Approach of the value of a rent when non-central moments of the capitalization factor are known: an R application with interest rates following normal and beta distributions


Ratio Mathematica                                                Volume 37, 2019, pp. 49-60 

 

 
 

49 

 

 

 

 

Mahgoub Transform on Boehmians 
 

 

Yogesh Khandelwal* 

Priti Chaudhary† 
 

 

 

Abstract 

Boehmian’s space is established utilizing an algebraic way that approximates 

identities or delta sequences and appropriate convolution. The space of 

distributions can be related to the proper subspace. In this paper, firstly we 

establish the appropriate Boehmian space, on which the Mahgoub 

Transformation can be described& function space K can be embedded. We 

add to more in this, our definitions enhance Mahgoub transform to 

progressively wide spaces. We additionally explain the functional axioms of 

Mahgoub transform on Boehmians. Lastly toward the finishing of topic, we 

analyze with specify axioms and properties for continuity and the enlarged 

Mahgoub transform, also its inverse regards to ∆-convergence and δ. 
 

Keywords: Mahgoub Transform; The Space 𝔹(𝔛); The Space𝔹(𝔛𝔐); 
Boehmian Spaces. 

 

2010 AMS subject classification: 44A99; 44A40;46F99; 20C20.‡ 
 

 
 

 
* Department of Mathematics, Jaipur National University, Jaipur, Rajasthan (India); 

yogeshmaths81@gmail.com. 
† Department of Mathematics, Jaipur National University, Jaipur, Rajasthan (India). 
‡ Received on July 3rd, 2019. Accepted on August 9th, 2019. Published on December 31th, 

2019. doi: 10.23755/rm.v37i0.468. ISSN: 1592-7415. eISSN: 2282-8214. ©Khandelwal and 

Chaudhary. This paper is published under the CC-BY licence agreement. 



Yogesh Khandelwal and Priti Chaudhary 

50 

 

1. Introduction 

The Mahgoub transform [8] which is denoted by the operator 𝔐(. )  and 
Mahgoub transform of 𝔗(𝓉∗) is defined by: 

𝔐( 𝔗(𝓉∗)) = 𝔼(𝜗) =  𝜗 ∫ 𝔗(𝓉∗)
∞

0

𝑒−𝜗𝓉
∗
𝑑𝓉 ∗, 𝓉 ∗ ≥ 0,               (1.1) 

and                                    𝜌1 ≤ 𝜗 ≤ 𝜌2. 
In a set ; 

𝔸 =  {𝔗(𝓉∗): ∃𝕄, 𝜌1, 𝜌2 > 0 . |𝔗(𝓉
∗)| < 𝕄𝑒

|𝓉 ∗|

𝜌𝑗 }    ,                            (1.2) 

where  𝜌1 and 𝜌2(may be finite or infinite), the constant 𝕄 must be finite. 
An existence’s Mahgoub transform of 𝔗(𝓉∗)  is essential for  𝓉∗ ≥ 0, a piece 
wise continuous and of exponential order is required, else it does not exist. 

 

Convolution Theorem For Mahgoub Transform [9-11]:  

If 𝔐( 𝔗(𝓉 ∗)) = 𝔼(𝜗)  and 𝔐( 𝔓(𝓉∗)) = 𝕎(𝜗) then 

𝔐(𝔗(𝓉∗) ⋆ 𝔓(𝓉 ∗)) =
1

ϑ
𝔐(𝔗(𝓉 ∗))𝔐(𝔓(𝓉∗)) =

1

ϑ
𝔼(ϑ)𝕎(ϑ)       (1.3) 

 

Linearity Property Of Mahgoub Transform: 

If       𝔐( 𝔗(𝓉∗)) = 𝔼(𝜗),  𝔐( 𝔓(𝓉∗)) = 𝕎(𝜗) then 
𝔐{𝔞𝔗(𝓉∗) +  𝔟 𝔓(𝓉∗)} = 𝔞 𝔐(𝔗(𝓉 ∗)) + 𝔟 𝔐(𝔓(𝓉∗))               (1.4) 

 

2. Boehmian Space 

Boehmians was first developed as a generalization’s standard mikusinski 

operators [2]. The formation necessary for Boehmians satisfying the following  

axioms. 

i. a non empty set 𝔄; 
ii. a semi group(𝜑,⊛) which is commutative; 

iii. ⊗: 𝔄 × 𝜑 → 𝔄  s.t.∀ ξ ∈ 𝔄 and 𝜂1, 𝜂2 ∈ 𝜑 , ξ ⊗ (𝜂1 ⊛ 𝜂2) =
(ξ ⊗ 𝜂1) ⊗ 𝜂2; 
iv. a collection ∆ ⊂ 𝜑𝑁such that 

a) Ifξ1, ξ2 ∈ 𝔄 , (𝜂𝑛) ∈ ∆ , ξ1 ⊗ 𝜂𝑛 = ξ2 ⊗ 𝜂𝑛  ∀ 𝑛then ξ1 = ξ2; 
b) If (𝜂𝑛), (𝜏𝑛) ∈ ∆,   then (𝜂𝑛 ⊗ 𝜏𝑛) ∈ ∆. where elements of ∆ are 

known as delta sequences. 

