# On the Favard-Type Theorem and the Jackson-Type Theorem (II)

Authors: Hee Sun Jung; Ryozi Sakai; Noriaki Suzuki
Journal: ISRN Applied Mathematics
 (2011)

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## Abstract

LetR=(-∞,∞), and let Q∈ℂ1:ℝ→[0,∞) be an even function. We consider the exponential weights w(x)=e-Q(x), x∈ℝ. In this paper we investigate the relations between the Favard-type inequality and the Jackson-type inequality. We also discuss the equivalence of two K-functionals and the modulus of smoothness.


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## Body

## 1. Introduction and Preliminaries

Letℝ=(-∞,∞), and let Q∈ℂ1:ℝ→[0,∞) be an even function. We consider the weights w(x):=exp⁡(-Q(x)) satisfying for all n=0,1,2,…,(1.1)∫0∞xnw(x)dx<∞.
First we need the following definition from [1]. We say that f:ℝ→[0,∞) is quasi-increasing if there exists C>0 such that f(x)⩽Cf(y), 0<x<y.Definition 1.1.
One defines,w=exp⁡(-Q)∈ℱ(C2+)as follows: LetQ:ℝ→[0,∞)be continuous and an even function, and satisfy the following properties:(a)
Q′(x)is continuous inℝ, withQ(0)=0;(b)
Q′′(x)exists and is positive inℝ∖{0};(c)
(1.2)lim⁡x→∞Q(x)=∞;(d)
the function(1.3)T(x):=xQ′(x)Q(x),x≠0
is quasi-increasing in (0,∞) with T(x)⩾Λ>1, x∈ℝ∖{0};(e)
there existsC1>0 such that
(1.4)Q′′(x)|Q′(x)|⩽C1|Q′(x)|Q(x),a.e.x∈R∖{0}.
Moreover, there also exists a compact subinterval J(∋0) of ℝ, and C2>0 such that
(1.5)Q′′(x)|Q′(x)|⩾C2|Q′(x)|Q(x),a.e.x∈R∖J.Example 1.2.
Letw=exp⁡(-Q)∈ℱ(C2+).(1)
IfT(x) is bounded, then we call the weight w=exp⁡(-Q) the Freud-type weight. The following example is the Freud-type weight:
(1.6)w(x)=exp⁡(-|x|α),α>1.(2)
IfT(x) is unbounded, then we call the weight w=exp⁡(-Q) the Erdös-type weight. The following examples give the Erdős-type weight w=exp⁡(-Q).
(i)
([1, Example  1.2], [2, Theorem  3.1]). For α>1, ℓ=1,2,3,…,(1.7)Q(x)=Ql,α(x)=exp⁡l(|x|α)-exp⁡l(0),whereexp⁡ℓ(x)=exp⁡(exp⁡(exp⁡…exp⁡x)…),(ℓtimes). More generally, we define for α+u>1, α⩾0, u⩾0, and l⩾1,
(1.8)Ql,α,u(x):=|x|u(exp⁡l(|x|α)-α*exp⁡l(0)),whereα*=0 if α=0, otherwise α*=1.(ii)
Forα>1, Q(x)=Qα(x):=(1+|x|)|x|α-1.We need the Mhaskar-Rakhmanov-Saff numbersax defined by (1.9)x=2π∫01axuQ′(axu)(1-u2)1/2du,x>0.
If f:ℝ→ℝ is measurable, we define (1.10)‖fw‖Lp(R):={(∫-∞∞|f(t)w(t)|pdt)1/p,if0<p<∞,esssup⁡t∈R|f(t)w(t)|,ifp=∞,
where if p=∞, then we suppose that f is continuous on ℝ and (1.11)lim⁡|x|→∞|f(x)w(x)|=0.
The class of all functions f for which ∥fw∥Lp(ℝ)<∞ will be denoted by Lp,w(ℝ), with the usual understanding that two functions are identified if they are equal almost everywhere. For f∈Lp,w(ℝ)(0<p⩽∞), the degree of weighted polynomial approximation is defined by (1.12)Ep,n(w;f):=inf⁡P∈Pn‖w(f-P)‖Lp(R),
where 𝒫n denotes the class of all polynomials of degree at most n. There are various estimates for this degree. Among them, in our previous article [3], we discuss the relation between the Favard-type inequality, the Jackson-type inequality, and the estimates by two K-functionals 𝒦r,p and 𝒦̃r,p. We recall and summarize them in Section 2. In Section 3, we will discuss the equivalence of 𝒦̃r,p and modulus of smoothness ωr,p. As the result, we see 𝒦r,p~𝒦̃r,p for a weight in a subclass of ℱ(C2+). For any nonzero real valued functions f(x) and g(x), we write f(x)~g(x) if there exists a constant C>0 independent of x such that for all x(1.13)(1C)f(x)⩽g(x)⩽Cf(x).
