Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

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Generalized Modulus of Smoothness and 𝑲 − 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒂𝒍. 
 

Bushra Kadhum Awaad 1*, Eman Samer Bhaya 2 

1Department of Mathematics, College of Education for Pure Sciences, University of Babylon, Babyel, Iraq. 

2Department of Mathematics, College of Education for Pure Sciences, University of Babylon, Babyel, Iraq. 

2Department of Mathematics, College of Education, Al-  Zahraa  University for Women, Karbala, Iraq. 

bushra.k@uokerbala.edu..iq 

emanbhaya@itnet.uobabylon.edu.iq 

emanbhaya@alzahraa.edu.iq 

 

Article History: 

Received: 09-04-2024 

Revised: 22-05-2024 

Accepted: 09-06-2024 

Abstract 

Many articles introduced about direct and inverse theorems interims of ordinary modulus 

of smoothness andK-functional. Here we shall define a generalized modulus of smoothness 

andK-functional, then we prove they are equivalent. 

Keywords: K—functional, Modulus etc. 

 

1. Introduction 

Many articles such us[1], [3], [6], [8],[9],[10] introduced approximation theorems interims of the 

ordinary modulus of smoothness, with ordinary symmetric difference. Here we shall define new 

symmetric difference, then we use it to obtain anew modulus of smoothness and  𝐾 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙, in 

anew quasi normed spaces, call it 𝐿𝑝,𝛽 . let us define 𝐿𝑝 ,0 < 𝑝 < 1 , an [−1,1] as the spaces of all 

measurable functions satisfies : 

‖𝑓‖𝑝   =    (∫ |𝑓|𝑝1

−1
)

1
𝑝

 < ∞ .        [9]                                                                                     

Define  𝐿𝑝,𝛽 , measurable function spaces of functions  𝑓  with domain     [−1,1] . 

                      √(1 − 𝑥)𝛽  𝑓 ∈ 𝐿𝑃 , 

It mean                                                                                         

                      ‖𝑓‖𝑝,𝛽 =   (∫ (|𝑓(𝑥)(1 − 𝑥)
𝛽
2 )|

𝑝

𝑑𝑥
1

−1
)

1
𝑝

 , 

=  ‖𝑓   (1 − 𝑥)
𝛽
2 ‖

𝑝
. 

Let   𝐸𝑛(𝑓)𝑝,𝛽 , best approximation error of  𝑓  in 𝐿𝑝,𝛽 ,using algebraic < 𝑛 in polynomials of degree  

 𝐿𝑝,𝛽 , and 

𝐸𝑛(𝑓)𝑝,𝛽    =     inf
𝑝𝑛∈𝐼𝑃𝑛

   ‖𝑓 − 𝑝𝑛‖ , [8] 

Where,  

ǁℙ𝑛 is     𝑛 − 1, degree algebraic polynomials spaces . 



Communications on Applied Nonlinear Analysis 

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Let us define the operator  

𝐷𝑥,𝜈,𝜇  =   (√1 − 𝑥)
−𝜈

(1 + 𝑥)−𝜇 𝑑

𝑑𝑥
(√(1 − 𝑥))

𝜈+1

(√(1 + 𝑥))
𝜇+1 𝑑

𝑑𝑥
 .   [4] 

Peetre     𝐾 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙 on 𝐿𝑝,𝛼 and 𝐼𝑃𝑛is defined   by  

𝐾(𝑓, 𝛿)𝑝,𝛼 = inf
𝑔∈𝐼𝑃𝑛

(‖𝑓  −   𝑔  ‖𝑝,𝛼 + 𝛿2‖  𝐷𝑥,2,2𝑔(𝑥)  ‖
𝑝,𝛼

), [4]. 

Let us introduce usual smoothness modulus. 

