# Sharp constant of Hardy operators corresponding to general positive measures

## Abstract

We investigate a new kind of Hardy operator $$H_{\mu}$$ with respect to arbitrary positive measures μ and prove that $$H_{\mu}$$ is bounded on $$L^{p}(d\mu)$$ with an upper constant $$p/(p-1)$$. Moreover, we characterize a sufficient condition about the measure which makes $$p/(p-1)$$ to be the $$L^{p}$$-norm of $$H_{\mu}$$.

## 1 Introduction

Let μ be a positive measure on $$[0,\infty)$$ and f be a nonnegative μ-measurable function. Define Hardy operator with respect to the measure μ by

$$H_{\mu}f(x)=\frac{1}{\mu([0,x])} \int_{[0,x]}f(t)\,d\mu(t),$$
(1)

if $$0<\mu{([0,x])}<\infty$$, and set $$H_{\mu}f(x)=0$$, if $$\mu{([0,x])}=0$$ or ∞.

Observe that if μ is Lebesgue measure, then $$H_{\mu}$$ becomes the classical Hardy operator

$$Hf(x)=\frac{1}{x} \int_{0}^{x}f(t)\,dt,$$
(2)

and if $$\mu=\sum_{k=1}^{\infty}\delta_{k}$$, then $$H_{\mu}$$ becomes the discrete Hardy operator

$$\mathcal{H}f(k)=\frac{f(1)+\cdots+f(k)}{k}.$$

For $$1< p<\infty$$, reference [1] showed that the two operators are bounded on $$L^{p}$$ and $$l^{p}$$ respectively. Moreover, for both, the best constants are $$p/(p-1)$$ and the maximizing functions do not exist. We refer the reader to [26] for the background material and further references.

Hardy operator has a close relationship with Hardy–Littlewood maximal operator. From the point of rearrangement, Hf is equivalent to Mf (see reference [7]). In reference [8], Grafakos considered the $$L^{p}$$-boundedness for the maximal functions associated with general measures. In this paper, we shall discuss the sharp problems about $$H_{\mu}$$. We will show that the operator $$H_{\mu}$$ is bounded on $$L^{p}(d\mu)$$ with an upper bound no more than $$p/(p-1)$$. Furthermore, we will characterize a sufficient condition about μ such that $$\Vert H_{\mu} \Vert _{L^{p}\rightarrow L^{p}}=p/(p-1)$$.

From the definition about $$H_{\mu}$$, it is not necessary to consider the points x such that $$\mu([0,x])=0$$ or ∞. Therefore, we let

$$a=\inf\bigl\{ x:\mu\bigl([0,x]\bigr)>0\bigr\} ,$$

and

$$b= \textstyle\begin{cases} \infty& \mbox{if } B= \emptyset,\\ \inf B& \mbox{if } B\neq\emptyset, \end{cases}$$

where B denotes the set $$\{x:\mu([0,x])=\infty \mbox{ or } \mu([x,\infty))=0\}$$. Then we call that the measure μ is supported in the interval $$[a,b]$$.

For the case of weak type inequality, the best constant from $$L^{p}(d\mu)$$ to $$L^{p,\infty}(d\mu)$$ is always 1.

### Theorem 1.1

Let μ be a positive measure on $$[0,\infty]$$ and $$1\leq p<\infty$$. Then we have

$$\Vert H_{\mu} \Vert _{L^{p}(d\mu)\rightarrow L^{p,\infty}(d\mu)}=1.$$

### Theorem 1.2

Suppose that μ is supported in $$[a,b]$$ and $$f\in L^{p}(d\mu)$$ with $$1< p<\infty$$. For $$f\neq0$$, define

$$\mathcal{R}_{\mu}(f)=\frac{ \Vert H_{\mu}f \Vert _{L^{p}(d\mu)}}{ \Vert f \Vert _{L^{p}(d\mu)}}.$$

Then the following statements hold:

1. (i)

$$\Vert H_{\mu}f \Vert _{L^{p}(d\mu)}\leq\frac{p}{p-1} \Vert f \Vert _{L^{p}(d\mu)}$$ holds for arbitrary positive measure μ.

2. (ii)

There exists no function f such that $$\mathcal{R}_{\mu}(f)=\frac{p}{p-1}$$ holds.

### Theorem 1.3

If μ satisfies one of the following conditions:

Condition 1. $$\mu([a,b])=\infty$$ and

$$\lim_{x\rightarrow b}\frac{\mu([a,x])}{\mu([a,x))}=1;$$

Condition 2. $$\{a\}$$ is not an atom of μ, and

$$\lim_{x\rightarrow a}\frac{\mu([a,x])}{\mu([a,x))}=1,$$

then we have

$$\sup_{f\in L^{p}(d\mu),f\neq0}\mathcal{R}_{\mu}(f)=\frac{p}{p-1}.$$

We remark that there indeed exist some measures so that

\begin{aligned} \sup_{f\in L^{p}(d\mu),f\neq0}\mathcal{R}_{\mu}(f)< \frac{p}{p-1}. \end{aligned}
(3)

For example, it is easy to know that the Dirac measure $$\delta_{0}$$ satisfies inequality (3). In this paper, we will give some more complex counterexamples.

## 2 Preliminary and lemmas

In the study of sharp problems, the rearrangement of function is a very useful tool. Let

$$d_{f}(s)=\mu\bigl(\bigl\{ \vert f \vert >s\bigr\} \bigr).$$

Then the rearrangement of f is defined by

$$f^{*}(t)=\inf\bigl\{ s>0:d_{f}(s)\leq t\bigr\} .$$

By the properties of the rearrangement, we can easily have

$$\Vert f \Vert _{L^{p}(d\mu)}= \bigl\Vert f^{*} \bigr\Vert _{L^{p}(dm)}.$$

We refer the reader to [9] for more properties of rearrangement. In reference [1], Hardy gave the following result.

### Lemma 2.1

(G.H. Hardy and J.E. Littlewood)

Let $$(X,\mu)$$ be a measurable space. If $$f, g\in\mathcal{M}(X,\mu)$$, then

$$\int_{X} \vert fg \vert \,d\mu\leq \int_{0}^{\infty}f^{*}(t)g^{*}(t)\,dt$$

holds.

Moreover, the theory of rearrangement plays an important role in proving the existence of maximizing function. This is because of the following lemma introduced by Lieb [10].

