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Journal of Function Spaces and Applications
Volume 2012 (2012), Article ID 825240, 9 pages
doi:10.1155/2012/825240
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.

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.

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 𝐑 𝑛 ( 𝑛 2 ) the 𝑛 -dimensional Euclidean space. A point in 𝐑 𝑛 is denoted by 𝑃 = ( 𝑋 , 𝑥 𝑛 ) , where 𝑋 = ( 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 1 ) . The Euclidean distance of two points 𝑃 and 𝑄 in 𝐑 𝑛 is denoted by | 𝑃 𝑄 | . Also | 𝑃 𝑂 | with the origin 𝑂 of 𝐑 𝑛 is simply denoted by | 𝑃 | . The boundary, the closure, and the complement of a set 𝐒 in 𝐑 𝑛 are denoted by 𝜕 𝐒 , 𝐒 , and 𝐒 𝑐 , respectively.

We introduce a system of spherical coordinates ( 𝑟 , Θ ) ,   Θ = ( 𝜃 1 , 𝜃 2 , , 𝜃 𝑛 1 ) , in 𝐑 𝑛 which are related to cartesian coordinates ( 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 1 , 𝑥 𝑛 ) by 𝑥 𝑛 = 𝑟 c o s 𝜃 1 .

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

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

By 𝐶 𝑛 ( Ω ) , we denote the set 𝐑 + × Ω in 𝐑 𝑛 with the domain Ω on 𝐒 𝑛 1 . We call it a cone. Then 𝑇 𝑛 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 𝐶 𝑛 ( Ω ; 𝐼 ) and 𝑆 𝑛 ( Ω ; 𝐼 ) . By 𝑆 𝑛 ( Ω ) we denote 𝑆 𝑛 ( Ω ; ( 0 , + ) ) which is 𝜕 𝐶 𝑛 ( Ω ) { 𝑂 } .

Let Ω be a domain on 𝐒 𝑛 1 with smooth boundary. Consider the Dirichlet problem: Λ 𝑛 + 𝜆 𝜑 = 0 o n Ω , 𝜑 = 0 o n 𝜕 Ω , ( 1 . 1 ) where Λ 𝑛 is the spherical part of the Laplace operator Δ 𝑛 Δ 𝑛 = 𝑛 1 𝑟 𝜕 + 𝜕 𝜕 𝑟 2 𝜕 𝑟 2 + Λ 𝑛 𝑟 2 . ( 1 . 2 ) We denote the least positive eigenvalue of this boundary value problem by 𝜆 Ω and the normalized positive eigenfunction corresponding to 𝜆 Ω by 𝜑 Ω ( Θ ) , Ω 𝜑 2 Ω ( Θ ) 𝑑 𝜎 Θ = 1 , ( 1 . 3 ) where 𝑑 𝜎 Θ is the surface area on 𝑆 𝑛 1 . We denote the solutions of the equation 𝑡 2 + ( 𝑛 2 ) 𝑡 𝜆 Ω = 0 by 𝛼 Ω , 𝛽 Ω ( 𝛼 Ω , 𝛽 Ω > 0 ). If Ω = 𝐒 + 𝑛 1 , then 𝛼 Ω = 1 , 𝛽 Ω = 𝑛 1 , and 𝜑 1 ( Θ ) = ( 2 𝑛 𝑠 𝑛 1 ) 1 / 2 c o s 𝜃 1 , where 𝑠 𝑛 is the surface area 2 𝜋 𝑛 / 2 ( Γ ( 𝑛 / 2 ) ) 1 of 𝐒 1 .

To simplify our consideration in the following, we will assume that if 𝑛 3 , then Ω is a 𝐶 2 , 𝛼 -domain ( 0 < 𝛼 < 1 ) on 𝐒 𝑛 1 surrounded by a finite number of mutually disjoint closed hypersurfaces (e.g., see ([1], pages 88-89) for the definition of 𝐶 2 , 𝛼 -domain). Then by modifying Miranda’s method ([2], pages 7-8), we can prove the following inequality: 𝜑 Ω ( Θ ) d i s t ( Θ , 𝜕 Ω ) ( Θ Ω ) . ( 1 . 4 )

For any ( 1 , Θ ) Ω , we have (see [3]) 𝜑 Ω ( Θ ) d i s t ( 1 , Θ ) , 𝜕 𝐶 𝑛 ( Ω ) , ( 1 . 5 ) which yields that 𝛿 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) , ( 1 . 6 ) where 𝛿 ( 𝑃 ) = d i s t ( 𝑃 , 𝜕 𝐶 𝑛 ( Ω ) ) and 𝑃 = ( 𝑟 , Θ ) 𝐶 𝑛 ( Ω ) .

