# The Approximation of a Modified Baskakov Operator

Authors: Ma Yingdian; Wang Weimeng
Journal: Journal of Mathematics
 (2023)

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## Abstract

Because of their simple form and high quality, linear arithmetical operators, particularly linear positive operators, are very popular. The number of in-depth studies on linear operator approximation is extensive, and the majority of the published material falls into four categories: types and structures of operators, approximation of operators, order of approximation and the converse theorem of operators, and operator saturation. The Baskakov operator will be transformed in the same way to explore its approximation characteristics as well as the approximation theorem and converse theorem.


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## Body

## 1. Introduction and Background

### 1.1. Background

The famous Korovkin theorem [1], back in 1952, was a great boost to the studies of the positive linear arithmetic approximation. The results of the studies focusing on the problems with the order of approximation and the saturation of linear arithmetic were concluded in Devore’s book in 1972 [2]. In that same year, Berens and Lorentz [3] solved the converse theorem of the approximation problem of the Bernstein operator, and the converse theorem and its equivalence theorems have become a research hotspot since then. The subsequent development of the modulus of smoothness and the Korovkin-functionals, as well as some other qualified linear arithmetic, drew a large number of research studies and applications. Ditzian and Totik’s paper from 1987 [4] detailed the key findings.Many academics have developed tweaks and revisions to various well-known operators to speed up the approximation process [5]. Integral transform, generalisation, weighting, combinatorics, and applying probability approaches are some of the methods for making such alterations and corrections [6].We know that the Bernstein operator is linear, so there is(1)Bnei;x=xi,i=0,1.Here,ei=ti. However, the Bernstein Operator does not have its retention with quadratic function,(2)Bne2;x=x2+x1−xn≠e2.In order to keep the quadratic function constant, a transformation was applied to the Bernstein operator in the reference literature [7].(3)Bn∗f,x=∑k=0nfknnkrn∗xk1−rn∗xn−k.Here,(4)rn∗x=x2,n=1,−12n−1+nn−1x2+14n−12,n=2,3,⋯.Thus,(5)Bn∗ei;x=xi,i=0,2,Bn∗e1;x=rn∗x.Other literature [8, 9] have done similar transformations to the Meyer-König and Zeller operators; we will perform the same transformation to the Baskakov operator to study its approximation properties, the approximation theorem, and the converse theorem.


### 1.2. Modulus of Smoothness and theK-Functional

Assumef∈Lp0,+∞1≤P≤+∞ (asquare-integrable function on 0,+∞, written as L∞0,+∞≜C0,+∞), ϕx=x1+x, 0≤λ≤1, so the Rth order modulus of smoothness is defined as(6)ωϕλrf,tp=sup0<h≤tΔhϕλrfp.With(7)Δhφλrfx=∑k=0r−1krkfx+r2−khφλx,x±r2hφλx∈0,∞,0,other.Whenλ=0 in specific, it is called the classical modulus of smoothness; and when λ= 1, it is called the Ditzian–Totik modulus of smoothness. The Rth order K-functional of f is defined as(8)Kϕλrf,tp=infgr−1∈AClocf−gp+trφrλgrp.Here,A.Cloc is defined to be a sum set of absolute continuous functions on 0,+∞. Based on our reference literature [10], we know Kφλrf,trp is equivalent to ωϕλrf,tp, which means there exists a constant c>0,t0>0, so that when 0<t<t0:(9)C−1ωϕλrf,tp≤Kφλrf,trp≤Cωϕλrf,tp.Note: whenp=∞, we use ωφλrf,t∞=ωφλrf,t; Kφλrf,t∞=Kφλrf,t; f∞=f for convenience.


### 1.3. Baskakov Operator

Forf∈C0,∞, the famous Baskakov operator is defined as(10)Vnf,x=∑k=0∞vn,kxfkn.Wherex∈0,∞, vn,kx  =  n+k−1kxk1+x−n−k. Calculation shows(11)Vn1,x=1;Vnt,x=x;Vnt2,x=x2+ϕ2xn.This demonstrates that the Baskakov operator preserves the function’s linearity but not its quadratic function [11]. There have been several in-depth studies on the approximation of the Baskakov operator, the results of which are documented in the literature [12–19]. When we apply the procedure described in [2] to the Baskakov operator, we can easily get an approximation result using the first-order classical modulus of smoothness as follows:Theorem 1.
Ifδ>0,f∈C0,∞, then(12)Vnf,x−fx≤ωf,δ1+1δx1+xn,x∈0,∞.Literature [19] used the generalized modulus of smoothness ωϕλ2f,t to study the approximation theorem and its reverse theorem of Vnf,x and has gottenTheorem 2.
Letf∈CB0,+∞, 0≤λ≤1, then(13)Vnf,x−fx≤Cωφλ2f;n−1/2φ1−λx,x∈0,∞.Theorem 3.
Letf∈CB0,+∞, 0≤λ≤1; 0≤α≤1, then(14)Vnf,x−fx≤Cn−1/2φ1−λxα⟹ωφλ2f;t=Otα.Here, Theorems2 and 3 have unified the results of the classic modulus of smoothness and the Ditzian–Totik modulus correspondingly.


### 1.4. The Structure of the Modified Baskakov Operator

Literature [4] has addressed the linear combination to increase the order of approximation of the Baskakov operator, and this thesis will execute the following adjustment to the Baskakov operator:(15)Vn∗f,x≕Vnf,rn∗x=∑k=0∞vn,krn∗xfkn,where vn,kx keeps the same,(16)rn∗x=x2,n=1,−12n+1+nn+1x2+14n+12,n=2,3,⋯.Obviously,rn∗x≥0,x∈0,+∞n=1,2,...... Here we know that Vn∗ is linear positive operator on C0,∞.(17)Vn∗eix=xi,i=0,2whileVn∗e1,x=rn∗x.This means this operator remains constant to function 1 andx2.


