ON BOUNDEDNESS OF THE GENERALIZED RIESZ POTENTIAL IN LOCAL MORREY-TYPE SPACES

Abstract

For all admissible values of the numerical parameters sharp sufficient conditions on the functional parameters are obtained ensuring the boundedness of the generalized Riesz potential from one general local Morrey-type space to another one, which, for a certain range of the numerical parameters, coincide with the necessary ones.

Introduction

In this paper, we study the boundedness from one general local Morrey-type space to another one of the generalized Riesz potential

$$\begin{aligned} \left( I_{\rho (\cdot )}f\right) (x) = \int _{\mathbb {R}^{n}}^{}\rho (\vert x-y\vert )f(y) dy,\ x\in \mathbb {R}^{n}, \end{aligned}$$

under certain assumptions on the kernel $\rho$. Our aim is to generalize the results obtained in [6] for the case of the classical Riesz potential $I_{\alpha },$ in which $\rho (t)=t^{\alpha -n}$, $t>0$, $0<\alpha <n$. Some of the presented results were stated without proofs in [7]. Let $F,G : A\times B \longrightarrow [0,\infty ] .$ Throughout this paper we say that F is dominated by G uniformly in $x\in A$ and write

$$\begin{aligned} F \lesssim G\ \text {uniformly\ in}\ x \in A \end{aligned}$$

if there exists $c(B)>0$ such that

$$\begin{aligned} F(x,y)\le c(B) G(x,y) \ \text {for all} \ x \in A. \end{aligned}$$

(So c(B) is independent of $x \in A,$ but may depend on $y \in B$.) Respectively, we say that F dominates G uniformly in $x\in A$ and write

$$\begin{aligned} F \gtrsim G\ \text {uniformly\ in}\ x \in A \end{aligned}$$

if there exists $c(B)>0$ such that

$$\begin{aligned} F(x,y)\ge c(B)G(x,y) \ \text {for all} \ x \in A. \end{aligned}$$

We also say that F is equivalent to G uniformly in $x\in A$ and write

$$\begin{aligned} F \thickapprox G\ \text {uniformly\ in}\ x \in A \end{aligned}$$

if F and G dominate each other uniformly in $x\in A.$

Definitions and basic properties of general Morrey-type spaces

In this section we recall basic facts of the theory of general Morrey-type spaces.

Let, for a Lebesgue measurable set $\Omega \subset \mathbb {R}^{n},$ $\mathfrak {M}(\Omega )$ denote the space of all functions $f: \Omega \longrightarrow \mathbb {C}$ Lebesgue measurable on $\Omega ,$ and $\mathfrak {M^{+}}(\Omega )$ denote the subset of $\mathfrak {M}(\Omega )$ of all non-negative functions.

Definition 2.1

Let $0<p,\theta \le \infty$ and let $w \in \mathfrak {M^{+}}\big ((0,\infty )\big )$ be not equivalent to 0. We denote by $LM_{p\theta ,w (\cdot )}$ the local Morrey-type space, the space of all functions $f\in L_{p}^{ \text {loc}}(\mathbb {R}^{n})$ with finite quasinorm

$$\begin{aligned} \Vert f\Vert _{LM_{p\theta ,w (\cdot )}}\equiv \Vert f\Vert _{LM_{p\theta ,w (\cdot )}(\mathbb {R}^{n})}=\Vert w(r)\Vert f\Vert _{L_{p}(B(0,r))} \Vert _{L_{\theta }(0,\infty )}. \end{aligned}$$

Definition 2.2

Let $0<p,\theta \le \infty$. We denote by $\Omega _{\theta }$ the set of all functions $w \in \mathfrak {M^{+}}((0,\infty ))$ which are not equivalent to 0 and such that

$$\begin{aligned} \Vert w\Vert _{L_{\theta }(t,\infty )}<\infty \end{aligned}$$

(2.1)

for some $t>0$ .

Lemma 2.1

[3,4,5]. Let $0<p$, $\theta \le \infty$ and let w be a non-negative Lebesgue measurable function on $(0, \infty )$, which is not equivalent to 0.

Then the space $LM_{p\theta , w(\cdot )}$ is nontrivial if and only if $w \in \Omega _{\theta }$.

In the sequel, keeping in mind Lemma 2.1, we always assume that $w \in \Omega _{\theta }$ for the local Morrey-type spaces $LM_{p\theta ,w(\cdot )}$.

It is well known that the spaces $LM_{p \theta , w(\cdot )}$ are Banach spaces if $1\le p$, $\theta \le \infty$ and are quasi-Banach spaces if $0< p <1$, or $0<\theta < 1$, or both $0<p$, $\theta <1$.

For further properties of general Morrey-type spaces, operators acting in such spaces and applications see, for example, survey papers [1, 2, 12,13,14,15,16] and references therein.

$L_{p}$ and $WL_{p}$ estimates of generalized Riesz potentials over balls

Let $n \in \mathbb {N}, 0<p< \infty , \Omega \subset \mathbb {R}^{n}$ be a Lebesgue measurable set. Recall that the weak $L_{p}$-space $WL_{p}(\Omega )$ is the space of all Lebesgue measurable functions $f: \Omega \rightarrow \mathbb {C}$ for which

$$\begin{aligned} \Vert f \Vert _{ WL_{p}(\Omega )} = \sup \limits _{ t\ge 0} t \left| \left\{ y \in \Omega : \vert f(y)\vert > t \right\} \right| ^{\frac{1}{p}} < \infty . \end{aligned}$$

Here $\vert G \vert$ denotes the Lebesgue measure of a set $G \subset \mathbb {R}^{n}.$

Remark 3.1

Let $M > 0.$ It follows directly from the above definition that, if f is equivalent to M on $\Omega$, then

$$\begin{aligned} \Vert M \Vert _{ WL_{p}(\Omega )} = \Vert M \Vert _{L_{p}(\Omega )} = M\vert \Omega \vert ^{\frac{1}{p}} \end{aligned}$$

and if $f : \Omega \rightarrow \mathbb {R}$ is such that $f \geqslant M$ almost everywhere on $\Omega$, then

$$\begin{aligned} \Vert f \Vert _{WL_{p}(\Omega )} \ge M \vert \Omega \vert ^{\frac{1}{p}}. \end{aligned}$$

Definition 3.1

Let $n \in \mathbb {N}.$ We say that $\rho \in S_{n}$ if $\rho \in \mathfrak {M^{+}}((0,\infty ))$ and

1) ${\int _{0}^{r}}\rho (t)t^{n-1}dt < \infty$ for all $r > 0,$

2) for some $c_{1}, c_{2}> 0$,

$$\begin{aligned} c_{1}\rho (t)\le \rho (s)\le c_{2}\rho (t) \end{aligned}$$

for all $s,t>0$ satisfying the inequality $\frac{t}{2}\le s\le 2t.$

Remark 3.2

Let $x \in \mathbb {R}^{n}$, $r>0$, $y \in B(x,r)$, $z \in$ $^{c}B(x,2r).$ Then

$$\begin{aligned} \vert y-z \vert \le \vert y-x \vert + \vert x-z \vert \le r + \vert x-z \vert \le \frac{1}{2} \vert x-z \vert + \vert x-z \vert <2 \vert x-z \vert \end{aligned}$$

and

$$\begin{aligned} \vert y-z \vert \ge \vert x-z \vert - \vert y-x \vert \ge \vert x-z \vert -r \ge \vert x-z \vert - \frac{1}{2} \vert x-z \vert = \frac{1}{2} \vert x-z \vert . \end{aligned}$$

Let $s= \vert y-z\vert$ and $t= \vert x-z\vert .$ Then by Condition 2) of Definition 3.1 we have

$\text {3)} \ \ c_{1}\rho (\vert x-z \vert )\le \rho (\vert y-z \vert )\le c_{2}\rho (\vert x-z \vert )$

for all $x \in \mathbb {R}^{n}$, $r>0$, $y \in B(x,r)$, $z \in$ $^{c}B(x,2r).$

Remark 3.3

If functions $\rho _{1},\rho _{2} : (0,\infty )\rightarrow (0,\infty )$ satisfy Condition 2) of Definition 3.1 with $c_{11}$, $c_{12}>0$, $c_{21}$, $c_{22}>0$, respectively, then the product $\rho _{1}\rho _{2}$ satisfies Condition 2) with $c_{1}=c_{11} c_{21}$ and $c_{2}=c_{12} c_{22}$. Indeed, it suffices to multiply the inequalities

$$\begin{aligned} c_{i1}\ \rho _{i}(t)\le \rho _{i} (s)\le c_{i2} \rho _{i}(t), \end{aligned}$$

where $s, t>0$, $\frac{t}{2}\le s \le 2t$ and $i=1,2$.

Remark 3.4

If a function $\rho _{1} : (0,1]\rightarrow (0,\infty )$ satisfies Condition 2) of Definition 3.1 with $c_{11}$, $c_{12}>0$ and a function $\rho _{2} : (1,\infty ) \rightarrow (0,\infty )$ satisfies Condition 2) of Definition 3.1 with $c_{21}$, $c_{22}>0$, then the function

$$\begin{aligned} \rho (t) = \left\{ \begin{array}{ll} \rho _{1}(t),\ 0<t\le 1, \\ \rho _{2}(t),\ 1<t< \infty \end{array}\right. \end{aligned}$$

satisfies Condition 2) of Definition 3.1 with certain $c_{1}, c_{2}>0$.

Indeed, let $s,t>0$, $\frac{t}{2}\le s\le 2t$. If $t\le \frac{1}{2}$, then $s\le 1$, $\rho (s)=\rho _{1}(s)$, $\rho (t)=\rho _{1}(t)$ and Condition 2) of Definition 3.1 is satisfied with $c_{11}, c_{12}>0$. Respectively, if $t>2$, then $s>1$, $\rho (s)=\rho _{2}(s)$, $\rho (t)=\rho _{2}(t)$ and Condition 2) of Definition 3.1 is satisfied with $c_{21}, c_{22}>0$. Let $\frac{1}{2}<t\le 2$, then $\frac{1}{4}<s\le 4$,

$$\begin{aligned} \min \{c_{11},c_{21}\} \le \left\{ \begin{array}{ll} c_{11},\ \frac{1}{4}<s\le 1, \\ c_{21},\ 1<s<4 \end{array}\right. \le \rho (s)\le \left\{ \begin{array}{ll} c_{12},\ \frac{1}{4}<s\le 1, \\ c_{22},\ 1<s<4 \end{array}\right. \le \max \{c_{12},c_{22}\} , \end{aligned}$$

$$\begin{aligned} \min \{c_{11},c_{21}\} \le \left\{ \begin{array}{ll} c_{11},\ \frac{1}{2}<t\le 1, \\ c_{21},\ 1<t\le 2 \end{array}\right. \le \rho (t)\le \left\{ \begin{array}{ll} c_{12},\ \frac{1}{2}<t\le 1, \\ c_{22},\ 1<t\le 2 \end{array}\right. \le \max \{c_{12},c_{22}\} , \end{aligned}$$

hence

$$\begin{aligned} \min \{c_{11},c_{21}\} (\max \{c_{12},c_{22}\})^{-1}\rho (t)\le \rho (s)\le \max \{c_{12},c_{22}\} (\min \{c_{11},c_{21}\})^{-1}\rho (t). \end{aligned}$$

So, for $\frac{1}{2}<t\le 2$, $\frac{t}{2}\le s\le 2t$, Condition 2) of Definition 3.1 is satisfied with

$$\begin{aligned} c_{1}=\min \{c_{11},c_{21}\}\max \{c_{12},c_{22}\}^{-1}, \ c_{2}= \max \{c_{12},c_{22}\} \min \{c_{11},c_{21}\}^{-1}. \end{aligned}$$

The above inequality is satisfied for all $s,t>0$, $\frac{t}{2}\le s\le 2t$, because $c_{11}\le 1\le c_{12}$, $c_{21}\le 1\le c_{22}$, hence $c_{1} \le c_{11}$, $c_{1} \le c_{21}$, $c_{12} \le c_{2}$, $c_{22} \le c_{2}$.

Remark 3.5

If a function $\rho : (0,\infty ) \rightarrow (0,\infty )$ satisfies Condition 2) of Definition 3.1, then for any $\gamma >0$ $c_{1\gamma }, c_{2\gamma } >0$ such that for any $t>0$

$$\begin{aligned} c_{1\gamma }\rho (t)\le \rho (\gamma t) \le c_{2\gamma } \rho (t). \end{aligned}$$

Indeed, if $\frac{1}{2}\le \gamma \le 2$, then this inequality follows by Condition 2) with $s=\gamma t$. Let $2 \le \gamma < 4$, then by Condition 2) with $t=2\tau$, $\tau >0$ and $s=\gamma \tau$ and the above inequality with $t=\tau$ and $\gamma =2$

$$\begin{aligned} {c_{1}^{2}}\rho (\tau ) \le c_{1}\rho (2\tau ) \le \rho (\gamma \tau ) \le c_{2}\rho (2\tau ) \le {c_{2}^{2}}\rho (\tau ),\ \tau >0, \end{aligned}$$

because $\frac{1}{2} (2\tau ) \le \gamma \tau \le 2 (2\tau )$.

Next, let $4 \le \gamma < 8$ by Condition 2) with $t = 4\tau , \tau >0, s= \gamma \tau ,$

$$\begin{aligned} \frac{1}{2} (4\tau ) \le \gamma \tau \le 2(4\tau ), \end{aligned}$$

then

$$\begin{aligned} {c^{3}_{1}}\rho (\tau ) \le {c^{2}_{1}}\rho (2\tau ) \le c_{1}\rho (4\tau )\le \rho (\gamma \tau )\le c_{2}\rho (4\tau ) \le {c^{2}_{2}}\rho (2\tau )\le {c^{3}_{2}}\rho (\tau ), \ \tau >0. \end{aligned}$$

Furthermore, if $2^{k-1}\le \gamma <2^{k}$, $k\in \mathbb {N}$, $k>3 \Leftrightarrow k=\left[ \frac{\ln \gamma }{\ln 2}\right] +1$, then

$$\begin{aligned} {c^{k}_{1}}\rho (\tau )\le c^{k-1}_{1}\rho (2\tau )\le ...\le \rho (\gamma \tau )\le {c^{2}_{2}}\rho (2\tau )\le ...\le {c^{k}_{2}}\rho (\tau ). \end{aligned}$$

The argument for the case $0<\gamma <\frac{1}{2}$ is similar.

Example 1

Let $\rho (t)=t^{\alpha -n}, t>0, 0<\alpha < n.$ Then $\rho \in S_{n}$ because

$$\begin{aligned} {\int _{0}^{r}} t^{\alpha -n+n-1}dt< \infty \ \text { for all}\ \ 0<r<\infty \ \Leftrightarrow \alpha > 0 \end{aligned}$$

and, for any $s,t >0$ for which $\frac{t}{2} \le s\le 2t$,

$$\begin{aligned} 2^{\alpha -n}\rho (s)= 2^{\alpha -n}s^{\alpha -n}= (2s)^{\alpha -n} \end{aligned}$$

$$\begin{aligned} \le \rho (t)= t^{\alpha -n}\le \left( \frac{s}{2}\right) ^{\alpha -n}= 2^{n-\alpha }s^{\alpha -n}= 2^{n-\alpha }\rho (s). \end{aligned}$$

Example 2

Let $0<\alpha <n$, $-\infty<\beta _{1},\beta _{2}<\infty$,

$$\begin{aligned} \Psi _{\beta _{1},\beta _{2}}(t) = \left\{ \begin{array}{ll} (1+\vert \ln t\vert )^{\beta _{1}},\ 0<t\le 1, \\ (1+\vert \ln t\vert )^{\beta _{2}},\ 1<t< \infty \end{array}\right. \end{aligned}$$

and $\rho (t) =t^{\alpha -n}\Psi _{\beta _{1}\beta _{2}}(t)$, $t>0$. Then $\rho \in S_{n}$.

Since the function $\rho$ is continuous on $(0,\infty )$ it suffices to prove Condition 1) of Definition 3.1 for $r=1.$ Recall that for any $\varepsilon >0$ there exists $C_{\varepsilon }\ge 1$ such that

$$\begin{aligned} 1+\vert \ln t\vert \le C_{\varepsilon }t^{-\varepsilon } \end{aligned}$$

for all $t \in (0,1].$ If $\beta _{1}\le 0,$ then

$$\begin{aligned} {\int _{0}^{1}}\rho (t)t^{n-1}dt = {\int _{0}^{1}}t^{\alpha -1}(1+\vert \ln t\vert )^{\beta _{1}}dt \le {\int _{0}^{1}} t^{\alpha -1}dt <\infty . \end{aligned}$$

If $\beta _{1}>0,$ then

$$\begin{aligned} {\int _{0}^{1}}\rho (t)t^{n-1}dt = {\int _{0}^{1}}t^{\alpha -1}(1+\vert \ln t\vert )^{\beta _{1}}dt \le C_{\varepsilon }^{\beta _{1}} {\int _{0}^{1}} t^{\alpha -\varepsilon \beta _{1}-1}dt <\infty \end{aligned}$$

if we choose $\varepsilon >0$ to be such that $\varepsilon \beta _{1}<\alpha .$

As for Condition 2) of Definition 3.1, by Remark 3.3 and Example 1, it suffices to prove that the function $\varphi _{\beta _{1},\beta _{2}}$ satisfies this condition.