Let 𝐺 Ω ( 𝑃 , 𝑄 ) ( 𝑃 = ( 𝑟 , Θ ) , 𝑄 = ( 𝑡 , Φ ) 𝐶 𝑛 ( Ω ) ) be the Green function of 𝐶 𝑛 ( Ω ) . We define the Poisson kernel 𝐾 Ω ( 𝑃 , 𝑄 ) by 𝐾 Ω 1 ( 𝑃 , 𝑄 ) = 𝑐 𝑛 𝜕 𝜕 𝑛 𝑄 𝐺 Ω ( 𝑃 , 𝑄 ) , ( 1 . 7 ) where 𝑐 𝑛 = ( 2 𝜋 𝑛 = 2 , 𝑛 2 ) 𝑠 𝑛 𝑛 3 , ( 1 . 8 ) 𝑄 𝑆 𝑛 ( Ω ) and 𝜕 / 𝜕 𝑛 𝑄 denotes the differentiation at 𝑄 along the inward normal into 𝐶 𝑛 ( Ω ) .

In this paper, we consider functions 𝑓 𝐿 𝑝 ( 𝜕 𝐶 𝑛 ( Ω ) ) , where 1 𝑝 < . Then the Poisson integral 𝑊 Ω 𝑓 ( 𝑃 ) ( 𝑃 𝐶 𝑛 ( Ω ) ) is defined by 𝑊 Ω 𝑓 ( 𝑃 ) = 𝑆 𝑛 ( Ω ) 𝐾 Ω ( 𝑃 , 𝑄 ) 𝑓 ( 𝑄 ) 𝑑 𝜎 𝑄 , ( 1 . 9 ) where 𝑑 𝜎 𝑄 is the surface area element on 𝑆 𝑛 ( Ω ) .

Remark 1.1. Let Ω = 𝐒 + 𝑛 1 . Then 𝐺 𝐒 + 𝑛 1 | | ( 𝑃 , 𝑄 ) = l o g 𝑃 𝑄 | | | | | | | | | | l o g 𝑃 𝑄 𝑛 = 2 , 𝑃 𝑄 2 𝑛 | | 𝑃 𝑄 | | 2 𝑛 𝑛 3 , ( 1 . 1 0 ) where 𝑄 = ( 𝑌 , 𝑦 𝑛 ) , that is, 𝑄 is the mirror image of 𝑄 = ( 𝑌 , 𝑦 𝑛 ) with respect to 𝜕 𝑇 𝑛 . Hence, for the two points 𝑃 = ( 𝑋 , 𝑥 𝑛 ) 𝑇 𝑛 and 𝑄 = ( 𝑌 , 𝑦 𝑛 ) 𝜕 𝑇 𝑛 , we have 𝑐 𝑛 𝐾 𝐒 + 𝑛 1 𝜕 ( 𝑃 , 𝑄 ) = 𝜕 𝑛 𝑄 𝐺 𝐒 + 𝑛 1 2 | | | | ( 𝑃 , 𝑄 ) = 𝑃 𝑄 2 𝑥 𝑛 | | | | 𝑛 = 2 , 2 ( 𝑛 2 ) 𝑃 𝑄 𝑛 𝑥 𝑛 𝑛 3 . ( 1 . 1 1 )

We fix an open, nonempty, and bounded set 𝐺 ( Ω ) 𝜕 𝐶 𝑛 ( Ω ) . In 𝐶 𝑛 ( Ω ) , we normalise the extension, with respect to 𝐺 ( Ω ) , by 𝒫 Ω 𝑊 𝑓 ( 𝑃 ) = Ω 𝑓 ( 𝑃 ) 𝑊 Ω 𝜒 𝐺 ( Ω ) ( , 𝑃 ) ( 1 . 1 2 ) where 𝜒 𝐺 ( Ω ) denotes the characteristic function of 𝐺 ( Ω ) .