### 1.5. Conclusion

For the modified Baskakov operatorVn∗f,x, we have the following conclusions:Theorem 4.
Assumef∈C0,∞,and there exists tMf>0,l≥0, which let(18)ft≤Mf1+t2tl+11/2,t∈0,∞.So,(19)limn⟶∞Vn∗f−fCD=0,where D=α,β, ⊂0,∞ (any closed interval).Theorem 5.
Assumef∈C0,∞,x∈0,∞, δ>0, so(20)Vn∗f,x−fx≤ωf;δ1+1δ2xx−rn∗x.Wherern∗x is the same with expression (16)Theorem 6.
Assumef∈C0,∞, 0≤λ≤1, so(21)≤CVn∗f,x−fx≤Cωϕλ1f,n−1/2ϕ1−λx.Theorem 7.
Assumef∈C0,∞, 0≤λ≤1, 0<α<1, so(22)Vn∗f,x−fx=On−1/2ϕ1−λxα⟺ωϕλ1f,t=Otα.The approximation features of the modified Baskakov operator will be discussed in Chapter II (Theorems4 and 5). The approximation theorem and its reverse theorem will be discussed in Chapter III (Theorems 6 and 7).Let us restate the important notations that will be used:(1)
A=OB means there is constant C that makes A≤CB;(2)
ϕx=x1+x;(3)
fp =: ∫01fxpdx1/p, f∞=supx∈0,+∞fx;(4)
N means the positive integer set, N0=N∪0;(5)
A.Cloc is defined as an absolute continuous function space on 0,+∞;(6)
C means a constant, it refers to different values according to the situation.


## 1.1. Background

The famous Korovkin theorem [1], back in 1952, was a great boost to the studies of the positive linear arithmetic approximation. The results of the studies focusing on the problems with the order of approximation and the saturation of linear arithmetic were concluded in Devore’s book in 1972 [2]. In that same year, Berens and Lorentz [3] solved the converse theorem of the approximation problem of the Bernstein operator, and the converse theorem and its equivalence theorems have become a research hotspot since then. The subsequent development of the modulus of smoothness and the Korovkin-functionals, as well as some other qualified linear arithmetic, drew a large number of research studies and applications. Ditzian and Totik’s paper from 1987 [4] detailed the key findings.Many academics have developed tweaks and revisions to various well-known operators to speed up the approximation process [5]. Integral transform, generalisation, weighting, combinatorics, and applying probability approaches are some of the methods for making such alterations and corrections [6].We know that the Bernstein operator is linear, so there is(1)Bnei;x=xi,i=0,1.Here,ei=ti. However, the Bernstein Operator does not have its retention with quadratic function,(2)Bne2;x=x2+x1−xn≠e2.In order to keep the quadratic function constant, a transformation was applied to the Bernstein operator in the reference literature [7].(3)Bn∗f,x=∑k=0nfknnkrn∗xk1−rn∗xn−k.Here,(4)rn∗x=x2,n=1,−12n−1+nn−1x2+14n−12,n=2,3,⋯.Thus,(5)Bn∗ei;x=xi,i=0,2,Bn∗e1;x=rn∗x.Other literature [8, 9] have done similar transformations to the Meyer-König and Zeller operators; we will perform the same transformation to the Baskakov operator to study its approximation properties, the approximation theorem, and the converse theorem.


## 1.2. Modulus of Smoothness and theK-Functional

Assumef∈Lp0,+∞1≤P≤+∞ (asquare-integrable function on 0,+∞, written as L∞0,+∞≜C0,+∞), ϕx=x1+x, 0≤λ≤1, so the Rth order modulus of smoothness is defined as(6)ωϕλrf,tp=sup0<h≤tΔhϕλrfp.With(7)Δhφλrfx=∑k=0r−1krkfx+r2−khφλx,x±r2hφλx∈0,∞,0,other.Whenλ=0 in specific, it is called the classical modulus of smoothness; and when λ= 1, it is called the Ditzian–Totik modulus of smoothness. The Rth order K-functional of f is defined as(8)Kϕλrf,tp=infgr−1∈AClocf−gp+trφrλgrp.Here,A.Cloc is defined to be a sum set of absolute continuous functions on 0,+∞. Based on our reference literature [10], we know Kφλrf,trp is equivalent to ωϕλrf,tp, which means there exists a constant c>0,t0>0, so that when 0<t<t0:(9)C−1ωϕλrf,tp≤Kφλrf,trp≤Cωϕλrf,tp.Note: whenp=∞, we use ωφλrf,t∞=ωφλrf,t; Kφλrf,t∞=Kφλrf,t; f∞=f for convenience.


## 1.3. Baskakov Operator

Forf∈C0,∞, the famous Baskakov operator is defined as(10)Vnf,x=∑k=0∞vn,kxfkn.Wherex∈0,∞, vn,kx  =  n+k−1kxk1+x−n−k. Calculation shows(11)Vn1,x=1;Vnt,x=x;Vnt2,x=x2+ϕ2xn.This demonstrates that the Baskakov operator preserves the function’s linearity but not its quadratic function [11]. There have been several in-depth studies on the approximation of the Baskakov operator, the results of which are documented in the literature [12–19]. When we apply the procedure described in [2] to the Baskakov operator, we can easily get an approximation result using the first-order classical modulus of smoothness as follows:Theorem 1.
Ifδ>0,f∈C0,∞, then(12)Vnf,x−fx≤ωf,δ1+1δx1+xn,x∈0,∞.Literature [19] used the generalized modulus of smoothness ωϕλ2f,t to study the approximation theorem and its reverse theorem of Vnf,x and has gottenTheorem 2.
Letf∈CB0,+∞, 0≤λ≤1, then(13)Vnf,x−fx≤Cωφλ2f;n−1/2φ1−λx,x∈0,∞.Theorem 3.
Letf∈CB0,+∞, 0≤λ≤1; 0≤α≤1, then(14)Vnf,x−fx≤Cn−1/2φ1−λxα⟹ωφλ2f;t=Otα.Here, Theorems2 and 3 have unified the results of the classic modulus of smoothness and the Ditzian–Totik modulus correspondingly.