Let $s,t>0$ and $\frac{t}{2}\le s\le 2t.$ If $s>1,$ hence $t>\frac{1}{2},$ then

$$\begin{aligned} 1+\vert \ln s\vert \le 1+\ln 2t = 1+\ln 2+\ln t\le 2(1+\vert \ln t\vert ) \end{aligned}$$

and

$$\begin{aligned} 1+\vert \ln t\vert \le 1+\ln 2 <2(1+\vert \ln s\vert ), \end{aligned}$$

hence

$$\begin{aligned} \frac{1}{2}(1+\vert \ln t\vert )\le 1+\vert \ln s\vert \le 2 (1+\vert \ln t\vert ). \end{aligned}$$

For similar reasons this inequality also holds if $s\le 1,$ hence $t\le 2.$ Therefore, for $s\le 1$

$$\begin{aligned} 2^{-\vert \beta _{1}\vert }(1+\vert \ln t\vert )^{\beta _{1}}\le (1+\vert \ln s\vert )^{\beta _{1}}\le 2^{\vert \beta _{1}\vert }(1+\vert \ln t\vert )^{\beta _{1}}. \end{aligned}$$

If $t \le 1,$ this means that

$$\begin{aligned} 2^{-\vert \beta _{1}\vert }\Psi _{\beta _{1},\beta _{2}}(t)\le \Psi _{\beta _{1},\beta _{2}}(s)\le 2^{\vert \beta _{1}\vert } \Psi _{\beta _{1},\beta _{2}}(t). \end{aligned}$$

If $1<t\le 2,$ then

$$\begin{aligned} 2^{-\vert \beta _{1}\vert - \vert \beta _{2}\vert }\le (1+\vert \ln t\vert )^{\beta _{1}} (1+\vert \ln t\vert )^{-\beta _{2}}\le 2^{\vert \beta _{1}\vert +\vert \beta _{2}\vert }, \end{aligned}$$

hence

$$\begin{aligned} 2^{-2\vert \beta _{1}\vert -\vert \beta _{2}\vert }(1+\vert \ln t\vert )^{\beta _{2}}\le (1+\vert \ln s\vert )^{\beta _{1}}\le 2^{2\vert \beta _{1}\vert +\vert \beta _{2}\vert }(1+\vert \ln t\vert )^{\beta _{2}} \end{aligned}$$

which means that

$$\begin{aligned} 2^{-2\vert \beta _{1}\vert -\vert \beta _{2}\vert }\Psi _{\beta _{1},\beta _{2}}(t)\le \Psi _{\beta _{1},\beta _{2}}(s)\le 2^{2\vert \beta _{1}\vert +\vert \beta _{2}\vert }\Psi _{\beta _{1},\beta _{2}}(t). \end{aligned}$$

So, this inequality holds for all $s\le 1$ and all t satisfying the inequality $\frac{t}{2} \le s\le 2t.$

For similar reasons, for all $s>1$ and all t satisfying the inequality $\frac{t}{2} \le s\le 2t$

$$\begin{aligned} 2^{-\vert \beta _{1}\vert -2\vert \beta _{2}\vert }\Psi _{\beta _{1},\beta _{2}}(t)\le \Psi _{\beta _{1},\beta _{2}}(s)\le 2^{\vert \beta _{1}\vert +2\vert \beta _{2}\vert }\Psi _{\beta _{1},\beta _{2}}(t). \end{aligned}$$

Example 3

Let $0<\alpha <n$, $-\infty<\beta <\infty$ and

$$\begin{aligned} \rho (t) = \left\{ \begin{array}{ll} t^{\alpha -n} ,\ 0<t\le 1, \\ t^{\beta },\ 1<t< \infty . \end{array}\right. \end{aligned}$$

Then $\rho \in S_{n}$. Indeed, Condition 1) is clearly satisfied. As for Condition 2), it is also satisfied by Remark 3.4, because $\rho _{1}(t)=t^{\alpha -n}$ satisfies Condition 2) on (0, 1] by Example 1 and $\rho _{2}(t) =t^{\beta }$ satisfies Condition 2) on $(1,\infty )$ by an argument similar to that of the proof of Example 1.

Note that $t^{\beta }$ can be replaced by any function $\rho (t)$ satisfying Condition 2) on $(1,\infty )$.

Definition 3.2

For $\rho \in S_{n}$ and $f \in \mathfrak {M}(\mathbb {R}^{n})$

$$\begin{aligned} I_{\rho (\cdot )}f(x) = \int _{\mathbb {R}^{n}}^{}\rho (\vert x-y\vert )f(y) dy, \end{aligned}$$

$$\begin{aligned} \overline{ I}_{\rho (\cdot ),r}f(x)= \int _{^{c}B(x,r) }^{}\rho (\vert x-y\vert )f(y) dy \end{aligned}$$

and

$$\begin{aligned} \underline{I}_{\rho (\cdot ),r}f(x)= \int _{B(x,r) }^{}\rho (\vert x-y\vert )f(y) dy. \end{aligned}$$

Lemma 3.1

Let $n \in \mathbb {N},\rho \in S_{n},$ $0 < p\le \infty , x \in \mathbb {R}^{n}, r>0, f \in \mathfrak {M}(\mathbb {R}^{n}).$ Then¹$G= r^{\frac{n}{p}} \left( \overline{I}_{\rho (\cdot ), 2r} \vert f \vert \right) (x).$

$$\begin{aligned} \Vert I_{\rho (.)}\vert f\vert \Vert _{WL_{p}(B(x,r))}\gtrsim r^{\frac{n}{p}} \left( \overline{I}_{\rho (\cdot ), 2r} \vert f\vert \right) (x) \end{aligned}$$

(3.1)

uniformly in $x \in \mathbb {R}^{n}$, $r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$

Proof

By Property 3) of Remark 3.2 for any $y \in B(x,r)$

$$\begin{aligned} I_{\rho (.)}\vert f\vert (y) \ge \int _{^{c}B(x,2r) }^{}\rho (\vert y-z\vert )\vert f(z)\vert dz \end{aligned}$$

$$\begin{aligned} \ge c1 \int _{^{c}B(x,2r) }^{}\rho (\vert x-z\vert )\vert f(z)\vert dz, \end{aligned}$$

hence, by Remark 3.1, we get

$$\begin{aligned} \Vert I_{\rho (\cdot )}|f| \Vert _{WL_{p}(B(x,r))}\ge c_{1} \left( v_{n}r^{n}\right) ^{\frac{1}{p}}\int _{^{c}B(x,2r) }^{}\rho (\vert x-z\vert )\vert f(z)\vert dz \end{aligned}$$

$$\begin{aligned} = c_{1} v_{n}^{\frac{1}{p}}r^{\frac{n}{p}} \overline{I}_{\rho (\cdot ), 2r} \vert f\vert (x) \end{aligned}$$

for all $x \in \mathbb {R}^{n}, r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$ Here $v_{n}$ is the volume of the unit ball in $\mathbb {R}^{n}$.

Lemma 3.2

Let $n \in \mathbb {N},\rho \in S_{n},$ $0 < p\le \infty , x \in \mathbb {R}^{n}, r>0$, $f \in \mathfrak {M}(\mathbb {R}^{n}).$ Then

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{L_{p}(B(x,r))}\thickapprox \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{L_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (\cdot ), 2r} \vert f\vert \right) (x) \end{aligned}$$

(3.2)

and

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{WL_{p}(B(x,r))}\thickapprox \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{WL_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (\cdot ), 2r} \vert f\vert \right) (x) \end{aligned}$$

(3.3)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$

Proof

According to the properties² of the spaces $L_{p}(\Omega )$ and $WL_{p}(\Omega )$

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{L_{p}(B(x,r))} \le 2^{(\frac{1}{p}-1)_{+}}\left( \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{L_{p}(B(x,r))} + \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{^{c}B(x,2r)}}) \Vert _{L_{p}(B(x,r))}\right) , \end{aligned}$$

(3.4)

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{WL_{p}(B(x,r))} \le 2^{\frac{1}{p}} \Big ( \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{WL_{p}(B(x,r))} + \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{^{c}B(x,2r)}}) \Vert _{WL_{p}(B(x,r))}\Big ) . \end{aligned}$$

(3.5)

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{L_{p}(B(x,r))} \ge \frac{1}{2}\Big ( \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{L_{p}(B(x,r))} + \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{^{c}B(x,2r)}} ) \Vert _{L_{p}(B(x,r))}\Big ), \end{aligned}$$

(3.6)

and

$$\begin{aligned} \Vert I_{\rho (\cdot )}\vert f\vert \Vert _{L_{p}(B(x,r))} \Vert I_{\rho (\cdot )} \vert f\vert \Vert _{WL_{p}(B(x,r))} \ \ge \ \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{B(x,2r)}} ) \Vert _{WL_{p}(B(x,r))}. \end{aligned}$$

(3.7)

Next,

$$\begin{aligned} \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{^{c}B(x,2r)}} ) \Vert _{WL_{p}(B(x,r))} \ \le \ \Vert I_{\rho (\cdot )}(\vert f\vert \chi _{_{^{c}B(x,2r)}}) \Vert _{L_{p}(B(x,r))} \end{aligned}$$

$$\begin{aligned} = \left[ \int _{B(x,r)}^{} \left( \int _{^{c}B(x,2r)}^{}\rho (\vert y-z\vert )\vert f(z)\vert dz\right) ^{p} dy\right] ^{\frac{1}{p}} = J. \end{aligned}$$

By Condition 3) of Remark 3.2

$$\begin{aligned} J \le _{C2} \left[ \int _{B(x,r)}^{} \left( \int _{^{c}B(x,2r)}^{}\rho (\vert x-z\vert )\vert f(z)\vert dz\right) ^{p} dy\right] ^{\frac{1}{p}}=_{C2}v_{n}^{\frac{1}{p}}r^{\frac{n}{p}}\left( \overline{I}_{\rho (.), 2r} \vert f\vert \right) (x). \end{aligned}$$

Thus, by (3.4) and (3.5)

$$\begin{aligned} \Vert I_{\rho (.)}\vert f\vert \Vert _{L_{p}(B(x,r))} \lesssim \ \Vert I_{\rho (.)}(\vert f\vert \chi _{_{B(x,2r)}} ) \Vert _{L_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (.), 2r} \vert f\vert \right) (x), \end{aligned}$$

and

$$\begin{aligned} \Vert I_{\rho (.)}\vert f\vert \Vert _{WL_{p}(B(x,r))} \lesssim \ \Vert I_{\rho (.)}(\vert f\vert \chi _{_{B(x,2r)}} ) \Vert _{WL_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (.), 2r} \vert f\vert \right) (x), \end{aligned}$$

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$

Also, by Condition 3) of Remark 3.2

$$\begin{aligned} J \ge _{C1} \left[ \int _{B(x,r)}^{} \left( \int _{^{c}B(x,2r)}^{}\rho (\vert x-z\vert )\vert f(z)\vert dz\right) ^{p} dy\right] ^{\frac{1}{p}}=_{C1}^{-1}v_{n}^{\frac{1}{p}}r^{\frac{n}{p}}\left( \overline{I}_{\rho (.), 2r} |f|\right) (x). \end{aligned}$$

Hence, by (3.6)

$$\begin{aligned} \Vert I_{\rho (.)}\vert f\vert \Vert _{L_{p}(B(x,r))} \gtrsim \Vert I_{\rho (.)}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{L_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (.), 2r} \vert f \vert \right) (x), \end{aligned}$$

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$

Finally, by adding inequalities (3.1) and (3.7), we get that

$$\begin{aligned} \Vert I_{\rho (.)}\vert f\vert \Vert _{WL_{p}(B(x,r))} \gtrsim \Vert I_{\rho (.)}(\vert f\vert \chi _{_{B(x,2r)}}) \Vert _{WL_{p}(B(x,r))} + r^{\frac{n}{p}} \left( \overline{I}_{\rho (.), 2r} \vert f\vert \right) (x), \end{aligned}$$

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n}).$

Lemma 3.3

Let, for $n \in \mathbb {N},\rho \in S_{n}$, $1\le p_{1}<p_{2}<\infty$, and

$$\begin{aligned} \varphi _{n,\rho , p_{1},p_{2}}(t) =\rho (t) t^{n\left( \frac{1}{p_{1}^{\prime }}+\frac{1}{p_{2}}\right) }, \ t>0, \end{aligned}$$

(3.8)

where $p_{1}^{\prime }$ is the conjugate number to $p_{1}.$

1. Assume that $\rho$ is such that the function $\varphi _{n, \rho , p_{1}, p_{2}}$ is almost non-decreasing on $(0,\infty )$, that is for some $c>0$

$$\begin{aligned} \varphi _{n,\rho , p_{1},p_{2}}(t_{1}) \le c\varphi _{n,\rho , p_{1},p_{2}}(t_{2}) \ \text{for all} \ 0 < t_{1} < t_{2} < \infty . \end{aligned}$$

(3.9)

If $1<p_{1}<p_{2}<\infty$, then

$$\begin{aligned} \Vert I_{\rho (\cdot )}\left( \vert f\vert \chi _{_{B(x,2r)}}\right) \Vert _{L_{p_{2}}(B(x,r))}\ \lesssim \ \varphi _{n,\rho , p_{1},p_{2}}(r) \Vert f\Vert _{L_{p_{1}}(B(x,2r))} \end{aligned}$$

(3.10)

uniformly in $x\in \mathbb {R}^{n}$, $r>0$ and $f\in L_{p_{1}}(B(x,2r))$.

Also, if $p_{1}=1$, $1< p_{2} < \infty$, then

$$\begin{aligned} \Vert I_{\rho (\cdot )}\left( \vert f\vert \chi _{_{B(x,2r)}}\right) \Vert _{WL_{p_{2}}(B(x,r))}\ \lesssim \ \varphi _{n,\rho , p_{1},p_{2}}(r) \Vert f\Vert _{L_{p_{1}}(B(x,2r))} \end{aligned}$$

(3.11)

uniformly in $x\in \mathbb {R}^{n}$, $r>0$ and $f\in L_{1}(B(x,2r))$.

2. Inequality (3.10) also holds for $p_{1}=1$ and $1\le p_{2}\le \infty$ under the assumption

$$\begin{aligned} \Vert \rho (t)t^{\frac{n-1}{p_{2}}} \Vert _{L_{p_{2}}(0,r)}\lesssim \rho (r)r^{\frac{n}{p_{2}}} \end{aligned}$$

(3.12)

uniformly in $r >0,$ replacing assumption (3.9).

3. Inequality (3.10) and hence inequality (3.11) also holds for $1\le p_{1}< \infty$ and $0 <p_{2}\le p_{1}$ under the assumption

$$\begin{aligned} \Vert \rho (t)t^{n-1} \Vert _{L_{1}(0,r)}\lesssim \rho (r)r^{n} \end{aligned}$$

(3.13)

uniformly in $r >0,$ replacing assumption (3.9).

4. Condition (3.13) is necessary for the validity of inequalities (3.10) and (3.11) for all $0<p_{1},p_{2} \le \infty$.

5. Condition (3.13) is necessary and sufficient for the validity of inequalities (3.10) and (3.11) for all $1\le p_{1}< \infty$ and $0<p_{2}\le p_{1}$.

Proof

1. Let $1\le p_{1}<p_{2}<\infty$, $x\in \mathbb {R}^{n}$, $r>0$, $z\in B(x,r)$ and $f\in L_{p_{1}}(B(x,2r))$. Then

$$\begin{aligned} \Big (I_{\rho (\cdot )}\left( \vert f \vert \chi _{_{B(x,2r)}} \right) \Big )(z)=\int \limits _{\mathbb {R}^{n}}\rho ( \vert z-y \vert ) \vert f(y) \vert \chi _{_{B(x,2r)}} (y)dy \end{aligned}$$

$$\begin{aligned} =\int \limits _{B(x,2r)} \rho \left( \vert z-y \vert \right) \vert z-y \vert ^{n\left( 1-\frac{1}{p_{1}}+\frac{1}{p_{2}}\right) } \ \frac{\vert f(y) \vert \chi _{_{B(x,2r)}}(y)}{ \vert z-y \vert ^{n-n(\frac{1}{p_{1}}-\frac{1}{p_{2}})}}dy \end{aligned}$$

$$\begin{aligned} \le \Big (\underset{\underset{y\in B(x,2r)}{z\in B(x,r)}}{\sup } \varphi _{n,\rho , p_{1},p_{2}}(\vert z-y \vert ) \Big ) \Big (I_{n\left( \frac{1}{p_{1}}-\frac{1}{p_{2}}\right) }\left( \vert f \vert \chi _{_{B(x,2r)}}\right) \Big )(z) . \end{aligned}$$

Since the function $\varphi _{n,\rho , p_{1},p_{2}}$ is almost non-decreasing on $(0,\infty )$ and, for $z\in B(x,r)$, $y\in B(x,2r)$,

$$\begin{aligned} \vert z-y \vert \le \vert z-x \vert + \vert x-y \vert \le r+2r=3r, \end{aligned}$$

we have by (3.9)

$$\begin{aligned} \underset{\underset{y\in B(x,2r)}{z\in B(x,r)}}{\sup } \varphi _{n,\rho , p_{1},p_{2}}(\vert z-y \vert ) \le c \varphi _{n,\rho , p_{1},p_{2}}(3r)=c3^{n(1-\frac{1}{p_{1}}+\frac{1}{p_{2}})}\rho (3r)r^{n(1-\frac{1}{p_{1}}+\frac{1}{p_{2}})}. \end{aligned}$$

By Remark 3.5 with $\gamma =3$ it follows that

$$\begin{aligned} \rho (3r) \le c_{23}\rho (r),\ r>0. \end{aligned}$$

Hence,

$$\begin{aligned} \underset{\underset{y\in B(x,2r)}{z\in B(x,r)}}{\sup } \varphi _{n,\rho , p_{1},p_{2}}(\vert z-y \vert ) \le c\ c_{23} \ 3^{n(1-\frac{1}{p_{1}}+\frac{1}{p_{2}})} \varphi _{n,\rho , p_{1},p_{2}}(r) \end{aligned}$$

and

$$\begin{aligned} \Big (I_{\rho (\cdot )}\left( \vert f \vert \chi _{_{B(x,2r)}} \right) \Big )(z)\lesssim \varphi _{n,\rho , p_{1},p_{2}}(r) \Big (I_{n\left( \frac{1}{p_{1}}-\frac{1}{p_{2}}\right) }\left( \vert f \vert \chi _{_{B(x,2r)}} \right) \Big )(z) \end{aligned}$$

(3.14)

uniformly in $x\in \mathbb {R}^{n}$, $r>0$, $z\in B(x,r)$ and $f\in L_{p_{1}}(B(x,2r))$.

