Abstract

Let R=(,), and let 𝑄1[0,) be an even function. We consider the exponential weights 𝑤(𝑥)=𝑒𝑄(𝑥), 𝑥. 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.

1. Introduction and Preliminaries

Let =(,), and let 𝑄1[0,) be an even function. We consider the weights 𝑤(𝑥)=exp(𝑄(𝑥)) satisfying for all 𝑛=0,1,2,,0𝑥𝑛𝑤(𝑥)𝑑𝑥<.(1.1) First we need the following definition from [1]. We say that 𝑓[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)(𝐶2+) as follows: Let 𝑄[0,) be continuous and an even function, and satisfy the following properties: (a)𝑄(𝑥) is continuous in , with 𝑄(0)=0;(b)𝑄(𝑥) exists and is positive in {0};(c)lim𝑥𝑄(𝑥)=;(1.2)(d)the function 𝑇(𝑥)=𝑥𝑄(𝑥)𝑄(𝑥),𝑥0(1.3) is quasi-increasing in (0,) with 𝑇(𝑥)Λ>1, 𝑥{0};(e)there exists 𝐶1>0 such that 𝑄(𝑥)||𝑄||(𝑥)𝐶1||𝑄||(𝑥)𝑄(𝑥),a.e.𝑥{0}.(1.4) Moreover, there also exists a compact subinterval 𝐽(0) of , and 𝐶2>0 such that 𝑄(𝑥)||𝑄||(𝑥)𝐶2||𝑄||(𝑥)𝑄(𝑥),a.e.𝑥𝐽.(1.5)

Example 1.2. Let 𝑤=exp(𝑄)(𝐶2+).(1) If 𝑇(𝑥) is bounded, then we call the weight 𝑤=exp(𝑄) the Freud-type weight. The following example is the Freud-type weight: 𝑤(𝑥)=exp(|𝑥|𝛼),𝛼>1.(1.6)(2) If 𝑇(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, =1,2,3,,𝑄(𝑥)=𝑄,𝛼(𝑥)=exp(|𝑥|𝛼)exp𝑙(0),(1.7)where exp(𝑥)=exp(exp(expexp𝑥)),(times). More generally, we define for 𝛼+𝑢>1, 𝛼0, 𝑢0, and 𝑙1, 𝑄𝑙,𝛼,𝑢(𝑥)=|𝑥|𝑢exp𝑙(|𝑥|𝛼)𝛼exp𝑙(0),(1.8)where 𝛼=0 if 𝛼=0, otherwise 𝛼=1.(ii) For 𝛼>1, 𝑄(𝑥)=𝑄𝛼(𝑥)=(1+|𝑥|)|𝑥|𝛼1.

We need the Mhaskar-Rakhmanov-Saff numbers 𝑎𝑥 defined by 2𝑥=𝜋10𝑎𝑥𝑢𝑄𝑎𝑥𝑢1𝑢21/2𝑑𝑢,𝑥>0.(1.9) If 𝑓 is measurable, we define 𝑓𝑤𝐿𝑝()=||||𝑓(𝑡)𝑤(𝑡)𝑝𝑑𝑡1/𝑝,if0<𝑝<,esssupt||||𝑓(𝑡)𝑤(𝑡),if𝑝=,(1.10) where if 𝑝=, then we suppose that 𝑓 is continuous on and lim|𝑥|||||𝑓(𝑥)𝑤(𝑥)=0.(1.11) The class of all functions 𝑓 for which 𝑓𝑤𝐿𝑝()< will be denoted by 𝐿𝑝,𝑤(), with the usual understanding that two functions are identified if they are equal almost everywhere. For 𝑓𝐿𝑝,𝑤()(0<𝑝), the degree of weighted polynomial approximation is defined by 𝐸𝑝,𝑛(𝑤;𝑓)=inf𝑃𝒫𝑛𝑤(𝑓𝑃)𝐿𝑝(),(1.12) 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 𝒦𝑟,𝑝. We recall and summarize them in Section 2. In Section 3, we will discuss the equivalence of 𝒦𝑟,𝑝 and modulus of smoothness 𝜔𝑟,𝑝. As the result, we see 𝒦𝑟,𝑝𝒦𝑟,𝑝 for a weight in a subclass of (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 𝑥1𝐶𝑓(𝑥)𝑔(𝑥)𝐶𝑓(𝑥).(1.13) Similarly, for any positive numbers {𝑐𝑛}𝑛=1 and {𝑑𝑛}𝑛=1 we define 𝑐𝑛𝑑𝑛. 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

Let 𝑟1 be an integer (in this paper, we suppose that 𝑟 (or 𝑠) is an integer). The 𝑟th order 𝐾-functional of a function 𝑓𝐿𝑝,𝑤() is defined by the formula 𝒦𝑟,𝑝(𝑤,𝑓,𝛿)=inf𝑤(𝑓𝑔)𝐿𝑝()+𝛿𝑟𝑤𝑔(𝑟)𝐿𝑝(),(2.1) for 𝛿>0, where the infimum is taken over all 𝑟1 times continuously differentiable 𝑔 such that 𝑔(𝑟1) is absolutely continuous, and 𝑔(𝑟)𝐿𝑝,𝑤() (this means 𝑤𝑔(𝑟)𝐿𝑝()). Using this 𝑟th order 𝐾-functional, we can estimate the order of 𝐸𝑝,𝑛(𝑤;𝑓).

