(1)https://doi.org/10.4134/BKMS.b180836 pISSN eISSN ESTIMATES FOR RIESZ TRANSFORMS ASSOCIATED WITH SCHR ¨ODINGER TYPE OPERATORS Yueshan Wang Abstract

https://doi.org/10.4134/BKMS.b180836 pISSN: 1015-8634 / eISSN: 2234-3016

ESTIMATES FOR RIESZ TRANSFORMS ASSOCIATED WITH SCHR ¨ODINGER TYPE OPERATORS

Yueshan Wang

Abstract. Let L2 = (−∆)²+ V² be the Schr¨odinger type operator, where nonnegative potential V belongs to the reverse H¨older class RHs, s > n/2. In this paper, we consider the operator Tα,β = V^2αL^−β₂ and its conjugate T_α,β^∗ , where 0 < α ≤ β ≤ 1. We establish the (L^p, L^q)- boundedness of operator Tα,β and T_α,β^∗ , respectively, we also show that Tα,βis bounded from Hardy type space H_L¹

2(Rⁿ) into L^p2(Rⁿ) and T_α,β^∗ is bounded from L^p1(Rⁿ) into BM O type space BM OL₁(Rⁿ), where p1= _4(β−α)ⁿ , p2=_{n−4(β−α)}ⁿ .

1. Introduction

For 1 < s < ∞, a nonnegative locally L^s-integrable function V is said to belong to RH_s if there exists a constant C > 0 such that the reverse H¨older inequality

 1

|B|

Z

B

V (y)^sdy

1/s

≤ C

|B|

Z

B

V (y)dy holds for every ball B ⊂ Rⁿ.

Consider the Schr¨odinger type operators

Lj= (−∆)^j+ V^j (j = 1, 2) on Rⁿ, n ≥ 2j + 1,

where ∆ is the Laplacian on Rⁿ and the potential V belongs to the reverse H¨older class RHs for s > n/2. Note that L2 is biharmonic operator when V = 0. The boundedness of the operator V^αL^−β₁ has been well studied in some spaces (see [7], [8], [9], [13]). In this paper, we concentrate on the boundedness of the operators Tα,β = V^2αL^−β₂ and its conjugate T_α,β^∗ = L^−β₂ V^2αfor 0 < α ≤ β ≤ 1.

The operator Tα,β has been studied under the condition V ∈ RHs for s >

n/2. Zhong [14] and Sugano [13] showed the L^p-boundedness of T1,1, Chen et

Received September 5, 2018; Revised April 18, 2019; Accepted July 4, 2019.

2010 Mathematics Subject Classification. Primary 42B35, 35J10.

Key words and phrases. Riesz transform, Schr¨odinger operator, Hardy space, BMO.

c

2019 Korean Mathematical Society 1117

al. [2] established the L^p-boundedness of T_1/2,1/2. In this paper, we establish the following (L^p, L^q)-boundedness.

Theorem 1.1. Suppose V ∈ RH_sfor some s > n/2, 0 < α ≤ β ≤ 1.

(i) If _2α^s
0

< p < _4(β−α)ⁿ and ¹_q = ¹_p−^4(β−α)_n , then kT_α,β^∗ (f )k_Lq(Rⁿ)≤ Ckf kL^p(Rⁿ); (ii) If 1 < p < _2α ¹

s+^4(β−α)_n and ¹_q = ¹_p−^4(β−α)_n , then kT_α,β(f )k_Lq(Rⁿ)≤ Ckf k_Lp(Rⁿ).

Let us recall the concept of Hardy space related to Schr¨odinger type oper- ator. The Schr¨odinger type operators L_j (j = 1, 2) generate C₀ semigroups {e^−tLj}t>0. The maximal functions with respect to the semigroups {e^−tLj}t>0

are given by

M^Ljf (x) = sup

t>0

|e^−tLjf (x)|.

