# On Approximation by Algebraic Version of the Trigonometric Jackson IntegralsGs,n in Weighted Integral Metric

Authors: Teodora Zapryanova
Journal: Journal of Operators
 (2013)

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## Abstract

We characterize the errors of the algebraic version of trigonometric Jackson integralsGs,n in weighted integral metric. We prove direct and strong converse theorem in terms of the weighted K-functional.


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## Body

## 1. Introduction

We study linear approximation process together with the characterization of the rate of convergence of the algebraic version of the trigonometric Jackson integralsGs,n defined by
(1)Gs,n(f,x)=∫-ππf(cos(arccosx+v))Ks,n(v)dv,
where
(2)Ks,n(v)=μs,n(sin(nv/2)sin(v/2))2s,∫-ππKs,n(v)dv=1.
We established the equivalence in uniform norm for the approximation error of the operator Gs,n, the values of proper K-functional, and modulus of smoothness (see [1, 2]):
(3)∥f-Gs,nf∥~K(f,1n2;C[-1,1],C2,H)~Ω2(f,1n).
For every f∈C[-1,1] and t>0 the K-functional is defined by
(4)K(f,t;C[-1,1],C2,H):=inf{∥f-g∥+t∥Hg∥:g∈C2},
where the differential operator
(5)H:=(H1)2,(H1g)(x):=1-x2ddxg(x).
The modulus
(6)Ω2(f,t):=sup0<h≤tsup-1≤x≤1|f(cos(arccosx+h))+f(cos(arccosx-h))-2f(x)|.
The expression A(n)~B(n) means that there exists a positive constant c independent of n such that (1/c)B(n)≤A(n)≤cB(n). The equivalence ∥f-Gs,nf∥~K(f,1/n2;C[-1,1],C2,H) consists of a direct inequality and strong converse inequality of type A in the sense of [3]. Ditzian and Ivanov show that a converse inequality follows from several inequalities of Bernstein and Voronovskaya type. We apply their method.LetLp(u)[-1,1] be the weighted Lp space with the norm
(7)∥f∥p,u=∥f∥Lp(u)[-1,1]∥f∥p,u=(∫-11|u(x)f(x)|pdx)1/p∥f∥p,u=(∫-11|(1-x2)-1/2pf(x)|pdx)1/p,u(x)=(1-x2)-1/2p.
The approximation error ∥f-Gs,nf∥p,u of Gs,n in Lp(u)[-1,1] will be compared with the K-functional, which for every f∈Lp(u)[-1,1] and t>0 is defined by
(8)K(f,t;Lp(u)[-1,1],C2,H):=inf{∥f-g∥p,u+t∥Hg∥p,u:g∈C2}.
Our main result states the following.Theorem 1.
For everyf∈Lp(u)[-1,1],1≤p<∞, and s,n∈ℕ,s≥3, one has
(9)∥f-Gs,nf∥Lp(u)[-1,1]~K(f,1n2;Lp(u)[-1,1],C2,H).In Section2 we state and prove some auxiliary lemmas. The proof of Theorem 1 is given in Section 3.


