# The Modification of Kernel Function and Its Applications

Authors: Tao Zhao
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

By virtue of the modified Riesz kernel introduced by Qiao (2012), we give the integral representations for solutions of the Neumann problems in a half space.


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## Body

## 1. Introduction and Main Results

LetR and R+ be the sets of all real numbers and of all positive real numbers, respectively. Let Rn  (n≥3) denote the n-dimensional Euclidean space with points x=(x′,xn), where x′=(x1,x2,…,xn-1)∈Rn-1 and xn∈R. The boundary and closure of an open set Ω of Rn are denoted by ∂Ω and Ω¯, respectively. For x∈Rn and r>0, let Bn(x,r) denote the open ball with center at x and radius r in Rn. Let Bn(r)=Bn(O,r).The upper half space is the setH={(x′,xn)∈Rn:xn>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 Rn with Rn-1×R and Rn-1 with Rn-1×{0}, writing typical points x, y∈Rn as x=(x′,xn), y=(y′,yn), where y′=(y1,y2,…,yn-1)∈Rn-1. Let θ be the angle between x and e^n, that is, xn=|x|cos⁡θ and 0≤θ<π/2, where e^n is the ith unit coordinate vector and e^n is normal to ∂H.We will say that a setE⊂H has a covering {rj,Rj} if there exists a sequence of balls {Bj} with centers in H such that E⊂∪j=1∞Bj, where rj is the radius of Bj and Rj is the distance between the origin and the center of Bj.For positive functionsg1 and g2, we say that g1≲g2 if g1≤Mg2 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 u+=max⁡{u,0}, [d] is the integer part of d, and d=[d]+{d}, where d is a positive real number.Given a continuous functionf on ∂H, we say that h is a solution of the Neumann problem on H with f, if h is a harmonic function on H and
(1)lim⁡x∈H,x→y′∂∂xnh(x)=f(y′)
for every point y′∈∂H.Forx∈Rn and y′∈Rn-1, consider the kernel function
(2)Kn(x,y′)=-βn|x-y′|n-2,
where βn=2/(n-2)σn and σn is the surface area of the n-dimensional unit sphere.The Neumann integral onH is defined by
(3)N[f](x)=∫∂H‍Kn(x,y′)f(y′)dy′,
where f is a continuous function on ∂H.The Neumann integralN[f](x) is a solution of the Neumann problem on H with f if (see [1, Theorem 1 and Remarks])
(4)∫∂H‍f(y′)(1+|y′|)n-2dy′<∞.In this paper, we consider functionsf satisfying
(5)∫∂H‍|f(y′)|p(1+|y′|)γdy′<∞
for 1≤p<∞ and γ∈R.Forp and α, we define the positive measure μ on Rn by
(6)dμ(y′)={|f(y′)|p|y′|-γdy′y′∈∂H(1,+∞),0Q∈Rn-∂H(1,+∞).
If f is a measurable function on ∂H satisfying (5), we remark that the total mass of μ is finite.Letϵ>0 and δ≥0. For each x∈Rn, the maximal function M(x;μ,δ) is defined by
