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# Growth estimates for modified Neumann integrals in a half-space

## Abstract

Our aim in this paper is to deal with the growth properties for modified Neumann integrals in a half-space of $R n$. 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 $R +$ be the sets of all real numbers and of all positive real numbers, respectively. Let $R n$ ($n≥3$) denote the n-dimensional Euclidean space with points $x=( x ′ , x n )$, where $x ′ =( x 1 , x 2 ,…, x n − 1 )∈ R n − 1$ and $x n ∈R$. The boundary and closure of an open set Ω of $R n$ are denoted by Ω and $Ω ¯$, respectively. For $x∈ R n$ and $r>0$, let $B n (x,r)$ denote the open ball with center at x and radius r in $R n$.

The upper half-space is the set $H={( x ′ , x n )∈ R n : x n >0}$, whose boundary is ∂H. For a set F, $F⊂ R + ∪{0}$, we denote ${x∈H;|x|∈F}$ and ${x∈∂H;|x|∈F}$ by HF and $∂HF$, respectively. We identify $R n$ with $R n − 1 ×R$ and $R n − 1$ with $R n − 1 ×{0}$, writing typical points $x,y∈ R n$ as $x=( x ′ , x n )$, $y=( y ′ , y n )$, where $y ′ =( y 1 , y 2 ,…, y n − 1 )∈ R n − 1$. Let θ be the angle between x and $e ˆ n$, i.e., $x n =|x|cosθ$ and $0≤θ<π/2$, where $e ˆ n$ is the i th unit coordinate vector and $e ˆ n$ is normal to ∂H.

We shall say that a set $E⊂H$ has a covering ${ r j , R j }$ if there exists a sequence of balls ${ B j }$ with centers in H such that $E⊂ ⋃ j = 0 ∞ B j$, where $r j$ is the radius of $B j$ and $R j$ is the distance between the origin and the center of $B j$.

For positive functions $g 1$ and $g 2$, we say that $g 1 ≲ g 2$ if $g 1 ≤M g 2$ 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, $[d]$ is the integer part of d and $d=[d]+{d}$, 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

$lim x ∈ H , x → y ′ ∂ ∂ x n h(x)=f ( y ′ )$

for every point $y ′ ∈∂H$.

For $x∈ R n$ and $y ′ ∈ R n − 1$, consider the kernel function

$K n ( x , y ′ ) =− β n | x − y ′ | n − 2 ,$

where $β n =2/(n−2) σ n$ and $σ n$ is the surface area of the n-dimensional unit sphere. It has the expression

$K n ( x , y ′ ) = ∑ k = 0 ∞ | x | k | y | n + k − 2 C k n − 2 2 ( x ⋅ y ′ | x | | y ′ | ) ,$

where $C k n 2 (t)$ is the ultraspherical (Gegenbauer) polynomials . The series converges for $| y ′ |>|x|$, and each term in it is a harmonic function of x.

The Neumann integral is defined by

$N[f](x)= ∫ ∂ H K n ( x , y ′ ) f ( y ′ ) d y ′ ,$

where f is a continuous function on ∂H, $α n =2/n σ n$ and $σ n = π n 2 /Γ(1+ n 2 )$ is the volume of the unit n-ball.

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

$∫ ∂ H f ( y ′ ) ( 1 + | y ′ | ) n − 2 d y ′ <∞.$

In this paper, we consider functions f satisfying

$∫ ∂ H | f ( y ′ ) | p ( 1 + | y ′ | ) n + α − 2 d y ′ <∞$
(1.1)

for $1≤p<∞$ and $α∈R$.

For this p and α, we define the positive measure μ on $R n$ by

$dμ ( y ′ ) = { | f ( y ′ ) | p | y ′ | − n − α + 2 d y ′ , y ′ ∈ ∂ H ( 1 , + ∞ ) , 0 , Q ∈ R n − ∂ H ( 1 , + ∞ ) .$

If f is a measurable function on ∂H satisfying (1.1), we remark that the total mass of μ is finite.

Let $ϵ>0$ and $δ≥0$. For each $x∈ R n$, the maximal function $M(x;μ,δ)$ is defined by

$M(x;μ,δ)= sup 0 < ρ < | x | 2 μ ( B n ( x , r ) ) ρ δ .$

The set ${x∈ R n ;M(x;μ,δ)>ϵ}$ is denoted by $E(ϵ;μ,δ)$.