## 1.4. The Structure of the Modified Baskakov Operator

Literature [4] has addressed the linear combination to increase the order of approximation of the Baskakov operator, and this thesis will execute the following adjustment to the Baskakov operator:(15)Vn∗f,x≕Vnf,rn∗x=∑k=0∞vn,krn∗xfkn,where vn,kx keeps the same,(16)rn∗x=x2,n=1,−12n+1+nn+1x2+14n+12,n=2,3,⋯.Obviously,rn∗x≥0,x∈0,+∞n=1,2,...... Here we know that Vn∗ is linear positive operator on C0,∞.(17)Vn∗eix=xi,i=0,2whileVn∗e1,x=rn∗x.This means this operator remains constant to function 1 andx2.


## 1.5. Conclusion

For the modified Baskakov operatorVn∗f,x, we have the following conclusions:Theorem 4.
Assumef∈C0,∞,and there exists tMf>0,l≥0, which let(18)ft≤Mf1+t2tl+11/2,t∈0,∞.So,(19)limn⟶∞Vn∗f−fCD=0,where D=α,β, ⊂0,∞ (any closed interval).Theorem 5.
Assumef∈C0,∞,x∈0,∞, δ>0, so(20)Vn∗f,x−fx≤ωf;δ1+1δ2xx−rn∗x.Wherern∗x is the same with expression (16)Theorem 6.
Assumef∈C0,∞, 0≤λ≤1, so(21)≤CVn∗f,x−fx≤Cωϕλ1f,n−1/2ϕ1−λx.Theorem 7.
Assumef∈C0,∞, 0≤λ≤1, 0<α<1, so(22)Vn∗f,x−fx=On−1/2ϕ1−λxα⟺ωϕλ1f,t=Otα.The approximation features of the modified Baskakov operator will be discussed in Chapter II (Theorems4 and 5). The approximation theorem and its reverse theorem will be discussed in Chapter III (Theorems 6 and 7).Let us restate the important notations that will be used:(1)
A=OB means there is constant C that makes A≤CB;(2)
ϕx=x1+x;(3)
fp =: ∫01fxpdx1/p, f∞=supx∈0,+∞fx;(4)
N means the positive integer set, N0=N∪0;(5)
A.Cloc is defined as an absolute continuous function space on 0,+∞;(6)
C means a constant, it refers to different values according to the situation.


## 2. The Approximation of the Modified Baskakov Operator

First of all, let us introduce the Korovkin theorem built by Ditzian for the local convergence of linear positive operators. LetT−∞,+∞ or 0,∞, μt∈CT and μt≥1t∈T. Written as follows:(23)CμT=f∈CT|ft≤Mf1+t2μt,t∈T,where Mf is a positive constant related to f.Theorem 8 (see [20]).
LetLnn∈N be a linear positive operator sequence on CT, and D=α,β⊂T applies to the following conditions:(1)
Fork=0,1,2, there’s limn⟶∞Lnei−eiCD=0,(2)
limn⟶∞maxx∈DLnt−x2μt,x=0.Then, for everyf∈CμT, there is(24)limn⟶∞Lnf−fCD=0.Apply this theorem to operatorVn∗, and we can getTheorem 9.
Letf∈C0,∞, and there exists Mf>0,l≥0 that makes: ft≤Mf1+t2tl+11/2t∈0,∞,
Thus, there is(25)limn⟶∞Vn∗f−fCD=0,where D=α,β, ⊂0,∞ (any closed interval).Proof.
Maker∈N, then there is(26)Vn∗er+1;x=∑k=0∞knr+1vn,krn∗x=∑k=0∞kn+1nrn+knvn,krn∗xrn∗x1+rn∗x=rn∗x1+rn∗x∑j=0rrj1nr−jVn∗ej;x+rn∗x1+rn∗x∑j=0rrj1nr−jVn∗ej+1;x.
Therefore, a recurrence relation exists(27)Vn∗er+1;x=rn∗xnrVn∗e0;x+rn∗x∑j=1rrj1nr−j+rj−11nr−j+1Vn∗ej;x,Vn∗e0;x=1;Vn∗e1;x=rn∗x;Vn∗e2;x=x2.
Therefore,(28)Vn∗e3;x=rn∗xx2+2x2+rn∗xn+1+rn∗xn2,Vn∗e4;x=rn∗2xx2+rn∗xn3x2+5x2+2rn∗xrn∗x+rn∗xn23x2+22+3x2rn∗x+7rn∗2x+rn∗xn31+4rn∗x+3rn∗2x.
Written as(29)Tn,rx=Vn∗t−xr;xr=0,1,⋯.
So,(30)Tn,0x=1;Tn,1x=rn∗x−x;Tn,2x=2xx−rn∗x;Tn,3x=−4x3+4x2rn∗x+2rn∗xx2+rn∗xn+rn∗x1+rn∗xn2,Tn,4x=7x4−8x3rn∗x+x2rn∗2x+rn∗xn−8x3+3x2+5x2−8x+2rn∗xrn∗x+rn∗xn23x2−4x+22−2x+3x2rn∗x+7rn∗2x.
Whenx∈D=α,β, there is(31)rn∗x=x−1+x2n+4x2+4x+18xn2+o1n2.
Substitute this back intoTn,4x, we get(32)Tn,4x=9/4x4+6x3+3x2n2+o1n2.
Therefore,(33)Tn,4CD=maxx∈DVn∗t−x4;x≤MDn2,where MD is a positive constant dependent on D=α,β. Please note that(34)rn∗x−x≤1+xn+1.
So fork=0,1, there is(35)limn⟶∞Vn∗ek−ekCD=0,limn⟶∞Vn∗ek−ekCD=0.
Take the following values:μt=tl+11/2,t∈0,∞ (l is any determinate, nonnegative integer), then for any x∈D, based on the Schwarz inequality, we can get:(36)Vn∗t−x2μt;x≤Vn∗t−x4;x1/2Vn∗tl+1;x1/2≤MDn21/2MlD+11/2.
Thus,(37)limn⟶∞maxx∈DVn∗t−x2μt;x=0.
From Theorem8, we know that for f∈C0,∞ and ft≤Mf1+t2tl+11/2t∈0,∞, there is(38)limn⟶∞Vn∗f−fCD=0.
Theorem proven.
Now assumef∈C0,∞, then for any positive integer δ, there is(39)Vn∗f,x−fx≤Vn∗ft−fx;x+fxVn∗1;x−1≤Vn∗ωf,t−x;x≤ωf,δVn∗1+t−xδ;x≤ωf,δVn∗1;x+1δVn∗t−x;x≤ωf,δ1+1δVn∗1/2t−x2;x≤ωf,δ1+1δ2xx−Vn∗e1x.
Note thatVn∗e1x=rn∗x, thus we haveTheorem 10.
Assumef∈C0,∞,x∈0,∞, δ>0, so(40)Vn∗f;x−fx≤ωf;δ1+1δ2xx−rn∗x.Note: because(41)2xx−rn∗x≤x1+xn.Based on this, we can tell the operatorVn∗ has a better performance in the order of approximation than the classic Baskakov operator Vn. In fact,(42)2xx−rn∗x≤x1+xn,is equivalent with(43)rn∗x≥2n−12nx−12n,because(44)nn+1x2+14n+12−2n−124n2x2=3n−14n2n+1x2+14n+12≥0,which is(45)nn+1x2+14n+12≥2n−12nx.Thus, we can get(46)rn∗x≥2n−12nx−12n,x∈0,∞.