Theorem 2.1 (see [3]). Let 𝑤=exp(𝑄)(𝐶2+). Let 𝑟1, 𝑠0 be integers, and let 1𝑝. Let 𝑓 be 𝑠1 times continuously differentiable, 𝑓(𝑠1) be absolutely continuous on each compact interval, and 𝑓(𝑠)𝐿𝑝,𝑤() (when 𝑠=0, these assumptions state merely that 𝑓𝐿𝑝,𝑤()). Then, for every integer 𝑛𝑟+𝑠, 𝐸𝑝,𝑛(𝑎𝑤;𝑓)𝐶𝑛𝑛𝑠𝒦𝑟,𝑝𝑤,𝑓(𝑠),𝑎𝑛𝑛.(2.2)

Theorem 2.1 was shown by using the following Favard-type inequalities (see [3]).

Theorem 2.2 ([4, Corollary 8]). Let 𝑤=exp(𝑄)(𝐶2+). Let 𝑓 be 𝑠1 times continuously differentiable, and let 𝑓(𝑠1) for some integer 𝑠1 be absolutely continuous on each compact interval. Let 1𝑝 and 𝑓(𝑠)𝐿𝑝,𝑤(). Then one has 𝐸𝑝,𝑛(𝑎𝑤;𝑓)𝐶𝑛𝑛𝑠𝑤𝑓(𝑠)𝐿𝑝(),(2.3) equivalently, 𝐸𝑝,𝑛(𝑎𝑤;𝑓)𝐶𝑛𝑛𝑠𝐸𝑝,𝑛𝑠𝑤;𝑓(𝑠).(2.4)

Remark 2.3. (1)([4, Remark 11]) Let 𝑤(𝐶2+) and let 0<𝑝<1. Then there exists a constant 𝐶0 such that for every absolutely continuous function 𝑓 with 𝑓𝐿,𝑤 and 𝑤𝑓𝐿𝑝(), and for every 𝑛, we have 𝐸𝑝,𝑛(𝑤;𝑓)𝐶0𝑎𝑛𝑛𝑤𝑓𝐿()+𝑤𝑓𝐿𝑝().(2.5)
(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(𝐶2+),1𝑝 and 1/𝑝𝛽1. Let 𝑔 be absolutely continuous with |𝑔|𝛽𝐿𝑝,𝑤() (and for 𝑝=, we require 𝑔 to be continuous, and 𝑔𝑤 to vanish at ±). Let 𝑓𝐿𝑝,𝑤() and set 𝑓𝑔,𝛽(𝑥)=sup𝑦||||𝑓(𝑥)𝑓(𝑦)||||𝑔(𝑥)𝑔(𝑦)𝛽.(2.6) Then we define 𝐅𝑔,𝛽(𝑤,𝑝)=𝑓𝑓𝐿𝑝,𝑤𝑓(),𝑔,𝛽𝐿()<.(2.7)

Now, we have the following theorem:

Theorem see ([3, Theorem 4.2]). Let 1𝑝, 1/𝑝𝛽1, and let 𝑓𝐅𝑔,𝛽(𝑤,𝑝). Then one has 𝐸𝑝,𝑛𝑎(𝑤;𝑓)𝐶𝑛𝑛𝛽𝑤||𝑔||𝛽𝐿𝑝()𝑓𝑔,𝛽𝐿().(2.8)

We define the following class of weights from [5, Definition  1.1].

Definition 2.5. Let 𝑤(𝑥)=exp(𝑄(𝑥)), where 𝑄 is even, continuous, and 𝑄 is positive in (0,). Then one writes 𝑤=exp(𝑄)1, if the following are satisfied: (a)𝑥𝑄(𝑥) is strictly increasing in (0,) with lim𝑥0+𝑥𝑄(𝑥)=0;(b)the function𝑇(𝑥)=𝑥𝑄(𝑥)𝑄(𝑥)(2.9)is quasi-increasing in (𝐶,) for some 𝐶>0 and lim𝑥𝑇(𝑥)=;(c)assume𝑦𝑄(𝑦)𝑥𝑄(𝑥)𝐶1𝑄(𝑦)𝑄(𝑥)𝐶3,𝑦𝑥𝐶2,(2.10)for some positive constants 𝐶1,𝐶2, and 𝐶3.

Remark 2.6. Let 𝑤(𝑥)=exp(𝑄(𝑥))(𝐶2+) and 𝑇(𝑥) is unbounded, then we see 𝑤1. In fact, we see this as follows.
From Definition 1.1(e) and (d), we have for 𝑦𝑥>0, 𝑄(𝑦)𝑄(𝑥)=exp𝑦𝑥𝑄(𝑡)𝑄,𝐶(𝑡)𝑑𝑡exp1𝑦𝑥𝑄(𝑡)=𝑄(𝑡)𝑑𝑡𝑄(𝑦)𝑄(𝑥)𝐶1,𝑦𝑥=exp𝑦𝑥1𝑡1𝑑𝑡expΛ𝑦𝑥𝑄(𝑡)=𝑄(𝑡)𝑑𝑡𝑄(𝑦)𝑄(𝑥)1/Λ.(2.11) Therefore we obtain (c) in Definition 2.5 for 𝐶3=𝐶1+1/Λ.

