Mathematical Statistician and Engineering Applications

2326-9865

Vol. 71 No. 3s3 (2022) 1 http://philstat.org.ph

On Approximation of Functions by Algebraic Polynomials in the Weighted Quasi Normed Space

Nada Zuhair Abd AL-Sada

Iraq Diwaniyah, Qadisiyah University

- Al , College of Education ,

Mathematics

nada.abdalsada@qu.edu.iq

Issue: Special Issue on Mathematical Computation in Combinatorics and Graph Theory in Mathematical Statistician and Engineering Applications

Article Info

Page Number: 01 - 24 Publication Issue:

Vol 71 No. 3s3 (2022)

Article Received: 30 April 2022 Revised: 22 May 2022

Accepted: 25 June 2022 Publication: 02 August 2022

Abstract

In this paper , we discuss the approximation of weighted function in the weighted quasi normed space by algebraic polynomials. This approximation of the functions is found to the weight quasi normed space by using Jacobi function ,and the zero value is excluded , because it gives the space without weighted .

Two results were obtained in this research, each of which depends on a constant with certain values, One of these constants, it is restricted to several values, while the other result showed another restriction in the constant.

Keywords: -Approximation Theory, Algebraic polynomial, Jacobi weight, Heaviside function, quasi normed space.

Subject Classification: Mathematics subject classification 2010:41A10,41A50

1.Introduction and Main Result

In this paper we are study approximation of function F in the weighted quasi normed space L_ω,p(Ι), 1 ≤ p < ∞. Where we will study space in an interval Ι = [−𝔟, 𝔟], 𝔟 =positive integer and the positive weighted function ω(x) on Ι, its called Jacobi weight as in [1-3] ,ω(x) = (1 − x)^a1(1 + x)^a2, a₁, a₂ > −𝑏, x ∈ Ι ⊂ R.

Characteristic for L_ω,p(Ι), are the inequalities of Holder and Minkowski as in [4,5]. The study of approximation in this paper will be by sing algebraic polynomial in ∏_m, 0 < 𝑚 ≤¹

h,¹

h∈ R⁺(positive integer).The space L_ω,p(Ι), consisting of all functions F such that F: Ι → R and the function F is defined as

‖F‖_Lω,p(Ι) = (∫_Ι|F(x)|^pω(x)dx)

1

p < ∞, 𝑥 ∈ 𝛪, 1 ≤ 𝑝 < ∞. If a₁ = a₂ = 0 then L_ω,p(Ι) = L_p(Ι), and ‖F‖_Lp(Ι)= (∫_Ι|F(x)|^pdx)

1

p < ∞, 𝑥 ∈ 𝛪, 1 ≤ 𝑝 < ∞.

So we will exclude zero values to a₁ and a₂.Let the space W_ω,p^(k+v)(Ι), k > 0, 𝑣 ∈ (0, 𝑏], it is the set of all functions F is defined as {F: F ∈ L_p; ∃F^(k) ∈ L_p, ∀h ∈ (0, b) ∋ ‖F^(k)(x + h) −

F^(k)(x)‖_L

ω,p[−b,b−h]≤ h^v}. Let a₁, a₂ > −𝑏, 1 ≤ 𝑝 < ∞, 𝑘 > 0, 𝑣 ∈ (0, b], we will know the class

2326-9865

Vol. 71 No. 3s3 (2022) 2 http://philstat.org.ph

of function as EW_ω,p^(k+v)(Ι) = {f ∈ L_ω,p(Ι): ∃F^(k) ∈ L_ω,p(Ι), ∀h ∈ (0, b) ∋ (∫_−b+h^b |F^(k)(x) − F^(k)(x − h)|^pω(x)dx)

1

p ≤ Eh^v} (1)

The Heaviside function as in [6,7], was used in the search, may be defined as:

1) A piecewise function H(x) = {1, x > 0 0, x ≤ 0. 2) An indicator function H(x) = 1_x>0.

The Dirac delta function as in [8-10] is the derivative of the Heaviside function δ_a(x) = ^d

dxH(x), hence the Heaviside function can be considered to be the integral of the Dirac delta function. This is some time written as H(x) = ∫_−∞^x δ_a(u)du.

Theorem A : Let F ∈ W_ω,p^(k+v)(Ι), then there exist a polynomial Q_m of degree ≤ m, 0 < 𝑚 ≤¹

h, m depends on (a₁, a₂ , p, v, b), such that :

1) ∀¹

h≤ ¹

h₁ there exist a constant 𝒞 = 𝒞(a₁, a₂ , p, b, v, k, α, α₁) satisfy

‖ ^F−Qm

(φ+h)^k+v‖

L_ω,p(Ι)

≤ 𝒞h^k+v+

2 pLn

1 p1

h.

Such that a₁, a₂ > −𝑏, 1 ≤ 𝑝 < ∞, 𝑣 ∈ (0, 𝔟], k > 0, 𝛼 ∈ (0,1), α₁ = Lnα ∈ R⁺ and 𝔟 positive integer.

2) ∀¹

h≤ ¹

h₁, there exist a constant 𝒞 depends on a₁, a₂, p, b, v, k, μ satisfy

‖ F − Q_m (φ + h)^k+μ‖

L_ω,p(Ι)

≤ 𝒞h^k+v

𝒞 = 𝒞(a₁, a₂ , p, b, v, k, μ, s),Such that a₁, a₂ > −𝑏, 1 ≤ 𝑝 < ∞, 𝑣 ∈ (0, 𝔟], k > 0, 𝔟 positive integer and s is the integer function of the greatest integer.

2. Notation

The following notation is used in this paper:

Let x_i= bcosγ_i, i = 1, … , Log_α¹

h,¹

h positive integer 0 < 𝛼 < 1, x_i < x_i+1, it's clear to see that bsin γ_i= √1 − cos²γ_i = √1 − x_i², Then |Ι| = x_i+1− x_i = b cosγ_i+1− b cosγ_i = bh_i =^bhKi

αⁱ , (K_i is a maximum of i ). Let 𝕁_N is a partition to the an interval Ι = [−𝔟, 𝔟] such that: 𝕁_N = {𝒿_𝓈, 𝓈 = 0, … , N | − 𝔟 = 𝒿₀ < 𝒿₁ < ⋯ < 𝒿_N−1 < 𝒿_N = 𝔟}, and Δ¹

h

(x) = hφ(x) + h when φ(x) =

√1 − x² , x ∈ Ι. Suppose t_i the sets of points which is belong to an interval Ι and satisfy the followings :

1. t_i = 0, t_i+1 = t_i+^bhKi

α_i

2326-9865

Vol. 71 No. 3s3 (2022) 3 http://philstat.org.ph

2. α⁻ⁱ < √1 − t_i²

3. √1 − t_i+1² = √1 − (t_i+^bhKi

αⁱ )² < α⁻ⁱ, i = 1, … , Log_α¹

h It can be pointed out that t_k+1 = t_Log

α1

h+1 = b, t_−i = −t_i, and let A_k = [t_k, t_k+1]⋃[t_−(k+1), t_−k], k ≥ 0, ∀x ∈ A_k.

