# Weighted Simultaneous Approximation of the Linear Combinations of Baskakov Operators

Authors: Xiao-jun Tang; Xin-chang Wang; Hua Yi
Journal: Complexity
 (2020)

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## Abstract

In this paper, the approximation property of the linear combinations of Baskakov operators is investigated by using a Jacobi weight function. Moreover, both the positive and inverse theorems for the weighted simultaneous approximation of the linear combinations of Baskakov operators are proved.


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## Body

## 1. Introduction

The functional analysis methods are widely used to study the approximation theory since the 20th century. Due to the combination of functional analysis methods and classical analysis techniques, such approximation theories developed quickly and had formed a theoretical system.It is well known that the positive and inverse theories are the most significant problems in the operator approximation theorem. In 1972, some pioneering works had been done by H. Berens and G. Lorentz [1], which lead to a hot topic in related research fields. In the meantime, many kinds of approximation tools had been proposed and widely used in practice, such as smooth modulus and K-functional. In particular, the smooth modulus introduced by Ditzian [2] in 1994 contains the related results by using the classic smooth modulus and the Ditzian-Totik modulus. Moreover, such kinds of smooth modulus are usually used to construct an optimal polynomial to approximate a complicated function [3–5].For anyf∈C0,+∞, the corresponding Baskakov operator [6] is defined as(1)Vnf,x=∑k=0∞fknvn,kx,where Vn,kx=n+k−1kxk1+x−n−k, k=1,2,…,n. Linear combination of the Baskakov operator [7] is defined as follows:(2)Vn,kf,r,x=∑i=0r−1CinVnif,x,where ni and Cin are related constants which satisfy the following conditions:(3)an=n0<n1<…<nr−1≤Cn,b∑i=0r−1Cin≤C,c∑i=0r−1Cin=1,d∑i=0r−1Cinni−ρ=0,ρ=1,2,…,r−1.Letφx=x1+x,0<λ<1,δnx=φx+1/n,wx=xa1+x−b0<a<1,b≥0,⋅ be the supremum or infimum norm, fw=wf∞+f0,(4)ωφλrf,tw=sup0<h≤twxΔhφλxrfx,where(5)Δhφλxrfx=∑k=0r−1krkfx+r2−khφrx,x≥rhφλx20,0<x<rhφλx2,.Hence, the weighted K-functional can be defined as follows:(6)Kr,φλf,trw=infwf−g+trwφλgr:gr−1∈AC.loc,and the corrected weighted K-functional is defined as(7)K¯r,φλf,trw=infwf−g+trwφλgr+tr/1−λ/2wgr:gr−1∈AC.loc.According to [8], one can obtain the relationship between the smooth modulus and K-functional(8)ωφrrf,tw∼Kr,φλf,trw∼K¯r,φλf,trw.Note thatfs∈C0,∞ and s∈N, and we have(9)Vnsf,x=n+s−1!n−1!∑k=0∞∫​01/n…∫01/nfskn+∑i=1suidu1…dusvn+s,kx.Letg∈C0,∞ and s∈N, and introducing an auxiliary operator(10)Vn,sg,x=n+s−1!n−1!∑k=0∞∫​01/n…∫01/ngkn+∑i=1suidu1…dusvn+s,kx,then the linear combination of the auxiliary operator can be defined as follows:(11)Vn,sg,r,x=∑i=0r−1CixVn,sg,x,where ni and Cix are related constants which satisfy (3). Note that Vn,s is a bounded operator, and Vn,s1,x=0. Assuming fs∈C0,+∞, we have(12)Vn,sfs,x=Vnsf,x,(13)Vn,sfs,r,x=Vnsf,r,x.The Baskakov operator has been studied in many research studies by using a lot of deep methods [9–13]. In previous work [14, 15], some pointwise results of the Baskakov operator with weighted approximate were obtained.From the literature [16], some preexisting results are given as follows.Theorem 1.
Lets∈N,0≤λ≤1, fs∈0,∞, then we have(14)Vnsf,x−fs=Oωφλrfs,n−1/2δn1−λx.Theorem 2.
Lets∈N,0≤λ≤1,x∈0,∞, fs∈0,∞,s≤α≤s+r, then we have(15)Vnsf,x−fs=On−1/2δn1−λxα−s⟺ωφλrfs,t=Otα−s.According to [17], the following theorems hold.Theorem 3.
Lets∈N,0≤λ≤1, fs∈0,∞, then we obtain(16)wxVnsf,x−fs=Oωφλrfs,n−1/2δn1−λxw.Theorem 4.
Lets∈N,0≤λ≤1,x∈0,∞, fs∈0,∞,s≤α≤s+r, then we have(17)wxVnsf,x−fs=On−1/2δn1−λxα−s⟺ωφλrfs,tw=Otα−s.According to [18], we get the following theorems.Theorem 5.
Letr,s∈N,0≤λ≤1, fs∈0,∞, then we have(18)Vnsf,r,x−fs=Oωφλrfs,n−1/2δn1−λx.Theorem 6.
Letr,s∈N,0≤λ≤1,x∈0,∞, fs∈0,∞,s≤α≤s+r, then we get(19)Vnsf,r,x−fs=On−1/2δn1−λxα−s⟺ωφλrfs,t=Otα−s.Although there are many research studies about Baskakov operator approximation [19], we can further improve these theories. On the basis of the aforementioned research literature, we firstly applied the equivalence relation between the weighted smooth modulus and K-functional to explore the simultaneous approximation of the linear combination of the Baskakov operator with Jacobi weight. And we obtained both the positive and inverse theorems for the weighted simultaneous approximation of the linear combinations of Baskakov operators. Thereby, we unite and expand the results about the existing smooth modulus ω2f,t ωφ2f,t, ωφ2f,tw, and s=0, where C is a constant.Then, we will introduce some concepts and properties in Section2. In Section 3, we will prove the pros-theorem of simultaneous approximation of the linear combination of the Baskakov operator. The proof of the cons-theorem will be given in Section 4. Finally, we will analyze the equivalence theorem of approximation.