Similarly, for any positive numbers {cn}n=1∞ and {dn}n=1∞ we define cn~dn. Throughout this paper C,C1,…,c,c1,… denote positive constants independent of n,x,t, or Pn(x). The same symbols do not necessarily denote the same constants occurrences.


## 2. Known Results and Summarization

Letr⩾1 be an integer (in this paper, we suppose that r (or s) is an integer). The rth order K-functional of a function f∈Lp,w(ℝ) is defined by the formula (2.1)Kr,p(w,f,δ):=inf⁡{‖w(f-g)‖Lp(R)+δr‖wg(r)‖Lp(R)},
for δ>0, where the infimum is taken over all r-1 times continuously differentiable g such that g(r-1) is absolutely continuous, and g(r)∈Lp,w(ℝ) (this means wg(r)∈Lp(ℝ)). Using this rth order K-functional, we can estimate the order of Ep,n(w;f).Theorem 2.1 (see [3]).
Letw=exp⁡(-Q)∈ℱ(C2+). Let  r⩾1,  s⩾0 be integers, and let  1⩽p⩽∞. Let  f be  s-1 times continuously differentiable,  f(s-1) be absolutely continuous on each compact interval, and  f(s)∈Lp,w(ℝ) (when  s=0, these assumptions state merely that  f∈Lp,w(ℝ)). Then, for every integer  n⩾r+s,
(2.2)Ep,n(w;f)⩽C(ann)sKr,p(w,f(s),ann).Theorem2.1 was shown by using the following Favard-type inequalities (see [3]).Theorem 2.2 ([4, Corollary 8]).
Letw=exp⁡(-Q)∈ℱ(C2+). Let  f be  s-1 times continuously differentiable, and let  f(s-1) for some integer  s⩾1 be absolutely continuous on each compact interval. Let  1⩽p⩽∞ and  f(s)∈Lp,w(ℝ). Then one has
(2.3)Ep,n(w;f)⩽C(ann)s‖wf(s)‖Lp(R),
equivalently,
(2.4)Ep,n(w;f)⩽C(ann)sEp,n-s(w;f(s)).Remark 2.3.
(1)
([4, Remark 11]) Let w∈ℱ(C2+) and let 0<p<1. Then there exists a constant C0 such that for every absolutely continuous function f with f′∈L∞,w and w′f∈Lp(ℝ), and for every n∈ℕ, we have
(2.5)Ep,n(w;f)⩽C0(ann){‖wf′‖L∞(R)+‖w′f‖Lp(R)}.
(2)
([3, Section 4]) As a by-product of the method of the proof for Theorem 2.2, we can obtain the degree of functions which satisfy the Hölder-Lipschitz condition. Let w∈ℱ(C2+),1⩽p⩽∞ and 1/p⩽β⩽1. Let g be absolutely continuous with |g′|β∈Lp,w(ℝ) (and for p=∞, we require g to be continuous, and gw to vanish at ±∞). Let f∈Lp,w(ℝ) and set
(2.6)fg,β*(x):=sup⁡y∈R|f(x)-f(y)||g(x)-g(y)|β.
Then we define
(2.7)Fg,β(w,p):={f∣f∈Lp,w(R),‖fg,β*‖L∞(R)<∞}.Now, we have the following theorem:Theorem see ([3, Theorem 4.2]).
Let1⩽p⩽∞,  1/p⩽β⩽1, and let  f∈Fg,β(w,p). Then one has
(2.8)Ep,n(w;f)⩽C(ann)β‖w|g′|β‖Lp(R)‖fg,β*‖L∞(R).We define the following class of weights from [5, Definition  1.1].Definition 2.5.
Letw(x):=exp⁡(-Q(x)), whereQ:ℝ→ℝis even, continuous, andQ′is positive in(0,∞). Then one writesw=exp⁡(-Q)∈ℰ1, if the following are satisfied:(a)
xQ′(x)is strictly increasing in(0,∞)withlim⁡x→0+xQ′(x)=0;(b)
the function(2.9)T(x):=xQ′(x)Q(x)is quasi-increasing in(C,∞)for someC>0andlim⁡x→∞T(x)=∞;(c)
assume(2.10)yQ′(y)xQ′(x)⩽C1(Q(y)Q(x))C3,y⩾x⩾C2,for some positive constantsC1,C2,
andC3.Remark 2.6.