Consider 

ℋ = {(ξ𝑛, 𝜂𝑛 ): ξ𝑛 ∈ 𝔄 , (𝜂𝑛 ) ∈ ∆, ξ𝑛 ⊗ 𝜂𝑚 = ξ𝑚 ⊗ 𝜂𝑛 ∀ 𝑚, 𝑛 ∈ 𝑁},                  
Now if (ξ𝑛, 𝜂𝑛 ), (∅𝑛, 𝜏𝑛 ) ∈ ℋ then ξ𝑛 ⊗ 𝜏𝑚 = ∅𝑚 ⊗ 𝜂𝑛 , ∀ 𝑚, 𝑛 ∈ 𝑁. 



 An Reckon of Mahgoub Transform for Boehmians  
 

51 

 

We say that (ξ𝑛 , 𝜂𝑛 )~(∅𝑛, 𝜏𝑛). where ~ is an equivalence relation inℋ.The 
Set of equivalence classes in ℋ is denoted asℌ. Elements of ℌ are said to be 
Boehmians. 

We assume that there is a canonical embedding between ℌ and𝔄, expressed 

as   ξ →
ξ𝑛⊗𝜂𝑛

𝜂𝑛
 ,where⊗ can also be extended in 

                ℌ × 𝔄 by 
ξ𝑛

𝜂𝑛
⊗ 𝜏 =

ξ𝑛⊗𝜏 

𝜂𝑛
 . 

In ℌ, there are two types of convergence is given by 
i. if ℶ𝑛 → ℶ as 𝑛 → ∞which belongs to𝔄,𝓀 ∈ 𝜑 is any fixed 

element, then  ℶ𝑛 ⊗ 𝓀 → ℶ ⊗  𝓀 as 𝑛 → ∞in 𝔄. 
ii. if ℶ𝑛 → ℶas 𝑛 → ∞ in 𝔄 and 𝜆𝑛 ∈ ∆  then ℶ𝑛 ⊗ 𝜆𝑛 → ℶas 𝑛 →
∞in 𝔄. 

An operation ⊗can be extended inℌ × 𝜑 as per condition: 

If [
ℶ𝑛

𝜂𝑛
] ∈ ℌand 𝓀 ∈ 𝜑 then [

ℶ𝑛

𝜂𝑛
] ⊗ 𝓀 = [

ℶ𝑛⊗𝓀

𝜂𝑛
]. 

Now convergence in ℌas following: 
 

1. A sequence (𝜍𝑛)in ℌis called𝛿– convergent to 𝜍 in ℌ, i.e. 

𝜍𝑛
𝛿
→ 𝜍 if ∃ (𝜂𝑛) ∈ Δ  such that(𝜍𝑛 ⊗ 𝜂𝑛 ), (𝜍 ⊗ 𝜂𝑛 ) ∈ 𝔄, ∀ 𝑛 ∈ 𝑁 and 

 (𝜍𝑛 ⊗ 𝜂𝓀 ) → (𝜍 ⊗ 𝜂𝓀) as 𝑛 → ∞ in 𝔄, ∀ 𝓀, 𝑛 ∈ 𝑁. 

2. A sequence (𝜍𝑛) in ℌ is said to be ∆ convergent to 𝜍 in ℌ i.e. 𝜍𝑛
∆
→ 𝜍 , if 

∃(𝜂𝑛) ∈ Δ such that(𝜍𝑛 − 𝜍) → 0 as 𝑛 → ∞ which belongs to 𝔄 . 
For more details, see [3-6]. 

 

 

3. The Boehmian Space 𝔹(𝖃): 

Denoted by 𝔖+(ℝ) and 𝒞0+
∞ (ℝ)are the space’s smooth function over ℝ and the 

Schwarz space’s test function’s compact support over ℝ+ where ℝ+ = (0, ∞) 

respectively. We have found vital results for the structure of  Boehmian space 

𝔹(𝔛)where 𝔛 = (𝔖+, 𝒞0+
∞ , ∆+). 

Lemma 3.1: 

1) If 𝔡1, 𝔡2 ∈ 𝒞0+
∞ (ℝ)then 𝔡1 ⋆ 𝔡2 ∈ 𝒞0+

∞ (ℝ)(Closure). 
 

2) If 𝔉1, 𝔉2 ∈ 𝔖+(ℝ)and 𝔡1 ∈ 𝒞0+
∞ (ℝ)then 

(𝔉1 +  𝔉2) ⋆ 𝔡1 = 𝔉1 ⋆ 𝔡1 + 𝔉2 ⋆ 𝔡1 (Distributive). 
3) 𝔡1 ⋆ 𝔡2 = 𝔡2 ⋆ 𝔡1∀𝔡1, 𝔡2 ∈ 𝒞0+

∞ (ℝ) (Commutative). 
 



Yogesh Khandelwal and Priti Chaudhary 

52 

 

4) If 𝔉 ∈ 𝔖+(ℝ),  𝔡1, 𝔡2 ∈ 𝒞0+
∞ (ℝ)then (𝔉 ⋆ 𝔡1) ⋆ 𝔡2 = 𝔉 ⋆

(𝔡1 ⋆ 𝔡2)(Associative). 
 

Definition3.2: A sequence (𝜂𝑛) of function from 𝒞0+
∞ (ℝ)is said to be in∆+. 

