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Article

Exchange Formulae for the Stieltjes–Poisson Transform over Weighted Lebesgue Spaces

by
Hari M. Srivastava
1,2,3,4,5,6,
Emilio Ramón Negrín
7,8 and
Jeetendrasingh Maan
9,*
1
Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8P 5C2, Canada
2
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 406040, Taiwan
3
Center for Converging Humanities, Kyung Hee University, Seoul 02447, Republic of Korea
4
Department of Applied Mathematics, Chung Yuan Christian University, Taoyuan City 320314, Taiwan
5
Department of Mathematics and Informatics, Azerbaijan University, Baku 1007, Azerbaijan
6
Section of Mathematics, International Telematic University Uninettuno, 00186 Roma, Italy
7
Departamento de Análisis Matemático, Universidad de La Laguna (ULL), ES-38271 La Laguna, Spain
8
Instituto de Matemáticas y Aplicaciones (IMAULL), ULL Campus de Anchieta, Universidad de La Laguna (ULL), ES-38271 La Laguna, Spain
9
Department of Mathematics and Scientific Computing, National Institute of Technology, Hamirpur 177005, India
*
Author to whom correspondence should be addressed.
Axioms 2024, 13(11), 748; https://doi.org/10.3390/axioms13110748
Submission received: 4 September 2024 / Revised: 19 October 2024 / Accepted: 27 October 2024 / Published: 30 October 2024
(This article belongs to the Special Issue Advanced Approximation Techniques and Their Applications, 2nd Edition)

Abstract

:
This paper aims to develop exchange formulae for the Stieltjes–Poisson transform by using Mellin-type convolutions in the context of weighted Lebesgue spaces. A key result is the introduction of bilinear and continuous Mellin-type convolutions, expanding the scope of the analysis to include the space of weighted L 1 functions and the space of continuous functions vanishing at infinity.

1. Introduction and Preliminaries

The Stieltjes transform is represented by the equation
f ( x ) = R + d α ( t ) x + t = lim T + 0 T d α ( t ) x + t ,
where x > 0 and R + = ( 0 , ) . We consider that the function α ( t ) is a bounded variation of ( 0 , T ) for any positive value of T and the limit (1) exists. If α ( t ) is the integral of a function ϕ ( t ) , we arrive at the special case
f ( x ) = R + ϕ ( t ) x + t d t , x > 0 .
The Stieltjes transform, as discussed in prior works [1,2,3], results from iterating two Laplace transforms [4]. The characteristics of the integral (2) are intimately connected to those of the Laplace integral, as highlighted in the aforementioned studies. The Stieltjes transform was initially proposed in reference to the semi-infinite interval moment problem by T. S. Stieltjes [5]. Since then, it has been investigated and found to be useful in a number of areas, including operator theory, probability, continuous fractions, engineering, mathematical physics, image processing and signal processing, to name a few. On the other hand, several aspects of the generalized Stieltjes transform were investigated in [6,7,8] and in other relevant publications on the topic. The Stieltjes transform and its generalization have been applied to transforms of distributions in addition to classical functions [9,10]. The Stieltjes transform analysis for Boehmians over specific generalized function spaces was extensively studied in [11]. Recent research on the generalized Stieltjes transform’s properties over weighted Lebesgue spaces and compact support distributions may be found in [12]. A case dealing with exchange formulae for the generalized Stieltjes transform was analyzed by the authors of [13].
Widder [14] offered a comprehensive explanation of the integral transform
f ( x ) = R + t ϕ ( t ) x 2 + t 2 d t , x > 0 ,
where, through an exponential change in the variable, the expression transforms into a convolution transform with a kernel belonging to the general class discussed by Hirschman and Widder [15]. Notably, the kernel in the transform (3) is identical to the one found in the Poisson integral, which represents a harmonic function in a half-plane, as noted by Widder. He derived several inversion formulas for (3) based on this connection and subsequently applied his results to the study of harmonic functions.
Definition 1.
Let α , β R with α 0 . For the function f L 1 ( R + , t β α d t ) , the Stieltjes–Poisson transform of f is defined as follows:
( ( S P ) α , β f ) ( x ) = R + t β x α + t α f ( t ) d t , x > 0 .
When α = 1 and β = 0 , the result is the Stieltjes transform
( ( S P ) 1 , 0 f ) ( x ) = R + 1 x + t f ( t ) d t , x > 0 ,
studied by Widder [3] and Goldberg [16] (amongst others).
Additionally, when α = 2 and β = 1 , the result is the Poisson transform
( ( S P ) 2 , 1 f ) ( x ) = R + t x 2 + t 2 f ( t ) d t , x > 0 ,
studied in [14,16,17,18] (amongst others).
In this paper, we explore the Stieltjes–Poisson transform applied to the space L 1 ( R + , t β α d t ) focusing on the weighted integrable functions defined on R + .
Standard norms and notations:
  • The norm defined as f = ess sup x R + | f ( x ) | characterizes the space L ( R + ) , which consists of measurable functions that are essentially bounded on the positive real numbers R + .
  • The expression f 1 , t δ d t = R + | f ( t ) | t δ d t defines a norm on the space L 1 ( R + , t δ d t ) . This space consists of functions that are integrable with respect to the weight t δ on the positive real numbers R + , where δ R .
  • The notation C c ( R + ) represents the vector space of all continuous complex-valued functions on R + that have compact support.
  • The notation C 0 ( R + ) represents the vector space of all continuous complex-valued functions on R + vanishing at infinity, provided with the norm f = sup x R + | f ( x ) | .
Definition 2.
For a fixed γ R , we consider the next products
( f γ ψ ) ( t ) = R + f t s ψ ( s ) s γ d s , t > 0 ,
and
( f γ ψ ) ( t ) = R + f s ψ ( t s ) s γ d s , t > 0 ,
provided that these integrals are well defined.
We note that 1 = represents the standard Mellin convolution [19], while 0 = corresponds to the second kind of Mellin-type convolution [7] [(7.4.4), p. 218].
This paper is organized into three sections. Section 1 introduces the Stieltjes–Poisson transform, starting with several definitions and notations that will be referenced later. Section 2 examines the properties of the product γ over the space of weighted L 1 functions and the space of continuous functions vanishing at infinity and an exchange formula for the Stieltjes–Poisson transform using γ . Finally, Section 3 addresses the properties of the product γ over the space of weighted L 1 functions and the space of continuous functions vanishing at infinity and exchange formulae for the Stieltjes–Poisson transform via γ .