ῶ1  (𝑓, 𝛿)𝑝,𝛽   =   sup
|𝑡|≤𝛿

‖𝜏𝑡 (𝑓)   −   𝑓‖𝑝,𝛼  .      (1) 

Where, 

𝜏𝑡 (𝑓)   =     
1

𝜋(1−𝑥2) cos4𝑡

2

 ∫ (2(√1 − 𝑥2   cos 𝑡 + 𝑥 sin 𝑡 cos 𝜑 + √1 − 𝑥2  (1 − cos 𝑡) sin2 𝜑)
2

−
𝜋

0

1 + (𝑥 cos 𝑡 − √1 − 𝑥2   sin 𝑡 cos 𝜑)
2

)
⬚

𝑓(𝑥 cos 𝑡 − √1 − 𝑥2  sin 𝑡 cos 𝜑) 𝑑𝜑 .   

Let  𝑓  ∈ 𝐿𝑝,𝛼, .    put    𝑦 = cos 𝑡, 𝑧 = cos 𝜑 , in the  𝜏𝑡 (𝑓),  let 𝜏𝑦 (𝑓). Then 

    𝜏𝑡 (𝑓)    =     
4

𝜋(1−𝑥2)(1+𝑦)2   ∫ 𝐵𝑦(𝑥, 𝑧, 𝑅)
1

−1
𝑓(𝑅)   

𝑑𝑧

√1−𝑧2
 ,         [1]. 

where , 

𝑅  =   𝑥𝑦 − 𝑧√1 − 𝑥2 √1 − 𝑦2, 

𝐵𝑦(𝑥, 𝑧, 𝑅)   =  2 (√1 − 𝑥2 𝑦 + 𝑧𝑥√1 − 𝑦2 + √1 − 𝑥2 (1 − 𝑦)(1 − 𝑧2))
2

− (1 − 𝑅2).  

2. Auxiliary Results 

To prove our main theorem we need some results, that we make use of them in our proof. Let us 

begin with                                                   Lemma .2.1.                                                                                                         

For   𝑧 = cos 𝜑  and  𝑅 =   𝑥 cos 𝑡 − 𝑧√1 − 𝑥2 sin 𝑡, 𝑓 ∈ 𝐿𝑝,𝛽[−1,1], we get 

 (2.1)                   ‖  
1

1 − 𝑥2
∫ (1 − 𝑅2)|𝑓(𝑅)|

𝑑𝑧

√1 − 𝑧2

1

−1

‖
𝑝,𝛽

≤   ∁ |𝑓| 

Proof                                                                                                                    

Using simple calculating of integrals we obtain the result in (2.1).  

Lemma.2.2. [1] 

|𝐵𝑦(𝑥, 𝑧, 𝑅)|   ≤   19(1 − 𝑅2) 

Where,  

𝑅   =  𝑥 cos 𝑡 − 𝑧√1 − 𝑥2 sin 𝑡, 𝑧 = cos 𝜑 and    𝑥  = cos 𝜃1.                                                          

Lemma. 2.3.   



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If    𝑓 ∈  𝐿𝑝,𝛽[−1,1], we get  

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽     ≤  
19∁(𝑝)

𝜋 cos4 𝑡
2

  ‖𝑓‖𝑝,𝛽 , 

Where  𝑐 > 0. 

Proof.  

      ‖    𝜏𝑡 (𝑓)‖𝑝,𝛽 =    ‖
1

𝜋 cos4𝑡

2

 
1

1−𝑥2 ∫ 𝐵cos 𝑡(𝑥, 𝑧, 𝑅)
1

−1
𝑓(𝑅)

𝑑𝑧

√1−𝑧2
‖

𝑝,𝛽

 

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽   =   
1

𝜋 cos4 𝑡
2

‖
1

1 − 𝑥2
∫ 𝐵cos 𝑡(𝑥, 𝑧, 𝑅)

1

−1

𝑓(𝑅)
𝑑𝑧

√1 − 𝑧2
‖

𝑝,𝛽

 

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽   =   
1

𝜋 cos4𝑡

2

(∫ |
1

1−𝑥2 ∫ ∫ 𝐵cos 𝑡(𝑥, 𝑧, 𝑅)
1

−1
𝑓(𝑅)

𝑑𝑧

√1−𝑧2

1

−1
|

1

−1

𝑝
(1 − 𝑥)

𝛽

2 𝑑𝑥)

1

𝑝

. 