## 2. Concepts and Properties

The weight functionwx that we used has the following properties.Property 1.
Let0<a<1,0<b,x≤0,t≤u≤x, then we have(20)w−1u≤Ct−a1+xb+x−a1+tb,Ωrf,tw=sup0<h≤twΔhrfrh1/1−λ,∞,0<λ<1,Ωφλrf,tw=sup0<h≤twΔhφλrfrh1/1−λ,∞,0<λ<1.
Moreover, weighted smooth modulus and weighted main part smooth modulus have the following connection.Property 2.
ω
r
f
,
t
1
/
1
−
β
w
≤
C
ω
φ
r
f
,
t
w
,
β
<
1.(21)C−1Ωφλrf,tw≤ωφλrf,tw≤C∫0tΩφλrf,tττdτ,Ωφrf,tw≤Crr∫t1Ωφr+1f,τwτr+1dτ+wf.
In order to prove the cons-theorem, we need to introduce a new K-functional and put some symbols firstly. Let0<λ<1,0<α<s, then(22)f0=supx∈0,∞wxφα−sλ−1xfx,fr=supx∈0,∞wxφr+α−sλ−1xfrx,C0=ffx∈C0,∞,f0≤+∞,Cr=ffx∈C0,fr−1∈C0,∞,fr≤+∞.
A new K-functional is defined:(23)Kλα−s=inffs−g0+trgr/2,and a new smooth modulus is defined:(24)ωφλ2f,tα=sup0<h≤tφαλ−1xΔhφλ2fx,x±hφλx>0.


