An Integral Version of Hardy's Inequality

TL;DR

This paper introduces a generalized form of Hardy's inequality applicable to measure spaces with a measurable function replacing the absolute value or distance, broadening the inequality's scope.

Contribution

It presents an integral version of Hardy's inequality that extends classical results to more general measure spaces and functions.

Findings

Generalizes Hardy's inequality to measure spaces with measurable functions

Replaces absolute value or distance with a measurable function in the inequality

Applicable to metric spaces with a distance function from a point

Abstract

In this note we present a version of Hardy's inequality on a measure space $(X,\mu)$ endowed with a measurable function $N\colon X\to \mathbb R$ which replaces the absolute value on $\mathbb R$ or $\mathbb R^n$, and, more generally, the distance function from a given point when $X$ is a metric space.

Equations

$$\sum_{n=1}^{\infty}\mathopen{}\mathclose{{}\left(\frac{1}{n}\sum_{k=1}^{n}a_{k}}\right)^{p}\leqslant\mathopen{}\mathclose{{}\left(\frac{p}{p-1}}\right)^{p}\sum_{n=1}^{\infty}a_{n}^{p}$$

$$\int_{0}^{+\infty}\mathopen{}\mathclose{{}\left(\frac{1}{x}\int_{0}^{x}f(t)\,\mathrm{d}t}\right)^{p}\,\mathrm{d}x\leqslant\mathopen{}\mathclose{{}\left(\frac{p}{p-1}}\right)^{p}\int_{0}^{+\infty}f(x)^{p}\,\mathrm{d}x,$$

$$\int_{0}^{+\infty}\mathopen{}\mathclose{{}\left(\int_{0}^{x}f(t)\,\mathrm{d}t}\right)^{p}\,\mathrm{d}\mu(x)\leqslant C^{p}\int_{0}^{+\infty}f(x)^{p}\,\mathrm{d}\nu(x)$$

$$B\coloneqq\sup_{r>0}\mu([r,+\infty))^{1/p}\mathopen{}\mathclose{{}\left(\int_{0}^{r}\psi^{-p^{\prime}/p}(x)\,\mathrm{d}x}\right)^{1/p^{\prime}}<\infty$$

$$\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{N(x)<r}f(x)\,\mathrm{d}\mu_{1}(x)}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}\leqslant C\mathopen{}\mathclose{{}\left(\int_{X}f(x)^{p}\,\mathrm{d}\mu_{3}(x)}\right)^{1/p}$$

$$B\coloneqq\sup_{r\in\mathds{R}}\mu_{2}([r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)<r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}$$

$$\sup_{r\in\mathds{R}}\mu_{2}([r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)<r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}=\sup_{r\in\mathds{R}}\mu_{2}((r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)\leqslant r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}$$

$$B^{\prime}\coloneqq\sup_{r\in\mathds{R}}\mu_{2}([r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)\leqslant r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}$$

$$\sup_{r\in\mathds{R}}\mu_{2}((r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)<r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}$$

$$\mathopen{}\mathclose{{}\left(\int_{X}\mathopen{}\mathclose{{}\left(\int_{N(x)<N(y)}f(x)\,\mathrm{d}\mu_{1}(x)}\right)^{q}\,\mathrm{d}\nu(y)}\right)^{1/q}\leqslant C\mathopen{}\mathclose{{}\left(\int_{X}f(x)^{p}\,\mathrm{d}\mu_{3}(x)}\right)^{1/p}$$

$$\int_{0}^{+\infty}\mathopen{}\mathclose{{}\left(\int_{x}^{+\infty}f(t)\,\mathrm{d}t}\right)^{p}\,\mathrm{d}\mu(x)\leqslant C^{p}\int_{0}^{+\infty}f(x)^{p}\,\mathrm{d}\nu(x)$$

$$\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{N(x)\leqslant r}f(x)\,\mathrm{d}\mu_{1}(x)}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}\leqslant C\mathopen{}\mathclose{{}\left(\int_{X}f(x)^{p}\,\mathrm{d}\mu_{3}(x)}\right)^{1/p}$$

$$B^{\prime}\coloneqq\sup_{r\in\mathds{R}}\mu_{2}([r,+\infty))^{1/q}\mathopen{}\mathclose{{}\left(\int_{N(x)\leqslant r}\psi^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}}\right)^{1/p^{\prime}}$$

$$(\phi \circ f{)}^{\prime }=({\phi }^{\prime }\circ f)\cdot f^{\prime }.$$

$$\begin{split}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{-\infty}^{r}f\,\mathrm{d}\mu_{1}}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{p/q}&\leqslant\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r_{2}}\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r}(fh)^{p}\,\mathrm{d}\mu_{1}}\right)^{q/p}\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r}h^{-p^{\prime}}\,\mathrm{d}\mu_{1}}\right)^{q/p^{\prime}}\,\mathrm{d}\mu_{2}(r)}\right)^{p/q}\\ &\leqslant\int_{r_{1}}^{r_{2}}(fh)^{p}(s)\mathopen{}\mathclose{{}\left(\int_{s}^{r_{2}}\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r}h^{-p^{\prime}}\,\mathrm{d}\mu_{1}}\right)^{q/p^{\prime}}\,\mathrm{d}\mu_{2}(r)}\right)^{p/q}\,\mathrm{d}\mu_{1}(s).\end{split}$$

$${\int }_{r_1}^r{h}^{-p^{\prime }}\mathrm{d}{\mu }_1={\int }_{r_1}^r{\mu }_1((r_1,s){)}^{-1/p}\mathrm{d}{\mu }_1(s)=p^{\prime }{\mu }_1((r_1,r){)}^{1/p^{\prime }}⩽B{p}^{\prime }{\mu }_2([r,+\infty ){)}^{-1/q}$$

