1 Introduction

The Hartley transform is a linear operator defined for a suitable function \(\psi (x)\) as follows:

$$\begin{aligned} (\mathscr {H}\psi )(\lambda ) = \frac{1}{\sqrt{2\pi }} \int _{\mathbb {R}} \psi (x) \;\textrm{cas}(\lambda x) dx, \end{aligned}$$
(1.1)

where \(\textrm{cas}(x)\) is the cas function, defined as:

$$\begin{aligned} \textrm{cas}(x) = \sum _{n=0}^\infty \frac{(-1)^{\left( {\begin{array}{c}n+1\\ 2\end{array}}\right) }}{n!}x^n, \end{aligned}$$
(1.2)

with \(\left( {\begin{array}{c}n\\ 2\end{array}}\right) =\frac{n(n-1)}{2}\) being the binomial coefficient. The \(\textrm{cas}\) function (1.2) can be seen as a generalization of the exponential function \(\exp \). This becomes evident when considering that the cas function is the unique \(C^\infty \)-solution to the differential-reflection problem [1]:

$$\begin{aligned} \left\{ \begin{array}{l l} R\partial _x u(x) = \lambda u(x), \\ u(0) = 0. \end{array} \right. \end{aligned}$$
(1.3)

Here, \(\partial _x\) represents the first-order derivative, and R is the reflection operator acting on functions f(x) as:

$$\begin{aligned} (Rf)(x) = f(-x). \end{aligned}$$
(1.4)

The Hartley transform shares several essential properties with the Fourier transform, including linearity, invertibility, and Parseval’s identity. These transforms find extensive applications across various fields of mathematics, physics, and engineering, such as signal processing, data analysis, and number theory [2,3,4,5,6,7].

Building upon the foundational work presented in [1], it was revealed that the harmonic oscillator, traditionally analyzed through the Fourier transform, can seamlessly extend its reach into the domain of supersymmetric quantum mechanics by embracing the Hartley transform. While both transforms are valuable tools, the Hartley transform offers distinct advantages, such as being real-valued and symmetric, making it particularly well-suited for specific tasks like image processing. Nevertheless, the Fourier transform enjoys more widespread usage, boasts a more elaborate theoretical framework, and finds application in a broader range of scenarios.

In the existing body of literature, several generalizations of the operator \(R\partial _x\) have been thoroughly investigated. The most comprehensive self-adjoint first-order differential Dunkl operator, which preserves the space of polynomials of a specific degree, was originally introduced by Vinet and Zhedanov in reference [8]. This operator is expressed as follows:

$$\begin{aligned} L^{(\alpha ,\beta ,c)}=\frac{2(x-1)(x+c)}{x} \partial _x R+ \frac{(\alpha +\beta +1)x^2 +(c\alpha -\beta )x + c}{x^{2}}(R-I). \end{aligned}$$
(1.5)

A related operator of interest, presented in [9, 10], is a limiting case derived from \(L^{\alpha ,\beta ,c}\). This case introduces a one-parameter extension of the Hartley transform by considering the following first-order difference-differential operator:

$$\begin{aligned} \mathscr {L}_\alpha = R\left( \partial _x + \frac{\alpha }{x}\right) + \frac{\alpha }{x}, \quad \alpha \ge 0. \end{aligned}$$
(1.6)

In [9], it was demonstrated that for \(\lambda \in \mathbb {C}\), the problem stated as:

$$\begin{aligned} \left\{ \begin{array}{l l} \mathscr {L}_\alpha y(x) = \lambda y(x) \\ y(0) = 1, \end{array} \right. \end{aligned}$$
(1.7)

possesses a unique analytic solution given by:

$$\begin{aligned} \mathcal {J}_\lambda (x;\alpha ) = \mathscr {J}_{\alpha -1/2}(\lambda x) + \frac{\lambda x}{2\alpha +1} \mathscr {J}_{\alpha +1/2}(\lambda x), \end{aligned}$$
(1.8)

where \(\mathscr {J}_\alpha (x)\) represents the normalized Bessel function, defined by

$$\begin{aligned} \mathscr {J}_\alpha (x)&:=\sum _{n=0}^{\infty }\frac{(-1)^n}{n!(\alpha +1)_n}(\frac{x}{2})^{2n}, \quad \alpha >-1. \end{aligned}$$

Consequently, the function \(\mathcal {J}_\lambda (x;\alpha )\) represents a one-parameter extension of the "cas" function. To emphasize this extension, it is referred to as the Hartley–Bessel function. It is well-known that for \(\alpha > -\frac{1}{2}\), the normalized Bessel function \(\mathscr {J}_\alpha (x)\) satisfies the product formula [11]:

$$\begin{aligned} \mathscr {J}_\alpha (\gamma r)\mathscr {J}_\alpha (\gamma s) = \int _0^\infty \mathscr {J}_\alpha (\gamma t) \omega _\alpha (r,s,t) t^{2\alpha +1} dt, \quad \gamma \in \mathbb {C}, \end{aligned}$$
(1.9)

where the kernel \(w_\alpha (r,s,t)\) is given by:

$$\begin{aligned} w_\alpha (r, s, t) = 2^{2\alpha -2} \frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha ) \Gamma (\frac{1}{2})} \frac{\Delta ^{2\alpha -2}(r, s, t)}{(rst)^{2\alpha -1}}, \end{aligned}$$

with \(\Delta (r, s, t)\) representing the area of a triangle with sides of length r, s, and t, if such a triangle exists; otherwise, \(\Delta (r, s, t)\) is zero. Utilizing Heron’s formula for the area of a triangle [12, §1], we obtain:

$$\begin{aligned} \Delta (r, s, t) = \frac{1}{4} \sqrt{(r + s + t)(r + s - t)(r - s + t)(-r + s + t)}. \end{aligned}$$
(1.10)

