Optimal relationships between $L^p$-norms for the Hardy operator and its dual

Abstract

We obtain sharp two-sided inequalities between $L^p$-norms $(1<p<\infty )$ of functions $\textit{Hf}$ and $H^*f$, where $H$ is the Hardy operator, $H^*$ is its dual, and $f$ is a nonnegative measurable function on $(0,\infty ).$ In an equivalent form, it gives sharp constants in the two-sided relationships between $L^p$-norms of functions $H\varphi -\varphi$ and $\varphi$, where $\varphi$ is a nonnegative nonincreasing function on $(0,+\infty )$ with $\varphi (+\infty )=0.$ In particular, it provides an alternative proof of a result obtained by Kruglyak and Setterqvist (Proc Am Math Soc 136:2005–2013, 2008) for $p=2k \,\,(k\in \mathbb N )$ and by Boza and Soria (J Funct Anal 260:1020–1028, 2011) for all $p\ge 2$, and gives a sharp version of this result for $1<p<2$.

1 Introduction and main results

Denote by $\mathcal M ^+(\mathbb R _+)$ the class of all nonnegative measurable functions on $\mathbb R _+\equiv (0,+\infty ).$ Let $f\in \mathcal M ^+(\mathbb R _+).$ Set

$$\begin{aligned} Hf(x)=\frac{1}{x}\int \limits _0^x f(t)\,\text{ d}t \end{aligned}$$

and

$$\begin{aligned} H^*f(x)=\int \limits _x^\infty \frac{f(t)}{t}\,\text{ d}t. \end{aligned}$$

These equalities define the classical Hardy operator $H$ and its dual operator $H^*$. By Hardy’s inequalities [5, Ch. 9], these operators are bounded in $L^p(\mathbb R _+)$ for any $1<p<\infty .$ Furthermore, it is easy to show that for any $f\in \mathcal M ^+(\mathbb R _+)$ and any $1<p<\infty$, the $L^p$-norms of $Hf$ and $H^*f$ are equivalent. Indeed, let $f\in \mathcal M ^+(\mathbb R _+).$ By Fubini’s theorem,

$$\begin{aligned} Hf(x)=\frac{1}{x}\int \limits _0^x \text{ d}t\int \limits _t^x\frac{f(u)}{u}\,\text{ d}u\le \frac{1}{x}\int \limits _0^x H^*f(t)\,\text{ d}t. \end{aligned}$$

On the other hand, Fubini’s theorem gives that

$$\begin{aligned} H^*f(x)=\int \limits _x^\infty \frac{\text{ d}u}{u^2}\int \limits _x^u f(t)\,\text{ d}t\le \int \limits _x^\infty \frac{Hf(u)}{u}\,\text{ d}u. \end{aligned}$$

Using these estimates and applying Hardy’s inequalities [5, pp. 240, 244], we obtain that

$$\begin{aligned} \frac{1}{p^{\prime }}||Hf||_p\le ||H^*f||_p\le p||Hf||_p\quad \text{ for} \quad 1<p<\infty \end{aligned}$$

(1.1)

(as usual, $p^{\prime }=p/(p-1)).$

However, the constants in (1.1) are not optimal. The objective of this paper is to find optimal constants. Our main result is the following theorem.

Theorem 1.1

Let $f\in \mathcal M ^+(\mathbb R _+)$ and let $1<p<\infty .$ Then,

$$\begin{aligned} (p-1)||Hf||_p \le ||H^*f||_p\le (p-1)^{1/p}||Hf||_p \end{aligned}$$

(1.2)

if $1<p\le 2$, and

$$\begin{aligned} (p-1)^{1/p}||Hf||_p \le ||H^*f||_p\le (p-1)||Hf||_p \end{aligned}$$

(1.3)

if $2\le p<\infty .$ All constants in (1.2) and (1.3) are the best possible.

