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Journal of Function Spaces and Applications
Volume 2012 (2012), Article ID 825240, 9 pages
Research Article

Nontangential Limits for Modified Poisson Integrals of Boundary Functions in a Cone

Department of Mathematics and Information Science, Henan University of Economics and Law, Zhengzhou 450002, China

Received 17 May 2012; Accepted 8 July 2012

Academic Editor: Dachun Yang

Copyright © 2012 Lei Qiao. 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.


Our aim in this paper is to deal with non-tangential limits for modified Poisson integrals of boundary functions in a cone, which generalized results obtained by Brundin and Mizuta-Shimomura.

1. Introduction and Main Results

Let 𝐑 and 𝐑+ be the set of all real numbers and the set of all positive real numbers, respectively. We denote by 𝐑𝑛(𝑛2) the 𝑛-dimensional Euclidean space. A point in 𝐑𝑛 is denoted by 𝑃=(𝑋,𝑥𝑛), where 𝑋=(𝑥1,𝑥2,,𝑥𝑛1). The Euclidean distance of two points 𝑃 and 𝑄 in 𝐑𝑛 is denoted by |𝑃𝑄|. Also |𝑃𝑂| with the origin 𝑂 of 𝐑𝑛 is simply denoted by |𝑃|. The boundary, the closure, and the complement of a set 𝐒 in 𝐑𝑛 are denoted by 𝜕𝐒, 𝐒, and 𝐒𝑐, respectively.

We introduce a system of spherical coordinates (𝑟,Θ),  Θ=(𝜃1,𝜃2,,𝜃𝑛1), in 𝐑𝑛 which are related to cartesian coordinates (𝑥1,𝑥2,,𝑥𝑛1,𝑥𝑛) by 𝑥𝑛=𝑟cos𝜃1.

For positive functions 1 and 2, we say that 12 if 1𝑀2 for some constant 𝑀>0. If 12 and 21, we say that 12.

For 𝑃𝐑𝑛 and 𝑅>0, let 𝐵(𝑃,𝑅) denote the open ball with center at 𝑃 and radius 𝑅 in 𝐑𝑛. The unit sphere and the upper half unit sphere are denoted by 𝐒𝑛1 and 𝐒+𝑛1, respectively. For simplicity, a point (1,Θ) on 𝐒𝑛1 and the set {Θ;(1,Θ)Ω} for a set Ω, Ω𝐒𝑛1 are often identified with Θ and Ω, respectively. For two sets Ξ𝐑+ and Ω𝐒𝑛1, the set {(𝑟,Θ)𝐑𝑛;𝑟Ξ,(1,Θ)Ω} in 𝐑𝑛 is simply denoted by Ξ×Ω. In particular, the half space 𝐑+×𝐒+𝑛1={(𝑋,𝑥𝑛)𝐑𝑛;𝑥𝑛>0} will be denoted by 𝐓𝑛.

By 𝐶𝑛(Ω), we denote the set 𝐑+×Ω in 𝐑𝑛 with the domain Ω on 𝐒𝑛1. We call it a cone. Then 𝑇𝑛 is a special cone obtained by putting Ω=𝐒+𝑛1. We denote the sets 𝐼×Ω and 𝐼×𝜕Ω with an interval on 𝐑 by 𝐶𝑛(Ω;𝐼) and 𝑆𝑛(Ω;𝐼). By 𝑆𝑛(Ω) we denote 𝑆𝑛(Ω;(0,+)) which is 𝜕𝐶𝑛(Ω){𝑂}.

Let Ω be a domain on 𝐒𝑛1 with smooth boundary. Consider the Dirichlet problem: Λ𝑛+𝜆𝜑=0onΩ,𝜑=0on𝜕Ω,(1.1) where Λ𝑛 is the spherical part of the Laplace operator Δ𝑛Δ𝑛=𝑛1𝑟𝜕+𝜕𝜕𝑟2𝜕𝑟2+Λ𝑛𝑟2.(1.2) We denote the least positive eigenvalue of this boundary value problem by 𝜆Ω and the normalized positive eigenfunction corresponding to 𝜆Ω by 𝜑Ω(Θ), Ω𝜑2Ω(Θ)𝑑𝜎Θ=1,(1.3) where 𝑑𝜎Θ is the surface area on 𝑆𝑛1. We denote the solutions of the equation 𝑡2+(𝑛2)𝑡𝜆Ω=0 by 𝛼Ω,𝛽Ω (𝛼Ω,𝛽Ω>0). If Ω=𝐒+𝑛1, then 𝛼Ω=1,𝛽Ω=𝑛1, and 𝜑1(Θ)=(2𝑛𝑠𝑛1)1/2cos𝜃1, where 𝑠𝑛 is the surface area 2𝜋𝑛/2(Γ(𝑛/2))1 of 𝐒1.