## 3. Pros-Theorem of Simultaneous Approximation of the Linear Combination of the Baskakov Operator

### 3.1. Auxiliary Lemmas

Lemma 1 (see [18]).
Lets∈N,r∈n,m∈n, then we have(25)Vn,st−xm,r,x=0,m=1,2,…,r−1,(26)Vn,st−x2m,x≤Cn−mδn2mx.Lemma 2 (see [15]).
Ifa≥0,b∈R, then(27)∑k=1∞vn,kxkn−a1+knb≤Cx−a1+xb.According to Lemma2, we obtain the following corollaries.Corollary 1.
Whena≥0,b∈R, we have(28)Vn,st−a,x≤Cx−a,(29)Vn,s1+tb,x≤C1+xb.Lemma 3 (see [17]).
Iff∈C0,∞,s∈N, then(30)Vn,sf,xw≤Cfw.Corollary 2.
Iff∈C0,∞,s∈N, then(31)Vn,sf,r,xw≤Cfw.Proof.
It follows from (30) that we get(32)wxVn,sf,r,x=wx∑i=0r−1CiVni,sf,x≤∑i=0r−1CiwxVni,sf,x≤Cfw.Lemma 4.
(1)
Ift≤u≤x,x∈EnC0,1/n, then(33)t−ur−1δnru≤t−xr−1δnrx.(2)
Ift≤u≤x, then(34)t−ur−1φru≤t−xr−1xr/211+xr/2+11+tr/2.Proof.
(1)
Whenx<t, the conclusion is established obviously.For the case ofx>t, let fx=x/δn2x, then we have(35)f′x=δn2x−nδnxx+1/x+x/x+1δn4x=1/n−xx/x+1δn3x.Forx<1/n, f′x>0, so when x in 0,1/n, f increases. Hence, we have u/δn2u<x/δn2x; for x>u, we obtain u2/δn2u<x2/δn2x,ur−2/δnr−2u<xr−2/δnr−2x, and then ur−1/δnru<xr−1/δnrx, combined with inequality t−ux<t−xu; then, we get(36)t−ur−1δnru<t−xr−1δnrx.(2)
Due to1/1+ur/2≤1/1+xr/2+1/1+tr/2, thereby we have(37)ur/2φru≤xr/2xr/211+xr/2+11+tr/2,and combined with inequality t−ux<t−xu, then we get(38)t−ur/2φru≤t−xr/2xr/211+xr/2+11+tr/2,and then combined with inequality t−u<t−x, the proposition is established.Lemma 5.
Letg∈C0,∞, 0≤λ≤1, Rrg,t,x=∫xtt−ur−1grudu,r∈N, then we have(39)wxVn,sRrg,t,x,x≤Cn−r/2δn1−λrxφλrgrw+n−r/2δn1−λrxr/1−λ/2grw.Proof.
(1)
Ift≤u≤x, and x∈EnC0,1/n using Lemma 4, we get(40)wxVn,sRrg,t,x,x=wxVn,s∫xtt−ur−1grudu,x≤CwxδnλrgrwVn,s∫xtt−ur−1δn−λruw−1udu,x≤Cwxδn−λrxδnλrgrwVn,st−xr−1∫xtw−1udu,x≤Cwxδn−λrxδnλrgrwVn,st−xrt−a1+xb+x−a1+tb,x≤Cwxδn−λrxδnλrgrwVn,st−xrt−a1+xb,x+Vn,st−xrx−a1+tb,x≤Cn−r/2δn1−λrxδnλrgrw.In the proof process above, we used the Hőlder inequality and the conclusion of formula (26). i.e.,(41)Vn,st−xrt−a1+xb,x≤CVn,st−x2r,x1/2Vn,st−a1+xb,x1/2≤CVn,st−x2r,x1/2Vn,st−4a,x1/4Vn,s1+x−4a,x1/4≤Cn−r/2δn1−λrxx−a1+xb.Similarly, we obtain(42)Vn,st−xrx−a1+tb,x≤Cn−r/2δn1−λrxx−a1+xb.(2)
Whenx∈1/n,∞, by using the Hőlder inequality, (26), (28), and (29) and the conclusion of [15], we can get(43)wxVn,sRrg,t,x,x=wxVn,s∫xtt−ur−1grudu,x≤CwxδnλrgrwVn,s∫xtt−ur−1δn−λruw−1udu,x≤CwxδnλrgrwVn,sx−tr1xrλ/211+xrλ/2+11+trλ/2t−a1+xb+x−a1+tb,x≤CwxδnλrgrwVn,sx−trφrλxt−a1+xb+x−a1+tb,x+Vn,sx−trx1+trλ/2t−a1+xb+x−a1+tb,x≤Cn−r/2δn1−λrxδnλrgrw.Therefore,(44)Cn−r/2δn1−λrxδnλrgrw≤Cn−r/2δn1−λrxφλrgrw+n−λr/2grw≤Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw,and the proposition is established.