Clearly, the problem on relationships between various norms of Hardy operator and its dual is of independent interest (cf. [4]). At the same time, this problem has an equivalent formulation in terms of the difference operator $H\varphi -\varphi .$

Let $\varphi$ be a nonincreasing and nonnegative function on $\mathbb R _+$ such that $\varphi (+\infty )=0.$ The quantity $H\varphi -\varphi$ plays an important role in the analysis (see [2–4, 6, 7] and references therein). It is well known that the norms $||H\varphi -\varphi ||_p$ and $||\varphi ||_p$ ($1<p<\infty$) are equivalent (see [1, p. 384]). However, the sharp constant is known only in the following inequality.

Let $\varphi$ be a nonincreasing and nonnegative function on $\mathbb R _+$. Then, for any $p\ge 2$

$$\begin{aligned} ||H\varphi -\varphi ||_p\le (p-1)^{-1/p}||\varphi ||_p, \end{aligned}$$

(1.4)

and the constant is optimal.

This result was obtained in [7] for $p=2k\,\, (k\in \mathbb N )$ and in [2] for all $p\ge 2$ (we observe that (1.4) is a special case of the inequality proved in [2] for weighted $L^p$-norms).

We shall show that inequality (1.4) is equivalent to the first inequality in (1.3):

$$\begin{aligned} ||Hf||_p \le (p-1)^{-1/p}||H^*f||_p, \quad 2\le p<\infty . \end{aligned}$$

(1.5)

Thus, (1.5) can be derived from (1.4). However, below we give a simple direct proof of (1.5). Moreover, Theorem 1.1 has the following equivalent form.

Theorom 1.2

Let $\varphi$ be a nonincreasing and nonnegative function on $\mathbb R _+$ such that $\varphi (+\infty )=0$ and let $1<p<\infty .$ Then,

$$\begin{aligned} (p-1)||H\varphi -\varphi ||_p \le ||\varphi ||_p\le (p-1)^{1/p}||H\varphi -\varphi ||_p \end{aligned}$$

(1.6)

if $1<p\le 2$, and

$$\begin{aligned} (p-1)^{1/p}||H\varphi -\varphi ||_p \le ||\varphi ||_p\le (p-1)||H\varphi -\varphi ||_p \end{aligned}$$

(1.7)

if $2\le p<\infty .$ All constants in (1.6) and (1.7) are the best possible.

2 Proofs of main results

Proof of Theorem 1.1. Taking into account (1.1), we may assume that $Hf$ and $H^*f$ belong to $L^p(\mathbb R _+).$ We may also assume that $f(x)>0$ for all $x\in \mathbb R _+.$ Denote

$$\begin{aligned} I_p=\int \limits _0^\infty \left(\frac{1}{x}\int \limits _0^x f(t)\,\text{ d}t\right)^p\,\text{ d}x. \end{aligned}$$

Since $Hf\in L^p(\mathbb R _+),$ we have

$$\begin{aligned} Hf(x)=o(x^{-1/p})\quad \text{ as}\quad x\rightarrow 0+\quad \text{ or}\quad x\rightarrow +\infty . \end{aligned}$$

Thus, integrating by parts, we obtain

$$\begin{aligned} I_p=p^{\prime }\int \limits _0^{\infty } x^{1-p}f(x)\left(\int \limits _0^x f(t)\,\text{ d}t\right)^{p-1}\,\text{ d}x. \end{aligned}$$

(2.1)

Further, set

$$\begin{aligned} I_p^*=\int \limits _0^\infty \left(\int \limits _t^\infty \frac{f(x)}{x}\,\text{ d}x\right)^p\,\text{ d}t. \end{aligned}$$

(2.2)

First, we shall prove that

$$\begin{aligned} (p-1)I_p\le I_p^*\quad \text{ if}\quad 2\le p<\infty \end{aligned}$$

(2.3)

and

$$\begin{aligned} I_p^*\le (p-1)I_p \quad \text{ if}\quad 1< p\le 2. \end{aligned}$$

(2.4)

Set

$$\begin{aligned} \Phi (t,x)=\int \limits _t^x\frac{f(u)}{u}\,\text{ d}u, \quad 0<t\le x, \end{aligned}$$

and $G(t,x)= \Phi (t,x)^p.$ Since $G(t,t)=0,$ we have

$$\begin{aligned} \left(\int \limits _t^\infty \frac{f(x)}{x}\,\text{ d}x\right)^p = \int \limits _t^\infty G_x^{\prime }(t,x)\,\text{ d}x =p \int \limits _t^\infty \frac{f(x)}{x} \Phi (t,x)^{p-1}\,\text{ d}x. \end{aligned}$$

