Sharp bounds for Hardy type operators on higher-dimensional product spaces

Abstract

A new class of Hardy type operator defined on a higher-dimensional product space is discussed. It includes two different kinds of the classical Hardy operators. In addition, we also consider the fractional Hardy operator ${\mathbb{H}}_{\beta }$. The bound of operator ${\mathbb{H}}_{\beta }$ from $L^p$ to $L^q$ is explicitly worked out. Especially, the bound of operator ${\mathbb{H}}_{\beta }$ from $L^1$ to $L^{\frac{n}{n-\beta },\mathrm{\infty }}$ is sharp.

MSC:26D10, 26D15, 42B99.

1 Introduction

The most fundamental averaging operator is the Hardy operator defined by

$$H(f)(x)=\frac{1}{x}{\int }_0^{x}f(t)dt,$$

where the function f is a nonnegative integrable on ${\mathbb{R}}_{+}=(0,\mathrm{\infty })$ and $x>0$. A classical inequality, due to Hardy [1], states that

$${\parallel H(f)\parallel }_{L^p}\le \frac{p}{p-1}{\parallel f\parallel }_{L^p}$$

holds for $1<p<\mathrm{\infty }$, and the constant $\frac{p}{p-1}$ is best possible.

For the multidimensional case $n\ge 2$, generally speaking, there exist two different definitions. One is the rectangle averaging operator defined by

$${\mathfrak{R}}_n(f)(x_1,\dots ,x_n)=\frac{1}{x_1\cdots x_n}{\int }_0^{x_1}\cdots {\int }_0^{x_n}f(t_1,\dots ,t_n)d{t}_1\cdots d{t}_n,$$

(1)

where the function f is a nonnegative measurable function on $G={(0,\mathrm{\infty })}^n$ and $x_i>0$, $i=1,2,\dots ,n$.

Another definition is the spherical averaging operator, which was given by Christ and Grafakos in [2] as follows:

$$\mathcal{H}(f)(x)=\frac{1}{|B(0,|x|)|}{\int }_{|y|<|x|}f(y)dy,x\in {\mathbb{R}}^n\mathrm{\setminus }\{0\},$$

(2)

where f is a nonnegative measurable function on ${\mathbb{R}}^n$.

The boundedness of operator ${\mathfrak{R}}_n$ are discussed in many papers (cf. [3–6]). ${\parallel {\mathfrak{R}}_n\parallel }_{L^p\to L^p}$, the norm of ${\mathfrak{R}}_n$, is ${(\frac{p}{p-1})}^n$ and obviously depends on the dimension of the space. However, ${\parallel \mathcal{H}\parallel }_{L^p\to L^p}$, the norm of ℋ, is still $\frac{p}{p-1}$, and does not depend on the dimension of the space. The reason of generating the difference of the two kind of operators is, roughly speaking, that each variable can independently dilate by itself in the operator ${\mathfrak{R}}_n$, nevertheless, for the operator ℋ, all variables dilate by the same scale simultaneously. Generally speaking, the spherical averaging operator has better properties than the rectangle averaging operator does, such as the Hardy-Littlewood maximal function. A detailed account of the history of the topic can be found in the book [7]; see also Kufner and Persson’s book [8].

For the operator ${\mathfrak{R}}_n$, we note that every variable is defined on the one-dimensional space. In this paper, we shall extent the definition of ${\mathfrak{R}}_n$ so that every variable is spherical average defined on a higher-dimensional space.

Next, we will give the definition of Hardy type operator on higher-dimensional product spaces as follows and discuss the corresponding properties.