Letw(x):=exp⁡(-Q(x))∈ℱ(C2+) and T(x) is unbounded, then we see w∈ℰ1. In fact, we see this as follows.
From Definition1.1(e) and (d), we have for y⩾x>0,
(2.11)Q′(y)Q′(x)=exp⁡(∫xyQ′′(t)Q′(t)dt),⩽exp⁡(C1∫xyQ′(t)Q(t)dt)=(Q(y)Q(x))C1,yx=exp⁡(∫xy1tdt)⩽exp⁡(1Λ∫xyQ′(t)Q(t)dt)=(Q(y)Q(x))1/Λ.
Therefore we obtain (c) in Definition 2.5 for C3=C1+1/Λ.Iff:ℝ→ℝ, and h>0, then we define the differences of f inductively by the formula (2.12)Δh0f(x)=f(x),Δh1f(x)=Δhf(x)=f(x+h2)-f(x-h2),Δhkf(x)=Δhk-1(Δhf)(x)=∑j=0k(-1)j(kj)f(x+kh2-jh),k=2,3,..,x∈R.
We set (2.13)σ(t):=inf⁡{au:auu⩽t},t>0,Φt(x):=|1-|x|σ(t)|+T(σ(t))-1/2,x∈R.
By [6], when w=exp⁡(-Q) is the Erdős-type weight we define for f∈Lp,w(ℝ),0<p⩽∞, (2.14)ωr,p(w,f,t):=sup⁡0<h⩽t‖wΔhΦt(x)r(f)‖Lp(|x|⩽σ(2t))+inf⁡R∈Pr-1‖w(x)(f-R)(x)‖Lp(|x|⩾σ(4t)).
If w=exp⁡(-Q) is the Freud-type weights, then we define (2.15)ωr,p(w,f,t):=sup⁡0<h⩽t‖wΔhr(f)‖Lp(|x|⩽σ(h))+inf⁡R∈Pr-1‖w(x)(f-R)(x)‖Lp(|x|⩾σ(t)).The following Jackson-type inequality is known.Theorem 2.7 (see [5, Theorem  1.2], [6, Corollary  1.4]).
Letw∈ℰ1. Let  0<p⩽∞. Then for  f:ℝ→ℝ for which  f∈Lp,w(ℝ) (and for  p=∞, we require  f to be continuous, and  fw to vanish at  ±∞), one has for  n⩾C3,
(2.16)Ep,n(f;w)⩽C1ωr,p(w,f,C2ann),
where Cj, j=1,2,3, do not depend on f and n.We also consider the following class of weights which are called the Freud weights.Definition 2.8 ([7, Definition 3.3]).
Letw=exp⁡(-Q), whereQ:ℝ→ℝis even, andQ′exists and is positive on(0,∞). Moreover, assume thatxQ′(x)is strictly increasing, with right limit0at0, and for someλ,A,B>1,C>0,(2.17)A⩽Q′(λx)Q′(x)⩽B,x⩾C.
Then we call w Freud weight, and write w∈ℱ*.Remark 2.9.
Letw=exp⁡(-Q)∈ℱ(C2+). Then we see the following.
(1)
IfT(x)=(xQ′(x))/Q(x) is bounded, then we say that w=exp⁡(-Q) is the Freud-type weight, and we write w∈ℱ. Then we see ℱ*∩ℱ(C2+)=ℱ. In fact, when w∈ℱ, 1<ν and x>0 large enough, by Definition 1.1(e),
(2.18)Q′(νx)Q′(x)=exp⁡(∫xνxQ′′(u)Q′(u)du)⩽exp⁡(C1∫xνxQ′(u)Q(u)du),
and then since T(x) is bounded, there exists C>0 such that
(2.19)Q′(νx)Q′(x)⩽exp⁡(C1C∫xνx1udu)=νC1C.
Similarly,
(2.20)Q′(νx)Q′(x)=exp⁡(∫xνxQ′′(u)Q′(u)du)⩾exp⁡(C2∫xνxQ′(u)Q(u)du)⩾exp⁡(C2Λ∫xνx1udu)=νC2Λ.
Therefore, if we take λ:=ν>1 large enough, then we have (2.17).
Conversely, to showℱ*∩ℱ(C2+)⊂ℱ we suppose that there exists w=exp⁡(-Q)∈ℱ(C2+) such that T(x)=(xQ′(x))/Q(x) is unbounded. Then since T(x) is quasi-increasing in (0,∞), we see that T(x)→∞ as x→∞. So, for any M>0 there exists L>0 (large enough) such that T(x)>M for x>L. Therefore, we have for x>L and any λ>1,
(2.21)Q′(λx)Q′(x)=exp⁡(∫xλxQ′′(u)Q′(u)du)⩾exp⁡(C3∫xλxQ′(u)Q(u)du)=exp⁡(C3∫xλxT(u)udu)=λC3M,
where C3>0 is a constant, that is, (1) does not hold. Hence we have ℱ*∩ℱ(C2+)⊂ℱ. Consequently, we have ℱ=ℱ*∩ℱ(C2+).