## 2. Auxiliary Lemmas

Lemma 2.
Letf be summable on ℝ and g∈Lp[-π,π]2π—periodic function, 1≤p<∞. Then the following holds true:
(10)∥f*g∥Lp[-π,π]≤∥f∥L1(-∞,∞)∥g∥Lp[-π,π].Proof.
Using the Minkowski inequality we have(11)∥f*g∥Lp[-π,π]=∥∫-∞∞f(x-y)g(y)dy∥Lp[-π,π]=∥∫-∞∞g(x-y)f(y)dy∥Lp[-π,π]≤(∫-ππ|∫-∞∞|g(x-y)f(y)|dy|pdx)1/p≤∫-∞∞(∫-ππ|g(x-y)|p|f(y)|pdx)1/pdy=∫-∞∞(|f(y)|p∫-ππ|g(x-y)|pdx)1/pdy=∫-∞∞|f(y)|(∫-ππ|g(x-y)|pdx)1/pdy(asgis2πperiodic function)=∫-∞∞|f(y)|∥g∥Lp[-π,π]dy=∥g∥Lp[-π,π]∥f∥L1(-∞,∞).Lemma 3.
For2π periodic integrable on [-π,π] function g and every v≠0 one has
(12)∥1v∫zz+v|g(ξ)|dξ∥L1[-π,π]=∥g∥L1[-π,π].Proof.
We consider the functiong(ξ+z). By the Fubini theorem we have
(13)∫-ππ∫0v|g(ξ+z)|dξdz=∫0v∫-ππ|g(ξ+z)|dzdξ.
As∫0v|g(ξ+z)|dξ=∫zz+v|g(x)|dx we obtain that the integral on the left-hand side of (13) is equal to
(14)v∫-ππ(1v∫zz+v|g(ξ)|dξ)dz.
The integral on the right-hand side of (13) is
(15)∫0v∫-ππ|g(ξ+z)|dzdξ=∫0v[∫-π+ξπ+ξ|g(x)|dx]dξ(asgis2πperiodic function)=∫0v[∫-ππ|g(x)|dx]dξ=v∫-ππ|g(x)|dx.
From (13), (14), and (15) we obtain ∫-ππ((1/v)∫zz+v|g(ξ)|dξ)dz=∫-ππ|g(x)|dx=∥g∥L1[-π,π]. If ∫zz+v|g(ξ)|dξ≠0 then the sign of ∫zz+v|g(ξ)|dξ is constant and coincides with the sign of v. Therefore
(16)∫-ππ(1v∫zz+v|g(ξ)|dξ)dz=∫-ππ|1v∫zz+v|g(ξ)|dξ|dz=∥1v∫zz+v|g(ξ)|dξ∥L1[-π,π],
which proves the lemma.Lemma 4.
For2π periodic function g∈Lp[-π,π],1≤p<∞ and every v≠0 one has
(17)∥1v∫zz+v|g(ξ)|dξ∥Lp[-π,π]≤∥g∥Lp[-π,π].Proof.
Forp=1 Lemma 4 is fulfilled as equality. Let 1<p<∞. Using the Hölder inequality we get
(18)|∫zz+v|g(ξ)|dξ|≤|∫zz+v|g(ξ)|pdξ|1/p|∫zz+v1qdξ|1/q(1p+1q=1)=|v|1/q|∫zz+v|g(ξ)|pdξ|1/p.
Therefore
(19)|1v∫zz+v|g(ξ)|dξ|p≤|v|p/q|v|p|∫zz+v|g(ξ)|pdξ|=1|v||∫zz+v|g(ξ)|pdξ|=1v∫zz+v|g(ξ)|pdξ=|1v∫zz+v|g(ξ)|pdξ|.
Then
(20)∫-ππ|1v∫zz+v|g(ξ)|dξ|pdz≤∫-ππ|1v∫zz+v|g(ξ)|pdξ|dz.
We apply Lemma 3 to |g(ξ)|p to obtain
(21)∫-ππ|1v∫zz+v|g(ξ)|pdξ|dz=∫-ππ|g(ξ)|pdξ.
Hence
(22)∫-ππ|1v∫zz+v|g(ξ)|dξ|pdz≤∫-ππ|g(ξ)|pdξ.
Thus
(23)∥1v∫zz+v|g(ξ)|dξ∥Lp[-π,π]≤∥g∥Lp[-π,π].Definition 5.
