Sharp constant of Hardy operators corresponding to general positive measures

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Introduction
Let μ be a positive measure on [0, ∞) and f be a nonnegative μ-measurable function. Define Hardy operator with respect to the measure μ by Observe that if μ is Lebesgue measure, then H μ becomes the classical Hardy operator and if μ = ∞ k=1 δ k , then H μ becomes the discrete Hardy operator For 1 < p < ∞, 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 μ . We will show that the operator H μ is bounded on L p (dμ) with an upper bound no more than p/(p -1). Furthermore, we will characterize a sufficient condition about μ such that H μ L p →L p = p/(p -1). From the definition about H μ , it is not necessary to consider the points x such that μ([0, x]) = 0 or ∞. Therefore, we let For the case of weak type inequality, the best constant from L p (dμ) to L p,∞ (dμ) is always 1.
Then the following statements hold: We remark that there indeed exist some measures so that For example, it is easy to know that the Dirac measure δ 0 satisfies inequality (3). In this paper, we will give some more complex counterexamples.

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. 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, , μ) and (M , , μ ) are two measure spaces. Let X and Y be L p (M, , μ) and L q (M , , μ ) with 1 ≤ p ≤ q < ∞. Let A be a bounded linear operator from X to Y . For f ∈ X with f = 0, set Let {f j } be a uniform norm-bounded maximizing sequence for N, and assume that f j → f = 0 and that A(f j ) → A(f ) pointwise almost everywhere. Then f maximizes, i.e., R(f ) = N .

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 μ increases as x → ∞. It follows from Lemma 2.1 and the definition of H μ that Let Note that f * decreases, so we easily have that Hf * decreases as well. If we take Thus, we can obtain that We conclude that where | · | denotes the Lebesgue measure. It follows from inequalities (4) and (5) that Since f * ∈ 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).
It is easy to obtain The proof is completed.

L p -boundedness of the operator H μ 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 R μ (f ) = p p-1 . The proof of the result (ii) of Theorem 1.2 is completed.

A characterization of the measure μ which ensures sup f =0 R μ (f ) = p/(p -1)
In this section, we try to characterize the measure μ which ensures sup f =0 R μ (f ) = p/(p -1). We regard μ as a complete atom measure by giving an appropriate partition on [0, ∞]. We first present a partition on [0, ∞] by the following two lemmas. For k = 2, we let For k > 2, we let Thus, {I k } obviously constitutes a partition of [0, ∞]. We first show that By our construction, for any x > x k+1 , it follows that Thus the property of measure implies that To complete the proof, it remains to show that This is equivalent to prove that, for any > 0, there is an integer N > 0 such that In order to prove this result, we divide the set Z + \ {1} into two parts: and By definition (16), if k ∈ G , then we have We discuss the problem in two cases: Case I. G is not a finite set. Case II. G is a finite set.
If G is not a finite set, then by equality lim x→∞ μ({x}) μ([0,x]) = 0, there exists an integer N ∈ G such that, for any k ≥ N , Thus if k > N and k ∈ G , then by inequalities (17) and (18), we have On the other hand, if k > N and k ∈ F , since G is not a finite integer and N ∈ G , we can find a series of integers k 0 , k 0 + 1, . . . , k, such that k 0 ∈ G , and By the definition of F and inequality (14), we can conclude that if i ∈ F , then It immediately implies from inequality (20) that (21) Thus, by inequality (21), we have Since k 0 ∈ G , inequalities (14) and (20) imply If (1 + ) k-k 0 > 2, by inequality (22), we have If (1 + ) k-k 0 ≤ 2, by inequality (23), we have At last, we conclude that if k > N and k ∈ F , then The proof of Case I is complete. If G is a finite set, then we can find an integer k 0 such that k ∈ F for k > k 0 . Then, by inequality (22), we can find a big enough integer N such that The proof is completed.
Proof Without loss of generality, suppose If μ({1}) < 1, then we set k 0 = 0. If μ({1}) ≥ 1, 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 ≥ 1. 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→∞ x i = 0. It is easy to see that ( To prove this partition satisfying the requirement of the lemma, we define two integer sets: where is an arbitrary positive real number. Since lim x→0 It is easy to find an integer N such that for any integer i > N . Thus, by the construction of G , if i > N and i ∈ G , we have If i ∈ F , 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 ∈ F , there is Since 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 μ T which is supported in [0, ∞] 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, ∞]. We first consider Condition 1.
For any > 0, if we can find a function f such that R(f ) ≥ p p-1 -O( ), then the proof is completed.
By the property of the partition, there exists an integer N satisfying for k ≥ N . This inequality is equivalent to First we estimate the norm of f Next, we estimate the value of H μ f (x). When k ≥ N and x k < x ≤ x k+1 , we have By inequality (33), we have From this result and inequalities (32) and (34), we can get Since is arbitrary, it is easy to imply sup f =0 R(f ) = 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 ], . . . , (x k+1 , x k ], . . .