On L^p-Boundedness Properties and Parseval–Goldstein-Type Theorems for a Lebedev-Type Index Transform

Abstract

This paper investigates Parseval–Goldstein-type relations for a Lebedev-type index transform and examines its behavior in weighted Lebesgue spaces. Key results on $L^p$-boundedness establish conditions that support these relations. This contributes to understanding the functional framework of Lebedev-type index transforms in mathematical analysis.

1. Introduction and Preliminaries

Integral transforms play a central role in mathematical analysis and its applications, providing powerful tools for solving differential equations, analyzing asymptotic behaviors, and addressing boundary value problems in physics and engineering. Among the various classes of integral transforms, index transforms are distinguished by the presence of a kernel dependent on an index parameter, which adds flexibility and complexity to their theoretical structure and practical applications.

One of the most studied index transforms is the Kontorovich–Lebedev transform, introduced by M. I. Kontorovich and N. N. Lebedev in 1938. It is defined for a suitable complex-valued function f on ${\mathbb{R}}_{+}$ by the integral [1]

$$\begin{array}{c}\hfill \left(\mathcal{F}f\right)\left(\tau \right)={\int }_0^{\infty }f\left(x\right)K_{i\tau }\left(x\right)dx,\tau >0,\end{array}$$

(1)

where $K_{i\tau }\left(x\right)$ is the modified Bessel function of the third kind, commonly known as the Macdonald function. This transform has been extensively applied in fields such as diffraction theory and electrodynamics, where its kernel’s asymptotic properties and boundedness play a crucial role.

Building upon the foundational work on the Kontorovich–Lebedev transform, a Lebedev-type index transform was introduced in [2] employing the squared kernel $K_{i\tau }^2\left(x\right)$. This Lebedev-type index transform of a suitable complex-valued function f defined on ${\mathbb{R}}_{+}$ is formally given by the integral operator [2]

$$\begin{array}{c}\hfill \left(\mathcal{L}f\right)\left(\tau \right)={\int }_0^{\infty }f\left(x\right)K_{i\tau }^2\left(x\right)dx,\tau >0,\end{array}$$

(2)

(cf. also [1], Formula (1.236), p.37).

From [3] (Chapter 7), one has

$$\begin{array}{c}\hfill K_{i\tau }\left(x\right)={\displaystyle \frac{\pi }{2sin\left(i\tau \pi \right)}}\left[I_{-i\tau }\left(x\right)-I_{i\tau }\left(x\right)\right],\end{array}$$

where

$$\begin{array}{c}\hfill I_{i\tau }\left(x\right)=\sum _{0\le n\le \infty }{\displaystyle \frac{{\left(\frac{x}{2}\right)}^{2n+i\tau }}{n!\mathrm{\Gamma }(i\tau +n+1)}}.\end{array}$$

The Lebedev-type index transform (2) shares similarities with the Kontorovich–Lebedev transform but introduces additional challenges due to the structure of its kernel. For instance, the squared term amplifies the influence of the Macdonald function’s index dependency, affecting the transform’s convergence, asymptotics, and inversion properties. While the Kontorovich–Lebedev transform has been extensively studied, the Lebedev-type index transform (2) remains less explored, leaving significant theoretical gaps in understanding its functional spaces, inversion formulas, and analytical properties.

The inversion of (2) is given by [2]

$$\begin{array}{c}\hfill f\left(x\right)={\displaystyle \frac{2i}{\pi }}{\int }_0^{\infty }\tau \left(\mathcal{L}f\right)\left(\tau \right){\displaystyle \frac{d}{dx}}{I}_{i\tau }^2\left(x\right)d\tau ,x>0,\end{array}$$

(3)

(cf. also [1], Formula (1.236), p.37)

We begin by recalling, as noted in [3] (p. 82, Entry 21), that

$$\begin{array}{c}\hfill K_{i\tau }\left(x\right)={\int }_0^{\infty }{e}^{-xcosht}cos\left(\tau t\right)dt,x>0,\tau \in \mathbb{R}.\end{array}$$