$$\begin{split}\int_{s}^{r_{2}}\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r}h^{-p^{\prime}}\,\mathrm{d}\mu_{1}}\right)^{q/p^{\prime}}\,\mathrm{d}\mu_{2}(r)&\leqslant(Bp^{\prime})^{q/p^{\prime}}\int_{s}^{r_{2}}\mu_{2}((r,r_{2}))^{-1/p^{\prime}}\,\mathrm{d}\mu_{2}(r)\\ &=(Bp^{\prime})^{q/p^{\prime}}p\mu_{2}((s,r_{2}))^{1/p}\end{split}$$

$$\int_{r_{1}}^{r_{2}}(fh)^{p}(s)\mathopen{}\mathclose{{}\left(\int_{s}^{r_{2}}\mathopen{}\mathclose{{}\left(\int_{r_{1}}^{r}h^{-p^{\prime}}\,\mathrm{d}\mu_{1}}\right)^{q/p^{\prime}}\,\mathrm{d}\mu_{2}(r)}\right)^{p/q}\,\mathrm{d}\mu_{1}(s)\leqslant(Bp^{\prime})^{p/p^{\prime}}p^{p/q}B\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1}$$

$$\begin{split}\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left(\int_{-\infty}^{r}f\,\mathrm{d}\mu_{1}}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}&=\sup_{g\in\mathcal{G}}\int_{\mathds{R}}\int_{-\infty}^{r}f\,\mathrm{d}\mu_{1}\,g(r)\,\mathrm{d}\mu_{2}(r)\\ &=\sup_{g\in\mathcal{G}}\int_{\mathds{R}}\int_{s}^{+\infty}g\,\mathrm{d}\mu_{2}\,f(s)\,\mathrm{d}\mu_{1}(s)\\ &\leqslant\sup_{g\in\mathcal{G}}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{s}^{+\infty}g\,\mathrm{d}\mu_{2}}\right)^{p^{\prime}}\,\mathrm{d}\mu_{1}(s)}\right)^{1/p^{\prime}}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p}\\ &\leqslant(q^{\prime})^{1/p^{\prime}}q^{1/q}B\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p},\end{split}$$

$${\mu }_{1,n}:={\mu }_1+\sum _{r\in \text{mathds}R}(2^n{\chi }_{[r,r+2^{-n}]}\cdot {\mathcal{H}}^1-{\delta }_r){\mu }_1(\left\{r\right\})$$

$${\mu }_{1,n}((-\infty ,r))={\mu }_1((-\infty ,r))+\sum _{r^{\prime }<r}{\mu }_1(\left\{r^{\prime }\right\})(\min (1,2^n(r-r^{\prime }))-1)⩽{\mu }_1((-\infty ,r))$$

$$\begin{split}\mathopen{}\mathclose{{}\left\lvert\langle\chi_{(-\infty,r)}\cdot\mu_{1}-\chi_{(-\infty,r)}\cdot\mu_{1,n},\varphi\rangle}\right\rvert\leqslant\sum_{r^{\prime}<r}\mu_{1}(\Set{r^{\prime}})\mathopen{}\mathclose{{}\left\lvert 2^{n}\int_{r^{\prime}}^{\min(r,r^{\prime}+2^{-n})}\varphi(r^{\prime\prime})\,\mathrm{d}r^{\prime\prime}-\varphi(r^{\prime})}\right\rvert.\end{split}$$

$$\mathopen{}\mathclose{{}\left\lvert 2^{n}\int_{r^{\prime}}^{\min(r,r^{\prime}+2^{-n})}\varphi(r^{\prime\prime})\,\mathrm{d}r^{\prime\prime}-\varphi(r^{\prime})}\right\rvert\leqslant 2\lVert\varphi\rVert_{L^{\infty}(\mathds{R})}$$

$$\lim_{n\to\infty}\mathopen{}\mathclose{{}\left\lvert\langle\chi_{(-\infty,r)}\cdot\mu_{1}-\chi_{(-\infty,r)}\cdot\mu_{1,n},\varphi\rangle}\right\rvert=0,$$

$$\begin{split}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1}}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}&\leqslant\liminf_{n\to\infty}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1,n}}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}\\ &\leqslant C_{B}\liminf_{n\to\infty}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1,n}}\right)^{1/p}\\ &=C_{B}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p}\end{split}$$

$${\mu }_{2,n}:={\mu }_2+\sum _{r\in \text{mathds}R}(2^n{\chi }_{[r-2^{-n},r]}\cdot {\mathcal{H}}^1-{\delta }_r){\mu }_2(\left\{r\right\}),$$

$$\begin{split}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1}}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}&=\lim_{n\to\infty}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1}}\right)^{q}\,\mathrm{d}\mu_{2,n}(r)}\right)^{1/q}\\ &\leqslant C_{B}\mathopen{}\mathclose{{}\left(\int_{X}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p}.\end{split}$$