Set(24)Y={g∈C[-1,1]:H1g∈C[-1,1],Hg∈C[-1,1],H1g(±1)=0}.Z={g∈Y:H13g∈C[-1,1],H2g∈C[-1,1],H13g(±1)=0}.Lemma 6 ([2, page 402]).
LetY be the space from Definition 5, g∈Y, and g~(σ):=g(cosσ). Then g~∈C2(ℝ) and g~′′(σ)=Hg(cosσ) for σ∈ℝ.Lemma 7.
LetY be the space from Definition 5. Then for every f∈Lp(u)[-1,1] and t>0, one has
(25)K(f,t;Lp(u)[-1,1],Y,H)=K(f,t;Lp(u)[-1,1],C2,H).Proof.
FromC2⊂Y we see that K(f,t;Lp(u)[-1,1],Y,H)≤K(f,t;Lp(u)[-1,1],C2,H). In order to prove K(f,t;Lp(u)[-1,1],C2,H)≤K(f,t;Lp(u)[-1,1],Y,H) it is sufficient to show (see Lemma  2 in [4, page 116]) that for every g∈Y and ɛ>0 there exists G∈C2 such that ∥G-g∥Lp(u)[-1,1]≤ɛ and ∥HG∥Lp(u)[-1,1]≤∥Hg∥Lp(u)[-1,1]+ɛ. Let g(x)∈Y. We put x=cosσ and consider g~(σ):=g(cosσ). Since g(x)∈Y,g~(s)∈C2 (see Lemma 6). We use the Jackson integrals of the following type:
(26)Jn(g~,σ):=∫-ππg~(σ+v)K1s,n(v)dv=∫-ππg~(v)K1s,n(σ-v)dv,
where K1s,n(v):=λs,n(sin(mv/2)/sin(v/2))2s, ∫-ππK1s,n(v)dv=1,n,s>0,m:=[n/s]+1.
Since(1/2m)(sin(mv/2)/sin(v/2))2=1/2+∑k=0m-1(1-k/m)coskv, it follows that K1s,n is an even nonnegative trigonometric polynomial of degree ≤n.Jn(g~,σ) is a trigonometric polynomial of degree ≤n, which is even as g~ is even. From Jackson theorem (see [5, Chapter 7, Theorem 2.2]) we get
(27)∥g~-Jn(g~)∥Lp[0,π]≤cω2(g~,1n)Lp[0,π]=O(1n2).
By the substitution σ=arccosx in Jn(g~,σ) we obtain an algebraic polynomial, which is desired function from C2. We set
(28)G(x)=Jn(g~,arccosx).
From ∥g-G∥Lp(u)[-1,1]=∥g~-Jn(g~)∥Lp[0,π] and (27) we get
(29)∥g-G∥Lp(u)[-1,1]≤cω2(g~,1n)Lp[0,π]=O(1n2).
From (26) we get
(30)d2dσ2Jn(g~,σ)=∫-ππg~′′(σ+v)K1s,n(v)dvd2dσ2Jn(g~,σ)=∫-ππg~′′(v)K1s,n(σ-v)dvd2dσ2Jn(g~,σ)=Jn(g~′′,σ).
Using the Jackson theorem we get
(31)∥g~′′-Jn(g~′′)∥Lp[0,π]≤cω2(g~′′,1n)Lp[0,π].
Since (Hg)(x)=(d2/dσ2)g~(σ) and (HG)(x)=(d2/dσ2)Jn(g~,σ) (31) implies
(32)∥Hg-HG∥Lp(u)[-1,1]≤cω2(g~′′,1n)Lp[0,π].
For given ɛ>0 we choose n such that (1+c)ω2(g~′′,1/n)Lp[0,π]<ɛ and (1+c)ω2(g~,1/n)Lp[0,π]<ɛ to obtain
(33)∥HG∥Lp(u)[-1,1]<∥Hg∥Lp(u)[-1,1]+ɛ,∥G-g∥Lp(u)[-1,1]<ɛ.
This completes the proof of the lemma.