This result directly leads to the following inequality, providing an immediate bound under the given conditions:

$$\begin{array}{c}\hfill |K_{i\tau }\left(x\right)|\le {\int }_0^{\infty }{e}^{-xcosht}dt=K_0\left(x\right),x>0,\tau \in \mathbb{R}.\end{array}$$

(4)

Notice that

$$\begin{array}{c}\hfill K_{i\tau }^2\left(x\right)\le K_0^2\left(x\right),x>0,\tau \in \mathbb{R}.\end{array}$$

(5)

Moreover, based on results from [4] (p. 172, Entry 3) and [4] (p. 173, Entry 4), it follows that

$$\begin{array}{ccc}\hfill K_0^2\left(x\right)& \sim & {ln}^2\left({\displaystyle \frac{Cx}{2}}\right)\mathrm{as}x\to +0,\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill K_0^2\left(x\right)& \sim & {\displaystyle \frac{\pi }{4}}{\displaystyle \frac{e^{-2x}}{x}}\mathrm{as}x\to +\infty ,\hfill \end{array}$$

(7)

where $C=e^{\gamma }$ and $\gamma =0.5772\cdots$ is the Euler–Mascheroni constant.

Observe that $K_0^2\left(x\right)$ is unbounded in $(0,\infty )$.

Set

$$\begin{array}{c}\hfill \left({\mathcal{L}}^{\ast }g\right)\left(x\right)={\int }_0^{\infty }g\left(\tau \right)K_{i\tau }^2\left(x\right)d\tau ,x>0.\end{array}$$

(8)

For $g\in L^1\left({\mathbb{R}}_{+}\right)$, the integral defined in (8) converges for each $x>0$. This is due to the fact that

$$\begin{array}{c}\hfill |\left({\mathcal{L}}^{\ast }g\right)\left(x\right)|\le {\int }_0^{\infty }\left|g\left(\tau \right)\right|d\tau ·K_0^2\left(x\right)<\infty ,\mathrm{for} \mathrm{each}x>0.\end{array}$$

Let us consider the case for each $\tau >0$

$$\begin{array}{c}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|\le {\int }_0^{\infty }\left|f\left(x\right)\right|K_0^2\left(x\right)dx<\infty ;\end{array}$$

thus, for every function f belonging to the space $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$, the integral given by (2) converges for all $\tau >0$.

Parseval–Goldstein relations in integral transforms provide a vital link between the norms of a function in its original domain and in the transformed domain, offering insights into energy conservation and functional equivalence across spaces [5,6,7,8,9]. Analyzing $L^p$-boundedness further ensures stability and broad applicability of these transforms in various functional and weighted spaces, making it fundamental for both theoretical insights and practical applications [1]. The present article deals with the study of Parseval–Goldstein-type relations and $L^p$-boundedness properties for the Lebedev-type index transforms (2) and (8).

The structure of this article is organized as follows: Section 1 introduces the definitions of the Lebedev-type index transform (2) and presents foundational results that will be referenced throughout this paper. Section 2 examines the continuity properties of the Lebedev-type index transform (2) and its adjoint within Lebesgue spaces, alongside establishing Parseval–Goldstein-type relations. Section 3 investigates the $L^p$-boundedness properties of the Lebedev-type index transform (2) in various Lebesgue spaces. Section 4 further elaborates on these $L^p$-boundedness properties, ultimately leading to the derivation of Parseval–Goldstein-type theorems for the Lebedev-type index transform (2). Finally, Section 5 offers concluding remarks and discusses potential avenues for future research.

2. Parseval–Goldstein-Type Relations for the Lebedev-Type Index Transform

In this section, we obtain continuity properties over weighted Lebesgue spaces and Parseval–Goldstein relations for the transforms $\mathcal{L}$ and ${\mathcal{L}}^{\ast }$ given by (2) and (8), respectively.

2.1. The $\mathcal{L}$ Transform over the Space $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$

Proposition 1.

The Lebedev-type index transform $\mathcal{L}$ defined by (2) is a bounded linear operator from $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$. For any function $f\in L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$, we have

$$\begin{array}{c}\hfill {\parallel \mathcal{L}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}\le {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)},\end{array}$$

indicating that $\mathcal{L}f$ is a continuous function on ${\mathbb{R}}_{+}$. Furthermore, the Lebedev-type index transform $\mathcal{L}$ is a continuous map from $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$ to the Banach space of bounded continuous functions on ${\mathbb{R}}_{+}$.