𝟑.Auxiliary Result Lemma 3.1.‖𝐹¹

ℎ

− 𝐹_ℎ(𝐹)‖

𝐿_𝜔,𝑝(𝐼₁)

≤ 𝒞𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑏),

such that 𝑎₁, 𝑎₂ > −𝑏, 𝑏 =positive integer, and 𝐼₁ = [−(𝑏 − ℎ), 𝑏 − ℎ]

Proof: Let 𝐴 = ‖𝐹1 ℎ

− 𝐹_ℎ(𝐹)‖

𝐿_{𝜔,𝑝(𝐼1)}

= (∫ |𝐹1 ℎ

(𝑥) − 𝐹_ℎ(𝐹, 𝑥)|

𝑝

𝜔(𝑥)𝑑𝑥

𝑏−ℎ

−(𝑏−ℎ) )

1 𝑝

Let ∀𝐹_ℎ(𝐹, 𝑥) ∈ 𝐸𝑊_𝜔,𝑝^(𝑘+𝑣)(𝛪), defined

𝐹_ℎ(𝐹, 𝑥) = ℎ ∫ 𝐹(𝑥^∗)𝑑𝑥^∗, 𝑥 ∈ [−1 ℎ,1

ℎ− 1)

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

Then 𝐴 = (∑ ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|ℎ ∫_𝑥^𝑥+ℎ𝐹(𝑥^∗)𝑑𝑥^∗− ℎ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}𝐹(𝑥^∗)𝑑𝑥^∗|^𝑝𝜔(𝑥)𝑑𝑥

𝑏−2ℎ 𝑏ℎ

𝑘=^ℎ−𝑏_ℎ𝑏 )

1 𝑝

𝐴^𝑝 ≤ 𝒞 ∑ ∫ (ℎ^𝑝∫^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗ + ℎ) − 𝐹(𝑥^∗)|^𝑝

𝑘𝑏ℎ 𝜔(𝑥)𝑑𝑥^∗𝑑𝑥)

𝑘𝑏ℎ+ℎ 𝑘𝑏ℎ 𝑏−2ℎ

𝑏ℎ 𝑘=^ℎ−𝑏_ℎ𝑏

≤ 𝒞 ∑ ℎ^𝑝 ∫ |𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

𝑑𝑥^∗ ∫ 𝜔(𝑥)

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

𝑑𝑥

𝑏−2ℎ 𝑏ℎ

𝑘=ℎ−𝑏 ℎ𝑏

Let 𝐿^𝑝₁ = ∑ ℎ^𝑝∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗ + ℎ) − 𝐹(𝑥^∗)|^𝑝𝑑𝑥^∗∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}𝜔(𝑥)𝑑𝑥

−𝑏 𝑏ℎ 𝑘=^ℎ−𝑏_ℎ𝑏

And 𝐿^𝑝₂ = ∑ ℎ^𝑝∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝𝑑𝑥^∗∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}𝜔(𝑥)𝑑𝑥

𝑏−2ℎ 𝑏ℎ

𝑘=0 . We have

(1 − 𝑥)^𝑎1𝑚𝑖𝑛{𝐵₁^−𝑎2, 𝐵₁^𝑎2} ≤ 𝜔(𝑥) ≤ (1 − 𝑥)^𝑎1𝑀𝑎𝑥{B₁^−𝑎2, 𝐵₁^𝑎2}, 𝑥 ∈ [0, 𝑏) If 𝑎₁ < 0 then we get

2326-9865

Vol. 71 No. 3s3 (2022) 4 http://philstat.org.ph

𝐿^𝑝₂ ≤ 𝒞(𝑎₂)ℎ^𝑝∑ ℎ^𝑝∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝(1 − 𝑥^∗)^𝑎1𝑑𝑥^∗(∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}(1 − 𝑏𝑘ℎ)^−𝑎1(1 −

𝑏−2ℎ 𝑏ℎ 𝑘=0

𝑥)^𝑎1𝑑𝑥) . Let 𝑇_𝑘² = ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}(1 − 𝑏𝑘ℎ)^−𝑎1(1 − 𝑥)^𝑎1𝑑𝑥. Hence𝑇_𝑘² ≤ 𝒞(𝑎₁, 𝑏), then 𝐿^𝑝₂ ≤ 𝒞(𝑎₁, 𝑎₂, 𝑏)ℎ^𝑝∑ ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝𝜔(𝑥^∗)𝑑𝑥^∗

𝑏−2ℎ 𝑏ℎ

𝑘=0 (2)

If 𝑎₁ ≥ 0, then by the same way we get the same estimate. Now, for 𝐿^𝑝₁

(1 + 𝑥)^𝑎2𝑚𝑖𝑛{𝐵₂^−𝑎1, 𝐵₂^𝑎1} ≤ 𝜔(𝑥) ≤ (1 + 𝑥)^𝑎2𝑀𝑎𝑥{𝐵₂^−𝑎1, 𝐵₂^𝑎1}. ∀𝑎₂ ≥ 0

𝐿^𝑝₁ ≤ 𝒞(𝑎₁)ℎ^𝑝 ∑ ∫ |𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝 (1

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

−𝑏

𝑘=𝑏−2ℎ 𝑏ℎ

+ 𝑥^∗)^𝑎2𝑑𝑥^∗( ∫ (1 + 𝑏𝑘ℎ + ℎ)^−𝑎2(1 + 𝑥)^𝑎2

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

𝑑𝑥)

Let 𝑇_𝑘¹ = ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}(1 + 𝑏𝑘ℎ + ℎ)^−𝑎2(1 + 𝑥)^𝑎2𝑑.Hence 𝑇_𝑘¹ ≤ 𝒞(𝑎₂, 𝑏), then 𝐿^𝑝₁ ≤ 𝒞(𝑎₁, 𝑎₂, 𝑏)ℎ^𝑝∑^−𝑏 ∫_𝑘𝑏ℎ^{𝑘𝑏ℎ+ℎ}|𝐹(𝑥^∗+ ℎ) − 𝐹(𝑥^∗)|^𝑝𝜔(𝑥^∗)𝑑𝑥^∗

𝑘=^ℎ−𝑏_ℎ𝑏 (3)

There for we get from (2) and (3) and by using definition (1) 𝐴^𝑝 ≤ 𝐿^𝑝₁+ 𝐿^𝑝₂

‖𝐹1 ℎ

− 𝐹_ℎ(𝐹)‖

𝑝

𝐿_𝜔,𝑝(𝐼₁)

≤ 𝒞(𝑎₁, 𝑎₂, 𝑏)(𝐸ℎ^𝑣 )^𝑝

‖𝐹¹

ℎ

− 𝐹_ℎ(𝐹)‖

𝐿_𝜔,𝑝(𝐼₁)

≤ 𝒞𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑏), such that 𝑎₁, 𝑎₂ > −𝑏, 𝑏 =positive integer.

