# Estimates of the Fundamental Solution for Higher Order Schrödinger Type Operators and Their Applications

Authors: Satoko Sugano
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We consider higher order Schrödinger type operators with nonnegative potentials. We assume that the potential belongs to the reverse Hölder class which includes nonnegative polynomials. We establish estimates of the fundamental solution and showLp boundedness of some Schrödinger type operators.


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## Body

## 1. Introduction

LetV(x) be a nonnegative potential and consider the Schrödinger type operators Hk=(-Δ)k+Vk on Rn, where k is a positive integer and n≥2k+1. When V is a nonnegative polynomial, Zhong proved estimates of the fundamental solution for H1 and H2 and showed some estimates for H1 and H2 (see [1]). More precisely, he showed the Lp boundedness of the operators VH1-1, V1/2∇H1-1, and V2-j/2∇jH2-1, where j=0,1,2,3,4. He also proved that the operators ∇2H1-1 and ∇4H2-1 are Calderón-Zygmund operators. Recently, in [2], the authors showed the Lp boundedness of the operators VkHk-1, ∇2kHk-1, and ∇kHk-1/2 for nonnegative polynomial potential  V.For the potentialV which belongs to the reverse Hölder class, which includes nonnegative polynomials, Shen generalized Zhong's results on H1 (see [3]). Actually, he established estimates of the fundamental solution for H1 and showed the Lp estimates for the operators VH1-1, V1/2∇H1-1, ∇2H1-1, and so on. For the operator H1 with reverse Hölder class potentials, further results have been investigated by many researchers. See [4–7], for example. For the operator H2 with reverse Hölder class potentials, in [8], the author established estimates of the fundamental solution for H2 and showed the Lp boundedness of the operators V2-j/2∇jH2-1, where j=0,1,2,3,4. In [8], the author also showed that the operator ∇4H2-1 is a Calderón-Zygmund operator. Recently, in [9], the authors showed Lp boundedness of the operator ∇2H2-1/2 and that the operator ∇2H2-1/2 is a Calderón-Zygmund operator.In this paper, we studyH2m=(-Δ)2m+V2m with reverse Hölder class potential V on Rn, where m is a nonnegative integer and n≥2m+1+1. We establish estimates of the fundamental solution for H2m and show the Lp boundedness of the operators V2m-j/2∇jH2m-1, where j is an integer satisfying 0≤j≤2m+1.As mentioned above, in [2], the authors proved some results on Hk=(-Δ)k+Vk, where V is a nonnegative polynomial and k is an integer, k≥3. They proved their results by making use of [2, Lemma 3.3 and Corollary 3.1] which have been proved only for nonnegative polynomial  V. In this paper, our strategy is different from the one in [2], since the question whether the above two results can be proved for reverse Hölder class potentials V is yet to be settled. The purpose of this paper is to show some results on Hk=(-Δ)k+Vk with potential  V which belongs to the reverse Hölder class, which includes nonnegative polynomials. However, our results are only for k=2m, where m is an integer, m≥2.We recall the definitions of the reverse Hölder class (e.g., [3]). We denote by Br(x) the ball centered at x with radius r.Definition 1 (reverse Hölder class).
LetV≥0.
(
1
) For 1<p<∞ one says that  V∈(RH)p  if V∈Llocp(Rn) and there exists a positive constant C such that
(1)(1|Br(x)|∫Br(x)‍V(y)pdy)1/p≤C|Br(x)|∫Br(x)‍V(y)dy
holds for every x∈Rn and 0<r<∞.
(
2
) One  says that  V∈(RH)∞ if V∈Lloc∞(Rn) and there exists a positive constant C such that
(2)∥V∥L∞(Br(x))≤C|Br(x)|∫Br(x)‍V(y)dy
holds for every x∈Rn and 0<r<∞.Remark 2.
(
1
) If P(x) is a polynomial and α>0, then V(x)=|P(x)|α belongs to (RH)∞ (see [10, page 146]).
(
2
) For 1<p<∞, it is easy to see (RH)∞⊂(RH)p.Definition 3 (see [3, Definition 1.3]).