Thus, by Fubini’s theorem,

$$\begin{aligned} I_p^*&= p\int \limits _0^\infty \int \limits _t^\infty \frac{f(x)}{x} \Phi (t,x)^{p-1}\,\text{ d}x\,\text{ d}t\nonumber \\&= p\int \limits _0^\infty \frac{f(x)}{x}\int \limits _0^x \Phi (t,x)^{p-1}\,\text{ d}t\text{ d}x. \end{aligned}$$

(2.5)

On the other hand, Fubini’s theorem gives that

$$\begin{aligned} \int \limits _0^x f(t)\,\text{ d}t=\int \limits _0^x \Phi (t,x)\,\text{ d}t. \end{aligned}$$

Hence, by (2.1),

$$\begin{aligned} I_p=p^{\prime }\int \limits _0^\infty x^{1-p}f(x)\left(\int \limits _0^x \Phi (t,x)\,\text{ d}t\right)^{p-1}\,\text{ d}x. \end{aligned}$$

(2.6)

Comparing (2.1) with (2.2), we see that $I_2=I_2^*.$ In what follows, we assume that $p\not =2.$

Let $p>2.$ Then, by Hölder’s inequality

$$\begin{aligned} \left(\int \limits _0^x \Phi (t,x)\,\text{ d}t\right)^{p-1}\le x^{p-2}\int \limits _0^x \Phi (t,x)^{p-1}\,\text{ d}t. \end{aligned}$$

Thus, by (2.5) and (2.6),

$$\begin{aligned} I_p\le p^{\prime }\int \limits _0^\infty \frac{f(x)}{x}\int \limits _0^x \Phi (t,x)^{p-1}\,\text{ d}t\text{ d}x=\frac{I^*_p}{p-1}, \end{aligned}$$

and we obtain (2.3).

Let now $1<p< 2.$ Applying Hölder’s inequality, we get

$$\begin{aligned} \int \limits _0^x \Phi (t,x)^{p-1}\,\text{ d}t\le x^{2-p}\left(\int \limits _0^x \Phi (t,x)\,\text{ d}t\right)^{p-1}. \end{aligned}$$

Thus, by (2.5) and (2.6),

$$\begin{aligned} I_p^*\le p\int \limits _0^\infty x^{1-p}f(x)\left(\int \limits _0^x \Phi (t,x)\,\text{ d}t\right)^{p-1}\,\text{ d}x=(p-1)I_p, \end{aligned}$$

and we obtain (2.4).

Inequalities (2.3) and (2.4) imply the first inequality in (1.3) and the second inequality in (1.2), respectively.

Now, we shall show that

$$\begin{aligned} I_p^*\le (p-1)^pI_p\quad \text{ if}\quad 2< p<\infty \end{aligned}$$

(2.7)

and

$$\begin{aligned} (p-1)^pI_p\le I_p^* \quad \text{ if}\quad 1< p<2. \end{aligned}$$

(2.8)

Observe that by our assumption ($f>0$ and $H^*f\in L^p(\mathbb R _+))$,

$$\begin{aligned} 0<\int \limits _t^\infty \frac{f(x)}{x}\,\text{ d}x <\infty \quad \text{ for} \text{ all} \quad t>0. \end{aligned}$$

Thus, for any $q>0$, we have

$$\begin{aligned} \left(\int \limits _t^\infty \frac{f(x)}{x}\,\text{ d}x\right)^q=q\int \limits _t^\infty \frac{f(x)}{x}\left(\int \limits _x^\infty \frac{f(u)}{u}\,\text{ d}u\right)^{q-1}\,\text{ d}x. \end{aligned}$$

(2.9)