Let Γ ( Ω , 𝜁 ) = 𝑃 = ( 𝑟 , Θ ) 𝐶 𝑛 | | | | ( Ω ) ( 𝑟 , Θ ) 𝜁 𝛿 ( 𝑃 ) ( 1 . 1 3 ) be a nontangential cone in 𝐶 𝑛 ( Ω ) with vertex 𝜁 𝜕 𝐶 𝑛 ( Ω ) .

We define 𝑝 1 ( 𝑓 , 𝑙 , 𝑃 ) = 𝑙 𝑛 1 𝐵 ( 𝑃 , 𝑙 ) | | | | 𝑓 ( 𝑄 ) 𝑝 𝑑 𝜎 𝑄 1 / 𝑝 , 𝔼 𝑝 𝑓 ( 𝐺 ( Ω ) ) = 𝑃 𝐺 ( Ω ) 𝑝 . ( 𝑓 𝑓 ( 𝑃 ) , 𝑙 , 𝑃 ) 0 a s 𝑙 0 ( 1 . 1 4 )

Note that, if 𝑓 𝐿 𝑝 ( 𝜕 𝐶 𝑛 ( Ω ) ) , then | 𝐺 ( Ω ) 𝔼 𝑝 𝑓 ( 𝐺 ( Ω ) ) | = 0 (a.e. point is a Lebesgue point).

In 𝑇 𝑛 , 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 ) , 𝒫 𝐒 + 𝑛 1 𝑓 ( 𝑃 ) 𝑓 ( 𝜁 ) (see Remark 1.1 for the definition of 𝒫 𝐒 + 𝑛 1 𝑓 ( 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 any 𝜁 𝔼 𝑝 𝑓 ( 𝐺 ( Ω ) ) (in particular, for a.e. 𝜁 𝐺 ( Ω ) ) one has that 𝒫 Ω 𝑓 ( 𝑃 ) 𝑓 ( 𝜁 ) as 𝑃 𝜁 along Γ ( Ω , 𝜁 ) .

2. Some Lemmas

Lemma 2.1. One has 𝐾 Ω ( 𝑃 , 𝑄 ) 𝑟 𝛽 Ω 𝑡 𝛼 Ω 1 𝜑 Ω ( Θ ) , r e s p . 𝐾 Ω ( 𝑃 , 𝑄 ) 𝑟 𝛼 Ω 𝑡 𝛽 Ω 1 𝜑 Ω , ( Θ ) ( 2 . 1 ) for any 𝑃 = ( 𝑟 , Θ ) 𝐶 𝑛 ( Ω ) and any 𝑄 = ( 𝑡 , Φ ) 𝑆 𝑛 ( Ω ) satisfying 0 < 𝑡 / 𝑟 4 / 5 ( r e s p . 0 < 𝑟 / 𝑡 4 / 5 ) ; 𝐾 Ω ( 𝑃 , 𝑄 ) 𝑟 𝜑 Ω ( Θ ) | | | | 𝑃 𝑄 𝑛 , ( 2 . 2 ) for any 𝑃 = ( 𝑟 , Θ ) 𝐶 𝑛 ( Ω ) and any 𝑄 = ( 𝑡 , Φ ) 𝑆 𝑛 ( Ω ; ( 4 𝑟 / 5 , 5 𝑟 / 4 ) ) .

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

Lemma 2.2. One has 𝑊 Ω 1 ( 𝑃 ) = 𝑂 ( 1 ) a s 𝑃 𝜁 𝐺 . ( 2 . 3 )

