Abstract

We introduce a new space, space, of several real variables with nondecreasing functions . By giving basic properties of the weighted function , by establishing a Stegenga-type estimate, and by introducing the -Carleson measure on , we obtain various characterizations of space.

1. Introduction

Recall that a locally integrable function belongs to if where denotes a cube in with edges parallel to the coordinate axes and denotes the Lebesgue measure of and Via the John-Nirenberg inequality [1], one can show an equivalent condition of as follows:

C. Fefferman’s famous equation, , describes a deep relation between and the Hardy space (cf. [2, 3]). This leads quite naturally to increased study of these functions from the point of real variable theory and complex function theory views in the recent fifty years. See [2–9] for more results about space.

As a generalization of , the space , , introduced by Essén et al. in [10], is defined to be the class of all locally integrable functions such that where denotes the edge length of the cube .

It is easy to see that is always a subclass of and by choosing . Moreover, we know by [10] that if and only if . Also, we see that is trivial (containing a.e. constant functions only) if and only if and , , is trivial if and only if .

In this paper, we introduce and develop a more general space of several real variables, which can be viewed as an extension and improvement of spaces as well as . A theory of spaces on unit disc has been extensively studied for recent years in the context of a wide class of function spaces; see, for example, [11–15]. Motivated by the theory of analytic spaces, we define the following.

Definition 1. Let be a nondecreasing function. A function is said to belong to the space if
If we take , for , then . Modulo constants, is a Banach space under the norm defined in (5).
Our paper is organized as follows.
In Section 2, we investigate the relationship between and and give a sufficient and necessary condition for space which is nontrivial.
In Section 3, we give several results about the weight function on which obviously depends. In the study of , the auxiliary function defined by still works well as the analytic spaces.
In Section 4, we define the -Carleson measure on . By establishing a Stegenga-type estimate, we obtain a characterization of spaces in terms of the -Carleson measure.
Throughout this note, means that there is a positive constant such that . Moreover, if both and hold, then one says that . For the convenience of calculation, in this paper, we always assume that is nondecreasing and .

2. Basic Properties of

Our first observation is that is invariant under the conformal mappings and rotations; that is, for any conformal map , and , or any rotation for an orthogonal matrix of order , hold for any .

We say that the space is trivial if contains only a.e. constant functions. To discuss this problem, we recall the space , the class of all functions , with where . By [16], we know that Thus is not trivial for . However, is trivial.

For any cube , if , then . For , means that the cube has the same center as and the edge length . If and , then . By the change of variable, if and only if

Theorem 2. The following statements are true. (a). Moreover, is never trivial.(b)For , is not trivial if and only if Moreover, if (11) holds, then

Proof. (a) For any cube of and , we have By assumption on we have Hence, . is not trivial and so is .
(b) Necessity. It is enough to show that if then is trivial. We will prove the necessity by two steps.
Step 1. If and is nonconstant, we may assume that is real. Then there exists a point such that . By the Householder reflector [17, p. 71], there exists an orthogonal matrix ,  , such that which gives Denote by the transpose of the matrix . Set . Since , there exists a point such that . Write for convenience as follows: Consequently, Similarly, Thus Note that . Then there exist a positive constant and a small cube centered at on which and ,  . Define If and , using the mean value theorem, we get Thus If , then . Hence It means that is not an element of . On the other hand, since is invariant under rotations. This is a contraction.
Step 2. Note that is conformal invariant. By Minkowski’s inequality, if and , then and where In particular, if , the class of smooth functions with compact support, then . Thus is constant by Step  1. By [10], there exits a sequence with , , and shrinking to such that a.e. It follows that is constant a.e. Thus, we complete the proof of necessity.
Sufficiency. Give a cube and suppose . By we have It follows by Minkowski’s inequality that Thus, and is not trivial.

Theorem 3. The space is a subset of . Furthermore, (a)if , then ;(b)if , then, for all , .

Proof. Let . For any cube and , if is small enough, we know that the Lebesgue measure of the set is bigger than . Since is nondecreasing, Consequently, For a small enough , we obtain Thus .
(a) Note that For a cube and for every with , Therefore, Thus , and this deduces that .
(b) Consider the Schwartz space consisting of all those functions on such that for all multi-indices and of nonnegative integers, where and .
Let be a fixed function such that the Fourier transform of has support in the unit ball and on the cube . Let be a sequence of real numbers and define where is the first coordinate of . By [10], if and only if .
Suppose that . Set for . We know that . Choosing , we have Since for and , Since , we have . Hence, Writing , , we obtain

For any cube of , when , we have that . By the definition, the space depends on when . In fact, depends only on when is near origin, which can be found by the following theorem. Here the proof of the theorem is left to the reader.

Theorem 4. The following statements are true.(a)Suppose that for some . One defines . Then .(b)Let Then .

Remark 5. If , we have . Since is trivial for , we only pay attention to the case .

3. Weighted Functions

The characterization of depends on the properties of the weight function obviously. In this section we give several results about the weight functions that are needed for the next section.

In the analytic spaces, the auxiliary function plays a key role; see [12, 14, 15], for example.

To study spaces, we need some more constraints on as follows: Note that (46) implies the following two conditions:

In particular, if we choose , then condition (47) holds if and only if ; condition (48) holds if and only if ; and condition (49) holds if and only if .

Lemma 6. Let satisfy for some . Then one can find another nonnegative weight function with such that , and the new weight function has the following properties.(a) is nondecreasing on .(b) on , and thus satisfies condition (50).(c) on .(d) is differentiable (up to any given order) on .(e)For some small enough , is nonincreasing on . Consequently, is also nonincreasing on .