The aim of this paper is to establish a product formula for the Hartley–Bessel function \(\mathcal {J}_\lambda (x,\alpha )\), akin to the product formula (1.9) for the normalized Bessel function. Specifically, we establish the product formula in the form:

$$\begin{aligned} \mathcal {J}_\lambda (r;\alpha )\mathcal {J}_\lambda (s;\alpha )=\int _{\mathbb {R}}\mathcal {J}_\lambda (t;\alpha ) \gamma ^{\alpha }_{r,s}(dt). \end{aligned}$$
(1.11)

Here, \(\gamma ^\alpha _{r,s}(dt)\) represents an explicit real-valued measure with compact support on \(\mathbb {R}\), uniformly bounded for r and s within \(\mathbb {R}\) but may not always be positive. The derived product formula is of paramount importance as it provides the basis for defining the corresponding translation operators:

$$\begin{aligned} \tau ^r_\alpha f(s):=\int _{\mathbb {R}}f(t)\gamma ^\alpha _{r,s}(dt). \end{aligned}$$
(1.12)

These translation operators, in turn, serve as the foundation for introducing a novel convolution structure, defined for suitable functions f and g, by

$$\begin{aligned} (f*_\alpha g)(s):=\frac{1}{2^{\alpha -1/2}\Gamma (\alpha +1/2)}\int _{\mathbb {R}}f(t)\tau ^s_\alpha g(t)|t|^{2\alpha }dt. \end{aligned}$$
(1.13)

In the literature, several attempts have been made to extend the product formula (1.9) for the normalized Bessel function. Rosler [12] derived a product formula for Dunkl-Bessel functions, leading to the structure of signed hypergroups in the real line. Similarly, Salen Ben Saïd [13] established a similar formula for \(\alpha \)-deformed Dunkl-Bessel functions. More recently, Boubatra et al. [14] established a product formula for generalized deformed Dunkl-Bessel functions. Additionally, refinements have been achieved by Amir Bechir [15].

In the forthcoming sections:

In Sect. 2, we revisit essential properties of Hartley–Bessel functions, laying the groundwork for our study. Section 3 introduces our main result. In Sect. 4, we explore the properties of the measure \(\gamma _\alpha (dt)\) and conduct a comprehensive examination of the translation operator and convolution product. Section 5 furnishes a detailed proof of our main result.

2 The Hartley–Bessel function

In this section, we will begin by offering a brief overview of the Hartley–Bessel function. We will then take a closer look at the motivation behind the Hartley–Bessel function and explore its derivation, along with some of its properties. This exploration will involve a detailed examination of the limit transition from the \(-1\) little Jacobi polynomials.

Let \(\alpha \ge 0,\) and f, be a differentiable function \(\mathbb {R}.\) The Hartely-Bessel derivative \(\mathscr {L}_\alpha f(x)\) is defined by

$$\begin{aligned} \mathscr {L}_\alpha f (x)=\left\{ \begin{array}{l l} f'(-x)+ \frac{\alpha }{x}\big \{f(x)-f(-x)\big \},\quad \text{ if }\quad x\ne 0,\\ (2\alpha +1) f'(0) \quad \text{ if }\quad x=0.\end{array} \right. \end{aligned}$$
(2.1)

For each \(\lambda \in \mathbb {C}\), the following problem [1]

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\mathscr {L}_\alpha u(x) =\lambda u(x),\\ &{}\displaystyle u(0)=1. \end{array}\right. } \end{aligned}$$

admits a unique \(C^\infty \) solution on \(\mathbb {R}\), denoted by \(\mathcal {J}_\alpha (x)\) given by

$$\begin{aligned} \mathcal {J}_\alpha (x):=\mathscr {J}_{\alpha -1/2}(x)+\frac{x}{2\alpha +1}\mathscr {J}_{\alpha +1/2}(x). \end{aligned}$$
(2.2)

where \(\mathscr {J}_{\alpha }\) the normalized Bessel functions is defined by [16, §4 ]

$$\begin{aligned} \mathscr {J}_\alpha (x)&:=\sum _{n=0}^{\infty }\frac{(-1)^n}{n!(\alpha +1)_n}(\frac{x}{2})^{2n}, \alpha >-1. \end{aligned}$$

2.1 Limit from \(-1\) little Jacobi polynomials

Recall the \(-1\) little Jacobi polynomials \(J_{n}(x;\alpha ,\beta )\), defined as follows

(2.3)

These polynomials were originally introduced by Luc Vinet and Zhedanov in [17]. The \(-1\) little Jacobi polynomials \(J_{n}(x;\alpha ,\beta )\) serve as the eigenfunctions for the following operator:

$$\begin{aligned} Y^{(\alpha ,\beta )} J_{n}(x;\alpha ,\beta ) = \lambda _n J_{n}(x;\alpha ,\beta ), \end{aligned}$$
(2.4)

where \(Y^{(\alpha ,\beta )}\) is defined as:

$$\begin{aligned} Y^{(\alpha ,\beta )} = 2(1-x) \partial _x R + (\alpha +\beta +1 -\alpha x^{-1})(1-R). \end{aligned}$$
(2.5)

The eigenvalues \(\lambda _n\) are determined by the following conditions:

$$\begin{aligned} \lambda _n = {\left\{ \begin{array}{ll} -2n, &{} \text {if }n \text { is even},\\ -(\alpha +\beta +n+1), &{} \text {if }n\text { is odd}. \end{array}\right. } \end{aligned}$$
(2.6)