## 3. Theorem and the Converse Theorem of the Modified Baskakov Operator

### 3.1. Theorem of Approximation

To build the approximation theorem of the operator, we need to introduce a few following theorems first:Lemma 1.
Assumen∈N, then Vn∗t−x2,x≤4n−1ϕ2x.Proof.
We note that(47)Vn∗t−x2,x=Vn∗t2,x−2xVn∗t,x+x2Vn∗1,x=2x−−rn∗x.
Whenn=1, there is(48)x−rn∗x=x−x2≤x≤1+x,when n≥2, there is(49)x−rn∗x=x+12n+1−nn+1x2+14n+1≤x2+xn+1x≤21+xn.
Lemma proven.Lemma 2.
Assume g∈ A.C.loc, then(50)Vn∗R1g,t,x,x≤Cn−1/2ϕ1−λxϕλg′,where R1g,t,x = ∫xtg′udu.Proof.
From(51)R1g,t,x=∫xtg′udu≤∫xtg′udu≤φλg′∫xtφ−λudu.
we can get(52)Vn∗R1g,t,x;x≤Vn∗R1g,t,x,x≤ϕλg′Vn∗∫xtφ−λudu;x≤ϕλg′∑k=0∞vn,krn∗x∫xk/nϕ−λudu.
Now we estimate∫xk/nϕ−λudu
Whenk=0, there is(53)∫x0ϕ−λudu=∫0xu−λ/21+u−λ/2du=ϕ−λx∫0xx1+xu1+uλ/2du≤ϕ−λx∫0xxuλdu≤φ−λxxλ∫0xu−λdu≤Cφ−λxxλx−λ+1=Cφ−λxx.
Whenk≥1, there is(54)∫xk/nϕ−λudu=∫xk/nu−λ/21+u−λ/2du≤1+x−λ/2+1+kn−λ/2∫xk/nu−λ/2du≤1+x−λ/2+1+kn−λ/2∫xk/nu−1/2duλ∫xk/n1du1−λ≤C1+x−λ2+1+kn−λ/2kn−xλkn−x1−λ≤C1+x−λ/2+1+kn−λ/2k/n−xxλ/2≤Ckn−xφ−λx+k/n−x1+k/nλ/2xλ/2.
Therefore,(55)Vn∗R1g,t,x;x≤Cφλg′vn,0rn∗xxφλx+∑k=1∞k/n−xφλx+k/n−x1+k/nλ/2xλ/2vn,krn∗x≤Cφλg′∑k=0∞vn,krn∗xk/n−xφλx+∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗xCφλg′Vn∗t−x;xφλx+∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x.
The first item within the bracket in the above equation is as follows:(56)Vn∗t−x;xϕ−λx≤ϕ−λxVn∗1/2t−x2;xVn∗1/21;x≤ϕ−λxVn∗1/2t−x2;x≤Cn−1/2φ1−λx.
The second item within the bracket in the above equation is as follows:(57)∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x≤∑k=1∞vn,krn∗x1+k/n−λ1/2x−λ/2∑k=1∞vn,krn∗xk/n−x21/2≤x−λ/2∑k=1∞vn,krn∗xk/n−x21/2∑k=0∞vn,krn∗x1+k/n−1λ/2,and(58)∑k=0∞vn,krn∗x1+kn−1=11+rn∗x∑k=0∞vn−1,krn∗xnn+k−1n−1n+k≤C1+rn∗x.
Whenx≥1,(59)1+x1+rn∗x=1+x1−1/2n+1+n/n+1x2+1/4n+12≤1+xxn/n+1≤22.
When0≤x<1,(60)1+x1+rn∗x≤21−1/2n+1+1/4n+12=2.
Thus, we have(61)11+rn∗x≤C1+x.
Substitute back to (58), we get(62)∑k=0∞vn,krn∗x1+kn−1≤C1+x.
Substitute equation (62) and the result of Lemma 1 into (57), we get(63)∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x≤Cx−λ/2n−1/2ϕx1+x−λ/2≤Cn−1/2ϕ1−λx.
Substitute this equation and (56) back into (55) we get(64)Vn∗R1g,t,x,x≤Cn−1/2ϕ1−λxϕλg′.
Thus, Lemma2 is proven.
With Lemma1 and Lemma 2, we can build the approximation theorem of operator Vn∗ under the generalized modulus of smoothness as follows:Theorem 11.
Assumef∈C0,∞, 0≤λ≤1, then(65)Vn∗f;x−fx≤Cωφλf,n−1/2φ1−λx≤C.Proof.
Based on the definition ofKϕλf,t, we can select the gn∈A.C.loc that makes(66)f−gn+n−1/2ϕ1−λxϕλgn′≤Cωφλf,n−1/2φ1−λx.
Apply Lemma2, we get(67)Vn∗f;x−fx≤Vn∗f−gn;x−fx−gnx+Vn∗gn;x−gnx≤Cf−gn+Vn∗R1gn,t,x;x≤Cf−gn+n−1/2φ1−λxφλgn′≤Cωϕλ1f,n−1/2ϕ1−λx.
Theorem proven.