Proof. Write 𝑊 Ω 1 ( 𝑃 ) = 𝐸 1 + 𝐸 2 + 𝐸 3 = 𝑈 1 ( 𝑃 ) + 𝑈 2 ( 𝑃 ) + 𝑈 3 ( 𝑃 ) , ( 2 . 4 ) where 𝐸 1 = 𝑆 𝑛 4 Ω ; 0 , 5 𝑟 , 𝐸 2 = 𝑆 𝑛 5 Ω ; 4 𝑟 , , 𝐸 3 = 𝑆 𝑛 4 Ω ; 5 5 𝑟 , 4 𝑟 . ( 2 . 5 )
By (2.1), we have the following estimates 𝑈 1 ( 𝑃 ) 𝑟 𝛽 Ω 𝜑 Ω ( Θ ) 𝐸 1 𝑡 𝛼 Ω 1 𝑑 𝜎 𝑄 𝑠 𝑛 𝛽 Ω 4 5 𝛽 Ω 𝜑 Ω 𝑈 ( Θ ) , ( 2 . 6 ) 2 𝑠 ( 𝑃 ) 𝑛 𝛼 Ω 4 5 𝛼 Ω 𝜑 Ω ( Θ ) . ( 2 . 7 )
Next, we will estimate 𝑈 3 ( 𝑃 ) . Take a sufficiently small positive number 𝑘 such that 𝐸 3 𝑃 = ( 𝑟 , Θ ) Λ ( 𝑘 ) 𝐵 1 𝑃 , 2 𝑟 , ( 2 . 8 ) where Λ ( 𝑘 ) = 𝑃 = ( 𝑟 , Θ ) 𝐶 𝑛 ( Ω ) ; i n f 𝑧 𝜕 Ω | | | | ( 1 , Θ ) ( 1 , 𝑧 ) < 𝑘 , 0 < 𝑟 < . ( 2 . 9 )
Since 𝑃 𝜁 𝐺 , we only consider the case 𝑃 Λ ( 𝑘 ) . Now put 𝐻 𝑖 ( 𝑃 ) = 𝑄 𝐸 3 ; 2 𝑖 1 𝛿 | | | | ( 𝑃 ) 𝑃 𝑄 < 2 𝑖 𝛿 . ( 𝑃 ) ( 2 . 1 0 )
Since 𝑆 𝑛 ( Ω ) { 𝑄 𝐑 𝑛 | 𝑃 𝑄 | < 𝛿 ( 𝑃 ) } = Ø , we have 𝑈 3 ( 𝑃 ) 𝑖 ( 𝑃 ) 𝑖 = 0 𝐻 𝑖 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) | | | | 𝑃 𝑄 𝑛 𝑑 𝜎 𝑄 , ( 2 . 1 1 ) where 𝑖 ( 𝑃 ) is a positive integer satisfying 2 𝑖 ( 𝑃 ) 1 𝛿 ( 𝑃 ) 𝑟 / 2 < 2 𝑖 ( 𝑃 ) 𝛿 ( 𝑃 ) .
By (1.6) we have 𝐻 𝑖 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) | | | | 𝑃 𝑄 𝑛 𝑑 𝜎 𝑄 𝑟 𝜑 Ω ( Θ ) 𝐻 𝑖 ( 𝑃 ) 1 𝛿 ( 𝑃 ) 𝑑 𝜎 𝑄 = 𝑟 𝜑 Ω ( Θ ) 𝑠 𝛿 ( 𝑃 ) 𝑛 2 𝑖 ( 𝑃 ) 𝑠 𝑛 2 𝑖 ( 𝑃 ) ( 2 . 1 2 ) for 𝑖 = 0 , 1 , 2 , , 𝑖 ( 𝑃 ) .
So 𝑈 3 ( 𝑃 ) 𝑂 ( 1 ) . ( 2 . 1 3 )
Combining (2.6)–(2.13), Lemma 2.2 is proved.

Lemma 2.3. One has 𝑊 Ω 𝜒 𝐺 ( Ω ) ( 𝑃 ) = 𝑊 Ω 1 ( 𝑃 ) + 𝑂 ( 1 ) a s 𝑃 𝜁 𝐺 ( Ω ) . ( 2 . 1 4 )