From equation (2.3), we can easily derive the following limits as n tends to infinity:

$$\begin{aligned} \lim _{n\rightarrow \infty } J_{2n}\left( \frac{\lambda x}{2n},\alpha -\frac{1}{2}, \beta \right)&= \mathcal {J}_{\lambda }(x,\alpha ), \ \lim _{n\rightarrow \infty } J_{2n+1}\left( \frac{\lambda x}{2n},\alpha -\frac{1}{2}, \beta \right) \\&=\mathcal {J}_{\lambda }(-x,\alpha ). \end{aligned}$$

Starting from equation (2.4), we can obtain another equation in the limit as \(n \rightarrow \infty \), which involves the rescaling of variables \(x\rightarrow \lambda x/2n\). This leads to the equation:

$$\begin{aligned} \mathscr {L}_\alpha \mathcal {J}_\lambda (\pm x;\alpha ) = \pm \lambda \mathcal {J}_\lambda (\pm x;\alpha ). \end{aligned}$$
(2.7)

2.2 Sonine’s integral

Observe that for \(\alpha \ge 0\) and \(\lambda \in \mathbb {C}\), we can express the Hartley–Bessel function more succinctly as:

$$\begin{aligned} \mathcal {J}_\lambda (z,\alpha ) = \sum _{n=0}^\infty \frac{(-1)^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }}{2^n\lfloor n/2\rfloor ! (\alpha +\frac{1}{2})_{\lceil n/2\rceil }}\lambda ^n z^n,\quad z\in \mathbb {C}. \end{aligned}$$
(2.8)

Here, \(\lfloor x\rfloor \) denotes the floor function (the greatest integer less than or equal to x), and \(\lceil x \rceil \) denotes the ceiling function (the least integer greater than or equal to x).

We have some straightforward elementary identities involving the Pochhammer symbol and ceiling functions:

$$\begin{aligned} \begin{aligned}&(a)_{\lceil \frac{n+1}{2}\rceil } = a (a+1)_{\lfloor \frac{n}{2}\rfloor },\\&(a)_n = (a)_{\lfloor \frac{n}{2}\rfloor }(a+\lfloor \frac{n}{2}\rfloor )_{\lceil \frac{n}{2}\rceil },\\&(a)_n = (a)_{\lceil \frac{n}{2}\rceil }(a+\lceil \frac{n}{2}\rceil )_{\lfloor \frac{n}{2}\rfloor },\\&\lceil \frac{n+1}{2}\rceil ! = \lceil \frac{n+1}{2}\rceil \lfloor \frac{n}{2}\rfloor !. \end{aligned} \end{aligned}$$

Theorem 2.1

Let \(\alpha ,\beta > 0\) such that \(\beta >\alpha .\) Then

$$\begin{aligned} \mathcal {J}_\lambda (x;\beta )=\frac{\Gamma (\beta +\frac{1}{2})}{\Gamma (\beta -\alpha )\Gamma (\alpha +\frac{1}{2})}\int _{-1}^{1}(1-t)^{\beta -\alpha -1} (1+t)^{\beta -\alpha }\mathcal {J}_\lambda (xt;\alpha ) |t|^{2\alpha }dt. \end{aligned}$$

Proof

Starting from (1.8), we have

$$\begin{aligned} \int _{-1}^{1}(1-t)^{\alpha -\frac{1}{2}} (1+t)^{\alpha +\frac{1}{2}}\mathcal {J}_\lambda (x;\alpha ) |t|^{2\alpha }dt = \sum _{n=0}^{\infty } \frac{(-1)^{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }}{2^n\lfloor n/2\rfloor ! (\alpha +\frac{1}{2})_{\lceil n/2\rceil }}\lambda ^n x^n I_n(\alpha ,\beta ) \end{aligned}$$
(2.9)

where

$$\begin{aligned} I_n(\alpha ,\beta )=\int _{-1}^{1}t^n(1-t)^{\beta -\alpha -1} (1+t)^{\beta -\alpha } |t|^{2\alpha }dt. \end{aligned}$$

Let’s compute the terms \(I_{2n}(\alpha ,\beta )\) and \(I_{2n+1}(\alpha ,\beta )\) using the Beta integral formula:

$$\begin{aligned} \int _{0}^{1}t^{\alpha -1}(1-t)^{\beta -1} dt =\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}, \quad \alpha , \beta >0. \end{aligned}$$

After a straightforward computation, we obtain

$$\begin{aligned} I_n(\alpha ,\beta )&=\frac{\Gamma (\lceil \tfrac{n}{2}\rceil +\alpha +\frac{1}{2})\Gamma (\beta -\alpha )}{\Gamma (\lceil \tfrac{n}{2}\rceil +\beta +\frac{1}{2})}\\&=\frac{\Gamma (\alpha +\frac{1}{2})\Gamma (\beta -\alpha )}{\Gamma (\beta +\frac{1}{2})} \frac{(\alpha +\frac{1}{2})_{\lceil \tfrac{n}{2}\rceil }}{ (\beta +\frac{1}{2})_{\lceil \tfrac{n}{2}\rceil }}. \end{aligned}$$

Inserting this expression into (2.9), we obtain the desired result. \(\square \)

Corollary 2.2

Let \(\alpha >0.\) The function \(\mathcal {J}_\lambda (z;\alpha )\) possesses the Laplace integral representation:

$$\begin{aligned} \mathcal {J}_\lambda (x;\alpha )=\displaystyle \frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha )}\sqrt{\frac{2}{\pi }}\int _{-1}^{1}(1-t)^{\alpha -\frac{1}{2}} (1+t)^{\alpha +\frac{1}{2}}\sin (\lambda xt+\frac{\pi }{4}) dt. \end{aligned}$$