2. Properties of the Mellin-Type Convolution γ and Exchange Formula

This section investigates the properties of the product γ within the space of weighted L 1 functions and continuous functions that vanish at infinity, along with an exchange formula for the Stieltjes–Poisson transform using γ .
Lemma 1.
Let δ R . For a function f L 1 ( R + , t δ d t ) and a function ψ C c ( R + ) , it follows that f γ ψ L 1 ( R + , t δ d t ) and
f γ ψ 1 , t δ d t f 1 , t δ d t R + | ψ ( s ) | s γ + δ + 1 d s < .
Proof. 
We initially note that the integral mentioned above is valid for the given functions f and ψ . Additionally, we observe that if C = R + | ψ ( s ) | s γ + δ + 1 d s , then C < , as ψ C c ( R + ) . Therefore,
f γ ψ 1 , t δ d t = R + | ( f γ ψ ) ( t ) | t δ d t R + R + f t s ψ ( s ) s γ d s t δ d t R + R + f t s t δ d t | ψ ( s ) | s γ d s ( according to Fubini s theorem ) = R + R + | f ( u ) | u δ d u | ψ ( s ) | s γ + δ + 1 d s ( by making a substitution of variables u = t s ) = f 1 , t δ d t R + | ψ ( s ) | s γ + δ + 1 d s = C f 1 , t δ d t < .
This concludes the proof of Lemma 1. □
Remark 1.
Lemma 1 yielding to the Mellin-type convolution γ is a continuous operation in a specific sense. In fact, let δ R . If the sequence ( f n ) converges to f in L 1 ( R + , t δ d t ) as n and ψ C c ( R + ) , it follows that the sequence ( f n γ ψ ) converges to f γ ψ in L 1 ( R + , t δ d t ) as n .
The Mellin-type convolution γ is a bilinear operation in the following sense:
  • if f , g L 1 ( R + , t δ d t ) , ϕ , ψ C c ( R + ) and α , β C , then
( α f + β g ) γ ϕ = α ( f γ ϕ ) + β ( g γ ϕ )
and
f γ ( α ϕ + β ψ ) = α ( f γ ϕ ) + β ( f γ ψ ) .
Also, one obtains the following:
Proposition 1.
Let δ R . For a function f L 1 ( R + , t δ d t ) , and for functions ϕ , ψ C c ( R + ) , we have
(i)
( f γ ϕ ) γ ψ = f γ ( ϕ ψ ) ,
(ii)
f γ ( ϕ γ ψ ) = ( f γ ϕ ) 2 γ + 1 ψ .
Proof. 
( i ) Note that the left-hand side of Equation (5) is valid according to Lemma 1. Similarly, the right-hand side of (5) is also valid, as established by Lemma 1, and because ϕ ψ C c ( R + ) .
Now, for t > 0 ,
( ( f γ ϕ ) γ ψ ) ( t ) = R + ( f γ ϕ ) t s ψ ( s ) s γ d s = R + R + f t s u ϕ ( u ) u γ d u ψ ( s ) s γ d s = R + R + f t v ϕ v s v γ s γ d v s ψ ( s ) s γ d s ( by making a substitution of variables v = u s ) = R + f t v R + ϕ v s ψ ( s ) d s s v γ d v ( according to Fubini s theorem ) = R + f t v ( ϕ ψ ) ( v ) v γ d v = ( f γ ( ϕ ψ ) ) ( t ) .
(ii) Note that, based on Lemma 1, and given that ϕ γ ψ C c ( R + ) , the left-hand side of Equation (6) is well-defined. Similarly, from Lemma 1, we can confirm that the right-hand side of Equation (6) is also well-defined. Now, for t > 0 ,
( f γ ( ϕ γ ψ ) ) ( t ) = R + f t s ( ϕ γ ψ ) ( s ) s γ d s = R + f t s R + ϕ s u ψ ( u ) u γ d u s γ d s = R + ψ ( u ) R + f t s ϕ s u s γ d s u γ d u ( according to Fubini s theorem ) = R + ψ ( u ) R + f t v u ϕ ( v ) v γ u γ u d v u γ d u ( by making a substitution of variables v = s u ) = R + ψ ( u ) ( f γ ϕ ) t u u 2 γ + 1 d u = ( ( f γ ϕ ) 2 γ + 1 ψ ) ( t ) .
This concludes the proof of Proposition 1. □
The exchange formula for the Stieltjes–Poisson transform using γ is given as follows
Theorem 1.
(Exchange formula). Let α , β R , α 0 . If f L 1 ( R + , t β α d t ) , ψ C c ( R + ) then
( S P ) α , β ( f γ ψ ) = ( ( S P ) α , β f ) γ + β α + 1 ψ .
Proof. 
First, based on Lemma 1, it follows that f γ ψ L 1 ( R + , t β α d t ) . Therefore, the left-hand side of (7) is well defined. Additionally, referencing [20] (Theorem 2.2), we obtain the inequality ( S P ) α , β f f 1 , t β α d t . Furthermore, for every x > 0