Proof. 

Let ${\tau }_0>0$ be arbitrary. Since the mapping $\tau \to K_{i\tau }^2\left(x\right)$ is continuous for each fixed $x>0$, we have

$$\begin{array}{c}\hfill K_{i\tau }^2\left(x\right)\to K_{i{\tau }_0}^2\left(x\right)as\tau \to {\tau }_0.\end{array}$$

Also, we have that $\left(K_{i\tau }^2\left(x\right)-K_{i{\tau }_0}^2\left(x\right)\right)\left|f\left(x\right)\right|$ is dominated by the integrable function $2K_0^2\left(x\right)\left|f\left(x\right)\right|$. Consequently, by applying the dominated convergence theorem, we can conclude that

$$\begin{array}{c}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)-\left(\mathcal{L}f\right)\left({\tau }_0\right)\right|\le {\int }_0^{\infty }\left(K_{i\tau }^2\left(x\right)-K_{i{\tau }_0}^2\left(x\right)\right)\left|f\left(x\right)\right|dx\to 0,as\tau \to {\tau }_0.\end{array}$$

Consequently, $\mathcal{L}f$ is a continuous function on ${\mathbb{R}}_{+}$.

Since for each $\tau >0$,

$$\begin{array}{ccc}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|& \le & {\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(x\right)\right|dx\hfill \\ & \le & {\int }_0^{\infty }{K}_0^2\left(x\right)\left|f\left(x\right)\right|dx={\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)},\hfill \end{array}$$

(9)

one has that $\mathcal{L}f$ is a bounded function.

The linearity of the integral operator ensures that the $\mathcal{L}$ transform is linear. Also, from (9), we get that ${\parallel \mathcal{L}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}\le {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}$; hence, the mapping $\mathcal{L}:L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)\to L^{\infty }\left({\mathbb{R}}_{+}\right)$ is a continuous linear map. □

Proposition 2.

The Lebedev-type index transform (2) acts as a bounded linear operator mapping from $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$ into $L^q({\mathbb{R}}_{+},\rho \left(x\right)dx)$, $0<q<\infty$, under the conditions that $\rho >0$ a.e. on ${\mathbb{R}}_{+}$ and that the integral ${\int }_0^{\infty }\rho \left(x\right)dx<\infty$.

Proof. 

Note that according to (9) for each $\tau >0$

$$\begin{array}{ccc}\hfill \left|\right(\mathcal{L}f\left)\right(\tau \left)\right|& \le & {\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(x\right)\right|dx\hfill \\ & \le & {\int }_0^{\infty }{K}_0^2\left(x\right)\left|f\left(x\right)\right|dx={\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}.\hfill \end{array}$$

Then, for $0<q<\infty$, the following holds:

$$\begin{array}{c}\hfill {\left({\int }_0^{\infty }{\left|\left(\mathcal{L}f\right)\left(x\right)\right|}^q\rho \left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}\le {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}{\left({\int }_0^{\infty }\rho \left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}<\infty .\end{array}$$

□

Example 1.

The following are examples of weights ρ for Proposition 2:

(i)

$\rho \left(x\right)={(1+x)}^s, for s<-1.$

(ii)

$\rho \left(x\right)=e^{sx}, for s<0.$

2.2. The Transform ${\mathcal{L}}^{\ast }$ over the Space $L^1\left({\mathbb{R}}_{+}\right)$

Proposition 3.

The ${\mathcal{L}}^{\ast }$ is defined as a bounded linear operator mapping from $L^1\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},\rho \left(x\right)dx)$, for $0<q<\infty$, under the conditions that the weight function $\rho >0$ a.e. on ${\mathbb{R}}_{+}$ and that $K_0^2\left(x\right)\in L^q({\mathbb{R}}_{+},\rho \left(x\right)dx)$.

Proof. 