Corollary 2.3

For \(\alpha >0,\) we have for \(\lambda , x \in \mathbb {R}\)

$$|\mathcal {J}_\lambda (x;\alpha )|\le \sqrt{2}.$$

3 Statement of main result

For \(r, s, t>0,\) we define

$$\begin{aligned} \omega _\alpha (r, s, t) {:}{=}2^{2\alpha -2}\frac{\Gamma (\alpha +\frac{1}{2})}{\Gamma (\alpha ) \Gamma (\frac{1}{2})}\frac{\Delta ^{2\alpha -2}(r, s, t)}{(rst)^{2\alpha -1}}, \end{aligned}$$
(3.1)

where \(\Delta (r, s, t)\) is the area of a triangle with sides of length r, s, t if such a triangle exists; otherwise, \(\Delta (r, s, t)\) is zero. Using Heron’s formula (1.10)

$$\begin{aligned} \Delta (r, s, t) = \frac{1}{4}((r + s)^2 - t^2)^{1/2}(t^2-(r - s)^2)^{1/2}. \end{aligned}$$
(3.2)

Substituting (3.2) in (3.1) to get

$$\begin{aligned} \omega _\alpha (r, s, t) = \frac{2\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \frac{\big [(t^2 - (r - s)^2)((r + s)^2 - t^2))\big ] ^{\alpha -1}}{(2rst)^{2\alpha -1}}, \quad |r-s|\le t\le r+s. \end{aligned}$$
(3.3)

Furthermore, we set

$$\begin{aligned} \gamma _\alpha (r, s, t) {:}{=}\frac{1}{2}\omega _\alpha (|r|, |s|, |t|)\sigma (r, s, t) \quad \text {for all } r, s, t \in \mathbb {R}^*, \end{aligned}$$

where \(\sigma (r, s, t)\) is defined as follows:

$$\begin{aligned} \sigma (r, s, t)&= 1 + \frac{r^2 + s^2 - t^2}{2rs} \\&\quad + \frac{t^2 + s^2 - r^2}{2ts} + \frac{r^2 + t^2 - s^2}{2rt} \\&= 2 + \frac{(r + s - t)(t + r - s)(t - r + s)}{2rst}. \end{aligned}$$

The following obvious properties will play an important role in what follows:

  1. (i)

    The mapping \((r,s,t)\rightarrow \sigma (r,s,t)\) is homogeneous of degree 0, that is

    $$\sigma (\lambda r,\lambda s,\lambda t)=\sigma (r,s,t).$$
  2. (ii)

    For \(r,s,t\in \mathbb {R}^*,\) we have

    $$\begin{aligned} |\sigma (r,s,t)|\le 4.\end{aligned}$$
    (3.4)
  3. (iii)

    The kernel \(\gamma _\alpha \) satisfies the following symmetries:

    $$\begin{aligned} \gamma _\alpha (r, s, t) = {\left\{ \begin{array}{ll} \gamma _\alpha (s, r, t), \\ \gamma _\alpha (-r, -s, -t), \\ \gamma _\alpha (-t, s, -r) = \gamma _\alpha (r, -t, -s). \end{array}\right. } \end{aligned}$$
    (3.5)

The following theorem presents our main result, with its proof provided in Sect. 5.

Theorem 3.1

For \(\alpha > 0\), the Hartley–Bessel function \(\mathcal {J}_\lambda (r, \alpha )\) satisfies the following product formula:

$$\begin{aligned} \mathcal {J}_\lambda (r, \alpha )\mathcal {J}_\lambda (s, \alpha ) = \int _{\mathbb {R}}\mathcal {J}_\lambda (t, \alpha ) \gamma _\alpha (r, s, t)|t|^{2\alpha } dt, \quad \text {for } r, s \in \mathbb {R}^*. \end{aligned}$$

Definition 3.2

If f is a Lebesgue measurable locally integrable on \(\mathbb {R},\) we define the generalized translation function \(\tau ^r_{\alpha }f(s)\) by

$$\begin{aligned} \tau ^r_{\alpha }f(s):=\int _{\mathbb {R}}f(t)\gamma ^\alpha _{r,s}(dt), \end{aligned}$$
(3.6)

where the measure \(\gamma ^\alpha _{r,s}(dt)\) is given by

$$\begin{aligned} \gamma ^\alpha _{r,s}(dt) = {\left\{ \begin{array}{ll} \gamma _\alpha (r,s,t) |t|^{2\alpha } dt &{}\text {if }r,s\!\in \!\mathbb {R}^*,\\ d\delta _s(t) &{}\text {if }r\!=\!0,\\ d\delta _r(t) &{}\text {if }s\!=\!0. \end{array}\right. } \end{aligned}$$
(3.7)

The measure \(\gamma _{r,s}^\alpha \) is supported in \(||r| - |s|| \le |t| \le |r| + |s|\). This property can be intuitively understood: when both r and s are non-zero, \(t \in \text {supp}(\gamma ^\alpha _{r,s})\) if and only if |r|, |s|, and |t| can form the sides of a (possibly degenerate) triangle.

4 Convolution structure

We denote by

  • \(C_0(\mathbb {R})\) as the space of continuous functions on \(\mathbb {R}\) that tend to 0 as x approaches infinity.

  • \(S(\mathbb {R})\) as the Schwartz space.

  • \(L^p_{\alpha }(\mathbb {R})\), \(1 \le p \le \infty \), as the space of measurable functions on \(\mathbb {R}\) equipped with the norm

    $$\begin{aligned} \Vert f\Vert _{p,\alpha }= {\left\{ \begin{array}{ll} \left( \int _{\mathbb {R}} |f(t)\mu _{\alpha }(dt)|\right) ^{1/p}, &{} \text {if } 1 \le p < \infty , \\ \text {ess} \sup _{x\in \mathbb {R}} |f(x)|, &{} \text {if } p=\infty . \end{array}\right. } \end{aligned}$$
    (4.1)

    Here,

    $$\begin{aligned} \mu _{\alpha }(dt)=\frac{1}{2^{\alpha +1/2}\Gamma (\alpha +1/2)}|t|^{2\alpha }dt. \end{aligned}$$
    (4.2)

Let \(\alpha \ge 0.\) The Hartley–Bessel transform of a function \(f\in L^1_{\alpha }(\mathbb {R})\) is defined by

$$\begin{aligned} (\mathscr {H}_\alpha f)(\lambda ) = \int _{\mathbb {R}} f(t) \mathcal {J}_{\lambda }(t,\alpha ) \mu _{\alpha }(dt),\quad \lambda \in \mathbb {R}. \end{aligned}$$
(4.3)

Theorem 4.1

[9] Let \(\alpha > 0\). The following statements hold:

  1. 1.