### 3.2. The Reverse Theorem of Approximation

Let us now construct the opposite theorem of approximation of the operator’s equivalent formVn∗. First and foremost, certain important theorems must be introduced in order to prepare.Lemma 3.
Assumef ∈ A.C.loc, 0≤λ≤1, then(68)φλxVn∗′f,x≤Cϕλf′n≥2.Proof.
Whenn≥2,(69)vn,k′x=−nvn+1,0x,k=0,nvn+1,k−1x−vn+1,kx,k≥1,rn∗x′=n/n+1xn/n+1x2+1/4n+12.
Thus, there is(70)Vn∗′f,x=n∑k=1∞fknvn+1,k−1rn∗x−∑k=0∞vn+1,krn∗xfkn=rn∗x′=n∑k=0∞fk+1nvn+1,krn∗x−∑k=0∞fknvn+1,krn∗xrn∗x′=n∑k=0∞fk+1n−fknvn+1,krn∗xn/n+1xn/n+1x2+1/4n+12.
Now we estimatefk+1/n−fk/n
Whenk=0,(71)f1n−f0=∫01/nf′udu≤∫01/nf′udu≤ϕλf′∫01/nϕ−λudu≤ϕλf′∫01/nu−λ/2du≤n−1+λ/2ϕλf′.
Whenk≥1,(72)fk+1n−fkn=∫01/nf′kn+udu≤∫01/nf′kn+udu≤ϕλf′∫01/nϕ−λkn+udu≤ϕλf′1nkn−λ/21+kn−λ/2.
So we can get(73)φλxVn∗′f,x≤ϕλxn∑k=0∞fk+1n−fknvn+1,krn∗x·n/n+1xn/n+1x2+1/4n+12≤Cϕλf′ϕλx∑k=1∞nkλ/2nn+kλ/2vn+1,krn∗x+nλ/2vn+1,0rn∗x·n/n+1xn/n+1x2+1/4n+12.
For the summarized equation in the square bracket above, based on the Holder inequality, we get(74)∑k=1∞nkλ/2nn+kλ/2vn+1,krn∗x≤∑k=1∞n2kk+nvn+1,krn∗xλ/2=ϕ−λrn∗x∑k=1∞n2kn+kn+kk+1nn−1vn−1,k+1rn∗xλ/2≤Cϕ−λrn∗x∑k=1∞vn−1,k+1rn∗xλ/2≤Cϕ−λrn∗x.
For the second item in the above square bracket, becausevn+1,0rn∗x ≤ 1, so there is(75)nλ/2vn+1,0rn∗x≤nvn+1,0rn∗xλ/2=φ−λrn∗xnn−1vn−1,1rn∗xλ/2≤Cϕ−λrn∗x.
From equations (73), (74), and (75), we get(76)φλxVn∗′f,x≤Cφλfφλxφ−λrn∗x·n/n+1xn/n+1x2+1/4n+12.
Because whenn≥2, 0≤rn∗x≤x, so,(77)rn∗x=−12n+1+nn+1x2+14n+12≥n/n+1x22n/n+1x2+1/4n+12.
Thus,(78)ϕλxϕ−λrn∗xn/n+1xn/n+1x2+1/4n+12=xλ/21+xλ/2rn∗x−λ/21+rn∗x−λ/2·n/n+1xn/n+1x2+1/4n+12=1+x1+rn∗xλ/2xrn∗xλ/2n/n+1xn/n+1x2+1/4n+12.
From Lemma2, we have(79)1+x1+rn∗x≤C.
Whenx>1/2n+1,(80)xrn∗x≤2n/n+1x2+1/4n+12n/n+1x≤C.
For anyx∈0,∞,(81)n/n+1xn/n+1x2+1/4n+12≤C.
So whenx>1/2n+1, there is(82)xrn∗λ/2n/n+1xn/n+1x2+1/4n+12≤C.
Whenx≤1/2n+1, based on equation (77), there is(83)xrn∗xλ/2n/n+1xn/n+1x2+1/4n+12≤Cx1−λ/2nn+1x2+14n+12λ/2−1≤Cx1−λ/2nn+1x2+x2λ/2−1=Cx1−λ/2nn+1+1λ/2−1xλ/2−1≤C.
After all, we get(84)ϕλxϕ−λrn∗xn/n+1xn/n+1x2+1/4n+12≤C.
In summary,(85)φλxVn∗′f,x≤Cϕλf′n≥2.
Lemma3 proven.Lemma 4.
Assumef∈C0,∞, 0≤λ≤1,(86)Vn∗′f,x≤Cn1/2ϕ−1xf.Proof.
Use(87)vn,k′x=nϕ−2xkn−xvn,kx.
We can get(88)ϕλxVn∗′f,x=φλx∑k=0∞nφ−2rn∗xkn−rn∗xvn,krn∗xfknrn∗′x≤ϕλ−1xfnϕxϕ−2rn∗x∑k=0∞kn−rn∗xvn,krn∗xn/n+1xn/n+1x2+1/4n+12≤Cϕλ−1xfnϕxϕ−2rn∗x∑k=0∞kn−rn∗x2vn,krn∗x1/2n/n+1xn/n+1x2+1/4n+12≤Cnϕλ−1xfϕxϕ−2rn∗xn−1/2ϕrn∗xn/n+1xn/n+1x2+1/4n+12≤Cn1/2ϕλ−1xfϕxϕ−1rn∗xn/n+1xn/n+1x2+1/4n+12≤Cn1/2ϕλ−1xf.
The last inequality has used the result of (84), and now Lemma 4 is proven.Lemma 5.
For0<t<1/8, x≥t/2, 0≤β≤1, there is(89)∫−t/2t/2φ−βx+udu≤4tφ−βx.Proof.
Whenβ=0, it is clear that the equation holds. Now we move to the proof for β=1. For 0<β<1, the Holder inequality gives us the following:
Whenx≥t/2, −t/2≤u≤t/2, there is(90)121+x≤1+x+u,so(91)∫−t/2t/2φ−1x+udu≤21+x∫−t/2t/21x+udu=41+xx+t2−x−t2≤41+xtx+t/2+x−t/2≤4tφ−1x.Theorem 12.
Assumef∈C0,∞, 0≤λ≤1, 0<α<1,(92)Vn∗f,x−fx≤Cn−1/2ϕ1−λxα⟺ωϕλ1f,t=Οtα.Proof.
Theorem11 gives us its sufficiency directly, so now we try to prove its necessity. Since ϕα1−λx is a convex function, we can get(93)Δtφλ1fx≤Δtφλ1fx−Vn∗f,x+Δtϕλ1Vn∗f,x≤∑j=011jVn∗f,x+j−12tϕλx−fx+j−12tϕλxΔtϕλ1Vn∗f,x≤∑j=011jn−12ϕ1−λx+j−12tϕλxα+Δtϕλ1Vn∗f,x≤Cn−1/2ϕ1−λxα+Δtϕλ1Vn∗f,x.
We can select agn that satisfies(94)f−gn+n−1/2ϕ1−λxϕλgn′≤Cωφλf,n−1/2φ1−λx.
Apply, Lemmas3–5(95)Δtϕλ1Vn∗f,x=∫−tφλx/2tφλx/2Vn∗′f,x+udu≤∫−tφλx/2tφλx/2Vn∗′f−gn,x+udu+∫−tφλx/2tφλx/2Vn∗′gn,x+udu≤Cn1/2f−gn∫−tφλx/2tφλx/2ϕ−1x+udu+Cφλgn′∫−tφλx/2tφλx/2ϕ−λx+udu≤Ctn1/2φλ−1xf−gn+tφλgn′≤Ctn1/2ϕλ−1xf−gn+n−1/2ϕ1−λxϕλgn′≤Ctn1/2ϕλ−1xωϕλ1f,n−1/2ϕ1−λx.
From equations (93) and (95), we have(96)Δtϕλx1fx≤Cn−1/2φ1−λxα+tn−1/2φ1−λxωφλ1f,n−1/2φ1−λx.
For any fixed0<δ<1/8 and all x∈0,∞, use a sufficiently great n that makes the following equation stand:(97)n−1/2ϕ1−λx≤δ≤2n−1/2ϕ1−λx.
Now we have(98)Δtϕλx1fx≤Cδα+tδωφλ1f,δ.
Thus,(99)ωϕλ1f,t≤Cδα+tδωφλ1f,δ
Use lemma Lorenz-Hermann (literature [20] P199), then we can get(100)ωϕλ1f,t=Οtα.
Theorem12 proven.


