Boundedness of Bessel–Riesz Operator in Variable Lebesgue Measure Spaces

Abstract

In this manuscript, we establish the boundedness of the Bessel–Riesz operator $I_{\alpha ,\gamma }f$ in variable Lebesgue spaces $L^{p(·)}$. We prove that $I_{\alpha ,\gamma }f$ is bounded from $L^{p(·)}$ to $L^{p(·)}$ and from $L^{p(·)}$ to $L^{q(·)}$. We explore various scenarios for the boundedness of $I_{\alpha ,\gamma }f$ under general conditions, including constraints on the Hardy–Littlewood maximal operator $\mathcal{M}$. To prove these results, we employ the boundedness of $\mathcal{M}$, along with Hölder’s inequality and classical dyadic decomposition techniques. Our findings unify and generalize previous results in classical Lebesgue spaces. In some cases, the results are new even for constant exponents in Lebesgue spaces.

1. Introduction

Harmonic analysis is one of the most influential branches of mathematics, where the study of the boundedness of operators plays a central role. Many classical and modern results in this field revolve around the boundedness properties of integral and maximal operators in various function spaces. Among these operators, the Hardy–Littlewood maximal operator is particularly significant due to its ability to control singular integrals, providing a fundamental tool for harmonic analysis. Since the 1970s, weighted norm inequalities for the Hardy–Littlewood maximal operator $\mathcal{M}$ have been extensively studied, with significant contributions from Muckenhoupt [1] and Sawyer [2,3], laying the groundwork for weighted analysis in Lebesgue spaces.

In [4], the authors studied the harmonic analysis of Lebesgue spaces with variable exponents $L^{p(.)}$, establishing sufficient conditions for the maximal operator $\mathcal{M}$ to be bounded on $L^{p(.)}$ spaces. Furthermore, they highlighted that the study of the boundedness of operators in variable Lebesgue spaces is a crucial topic in harmonic analysis. In the classical $L^p$ spaces, the norm convergence of an approximate identity is relatively straightforward to prove, whereas pointwise convergence requires a more sophisticated argument using the maximal operator. In contrast, for variable Lebesgue spaces, this situation is reversed: pointwise convergence is an immediate consequence of classical results, while norm convergence requires the boundedness of the maximal operator.

Another class of operators that has attracted substantial attention in recent years is the family of Bessel–Riesz operators. These operators are closely related to the study of fractional integrals and have been analyzed extensively in Lebesgue and Morrey spaces. Kurata et al. [5] investigated the boundedness of integral operators with Bessel–Riesz kernels in generalized Morrey spaces and explored their connections to the Schrödinger operator. In [6,7], Idris et al. extended the study of the Bessel–Riesz operator and generalized Bessel–Riesz operator to Morrey spaces in Euclidean spaces, while Saba et al. [8,9] further examined their behavior in Lebesgue and Morrey spaces within measure metric spaces. These studies highlight the growing interest in understanding the boundedness of integral operators beyond the classical Euclidean setting.

Previous research on the Bessel–Riesz operator has primarily concentrated on classical Lebesgue spaces. Building upon the work of David Cruz-Uribe [4], who extended the boundedness of the Hardy–Littlewood maximal operator $\mathcal{M}$ to variable Lebesgue spaces $L^{p(·)}$, we explore the boundedness of the Bessel–Riesz operator $I_{\alpha ,\gamma }f$ in these spaces. Unlike classical cases, our results depend on specific conditions, which is why we consider different scenarios throughout this manuscript. The structure of this paper is as follows. In Section 1 and Section 2, we provide the necessary background on variable Lebesgue spaces and the Bessel–Riesz operator, which serve as the foundation for our main results. Section 3 is dedicated to formulating and proving the key results concerning the boundedness of the Bessel–Riesz operator $I_{\alpha ,\gamma }f$ in variable Lebesgue spaces $L^{p(·)}$. Specifically, in Theorem 6, we establish the boundedness of $I_{\alpha ,\gamma }f$ from $L^{p(·)}$ to $L^{p(·)}$, while, in Theorems 7 and 8, we prove its boundedness from $L^{p(·)}$ to $L^{q(·)}$. Finally, we conclude the manuscript with remarks in Section 4.

2. Preliminaries

Throughout the paper, $\Omega$ will be a subset of ${\mathbb{R}}^n$.

Given a set $\Omega$ and $\mathcal{P}(\Omega )$, denote the collection of the Lebesgue measurable functions $p(.):\Omega \to [1,\infty ]$. The elements of $\mathcal{P}(\Omega )$ are called exponent functions. Given any $p(.)\in \mathcal{P}(\Omega )$ a set $E\subseteq \Omega$, we suppose ${\displaystyle p_{-}\left(E\right)=\mathrm{ess}\underset{u\in E}{\inf }p\left(u\right)}$ and ${\displaystyle p_{+}\left(E\right)=\mathrm{ess}\underset{u\in E}{\sup }p\left(u\right).}$ If there is no confusion in the domain, we shall simply denote $p_{-}$ and $p_{+}$. Given any exponent function $p(·)\in \mathcal{P}(\Omega )$, the conjugate exponent function $p^{\prime }(·)$ of $p(·)$ is defined as ${\displaystyle \frac{1}{p\left(u\right)}}+{\displaystyle \frac{1}{p^{\prime }\left(u\right)}}=1.$ Further, ${\left(p^{\prime }(·)\right)}_{+}=(p_{-}{)}^{\prime },{\left(p^{\prime }(·)\right)}_{-}={\left(p_{+}\right)}^{\prime }.$ The following results on variable Lebesgue spaces $L^{p(.)}$ have been taken from [4].

Definition 1.

Let $p(.)\in \mathcal{P}\left(\mathbb{R}\right)$ be an exponent function with $p_{+}<\infty ,$ and let h be a Lebesgue measurable function; we define the modular functional related to $p(.)$ as

$$\begin{array}{c}\hfill {\eta }_{p(·)}\left(h\right)={\int }_{\mathbb{R}}{\left|h\left(u\right)\right|}^{p\left(u\right)}du.\end{array}$$

(1)

We shall write simply ${\eta }_{p(.)}\left(h\right)$ or $\eta \left(h\right)$ for the modular functional.

Proposition 1.

Suppose that $\Omega \subseteq {\mathbb{R}}^n$ and an exponent function $p(·)\in \mathcal{P}(\Omega )$. Then, the following properties hold for the modular functional η.

• For a function h, we have $\eta \left(h\right)\ge 0$ and also $\eta \left(h\right)=\eta \left(\right|h\left|\right)$.
• $\eta \left(h\right)=0$⇔$h=0$ for a.e. $u\in \Omega$.
• If $\eta \left(h\right)<\infty$, then $h<\infty$ for a.e. $u\in \Omega$.
• η is convex, i.e., for $a,b>0$ with $a+b=1$, we have

  $$\eta (ag+bh)\le a\eta \left(g\right)+b\eta \left(h\right).$$
• η is order-preserving, i.e., if $\left|g\right(u\left)\right|\ge \left|h\right(u\left)\right|$ a.e., then $\eta \left(g\right)\ge \eta \left(h\right)$.
• If, for some $\xi >0$, $\eta (h/\xi )<\infty$, then the function $\nu \mapsto \eta (h/\nu )$ is continuous and decreasing on $[\xi ,\infty )$. Furthermore, $\eta (h/\nu )\to 0$ as $\nu \to \infty$.

Definition 2.

Let $p(.)\in \mathcal{P}(\Omega )$, and we denote a variable Lebesgue space by $L^{p(.)}(\Omega )$ and define the set of all functions h that are Lebesgue-measurable and $\eta (h/\lambda )<\infty$ for some $\lambda >0$, i.e.,

$$\begin{array}{c}\hfill L^{p(.)}(\Omega )=\{h:{\eta }_{p(.)}(h/\lambda )<\infty ,\mathit{for}\mathit{some}\lambda >0\}.\end{array}$$

(2)

Definition 3.

