Sharp constant of Hardy operators corresponding to general positive measures

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Research

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 \(H_{\mu}\).

Keywords

Hardy operator Best constant Sharp problem 

MSC

42B20 42B35 

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 [2, 3, 4, 5, 6] 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

Letμbe 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 thatμis 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 measureμ.

     
  2. (ii)

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

     

Theorem 1.3

Ifμsatisfies 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 ofμ, 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. LetXandYbe\(L^{p}(M,\Sigma,\mu)\)and\(L^{q}(M',\Sigma',\mu')\)with\(1\leq p\leq q<\infty\). LetAbe a bounded linear operator fromXtoY. 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 forN, and assume that\(f_{j}\rightarrow f\neq0\)and that\(A(f_{j})\rightarrow A(f)\)pointwise almost everywhere. Thenfmaximizes, 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

Letμbe 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 thatμis 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

Supposeμis 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

Supposeμis 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. □

Notes

Authors’ contributions

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

Competing interests

The authors declare that they have no competing interests.

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Authors and Affiliations

  1. 1.Department of Mathematics and PhysicsShijiazhuang Tiedao UniversityShijiazhuangP.R. China
  2. 2.School of Mathematical SciencesUniversity of the Chinese Academy of SciencesBeijingP.R. China

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