Lemma 3.2.‖𝐹 − 𝐹1 ℎ

‖

𝐿_𝜔,𝑝(𝐼₂)

≤ 𝒞𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑏),

such that 𝑎₁, 𝑎₂ > −𝑏, 𝑏 =positive integer, and 𝐼₂ = [−(𝑏 − ℎ), 𝑏 − ℎ]

Proof: Let 𝐵 = ‖𝐹 − 𝐹1 ℎ

‖

𝐿_{𝜔,𝑝(𝐼2)}

= (∫ |𝐹 − 𝐹1 ℎ

(𝑥)|

𝑝

𝜔(𝑥)𝑑𝑥

𝑏−ℎ

−(𝑏−ℎ) )

1 𝑝

By using the same method in Lemma 3.1. then

2326-9865

Vol. 71 No. 3s3 (2022) 5 http://philstat.org.ph

𝐵^𝑝 ≤ 𝒞(𝑎₂)ℎ^𝑝 ∑ ℎ^𝑝 ∫ |𝐹(𝑥^∗) − 𝐹(𝑥^∗− ℎ)|^𝑝

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

𝑑𝑥^∗( ∫ 𝜔(𝑥)

𝑘𝑏ℎ+ℎ

𝑘𝑏ℎ

𝑑𝑥)

𝑏−2ℎ 𝑏ℎ

𝑘=ℎ−𝑏 𝑏ℎ

Hence ‖𝐹 − 𝐹1 ℎ

‖

𝐿_𝜔,𝑝(𝐼₂)

≤ 𝒞𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑏), such that 𝑎₁, 𝑎₂ > −𝑏, 𝑏 =positive integer.

Lemma 3.3.∫ ∫ |𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑥𝑑𝑦 ≤ 𝒞𝐸^𝑝ℎ^(𝑣+

1 𝑝)𝑝 𝑏

𝑏−ℎ 𝑏 𝑏−ℎ

𝒞 = 𝒞(𝑣, 𝑝), 1 ≤ 𝑝 < ∞, 𝑣 ∈ (0, 𝑏]

Proof: Let 𝑇 = ∫_𝑏−ℎ^𝑏 ∫_𝑏−ℎ^𝑏 |𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑥𝑑𝑦

We will now make a s substitution for the internal integration, put 𝑡 = 𝑥 − 𝑦, then we get

𝑇 = ∫ ∫ |𝐹(𝑡 + 𝑦) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑦𝑑𝑡

𝑏−𝑦

𝑡=𝑏−ℎ−𝑦 𝑏

𝑦=𝑏−ℎ

+ ∫ ∫ |𝐹(𝑡 + 𝑦) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑦𝑑𝑡

𝑏−𝑡

𝑏−ℎ ℎ

0

Let 𝑣 = −𝑡, 𝑑𝑣 = −𝑑𝑡

𝑇 ≤ ∫ ∫ |𝐹(𝑦) − 𝐹(𝑦 − 𝑣)|^𝑝𝜔(𝑦)𝑑𝑦𝑑𝑣

𝑏

𝑏−ℎ+𝑣 ℎ

0

+ ∫ ∫ |𝐹(𝑡 + 𝑦) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑦𝑑𝑡

𝑏−𝑡

−𝑏 ℎ

0

By definition (1) when 𝑡 ∈ (0, 𝑏)

𝑇 ≤ ∫ ∫₀^ℎ _−𝑏+𝑣^𝑏 |𝐹(𝑦) − 𝐹(𝑦 − 𝑣)|^𝑝𝜔(𝑦)𝑑𝑦𝑑𝑣+ ∫ (𝐸𝑡₀^{ℎ 𝑣})^𝑝𝑑𝑡

Also by definition (1) when 𝑣 ∈ (0, 𝑣), −𝑏 + 𝑣 ≤ 𝑦 ≤ 𝑏, (𝑘 = 0), then

∫ |𝐹(𝑦) − 𝐹(𝑦 − 𝑣)|^𝑝𝜔(𝑦)𝑑𝑦

𝑏

−𝑏+𝑣

≤ (𝐸𝑡^𝑣)^𝑝

Hence 𝑇 ≤ 2𝐸^𝑝∫ 𝑡₀^{ℎ 𝑣𝑝}𝑑𝑡, hence

∫_𝑏−ℎ^ℎ ∫_𝑏−ℎ^𝑏 |𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝑑𝑥 𝜔(𝑦)𝑑𝑦 ≤ ²

𝑣𝑝+1𝐸^𝑝 ℎ^(𝑣+

1 𝑝)𝑝

=𝒞(𝑣, 𝑝)𝐸^𝑝ℎ^(𝑣+

1 𝑝)𝑝

, 1 ≤ 𝑝 < ∞, 𝑣 ∈ (0, 𝑏].

Lemma 3.4. ∫ ∫ |𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝑑𝑥𝜔(𝑦)𝑑𝑦 ≤ 𝒞𝐸^𝑝ℎ^(𝑣+

1 𝑝)𝑝

−𝑏+ℎ

−𝑏

−𝑏+ℎ

−𝑏 ,𝒞 = 𝒞(𝑣, 𝑝), 1 ≤ 𝑝 < ∞, 𝑣 ∈

(0, 𝑏]

2326-9865

Vol. 71 No. 3s3 (2022) 6 http://philstat.org.ph

Proof: By using the same way in Lemma (3.3) we get the above result.

Theorem 3.5. Let 𝐹 ∈ 𝐸𝑊_𝜔,𝑝^(𝑣)(𝛪), 1 < 𝑝 < ∞, then

‖𝐹 − 𝐹_ℎ(𝐹)‖_{𝐿𝜔,𝑝(𝛪)} ≤ 𝒞𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑣, 𝑝) Such that 𝑎₁, 𝑎₂ > −𝑏, 𝑣 ∈ (0, 𝑏], 1 < 𝑝 < ∞.