### 3.2. Positive Theorem

Theorem 7.
Lets∈N,0≤λ≤1,fs∈C0,∞, s≤α≤s+r, then we get(45)wxVnsf,r,x−fsx=Oωφλrfs,n−1/2δn1−λxw.Proof.
Forgr−1∈C0,∞, let gt Taylor expand in x,(46)gt=gx+t−xg′x+⋯+t−xr−1r−1!gr−1x+1r−1!∫xtt−ur−1grudu.
It follows from (25) that we obtain(47)wxVn,sg,r,x−gx=wxVn,s1r−1!∫xtt−ur−1grudu,r,x≤∑i=0r−1CiwxVn,s1r−1!∫xtt−ur−1grudu,x≤Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw.
Then, forfs∈C0,∞ and all of gr−1∈A.C.loc, we have(48)wxVnsf,r,x−fsx≤wxVn,sf−g,r,x+wxfsx−gx+wxVn,sf,r,x−gx≤Cfsx−gxw+Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw≤K¯r,φλfs,n−1/2δn1−λxrw≤Cωφλrfs,n−1/2δn1−λxw.Theorem 8.
Lets∈N,0≤λ≤1,x∈0,∞,fs∈C0,∞,s≤α≤s+r, and if ωφλrfs,tw=Otα−s, then we have(49)wxVnsf,r,x−fsx=On−1/2δn1−λxα−s.Proof.
From Theorem7, we obtain(50)wxVnsf,r,x−fsx=Oωφλrfs,n−1/2δn1−λxw=On−1/2δn1−λxα−s.


## 3.1. Auxiliary Lemmas

Lemma 1 (see [18]).
Lets∈N,r∈n,m∈n, then we have(25)Vn,st−xm,r,x=0,m=1,2,…,r−1,(26)Vn,st−x2m,x≤Cn−mδn2mx.Lemma 2 (see [15]).
Ifa≥0,b∈R, then(27)∑k=1∞vn,kxkn−a1+knb≤Cx−a1+xb.According to Lemma2, we obtain the following corollaries.Corollary 1.
Whena≥0,b∈R, we have(28)Vn,st−a,x≤Cx−a,(29)Vn,s1+tb,x≤C1+xb.Lemma 3 (see [17]).
Iff∈C0,∞,s∈N, then(30)Vn,sf,xw≤Cfw.Corollary 2.
Iff∈C0,∞,s∈N, then(31)Vn,sf,r,xw≤Cfw.Proof.
It follows from (30) that we get(32)wxVn,sf,r,x=wx∑i=0r−1CiVni,sf,x≤∑i=0r−1CiwxVni,sf,x≤Cfw.Lemma 4.
(1)
Ift≤u≤x,x∈EnC0,1/n, then(33)t−ur−1δnru≤t−xr−1δnrx.(2)
Ift≤u≤x, then(34)t−ur−1φru≤t−xr−1xr/211+xr/2+11+tr/2.Proof.
(1)
Whenx<t, the conclusion is established obviously.For the case ofx>t, let fx=x/δn2x, then we have(35)f′x=δn2x−nδnxx+1/x+x/x+1δn4x=1/n−xx/x+1δn3x.Forx<1/n, f′x>0, so when x in 0,1/n, f increases. Hence, we have u/δn2u<x/δn2x; for x>u, we obtain u2/δn2u<x2/δn2x,ur−2/δnr−2u<xr−2/δnr−2x, and then ur−1/δnru<xr−1/δnrx, combined with inequality t−ux<t−xu; then, we get(36)t−ur−1δnru<t−xr−1δnrx.(2)
Due to1/1+ur/2≤1/1+xr/2+1/1+tr/2, thereby we have(37)ur/2φru≤xr/2xr/211+xr/2+11+tr/2,and combined with inequality t−ux<t−xu, then we get(38)t−ur/2φru≤t−xr/2xr/211+xr/2+11+tr/2,and then combined with inequality t−u<t−x, the proposition is established.Lemma 5.
Letg∈C0,∞, 0≤λ≤1, Rrg,t,x=∫xtt−ur−1grudu,r∈N, then we have(39)wxVn,sRrg,t,x,x≤Cn−r/2δn1−λrxφλrgrw+n−r/2δn1−λrxr/1−λ/2grw.Proof.
(1)
Ift≤u≤x, and x∈EnC0,1/n using Lemma 4, we get(40)wxVn,sRrg,t,x,x=wxVn,s∫xtt−ur−1grudu,x≤CwxδnλrgrwVn,s∫xtt−ur−1δn−λruw−1udu,x≤Cwxδn−λrxδnλrgrwVn,st−xr−1∫xtw−1udu,x≤Cwxδn−λrxδnλrgrwVn,st−xrt−a1+xb+x−a1+tb,x≤Cwxδn−λrxδnλrgrwVn,st−xrt−a1+xb,x+Vn,st−xrx−a1+tb,x≤Cn−r/2δn1−λrxδnλrgrw.In the proof process above, we used the Hőlder inequality and the conclusion of formula (26). i.e.,(41)Vn,st−xrt−a1+xb,x≤CVn,st−x2r,x1/2Vn,st−a1+xb,x1/2≤CVn,st−x2r,x1/2Vn,st−4a,x1/4Vn,s1+x−4a,x1/4≤Cn−r/2δn1−λrxx−a1+xb.Similarly, we obtain(42)Vn,st−xrx−a1+tb,x≤Cn−r/2δn1−λrxx−a1+xb.(2)
Whenx∈1/n,∞, by using the Hőlder inequality, (26), (28), and (29) and the conclusion of [15], we can get(43)wxVn,sRrg,t,x,x=wxVn,s∫xtt−ur−1grudu,x≤CwxδnλrgrwVn,s∫xtt−ur−1δn−λruw−1udu,x≤CwxδnλrgrwVn,sx−tr1xrλ/211+xrλ/2+11+trλ/2t−a1+xb+x−a1+tb,x≤CwxδnλrgrwVn,sx−trφrλxt−a1+xb+x−a1+tb,x+Vn,sx−trx1+trλ/2t−a1+xb+x−a1+tb,x≤Cn−r/2δn1−λrxδnλrgrw.Therefore,(44)Cn−r/2δn1−λrxδnλrgrw≤Cn−r/2δn1−λrxφλrgrw+n−λr/2grw≤Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw,and the proposition is established.