To obtain the Neumann solution for the boundary data f, as in , we use the following modified kernel function defined by

$L n , m ( x , y ′ ) = { − β n ∑ k = 0 m − 1 | x | k | y | n + k − 2 C k n − 2 2 ( x ⋅ y ′ | x | | y ′ | ) , | y ′ | ≥ 1 m ≥ 1 , 0 , | y ′ | < 1 m ≥ 1 , 0 , m = 0$

for a non-negative integer m.

For $x∈ R n$ and $y ′ ∈ R n − 1$, the generalized Neumann kernel is defined by

$K n , m ( x , y ′ ) = K n ( x , y ′ ) − L n , m ( x , y ′ ) (m≥0).$

Since $|x | k C k n − 2 2 ( x ⋅ y ′ | x | | y ′ | )$ ($k≥0$) is harmonic in H (see ), $K n , m (⋅, y ′ )$ is also harmonic in H for any fixed $y ′ ∈∂H$. Also, $K n , m (x, y ′ )$ will be of order $| y ′ | − ( n + m − 2 )$ as $y ′ →∞$ (see [, Theorem D]).

Put

$N m [f](x)= ∫ ∂ H K n , m ( x , y ′ ) f ( y ′ ) d y ′ ,$

where f is a continuous function on ∂H. Here, note that $N 0 [f](x)$ is nothing but the Neumann integral $N[f](x)$.

The following result is due to Siegel and Talvila (see [, Corollary 2.1]). For similar results with respect to the Schrödinger operator in a half-space, we refer readers to papers by Su (see ).

Theorem A If f is a continuous function on ∂H satisfying (1.1) with $p=1$ and $α=m$, then

$lim | x | → ∞ , x ∈ H N m [f](x)=o ( | x | m sec n − 2 θ ) .$
(1.2)

The next result deals with a type of uniqueness of solutions for the Neumann problem on H (see [, 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

$∫ ∂ H | f ( y ′ ) | ( 1 + | y ′ | ) n + m − 2 d y ′ <∞,$

and h is a solution of the Neumann problem on H with f such that

$lim | x | → ∞ , x ∈ H h + (x)=o ( | x | l + m ) ,$

then

$h(x)= N m [f](x)+Π ( x ′ ) + ∑ j = 1 [ l + m 2 ] ( − 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x ′ )$

for any $x=( x ′ , x n )∈H$, where $h + (x)$ is the positive part of h,

$Δ j = ( ∂ 2 ∂ x 1 2 + ∂ 2 ∂ x 2 2 + ⋯ + ∂ 2 ∂ x n − 1 2 ) (j=1,2…)$

and $Π( x ′ )$ is a polynomial of $x ′ ∈ R n − 1$ of degree less than $l+m$.

Our first aim is to be concerned with the growth property of $N m [f]$ at infinity and establish the following theorem.

Theorem 1 Let $1≤p<∞$, $0≤β≤(n−2)p$, $n+α−2>−(n−1)(p−1)$ and

If f is a measurable function on satisfying (1.1), then there exists a covering ${ r j , R j }$ of $E(ϵ;μ,(n−2)p−β)$ (H) satisfying

$∑ j = 0 ∞ ( r j R j ) ( n − 2 ) p − β <∞$
(1.3)

such that

$lim | x | → ∞ , x ∈ H − E ( ϵ ; μ , ( n − 2 ) p − β ) N m [f](x)=o ( | x | 1 + α − 1 p sec β p θ ) .$
(1.4)

Corollary 1 Let $1, $n+α−2>−(n−1)(p−1)$ and

$1− 1 − α p

If f is a measurable function on ∂H satisfying (1.1), then

$lim | x | → ∞ , x ∈ H N m [f](x)=o ( | x | 1 + α − 1 p sec n − 2 θ ) .$
(1.5)

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 $N m [f]$ is a solution of the Neumann problem on H with f and (1.4) holds, where the exceptional set $E(ϵ;μ,(n−2)p−β)$ (H) has a covering ${ r j , R j }$ satisfying (1.3).

Remark In the case $p=1$, $α=m$ and $β=n−2$, then (1.3) is a finite sum and the set $E(ϵ;μ,0)$ 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 $1≤p<∞$, $n+α−2>−(n−1)(p−1)$ and

If f is a continuous function on ∂H satisfying (1.1), then the function $N m [f]$ 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 $p=1$ and $α=m$.