Proof: Since the an interval 𝛪 = [−𝔟, 𝔟], can be written as

[−𝔟, 𝔟] = [−𝔟, −𝔟 + ℎ]⋃[−𝔟 + ℎ, 𝔟 − ℎ]⋃[𝔟 − ℎ, 𝔟], and we have

|𝐹(𝑥) − 𝐹_ℎ(𝐹, 𝑥)| ≤ 𝒞 |𝐹(𝑥) − 𝐹1 ℎ

(𝑥)| + 𝒞 |𝐹1 ℎ

(𝑥) − 𝐹_ℎ(𝐹, 𝑥)|

Now by using Holder inequality if 𝑝 ≥ 1, [10],and by take 𝐿_𝜔,𝑝- of both sides, we get

‖𝐹 − 𝐹_ℎ(𝐹)‖_{𝐿𝜔,𝑝(𝛪)}

≤ 𝒞 ∫ ∫ |𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑥𝑑𝑦 + ( ∫ |𝐹1 ℎ

(𝑥) − 𝐹_ℎ(𝐹, 𝑥)|

𝑝

𝜔(𝑥)𝑑𝑥

𝑏−ℎ

−(𝑏−ℎ)

)

1

−𝑏+ℎ 𝑝

−𝑏

−𝑏+ℎ

−𝑏

+ ( ∫ ∫|𝐹(𝑥) − 𝐹(𝑦)|^𝑝𝜔(𝑦)𝑑𝑥

𝔟

𝔟−ℎ 𝑏

𝔟−ℎ

𝑑𝑦)

Now, by using Lemmas (3.1),(3.3) and (3.4) we get

‖𝐹 − 𝐹_ℎ(𝐹)‖_{𝐿𝜔,𝑝(𝛪)} ≤ 𝒞(𝑣, 𝑝)𝐸^𝑝ℎ^(𝑣+

1 p)𝑝

+ 𝒞(𝑎₁, 𝑎₂, 𝔟)𝐸ℎ^𝑣 + 𝒞(𝑣, 𝑝)𝐸^𝑝ℎ^(𝑣+

1 𝑝)𝑝

=𝒞(𝑎₁, 𝑎₂, 𝑝, 𝔟, 𝑣)𝐸ℎ^𝑣, 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝑣, 𝑝)

Such that 𝑎₁, 𝑎₂ > −𝑏, 𝑣 ∈ (0, 𝑏], 1 < 𝑝 < ∞, 𝑏 =positive integer.

Theorem 3.6. Let 𝐹 ∈ 𝐸𝑊_𝜔,𝑝^(𝑘+𝑣)(𝛪), 𝑘 = 0, 1 < 𝑝 < ∞, then

‖𝐹 − 𝐹_ℎ(𝐹)

∆1 ℎ

‖

𝐿_𝜔,𝑝(𝛪)

≤ 𝒞ℎ

2 𝑝 (𝐿𝑛1

ℎ)

1

𝑝, 𝒞 = 𝒞(𝛼, 𝛼₁, 𝑣, 𝑝, 𝔟)

Such that 𝛼 ∈ (0,1), 𝛼₁ ∈ 𝑅⁺, 𝑣 ∈ (0, 𝑏], 1 ≤ 𝑝 < ∞, 𝑏 =positive integer.

Proof: We have 𝐴_𝑘 ⊂ [−𝔟, 𝔟], 𝑘 ≥ 0 then

‖𝐹 − 𝐹_ℎ(𝐹)

∆1 ℎ

‖

𝐿_𝜔,𝑝(𝛪) 𝑝

= ∫ |𝐹(𝑥) − 𝐹_ℎ(𝐹, 𝑥)

∆1 ℎ

(𝑥) |

𝑝

𝜔(𝑥)𝑑𝑥

𝔟

−𝔟

2326-9865

Vol. 71 No. 3s3 (2022) 7 http://philstat.org.ph

= ∑ ∫_𝐴

𝑘 𝐿𝑜𝑔_𝛼1

ℎ

𝑘=0

|𝐹(𝑥) − 𝐹_ℎ(𝐹, 𝑥)

∆1 ℎ

(𝑥) |

𝑝

𝜔(𝑥)𝑑𝑥

By using Theorem 3.5 and ∀ 𝐹 ∈ 𝑊_𝜔,𝑝^(𝑣)(𝛪), 𝐹_ℎ𝑖(𝐹, 𝑥) = 𝐹ℎ 𝛼𝑘

(𝐹, 𝑥)

∫_𝐴

𝑘|𝐹(𝑥) − 𝐹ℎ 𝛼^𝑘

(𝐹, 𝑥)|

𝑝

𝜔(𝑥)𝑑𝑥 ≤ ∫_𝛪|𝐹(𝑥) − 𝐹_ℎ(𝐹, 𝑥)|^𝑝𝜔(𝑥)𝑑𝑥 ≤ 𝒞(𝑎₁, 𝑎₂, 𝑣, 𝑝) (^ℎ

𝛼^𝑘)^𝑣𝑝. Now,for 𝑘 < 𝐿𝑜𝑔_𝛼¹

ℎ, on an interval 𝐴_𝑘, we have 𝛼^𝑘 < 𝛼^𝑘+1, hence by property (2) we get 𝛼^−(𝑘+1)< 𝛼^−𝑘 < √1 − 𝑡_𝑘²

< √1 − 𝑡_𝑘+1² < √1 − 𝑥² (𝛼^𝑘+1)^𝑣(√1 − 𝑥²+ ℎ)^𝑣 > 1 1

ℎ^𝑣

(√1 − 𝑥² + ℎ) < (𝛼^𝑘+1)^𝑣

ℎ^𝑣 ⇒ 1

(ℎ𝜑(𝑥) + ℎ²)^𝑣 < (𝛼^𝑘+1 ℎ )

𝑣

Hence for 𝑘 = 𝐿𝑜𝑔_𝛼¹

ℎ, we get

‖𝐹 − 𝐹_ℎ(𝐹)

∆1 ℎ

‖

𝐿_𝜔,𝑝(𝛪) 𝑝

≤ 𝒞(𝑣, 𝑝, 𝛼)|𝛪| = 𝒞(𝛼, 𝑣, 𝑝)ℎ_𝑖 = 𝒞 (𝑏ℎ𝑘_𝑖 𝛼^𝑖 )

=𝒞 (^{ℎ𝐿𝑜𝑔𝛼}

1 ℎ 𝛼^{𝐿𝑜𝑔𝛼}

1 ℎ

) = 𝒞(𝛼, 𝑝, 𝑣, 𝔟)ℎ²(^𝐿n

1 ℎ

𝐿𝑛𝛼), let 𝐿𝑛𝛼 = 𝛼₁ ∈ 𝑅⁺

‖𝐹 − 𝐹_ℎ(𝐹, . )

∆1 ℎ

(. ) ‖

𝐿_𝜔,𝑝(𝛪)

≤ 𝒞ℎ

2

𝑝 (𝐿𝑛1 ℎ)

1

𝑝, 𝒞 = 𝒞(𝛼, 𝛼₁, , 𝑣, 𝑝, 𝔟)

Such that 𝛼 ∈ (0,1), 𝛼₁ ∈ 𝑅⁺, 𝑣 ∈ (0, 𝑏], 1 ≤ 𝑝 < ∞, 𝑏 =positive integer.