If 

∆+
1 : ∫ 𝜂𝑛 (𝜉)𝑑𝜉 = 1.

ℝ+
 

∆+
2 : ∫ |𝜂𝑛(𝜉)|𝑑𝜉 ≤ 𝑚,ℝ+ where 𝑚 is a positive integer; 

 

∆+
3 ∶ 𝑆𝑢𝑝𝑝 𝜂𝑛 (𝜉) ⊂ (0, 𝜖𝑛),         𝜖𝑛 → 0   as 𝑛 → ∞. 

 

i.e.(𝜂𝑛) shrink to zero as 𝑛 → ∞.every  member of ∆+ is known as an 
approximation identity or a delta sequences. In all manners delta sequences 

arise in numerous parts of Mathematics, however likely the very important 

application are those in the presupposition’s generalized functions. The 

fundamental application of delta sequence is the regularization’s established 

functions and ahead we can be utilized to characterize the convolution product 

and its established functions. 

 

Lemma 3.3: If(𝜂𝑛),(𝜏𝑛)  ∈ ∆+, then  𝑠𝑢𝑝𝑝(𝜂𝑛 ⋆ 𝜏𝑛) ⊂ 𝑠𝑢𝑝𝑝𝜂𝑛 + 𝑠𝑢𝑝𝑝𝜏𝑛 . 

Lemma 3.4: If 𝔡1, 𝔡2 ∈ 𝒞0+
∞ (ℝ) then so is 𝔡1 ⋆ 𝔡2 and ∫ |𝔡1 ⋆ 𝔡2| ≤ℝ+

∫ |𝔡1|ℝ+ ∫ |𝔡2|ℝ+ . 

 

Theorem 3.5: Let𝔉1, 𝔉2 ∈ 𝔖+(ℝ) and (𝜂𝑛 ) ∈ ∆+ such that  
𝔉1 ⋆ 𝜂𝑛 = 𝔉2 ⋆ 𝜂𝑛 . 

where 𝑛 = 1,2,3, …, then 𝔉1 = 𝔉2   in  𝔖+(ℝ). 
 

Proof: To prove that  𝔉1 ⋆ 𝜂𝑛 = 𝔉1.  
Let K be a compact support accommodating the 𝑠𝑢𝑝𝑝𝜂𝑛 for each𝑛 ∈ 𝑁. By 

using   ∆+
1 , we write  

|𝜉
𝑘

𝐷𝑚(𝔉1 ⋆ 𝜂𝑛 − 𝔉1)(𝜉)|

≤ ∫ |𝜂𝑛(𝜏)| |𝜉
𝑘

𝐷𝑚(𝔉1(𝜉 − 𝜏) − 𝔉1(𝜉))| 𝑑𝜏 
𝐾

                   (3.1) 

The mapping 𝜏 → 𝔉1
𝜏 where𝔉1

𝜏 = 𝔉1(𝜉 − 𝜏), is Uniformly continuous 
from ℝ+ → ℝ+. By using axiom∆+

3 that 𝑠𝑢𝑝𝑝𝜂𝑛 → 0 as  𝑛 → ∞, now we 
choose 𝑟 > 0;𝑠𝑢𝑝𝑝𝜂𝑛 ⊆ [0, 𝑟] for large 𝑛 and 𝜏 < 𝑟, that is 

|𝔉1(𝜉 − 𝜏) − 𝔉1(𝜉)| = |𝔉1
𝜏 − 𝔉1| <

𝜖𝑛
𝑀

                                                           (3.2) 

Hence   using ∆+
2  and  Eq’s. (3.2), (3.1) we get  



 An Reckon of Mahgoub Transform for Boehmians  
 

53 

 

|𝜉
𝑘

𝐷𝑚(𝔉1 ⋆ 𝜂𝑛 − 𝔉1)(𝜉)| < 𝜖𝑛 → 0 as 𝑛 → ∞.  
Thus   𝔉1 ⋆ 𝜂𝑛 →  𝔉1 in𝔖+(ℝ). Similarly,  
we prove that 𝔉2 ⋆ 𝜂𝑛 →

𝔉2in  𝔖+(ℝ)                                                                 ⌂ 
 

Theorem 3.6:if  𝔉𝑛 → 𝔉in 𝔖+(ℝ)as𝑛 → ∞ and 𝔡 ∈
𝒞0+

∞  (ℝ) then  lim
𝑛→∞

𝔉𝑛 ⋆ 𝔡 = 𝔉 ⋆ 𝔡. 

 

Proof: Using Theorem we get 

|𝜉
𝑘

𝐷𝑚 ((𝔉𝑛 ⋆ 𝔡) − (𝔉 ⋆ 𝔡))(𝜉)| = |𝜉
𝑘

(𝐷𝑚(𝔉𝑛 − 𝔉) ⋆ 𝔡)(𝜉)| (3.3) 

The equation follows from [3] 
       𝐷𝑚𝔉 ⋆ 𝔡 = 𝐷𝑚𝔉 ⋆ 𝔡 = 𝔉 ⋆ 𝐷𝑚𝔡          

for all     𝔡 ∈ 𝒞0+
∞  (ℝ), we have  

|𝜉
𝑘

𝐷𝑚 ((𝔉𝑛 ⋆ 𝔡) − (𝔉 ⋆ 𝔡))(𝜉)| ≤ ∫ 𝜉
𝑘

|𝐷𝑚(𝔉𝑛 − 𝔉)(𝜉 − 𝜏)||𝔡(𝜏)|𝑑𝜏
𝐾

 

≤ 𝑀𝛾𝑘 (𝔉𝑛 − 𝔉)for some constant M→ 0  as   𝑛 → ∞.                                 ⌂ 
 

Theorem 3.7:In 𝔖+(ℝ), Let lim
𝑛→∞

𝔉𝑛 = 𝔉   and 

         (𝜂𝑛 ) ∈ ∆+⇒ lim
𝑛→∞

𝔉𝑛 ⋆ 𝜂𝑛 = 𝔉. 