### Lemma 2.2

Suppose that $$(M, \Sigma, \mu)$$ and $$(M',\Sigma',\mu')$$ are two measure spaces. Let X and Y be $$L^{p}(M,\Sigma,\mu)$$ and $$L^{q}(M',\Sigma',\mu')$$ with $$1\leq p\leq q<\infty$$. Let A be a bounded linear operator from X to Y. For $$f\in X$$ with $$f\neq0$$, set

$$\mathcal{R}(f)=\frac{ \Vert Af \Vert _{Y}}{ \Vert f \Vert _{X}}$$

and

$$N=\sup\bigl\{ \mathcal{R}(f):f\neq0\bigr\} .$$

Let $$\{f_{j}\}$$ be a uniform norm-bounded maximizing sequence for N, and assume that $$f_{j}\rightarrow f\neq0$$ and that $$A(f_{j})\rightarrow A(f)$$ pointwise almost everywhere. Then f maximizes, i.e., $$\mathcal{R}(f)=N$$.

## 3 The boundedness of weak-$$L^{p}$$

In this section, we first prove Theorem 1.1. For the sake of clarity, we define a function as

$$F_{\mu}(x):=\mu\bigl([0,x]\bigr).$$

Obviously $$F_{\mu}$$ increases as $$x\rightarrow\infty$$. It follows from Lemma 2.1 and the definition of $$H_{\mu}$$ that

\begin{aligned} H_{\mu}f(x)&=\frac{1}{\mu([0,x])} \int_{[0,x]}f(t)\,d\mu(t) \\ &\leq \frac{1}{F_{\mu}(x)} \int_{[0,F_{\mu}(x)]}f^{*}(t)\,dt \\ &=Hf^{*} \bigl(F_{\mu}(x) \bigr). \end{aligned}
(4)

Let

$$E_{\mu}^{f^{*}}(\lambda):= \bigl\{ x:Hf^{*}\bigl(F_{\mu}(x) \bigr)>\lambda \bigr\} .$$

Note that $$f^{*}$$ decreases, so we easily have that $$Hf^{*}$$ decreases as well. If we take

$$x_{0}=\sup\bigl\{ x:Hf^{*}\bigl(F_{\mu}(x)\bigr)>\lambda\bigr\} ,$$

then it implies that

$$E_{\mu}^{f^{*}}(\lambda)=[0,x_{0}).$$

Thus, we can obtain that

$$\bigl\{ x:Hf^{*}(x)>\lambda\bigr\} \supset\bigl[0,F_{\mu}(x_{0})\bigr).$$

We conclude that

$$\mu \bigl(\bigl\{ x:Hf^{*}\bigl(F_{\mu}(x)\bigr)>\lambda \bigr\} \bigr)\leq F_{\mu}(x_{0}) \leq \bigl\vert \bigl\{ x:Hf^{*}(x)>\lambda\bigr\} \bigr\vert ,$$
(5)

where $$\vert \cdot \vert$$ denotes the Lebesgue measure. It follows from inequalities (4) and (5) that

\begin{aligned} \frac{\sup_{\lambda>0}\lambda \mu(\{x:H_{\mu}f(x)>\lambda\})^{\frac{1}{p}}}{ \Vert f \Vert _{L^{p}(d\mu)}}&\leq \frac{\sup_{\lambda>0}\lambda \mu(\{x:Hf^{*}(F_{\mu}(x))>\lambda\})^{\frac{1}{p}}}{ \Vert f^{*} \Vert _{L^{p}(dm)}} \\ &\leq \frac{\sup_{\lambda>0}\lambda \vert \{x:Hf^{*}(x)>\lambda\} \vert ^{\frac{1}{p}}}{ \Vert f^{*} \Vert _{L^{p}(dm)}}. \end{aligned}
(6)

Since $$f^{*}\in L^{p}(dm)$$, by Hölder’s inequality, we have that

$$Hf^{*}(x)=\frac{1}{x} \int_{0}^{x}f^{*}(t)\,dt\leq \biggl(\frac{1}{x} \int_{0}^{x} \bigl\vert f^{*}(t) \bigr\vert ^{p}\,dt \biggr)^{\frac{1}{p}} \leq x^{-\frac{1}{p}} \bigl\Vert f^{*} \bigr\Vert _{L^{p}(dm)}.$$
(7)

Thus it is obvious to obtain that

$$\bigl\vert \bigl\{ x:Hf^{*}(x)>\lambda\bigr\} \bigr\vert \leq \bigl\vert \bigl\{ x:x^{-\frac{1}{p}} \bigl\Vert f^{*} \bigr\Vert _{L^{p}(dm)}>\lambda \bigr\} \bigr\vert = \frac{ \Vert f^{*} \Vert _{L^{p}(dm)}^{p}}{\lambda^{p}}.$$
(8)

From inequality (6) and inequality (8), we have

$$\frac{\sup_{\lambda>0}\lambda \mu(\{x:H_{\mu}f(x)>\lambda\})^{\frac{1}{p}}}{ \Vert f \Vert _{L^{p}(d\mu)}}\le1.$$

That is,

$$\frac{ \Vert H_{\mu}f \Vert _{L^{p,\infty}(d\mu)}}{ \Vert f \Vert _{L^{p}(d\mu)}}\le1$$
(9)

holds. This is equivalent to

$$\Vert H_{\mu} \Vert _{L^{p}(d\mu)\rightarrow L^{p,\infty}(d\mu)}\leq1.$$
(10)

Next it suffices to show that the constant 1 is sharp for inequality (10).

Take $$0\leq x_{1}< x_{2}<\infty$$ such that $$0<\mu ([x_{1},x_{2}] )<\infty$$. Let $$g=\chi_{[x_{1},x_{2}]}$$. It is easy to obtain

$$\Vert H_{\mu}g \Vert _{L^{p,\infty}(d\mu)}= \Vert g \Vert _{L^{p}(d\mu)}.$$

The proof is completed.

## 4 $$L^{p}$$-boundedness of the operator $$H_{\mu}$$ with upper bound $$p/(p-1)$$

Now we will show the results (i) and (ii) of Theorem 1.2.