Lemma 3.7. For all 𝑥 ∈ 𝛪 = [−𝔟, 𝔟], and [𝑥_𝑖, 𝑥_𝑖+1] ⊂ [𝑡_𝑘−1, 𝑡_𝑘] we have

∫ 𝜔(𝑥)𝑑𝑥 ≤ 𝒞ℎ𝑠𝑖𝑛^2𝑎1+1

𝑥_𝑖+1

𝑥_𝑖

𝛾𝑖,1 ℎ

2326-9865

Vol. 71 No. 3s3 (2022) 8 http://philstat.org.ph

Where 𝜔(𝑥) the Jacobi weights, and 𝒞 = 𝒞(𝑎₁, 𝑎₂, 𝔟), such that 𝑎₁, 𝑎₂ > −𝑏, 𝔟 =positive integer.

Proof: i) If 𝑎₁ < 0, the proof will be discussed in case 𝑎₁ < 0, since [𝑥_𝑖, 𝑥_𝑖+1] ⊂ [𝑡_𝑘−1, 𝑡_𝑘], hence

∫ (1 − 𝑥)^𝑎1(1 + 𝑥)^𝑎2𝑑𝑥 ≤ 𝒞(𝑎₂) ∫ (1 − 𝑥)^𝑎1𝑑𝑥

𝑡_𝑘

𝑡_𝑘− ℎ 𝛼^𝑖−1 𝑥_𝑖+1

𝑥_𝑖

1) If 𝑡_𝑘 < 𝑏 (𝑘 ≤ 𝐿𝑜𝑔_𝛼¹

ℎ ), then

𝒞(𝑎₂) ∫_𝑡^𝑡𝑘 (1 − 𝑥)^𝑎1 𝑑𝑥 ≤ 𝒞(𝑎₁, 𝑎₂, 𝔟) (𝑡_𝑘− 𝑡_𝑘−1) (√1 − 𝑡_𝑘−1² )^2𝑎1

𝑘−ℎ𝛼^{−(𝑖−1)}

𝛥𝑡_𝑘−1= 𝑡_𝑘− 𝑡_𝑘−1= ℎ𝛼^{−(𝑘−1)}, and by property (2) we have 𝛼^{−(𝑘−1)}< √1 − 𝑡_𝑘−1² , hence 𝒞(𝑎₂) ∫_𝑡^𝑡𝑘 (1 − 𝑥)^𝑎1𝑑𝑥

𝑘−ℎ𝛼^{−(𝑖−1)} ≤ 𝒞(𝑎₁, 𝑎₂, 𝔟)ℎ (√1 − 𝑡_𝑘−1² )^2𝑎1+1. It is clear that 𝑐𝑜𝑠 𝛾_𝑖,¹

ℎ

= 𝑡_𝑘−1, hence 𝑠𝑖𝑛²𝛾_𝑖,¹

ℎ

= 1 − 𝑐𝑜𝑠²𝛾_𝑖,¹

ℎ

, hence 𝛾_𝑖,¹

ℎ

= √1 − 𝑡_𝑘−1² , then we get

∫_𝑥^𝑥𝑖+1𝜔(𝑥)𝑑𝑥 ≤ 𝒞(𝑎₁, 𝑎₂, 𝑏)ℎ 𝑠𝑖𝑛^2𝑎1+1

𝑖 𝛾_𝑖,1

ℎ

. 2) If 𝑡_𝑘 = 𝑏 ( 𝑘 = 𝐿𝑜𝑔_𝛼¹

ℎ+ 1), then

∫ 𝜔(𝑥)𝑑𝑥 ≤ 𝒞(𝑎₁) ∫ (1 − 𝑥)^𝑎1𝑑𝑥

𝔟

𝔟− ℎ 𝛼^{𝐿𝑜𝑔𝛼}

1 ℎ 𝑥_𝑖+1

𝑥_𝑖

Let 1 − 𝑥 = 𝑦, 𝑑𝑦 = −𝑑𝑥

∫ 𝜔(𝑥)𝑑𝑥 ≤ −𝒞(𝑎₂, 𝔟) ∫ 𝑦^𝑎1

0

−ℎ²

𝑑𝑦 ≤

𝑥_𝑖+1

𝑥_𝑖

𝒞(𝑎₁, 𝑎₂, 𝔟)ℎℎ^2𝑎1+1

We have from property (1) and (2) 1

𝛼^(𝑖−1) < √1 − 𝑡_𝑖−1² = √1 − (𝑡_𝑖− ℎ 𝛼^𝑖−1)

2

, (𝑡_𝑘= 𝔟)

1 𝛼^{𝐿𝑜𝑔𝛼}

1 ℎ

< √1 − (𝔟 − ^ℎ

𝛼^{𝐿𝑜𝑔𝛼} 1 ℎ

)

2

, (𝑘 = 𝐿𝑜𝑔_𝛼¹

ℎ+ 1)hence ℎ ≤ 𝑠𝑖𝑛𝛾_𝑖,¹

ℎ

(4)

∫ 𝜔(𝑥)𝑑𝑥 ≤ 𝒞(𝑎₁, 𝑎₂, 𝔟)ℎ𝑠𝑖𝑛^2𝑎1+1

𝑥_𝑖+1

𝑥_𝑖

𝛾𝑖,1 ℎ

ii) If 𝑎₁ ≥ 0, we take this case if 𝑡_𝑘 < 𝑏 and if 𝑡_𝑘 = 𝔟, and by the same way in (i) we get

2326-9865

Vol. 71 No. 3s3 (2022) 9 http://philstat.org.ph

∫_𝑥^𝑥𝑖+1𝜔(𝑥)𝑑𝑥 ≤ 𝒞(𝑎₁, 𝑎₂, 𝔟)ℎ𝑠𝑖𝑛^2𝑎1+1

𝑖 𝛾_𝑖,1

ℎ

.

