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

Authors: Lei Qiao
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

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.


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## Body

## 1. Introduction and Main Results

LetR and R+ be the set of all real numbers and the set of all positive real numbers, respectively. We denote by Rn(n≥2) the n-dimensional Euclidean space. A point in Rn is denoted by P=(X,xn), where X=(x1,x2,…,xn-1). The Euclidean distance of two points P and Q in Rn is denoted by |P-Q|. Also |P-O| 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,…,θn-1), in Rn which are related to cartesian coordinates (x1,x2,…,xn-1,xn) by xn=rcosθ1.For positive functionsh1 and h2, we say that h1≲h2 if h1≤Mh2 for some constant M>0. If h1≲h2 and h2≲h1, we say that h1≈h2.ForP∈Rn 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 Sn-1 and S+n-1, respectively. For simplicity, a point (1,Θ) on Sn-1 and the set {Θ;(1,Θ)∈Ω} for a set Ω, Ω⊂Sn-1 are often identified with Θ and Ω, respectively. For two sets Ξ⊂R+ and Ω⊂Sn-1, the set {(r,Θ)∈Rn;r∈Ξ,(1,Θ)∈Ω} in Rn is simply denoted by Ξ×Ω. In particular, the half space R+×S+n-1={(X,xn)∈Rn;xn>0} will be denoted by Tn.ByCn(Ω), we denote the set R+×Ω in Rn with the domain Ω on Sn-1. We call it a cone. Then Tn is a special cone obtained by putting Ω=S+n-1. 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}.LetΩ be a domain on Sn-1 with smooth boundary. Consider the Dirichlet problem:
(1.1)(Λn+λ)φ=0on⁡Ω,φ=0on⁡∂Ω,
where Λn is the spherical part of the Laplace operator Δn:(1.2)Δn=n-1r∂∂r+∂2∂r2+Λnr2.
We denote the least positive eigenvalue of this boundary value problem by λΩ and the normalized positive eigenfunction corresponding to λΩ by φΩ(Θ),
(1.3)∫Ω‍φΩ2(Θ)dσΘ=1,
where dσΘ is the surface area on Sn-1. We denote the solutions of the equation t2+(n-2)t-λΩ=0 by αΩ,-βΩ (αΩ,βΩ>0). If Ω=S+n-1, then αΩ=1,βΩ=n-1, and φ1(Θ)=(2nsn-1)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 ifn≥3, then Ω is a C2,α-domain (0<α<1) on Sn-1 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:
(1.4)φΩ(Θ)≈dist⁡(Θ,∂Ω)(Θ∈Ω).For any(1,Θ)∈Ω, we have (see [3])
(1.5)φΩ(Θ)≈dist⁡((1,Θ),∂Cn(Ω)),
which yields that
(1.6)δ(P)≈rφΩ(Θ),
where δ(P)=dist(P,∂Cn(Ω)) and P=(r,Θ)∈Cn(Ω).LetGΩ(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
(1.7)KΩ(P,Q)=1cn∂∂nQGΩ(P,Q),
where
(1.8)cn={2πn=2,(n-2)snn≥3,Q∈Sn(Ω) and ∂/∂nQ denotes the differentiation at Q along the inward normal into Cn(Ω).In this paper, we consider functionsf∈Lp(∂Cn(Ω)), where 1≤p<∞. Then the Poisson integral WΩf(P)(P∈Cn(Ω)) is defined by
(1.9)WΩf(P)=∫Sn(Ω)‍KΩ(P,Q)f(Q)dσQ,
where dσQ is the surface area element on Sn(Ω).Remark 1.1.
LetΩ=S+n-1. Then
(1.10)GS+n-1(P,Q)={log⁡|P-Q*|-log⁡|P-Q|n=2,|P-Q|2-n-|P-Q*|2-nn≥3,
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
(1.11)cnKS+n-1(P,Q)=∂∂nQGS+n-1(P,Q)={2|P-Q|-2xnn=2,2(n-2)|P-Q|-nxnn≥3.We fix an open, nonempty, and bounded setG(Ω)⊂∂Cn(Ω). In Cn(Ω), we normalise the extension, with respect to G(Ω), by
(1.12)PIΩf(P)=WΩf(P)WΩχG(Ω)(P),
where χG(Ω) denotes the characteristic function of G(Ω).Let(1.13)Γ(Ω,ζ)={P=(r,Θ)∈Cn(Ω):|(r,Θ)-ζ|≲δ(P)}
be a nontangential cone in Cn(Ω) with vertex ζ∈∂Cn(Ω).We define(1.14)ℵp(f,l,P)=(1ln-1∫B(P,l)‍|f(Q)|pdσQ)1/p,Efp(G(Ω))={P∈G(Ω):ℵp(f-f(P),l,P)⟶0asl⟶0}.Note that, iff∈Lp(∂Cn(Ω)), then |G(Ω)∖𝔼fp(G(Ω))|=0 (a.e. point is a Lebesgue point).InTn, 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.ζ∈G(S+n-1), 𝒫ℐS+n-1f(P)→f(ζ) (see Remark 1.1 for the definition of 𝒫ℐS+n-1f(P)) as P→ζ along Γ(S+n-1,ζ).Our aim is to generalize Theorem A to the conical case.Theorem 1.2.
For anyζ∈𝔼fp(G(Ω)) (in particular, for a.e. ζ∈G(Ω)) one has that 𝒫ℐΩf(P)→f(ζ) as P→ζ along Γ(Ω,ζ).