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽     ≤
1

𝜋 cos4𝑡

2

   (∫ (
1

1−𝑥2 ∫ |𝐵cos 𝑡(𝑥, 𝑧, 𝑅)||𝑓(𝑅)|
1

−1

𝑑𝑧

√1−𝑧2
)

𝑝
(1 − 𝑥)

𝛽

2  
1

−1
𝑑𝑥)

1

𝑝

. 

Using Lemma (2.2), we get  

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽   ≤
1

𝜋 cos4𝑡

2

(∫ (
1

1−𝑥2 ∫ 19(1 − 𝑅2)
1

−1
|𝑓(𝑅)|

𝑑𝑧

√1−𝑧2
)

𝑝1

−1
(1 − 𝑥)

𝛽

2 𝑑𝑥)

1

𝑝

. 

Using Lemma (2.1) ,to obtain  

‖    𝜏𝑡 (𝑓)‖𝑝,𝛽  ≤  
19∁(𝑝)

𝜋 cos4 𝑡
2

‖𝑓‖𝑝,𝛽 . 

Lemma.2.4. [2].                                                                                                             

If 𝑦 ∈ (−1,1),  and 𝑥 ∈ (−1,1) ,    for almost every then  

𝜏𝑦 (𝐷𝑥,2,2𝑓) =  𝐷𝑥,2,2𝜏𝑦(𝑓). 

Lemma.2.5.[3].                                                                                                    

let   𝑓 ˊ be an  absolutely continuous with domain  [𝑎, 𝑏] ⊂ (−1,1) ,with 𝐷𝑥,2,2𝑓 ∈ 𝐿1,2. , So  almost 

every 𝑥 ∈ (−1,1) and every 𝑦 ∈ (−1,1) 

𝜏𝑦 (𝑓) − 𝑓 = ∫ (1 − 𝜈)−1(1 − 𝜈)−5 ∫ (1 + 𝑢)4𝜈

1

𝑦

1
𝜏𝑢(𝐷𝑥,2,2𝑓, 𝑥)𝑑𝑢𝑑𝜈 . 

𝜏𝑦 (𝑓) − 𝜏0 (𝑓) = − ∫ (1 − 𝜈)−1(1 − 𝜈)−5 ∫ (1 + 𝑢)4−1

𝜈

𝑦

1
𝜏𝑢(𝐷𝑥,2,2𝑓, 𝑥)𝑑𝑢𝑑𝜈 . 

Lemma.2.6.[3].                                                                                                   

let   𝑓 ˊ be  absolutely continuous with domain  [𝑎, 𝑏] ⊂ (−1,1) ,with 𝐷𝑥,2,2𝑓(𝑥) ∈ 𝐿1,2. So   almost 

every  𝑥 ∈ (−1,1) and every   𝑡 ∈ (−𝜋, 𝜋). 



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𝜏𝑡
̭
 (𝑓) − 𝑓 = ∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑓, 𝑥)  sin
𝑢

2

𝜈

0

𝑡

0
 (cos

𝑢

2
)

9

𝑑𝑢𝑑𝜈 . 

𝜏𝑡 (𝑓)  −  𝜏𝜋

2
 (𝑓)   = − ∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑓, 𝑥)
𝜋

𝜈

𝑡
𝜋

2

 sin
𝑢

2
(cos

𝑢

2
)

9

 𝑑𝑢𝑑𝜈 . 

Lemma.2.7.                                                                                                              

If  𝑓 ∈ 𝐿𝑝,𝛽 , then 

                             𝐸𝑛(𝑓)𝑝,𝛽  ≤ ∁(𝑝)
1

𝑛2
 ‖𝐷(𝑓)‖𝑝,𝛽 . 