 

Proof: By the hypothesis of the Theorem 3.5, we get lim
𝑛→∞

𝔉𝑛 ⋆ 𝜂𝑛 = 𝔉𝑛 →

𝔉as𝑛 → ∞. 
Hence , we arrive, 

lim
𝑛→∞

𝔉𝑛 ⋆ 𝜂𝑛 = 𝔉 as  𝑛

→ ∞.                                                                                                  ⌂ 

The Canonical embedding between𝔹(𝔛)and𝔖+(ℝ), defined as𝜉 → [
𝜉⋆𝜂

𝑛

𝜂
𝑛

]. 

The extension of ⋆ to 𝔹(𝔛) × 𝔖+(ℝ)  is given by  [
ξ𝑛

𝜂𝑛
] ⋆ 𝜏 = [

ξ𝑛⋆𝜏

𝜂𝑛
].  

Convergence in 𝔹(𝔛)is followed: 
𝜹– Convergence: A sequence(𝜍𝑛)in 𝔹(𝔛)is called𝛿– convergent to 𝜍 in 

𝔹(𝔛)denoted by 𝜍𝑛
𝛿
→ 𝜍 if ∃ (𝜂𝑛) ∈ Δ  such that(𝜍𝑛 ⋆ 𝜂𝑛 ), (𝜍 ⋆ 𝜂𝑛) ∈

𝔖+(ℝ), ∀ 𝑛 ∈ 𝑁 and (𝜍𝑛 ⋆ 𝜂𝓀 ) → (𝜍 ⋆ 𝜂𝓀 )as𝑛 → ∞ in 𝔖+(ℝ), ∀ 𝓀, 𝑛 ∈ 𝑁. 
𝚫+ − Convergence:A sequence (𝜍𝑛)in 𝔹(𝔛)is said to be Δ+ − convergent 

to 𝜍  in 𝔹(𝔛)i.e.𝜍𝑛
∆
→ 𝜍 , if ∃(𝜂𝑛) ∈ Δ+such that(𝜍𝑛 − 𝜍) ⊗ 𝜂𝑛 ∈ 𝔖+(ℝ)∀ 𝑛 ∈

𝑁 and (𝜍𝑛 − 𝜍) ⊗ 𝜂𝑛 → 0 as 𝑛 → ∞in 𝔖+(ℝ). 



Yogesh Khandelwal and Priti Chaudhary 

54 

 

Theorem 3.8:Define   𝔉 → [
𝔉⋆𝜂𝑛

𝜂𝑛
] is continuous mapping which is 

embedding from  𝔖+(ℝ) into 𝔹(𝔛). 
Proof: To show: The mapping is one - one. 

We have [
𝔉1⋆𝜂𝑛

𝜂𝑛
] = [

𝔉2⋆𝜏𝑛

𝜏𝑛
], then 

(𝔉1 ⋆ 𝜂𝑛 ) ⋆ 𝜏𝑚 = (𝔉2 ⋆ 𝜏𝑚) ⋆ 𝜂𝑛  , 𝑚, 𝑛 ∈ 𝑁. 
∵ (𝜏𝑛), (𝜂𝑛) ∈ Δ+, 𝔉1 ⋆ (𝜂𝑚 ⋆ 𝜏𝑛) = 𝔉2 ⋆ (𝜏𝑛 ⋆ 𝜂𝑚) = 𝔉2 ⋆ (𝜂𝑚 ⋆ 𝜏𝑛). 
Using Theorem 3.5, we get 𝔉1 = 𝔉2. 
To prove: The mapping is continuous. 

Let 𝔉𝑛 → 0 in𝔖+(ℝ)as 𝑛 → ∞. Then we have[
𝔉𝑛⋆𝜂𝑚

𝜂𝑚
]

𝛿
→ 0as 𝑛 → ∞. 

From the Theorem 3.5, [
𝔉𝑛⋆𝜂𝑚

𝜂𝑚
] ⋆ 𝜂𝑚 = 𝔉𝑛 ⋆ 𝜂𝑚 → 0as 𝑛 → ∞.          ⌂ 

 

Theorem3.9: Let 𝔡 ∈ 𝒞0+
∞  (ℝ)  and 𝔉 ∈  𝔖+(ℝ)⇒𝔐(𝔉 ⋆ 𝔡)(𝜉) =

1

𝜉
𝔉𝔐(𝜉)𝔡𝔐(𝜉). 

