Edelweiss Applied Science and Technology 
ISSN: 2576-8484 
Vol. 8, No. 6, 4910-4919 
2024 
Publisher: Learning Gate 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

© 2024 by the authors; licensee Learning Gate 
* Correspondence:  nadiah.a@uokerbala.edu.iq 

 
 
 
 
 

Fractional integral Ostrowski inequality on L_P,0<P<∞ spaces 

 
Nadia Abed Habeeb1*, Eman Samir Bhaya2 
1College of Engineering, Kerbala University, Iraq; nadiah.a@uokerbala.edu.iq (N.A.H.). 
2College of Education, Al-Zahraa University for women, Iraq; eman.bhaya@alzahraa.edu.iq (E.S.B.). 

 

 
Abstract: Recently we proved a type of Ostrowski inequality in terms of the quasi norm of the first 
derivative. Here we generalize this inequality type for fractional integral. 
Keywords: Convex function, Fractional integral, Holder inequality, L_pspace. 
Classification: 60E15; 39B72. 

 
1. Introduction  

Firstly, let us recall and introduce some definitions notations, that we need in our work. 

The function 𝑓: [𝑎, 𝑏] ⊂ 𝑅 → 𝑅 is said to be convex if the following inequality holds 

𝑓(𝑡𝑥 + (1 − 𝑡)𝑦) ≤ 𝑡𝑓(𝑥) + (1 − 𝑡)𝑓(𝑦  [4] 

For all 𝑥, 𝑦 ∈ [𝑎, 𝑏] and 𝑡 ∈ [0,1], we say that f is concave if(−𝑓) is convex. 

𝑅∝is the set of real numbers, 𝑅∝ = 𝑄∝ ∪ 𝐽∝,where 𝑄∝ is the 𝛼 −type set of the rational numbers is 
defined as the set 

{𝑚∝ = (
𝑃

𝑞
)
𝛼
, 𝑝, 𝑞𝜖𝑍, 𝑞 ≠ 0}, 

 also 𝐽∝ is the  𝛼 −type set of the irrational numbers is defined as the set  

{𝑚∝ ≠ (
𝑃

𝑞
)
𝛼

, 𝑝, 𝑞𝜖𝑍, 𝑞 ≠ 0}. 

 The local fractional derivative of 𝑓(𝑥) of order 𝛼 at 𝑋 = 𝑋° is defined as: 

𝑓𝛼(𝑋°) =
𝑑𝛼𝑓(𝑥)

𝑑𝑥𝛼
|
𝑋=𝑋°

= lim
𝑋→𝑋°

∆𝛼(𝑓(𝑋) − 𝑓(𝑋°))

(𝑋 − 𝑋°)
 

Where ∆𝛼(𝑓(𝑋) − 𝑓(𝑋°)) ≅ Γ(𝛼 + 1)(𝑓(𝑋) − 𝑓(𝑋°)). 

If there exists 𝑓(𝑘+1)𝛼(𝑥) = 𝐷𝑥
𝛼…𝐷𝑥

𝛼𝑓(𝑥) mm times for any 𝑋 ∈ 𝐼 ⊆ 𝑅,then we denoted 𝑓 ∈
𝐷(𝑘+1)𝛼(𝐼), where 𝑘 = 0,1,…      [5] 

Anon –differentiable function 𝑓: 𝑅 → 𝑅𝛼, 𝑋 → 𝑓(𝑥) is called to be local fractional continuous at 𝑋°.If 

for any ∈> 0, there exists 𝛿 > 0, such that |𝑓(𝑥) − 𝑓(𝑋°)| <∈
∝ holds for |𝑋 − 𝑋°| < 𝛿, where ∈, 𝛿 ∈

𝑅. If 𝑓(𝑥) is local continuous on the interval (𝑎, 𝑏).We denote 𝑓(𝑥) ∈ 𝐶𝛼(𝑎, 𝑏). [5] 

