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Journal of Function Spaces
Volume 2014 (2014), Article ID 274521, 14 pages
http://dx.doi.org/10.1155/2014/274521
Research Article

The Boundedness of Intrinsic Square Functions on the Weighted Herz Spaces

College of Mathematics and Econometrics, Hunan University, Changsha 410082, China

Received 23 September 2013; Accepted 4 November 2013; Published 12 January 2014

Academic Editor: Dachun Yang

Copyright © 2014 Hua Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We will obtain the strong type and weak type estimates of intrinsic square functions including the Lusin area integral, Littlewood-Paley -function, and -function on the weighted Herz spaces with general weights.

1. Introduction and Main Results

Let and . The classical square function (Lusin area integral) is a familiar object. If is the Poisson integral of , where denotes the Poisson kernel in , then we define the classical square function (Lusin area integral) by (see [1, 2]) where denotes the usual cone of aperture one: Similarly, we can define a cone of aperture for any : and corresponding square function The Littlewood-Paley -function (could be viewed as a “zero-aperture” version of ) and the -function (could be viewed as an “infinite aperture” version of ) are defined, respectively, by (see, e.g., [3, 4])

The modern (real-variable) variant of can be defined in the following way (here we drop the subscript if ). Let be real, radial and have support contained in , and let . The continuous square function is defined by (see, e.g., [5, 6])

In 2007, Wilson [7] introduced a new square function called intrinsic square function which is universal in a sense (see also [8]). This function is independent of any particular kernel , and it dominates pointwise all the above-defined square functions. On the other hand, it is not essentially larger than any particular . For , let be the family of functions defined on such that has support containing in , , and for all , For and , we set Then we define the intrinsic square function of (of order ) by the following formula: We can also define varying-aperture versions of by the formula The intrinsic Littlewood-Paley -function and the intrinsic -function will be given, respectively, by

In [8], Wilson showed the following weighted boundedness of the intrinsic square functions.

Theorem A. Let , , and . Then there exists a constant independent of such that

Moreover, in [9], Lerner obtained sharp norm inequalities for the intrinsic square functions in terms of the characteristic constant of for all . For further discussions about the boundedness of intrinsic square functions on various function spaces, we refer the readers to [1017].

Before stating our main results, let us first recall some definitions about the weighted Herz and weak Herz spaces. For more information about these spaces, one can see [1822] and the references therein. Let and let for any . Denote for , if , and , where is the characteristic function of the set . For any given weight function on and , we denote by the space of all functions satisfying

Definition 1 (see [19]). Let , , and , and let , be two weight functions on .(a)The homogeneous weighted Herz space is defined by where (b)The nonhomogeneous weighted Herz space is defined by where

For any , , and any measurable function on , we set . Let for and .

Definition 2 (see [21]). Let , , and , and let , be two weight functions on .(c)A measurable function on is said to belong to the homogeneous weighted weak Herz space if (d)A measurable function on is said to belong to the nonhomogeneous weighted weak Herz space if

Obviously, if , then for any . We also have when and , where

Thus, weighted (weak) Herz spaces are generalizations of the weighted (weak) Lebesgue spaces. The main purpose of this paper is to consider the boundedness of intrinsic square functions on weighted Herz spaces with weights. At the extreme case, we will also prove that these operators are bounded from the weighted Herz spaces to the weighted weak Herz spaces. Our main results in the paper are formulated as follows.

Theorem 3. Let , , , , and . Then is bounded on provided that and satisfy either of the following:(i) , , and ;(ii) , , , and .

Theorem 4. Let , , , , and . If , , and , then is bounded from into .

Theorem 5. Let , , , , and . If , then is bounded on provided that and satisfy either of the following:(i) , , and ;(ii) , , , and .

Theorem 6. Let , , , , and . If , , , and , then is bounded from into .

In [7], Wilson also showed that for any , the functions and are pointwise comparable, with comparability constants depending only on and . Thus, as direct consequences of Theorems 3 and 4, we obtain the following.

Corollary 7. Let , , , , and . Then is bounded on provided that and satisfy either of the following:(i) , , and ;(ii) , , , and .

Corollary 8. Let , , , , and . If , , and , then is bounded from into .