 

4. TheBoehmian Space 𝔹(𝔛𝔐) 

We delineate Boehmian space as ensues. Let 𝔖+(ℝ) be the space’s 
immediately decreasing function [3]. We have 

 

𝒞0+
∞𝔐(ℝ) = {𝔡𝔐: ∀𝔡 ∈ 𝒞0+

∞  (ℝ)}                                 (4.1) 

 
here  𝔡𝔐 express the Mahgoub transform of  𝔡   and also 

characterize  𝔉∎𝔡𝔐 by 

(𝔉∎𝔡𝔐)(𝜉) =
1

𝜉
𝔉(𝜉)𝔡𝔐(𝜉)                                       (4.2) 

 

Lemma 4.1 Let 𝔉 ∈ 𝔖+(ℝ),𝔡
𝔐 ∈ 𝒞0+

∞𝔐(ℝ)⇒𝔉∎𝔡𝔐 ∈ 𝔖+(ℝ). 

Proof. Let 𝔉 ∈ 𝔖+(ℝ), 𝔡
𝔐 ∈ 𝒞0+

∞𝔐(ℝ),  by Leibnitz’ Theorem and applying 
the definition of  𝔖+(ℝ), we found   

|𝜉
𝑘

𝐷𝜉
𝑚(𝔉∎𝔡𝔐)(𝜉)| ≤ |𝜉

𝑘
∑ 𝐷𝑚−𝑗 (

1

𝜉
𝔉(𝜉)) 𝐷𝑗 𝔡𝔐(𝜉)

𝑚

𝑗=1

| 

                               ≤ ∑ |𝜉
𝑘

𝐷𝑚−𝑗 (
1

𝜉
𝔉(𝜉))|

𝑚

𝑗=1

|𝐷𝑗 𝔡𝔐(𝜉)| 

                                 = ∑ |𝜉
𝑘

𝐷𝑚−𝑗 𝔉1(𝜉)|
𝑚
𝑗=1 |𝜗 ∫ 𝔡(𝜏)𝑒

−
𝜏

𝜗𝑑𝜏
𝐾

| 



 An Reckon of Mahgoub Transform for Boehmians  
 

55 

 

Where 𝔉1(𝜉) =
1

𝜉
𝔉(𝜉) ∈ 𝔖+(ℝ) and K is a compact support        

accommodating the  𝑠𝑢𝑝𝑝𝑢(𝜏). 

|𝜉
𝑘

𝐷𝜉
𝑚(𝔉∎𝔡𝔐)(𝜉)| ≤ 𝑀𝛾𝑘,𝑚−𝑗 (𝔉1) < ∞, 

for some  positive constant M.                                                                     ⌂ 
 

Lemma 4.2   A mapping    𝔖+ × 𝒞0+
∞𝔐 → 𝔖+ is defined by 

(𝔉, 𝔡𝔐) → 𝔉∎𝔡𝔐 

Satisfying the following axioms: 

(1) If 𝔡1
𝔐, 𝔡2

𝔐 ∈ 𝒞0+
∞𝔐(ℝ), then 𝔡1 

𝔐∎𝔡2
𝔐  ∈ 𝒞0+

∞𝔐(ℝ). 
 

(2) If   𝔉1, 𝔉2 ∈ 𝔖+(ℝ), 𝔡
𝔐 ∈ 𝒞0+

∞𝔐(ℝ), then (𝔉1+ 𝔉2)∎𝔡
𝔐 =

𝔉1∎𝔡
𝔐 + 𝔉2∎𝔡

𝔐. 
 

(3) For  𝔡1
𝔐, 𝔡2

𝔐 ∈ 𝒞0+
∞𝔐(ℝ), 𝔡1

𝔐∎𝔡2
𝔐 = 𝔡2

𝔐∎𝔡1
𝔐. 

 

(4) For  𝔉 ∈ 𝔖+(ℝ),𝔡1 
𝔐, 𝔡2

𝔐 ∈ 𝒞0+
∞𝔐 (ℝ)then(𝔉∎𝔡1 

𝔐)∎𝔡2
𝔐 =

𝔉∎(𝔡1 
𝔐∎𝔡2

𝔐). 

Proof .Axioms of above lemma as follows: 

(1) Let 𝔡1, 𝔡2 ∈ 𝒞0+
∞  (ℝ) then 𝔡1∎𝔡2 ∈ 𝒞0+

∞ (ℝ). 

⇒ (𝔡1∎𝔡2)
𝔐 ∈ 𝒞0+

∞𝔐(ℝ) 

By using Theorem (3.9) implies  𝔡1 
𝔐∎𝔡2

𝔐 ∈ 𝒞0+
∞𝔐 (ℝ). 

 (2) Proof is straightforward. 