### Proof

Following the proof of (5), we obtain

$$\int_{[0,\infty]}\bigl(f\bigl(\mu\bigl([0,x]\bigr)\bigr) \bigr)^{p}\,d\mu(x)\leq \int_{[0,\infty]}f^{p}(x)\,dx.$$
(11)

By inequality (11), we conclude that

\begin{aligned} \Vert H_{\mu}f \Vert _{L^{p}(\mathbb {R}_{+},d\mu)}&= \biggl( \int_{\mathbb {R}_{+}} \biggl\vert \frac{1}{\mu([0,x])} \int_{[0,x]}f(t)\,d\mu(t) \biggr\vert ^{p}\,d\mu(x) \biggr)^{\frac{1}{p}} \\ &\leq \biggl( \int_{\mathbb {R}_{+}} \biggl\vert \frac{1}{F_{\mu}(x)} \int_{[0,F_{\mu}(x)]}f^{*}(t)\,dt \biggr\vert ^{p}\,d\mu (x) \biggr)^{\frac{1}{p}} \\ &= \biggl( \int_{\mathbb {R}_{+}} \bigl\vert Hf^{*}\bigl(F_{\mu}(x)\bigr) \bigr\vert ^{p}\,d\mu(x) \biggr)^{\frac{1}{p}} \\ &\leq \biggl( \int_{\mathbb {R}_{+}} \bigl\vert Hf^{*}(x) \bigr\vert ^{p}\,dx \biggr)^{\frac{1}{p}}. \end{aligned}
(12)

It follows from the inequality of classical Hardy operator that

$$\biggl( \int_{\mathbb {R}_{+}} \bigl\vert Hf^{*}(x) \bigr\vert ^{p}\,dx \biggr)^{\frac{1}{p}} \leq\frac{p}{p-1} \bigl\Vert f^{*} \bigr\Vert _{L^{p} (dm)} =\frac{p}{p-1} \Vert f \Vert _{L^{p}(d\mu)}.$$
(13)

Combining inequality (12) with inequality (13), we have

$$\Vert H_{\mu}f \Vert _{L^{p}(\mathbb {R}_{+},d\mu)}\leq\frac{p}{p-1} \Vert f \Vert _{L^{p}(d\mu)}.$$

Since the sharp function for the classical Hardy operator does not exist, it is easy to know from inequality (12) that there exists no function f such that $$\mathcal{R}_{\mu}(f)=\frac{p}{p-1}$$. The proof of the result (ii) of Theorem 1.2 is completed. □

## 5 A characterization of the measure μ which ensures $$\sup_{f\neq0}\mathcal{R}_{\mu}(f)=p/(p-1)$$

In this section, we try to characterize the measure μ which ensures $$\sup_{f\neq0}\mathcal{R}_{\mu}(f)=p/(p-1)$$. We regard μ as a complete atom measure by giving an appropriate partition on $$[0,\infty]$$. We first present a partition on $$[0,\infty]$$ by the following two lemmas.

### Lemma 5.1

Let μ be a positive measure that is supported on $$[0,\infty]$$. If $$\mu([0,\infty])=\infty$$ and

$$\lim_{x\rightarrow\infty}\frac{\mu(\{x\})}{\mu([0,x])}=0,$$

then there exists a partition on $$[0,\infty]$$ as

$$I_{0}=[0,x_{1}],\qquad I_{1}=(x_{1},x_{2}], \ldots,\qquad I_{k}=(x_{k},x_{k+1}], \ldots,$$

such that

$$\mu(I_{k+1})\geq\mu(I_{k}),$$

and

$$\lim_{k\rightarrow\infty}\frac{\mu({I_{k}})}{\mu([0,x_{k+1}])}=0.$$

### Proof

Let $$x_{1}$$ be any positive number. Denote $$I_{0}=[0,x_{1}]$$. Since μ is supported on $$[0,\infty]$$, we have

$$\mu(I_{0})>0.$$

For $$k=2$$, we let

$$x_{2}=\inf\bigl\{ x:\mu \bigl((x_{1},x] \bigr)\geq\mu \bigl([0,x_{1}]\bigr)\bigr\} .$$

For $$k>2$$, we let

$$x_{k}=\inf\bigl\{ x:\mu \bigl((x_{k-1},x] \bigr)\geq\mu \bigl((x_{k-2},x_{k-1}]\bigr)\bigr\} .$$

Denote $$I_{k}=[x_{k-1},x_{k}]$$ with $$k=2,3,\ldots$$ . Since $$\mu([0,\infty])=\infty$$, we easily have

$$\lim_{k\rightarrow\infty}x_{k}=\infty.$$

Thus, $$\{I_{k}\}$$ obviously constitutes a partition of $$[0,\infty]$$.

We first show that

$$\mu(I_{k})\geq\mu(I_{k-1})$$

and

\begin{aligned} \mu\bigl((x_{k},x_{k+1})\bigr)\leq \mu(I_{k-1}). \end{aligned}
(14)

By our construction, for any $$x>x_{k+1}$$, it follows that

$$\mu\bigl((x_{k},x]\bigr)\geq\mu\bigl((I_{k-1})\bigr).$$

Thus the property of measure implies that

$$\mu(I_{k}) =\mathop{\lim_{x>x_{k+1}}}_{ x\rightarrow x_{k+1}}\mu\bigl((x_{k},x]\bigr)\geq \mu(I_{k-1}).$$

Moreover, if $$x_{k}< x< x_{k+1}$$, then $$\mu([x_{k},x])<\mu(I_{k-1})$$. Thus, it follows that

$$\mu\bigl((x_{k},x_{k+1})\bigr)=\mathop{\lim_{x< x_{k+1} }}_{ x\rightarrow x_{k+1}} \mu\bigl((x_{k},x]\bigr)\leq\mu(I_{k-1}).$$

To complete the proof, it remains to show that

$$\lim_{k\rightarrow\infty}\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}=0.$$

This is equivalent to prove that, for any $$\epsilon>0$$, there is an integer $$N>0$$ such that

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq2\epsilon$$

holds for $$k\geq N$$.

In order to prove this result, we divide the set $$\mathbb {Z}^{+}\setminus\{1\}$$ into two parts:

$$F_{\epsilon} := \biggl\{ k\in\mathbb{Z}:k\geq2, \frac{\mu(\{x_{k}\})}{\mu ((x_{k-1},x_{k}))}< \epsilon \biggr\}$$
(15)

and

$$G_{\epsilon}:= \biggl\{ k\in\mathbb{Z}:k\geq2, \frac{\mu(\{x_{k}\} )}{\mu((x_{k-1},x_{k}))}\geq\epsilon \biggr\} .$$
(16)

By definition (16), if $$k\in G_{\epsilon}$$, then we have

\begin{aligned} \mu(I_{k-1})\leq \biggl(1+\frac{1}{\epsilon} \biggr)\mu \bigl(\{x_{k}\}\bigr). \end{aligned}
(17)

We discuss the problem in two cases:

Case I. $$G_{\epsilon}$$ is not a finite set.

Case II. $$G_{\epsilon}$$ is a finite set.