Next, we apply the well-known inequalities for the Riesz potential. If $1<p_{1}<p_{2}<\infty$, then

$$\begin{aligned} \Vert I_{n\left( \frac{1}{p_{1}}-\frac{1}{p_{2}}\right) }|f|\Vert _{L_{p_{2}}(\mathbb {R}^{n})}\lesssim \Vert f\Vert _{L_{p_{1}}(\mathbb {R}^{n})} \end{aligned}$$

(3.15)

uniformly in $f\in L_{p_{1}}(\mathbb {R}^{n})$. Also, if $1<p_{2}<\infty$, then

$$\begin{aligned} \Vert I_{n\left( 1-\frac{1}{p_{2}}\right) }|f|\Vert _{WL_{q}(\mathbb {R}^{n})}\lesssim \Vert f\Vert _{L_{1}(\mathbb {R}^{n})} \end{aligned}$$

(3.16)

uniformly in $L_{1}(\mathbb {R}^{n})$.

If $1<p_{1}<p_{2}<\infty$, then, by (3.14) and (3.15), it follows that

$$\begin{aligned} \Vert I_{\rho (\cdot )}\left( \vert f \vert \chi _{_{B(x,2r)}} \right) \Vert _{L_{p_{2}}(\mathbb {R}^{n})} \lesssim \varphi _{n,\rho , p_{1},p_{2}}(r)\ \Vert \ |f|\chi _{_{B(x,2r)}} \Vert _{L_{p_{1}}(\mathbb {R}^{n})} \end{aligned}$$

$$\begin{aligned} =\varphi _{n,\rho , p_{1},p_{2}}(r) \Vert f\Vert _{L_{p_{1}}(B(x,2r))} \end{aligned}$$

uniformly in $x\in \mathbb {R}^{n}$, $r>0$ and $f\in L_{p_{1}}(B(x,2r))$, which is inequality (3.10).

Respectively, if $1<p_{2}<\infty$, then by (3.14) and (3.16) there follows inequality (3.11).

2. If $p_{1}=1$ and $1\le p_{2}\le \infty ,$ then, by applying Young’s inequality for truncated convolutions, we get

$$\begin{aligned} \parallel I_{\rho (.)} (\vert f \vert \chi _{_{B(x,2r)}} )\parallel _{L_{p_{2}}(B(x,r))}= \left\| \int _{B(x,2r) }^{} \rho (|.-y|) \vert f(y) \vert dy\right\| _{L_{p_{2}}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \le \parallel \rho \parallel _{L_{p_{2}}(B(x,r)-B(x,2r))}\parallel f\parallel _{L_{1}(B(x,2r))} \end{aligned}$$

$$\begin{aligned} = \parallel \rho \parallel _{L_{p_{2}}(B(0,3r))}\parallel f\parallel _{L_{1}(B(x,2r))} \end{aligned}$$

$$\begin{aligned} = \sigma _{n}^{\frac{1}{p_{2}}}\parallel \rho (t)t^{\frac{n-1}{p_{2}}} \parallel _{L_{p_{2}}(0,3r)}\parallel f\parallel _{L_{1}(B(x,2r))}, \end{aligned}$$

where $\sigma _{n}=nv_{n}$ is the surface area of the unit ball in $\mathbb {R}^{n}.$

By inequality (3.12) and Remark 3.5 with $\gamma = 3$ it follows that

$$\begin{aligned} \parallel \rho (t)t^{\frac{n-1}{p_{2}}} \parallel _{L_{p_{2}}(0,3r)}\lesssim \rho (3r)r^{\frac{n}{p_{2}}}\lesssim \rho (r)r^{\frac{n}{p_{2}}}=\varphi _{n,\rho ,1,p_{2}}(r), \end{aligned}$$

which implies inequality (3.10).

3. If $1\le p_{2}=p_{1}< \infty ,$ then similarly to Step 2

$$\begin{aligned} \parallel I_{\rho (.)} (\vert f \vert \chi _{_{B(x,2r)}} )\parallel _{L_{p_{1}}(B(x,r))}\le \parallel \rho \parallel _{L_{1}(B(x,r)-B(x,2r))}\parallel f\parallel _{L_{p_{1}}(B(x,2r))} \end{aligned}$$

$$\begin{aligned} =\sigma _{n}\parallel \rho (t)t^{n-1} \parallel _{L_{1}(0,3r)} \parallel f\parallel _{L_{p_{1}}(B(x,2r))}. \end{aligned}$$

By inequality (3.13) and Remark 3.5 with $\gamma =3$ it follows that

$$\begin{aligned} \parallel \rho (t)t^{n-1} \parallel _{L_{1}(0,3r)} \lesssim \rho (3r)r^{n}\lesssim \rho (r)r^{n} =\varphi _{n,\rho ,p_{1}, p_{1}}(r), \end{aligned}$$

which implies inequality (3.10).

If $0<p_{2}< p_{1}, 1\le p_{1}<\infty ,$ then, by applying Hölder’s inequality and inequality (3.10) with $p_{2}=p_{1}$ we get

$$\begin{aligned} \parallel I_{\rho (.)} ( \vert f \vert \chi _{_{B(x,2r)}} )\parallel _{L_{p_{2}}(B(x,r))}\le (v_{n}r^{n})^{\frac{1}{p_{2}}-\frac{1}{p_{1}}}\parallel I_{\rho (.)} (\vert f \vert \chi _{_{B(x,2r)}} )\parallel _{L_{p_{1}}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \lesssim \rho (r)r^{n\left( \frac{1}{p_{1}^{\prime }}+\frac{1}{p_{2}}\right) }\parallel f\parallel _{L_{p_{1}}(B(x,2r))}= \varphi _{n, \rho , p_{1}, p_{2}} (r)\parallel f\parallel _{L_{p_{1}}(B(x,2r))} , \end{aligned}$$

which is inequality (3.10).

4. Assume that for some $0<p_{1},p_{2}\le \infty$ inequality (3.10) and (3.11) is satisfied. If inequality (3.10) is satified, then inequality (3.11) is also satisfied. Take in this inequality $x=0$ and $f\equiv 1$. Then for any $y\in B(0,r)$ we have $B(y,2r) \supset B(0,r)$ and

$$\begin{aligned} I_{\rho (.)} \big (\chi _{_{B(0,2r)}}\big )(y)= \int _{B(0,2r)}\rho (y-z)dz=\int _{B(y,2r)}\rho (u)du \end{aligned}$$

$$\begin{aligned} \ge \int _{B(0,r)}\rho (u) du =\sigma _{n} {\int _{0}^{r}} \rho (t) t^{n-1}dt. \end{aligned}$$

By Remark 3.1

$$\begin{aligned} \left\| I_{\rho (.)} \big (\chi _{_{B(0,2r)}}\big )\right\| _{WL_{p_{2}}(B(0,r))}\gtrsim r^{\frac{n}{p_{2}}}{\int _{0}^{r}}\rho (t) t^{n-1}dt \end{aligned}$$

and $\left\| \chi _{_{B(0,2r)}}\right\| _{L_{p_{2}}(B(0,r))}= (\sigma _{n} r^{n})^{\frac{1}{p_{1}}}$, hence by (3.11)

$$\begin{aligned} r^{\frac{n}{p_{2}}}{\int _{0}^{r}}\rho (t) t^{n-1}dt\lesssim \rho (r) r^{n\left( \frac{1}{p'_{1}}+\frac{1}{p_{2}}\right) }r^{\frac{n}{p_{1}}} \end{aligned}$$

uniformly in $r>0$, which implies inequality (3.13).

5. The last statement of Lemma 3.3 follows by Steps 3 and 4.

Remark 3.6

If $1 \le p_{1}<p_{2}< \infty$ and

$$\begin{aligned} \lim \limits _{t \rightarrow 0^{+}} \varphi _{n, \rho , p_{1}, p_{2}}(t)= 0, \end{aligned}$$

(3.17)

then inequality (3.10) for $1<p_{1}<p_{2}<\infty$ and inequality (3.11) for $1<p_{2}<\infty$ cannot hold for any $f \in L_{p_{1}}{(\mathbb {R}^{n})}$, $f \in L_{1}(\mathbb {R}^{n})$, respectively, not equivalent to 0 on $\mathbb {R}^{n}$. Indeed, by passing to the limit as $r \rightarrow \infty ,$ in (3.10) and (3.11), it follows that $\parallel I_{\rho (.)}|f| \parallel _{L_{p_{2}}(\mathbb {R}^{n})}= 0$, $\parallel I_{\rho (.)}|f| \parallel _{WL_{p_{2}}(\mathbb {R}^{n})}= 0,$ hence, f is equivalent to 0 on $\mathbb {R}^{n}.$

If $\rho (t)= t^{\alpha - n}$, $0<\alpha <n$ and $1\le p_{1}<p_{2},$ then Condition (3.9) reduces to the condition $\alpha \ge n(\frac{1}{p_{1}}-\frac{1}{p_{2}}),$ Condition (3.12) reduces to the condition $\alpha > n(1-1/{p_{2}})$ and Condition (3.13) is satisfied since $\alpha >0.$

Corollary 3.1

Let the assumptions of Lemma 3.3 for $n, \rho , p_{1}, p_{2}$ and the function $\varphi _{n,\rho , p_{1}, p_{2}}$ be satisfied. Then, for $1<p_{1}<p_{2}< \infty ,$ for $p_{1}=1, 1<p_{2}\le \infty ,$ and for $1\le p_{1}< \infty , 0<p_{2}\le p_{1}$

$$\begin{aligned} \parallel I_{\rho (.)}\vert f\vert \parallel _{L_{p_{2}}(B(x,r))}\lesssim \varphi _{n,\rho , p_{1}, p_{2}}(r)\parallel f\parallel _{L_{p_{1}}(B(x,2r))} + r^{\frac{n}{p_{2}}} \left( \overline{I}_{\rho (.), 2r} \vert f \vert \right) (x) \end{aligned}$$

(3.18)

uniformly in $x \in \mathbb {R}^{n}$, $r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n})\bigcap L_{p_{1}}(B(x,2r))$ and, for $1<p_{2}<\infty$

$$\begin{aligned} \parallel I_{\rho (.)}\vert f\vert \parallel _{WL_{p_{2}}(B(x,r))} \lesssim \varphi _{n,\rho , 1, p_{2}}(r)\parallel f\parallel _{L_{1}(B(x,2r))} + r^{\frac{n}{p_{2}}} \left( \overline{I}_{\rho (.), 2r} \vert f \vert \right) (x) \end{aligned}$$

(3.19)

uniformly in $x \in \mathbb {R}^{n}$, $r>0$ and $f \in \mathfrak {M}(\mathbb {R}^{n})\bigcap L_{1}(B(x,2r)).$

Lemma 3.4

Let $f \in \mathfrak {M}(\mathbb {R}^{n})$ be a non-negative function and $\rho$ be a non-negative, non-increasing, continuously differentiable function on $(0,\infty )$ such that $\lim _{t\rightarrow +\infty }\rho (t)=0.$ Then for any $x \in \mathbb {R}^{n}$ and $r>0$

$$\begin{aligned} \int _{c_{B(x,r)}}^{}\rho (\vert x-y\vert )f(y)dy = \int _{r}^{\infty }\left( \int _{B(x,t)\setminus B(x,r)}^{ }f(y)dy\right) \vert \rho ^{\prime }(t)\vert dt. \end{aligned}$$

(3.20)

For $\rho (t)=t^{- \beta }$, $t>0$, $\beta >0$ this lemma was proved in [3] (Lemma 3).

Proof

By applying the Fubini theorem we get

$$\begin{aligned} \int _{c_{B(x,r)}}^{}\rho (\vert x-y \vert )f(y)dy= & {} \int _{y : \vert x-y\vert>r}^{}\rho (\vert x-y \vert )f(y)dy \\= & {} \int _{y : \vert x-y\vert>r}^{}\left( \int _{\vert x-y\vert }^{\infty }\vert \rho ^{\prime }(t)\vert dt\right) f(y)dy \\= & {} \int _{r}^{\infty }\left( \int _{y : \vert x-y\vert >r, \vert x-y\vert <t }^{ }\vert \rho ^{\prime }(t)\vert f(y)dy\right) dt \\= & {} \int _{r}^{\infty }\left( \int _{B(x,t)\setminus B(x,r)}^{} f(y)dy \right) \vert \rho ^{\prime }(t)\vert dt. \end{aligned}$$

Definition 3.3

Let $n \in \mathbb {N}, 1 \le p_{1}< \infty , 0<p_{2}\le \infty .$ Then $\rho \in S_{n, p_{1}, p_{2}}$ if $\rho \in \mathfrak {M^{+}}((0,\infty ))$ and there exists a positive non-increasing continuously differentiable function $\tilde{\rho }: (0,\infty ) \rightarrow (0,\infty )$ such that

1) $\tilde{\rho }(t)\approx \rho (t)$ uniformly in $t>0,$

$2)\ \lim \limits _{t \rightarrow \infty } \tilde{\rho }(t)=0,$

$3)\ \tilde{\rho } \in S_{n},$

$4)\ {\int \limits _{0}^{1}} \tilde{\rho }(t)t^{\frac{n}{p_{1}^{\prime }}-1}dt= \infty ,$ $\int \limits _{1}^{\infty } \tilde{\rho }(t)t^{\frac{n}{p_{1}^{\prime }}-1}dt< \infty ,$

$5)\ \vert \tilde{\rho }^{\prime }(t)\vert t \gtrsim \tilde{\rho }(t)$ uniformly in $t>0,$

6) if $0< p_{2}\le p_{1}$ and $1\le p_{1}< \infty$, then

$$\begin{aligned} {\int _{0}^{r}} \tilde{\rho }(t)t^{n-1} dt \lesssim \tilde{\rho }(r)r^{n} \end{aligned}$$

uniformly in $r>0$,

7) if $1\le p_{1}<p_{2}<\infty$, then the function $\phi _{n,\tilde{\rho },p_{1}, p_{2}}(t)= \tilde{\rho }(t)t^{n\big (\frac{1}{p_{1}'}+\frac{1}{p_{2}}\big )}$ is almost non-decreasing on $(0,\infty )$.

Moreover, if $p_{1}=1$ and $1<p_{2}<\infty$, then $\rho \in \tilde{S}_{n,1,p_{2}}$ if there exists a positive non-increasing continuously differentiable function $\tilde{\rho }:(0,\infty )\rightarrow (0,\infty )$ such that Conditions 1)–3) and 5) are satisfied, Conditions 4) and 6) are satisfied for $p_{1}=1$ and instead of Condition 7) the following condition is satisfied

8)

$$\begin{aligned} \big \Vert \tilde{\rho }(t)t^{\frac{n-1}{p_{2}}} \big \Vert _{L_{p_{2}}(0,r)} \lesssim \tilde{\rho }(r)r^{\frac{n}{p_{2}}} \end{aligned}$$

uniformly in $r>0.$

Remark 3.7

For the Riesz potential $\rho (t) =t^{\alpha -n}$, $0<\alpha <n$, $\tilde{\rho }=\rho$. Condition 2) is satisfied because $\alpha <n$, Condition 3) is satisfied because $\alpha >0$, Condition 4) is satisfied if $p_{1}<\infty$ and $\alpha <\frac{n}{p_{1}}$, Condition 5) is obviously satisfied, Condition 6) is satisfied because $\alpha > 0,$ Condition 7) is satisfied if $\alpha \ge n(1/{p_{1}}-1/{p_{2}},$ Condition 8) is satisfied if $\alpha > n(1-1/{p_{2}}).$

Remark 3.8

Clearly, due to Condition 1), Conditions 2)–4), 6)–8) are equivalent to the conditions, obtained from them by replacing $\tilde{\rho }$ by ${\rho }$

Theorem 3.1

Let $n \in \mathbb {N}, 1 \le p_{1}<\infty , 0<p_{2} \le \infty$.