Suppose that $p(.)\in \mathcal{P}(\Omega )$ and let h be a measurable function; then, the following formula defines a norm on $L^{p(.)}(\Omega ),$ making it a normed space.

$$\begin{array}{c}\hfill {\left|\right|h\left|\right|}_{L_{p(.)}}=\inf \{\lambda >0:\eta (h/\lambda )\le 1\}.\end{array}$$

(3)

Proposition 2.

Suppose that $\Omega \subseteq {\mathbb{R}}^n$ and let $p(·)$ be any exponent function from the collection $\mathcal{P}(\Omega ).$ If $p_{+}<\infty ,$ then ∀$\kappa \ge 1,$ and we have ${\kappa }^{p_{-}}\eta \left(h\right)\le \eta \left(\kappa h\right)\le {\kappa }^{p_{+}}\eta \left(h\right).$ Moreover, for $0<\kappa <1$, the reverse inequalities hold true.

Proposition 3.

Suppose that $\Omega \subseteq {\mathbb{R}}^n$ and let $p(·)$ be any exponent function from the collection $\mathcal{P}(\Omega );$ suppose that $h\in L^{p(·)}(\Omega )$ such that ${\parallel h\parallel }_{L^{p(·)}}>0.$ In this case, we have that $\eta \left(h/{\parallel h\parallel }_{L^{p(·)}}\right)\le 1.$ Moreover, if $p_{+}<\infty ,$ then $\eta \left(h/{\parallel h\parallel }_{L^{p(·)}}\right)=1,$ for all non-trivial $h\in L^{p(·)}(\Omega ).$

Corollary 1.

Let $p(·)$ be any exponent function from the collection $\mathcal{P}(\Omega )$ and $p_{+}<\infty .$ Let ${\parallel h\parallel }_{p(·)}>1$, and we have, in this case, $\eta {\left(h\right)}^{1/p_{+}}\le {\parallel h\parallel }_{p(·)}\le \eta {\left(h\right)}^{1/p_{-}}.$ Moreover, for ${0<\parallel h\parallel }_{p(·)}\le 1$, and we have $\eta {\left(h\right)}^{1/p_{-}}\le {\parallel h\parallel }_{p(·)}\le \eta {\left(h\right)}^{1/p_{+}}.$ Further, if the exponent function $p(·)$ is a constant, then $p_{-}=p_{+}$, and these inequalities reduce to the following identity:

$$\begin{array}{c}\hfill {\parallel h\parallel }_p={\left({\int }_{\Omega }{\left|h\left(u\right)\right|}^{p}du\right)}^{1/p},\end{array}$$

which is the norm on classical Lebesgue spaces $L^p(\Omega )$.

Theorem 1.

[Hölder’s Inequality] Let $\Omega \subseteq {\mathbb{R}}^n$ and $p(·)$ be any exponent function from the collection $\mathcal{P}(\Omega )$. Then, ∀$h_1\in L^{p(·)}(\Omega )$, $h_2\in L^{p^{\prime }(·)}(\Omega )$; we have $h_1{h}_2\in L^1(\Omega )$ and

$${\int }_{\Omega }|h_1\left(u\right)h_2\left(u\right)|du\le K_{p(·)}\parallel h_1{\parallel }_{p(·)}{\parallel h_2\parallel }_{p^{\prime }(·)},$$

where $p^{\prime }(·)$ denotes the conjugate exponent function of $p(·)$ and $K_{p(·)}$, a constant that depends on $p(·)$.

Definition 4.

Given any $\Omega \subseteq {\mathbb{R}}^n$ and a measurable function that maps $p(·):\Omega \to \mathbb{R}$, the exponent function $p(·)$ is said to be locally log-Hölder continuous and is denoted as $p(·)\in L{H}_0(\Omega )$, if we have a constant $C_0$ and ∀$u_1,u_2\in \Omega$ with $|u_1-u_2| <{\displaystyle \frac{1}{2}}$,

$$\begin{array}{c}\hfill |p\left(u_1\right)-p\left(u_2\right)|\le {\displaystyle \frac{C_0}{-\log \left(\right|u_1-u_2\left|\right)}}.\end{array}$$

$p(·)$ is said to be log-Hölder continuous at infinity and is denoted as $p(·)\in L{H}_{\infty }(\Omega )$, if we have constants $C_{\infty }$, $p_{\infty }$ and ∀$u\in \Omega$,

$$|p\left(u\right)-p_{\infty }|\le {\displaystyle \frac{C_{\infty }}{\log (e+|u\left|\right)}}.$$

When $p(·)$ is log-Hölder continuous both locally and at infinity, we denote this as $p(·)\in LH(\Omega )$.

Definition 5.

Suppose that we have a function $h\in L_{loc}^1\left({\mathbb{R}}^n\right)$; the Hardy–Littlewood maximal operator of h is denoted by $\mathcal{M}h$ and is defined as

$$\mathcal{M}h\left(u\right)=\underset{Q\ni u}{\sup }{\displaystyle \frac{1}{\left|Q\right|}}{\int }_Q\left|h\left(v\right)\right|dv,u\in {\mathbb{R}}^n$$

(4)

Here, the supremum is calculated over all such cubes $Q\subset {\mathbb{R}}^n$ with sides parallel to the coordinate axes and containing $‘u’$.

The characterization of the boundedness of the Hardy–Littlewood maximal operator $\mathcal{M}$ in weighted Lebesgue spaces $L_w^p$ was addressed by Muckenhoupt [1] and further developed by Sawyer [2,3] in 1970.

Theorem 2.

Let $1<p<\infty .$ Suppose that $w_1$ and $w_2$ are non-negative measurable functions. Then, the following are equivalent.

• There exists $C>0$ such that, for all $h\in L^p\left(w_2\right),$

  $$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^n}{\left(\mathcal{M}h\left(u\right)\right)}^p{w}_1\left(u\right)du\le C{\int }_{{\mathbb{R}}^n}{\left(h\left(u\right)\right)}^p{w}_2\left(u\right)du.\end{array}$$
• The pair $(w_1,w_2)$ satisfies the following inequality—that is, there exists $C>0$ such that

  $$\begin{array}{c}\hfill {\int }_Q{\left(\mathcal{M}\delta {\chi }_Q\left(u\right)\right)}^p{w}_1\left(u\right)du\le C{\int }_Q\delta \left(u\right)du<\infty ,\end{array}$$
  ∀ cubes Q, where $\delta =w_2^{1-p^{\prime }}$ and $p^{\prime }$ is the conjugate exponent of $p.$
  If $w_1=w_2$, then $\left(1\right)$ and $\left(2\right)$ are equivalent to the following.
• w satisfies the following inequality—that is, there exists $C>0$ such that, for all cubes $Q,$

  $$\begin{array}{c}\hfill {\left({\displaystyle \frac{1}{\left|Q\right|}}{\int }_{Q}w\right)}^{1/p}{\left({\displaystyle \frac{1}{\left|Q\right|}}{\int }_Q{w}^{1-p^{\prime }}\right)}^{1/p^{\prime }}\le C.\end{array}$$

As in the continuation of the above results, the boundedness of the maximal operator $\mathcal{M}$ is extended to $L^{p(.)}$ spaces in [4].

Theorem 3.

Let there be any exponent function $p(.)$ from the collection $\mathcal{P}\left({\mathbb{R}}^n\right)$. Suppose $1/p(·)\in LH\left({\mathbb{R}}^n\right)$. Then, the following inequality holds:

$$\parallel \beta {\chi }_{\{u:\mathcal{M}h(u)>\beta \}}{\parallel }_{L^{p(·)}}\le C{\parallel h\parallel }_{L^{p(·)}}.$$

Furthermore, if $p_{-}>1$, it follows that

$${\parallel \mathcal{M}h\parallel }_{L^{p(·)}}\le C{\parallel h\parallel }_{L^{p(·)}}.$$

(5)

Here, the constant C depends on n, the log-Hölder continuity of $1/p(·)$, and the values of $p_{-}$ and $p_{\infty }$.