Note that for each $x>0$,

$$\begin{array}{ccc}\hfill \left|\left({\mathcal{L}}^{\ast }f\right)\left(x\right)\right|& \le & {\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(\tau \right)\right|d\tau \hfill \\ & \le & K_0^2\left(x\right){\int }_0^{\infty }\left|f\left(\tau \right)\right|d\tau .\hfill \end{array}$$

Then, for $0<q<\infty$, the following holds:

$$\begin{array}{c}\hfill {\left({\int }_0^{\infty }{\left|\left({\mathcal{L}}^{\ast }f\right)\left(x\right)\right|}^q\rho \left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}\le {\parallel f\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2q}\rho \left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}<\infty .\end{array}$$

□

Example 2.

The following are examples of weights ρ for Proposition 3:

(i)

$\rho \left(x\right)=x^s,\mathit{for} s>-1.$

(ii)

$\rho \left(x\right)={(1+x)}^s, for all reals.$

(iii)

$\rho \left(x\right)=e^{sx},for s<2q;\mathrm{and} s=2qbeingq>1.$

2.3. Parseval–Goldstein-Type Theorems

Theorem 1.

If $f\in L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$ and $g\in L^1\left({\mathbb{R}}_{+}\right)$, then the following Parseval–Goldstein-type relation holds:

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathcal{L}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathcal{L}}^{\ast }g\right)\left(x\right)dx.\end{array}$$

(10)

Proof. 

In particular, for each $\tau >0$,

$$\begin{array}{c}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|\le {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}.\end{array}$$

Therefore,

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left|\left(\mathcal{L}f\right)\left(\tau \right)\right|\left|g\left(\tau \right)\right|d\tau \le {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\end{array}$$

Furthermore, for each $x>0$,

$$\begin{array}{c}\hfill |\left({\mathcal{L}}^{\ast }g\right)\left(x\right)|\le {\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)|g\left(\tau \right)|d\tau \le K_0^2\left(x\right){\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\end{array}$$

Then,

$$\begin{array}{ccc}\hfill {\int }_0^{\infty }\left|f\left(x\right)\right||\left({\mathcal{L}}^{\ast }g\right)\left(x\right)|dx& \le & {\int }_0^{\infty }\left|f\left(x\right)\right|K_0^2\left(x\right)dx{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}\hfill \\ & =& {\parallel f\parallel }_{L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)}{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\hfill \end{array}$$

Using Fubini’s theorem, we can derive the relationship expressed in Equation (10). □

Remark 1.

From this result, the transform ${\mathcal{L}}^{\ast }$ is identified as the adjoint of the Lebedev-type index transform $\mathcal{L}$ over $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$.

3. $L^p$-Boundedness Properties for the Lebedev-Type Index Transform

In this section, we obtain continuity properties over weighted Lebesgue spaces for the transforms $\mathcal{L}$ given by (2).

Theorem 2.

Let $1<p<\infty$ and $0<q<\infty$. Then,

(i)

For $\gamma <-1$ and all q, $0<q<\infty$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$. Furthermore, for all $\gamma \in \mathbb{R}$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$;

(ii)

For $-2p\le \alpha <0$ and all q, $0<q<\infty$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^q({\mathbb{R}}_{+},e^{\alpha x}dx)$. Furthermore, for $\alpha \ge -2p$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

Proof. 

(i)

Using Hölder’s inequality, we arrive at

$$\begin{array}{ccc}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|& =& \left|{\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)f\left(x\right)dx\right|\hfill \\ \hfill & \le & {\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(x\right)\right|dx\hfill \\ & =& {\int }_0^{\infty }\left|f\left(x\right)\right|{(1+x)}^{{\displaystyle \frac{\gamma }{p}}}{K}_{i\tau }^2\left(x\right){(1+x)}^{-{\displaystyle \frac{\gamma }{p}}}dx\hfill \\ \hfill & \le & {\left({\int }_0^{\infty }{\left|f\left(x\right)\right|}^p{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill & ·& {\left({\int }_0^{\infty }{\left(K_{i\tau }\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ \hfill & =& {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}·{\left({\int }_0^{\infty }{\left(K_{i\tau }\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}},\hfill \end{array}$$