( ( S P ) α , β f ) γ + β α + 1 ψ ( x ) = R + ( ( S P ) α , β f ) x s ψ ( s ) s γ + β α + 1 d s ( S P ) α , β f R + | ψ ( s ) | s γ + β α + 1 d s f 1 , t β α d t R + | ψ ( s ) | s γ + β α + 1 d s < ,
then the right-hand side of (7) makes sense.
Now, for x > 0 , we have
( S P ) α , β ( f γ ψ ) ( x ) = R + t β x α + t α ( f γ ψ ) ( t ) d t = R + t β x α + t α 0 f t s ψ ( s ) s γ d s d t = R + R + t β x α + t α f t s d t ψ ( s ) s γ d s ( according to Fubini s theorem ) = R + R + ( u s ) β x α + ( u s ) α f ( u ) s d u ψ ( s ) s γ d s ( by making a substitution of variables u = t s ) = R + R + ( u ) β x s α + u α f ( u ) d u ψ ( s ) s γ + β α + 1 d s = R + ( ( S P ) α , β f ) x s ψ ( s ) s γ + β α + 1 d s = ( ( S P ) α , β f ) γ + β α + 1 ψ ( x ) .
This concludes the proof of Theorem 1. □
From this Theorem and taking β = α 1 , α 0 , one obtains
Corollary 1.
Let α 0 . If f L 1 ( R + , t 1 d t ) and ψ C c ( R + ) , then
(i)
( S P ) α , α 1 ( f γ ψ ) = ( ( S P ) α , α 1 f ) γ ψ ,
(ii)
( S P ) α , α 1 ( f ψ ) = ( ( S P ) α , α 1 f ) ψ .
Remark 2.
Observe that the Stieltjes transform ( S P ) 1 , 0 and the Poisson transform ( S P ) 2 , 1 are particular cases of the transform ( S P ) α , α 1 considered in Corollary 1.
The next result holds as follows:
Lemma 2.
If f C 0 ( R + ) and ψ C c ( R + ) then f γ ψ C 0 ( R + ) and f γ ψ f R + | ψ ( s ) | s γ d s < .
Proof. 
We find a closed and bounded interval [ a , b ] R + such that the support of ψ [ a , b ] . Let t n t as n in R + . As f is a continuous function on R + , we have
f ( t n s ) ψ ( s ) s γ f ( t s ) ψ ( s ) s γ for each s R + .
Since
f ( t n s ) ψ ( s ) s γ f ψ ( s ) s γ
and ψ ( s ) s γ is an integrable function on R + , we can apply the dominated convergence theorem, and hence
lim n ( f γ ψ ) ( t n ) = lim n R + f ( t n s ) ψ ( s ) s γ d s = R + f ( t s ) ψ ( s ) s γ d s = ( f γ ψ ) ( t ) .
Thus, f γ ψ is a continuous function on R + .
Let ϵ > 0 be arbitrary. Since f vanishes at infinity, we can find a closed and bounded interval [ c , d ] R + such that
| f ( u ) | ϵ C + 1 , for all u [ c , d ] ,
where C = R + | ψ ( s ) | s γ d s .
Since t s [ c , d ] , s [ a , b ] then t [ a c , b d ] , we find that if t [ a c , b d ] and s [ a , b ] then t s [ c , d ] . Then, | f ( t s ) |   ϵ C + 1 , for t [ a c , b d ] . Thus, for t [ a c , b d ] , we obtain
| ( f γ ψ ) ( t ) | R + f ( t s ) | ψ ( s ) | s γ d s ϵ C + 1 · C ϵ .
Thus, f γ ψ C 0 ( R + ) .
By direct computation, we obtain
f γ ψ R + f ( t s ) | ψ ( s ) | s γ d s f R + | ψ ( s ) | s γ d s < .
This concludes the proof of Lemma 2. □
Remark 3.
From this Lemma, it can be seen that the Mellin-type convolution γ is a continuous operation in other specific senses different to that obtained in Remark 1. In fact, if the sequence ( f n ) converges to f in C 0 ( R + ) as n and ψ C c ( R + ) , it follows that the sequence ( f n γ ψ ) converges to f γ ψ in C 0 ( R + ) as n .
Observe that from Lemma 2, the Mellin-type convolution γ is a bilinear operation in the following sense:
If f , g C 0 ( R + ) , ϕ , ψ C c ( R + ) and α , β C , then
( α f + β g ) γ ϕ = α ( f γ ϕ ) + β ( g γ ϕ )
and
f γ ( α ϕ + β ψ ) = α ( f γ ϕ ) + β ( f γ ψ ) .
Again using Lemma 2, one can obtain a corresponding result to Proposition 1.
Indeed:
Proposition 2.
For a function f C 0 ( R + ) , and for functions ϕ , ψ C c ( R + ) , we have
(i)
( f γ ϕ ) γ ψ = f γ ( ϕ ψ ) ,
(ii)
f γ ( ϕ γ ψ ) = ( f γ ϕ ) 2 γ + 1 ψ .