Definition 1.1 Let $m\in \mathbb{N}$, $n_i\in \mathbb{N}$, $x_i\in {\mathbb{R}}^{n_i}$, $1\le i\le m$, and f be a nonnegative measurable function on ${\mathbb{R}}^{n_1}\times {\mathbb{R}}^{n_2}\times \cdots \times {\mathbb{R}}^{n_m}$. The Hardy type operator is defined by

$${\mathcal{H}}_m(f)(x)=\left(\prod _{i=1}^m\frac{1}{|B(0,|x_i|)|}\right){\int }_{|y_1|<|x_1|}\cdots {\int }_{|y_m|<|x_m|}f(y_1,\dots ,y_m)d{y}_1\cdots d{y}_m,$$

(3)

where $x=(x_1,x_2,\dots ,x_m)\in {\mathbb{R}}^{n_1}\times {\mathbb{R}}^{n_2}\times \cdots \times {\mathbb{R}}^{n_m}$ with ${\prod }_{i=1}^m|x_i|\ne 0$.

Our first aim in this paper is to provide transparent treatments of multivariate inequalities of higher-dimensional Hardy type both for the rectangle and for the ball case. Our results subsume those of [9] and [6]. In fact, if $m=1$, then the operator ${\mathcal{H}}_m$ will become ℋ defined by (2); if $n_1=n_2=\cdots =n_m=1$, then ${\mathcal{H}}_m$ will become ${\mathfrak{R}}_n$ defined by (1). Consequently, the operator ${\mathcal{H}}_m$ includes both ${\mathfrak{R}}_n$ and ℋ. It is much significant to discuss the properties of ${\mathcal{H}}_m$.

Our second aim is to consider the fractional Hardy operator on the Lebesgue spaces. Recall that, for a nonnegative measurable function f on ${\mathbb{R}}^n$, the n-dimensional fractional Hardy operator ${\mathbb{H}}_{\beta }$ with spherical mean is defined by

$${\mathbb{H}}_{\beta }(f)(x)=\frac{1}{|B(0,|x{|)|}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}f(y)dy,x\in {\mathbb{R}}^n\mathrm{\setminus }\{0\},$$

(4)

where $0\le \beta <n$ (cf. [10]). Clearly, ${\mathbb{H}}_{\beta }(|f|)(x)\le C{M}_{\beta }(f)(x)$, where $M_{\beta }$ is the fractional Hardy-Littlewood maximal function defined by

$$M_{\beta }(f)(x)=\underset{r>0}{sup}\frac{1}{{|B(x,r)|}^{1-\frac{\beta }{n}}}{\int }_{|y-x|<r}|f(y)|dy,$$

where f is a measurable function.

For $0\le \beta <n$, $1<p\le \frac{n}{\beta }$, and $\frac{1}{p}-\frac{1}{q}=\frac{\beta }{n}$, the following two statements (a) and (b) are well-known (cf. [11]).

• (a) If $f\in L^p({\mathbb{R}}^n)$, then

  $${\parallel M_{\beta }(f)\parallel }_{L^q}\le C{\parallel f\parallel }_{L^p};$$

• (b) if $f\in L^1({\mathbb{R}}^n)$, then, for any $\lambda >0$,

  $$\left|\{x\in {\mathbb{R}}^n:M_{\beta }(f)(x)>\lambda \}\right|\le {\left(\frac{C}{\lambda }{\parallel f\parallel }_{L^1}\right)}^{\frac{n}{n-\beta }}.$$

However the constant C, the bound of operator $M_{\beta }$, were not given explicit expression of depending on the parameters p, q and β. In this paper, the bounds of the operator ${\mathbb{H}}_{\beta }$ from $L^p$ to $L^q$ and from $L^1$ to $L^{\frac{n}{n-\beta },\mathrm{\infty }}$ are explicitly worked out. Furthermore, we will show that the constant 1 is the bound of operator ${\mathbb{H}}_{\beta }$ from $L^1$ to $L^{\frac{n}{n-\beta },\mathrm{\infty }}$ and is beat possible, that is,

$${\parallel {\mathbb{H}}_{\beta }\parallel }_{L^1\to L^{\frac{n}{n-\beta },\mathrm{\infty }}}=1.$$

Throughout the paper, we use the following notation. The set $B(0,|x|)$ denotes a ball with center at the original point and radius $|x|$, and $|B(0,|x|)|$ denotes the volume of the ball $B(0,|x|)$; ${(B(0,|x|))}^c={\mathbb{R}}^n\setminus B(0,|x|)$.