(2)
IfT(x)=(xQ′(x))/Q(x) is bounded, then for x⩾1, there exists c,C>0 such that
(2.22)xΛ⩽Q(x)⩽Cxc.Theorem 2.10  2.10 ([7, Theorem 3.5]).
Letw∈ℱ*. Let  0<p⩽∞. Then for  f:ℝ→ℝ for which  f∈Lp,w(ℝ) (and for  p=∞, one requires  f to be continuous, and  fw to vanish at  ±∞), one has for  n⩾C3,
(2.23)Ep,n(f;w)⩽C1ωr,p(w,f,C2ann),
where Cj, j=1,2,3, do not depend on f and n.Damelin [6] introduces the following K-functional: let f∈Lp,w(ℝ), 0<p⩽∞ and r⩾1 be an integer, then we define (2.24)K̃r,p(w,f,δ):=inf⁡P∈Pn{‖w(f-P)‖Lp(R)+δr‖wP(r)‖Lp(R)},
where δ>0 are chosen in advance and (2.25)n=n(δ):=inf⁡{k:akk⩽δ}.Then Damelin gives the following.Theorem 2.11 ([6, Theorem 1.3 (b)]).
Letw∈ℰ1,  r⩾1,  0<p⩽∞, and let  fw∈Lp(ℝ) (and for  p=∞, one requires  f to be continuous, and  fw to vanish at  ±∞). Then one has
(2.26)ωr,p(w,f,t)~K̃r,p(w,f,δ).For the Freud-type weights we have also the followings.Theorem 2.12 ([7, Theorem 3.9, 3.10]).
Letw∈ℱ*,  r⩾1, 0<p⩽∞, and let  fw∈Lp(ℝ) (and for  p=∞, we require  f to be continuous, and  fw to vanish at  ±∞). Then we have
(2.27)Kr,p(w,f,δ)~ωr,p(w,f,t)~K̃r,p(w,f,δ).Forw∈ℱ(C2+), we see easily that (2.28)Kr,p(w,f,δ)⩽K̃r,p(w,f,δ).So, from Theorem2.1 we obtain the following corollary.Corollary 2.13.
Letw=exp⁡(-Q)∈ℱ(C2+). Let  r⩾1,s⩾0 be integers, and let  1⩽p⩽∞. Let  f be  s-1 times continuously differentiable,  f(s-1) be absolutely continuous, and  f(s)∈Lp,w(ℝ) (when  s=0, these assumptions state merely that  f∈Lp,w(ℝ)). Then, for every integer  n⩾r+s,
(2.29)Ep,n(w;f)⩽C(ann)sK̃r,p(w,f(s),ann).The main theme in [3] is to summarize the above theorems. Let w∈ℱ(C2+),r⩾1 be an integer, and let 1⩽p⩽∞. We have the following succession of the theorems. We use the constant Ci>0,i=1,2,3,… which do not depend on f and n.(a)
[Theorem2.7 with r=1 (the Erdős-type case)], [Theorem 2.10 with r=1 (the Freud case)]:letf:ℝ→ℝ and if 1⩽p<∞, assume that fw∈Lp(ℝ). If p=∞, assume in addition that f is continuous and that fw has limit 0 at ±∞. Then we have
(2.30)Ep,n(w;f)⩽C1ω1,p(w,f,C2ann).(b)
[Theorem2.2]:letf be s-1 times continuously differentiable, and let for some integer s⩾1, f(s-1)(x) be absolutely continuous on each compact interval. Let 1⩽p⩽∞ and f(s)∈Lp,w(ℝ). Then we have
(2.31)Ep,n(w;f)⩽C3(ann)s‖wf(s)‖Lp(R),
equivalently,
(2.32)Ep,n(w;f)⩽C4(ann)sEp,n(w;f(s)).(c)
[Theorem2.1]:lets⩾0 be an integer, and let f be s-1 times continuously differentiable, f(s-1) be absolutely continuous, and f(s)∈Lp,w(ℝ) (when s=0, these assumptions state merely that f∈Lp,w(ℝ)). Then, for every integer r⩾1 and n⩾r+s,(2.33)Ep,n(w;f)⩽C5(ann)sKr,p(w,f(s),ann).(d)
[Theorem2.7, 2.11 (the Erdős-type case)], [Theorem 2.10, 2.12 (the Freud case)]:forf:ℝ→ℝ with f∈Lp,w(ℝ) (and for p=∞ we require f to be continuous, and fw to vanish at ±∞), for every n large enough and for every integer r⩾1 we have(2.34)Ep,n(w;f)⩽C6K̃r,p(w,f,ann).Consequently we find an interesting fact as follows:(b) is shown by the use of (a) (see [4]). Using (b), we proved (c). If we use [6, Theorem 1.3 (b)] and [7, Theorems 3.9 and 3.10], we see easily that (c) means (d) with Theorem 2.11. Trivially, we have (a) from (d).