Applying this equality with $q=p$ in (2.2) and using Fubini’s theorem, we obtain

$$\begin{aligned} I_p^*=p\int \limits _0^\infty f(x)\left(\int \limits _x^\infty \frac{f(u)}{u}\,\text{ d}u\right)^{p-1}\,\text{ d}x. \end{aligned}$$

(2.10)

Further, apply (2.9) for $q=p-1$ and use again Fubini’s theorem. This gives

$$\begin{aligned} I_p^*&= p(p-1)\int \limits _0^\infty f(x)\int \limits _x^\infty \frac{f(u)}{u}\left(\int \limits _u^\infty \frac{f(v)}{v}\,\text{ d}v\right)^{p-2}\,\text{ d}u\,\text{ d}x\\&= p(p-1)\int \limits _0^\infty \frac{f(u)}{u}\left(\int \limits _u^\infty \frac{f(v)}{v}\,\text{ d}v\right)^{p-2}\int \limits _0^u f(x)\,\text{ d}x\,\text{ d}u. \end{aligned}$$

Set

$$\begin{aligned} \varphi (u)=\frac{f(u)^{1/(p-1)}}{u}\int \limits _0^u f(x)\,\text{ d}x \end{aligned}$$

and

$$\begin{aligned} \psi (u)=f(u)^{(p-2)/(p-1)}\left(\int \limits _u^\infty \frac{f(x)}{x}\,\text{ d}x\right)^{p-2} \end{aligned}$$

(recall that $f>0$). Then, we have

$$\begin{aligned} I_p^*=p(p-1)\int \limits _0^\infty \varphi (u)\psi (u)\,\text{ d}u. \end{aligned}$$

(2.11)

Furthermore, by (2.1),

$$\begin{aligned} \int \limits _0^\infty \varphi (u)^{p-1}\,\text{ d}u=\int \limits _0^\infty \frac{f(u)}{u^{p-1}}\left(\int \limits _0^u f(x)\,\text{ d}x\right)^{p-1}\,\text{ d}u= \frac{I_p}{p^{\prime }}, \end{aligned}$$

(2.12)

and by (2.10),

$$\begin{aligned} \int \limits _0^\infty \psi (u)^{(p-1)/(p-2)}\,\text{ d}u= \int \limits _0^\infty f(u)\left(\int \limits _u^\infty \frac{f(x)}{x}\,\text{ d}x\right)^{p-1}\,\text{ d}u=\frac{I_p^*}{p} \end{aligned}$$

(2.13)

for any $p>1, \, p\not =2.$

Let $p>2.$ Applying in (2.11) Hölder’s inequality with the exponent $p-1$ and taking into account equalities (2.12) and (2.13), we obtain

$$\begin{aligned} I_p^*\le p(p-1)\left(\frac{I_p}{p^{\prime }}\right)^{1/(p-1)}\left(\frac{I_p^*}{p}\right)^{(p-2)/(p-1)}. \end{aligned}$$

This implies (2.7), which is the second inequality in (1.3).

Let now $1< p<2.$ Applying in (2.11) Hölder’s inequality with the exponent $p-1\in (0,1)$ (see [5, p. 140]), and using equalities (2.12) and (2.13), we get

$$\begin{aligned} I_p^*\ge p(p-1)\left(\frac{I_p}{p^{\prime }}\right)^{1/(p-1)}\left(\frac{I_p^*}{p}\right)^{(p-2)/(p-1)}. \end{aligned}$$

Thus,

$$\begin{aligned} (I_p^*)^{1/(p-1)}\ge (p-1)^{p/(p-1)}I_p^{1/(p-1)}. \end{aligned}$$

This implies (2.8), which is the first inequality in (1.2).