    If \(f \in L^1_{\alpha }(\mathbb {R})\), then \(\mathscr {H}_\alpha f\in C_0(\mathbb {R})\). Moreover, \(\Vert \mathscr {H}_\alpha f\Vert _{\infty } \le \sqrt{2}\Vert f\Vert _{1,\alpha }\).

  2. 2.

    If \(f \in L^1_{\alpha }(\mathbb {R})\cap L^2_{\alpha }(\mathbb {R})\), then \(\mathscr {H}_\alpha f \in L^2_{\alpha }(\mathbb {R})\) and \(\Vert \mathscr {H}_\alpha f\Vert _{2,\alpha } = \Vert f\Vert _{2,\alpha }\).

  3. 3.

    There exists a unique isometry in \(L^2_{\alpha }(\mathbb {R})\) that coincides with \(\mathscr {H}_\alpha \) when restricted to \(L^1_{\alpha }(\mathbb {R})\cap L^2_{\alpha }(\mathbb {R})\).

Theorem 4.2

For all \(f\in S(\mathbb {R})\) and for all \(r, s\in \mathbb {R}\), the following properties hold:

  1. 1.

    \(\tau ^r_{\alpha }f(s)=\tau ^s_{\alpha }f(r)\).

  2. 2.

    \(\tau ^0_{\alpha }f(s)=f(s)\).

  3. 3.

    \(\tau ^r_{\alpha }\tau ^s_{\alpha }f=\tau ^s_{\alpha }\tau ^r_{\alpha }f\).

  4. 4.

    \(\mathscr {H}_{\alpha } \tau ^r_{\alpha }f(\lambda )=\mathcal {J}_\lambda (r,\alpha )\mathscr {H}_{\alpha } f(\lambda )\).

  5. 5.

    \(\tau ^r_{\alpha }\mathscr {L}_{\alpha }=\mathscr {L}_{\alpha }\tau ^r_{\alpha }\).

Proof

  1. 1.

    Property 1 is a direct consequence of the symmetry property \(\gamma _\alpha (r,s,t)=\gamma _\alpha (s,r,t)\).

  2. 2.

    Property 2 directly follows from the fact that \(\mathcal {J}_\lambda (0,\alpha )=1\).

  3. 3.

    Property 3 can be derived from property 1.

  4. 4.

    Let \(f \in S(\mathbb {R})\), then from Definition 3.2 and Fubini’s theorem, we obtain

    $$\begin{aligned} \mathscr {H}_{\alpha } \tau ^r_{\alpha }f(\lambda )&=\int _{\mathbb {R}}\tau _\alpha ^rf(s)\mathcal {J}_\lambda (s,\alpha )\mu _\alpha (ds)\\ {}&=\int _{\mathbb {R}}\int _{\mathbb {R}}\gamma _\alpha (r,s,t)f(t)\mathcal {J}_\lambda (s,\alpha )|t|^{2\alpha }dt\mu _\alpha (ds)\\ {}&=\int _{\mathbb {R}}(\int _{\mathbb {R}}\gamma _\alpha (r,s,t)\mathcal {J}_\lambda (s,\alpha )|s|^{2\alpha }ds)f(t)\mu _\alpha (dt) \\ {}&=\mathcal {J}_\lambda (r,\alpha )\int _{\mathbb {R}}f(t)\mathcal {J}_\lambda (t,\alpha )\mu _\alpha (dt)\\ {}&= \mathcal {J}_\lambda (r,\alpha )\mathscr {H}_{\alpha } f(\lambda ). \end{aligned}$$

    Thus proves \(\mathscr {H}_{\alpha } \tau ^r_{\alpha }f(\lambda ) =\mathcal {J}_\lambda (r,\alpha )\mathscr {H}_{\alpha } f(\lambda ).\)

  5. 5.

    Given that f and g belong to \(S(\mathbb {R})\), we have the equality

    $$\begin{aligned} \int _{\mathbb {R}}\mathscr {L}_{\alpha }f(t)g(t)\mu _\alpha (dt)= \int _{\mathbb {R}}f(t)\mathscr {L}_{\alpha }g(t)\mu _\alpha (dt). \end{aligned}$$

    Utilizing Property 3 and (1.7), we can derive the following relations:

    $$\begin{aligned} \mathscr {H}_{\alpha } \tau ^r_{\alpha }\mathscr {L}_{\alpha }f(\lambda )&=\mathcal {J}_\lambda (r,\alpha )\mathscr {H}_{\alpha } \mathscr {L}_{\alpha }f(\lambda )\\&=\lambda \mathcal {J}_\lambda (r,\alpha )\mathscr {H}_{\alpha } f(\lambda )\\&=\mathscr {H}_{\alpha } \mathscr {L}_{\alpha }\tau ^r_{\alpha }f(\lambda ). \end{aligned}$$

    This relation follows by the injectivity of the Hartley–Bessel Hankel transform.