(7)M(x;μ,δ)=sup⁡0<ρ<|x|/2μ(Bn(x,ρ))ρδ.The set{x∈Rn;M(x;μ,δ)|x|δ>ϵ} is denoted by E(ϵ;μ,δ).To obtain the Neumann solution for the boundary dataf on H, as in [2, 3], we use the following modified Riesz kernel defined by
(8)Ln,m(x,y′)={-βn∑k=0m-1|x|k|y|n+k-2Ck(n-2)/2(x·y′|x||y′|)|y′|≥1,m≥1,0|y′|<1,m≥1,0m=0,
where m is a nonnegative integer.Forx∈Rn and y′∈Rn-1, the generalized Neumann kernel is defined by
(9)Kn,m(x,y′)=Kn(x,y′)-Ln,m(x,y′)(m≥0).Put(10)Nm[f](x)=∫∂H‍Kn,m(x,y′)f(y′)dy′,
where f is continuous function on ∂H. Here note that N0[f](x) is nothing but the Neumann integral N[f](x).The following result is due to Su (see [4]).Theorem A.
Iff is a continuous function on ∂H satisfying (5) with p=1 and α=m, then
(11)lim⁡|x|→∞,x∈HNm[f](x)=o(|x|msecn-2θ).Our first aim is to be concerned with the growth property ofNm[f] at infinity in a half space and establish the following theorem.Theorem 1.
Let1≤p<∞, 0≤β≤(n-2)p, γ>-(n-1)(p-1) and
(12)1-n-γ-1p<m<2-n-γ-1pifp>1,γ-n+2≤m<γ-n+3ifp=1.
If f is a measurable function on ∂H satisfying (5), then there exists a covering {rj,Rj} of E(ϵ;μ,(n-2)p-β)(⊂H) satisfying
(13)∑j=0∞‍(rjRj)(n-2)p-β<∞
such that
(14)lim⁡|x|→∞,x∈H-E(ϵ;μ,(n-2)p-β)Nm[f](x)=o(|x|1+((γ-n+1)/p)).Remark 2.
In the case thatp=1, α=m, and β=n-2, then (13) is a finite sum and the set E(ϵ;μ,0) is a bounded set. So (14) holds in H. That is to say, (11) holds. This is just the result of Theorem A.Corollary 3.
Let1<p<∞, n+α-2>-(n-1)(p-1) and
(15)1-n-γ-1p<m<2-n-γ-1p.
If f is a measurable function on ∂H satisfying (5), then
(16)lim⁡|x|→∞,x∈HNm[f](x)=o(|x|1+((γ-n+1)/p)).As an application of Theorem1, we now show the solution of the Neumann problem with continuous data on H. About the solutions of the Dirichlet problem with respect to the Schrödinger operator in a half space, we refer readers to the paper by Su (see [5]).Theorem 4.
Letp, β, α, and m be defined as in Theorem 1. If f is a continuous function on ∂H satisfying (5), then the function Nm[f] is a solution of the Neumann problem on H with f and (14) holds, where the exceptional set E(ϵ;μ,(n-2)p-β)(⊂H) has a covering {rj,Rj} satisfying (13).Finally we have the following result.Theorem 5.
Let1≤p<∞, α>1-p, l be a positive integer and
(17)1-n-γ-1p<m<2-n-γ-1pifp>1,α≤m<α+1ifp=1.
If f is a continuous function on ∂H satisfying (5) and h is a solution of the Neumann problem on H with f such that
(18)lim⁡|x|→∞,x∈Hh+(x)=o(|x|l+[1+((γ-n+1)/p)]),
then
(19)h(x)=Nm[f](x)+Π(x′)+∑j=1[l+[1+((γ-n+1)/p)]/2]‍(-1)j(2j)!xn2jΔjΠ(x′)
for any x=(x′,xn)∈H, where
(20)Δj=(∂2∂x12+∂2∂x22+⋯+∂2∂xn-12)(j=1,2,…)
and Π(x′) is a polynomial of x′∈Rn-1 of degree less than l+[1+((γ-n+1)/p)].