$$\begin{split}\mathopen{}\mathclose{{}\left(\mathopen{}\mathclose{{}\left(\int_{N(x)<r}f(x)\,\mathrm{d}\mu_{1}(x)}\right)^{q}\,\mathrm{d}\mu_{2}(r)}\right)^{1/q}&=\sup_{g\in\mathcal{G}}\int_{\mathds{R}}\int_{N(x)<r}f(x)\,\mathrm{d}\mu_{1}(x)\,g(r)\,\mathrm{d}\mu_{2}(r)\\ &=\sup_{g\in\mathcal{G}}\int_{X}\int_{N(x)}^{+\infty}g\,\mathrm{d}\mu_{2}\,f(x)\,\mathrm{d}\mu_{1}(x)\\ &\leqslant\sup_{g\in\mathcal{G}}\mathopen{}\mathclose{{}\left(\int_{X}\mathopen{}\mathclose{{}\left(\int_{N(x)}^{+\infty}g\,\mathrm{d}\mu_{2}}\right)^{p^{\prime}}\,\mathrm{d}\mu_{1}(x)}\right)^{1/p^{\prime}}\mathopen{}\mathclose{{}\left(\int_{X}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p}\\ &=\sup_{g\in\mathcal{G}}\mathopen{}\mathclose{{}\left(\int_{\mathds{R}}\mathopen{}\mathclose{{}\left(\int_{s}^{+\infty}g\,\mathrm{d}\mu_{2}}\right)^{p^{\prime}}\,\mathrm{d}N_{*}(\mu_{1})(s)}\right)^{1/p^{\prime}}\mathopen{}\mathclose{{}\left(\int_{X}f^{p}\,\mathrm{d}\mu_{1}}\right)^{1/p}\\ &\leqslant C\int_{\mathds{R}}f^{p}\,\mathrm{d}\mu_{1},\end{split}$$

$${\mu }_{1,n}:=[({\chi }_{[r_{1,n},r_{2,n}]}\circ N){\psi }^{-p^{\prime }/p}]\cdot {\mu }_1$$

$${\mu }_{2,n}:={\chi }_{[r_{1,n}^{\prime },r_{2,n}^{\prime }]}\cdot {\mu }_2.$$

Full text

*Math Subject Classification 2020 *

The author is a member of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM). The author was partially funded by the INdAM-GNAMPA Project CUP_E55F22000270001.

1. Introduction

In [10, 13], Hardy’s inequality was first proved in its discrete version

[TABLE]

for every $p\in(1,\infty)$ and for every positive sequence $(a_{n})$. The continuous version

[TABLE]

for every positive measurable function $f$ on $[0,\infty)$ and for every $p>1$, was later proved in [11]. Since then, Hardy’s inequality has been generalized and extended to several contexts. On the one hand, the function $1/x$ has been replaced with more general weights in a series of works (cf., e.g., [20, 21, 15]), finally reaching a characterization of the positive Radon measures $\mu,\nu$ for which there is a $C\geqslant 0$ such that

[TABLE]

for every positive Borel measurable function $f$ on $\mathds{R}$. Explicitly, there is a finite constant $C\geqslant 0$ as above if and only if

[TABLE]

(interpreting \big{(}\int_{0}^{r}\psi^{-p^{\prime}/p}(x)\,\mathrm{d}x\big{)}^{1/p^{\prime}} as $\lVert\chi_{(0,r)}\psi^{-1/p}\rVert_{L^{p^{\prime}}(\mathds{R})}$ when $p=1$), where $\psi$ is the density of the absolutely continuous part of $\nu$ with respect to Lebesgue measure, in which case the minimal constant $C$ such that (1) holds satisfies $B\leqslant C\leqslant(p^{\prime})^{1/p^{\prime}}p^{1/p}B$, with the conventions $0^{0}=0$ and $\infty^{0}=1$. A similar characterization was also provided for $p=\infty$, with suitable modifications.

On the other hand, $\mathds{R}$ has been replaced with more general spaces, from $\mathds{R}^{n}$, replacing the segments $[0,r]$ with the balls $B(0,r)$, to general metric spaces, with similar substitutions, cf. [18, 19]. These latter papers provide a very general version of Hardy’s inequality on metric spaces based on a suitable ‘polar decomposition’. On the contrary, we shall use a duality method in order to essentially extend the results of [15]. For more detailed accounts on the available literature on Hardy’s inequalities (including several equivalent formulations of the finiteness of $B$, in the above notation), cf., e.g., [2, 16, 5, 17, 8, 6, 7, 9, 14, 3, 12, 18, 19] and the references therein.

In this note, we shall prove the following general version of Hardy’s inequality.

Theorem 1**.**

Let $(X,\mathfrak{M})$ be a measurable space, let $\mu_{2}$ be a positive Borel measure on $\mathds{R}$, and let $\mu_{1},\mu_{3}$ be two positive measures on $(X,\mathfrak{M})$, with $\mu_{1}$ semi-finite. Take $p,q\in[1,\infty]$ with $p\leqslant q$, and assume that the absolutely continuous part of $\mu_{3}$ with respect to $\mu_{1}$ has a density $\psi\colon X\to[0,+\infty)$ with respecto to $\mu_{1}$.111This is certainly the case if $\mu_{1}$ and $\mu_{3}$ are both Radon or $\sigma$-finite. One may simply require $\mu_{1}$ to be Radon or $\sigma$-finite if one allows $\psi$ to take also the value $+\infty$, but in this case the assertion is still equivalent to Theorem 1, applied to $X\setminus\psi^{-1}(+\infty)$. In addition, let $N\colon X\to\mathds{R}$ be a $\mathfrak{M}$-measurable function. Then, there is a constant $C>0$ such that

[TABLE]

(with the usual modifications when $p$ or $q$ is $\infty$), for every positive $\mathfrak{M}$-measurable function $f$ on $X$, if and only if

[TABLE]

(with the usual modification when $p=1$ and the conventions $0^{0}=0$ and $\infty^{0}=1$) is finite. The minimal $C$ for which (2) holds satisfies the inequalities $B\leqslant C\leqslant\min((p^{\prime})^{1/p^{\prime}}p^{1/q},(q^{\prime})^{1/p^{\prime}}q^{1/q})B$.