Proof.                                                                                                                    

By using   𝐸𝑛(𝑓)𝑝,𝛽   ≤  ‖𝑓 − 𝑄‖𝑝,𝛽 ,                                                                                 

where 𝑄 is algebraic polynomial of degree less than   𝑛 − 1,                                                                                                                                   

and                                                                                                                               

𝑄(𝑥) =
1

𝛾𝑚
∫ 𝑇2;cos 𝑡(𝑓, 𝑥)

𝜋

0
(

sin
𝑚𝑡

2

sin
𝑡

2

)

2𝑞+4

sin5 𝑡𝑑𝑡                                                   

where 

                            𝛾𝑚 = ∫ (
sin

𝑚𝑡

2

sin
𝑡

2

)

2𝑞+4

sin5 𝑡𝑑𝑡
𝜋

0
 , 

Then  

𝐸𝑛(𝑓)𝑝,𝛽  ≤ ‖𝑓 −
1

𝛾𝑚
∫ 𝑇2;cos 𝑡(𝑓, 𝑥)

𝜋

0

(
sin

𝑚𝑡
2

sin
𝑡
2

)

2𝑞+4

sin5 𝑡𝑑𝑡‖

𝑝,𝛽

. 

= ‖
1

𝛾𝑚
∫ (

sin
𝑚𝑡

2

sin
𝑡

2

)

2𝑞+4

sin5 𝑡𝑑𝑡
𝜋

0
 𝑓(𝑥) −  

1

𝛾𝑚
∫ 𝑇(𝑓) (

sin
𝑚𝑡

2

sin
𝑡

2

)

2𝑞+4

sin5 𝑡𝑑𝑡
𝜋

0
‖

𝑝,𝛽

  

𝐸𝑛(𝑓)𝑝,𝛽 ≤ ∁(𝑝)‖𝑓 − 𝑇‖𝑝,𝛽. 

Using the 

 ‖𝜏𝑡 (𝑔)  −  𝑔 ‖𝑝,𝛽  ≤  ∁(𝑝)
1

cos4𝑡

2

 𝑡2 ‖𝐷𝑥,2,2𝑔 ‖
𝑝,𝛽

,[2]. 

To obtain  

𝐸𝑛(𝑓)𝑝,𝛽  ≤  ∁(𝑝) 
1

𝑛2
  ‖𝐷(𝑓)‖𝑝,𝛽 . 

3. The Main Result 

After we define our new modulus of smoothness, and 𝑘 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑎𝑙. we shall relate then to others. 

That is we shall introduce the theorem:                                                                                        

 



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Theorem.3.1.                                                                                                                  

For 𝑓 ∈ 𝐿𝑝,𝛽  ,0 < 𝑝 < 1 , we have                                            

  ∁1(𝑝) 𝐾(𝑓, 𝛿)𝑝,𝛽   ≤ ῶ(𝑓, 𝛿)𝑝,𝛽  ≤ ∁2(𝑝)
1

cos4𝛿

2

  𝐾(𝑓, 𝛿)𝑝,𝛽  .                           

Where, ∁1(𝑝)  and  ∁2(𝑝)  are positive constants depending on p only.    

Proof .                                                                                                             

Let ℙ𝑛be the   set of 𝑛 th degree algebraic  polynomials we shall prove  (3.1) 

  ‖𝜏𝑡 (𝑔) − 𝑔‖𝑝,𝛽  ≤  ∁(𝑝)
1

cos4 𝑡
2

 𝑡2 ‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

, 

Where,  ∁(𝑝)  is constant depend on p only. 

If  0 < 𝑡 ≤
𝜋

2
 , then lemma.2.6.implies   

𝐼1  =  ‖𝜏𝑡 (𝑔) − 𝑔‖𝑝,𝛽. 

= ‖∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑔, 𝑥)  sin
𝑢

2

𝜈

0

𝑡

0
 (cos

𝑢

2
)

9

𝑑𝑢𝑑𝜈‖
𝑝,𝛽

. 