Remark 1.

Suppose that $p_{+}<\infty$; in this case, the hypothesis ${\displaystyle \frac{1}{p(·)}}\in LH(\Omega )$ is equivalent to assuming $p(·)\in LH(\Omega )$.

On the other hand, similar to the maximal operators, several other integral operators have been a significant focus of research for the past fifty years. Among these operators, the Bessel–Riesz operator is particularly notable. Therefore, we now shift our discussion to the Bessel–Riesz operator.

Definition 6.

Let $0<\gamma <\infty$ and $0<\alpha <n$. The Bessel–Riesz operator is denoted as $I_{\alpha ,\gamma }h$ and is defined through the convolution of a function $h\in L_{\mathrm{loc}}^{p(.)}\left({\mathbb{R}}^n\right)$ with the Bessel–Riesz kernel as

$$\begin{array}{c}\hfill I_{\alpha ,\gamma }\ast h\left(u\right)={\int }_{{\mathbb{R}}^n}{K}_{\alpha ,\gamma }(u-v)h\left(v\right)dv,\end{array}$$

(6)

where $K_{\alpha ,\gamma }\left(u\right)={\displaystyle \frac{{\left|u\right|}^{\alpha -n}}{{(1+|u\left|\right)}^{\gamma }}},u\in {\mathbb{R}}^n\setminus \left\{0\right\}$ and zero when $u=0.$

Note that the kernel $K_{\alpha ,\gamma }$ is the product of two distinct kernels, i.e., $K_{\alpha ,\gamma }\left(u\right)=K_{\gamma }\left(u\right)K_{\alpha }\left(u\right),$∀$u\in {\mathbb{R}}^n.$ According to [10], $K_{\gamma }={\displaystyle \frac{1}{{(1+|u\left|\right)}^{\gamma }}}$ and $K_{\alpha }={\left|u\right|}^{\alpha -n}$ are recognized as the Bessel kernel and the Riesz kernel, respectively. Thus, the kernel $K_{\alpha ,\gamma }$ is referred to as the Bessel–Riesz kernel, and the corresponding operator $I_{\alpha ,\gamma }$ is known as the Bessel–Riesz operator. Significant research has been carried out on the boundedness of this singular integral operator $I_{\alpha ,\gamma }f$ under various conditions. For example, in classical Lebesgue spaces, the boundedness of $I_{\alpha ,\gamma }f$ can be directly established by Young’s inequality, as shown in [6,11].

Lemma 1

(Young’s inequality for convolution). Suppose that $q,s,t\in [1,\infty ]$ and ${\displaystyle \frac{1}{t}}={\displaystyle \frac{1}{q}}+{\displaystyle \frac{1}{s}}-1.$ If $h_1\in L^q\left(\mathbb{R}\right)$ and $h_2\in L^s\left(\mathbb{R}\right)$, then $h_1\ast h_2\in L^t\left(\mathbb{R}\right),$ such that

$$\begin{array}{c}\hfill \parallel h_1\ast h_2{\parallel }_{L^t\left(\mathbb{R}\right)}\le \parallel h_1{\parallel }_{L^q\left(\mathbb{R}\right)}{\parallel h_2\parallel }_{L^s\left(\mathbb{R}\right)}.\end{array}$$

(7)

Further results on the boundedness of the Bessel–Riesz operator in classical Lebesgue spaces for kernel $K_{\alpha ,\beta ,\gamma }\left(u\right)={\displaystyle \frac{{\left|u\right|}^{\alpha -n}}{{(1+|u|}^{\beta }{)}^{\gamma }}},u\in {\mathbb{R}}^n\setminus \left\{0\right\}$ and zero when $u=0$ are provided in [11].

Lemma 2.

Let $0<\alpha <1,0<\beta <\infty$ and $0<\gamma <\infty ;$ then, $K_{\alpha ,\beta ,\gamma }\in L^s\left(\mathbb{R}\right),$ where ${\displaystyle \frac{1}{\beta \gamma +1-\alpha }}<s<{\displaystyle \frac{1}{1-\alpha }}.$

Theorem 4.

Let $0<\alpha <1,0<\beta <\infty$ and $0<\gamma <\infty .$ Assume that the kernel $K_{\alpha ,\beta ,\gamma }\in L^1\left(\mathbb{R}\right)$ and $h\in L^p\left(\mathbb{R}\right)$ for $1<p<\infty ;$ then, there exists a positive constant c such that

$$\parallel I_{\alpha ,\beta ,\gamma }{h\parallel }_{L^p\left(\mathbb{R}\right)}\le c\parallel K_{\alpha ,\beta ,\gamma }{\parallel }_{L^1\left(\mathbb{R}\right)}{\parallel h\parallel }_{L^p\left(\mathbb{R}\right)}.$$

Theorem 5.

Let $0<\alpha <1,0<\beta <\infty$ and $0<\gamma <\infty .$ Assume that the kernel $K_{\alpha ,\beta ,\gamma }\in L^s\left(\mathbb{R}\right)$ with ${\displaystyle \frac{1}{\beta \gamma +1-\alpha }}<s<{\displaystyle \frac{1}{1-\alpha }}$ and $h\in L^p\left(\mathbb{R}\right)$ for $1<p<{\displaystyle \frac{s-1}{s}};$ then, there exists a positive constant $C_0,$ such that

$$\parallel I_{\alpha ,\beta ,\gamma }{h\parallel }_{L^q\left(\mathbb{R}\right)}\le C_0\parallel K_{\alpha ,\beta ,\gamma }{\parallel }_{L^s\left(\mathbb{R}\right)}{\parallel h\parallel }_{L^p\left(\mathbb{R}\right)},$$

where ${\displaystyle \frac{1}{q}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{s}}-1.$

However, for variable Lebesgue spaces, the scenario is different. Young’s inequality for convolution does not hold true in such spaces, as highlighted in [4]. The boundedness of the Bessel–Riesz operator $I_{\alpha ,\gamma }f$ in variable Lebesgue spaces $L^{p(.)}$ is the main topic discussed in this manuscript. We use the boundedness of the maximal operator $\mathcal{M}$ in $L^{p(.)}$ spaces, Holder’s inequality, and the traditional dyadic decomposition technique to analyze the boundedness of $I_{\alpha ,\gamma }f$ in these spaces. In Lemma 3, we discuss the membership of the Bessel–Riesz kernel in Lebesgue spaces with variable exponents $L^{p(.)}$, while Lemmas 4–7 provide useful estimations that will be applied in our main results. Furthermore, in Theorem 6, we establish the boundedness of the Bessel–Riesz operator $I_{\alpha ,\gamma }f$ from $L^{p(.)}$ to $L^{p(.)}$. In Theorem 7, we establish the boundedness of $I_{\alpha ,\gamma }f$ from $L^{p(.)}$ to $L^{q(.)}$ without any constraint on the maximal function. Theorem 8 presents the results in a broader context, examining the boundedness of the Bessel–Riesz operator from $L^{p(.)}$ to $L^{q(.)}$ under some constraints on the maximal function. Additionally, we discuss how our results are more generalized in nature and show how the corresponding results in classical Lebesgue spaces are special cases of our results. Moreover, in certain scenarios, our results even cover cases that do not exist in the literature for the classical framework.

3. Main Results

Suppose that $f\in L^{p(.)}\left(\mathbb{R}\right)$. The Bessel–Riesz operator is defined as

$$\begin{array}{c}\hfill I_{\alpha ,\gamma }f\left(u\right):=(K_{\alpha ,\gamma }\ast f)\left(u\right)={\int }_{\mathbb{R}}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{[1+|u-v\left|\right]}^{\gamma }}}dv;u\in \mathbb{R}.\end{array}$$

We now present and prove our results concerning the boundedness of the Bessel–Riesz operator. Throughout this section, we assume that $1<p_{-}\le p_{+}<\infty ,q_{+}<\infty$ and $s_{+}<\infty .$

Theorem 6.