Some comments are in order. First of all, we say that a measure $\mu$ on $(X,\mathfrak{M})$ is semi-finite if every $\mathfrak{M}$-measurable subset $E$ of $X$ with $\mu(E)=+\infty$ contains a $\mathfrak{M}$-measurable subset $F$ with $0<\mu(F)<+\infty$. This is readily seen to be equivalent to the fact that $\mu(E)$ is the least upper bound of the $\mu(F)$, as $F$ runs through the set of $\mathfrak{M}$-measurable subsets of $E$ with finite measure, for every $\mathfrak{M}$-measurable subset $E$ of $X$. Thus, $\mu$ is semi-finite if it satisfies some kind of inner regularity with respect to $\mathfrak{M}$-measurable sets with finite measure. In particular, bounded and $\sigma$-finite measures are semi-finite.

Furthermore, given a positive $\mathfrak{M}$-measurable function $\psi$, we denote by $\psi\cdot\mu$ the measure such that $(\psi\cdot\mu)(E)=\int_{E}\psi\,\mathrm{d}\mu$ for every $\mathfrak{M}$-measurable subset $E$ of $X$, in which case we say that $\psi$ is a density of $\psi\cdot\mu$ with respect to $\mu$. We remark that, when we say ‘positive’, we mean ‘$\geqslant 0$’, so that a positive function is a function which is $\geqslant 0$ everywhere.

Let us now briefly comment on the assumptions of Theorem 1. Observe first that, if $\mu_{1}$ is not assumed to be semi-finite, then the inequality $B\leqslant C$ may fail (and the existence of $C$ may fail to imply the finiteness of $B$). For example, if $\mu_{1}(E)=+\infty$ for every non-empty element of $\mathfrak{M}$, and if $\mu_{3}=\mu_{1}$ (so that one may take $\psi=1$), then (2) holds trivially for every positive $\mathfrak{M}$-measurable function $f$ on $X$ with any non-zero $C$, but $B$ is infinite unless $\mu_{2}=0$ on $(\inf N(X),+\infty)$.

Furthermore, notice that

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since the two sides are the least upper bound of a left-continuous and a right-continuous function which coincide on the complement of a countable set (if $q<\infty$ and $p>1$; the other cases require a slightly different treatment). On the other hand,

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may be strictly larger than $B$ (and is crucial in Theorem 2), while

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may be strictly smaller than $B$.

Finally, observe that, if there is a measure $\nu$ on $(X,\mathfrak{M})$ such that $N_{*}(\nu)=\mu_{2}$, that is, $\mu_{2}(A)=\nu(N^{-1}(A))$ for every Borel subset $A$ of $\mathds{R}$, then (2) may be written in the equivalent way

[TABLE]

for every positive $\mathfrak{M}$-measurable function $f$ on $X$, since the function $r\mapsto\int_{N(x)<r}f(x)\,\mathrm{d}\mu_{1}(x)$ is increasing, hence Borel measurable. Nonetheless, even if $N(X)=\mathds{R}$, there may be positive Borel measures on $\mathds{R}$ which are not images of a positive measure on $(X,\mathfrak{M})$222For example, if $X=\mathds{R}$ and $\mathfrak{M}$ is the set of subsets of $X$, then $N_{*}(\nu)$ is the restriction of $\nu$ to the Borel $\sigma$-algebra of $\mathds{R}$ and it seems unlikely that every Borel measure $\mu_{2}$ on $\mathds{R}$ may be written in the form $N_{*}(\nu)$ for some measure $\nu$ on $(X,\mathfrak{M})$. Under the continuum hypothesis, $\mathds{R}$ is ulamian, so that every $\nu$ as above must be atomic, that is, $\nu=\sum_{x\in\mathds{R}}f(x)\delta_{x}$ for some $f\colon X\to[0,+\infty]$, so that only the atomic measures on $\mathds{R}$ may be of the form $N_{*}(\nu)$.

Let us now compare Theorem 1 with some results in the literature. First of all, the discrete version of Hardy’s inequality may be obtained from Theorem 1 choosing $X=1+\mathds{N}=\Set{1,2,3,\dots}$, $\mu_{1}=\mu_{3}$ as the counting measure on $X$, $\mu_{2}=\sum_{n=1}^{\infty}\frac{1}{n^{p}}\delta_{n+1}$, and letting $N$ be the natural inclusion of $X$ in $\mathds{R}$ (nonetheless, the discrete version of Hardy’s inequality is most naturally obtained as a particular case of Theorem 2 below). Analogously, the continuous version of Hardy’s inequaliy may be obtained choosing $X=(0,\infty)$, $\mu_{1}=\mu_{3}$ as Lebesgue measure, $\mu_{2}=\chi_{(0,+\infty)}\frac{1}{(\,\cdot\,)^{p}}\cdot\mathcal{H}^{1}$, where $\mathcal{H}^{1}$ denotes the $1$-dimensional Hausdorff measure, that is, Lebesgue measure, and letting $N$ be the natural inclusion of $X$ in $\mathds{R}$. The general version (1) of Hardy’s inequality on $[0,\infty)$ may be obtained in a similar way, as well as several other versions of Hardy’s inequality on $\mathds{R}^{n}$ on in more general metric spaces (including [18]).

Furthermore, other interesting examples may be studied by means of Theorem 1. For instance, Theorem 1 may be applied to every Riemannian manifold (complete or not), endowed with the canonical volume form, and even to every sub-Riemannian manifold, endowed with a suitable measure (such as Popp’s measure, when defined). Once the behaviour of the volume of the balls centred at a fixed point near [math] and near $\infty$ is understood, one may find estimates for $B$ and then sufficient conditions for (2) to hold with a certain $C$. Besides that, Thereom 1 applies also to discrete spaces, as remarked earlier, even though Theorem 2 below applies more naturally, in this latter context.