(1) LetV∈(RH)n/2 and V≢0. Then it is well known that there exists a positive ϵ such that V∈(RH)n/2+ϵ (see [11, Lemma 2]). Then the function ρ(x,V) is well defined by
(3)1ρ(x,V)=sup{r>0:r2|Br(x)|∫Br(x)‍V(y)dy≤1}
and satisfies 0<ρ(x,V)<∞ for every x∈Rn.One recalls that(4)r2|Br(x)|∫Br(x)‍V(y)dy=1forr=1ρ(x,V)
(see [3, page 519]).Remark 4.
(
1
) If V∈(RH)p, p≥n/2, then there exists a positive constant C such that
(5)(1|Br(x)|∫Br(x)‍V(y)pdy)1/p≤Cρ(x,V)2
(cf. [3, Lemma 1.8] and [5, Lemma 2.2 (a)]).
(
2
) If V∈(RH)∞ then there exists a positive constant C such that V(x)≤Cρ(x,V)2 (see [3, Remark 2.9]).
Letα=(α1,…,αn) denote the multi-index with αi∈N, i∈N, and  1≤i≤n. Define xα=x1α1⋯xnαn and ∇j=∂α=∂|α|/(∂x1α1⋯∂xnαn) for j=|α|=α1+⋯+αn. For any positive integer j and a function u∈Cj(Rn), denote that  ∇ju(x)=(∂αu(x):|α|=j) and |∇ju(x)|2=∑|α|=j‍|∂αu(x)|2. We denote by ΓHk(x,y) the fundamental solution for Hk. The operator Hk-1 is the integral operator with ΓHk(x,y) as its kernel.Now we state our theorems.Theorem 5.
Letj, m, and n be integers, m≥0, n≥2m+1+1, and 0≤j≤2m+1-1. Suppose that V∈(RH)q0 for some q0≥2mn/(2m+1-j). Then there exist positive constants Cj such that
(6)∥V2m-j/2∇jH2m-1f∥Lp(Rn)≤Cj∥f∥Lp(Rn),
where 1≤p≤2q0/(2m+1-j).In Theorem5, the case m=0 was shown in [3, Theorems 2.7 and 4.13].Remark 6.
In Theorem5, for the case m=j=0, inequality (6) was shown in [3, Remark 2.9] under the assumption V∈(RH)∞. If we take the limit q0→∞, then class (RH)q0 becomes (RH)∞ and V∈(RH)∞ implies “V∈(RH)n/2 and V(x)≤Cρ(x,V)2.” See [8, Remark 3] and also [8, Theorem 1].Remark 7.
We assume “V∈(RH)n/2 and V(x)≤Cρ(x,V)2.” Then inequality (6) is true for 1≤p≤∞. See Section 7 for the details. We also remark that the inequality (6) is true for j=2m+1 and 1<p<∞ under the assumption “V∈(RH)n/2 and V(x)≤Cρ(x,V)2.” See [3, Remark 2.9] and also [8, Theorem 1 (2)].Theorem 8.
Letj, m, and  n be integers, m≥0, n≥2m+1+1, and 1≤j≤2m+1-1. Suppose that V∈(RH)q0 for some q0, n/2≤q0<2mn/(2m+1-j). Then there exist positive constants Cj such that (6) holds, where 1≤p≤p0 and 1/p0=(2m+2-j)/(2q0)-(2m+1-j)/n.To prove Theorems5 and 8 we need the estimates of the fundamental solution (Theorems 9 and 10). The following Theorem 9 generalizes the results in [2, Theorem 3.1] to the operator H2m with potential  V which belongs to the reverse Hölder class.Theorem 9.
Letm and n be integers, m≥0, and n≥2m+1+1. Suppose that  V∈(RH)n/2. Then for any positive integer N there exists a positive constant CN such that
(7)(0≤)ΓH2m(x,y)≤CN{1+ρ(x,V)|x-y|}N·1|x-y|n-2m+1.In Theorem9, the cases m=0 and m=1 were shown in [3, Theorem 2.7] and [8, Theorem 2], respectively. We prove Theorem 9 by induction; that is, we assume that Theorem 9 is true for m=l and show the case m=l+1. We also prove the following theorem which states derivative estimates of the fundamental solution.Theorem 10.
Letj, m, and n be integers, m≥0, n≥2m+1+1, and 1≤j≤2m+1-1. Suppose that  V∈(RH)2mn/(2m+1-j). Then for any positive integer N there exists a positive constant CN such that
(8)|∇xjΓH2m(x,y)|≤CN{1+ρ(x,V)|x-y|}N·1|x-y|n-2m+1+j.In Theorem10, the cases m=0 and m=1 were shown in [3, page 537] and [8, Theorem 6], respectively.The plan of this paper is as follows. In Section2, we describe some lemmas needed later. In Section 3, we assume that Theorem 9 is true for m=l and show some estimates for H2l which are needed to prove the case m=l+1 in Theorem 9. In Section 4, we prove the case m=l+1 in Theorem 9. Section 5 is devoted to proof of Theorem 10. In Section 6, we prove Theorems 5 and 8. Finally, in Section 7, we state some remarks. Throughout this paper the letter C stands for a constant not necessarily the same at each occurrence.