To simplify our consideration in the following, we will assume that if 𝑛3, then Ω is a 𝐶2,𝛼-domain (0<𝛼<1) on 𝐒𝑛1 surrounded by a finite number of mutually disjoint closed hypersurfaces (e.g., see ([1], pages 88-89) for the definition of 𝐶2,𝛼-domain). Then by modifying Miranda’s method ([2], pages 7-8), we can prove the following inequality: 𝜑Ω(Θ)dist(Θ,𝜕Ω)(ΘΩ).(1.4)

For any (1,Θ)Ω, we have (see [3]) 𝜑Ω(Θ)dist(1,Θ),𝜕𝐶𝑛(Ω),(1.5) which yields that 𝛿(𝑃)𝑟𝜑Ω(Θ),(1.6) where 𝛿(𝑃)=dist(𝑃,𝜕𝐶𝑛(Ω)) and 𝑃=(𝑟,Θ)𝐶𝑛(Ω).

Let 𝐺Ω(𝑃,𝑄)(𝑃=(𝑟,Θ),𝑄=(𝑡,Φ)𝐶𝑛(Ω)) be the Green function of 𝐶𝑛(Ω). We define the Poisson kernel 𝐾Ω(𝑃,𝑄) by 𝐾Ω1(𝑃,𝑄)=𝑐𝑛𝜕𝜕𝑛𝑄𝐺Ω(𝑃,𝑄),(1.7) where 𝑐𝑛=(2𝜋𝑛=2,𝑛2)𝑠𝑛𝑛3,(1.8)𝑄𝑆𝑛(Ω) and 𝜕/𝜕𝑛𝑄 denotes the differentiation at 𝑄 along the inward normal into 𝐶𝑛(Ω).

In this paper, we consider functions 𝑓𝐿𝑝(𝜕𝐶𝑛(Ω)), where 1𝑝<. Then the Poisson integral 𝑊Ω𝑓(𝑃)(𝑃𝐶𝑛(Ω)) is defined by 𝑊Ω𝑓(𝑃)=𝑆𝑛(Ω)𝐾Ω(𝑃,𝑄)𝑓(𝑄)𝑑𝜎𝑄,(1.9) where 𝑑𝜎𝑄 is the surface area element on 𝑆𝑛(Ω).

Remark 1.1. Let Ω=𝐒+𝑛1. Then 𝐺𝐒+𝑛1||(𝑃,𝑄)=log𝑃𝑄||||||||||log𝑃𝑄𝑛=2,𝑃𝑄2𝑛||𝑃𝑄||2𝑛𝑛3,(1.10) where 𝑄=(𝑌,𝑦𝑛), that is, 𝑄 is the mirror image of 𝑄=(𝑌,𝑦𝑛) with respect to 𝜕𝑇𝑛. Hence, for the two points 𝑃=(𝑋,𝑥𝑛)𝑇𝑛 and 𝑄=(𝑌,𝑦𝑛)𝜕𝑇𝑛, we have 𝑐𝑛𝐾𝐒+𝑛1𝜕(𝑃,𝑄)=𝜕𝑛𝑄𝐺𝐒+𝑛12||||(𝑃,𝑄)=𝑃𝑄2𝑥𝑛||||𝑛=2,2(𝑛2)𝑃𝑄𝑛𝑥𝑛𝑛3.(1.11)

We fix an open, nonempty, and bounded set 𝐺(Ω)𝜕𝐶𝑛(Ω). In 𝐶𝑛(Ω), we normalise the extension, with respect to 𝐺(Ω), by 𝒫Ω𝑊𝑓(𝑃)=Ω𝑓(𝑃)𝑊Ω𝜒𝐺(Ω)(,𝑃)(1.12) where 𝜒𝐺(Ω) denotes the characteristic function of 𝐺(Ω).

Let Γ(Ω,𝜁)=𝑃=(𝑟,Θ)𝐶𝑛||||(Ω)(𝑟,Θ)𝜁𝛿(𝑃)(1.13) be a nontangential cone in 𝐶𝑛(Ω) with vertex 𝜁𝜕𝐶𝑛(Ω).