## 3. Equivalence ofK-Functional and Modulus of Smoothness

In this section we consider a subclass ofℱ(C2+).Assumption 3.1.
Letr⩾2 be an integer. Let w(x)=exp⁡(-Q(x))∈ℱ(C2+). Then we suppose that Q∈Cr-1(ℝ), Q(r)(x) exists in ℝ∖{0}, furthermore, the following inequalities hold
(3.1)|Q(j)(x)|⩽Cj|Q(j-1)(x)||Q′(x)Q(x)|,j=1,2,…,r,
where for j=r we suppose that (3.1) hold almost everywhere on x∈ℝ∖{0}. Then we write w∈ℱ(Cr+).Remark 3.2.
Example1.2, (i), (ii) satisfy Assumption 3.1, moreover Q(x)=x2m, m: a positive integer, satisfies Assumption 3.1.In Section2, we know (3.2)Kr,p(w,f,t)⩽K̃r,p(w,f,t)~ωr,p(w,f,t).
The equivalences mentioned in the last of Section 2 give a certain suggestion, that is, (3.3)Kr,p(w,f,t)~ωr,p(w,f,t).
In fact, this is true.Theorem 3.3.
Letr⩾2 be a positive integer, and let  w∈ℱ(Cr+). Let  f∈Lp,w(ℝ). Then one has (3.3). Similarly, when  w∈ℱ(C2+) and  r=1, one has (3.3).To prove Theorem3.3, we need some lemmas.Lemma 3.4.
Letr⩾1 be an integer. If  f(r-1)(x) is absolutely continuous on  ℝ, then for  k=1,2,…,r, one has the following representation:
(3.4)ΔΦt(x)hkf(x)=(Φt(x))k{∫-h/2h/2}kf(k)(x+Φt(x)(u1+u2+⋯+uk))du1du2⋯duk.Proof.
(3.5)∫h/2h/2f(1)(x+Φt(x)u1)du1=(Φt(x))-1∫x-Φt(x)h/2x+Φt(x)h/2f(1)(v1)dv1=(Φt(x))-1∆Φt(x)hf(x).
Therefore, for r=1 we have (3.4). For some k⩾1 we suppose (3.4). Then we have
(3.6){∫-h/2h/2}k+1f(k+1)(u)du1du2⋯duk+1=∫-h/2h/2[{∫-h/2h/2}k(f′)(k)(ũ)du1du2⋯duk]duk+1=(Φt(x))-k∫-h/2h/2ΔΦt(x)hkf′(x+Φt(x)uk+1)duk+1=(Φt(x))-(k+1)ΔΦt(x)hk(ΔΦt(x)hf(x))=(Φt(x))-(k+1)ΔΦt(x)hk+1f(x),
where
(3.7)u=u(x;u1,…,uk+1)=x+Φt(x)(u1+u2+⋯+uk+1),ũ=ũ(x;u1,…,uk+1)=x+Φt(x)uk+1+Φt(x)(u1+u2+⋯+uk).
Hence we have (3.4) for r=1,2,3,….Lemma 3.5 (see [1, Lemma  3.4 (3.17)]).
Uniformly fort>0, one has
(3.8)Q′(at)~tT(at)at.Lemma 3.6.
Letw∈ℱ(Cr+) and  r⩾2 be an integer. Then for any integer  k,  1⩽k⩽r-1, there exist  c1,c2>0 and  A>0 such that for  |x|⩾A,
(3.9)c1⩽(-1)kw(k)(x)(1+Q′(x)2)k/2w(x)⩽c2.
Also, there exist c3,c4>0 and B>0 such that for |x|⩾B,
(3.10)c3⩽(-1)k(w-1)(k)(x)(1+Q′(x)2)k/2w-1(x)⩽c4.
Furthermore, for k=r, (3.9) and (3.10) hold almost everywhere on |x|⩾A and |x|⩾B, respectively. When w∈ℱ(C2+) and r=1, one also has (3.9) and (3.10).Proof.