1. If $1<p_{1}<p_{2}<\infty$ or $1 \le p_{1}< \infty$ and $0< p_{2}\le p_{1}$, and $\rho \in S_{n,p_{1}.p_{2}},$ then

$$\begin{aligned} \parallel I_{\rho (.)}(\vert f\vert )\parallel _{L_{p_{2}}(B(x,r))} \lesssim r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \rho (t)t^{\frac{n}{p_{1}^{\prime }}-1}\parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \end{aligned}$$

(3.21)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{p_{1}}^{\text {loc}}(\mathbb {R}^{n}).$

2. If $p_{1}=1$ and $0<p_{2}<\infty ,$ and $\rho \in \tilde{S}_{n,1,p_{2}},$ then

$$\begin{aligned} \parallel I_{\rho (.)}\vert f \vert \parallel _{WL_{p_{2}}(B(x,r))} \thickapprox \parallel I_{\rho (.)}\vert f \vert \parallel _{L_{p_{2}}(B(x,r))}\thickapprox r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \rho (t)t^{-1} \parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.22)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{1}^{\text {loc}}(\mathbb {R}^{n}).$

3. If $p_{1}=1$ and $1<p_{2}<\infty ,$ and $\rho \in S_{n,1,p_{2}},$ then

$$\begin{aligned} \parallel I_{\rho (.)}\vert f \vert \parallel _{WL_{p_{2}}(B(x,r))} \thickapprox r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \rho (t)t^{-1} \parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.23)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{1}^{\text {loc}}(\mathbb {R}^{n}).$

For $\rho (t) = t^{\alpha -n}$, $0<\alpha <n,$ hence for the Riesz potential $I_{\alpha }$, this theorem is proved in [6] (Lemma 3.6 and Theorem 3.9).

Remark 3.9

If $p_{1}=1$ and $0<p_{2}<\infty$ and $\rho \in \tilde{S}_{n,1 ,p_{2}},$ equivalence (3.22) holds if Condition 6) of Definition 3.3 for $0 < p_{2} \le 1$ and Condition 7) of Definition 3.3 for $1< p_{2} < \infty$ are satisfied. If $\rho (t)=t^{\alpha -n}, 0<\alpha <n,$ then this means that $n(1-\frac{1}{p_{2}})_{+}<\alpha <n.$

If $p_{1}=1$ and $1<p_{2}<\infty$ and $\rho \in S_{n,1 ,p_{2}}$, equivalence (3.23) holds if Condition 8) of Definition 3.3 is satisfied. If $\rho (t)=t^{\alpha -n}$, $0<\alpha <n,$ then this means that $n(1-\frac{1}{p_{2}})\le \alpha <n.$

In particular, for $\alpha =n(1-1/{p_{2}})$ Statement 3 of Theorem 3.1 holds, while, as proved in Lemma 3.3 of [6], Statement 2 does not hold.

Proof

Step 1 (proof of (3.21)). We apply equivalence (3.2).

1.1a. To the first summand in the right-hand side of (3.2) we apply Condition 2) of Definition 3.1 with $\rho$ replaced by $\tilde{\rho }, s$ by r, t by 2r and inequality (3.10) and get that

$$\begin{aligned}&r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \rho (t)t^{ \frac{n}{p_{1}^{\prime }}-1}\parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \\&\gtrsim r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \tilde{\rho }(t)t^{ \frac{n}{p_{1}^{\prime }}-1} \parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \\&\ge r^{\frac{n}{p_{2}}} \int _{r}^{2r} \tilde{\rho }(t) t^{\frac{n}{p_{1}^{\prime }}-1} \parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \\&\ge r^{\frac{n}{p_{2}}} \tilde{\rho }(2r)\left( \int _{r}^{2r} t^{ \frac{n}{p_{1}^{\prime }}-1}dt \right) \parallel f\parallel _{L_{p_{1}}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \gtrsim \tilde{\rho }(r)r^{\frac{n}{p_{1}^{\prime }}+\frac{n}{p_{2}}} \parallel f\parallel _{L_{p_{1}}(B(x,r))}= \varphi _{n,\tilde{\rho },p_{1}, p_{2}}(r)\parallel f\parallel _{L_{p_{1}}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \gtrsim \parallel I_{\tilde{\rho }(.)} (|f|\chi _{_{B(x,2r)}} )\parallel _{L_{p_{2}}(B(x,r))}\gtrsim \parallel I_{\rho (.)} (\vert f\vert \chi _{_{B(x,2r)}} )\parallel _{L_{p_{2}}(B(x,r))} \end{aligned}$$

(3.24)

uniformly in $x \in \mathbb {R}^{n},$ $r>0$ and $f \in L^{\text {loc}}_{1}( \mathbb {R}^{n}).$

1.1b (second proof of inequality (3.24)). To the first summand in the right-hand side of (3.2) we apply inequality (3.10) and Condition 2) of Definition 3.1 with $\rho$ replaced by $\tilde{\rho }$, firstly, s replaced by r and t by 2r,  secondly, s replaced by $\frac{\tau }{2}$ and t by $\tau ,$ and we get that

$$\begin{aligned} \parallel I_{\rho (.)} (\vert f\vert \chi _{B(x,2r)})\parallel _{L_{p_{2}}(B(x,r))}\lesssim \parallel I_{\tilde{\rho }(.)} (\vert f\vert \chi _{B(x,2r)})\parallel _{L_{p_{2}}(B(x,r))} \end{aligned}$$

$$\begin{aligned}\lesssim & {} \tilde{\rho }(r)r^{n\Big (\frac{1}{p^{\prime }_{1}}+\frac{1}{p_{2}}\Big )} \parallel f\parallel _{L_{p_{1}}(B(x,2r))} \lesssim\; \tilde{\rho }(2r)r^{n\Big (\frac{1}{p^{\prime }_{1}}+\frac{1}{p_{2}}\Big )} \parallel f\parallel _{L_{p_{1}}(B(x,2r))}\\\lesssim & {} r^{\frac{n}{p_{2}}}\left( \int _{2r}^{4r}\tilde{\rho }(t)t^{\frac{n}{p^{\prime }_{1}}-1}dt\right) \parallel f\parallel _{L_{p_{1}}(B(x,2r))} \lesssim\; r^{\frac{n}{p_{2}}} \int _{2r}^{\infty }\tilde{\rho }(t)t^{\frac{n}{p^{\prime }_{1}}-1}\parallel f\parallel _{L_{p_{1}}(B(x,2t))}dt \\= & {} 2^{-\frac{n}{p^{\prime }_{1}}}r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\tilde{\rho }\Big (\frac{\tau }{2}\Big )\tau ^{\frac{n}{p^{\prime }_{1}}-1}\parallel f\parallel _{L_{p_{1}}(B(x,\tau ))}d\tau \;\lesssim\; r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\tilde{\rho }(\tau )^{\frac{n}{p^{\prime }_{1}}-1}\parallel f\parallel _{L_{p_{1}}(B(x,\tau ))}d\tau \\\lesssim & {} r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\rho (\tau )^{\frac{n}{p^{\prime }_{1}}-1}\parallel f\parallel _{L_{p_{1}}(B(x,\tau ))}d\tau \end{aligned}$$

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{p_{1}}^{\text {loc}}(\mathbb {R}^{n}).$

1.2a. In order to estimate the second summand in the right-hand side of (3.2) we obtain an estimate for $(\overline{I}_{\rho (.),r}|f|)(x).$ First we note that

$$\begin{aligned} \int _{2r}^{\infty } \tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt= \sum _{k=1}^{\infty } \int _{2^{k}r}^{2^{k+1}r } \tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

$$\begin{aligned} \ge \sum _{k=1}^{\infty }\tilde{\rho }(2^{k}r) \parallel f\parallel _{L_{1}(B(x,2^{k}r))}\int _{2^{k}r}^{2^{k+1}r }\frac{dt}{t} = \ln 2 \sum _{k=1}^{\infty }\tilde{\rho }(2^{k}r) \parallel f\parallel _{L_{1}(B(x,2^{k}r))} \end{aligned}$$

$$\begin{aligned} \ge \ln 2 \sum _{k=1}^{\infty }\tilde{\rho }(2^{k}r) \parallel f\parallel _{L_{1}(B(x,2^{k}r)\setminus B(x,2^{k-1}r))} \end{aligned}$$

Since by Condition 2) of Definition 3.1 $\tilde{\rho }(2^{k}r)\ge c_{1}\tilde{\rho }(2^{k-1}r)$ and $\vert x-y \vert \ge 2^{k-1}r$ for all $y \in B(x,2^{k}r)\setminus B(x,2^{k-1}r)$, it follows that

$$\begin{aligned} \int _{2r}^{\infty } \tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \gtrsim \sum _{k=1}^{\infty }\int _{B(x,2^{k}r)\setminus B(x,2^{k-1}r)}^{}\tilde{\rho }(2^{k-1}r)\vert f(y)\vert dy \end{aligned}$$

$$\begin{aligned} \ge \sum _{k=1}^{\infty }\int _{B(x,2^{k}r)\setminus B(x,2^{k-1}r)}^{}\tilde{\rho }(\vert x-y\vert )\vert f(y)\vert dy = \int _{^{c}B(x,r)}^{}\tilde{\rho }(\vert x-y\vert )\vert f(y)\vert dy = (\overline{I}_{\tilde{\rho }(.),r}\vert f\vert )(x). \end{aligned}$$

So,

$$\begin{aligned} (\overline{I}_{\tilde{\rho }(.),r}\vert f\vert )(x) \lesssim \int _{2r}^{\infty } \tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.25)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{1}^{loc}(\mathbb {R}^{n}).$

Hence, by applying Hölder’s inequality, we get that

$$\begin{aligned} r^{\frac{n}{p_{2}}} \left( \overline{I}_{\rho (.), 2r} \vert f\vert \right) (x) \lesssim r^{\frac{n}{p_{2}}} \left( \overline{I}_{ \tilde{\rho }(.), 2r} \vert f\vert \right) (x) \lesssim r^{\frac{n}{p_{2}}}\int _{4r}^{\infty }\tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt\end{aligned}$$

$$\begin{aligned} \lesssim r^{\frac{n}{p_{2}}}\int _{4r}^{\infty }\tilde{\rho }(t)t^{\frac{n}{p_{1}^{\prime }}-1}\parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \lesssim r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\rho (t)t^{\frac{n}{p_{1}^{\prime }}-1}\parallel f\parallel _{L_{p_{1}}(B(x,t))}dt \end{aligned}$$

(3.26)

uniformly in $x \in \mathbb {R}^{n}, r>0$ and $f \in L_{1}^{loc}(\mathbb {R}^{n}).$

Step 2 (proof of equivalences (3.22) and (3.23) for $p_{1}=1$). We apply equivalences (3.2) and (3.3).

2.1a. Under the assumption $\tilde{\rho } \in \tilde{S}_{n,1,p_{2}}$ inequality (3.12) holds for $p_{1}=1$ and $1\le p_{2}\le \infty$ and inequality (3.13) holds for $p_{1}=1$ and $0<p_{2}\le 1,$ therefore, Lemma 3.3 ensures the validity of inequality (3.10) with $p_{1}=1$ and we have

$$\begin{aligned} r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\rho (t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))} dt\gtrsim r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

$$\begin{aligned} \gtrsim \varphi _{n, \tilde{\rho },1,p_{2}}(r)\parallel f\parallel _{L_{1}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \gtrsim \parallel I_{\rho (.)}(\vert f\vert \chi _{B(x,2r)}) \parallel _{L_{p_{2}}(B(x,r))}\ge \parallel I_{\rho (.)}(\vert f\vert \chi _{B(x,2r)}) \parallel _{WL_{p_{2}}(B(x,r))} \end{aligned}$$

(3.27)

uniformly in $x\in \mathbb {R}^{n}, \ r>0$ and $f \in L^{\text {loc}}_{1}( \mathbb {R}^{n})$ and we get the estimate above for the first summand in (3.22).

2.1b. If $\rho \in S_{n,1,p_{2}},$ then condition (3.9) holds for $p_{1}=1$ and $1<p_{2}< \infty ,$ therefore, Lemma 3.3 ensures the validity of inequality (3.11) and we, respectively, have

$$\begin{aligned} r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\rho (t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))} dt \end{aligned}$$

$$\begin{aligned} \gtrsim r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt\ge \varphi _{n, \tilde{\rho },1,p_{2}}(r)\parallel f\parallel _{L_{1}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \gtrsim \parallel I_{\rho (.)}(\vert f\vert \chi _{B(x,2r)}) \parallel _{WL_{p_{2}}(B(x,r))} \end{aligned}$$

(3.28)

uniformly in $x\in \mathbb {R}^{n}, \ r>0$ and $f \in L^{\text {loc}}_{1}( \mathbb {R}^{n})$ and we get the estimate above for the first summand in (3.23).

2.2a. If $\tilde{\rho }\in \tilde{S}_{n,1,p_{2}}$ and $1\le p_{2}\le \infty$, to the second summand in the right-hand side of (3.2) we apply inequality (3.25). Equivalence (3.2) and inequalities (3.25) and (3.27) imply that

$$\begin{aligned} \parallel I_{\rho (.)}\vert f\vert \parallel _{WL_{p_{2}}(B(x,r))} \le \parallel I_{\rho (.)}\vert f\vert \parallel _{L_{p_{2}}(B(x,r))}\lesssim r^{\frac{n}{p_{2}}}\ \int _{r}^{\infty }\rho (t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.29)

uniformly in $x \in \mathbb {R}^{n},$ $r>0$ and $f \in L^{\text {loc}}_{1}(\mathbb {R}^{n}).$

2.2b. Accordingly, if $\tilde{\rho } \in S_{n,1,p_{2}},$ $p_{1}=1$ and $1<p_{2}<\infty ,$ then equivalence (3.3) and inequalities (3.25) and (3.28) imply that

$$\begin{aligned} \parallel I_{\rho (.)}\vert f\vert \parallel _{WL_{p_{2}}(B(x,r))} \lesssim r^{\frac{n}{p_{2}}}\ \int _{r}^{\infty }\rho (t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.30)

uniformly in $x \in \mathbb {R}^{n},$ $r>0$ and $f \in L^{\text {loc}}_{1}(\mathbb {R}^{n}).$

2.3. For all $y \in B(x,r)$ we have $\vert y-z \vert \le 2r$ if $z \in B(x,r)$ and $\vert y-z\vert \le 2 \vert x-z \vert$ if $z \in ^{c}B(x,r)$ (since $\vert y-z \vert \le \vert y-x \vert + \vert x-z\vert \le r+ \vert x-z\vert \le 2 \vert x-z \vert ).$ Therefore, since $\tilde{\rho }$ is non-increasing

$$\begin{aligned} I_{\tilde{\rho }}(.)(\vert f\vert )(y)= \int _{B(x,r) }^{}\tilde{\rho }(\vert y-z \vert )\vert f(z) \vert dz + \int _{^{c}B(x,r)}^{}\tilde{\rho }(\vert y-z\vert )\vert f(z)\vert dz \end{aligned}$$

$$\begin{aligned} \ge \tilde{\rho }(2r) \int _{B(x,r)}^{}\vert f(z)\vert dz+ \int _{^{c}B(x,r)}^{}\tilde{\rho }(2\vert x-z\vert )\vert f(z)\vert dz. \end{aligned}$$

By Condition 2) of Definition 3.1 with $s=2t$ we have $\tilde{\rho }(2t)\ge c_{1}\tilde{\rho }(z)$ for any $t>0,$ hence, by equality (3.18) uniformly in $x \in \mathbb {R}^{n},$ $r>0, \ y \in B(x,r)$ and $f \in L^{\text {loc}}_{1}(\mathbb {R}^{n})$

$$\begin{aligned} I_{\rho (.)}(\vert f\vert )(y)\gtrsim I_{\tilde{\rho }(.)}(\vert f)(y) \ge c_{1} \left( \tilde{\rho }(r) \int _{B(x,r)}^{}\vert f(z)\vert dz+ \int _{^{c}B(x,r)}^{} \tilde{\rho }(\vert x-z \vert )\vert f(z)\vert dz\right) \end{aligned}$$

$$\begin{aligned} =c_{1} \left( \tilde{\rho }(r) \int _{B(x,r)}^{}\vert f(z)\vert dz+ \int _{ r}^{\infty } \left( \int _{B(x,t)\setminus B(x,r)}^{}\vert f(z)\vert dz\right) \vert (\tilde{\rho })'(t)\vert dt\right) \end{aligned}$$

$$\begin{aligned} =c_{1} \Big ( \tilde{\rho }(r) \int _{B(x,r)}^{}\vert f(z)\vert dz+ \int _{ r}^{\infty } \left( \int _{B(x,t)}^{}\vert f(z)\vert dz\right) \vert (\tilde{\rho })' (t)\vert dt - \left( \int _{ r}^{\infty }\vert (\tilde{\rho })' (t)\vert dt\right) \int _{B(x,r)}^{}\vert f(z)\vert dz\Big ) \end{aligned}$$

$$\begin{aligned} = c_{1} \int _{ r}^{\infty }\left( \int _{B(x,t)}^{}\vert f(z)\vert dz\right) \vert (\tilde{\rho })' (t)\vert dt. \end{aligned}$$

Consequently, by Remark 3.1 and by Condition 5) of Definition 3.3

$$\begin{aligned} \parallel I_{\rho (.)}\vert f\vert \parallel _{L_{p_{2}}(B(x,r))}\ge \parallel I_{\rho (.)}\vert f\vert \parallel _{WL_{p_{2}}(B(x,r))}\gtrsim \parallel I_{\tilde{\rho }}\vert f\vert \parallel _{WL_{p_{2}}(B(x,r))} \end{aligned}$$

$$\begin{aligned} \ge c_{1}v_{n}^{\frac{n}{p_{2}}}r^{\frac{n}{p_{2}}}\int _{ r}^{\infty }\left( \int _{B(x,t)}^{}\vert f(z)\vert dz\right) \vert (\tilde{\rho })'(t)\vert dt \end{aligned}$$

$$\begin{aligned} \gtrsim r^{\frac{n}{p_{2}}} \int _{r}^{\infty } \tilde{\rho }(t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \gtrsim r^{\frac{n}{p_{2}}}\int _{r}^{\infty } \rho (t)t^{-1}\parallel f\parallel _{L_{1}(B(x,t))}dt \end{aligned}$$

(3.31)

uniformly in $x \in \mathbb {R}^{n},$ $r>0$ and $f \in L^{\text {loc}}_{1}(\mathbb {R}^{n}).$

Step 3. Statement 1 of Theorem 3.1 follows by inequalities (3.24) and (3.26). Statement 2 follows by inequalities (3.25), (3.27) and (3.31). Statement 3 follows by inequalities (3.25), (3.28) and (3.31).