In addition, the so-called conjugate (or dual) Hardy’s inequality (which we now express on $[0,\infty)$ for the sake of simplicity)

[TABLE]

may be obtained as a particular case of (2) choosing $N(x)=-x$.

Notice that we do not consider the cases in which $\min(p,q)<1$ or $p<q$ (cf., e.g., [16, 17, 19]). Indeed, the former cannot be studied with duality methods; the latter is not considered for the sake of simplicity.

Concerning the vast wealth of equivalent formulations of the finiteness of $B$ (cf., e.g., [8, 9]), we observe that, at least when $\mu_{2}$ and $N_{*}(\mu_{1})$ are $\sigma$-finite and absolutely continuous with respect to Lebesgue measure, no new proofs are necessary, as these equivalent conditions are formulated in terms of the left-continuous monotone functions $r\mapsto\mu_{2}([r,+\infty))$ and $r\mapsto\int_{N(x)<r}\psi^{-p^{\prime}/p}(x)\,\mathrm{d}\mu_{1}(x)$ and their derivatives $-\mu_{2}$ and $N_{*}(\psi^{-p^{\prime}/p}\cdot\mu_{1})$ (for $p>1$). For more general measures, suitable extensions may be in order. Nonetheless, once these equivalences are formulated for general pairs of left-continuous decresing and increasing functions, they immediately apply to this more general context.

We shall also prove the following variant of Theorem 1.

Theorem 2**.**

Keep the hypotheses and the notation of Theorem 1. Then, there is a constant $C^{\prime}>0$ such that

[TABLE]

(with the usual modifications when $p$ or $q$ is $\infty$), for every positive $\mathfrak{M}$-measurable function $f$ on $X$, if and only if

[TABLE]

(with the usual modification when $p=1$) is finite. The minimal $C^{\prime}$ for which (4) holds satisfies the inequalities $B^{\prime}\leqslant C^{\prime}\leqslant\min((p^{\prime})^{1/p^{\prime}}p^{1/q},(q^{\prime})^{1/p^{\prime}}q^{1/q})B^{\prime}$.

2. Proof of Theorems 1 and 2

We first need to recall a simple lemma (cf. [1, Theorem 3.96] for a proof of a much more general version of this result). We restrict our attention to diffuse measures (that is, measures $\mu$ such that $\mu(\Set{x})=0$ for every $x$) since this additional assumption simplifies the statement and this case will be sufficient for our purposes.

Lemma 3**.**

Let $f$ be a continuous function on an open subset $I$ of $\mathds{R}$, and assume that its distributional derivative $f^{\prime}$ is a (Radon) measure. Let $J$ be an open subset of $\mathds{R}$ which contains $f(I)$, and let $\varphi$ be a function of class $C^{1}$ on $J$. Then,

[TABLE]

Recall that, for a function $f$ as in the statement of Lemma 3, $f(b)-f(a)=f^{\prime}([a,b])$ for every $a,b\in I$ such that $a<b$ and $[a,b]\subseteq I$, so that the conclusion of Lemma 3 may be rephrased saying that $\varphi(f(b))-\varphi(f(a))=\int_{a}^{b}(\varphi^{\prime}\circ f)\,\mathrm{d}f^{\prime}$ for every $a,b\in I$ such that $a<b$ and $[a,b]\subseteq I$. In particular, saying that $f$ is continuous is essentially equivalent to saying that $f^{\prime}$ is diffuse.

Notice that [1, Theorem 3.96] requires $\varphi$ to be Lipschitz since the authors want to deduce that the measure $(\varphi\circ f)^{\prime}$ is bounded (assuming $f^{\prime}$ bounded as well). If we do not require this fact and apply the cited result locally, we obtain Lemma 3.

We may now pass to the proof of Theorem 1.

Proof of Theorem 1..

Throughout the proof, we shall assume that $p,q\in(1,\infty)$. We shall leave to the reader the simple modifications which are needed to deal with the remaining cases. We shall set $C_{B}=\min((p^{\prime})^{1/p^{\prime}}p^{1/q},(q^{\prime})^{1/p^{\prime}}q^{1/q})$.

Step I. Assume that: $B$ is finite; $X=\mathds{R}$; $\mathfrak{M}$ is the Borel $\sigma$-algebra of $\mathds{R}$; $N(r)=r$ for every $r\in\mathds{R}$; $\mu_{1},\mu_{2}$ are diffuse Radon measures with compact support on $\mathds{R}$; $\mu_{3}=\mu_{1}$ (so that $\psi=1$). Set $h(r)\coloneqq\mu_{1}((-\infty,r))^{1/(pp^{\prime})}$ for every $R\in\mathds{R}$. Observe that (2) holds trivially with $C=0$ if $\mu_{1}=0$, that is, if $h=0$, so that we may assume that $h\neq 0$. Let $r_{1}$ be the maximum of the $r$ such that $h(r)=0$. Analogously, we may assume that $\mu_{2}\neq 0$, in which case we define $r_{2}$ as the minimum of the $r$ such that $\mu_{2}((r,+\infty))=0$. We may assume that $r_{1}<r_{2}$, for otherwise the assertion is trivial.

Then, take a positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$, and observe that, by Hölder’s and Minkowski’s integral inequality,

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Then, observe that, for every $r>r_{1}$,

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by Lemma 3. In addition,

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again by Lemma 3. Thus,

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by the definition of $B$. Thus, (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$ with $C=(p^{\prime})^{1/p^{\prime}}p^{1/q}B$.