Lemma 3.8. For 𝑥 ∈ [𝑥_𝑖, 𝑥_𝑖+1] and 𝛾_𝑖 = 𝑐𝑜𝑠⁻¹𝑥_𝑖, we have

(𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|)^𝑝≤ 1 + 𝒞(𝓂, 𝛼, 𝑝, 𝑞)𝛼^−𝑘𝑝ℎ^{2𝓂−12 +1+𝑝} Where 𝜒_𝑗(𝑥) = {1 , 𝑥 ≥ 𝑥_𝑖

0 𝑥 < 𝑥_𝑖, 𝛼 ∈ (0,1), 𝓂 ≥ 2, 1 < 𝑝 < ∞ Proof: By reference [11], then we get when ℵ_𝑗(𝑥) = 1, 𝑥 ≥ 𝑥_𝑖

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=`

)

𝑝

≤ (∑ (1 + 𝒞 2𝓂 − 1ℎ

𝑗,1 ℎ

𝛹𝑗,1 ℎ

2𝓂−1(𝑥))

𝑁

𝑗=`

)

𝑝

≤ (∑ (1 + 𝒞(𝓂)ℎ

𝑗,1 ℎ

(

𝛥1 ℎ

(𝑥_𝑗) 𝛥1

ℎ

(𝑥_𝑗) + |𝑐𝑜𝑠⁻¹𝑥 − 𝑐𝑜𝑠⁻¹𝑥_𝑗|)

2𝓂−1

)

𝑁

𝑗=`

)

𝑝

Since 𝛥1 ℎ

(𝑥_𝑗) = 𝑥_𝑗+1− 𝑥_𝑗 = ℎ_𝑗 = ^ℎ

𝛼^𝑘

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=`

)

𝑝

≤ ( ∑

(

1 + 𝒞(𝓂)ℎ𝛼^−𝑘( 1

1 +|𝑐𝑜𝑠⁻¹𝑥 − 𝑐𝑜𝑠⁻¹𝑥_𝑗| ℎ𝛼^−𝑘

)

2𝓂−1

)

𝑁

𝑗=` )

𝑝

Let 𝛾 = 𝑐𝑜𝑠⁻¹𝑥 and 𝛾_𝑗,1 ℎ

= 𝑐𝑜𝑠⁻¹𝑥_𝑗, and we have 𝛾 ∈ [𝛾_𝑖+1,1 ℎ

, 𝛾_𝑖,1 ℎ

], since

|𝛾𝑖,1 ℎ

− 𝛾𝑗,1 ℎ

| ≤ |𝛾

𝑖,1 ℎ

− 𝛾| + |𝛾 − 𝛾

𝑗,1 ℎ

|

≤ (1 +¹

ℎ|𝛾_𝑖,¹

ℎ

− 𝛾_𝑖+1,¹

ℎ

| (1 +¹

ℎ|𝛾 − 𝛾_𝑗,¹

ℎ

|), then 1 +1

ℎ |𝛾

𝑖,1 ℎ

− 𝛾𝑗,1 ℎ

| 1 +1

ℎ |𝛾 − 𝛾𝑗,1 ℎ

|

≤ 1 + 1 ℎ|𝛾

𝑖,1 ℎ

− 𝛾|

1 +1

ℎ |𝛾 − 𝛾𝑗,1 ℎ

|

≤ 1 +1 ℎ |𝛾_𝑖,1

ℎ

, 𝛾_𝑖+1,1 ℎ

|

Therefor

2326-9865

Vol. 71 No. 3s3 (2022) 10 http://philstat.org.ph

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=`

)

𝑝

≤ ( ∑

(

1 + 𝒞(𝓂)ℎ𝛼^−𝑘( 1 1 +𝛼^𝑘

ℎ |𝛾𝑖,1 ℎ

− 𝛾𝑗,1 ℎ

| )

2𝓂−1

)

𝑁

j=1 )

𝑝

By Holder inequality

(∑ ( ¹

1+^𝛼𝑘 ℎ|𝛾

𝑖,1 ℎ

−𝛾𝑗,1 ℎ

|

)

2𝓂−1

𝑁𝑗=1 )

𝑝

= (

∑ ¹

(1+^𝛼𝑘 ℎ|𝛾

𝑖,1 ℎ

−𝛾𝑗,1 ℎ

|) 2𝓂−1

2

𝑁𝑗=1 . ¹

(1+^𝛼𝑘 ℎ|𝛾

𝑖,1 ℎ

−𝛾𝑗,1 ℎ

|) 2𝓂−1

2

)

𝑝

≤ (

∑ 1

(1 +𝛼^𝑘 ℎ |𝛾𝑖,1

ℎ

− 𝛾𝑗,1 ℎ

|)

2𝓂−1 2 𝑝 𝑁

𝑗=1

) .

((

∑ 1

(1 +𝛼^𝑘 ℎ |𝛾𝑖,1

ℎ

− 𝛾𝑗,1 ℎ

|)

2𝓂−1 2 𝑞 𝑁

𝑗=1

))

𝑝 𝑞

∀𝑗 ∈ 𝕀_𝑘 = {𝑗: [𝑥_𝑗, 𝑥_j+1] ⊂ 𝐴_𝑘}, hence

(

∑ 1

(1 +𝛼^𝑘 ℎ |𝛾𝑖,1

ℎ

− 𝛾𝑗,1 ℎ

|)

2𝓂−1 2 𝑞 𝑁

𝑗=1

)

= ∑ ∑ 1

(1 +𝛼^𝑘 ℎ |𝛾𝑖,1

ℎ

− 𝛾𝑗,1 ℎ

|)

2𝓂−1 2 𝑞 𝑗∈𝕀_𝑘

𝐿𝑜𝑔_𝛼1 ℎ

𝑘=0

≤ 𝒞(𝓂, 𝑞, 𝛼), 𝛼 ∈ (0,1), 𝓂 ≥ 2, 𝑞 > 1 By using the property (3) for 𝑘

1 − (𝑡_𝑘+ ¹

ℎ𝛼^𝑘−1)² < ¹

𝛼^2𝑘

1 − 𝑡_𝑘² < 𝛼^−2𝑘(1 +^{2𝛼𝛼𝑘𝑡𝑘}

ℎ +^𝛼2

ℎ²) 1 − 𝑡_𝑘² < 𝛼^−2𝑘+2 (5)