3. Properties of the Mellin-Type Convolution γ and Exchange Formulae

This section explores the properties of the product γ in the context of weighted L 1 functions and continuous functions that vanish at infinity, along with the exchange formulae for the Stieltjes–Poisson transform using γ .
Lemma 3.
Let δ R . For a function f L 1 ( R + , t δ d t ) and a function ψ C c ( R + ) , then
(i)
f 1 + 2 δ ψ L 1 ( R + , t δ d t ) and f 1 + 2 δ ψ 1 , t δ d t f 1 , t δ d t R + | ψ ( s ) | s δ d s < ,
(ii)
ψ γ f L 1 ( R + , t δ d t ) and ψ γ f 1 , t δ d t f 1 , t δ d t R + | ψ ( s ) | s γ δ 1 d s < .
Proof. 
(i) We initially note that the integral mentioned above is valid for the given functions f and ψ . Also,
f 1 + 2 δ ψ 1 , t δ d t = R + | ( f 1 + 2 δ ψ ) ( t ) | t δ d t R + R + f s ψ ( t s ) s 1 + 2 δ d s t δ d t R + R + | ψ ( t s ) | t δ d t f s s 1 + 2 δ d s ( according to Fubini s theorem ) = R + R + | ψ ( u ) | u s δ 1 s d u | f ( s ) | s 1 + 2 δ d s ( by making a substitution of variables u = t s ) = R + | ψ ( u ) | u δ d u R + | f ( s ) | s δ d s = C f 1 , t δ d t < , w h e r e C = R + | ψ ( u ) | u δ d u < .
(ii) To ensure the integrity of the integral for the functions f and ψ , we start by noting that the integral is defined under these conditions. Additionally, let C = R + | ψ ( s ) | s γ δ 1 d s . Given that ψ C c ( R + ) , we can conclude that C < . This leads us to the following observations:
ψ γ f 1 , t δ d t = R + | ( ψ γ f ) ( t ) | t δ d t R + R + ψ ( s ) f ( t s ) s γ d s t δ d t R + R + f t s t δ d t | ψ ( s ) | s γ d s ( according to Fubini s theorem ) = R + R + | f ( u ) | u s δ 1 s d u | ψ ( s ) | s γ d s ( by making a substitution of variables u = t s ) = f 1 , t δ d t R + | ψ ( s ) | s γ δ 1 d s = C f 1 , t δ d t < .
This concludes the proof of Lemma 3. □
Remark 4.
Lemma 3 yielding to the Mellin-type convolution γ is a continuous operation in a specific sense. In fact, let δ R . If a sequence ( f n ) converges to f in L 1 ( R + , t δ d t ) as n and ψ C c ( R + ) , then:
(i)
the sequence ( f n 1 + 2 δ ψ ) converges to f 1 + 2 δ ψ in L 1 ( R + , t δ d t ) as n ,
(ii)
the sequence ( ψ γ f n ) converges to ψ γ f in L 1 ( R + , t δ d t ) as n .
The Mellin-type convolution γ is a bilinear operation in the following sense:
  • If f , g L 1 ( R + , t δ d t ) , ϕ , ψ C c ( R + ) and α , β C , then
( α f + β g ) 1 + 2 δ ϕ = α ( f 1 + 2 δ ϕ ) + β ( g 1 + 2 δ ϕ )
and
f 1 + 2 δ ( α ϕ + β ψ ) = α ( f 1 + 2 δ ϕ ) + β ( f 1 + 2 δ ψ ) .
Also,
( α ϕ + β ψ ) γ f = α ( ϕ γ f ) + β ( ψ γ f )
and
ϕ γ ( α f + β g ) = α ( ϕ γ f ) + β ( ϕ γ g ) .
Proposition 3.
Let δ R . For a function f L 1 ( R + , t δ d t ) , and for functions ϕ , ψ C c ( R + ) , we have
(i)
( f 1 + 2 δ ϕ ) 1 + 2 δ ψ = f ( ϕ 1 + 2 δ ψ ) ,
(ii)
f 1 + 2 δ ( ϕ 1 + 2 δ ψ ) = ϕ 1 + 2 δ ( f 1 + 2 δ ψ ) ,
(iii)
( ψ γ ϕ ) γ f = ψ ( ϕ γ f ) ,
(iv)
ψ γ ( ϕ γ f ) = ( ϕ ψ ) γ f .
Proof. 
( i ) Note that, according to Lemma 3(i), the left side of (8) is well defined. Similarly, the right side of (8) is well defined based on Lemma 1 and the fact that ϕ 1 + 2 δ ψ C c ( R + ) . Now, for t > 0