2. Weights

The classical weight theory was first introduced by Muckenhoupt in the study of weighted boundedness of Hardy-Littlewood maximal functions in [23]. A weight is a nonnegative, locally integrable function on , that denotes the ball with the center and radius . For any ball and , denotes the ball concentric with whose radius is times as long. For a given weight function and a measurable set , we also denote the Lebesgue measure of by and set weighted measure . We say that is in the Muckenhoupt class with , if there exists a constant such that for every ball , For the endpoint case , , if where is a positive constant which is independent of the choice of . The smallest value of such that the above inequalities hold is called the characteristic constant of and denoted by . If there exist two constants and such that the following reverse Hölder inequality holds: then we say that satisfies the reverse Hölder condition of order and write . It is well known that if with , then for all . Moreover, if with , then there exists such that .

The following properties for weights will be repeatedly used in this paper.

Lemma 9 (see [24]). Let with . Then, for any ball , there exists an absolute constant such that In general, for any , one has where does not depend on nor on .

Lemma 10 (see [1, 24]). Let , , and . Then there exist constants such that for any measurable subset of a ball .

Throughout this paper, always denotes a positive constant which is independent of the main parameters involved but may vary from line to line.

3. Proofs of Theorems 3 and 4

Proof of Theorem 3. We only need to show the theorem for the homogeneous case because the proof of the nonhomogeneous result is similar and so is omitted here. Let . Following [25], for any , we decompose as Since () is a sublinear operator, then we can write Since and , then . By Theorem A and Lemma 9, we have For the term , we first use Minkowski's inequality to derive For any , , and , we have For any , , and with , by a direct computation, we can easily see that Thus, by using inequality (31) and Minkowski's inequality, we deduce Denote the conjugate exponent of by . Applying Hölder's inequality and the condition, we can deduce that Substituting inequality (34) into (33), we thus obtain Here, we will consider two cases. For the case of , using the well-known inequality and changing the order of summation, we find that Moreover, it follows immediately from Lemma 9 that Since when and for , by Lemma 10, we can get Therefore where the last inequality holds since . On the other hand, for the case of , we will use Hölder's inequality to obtain Using the same arguments as above, we can also prove the following estimates under the assumption that : Hence Summarizing the above estimates for the term , we obtain that for every , Let us now turn to estimate the last term . In this case, for any , , and with , it is easy to check that Then it follows from inequality (31) and Minkowski's inequality that
This estimate together with (34) implies that Hence Now we will consider the following two cases again. For the case of , by using the inequality and changing the order of summation, we obtain Since , then there exists such that for . Thus, by Lemma 10 again, we can get where . Therefore, we have where in the last inequality we have used the fact that under our assumption (i) or (ii). On the other hand, for the case of , an application of Hölder's inequality gives us that By using the same arguments as for , we are able to prove that the following two series are convergent by an absolute constant under the assumption (i) or (ii): Consequently From the above discussions for the term , we know that, for any , Summing up the above estimates for , , and , we complete the proof of Theorem 3.

Proof of Theorem 4. Let . For any , as in the proof of Theorem 3, we will split into three parts: Then, for any given , we have Applying Chebyshev’s inequality, Theorem A, and Lemma 9, we obtain For any , it follows from inequalities (33) and (34) that By using Lemma 9, inequality (38), and the fact that , we deduce that Moreover, since , then we have that, for any , Set . If , then the inequality holds trivially. Now we suppose that . First it is easy to verify that . Then, for any fixed , we are able to find a maximal positive integer such that Hence Because , then by Lemma 10, with the same notations as in (49), we can get Therefore On the other hand, it follows from inequalities (34) and (45) that In the present situation, since with , then it follows from inequality (49) that Furthermore, recalling that , then, for any , we have Repeating the arguments used for the term , we can also obtain Combining the above estimates for , , and and then taking the supremum over all , we finish the proof of Theorem 4.

4. Proofs of Theorems 5 and 6

In order to prove the main theorems of this section, let us first establish the following results.

Proposition 11. Let , , and with . Then, for any , one has

Proof. Since , then, by Lemma 9, we know that for any , Therefore Taking square-roots on both sides of the above inequality, we are done.

Proposition 12. Let , , and with . Then, for any , one has

Proof. For any and , it is easy to see that Since , then, by duality, we have For , we denote the weighted maximal operator by ; that is, where the supremum is taken over all balls which contain . Then, by Lemma 9, we can get