## 3.2. Positive Theorem

Theorem 7.
Lets∈N,0≤λ≤1,fs∈C0,∞, s≤α≤s+r, then we get(45)wxVnsf,r,x−fsx=Oωφλrfs,n−1/2δn1−λxw.Proof.
Forgr−1∈C0,∞, let gt Taylor expand in x,(46)gt=gx+t−xg′x+⋯+t−xr−1r−1!gr−1x+1r−1!∫xtt−ur−1grudu.
It follows from (25) that we obtain(47)wxVn,sg,r,x−gx=wxVn,s1r−1!∫xtt−ur−1grudu,r,x≤∑i=0r−1CiwxVn,s1r−1!∫xtt−ur−1grudu,x≤Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw.
Then, forfs∈C0,∞ and all of gr−1∈A.C.loc, we have(48)wxVnsf,r,x−fsx≤wxVn,sf−g,r,x+wxfsx−gx+wxVn,sf,r,x−gx≤Cfsx−gxw+Cn−r/2δn1−λrxφλrgrw+n−λr/2δn1−λrxr/1−λ/2grw≤K¯r,φλfs,n−1/2δn1−λxrw≤Cωφλrfs,n−1/2δn1−λxw.Theorem 8.
Lets∈N,0≤λ≤1,x∈0,∞,fs∈C0,∞,s≤α≤s+r, and if ωφλrfs,tw=Otα−s, then we have(49)wxVnsf,r,x−fsx=On−1/2δn1−λxα−s.Proof.
From Theorem7, we obtain(50)wxVnsf,r,x−fsx=Oωφλrfs,n−1/2δn1−λxw=On−1/2δn1−λxα−s.