( ( f 1 + 2 δ ϕ ) 1 + 2 δ ψ ) ( t ) = R + ( f 1 + 2 δ ϕ ) s ψ ( t s ) s 1 + 2 δ d s = R + R + f u ϕ ( u s ) u 1 + 2 δ d u ψ ( t s ) s 1 + 2 δ d s = R + f u R + ϕ ( u s ) ψ ( t s ) s 1 + 2 δ d s u 1 + 2 δ d u ( according to Fubini s theorem ) = R + f u R + ϕ ( v ) ψ ( t u v ) v u 1 + 2 δ 1 u d v u 1 + 2 δ d u ( by making a substitution of variables v = u s ) = R + ( ϕ 1 + 2 δ ψ ) t u f ( u ) d u u = ( f ( ϕ 1 + 2 δ ψ ) ) ( t ) .
(ii) Note that, based on Lemma 3(i), and given that ϕ 1 + 2 δ ψ C c ( R + ) , the left-hand side of Equation (9) is well defined. Furthermore, referring to Lemma 3(i) and (ii), we see that the right-hand side of (9) is also well defined. Now, for t > 0
( f 1 + 2 δ ( ϕ 1 + 2 δ ψ ) ) ( t ) = R + f ( s ) ( ϕ 1 + 2 δ ψ ) ( t s ) s 1 + 2 δ d s = R + f ( s ) R + ϕ ( u ) ψ ( t s u ) u 1 + 2 δ d u s 1 + 2 δ d s = R + ϕ ( u ) R + f ( s ) ψ ( t s u ) s 1 + 2 δ d s u 1 + 2 δ d u ( according to Fubini s theorem ) = R + ϕ ( u ) ( f 1 + 2 δ ψ ) ( t u ) u 1 + 2 δ d u = ( ϕ 1 + 2 δ ( f 1 + 2 δ ψ ) ) ( t ) .
(iii) Note that based on Lemma 3(ii) and the fact that ψ γ ϕ C c ( R + ) , the left-hand side of (10) is well defined. Additionally, using Lemma 3(ii) in conjunction with Lemma 1, we can conclude that the right-hand side of (10) is also well defined. Now, consider t > 0
( ( ψ γ ϕ ) γ f ) ( t ) = R + ( ψ γ ϕ ) ( s ) f ( s t ) s γ d s = R + R + ψ ( u ) ϕ ( u s ) u γ d u f ( s t ) s γ d s = R + ψ ( u ) R + ϕ ( u s ) f ( s t ) s γ d s u γ d u ( according to Fubini s theorem ) = R + ψ ( u ) R + ϕ ( v ) f v u t v γ u γ 1 u d v u γ d u ( by making a substitution of variables v = u s ) = R + ψ ( u ) ( ϕ γ f ) t u 1 u d u = ( ψ ( ϕ γ f ) ) ( t ) .
(iv) Note that according to Lemma 3(ii), the left-hand side of Equation (11) is well defined. Similarly, the right-hand side of (11) is also well defined, as indicated by Lemma 3(ii) and the fact that ϕ ψ C c ( R + ) . Now, let us consider t > 0
( ψ γ ( ϕ γ f ) ) ( t ) = R + ψ ( s ) ( ϕ γ f ) ( t s ) s γ d s = R + R + ϕ ( u ) f u t s u γ d u ψ ( s ) s γ d s = R + R + f t v ϕ v s v γ s γ d v s ψ ( s ) s γ d s ( by making a substitution of variables v = u s ) = R + f t v R + ϕ v s ψ ( s ) d s s v γ d v ( according to Fubini s theorem ) = R + f t v ( ϕ ψ ) ( v ) v γ d v = ( ( ϕ ψ ) γ f ) ( t ) .
This concludes the proof of Proposition 3. □
The exchange formulae for the Stieltjes–Poisson transform using γ are given as follows:
Theorem 2.
(Exchange formulae). Let α , β R , α 0 . If f L 1 ( R + , t β α d t ) and ψ C c ( R + ) , then we have the following result:
(i)
( S P ) α , β ( f 1 + 2 ( β α ) ψ ) = f β α ( ( S P ) α , β ψ ) ,
(ii)
( S P ) α , β ( ψ γ f ) = ψ γ + α β 1 ( ( S P ) α , β f ) .
Proof. 
(i) Consider f L 1 ( R + , t β α d t ) and ψ C c ( R + ) . By applying Lemma 3(i), we can deduce that f 1 + 2 ( β α ) ψ L 1 ( R + , t β α d t ) . This confirms the existence of the left-hand side of Equation (12). Furthermore, the right-hand side of (12) also exists, as it holds true for every x > 0