If 𝑓, and >0, then we define the differences of 𝑓 inductively by the formula Δ0Δ𝑓(𝑥)=𝑓(𝑥),1𝑓(𝑥)=Δ𝑓(𝑥)=𝑓𝑥+2𝑓𝑥2,Δ𝑘𝑓(𝑥)=Δ𝑘1Δ𝑓(𝑥)=𝑘𝑗=0(1)𝑗𝑘𝑗𝑓𝑥+𝑘2𝑗,𝑘=2,3,..,𝑥.(2.12) We set 𝑎𝜎(𝑡)=inf𝑢𝑎𝑢𝑢Φ𝑡,𝑡>0,𝑡(𝑥)=||||1|𝑥|𝜎||||(𝑡)+𝑇(𝜎(𝑡))1/2,𝑥.(2.13) By [6], when 𝑤=exp(𝑄) is the Erdős-type weight we define for 𝑓𝐿𝑝,𝑤(),0<𝑝, 𝜔𝑟,𝑝(𝑤,𝑓,𝑡)=sup0<𝑡𝑤Δ𝑟Φ𝑡(𝑥)(𝑓)𝐿𝑝(|𝑥|𝜎(2𝑡))+inf𝑅𝒫𝑟1𝑤(𝑥)(𝑓𝑅)(𝑥)𝐿𝑝(|𝑥|𝜎(4𝑡)).(2.14) If 𝑤=exp(𝑄) is the Freud-type weights, then we define 𝜔𝑟,𝑝(𝑤,𝑓,𝑡)=sup0<𝑡𝑤Δ𝑟(𝑓)𝐿𝑝(|𝑥|𝜎())+inf𝑅𝒫𝑟1𝑤(𝑥)(𝑓𝑅)(𝑥)𝐿𝑝(|𝑥|𝜎(𝑡)).(2.15)

The following Jackson-type inequality is known.

Theorem 2.7 (see [5, Theorem  1.2], [6, Corollary  1.4]). Let 𝑤1. Let 0<𝑝. Then for 𝑓 for which 𝑓𝐿𝑝,𝑤() (and for 𝑝=, we require 𝑓 to be continuous, and 𝑓𝑤 to vanish at ±), one has for 𝑛𝐶3, 𝐸𝑝,𝑛(𝑓;𝑤)𝐶1𝜔𝑟,𝑝𝑤,𝑓,𝐶2𝑎𝑛𝑛,(2.16) 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]). Let 𝑤=exp(𝑄), where 𝑄 is even, and 𝑄 exists and is positive on (0,). Moreover, assume that 𝑥𝑄(𝑥) is strictly increasing, with right limit 0 at 0, and for some 𝜆,𝐴,𝐵>1, 𝐶>0, 𝑄𝐴(𝜆𝑥)𝑄(𝑥)𝐵,𝑥𝐶.(2.17) Then we call 𝑤 Freud weight, and write 𝑤.

Remark 2.9. Let 𝑤=exp(𝑄)(𝐶2+). Then we see the following.
(1)If 𝑇(𝑥)=(𝑥𝑄(𝑥))/𝑄(𝑥) is bounded, then we say that 𝑤=exp(𝑄) is the Freud-type weight, and we write 𝑤. Then we see (𝐶2+)=. In fact, when 𝑤, 1<𝜈 and 𝑥>0 large enough, by Definition 1.1(e), 𝑄(𝜈𝑥)𝑄(𝑥)=exp𝑥𝜈𝑥𝑄(𝑢)𝑄𝐶(𝑢)𝑑𝑢exp1𝑥𝜈𝑥𝑄(𝑢)𝑄(𝑢)𝑑𝑢,(2.18) and then since 𝑇(𝑥) is bounded, there exists 𝐶>0 such that 𝑄(𝜈𝑥)𝑄𝐶(𝑥)exp1𝐶𝑥𝜈𝑥1𝑢𝑑𝑢=𝜈𝐶1𝐶.(2.19) Similarly, 𝑄(𝜈𝑥)𝑄(𝑥)=exp𝑥𝜈𝑥𝑄(𝑢)𝑄𝐶(𝑢)𝑑𝑢exp2𝑥𝜈𝑥𝑄(𝑢)𝐶𝑄(𝑢)𝑑𝑢exp2Λ𝑥𝜈𝑥1𝑢𝑑𝑢=𝜈𝐶2Λ.(2.20) Therefore, if we take 𝜆=𝜈>1 large enough, then we have (2.17).
Conversely, to show (𝐶2+) we suppose that there exists 𝑤=exp(𝑄)(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, 𝑄(𝜆𝑥)𝑄(𝑥)=exp𝑥𝜆𝑥𝑄(𝑢)𝑄𝐶(𝑢)𝑑𝑢exp3𝑥𝜆𝑥𝑄(𝑢)𝐶𝑄(𝑢)𝑑𝑢=exp3𝑥𝜆𝑥𝑇(𝑢)𝑢𝑑𝑢=𝜆𝐶3𝑀,(2.21) where 𝐶3>0 is a constant, that is, (1) does not hold. Hence we have (𝐶2+). Consequently, we have =(𝐶2+).
(2)If 𝑇(𝑥)=(𝑥𝑄(𝑥))/𝑄(𝑥) is bounded, then for 𝑥1, there exists 𝑐,𝐶>0 such that 𝑥Λ𝑄(𝑥)𝐶𝑥𝑐.(2.22)

Theorem 2.10  2.10 ([7, Theorem 3.5]). Let 𝑤. Let 0<𝑝. Then for 𝑓 for which 𝑓𝐿𝑝,𝑤() (and for 𝑝=, one requires 𝑓 to be continuous, and 𝑓𝑤 to vanish at ±), one has for 𝑛𝐶3, 𝐸𝑝,𝑛(𝑓;𝑤)𝐶1𝜔𝑟,𝑝𝑤,𝑓,𝐶2𝑎𝑛𝑛,(2.23) where 𝐶𝑗, 𝑗=1,2,3, do not depend on 𝑓 and 𝑛.