We call 𝑓 is fractional integrable if  

1

Γ(1 + 𝛼)
∫ 𝑓(𝑥)(𝑑𝑡)𝛼 =
𝑏

𝑎

1

Γ(1 + 𝛼)
lim
∆𝑡→0

∑𝑓(𝑡𝑗)

𝑛−1

𝑗=0

(∆𝑡𝑗)
𝛼
< ∞    (1) 

and the fractional integrable defined by: 

𝑎𝐼𝑏
∝𝑓(𝑥) =

1

Γ(1 + 𝛼)
∫ 𝑓(𝑥)(𝑑𝑡)𝛼
𝑏

𝑎

=
1

Γ(1 + 𝛼)
lim
∆𝑡→0

∑𝑓(𝑡𝑗)

𝑛−1

𝑗=0

(∆𝑡𝑗)
𝛼

 

With ∆𝑡𝑗 = 𝑡𝑗+1 − 𝑡,and ∆𝑡 = 𝑚𝑎𝑥{∆𝑡1, ∆𝑡2, … , ∆𝑡𝑛−1},where [𝑡𝑗, 𝑡𝑗+1], 



4911 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

𝑗 = 0,… , 𝑛 − 1, and 𝑎 = 𝑡0 < 𝑡1 < ⋯ < 𝑡𝑛−1 = 𝑏 is a partition of interval [𝑎, 𝑏].Here it follows that 

𝑎𝐼𝑏
∝𝑓(𝑥) = 0 if 𝑎 = 𝑏 and 𝑎𝐼𝑏

∝𝑓(𝑥) = −𝑏 𝑎𝐼𝑏
∝𝑓(𝑥) if 𝑎 < 𝑏. 

Let 𝑓: 𝐼 ⊂ 𝑅 → 𝑅𝛼 .For any 𝑋1,𝑋2 ∈ 𝐼and 𝜆 ∈ [0,1], if the following inequality 𝑓(𝜆𝑋1) + (1 − 𝜆)𝑋2 ≤

𝜆𝛼𝑓(𝑋1) + (1 − 𝜆)
𝛼𝑓(𝑋2) holds, then f is called a generalized convex function on I. 

Now, let us introduce our LP,∝ space for 0 < 𝑃 < ∞. 
Let us define the fractional integrable quasi normed space as: 

LP,∝[𝑎, 𝑏] = {𝑓: [𝑎, 𝑏] → 𝑅: ‖𝑓‖𝑃,𝛼 = (∫ |𝑓(𝑥)|𝑝
𝑏

𝑎

(𝑑𝑥)𝛼)

1

𝑃

< ∞} 

and ‖. ‖𝑃,𝛼 is a fractional LP integrable norm. 

 
2. Auxilary Results 
Lemma 2.1 [6]: 

dαf(x)

dxα
=

Γ(1 + 𝑘𝛼)

Γ(1 + (𝑘 − 1)𝛼)
𝑋(𝑘−1)𝛼 

1

Γ(1 + 𝛼)
∫𝑋𝑘𝛼(𝑑𝑥)𝛼 =

𝑏

𝑎

Γ(1 + 𝑘𝛼)

Γ(1 + (𝑘 + 1)𝛼)
(𝑏(𝑘+1)𝛼 − 𝑎(𝑘+1)𝛼), 𝑘𝜖𝑅. 

Lemma 2.2 [1]: Generalized Holder,s inequality 

Let 𝑓, 𝑔 ∈ 𝐶𝛼[𝑎, 𝑏], 𝑝, 𝑞 > 1, with 
1

𝑝
+
1

𝑞
= 1, then 

1

Γ(1 + 𝛼)
∫|𝑓(𝑥)𝑔(𝑥)|(𝑑𝑥)𝛼 ≤ (

1

Γ(1 + 𝛼)
∫|𝑓(𝑥)|𝑝(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑝𝑏

𝑎

. (
1

Γ(1 + 𝛼)
∫|𝑔(𝑥)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

. 

