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Sharp constant of Hardy operators corresponding to general positive measures
Journal of Inequalities and Applications volume 2018, Article number: 78 (2018)
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
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
and if \(\mu=\sum_{k=1}^{\infty}\delta_{k}\), then \(H_{\mu}\) becomes the discrete Hardy operator
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 [2–6] 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
and
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
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
Then the following statements hold:
-
(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.
-
(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
Condition 2. \(\{a\}\) is not an atom of μ, and
then we have
We remark that there indeed exist some measures so that
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
Then the rearrangement of f is defined by
By the properties of the rearrangement, we can easily have
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
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
and
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
Obviously \(F_{\mu}\) increases as \(x\rightarrow\infty\). It follows from Lemma 2.1 and the definition of \(H_{\mu}\) that
Let
Note that \(f^{*}\) decreases, so we easily have that \(Hf^{*}\) decreases as well. If we take
then it implies that
Thus, we can obtain that
We conclude that
where \(\vert \cdot \vert \) denotes the Lebesgue measure. It follows from inequalities (4) and (5) that
Since \(f^{*}\in L^{p}(dm)\), by Hölder’s inequality, we have that
Thus it is obvious to obtain that
From inequality (6) and inequality (8), we have
That is,
holds. This is equivalent to
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
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
By inequality (11), we conclude that
It follows from the inequality of classical Hardy operator that
Combining inequality (12) with inequality (13), we have
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
then there exists a partition on \([0,\infty]\) as
such that
and
Proof
Let \(x_{1}\) be any positive number. Denote \(I_{0}=[0,x_{1}]\). Since μ is supported on \([0,\infty]\), we have
For \(k=2\), we let
For \(k>2\), we let
Denote \(I_{k}=[x_{k-1},x_{k}]\) with \(k=2,3,\ldots \) . Since \(\mu([0,\infty])=\infty\), we easily have
Thus, \(\{I_{k}\}\) obviously constitutes a partition of \([0,\infty]\).
We first show that
and
By our construction, for any \(x>x_{k+1}\), it follows that
Thus the property of measure implies that
Moreover, if \(x_{k}< x< x_{k+1}\), then \(\mu([x_{k},x])<\mu(I_{k-1})\). Thus, it follows that
To complete the proof, it remains to show that
This is equivalent to prove that, for any \(\epsilon>0\), there is an integer \(N>0\) such that
holds for \(k\geq N\).
In order to prove this result, we divide the set \(\mathbb {Z}^{+}\setminus\{1\}\) into two parts:
and
By definition (16), if \(k\in G_{\epsilon}\), then we have
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 \(k\geq N\),
Thus if \(k>N\) and \(k\in G_{\epsilon}\), then by inequalities (17) and (18), we have
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
By the definition of \(F_{\epsilon}\) and inequality (14), we can conclude that if \(i\in F_{\epsilon}\), then
It immediately implies from inequality (20) that
Thus, by inequality (21), we have
Since \(k_{0}\in G_{\epsilon}\), inequalities (14) and (20) imply
If \((1+\epsilon)^{k-k_{0}}> 2\), by inequality (22), we have
If \((1+\epsilon)^{k-k_{0}}\leq2\), by inequality (23), we have
At last, we conclude that if \(k>N\) and \(k\in F_{\epsilon}\), then
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
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 \((0,1]\),
such that \(\lim_{k\rightarrow\infty}x_{k}=0\) and
Proof
Without loss of generality, suppose
If \(\mu(\{1\})<1\), then we set \(k_{0}=0\). If \(\mu(\{1\})\geq1\), then we set
It is easy to see that
Then we can find a positive real number \(x_{1}<1\) such that
Proceeding in this way, we set
and
for \(i\geq1\). By (27), (25), and (26), we can conclude
It is easy to see that \(x_{i}>x_{i+1}\) and
Thus we have \(\lim_{i\rightarrow\infty}x_{i}=0\). It is easy to see that
divide \((0,1]\). It can be implied from inequality (27) that
To prove this partition satisfying the requirement of the lemma, we define two integer sets:
and
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
for any integer \(i>N\). Thus, by the construction of \(G_{\epsilon}\), if \(i>N\) and \(i\in G_{\epsilon}\), we have
If \(i\in F_{\epsilon}\), then we have
By inequalities (28) and (29), we have
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
Since ϵ is an arbitrary real number, we have
The proof is completed. □
After finishing our preparations, we can give the proof of the result (iii) of the main theorem.
Proof
Let
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
Then it is easy to get
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
such that
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
for \(k\geq N\). This inequality is equivalent to
Let
First we estimate the norm of \(f_{\epsilon}\)
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
Set
and
Then we have
By inequality (33), we have
From this result and inequalities (32) and (34), we can get
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
such that
Then, for any \(\epsilon>0\), there is a sufficiently large integer N such that
for \(k\geq N\).
Thus, we have
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
It follows from inequality (36) that
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
then we have
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
which is equivalent to \(f_{n}^{*}(R)\leq CR^{-\frac{1}{p}}\). Thus, by the control convergence theorem,
However, by inequality (12), we have
it obviously shows that \(\{f_{n}^{*}\}\) is a maximizing sequence for H, i.e.,
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
then there exists a series of functions \(\{f_{k}\}\), \(k=1,2,\ldots\) , and \(f_{k}(x_{0})=0\) such that
Proof
It is obvious that we can assume there exists a series of functions \(g_{k}\), \(g_{k}(x_{0})=1\), such that
Let
Then we have
By equality (39), we can get
On the other hand, it is easy to obtain
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
□
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
Let \(A=\mu(\{a\})\), \(B=\mu(\mathbb {R}_{+})\), and \(\mu_{1}=\mu-A\delta_{a}\). Then we have
and
By inequalities (42) and (43), we obtain
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
and
By inequalities (44), (45), and Minkowski’s inequality, it follows
It is easy to get \(\frac{\lambda-1}{\lambda-\lambda^{\frac{1}{p}}}<\frac{p}{p-1}\). The proof is completed. □
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Funding
The research was supported by the National Natural Foundation of China (Grant Nos. 11471039, 11271162).
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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
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DOI: https://doi.org/10.1186/s13660-018-1660-8