#### Abstract

By virtue of the modified Riesz kernel introduced by Qiao (2012), we give the integral representations for solutions of the Neumann problems in a half space.

#### 1. Introduction and Main Results

Let and be the sets of all real numbers and of all positive real numbers, respectively. Let    denote the -dimensional Euclidean space with points , where and . The boundary and closure of an open set of are denoted by and , respectively. For and , let denote the open ball with center at and radius in . Let .

The upper half space is the set , whose boundary is . For a set , , we denote and by and , respectively. We identify with and with , writing typical points , as , , where . Let be the angle between and , that is, and , where is the th unit coordinate vector and is normal to .

We will say that a set has a covering if there exists a sequence of balls with centers in such that , where is the radius of and is the distance between the origin and the center of .

For positive functions and , we say that if for some positive constant . Throughout this paper, let denote various constants independent of the variables in question. Further, we use the standard notations , is the integer part of , and , where is a positive real number.

Given a continuous function on , we say that is a solution of the Neumann problem on with , if is a harmonic function on and for every point .

For and , consider the kernel function where and is the surface area of the -dimensional unit sphere.

The Neumann integral on is defined by where is a continuous function on .

The Neumann integral is a solution of the Neumann problem on with if (see [1, Theorem 1 and Remarks])

In this paper, we consider functions satisfying for and .

For and , we define the positive measure on by If is a measurable function on satisfying (5), we remark that the total mass of is finite.

Let and . For each , the maximal function is defined by

The set is denoted by .

To obtain the Neumann solution for the boundary data on , as in [2, 3], we use the following modified Riesz kernel defined by where is a nonnegative integer.

For and , the generalized Neumann kernel is defined by

Put where is continuous function on . Here note that is nothing but the Neumann integral .

The following result is due to Su (see [4]).

Theorem A. If is a continuous function on satisfying (5) with and , then

Our first aim is to be concerned with the growth property of at infinity in a half space and establish the following theorem.

Theorem 1. Let , , and If is a measurable function on satisfying (5), then there exists a covering of satisfying such that

Remark 2. In the case that , , and , then (13) is a finite sum and the set is a bounded set. So (14) holds in . That is to say, (11) holds. This is just the result of Theorem A.

Corollary 3. Let , and If is a measurable function on satisfying (5), then

As an application of Theorem 1, we now show the solution of the Neumann problem with continuous data on . About the solutions of the Dirichlet problem with respect to the Schrödinger operator in a half space, we refer readers to the paper by Su (see [5]).

Theorem 4. Let , , , and be defined as in Theorem 1. If is a continuous function on satisfying (5), then the function is a solution of the Neumann problem on with and (14) holds, where the exceptional set has a covering satisfying (13).

Finally we have the following result.

Theorem 5. Let , , be a positive integer and If is a continuous function on satisfying (5) and is a solution of the Neumann problem on with such that then for any , where and is a polynomial of of degree less than .

#### 2. Lemmas

In our discussions, the following estimates for the kernel function are fundamental (see [6, Lemma 4.2] and [3, Lemmas 2.1 and 2.4]).

Lemma 6. (1) If , then .
(2) If , then .
(3) If , then .
(4) If and , then .

The following Lemma is due to Qiao (see [3]).

Lemma 7. If , , and is a positive measure in satisfying , then has a covering    such that

Lemma 8. Let , , , and be defined as in Theorem 1. If is a local integral and upper semicontinuous function on satisfying (5), then for any fixed point .

Proof. Let be any fixed point an and let be any positive number. Take a positive number , , such that for any .
By Lemma 6(4) and (5), we can choose a number , , such that for any .
Put
Since for any and , we have for any .
Since for any , we observe that
Finally (23), (27), and (29) yield
From Lemma 6(4) we obtain for any .
These and (24) yield

Now the conclusion immediately follows.

Lemma 9 (see [1, Lemma 1]). If is a harmonic polynomial of of degree and vanishes on , then there exists a polynomial of degree such that

#### 3. Proof of Theorem 1

We prove only the case ; the proof of the case is similar.

For any , there exists such that

Take any point such that and write where

First note that so that

If and , then . By Lemma 6(1), (34), and Hölder inequality, we have

Put where

If , then

Moreover, by (34) and (39) we get

That is,

By Lemma 6(3), (34), and Hölder inequality, we have

If , then . We obtain Lemma 6(4), (34), and Hölder inequality:

Finally, we will estimate . Take a sufficiently small positive number such that for any , where and divide into two sets and .

If , then there exists a positive number such that for any , and hence which is similar to the estimate of .

We will consider the case . Now put where .

Since , we have where is a positive integer satisfying .

Similar to the estimate of we obtain for .

Since , we have for and

So

Combining (38) and (44)–(54), we obtain that if is sufficiently large and is a sufficiently small number, then as , where . Finally, there exists an additional finite ball covering , which, together with Lemma 7, gives the conclusion of Theorem 1.

#### 4. Proof of Theorem 4

For any fixed , take a number satisfying . If , then . By (5), Lemma 6(4), and Hölder inequality, we have Hence, is absolutely convergent and finite for any . Thus, is harmonic on .

To prove for any point , we only need to apply Lemma 8 to and .

We complete the proof of Theorem 4.

#### 5. Proof of Theorem 5

Consider the function . Then it follows from Theorems 4 and 5 that is a solution of the Neumann problem on with and it is an even function of (see [1, page 92]).

Sincefor any , we havefrom Theorem 4.

Moreover, (18) gives that

This implies that is a polynomial of degree less than (see [7, Appendix]), which gives the conclusion of Theorem 5 from Lemma 9.

#### Conflict of Interests

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

#### Acknowledgment

This work was supported by the National Natural Science Foundation of China under Grant nos. 11301140 and U1304102.