Generalized Riesz potential and Hardy operator

Let $\mathfrak {M}^{+}((0,\infty ))$ be the subset of $\mathfrak {M}((0,\infty ))$ consisting of all non-negative functions on $(0,\infty ).$ We denote by $\mathfrak {M}^{+}((0,\infty );\downarrow )$ the cone of all functions in $\mathfrak {M}^{+}((0,\infty ))$ which are non-increasing on $(0,\infty )$ and we set

$$\begin{aligned} \mathbb {A} = \left\{ \varphi \in \mathfrak {M}^{+}((0,\infty );\downarrow ): \lim \limits _{t \rightarrow \infty } \varphi (t)=0 \right\} . \end{aligned}$$

Let H be the Hardy operator

$$\begin{aligned} \left( Hg\right) (t) := {\int _{0}^{t}}g(r)dr, \ 0< t<\infty \end{aligned}$$

Moreover, let, for $0<p\le \infty$ and a Lebesgue measurable function $v:(0,\infty )\rightarrow [0,\infty )$, $L_{p_{1}v(\cdot )}(0,\infty )$ denote the space of all Lebesgue measurable functions $f:(0,\infty )\rightarrow \mathbb {C}$ for which

$$\begin{aligned} \Vert f\Vert _{L_{ p,v(\cdot ) }(0,\infty )}= \Vert fv\Vert _{L_{p}(0,\infty )}<\infty . \end{aligned}$$

Theorem 4.1

Let $n \in \mathbb {N}$, $1\le p_{1}< \infty ,$ $0<p_{2}\le \infty ,$ $0<\theta _{2}\le \infty$, $w_{2} \in \Omega _{\theta _{2}}$, $\rho$ be a positive continuous function on $(0,\infty )$ and

$$\begin{aligned} \mu _{n,\rho ,p_{1}}(r)= \frac{\int _{r}^{\infty }\rho (t)t^{\frac{n}{p^{\prime }_{1}}-1}dt}{\int _{1}^{\infty }\rho (t)t^{\frac{n}{p^{\prime }_{1}}-1}dt } ,\ r>0, \end{aligned}$$

(4.1)

$$\begin{aligned} v_{2}(r)= w_{2}\left( \mu ^{(-1)}_{n,\rho ,p_{1}}(r)\right) \left( \mu ^{(-1)}_{n,\rho ,p_{1}}(r)\right) ^{\frac{n}{p_{2}}}\vert ( \mu ^{(-1)}_{n,\rho ,p_{1}}(r))^{\prime }\vert ^{\frac{1}{\theta _{2}}}, \ r>0, \end{aligned}$$

(4.2)

$$\begin{aligned} g_{n,\rho ,p_{1}}(t)= \Vert f\Vert _{L_{p_{1}}\left( B(0,\mu _{n,\rho ,p_{1}}^{(-1)}(t) )\right) }, \ t>0. \end{aligned}$$

(4.3)

1. If $1<p_{1}<p_{2}<\infty$ or $1\le p_{1} <\infty$ and $0< p_{2}\le p_{1}$, and $\rho \in S_{n,p_{1}.p_{2}}$, then

$$\begin{aligned} \Vert I_{\rho (\cdot )} f \Vert _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\lesssim \Vert Hg_{n,\rho ,p_{1}} \Vert _{L_{\theta _{2},v_{2}(\cdot )}(0,\infty )} \end{aligned}$$

(4.4)

uniformly in $f\in \mathfrak {M} (\mathbb {R}^{n})$.

2. If $p_{1}= 1$ and $0<p_{2}<\infty$, and $\rho \in \tilde{S}_{n,1,p_{2}}$, then

$$\begin{aligned} \Vert I_{\rho (\cdot )} f \Vert _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\approx \Vert Hg_{n,\rho , 1} \Vert _{L_{\theta _{2},v_{2}(\cdot )}(0,\infty )} \end{aligned}$$

(4.5)

uniformly in all non-negative functions $f\in \mathfrak {M} (\mathbb {R}^{n})$.

3. If $p_{1}=1$ and $1<p_{2}<\infty$, and $\rho \in S_{n,1,p_{2}}$, then

$$\begin{aligned} \Vert I_{\rho (\cdot )} f \Vert _{WLM_{p_{2}\theta _{2},w_{2}(\cdot )}}\approx \Vert Hg_{n,\rho , 1} \Vert _{L_{\theta _{2},v_{2}(\cdot )}(0,\infty )} \end{aligned}$$

(4.6)

uniformly in all non-negative functions $f\in \mathfrak {M} (\mathbb {R}^{n})$.

Remark 4.1

If $\rho (t)=t^{\alpha -n}$, $1 \le p_{1}<\infty ,$ $0< p_{2}<\infty ,$ $0<\alpha <\frac{n}{p_{1}},$ then $\mu _{n,\rho , p_{1}}(r)=r^{-\sigma },$ where $\sigma =\frac{n}{p_{1}}-\alpha ,$ $\mu _{n,\rho , p_{1}}^{(-1)}(r)=r^{- \frac{1}{\sigma }},$ $v_{2}(r)= \sigma ^{-\frac{1}{\theta _{2}}}w_{2}(r^{-\frac{1}{\sigma }})r^{-\frac{n}{\sigma p_{2}}-\frac{1}{\sigma \theta _{2}}-\frac{1}{\theta _{2}}},$ $g_{n,\rho , p_{1}}(t)$ $= \parallel f\parallel _{L_{p_{1}}(B(0,t^{-\frac{1}{\sigma }}))}$ and Lemma 4.1 takes the form of Lemma 4.1 in [6].

Remark 4.2

Due to Condition 4) of Definition 3.3

$$\begin{aligned} \lim \limits _{r \rightarrow 0^{+}}\mu _{n,\rho , p_{1}}(r)=\lim _{r\rightarrow 0^{+}} \mu _{n,\tilde{\rho }, p_{1}} (r) =\infty , \lim _{r\rightarrow \infty }\mu _{n,\rho ,p_{1}} (r)= \lim _{r\rightarrow \infty } \mu _{n,\tilde{\rho },p_{1}} (r)=0 \end{aligned}$$

(4.7)

and $\mu _{n,\rho , p_{1}}$ is a strictly decreasing continuously differentiable function on $(0,\infty ).$ Moreover,

$$\begin{aligned} \lim \limits _{r\rightarrow 0^{+}} \mu _{n,\rho , p_{1}}^{(-1)}(r) = \infty ,\ \lim \limits _{r\rightarrow \infty } \mu _{n,\rho , p_{1}}^{(-1)}(r) = 0. \end{aligned}$$

(4.8)

Proof

1. Let $1< p_{1}< p_{2} < \infty$ or $1 \le p_{1} \le \infty$ and $0 < p_{2} \le p_{1}$, and $\rho \in S_{n,p_{1},p_{2}}$. By inequality (3.21) we have $\Big (c=\left( \int _{1}^{\infty } \rho (t)t^{\frac{n}{p'_{1}}-1}dt \right) ^{-1}\Big )$

$$\begin{aligned} \Vert I_{\rho (\cdot )} f \Vert _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\lesssim \Big \Vert w_{2}(r)r^{\frac{n}{p_{2}}}\int _{r}^{\infty }\rho (t) t^{ \frac{n}{p'_{1}}-1} \Vert f\Vert _{L_{p_{1}}(B(0,t))}dt\Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} =c \Big \Vert w_{2}(r)r^{\frac{n}{p_{2}}}\left( -\int _{r}^{\infty } \Vert f\Vert _{L_{p_{1}}(B(0,t))}d \mu _{n,\rho , p_{1}}(t)\right) \Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned}=c\left( \mu _{n,\rho , p_{1}}(t) =\tau ,\ t=\mu _{n,\rho , p_{1}}^{(-1)}(\tau ),\ \lim \limits _{r\rightarrow \infty } \mu _{n,\rho , p_{1}}(t)=0 \right) \end{aligned}$$

$$\begin{aligned} = c\Big \Vert w_{2}(r)r^{\frac{n}{p_{2}}} \int \limits ^{ \mu _{n,\rho ,p_{1}}(r) }_{0} \Vert f\Vert _{L_{p_{1}}\left( B(0,\mu _{n,\rho , p_{1}}^{(-1)}(\tau ) )\right) } d \tau \Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} = c\Big (\mu _{n,\rho , p_{1}}(r)=u,\ r= \mu _{n,\rho , p_{1}}^{(-1)}(u),\ \lim \limits _{r\rightarrow 0} \mu _{n,\rho , p_{1}}(r)=\infty ,\ \lim \limits _{r\rightarrow \infty } \mu _{n,\rho , p_{1}}(r)=0\Big ) \end{aligned}$$

$$\begin{aligned} =c\Big \Vert w_{2}\left( \mu _{n,\rho , p_{1}}^{(-1)}(u) \right) \left( \mu _{n,\rho , p_{1}}^{(-1)}(u) \right) ^{\frac{n}{p_{2}}} \left| \left( \mu _{n,\rho , p_{1}}^{(-1)}(u) \right) ^{\prime }\right| ^{\frac{1}{\theta _{2}}}\int \limits _{0}^{u} \Vert f\Vert _{L_{p_{1}}\left( B (0, \mu _{n,\rho , p_{1}}^{(-1)}(\tau )) \right) } d\tau \Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} =c \left\| v_{2}(u) \left( Hg_{n,\rho , p_{1}}\right) (u)\right\| _{L_{\theta _{2}}(0,\infty )}= c\Vert Hg_{n,\rho , p_{1}}\Vert _{L_{\theta _{2},v_{2}(\cdot ),}(0,\infty )} \end{aligned}$$

uniformly in $f\in \mathfrak {M} (\mathbb {R}^{n})$.

2. Let $p_{1} =1$ and $0 < p_{2} \le 1$, and $\rho \in \tilde{S}_{n,p_{1},p_{2}}$. By inequality (3.22) we have similarly to Step 1

$$\begin{aligned} \Big \Vert I_{\rho (\cdot )} f \Vert _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\approx \Vert w_{2}(r)r^{ \frac{n}{p_{2}}}\int _{r}^{\infty }\rho (t) t^{ -1} \Vert f\Vert _{L_{1}(B(0,t))}\Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} = c \Big \Vert w_{2}(r)r^{ \frac{n}{p_{2}}}\left( -\int _{r}^{\infty } \Vert f\Vert _{L_{1}(B(0,t))}d \mu _{n,\rho , 1}(t)\right) \Big \Vert _{L_{\theta _{2}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} =c\left\| v_{2}(u) \left( Hg_{n,1}\right) (u)\right\| _{L_{\theta _{2}}(0,\infty )}= c\Vert Hg_{n,\rho , 1}\Vert _{L_{\theta _{2},v_{2}(\cdot ),}(0,\infty )} \end{aligned}$$

uniformly in all non-negative functions $f\in \mathfrak {M} (\mathbb {R}^{n})$.

3. Let $p_{1} = 1$ and $1< p_{2} < \infty$, and $\rho \in S_{n,1,p_{2}}$. Equivalence (4.6) follows similarly to Step 2 from equivalence (3.23). $\square$

Theorem 4.2

Assume that $n \in \mathbb {N}$, $1\le p_{1}< \infty$, $0<p_{2}\le \infty$, $0<\theta _{1}$, $\theta _{2}\le \infty$, $w_{1}\in \Omega _{\theta _{1}}$, $w_{2}\in \Omega _{\theta _{2}}$, $\mu _{n,\rho , p_{1}}$ is defined by formula (4.1),

$$\begin{aligned} v_{1}(r) = w_{1}\left( \mu ^{(-1)}_{n,\rho , p_{1}}(r)\right) \big \vert \left( \mu ^{(-1)}_{n,\rho , p_{1}}(r)\right) ' \big \vert ^{\frac{1}{\theta _{1}}}, \ r>0, \end{aligned}$$

(4.9)

and $v_{2}$ is defined by formula (4.2).

1. Let $1<p_{1}<p_{2}< \infty$ or $1\le p_{1}< \infty$ and $0<p_{2}\le p_{1}$, and $\rho \in S_{n,p_{1}, p_{2}}$. If the operator H is bounded from $L_{\theta _{1}, v_{1}(\cdot )}(0,\infty )$ to $L_{\theta _{2}, v_{2}(\cdot )}(0,\infty )$ on the cone $\mathbb {A}$, that is

$$\begin{aligned} \Vert Hg\Vert _{L_{\theta _{2}, v_{2}(\cdot )}(0,\infty )}\lesssim \Vert g\Vert _{L_{\theta _{1}, v_{1}(\cdot )} (0,\infty ) } \end{aligned}$$

(4.10)

uniformly in $g\in \mathbb {A}$, then the operator $I_{\rho (\cdot )}$ is bounded from $LM_{p_{1} \theta _{1},w_{1}(\cdot )}$ to $LM_{p_{2} \theta _{2},w_{2}(\cdot )}$.

2. Let $p_{1}=1$, $0<p_{2}<\infty$ and $\rho \in \tilde{S}_{n,1,p_{2}}$. Then the operator $I_{\rho (\cdot )}$ is bounded from $LM_{1 \theta _{1},w_{1}(\cdot )}$ to $LM_{p_{2} \theta _{2},w_{2}(\cdot )}$ if and only if the operator H is bounded from $L_{ \theta _{1},v_{1}(\cdot )}(0, \infty )$ to $L_{ \theta _{2},v_{2}(\cdot )}(0, \infty )$ on the cone $\mathbb {A}$.

3. Let $p_{1}=1$, $1<p_{2}<\infty$ and $\rho \in S_{n,1,p_{2}}$. Then the operator $I_{\rho (\cdot )}$ is bounded from $LM_{p_{1} \theta _{1},w_{1}(\cdot )}$ to $WLM_{p_{2} \theta _{2},w_{2}(\cdot )}$ if and only if the operator H is bounded from $L_{ \theta _{1},v_{1}(\cdot )}(0, \infty )$ to $L_{ \theta _{2},v_{2}(\cdot )}(0, \infty )$ on the cone $\mathbb {A}$.