Let $0<\alpha <1$, $0<\gamma <\infty$ with $\alpha <\gamma$ and $p(.)\in LH\left({\mathbb{R}}_{+}\right)$; then, the Bessel–Riesz operator $I_{\alpha ,\gamma }f:L^{p(·)}\left({\mathbb{R}}_{+}\right)\to L^{p(·)}\left({\mathbb{R}}_{+}\right)$ is bounded.

Proof. 

Since $p_{+}<\infty$ and $p(.)\in LH\left({\mathbb{R}}_{+}\right)$, there exists a constant $c_0>0$ such that ${\parallel \mathcal{M}f\parallel }_{L^{p(.)}}\le c_0{\parallel f\parallel }_{L^{p(.)}}.$ Now, for any $u\in {\mathbb{R}}_{+}$ and $f\in L^{p(·)}$, we have

$$\begin{array}{cc}\hfill I_{\alpha ,\gamma }f\left(u\right)& =(K_{\alpha ,\gamma }\ast f)\left(u\right)={\int }_{\mathbb{R}}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{[1+|u-v\left|\right]}^{\gamma }}}dv\hfill \\ \hfill & ={\int }_{|u-v|<1}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{[1+|u-v\left|\right]}^{\gamma }}}dv+{\int }_{|u-v|\ge 1}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{[1+|u-v\left|\right]}^{\gamma }}}dv=I_1\left(u\right)+I_2\left(u\right).\hfill \end{array}$$

To estimate $I_1\left(u\right)$, we proceed as follows:

$$\begin{array}{cc}\hfill |I_1\left(u\right)|& \le {\int }_{|u-v|<1}{\displaystyle \frac{{|u-v|}^{\alpha -1}\left|f\left(v\right)\right|}{{[1+|u-v\left|\right]}^{\gamma }}}dv\hfill \\ \hfill & \le \sum _{k=-\infty }^{-1}{\int }_{2^k\le |u-v|<2^{k+1}}{\displaystyle \frac{{|u-v|}^{\alpha -1}\left|f\left(v\right)\right|}{{[1+|u-v\left|\right]}^{\gamma }}}dv\hfill \\ \hfill & \le \sum _{k=-\infty }^{-1}{\displaystyle \frac{{\left(2^k\right)}^{\alpha -1}}{{[1+\left(2^k\right)]}^{\gamma }}}{\int }_{2^k\le |u-v|<2^{k+1}}\left|f\left(v\right)\right|dv\hfill \\ \hfill & \le \mathcal{M}f\left(u\right)\sum _{k=-\infty }^{-1}{\displaystyle \frac{{\left(2^k\right)}^{\alpha }}{{(1+2^k)}^{\gamma }}}\hfill \\ \hfill & \le \mathcal{M}f\left(u\right)\sum _{k=-\infty }^{-1}{2}^{\alpha k}=C_1\mathcal{M}f\left(u\right).\hfill \end{array}$$

(8)

Now, we estimate $I_2\left(u\right),$ as follows:

$$\begin{array}{cc}\hfill |I_2\left(u\right)|& \le {\int }_{|u-v|\ge 1}{\displaystyle \frac{{|u-v|}^{\alpha -1}\left|f\left(v\right)\right|}{{[1+|u-v\left|\right]}^{\gamma }}}dv\hfill \\ \hfill & =\sum _{k=0}^{\infty }{\int }_{2^k\le |u-v|<2^{k+1}}{\displaystyle \frac{{|u-v|}^{\alpha -1}\left|f\left(v\right)\right|}{{[1+|u-v\left|\right]}^{\gamma }}}dv\hfill \\ \hfill & \le \sum _{k=0}^{\infty }{\displaystyle \frac{{\left(2^k\right)}^{\alpha -1}}{{[1+\left(2^k\right)]}^{\gamma }}}{\int }_{2^k\le |u-v|<2^{k+1}}\left|f\left(v\right)\right|dv\hfill \\ \hfill & \le \mathcal{M}f\left(u\right)\sum _{k=0}^{\infty }{\displaystyle \frac{{\left(2^k\right)}^{\alpha }}{{[1+\left(2^k\right)]}^{\gamma }}}\le \mathcal{M}f\left(u\right)\sum _{k=0}^{\infty }{2}^{k(\alpha -\gamma )}=C_2\mathcal{M}f\left(u\right)\hfill \end{array}$$

(9)

Combining estimates (8) and (9), we obtain

$$\begin{array}{cc}\hfill |I_{\alpha ,\gamma }f\left(u\right)|\le |I_1\left(u\right)| + |I_2\left(u\right)|& \le C_3\mathcal{M}f\left(u\right).\hfill \end{array}$$

For any $\lambda \in \{\lambda >0;{\eta }_{p(.)}(\mathcal{M}f/\lambda )\le 1\}$, we have $\left|{\displaystyle \frac{I_{\alpha ,\gamma }f\left(u\right)/C_3}{\lambda }}\right|\le {\displaystyle \frac{\left|\mathcal{M}f\right(u\left)\right|}{\lambda }}.$ Further, by definition of the modular, we obtain ${\eta }_{p(.)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_3}{\lambda }}\right)\le {\eta }_{p(.)}\left({\displaystyle \frac{\mathcal{M}f}{\lambda }}\right)\le 1.$ Thus, $\lambda \in \{\lambda >0;{\eta }_{p(.)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_3}{\lambda }}\right)\le 1\}$. Therefore, $\parallel I_{\alpha ,\gamma }f/C_3{\parallel }_{L^{p(.)}}\le {\parallel \mathcal{M}f\parallel }_{L^{p(.)}}.$ Now, by following the boundedness of the maximal operator, we obtain

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{p(.)}}\le C_4{\parallel f\parallel }_{L^{p(.)}}.$$

□

Lemma 3.

Let $0<\alpha <1$ and $0<\gamma <\infty$; then, $K_{\alpha ,\gamma }\in L^{s(.)}\left({\mathbb{R}}_{+}\right)$ for

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\gamma +1-\alpha }}<s_{-}\le s_{+}<{\displaystyle \frac{1}{1-\alpha }}.\end{array}$$

Proof. 

By definition of the modular, we have

$${\eta }_{s(.)}(K_{\alpha ,\gamma }/\lambda )={\int }_{{\mathbb{R}}_{+}}{\left|{\displaystyle \frac{K_{\alpha ,\gamma }\left(t\right)}{\lambda }}\right|}^{s\left(t\right)}dt.$$

For $0<\lambda <1$, we can write

$$\begin{array}{cc}\hfill {\eta }_{s(.)}(K_{\alpha ,\gamma }/\lambda )& \le {\displaystyle \frac{1}{{\lambda }^{s_{+}}}}{\int }_{{\mathbb{R}}_{+}}{\left|K_{\alpha ,\gamma }\left(t\right)\right|}^{s\left(t\right)}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{{\lambda }^{s_{+}}}}\left({\int }_0^1{\displaystyle \frac{t^{(\alpha -1)s\left(t\right)}}{{(1+t)}^{\gamma s\left(t\right)}}}dt+{\int }_1^{\infty }{\displaystyle \frac{t^{(\alpha -1)s\left(t\right)}}{{(1+t)}^{\gamma s\left(t\right)}}}dt\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{{\lambda }^{s_{+}}}}\left({\int }_0^1{t}^{(\alpha -1)s\left(t\right)}dt+{\int }_1^{\infty }{\displaystyle \frac{t^{(\alpha -1)s\left(t\right)}}{t^{\gamma s\left(t\right)}}}dt\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{{\lambda }^{s_{+}}}}\left({\int }_0^1{t}^{(\alpha -1)s_{+}}dt+{\int }_1^{\infty }{t}^{(\alpha -1-\gamma )s_{-}}dt\right).\hfill \end{array}$$

By adhering to the conditions on $s_{-}$ and $s_{+}$, both integrals on the right-hand side are finite. Therefore, we have

$${\eta }_{s(·)}\left({\displaystyle \frac{K_{\alpha ,\gamma }}{\lambda }}\right)<\infty ,\mathrm{for}0<\lambda <1.$$

Hence, $K_{\alpha ,\gamma }\in L^{s(·)}\left({\mathbb{R}}_{+}\right)$. A similar expression holds for $\lambda \ge 1$, with the only difference being the factor $1/{\lambda }^{s_{-}}$ replacing $1/{\lambda }^{s_{+}}$. □

Lemma 4.