Proof. In fact, we only need to prove 𝑈 4 ( 𝑃 ) = 𝑆 𝑛 ( Ω ) 𝐺 ( Ω ) 𝐾 Ω ( 𝑃 , 𝑄 ) 𝑑 𝜎 𝑄 𝑂 ( 1 ) . ( 2 . 1 5 )
Write 𝑈 4 ( 𝑃 ) = ( 𝑆 𝑛 ( Ω ) 𝐺 ( Ω ) ) 𝐸 1 + ( 𝑆 𝑛 ( Ω ) 𝐺 ( Ω ) ) 𝐸 2 + ( 𝑆 𝑛 ( Ω ) 𝐺 ( Ω ) ) 𝐸 3 = 𝑈 5 ( 𝑃 ) + 𝑈 6 ( 𝑃 ) + 𝑈 7 ( 𝑃 ) , ( 2 . 1 6 ) where 𝐸 1 , 𝐸 2 , and 𝐸 3 are sets on 𝑆 𝑛 ( Ω ) used in Lemma 2.2.
Obviously, 𝑈 5 ( 𝑃 ) 𝑈 1 𝑈 ( 𝑃 ) 𝑂 ( 1 ) , ( 2 . 1 7 ) 6 ( 𝑃 ) 𝑈 2 ( 𝑃 ) 𝑂 ( 1 ) . ( 2 . 1 8 )
Further, we have by (2.2) 𝑈 7 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) ( 𝑆 𝑛 ( Ω ) 𝐺 ( Ω ) ) 𝐸 3 1 | | | | 𝑃 𝑄 𝑛 𝑑 𝜎 𝑄 𝑠 𝑛 𝑑 | | 𝜁 | | 𝜑 Ω ( Θ ) ( 𝑃 𝜁 𝐺 ( Ω ) ) , ( 2 . 1 9 ) where 𝑑 = i n f 𝑄 𝜕 𝐶 𝑛 ( Ω ) 𝐺 ( Ω ) | | | | . 𝑄 𝜁 ( 2 . 2 0 )
Combining (2.17)–(2.19), (2.15) holds which gives the conclusion.

3. Proof of the Theorem 1.2

As 𝑃 𝜁 𝐺 ( Ω ) , 𝑊 Ω 𝜒 𝐺 ( Ω ) ( 𝑃 ) = 𝑂 ( 1 ) from Lemmas 2.2 and 2.3.

Now, let 𝑓 𝐿 𝑝 ( 𝜕 𝐶 𝑛 ( Ω ) ) and 𝜁 𝔼 𝑝 𝑓 ( 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 𝑊 Ω 𝑓 ( 𝑃 ) = 𝐸 1 + 𝐸 2 + 𝐸 3 𝐵 ( 𝜁 , 2 𝑠 ) + 𝐸 3 𝐵 𝑐 ( 𝜁 , 2 𝑠 ) = 𝑉 1 𝑓 ( 𝑃 ) + 𝑉 2 𝑓 ( 𝑃 ) + 𝑉 3 𝑓 ( 𝑃 ) + 𝑉 4 𝑓 ( 𝑃 ) , ( 3 . 1 ) where 𝐸 1 , 𝐸 2 , and 𝐸 3 are sets on 𝑆 𝑛 ( Ω ) used in Lemma 2.2.

By using Hölder’s inequality, (2.1), we have the following estimates | | 𝑉 1 | | 𝑓 ( 𝑃 ) 𝑟 𝛽 Ω 𝜑 Ω ( Θ ) 𝐸 1 𝑡 𝛼 Ω 1 𝑓 ( 𝑄 ) 𝑑 𝜎 𝑄 𝑟 ( 1 𝑛 ) / 𝑝 𝑓 𝑝 , | | 𝑉 2 | | 𝑓 ( 𝑃 ) 𝑟 ( 1 𝑛 ) / 𝑝 𝑓 𝑝 . ( 3 . 2 )

Similar to the estimate of 𝑈 3 ( 𝑃 ) in Lemma 2.2, we only consider the following inequality by (1.6) 𝐻 𝑖 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) | | | | 𝑃 𝑄 𝑛 𝑑 𝜎 𝑄 𝑟 𝜑 Ω ( Θ ) 𝐻 𝑖 ( 𝑃 ) 1 2 𝑖 1 𝛿 ( 𝑃 ) 𝑛 𝑑 𝜎 𝑄 𝑟 𝛼 Ω 𝜑 Ω ( Θ ) 𝐸 2 𝑡 𝛽 Ω 1 | | 𝑓 | | ( 𝑄 ) 𝑑 𝜎 𝑄 𝑟 ( 1 𝑛 ) / 𝑝 𝑓 𝑝 ( 3 . 3 ) for 𝑖 = 0 , 1 , 2 , , 𝑖 ( 𝑃 ) , which is similar to the estimate of 𝑉 2 𝑓 ( 𝑃 ) .