Lemma 2.3 [1]: in 𝐿𝑃-space if If p < q, then  

(∑ |𝑥𝑖|
𝑞

∞

𝑖=1
)

1

𝑞

≤ (∑ |𝑥𝑖|
𝑝

∞

𝑖=1
)

1

𝑝

. 

Lemma 2.4 [2]: Generalized Montgomery inequality 

Let 𝐼 ⊂ 𝑅 be an interval,𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼 (𝐼∘is the interior of I) such that 𝑓 is ∝ −integrable for 

𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏.Then we have the identity 

𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥) =

1

Γ(1 + 𝛼)
∫ 𝑝(𝑥, 𝑡)𝑓(𝑥)𝛼(𝑑𝑡)𝛼
𝑏

𝑎

 

Where  

𝑝(𝑥, 𝑡) = {
(𝑡 − 𝑎)𝛼 , 𝑡 ∈ [𝑎, 𝑥] 

(𝑡 − 𝑏)𝛼 , 𝑡 ∈ [𝑥, 𝑏]
 

Lemma 2.5 [6]: A second type generalized Montgomery 
                                        inequality   

Let 𝐼 ⊂ 𝑅 be an interval,𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼(𝐼∘is the interior of I) such that 𝑓 ∈ 𝐷𝛼(𝐼
∘)  and 𝑓 is ∝

−integrable with  𝑎 < 𝑏 ,then 

(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥) =

1

Γ(1 + 𝛼)
∫ 𝑝(𝑥, 𝑡)𝑓(𝑥)𝛼(𝑑𝑡)𝛼
𝑏

𝑎

 . 

Where  



4912 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

𝑝(𝑥, 𝑡) =

{
 
 

 
 𝑡 − (𝑎 + ℎ (

𝑏 − 𝑎

2
))

∝

, 𝑡 ∈ [𝑎, 𝑥] 

𝑡 − (𝑏 − ℎ (
𝑏 − 𝑎

2
))

∝

, 𝑡 ∈ [𝑥, 𝑏]

 . 

Where ℎ ∈ [0,1] and 𝑎 + ℎ (
𝑏−𝑎

2
) ≤ 𝑥 ≤ 𝑏 − ℎ (

𝑏−𝑎

2
). 

 
3. Main Results 

Let us now introduce our main results. We use two kinds of generalized Montgomery identity to 
prove types generalized Ostrowski Theorems. 
Theorem 3.1: 

If 𝑓 ⊂ 𝑅, 𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼 be a map [𝑎, 𝑏] ⊂ 𝐼∘and 𝑓 ∈ 𝐿𝑃,𝛼[𝑎, 𝑏].  
Then 

 (1) |𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

                                                                              1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

(2) |𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
𝐶(𝑝)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,            0 < 𝑝 < 1. 

Where, 𝐼∘ is the interior of the interval I. 
Proof: According to p, let us divide our proof into two cases. 

Case1: 1 ≤ 𝑝 ≤ ∞ 
by using the generalized Holder,s inequality described in Lemma(2.2) , we get 

1

Γ(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

≤
1

(𝑏 − 𝑎)𝛼
(

1

Γ(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

(
1

Γ(1 + 𝛼)
∫|𝑓(𝑥)𝛼|𝑝(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑝

 

                      1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1                                             (2) 

Let us calculate (
1

Γ(1+𝛼)
∫ |𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎
)

1

𝑞
 , we have 

(
1

Γ(1 + 𝛼)
∫|(𝑡 − 𝑎)|𝑞(𝑑𝑡)𝛼
𝑥

𝑎

)

1

𝑞

+ (
1

Γ(1 + 𝛼)
∫|(𝑡 − 𝑏)|𝑞(𝑑𝑡)𝛼
𝑏

𝑥

)

1

𝑞

 

       = (𝐼1 + 𝐼2)
1

𝑞                                                                                            (3) 
By using Lemma 2,1, 

𝐼1 =
1

Γ(1 + 𝛼)
∫|(𝑡 − 𝑎)|𝑞(𝑑𝑡)𝛼
𝑥

𝑎

 