Step II. Assume that: $B$ is finite; $X=\mathds{R}$; $\mathfrak{M}$ is the Borel $\sigma$-algebra of $\mathds{R}$; $N(r)=r$ for every $r\in\mathds{R}$; $\mu_{1},\mu_{2}$ are diffuse Radon measures with compact support on $\mathds{R}$; $\mu_{3}=\mu_{1}$. We proceed by duality, using step I. Define $\mathcal{G}$ as the set of positive $\mathfrak{M}$-measurable functions $g$ on $\mathds{R}$ such that $\lVert g\rVert_{L^{q^{\prime}}(\mu_{2})}\leqslant 1$. Take a positive $p$-th power $\mu_{1}$-integrable function $f$ on $\mathds{R}$, and observe that

[TABLE]

where the first equality follows from the fact that $\mu_{2}$ is bounded, the second equality follows from Tonelli’s theorem, since both $\mu_{1}$ and $\mu_{2}$ are bounded, the first inequality follows from Hölder’s inequality, and the last inequality follows from step I, applied to $\check{\mu}_{2}$, $\check{\mu}_{1}$, $\check{g}$, $q^{\prime}$, and $p^{\prime}$ instead of $\mu_{1}$, $\mu_{2}$, $f$, $p$, and $q$, where $\check{\;}$ denotes the reflection $r\mapsto-r$ (taking into account (3)). Thus, (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$ with $C=C_{B}$ (the case in which $f\not\in L^{p}(\mu_{1})$ being trivial).

Step III. Assume that: $B$ is finite; $X=\mathds{R}$; $\mathfrak{M}$ is the Borel $\sigma$-algebra of $\mathds{R}$; $N(r)=r$ for every $r\in\mathds{R}$; $\mu_{1},\mu_{2}$ are Radon measures with compact support on $\mathds{R}$, with $\mu_{2}$ diffuse; $\mu_{3}=\mu_{1}$. Define

[TABLE]

for every $n\in\mathds{N}$, so that $(\mu_{1,n})$ is a sequence of diffuse positive Radon measures, and

[TABLE]

for every $r\in\mathds{R}$ and for every $n\in\mathds{N}$. Let us prove that $\chi_{(-\infty,r)}\cdot\mu_{1,n}$ converges vaguely (that is, in the weak dual of $C_{c}(\mathds{R})$) to $\chi_{(-\infty,r)}\cdot\mu_{1}$ for every $r\in\mathds{R}$. Indeed, for every $\varphi\in C_{c}(\mathds{R})$,

[TABLE]

Now, it is clear that $2^{n}\int_{r^{\prime}}^{\min(r,r^{\prime}+2^{-n})}\varphi(r^{\prime\prime})\,\mathrm{d}r^{\prime\prime}$ converges to $\varphi(r^{\prime})$ for every $r^{\prime}<r$, that

[TABLE]

for every $r^{\prime}<r$, and that $\sum_{r^{\prime}\in(-\infty,r)}\mu_{1}(\Set{r})\leqslant\mu_{1}(\mathds{R})<\infty$. Then, the dominated convergence theorem shows that

[TABLE]

whence our claim by the arbitrariness of $\varphi$ and $r$. In particular, taking $r\geqslant\max\operatorname{Supp}\mu_{1}+1$, we see that $\mu_{1,n}$ converges vaguely to $\mu_{1}$.

Then, for every positive $f\in C_{c}(\mathds{R})$, steps I and II show that

[TABLE]

by Fatou’s lemma. Thus, (2) holds for $f$. Now, if $f\colon\mathds{R}\to[0,+\infty]$ is lower semi-continuous, then $f$ is the pointwise limit of an increasing sequence of elements of $C_{c}(\mathds{R})$, so that (2) is established for $f$ by monotone convergence. Analogously, if $f$ is a $p$-th power $\mu_{1}$-integrable positive function, then there is a decresing sequence of positive $p$-th power $\mu_{1}$-integrable lower semicontinuous functions which converge $\mu_{1}$-almost everywhere to $f$, so that (2) is established for $f$ by monotone convergence and Fatou’s lemma. Then (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$ (the case in which $f\not\in L^{p}(\mu_{1})$ being trivial).

Step IV. Assume that: $B$ is finite; $X=\mathds{R}$; $\mathfrak{M}$ is the Borel $\sigma$-algebra of $\mathds{R}$; $N(r)=r$ for every $r\in\mathds{R}$; $\mu_{1},\mu_{2}$ are Radon measures with compact support on $\mathds{R}$; $\mu_{3}=\mu_{1}$. Define

[TABLE]

so that $\mu_{2,n}$ is a sequence of diffuse positive Radon measures such that $\mu_{2,n}([r,+\infty))\leqslant\mu_{2}([r,+\infty))$ for every $r\in\mathds{R}$ and for every $n\in\mathds{N}$, and such that $\int_{\mathds{R}}f\,\mathrm{d}\mu_{2,n}$ converges to $\int_{\mathds{R}}f\,\mathrm{d}\mu_{2}$ for every positive left-continuous function $f$ on $\mathds{R}$ (the proof is analogous to that of step III). Now, observe that, for every positive $p$-th power $\mu_{1}$-integrable function $f$ on $\mathds{R}$, the increasing function $r\mapsto\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1}$ is left-continuous and bounded (by $\mu_{1}(\mathds{R})^{1/p^{\prime}}\lVert f\rVert_{L^{p}(\mu_{1})}$). Thus, by means of step III we see that

[TABLE]

Thus, (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$ (the case in which $f\not\in L^{p}(\mu_{1})$ being trivial).