## 2. Preliminaries

In this section, we describe some lemmas needed later. First, we remark thatV∈(RH)q, q>1, implies that V(x)dx is a doubling measure; that is, there exists a positive constant C0 such that
(9)∫B2r(x)‍V(y)dy≤C0∫Br(x)‍V(y)dy
(see [3, page 518] and also [5, Remark 1.2 (1)]).Lemma 11 (see [3, Lemma 1.2]).
Assume thatV∈(RH)q for some q>n/2. Then there exists a positive constant C such that, for 0<r<R<∞,
(10)r2|Br(x)|∫Br(x)‍V(y)dy≤C(Rr)n/q-2R2|BR(x)|∫BR(x)‍V(y)dy.Lemma 12 (see [3, Lemma 1.4 (b), (c)]).
(
1
) Suppose that  V∈(RH)n/2. Then there exist positive constants C and k0 such that, for x,y∈Rn,
(11)ρ(y,V)≤C{1+ρ(x,V)|x-y|}k0ρ(x,V).(2) Suppose V∈(RH)n/2. Then there exist positive constants C′ and k0 such that, for x, y∈Rn,
(12)ρ(y,V)≥C′ρ(x,V){1+ρ(x,V)|x-y|}k0/(k0+1).Lemma 13 (Caccioppoli type inequality).
Leta, j, and  k be integers, k≥2, 1≤a≤k/2, and 1≤j≤2a. Assume that  (-Δ)ku+Vku=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(13)∫BR/2(x0)‍|Δk-au(x)|2dx+∫BR/2(x0)‍V(x)k|u(x)||Δk-2au(x)|dx≤C∑j=12a‍1R2j∫BR(x0)‍|∇2a-j(Δk-2au)(x)|2dx.For readers’ convenience, we give the proof of Lemma13 at the end of this section. If k is an even number, then letting a=k/2 in Lemma 13, we have the following.Corollary 14 (see [2, Lemma 3.5]).
Letj be an integer, k an even number, k≥2, and  0≤j≤k-1. Assume that  (-Δ)ku+Vku=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(14)∫BR/2(x0)‍|Δk/2u(x)|2dx+∫BR/2(x0)‍V(x)k|u(x)|2dx≤C∑j=0k-1‍1R2(k-j)∫BR(x0)‍|∇ju(x)|2dx.Lemma 15 (Caccioppoli type inequality).
Leta, j, and k be integers, k≥1, 1≤a≤(k+1)/2, and 1≤j≤2a-1. Assume that  (-Δ)ku+Vku=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(15)∫BR/2(x0)‍|∇(Δk-au)(x)|2dx+∫BR/2(x0)‍V(x)k|u(x)||Δk-2a+1u(x)|dx≤C∑j=12a-1‍1R2j∫BR(x0)‍|∇2a-j-1(Δk-2a+1u)(x)|2dx.Ifk is an odd number, then letting a=(k+1)/2 in Lemma 15, we have the following.Corollary 16 (see [2, Lemma 3.5]).
Letj be an integer, k an odd number, k≥1, and  0≤j≤k-1. Assume that  (-Δ)ku+Vku=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(16)∫BR/2(x0)‍|∇(Δ(k-1)/2u)(x)|2dx+∫BR/2(x0)‍V(x)k|u(x)|2dx≤C∑j=0k-1‍1R2(k-j)∫BR(x0)‍|∇ju(x)|2dx.Lemma 17.
Letj, k, and  n be integers, k≥1, n≥2k+1, and 0≤j≤k-1. Assume that  (-Δ)ku+Vku=0, u≥0, in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(17)supy∈BR/2(x0)|u(y)|≤C∑j=0k-1‍Rj(1|BR(x0)|∫BR(x0)‍|∇ju(x)|2dx)1/2.In Lemma17, the cases k=1 and k=2 have been proved (see [3, page 523] and [1, Corollary 5.6]).Proof of Lemma17.
We proceed following the proof of [1, Corollary 5.6]. We prove Lemma 17 for u which satisfies (-Δ)ku+Vku=0, u≥0, in BR(x0). Let a and b be nonnegative integers and choose η∈C0∞(BR/4(x0)) such that η≡1 on BR/8(x0) and |∇b(Δaη)|≤C/R2a+b. Let ΓHk-1,0(x,y)=C|x-y|2(k-1)-n be the fundamental solution for (-Δ)k-1. Note that
(18)Δk-1(u(y)η(y))=∑a,b‍(2(k-1)2a+b)∇b(Δk-1-a-bu)(y)·∇b(Δaη)(y),
where
(19)∇b(Δk-1-a-bu)(y)·∇b(Δaη)(y)=∑i1,i2,…,ik-1=1n‍∂b∂yi1∂yi2⋯∂yib×(∂2∂yi12(∂2∂yi22(⋯(∂2∂yik-1-a-b2u)⋯)))·∂b∂yi1∂yi2⋯∂yib(∂2∂yi12(∂2∂yi22(⋯(∂2∂yia2η)⋯)))
and the summation is taken over all integers a and b satisfying a≥0, b≥0, and 0≤a+b≤k-1. Then, integrating by parts, for x∈BR/4(x0) we have
(20)u(x)η(x)=∫Rn‍ΓHk-1,0(x,y)(-Δ)k-1(uη)(y)dy=∫Rn‍ΓHk-1,0(x,y)×∑a,b‍{(-1)k-1(2(k-1)2a+b)∇b(Δk-1-a-bu)(y)×{∑a,b‍·∇b(Δaη)(y)(2(k-1)2a+b)}dy=∫Rn‍ΓHk-1,0(x,y)(-Δ)k-1u(y)η(y)dy+C∫Rn‍∑(a,b)≠(0,0){∇b(Δk-1-a-bΓHk-1,0(x,y)){+C∑(a,b)≠(0,0)∫Rn‍·∇b(Δaη(y))u(y)∇b(Δk-1-a-bΓHk-1,0(x,y))}dy.