(11)

which leads us to the following inequality:

$$\begin{array}{ccc}\hfill & & {\int }_0^{\infty }{\left|\left(\mathcal{L}f\right)\left(\tau \right)\right|}^q{(1+\tau )}^{\gamma }d\tau \hfill \\ \hfill & \le & {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}^q{\int }_0^{\infty }{\left({\int }_0^{\infty }{\left(K_{i\tau }\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}{(1+\tau )}^{\gamma }d\tau \hfill \\ & \le & {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}^q{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}·{\int }_0^{\infty }{(1+\tau )}^{\gamma }d\tau \hfill \\ \hfill & \le & {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}^q{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}·{(-1-\gamma )}^{-1},\hfill \end{array}$$

(12)

where $p+p^{\prime }=p{p}^{\prime }$ and $\gamma <-1$. Therefore, we have

$$\begin{array}{ccc}\hfill {\parallel \mathcal{L}f\parallel }_{L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}& \le & {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & ·& {(-1-\gamma )}^{{\displaystyle \frac{-1}{q}}}.\hfill \end{array}$$

(13)

In light of (6) and (7), the above integral converges. So, clearly, we have

$${\parallel \mathcal{L}f\parallel }_{L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}\le C{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)},$$

for a certain real constant C depending on p, q and $\gamma$. Consequently, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$.

Thus, analogously, we find that

$$\begin{array}{ccc}& & \underset{\tau \in {\mathbb{R}}_{+}}{ess\; sup}\left\{\right|\left(\mathcal{L}f\right)\left(\tau \right)\left|\right\}\hfill \\ & & \le {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}·\underset{\tau \in {\mathbb{R}}_{+}}{ess\; sup}\left\{{\left({\int }_0^{\infty }{\left(K_{i\tau }\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\right\}\hfill \\ & & \le {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}·{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{\gamma p^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}.\hfill \end{array}$$

We next observe that, by virtue of (6) and (7), the above integral converges for all $\gamma \in \mathbb{R}$. So, clearly, we have

$${\parallel \mathcal{L}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}\le C{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)},$$

for a certain real constant C depending on p and $\gamma$. Consequently, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

(ii)

Following the same technique as the above for $-2p\le \alpha <0$ and all q, $0<q<\infty$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^q({\mathbb{R}}_{+},e^{\alpha x}dx)$. Furthermore, for $\alpha \ge -2p$, the Lebedev-type index transform (2) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

□

Theorem 3.

For any q, $0<q<\infty$, and $\gamma <-1$, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$. Also, for $\alpha <0$, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},e^{\alpha x}dx)$. Moreover, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

Proof. 

Observe that

$$\begin{array}{c}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|={\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(x\right)\right|dx\le {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}·{\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)dx,\end{array}$$

(14)

so that, for any $0<q<\infty$, we get

$$\begin{array}{c}\hfill {\left|\left(\mathcal{L}f\right)\left(\tau \right)\right|}^q\le {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}^q·{\left({\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)dx\right)}^q.\end{array}$$

(15)

We thus find that

$$\begin{array}{c}\hfill {\int }_0^{\infty }{\left|\left(\mathcal{L}f\right)\left(\tau \right)\right|}^q{(1+\tau )}^{\gamma }d\tau \le {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}^q·{\int }_0^{\infty }{\left({\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)dx\right)}^q{(1+\tau )}^{\gamma }d\tau .\end{array}$$

(16)

and, therefore, that

$$\begin{array}{c}\hfill {\parallel \mathcal{L}f\parallel }_{L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}\le {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}·{\left({\int }_0^{\infty }{\left({\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)dx\right)}^q{(1+\tau )}^{\gamma }d\tau \right)}^{{\displaystyle \frac{1}{q}}}.\end{array}$$

(17)

Moreover, in view of (5), we have

$$\begin{array}{ccc}\hfill & & {\parallel \mathcal{L}f\parallel }_{L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}\hfill \\ & & \le {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}·{\left({\int }_0^{\infty }{K}_0^2\left(x\right)dx\right)}^2{\left({\int }_0^{\infty }{(1+\tau )}^{\gamma }d\tau \right)}^{{\displaystyle \frac{1}{q}}}\hfill \\ \hfill & & \le {\displaystyle \frac{A}{{(-1-\gamma )}^q}}·{\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)},\gamma <-1,\mathrm{for} \mathrm{some}A>0.\hfill \end{array}$$

(18)

Hence, we have

$${\parallel \mathcal{L}f\parallel }_{L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}\le C{\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)},$$

for a certain real constant C depending on q and $\gamma$. Consequently, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$.