Let $K_{\alpha ,\gamma }\in L^{s(.)}\left({\mathbb{R}}_{+}\right)$; then, there exists an integer $N_r$ depending upon $r,$ such that, for any $k<N_r,$

$${\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)\ge C_1{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{-}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}$$

(10)

and, for any $k\ge N_r,$

$${\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)\ge C_2{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{+}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}.$$

(11)

Proof. 

By definition of the modular, we have

$$\begin{array}{cc}\hfill {\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)& ={\int }_{{\mathbb{R}}_{+}}{\left({\displaystyle \frac{t^{(\alpha -1)}}{{(1+t)}^{\gamma }}}\right)}^{s\left(t\right)}dt\hfill \\ \hfill & =\sum _{k\in \mathbb{Z}}{\int }_{2^{k}r\le t<2^{k+1}r}{\displaystyle \frac{t^{(\alpha -1)s\left(t\right)}}{{(1+t)}^{\gamma s\left(t\right)}}}dt\hfill \\ \hfill & \ge \sum _{k\in \mathbb{Z}}{\int }_{2^{k}r\le t<2^{k+1}r}{\displaystyle \frac{t^{(\alpha -1)s\left(t\right)}}{{(1+t)}^{\gamma s_{+}}}}dt\hfill \\ \hfill & \ge \sum _{k\in \mathbb{Z}}{\displaystyle \frac{1}{{(1+2^{k+1}r)}^{\gamma s_{+}}}}{\int }_{2^{k}r\le t<2^{k+1}r}{t}^{(\alpha -1)s\left(t\right)}dt,\hfill \end{array}$$

and there exists an integer $N_r$ depending upon r such that $2^{N_r}r\ge 1,$ and $2^{N_r-1}r<1;$ surely, $N_r$ will be a positive integer if $r<1$ and a negative integer if $r>1$ and $N_r=0$ if $r=1.$ Thus, for $0<\alpha <1,$ we have

$$\begin{array}{cc}\hfill {\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)& \ge {\displaystyle \frac{1}{2^{\gamma s_{+}}}}(\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{1}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}{\int }_{2^{k}r\le t<2^{k+1}r}{t}^{(\alpha -1)s\left(t\right)}dt\hfill \\ \hfill & +\sum _{k=N_r}^{\infty }{\displaystyle \frac{1}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}{\int }_{2^{k}r\le t<2^{k+1}r}{t}^{(\alpha -1)s\left(t\right)}dt)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2^{\gamma s_{+}}}}(\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{1}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}{\int }_{2^{k}r\le t<2^{k+1}r}{t}^{(\alpha -1)s_{-}}dt\hfill \\ \hfill & +\sum _{k=N_r}^{\infty }{\displaystyle \frac{1}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}{\int }_{2^{k}r\le t<2^{k+1}r}{t}^{(\alpha -1)s_{+}}dt)\hfill \\ \hfill & =C_1\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{-}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}+C_2\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{+}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}},\hfill \end{array}$$

where $C_1={\displaystyle \frac{2^{(\alpha -1)s_{-}+1}-1}{2^{\gamma s_{+}}((\alpha -1)s_{-}+1)}}>0$ and $C_2={\displaystyle \frac{2^{(\alpha -1)s_{+}+1}-1}{2^{\gamma s_{+}}((\alpha -1)s_{+}+1)}}>0.$ Thus, for any $k<N_r,$ we have

$${\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)\ge C_1\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{-}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}\ge C_1{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{-}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}.$$

Similarly, for any $k\ge N_r,$ we have

$${\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right)\ge C_2\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{+}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}\ge C_2{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)s_{+}+1}}{{\left(1+2^{k}r\right)}^{\gamma s_{+}}}}.$$

□

Lemma 5.

Let $K_{\alpha ,\gamma }\in L^{s(.)}\left({\mathbb{R}}_{+}\right)$ and $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1;$ then, there exists an integer $N_r$ depending upon $r,$ such that, for any $k<N_r,$

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{1/s_{+}}\ge C_1{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)\frac{s_{-}}{s_{+}}+\frac{1}{s_{+}}}}{{\left(1+2^{k}r\right)}^{\gamma }}}$$

(12)

and, for any $k\ge N_r,$

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{1/s_{+}}\ge C_2{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)+\frac{1}{s_{+}}}}{{\left(1+2^{k}r\right)}^{\gamma }}}.$$

(13)

Proof. 

Since $s_{+}<\infty ,$ and $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1,$ by Propositions 2 and 3, we obtain

$$1={\eta }_{s(·)}\left(K_{\alpha ,\gamma }/{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}\right)\ge {\displaystyle \frac{1}{\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{s_{-}}}}{\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right),$$

or

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{s_{-}}\ge {\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right),$$

and clearly

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}\ge \parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{s_{-}}\ge {\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right),$$

and we can write

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{1/s_{+}}\ge {\eta }_{s(·)}{\left(K_{\alpha ,\gamma }\right)}^{1/s_{+}}.$$

Thus, by following Lemma 4, we obtain the required results. □

Lemma 6.

Let $K_{\alpha ,\gamma }\in L^{s(.)}\left({\mathbb{R}}_{+}\right)$ and $\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}\ge 1;$ then, there exists an integer $N_r$ depending upon $r,$ such that, for any $k<N_r,$

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}\ge C_1{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)\frac{s_{-}}{s_{+}}+\frac{1}{s_{+}}}}{{\left(1+2^{k}r\right)}^{\gamma }}}$$

(14)

and, for any $k\ge N_r,$

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}\ge C_2{\displaystyle \frac{{\left(2^{k}r\right)}^{(\alpha -1)+\frac{1}{s_{+}}}}{{\left(1+2^{k}r\right)}^{\gamma }}}.$$

(15)

Proof. 

Since $s_{+}<\infty ,$ and $\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}\ge 1,$ by Propositions 2 and 3, we obtain

$$1={\eta }_{s(·)}\left(K_{\alpha ,\gamma }/{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}\right)\ge {\displaystyle \frac{1}{\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{s_{+}}}}{\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right),$$

or

$$\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{s_{+}}\ge {\eta }_{s(·)}\left(K_{\alpha ,\gamma }\right);$$

thus, by following Lemma 4, we obtain the required results. □

Lemma 7.

For any $u\in {\mathbb{R}}_{+}$ and $f\in L^{p(·)}\left({\mathbb{R}}_{+}\right),$ the following estimation holds for the Bessel–Riesz operator:

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_4\left(\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}+{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\right).$$

(16)

Proof. 

