr(t)|
> c, t≥tn,
which contradicts (). Sufficiency is proved.
We will prove the necessity now. limn→∞ϕr,n,f(∞) =δ > . Assume the converse, let
limt→∞|Pfr((tt))|=∞. Then there exists a numberM> such that for allt>Mthe inequality
f(t) |Pr(t)|
>
δ
holds; this is equivalent to
Pr(t)<δf(t).
In view of Lemma , for alln, the following inequality holds:
M
Pr,n,f(t)dt≤
M
Pr–(t)dt. ()
Choosenso big that
n·f(M) >
M
Pr–(t)dt. ()
Choosemsuch that theg-splinePr,m,f(t) has at leastn+ oscillation points (this is possible due to Lemma ). Denote by ≤a<a<· · ·<an+ the first oscillation points of the g-splinePr,m,f(t). Then, in virtue of (), and by the facts thatfis non-increasing function,
M
Pr,m,f = M|Pr,m,f(t)|dtand (), we get thatan>M. Thusan+>an>Mare oscillation points of theg-splinePr,m,f(t), and we get
Pr,m,f(an+) –Pr,m,f(an)=ϕr,m,f(∞)·
f(an+) +f(an)
≥δ·f(an+) +f(an)
>δf(an)
>Pr(an)=
∞
an
Pr–(t)dt.
On the other hand, in view of Lemma ,
Pr,m,f(an+) –Pr,m,f(an)≤
an+
an
Pr,m,f(t)dt
≤
an+
an
Pr–(t)dt
<
∞
an
Pr–(t)dt.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors have made an equal contribution to the paper, have read and approved the final manuscript.
Acknowledgements
The authors would like to thank the referees for their valuable remarks and suggestions.
Received: 19 December 2014 Accepted: 16 September 2015
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