Following the same technique as the above for $\alpha <0$, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},e^{\alpha x}dx)$.

Also, in view of (5), we have

$$\begin{array}{ccc}\hfill {\parallel \mathcal{L}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}& \le & {\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)}·{\int }_0^{\infty }{K}_0^2\left(x\right)dx\hfill \\ \hfill & \le & A·{\parallel f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)},\mathrm{for} \mathrm{some}A>0.\hfill \end{array}$$

(19)

Therefore, the Lebedev-type index transform (2) is bounded from $L^{\infty }\left({\mathbb{R}}_{+}\right)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$. □

Remark 2.

Since $K_0$ is unbounded in $(0,\infty )$, we do not know whether or not Theorem 3.1 can be extended to the case $p=1$.

4. Weighted $L^p$ Inequalities for the Lebedev-Type Index Transform

In this section, we obtain continuity properties over weighted Lebesgue spaces for the transform ${\mathcal{L}}^{\ast }$ given by (8). We also establish new Parseval–Goldstein relations.

Proposition 4.

Assume $1<p<\infty$, $p+p^{\prime }=p{p}^{\prime }$. Then,

(i)

For $\gamma >p-1$, the ${\mathcal{L}}^{\ast }$ given by (8) is bounded from $L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$ into $L^{p^{\prime }}({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$;

(ii)

For $0<\alpha \le 2p^{\prime }$, the ${\mathcal{L}}^{\ast }$ given by (8) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^{p^{\prime }}({\mathbb{R}}_{+},e^{\alpha x}dx)$.

Proof. 

(i)

Let $g\in L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$. From (8), one has

$$\begin{array}{c}\hfill |\left({\mathcal{L}}^{\ast }g\right)\left(x\right)|\le {\int }_0^{\infty }\left|g\left(\tau \right)\right|{(1+\tau )}^{{\displaystyle \frac{\gamma }{p}}}{K}_{i\tau }^2\left(x\right){(1+\tau )}^{-{\displaystyle \frac{\gamma }{p}}}d\tau .\end{array}$$

(20)

Using Hölder’s inequality, the right side of (20) is less than or equal to

$$\begin{array}{ccc}& & {\left({\int }_0^{\infty }{\left|g\left(\tau \right)\right|}^p{(1+\tau )}^{\gamma }d\tau \right)}^{{\displaystyle \frac{1}{p}}}{\left({\int }_0^{\infty }{\left(K_{i\tau }\left(x\right)\right)}^{2p^{\prime }}{(1+\tau )}^{{\displaystyle \frac{-p^{\prime }\gamma }{p}}}d\tau \right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & & \le K_0^2\left(x\right){\left({\int }_0^{\infty }{(1+\tau )}^{{\displaystyle \frac{-p^{\prime }\gamma }{p}}}d\tau \right)}^{{\displaystyle \frac{1}{p^{\prime }}}}{\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}.\hfill \end{array}$$

Now, since ${\int }_0^{\infty }{(1+\tau )}^{{\displaystyle \frac{-p^{\prime }\gamma }{p}}}d\tau <\infty$, we have

$$\begin{array}{ccc}& & {\left({\int }_0^{\infty }{\left|\left({\mathcal{L}}^{\ast }g\right)\left(x\right)\right|}^{p^{\prime }}{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & & \le {C\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}·{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}},\hfill \end{array}$$

for certain constant $C>0$. From this inequality and taking into account that ${\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{\gamma }dx<\infty$, it follows that

$$\parallel {\mathcal{L}}^{\ast }{g\parallel }_{L^{p^{\prime }}({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}\le M·{\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)},$$

for certain constant $M>0$.