f β α ( ( S P ) α , β ψ ) ( x ) = R + f ( s ) ( ( S P ) α , β f ) x s s β α d s ( S P ) α , β f R + | f ( s ) | s β α d s f 1 , t β α d t 2 ,
using ( S P ) α , β f f 1 , t β α d t [20] (Theorem 2.2).
Now, for x > 0 , we have
( S P ) α , β ( f 1 + 2 ( β α ) ψ ) ( x ) = R + t β x α + t α ( f 1 + 2 ( β α ) ψ ) ( t ) d t = R + t β x α + t α R + f s ψ ( t s ) s 1 + 2 ( β α ) d s d t = R + R + t β x α + t α ψ ( t s ) d t f s s 1 + 2 ( β α ) d s ( according to Fubini s theorem ) = R + R + u s β x α + u s α ψ ( u ) 1 s d u f ( s ) s 1 + 2 ( β α ) d s ( by making a substitution of variables u = t s ) = R + R + u β ( x s ) α + u α ψ ( u ) d u f ( s ) s β α d s = R + ( ( S P ) α , β ψ ) x s f ( s ) s β α d s = f β α ( ( S P ) α , β ψ ) ( x ) .
(ii) Initially, based on Lemma 3(ii), it can be concluded that ψ γ f L 1 ( R + , t δ d t ) . This implies that the left-hand side of (13) is well defined. Additionally, as stated in [20] (Theorem 2.2), we find that ( S P ) α , β f f 1 , t β α d t . Thus, for every x > 0
ψ γ + α β 1 ( ( S P ) α , β f ) ( x ) = R + ψ ( s ) ( ( S P ) α , β f ) x s s γ + α β 1 d s ( S P ) α , β f R + | ψ ( s ) | s γ + α β 1 d s f 1 , t β α d t R + | ψ ( s ) | s γ + α β 1 d s < ,
then the right-hand side of (13) makes sense.
Now, for x > 0 , we have
( S P ) α , β ( ψ γ f ) ( x ) = R + t β x α + t α ( ψ γ f ) ( t ) d t = R + t β x α + t α R + ψ ( s ) f t s s γ d s d t = R + R + t β x α + t α f t s d t ψ ( s ) s γ d s ( according to Fubini s theorem ) = R + R + u s β x α + u s α f ( u ) 1 s d u ψ ( s ) s γ d s ( by making a substitution of variables u = t s ) = R + R + u β ( x s ) α + u α f ( u ) d u ψ ( s ) s γ + α β 1 d s = R + ψ ( s ) ( ( S P ) α , β f ) x s s γ + α β 1 d s = ψ γ + α β 1 ( ( S P ) α , β f ) ( x ) .
This concludes the proof of Theorem 2. □
From this Theorem and taking β = α 1 , α 0 , one obtains
Corollary 2.
Let α 0 . If f L 1 ( R + , t 1 d t ) and ψ C c ( R + ) , then
(i)
( S P ) α , α 1 ( f 1 ψ ) = f 1 ( ( S P ) α , α 1 ψ ) ,
(ii)
( S P ) α , α 1 ( ψ γ f ) = ψ γ ( ( S P ) α , α 1 f ) ,
(iii)
( S P ) α , α 1 ( ψ f ) = ψ ( ( S P ) α , α 1 f ) ,
Remark 5.
Observe that the Stieltjes transform ( S P ) 1 , 0 and the Poisson transform ( S P ) 2 , 1 are particular cases of the transform ( S P ) α , α 1 considered in Corollary 2.
The next result holds as follows:
Lemma 4.
If f C 0 ( R + ) and ψ C c ( R + ) , then
(i)
f 1 ψ C 0 ( R + ) a n d f 1 ψ f R + | ψ ( s ) | s 1 d s < ,
(ii)
ψ γ f C 0 ( R + ) a n d ψ γ f f R + | ψ ( s ) | s γ d s < .
Proof. 
(i) We find a closed and bounded interval [ a , b ] R + such that the support of ψ [ a , b ] . Let t n t as n in R + . As f is a continuous function on R + , we have
f ( s t n ) ψ ( s ) s 1 f ( s t ) ψ ( s ) s 1 f o r e a c h s R + .
Since
f ( s t n ) ψ ( s ) s 1 f ψ ( s ) s 1
and ψ ( s ) s 1 is an integrable function on R + , we can apply the dominated convergence theorem, and hence
lim n ( f 1 ψ ) ( t n ) = lim n R + f ( s ) ψ ( t n s ) s 1 d s = R + f ( s t n ) ψ ( s ) s 1 d s = R + f ( s t ) ψ ( s ) s 1 d s = R + f ( s ) ψ ( s t ) s 1 d s = ( f 1 ψ ) ( t ) .