Applying lemma .2.3  [‖    𝜏𝑡 (𝑓)‖𝑝,𝛽   ≤  
19∁(𝑝)

𝜋 cos4𝑡

2

‖𝑓‖𝑝,𝛽] ,          we get  

𝐼1 ≤  ∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑔, )  sin
𝑢

2

𝜈

0

𝑡

0

 (cos
𝑢

2
)

9

𝑑𝑢𝑑𝜈. 

𝐼1 ≤ ∁(𝑝)‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫   sin
𝑢

2

𝜈

0

𝑡

0

 (cos
𝑢

2
)

5

𝑑𝑢𝑑𝜈. 

The inequality  

∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫   sin
𝑢

2

𝜈

0

𝑡

0

 (cos
𝑢

2
)

5

𝑑𝑢𝑑𝜈 ≤ ∁(𝑝)𝑡2. 

For   0 < 𝑡 ≤  
𝜋

2
, we obtain 

𝐼1  ≤ ∁(𝑝)𝑡2‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

 ≤  ∁(𝑝) 
1

cos4𝑡

2

 𝑡2 ‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

.  

If    
𝜋

2
 ≤ 𝑡 < 𝜋,    so by Lemma (2.6).                                       

[𝜏𝑡 (𝑓) − 𝜏𝜋

2
 (𝑓) = − ∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑓)
𝜋

𝜈

𝑡
𝜋

2

 sin
𝑢

2
(cos

𝑢

2
)

9

 𝑑𝑢𝑑𝜈.
 

]We get 

𝐼2 = ‖𝜏𝑡 (𝑓) − 𝜏𝜋

2
 (𝑓)‖

𝑝,𝛽
=  ‖∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑔, 𝑥)
𝜋

𝜈

𝑡
𝜋

2

 sin
𝑢

2
(cos

𝑢

2
)

9

 𝑑𝑢𝑑𝜈‖
𝑝,𝛽

.  



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Applying  Lemma. 2.3 [‖    𝜏𝑡 (𝑓)‖𝑝,𝛽  ≤  
19∁(𝑝)

𝜋 cos4𝑡

2

‖𝑓‖𝑝,𝛽] , we get  

𝐼2  ≤   ∁(𝑝)‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫  sin
𝑢

2

𝜋

𝜈

𝑡
𝜋

2

(cos
𝑢

2
)

5

 𝑑𝑢𝑑𝜈.  

Assume    
𝜋

2
 ≤ 𝑡 < 𝜋, so   

∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫  sin
𝑢

2

𝜋

𝜈

𝑡

𝜋
2

(cos
𝑢

2
)

5

 𝑑𝑢𝑑𝜈 ≤ ∁(𝑝)
1

cos4 𝑡
2

 , 

It the follows that  

(3.2)            𝐼2   ≤  ∁(𝑝)
1

cos4𝑡

2

 ‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

≤  ∁(𝑝)
1

cos4𝑡

2

𝑡2‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

. 

Since   

‖𝜏𝑡 (𝑔) − 𝑔‖𝑝,𝛽   ≤  ‖𝜏𝑡 (𝑔) − 𝜏𝜋
2

 
(𝑔)‖

𝑝,𝛽

+    ‖𝜏𝜋
2

 
(𝑔) − 𝑔‖

𝑝,𝛽

 

Using (3.2)and (3.1)we obtain,  for 

0< 𝑡 ≤
𝜋

2
 , that 

‖𝜏𝑡 (𝑔) − 𝑔‖𝑝,𝛽   ≤  ∁(𝑝) 
1

cos4 𝑡
2

𝑡2 ‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

. 

For   
𝜋

2
< 𝑡 ≤ 𝜋, 

we proved inequality (3.1) for   0 < 𝑡 ≤ 𝜋.   

𝜏cos 𝑡(𝑔)  =  𝜏cos −𝑡(𝑔, ), 

Let  𝑓 ∈ 𝐿𝑝,𝛽 and 0 ≤ |𝑡| ≤ 𝛿 < 𝜋. so for any 𝑔 ∈ 𝐿𝑝,𝛽 . 