Let
(21)|u(x0)|≤∫BR/4(x0)‍|(-Δ)k-1u(y)||x0-y|n-2(k-1)dy+CR2a+b∑(a,b)≠(0,0)‍∫BR/4(x0)∖BR/8(x0)‍|u(y)||x0-y|n-2a-bdy=I1+I2.
Then we have
(22)I2≤CRn∫BR/4(x0)‍|u(y)|dy≤C(1|BR(x0)|∫BR(x0)‍|u(y)|2dy)1/2.
It remains to estimate I1. Since u≥0, it follows that Δ((-Δ)k-1u)=-(-Δ)ku=Vku≥0. Hence (-Δ)k-1u is subharmonic. Then for all x∈BR/4(x0),
(23)(-Δ)k-1u(x)≤1|BR/4(x)|∫BR/4(x)‍(-Δ)k-1u(y)dy(-Δ)k-1u(x)≤C(1|BR/2(x0)|∫BR/2(x0)‍|Δk-1u(y)|2dy)1/2.
Then we have
(24)I1≤Csupy∈BR/4(x0)|(-Δ)k-1u(y)|∫BR/4(x0)‍dy|x0-y|n-2(k-1)≤CR2(k-1)supy∈BR/4(x0)|Δk-1u(y)|≤CR2k-2(1|BR/2(x0)|∫BR/2(x0)‍|Δk-1u(y)|2dy)1/2.
Then using Lemmas 13 and 15 repeatedly, we have
(25)I1≤CR2k-2-n/2(1R2∫BR(x0)‍|∇(Δk-2u)(x)|2dx(≤CR2k-2-n/2+1R4∫BR(x0)‍|Δk-2u(x)|2dx)1/2≤⋯≤C∑j=0k-1‍Rj(1Rn∫BR(x0)‍|∇ju(x)|2dx)1/2.
It follows that
(26)|u(x0)|≤C∑j=0k-1‍Rj(1|BR(x0)|∫BR(x0)‍|∇ju(x)|2dx)1/2.
From (26) we have, for all y∈BR/2(x0),
(27)|u(y)|≤C∑j=0k-1‍Rj(1|BR/4(y)|∫BR/4(y)‍|∇ju(x)|2dx)1/2.
Then we arrive at the desired estimate.Lemma 18.
Letj, k, and  n be integers, k≥1, n≥2k+1, and 1≤j≤2k-1. Suppose that  V∈(RH)q0 for some q0, n/2≤q0<kn/(2k-j). Assume also that (-Δ)ku+Vku=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(28)(∫BR/2(x0)‍|∇ju(x)|tdx)1/t≤CRkn/q0-2k{1+Rρ(x0,V)}2ksupy∈BR(x0)|u(y)|,
where 1/t=k/q0-(2k-j)/n.In Lemma18, the cases k=1 and k=2 were shown in [3, Lemma 4.6] and [8, Lemma 7], respectively.Proof of Lemma18.
We show Lemma18 by a method similar to the one used in the proof of [3, Lemma 4.6]. Let η∈C0∞(BR(x0)) such that η≡1 on B3R/4(x0) and |∇b(Δaη)|≤C/R2a+b, where a and b are nonnegative integers. We denote by ΓHk,0(x,y) the fundamental solution for (-Δ)k. Note that
(29)u(x)η(x)=∫Rn‍ΓHk,0(x,y){∑a,b‍(2k2a+b)-V(y)ku(y)η(y)+(-1)k{=∫Rn‍ΓHk,0(x,y)×∑a,b‍(2k2a+b)∇b(Δk-a-bu)(y){=∫Rn‍ΓHk,0(x,y)1111×·∇b(Δaη)(y)∑a,b‍(2k2a+b)}dy,
where the summation is taken over all integers a and b satisfying a≥0, b≥0, and 1≤a+b≤k. Then  integrating by parts, for x∈BR/2(x0), we have
(30)|∇ju(x)|≤C∫BR(x0)‍V(y)k|u(y)||η(y)||x-y|n-2k+jdy+CRn-2k+j+b+2(k-a-b)+2a+b∫BR(x0)‍|u(y)|dy≤Csupy∈BR(x0)|u(y)|·∫BR(x0)‍V(y)k|x-y|n-2k+jdy+CRn+j∫BR(x0)‍|u(y)|dy.
It then follows from the well-known theorem on fractional integrals that
(31)(∫BR/2(x0)‍|∇ju(x)|tdx)1/t≤Csupy∈BR(x0)|u(y)|(∫BR(x0)‍V(x)q0dx)k/q0+CRkn/q0-2ksupy∈BR(x0)|u(y)|≤CRkn/q0-2k{1+Rρ(x0,V)}2ksupy∈BR(x0)|u(y)|,
where 1/t=k/q0-(2k-j)/n and we have used Remark 4 (1).At the end of this section, we give the following.Proof of Lemma13.
Letl and m be nonnegative integers and choose η∈C0∞(BR(x0)) such that η≡1 on BR/2(x0) and |∇m(Δlη)|≤C/R2l+m. Multiplying (-Δ)ku+Vku=0 by ((-Δ)k-2au)¯ηk+1 and integrating over Rn by integrating by parts, we have
(32)∫BR/2(x0)‍|Δk-au(x)|2dx+∫BR/2(x0)‍V(x)k|u(x)||Δk-2au(x)|dx≤C∫BR(x0)‍∑l,m‍|Δk-au(x)||∇m(Δk-a-l-mu)(x)≤C∫BR(x0)‍∑l,m‍||Δk-au(x)|·∇m(Δlηk+1)(x)|dx,
where the summation is taken over all integers l and m satisfying l≥0, m≥0, and 1≤l+m≤a. Let ϵ be a positive real number which will be determined later. Then the right-hand side of (32) is bounded by
(33)C∫BR(x0)‍∑l,m‍ϵ|Δk-au(x)|·1ϵ·1R2l+m|∇m(Δk-a-l-mu)(x)|dx≤C∫BR(x0)‍∑l,m‍(ϵ|Δk-au(x)|2+1ϵ·1R2(2l+m)×|∇m(Δk-a-l-mu)(x)|2)dx.
Then choosing ϵ such that C(a-1)(a+2)ϵ=1, we arrive at the desired estimate.