## 3. Proof of Theorem1

The theorem will be proved if we show that∥f-Gs,nf∥p,u~K(f,1/n2;Lp(u)[-1,1],Y,H) (see Lemma 7). First we prove the converse result K(f,1/n2;Lp(u)[-1,1],Y,H)≤c∥f-Gs,nf∥Lp(u)[-1,1], which is a strong converse inequality of type A in the terminology of [3]. We utilize Theorems 3.1 and   4.1 in [3] with Qα=Gs,n, Df=Hf, Φ(f)=∥D2f∥Lp(u)[-1,1], and λ(n)=(1/2)∫-ππv2Ks,n(v)dv~n-2 for s≥2, λ1(n)=(1/3!)∫-ππv4Ks,n(v)dv~n-4 for s≥3. The result needed for inequality (3.3) from Theorem  3.1 in [3] with M=1 is given by
(34)∥Gs,nf∥Lp(u)[-1,1]≤∥f∥Lp(u)[-1,1].We setf(cos(arccosx+v)):=f(cos(z+v)):=f~(z+v),z=arccosx. To prove (34) we recall the representation
(35)(Gs,nf)(x)=∫-ππf(cos(arccosx+v))Ks,n(v)dv(Gs,nf)(x)=∫-ππf~(z+v)Ks,n(v)dv(Gs,nf)(x)=∫-ππf~(z-v)Ks,n(v)dv(Gs,nf)(x)=f~*K(z)(Gs,nf)(x)=K*f~(z)(Gs,nf)(x)=(G~s,nf~)(z),
where
(36)K(v)=K1(v)={μs,n(sin(nv/2)sin(v/2))2s,if|v|≤π,0,otherwise.
And the convolution is defined by
(37)F*G(x):=∫-∞∞F(x-v)G(v)dv.We have(38)∥Gs,nf∥Lp(u)[-1,1]={∫-11|(Gs,nf)(x)|p11-x2dx}1/p(x=cosz)={∫π0|(G~s,nf~)(z)|p11-cos2z(-sinz)dz}1/p={∫0π|(G~s,nf~)(z)|pdz}1/p=∥G~s,nf~∥Lp[0,π].Using Lemma2 as f~ and G~s,nf~ are even, 2π periodic functions, we estimate
(39)∥G~s,nf~∥Lp[0,π]=∥K*f~∥Lp[0,π]≤∥K∥L1(-∞,∞)∥f~∥Lp[0,π]=∥f~∥Lp[0,π]=∥f∥Lp(u)[-1,1]
to complete the proof of (34).We will show now that forf∈Z the following Voronovskaya-type estimate holds true:
(40)∥f-Gs,nf+λ(n)Df∥Lp(u)[-1,1]≤λ1(n)Φ(f),
with Φ(f)=∥f~(4)∥Lp[0,π]=∥H2f∥Lp(u)[-1,1], Df=Hf, and λ(n)=(1/2)∫-ππv2Ks,n(v)dv, λ1(n)=(1/3!)∫-ππv4Ks,n(v)dv. Inequality (40) will serve for (3.4) from Theorem  3.1 in [3]. We note that Gs,nmf∈Z as Gs,nf is an algebraic polynomial. Let f∈Z. From Z⊂Y and Lemma 6  f~∈C2(ℝ). We apply Lemma 6 for Hf to obtain that f~∈C4(ℝ). With λ(n)=(1/2)∫-ππv2Ks,n(v)dv we have
(41)f(x)-(Gs,nf)(x)+λ(n)Df(x)=f~(z)-(G~s,nf~)(z)+λ(n)f~′′(z)=∫-ππ[f~(z)-f~(z+v)+12v2f~′′(z)]Ks,n(v)dv.Expandingf~(z+v) by Taylor’s formula and using that ∫-ππvKs,n(v)dv=0, ∫-ππv3Ks,n(v)dv=0, we now have