We have ∀𝑗 ∈ 𝕀_𝑘, 𝛾_𝑘 = 𝑐𝑜𝑠⁻¹𝑥_𝑘 by inequality (5) then 𝑠𝑖𝑛𝛾𝑗,1

ℎ

< 𝑠𝑖𝑛𝛾

𝑘,1 ℎ

= √1 − 𝑡_𝑘² ≤ √𝛼²𝛼^−2𝑘= 𝛼𝛼^−𝑘 𝑠𝑖𝑛 𝛾_𝑗,1

ℎ

≤ 𝒞₂(𝛼)𝛼^−𝑘, 𝒞₂ > 0 (6) And from property (2) for 𝑘 ≥ 0

2326-9865

Vol. 71 No. 3s3 (2022) 11 http://philstat.org.ph 𝛼^−𝑘 < √1 − 𝑡_𝑘² = 𝑠𝑖𝑛𝛾_𝑗,¹

ℎ

, ∃ 𝒞₁ > 0 ∋ 𝒞₁ 𝛼^−𝑘 < 𝑠𝑖𝑛 𝛾_𝑗,¹

ℎ

(7) From the inequality (6) and (7) we get

𝒞₁ 𝛼^−𝑘 < 𝑠𝑖𝑛 𝛾_𝑗,¹

ℎ

≤ 𝒞₂(𝛼)𝛼^−𝑘, ∀𝒞₁, 𝒞₂ > 0 (8) Now from the inequality (4) we have ℎ ≤ 𝑠𝑖𝑛𝛾_𝑖,¹

ℎ

, and ℎ ≤ 𝑠𝑖𝑛𝛾_𝑖,1

ℎ

≤ 𝑠𝑖𝑛𝛾_𝑖,1 ℎ

+ |𝛾_𝑖,1 ℎ

− 𝛾_𝑗,1 ℎ

| ℎ − 𝑠𝑖𝑛𝛾_𝑗,¹

ℎ

≤ |𝛾_i,¹

ℎ

− 𝛾_𝑗,¹

ℎ

| 1 +^𝛼𝑘

ℎ (ℎ − 𝑠𝑖𝑛𝛾_𝑗,1 ℎ

) ≤ 1 +^𝛼𝑘

ℎ |𝛾_𝑖,1 ℎ

− 𝛾_𝑗,1 ℎ

| (9) From the inequality (8) we have 𝑠𝑖𝑛𝛾_𝑗,¹

ℎ

≤ 𝒞𝛼^−𝑘 𝛼^𝑘

ℎ 𝑠𝑖𝑛𝛾

𝑗,1 ℎ

≤ 𝒞 𝛼^𝑘 ℎ 𝛼^−𝑘 1 + 𝛼^𝑘−^𝛼𝑘

ℎ 𝑠𝑖𝑛𝛾_𝑗,¹

ℎ

> 1 + 𝛼^𝑘−^𝒞

ℎ (10)

By substantiating (10) in (9) we get 1 + 𝛼^𝑘−^𝒞

ℎ≤ 1 +^𝛼𝑘

ℎ |𝛾_𝑖,1 ℎ

− 𝛾_𝑗,1 ℎ

| , ∀ 𝑖, 𝑗

∑ ∑ 1

(1 +𝛼^𝑘 ℎ |𝛾𝑖,1

ℎ

− 𝛾𝑗,1 ℎ

|)

2𝓂−1

2 𝑝 < ∑ 1

(1 + 𝛼^𝑘−𝒞 ℎ)

2𝓂−1 2 𝑝

∞

𝑘=0 𝑗∈𝕀_𝑘

∞

𝑘=0

Hence

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=1

)

𝑝

≤ [1 + 𝒞(𝓂)ℎ^𝑝𝛼^−𝑘𝑝(𝒞(𝓂, 𝛼, 𝑞) ∑ (1 + 𝛼^𝑘−𝒞 ℎ)

−(2𝓂−1)

2 𝑝

∞

𝑘=0

)]

∑ (1 + 𝛼^𝑘−𝒞 ℎ)

−(2𝓂−1)

2 𝑝

∞

𝑘=0

= ∑ (ℎ + ℎ𝛼^𝑘− 𝒞

ℎ )

−(2𝓂−1)

2 𝑝

∞

𝑘=0

2326-9865

Vol. 71 No. 3s3 (2022) 12 http://philstat.org.ph

= (1 ℎ)

−(2𝓂−1)

2 𝑝

∑(ℎ + ℎ𝛼^𝑘− 𝒞)^{−(2𝓂−1)2 𝑝}

∞

𝑘=0

= (ℎ)^{(2𝓂−1)2 𝑝} (𝒞(𝓂, 𝑝, 𝛼)ℎ)

= 𝒞(𝓂, 𝑝, 𝛼)ℎ^{2𝓂−12 𝑝+1}, 𝛼 ∈ (0,1), 𝓂 ≥ 2, 1 ≤ 𝑝 < ∞

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=1

)

𝑝

≤ 1 + 𝒞(𝓂)ℎ^𝑝𝛼^−𝑘𝑝(𝒞(𝓂, 𝑝, 𝑞, 𝛼)ℎ^{2𝓂−12 𝑝+1})

(∑ 𝑠𝑢𝑝_{𝑥𝑖≤𝑥≤𝑥𝑖+1}|ℵ_𝑗(𝑥) − 𝛲_𝑗(𝑥)|

𝑁

𝑗=1

)

𝑝

≤ 1 + 𝒞(𝓂, 𝑝, 𝑞, 𝛼)ℎ^{2𝓂−12 𝑝+1+𝑝}𝛼^−𝑘𝑝

Lemma 3.9. ∀ 𝑗 ∈ 𝕀_𝑘, there exist a polynomial 𝒫_𝑗,1 ℎ

of degree ≤ 𝓂, 0 < 𝓂 ≤¹

ℎ such that

∑ |𝒫_𝑗,1 ℎ

|

𝑝

≤ 𝒞𝛼^{2𝑎1𝑘−(𝑣𝑝−1)𝑘}ℎ^𝑣𝑝−1,

𝑗∈𝕀_𝑘 , where

𝒫𝑗,1 ℎ

= 1

ℎ_𝑗−1 ∫ 𝐹(𝑡)𝑑𝑡, ℎ_𝑗−1 = 𝑥_𝑗− 𝑥_𝑗−1, 𝑗 > 0, 𝒞(𝑝) = 𝒞 > 0

𝑥_𝑗

𝑥_𝑗−1

Proof: ∀ 𝑗 ∈ 𝕀_𝑘 we have ∑ |𝒫_𝑗

𝑘,¹ ℎ

|

𝑝

= ∑ |¹

ℎ∫_𝑥^𝑥𝑗 𝐹(𝑡)𝑑𝑡

𝑗−1 |^𝑝

𝑗_𝑘∈𝕀_𝑘

𝑗_𝑘∈𝕀_𝑘 , since

[𝑥_𝑗−1, 𝑥_𝑗] ⊂ [𝑡_𝑘, 𝑡_𝑘+1]⋃[𝑡_−(𝑘+1), 𝑡_−𝑘]