## 3. Estimates forH2l

In this section, we assume that Theorem9 is true for m=l and show estimates for H2l which is needed to prove the case m=l+1 in Theorem 9.Lemma 19.
Letj, l, and n be integers, l≥0, and n≥2l+1+1. Suppose that  V∈(RH)n/2. Assume also that Theorem 9 is true for m=l. Then there exists a positive constant C such that
(34)∥ρ(·,V)2l+1H2l-1f∥Lp(Rn)≤C∥f∥Lp(Rn),
where 1≤p≤∞.Proof.
We can prove Lemma19 by the same way as in the proof of [3, Corollary 2.8]. It follows from the inductive assumption that
(35)∫Rn‍|ΓH2l(x,y)|dy≤C∫Rn‍dy{1+ρ(x,V)|x-y|}2l+1+1|x-y|n-2l+1=Cρ(x,V)2l+1.
If f∈Lp(Rn), then
(36)u(x)=H2l-1f(x)=∫Rn‍ΓH2l(x,y)f(y)dy.
Let p′=p/(p-1). Then using Hölder's inequality and (35) we have
(37)|u(x)|≤(∫Rn‍|ΓH2l(x,y)|dy)1/p′×(∫Rn‍|ΓH2l(x,y)||f(y)|pdy)1/p≤Cρ(x,V)2l+1/p′(∫Rn‍|ΓH2l(x,y)||f(y)|pdy)1/p.
It follows that
(38)∫Rn‍|ρ(x,V)2l+1u(x)|pdx≤C∫Rn‍|f(y)|p(∫Rn‍ρ(x,V)2l+1|ΓH2l(x,y)|dx)dy.
Finally, note that, by the inductive assumption and Lemma 12 (1),
(39)∫Rn‍ρ(x,V)2l+1|ΓH2l(x,y)|dx≤CCN∫Rn‍ρ(y,V)2l+1{1+ρ(y,V)|x-y|}N-2l+1k0|x-y|n-2l+1dx≤C
if we choose N=2l+1k0+2l+1+1. Then Lemma 19 follows.We also need the following lemma.Lemma 20.
Letj, l, and n be integers, l≥0, n≥2l+1+1, and 1≤j≤2l+1-1. Suppose that V∈(RH)q0 for some q0, n/2≤q0<2ln/(2l+1-j). Assume also that Theorem 9 is true for m=l. Then for 1≤p≤p0 there exists a positive constant C such that
(40)∥ρ(·,V)2l+1-j∇jH2l-1f∥Lp(Rn)≤C∥f∥Lp(Rn),
where 1/p0=2l/q0-(2l+1-j)/n.Proof.
We show Lemma20 by a method similar to the one used in the proof of [3, Theorem 4.13]. Suppose that V∈(RH)q0 for some q0≥n/2. Then V∈(RH)q1 for some q1, 2ln/(2l+1-j)>q1>q0. Let
(41)T2lf(x)=ρ(x,V)2l+1-j∫Rn‍∇xjΓH2l(x,y)f(y)dy.
The adjoint of T2l is given by
(42)T2l*f(x)=∫Rn‍∇yjΓH2l(y,x)ρ(y,V)2l+1-jf(y)dy.
By duality, it suffices to show that
(43)∥T2l*f∥Lp(Rn)≤C∥f∥Lp(Rn)forp0′≤p≤∞,
where 1/p0+1/p0′=1. Let r=1/ρ(x,V). We choose t and p1 such that 1/t=2l/q1-(2l+1-j)/n, 1/p1=1-2l/q1+(2l+1-j)/n. Thus 1/t+1/p1=1. By Hölder’s inequality we have
(44)|T2l*f(x)|≤∑i=-∞∞‍∫2i-1r<|y-x|≤2ir‍|∇yjΓH2l(y,x)|×ρ(y,V)2l+1-j|f(y)|dy≤∑i=-∞∞‍(∫2i-1r<|y-x|≤2ir‍{(y,V)2l+1-j|∇yjΓH2l(y,x)|≤∑i=-∞∞‍({∫2i-1r<|y-x|≤2ir‍×ρ(y,V)2l+1-j}tdy)1/t·(∫|y-x|≤2ir‍|f(y)|p1dy)1/p1.
Let a be a positive integer and ya∈Rn. It follows from Lemma 18, Lemma 12 (1), and the inductive assumption that
(45)(∫2i-1r<|y-x|≤2ir‍{|∇yjΓH2l(y,x)|ρ(y,V)2l+1-j}tdy)1/t≤∑a=1A‍(∫|z-ya|≤2i-1r|ya-x|=3·2i-2r‍{|∇zjΓH2l(z,x)|ρ(z,V)2l+1-j}tdz)1/t≤C(2i-1r)2ln/q1-2l+1{1+2irρ(x,V)}2l+1+(2l+1-j)k0×ρ(x,V)2l+1-j·∑a=1A‍supz∈B5·2i-3r(ya)|ΓH2l(z,x)|≤C(2ir)2ln/q1-2l+1(1+2i)2l+1+(2l+1-j)k0·1r2l+1-j·CN{1+ρ(x,V)|z-x|}N·∑a=1A‍supz∈B5·2i-3r(ya)1|z-x|n-2l+1≤CCN(2ir)2ln/q1-2l+1(1+2i)2l+1+(2l+1-j)k0r2l+1-j(1+2i-3)N(2i-3r)n-2l+1≤CCN(2ir)2ln/q1-nr2l+1-j(1+2i)N-2l+1-(2l+1-j)k0,
where A is a finite integer not depending on i and r. Thus
(46)|T2l*f(x)|≤CCN∑i=-∞∞‍(2i)2l+1-j(1+2i)N-2l+1-(2l+1-j)k0×{1(2ir)n∫B2ir(x)‍|f(y)|p1dy}1/p1≤C{M(|f|p1)(x)}1/p1,
where we choose N≥2l+2+(2l+1-j)k0-j+1 and M is the Hardy-Littlewood maximal operator. Hence it follows that
(47)∥T2l*f∥Lp(Rn)≤C∥f∥Lp(Rn)forp1<p≤∞.
Then (43) follows since p0′>p1.For the caseq0=n/2 in Lemma 20, if n≥2l+2+1, letting p=2, we have the following.Corollary 21.
Letj, l, and n be integers, l≥0, n≥2l+2+1, and 1≤j≤2l+1-1. Suppose that V∈(RH)n/2. Assume also that Theorem 9 is true for m=l. Then there exists a positive constant C such that
(48)∥ρ(·,V)2l+1-j∇jH2l-1f∥L2(Rn)≤C∥f∥L2(Rn).