Using Lemma 2.3 we get 

‖𝜏𝑡 (𝑓) − 𝑓‖𝑝,𝛽 ≤ ∁(𝑝)(‖𝜏𝑡 (𝑓 − 𝑔, 𝑥)‖𝑝,𝛽 + ‖𝜏𝑡 (𝑔) − 𝑔‖𝑝,𝛽 + ‖𝑔 − 𝑓‖𝑝,𝛽). 

‖𝜏𝑡 (𝑓) − 𝑓‖𝑝,𝛽 ≤ ∁(𝑝)
1

cos4𝑡

2

‖𝑓 − 𝑔‖𝑝,𝛽 + ‖𝜏𝑡(𝑔) − 𝑔‖𝑝,𝛽. 

By using(3.1), to get  

‖𝜏𝑡 (𝑓) − 𝑓‖𝑝,𝛽 ≤ ∁(𝑝)
1

cos4 𝑡
2

‖𝑓 − 𝑔‖𝑝,𝛽 +  𝑡2‖𝐷𝑥,2,2𝑔‖
𝑝,𝛽

. 

 To prove the second inequality, assume function  

𝑔𝛿   =
1

𝒦(𝛿)
∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝑓, 𝑥) sin
𝑢

2

𝜈

0

𝛿

0
(cos

𝑢

2
)

9

 𝑑𝑢𝑑𝜈, 



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Where 

𝒦(𝛿) = ∫ (sin
𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝑓) sin
𝑢

2

𝜈

0

𝛿

0

(cos
𝑢

2
)

9

 𝑑𝑢𝑑𝜈. 

Let   0 < 𝛿 <
𝜋

2
, then 

(3.3)                            ∁(𝑝)𝛿2 ≤ 𝒦(𝛿) ≤ ∁(𝑝)𝛿2. 

Applying Lemma.2.3, to get  ‖𝑔𝛿‖𝑝,𝛽  ≤

1

𝒦(𝛿)
∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ ‖𝜏𝑢(𝑓)‖𝑝,𝛽 sin
𝑢

2

𝜈

0

𝛿

0
(cos

𝑢

2
)

9

 𝑑𝑢𝑑𝜈 ≤ ∁(𝑝)
1

cos4𝛿

2

‖𝑓‖𝑝,𝛽 . 

That is 𝑔𝛿 ∈ 𝐿𝑝,𝛽 . 

Put      𝑔  = − ∫ (1 − 𝑦2)−3𝑥

0
∫ (1 − 𝑧2)21

𝑦
(𝑓(𝑧) −

𝑚1

𝑚2
) 𝑑𝑧𝑑𝑦, 

Where 

𝑚1 =  ∫ (1 − 𝑧2)2𝑓(𝑧)𝑑𝑧
1

−1
 , 𝑚2 = ∫ (1 − 𝑧2)21

−1
𝑑𝑧. 

                                    𝐷𝑥,2,2𝑔(𝑥) = 𝑓 −
𝑚1

𝑚2
 , 

𝐷𝑥,2,2𝑔(𝑥)  = (√1 − 𝑥)
−2

(1 + 𝑥)−2 𝑑𝑔

𝑑𝑥 
(√1 − 𝑥)

3
(√1 + 𝑥)

3 𝑑𝑔

𝑑𝑥
, 

We have  

𝑔𝛿   =
1

𝒦(𝛿)
∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫ 𝜏𝑢(𝐷𝑥,2,2𝑔, 𝑥) sin
𝑢

2

𝜈

0

𝛿

0

(cos
𝑢

2
)

9

 𝑑𝑢𝑑𝜈 +
𝑚1

𝑚2
. 

Using Lemma.2.6.we get                                                                                       

(3.4)                       𝑔𝛿   =   
1

𝒦(𝛿)
(𝜏𝛿(𝑔) − 𝑔) + 

𝑚1

𝑚2
. 