\(\square \)

Definition 4.3

We define the convolution product for well-defined functions, denoted as \(f*_\alpha g\), as follows:

$$\begin{aligned} (f*_{\alpha }g)(r) = \int _{\mathbb {R}} \tau _{\alpha }^{r}f(s)g(s)\mu _{\alpha }(ds). \end{aligned}$$
(4.4)

Theorem 4.4

Let \(\alpha > 0\). The following hold:

  1. (i)

    For \(1 \le p, q, r \le \infty \) such that \(\frac{1}{p} + \frac{1}{q} - 1 = \frac{1}{r}\) and for \(f \in L^p_{\alpha }(\mathbb {R})\) and \(g \in L^q_{\alpha }(\mathbb {R})\), \(f *_{\alpha } g\) is well defined in \(L^r_{\alpha }(\mathbb {R})\) and \(\Vert f*g\Vert _{r,\alpha }\le 4\Vert f\Vert _{p,\alpha }\Vert g\Vert _{q,\alpha }\).

  2. (ii)

    For \(1 \le p, q, r \le 2\) such that \(\frac{1}{p} + \frac{1}{q} - 1 = \frac{1}{r}\) and for \(f \in L^p_{\alpha }(\mathbb {R})\) and \(g \in L^q_{\alpha }(\mathbb {R})\), \(\mathscr {H}_\alpha (f*_{\alpha }g)=\mathscr {H}_\alpha f \mathscr {H}_\alpha g\). In particular, the convolution product \(*_{\alpha }\) is associative in \(L^1_{\alpha }(\mathbb {R})\).

Proof

For part (i), if \(r = \infty \), then \(1/p + 1/q = 1\). Hence, by Hölder’s inequality and equation (3.4), \(f * g\) exists, and

$$\begin{aligned} \Vert f * g\Vert _{\infty ,\alpha } \le 4\Vert f\Vert _{p,\alpha } \Vert g\Vert _{q,\alpha }. \end{aligned}$$

Assume \(r < \infty \), implying \(p, q \le r\). Let \(s = p(1 - 1/q) = 1 - p/r\), where \(0 \le s < 1\). We have

$$\begin{aligned} |(f * g)(x)|\le & {} \int _{\mathbb {R}} |f(y)| |\tau _\alpha ^x g(y)|^{1-s} |\tau _\alpha ^x g(y)|^{s} \mu _\alpha (dy)\\\le & {} \left( \int _{\mathbb {R}} |f(y)|^{q} |\tau _\alpha ^x g(y)|^{q(1-s)} \mu _\alpha (dy) \right) ^{1/q} \Vert \tau _\alpha ^x g|^s\Vert _{q',\alpha }. \end{aligned}$$

If \(s = 0\), then \(q = 1\). If \(s \ne 0\), then \(sq' = p\). In either case, taking the qth power, we obtain

$$\begin{aligned}{} & {} |f * g(x)|^q\\{} & {} \quad \le \int _{\mathbb {R}} |f(y)|^{q} |\tau _\alpha ^x g(y)|^{q(1-s)} \mu _\alpha (dy) \Vert \tau _\alpha ^x g|^{sq}\Vert _{q',\alpha }\\{} & {} \quad \le 4^{sq} \Vert g\Vert ^{sq}_{p,\alpha }\int _{\mathbb {R}} |f(y)|^{q} |\tau _\alpha ^x g(y)|^{q(1-s)} \mu _\alpha (dy). \end{aligned}$$

Thus, for \(t:= r/q\), applying the generalized Minkowski inequality yields

$$\begin{aligned} \Vert f*g\Vert ^q_{qt,\alpha }&=\Vert |f*g|^q\Vert _{t,\alpha }\\&\le 4^{sq} \Vert g\Vert ^{sq}_{p,\alpha }\int _{\mathbb {R}}\left( \int _{\mathbb {R}}|f(y)|^{qt} |\tau _\alpha ^x g(y)|^{qt(1-s)} \mu _\alpha (dx)\right) ^{1/t}\mu _\alpha (dy)\\&\le 4^q \Vert f\Vert _{q,\alpha }^q\Vert g\Vert _{p,\alpha }^q. \end{aligned}$$

Since \(qt = r\) and \((1 - s)r = p\), we have

$$\begin{aligned} \Vert f*g\Vert _{r,\alpha }\le 4 \Vert f\Vert _{q,\alpha }\Vert g\Vert _{p,\alpha }. \end{aligned}$$

This proves (i).

The assertion (ii) follows easily from part (i). \(\square \)

5 Proof of main result

The following formulas are crucial for our subsequent calculations.

Lemma 5.1

For \(r, s>0\) and \(\Re (\alpha ) > 0\), the following relations hold:

$$\begin{aligned} j_{\alpha -1/2}(r)j_{\alpha -1/2}(s)&= \frac{\Gamma (\alpha +\frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{-1}^{1}j_{\alpha -1/2}(\sqrt{r^2+s^2-2rst} ) (1-t^2)^{\alpha -1} dt, \end{aligned}$$
(5.1)
$$\begin{aligned} j_{\alpha +1/2}(r)j_{\alpha +1/2}(s)&= \frac{4\Gamma (\alpha +\frac{3}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{-1}^{1}j_{\alpha -1/2}(\sqrt{r^2+s^2-2rst} ) t(1-t^2)^{\alpha -1} dt, \end{aligned}$$
(5.2)
$$\begin{aligned} rj_{\alpha +1/2}(r)j_{\alpha -1/2}(s)&= \frac{\Gamma (\alpha +\frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{-1}^{1}j_{\alpha +1/2}(\sqrt{r^2+s^2-2rst} )(r-st) (1-t^2)^{\alpha -1} dt. \end{aligned}$$
(5.3)