We define 𝑝1(𝑓,𝑙,𝑃)=𝑙𝑛1𝐵(𝑃,𝑙)||||𝑓(𝑄)𝑝𝑑𝜎𝑄1/𝑝,𝔼𝑝𝑓(𝐺(Ω))=𝑃𝐺(Ω)𝑝.(𝑓𝑓(𝑃),𝑙,𝑃)0as𝑙0(1.14)

Note that, if 𝑓𝐿𝑝(𝜕𝐶𝑛(Ω)), then |𝐺(Ω)𝔼𝑝𝑓(𝐺(Ω))|=0 (a.e. point is a Lebesgue point).

In 𝑇𝑛, the following conclusion was proved by Brundin (see ([4], pages 11–16)) and Mizuta and Shimomura (see ([5], Theorem 3)), respectively. In the unit disc, about related results, we refer the readers to the papers by Sjögren (see [6, 7]), Rönning (see [8]), and Brundin (see [9]).

Theorem A. For a.e. 𝜁𝐺(𝐒+𝑛1), 𝒫𝐒+𝑛1𝑓(𝑃)𝑓(𝜁) (see Remark 1.1 for the definition of 𝒫𝐒+𝑛1𝑓(𝑃)) as 𝑃𝜁 along Γ(𝐒+𝑛1,𝜁).

Our aim is to generalize Theorem A to the conical case.

Theorem 1.2. For any 𝜁𝔼𝑝𝑓(𝐺(Ω)) (in particular, for a.e. 𝜁𝐺(Ω)) one has that 𝒫Ω𝑓(𝑃)𝑓(𝜁) as 𝑃𝜁 along Γ(Ω,𝜁).

2. Some Lemmas

Lemma 2.1. One has 𝐾Ω(𝑃,𝑄)𝑟𝛽Ω𝑡𝛼Ω1𝜑Ω(Θ),resp.𝐾Ω(𝑃,𝑄)𝑟𝛼Ω𝑡𝛽Ω1𝜑Ω,(Θ)(2.1) for any 𝑃=(𝑟,Θ)𝐶𝑛(Ω) and any 𝑄=(𝑡,Φ)𝑆𝑛(Ω) satisfying 0<𝑡/𝑟4/5(resp.0<𝑟/𝑡4/5); 𝐾Ω(𝑃,𝑄)𝑟𝜑Ω(Θ)||||𝑃𝑄𝑛,(2.2) for any 𝑃=(𝑟,Θ)𝐶𝑛(Ω) and any 𝑄=(𝑡,Φ)𝑆𝑛(Ω;(4𝑟/5,5𝑟/4)).

Proof. These immediately follow from ([10], Lemma 2), ([11], Lemma 4 and Remark), and (1.4).

Lemma 2.2. One has 𝑊Ω1(𝑃)=𝑂(1)as𝑃𝜁𝐺.(2.3)