2 Sharp bounds for the Hardy type operator on product space

Theorem 2.1 Let $1<p<\mathrm{\infty }$, $m\in \mathbb{N}$, $n_i\in \mathbb{N}$, $x_i\in {\mathbb{R}}^{n_i}$, $i=1,\dots ,m$. If $f\in L^p({\mathbb{R}}^{n_1}\times {\mathbb{R}}^{n_2}\times \cdots \times {\mathbb{R}}^{n_m},{|x|}^{\overrightarrow{\alpha }})$, where ${|x|}^{\overrightarrow{\alpha }}:={|x_1|}^{{\alpha }_1}{|x_2|}^{{\alpha }_2}\cdots {|x_m|}^{{\alpha }_m}$ and ${\alpha }_j<(p-1)n_j$, then the Hardy type operator ${\mathcal{H}}_m$ defined in (3) is bounded on $L^p({\mathbb{R}}^{n_1}\times {\mathbb{R}}^{n_2}\times \cdots \times {\mathbb{R}}^{n_m},{|x|}^{\overrightarrow{\alpha }})$, moreover, the norm of ${\mathcal{H}}_m$ can be obtained as follows:

$${\parallel {\mathcal{H}}_m\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})\to L^p({|x|}^{\overrightarrow{\alpha }})}=\prod _{j=1}^m\left(\frac{p}{p-1-\frac{{\alpha }_j}{n_j}}\right).$$

Proof of Theorem 2.1 We merely give the proof with the case $m=2$ for the sake of clarity in writing, and the same is true for the general case $m>2$. We adapt some ideas and methods used in [12].

Set

$$g_f(x_1,x_2)=\frac{1}{{\omega }_{n_1}}\frac{1}{{\omega }_{n_2}}{\int }_{S^{n_1-1}}{\int }_{S^{n_2-1}}f(|x_1|{\xi }_1,|x_2|{\xi }_2)d\sigma ({\xi }_1)d\sigma ({\xi }_2),$$

where ${\omega }_{n_i}=\frac{2{\pi }^{\frac{n_i}{2}}}{\mathrm{\Gamma }(\frac{n_i}{2})}$ and $x_i\in {\mathbb{R}}^{n_i}$, $i=1,2$. Obviously, g is a nonnegative radial function with respect to the variables $x_1$ and $x_2$, respectively. In the following, we briefly call this function is a radial function on product space.

It follows that ${\mathcal{H}}_2(g_f)(x_1,x_2)$ is equal to

Using the generalized Minkowski’s inequality and Hölder’s inequality, we conclude that ${\parallel g_f\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}$ is equal to

Thus, we conclude that the following inequality:

$$\frac{{\parallel {\mathcal{H}}_2(f)\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}{{\parallel f\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}\le \frac{{\parallel {\mathcal{H}}_2(g_f)\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}{{\parallel g_f\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}$$

holds provided that ${\parallel f\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}\ne 0$. In addition, clearly if f is a nonnegative radial function, then we have $g_f=f$. This means that the norm of the operator ${\mathcal{H}}_2$ is equal to the norm that ${\mathcal{H}}_2$ restricts to the set of nonnegative radial functions. Consequently, without loss of generality, it suffices to fulfil the proof of the theorem by assuming that f is a nonnegative radial function.

Substituting the variables $z_1=\frac{y_1}{|x_1|}$ and $z_2=\frac{y_2}{|x_2|}$, we have that ${\parallel {\mathcal{H}}_2(f)\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}$ equals

Using the generalized Minkowski’s inequality again and noting that f is a radial function with respect to the first variable and the second variable, respectively, we have that ${\parallel {\mathcal{H}}_2(f)\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}$ is not greater than

where ${\nu }_{n_i}=\frac{{\pi }^{n_i/2}}{\mathrm{\Gamma }(1+n_i/2)}$ is the volume of the unit ball in ${\mathbb{R}}^{n_i}$, $i=1,2$.