## 2. Some Lemmas

Lemma 2.1.
One has(2.1)KΩ(P,Q)≈r-βΩtαΩ-1φΩ(Θ),(resp.KΩ(P,Q)≈rαΩt-βΩ-1φΩ(Θ)),
for any P=(r,Θ)∈Cn(Ω) and any Q=(t,Φ)∈Sn(Ω) satisfying 0<t/r≤4/5(resp.0<r/t≤4/5);
(2.2)KΩ(P,Q)≈rφΩ(Θ)|P-Q|n,
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(2.3)WΩ1(P)=O(1)asP⟶ζ∈G.Proof.
Write(2.4)WΩ1(P)=∫E1‍+∫E2‍+∫E3‍=U1(P)+U2(P)+U3(P),
where
(2.5)E1=Sn(Ω;(0,45r]),E2=Sn(Ω;[54r,∞)),E3=Sn(Ω;(45r,54r)).
By (2.1), we have the following estimates
(2.6)U1(P)≈r-βΩφΩ(Θ)∫E1‍tαΩ-1dσQ≈snβΩ(45)βΩφΩ(Θ),(2.7)U2(P)≈snαΩ(45)αΩφΩ(Θ).
Next, we will estimateU3(P). Take a sufficiently small positive number k such that
(2.8)E3⊂⋃P=(r,Θ)∈Λ(k)‍B(P,12r),
where
(2.9)Λ(k)={P=(r,Θ)∈Cn(Ω);inf⁡z∈∂Ω|(1,Θ)-(1,z)|<k,0<r<∞}.
SinceP→ζ∈G, we only consider the case P∈Λ(k). Now put
(2.10)Hi(P)={Q∈E3;2i-1δ(P)≤|P-Q|<2iδ(P)}.
SinceSn(Ω)∩{Q∈Rn:|P-Q|<δ(P)}=⌀, we have
(2.11)U3(P)≈∑i=0i(P)‍∫Hi(P)‍rφΩ(Θ)|P-Q|ndσQ,
where i(P) is a positive integer satisfying 2i(P)-1δ(P)≤r/2<2i(P)δ(P).
By (1.6) we have
(2.12)∫Hi(P)‍rφΩ(Θ)|P-Q|ndσQ≈rφΩ(Θ)∫Hi(P)‍1δ(P)dσQ=rφΩ(Θ)δ(P)sn2i(P)≈sn2i(P)
for i=0,1,2,…,i(P).
So(2.13)U3(P)≈O(1).
Combining (2.6)–(2.13), Lemma 2.2 is proved.Lemma 2.3.
One has(2.14)WΩχG(Ω)(P)=WΩ1(P)+O(1)asP⟶ζ∈G(Ω).Proof.
In fact, we only need to prove(2.15)U4(P)=∫Sn(Ω)-G(Ω)‍KΩ(P,Q)dσQ≲O(1).
Write(2.16)U4(P)=∫(Sn(Ω)-G(Ω))∩E1‍+∫(Sn(Ω)-G(Ω))∩E2‍+∫(Sn(Ω)-G(Ω))∩E3‍=U5(P)+U6(P)+U7(P),
where E1, E2, and E3 are sets on Sn(Ω) used in Lemma 2.2.
Obviously,(2.17)U5(P)≲U1(P)≈O(1),(2.18)U6(P)≲U2(P)≈O(1).
Further, we have by (2.2)
(2.19)U7(P)≈rφΩ(Θ)∫(Sn(Ω)-G(Ω))∩E3‍1|P-Q|ndσQ≲snd|ζ|φΩ(Θ)(P⟶ζ∈G(Ω)),
where
(2.20)d=inf⁡Q∈∂Cn(Ω)-G(Ω)|Q-ζ|.
Combining (2.17)–(2.19), (2.15) holds which gives the conclusion.