By [1, 4], a function f ∈ L¹(Rⁿ) is said to be in H_L¹

j(Rⁿ) if the semigroup maximal function M^Ljf belongs to L¹(Rⁿ). The norms of such a function are defined by

kf k_H1

Lj(Rⁿ)= kM^Ljf k_L1(Rⁿ), j = 1, 2.

Theorem 1.1 in [1] showed that H_L¹

2(Rⁿ) = H_L¹

1(Rⁿ) with equivalent norms.

We consider the boundedness of Tα,β at the endpoint p = 1, and get the following result.

Theorem 1.2. Suppose V ∈ RHsfor some s > n/2, 0 < α < β ≤ 1. Then, kTα,β(f )k_Lp2(Rⁿ)≤ Ckf k_H1

L2(Rⁿ), where p2= _{n−4(β−α)}ⁿ .

The dual space of H_L¹

1(Rⁿ) is the BM O type space BM O_L1(Rⁿ) (see [3]).

Let f be a locally function on Rⁿand B = B(x, r). Set fB= _|B|¹ R

Bf (y)dy and f (B, V ) = fB if r < ρ(x); f (B, V ) = 0 if r ≥ ρ(x). We say f ∈ BM OL1(Rⁿ) if

kf k_{BM OL1(R}n)= sup

B

1

|B|

Z

B

|f (y) − f (B, V )|dy < ∞.

It follows from [3] that kf kBM O_L1(Rⁿ) is actually a norm which makes BM OL₁(Rⁿ) a Banach space. Due to H¹(Rⁿ) ⊂ H_L¹₁(Rⁿ), we conclude by duality that BM O_L1(Rⁿ) ⊂ BM O(Rⁿ).

By Theorem 1.2 and duality, we get:

Corollary 1.3. Suppose V ∈ RHs for s > ⁿ₂. Let 0 < α < β ≤ 1. Then kT_α,β^∗ (f )k_{BM OL1(R}n)≤ Ckf k_Lp1(Rⁿ),

where p1= _4(β−α)ⁿ .

2. Some preliminaries

As in [12], for a given potential V ∈ RHs with s ≥ n/2, we define the auxiliary function

ρ(x) = sup (

r > 0 : 1 rⁿ⁻²

Z

B(x,r)

V (y)dy ≤ 1 )

, x ∈ Rⁿ. It is well known that 0 < ρ(x) < ∞ for any x ∈ Rⁿ.

We recall some important properties concerning the auxiliary function which will play an important role to obtain the main results. Throughout this section we always assume V ∈ RHswith s > n/2.

Lemma 2.1 ([12]). For 0 < r < R < ∞, we have 1

rⁿ⁻² Z

B(x,r)

V (y)dy ≤ C R r

n/s−2

1 Rⁿ⁻²

Z

B(x,R)

V (y)dy.

Lemma 2.2 ([5]). There exists a constant l₀> 0 such that 1

rⁿ⁻² Z

B(x,r)

V (y)dy ≤ C

 1 + r

ρ(x)

l0

. Lemma 2.3 ([12]). There exists k0≥ 1 such that

C⁻¹ρ(x)



1 + |x − y|

ρ(x)

−k0

≤ ρ(y) ≤ Cρ(x)



1 + |x − y|

ρ(x)

_1+k0^k0

for all x, y ∈ Rⁿ.

It is easy to get the following result from Lemma 2.3.

Lemma 2.4. Let k ∈ N and x ∈ 2^k+1B(x₀, r) \ 2^kB(x₀, r). Then we have 1



1 + _ρ(x)^2kr^N ≤ C



1 + _ρ(x^2kr

0)

^N/(k0+1).

A ball B(x0, ρ(x0)) is called critical. Assume that Q = B(x0, ρ(x0)), for x ∈ Q, Lemma 2.3 tell us that ρ(x) ∼ ρ(y) if |x − y| < Cρ(x).

We give the estimates of fundamental solutions. Denote Γ_L2(x, y, λ) by the fundamental solution of L2+ λ, where λ ∈ [0, ∞). When λ = 0 it follows from Theorem 2 in [13] that for any positive integer N there exists a positive constant CN such that

0 ≤ ΓL₂(x, y, 0) .