Theorem 3 Let $1≤p<∞$, $α>1−p$, 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

$lim | x | → ∞ , x ∈ H h + (x)=o ( | x | l + [ 1 + α − 1 p ] ) ,$
(1.6)

then

$h(x)= N m [f](x)+Π ( x ′ ) + ∑ j = 1 [ l + [ 1 + α − 1 p ] 2 ] ( − 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x ′ )$
(1.7)

for any $x=( x ′ , x n )∈H$ and $Π( x ′ )$ is a polynomial of $x ′ ∈ R n − 1$ of degree less than $l+[1+ α − 1 p ]$.

## 2 Lemmas

In our discussions, the following estimates for the kernel function $K n , m (x, y ′ )$ are fundamental (see [, Lemma 4.2] and [, Lemmas 2.1 and 2.4]).

Lemma 1

1. (1)

If $1≤| y ′ |≤ | x | 2$, then $| K n , m (x, y ′ )|≲|x | m − 1 | y ′ | − n − m + 3$.

2. (2)

If $| x | 2 <| y ′ |≤ 3 2 |x|$, then $| K n , m (x, y ′ )|≲|x− y ′ | 2 − n$.

3. (3)

If $3 2 |x|<| y ′ |≤2|x|$, then $| K n , m (x, y ′ )|≲ x n 2 − n$.

4. (4)

If $| y ′ |≥2|x|$ and $| y ′ |≥1$, then $| K n , m (x, y ′ )|≲|x | m | y ′ | 2 − n − m$.

The following lemma is due to Qiao (see ).

Lemma 2 If $ϵ>0$, $η≥0$ and λ is a positive measure in $R n$ satisfying $λ( R n )<∞$, then $E(ϵ;λ,η)$ has a covering ${ r j , R j }$ ($j=1,2,…$) such that

$∑ j = 1 ∞ ( r j R j ) η <∞.$

Lemma 3 ([, 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

$lim sup x ∈ H , x → y ′ ∂ ∂ x n N m [f](x)≤f ( y ′ )$

for any fixed $y ′ ∈∂H$.

Lemma 4 ([, Lemma 1])

If $h(x)$ is a harmonic polynomial of $x=( x ′ , x n )∈H$ of degree m and $∂h/∂ x n$ vanishes on ∂H, then there exists a polynomial $Π( x ′ )$ of degree m such that

$h(x)= { Π ( x ′ ) + ∑ j = 1 [ m 2 ] ( − 1 ) j ( 2 j ) ! x n 2 j Δ j Π ( x ′ ) , m ⩾ 2 , Π ( x ′ ) , m = 0 , 1 .$

## 3 Proof of Theorem 1

For any $ϵ>0$, there exists $R ϵ >1$ such that

$∫ ∂ H ( R ϵ , ∞ ) | f ( y ′ ) | p ( 1 + | y ′ | ) n + α − 2 d y ′ <ϵ.$
(3.1)

Take any point $x∈H( R ϵ ,∞)−E(ϵ;μ,(n−2)p−β)$ such that $|x|>2 R ϵ$, and write

$N m [ f ] ( x ) = ( ∫ G 1 + ∫ G 2 + ∫ G 3 + ∫ G 4 + ∫ G 5 ) K n , m ( x , y ′ ) f ( y ′ ) d y ′ = U 1 ( x ) + U 2 ( x ) + U 3 ( x ) + U 4 ( x ) + U 5 ( x ) ,$

where

$G 1 = { y ′ ∈ ∂ H : | y ′ | ≤ 1 } , G 2 = { y ′ ∈ ∂ H : 1 < | y ′ | ≤ | x | 2 } , G 3 = { y ′ ∈ ∂ H : | x | 2 < | y ′ | ≤ 3 2 | x | } , G 4 = { y ′ ∈ ∂ H : 3 2 | x | < | y ′ | ≤ 2 | x | } G 5 = { y ′ ∈ ∂ H : | y ′ | ≥ 2 | x | } .$

First note that

$| U 1 ( x ) | ≲ ∫ G 1 | f ( y ′ ) | | x − y ′ | n − 2 d y ′ ≲ | x | 2 − n ∫ G 1 | f ( y ′ ) | d y ′ ,$

so that

$lim | x | → ∞ , x ∈ H |x | − 1 + 1 − α p U 1 (x)=0.$
(3.2)

If $m<2− 1 − α p$ and $1 p + 1 q =1$, then $(3−n−m+ n + α − 2 p )q+n−1>0$. By Lemma 1(1), (3.1) and the Hölder inequality, we have