Proof. By Theorem 4, we may assume that for . Since satisfies condition (50), we claim that If , the claim is true. If , the claim will be confirmed by showing For the case of , by (50), Taking in (50), we have Since , for , we have Thus Now, we prove In fact, if , , If , and , If , and , Hence, Therefore, our claim above has been confirmed. Let Then for .
(a) Fix and consider the difference Since is nondecreasing and nonnegative, we have
(b) Using the assumption that is nondecreasing again, we obtain for . On the other hand, Thus, on and .
(c) For any , we have Since is nondecreasing, .
(d) If we repeat the construction , then we can make the new weight function differentiable up to any desired order.
(e) Note that if is sufficiently small, then we have This means that is nonincreasing. The proof is complete.

The following result shows that there is no essential difference between (47) and (50).

Lemma 7. The following are equivalent. (a)Equation (50) holds for .(b)There exists a weight , comparable with , such that, for some small enough , is nonincreasing on .(c)Equation (47) holds for .

Proof. We assume that for .
is obvious by Lemma 6.
Suppose that (b) holds. We have We obtain that (50) holds for since is comparable with . It means that holds.
For, assume that (47) holds for . We claim that If , it is clear that and by (47) Thus, we have which gives the claim. We define It is easy to check that is nondecreasing. Since is nondecreasing, it follows that ,  . We note that, for , Thus, we obtain that For , we have Therefore, we get that on . Note that if is sufficiently small, then we have This means that is nonincreasing.
Suppose that is nonincreasing on . For , which gives Thus holds.

Lemma 8. Let satisfy (49). Then one can find another nonnegative weight function such that , and the new weight function has the following properties. (a) is nondecreasing on .(b) on , and thus satisfies condition (49).(c)For some small enough , is nondecreasing on . Consequently, is also nondecreasing on . Conversely, if is nondecreasing, for some , then (49) holds for .(d)

Proof. Assume that for . Define Note that (49) is a condition for . Then, consider the following.
(a) is obvious.
(b) For , On the other hand, since we always assume that , we obtain that Thus, on . For , clearly, . Therefore, on and we get that .
(c) If , for some small enough , If , is nondecreasing. Thus is nondecreasing on .
Conversely, if is nondecreasing, for some , then, for , which gives
(d) Note that is nondecreasing. For , we have The proof is complete.

We end this section by giving an example. Fix , and set Since we obtain that is not trivial and Moreover, a direct calculation shows that (46) and (47) hold for .

4. Carleson-Type Measures

Let be a cube of and let denote the upper half space based on . Define the Carleson box as follows: For and a positive Borel measure on , is said to be a -Carleson measure if for some and all cubes .

Denote by the distance of the point to the boundary . Also stands for the symmetric point of with respect to ; that is, if , then .

A positive Borel measure is said to be a -Carleson measure on , as a modification of -Carleson measure, provided

Clearly, if , then is a -Carleson measure on if and only if is a -Carleson measure on . Now, we give a characterization of -Carleson measure as follows.

Theorem 9. Let satisfy (48). Let be a positive Borel measure on . Then is a -Carleson measure if and only if

Proof (sufficiency). Let be a cube and take to be the center of the Carleson box . Then . If , then and hence Thus, if (95) holds, then is a -Carleson measure.
Necessity. For , let be the cube with center and edge length . Set to be the Carleson box for each positive integer . It is clear that Then Since is a -Carleson measure, This together with (48) yields

Let be a measurable function on satisfying Its Poisson integral is defined by where The gradient of is It is known that (101) holds for (see [9]).

The following main theorem generalizes the result of in [10].

Theorem 10. Suppose that (47) and (81) hold for . Let with (101). Then if and only if is a -Carleson measure.

In order to prove Theorem 10, we need the following Hardy-type inequalities.

Lemma 11 (see [18]). Let , , and . Assume that functions and are measurable and nonnegative in the interval . Then holds for all measurable functions if and only if holds for all measurable functions if and only if Here depends only on , , or .

The following Stegenga-type estimate will be used in the proof of Theorem 10.

Lemma 12. Suppose that (81) holds for and Let and be cubes in centered at with and let satisfy (101). Then, there is a constant independent of , , and , such that

Proof. Without loss of generality, we may assume that . Let be a function with such that on , , and Following Stegenga [19], we write Then we have for the corresponding Poisson integrals. Since is constant, it contributes nothing to the integral with the gradient square.
Note that We obtain Hence, A direct calculation shows that which gives Therefore, We write with and . Let Then Thus, Note that By Lemma 11, Since (81) holds for , by Lemma 11 again, Therefore, By Hölder’s inequality, Note that Hence, It follows that Since this deduces that We have Repeating the procedure above, we also can obtain Therefore, To estimate , we note that Thus We have Hence, To handle , note that We obtain Similarly, Moreover, if and  , then and, by , we have Combining the inequalities above, Lemma 12 is proved.

Proof of Theorem 10. We assume that for . By Lemma 7, satisfies (47) if and only if Sufficiency. Let be a cube and is the Poisson integral of . Note that By Minkowski’s inequality, we have Hence, The last inequality above holds by Lemma 11 since satisfies Thus, For , Similarly, we get Note that When , we employ Minkowski’s inequality to get Hence, By the triangle inequality, we get Thus .
Necessity. Note that Thus satisfies the conditions of Lemma 12. Let and be cubes in centered at with . Since , by [10, p. 590], we have Note that Lemma 12 gives The proof of Theorem 10 is complete.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors are supported by NSF of China (no. 11371234).