Therefore, it implies that

$${\parallel {\mathcal{H}}_2\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})\to L^p({|x|}^{\overrightarrow{\alpha }})}\le \prod _{j=1}^2\left(\frac{p}{p-1-{\alpha }_j/n_j}\right).$$

Next, we need to prove the converse inequality.

For the purpose of getting the sharp bound, we set

$$0<\epsilon <min\{1,\frac{(p_1-1)n}{p_2},\frac{(p_2-1)n}{p_2}\},$$

and define

$$f_{\epsilon }(x_1,x_2)={|x_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }{|x_2|}^{-\frac{n_2+{\alpha }_2}{p}+\epsilon }{\chi }_{\{|x_1|<1,|x_2|<1\}}(x_1,x_2).$$

It follows from the elementary calculation that ${\parallel f_{\epsilon }\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}$ is

We rewrite ${\mathcal{H}}_2(f_{\epsilon })$ as follows:

$$\begin{array}{rcl}{\mathcal{H}}_2(f_{\epsilon })(x_1,x_2)& =& \frac{1}{|B(0,|x_1|)||B(0,|x_2|)|}{\int }_{|y_1|<|x_1|}{\int }_{|y_2|<|x_2|}{f}_{\epsilon }(y_1,y_2)d{y}_{1}d{y}_2\\ =& \frac{1}{|B(0,1)||B(0,1)|}{\int }_{|z_1|<1}{\int }_{|z_2|<1}{f}_{\epsilon }(z_1|x_1|,z_2|x_2|)d{z}_{1}d{z}_2\\ =& \frac{{|x_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }{|x_2|}^{-\frac{n_2+{\alpha }_2}{p}+\epsilon }}{|B(0,1)||B(0,1)|}\\ \times {\int }_{\{|z_1|<1,|z_1|<1/|x_1|\}}{\int }_{\{|z_2|<1,|z_2|<1/|x_2|\}}{|z_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }{|z_2|}^{-\frac{n_2+{\alpha }_2}{p}+\epsilon }d{z}_{2}d{z}_1.\end{array}$$

Thus, we estimate the norm of ${\parallel {\mathcal{H}}_2(f_{\epsilon })\parallel }_{L^p}$ as follows:

$$\begin{array}{rcl}{\parallel {\mathcal{H}}_2(f_{\epsilon })\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}^p& =& \frac{1}{{({\nu }_{n_1}{\nu }_{n_2})}^p}{\int }_{{\mathbb{R}}^{n_1}}{\int }_{{\mathbb{R}}^{n_2}}|{\int }_{\{|z_1|<1,|z_1|<\frac{1}{|x_1|}\}}{\int }_{\{|z_2|<1,|z_2|<\frac{1}{|x_2|}\}}{|z_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }\\ \times {|z_2|}^{-\frac{n_2+{\alpha }_1}{p}+\epsilon }d{z}_{2}d{z}_1{|}^p{|x_1|}^{p\epsilon -n_1-{\alpha }_1}{|x_2|}^{p\epsilon -n_2-{\alpha }_2}{|x|}^{\overrightarrow{\alpha }}d{x}_{2}d{x}_1\\ \ge & \frac{1}{{({\nu }_{n_1}{\nu }_{n_2})}^p}{\int }_{|x_1|<1}{\int }_{|x_2|<1}|{\int }_{|z_1|<1}{\int }_{|z_2|<1}{|z_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }\\ \times {|z_2|}^{-\frac{n_2+{\alpha }_2}{p}+\epsilon }d{z}_{2}d{z}_1{|}^p{|x_1|}^{p\epsilon -n_1}{|x_2|}^{p\epsilon -n_2}d{x}_{2}d{x}_1\\ =& \frac{1}{{({\nu }_{n_1}{\nu }_{n_2})}^p}\frac{{\omega }_{n_1}}{p\epsilon }\frac{{\omega }_{n_2}}{p\epsilon }{\left({\int }_{|z_1|<1}{\int }_{|z_2|<1}{|z_1|}^{-\frac{n_1+{\alpha }_1}{p}+\epsilon }{|z_2|}^{-\frac{n_2+{\alpha }_2}{p}+\epsilon }d{z}_{2}d{z}_1\right)}^p\\ =& \frac{1}{{({\nu }_{n_1}{\nu }_{n_2})}^p}{\left(\frac{{\omega }_{n_1}}{n_1(1-1/p-{\alpha }_1/p{n}_1+\epsilon /n_1)}\right)}^p\\ \times {\left(\frac{{\omega }_{n_2}}{n_2(1-1/p-{\alpha }_2/p{n}_2+\epsilon /n_2)}\right)}^p{\parallel f_{\epsilon }\parallel }_{L^p}^p\\ =& {\left(\frac{p}{p-1-{\alpha }_1/n+p\epsilon /n_1}\frac{p}{p-1-{\alpha }_2/n+p\epsilon /n_2}\right)}^p{\parallel f_{\epsilon }\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}^p.\end{array}$$