Since, for any $u\in {\mathbb{R}}_{+}$ and $f\in L^{p(·)}\left(\mathbb{R}\right),$ by definition of the Bessel–Riesz operator, we have

$$I_{\alpha ,\gamma }f\left(u\right)={\int }_{{\mathbb{R}}_{+}}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{(1+|u-v\left|\right)}^{\gamma }}}dv=\sum _{k\in \mathbb{Z}}{\int }_{2^{k}r\le |u-v|<2^{k+1}r}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{(1+|u-v\left|\right)}^{\gamma }}}dv,$$

there exists an integer $N_r$ depending upon r such that $2^{N_r}r\ge 1$ and $2^{N_r-1}r<1.$ Thus, we have

$$\begin{array}{cc}\hfill I_{\alpha ,\gamma }f\left(u\right)& =\sum _{k=-\infty }^{N_r-1}{\int }_{2^{k}r\le |u-v|<2^{k+1}r}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{(1+|u-v\left|\right)}^{\gamma }}}dv+\sum _{k=N_r}^{\infty }{\int }_{2^{k}r\le |u-v|<2^{k+1}r}{\displaystyle \frac{{|u-v|}^{\alpha -1}f\left(v\right)}{{(1+|u-v\left|\right)}^{\gamma }}}dv\hfill \\ \hfill & =I_1\left(u\right)+I_2\left(u\right).\hfill \end{array}$$

(17)

Now, first, we estimate $I_1\left(u\right)$:

$$\begin{array}{cc}\hfill |I_1\left(u\right)|& \le \sum _{k=-\infty }^{N_r-1}{\int }_{2^{k}r\le |u-v|<2^{k+1}r}{\displaystyle \frac{{|u-v|}^{\alpha -1}\left|f\left(v\right)\right|}{{(1+|u-v\left|\right)}^{\gamma }}}dv\hfill \\ \hfill |I_1\left(u\right)|& \le \sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1}}{{(1+2^{k}r)}^{\gamma }}}{\int }_{2^{k}r\le |u-v|<2^{k+1}r}\left|f\left(v\right)\right|dv\hfill \\ \hfill & =\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}.\hfill \end{array}$$

(18)

To estimate $I_2\left(u\right)$, we proceed as follows:

$$\begin{array}{cc}\hfill |I_2\left(u\right)|& \le \sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1}}{{(1+2^{k}r)}^{\gamma }}}{\int }_{2^{k}r\le |u-v|<2^{k+1}r}\left|f\left(v\right)\right|dv.\hfill \end{array}$$

By using Hölder’s inequality, we obtain

$$\begin{array}{cc}\hfill |I_2\left(u\right)|& \le C_3\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1}}{{(1+2^{k}r)}^{\gamma }}}{\parallel f\parallel }_{L^{p(.)}[u+2^{k}r,u+2^{k+1}r)}{\parallel {\chi }_{[u+2^{k}r,u+2^{k+1}r)}\parallel }_{L^{p^{\prime }(.)}(u+2^{k}r,u+2^{k+1}r)}\hfill \\ \hfill & \le C_3\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1}}{{(1+2^{k}r)}^{\gamma }}}{\parallel f\parallel }_{L^{p(.)}}{\parallel {\chi }_{[u+2^{k}r,u+2^{k+1}r)}\parallel }_{L^{p^{\prime }(.)[u+2^{k}r,u+2^{k+1}r)}}.\hfill \end{array}$$

(19)

By simplifying and using ${\left(p^{\prime }\right)}_{-}=(p_{+}{)}^{\prime }$, we obtain

$$\begin{array}{cc}\hfill |I_2\left(u\right)|& \le C_3{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1}{\left(2^{k}r\right)}^{\frac{1}{{\left(p_{+}\right)}^{\prime }}}}{{(1+2^{k}r)}^{\gamma }}}\hfill \\ \hfill & =C_3{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}.\hfill \end{array}$$

(20)

Thus, by combining (18) and (20), we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_4\left(\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}+{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\right).$$

□

Theorem 7.

Let $p(·),q(·),s(·)\in \mathcal{P}\left({\mathbb{R}}_{+}\right),$$0<\alpha <1$ and $0<\gamma <\infty$ with Range $\left(s(·)\right)\subset \left({\displaystyle \frac{1}{\gamma +1-\alpha }},{\displaystyle \frac{1}{1-\alpha }}\right).$ If there exists a constant $c>0$ such that ${\displaystyle \frac{1}{q_{+}}}\le c\left({\displaystyle \frac{1}{p_{+}}}+{\displaystyle \frac{1}{s_{+}}}-1\right)$ and ${\displaystyle \frac{q(·)}{p(·)}}=1+{\displaystyle \frac{c{q}_{+}(\alpha (s_{+}-s_{-})+s_{-}-1)}{s_{+}}}$ a.e. with $p(.)\in LH\left({\mathbb{R}}_{+}\right)$, then the Bessel–Riesz operator $I_{\alpha ,\gamma }:L^{p(·)}\to L^{q(·)}$ is bounded. Further, for any $f\in L^{p(.)}$, there exists a constant $C>0$, such that, for $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1,$ we obtain

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(.)}}\le C\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}^{1/s_{+}}{\parallel f\parallel }_{L^{p(.)}}$$

(21)

and, for $\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}\ge 1,$ we obtain

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(.)}}\le C\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}{\parallel f\parallel }_{L^{p(.)}}.$$

(22)

Proof. 

Since $p_{+}<\infty$ and $p(·)\in LH\left({\mathbb{R}}_{+}\right),$ the Hardy–Littlewood maximal operator $\mathcal{M}$ is bounded, and there exists a constant $C_0>0$ such that, for any $f\in L^{p(·)}\left({\mathbb{R}}_{+}\right),$ we have ${\parallel \mathcal{M}f\parallel }_{L^{p(·)}}\le C_0{\parallel f\parallel }_{L^{p(·)}}.$ We can write ${\parallel \mathcal{M}f\parallel }_{L^{p(·)}}\le {\parallel C_{0}f\parallel }_{L^{p(·)}},$ further since variable Lebesgue space $L^{p(·)}$ is a vector space, and we also have $C_{0}f\in L^{p(·)}.$ Thus, without loss of generality, we can choose any fixed $f\in L^{p(·)}$ such that ${\parallel \mathcal{M}f\parallel }_{L^{p(·)}}\le {\parallel f\parallel }_{L^{p(·)}}.$ Further, by following the definition of the norm in $L^{p(·)},$ we obtain

$${\parallel f\parallel }_{L^{p(·)}}\in \{\lambda >0:{\eta }_{p(·)}\left(\mathcal{M}f/\lambda \right)\le 1\}.$$

(23)

The case is trivial for $f=0$.

Now, for any $u\in {\mathbb{R}}_{+}$ and any non-zero function $f\in L^{p(·)},$ by following Lemma 7, we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_4\left(\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}+{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\right).$$

(24)

Now, before we proceed further, first, we derive the following expression, which we shall use later:

$$\begin{array}{c}\hfill A=r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}.\end{array}$$

(25)

We choose $r={\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{-{\displaystyle \frac{p\left(u\right)c{q}_{+}}{q\left(u\right)}}},$ where ${\parallel f\parallel }_{L^{p(·)}}\ne 0.$ Then, from (25), it follows that

$$\begin{array}{cc}\hfill A& ={\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{-c{q}_{+}{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(.)}}\hfill \\ \hfill & ={\displaystyle \frac{{\parallel f\parallel }_{L^{p(·)}}^{\frac{bp\left(u\right)c{q}_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}+{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)c{q}_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)c{q}_{+}}{q\left(u\right)s_{+}}-1}{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}},\hfill \end{array}$$

where $b=\alpha (s_{+}-s_{-})+s_{-}-1$, and we have ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}=1+{\displaystyle \frac{bc{q}_{+}}{s_{+}}}.$ Hence, ${\displaystyle \frac{bp\left(u\right)c{q}_{+}}{q\left(u\right)s_{+}}}+{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}-1=0,$ and we obtain

$$\begin{array}{cc}\hfill A& =2{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(·)}}.\hfill \end{array}$$

Hence,

$$r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}=2{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(·)}}.$$

(26)

Now, we discuss two cases depending upon the norm of the Bessel–Riesz kernel.