## 4. Proof of Theorem9 (Casem=l+1)

In this section, we assume that Theorem9 is true for m=l and prove the case m=l+1 in Theorem 9. It follows easily from the following lemma.Lemma 22.
Letj, l, and n be integers, l≥0, and n≥2l+2+1. Suppose that V∈(RH)n/2. Assume also that (-Δ)2l+1u+V2l+1u=0 in BR(x0) for some x0∈Rn and Theorem 9 is true for m=l. Then for any positive integer N there exists a positive constant CN such that
(49)supy∈BR/2(x0)|u(y)|≤CN{1+Rρ(x0,V)}Nsupy∈BR(x0)|u(y)|.Assuming Lemma22 for the moment, we giveProof of Theorem9  (Casem=l+1).
Fixx0, y0∈Rn and put R=|x0-y0|. Then u(x)=ΓH2l+1(x,y0) is a solution of (-Δ)2l+1u+V2l+1u=0 on BR/2(x0). Using the estimate 0≤ΓH2l+1(x,y)≤C|x-y|2l+2-n and (49), we arrive at the desired estimate.To prove Lemma22, we need some lemmas.Lemma 23.
Letj, l, and n be integers, l≥0, n≥2l+2+1, and 0≤j≤2l+1-1. Suppose that V∈(RH)n/2. Assume also that Theorem 9 is true for m=l. Then there exists a positive constant C such that
(50)∑j=02l+1-1‍∫Rn‍ρ(x,V)2(2l+1-j)|∇ju(x)|2dx≤C(∫Rn‍|Δ2lu(x)|2dx+∫Rn‍V(x)2l+1|u(x)|2dx),
where u∈C0∞(Rn).Proof.
By the casep=2 in Lemma 19 and Corollary 21, we have
(51)∑j=02l+1-1‍∫Rn‍ρ(x,V)2(2l+1-j)|∇ju(x)|2dx≤C∫Rn‍|{(-Δ)2l+V2l}u(x)|2dx≤C(∫Rn‍|Δ2lu(x)|2dx+∫Rn‍V(x)2l+1|u(x)|2dx).Ifn≥2l+2+1 and 1≤j≤2l+1-1, then letting k=2l+1 and q0=n/2 in Lemma 18 we have the following.Corollary 24.
Letj, l, and n be integers, l≥0, n≥2l+2+1, and 1≤j≤2l+1-1. Suppose that V∈(RH)n/2. Assume also that (-Δ)2l+1u+V2l+1u=0 in BR(x0) for some x0∈Rn. Then there exists a positive constant C such that
(52)(1|BR/2(x0)|∫BR/2(x0)‍|∇ju(x)|2dx)1/2≤C{1+Rρ(x0,V)}2l+2Rjsupy∈BR(x0)|u(y)|.Now we are ready to give the following.Proof of Lemma22.
Letη∈C0∞(BR/2(x0)) such that η≡1 on BR/4(x0), |∇jη|≤C/Rj, where j is an integer satisfying 1≤j≤2l+1-1. Applying Lemma 23 to uη and using Corollary 14 we have
(53)∑j=02l+1-1‍∫BR/4(x0)‍ρ(x,V)2(2l+1-j)|∇ju(x)|2dx≤C∑j=02l+1-1‍1R2(2l+1-j)∫BR(x0)‍|∇ju(x)|2dx.
From Lemma 12 (2) it follows for each integer j satisfying 1≤j≤2l+1-1 that
(54)∫BR/4(x0)‍|∇ju(x)|2dx≤C{1+Rρ(x0,V)}2(2l+1-j)k0/(k0+1){Rρ(x0,V)}2(2l+1-j)·R2l+2R2j·∑i=02l+1-1‍R2iR2l+2∫BR(x0)‍|∇iu(x)|2dx.
Then we have
(55)∑j=02l+1-1‍Rj(1|BR/4(x0)|∫BR/4(x0)‍|∇ju(x)|2dx)1/2≤C{1+Rρ(x0,V)}1/(k0+1)∑i=02l+1-1‍Ri·(1|BR(x0)|∫BR(x0)‍|∇iu(x)|2dx)1/2.
Repeating the above argument, for any positive integer K we have
(56)∑j=02l+1-1‍Rj(1|BR/4K(x0)|∫BR/4K(x0)‍|∇ju(x)|2dx)1/2≤CK{1+Rρ(x0,V)}K/(k0+1)∑j=02l+1-1‍Rj≤·(1|BR(x0)|∫BR(x0)‍|∇ju(x)|2dx)1/2.
Then using Lemma 17, Corollary 24, and estimate (56), we arrive at the desired estimate.