If $$G_{\epsilon}$$ is not a finite set, then by equality $$\lim_{x\rightarrow\infty}\frac{\mu(\{x\})}{\mu([0,x])}=0$$, there exists an integer $$N\in G_{\epsilon}$$ such that, for any $$k\geq N$$,

\begin{aligned} \frac{\mu(\{x_{k}\})}{\mu([0,x_{k}])}< \frac{\epsilon^{2}}{1+\epsilon}. \end{aligned}
(18)

Thus if $$k>N$$ and $$k\in G_{\epsilon}$$, then by inequalities (17) and (18), we have

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq\epsilon.$$
(19)

On the other hand, if $$k> N$$ and $$k\in F_{\epsilon}$$, since $$G_{\epsilon}$$ is not a finite integer and $$N\in G_{\epsilon}$$, we can find a series of integers $$k_{0}$$, $$k_{0}+1$$, …, k, such that $$k_{0}\in G_{\epsilon}$$, and

$$k_{0}+1,\ldots,k \in F_{\epsilon}.$$

By the definition of $$F_{\epsilon}$$ and inequality (14), we can conclude that if $$i\in F_{\epsilon}$$, then

\begin{aligned} \mu\bigl((x_{i-1},x_{i}]\bigr)&=\mu\bigl((x_{i-1},x_{i}) \bigr)+\mu\bigl(\{x_{i}\}\bigr) \\ &\leq(1+\epsilon)\mu\bigl((x_{i-1},x_{i})\bigr) \\ &\leq(1+\epsilon)\mu\bigl((x_{i-2},x_{i-1}]\bigr). \end{aligned}
(20)

It immediately implies from inequality (20) that

\begin{aligned} \mu\bigl((x_{k_{0}},x_{k}]\bigr)&=\sum _{i=k_{0}+1}^{k}\mu\bigl((x_{i-1},x_{i}]\bigr) \\ &\geq\sum_{i=k_{0}+1}^{k}(1+ \epsilon)^{i-k}\mu \bigl((x_{k-1},x_{k}]\bigr) \\ &=\mu\bigl((x_{k-1},x_{k}]\bigr)\frac{1- (\frac{1}{1+\epsilon} )^{k-k_{0}}}{1-\frac{1}{1+\epsilon}}. \end{aligned}
(21)

Thus, by inequality (21), we have

\begin{aligned} \frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}&\leq\frac{\mu ((x_{k-1},x_{k}])}{\mu((x_{k_{0}-1},x_{k}])} \leq\frac{1-\frac{1}{1+\epsilon}}{1- (\frac{1}{1+\epsilon } )^{k-k_{0}}} \\ &\leq\frac{\epsilon}{1- (\frac{1}{1+\epsilon} )^{k-k_{0}}}. \end{aligned}
(22)

Since $$k_{0}\in G_{\epsilon}$$, inequalities (14) and (20) imply

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq(1+\epsilon)^{k-k_{0}} \frac {\mu({I_{k_{0}-1}})}{\mu([0,x_{k}])}\leq(1+\epsilon)^{k-k_{0}}\epsilon.$$
(23)

If $$(1+\epsilon)^{k-k_{0}}> 2$$, by inequality (22), we have

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq2\epsilon.$$

If $$(1+\epsilon)^{k-k_{0}}\leq2$$, by inequality (23), we have

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq2\epsilon.$$

At last, we conclude that if $$k>N$$ and $$k\in F_{\epsilon}$$, then

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq2\epsilon.$$
(24)

The proof of Case I is complete.

If $$G_{\epsilon}$$ is a finite set, then we can find an integer $$k_{0}$$ such that $$k\in F_{\epsilon}$$ for $$k>k_{0}$$. Then, by inequality (22), we can find a big enough integer N such that

$$\frac{\mu({I_{k-1}})}{\mu([0,x_{k}])}\leq2\epsilon$$

if $$k\geq N$$. The proof is completed. □

### Lemma 5.2

Suppose that μ is supported in $$[0,\infty]$$. If $$\mu(\{0\})=0$$ and $$\lim_{x\rightarrow0}\frac{\mu([0,x])}{\mu([0,x))}=1$$, then there exists a partition on $$(0,1]$$,

$$(x_{1},1], (x_{2},x_{1}], \ldots, (x_{k},x_{k-1}], \ldots,$$

such that $$\lim_{k\rightarrow\infty}x_{k}=0$$ and

$$\lim_{k\rightarrow\infty}\frac{\mu((x_{k},x_{k-1}])}{\mu([0,x_{k-1}])}=0.$$

### Proof

Without loss of generality, suppose

$$\mu\bigl([0,1]\bigr)=\sum_{k=1}^{\infty} \frac{1}{k^{2}}.$$

If $$\mu(\{1\})<1$$, then we set $$k_{0}=0$$. If $$\mu(\{1\})\geq1$$, then we set

$$k_{0}=\max \Biggl\{ m:\sum_{k=1}^{m} \frac{1}{k^{2}}\leq\mu\bigl(\{1\}\bigr) \Biggr\} .$$

It is easy to see that

$$\sum_{k=1}^{k_{0}+1}\frac{1}{k^{2}}>\mu\bigl( \{1\}\bigr).$$

Then we can find a positive real number $$x_{1}<1$$ such that

$$x_{1}=\sup \Biggl\{ x:\mu\bigl((x,1]\bigr)\geq\sum_{k=1}^{k_{0}+1} \frac{1}{k^{2}} \Biggr\} .$$

Proceeding in this way, we set

\begin{aligned} k_{i}=\max \Biggl\{ m:\sum_{k=1}^{m} \frac{1}{k^{2}}\leq\mu\bigl([x_{i},1]\bigr) \Biggr\} \end{aligned}
(25)

and

\begin{aligned} x_{i+1}=\sup \Biggl\{ x:\mu\bigl((x,1]\bigr)\geq\sum _{k=1}^{k_{i}+1}\frac{1}{k^{2}} \Biggr\} \end{aligned}
(26)

for $$i\geq1$$. By (27), (25), and (26), we can conclude

\begin{aligned} \sum_{k=1}^{k_{i}} \frac{1}{k^{2}}\leq\mu\bigl([x_{i},1]\bigr) \leq\mu\bigl((x_{i+1},1]\bigr) \leq\sum_{k=1}^{k_{i}+1}\frac{1}{k^{2}}\leq\mu \bigl([x_{i+1},1]\bigr). \end{aligned}
(27)

It is easy to see that $$x_{i}>x_{i+1}$$ and

$$\lim_{i\rightarrow\infty}\mu\bigl([x_{i},1]\bigr)\geq\lim _{i\rightarrow \infty}\sum_{k=1}^{k_{i}} \frac{1}{k^{2}}=\mu\bigl((0,1]\bigr).$$

Thus we have $$\lim_{i\rightarrow\infty}x_{i}=0$$. It is easy to see that

$$(x_{1},1], (x_{2},x_{1}], \ldots, (x_{k},x_{k-1}], \ldots,$$

divide $$(0,1]$$. It can be implied from inequality (27) that

$$\mu\bigl([x_{i},1]\bigr)+\frac{1}{(k_{i}+1)^{2}}\geq\sum _{k=1}^{k_{i}+1}\frac{1}{k^{2}}\geq \mu\bigl((x_{i+1},1]\bigr).$$
(28)