$| U 2 ( x ) | ≲ | x | m − 1 ∫ G 2 | y ′ | − n − m + 3 | f ( y ′ ) | d y ′ ≲ | x | m − 1 ( ∫ G 2 | f ( y ′ ) | p | y ′ | n + α − 2 d y ′ ) 1 p ( ∫ G 2 | y ′ | ( − n − m + 3 + n + α − 2 p ) q d y ′ ) 1 q ≲ | x | 1 − 1 − α p ( ∫ G 2 | f ( y ′ ) | p | y ′ | n + α − 2 d y ′ ) 1 p .$
(3.3)

Put

$U 2 (x)= U 21 (x)+ U 22 (x),$

where

$U 21 ( x ) = ∫ G 2 ∩ B n − 1 ( R ϵ ) K n , m ( x , y ′ ) f ( y ′ ) d y ′ , U 22 ( x ) = ∫ G 2 ∖ B n − 1 ( R ϵ ) K n , m ( x , y ′ ) f ( y ′ ) d y ′ .$

If $|x|≥2 R ϵ$, then

$| U 21 (x)|≲ R ϵ 2 − m − 1 − α p |x | m − 1 .$

Moreover, by (3.1) and (3.3), we get

$| U 22 (x)|≲ϵ|x | 1 − 1 − α p .$

That is,

$| U 2 (x)|≲ϵ|x | 1 − 1 − α p .$
(3.4)

By Lemma 1(3), (3.1) and the Hölder inequality, we have

$| U 4 (x)|≲ϵ x n 2 − n |x | n − 1 − 1 − α p .$
(3.5)

If $m>1− 1 − α p$, then $(2−n−m+ n + α − 2 p )q+n−1<0$. We obtain, by Lemma 1(4), (3.1) and the Hölder inequality,

$| U 5 ( x ) | ≲ | x | m ∫ G 5 | y ′ | − n − m + 2 | f ( y ′ ) | d y ′ ≲ | x | m ( ∫ G 5 | f ( y ′ ) | p | y ′ | n + α − 2 d y ′ ) 1 p ( ∫ G 5 | y ′ | ( − n − m + 2 + n + α − 2 p ) q d y ′ ) 1 q ≲ ϵ | x | 1 − 1 − α p .$
(3.6)

Finally, we shall estimate $U 3 (x)$. Take a sufficiently small positive number b such that $∂H[ | x | 2 , 3 2 |x|]⊂B(x, | x | 2 )$ for any $x∈Π(b)$, where

$Π(b)= { x ∈ H ; inf y ′ ∈ ∂ H | x | x | − y ′ | y ′ | | < b }$

and divide H into two sets $Π(b)$ and $H−Π(b)$.

If $x∈H−Π(b)$, then there exists a positive number $b ′$ such that $|x− y ′ |≥ b ′ |x|$ for any $y ′ ∈∂H$, and hence

$| U 3 ( x ) | ≲ ∫ G 3 | y ′ | 2 − n | f ( y ′ ) | d y ′ ≲ | x | m ∫ G 3 | y ′ | 2 − n − m | f ( y ′ ) | d y ′ ≲ ϵ | x | 1 − 1 − α p ,$

which is similar to the estimate of $U 5 (x)$.

We shall consider the case $x∈Π(b)$. Now put

$H i (x)= { y ′ ∈ ∂ H [ | x | 2 , 3 2 | x | ] ; 2 i − 1 δ ( x ) ≤ | x − y ′ | < 2 i δ ( x ) } ,$

where $δ(x)= inf y ′ ∈ H |x− y ′ |$.

Since $∂H∩{ y ′ ∈ R n − 1 :|x− y ′ |<δ(x)}=∅$, we have

$U 3 (x)= ∑ i = 1 i ( x ) ∫ H i ( x ) | g ( y ′ ) | | x − y ′ | n − 2 d y ′ ,$

where $i(x)$ is a positive integer satisfying $2 i ( x ) − 1 δ(x)≤ | x | 2 < 2 i ( x ) δ(x)$.

Similar to the estimate of $U 5 (x)$, we obtain

$∫ H i ( x ) | g ( y ′ ) | | x − y ′ | n − 2 d y ′ ≲ ∫ H i ( x ) | g ( y ′ ) | { 2 i − 1 δ ( x ) } n − 2 d y ′ ≲ δ ( x ) β − ( n − 2 ) p p ∫ H i ( x ) δ ( x ) ( n − 2 ) p − β p − n + 2 | g ( y ′ ) | d y ′ ≲ cos − β p θ δ ( x ) β − ( n − 2 ) p p ∫ H i ( x ) | x | − β p | g ( y ′ ) | d y ′ ≲ | x | n − 2 − β p cos − β p θ δ ( x ) β − ( n − 2 ) p p ∫ H i ( x ) | y ′ | 2 − n | g ( y ′ ) | d y ′ ≲ | x | n − 1 + α − β − 1 p cos − β p θ ( μ ( H i ( x ) ) 2 i δ ( x ) ( n − 2 ) p − β ) 1 p$

for $i=0,1,2,…,i(x)$.