Damelin [6] introduces the following 𝐾-functional: let 𝑓𝐿𝑝,𝑤(), 0<𝑝 and 𝑟1 be an integer, then we define 𝒦𝑟,𝑝(𝑤,𝑓,𝛿)=inf𝑃𝒫𝑛𝑤(𝑓𝑃)𝐿𝑝()+𝛿𝑟𝑤𝑃(𝑟)𝐿𝑝(),(2.24) where 𝛿>0 are chosen in advance and 𝑎𝑛=𝑛(𝛿)=inf𝑘𝑘𝑘𝛿.(2.25)

Then Damelin gives the following.

Theorem 2.11 ([6, Theorem 1.3 (b)]). Let 𝑤1, 𝑟1, 0<𝑝, and let 𝑓𝑤𝐿𝑝() (and for 𝑝=, one requires 𝑓 to be continuous, and 𝑓𝑤 to vanish at ±). Then one has 𝜔𝑟,𝑝𝒦(𝑤,𝑓,𝑡)𝑟,𝑝(𝑤,𝑓,𝛿).(2.26)

For the Freud-type weights we have also the followings.

Theorem 2.12 ([7, Theorem 3.9, 3.10]). Let 𝑤, 𝑟1, 0<𝑝, and let 𝑓𝑤𝐿𝑝() (and for 𝑝=, we require 𝑓 to be continuous, and 𝑓𝑤 to vanish at ±). Then we have 𝒦𝑟,𝑝(𝑤,𝑓,𝛿)𝜔𝑟,𝑝𝒦(𝑤,𝑓,𝑡)𝑟,𝑝(𝑤,𝑓,𝛿).(2.27)

For 𝑤(𝐶2+), we see easily that 𝒦𝑟,𝑝𝒦(𝑤,𝑓,𝛿)𝑟,𝑝(𝑤,𝑓,𝛿).(2.28)

So, from Theorem 2.1 we obtain the following corollary.

Corollary 2.13. Let 𝑤=exp(𝑄)(𝐶2+). Let 𝑟1,𝑠0 be integers, and let 1𝑝. Let 𝑓 be 𝑠1 times continuously differentiable, 𝑓(𝑠1) be absolutely continuous, and 𝑓(𝑠)𝐿𝑝,𝑤() (when 𝑠=0, these assumptions state merely that 𝑓𝐿𝑝,𝑤()). Then, for every integer 𝑛𝑟+𝑠, 𝐸𝑝,𝑛(𝑎𝑤;𝑓)𝐶𝑛𝑛𝑠𝒦𝑟,𝑝𝑤,𝑓(𝑠),𝑎𝑛𝑛.(2.29)

The main theme in [3] is to summarize the above theorems. Let 𝑤(𝐶2+),𝑟1 be an integer, and let 1𝑝. 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)[Theorem 2.7 with 𝑟=1 (the Erdős-type case)], [Theorem 2.10 with 𝑟=1 (the Freud case)]:let 𝑓 and if 1𝑝<, assume that 𝑓𝑤𝐿𝑝(). If 𝑝=, assume in addition that 𝑓 is continuous and that 𝑓𝑤 has limit 0 at ±. Then we have 𝐸𝑝,𝑛(𝑤;𝑓)𝐶1𝜔1,𝑝𝑤,𝑓,𝐶2𝑎𝑛𝑛.(2.30)(b) [Theorem 2.2]: let 𝑓 be 𝑠1 times continuously differentiable, and let for some integer 𝑠1, 𝑓(𝑠1)(𝑥) be absolutely continuous on each compact interval. Let 1𝑝 and 𝑓(𝑠)𝐿𝑝,𝑤(). Then we have 𝐸𝑝,𝑛(𝑤;𝑓)𝐶3𝑎𝑛𝑛𝑠𝑤𝑓(𝑠)𝐿𝑝(),(2.31) equivalently, 𝐸𝑝,𝑛(𝑤;𝑓)𝐶4𝑎𝑛𝑛𝑠𝐸𝑝,𝑛𝑤;𝑓(𝑠).(2.32)(c) [Theorem 2.1]:let 𝑠0 be an integer, and let 𝑓 be 𝑠1 times continuously differentiable, 𝑓(𝑠1) be absolutely continuous, and 𝑓(𝑠)𝐿𝑝,𝑤() (when 𝑠=0, these assumptions state merely that 𝑓𝐿𝑝,𝑤()). Then, for every integer 𝑟1 and 𝑛𝑟+𝑠, 𝐸𝑝,𝑛(𝑤;𝑓)𝐶5𝑎𝑛𝑛𝑠𝒦𝑟,𝑝𝑤,𝑓(𝑠),𝑎𝑛𝑛.(2.33)(d) [Theorem 2.7, 2.11 (the Erdős-type case)], [Theorem 2.10, 2.12 (the Freud case)]:for 𝑓 with 𝑓𝐿𝑝,𝑤() (and for 𝑝= we require 𝑓 to be continuous, and 𝑓𝑤 to vanish at ±), for every 𝑛 large enough and for every integer 𝑟1 we have 𝐸𝑝,𝑛(𝑤;𝑓)𝐶6𝒦𝑟,𝑝𝑎𝑤,𝑓,𝑛𝑛.(2.34)

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 of 𝐾-Functional and Modulus of Smoothness

In this section we consider a subclass of (𝐶2+).