## 2. Lemmas

In our discussions, the following estimates for the kernel functionKn,m(x,y′) are fundamental (see [6, Lemma 4.2] and [3, Lemmas 2.1 and 2.4]).Lemma 6.
(1) If 1≤|y′|≤|x|/2, then |Kn,m(x,y′)|≲|x|m-1|y′|-n-m+3.
(2) If |x|/2<|y′|≤(3/2)|x|, then |Kn,m(x,y′)|≲|x-y′|2-n.
(3) If (3/2)|x|<|y′|≤2|x|, then |Kn,m(x,y′)|≲xn2-n.
(4) If |y′|≥2|x| and |y′|≥1, then |Kn,m(x,y′)|≲|x|m|y′|2-n-m.The following Lemma is due to Qiao (see [3]).Lemma 7.
Ifϵ>0, η≥0, and λ is a positive measure in Rn satisfying λ(Rn)<∞, then E(ϵ;λ,η) has a covering {rj,Rj}  (j=1,2,…) such that
(21)∑j=1∞‍(rjRj)η<∞.Lemma 8.
Letp, β, α, and m be defined as in Theorem 1. If f is a local integral and upper semicontinuous function on ∂H satisfying (5), then
(22)limsup⁡x∈H,x→y′∂∂xnNm[f](x)≤f(y′),
for any fixed point y′∈∂H.Proof.
Lety* be any fixed point an ∂H and let ϵ be any positive number. Take a positive number δ, δ<1, such that
(23)f(y)<f(y*)+ϵ,
for any y∈Bn-1(y*,δ).
By Lemma6(4) and (5), we can choose a number R*, R*>2(|y*|+1), such that
(24)∫∂H∖Bn-1(R*)|∂∂xnKn,m(x,y′)||f(y′)|dy′<ϵ,
for any x∈∂H∩Bn-1(y*,δ).
Put(25)Λ1(x)=∫Bn-1(R*)∂∂xnKn,0(x,y′)f(y′)dy′,Λ2(x)=-∫Bn-1(R*)∂∂xnLn,m(x,y′)f(y′)dy′.
Since(26)∂∂xnKn,0(x,y′)=2xnσn1|x-y′|n,
for any x=(x′,xn)∈H and y′∈∂H, we have
(27)|∫Bn-1(R*)∖Bn-1(y*,δ)∂∂xnKn,0(x,y′)f(y′)dy′|≲xn(δ2)-n∫Bn-1(R*)∖Bn-1(y*,δ)f(y′)dy′
for any x∈H∩Bn(y*,δ/2).
Since(28)1-∫Bn-1(y*,δ)∂∂xnKn,0(x,y′)dy′=∫∂H∖Bn-1(y*,δ)∂∂xnKn,0(x,y′)dy′=2xnσn∫∂H∖Bn-1(y*,δ)1|x-y′|ndy′,
for any x∈H, we observe that
(29)limsup⁡x∈H,x→y*∫Bn-1(y*,δ)∂∂xnKn,0(x,y′)dy′=1.
Finally (23), (27), and (29) yield
(30)lim⁡x∈H,x→y*Λ1(x)≤f(y*)+ϵ.
From Lemma6(4) we obtain
(31)|Λ2(x)|≲∫BN-1(R*)xn|f(y′)|dy′≲xn
for any x∈H∩Bn-1(y*,δ).
These and (24) yield
(32)limsup⁡x∈H,x→y*∂∂xnNm[f](x)=limsup⁡x∈H,x→y*∫∂H∂∂xnKn,m(x,y′)f(y′)dy′=limsup⁡x∈H,x→y*(∫∂H∖Bn-1(R*)∂∂xnKn,m(x,y′)f(y′)dy′Λ1(x)+Λ2(x)+∫∂H∖Bn-1(R*)∂∂xnKn,m(x,y′)f(y′)dy′)≤f(y*)+2ϵ.Now the conclusion immediately follows.Lemma 9 (see [1, Lemma 1]).
Ifh(x) is a harmonic polynomial of x=(x′,xn)∈H of degree m and ∂h/∂xn vanishes on ∂H, then there exists a polynomial Π(x′) of degree m such that
(33)h(x)={Π(x′)+∑j=1[m/2](-1)j(2j)!xn2jΔjΠ(x′)ifm≥2,Π(x′)ifm=0,1.