Therefore, it implies that

$$\frac{{\parallel {\mathcal{H}}_2(f_{\epsilon })\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}{{\parallel f_{\epsilon }\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})}}\ge \frac{p}{p-1-{\alpha }_1/n+p\epsilon /n_1}\cdot \frac{p}{p-1-{\alpha }_2/n+p\epsilon /n_2}.$$

Consequently, using the definition of the norm of the operator and letting $\epsilon \to 0$, we conclude that

$${\parallel {\mathcal{H}}_2\parallel }_{L^p({|x|}^{\overrightarrow{\alpha }})\to L^p({|x|}^{\overrightarrow{\alpha }})}\ge \prod _{j=1}^2\left(\frac{p}{p-1-{\alpha }_j/n_j}\right).$$

This finishes the proof of the theorem. □

3 Explicit bounds for the fractional Hardy operator

Theorem 3.1 Suppose that $0\le \beta <n$, $1<p\le \frac{n}{\beta }$ and $\frac{1}{p}-\frac{1}{q}=\frac{\beta }{n}$.

• (i) If $f\in L^p({\mathbb{R}}^n)$, then we have

  $${\parallel {\mathbb{H}}_{\beta }(f)\parallel }_{L^q}\le C{\parallel f\parallel }_{L^p},$$

where

$${\left(\frac{p}{q}\right)}^{1/q}{\left(\frac{p}{p-1}\right)}^{1/q}{\left(\frac{q}{q-1}\right)}^{1-1/q}{(1-\frac{p}{q})}^{1/p-1/q}\le C\le {\left(\frac{p}{p-1}\right)}^{\frac{p}{q}}.$$

• (ii) If $f\in L^1({\mathbb{R}}^n)$, then for any $\lambda >0$,

  $$\left|\{x\in {\mathbb{R}}^n:|{\mathbb{H}}_{\beta }(f)(x)|>\lambda \}\right|\le {\left(\frac{1}{\lambda }{\parallel f\parallel }_{L^1}\right)}^{\frac{n}{n-\beta }}.$$

Moreover,

$${\parallel {\mathbb{H}}_{\beta }\parallel }_{L^1\to L^{\frac{n}{n-\beta },\mathrm{\infty }}}=1.$$

Proof of (i) of Theorem 3.1 Set ${\omega }_n=2{\pi }^{\frac{n}{2}}/\mathrm{\Gamma }(\frac{n}{2})$ as above. Let

$$g_f(y)=\frac{1}{{\omega }_n}{\int }_{|\xi |=1}f(|y|\xi )d\xi ,y\in {\mathbb{R}}^n.$$

Clearly, $g_f$ is a radial function.

$$\begin{array}{rcl}{\mathbb{H}}_{\beta }(g_f)(x)& =& \frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}{g}_f(y)dy\\ =& \frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}\left(\frac{1}{{\omega }_n}{\int }_{|\xi |=1}f(|y|\xi )d\xi \right)dy\\ =& \frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|\xi |=1}\left(\frac{1}{{\omega }_n}{\int }_{|y|<|x|}f(|y|\xi )dy\right)d\xi \\ =& \frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}f(y)dy\\ =& {\mathbb{H}}_{\beta }(f)(x).\end{array}$$