Assumption 3.1. Let 𝑟2 be an integer. Let 𝑤(𝑥)=exp(𝑄(𝑥))(𝐶2+). Then we suppose that 𝑄𝐶𝑟1(), 𝑄(𝑟)(𝑥) exists in {0}, furthermore, the following inequalities hold ||𝑄(𝑗)||(𝑥)𝐶𝑗||𝑄(𝑗1)||||||(𝑥)𝑄(𝑥)||||𝑄(𝑥),𝑗=1,2,,𝑟,(3.1) where for 𝑗=𝑟 we suppose that (3.1) hold almost everywhere on 𝑥{0}. Then we write 𝑤(𝐶𝑟+).

Remark 3.2. Example 1.2, (i), (ii) satisfy Assumption 3.1, moreover 𝑄(𝑥)=𝑥2𝑚, 𝑚: a positive integer, satisfies Assumption 3.1.

In Section 2, we know 𝒦𝑟,𝑝𝒦(𝑤,𝑓,𝑡)𝑟,𝑝(𝑤,𝑓,𝑡)𝜔𝑟,𝑝(𝑤,𝑓,𝑡).(3.2) The equivalences mentioned in the last of Section 2 give a certain suggestion, that is, 𝒦𝑟,𝑝(𝑤,𝑓,𝑡)𝜔𝑟,𝑝(𝑤,𝑓,𝑡).(3.3) In fact, this is true.

Theorem 3.3. Let 𝑟2 be a positive integer, and let 𝑤(𝐶𝑟+). Let 𝑓𝐿𝑝,𝑤(). Then one has (3.3). Similarly, when 𝑤(𝐶2+) and 𝑟=1, one has (3.3).

To prove Theorem 3.3, we need some lemmas.

Lemma 3.4. Let 𝑟1 be an integer. If 𝑓(𝑟1)(𝑥) is absolutely continuous on , then for 𝑘=1,2,,𝑟, one has the following representation: Δ𝑘Φ𝑡(𝑥)Φ𝑓(𝑥)=𝑡(𝑥)𝑘/2/2𝑘𝑓(𝑘)𝑥+Φ𝑡(𝑢𝑥)1+𝑢2++𝑢𝑘𝑑𝑢1𝑑𝑢2𝑑𝑢𝑘.(3.4)

Proof. /2/2𝑓(1)𝑥+Φ𝑡(𝑥)𝑢1𝑑𝑢1=Φ𝑡(𝑥)1𝑥+Φ𝑡(𝑥)/2𝑥Φ𝑡(𝑥)/2𝑓(1)𝑣1𝑑𝑣1=Φ𝑡(𝑥)1Φ𝑡(𝑥)𝑓(𝑥).(3.5) Therefore, for 𝑟=1 we have (3.4). For some 𝑘1 we suppose (3.4). Then we have /2/2𝑘+1𝑓(𝑘+1)(𝑢)𝑑𝑢1𝑑𝑢2𝑑𝑢𝑘+1=/2/2/2/2𝑘𝑓(𝑘)(̃𝑢)𝑑𝑢1𝑑𝑢2𝑑𝑢𝑘𝑑𝑢𝑘+1=Φ𝑡(𝑥)𝑘/2/2Δ𝑘Φ𝑡(𝑥)𝑓𝑥+Φ𝑡(𝑥)𝑢𝑘+1𝑑𝑢𝑘+1=Φ𝑡(𝑥)(𝑘+1)Δ𝑘Φ𝑡(𝑥)ΔΦ𝑡(𝑥)𝑓=Φ(𝑥)𝑡(𝑥)(𝑘+1)ΔΦ𝑘+1𝑡(𝑥)𝑓(𝑥),(3.6) where 𝑢=𝑢𝑥;𝑢1,,𝑢𝑘+1=𝑥+Φ𝑡𝑢(𝑥)1+𝑢2++𝑢𝑘+1,̃𝑢=̃𝑢𝑥;𝑢1,,𝑢𝑘+1=𝑥+Φ𝑡(𝑥)𝑢𝑘+1+Φ𝑡(𝑢𝑥)1+𝑢2++𝑢𝑘.(3.7) Hence we have (3.4) for 𝑟=1,2,3,.

Lemma 3.5 (see [1, Lemma  3.4 (3.17)]). Uniformly for 𝑡>0, one has 𝑄𝑎𝑡𝑡𝑇𝑎𝑡𝑎𝑡.(3.8)

Lemma 3.6. Let 𝑤(𝐶𝑟+) and 𝑟2 be an integer. Then for any integer 𝑘, 1𝑘𝑟1, there exist 𝑐1,𝑐2>0 and 𝐴>0 such that for |𝑥|𝐴, 𝑐1(1)𝑘𝑤(𝑘)(𝑥)1+𝑄(𝑥)2𝑘/2𝑤(𝑥)𝑐2.(3.9) Also, there exist 𝑐3,𝑐4>0 and 𝐵>0 such that for |𝑥|𝐵, 𝑐3(1)𝑘𝑤1(𝑘)(𝑥)1+𝑄(𝑥)2𝑘/2𝑤1(𝑥)𝑐4.(3.10) Furthermore, for 𝑘=𝑟, (3.9) and (3.10) hold almost everywhere on |𝑥|𝐴 and |𝑥|𝐵, respectively. When 𝑤(𝐶2+) and 𝑟=1, one also has (3.9) and (3.10).