Thus, f 1 ψ is a continuous function on R + .
Let ϵ > 0 be arbitrary. Since f vanishes at infinity, we can find a closed and bounded interval [ c , d ] R + such that
| f ( u ) | ϵ C + 1 , for all u [ c , d ] ,
where C = R + | ψ ( s ) | s 1 d s .
Since s t [ c , d ] , s [ a , b ] then t [ a d , b c ] ; we find that if t [ a d , b c ] and s [ a , b ] then s t [ c , d ] . Then, | f ( s t ) |   ϵ C + 1 , for t [ a d , b c ] . Thus, for t [ a d , b c ] one obtains
| ( f 1 ψ ) ( t ) | R + f ( s t ) | ψ ( s ) | s 1 d s ϵ C + 1 · C ϵ .
Thus, f 1 ψ C 0 ( R + ) .
By direct computation, we obtain
f 1 ψ R + f ( s t ) | ψ ( s ) | s 1 d s f R + | ψ ( s ) | s 1 d s < .
Therefore, (i) holds.
(ii) We find a closed and bounded interval [ a , b ] R + such that the support of ψ [ a , b ] . Let t n t as n in R + . As f is a continuous function on R + , we have
f ( t n s ) ψ ( s ) s γ f ( t s ) ψ ( s ) s γ f o r e a c h s R + .
Since
f ( t n s ) ψ ( s ) s γ f ψ ( s ) s γ
and ψ ( s ) s γ is an integrable function on R + , we can apply the dominated convergence theorem, and hence
lim n ( ψ γ f ) ( t n ) = lim n R + ψ ( s ) f ( t n s ) s γ d s = R + ψ ( s ) f ( s t ) s γ d s = ( ψ γ f ) ( t ) .
Thus, ψ γ f is a continuous function on R + .
Let ϵ > 0 be arbitrary. Since f vanishes at infinity, we can find a closed and bounded interval [ c , d ] R + such that
| f ( u ) | ϵ C + 1 , for all u [ c , d ] ,
where C = R + | ψ ( s ) | s γ d s .
Since t s [ c , d ] , s [ a , b ] then t [ c b , d a ] , we find that if t [ c b , d a ] and s [ a , b ] then t s [ c , d ] . Then, | f ( s t ) |   ϵ C + 1 , for t [ c b , d a ] . Thus, for t [ c b , d a ] we obtain
| ( ψ γ f ) ( t ) | R + | ψ ( s ) | f ( t s ) s γ d s ϵ C + 1 · C ϵ .
Thus, ψ γ f C 0 ( R + ) .
By direct computation, we obtain
ψ γ f R + | ψ ( s ) | f ( t s ) s γ d s f R + | ψ ( s ) | s γ d s < .
Therefore, ( ii ) holds.
This concludes the proof of Lemma 4. □
Remark 6.
From this Lemma, one can see that the Mellin-type convolution γ is a continuous operation in another specific sense different to that obtained in Remark 4. In fact, if the sequence ( f n ) converges to f in C 0 ( R + ) as n and ψ C c ( R + ) , then:
(i)
the sequence ( f n 1 ψ ) converges to f 1 ψ in C 0 ( R + ) as n ,
(ii)
the sequence ( ψ γ f n ) converges to ψ γ f in C 0 ( R + ) as n .
Observe that from Lemma 4, the Mellin-type convolution denoted as γ represents a bilinear operation characterized by the following properties:
If f , g C 0 ( R + ) , ϕ , ψ C c ( R + ) and α , β C , then
( α f + β g ) 1 ϕ = α ( f 1 ϕ ) + β ( g 1 ϕ )
and
f 1 ( α ϕ + β ψ ) = α ( f 1 ϕ ) + β ( f 1 ψ ) .
Also,
( α ϕ + β ψ ) γ f = α ( ϕ γ f ) + β ( ψ γ f )
and
ϕ γ ( α f + β g ) = α ( ϕ γ f ) + β ( ϕ γ g ) .
Also, again using Lemma 4, one obtains a corresponding result to Proposition 3.
Indeed
Proposition 4.
For a function f C 0 ( R + ) , and for functions ϕ , ψ C c ( R + ) , we have
(i)
( f 1 ϕ ) 1 ψ = f ( ϕ 1 ψ ) ,
(ii)
f 1 ( ϕ 1 ψ ) = ϕ 1 ( f 1 ψ ) ,
(iii)
( ψ γ ϕ ) γ f = ψ ( ϕ γ f ) ,
(iv)
ψ γ ( ϕ γ f ) = ( ϕ ψ ) γ f .