Proof. Write 𝑊Ω1(𝑃)=𝐸1+𝐸2+𝐸3=𝑈1(𝑃)+𝑈2(𝑃)+𝑈3(𝑃),(2.4) where 𝐸1=𝑆𝑛4Ω;0,5𝑟,𝐸2=𝑆𝑛5Ω;4𝑟,,𝐸3=𝑆𝑛4Ω;55𝑟,4𝑟.(2.5)
By (2.1), we have the following estimates 𝑈1(𝑃)𝑟𝛽Ω𝜑Ω(Θ)𝐸1𝑡𝛼Ω1𝑑𝜎𝑄𝑠𝑛𝛽Ω45𝛽Ω𝜑Ω𝑈(Θ),(2.6)2𝑠(𝑃)𝑛𝛼Ω45𝛼Ω𝜑Ω(Θ).(2.7)
Next, we will estimate 𝑈3(𝑃). Take a sufficiently small positive number 𝑘 such that 𝐸3𝑃=(𝑟,Θ)Λ(𝑘)𝐵1𝑃,2𝑟,(2.8) where Λ(𝑘)=𝑃=(𝑟,Θ)𝐶𝑛(Ω);inf𝑧𝜕Ω||||(1,Θ)(1,𝑧)<𝑘,0<𝑟<.(2.9)
Since 𝑃𝜁𝐺, we only consider the case 𝑃Λ(𝑘). Now put 𝐻𝑖(𝑃)=𝑄𝐸3;2𝑖1𝛿||||(𝑃)𝑃𝑄<2𝑖𝛿.(𝑃)(2.10)
Since 𝑆𝑛(Ω){𝑄𝐑𝑛|𝑃𝑄|<𝛿(𝑃)}=Ø, we have 𝑈3(𝑃)𝑖(𝑃)𝑖=0𝐻𝑖(𝑃)𝑟𝜑Ω(Θ)||||𝑃𝑄𝑛𝑑𝜎𝑄,(2.11) where 𝑖(𝑃) is a positive integer satisfying 2𝑖(𝑃)1𝛿(𝑃)𝑟/2<2𝑖(𝑃)𝛿(𝑃).
By (1.6) we have 𝐻𝑖(𝑃)𝑟𝜑Ω(Θ)||||𝑃𝑄𝑛𝑑𝜎𝑄𝑟𝜑Ω(Θ)𝐻𝑖(𝑃)1𝛿(𝑃)𝑑𝜎𝑄=𝑟𝜑Ω(Θ)𝑠𝛿(𝑃)𝑛2𝑖(𝑃)𝑠𝑛2𝑖(𝑃)(2.12) for 𝑖=0,1,2,,𝑖(𝑃).
So 𝑈3(𝑃)𝑂(1).(2.13)
Combining (2.6)–(2.13), Lemma 2.2 is proved.

Lemma 2.3. One has 𝑊Ω𝜒𝐺(Ω)(𝑃)=𝑊Ω1(𝑃)+𝑂(1)as𝑃𝜁𝐺(Ω).(2.14)

Proof. In fact, we only need to prove 𝑈4(𝑃)=𝑆𝑛(Ω)𝐺(Ω)𝐾Ω(𝑃,𝑄)𝑑𝜎𝑄𝑂(1).(2.15)
Write 𝑈4(𝑃)=(𝑆𝑛(Ω)𝐺(Ω))𝐸1+(𝑆𝑛(Ω)𝐺(Ω))𝐸2+(𝑆𝑛(Ω)𝐺(Ω))𝐸3=𝑈5(𝑃)+𝑈6(𝑃)+𝑈7(𝑃),(2.16) where 𝐸1, 𝐸2, and 𝐸3 are sets on 𝑆𝑛(Ω) used in Lemma 2.2.
Obviously, 𝑈5(𝑃)𝑈1𝑈(𝑃)𝑂(1),(2.17)6(𝑃)𝑈2(𝑃)𝑂(1).(2.18)
Further, we have by (2.2) 𝑈7(𝑃)𝑟𝜑Ω(Θ)(𝑆𝑛(Ω)𝐺(Ω))𝐸31||||𝑃𝑄𝑛𝑑𝜎𝑄𝑠𝑛𝑑||𝜁||𝜑Ω(Θ)(𝑃𝜁𝐺(Ω)),(2.19) where 𝑑=inf𝑄𝜕𝐶𝑛(Ω)𝐺(Ω)||||.𝑄𝜁(2.20)
Combining (2.17)–(2.19), (2.15) holds which gives the conclusion.

3. Proof of the Theorem 1.2

As 𝑃𝜁𝐺(Ω), 𝑊Ω𝜒𝐺(Ω)(𝑃)=𝑂(1) from Lemmas 2.2 and 2.3.

Now, let 𝑓𝐿𝑝(𝜕𝐶𝑛(Ω)) and 𝜁𝔼𝑝𝑓(𝐺(Ω)) be given. We may, without loss of generality, assume that 𝑓(𝜁)=0. Furthermore, we assume that 𝑃=(𝑟,Θ)Γ(Ω,𝜁). For short, let 𝑠=|(𝑟,Θ)𝜁|. We write 𝑊Ω𝑓(𝑃)=𝐸1+𝐸2+𝐸3𝐵(𝜁,2𝑠)+𝐸3𝐵𝑐(𝜁,2𝑠)=𝑉1𝑓(𝑃)+𝑉2𝑓(𝑃)+𝑉3𝑓(𝑃)+𝑉4𝑓(𝑃),(3.1) where 𝐸1, 𝐸2, and 𝐸3 are sets on 𝑆𝑛(Ω) used in Lemma 2.2.