First, we will see that forμ=±1 the following equations hold:
(3.11)lim⁡|x|→(wμ)(k)(x)Q′(x)kwμ(x)=(-μ)k,k=1,…,r-1.
For k=r we take Q̃(x) as follows;
(3.12)Q̃(j)(x)=Q(j)(x),x∈R,j=1,2,…,r-1,Q̃(r)(x)=Q(r)(x),a.e.x∈R∖{0},
and for all j=1,2,…,r, Q̃(x) satisfies (3.1) for all x≠0. Then we obtain that exchanging Q with Q̃, (3.11) also holds.
Letμ=1, and let 1⩽k⩽r-1.
(3.13)w′(x)=-Q′(x)w(x),w′′(x)=(-Q′′(x)+Q′(x)2)w(x),w′′′(x)=(-Q′′′(x)+3Q′(x)Q′′(x)-Q′(x)3)w(x),
and we continue this manner, so
(3.14)w(k)(x)=(∑i1+2i2+⋯+kik=k,i1<kci1,i2,...,ikQ′(x)i1Q′′(x)i2⋯Q(k)(x)ik+(-1)kQ′(x)k)w(x),
where ci1,i2,...,ik are coefficients. Here, from (3.1), for |x|⩾A>0 large enough and ij≠0, 2⩽j⩽k,(3.15)|Q(j)(x)|ij⩽Cj(|Q′(x)Q(x)|j-1|Q′(x)|)ij=Cj(|Q′(x)|jQ(x)j-1)ij.
Hence, from (3.14) and (3.15) we have
(3.16)|lim⁡|x|→∞w(k)(x)Q′(x)kw(x)-(-1)k|⩽Clim⁡|x|→∞1Q(x)=0,k=1,…,r-1,
where C is a positive constant. Therefore, we have (3.11) with μ=1. Similarly, for μ=-1,
(3.17)lim⁡|x|→∞(w-1)(k)(x)Q′(x)kw-1(x)=1,k=1,…,r-1.
Therefore, we have (3.11) for k=1,2,…,r-1, and hence we also have (3.9) and (3.10) for |x| large enough. If in (3.11), we replace Q with Q̃, then repeating the above proof we also obtain (3.11), so for k=r we conclude (3.9) and (3.10) with a.e.x large enough, that is,
(3.18)c1⩽(-1)rw(r)(x)(1+Q′(x)2)r/2w(x)⩽c2,a.e.|x|⩾A,c3⩽(-1)r(w-1)(r)(x)(1+Q′(x)2)r/2w-1(x)⩽c4,a.e.|x|⩾B.
For r=1, the lemma is trivial.Lemma 3.7.
Letr⩾1 be an integer. There exists  C>0 such that for every integer  k=1,2,…,r,
(3.19)|Q′(x)|kw(x)⟶0,as|x|⟶∞,(3.20)w-1(x)(1+Q′(x)2)(k-1)/2∫|x|∞(1+Q′(t)2)k/2w(t)dt⩽C,x∈R,
and for x∈ℝ,
(3.21)(1+Q′(x)2)k/2w(x)∫0|x|(1+Q′(t)2)-(k-1)/2w-1(t)dt⩽C,x∈R.Proof.
From Definition1.1(e) we see that there exist C>0 and λ>0 such that
(3.22)|Q′(x)|⩽CQ(x)λ,x∈R,
so we have
(3.23)|Q′(x)|kw(x)⩽CkQ(x)λkw(x)⟶0,as|x|⟶∞.
Hence, we have (3.19). From (3.9) with k-1, we see that for |x|⩾A>0 and constants c1,c2,c3>0,
(3.24)w-1(x)(1+Q′(x)2)(k-1)/2∫|x|∞(1+Q′(t)2)k/2w(t)dt⩽c1w-1(x)(1+Q′(x)2)(k-1)/2∫|x|∞(-1)kw(k)(t)dt⩽c2w-1(x)(1+Q′(x)2)(k-1)/2(-1)k-1w(k-1)(x)⩽c3.
Especially, if we use (3.24) with |x|=A, then we have
(3.25)w-1(A)(1+Q′(A)2)(k-1)/2∫A∞(1+Q′(t)2)k/2w(t)dt⩽c3.
Then, for |x|⩽A there exists a constant C=C(A) such that
(3.26)w-1(x)(1+Q′(x)2)(k-1)/2⩽Cw-1(A)(1+Q′(A)2)(k-1)/2.
In fact, for |x|⩽A we have
(3.27)1+Q′(A)2⩽(1+Q′(A)2)(1+Q′(x)2),
that is,
(3.28)(1+Q′(A)2)(k-1)/2⩽C(A)(1+Q′(x)2)(k-1)/2,
where C(A)=(1+Q′(A)2)(k-1)/2. Hence we have (3.26).