Remark 4.3

If we put $r=\mu _{n_{1},\rho , p_{1}}(t)$ in (4.9), then, taking into account that $\big ( \mu ^{(-1)}_{n_{1},\rho , p_{1}}(r) \big )'= \Big (\mu '_{n_{1},\rho , p_{1}}\big (\mu ^{(-1)}_{n_{1},\rho , p_{1}}(r)\big ) \Big )^{-1}$, $r>0$, we get

$$\begin{aligned} v_{1}\big ( \mu _{n_{1},\rho , p_{1}} (t)\big ) =w_{1}(t) \left| \mu '_{n_{1},\rho , p_{1}}(t) \right| ^{-\frac{1}{\theta _{1}}}, \ t>0 \end{aligned}$$

(4.11)

and

$$\begin{aligned} v_{2}\big ( \mu _{n_{1},\rho , p_{1}}(t) \big )=w_{2}(t)t^{\frac{n}{p_{2}}}\left| \mu '_{n,\rho , p_{1}}(t) \right| ^{-\frac{1}{\theta _{1}}}, \ t>0, \end{aligned}$$

(4.12)

Lemma 4.1

Assume that $0<p_{1}, p_{2}, \theta _{1}, \theta _{2} \le \infty$, $w_{1}\in \Omega _{\theta _{1}}$, $\rho$ is a positive continuous function on $(0, \infty )$ such that

$$\begin{aligned} {\int _{0}^{1}}\rho (t) t^{\frac{n}{p'_{1}}-1}dt=\infty , \ \int _{1}^{\infty }\rho (t) t^{\frac{n}{p'_{1}}-1}dt<\infty , \end{aligned}$$

(4.13)

and the functions $\mu _{n,\rho , p_{1}}$, $g_{n,\rho , p_{1}}$, $v_{1}$, $v_{2}$ are defined by formulas (4.1), (4.3), (4.9), (4.2), respectively. Then

$$\begin{aligned} \Vert g_{n,\rho , p_{1}}\Vert _{L_{\theta _{1},v_{1}(\cdot )}}= \Vert f\Vert _{LM_{p_{1}\theta _{1},w_{1}(\cdot )}} \end{aligned}$$

(4.14)

for all $f \in LM_{p_{1}\theta _{1},w_{1}(\cdot )}$, for any measurable function $\phi _{1}:(0,\infty )\rightarrow (0, \infty )$ and $t>0$

$$\begin{aligned} \Vert v_{1}\phi _{1}\Vert _{L_{\theta _{1} }(0,t)}= \Vert w_{1}(r)\phi _{1}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\theta _{1} }(\mu ^{(-1)}_{n,\rho ,p_{1}}(t),\infty )}, \end{aligned}$$

(4.15)

$$\begin{aligned} \Vert v_{1}\phi _{1}\Vert _{L_{\theta _{1} }(t, \infty )}= \Vert w_{1}(r)\phi _{1}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\theta _{1} }(0,\mu ^{(-1)}_{n,\rho ,p_{1}}(t))}, \end{aligned}$$

(4.16)

(4.16')

$$\begin{aligned} \Vert v_{1}\Vert _{L_{\theta _{1} }(0,\infty )}=\left\| w_{1}\right\| _{L_{\theta _{1}} (0,\infty )}, \end{aligned}$$

(4.17)

and for any measurable function $\phi _{2}:(0,\infty )\rightarrow (0, \infty )$ and $t>0$

$$\begin{aligned} \Vert v_{2}\phi _{2}\Vert _{L_{\theta _{2} }(0,t)}= \Vert w_{2}(r)r^{\frac{n}{p_{2}}}\phi _{2}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\theta _{2} }(\mu ^{(-1)}_{n,\rho ,p_{1}}(t),\infty )}, \end{aligned}$$

(4.18)

$$\begin{aligned} \Vert v_{2}\phi _{2}\Vert _{L_{\theta _{2} }(t,\infty )}= \Vert w_{2}(r)r^{\frac{n}{p_{2}}}\phi _{2}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\theta _{2} }(0,\mu ^{(-1)}_{n,\rho ,p_{1}}(t))}, \end{aligned}$$

(4.19)

(4.19')

$$\begin{aligned} \Vert v_{2}\Vert _{L_{\theta _{2} }(0,\infty )}=\left\| w_{2}(r)r^{\frac{n}{p_{2}}}\right\| _{L_{\theta _{2}} (0,\infty )}. \end{aligned}$$

(4.20)

Proof

1. Indeed,

$$\begin{aligned} \Vert g_{n,\rho , p_{1}}\Vert _{L_{\theta _{1},v_{1}(\cdot )}}= \left\| v_{1}(t)\Vert f\Vert _{L_{p_{1}}(B(0,\mu ^{(-1)}_{n,\rho ,p_{1}}(t)))} \right\| _{L_{\theta _{1}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} =\left( \mu ^{(-1)}_{n,\rho ,p_{1}}(t) =r, t=\mu _{n,\rho ,p_{1}}(r), \ {\text {by\ (4.13) }} \lim \limits _{r\rightarrow 0^{+}} \mu _{n,\rho , p_{1}}^{(-1)}(t) = \infty , \lim \limits _{n\rightarrow \infty } \mu _{n,\rho , p_{1}}^{(-1)}(t) = 0 \right) \end{aligned}$$

$$\begin{aligned} = \left\| v_{1} \left( \mu _{n,\rho ,p_{1}}(r) \right) \vert \mu '_{n,\rho ,p_{1}}(r)\vert ^{\frac{1}{\theta _{1}}} \Vert f\Vert _{L_{p_{1}}(B(0,r))}\right\| _{L_{\theta _{1}}(0,\infty )} \end{aligned}$$

$$\begin{aligned} = \left\| w_{1}(r) \Vert f\Vert _{L_{p_{1}}(B(0,r))}\right\| _{L_{\theta _{1}}(0,\infty )}= \Vert f\Vert _{LM_{p_{1}\theta _{1},w_{1}(\cdot )}}. \end{aligned}$$

(We have used equality (4.11)).

2. Note that, by (3.10) and (4.7), for $0<\theta _{1}<\infty$, $t>0$

$$\begin{aligned} \Vert v_{1}\phi _{1}\Vert _{L_{\theta _{1}}(0,t)}= \left( -{\int _{0}^{t}} w_{1} \left( \mu _{n,\rho , p_{1}}^{(-1)}(s)\right) ^{\theta _{1}}\phi _{1}(s)^{\theta _{1}} \left( \mu _{n,\rho , p_{1}}^{(-1)}(s)\right) ' ds\right) ^{\frac{1}{\theta _{1}}} \end{aligned}$$

$$\begin{aligned} =\left( \mu _{n,\rho , p_{1}}^{(-1)}(s)=r\right) =\left( \int _{\mu _{n,\rho , p_{1}}^{(-1)}(t)}^{\infty }w_{1} (r)^{\theta _{1}}\phi _{1}(\mu _{n,\rho ,p_{1}}(r))^{\theta _{1}} dr\right) ^{\frac{1}{\theta _{1}}} \end{aligned}$$

$$\begin{aligned} =\Vert w_{1}(r)\phi _{1}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\theta _{1}}(\mu _{n,\rho , p_{1}}^{(-1)}(t),\infty )} \end{aligned}$$

and, similarly

$$\begin{aligned} \Vert v_{1}\Vert _{L_{\theta _{1}}(t,\infty )}=\Vert w_{1}(r)\phi _{1}(\mu _{n,\rho ,p_{1}}(r)) \Vert _{L_{\theta _{1}}(0,\mu _{n,\rho , p_{1}}^{(-1)}(t))}. \end{aligned}$$

These equalities also hold for $\theta _{1}=\infty$, because, for example, by (4.9)

$$\begin{aligned} \Vert v_{1}\phi _{1}\Vert _{L_{\infty }(0,t)}= \left\| w_{1}\left( \mu _{n,\rho , p_{1}}^{(-1)} (s)\right) \phi _{1}(s)\right\| _{L_{\infty }(0,t)} \end{aligned}$$

$$=\underset{0<s<t}{ess\,sup}\vert w_1\left(\mu_{n,\rho,p_1}^{(-1)}(s)\right)\phi_1(s)\vert$$

$$\begin{aligned} =\Big (\mu _{n,\rho , p_{1}}^{(-1)} (s)= r, \mu _{n,\rho , p_{1}}^{(-1)} ((0,t))= \left( \mu ^{(-1)}_{n,\rho ,p_{1}},\infty \right) \Big ) \end{aligned}$$

$$=\underset{\mu_{n,\rho,p_1}^{(-1)}(t)<r<\infty}{\text{ess}\,\text{sup}}\;\vert w_1\left(r\right)\phi_1\left(\mu_{n,\rho,p_1}(r)\right)\vert =\Vert w_{1}(r)\phi _{1}(\mu _{n,\rho ,p_{1}}(r))\Vert _{L_{\infty }\left( \mu _{n,\rho , p_{1}}^{(-1)}(t),\infty \right) }.$$

Similarly, by (4.2), (4.7) and (4.12) for $0<\theta _{2}\le \infty$, $t>0$

$$\begin{aligned} \Vert v_{2}\Vert _{L_{\theta _{2}}(0,t)}= \left\| w_{2} (r)r^{\frac{n}{p_{2}}} \right\| _{L_{\theta _{2}}\left( \mu _{n,\rho , p_{1}}^{(-1)}(t), \infty \right) }\end{aligned}$$

and

$$\begin{aligned} \Vert v_{2}\Vert _{L_{\theta _{2}}(t, \infty )}= \left\| w_{2} (r)r^{\frac{n}{p_{2}}} \right\| _{L_{\theta _{2}}\left( 0,\mu _{n,\rho , p_{1}}^{(-1)}(t)\right) }. \end{aligned}$$

The proved equalities imply equalities (4.17) and (4.20) by passing to the limit as $t\rightarrow 0^{+}$ or $t\rightarrow +\infty$. $\square$

Proof of Theorem 4.2

1. Let $1< p_{1}< p_{2} < \infty$ or $1 \le p_{1} < \infty$ and $0 < p_{2} \le p_{1}$, and $\rho \in S_{n,p_{1},p_{2}}$. Assume that the operator H is bounded from $L_{\theta _{1},v_{1}(\cdot )}(0,\infty )$ to $L_{\theta _{2},v_{2}(\cdot ) }(0,\infty )$ on the cone $\mathbb {A}$. Since $g_{n,\rho ,p_{1}}\in \mathbb {A}$, by Statement 1 of Theorem 4.1 and formula (4.14) we have

$$\begin{aligned} \left\| I_{\rho (\cdot )}f\right\| _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\lesssim \left\| Hg_{n,\rho ,p_{1}}\right\| _{ L_{\theta _{2},v_{2}(\cdot )}(0,\infty )} \lesssim \left\| g_{n,\rho ,p_{1}}\right\| _{ L_{\theta _{1},v_{1}(\cdot )}(0,\infty )} = \Vert f\Vert _{LM_{p_{1}\theta _{1},w_{1}(\cdot )}} \ \end{aligned}$$

(4.21)

uniformly in $f\in LM_{p_{1}\theta _{1},w_{1}(\cdot )}$, hence the operator $I_{\rho (\cdot )}$ is bounded from $LM_{p_{1}\theta _{1},w_{1}(\cdot )}$ to $LM_{p_{2}\theta _{2},w_{2}(\cdot )}$.

2.1. Let $p_{1}=1$, $0<p_{2}<\infty$ and $\rho \in \tilde{S}_{n,1,p_{2}}$. If the operator H is bounded from $L_{\theta _{1},v_{1}(\cdot )(0,\infty )}$ to $L_{\theta _{2} ,v_{2}(\cdot )}(0,\infty )$ on the cone $\mathbb {A}$, then by Statement 2 of Theorem 4.1 as in Step 1 it follows that the operator $I_{\rho (\cdot )}$ is bounded from $LM_{p_{1}\theta _{1},w_{1}(\cdot )}$ to $LM_{p_{2}\theta _{2},w_{2}(\cdot )}$.

2.2. Assume that $I_{\rho (\cdot )}$ is bounded from $LM_{1\theta _{1},w_{1}(\cdot )}$ to $LM_{p_{2}\theta _{2},w_{2}(\cdot )}$. Then by Statement 2 of Theorem 4.1 and formula (4.14) with $p_{1}=1$

$$\begin{aligned} \left\| H_{g_{n,\rho ,1}} \right\| _{L_{\theta _{2},v_{2}(\cdot )}(0,\infty )} \approx \left\| I_{\rho (\cdot )}f\right\| _{LM_{p_{2}\theta _{2},w_{2}(\cdot )}}\lesssim \left\| f\right\| _{LM_{1\theta _{1},w_{1}(\cdot )}}= \left\| g_{n,\rho ,1}\right\| _{L_{\theta _{1},v_{1}(\cdot )}} \end{aligned}$$

(4.22)

uniformly in all non-negative functions $f \in L_{1}^{\text {loc}} (\mathbb {R}^{n})$.

Let $g\in \mathbb {A}$ be continuously differentiable on $(0,\infty )$. Consider the positive Lebesgue measurable function h on $(0,\infty )$ defined uniquely up to equivalence by the equality

$$\begin{aligned} g(t)= \left\| h(\vert \cdot \vert )\right\| _{L_{1}(B(0,\mu _{n,\rho ,1}^{(-1)}(t)))}=\sigma _{n} \int _{0}^{\mu _{n,\rho ,1}^{(-1)}(t)} h (\tau )\tau ^{n-1}d \tau ,\ t>0 \end{aligned}$$

$$\begin{aligned} \Leftrightarrow g'(t)= \sigma _{n}\big ( \mu _{n,\rho ,1}^{(-1)}(t) \big )' h \big ( \mu _{n,\rho ,1}^{(-1)}(t) \big )\big ( \mu _{n,\rho ,1}^{(-1)}(t) \big )^{n-1} , t>0 \end{aligned}$$

$$\begin{aligned} \Leftrightarrow h(r) =\sigma ^{-1}_{n} g'(\mu _{n,\rho ,1}(r))\mu '_{n,\rho ,1}(r)r^{1-n},r>0. \end{aligned}$$

If we take in (4.22) $f(x)=h(|x|)$, then by (4.3) $\Vert f\Vert _{L_{1}(B(0,\mu _{n,\rho ,1}^{(-1)}(t)))}=g(t)$ and by (4.21)

$$\begin{aligned} \Vert Hg\Vert _{L_{\theta _{2},v_{2}(\cdot )}(0, \infty )}\lesssim \Vert g\Vert _{L_{\theta _{1},v_{1}(\cdot )}} \end{aligned}$$

(4.23)

uniformly in all $g\in \mathbb {A}$ which are continuously differentiable on $(0,\infty )$.

Finally, if g is an arbitrary function in $\mathbb {A}$, then there exist functions $g_{k}\in \mathbb {A}$, $k\in \mathbb {N}$, which are continuously differentiable on $(0,\infty )$ and such $g_{k}\le g_{k+1}$, $k\in \mathbb {N}$, and $\lim \limits _{k\rightarrow \infty }g_{k}=g$ on $(0,\infty )$. Therefore, by passing to the limit in (4.23), with g replaced by $g_{k}$, as $k \rightarrow \infty$, it follows that (4.23) holds for all $g\in \mathbb {A}$.

3. Let $p_{1}=1$, $1<p_{2}<\infty$ and $\rho \in S_{n,1,p_{2}}$. In this case the argument is similar to that of Step 2, only Statement 3 of Theorem 4.1 should be used and the space $LM_{p_{2}\theta _{2},w_{2}(\cdot )}$ should be replaced by the space $WLM_{p_{2}\theta _{2},w_{2}(\cdot )}$. $\square$

Hardy’s inequality on the cone of monotonic functions

Necessary and sufficient conditions for the boundedness of the operator H from one weighted Lebesgue space $L_{\theta _{1},v_{1}(\cdot )}(0,\infty )$ to another one $L_{\theta _{2},v_{2}(\cdot )}(0,\infty )$ on the cone $\mathbb {A}$ are knows for all values of $0<\theta _{1}$, $\theta _{2}\le \infty$. We present below these results which are corollaries of more general statements contained in the survey by A. Gogatishvili and V.D. Stepanov [10] (Theorems 2.5, 3.15 and 3.16). See also the survey by M.L. Goldman [11].

Theorem 5.1

Let $0<\theta _{1}$, $\theta _{2}\le \infty$, $v_{1}$, $v_{2}$ be non-negative Lebesgue measurable functions on $(0,\infty )$. Then the operator H is bounded from $L_{\theta _{1},v_{1}(\cdot )}(0,\infty )$ to $L_{\theta _{2},v_{2}(\cdot )}(0,\infty )$ on the cone $\mathbb {A}$ if and only if

(a) if $1<\theta _{1}\le \theta _{2}<\infty$, then

$$\begin{aligned} A_{11}:=\sup \limits _{t>0}\left( {\int _{0}^{t}} s^{\theta _{2}} v_{2}^{\theta _{2}}(s) ds\right) ^{\frac{1}{\theta _{2}}} \left( {\int _{0}^{t}}v_{1}^{\theta _{1}}(s)ds\right) ^{-\frac{1}{\theta _{1}}} <\infty , \end{aligned}$$

and

$$\begin{aligned} A_{12} : = \sup \limits _{t>0}\Bigg (\int ^{\infty } _{t} v_{2}^{\theta _{2}}(\tau )d\tau \Bigg )^{\frac{1}{\theta _{2}}} \Bigg (\int ^{t} _{0} z^{\theta _{1}'} \Big (\int ^{z} _{0} v_{1}^{\theta _{1}}(s)ds \Big )^{-\theta _{1}'}v_{1}^{\theta _{1}}(z)dz\Bigg )^{\frac{1}{\theta _{1}'}}<\infty , \end{aligned}$$

(b) if $0<\theta _{1}\le 1$, $\theta _{1} \le \theta _{2} <\infty$, then

$$\begin{aligned} A_{2} : = \sup \limits _{t>0}\Bigg ({\int ^{t}_{0}} v_{1}^{\theta _{1}}(s)ds\Bigg )^{-\frac{1}{\theta _{1}}} \Bigg (\int ^{\infty }_{0} \min \{s,t\}^{\theta _{2}} v_{2}^{\theta _{2}}(s)ds \Bigg )^{\frac{1}{\theta _{2}}} <\infty , \end{aligned}$$

(c) if $1<\theta _{1}<\infty$, $0<\theta _{2}<\theta _{1}<\infty$, then

$$\begin{aligned} A_{31} : = \Bigg ( \int ^{\infty }_{0}\Bigg (\int ^{t} _{0} v_{1}^{\theta _{1}}(\tau )d\tau \Bigg )^{-\frac{r}{\theta _{1}}}\Bigg (\int ^{t} _{0}s^{\theta _{2}}v_{2}^{\theta _{2}}(s)ds \Bigg )^{\frac{r}{\theta _{1}}} t^{\theta _{2}}v_{2}^{\theta _{2}}(t)dt \Bigg )^{\frac{1}{r}}<\infty , \end{aligned}$$

and

$$\begin{aligned} A_{32} := \Bigg (\int ^{\infty } _{0}\Bigg (\int ^{\infty } _{t} v_{2}^{\theta _{2}} (s)ds\Bigg )^{\frac{r}{\theta _{1}}}\Bigg (\int ^{t} _{0}s^{\theta _{1}'} \Big (\int ^{s} _{0} v_{1}^{\theta _{1}} (\tau )d\tau \Big )^{-\theta _{1}'}v_{1}^{\theta _{1}}(s)ds \Bigg )^{\frac{r}{\theta _{1}'}} v_{2}^{\theta _{2}}(t)dt \Bigg )^{\frac{1}{r}}<\infty , \end{aligned}$$

where $\frac{1}{r}=\frac{1}{\theta _{2}}-\frac{1}{\theta _{1}}$,

(d) if $0<\theta _{2}<\theta _{1}\le 1$, then $A_{41} := A_{31}<\infty$ and

$$\begin{aligned} A_{42}: = \Bigg (\int ^{\infty } _{0} \Bigg ( \sup \limits _{0< s\le t} s^{\theta _{1}} \Bigg ({\int ^{s}_{0}}v_{1}^{\theta _{1}}(\tau )d\tau \Bigg )^{-1} \Bigg )^{\frac{r}{\theta _{1}}} \Bigg (\int ^{\infty } _{t} v_{2}^{\theta _{2}} (z)dz \Bigg )^{^{\frac{r}{\theta _{1}}}} v_{2}^{\theta _{2}}(t)dt \Bigg )^{\frac{1}{r}}<\infty , \end{aligned}$$