= ΓL₂(x, y) ≤ C_N



1 +^|x−y|_ρ(x) ^N 1

|x − y|ⁿ⁻⁴. When λ 6= 0, we have the following result [10].

Lemma 2.5. Let 0 < λ < ∞. For any positive integer N , there exists a positive constant CN such that

0 ≤ ΓL2(x, y, λ) ≤ C_N



1 + λ¹²|x − y|²^N

1 +^|x−y|_ρ(x) ^N 1

|x − y|ⁿ⁻⁴. Lemma 2.6 ([11]). Assume that (L₂+λ)u = 0 in B(x₀, 2R) for some x₀∈ Rⁿ. Then there exists a k₀⁰ such that

Z

B(x₀,R)

|∇u|^tdx

!^1/t

≤ CR^2n/s−4

 1 + R

ρ(x0)

k⁰₀

sup

B(x0,2R)

|u|, where 1/t = 2/s − 3/n.

Let K_β be the kernel of operator L^−β₂ . We have the following estimates for K_β(x, y).

Lemma 2.7. Suppose V ∈ RHsfor s > ⁿ₂, 0 < β ≤ 1.

(i) For every N > 0, there exists a constant C such that

|K_β(x, y)| ≤ C



1 +^|x−y|_ρ(x) N

1

|x − y|^n−4β;

(ii) For every N > 0, and any 0 < h < |x − y|/16, we have

|Kβ(x, y + h) − Kβ(x, y)| ≤ C



1 + ^|x−y|_ρ(x)N

h^δ

|x − y|^n−4β+δ, where δ = 4 − 2n/s.

Proof. (i) When β = 1, we have

Kβ(x, y) = ΓL2(x, y, 0) ≤ C_N



1 + ^|x−y|_ρ(x)^N 1

|x − y|ⁿ⁻⁴. For 0 < β < 1, by the functional calculus, we may write

L^−β₂ =sin πβ π

Z ∞ 0

λ^−β(L2+ λ)⁻¹dλ.

Let f ∈ C₀^∞. From (L2+ λ)⁻¹f (x) =R

RⁿΓ_L2(x, y, λ)f (y)dy, it follows that L^−β₂ (f )(x) =

Z

Rⁿ

Kβ(x, y)f (y)dy, where

K_β(x, y) = sin πβ π

Z ∞ 0

λ^−βΓ_L2(x, y, λ)dλ.

Note that

Z ∞ 0

λ^−β 1

1 + λ¹²|x − y|²
Ndλ ≤ C|x − y|^4β−4 for 0 < β < 1. So, by Lemma 2.5 we get

|K_β(x, y)| ≤ C



1 +^|x−y|_ρ(x) N

1

|x − y|^n−4β.

(ii) For β = 1, by Lemma 3.3 in [2] we get (ii). When 0 < β < 1, by the functional calculus, we have

Kβ(x, y + h) − Kβ(x, y) = sin βπ π

Z ∞ 0

λ^−β ΓL₂(x, y + h, λ) − ΓL₂(x, y, λ)
dλ.

We need to estimate |Γ_L2(x, y + h, λ) − Γ_L2(x, y, λ)| in advance.

Fix x, y ∈ Rⁿ and let R = |x − y|/8, 1/t = 1/s − 3/n. It follows from the Morrey embedding theorem (see [6]) and Lemma 2.6 that

|ΓL2(x, y + h, λ) − Γ_L2(x, y, λ)|

≤ Ch^1−n/t

 Z

B(y,R)

|∇yΓ_L2(x, z, λ)|^tdu

1/t

≤ C h R

^1−n/t 1 + R

ρ(y)

^k00

sup

z∈B(y,2R)

Γ_L2(x, z, λ)

≤ C h R

^1−n/t 1 + R

ρ(x)

^k

00 0

sup

z∈B(y,2R)

Γ_L2(x, z, λ).