Using the generalized Minkowski inequality and Hölder’s inequality, we have that

$$\begin{array}{rcl}{\parallel g_f\parallel }_{L^p}& =& {\left({\int }_{{\mathbb{R}}^n}{|g_f(x)|}^{p}dx\right)}^{\frac{1}{p}}\\ \le & \frac{1}{{\omega }_n}{\int }_{|\xi |=1}{\left({\int }_{{\mathbb{R}}^n}{\left|f(|x|\xi )\right|}^{p}dx\right)}^{\frac{1}{p}}d\xi \\ =& \frac{{\omega }_n^{1/p}}{{\omega }_n}{\int }_{|\xi |=1}{\left({\int }_0^{\mathrm{\infty }}{|f(t\xi )|}^p{t}^{n-1}dt\right)}^{1/p}d\xi \\ \le & \frac{{\omega }_n^{1/p}}{{\omega }_n}{\left({\int }_{|\xi |=1}{\int }_0^{\mathrm{\infty }}{|f(t\xi )|}^p{t}^{n-1}dtd\xi \right)}^{1/p}{\left({\int }_{|\xi |=1}d\xi \right)}^{1-1/p}\\ =& {\parallel f\parallel }_{L^p}.\end{array}$$

Therefore, we have that

$$\frac{{\parallel {\mathbb{H}}_{\beta }(f)\parallel }_{L^q}}{{\parallel f\parallel }_{L^p}}\le \frac{{\parallel {\mathbb{H}}_{\beta }(g_f)\parallel }_{L^q}}{{\parallel g_f\parallel }_{L^p}}.$$

As above, this means that the norm of the operator ${\mathbb{H}}_{\beta }$ from $L^p$ to $L^q$ is equal to the norm that ${\mathbb{H}}_{\beta }$ restricts to radial functions. Consequently, without loss of generality, it suffices to carry out the proof of the theorem by assuming that f is a radial function.

A simple estimate implies that ${\parallel {\mathbb{H}}_{\beta }(f)\parallel }_{L^q}$ equals

where we use the well-known consequence that ℋ is bounded on $L^p({\mathbb{R}}^n)$ with the sharp bound $\frac{p}{p-1}$ and the following relationship:

$$(q-p)(1-\frac{1}{p})+p+(\frac{\beta }{n}-1)q=0.$$

To obtain a better lower bound of ${\parallel \mathcal{H}\parallel }_{L^p\to L^q}$, we can take $f_0(x)={|x|}^{\frac{n{C}_1}{p}}{\chi }_{\{|x|<1\}}(x)$, $C_1>-1$. Then

$${\parallel f_0\parallel }_{L^p}^p=\frac{{\nu }_n}{C_1+1}.$$

We have

$$\begin{array}{rcl}{\parallel {\mathbb{H}}_{\beta }(f_0)\parallel }_{L^q}^q& =& {\int }_{|x|>1}{\left(\frac{1}{{({\nu }_n{|x|}^n)}^{(1-\beta /n)}}{\int }_{|y|<1}{|y|}^{\frac{n{C}_1}{p}}dy\right)}^{q}dx\\ +{\int }_{|x|\le 1}{\left(\frac{1}{{({\nu }_n{|x|}^n)}^{(1-\beta /n)}}{\int }_{|y|<|x|}{|y|}^{\frac{n{C}_1}{p}}dy\right)}^{q}dx\\ =& \frac{{\omega }_n^q}{{\nu }_n^{(1-\frac{\beta }{n})q}{(n{C}_1/p+n)}^q}({\int }_{|x|>1}{|x|}^{(\beta -n)q}dx+{\int }_{|x|\le 1}{|x|}^{(\beta +\frac{n{C}_1}{p})q}dx)\\ =& \frac{{\omega }_n^{(1+q)}}{{\nu }_n^{(1-\frac{\beta }{n})q}{(n{C}_1/p+n)}^q}\frac{p}{qn}(\frac{1}{1+C_1}+\frac{1}{p-1}).\end{array}$$