By using Hölder’s inequality, (2.1), we have the following estimates ||𝑉1||𝑓(𝑃)𝑟𝛽Ω𝜑Ω(Θ)𝐸1𝑡𝛼Ω1𝑓(𝑄)𝑑𝜎𝑄𝑟(1𝑛)/𝑝𝑓𝑝,||𝑉2||𝑓(𝑃)𝑟(1𝑛)/𝑝𝑓𝑝.(3.2)

Similar to the estimate of 𝑈3(𝑃) in Lemma 2.2, we only consider the following inequality by (1.6) 𝐻𝑖(𝑃)𝑟𝜑Ω(Θ)||||𝑃𝑄𝑛𝑑𝜎𝑄𝑟𝜑Ω(Θ)𝐻𝑖(𝑃)12𝑖1𝛿(𝑃)𝑛𝑑𝜎𝑄𝑟𝛼Ω𝜑Ω(Θ)𝐸2𝑡𝛽Ω1||𝑓||(𝑄)𝑑𝜎𝑄𝑟(1𝑛)/𝑝𝑓𝑝(3.3) for 𝑖=0,1,2,,𝑖(𝑃), which is similar to the estimate of 𝑉2𝑓(𝑃).

So ||𝑉3||𝑓(𝑃)𝑟(1𝑛)/𝑝𝑓𝑝.(3.4)

Notice that |𝑃𝑄|>(1/2)|𝜁𝑄| in the case 𝑄𝐸3𝐵𝑐(𝜁,2𝑠). By (1.6) and (2.2), we have ||𝑉4||𝑓(𝑃)𝛿(𝑃)𝐸3𝐵𝑐(𝜁,2𝑠)||||𝑓(𝑄)||||𝑃𝑄𝑛𝑑𝜎𝑄𝛿(𝑃)𝑖=1𝐸3(𝐵(𝜁,2𝑖+1𝑠)𝐵(𝜁,2𝑖𝑠))||𝑓||(𝑄)||||𝜁𝑄𝑛𝑑𝜎𝑄𝛿(𝑃)𝑖=112𝑖𝑠𝑛𝐸3𝐵(𝜁,2𝑖+1𝑠)||||𝑓(𝑄)𝑑𝜎𝑄𝛿(𝑃)𝑖=11𝑓,2𝑖+1𝑠,𝜁𝛿(𝑃)𝑖=12𝑖+2𝑠2𝑖+1𝑠1(𝑓,𝑙,𝜁)𝑙𝑑𝑙𝛿(𝑃)𝑠1(𝑓,𝑙,𝜁)𝑙𝑑𝑙𝛿(𝑃)𝛿(𝑃)1(𝑓,𝑙,𝜁)𝑙𝑑𝑙.(3.5)

Thus, it follows that ||𝒫Ω||1𝑓(𝑃)𝑂||𝑉(1)1||+||𝑉𝑓(𝑃)2||+||𝑉𝑓(𝑃)3||+||𝑉𝑓(𝑃)4||𝑓(𝑃)𝑟(1𝑛)/𝑝𝑓𝑝+𝛿(𝑃)𝛿(𝑃)1(𝑓,𝑙,𝜁)𝑙𝑑𝑙.(3.6)

Using the fact that 𝑠𝛿(𝑃)𝑟𝜑Ω(Θ), we get ||𝒫Ω||𝑓(𝑃)1(𝑓,2𝑠,𝜁)+𝛿(𝑃)𝛿(𝑃)1(𝑓,𝑙,𝜁)𝑙𝑑𝑙.(3.7)

It is clear that 𝛿(𝑃)1(𝑓,𝑙,𝜁)𝑙𝑑𝑙(3.8) is a convergent integral, since 1(𝑓,l,𝜁)𝑙𝑠1𝑛𝑠𝑛/𝑞𝑓𝑝𝑠1(𝑛/𝑝)𝑓𝑝(3.9) from the Hölder’s inequality.

Now, as 𝛿(𝑃)0, we also have 𝑠0. Since 𝑓(𝜁)=0 and since we have assumed that 𝜁𝔼𝑝𝑓(𝐺(Ω)) (and thus that 𝜁𝔼1𝑓(𝐺(Ω))), it follows that 𝒫Ω𝑓(𝑃)0=𝑓(𝜁) as 𝑃=(𝑟,Θ)𝜁 along Γ(Ω,𝜁). This concludes the proof.


This paper is supported by SRFDP (no. 20100003110004) and NSF of China (no. 11071020).


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