To prove this partition satisfying the requirement of the lemma, we define two integer sets:

$$F_{\epsilon}= \biggl\{ k\geq1:\frac{\mu(\{x_{k}\})}{\mu ((x_{k+1},x_{k}])}< \epsilon \biggr\}$$

and

$$G_{\epsilon}= \biggl\{ k\geq1:\frac{\mu(\{x_{k}\})}{\mu ((x_{k+1},x_{k}])}\geq\epsilon \biggr\} ,$$

where ϵ is an arbitrary positive real number. Since $$\lim_{x\rightarrow0}\frac{\mu([0,x])}{\mu([0,x))}=1$$, we have $$\lim_{x\rightarrow0}\frac{\mu(\{x\})}{\mu([0,x))}=0$$. It is easy to find an integer N such that

$$\frac{\mu(\{x_{i-1}\})}{\mu([0,x_{i-1}])}< 2\epsilon^{2}$$

for any integer $$i>N$$. Thus, by the construction of $$G_{\epsilon}$$, if $$i>N$$ and $$i\in G_{\epsilon}$$, we have

$$\frac{\mu((x_{i},x_{i-1}])}{\mu([0,x_{i-1}])}< 2\epsilon.$$

If $$i\in F_{\epsilon}$$, then we have

$$\mu\bigl((x_{i+1},x_{i}]\bigr)\leq\frac{1}{1-\epsilon}\mu \bigl((x_{i+1},x_{i})\bigr).$$
(29)

By inequalities (28) and (29), we have

\begin{aligned} \frac{\mu((x_{i+1},x_{i}])}{\mu([0,x_{i}])}&\leq\frac{1}{1-\epsilon }\frac{\mu((x_{i+1},x_{i}))}{\mu([0,x_{i}])} \\ &= \frac{1}{1-\epsilon}\frac{\mu((x_{i+1},1])-\mu([x_{i},1])}{\mu ((0,1])-\mu((x_{i},1])} \\ &\leq \frac{1}{1-\epsilon}\frac{1/(k_{i}+1)^{2}}{\sum_{k=k_{i}+2}^{\infty}1/k^{2}}. \end{aligned}

Thus we can find a sufficiently large integer which is still denoted by N such that, for any integer $$i>N$$ and $$i\in F_{\epsilon}$$, there is

$$\frac{\mu((x_{i},x_{i-1}])}{\mu([0,x_{i-1}])}< 2\epsilon.$$

Since ϵ is an arbitrary real number, we have

$$\lim_{k\rightarrow\infty}\frac{\mu((x_{k},x_{k-1}])}{\mu([0,x_{k-1}])}=0.$$

The proof is completed. □

After finishing our preparations, we can give the proof of the result (iii) of the main theorem.

### Proof

Let

$${ } T_{a,b}(x)=\textstyle\begin{cases} \tan (\frac{\pi}{2} (\frac{x-a}{b-a} ) ),& 0< b< \infty;\\ x-a,& b=\infty. \end{cases}$$
(30)

By equality (30), we can obtain a new measure denoted by $$\mu_{T}$$ which is supported in $$[0,\infty]$$ so that, for any open interval $$(x,y)$$, we have

$$\mu_{T}\bigl((x,y)\bigr)=\mu\bigl(\bigl(T_{a,b}^{-1}(x),T_{a,b}^{-1}(y) \bigr)\bigr).$$

Then it is easy to get

$$\sup\bigl\{ \mathcal{R}_{\mu}f \bigl\vert f\in L^{p}(d\mu) \bigr\} =\sup\bigl\{ \mathcal{R}_{\mu _{T}}f \bigr\vert f\in L^{p}(d \mu_{T})\bigr\} .$$

Thus it is enough to assume that the measure μ is supported in $$[0,\infty]$$.

We first consider Condition 1.

By Lemma 5.1, we can divide $$\mathbb {R}^{+}$$ into a series of intervals

$$[0,x_{1}],(x_{1},x_{2}], \ldots, (x_{k},x_{k+1}], \ldots,$$

such that

$$\lim_{k\rightarrow\infty}\frac{\mu((x_{k},x_{k+1}])}{\mu([0,x_{k}])}=0.$$

For any $$\epsilon>0$$, if we can find a function $$f_{\epsilon}$$ such that $$\mathcal{R}(f_{\epsilon})\geq\frac{p}{p-1}-O(\epsilon)$$, then the proof is completed.

By the property of the partition, there exists an integer N satisfying

$$\frac{\mu((x_{k},x_{k+1}])}{\mu([0,x_{k}])}< \epsilon$$

for $$k\geq N$$. This inequality is equivalent to

$$\frac{\mu([0,x_{k+1}])}{\mu([0,x_{k}])}< 1+\epsilon.$$
(31)

Let

$$f_{\epsilon}=\sum_{k=N}^{\infty}\mu \bigl([0,x_{k+1}]\bigr)^{-\frac{1}{p}-\epsilon} \chi_{(x_{k},x_{k+1}]}.$$

First we estimate the norm of $$f_{\epsilon}$$

\begin{aligned} \Vert f_{\epsilon} \Vert _{L^{p}(d\mu)}&= \Biggl(\sum _{k=N}^{\infty}\mu \bigl([0,x_{k+1}] \bigr)^{-1-p\epsilon} \mu\bigl((x_{k},x_{k+1}]\bigr)\Biggr)^{\frac{1}{p}} \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon}\Biggl(\sum _{k=N}^{\infty}\mu\bigl([0,x_{k}] \bigr)^{-1-p\epsilon} \mu\bigl((x_{k},x_{k+1}]\bigr) \Biggr)^{\frac{1}{p}} \\ &= \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon} \Biggl(\sum _{k=N}^{\infty} \int_{\mu([0,x_{k}])}^{\mu ([0,x_{k+1}])}\mu\bigl([0,x_{k}] \bigr)^{-1-p\epsilon}\,dt \Biggr)^{\frac{1}{p}} \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon} \biggl( \int_{\mu([0,x_{N}])}^{\infty} t^{-1-p\epsilon}\,dt \biggr)^{\frac{1}{p}} \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon} \biggl( \frac{1}{p\epsilon} \biggr)^{\frac{1}{p}} \mu\bigl([0,x_{N}] \bigr)^{-\epsilon}. \end{aligned}
(32)