Note z ∈ B(y, 2R), |x − y| = 8R, then 6R ≤ |z − x| ≤ 10R. By Lemma 2.5 we get

ΓL₂(x, z, λ) ≤ C



1 + λ¹²|z − x|²^N

1 + ^|z−x|_ρ(x) ^N 1

|z − x|ⁿ⁻⁴

≤ C

1 + λ¹²R²
N

1 +_ρ(x)^R ^N 1 Rⁿ⁻⁴. Then

|Γ_L2(x, y + h, λ) − Γ_L2(x, y, λ)|

≤ C

1 + λ¹²R²
N

1 + _ρ(x)^R ^N h^δ R^n−4+δ

≤ Ch^δ

1 + λ¹²|x − y|^2
N

1 +^|x−y|_ρ(x) N

1

|x − y|^n−4+δ.

So

|Kβ(x, y + h) − Kβ(x, y)|

≤ C Z ∞

0

λ^−β

Γ_L2(x, y, λ) − Γ_L2(x, x0, λ) dλ

≤ C



1 +^|x−y|_ρ(x) N

h^δ

|x − y|^n−4+δ Z ∞

0

λ^−β 1

1 + λ¹²|x − y|²
Ndλ

≤ C



1 +^|x−y|_ρ(x) ^N

h^δ

|x − y|^n−4β+δ.

 Let f ∈ L^q_loc(Rⁿ). Denote |B| by the Lebesgue measure of the ball B ⊂ Rⁿ. The fractional Hardy-Littlewood maximal function Mσ,γ is defined by

Mσ,γ(f )(x) = sup

 1

|B|^1−σγn Z

B

|f (y)|^γdy

^1/γ . Lemma 2.8. Suppose 1 < γ < p < ⁿ_σ, and ¹_q = ¹_p−^σ_n. Then

kMσ,γf k_Lq(Rⁿ)≤ Ckf kL^p(Rⁿ). 3. (L^p, L^q)-estimates In this section, we prove Theorem 1.1. By Lemma 2.5,

T_α,β^∗ f (x) = Z

Rⁿ

Kβ(y, x)V (y)^2αf (y)dy, where

|Kβ(y, x)| ≤ CN



1 +^|x−y|_ρ(y) ^N 1

|x − y|^n−4β. We will prove part (i) and part (ii) follows by duality.

Let r = ρ(x). Then by Lemma 2.4 we have

|T_α,β^∗ f (x)| = Z

Rⁿ

|Kβ(y, x)|V (y)^2α|f (y)|dy

≤ C Z

Rⁿ

V (y)^2α 1 +^|x−y|_ρ(y)
^N

|f (y)|

|x − y|^n−4βdy

≤ C

∞

X

k=−∞

(2^kr)^4β (1 + 2^k)^N/(k0+1)

1 (2^kr)ⁿ

Z

|x−y|≤2^kr

V (y)^2α|f (y)|dy.

By H¨older inequality and V ∈ RHs, s > n/2, we have 1

(2^kr)ⁿ Z

|x−y|≤2^kr

V (y)^2α|f (y)|dy

≤ C 1 (2^kr)ⁿ

Z

|x−y|≤2^kr

V (y)dy

!^2α

× 1

(2^kr)ⁿ Z

|x−y|≤2^kr

|f (y)|^(2αs)0dy

!1/(_2α^s)⁰

. For k ≥ 1, by Lemma 2.2, there exists l0> 0 such that

1 (2^kr)ⁿ

Z

|x−y|≤2^kr

V (y)dy

!^2α

≤ C(2^kr)^−4α 1 (2^kr)ⁿ⁻²

Z

|x−y|≤2^kr

V (y)dy

!2α

≤ C(2^kr)^−4α(1 + 2^k)^2l0α≤ C(2^kr)^−4α2^2l0αk. For k ≤ 0, by Lemma 2.1, we have

1 (2^kr)ⁿ

Z

|x−y|≤2^kr

V (y)dy

!2α

≤ C(2^kr)^−4α

 r 2^kr

^(n/s−2)2α 1 rⁿ⁻²

Z

|x−y|≤r

V (y)dy

^2α

≤ C(2^kr)^−4α2^{2αk(2−n/s)}. Taking N > 2(k₀+ 1)l₀α, then

|T_α,β^∗ f (x)| ≤ C 1

(2^kr)^{n−4(β−α)(2αs )0} Z

|x−y|≤2^kr

|f (y)|^(2αs)0dy

!^1/(_2α^s)0

×

∞

X

k=1

1 2^{k(k0+1N −2l0α)}

+

0

X

k=−∞

2^{2αk(2−n/s)}

!