Proof. First, we will see that for 𝜇=±1 the following equations hold: lim|𝑥|(𝑤𝜇)(𝑘)(𝑥)𝑄(𝑥)𝑘𝑤𝜇(𝑥)=(𝜇)𝑘,𝑘=1,,𝑟1.(3.11) For 𝑘=𝑟 we take 𝑄(𝑥) as follows; 𝑄(𝑗)(𝑥)=𝑄(𝑗)𝑄(𝑥),𝑥,𝑗=1,2,,𝑟1,(𝑟)(𝑥)=𝑄(𝑟)(𝑥),a.e.𝑥{0},(3.12) and for all 𝑗=1,2,,𝑟, 𝑄(𝑥) satisfies (3.1) for all 𝑥0. Then we obtain that exchanging 𝑄 with 𝑄, (3.11) also holds.
Let 𝜇=1, and let 1𝑘𝑟1. 𝑤(𝑥)=𝑄𝑤(𝑥)𝑤(𝑥),(𝑥)=𝑄(𝑥)+𝑄(𝑥)2𝑤𝑤(𝑥),(𝑥)=𝑄(𝑥)+3𝑄(𝑥)𝑄(𝑥)𝑄(𝑥)3𝑤(𝑥),(3.13) and we continue this manner, so 𝑤(𝑘)(𝑥)=𝑖1+2𝑖2++𝑘𝑖𝑘𝑖=𝑘,1<𝑘𝑐𝑖1,𝑖2,...,𝑖𝑘𝑄(𝑥)𝑖1𝑄(𝑥)𝑖2𝑄(𝑘)(𝑥)𝑖𝑘+(1)𝑘𝑄(𝑥)𝑘𝑤(𝑥),(3.14) where 𝑐𝑖1,𝑖2,...,𝑖𝑘 are coefficients. Here, from (3.1), for |𝑥|𝐴>0 large enough and 𝑖𝑗0, 2𝑗𝑘,||𝑄(𝑗)||(𝑥)𝑖𝑗𝐶𝑗||||𝑄(𝑥)||||𝑄(𝑥)𝑗1||𝑄||(𝑥)𝑖𝑗=𝐶𝑗||𝑄||(𝑥)𝑗𝑄(𝑥)𝑗1𝑖𝑗.(3.15) Hence, from (3.14) and (3.15) we have ||||lim|𝑥|𝑤(𝑘)(𝑥)𝑄(𝑥)𝑘𝑤(𝑥)(1)𝑘||||𝐶lim|𝑥|1𝑄(𝑥)=0,𝑘=1,,𝑟1,(3.16) where 𝐶 is a positive constant. Therefore, we have (3.11) with 𝜇=1. Similarly, for 𝜇=1, lim|𝑥|𝑤1(𝑘)(𝑥)𝑄(𝑥)𝑘𝑤1(𝑥)=1,𝑘=1,,𝑟1.(3.17) Therefore, we have (3.11) for 𝑘=1,2,,𝑟1, and hence we also have (3.9) and (3.10) for |𝑥| large enough. If in (3.11), we replace 𝑄 with 𝑄, 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, 𝑐1(1)𝑟𝑤(𝑟)(𝑥)1+𝑄(𝑥)2𝑟/2𝑤(𝑥)𝑐2𝑐,a.e.|𝑥|𝐴,3(1)𝑟𝑤1(𝑟)(𝑥)1+𝑄(𝑥)2𝑟/2𝑤1(𝑥)𝑐4,a.e.|𝑥|𝐵.(3.18) For 𝑟=1, the lemma is trivial.

Lemma 3.7. Let 𝑟1 be an integer. There exists 𝐶>0 such that for every integer 𝑘=1,2,,𝑟, ||𝑄||(𝑥)𝑘𝑤(𝑥)0,𝑎𝑠|𝑥|,(3.19)𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2|𝑥|1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝐶,𝑥,(3.20) and for 𝑥, 1+𝑄(𝑥)2𝑘/2𝑤(𝑥)0|𝑥|1+𝑄(𝑡)2(𝑘1)/2𝑤1(𝑡)𝑑𝑡𝐶,𝑥.(3.21)