## 3.1. Theorem of Approximation

To build the approximation theorem of the operator, we need to introduce a few following theorems first:Lemma 1.
Assumen∈N, then Vn∗t−x2,x≤4n−1ϕ2x.Proof.
We note that(47)Vn∗t−x2,x=Vn∗t2,x−2xVn∗t,x+x2Vn∗1,x=2x−−rn∗x.
Whenn=1, there is(48)x−rn∗x=x−x2≤x≤1+x,when n≥2, there is(49)x−rn∗x=x+12n+1−nn+1x2+14n+1≤x2+xn+1x≤21+xn.
Lemma proven.Lemma 2.
Assume g∈ A.C.loc, then(50)Vn∗R1g,t,x,x≤Cn−1/2ϕ1−λxϕλg′,where R1g,t,x = ∫xtg′udu.Proof.
From(51)R1g,t,x=∫xtg′udu≤∫xtg′udu≤φλg′∫xtφ−λudu.
we can get(52)Vn∗R1g,t,x;x≤Vn∗R1g,t,x,x≤ϕλg′Vn∗∫xtφ−λudu;x≤ϕλg′∑k=0∞vn,krn∗x∫xk/nϕ−λudu.
Now we estimate∫xk/nϕ−λudu
Whenk=0, there is(53)∫x0ϕ−λudu=∫0xu−λ/21+u−λ/2du=ϕ−λx∫0xx1+xu1+uλ/2du≤ϕ−λx∫0xxuλdu≤φ−λxxλ∫0xu−λdu≤Cφ−λxxλx−λ+1=Cφ−λxx.
Whenk≥1, there is(54)∫xk/nϕ−λudu=∫xk/nu−λ/21+u−λ/2du≤1+x−λ/2+1+kn−λ/2∫xk/nu−λ/2du≤1+x−λ/2+1+kn−λ/2∫xk/nu−1/2duλ∫xk/n1du1−λ≤C1+x−λ2+1+kn−λ/2kn−xλkn−x1−λ≤C1+x−λ/2+1+kn−λ/2k/n−xxλ/2≤Ckn−xφ−λx+k/n−x1+k/nλ/2xλ/2.
Therefore,(55)Vn∗R1g,t,x;x≤Cφλg′vn,0rn∗xxφλx+∑k=1∞k/n−xφλx+k/n−x1+k/nλ/2xλ/2vn,krn∗x≤Cφλg′∑k=0∞vn,krn∗xk/n−xφλx+∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗xCφλg′Vn∗t−x;xφλx+∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x.
The first item within the bracket in the above equation is as follows:(56)Vn∗t−x;xϕ−λx≤ϕ−λxVn∗1/2t−x2;xVn∗1/21;x≤ϕ−λxVn∗1/2t−x2;x≤Cn−1/2φ1−λx.
The second item within the bracket in the above equation is as follows:(57)∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x≤∑k=1∞vn,krn∗x1+k/n−λ1/2x−λ/2∑k=1∞vn,krn∗xk/n−x21/2≤x−λ/2∑k=1∞vn,krn∗xk/n−x21/2∑k=0∞vn,krn∗x1+k/n−1λ/2,and(58)∑k=0∞vn,krn∗x1+kn−1=11+rn∗x∑k=0∞vn−1,krn∗xnn+k−1n−1n+k≤C1+rn∗x.
Whenx≥1,(59)1+x1+rn∗x=1+x1−1/2n+1+n/n+1x2+1/4n+12≤1+xxn/n+1≤22.
When0≤x<1,(60)1+x1+rn∗x≤21−1/2n+1+1/4n+12=2.
Thus, we have(61)11+rn∗x≤C1+x.
Substitute back to (58), we get(62)∑k=0∞vn,krn∗x1+kn−1≤C1+x.
Substitute equation (62) and the result of Lemma 1 into (57), we get(63)∑k=1∞k/n−x1+k/nλ/2xλ/2vn,krn∗x≤Cx−λ/2n−1/2ϕx1+x−λ/2≤Cn−1/2ϕ1−λx.
Substitute this equation and (56) back into (55) we get(64)Vn∗R1g,t,x,x≤Cn−1/2ϕ1−λxϕλg′.
Thus, Lemma2 is proven.
With Lemma1 and Lemma 2, we can build the approximation theorem of operator Vn∗ under the generalized modulus of smoothness as follows:Theorem 11.
Assumef∈C0,∞, 0≤λ≤1, then(65)Vn∗f;x−fx≤Cωφλf,n−1/2φ1−λx≤C.Proof.
Based on the definition ofKϕλf,t, we can select the gn∈A.C.loc that makes(66)f−gn+n−1/2ϕ1−λxϕλgn′≤Cωφλf,n−1/2φ1−λx.
Apply Lemma2, we get(67)Vn∗f;x−fx≤Vn∗f−gn;x−fx−gnx+Vn∗gn;x−gnx≤Cf−gn+Vn∗R1gn,t,x;x≤Cf−gn+n−1/2φ1−λxφλgn′≤Cωϕλ1f,n−1/2ϕ1−λx.
Theorem proven.