(e) if $0<\theta _{1}\le 1$, $\theta _{2}=\infty$, then

$$\begin{aligned} A_{5} : = \sup \limits _{t>0} \Bigg ({\int ^{t}_{0}} v_{1}^{\theta _{1}}(s)ds\Bigg )^{-\frac{1}{\theta _{1}}} \Bigg (\text {ess} \sup \limits _{y >0 \ \ \ \ } \min (t,y) v_{2}(y)\Bigg ) <\infty , \end{aligned}$$

(f) if $1<\theta _{1}<\infty$, $\theta _{2}=\infty$, then

$$\begin{aligned} A_{6} : = \text {ess} \sup \limits _{s>0 \ \ \ \ } v_{2}(s)\Bigg ({\int ^{s}_{0}} \Bigg ({\int ^{s}_{t}}\Bigg ({\int ^{y}_{0}} v_{1}^{\theta _{1}} (z) dz\Bigg )^{-1}dy \Bigg )^{\theta '_{1}} v_{1}^{\theta _{1}}(t)dt \Bigg )^{\frac{1}{\theta _{1}'}}<\infty , \end{aligned}$$

(g) if $\theta _{1}=\infty$, $0<\theta _{2}<\infty$, then

$$\begin{aligned} A_{7}:=\Bigg (\int ^{\infty }_{0}\Bigg (v_{2}(x){\int ^{x}_{0}} \frac{dy}{\text {ess} \sup \limits _{0<z<y \ \ \ \ } v_{1} (z)}\Bigg )^{\theta _{2}}dx\Bigg )^{\frac{1}{\theta _{2}}} <\infty , \end{aligned}$$

(h) if $\theta _{1}=\theta _{2}=\infty$, then

$$\begin{aligned} A_{8} : = \text {ess} \sup \limits _{x>0 \ \ \ \ } \Bigg (v_{2}(x){\int ^{x}_{0}} \frac{dy}{\text {ess} \sup \limits _{0<z<y \ \ \ \ } v_{1} (z)}\Bigg ) <\infty . \end{aligned}$$

Conditions, ensuring boundedness of the generalized Riesz potential

In order to obtain sufficient conditions and necessary and sufficient conditions for $p_{1}=1$ on the weight functions $w_{1}$, $w_{2}$ ensuring the boundedness of $I_{\rho (.)}$ from $LM_{p_{1},\theta _{1},w_{1}(.)}$ to $LM_{p_{2},\theta _{2},w_{2}(.)}$ for $1<p_{1}<p_{2}<\infty$ or $1\le p_{1}<\infty$, it, clearly, suffices to apply Theorems 4.2 and 5.1.

In order to make these conditions have a simpler form we shall carry out certain changes of variables and apply equalities (4.15)–(4.20) proved in Lemma 4.1.

Theorem 6.1

Assume that $0<\theta _{1},\theta _{2}\le \infty , w_{1} \in \Omega _{\theta _{1}}, w_{2} \in \Omega _{ \theta _{2}}$.

1. Let $1< p_{1}< p_{2}<\infty$ or $1 \le p_{1}< \infty$ and $0<p_{2}\le \infty ,$ and $\rho \in S_{n, p_{1},p_{2}}.$ Then the operator $I_{\rho (.)}$ is bounded from $LM_{p_{1},{ \theta _{1}}, w_{1}(\cdot ) }$ to $LM_{p_{2},{\theta _{2}},w_{2}(\cdot ) }$ if the following conditions are satistied.

(a) If $1<\theta _{1}\le \theta _{2}<\infty ,$ then

$$\begin{aligned} B_{11} := \sup _{t>0}\left( \int _{t}^{\infty }w_{2,n,p_{2} }^{\theta _{2}}(x)\mu ^{\theta _{2}}_{n,\rho ,p_{1}}(x) dx\right) ^{\frac{1}{\theta _{2}}} \left( \int _{t}^{\infty } w_{1}^{\theta _{1}} (x)dx \right) ^{-\frac{1}{\theta _{1}}} < \infty , \end{aligned}$$

and

$$\begin{aligned} B_{12} := \sup _{t>0}\left( {\int _{0}^{t}}w_{2,n,p_{2} }^{\theta _{2}}(x) dx\right) ^{\frac{1}{\theta _{2}}} \left( \int _{t}^{\infty } \frac{w_{1}^{\theta _{1}} (x)\mu ^{\theta _{1}^{\prime }}_{n,\rho ,p_{1}}(x)}{(\int _{x}^{\infty }w_{1}^{\theta _{1}}(s)ds)^{\theta _{1}^{\prime }}} dx \right) ^{\frac{1}{\theta '_{1}}} < \infty . \end{aligned}$$

where $w_{2,n,p_{2} }(x) =w_{2}(x)x^{\frac{n}{p_{2}}}$.

(b) If $0<\theta _{1}\le 1$, $\theta _{1}\le \theta _{2}<\infty ,$ then

$$\begin{aligned} B_{2} := \sup _{t>0} \left( \int _{t}^{\infty } w_{1}^{\theta _{1}} (x)dx \right) ^{-\frac{1}{\theta _{1}}} \left( \int _{0}^{\infty }w_{2 ,n,p_{2} }^{\theta _{2}}(x) \min \{ \mu _{n,\rho ,p_{1}}(t), \mu _{n,\rho ,p_{1}}(x) \}^{\theta _{2}} dx\right) ^{\frac{1}{\theta _{2}}} < \infty . \end{aligned}$$

(c) If $1<\theta _{1}< \infty , 0<\theta _{2}< \theta _{1}< \infty ,$ then

$$\begin{aligned} B_{31} := \left( \int _{0}^{\infty } \left( \frac{\int _{t}^{\infty }w_{2,n,p_{2} }^{\theta _{2}}(x)\mu ^{\theta _{2}}_{n,\rho ,p_{1}}(x) dx}{\int _{t}^{\infty }w_{1}^{\theta _{1}}(x)dx} \right) ^{\frac{r}{\theta _{1}} } w_{2,n,p_{2} }^{\theta _{2}}(t)\mu ^{\theta _{2}}_{n,\rho ,p_{1}}(t) dt\right) ^{\frac{1}{r} }< \infty , \end{aligned}$$

and

$$\begin{aligned} B_{32} := \Bigg (\int \limits ^{\infty }_{0} \Bigg (\int \limits ^{\infty }_{z} w_{2 ,n,p_{2} }^{\theta _{2}} (x) dx\Bigg )^{\frac{r}{\theta _{1}} }\Bigg (\int \limits ^{\infty } _{ z}\frac{\mu ^{\theta _{1}'}_{n,\rho ,p_{1}}(\tau )w_{1}^{\theta _{1}}(\tau )}{ \Big (\int \limits ^{\infty } _{ \tau } w_{1}^{\theta _{1}} (u)du \Big )^{\theta _{1}'}}d\tau \Bigg )^{ \frac{r}{\theta '_{1}} } w_{2 ,n,p_{2} }^{\theta _{2}}(z) dz \Bigg )^{ \frac{1}{r}} <\infty , \end{aligned}$$

where $\frac{1}{r}=\frac{1}{\theta _{2}}-\frac{1}{\theta _{1}}$.

(d) If $0<\theta _{2}< \theta _{1}\le 1,$ then $B_{41}: =B_{31}<\infty$ and

$$\begin{aligned} B_{42}:= \Bigg (\int \limits ^{\infty }_{0} \sup \limits _{y<z<\infty } \mu _{n,\rho ,p_{1}}(z)^{r} \Bigg (\int \limits ^{\infty }_{ z} w_{1}^{\theta _{1}}(s)ds\Bigg )^{-\frac{r}{\theta _{1}} } \Bigg (\int \limits ^{y}_{ 0} w_{2 ,n,p_{2} }^{\theta _{2}}(x) dx\Bigg )^{\frac{r}{\theta _{1}} } w_{2 ,n,p_{2} }^{\theta _{2}}(y)dy\Bigg )^{\frac{1}{r} } <\infty . \end{aligned}$$

(e) If $0<\theta _{1}\le 1,\theta _{2}=\infty ,$ then

$$\begin{aligned} B_{5}:= \sup \limits _{ t>0 } \Bigg (\int \limits ^{\infty }_{t} w_{1}^{\theta _{1} } (x) dx\Bigg )^{-\frac{1}{\theta _{1}} } \text {ess}\sup _{x>0} w_{2,n,p_{2} } (x) \min \{ \mu _{n,\rho ,p_{_{1}}} (t), \mu _{n,\rho ,p_{_{1}}} (x) \}<\infty . \end{aligned}$$

(f) If $1<\theta _{1}< \infty ,\theta _{2}=\infty ,$ then

$$\begin{aligned} B_{6}:= \text {ess}\sup _{z>0}w_{2 ,n,p_{2} }(z) \Bigg (\int \limits ^{\infty }_{ z}\Bigg ( \int \limits ^{ z}_{ x} \Bigg (\int \limits _{u}^{\infty } w^{\theta _{1}}_{1}(s)ds \Bigg )^{-1} \rho (u) u^{\frac{n}{p'_{1}}-1}du \Bigg )^{\theta '_{1}}w_{1}^{\theta _{1}}(x)dx\Bigg )^{ \frac{1}{\theta '_{1}}} < \infty . \end{aligned}$$

(g) If $\theta _{1}=\infty$, $0<\theta _{2}<\infty$, then

$$\begin{aligned} B_{7} : = \Bigg (\int ^{\infty }_{0}\Bigg ( w_{2 ,n,p_{2} } (\tau ) \int _{\tau }^{\infty }\frac{ \rho (z) z^{\frac{n}{p'_{1}}-1} dz}{\text {ess}\sup \limits _{z<s<\infty \ \ } w_{1}(s)}\Bigg )^{\theta _{2}} d \tau \Bigg )^{\frac{1}{\theta _{2}}}< \infty . \end{aligned}$$

(h) If $\theta _{1}=\theta _{2}=\infty ,$ then

$$\begin{aligned} B_{8} : = \text {ess}\sup _{\tau >0 \ \ } \Bigg ( \int _{\tau }^{\infty }\frac{\rho (z) z^{\frac{n}{p'_{1}}-1} dz}{\text {ess}\sup \limits _{z<s<\infty \ \ } w_{1}(s)}\Bigg ) w_{2,n,p_{2} } (\tau ) < \infty . \end{aligned}$$

2. Let $p_{1}=1, 0<p_{2}<\infty$ and $\rho \in \tilde{S}_{n,1,p_{2}}.$ Then the operator $I_{\rho (.)}$ is bounded from $LM_{1,\theta _{1},w_{1}(.)}$ to $LM_{p_{2},\theta _{2},w_{2}(.)}$ if and only if conditions (a) - (h) with $p_{1}=1$ are satisfied.

3. Let $p_{1}=1, 1<p_{2}<\infty$ and $\rho \in {S}_{n,1,p_{2}}.$ Then the operator $I_{\rho (.)}$ is bounded from $LM_{1,\theta _{1},w_{1}(.)}$ to $WLM_{p_{2},\theta _{2},w_{2}(.)}$ if and only if conditions (a) - (h) with $p_{1}=1$ are satisfied.

Proof

It sufficies to prove by using Lemma 4.1 and the appropriate changes of variables that, for $\mu _{n,\rho ,p_{1}}$, $v_{1}$ and $v_{2}$ defined by formulas (4.1), (4.9), (4.2), respectively, $A_{11}=B_{11}$. $A_{12}=B_{12}$, $A_{2}=B_{2}$, $A_{31}=B_{31}$, $A_{32}=B_{32}$, $A_{41}=B_{41}$, $A_{5}=B_{5}$ and $A_{6}= c B_{6}$, $A_{7}= c B_{7}$, $A_{8}= c B_{8}$, where $c=\Big (\int \limits _{1}^{\infty } \rho (t)t^{\frac{n}{p'_{1}}-1}dt \Big )^{-1}$.

(a) If $1<\theta _{1}\le \theta _{2}<\infty$, then according to Theorem 5.1 and by (4.15) with $\phi _{1}\equiv 1$ and (4.18) with $\phi _{2}(s)\equiv s$, we get

$$\begin{aligned} A_{11} = \sup \limits _{t>0} \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) } w_{2,n,p_{2} }^{\theta _{2}}(r) \mu ^{\theta _{2}}_{n,\rho ,p_{1}}(r)dr \Bigg )^{\frac{1}{\theta _{2}}} \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) } w_{1}^{\theta _{1}}(r)dr \Bigg )^{- \frac{1}{\theta _{1}}} \end{aligned}$$

$$\begin{aligned} \left( \mu ^{(-1)}_{n,\rho ,p_{1}}(t)= \tau \right) \end{aligned}$$

$$\begin{aligned} = \sup \limits _{\tau >0} \Bigg (\int \limits ^{\infty }_{ \tau } w_{2,n,p_{2} }^{\theta _{2}}(r) \mu ^{\theta _{2}}_{n,\rho ,p_{1}}(r)dr \Bigg )^{\frac{1}{\theta _{2}}} \Bigg (\int \limits ^{\infty }_{ \tau } w_{1}^{\theta _{1}}(r)dr \Bigg )^{- \frac{1}{\theta _{1}}} =B_{11 } \end{aligned}$$

and

$$\begin{aligned} A_{12} = \sup \limits _{t>0} \Bigg (\int \limits _{0}^{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) } w_{2,n,p_{2} }^{\theta _{2}}(x) dx \Bigg )^{\frac{1}{\theta _{2}}} \Bigg ({\int \limits ^{t}_{0}} z^{\theta '_{1}} \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(z) } w_{1}^{\theta _{1}}(r)dr \Bigg )^{-\theta '_{1}} w_{1}^{\theta _{1}}(z)dz \Bigg )^{\frac{1}{\theta '_{1}}} . \end{aligned}$$

Next, by (4.16) with

$$\begin{aligned} \phi _{1}(z)=\Bigg (z^{\theta '_{1}} \Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(z)}^{\infty } w_{1}^{\theta _{1}}(r)dr \Bigg )^{ -\theta '_{1}} \Bigg ) ^{\frac{1}{\theta _{1}}}, \end{aligned}$$

we get

$$\begin{aligned} A_{12} = \sup \limits _{t>0} \Bigg (\int \limits ^{\mu ^{(-1)}_{n,\rho ,p_{1}}(t) }_{ 0 } w_{2 ,n,p_{2} }^{\theta _{2}}(x) dx \Bigg )^{\frac{1}{\theta _{2}}} \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) } \mu ^{\theta _{1}'}_{n,\rho ,p_{1}}(x) \Bigg (\int \limits ^{\infty }_{ x} w_{1}^{\theta _{1}}(r)dr \Bigg )^{- \theta '_{1}} w_{1}^{\theta _{1}}(x)dx \Bigg )^{\frac{1}{\theta _{1}'}} \end{aligned}$$

$$\begin{aligned} \left( \mu ^{(-1)}_{n,\rho ,p_{1}}(t)= \tau \right) \end{aligned}$$

$$\begin{aligned} = \sup \limits _{\tau >0} \Bigg (\int \limits ^{\tau }_{ 0 } w_{2}^{\theta _{2}}(x)x^{\frac{n \theta _{2}}{p_{2} } } dx \Bigg )^{\frac{1}{\theta _{2}}} \Bigg (\int \limits ^{\infty }_{ \tau } \mu ^{\theta _{1}'}_{n,\rho ,p_{1}}(x) \Bigg (\int \limits ^{\infty }_{ x} w_{1}^{\theta _{1}}(r)dr \Bigg )^{ -\theta '_{1}} w_{1}^{\theta _{1}}(x)dx \Bigg )^{ \frac{1}{\theta _{1}'}} =B_{12} . \end{aligned}$$