Next, we estimate the value of $$H_{\mu}f_{\epsilon}(x)$$. When $$k\geq N$$ and $$x_{k}< x\leq x_{k+1}$$, we have

\begin{aligned} H_{\mu}f_{\epsilon}(x)&=\frac{1}{\mu([0,x])} \int_{[0,x]}f_{\epsilon }(t)\,d\mu(t) \\ &\geq\frac{1}{\mu([0,x_{k+1}])} \int_{[0,x_{k}]}f_{\epsilon}(t)\,d\mu (t) \\ &=\frac{1}{\mu([0,x_{k+1}])}\sum_{i=N}^{k-1}\mu \bigl([0,x_{i+1}]\bigr)^{-\frac{1}{p}-\epsilon} \mu\bigl((x_{i},x_{i+1}]\bigr) \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon}\frac{1}{\mu([0,x_{k+1}])} \sum _{i=N}^{k-1}\mu\bigl([0,x_{i}] \bigr)^{-\frac{1}{p}-\epsilon} \mu\bigl((x_{i},x_{i+1}]\bigr) \\ &= \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon}\frac{1}{\mu([0,x_{k+1}])} \sum _{i=N}^{k-1} \int_{\mu([0,x_{i}])}^{\mu([0,x_{i+1}])} \mu\bigl([0,x_{i}] \bigr)^{-\frac{1}{p}-\epsilon}\,dt \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon }\frac{1}{\mu([0,x_{k}])} \int_{\mu([0,x_{N}])}^{\mu([0,x_{k}])}t^{-\frac{1}{p}-\epsilon }\,dt \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon }\frac{1}{1-\frac{1}{p}-\epsilon} \biggl(\mu\bigl([0,x_{k}]\bigr)^{-\frac{1}{p}-\epsilon}-\frac{\mu ([0,x_{N}])^{1-\frac{1}{p}-\epsilon}}{ \mu([0,x_{k}])} \biggr). \end{aligned}
(33)

Set

\begin{aligned} f_{\epsilon}^{(1)}&= \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon}\frac {1}{1-\frac{1}{p}-\epsilon} \sum_{k=N}^{\infty} \mu\bigl([0,x_{k}]\bigr)^{-\frac{1}{p}-\epsilon}\chi _{(x_{k},x_{k+1}]} \\ &= \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon}\frac {1}{1-\frac{1}{p}-\epsilon} f_{\epsilon} \end{aligned}

and

$$f_{\epsilon}^{(2)}= \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon} \frac {1}{1-\frac{1}{p}-\epsilon} \sum_{k=N}^{\infty} \frac{\mu([0,x_{N}])^{1-\frac{1}{p}-\epsilon}}{\mu ([0,x_{k}])}\chi_{(x_{k},x_{k+1}]}.$$

Then we have

\begin{aligned} \bigl\Vert f_{\epsilon}^{(2)} \bigr\Vert _{L^{p}(d\mu)}&= \biggl(\frac{1}{1+\epsilon } \biggr)^{1+\frac{1}{p}+\epsilon} \frac{1}{1-\frac{1}{p}-\epsilon} \mu\bigl([0,x_{N}]\bigr)^{1-\frac{1}{p}-\epsilon} \Biggl(\sum _{k=N}^{\infty}\mu\bigl([0,x_{k}] \bigr)^{-p}\mu\bigl((x_{k},x_{k+1}]\bigr) \Biggr)^{\frac{1}{p}} \\ &\leq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon}\frac {1}{1-\frac{1}{p}-\epsilon} \mu \bigl([0,x_{N}]\bigr)^{1-\frac{1}{p}-\epsilon} \Biggl(\sum _{k=N}^{\infty}\mu\bigl([0,x_{k+1}] \bigr)^{-p}\mu \bigl((x_{k},x_{k+1}]\bigr) \Biggr)^{\frac{1}{p}} \\ &\leq \biggl(\frac{1}{1+\epsilon} \biggr)^{\frac{1}{p}+\epsilon}\frac {1}{1-\frac{1}{p}-\epsilon} \biggl(\frac{1}{p-1} \biggr)^{\frac{1}{p}} \mu\bigl([0,x_{N}] \bigr)^{-\epsilon}. \end{aligned}
(34)

By inequality (33), we have

$$\Vert H_{\mu}f_{\epsilon} \Vert _{L^{p}(d\mu)}\geq \bigl\Vert f_{\epsilon}^{(1)} \bigr\Vert _{L^{p}(d\mu)} - \bigl\Vert f_{\epsilon}^{(2)} \bigr\Vert _{L^{p}(d\mu)}.$$

From this result and inequalities (32) and (34), we can get

\begin{aligned} \mathcal{R}(f_{\epsilon}) &\geq\frac{ \Vert f_{\epsilon}^{(1)} \Vert _{L^{p}(d\mu)} - \Vert f_{\epsilon}^{(2)} \Vert _{L^{p}(d\mu)}}{ \Vert f_{\epsilon} \Vert _{L^{p}(d\mu )}} \\ &= \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon} \frac{1}{1-\frac{1}{p}-\epsilon}- \frac{ \Vert f_{\epsilon}^{(2)} \Vert _{L^{p}(d\mu)}}{ \Vert f_{\epsilon} \Vert _{L^{p}(d\mu)}} \\ &\geq \biggl(\frac{1}{1+\epsilon} \biggr)^{1+\frac{1}{p}+\epsilon} \frac{1}{1-\frac{1}{p}-\epsilon}- \frac{ (\frac{1}{1+\epsilon} )^{\frac{1}{p}+\epsilon}\frac {1}{1-\frac{1}{p}-\epsilon} (\frac{1}{p-1} )^{\frac{1}{p}}}{ (\frac{1}{1+\epsilon } )^{\frac{1}{p}+\epsilon} (\frac{1}{p\epsilon} )^{\frac{1}{p}}}. \end{aligned}
(35)

Since ϵ is arbitrary, it is easy to imply $$\sup_{f\neq 0}\mathcal{R}(f)=\frac{p}{p-1}$$.

To prove condition (ii), by Lemma 5.2, we can part the intervals $$(0,1]$$ to

$$(x_{1},1], (x_{2},x_{1}], \ldots, (x_{k+1},x_{k}], \ldots$$

such that

$$\lim_{k\rightarrow\infty}\frac{\mu((x_{k+1},x_{k}])}{\mu((0,x_{k}])}=0.$$

Then, for any $$\epsilon>0$$, there is a sufficiently large integer N such that

$$\frac{\mu((x_{k+1},x_{k}])}{\mu((0,x_{k}])}< \epsilon$$

for $$k\geq N$$.