## 4. Cons-Theorem of Simultaneous Approximation of the Linear Combination of the Baskakov Operator

### 4.1. Base Lemmas

Lemma 6.
If0≤λ≤1,r∈N,s≤α≤s+r,f∈C0, we have(51)Vn,sf,r,xr≤Cnr/2f0.Proof.
(1)
Whenx∈0,1/n, we have φx≤1/n, and then(52)Vn,srf,x≤Cnr+s∑k=0∞vn+s+r,kx∫​01/n…∫01/n∑j=0r−1rrjfk+rn+∑i=1suidu1…dus.Firstly, considering the first term of∑j=0r−1rrjfk+r−j/n+∑​i=1suidu1…dus, the following part is similar, and by using the methods of [17] and the Hǒlder inequality, we get(53)nr+s∑k=0∞vn+s+r,kx∫​01/n…∫​01/n∑j=0r−1rrjfk+rn+∑i=1suidu1…dus≤Cf0nr∑k=0∞vn+s,kxk+r/nα−s1−λ/2−a1+k+rnα−s1−λ/2+b≤Cnrnα−s1−λ/2w−1xf0.Therefore,(54)Vn,sf,xr≤Cwxφr+α−sλ−1nrnα−s1−λ/2w−1xf0≤Cnr/2f0.(2)
Whenx∈1/n,∞, similar to the proof of [8], we have(55)Vn,srf,x≤Cf0ns∑i=0rn+sφ2xr+i/2∑k=0∞vn+s,kkn+s−xi⋅∫​01/n…∫01/nw−1φα−s1−λkn+∑j=0sujdu1…dus≡I1+I2,where I1 is the situation when k=0, and I2 is the last situations. Let p∈N, s.t. i/2p<1−α−s1−λ/2, by the Hǒlder inequality, then we have(56)I2≤Cf0∑i=0rn+sφ2xr+i/2∑k=1∞vn+s,kxkn+s−xiw−1φα−s1−λkn≤Cf0∑i=0rnφ2xr+i/2∑k=1∞vn+s,kxφ2knα−s1−λ/2⋅∑k=1∞vn+s,kxkn+s−x2pi/2p∑k=1∞vn+s,kxwkn−1/1−α−s1−λ/2−i/2p1−α−s1−λ/2−i/2p≤Cf0nr/2w−1xφ−r+α−s1−λxI1≤Cf0∑i=0rn+sφ2xr+i/2ns1+x−n+sxi∫​01/n…∫01/nw−1φα−s1−λ∑j=0sujdu1…dus≤Cf0nr/2w−1xφ−r+α−s1−λx.
Then, we obtainVn,sf,xr≤Cnr/2f0; therefore,(57)Vn,sf,r,xr≤Vn,sf,xr≤Cnr/2f0.Lemma 7.
Supposing that0≤λ≤1,r∈N,s≤α≤s+r, we get(58)Vn,sf,r,xr≤Cfr.Proof.
Supposing thatFx satisfies Fsx=fx, and then by reference [8], we have(59)wxφr+α−s1−λxVn,srf,x≤wxφr+α−s1−λxnsn−1!n+s−1!Vnr+sF,x≤Cnr+swxφr+α−s1−λ∑k=0∞Δ⟶1/nr+sFknvn+r+s,kx,where Δ⟶1/nr+sFk/n=∑j=0r+s−1jr+sjFk+r+s−j/n, and when k≥1, we get(60)nr+sΔ⟶1/nr+sFkn≤supk/n≤θ≤k+r+s/nFr+sx≤Ckn−a−r+α−sλ−1/21+knb−r+α−sλ−1/2fr.
Whenk=0, using the method of [8], we can obtain(61)nr+sΔ⟶1/nr+sF0≤Cw−1xn−r+α−sλ−1/2fr.
So, using the Hőlder inequality and Lemma2, we get(62)wxφr+α−sλ−1xVn,srf,x≤Cwxφr+α−sλ−1xfr∑k=0∞kn−a−r+α−sλ−1/21+knb−r+α−sλ−1/2vn+r+s,kx≤Cfr,but Vn,sf,r,xr≤CVn,sf,xr≤Cfr, and thereby the conclusion is established.Lemma 8 (see [20]).
Supposingr∈N,0≤λ≤1,0<h<1/2λr,x>rhφλx/2, then we have(63)wx∫​−hφλx/2hφλx/2…∫−hφλx/2hφλx/2φ−rx+∑i=0ruiw−1x+∑i=0ruidu1…dur≤Chr.