4. Final Observations and Conclusions

This paper develops exchange formulae for the Stieltjes–Poisson transform using Mellin-type convolutions in weighted Lebesgue spaces. A key contribution of this paper is the inclusion of a more general Mellin-type convolution product, which is bilinear and continuous. While a broader exchange formula was proven in [21], this paper provides novel insights, particularly with the incorporation of the space of continuous functions vanishing at infinity. This inclusion enlarges the applicability of the results, offering a more versatile framework for understanding the Stieltjes-Poisson transform.
In conclusion, this research expands upon existing work by providing new exchange formulae for the Stieltjes–Poisson transform and extending the function spaces considered, thereby opening new avenues for further exploration and application in the field of harmonic analysis and related areas. The results obtained in this paper can be used as a foundation for future studies on the interaction between Mellin-type convolutions and other integral transforms in weighted function spaces.

Author Contributions

Methodology, E.R.N.; Validation, E.R.N. and J.M.; Formal analysis, E.R.N. and J.M.; Investigation, J.M.; Writing—original draft, J.M.; Writing—review & editing, H.M.S. and E.R.N.; Supervision, H.M.S. and E.R.N. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflicts of interest.

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Srivastava, H.M.; Negrín, E.R.; Maan, J. Exchange Formulae for the Stieltjes–Poisson Transform over Weighted Lebesgue Spaces. Axioms 2024, 13, 748. https://doi.org/10.3390/axioms13110748

AMA Style

Srivastava HM, Negrín ER, Maan J. Exchange Formulae for the Stieltjes–Poisson Transform over Weighted Lebesgue Spaces. Axioms. 2024; 13(11):748. https://doi.org/10.3390/axioms13110748

Chicago/Turabian Style

Srivastava, Hari M., Emilio Ramón Negrín, and Jeetendrasingh Maan. 2024. "Exchange Formulae for the Stieltjes–Poisson Transform over Weighted Lebesgue Spaces" Axioms 13, no. 11: 748. https://doi.org/10.3390/axioms13110748

APA Style

Srivastava, H. M., Negrín, E. R., & Maan, J. (2024). Exchange Formulae for the Stieltjes–Poisson Transform over Weighted Lebesgue Spaces. Axioms, 13(11), 748. https://doi.org/10.3390/axioms13110748

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