It remains to show that the constants in (1.2) and (1.3) are optimal. First, set $f_\varepsilon (x)=\chi _{[1,1+\varepsilon ]}(x)\,\, (\varepsilon >0).$ Then,

$$\begin{aligned} ||Hf_\varepsilon ||_p^p =\int \limits _1^{1+\varepsilon } x^{-p}(x-1)^p\,\text{ d}x + \varepsilon ^p\int \limits _{1+\varepsilon }^\infty x^{-p}\,\text{ d}x. \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\varepsilon ^p(1+\varepsilon )^{1-p}}{p-1}\le ||Hf_\varepsilon ||_p^p\le \frac{\varepsilon ^p(1+\varepsilon )^{1-p}}{p-1} +\varepsilon ^{p+1}. \end{aligned}$$

Further,

$$\begin{aligned} ||H^*f_\varepsilon ||_p^p&= \int \limits _0^1\left(\int \limits _1^{1+\varepsilon }\frac{\text{ d}t}{t}\right)^p\,\text{ d}x +\int \limits _1^{1+\varepsilon }\left(\int \limits _x^{1+\varepsilon }\frac{\text{ d}t}{t}\right)^p\,\text{ d}x\\&= (\ln (1+\varepsilon ))^p+\int \limits _1^{1+\varepsilon }\left(\ln \frac{1+\varepsilon }{x}\right)^p\,\text{ d}x. \end{aligned}$$

Thus,

$$\begin{aligned} (\ln (1+\varepsilon ))^p\le ||H^*f_\varepsilon ||_p^p\le (\ln (1+\varepsilon ))^p(1 +\varepsilon ). \end{aligned}$$

Using these estimates, we obtain that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0+}\frac{||Hf_\varepsilon ||_p}{||H^*f_\varepsilon ||_p}=(p-1)^{-1/p}. \end{aligned}$$

It follows that the constants in the right-hand side of (1.2) and the left-hand side of (1.3) cannot be improved.

Let $1<p<2.$ Set $f_\varepsilon (x)=x^{\varepsilon -1/p}\chi _{[0,1]}(x)\,\, (0<\varepsilon <1/p).$ Then,

$$\begin{aligned} ||Hf_\varepsilon ||_p^p\ge \int \limits _0^1\left(\frac{1}{x}\int \limits _0^x t^{\varepsilon -1/p}\,\text{ d}t\right)^p\,\text{ d}x=\frac{p^p}{\varepsilon p(p-1+\varepsilon p)^p}. \end{aligned}$$

On the other hand,

$$\begin{aligned} ||H^*f_\varepsilon ||_p^p\le \left(\frac{1}{p}-\varepsilon \right)^{-p}\int \limits _0^1 x^{(\varepsilon -1/p)p}\,\text{ d}x=\frac{p^p}{\varepsilon p(1-\varepsilon p)^p}. \end{aligned}$$

Hence,

$$\begin{aligned} \mathop {\underline{\text{ lim}}}\limits _{\varepsilon \rightarrow 0+}\frac{||Hf_\varepsilon ||_p}{||H^*f_\varepsilon ||_p}\ge \frac{1}{p-1}. \end{aligned}$$

This implies that the constant in the left-hand side of (1.2) is optimal.

Let now $p>2.$ Set $f_\varepsilon (x)=x^{-\varepsilon -1/p}\chi _{[1,+\infty )}(x)\,\, (0<\varepsilon <1/p^{\prime }).$ Then

$$\begin{aligned} ||H^*f_\varepsilon ||_p^p\ge \int \limits _1^\infty \left(\int \limits _x^\infty \frac{\text{ d}t}{t^{1+1/p+\varepsilon }}\right)^p\,\text{ d}x=\frac{p^p}{\varepsilon p(1+\varepsilon p)^p} \end{aligned}$$

and

$$\begin{aligned} ||Hf_\varepsilon ||_p^p\le \int \limits _1^\infty \left(\frac{1}{x}\int \limits _0^x\frac{\text{ d}t}{t^{1/p+\varepsilon }}\right)^p\,\text{ d}x=\frac{p^p}{\varepsilon p(p-1-\varepsilon p)^p}. \end{aligned}$$

Thus,

$$\begin{aligned} \mathop {\underline{\text{ lim}}}\limits _{\varepsilon \rightarrow 0+}\frac{||H^*f_\varepsilon ||_p}{||Hf_\varepsilon ||_p}\ge p-1. \end{aligned}$$

This shows that the constant in the right-hand side of (1.3) is the best possible. The proof is completed.