Thus, we have

$$\frac{\mu((0,x_{k+1}])}{\mu((0,x_{k}])}\geq1-\epsilon.$$

Let $$f_{\epsilon}=\sum_{k=N}^{\infty}\mu((0,x_{k}])^{-\frac {1}{p}+\epsilon}\chi_{(x_{k+1},x_{k}]}$$. Then, for $$x_{k+1}< x\leq x_{k}$$ and $$k\geq N$$, we have

\begin{aligned} H_{\mu}f_{\epsilon}(x)&=\frac{1}{\mu((0,x])} \int_{(0,x]}f_{\epsilon }(t)\,d\mu(t) \\ &\geq \frac{1}{\mu((0,x_{k}])} \int _{(0,x_{k+1}]}f_{\epsilon}(t)\,d\mu(t) \\ &=\frac{1}{\mu((0,x_{k}])}\sum_{i=k+1}^{\infty} \mu\bigl((0,x_{i}]\bigr)^{-\frac{1}{p}+\epsilon} \mu\bigl((x_{i+1},x_{i}]\bigr) \\ &\geq\frac{1}{\mu((0,x_{k}])}\frac{\mu((0,x_{k+1}])^{1-\frac{1}{p}+\epsilon}}{1-\frac{1}{p}+\epsilon} \\ &\geq\frac{(1-\epsilon)^{1-\frac{1}{p}+\epsilon}}{1-\frac{1}{p}+\epsilon }\mu\bigl((0,x_{k}]\bigr)^{-\frac{1}{p}+\epsilon} \end{aligned}
(36)
\begin{aligned} &=\frac{(1-\epsilon)^{1-\frac{1}{p}+\epsilon}}{1-\frac{1}{p}+\epsilon }f_{\epsilon}(x). \end{aligned}
(37)

It follows from inequality (36) that

$$\mathcal{R}(f_{\epsilon})\geq\frac{(1-\epsilon)^{1-\frac{1}{p}+\epsilon}}{1-\frac{1}{p}+\epsilon}.$$

Because ϵ is arbitrary, it is easy to know $$\sup_{f}\mathcal {R}(f)\geq\frac{p}{p-1}$$. The proof of the suffice part of Theorem 1.2 is then completed. □

## 6 Counterexample

In this section we give some counterexamples that make $$\sup_{f\neq 0,f\in L^{p}}\mathcal{R}(f)< p/(p-1)$$. The following two lemmas tell us that we can limit our discussion to a special function set.

### Lemma 6.1

Suppose μ is a positive measure on $$\mathbb {R}_{+}$$ and it has an atom $$x_{0}$$. If $$\{f_{n}\}$$, $$n=1,2,\ldots$$ , is a series of functions satisfying $$f_{n}(x_{0})=1$$ and

$$\lim_{n\rightarrow\infty}\mathcal{R}_{\mu}(f_{n})= \frac{p}{p-1},$$

then we have

$$\lim_{n\rightarrow\infty} \Vert f_{n} \Vert _{L^{p}(d\mu)}= \infty.$$

### Proof

Without loss of generality, we assume that $$\mu(\{x_{0}\})=1$$. If the assertion does not hold, then we can assume that there exists a constant C satisfying $$\Vert f_{n} \Vert _{L^{p}(d\mu)}\leq C$$. Let $$f_{n}^{*}$$ be the decreasing rearrangement of $$f_{n}$$, then it is easy to get $$\Vert f_{n}^{*} \Vert _{L^{p}(dm)}\leq C$$ and $$f_{n}^{*}(1)\geq1$$. Thus we have $$f^{*}(x)\geq1$$ for $$0< x\leq1$$. By Helly’s theorem, we can assume $$\lim_{n\rightarrow\infty} f^{*}_{n}=f^{*}$$ almost everywhere. Since $$f_{n}^{*}$$ is decreasing, we have

$$C^{p}\geq \bigl\Vert f_{n}^{*} \bigr\Vert _{L^{p}(dm)}^{p}\geq \int_{[0,R]} \bigl\vert f_{n}^{*}(t) \bigr\vert ^{p}\,dt\geq R \bigl\vert f_{n}^{*}(R) \bigr\vert ^{p},$$

which is equivalent to $$f_{n}^{*}(R)\leq CR^{-\frac{1}{p}}$$. Thus, by the control convergence theorem,

$$\lim_{n\rightarrow\infty}Hf_{n}^{*}(x)= \lim _{n\rightarrow\infty}\frac{1}{x} \int_{[0,x]}f_{n}^{*}(t)\,dt=Hf^{*}(x).$$
(38)

However, by inequality (12), we have

$$\mathcal{R}_{m}\bigl(f_{n}^{*}\bigr)\geq \mathcal{R}_{\mu}(f_{n}),$$

it obviously shows that $$\{f_{n}^{*}\}$$ is a maximizing sequence for H, i.e.,

$$\lim_{n\rightarrow\infty}\mathcal{R}\bigl(f_{n}^{*}\bigr)= \frac{p}{p-1}.$$

By $$\lim_{n\rightarrow\infty} f^{*}_{n}=f^{*}$$ and equality (38), using Lemma 2.2, we get $$\mathcal{R}_{m}(f^{*})=\frac{p}{p-1}$$, which contradicts the result about Hardy operator we have known. The proof is completed. □

### Lemma 6.2

Suppose μ is a positive measure on $$\mathbb {R}_{+}$$ and it has an atom $$x_{0}$$. If

$$\sup\bigl\{ \mathcal{R}_{\mu}f| f\in L^{p}(d\mu)\bigr\} = \frac{p}{p-1},$$

then there exists a series of functions $$\{f_{k}\}$$, $$k=1,2,\ldots$$ , and $$f_{k}(x_{0})=0$$ such that

$$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu}(f_{k})= \frac{p}{p-1}.$$

### Proof

It is obvious that we can assume there exists a series of functions $$g_{k}$$, $$g_{k}(x_{0})=1$$, such that

$$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu}(g_{k})= \frac{p}{p-1}.$$

Let

$$f_{k}(x)=\textstyle\begin{cases} g_{k}(x),&x\neq x_{0},\\ 0,&x=x_{0}. \end{cases}$$

Then we have

$$H_{\mu}f_{k}(x)=\textstyle\begin{cases} H_{\mu}g_{k}(x),&x< x_{0};\\ H_{\mu}g_{k}(x)-\mu(\{x_{0}\})/\mu([0,x]),&x\geq x_{0}. \end{cases}$$
(39)

By equality (39), we can get

$$\bigl\Vert H_{\mu}(f_{k}) \bigr\Vert _{L^{p}}\geq \bigl\Vert H_{\mu}(g_{k}) \bigr\Vert _{L^{p}}- \biggl\Vert \frac{\mu(\{x_{0}\})}{\mu([0,\cdot])}\chi_{[x_{0},\infty ]} \biggr\Vert _{L^{p}}.$$
(40)

On the other hand, it is easy to obtain

$$\Vert f_{k} \Vert _{L^{p}}\leq \Vert g_{k} \Vert +\mu\bigl(\{x_{0}\}\bigr)^{\frac{1}{p}}.$$
(41)

By Lemma 6.1, we know that $$\lim_{k\rightarrow\infty} \Vert g_{k} \Vert _{L^{p}(d\mu)}=\infty$$ and $$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu}(g_{k})=p/(p-1)$$. By this result, together with inequalities (40) and (41), we can have

$$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu}(f_{k})= \frac{p}{p-1}.$$

□

Now we can give some counterexamples.