4913 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

 =
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑥(𝑞+1)𝛼

− 𝑎(𝑞+1)𝛼)                                                                                                           (4) 

𝐼2 =
1

Γ(1 + 𝛼)
∫|(𝑡 − 𝑏)|𝑞(𝑑𝑡)𝛼
𝑏

𝑥

=
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑏(𝑞+1)𝛼 − 𝑥(𝑞+1)𝛼)                                               (5) 

Put (4) and(5) in (3), we get  

(
1

Γ(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

 

= (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑏(𝑞+1)𝛼 − 𝑎(𝑞+1)𝛼))

1

𝑞

 

≤ (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
)

1

𝑞

(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞                                                    (6) 

Since (
1

Γ(1+𝛼)
∫ |𝑓(𝑥)𝛼|𝑝(𝑑𝑥)𝛼
𝑏

𝑎
)

1

𝑝
= (

1

Γ(1+𝛼)
)

1

𝑝 ‖𝑓𝛼‖𝑝                       (7) 

Put (6) and (7) in (2), we get 

1

Γ(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼 ≤
𝑏

𝑎

 

(𝛤(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞

(𝑏 − 𝑎)𝛼(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

                                                               1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1   

Now by using Lemma 2.4,we get 

|𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(𝛤(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
𝛼

𝑞

(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝. 

The first case is proved. 

Case2: 0 < 𝑃 < 1 
By using the generalized Holder,s inequality described in Lemma2.2 , we obtain. 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

≤
1

(𝑏 − 𝑎)𝛼
(

1

𝛤(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

(
1

𝛤(1 + 𝛼)
∫|𝑓(𝑥)𝛼|ℎ(𝑑𝑥)𝛼
𝑏

𝑎

)

1

ℎ

 

          1 ≤ 𝑝, 𝑞 ≤ ∞,
1

ℎ
+
1

𝑞
= 1  

by using definition of the fractional integral in(1) and (6) 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

 



4914 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

≤ (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
)

1

𝑞

(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞 × (
1

𝛤(1 + 𝛼)
∑|𝑓𝛼(𝑡𝑖)|

ℎ(∆𝑡𝑖)
𝛼

𝑛

𝑖=1

)

1

ℎ

 

Where 0 < 𝑝 < 1 and by using (1), we get  

1

Γ(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼 ≤
𝑏

𝑎

 

𝐶(𝑝)(𝛤(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
𝛼

𝑞

(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,   0 < 𝑝 < 1. 

Now by using Lemma 2.4,we get 

|𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
𝐶(𝑝)(𝛤(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
𝛼

𝑞

(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,   0 < 𝑝 < 1. 

The second case is proved. 
Theorem 3.2: 

If 𝑓 ⊂ 𝑅, 𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼 be a map [𝑎, 𝑏] ⊂ 𝐼∘, 𝑓 ∈ 𝐿𝑃,𝛼[𝑎, 𝑏].  
Then 

 (1) |(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)|

≤
(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

                                                                       1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

 

 (2) |(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
𝐶(𝑝)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝     ,0 < 𝑝 < 1. 

 
 
Proof:  
We take two cases to prove our Theorem. 

Case1: 1 ≤ 𝑝 ≤ ∞ 
By using Lemma 2.5 , we obtain. 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

=
1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

 

By using the generalized Holder,s inequality described in Lemma2.1 , we obtain. 