Now, by (3.26), if |x|⩽A, then we have
(3.29)w-1(x)(1+Q′(x)2)(k-1)/2∫A∞(1+Q′(t)2)k/2w(t)dt⩽c4(A).
When |x|<A, we see easily that
(3.30)w-1(x)(1+Q′(x)2)(k-1)/2∫|x|A(1+Q′(t)2)k/2w(t)dt⩽c5=c5(A).
Hence, with (3.29) we have for |x|⩽A,
(3.31)w-1(x)(1+Q′(x)2)(k-1)/2∫|x|∞(1+Q′(t)2)k/2w(t)dt⩽c6(A).
Consequently, from (3.24) and (3.31) we have (3.20). We need to show (3.21). For |x| large enough, we see
(3.32)Q′′(x)~Q′(x)2Q(x),a.e.x∈R,
(see Definition 1.1(e)) so we can select |x|⩾A>0 large enough such that
(3.33)Q′′(x)<12kQ′(x)2,a.e.x∈R.
We show (3.21) for ∫A|x|, |x|⩾A>0. For |x|⩾A, we have by (3.33),
(3.34)I:=∫A|x|w-1(t)Q′(t)k-1dt=∫A|x|Q′(t)w-1(t)Q′(t)kdt⩽w-1(x)Q′(|x|)k+k∫A|x|w-1(t)Q′′(t)Q′(t)k+1dt⩽w-1(x)Q′(|x|)k+12I.
Hence, we have for a certain constant c7>0,
(3.35)(1+Q′(x)2)k/2w(x)∫A|x|w-1(t)(1+Q′(t)2)(k-1)/2dt⩽c7.
For ∫0A we have (3.21) by the use of (3.19). In fact, there exists c8=c8(A) such that
(3.36)(1+Q′(x)2)k/2w(x)∫0Aw-1(t)(1+Q′(t)2)(k-1)/2dt⩽c8.
So, with (3.35) we conclude (3.21).To prove Theorem3.3 we further need the following lemma.Lemma 3.8.
Forr=1, one lets  w∈ℱ(C2+), and for integer  r⩾2 one lets  w∈ℱ(Cr+). Let  1⩽p⩽∞, and  1⩽k⩽r be an integer. If  g:ℝ→ℝ is absolutely continuous,  g(0)=0, and  |Q′(x)|k-1g′∈Lp,w(ℝ), then
(3.37)‖(1+Q′(x)2)k/2wg‖Lp(R)⩽C‖(1+Q′(x)2)(k-1)/2wg′‖Lp(R).Proof.
We will prove (3.37) for p=1 and p=∞, and then we use the Riesz-Thorin interpolation theorem. Let
(3.38)ψ(t):=(1+Q′(t)2)(k-1)/2w(t)g′(t),t∈R.
Then for almost all x⩾0,
(3.39)|g(x)|+|g(-x)|⩽∫0xw-1(t)(1+Q′(t)2)(k-1)/2{|ψ(t)|+|ψ(-t)|}dt.
Denoting
(3.40)Ψ(t):=|ψ(t)|+|ψ(-t)|,
we get
(3.41)‖(1+Q′2)k/2wg‖L1(R)=∫0∞(1+Q′(x)2)k/2w(x){|g(x)|+|g(-x)|}dx⩽∫0∞(1+Q′(x)2)k/2w(x)∫0|x|w-1(t)(1+Q′(t)2)(k-1)/2Ψ(t)dtdx=∫0∞{w-1(x)(1+Q′(x)2)(k-1)/2∫|x|∞(1+Q′(t)2)k/2w(t)dt}Ψ(x)dx,
by changing of the integral order. Hence, from (3.20) we have
(3.42)‖(1+Q′2)k/2wg‖L1(R)⩽C∫0∞Ψ(x)dx=‖ψ‖L1(R).
By the definition of (3.38), we have (3.37) for p=1.
Next, we show (3.37) for p=∞. From (3.39) we see that(3.43)|(1+Q′(x)2)k/2w(x)g(x)|⩽(1+Q′(x)2)k/2w(x){|g(x)|+|g(-x)|}⩽(1+Q′(x)2)k/2w(x)∫0|x|w-1(t)(1+Q′(t)2)(k-1)/2{|ψ(t)|+|ψ(-t)|}dt⩽2‖ψ‖L∞(R)(1+Q′(x)2)k/2w(x)∫0|x|w-1(t)(1+Q′(t)2)(k-1)/2dt⩽C‖ψ‖L∞(R),
by (3.21). So we have (3.37) for p=∞. Let ϕ∈L1(ℝ)∩L∞(ℝ)⊂Lp(ℝ), 1⩽p⩽∞, then we set
(3.44)g(x):=∫0xϕ(t)(1+Q′(t)2)(k-1)/2w(t)dt.