### Example 6.3

Suppose that μ is supported in $$[a,b)$$, $$\mu(\{a\})>0$$, and $$\mu (\mathbb {R}_{+})<\infty$$. Then $$\sup_{f\neq0}\mathcal{R}_{\mu}(f)< p/(p-1)$$.

### Proof

Suppose that the result is not valid. By Lemma 6.2, we can find a series of functions $$\{f_{k}\}$$, $$f_{k}(a)=0$$, such that

$$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu}(f_{k})= \frac{p}{p-1}.$$

Let $$A=\mu(\{a\})$$, $$B=\mu(\mathbb {R}_{+})$$, and $$\mu_{1}=\mu-A\delta_{a}$$. Then we have

$$H_{\mu}f_{k}(x)=\frac{\mu_{1}([0,x])}{\mu([0,x])} \frac{1}{\mu_{1}([0,x])} \int_{[0,x]}f_{k}\,d\mu_{1}\leq \frac{B-A}{B}H_{\mu_{1}}f_{k}(x)$$
(42)

and

$$\Vert f_{k} \Vert _{L^{p}(d\mu)}= \Vert f_{k} \Vert _{L^{p}(d\mu_{1})}.$$
(43)

By inequalities (42) and (43), we obtain

$$\mathcal{R}_{\mu}(f_{k})\leq\frac{B-A}{B} \mathcal{R}_{\mu_{1}}(f_{k}) \leq\frac{B-A}{B} \frac{p}{p-1}.$$

It contradicts with $$\lim_{k\rightarrow\infty}\mathcal{R}_{\mu }(f_{k})=p/(p-1)$$. Then the counterexample is valid. □

### Example 6.4

If $$\mu=\sum_{k=-\infty}^{\infty}\lambda^{k} \delta_{\lambda^{k}}$$ with $$\lambda>1$$, then $$\sup_{f\in L^{p}(d\mu)}\mathcal{R}f<\frac{p}{p-1}$$.

### Proof

By the definition of μ, we have

\begin{aligned} H_{\mu}f(\lambda_{k})&= \frac{1}{\mu([0,\lambda_{k}])} \int _{[0,\lambda_{k}]}f(t)\,d\mu(t) \\ &=\frac{(\lambda-1)\sum_{i=-\infty}^{k}\lambda^{i} f(\lambda ^{i})}{\lambda^{k+1}} \\ &=\frac{\lambda-1}{\lambda}\sum_{i=-\infty}^{0} \lambda ^{i}f\bigl(\lambda^{i+k}\bigr) \end{aligned}
(44)

and

\begin{aligned} \bigl\Vert f\bigl(\lambda^{i}\cdot\bigr) \bigr\Vert _{L^{p}(d\mu)} &= \Biggl(\sum_{k=-\infty}^{\infty} \bigl\vert f\bigl(\lambda^{i+k}\bigr) \bigr\vert ^{p} \lambda ^{k} \Biggr)^{\frac{1}{p}} \\ &= \Biggl(\sum_{k=-\infty}^{\infty} \bigl\vert f\bigl(\lambda^{k}\bigr) \bigr\vert ^{p}\lambda ^{k-i} \Biggr)^{\frac{1}{p}} \\ &=\lambda^{-\frac{i}{p}} \Vert f \Vert _{L^{p}(d\mu)}. \end{aligned}
(45)

By inequalities (44), (45), and Minkowski’s inequality, it follows

\begin{aligned} \Vert H_{\mu}f \Vert _{L^{p}(d\mu)}&= \Biggl\Vert \frac{\lambda-1}{\lambda}\sum_{i=-\infty}^{0}\lambda ^{i}f\bigl(\lambda^{i}\cdot\bigr) \Biggr\Vert _{L^{p}(d\mu)} \\ &\leq \frac{\lambda-1}{\lambda}\sum_{i=-\infty}^{0} \lambda^{i} \bigl\Vert f\bigl(\lambda^{i}\cdot\bigr) \bigr\Vert _{L^{p}(d\mu)} \\ &=\frac{\lambda-1}{\lambda}\sum_{i=-\infty}^{0} \lambda^{i-\frac{i}{p}} \Vert f \Vert _{L^{p}(d\mu)} \\ &=\frac{\lambda-1}{\lambda-\lambda^{\frac{1}{p}}} \Vert f \Vert _{L^{p}(d\mu)}. \end{aligned}
(46)

It is easy to get $$\frac{\lambda-1}{\lambda-\lambda^{\frac{1}{p}}}<\frac{p}{p-1}$$. The proof is completed. □

## References

1. Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, Cambridge (1952)

2. Komori, Y.: Notes on commutators of Hardy operators. Int. J. Pure Appl. Math. 7, 329–334 (2003)

3. Lu, S., Yan, D., Zhao, F.: Sharp bounds for Hardy type operators on higher-dimensional product spaces. J. Inequal. Appl. 2013, 148 (2013)

4. Muckenhoupt, B.: Hardy’s inequality with weights. Stud. Math. 44(1), 31–38 (1972)

5. Shunchao, L., Jian, W.: Commutators of Hardy operators. J. Math. Anal. Appl. 274(2), 626–644 (2002)

6. Sinnamon, G.: Operators of Lebesgue Spaces with General Measures. McMaster University, Toronto (1987)

7. Bennett, C., Sharpley, R.C.: Interpolation of Operators, vol. 129. Academic Press, San Diego (1988)

8. Grafakos, L., Kinnunen, J.: Sharp inequalities for maximal functions associated with general measures. Proc. R. Soc. Edinb., Sect. A, Math. 128(4), 717–723 (1998)

9. Grafakos, L.: Classical and Modern Fourier Analysis. Prentice Hall, New York (2004)

10. Lieb, E.H.: Sharp constants in the Hardy–Littlewood–Sobolev and related inequalities. Ann. Math. 118, 349–374 (1983)

## Funding

The research was supported by the National Natural Foundation of China (Grant Nos. 11471039, 11271162).

## Author information

Authors

### Contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Xudong Nie.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests.

### Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Rights and permissions

Reprints and permissions

Nie, X., Yan, D. Sharp constant of Hardy operators corresponding to general positive measures. J Inequal Appl 2018, 78 (2018). https://doi.org/10.1186/s13660-018-1660-8