Since $x∉E(ϵ;μ,(n−2)p−β)$, we have

$μ ( H i ( x ) ) { 2 i δ ( x ) } ( n − 2 ) p − β ≲ μ ( B n − 1 ( x , 2 i δ ( x ) ) ) { 2 i δ ( x ) } ( n − 2 ) p − β ≲M ( x ; μ , ( n − 2 ) p − β ) ≲ϵ|x | β − ( n − 2 ) p$

for $i=0,1,2,…,i(x)−1$ and

$μ ( H i ( x ) ( x ) ) { 2 i δ ( x ) } ( n − 2 ) p − β ≲ μ ( B n − 1 ( x , | x | 2 ) ) ( | x | 2 ) ( n − 2 ) p − β ≲ϵ|x | β − ( n − 2 ) p .$

So

$| U 3 (x)|≲ϵ|x | 1 + α − 1 p sec β p θ.$
(3.7)

Combining (3.2), (3.4)-(3.7), we obtain that if $R ϵ$ is sufficiently large and ϵ is a sufficiently small number, then $N m [f](x)=o(|x | 1 + α − 1 p sec β p θ)$ as $|x|→∞$, where $x∈H( R ϵ ,+∞)−E(ϵ;μ,(n−2)p−β)$. Finally, there exists an additional finite ball $B 0$ covering $H(0, R ϵ ]$, which together with Lemma 2, gives the conclusion of Theorem 1.

## 4 Proof of Theorem 2

For any fixed $x∈H$, take a number R satisfying $R>max{1,2|x|}$. If $m> 1 − α p$, then $(2−n−m+ n + α − 2 p )q+n−1<0$. By (1.1), Lemma 1(4) and the Hölder inequality, we have

$∫ ∂ H ( R , ∞ ) | K n , m ( x , y ′ ) | | f ( y ′ ) | d y ′ ≲ | x | m ∫ ∂ H ( R , ∞ ) | y ′ | 2 − n − m | f ( y ′ ) | d y ′ ≲ | x | m ( ∫ ∂ H ( R , ∞ ) | f ( y ′ ) | p | y ′ | n + α − 2 d y ′ ) 1 p ( ∫ ∂ H ( R , ∞ ) | y ′ | ( − n − m + 2 + n + α − 2 p ) q d y ′ ) 1 q < ∞ .$

Hence $N m [f](x)$ is absolutely convergent and finite for any $x∈H$. Thus $N m [f](x)$ is harmonic on H.

To prove

$lim x → y ′ , x ∈ H ∂ ∂ x n N m [f](x)=f ( y ′ )$

for any point $y ′ ∈∂H$, we only need to apply Lemma 3 to $f(y)$ and $−f(y)$.

We complete the proof of Theorem 2.

## 5 Proof of Theorem 3

Consider the function $h ′ (x)=h(x)− N m [f](x)$. Then it follows from Theorems 2 and 3 that $h ′ (x)$ is a solution of the Neumann problem on H with f and it is an even function of $x n$ (see [, p.92]).

Since

$0≤ { h − N m [ f ] } + (x)≤ h + (x)+ { N m [ f ] } − (x)$

for any $x∈H$, and

$lim | x | → ∞ , x ∈ H N m [f](x)=o ( | x | 1 + α − 1 p )$

from Theorem 2.

Moreover, (1.6) gives that

$lim | x | → ∞ , x ∈ H ( h − N m [ f ] ) (x)=o ( | x | l + [ 1 + α − 1 p ] ) .$

This implies that $h ′ (x)$ is a polynomial of degree less than $l+[1+ α − 1 p ]$ (see [, Appendix]), which gives the conclusion of Theorem 3 from Lemma 4.

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## Acknowledgements

The authors are thankful to the referees for their helpful suggestions and necessary corrections in the completion of this paper.

## Author information

Correspondence to Yudong Ren.

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

### Authors’ contributions

All authors contributed equally to the manuscript and read and approved the final manuscript.

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