Remark 2.1

We emphasize that in Theorem 1.1, we do not assume that $f$ belongs to $L^p(\mathbb R _+).$ It is clear that the condition $Hf\in L^p(\mathbb R _+)$ does not imply that $f\in L^p(\mathbb R _+).$ For example, let $f(x)=|x-1|^{-1/p}\chi _{[1,2]}(x),\,p>1.$ Then,

$$\begin{aligned} Hf(x)=0\quad \text{ for}\quad x\in [0,1]\quad \text{ and}\quad Hf(x)\le \frac{p^{\prime }}{x}\quad \text{ for}\quad x\ge 1. \end{aligned}$$

Thus, $Hf\in L^p(\mathbb R _+)$, but $f\not \in L^p(\mathbb R _+).$

Now, we shall show that Theorems 1.1 and 1.2 are equivalent. First, we observe that without the loss of generality, we may assume that a function $\varphi$ in Theorem 1.2 is locally absolutely continuous on $\mathbb R _+.$ Indeed, let $\varphi$ be a nonincreasing and nonnegative function on $\mathbb R _+$ such that $\varphi (+\infty )=0.$ Set

$$\begin{aligned} \varphi _n(x)=n\int \limits _x^{x+1/n} \varphi (t)\,\text{ d}t \quad (n\in \mathbb N ). \end{aligned}$$

Then, functions $\varphi _n$ are nonincreasing, nonnegative, and locally absolutely continuous on $\mathbb R _+.$ Besides, the sequence $\{\varphi _n(x)\}$ increases for any $x\in \mathbb R _+$ and converges to $\varphi (x)$ at every point of continuity of $\varphi .$ By the monotone convergence theorem, $H\varphi _n(x)\rightarrow H\varphi (x)$ as $n\rightarrow \infty$ for any $x\in \mathbb R _+$, and $||\varphi _n||_p\rightarrow ||\varphi ||_p.$ Furthermore, in Theorem 1.2, we may assume that $\varphi \in L^p(\mathbb R _+)$ (in conditions of this theorem, the norms $||H\varphi -\varphi ||_p$ and $||\varphi ||_p$ are equivalent [1, p. 384]). Using this assumption, Hardy’s inequality, and the dominated convergence theorem, we obtain that $||H\varphi _n-\varphi _n||_p\rightarrow ||H\varphi -\varphi ||_p$.

Let $\varphi$ be a nonincreasing, nonnegative, and locally absolutely continuous function on $\mathbb R _+$ such that $\varphi (+\infty )=0.$ Then,

$$\begin{aligned} H\varphi (x)-\varphi (x)&= \frac{1}{x}\int \limits _0^x[\varphi (t)-\varphi (x)]\,\text{ d}t\\&= \frac{1}{x}\int \limits _0^x\int \limits _t^x|\varphi ^{\prime }(u)|\,\text{ d}u\text{ d}t=\frac{1}{x}\int \limits _0^x u|\varphi ^{\prime }(u)|\,\text{ d}u. \end{aligned}$$

Set $u|\varphi ^{\prime }(u)|=f(u).$ Since $\varphi (+\infty )=0,$ we have

$$\begin{aligned} \varphi (x)=\int \limits _x^\infty |\varphi ^{\prime }(u)|\,\text{ d}u=\int \limits _x^\infty \frac{f(u)}{u}\,\text{ d}u. \end{aligned}$$

Thus,

$$\begin{aligned} H\varphi (x)-\varphi (x)=\frac{1}{x}\int \limits _0^x f(u)\,\text{ d}u = Hf(x) \end{aligned}$$

(2.14)

and

$$\begin{aligned} \varphi (x)=\int \limits _x^\infty \frac{f(u)}{u}\,\text{ d}u=H^*f(x). \end{aligned}$$

(2.15)

Conversely, if $f\in \mathcal M ^+(\mathbb R _+)$ and

$$\begin{aligned} \int \limits _0^x f(u)\,\text{ d}u<\infty \quad \text{ for} \text{ any}\quad x>0, \end{aligned}$$

we define $\varphi$ by (2.15) and then we have equality (2.14). These arguments show the equivalence of Theorems 1.1 and 1.2.