## 3.2. The Reverse Theorem of Approximation

Let us now construct the opposite theorem of approximation of the operator’s equivalent formVn∗. First and foremost, certain important theorems must be introduced in order to prepare.Lemma 3.
Assumef ∈ A.C.loc, 0≤λ≤1, then(68)φλxVn∗′f,x≤Cϕλf′n≥2.Proof.
Whenn≥2,(69)vn,k′x=−nvn+1,0x,k=0,nvn+1,k−1x−vn+1,kx,k≥1,rn∗x′=n/n+1xn/n+1x2+1/4n+12.
Thus, there is(70)Vn∗′f,x=n∑k=1∞fknvn+1,k−1rn∗x−∑k=0∞vn+1,krn∗xfkn=rn∗x′=n∑k=0∞fk+1nvn+1,krn∗x−∑k=0∞fknvn+1,krn∗xrn∗x′=n∑k=0∞fk+1n−fknvn+1,krn∗xn/n+1xn/n+1x2+1/4n+12.
Now we estimatefk+1/n−fk/n
Whenk=0,(71)f1n−f0=∫01/nf′udu≤∫01/nf′udu≤ϕλf′∫01/nϕ−λudu≤ϕλf′∫01/nu−λ/2du≤n−1+λ/2ϕλf′.
Whenk≥1,(72)fk+1n−fkn=∫01/nf′kn+udu≤∫01/nf′kn+udu≤ϕλf′∫01/nϕ−λkn+udu≤ϕλf′1nkn−λ/21+kn−λ/2.
So we can get(73)φλxVn∗′f,x≤ϕλxn∑k=0∞fk+1n−fknvn+1,krn∗x·n/n+1xn/n+1x2+1/4n+12≤Cϕλf′ϕλx∑k=1∞nkλ/2nn+kλ/2vn+1,krn∗x+nλ/2vn+1,0rn∗x·n/n+1xn/n+1x2+1/4n+12.
For the summarized equation in the square bracket above, based on the Holder inequality, we get(74)∑k=1∞nkλ/2nn+kλ/2vn+1,krn∗x≤∑k=1∞n2kk+nvn+1,krn∗xλ/2=ϕ−λrn∗x∑k=1∞n2kn+kn+kk+1nn−1vn−1,k+1rn∗xλ/2≤Cϕ−λrn∗x∑k=1∞vn−1,k+1rn∗xλ/2≤Cϕ−λrn∗x.
For the second item in the above square bracket, becausevn+1,0rn∗x ≤ 1, so there is(75)nλ/2vn+1,0rn∗x≤nvn+1,0rn∗xλ/2=φ−λrn∗xnn−1vn−1,1rn∗xλ/2≤Cϕ−λrn∗x.
From equations (73), (74), and (75), we get(76)φλxVn∗′f,x≤Cφλfφλxφ−λrn∗x·n/n+1xn/n+1x2+1/4n+12.
Because whenn≥2, 0≤rn∗x≤x, so,(77)rn∗x=−12n+1+nn+1x2+14n+12≥n/n+1x22n/n+1x2+1/4n+12.
Thus,(78)ϕλxϕ−λrn∗xn/n+1xn/n+1x2+1/4n+12=xλ/21+xλ/2rn∗x−λ/21+rn∗x−λ/2·n/n+1xn/n+1x2+1/4n+12=1+x1+rn∗xλ/2xrn∗xλ/2n/n+1xn/n+1x2+1/4n+12.
From Lemma2, we have(79)1+x1+rn∗x≤C.
Whenx>1/2n+1,(80)xrn∗x≤2n/n+1x2+1/4n+12n/n+1x≤C.
For anyx∈0,∞,(81)n/n+1xn/n+1x2+1/4n+12≤C.
So whenx>1/2n+1, there is(82)xrn∗λ/2n/n+1xn/n+1x2+1/4n+12≤C.
Whenx≤1/2n+1, based on equation (77), there is(83)xrn∗xλ/2n/n+1xn/n+1x2+1/4n+12≤Cx1−λ/2nn+1x2+14n+12λ/2−1≤Cx1−λ/2nn+1x2+x2λ/2−1=Cx1−λ/2nn+1+1λ/2−1xλ/2−1≤C.
After all, we get(84)ϕλxϕ−λrn∗xn/n+1xn/n+1x2+1/4n+12≤C.