 (3) We have  

(𝔡1 
𝔐∎𝔡2

𝔐)(𝜉) =
1

𝜉
𝔡1 

𝔐(𝜉)𝔡2
𝔐 (𝜉) 

                             =
1

𝜉
𝔡2 

𝔐(𝜉)𝔡1
𝔐(𝜉) 

                            = (𝔡2
𝔐∎𝔡1

𝔐)(𝜉). 
(𝔡1 

𝔐∎𝔡2
𝔐) = (𝔡2

𝔐∎𝔡1
𝔐) 

(4)Let 𝔉 ∈ 𝔖+(ℝ)and 𝔡1 
𝔐, 𝔡2

𝔐 ∈ 𝒞0+
∞𝔐(ℝ), then  

((𝔉∎𝔡1 
𝔐)∎𝔡2

𝔐) (𝜉) =
1

𝜉
(𝔉∎𝔡1 

𝔐)(𝜉)𝔡2
𝔐 (𝜉) 

       =
1

𝜉
{

1

𝜉
𝔉(𝜉)𝔡1

𝔐 (𝜉) } 𝔡2
𝔐(𝜉) 

=
1

𝜉
𝔉(𝜉)

1

𝜉
𝔡1

𝔐 (𝜉) 𝔡2
𝔐(𝜉) 



Yogesh Khandelwal and Priti Chaudhary 

56 

 

                                                =
1

𝜉
𝔉(𝜉)(𝔡1 

𝔐∎𝔡2
𝔐)(𝜉) 

                                                = (𝔉∎(𝔡1 
𝔐∎𝔡2

𝔐)) (𝜉), 

(𝔉∎𝔡1 
𝔐)∎𝔡2

𝔐 = 𝔉∎(𝔡1 
𝔐∎𝔡2

𝔐)                                        ⌂ 

 

Denote by∆+
𝔐  where∆+

𝔐 is the collection of all Mahgoub transform’s delta 

sequence in∆+.  i.e., 

∆+
𝔐 =  {(𝜂𝑛

𝔐): (𝜂𝑛) ∈ ∆+, ∀ 𝑛 ∈ 𝑁}.                                      (4.3) 

Lemma 4.3Let  𝔉1, 𝔉2 ∈ 𝔖+(ℝ), (𝜂𝑛
𝔐) ∈ ∆+

𝔐   such that  

            𝔉1∎𝜂𝑛
𝔐 = 𝔉2∎𝜂𝑛

𝔐 , ∀𝑛, then 𝔉1 = 𝔉2 in 𝔖+(ℝ). 
Proof. Let𝔉1, 𝔉2 ∈ 𝔖+(ℝ), (𝜂𝑛

𝔐) ∈ ∆+
𝔐. Since 𝔉1∎𝜂𝑛

𝔐 = 𝔉2∎𝜂𝑛
𝔐, using 

Eq.(4.2)  

⇒
1

𝜉
𝔉1(𝜉)𝜂𝑛

𝔐(𝜉) =
1

𝜉
𝔉2(𝜉)𝜂𝑛

𝔐(𝜉) 

Hence 𝔉1(𝜉) = 𝔉2(𝜉) for all𝜉.                                                     ⌂ 
 

Lemma 4.4 For all (𝜏𝑛), (𝜂𝑛 ) ∈ Δ+, (𝜂𝑛
𝔐∎𝜏𝑛

𝔐
) ∈ ∆+

𝔐.  

Proof. Since(𝜏𝑛), (𝜂𝑛) ∈ Δ+,𝜂𝑛 ⋆ 𝜏𝑛 ∈ Δ+, ∀𝑛 hence from Theorem 3.9, we 
get 

𝔐(𝜂𝑛 ⋆ 𝜏𝑛)(𝜉) =
1

𝜉
𝜂𝑛

𝔐(𝜉)𝜏𝑛
𝔐(𝜉) = 𝜂𝑛

𝔐∎𝜏𝑛
𝔐 ∈ ∆+

𝔐, for each 𝑛.      ⌂ 

 

Lemma 4.5  Let lim
𝑛→∞

𝔉𝑛 = 𝔉in 𝔖+(ℝ), 𝔡
𝔐 ∈ 𝒞0+

∞𝔐(ℝ) then 𝔉𝑛∎𝔡
𝔐 →

𝔉∎𝔡𝔐in 𝔖+(ℝ). 
Proof.we know that 𝔡𝔐 is bounded in 𝒞0+

∞𝔐(ℝ) we have  

(𝔉𝑛∎𝔡
𝔐)(𝜉) →

1

𝜉
𝔉(𝜉)𝔡𝔐(𝜉) 

→ (𝔉∎𝔡𝔐)(𝜉). 
Hence    𝔉𝑛∎𝔡

𝔐 → 𝔉∎𝔡𝔐.                                                                            ⌂ 
 

Lemma 4.6 Let lim
𝑛→∞

𝔉𝑛 = 𝔉 in  𝔖+(ℝ), (𝜂𝑛
𝔐) ∈ ∆+

𝔐 then 

 𝔉𝑛 ∎𝜂𝑛
𝔐 →  𝔉 in 𝔖+(ℝ).  

Proof.Let (𝜂𝑛) ∈ Δ+, 𝜂𝑛
𝔐(𝜉) → 𝜉 is uniformly on compact subsets of ℝ+. 

Hence  

|𝜉
𝑘

𝐷𝜉
𝑚(𝔉𝑛∎𝜂𝑛

𝔐 − 𝔉)(𝜉)| = |𝜉
𝑘

𝐷𝜉
𝑚 (

1

𝜉
𝔉𝑛 (𝜉)𝜂𝑛

𝔐(𝜉)) − 𝔉(𝜉)| 

→ |𝜉
𝑘

𝐷𝜉
𝑚(𝔉𝑛 − 𝔉)(𝜉)|    as      𝑛 → ∞ 

Thus |𝜉
𝑘

𝐷𝜉
𝑚(𝔉𝑛 ∎𝜂𝑛

𝔐 − 𝔉)(𝜉)| → 0  as 𝑛 → ∞.  