Step V. Assume that: $B$ is finite; $\mu_{1},\mu_{2}$ are bounded; $\mu_{1}=\mu_{3}$; $N_{*}(\mu_{1})$ and $\mu_{2}$ are compactly supported. Notice that every bounded Borel measure on $\mathds{R}$ is a Radon measure, so that both $N_{*}(\mu_{1})$ and $\mu_{2}$ are Radon measures. We proceed by duality as in step II. Define $\mathcal{G}$ as the set of positive Borel measurable functions $g$ on $\mathds{R}$ such that $\lVert g\rVert_{L^{q^{\prime}}(\mu_{2})}\leqslant 1$. Take a positive $f\in L^{p}(\mu_{1})$, and observe that

[TABLE]

where each step is justified as in step II, except for the last equality, which holds since the function $s\mapsto\int_{s}^{+\infty}g\,\mathrm{d}\mu_{2}$ is decreasing, hence Borel measurable. Thus, (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$ (the case in which $f\not\in L^{p}(\mu_{1})$ being trivial).

Step VI. Assume that $B$ is finite. If $\mu_{1}(N^{-1}((-\infty,r)))=0$ for some $r\in\mathds{R}$, let $r_{1}$ be the maximum of these $r$. Set $r_{1}\coloneqq-\infty$ otherwise. Analogously, if $\mu_{2}([r,+\infty))=0$ for some $r\in\mathds{R}$, let $r_{2}$ be the greatest lower bound of these $r$. Set $r_{2}\coloneqq+\infty$ otherwise. Then, the finiteness of $B$ implies that, for every $r\in(r_{1},r_{2})$, both $\mu_{2}([r,+\infty))$ and $\int_{N(x)<r}\psi(x)^{-p^{\prime}/p}\,\mathrm{d}\mu_{1}(x)$ are finite and non-zero. Notice that, if $r_{2}\leqslant r_{1}$, then (2) trivially holds (with $C=0$) for every positive $\mathfrak{M}$-measurable function $f$ on $X$, since $\int_{N(y)<r}f(y)\,\mathrm{d}\mu_{1}=0$ for every $r\in(-\infty,r_{2}]$, hence for $\mu_{2}$-almost every $r\in\mathds{R}$. Then, we may assume that $r_{1}<r_{2}$.

Define two sequences $\mu_{1,n}$ and $\mu_{2,n}$ of bounded Radon measures on $X$ and $\mathds{R}$, respectively, as follows. Let $(r_{1,n})$ be a strictly decreasing sequence which converges to $r_{1}-1$, and let $(r_{2,n})$ be a strictly increasing sequence which converges to $r_{2}$, and define

[TABLE]

for every $n\in\mathds{N}$. In addition, let $(r^{\prime}_{2,n})$ be a strictly increasing sequence which converges to $r_{2}+1$, and let $(r^{\prime}_{1,n})$ be a strictly decreasing sequence which converges to $r_{1}$, and define

[TABLE]

Then, the preceding considerations show that $\mu_{1,n}$ and $\mu_{2,n}$ are bounded inner regular measures for every $n\in\mathds{N}$. In addition, $N_{*}(\mu_{1,n})$ and $\mu_{2,n}$ are compactly supported for every $n\in\mathds{N}$. Now, take a positive $\mathfrak{M}$-measurable function $f$ on $X$, so that $f\psi^{p^{\prime}/p}$ is $\mu_{1,n}$-measurable for every $n\in\mathds{N}$. Then, by monotone convergence,

[TABLE]

Thus, (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$.

Step VII. Assume that there is a finite constant $C$ such that (2) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$. Observe that (2) holds with $\psi\cdot\mu_{1}$ in place of $\mu_{3}$ as one sees taking $f=0$ on a $\mu_{1}$-negligible Borel subset of $X$ where $\mu_{3}-\psi\cdot\mu_{1}$ is concentrated. Therefore, we may assume that $\mu_{3}=\psi\cdot\mu_{1}$. Take $r\in\mathds{R}$. If $E$ is an $\mathfrak{M}$-measurable subset of $N^{-1}((-\infty,r))$ such that $\mu_{3}(E)=0$, then applying (2) with $f=\chi_{E}$ we see that

[TABLE]

If $\mu_{1}(E)>0$, this implies that

[TABLE]

so that

[TABLE]

Therefore, if either $\mu_{1}(N^{-1}((-\infty,r)))=0$ or if $N^{-1}((-\infty,r))$ contains an $\mathfrak{M}$-measurable subset $E$ with $\mu_{1}(E)>0$ and $\mu_{3}(E)=0$, then

[TABLE]

Then, assume that $\mu_{1}(N^{-1}((-\infty,r)))>0$ and that $\mu_{3}(E)>0$ for every $\mathfrak{M}$-measurable subset $E$ of $N^{-1}((-\infty,r))$ with $\mu_{1}(E)>0$. In other words, we assume that $\psi(x)>0$ for $\mu_{1}$-almost every $x\in N^{-1}((-\infty,r))$, and that $\mu_{1}(N^{-1}((-\infty,r)))>0$. Since $\mu_{1}$ is semi-finite, also $\psi^{-p^{\prime}/p}\cdot\mu_{1}$ is semi-finite (on $N^{-1}((-\infty,r))$), so that $(\psi^{-p^{\prime}/p}\cdot\mu_{1})(N^{-1}((-\infty,r)))$ is the least upper bound of the $(\psi^{-p^{\prime}/p}\cdot\mu_{1})(E)$, as $E$ runs through the set of $\mathfrak{M}$-measurable subsets of $N^{-1}((-\infty,r))$ such that $0<(\psi^{-p^{\prime}/p}\cdot\mu_{1})(E)<\infty$. Then, fix such an $E$, and choose $f=\chi_{E}\psi^{-p^{\prime}/p}$ in (2), so that

[TABLE]

Therefore,

[TABLE]

The arbitrariness of $E$ then shows that

[TABLE]

By the arbitrariness of $r$, we have thus proved that $B\leqslant C$. ∎

Lemma 4**.**

Let $f,g\colon\mathds{R}\to[0,+\infty)$ be two functions, with $f$ decreasing and left-continuous, $g$ increasing and right-continuous, and $fg$ compactly supported. Then,

[TABLE]

Proof.