Applying the operator 𝐷𝑥,2,2𝑔  for (3.4) and Lemm2.4, 

(𝜏𝑦(𝐷𝑥,2,2𝑓, ) =  𝐷𝑥,2,2𝜏𝑦(𝑓). ), 

we get  

𝐷𝑥,2,2𝑔𝛿(𝑥), 𝑔𝛿(𝑥)   =
1

𝒦(𝛿)
(𝐷𝑥,2,2𝑔(𝜏𝛿)  −  𝐷𝑥,2,2𝑔(𝑥)) +

𝑚1

𝑚2
. 

𝐷𝑥,2,2𝑔𝛿(𝑥)   =
1

𝒦(𝛿)
(𝜏𝛿(𝐷𝑥,2,2𝑔, 𝑥) − 𝑓) −

𝑚1

𝑚2
+

𝑚1

𝑚2
.                                      

(3.5)                  𝐷𝑥,2,2𝑔𝛿(𝑥)  =
1

𝒦(𝛿)
  (𝜏𝛿(𝑓) − 𝑓). 

 Therefore, by Lemmas. 2.3. and  2.4  ,also 𝑔𝛿 ∈ 𝐿𝑝,𝛽. 

Then using (3.5) equality and inequality  (3.3) , we get                             



Communications on Applied Nonlinear Analysis 

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125 
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(3.6)                ‖𝐷𝑥,2,2𝑔𝛿 (𝑥)‖
𝑝,𝛽

 ≤   ∁(𝑝)
1

𝛿2
‖𝜏𝛿(𝑓) − 𝑓‖𝑝,𝛽 . 

Then using(3.6), to get                                                                                  

(3.7)                     ‖𝐷𝑥,2,2𝑔𝛿(𝑥)‖
𝑝,𝛽

 ≤  ∁(𝑝)
1

𝛿2
ῶ(𝑓, 𝛿)𝑝,𝛽. 

And                                                                                                                    

(3.8)                        ‖𝑓 − 𝑔𝛿‖𝑝,𝛽

≤
1

𝒦(𝛿)
∫ (sin

𝜈

2
)

−1

(cos
𝜈

2
)

−9

∫‖𝑓 − 𝜏𝑢(𝑓)‖𝑝,𝛽 sin
𝑢

2

𝜈

0

𝛿

0

(cos
𝑢

2
)

9

 𝑑𝑢𝑑𝜈 ≤ ῶ(𝑓, 𝛿)𝑝,𝛽 . 

For 0 < 𝛿 ≤
𝜋

2
, we have proved that  

𝐼(𝛿) = ‖𝑓 − 𝑔𝛿‖𝑝,𝛽 + 𝛿2‖𝐷𝑥,2,2𝑔𝛿(𝑥)‖
𝑝,𝛽

≤ ∁(𝑝)ῶ(𝑓, 𝛿)𝑝,𝛽 . 

For  
𝜋

2
< 𝛿 ≤ 𝜋 , we get   𝛿2 < 𝜋2. 1 , and,  1 <

𝜋

2
,  

𝐾(𝑓, 𝛿)𝑝,𝛽 ≤ 𝜋2 (‖𝑓 − 𝑔1‖𝑝,𝛽 + 12‖𝐷𝑥,2,2𝑔1(𝑥)‖
𝑝,𝛽

) 

Using (3.7) 𝑎𝑛𝑑(3.8),to obtain  

 

𝐾(𝑓, 𝛿)𝑝,𝛽 ≤ ∁(𝑝) ῶ(𝑓, 1)𝑝,𝛽 ≤ ῶ(𝑓, 1)𝑝,𝛽 . 

Conclusions 

K functional and modulus of smoothness are equivalent on weighted spaces. 

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smoothness, East J.Approx.4(1998),no.2,217-241.MR 1638345. 

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ISSN: 1074-133X 

Vol 31 No. 3s (2024) 

 

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