## 3. Proof of Theorem1

We prove only the casep>1; the proof of the case p=1 is similar.For anyϵ>0, there exists Rϵ>1 such that
(34)∫∂H(Rϵ,∞)|f(y′)|p(1+|y′|)n+α-2dy′<ϵ.Take any pointx∈H(Rϵ,∞)-E(ϵ;μ,(n-2)p-β) such that |x|>2Rϵ and write
(35)Nm[f](x)=(∫G1‍+∫G2‍+∫G3‍+∫G4‍+∫G5‍)Kn,m(x,y′)f(y′)dy′=U1(x)+U2(x)+U3(x)+U4(x)+U5(x),
where
(36)G1={y′∈∂H:|y′|≤1},G2={y′∈∂H:1<|y′|≤|x|2},G3={y′∈∂H:|x|2<|y′|≤32|x|},G4={y′∈∂H:32|x|<|y′|≤2|x|},G5={y′∈∂H:|y′|≥2|x|}.First note that(37)|U1(x)|≲∫G1|f(y′)||x-y′|n-2dy′≲|x|2-n∫G1‍|f(y′)|dy′,
so that
(38)lim⁡|x|→∞,x∈H|x|-1+((n-γ-1)/p)U1(x)=0.Ifm<2-((n-γ-1)/p) and 1/p+1/q=1, then (3-n-m+((n+α-2)/p))q+n-1>0. By Lemma 6(1), (34), and Hölder inequality, we have
(39)|U2(x)|≲|x|m-1∫G2‍|y′|-n-m+3|f(y′)|dy′≲|x|m-1(∫G2‍|f(y′)|p|y′|n+α-2dy′)1/p×(∫G2|y′|(-n-m+3+((n+α-2)/p))qdy′)1/q≲|x|1-((n-γ-1)/p)(∫G2‍|f(y′)|p|y′|n+α-2dy′)1/p.Put(40)U2(x)=U21(x)+U22(x),
where
(41)U21(x)=∫G2∩Bn-1(Rϵ)Kn,m(x,y′)f(y′)dy′,U22(x)=∫G2∖Bn-1(Rϵ)‍Kn,m(x,y′)f(y′)dy′.If|x|≥2Rϵ, then
(42)|U21(x)|≲Rϵ2-m-((n-γ-1)/p)|x|m-1.Moreover, by (34) and (39) we get
(43)|U22(x)|≲ϵ|x|1-((n-γ-1)/p).That is,(44)|U2(x)|≲ϵ|x|1-((n-γ-1)/p).By Lemma6(3), (34), and Hölder inequality, we have
(45)|U4(x)|≲ϵxn2-n|x|n-1-((n-γ-1)/p).Ifm>1-((n-γ-1)/p), then (2-n-m+((n+α-2)/p))q+n-1<0. We obtain Lemma 6(4), (34), and Hölder inequality:
(46)|U5(x)|≲|x|m∫G5|y′|-n-m+2|f(y′)|dy′≲|x|m(∫G5‍|f(y′)|p|y′|n+α-2dy′)1/p×(∫G5‍|y′|(-n-m+2+((n+α-2)/p))qdy′)1/q≲ϵ|x|1-((n-γ-1)/p).Finally, we will estimateU3(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
(47)Π(b)={x∈H;inf⁡y′∈∂H|x|x|-y′|y′||<b},
and divide H into two sets Π(b) and H-Π(b).Ifx∈H-Π(b), then there exists a positive number b′ such that |x-y′|≥b′|x| for any y′∈∂H, and hence
(48)|U3(x)|≲∫G3|y′|2-n|f(y′)|dy′≲|x|m∫G3|y′|2-n-m|f(y′)|dy′≲ϵ|x|1-((n-γ-1)/p),
which is similar to the estimate of U5(x).We will consider the casex∈Π(b). Now put
(49)Hi(x)={y′∈∂H[|x|2,32|x|];2i-1δ(x)≤|x-y′|<2iδ(x)y′∈∂H[|x|2,32|x|]},
where δ(x)=inf⁡y′∈H⁡|x-y′|.Since∂H∩{y′∈Rn-1:|x-y′|<δ(x)}=∅, we have
(50)U3(x)=∑i=1i(x)∫Hi(x)|g(y′)||x-y′|n-2dy′,
where i(x) is a positive integer satisfying 2i(x)-1δ(x)≤|x|/2<2i(x)δ(x).Similar to the estimate ofU5(x) we obtain
(51)∫Hi(x)‍|g(y′)||x-y′|n-2dy′≲∫Hi(x)|g(y′)|{2i-1δ(x)}n-2dy′≲δ(x)(β-(n-2)p)/p∫Hi(x)δ(x)(((n-2)p-β)/p)-n+2|g(y′)|dy′≲cos⁡-β/pθδ(x)(β-(n-2)p)/p∫Hi(x)|x|-β/p|g(y′)|dy′≲|x|n-2-(β/p)cos⁡-β/pθδ(x)(β-(n-2)p)/p×∫Hi(x)‍|y′|2-n|g(y′)|dy′≲|x|n-1+((α-β-1)/p)(μ(Hi(x))2iδ(x)(n-2)p-β)1/p
for i=0,1,2,…,i(x).Sincex∉E(ϵ;μ,(n-2)p-β), we have
(52)μ(Hi(x)){2iδ(x)}(n-2)p-β≲μ(Bn-1(x,2iδ(x))){2iδ(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
(53)μ(Hi(x)(x)){2iδ(x)}(n-2)p-β≲μ(Bn-1(x,|x|/2))(|x|/2)(n-2)p-β≲ϵ|x|β-(n-2)p.So(54)|U3(x)|≲ϵ|x|1+((γ-n+1)/p).Combining (38) and (44)–(54), we obtain that if Rϵ is sufficiently large and ϵ is a sufficiently small number, then Nm[f](x)=o(|x|1+((γ-n+1)/p)) as |x|→∞, where x∈H(Rϵ,+∞)-E(ϵ;μ,(n-2)p-β). Finally, there exists an additional finite ball B0 covering H(0,Rϵ], which, together with Lemma 7, gives the conclusion of Theorem 1.