## 3. Proof of the Theorem1.2

AsP→ζ∈G(Ω), WΩχG(Ω)(P)=O(1) from Lemmas 2.2 and 2.3.Now, letf∈Lp(∂Cn(Ω)) and ζ∈𝔼fp(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
(3.1)WΩf(P)=∫E1‍+∫E2‍+∫E3∩B(ζ,2s)‍+∫E3∩Bc(ζ,2s)‍=V1f(P)+V2f(P)+V3f(P)+V4f(P),
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
(3.2)|V1f(P)|≲r-βΩφΩ(Θ)∫E1‍tαΩ-1f(Q)dσQ≲r(1-n)/p‖f‖p,|V2f(P)|≲r(1-n)/p‖f‖p.Similar to the estimate ofU3(P) in Lemma 2.2, we only consider the following inequality by (1.6)
(3.3)∫Hi(P)‍rφΩ(Θ)|P-Q|ndσQ≈rφΩ(Θ)∫Hi(P)‍1{2i-1δ(P)}ndσQ≲rαΩφΩ(Θ)∫E2‍t-βΩ-1|f(Q)|dσQ≲r(1-n)/p‖f‖p
for i=0,1,2,…,i(P), which is similar to the estimate of V2f(P).So(3.4)|V3f(P)|≲r(1-n)/p‖f‖p.Notice that|P-Q|>(1/2)|ζ-Q| in the case Q∈E3∩Bc(ζ,2s). By (1.6) and (2.2), we have
(3.5)|V4f(P)|≲δ(P)∫E3∩Bc(ζ,2s)‍|f(Q)||P-Q|ndσQ≲δ(P)∑i=1∞‍∫E3∩(B(ζ,2i+1s)∖B(ζ,2is))‍|f(Q)||ζ-Q|ndσQ≲δ(P)∑i=1∞‍(12is)n∫E3∩B(ζ,2i+1s)‍|f(Q)|dσQ≲δ(P)∑i=1∞‍ℵ1(f,2i+1s,ζ)≲δ(P)∑i=1∞‍∫2i+1s2i+2s‍ℵ1(f,l,ζ)ldl≲δ(P)∫s∞‍ℵ1(f,l,ζ)ldl≲δ(P)∫δ(P)∞‍ℵ1(f,l,ζ)ldl.Thus, it follows that(3.6)|PIΩf(P)|≲1O(1)[|V1f(P)|+|V2f(P)|+|V3f(P)|+|V4f(P)|]≲r(1-n)/p‖f‖p+δ(P)∫δ(P)∞‍ℵ1(f,l,ζ)ldl.Using the fact thats≲δ(P)≲rφΩ(Θ), we get
(3.7)|PIΩf(P)|≲ℵ1(f,2s,ζ)+δ(P)∫δ(P)∞‍ℵ1(f,l,ζ)ldl.It is clear that(3.8)∫δ(P)∞ℵ1(f,l,ζ)ldl
is a convergent integral, since
(3.9)ℵ1(f,l,ζ)l≲s-1-nsn/q‖f‖p≲s-1-(n/p)‖f‖p
from the Hölder’s inequality.Now, asδ(P)→0, we also have s→0. Since f(ζ)=0 and since we have assumed that ζ∈𝔼fp(G(Ω)) (and thus that ζ∈𝔼f1(G(Ω))), it follows that 𝒫ℐΩf(P)→0=f(ζ) as P=(r,Θ)→ζ along Γ(Ω,ζ). This concludes the proof.



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*Source: 825240-2012-08-01.xml*