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Growth estimates for modified Neumann integrals in a half-space
Journal of Inequalities and Applications volume 2013, Article number: 572 (2013)
Abstract
Our aim in this paper is to deal with the growth properties for modified Neumann integrals in a half-space of . As an application, the solutions of Neumann problems in it for a slowly growing continuous function are also given.
1 Introduction and main results
Let R and be the sets of all real numbers and of all positive real numbers, respectively. Let () denote the n-dimensional Euclidean space with points , where and . The boundary and closure of an open set Ω of are denoted by ∂ Ω and , respectively. For and , let denote the open ball with center at x and radius r in .
The upper half-space is the set , whose boundary is ∂H. For a set F, , we denote and by HF and , respectively. We identify with and with , writing typical points as , , where . Let θ be the angle between x and , i.e., and , where is the i th unit coordinate vector and is normal to ∂H.
We shall say that a set has a covering if there exists a sequence of balls with centers in H 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 M. Throughout this paper, let M denote various constants independent of the variables in question. Further, we use the standard notations, is the integer part of d and , where d is a positive real number.
Given a continuous function f in ∂H, we say that h is a solution of the Neumann problem in H with f, if h is a harmonic function in H and
for every point .
For and , consider the kernel function
where and is the surface area of the n-dimensional unit sphere. It has the expression
where is the ultraspherical (Gegenbauer) polynomials [1]. The series converges for , and each term in it is a harmonic function of x.
The Neumann integral is defined by
where f is a continuous function on ∂H, and is the volume of the unit n-ball.
The Neumann integral is a solution of the Neumann problem on H with f if (see [[2], Theorem 1 and Remarks])
In this paper, we consider functions f satisfying
for and .
For this p and α, we define the positive measure μ on by
If f is a measurable function on ∂H satisfying (1.1), we remark that the total mass of μ 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 f, as in [3–6], we use the following modified kernel function defined by
for a non-negative integer m.
For and , the generalized Neumann kernel is defined by
Since () is harmonic in H (see [4]), is also harmonic in H for any fixed . Also, will be of order as (see [[7], Theorem D]).
Put
where f is a continuous function on ∂H. Here, note that is nothing but the Neumann integral .
The following result is due to Siegel and Talvila (see [[5], Corollary 2.1]). For similar results with respect to the Schrödinger operator in a half-space, we refer readers to papers by Su (see [8]).
Theorem A If f is a continuous function on ∂H satisfying (1.1) with and , then
The next result deals with a type of uniqueness of solutions for the Neumann problem on H (see [[9], Theorem 3]).
Theorem B Let l be a positive integer and m be a non-negative integer. If f is a continuous function on ∂H satisfying
and h is a solution of the Neumann problem on H with f such that
then
for any , where is the positive part of h,
and is a polynomial of of degree less than .
Our first aim is to be concerned with the growth property of at infinity and establish the following theorem.
Theorem 1 Let , , and
If f is a measurable function on ∂ satisfying (1.1), then there exists a covering of (⊂H) satisfying
such that
Corollary 1 Let , and
If f is a measurable function on ∂H satisfying (1.1), then
As an application of Theorem 1, we now show the solution of the Neumann problem with continuous data on H.
Theorem 2 Let p, β, α and m be defined as in Theorem 1. If f is a continuous function on ∂H satisfying (1.1), then the function is a solution of the Neumann problem on H with f and (1.4) holds, where the exceptional set (⊂H) has a covering satisfying (1.3).
Remark In the case , and , then (1.3) is a finite sum and the set is a bounded set. So (1.4) holds in H. That is to say, (1.2) holds. This is just the result of Theorem A.
Corollary 2 Let , and
If f is a continuous function on ∂H satisfying (1.1), then the function is a solution of the Neumann problem on H with f and (1.5) holds.
The following result extends Theorem B, which is our result in the case and .
Theorem 3 Let , , l be a positive integer and
If f is a continuous function on ∂H satisfying (1.1) and h is a solution of the Neumann problem on H with f such that
then
for any and is a polynomial of of degree less than .
2 Lemmas
In our discussions, the following estimates for the kernel function are fundamental (see [[10], Lemma 4.2] and [[4], Lemmas 2.1 and 2.4]).
Lemma 1
-
(1)
If , then .
-
(2)
If , then .
-
(3)
If , then .
-
(4)
If and , then .
The following lemma is due to Qiao (see [4]).
Lemma 2 If , and λ is a positive measure in satisfying , then has a covering () such that
Lemma 3 ([[9], Lemma 4])
Let p, β, α and m be defined as in Theorem 1. If f is a locally integral and upper semi-continuous function on ∂H satisfying (1.1), then
for any fixed .
Lemma 4 ([[2], Lemma 1])
If is a harmonic polynomial of of degree m and vanishes on ∂H, then there exists a polynomial of degree m such that
3 Proof of Theorem 1
For any , there exists such that
Take any point such that , and write
where
First note that
so that
If and , then . By Lemma 1(1), (3.1) and the Hölder inequality, we have
Put
where
If , then
Moreover, by (3.1) and (3.3), we get
That is,
By Lemma 1(3), (3.1) and the Hölder inequality, we have
If , then . We obtain, by Lemma 1(4), (3.1) and the Hölder inequality,
Finally, we shall estimate . Take a sufficiently small positive number b such that for any , where
and divide H 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 shall 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 (3.2), (3.4)-(3.7), we obtain that if is sufficiently large and ϵ is a sufficiently small number, then as , where . Finally, there exists an additional finite ball covering , which together with Lemma 2, gives the conclusion of Theorem 1.
4 Proof of Theorem 2
For any fixed , take a number R satisfying . If , then . By (1.1), Lemma 1(4) and the Hölder inequality, we have
Hence is absolutely convergent and finite for any . Thus is harmonic on H.
To prove
for any point , we only need to apply Lemma 3 to and .
We complete the proof of Theorem 2.
5 Proof of Theorem 3
Consider the function . Then it follows from Theorems 2 and 3 that is a solution of the Neumann problem on H with f and it is an even function of (see [[2], p.92]).
Since
for any , and
from Theorem 2.
Moreover, (1.6) gives that
This implies that is a polynomial of degree less than (see [[11], Appendix]), which gives the conclusion of Theorem 3 from Lemma 4.
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The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper.
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Ren, Y., Yang, P. Growth estimates for modified Neumann integrals in a half-space. J Inequal Appl 2013, 572 (2013). https://doi.org/10.1186/1029-242X-2013-572
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DOI: https://doi.org/10.1186/1029-242X-2013-572