 An Reckon of Mahgoub Transform for Boehmians  
 

57 

 

This yield      𝔉𝑛 ∎𝜂𝑛
𝔐 →  𝔉 in𝔖+(ℝ).                                                          ⌂ 

 

Lemma 4.7 Define𝔉 → [
𝔉∎𝜂𝑛

𝔐

𝜂𝑛
𝔐 ] is a continuous mapping which is 

embedding from   𝔖+(ℝ)  into   𝔹(𝔛
𝔐).                                                             (4.4) 

Proof. Let 
𝔉∎𝜂𝑛

𝔐

𝜂𝑛
𝔐 is a quotient of sequences where𝔉 ∈ 𝔖+(ℝ), 𝜂𝑛

𝔐 ∈ ∆+
𝔐. We 

have (𝔉∎𝜂𝑛
𝔐)∎𝜂𝑚

𝔐 = 𝔉∎(𝜂𝑚
𝔐∎𝜂𝑛

𝔐).We show that the map (4.3) is  
one - to - one.  

Let[
𝔉1∎𝜂𝑛

𝔐

𝜂𝑛
𝔐 ] = [

𝔉2∎𝜏𝑛
𝔐

𝜏𝑛
𝔐 ], then(𝔉1∎𝜂𝑛

𝔐)∎𝜏𝑚
𝔐 = (𝔉2∎𝜏𝑚

𝔐)∎𝜂𝑛
𝔐 , 𝑚, 𝑛 ∈ 𝑁. 

Now using of Lemma 4.2& 4.3, we get 𝔉1 = 𝔉2.  
To establish the continuity of Eq.(4.4), let 𝔉𝑛 → 0as 𝑛 → ∞in 𝔖+(ℝ). Then 

𝔉𝑛 ∎𝜂𝑛
𝔐 → 0  as 𝑛 → ∞by Lemma 4.6, and hence 

[
𝔉∎𝜂𝑛

𝔐

𝜂𝑛
𝔐 ] → 0,as   𝑛 → ∞  in  𝔹(𝔛

𝔐).                                                    ⌂ 

 

 

5. The Mahgoub transform of Bohemians  

Let ℌ = [
𝔉𝑛

𝜂𝑛
] ∈ 𝔹(𝔛), then we delineate the Mahgoub transform of  ℌ by 

the relation  

ℌ1
𝔐 =  [

𝔉𝑛
𝔐

𝜂𝑛
𝔐    ]in  𝔹(𝔛

𝔐).                                                      (5.1) 

 

Theorem 5.1 ℌ1
𝔐: 𝔹(𝔛) → 𝔹(𝔛𝔐) is well defined. 

Proof. Let ℌ1 = ℌ2 ∈ 𝔹(𝔛), where ℌ1 = [
𝔉𝑛

𝜂𝑛
] , ℌ2 = [

𝑔𝑛

𝜏𝑛
]Then the concept 

of quotients yields  𝔉𝑛 ⋆ 𝜏𝑚 =  𝑔𝑚 ⋆ 𝜂𝑛 . Applying Theorem 3.9, we get 
1

𝜉
𝔉𝑛

𝔐(𝜉)𝜏𝑚
𝔐(𝜉) =

1

𝜉
𝑔𝑚

𝔐(𝜉)𝜂𝑛
𝔐(𝜉), 

𝑖. 𝑒. 𝔉𝑛
𝔐∎𝜏𝑚

𝔐 = 𝑔𝑚
𝔐∎𝜂𝑛

𝔐 ⇒
𝑓𝑛

𝔐

𝜂𝑛
𝔐 ~

𝑔𝑛
𝔐

𝜏𝑛
𝔐 . Thus ℌ1

𝔐 = ℌ2
𝔐.                            ⌂ 

 

Theorem 5.2  ℌ𝔐: 𝔹(𝔛) → 𝔹(𝔛𝔐) is continuous regards to 𝛿-convergence. 

Proof. Let ℌ𝑛 → 0 in 𝔹(𝔛)as 𝑛 → ∞. using [4] we get,ℌ𝑛 = [
𝔉𝑛,𝓀

𝜂𝓀
] 

and 𝔉𝑛,𝓀 → 0 in 𝔖+(ℝ) as 𝑛 → ∞ in 𝔖+(ℝ). Now we apply the Mahgoub 

transform to both sides revenue   𝔉𝑛,𝑘
𝔐 →  0  as 𝑛 → ∞. Hence  

ℌ𝑛
𝔐 = [

𝔉𝑛,𝓀
𝔐

𝜂𝓀
𝔐 ] → 0   as 𝑛 → ∞  in  𝔹(𝔛

𝔐).                            ⌂ 

 

 



Yogesh Khandelwal and Priti Chaudhary 

58 

 

Theorem 5.3   ℌ𝔐 ∶ 𝔹(𝔛) → 𝔹(𝔛𝔐)is  one-to-one mapping. 