So we have
(3.45)(1+Q′(x)2)(k-1)/2w(x)g′(x)=ϕ(x)∈L1(R)∩L∞(R)⊂Lp(R),1⩽p⩽∞.
Since L1(ℝ)∩L∞(ℝ) is dense in Lp(ℝ), 1⩽p⩽∞, using the Riesz-Thorin interpolation theorem for the linear operator,
(3.46)ϕ⟼(1+Q′(x)2)k/2w(x)∫0xw-1(t)(1+Q′(t)2)(k-1)/2ϕ(t)dt,ϕ∈Lp(R),
we have the result.Corollary 3.9.
Forr=1, one lets  w∈ℱ(C2+), and for integer  r⩾2 one lets   w∈ℱ(Cr+). Let  1⩽p⩽∞, and  1⩽k be an integer. If  g:ℝ→ℝ is absolutely continuous on  ℝ,  g(j)(0)=0,  j=0,…,k-1, and  g(k)∈Lp,w(ℝ), then
(3.47)‖(1+Q′(x)2)k/2wg‖Lp(R)⩽C‖wg(k)‖Lp(R).Lemma 3.10 ([4, Lemma  7]).
Letw∈ℱ(C2+). For a certain constant  C1>0, let  u satisfy
(3.48)0<C1⩽u,t=auu.
Then there exists a constant C2>1 such that for every x and y which satisfy |x|⩽σ(2t) and |x-y|⩽tΦt(x)/2, one has
(3.49)1C2w(y)⩽w(x)⩽C2w(y).Proof of Theorem3.3.
For a givenε>0, we can select g, where g(r-1)(x) is absolutely continuous on ℝ, and g(r)∈Lp,w(ℝ) such that
(3.50)‖w(f-g)‖Lp(R)+tr‖wg(r)‖Lp(R)-ε⩽Kr,p(w,f,t).
Let f=h+g, where h∈Lp,w(ℝ). Then we have
(3.51)ωr,p(h,t)⩽C‖wh‖Lp(R).
Let 0<s⩽t/r. From Lemma 3.4, 3.10 and the Hőlder-Minkowski inequality, we have
(3.52)‖wΔsΦt(x)rg(x)‖Lp(|x|⩽σ(2t)),C‖w(x){∫-s/2s/2}r|g(r)(u)|du1du2…dur‖Lp(|x|⩽σ(2t))(noteΦt(x)⩽C)⩽C‖{∫-s/2s/2}r|(wg(r))(u)|du1du2⋯dur‖Lp(|x|⩽σ(2t))⩽C[{∫-s/2s/2}r‖(wg(r))(u)‖Lp(|x|⩽σ(2t))du1du2⋯dur]⩽Csr‖wg(r)‖Lp(R)⩽Ctr‖wg(r)‖Lp(R),
where u=x+Φt(x)(u1+u2+⋯+ur). We estimate
(3.53)inf⁡P∈Pr-1‖(g(x)-P(x))w(x)‖Lp(|x|⩾σ(4t)).
Then we may suppose
(3.54)g(j)(0)-P(j)(0)=0,j=0,1,…,r-1.
Using Lemma 3.5, we see
(3.55)|Q′(σ(4t))|-1⩽Ct,
(see [7, page 12]). By Corollary 3.9 with k=r and (3.55) we have
(3.56)‖w(g-P)‖Lp(|x|⩾σ(4t))⩽|Q′(σ(4t))|-r‖(1+Q′(x)2)r/2w(g-P)‖Lp(|x|⩾σ(4t))⩽Ctr‖wg(r)‖Lp(R).
Hence, from (3.51), (3.52), and (3.56) we have
(3.57)ωr,p(w,f,t)-Cε⩽C(‖wh‖Lp(R)+tr‖wg(r)‖Lp(R)-ε)⩽CKr,p(w,f,t).
Consequently, we have
(3.58)ωr,p(w,f,t)⩽CKr,p(w,f,t).
Therefore, from (3.2) we have (3.3).Corollary 3.11.
Forr=1, one lets  w∈ℱ(C2+), and for integer  r⩾2 one lets  w∈ℱ(Cr+). Let  f∈Lp,w(ℝ). Then one has
(3.59)ωr,p(w,f,t)~Kr,p(w,f,t)~K̃r,p(w,f,t).



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*Source: 725638-2011-12-14.xml*