(b) If $0<\theta _{1}\le 1$, $\theta _{1}\le \theta _{2}<\infty$, then by (4.15) with $\phi _{1} \equiv 1$ and by (4.19′) with $\phi _{2}(s)=\min \{t,s\}$, $s>0$, we get

$$\begin{aligned} A_{2}=\sup \limits _{t>0} \bigg ( \int ^{\infty }_{ \mu ^{-1}_{n,\rho ,p_{1}}(t) }w_{1}^{\theta _{1}}(r)dr\bigg )^{-\frac{1}{\theta _{1}} } \bigg ( \int ^{\infty }_{0} w_{2,n,p_{2}} ^{\theta _{2}} (r ) \min \{t,\mu _{n,\rho ,p_{1}}(r) \}^{\theta _{2}} dr \bigg )^{\frac{1}{\theta _{2}}} \end{aligned}$$

$$\begin{aligned} ( \mu ^{-1}_{n,\rho ,p_{1}}(t)=\tau ) \end{aligned}$$

$$\begin{aligned} =\sup \limits _{\tau >0} \bigg ( \int ^{\infty }_{\tau }w_{1}^{\theta _{1}}(r)dr\bigg )^{-\frac{1}{\theta _{1}} } \bigg ( \int ^{\infty }_{0} w_{2,n,p_{2} }^{\theta _{2}} (r ) \min \{ \mu _{n,\rho ,p_{1}}(\tau ) ,\mu _{n,\rho ,p_{1}}(r) \}^{\theta _{2}} dr \bigg )^{\frac{1}{\theta _{2}}} = B_{2} . \end{aligned}$$

(c) If $1<\theta _{1}<\infty$, $0<\theta _{2}<\theta _{1}<\infty$, then by (4.15) with $\phi _{1}\equiv 1$, (4.18) with $\phi _{s}=s$ and by (4.19’) we get

$$\begin{aligned} A_{31} = \Bigg (\int ^{\infty }\limits _{0}\Bigg (\int \limits ^{\infty }_{\mu ^{(-1)}_{n,\rho ,p_{1}(t)}} w_{1}^{\theta _{1}}(x) dx \Bigg )^{\frac{r}{\theta _{1}} }\Bigg (\int \limits ^{\infty }_{\mu ^{(-1)}_{n,\rho ,p_{1}(t)}} w_{2,n,p_{2} }^{\theta _{2}} (x) \mu ^{\theta _{2}}_{n,\rho ,p_{1}}(x)dx \Bigg )^{\frac{r}{\theta _{1}} } t^{\theta _{2}} v_{2}^{\theta _{2}}(t) dt\Bigg )^{\frac{1}{r} } \end{aligned}$$

$$\begin{aligned} =\Bigg (\int \limits ^{\infty }_{0}\Bigg (\int \limits ^{\infty }_{z}w_{1}^{\theta _{1}}(x) dx \Bigg )^{\frac{r}{\theta _{1}} }\Bigg (\int \limits ^{\infty }_{z}w_{2,n,p_{2} }^{\theta _{2}}(x) \mu ^{\theta _{2}}_{n,\rho ,p_{1}}(x)dx\Bigg )^{\frac{r}{\theta _{1}} }\mu ^{\theta _{2}}_{n,\rho ,p_{1}}(z) w_{2,n,p_{2} }^{\theta _{2}} (z) dz\Bigg )^{\frac{1}{r} } = B_{31} \end{aligned}$$

and

$$\begin{aligned} A_{32} = \Bigg (\int \limits ^{\infty }_{0} \Bigg (\int \limits ^{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t)}_{0} w_{2 ,n,p_{2} }^{\theta _{2}} (x) dx\Bigg )^{ \frac{r}{\theta _{1}} } \Bigg (\int \limits ^{t} _{0}\frac{s^{\theta _{1}'} }{ \Big (\int \limits ^{\infty } _{ \mu ^{(-1)}_{n,\rho ,p_{1}}(s) } w_{1}^{\theta _{1}} (u)du \Big )^{\theta _{1}'}}v_{1}^{\theta _{1}}(s)ds \Bigg )^{\frac{r}{\theta '_{1}} } v_{2}^{\theta _{2}} (t)dt \Bigg )^{\frac{1}{r} } \end{aligned}$$

$$\begin{aligned} = \Bigg (\int \limits ^{\infty }_{0} \Bigg (\int \limits ^{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t)}_{0} w_{2,n,p_{2} }^{\theta _{2}} (x) dx\Bigg )^{\frac{r}{\theta _{1}} } \Bigg (\int \limits ^{\infty } _{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) }\frac{\mu ^{\theta _{1}'}_{n,\rho ,p_{1}}(\tau )w_{1}^{\theta _{1}}(\tau )}{ \Big (\int \limits ^{\infty } _{ \tau } w_{1}^{\theta _{1}} (u)du \Big )^{\theta _{1}'}}d\tau \Bigg )^{\frac{r}{\theta '_{1}} } v_{2}^{\theta _{2}} (t)dt \Bigg )^{\frac{1}{r} } \end{aligned}$$

$$\begin{aligned} = \Bigg (\int \limits ^{\infty }_{0} \Bigg (\int \limits ^{\infty }_{z} w_{2 ,n,p_{2} }^{\theta _{2}} (x) dx\Bigg )^{\frac{r}{\theta _{1}} }\Bigg (\int \limits ^{\infty } _{ z}\frac{\mu ^{\theta _{1}'}_{n,\rho ,p_{1}}(\tau )w_{1}^{\theta _{1}}(\tau )}{ \Big (\int \limits ^{\infty } _{ \tau } w_{1}^{\theta _{1}} (u)du \Big )^{\theta _{1}'}}d\tau \Bigg )^{\frac{r}{\theta '_{1}} } w_{2 ,n,p_{2} }^{\theta _{2}}(z) dz \Bigg )^{\frac{1}{r} }= B_{32}. \end{aligned}$$

(d) If $0<\theta _{2}<\theta _{1}\le 1$, then by (4.15) with $\phi _{1}\equiv 1$, by (4.19) with $\phi _{2}\equiv 1$, we get

$$\begin{aligned} A_{41} = \Bigg (\int \limits ^{\infty }_{0}\sup \limits _{ 0<\tau <t }\tau ^{ r}\Bigg (\int \limits ^{\infty }_{\mu ^{(-1)}_{n,\rho ,p_{1}}(\tau )} w_{1}^{\theta _{1}}(u)du\Bigg )^{\frac{r}{\theta _{1}} }\Bigg (\int \limits ^{\mu ^{(-1)}_{n,\rho ,p_{1}}(t) }_{0}w_{2 ,n,p_{2} }^{\theta _{2}}(x)dx\Bigg )^{\frac{r}{\theta _{1}} } v_{2}^{\theta _{2}}(t)dt\Bigg )^{\frac{1}{r} } \end{aligned}$$

$$\begin{aligned} =\Bigg (\int \limits ^{\infty }_{0} \sup \limits _{ 0<\tau < \mu _{n,\rho ,p_{1}}(y) } \tau ^{r}\Bigg (\int \limits ^{\infty }_{\mu ^{(-1)}_{n,\rho ,p_{1}}(\tau ) } w_{1}^{\theta _{1}}(s)ds\Bigg )^{\frac{r}{\theta _{1}} }\Bigg ({\int \limits ^{y}_{0}}w_{2 ,n,p_{2} }^{\theta _{2}}(x) dx\Bigg )^{\frac{r}{\theta _{1}} } w_{2 ,n,p_{2} }^{\theta _{2}}(y)dy\Bigg )^{\frac{1}{r} }. \end{aligned}$$

$$\begin{aligned} \left( \tau =\mu ^{(-1)}_{n,\rho ,p_{1}}(z)\right) \end{aligned}$$

$$\begin{aligned} =\Bigg ( \int \limits ^{\infty }_{0} \text {ess} \sup \limits_{ y < z < \infty } \mu _{n,\rho ,p_{1}}(z)^{r} \Bigg (\int \limits ^{\infty }_{ z} w_{1}^{\theta _{1}}(s)ds\Bigg )^{ \frac{r}{\theta _{1}} } \Bigg (\int \limits ^{y}_{ 0} w_{2 ,n,p_{2} }^{\theta _{2}}(x) dx \Bigg ) ^{\frac{r}{\theta _{1}} } \mu _{n,\rho ,p_{1}}(x)^{\theta _{2}}(y)dy \Bigg )^{ \frac{1}{r} } = B_{42}.\end{aligned}$$

(e) If $0<\theta _{1} \le 1$, $\theta _{2}=\infty$, then by (4.15) with $\phi _{1}\equiv 1$ and (4.19’) with $\phi (y) =\min \{t,y\}$, $y>0$, we get

$$\begin{aligned} A_{5} =\sup \limits _{t>0} \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(t) }w^{\theta _{1}}_{1}(x)dx\Bigg )^{- \frac{1}{\theta _{1}}}\text {ess}\sup \limits _{x>0\ \ \ \ } w_{2 ,n,p_{2} }(x) \min \lbrace t, \mu _{n,\rho ,p_{1}}(x)\rbrace \end{aligned}$$

$$\begin{aligned} (\mu ^{(-1)}_{n,\rho ,p_{1}}(t) =\tau ) \end{aligned}$$

$$\begin{aligned} =B_{42}. =\sup \limits _{\tau>0} \Bigg (\int \limits ^{\infty }_{\tau } w^{\theta _{1}}_{1}(x)dx\Bigg )^{- \frac{1}{\theta _{1}}}\text {ess}\sup \limits _{x>0\ \ \ \ } w_{2 ,n,p_{2} }(x) \min \lbrace \mu _{n,\rho ,p_{1}}(\tau ), \mu _{n,\rho ,p_{1}}(x) \rbrace =B_{5}. \end{aligned}$$

(f) If $1<\theta _{1}<\infty$, $\theta _{2}=\infty$, then by (4.15) with $\phi _{1}\equiv 1$ and by the same formula with

$$\begin{aligned} \phi _{1}(t) = \Bigg ( {\int \limits ^{s}_{t}} \Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(y) }^{\infty } w^{\theta _{1}}_{1}(s)ds\Bigg )^{-1} dy\Bigg )^{ \frac{\theta '_{1}}{\theta _{1}} } \end{aligned}$$

and, finally, by (4.19’) with $\theta _{2}=\infty$ and

$$\begin{aligned} \phi _{2}(s) = \Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(s)} \Bigg ( \int \limits ^{s}_{ \mu _{n,\rho ,p_{1}}(r)} \Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(y) }^{\infty } w^{\theta _{1}}_{1}(s)ds\Bigg )^{-1} dy\Bigg )^{\theta '_{1}} w_{1}^{\theta _{1}} (r)dr \Bigg )^{\frac{1}{\theta _{1}'}} \end{aligned}$$

we get

$$\begin{aligned} A_{6} = \text {ess}\sup _{s>0}v_{2}(s)\Bigg ({\int \limits ^{s}_{0}}\Bigg ( {\int \limits ^{s}_{t}} \Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(y) }^{\infty } w^{\theta _{1}}_{1}(s)ds\Bigg )^{-1} dy\Bigg )^{\theta '_{1}} v_{1}^{\theta _{1}}(t)dt\Bigg )^{ \frac{1}{\theta '_{1}}} \end{aligned}$$

$$\begin{aligned} =\text {ess}\sup _{s>0}v_{2}(s)\Bigg (\int \limits ^{\infty }_{ \mu ^{(-1)}_{n,\rho ,p_{1}}(s)}\Bigg ( \int \limits ^{s}_{\mu _{n,\rho ,p_{1}}(r) }\Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(y) }^{\infty } w^{\theta _{1}}_{1}(s)ds\Bigg )^{-1} dy\Bigg )^{\theta '_{1}} w_{1}^{\theta _{1}}(r)dr\Bigg )^{ \frac{1}{\theta '_{1}}} \end{aligned}$$

$$\begin{aligned} =\text {ess}\sup _{z>0}w_{2}(z)z^{\frac{n}{p_{2}}}\Bigg (\int \limits ^{\infty }_{ z}\Bigg ( \int \limits ^{\mu _{n,\rho ,p_{1}}(z) }_{\mu _{n,\rho ,p_{1}}(r) } \Bigg (\int \limits _{\mu ^{(-1)}_{n,\rho ,p_{1}}(y) }^{\infty } w^{\theta _{1}}_{1}(s)ds \Bigg )^{-1} dy\Bigg )^{\theta '_{1}} w_{1}^{\theta _{1}}(r)dr\Bigg )^{ \frac{1}{\theta '_{1}}} \end{aligned}$$

$$\begin{aligned} (y=\mu _{n,\rho ,p_{1}}(u)) \end{aligned}$$

$$\begin{aligned} =c B_{6}, =\text {ess}\sup \limits _{z>0\ \ \ \ } w_{2 ,n,p_{2} }(z)\Bigg (\int \limits ^{\infty }_{ z}\Bigg (\int \limits ^{ z}_{ r} \Bigg (\int \limits ^{\infty }_{ u} w_{1}(s) ^{\theta _{1}} ds \Bigg )^{-1} \vert ( \mu _{n,\rho ,p_{1}} (u))'\vert du \Bigg )^{\theta '_{1}} w_{1}^{\theta _{1}}(r)dr \Bigg )^{\frac{1}{\theta '_{1}}} \end{aligned}$$

where

$$\begin{aligned} c= \Bigg (\int \limits _{1}^{\infty } \rho (t)t^{\frac{n}{p'_{1}}-1}dt \Bigg )^{-1}. \end{aligned}$$

(g) If $\theta _{1}=\infty$, $0<\theta _{2}<\infty$, then by using formula (4.15) with $\phi _{1}\equiv 1$ and formula (4.19’) with $\phi _{2}(x)={\int \limits _{0}^{x}} \ \ \ \frac{dy}{ \text {ess} \sup \limits _{r> \mu ^{(-1)}_{n,\rho ,p_{1}}(y) \ \ \ \ \ } w_{1}(r) }$ we get

$$\begin{aligned} A_{7} = \Bigg (\int \limits ^{\infty }_{0} \Bigg (v_{2}(x){\int \limits ^{x}_{0}} \ \ \frac{dy}{\text {ess} \sup \limits _{r>\mu ^{(-1)}_{n,\rho ,p_{1}}(y) \ \ \ \ \ } w_{1}(r)} \Bigg )^{\theta _{2}}dx \Bigg )^{\frac{1}{\theta _{2}}} \end{aligned}$$

$$\begin{aligned} = \Bigg (\int \limits ^{\infty }_{0} \Bigg ( w_{2,n,p_{2}}(\tau ) \int \limits ^{\mu _{n,\rho ,p_{1}} (\tau ) }_{0} \ \frac{dy}{\text {ess} \sup \limits _{r>\mu ^{(-1)}_{n,\rho ,p_{1}}(y) \ \ \ \ \ } w_{1}(r)} \Bigg )^{\theta _{2}}d\tau \Bigg )^{\frac{1}{\theta _{2}}} \end{aligned}$$

$$\begin{aligned} \left( \mu ^{(-1)}_{n,\rho ,p_{1}}(y)=z \right) \end{aligned}$$

$$\begin{aligned} = c \Bigg (\int \limits ^{\infty }_{0} \Bigg ( w_{2,n,p_{2}}(\tau ) \int \limits ^{\infty }_{\tau } \frac{ \rho (z) z^{\frac{n}{p'_{1}}-1} }{\text {ess} \sup \limits _{ z<r<\infty \ \ \ \ \ } w_{1}(r)} dz\Bigg )^{\theta _{2}}d\tau \Bigg )^{\frac{1}{\theta _{2}}}= c B_{7}. \end{aligned}$$

(h) If $\theta _{1}=\theta _{2}=\infty$, then in $A_{8}$ according to (4.10) and (4.2)

$$\begin{aligned} v_{1}(x)=w_{1}\Big (\mu ^{(-1)}_{n,\rho ,p_{1}}(x)\Big ),\ v_{2}(x)=w_{2}\Big (\mu ^{(-1)}_{n,\rho ,p_{1}}(x)\Big )\Big (\mu ^{(-1)}_{n,\rho ,p_{1}}(x)\Big )^{\frac{n}{p_{2}}} ,x>0. \end{aligned}$$

By using formula (4.15) with $\psi _{1}\equiv 1$ and by carrying out the following changes of variables: $y=\mu _{n,\rho ,p_{1}}(z)$, $z>0$, and, finally, $t=\mu _{n,\rho ,p_{1}}(\tau )$, $\tau >0$, we get

$$\begin{aligned} A_{8} = \text {ess}\sup _{t>0 \ \ } \Bigg ( {\int _{0}^{t}}\frac{ dy }{\text {ess}\sup \limits _{s>\mu ^{(-1)}_{n,\rho ,p_{1}}(y)} w_{1}(s)}\Bigg ) w_{2}\Big (\mu ^{(-1)}_{n,\rho ,p_{1}}(t)\Big )\Big (\mu ^{(-1)}_{n,\rho ,p_{1}}(t)\Big )^{\frac{n}{p_{2}}} \end{aligned}$$

$$\begin{aligned} = c \ \text {ess}\sup _{t>0 \ \ } \Bigg ( \int _{\mu ^{(-1)}_{n,\rho ,p_{2}}(t)}^{\infty }\frac{\rho (z) z^{\frac{n}{p'_{1}}-1} }{\text {ess}\sup \limits _{s>z\ \ } w_{1}(s)}dz\Bigg ) w_{2}\Big (\mu ^{(-1)}_{n,\rho ,p_{2}}(t)\Big )\Big (\mu ^{(-1)}_{n,\rho ,p_{2}}(t)\Big )^{\frac{n}{p_{2}}} \end{aligned}$$

$$\begin{aligned} = c \ \text {ess}\sup _{\tau >0 \ \ } \Bigg ( \int _{\tau }^{\infty }\frac{ \rho (z) z^{\frac{n}{p'_{1}}-1} }{\text {ess}\sup \limits _{z<s<\infty \ \ } w_{1}(s)}dz\Bigg ) w_{2,n,p_{2}}(\tau ) = c B_{8}. \end{aligned}$$

$\square$