So, we have

$$\begin{array}{rcl}\frac{{\parallel {\mathbb{H}}_{\beta }(f_0)\parallel }_{L^q}}{{\parallel f_0\parallel }_{L^p}}& =& {\left(\frac{{\omega }_n^{(1+q)}}{{\nu }_n^{(1-\beta /n)q}{(n{C}_1/p+n)}^q}\frac{p}{qn}(\frac{1}{1+C_1}+\frac{1}{p-1})\right)}^{\frac{1}{q}}{\left(\frac{C_1+1}{{\nu }_n}\right)}^{\frac{1}{p}}\\ =& {\left(\frac{p}{q}\right)}^{1/q}\frac{{(1+C_1)}^{1/p}}{1+C_1/p}{(\frac{1}{1+C_1}+\frac{1}{p-1})}^{\frac{1}{q}}\\ =& {\left(\frac{p}{q}\right)}^{1/q}{\left(\frac{p}{p-1}\right)}^{1/q}\frac{{(1+C_1)}^{\frac{1}{p}-\frac{1}{q}}}{{(1+C_1/p)}^{1-\frac{1}{q}}}.\end{array}$$

(5)

Let $g(t)={(1+t)}^{\frac{1}{p}-\frac{1}{q}}{(1+t/p)}^{-(1-\frac{1}{q})}$, $t>-1$. Taking $g^{\prime }(t^{\ast })=0$, we get $t^{\ast }=-p/q$, where $g^{\prime }$ is the derivative of the function g. It can be easily verified that the function $g(t)$ defined over the open interval $(-1,\mathrm{\infty })$ has a unique global maximum at the point $t^{\ast }$. Taking $C_1=-p/q$ together with inequality (5), we get

$${\parallel {\mathbb{H}}_{\beta }\parallel }_{L^p\to L^q}\ge {\left(\frac{p}{q}\right)}^{1/q}{\left(\frac{p}{p-1}\right)}^{1/q}{\left(\frac{q}{q-1}\right)}^{1-1/q}{(1-\frac{p}{q})}^{1/p-1/q}.$$

(6)

 □

Proof of (ii) of Theorem 3.1

Since

$${\mathbb{H}}_{\beta }(f)(x)=\frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}|f(y)|dy\le \frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\parallel f\parallel }_{L^1},$$

we have that

$$\begin{array}{rcl}\left|\{x\in {\mathbb{R}}^n:|{\mathbb{H}}_{\beta }(f)(x)|>\lambda \}\right|& \le & \left|\{x\in {\mathbb{R}}^n:\frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\parallel f\parallel }_{L^1}>\lambda \}\right|\\ =& \left|\{x\in {\mathbb{R}}^n:|x|<{\left(\frac{{\parallel f\parallel }_{L^1}}{\lambda {\nu }_n^{1-\frac{\beta }{n}}}\right)}^{\frac{1}{n-\beta }}\}\right|\\ =& {\left(\frac{{\parallel f\parallel }_{L^1}}{\lambda }\right)}^{\frac{n}{n-\beta }}.\end{array}$$

Next, we will show that the constant 1 is a sharp bound by constructing a suitable function. In fact, set

$$g(x)={\chi }_{|x|<1}(x).$$

we have

$${\parallel g\parallel }_{L^1}={\nu }_n.$$

It follows that

$${\mathbb{H}}_{\beta }(g)(x)=\frac{1}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}{\int }_{|y|<|x|}{\chi }_{|y|<1}(y)dy=\frac{|\{y:|y|<min\{|x|,1\}\}|}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}.$$

We assert that ${\mathbb{H}}_{\beta }(g)(x)\le {\nu }_n^{\frac{\beta }{n}}$ for all $x\in {\mathbb{R}}^n$. We rewrite

$$\{x\in {\mathbb{R}}^n:{\mathbb{H}}_{\beta }(g)(x)>{\nu }_n^{\frac{\beta }{n}}\}:=A\cup B,$$

where

Next, we estimate two sets A and B, respectively.