Proof. Let  ℌ1
𝔐 = [

𝔉𝑛
𝔐

𝜂𝑛
𝔐] = [

𝑔𝑛
𝔐

𝜏𝑛
𝔐 ] = ℌ2

𝔐,then 𝔉𝑛
𝔐∎𝜏𝑚

𝔐 = 𝑔𝑚
𝔐∎𝜂𝑛

𝔐 .  

Hence   

(𝔉𝑛 ⋆ 𝜏𝑚)
𝔐 = (𝑔𝑚 ⋆ 𝜂𝑛 )

𝔐. 
Since the Mahgoub transform is one to one, we get 𝔉𝑛 ⋆ 𝜏𝑚 = 𝑔𝑚 ⋆

𝜂𝑛 .Thus  
𝔉𝑛
𝜂𝑛

~
𝑔𝑛
𝜏𝑛

 . 

Hence [
𝔉𝑛

𝜂𝑛
] = ℌ1 = [

𝑔𝑛

𝜏𝑛
] = ℌ2.                                                                 ⌂ 

 

Theorem 5.4 Let ℌ1, ℌ2 ∈ 𝔹(𝔛), then  
(1) (ℌ1 + ℌ2)

𝔐 = ℌ1
𝔐 + ℌ2

𝔐; 
(2) (𝓀ℌ)𝔐 = 𝓀 ℌ𝔐 , 𝓀 ∈ ℂ . 

 

Theorem5.5 ℌ𝔐 ∶ 𝔹(𝔛) →  𝔹(𝔛𝔐) is continuous regards to ∆+ -
convergence. 

Proof. Let ℌ𝑛
∆
→ ℌ  in   𝔹(𝔛) as  𝑛 → ∞ Then ∃𝔉𝑛 → 0 𝔖+(ℝ) and  (𝜂𝑛) ∈

∆+  such that (ℌ𝑛 − ℌ) ∗ 𝜂𝑛 = [
𝔉𝑛∗𝜂𝓀

𝜂𝓀
] and 𝔉𝑛 → 0 as 𝑛 → ∞.Applying in 

Eq.(5.1) , we get  

𝔐((ℌ𝑛 − ℌ) ∗ 𝜂𝑛 ) = [
𝔐(𝔉𝑛 ∗ 𝜂𝓀)

𝜂𝓀
𝔐

]. 

Hence we have 𝔐((ℌ𝑛 − ℌ) ∗ 𝜂𝑛 ) =  [
𝔉𝑛

𝔐𝜂𝓀
𝔐

𝜉𝜂𝓀
𝔐 ] → 0 as 𝑛 → ∞ in 

𝔹(𝔛𝔐). therefore   

𝔐((ℌ𝑛 − ℌ) ∗ 𝜂𝑛 ) =
1

𝜉
(ℌ𝑛

𝔐 − ℌ𝔐)𝜂𝑛
𝔐 → 0 as 𝑛 → ∞.             ⌂ 

 

Theorem5.6  Let ℌ𝔐 ∶ 𝔹(𝔛) → 𝔹(𝔛𝔐) is onto. 

Proof. Let  [
𝔉𝑛

𝔐

𝜂𝑛
𝔐] ∈ 𝔹(𝔛

𝔐) be arbitrary then 𝔉𝑛
𝔐 ∎𝜂𝑚

𝔐 = 𝔉𝑚
𝔐 ∎𝜂𝑛

𝔐  for each 

𝑚, 𝑛 ∈ 𝑁.Then 

 𝔉𝑛 ⋆ 𝜂𝑚 = 𝔉𝑚 ⋆ 𝜂𝑛.That is, 
𝔉𝑛  

𝜂𝑛 
is the corresponding quotient of sequences 

of  
𝔉𝑛

𝔐

𝜂𝑛
𝔐.   Thus    

𝔉𝑛

𝜂𝑛
∈ 𝔹(𝔛)  is such that 𝔐 [

𝔉𝑛

𝜂𝑛
] = [

𝔉𝑛
𝔐

𝜂𝑛
𝔐 ]    in   𝔹(𝔛

𝔐). 

Let  ℌ𝔐 = [
𝔉𝑛

𝔐

𝜂𝑛
𝔐] ∈ 𝔹(𝔛

𝔐), then we express the inverse of Mahgoub 

transform of  ℌ𝔐 given by 



 An Reckon of Mahgoub Transform for Boehmians  
 

59 

 

ℌ𝔐
−1

=  [
𝔉𝑛

𝜂𝑛
] in the space 𝔹(𝔛).                                                              ⌂ 

 

Theorem5.7 Let [
𝔉𝑛

𝔐

𝜂𝑛
𝔐] ∈ 𝔹(𝔛

𝔐) and   𝔡𝔐 ∈ 𝒞0+
∞𝔐 (ℝ), 𝔡 ∈ 𝒞0+

∞  (ℝ).  

ℌ ([
𝔉𝑛

𝜂𝑛
] ⋆ 𝔡) = [

𝔉𝑛
𝔐

𝜂𝑛
𝔐] ∎𝔡

𝔐andℌ𝔐
−1

([
𝔉𝑛

𝔐

𝜂𝑛
𝔐] ∎𝔡

𝔐) =  [
𝔉𝑛

𝜂𝑛
] ⋆ 𝔡. 

We can easily proof from the definitions. 

 

Conflict of Interests 

The authors declare that there is no conflict of interests regarding the 

publication of this paper. 

 

 

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60 

 

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