So | | 𝑉 3 | | 𝑓 ( 𝑃 ) 𝑟 ( 1 𝑛 ) / 𝑝 𝑓 𝑝 . ( 3 . 4 )

Notice that | 𝑃 𝑄 | > ( 1 / 2 ) | 𝜁 𝑄 | in the case 𝑄 𝐸 3 𝐵 𝑐 ( 𝜁 , 2 𝑠 ) . By (1.6) and (2.2), we have | | 𝑉 4 | | 𝑓 ( 𝑃 ) 𝛿 ( 𝑃 ) 𝐸 3 𝐵 𝑐 ( 𝜁 , 2 𝑠 ) | | | | 𝑓 ( 𝑄 ) | | | | 𝑃 𝑄 𝑛 𝑑 𝜎 𝑄 𝛿 ( 𝑃 ) 𝑖 = 1 𝐸 3 ( 𝐵 ( 𝜁 , 2 𝑖 + 1 𝑠 ) 𝐵 ( 𝜁 , 2 𝑖 𝑠 ) ) | | 𝑓 | | ( 𝑄 ) | | | | 𝜁 𝑄 𝑛 𝑑 𝜎 𝑄 𝛿 ( 𝑃 ) 𝑖 = 1 1 2 𝑖 𝑠 𝑛 𝐸 3 𝐵 ( 𝜁 , 2 𝑖 + 1 𝑠 ) | | | | 𝑓 ( 𝑄 ) 𝑑 𝜎 𝑄 𝛿 ( 𝑃 ) 𝑖 = 1 1 𝑓 , 2 𝑖 + 1 𝑠 , 𝜁 𝛿 ( 𝑃 ) 𝑖 = 1 2 𝑖 + 2 𝑠 2 𝑖 + 1 𝑠 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 𝛿 ( 𝑃 ) 𝑠 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 𝛿 ( 𝑃 ) 𝛿 ( 𝑃 ) 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 . ( 3 . 5 )

Thus, it follows that | | 𝒫 Ω | | 1 𝑓 ( 𝑃 ) 𝑂 | | 𝑉 ( 1 ) 1 | | + | | 𝑉 𝑓 ( 𝑃 ) 2 | | + | | 𝑉 𝑓 ( 𝑃 ) 3 | | + | | 𝑉 𝑓 ( 𝑃 ) 4 | | 𝑓 ( 𝑃 ) 𝑟 ( 1 𝑛 ) / 𝑝 𝑓 𝑝 + 𝛿 ( 𝑃 ) 𝛿 ( 𝑃 ) 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 . ( 3 . 6 )

Using the fact that 𝑠 𝛿 ( 𝑃 ) 𝑟 𝜑 Ω ( Θ ) , we get | | 𝒫 Ω | | 𝑓 ( 𝑃 ) 1 ( 𝑓 , 2 𝑠 , 𝜁 ) + 𝛿 ( 𝑃 ) 𝛿 ( 𝑃 ) 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 . ( 3 . 7 )

It is clear that 𝛿 ( 𝑃 ) 1 ( 𝑓 , 𝑙 , 𝜁 ) 𝑙 𝑑 𝑙 ( 3 . 8 ) is a convergent integral, since 1 ( 𝑓 , l , 𝜁 ) 𝑙 𝑠 1 𝑛 𝑠 𝑛 / 𝑞 𝑓 𝑝 𝑠 1 ( 𝑛 / 𝑝 ) 𝑓 𝑝 ( 3 . 9 ) from the Hölder’s inequality.

Now, as 𝛿 ( 𝑃 ) 0 , we also have 𝑠 0 . Since 𝑓 ( 𝜁 ) = 0 and since we have assumed that 𝜁 𝔼 𝑝 𝑓 ( 𝐺 ( Ω ) ) (and thus that 𝜁 𝔼 1 𝑓 ( 𝐺 ( Ω ) ) ), it follows that 𝒫 Ω 𝑓 ( 𝑃 ) 0 = 𝑓 ( 𝜁 ) as 𝑃 = ( 𝑟 , Θ ) 𝜁 along Γ ( Ω , 𝜁 ) . This concludes the proof.

Acknowledgments

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

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