(42)f(x)-(Gs,nf)(x)+λ(n)Df(x)=-∫-ππ∫zz+v13!f~(4)(ξ)(z+v-ξ)3dξKs,n(v)dv.We recall thatf~(4)(ξ)=(H2f)(cosξ). As for ξ∈[z,z+v], z+v-ξ∈[0,v], and the sign of ∫zz+v|f~(4)(ξ)|dξ is constant and coincides with the sign of v (if the integral is not zero); via Minkowski’s inequality, we get
(43)∥f-Gs,nf+λ(n)Df∥Lp(u)[-1,1]=∥f~-G~s,nf~+λ(n)f~′′∥Lp[0,π]={∫-ππ∫zz+v13!f~(4)(ξ)∫0π|∫-ππ∫zz+v13!f~(4)(ξ)×(z+v-ξ)3dξKs,n(v)dv∫-ππ∫zz+v13!f~(4)(ξ)|pdz}1/p≤{|∫zz+v|f~(4)(ξ)(z+v-ξ)3|dξ|dv)p∫0π(∫-ππKs,n(v)13!×|∫zz+v|f~(4)(ξ)(z+v-ξ)3|dξ|dv)pdz}1/p≤{∫0π(∫-ππKs,n(v)13!×|∫zz+v|f~(4)(ξ)||v3|dξ|dv)pdz}1/p={∫0π(∫-ππKs,n(v)|v3|3!|∫zz+v|f~(4)(ξ)|dξ|dv)pdz}1/p=13!{∫0π(∫-ππKs,n(v)v41|v|×|∫zz+v|f~(4)(ξ)|dξ|dv)pdz}1/p=13!{∫0π(∫-ππv4Ks,n(v)1v∫zz+v|f~(4)(ξ)|dξdv︸φ(z,v))p×1v∫zz+v|f~(4)(ξ)|dξ︸φ(z,v)dv)pdz}1/p=13!{∫0π(∫-ππv4Ks,n(v)φ(z,v)dv)pdz}1/p≤13!∫-ππ{∫0π[v4Ks,n(v)φ(z,v)]pdz}1/pdv=13!∫-ππv4Ks,n(v){∫0π[φ(z,v)]pdz}1/pdv=13!∫-ππv4Ks,n(v){∫0π|φ(z,v)|pdz}1/pdv=13!∫-ππv4Ks,n(v)∥1v∫zz+v|f~(4)(ξ)|dξ∥Lp[0,π]dv≤13!∫-ππv4Ks,n(v)∥f~(4)∥Lp[0,π]dv=λ1(n)Φ(f).We apply Lemma4 (as f~(4) is even 2π periodic function) to obtain ∥(1/v)∫zz+v|f~(4)(ξ)|dξ∥Lp[0,π] ≤ ∥f~(4)∥Lp[0,π]. This establishes (40).To obtain results corresponding to(3.5) and (3.6) from Theorem  3.1 in [3] we need a weighted Bernstein-type inequality for the power of the operator Gs,n like (6.10) in [3].We use representations(44)(Gs,nf)(x)=∫-ππf~(z-v)Ks,n(v)dv(Gs,nf)(x)=f~*K(z)=K*f~(z)=(G~s,nf~)(z),Gs,n(Gs,nf)(x)=K*K*f~(z)=K2*f~(z),…(Gs,nmf)(x)=K*Km-1*f~(z)=Km*f~(z)(Km=K*Km-1form=2,3,4,…)to  obtainH1Gs,nmf=-{K′}*Km-1*f~,
where
(45){K′}(v)={μs,nddv[(sin(nv/2)sin(v/2))2s],if|v|≤π,0,otherwise.We estimate the action ofH1 on the mth degree Gs,nm of the operator. Using Lemma 2 we have
(46)∥H1Gs,nmf∥Lp(u)[-1,1]=∥{K′}*Km-1*f~∥Lp[0,π]≤∥{K′}*Km-1∥L1(-∞,∞)∥f~∥Lp[0,π]=∥{K′}*Km-1∥L1(-∞,∞)∥f∥Lp(u)[-1,1].We have already proved earlier (see Assertion  1.2 in [1]) that
(47)∥{K′}*Km-1∥L1(-∞,∞)≤cnm
and therefore
(48)∥H1Gs,nmf∥Lp(u)[-1,1]≤cnm∥f∥Lp(u)[-1,1].AsGs,n commutes with the operator H1 using estimation (48) we now observe that