## 5. Proof of Theorem10

In this section, we prove Theorem10 which states derivative estimates of the fundamental solution for H2m. We arrive at Theorem 10 combining Lemmas 25 with 22.Lemma 25.
Letj, m, and n be integers, m≥0, n≥2m+1+1, and 1≤j≤2m+1-1. Suppose that  V∈(RH)2mn/(2m+1-j). Assume also that (-Δ)2mu+V2mu=0 in BR(x0) for some x0∈Rn. Then there exist positive constants Cj and Cj′ such that
(57)supy∈BR/2(x0)|∇ju(y)|≤Cj{1+Rρ(x0,V)}Cj′Rjsupy∈BR(x0)|u(y)|.Proof.
Letη∈C0∞(BR(x0)) such that η≡1 on B3R/4(x0) and |∇b(Δaη)|≤C/R2a+b, where a and b are nonnegative integers. We use (29); integrating by parts, we have
(58)|∇ju(x0)|≤C∫BR(x0)‍V(y)2m|u(y)||x0-y|n-2m+1+jdy+CRn+j∫BR(x0)‍|u(y)|dy.
Since V∈(RH)2mn/(2m+1-j), it follows that V∈(RH)q for some q>2mn/(2m+1-j). We choose r such that 2m/q+1/r=1 and r>1. By Hölder’s inequality we have
(59)|∇ju(x0)|≤CRn(1Rn∫BR(x0)‍V(y)qdy)2m/q·(1Rn∫BR(x0)‍dy|x0-y|(n-2m+1+j)r)1/rsupy∈BR(x0)|u(y)|+CRn+j∫BR(x0)‍|u(y)|dy≤C{1+Rρ(x0,V)}2m+1Rjsupy∈BR(x0)|u(y)|,
where we have used Remark 4 (1). From (59) we have, for all y∈BR/2(x0),
(60)|∇ju(y)|≤C{1+Rρ(y,V)}2m+1Rjsupx∈BR/4(y)|u(x)|.
Using Lemma 12 (1), we have
(61)supy∈BR/2(x0)|∇ju(y)|≤C{1+Rρ(x0,V)}2m+1(k0+1)Rjsupy∈BR(x0)|u(y)|.
Then the proof is complete.


## 6. Proofs of Theorems5 and 8

In this section, we prove Theorems5 and 8.Proof of Theorem5.
We assume thatV∈(RH)q0 for some q0≥2mn/(2m+1-j). Then V∈(RH)q1 for some q1>q0. Let
(62)T2mf(x)=V(x)(2m+1-j)/2∫Rn‍∇xjΓH2m(x,y)f(y)dy.
The adjoint of T2m is given by
(63)T2m*f(x)=∫Rn‍∇yjΓH2m(y,x)V(y)(2m+1-j)/2f(y)dy.
Let p0=2q0/(2m+1-j). By duality, it suffices to show that
(64)∥T2m*f∥Lp(Rn)≤C∥f∥Lp(Rn)forp0′≤p≤∞,
where 1/p0+1/p0′=1. Let r=1/ρ(x,V). By Theorems 9 and 10 and Hölder's inequality we have
(65)|T2m*f(x)|≤∑i=-∞∞‍∫2i-1r<|y-x|≤2ir‍|∇yjΓH2m(y,x)|×V(y)(2m+1-j)/2|f(y)|dy≤CCN∑i=-∞∞‍1(1+2i)N·1(2ir)n-2m+1+j×∫B2ir(x)‍V(y)(2m+1-j)/2|f(y)|dy≤CCN∑i=-∞∞‍(2ir)2m+1-j(1+2i)N×{1(2ir)n∫B2ir(x)‍V(y)q1dy}(2m+1-j)/(2q1)·{1(2ir)n∫B2ir(x)‍|f(y)|p1dy}1/p1,
where 1/p1=1-(2m+1-j)/(2q1). Since V∈(RH)q1 we have
(66)|T2m*f(x)|≤CCN{M(|f|p1)(x)}1/p1·∑i=-∞∞‍1(1+2i)N{1(2ir)n-2∫B2ir(x)‍V(y)dy}(2m+1-j)/2.
For the case j≤0, by Lemma 11 and (4), we have
(67)1(2ir)n-2∫B2ir(x)‍V(y)dy≤C(r2ir)n/q1-2·1rn-2∫Br(x)‍V(y)dy=C(2i)2-n/q1.
For the case j≥1, by the doubling condition (9) and (4), we have
(68)1(2ir)n-2∫B2ir(x)‍V(y)dy≤22i·2-inC0i·1rn-2∫Br(x)‍V(y)dy≤C(2i)k1,
where k1=2-n+log2C0. Hence, choosing N≥k1(2m+1-j)/2+1, we obtain
(69)|T2m*f(x)|≤C{M(|f|p1)(x)}1/p1.
It follows that
(70)∥T2m*f∥Lp(Rn)≤C∥f∥Lp(Rn)forp1<p≤∞.
Then (64) follows since p1={2q1/(2m+1-j)}′<{2q0/(2m+1-j)}′=p0′.Proof of Theorem8.
We assume thatV∈(RH)q0 for some q0, n/2≤q0<2mn/(2m+1-j). Then V∈(RH)q1 for some q1, 2mn/(2m+1-j)>q1>q0. We use (62) and (63) again and it suffices to show (64) for 1/p0=(2m+2-j)/(2q0)-(2m+1-j)/n and 1/p0+1/p0′=1. We choose t and p1 such that 1/t=2m/q1-(2m+1-j)/n and 1/p1=1-(2m+2-j)/(2q1)+(2m+1-j)/n. Thus,
(71)1t+2m+1-j2q1+1p1=1.
Let r=1/ρ(x,V). By Hölder’s inequality we have
(72)|T2m*f(x)|≤∑i=-∞∞‍(∫2i-1r<|y-x|≤2ir‍|∇yjΓH2m(y,x)|tdy)1/t·(∫|y-x|≤2ir‍V(y)q1dy)(2m+1-j)/(2q1)×(∫|y-x|≤2ir‍|f(y)|p1dy)1/p1.
It follows from Lemma 18 and Theorem 9 that
(73)(∫2i-1r<|y-x|≤2ir‍|∇yjΓH2m(y,x)|tdy)1/t≤CCN(2ir)2mn/q1-n(1+2i)N.
Thus, as in the proof of Theorem 5, we have
(74)|T2m*f(x)|≤CCN{M(|f|p1)(x)}1/p1·∑i=-∞∞‍1(1+2i)N{1(2ir)n-2∫B2ir(x)‍V(y)dy}(2m+1-j)/2≤C{M(|f|p1)(x)}1/p1.
Then (64) follows since p0′>p1.