Proof

Formulas (5.1) and (4.1) can be derived directly by applying suitable substitutions from the following integral formula [18, formula (1) 7.334]:

$$\begin{aligned} \int _{0}^{\pi }C_n^{\nu }(\cos \theta ) \frac{J_\nu (\varpi )}{\varpi ^{\nu }} \sin ^{2\nu } \theta d\theta = \frac{\pi \Gamma (2\nu +n)}{2^{\nu -1}n!\Gamma (\nu )} \frac{J_{\nu +n}(r)}{r^\nu } \frac{J_{\nu +n}(s)}{s^\nu }, \end{aligned}$$

Here, the Gegenbauer polynomials \(C_{n}^{\nu }(t)\) are defined as per [16, (18.5.7)]:

$$\begin{aligned} C_n^{\nu }(t)&= \frac{\Gamma (n+2\nu )}{\Gamma (2\nu )n!} _2F_1\left( \begin{matrix}-n, n+2\nu \\ \nu +\frac{1}{2}\end{matrix};{\frac{1-t}{2}}\right) . \end{aligned}$$

It’s important to note that \(C^{\alpha }_{0}(t )=1\) and \(C^{\alpha }_{1}(t)=2\alpha t.\) Formula (5.3), on the other hand, follows from differentiation with respect to \(r\) in (5.1), utilizing the identity:

$$\begin{aligned} j_{\alpha -1/2}'(r)=-\frac{r}{2\alpha +1} j_{\alpha +1/2}(r). \end{aligned}$$

\(\square \)

Lemma 5.2

Let \(r, s \in \mathbb {R}^*\) and f be a bounded continuous function on \(\mathbb {R}.\)

  • If f is an even function, then

    $$\begin{aligned} \int _{\mathbb {R}} f(t)\gamma ^{\alpha }_{r,s}(dt) = \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{||r|-|s||}^{|r|+|s|} f(t)\gamma ^{e,\alpha }_{r,s}(dt), \end{aligned}$$
    (5.4)

    where

    $$\begin{aligned} \gamma ^{e,\alpha }_{r,s}(dt) ={\left\{ \begin{array}{ll} \frac{\big [(t^2 - (|r| - |s|)^2)\big ]^{\alpha -1}\big [((|r| + |s|)^2 - t^2)) \big ]^{\alpha }}{|2rs|^{2\alpha }}t dt &{}\text {if }rs>0,\\ \\ \frac{\big [(t^2 - (|r| - |s|)^2)\big ]^{\alpha }\big [((|r| + |s|)^2 - t^2)) \big ]^{\alpha -1}}{|2rs|^{2\alpha }}t dt &{}\text {if }rs<0. \end{array}\right. } \end{aligned}$$
    (5.5)

  • If f is an odd function, then

    $$\begin{aligned} \int _{\mathbb {R}} f(t)\gamma ^{\alpha }_{r,s}(dt) =\frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{||r|-|s||}^{|r|+|s|}f(t)\gamma ^{o,\alpha }_{r,s}(dt), \end{aligned}$$
    (5.6)

    where

    $$\begin{aligned} \gamma ^{o,\alpha }_{r,s}(dt) =(r+s){\left\{ \begin{array}{ll} \frac{\big [(t^2 - (|r| - |s|)^2)\big ]^{\alpha }\big [((|r| + |s|)^2 - t^2)) \big ]^{\alpha -1}}{|2rs|^{2\alpha }} dt &{}\text {if }rs>0,\\ \\ \frac{\big [(t^2 - (|r| - |s|)^2)\big ]^{\alpha -1}\big [((|r| + |s|)^2 - t^2)) \big ]^{\alpha }}{|2rs|^{2\alpha }} dt &{}\text {if }rs<0. \end{array}\right. } \end{aligned}$$
    (5.7)

Proof

For fixed \(r, s \in \mathbb {R}^*,\) observe that the function \(t \rightarrow \sigma (r,s,t)\) can be decomposed into even and odd parts as follows:

$$\begin{aligned} \sigma (r,s,t) = \sigma _e(r,s,t) + \sigma _o(r,s,t), \end{aligned}$$
(5.8)

where

$$\begin{aligned} \sigma _e(r,s,t)&= \frac{(r + s)^2 - t^2}{2rs} \quad \text {if} \quad r,s\ne 0 \\&= {\left\{ \begin{array}{ll} \frac{(|r| + |s|)^2 - t^2}{2rs} &{} \text {if} \quad rs > 0, \\ \frac{t^2 - (|r| - |s|)^2}{2|rs|} &{} \text {if} \quad rs < 0, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \sigma _o(r,s,t)&= \frac{r + s}{t} \cdot \frac{t^2 - (r - s)^2}{2rs} \quad \text {if} \quad r,s \ne 0 \\&= \frac{r+s}{t} \cdot {\left\{ \begin{array}{ll} \frac{t^2-(|r| - |s|)^2}{2rs} &{} \text {if} \quad rs > 0, \\ \frac{(|r| + |s|)^2-t^2}{2|rs|} &{} \text {if} \quad rs < 0. \end{array}\right. } \end{aligned}$$

Taking this decomposition into account, we can straightforwardly derive the result. \(\square \)

Lemma 5.3

Let \(\alpha >0\) and \(r, s\in \mathbb {R}^*.\) The following relationships hold:

$$\begin{aligned}&\frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} j_{\alpha -1/2}(t)\gamma ^{e,\alpha }_{r,s}(dt) = j_{\alpha -1/2}(r)j_{\alpha -1/2}(s) + \frac{rs}{(2\alpha +1)^2}j_{\alpha +1/2}(r)j_{\alpha +1/2}(s), \\&\frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} t j_{\alpha +1/2}(t)\gamma ^{o \alpha }_{r,s}(dt) = r j_{\alpha +1/2}(r) j_{\alpha -1/2}(s) + s j_{\alpha -1/2}(r)j_{\alpha +1/2}(s). \end{aligned}$$