4915 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
1

(𝑏 − 𝑎)𝛼
(

1

𝛤(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

(
1

𝛤(1 + 𝛼)
∫|𝑓(𝑥)𝛼|ℎ(𝑑𝑥)𝛼
𝑏

𝑎

)

1

ℎ

 

                         1 ≤ 𝑝, 𝑞 ≤ ∞,
1

ℎ
+
1

𝑞
= 1                                                 (8) 

Let us calculate (
1

Γ(1+𝛼)
∫ |𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎
)

1

𝑞
 , we have 

(
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑎 + ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑥

𝑎

)

1

𝑞

+(
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑏 − ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑏

𝑥

)

1

𝑞

 

          = (𝑀1 +𝑀2)
1

𝑞                                                                                                                                   (9) 

𝑀1 =
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑎 + ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑥

𝑎

 

 =
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑎 + ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼 +
𝑎+ℎ(

𝑏−𝑎

2
)

𝑎

 

1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑎 + ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑥

𝑎+ℎ(
𝑏−𝑎

2
)

 

By using Lemma 2,1 
 
 

𝑀1 =
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
[((𝑎 + ℎ (

𝑏 − 𝑎

2
))

(𝑞+1)𝛼

− 𝑎(𝑞+1)𝛼) 

+(𝑥(𝑞+1)𝛼 − (𝑎 + ℎ (
𝑏 − 𝑎

2
))

(𝑞+1)𝛼

)]     

    =
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑥(𝑞+1)𝛼 − 𝑎(𝑞+1)𝛼)                                         (10) 

also, 

𝑀2 =
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑏 − ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑏

𝑥

 

=
1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑏 − ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼 +
𝑏−ℎ(

𝑏−𝑎

2
)

𝑎

 

1

Γ(1 + 𝛼)
∫ |𝑡 − (𝑏 − ℎ (

𝑏 − 𝑎

2
))|

𝛼𝑞

(𝑑𝑡)𝛼
𝑥

𝑎+ℎ(
𝑏−𝑎

2
)

 



4916 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

𝑀2 =
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
[((𝑏 − ℎ (

𝑏 − 𝑎

2
))

(𝑞+1)𝛼

− 𝑥(𝑞+1)𝛼) 

+(𝑥(𝑞+1)𝛼 − (𝑏 − ℎ (
𝑏 − 𝑎

2
))

(𝑞+1)𝛼

)]              

=
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑏(𝑞+1)𝛼 − 𝑥(𝑞+1)𝛼)                                   (11) 

Put (10) and(11) in (9), we get 

(
1

Γ(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

 

= [
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑥(𝑞+1)𝛼 − 𝑎(𝑞+1)𝛼) + (𝑏(𝑞+1)𝛼 − 𝑥(𝑞+1)𝛼)]

1

𝑞

 

 

= (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
(𝑏(𝑞+1)𝛼 − 𝑎(𝑞+1)𝛼))

1

𝑞

≤ (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
)

1

𝑞

(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞                           (12) 

 
Put (7) and (12) in (8), we get 
 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
1

(𝑏 − 𝑎)𝛼
(

(𝛤(1 + 𝛼𝑞))

𝛤(1 + (𝑞 + 1)𝛼)
(𝑏(𝑞+1)𝛼 − 𝑎(𝑞+1)𝛼))

1

𝑞

× (
1

𝛤(1 + 𝛼)
)

1

𝑝

‖𝑓𝛼‖𝑝 .  

This implies, 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
(𝛤(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
𝛼

𝑞

(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1 

         The first case is proved. 

Case2: 0 < 𝑃 < 1 
 By using the generalized Holder,s inequality described in Lemma2.2 , we obtain. 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

≤
1

(𝑏 − 𝑎)𝛼
(

1

𝛤(1 + 𝛼)
∫|𝑝(𝑥, 𝑡)|𝑞(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝑞

(
1

𝛤(1 + 𝛼)
∫|𝑓(𝑥)𝛼|𝐿(𝑑𝑥)𝛼
𝑏

𝑎

)

1

𝐿

 



4917 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝐿
+
1

𝑞
= 1  

by using (1) and using(12) implies 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

 

≤ (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
)

1

𝑞

(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞 × (
1

𝛤(1 + 𝛼)
∑|𝑓𝛼(𝑡𝑖)|

𝐿(∆𝑡𝑖)
𝛼

𝑛

𝑖=1

)

1

𝐿

 

Where 0 < 𝑝 < 1 , by using Lemma 2.3,we get 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

 

≤ (
Γ(1 + 𝛼𝑞)