(49)∥D2Gs,n4mf∥Lp(u)[-1,1]=∥H2Gs,n4mf∥Lp(u)[-1,1]∥D2Gs,n4mf∥Lp(u)[-1,1]≤cn2m-1∥HGs,n2mf∥Lp(u)[-1,1]∥D2Gs,n4mf∥Lp(u)[-1,1]=Aλ(n)λ1(n)∥DGs,n2mf∥Lp(u)[-1,1],(50)∥DGs,n2mf∥Lp(u)[-1,1]=∥HGs,n2mf∥Lp(u)[-1,1]∥DGs,n2mf∥Lp(u)[-1,1]=∥H1Gs,nmH1Gs,nmf∥Lp(u)[-1,1]∥DGs,n2mf∥Lp(u)[-1,1]≤cnm-1/2∥H1Gs,nmf∥Lp(u)[-1,1]∥DGs,n2mf∥Lp(u)[-1,1]≤cn2m-1∥f∥Lp(u)[-1,1]∥DGs,n2mf∥Lp(u)[-1,1]=B1λ(n)∥f∥Lp(u)[-1,1].Estimations (49) and (50) correspond to (3.5) and (3.6) from Theorem  3.1 in [3]. To match the conditions of Theorem  4.1 in [3] we need
(51)Aλ(n)λ1(n)=cn2m-1
to be satisfied with A<1 or
(52)A=cm-1<1
which is true for large m. From (34), (40), (49), and (50) we obtain
(53)K(f,1n2;Lp(u)[-1,1],Y,H)≤c∥f-Gs,nf∥Lp(u)[-1,1].To prove the direct result we need inequality (34) and
(54)∥g-Gs,ng∥Lp(u)[-1,1]≤c1n2∥Hg∥Lp(u)[-1,1]forg∈Y
to be satisfied (see Theorem  3.4 in [3]).Letg∈Y. We set g(cos(arccosx+v)):=g(cos(z+v)):=g~(z+v),z=arccosx. We have
(55)(Gs,ng-g)(x)=∫-ππ[g~(z+v)-g~(z)]Ks,n(v)dv.Expandingg~(z+v) by Taylor’s formula (by Lemma 6  g~∈C2(ℝ))
(56)g~(z+v)-g~(z)=vg~′(z)+∫zz+vg~′′(ξ)(z+v-ξ)dξ,
and using that ∫-ππvKs,n(v)dv=0 we have
(57)(Gs,ng-g)(x)=∫-ππ∫zz+vg~′′(ξ)(z+v-ξ)dξKs,n(v)dv.We recall thatg~′′(ξ)=(Hg)(cosξ). As for ξ∈[z,z+v], z+v-ξ∈[0,v], and the sign of ∫zz+v|g~′′(ξ)|dξ is constant and coincides with the sign of v (if the integral is not zero); we get
(58)∥Gs,ng-g∥Lp(u)[-1,1]=∥G~s,ng~-g~∥Lp[0,π]={∫0π|∫-ππ∫zz+vg~′′(ξ)(z+v-ξ)dξKs,n(v)dv|pdz}1/p≤{∫0π(∫-ππ|v||∫zz+v|g~′′(ξ)|dξ|Ks,n(v)dv)pdz}1/p={∫0π(∫-ππv2Ks,n(v)1v∫zz+v|g~′′(ξ)|dξ︸ψ(z,v)dv)pdz}1/p={∫0π(∫-ππv2Ks,n(v)ψ(z,v)dv)pdz}1/p.Using the Minkowski inequality we have(59){∫0π(∫-ππv2Ks,n(v)ψ(z,v)dv)pdz}1/p≤∫-ππ{∫0π[v2Ks,n(v)ψ(z,v)]pdz}1/pdv=∫-ππv2Ks,n(v){∫0π[ψ(z,v)]pdz}1/pdv=∫-ππv2Ks,n(v)∥1v∫zz+v|g~′′(ξ)|dξ∥Lp[0,π]dv≤∫-ππv2Ks,n(v)∥g~′′∥Lp[0,π]dv≤c1n2∥g~′′∥Lp[0,π]=c1n2∥Hg∥Lp(u)[-1,1].We applied Lemma4 (as g~′′ is even 2π periodic function) to obtain ∥(1/v)∫zz+v|g~′′(ξ)|dξ∥Lp[0,π] ≤ ∥g~′′∥Lp[0,π]. The proof of the direct result
(60)∥f-Gs,nf∥Lp(u)[-1,1]≤cK(f,1/n2;Lp(u)[-1,1],C2,H)
is completed. This concludes the proof of Theorem 1.



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