Proof

Let \(t=\sqrt{r^2+s^2-2|rs|u}\). Then we have:

$$\begin{aligned} u&= \frac{r^2+s^2-t^2}{2|rs|}, \\ tdt&=- |rs|u du, \\ \frac{t^2-(|r|-|s|)^2}{2|rs|}&= 1-u, \\ \frac{(|r|+|s|)^2-t^2}{2|rs|}&= 1+u. \end{aligned}$$

Therefore,

$$\begin{aligned} \gamma ^{e,\alpha }_{r,s}(dt)=-(1-u^2)^{\alpha -1} (1+{{\,\textrm{sign}\,}}(rs)u ) du, \end{aligned}$$
(5.9)

where

$$\begin{aligned} \text {sign}(x) = {\left\{ \begin{array}{ll} 1 &{} \text {if } x > 0, \\ -1 &{} \text {if } x < 0. \end{array}\right. } \end{aligned}$$

By Lemma 5.2, we have:

$$\begin{aligned}&\frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} j_{\alpha -1/2}(t)\gamma ^{e,\alpha }_{r,s}(dt) \\&= \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{||r|-|s||}^{|r|+|s|} j_{\alpha -1/2}( t)\gamma ^{e,\alpha }_{r,s}(dt) \\&= \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )} \int _{-1}^{1} j_{\alpha -1/2}( \sqrt{r^2+s^2-2|rs|u} ) \\&\qquad \times (1-u^2)^{\alpha -1}(1+{{\,\textrm{sign}\,}}(rs)u) du. \end{aligned}$$

Using formulas (5.1) and (5.2), we get:

$$\begin{aligned} \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} j_{\alpha -1/2}(t)\gamma ^{e,\alpha }_{r,s}(dt)&= j_{\alpha -1/2}(r)j_{\alpha -1/2}(s)\\&\quad + \frac{rs}{(2\alpha +1)^2}j_{\alpha +1/2}(r)j_{\alpha +1/2}(s). \end{aligned}$$

Since

$$\sigma _o(r,s,t)=\frac{r+s}{t}(1-\text {sign}(rs)u),$$

we have:

$$\begin{aligned} \gamma ^{o,\alpha }_{r,s}(dt)=-(r+s)(1-u^2)^{\alpha -1} \frac{1-{{\,\textrm{sign}\,}}(rs) u}{\sqrt{r^2+s^2-2|rs|u}} du. \end{aligned}$$

By Lemma 3.1, and formula (5.3), and the following decomposition:

$$(r+s)(1-{{\,\textrm{sign}\,}}(rs)u)=r-s{{\,\textrm{sign}\,}}(rs)u + s-r{{\,\textrm{sign}\,}}(rs)u,$$

we obtain:

$$\begin{aligned} \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} tj_{\alpha +1/2}(t)\gamma ^{o \alpha }_{r,s}(dt)= rj_{\alpha +1/2}(r)j_{\alpha -1/2}(s)+ sj_{\alpha -1/2}(r)j_{\alpha +1/2}(s) \end{aligned}$$

This completes the proof. \(\square \)

We shall now establish Theorem 3.1.

Proof

First, let’s consider the case where \(\lambda > 0\) and \(r, s \in \mathbb {R}\). Using the explicit expression (1.8) for the kernel \(\mathcal {J}_\lambda (\cdot , \alpha )\), we can write:

$$\begin{aligned} \mathcal {J}_\lambda (r, \alpha ) \mathcal {J}_\lambda (s, \alpha )&= j_{\alpha -1/2}(\lambda r) j_{\alpha -1/2}(\lambda s) \\&\quad + \frac{\lambda ^2 rs}{(2\alpha +1)^2} j_{\alpha +1/2}(\lambda r) j_{\alpha +1/2}(\lambda s) \\&\quad + \frac{\lambda r}{2\alpha +1} j_{\alpha +1/2}(\lambda r) j_{\alpha -1/2}(\lambda s) \\&\quad + \frac{\lambda s}{2\alpha +1} j_{\alpha -1/2}(\lambda r) j_{\alpha +1/2}(\lambda s). \end{aligned}$$

Now, by applying Lemma 5.2, we obtain:

$$\begin{aligned} \mathcal {J}_\lambda (r, \alpha ) \mathcal {J}_\lambda (s, \alpha )&= \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} j_{\alpha -1/2}(\lambda t) \gamma ^{e,\alpha }_{r,s}(dt) \\&\quad + \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} \frac{\lambda t}{2\alpha +1} j_{\alpha +1/2}(\lambda t) \gamma ^{o,\alpha }_{r,s}(dt) \\&= \frac{\Gamma (\alpha + \frac{1}{2})}{\sqrt{\pi }\Gamma (\alpha )}\int _{\mathbb {R}} \mathcal {J}_{\lambda }(t,\alpha ) \gamma ^{\alpha }_{r,s}(dt). \end{aligned}$$

The symmetry relations (3.5) complete the proof. \(\square \)

6 Conclusion

In conclusion, this paper presents a comprehensive investigation into the one-parameter extension of the Hartley kernel, embodied in the Hartley–Bessel function. By exploring its derivation from the \(-1\) little Jacobi polynomials, we elucidate its intrinsic properties, including its emergence as an eigenfunction of a first-order difference-differential operator and its Sonin integral-type representation.

The primary contribution of this work lies in establishing a novel product formula for the Hartley–Bessel function. This formula, akin to its counterpart for the normalized Bessel function and Dunkl-Bessel function, paves the way for the development of innovative translation and convolution structures on the real line. Through the rigorous derivation and analysis of this product formula, we introduce a non-positivity-preserving convolution structure, enhancing the mathematical framework for applications in various fields.

Building upon foundational results in the literature, we advance the existing body of knowledge by providing a deeper understanding of the Hartley–Bessel function and its implications. The significance of our findings is underscored by their potential impact in harmonic analysis.