Γ(1 + (𝑞 + 1)𝛼)
)

1

𝑞

(𝑏 − 𝑎)
(𝑞+1)𝛼

𝑞 × (
1

𝛤(1 + 𝛼)
∑|𝑓𝛼(𝑡𝑖)|

𝑃(∆𝑡𝑖)
𝛼

𝑛

𝑖=1

)

1

𝑃

 

By using (1), we get 

1

𝛤(1 + 𝛼)(𝑏 − 𝑎)𝛼
∫ |𝑝(𝑥, 𝑡)||𝑓(𝑥)𝛼| (𝑑𝑡)𝛼
𝑏

𝑎

 

≤
𝐶(𝑝)(𝛤(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
𝛼

𝑞

(𝛤(1 + (𝑞 + 1)𝛼))
1

𝑞(𝛤(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,   0 < 𝑝 < 1. 

Now by using Lemma 2.5, we get 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

    ≤
𝐶(𝑝)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,               0 < 𝑝 < 1. 

The second case is proved. 
Corollary3.3: 

If 𝑓 ⊂ 𝑅, 𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼 be a map [𝑎, 𝑏] ⊂ 𝐼∘and 𝑓 ∈ 𝐿𝑃,𝛼[𝑎, 𝑏].  
Then 

 (1) |𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

                                                                              1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

(2) |𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
𝐶(𝑝)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝     ,0 < 𝑝 < 1. 

Proof:  
By usingTheorem 3.2, we  get 
Case 1: 



4918 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

If ℎ = 0, we get 

|𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,    1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

Case 2: 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

𝐶(𝑃)(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

If ℎ = 0, we get 

|𝑓(𝑥) −
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

𝐶(𝑃)(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝,   0 < 𝑃 < 1. 

Corollary 3.4: 

If 𝑓 ⊂ 𝑅, 𝑓: 𝐼∘ ⊂ 𝑅 → 𝑅𝛼 be a map [𝑎, 𝑏] ⊂ 𝐼∘and 𝑓 ∈ 𝐿𝑃,𝛼[𝑎, 𝑏].  
Then 

 (1) |𝑓(𝑥) +
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)|    

≤
(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝   , 1 ≤ 𝑝, 𝑞 ≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

(2) |𝑓(𝑥) +
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| 

≤
𝐶(𝑝)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝     ,0 < 𝑝 < 1. 

Proof:  
By using Theorem 3.2, we  get 
Case 1: 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

If ℎ = 1, we get 

|𝑓(𝑥) +
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 1 ≤ 𝑝, 𝑞

≤ ∞,
1

𝑝
+
1

𝑞
= 1. 

 
 
Case 2: 



4919 

 

 
Edelweiss Applied Science and Technology 
ISSN: 2576-8484 

Vol. 8, No. 6: 4910-4919, 2024 
DOI: 10.55214/25768484.v8i6.3054 
© 2024 by the authors; licensee Learning Gate 

 

|(𝐼 − ℎ)𝛼𝑓(𝑥) + ℎ𝛼
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)| ≤

𝐶(𝑃)(Γ(1 + 𝛼𝑞))
1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝, 

If ℎ = 1, we get 

|𝑓(𝑥) +
𝑓(𝑎) + 𝑓(𝑏)

2𝛼
−
Γ(1 + 𝛼)

(𝑏 − 𝑎)𝛼
𝑎𝐼𝑏
∝𝑓(𝑥)|            

≤
𝐶(𝑃)(Γ(1 + 𝛼𝑞))

1

𝑞(𝑏 − 𝑎)
∝

𝑞

(Γ(1 + (𝑞 + 1)𝛼))
1

𝑞(Γ(1 + 𝛼))
1

𝑝

‖𝑓𝛼‖𝑝                         0 < 𝑃 < 1. 

 
Copyright:  
© 2024 by the authors. This article is an open access article distributed under the terms and conditions 
of the Creative Commons Attribution (CC BY) license (https://creativecommons.org/licenses/by/4.0/). 
 
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