(ii)

Following the same technique as the above for $0<\alpha \le 2p^{\prime }$, the ${\mathcal{L}}^{\ast }$ given by (8) is bounded from $L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$ into $L^{p^{\prime }}({\mathbb{R}}_{+},e^{\alpha x}dx)$.

□

Proposition 5.

Assume $1<p<\infty$, $p+p^{\prime }=p{p}^{\prime }$. Then,

(i)

For $\gamma >p-1$, $f,g\in L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx);$

or

(ii)

For $0<\alpha \le 2p^{\prime }$, $f,g\in L^p({\mathbb{R}}_{+},e^{\alpha x}dx);$

the following Parseval–Goldstein-type relation holds:

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathcal{L}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathcal{L}}^{\ast }g\right)\left(x\right)dx.\end{array}$$

(21)

Proof. 

(i)

For $f\in L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)$, we have

$$\begin{array}{c}\hfill \left|\left(\mathcal{L}f\right)\left(\tau \right)\right|={\int }_0^{\infty }{K}_{i\tau }^2\left(x\right)\left|f\left(x\right)\right|dx={\int }_0^{\infty }{K}_{i\tau }^2\left(x\right){(1+x)}^{-{\displaystyle \frac{\gamma }{p}}}\left|f\left(x\right)\right|{(1+x)}^{{\displaystyle \frac{\gamma }{p}}}dx.\end{array}$$

Using Hölder’s inequality and relation (5), we have

$$\begin{array}{ccc}& & {\left({\int }_0^{\infty }{\left|f\left(x\right)\right|}^p{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p}}}·{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{p^{\prime }\gamma }{p}}}\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & & \le {\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}·{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & & \le M^{\ast }{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)},\hfill \end{array}$$

where $M^{\ast }>0$ is some constant. Therefore,

$$\begin{array}{ccc}& & {\int }_0^{\infty }\left|\left(\mathcal{L}f\right)\left(x\right)\right|\left|g\left(x\right)\right|dx\hfill \\ & & \le M^{\ast }{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\int }_0^{\infty }\left|g\left(x\right)\right|dx\hfill \\ & & =M^{\ast }{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\int }_0^{\infty }{(1+x)}^{-{\displaystyle \frac{\gamma }{p}}}\left|g\left(x\right)\right|{(1+x)}^{{\displaystyle \frac{\gamma }{p}}}dx\hfill \\ & & \le M^{\ast }{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\left({\int }_0^{\infty }{(1+x)}^{-{\displaystyle \frac{p^{\prime }\gamma }{p}}}dx\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}{\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}.\hfill \end{array}$$

On the other hand,

$$\begin{array}{ccc}& & {\int }_0^{\infty }f\left(x\right)\left({\mathcal{L}}^{\ast }g\right)\left(x\right)dx\hfill \\ & & \le {C\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\left({\int }_0^{\infty }{\left|f\left(x\right)\right|}^p{(1+x)}^{\gamma }dx\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ & & ·{\left({\int }_0^{\infty }{\left(K_0\left(x\right)\right)}^{2p^{\prime }}{(1+x)}^{-{\displaystyle \frac{p^{\prime }\gamma }{p}}}\right)}^{{\displaystyle \frac{1}{p^{\prime }}}}\hfill \\ & & \le M·{\parallel f\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)}{\parallel g\parallel }_{L^p({\mathbb{R}}_{+},{(1+x)}^{\gamma }dx)},\hfill \end{array}$$

for some constants C and M.

Finally, using Fubini’s theorem, we obtain the relation (21).

(ii)

Following the same technique as the above for $0<\alpha \le 2p^{\prime }$, $f,g\in L^p({\mathbb{R}}_{+},e^{\alpha x}dx)$, we obtain the relation (21).

□

Remark 3.

Since $K_0$ is unbounded in $(0,\infty )$, we do not know whether or not Proposition 4.1 and Proposition 4.2 can be extended to the case $p=1$.

5. Final Observations and Conclusions

This research has successfully demonstrated $L^p$-boundedness properties and Parseval–Goldstein-type relations for the Lebedev-type index transform given by (2), enhancing our understanding of its continuity in weighted Lebesgue spaces. These findings provide a robust framework applicable to various index integral transforms [1].