## 4. Proof of Theorem4

For any fixedx∈H, take a number R satisfying R>max⁡{1,2|x|}. If m>(n-γ-1)/p, then (2-n-m+((n+α-2)/p))q+n-1<0. By (5), Lemma 6(4), and Hölder inequality, we have
(55)∫∂H(R,∞)|Kn,m(x,y′)||f(y′)|dy′≲|x|m∫∂H(R,∞)|y′|2-n-m|f(y′)|dy′≲|x|m(∫∂H(R,∞)|f(y′)|p|y′|n+α-2dy′)1/p×(∫∂H(R,∞)|y′|(-n-m+2+((n+α-2)/p))qdy′)1/q<∞.
Hence, Nm[f](x) is absolutely convergent and finite for any x∈H. Thus, Nm[f](x) is harmonic on H.To prove(56)lim⁡x→y′,x∈H∂∂xnNm[f](x)=f(y′)
for any point y′∈∂H, we only need to apply Lemma 8 to f(y) and -f(y).We complete the proof of Theorem4.


## 5. Proof of Theorem5

Consider the functionh′(x)=h(x)-Nm[f](x). Then it follows from Theorems 4 and 5 that h′(x) is a solution of the Neumann problem on H with f and it is an even function of xn (see [1, page 92]).Since(57)0≤{h-Nm[f]}+(x)≤h+(x)+{Nm[f]}-(x)for any x∈H, we have(58)lim⁡|x|→∞,x∈HNm[f](x)=o(|x|1+((γ-n+1)/p))from Theorem 4.Moreover, (18) gives that
(59)lim⁡|x|→∞,x∈H(h-Nm[f])(x)=o(|x|l+[1+((γ-n+1)/p)]).This implies thath′(x) is a polynomial of degree less than l+[1+((γ-n+1)/p)] (see [7, Appendix]), which gives the conclusion of Theorem 5 from Lemma 9.



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