∑ |𝒫𝑗_𝑘,1 ℎ

|

𝑝

≤ 𝒞(𝑝)𝛼^𝑘

ℎ || ∫ 𝐹(𝑡) + ∫ 𝐹(𝑡) + ∫ 𝐹(𝑡)

t_𝑘

𝑡_𝑘−ℎ 𝛼^𝑘 𝑡_𝑘−ℎ

𝛼^𝑘

𝑡_𝑘− ℎ 𝛼^𝑘−1 𝑡_𝑘+ℎ

𝛼^𝑘

𝑡_𝑘

||

𝑝

𝑑𝑡

𝑗_𝑘∈𝕀_𝑘

≤ 𝒞(𝑝)𝛼^𝑘 ℎ

(

∫ |𝐹(𝑡)| + ∫ |𝐹(𝑡)|

𝑡_𝑘

𝑡_𝑘− ℎ 𝛼^𝑘−1 𝑡_𝑘+ℎ

𝛼^𝑘

𝑡_𝑘

)

𝑝

𝑑𝑡

≤ 𝒞(𝑝) (𝛼^𝑘 ℎ )

𝑝

∫ |𝐹 (𝑡 + ℎ

𝛼^𝑘) − 𝐹(𝑡)|

𝑡_𝑘 𝑝

𝑡_𝑘− ℎ 𝛼^𝑘−1

𝑑𝑡

2326-9865

Vol. 71 No. 3s3 (2022) 13 http://philstat.org.ph

≤ 𝒞(𝑝) (𝛼^𝑘 ℎ)

𝑝

𝑠𝑖𝑛^−2𝑎1𝛾

𝑗,1 ℎ

∫ |𝐹 (𝑡 + ℎ

𝛼^𝑘) − 𝐹(𝑡)|

𝑝

𝜔(𝑡)

𝑡_𝑘

𝑡_𝑘− ℎ 𝛼^𝑘−1

𝑑𝑡

𝑠𝑖𝑛^−2𝑎1𝛾_𝑗,¹

ℎ

< 𝛼^2𝑎1𝑘, by using inequality (8) [𝑡_𝑘− ℎ

𝛼^𝑘−1, 𝑡_𝑘] ⊂ [𝑡_𝑘− ℎ

α^𝑘−1, 𝑡_𝑘+1− ℎ 𝛼^𝑘]

∑ |𝒫𝑗_𝑘,1 ℎ

|

𝑝

≤ 𝒞(𝑝)𝛼^𝑘

ℎ 𝑠𝑖𝑛^−2𝑎1𝛾

𝑗,1 ℎ

∫ |𝐹 (𝑡 + ℎ

𝛼^𝑘) − 𝐹(𝑡)|

𝑝

𝜔(𝑡)

𝑡_𝑘

𝑡_𝑘− ℎ 𝛼^𝑘−1

𝑑𝑡

𝑗_𝑘∈𝕀_𝑘

≤ 𝒞(𝑝)𝛼^𝑘

ℎ 𝛼^2𝑎1𝑘 ∫ |𝐹 (𝑡 + ℎ

𝛼^𝑘) − 𝐹(𝑡)|

𝑝

𝜔(𝑡)

𝔟−ℎ 𝛼^𝑘

−𝔟

𝑑𝑡

≤ 𝒞(𝑝)^𝛼𝑘

ℎ 𝛼^2𝑎1𝑘(^ℎ

𝛼^𝑘)^𝑣𝑝 = 𝒞(𝑝)𝛼^{𝑘−𝑘𝑣𝑝+2𝑘𝑎1}ℎ^𝑣𝑝−1

∑ |𝒫_𝑗,1 ℎ

|

𝑝

≤ 𝒞𝛼^{2𝑎1𝑘−(v𝑝−1)𝑘}ℎ^𝑣𝑝−1

𝑗∈𝕀_𝑘

Theorem 3.10. ∀ 𝐹 ∈ 𝑊_𝜔,𝑝^(𝑣)(𝛪), 𝑘 = 0, 1 < 𝑝 < ∞, there exist a polynomial 𝒫1 ℎ

(𝑥), of degree ≤ 𝓂, 0 < 𝓂 ≤¹

ℎ (¹

ℎ∈ 𝑅⁺)(𝓂 which is depended on 𝑎₁, 𝑎₂, 𝑣, 𝑝) such that

‖ 𝐹1

ℎ

(𝐹) − 𝒫_𝓂 𝛥₁

ℎ

𝑣 ‖

𝐿_𝜔,𝑝(𝛪)

≤ 𝒞ℎ

2

𝑝 (𝐿𝑛1 ℎ)

1

𝑝, 𝒞 = 𝒞(𝛼, 𝛼₁, 𝑎₂, 𝑣, 𝑝, 𝔟)

Such that 𝛼 ∈ (0,1), 𝛼₂ ∈ 𝑅⁺, 𝑣 ∈ (0, 𝑏], 1 ≤ 𝑝 < ∞, 𝑏 =positive integer.

Proof: ∀ 𝑥 ∈ [−𝔟, 𝔟], 𝑛 ∈ 𝑁, by lemma (3.9) there is a polynomial 𝒫¹

ℎ

(𝑥), of degree ≤ 𝓂, 0 <

𝓂 ≤ ¹

ℎ (¹

ℎ∈ 𝑅⁺), have the form

𝒫_𝓂(𝑥) = 𝒫

0,1 ℎ

(𝑥) + ∑ 𝒫_𝑥𝑗(𝑥)𝒫

𝑗,1 ℎ

(𝑥)

𝑁

𝑗=1

𝒫_𝑗,¹

ℎ

(𝑥), ia a polynomial of degree ≤ 𝓂, 0 < 𝓂 ≤¹_ℎ, such that

𝒫𝑗,1 ℎ

(𝑥) = 1

ℎ_𝑗 ∫ 𝐹(𝑡)𝑑𝑡, ℎ_𝑗 = 𝑥_𝑗+1− 𝑥_𝑗

𝑥_𝑗+1

𝑥_𝑗−1

, 𝑗 = 1, … , 𝑁