This is the case of the Lebedev-type index transform given by

$$\begin{array}{c}\hfill {\int }_0^{\infty }{\left|K_{{\displaystyle \frac{\alpha +i\tau }{2}}}\left(x\right)\right|}^{2}f\left(x\right)dx,\tau \in \mathbb{R},\end{array}$$

(22)

for $\alpha \in \mathbb{R}$, which was analyzed in [10].

Observe that

$$\begin{array}{c}\hfill {\left|K_{{\displaystyle \frac{\alpha +i\tau }{2}}}\left(x\right)\right|}^2\le K_{{\displaystyle \frac{\alpha }{2}}}^2\left(x\right),x>0,\tau \in \mathbb{R},\end{array}$$

for each $\alpha \in \mathbb{R}$.

Thus, for the transform (22) in Section 2, one considers the space $L^1({\mathbb{R}}_{+},K_{{\displaystyle \frac{\alpha }{2}}}^2\left(x\right)dx)$ instead of the space $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$.

Also, one considers the Lebedev-Skalskaya-type index transform

$$\begin{array}{c}\hfill \underset{0}{\overset{\infty }{\int }}f\left(x\right)\mathfrak{R}\left[K_{\alpha +i\tau }\left(x\right)K_{\beta +i\tau }\left(x\right)\right]dx,\tau >0,\end{array}$$

(23)

where we mean $\alpha +\beta =1$, (cf. [1], Formula (7.106), p.221).

Observe that

$$\begin{array}{c}\hfill \left|\mathfrak{R}\left[K_{\alpha +i\tau }\left(x\right)K_{\beta +i\tau }\left(x\right)\right]\right|\le \left|K_{\alpha +i\tau }\left(x\right)K_{\beta +i\tau }\left(x\right)\right|\le K_{\alpha }\left(x\right)K_{\beta }\left(x\right),x>0,\tau \in \mathbb{R}.\end{array}$$

Thus, for the transform (23) in Section 2, one considers the space $L^1\left({\mathbb{R}}_{+},K_{\alpha }\left(x\right)K_{\beta }\left(x\right)dx\right)$ instead of the space $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$.

An interesting particular case of the transform (23) is obtained by taking $\alpha =\beta ={\displaystyle \frac{1}{2}}$, namely

$$\begin{array}{c}\hfill \underset{0}{\overset{\infty }{\int }}f\left(x\right)\mathfrak{R}\left[K_{{\displaystyle \frac{1}{2}}+i\tau }^2\left(x\right)\right]dx,\tau >0.\end{array}$$

(24)

For this particular case, we have

$$\begin{array}{c}\hfill \left|\mathfrak{R}\left[K_{{\displaystyle \frac{1}{2}}+i\tau }^2\left(x\right)\right]\right|\le \left|K_{{\displaystyle \frac{1}{2}}+i\tau }^2\left(x\right)\right|\le K_{{\displaystyle \frac{1}{2}}}^2\left(x\right)={\displaystyle \frac{\pi }{2}}{\displaystyle \frac{e^{-2x}}{x}},x>0,\tau \in \mathbb{R}.\end{array}$$

Thus, for the transform (24) in Section 2, one considers the space $L^1\left({\mathbb{R}}_{+},{\displaystyle \frac{e^{-2x}}{x}}dx\right)$ instead of the space $L^1({\mathbb{R}}_{+},K_0^2\left(x\right)dx)$.

Similarly, this approach one can spread on the $\mathfrak{F}$-type Lebedev–Skalskaya-type index transform where $\alpha +\beta =1$

$$\begin{array}{c}\hfill \underset{0}{\overset{\infty }{\int }}f\left(x\right)\mathfrak{F}\left[K_{\alpha +i\tau }\left(x\right)K_{\beta +i\tau }\left(x\right)\right]dx,\tau >0.\end{array}$$

(25)

By extending our results, future investigations can explore similar boundedness and continuity features in related transforms, potentially revealing new insights in mathematical analysis. Overall, this work contributes significantly to the theoretical underpinnings of integral transforms.