Case 1. If $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1,$ then, by following (12), we have

$$\begin{array}{cc}\hfill \mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}& \le C_1\mathcal{M}f\left(u\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{\left(2^{k}r\right)}^{(\alpha -1)\frac{s_{-}}{s_{+}}+\frac{1}{s_{+}}}}}\hfill \\ \hfill & =C_1{r}^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\sum _{k=-\infty }^{N_r-1}{2}^{{\displaystyle \frac{k}{s_{+}}}(\alpha (s_{+}-s_{-})+s_{-}-1)}\hfill \\ \hfill & \le C_2{r}^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}.\hfill \end{array}$$

(27)

Further, by following (13) and the fact that $1-{\displaystyle \frac{1}{p_{+}}}-{\displaystyle \frac{1}{s_{+}}}\le -{\displaystyle \frac{1}{c{q}_{+}}}$, we obtain

$$\begin{array}{cc}\hfill {\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}& \le C_3{\parallel f\parallel }_{L^{p(.)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{(2^{k}r{)}^{\alpha -\frac{1}{p_{+}}}}{{\left(2^{k}r\right)}^{(\alpha -1)+\frac{1}{s_{+}}}}}\hfill \\ \hfill & =C_3{r}^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\sum _{k=N_r}^{\infty }{\left(2^k\right)}^{-{\displaystyle \frac{1}{c{q}_{+}}}}\hfill \\ \hfill & \le C_4{r}^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}.\hfill \end{array}$$

(28)

By (24), (27) and (28), we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_5\left(r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}.$$

(29)

Thus, by following (26), we obtain

$${\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}},$$

or

$${\left({\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{q\left(u\right)}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{p\left(u\right)};$$

thus,

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_5{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right).$$

Consequently, we have

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le 1.$$

Thus,

$${\parallel f\parallel }_{L^{p(·)}}\in \{\lambda >0:{\eta }_{q(·)}\left(\left(I_{\alpha ,\gamma }f/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\right)/\lambda \right)\le 1\}.$$

Hence,

$$\parallel {\displaystyle \frac{I_{\alpha ,\gamma }f}{C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}}{\parallel }_{L^{q(·)}}\le {\parallel f\parallel }_{L^{p(·)}},$$

or

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(·)}}\le C_6\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{1/s_{+}}{\parallel f\parallel }_{L^{p(·)}}.$$

Case 2. If $\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}\ge 1,$ by following (14) and using a similar procedure as in Case 1, we obtain

$$\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}\le C_2{r}^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

(30)

Further, by following (15), we obtain

$${\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\le C_4{r}^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

(31)

By (24), (30) and (31), we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_6\left(r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{c{q}_{+}}}}{\parallel f\parallel }_{L^{p(.)}}\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

(32)

Now, by following (26), we obtain

$${\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}{{\parallel f\parallel }_{L^{p(·)}}}}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}};$$

thus,

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_6{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right).$$

Consequently, we have

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(·)}}\le C_6{\parallel f\parallel }_{L^{p(·)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

□

Corollary 2.

If all exponent functions in Theorem 6 become constants, such that $q(·)=q_{-}=q_{+}=q,$$p(·)=p_{-}=p_{+}=p$ and $s(·)=s_{-}=s_{+}=s,$ with $c=1$, the expression ${\displaystyle \frac{1}{q_{+}}}\le c\left({\displaystyle \frac{1}{p_{+}}}+{\displaystyle \frac{1}{s_{+}}}-1\right)$ becomes ${\displaystyle \frac{1}{q}}\le {\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{s}}-1.$ Further, the relation ${\displaystyle \frac{q(·)}{p(·)}}=1+{\displaystyle \frac{c{q}_{+}(\alpha (s_{+}-s_{-})s_{-}-1)}{s_{+}}}$ becomes ${\displaystyle \frac{1}{q}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{s}}-1.$ Thus, by combining both, we obtain ${\displaystyle \frac{1}{q}}={\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{s}}-1.$ In such a situation, our results coincide with the results given in [6,11] for classical Lebesgue spaces and proven by using Young’s inequality.

Remark 2.

Since Theorem 7 holds only when ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}=1+{\displaystyle \frac{b{q}_{+}}{s_{+}}}$ a.e. and ${\displaystyle \frac{1}{q_{+}}}\le c\left({\displaystyle \frac{1}{p_{+}}}+{\displaystyle \frac{1}{s_{+}}}-1\right)$, the result does not hold if the relation ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}=1+{\displaystyle \frac{b{q}_{+}}{s_{+}}}$ a.e. fails for $u\in {\mathbb{R}}_{+}$. Therefore, in Theorem 8, we discuss the cases where the boundedness holds for all possible scenarios of the exponent functions. However, in this case, additional conditions on the maximal function in terms of the norm ${\parallel f\parallel }_{L^{p(·)}}$ are required, which were not needed in Theorem 7. Furthermore, Theorem 8 does not cover Theorem 7 because we impose conditions on the maximal function in Theorem 8, while Theorem 7 does not depend on such conditions. In the case of equality, Theorem 7 becomes a special case of Theorem 8.

Definition 7.

Let $0<\alpha <1$; then, the pair of exponent functions $\left(q\right(·),p(·\left)\right)$ is said to satisfy the boundedness property if, for any $u\in A=\{u\in {\mathbb{R}}_{+}:\mathcal{M}f\left(u\right)\le \parallel f{\parallel }_{L^{p(·)}}\}$, the exponents satisfy ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}\ge 1+{\displaystyle \frac{q_{+}(\alpha (s_{+}-s_{-})+s_{-}-1)}{s_{+}}}$ a.e. and, for any $u\in B=\{u\in {\mathbb{R}}_{+}:\mathcal{M}f\left(u\right)>\parallel f{\parallel }_{L^{p(·)}}\}$, the exponents satisfy ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}<1+{\displaystyle \frac{q_{+}(\alpha (s_{+}-s_{-})+s_{-}-1)}{s_{+}}}$ a.e.

Theorem 8.

Let $0<\gamma <\infty$, and let the pair $\left(q\right(·),p(·\left)\right)$ satisfy the boundedness property and ${\displaystyle \frac{1}{\gamma +1-\alpha }}<s_{-}\le s_{+}<{\displaystyle \frac{1}{1-\alpha }}$, ${\displaystyle \frac{1}{q_{+}}}={\displaystyle \frac{1}{p_{+}}}+{\displaystyle \frac{1}{s_{+}}}-1$ with $p(.)\in LH\left(\mathbb{R}\right)$; then, the Bessel–Riesz operator $I_{\alpha ,\gamma }:L^{p(·)}\to L^{q(·)}$ is bounded. Further, for any $f\in L^{p(.)}$, there exists a constant $C>0$, such that, for $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1,$ we have

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(.)}}\le C\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}^{1/s_{+}}{\parallel f\parallel }_{L^{p(.)}}$$

(33)

and, for $\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}\ge 1,$ we have

$$\parallel I_{\alpha ,\gamma }{f\parallel }_{L^{q(.)}}\le C\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}{\parallel f\parallel }_{L^{p(.)}}.$$

(34)

Proof. 

The case is trivial for $f=0$.