Proof. From Definition 1.1(e) we see that there exist 𝐶>0 and 𝜆>0 such that ||𝑄||(𝑥)𝐶𝑄(𝑥)𝜆,𝑥,(3.22) so we have ||𝑄||(𝑥)𝑘𝑤(𝑥)𝐶𝑘𝑄(𝑥)𝜆𝑘𝑤(𝑥)0,as|𝑥|.(3.23) Hence, we have (3.19). From (3.9) with 𝑘1, we see that for |𝑥|𝐴>0 and constants 𝑐1,𝑐2,𝑐3>0, 𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2|𝑥|1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝑐1𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2|𝑥|(1)𝑘𝑤(𝑘)(𝑡)𝑑𝑡𝑐2𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2(1)𝑘1𝑤(𝑘1)(𝑥)𝑐3.(3.24) Especially, if we use (3.24) with |𝑥|=𝐴, then we have 𝑤1(𝐴)1+𝑄(𝐴)2(𝑘1)/2𝐴1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝑐3.(3.25) Then, for |𝑥|𝐴 there exists a constant 𝐶=𝐶(𝐴) such that 𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2𝑤𝐶1(𝐴)1+𝑄(𝐴)2(𝑘1)/2.(3.26) In fact, for |𝑥|𝐴 we have 1+𝑄(𝐴)21+𝑄(𝐴)21+𝑄(𝑥)2,(3.27) that is, 1+𝑄(𝐴)2(𝑘1)/2𝐶(𝐴)1+𝑄(𝑥)2(𝑘1)/2,(3.28) where 𝐶(𝐴)=(1+𝑄(𝐴)2)(𝑘1)/2. Hence we have (3.26).
Now, by (3.26), if |𝑥|𝐴, then we have 𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2𝐴1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝑐4(𝐴).(3.29) When |𝑥|<𝐴, we see easily that 𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2𝐴|𝑥|1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝑐5=𝑐5(𝐴).(3.30) Hence, with (3.29) we have for |𝑥|𝐴, 𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2|𝑥|1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡𝑐6(𝐴).(3.31) Consequently, from (3.24) and (3.31) we have (3.20). We need to show (3.21). For |𝑥| large enough, we see 𝑄𝑄(𝑥)(𝑥)2𝑄(𝑥),a.e.𝑥,(3.32) (see Definition 1.1(e)) so we can select |𝑥|𝐴>0 large enough such that 𝑄1(𝑥)<𝑄2𝑘(𝑥)2,a.e.𝑥.(3.33)
We show (3.21) for 𝐴|𝑥|, |𝑥|𝐴>0. For |𝑥|𝐴, we have by (3.33), 𝐼=𝐴|𝑥|𝑤1(𝑡)𝑄(𝑡)𝑘1𝑑𝑡=𝐴|𝑥|𝑄(𝑡)𝑤1(𝑡)𝑄(𝑡)𝑘𝑤𝑑𝑡1(𝑥)𝑄(|𝑥|)𝑘+𝑘𝐴|𝑥|𝑤1𝑄(𝑡)(𝑡)𝑄(𝑡)𝑘+1𝑤𝑑𝑡1(𝑥)𝑄(|𝑥|)𝑘+12𝐼.(3.34) Hence, we have for a certain constant 𝑐7>0, 1+𝑄(𝑥)2𝑘/2𝑤(𝑥)𝐴|𝑥|𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2𝑑𝑡𝑐7.(3.35) For 𝐴0 we have (3.21) by the use of (3.19). In fact, there exists 𝑐8=𝑐8(𝐴) such that 1+𝑄(𝑥)2𝑘/2𝑤(𝑥)𝐴0𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2𝑑𝑡𝑐8.(3.36) So, with (3.35) we conclude (3.21).

To prove Theorem 3.3 we further need the following lemma.

Lemma 3.8. For 𝑟=1, one lets 𝑤(𝐶2+), and for integer 𝑟2 one lets 𝑤(𝐶𝑟+). Let 1𝑝, and 1𝑘𝑟 be an integer. If 𝑔 is absolutely continuous, 𝑔(0)=0, and |𝑄(𝑥)|𝑘1𝑔𝐿𝑝,𝑤(), then 1+𝑄(𝑥)2𝑘/2𝑤𝑔𝐿𝑝()𝐶1+𝑄(𝑥)2(𝑘1)/2𝑤𝑔𝐿𝑝().(3.37)

Proof. We will prove (3.37) for 𝑝=1 and 𝑝=, and then we use the Riesz-Thorin interpolation theorem. Let 𝜓(𝑡)=1+𝑄(𝑡)2(𝑘1)/2𝑤(𝑡)𝑔(𝑡),𝑡.(3.38) Then for almost all 𝑥0, ||||+||||𝑔(𝑥)𝑔(𝑥)𝑥0𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2||||+||||𝜓(𝑡)𝜓(𝑡)𝑑𝑡.(3.39) Denoting ||||+||||Ψ(𝑡)=𝜓(𝑡)𝜓(𝑡),(3.40) we get 1+𝑄2𝑘/2𝑤𝑔𝐿1()=01+𝑄(𝑥)2𝑘/2||||+||||𝑤(𝑥)𝑔(𝑥)𝑔(𝑥)𝑑𝑥01+𝑄(𝑥)2𝑘/2𝑤(𝑥)0|𝑥|𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2=Ψ(𝑡)𝑑𝑡𝑑𝑥0𝑤1(𝑥)1+𝑄(𝑥)2(𝑘1)/2|𝑥|1+𝑄(𝑡)2𝑘/2𝑤(𝑡)𝑑𝑡Ψ(𝑥)𝑑𝑥,(3.41) by changing of the integral order. Hence, from (3.20) we have (1+𝑄2)𝑘/2𝑤𝑔𝐿1()𝐶0Ψ(𝑥)𝑑𝑥=𝜓𝐿1().(3.42) 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|||1+𝑄(𝑥)2𝑘/2|||𝑤(𝑥)𝑔(𝑥)1+𝑄(𝑥)2𝑘/2||||+||||𝑤(𝑥)𝑔(𝑥)𝑔(𝑥)1+𝑄(𝑥)2𝑘/2𝑤(𝑥)0|𝑥|𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2||||+||||𝜓(𝑡)𝜓(𝑡)𝑑𝑡2𝜓𝐿()1+𝑄(𝑥)2𝑘/2𝑤(𝑥)0|𝑥|𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2𝑑𝑡𝐶𝜓𝐿(),(3.43) by (3.21). So we have (3.37) for 𝑝=. Let 𝜙𝐿1()𝐿()𝐿𝑝(), 1𝑝, then we set 𝑔(𝑥)=𝑥0𝜙(𝑡)1+𝑄(𝑡)2(𝑘1)/2𝑤(𝑡)𝑑𝑡.(3.44) So we have 1+𝑄(𝑥)2(𝑘1)/2𝑤(𝑥)𝑔(𝑥)=𝜙(𝑥)𝐿1()𝐿()𝐿𝑝(),1𝑝.(3.45) Since 𝐿1()𝐿() is dense in 𝐿𝑝(), 1𝑝, using the Riesz-Thorin interpolation theorem for the linear operator, 𝜙1+𝑄(𝑥)2𝑘/2𝑤(𝑥)𝑥0𝑤1(𝑡)1+𝑄(𝑡)2(𝑘1)/2𝜙(𝑡)𝑑𝑡,𝜙𝐿𝑝(),(3.46) we have the result.