If $x\in A$, then $|x|<1$. It follows that

$${\mathbb{H}}_{\beta }(g)(x)\le \frac{|\{y:|y|<|x|\}|}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}={\left({\nu }_n{|x|}^n\right)}^{\frac{\beta }{n}}<{\nu }_n^{\frac{\beta }{n}}.$$

If $x\in B$, then $|x|\ge 1$. It follows that

$${\mathbb{H}}_{\beta }(g)(x)\le \frac{|\{y:|y|<1\}|}{{({\nu }_n{|x|}^n)}^{1-\frac{\beta }{n}}}\le {\nu }_n^{\frac{\beta }{n}}.$$

Consequently, we have $A=B=\mathrm{\varnothing }$. This implies that ${\mathbb{H}}_{\beta }(g)(x)\le {\nu }_n^{\frac{\beta }{n}}$.

For any $0<\lambda <{\nu }_n^{\frac{\beta }{n}}$, we conclude that

(7)

If there exists a constant C such that

$$\left|\{x\in {\mathbb{R}}^n:|{\mathbb{H}}_{\beta }(f)(x)|>\lambda \}\right|\le {\left(\frac{C{\parallel f\parallel }_{L^1}}{\lambda }\right)}^{\frac{n}{n-\beta }}$$

holds for all $f\in L^1({\mathbb{R}}^n)$. Then we can choose that

$$f(x)={\chi }_{|x|<1}(x).$$

It follows from equality (7) that

$${\left(\frac{{\nu }_n}{\lambda }\right)}^{\frac{n}{n-\beta }}-{\lambda }^{\frac{n}{\beta }}\le {\left(\frac{C{\nu }_n}{\lambda }\right)}^{\frac{n}{n-\beta }}$$

always holds for every $0<\lambda <{\nu }_n^{\frac{\beta }{n}}$. Letting $\lambda \to 0^{+}$, this forces that $C\ge 1$. This means that the constant 1 is sharp. □

At the end of this paper, we revisit the lower bound of ${\parallel {\mathbb{H}}_{\beta }\parallel }_{L^p\to L^q}$. Using L’Hospital’s rule, we obtain that

$$\underset{q\to p^{+}}{lim}{(1-\frac{p}{q})}^{1/p-1/q}=1.$$

This implies that

$${\parallel \mathcal{H}\parallel }_{L^p\to L^p}=\frac{p}{p-1}.$$

Remark 3.1 If $\beta =0$, the operator ${\mathbb{H}}_{\beta }$ is reduced to the classical Hardy operator ℋ. In addition, in order to study the endpoint estimate for Hardy operator, we in [13] modified the definition of (2) as follows:

$$\tilde{\mathcal{H}}(f)(x)=\frac{1}{|B(0,|x|)|}{\int }_{|y|<|x|}f(y)dy,x\in {\mathbb{R}}^n\mathrm{\setminus }\{0\},$$

where f is a measurable function on ${\mathbb{R}}^n$. In fact, the operators $\tilde{\mathcal{H}}$ and ℋ enjoy the same boundedness property for $L^p\to L^p$ and $L^1\to L^{1,\mathrm{\infty }}$, but they do not have the same property involving the Hardy space $H^1$. For instance, the operator $\tilde{\mathcal{H}}$ is bounded from $H^1$ to $L^1$; but the operator ℋ is not since ℋ is nonnegative (cf. [13]).

The well-known fact is that ${\parallel \mathcal{H}\parallel }_{L^p\to L^p}=\frac{p}{p-1}$ (cf. [2]) and ${\parallel \tilde{\mathcal{H}}\parallel }_{L^1\to L^{1,\mathrm{\infty }}}={\parallel \mathcal{H}\parallel }_{L^1\to L^{1,\mathrm{\infty }}}=1$ (cf. [13]). On account of these facts, we guess that the lower bound of (i) in Theorem 3.1 is sharp.