≤ CM_4(β−α),(^s

2α)⁰(f )(x).

By Lemma 2.6, we get

kT_α,β^∗ (f )k_Lq(Rⁿ)≤ Ckf k_Lp(Rⁿ), where _2α^s
0

< p < _4(β−α)ⁿ , and ¹_q = ¹_p−^4(β−α)_n . This finishes the proof of (i).

By duality, we get

kT_α,β(f )k_Lq(Rⁿ)≤ Ckf k_Lp(Rⁿ)

for _{n−4(β−α)}ⁿ < q < _2α^s , and ¹_q = ¹_p −^4(β−α)_n . These conditions are equivalent to

1 < p < 1

2α

s +^4(β−α)_n

and 1 q =1

p−4(β − α)

n .

This completes the proof of Theorem 1.1.

4. (H_L¹

2, L^p2)-estimates Because of H_L¹

2(Rⁿ) = H_L¹

1(Rⁿ) and their norms are equivalent, we only need to show that the operator Tα,β maps the Hardy space H_L¹₁(Rⁿ) contin- uously into L^p2(Rⁿ), where p2= _{n−4(β−α)}ⁿ . Firstly, we review the concept of (1, q)_ρ-atom.

Let 1 < q ≤ ∞. A measurable function a is called a (1, q)_ρ-atom associated with the ball B(x, r) if r < ρ(x) and the following conditions hold:

(i) supp a ⊂ B(x, r) for some x ∈ Rⁿ and r > 0;

(ii) kak_Lq(Rⁿ)≤ |B(x, r)|^1/q−1; (iii) if r < ρ(x)/4, thenR

B(x,r)a(x)dx = 0.

By [4], the Hardy space H_L¹

1(Rⁿ) admits the following atomic decomposition:

Lemma 4.1. f ∈ H_L¹

1(Rⁿ) if and only if f can be written as f =P

jλjaj, where a_j are (1, q)_ρ-atoms and P

j|λ_j| < ∞. Moreover kf k_H1

L1 ∼ inf

 X

j

|λ_j|

 ,

where the infimum is taken over all atomic decompositions of f into H_L¹

1-atoms.

We choose q1 and q2such that

1 < q1< 1

2α

s +^4(β−α)_n

and 1

q2

= 1 q1

−4(β − α)

n .

By Lemma 4.1, we only need to prove

kTα,βak_Lp2(Rⁿ)≤ C

holds for any (1, q1)ρ-atom a, where C > 0 is independent of a.

Assume that suppa ⊂ B(x0, r), r < ρ(x0). Then

kT_α,β(a)k_Lp2(Rⁿ)≤ kχ_16BT_α,β(a)k_Lp2(Rⁿ)+ kχ_(16B)cT_α,β(a)k_Lp2(Rⁿ)= I₁+ I₂. By H¨older inequality, Theorem 1.1 and _p¹

2 = 1 −^4(β−α)_n , we have I₁= kχ_16BT_α,β(a)k_Lp2(Rⁿ)

≤ C|16B|^p21−q21

Z

Rⁿ

|T_α,βa(x)|^q2dx

^1/q2

≤ C|16B|^p21−q21

Z

B

|a(x)|^q1dx

^1/q1

≤ C|16B|^p21−q21 |B|^q11−1≤ C.

We divided into two case for the estimate of I₂: r ≥ ρ(x₀)/4 and r < ρ(x₀)/4.

Case I: r ≥ ρ(x0)/4. In this case, we have r ∼ ρ(x0). Let Ek = 2^kB \ 2^k−1B.