Corollary 3.9. For 𝑟=1, one lets 𝑤(𝐶2+), and for integer 𝑟2 one lets 𝑤(𝐶𝑟+). Let 1𝑝, and 1𝑘 be an integer. If 𝑔 is absolutely continuous on , 𝑔(𝑗)(0)=0, 𝑗=0,,𝑘1, and 𝑔(𝑘)𝐿𝑝,𝑤(), then 1+𝑄(𝑥)2𝑘/2𝑤𝑔𝐿𝑝()𝐶𝑤𝑔(𝑘)𝐿𝑝().(3.47)

Lemma 3.10 ([4, Lemma  7]). Let 𝑤(𝐶2+). For a certain constant 𝐶1>0, let 𝑢 satisfy 0<𝐶1𝑎𝑢,𝑡=𝑢𝑢.(3.48) Then there exists a constant 𝐶2>1 such that for every 𝑥 and 𝑦 which satisfy |𝑥|𝜎(2𝑡) and |𝑥𝑦|𝑡Φ𝑡(𝑥)/2, one has 1𝐶2𝑤(𝑦)𝑤(𝑥)𝐶2𝑤(𝑦).(3.49)

Proof of Theorem 3.3. For a given 𝜀>0, we can select 𝑔, where 𝑔(𝑟1)(𝑥) is absolutely continuous on , and 𝑔(𝑟)𝐿𝑝,𝑤() such that 𝑤(𝑓𝑔)𝐿𝑝()+𝑡𝑟𝑤𝑔(𝑟)𝐿𝑝()𝜀𝒦𝑟,𝑝(𝑤,𝑓,𝑡).(3.50) Let 𝑓=+𝑔, where 𝐿𝑝,𝑤(). Then we have 𝜔𝑟,𝑝(,𝑡)𝐶𝑤𝐿𝑝().(3.51) Let 0<𝑠𝑡/𝑟. From Lemma 3.4, 3.10 and the Hőlder-Minkowski inequality, we have 𝑤Δ𝑟𝑠Φ𝑡(𝑥)𝑔(𝑥)𝐿𝑝(|𝑥|𝜎(2𝑡)),𝐶𝑤(𝑥)𝑠/2𝑠/2𝑟||𝑔(𝑟)||(𝑢)𝑑𝑢1𝑑𝑢2𝑑𝑢𝑟𝐿𝑝(|𝑥|𝜎(2𝑡))noteΦ𝑡(𝑥)𝐶𝐶𝑠/2𝑠/2𝑟||𝑤𝑔(𝑟)||(𝑢)𝑑𝑢1𝑑𝑢2𝑑𝑢𝑟𝐿𝑝(|𝑥|𝜎(2𝑡))𝐶𝑠/2𝑠/2𝑟(𝑤𝑔(𝑟))(𝑢)𝐿𝑝(|𝑥|𝜎(2𝑡))𝑑𝑢1𝑑𝑢2𝑑𝑢𝑟𝐶𝑠𝑟𝑤𝑔(𝑟)𝐿𝑝()𝐶𝑡𝑟𝑤𝑔(𝑟)𝐿𝑝(),(3.52) where 𝑢=𝑥+Φ𝑡(𝑥)(𝑢1+𝑢2++𝑢𝑟). We estimate inf𝑃𝒫𝑟1(𝑔(𝑥)𝑃(𝑥))𝑤(𝑥)𝐿𝑝(|𝑥|𝜎(4𝑡)).(3.53) Then we may suppose 𝑔(𝑗)(0)𝑃(𝑗)(0)=0,𝑗=0,1,,𝑟1.(3.54) Using Lemma 3.5, we see ||𝑄||(𝜎(4𝑡))1𝐶𝑡,(3.55) (see [7, page 12]). By Corollary 3.9 with 𝑘=𝑟 and (3.55) we have 𝑤(𝑔𝑃)𝐿𝑝(|𝑥|𝜎(4𝑡))||𝑄||(𝜎(4𝑡))𝑟1+𝑄(𝑥)2𝑟/2𝑤(𝑔𝑃)𝐿𝑝(|𝑥|𝜎(4𝑡))𝐶𝑡𝑟𝑤𝑔(𝑟)𝐿𝑝().(3.56) Hence, from (3.51), (3.52), and (3.56) we have 𝜔𝑟,𝑝(𝑤,𝑓,𝑡)𝐶𝜀𝐶𝑤𝐿𝑝()+𝑡𝑟𝑤𝑔(𝑟)𝐿𝑝()𝜀𝐶𝒦𝑟,𝑝(𝑤,𝑓,𝑡).(3.57) Consequently, we have 𝜔𝑟,𝑝(𝑤,𝑓,𝑡)𝐶𝒦𝑟,𝑝(𝑤,𝑓,𝑡).(3.58) Therefore, from (3.2) we have (3.3).

Corollary 3.11. For 𝑟=1, one lets 𝑤(𝐶2+), and for integer 𝑟2 one lets 𝑤(𝐶𝑟+). Let 𝑓𝐿𝑝,𝑤(). Then one has 𝜔𝑟,𝑝(𝑤,𝑓,𝑡)𝒦𝑟,𝑝𝒦(𝑤,𝑓,𝑡)𝑟,𝑝(𝑤,𝑓,𝑡).(3.59)

Acknowledgments

The authors thank the referees for many kind suggestions and comments. H. S. Jung was supported by SEOK CHUN Research Fund, Sungkyunkwan University, 2010.