In summary,(85)φλxVn∗′f,x≤Cϕλf′n≥2.
Lemma3 proven.Lemma 4.
Assumef∈C0,∞, 0≤λ≤1,(86)Vn∗′f,x≤Cn1/2ϕ−1xf.Proof.
Use(87)vn,k′x=nϕ−2xkn−xvn,kx.
We can get(88)ϕλxVn∗′f,x=φλx∑k=0∞nφ−2rn∗xkn−rn∗xvn,krn∗xfknrn∗′x≤ϕλ−1xfnϕxϕ−2rn∗x∑k=0∞kn−rn∗xvn,krn∗xn/n+1xn/n+1x2+1/4n+12≤Cϕλ−1xfnϕxϕ−2rn∗x∑k=0∞kn−rn∗x2vn,krn∗x1/2n/n+1xn/n+1x2+1/4n+12≤Cnϕλ−1xfϕxϕ−2rn∗xn−1/2ϕrn∗xn/n+1xn/n+1x2+1/4n+12≤Cn1/2ϕλ−1xfϕxϕ−1rn∗xn/n+1xn/n+1x2+1/4n+12≤Cn1/2ϕλ−1xf.
The last inequality has used the result of (84), and now Lemma 4 is proven.Lemma 5.
For0<t<1/8, x≥t/2, 0≤β≤1, there is(89)∫−t/2t/2φ−βx+udu≤4tφ−βx.Proof.
Whenβ=0, it is clear that the equation holds. Now we move to the proof for β=1. For 0<β<1, the Holder inequality gives us the following:
Whenx≥t/2, −t/2≤u≤t/2, there is(90)121+x≤1+x+u,so(91)∫−t/2t/2φ−1x+udu≤21+x∫−t/2t/21x+udu=41+xx+t2−x−t2≤41+xtx+t/2+x−t/2≤4tφ−1x.Theorem 12.
Assumef∈C0,∞, 0≤λ≤1, 0<α<1,(92)Vn∗f,x−fx≤Cn−1/2ϕ1−λxα⟺ωϕλ1f,t=Οtα.Proof.
Theorem11 gives us its sufficiency directly, so now we try to prove its necessity. Since ϕα1−λx is a convex function, we can get(93)Δtφλ1fx≤Δtφλ1fx−Vn∗f,x+Δtϕλ1Vn∗f,x≤∑j=011jVn∗f,x+j−12tϕλx−fx+j−12tϕλxΔtϕλ1Vn∗f,x≤∑j=011jn−12ϕ1−λx+j−12tϕλxα+Δtϕλ1Vn∗f,x≤Cn−1/2ϕ1−λxα+Δtϕλ1Vn∗f,x.
We can select agn that satisfies(94)f−gn+n−1/2ϕ1−λxϕλgn′≤Cωφλf,n−1/2φ1−λx.
Apply, Lemmas3–5(95)Δtϕλ1Vn∗f,x=∫−tφλx/2tφλx/2Vn∗′f,x+udu≤∫−tφλx/2tφλx/2Vn∗′f−gn,x+udu+∫−tφλx/2tφλx/2Vn∗′gn,x+udu≤Cn1/2f−gn∫−tφλx/2tφλx/2ϕ−1x+udu+Cφλgn′∫−tφλx/2tφλx/2ϕ−λx+udu≤Ctn1/2φλ−1xf−gn+tφλgn′≤Ctn1/2ϕλ−1xf−gn+n−1/2ϕ1−λxϕλgn′≤Ctn1/2ϕλ−1xωϕλ1f,n−1/2ϕ1−λx.
From equations (93) and (95), we have(96)Δtϕλx1fx≤Cn−1/2φ1−λxα+tn−1/2φ1−λxωφλ1f,n−1/2φ1−λx.
For any fixed0<δ<1/8 and all x∈0,∞, use a sufficiently great n that makes the following equation stand:(97)n−1/2ϕ1−λx≤δ≤2n−1/2ϕ1−λx.
Now we have(98)Δtϕλx1fx≤Cδα+tδωφλ1f,δ.
Thus,(99)ωϕλ1f,t≤Cδα+tδωφλ1f,δ
Use lemma Lorenz-Hermann (literature [20] P199), then we can get(100)ωϕλ1f,t=Οtα.
Theorem12 proven.



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