By Lemma 2.4 and Lemma 2.5, we get I2=

Z

(16B)^c

|Tα,βa(x)|^p2dx

!1/p2

≤ C

∞

X

k=5

Z

Ek

V (x)^2αp2

Z

B

|Kβ(x, y)a(y)|dy

^p2

dx

!1/p2

≤ C





∞

X

k=5

Z

Ek

V (x)^2αp2



 Z

B

|a(y)|dy 1 +^|x−y|_ρ(x)
N

|x − y|^n−4α





p₂

dx





1/p₂

≤ C





∞

X

k=5

1 1 +_ρ(x^2kr

0)

^{N p}2/(k₀+1)

(2^kr)^4βp2 (2^kr)^np2

Z

2^kB

V (x)^2αp2dx





1/p₂

× Z

B

|a(y)|dy.

Notice

p2= n

n − 4(β − α) < n 4α < s

2α. By H¨older inequality and V ∈ RHs, we have

1

|2^kB|

Z

2^kB

V (x)^2αp2dx

≤ C

 1

|2^kB|

Z

2^kB

V (x)^sdx

2αp2/s

≤ C

 1

|2^kB|

Z

2^kB

V (x)dx

2αp2

. Then, by Lemma 2.2 we get

1

|2^kB|

Z

2^kB

V (x)^2αp2dx ≤ C(2^kr)^−4αp2



1 + 2^kr ρ(x0)

^2l0αp2

. Due to a is a (1, q1)ρ-atom, by H¨older inequality we get

Z

B

|a(y)|dy ≤ C|B|^1−q11

Z

B

|a(y)|^q1dy

1/q1

≤ C.

Then, we have

I2≤ C

∞

X

k=5

(2^kr)^{n−p2(n−4(β−α))}

 1 +_ρ(x^2kr

0)

_k0+1^{N p2}−2l0αp2

!^1/p2

.

Noticing r ∼ ρ(x₀), and p₂=_{n−4(β−α)}ⁿ , we get

I₂≤ C

∞

X

k=5

1 (2^k)^{k0+1N p2−2l0αp2}

!^1/p2

.

Taking _k^N

0+1 − 2αl0> 0, we obtain I2≤ C.

Case II: r < ρ(x₀)/4. By the vanishing condition of a, Lemma 4.1 and Lemma 2.4 we have

I2≤ C Z

(16B)^c

V (x)^2αp2

 Z

B

|(Kβ(x, y) − Kβ(x, x0))a(y)|dy

p2

dx

!^1/p2

≤ C

∞

X

k=5

Z

E_k

V (x)^2αp2 Z

B

|a(y)|



1 + ^|x−y|_ρ(x)^N

|y − x0|^δ

|x − y|^n+δ−4βdy

!^p2

dx

!^1/p2

≤ C

∞

X

k=5

1



1 + _ρ(x^2kr

0)

_k0+1^{N p2}

r^δp2 (2^kr)^{(n+δ−4β)p2}

Z

2^kB

V (x)^2αp2dx

!^1/p2

× Z

B

|a(y)|dy.

Note that 1

|2^kB|

Z

2^kB

V (x)^2αp2dx ≤ C(2^kr)^−4αp2



1 + 2^kr ρ(x0)

^2l0αp2

, Z

B

|a(y)|dy ≤ C,

p2=_{n−4(β−α)}ⁿ , and δ > 0. Then, taking N ≥ 2l0α(k0+ 1), we get

I2≤ C

∞

X

k=5

1

 1 +_ρ(x^2kr

0)

_k0+1^{N p2}−2l0αp₂

1

(2^kr)(n−4(β−α))p₂−n

r^δp2 (2^kr)^δp2

!1/p2

≤ C

∞

X

k=5

1 2^kδp2

!^1/p2

≤ C.

This completes the proof of Theorem 1.2.

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Yueshan Wang

Department of Mathematics Jiaozuo University

Jiaozuo 454003, P. R. China Email address: wangys1962@163.com