## 7. Remarks

We show estimate (57) under another assumption.Lemma 26.
Letk, l, and n be integers, m≥0, n≥2m+1+1, and 1≤j≤2m+1-1. Suppose that V∈(RH)n/2 and there exists a positive constant C such that V(x)≤Cρ(x,V)2. Assume also that (-Δ)2mu+V2mu=0 in BR(x0) for some x0∈Rn. Then there exist constants Cj and Cj′ such that
(75)supy∈BR/2(x0)|∇ju(y)|≤Cj{1+Rρ(x0,V)}Cj′Rjsupy∈BR(x0)|u(y)|.Proof.
We use (58). From Lemma 12 (1) we have
(76)|∇ju(x0)|≤C{1+Rρ(x0,V)}2m+1k0×ρ(x0,V)2m+1R2m+1-jsupy∈BR(x0)|u(y)|+CRj(1Rn∫BR(x0)‍|u(x)|2dx)1/2≤C{1+Rρ(x0,V)}2m+1(k0+1)Rjsupy∈BR(x0)|u(y)|.
Then by the same argument as in the proof of Lemma 25, we arrive at the desired estimate.The following theorem immediately follows from Lemmas22 and 26.Theorem 27.
Letm and n be integers, m≥0, n≥2m+1+1, and 1≤j≤2m+1+1. Suppose V∈(RH)n/2 and there exists a positive constant C such that V(x)≤Cρ(x,V)2. Then for any positive integer N there exists a constant CN such that
(77)|∇xjΓH2m(x,y)|≤CN{1+ρ(x,V)|x-y|}N·1|x-y|n-2m+1+j.Using Theorem27 we can prove the following theorem by the same way as the proof of Lemma 19.Theorem 28.
Letj, m, and n be integers, m≥0, n≥2m+1+1, and 0≤j≤2m+1-1. Suppose that V∈(RH)n/2 and there exists a positive constant C such that V(x)≤Cρ(x,V)2. Then there exist positive constants Cj such that (6) holds, where 1≤p≤∞.We obtain the following pointwise estimate stated with the Hardy-Littlewood maximal operator.Theorem 29.
Letj, m, and n be integers, m≥0, n≥2m+1+1, and 0≤j≤2m+1-1. Suppose that V∈(RH)n/2 and there exists a positive constant C such that V(x)≤Cρ(x,V)2. Then there exists a constant C′ such that
(78)|V2m-j/2∇jH2m-1f(x)|≤C′Mf(x).In Theorem29, the cases m=0 and m=1 were shown in [5, Theorem 1.3] and [8, Lemma 10], respectively.Proof of Theorem29.
Letr=1/ρ(x,V). Since V(x)≤Cρ(x,V)2, it follows from Theorems 9 and 27 that
(79)|V(x)2m-j/2∇jH2m-1f(x)|≤CCN∫Rn‍ρ(x,V)2m+1-j|f(y)|{1+ρ(x,V)|x-y|}N|x-y|n-2m+1+jdy≤CCN∑i=-∞∞‍∫2i-1r<|x-y|≤2ir‍×|f(y)|dyr2m+1-j(1+r-1|x-y|)N|x-y|n-2m+1+j≤CCN∑i=-∞∞‍(2i-1)2m+1-j(1+2i-1)N·1(2i-1r)n∫|x-y|≤2ir‍|f(y)|dy≤CCN∑i=-∞∞‍(2i)2m+1-j(1+2i)NMf(x).
Then choosing N≥2m+1-j+1 we arrive at the desired estimate.Remark 30 (see [5, Remark 1.6]).
Once we obtain the estimate (78), we can obtain boundedness of the operator V2m-j/2∇jH2m-1 on any space on which M is bounded, for example, Morrey spaces (see [12]).



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