We may assume that $f,g\neq 0$, for otherwise the assertion is trivial. Take $a,b\in\mathds{R}$ so that $f(b-\varepsilon)>0=f(b+\varepsilon)$ $g(a-\varepsilon)=0<g(b+\varepsilon)$ for every $\varepsilon>0$: this is possible since$f,g\neq 0$ and $fg$ has compact support. In addition, the assertion is trivial if $b<a$, so that we may assume that $a\leqslant b$. Observe that, since $f(r-\varepsilon)\to f(r)$ for every $r\in\mathds{R}$, and for $\varepsilon\to 0^{+}$, thanks to the left continuity of $f$, it is clear that

[TABLE]

Conversely, let $(\varepsilon_{k})$ be a sequence in $(0,+\infty)$ such that $\varepsilon_{k}\to 0$ and

[TABLE]

Take $\eta>0$. For every $k\in\mathds{N}$, take $r_{k}\in[a,b+\varepsilon_{k}]$ so that $\sup_{r\in\mathds{R}}f(r-\varepsilon_{k})g(r)\leqslant f(r_{k}-\varepsilon_{k})g(r_{k})+\eta$. Up to a subsequence, we may then assume that $(r_{k})$ is a monotone sequence; let $\bar{r}\in[a,b]$ be its limit. If $(r_{k})$ is increasing, then, $f(r_{k}-\varepsilon_{k})\to f(\bar{r})$ since $f$ is left-continuous, and $g(r_{k})\leqslant g(\bar{r})$ since $g$ is increasing, so that

[TABLE]

If, otherwise, $(r_{k})$ is decreasing, then $f(r_{k}-\varepsilon_{k})\leqslant f(\bar{r}-\varepsilon_{k})\to f(\bar{r})$ since $f$ is decreasing and left-continuous, and $g(r_{k})\to g(\bar{r})$ since $g$ is right-continuous, so that

[TABLE]

By the arbitrariness of $\eta$, this implies that

[TABLE]

whence the result. ∎

Proof of Theorem 2.

As in the proof of Thereom 1, we shall assume, for simplicity, $p,q\in(1,\infty)$, and leave to the reader the simple modifications needed to deal with the remaining cases.

Step I. Assume first that: $B^{\prime}$ is finite; $X=\mathds{R}$; $\mathfrak{M}$ is the Borel $\sigma$-algebra of $\mathds{R}$; $N(r)=r$ for every $r\in\mathds{R}$; $\mu_{1}$ and $\mu_{2}$ are bounded and compactly supported; $\mu_{1}=\mu_{3}$. Then, the assertion follows from Theorem 1 if $\mu_{1}(\Set{r})\mu_{2}(\Set{r})=0$ for every $r\in\mathds{R}$, since, in this case, for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$, one has $\int_{(-\infty,r)}f\,\mathrm{d}\mu_{1}=\int_{(-\infty,r]}f\,\mathrm{d}\mu_{1}$ for $\mu_{2}$-almost every $r\in\mathds{R}$, and

[TABLE]

since $\mu_{2}([\,\cdot\,,+\infty))$ is continuous at $r$ for every $r\in\mathds{R}$ such that $\mu_{1}(\Set{r})>0$. In the general case, observe that, since both $\mu_{1}$ and $\mu_{2}$ have at most countably many atoms, there is a sequence $(\varepsilon_{k})_{k\in\mathds{N}}$ of elements of $(0,+\infty)$ such that $\varepsilon_{k}\to 0$ and such that $\mu_{1}(\Set{r})\mu_{2}(\Set{r-\varepsilon_{k}})=0$ for every $r\in\mathds{R}$. Then, by Lemma 4

[TABLE]

so that, for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$,

[TABLE]

by the right-continuity of the function $r\mapsto\int_{(-\infty,r]}f\,\mathrm{d}\mu_{1}$, Fatou’s lemma, and the previous remarks, with $C^{\prime}=\min((p^{\prime})^{1/p^{\prime}}p^{1/q},(q^{\prime})^{1/p^{\prime}}q^{1/q})B^{\prime}$. Thus, (4) holds for every positive $\mathfrak{M}$-measurable function $f$ on $\mathds{R}$.

Step II. Now, assume that $B^{\prime}$ is finite, and let us prove that (4) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$, with $C^{\prime}=\min((p^{\prime})^{1/p^{\prime}}p^{1/q},(q^{\prime})^{1/p^{\prime}}q^{1/q})B^{\prime}$. The proof proceeds as in steps V and VI of the proof of Theorem 1, with the following modification: in the proof of step VI, one chooses $r_{2,n}=r_{2}$ if $r_{2}\in\mathds{R}$ and $\mu_{2}(\Set{r_{2}})>0$ (in which case the finiteness of $B^{\prime}$ implies that $\mu_{1}(N^{-1}((-\infty,r_{2}])))$ is finite), and, analogously, $r^{\prime}_{1,n}=r_{1}$ if $r_{1}\in\mathds{R}$ and $\mu_{1}(N^{-1}(\Set{r_{1}}))>0$ (in which case the finiteness of $B^{\prime}$ implies that $\mu_{2}([r_{1},+\infty))$ is finite).

Step III. Finally, assume that there is a finite $C^{\prime}$ for which (4) holds for every positive $\mathfrak{M}$-measurable function $f$ on $X$, and let us prove that $B^{\prime}\leqslant C^{\prime}$. The proof proceeds as in step VII of the proof of Theorem 1, simply replacing $(-\infty,r)$ with $(-\infty,r]$ (and $N(x)<r$ with $N(x)\leqslant r$) everywhere. ∎