### 4.2. Cons-Theorem

Theorem 9.
Supposings∈N,0≤λ≤1,x∈0,∞,fs∈0,∞,s≤α≤s+r, then we have(64)wxVnsf,r,x−fs=On−1/2δn1−λxα−s⟺ωφλrfs,t=Otα−s.Proof.
By the definition ofKλα−sf,trw, selecting a suitable g, which satisfies the following formula(65)fs−g0+n−r/2gr≤2Kλα−sf,n−r/2w,by the known condition(66)wxVnsfs,r,x−fs=On−1/2δn1−λxα−s,we have(67)wxφα−sλ−1xVn,sfs,r,x−fsx≤Cn−α−s/2⟹Vnsfs,r,x−fsx0≤Cn−α−s/2.
By using Lemmas6 and 7, we obtain(68)Kλα−sfs,trw≤Vn,sfs,r,x−fsx0+trVn,sfs,r,xr≤Cn−α−s/2+trVn,sfs−g,r,xr+Vn,sg,r,xr≤Cn−α−s/2+trn−r/2fs−g0+trgr≤Cn−α−s/2+trn−r/2Kλα−sf,n−r/2w,by using the Berens-Lorentz lemma [8], we get Kλα−sfs,trw≤Ctα−s. Especially, when λ=1, we can obtain K1α−sfs,trw=Kφfs,trw≤Ctα−s, i.e., ωφrfs,t=Otα−s. By contrast, when 0≤λ≤1,x≥rhφλx/2,x∈rh1/1−λ,∞, for f∈Cλ,w, by using Lemma 8, we get(69)wxΔhφλxrfsx≤wxΔhφλxrfs−gx+wxΔhφλxrgx=wx∑k=0r−1krkfs−gx+r−khφλx+wx∫​−rhφλx/2rhφλx/2…∫−rhφλx/2rhφλx/2grx+∑i=0ruidu1…dur≤Cφ−α−sλ−1xfs−g0+grhrφrλ−1x≤Cφ−α−sλ−1xKλα−sfs,hrφrλ−1xw≤Chα−s.
Then,Ωφλrfs,tw=Otα−s, and thereby(70)ωφλrfs,tw≤C∫0tΩφλrfs,τwτdτ≤C∫0tτα−s−1dτ≤Ctα−s.


## 4.1. Base Lemmas

Lemma 6.
If0≤λ≤1,r∈N,s≤α≤s+r,f∈C0, we have(51)Vn,sf,r,xr≤Cnr/2f0.Proof.
(1)
Whenx∈0,1/n, we have φx≤1/n, and then(52)Vn,srf,x≤Cnr+s∑k=0∞vn+s+r,kx∫​01/n…∫01/n∑j=0r−1rrjfk+rn+∑i=1suidu1…dus.Firstly, considering the first term of∑j=0r−1rrjfk+r−j/n+∑​i=1suidu1…dus, the following part is similar, and by using the methods of [17] and the Hǒlder inequality, we get(53)nr+s∑k=0∞vn+s+r,kx∫​01/n…∫​01/n∑j=0r−1rrjfk+rn+∑i=1suidu1…dus≤Cf0nr∑k=0∞vn+s,kxk+r/nα−s1−λ/2−a1+k+rnα−s1−λ/2+b≤Cnrnα−s1−λ/2w−1xf0.Therefore,(54)Vn,sf,xr≤Cwxφr+α−sλ−1nrnα−s1−λ/2w−1xf0≤Cnr/2f0.(2)
Whenx∈1/n,∞, similar to the proof of [8], we have(55)Vn,srf,x≤Cf0ns∑i=0rn+sφ2xr+i/2∑k=0∞vn+s,kkn+s−xi⋅∫​01/n…∫01/nw−1φα−s1−λkn+∑j=0sujdu1…dus≡I1+I2,where I1 is the situation when k=0, and I2 is the last situations. Let p∈N, s.t. i/2p<1−α−s1−λ/2, by the Hǒlder inequality, then we have(56)I2≤Cf0∑i=0rn+sφ2xr+i/2∑k=1∞vn+s,kxkn+s−xiw−1φα−s1−λkn≤Cf0∑i=0rnφ2xr+i/2∑k=1∞vn+s,kxφ2knα−s1−λ/2⋅∑k=1∞vn+s,kxkn+s−x2pi/2p∑k=1∞vn+s,kxwkn−1/1−α−s1−λ/2−i/2p1−α−s1−λ/2−i/2p≤Cf0nr/2w−1xφ−r+α−s1−λxI1≤Cf0∑i=0rn+sφ2xr+i/2ns1+x−n+sxi∫​01/n…∫01/nw−1φα−s1−λ∑j=0sujdu1…dus≤Cf0nr/2w−1xφ−r+α−s1−λx.
Then, we obtainVn,sf,xr≤Cnr/2f0; therefore,(57)Vn,sf,r,xr≤Vn,sf,xr≤Cnr/2f0.Lemma 7.
Supposing that0≤λ≤1,r∈N,s≤α≤s+r, we get(58)Vn,sf,r,xr≤Cfr.Proof.
Supposing thatFx satisfies Fsx=fx, and then by reference [8], we have(59)wxφr+α−s1−λxVn,srf,x≤wxφr+α−s1−λxnsn−1!n+s−1!Vnr+sF,x≤Cnr+swxφr+α−s1−λ∑k=0∞Δ⟶1/nr+sFknvn+r+s,kx,where Δ⟶1/nr+sFk/n=∑j=0r+s−1jr+sjFk+r+s−j/n, and when k≥1, we get(60)nr+sΔ⟶1/nr+sFkn≤supk/n≤θ≤k+r+s/nFr+sx≤Ckn−a−r+α−sλ−1/21+knb−r+α−sλ−1/2fr.
Whenk=0, using the method of [8], we can obtain(61)nr+sΔ⟶1/nr+sF0≤Cw−1xn−r+α−sλ−1/2fr.
So, using the Hőlder inequality and Lemma2, we get(62)wxφr+α−sλ−1xVn,srf,x≤Cwxφr+α−sλ−1xfr∑k=0∞kn−a−r+α−sλ−1/21+knb−r+α−sλ−1/2vn+r+s,kx≤Cfr,but Vn,sf,r,xr≤CVn,sf,xr≤Cfr, and thereby the conclusion is established.Lemma 8 (see [20]).
Supposingr∈N,0≤λ≤1,0<h<1/2λr,x>rhφλx/2, then we have(63)wx∫​−hφλx/2hφλx/2…∫−hφλx/2hφλx/2φ−rx+∑i=0ruiw−1x+∑i=0ruidu1…dur≤Chr.