Now, for any $u\in {\mathbb{R}}_{+}$ and any non-zero function $f\in L^{p(·)},$ by Lemma 7, we have

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_4\left(\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}+{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -\frac{1}{p_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\right).$$

Thus, by using ${\displaystyle \frac{1}{q_{+}}}={\displaystyle \frac{1}{p_{+}}}+{\displaystyle \frac{1}{s_{+}}}-1,$ we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_4\left(\mathcal{M}f\left(u\right)\sum _{k=-\infty }^{N_r-1}{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha }}{{(1+2^{k}r)}^{\gamma }}}+{\parallel f\parallel }_{L^{p(.)}}\sum _{k=N_r}^{\infty }{\displaystyle \frac{{\left(2^{k}r\right)}^{\alpha -1+\frac{1}{s_{+}}-\frac{1}{q_{+}}}}{{(1+2^{k}r)}^{\gamma }}}\right).$$

(35)

Now, to proceed further, first, we simplify the following for $r={\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{-q_{+}{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}},$ where ${\parallel f\parallel }_{L^{p(·)}}\ne 0,$ so

$$\begin{array}{cc}\hfill T& =r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{q_{+}}}}{\parallel f\parallel }_{L^{p(.)}}\hfill \\ \hfill & ={\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{-q_{+}{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(.)}}\hfill \\ \hfill & ={\displaystyle \frac{{\parallel f\parallel }_{L^{p(·)}}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}-1}}}+{\displaystyle \frac{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{p\left(u\right)}{q\left(u\right)}}}{{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}}\hfill \\ \hfill & ={\displaystyle \frac{{\parallel f\parallel }_{L^{p(·)}}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}+{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}-1}{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}},\hfill \end{array}$$

(36)

where $b=\alpha (s_{+}-s_{-})+s_{-}-1.$ For any $u\in {\mathbb{R}}_{+},$ we discuss the following two possibilities. If $u\in A,$ then ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}\ge 1+{\displaystyle \frac{q_{+}(\alpha (s_{+}-s_{-})+s_{-}-1)}{s_{+}}}=1+{\displaystyle \frac{b{q}_{+}}{s_{+}}}$ a.e. This implies that ${\displaystyle \frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}}+{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}-1\le 0$ a.e. and we obtain from (36)

$$\begin{array}{cc}\hfill T& \le {\displaystyle \frac{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}+{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}-1}{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}}\hfill \\ \hfill & =2{\displaystyle \frac{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}-1}{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}}\hfill \\ \hfill & =2{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(·)}}.\hfill \end{array}$$

(37)

If $u\in B$, then ${\displaystyle \frac{q\left(u\right)}{p\left(u\right)}}<1+{\displaystyle \frac{q_{+}(\alpha (s_{+}-s_{-})+s_{-}-1)}{s_{+}}}=1+{\displaystyle \frac{b{q}_{+}}{s_{+}}}$ a.e.; thus, ${\displaystyle \frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}}+{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}-1>0$ a.e. and we obtain from (36)

$$\begin{array}{cc}\hfill T& \le {\displaystyle \frac{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}+{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}+\frac{p\left(u\right)}{q\left(u\right)}-1}}{{\left(\mathcal{M}f\left(u\right)\right)}^{\frac{bp\left(u\right)q_{+}}{q\left(u\right)s_{+}}-1}{\parallel f\parallel }_{L^{p(·)}}^{\frac{p\left(u\right)}{q\left(u\right)}-1}}}\hfill \\ \hfill & =2{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(·)}}.\hfill \end{array}$$

(38)

Thus, following from (37) and (38), for any $u\in {\mathbb{R}}_{+},$ we have

$$T\le 2{\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}}{\parallel f\parallel }_{L^{p(·)}}.$$

(39)

Case 1. If $0<\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(.)}}<1,$ then, by Lemma 5 and relation (35), we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_7\left(r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{q_{+}}}}{\parallel f\parallel }_{L^{p(.)}}\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}.$$

By following (38), we obtain, for any $u\in R_{+},$

$${\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_8{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}},$$

or

$${\left({\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_8{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{q\left(u\right)}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{p\left(u\right)};$$

thus,

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_8{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right).$$

Since, by the boundedness of the maximal function, we have ${\parallel f\parallel }_{L^{p(·)}}\in \{\lambda >0:{\eta }_{p(·)}\left(\mathcal{M}f/\lambda \right)\le 1\},$ we obtain

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_8{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le 1.$$

Thus,

$${\parallel f\parallel }_{L^{p(·)}}\in \{\lambda >0:{\eta }_{q(·)}\left(\left(I_{\alpha ,\gamma }f\left(u\right)/C_8{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}\right)/\lambda \right)\le 1\}.$$

Hence,

$${\parallel I_{\alpha ,\gamma }f\parallel }_{L^{q(·)}}/C_8\parallel K_{\alpha ,\gamma }{\parallel }_{L^{s(·)}}^{1/s_{+}}\le {\parallel f\parallel }_{L^{p(·)}},$$

or

$${\parallel I_{\alpha ,\gamma }f\parallel }_{L^{q(·)}}\le C_8{\parallel f\parallel }_{L^{p(·)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}^{1/s_{+}}.$$

Case 2. If ${\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(.)}}\ge 1,$ by Lemma 6 and relation (35), and then by the convergence of the geometric series, we obtain

$$|I_{\alpha ,\gamma }f\left(u\right)|\le C_9\left(r^{{\displaystyle \frac{\alpha (s_{+}-s_{-})+s_{-}-1}{s_{+}}}}\mathcal{M}f\left(u\right)+r^{-{\displaystyle \frac{1}{q_{+}}}}{\parallel f\parallel }_{L^{p(.)}}\right){\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

By using the value of $T,$ from (38) for any $u\in R_{+},$ we have

$${\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_{10}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}{{\parallel f\parallel }_{L^{p(·)}}}}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{{\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}},$$

or

$${\left({\displaystyle \frac{|I_{\alpha ,\gamma }f\left(u\right)|/C_{10}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{q\left(u\right)}\le {\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{p\left(u\right)}.$$

Thus,

$${\eta }_{q(·)}\left({\displaystyle \frac{I_{\alpha ,\gamma }f/C_{10}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le {\eta }_{p(·)}\left({\displaystyle \frac{\mathcal{M}f}{{\parallel f\parallel }_{L^{p(·)}}}}\right)\le 1.$$

Thus,

$${\parallel f\parallel }_{L^{p(·)}}\in \left\{\lambda >0:{\eta }_{q(·)}\left(\left(I_{\alpha ,\gamma }f\left(u\right)/C_{10}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}\right)/\lambda \right)\le 1\right\}.$$

Hence,

$$\parallel {\displaystyle \frac{I_{\alpha ,\gamma }f\left(u\right)}{C_{10}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}}}{\parallel }_{L^{q(·)}}\le {\parallel f\parallel }_{L^{p(·)}},$$

or

$${\parallel I_{\alpha ,\gamma }f\parallel }_{L^{q(·)}}\le C_{10}{\parallel f\parallel }_{L^{p(·)}}{\parallel K_{\alpha ,\gamma }\parallel }_{L^{s(·)}}.$$

□

Remark 3.

If we choose $r={\left({\displaystyle \frac{\mathcal{M}f\left(u\right)}{{\parallel f\parallel }_{L^{p(·)}}}}\right)}^{\left({\displaystyle \frac{p\left(u\right)}{q\left(u\right)}}-1\right){\displaystyle \frac{s_{+}}{\alpha (s_{+}-s_{-})+s_{-}-1}}}$ to estimate T in Theorem 8, then, by following a similar procedure, we obtain the same results.

Remark 4.

When $p(·)=q(·)$, it follows from Theorem 8 that $\alpha (s_{+}-s_{-})=1-s_{-}$. This relation is valid only when $s_{-}=1$, which implies $s_{+}=s_{-}$ and thus $s(·)$ is constant. However, for $s_{-}>1$, Theorem 7 does not hold in the case $p(·)=q(·)$. Therefore, Theorem 7 is not covered by Theorem 8. Additionally, in Theorem 8, we require additional conditions on $\mathcal{M}f\left(u\right)$.

4. Conclusions

In this manuscript, we present significant findings in harmonic analysis concerning the boundedness of the Bessel–Riesz operator in variable Lebesgue spaces, a cutting-edge topic in the field. Our approach to investigating this boundedness is based on standard techniques in analysis, relying on the boundedness of the Hardy–Littlewood maximal operator, Hölder’s inequality, and the classical dyadic decomposition method. Moreover, we explore different scenarios, and our results are more general than those already established in classical cases.