## 4.2. Cons-Theorem

Theorem 9.
Supposings∈N,0≤λ≤1,x∈0,∞,fs∈0,∞,s≤α≤s+r, then we have(64)wxVnsf,r,x−fs=On−1/2δn1−λxα−s⟺ωφλrfs,t=Otα−s.Proof.
By the definition ofKλα−sf,trw, selecting a suitable g, which satisfies the following formula(65)fs−g0+n−r/2gr≤2Kλα−sf,n−r/2w,by the known condition(66)wxVnsfs,r,x−fs=On−1/2δn1−λxα−s,we have(67)wxφα−sλ−1xVn,sfs,r,x−fsx≤Cn−α−s/2⟹Vnsfs,r,x−fsx0≤Cn−α−s/2.
By using Lemmas6 and 7, we obtain(68)Kλα−sfs,trw≤Vn,sfs,r,x−fsx0+trVn,sfs,r,xr≤Cn−α−s/2+trVn,sfs−g,r,xr+Vn,sg,r,xr≤Cn−α−s/2+trn−r/2fs−g0+trgr≤Cn−α−s/2+trn−r/2Kλα−sf,n−r/2w,by using the Berens-Lorentz lemma [8], we get Kλα−sfs,trw≤Ctα−s. Especially, when λ=1, we can obtain K1α−sfs,trw=Kφfs,trw≤Ctα−s, i.e., ωφrfs,t=Otα−s. By contrast, when 0≤λ≤1,x≥rhφλx/2,x∈rh1/1−λ,∞, for f∈Cλ,w, by using Lemma 8, we get(69)wxΔhφλxrfsx≤wxΔhφλxrfs−gx+wxΔhφλxrgx=wx∑k=0r−1krkfs−gx+r−khφλx+wx∫​−rhφλx/2rhφλx/2…∫−rhφλx/2rhφλx/2grx+∑i=0ruidu1…dur≤Cφ−α−sλ−1xfs−g0+grhrφrλ−1x≤Cφ−α−sλ−1xKλα−sfs,hrφrλ−1xw≤Chα−s.
Then,Ωφλrfs,tw=Otα−s, and thereby(70)ωφλrfs,tw≤C∫0tΩφλrfs,τwτdτ≤C∫0tτα−s−1dτ≤Ctα−s.


## 5. Equivalence Theorem

Theorem 10.
Supposings∈N,0≤λ≤1,x∈0,∞,fs∈C0,∞,s≤α≤s+r, then the following propositions are equivalent:(71)1wxVnsf,r,x−fs=On−1/2δn1−λxα−s,(72)2ωφλrfs,tw=Otα−s,(73)3ωφλr+sfs,tw=Otα.Proof.
It follows from Theorem8, and we know (71)⟹(72) is established. And from Theorem 9, (72)⟹(71) is also established. Finally, by reference [8], we know (72)⟺(73) is established, too.



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